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0 # Current value depends on the amount of charge moving past a point in a unit of time? Updated: 9/17/2023 Wiki User 13y ago Be notified when an answer is posted Earn +20 pts Q: Current value depends on the amount of charge moving past a point in a unit of time? Submit Still have questions? Related questions ### What is the rate at which electrons do work? Current. The flow of electrons is the flow of a moving charge. The rate of flow is current (the amount of charge that flows in a set time). The equation is: I = Qt Hope this helps. a current ### Why does a current carrying wire produce magnetic field but not electric field-whereas moving charge produces both? There's something seriously wrong with the question's hypotheses.Current is moving charge, and moving charge is current. ### What is the definition of current-? Current is known as the flow of an electrical charge. This is carried by moving electrons in a wire. ### Suppose you charge a pocket comb by friction and then walk around the room carrying the comb Are you producing an electric current? Yes indeed, a current is just a moving charge. Even if the charge is on a comb, and you are physically moving the comb around the place. ### What does moving electricale charge produce? It produces a magnetic field. ### What surrounds a moving electrical charge? it depends on what kind of charge it is if its a positive charge then protons are surrounding it if it is a negative charge then electrons are surrounding it does that make sense? ### What do moving electrical charges create? -- Electric charge that's moving is the definition of electric current.-- It creates a magnetic field in its neighborhood. ### Can you have current without voltage? Sure - an unconnected power supply of any kind - a battery is + at one end and - at the other. ### What is the amount of electrons moving past a certain point on a wire? It's called current. ### How do you calculate a starting current? The formula for the electric current can be given as I= Qt where Q refers to coulomb charge and t= amount of time in second . By applying this formula the amount of current passing through the conductor can be known foor any instant of time. For any conductor the amount of charge is always constant. Current measures the flow of moving charge per unit time. Therefore the formula for current is I=Q/t as the unit of current is C/s (Coulomb's per sec) or amps (A). ### In general how much do a moving company charge for a local move? It depends. Get several moving quotes from local movers and make a comparison.

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https://zims-en.kiwix.campusafrica.gos.orange.com/wikipedia_en_all_nopic/A/Sequential_proportional_approval_voting

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# Sequential proportional approval voting Sequential proportional approval voting (SPAV) or reweighted approval voting (RAV) is an electoral system that extends the concept of approval voting to a multiple winner election. Proposed by Danish statistician Thorvald N. Thiele in the early 1900s,[1] it was used (with adaptations for party lists) in Sweden for a short period after 1909.[2] ## Description This system converts AV into a multi-round rule,[3] selecting a candidate in each round and then reweighing the approvals for the subsequent rounds. The first candidate elected is the AV winner (w1). The value of all ballots that approve of w1 are reduced in value from 1 to 1/2 and the approval scores recalculated. Next, the unelected candidate who has the highest approval score is elected (w2). Then the value of ballots that approve of both w1 and w2 are reduced in value to 1/3, and the value of all ballots that approve of either w1 or w2 but not both are reduced in value to 1/2.[4] At each stage, the unelected candidate with the highest approval score is elected. Then the value of each voter’s ballot is set at 1/(1+m) where m is the number of candidates approved on that ballot who were already elected, until the required number of candidates is elected. The system disadvantages minority groups who share some preferences with the majority. In terms of tactical voting, it is therefore highly desirable to withhold approval from candidates who are likely to be elected in any case, as with cumulative voting and the single non-transferable vote. It is however a much computationally simpler algorithm than proportional approval voting, permitting votes to be counted either by hand or by computer, rather than requiring a computer to determine the outcome of all but the simplest elections.[5]

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https://aic-components.com/power-equipment/how-can-i-reduce-the-amperage-of-a-transformer.html

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# How can I reduce the amperage of a transformer? Contents ## What happens if amperage is too high? As long as the correct voltage is used, a device will draw only the amperage it needs, meaning there will not be “too many amps”. If an incorrect voltage is used — say a higher voltage than the device is rated to accept — then yes, too many amps may be drawn and the device can be damaged. ## Does amperage change in a transformer? Transformers change electricity from one voltage level to another. But changing the voltage does not change the power. … So when a transformer increases voltage, it decreases current. Likewise, if it decreases voltage, it increases current. ## Does increasing voltage reduce amperage? In a circuit, cutting the resistance by half and leaving the voltage unchanged will double the amperage across the circuit. If the circuit’s resistance remains unchanged, the amperage in a circuit can be increased by increasing the voltage. ## How do transformers reduce current? A transformer is a device that can change the voltage of an alternating current (ac). … A step-up transformer has more turns of wire on its secondary coil than it does on its primary coil. Transformers will only work with an alternating current (ac) input. This transformer steps up the voltage by reducing the current. ## How do you determine the maximum amps of a transformer? The full load current I(A) in amps is equal to 1000 times of transformer rating S(kVA) in kVA divided by the multiplication of root 3 times of line to line voltage V(V) in volts. THIS IS IMPORTANT:  How long should you warm up on a treadmill? ## How can the current of a transformer be increased? Similarly, when the voltage decreases in a step-down transformer, the current increases proportionately. Thus, if the voltage is cut in half, the current doubles. In other words, power equals voltage times current. A transformer transfers power from the primary coil to the secondary coil.

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Number System and Codes Presentation on theme: "Number System and Codes"— Presentation transcript: Number System and Codes Chapter 3 Number System and Codes Decimal and Binary Numbers Decimal and Binary Numbers Converting Decimal to Binary Sum of powers of 2 Converting Decimal to Binary Repeated Division Binary Numbers and Computers Representing Integers with binary Some of challenges:- Integers can be positive or negative Each integer should have a unique representation The addition and subtraction should be efficient. Representing a positive numbers Representing a negative numbers using Sign-Magnitude notation -5 = bits sign-manitude -55= bits sign-magnitude 1’s Complement The 1’s complement representation of the positive number is the same as sign-magnitude. +84 = 1’s Complement The 1’s complement representation of the negative number uses the following rule:- Subtract the magnitude from 2n-1 For example: -36 = ??? +36 = 1’s Complement Example :- - 57 +57 = -57 = Converting to decimal format 2’s Complement For negative numbers:- Subtract the magnitude from 2n. Or Add 1 to the 1’s complement Example Convert to decimal value Positive values:- = +89 Negative values Two's Complement Arithmetic Adding Positive Integers in 2's Complement Form Adding Positive and Negative Integers in 2's Complement Form Adding Positive and Negative Integers in 2's Complement Form Subtraction of Positive and Negative Integers Digital Codes Binary Coded Decimal (BCD) BCD BCD 4221 Code Gray Code In pure binary coding or 8421 BCD then counting from 7 (0111) to 8 (1000) requires 4 bits to be changed simultaneously. Gray coding avoids this since only one bit changes between subsequent numbers Binary –to-Gray Code Conversion Gray –to-Binary Conversion Gray –to-Binary Conversion The Excess-3- Code Parity The method of parity is widely used as a method of error detection. Extar bit known as parity is added to data word The new data word is then transmitted. Two systems are used: Even parity: the number of 1’s must be even. Odd parity: the number of 1’s must be odd. Parity Example: Odd parity Even Parity 110010 110011 11001 111101 111100 11110 110001 110000 11000

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https://pypi.org/project/solarsystem/

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Our Solar System. Planet's positions, Sun's and Moon' s position and rise/set # SOLARSYSTEM Our Solar System consists of: • our Star, the Sun • 8 Planets: Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus and Neptune • Some dwarf Planets: Pluto, Ceres, Eris, (Haumea, Makemake, Quaoar, Sedna, Orcus, 2007 OR10, not yet included here) • Some Centaurs: Chiron (onlyone included here) • Many moons orbiting planets. Our Moon (Selene in Greek or Luna in Latin) is orbiting Earth solarsystem is a python library for finding position of planets around Sun or around Earth. Also with solarsystem we can find positions around Sun/Earth of dwarf planets (only 3 planets so far) and Chiron Centaur and position of moon around Earth Furthermore we compute sunrise/sunset, moonrise/moonset and moon phase for given place (geocoordinates). Except all computation above with this library a set of usefull functions are included with which we can convert between coordinate systems: • Transform spherical to rectangular projection. • Transform rectangular to spherical projection. • Transform ecliptic to equatorial projection. • Transform equatorial to ecliptic projection. • Transform eclipitc to spherical projection. • Transform spherical to eclipitc projection. ## Quick start import solarsystem Initialize class H = solarsystem.Heliocentric(year=2020, month=1, day=1, hour=12, minute=0 ) Compute position of planets around sun planets_dict=H.planets() print('Planet',' \t','Longitude',' \t','Latitude',' \t','Distance in AU') for planet in planets_dict: pos=planets_dict[planet] print(planet,' \t',round(pos[0],2),' \t',round(pos[1],2),' \t',round(pos[2],2)) # Planet Longitude Latitude Distance in AU # Mercury 263.83 -4.06 0.47 # Venus 5.23 -3.22 0.73 # Earth 100.53 0.0 0.98 # Mars 214.38 0.49 1.59 # Jupiter 276.1 0.1 5.23 # Saturn 292.51 0.05 10.05 # Uranus 35.35 359.52 19.81 # Neptune 348.02 -1.04 29.91 # Pluto 292.75 359.33 33.88 # Ceres 290.87 -5.4 2.92 # Chiron 4.33 2.94 18.81 # Eris 23.55 -11.74 96.0 ## Examples - Use Cases • Plot planets around Sun, watch where planets are around Sun • Get the Geocentric positions of Sun, planets, nano planets, our Moon and 1 Centaur • Time of sun rise and set within each day Time of moon rise and set within each day Moon phase - percent of illumination ## Documentation The full documentation is available at solarsystem.readthedocs.io Alternatively you can build documentation: install sphinx Go to docs/ directory cd docs Build html files make html Open _build/html/index.html in browser. ## Installation install from Pypi: pip install solarsystem pip install git+https://github.com/IoannisNasios/solarsystem ## Requirements No requirements, no additional libraries needs to be installed. Exception: example notebook Solar System Live, matplotlib is needed in order to view the plot ## Python versions • solarsystem is tested and runs normal for python versions 3.4+ and 2.7 • running solarsystem on previous python versions should also run but use with caution. ## Citing If you find this library useful, please consider citing: @misc{Nasios:2020, Author = {Ioannis Nasios}, Title = {solarsystem}, Year = {2020}, Publisher = {GitHub}, Journal = {GitHub repository}, Howpublished = {\url{https://github.com/IoannisNasios/solarsystem}} } ## Project details Uploaded source

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https://news.ycombinator.com/item?id=12528144

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Music theory for nerds (eev.ee) 766 points by hardmath123 on Sept 19, 2016 | hide | past | web | favorite | 378 comments I'd like to elaborate a bit, in the same "for nerds" manner, on where Eevee seems to get lost a bit with scales and notation. He (she? not sure) calls the A minor and C major scales the same, because they contain the same notes. That's not an odd thought, but it's like calling sine and cosine the same because both functions contain the same set of values, in the same order.The difference is phase. Basically, scales aren't just an ordered set of notes, they also have a starting point. This note, the note the scale is named after, is often the last note of a tune played in that scale, and often the first too (especially for more poppy tunes). So if you play Für Elise in C major, like Eevee suggests, the entire melody will be pitched 3 semitones higher than playing Für Elise in A major. And it'll sound awkward because you're supposed to play it on a minor scale.Once you understand this, the whole notation thing makes a lot more sense as well. Even though the key signature for a major and its relative minor are the same, that is only part of the story.When you see that A Minor has the same key signature as C Major, what you are seeing is the "natural minor" scale. This is simply the notes implied by the key signature, and so is identical (but down a third) from the relative major. If you play purely in the natural minor, you are really closer to the Aeolian mode than a minor key.There is also the "harmonic minor" and "melodic minor" which, in common practice period, are much more commonly used than natural minor. Harmonic minor has a raised seventh. This makes the dominant (V) triad a major triad which increases its need to resolve to the tonic (I). It is used for harmonies (obviously based on the name) for this reason, but it makes the step from 6 to 7 an augmented second (i.e. minor third) so its isn't used melodically very often. Which is why there is also a Melodic Minor.Melodic minor is tricky because it varies depending on whether the melody is ascending or descending. Descending is easy, it is identical to natural minor. Ascending is similar to the harmonic minor with its raised seventh (creating what is known as a leading tone, i.e. a half step below the tonic that wants to resolve up to the tonic.) But the ascending scale also has a raised sixth, which eliminates the augmented second between 6 and 7. I've never quite understood why you learn a melodic minor as being different depending on if it's ascending or descending.Are there examples of this in music? Or is it just something invented to test young pianists in exams? That's only in classical music. Jazzers play "pure" melodic minor both up and down. Thank you for this. I learned to write natural, harmonic, and melodic minor scales when I was a kid, but I never knew this explanation. It makes perfect sense! > they also have a starting pointIndeed. For anyone that is not sure about that, take a piano (like this virtual one[1]) and play a scale of all the white notes starting with C.The play a scale of all those same white notes, but starting with A.You should immediately notice that the second version is in a minor key (sounds "sadder"). Awesome example! Folks, do this if this stuff is new to you, it's beautifully enlightening. Thanks, that's an explanation I haven't heard that way before, but that makes sense to me.Follow-up question: how are phase of the scale and the piece of music synchronized? When I think in terms of a wave, I could Fourier-Transform it into sines or in cosines, or any other phase-shifted variant (a * sin(nx + const.))?Is it always the first note of the piece of music that "anchors" the piece in its scale?Does that question even make sense? :-) Nice question, I think I can answer (although you'll soon notice that my understanding of this subject is limited as well)First of all, this is a mathematically undecidable problem, as in that there exist tunes that could be in multiple scales. The extreme example is a melody that consists of only a single note. Ridiculous, but as a fellow nerd I'm sure you can see how I could call a single note a melody. So we have to lose the phase metaphor here a little bit.Most songs "anchor" with the last note. If you really can't tell whether a song is in major or minor (songs in major are a bit happier, jollier; songs in minor a bit sadder and more melancholic), going for the very last note (or the last note of the chorus, if the song a chorus) is a very good bet. Find the core melody of the song, the thing everything is hung up on, the last note of that core melody is the ground note of the scale. This is a really safe bet.There are, of course, exceptions, and then I think you'll need to do some mathematical analysis to find out which key the notes in the melody match the best to. I'm not 100% sure how this is defined but I'm sure people have researched it.All that said, musicians don't really have this problem. After all, hardly any of this stuff was designed. The only genius piece of engineering in all this was the discovery that if you choose to 12 notes in an octave (and not 8 or 15 or whatever), you have a very versatile instrument that can play almost any melody at any pitch in a way that sounds pretty good to human ears (pretty good, not perfect, because of the all the ≈'s in the "Intervals" section of Eevee's article).But the rest was discovered, not designed, just by fooling around. The analysis came after the music.Musicians just start with a scale and then make the music they compose fit. This comes pretty natural to you with a bit of practice; most people have a pretty decent innate ability to hear which melodies "match" with a chord and which chords "match" with a previous chord. In all honesty, once you've understood things this far, I'd recommend fooling around with the easiest instrument or tool you know, rather than diving even deeper in to the mathematics of things :-) > The extreme example is a melody that consists of only a single note. Ridiculous, but as a fellow nerd I'm sure you can see how I could call a single note a melody.I got your single note melody right here: I was half expecting the One Note Samba (which is sort of the pathological case for the key-finding heuristic skrebbel describes). This gets pretty complicated and subjective.Strictly speaking, a "note" cannot define a key. To determine a key, you really need at least 2 intervals, wherein interval a somehow resolves to interval b. Keys are just a mapping of relationships; how does each of the 12 notes relate to each other? If you just play a C major chord, you aren't really playing in the key of CMaj, you're just playing a CMaj chord.For example: Play B+G, then C+G. This is (arguably) C major. The B note resolves into the C note. That's the 7th moving into the Tonic (#1). This is an example of a V-I resolution, the strongest possible. If you'd like more examples, look up "cadences". There are 'principles' for determining the key, but they should not be understood as proofs.I think you're also asking if/how to change keys. This is not only possible, but often highly encouraged. There are two extremely common key changes in pop music: 1) Up one step: This is incredibly common and can be heard in "I Will Always Love You" (album version, right after the 3 minute mark when she starts singing the chorus again). Once you start hearing this, you can't stop. It's all over the place. If we're in the key of C, we'll just play a D chord with a lot of confidence, typically after a G chord. 2) Minor to Major (or reverse, moving from sine to cosine in the analogy): This is much more subtle than the previous example. Typically you'll recognize this as a change in 'mood' or 'feeling'. You'll find a lot of Am-Em changes, then the chorus will be a lot of C-G changes. The tonic changes from Am to C, even though all the chords and notes being used are still the same (although the duration of said chords will probably be different).Most "classical" music moves around keys pretty frequently, in extreme cases multiple times within a measure. And then there are chords/sections in which the key is debatable if not indiscernible (I would argue that the intro to Smashing Pumpkins "1979" is changing keys every 2 measures, but I think there's also a strong argument for it being in a single key)Thinking about keys in terms of phases is a good basic explanation, but ultimately phase is much easier to measure. Perhaps I see it this way because my understanding of music is much better than my understanding of physics. Yes, I was trying to explain some theory to my sister and she had it in her head that "black notes" on the piano - or anything we call a sharp or flat - were "minor". It took a lot of explaining to get her to realize that a single note on it's own is in no particular key and no particular mode (scale). That's funny because B+G, C+G to me says G major: a major third, then a fourth. Is the ear guided to keys by inversion, some inversions more natural or root-y than others?I've been composing pop music for a long time without knowing stuff like this. I can see the G major argument. That would be a I/3 into IV. My professor would have said that answer is wrong due to voice leading, the 7 to 1 is very powerful. And the last chord is C, in root, meaning that's where you have resolved. Had the second interval been G+B, it would make the GMaj argument stronger. Ultimately, it's really context. You're really just asking about cadences, very roughly translated means "how chords resolve". B+G is a minor sixth, not a major third. If you invert it, it's a major third.B+G followed by C+G is most likely going to sound like a cadence in C Major. You could claim it is G major only if you considered it unresolved transition to the subdominant. >Is it always the first note of the piece of music that "anchors" the piece in its scale?No it isn't. If you put any piece of music that is harmonious through a device that analyses the frequencies you'll find three main notes and those give you the key. In C-major for example the main notes are C, E and G. The section will be in C even if the first note is E for example.You can do this with software. There's a nice video of this here https://youtu.be/xVmkIznjUPE?t=24sor you can upload a recording to get the chords https://chordify.net/You can indeed fourier transform the music. I imagine the above software does so. My teacher always said "music is about tension and the resolution of tension". So in A minor if you play tense chords and then resolve them to A minor, you're in A minor. If you resolve them to C, you're in C major. The phase of a sound wave doesn't play a role in the context of compositions or scales. It does of course play a role in audio in general though - in recording, and analog and digital signal processing etc. they're not talking about phase of the sound wave but "phase" of the scale notes as an alternative way of describing what music theory calls modes It's not even just that. They are only "the same" with equal temperament. If you use proper intervals then they are quite different. I'm not sure this applies here. Equal temperament explains how an interval is in tune and does not determine what key a song is in. Torrent-of-ions is correct. Equal temperament is a compromise in tuning (assigning pitches to notes) which makes the most keys sound ok. The intervals aren't perfect, but close enough. If you tuned a piano for perfect intervals in one key, those pitches would be too much off for another key. Yes, equal temperament is basically a compromise of tuning methods. While it can help ensure that a piece which modulates into different keys is in tune (as close as possible), equal temperament cannot tell the key that is being played. > He (she? not sure)FTR the pronoun where gender is ambiguous is 'they' (It's also sometimes explicitly requested by nonbinary people). From eevee's Twitter:> she/they/heI have a similar relationship with pronouns. My own perception of my gender jumps around so much that I don't even bother trying to figure it out. It usually leans in one direction, but I don't have many strong feelings about it. > It completely obscures the relationship between the pitches, though.It doesn't actually obscure the relationship between the notes -- it makes them clearer. For example, I see the notes C, E, and G on some sheet music, maybe with some accidentals on some of those notes. I know that I'm therefore supposed to play a C triad. Now, there are multiple kinds of triads, but once I know I'm supposed to play a triad, it's easy to use context to pick out which one I need (major, minor, diminished, augmented -- usually one of the first two). If I were supposed to play a C# major triad, though, and the written notes were (C#-F-G#) as opposed to what they should be (C#-E#-G#) then that's confusing because it looks like I should be playing an arpeggiated sus4 of some kind. So the written nature of scales on the staff engenders an understanding of the relationship between the notes. Basically we write things the way we do so that the people reading the music can more efficiently pattern-match.> C major is identical to A minor, and I don’t understand why we need both.They're not identical. C major and A minor have the same notes in their respective scales. But we say that a piece is in the key of C major when it resolves to the a C major chord at the end, and we say a piece is in the key of A minor when it resolves to an A minor chord at the end -- an important concept for reasoning about how a piece is supposed to be performed.> C minor: C D D# F G G# A# CEb, Ab, and Bb, not D#, G#, and A#.> This has got to be some of the worst jargon and notation for anything, ever.It's really not. Keep practicing. It makes sense, I promise.I hear a lot of people -- usually people who have not been studying music for very long -- insist that the system would be more logical if there were no accidentals and there were 12 notes with distinct names and the staff had a bunch more lines on it. I've never bought it. The notation of music isn't arbitrary, it's informed by experience and it works. Reading this over again, it seems like the author's not yet fully wrapped their mind around "big picture" stuff like keys, chords, basic compositional structure, etc. It's an... interesting choice to write a piece proclaiming that musical notation is absurd when you're only a beginner. Is there an article/book/video anyone can recommend that explains the advantages of modern staff notation in terms that beginners can understand?In particular, I think everyone would agree that it's it's much harder for beginners to hunt and peck sheet music in modern staff notation than it is for a novice typist to type words on a QWERTY keyboard. (At least when the sheet music is written in any key except C major, since the configuration of black/white piano keys corresponds to C major.)This prompts beginners to ask, "why is this unnecessarily hard?"The general advice I see (especially in the comment thread here) is to just spend years practicing and then you'll "get it;" you won't just learn how to play, but you'll understand why staff notation is awesome.Can the beginner's question be answered, except by saying, "uh, trust me, it's great, just keep practicing"? I don't think you can explain it to someone who can't play music, because it's a notation for music. You could teach someone to play an instrument the old-fashioned way, person-to-person without any sheets, and once they understood music (i.e. what patterns of notes sound good or don't) ask them to come up with a way to write it down, and they'd come up with something much like staff notation. But that's a very labour-intensive way to learn.Hunt-and-peck is always going to be hard on an instrument that supports multiple modes, because each mode only uses some of the modes. At my school music classes were taught on the xylophone/glockenspiel, which you configure for your piece at the start (putting the correct bars on for the mode/key you're working in), which is probably easier for beginners, but it's not a popular instrument (and nor is the harp, which is the only other example I can think of of that approach). I think a lot of the favoring of staff notation comes down to what the instrument does well. A majority of acoustic instruments are monophonic(voice, wind, brass, many strings) and have some form of linearity in pitch, and the staff accommodates them optimally - the beginner's sheet music always starts out by accompanying the staff with fingerings for each note, to help you get some confidence that you are actually playing right notes, and then after that it's pretty straightforward to extend your knowlege to read more notes.Chord structure just isn't even touched in most of these instruments, at least in classical performance, because they aren't capable of polyphony! That only becomes a topic as one starts moving into, e.g., jazz improvisation, where knowing how to recognize and play around a root is critical. Piano is exceptional here, since finger independence and chording practically define that instrument.On the other hand, guitar tabs are a Big Thing in part because guitar chords transpose very well, so you don't have to learn too many unique fingerings to start accessing the others - and then if the song is defined as a sung melody plus rhythmic chord backing, as a lot of popular songs are, you don't need more than a tab and hearing the tune once to have a shot at covering it competently.Guitar's 2D layout favors isomorphism - this is what gives it this extra power to transpose. Guitar is conventionally not tuned isomorphically, but some forms of accordion like bayan, and alternative keyboard layouts such as Wicki-Hayden, Harmonic Table or Janko, are fully isomorphic. In that case, you don't have to mode set or learn any unique scales or chord fingering per key: Learn any set of intervals(chords or scales) and you can reuse them in every key by moving your fingers over. This is wonderful for learning theory and doing composition, but it doesn't make the instruments a 1-to-1 replacement for similar instruments as in performance the distances are usually much smaller, fingerings can get tangled, it can be harder to find your place or maintain tempo, etc. > Can the beginner's question be answered, except by saying, "uh, trust me, it's great, just keep practicing"?Basically: not really.Musical notation... notates a bunch of stuff, not only pitches.Unless you have a basic understanding of all the factors involved in reading and writing music, you won't be able to understand the choices fully, and you'll have to accept them.TLDR: music notation needs to express the five qualities of sound: pitch, duration, loudness, intention and timbre. You can't just look at pitch. I'm working on a blog post now which I hope will answer that question (at least, in very general terms). The way to learn the logic behind the scales in my opinion is to learn the theory of the early Western music, up to the Baroque era, that was around while the staff was being developed. That is: learn the seven modes of Western music (https://en.wikipedia.org/wiki/Mode_(music)).Each mode is a diatonic scale (https://en.wikipedia.org/wiki/Diatonic_scale). Each mode has a particular combination of whole steps and half steps. Each mode has a "natural" key which can be represented in modern staff notation by all white keys.In Gregorian chant, there are rarely any accidentals. Even by the Baroque era, while there are more accidentals, I would say the music is still very diatonic in nature.Early music was more oriented towards just intonation (the whole integer ratio harmony mentioned in the article, which is what I would consider the more "natural" way of harmony). With just intonation, however, you can't just shift to any random key on the fly. Some keys sound nice and related. Some keys sound awful and horrible.The article author made a statement: "If your music mostly relies on the seven notes from a particular scale, then it’s more compact to only have room for seven notes in your sheet music, and adjust the meaning of those notes when necessary… right?" I think that's an absolutely a correct way of framing early Western classical music, yes.What makes staff notation harder probably is the corresponding rise in chromatic music.The rise of 12TET instruments (equal temperament) like the piano, and the corresponding development of chromatic instruments (compare: the natural horn used in the Baroque era vs. the modern valved horn) allows for this sort of modulation at the cost of some chords being slightly out of tune. Modulating key signature all over the place is now possible. Yes, the author should realize that some composers do change the key in the middle of a song to change the mood. (For a nice pop example, here's the Temptations' "My Girl", which changes keys midway through -- https://www.youtube.com/watch?v=6IUG-9jZD-g)Incidentally, the diatonic nature of pre-19th century classical music is the answer to the author's sharp / flat question as well. (http://jtauber.com/blog/2006/11/17/why_a-sharp_is_not_b-flat...)This shift to chromaticism really isn't reflected in staff notation. In fact, I'm aware that some 20th century composers abandoned the use of key notation, ledger lines, etc. altogether, perhaps in part for this sort of reason. Yes, I almost mentioned this point about A# and Bb on his blog but didn't. He gives an example of an A# major scale and it becomes immediately obvious that something's wrong. In order to write such a scale you either have to skip named notes (A#, C, D.. what happened to B?) or have a key signature that contains double sharps. That would be the "theoretically correct" way to do it but there are no key signatures with double sharps. So the correct name for the scale is B flat, which follows the diatonic major scale pattern of "wwhwwwh" and does not skip named notes.But from his point of view (as any beginner) this is very confusing. They sound the same and they use the same physical keys on the keyboard, so what's the difference? There is no good answer to that except it satisfies the theory (the musical "rules") of western music. I don't think the author was necessarily proclaiming that it's absurd, only that it seems absurd from his/her beginner perspective. He/she was very up front about not having much experience, so I think this is totally reasonable (and expected) If a beginner can't immediately grasp musical notation, isn't that evidence that it is, in fact, absurd? A beginner can grasp musical notation; its rules are simple and consistent, and the conventions are well-documented and reasonably straightforward. This blog post should be evidence of that: the beginner who wrote it clearly has the basics of notation down, and there are no inaccuracies at least with regard to the notation.A beginner can't grasp the motivations behind the design of musical notation, and I don't expect a beginner to grasp the motivations behind the design of any complex system. That does not make the system absurd. Maybe the notation is optimized for efficiency for experts, at the cost of beginner-friendliness. Maybe it's representing something genuinely hard. Absurdity is possible but not the only possible reason. Ah, the hipster programmer attitude to learning, based on the idea that because something is:a) hard (it takes more than a day to understand the complexities)b) incredibly oldthe community should declare it obsolete and immediately begin work on a replacement (preferably based on JavaScript). If a beginner can't immediately grasp mathematical notation, or a programming language - or a native language for that matter - is that evidence that it is, in fact, absurd? The musical notation has evolved (toward some local optimum) to be useful for practicing musicians and composers, at the expense of having a learning curve. I'm curious, did you immediately grasp the syntax of every programming language you've ever learned? Programming languages are a lot easier to grasp. ;)The oldest programming languages are only 50 years old, whereas the oldest music notation is at least 4000 years old.Most programming lanuguages haven't crossed any spoken languages, but modern music notation & terminology has been heavily developed by countries all over Asia and Europe.This is a big reason why musical notation seems so weird at first, especially to engineers, because it is a legacy that comes from a different time, a different context, in a different language. The people who developed musical notation had different math, different logic and different musical motivations than we have now.Think about this for a while and it starts to feel like a miracle that musical notation works at all, not to mention how well it works.Programming languages were developed by people nearer to us in every way, and made to be logical and simple, so it makes sense that they're easier to grasp quickly. Can't speak for the parent, but as far as I can recall, I did. Well, maybe not literally immediately, but to me, the syntax was generally the easiest part of learning a programming language. (It probably helps that they are often very similar to each other.)The only things that were hard were forgetting to write semicolons after statements in C (I have previously mostly written Pascal), and C declarators — but even the latter were easy after I learned that "declaration reflects use". Yes, but did you learn your first programming language immediately? Because that's closer to the author's situation. Well, in our programming world new languages seem to come out every week, trying to find better ways to write programs. By this standard a new way to encode music is long overdue. In that metaphor, coming up with a new way to encode music is like releasing a new programming language and then asking everyone to write their own compiler for it. Not impossible, but it would have to be a hell of an encoding! Maybe not as many as programming languages, but new notations do appear time to time. It's just that none has ever got as popular as the traditional notation, I guess. So if you can't read Italian after a half-arsed effort to learn it, is this proof Italian is broken? Not as badly as English is broken.I mean, it's generally agreed that English as a language has accrued so much historical baggage that it's shed any elegance or coherence it may or may not have once had. It would honestly be surprising to me if the same thing didn't happen to musical language. > it's generally agreedBuy people who know nothing about how languages and linguistics work, I can only assume. well a beginner can't really grasp shorthand or kanji either, right? A beginner not being able to get things means that the learning curve is higher.Most musical notation is meant for non-beginners, right? And a lot of the beginner confusion is from features that are useful for advanced users. So there's a tradeoff.There might, of course, be changes to be had. But a lot of beginner confusion is because the higher-level abstractions are needed.A good example in mathematical notation is order of operations. Why not just do left to right? Or just going with Polish Notation?Standard order of operations work well for people working with large formula, because they allow you to write many common things without many parentheses. But mandatory parentheses + left-to-right only would be easier for beginners. If you are mathematically literate but a beginner in music most texts on music won't tell you the simple stuff in here. And most music students can't tell you what an octave means (i.e. frequency doubles) besides something wooly like "(a C an octave up from another C) is the same note but a different pitch".This text is what it claims to be: music theory for nerds (just not necessarily music nerds!) Yes, it's like some of my students who wonder why I won't let them use a GOTO in their programming assignments.I tell them that, if they are using assembly, then it is fair game. It can also be worth it, in very specific limited circumstances, when you're emulating constructs not present in the target language.Emulating a multi-level break statement and error handlers come to mind as examples. Not to mention that A minor is generally considered to contain two additional notes compared to the A natural minor scale (AKA the A Aeolian mode). Well, two _or_ one. The melodic (two) and harmonic (one) minor scales. Sort of. They are separate scales, but if you compose (classical-style) music in a minor key, you don't typically pick one of those scales. The chords that appear might include any of those notes, even within individual phrases.In my totally non-expert observation, contemporary music often is more modal, in the sense that it probably would just pick one of those scales (particularly natural minor) and stick to it. In jazz A minor generally implies A Dorian Yeah, in my reading, it never seems that there was a compelling reason to chose Aeolian vs. Dorian as the definitive "minor". I think it's largely because common practice evolved to much more frequently approach the tonic from below instead of above. That would favor harmonic minor over Dorian. I suspect that this is because it allows for very similar cadences as major keys. I think a lot of the time (in a jazz context) the chords are approached as almost individual elements or as part of a small "local" progression instead of in the context of the global key of the song, since the key could be changing as often as every couple bars. In the case of a common minor progression: B-E7alt-Am7 the major third of the dominant chord is the same note as the major 7 in the harmonic minor scale of the target minor chord, which would be one option for playing straight through that progression, but you'll often find that players will use either A harmonic minor, diminished, E altered, G# whole tone, etc. over the V7 and resolve to dorian or aeolian or something with a b7. It doesn't make much sense to analyze that progression in a classical context since those chords aren't entirely contained within any single key, and you end up something that looks like a key change, or has B as a secondary dominant implying a key change to the E7, where a jazz player would immediately recognize it as a minor 2-5-1 That's wild. Jazz blows my mind!I've heard it said that it's fundamentally actually very tonal, in the sense that the underlying progressions are typically 2-5-1. Which has always been hard for me to imagine :), but I guess it simply dispenses with the slavish dedication to a particular scale of the tonic. "A Geometry of Music", which I've linked to elsewhere in this thread, digs into jazz practice a lot. > we say that a piece is in the key of C major when it resolves to the a C major chord at the endCan you ELI5 that for me, please? I'm very interested to have a better understanding of that concept. Broadly speaking, major chords sound bright and happy. Minor chords sound dark and sad.The C major scale consists of C-D-E-F-G-A-B-C. The A-minor scale consists of A-B-C-D-E-F-G-A. Those are the same notes, but if you play each of those patterns on a keyboard, the first one sounds happy, the second one sounds sad.A C-major chord (technically, triad) is made up of C-E-G, the first, third, and fifth notes of the scale. Again, sounds happy. The A-minor triad is A-C-E ... sad.But if you play a melody -- that is, one note at at a time -- it isn't always clear whether it's happy or sad. In most western music, though (including virtually all pre-1900 classical music and the vast majority of modern pop), the piece will end ("resolve") with a clearer "happy" or "sad" type of chord. That final chord is what determines the key.(In a huge amount of classical and popular music, the final chord is the same as the opening chord, but not always. When they're different, the final chord tells you the key.) > the final chord tells you the keyThat makes it sound like a definitive rule, but it's just a common convention. It is just a convention, not a rule. It is very common in classical music for a piece in a minor key to end on a major chord (called a Picardy 3rd) or in some cases to end on the dominate 5th - which is a major chord. The latter though, isn't usually the absolute end of a piece because it leaves you hanging (rather like ending "Happy Birthday" on the word "to" - and leaving off the "you"). Resolving to a chord just means that the music sounds "complete" when it hits that chord. In other words, it wouldn't sound weird if the song ended right there.For example, if you hear "Happy Birthday" played in C, the final chord is C major, and the song sounds done. If you heard it with the final chord changed to something else, it would sound like it was leading somewhere, and you'd expect another verse or a bridge or something to follow. Just as it can be said that music requires at least 3 consecutive notes or beats to have a tempo and a rhythm, it can also be said that it requires a "cadence" which at bare minimum is two consecutive pitched notes but usually at least three.C-G-C is a simple 'harmonic' cadence. It is, for example, the basis of oom-pah music (think military marches which repeat C-G in the bass with a melody on top, and end on C when the melody is done).There is a form of musical analysis (Schenkerian) that can be used to show that nearly all music has the basic form C-G-C, which is usually expressed as I-V-I (chord I is C major and chord V is the 5th chord of C major, which is G major). It is of course nowhere near as simple as I've expressed here!So resolution is basically the cadence of the piece's harmony, much like a story has beginning, middle and end, so does harmony.The reason chord V and chord I are so powerfully related is that the 3rd note of V is the leading note of chord I and the 7th note of V is the 4th of chord I. This is actually significant to understand because it is how one can see the purest relationship between harmony and melody. In early music (eg medieval plainsong) the fundamentals of more complex harmony were first developed from melody.In the case of C major with two voices, the notes would provide resolution from notes F and B to notes E and C. The FB is a tritone, which is dissonant and wants to be resolved in the human ear. The EC is the major triad of C major, which is harmonious and consonant and resolves the dissonance of FB. Many musicians much better than me are surprised at how I can play a song just by hearing it on the radio. My breakthrough came from understanding music was realizing that the real “meaning” of a note lies in its position relative to the tonic note (e,g, I-II-II, etc, also written do-re-mi). Suddenly, almost all of the clutter was removed, and the problem became manageable.Let's consider the three-note tune “do, re, mi”. If that tune were played in the key of C, it would become C-D-E. If it were played in G, it would become G-A-B. But in either case, it's the same tune but with each frequency increased by the same percentage.Trying to understand music by understanding the letters is like trying to read in a world where every article has been enciphered into a different “key”: e.g., the word "cab" in “the key of A” (the alphabet we normally use) would be written as "dbc" if the article were written in “the key of B”. In the latter case, you could discern meaning only once you realised that the letter “d” represented the third letter of the alphabet. There's nothing meaningful about a “d” but there is something meaningful about a “4th letter of the alphabet”.Once you start to “decipher” all music into I, II, III, IV, V, etc., the complexity becomes manageable. You can start to learn to recognize the sound of a III note, or of a VI minor chord. After all, there are only eight notes in the major scale. I'm not a good musician by any means, never had music education apart from the primary school which was abysmal. I can't recognise pure tones (just the intervals). I still can play any song I hear on guitar or keyboard "good enough" so that people have fun singing to it.The huge reveal to me was the same - notes doesn't matter - the intervals make the song recognizable. People change notes all the time when singing (jump octaves, start again lower to adjust to others, etc).So on amateur level it's really just starting on random place on keyboard and guessing which note will sound "right" after that. Everybody hear if the next note is higher or lover, so it's just "was that +1, +2, or +3?" Usually you can guess, if not - start again. Very easy and makes playing instruments so fun.I never understood why they bother kids with these complicated drawings and hashes and be-mols, if they could've just wrote all songs as "start at this note, and jump by +2, +3, -5, ...". Some notation systems do exactly that. For example, the notation used for byzantine chanting: http://www.byzantinechant.org/notation/Table%20of%20Byzantin... Many (but not all, or even the majority) music classrooms around the world teach students fluency in solfege (do-re-mi), and everyone I've spoken to about it agrees it helps out a lot in the way you describe. That's interesting from a personal perspective. I was taught music in the UK where that is specifically not used (or at least wasn't when I learnt).I took a year out of my Music degree to attend the Sorbonne. French music education places a heavy emphasis on solfege. When I started going to their undergraduate classes it was immediately apparent that the level was several years behind that of the UK (in classes for composition and orchestration most noticeably). To attend classes dealing with similar material to what I was used to as a UK undergraduate, I was attending Post-grad courses. Having just completed my first year on a UK BMus course it was quite an eye opener to see 19yr olds learning material I had been taught at 16. I wonder if the UK system is better at sieving natural talent whereas the French system is better at teaching? Personally I hated music (in the UK) because we never seemed to get taught anything, we we largely expected to just know things or magically learn through awkward repetition.Knowing what I know now, I think there are a lot of ways we could have practised music early on that would have helped those of us not born with perfect absolute pitch. Most people have perfect relative pitch (afaik I fall into this group). Perfect absolute pitch and tone-deafness are both quite rare.It sounds like the French system is optimised for the majority, forcing everyone to practise interval differentiation, including those who don't need it (and the small minority who will never be able to do it). Without trying to cite what currently happens in the UK, I think it is fair to say that kids are handed instruments like the recorder at a fairly early age, and those that do well are encouraged to progress. That's a sieving process for sure.When they get to secondary school, music as a subject is most often just a minor inconvenience in the curriculum to most pupils, and those who have ability are pushed into learning flute/violin etc (at additional cost to themselves and outside of the timetable). In the course of going through the grades of music (performance exams), kids are taught aural skills and theory (grade 5 theory is required to take higher performance grades). This results in a select group of instrumentalists that have learned intervals, harmony and scales practically. Whether any of those skills are useful to a non musician is debatable, so one could say that it is the most efficient way of getting a rounded skill set into the brain of a musical 15-16 yr old.The French system would, I agree, produce a broader spectrum of musically able people, but in practice it results in a lower level of specific and important knowledge. The UK system produces more complete performers whereas I would say the French system has large gaps which then get filled in at degree level.Perhaps things are greatly different these days. I know for example that studio production is an option for A level music, and there is absolutely no musical theory knowledge required to produce a studio track. I wonder if A level students are even taught basic 4 part harmony any more.Music, to an individual, has always been a matter of ability, discipline and perseverance to practice. In education, the solution to nourishing those qualities is never going to be perfect. I do recall the French students I was with were quite annoyed that my education was years ahead of theirs, but with perspective, I'm sure it didn't really matter then, and it surely doesn't now. It's also interesting to compare and contrast the apparent results. The UK has a fairly long history of producing a far above average number of world-class musicians. France on the other hand seems to have a broader musical culture, or at least, so it seems in Paris during the Fête de la Musique. I'm a very good musician and have never thought about it exactly that way before but I have to say, that's a very good explanation and way of thinking about it. I think it's a bit abstract for someone trying to learn to read music, but it's absolutely correct. The gist is correct, but it's worth noting that, for example, if you played do-re-mi in E it would do E-F#-G#, not E-F-G, to preserve tone/semitone order. That's where the key signature comes in. And that shows how WTF the musical notation is. o.O why so? And what on earth do you propose? C is 0. Each half-tone up is +1.If you really want to keep it concise you can write it as base-12 numbers.`````` C1 C1# D1 D1# E1 F1 F1# G1 G1# A1 A1# B1 C2 C2# D2 D2# E2 F2 F2# G2 G2# A2 A2# B2 ... 00 01 02 03 04 05 06 07 08 09 0a 0b 10 11 12 13 14 15 16 17 18 19 1a 1b ... `````` First number is octave. Second number is half-tone in that octave. Translating is just mechanical addition.EDIT: on second thought making it base-12 just to save some space makes no sense, people are good with base-10, just keep the numbers. It looks better on paper for an engineer, but not for someone who actually plays an instrument and reads the notation. With most western music, not all 12 pitches in an octave are used most of the time, but only a subset determined by the key and scale. Although the currently used notation may look weird for a newbie, it takes just a quick look at the key signature and you know which pitches will be used in a piece of music. When you know the scale (and practicing scales is just a standard part of learning), then "decoding" a note by counting tones is much easier than counting individual semitones (12 seems just too many). After a little practice you get it intuitively and you really don't count; you just know where each tone (or chord) is in a given scale and what function it has. And then when you suddenly see an additional flat or sharp symbol before a note, you know that this is an out-of-scale note, so it is also easier to play it. Disclaimer: I'm an engineer. That completely misses the point. Without wanting to sound brash, you know jack about music and you should act accordingly, that is, don't spout ill informed suggestions when you clearly don't play an instrument or studied this mater well enough to give an informed opinion.Scales have 7 notes, not 12. A musician plays music in a scale, they aren't a computer outputting pitches, they are a person playing notes. Music notation has a reason to be this way: notes in the scale don't have flats or sharps next to them, accidentals do. Reducing everything to a number describing absolute pitches is the right thing to do for a computer to play (see midi). It's not the way to go for a person that actually has to understand the logic and patterns in the music. What would D Flat be, 0.5? what about D Double flat? Db would be 01. Dbb would be 00. (or 11 and 10 for the octave)The point ajuc is making is that the flat-sharp accidentals aren't used or needed at all if you just assign numbers to each tone. There's no concept of flat or sharp, unless you want to deal with microtones. The letters and flats/sharps give you key/value over frequencies, which is better than just a numeric index over a chromatic scale. Working with the keys allows for the same abstractions to be used with all 12 keys at the same time, on the same staff. I'm not really arguing one way or the other, just pointing out that they're functionally equivalent. It's two different maps keying to the same set of frequencies. The BASE12 system described above would allow for keys as well, just with a different notation. Instead of flats/sharps marked next to the clef in traditional sheet music, the BASE12 system could start each line with a list of 'prohibited' notes. For example, when indicating that a piece is in the key of Gm, the staff in BASE12 could start with : [01,04,06,08,0b]. This indicates that the majority of the song will be made of the notes 7,9,b,0,2,3, and 5, and serves the same purpose as having two flats next to the treble clef, one on the middle line and one in the top space. Which of these systems would be easier for humans to grok is up for debate. I personally don't think either one is better.Thinking about it, the traditional notation is just mapping to an octal system, with the key accidentals acting as modifiers to the map and the base-8 values being displayed graphically as vertical position on the staff. You need to stop thinking about things in engineer terms. The scale going 7,9,b,0,2,3 instead of G,A,B,C,D,E,F with B and E flat may be functionally identical, but one is sure a better representation of the actual patterns in the music than simply referring to a number proportional to the log(frequency). I started music with classical piano and only recently started transposing instrument (bamboo flute), and am trying to get used to movable do solfege. I'm still easily confused when key change occurs or borrowed chord appears. (For the latter there are limited patterns so I should remember them, I guess.) @karb: mind if I shoot some questions about the "decipher” all music into I, II, III, IV, V" part in email? Thanks. Indeed, the scales are isomorphic. This has got to be some of the worst jargon and notation for anything, ever.Indeed. I'm a musician, and something non-musicians often ask (especially techies, it seems) is why we use such an archaic notation system.The reason is simply that a certain number of musicians have developed the skill of sight reading which is the ability to perform a composition directly from a written sheet, with little or no rehearsal. Those players, myself included, can't quite explain how we do it, and aren't going to learn a new notation system. It's not merely sight reading that relies on the existing notation, jargon, and theory.Every element of Western music builds on the same building blocks. Western harmony, as taught, relies on understanding chromatic and diatonic harmony; the author did a nice job figuring out and explaining the compromises inherent in equal temperament (without knowing about how ratios and tuning worked before equal temperament; people didn't always fudge the ratios to make them work out the same across all keys), which is cool. But, he still doesn't actually know much about western musical harmony, as evidenced by the assertion that C Major and A minor are "the same", because they share a key signature and the same notes. Had he known that equal temperament is a moden-ish invention, he might have also figured that C Major and A minor actually have (slightly) different notes if your instrument is tuned specifically for that key rather than with equal temperament.In short, he's just not done learning yet. Most musicians and composers never really are, as it is a vast subject. And, what he's calling "music theory" is really more "equal temperament tuning math" with very little theory.Not to say it's not interesting and well-presented. It may be useful for non-musical nerds to see it presented in this way. Very important historical details that are missing completely from all music theory textbooks I have had the misfortune to read. They only give the rules, without explaining the reasons. Also, imagine using notation to write down, say, electonic music. Drums? Maybe, but which notation? Filters, effects? ugh As a percussionist, yes, musical notation is used for drums. However, it ends up looking rather like a step sequencer with some standard notational effects (staccato/legato, rests or varying lengths instead of the absence of markings, etc.) This helped me out quite a bit later when I started getting into digital music. This is true for me. I can go from notation to my fingers, faster than I can go from notation to singing a tune. It's a hyper specialized skill. I can sight-read treble clef on flute, and bass clef on the double bass, but it's hard for me to sight-read treble clef music on the double bass, even though the transposition of a couple octaves is no problem at all. You can almost guess what instrument the composer played because of that bias. If you see a violin piece with 6 flats, it's a pretty good guess that the composer didn't play violin. > The reason is simply that a certain number of musicians have developed the skill of sight reading ...It's moer than just that though. The inertia to overcome also includes pretty much all sheet music ever printed, all the schools that have adopted the current system, all the educational materials... etc. etc. the list goes on.In addition to all of that, the new system needs to convey most of (if not all) the information that the current system has. I have seen a few attempts at improvement, but they all fell short of the current system, which let's face it has been in development for several hundred years. Once we get off printed paper, computerized notes should solve much of the inertia. problemDisplaying a song in whatever notation the musician/reader prefers should be as simple as setting a preference on your hyperPad.That still leaves the difficult task of figuring out better notations, but it can be done incrementally. You don't need to convert the whole world or the entire notation at once. > Once we get off printed paper, computerized notes should solve much of the inertia. problemI don't know...ebooks still look a lot like treebooks.The one example I've seen of what you're describing is I've seen jazz musicians play off ipads. This give them access to very large catalogs without lugging around giant binders, and also they can transpose their sheets into any key.(That's more important than it might sound at first, because different instruments "play" in different keys -- if a pianist thinks the piece is in C major, the clarinetist thinks the piece is in Bb major. So now you don't have to have separate books for the different instruments in the band.)But the music as written down in these electronic fakebooks is a lot less complex than your average classical piece, so it's a much more tractable problem. I have those electronic materials. They come in two varieties. There's something called iRealB, which contains thousands of tunes, but is just chord changes due to a quirk of copyright law -- the melody and lyrics are copyrighted but not the harmony. That's the one where you can transpose tunes. But you don't get the melody.Then there are large PDFs where somebody ran the old paper fake books through a scanner, but they are just images and are not in an actual computer readable format, so they can't be transposed.So the computer has not solved the notation problem, yet.Myself, I've memorized most of the standard jazz repertoire.In fact I suspect that most of the written music repertoire that exists today will never be translated into computer readable form, because it's just too much work. And not enough new music is being composed to form a critical mass around some new notation system or computer format. When somebody composes a tune (I play in one band that does original jazz compositions), they send out a PDF.Maybe software will eventually automate the process reliably enough to be useful. Notation would tend to be one of the easiest things for which to develop automatic recognition / parsing? It should be, but there's the problem of demand. Somebody has to be motivated to do it, meaning that they are probably equally passionate about music and image analysis.Also, a lot of the written stuff is handwritten, not typeset. So it's a subset of the handwriting recognition problem.It's pretty bad for the bassist, since the bass part is usually the second to last part to be copied, meaning that the copyist was probably drunk. ;-) Just a nitpick, but the transposition is the other way around. Concert B flat is C for B flat instruments. Argh. In my defense, I play a C instrument. :) >why we use such an archaic notation system.The answer is, because it works, and nobody has managed to propose a better one. > nobody has managed to propose a better oneA new notation system will need to be a lot better to justify the change, because there is also a lot of value in compatibility with everything that already exists.I'm not sure a sufficiently better system exists, because as you say, the traditional notation works. It has its quirks and rough corners, but music is complicated enough that any system would probably have similar imperfections. One attempt was Hummingbird. http://www.hummingbirdnotation.com Which in durations are represented spacially, which in my opinion has two negative effects.The first is that spacial recognition takes more effort than symbol recognition, because it's comparative.The second, and more important being that complex sequences of notes will be very dense on the page, and simple sequences of notes will take up a lot of space on the page, so suddenly there is a tradeoff between having sheet music that doesn't take 10 pages, and having enough space to represent hemi-demi-semiquaver sequences when they inevitably appear somewhere. Yeah wtf. Like I don't think current notation is perfect by any means (and I am a PLer, I love new representations) but this and every other replacement I've seen blatantly sucks. ....As with so many things, know what your disrupting![In this case, I'd like to say go apprentice engraving if you are 100% serious.] With Hummingbird's inclusion of a trailing line as part of the indicator of note duration you also have to read ahead in the score and then jump back if you want to use the trailing line to identify the duration, which is a lot harder than just identifying by local information in the form of a set of note tails and whether or not a note is filled in. While there are additional parts to the glyphs for half and whole notes, these aren't the dominant part of the symbol.Another question is whether or not the length of a trailing line is absolute or relative to the bar itself.On the subject of filled notes, Hummingbird is also conveying a lot of information that for performance of a score is useless. Note letters (A-G) aren't actually important for performance, only the action or position that they map to for each instrument. No musician parses a score and translates each note to a letter and then each letter to an action, instead going directly from note to action.Essentially, telling you the note letter with a glyph shape on top of the position on the stave is adding noise to the signal.I'll admit I'm coming at it from a position where I'm perfectly comfortable with traditional notation, so part of the reason that it appears difficult is simply because it's unfamiliar, however the terseness of traditional notation and ability to read in one "parse" without forward- and back-skipping seems to give it the advantage.That and over 300 years of existing music, too ;) Your second point seems pretty valid, and to address your first point:"There are multiple cues to the same information. Everything has both a symbol and spatial element, for all kinds of thinkers."Indeed, in addition to the spatial length of the notes, there is also a symbol next to the notes denoting their life. You can see this on the linked page, in the second section. (Next to "Intuitive.") Try writing hummingbird notation by hand though! Their pages claims: "It’s quick and easy enough to write with an unsharpened pencil. You can scrawl it on a napkin in a pinch."I don't know if it's true or not, as I view Hummingbird to be fixing things that aren't broken (is the difference between a whole note and a half that hard to suss?) and doesn't fix the things that are. I think the more difficult part writing by hand would be the redundancies -- not only do I have to know which line to put the note on, I have to know which symbol to draw. Which could be tough, as it forces the composer to be consciously aware that e.g. "this note is D#" rather than "this note should be two steps above the previous note in the current key". The thing I most want to change is how time signatures are notated: I'd like to specify branching factor all the way down the rhythm tree. That works, and would be an improvement, so long as your rhythms are based on constant integer subdivisions.It's possible (and fun) to play music where the rhythmic structure changes smoothly and continuously. But it's bloody impossible to notate and very difficult to orally communicate, so music cultures that depend on notation or oral communication have left this territory largely unexplored.The handful of classical composers that have attempted this (Steve Reich, Brian Current, etc) have either abandoned western notation (Reich) or hacked on their own bespoke glyphs with their own situation-specific explanations (Current).Electronic musicians can easily explore this space by writing their own software (Autechre, your humble author, etc). When your musicians are mechanical, you can explore all sorts of otherwise impractical permutations of theory.Most delightful, though, are the non-western cultures that communicate musical ideas entirely without written notation or spoken language, and instead communicate musical ideas through play (Indonesian Gamelan, Australian Aborigines, etc). You get the ineffable human qualities that make music most beautiful, and the freedom to explore structural spaces that are difficult to capture with discrete/unitized/quantized notation and language. > But it's bloody impossible to notate and very difficult to orally communicateIt's really easy to orally communicate. You can just sing it!(I know what you mean, is it's difficult to describe using a computer keyboard. Using a pen and paper it is really easy, given that you can just draw notes and time signature changes in the margin.) I'd like to specify branching factor all the way down the rhythm tree.I'm not really sure what this means, but it sounds interesting. By default the next note is half as long as the previous. But sometimes that's inconvenient. For example 9/8 is usually 3-3-2-2-2-2-2-2-2... A bar is devided into 3 dotted quarters and a dotted quarter into 3 8ths, and anything further down is split in half as is normal. Oh that sounds great, but very difficult to read/play Sometimes when there's a bunch of triplets the editor will just write "simile" and drop the triplet notation.This could work like that and only the time signature would need a brand new notation, so no I don't think it would be super hard to read. Triplets are tough to play haha And then most musicians will have to learn both notation systems. Have you played with trackers?2d grid, time is down, instruments(channels) are right, each cell can have a note or "stop this channel". There also can be modifiers in each cell (louder, start tremolo, slide this note into the next, etc).It's most applicable to keyboard music, I think. Sight-reading this without rehaersal would be trivial, right? For better or worse, the only way to guess if this notation can be sight-read, is for someone to train themselves to do it. And to make matters worse, learning to sight-read for adults is so hard that it's virtually prohibitive. Every musician I know who can sight-read fluently, learned it as a kid.So it's virtually impossible to try out a new notation system.But my impression is that this would be phenomenally hard to read, especially in a live performance situation where your attention is divided between the sheet music and other stuff. If I were staring at a solid grid of text, and were to glance away for a split second, I'd be lost. Part of sight-reading for me is being able to read ahead by a few notes or even a few bars.It may also be that conventional notation displays a lot more density on a single page, because a 16th note takes up no more space than a whole note. The main problem I can see with that sort of system is that it probably would get quite unwieldy for decently complex chord patterns (try scoring a chord pattern up the keyboard, say, like what's at the beginning of Tchaikovsky's Piano Concerto 1 for a start -- http://imslp.org/wiki/File:PMLP02744-Tchaikovsky-Op23v1FSmuz...).For other styles I think it would be pretty good; although the score won't be as compact, it might be easier to understand a "fake book" scored piece (ala what's used a lot in jazz) written this way, let alone pop (which often can be represented with a melody line and Roman numeral chords). It's a lot like why the QWERTY keyboard remains dominant. Yeah, it would be really hard to change all the millions of documents written with QWERTY. It'd still probably be hard to convert thousands of existing computer keyboards to anything else. Every keyboard at every school, office, home, etc. And it'd be harder still to convert billions of existing people to another layout. Inertia's hard to overcome, even if another keyboard layout or music notation system had ever been convincingly proven superior than the respective dominant ones. No, it's not the same. Current music notation is a result of long evolution, unlike QWERTY. Try to invent usable alternative - then you will appreciate the convenience of current notation. What are you talking there? Qwerty is indeed not that old as the musical notation but it is ~150 years old, and it was invented for the typing machines in order to avoid jams, so it does have some history and evolution [0] "In order to avoid jams" is a popular story but it's credibility has been challenged [1]. It may be worth noted that Qwerty was made before people used all fingers and touch-typed---they were likely to use just two fingers and typing speed might be quite slow in today's standard.[1] http://gizmodo.com/qwertys-origin-story-is-a-big-fat-lie-493... (This is just a first one I found in English, but Prof. Yasuoka referred in the article publishes quite a few articles about early typewriters in Japanese, in which he lists several evidences that "jamming" wasn't the reason.) This has got to be some of the worst jargon and notation for anything, ever.I'd like to argue that as inefficient as it seems, it's really a reflection of how broad and complex music actually is. This notation and organization system is all a perspective or reference to take when actually trying to comprehend/play music. It's not a fixed set of rules and there are exceptions everywhere.It's one of those systems where you learn the rules only to know when you're breaking them. So by all means learn to do this, but don't get hung up on real-life deviations.It actually works well for people performing together, especially when led by a conductor. I think you sum it up pretty well. As a techie who has never understood music, and who is fascinated by watching my own kids learn and understand music theory (something I was never exposed to as a child), I found this an interesting article.Clearly history, culture, notation and reproduction technology have all conspired to produce a certain flawed, but accepted, jargon that you just have to bite the bullet and learn. If we could start again using colours, numbers, augmented reality, etc., then it seems obvious we could come up with a better system, but that would be tantamount to proscribing a new alphabet, or telling everyone to start using base-16. "2. Learn your intervals, and learn Solfege. If you're a musician, you already know these things in principal, even if you don't know the words or the terminology. In particular, you should be able to "hear" the distance between Do-Mi-Sol without too much difficulty, because you hear those distances in music a lot."I believe the Solfege named intervals (Do-Re-Mi-Fa-So-La-Ti-Do) are only taught as a historical oddity these days (or maybe used more in classical training?). All of my musical instruction, at the high school and college level, used numeric interval names (1-2-3-4-5-6-7-8). Most of the serious musicians I've played with also used the numeric scale rather than Do-Re-Mi. We learned how to sing Do-Re-Mi, "Just in case", but we never used it.Were you taught in the US, or somewhere else? Maybe it is a regional thing. I was taught in the US, and Solfege was used prominently in my voice classes, but rarely in my theory and piano classes. I think it's a system that works well when you don't have to think about the actual note name that you're singing, as it describes intervals very well, but relates somewhat poorly to pitch.I grew up attending a Church of Christ, which used full congregational singing with four part harmony. Our songbooks used a variation on music notation that used shape notes, and the shapes corresponded to Solfege. If you knew your musical intervals, you could completley ignore the key signature, because a Do was always drawn as a triangle, a Sol was a circle, a La was a square, etc etc. This was especially handy because the song leaders were always men, and could not always sing as high as the written music required. They picked whatever key they could sing comfortably, and the congregation adjusted to them. Drove the music majors in the audience nuts. :Dhttps://en.wikipedia.org/wiki/Shape_noteGrowing up with that system meant that Solfege was simply the easiest system I had to understand music. To this day, I struggle with pieces in unusual modes, and with passages that modulate their key and make use of unusual progressions, because it breaks down my innate understanding of music and requires me to think in a different way. Shape note seems really clever! But, I can also see where it would break down in many cases. I learned music predominantly in a Jazz context. That'd be very tricky to use for jazz...especially the various modal types of jazz. Solfege/solfège is still used prominently in France, or at least it was ~25 years ago. I took music lessons when I lived in France as a teenager. My piano instructor was very confused by the fact that I spoke good French, was at a solid intermediate level at the piano, and yet I could not at all follow his solfège commands (like you say, I had only used numeric intervals back home, although teachers usually called the note by its alphabetic name, like most here I think).I tried reviewing in my head each week before class what I remembered from the Sound of Music, but made the mistake of thinking that Do-Re-Mi etc. was a static C-D-E instead of realizing that my music instructor was simply describing intervals depending on the key we were in.In the present day, I would just look it up online, but back in 1991, in a small city in northern France, I didn't have that privilege. It took me several months of twice-weekly instruction before I finally figured out that he was using solfège for intervals, I'm embarrassed to say (to my instructor's frustration and confusion). I'm wincing even now when I think of it.I asked around at the time and was told it was pretty universal to use solfège there.I do think using solfège to indicate notes is a much better system for students, since it emphasizes the importance of intervals and keys. It's probably harder at the beginning that just learning static A-B-etc., but worth it. Solfege is definitely still in strong use, especially in public school choirs. They are convinced that it helps in sight-reading competitions (yes, sight reading is one of several areas in which a choir can compete).Having said that, I absolutely hate solfege. But my bachelor's was in piano performance, not vocal.Interesting note: quite a few countries use a "fixed Do" system rather than "A-B-C". It is quite confusing (and humorous to observe) when a solfege disciple tries to sing with a fixed Do native. Fixed Do sounds awful! I just spent some bit of time reading the wiki on Solfege, as I realized I have a limited view of it. Interestingly, there are additional syllables beyond the 7 I learned! There are also syllables for flattened and raised notes, which is really nice. I have always been bugged by saying "flat three" or "minor three" when singing intervals, and it's hard to make the voice actually make it minor (for me) because of the muscle memory for three being so firmly set.So, I may have to somewhat rethink my dismissal of the solfege, at least for singing intervals. Unless there's a secret system for singing with intervals by number that accommodates accidentals. As a computery person, what bothers me the most is the 1-based indexing of intervals. I get that music theory predates zero, but it makes it incredibly frustrating to work with (e.g. add a third to a fourth and you get a... sixth). When do you need to do math like that with intervals?As a computer nerd myself, I can understand the argument for 0-based indexing in this case, but I don't recall ever being stumped by it being 1-based. When would you need to add a 3rd and a 4th to get a 6th? Harmonic theory doesn't use addition like that. e.g., playing a 6th is not the same as playing a third and a fourth. So, why do that kind of math with intervals? Whenever you play three notes in sequence, no? I'll read a passage and think "tonic, up a third, up a fifth so that puts me up to the tonic again... nope." Huh. That's interesting. In a "people's brains think surprisingly differently sometimes and we rarely think about those differences when the resulting behaviors look the same" kinda way.I mean, I guess that's not so foreign...But, I tend to think of it as pulling out the notes I need from the scale, and not actually counting up to them. e.g. in my brain I'm grabbing the third and the octave (well 7th, if you've got a third and then the fifth of that third, which I guess is why you're preferring 0-based) that are already there...not climbing up them to find there's the tonic there. I mean, I can see that it's a fifth interval if I go from E to B (in C), but unless I'm building a chord on E, I don't care..it's either the 7 in C, and I'm not so much thinking of its relation to E as I play it, and it's a phrase in C, with maybe an Em chord (either implicit or explicit) underneath; or I'm playing jazz, or some other very chord-based music, and I want my phrase to be relative to the chord we're currently playing (so we're inside that Em, and the key is less relevant).Sight reading is different, as well, in my brain, but, I think it even bypasses the intervals to some degree and is just distances and shapes and an awareness of the key I'm in. I don't read much these days, but I recall it working best (or at least fastest and most accurately) when most of the theory was turned off in my brain and I just let the shape of the notes (their distance from each other) guide me. But, I feel like it's only in improvising and composition where one would be doing any sort of interval math. But, maybe I'm wrong.When are you doing this kind of math? When reading, improvising, playing memorized pieces, or composing? The main context I was thinking of was learning a new piece, particularly the initial read-through of something I haven't heard - I sometimes try to think what it "should" sound like ahead of playing it. > (e.g. add a third to a fourth and you get a... sixth)Better:In equal temperament, log base two: add 4/12 to 5/12 and you get 9/12.In ratios (depending on tuning, these could be loose approximations): multiply 5/4 by 4/3 and you get 5/3. Yes, I know. But having the log base 2^(1/7)-ish built in is useful. If we could just subtract 1 from all the numbers (i.e. what we call a fifth should be 4) then we would have a measure that actually made sense. Eh. Calling it a “fifth” is making it clear that the label is an ordinal number: first second third fourth fifth ...You need to think of it as “if the bottom note was the first note of a scale, where in the scale would the top note be?”As with many questions of indexing, off by one errors are tricky.It’s a system that confuses names for notes in a scale with names for intervals between notes. You’d rather they called them by cardinal numbers representing some kind of “distance”, instead of a count starting at one.But that would be applying a later mathematical understanding on the earlier system. If that’s what you want, you should just use a log scale and count twelfth roots of two.Ideally we’d switch all our indexing to start at zero, and use half-open intervals everywhere. Start at 0 AD, call the ground floor of a building “0”, start spreadsheets with row 0, switch Matlab to index from 0, et cetera. This is pretty unlikely to happen though. > If that’s what you want, you should just use a log scale and count twelfth roots of two.I don't want to count in twelfths, I want to count up the scale.> call the ground floor of a building “0”I'm a Brit, we do that here already. The intervals in a 7-note scale inherently don’t add up like that, because they’re not based on even divisions. So regardless we need to have 12 different named intervals for various numbers of semitones:For instance, “minor second”, “major second”, “minor third”, “major third”, “perfect fourth”, “augmented fourth”, “perfect fifth”, “minor sixth”, “major sixth”, “minor seventh”, “major seventh”, “octave”.Reducing all those ordinal numbers by one really doesn’t help all that much. You still have to remember how the “minor” and “major” labels interact for every interval in the scale, and remember that sometimes the interval between the same two notes is given multiple names depending on the key, etc., which is all horribly confusing mess. > Reducing all those ordinal numbers by one really doesn’t help all that much. You still have to remember how the “minor” and “major” labels interact for every interval in the scale, which is a horribly confusing mess.Those interactions are pretty intuitive. Where defined, major + minor = perfect (considering an octave as perfect), perfect + major/minor = major/minor. As long as you remember which notes exist, you can't get it wrong, so you'll never get confused by a piece of arithmetic in an actual piece. They’re not remotely “intuitive”. They only make sense to someone with years of training.If instead you used digits from –5 to 6, using arithmetic mod 12, it becomes obvious that e.g.:`````` -2 + -3 = -5 4 + 3 = -5 5 + 5 = -2 -5 + -1 = 6 4 + -3 = 1 etc. `````` The “perfect” intervals are just ±5 (ratios very close to 3:2 and 4:3). The “major” intervals are –3, –1, 2, 4 (approx. ratios of 5:3, 15:8, 9:8, 5:4). The “minor” intervals are –4, –2, 1, 3 (approx. ratios of 8:5, 16:9, 16:15, 6:5).Then it’s easy to see that your “major + minor = perfect” formula only works for some intervals, Etc. Overall the simple heuristics are more obfuscatory than helpful IMO. > Then it’s easy to see that your “major + minor = perfect” formula only works for some intervals, Etc.Where does it go wrong? Do those cases come up in practice?Counting up and down the scale is a core use case for a notation for intervals. It absolutely needs to be well-supported. A 12-semitone approach is never going to match the usability of even the existing system. >A 12-semitone approach is never going to match the usability of even the existing systemI am obliged to point out that a 12-semitone approach is in fact part of the "existing system" (see: pitch-class set theory).(Mind you, I of course think its usefulness is overrated, because I think the "atonal" repertory is tonal.) Nearly all aural skills classes (learning to hear/sing music) for music majors use solfege or something like it. Some people use scale degree numbers instead (so a IV chord in a major key is "4-6-1" rather than "fa-la-do"), but the concept is still very useful. What I'm questioning is the popularity of Solfege vs numeric interval names.My music classes of ~20 years ago treated Solfege as being of historic interest, but not particularly common. While numeric intervals were used daily. It came up somewhat more in sight singing and vocal training than in any of the instrument or theory oriented classes. But, I think it was mostly students who were used to it using it rather than instructors teaching it.But, maybe I just so strongly preferred numbers that I immediately discarded any instruction involving Solfege as being silly and a waste of my time.Still, I can only recall seeing numbers (and Roman numerals) in writings on theory and such. Fair enough. It's still pretty common...it's hard to pin down any numbers exactly, but I'd guess it's probably half and half for solfege vs. other systems. We teach solfege at my school (although I prefer numbers myself). And you're right that it's used mostly in sight-singing/aural-skills classes; I mention solfege much less often in my written theory classes. Exact opposite for me. I learned do-re-me as a kid and then ever saw it again, until I started reading about music theory recently. Then I saw it a lot - e.g. voice leading rules saying that a voice must start from Do and end on a ti-do step, etc. Doesn't everyone learn Do-Re-Mi from "The Sound Of Music"? Well, sure, but what's that got to do with actual music instruction and how musicians talk about music? Movies aren't always entirely accurate representations of the world, particularly on highly technical topics.Knowing solfege because you heard it in a movie and using it on a daily basis in the process of teaching or making music are independent concepts.Anyway, conversation here has brought it to my attention that it is still pretty common, there are some areas where it is useful (maybe even better than numbers, as in the singing and vocal training area; I personally have recognized the limitation of numbers when singing minor notes, for example), and that my own experience was only partly representative of music pedagogy in the US and elsewhere. That said, when I'm teaching people about music, I still plan to only use numbers...the areas where solfege would be useful are pretty advanced, and require more than watching Sound of Music to understand. > 2. Learn your intervals> 4c. Get a good feel for common jumpsi think that's why the current system works so well. most musicians have intervals burned into their muscles. to use an excel reference, reading R1C1 from the staves is much faster than reading A1, because you can read R1C1 from any line in the staves. Sounds like vim! Sounds like... And is it? ;)In terminal run vimtutor. The sound will change.Does anyone know of a similar tutor for sheet music? In case anyone is searching for a really smart, modern method for learning music theory, this is it: https://www.hooktheory.comThe author devised his own system for visually representing notes, it makes it much easier to understand things like scale degrees and relative notation (and thus the theory around famous harmonies, melodies, etc).I think music tools are in desperate need for improvement... Starting with notation, which is still a bit akin to forcing programmers to go straight to Assembly. Little is gained from it as most people just completely give up and then go on to live the rest of their musical lives "in the dark", without knowing how to read and write at all. This can actually be good for some but I'm sure it hinders the creativity of a lot more.I think we also need way better digital instruments... That make it easier to stay on scale (or to modulate, etc -- whatever the mood is), for instance, allowing people to just play away which is what actually matters.I've spent countless hours of my life learning scales on several different instruments and think a lot of that was wasteful. More often than not I'm just trying to stay in a given key anyway, nothing fancy...Instruments really need better interfaces :) No, instruments really don't. The point of conventional instruments is that once you learn them - which takes years - you can instantly express almost any musical idea using all the possible degrees of freedom available on that instrument.With something like Ableton Push, you're one step removed from the sound generation, because you're triggering automata with a very limited expressive repertoire. (With Push, it's often just a triggered sample, which has almost no expressive potential at all.)You can change keys instantly on a piano. You can play any chord you can get your fingers around, in any inversion, using any voicing, with fine control of the relative level of each note in the chord.With button controllers the best you'll get is one chord per button with no fine shading of levels, no control over inversions or voicings, and so on.It's absolutely fine to make music like this, but it's not fine to demand that all music be made like this.Controllers like Push are good for performing effects - filter sweeps, and such - which aren't possible on a keyboard. But that's a different skill to learning scales, and much more expressively limited.Electronic art forms generally are more rigid and less expressive than non-mechanised media. In theory you should be able to do more, but in practice no one has cracked the problem of building high-bandwidth expressive automata that are as physically responsive and open as traditional instruments/media.Aesthetically, that can be a problem. A lot of machine-assisted art is either chaotic and formless, or formulaic and repetitive. The best classical music and classical performance lives in an expressive and creative sweet spot between those extremes, and it's incredibly hard to hit that spot with machine assistance. You seem to be pegged on what current controllers can do... And that's exactly what I am saying: they often suck!But they can improve and I am confident they will. When I am learning a brand new instrument I can literally feel my brain knowing exactly what I want to do way before my fingers/mouth/feet are able to perform the task at hand. How is this not an interface problem?With button controllers the best you'll get is one chord per button with no fine shading of levels, no control over inversions or voicings, and so on.No way. If you don't have to be memorizing stupid things such as "where is the minor 7th again on this one particular instrument?" maybe you could use your free mental cycles (and fingers, feet, mouth) to control that instead... And who knows, maybe you could now do 4-5 inversions in the same amount of time it would take you to do a single one on a piano. Or maybe you can do inversions way more effortlessly on another instrument and focus on really nailing the vibrato.It's absolutely fine to make music like this, but it's not fine to demand that all music be made like this.I never said this, I'm just saying that a lot more can be done with a lot less effort if instrument/controller interfaces improve. Once you actually practice a physical instrument for a reasonable amount of time things like the concept of "memorizing where the minor 7th is" quickly become non issues - the only memorization involved is that of your muscles i.e. the cognitive load is essentially nonexistent. Involving more parts of your body than your cognition is one of the joys of playing a physical instrument, versus pressing a button and thinking a lot. I play several! But of course I'm not proficient in all of them, which is the whole point. There isn't an "universal controller" that is expressive enough across a variety of timber types... Yet if that existed one could master one interface and do a lot more musically with that acquired skill. "The point of conventional instruments is that once you learn them - which takes years - you can instantly express almost any musical idea using all the possible degrees of freedom available on that instrument."But the degrees of freedom of conventional instruments are severely limited compared to what is possible.Let us also recall that every "conventional instrument" was at one time not only unconventional, but even radically new. The piano, is itself only a few hundred years old. I'm sure when it was invented there were some people who argued against its use and that one should instead stay with "conventional instruments", which then did not then include the piano.I strongly recommend a talk[1] by Jordan Rudess, who is widely considered to be one of the greatest living keyboard players.In this talk, Rudess discusses and vividly demonstrates the greatly expanded possibilities that innovative keyboards bring to the table.Novel instruments that somewhat resemble conventional instruments like the keyboard are only the tip of the iceberg of music interface possibility, however. There are plenty of novel music expression technologies that don't have even the remotest resemblance to conventional instruments, and allow ways of expression that were hardly imaginable a hundred years ago. Things like whole body position tracking, which allows you to make music through dance.Of course, mature musicians like Rudess who've spent their entire lives learning and practicing on traditional instruments will be unlikely to switch to something radically different, as they'll be starting from ground zero on those instruments. But others with less to lose will be more open to learning something completely new.It's impossible to tell which novel instrument will become the conventional instrument of tomorrow, but it's very likely some will, because that's how we got all of the conventional instruments of today. Let us also recall that every "conventional instrument" was at one time not only unconventional, but even radically new.Amen.Thanks for sharing the talk! I'm a proud owner of one of those keyboards he is playing, a ROLI Seaboard. It is indeed an amazingly expressive, fantastic product This is utter BS, you can play the Push as if it were a piano and the sound design options are endless.Saying electronic music is either too chaotic or too repetitive is not only entirely subjective but completely impossible for you to say. Artists like Kiasmos or, famously, Aphex Twin, just to name a couple amongst hundreds, make music that can be neither repetitive or chaotic, for example. > I think we also need way better digital instruments... That make it easier to stay on scale (or to modulate, etc -- whatever the mood is), for instance, allowing people to just play away which is what actually matters.I'm not sure what new stuff has come out in the last few years, but Ableton Push is exactly what you're describing. It's a grid where you can select a key, scale, and tone, and then you can apply effects in series / parallel. The notes in the scale light up. There's a bunch of other stuff you can do as well. Thanks! I've played with Push, it's pretty good... but as someone commented above it's fairly limited in many ways too.This is an interesting piece of software I have used before: http://autotheory.net. It simply translates incoming midi data so you can use whatever controller/instrument you're already familiar with. The creator is nice and responsive, he often attends shows like AES and NAMM. Thanks for the recommendation, I just bought the Hook Theory book and it looks quite good. I've been searching for something like that for a while! I've not read the book but the tool is amazing. I believe the authors have been working on a HTML rewrite for some years now (currently it's Flash). The whole site design got redone a month or two ago, so there may be more changes on the way. > In case anyone is searching for a really smart, modern method for learning music theory, this is it: https://www.hooktheory.comI really want to buy the books but they're all DRM'd :-(.Here's hoping someone from there reads this page and releases it as an epub... I've skimmed a lot of articles on music "theory" but none of them provide anything like what I'm looking for. A music theory should explain:1. Why do we like pieces when played forward but not backward or inverted?2. Why do certain sounds evoke certain emotions?3. How could you write a program to pick out music that people find especially good (versus music that has surface similarities)?In other words, why does a particular sequence of sounds A, B, C lead to a mental state M that has particular internal qualities? > 1. Why do we like pieces when played forward but not backward or inverted?Why do we like text when read forward, but not backward or inverted?There are, of course, works that are palindromic or otherwise written to be read/heard backwards, but most of the time that kind of global transformation tends to ruin the "spelling"/"narrative".> 2. Why do certain sounds evoke certain emotions?Just like text, evoking emotions needs some sort of narrative. A story isn't a single fact or statement (or a single sound); it's about how those facts (or sounds) flow or change.In music you might hear a brief bit of new melody that foreshadows something big later in the song. A clear rhythm or melody might be repeated to get the listener to follow along only to have it cut short at a key moment to deny the obvious resolution (similar to a melodrama that suddenly reveals a new twist in the plot as a cliffhanger).It's the story you tell that matters, and it takes a skilled composer to put sounds together to make a song emotionally evocative. The song that is mostly a 16 bar loop probably sounds boring (but not always!), while the song that introduces the same 16 bars and then plays with variations of it to create an initial conflict, rising action, and a climax is probably a lot more interesting. An obvious example might be Mozart playing Salieri's march in Amadeus[1]. It's not just that he embellished the simple march; Mozart adds a lot of variations that culminate at a comic ending. Actually, great composers such as Beethoven and Bach and Chopin had very definite ideas about what emotions are evoked by certain keys. They even argued about it with their peers. Music is not something that is reducible to mere quanta and waves and frequency. You all are missing the human part. Sorry, but it's true. > what emotions are evoked by certain keysYes, choice of key is one of the tropes that is useful when composing a song's "plot".> Music is not something that is reducible to mere quanta and waves and frequency.That's my point; interesting aspects of a song are not derived from specific sounds (and their frequency/etc). Those are the atoms that can be used to create the larger plot.While it is possible to reduce music to the frequency and timing of its atomic structure, it's similar to analyzing the phonetics of speech or the glyphs of text in isolation. A low level perspective may be useful, but misses the larger structure we call a "song" or "essay". That, and they didn't all necessarily use a pure Equal Temperament, either. Different temperaments can give more distinct feelings to certain keys more so than the modern equal temperament. (Note: I used to tune pianos.) Music is waves and frequency. Music appreciation is what you are describing. And appreciation is very dependant on culture. That is why Bach is not (as) appreciated in certain cultures.Just like photography. Why is one photograph more meaningful than another? it has nothing to do with photography, per se, it has everything to do with the culture of the person doing the appreciation.There is a link between the two, between creation and appreciation, and those who understand it generally fare better. But it is not required to be a musician, or a photographer or a poet or anything really. > Music is waves and frequency.Sound is waves and frequency. Music is a collection of sounds arranged in a specific sequence.> Music appreciation is what you are describing. And appreciation is very dependant on culture.Music relies on various "tropes" to construct a narrative. This includes the choice of key/scale (or none at all), ideas about timing and harmony, etc. These "standard parts" of music are usually from the local culture, just like how a play or movie will use standard character archetypes ("tropes") that are culturally derived. "music is waves and frequency" in the same way that "spoken language is waves and frequency"---not very usefully. I think bringing in" appreciation" muddies the waters. I was just responding to the parent who claimed that music was not "waves and frequency".My point is music is (mostly) independent of its appreciation. Machines can, and do, make music based entirely on the theory of music. Music theory is an accepted and used term of art for the category of things that this article talks about. There are courses, books, university departments, and degree programmes that use the term. Nobody is going to stop using it because you skimmed it but it doesn't describe some other thing that you think it should.Some examples: Grand parent is saying he wants an article that covers the "why does music sound good" part of music theory, something I want too. I think most people have a basic grasp of notes, scales and that a middle C is air oscillating at 440 cycles per second. How can you make something sound good is the interesting part of music theory for me and I still haven't found a good intro to it! > How can you make something sound good is the interesting part of music theory for me and I still haven't found a good intro to it!As someone who majored in music composition, I have a very simple answer. I'd have some sort of idea in my head of what I wanted the music to sound like (or the emotion to evoke). Then I'd fiddly around for quite some time, discarding the things that didn't meet my criteria.That's sort of a glib answer, but the fact is that no one really knows exactly why certain things evoke certain emotions, even though most composers understand various building block ideas like "odd meters like 5/8 and 7/8 generally evoke intensity and tension" or "brass chorale in a major key sounds triumphant" or "gong crescendo roll is scary". And of course, even then, we could find counter-examples for every one of those things.Also, all music theory will tell you is why something in some piece of historical music sounded the way people expected it to sound at the time. It will definitely not tell you how to write good original music (though it may be a good guide on how to imitate past composers if that's useful for what you're trying to do). > like "odd meters like 5/8 and 7/8 generally evoke intensity and tension" or "brass chorale in a major key sounds triumphant" or "gong crescendo roll is scary"Can you recommend any books that teach these sorts of general rules, or the emotive feeling generally associated with different keys and modes? AFAIK there are no such books. As for whether certain keys evoke certain emotions, that's highly debated other than "major happy, minor sad (other modes weird)".The way I learned about how composition worked was mostly two things. One, listening to lots of music, ideally with the score in front of me so I could zoom in on some particular bit I really liked. Two, writing music and seeing how it turned out in practice. How can you make something sound good is the interesting part of music theoryYou need something like this: https://www.amazon.com/Alfreds-Essentials-Music-Theory-Self-...It's not something you're going to learn in an afternoon or a weekend, it's hard work and Beethoven was still working on it at the end of his life.Last month I spent an evening analyzing and discussing a passage of Rachmaninoff, trying to understand how he knew how to write a certain sequence. That may be a very good book, but I'll be honest, it doesn't look like a very gentle introduction. It's not :(. Although the average person could probably get a lot of good high-level info by browsing it.I haven't yet come across anything that is a good gentle intro. Most resources that approach music and math make the mistake of treating music theory like the law, without any rationale for it provided. Music history textbooks typically give a lot more context of how our music theories emerged, but they don't talk about why that might be, based on acoustics, psychoacoustic, and math.Maybe one day, I'll write the comprehensive intro I wish I'd had. Does that book provide anything helpful to someone who already understand music theory? Oh most definitely! I'd describe it as an extension of what's taught in the standard music theory curriculum. It makes the very ambitious claim of developing a framework that can be used to analyze all tonal music, from the renaissance to the present. middle C isn't 440hz. that's generally an A. Good point, not sure where I got that from then. Record updated. When we have discordant sound (e.g. a collection of plucked strings where the fundamental frequency ratios are not in nice simple ratios, so that none of the overtones align), there’s a great deal of complexity to the sound, and you get interference patterns between them, similar to moiré patterns in images.This causes “musical tension”.Some types of discord are mild, and cause a bit of mild annoyance or “sadness” in the sound. Other types are aggressive and cause serious anxiety.When you return most of the sounds to be in harmony, that tension is relieved. This causes a more positive emotional response. The greater the former tension, the more satisfying the release.Imagine you’re in a crowd of applauding people, each clapping at a different rate, so that the sound is like a cacophony. Your brain can’t make out any pattern except a wave of sound. Now imagine the people start clapping in rhythmic unison, with some kind of structure. Suddenly your brain can make sense of the pattern. Music theory can't explain why a piece is designed in some way; it explains what patterns can be found within an existing piece. Designing any aesthetic is primarily about how patterns are prioritized, associated to other parts of culture and turned into conventional tropes or motives. As we get new genres of music the pattern languages tend to change. (The idea of music as "universal language" is only true in a basic sense of what things our ears and brains can comprehend and how we would perceive them in an ungrounded state. In the details, cultural differences will definitely matter.)So, music theory "catches up" to the pattern language by associating it to human natural language, but it doesn't say why. I concur with the "music appreciation" recommendation for learning the whys. When you get deep into analysis of a work, all sorts of angles can be found to correlate "the thing in the work" with "the reason and context of its creation". For one song, maybe it's the lyrical content that is important. For another, it's about rhythm, or dynamics. The artist's life at that moment, sociopolitical context, and newly available technology are often considered as factors. In a complete work, these elements blend such that it can't be reduced to a singular "this word or phrase is definitely all this thing is" - analysis highlights parts of an experience that can't be fully conveyed in a different form, rather than trying to "spoil" or "solve" its mysteries. An aesthetic just seems like a unconscious favorable reaction to stimuli based on genetics and culture. Why our genes and culture have favored certain forms, I suspect there isn't a way to reduce it to something satisfactory.I used to wonder why I felt good when looking at sunsets, landscapes and clouds. But I figured that our aesthetics probably evolved in response to what was around us. Happier people probably survive better.The way we hear music seems like such an aesthetic, as it might've occurred within ancient cultures as a form of play and release. I suspect that our random genetic hunger towards different aesthetics might have created an incredible developmental feedback loop. I'm not sure where I'm going with this incredibly complicated topic, but I have a lot of very unrefined thoughts about them, that are probably overly-reductive and wrong. You are looking for a "theory of music", as opposed to "music theory", ie an explanatory or scientific theory.I offer my own efforts in this direction at http://whatismusic.info/. Thanks! This sounds like good terminology and your writing attempts to answer the question in ways I haven't seen elsewhere.Also I think the fact that songs can get stuck in your head suggests some kind of mental reinforcement exercise for patterns over time. Those qualities aren't universal in people and as such what you are describing is more of a study of culture than musical theory. You would probably enjoy a music appreciation course, and possibly one taught by a philosophical professor. Music theory describes music.Your questions are very interesting, but theories answering them would describe humans. Not an article, but I think this talk hits a lot of interesting points: Answer to question 1 is simple: good resolution is a semitone up, but scale step down (normally, whole tone). No symmetry here. If you play backwards, you won't get resolutions - it will sound as nonsense. Harmony Explained: Progress Towards A Scientific Theory of Music 3. How could you write a program to pick out music that people find especially good (versus music that has surface similarities)?Be an artist. I don't think really we know the answers to those questions. > 1. Why do we like pieces when played forward but not backward or inverted?Would you like a movie played backward?> 2. Why do certain sounds evoke certain emotions?Large part of this boils down to if the waves representing the sounds meet at zeroes or not.> 3. How could you write a program to pick out music that people find especially good (versus music that has surface similarities)?I think that this is currently impossible. The music composition search space is actually extremely large, larger than say the search space of Go. You can restrict the search space quite a bit but it's still large.> In other words, why does a particular sequence of sounds A, B, C lead to a mental state M that has particular internal qualities?Think of it as design. It's the same sort of problem. If anyone is interested in learning to read music, I've been slowly building a tool to practice: http://sightreading.training/source is here (built in react, es6): https://github.com/leafo/mursicjsIt's still lacking a lot (like rhythm), but the different generators definitely give my brain a workout. It works best if you hook up a midi keyboard. As someone self-teaching themselves piano, this is really fantastic. I already play drums but have no notion of what notes are what on a staff and piano keys and have been slowly teaching myself. The MIDI keyboard support really makes this stand out. I've been teaching myself piano after someone gave me one.Here's a great set of online lessons [1] (not free, costs about \$20 per month if you put in your email address). The guy is really talented, and teaches non-classical stuff like pop, boogie woogie, etc. I'm super inspired by it!I also found this [2] which does have free lessons and also looks good. If you want a series of books that constitute "music theory for nerds" -- building up music theory from a solid foundation of acoustics, and math -- try "Musimathics" by Gareth Loy. It is a great read.It takes very little for granted. Now, sometimes you have to say "this is just the way it turned out" to explain Western music, but the best way to do so is to show some other ways it could have turned out, and show their role in non-Western music. Musimathics does that often. On the science side of things, Vi Hart put together just an excellent video that goes over harmonics, the overtone series, and why 440 Hz and 880 Hz sound so "indescribably similar" to this blog author. This is great, but I would recommend Robert Greenberg's "How to Listen to and Understand Great Music". He goes through the history of western music in a way that makes it clear why Amin != Cmaj, and other questions that the OP has. Yes, sheet music is crap, and he explains how it evolved to be the way it is, after which you'll be much more forgiving. He's a great speaker who obviously knows the material inside and out. The explanation of the origin of major scale in the article is pure voodoo. Minor third is not a simple fraction - is that the reason to exclude it from the scale? How do you explain minor scale then? Maybe it should be excluded, too?Here're my thoughts on the subject.For some reason no one can explain, Western music settled on a system of 12 tones with equal temperament, This system emerged as a result of long evolution of Western music, and experimentally proven to be very rich in possibilities.Scales used in Western music (of which jazz is a part of) are built on two simple principles: 1) interval between adjacent notes of the scale is either tone or semitone 2) there's no two semitones in a row.It's easy to check that all scales that satisfy these 2 rules are:major scale and its modes (7-note scales; 7 modes)melodic minor scale and its modes (7-note scales; 7 modes)diminished scale and its modes (8-note scales; 2 modes)whole-tone scale (6-note scale, single mode).(Whole tone scale is not used very often, except by T.Monk)But even after we "explain" scales, we need to figure out how to use them, what their role is, what the properties of each mode are. There's no hard science behind this, the properties just "emerge", and you have to experience them - theorizing is not of much help, math formulas don't explain anything, just lead to confusion.In short: you have to play AND think; thinking alone won't help. It's an experimental subject.Edit: forgot to say: scale is a very useful notion, but in some contexts, it's more convenient to think in terms of triads and interpolation. I know this all sounds hand-wavy, and it is! Unfortunately, without piano, it's impossible to to illustrate what it all means. The subject doesn't easily lend itself to verbalization. I agree very much with your post's thesis (you have to play AND think; thinking alone won't help. It's an experimental subject.) Just noticed one thing:> For some reason no one can explain, Western music settled on a system of 12 tones with equal temperamentIt doesn't seems surprising to me. If you start from a pitch and go upwards in both octaves and perfect fifths (2:1 and 3:2, the two most fundamental intervals), the perfect fifth sequence will land on 11 distinct tones before (nearly) meeting the octave sequence. Mathematically, (3/2)^12 ≈ 2^7.So 12 semitones works out nicely because you can follow perfect fifths out in any direction as far as you want and never go outside the set of semitones. And most of the small-ratio'd intervals can be represented with pairs of notes inside this set. Interesting idea indeed. I need to think about it.Edit after thinking: still, it doesn't explain the number 12 IMO. It could be 17 or something else. Probably, it's a long chain of coincidences at play: Western music settled on 7-note scales long time ago (long before equal temperament was invented), and we should start looking for explanations from here.Another edit: one of the important coincidences is that number 12 makes possible the existence of diminished scale, which serves as a "universal glue" due to 2 tritones. (There's not enough space here to elaborate, but you probably know what I mean). And maybe tritone itself is one of factors leading to number 12. The number 12 is just a coincidence, that 3^12 = 531441 ≈ 524288 = 2^19This means that 3/2 ≈ 2^(7/12) [accurate to about 0.1%]And also 4/3 ≈ 2^(5/12) [also accurate to about 0.1%]And also 9/8 = (3/2) / (4/3) ≈ 2^(2/12) [accurate to about 0.2%; putting two factors of 3 in makes the approximation only half as good]You also get another nice coincidence, that 5^3 = 125 ≈ 128 = 2^7This means that 5/4 ≈ 2^(4/12) [accurate to about 1%]There are some people who have written music in a 41-note equal tempered scale, because then you get an even better approximation to the 3/2 ratio:3^41 = 36472996377170786403 ≈ 36893488147419103232 = 2^65(3/2) ≈ 2^(24/41) [accurate to about 0.03%] If you start off from assuming that the Do-Sol (fifth, 3:2) harmony is a "pleasing" one, and also the Do-Mi (third, 5:4) one, you can create new "mostly pleasing" harmonies by for example taking the fifth of a fifth (9:4, which can be transposed an octave to get 9:8, which is Re or a second), and doing similar things (you can also do things like finding the note whose fifth is Do, which is 2:3, or 4:3, a fourth or Fa).Repeat this process and you start getting a bunch of notes which fall on the 7-note scale. In the blog post the major seventh is listed as 17:9, but by this method you get a 16:9. Basically the same thing.At this stage, you may notice that the notes are roughly equidistant, except for Mi-Fa and Ti-Do, which are at ~half the distance. This is the first hint of the 12-note scale. We could have stopped earlier in the notemaking process and had a 6-note or a 5-note scale or whatever, but it wouldn't be so equidistant.Now pick each note, and build an octave from it. The new notes created will invariably be very close to existing notes, or very close to the midpoint between existing notes. This gets us the 12 note scale (5 midpoints + 7 notes, the aforementioned half-step notes don't have midpoints), if you choose a canonical note for each part. The number 12 just happens to be the number where simple harmonic ratios can get you a mostly-equidistant scale.At this stage, different music systems do different things.One kind of Chinese scale uses a 2:3 ratio and generates ratios involving these numbers that form a 12-note (roughly equidistant) division.Indian music does something similar, though it instead generates a 22-note scale, where many of the 12-note scale notes have two forms. It is rare that a given piece of music will use both forms of the same note.Western music goes ahead and invents the piano, realizes that the piano is hard to tune/transpose, and settles on the twelfth-root-of-two stuff so that transposing becomes dead easy. A bit better modification of the argument: continue cycle of 5th. After 12 steps, you get (3/2)^12=129.7, which is really close to 128=2^7 (whole number of octaves). That's where 12 steps come from!And from here, the natural idea follows: what if we take not exactly 3/2 for fifth, but value x such that x^12 is exactly equal to 128? This leads to equal temperament.Yeah, that might be it! (Not sure that it's true historically though). I think that there are several thing that brought us to 12. First it is very easy to divide, this is why we use 12 hours clocks and between 2 octave the ear is able to distinguish 1/12 of an octave as 2 different sounds, it might be possible to do better but it would be unpractical because it would be more difficult to found chords that sound good. Arabic music use quater tones (24 quater tones per octave). > Minor third is not a simple fraction - is that the reason to exclude it from the scale?The minor third is a 6:5 ratio. Does that change your mind?Where do your tones and semitones come from? You've just rejected the explanation for why we have a 12-tone scale (it is a local optimum of tuning, closed under the operation of transposition, that satisfies lots of nice ratios), so what do you propose instead? Saying "no one can explain" is a cop-out.To follow up: Major and minor scales are 7-tone subsets of the 12-tone scale that were discovered first. They allow for many of the same possibilities, but they're not closed under transposition, which is why we now have the 12-tone scale that includes all of them. Please read wikipedia article on temperament. 7-note scales were "discovered" long before equal temperament was introduced. Equal temperament is a relatively recent invention, initially was very controversial, and remained so for a long time. In other cultures, there's still a variety of scales and temperaments that have nothing to do with 7-note scales at all. (Jazz uses variety of scales, too - some even with 2 semitones in a row). There's only one thing people seem to agree on: the role of fifth (3/2). This is exactly what I'm telling you. Why are you being condescending to me about it? You're the one claiming that the semitones of the 12-tone scale are some sort of fundamental axiom.Why didn't you know where minor thirds come from?Why do you seem to be denying that simple harmonic ratios are where harmony comes from? (EDIT: he's not, it's fine)Why are you promoting a theory of music with absurd axioms, which manage to explain the octatonic scale (which almost nobody uses) better than the pentatonic scale (which almost everybody uses)? That's not a sign of a good theory. Please read the part of the article starting with "The twelfth root of two may be irrational, but it turns out to almost create several nice ratios". The author excludes minor third from the set of "nice ratios". I referred to this while calling it a voodoo. A simplified explanation is not voodoo.Yes, he chose to list only the intervals in the major scale. Showing the intervals in the major and minor scale at the same time would have been very confusing.I understand now that you weren't trying to say that minor thirds were a hole in the theory he presented, but that they were a hole in the way he presented it. I took issue when I thought you were criticizing the theory itself, because what he presented is a simplified view of generally accepted foundations of harmony. I apologize for mis-characterizing your position there.There is more that could have been said -- and it sounds like we agree on what could have been said -- but I think it's not necessary to go into all the details of the construction of scales, including temperament, just to write a blog post. Describing temperament accurately, without handwaving, requires a book. Dude, if you can't hear it, it doesn't matter. You'll never listen to Coletrane's "Night Train" and have your head explode when you realize he's jumping from Dorian to Mixolodian to whatever the fuck was that!? BTW, please read something like "Musimathics", it'll be better than trying to learn music theory from Wikipedia. I'm not learning it from wikipedia. I'm playing for 50+ years. Best resource is Jazz Piano Book by M.Levine. (I read lots of others, but... everything I learned, I learned from this one). Okay, yeah. I bristled at your suggestion that I should go learn temperament from Wikipedia, when I was making an effort to not bring up temperament to keep my reply focused.I thought you didn't understand harmony, you thought I didn't understand temperament, I was talking from a historical point of view, you were talking from a jazz point of view.We might disagree on two kind of minor points:* How this blog post should have been written* I'm uncomfortable with describing fundamentals of music from a jazz point of view, because that seems to me to be putting the effect before the cause. I think you're being unfair on the whole tone there, it's the intro to 'Take the A Train', and people will often substitute it in a dominant chord when improvising. "C major is identical to A minor, and I don’t understand why we need both."Cringe. If I understand nothing of some subject, I would do well to just shut my mouth about it. In fairness, the author said " I don’t understand why we need both", and not "We don't need both". >I suppose it’s possible to change the sound of an entire piece of music just by changing the key signature, but does anyone actually do that?>How would that work for music that also uses notes outside the scale? These seem more like questions of composition, which I definitely don’t know anything about.From wikipedia:Although transpositions are usually written out, musicians are occasionally asked to transpose music "at sight", that is, to read the music in one key while playing in another. Musicians who play transposing instruments sometimes have to do this (for example when encountering an unusual transposition, such as clarinet in C), as well as singers' accompanists, since singers sometimes request a different key than the one printed in the music to better fit their vocal range (although many, but not all, songs are printed in editions for high, medium, and low voice).There are three basic techniques for teaching sight transposition: interval, clef, and numbers ... I don't think the author was referring to transposition there. I think that was about leaving the notes on the staff the same but just changing the key signature. This is done sometimes as a novelty, and the best example is changing something from a major key to a minor key or vice versa.The article has a link to a recording of Für Elise in a major key [1], and there are many similar renditions of other pieces around. You could in principle do this with any of the seven modes, not just major and minor.As for notes outside the scale, it seems like these key-signature-changing compositions typically keep them the same (like the D# in Für Elise). The author and anyone else who understands correctly that traditional music notation is shitty in lots of ways and is trying to understand how music really works should go get the book "Music and Memory: An Introduction" by Bob Snyder. It uses no music notation and explains music in terms of psychological principles of perception of time. It's not a complete theory of everything, but it shows the way you should be understanding the nature of music.Beyond that, check out Sweet Anticipation by David Huron, and Tuning Timber Spectrum Scale by William Sethares (and his other Rhythms and Transforms, see http://sethares.engr.wisc.edu/ for web versions of first chapter of each). These sorts of resources are where real understanding of music comes from. Not from the "theory" stuff us music professionals had to deal with that fails to explain anything well. Well, the first figure: frequency is NOT the period, and amplitude is not that. I guess there are lots of errors in the text if the first figure is completely wrong. This source is a bit wordy. Let’s summarize:The core idea of music made with harmonic sounds is that “notes” with frequencies at small-integer ratios will “harmonize”. Harmonic sounds means something like a vibrating string where the vibrations are integer multiples of some fundamental frequency, because other non-integer-multiple vibrations are damped out by the fixture of the string at two points. Different (non-harmonic) types of sounds often sound better with a different sort of scale, for details see this book http://sethares.engr.wisc.edu/ttss.html* * *The “octave”, 2:1, is the simplest whole-number ratio, and makes many of the vibrations in two notes in such frequency ratio align with each-other, to the point that two harmonic sounds exactly an octave apart almost sound like the same sound.Other simple ratios like 3:1, 4:3, 5:4, etc. also “harmonize”, with (not quite as) many aligned overtones.The core idea of the 12-note musical scale (pretty much regardless of specific tuning) is the approximation:3^12 = 531441 ≈ 524288 = 2^193/2 ≈ 2^(7/12) [this is accurate to about 0.1%]Or another way to say this: 7/12 of “doubling” on a log scale is very nearly “three-to-two”. Musicians call this ratio a “perfect fifth”.In the case of equal temperament, an octave is split into 12 precisely equal steps (on a log scale), each one the 12th root of 2.There’s one other nice approximation to take advantage of:5^3 = 125 ≈ 128 = 2^75/4 ≈ 2^(4/12) [this approximation is only accurate to about 1%]Musicians call this ratio a “major third”.* * *Even outside music, these approximations can be useful for doing approximate computations.If only our society switched from decimal to “duodecimal” numerals, it would be very natural to use logarithms base two, notated with “duodecimal” fractions.If you have a number expressed in log base two, and you use duodecimal notation, approximately multiplying or dividing by 2, 3, 4, 5, 6, 8, 9, 10, 12, ... is very easy using addition/subtraction of easy-to-remember multiples of 2^(1/12).Unfortunately our society instead has slide rules and measurement scales (decibels, etc.) which are all built around logarithms base ten, and decimal notation. The article says that the "human ear loves ratios", but doesn't dig deeper into why. Here's my two cents.First of all, let's focus on harmony (notes played at the same time) as opposed to melody (notes played one after another). What sounds good in a melody is quite culture-dependent, but there are reasons why harmony is more universal.Second, let's focus on sounds that are produced by something long and narrow. In a guitar, violin, or piano it's a string, and in a flute it's a column of air. The physics of vibrations goes so that in such a case the sound is composed of harmonics: sine waves of frequencies f, 2f, 3f, 4f, ... If the shape is different (say, a circular membrane of a drum), then this may not apply.Suppose we add a second sound, whose fundamental frequency is, say, 3/2 f. This means that its harmonics are 1.5f, 3f, 4.5f, 6f, 7.5f, 9f, ... Half of these (3f, 6f, ...) coincide with the harmonics of the first sound, so the sounds "reinforce" each other. More generally, if the ratio of the frequencies is p/q for some integers p and q, then there will be overlap in the harmonics. And the smaller p and q are, the more overlap there will be. An entire class of music, Carnatic music, has existed for thousands of years with millions of listeners that is based on melody alone.For a drum, interestingly, the fundamental vibration modes are all Bessel functions. The really wierd thing is that intervals such as a fifth, major and minor thirds, and sevenths occur regularly in bird song, whale song, and a bunch of other non-human sounds. I assume that evolution favours constructive interference as the communication will generally travel further, but I also feel that human music is in some way influenced by our pre-lingual history.In some way our brains are hard wired to appreciate and recognise these intervals, and to infer certain emotions from them. Melody in the form of the pentatonic scale is much more universal across cultures, though. Use of non-trivial harmony is significantly less widespread, it's usually no more than a melody played over a single root tone or chord. South-Asian classical and folk music is one example in which Western-style harmony is not used at all. You're right, of course. My wording is was bit poor.What I should have said that harmony is less ad hoc; it has less "degrees of freedom".With regards to melody, there are tons of tuning systems that are quite close to the usual twelve-tone equal temperament. It would be hard to give a convincing argument that one of these sounds better than all others.Contrast this to the system of harmony where the basic principle is that ratios of small integers sound good together. This is not the only possible system of harmony, but it does seem to represent some kind of local optimum. And this makes it more amenable to the kind of purely theoretical reasoning that the article is trying to do. More Search:

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# Chapter 07 dimensioning Chapter 7 Dimensioning TOPICS Introduction Dimensioning components Dimensioning object’ s features Placement of dimensions. Introduction ENGINEERING DESIGN PROCESS RESULT Design a part Sketches of ideas Create drawings Multiview Drawing Dimensioning Manufacture TRANSFERRED INFORMATION Shape 1. Size, Location 2. Non-graphic information DEFINITION Dimensioning is the process of specifying part’ s information by using of figures, symbols and notes. This information are such as: 1. Sizes and locations of features 2. Material’s type 3. Number required 4. Kind of surface finish 5. Manufacturing process 6. Size and geometric tolerances This course DIMENSIONING SYSTEM 1. Metric system : ISO and JIS standards Examples 32, 32.5, 32.55, 0.5 (not .5) etc. 2. Decimal-inch system Examples 0.25 (not .25), 5.375 etc. 3. Fractional-inch system 3 1 , Examples 5 4 8 etc. This course Dimensioning Components DIMENSIONING COMPONENTS Extension lines Dimension lines Drawn with 4H pencil Dimension figures Notes : - local note - general note Lettered with 2H pencil. EXTENSION LINES indicate the location on the object’s features that are dimensioned. DIMENSION LINES indicate the direction and extent of a dimension, and inscribe dimension figures. 27 o 123 13 10 43 indicate details of the feature with a local note. 27 10 Drill, 2 Holes R16 o 123 13 10 43 Recommended Practices EXTENSION LINES Leave a visible gap (≈ 1 mm) from a view and start drawing an extension line. Extend the lines beyond the (last) dimension line 1-2 mm. COMMON MISTAKE Visible gap EXTENSION LINES Do not break the lines as they cross object lines. COMMON MISTAKE Continuous DIMENSION LINES Dimension lines should not be spaced too close to each other and to the view. 34 11 35 16 Leave a space at least 2 times of a letter height. Leave a space at least 1 time of a letter height. DIMENSION FIGURES The height of figures is suggested to be 2.5~3 mm. Place the numbers at about 1 mm above dimension line and between extension lines. 11 34 11 34 COMMON MISTAKE DIMENSION FIGURES When there is not enough space for figure or arrows, put it outside either of the extension lines. Not enough space for figures 16.25 16.25 Not enough space for arrows 1 1 1 or DIMENSION FIGURES : UNITS The JIS and ISO standards adopt the unit of Length dimension in millimeters without specifying a unit symbol “mm”. Angular dimension in degree with a symbol “o” place behind the figures (and if necessary minutes and seconds may be used together). DIMENSION FIGURES : ORIENTATION 1. Aligned method The dimension figures are placed so that they are readable from the bottom and right side of the drawing. 2. Unidirectional method The dimension figures are placed so that they can be read from the bottom of the drawing. Do not use both system on the same drawing or on the same series of drawing (JIS Z8317) EXAMPLE : Dimension of length using aligned method. 30 30 30 30 30 30 30 30 EXAMPLE : Dimension of length using unidirectional method. 30 30 30 30 30 30 30 30 EXAMPLE : Dimension of angle using aligned method. 45o 45 o o 45 45 o 45o 45o o 45 45o EXAMPLE : Dimension of angle using unidirectional method. 45o 45o 45o 45o 45o 45o 45o 45o LOCAL NOTES Place the notes near to the feature which they apply, and should be placed outside the view. 10 Drill ≈ 10mm 10 Drill Too far 10 Drill COMMON MISTAKE Dimensioning Practices ### Tài liệu bạn tìm kiếm đã sẵn sàng tải về Tải bản đầy đủ ngay ×

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(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff) $a + b < \min(a,b) \Leftrightarrow a < 0 \and b < 0$ If both a and b are negative, $a + b < min(a, b)$. For example $a = -5$ $b = -7$ $a+ b = -5 + (-7) = -12$ $min(-5, -7) = -7$

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# 100 Lockers. Riddle 15 Lvl Easy? Suppose you're in a hallway lined with 100 closed lockers. You begin by opening every locker. Then you close every second locker. Then you go to every third locker and open it (if it's closed) or close it (if it's open). Let's call this action toggling a locker. Continue toggling every nth locker on pass number n. After 100 passes, where you toggle only locker #100, how many lockers are open? Nth means "ninth" by the way :3 My apologies, as someone pointed out to me, nth doesn't mean ninth, it's actually like a variable. >.< I had to go back and look it up to make sure. So pay the first update no mind. 0|0 1|5 • After quite some thinking i got it. Here's what happens. First, let's agree on the fact that prime numbers will be closed. Because the only factors they have are 1 and the number itself. So, if door opens at 1 , it has to close at the number itself. Secondly, all other numbers will have factor , which can be grouped into pairs , for example take number 24, all possible factors are , 1,2,3,4,6,8,12,24 which can be grouped as , 1x24 , 2x12, 3x8, 4x6. If door opens at 1 it will go like open close open close open close open close. It closes at the number itself. But the exception lies in perfect squares where you have 5x5, 6x6 where u cannot account for 5 or 6 two times, therefore the door remains open. 1,4,9,16,25,36,49,64,81,100 will remain open. 1|0 0|0 • and nth doesn't mean ninth , it means any general number satisfying the equation. It's like a variable. • Show All • ty for MHO • You're welcome

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10K - views # Nrich - answers to mode, median and mean questions. ## Nrich - answers to mode, median and mean questions. Download Presentation - The PPT/PDF document "Nrich - answers to mode, median and mean..." is the property of its rightful owner. Permission is granted to download and print the materials on this web site for personal, non-commercial use only, and to display it on your personal computer provided you do not modify the materials and that you retain all copyright notices contained in the materials. By downloading content from our website, you accept the terms of this agreement. ## Presentation on theme: "Nrich - answers to mode, median and mean questions."— Presentation transcript: Slide1 Nrich- answers to mode, median and mean questions. By Joshua and Matthew Slide2 Question 1 One of the answers to the question can you find 5 sets of positive numbers that make the mode < median < mean is …. 2,2,5,10,11- We got this answer because we thought that the first 2 numbers had to be identical and small because that would give us a small number as the mode. We then chose the median as 5 and then we thought that the mean had to be more than 5, so we chose 6! Next we did 6 x 5 which equals 30 so then we realised that all of the numbers added together must equal 30. 2+2+5 = 9 and 30-9 = 21 and then we picked 2 numbers which were different that equalled 21. This makes the Mode 2 Median 5 Mean 6 This question can also be completed only using 4 numbers. If you use the numbers 1,1,3,35 then the MODE = 1 MEDIAN = 2 MEAN = 10 Slide3 Question 2 One of the answers to the question can you find 5 sets of positive numbers that make the mode < mean < median is …. 2,2,6,7,8- We got this answer by first of all picking the first 2 numbers identical and small- the same as last question. We then picked the median as a reasonably big number, so we picked 6. After that we picked two numbers that were bigger than the median, but only just otherwise the mean would become bigger than the median. This makes the Mode 2 Mean 5 Median 6 Slide4 Question 3 For this question we couldn’t find an answer to the question One of the answers to the question can you find 5 sets of positive numbers that make the mean < mode < median only using 5 numbers. We couldn’t find an answer because the median (The 3 rd number) has to be bigger than the mode which has to be the first 2 numbers. After that you have to put 2 numbers bigger than the median into the sequence which will automatically make the mean bigger than the mode. Slide5 Question 4 One of the answers to the question can you find 5 sets of positive numbers that make the mean < median < mode is …. 3,4,7,8,8- We got this answer by making the 4 th and 5th numbers identical and slightly bigger than the median number which we chose as 7. This ensured that the mode would be bigger than the median as long as the first 2 numbers were different to any other in the sequence. We tried to make the mean 6 which would make the numbers total to 30. The 3 numbers we already had equalled 23 so we realised that the remaining 2 numbers must = 7, we chose these numbers as 4 and 3. This makes the …. Mean 6 Median 7 Mode 8 This question can also be completed using only 4 numbers. If you use the numbers 2,4,6,6 then the MEAN = 4.5 MEDIAN = 5 MODE = 6 Slide6 Question 5 One of the answers to the question can you find 5 sets of positive numbers that make the median < mode < mean. We think we couldn’t find the answer because the 3 rd number in the sequence has to be smaller than the mean and the mode. This means that the 2 numbers after the median have to be big and the same, because these 2 numbers are the biggest this means that it is impossible for the mean to be higher than the mode. Slide7 Question 6 One of the answers to the question can you find 5 sets of positive numbers that make the median < mean < mode is …. 4,5,6,10,10- We got this answer by first of all selecting a biggish number which we would have as the mode, we chose 10. After this we had to make sure the mean was bigger than the median so we worked out if the mean was to be 7, the 3 numbers would have to add up to 15 but the biggest of the numbers could only be up to 6. Then we found that 6+5+4= 15 so we used them numbers. mode 7 mean 10 Slide8 Let’s do the same using 6 numbers Mode < Median < Mean 3,3,6,8,12,16 Mode= 3 Median= 7 Mean= 8 Mode < Mean < Median 1,1,10,12,14,16 Mode= 1 Mean= 9 Median= 11 Mean < Mode < Median 1,100,100,101,102,103 Mean = 84.5 Mode= 100 Median= 100.5 Mean < Median < Mode 3,4,5,6,6,6 Mean= 5 Median= 5.5 Mode= 6 Median < Mode < Mean 3,4,5,7,7,34 Median= 6 Mode=7 Mean=10 Median < Mean < Mode 4,5,6,7,10,10 Median= 6.5 Mean= 7 Mode= 10 Slide9

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# paintingmississauga.com ## How To Divide Mixed Numbers how to divide mixed numbers save resource. how to divide mixed numbers mixed number division worksheet how to divide fractions math subtracting dividing mixed numbers worksheet number multiplication mixed number division. how to divide mixed numbers snapshot image of multiplying mixed numbers worksheets 1 and 2. how to divide mixed numbers course 1 5 8 multiplying mixed numbers. how to divide mixed numbers adding subtracting multiplying and dividing mixed numbers worksheet how to divide fractions with whole numbers math. how to divide mixed numbers image titled divide mixed fractions step 1. how to divide mixed numbers how to add mixed fraction adding fractions with like denominators. how to divide mixed numbers multiplying mixed numbers and fractions math worksheets math crush fractions obtain multiply mixed numbers fractions and. how to divide mixed numbers cover image. how to divide mixed numbers multiplying mixed numbers fractions 4 multiplying mixed numbers. how to divide mixed numbers multiplication of mixed numbers worksheet fraction multiplying mixed numbers by whole numbers worksheets. how to divide mixed numbers dividing mixed numbers calculator math dividing simplifying fractions math playground duck life. how to divide mixed numbers example 3 dividing mixed numbers a 3 6 multiply by the. how to divide mixed numbers multiply mixed numbers by mixed numbers using visual representations. how to divide mixed numbers dividing mixed fractions with whole numbers math dividing fractions by whole numbers worksheet awesome math pics. how to divide mixed numbers multiplying and dividing mixed numbers worksheet dividing mixed numbers worksheet multiplication of adding subtracting and multiplying. how to divide mixed numbers multiplying and dividing mixed numbers. how to divide mixed numbers image titled divide mixed fractions step 7. how to divide mixed numbers dividing mixed fractions and whole numbers math divide mixed number fraction using model dividing fractions free. how to divide mixed numbers. how to divide mixed numbers divide mixed fractions how to divide mixed fractions dividing and division worksheets math how to divide divide mixed. how to divide mixed numbers fraction word problems worksheets from multiplying mixed numbers improper fractions grade with answers division.

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Share Explore BrainMass # Description of Regression problem Mr. James McWhinney, president of Daniel-James Financial Services, believes there is a relationship between the number of client contacts and the dollar amount of sales. To document this assertion, Mr. McWhinney gathered the following sample information. The X column indicates the number of client contacts last month, and the Y column shows the value of sales (\$ thousands) last month for each client sampled. 1. Determine the regression equation. 2. Determine the estimated sales if 40 contacts are made. #### Solution Summary This posting contains the solution to the regression problem stated. It is solved using an attached Excel file. \$2.19

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# Difference between revisions of "Algebraic operation" $n$-ary operation, on a set $A$ A mapping $$\omega\colon A^n\to A$$ of the $n$-th Cartesian power of the set $A$ into the set $A$ itself. The number $n$ is known as the arity of the algebraic operation. Historically, the concepts of binary $(n=2)$ and unary ($n=1$) operations were the first to be considered. Nullary $(n=0)$ operations are fixed elements of the set $A$; they are also known as distinguished elements or constants. In the 20th century the concept of an infinitary operation appeared, i.e. a mapping $\omega\colon A^\alpha\to A$, where $\alpha$ is an arbitrary cardinal number. A set with a system of algebraic operations defined on it is called a universal algebra. The study of infinitary operations actually started in the late 1950s [a1]. A nullary operation is also called a noughtary operation [a2]. #### References [a1] Józef Słomiński, "The theory of abstract algebras with infinitary operations" Rozprawy Mat. , 18 (1959). Zbl 0178.34104 [a2] P.M. Cohn, "Universal algebra" (rev.ed.), Reidel (1981) pp. 13–14. ISBN 90-277-1213-1 Zbl 0461.08001 How to Cite This Entry: Algebraic operation. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Algebraic_operation&oldid=39753 This article was adapted from an original article by T.M. Baranovich (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article

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BETA This is a BETA experience. You may opt-out by clicking here # \$500 Million Jackpot: Calculating Your Odds Following Mega Millions logo (Photo credit: Wikipedia) The Mega Millions lottery drawing Friday night will deliver a jackpot estimated at \$540 million. Your odds of picking the winning combo are 175.7 million to one. Divide one number by the other. That means a ticket is worth \$3.07, right? Maybe you should buy as many as you can? Not so fast. There’s a little more to the math here. You have to allow for not only the long odds that you’ll pick the right numbers but also the short odds that you will be sharing that jackpot. It turns out that each ticket has a value a bit shy of the one dollar you pay for it. But at least you’re doing a lot better than you would in a lottery that doesn’t have a carryover from the previous drawing. If you like to buy lottery tickets, now is a good time. I’ll explain how you calculate odds, and how to improve them slightly. I’ll also give you some sad news about the time value of money and the taxation of gamblers. The jackpot is a giant one because no one picked the numbers to win the \$363 million pot offered in Tuesday’s drawing. That sum is added to a percentage of the take from ticket sales over the next 72 hours to arrive at the Mar. 30 pot. The frenzy of ticket buying that began Wednesday leads to an official projection of a \$540 million pot. I would bet that the operators are lowballing here. My speculative guess: a pot near \$600 million, with just under 620 million tickets sold. That would be more than triple the number sold for the Tuesday draw. You can buy tickets in 42 states plus the District of Columbia. Biggest jackpot so far: \$390 million in March 2007, shared by two winning tickets. To win the big one you have to pick which five numbers between 1 and 56 will be pulled, and then pick a second number between 1 and 46. The order of the first five is not important and there are no repeats. Here’s how you get the odds: multiply 56x55x54x53x52, then divide by 5x4x3x2 to get 3.82 million. Now multiply that by 46 to get 175.7 million. If you get the five-number combo but not the 1-to-46 play, then you win a flat \$250,000 (except in California, where it gets more complicated). Lesser matches have smaller payoffs, ranging down to \$2. If you knew no one else could play, it would make sense to buy a ticket. In fact, you should buy all the numbers. You’d invest \$176 million and haul in almost triple that. Alas, other people have heard about this jackpot. So what are the odds on sharing? In the past decade there have been 13 shared jackpots. Three jackpot winners neglected to claim their prizes; they were from Brooklyn, Queens and an unidentified place in New York State. History, though, is an imperfect  judge of odds. Here’s how to do it with a calculator. Call the number of tickets sold T. Call that 175.7 million figure X. Use ^ as the sign for exponents (2^3, for example, means 2x2x2). Call the base of natural logarithms e (it’s roughly 2.718). If you did win, what are the odds that no one else won? It equals (1-1/X)^T. That’s a mess, but a very close approximation is e^(-T/X). Comes to 0.03 on my calculator when I assume that T = 616 million. Translation: If you do win, there’s only a 3% chance you’ll have the big prize to yourself. Odds there will be exactly one other winner? Take the 0.03 and multiply by T/X. I get 0.105. So there's a 10.5% chance that if you win, you will be splitting the pot in two. Odds for two other winners: take the last number you had, multiply by T/X again and divide by 2. Now you have 0.184. For three other winners: previous number, multiplied by T/X and divided by 3. Now you're up to 0.216. This could go on, but the numbers start to get pretty small. There’s only a 1.7% chance that if you win you’ll be splitting the prize 9 ways. Add it all up, and you find that a \$600 million pot with 616 million tickets competing makes each chance on the big prize worth 94 cents. Now there’s the time value problem. Turns out you don’t get the prize right away. You have to take it in installments. It reminds one of that joke from Soviet days: Did you hear about the \$1 million Russian lottery? You win \$1 a year for 1 million years. The Mega Millions deal isn’t quite that bad, but if you want your money right away there’s a 28% haircut. So your 94 cents shrinks to 68 cents. The smaller prizes are worth something, somewhere around 18 cents. Add it up: Each one-dollar ticket is worth not even a dollar. Now there’s the little problem of taxes. Big wins (\$10,000 and up) are taxable. The smaller ones (\$150 and down) are not, because you are entitled to deduct gambling losses up to the amount of gambling wins, and if you win \$150 you probably spent more than \$150 on tickets. If you live in New York, you have a high probability of being subject to federal, state and city taxes, and a federal alternative minimum tax for good measure. You also have, as noted above, a high probability of being too dumb to collect. The tax collectors don’t stop with income tax. This brings us to the curious case of the Waffle House waitress, decided a few weeks ago in U.S. Tax Court. A patron of the Grand Bay, Ala. Waffle House decided to leave behind a Florida lottery ticket. He evidently either didn’t realize that the lottery had been drawn the night before or just didn’t check the numbers. Had he done so, he would have realized that he was leaving a \$5 million tip. The waitress, being a good soul, transferred the ticket to a corporation in which relatives were shareholders. There was an unwritten promise in her family, she said, to share winnings. The IRS said she was just being nice and hit her up for a \$771,570 gift tax. This would be on top of any income tax being paid by her or the relatives. The judge sided largely with the IRS, although he cut the tax bill a little on the theory that, at the time of the ticket transfer, there was some risk that fellow restaurant workers would horn in on the jackpot. They sued her but she won that battle. So, how do you improve your expected payoff in Mega Millions? Pick numbers between 32 and 56. They are no more likely to be picked than smaller numbers, but they are less likely to be shared. That’s because a lot of buyers play hunches that involve birthdays. Even if you don't win, get help with your taxes from The  Forbes 2012 Tax Guide

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https://pythonstars.com/what-is-quick-sort-algorithm-how-to-use-it-in-python/

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Quick sort algorithm is a popular sorting algorithm that uses the divide and conquer strategy to sort an array of elements. It works by selecting a pivot element and partitioning the array around the pivot such that all elements on the left of the pivot are smaller than the pivot, and all elements on the right are larger. The partitioning is done recursively until the entire array is sorted. In this article, we will explain in detail what quick sort algorithm is, how it works, and provide examples of implementing the algorithm using Python code. ## How Quick Sort Algorithm Works Quick sort algorithm follows the divide and conquer strategy, which involves the following steps: 1. Select a pivot element from the array. 2. Partition the array around the pivot such that all elements on the left of the pivot are smaller than the pivot, and all elements on the right are larger. 3. Recursively apply the above two steps to the sub-arrays until the entire array is sorted. The pivot element can be selected using different techniques, such as selecting the first or last element, selecting a random element, or selecting the median element. The partitioning is done by comparing each element to the pivot and placing it on the left or right of the pivot accordingly. ## Python Code Implementation Here is an example of implementing quick sort algorithm in Python: ``````def quick_sort(arr): if len(arr) <= 1: return arr else: pivot = arr[0] left_half = [x for x in arr[1:] if x < pivot] right_half = [x for x in arr[1:] if x >= pivot] return quick_sort(left_half) + [pivot] + quick_sort(right_half) `````` The above code implements quick sort algorithm using the `quick_sort` function, which takes an array `arr` as input and sorts it in ascending order. The function first checks if the length of the array is less than or equal to one. If it is, the function returns the array as it is already sorted. Otherwise, the function selects the first element as the pivot and partitions the array around the pivot. Then, the function recursively calls itself on the left and right sub-arrays and combines the sorted sub-arrays with the pivot element to produce the final sorted array. Quick sort algorithm has the following advantages: • Efficient for large data sets due to its time complexity of O(n log n) • In-place sorting algorithm, meaning that it does not require additional memory to sort the array • Faster in practice than other sorting algorithms for small data sets However, quick sort algorithm has the following disadvantages: • Not stable, meaning that it does not preserve the order of equal elements in the array • Worst-case time complexity of O(n^2) when the pivot is chosen poorly ## Conclusion Quick sort algorithm is a popular sorting algorithm that uses the divide and conquer strategy to sort an array of elements. It is efficient for sorting large data sets and is an in-place sorting algorithm. In this article, we have explained in detail what quick sort algorithm is, how it works, and provided examples of implementing the algorithm using Python code. For complete list of topic on DATA STRUCTURE AND ALGORITHM click hear ## FAQs 1. What is the time complexity of quick sort algorithm? • The time complexity of quick sort algorithm is O(n log n) on average and O(n^2) in the worst case. 1. Is quick sort a stable sorting algorithm? • No, quick sort is not a stable sorting algorithm as it does not preserve the order of equal elements in the array.

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jGuru Forums Unary Minus 2 posts in topic Flat View TOPIC ACTIONS: Posted By:   Quirin_Meyer Posted On:   Tuesday, November 11, 2003 05:09 AM how do i realize a *real* unary minus i have a problem with unary minus. i use standart grammatics for arithmetic expressions found all over the place, but they all seem to have the same problem (or i guess, i only have) - they are not parsing unary operators correctly here the grammar: primary_expr : constant | (LPAREN! expr RPAREN! ) ; sign_expr : (MINUS)? primary_expr ; mul_expr : sign_expr (( STAR^ | DIV^) sign_expr)* ; expr : mul_expr (( PLUS^ | MINUS^) mul_expr)* ; constant : NUMBER ; -2+3 is a correct expression according to those grammars but   More>> how do i realize a *real* unary minus i have a problem with unary minus. i use standart grammatics for arithmetic expressions found all over the place, but they all seem to have the same problem (or i guess, i only have) - they are not parsing unary operators correctly here the grammar: ``` primary_expr : constant | (LPAREN! expr RPAREN! ) ; sign_expr : (MINUS)? primary_expr ; mul_expr : sign_expr (( STAR^ | DIV^) sign_expr)* ; expr : mul_expr (( PLUS^ | MINUS^) mul_expr)* ; constant : NUMBER ; ``` -2+3 is a correct expression according to those grammars but -2+-3 is one as well but doesn't make much sense, even to a second grader ;-) and even worse: this word (-2+-3) gets accepted and produces (+ (-2) (-3)) (in lisp notation) could anyone possible help me to change that grammar so that expressions such as the above do not get accepted by the parser. thanks a lot. q    <<Less Re[2]: Unary Minus Posted By:   Quirin_Meyer Posted On:   Sunday, November 16, 2003 08:52 AM thanks a lot.. that looks almost too good. but i miss the sign_expr production.. q Re: Unary Minus Posted By:   Sebastian_Kaliszewski Posted On:   Thursday, November 13, 2003 12:29 PM The following `primary_expr : constant | (LPAREN! expr RPAREN! ) ;mul_expr : sign_expr (( STAR^ | DIV^) sign_expr)* ;expr : (MINUS^)? mul_expr (( PLUS^ | MINUS^) mul_expr)* ;constant : NUMBER ;` does the job, but notice that now unary minus is just a special case of binary sumation/difference operators (has the same priority) -- this is the way it's introduced at school. Is it what you want? rgds Sebastian

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https://kurser.ku.dk/course/ndaa09009u/

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# NDAA09009U Numerical Optimisation (NO) Volume 2021/2022 Education MSc Programme in Bioinformatics MSc Programme in Computer Science MSc Programme in Physics MSc Programme in Statistics Content Numerical optimisation is a useful computer tool in many disciplines like image processing, computer vision, machine learning, bioinformatics, eScience, scientific computing and computational physics, computer animation and many more. A wide range of problems can be solved using numerical optimisation like; inverse kinematics in robotics, image segmentation and registration in medical imaging, protein folding in computational biology, stock portfolio optimisation, motion planning and many more. This course will build up a toolbox of numerical optimisation methods which the student can use when building solutions in his or her future studies. Therefore this course is an ideal supplement for students coming from many different fields of science. This course teaches the basic theory of numerical optimisation methods. The focus is on deep learning, and how the methods covered during the course works. Both on a theoretical level that goes into deriving the math but also on an implementational level focusing on computer science and good programming practice. There will be weekly programming exercises where students will implement the algorithms and methods introduced from theory on their own case-study problems like computing the motion of a robot hand or fitting a model to highly non-linear data or similar problems. The topics covered during the course are: • First-order optimality conditions, Karush-Kuhn-Tucker Conditions, Taylors Theorem, Mean Value Theorem. • Nonlinear Equation Solving: Newtons Method, etc. • Linear Search Methods: Newton Methods, Quasi-Newton Methods, etc. • Trust Region Methods: Levenberg-Marquardt, Dog leg method, etc. • Linear Least-squares fitting, Regression Problems, Normal Equations, etc. • And many more... Learning Outcome Knowledge of • The theory of gradient descent method • The theory of Newton and Quasi-Newton Methods • The theory of Thrust Region Methods • The theory of quadratic programming problems • First-order optimality conditions (KKT conditions) Skills in • Applying numerical optimisation problems to solve unconstrained and constrained minimisation problems and nonlinear root search problems • Reformulating one problem type into another form - i.e. reformulating root search into minimisation and vice versa • Implementing and testing numerical optimisation methods Competences to • Evaluate which numerical optimisation methods are best suited for solving a given optimisation problem • Understand the implications of theoretical theorems and being able to analyse real problems on that basis See Absalon when the course is set up. We do not impose any programming language, however Python will be the programming language of choice and it is expected that students know how to install and use Python, Numpy and Matplotlib by themselves. It is also expected that students know what matrices and vectors are and that students are able to differentiate vector functions. Theorems like fundamental theorem of calculus, mean value theorem or Taylor's theorem will be touched upon during the course. The inquisitive students may find more in depth knowledge from Chapters 2, 3, 5, 6 and 13 of R. A. Adams, Calculus, 3rd ed. Addison Wesley. Academic qualifications equivalent to a BSc degree is recommended. Mixture of study groups and project group work with hand-ins (flipped classroom). • Category • Hours • Preparation • 50 • Exercises • 72 • Project work • 84 • Total • 206 Written Oral Individual Continuous feedback during the course of the semester Credit 7,5 ECTS Type of assessment Continuous assessment The assessment is based on 5-7 written assignments (must be individually approved). The student acting as opponent in respect of fellow students’ work is also considered as part of the continuous assessment. The final grade is based on an overall assessment. Aid All aids allowed Marking scale

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View the step-by-step solution to: 2-2 Three-point Estimate Problem (a) A project requires the use of structural steel in several tasks over the 12 month period expected to complete... 2-2 Three-point Estimate Problem (a) A project requires the use of structural steel in several tasks over the 12 month period expected to complete the work. Based on recent experience the most likely price for the material is \$0.30/lb. There are, however, price fluctuations over time based on suppliers used and structural steel availability. The most optimistic price estimate is \$0.24/lb., and the most pessimistic estimate is \$0.62/lb. Use the three-point estimating process to determine the expected price of the material. (b) In addition to price fluctuations, it is also uncertain how much of structural steel will be required for the project. Design uncertainties and waste at the project site will affect the amount of structural steel actually needed. The most likely amount required is 40 tons. However, as little as 37 tons and as much as 49 tons might be required. One ton = 2000 lbs. Use the three-point estimating process to determine to determine the expected amount of structural steel needed for the project? (c) Using the estimates from 1 & 2, what is the expected cost for structural steel over the life of the project? (d) Using the "Complex" three-point method, what is the expected cost for structural steel over the life of the project? Expected price = \$0.34 per pound Expected usage = 41 tons Expected total cost (simple method) = \$28,153 (\$27,880 using rounded numbers) Expected total cost (complex method) = \$31,856 (\$31,673 using rounded numbers) This question was asked on May 14, 2011. Why Join Course Hero? Course Hero has all the homework and study help you need to succeed! We’ve got course-specific notes, study guides, and practice tests along with expert tutors and customizable flashcards—available anywhere, anytime. - Educational Resources • - Study Documents Find the best study resources around, tagged to your specific courses. Share your own to gain free Course Hero access or to earn money with our Marketplace. Browse Documents

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DIY Paper Hexahexaflexagon 411 4 1 You may be wondering to yourself, "What is a hexahexaflexagon?" A hexahexaflexaon is a fun, hexagonal paper toy that has many faces that you can flex to reveal. This instructable will walk you through how to make your own! Supplies: All you need is: • a roll of adding machine tape, 2 1/4" or 57.15 mm wide • a glue stick • a mechanical pencil • a metric ruler • Optional - coloring medium of your choice Teacher Notes Teachers! Did you use this instructable in your classroom? Add a Teacher Note to share how you incorporated it into your lesson. Step 1: Marking the Lines To start, along one edge of the paper mark every 66 mm with your pencil until you have 10 marks. In the pictures I have put the marks in bold. From the corner, measure and mark 66 mm with your ruler diagonally to the other side of the paper. Cut on this line so you have an angle on the end. Then mark every 66 mm on that side until you have 10 marks. Using your ruler, you should now be able to connect these lines bridging across sides to form 19 perfect equilateral triangles. Cut at the last diagonal line. Step 2: Numbering the Triangles Using your pencil, lightly number the side still facing up by numbering 1-2-3 six times and leaving the last triangle blank as shown. Then flip the paper over and leave the first one blank and number in the pattern 4-4-5-5-6-6 three times as shown. You will have hold the paper up to the light to look through it to see where to number the back, although you don't have to be exact as you will erase them later. Step 3: Folding With the 1-2-3 side facing up, fold on the lines so that on the 4-5-6 side, the same numbers next to each other match up. You should end up with the flattened spiral in the first picture and have a blank triangle on each end. To fold into a hexagon, pay close attention to the second picture. Starting from the right end, fold the two 1s right next two each other, three from the right, down as shown. Then, going down, fold the other two 1s up and lift the two over the three so it is a hexagon with a 1 sticking of to the side. Step 4: Gluing Flip the hexagon over. There should be two blanks right next to each other. Put glue on one and press them together exactly on top of each other. Step 5: Flexing To flex it, push the sides in so you get a sort of three pointed star as shown in the third picture. Then pull the middle apart to form a completely new hexagon. If you did it right, there should be a different number on each of the triangles. If a side won't open, shift to a different corner and do it there instead. Step 6: (Optional) Decorating To make this look a little cooler, you can erase the numbers and draw a design on a side. Then flex it and do that side too until you have done all six sides. I did a planets theme but you can do anything you want. ***WARNING*** Be careful to make a design that will work in any orientation of the triangles because when it comes back up, it will not be in the same orientation. Step 7: Have Fun! Now you have your hexahexaflexagon! Have fun playing with it! Participated in the 90 13K 2 146 14K 97 8.4K Discussions Cool! Do you have any video of it in action?

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# Another Matrix Order Question :) This topic is 2381 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic. ## Recommended Posts I'm trying to get my head around row/column major matrices and row/column vectors. Mostly, I'm using the following for reference: http://www.j3d.org/matrix_faq/matrfaq_latest.html So, trying to write a 4x4 matrix class, I have a class with an array of values: float m_Values[16]; Now, I'm going to be using column vectors, so: 1) Am I right in saying the following: Multiplication should be written like so: Matrix * Vector Combining rotation matrices should be written (for example): Projection * View * World 2) Row/Column major matrices refers to the order the data is stored in memory, NOT whether or not the translation component should be down the right hand side / along the bottom. The translation is down the right hand side of the matrix when using column vectors, and across the bottom when using row vectors. Is this correct? 3) OpenGL uses column vectors, and uses column major matrices, and we want to upload the matrices to OpenGL using glUniformMatrix4fv I'm pretty sure this is right, so does that mean that the translation component should be in m_Values[12], m_Values[13] and m_Values[14]? 4) When multiplying two matrices together, the operator* should have something like the following final.m_Values[0] = (this.m_Values[0] * other.m_Values[0]) + (this.m_Values[1] * other.m_Values[4]) + (this.m_Values[2] * other.m_Values[8]) + (this.m_Values[3] * other.m_Values[12]); final.m_Values[1] = (this.m_Values[0] * other.m_Values[1]) + (this.m_Values[1] * other.m_Values[5]) + (this.m_Values[2] * other.m_Values[9]) + (this.m_Values[3] * other.m_Values[13]); final.m_Values[2] = (this.m_Values[0] * other.m_Values[2]) + (this.m_Values[1] * other.m_Values[6]) + (this.m_Values[2] * other.m_Values[10]) + (this.m_Values[3] * other.m_Values[14]); I'm not sure if this order is right, or if I need to multiply a different way because the matrices are stored in column major order in memory. Can anyone clarify if this is the right way or not? 5) The Matrix Quaternion FAQ I linked to at the top seems to be using column vectors, so vectors are multiplied on the right, as shown in Q13 of the FAQ. Q35 appears to suggest that matrices should be multiplied the other way though. Q35 states that in order to get an X rotation, followed by a Y rotation, followed by a Z rotation, you multiply: M = X.Y.Z But I thought that, because it is using column vectors, the first transformation should be at the right hand side. I.e.: M = Z.Y.X In fact, looking at the function calls listed at the start of Q36, it seems to be multiplying in the order I'd think was correct (i.e. Z * Y * X), but still goes on to state that M = X.Y.Z. Are Qs 35 & 36 wrong, or is it just a difference in notation for multiplying matrices instead of vectors (as Q13 says that V' = M.V, which seems like the right way round to me). Or is it just that I'm very confused? ;) Hopefully someone can help me out on this, it would be very useful. Thanks 1. 1 Rutin 29 2. 2 3. 3 4. 4 5. 5 • 13 • 13 • 11 • 10 • 14 • ### Forum Statistics • Total Topics 632961 • Total Posts 3009493 • ### Who's Online (See full list) There are no registered users currently online ×

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# How to solve this seemingly simple system of algebraic equations? I mean $$\text{z1}+\text{z2}+\text{z3}+\text{z4}=0\land | \text{z1}| =| \text{z2}| \land | \text{z3}| =| \text{z2}| \land | \text{z3}| =| \text{z4}|$$ in $$\text{z1},\text{z2},\text{z3},\text{z4}$$ over the complexes. Here are my unsuccessful trials in 13.2 on Windows 10. (i) Solve[z1 + z2 + z3 + z4 == 0 && Abs[z1] == Abs[z2] && Abs[z3] == Abs[z2] && Abs[z3] == Abs[z4], {z1, z2, z3,z4}, Complexes] Solve::ifun: Inverse functions are being used by Solve, so some solutions may not be found; use Reduce for complete solution information. Solve::svars: Equations may not give solutions for all "solve" variables. {{z2->z1,z3->-z1,z4->-z1},{z2->-z1,z3->z1,z4->-z1},{z2->-z1,z3->-z1,z4->z1},{z1->0,z2->0,z3->0,z4->0}} Most of the solutions are lost. (ii) The command Reduce[z1 + z2 + z3 + z4 == 0 && Abs[z1] == Abs[z2] && Abs[z3] == Abs[z2] && Abs[z3] == Abs[z4], {z1, z2, z3, z4}, Complexes] almost crashes my comp: "Kernel connection is lost". (iii) With the additional constraint Abs[z1]==1, Solve[z1 + z2 + z3 + z4 == 0 && Abs[z1] == Abs[z2] && Abs[z3] == Abs[z2] && Abs[z3] == Abs[z4] && Abs[z1] == 1, {z1, z2, z3, z4}, Complexes] Solve::ifun: Inverse functions are being used by Solve, so some solutions may not be found; use Reduce for complete solution information. {{z1 -> 1, z2 -> 1, z3 -> -1, z4 -> -1}, {z1 -> 1, z2 -> -1, z3 -> 1, z4 -> -1}, {z1 -> -1, z2 -> 1, z3 -> 1, z4 -> -1}, {z1 -> 1, z2 -> -1, z3 -> -1, z4 -> 1}, {z1 -> -1, z2 -> 1, z3 -> -1, z4 -> 1}, {z1 -> -1, z2 -> -1, z3 -> 1, z4 -> 1}} Most of the solutions are lost. (iv) Reduce[z1 + z2 + z3 + z4 == 0 && Abs[z1] == Abs[z2] && Abs[z3] == Abs[z2] && Abs[z3] == Abs[z4] && Abs[z1] == 1, {z1, z2, z3, z4}, Complexes] is running without any response on my comp for hours. The resources of my comp are not exhausted. Likely an infinite loop is formed. (v) The switching to the reals by Reduce[x1 + x2 + x3 + x4 == 0 && y1 + y2 + y3 + y4 == 0 && x1^2 + y1^2 == x2^2 + y2^2 && x3^2 + y3^2 == x2^2 + y2^2 && x3^2 + y3^2 == x4^2 + y4^2 && x1^2 + y1^2 == 1, {x1, x2, x3, x4, y1,y2, y3, y4},Reals] does not help. The command is running without any response on my comp for hours. The resources of my comp are not exhausted. Likely an infinite loop is created. • If the absolute values of all the complexes are the same, it should help to reformulate the problem as z1 = r Exp[I phi1] etc. and impose that r >= 0. Commented Apr 5, 2023 at 15:28 • As I know it, the geometric interpretation of the solutions is the following: z1, z2, z3, z4 are vertices of a rectangle (maybe, a degenerate one) in the complex plane centered at the origin Commented Apr 5, 2023 at 17:21 • @SjoerdSmit, you can go even further by taking r==1 Commented Apr 5, 2023 at 19:35 • This is a 2-parameter family if we restrict length to be 1. Commented Apr 5, 2023 at 22:58 • The geometric interpretation is more general: z1,z2,z3,z4 form a rhombus! Commented Apr 7, 2023 at 15:42 By taking z1==1, you can find a solution set that you can scale by any complex constant. Decomposing into real and imaginary components I have eqn = 1 + c1 + c2 + c3 == 0 && s1 + s2 + s3 == 0 && c1^2 + s1^2 == c2^2 + s2^2 == c3^2 + s3^2 == 1; This can be solved easily using Reduce[eqn, Reals] Alternatively, you find a few interesting values, e.g. FindInstance[1 + c1 + c2 + c3 == 0 && s1 + s2 + s3 == 0 && c1^2 + s1^2 == c2^2 + s2^2 == c3^2 + s3^2 == 1, {c1, c2, c3, s1, s2, s3}, Reals, 8] • Thank you. I find this approach advanced. Sum spots on the Sun. (i) "By taking z1==1, you can find a solution set that you can scale by any complex constant" should be elaborated. Commented Apr 6, 2023 at 3:04 • (ii) Your result is a=((c3 == 0 && ((c2 == 0 && ((s2 == 1 && 1 + s3 == 0) || (1 + s2 == 0 && s3 == 1))) || (1 + c2 == 0 && s2 == 0 && (s3 == 1 || 1 + s3 == 0)))) || (s3 == 0 && ((1 + c3 == 0 && ((c2 == 0 && (1 + s2 == 0 || s2 == 1)) || (s2 == 0 && (c2 == 1 || 1 + c2 == 0)) || ((-1 < s2 < 0 || 0 < s2 < 1) && (Sqrt[1 - s2^2] == c2 || c2 + Sqrt[1 - s2^2] == 0)))) || Commented Apr 6, 2023 at 3:06 • (1 + c2 == 0 && c3 == 1 && s2 == 0))) || ((-1 < s3 < 0 || 0 < s3 < 1) && ((s2 == 0 && c2 + Sqrt[c3^2 + s3^2] == 0 && (Sqrt[1 - s3^2] == c3 || c3 + Sqrt[1 - s3^2] == 0)) || (s2 + s3 == 0 && ((c3 + Sqrt[1 - s3^2] == 0 && Sqrt[c3^2 - s2^2 + s3^2] == c2) || (Sqrt[1 - s3^2] == c3 && c2 + Sqrt[c3^2 - s2^2 + s3^2] == 0)))))) && s1 + s2 + s3 == 0 && 1 + c1 + c2 + c3 == 0 . This description leaves much to be desired. For example a /. {c3 -> 1/2, s1 -> 1/4, s3 -> -1/4} results in False. Commented Apr 6, 2023 at 3:11 • (iii) I repeat "As I know it, the geometric interpretation of the solutions is the following: z1, z2, z3, z4 are the vertices of a rectangle (maybe, a degenerate one) in the complex plane centered at the origin". Commented Apr 6, 2023 at 3:18

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# How do I set the point to rotate at on SPROT? BlackDrag0n23Created: Essentially, how do set the origin point of a sprite? Use SPHOME `SPHOME ID,X,Y` For example, if you want to set it to the middle of a 16x16 sprite, you would use SPHOME ID,8,8 This also sets the point that is controlled by SPOFS, and the center for scaling. Why thank you kind sir. What if I wanted to make an animal rotate and move in that direction? What if I wanted to make an animal rotate and move in that direction? From the sound of it, I believe you would just need a simple loop for it: ```ACLS X=100 Y=50 SPSET 1,[animal sprite number] SPHOME 1,8,8 WHILE 1 VSYNC 1 SPOFS 1,X,Y SPROT 1,360 INC X,1 WEND ``` Unfortunately, I don’t know the syntax for SPROT very well, so you may have to change that slightly. What if I wanted to make an animal rotate and move in that direction? From the sound of it, I believe you would just need a simple loop for it: ```ACLS X=100 Y=50 SPSET 1,[animal sprite number] SPHOME 1,8,8 WHILE 1 VSYNC 1 SPOFS 1,X,Y SPROT 1 INC X,1 WEND ``` Unfortunately, I don’t know the syntax for SPROT very well, so you may have to change that slightly. Thank you! What if I wanted to make an animal rotate and move in that direction? From the sound of it, I believe you would just need a simple loop for it: ```ACLS X=100 Y=50 SPSET 1,[animal sprite number] SPHOME 1,8,8 WHILE 1 VSYNC 1 SPOFS 1,X,Y SPROT 1 INC X,1 WEND ``` Unfortunately, I don’t know the syntax for SPROT very well, so you may have to change that slightly. Thank you! i'm having trouble. I don't quiet know what i'm doing wrong. the fish either goes in a random direction or not at a;; snip the fish either goes in a random direction or not at a;; Ever heard of the edit button? or Post checking? snip the fish either goes in a random direction or not at a;; Ever heard of the edit button? or Post checking? Sorry didin't think that someone would be watching every single thing I post. XD i'm having trouble. I don't quiet know what i'm doing wrong. the fish either goes in a random direction or not at a;; Sorry, it may have been a problem with my first post of the code. I’ve fixed up my code to make the sprite continue rotating instead of just going to a single specified angle: ```ACLS X=100 Y=50 SPSET 1,[animal sprite number] SPHOME 1,8,8 WHILE 1 VSYNC 1 SPOFS 1,X,Y SPROT 1,R INC X,1 INC R,2 IF R>360 THEN R=0 WEND ``` You can make the sprite spin faster by increasing the number that R is raised by in INC R,2. You can do you same for moving the sprite to the right by increasing the number that X is increased by.

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# In an integer program, how I can force a binary variable to equal 1 if some condition holds? Suppose we have a binary or continuous variable $$x$$, a binary variable $$y$$, and a constant $$b$$, and we want to enforce a relationship like If $$x \gtreqless b$$, then $$y = 1$$. How can we write this using one or more linear constraints? If $$x$$ is binary: Then the "if" condition really means either "$$x = 0$$" or "$$x=1$$". To enforce "if $$x=0$$ then $$y=1$$": use $$y \ge 1-x.$$ To enforce "if $$x=1$$ then $$y=1$$": use $$y \ge x.$$ If you want to require that $$y=1$$ if and only if the condition holds, then replace the $$\ge$$s above with $$=$$s. If $$x$$ is continuous: In this case, numerical inaccuracy might produce errors, so be prepared for $$y$$ to be set incorrectly if $$x$$ is close to but on the “wrong” side of $$b$$. To avoid this, you can increase or decrease $$b$$ a little bit to provide some buffer. To enforce "if $$x < b$$ then $$y=1$$": $$b - x \le My,$$ where $$M$$ is a large constant. The logic is that if $$b - x > 0$$, then $$y$$ must equal 1, and otherwise it may equal 0. To enforce "if $$x > b$$ then $$y=1$$": $$x - b \le My,$$ with similar logic as above. To enforce "if $$x = b$$ then $$y=1$$": This one is tricky. I'm not sure my approach is the easiest. (Anyone have a better solution?) We really can't check for $$x=b$$, but we can check for $$b-\delta \le x \le b+\delta$$ for some small $$\delta > 0$$. To do this, we introduce two new binary decision variables. Let $$z_1$$ be a binary variable that equals 1 if $$x > b - \delta$$, equals 0 if $$x < b - \delta$$, and could equal either if $$x = b - \delta$$. Enforce this definition by adding the following constraints: \begin{alignat}{2} Mz_1 & \ge x - b + \delta\tag1 \\ M(1-z_1) & \ge b - x - \delta\tag2 \end{alignat} The logic is: • If $$x > b - \delta$$, then (1) forces $$z_1=1$$ and (2) has no effect. • If $$x < b - \delta$$, then (2) forces $$z_1=0$$ and (1) has no effect. • If $$x = b - \delta$$, then (1) and (2) have no effect; $$z_1$$ could equal either 0 or 1. Next, introduce a second binary variable $$z_2$$, which equals 1 if $$x < b + \delta$$, equals 0 if $$x > b + \delta$$, and could equal either if $$x = b + \delta$$. Introduce the following constraints: \begin{alignat}{2} Mz_2 & \ge b - x + \delta\tag3 \\ M(1-z_2) & \ge x - b - \delta\tag4 \end{alignat} The logic is similar: • If $$x < b + \delta$$, then (3) forces $$z_2=1$$ and (4) has no effect. • If $$x > b + \delta$$, then (4) forces $$z_2=0$$ and (3) has no effect. • If $$x = b + \delta$$, then (3) and (4) have no effect; $$z_2$$ could equal either 0 or 1. From constraints (1)-(4), we can say that if $$z_1=z_2=1$$, then $$b - \delta \le x \le b + \delta$$. Therefore, we can enforce "if $$b - \delta \le x \le b + \delta$$ then $$y=1$$" using: $$y \ge z_1 + z_2 - 1.$$ Note: If your model is relatively large, i.e., it takes a non-negligible amount of time to solve, then you need to be careful with big-$$M$$-type formulations. In particular, you want $$M$$ to be as small as possible while still enforcing the logic of the constraints above. • In the first part where $x$ is binary, what if I have 2 or more binary variable that leads to the decision of $y$ like $x$ and $z$ when $x=1 \wedge z = 1$ then $y = 1$ – ooo Jan 28, 2020 at 21:22 • Then you'd have to formulate separate constraints in which you enforce the definition of $y$ and then use $y$ as described above. Jan 29, 2020 at 14:33 • I didn't get it. – ooo Jan 29, 2020 at 19:38 • Lets say I have 4 binary variable $a,b,c,d$ if all $a,b,c,d = 1$ the $x = 1$ else $x =0$, then can I write $x \ge 3 - a+b+c+d$ – ooo Jan 30, 2020 at 13:57 • @LarrySnyder610: how small could we set $\delta$ to? Jun 4, 2020 at 17:39 Rather than linearising the logical constraint, I would try the logical constraints built in a solver. Gurobi and SCIP both have indicator constraints. My colleague works with these a lot and he’s finding the indicator constraints in Gurobi perform worse than big-M. He’s in contact with the Gurobi developers so I might be able to get more info if there’s interest. • I've never tried those; that's a good suggestion. More info would certainly be welcome (maybe in a new Q&A). May 31, 2019 at 15:36 To model $$x=b \implies y=1$$, where $$L \le x \le U$$, you can do the following: \begin{align} L y^- + b y + (b+\delta)y^+ \le x &\le (b-\delta) y^- + b y + U y^+\\ y^- + y + y^+ &= 1 \\ y^-, y, y^+ &\in \{0,1\} \end{align} In fact, this formulation also enforces the converse $$y=1 \implies x=b$$. • If we assume $L=b-\delta$ and $U=b+\delta$, then the first constraint is essentially: $L y^- + b y + Uy^+ \le x \le L y^- + b y + U y^+$ or $x = L y^- + b y + U y^+$. – EhsanK Jan 17, 2020 at 1:40 • True, but the idea here is that $\delta>0$ is small, so those assumptions on $L$ and $U$ would imply that $x$ is essentially constant. The more useful setting would be when $L\ll b\ll U$. Jan 17, 2020 at 1:48

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In the conflict between the israelis and the palestinians, : GMAT Sentence Correction (SC) Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack It is currently 22 Jan 2017, 00:26 GMAT Club Daily Prep Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email. Customized for You we will pick new questions that match your level based on your Timer History Track every week, we’ll send you an estimated GMAT score based on your performance Practice Pays we will pick new questions that match your level based on your Timer History Events & Promotions Events & Promotions in June Open Detailed Calendar In the conflict between the israelis and the palestinians, Author Message TAGS: Hide Tags Intern Joined: 24 Aug 2006 Posts: 8 Followers: 0 Kudos [?]: 1 [0], given: 0 In the conflict between the israelis and the palestinians, [#permalink] Show Tags 31 Aug 2006, 05:02 1 This post was BOOKMARKED 00:00 Difficulty: (N/A) Question Stats: 80% (01:50) correct 20% (01:23) wrong based on 45 sessions HideShow timer Statistics In the conflict between the israelis and the palestinians, the refusal of each side to acknowledge each other as a legitimate national movement is closer to the heart of the problem than is any other issue. a- the refusal of each side to acknowledge each other as a legitimate national movement is closer to the heart of the problem than b- that the refusal of each side to acknowledge another as a legitimate national movement is closer to the heart of the problem as c- the refusal of each side to acknowledge another as a legitimate national movement is closer to the heart of the problem than d- that the refusal of each side to acknowledge each other as a legitimate national movement is closer to the heart of the problem than e- the refusal of each side to acknowledge each other as a legitimate national movement is closer to the heart of the problem as Edited for clarity If you have any questions New! Manager Joined: 09 Aug 2005 Posts: 72 Followers: 1 Kudos [?]: 2 [1] , given: 0 Show Tags 31 Aug 2006, 05:26 1 KUDOS I'd go with A. The clause cant begin with "that", so it brings us down to A, C, E. It has to end with "than", not "as" since there is no as (adj) as clause. This would require the phrase "as close as" which none of them have. Thus we are down to A and C. C has "acknowledge another" which is wrong. So its A. ( Actually as far as I'm concerned even " each side to acknowledge each other" is wrong, it should be "each side to acknowledge the other". ) MG Senior Manager Joined: 15 Aug 2004 Posts: 329 Followers: 1 Kudos [?]: 8 [0], given: 0 Show Tags 31 Aug 2006, 05:33 Should be A... Each Other - When focussing on two things Each Another - When focussing on more than two... Senior Manager Joined: 05 Jun 2005 Posts: 454 Followers: 1 Kudos [?]: 41 [0], given: 0 SC --> each other x another [#permalink] Show Tags 31 Aug 2006, 07:17 ps_dahiya wrote: sumitsarkar82 wrote: Should be A... Each Other - When focussing on two things Each Another - When focussing on more than two... A for the same reasoning. A for the same reason. C is wrong because another is used with more than 2 things, so it would have been "one another" Since only 2 things, we like each other Current Student Joined: 29 Jan 2005 Posts: 5238 Followers: 25 Kudos [?]: 378 [0], given: 0 Re: SC --> each other x another [#permalink] Show Tags 31 Aug 2006, 07:19 uvs_mba wrote: ps_dahiya wrote: sumitsarkar82 wrote: Should be A... Each Other - When focussing on two things One Another - When focussing on more than two... A for the same reasoning. A for the same reason. C is wrong because another is used with more than 2 things, so it would have been "one another" Since only 2 things, we like each other Should be one anotherSumit Director Joined: 17 Jul 2006 Posts: 714 Followers: 1 Kudos [?]: 12 [0], given: 0 Show Tags 31 Aug 2006, 13:03 Matt you pointed right. I got really confused. one another when we focus on more than 2 things each other for 2 things. VP Joined: 07 Nov 2005 Posts: 1131 Location: India Followers: 5 Kudos [?]: 41 [0], given: 1 Show Tags 31 Aug 2006, 21:11 sumitsarkar82 wrote: Should be A... Each Other - When focussing on two things Each Another - When focussing on more than two... Agree with A. I think this construction would also be correct : the refusal of each side to acknowledge the other as a legitimate _________________ Trying hard to conquer Quant. Director Joined: 29 Nov 2012 Posts: 898 Followers: 14 Kudos [?]: 1045 [0], given: 543 Re: In the conflict between the israelis and the palestinians, [#permalink] Show Tags 06 Jun 2013, 01:10 Can someone please explain this question! Would be grateful if I could get the usage of The other another each other as the question tests this concept. _________________ Click +1 Kudos if my post helped... Amazing Free video explanation for all Quant questions from OG 13 and much more http://www.gmatquantum.com/og13th/ GMAT Prep software What if scenarios http://gmatclub.com/forum/gmat-prep-software-analysis-and-what-if-scenarios-146146.html VP Status: Far, far away! Joined: 02 Sep 2012 Posts: 1123 Location: Italy Concentration: Finance, Entrepreneurship GPA: 3.8 Followers: 181 Kudos [?]: 1965 [2] , given: 219 Re: In the conflict between the israelis and the palestinians, [#permalink] Show Tags 06 Jun 2013, 04:48 2 KUDOS fozzzy wrote: Can someone please explain this question! Would be grateful if I could get the usage of The other another each other as the question tests this concept. In the conflict between the israelis and the palestinians, the refusal of each side to acknowledge each other as a legitimate national movement is closer to the heart of the problem than is any other issue. The first split, before considering "other, another", is "than" VS "as". Because each sentence uses "closer" => "than" is required. Out B and E Secondly we can eliminate D, because has a "that" at the beginning. Out D as well a- the refusal of each side to acknowledge each other as a legitimate national movement is closer to the heart of the problem than c- the refusal of each side to acknowledge another as a legitimate national movement is closer to the heart of the problem than Here is the "each other" VS "another" problem. The solution relies in the meaning of the sentence. the refusal of each side to acknowledge each other (side) This expression means that I (side A) refuse to acknowledge side B as legit. The rule of "each" is this one: it limits the "range". The text talks about two sides that refuse to acknowledge each other. Side A with side B and the opposite. In C however the refusal of each side to acknowledge another (side), here the intended meaning could vary. I (side A) refuse to acknowledge another side (that could be side B or not). I hope that the expression "limit the range" makes more sense now. _________________ It is beyond a doubt that all our knowledge that begins with experience. Kant , Critique of Pure Reason Tips and tricks: Inequalities , Mixture | Review: MGMAT workshop Strategy: SmartGMAT v1.0 | Questions: Verbal challenge SC I-II- CR New SC set out !! , My Quant Rules for Posting in the Verbal Forum - Rules for Posting in the Quant Forum[/size][/color][/b] Senior Manager Joined: 21 Jan 2010 Posts: 344 Followers: 3 Kudos [?]: 177 [0], given: 12 Re: In the conflict between the israelis and the palestinians, [#permalink] Show Tags 06 Jun 2013, 12:42 fozzzy wrote: Can someone please explain this question! Would be grateful if I could get the usage of The other another each other as the question tests this concept. each other means they are talking about themselves. No 3rd party involved. Another adds the ambiguity here. Does it mean they don't recognize a 3rd party or each other ? GMAT Club Legend Joined: 01 Oct 2013 Posts: 10537 Followers: 919 Kudos [?]: 203 [0], given: 0 Re: In the conflict between the israelis and the palestinians, [#permalink] Show Tags 09 Sep 2014, 19:39 Hello from the GMAT Club VerbalBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. Intern Joined: 12 Feb 2013 Posts: 25 Location: India Concentration: Strategy, Entrepreneurship GMAT 1: 780 Q51 V44 GPA: 3.85 Followers: 1 Kudos [?]: 3 [0], given: 58 Re: In the conflict between the israelis and the palestinians, [#permalink] Show Tags 19 Jan 2016, 21:46 Each Other Vs One another "each other" is for two things; "one another" for more than two. If you have trouble "getting this" suggest you make a flashcard. Idioms can be troublesome. Also arrive at 2/3 or 3/2split and then look for differences between the narrowed answer choices. Between A,E go with A as closer than correctly conveys the intended meaning. Not understanding the intended meaning is sacrilegious when going about SC. Pronoun ambiguities are easier to spot while you're comprehending the question as well, you so can kill two birds with one stone. VP Joined: 08 Jun 2010 Posts: 1401 Followers: 3 Kudos [?]: 114 [0], given: 813 Re: In the conflict between the israelis and the palestinians, [#permalink] Show Tags 02 Feb 2016, 05:53 PedroPedro wrote: In the conflict between the israelis and the palestinians, the refusal of each side to acknowledge each other as a legitimate national movement is closer to the heart of the problem than is any other issue. a- the refusal of each side to acknowledge each other as a legitimate national movement is closer to the heart of the problem than b- that the refusal of each side to acknowledge another as a legitimate national movement is closer to the heart of the problem as c- the refusal of each side to acknowledge another as a legitimate national movement is closer to the heart of the problem than d- that the refusal of each side to acknowledge each other as a legitimate national movement is closer to the heart of the problem than e- the refusal of each side to acknowledge each other as a legitimate national movement is closer to the heart of the problem as Edited for clarity I think differently th refusal of each side " need "to acknowlege THE OTHER". it is not logic to say that " the refusal of each side to ackowlege another" gmat use simple and basic grammar rule to make illogic combination and force us to realize those illogic combinations. practive realizing illogic combination is important Re: In the conflict between the israelis and the palestinians,   [#permalink] 02 Feb 2016, 05:53 Similar topics Replies Last post Similar Topics: 28 Out of the public's interest in the details of and conflicts 28 22 May 2012, 08:17 56 Out of the publc's interest in the details of and conflicts 30 27 Jan 2011, 05:50 6 Out of the publics interest in the details of and conflicts 35 10 Jan 2010, 15:45 1000 SC key conflict with OG 0 10 Jul 2009, 06:35 Out of the public's interest in the details of and conflicts 2 01 Sep 2007, 22:45 Display posts from previous: Sort by

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The OEIS is supported by the many generous donors to the OEIS Foundation. Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 60th year, we have over 367,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”). Other ways to Give Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!) A091420 Square roots of A091419. 3 %I #9 Sep 29 2019 02:52:40 %S 1,1,2,1,3,1,4,3,2,5,1,6,3,3,5,1,8,3,9,3,10,5,11,2,1,12,3,9,13,14,3,5, %T 15,11,3,1,16,4,17,3,8,9,5,13,19,3,20,5,21,15,22,11,3,1,23,24,17,2,25, %U 3,9,5,26,13,19,12,27,28,3,5,29,8,13,21,30,15,31,11,3,1,23,33,34,17,35 %N Square roots of A091419. %H Amiram Eldar, <a href="/A091420/b091420.txt">Table of n, a(n) for n = 1..10000</a> %F a(n) = sqrt(A091419(n)). - _Amiram Eldar_, Sep 29 2019 %Y Cf. A089653, A091419. %K nonn %O 1,3 %A _Ray Chandler_, Jan 05 2004 Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recents The OEIS Community | Maintained by The OEIS Foundation Inc. Last modified December 3 01:44 EST 2023. Contains 367529 sequences. (Running on oeis4.)

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Algebra Properties 5 terms a + 0 = a 7 x 1 = 7 9 x 0 = 0 1/3 x 3 = 1 Subs If n = 15 then 3n = 3 x 15

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+1(978)310-4246 [email protected] Select Page Question Description I’m working on a mechanical engineering project and need a sample draft to help me learn. Hello, I need a lab manual for a topic from the mentioned topics in the assignment description, please follow the guidelines and write everything according to the rubric. Please choose the topic that you think would be suitable for you. Attached are both the description and a sample lab manual from class. please include equations, tables, and everything needed. Thanks EML 3303 – Term Project Goal Design a brief lab session (similar to the ones you have taken throughout this course) and write the lab manual. The project is to be completed in groups, same groups as in labs. You will submit one thing: a) A written document (pdf) in the form of a lab manual: no more than 5-6 pages approximately. Due date Friday, December 3rd by midnight (11:59PM) 2021. What to do: Think of a physical variable/material property that is typically measured in mechanical engineering and design a brief laboratory session that shows how to conduct measurement(s) for such variable. IMPORTANT: you cannot choose a physical variable that can be directly measured with an instrument. It has to be an indirect measurement. For example, you cannot choose mass as the physical variable since a scale can directly provide that measurement. Instead you could choose density and measure mass and volume in order to find density. Once you pick a physical variable/material property you will: – Choose real measuring devices/instruments/sensors to carry out the measurement you picked. – Explain how the measuring devices/instruments/sensors work. – Describe in detail step by step how to use the measuring devices/instruments/sensors to eventually obtain the desired measurement. Written document structure The written document you submit should have the following sections: Cover page: Project title, Authors names. 1 – Introduction: goal of the lab session and context for the measurement. Relate it to real applications. Where does this measurement occur? Who uses it? What is it used for? Why is it important? 2 – Theory: here you will present the measuring devices/instruments/sensors needed to conduct your measurement and the fundamental principles behind such devices. Choose real existing measuring devices/instruments/sensors and provide the manufacturer/seller and model numbers. Also, you will include any equations required for your measurement and explain them briefly and clearly. 3 – Procedure: explain in full detail, step by step, how to use the measuring devices/instruments/sensors to conduct your measurement. Use a bulleted list for the procedure. 4 – Uncertainty: list the sources of uncertainty for the measuring devices/instruments/sensors you decided to use and provide the formulas to be used to calculate the uncertainty of the physical variable/material property you chose for this project. You are not required to use numerical values, although you may choose some hypothetical ones if you would like. Ideas (here are some ideas that you can borrow or you can come up with your own) • • • • • • • • • • • • • • • • • • • • • • Measuring the drag coefficient of spherical objects. Measuring the flow coefficient (Ko) of a differential pressure flow meter. Measuring the Young’s Modulus of a metal. Measuring material hardness with the Vickers and Brinell methods. Measuring distance with optical techniques. Measuring sound wave attenuation through a solid material. Measuring the thermal conductivity of a material. Measuring the specific heat of a material. Measuring the density of an ideal gas. Measuring solar irradiance on a flat surface. Measuring the Poisson’s ratio of a solid material. Measuring the resistivity of a material. Measuring the electrical conductivity of a metal. Measuring corrosion. Measuring surface tension in a liquid. Measuring the Seebeck coefficient of a thermoelectric material. Measuring the Shear Modulus. Measuring material toughness. Measuring the thermal expansion coefficient of a metal. Measuring fracture toughness against surface cracks. Measuring magnetic permeability. Measuring the speed of sound in liquids with ultrasonic transducers. Lab 2 Pneumatic Cylinder Objective Introduction to: • Pneumatic Cylinders • Propagation of Uncertainty Introduction A pneumatic cylinder converts energy stored in a pressurized gas into mechanical work. The high pressure air exerts a large force on one side of the piston and the low pressure air exerts a smaller force on the other side of the piston, resulting in a net force given by equation 1a (pressures are gage pressure, not absolute pressure). This force causes the piston (and the attached rod) to extend or retract, depending on the direction of the net force. If the low pressure side is at atmospheric pressure, then equation 1a becomes 1b, which is the case in this lab. 𝑭 = 𝑷𝒄𝒂𝒑 ∗ 𝑨𝒄𝒂𝒑 − 𝑷𝒓𝒐𝒅 ∗ 𝑨𝒓𝒐𝒅 𝑭=𝑷∗𝑨 (1a) (1b) The terminology used in pneumatic cylinders refers to the larger area (left side of the piston in figure 1) as the “cap” side of the piston and the smaller piston area (right side of the piston in figure 1) as the “rod” side of the piston. Notice that the rod side area is smaller than the cap side area. Also, note that Arod in this case does not refer to the area of the rod, but rather, the area of the rod side of the piston. Figure 1 depicts the anatomy of a typical pneumatic cylinder. These devices are also known as linear actuators because they produce linear motion. Figure 1: Anatomy of a typical air cylinder The advantages of using a pneumatic cylinder: • Air cylinders are relatively quiet in operation • Air cylinders are clean – there is no hazardous fluid that can leak into the surroundings • Initial cost is low The disadvantages of using a pneumatic cylinder: • Can lack precision if the load on the cylinder is compressing the gas as the cylinder actuates (i.e. extending or retracting in vertical position) • Relatively poor speed control • The long term cost of the increased energy required run the air compressor Single-acting cylinders (SAC) – Air cylinders which have only one air supply to push the piston only one way. A SAC type air cylinder usually has an internal compression spring resetting the piston after actuation. Double-acting cylinders (DAC) – Air cylinders that have air supply ports on both ends. This type of air cylinder is more controllable (position and speed) by controlling the differential pressure on both air supply ports. One-sided cylinder – A cylinder with a rod extending out in only one direction. This is what is being used in this lab. Two-sided cylinder – A cylinder with a rod extending out of each side. Typically the rod diameter is equal on both sides, resulting in simplified analysis because both sides of the piston will have the same area. Bore – Refers to the diameter of the cylinder barrel in which the piston travels. This dimension will be used to calculate the piston area. Position Sensors (Switches) – The cylinder’s speed and position can be controlled by having a magnetized piston traveling inside the cylinder barrel. The position of the magnetized barrel can be sensed by position sensors (switches) attached to the outside of the cylinder barrel via the Hall effect. Tools and Material • • • • • 1 Air cylinder setup with two On/Off switches 1 Air regulator/filter 1 Dissected air cylinder (one per class) 1 load scale (aka. fish scale) 1 Vernier caliper Procedure 1. Make sure that all laboratory tools are present and properly working. 2. With a caliper, measure the inside diameter of the provided pneumatic cylinder. A dissected air cylinder will be provided so that the inside diameter is accessible. Each member of the group should perform this measurement twice. Record your measurements in the table provided in the back of this laboratory. 3. With a caliper, measure the diameter of the air cylinder rod. Each member of the group should perform this measurement twice. Record your measurements in the table provided in the back of this laboratory. 4. Familiarize yourselves with the set-up, valves, pressure regulator, and gages. Be sure you know which direction to turn the valve switches (A and B) to turn supply air pressure on and off. Caution: While under pressure the cylinder can snap quickly and cause minor injuries so take caution when switching air supply switches! 5. Record the force exerted by the extending air cylinder at approximately 40 psi. Without adjusting the regulator, record the force exerted by the retracting air cylinder at the same pressure. Carefully monitor regulator readings and allow it to settle to steady state before recording the data. 6. Repeat step 5 for each pressure setting required. The pressure should range from approximately 5 psi up to 40 psi. It is important that you leave the pressure regulator setting unchanged while measuring the extension and retraction force because it would be difficult to achieve the identical pressure after adjusting the regulator. Calculations 1. Plot cylinder performance (do in lab) Compare the extending and retracting forces, that is, compare measured forces to the forces calculated (theoretical) from your area and pressure measurements. For each force direction (extension and retraction), plot graphs with force on the vertical axis and pressure on the horizontal axis, clearly indicating units. Make sure your data looks reasonable. 2. Find the uncertainty in the cylinder extension force measurement (do in lab) From your caliper measurements, estimate the error in surface area. Also estimate the error in pressure from your experience in setting the pressure at the regulator. Finally, use Equation 1b with equation 2, 3, and 4 to estimate the error in your force calculation. Be sure to check your units, whether you use English or SI units. The uncertainty of the extending or retracting cylinder force, can be found with Equation 2. This provides the ‘most likely error’. 4 1. 1. 𝑢. = /012 𝑢2 3 + 016 𝑢6 3 1. 12 = 𝐴 and 1. 16 4 (2) =𝑃 (3),(4) Substituting Equations 1, 3, and 4 into Equation 2 and rearranging gives Equation 5. 9: . 4 9 9 = /0 2;3 + 0 6? = 𝜋𝑑 4 /4 (For Extension Only) (6) Applying the same method used to get Equation 5, we get the uncertainty in the area to be 16 4 𝑢6 = /01E 𝑢E 3 = /02 GE H 4 𝑢E 3 = 2 GE H |𝑢E | (7) 𝑢E is the uncertainty of the cylinder inner diameter. This 𝑢6 can then be used to solve for 𝑢. (Equation 5). These equations are for the extension when the pressure acts on the entire area. Looking at equation 5, you can see that the uncertainty will change depending on the value of F and P, so your uncertainty will be different for each measurement – calculate the uncertainty of the calculated force for each pressure on extension mode only. Check to see if the measured force is within the bounds of uncertainty of the calculated force. If not, what do you think could account for the differences? 3. Uncertainty on retraction force The equations change for retraction when the area of the rod must be subtracted. Derive the equations needed to find the uncertainty on the retraction force and present them in the appendix. No numerical values required. Lab Report This report is to be completed in groups (one printed report per group). The report should also include the following graphs as part of the results: 1. One scatter plot of the measured force for extension and retraction VS air supply pressure (2 sets of points on plot) 2. One scatter plot of the calculated force for extension and retraction VS air supply pressure (2 sets of points on plot) 3. One scatter plot of the measured force for extension and the calculated force ± Force Uncertainty (𝑢. ) VS air supply pressure. (two lines, one with Error bars) Figure 2 presents a general sample plot formatted properly in Microsoft Excel. Pneumatic Cylinder Extension Force vs. Supply Pressure Force (lbs) 150 100 Measured force 50 Calculated force 0 0 20 40 Pressure (psi) 60 Figure 2: Example of cylinder extension vs supply pressure plot created in Microsoft Excel. As always, normal formatting rules for any graphs and tables apply (axis labels, caption, table title, etc). Remember to still include tables with the calculated values. Include a discussion of observations and trends for each graph and discuss what would need to be done to reduce the error in the force measurement and calculation. Review the lab report formatting requirements and pay attention. They are not suggestions. Upon a reasonable effort to adhere to them, you will receive a zero on your lab report with no opportunity to resubmit and with no detailed feedback. Data Sheet Pressure Regulator Setting (_____)* Cylinder Extension Force (______)* Cylinder Retraction Force (______)* *Record units used Cylinder Inside Diameter (_______)* Lab Partner Measurement 1 Measurement 2 Cylinder Rod Diameter (_______)* Measurement 1 1 2 3 4 *Record units used Measurement 2 Appendix: Why do we use gage pressure instead of absolute pressure in the above analysis? The absolute pressure ultimately is what determines the force. Consider all the forces acting on the piston-rod assembly in the direction of motion: 1) the absolute pressure in the cap side acting on the cap side area 2) the absolute pressure in the rod side acting on the rod side area 3) the absolute pressure (atmospheric) acting on the tip of the rod 4) friction between the cylinder wall and the piston. Conveniently, the friction force goes to zero as the piston velocity goes to zero, leaving the other 3 forces. The resulting force is given below 𝐹 = (𝑃L,=>? + 𝑃>NOPQ?RSTU= )𝐴=>? − (𝑃L,TPW + 𝑃>NOPQ?RSTU= )𝐴TPW − 𝑃>NOPQ?RSTU= ∗ 𝐴TPW NU? But notice that 𝐴=>? = 𝐴TPW + 𝐴TPW NU? so that the above equation simplifies as follows: 𝐹 = 𝑃L,=>? 𝐴=>? + 𝑃>NOPQ?RSTU= 𝐴=>? − 𝑃L,TPW 𝐴TPW − 𝑃>NOPQ?RSTU= 𝐴TPW − 𝑃>NOPQ?RSTU= ∗ 𝐴TPW NU? 𝐹 = 𝑃L,=>? 𝐴=>? + 𝑃>NOPQ?RSTU= 𝐴=>? − 𝑃L,TPW 𝐴TPW − 𝑃>NOPQ?RSTU= (𝐴TPW + 𝐴TPW NU? ) 𝐹 = 𝑃L,=>? 𝐴=>? + 𝑃>NOPQ?RSTU= 𝐴=>? − 𝑃L,TPW 𝐴TPW − 𝑃>NOPQ?RSTU= (𝐴=>? ) 𝐹 = 𝑃L,=>? 𝐴=>? − 𝑃L,TPW 𝐴TPW

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# AVERAGE Examples 1. A batsman has a certain average runs for 16 innings. In the 17th inning he made a score of 85 runs thereby his average is increased by 3. What is his average after 17th inning?Examples Sol: The average for 17th inning has been increased by 3. The total increase in the runs for 17th inning = 17 × 3 = 51 But the batsman scores 85. Average runs in his 16th innings = 85 – 51 = 34. Hence the average of runs after 17th innings = 34  + 3 = 37 2.  A man has 7 children. When their average age was 12 years, the child who was 6 years of age, died. What was the average age of surviving children 5 years after the death of the above child? Sol: Average age of 7 children  = 12 years Total age of 6 children  =  12 × 7 = 84 years Total age of 6 children after the death of a child aged 6 years = 84 – 6 = 78 Hence   the average age of the surviving children = After 5 yrs. = 13 + 5 = 18 yrs. 3.  If the weights of 5 students of a class are 49.6 kg, 39.8 kg, 45.2 kg and 24.6 kg respectively then what is their average weight? Sol: Total weight of 5 students = 49.6 + 39.8 + 40.8 + 45.2 + 24.6 = 200 kg. ∴    Their average weight = = 40 kg. 4.  The average temperature for Monday, Tuesday and Wednesday was 36oC. The average temperature for Tuesday, Wednesday and Thursday was. If the temperature for Thursday was 37oC, what was the temperature on Monday? Sol:   Average temperature for Monday, Tuesday and Wednesday = 36oC Total temperature for Monday, Tuesday and Wednesday = 36 × 3 = 108oC ∴   Average temperature for Tuesday, Wednesday and Thursday = 38oC ∴    Total temperature for Tuesday, Wednesday and Thursday = 38 × 3 = 114oC ∴    Total temperature for Tuesday and Wednesday only = 114 – 37 =  77oC ∴    Temperature for  Monday only = 108 – 77 = 31oC 5.  A train covers the first 16 km at a speed of 20 km per hour another 20 km at 40 km per hour and the last 10 km at 15 km per hour. Find the average speed for the entire journey. 6.  A vehicle travels from A to B at the speed of 40 km/hr, but from B to A at the speed of 60km/hr. what is its average speed during the whole journey? Sol: Let the distance from A to B be x km 7.  The average age of a class of 40 boys is 16.95 years. A new boy joins the class and the average age now is 17 years. What is the age of the new boy? Sol: The average age of 40 boys  = 16.95 years Total are of 40 boys  = 16.95 × 40 = 678 years The average age of 41 boys  = 17 years Total age of 41 boys  = 17 × 41 = 697 years Age of the new boy  = 697 – 678 = 19 years The average of the given quantities is calculated by dividing the sum of the quantities by their number. If the average of ‘x’ quantities is ‘a’ and average of ‘y’ quantities is ‘b’, then average of x and y is Weighted Average Let x1, x2, x3, …..xn be the quantities and w1, w2, w3, ….wn be the weights attached to them then Weighted Average SOLVED EXAMPLES Example 1. A cricketer has completed 18 innings and his average is 26.5 runs. How many runs must he make in his next innings so as to raise his average to 27? (1) 36           (2)46            (3)38            (4) 40            (5) None of these Explanation: (1)   Formula = n(y – x) + y n = 18 x = 265 y = 27 = 18(27 – 26.5) + 27 = 9 + 7 = 36 Example 2. The average of 13 numbers is 30. The average of 1st 7 of these numbers is 32 and the last 7 of these numbers is 22. Find the 7th number. (1) 15         (2) 16          (3) 12          (4) 20          (5) None of these Explanation: (3)    Total number = 13 × 30 = 390 1st 7 numbers = 7 × 32 = 224 Last 7 number = 7 × 22 = 154 7th number = 390 – (224 + 154) = 12 Example 3. The average of marks obtained by 65 candidates in a certain examination is 25. If the average marks of passed candidates is 27 and that of the failed candidates is 14, what is the no. of candidates who passed the examination? (1) 53            (2) 64            (3) 55            (4) 70            (5) None of these Explanation: (3)  Passed candidates = x Failed candidates = 65 – x Total marks ∴   25  ×  65 = 27x + (65 – x)14 ⇒ 1625 = 27x + 910 – 14x ⇒ 13x = 715 ⇒ x = 55 Example 4. In a class there are 30 students whose average is decreased by6 months, when 4 students aged 16, 17, 18 and 19 years respectively are replaced by the some no. of students. Find the average of the new students. (1) 16.25               (2) 15.25               (3) 16               (4) 15.75               (5) None of these Example 5. The average age of 11 persons in a committee is increased by 2 years when three men aged 32 years, 33 years and 34 years are substituted by three women. Find the average age of these three women. (1) 40 1/3        (2) 40          (3) 41 1/3        (4) 41 years          (5) None of these Example 6. The average of 40 numbers is 405. If each of the numbers is divided by 15. Find the average of new set of numbers. (1) 27          (2) 28          (3) 21          (4) 26          (5) None of these Explanation: 405/15 = 27 Example 7. The average age of 80 boys in a class is 15. The average age of a group of 15 boys in the class is 16 and the average age of another 25 boys in the class is 14. What is the average age of remaining boys in the class. (approx.) (1) 20          (2) 15          (3) 18          (4) 21          (5) None of these Explanation: (2)    Remaining boys average age = x ⇒ 15 × 16 + 25 × 14 + 40x = 80 × 15 ⇒ 240 + 350 + 40x = 1200 ⇒ 40x = 1200 – 590 x = 610/40 x = 15. 25 Example 8. A batsman in his 12th innings makes a score of 63 runs and there by increases his average score by ‘2’. The average of his score after 12th innings is. (1) 39        (2) 41         (3) 37         (4) 45         (5) None of these Explanation: (2)    Average score = x till 11 innings Total runs = 11x Total runs after 12th inning = 11x + 63 Average after 12th inning = ( x + 2) Total run = 12 ( x + 2) 12 ( x + 2) = 11x + 63 ⇒ x = 63 – 12 × 2 = 39 Average of his score = 39 + 2 = 41 Example 9. A certain amount was to be distributed among the A, B and C in the ratio 2 : 3 : 4 but was erroneously distributed in the ratio 7 : 2 : 5 as a result of this ‘B’ received Rs. 40 less. What is the actual? (1) 210         (2) 270         (3) 230         (4) 280         (5) None of these Example 10. Two number are respectively 20% and 50% more than a third number. The ratio of the two numbers is. (1) 2 : 5         (2) 4 : 5        (3) 3 : 5        (4) 6 : 7        (5) None of these PRACTICE EXERCISE 1. The average of marks obtained by 120 candidates in a certain examination is 35. If the average marks of passed candidates is 39 and that  of the failed candidates is 15. What is the number of candidates who passed the examination? (1) 90        (2) 85         (3) 100         (4) 120         (5) None of these Ans: 100 2. The average salary of all workers in workshop is 8000. The average salary of 7-technicians is Rs. 12000 and the average salary of the rest is Rs. 6000. The total number of workers in the work shop is. (1) 20         (2) 21         (3) 23         (4) 22         (5) None of these Ans: 21 3. In a competitive examination, the average marks obtained was 45. It was later discovered that there was some error in computerization and the marks of 90 candidates had to be changed from 80 to 50, and the average come down to 40 marks. The total number of candidates appeared in the examination. (1) 500         (2) 600         (3) 540         (4) 580         (5) None of these Ans: 540 4. The average weight of A, B, C is 84 kg. If D joins the group the average weight of the group becomes 80 kg. If another man ‘E’ who weights is 3kg more than D replace A, then average of B, C, D and E becomes ‘79’ kg. What is the weight of ‘A’ (1) 64        (2) 72         (3) 75         (4) 100         (5) None of these Ans: 75 5. 1/3rd of certain journey is covered at the rate of 25kmph. 1/4th at the rate of 30 kmph and the respect 50 kmph. Find the average speed for the hole journey . (1) 33.33 kmph         (2) 36 kmph         (3) 42 kmph         (4) 27 kmph         (5) None of these Ans: 33.33 kmph 6. Deepak’s opinion, his weight is greater than 65 kg but less than 72 kg. His brother does not agree with Deepak and he thinks that Deepak’s weight is greater than 60 kg but less than 70 kg. His mother’s view is that his weight cannot be greater than 68 kg. If all are them are correct in their estimation, what is the average of different probable weights of Deepak. (1) 67 kg         (2) 68kg         (3) 69 kg         (4) 70 kg         (5) None of these Ans: 67 kg 7. The average of marks obtained by 120 candidates in a certain examination is 35. If the average marks of passed candidates is 39 and that of the failed candidates is 15. What is the number of candidates who passed the examination? (1) 90         (2) 85          (3) 100          (4) 120          (5) None of these Ans: 100 8. The average age of Raju & Aditya is 35 years. If Raghu replaces Raju. The average age becomes 32 years and if Raghu replaces Adithya then the average becomes 32 years. If the average age of Naveen and Ramu behalf of the average age of Raju, Aditya and Raghu. Then the average age of all the people-? (1) 28          (2) 32          (3) 35          (4) 30          (5) None of these Ans: 28 9. The average salary is being paid to all it’s employees by a company is 15,500. The average salary of the senior employees is 18000 per month and the average salary of the junior employees is 12000 per month. If there are only two levels of employees viz junior and senior levels, then what fraction of the total employees is the too junior level employees are (1) 7/5    (2) 7/12       (3) 5/12       (4) 5/7         (5) None of these Ans: 7/12 10. The average age of ‘8’ persons in a committee is increased by 2 years when two men aged 35 years and 45 years are substituted by two women. Find the average age of these two women. (1) 48           (2) 36          (3) 42          (4) 29          (5) None of these Ans:  48 The average of the given quantities is calculated by dividing the sum of the quantities by their number. Average = Sum of the quantities / Number of quantities Weighted average: Let x1, x2, x3, ...... xn be the quantities and w1, w2, w3, ...... wn be the weights attached to them then Weighted Average = w1x1 + w2x2 + w3x3 + ........ + wnxn / w1 + w2 + w3 + ........ + wn MODEL QUESTIONS 1. The average of seven numbers is 18. The average of first three numbers is 14 and the average of last three numbers is 19. What is the middle number? A) 9           B) 27           C) 32           D) 35 2. The average of 17 results is 60. If the average of first 9 results is 57 and that of the last 9 results is 65. What is the value of 9th result? (SSC-CGL – 2017) A) 39         B) 78         C) 117         D) 156 3. In a class, the average age of 23 boys is 12 years and the average age of 17 girls is 11 years. What is the average age of the whole class? A) 11.5 years      B) 11.245 years     C) 11.575 years      D) 11.525 years 4. In a class, the average score of thirty students on a test is 69. Later it was found that the score of one of the students was wrongly read as 88 instead of 58. The actual average score is (SSC-CGL – 2019) A) 71           B) 78           C) 65           D) 68 5. The average of 23 numbers is 28. Later, it was pointed out that value of one the numbers was wrongly taken as 78 instead of 87. What is the correct average? A) 31.25     B) 27.15     C) 28.39     D) 32.65 6. Out of 6 numbers the sum of the first 5 numbers is 7 times the 6th number. If their average is 136, then the 6th number is.... (SSC-CGL – 2019) A) 102         B) 96         C) 84         D) 116 7. A cricketer’s average for 11 matches is 25 runs. In the next four matches he scored 72, 34, 51 and 48 runs. Find his new average for the 15 matches? A) 39          B) 32          C) 36          D) 28 8. The average age of a class of 24 students is 11 years. If the teacher’s age is also included, the average increases by one year. Find the age of the teacher? A) 35           B) 39           C) 34           D) 36 9. The average age of six members of a family is 20 years. If the age of the servant is included, then the average age increase by 25%. What is the age of the servant (in years)? (SSC-CGL – 2017) A) 30           B) 55           C) 50           D) 35 10. The average weight of 8 persons increases by 1.5 kg when a person weighting 65 kg is replaced by a new person. What could be the weight of the new person? A) 55 kg     B) 57.5 kg     C) 75 kg    D) 77 kg 11. The average of twelve numbers is 45.5. The average of the first four numbers is 41.5 and that of the next five numbers is 48. The 10th number is 4 more than the 11th number and 9 more than the 12th number. What is the average of the 10th and 12th number? (SSC-CGL – 2019) A) 46.5         B) 46         C) 47.8         D) 47 12. 5 kg sugar is purchased at Rs.10 per kg, 6kg at Rs.11 and 9kg at Rs.12 per kg. What is the weighted average price of sugar per kg? A) 8.8        B) 10.8        C) 11.2        D) 11.8 13. The average of 12 numbers is 9. If each number is multiplied by 2 and added to 3, the average of the new set of numbers is (SSC-CGL – 2016) A) 9           B) 18           C) 21           D) 27 14. In a class, average height of all students is 'a' cms. Among them, average height of 10 students is 'b' cms and the average height of the remaining students is 'c' cms. Find the number of students in the class. (Here a > c and b > c) (SSC-CGL – 2016) 15. The average weight of some students in a class is 68.5 kg. If four new students of 72.2 kg, 70.8 kg, 70.3kg and 66.7 kg are enrolled in the class, the average weight of students increases to 300 g. Initially how many students were there in the class? (SSC-CGL – 2018) A) 11          B) 26          C) 21          D) 16 ANSWERS: 1-B; 2-B; 3-C; 4-D; 5-C; 6-A; 7-B; 8-D; 9-B; 10-D; 11-A; 12-C; 13-C; 14-D; 15-D. Explanations 1. The total of seven numbers = 7 × 18 = 126 The total of first 3 and last 3 numbers is = 3 × 14 + 3 × 19 = 99 ∴ Middle number is (126 − 99) = 27 Shortcut: 3 × (−4) + 3 × (+1) = −12 + 3 = −9 ∴ Middle number = 18 + 9 = 27 2. The total of 17 numbers = 17 × 60 = 1020 The total of first 9 and last 9 numbers is = 9 × 57 + 9 × 65 = 1098 ∴ Middle number is (1098 − 1020) = 78 Shortcut:  9 × (−3) + 9 × (+5) = −27 + 45 = +18 ∴ Middle number = 60 + 18 = 78 3. Total age of 40 students = (23 × 12 + 17 × 11) = 463 Average age = 463/40 = 11.575 years Shortcut: Let all the students average be 11 years Extra age by all boys = 23 × 1 yr. = 23 years Extra average for all 40 students = 23/40 = 0.575 ∴ Actual Average = 11 + 0.575 = 11.575 4. The Correct Average Shortcut: Additional score taken = 88 − 58 = 30 Additional average = 30/30 = 1 ∴ New average is 69 − 1 = 68 5. Correct Average Shortcut: Value less taken = 87 − 78 = 9 Average less taken = 9/23 = 0.39 ∴ New average is 28 + 0.39 = 28.39 6. Let the 6th number be ‘X’ ∴ 7x = (136 × 6) − x ⇒ x = 102 7. Total runs in 11 matches = 11 × 25 = 275 runs Total runs in the next 4 matches = 72 + 34 + 51 + 48 = 205 runs = 32 runs Shortcut: Let the average of all 15 matches be 25 Extra runs in 4 matches = (72 − 25) + (34 − 25) + (51 − 25) + (48 − 25) = 105 Extra Average = 105 /15 = 7 ∴ New Average = 25 + 7 = 32 8. Total of 24 students is 24 × 11 = 264 years Total age with teacher is 25 × 12 = 300 years ∴ Teacher’s age is 300 − 264 = 36 years Shortcut: Teacher’s age = 11 + (25 × 1) = 36 years 9. Total age of the family is 6 × 20 = 120 years Average age after servant’s inclusion = 125% of 20 = 25 years Total age with servant = 7 × 25 = 175 years ∴ Servant’s age is 175 − 120 = 55 years Shortcut: Servant’s age = 20 + (7 × 5) = 55 years 10. Let the average weight of 8 persons be ‘x’ years and the age of the new person be ‘y’ years Shortcut: 65 + 8 × 1.5 = 77 kg 11. The total of 12 numbers = 12 × 45.5 = 546 The total of first 4 and next 5 numbers is = 4 × 41.5 + 5 × 48 = 406 Total of 9th, 10th & 11th numbers = 546 − 406 = 140 Let the 10th number be x ∴ x + x − 4 + x − 9 = 140 ⇒ x = 51 Shortcut: 4 × (− 4) + 5 × (+ 2.5) = − 16 + 12.5 = − 3.5 ∴ Total of 9th, 10th & 11th numbers = (45.5 × 3) + 3.5 = 140 Let the 10th number be x ∴ x + x − 4 + x − 9 = 140 ⇒ x = 51 = 46.5 12. Weighted average Shortcut: Let the assumed average be Rs. 10 13. New average Shortcut: New average = (9 × 2) + 3 = 21 14. Let the number of students be ‘x’ 15. Let the number of students be ‘x’ ⇒ 68.5x + 280 = 68.8(x + 4) ⇒ 68.5x + 280 = 68.8x + 275.2 ⇒ 0.3x = 4.8 Posted Date : 20-02-2021 గమనిక : ప్రతిభ.ఈనాడు.నెట్‌లో కనిపించే వ్యాపార ప్రకటనలు వివిధ దేశాల్లోని వ్యాపారులు, సంస్థల నుంచి వస్తాయి. మరి కొన్ని ప్రకటనలు పాఠకుల అభిరుచి మేరకు కృత్రిమ మేధస్సు సాంకేతికత సాయంతో ప్రదర్శితమవుతుంటాయి. ఆ ప్రకటనల్లోని ఉత్పత్తులను లేదా సేవలను పాఠకులు స్వయంగా విచారించుకొని, జాగ్రత్తగా పరిశీలించి కొనుక్కోవాలి లేదా వినియోగించుకోవాలి. వాటి నాణ్యత లేదా లోపాలతో ఈనాడు యాజమాన్యానికి ఎలాంటి సంబంధం లేదు. ఈ విషయంలో ఉత్తర ప్రత్యుత్తరాలకు, ఈ-మెయిల్స్ కి, ఇంకా ఇతర రూపాల్లో సమాచార మార్పిడికి తావు లేదు. ఫిర్యాదులు స్వీకరించడం కుదరదు. పాఠకులు గమనించి, సహకరించాలని మనవి.

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{[ promptMessage ]} Bookmark it {[ promptMessage ]} 121homework 5 # 121homework 5 - Felder Jacob Homework 6 Due 9:00 pm Inst... This preview shows pages 1–3. Sign up to view the full content. Felder, Jacob – Homework 6 – Due: Oct 17 2006, 9:00 pm – Inst: Vitaly 1 This print-out should have 23 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 4) 10 points An air-filled capacitor consists of two parallel plates, each with an area of 6 . 6 cm 2 , sepa- rated by a distance 3 mm . A 23 V potential difference is applied to these plates. The permittivity of a vacuum is 8 . 85419 × 10 - 12 C 2 / N · m 2 . 1 pF is equal to 10 - 12 F . The magnitude of the electric field between the plates is 1. E = 1 V d . 2. E = d V 2 . 3. E = 1 ( V d ) 2 . 4. E = V d . correct 5. E = V d . 6. E = d V . 7. E = V d 2 . 8. E = ( V d ) 2 . 9. None of these Explanation: Since E is constant between the plates, V = Z ~ E · d ~ l = E d E = V d . 002 (part 2 of 4) 10 points The magnitude of the surface charge density on each plate is 1. σ = ² 0 d V 2 . 2. σ = ² 0 V d 3. σ = ² 0 V d . correct 4. None of these 5. σ = ² 0 ( V d ) 2 . 6. σ = ² 0 ( V d ) 2 . 7. σ = ² 0 d V . 8. σ = ² 0 V d . 9. σ = ² 0 V d 2 . Explanation: Use Gauss’s Law. We find that a pillbox of cross section S which sticks through the sur- face on one of the plates encloses charge σ S. The flux through the pillbox is only through the top, so the total flux is E S. Gauss’ Law gives σ = ² 0 E = ² 0 V d Alternatively, we could just recall this result for an infinite conducting plate (meaning we neglect edge effects) and apply it. 003 (part 3 of 4) 10 points This preview has intentionally blurred sections. Sign up to view the full version. View Full Document Felder, Jacob – Homework 6 – Due: Oct 17 2006, 9:00 pm – Inst: Vitaly 2 Calculate the capacitance. Correct answer: 1 . 94792 pF. Explanation: Let : A = 0 . 00066 m 2 , d = 0 . 003 m , V = 23 V , and ² 0 = 8 . 85419 × 10 - 12 C 2 / N · m 2 . The capacitance is given by C = ² 0 A d = 8 . 85419 × 10 - 12 C 2 / N · m 2 × 0 . 00066 m 2 0 . 003 m = 1 . 94792 × 10 - 12 F = 1 . 94792 pF . 004 (part 4 of 4) 10 points Calculate plate charge; i.e. , the magnitude of the charge on each plate. Correct answer: 44 . 8022 pC. Explanation: The charge Q on one of the plates is simply Q = C V = (1 . 94792 × 10 - 12 F) (23 V) = 4 . 48022 × 10 - 11 C = 44 . 8022 pC . keywords: 005 (part 1 of 2) 10 points An isolated conducting sphere can be consid- ered as one element of a capacitor (the other element being a concentric sphere of infinite radius). If k = 1 4 π ² 0 , the capacitance of the system is C , and the charge on the sphere is Q , what is the radius of the sphere? 1. r = C k 2. r = k C Q 3. r = Q 2 C k 4. r = k Q C 5. r = k Q 6. r = k C correct Explanation: If the radius of the sphere is r , then the potential difference for the charged sphere is V = k Q r and for a capacitor, C = Q V = r k . Thus r = k C . 006 (part 2 of 2) 10 points The Coulomb constant is 8 . 98755 × 10 9 N · m 2 / C 2 . If the potential on the surface of the sphere is 2603 V and the capacitance is 6 . 2 × 10 - 11 F, what is the surface charge density? This is the end of the preview. Sign up to access the rest of the document. {[ snackBarMessage ]} ### Page1 / 8 121homework 5 - Felder Jacob Homework 6 Due 9:00 pm Inst... This preview shows document pages 1 - 3. Sign up to view the full document. View Full Document Ask a homework question - tutors are online

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# Lesson Seventeen ## Telling Time have, know, be, tell Click for the audio. Questions: Answers: Do you have the time? Sure, it’s 6:00. Do you know what time it is? Sorry, I’m not wearing a watch. What time is it? 5:50 Can you tell me what time it is, please? It’s a quarter after two. Watch this video: Important vocabulary • o’clock = :00 ( 7:00 = seven o’clock ) • a quarter after = 15 minutes after the hour ( 3:15 = It’s a quarter after three.) • a quarter past = 15 minutes after the hour (3:15 = It’s a quarter past three.) • a quarter before = 15 minutes before the hour( 3:45 = It’s a quarter before four ) • a quarter to = 15 minutes before the hour (3:45 = It’s a quarter to four.) • half past = 30 minutes after the hour ( 11:30 = It’s half past eleven) • thirty = 30 minutes after the hour. (11:30 = It’s eleven thirty.) The easiest way to tell someone the time is to use a digital format. For example, when someone asks you what time it is, you can say, “It’s 5:30.” Instead of, “It’s half past five.” “It’s 5:13.” = (five thirteen) or “It’s 5:02.” (five o two)* or “It’s 5:50.” (five fifty) *Note: 0 is pronounced “O” not “zero.” Giving the Date Question: Answers: What’s today? or What day is it today? Today’s Tuesday, June 6. (June sixth) What’s the date? It’s June 6. or It’s the 6th of June. There’s a difference between “day” and “date • day: Sunday, Monday, Tuesday, etc. • date: June 6 When someone asks you the date, it’s not necessary to give the year. For example: • What’s the date? —-> It’s June 6.  (Not It’s June 6, 2005) Make sure you use ordinal numbers when you say the date. You can listen to the way I say the date every day on my blog. When someone asks about time in the future or the past, use the year. • When were you born? —-> March 25, 1965 When did you arrive in the U.S.? —-> In 2012. • When is the next election? —-> In November of 2020.

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# EE 221 AC Circuit Power Analysis ```EE 221 AC Circuit Power Analysis Instantaneous and average power RMS value Apparent power and power factor Complex power Instantaneous Power ƒ Product of time-domain voltage and time-domain current p(t) = v(t) i(t) ƒ Determine maximum value ƒ Transients As the transient dies out, the operation. Since the only source remaining in the circuit is dc, the inductor eventually acts as a short circuit absorbing zero power. Power due to Sinusoidal Excitation ƒ v(t) = Vm cos(ωt+θ) and i(t) = Im cos(ωt +ϕ) ƒ p(t) = Vm Im cos(ωt+θ) cos(ωt +ϕ) = = ½ Vm Im cos(θ - ϕ) + ½ Vm Im cos(2ωt + θ + ϕ) p(t) = ½ Vm Im cos(θ - ϕ) + trigonometric identity ½ Vm Im cos(2ωt + θ + ϕ) ƒ instantaneous ƒ constant ƒ periodic ƒ two parts ƒ independent of t ƒ period is ½T ƒ average ƒ the "average" ƒ average is zero ƒ periodic ƒ wanted ƒ unwanted ƒ active or real Power due to Sinusoidal Excitation p(t) = ½ Vm Im cos(θ - ϕ) + ½ Vm Im cos(2ωt + θ + ϕ) • voltage V = 4∠0° V, impedance Z = 2∠60° Ω, ω = π/ 6 rad/s • I = 2∠-60° A • p(t) = 2 + 4 cos(π / 3 - 60°) W Power due to Sinusoidal Excitation p(t) = ½ Vm Im cos(θ - ϕ) ½ Vm Im cos(2ωt + θ + ϕ) + Example 1: Average power delivered to resistor T 2 PR = ½ Vm Im cos(θ - ϕ) = ½ Vm Im cos(0) 1.5 = ½ Vm Im = ½ Vm2 / R =1W 0.5 v(t), i(t), p(t) = ½ R Im2 1 0 -0.5 -1 V = 2∠-45° V R = 2Ω -1.5 -2 0 0.2 0.4 0.6 0.8 1 time (s) 1.2 1.4 1.6 1.8 2 Power due to Sinusoidal Excitation p(t) = ½ Vm Im cos(θ - ϕ) ½ Vm Im cos(2ωt + θ + ϕ) + Example 2: Average power delivered to purely reactive elements T 2 PX = ½ Vm Im cos(θ - ϕ) = ½ Vm Im cos(±90°) = 0 1.5 1 v(t), i(t), p(t) 0.5 0 -0.5 -1 V = 2∠-45° V X = j2Ω -1.5 -2 0 0.2 0.4 0.6 0.8 1 time (s) 1.2 1.4 1.6 1.8 2 Power due to Sinusoidal Excitation p(t) = ½ Vm Im cos(θ - ϕ) Example 3: + ½ Vm Im cos(2ωt + θ + ϕ) Voltage across impedance (V = 100∠25o V and Z = 50∠55o Ω) Determine active power absorbed. I =V/Z = 2 ∠25-55° V = 2 ∠-30° = 1.7 - j A 25o -30o Z = 28.7 + j41.9 Ω P = ½ 22 28.7 = ½ (1.72+12) 28.7 = 57.4 W I Power due to Sinusoidal Excitation p(t) = ½ Vm Im cos(θ - ϕ) Example 4: + ½ Vm Im cos(2ωt + θ + ϕ) (Chapter 11, Problem 8.) In the circuit shown in Fig. 11.27, find the average power being (a) dissipated in the 3-Ω resistor; (b) generated by the source. ZR = 3 + 1 = 3 + 1 + j3 = 4 + j3 Ω 0.1 − j 0.3 Ignore 30° on Vs , I R = 5 (a) 2 + j5 5 29 , IR = 6 + j8 10 2 P3 Ω 1  5 29  =   × 3 = 10.875 W 2  10  (b) (2 + j5)(4 + j3) = 13.463∠51.94° V 6 + j8 1 = ×13.463× 5cos 51.94° = 20.75 W 2 Vs = 5∠0° ∴Ps,gen Maximum Power Transfer A simple loop circuit used to illustrate the derivation of the maximum power transfer theorem as it applies to circuits operating in the sinusoidal steady state. Zth = Rth + j Xth ZL = RL + j XL ZL = Zth* RL = Rth and XL = - Xth Maximum Power Transfer Example: (Chapter 11, Problem 12.) For the circuit of Fig. 11.30; (a) what value of ZL will absorb a maximum average power? (b) What is the value of this maximum power? −10 j = 107.33∠ − 116.57° V 10 + j 5 − j10 (10 + j15) Zth = = 8 − j14 Ω 10 + j 5 Vth = 120 (a) ZL = 8 + j14 Ω (b) IL = 107.33∠ − 116.57° ∴ 16 PL ,max 1 = 2 2  107.33    × 8 = 180 W  16  Effective Values • Measure for sinusoidal voltages and currents • Power outlets: 60 Hz, "voltage of 115V" • Not the mean of T (T/2) • Not the Amplitude ( 2 115V) • Measure of effectiveness of a source in delivering • Effective value of periodic current • is equal to the DC value that • delivers the same average power to resistor • i(t) R → p(t) → PR and compare to IDC R Effective Values Mathematical expression T T R 2 1 2 2 P = ∫ i Rdt = ∫ i dt =I DC R = I eff2 R T 0 T 0 T I eff = 1 2 i Rdt ∫ T 0 • (Square) root of the mean of the square current • rms value • Defined for all periodic signals Effective Values - Sinusoids ƒ i(t) = Im cos(ωt +ϕ) with a period of T = 2π/ω I eff T ω 1 2 2 = I m cos (ωt + ϕ )dt = I m ∫ 2π T 0 I eff 2π / ω ∫ 0 1 1   2 + 2 cos(2ωt + 2ϕ ) dt   Im = 2 • real quantity • independent of phase angle • equal to 0.707 the amplitude • example: 2∠-30o A delivers the same as IDC = 1A RMS value to compute average power ƒ In general P = ½ Vm Im cos(θ - ϕ) = Veff Ieff cos(θ - ϕ) ƒ For resistors P = Veff Ieff = Veff2 / R = Ieff2 R ƒ Note: ƒ We can use amplitude or rms value ƒ Use V and V rms to designate voltages RMS value to compute average power ƒ Example: (Chapter 11, Problem 30.) The series combination of a 1-kΩ resistor and a 2-H inductor must not dissipate more than 250 mW of power at any instant. Assuming a sinusoidal current with ω =500 rad/s, what is the largest rms current that can be tolerated? The peak instantaneous power is 250 mW. The combination of elements yields Z = 1000 + j1000 Ω = 1414 ∠45o Ω. Vm ∠0 Vm ∠ − 45o A and Vm = 1414 Im. Arbitrarily designate V = Vm ∠0 , so that I = = 1414 Z We may write p(t) = ½ Vm Im cos φ + ½ Vm Im cos (2ωt + φ) where φ = the angle of the current (-45o). This function has a maximum value of ½ VmIm cos φ + ½ VmIm. Thus, 0.250 = ½ VmIm (1 + cos φ) = ½ (1414) Im2 (1.707) and Im = 14.39 mA. In terms of rms current, the largest rms current permitted is 14.39m / 2 = 10.18 mA rms. Apparent Power P = ½ Vm Im cos(θ - ϕ) = Veff Ieff cos(θ - ϕ) In case of direct current we would use voltage times current: S = Veff Ieff • This is not the average power • Is the "apparent" power (S or AP) • Measured in volt-ampere or VA • (rather than W to avoid confusion) • Magnitude of S is always greater or equal to magnitude of P: |S| ≥ |P| Power Factor Defined as the ratio of average (real) power to apparent power PF = P / S = P / (Veff Ieff) In the sinusoidal case the power factor is PF = cos(θ - ϕ) • θ - ϕ is the angle the voltage leads the current: PF angle • note: • Purely resistive load has PF = 1 • Purely reactive load has PF = 0 • PF = 0.5 means a phase angle of ±60° • Resolve ambiguity Apparent Power and Power Factor Example: (Chapter 11, Problem 30.) (a) Find the power factor at which the source in the circuit of Fig. 11.42 is operating. (b) Find the average power being supplied by the source. (c) What size capacitor should be placed in parallel with the source to cause its power factor to be unity? (d) Verify your 120 = 9.214∠ − 26.25° A rms j192 4+ 12 + j16 ∴ PFs = cos 26.25 = 0.8969 lag (a) I s = (b) Ps = 120 × 9.214 × 0.8969 = 991.7W j 48 1 (192 + j144) = 4+ 3 + j4 25 11.68 − j 5.76 ∴ Z L = 11.68 + j 5.76 Ω, YL = 11.682 + 5.762 j 5.76 , C = 90.09 µ F ∴ j120π C = 11.682 + 5.762 ZL = 4 + (c) Apparent Power and Power Factor (Chapter 11, Problem 30.) (d) Examine simulation output file (AC sweep at 60Hz). FREQ VM(\$N_0003,0) VP(\$N_0003,0) 6.000E+01 1.200E+02 0.000E+00 FREQ IM(V_PRINT1) IP(V_PRINT1) 6.000E+01 9.215E+00 -2.625E+01 (a) and (b) are correct Next, add a 90.09-µF capacitor in parallel with the source: FREQ IM(V_PRINT1) IP(V_PRINT1) 6.000E+01 8.264E+00 -9.774E-05 (c) is correct (-9.8×10-5 degrees is essentially zero, for unity PF). Complex Power • Simplifies power calculations • We had p(t) = Veff Ieff cos(θ - ϕ) + ½ Vm Im cos(2ωt + θ + ϕ) • Where the average (real) power P = Veff Ieff cos(θ - ϕ) • Using complex nomenclature P = Veff Ieff Re{ej(θ - ϕ)} = Veff Re{ej(θ)} Ieff Re{e-jϕ)} phasor voltage • Hence P = Re{Veff I*eff} • Define S = Veff I*eff complex conjugate of phasor current Complex Power • S = Veff I*eff can be written as S = Veff Ieff ej(θ - ϕ) = P + jQ magnitude equals apparent power |S| = S PF angle • Reactive power Q average (real) power reactive power • Imaginary part of complex power • Dimensions are those of P, S, AP=S (|S|) • Avoiding confusion by using volt-ampere-reactive or VAr • Q = Veff Ieff sin(θ - ϕ) • Physical interpretation: • Time rate of energy flow back&forth between source and reactive loads • Reactive components charge and discharge at 2ω (→ current flows) Complex Power (Example) p(t) = Veff Ieff cos(θ - ϕ) + Veff Ieff cos(2ωt + θ + ϕ) |S| P • voltage V = 4∠0o V, impedance Z = 2∠60o Ω, I = 2∠-60o A, ω = π/ 6 rad/s • p(t) = 2 + 4 cos(π / 3 - 60°) W • voltage V = 2.83∠0o Vrms, I = 2∠-60o Arms • S = P + jQ = V I* = 2W + j 3.46VAr = 4 ∠60o VA Q Power Triangle • Commonly used graphical representation of S = P + jQ = V I* = V (V/Z)* = V V* / Z* = |V|2 / Z* = |I|2 Z • Useful relationships: P = |S| cos(Θ), Q = |S| sin(Θ), Θ = power angle = tan-1(Q/P) Q = P tan(Θ) • 1st - power factor is lagging, inductive load Im |S| • Need only two quantities to find third S Q Re P (4 ∠60o VA = 2W + j 3.46VAr) Power and Phasors • Another interpretation of active and reactive power components • Current components • In phase with voltage - Ieff cos(θ - ϕ) • 90° out of phase (quadrature component) - Ieff sin(θ - ϕ) • Multiplied by |V| results in P and Q Veff Im Ieff cos(θ - ϕ) Ieff sin|θ - ϕ| θ-ϕ Ieff Re Power and Phasors Example: (Chapter 11, Problem 42.) Both sources are operating at the same frequency. Find the complex power generated by each source and the complex power absorbed by each passive circuit element. Vx − 100 V V − j100 + x + x =0 6 + j4 − j10 5  1  100 ∴ Vx  + j 0.1 + 0.2  = + j 20 6 + j 4 6 + j 4   ∴ Vx = 53.35− ∠42.66° V 100 − 53.35− ∠42.66° ∴ I1 = = 9.806∠ − 64.44° A 6 + j4 1 ∴ S1. gen = × 100 × 9.806∠64.44° = 211.5 + j 4423VA . 2 Power and Phasors Example: (Chapter 11, Problem 42.) Both sources are operating at the same frequency. Find the complex power generated by each source and the complex power absorbed by each passive circuit element. S 6, abs = 1 × 6 × 9.806 2 = 288.5 + j 0 VA 2 1 ( j 4) 9.806 2 = 0 + j192.3 VA 2 j100 − 53.35 − ∠ 42.66 ° = 14.99 ∠121.6 °, I2 = 5 1 S 5 abs = × 5 × 14.99 2 = 561.5 + j 0 VA 2 1 S 2, gen = ( j100)14.99 ∠ − 121.57 ° = 638.4 − j 392.3 VA 2 2 1  53.35  -90 0 VA S − j10, abs =   ( − j10) = 0 − j142.3 VA = 142.3 ∠ 2  10  S j 4, abs = Power Measurement • Average PF is used to adjust consumer's bill (industry has to pay for unwanted losses) • Complex power delivered to individual loads • equals their sum • no matter how loads are connected • S = V I* = V (I1 + I2)* = V (I*1 + I*2) = VI*1 + VI*2 Power Factor Correction • Large industrial consumers pay penalty when PF < 0.85 • Caused by inductive loads (motors) • Why: Causes increased • Device ratings • Transmission and distribution losses • Example: • \$0.22/kVAr above 62% of average (real) power demand • S = P + jQ = P + j0.62P = P (1+j0.62) = P (1.77∠31.8°) • Targets cos(31.8°) = 0.85 (pay penalty when PF < 0.85) • Use compensation capacitors in parallel with load Power Factor Correction • Value of capacitance P(tan θ old − tan θ new ) C= 2 ωVrms corrects the PF angle from old to new at the specified frequency and voltage • Derived from ∆Q = V2rms / Xc = V2rms ωC (Qold - Qnew) / V2rms = ωC P(tan θold - tan θnew) / V2rms = ωC Complex Power Example: Source of 115 Vrms supplies two loads. Loads are connected in parallel: 7kW / 3 kVAr and 4kVA at 0.85 pf lagging. Find the pf of the equivalent load as seen from the input terminal. S1 = 7000 + j 3000 S2 = 4k [cos(θ) + j sin(θ)] = 4k [0.85 + j sin(cos-1(0.85))] = 3400 + j 2107 S = S1 + S2 = 10400 + j 5107 = 11586 ∠26.15° θ = tan-1(Q / P) = tan-1(5107 / 10400) = 26.15° cos(θ) = cos 26.15 ° = 0.8976 Z = |V|2 / |S| = 1152 / 11586 = 1.142Ω Z = 1.142Ω ∠26.15° = 1.02 + j 0.5 Ω (remember: S = |V|2 / Z*) ``` Electronics 23 Cards Mobile computers 37 Cards

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The overall goal is that the student should be able to use digital and microprocessor based technology in the design of a product. The student should also be able to dimension the drive and the size of a motor in an application. ### Choose semester and course offering Choose semester and course offering to see information from the correct course syllabus and course offering. ## Content and learning outcomes ### Course contents Electrical circuits: DC, AC and transients. Analogy between electrical and mechanical quantities. Electrical measurements and analog circuits: Measuring with multimeter and oscilloscope. Use of LabVIEW. The OP-amplifier model and how it is used in amplifier circuits and as a comarator. Use of filters to pass or block diferent frequency ranges. Digital electronics and microcontrollers: Transistors in switched applications. Analysis and synthesis of combinatorical and sequence cirquits. The functionality of a microprocessor and a microcontroller. Use of microcontrollers in simple applications. Analog cirquits for signalcondition of sensorsignals before ADC (analog to digital conversion). Examples of sensors such as encoders and strain gauges. Electrical motordrives: Single- and three- phase systems. Theory and properties of DC machines and PM synchronous machines. Principles for speedcontrol of electrical machines. Mechanical and thermal transients in electrical machines. Choice of machine size for time varying mechanical loads. Power electronics and drive units for machines. calculation of the required voltage and current for a motordrive. Sustainable development: Electric and hybrid cars. Calculation of quantities such as e.g. energy, power, force, velocity, acceleration, current and voltage in different parts of a electric or hybrid car under different conditions such as acceleration or regenerative braking. Dimension of energystorages such as batteries and capacitors (ultracap). ### Intended learning outcomes After the course the student will be able to - Analyze the conditions in simple circuits such as DC, AC and transient events of the first order. - Choose an electric motor to a mechanical load whose torque varies in time. - Calculate the speed, torque, power, current and voltage in different parts of an electric motor drive (consisting of mechanical load, electric motor and power supply) at constant speed and also during acceleration and braking. - With given cooling conditions estimate the temperature of an electric motor for some time after a known load is applied. - Able to explain the problems and possibilities of electricity and / or hybrid operation compared to other technologies for the propulsion of cars viewed from a sustainability perspective. - Describe and perform basic calculations on different powertrain concepts for electric and hybrid cars. - Designing an energy storage in an electric or hybrid vehicle to achieve the desired performance such as range. With energy storage means in this context mainly batteries and / or ultra-capacitors. - Use a microcontroller to solve simple tasks such as controlling the voltage to an electric motor. - Describe a system using a state diagram and write a program to control such a system. - apply the  OP-amplifier model to dimension and analyze basic circuits. - dimension and analyze simple filters. - Design a digital design to solve a combinatorial problem. - Estimate the deviations in the measurement results. - Connect simple electrical circuits. - Connect electric measuring instruments such as multimeter and oscilloscopes to simple electrical circuits. Performing measurements with these instruments. - Experimentally determine the current-voltage-characteristics of a device or component. - Solve simple problems and show the solution function by performing an experiment. - Give a short oral presentation about the outcome of an experiment or a laboratory exercise. - Translate the substance technical terms into English. - Work constructively in a group of 2-3 persons with laboratory and experimenta. ### Course Disposition No information inserted ## Literature and preparations ### Specific prerequisites SF1624 Algebra and Geometry, SF1625 Calculus in One Variable and SF1626 Calculus in Several Variable ### Recommended prerequisites No information inserted ### Equipment No information inserted ### Literature Will be announced at the beginning of the course. ## Examination and completion If the course is discontinued, students may request to be examined during the following two academic years. A, B, C, D, E, FX, F ### Examination • INL1 - Assignments, 3,0 hp, betygsskala: P, F • LAB1 - Laboratory Work, 3,0 hp, betygsskala: P, F • TEN1 - Written examination, 3,0 hp, betygsskala: A, B, C, D, E, FX, F Based on recommendation from KTH’s coordinator for disabilities, the examiner will decide how to adapt an examination for students with documented disability. The examiner may apply another examination format when re-examining individual students. ### Opportunity to complete the requirements via supplementary examination No information inserted ### Opportunity to raise an approved grade via renewed examination No information inserted ### Ethical approach • All members of a group are responsible for the group's work. • In any assessment, every student shall honestly disclose any help received and sources used. • In an oral assessment, every student shall be able to present and answer questions about the entire assignment and solution. ## Further information ### Course web Further information about the course can be found on the Course web at the link below. Information on the Course web will later be moved to this site. Course web MF1016 ### Offered by ITM/Machine Design Technology First cycle

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{[ promptMessage ]} Bookmark it {[ promptMessage ]} # 3 advanced formulas - formula results in the priority of... This preview shows pages 1–5. Sign up to view the full content. ADVANCED FINANCIAL MODELING FORMULAS Optional Debt Prepayment / Waterfall Formula =MAX(MIN(D151-E218,E\$241- SUM(E\$222:E224)),0)*\$D\$225 YES / NO TRIGGER = \$D\$225 If \$D\$225= 0 or “NO” there is no prepayment and the value reverts to 0 If \$D\$225 = 1 or “YES” there is prepayment and the value = the output of the min / max function This preview has intentionally blurred sections. Sign up to view the full version. View Full Document MIN / MAX FUNCTIONS =MAX(MIN(D151-E218,E\$241- SUM(E\$222:E224)),0)*\$D\$225 The amount of optional debt retirement cannot: 1) Be a negative payment - Achieved using a MAX function 2) Exceed the Debt balance - Achieved using a MIN function Further detail on MIN function MIN(D151-E218,E\$241- SUM(E\$222:E224)) D151 – E218 = Revolver Balance – Required Revolver Pmt Required Revolver Payment is 0 in this case. E\$241 – SUM(E\$222:E224) Where: E\$241 = Cash Available for Debt Repayment + Existing Excess Cash SUM(E\$222:E224) = Required Debt Payments + Previous Optional Prepayments If you highlight the Senior Sub cell this This preview has intentionally blurred sections. Sign up to view the full version. View Full Document This is the end of the preview. Sign up to access the rest of the document. Unformatted text preview: formula results in the priority of paying off the Revolver, then Term Loan, and then the Senior Sub Notes. (Waterfall formula) Average Interest Expense Calculation =\$C104*IF(B7=1, (D160+E160)/2,D160) If average interest formula trigger \$C104 is turned on then multiply the interest rate by an average of this period’s debt balance and the prior period’s debt balance… Else Multiply the interest rate by last period’s debt balance. Calculation Of Fully Diluted Shares Outstanding Using Options Sample Formula: Diluted Shrs =E7+IF(E6>E9,E8-E8*E9/E6,0) Logic If the options are out of the money use current shares outstanding (E7)…else use current shares outstanding (E7) + new net shares Where net new shares = number of vested options (E8) – amount of shares the firm could buyback with the exercise price proceeds (E8*E9/E6)... View Full Document {[ snackBarMessage ]}

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# Math - Trigonometry Solve for x, correct to one decimal place. It`s a right angle triangle. The height for the triangle is 4 cm with a 48° angle. The bottom of the triangle is an x. I got the answer 3.3 cm, is this correct? Any help would be appreciated. Thanks! 1. 0 2. 0 3. 12 1. Did you use tan48 = x/4 ? and then solve for x. 1. 0 2. 0 posted by John 2. Hello John, I was thinking of doing 4tan48° on my calculator, and the answer i got was 4.4 cm, but i wasn't sure if this was correct or not. Is this right? Thank you. 1. 0 2. 0 posted by Ollie 3. Yes. You have rounded. But it is correct : ) 1. 0 2. 0 posted by John 4. Thank you so much for your help, John. :) 1. 0 2. 0 posted by Ollie ## Similar Questions 1. ### Math Find, correct to one decimal place, the size of the smallest angle of the triangle which has sides of length 3,5,7 asked by Anonymous on October 17, 2012 2. ### Trigonometry Find the distance from the ladder to the base of the slide, marked with an x in the diagram. Give your answer accurate to one decimal place. The height of the right angle triangle is 4 m, the hypotenuse is 7 m and the missing asked by Michael on February 22, 2018 3. ### 5 Pre-Calculus Questions 1. What is the value of x in the right triangle below? If needed, round your answer to two decimal places. 32 angle, 15 inches, find the x in base… 2. A 10-foot ladder is leaning up against the side of a building so that the top asked by Anonymous on June 30, 2011 how do you find the isoperimetric quotient of an n sided figure? my side lengths are 5cm i have split it in to triangles so 360/n is the central angle then to work out area of singular triangle its 1/2 base times height height is asked by Spirit on October 12, 2006 5. ### math Isosceles Triangle ABC is inscribed in a circle with radius 12cm. Aide AB is a diameter. Determine the perimeter of the triangle, correct to one decimal place. asked by Amy on July 27, 2009 6. ### Math A solid S is made up of a cylindrical part and conical part. The height of the Soild is 4.5m. The common radius of the cylindrical part and the conical part is 0.9m. The height of the conical part is 1.4m. a. Calculate the asked by Kd on September 7, 2018 7. ### Math A solid S is made up of a cylindrical part and conical part. The height of the Soild is 4.5m. The common radius of the cylindrical part and the conical part is 0.9m. The height of the conical part is 1.4m. a. Calculate the asked by Kd on September 4, 2018 8. ### math In a right triangle ABC, C = 90°, b = 380.1 inches and c = 589.1 inches. Solve for all the missing parts using the given information. (Round each answer to one decimal place for side and to two decimal places for angles.) A = ° asked by Anonymous on February 10, 2012 9. ### maths a cube has a lenght of 16cmby12cm height of 10cm with angles z,q,p,w,s,v,r and x.find the angle between z,q and the base p,q,r,s,.2.the angle between zq and the vertical plane p,w,z,s to 1 decimal place.the angle between plane asked by golda on April 14, 2016 10. ### mathematics the vertices of angle PQR are P(3,0),Q(9,-2) and R(9,8)find the area of angle PQR to one decimal place and state the type of triangle asked by bright ulubi on March 19, 2016 More Similar Questions

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## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition) $5\left( 7a+11c+1 \right)$ The $GCF$ of the given expression, $35a+55c+5 ,$ is $5,$ since it is the highest number that can evenly divide (no remainder) all the terms. Factoring the $GCF$ results to \begin{array}{l}\require{cancel} 5\left( \dfrac{35a}{5}+\dfrac{55c}{5}+\dfrac{5}{5} \right) \\\\= 5\left( 7a+11c+1 \right) .\end{array}

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https://www.mechanicaltutorial.com/ac-circuit-objective-questions-and-answers-02-03

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# AC Circuit - R-L-C Circuit objective questions (mcq) and answers 11.  A series R-L-C circuit draws a current at a leading power factor at A. More than resonant frequency B. Less than resonant frequency C. Resonant frequency D. None of the above B.Less than resonant frequency 12.  In a series R-L-C circuit at resonance A. wLC = 1 B. wL2C2 = 1 C. w2LC = 1 D. w2L2C = 1 C. w2LC = 1 13.  The power factor of a series R-L-C circuit at its half-power points is A. Unity B. Lagging 14.  The dynamic impedance of an R-L and C parallel circuit at resonance is........ohm. A. R/LC B. C/LR C. LC/ R D. L/ CR D. L/ CR 15.  In an RLC circuit, supplied from an AC source, the reactive power is proportional to the A. Average energy stored in the electric field B. Average energy stored in the magnetic field C. Some of the average energy stored in the electric field and that stored in the magnetic field D. Difference between the average energy stored in the electric field and that stored in the magnetic field

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http://blade.nagaokaut.ac.jp/cgi-bin/scat.rb/ruby/ruby-talk/129926

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```I made one guess - count of possible wins of the first and second player on the 4x4 board is in direct proportion to count of their wins on the 4x2 board (due to the symmetry of the 4x4 board and possible moves). I have not managed to prove it mathematically, so my program below may be totally wrong... :-) ############################################################### #!/usr/bin/ruby -w NEW_GAME = 0b0000_0000 END_GAME = 0b1111_1111 POSSIBLE_MOVES = [ 1, 2, 3, 4, 6, 7, 8, 12, 14, 15, 16, 17, 32, 34, 48, 64, 68, 96, 112, 128, 136, 192, 224, 240 ] # wins_of_first and wins_of_second \$f = \$s = 0 # last_move_by - true for second player # false for first player def play( state, last_move_by, possible_moves ) possible_moves.delete_if { |m| state & m != 0 } if state != END_GAME possible_moves.each do |m| play( state | m, ! last_move_by, possible_moves.clone ) end elsif last_move_by # last move was by second player \$f += 1 else \$s += 1 end end play( NEW_GAME, true, POSSIBLE_MOVES ) puts "Wins of first == #{\$f}\nWins of second == #{\$s}", "#{\$f < \$s ? 'First' : 'Second'} player is bounded to win" ```

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https://molecularsciences.org/content/understanding-iteration-and-recursion-in-python-a-comparative-analysis/

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Understanding Iteration and Recursion in Python: A Comparative Analysis In programming, iteration and recursion are two fundamental approaches for executing repetitive tasks. Both methods have their unique applications, advantages, and performance implications. This article aims to elucidate the differences between iteration and recursion, supplemented by Python examples and benchmark results. Iteration in Python Iteration refers to the technique of executing a block of code repeatedly through loops such as `for` or `while`. It’s a straightforward approach where the number of repetitions is determined by the loop’s control conditions. Example: Calculating Factorial Using Iteration ``````def factorial_iterative(n): result = 1 for i in range(1, n + 1): result *= i return result`````` In this example, the `factorial_iterative` function calculates the factorial of a number by iteratively multiplying the numbers from 1 to `n`. Recursion in Python Recursion, on the other hand, involves a function calling itself with a modified parameter until a base condition is met. It’s a more elegant approach, often resulting in cleaner and more readable code. Example: Calculating Factorial Using Recursion ``````def factorial_recursive(n): if n == 1: return 1 else: return n * factorial_recursive(n - 1)`````` Here, the `factorial_recursive` function calls itself with `n - 1` until it reaches the base case where `n` equals 1. Performance Benchmarking When it comes to performance, iteration and recursion can vary significantly. Iteration is generally more efficient in Python due to less overhead. Recursion can lead to a stack overflow if the depth of recursion is too great, and it also involves additional overhead due to function calls. Benchmark results indicate that for simple tasks like calculating a factorial, iteration is faster and consumes less memory compared to recursion. However, recursion can be more intuitive and easier to implement for problems that have a natural recursive structure, such as tree traversals or algorithms like quicksort and mergesort. Conclusion Choosing between iteration and recursion depends on the specific problem at hand, the limitations of the programming language, and the programmer’s preference for code readability and maintainability. In Python, it’s essential to be mindful of the recursion depth limit and the potential for stack overflow errors. For tasks that require optimal performance and efficiency, iteration is the preferred method. For problems inherently recursive in nature, recursion provides a more straightforward and elegant solution, albeit with a potential cost in performance. In summary, both iteration and recursion are valuable tools in a programmer’s toolkit, and understanding when and how to use each can lead to more effective and efficient programming practices.

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# ChemSiddiqui Senior Members 378 ## Posts posted by ChemSiddiqui ### Quantum Physics dont worry, pay attention to word 'Quantum', look it up in dicitionary if you dont already know. I am a chemistry student and I never did physic before starting university and trust me you do a lot of quantum mechanics in physcial chemistry. Why, well atoms, molecules are very very tiny and our everyday observation give us a misleading picture in general so we use quantum mechanics to help us understand what is possible and what isnt even if the impossible may actually be possible in the real 'quantum' world. hope this helps you to think about it, also there are some books that will also help you with your essay. maybe google 'quantum mechanics and real world' who know something interesting might show up. good luck. ### math problem! For a start: I think U is not an operator but a terrible mess . Do you know what U means? I don't. Is it really the operator as it is given in the question/book or did you rearrange some letters? E.g. if U was $\frac{\sin \theta}{\cos \theta} \frac{d}{d\theta}$ then f would obviously be an eigenfuntion. You know what atheist, you got it right. The way the question was written down I didnt understand it . U is exactly as you suggested; $\frac{\sin \theta}{\cos \theta} \frac{d}{d\theta}$ but I am not sure yet if it is an eigenfunction, so I will do the calculations but I know how to do that. Thanks a lot. Merged post follows: Consecutive posts merged just one more question. Its the double derivative question; $\frac {d^2}{dx^2} e^\frac{-x^2}{2}$ and this is what I did, frst taking the first derivate: $\frac {d}{dx} e^\frac{-x^2}{2} = \frac{-1}{2}e^\frac{-x^2}{2}$ Now taking the second derivative; $\frac {d^2}{dx^2} (\frac{-1}{2}e^\frac{-x^2}{2})=\frac{-x}{4} e^\frac{-x^2}{2}$ could anybody confirm id this is right or not? any help most appreciated! ### math problem! well this is why I am slighty confused. The question goes to operate wavefunction by an operator to check if the wavefunction is an eigenfunction of the operator. The operator is $U = \frac{d sin(\theta)}{d(\theta)cos(\theta)}$ and the function is $f(\theta, \phi) = sin(\theta)sin(\phi)$ Now I know that I will have to use the product rule , but I dont know how, becuase the operator is very strange. Its like its saying that diffentiate the function f by $\frac{d sin(\theta)}{d(\theta)cos(\theta)}$ Maybe i havent understood the question properly? ### math problem! Hi everyone, I have been trying to practice calculus for the benfit of my quantum mechanics studies and I am stuck with a problem so thought you people might have some advice to give.Here goes; $\frac{d sin\theta}{d(\theta)cos(\theta)} sin(\theta)sin(\phi)$ Now we have to operate the $\frac{d sin(\theta)}{d(\theta)cos(\theta)}$ onto function $sin(\theta)sin(\phi)$. The operator can expressed into $cos^2(\theta)-sin^2(\theta)$ after diffentiation. Now my question is that can I simply multiply this by $sin(\theta)sin(\phi)$ or will this be wrong? It would have been much easier if the expression were $\frac{d}{d(\theta)} sin(\theta)sin(\phi)$ i.e. to diffentiatie the function?! I'll be grateful for your help! ### 2 questions! I was just wondering about the reaction of soft-acids and soft bases. Like generic acid/base reaction give a salt and water/neutral species, does soft acid and soft bases give a salt or they would give an adduct/complex? Lets say that the soft base is NMe3 and soft acid is a diradical oxgen molecule. Is this similar to the reaction between ammonia and diradical oxygen which goes like this 4NH3 + 5O2 -------> 4NO + 6H2O? Second question is more of a thought really. Why do we want to look for electrons when we look at an atom. I am asking this question as a quantum mechanics student. Does position or state of an electron give us all the information about a whole system? I am thinking that once we know which state the electron is in we can determine the reactivity of the system? any help, thoughts or comments are appreciated. you are right hermanntrude. thanks for correcting me! hmm, I think we could do the acid-base reaction with it if there is a sp3 hybrid orbital on B and it is empty. But I am not sure if there be one on the B atom? hey everyone, I was reading the inorganic book today and i have got a question. say we have an amine-borane adduct lets take (CH3)3CH2N:BH3, can it undergo redox or acid-base reactions? I mean the adduct doesnt act as a base because the empty 2p orbital on B is filled when Nitrogen in (CH3)3CNH2 datively forms a bond with BH3 so I am thinking that it cant have undergo any acid-base reaction. Can anyone tell me if I am thinking rightly here. I will be grateful for any direction. ### say hello Welcome to SFN alan, hope your stay here is an enjoyable and informative one. all the best. ### Birthday Gifts for a Scientist When I was reading a blog today, I saw a post about a sweater which had the complete periodic table on it. The only problem is that it was an ugly colour. It's my housemate's birthday coming up, and he's doing a PhD in genetics. Does anyone know of any sciencey/geeky gifts online that I could get him along these lines? Also, if I saw something cool enough, I would probably get something for myself. Oh, and I should probably note that he already has just about every one of those Giant Microbe toys! Many thanks! I bought myself a periodic table t-shirt which looks really cool, but its not the kind of thing you'd like to wear when going out or as a causual wear. But, if your friend like chemistry stuff i suggest buy him/her a molecular modelling set. hope this helps a bit. ### trailer cost belittled? thank Mr Skeptic, I googled the prices and without the things you stated in your reply it seems the cost can be as much as 300m GBR pounds. ### trailer cost belittled? Hi there everyone, its been a long time since i posted;kinda taking a rest from the forums and trying to enjoy the holidays. I was just chatting with someone the other day and he told me that you can actually build a trailer for 300 GBR pounds. Although i was not convinced I couldnt help but ask anyone here if that is possible. lets say its a wooden trailer. is this possible to build it that cheap? feel free to tell me of your experiences. Thanks. ### How long do you want to live for? For me when I find what I am looking in life ,I wouldnt complain God if die. ### RIP Michael Jackson He was the first ever person I heard to when I started listening to music. Liked him then but never really became a fan, though i was quite unaware about the allegations(I need not mention) leveled against him. I believe anyone who dies deserve some respect if they hadnt really killed anyone or done something of the sort ,humans do have false afterall. God knows him the best so let He be the judge so all I will say is RIP MJ. ### why do some metals wet glass? I think even though what you said above may be correct, if the adhesive forces between the glass and liquid are dominant than cohesive forces in the liquid itself then some of the liquid will stick to the wall and make it wet, but if the cohesive forces (attractive forces b/w like molecules) in liquid are dominant than the adhesive forces b/w wall of glass and the liquid as is the case with mercury, surface tension will make the liquid to occupy the minimun surface area and hence will not allow it to stick to the wall. ### QM problem! So I can use the classical formula for kinetic energy as used in my calculation above? Even if its a quantum mechanics question. ### why do some metals wet glass? mercury is a liquid metal (only liquid metal), it has mettalic bonding and surface tension is great in mercury so Cohesive forces in mercury are much more dominant than the adhesive forces with the wall of the glass. so mercury form 'bead' rather than stick to the wall of glass and make it wet. ### QM problem! what I think is that if we find the energy of electron in 1D box, by assuming that the length of box is $10^{-10} m$ and n=1 (lowest energy of the system) then we can find the momentum but we we would have to use the classical foruma relating kinetic energy with momentum; $E= \frac {p^2}{2m}$ if the use of the above forumla is acceptable ( i say acceptable because then it will be switching from QM to classical mechanics), then we can find mommentum and then use the de broglie relation to get the wavelegnth? BTW just did the calculation assuming what i said above is correct, and the de broglie comes out $1.99*10^{-10} m$ ### QM problem! hey everyone, I am doing QM, trying to solve as much problems there are to get a strong hold of the idea. here one question which i cant understand, i think that there is not enough information to solve this one; Assuming that a chemical bond can be modelled as a one-dimensional box,estimate the order of magnitude of the de Broglie wavelength of an electron in a typical chemical bond. do you think we have enough information in the question to solve this. The de broglie wavelength can be calculated using $lambda = \frac {h}{p}$ where p is momentum. how do you calculate momentum? I dont get it. Could anyone help. thanks!. ### Am I doing right? thanks a lot for verifing that. Just one little question aslo from past papers but a trivial one; The typical bond lengths of C-H, C-C and C(triple bond)C bonds are, respectively,110 pm, 150 pm and 120 pm. Use this information to obtain a reasonable length for a 1-D box that you can use to model the states of the pi electrons of butadiyne. Butadiyne , H-C(tiple bond)C-C(triple bond)C-H. there are 2 C(triple)C bonds, 2 C-H bonds and 1 C-C bond in the molecule. To get the ength of the box we add all the lengths togther; so 2(110) + (150) + 2(120) = 610 pm. Do you think this is right or must I have to take the average? Now I am able to solve most of the problems of particle in a box, QM actually isnt that bad I am starting to like it. ### Am I doing right? Well actually i did come up with that expression myself, but it was the book that made the distiction between energy of the system with equal lengths in all direction and one where the length are not equal, so I knew that this expression was for a box of equal legths! I will do the calculation now and see what I can get for an answer. I will post the answer here. Thanks for the help along the way. Merged post follows: Consecutive posts merged When we plug in the numbers in the expression $\frac{H^2}{8mL^2}$ for the mass of electron which is $me = 9.109*10^{-31}$ and $L= 6.626 * 10^{-9}$ we get the $1.37 *10^{-21}$ we can now write; $E = 1.37*10^{-21}{(n^2x + n^2y + n^2z)}$ using different combination of $nx, ny, nz$; first for lowest energy of the system $nx =1, ny=1, nz=1$ $E = 4.11*10^{-21}$ which is lower than the given energy, so this combintation is one of the required ones. Becuase we there is only one way to arrange the energy in 111, the degenercy is 1. for $nx=1, ny= 1, nz=2$ , $E = 8.22*10{-21}$ which again has lower energy value than the one given in question. But, now the degenrecy is 3 becuase there are three ways to arrange the energy i.e. 112, 121, 211. for $nx =1, ny=2, nz=1$, $E = 1.23*10^{-20}$. again this combination meets the condition refered to in the question, and the degenercy is 3. An electron is trapped in a three-dimensional box with edge lengths Lx=Ly=Lz= 6.626 x 10^(-9) m. Among the energy levels lying below E = 1.35 x10^(-20) J, how many are degenerate? What is the degeneracy of each one of them? There are 3 degenerate states and their degenercies are 1,3 and 3. ### Am I doing right? Is that the same n or is it three different n? right, now, we want to find 'n' and it is not neccessary that they are same. They will be same when we know that the translation of electron at a particular direction has the same energy as the other direction e.g. $nx = ny$ all we know is that the lengths are equal so the energy of the system can be equal to; $E = \frac {h^2}{8mL^2}{(n^2x + n^2y + n^2z)}$ I got the above expression from a physical chemistry book which says that when the lengths a=b=c then the above expression is applicable for energy of the system. we could use $n=1$ for $nx,ny,nz$ to get the lowest energy of the system and see if that energy equals the energy given in the question, if it is lower than the provided energy we have one state which has lower energy than asked in the question. how am i doing uptil here? ### Am I doing right? Ok here goes; the question is about 'a particle in 3D box'. The hamitonian operator for it is; now because the total enegy is equal to sum of energies in the 3 directions we have; $E(x,y,z) = Ex + Ey + Ez$ where $Ex = \frac {h^2n^2}{8mL^2x}, Ey = \frac {h^2n^2}{8mL^2y} , Ez = \frac {h^2n^2}{8mL^2z}$ so total energy is; $Ex,y,z = \frac {h^2}{8m}{(\frac{n^2}{L^2x} +\frac{n^2}{L^2y} + \frac{n^2}{L^2z})}$ so now the schrodingers equation will be; $H\psi_(x,y,z) = E\psi_(x,y,z)$ and $-\frac{\hbar^2}{2m} \left( \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} +\frac{\partial^2}{\partial z^2} \right) = \frac {h^2}{8m}{(\frac{n^2}{L^2x} +\frac{n^2}{L^2y} + \frac{n^2}{L^2z})}$ the total energy provided is $E = 1.35 * 10^{-20}$, so to find the value for n we put E equal to the energy provided. $\frac {h^2}{8m}{(\frac{n^2}{L^2x} +\frac{n^2}{L^2y} + \frac{n^2}{L^2z})} = 1.35*10^{-20} J$ by what we do after I cant seem to figure out, we cant possible take the LCM of the terms in bracket on the left hand side of the equation and take the denominator to the other side of the equation? note; by trying to use LaTex, i can appreciate now that it takes ages and is a monotonious work to write all that stuff dowm. grateful for your help!. ### Am I doing right? thanks I got it. I might not be using it immediately; takes a bit of time. If you have ever worked on maple then I think its very easy to use. Except that a few commands will need some learning!. sorry it will be really frustrating for you I am sure, but what do you think of my reasons to use the expressions and in your opinion whats the answer! ### Am I doing right? Yes. Two natural questions: 1) Why? 2) Why would the 3D case with 3 equal-length sides not be Enx,ny,ny = h^2/8m ( (nx/L)^2 + (ny/L)^2 + (nz/L)^2 ), then? . 1).well, we are considering a particle in a 3-D box with different lengths. So the particle will have different energies in different directions. 2). Well i tried to solve it with that expression and am reduced to; E8m/h^2 = ((nx+ny+nz/Lx,y,z,^2)) here the algebra seems a bit complicated because on the right hand of equation you have a numerator which is a sum and demoniator cant possibly be taken to the left or can it? Any way, even if you break that rule you still get the same answer i.e. n= approx 3. So i guess we could use the energy forumla from 1-D box?. I too have atkins 'physical chemistry', although it is recommended for next year I am trying to use it so solve that problem. How can I use the LaTex, are there any instructions here in sfn. Thanks a lot. ×

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https://scienceoxygen.com/how-do-you-calculate-freezing-point-depression/

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# How do you calculate freezing point depression? 1. Step 1: Calculate the freezing point depression of benzene. Tf = (Freezing point of pure solvent) – (Freezing point of solution) 2. Step 2 : Calculate the molal concentration of the solution. molality = moles of solute / kg of solvent. 3. Step 3: Calculate Kf of the solution. Tf = (Kf) (m) ## What is freezing point depression in chemistry? Freezing Point Depression. Freezing Point Depression. The freezing point of a solution is less than the freezing point of the pure solvent. This means that a solution must be cooled to a lower temperature than the pure solvent in order for freezing to occur. ## How do you calculate KF in chemistry? Divide the freezing point depression by the molal concentration so you have: Kf = delta Tf / cm. Insert the values for delta Tf and cm. For instance, if you have a solution with a molality of 0.455 which freezes at 3.17 degrees Celsius, then Kf would equal 3.17 divided by 0.455 or 6.96 degrees Celsius. ## What is the freezing point depression of NaCl? Commonly used sodium chloride can depress the freezing point of water to about −21 °C (−6 °F). If the road surface temperature is lower, NaCl becomes ineffective and other salts are used, such as calcium chloride, magnesium chloride or a mixture of many. ## How do you calculate freezing point depression and boiling point elevation? Multiply the original molality (m) of the solution by the number of particles formed when the solution dissolves. This will give you the total concentration of particles dissolved. Compare these values. The higher total concentration will result in a higher boiling point and a lower freezing point. ## How do you determine freezing point? Insert the thermometer in the slush, before the one you’re measuring turns completely liquid. Leave the thermometer in there until the point when it becomes all liquid. Write down the temperature when that happens. Make sure the thermometer you are using reads below 0 degree C. ## What is freezing point depression of water? The proportionality constant, Kf, is called the molal freezing-point depression constant. It is a constant that is equal to the change in the freezing point for a 1-molal solution of a nonvolatile molecular solute. For water, the value of Kf is −1.86oC/m. ## What is the value of KF? as KCN dissociates completely, the depression in freezing point of water: Kf=mΔT=2×0. 18920. ## How do you calculate the KF value of water? The value of Kf for water is 1.86^o , calculated from glucose solution. ## How do you calculate freezing point depression with molality? The freezing point depression ∆T = KF·m where KF is the molal freezing point depression constant and m is the molality of the solute. Rearrangement gives: mol solute = (m) x (kg solvent) where kg of solvent is the mass of the solvent (lauric acid) in the mixture. This gives the moles of the solute. ## What formula is used to calculate KF form factor? kf = dT / (m x i).. if your using the value of 1 for i for NaCl instead of 2, then Kf will be 2x’s too large. so you need to divide by 2. ## Is freezing point depression always negative? The change in the freezing point is proportional to the amount of solute added. This phenomenon is called freezing point depression. The change in the freezing point is defined as: ∆Tf = Tf,solution − Tf,solvent. ∆Tf is negative because the temperature of the solution is lower than that of the pure solvent. ## What is the freezing point depression of cacl2? Therefore the freezing point depressed by 5.58°C. It was important to know that the solute, calcium chloride was ionic, otherwise our van’t Hoff factor would have been 1 for a molecular solute and changed our calculations significantly. ## How do you calculate the freezing point of salt water? Calculate the freezing point depression of the saltwater solutions in glasses A, B, C, and D using the formula: ∆T = – (Kf) (m) (i) = (1.86)(m)(2). This is also the freezing point of the salt- water solution in °C. If you wish to see this freezing point in °F, convert using the formula: °F = (1.8*°C) + 32. ## What is the unit of KF? Kf is a constant for a given solvent. Kf is called the molal freezing point depression constant and represents how many degrees the freezing point of the solvent will change when 1.00 mole of a nonvolatile nonionizing (nondissociating) solute dissolves in one kilogram of solvent. ## How can you determine the value of KF and KB? The value of Kb or Kf depends only on the type of solvent and not solute dissolved in it. Here, Kb and Kf are molal elevation in boiling point constant and molal depression in freezing point constant respectively. They are characteristic of solvent and independent of the solute. Need a fast expert’s response? ## How do you calculate depression? 1. Scores below 7 generally represent the absence or remission of depression. 2. Scores between 7-17 represent mild depression. 3. Scores between 18-24 represent moderate depression. 4. Scores 25 and above represent severe depression. ## Why do we calculate form factor? It identifies the ratio of the direct current of equal power relative to the given alternating current. ## How do you calculate form factor? Another common way to calculate Building’s Form Factor is Surface to Volume Ratio (SVR). SVR is the ratio between the building’s envelope area (EA) and it’s volume (V), it is calculated as SVR = EA / V. ## Which factors affect the freezing point depression of a solution? The molar freezing point depression constant, the morality of the solution and the van Hoff factor of the salute. These are the three things that determine the freezing point depression, DELTA. ## Will the depression in freezing point be same or different? Solution : The depression in freezing point will be same in both the solutions because both are non-electrolytes and give same number of solute particles. ## Which has highest depression in freezing point? Depression in freezing point is a colligative property which depends on number of particles. Among given choices K2SO4 gives maximum number of ions, so it will have maximum depression in freezing point.

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You are on page 1of 39 Manual de Formao 2013 MANUAL DE EXCEL 2007 ORIENTAES METODOLGICAS DE ORIENTAO ................................................................................ 4 OBJETIVO GERAL.................................................................................................................................... 4 REFERNCIAS A ENDEREOS RELATIVOS E ABSOLUTOS ......................................................................... 5 REFERENCIAS A OUTRAS FOLHAS E FICHEIROS ....................................................................................... 8 FUNES DE PROCURA E REFERNCIA ................................................................................................. 10 PROCV (VLOOKUP) ........................................................................................................................... 10 PROCH (HLOOKUP) ........................................................................................................................... 12 FUNES LGICAS ............................................................................................................................... 13 SE (IF) ............................................................................................................................................... 13 E (AND) ............................................................................................................................................ 14 OU (OR) ............................................................................................................................................ 15 FUNES DE MATEMTICA ................................................................................................................. 17 SOMA.SE (SUMIF) ............................................................................................................................. 17 FUNES DE ESTATSTICA .................................................................................................................... 19 CONTAR.SE....................................................................................................................................... 19 CONTAR.VAZIO................................................................................................................................. 19 FUNES FINANCEIRAS ....................................................................................................................... 20 PGTO (PMT) ..................................................................................................................................... 20 FUNES DE DATA E HORA.................................................................................................................. 22 AGORA (NOW).................................................................................................................................. 22 DIAS360 (DAYS360) .......................................................................................................................... 22 HOJE (TODAY)................................................................................................................................... 22 DIATRABALHO (NETWORKDAYS) ....................................................................................................... 22 FORMATAO, PROTEO DE DADOS E FORMULRIOS...................................................................... 23 FORMATAO CONDICIONAL .............................................................................................................. 23 FORMATAO RPIDA ............................................................................................................................ 24 FORMATAO AVANADA ....................................................................................................................... 24 LISTAS PENDENTES............................................................................................................................... 26 PROTEO DE FOLHAS E LIVROS .......................................................................................................... 28 FORMULRIOS..................................................................................................................................... 30 CRIAR UM FORMULRIO DE DADOS ............................................................................................................ 31 TABELAS DINMICAS ........................................................................................................................... 32 IMPORTAO DE DADOS EXTERNOS............................................................................................................ 32 CRIAR UM RELATRIO DE TABELA DINMICA OU GRFICO DINMICO .................................................................. 32 UTILIZAR DADOS EXTERNOS ..................................................................................................................... 33 GRFICOS DINMICOS ............................................................................................................................ 34 CRIAR UM RELATRIO DE GRFICO DINMICO A PARTIR DE UM RELATRIO DE TABELA DINMICA EXISTENTE ................. 35 MACROS E EDIO VBA ....................................................................................................................... 35 MANUAL DE EXCEL 2007 O QUE UMA MACRO ........................................................................................................................... 35 GRAVAR E EXECUTAR UMA MACRO..................................................................................................... 36 O EDITOR DE VISUAL BASIC.................................................................................................................. 38 BIBLIOGRAFIA: ..................................................................................................................................... 39 Orientaes Metodolgicas de Orientao Este Manual foi elaborado com o objetivo de apoiar as aes de formao presenciais asseguradas pelo Departamento de Desenvolvimento e Formao. Visa facultar aos participantes das referidas aes um documento facilitador da aprendizagem nas ferramentas abordadas pelo mesmo. Ficar ao critrio do formador a forma como introduzir esta documentao nas sesses de formao. Objetivo Geral Atravs desta formao os formandos tero acesso a diversas explicaes sobre folhas de clculo e o aplicativo Excel, dotando-os dos conhecimentos para um bom desempenho profissional. Ser-lhes-o facultadas informaes sobre como trabalhar com diferentes livros simultaneamente, proteger informao, aplicar funes complexas, gerir informao atravs de tabelas dinmicas, bem como uma breve introduo elaborao de macros para automatizao de tarefas. MANUAL DE EXCEL 2007 Referncias a endereos relativos e absolutos A utilizao de endereos nas frmulas, em vez de valores constantes, permite que, sempre que exista alterao nos valores que influenciam a frmula, o resultado apresentado seja automaticamente atualizado. Dizemos ento que a frmula possui endereos relativos, porque ao ser copiada para as linhas seguintes, os seus endereos alteram-se, adaptando-se s novas coordenadas e fazendo com que o utilizador no tenha de digitar vrias frmulas idnticas. Nem sempre possvel utilizar a cpia das frmulas e a sua consequente adaptao a novas coordenadas, de forma directa. Na figura seguinte, na coluna E pretende-se calcular o valor total com 5% de desconto a pronto pagamento (P.P.) MANUAL DE EXCEL 2007 Se tentar copiar esta frmula para a linha seguinte ir verificar que o valor difere do que realmente deveria resultar do clculo. Isto porque a adaptao feita frmula copiada para a clula E6 deu origem frmula =D6*(1-E3), quando na realidade deveria ser =D6*(1-E2). Pretende-se, ento, que exista uma adaptao s novas linhas, mas a multiplicao seja sempre feita pela clula E2. Deve-se, para isso, indic-la como um endereo absoluto, pelo que a frmula correcta ser: =D5*(1-\$E\$2). MANUAL DE EXCEL 2007 O endereo absoluto difere do relativo, devido aos smbolos \$. A colocao destes smbolos pode ser feita atravs da simples digitao ou pressionando a tecla F4. MANUAL DE EXCEL 2007 Referencias a outras folhas e ficheiros Nas imagens seguintes temos duas folhas, na 1 folha (Preo Unit) temos os preos por Kg da fruta, na 2 folha (Preo Quant) necessitamos efectuar os calculos de acordo com o preo por Kg que nos indicado na 1 folha. Iremos estabelecer uma referencia ou ligaao entre as duas folhas que nos permita calcular os Kg da fruta pelo preo unitario. De salientar que este procedimento igual sendo CML/DMRH/DEPARTAMENTO DE DESENVOLVIMENTO E FORMAO - Joo Martins MANUAL DE EXCEL 2007 aplicado entre folhas do mesmo livro ou entre livros de ficheiros do Excel diferentes desde que estes ultimos se encontrem na mesma localizaao fisica. Tambem possivel executar em ficheiros com localizaoes diferentes, exemplo, entre 2 computadores distantes, desde que essa ligaao no seja quebrada, seja pela alteraao do nome do ficheiro ou pela alteraao da sua localizaao. Voltando ao exemplo apresentado, posicionando o rato na celula D3 da folha Preo Quant iremos executar a seguinte ligaao: (Sinal de igual) ='Preo Unit'!D3*'Preo Quant'!B3 Ou seja, iremos indicar que pretendemos multiplicar o valor por Kg das Maas que de 0,35 pelos 9 Kg que constam na clula B9 da folha Preo Quant . MANUAL DE EXCEL 2007 Funes de procura e referncia PROCV (VLOOKUP) Localiza um valor especfico na primeira coluna da esquerda numa matriz (tabela). Por outras palavras, esta funo permite a extraco de um valor de uma tabela com base num valor de pesquisa. Sintaxe da Funo PROCV(valor_proc; matriz_tabela; nm_ndice_coluna;procurar_intervalo) Valor_proc Valor a ser procurado na primeira coluna da tabela (matriz_tabela). Pode ser um nmero, uma referncia ou um texto. Matriz_tabela Tabela onde os dados so pesquisados. Os valores da primeira coluna da tabela (coluna onde feita a pesquisa do Valor_proc) devem estar ordenados por ordem ascendente. Nm_ndice_coluna o nmero da coluna de onde o valor correspondente (ao da coluna de pesquisa, onde actua o parmetro Valor_proc) extrado. Localizar_intervalo um valor lgico que indica se a pesquisa deve ser exacta ou por aproximao. Ou seja, se o parmetro for omitido a funo localiza um valor aproximado. Caso o parmetro seja falso (valor 0) a funo localiza o valor exacto. Exemplo de utilizao da funo PROCV. Esta funo permite pesquisar o contedo de uma determinada clula, na 1 coluna de uma tabela especificada, devolvendo em seguida um valor correspondente, numa outra coluna indicada, da mesma tabela. 10 MANUAL DE EXCEL 2007 No exemplo pretendemos introduzir um cdigo nas clulas A9, A10 e A11 e obter automticamente a respectiva descrio e o correspondente preo unitrio. Para tal iremos utilizar a funo Procv nas clulas B9:C11. No exemplo podemos ver a funo aplicada em B9 que seria copiada para B10 e B11. NOTA: Ao utilizarmos a funo Procv necessitamos de ter em ateno os seguintes factores: 1. A pesquisa feita sempre na 1 coluna da tabela. 2. A 1 coluna da tabela, onde feita a pesquisa, deve estar ordenada por ordem crescente. 3. Ao seleccionarmos a tabela no devemos abranger a linha de cabealho, se esta existir. Aqui na janela da funo podemos ver a frmula que estaria na clula C9 e posteriormente copiada para C10 e C11. 11 MANUAL DE EXCEL 2007 Vemos que esta funo tem 4 partes: 1. Indicar a clula onde ser introduzido o valor a pesquisar (Valor_proc). 2. Indicar a rea da tabela onde ser efectuada a pesquisa (Matriz_tabela). 3. Indicar o n da coluna onde est o resultado que pretendemos visualizar (Num_ndice_coluna). 4. (Procurar_intervalo) Este ltimo opcional e serve para indicar um Valor Lgico: 1 ou Verdadeiro, faz com que no caso de o valor introduzido na clula especificada em Valor_proc no existir seja colocado outro semelhante. 0 ou Falso s ser pesquisado um valor existente de facto na 1 coluna da tabela, caso contrrio surgir a indicao #N/D (Not Detected). PROCH (HLOOKUP) Esta funo basicamente idntica a Procv apenas a pesquisa, do contedo de uma determinada clula, realizada na 1 linha de uma tabela especificada, devolvendo em seguida um valor correspondente, numa outra linha indicada, da mesma tabela. 12 MANUAL DE EXCEL 2007 Funes lgicas SE (IF) Com base numa ou mais condies, devolve um valor se o teste condio for Verdadeiro ou outro se o teste for Falso. Sintaxe da Funo SE(teste_lgico; valor_se_verdadeiro; valor_se_falso) Teste_lgico Condio ou condies que se pretendem testar. Pode ser um valor ou uma expresso. Quando existe a necessidade de combinar vrias condies recorre-se s funes lgicas E(AND) e OU(OR). Exemplo de utilizao da funo SE. Na coluna Resultado pretendemos obter uma informao referente ao aproveitamento do aluno. Para tal iremos verificar se a nota em B2 superior ou igual a 10 e se tal se verificar surgir o texto Aprovado, caso contrrio surgir o texto Reprovado. 13 MANUAL DE EXCEL 2007 NOTA: Poderamos verificar se a nota era negativa, nesse caso a funo ficaria da seguinte forma: SE(B2<10;Reprovado;Aprovado) E (AND) As seguintes frases sero VERDADE ou MENTIRA? Tenho dois olhos E duas orelhas E nasci no planeta Terra. ou tem dois olhos E cinco orelhas E nasceu no planeta Terra. Resumindo: a funo E() para ser Verdadeira precisa que todas as opes sejam verdadeiras. E a funo ser assim: E(2+2=4;3+3=6) Portanto: verdade que 2+2=4 E tambm verdade que 3+3=6? Logo: E(10<100;5*5=25) E(10+90=100;5*5=25;4+4=8;10+1=175,4) A primeira verdade e a segunda falsa. Na prtica usamos o E() acoplado ao SE() CML/DMRH/DEPARTAMENTO DE DESENVOLVIMENTO E FORMAO - Joo Martins 14 MANUAL DE EXCEL 2007 Na frase: tem dois olhos E duas orelhas E nasceu no planeta Terra? Vamos transformar isto no SE() combinado com E() SE(E( tem dois olhos;duas orelhas; nasceu no planeta Terra) ganha 100; seno ganha 0,00) (O Excel l da seguinte forma: SE verdade que tem dois olhos E tambm verdade que tem duas orelhas E tambm verdade que nasceu no planeta Terra ganha 100; seno for verdade ganha 0,00) Portanto: SE(E(10+90=100;5*5=25;2+2=4);100;0) - Resposta: 100 SE(E(10+90=100;5*5=80;2+2=4);100;0) - Resposta: 0 OU (OR) Da mesma forma utilizando a funo OU OU sabe ler, OU est a estudar, OU respira. OU respira, OU s fala chines, OU nasceu na Lua. Resumindo: na funo OU() basta que uma condio seja preenchida. Outra com OU() OU(=Marciano; >120 anos; =terrorista) Juntando o SE( ) com o OU() Na folha abaixo observe a funo: =SE(OU(A1>5;B1>5);"Certo";"Errado") escrita na barra de frmulas. Vamos ler: SE(OU(A1>5 ou B1>5); escreva Certo;seno escreva "Errado") Observe a figura a seguir e os resultados. Como j vimos, na funo OU se apenas um dos valores, ou todos, forem Verdadeiros a funo ser Verdadeira e retornar o primeiro valor. 15 MANUAL DE EXCEL 2007 Numa Escola queremos atravs do Excel colocar ao lado das notas as expresses Aprovado - Em recuperao - Reprovado At aqui vimos a funo SE com apenas duas possibilidades. Ento, como fazer com trs? Iremos colocar uma Funo SE() dentro de outra Funo SE() Ficar assim: = SE(Nota>=5; "Aprovado";SE(Nota>=4;"Recuperao";"Reprovado")) Explicando melhor: Se a Nota for maior ou igual a 5 a resposta ser "Aprovado" se no for Camos na segunda funo SE (em azul) que tambm tem duas possibilidades - Recuperao ou Reprovado. 16 MANUAL DE EXCEL 2007 Ficar assim: Funes de Matemtica SOMA.SE (SUMIF) Soma um intervalo de clulas mediante uma condio estabelecida. Sintaxe da Funo SOMA.SE(intervalo; critrios; intervalo_soma) intervalo Intervalo de clulas para a validao do critrio. Este intervalo deve ter um comprimento de clulas igual ao intervalo_soma. critrios Condio ou condies que iro limitar a soma. A condio ser verficada no intervalo (primeiro argumento desta funo). Exemplo de utilizao da funo Soma.Se Pretendemos saber quanto foi pago a cada vendedor em comisses. Para tal, iremos efectuar uma soma de todas as comisses por vendedor 17 MANUAL DE EXCEL 2007 De acordo com o exemplo, vemos que a funo Soma.Se composta por 3 partes; rea onde iremos pesquisar (Intervalo) o nome do vendedor que nos interessa (Critrios) e a rea que contm os valores a somar (Intervalo_soma). 18 MANUAL DE EXCEL 2007 Funes de Estatstica CONTAR.SE Possui uma folha onde tem o nome dos alunos e as suas mdias. E deseja saber quantos alunos tiraram mdias maior e igual a 9. Veja o exemplo: CONTAR.VAZIO Conta as clulas que esto vazias. Exemplo: gostaria de saber quantos alunos esto sem mdia. 19 MANUAL DE EXCEL 2007 Funes financeiras PGTO (PMT) Calcula o valor da prestao para um dado capital com base numa determinada taxa de juro e por um perodo especfico. Sintaxe da Funo PGTO(taxa; nper; va; vf; tipo) Taxa Taxa de juro. Se a prestao a calcular for mensal, ento a taxa de juro deve ser mensal. Nper Nmero de perodos. Podem ser anos, meses, trimestres ou outra periodicidade. Va Valor actual. Valor do capital inicial. Vf Valor futuro ou o saldo em dinheiro que deseja obter aps o ultimo pagamento ter sido efectuado, 0 (zero), se omisso. Tipo valor logico; pagamento no inicio do perodo =1; pagamento no final do perodo = 0 ou omisso. Exemplo de utilizao da funo Pgto: Pretendemos saber quanto ficaremos a pagar, mensalmente, se adquirirmos uma habitao de 54.000 Euros, ficando em dvida 46.500 Euros a serem pagos em 30 anos a uma taxa anual de 9%. 20 MANUAL DE EXCEL 2007 Ao aplicar a funo, conforme est na clula B7, deveremos introduzir os seguintes dados. Taxa = B4/12, onde B4 representa a taxa de juro anual, mas como pretendemos calcular a prestao mensal, essa taxa ter que ser mensal. Nper = B5*12, onde B5 representa os anos, obtendo assim o n total de pagamentos. Va = -B6, representa o valor em dvida (o sinal de opcional, depende apenas da utilizao que pretendemos dar ao aplicarmos a funo Pgto. Vf = Saldo que pretende obter aps ltimo pagamento (0 se omisso). Tipo = Valor lgico: 1 pagamento no incio do perodo 0 ou omisso, pagamento no final do perodo. 21 Funes de data e hora AGORA (NOW) Devolve a data actual, ou seja, a data do sistema indicando a data e a hora do dia. Sintaxe da Funo AGORA() DIAS360 (DAYS360) Devolve o nmero de dias entre duas datas com base num ano de 360 dias (doze meses de 30 dias). Para calcular a diferena entre duas datas baseadas num ano de 365 dias, basta uma simples subtraco entre as mesmas. Sintaxe da Funo DIAS360(data_inicial; data_final; mtodo) HOJE (TODAY) Devolve a data actual. Sintaxe da Funo HOJE() DIATRABALHO (NETWORKDAYS) Devolve o nmero de dias teis compreendido entre duas datas. Sintaxe da Funo DIASTRABALHO(data_inicial; data_final; feriados) Data_inicial Data de incio. Data_final Data de fim. Feriados Datas dos feriados compreendidos no intervalo de datas especificado. 22 Formatao, proteo de dados e Formulrios Formatao condicional A formatao condicional ajuda a responder a questes especficas sobre dados. Pode aplicar a formatao condicional a um intervalo de clulas, a uma tabela do Excel ou a um relatrio de Tabela Dinmica. H diferenas importantes que devem ser entendidas ao usar formatao condicional num relatrio de Tabela Dinmica. Os benefcios da formatao condicional ao analisar dados, ajuda a questes como: Onde esto as excees num resumo de lucros nos ltimos cinco anos? Quais so as tendncias numa pesquisa de opinio de marketing ao longo dos ltimos dois anos? Quem vendeu mais do que 50.000 este ms? Qual a distribuio etria dos empregados? Quais so os alunos com melhor e pior desempenho na classe? A formatao condicional ajuda a responder essas questes tornando mais fcil destacar clulas ou intervalos de clulas, enfatizar valores no-usuais e visualizar dados usando barras de dados, escalas de cores e conjuntos de cones. Um formato condicional altera a aparncia de um intervalo de clulas com base numa condio (ou critrio). Se a condio for verdadeira, o intervalo de clulas ser formatado com base nessa condio; se a condio for falsa, o intervalo de clulas no ser formatado com base nessa condio. Observao Quando criar um formato condicional, poder fazer referncia a outras clulas na mesma folha, mas no poder fazer referncia a clulas em outras folhas na mesma livro, ou utilizar referncias externas a outra livro. Escalas de cores so Menus visuais que ajudam a entender a distribuio e a variao de dados. Uma escala de duas cores ajuda a comparar um intervalo de clulas usando uma gradao de duas cores. O tom da cor representa valores maiores ou menores. Por exemplo, numa escala de cores verde e vermelha, pode especificar que clulas de valores mais altos tenham uma cor mais verde e clulas de valores mais baixos tenham uma cor mais vermelha. Problema: no vejo a minha formatao condicional para nenhuma clula no intervalo. Se uma ou mais clulas do intervalo, contiver uma frmula que retorna erro, a formatao condicional no ser aplicada ao intervalo inteiro. Para garantir que a formatao condicional seja aplicada ao intervalo inteiro, use uma funo ou SEERRO para retornar um valor diferente de erro. CML/DMRH/DEPARTAMENTO DE DESENVOLVIMENTO E FORMAO - Joo Martins 23 MANUAL DE EXCEL 2007 Formatao rpida 1. Selecione uma ou mais clulas num intervalo, uma tabela ou um relatrio de Tabela Dinmica. 2. Na Menu Pgina Inicial, no grupo Estilos, clique na seta ao lado de Formatao Condicional e, em seguida, clique em Escalas de Cor. 3. Selecione uma escala de duas cores. Dica Passe o rato sobre os cones de escala de cores para ver qual deles corresponde a uma escala de duas cores. A cor da parte superior representa valores maiores; e a da parte inferior, valores menores. Dica possvel alterar o mtodo de escopo de campos na rea Valores de um relatrio de Tabela Dinmica usando o boto de opo Aplicar regra de formatao a. 1. Selecione uma ou mais clulas num intervalo, uma tabela ou um relatrio de Tabela Dinmica. 2. Na Menu Pgina Inicial, no grupo Estilos, clique na seta ao lado de Formatao Condicional e, em seguida, clique em Gerir Regras. A caixa de dilogo Gestor de Regras de Formatao Condicional ser exibida. 3. Siga um destes procedimentos: Para adicionar um formato condicional, clique em Nova Regra. A caixa de dilogo Nova Regra de Formatao ser exibida. Para alterar um formato condicional, siga este procedimento: 1. Verifique se a folha, a tabela ou o relatrio de Tabela Dinmica apropriado est selecionado na caixa de listagem Mostrar regras de formatao para. 2. Como opo, altere o intervalo de clulas clicando em Recolher Caixa de Dilogo na caixa Aplica-se a para ocultar temporariamente a caixa de dilogo, selecionando o novo intervalo de clulas na folha e, em seguida, selecionando Expandir Caixa de Dilogo . 3. Selecione a regra e, em seguida, clique em Editar regra. A caixa de dilogo Editar Regra de Formatao ser exibida. CML/DMRH/DEPARTAMENTO DE DESENVOLVIMENTO E FORMAO - Joo Martins 24 MANUAL DE EXCEL 2007 4. Em Aplicar Regra a, altere opcionalmente o escopo dos campos na rea Valores de um relatrio de Tabela Dinmica por: Seleo, clique em Apenas estas clulas. Campo de correspondncia, clique em Todas as clulas de <campo de valor> com os mesmos campos. Campo de valor, clique em Todas as clulas de <campo de valor>. 5. Em Selecione um Tipo de Regra, clique em Formatar todas as clulas com base nos valores. 6. Em Edite a Descrio da Regra, na caixa de listagem Formatar Estilo, selecione Escala Bicolor. 7. Selecione um Tipo Mnimo e Mximo. Siga um destes procedimentos: Formatar valor mais alto e valor mais baixo Selecione Valor Mais Baixo e Valor Mais Alto. Neste caso , no digita um Valor para Mnimo e Mximo. Formatar um valor numrico, de data ou hora Selecione Nmero e digite um Valor Mnimo e Mximo. Formatar uma percentagem Selecione Percentagem e digite um Valor para Mnimo e Mximo. Os valores vlidos esto entre 0 (zero) e 100. No digite o sinal de percentagem. Use uma percentagem quando desejar visualizar todos os valores proporcionalmente porque a distribuio de valores proporcional. Selecione Percentil e digite um Valor para Mnimo e Os percentuais vlidos esto entre 0 (zero) e 100. no pode usar um percentual se o intervalo de clulas contiver mais de 8.191 pontos de dados. Use um percentual quando desejar visualizar um grupo de valores elevados (como os 20 percentuais ) numa proporo de cores e valores baixos (como os 20 percentuais) em outra proporo de cores, porque eles representam valores extremos que podem distorcer a visualizao de seus dados. Formatar um resultado de frmula Selecione Frmula e digite um Valor Mnimo e Mximo. A frmula deve retornar um nmero, uma data ou uma hora. Inicie a frmula com um sinal de igual (=). Frmulas invlidas resultam em nenhuma formatao aplicada. Convm testar a frmula na folha para assegurar que ela no retorne um valor de erro. Valores Mnimo e Mximo so os valores mnimo e mximo do intervalo de clulas. Verifique se o valor Mnimo menor do que o valor Mximo. CML/DMRH/DEPARTAMENTO DE DESENVOLVIMENTO E FORMAO - Joo Martins 25 MANUAL DE EXCEL 2007 pode escolher um Tipo diferente para Mnimo e Mximo. Por exemplo, possvel escolher um Nmero para Mnimo e um Percentual para Mximo. 8. Para escolher uma escala de cores para Mnimo e Mximo, clique em Cor para cada um e, em seguida, selecione uma cor. Se desejar escolher cores adicionais ou criar uma cor personalizada, clique em Mais Cores. A escala de cores que selecionar ser exibida na caixa Visualizao. Listas pendentes Listas pendentes so criadas a partir de um intervalo de clulas. Para simplificar a introduo de dados ou limitar as entradas a itens especficos que defina, pode criar uma lista pendente de entradas vlidas que compilada noutra localizao do livro. Quando se cria uma lista pendente para uma clula, aparece uma seta nessa clula. Para introduzir informaes nessa clula, clique na seta e clique na entrada pretendida. Para criar uma lista pendente a partir de um intervalo de clulas, utilize o comando Validao de Dados no grupo Ferramentas de Dados do separador Dados. Para criar uma lista de entradas vlidas para a lista pendente, introduza as entradas numa coluna ou linha, sem clulas em branco. Por exemplo: A 1 2 3 4 Vendas Finanas Investigao e Desenvolvimento MIS 26 MANUAL DE EXCEL 2007 NOTA possvel que pretenda ordenar os dados pela ordem em que pretende que apaream na lista pendente. Se pretender utilizar outra folha de clculo, escreva a lista nessa folha de clculo e, em seguida, defina um nome para a lista. Para definir um nome: Seleccione a clula, o intervalo de clulas ou as seleces no adjacentes s quais pretende atribuir nomes. Escreva o nome das clulas; por exemplo, DepartVlidos. Prima ENTER. NOTA No pode atribuir um nome a uma clula enquanto estiver a alterar o respectivo contedo. Clique na caixa Nome na extremidade esquerda da barra de frmulas. apresentada a caixa de dilogo Validao de Dados. Clique no separador Definies. Na caixa Por, clique em Lista. Para especificar a localizao da lista de entradas vlidas, efectue um dos seguintes procedimentos: Se a lista estiver na folha de clculo actual, introduza uma referncia lista na caixa Origem. Se a lista se encontrar noutra folha de clculo, introduza o nome que definiu para a lista na caixaOrigem. Em ambos os casos, certifique-se de que insere um sinal de igual (=) antes da referncia ou do nome. Por exemplo, introduza =DepartVlidos. Certifique-se de que a caixa de verificao Lista pendente na clula est seleccionada. Para especificar se a clula pode ser deixada em branco, marque ou desmarque a caixa de verificao Ignorar clulas em branco. Em opo, defina uma mensagem de entrada a ser apresentada ao clicar na caixa. 27 MANUAL DE EXCEL 2007 Para visualizar uma mensagem de entrada: Clique no separador Mensagem de Entrada. Certifique-se de que a caixa de verificao Mostrar mensagem de entrada ao seleccionar clula est seleccionada. Introduza o ttulo e o texto para a mensagem (at 225 caracteres). Proteo de folhas e livros Para impedir que, por acidente ou deliberadamente, um utilizador altere, mova ou exclua dados importantes, pode proteger determinados elementos de uma folha ou livro, com ou snuma senha. Para proteger os dados de uma folha: a) Selecione a folha que deseja proteger. b) Para desbloquear clulas ou intervalos que deseja liberar para outros utilizadores alterarem, faa o seguinte: a. Selecione cada clula ou intervalo que deseja desbloquear. b. Na Menu Incio, no grupo Clulas, clique em Formatar e, em seguida, clique em Formatar Clulas. a. Na Menu Proteo, limpe a caixa de seleo Bloqueada e clique em OK. b) Para ocultar frmulas que no deseja que fiquem visveis, faa o seguinte: 28 MANUAL DE EXCEL 2007 b. Na Menu Propriedades, desmarque a caixa de seleo Bloqueado e desmarque a caixa de seleo Bloquear texto, se houver uma. f) OBSERVAO Para permitir que os utilizadores utilizem controles ou botes e cliquem neles, no necessrio desbloque-los. Pode desbloquear os grficos incorporados, as caixas de texto e os outros objetos criados com as ferramentas de desenho para que os utilizadores possam modific-los. g) Na Menu Reviso, no grupo Alteraes, clique em Proteger Folha. h) 29 MANUAL DE EXCEL 2007 i) Na lista Permitir a todos os utilizadores desta livro, selecione os elementos que deseja que os utilizadores possam alterar. Na caixa Senha para desproteger a folha, digite uma senha para a folha, clique em OK e digite novamente a senha para confirm-la. OBSERVAO A senha opcional. Se no fornecer uma senha, qualquer utilizador poder desproteger a folha e alterar os elementos protegidos. Certifique-se de escolher uma senha que seja fcil de lembrar j que se a perder no poder aceder aos elementos protegidos na folha. Formulrios 30 MANUAL DE EXCEL 2007 Criar um formulrio de dados 1. Se necessrio, adicione um cabealho de coluna a cada coluna no intervalo ou tabela. O Excel usa esses cabealhos de modo a criar rtulos para cada campo do formulrio. Importante Verifique se no h linhas em branco no intervalo de dados. 2. Clique numa clula no intervalo ou tabela ao qual deseja adicionar o formulrio. 3. Para adicionar o boto Formulrio Barra de Ferramentas de Acesso Rpido, faa o seguinte: 1. Clique na seta ao lado de Barra de Ferramentas de Acesso Rpido e clique em Mais Comandos. 2. Na caixa Escolher comandos em, clique em Todos os Comandos e selecione o boto Formulrio na lista. 3. Clique em Adicionar e em OK. 4. Na Barra de Ferramentas de Acesso Rpido, clique em Formulrio . 31 MANUAL DE EXCEL 2007 Tabelas dinmicas Mtodos para obteno de dados externos: Criar um relatrio de tabela dinmica ou grfico dinmico Para criar um relatrio de tabela dinmica ou de grfico dinmico, deve ligar-se fonte de dados e inserir o local do relatrio. 1. Selecione uma clula em um intervalo de clulas ou coloque o ponto de insero dentro da tabela do Microsoft Office Excel . Certifique-se de que o intervalo de clulas tenha ttulos de coluna. 2. Selecione o tipo de relatrio a ser gerado seguindo um destes procedimentos: Para criar um relatrio de Tabela Dinmica, na Menu Inserir, no grupo Tabelas, clique em Tabela Dinmica e em Tabela Dinmica. 32 MANUAL DE EXCEL 2007 o Excel exibe a caixa de dilogo Criar Tabela Dinmica. Para criar um relatrio de tabela dinmica e de grfico dinmico, na Menu Inserir, no grupo Tabelas, clique em Tabela Dinmica e, em seguida, clique em Grfico Dinmico. O Excel exibe a caixa de dilogo Criar Tabela Dinmica com Grfico Dinmico. 3. Selecione uma fonte de dados seguindo um destes procedimentos: Selecionar os dados que deseja analisar 1. Clique em Selecionar uma tabela ou um intervalo. 2. Digite o intervalo de clulas ou a referncia do nome da tabela, como =LucrosTrimestrais, na caixa Tabela/Intervalo. Se tiver selecionado uma clula em um intervalo de clulas ou se o ponto de insero estava numa tabela antes de iniciar o assistente, o Excel exibir o intervalo de clulas ou a referncia do nome da tabelas na caixa Tabela/Intervalo. Como alternativa, para selecionar um intervalo de clulas ou uma tabela, clique em Recolher Caixa de Dilogo para ocultar temporariamente a caixa de dilogo. Selecione o intervalo na . folha e pressione Expandir Caixa de Dilogo Dica Considere o uso de uma referncia de nome de tabela no lugar de um intervalo de clulas, uma vez que as linhas adicionadas a uma tabela so automaticamente inclusas no relatrio de Tabela Dinmica quando atualiza os dados. Observao Se o intervalo estiver em outra folha na mesma pasta de trabalho ou em outra pasta de trabalho, digite o nome da pasta de trabalho e da folha usando a seguinte sintaxe: ([nomedapastadetrabalho]nomedafolha!intervalo). 1. Clique em Usar uma fonte de dados externa. 2. Clique em Escolher Conexo. O Excel exibe a caixa de dilogo Conexes Existentes. 3. Na caixa de listagem suspensa Mostrar na parte superior da caixa de dilogo, selecione a categoria de conexes para a qual deseja escolher uma conexo ou selecione Todas as Conexes Existentes (que o padro). CML/DMRH/DEPARTAMENTO DE DESENVOLVIMENTO E FORMAO - Joo Martins 33 MANUAL DE EXCEL 2007 4. Selecione uma conexo a partir da caixa de listagem Selecionar uma Conexo e clique em Abrir. Observao Ao escolher uma conexo da categoria Conexes nesta Pasta de Trabalho, estar reutilizando ou compartilhando uma conexo existente. Se escolher uma conexo a partir das categorias Arquivos de conexo na rede ou Arquivos de conexo neste computador, o Excel copiar o arquivo de conexo na pasta de trabalho como uma nova conexo de pasta de trabalho e usar esse arquivo como a nova conexo para o relatrio de Tabela Dinmica. Para obter mais informaes, consulte Gerir conexes aos dados numa pasta de trabalho. 4. Especifique um local seguindo um destes procedimentos: Para colocar o relatrio de Tabela Dinmica numa nova folha comeando na clula A1, clique em Nova Folha. Para colocar o relatrio de tabela dinmica numa folha existente, selecione Folha Existente e especifique a primeira clula no intervalo de clulas onde deseja posicionar o relatrio de Tabela Dinmica. para ocultar temporariamente a . Como alternativa, clique em Recolher Caixa de Dilogo caixa de dilogo, selecione a clula inicial na folha e pressione Expandir Caixa de Dilogo 5. Clique em OK. O Excel adiciona um relatrio de Tabela Dinmica vazio ao local especificado e exibe a Lista de Campos da Tabela Dinmica para que possa adicionar campos, criar um layout e personalizar o relatrio de Tabela Dinmica. Para obter mais informaes, consulte Viso geral dos relatrios de Tabela Dinmica e de Grfico Dinmico e Criar e alterar o layout dos campos em um relatrio de Tabela Dinmica. Se criar um relatrio de Grfico Dinmico, o Excel criar um relatrio de Tabela Dinmica associado logo abaixo desse relatrio de Grfico Dinmico. Um relatrio de Grfico Dinmico e seu relatrio de Tabela Dinmica acompanhante devem estar sempre na mesma pasta de trabalho. Grficos dinmicos 34 MANUAL DE EXCEL 2007 Criar um relatrio de grfico dinmico a partir de um relatrio de tabela dinmica existente 1. Clique no relatrio de tabela dinmica. 2. Na Menu Inserir, no grupo Grficos, clique em um tipo de grfico. pode usar qualquer tipo de grfico exceto um grfico de disperso (xy), de bolha ou de aes. Macros e Edio VBA O que uma Macro Caso execute uma tarefa vrias vezes no Microsoft Excel, possvel automatiz-la com uma macro. Uma macro uma sequncia de comandos e funes armazenados em um mdulo do Visual Basic for Applications - VBA e pode ser executada sempre que voc precisar executar a tarefa. Quando voc grava uma macro, o Excel armazena informaes sobre cada etapa realizada medida que voc executa uma sequncia de comandos. Em seguida, voc executa a macro para repetir, ou "reproduzir", os comandos. Por exemplo, vamos supor que, seguidamente precisa formatar uma clula com Negrito, cor de fonte Vermelha, Itlico, Fonte Verdana de Tamanho 13 com quebra automtica de linha. Ao invs de ter que executar todos os comandos de formatao em cada clula, pode criar uma Macro que aplica todos os comandos de formatao. Aps criada a Macro, cada vez que tiver que aplicar o conjunto de comandos de formatao, basta executar a Macro, o que normalmente feito atravs da associao de uma combinao de teclas com a Macro, como por exemplo Ctrl+L. No nosso exemplo, cada vez que quiser formatar uma clula com os formatos descritos, basta clicar na clula e pressionar Ctrl+L. Mais fcil do que aplicar cada comando individualmente. 35 MANUAL DE EXCEL 2007 Antes de gravar uma macro, planeie as etapas e os comandos que deseja que a macro execute. Se cometer um erro durante a gravao da macro, as correes feitas tambm so gravadas. Ao gravar macros, o VBA armazena cada macro em um novo mdulo anexado a uma pasta de trabalho. Gravar e executar uma Macro Quando grava uma macro, o gravador de macro grava todas as etapas necessrias para concluir as aes a serem executadas por essa macro. A navegao na Faixa de Opes no includa nas etapas gravadas. Observao A Faixa de Opes um componente da Interface de utilizador do Microsoft Office Fluent. Primeiramente o menu Programador deve estar ativado, aparecendo na barra de menus, caso no esteja: 1. Se o Menu Programador no estiver disponvel, faa o seguinte: 1. Clique no Boto Microsoft Office Excel. e, em seguida, clique em Opes do 2. Na categoria Popular, em Opes principais para trabalhar com o Excel, marque a caixa de seleo Mostrar Separador Programador no Friso e clique em OK. 2. Para definir o nvel de segurana temporariamente e habilitar todas as macros, faa o seguinte: 1. Na Menu Programador, no grupo Cdigo, clique em Segurana de Macros. 36 MANUAL DE EXCEL 2007 CML/DMRH/DEPARTAMENTO DE DESENVOLVIMENTO E FORMAO - Joo Martins 37 MANUAL DE EXCEL 2007 6. Clique em OK para iniciar a gravao. 7. Execute as aes que deseja gravar. 8. Na Menu Programador, no grupo Cdigo, clique em Parar Gravao Dica tambm pode clicar em Parar Gravao . O Editor de Visual Basic O Editor de Visual Basic um ambiente de desenvolvimento que possibilita trabalhar com formulrios e objetos. Serve para que possamos atravs deste formulrios e objetos criar cdigo de programao para facilitar a criao de solues de uma maneira mais significativa. o ambiente de desenvolvimento e edio de macros disponibilizado nos aplicativos do Microsoft Office. Permite ainda Editar ou altera macros existentes ou previamente criadas. Esta disponvel no Menu Programador junto ao Boto de Macros. 38 Bibliografia:

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# Shigley's mechanical engineering design 9th edition solutions manual of 712 All materials on our website are shared by users. If you have any questions about copyright issues, please report us to resolve them. We are always happy to assist you. Description Text • Chapter 1 Problems 1-1 through 1-6 are for student research. No standard solutions are provided. 1-7 From Fig. 1-2, cost of grinding to  0.0005 in is 270%. Cost of turning to  0.003 in is 60%. Relative cost of grinding vs. turning = 270/60 = 4.5 times Ans. ______________________________________________________________________________ 1-8 CA = CB, 10 + 0.8 P = 60 + 0.8 P  0.005 P 2 P 2 = 50/0.005  P = 100 parts Ans. ______________________________________________________________________________ 1-9 Max. load = 1.10 P Min. area = (0.95)2A Min. strength = 0.85 S To offset the absolute uncertainties, the design factor, from Eq. (1-1) should be  2 1.10 1.43 . 0.85 0.95d n A  ns  ______________________________________________________________________________ 1-10 (a) X1 + X2:    1 2 1 1 2 2 1 2 1 2 1 2 error . x x X e X e e x x X X e e Ans             (b) X1  X2:       1 2 1 1 2 2 1 2 1 2 1 2 . x x X e X e e x x X X e e Ans            (c) X1 X2:   1 2 1 1 2 2 1 2 1 2 1 2 2 1 1 2 1 2 1 2 2 1 1 2 1 2 . x x X e X e e x x X X X e X e e e e eX e X e X X Ans X X                  Chapter 1 Solutions - Rev. B, Page 1/6 • (d) X1/X2: 1 1 1 1 1 1 2 2 2 2 2 2 1 2 2 1 1 1 2 1 2 2 2 2 1 2 1 1 1 1 1 2 2 2 2 1 2 1 1 11 1 then 1 1 1 1 Thus, . x X e X e X x X e X e X e e e X e e e 2 2 e X X e X X X X x X X e ee Ans x X X X X                                               X  ______________________________________________________________________________ 1-11 (a) x1 = 7 = 2.645 751 311 1 X1 = 2.64 (3 correct digits) x2 = 8 = 2.828 427 124 7 X2 = 2.82 (3 correct digits) x1 + x2 = 5.474 178 435 8 e1 = x1  X1 = 0.005 751 311 1 e2 = x2  X2 = 0.008 427 124 7 e = e1 + e2 = 0.014 178 435 8 Sum = x1 + x2 = X1 + X2 + e = 2.64 + 2.82 + 0.014 178 435 8 = 5.474 178 435 8 Checks (b) X1 = 2.65, X2 = 2.83 (3 digit significant numbers) e1 = x1  X1 =  0.004 248 688 9 e2 = x2  X2 =  0.001 572 875 3 e = e1 + e2 =  0.005 821 564 2 Sum = x1 + x2 = X1 + X2 + e = 2.65 +2.83  0.001 572 875 3 = 5.474 178 435 8 Checks ______________________________________________________________________________ 1-12     3 3 25 1016 1000 0.799 in . 2.5d S d A n d        ns Table A-17: d = 7 8 in Ans. Factor of safety:       3 3 7 8 25 10 3.29 . 16 1000 Sn A      ns ______________________________________________________________________________ 1-13 Eq. (1-5): R = 1 n i i R   = 0.98(0.96)0.94 = 0.88 Overall reliability = 88 percent Ans. ______________________________________________________________________________ Chapter 1 Solutions - Rev. B, Page 2/6 • 1-14 a = 1.500  0.001 in b = 2.000  0.003 in c = 3.000  0.004 in d = 6.520  0.010 in (a) d a b c   w = 6.520  1.5  2  3 = 0.020 in = 0.001 + 0.003 + 0.004 +0.010 = 0.018 allt w t w = 0.020  0.018 in Ans. (b) From part (a), wmin = 0.002 in. Thus, must add 0.008 in to d . Therefore, d = 6.520 + 0.008 = 6.528 in Ans. ______________________________________________________________________________ 1-15 V = xyz, and x = a   a, y = b   b, z = c   c, V abc    V a a b b c c abc bc a ac b ab c a b c b c a c a b a b c                            The higher order terms in  are negligible. Thus, V bc a ac b ab c      and, .V bc a ac b ab c a b c a b c Ans V abc a b c a b c                   For the numerical values given,   31.500 1.875 3.000 8.4375 inV     30.002 0.003 0.004 0.00427 0.00427 8.4375 0.036 in 1.500 1.875 3.000 V V V          V = 8.438  0.036 in3 Ans. ______________________________________________________________________________ Chapter 1 Solutions - Rev. B, Page 3/6 • 1-16 wmax = 0.05 in, wmin = 0.004 in 0.05 0.004 0.027 in 2  w = Thus,  w = 0.05  0.027 = 0.023 in, and then, w = 0.027  0.023 in. 0.027 0.042 1.5 1.569 in a b c a a       w = tw =  0.023 = tallt a + 0.002 + 0.005  ta = 0.016 in Thus, a = 1.569  0.016 in Ans. ______________________________________________________________________________ 1-17  2 3.734 2 0.139 4.012 ino iD D d      all 0.028 2 0.004 0.036 inoDt t    Do = 4.012  0.036 in Ans. ______________________________________________________________________________ 1-18 From O-Rings, Inc. (oringsusa.com), Di = 9.19  0.13 mm, d = 2.62  0.08 mm  2 9.19 2 2.62 14.43 mmo iD D d      all 0.13 2 0.08 0.29 mmoDt t    Do = 14.43  0.29 mm Ans. ______________________________________________________________________________ 1-19 From O-Rings, Inc. (oringsusa.com), Di = 34.52  0.30 mm, d = 3.53  0.10 mm  2 34.52 2 3.53 41.58 mmo iD D d      all 0.30 2 0.10 0.50 mmoDt t    Do = 41.58  0.50 mm Ans. ______________________________________________________________________________ Chapter 1 Solutions - Rev. B, Page 4/6 • 1-20 From O-Rings, Inc. (oringsusa.com), Di = 5.237  0.035 in, d = 0.103  0.003 in  2 5.237 2 0.103 5.443 ino iD D d      all 0.035 2 0.003 0.041 inoDt t    Do = 5.443  0.041 in Ans. ______________________________________________________________________________ 1-21 From O-Rings, Inc. (oringsusa.com), Di = 1.100  0.012 in, d = 0.210  0.005 in  2 1.100 2 0.210 1.520 ino iD D d      all 0.012 2 0.005 0.022 inoDt t    Do = 1.520  0.022 in Ans. ______________________________________________________________________________ 1-22 From Table A-2, (a)  = 150/6.89 = 21.8 kpsi Ans. (b) F = 2 /4.45 = 0.449 kip = 449 lbf Ans. (c) M = 150/0.113 = 1330 lbf  in = 1.33 kip  in Ans. (d) A = 1500/ 25.42 = 2.33 in2 Ans. (e) I = 750/2.544 = 18.0 in4 Ans. (f) E = 145/6.89 = 21.0 Mpsi Ans. (g) v = 75/1.61 = 46.6 mi/h Ans. (h) V = 1000/946 = 1.06 qt Ans. ______________________________________________________________________________ 1-23 From Table A-2, (a) l = 5(0.305) = 1.53 m Ans. (b)  = 90(6.89) = 620 MPa Ans. (c) p = 25(6.89) = 172 kPa Ans. Chapter 1 Solutions - Rev. B, Page 5/6 • Chapter 1 Solutions - Rev. B, Page 6/6 (d) Z =12(16.4) = 197 cm3 Ans. (e) w = 0.208(175) = 36.4 N/m Ans. (f)  = 0.001 89(25.4) = 0.0480 mm Ans. (g) v = 1200(0.0051) = 6.12 m/s Ans. (h)  = 0.002 15(1) = 0.002 15 mm/mm Ans. (i) V = 1830(25.43) = 30.0 (106) mm3 Ans. ______________________________________________________________________________ 1-24 (a)  = M /Z = 1770/0.934 = 1895 psi = 1.90 kpsi Ans. (b)  = F /A = 9440/23.8 = 397 psi Ans. (c) y =Fl3/3EI = 270(31.5)3/[3(30)106(0.154)] = 0.609 in Ans. (d)  = Tl /GJ = 9740(9.85)/[11.3(106)( /32)1.004] = 8.648(102) rad = 4.95 Ans. ______________________________________________________________________________ 1-25 (a)  =F / wt = 1000/[25(5)] = 8 MPa Ans. (b) I = bh3 /12 = 10(25)3/12 = 13.0(103) mm4 Ans. (c) I = d4/64 =  (25.4)4/64 = 20.4(103) mm4 Ans. (d)  =16T / d 3 = 16(25)103/[ (12.7)3] = 62.2 MPa Ans. ______________________________________________________________________________ 1-26 (a)  =F /A = 2 700/[ (0.750)2/4] = 6110 psi = 6.11 kpsi Ans. (b)  = 32Fa/ d 3 = 32(180)31.5/[ (1.25)3] = 29 570 psi = 29.6 kpsi Ans. (c) Z = (do4  di4)/(32 do) =  (1.504  1.004)/[32(1.50)] = 0.266 in3 Ans. (d) k = (d 4G)/(8D 3 N) = 0.06254(11.3)106/[8(0.760)3 32] = 1.53 lbf/in Ans. ______________________________________________________________________________ • Chapter 2 2-1 From Tables A-20, A-21, A-22, and A-24c, (a) UNS G10200 HR: Sut = 380 (55) MPa (kpsi), Syt = 210 (30) Mpa (kpsi) Ans. (b) SAE 1050 CD: Sut = 690 (100) MPa (kpsi), Syt = 580 (84) Mpa (kpsi) Ans. (c) AISI 1141 Q&T at 540C (1000F): Sut = 896 (130) MPa (kpsi), Syt = 765 (111) Mpa (kpsi) Ans. (d) 2024-T4: Sut = 446 (64.8) MPa (kpsi), Syt = 296 (43.0) Mpa (kpsi) Ans. (e) Ti-6Al-4V annealed: Sut = 900 (130) MPa (kpsi), Syt = 830 (120) Mpa (kpsi) Ans. ______________________________________________________________________________ 2-2 (a) Maximize yield strength: Q&T at 425C (800F) Ans. (b)Maximize elongation: Q&T at 650C (1200F) Ans. ______________________________________________________________________________ 2-3 Conversion of kN/m3 to kg/ m3 multiply by 1(103) / 9.81 = 102 AISI 1018 CD steel: Tables A-20 and A-5     3370 10 47.4 kN m/kg . 76.5 102 yS Ans     2011-T6 aluminum: Tables A-22 and A-5     3169 10 62.3 kN m/kg . 26.6 102 yS Ans     Ti-6Al-4V titanium: Tables A-24c and A-5     3830 10 187 kN m/kg . 43.4 102 yS Ans     ASTM No. 40 cast iron: Tables A-24a and A-5.Does not have a yield strength. Using the ultimate strength in tension      342.5 6.89 10 40.7 kN m/kg 70.6 102 utS Ans     ______________________________________________________________________________ 2-4 AISI 1018 CD steel: Table A-5     6 6 30.0 10 106 10 in . 0.282 E Ans    2011-T6 aluminum: Table A-5     6 6 10.4 10 106 10 in . 0.098 E Ans    Ti-6Al-6V titanium: Table A-5 Chapter 2 - Rev. D, Page 1/19 •     6 6 16.5 10 103 10 in . 0.160 E Ans    No. 40 cast iron: Table A-5     6 6 14.5 10 55.8 10 in . 0.260 E Ans    ______________________________________________________________________________ 2-5 22 (1 ) 2 E GG v E v G      From Table A-5 Steel:     30.0 2 11.5 0.304 . 2 11.5 v A    ns Aluminum:    10.4 2 3.90 0.333 . 2 3.90 v A    ns Beryllium copper:     18.0 2 7.0 0.286 . 2 7.0 v A    ns Gray cast iron:     14.5 2 6.0 0.208 . 2 6.0 v A    ns ______________________________________________________________________________ 2-6 (a) A0 =  (0.503)2/4,  = Pi / A0 For data in elastic range,  =  l / l0 =  l / 2 For data in plastic range, 0 0 0 0 0 1 1l l Al l l l l A        On the next two pages, the data and plots are presented. Figure (a) shows the linear part of the curve from data points 1-7. Figure (b) shows data points 1-12. Figure (c) shows the complete range. Note: The exact value of A0 is used without rounding off. (b) From Fig. (a) the slope of the line from a linear regression is E = 30.5 Mpsi Ans. From Fig. (b) the equation for the dotted offset line is found to be  = 30.5(106)  61 000 (1) The equation for the line between data points 8 and 9 is  = 7.60(105) + 42 900 (2) Chapter 2 - Rev. D, Page 2/19 • Solving Eqs. (1) and (2) simultaneously yields  = 45.6 kpsi which is the 0.2 percent offset yield strength. Thus, Sy = 45.6 kpsi Ans. The ultimate strength from Figure (c) is Su = 85.6 kpsi Ans. The reduction in area is given by Eq. (2-12) is    0 0 0.1987 0.1077100 100 45.8 % . 0.1987 fA AR Ans A      Data Point Pi l, Ai   1 0 0 0 0 2 1000 0.0004 0.00020 5032 3 2000 0.0006 0.00030 10065 4 3000 0.001 0.00050 15097 5 4000 0.0013 0.00065 20130 6 7000 0.0023 0.00115 35227 7 8400 0.0028 0.00140 42272 8 8800 0.0036 0.00180 44285 9 9200 0.0089 0.00445 46298 10 8800 0.1984 0.00158 44285 11 9200 0.1978 0.00461 46298 12 9100 0.1963 0.01229 45795 13 13200 0.1924 0.03281 66428 14 15200 0.1875 0.05980 76492 15 17000 0.1563 0.27136 85551 16 16400 0.1307 0.52037 82531 17 14800 0.1077 0.84506 74479 (a) Linear range Chapter 2 - Rev. D, Page 3/19 • (b) Offset yield (c) Complete range (c) The material is ductile since there is a large amount of deformation beyond yield. (d) The closest material to the values of Sy, Sut, and R is SAE 1045 HR with Sy = 45 kpsi, Sut = 82 kpsi, and R = 40 %. Ans. ______________________________________________________________________________ 2-7 To plot  true vs., the following equations are applied to the data. true P A   Eq. (2-4) Chapter 2 - Rev. D, Page 4/19 • 0 0 ln for 0 0.0028 in ln for 0.0028 in l l l A l A          where 2 2 0 (0.503) 0.1987 in 4 A   The results are summarized in the table below and plotted on the next page. The last 5 points of data are used to plot log  vs log  The curve fit gives m = 0.2306 log 0 = 5.1852  0 = 153.2 kpsi Ans. For 20% cold work, Eq. (2-14) and Eq. (2-17) give, A = A0 (1 – W) = 0.1987 (1 – 0.2) = 0.1590 in2 0 0.2306 0 0.1987ln ln 0.2231 0.1590 Eq. (2-18): 153.2(0.2231) 108.4 kpsi . Eq. (2-19), with 85.6 from Prob. 2-6, 85.6 107 kpsi . 1 1 0.2 m y u u u A A S A S SS Ans W                 ns P L A   true log  log true 0 0 0.198 713 0 0 1000 0.0004 0.198 713 0.000 2 5032.388 -3.699 01 3.701 774 2000 0.0006 0.198 713 0.000 3 10 064.78 -3.522 94 4.002 804 3000 0.001 0.198 713 0.000 5 15 097.17 -3.301 14 4.178 895 4000 0.0013 0.198 713 0.000 65 20 129.55 -3.187 23 4.303 834 7000 0.0023 0.198 713 0.001 149 35 226.72 -2.939 55 4.546 872 8400 0.0028 0.198 713 0.001 399 42 272.06 -2.854 18 4.626 053 8800 0.0036 0.198 4 0.001 575 44 354.84 -2.802 61 4.646 941 9200 0.0089 0.197 8 0.004 604 46 511.63 -2.336 85 4.667 562 9100 0.196 3 0.012 216 46 357.62 -1.913 05 4.666 121 13200 0.192 4 0.032 284 68 607.07 -1.491 01 4.836 369 15200 0.187 5 0.058 082 81 066.67 -1.235 96 4.908 842 17000 0.156 3 0.240 083 108 765.20 -0.619 64 5.036 49 16400 0.130 7 0.418 956 125 478.20 -0.377 83 5.098 568 14800 0.107 7 0.612 511 137 418.80 -0.212 89 5.138 046 Chapter 2 - Rev. D, Page 5/19 • ______________________________________________________________________________ 2-8 Tangent modulus at  = 0 is     6 3 5000 0 25 10 psi 0.2 10 0 E           Ans. At  = 20 kpsi Chapter 2 - Rev. D, Page 6/19 •         3 6 20 3 26 19 10 14.0 10 psi 1.5 1 10 E      Ans.  (10-3)  (kpsi) 0 0 0.20 5 0.44 10 0.80 16 1.0 19 1.5 26 2.0 32 2.8 40 3.4 46 4.0 49 5.0 54 ______________________________________________________________________________ 2-9 W = 0.20, (a) Before cold working: Annealed AISI 1018 steel. Table A-22, Sy = 32 kpsi, Su = 49.5 kpsi, 0 = 90.0 kpsi, m = 0.25, f = 1.05 After cold working: Eq. (2-16), u = m = 0.25 Eq. (2-14), 0 1 1 1.25 1 1 0.20i A A W      Eq. (2-17), 0ln ln1.25 0.223i u i A A      Eq. (2-18), S 93% increase Ans.  0.250 90 0.223 61.8 kpsi .my i     Ans Eq. (2-19), 49.5 61.9 kpsi . 1 1 0.20 u u SS A W       ns 25% increase Ans. (b) Before: 49.5 1.55 32 u y S S   After: 61.9 1.00 61.8 u y S S     Ans. Lost most of its ductility ______________________________________________________________________________ 2-10 W = 0.20, (a) Before cold working: AISI 1212 HR steel. Table A-22, Sy = 28 kpsi, Su = 61.5 kpsi, 0 = 110 kpsi, m = 0.24, f = 0.85 After cold working: Eq. (2-16), u = m = 0.24 Chapter 2 - Rev. D, Page 7/19 • Eq. (2-14), 0 1 1 1.25 1 1 0.20i A A W      Eq. (2-17), 0ln ln1.25 0.223i u i A A      Eq. (2-18), 174% increase Ans.  0.240 110 0.223 76.7 kpsi .my iS A     ns Eq. (2-19), 61.5 76.9 kpsi . 1 1 0.20 u u SS A W       ns 25% increase Ans. (b) Before: 61.5 2.20 28 u y S S   After: 76.9 1.00 76.7 u y S S     Ans. Lost most of its ductility ______________________________________________________________________________ 2-11 W = 0.20, (a) Before cold working: 2024-T4 aluminum alloy. Table A-22, Sy = 43.0 kpsi, Su = 64.8 kpsi, 0 = 100 kpsi, m = 0.15, f = 0.18 After cold working: Eq. (2-16), u = m = 0.15 Eq. (2-14), 0 1 1 1.25 1 1 0.20i A A W      Eq. (2-17), 0ln ln1.25 0.223i i A A f      Material fractures. Ans. ______________________________________________________________________________ 2-12 For HB = 275, Eq. (2-21), Su = 3.4(275) = 935 MPa Ans. ______________________________________________________________________________ 2-13 Gray cast iron, HB = 200. Eq. (2-22), Su = 0.23(200)  12.5 = 33.5 kpsi Ans. From Table A-24, this is probably ASTM No. 30 Gray cast iron Ans. ______________________________________________________________________________ 2-14 Eq. (2-21), 0.5HB = 100  HB = 200 Ans. ______________________________________________________________________________ Chapter 2 - Rev. D, Page 8/19 • 2-15 For the data given, converting HB to Su using Eq. (2-21) HB Su (kpsi) Su2 (kpsi) 230 115 13225 232 116 13456 232 116 13456 234 117 13689 235 117.5 13806.25 235 117.5 13806.25 235 117.5 13806.25 236 118 13924 236 118 13924 239 119.5 14280.25 Su = 1172 Su2 = 137373 1172 117.2 117 kpsi . 10 u u S S A N     ns Eq. (20-8),   10 2 2 2 1 137373 10 117.2 1.27 kpsi . 1 9u u u i S S NS s A N         ns ______________________________________________________________________________ 2-16 For the data given, converting HB to Su using Eq. (2-22) HB Su (kpsi) Su2 (kpsi) 230 40.4 1632.16 232 40.86 1669.54 232 40.86 1669.54 234 41.32 1707.342 235 41.55 1726.403 235 41.55 1726.403 235 41.55 1726.403 236 41.78 1745.568 236 41.78 1745.568 239 42.47 1803.701 Su = 414.12 Su2 =17152.63 Chapter 2 - Rev. D, Page 9/19 • 414.12 41.4 kpsi . 10 u u S S A N    ns Eq. (20-8),   10 2 2 2 1 17152.63 10 41.4 1.20 . 1 9u u u i S S NS s A N         ns ______________________________________________________________________________ 2-17 (a) 2 345.5 34.5 in lbf / in . 2(30)R u A  ns (b) P L A A0 / A – 1   = P/A0 0 0 0 0 1000 0.0004 0.0002 5 032.39 2000 0.0006 0.0003 10 064.78 3000 0.0010 0.0005 15 097.17 4000 0.0013 0.000 65 20 129.55 7000 0.0023 0.001 15 35 226.72 8400 0.0028 0.0014 42 272.06 8800 0.0036 0.0018 44 285.02 9200 0.0089 0.004 45 46 297.97 9100 0.1963 0.012 291 0.012 291 45 794.73 13200 0.1924 0.032 811 0.032 811 66 427.53 15200 0.1875 0.059 802 0.059 802 76 492.30 17000 0.1563 0.271 355 0.271 355 85 550.60 16400 0.1307 0.520 373 0.520 373 82 531.17 14800 0.1077 0.845 059 0.845 059 74 479.35 From the figures on the next page,         5 1 3 3 1 (43 000)(0.001 5) 45 000(0.004 45 0.001 5) 2 1 45 000 76 500 (0.059 8 0.004 45) 2 81 000 0.4 0.059 8 80 000 0.845 0.4 66.7 10 in lbf/in . T i i u A Ans                Chapter 2 - Rev. D, Page 10/19 • Chapter 2 - Rev. D, Page 11/19 • 2-18, 2-19 These problems are for student research. No standard solutions are provided. ______________________________________________________________________________ 2-20 Appropriate tables: Young’s modulus and Density (Table A-5)1020 HR and CD (Table A- 20), 1040 and 4140 (Table A-21), Aluminum (Table A-24), Titanium (Table A-24c) Appropriate equations: For diameter,   2 4 / 4 y y F F S d F A d S           Weight/length = A, Cost/length = \$/in = (\$/lbf) Weight/length, Deflection/length =  /L = F/(AE) With F = 100 kips = 100(103) lbf, Material  Young's  Modulus  Density   Yield  Strength  Cost/lbf Diameter Weight/  length  Cost/  length  Deflection/  length  units  Mpsi  lbf/in^3  kpsi  \$/lbf  in  lbf/in  \$/in  in/in                      1020 HR  30  0.282  30 \$0.27 2.060 0.9400 \$0.25  1.000E‐03 1020 CD  30  0.282  57 \$0.30 1.495 0.4947 \$0.15  1.900E‐03 1040  30  0.282  80 \$0.35 1.262 0.3525 \$0.12  2.667E‐03 4140  30  0.282  165 \$0.80 0.878 0.1709 \$0.14  5.500E‐03 Al  10.4  0.098  50 \$1.10 1.596 0.1960 \$0.22  4.808E‐03 Ti  16.5  0.16  120 \$7.00 1.030 0.1333 \$0.93  7.273E‐03 The selected materials with minimum values are shaded in the table above. Ans. ______________________________________________________________________________ 2-21 First, try to find the broad category of material (such as in Table A-5). Visual, magnetic, and scratch tests are fast and inexpensive, so should all be done. Results from these three would favor steel, cast iron, or maybe a less common ferrous material. The expectation would likely be hot-rolled steel. If it is desired to confirm this, either a weight or bending test could be done to check density or modulus of elasticity. The weight test is faster. From the measured weight of 7.95 lbf, the unit weight is determined to be 3 32 7.95 lbf 0.281 lbf/in 0.28 lbf/in [ (1 in) / 4](36 in) W Al     w which agrees well with the unit weight of 0.282 lbf/in3 reported in Table A-5 for carbon steel. Nickel steel and stainless steel have similar unit weights, but surface finish and darker coloring do not favor their selection. To select a likely specification from Table Chapter 2 - Rev. D, Page 12/19 • A-20, perform a Brinell hardness test, then use Eq. (2-21) to estimate an ultimate strength of . Assuming the material is hot-rolled due to the rough surface finish, appropriate choices from Table A-20 would be one of the higher carbon steels, such as hot-rolled AISI 1050, 1060, or 1080. Ans. 0.5 0.5(200) 100 kpsiu BS H   ______________________________________________________________________________ 2-22 First, try to find the broad category of material (such as in Table A-5). Visual, magnetic, and scratch tests are fast and inexpensive, so should all be done. Results from these three favor a softer, non-ferrous material like aluminum. If it is desired to confirm this, either a weight or bending test could be done to check density or modulus of elasticity. The weight test is faster. From the measured weight of 2.90 lbf, the unit weight is determined to be 3 32 2.9 lbf 0.103 lbf/in 0.10 lbf/in [ (1 in) / 4](36 in) W Al     w which agrees reasonably well with the unit weight of 0.098 lbf/in3 reported in Table A-5 for aluminum. No other materials come close to this unit weight, so the material is likely aluminum. Ans. ______________________________________________________________________________ 2-23 First, try to find the broad category of material (such as in Table A-5). Visual, magnetic, and scratch tests are fast and inexpensive, so should all be done. Results from these three favor a softer, non-ferrous copper-based material such as copper, brass, or bronze. To further distinguish the material, either a weight or bending test could be done to check density or modulus of elasticity. The weight test is faster. From the measured weight of 9 lbf, the unit weight is determined to be 3 32 9.0 lbf 0.318 lbf/in 0.32 lbf/in [ (1 in) / 4](36 in) W Al     w which agrees reasonably well with the unit weight of 0.322 lbf/in3 reported in Table A-5 for copper. Brass is not far off (0.309 lbf/in3), so the deflection test could be used to gain additional insight. From the measured deflection and utilizing the deflection equation for an end-loaded cantilever beam from Table A-9, Young’s modulus is determined to be     33 4 100 24 17.7 Mpsi 3 3 (1) 64 (17 / 32) FlE Iy     which agrees better with the modulus for copper (17.2 Mpsi) than with brass (15.4 Mpsi). The conclusion is that the material is likely copper. Ans. ______________________________________________________________________________ 2-24 and 2-25 These problems are for student research. No standard solutions are provided. ______________________________________________________________________________ Chapter 2 - Rev. D, Page 13/19 • 2-26 For strength,  = F/A = S  A = F/S For mass, m = Al = (F/S) l Thus, f 3(M ) =  /S , and maximize S/ ( = 1) In Fig. (2-19), draw lines parallel to S/ From the list of materials given, both aluminum alloy and high carbon heat treated steel are good candidates, having greater potential than tungsten carbide or polycarbonate. The higher strength aluminum alloys have a slightly greater potential. Other factors, such as cost or availability, may dictate which to choose. Ans. ______________________________________________________________________________ 2-27 For stiffness, k = AE/l  A = kl/E For mass, m = Al = (kl/E) l =kl2  /E Thus, f 3(M) =  /E , and maximize E/ ( = 1) In Fig. (2-16), draw lines parallel to E/ Chapter 2 - Rev. D, Page 14/19 • From the list of materials given, tungsten carbide (WC) is best, closely followed by aluminum alloys, and then followed by high carbon heat-treated steel. They are close enough that other factors, like cost or availability, would likely dictate the best choice. Polycarbonate polymer is clearly not a good choice compared to the other candidate materials. Ans. ______________________________________________________________________________ 2-28 For strength,  = Fl/Z = S (1) where Fl is the bending moment and Z is the section modulus [see Eq. (3-26b), p. 90 ]. The section modulus is strictly a function of the dimensions of the cross section and has the units in3 (ips) or m3 (SI). Thus, for a given cross section, Z =C (A)3/2, where C is a number. For example, for a circular cross section, C =   14   . Then, for strength, Eq. (1) is 2/3 3/2 Fl FlS A CA CS         (2) Chapter 2 - Rev. D, Page 15/19 • For mass, 2/3 2/3 5/3 2/3 Fl Fm Al l l CS C S                     Thus, f 3(M) =  /S 2/3, and maximize S 2/3/ ( = 2/3) In Fig. (2-19), draw lines parallel to S 2/3/ From the list of materials given, a higher strength aluminum alloy has the greatest potential, followed closely by high carbon heat-treated steel. Tungsten carbide is clearly not a good choice compared to the other candidate materials. .Ans. ______________________________________________________________________________ 2-29 Eq. (2-26), p. 65, applies to a circular cross section. However, for any cross section shape it can be shown that I = CA 2, where C is a constant. For example, consider a rectangular section of height h and width b, where for a given scaled shape, h = cb, where c is a Chapter 2 - Rev. D, Page 16/19 • constant. The moment of inertia is I = bh 3/12, and the area is A = bh. Then I = h(bh2)/12 = cb (bh2)/12 = (c/12)(bh)2 = CA 2, where C = c/12 (a constant). Thus, Eq. (2-27) becomes 1/23 3 klA CE        and Eq. (2-29) becomes 1/2 5/2 1/23 km Al l C E               Thus, minimize  3 1/2f M E   , or maximize 1/2EM   . From Fig. (2-16) From the list of materials given, aluminum alloys are clearly the best followed by steels and tungsten carbide. Polycarbonate polymer is not a good choice compared to the other candidate materials. Ans. ______________________________________________________________________________ 2-30 For stiffness, k = AE/l  A = kl/E For mass, m = Al = (kl/E) l =kl2  /E Chapter 2 - Rev. D, Page 17/19 • So, f 3(M) =  /E, and maximize E/ . Thus,  = 1. Ans. ______________________________________________________________________________ 2-31 For strength,  = F/A = S  A = F/S For mass, m = Al = (F/S) l So, f 3(M ) =  /S, and maximize S/ . Thus,  = 1. Ans. ______________________________________________________________________________ 2-32 Eq. (2-26), p. 65, applies to a circular cross section. However, for any cross section shape it can be shown that I = CA 2, where C is a constant. For example, consider a rectangular section of height h and width b, where for a given scaled shape, h = cb, where c is a constant. The moment of inertia is I = bh 3/12, and the area is A = bh. Then I = h(bh2)/12 = cb (bh2)/12 = (c/12)(bh)2 = CA 2, where C = c/12. Thus, Eq. (2-27) becomes 1/23 3 klA CE        and Eq. (2-29) becomes 1/2 5/2 1/23 km Al l C E               So, minimize  3 1/2f M E   , or maximize 1/2EM   . Thus,  = 1/2. Ans. ______________________________________________________________________________ 2-33 For strength,  = Fl/Z = S (1) where Fl is the bending moment and Z is the section modulus [see Eq. (3-26b), p. 90 ]. The section modulus is strictly a function of the dimensions of the cross section and has the units in3 (ips) or m3 (SI). Thus, for a given cross section, Z =C (A)3/2, where C is a number. For example, for a circular cross section, C =   14   . Then, for strength, Eq. (1) is 2/3 3/2 Fl FlS A CA CS         (2) For mass, 2/3 2/3 5/3 2/3 Fl Fm Al l l CS C S                     So, f 3(M) =  /S 2/3, and maximize S 2/3/. Thus,  = 2/3. Ans. ______________________________________________________________________________ 2-34 For stiffness, k=AE/l, or, A = kl/E. Chapter 2 - Rev. D, Page 18/19 • Chapter 2 - Rev. D, Page 19/19 Thus, m = Al = (kl/E )l = kl 2  /E. Then, M = E / and  = 1. From Fig. 2-16, lines parallel to E / for ductile materials include steel, titanium, molybdenum, aluminum alloys, and composites. For strength, S = F/A, or, A = F/S. Thus, m = Al = F/Sl = Fl  /S. Then, M = S/ and  = 1. From Fig. 2-19, lines parallel to S/ give for ductile materials, steel, aluminum alloys, nickel alloys, titanium, and composites. Common to both stiffness and strength are steel, titanium, aluminum alloys, and composites. Ans. • Chapter 3 3-1 0oM  18 6(100) 0BR   33.3 lbf .BR Ans 0yF  100 0o BR R   66.7 lbf .oR Ans 33.3 lbf .C BR R A  ns ______________________________________________________________________________ 3-2 Body AB: 0xF  Ax BxR R 0yF  Ay ByR R 0BM  (10) (10) 0Ay AxR R  Ax AyR R Body OAC: 0OM  (10) 100(30) 0AyR   300 lbf .AyR Ans 0xF  300 lbf .Ox AxR R A    ns 0yF  100 0Oy AyR R   200 lbf .OyR Ans  ______________________________________________________________________________ Chapter 3 - Rev. A, Page 1/100 • 3-3 0.8 1.39 kN . tan 30O R Ans  0.8 1.6 kN . sin 30A R Ans  ______________________________________________________________________________ 3-4 Step 1: Find RA & RE 4.5 7.794 m tan 30 0 9 7.794(400cos30 ) 4.5(400sin 30 ) 0 400 N . A E E h M R R Ans            2 2 0 400cos30 0 346.4 N 0 400 400sin 30 0 200 N 346.4 200 400 N . x Ax Ax y Ay Ay A F R R F R R R Ans                   Step 2: Find components of RC on link 4 and RD    4 4 0 400(4.5) 7.794 1.9 0 305.4 N . 0 305.4 N 0 ( ) 400 N C D D x Cx y Cy M R R Ans F R F R                Chapter 3 - Rev. A, Page 2/100 • Step 3: Find components of RC on link 2       2 2 2 0 305.4 346.4 0 41 N 0 200 N x Cx Cx y Cy F R R F R          ____________________________________________________________________________________________________________________ _ Chapter 3 - Rev. A, Page 3/100 • 3-5 0CM  11500 300(5) 1200(9) 0R    1 8.2 kN .R Ans 0yF  28.2 9 5 0R    2 5.8 kN .R Ans 1 8.2(300) 2460 N m .M Ans   2 2460 0.8(900) 1740 N m .M Ans    3 1740 5.8(300) 0 checks!M    _____________________________________________________________________________ 3-6 0yF  0 500 40(6) 740 lbf .R Ans   0 0M  0 500(8) 40(6)(17) 8080 lbf in .M Ans    1 8080 740(8) 2160 lbf in .M Ans      2 2160 240(6) 720 lbf in .M Ans      3 1720 (240)(6) 0 checks! 2 M     ______________________________________________________________________________ Chapter 3 - Rev. A, Page 4/100 • 3-7 0BM  12.2 1(2) 1(4) 0R    1 0.91 kN .R Ans  0yF  20.91 2 4 0R     2 6.91 kN .R Ans 1 0.91(1.2) 1.09 kN m .M Ans     2 1.09 2.91(1) 4 kN m .M Ans      3 4 4(1) 0 checks!M     ______________________________________________________________________________ 3-8 Break at the hinge at B Beam OB: From symmetry, 1 200 lbf .BR V Ans  Beam BD: 0DM  2200(12) (10) 40(10)(5) 0R   2 440 lbf .R Ans 0yF  3200 440 40(10) 0R     3 160 lbf .R Ans Chapter 3 - Rev. A, Page 5/100 • 1 200(4) 800 lbf in .M Ans   2 800 200(4) 0 checks at hingeM    3 800 200(6) 400 lbf in .M Ans     4 1400 (240)(6) 320 lbf in . 2 M Ans     5 1320 (160)(4) 0 checks! 2 M    ______________________________________________________________________________ 3-9 1 1 1 1 2 0 0 0 1 2 1 1 1 1 2 9 300 5 1200 1500 9 300 5 1200 1500 (1) 9 300 5 1200 1500 (2) q R x x x R x V R x x R x M R x x x R x                        1 At x = 1500+ V = M = 0. Applying Eqs. (1) and (2), 1 2 1 29 5 0 14R R R R       1 11500 9(1500 300) 5(1500 1200) 0 8.2 kN .R R A       2 14 8.2 5.8 kN . ns R Ans   0 300 : 8.2 kN, 8.2 N m 300 1200 : 8.2 9 0.8 kN 8.2 9( 300) 0.8 2700 N m 1200 1500 : 8.2 9 5 5.8 kN 8.2 9( 300 x V M x x V M x x x x V M x x                             ) 5( 1200) 5.8 8700 N mx x      Plots of V and M are the same as in Prob. 3-5. ______________________________________________________________________________ Chapter 3 - Rev. A, Page 6/100 • 3-10 1 2 1 0 0 0 0 1 0 1 1 0 0 1 2 2 0 0 500 8 40 14 40 20 500 8 40 14 40 20 (1) 500 8 20 14 20 20 (2) at 20 in, 0, Eqs. (1) and (2) give q R x M x x x x V R M x x x x M R x M x x x x V M R                                  0 0 2 0 0 0 500 40 20 14 0 740 lbf . (20) 500(20 8) 20(20 14) 0 8080 lbf in . R Ans R M M                Ans 0 8 : 740 lbf, 740 8080 lbf in 8 14 : 740 500 240 lbf 740 8080 500( 8) 240 4080 lbf in 14 20 : 740 500 40( 14) 40 800 lbf 740 8080 x V M x x V M x x x x V x x M x                               2 2500( 8) 20( 14) 20 800 8000 lbf inx x x x        Plots of V and M are the same as in Prob. 3-6. ______________________________________________________________________________ 3-11 1 1 1 1 1 2 0 0 0 1 2 1 1 1 1 2 2 1.2 2.2 4 3.2 2 1.2 2.2 4 3.2 (1) 2 1.2 2.2 4 3.2 (2) q R x x R x x V R x R x x M R x x R x x                         at x = 3.2+, V = M = 0. Applying Eqs. (1) and (2), Solving Eqs. (3) and (4) simultaneously, 1 2 1 2 1 2 1 2 2 4 0 6 (3) 3.2 2(2) (1) 0 3.2 4 (4) R R R R R R R R              R1 = -0.91 kN, R2 = 6.91 kN Ans. 0 1.2 : 0.91 kN, 0.91 kN m 1.2 2.2 : 0.91 2 2.91 kN 0.91 2( 1.2) 2.91 2.4 kN m 2.2 3.2 : 0.91 2 6.91 4 kN 0.91 2( x V M x x V M x x x x V M x x                                 1.2) 6.91( 2.2) 4 12.8 kN mx x     Plots of V and M are the same as in Prob. 3-7. ______________________________________________________________________________ Chapter 3 - Rev. A, Page 7/100 • 3-12 1 1 1 0 0 1 1 2 3 0 0 1 1 0 1 2 3 1 1 2 2 1 1 2 3 1 400 4 10 40 10 40 20 20 400 4 10 40 10 40 20 20 (1) 400 4 10 20 10 20 20 20 (2) 0 at 8 in 8 400( q R x x R x x x R x V R x R x x x R x M R x x R x x x R x M x R                                         18 4) 0 200 lbf .R Ans    at x = 20+, V =M = 0. Applying Eqs. (1) and (2), 2 3 2 3 2 2 2 200 400 40(10) 0 600 200(20) 400(16) (10) 20(10) 0 440 lbf . R R R R R R A               3 600 440 160 lbf . ns R Ans   0 4 : 200 lbf, 200 lbf in 4 10 : 200 400 200 lbf, 200 400( 4) 200 1600 lbf in 10 20 : 200 400 440 40( 10) 640 40 lbf 200 400( 4) x V M x x V M x x x x V x x M x x                                 2 2440( 10) 20 10 20 640x x x 4800 lbf inx        Plots of V and M are the same as in Prob. 3-8. ______________________________________________________________________________ 3-13 Solution depends upon the beam selected. ______________________________________________________________________________ 3-14 (a) Moment at center,     2 2 2 2 2 2 2 2 4 c c l a x l l lM l a a                      w wl At reaction, 2 2rM aw a = 2.25, l = 10 in, w = 100 lbf/in  2 100(10) 10 2.25 125 lbf in 2 4 100 2.25 253 lbf in . 2 c r M M Ans            (b) Optimal occurs when c rM M Chapter 3 - Rev. A, Page 8/100 • 2 2 20.25 0 2 4 2 l l aa a al l          w w Taking the positive root    2 21 4 0.25 2 1 0.207 .2 2 la l l l l A          ns for l = 10 in, w = 100 lbf, a = 0.207(10) = 2.07 in   2min 100 2 2.07 214 lbf inM    ______________________________________________________________________________ 3-15 (a) 20 10 5 kpsi 2 C   20 10 15 kpsi 2 CD   2 215 8 17 kpsiR    1 5 17 22 kpsi    2 5 17 12 kpsi     11 8tan 14.04 cw 2 15p          1 17 kpsi 45 14.04 30.96 ccws R         (b) 9 16 12.5 kpsi 2 C   16 9 3.5 kpsi 2 CD   2 25 3.5 6.10 kpsiR    1 12.5 6.1 18.6 kpsi    2 12.5 6.1 6.4 kpsi    11 5tan 27.5 ccw 2 3.5p          1 6.10 kpsi 45 27.5 17.5 cws R         Chapter 3 - Rev. A, Page 9/100 • (c) 2 2 1 2 24 10 17 kpsi 2 24 10 7 kpsi 2 7 6 9.22 kpsi 17 9.22 26.22 kpsi 17 9.22 7.78 kpsi C CD R                  11 790 tan 69.7 ccw 2 6p             1 9.22 kpsi 69.7 45 24.7 ccws R         (d) 2 2 1 2 12 22 5 kpsi 2 12 22 17 kpsi 2 17 12 20.81 kpsi 5 20.81 25.81 kpsi 5 20.81 15.81 kpsi C CD R                    11 1790 tan 72.39 cw 2 12p             Chapter 3 - Rev. A, Page 10/100 • 1 20.81 kpsi 72.39 45 27.39 cws R        ______________________________________________________________________________ Chapter 3 - Rev. A, Page 11/100 • 3-16 (a) 2 2 1 2 8 7 0.5 MPa 2 8 7 7.5 MPa 2 7.5 6 9.60 MPa 9.60 0.5 9.10 MPa 0.5 9.6 10.1 Mpa C CD R                      11 7.590 tan 70.67 cw 2 6p             1 9.60 MPa 70.67 45 25.67 cws R         (b) 2 2 1 2 9 6 1.5 MPa 2 9 6 7.5 MPa 2 7.5 3 8.078 MPa 1.5 8.078 9.58 MPa 1.5 8.078 6.58 MPa C CD R                   11 3tan 10.9 cw 2 7.5p          1 8.078 MPa 45 10.9 34.1 ccws R         Chapter 3 - Rev. A, Page 12/100 • (c) 2 2 1 2 12 4 4 MPa 2 12 4 8 MPa 2 8 7 10.63 MPa 4 10.63 14.63 MPa 4 10.63 6.63 MPa C CD R                   11 890 tan 69.4 ccw 2 7p             1 10.63 MPa 69.4 45 24.4 ccws R         (d) 2 2 1 2 6 5 0.5 MPa 2 6 5 5.5 MPa 2 5.5 8 9.71 MPa 0.5 9.71 10.21 MPa 0.5 9.71 9.21 MPa C CD R                   11 8tan 27.75 ccw 2 5.5p          1 9.71 MPa 45 27.75 17.25 cws R         ______________________________________________________________________________ Chapter 3 - Rev. A, Page 13/100 • 3-17 (a) 2 2 1 2 12 6 9 kpsi 2 12 6 3 kpsi 2 3 4 5 kpsi 5 9 14 kpsi 9 5 4 kpsi C CD R                  11 4tan 26.6 ccw 2 3p         1 5 kpsi 45 26.6 18.4 ccws R         (b) 2 2 1 2 30 10 10 kpsi 2 30 10 20 kpsi 2 20 10 22.36 kpsi 10 22.36 32.36 kpsi 10 22.36 12.36 kpsi C CD R                   11 10tan 13.28 ccw 2 20p          1 22.36 kpsi 45 13.28 31.72 cws R         Chapter 3 - Rev. A, Page 14/100 • (c) 2 2 1 2 10 18 4 kpsi 2 10 18 14 kpsi 2 14 9 16.64 kpsi 4 16.64 20.64 kpsi 4 16.64 12.64 kpsi C CD R                    11 1490 tan 73.63 cw 2 9p             1 16.64 kpsi 73.63 45 28.63 cws R        (d) 2 2 1 2 9 19 14 kpsi 2 19 9 5 kpsi 2 5 8 9.434 kpsi 14 9.43 23.43 kpsi 14 9.43 4.57 kpsi C CD R                  11 590 tan 61.0 cw 2 8p             1 9.34 kpsi 61 45 16 cws R         ______________________________________________________________________________ Chapter 3 - Rev. A, Page 15/100 • 3-18 (a) 2 2 1 2 3 80 30 55 MPa 2 80 30 25 MPa 2 25 20 32.02 MPa 0 MPa 55 32.02 22.98 23.0 MPa 55 32.0 87.0 MPa C CD R                            1 2 2 3 1 3 23 8711.5 MPa, 32.0 MPa, 43.5 MPa 2 2        (b) 2 2 1 2 3 30 60 15 MPa 2 60 30 45 MPa 2 45 30 54.1 MPa 15 54.1 39.1 MPa 0 MPa 15 54.1 69.1 MPa C CD R                        1 3 1 2 2 3 39.1 69.1 54.1 MPa 2 39.1 19.6 MPa 2 69.1 34.6 MPa 2           Chapter 3 - Rev. A, Page 16/100 • (c) 2 2 1 2 3 40 0 20 MPa 2 40 0 20 MPa 2 20 20 28.3 MPa 20 28.3 48.3 MPa 20 28.3 8.3 MPa 30 MPaz C CD R                        1 3 1 2 2 3 48.3 30 30 8.339.1 MPa, 28.3 MPa, 10.9 MPa 2 2         (d) 2 2 1 2 3 50 25 MPa 2 50 25 MPa 2 25 30 39.1 MPa 25 39.1 64.1 MPa 25 39.1 14.1 MPa 20 MPaz C CD R                      1 3 1 2 2 3 64.1 20 20 14.142.1 MPa, 39.1 MPa, 2.95 MPa 2 2         ______________________________________________________________________________ 3-19 (a) Since there are no shear stresses on the stress element, the stress element already represents principal stresses. 1 2 3 10 kpsi 0 kpsi 4 kpsi x y            1 3 1 2 2 3 10 ( 4) 7 kpsi 2 10 5 kpsi 2 0 ( 4) 2 kpsi 2              Chapter 3 - Rev. A, Page 17/100 • (b) 2 2 1 2 3 0 10 5 kpsi 2 10 0 5 kpsi 2 5 4 6.40 kpsi 5 6.40 11.40 kpsi 0 kpsi, 5 6.40 1.40 kpsi C CD R                     1 3 1 2 3 11.40 1.406.40 kpsi, 5.70 kpsi, 0.70 kpsi 2 2 R        (c) 2 2 1 2 3 2 8 5 kpsi 2 8 2 3 kpsi 2 3 4 5 kpsi 5 5 0 kpsi, 0 kpsi 5 5 10 kpsi C CD R                         1 3 1 2 2 3 10 5 kpsi, 0 kpsi, 5 kpsi 2       (d) 2 2 1 2 3 10 30 10 kpsi 2 10 30 20 kpsi 2 20 10 22.36 kpsi 10 22.36 12.36 kpsi 0 kpsi 10 22.36 32.36 kpsi C CD R                        1 3 1 2 2 3 12.36 32.3622.36 kpsi, 6.18 kpsi, 16.18 kpsi 2 2        ______________________________________________________________________________ Chapter 3 - Rev. A, Page 18/100 • 3-20 From Eq. (3-15), 3 2 2 2 2 2 2 3 ( 6 18 12) 6(18) ( 6)( 12) 18( 12) 9 6 ( 15) 6(18)( 12) 2(9)(6)( 15) ( 6)(6) 18( 15) ( 12)(9) 0 594 3186 0 2                                       Roots are: 21.04, 5.67, –26.71 kpsi Ans. 1 2 2 3 max 1 3 21.04 5.67 7.69 kpsi 2 5.67 26.71 16.19 kpsi 2 21.04 26.71 23.88 kpsi . 2 Ans               _____________________________________________________________________________ 3-21 From Eq. (3-15)       2 3 2 2 2 2 2 3 2 (20 0 20) 20(0) 20(20) 0(20) 40 20 2 0 20(0)(20) 2(40) 20 2 (0) 20 20 2 0(0) 20(40) 0 40 2 000 48 000 0 2                                  Roots are: 60, 20, –40 kpsi Ans. 1 2 2 3 max 1 3 60 20 20 kpsi 2 20 40 30 kpsi 2 60 40 50 kpsi . 2 Ans               _____________________________________________________________________________ Chapter 3 - Rev. A, Page 19/100 • 3-22 From Eq. (3-15)    2 23 2 2 2 2 2 3 2 (10 40 40) 10(40) 10(40) 40(40) 20 40 20 10(40)(40) 2(20)( 40)( 20) 10( 40) 40( 20) 40(20) 0 90 0                                 Roots are: 90, 0, 0 MPa Ans. 2 3 1 2 1 3 max 0 90 45 MPa . 2 Ans          _____________________________________________________________________________ 3-23        2 6 6 1 15000 33 950 psi 34.0 kpsi . 4 0.75 6033 950 0.0679 in . 30 10 0.0679 1130 10 1130 . 60 F Ans A FL L Ans AE E Ans L                    From Table A-5, v = 0.292     2 1 6 6 2 0.292(1130) 330 . 330 10 (0.75) 248 10 in . v A d d An ns s                   _____________________________________________________________________________ 3-24        2 6 6 1 3000 6790 psi 6.79 kpsi . 4 0.75 606790 0.0392 in . 10.4 10 0.0392 653 10 653 . 60 F Ans A FL L Ans AE E Ans L                   From Table A-5, v = 0.333     2 1 6 6 2 0.333(653) 217 . 217 10 (0.75) 163 10 in . v Ans d d Ans                   Chapter 3 - Rev. A, Page 20/100 • _____________________________________________________________________________ 3-25 2 0.0001 0.0001d d d d       From Table A-5, v = 0.326, E = 119 GPa           62 1 6 9 1 2 6 0.0001 306.7 10 0.326 and , so = 306.7 10 (119) 10 36.5 MPa 0.03 36.5 10 25 800 N 25.8 kN . 4 v FL F AE A E E L F A An                        s Sy = 70 MPa >  , so elastic deformation assumption is valid. _____________________________________________________________________________ 3-26  6 8(12)20 000 0.185 in . 10.4 10 FL L Ans AE E      _____________________________________________________________________________ 3-27     6 9 3140 10 0.00586 m 5.86 mm . 71.7 10 FL L Ans AE E       _____________________________________________________________________________ 3-28  6 10(12)15 000 0.173 in . 10.4 10 FL L Ans AE E      _____________________________________________________________________________ 3-29 With 0,z  solve the first two equations of Eq. (3-19) simulatenously. Place E on the left-hand side of both equations, and using Cramer’s rule,   2 2 1 1 1 1 1 x y xx y x E v E EE vE v v v v               yv Likewise, Chapter 3 - Rev. A, Page 21/100 •   21 y x y E v        From Table A-5, E = 207 GPa and ν = 0.292. Thus,               9 6 2 2 9 6 2 207 10 0.0019 0.292 0.000 72 10 382 MPa . 1 1 0.292 207 10 0.000 72 0.292 0.0019 10 37.4 MPa . 1 0.292 x y x y E v Ans v Ans                       _____________________________________________________________________________ 3-30 With 0,z  solve the first two equations of Eq. (3-19) simulatenously. Place E on the left-hand side of both equations, and using Cramer’s rule,   2 2 1 1 1 1 1 x y xx y x E v E EE vE v v v v               yv Likewise,   21 y x y E v        From Table A-5, E = 71.7 GPa and ν = 0.333. Thus,               9 6 2 2 9 6 2 71.7 10 0.0019 0.333 0.000 72 10 134 MPa . 1 1 0.333 71.7 10 0.000 72 0.333 0.0019 10 7.04 MPa . 1 0.333 x y x y E v Ans v Ans                       _____________________________________________________________________________ 3-31 (a) 1 max 1 c acR F M R a F l l    2 2 2 6 6 6 M ac bh lF F bh bh l ac       Ans. (b)          2 2 1 21( )( ) ( ) . ( )( ) m m m mm m m b b h h l lF s s s s Ans F a a c c s s      3-32 For equal stress, the model load varies by the square of the scale factor. _____________________________________________________________________________ Chapter 3 - Rev. A, Page 22/100 • 2 1 max /2 , 2 2 2 2x l l l lR M l         w w 8 l  ww(a) 2 2 2 2 2 6 6 3 4 . 8 4 3 M l Wl bhW A bh bh bh l      w ns (b) 2 2 2( / )( / )( / ) 1( )( ) . / m m m m m W b b h h s s s An W l l s      s 2 2 .m m ml ss s l s     w w w w Ans For equal stress, the model load w varies linearly with the scale factor. _____ _____________ -33 (a) Can solve by iteration or derive _ __________________________________________________________ 3 equations for the general case. Find maximum moment under wheel 3W . W W  at centroid of W’s T 3 3d A T l xR W l   Under wheel 3,   3 3 3 3 1 13 2 23 3 1 13 2 23A T l x d M R x W a W a W x W a W a l       For maximum,  3 33 3 3 3 0 2 2 TdM l dWl d x x dx l        Substitute into   2 3 3 1 14 T l d 3 2 23M M W W al      W a intersects the midpoint of the beam. For wheel i, This means the midpoint of 3d  2 1il dl d 1 , 2 4 ii T j ji j i ix M W W al     Note for wheel 1:  0j jiW a  1 2 3 4 104.4104.4, 26.1 kips 4T W W W W W      Wheel 1: 2 1 1 476 (1200 238)238 in, (104.4) 20128 kip in 2 4(1200) d M      Wheel 2: 238 84 154 ind    2 Chapter 3 - Rev. A, Page 23/100 • 2 2 max (1200 154) (104.4) 26.1(84) 21605 kip in . 4(1200) M M A     ns Check if all of the wheels are on the rail. (b) max 600 77 523 in .x Ans   (c) See above sketch. (d) Inner axles _____________________________________________________________________________ 3-34 (a) Let a = total area of entire envelope Let b = area of side notch            2 3 3 6 4 2 40(3)(25) 25 34 2150 mm 1 12 40 75 34 25 12 12 1.36 10 mm . a b A a b I I I I Ans           Dimensions in mm. (b) 2 2 2 0.375(1.875) 0.703 125 in 0.375(1.75) 0.656 25 in 2(0.703125) 0.656 25 2.0625 in a b A A A        3 4 3 4 2 2 1 2(0.703 125)(0.9375) 0.656 25(0.6875) 0.858 in . 2.0625 0.375(1.875) 0.206 in 12 1.75(0.375) 0.007 69 in 12 2 0.206 0.703 125(0.0795) 0.00769 0.656 25(0.1705) 0.448 in . a b y A I I 4 ns I Ans                   (c) Use two negative areas. 2 2 2 625 mm , 5625 mm , 10 000 mm 10 000 5625 625 3750 mm ; a b cA A A A        2 Chapter 3 - Rev. A, Page 24/100 •     1 3 4 3 6 4 3 6 4 6.25 mm, 50 mm, 50 mm 10 000(50) 5625(50) 625(6.25) 57.29 mm . 3750 100 57.29 42.71 mm . 50(12.5) 8138 mm 12 75(75) 2.637 10 mm 12 100(100) 8.333 10 in 12 a b c a b c y y y y Ans c Ans I I I                           2 26 2 6 1 6 4 1 8.333 10 10000(7.29) 2.637 10 5625 7.29 8138 625 57.29 6.25 4.29 10 in . I I Ans                  (d)       2 2 2 4 0.875 3.5 in 2.5 0.875 2.1875 in 5.6875 in 2.9375 3.5 1.25(2.1875) 2.288 in . 5.6875 a b a b A A A A A y Ans                  3 2 3 4 1 1(4) 0.875 3.5 2.9375 2.288 0.875 2.5 2.1875 2.288 1.25 12 12 5.20 in . I I Ans        2 _____________________________________________________________________________ 3-35  3 5 2 1 (20)(40) 1.067 10 mm 12 20(40) 800 mm I A     4 Mmax is at A. At the bottom of the section,  max 5 450 000(20) 84.3 MPa . 1.067 10 Mc Ans I     Due to V, max is between A and B at y = 0. max 3 3 3000 5.63 MPa . 2 2 800 V Ans A         _____________________________________________________________________________ Chapter 3 - Rev. A, Page 25/100 • 3-36 3 41 (1)(2) 0.6667 in 12 I   21(2) 2 inA   0oM  8 100(8)(12) 0AR   1200 lbfAR  1200 100(8) 400 lbfoR    is at A. At the top of the beam, maxM max 3200(0.5) 2400 psi . 0.6667 Mc Ans I     Due to V, max is at A, at y = 0. max 3 3 800 600 psi . 2 2 2 V Ans A         _____________________________________________________________________________ 3-37 3 41 (0.75)(2) 0.5 in 12 I   2(0.75)(2) 1.5 inA   0AM  15 1000(20) 0BR   1333.3 lbfBR  3000 1333.3 1000 2666.7 lbfAR     is at B. At the top of the beam, maxM max 5000(1) 10000 psi . 0.5 Mc Ans I     Due to V, max is between B and C at y = 0. max 3 3 1000 1000 psi . 2 2 1.5 V Ans A         _____________________________________________________________________________ Chapter 3 - Rev. A, Page 26/100 • 3-38   4 4 3 4(50) 306.796 10 mm 64 64 dI     2 2 2(50) 1963 mm 4 4 dA     0BM  6(300)(150) 200 0AR  1350 kNAR  6(300) 1350 450 kNBR    maxM is at A. At the top, max Mc I    Due to V, max is at A, at y = 0. 2 max 4 4 750 0.509 kN/mm 509 MPa . 3 3 1963 V Ans A          _____________________________________________________________________________ 3-39 2 2 max max max 2 8 8 8 Il l cM I cl     w w w (a) 448 in; Table A-8, 0.537 inl I         3 2 8 12 10 0.537 22.38 lbf/in . 1 48 Ans w (b)        3 360 in, 1 12 2 3 1 12 1.625 2.625 2.051 inl I  4        3 2 8 12 10 2.051 36.5 lbf/in . 1.5 60 Ans w (c)   460 in; Table A-6, 2 0.703 1.406 inl I   y = 0.717 in, cmax = 1.783 in       3 2 8 12 10 1.406 21.0 lbf/in . 1.783 60 Ans w (d) 460 in, Table A-7, 2.07 inl I         3 2 8 12 10 2.07 36.8 lbf/in . 1.5 60 Ans w _____________________________________________________________________________ Chapter 3 - Rev. A, Page 27/100 • 3-40      4 3 4 2 20.5 3.068 10 in , 0.5 0.1963 in64 4I A      Model (c)  3 max 500(0.5) 500(0.75 / 2) 218.75 lbf in 2 2 218.75(0.25) 3.068 10 17 825 psi 17.8 kpsi . 4 4 500 3400 psi 3.4 kpsi . 3 3 0.1963 M Mc I Ans V Ans A                 Model (d)  3 500(0.625) 312.5 lbf in 312.5(0.25) 3.068 10 25 464 psi 25.5 kpsi . M Mc I Ans           max 4 4 500 3400 psi 3.4 kpsi . 3 3 0.1963 V Ans A      Model (e)   3 max 500(0.4375) 218.75 lbf in 218.75(0.25) 3.068 10 17 825 psi 17.8 kpsi . 4 4 500 3400 psi 3.4 kpsi . 3 3 0.1963 M Mc I Ans V Ans A                _____________________________________________________________________________ 3-41 Chapter 3 - Rev. A, Page 28/100 •    4 4 212 1018 mm , 12 113.1 mm64 4I A 2     Model (c) 2 2 max 2000(6) 2000(9) 15 000 N mm 2 2 15 000(6) 1018 88.4 N/mm 88.4 MPa . 4 4 2000 23.6 N/mm 23.6 MPa . 3 3 113.1 M Mc I Ans V Ans A                    Model (d) 2 2000(12) 24 000 N mm 24 000(6) 1018 141.5 N/mm 141.5 MPa . M Mc I Ans          2 max 4 4 2000 23.6 N/mm 23.6 MPa . 3 3 113.1 V Ans A          Model (e) 2 2000(7.5) 15000 N mm 15000(6) 1018 88.4 N/mm 88.4 MPa . M Mc I Ans          2 max 4 4 2000 23.6 N/mm 23.6 MPa . 3 3 113.1 V Ans A          _____________________________________________________________________________  4 3 / 2 32 / 64 M dMc M I d d       3-42 (a) Chapter 3 - Rev. A, Page 29/100 • 3 3 32 32(218.75) 0.420 in . (30 000) Md A      ns (b) 2 / 4 V V A d     4 4( 500) 0.206 in . (15000) Vd Ans      (c)  2 4 4 3 3 / 4 V V A d     4 4 4 4(500) 0.238 in . 3 3 (15000) Vd A      ns ______________ __________________ ______________________________ _____________ _ _ 3-43 1 0 11 2 1 1 21 2 1 2 31 1 2 terms for terms for 2 terms for 2 6 p pq F x p x l x l x l a a p pV F p x l x l x l a a p p pM Fx x l x l x l a a                               terms for x > l + a = 0 At x ( ) , 0,l a V M    21 2 1 1 2 2 Fp p   2 31 1 2 1 2 2 0 (1) 2 6 ( )( ) 0 2 (2) 2 6 p pF p a a a a p a p p F l aF l a a p p a a                From (1) and (2) 1 22 2 2 2(3 2 ), (3 ) (3)F Fp l a p l a a a     From similar triang les 2 2 1 2 1 2 (4)apb a b p p p p p      Chapter 3 - Rev. A, Page 30/100 • Mmax occurs where V = 0 max 2x l a b   2 31 1 2 max 2 31 1 2 ( 2 ) ( 2 ) ( 2 ) 2 6 ( 2 ) ( 2 ) ( 2 ) 2 6 p p pM F l a b a b a b a p p pFl F a b a b a b a                   Normally Mmax =  Fl The fractional increase in the magnitude is    2 31 22 ( 2 ) 6 ( 2 ) (5)a b p p a a b      For example, consider F = 1500 lbf, a = 1.2 in, l = 1.5 in (3) 1( 2 )F a b p  Fl    1 2 2(1500) 3 1.5 2(1.2) 14 375 lbf/in 1.2 p        2 2 2(1500) 3 1.5 1.2 11 875 lbf/in 1.2 p       (4) b = 1.2(11 875)/(14 375 + 11 875) = 0.5429 in Substituting into (5) yields _____________________________________________________________________________ -44  = 0.036 89 or 3.7% higher than -Fl 3 Chapter 3 - Rev. A, Page 31/100 • 1 2 300(30)R  401800 6900 lbf 2 30 300(30) 101800 3900 lbf 2 30 3900 13 in 300 R a        MB = 1800(10) = 18 000 lbfin x = 27 in = (1/2)3900(13) = 25 350 lbfin   MB = 1800(10) = 18 000 lbfin x = 27 in = (1/2)3900(13) = 25 350 lbfin MM 3 4 1 3 4 2 0.5(3) 2.5(3) 1.5 in 6 1 (3)(1 ) 0.25 in 12 1 (1)(3 ) 2.25 in 12 y I I        Applying the parallel-axis theorem, (a) 20.25 3(1.5 0.5) 2.25 3zI          2 4 (2.5 1.5) 8.5 in  18000( 1.5)At 10 in, 1.5 in, 3176 psi 8.5 18000(2.5)At 10 in, 2.5 in, 5294 psi 8.5 25350( 1.5)At 27 in, 1.5 in, 4474 psi 8.5 At 27 in, 2.5 in, x x x x x y x y x y x y                              25350(2.5) 7456 psi 8.5    Max tension 5294 psi . Max compression 7456 psi . Ans Ans    aximum shear stress due to V is at B, at the neutral axis. (b) The m max 5100 lbfV    3 max 1.25(2.5)(1) 3.125 in 5100(3.125) 1875 psi . 8.5(1)V Q y A VQ Ans Ib         (c) There are three potentially critical locations for the maximum shear stress, all at x = 27 in: (i) at the top where the bending stress is maximum, (ii) at the neutral axis where Chapter 3 - Rev. A, Page 32/100 • the transverse shear is maximum, or (iii) in the web just above the flange where bending stress and shear stress are in their largest combination. For (i): The maximum bending stress was previously found to be 7456 psi, and the shear stress is zero. From Mohr’s circle, maxmax 7456 3728 psi 2 2      For (ii): The bending stress is zero, and the transverse shear stress was found previously to be 1875 psi. Thus, max = 1875 psi. For (iii): The bending stress at y = – 0.5 in is 18000( 0.5) 1059 psi 8.5x       The transverse shear stress is 3(1)(3)(1) 3.0 in 5100(3.0) 1800 psi 8.5(1) Q y A VQ Ib         From Mohr’s circle, 2 2 max 1059 1800 1876 psi 2         The critical location is at x = 27 in, at the top surface, where max = 3728 psi. Ans. _____________________________________________________________________________ 3-45 (a) L = 10 in. Element A:  34(1000)(10)(0.5) 10 101.9 kpsi( / 64)(1)A My I        , 0A A VQ Q 0 Ib      2 2 2 2 max 101.9 (0) 50.9 kpsi . 2 2 A A Ans               Element B: , 0 0B B My y I        32 3 34 0.54 4 1/12 in 3 2 6 6 r r rQ y A                Chapter 3 - Rev. A, Page 33/100 •  34(1000)(1/12) 10 1.698 kpsi( / 64)(1) (1)B VQ Ib      2 2 max 0 1.698 1.698 kpsi . 2 Ans        Element C:  34(1000)(10)(0.25) 10 50.93 kpsi( / 64)(1)C My I                  2 2 1 1 1 3/2 3/2 3/22 2 2 2 2 2 1 1 3/22 2 1 (2 ) 2 2 2 3 3 2 3 r r r y y y r y Q ydA y x dy y r y dy r y r r r y r y                     For C, y1 = r /2 =0.25 in  3/22 22 0.5 0.25 0.054133Q    in 3 2 2 2 2 12 2 2 0.5 0.25 0.866 inb x r y       34(1000)(0.05413) 10 1.273 kpsi( / 64)(1) (0.866)C VQ Ib      2 2 max 50.93 (1.273) 25.50 kpsi . 2 Ans        (b) Neglecting transverse shear stress: Element A: Since the transverse shear stress at point A is zero, there is no change. max 50.9 kpsi .Ans  % error 0% .Ans Element B: Since the only stress at point B is transverse shear stress, neglecting the transverse shear stress ignores the entire stress. 2 max 0 0 psi . 2 Ans       1.698 0% error *(100) 100% . 1.698 Ans      Chapter 3 - Rev. A, Page 34/100 • Element C: 2 max 50.93 25.47 kpsi . 2 Ans       25.50 25.47% error *(100) 0.12% . 25.50 Ans        (c) Repeating the process with different beam lengths produces the results in the table. Bending stress, kpsi) Transverse shear stress, kpsi) Max shear stress, max kpsi) Max shear stress, neglecting  max kpsi) % error L = 10 in A 102 0 50.9 50.9 0 B 0 1.70 1.70 0 100 C 50.9 1.27 25.50 25.47 0.12 L = 4 in A 40.7 0 20.4 20.4 0 B 0 1.70 1.70 0 100 C 20.4 1.27 10.26 10.19 0.77 L = 1 in A 10.2 0 5.09 5.09 0 B 0 1.70 1.70 0 100 C 5.09 1.27 2.85 2.55 10.6 L = 0.1in A 1.02 0 0.509 0.509 0 B 0 1.70 1.70 0 100 C 0.509 1.27 1.30 0.255 80.4 Discussion: The transverse shear stress is only significant in determining the critical stress element as the length of the cantilever beam becomes smaller. As this length decreases, bending stress reduces greatly and transverse shear stress stays the same. This causes the critical element location to go from being at point A, on the surface, to point B, in the center. The maximum shear stress is on the outer surface at point A for all cases except L = 0.1 in, where it is at point B at the center. When the critical stress element is at point A, there is no error from neglecting transverse shear stress, since it is zero at that location. Neglecting the transverse shear stress has extreme significance at the stress element at the center at point B, but that location is probably only of practical significance for very short beam lengths. _____________________________________________________________________________ Chapter 3 - Rev. A, Page 35/100 • 3-46 1 0 cR F l cM Fx x a l       2 2 max 66 6 0 . c l FxM bh bh Fcxh x lb        a Ans _____________________________________________________________________________ 3-47 From Problem 3-46, 1 , 0 cR F V x a l     max max 3 3 ( / ) 3 . 2 2 2 V c l F Fch A bh bh lb       ns From Problem 3-46, max 6( ) Fcxh x . lb  Sub in x = e and equate to h above. max max max 2 max 3 6 2 3 . 8 Fc Fce lb lb Fce A lb       ns _____________________________________________________________________________ 3-48 (a) x-z plane 20 1.5(0.5) 2(1.5)sin(30 )(2.25) (3)O zM R      2 1.375 kN .zR Ans 10 1.5 2(1.5)sin(30 ) 1.375z zF R       1 1.625 kN .zR Ans x-y plane 20 2(1.5)cos(30 )(2.25) (3)O yM R      2 1.949 kN .yR Ans 10 2(1.5) cos(30 ) 1.949y yF R      1 0.6491 kN .yR Ans Chapter 3 - Rev. A, Page 36/100 • (b) (c) The transverse shear and bending moments for most points of interest can readily be taken straight from the diagrams. For 1.5 < x < 3, the bending moment equations are parabolic, and are obtained by integrating the linear expressions for shear. For convenience, use a coordinate shift of x = x – 1.5. Then, for 0 < x < 1.5,     2 2 0.125 0.125 2 At 0, 0.9375 0.5 0.125 0.9375 z y z y y V x x M V dx x C x M C M x x                        2 2 1.949 0.6491 1.732 0.6491 1.125 1.732 0.6491 2 At 0, 0.9737 0.8662 0.125 0.9375 y z z z V x x M x x C x M C M x x                      By programming these bending moment equations, we can find My, Mz, and their vector combination at any point along the beam. The maximum combined bending moment is found to be at x = 1.79 m, where M = 1.433 kN·m. The table below shows values at key locations on the shear and bending moment diagrams. x (m) Vz (kN) Vy (kN) V (kN) My (kNm) Mz (kNm) M (kNm) 0 –1.625 0.6491 1.750 0 0 0 0.5 –1.625 0.6491 1.750 –0.8125 0.3246 0.8749 1.5 –0.1250 0.6491 0.6610 0.9375 0.9737 1.352 1.625 0 0.4327 0.4327 –0.9453 1.041 1.406 1.875 0.2500 0 0.2500 –0.9141 1.095 1.427 3 1.375 –1.949 2.385 0 0 0 Chapter 3 - Rev. A, Page 37/100 • (d) The bending stress is obtained from Eq. (3-27), y Az A x z y M zM y I I    The maximum tensile bending stress will be at point A in the cross section of Prob. 3-34 (a), where distances from the neutral axes for both bending moments will be maximum. At A, for Mz, yA = –37.5 mm, and for My, zA = –20 mm. 3 3 6 4 640(75) 34(25) 1.36(10 ) mm 1.36(10 ) m 12 12z I     4 3 3 5 4 725(40) 25(6)2 2.67(10 ) mm 2.67(10 ) m 12 12y I           4 It is apparent the maximum bending moment, and thus the maximum stress, will be in the parabolic section of the bending moment diagrams. Programming Eq. (3-27) with the bending moment equations previously derived, the maximum tensile bending stress is found at x = 1.77 m, where My = – 0.9408 kN·m, Mz = 1.075 kN·m, and x = 100.1 MPa. Ans. _____________________________________________________________________________ 3-49 (a) x-z plane 3 6000 (1000)(4) (10) 5 2O Oy M M     1842.6 lbf in .OyM Ans  3 60 (1000) 5 2z Oz F R     00 175.7 lbf .OzR Ans x-y plane 4 6000 (1000)(4) (10) 5 2O Oz M M      7442.5 lbf in .OzM Ans  4 60 (1000) 5 2y Oy F R     00 1224.3 lbf .OyR Ans Chapter 3 - Rev. A, Page 38/100 • (b) ( (c) 1/22 2( ) ( ) ( )y zV x V x V x    1/22 2( ) ( ) ( )y zM x M x M x    x (m) Vz (kN) Vy (kN) V (kN) My (kNm) Mz (kNm) M (kNm) 0 –175.7 1224.3 1237 –1842.6 –7442.6 7667 4 –175.7 1224.3 1237 –2545.4 –2545.4 3600 10 424.3 424.3 600 0 0 0 (d) The maximum tensile bending stress will be at the outer corner of the cross section in the positive y, negative z quadrant, where y = 1.5 in and z = –1 in. 3 3 42(3) (1.625)(2.625) 2.051 in 12 12z I    3 3 43(2) (2.625)(1.625) 1.601 in 12 12y I    At x = 0, using Eq. (3-27), yz x z y M zM y I I     ( 7442.6)(1.5) ( 1842.6)( 1) 6594 psi 2.051 1.601x        Check at x = 4 in, ( 2545.4)(1.5) ( 2545.4)( 1) 2706 psi 2.051 1.601x        The critical location is at x = 0, where x = 6594 psi. Ans. _____________________________________________________________________________ Chapter 3 - Rev. A, Page 39/100 • 3-50 The area within the wall median line, Am, is Square: 2( )mA b t  . From Eq. (3-45) 2 sq all all2 2( )mT A t b t t    Round: 2( ) /mA b t  4 2 rd all2 ( ) / 4T b t t   Ratio of Torques 2 sq all 2 rd all 2( ) 4 1.27 ( ) / 2 T b t t T b t t         Twist per unit length from Eq. (3-46) is all all 1 2 2 2 4 4 2 m m m m m m m TL A t L L LC GA t GA t G A A m m       Square: sq 2 4( ) ( ) b tC b t    Round: rd 2 2 ( ) 4( ( ) / 4 ( ) b t b tC C b t b t )        Ratio equals 1. Twists are the same. _____________________________________________________________________________ 3-51 (a) The area enclosed by the section median line is Am = (1  0.0625)2 = 0.8789 in2 and the length of the section median line is Lm = 4(1  0.0625) = 3.75 in. From Eq. (3-45), 2 2(0.8789)(0.0625)(12 000) 1318 lbf in .mT A t Ans    From Eq. (3-46),       1 2 6 2 (1318)(3.75) 36 0.0801 rad 4.59 . 4 4 11.5 10 (0.8789) 0.0625 m m TL ll A GA t        ns (b) The radius at the median line is rm = 0.125 + (0.5)(0.0625) = 0.15625 in. The area enclosed by the section median line is Am = (1  0.0625)2 – 4(0.15625)2 + 4(π /4)(0.15625)2 = 0.8579 in2. The length of the section median line is Lm = 4[1 – 0.0625 – 2(0.15625)] + 2π(0.15625) = 3.482 in. Chapter 3 - Rev. A, Page 40/100 • From Eq. (3-45), 2 2(0.8579)(0.0625)(12 000) 1287 lbf in .mT A t Ans    From Eq. (3-46),       1 2 6 2 (1287)(3.482) 36 0.0762 rad 4.37 . 4 4 11.5 10 (0.8579) 0.0625 m m TL ll A GA t        ns _____________________________________________________________________________ 3-52 3 1 1 3 3 3 i i i i i T GT GL c      i L c 3 31 1 2 3 1 . 3 i ii GT T T T L c Ans       From Eq. (3-47), G1c G and 1 are constant, therefore the largest shear stress occurs when c is a maximum. max 1 max .G c Ans  _____________________________________________________________________________ 3-53 (b) Solve part (b) first since the twist is needed for part (a).  max allow 12 6.89 82.7 MPa        6 max 1 9 max 82.7 10 0.348 rad/m . 79.3 10 (0.003) Ans Gc     (a)  9 331 1 1 1 0.348(79.3) 10 (0.020)(0.002 ) 1.47 N m . 3 3 GL cT A    ns     9 33 2 2 2 2 9 33 3 3 3 3 1 2 3 0.348(79.3) 10 (0.030)(0.003 ) 7.45 N m . 3 3 0.348(79.3) 10 (0)(0 ) 0 . 3 3 1.47 7.45 0 8.92 N m . GL cT A GL cT A T T T T Ans                  ns ns _____________________________________________________________________________ Chapter 3 - Rev. A, Page 41/100 • 3-54 (b) Solve part (b) first since the twist is needed for part (a).     3max 1 6 max 12000 8.35 10 rad/in . 11.5 10 (0.125) Ans Gc     (a)                       3 6 33 1 1 1 1 3 6 33 2 2 2 2 3 6 33 3 3 3 3 1 2 3 8.35 10 11.5 10 0.75 0.0625 5.86 lbf in . 3 3 8.35 10 11.5 10 1 0.125 62.52 lbf in . 3 3 8.35 10 11.5 10 0.625 0.0625 4.88 lbf in . 3 3 5.86 62.52 4 GL cT A GL cT A GL cT A T T T T                         .88 73.3 lbf in .Ans  ns ns ns _____________________________________________________________________________ 3-55 (b) Solve part (b) first since the twist is needed for part (a).  max allow 12 6.89 82.7 MPa        6 max 1 9 max 82.7 10 0.348 rad/m . 79.3 10 (0.003) Ans Gc     (a)  9 331 1 1 1 0.348(79.3) 10 (0.020)(0.002 ) 1.47 N m . 3 3 GL cT A    ns     9 33 2 2 2 2 9 33 3 3 3 3 1 2 3 0.348(79.3) 10 (0.030)(0.003 ) 7.45 N m . 3 3 0.348(79.3) 10 (0.025)(0.002 ) 1.84 N m . 3 3 1.47 7.45 1.84 10.8 N m . GL cT A GL cT A T T T T Ans                   ns ns _____________________________________________________________________________ 3-56 (a) From Eq. (3-40), with two 2-mm strips,        6 22 max max 80 10 0.030 0.002 3.08 N m 3 1.8 / ( / ) 3 1.8 / 0.030 / 0.002 2(3.08) 6.16 N m . bcT b c T Ans           Chapter 3 - Rev. A, Page 42/100 • From the table on p. 102, with b/c = 30/2 = 15, and has a value between 0.313 and 0.333. From Eq. (3-40), 1 0.321 3 1.8 / (30 / 2)     From Eq. (3-41),     3 3 9 3.08(0.3) 0.151 rad . 0.321 0.030 0.002 79.3 10 6.16 40.8 N m . 0.151t Tl Ans bc G Tk Ans           From Eq. (3-40), with a single 4-mm strip,        6 22 max max 80 10 0.030 0.004 11.9 N m . 3 1.8 / ( / ) 3 1.8 / 0.030 / 0.004 bcT A b c        ns Interpolating from the table on p. 102, with b/c = 30/4 = 7.5, 7.5 6 (0.307 0.299) 0.299 0.305 8 6       From Eq. (3-41)     3 3 9 11.9(0.3) 0.0769 rad . 0.305 0.030 0.004 79.3 10 11.9 155 N m . 0.0769t Tl Ans bc G Tk Ans           (b) From Eq. (3-47), with two 2-mm strips,     2 62 max 0.030 0.002 80 10 3.20 N m 3 3 2(3.20) 6.40 N m . LcT T Ans              3 3 9 3 3(3.20)(0.3) 0.151 rad . 0.030 0.002 79.3 10 6.40 0.151 42.4 N m .t Tl Ans Lc G k T Ans          From Eq. (3-47), with a single 4-mm strip,     2 62 max 0.030 0.004 80 10 12.8 N m . 3 3 LcT A    ns Chapter 3 - Rev. A, Page 43/100 •     3 3 9 3 3(12.8)(0.3) 0.0757 rad . 0.030 0.004 79.3 10 12.8 0.0757 169 N m .t Tl Ans Lc G k T Ans          The results for the spring constants when using Eq. (3-47) are slightly larger than when using Eq. (3-40) and Eq. (3-41) because the strips are not infinitesimally thin (i.e. b/c does not equal infinity). The spring constants when considering one solid strip are significantly larger (almost four times larger) than when considering two thin strips because two thin strips would be able to slip along the center plane. _____________________________________________________________________________ 3-57 (a) Obtain the torque from the given power and speed using Eq. (3-44). (40000)9.55 9.55 152.8 N m 2500 HT n     max 3 16Tr T J d          1 3 1 3 6 max 16 152.816 0.0223 m 22.3 mm . 70 10 Td A                 ns (b) (40000)9.55 9.55 1528 N m 250 HT n        1 3 6 16(1528) 0.0481 m 48.1 mm . 70 10 d A            ns _____________________________________________________________________________ 3-58 (a) Obtain the torque from the given power and speed using Eq. (3-42). 63025 63025(50) 1261 lbf in 2500 HT n     max 3 16Tr T J d       1 31 3 max 16 126116 0.685 in . (20000) Td A                ns (b) 63025 63025(50) 12610 lbf in 250 HT n     1 3 16(12610) 1.48 in . (20000) d A         ns _____________________________________________________________________________ Chapter 3 - Rev. A, Page 44/100 • 3-59     6 33max max 3 50 10 0.0316 265 N m 16 16 dT T d           Eq. (3-44),  3265(2000) 55.5 10 W 55.5 kW .9.55 9.55 TnH A    ns _____________________________________________________________________________ 3-60         3 6 3 3 4 9 4 16 110 10 0.020 173 N m 16 16 0.020 79.3 10 15 180 32 32(173) 1.89 m . T T d d Tl d Gl JG T l Ans                        _____________________________________________________________________________ 3-61        3 3 3 4 4 6 16 30 000 0.75 2485 lbf in 16 16 32 32(2485)(24) 0.167 rad 9.57 . 0.75 11.5 10 T T d d Tl Tl Ans JG d G                    _____________________________________________________________________________ 3-62 (a) 4 4 max max max max solid hollow ( ) 16 16 o o o o J d J d dT T r d r d           4 i     44 solid hollow 4 4 solid 36 % (100%) (100%) (100%) 65.6% . 40 i o T T dT A T d       ns (b)  2 2solid hollow, o oW kd W k d d  2i     22 solid hollow 2 2 solid 36 % (100%) (100%) (100%) 81.0% . 40 i o W W dW A W d       ns _____________________________________________________________________________ 3-63 (a)  444 maxmax max max solid hollow 16 16 d xdJ d JT T r d r d            4 4solid hollow 4 soli ( )% (100%) (100%) (100%) . d T T xdT x T d      Ans Chapter 3 - Rev. A, Page 45/100 • (b)   22 2solid hollow W kd W k d xd    2 2solid hollow 2 solid % (100%) (100%) (100%) . xdW WW x W d      Ans Plot %T and %W versus x. The value of greatest difference in percent reduction of weight and torque is 25% and occurs at 2 2x  . _____________________________________________________________________________ 3-64 (a)          46 344 2.8149 104200 2 120 10 32 0.70 dTc J dd d              1 34 2 6 2.8149 10 6.17 10 m 61.7 mm 120(10 ) d            d From Table A-17, the next preferred size is d = 80 mm. Ans. i = 0.7d = 56 mm. The next preferred size smaller is di = 50 mm Ans. (b)              4 4 44 4200 2 4200 0.050 2 30.8 MPa . 32 0.080 0.05032 i i dTc Ans J d d              _____________________________________________________________________________ Chapter 3 - Rev. A, Page 46/100 • 3-65 (1500)9.55 9.55 1433 N m 10 HT n          1 3 1 3 3 6 16 143316 16 = 0.045 m 45 mm 80 10 C C T Td d                    From Table A-17, select 50 mm. Ans. (a)        6 start 3 16 2 1433 117 10 Pa 117 MPa . 0.050 Ans     (b) Design activity _____________________________________________________________________________ 3-66     1 31 3 3 63 025 63 025(1) 7880 lbf in 8 16 788016 16 = 1.39 in 15 000CC HT n T Td d                    From Table A-17, select 1.40 in. Ans. _____________________________________________________________________________ 3-67 For a square cross section with side length b, and a circular section with diameter d, 2 2 square circular 4 2 A A b d b d      From Eq. (3-40) with b = c,   3 max 2 3 3square 1.8 1.8 23 3 (4.8) 6.896 / 1 T T T bc b c b d d                       3 T For the circular cross section,  max 3 3circular 16 5.093T T d d         3max square max circular 3 6.896 1.354 5.093 T d T d     The shear stress in the square cross section is 35.4% greater. Ans. (b) For the square cross section, from the table on p. 102, β = 0.141. From Eq. (3-41), Chapter 3 - Rev. A, Page 47/100 • square 43 4 11.50 0.141 2 Tl Tl Tl Tl bc G b G d G d G               4 For the circular cross section,   44 10.19 32rd Tl Tl Tl GJ d GG d      4 4 11.50 1.129 10.19 sq rd Tl d G Tl d G     The angle of twist in the square cross section is 12.9% greater. Ans. _____________________________________________________________________________ 3-68 (a)        1 2 2 1 2 2 2 2 1 0.15 0 (500 75)(4) 5 1700 0.15 5 1700 4.25 0 400 lbf . 0.15 400 60 lbf . T T T T T T T T Ans T Ans                 T s (b) 0 575(10) 460(28) (40) 178.25 178 lbf . 0 575 460 178.25 293.25 lbf . O C C O O M R R An F R R Ans                 (c) Chapter 3 - Rev. A, Page 48/100 • (d) The maximum bending moment is at x = 10 in, and is M = 2932.5 lbf·in. Since the shaft rotates, each stress element will experience both positive and negative bending stress as it moves from tension to compression. The torque transmitted through the shaft from A to B is T = (500  75)(4) = 1700 lbf·in. For a stress element on the outer surface where the bending stress and the torsional stress are both maximum,  3 3 32 2932.532 15 294 psi = 15.3 kpsi . (1.25) Mc M Ans I d        3 3 16 16(1700) 4433 psi = 4.43 kpsi . (1.25) Tr T Ans J d        (e)         2 2 2 2 1 2 1 2 2 2 2 2 max 15.3 15.3, 4.43 2 2 2 2 16.5 kpsi . 1.19 kpsi . 15.3 4.43 8.84 kpsi . 2 2 x x xy x xy Ans Ans Ans                                       _____________________________________________________________________________ 3-69 (a)             2 1 3 2 1 1 1 3 1 1 2 0.15 0 1800 270 (200) (125) 306 10 125 0.15 306 10 106.25 0 2880 N . 0.15 2880 432 N . T T T T T T T Ans T Ans                 T T (b) 0 3312(230) (510) 2070(810) 1794 N . 0 3312 1794 2070 3036 N . O C C y O O M R R Ans F R R Ans               (c) Chapter 3 - Rev. A, Page 49/100 • (d) The maximum bending moment is at x = 230 mm, and is M = –698.3 N·m. Since the shaft rotates, each stress element will experience both positive and negative bending stress as it moves from tension to compression. The torque transmitted through the shaft from A to B is T = (1800  270)(0.200) = 306 N·m. For a stress element on the outer surface where the bending stress and the torsional stress are both maximum,    33 3 32 698.332 263 10 Pa 263 MPa . (0.030) Mc M Ans I d          63 316 16(306) 57.7 10 Pa 57.7MPa .(0.030) Tr T Ans J d         (e)         2 2 2 2 1 2 1 2 2 2 2 2 max 263 263, 57.7 2 2 2 2 275 MPa . 12.1 MPa . 263 57.7 144 MPa . 2 2 x x xy x xy Ans Ans Ans                                       _____________________________________________________________________________ 3-70 (a)         2 1 2 1 1 1 1 1 2 0.15 0 300 50 (4) (3) 1000 0.15 (3) 1000 2.55 0 392.16 lbf . 0.15 392.16 58.82 lbf . T T T T T T T Ans T Ans                 T T (b) Chapter 3 - Rev. A, Page 50/100 • 0 450.98(16) (22) 327.99 lbf . 0 450.98 327.99 122.99 lbf . 0 350(8) (22) 127.27 lbf . 0 350 127.27 222.73 lbf . O y C z C z z O z O z O z C y C y y O y O y M R R Ans F R R Ans M R R Ans F R R Ans                            Chapter 3 - Rev. A, Page 51/100 • (c) (d) Combine the bending moments from both planes at A and B to find the critical location. 2 2 2 2 (983.92) ( 1781.84) 2035 lbf in (1967.84) ( 763.65) 2111 lbf in A B M M           The critical location is at B. The torque transmitted through the shaft from A to B is T = (300  50)(4) = 1000 lbf·in. For a stress element on the outer surface where the bending stress and the torsional stress are both maximum,  3 3 32 211132 21502 psi = 21.5 kpsi . (1) Mc M Ans I d        3 3 16 16(1000) 5093 psi = 5.09 kpsi . (1) Tr T Ans J d        (e)         2 2 2 2 1 2 1 2 2 2 2 2 max 21.5 21.5, 5.09 2 2 2 2 22.6 kpsi . 1.14 kpsi . 21.5 5.09 11.9 kpsi . 2 2 x x xy x xy Ans Ans Ans                                       _____________________________________________________________________________ Chapter 3 - Rev. A, Page 52/100 • 3-71 (a)         2 1 2 1 1 1 1 1 2 0.15 0 300 45 (125) (150) 31 875 0.15 (150) 31 875 127.5 0 250 N mm . 0.15 250 37.5 N mm . T T T T T T T T Ans T Ans                   T (b) o o o o 0 345sin 45 (300) 287.5(700) (850) 150.7 N . 0 345cos 45 287.5 150.7 107.2 N . 0 345sin 45 (300) (850) 86.10 N . 0 345cos 45 86.10 O y C z C z z O z O z O z C y C y y O y O y M R R Ans F R R Ans M R R Ans F R R                          157.9 N .Ans  (c) ( d ) F r o m t h e b e n ding moment diagrams, it is clear that the critical location is at A where both planes have the maximum bending moment. Combining the bending moments from the two planes,    2 247.37 32.16 57.26 N mM       Chapter 3 - Rev. A, Page 53/100 • The torque transmitted through the shaft from A to B is T = (300  45)(0.125) = 31.88 N·m. For a stress element on the outer surface where the bending stress and the torsional stress are both maximum,    63 3 32 57.2632 72.9 10 Pa 72.9 MPa . (0.020) Mc M Ans I d          63 316 16(31.88) 20.3 10 Pa 20.3 MPa .(0.020) Tr T Ans J d         (e)         2 2 2 2 1 2 1 2 2 2 2 2 max 72.9 72.9, 20.3 2 2 2 2 78.2 MPa . 5.27 MPa . 72.9 20.3 41.7 MPa . 2 2 x x xy x xy Ans Ans Ans                                       _____________________________________________________________________________ 3-72 (a) 0 300(cos 20º )(10) (cos 20º )(4) 750 lbf . B B T F F Ans       (b) 0 300(cos 20º )(16) 750(sin 20º )(39) (30) 183 lbf . 0 300(cos 20º ) 183 750(sin 20º ) 208 lbf . 0 300(sin 20º )(16) (30) 750(cos 20º )(39) 861 lbf . 0 300 O z C y C y y O y O y O y C z C z z O z M R R Ans F R R Ans M R R Ans F R                          (sin 20º ) 861 750(cos 20º ) 259 lbf .O zR Ans    Chapter 3 - Rev. A, Page 54/100 • (c) (d) Combine the bending moments from both planes at A and C to find the critical location. 2 2 2 2 ( 3336) ( 4149) 5324 lbf in ( 2308) ( 6343) 6750 lbf in A C M M             The critical location is at C. The torque transmitted through the shaft from A to B is . For a stress element on the outer surface where the bending stress and the torsional stress are both maximum,   300cos 20º 10 2819 lbf inT     3 3 32 675032 35 203 psi = 35.2 kpsi . (1.25) Mc M Ans I d        3 3 16 16(2819) 7351 psi = 7.35 kpsi . (1.25) Tr T Ans J d        (e)         2 2 2 2 1 2 1 2 2 2 2 2 max 35.2 35.2, 7.35 2 2 2 2 36.7 kpsi . 1.47 kpsi . 35.2 7.35 19.1 kpsi . 2 2 x x xy x xy Ans Ans Ans                                       _____________________________________________________________________________ Chapter 3 - Rev. A, Page 55/100 • 3-73 (a) 0 11000(cos 20º )(300) (cos 25º )(150) 22 810 N . B B T F F Ans      (b) 0 11 000(sin 20º )(400) 22 810(sin 25º )(750) (1050) 8319 N . O z C y C y M R R Ans        0 11000(sin 20º ) 22 810sin(25º ) 8319 5083 N . 0 11 000(cos 20º )(400) 22 810(cos 25º )(750) (1050) 10 830 N . 0 11 000(cos 20º ) 22 810(cos 25º ) 10 830 494 N . y O y O y O y C z C z z O z O z F R R Ans M R R Ans F R R Ans                      (c) (d) From the bending moment diagrams, it is clear that the critical location is at B where both planes have the maximum bending moment. Combining the bending moments from the two planes,    2 22496 3249 4097 N mM      The torque transmitted through the shaft from A to B is .   11000cos 20º 0.3 3101 N mT   For a stress element on the outer surface where the bending stress and the torsional stress are both maximum, Chapter 3 - Rev. A, Page 56/100 •    63 3 32 409732 333.9 10 Pa 333.9 MPa . (0.050) Mc M Ans I d          63 316 16(3101) 126.3 10 Pa 126.3 MPa .(0.050) Tr T Ans J d         (e)         2 2 2 2 1 2 1 2 2 2 2 2 max 333.9 333.9, 126.3 2 2 2 2 376 MPa . 42.4 MPa . 333.9 126.3 209 MPa . 2 2 x x xy x xy Ans Ans Ans                                       _____________________________________________________________________________ 3-74 (a)   6.13 3.8(92.8) 3.88(362.8) 0D xzM C     287.2 lbf .xC A ns ns   6.13 2.33(92.8) 3.88(362.8) 0C xzM D     194.4 lbf .xD A   3.80 (808) 500.9 lbf . 6.13D zx M C Ans       2.330 (808) 307.1 lbf . 6.13C zx M D A     ns (b) For DQC, let x, y, z correspond to the original  y, x, z axes. Chapter 3 - Rev. A, Page 57/100 • (c) The critical stress element is just to the right of Q, where the bending moment in both planes is maximum, and where the torsional and axial loads exist. 808(3.88) 3135 lbf inT    2 2669.2 1167 1345 lbf inM      3 3 16 16(3135) 11 070 psi . 1.13 T Ans d        3 3 32 32(1345) 9495 psi . 1.13b M Ans d           2 362.8 362 psi . ( / 4) 1.13a F Ans A         (d) The critical stress element will be where the bending stress and axial stress are both in compression. max 9495 362 9857 psi      2 2 max 9857 11 070 12118 psi 12.1 kpsi . 2 Ans         2 2 1 2 9857 9857, 11 070 2 2           1 7189 psi 7.19 kpsi .Ans   2 17 046 psi 17.0 kpsi .Ans     _____________________________________________________________________________ 3-75 (a)   0 6.13 3.8(46.6) 3.88(140) 0 D z x M C      ns  ns 117.5 lbf .xC A   0 6.13 2.33(46.6) 3.88(140) 0 C z x M D      70.9 lbf .xD A   3.80 (406) 251.7 lbf . 6.13D zx M C A       ns 2.330 (406) 154.3 lbf . 6.13C zx M D A     ns Chapter 3 - Rev. A, Page 58/100 • (b) For DQC, let x, y, z correspond to the original  y, x, z axes. (c) The critical stress element is just to the right of Q, where the bending moment in both planes is maximum, and where the torsional and axial loads exist. 406(3.88) 1575 lbf inT    2 2273.8 586.3 647.1 lbf inM      3 3 16 16(1575) 8021 psi . 1 T Ans d        3 3 32 32(647.1) 6591 psi . 1b M Ans d           2 140 178.3 psi . ( / 4) 1a F Ans A         (d) The critical stress element will be where the bending stress and axial stress are both in compression. max 6591 178.3 6769 psi      2 2 max 6769 8021 8706 psi 8.71 kpsi . 2 Ans         2 2 1 2 6769 6769, 8021 2 2           Chapter 3 - Rev. A, Page 59/100 • 1 5321 psi 5.32 kpsi .Ans   2 12090 psi 12.1 kpsi .Ans     _____________________________________________________________________________ 3-76   5.62(362.8) 1.3(92.8) 3 0B yzM A      639.4 lbfyA  Ans.   2.62(362.8) 1.3(92.8) 3 0A yzM B      276.6 lbfyB  Ans.   5.620 (808) 1513.7 lbf 3B zy M A     Ans.   2.620 (808) 705.7 lbf 3A zy M B     Ans. (b) (c) The critical stress element is just to the left of A, where the bending moment in both planes is maximum, and where the torsional and axial loads exist. Chapter 3 - Rev. A, Page 60/100 • 808(1.3) 1050 lbf inT     3 16(1050) 7847 psi . 0.88 Ans    2 2(829.8) (2117) 2274 lbf inM      3 3 32 32(2274) 33 990 psi . 0.88b M Ans d           2 92.8 153 psi . ( / 4) 0.88a F Ans A         (d) The critical stress will occur when the bending stress and axial stress are both in compression. max 33 990 153 34143 psi      2 2 max 34143 7847 18 789 psi 18.8 kpsi . 2 Ans         2 2 1 2 34143 34143, 7847 2 2           1 1717 psi 1.72 kpsi .Ans   2 35 860 psi 35.9 kpsi .Ans     _____________________________________________________________________________ 3-77 100 1600 N / 2 0.125 / 2t TF c            1600 tan 20 582.4 N 2 1600 0.250 2 200 N m 200 2667 N 2 0.150 2 n C t C F T F b TP a           0 450 582.4(325) 2667(75) 0 865.1 N A z Dy Dy M R R          0 450 1600(325)A DzyM R    0 865.1 582.4 2667y AyF R     0 1156 1600 z AzF R     1156 NDzR  2384 NAyR  444 NAzR  Chapter 3 - Rev. A, Page 61/100 • AB The maximum bending moment will either be at B or C. If this is not obvious, sketch the shear and bending moment diagrams. We will directly obtain the combined moments from each plane. 2 2 2 2 2 2 2 2 0.075 2384 444 181.9 N m 0.125 865.1 1156 180.5 N m y z y z B A A C D D M AB R R M CD R R             The stresses at B and C are almost identical, but the maximum stresses occur at B. Ans.         6 3 3 6 3 3 32 32(181.9) 68.6 10 Pa 68.6 MPa 0.030 16 16(200) 37.7 10 Pa 37.7 MPa 0.030 B B B B M d T d               2 2 2 2 max 68.6 68.6 37.7 85.3 MPa . 2 2 2 2 B B B Ans                  2 2 2 2 max 68.6 37.7 51.0 MPa . 2 2 B B Ans               _____________________________________________________________________________ 3-78 100 1600 N / 2 0.125 / 2t TF c            1600 tan 20 582.4 N 2 1600 0.250 2 200 N m 200 2667 N 2 0.150 2 n C t C F T F b TP a              0 450 582.4(325) 420.6 N 0 450 1600(325) 2667(75) 711.1 N 0 420.6 582.4 161.8 N 0 711.1 1600 2667 A Dy Dyz A Dz Dzy y Ay Ay z Az M R R M R R F R R F R                            1778 NAzR   Chapter 3 - Rev. A, Page 62/100 • The maximum bending moment will either be at B or C. If this is not obvious, sketch shear and bending moment diagrams. We will directly obtain the combined moments from each plane.  22 2 2 2 2 2 2 0.075 161.8 1778 133.9 N m 0.125 420.6 711.1 103.3 N m y z y z B A A C D D M AB R R M CD R R              The maximum stresses occur at B. Ans.         6 3 3 6 3 3 32 32(133.9) 50.5 10 Pa 50.5 MPa 0.030 16 16(200) 37.7 10 Pa 37.7 MPa 0.030 B B B B M d T d               2 2 2 2 max 50.5 50.5 37.7 70.6 MPa . 2 2 2 2 B B B Ans                  2 2 2 2 max 50.5 37.7 45.4 MPa . 2 2 B B Ans               _____________________________________________________________________________ 3-79 900 180 lbf / 2 10 / 2t TF c            180 tan 20 65.5 lbf 2 180 5 2 450 lbf in 450 150 lbf 2 6 2 n C t C F T F b TP a              0 20 65.5(14) 150(4) 75.9 lbf 0 20 180(14) 126 lbf 0 75.9 65.5 150 140 lbf 0 126 180 A Dy Dyz A Dz Dzy y Ay Ay z Az M R R M R R F R R F R                            54.0 lbfAzR  Chapter 3 - Rev. A, Page 63/100 • The maximum bending moment will either be at B or C. If this is not obvious, sketch shear and bending moment diagrams. We will directly obtain the combined moments from each plane. 2 2 2 2 2 2 2 2 4 140 54 600 lbf in 6 75.9 126 883 lbf in y z y z B A A C D D M AB R R M CD R R             The maximum stresses occur at C. Ans.     3 3 3 3 32 32(883) 3460 psi 1.375 16 16(450) 882 psi 1.375 C C C C M d T d             2 2 2 2 max 3460 3460 882 3670 psi . 2 2 2 2 C C C Ans                  2 2 2 2 max 3460 882 1940 psi . 2 2 C C Ans               _____________________________________________________________________________ 3-80 (a) Rod AB experiences constant torsion throughout its length, and maximum bending moment at the wall. Both torsional shear stress and bending stress will be maximum on the outer surface. The transverse shear will be very small compared to bending and torsion, due to the reasonably high length to diameter ratio, so it will not dominate the determination of the critical location. The critical stress element will be at the wall, at either the top (compression) or the bottom (tension) on the y axis. We will select the bottom element for this analysis. (b) Transverse shear is zero at the critical stress elements on the top and bottom surfaces.       34 3 / 2 32 8 20032 16 297 psi 16.3 kpsi / 64 1x M dMc M I d d                 34 3 / 2 16 5 20016 5093 psi 5.09 kpsi / 32 1xz T dTr T J d d           Chapter 3 - Rev. A, Page 64/100 • (c)         2 2 2 2 1 2 1 2 2 2 2 2 max 16.3 16.3, 5.09 2 2 2 2 17.8 kpsi . 1.46 kpsi . 16.3 5.09 9.61 kpsi . 2 2 x x xz x xz Ans Ans Ans                                       _____________________________________________________________________________ 3-81 (a) Rod AB experiences constant torsion throughout its length, and maximum bending moments at the wall in both planes of bending. Both torsional shear stress and bending stress will be maximum on the outer surface. The transverse shear will be very small compared to bending and torsion, due to the reasonably high length to diameter ratio, so it will not dominate the determination of the critical location. The critical stress element will be on the outer surface at the wall, with its critical location determined by the plane of the combined bending moments. My = – (100)(8) = – 800 lbf·in Mz = (175)(8) = 1400 lbf·in   2 2 tot 2 2 1 1 800 1400 1612 lbf in 800= tan tan 29.7º 1400 y z y z M M M M M                       The combined bending moment vector is at an angle of 29.7º CCW from the z axis. The critical bending stress location, and thus the critical stress element, will be ±90º from this vector, as shown. There are two equally critical stress elements, one in tension (119.7º CCW from the z axis) and the other in compression (60.3º CW from the z axis). We’ll continue the analysis with the element in tension. (b) Transverse shear is zero at the critical stress elements on the outer surfaces.       tottot tot 34 3 / 2 32 161232 16 420 psi 16.4 kpsi / 64 1x M dM c M I d d                 34 3 / 2 16 5 17516 4456 psi 4.46 kpsi / 32 1 T dTr T J d d           Chapter 3 - Rev. A, Page 65/100 • (c)     2 2 22 1 2 1 2 2 2 22 max 16.4 16.4, 4.46 2 2 2 2 17.5 kpsi . 1.13 kpsi . 16.4 4.46 9.33 kpsi . 2 2 x x x Ans Ans Ans                                       _____________________________________________________________________________ 3-82 (a) Rod AB experiences constant torsion and constant axial tension throughout its length, and maximum bending moments at the wall from both planes of bending. Both torsional shear stress and bending stress will be maximum on the outer surface. The transverse shear will be very small compared to bending and torsion, due to the reasonably high length to diameter ratio, so it will not dominate the determination of the critical location. The critical stress element will be on the outer surface at the wall, with its critical location determined by the plane of the combined bending moments. My = – (100)(8) – (75)(5) = – 1175 lbf·in Mz = (–200)(8) = –1600 lbf·in     2 2 tot 2 2 1 1 1175 1600 1985 lbf in 1175= tan tan 36.3º 1600 y z y z M M M M M                        The combined bending moment vector is at an angle of 36.3º CW from the negative z axis. The critical bending stress location will be ±90º from this vector, as shown. Since there is an axial stress in tension, the critical stress element will be where the bending is also in tension. The critical stress element is therefore on the outer surface at the wall, at an angle of 36.3º CW from the y axis. (b) Transverse shear is zero at the critical stress element on the outer surface.       tottot tot ,bend 34 3 / 2 32 198532 20220 psi 20.2 kpsi / 64 1x M dM c M I d d            ,axial 22 75 95.5 psi 0.1 kpsi / 4 1 / 4 x x x F F A d         , which is essentially negligible ,axial ,bend 20 220 95.5 20 316 psi 20.3 kpsix x x            33 16 5 20016 5093 psi 5.09 kpsi 1 Tr T J d         Chapter 3 - Rev. A, Page 66/100 • (c)     2 2 22 1 2 1 2 2 2 22 max 20.3 20.3, 5.09 2 2 2 2 21.5 kpsi . 1.20 kpsi . 20.3 5.09 11.4 kpsi . 2 2 x x x Ans Ans Ans                                       _____________________________________________________________________________ 3-83 T = (2)(200) = 400 lbf·in The maximum shear stress due to torsion occurs in the middle of the longest side of the rectangular cross section. From the table on p. 102, with b/c = 1.5/0.25 = 6,  = 0.299. From Eq. (3-40),     max 22 400 14 270 psi 14.3 kpsi . 0.299 1.5 0.25 T Ans bc       ____________________________________________________________________________ 3-84 (a) The cross section at A will experience bending, torsion, and transverse shear. Both torsional shear stress and bending stress will be maximum on the outer surface. The transverse shear will be very small compared to bending and torsion, due to the reasonably high length to diameter ratio, so it will not dominate the determination of the critical location. The critical stress element will be at either the top (compression) or the bottom (tension) on the y axis. We’ll select the bottom element for this analysis. (b) Transverse shear is zero at the critical stress elements on the top and bottom surfaces.       34 3 / 2 32 11 25032 28 011 psi 28.0 kpsi / 64 1x M dMc M I d d                 34 3 / 2 16 12 25016 15 279 psi 15.3 kpsi / 32 1xz T dTr T J d d           Chapter 3 - Rev. A, Page 67/100 • (c)         2 2 2 2 1 2 1 2 2 2 2 2 max 28.0 28.0, 15.3 2 2 2 2 34.7 kpsi . 6.7 kpsi . 28.0 15.3 20.7 kpsi . 2 2 x x xz x xz Ans Ans Ans                                       ____________________________________________________________________________ 3-85 (a) The cross section at A will experience bending, torsion, axial, and transverse shear. Both torsional shear stress and bending stress will be maximum on the outer surface. The transverse shear will be very small compared to bending and torsion, due to the reasonably high length to diameter ratio, so it will not dominate the determination of the critical location. The critical stress element will be on the outer surface, with its critical location determined by the plane of the combined bending moments. My = (300)(12) = 3600 lbf·in Mz = (250)(11) = 2750 lbf·in     2 2 tot 2 2 1 1 3600 2750 4530 lbf in 2750= tan tan 37.4º 3600 y z z y M M M M M                      The combined bending moment vector is at an angle of 37.4º CCW from the y axis. The critical bending stress location will be 90º CCW from this vector, where the tensile bending stress is additive with the tensile axial stress. The critical stress element is therefore on the outer surface, at an angle of 37.4º CCW from the z axis. (b)       tottot tot ,bend 34 3 / 2 32 453032 46142 psi 46.1 kpsi / 64 1x M dM c M I d d            ,axial 22 300 382 psi 0.382 kpsi / 4 1 / 4 x x x F F A d         ,axial ,bend 46142 382 46 524 psi 46.5 kpsix x x             33 16 12 25016 15 279 psi 15.3 kpsi 1 Tr T J d         Chapter 3 - Rev. A, Page 68/100 • (c)     2 2 22 1 2 1 2 2 2 22 max 46.5 46.5, 15.3 2 2 2 2 51.1 kpsi . 4.58 kpsi . 46.5 15.3 27.8 kpsi . 2 2 x x x Ans Ans Ans                                       ____________________________________________________________________________ 3-86 (a) The cross section at A will experience bending, torsion, axial, and transverse shear. Both torsional shear stress and bending stress will be maximum on the outer surface. The transverse shear will be very small compared to bending and torsion, due to the reasonably high length to diameter ratio, so it will not dominate the determination of the critical location. The critical stress element will be on the outer surface, with its critical location determined by the plane of the combined bending moments. My = (300)(12) – (–100)(11) = 4700 lbf·in Mz = (250)(11) = 2750 lbf·in     2 2 tot 2 2 1 1 4700 2750 5445 lbf in 2750= tan tan 30.3º 4700 y z z y M M M M M                      The combined bending moment vector is at an angle of 30.3º CCW from the y axis. The critical bending stress location will be 90º CCW from this vector, where the tensile bending stress is additive with the tensile axial stress. The critical stress element is therefore on the outer surface, at an angle of 30.3º CCW from the z axis. (b)       tottot tot ,bend 34 3 / 2 32 544532 55 462 psi 55.5 kpsi / 64 1x M dM c M I d d           Chapter 3 - Rev. A, Page 69/100 •  ,axial 22 300 382 psi 0.382 kpsi / 4 1 / 4 x x x F F A d         ,axial ,bend 55 462 382 55 844 psi 55.8 kpsix x x             33 16 12 25016 15 279 psi 15.3 kpsi 1 Tr T J d         (c)     2 2 22 1 2 1 2 2 2 22 max 55.8 55.8, 15.3 2 2 2 2 59.7 kpsi . 3.92 kpsi . 55.8 15.3 31.8 kpsi . 2 2 x x x Ans Ans Ans                                       ____________________________________________________________________________ 3-87 (a) The cross section at A will experience bending, torsion, and transverse shear. Both torsional shear stress and bending stress will be maximum on the outer surface, where the stress concentration will also be applicable. The transverse shear will be very small compared to bending and torsion, due to the reasonably high length to diameter ratio, so it will not dominate the determination of the critical location. The critical stress element will be at either the top (compression) or the bottom (tension) on the y axis. We’ll select the bottom element for this analysis. (b) Transverse shear is zero at the critical stress elements on the top and bottom surfaces. / 0.125 /1 0.125 / 1.5 /1 1.5 r d D d     Fig. A-15-8 ,torsion 1.39tK  Fig. A-15-9 ,bend 1.59tK      ,bend ,bend 33 32 11 25032 (1.59) 44 538 psi 44.5 kpsi 1x t t Mc MK K I d             ,torsion ,torsion 33 16 12 25016 (1.39) 21 238 psi 21.2 kpsi 1xz t t Tr TK K J d         Chapter 3 - Rev. A, Page 70/100 • (c)         2 2 2 2 1 2 1 2 2 2 2 2 max 44.5 44.5, 21.2 2 2 2 2 53.0 kpsi . 8.48 kpsi . 44.5 21.2 30.7 kpsi . 2 2 x x xz x xz Ans Ans Ans                                       ____________________________________________________________________________ 3-88 (a) The cross section at A will experience bending, torsion, axial, and transverse shear. Both torsional shear stress and bending stress will be maximum on the outer surface, where the stress concentration will also be applicable. The transverse shear will be very small compared to bending and torsion, due to the reasonably high length to diameter ratio, so it will not dominate the determination of the critical location. The critical stress element will be on the outer surface, with its critical location determined by the plane of the combined bending moments. My = (300)(12) = 3600 lbf·in Mz = (250)(11) = 2750 lbf·in     2 2 tot 2 2 1 1 3600 2750 4530 lbf in 2750= tan tan 37.4º 3600 y z z y M M M M M                      The combined bending moment vector is at an angle of 37.4º CCW from the y axis. The critical bending stress location will be 90º CCW from this vector, where the tensile bending stress is additive with the tensile axial stress. The critical stress element is therefore on the outer surface, at an angle of 37.4º CCW from the z axis. (b) / 0.125 /1 0.125 / 1.5 /1 1.5 r d D d     Fig. A-15-7 , 1.75t axialK  Fig. A-15-8 ,torsion 1.39tK  Fig. A-15-9 ,bend 1.59tK  Chapter 3 - Rev. A, Page 71/100 •    ,bend ,bend ,bend 33 32 453032 (1.59) 73 366 psi 73.4 kpsi 1x t t Mc MK K I d            ,axial ,axial 2 3001.75 668 psi 0.668 kpsi 1 / 4 x x t FK A       ,axial ,bend 73 366 668 74 034 psi 74.0 kpsix x x            ,torsion ,torsion 33 16 12 25016 (1.39) 21 238 psi 21.2 kpsi 1t t Tr TK K J d         (c)     2 2 22 1 2 1 2 2 2 22 max 74.0 74.0, 21.2 2 2 2 2 79.6 kpsi . 5.64 kpsi . 74.0 21.2 42.6 kpsi . 2 2 x x x Ans Ans Ans                                       ____________________________________________________________________________ 3-89 (a) The cross section at A will experience bending, torsion, axial, and transverse shear. Both torsional shear stress and bending stress will be maximum on the outer surface, where the stress concentration is also applicable. The transverse shear will be very small compared to bending and torsion, due to the reasonably high length to diameter ratio, so it will not dominate the determination of the critical location. The critical stress element will be on the outer surface, with its critical location determined by the plane of the combined bending moments. My = (300)(12) – (–100)(11) = 4700 lbf·in Mz = (250)(11) = 2750 lbf·in     2 2 tot 2 24700 2750 5445 lbf in y zM M M      1 1 2750= tan tan 30.3º 4700 z y M M                Chapter 3 - Rev. A, Page 72/100 • The combined bending moment vector is at an angle of 30.3º CCW from the y axis. The critical bending stress location will be 90º CCW from this vector, where the tensile bending stress is additive with the tensile axial stress. The critical stress element is therefore on the outer surface, at an angle of 30.3º CCW from the z axis. (b) / 0.125 /1 0.125 / 1.5 /1 1.5 r d D d     Fig. A-15-7 , 1.75t axialK  Fig. A-15-8 ,torsion 1.39tK  Fig. A-15-9 ,bend 1.59tK     ,bend ,bend ,bend 33 32 544532 (1.59) 88185 psi 88.2 kpsi 1x t t Mc MK K I d            ,axial ,axial 2 3001.75 668 psi 0.668 kpsi 1 / 4 x x t FK A       ,axial ,bend 88185 668 88 853 psi 88.9 kpsix x x            ,torsion ,torsion 33 16 12 25016 (1.39) 21 238 psi 21.2 kpsi 1t t Tr TK K J d         (c)     2 2 22 1 2 1 2 2 2 22 max 88.9 88.9, 21.2 2 2 2 2 93.7 kpsi . 4.80 kpsi . 88.9 21.2 49.2 kpsi . 2 2 x x x Ans Ans Ans                                       ____________________________________________________________________________ 3-90 (a) M = F(p / 4), c = p / 4, I = bh3 / 12, b =  dr nt, h = p / 2 Chapter 3 - Rev. A, Page 73/100 •        2 33 / 4 / 4 /12 16 / 2 /12 6 . b r t b r t F p pMc Fp I bh d n p F Ans d n p               (b) 2 2 4 / 4a r r F F F Ans. A d d           4 3 / 2 16 . / 32 r t r r T dTr T Ans J d d       (c) The bending stress causes compression in the x direction. The axial stress causes compression in the y direction. The torsional stress shears across the y face in the negative z direction. (d) Analyze the stress element from part (c) using the equations developed in parts (a) and (b).               2 2 3 3 1.5 0.25 1.25 in 6 15006 4584 psi = 4.584 kpsi 1.25 2 0.25 4 15004 = = 1222 psi = 1.222 kpsi 1.25 16 23516 = = 612.8 psi = 0.6128 kpsi 1.25 r x b r t y a r yz t r d d p F d n p F d T d                                       Use Eq. (3-15) for the three-dimensional stress element.          2 23 2 3 2 4.584 1.222 4.584 1.222 0.6128 4.584 0.6128 0 5.806 5.226 1.721 0                              The roots are at 0.2543, – 4.584, and –1.476. Thus, the ordered principal stresses are 1 = 0.2543 kpsi, 2 = –1.476 kpsi, and 3 = – 4.584 kpsi. Ans. From Eq. (3-16), the principal shear stresses are Chapter 3 - Rev. A, Page 74/100 •         1 2 1/2 2 3 2/3 1 3 1/3 0.2543 1.476 0.8652 kpsi . 2 2 1.476 4.584 1.554 kpsi . 2 2 0.2543 4.584 2.419 kpsi . 2 2 Ans Ans Ans                       ____________________________________________________________________________ 3-91 As shown in Fig. 3-32, the maximum stresses occur at the inside fiber where r = ri. Therefore, from Eq. (3-50) 2 2 ,max 2 2 2 2 2 2 2 2 2 ,max 2 2 2 1 . 1 . i i o t o i i o i i o i i i o r i o i i r p r r r r r rp Ans r r r p r p Ans r r r                            ______________________________________________________________________________ 3-92 If pi = 0, Eq. (3-49) becomes 2 2 2 2 2 2 2 2 2 2 2 / 1 o o i o o t o i o o i o i p r r r p r r r p r r r r r             The maximum tangential stress occurs at r = ri. So 2 ,max 2 2 2 .o ot o i p r Ans r r     For σr, we have 2 2 2 2 2 2 2 2 2 2 2 / 1 o o i o o r o i o o i o i p r r r p r r r p r r r r r            So σr = 0 at r = ri. Thus at r = ro 2 2 2 ,max 2 2 2 . o o i o r o o i o p r r r p Ans r r r          ______________________________________________________________________________ Chapter 3 - Rev. A, Page 75/100 • 3-93 The force due to the pressure on half of the sphere is resisted by the stress that is distributed around the center plane of the sphere. All planes are the same, so   2 av 1 2 / 4 ( ) . 4 i i t i p d pd Ans d t t          The radial stress on the inner surface of the shell is, 3 =  p Ans. ______________________________________________________________________________ 3-94 σt > σl > σr τmax = (σt − σr)/2 at r = ri 2 2 2 2 2 max 2 2 2 2 2 2 2 2 2 2 2 2 max2 2 1 1 1 2 3 2.75 (10 000) 1597 psi . 3 i i o i i o o i o i i o i i o i o i i o r p r r p r r p r r r r r r r r r rp Ans r                             ______________________________________________________________________________ 3-95 σt > σl > σr τmax = (σt − σr)/2 at r = ri   2 2 2 2 2 2 2 max 2 2 2 2 2 2 2 2 2 2 2 6 max 6 max 1 1 1 2 ( ) (25 4)10100 91.7 mm 25 10 100 91.7 8.3 mm . i i o i i o i i o o i o i i o i i o i i o i i i o o i r p r r p r r p r r p r r r r r r r r r r r pr r t r r Ans                                           ______________________________________________________________________________ 3-96 σt > σl > σr τmax = (σt − σr)/2 at r = ri 2 2 2 2 2 2 2 max 2 2 2 2 2 2 2 2 2 2 2 1 1 1 2 i i o i i o i i o o i o i i o i i o i i o i r p r r p r r p r r p r r r r r r r r r r r                              2 2 2 4 (500) 4129 psi . 4 3.75 Ans   ______________________________________________________________________________ 3-97 From Eq. (3-49) with pi = 0, Chapter 3 - Rev. A, Page 76/100 • 2 2 2 2 2 2 2 2 2 2 1 1 o o i t o i o o i r o i r p r r r r r p r r r r                   σt > σl > σr, and since σt and σr are negative, τmax = (σr − σt)/2 at r = ro 2 2 2 2 2 2 2 max 2 2 2 2 2 2 2 2 2 2 2 2 2 2 max2 2 1 1 1 2 3 2.75 (10 000) 1900 psi . 2.75 o o i o o i o o i i o o i o o i o o i o o i o i o i r p r r p r r p r r p r r r r r r r r r r r r rp Ans r                                     2 ______________________________________________________________________________ 3-98 From Eq. (3-49) with pi = 0, 2 2 2 2 2 2 2 2 2 2 1 1 o o i t o i o o i r o i r p r r r r r p r r r r                   σt > σl > σr, and since σt and σr are negative, τmax = (σr − σt)/2 at r = ro     2 2 2 2 2 2 2 max 2 2 2 2 2 2 2 2 2 2 6 max 6 max 1 1 1 2 25 10 100 92.8 mm ( ) 25 4 10 100 92.8 7.2 mm . o o i o o i o o i i o o i o o i o o i o o i i o o o i r p r r p r r p r r p r r r r r r r r r r r r r p t r r Ans                                           2 ______________________________________________________________________________ 3-99 From Eq. (3-49) with pi = 0, 2 2 2 2 2 2 2 2 2 2 1 1 o o i t o i o o i r o i r p r r r r r p r r r r                   σt > σl > σr, and since σt and σr are negative, τmax = (σr − σt)/2 at r = ro Chapter 3 - Rev. A, Page 77/100 • 2 2 2 2 2 2 2 max 2 2 2 2 2 2 2 2 2 2 2 2 2 1 1 1 2 3.75 (500) 3629 psi . 4 3.75 o o i o o i o o i i o o i o o i o o i o o i r p r r p r r p r r p r r r r r r r r r r r Ans                                 2 ______________________________________________________________________________ 3-100 From Table A-20, Sy=490 MPa From Eq. (3-49) with pi = 0, 2 2 2 2 21 o o i t o i r p r r r r          Maximum will occur at r = ri   2 22 22 ,max ,max 2 2 2 2 0.8( 490) 25 19( )2 82.8 MPa . 2 2(25 ) t o io o t o o i o r rr p p Ans r r r              ______________________________________________________________________________ 3-101 From Table A-20, Sy = 71 kpsi From Eq. (3-49) with pi = 0, 2 2 2 2 21 o o i t o i r p r r r r          Maximum will occur at r = ri     2 2 2 22 ,max ,max 2 2 2 2 0.8( 71) 1 0.752 12.4 kpsi . 2 2(1 ) t o io o t o o i o r rr p p Ans r r r               ______________________________________________________________________________ 3-102 From Table A-20, Sy=490 MPa From Eq. (3-50) 2 2 2 2 21 i i o t o i r p r r r r         Maximum will occur at r = ri     2 22 2 ,max 2 2 2 2 2 2 2 2 2 ,max 2 2 2 2 1 ( ) 0.8(490) (25 19 ) 105 MPa . (25 19 ) i o ii i o t o i i o i t o i i o i p r rr p r r r r r r r r p Ans r r                   ______________________________________________________________________________ Chapter 3 - Rev. A, Page 78/100 • 3-103 From Table A-20, Sy =71 MPa From Eq. (3-50) 2 2 2 2 21 i i o t o i r p r r r r         Maximum will occur at r = ri   2 2 2 2 ,max 2 2 2 2 2 2 2 2 2 ,max 2 2 2 2 ( )1 ( ) 0.8(71) (1 0.75 ) 15.9 ksi . (1 0.75 ) i i o i o i t o i i o i t o i i o i r p r p r r r r r r r r r p Ans r r                    ______________________________________________________________________________ 3-104 The longitudinal stress will be due to the weight of the vessel above the maximum stress point. From Table A-5, the unit weight of steel is s = 0.282 lbf/in3. The area of the wall is Awall = ( /4)(3602  358.52) = 846. 5 in2 The volume of the wall and dome are Vwall = Awall h = 846.5 (720) = 609.5 (103) in3 Vdome = (2 /3)(1803  179.253) = 152.0 (103) in3 The weight of the structure on the wall area at the tank bottom is W = s Vtotal = 0.282(609.5 +152.0) (103) = 214.7(103) lbf  3 wall 214.7 10 254 psi 846.5l W A        The maximum pressure will occur at the bottom of the tank, pi = water h. From Eq. (3-50) with ir r 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 1 ft 180 179.2562.4(55) 5708 5710 psi . 144 in 180 179.25 i i o o i t i o i i o i r p r r rp r r r r r Ans                                2 2 2 2 2 2 2 1 ft1 62.4(55) 23.8 psi . 144 in i i o r i o i i r p r p Ans r r r                    Note: These stresses are very idealized as the floor of the tank will restrict the values calculated. Chapter 3 - Rev. A, Page 79/100 • Since 1  2  3, 1 = t = 5708 psi, 2 = r =  24 psi and3 = l =  254 psi. From Eq. (3-16), 1 3 1 2 2 3 5708 254 2981 2980 psi 2 5708 24 2866 2870 psi . 2 24 254 115 psi 2 Ans                ______________________________________________________________________________ 3-105 Stresses from additional pressure are, Eq. (3-51),     2 2 250psi 50 179.25 5963 psi 180 179.25l     (r)50 psi =  50 psi Eq. (3-50)   2 2 2 250psi 180 179.2550 11 975 psi 180 179.25t     Adding these to the stresses found in Prob. 3-104 gives t = 5708 + 11 975 = 17683 psi = 17.7 kpsi Ans. r =  23.8  50 =  73.8 psi Ans. l =  254 + 5963 = 5709 psi Ans. Note: These stresses are very idealized as the floor of the tank will restrict the values calculated. From Eq. (3-16) 1 3 1 2 2 3 17 683 73.8 8879 psi 2 17 683 5709 5987 psi . 2 5709 23.8 2866 psi 2 Ans             ______________________________________________________________________________ 3-106 Since σt and σr are both positive and σt > σr  max max 2t  From Eq. (3-55), t is maximum at r = ri = 0.3125 in. The term Chapter 3 - Rev. A, Page 80/100 •   22 2 50003 0.282 3 0.292 82.42 lbf/in 8 386 60 8                         2 2 2 2 2 2max 0.3125 2.75 1 3(0.292)82.42 0.3125 2.75 0.3125 3 0.2920.3125 1260 psi t            max 1260 630 psi . 2 Ans   Radial stress: 2 2 2 2 2 2 i o r i o r rk r r r r          Maxima: 2 2 32 2 0 0.3125(2.75) 0.927 in i or i o r rd k r r r r dr r                  2 22 2 2max 0.3125 2.75 82.42 0.3125 2.75 0.927 0.927 490 psi . r Ans              2 ______________________________________________________________________________ 3-107  = 2 (2000)/60 = 209.4 rad/s,  = 3320 20 kg/m3,  = 0.24, ri = 0.01 m, ro = 0.125 m Using Eq. (3-55)     2 22 2 23 0.24 1 3(0.24)3320(209.4) 0.01 (0.125) (0.125) 0.01 (10) 8 3 0.24 1.85 MPa . t Ans  6               ______________________________________________________________________________ 3-108  = 2 (12 000)/60 = 1256.6 rad/s,         4 2 2 2 5 /16 6.749 10 lbf s / in 386 1 16 4 5 0.75      4 The maximum shear stress occurs at bore where max = t /2. From Eq. (3-55)   24 2 2 2 2 max 3 0.20 1 3(0.20)( ) 6.749(10 ) 1256.6 0.375 2.5 2.5 (0.375) 8 3 0.20 5360 psi t                 Chapter 3 - Rev. A, Page 81/100 • max = 5360 / 2 = 2680 psi Ans. ______________________________________________________________________________ 3-109  = 2 (3500)/60 = 366.5 rad/s, mass of blade = m = V = (0.282 / 386) [1.25(30)(0.125)] = 3.425(103) lbfs2/in F = (m/2)  2r = [3.425(103)/2]( 366.52)(7.5) = 1725 lbf Anom = (1.25  0.5)(1/8) = 0.093 75 in2 nom = F/ Anom = 1725/0.093 75 = 18 400 psi Ans. Note: Stress concentration Fig. A-15-1 gives Kt = 2.25 which increases σmax and fatigue. ______________________________________________________________________________ 3-110  = 0.292, E = 207 GPa, ri = 0, R = 25 mm, ro = 50 mm Eq. (3-57),   9 2 2 2 9 3 3 2 207(10 ) (0.05 0.025 )(0.025 0) 10 3.105(10 ) (1) 2(0.025) (0.05 0) p          where p is in MPa and  is in mm. Maximum interference, max 1 [50.042 50.000] 0.021 mm . 2 Ans    Minimum interference, min 1 [50.026 50.025] 0.0005 mm . 2 Ans    From Eq. (1) pmax = 3.105(103)(0.021) = 65.2 MPa Ans. pmin = 3.105(103)(0.0005) = 1.55 MPa Ans. ______________________________________________________________________________ 3-111  = 0.292, E = 30 Mpsi, ri = 0, R = 1 in, ro = 2 in Eq. (3-57), 6 2 2 2 7 3 2 30(10 ) (2 1 )(1 0) 1.125(10 ) (1) 2(1 ) (2 0) p          where p is in psi and  is in inches. Maximum interference, Chapter 3 - Rev. A, Page 82/100 • max 1 [2.0016 2.0000] 0.0008 in . 2 Ans    Minimum interference, min 1 [2.0010 2.0010] 0 . 2 Ans    From Eq. (1), pmax = 1.125(107)(0.0008) = 9 000 psi Ans. pmin = 1.125(107)(0) = 0 Ans. ______________________________________________________________________________ 3-112  = 0.292, E = 207 GPa, ri = 0, R = 25 mm, ro = 50 mm Eq. (3-57),   9 2 2 2 9 3 3 2 207(10 ) (0.05 0.025 )(0.025 0) 10 3.105(10 ) (1) 2(0.025) (0.05 0) p          where p is in MPa and  is in mm. Maximum interference, max 1 [50.059 50.000] 0.0295 mm . 2 Ans    Minimum interference, min 1 [50.043 50.025] 0.009 mm . 2 Ans    From Eq. (1) pmax = 3.105(103)(0.0295) = 91.6 MPa Ans. pmin = 3.105(103)(0.009) = 27.9 MPa Ans. ______________________________________________________________________________ 3-113  = 0.292, E = 30 Mpsi, ri = 0, R = 1 in, ro = 2 in Eq. (3-57), 6 2 2 2 7 3 2 30(10 ) (2 1 )(1 0) 1.125(10 ) (1) 2(1 ) (2 0) p          where p is in psi and  is in inches. Maximum interference, max 1 [2.0023 2.0000] 0.00115 in . 2 Ans    Minimum interference, Chapter 3 - Rev. A, Page 83/100 • min 1 [2.0017 2.0010] 0.00035 . 2 Ans    From Eq. (1), pmax = 1.125(107)(0.00115) = 12 940 psi Ans. pmin = 1.125(107)(0.00035) = 3 938 Ans. ______________________________________________________________________________ 3-114  = 0.292, E = 207 GPa, ri = 0, R = 25 mm, ro = 50 mm Eq. (3-57),   9 2 2 2 9 3 3 2 207(10 ) (0.05 0.025 )(0.025 0) 10 3.105(10 ) (1) 2(0.025) (0.05 0) p          where p is in MPa and  is in mm. Maximum interference, max 1 [50.086 50.000] 0.043 mm . 2 Ans    Minimum interference, min 1 [50.070 50.025] 0.0225 mm . 2 Ans    From Eq. (1) pmax = 3.105(103)(0.043) = 134 MPa Ans. pmin = 3.105(103)(0.0225) = 69.9 MPa Ans. ______________________________________________________________________________ 3-115  = 0.292, E = 30 Mpsi, ri = 0, R = 1 in, ro = 2 in Eq. (3-57), 6 2 2 2 7 3 2 30(10 ) (2 1 )(1 0) 1.125(10 ) (1) 2(1 ) (2 0) p          where p is in psi and  is in inches. Maximum interference, max 1 [2.0034 2.0000] 0.0017 in . 2 Ans    Minimum interference, min 1 [2.0028 2.0010] 0.0009 . 2 Ans    From Eq. (1), Chapter 3 - Rev. A, Page 84/100 • pmax = 1.125(107)(0.0017) = 19 130 psi Ans. pmin = 1.125(107)(0.0009) = 10 130 Ans. ______________________________________________________________________________ 3-116 From Table A-5, Ei = Eo = 30 Mpsi, i o ri = 0, R = 1 in, ro = 1.5 in The radial interference is  1 2.002 2.000 0.001in . 2 Ans    Eq. (3-57),             2 2 2 2 6 2 2 2 3 2 2 3 2 30 10 0.001 1.5 1 1 0 2 2 1 1.5 0 8333 psi 83.3 kpsi . o i o i r R R rEp R r r Ans                        The tangential stresses at the interface for the inner and outer members are given by Eqs. (3-58) and (3-59), respectively. 2 2 2 2 2 2 2 2 1 0( ) (8333) 8333 psi 8.33 kpsi . 1 0 i t i r R i R rp Ans R r               2 2 2 2 2 2 2 2 1.5 1( ) (8333) 21 670 psi 21.7 kpsi . 1.5 1 o t o r R o r Rp Ans r R           ______________________________________________________________________________ 3-117 From Table A-5, Ei = 30 Mpsi, Eo =14.5 Mpsi, i o   ri = 0, R = 1 in, ro = 1.5 in The radial interference is  1 2.002 2.000 0.001in . 2 Ans    Eq. (3-56),     2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 26 6 1 1 0.001 4599 psi . 1 1.5 1 1 1 01 0.211 0.292 1.5 1 1 014.5 10 30 10 o i o i o o i i p r R R rR E r R E R r p Ans                                           The tangential stresses at the interface for the inner and outer members are given by Eqs. (3-58) and (3-59), respectively. 2 2 2 2 2 2 2 2 1 0( ) (4599) 4599 psi . 1 0 i t i r R i R rp Ans R r             Chapter 3 - Rev. A, Page 85/100 • 2 2 2 2 2 2 2 2 1.5 1( ) (4599) 11960 psi . 1.5 1 o t o r R o r Rp Ans r R          ______________________________________________________________________________ 3-118 From Table A-5, Ei = Eo = 30 Mpsi, i o ri = 0, R = 0.5 in, ro = 1 in The minimum and maximum radial interferences are  min 1 1.002 1.002 0.000 in . 2 Ans     max 1 1.003 1.001 0.001in . 2 Ans    Since the minimum interference is zero, the minimum pressure and tangential stresses are zero. Ans. The maximum pressure is obtained from Eq. (3-57).             2 2 2 2 3 2 2 6 2 2 2 3 2 2 30 10 0.001 1 0.5 0.5 0 22 500 psi 2 0.5 1 0 o i o i r R R rEp R r r p Ans                  The maximum tangential stresses at the interface for the inner and outer members are given by Eqs. (3-58) and (3-59), respectively. 2 2 2 2 2 2 2 2 0.5 0( ) (22 500) 22 500 psi . 0.5 0 i t i r R i R rp Ans R r             2 2 2 2 2 2 2 2 1 0.5( ) (22 500) 37 500 psi . 1 0.5 o t o r R o r Rp Ans r R          ______________________________________________________________________________ 3-119 From Table A-5, Ei = 10.4 Mpsi, Eo =30 Mpsi, i o   ri = 0, R = 1 in, ro = 1.5 in The minimum and maximum radial interferences are min 1 [2.003 2.002] 0.0005 in . 2 Ans    max 1 [2.006 2.000] 0.003 in . 2 Ans    Eq. (3-56), Chapter 3 - Rev. A, Page 86/100 •       2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 26 6 6 1 1 1 1.5 1 1 1 01 0.292 0.333 1.5 1 1 030 10 10.4 10 6.229 10 psi . o i o i o o i i p r R R rR E r R E R r p p Ans                                                 6 6min min6.229 10 6.229 10 0.0005 3114.6 psi 3.11 kpsi .p Ans        6 6max max6.229 10 6.229 10 0.003 18 687 psi 18.7 kpsi .p Ans    The tangential stresses at the interface for the inner and outer members are given by Eqs. (3-58) and (3-59), respectively. Minimum interference: 2 2 2 2 min 2 2 2 2min 1 0( ) (3.11) 3.11 kpsi . 1 0 i t i i R rp Ans R r           2 2 2 2 min 2 2 2 2min 1.5 1( ) (3.11) 8.09 kpsi . 1.5 1 o t o o r Rp Ans r R        Maximum interference: 2 2 2 2 max 2 2 2 2max 1 0( ) (18.7) 18.7 kpsi . 1 0 i t i i R rp Ans R r           2 2 2 2 max 2 2 2 2max 1.5 1( ) (18.7) 48.6 kpsi . 1.5 1 o t o o r Rp Ans r R        ______________________________________________________________________________ 3-120 20 mm, 37.5 mm, 57.5 mmi od r r   From Table 3-4, for R = 10 mm, 37.5 10 47.5 mmcr      2 2 2 10 46.96772 mm 2 47.5 47.5 10 nr     47.5 46.96772 0.53228 mmc ne r r     46.9677 37.5 9.4677 mmi n ic r r     57.5 46.9677 10.5323 mmo o nc r r     2 2/ 4 (20) / 4 314.16 mmA d    2 4000(47.5) 190 000 N mmcM Fr    Using Eq. (3-65) for the bending stress, and combining with the axial stress, Chapter 3 - Rev. A, Page 87/100 • 4000 190 000(9.4677) 300 MPa . 314.16 314.16(0.53228)(37.5) 4000 190 000(10.5323) 195 MPa . 314.16 314.16(0.53228)(57.5) i i i o o o McF Ans A Aer McF Ans A Aer              ______________________________________________________________________________ 3-121 0.75 in, 1.25 in, 2.0 ini od r r   From Table 3-4, for R = 0.375 in, 1.25 0.375 1.625 incr      2 2 2 0.375 1.60307 in 2 1.625 1.625 0.375 nr     1.625 1.60307 0.02193 inc ne r r     1.60307 1.25 0.35307 ini n ic r r     2.0 1.60307 0.39693 ino o nc r r     2 2/ 4 (0.75) / 4 0.44179 inA d    2 750(1.625) 1218.8 lbf incM Fr    Using Eq. (3-65) for the bending stress, and combining with the axial stress, 750 1218.8(0.35307) 37 230 psi 37.2 kpsi . 0.44179 0.44179(0.02193)(1.25) 750 1218.8(0.39693) 23 269 psi 23.3 kpsi . 0.44179 0.44179(0.02193)(2.0) i i i o o o McF Ans A Aer McF Ans A Aer                 ______________________________________________________________________________ 3-122 6 mm, 10 mm, 16 mmi od r r   From Table 3-4, for R = 3 mm, 10 3 13 mmcr      2 2 2 3 12.82456 mm 2 13 13 3 nr     13 12.82456 0.17544 mmc ne r r     12.82456 10 2.82456 mmi n ic r r     16 12.82456 3.17544 mmo o nc r r     2 2/ 4 (6) / 4 28.2743 mmA d    2 300(13) 3900 N mmcM Fr    Using Eq. (3-65) for the bending stress, and combining with the axial stress, Chapter 3 - Rev. A, Page 88/100 • 300 3900(2.82456) 233 MPa . 28.2743 28.2743(0.17544)(10) 300 3900(3.17544) 145 MPa . 28.2743 28.2743(0.17544)(16) i i i o o o McF Ans A Aer McF Ans A Aer              ______________________________________________________________________________ 3-123 6 mm, 10 mm, 16 mmi od r r   From Table 3-4, for R = 3 mm, 10 3 13 mmcr      2 2 2 3 12.82456 mm 2 13 13 3 nr     13 12.82456 0.17544 mmc ne r r     12.82456 10 2.82456 mmi n ic r r     16 12.82456 3.17544 mmo o nc r r     2 2/ 4 (6) / 4 28.2743 mmA d    2 The angle  of the line of radius centers is     1 1/ 2 10 6 / 2sin sin 30 10 6 10 / 2 sin 300 10 6 / 2 sin 30 1950 N mm R d R d R M F R d                            Using Eq. (3-65) for the bending stress, and combining with the axial stress, sin 300sin 30 1950(2.82456) 116 MPa . 28.2743 28.2743(0.17544)(10) sin 300sin 30 1950(3.17544) 72.7 MPa . 28.2743 28.2743(0.17544)(16) i i i o o o McF Ans A Aer McF Ans A Aer                Note that the shear stress due to the shear force is zero at the surface. ______________________________________________________________________________ 3-124 0.25 in, 0.5 in, 0.75 ini od r r   From Table 3-4, for R = 0.125 in, 0.5 0.125 0.625 incr      2 2 2 0.125 0.618686 in 2 0.625 0.625 0.125 nr     0.625 0.618686 0.006314 inc ne r r     0.618686 0.5 0.118686 ini n ic r r     0.75 0.618686 0.131314 ino o nc r r     Chapter 3 - Rev. A, Page 89/100 • 2 2/ 4 (0.25) / 4 0.049087 inA d    2 75(0.625) 46.875 lbf incM Fr    Using Eq. (3-65) for the bending stress, and combining with the axial stress, 75 46.875(0.118686) 37 428 psi 37.4 kpsi . 0.049087 0.049087(0.006314)(0.5) 75 46.875(0.131314) 24 952 psi 25.0 kpsi . 0.049087 0.049087(0.006314)(0.75) i i i o o o McF Ans A Aer McF Ans A Aer                 ______________________________________________________________________________ 3-125 0.25 in, 0.5 in, 0.75 ini od r r   From Table 3-4, for R = 0.125 in, 0.5 0.125 0.625 incr      2 2 2 0.125 0.618686 in 2 0.625 0.625 0.125 nr     0.625 0.618686 0.006314 inc ne r r     0.618686 0.5 0.118686 ini n ic r r     0.75 0.618686 0.131314 ino o nc r r     2 2/ 4 (0.25) / 4 0.049087 inA d    2 The angle  of the line of radius centers is     1 1/ 2 0.5 0.25 / 2sin sin 30 0.5 0.25 0.5 / 2 sin 75 0.5 0.25 / 2 sin 30 23.44 lbf in R d R d R M F R d                            Using Eq. (3-65) for the bending stress, and combining with the axial stress, sin 75sin 30 23.44(0.118686) 18 716 psi 18.7 kpsi . 0.049087 0.049087(0.006314)(0.5) sin 75sin 30 23.44(0.131314) 12 478 psi 12.5 kpsi 0.049087 0.049087(0.006314)(0.75) i i i o o o McF Ans A Aer McF A Aer                   .Ans Note that the shear stress due to the shear force is zero at the surface. ______________________________________________________________________________ 3-126 (a)     3 3(4) 0.5(0.1094) 8021 psi 8.02 kpsi . (0.75) 0.1094 /12 Mc Ans I          (b) ri = 0.125 in, ro = ri + h = 0.125 + 0.1094 = 0.2344 in From Table 3-4, Chapter 3 - Rev. A, Page 90/100 • 0.125 (0.5)(0.1094) 0.1797 in 0.1094 0.174006 in ln(0.2344 / 0.125) 0.1797 0.174006 0.005694 in 0.174006 0.125 0.049006 in 0.2344 0.174006 0.060394 in 0.75(0.1094) 0.08205 c n c n i n i o o n r r e r r c r r c r r A bh                        2 in 3(4) 12 lbf inM      The negative sign on the bending moment is due to the sign convention shown in Fig. 3-34. Using Eq. (3-65), 12(0.049006) 10 070 psi 10.1 kpsi . 0.08205(0.005694)(0.125) 12(0.060394) 6618 psi 6.62 kpsi . 0.08205(0.005694)(0.2344) i i i o o o Mc Ans Aer Mc Ans Aer                 (c) 10.1 1.26 . 8.02 i iK Ans        6.62 0.825 . 8.02 o oK Ans      ______________________________________________________________________________ 3-127 (a)     3 3(4) 0.5(0.1406) 4856 psi 4.86 kpsi . (0.75) 0.1406 /12 Mc Ans I          (b) ri = 0.125 in, ro = ri + h = 0.125 + 0.1406 = 0.2656 in From Table 3-4, 0.125 (0.5)(0.1406) 0.1953 in 0.1406 0.186552 in ln(0.2656 / 0.125) 0.1953 0.186552 0.008748 in 0.186552 0.125 0.061552 in 0.2656 0.186552 0.079048 in 0.75(0.1406) 0.10545 c n c n i n i o o n r r e r r c r r c r r A bh                        2 in 3(4) 12 lbf inM      The negative sign on the bending moment is due to the sign convention shown in Fig. 3-34. Using Eq. (3-65), Chapter 3 - Rev. A, Page 91/100 • 12(0.061552) 6406 psi 6.41 kpsi . 0.10545(0.008748)(0.125) 12(0.079048) 3872 psi 3.87 kpsi . 0.10545(0.008748)(0.2656) i i i o o o Mc Ans Aer Mc Ans Aer                 (c) 6.41 1.32 . 4.86 i iK Ans        3.87 0.80 . 4.86 o oK Ans      ______________________________________________________________________________ 3-128 (a)     3 3(4) 0.5(0.1094) 8021 psi 8.02 kpsi . (0.75) 0.1094 /12 Mc Ans I          (b) ri = 0.25 in, ro = ri + h = 0.25 + 0.1094 = 0.3594 in From Table 3-4, 0.25 (0.5)(0.1094) 0.3047 in 0.1094 0.301398 in ln(0.3594 / 0.25) 0.3047 0.301398 0.003302 in 0.301398 0.25 0.051398 in 0.3594 0.301398 0.058002 in 0.75(0.1094) 0.08205 in c n c n i n i o o n r r e r r c r r c r r A bh                        2 3(4) 12 lbf inM      The negative sign on the bending moment is due to the sign convention shown in Fig. 3-34. Using Eq. (3-65), 12(0.051398) 9106 psi 9.11 kpsi . 0.08205(0.003302)(0.25) 12(0.058002) 7148 psi 7.15 kpsi . 0.08205(0.003302)(0.3594) i i i o o o Mc Ans Aer Mc Ans Aer                 (c) 9.11 1.14 . 8.02 i iK Ans        7.15 0.89 . 8.02 o oK Ans      ______________________________________________________________________________ 3-129 ri = 25 mm, ro = ri + h = 25 + 87 = 112 mm, rc = 25 + 87/2 = 68.5 mm The radius of the neutral axis is found from Eq. (3-63), given below. Chapter 3 - Rev. A, Page 92/100 •  /n Ar dA r   For a rectangular area with constant width b, the denominator is lno i r o r i rbdr b r r       Applying this equation over each of the four rectangular areas, 45 54.5 92 1129 ln 31 ln 31 ln 9 ln 16.3769 25 45 82.5 92 dA r                              22 20(9) 31(9.5) 949 mmA      949 57.9475 mm 16.3769/n Ar dA r     68.5 57.9475 10.5525 mmc ne r r     57.9475 25 32.9475 mmi n ic r r     112 57.9475 54.0525 mmo o nc r r     M = 150F2 = 150(3.2) = 480 kN·mm We need to find the forces transmitted through the section in order to determine the axial stress. It is not immediately obvious which plane should be used for resolving the axial versus shear directions. It is convenient to use the plane containing the reaction force at the bushing, which assumes its contribution resolves entirely into shear force. To find the angle of this plane, find the resultant of F1 and F2.   1 2 1 2 1 22 2 2.4cos60 3.2cos0 4.40 kN 2.4sin 60 3.2sin 0 2.08 kN 4.40 2.08 4.87 kN x x x y y y F F F F F F F                  This is the pin force on the lever which acts in a direction 1 1 2.08tan tan 25.3 4.40 y x F F       On the surface 25.3° from the horizontal, find the internal forces in the tangential and normal directions. Resolving F1 into components,     2.4cos 60 25.3 1.97 kN 2.4sin 60 25.3 1.37 kN t n F F           The transverse shear stress is zero at the inner and outer surfaces. Using Eq. (3-65) for the bending stress, and combining with the axial stress due to Fn, Chapter 3 - Rev. A, Page 93/100 •       3200 150 (32.9475)1370 64.6 MPa . 949 949(10.5525)(25) 3200 150 (54.0525)1370 21.7 MPa . 949 949(10.5525)(112) n i i i n o o o F Mc Ans A Aer F Mc Ans A Aer                  ______________________________________________________________________________ 3-130 ri = 2 in, ro = ri + h = 2 + 4 = 6 in, 2 0.5(4) 4 incr    2(6 2 0.75)(0.75) 2.4375 inA     Similar to Prob. 3-129, 3.625 60.75ln 0.75ln 0.682 920 in 2 4.375 dA r    2.4375 3.56923 in 0.682 920( / )n Ar dA r     4 3.56923 0.43077 inc ne r r     3.56923 2 1.56923 ini n ic r r     6 3.56923 2.43077 ino o nc r r     6000(4) 24 000 lbf incM Fr    Using Eq. (3-65) for the bending stress, and combining with the axial stress, 6000 24 000(1.56923) 20 396 psi 20.4 kpsi . 2.4375 2.4375(0.43077)(2) 6000 24 000(2.43077) 6 799 psi 6.80 kpsi . 2.4375 2.4375(0.43077)(6) i i i o o o McF Ans A Aer McF Ans A Aer                 ______________________________________________________________________________ 3-131 ri = 12 in, ro = ri + h = 12 + 3 = 15 in, rc = 12 + 3/2 = 13.5 in 3 3(1.5 )(0.75) 1.988 in 4 4 (1.5)(0.75) 3.534 I a b A ab 4          20(3 1.5) 90 kip inM     Since the radius is large compared to the cross section, assume Eq. 3-67 is applicable for the bending stress. Combining the bending stress and the axial stress, 20 90(1.5)(13.5) 82.1 kpsi . 3.534 (1.988)(12) i c i i Mc rF Ans A Ir       20 90(1.5)(13.5) 55.5 kpsi . 3.534 1.988(15) o c o o Mc rF Ans A Ir        ______________________________________________________________________________ Chapter 3 - Rev. A, Page 94/100 • 3-132 ri = 1.25 in, ro = ri + h = 1.25 + 0.5 + 1 + 0.5 = 3.25 in rc = (ri + ro) / 2 = (1.25 + 3.25)/2 = 2.25 in Ans. For outer rectangle, ln o i rdA b r r       For circle,     2 2 O2 2 O O , 2 c c A r A r dA r r r r                  2 2 O 2 ( )c c dA r r r r        Combine the integrals subtracting the circle from the rectangle  2 23.251.25ln 2 2.25 2.25 0.5 0.840 904 in1.25 dA r       2 21.25(2) (0.5 ) 1.714 60 in .A A   ns 1.71460 2.0390 in . 0.840904( / )n Ar A dA r     ns  2.25 2.0390 0.2110 in .c ne r r Ans     2.0390 1.25 0.7890 ini n ic r r     3.25 2.0390 1.2110 ino o nc r r     2000(4.5 1.25 0.5 0.5) 13 500 lbf inM      2000 13 500(0.7890) 20 720 psi = 20.7 kpsi . 1.7146 1.7146(0.2110)(1.25) i i i McF Ans A Aer       2000 13 500(1.2110) 12 738 psi 12.7 kpsi . 1.7146 1.7146(0.2110)(3.25) o o o McF Ans A Aer          ______________________________________________________________________________ 3-133 From Eq. (3-68),     1 3 2 1 3 1 3 2 13 8 2 1 E a KF F d                Use 0.292,  F in newtons, E in N/mm2 and d in mm, then 1/323 [(1 0.292 ) / 207 000] 0.03685 8 1/ 30 K          From Eq. (3-69), 1/3 1/3 1/3 max 2 1/3 2 2 2 3 3 3 3 352 MPa 2 2 ( ) 2 2 (0.03685) F F F Fp F a KF K         Chapter 3 - Rev. A, Page 95/100 • From Eq. (3-71), the maximum principal stress occurs on the surface where z = 0, and is equal to – pmax. 1/3 max max 352 MPa .z p F A      ns From Fig. 3-37, 1/3 max max0.3 106 MPa .p F A   ns ______________________________________________________________________________ 3-134 From Eq. (3-68),               2 2 1 1 2 2 3 1 2 2 2 3 1 13 8 1 1 1 0.292 207 000 1 0.333 717003 10 0.0990 mm 8 1 25 1 40 E EFa d d a                   From Eq. (3-69),    max 2 2 3 103 487.2 MPa 2 2 0.0990 Fp a     From Fig. 3-37, the maximum shear stress occurs at a depth of z = 0.48 a.  0.48 0.48 0.0990 0.0475 mm .z a A   ns The principal stresses are obtained from Eqs. (3-70) and (3-71) at a depth of z/a = 0.48.       1 1 2 2 1487.2 1 0.48 tan 1/ 0.48 1 0.333 101.3 MPa 2 1 0.48                    3 2 487.2 396.0 MPa 1 0.48      From Eq. (3-72),    1 3max 101.3 396.0 147.4 MPa . 2 2 Ans        Note that if a closer examination of the applicability of the depth assumption from Fig. 3- 37 is desired, implementing Eqs. (3-70), (3-71), and (3-72) on a spreadsheet will allow for calculating and plotting the stresses versus the depth for specific values of . For = 0.333 for aluminum, the maximum shear stress occurs at a depth of z = 0.492a with max = 0.3025 pmax. Chapter 3 - Rev. A, Page 96/100 • This gives max = 0.3025 pmax = (0.3025)(487.2) = 147.38 MPa. Even though the depth assumption was a little off, it did not have significant effect on the the maximum shear stress. ______________________________________________________________________________ 3-135 From the solution to Prob. 3-134, a = 0.0990 mm and pmax = 487.2 MPa. Assuming applicability of Fig. 3-37, the maximum shear stress occurs at a depth of z = 0.48 a = 0.0475 mm. Ans. The principal stresses are obtained from Eqs. (3-70) and (3-71) at a depth of z/a = 0.48.       1 1 2 2 1487.2 1 0.48 tan 1/ 0.48 1 0.292 92.09 MPa 2 1 0.48                    3 2 487.2 396.0 MPa 1 0.48      From Eq. (3-72),    1 3max 92.09 396.0 152.0 MPa . 2 2 Ans        Note that if a closer examination of the applicability of the depth assumption from Fig. 3- 37 is desired, implementing Eqs. (3-70), (3-71), and (3-72) on a spreadsheet will allow for calculating and plotting the stresses versus the depth for specific values of . For = 0.292 for steel, the maximum shear stress occurs at a depth of z = 0.478a with max = 0.3119 pmax. ______________________________________________________________________________ 3-136 From Eq. (3-68),         2 3 1 2 2 3 2 13 8 1 1 2 1 0.292 207 0003 20 0.1258 mm 8 1 30 1 EFa d d a               From Eq. (3-69),    max 2 2 3 203 603.4 MPa 2 2 0.1258 Fp a     From Fig. 3-37, the maximum shear stress occurs at a depth of  0.48 0.48 0.1258 0.0604 mm .z a A   ns Also from Fig. 3-37, the maximum shear stress is max max0.3 0.3(603.4) 181 MPa .p Ans    Chapter 3 - Rev. A, Page 97/100 • _____________ ____________________________________ _____________________________ -137 Aluminum Plate-Ball interface: From Eq. (3-68), 3                 3 1 2 2 6 2 6 3 1/33 8 1 1 1 0.292 30 10 1 0.333 10.4 103 3.517 10 in 8 1 1 1 a d d Fa F                      rom Eq. (3-69), 2 2 1 1 2 21 13 E EF      F    4 1/3max 22 3 1/3 3Fp   3 3.860 10 psi 2 2 3.517 10 F F a F        By examination of Eqs. (3-70), (3-71), and (3-72), it can be seen that the only difference in the maximum shear stress for the plate and the ball will be due to poisson’s ratio in Eq. (3-70). The larger poisson’s ratio will create the greater maximum shear stress, so the aluminum plate will be the critical element in this interface. Applying the equations for the aluminum plate,         4 1/3 1 1/3 1 2 13.86 10 1 0.48 tan 1/ 0.48 1 0.333 8025 psi 2 1 0.48 F F                     4 1/3 4 1/3 3 2 3.86 10 3.137 10 psi 1 0.48 F F      From Eq. (3-72),        1/3 4 1/3 4 1/31 3   max 8025 3.137 10 1.167 10 psi 2 2 F F F     omparing this stress to the allowable stress, and solving for F, C   3 20 000  4 5.03 lbf 1.167 10 F       able-Ball interface: From Eq. (3-68), T             2 6 2 6 3 1/33 1 0.292 30 10 1 0.211 14.5 103 3.306 10 in 8 1 1 1 Fa F                 From Eq. (3-69), Chapter 3 - Rev. A, Page 98/100 •    4 1/3max 22 3 1/3 3 3 4.369 10 psi 2 2 3.306 10 F Fp F a F          The steel ball has a higher poisson’s ratio than the cast iron table, so it will dominate.         4 1/3 1 1/3 1 2 14.369 10 1 0.48 tan 1/ 0.48 1 0.292 8258 psi 2 1 0.48 F F                     4 1/3 4 1/3 3 2 4.369 10 3.551 10 psi 1 0.48 F F      From Eq. (3-72),        1/3 4 1/3 4 1/31 3 max 8258 3.551 10 1.363 10 psi 2 2 F F F        Comparing this stress to the allowable stress, and solving for F,   3 4 20 000 3.16 lbf 1.363 10 F         The steel ball is critical, with F = 3.16 lbf. Ans. ______________________________________________________________________________ 3-138 v1 = 0.333, E1 = 10.4 Mpsi, l = 2 in, d1 = 1.25 in, v2 = 0.211, E2 = 14.5 Mpsi, d2 = –12 in. With b = KcF1/2           1 2 2 6 2 6 4 1 0.333 10.4 10 1 0.211 14.5 102 (2) 1/1.25 1/12 2.336 10 cK                  By examination of Eqs. (3-75), (3-76), and (3-77, it can be seen that the only difference in the maximum shear stress for the two materials will be due to poisson’s ratio in Eq. (3- 75). The larger poisson’s ratio will create the greater maximum shear stress, so the aluminum roller will be the critical element in this interface. Instead of applying these equations, we will assume the poisson’s ratio for aluminum of 0.333 is close enough to 0.3 to make Fig. 3-39 applicable. max max max 0.3 4000 13 300 psi p p 0.3     From Eq. (3-74), pmax = 2F / (bl ), so we have Chapter 3 - Rev. A, Page 99/100 • 1 2 max 1 2 2 2 c c F Fp lK F lK    So,   2 max 24 2 (2)(2.336) 10 (13 300) 2 95.3 lbf . clK pF Ans                   ______________________________________________________________________________ 3-139 v = 0.292, E = 30 Mpsi, l = 0.75 in, d1 = 2(0.47) = 0.94 in, d2 = 2(0.62) = 1.24 in. Eq. (3-73):       1 2 2 6 3 2 1 0.292 30 102(40) 1.052 10 in (0.75) 1/ 0.94 1/1.24 b              Eq. (3-74):      max 3 2 402 32 275 psi 32.3 kpsi . 1.052 10 0.75 Fp Ans bl       From Fig. 3-39, max max0.3 0.3(32 275)=9682.5 psi 9.68 kpsi .p Ans    ______________________________________________________________________________ 3-140 Use Eqs. (3-73) through (3-77). 1/22 2 1 1 2 2 1 2 (1 ) / (1 ) /2 (1/ ) (1/ ) E v EFb l d d           1/22 6 2 62(600) (1 0.292 ) / (30(10 )) (1 0.292 ) / (30(10 )) (2) 1/ 5 1/          0.007 631 inb  max 2 2(600) 25 028 psi (0.007 631)(2) Fp bl     Chapter 3 - Rev. A, Page 100/100 •    2 2 max 22 1 2 0.292 25 028 1 0.786 7102 psi 7.10 kpsi . x z zp b b Ans                     0.786       2 max 2 2 2 2 25 028 2 0.786 1 0.7861 4 646 psi 4.65 kpsi . y zbp bz b Ans                       2 21 2 1 2 0.786 z      max 2 2 25 028 19 677 psi 19.7 kpsi . 1 0.786 z p Ans z          21 b    max 4 646 19 677 7 516 psi 7.52 kpsi .y z Ans  2 2           ______________________________________________________________________________ 3-141 Use Eqs. (3-73) through (3-77).     1 22 21 1 2 2 1 2 1 12 1/ 1/ E EF l d d             b         1 2 2 30.211 100 10 /  2 31 0.292 207 10 12(2000) (40) 1/150 1                0.2583 mmb  max 2 2(2000) 123.2 MPa (0.2583)(40) Fp bl        2 2 max 22 1 2 0.292 123.2 1 0.786 35.0 MPa . x z zp b b Ans                   0.786       2 21 2 0.786 z            2 max 2 2 2 1 2 0.786 2 123.2 2 1 0.7861 22.9 MPa . y zbp bz b Ans                 max 2 2 123.2 96.9 MPa . 1 0.786 z p Ans z        21 b  Chapter 3 - Rev. A, Page 101/100 •   Chapter 3 - Rev. A, Page 102/100 max 22.9 96.9 37.0 MPa . 2 2 y z Ans           ______________________________________________________________________________ 3-142 Note to the Instructor: The first printing incorrectly had a width w = 1.25 mm instead of w = 1.25 in. The solution presented here reflects the correction which will be made in subsequent printings. Use Eqs. (3-73) through (3-77).     1 22 21 1 2 2 1 2 1 12 1/ 1/ E EF l d d            b           1 2 2 60.211 14.5 10 2 61 0.211 14.5 10 12(250) (1.25) 1/ 3 1/                0.007 095 inb  max 2 2(250) 17 946 psi (0.007 095)(1.25) Fp bl        2 2 max 22 1 2 0.211 17 946 1 0.786 0. 3 680 psi 3.68 kpsi . x z zp b b Ans                     786       2 max 2 2 2 1 2 0.786 2 17 946 2 0.786 1 0.7861 3 332 psi 3.33 kpsi . y zbp bz b Ans                       2 21 2 z      max 2 2 17 946 14109 psi 14.1 kpsi .z p Ans        2 1 0.7861 z b     max 3 332 14109 5 389 psi 5.39 kpsi . 2 2 y z Ans            ______________________________________________________________________________ • Chapter 4 4-1 For a torsion bar, kT = T/ = Fl/, and so  = Fl/kT. For a cantilever, kl = F/ , = F/kl. For the assembly, k = F/y, or, y = F/k = l +  Thus 2 T l F Fl Fy k k k    Solving for k 2 2 1 . 1 l T l T T l k kk A l k l k k k    ns ______________________________________________________________________________ 4-2 For a torsion bar, kT = T/ = Fl/, and so  = Fl/kT. For each cantilever, kl = F/l, l = F/kl, and,L = F/kL. For the assembly, k = F/y, or, y = F/k = l + l +L. Thus 2 T l F Fl F Fy k k k k     L Solving for k 2 2 1 . 1 1 L l T l L T L T l T l L k k kk A l k k l k k k k k k k      ns ______________________________________________________________________________ 4-3 (a) For a torsion bar, k =T/ =GJ/l. Two springs in parallel, with J =di 4/32, and d1 = d1 = d, 4 4 1 2 1 2 4 32 1 1 . (1) 32 J G J G d dk G x l x x l x Gd Ans x l x                 Deflection equation,     21 2 1results in (2) T l xT x JG JG T l x T x       From statics, T1 + T2 = T = 1500. Substitute Eq. (2) Chapter 4 - Rev B, Page 1/81 • 2 2 21500 1500 . (3) l x xT T T Ans x l          Substitute into Eq. (2) resulting in 1 1500 . (4) l xT An l s (b) From Eq. (1),      4 6 31 10.5 11.5 10 28.2 10 lbf in/rad . 32 5 10 5 k A        ns From Eq. (4), 1 10 51500 750 lbf in . 10 T Ans   From Eq. (3), 2 51500 750 lbf in . 10 T Ans   From either section,       3 3 3 16 150016 30.6 10 psi 30.6 kpsi . 0.5 i i T Ans d        ______________________________________________________________________________ 4-4 Deflection to be the same as Prob. 4-3 where T1 = 750 lbfin, l1 = l / 2 = 5 in, and d1 = 0.5 in  1 =  2 =           1 2 31 24 4 4 4 4 1 2 1 2 4 6 750 5 4 6 60 10 (1) 0.5 32 32 32 T T T T d dd G d G G        Or,  3 41 115 10 (2)T d  3 42 210 10 (3)T d Equal stress, 1 2 1 21 2 3 3 3 3 1 2 1 2 16 16 (4)T T T T d d d d          Divide Eq. (4) by the first two equations of Eq.(1) results in 1 2 3 3 1 2 2 1 1 2 4 4 1 2 1.5 (5)4 4 T T d d d dT T d d    Statics, T1 + T2 = 1500 (6) Substitute in Eqs. (2) and (3), with Eq. (5) gives     43 4 31 115 10 10 10 1.5 1500d d  Solving for d1 and substituting it back into Eq. (5) gives d1 = 0.388 8 in, d2 = 0.583 2 in Ans. Chapter 4 - Rev B, Page 2/81 • From Eqs. (2) and (3), T1 = 15(103)(0.388 8)4 = 343 lbfin Ans. T2 = 10(103)(0.583 2)4 = 1 157 lbfin Ans. Deflection of T is        1 1 1 4 6 1 343 4 0.053 18 rad / 32 0.388 8 11.5 10 T l J G      Spring constant is  3 1 1500 28.2 10 lbf in . 0.053 18 Tk Ans      The stress in d1 is      311 33 1 16 34316 29.7 10 psi 29.7 kpsi . 0.388 8 T Ans d        The stress in d1 is      322 33 2 16 115716 29.7 10 psi 29.7 kpsi . 0.583 2 T Ans d        ______________________________________________________________________________ 4-5 (a) Let the radii of the straight sections be r1 = d1 /2 and r2 = d2 /2. Let the angle of the taper be  where tan  = (r2  r1)/2. Thus, the radius in the taper as a function of x is r = r1 + x tan , and the area is A =  (r1 + x tan )2. The deflection of the tapered portion is       2 10 0 1 0 1 1 1 2 1 1 2 1 2 1 2 1 2 1 tan tantan 1 1 1 tan tan tan tan tan tan tan 4 . ll lF F dx Fdx AE E E r xr x F F E r r l E r r r rF F l Fl E r r E r r r r E Fl Ans d d E  2 1                                            (b) For section 1, 41 2 2 6 1 4 4(1000)(2) 3.40(10 ) in . (0.5 )(30)(10 ) Fl Fl Ans AE d E        For the tapered section, 46 1 2 4 4 1000(2) 2.26(10 ) in . (0.5)(0.75)(30)(10 ) Fl Ans d d E       For section 2, Chapter 4 - Rev B, Page 3/81 • 42 2 2 6 1 4 4(1000)(2) 1.51(10 ) in . (0.75 )(30)(10 ) Fl Fl Ans AE d E        ______________________________________________________________________________ 4-6 (a) Let the radii of the straight sections be r1 = d1 /2 and r2 = d2 /2. Let the angle of the taper be  where tan  = (r2  r1)/2. Thus, the radius in the taper as a function of x is r = r1 + x tan , and the polar second area moment is J = ( /2) (r1 + x tan )4. The angular deflection of the tapered portion is         4 3 0 0 1 1 0 33 3 1 11 2 23 3 3 3 1 1 2 22 1 2 1 3 3 3 3 3 3 1 2 2 1 1 2 1 2 2 1 2 1 3tan tan tan 2 1 1 2 1 1 3 tan 3 tantan tan 2 2 2 3 tan 3 3 32 3 l l lT T dx Tdx GJ G Gr x r x T T G r G r rr l r r r rr r r rT T l Tl G r r G r r r r G r r T                                                 3 2   2 21 1 2 2 3 3 1 2 . d d d dl Ans G d d   (b) The deflections, in degrees, are For section 1, 1 4 4 6 1 180 32 180 32(1500)(2) 180 2.44 deg . (0.5 )11.5(10 ) Tl Tl Ans GJ d G                           For the tapered section, 2 2 1 1 2 2 3 3 1 2 2 2 6 3 3 ( )32 180 3 (1500)(2) 0.5 (0.5)(0.75) 0.7532 180 1.14 deg . 3 11.5(10 )(0.5 )(.75 ) Tl d d d d Gd d Ans                        For section 2, 2 4 4 6 2 180 32 180 32(1500)(2) 180 0.481 deg . (0.75 )11.5(10 ) Tl Tl Ans GJ d G                           ______________________________________________________________________________ 4-7 The area and the elastic modulus remain constant, however the force changes with respect to x. From Table A-5 the unit weight of steel is  = 0.282 lbf/in3, and the elastic modulus is E = 30 Mpsi. Starting from the top of the cable (i.e. x = 0, at the top). F = (A)(lx) Chapter 4 - Rev B, Page 4/81 •   22 2 60 0 0.282 500(12)1( ) 0.169 in 2 2 2(30)10 l l l c o Fdx ll x dx lx x AE E E E                w From the weight at the bottom of the cable,  2 2 6 4(5000) 500(12)4 5.093 in (0.5 )30(10 )W Wl Wl AE d E        0.169 5.093 5.262 in .c W Ans       The percentage of total elongation due to the cable’s own weight 0.169 (100) 3.21% . 5.262 Ans ______________________________________________________________________________ 4-8 Fy = 0 = R 1  F  R 1 = F MA = 0 = M1  Fa  M1 = Fa VAB = F, MAB =F (x  a ), VBC = MBC = 0 Section AB:   2 1 1 2AB F xF x a dx ax C EI EI             (1) AB = 0 at x = 0  C1 = 0 2 3 22 6AB F x F x xy ax dx a EI EI                 2 2 C (2) yAB = 0 at x = 0  C2 = 0    2 3 . 6AB Fxy x a Ans EI   Section BC:   3 1 0 0BC dx CEI     From Eq. (1), at x = a (with C1 = 0), 2 2 ( ) 2 F a Faa a EI EI           2 = C3. Thus, 2 2BC Fa EI    2 2 42 2BC Fa Fay dx x C EI EI      (3) Chapter 4 - Rev B, Page 5/81 • From Eq. (2), at x = a (with C2 = 0), 3 2F a a 3 6 2 3 Fay a EI EI          . Thus, from Eq. (3) 2 3Fa Fa 3 4 42 3 6 Faa C C EI EI EI       Substitute into Eq. (3)    2 3 2 3 . 2 6 6BC Fa Fa Fay x a x EI EI EI      Ans maximum deflection occurs at x= l, The   2 max 3 . Fa 6 y a l Ans  EI MAB = R 1 x = Fx /2 : =  F /2, MBC = R 1 x  F ( x  l / 2) = F (l  x) /2 ______________________________________________________________________________ 4-9 MC = 0 = F (l /2)  R1 l  R1 = F /2 Fy = 0 = F /2 + R 2  F  R 2 = F /2 Break at 0  x  l /2: VAB = R 1 = F /2, Break at l /2  x  l VBC = R 1  F =  R 2 Section AB: 2 1 1 AB Fx   2 4 F xdx C EI EI   From symmetry, AB = 0 at x = l /2  2 2 1 1 2 0 4 1 lF FlC C EI EI            6 . Thus,   2 2 2 2F x Fl F x 4 4 16 16AB l EI EI EI      (1)   34x 2 2 2 2416 16 3AB F Fy x l dx l x C EI EI          Chapter 4 - Rev B, Page 6/81 • at x = 0  C2 = 0, and, yAB = 0   2 24 348AB Fxy x l EI   (2) is not given, because with symmetry, Eq. (2) can be used in this region. The maximum deflection occurs at x =l /2,  yBC 2 2 lF l 32 max 4 3 .48 2 48 Fly l Ans EI EI           4-10 From Table A-6, for each angle, I = 207 cm4. Thus, I = 2(207) (104) = 4.14(106) mm4 From Table A-9, use beam 2 with F = 2500 N, a = 2000 mm, and l = 3000 mm; and beam         ______________________________________________________________________________ 1-1 3 with w = 1 N/mm and l = 3000 mm. 2 4 max ( 3 ) Fa ly a l   w 6 8EI EI   2 4 3 6 3 2500(2000) (1)(3000)2000 3(3000) 6(207)10 (4.14)10 8(207)(10 )(4.14)(10 ) 25.4 mm .Ans      6 ) =  2500(2000)  [1(30002)/2] =  9.5(106) Nmm rom Table A-6, from id to upper surface is y = 29 mm. From centroid to bottom is compressive at the bottom of the beam at the wall. This stress is 2( / 2OM Fa l   w F centro surface is y = 29.0  100=  71 mm. The maximum stress 6 max 6 9.5(10 )( 71) 163 MPa . 4.14(10 ) My Ans I         ______________________________________________________________________________ Chapter 4 - Rev B, Page 7/81 • 4-11 14 10(450) (300) 465 lbf 20 20 6 10(450) (300) 285 lbf 20 20 O C R R       M1 = 465(6)12 = 33.48(103) lbfin M2 = 33.48(103) +15(4)12 = 34.20(103) lbfin 3maxmax 34.2 15 2.28 inM Z Z Z      For deflections, use beams 5 and 6 of Table A-9   2 3 21 2 10ft 3 2 2 2 6 6 4 4 [ ( / 2)] 2 6 2 2 48 450(72)(120) 300(240 )0.5 120 72 240 6(30)(10 ) (240) 48(30)(10 ) 12.60 in / 2 6.30 in x F a l l F ll ly a l EIl EI I I I I                       Select two 5 in-6.7 lbf/ft channels from Table A-7, I = 2(7.49) = 14.98 in4, Z =2(3.00) = 6.00 in3 midspan max 12.60 1 0.421 in 14.98 2 34.2 5.70 kpsi 6.00 y            ______________________________________________________________________________ 4-12 4 4(1.5 ) 0.2485 in 64 I   From Table A-9 by superposition of beams 6 and 7, at x = a = 15 in, with b = 24 in and l = 39 in 2 2 2 2 3 3[ ] (2 6 24 Fba ay a b l la a )l EIl EI       w 2 2 2 6 2 3 3 6 340(24)15 15 24 39 6(30)10 (0.2485)39 (150 /12)(15) 2(39)(15 ) 15 39 0.0978 in . 24(30)10 (0.2485) Ay Ans             At x = l /2 = 19.5 in Chapter 4 - Rev B, Page 8/81 • 2 2 2 3[ ( / 2)] ( / 2)2 2 6 2 2 24 2 2 Fa l l l l l l ly a l l EIl EI                               w 3 l     2 2 2 6 2 3 3 6 340(15)(19.5) 19.5 15 39 6(30)(10 )(0.2485)(39) (150 /12)(19.5) 2(39)(19.5 ) 19.5 39 0.1027 in . 24(30)(10 )(0.2485) y Ans             0.1027 0.0978% difference (100) 5.01% . 0.0978 Ans    ______________________________________________________________________________ 4-13  3 31 (6)(32 ) 16.384 10 mm12I   4 From Table A-9-10, beam 10 2 ( ) 3C Fay l a EI     2 26AB Faxy l x EIl   2 2( 3 6 ABdy Fa l x dx EIl   ) At x = 0, AB A dy dx  2 6 6A Fal Fal EIl EI    2 6O A Fa ly a EI     With both loads, 2 2 ( ) 6 3O Fa l Fay l a EI EI       2 2 3 3 400(300 )(3 2 ) 3(500) 2(300) 3.72 mm . 6 6(207)10 (16.384)10 Fa l a Ans EI         At midspan,     2 2 2 2 3 3 2 ( / 2) 3 3 400(300)(500 ) 1.11 mm . 6 2 24 24 207 10 16.384 10E Fa l l Faly l EIl EI              Ans _____________________________________________________________________________ 4-14 4 4(2 1.5 ) 0.5369 in 64 I 4   Chapter 4 - Rev B, Page 9/81 • From Table A-5, E = 10.4 Mpsi From Table A-9, beams 1 and 2, by superposition       3 23 2 6 6 200 4(12) 300 2(12) ( 3 ) 2(12) 3(4)(12) 3 6 3(10.4)10 (0.5369) 6(10.4)10 (0.5369) B A B F l F ay a l EI EI         1.94 in .By Ans  ______________________________________________________________________________ 4-15 From Table A-7, I = 2(1.85) = 3.70 in4 From Table A-5, E = 30.0 Mpsi From Table A-9, beams 1 and 3, by superposition   443 3 6 6 5 2(5 /12) (60 )( ) 150(60 ) 0.182 in . 3 8 3(30)10 (3.70) 8(30)10 (3.70) c A lFly Ans EI EI          w w ______________________________________________________________________________ 4-16 4 64 I d From Table A-5, 3207(10 ) MPaE  From Table A-9, beams 5 and 9, with FC = FA = F, by superposition 3 2 2 3 2 21(4 3 ) 2 (4 3 ) 48 24 48 B B B B F l Fay a l I F l Fa a l EI EI Ey                   3 2 3 3 4 1 550(1000 ) 2 375 (250) 4(250 ) 3(1000 ) 48(207)10 2 53.624 10 mm I 2       34 464 64 (53.624)10 32.3 mm .d I A      ns ______________________________________________________________________________ 4-17 From Table A-9, beams 8 (region BC for this beam with a = 0) and 10 (with a = a), by superposition         3 2 2 2 2 3 2 2 2 2 3 2 6 6 1 3 2 6 A AB A M Faxy x lx l x l x EIl EIl .M x lx l x Fax l x An EIl             s  3 2 2 2( )3 2 ( ) [( ) (36 6 A BC x l Md F x ly x lx l x x l x l a x l dx EIl EI               )] 2( )( ) [( ) (3 ) 6 6 AM l F x l ]x l x l a x l EI EI         Chapter 4 - Rev B, Page 10/81 •  2( ) ( ) (3 )6 A x l .M l F x l a x l Ans EI          ______________________________________________________________________________ 4-18 Note to the instructor: Beams with discontinuous loading are better solved using singularity functions. This eliminates matching the slopes and displacements at the discontinuity as is done in this solution.  1 10 22 2C a a .M R l a l a R l a Ans l              ww   2 2 20 22 2y a aF l a R a R l l        w ww .Ans     21 2 2 .V R 2 2AB a l a l a x a Ans l l         w wwx = wx = 2 2 .2BC aV R A l     w ns 2 2 122 2AB AB xM V dx l ax a x C l               w 210 at 0 0 2 .2AB AB M x C M al a lx A l           wx ns 2 2 22 2BC BC a aM V dx dx x C l l        w w 2 2 20 at ( ) .2 2BC BC a aM x l C M l x Ans l        w w  2 2 2 2 3 2 2 2 3 3 3 2 3 4 3 4 4 1 1 12 2 2 2 1 1 1 2 2 3 1 1 1 1 2 3 6 12 0 at 0 0 AB AB AB AB AB M xdx al a lx dx alx a x lx C EI EI l EI l y dx alx a x lx C dx EI l alx a x lx C x C EI l y x C   31 3                                             w w w w 2 2 2 5 2 3 2 4 3 2 3 5 3 1 1 1( ) 2 2 2 at 1 1 1 1 1 (1) 2 2 3 2 2 6 BC BC AB BC M a adx l x dx lx x C EI EI l EI l x a a aala a la C la a C C C EI l EI l                                              w w w w w 5 Chapter 4 - Rev B, Page 11/81 • 2 2 2 2 3 5 5 2 2 6 5 2 2 3 3 5 1 1 1 1 1 2 2 2 2 6 0 at 6 1 1 1 1 ( ) 2 2 6 3 BC BC BC BC a ay dx lx x C dx lx x C x EI l EI l a ly x l C C l ay lx x l C x l EI l                                            w w w w 6C      2 3 5 4 2 3 3 3 5 2 2 3 3 5 at 1 1 1 1 1 1 ( ) 2 3 6 12 2 2 6 3 3 4 ( ) (2) 24 AB BCy y x a aala a la C a la a l C a l l l aC a la l C a l l                          w w w Substituting (1) into (2) yields  2 2 2 5 424 aC a l    w l . Substituting this back into (2) gives   2 2 2 3 4 424 aC al a l    w l . Thus,  3 2 3 4 3 4 2 24 2 4 424ABy alx a x lx a lx a x a l xEIl      w  22 3 22 (2 ) 2 24AB xy ax l a lx a l a .Ans EIl         w  2 2 2 3 4 2 2 46 2 424BCy a lx a x a x a l x a l Ans.EIl     w This result is sufficient for yBC. However, this can be shown to be equivalent to  3 2 3 4 2 2 3 4 4 4 2 4 4 ( 24 24 ( ) . 24 BC BC AB y alx a x lx a l x a lx a x x a EIl EI y y x a Ans EI           w w w 4) by expanding this or by solving the problem using singularity functions. ______________________________________________________________________________ 4-19 The beam can be broken up into a uniform load w downward from points A to C and a uniform load upward from points A to B.         2 22 3 2 2 3 2 2 22 2 2 2 2 (2 ) 2 2 (2 ) 2 24 24 2 (2 ) 2 2 (2 ) 2 . 24 AB x xy bx l b lx b l b ax l a lx a l EIl EIl x bx l b b l b ax l a a l a Ans EIl                        w w w a       23 4 2 3 2 3 4 2 2 3 4 4 2 (2 ) 2 24 4 2 4 4 ( ) BCy bx l b lx b x l bEIl alx a x lx a l x a lx a x l x a Ans               w . Chapter 4 - Rev B, Page 12/81 • 3 2 3 4 2 2 3 4 4 3 2 3 4 2 2 3 4 4 4 4 4 2 4 4 ( ) 24 4 2 4 4 ( ) 24 ( ) ( ) . 24 CD AB y blx b x lx b l x b lx b x l x b EIl alx a x lx a l x a lx a x l x a EIl x b x a y Ans EI                           w w w  ______________________________________________________________________________ 4-20 Note to the instructor: See the note in the solution for Problem 4-18.   2 0 2 2 2y B B a aF R a R l a A l l        w ww .ns For region BC, isolate right-hand element of length (l + a  x)   2 , . 2AB A BC aV R V l a x An l        w w s   2 2, . 2 2AB A BC aM R x x M l a x Ans l         w w 2 2 14AB AB aEI M dx x C l     w  2 3 1 212AB aEIy x C x C l     w yAB = 0 at x = 0  C2 = 0  2 3 112AB aEIy x C x l    w yAB = 0 at x = l  2 1 12 a lC w      2 2 2 2 3 2 2 2 . 12 12 12 12AB AB a a l a x a xEIy x x l x y l x Ans l l EIl         w w w w 2  3 36BC BCEI M dx l a x      w C  4 3 424BCEIy l a x C x C      w yBC = 0 at x = l  4 4 3 4 4024 24 a aC l C C C l      w w 3 (1) AB = BC at x = l    2 2 3 2 3 34 12 6 6 a l a l aC C l       w w wa w a Substitute C3 into Eq. (1) gives   2 2 4 424 aC a l l a     w . Substitute back into yBC            2 4 2 4 4 2 4 1 24 6 24 6 4 . 24 BC ly l a x x l a EI l a x a l x l a a Ans EI                         w wa wa wa w l a Chapter 4 - Rev B, Page 13/81 • 4-21 Table A-9, beam 7, 1 2 100(10) 500 lbf 2 2 lR R   w            2 3 3 2 3 3 6 6 2 3 1002 2(10) 10 24 24 30 10 0.05 2.7778 10 20 1000 AB x xy lx x l x x EI x x x            w Slope:  2 3 36 424 AB AB d y lx x l d x EI     w At x = l,   3 2 3 36 4 24 24AB x l ll l l l EI EI       w w            33 3 6 100 10 10 2.7778 10 10 24 24(30)10 (0.05)BC AB x l ly x l x l x x EI            w From Prob. 4-20,         22 100 4 100 4 80 lbf 2 2(10) 4 480 lbf 2 2(10) 2 2(10)A B a aR R l a l l          w w              22 2 2 2 2 6 2 6 100 4 10 8.8889 10 100 12 12 30 10 0.05AB xa xy l x x x EIl       w x                    4 2 4 4 2 4 6 46 4 24 100 10 4 4 4 10 10 4 4 24 30 10 0.05 2.7778 10 14 896 9216 BCy l a x a l x l a aEI x x x x                            w Superposition, 500 80 420 lbf 500 480 980 lbf .A BR R A        ns        6 2 3 6 22.7778 10 20 1000 8.8889 10 100 .ABy x x x x x Ans            43 62.7778 10 10 2.7778 10 14 896 9216 .BCy x x x          Ans The deflection equations can be simplified further. However, they are sufficient for plotting. Using a spreadsheet, x 0 0.5 1 1.5 2 2.5 3 3.5 y 0.000000 -0.000939 -0.001845 -0.002690 -0.003449 -0.004102 -0.004632 -0.005027 x 4 4.5 5 5.5 6 6.5 7 7.5 y -0.005280 -0.005387 -0.005347 -0.005167 -0.004853 -0.004421 -0.003885 -0.003268 Chapter 4 - Rev B, Page 14/81 • x 8 8.5 9 9.5 10 10.5 11 11.5 y -0.002596 -0.001897 -0.001205 -0.000559 0.000000 0.000439 0.000775 0.001036 x 12 12.5 13 13.5 14 y 0.001244 0.001419 0.001575 0.001722 0.001867 ______________________________________________________________________________ 4-22 (a) Useful relations   3 33 4 6 48 1800 36 0.05832 in 48 48(30)10 F EIk y l klI E      From I = bh 3/12, and b = 10 h, then I = 5 h 4/6, or, 4 4 6 6(0.05832) 0.514 in 5 5 Ih    h is close to 1/2 in and 9/16 in, while b is close to 5.14 in. Changing the height drastically changes the spring rate, so changing the base will make finding a close solution easier. Trial and error was applied to find the combination of values from Table A-17 that yielded the closet desired spring rate. h (in) b (in) b/h k (lbf/in) 1/2 5 10 1608 1/2 5½ 11 1768 1/2 5¾ 11.5 1849 9/16 5 8.89 2289 9/16 4 7.11 1831 Chapter 4 - Rev B, Page 15/81 • h = ½ in, b = 5 ½ in should be selected because it results in a close spring rate and b/h is still reasonably close to 10. (b) 3 45.5(0.5) /12 0.05729 inI       3 33 6 ( / 4) 4 4(60)10 (0.05729) 1528 lbf 36 (0.25) (1528) 36 0.864 in . 48 48(30)10 (0.05729) Mc Fl c IF I I lc Fly A EI           ns ______________________________________________________________________________ 4-23 From the solutions to Prob. 3-68, 1 260 lbf and 400 lbfT T  4 4 41198 in(1.25) 0. 64 64 dI     From Table A-9, beam 6,     2 2 2 2 2 21 1 2 2 1 2 10in 2 2 2 6 2 2 2 6 ( ) ( ) 6 6 ( 575)(30)(10) 10 30 40 6(30)10 (0.1198)(40) 460(12)(10) 10 12 40 0.0332 in . 6(30)10 (0.1198)(40) A x Fb x F b xz x b l x b l EIl EIl Ans                       2 2 2 2 2 21 1 2 2 1 2 10in 10in 2 2 2 2 2 21 1 2 2 1 2 10in 2 2 2 6 ( ) ( ) 6 6 (3 ) (3 ) 6 6 (575)(30) 3 10 30 40 6(30)10 (0.1198)(40) 460(12) 6(30 A y x x x Fb x F b xd z d x b l x b l dx dx EIl EIl Fb F bx b l x b l EIl EIl                                             2 2 26 4 3 10 12 40 )10 (0.1198)(40) 6.02(10 ) rad .Ans      ______________________________________________________________________________ 4-24 From the solutions to Prob. 3-69, 1 22880 N and 432 NT T    4 4 3 4(30) 39.76 10 mm 64 64 dI     Chapter 4 - Rev B, Page 16/81 • The load in between the supports supplies an angle to the overhanging end of the beam. That angle is found by taking the derivative of the deflection from that load. From Table A-9, beams 6 (subscript 1) and 10 (subscript 2),    2 beam10beam6A BC ACy a y    (1)         1 1 2 2 2 2 21 1 1 1 2 21 1 1 2 6 3 6 6 6 BC C 2 x lx l F a l x F ad x a lx lx x a l dx EIl EIl F a l a EIl                        Equation (1) is thus        2 2 21 1 2 2 1 2 2 2 2 2 3 3 3 3 ( ) 6 3 3312(230) 2070(300 )510 230 300 510 300 6(207)10 (39.76)10 (510) 3(207)10 (39.76)10 7.99 mm . A F a F ay l a a l a EIl EI Ans            The slope at A, relative to the z axis is       2 2 2 2 21 1 2 1 2 2 2 21 1 2 1 2 2 2 2 21 1 2 1 2 2 2 3 3 ( )( ) ( ) (3 ) 6 6 3( ) 3 ( ) (3 ) 6 6 ( ) 3 2 6 6 3312(230) 510 2 6(207)10 (39.76)10 (510) A z x l a x l a F a F x ldl a x l a x l EIl dx EI F a Fl a x l a x l a x l EIl EI F a Fl a a la EIl EI                                      2 2 3 3 30 2070 3(300 ) 2(510)(300) 6(207)10 (39.76)10 0.0304 rad .Ans       ______________________________________________________________________________ 4-25 From the solutions to Prob. 3-70, 1 2392.16 lbf and 58.82 lbfT T  4 4 4(1) 0.049 09 in 64 64 dI     From Table A-9, beam 6, Chapter 4 - Rev B, Page 17/81 •    2 2 2 2 2 21 1 1 6 8in ( 350)(14)(8) 8 14 22 0.0452 in . 6 6(30)10 (0.049 09)(22)A x F b xy x b l Ans EIl             2 2 2 2 2 22 2 2 6 8in ( 450.98)(6)(8)( ) 8 6 22 0.0428 in . 6 6(30)10 (0.049 09)(22)A x F b xz x b l EIl            Ans The displacement magnitude is 2 2 2 20.0452 0.0428 0.0622 in .A Ay z Ans            11 2 2 2 2 2 21 1 1 1 1 1 2 2 2 6 (3 ) 6 6 ( 350)(14) 3 8 14 22 0.00242 rad . 6(30)10 (0.04909)(22) A z x ax a F b x F bd y d 1x b l a b ld x dx EIl EIl Ans                                  11 2 2 2 2 2 22 2 2 2 2 1 2 2 2 6 ( ) 3 6 6 (450.98)(6) 3 8 6 22 0.00356 rad . 6(30)10 (0.04909)(22) A y x ax a F b x F bd z d 2x b l a b ld x dx EIl EIl Ans                               The slope magnitude is  220.00242 0.00356 0.00430 rad .A Ans     ______________________________________________________________________________ 4-26 From the solutions to Prob. 3-71, 1 2250 N and 37.5 NT T  4 4 4(20) 7 854 mm 64 64 dI         o 1 1 2 2 2 2 2 2 1 3 300mm 345sin 45 (550)(300) ( ) 300 550 850 6 6(207)10 (7 854)(850) 1.60 mm . y A x F b x y x b l EIl Ans              2 2 2 2 2 21 1 2 2 1 2 300mm ( ) ( ) 6 6 z A x F b x F b xz x b l x b l EIl EIl                 o 2 2 2 3 2 2 2 3 345cos 45 (550)(300) 300 550 850 6(207)10 (7 854)(850) 287.5(150)(300) 300 150 850 0.650 mm . 6(207)10 (7 854)(850) Ans          The displacement magnitude is  22 2 21.60 0.650 1.73 mm .A Ay z Ans       Chapter 4 - Rev B, Page 18/81 •         1 1 1 1 1 12 2 2 2 2 2 1 1 o 2 2 2 3 (3 ) 6 6 345sin 45 (550) 3 300 550 850 0.00243 rad . 6(207)10 (7 854)(850) y y A z x a x a F b x F bd y d x b l a b l d x dx EIl EIl Ans                                 1               11 2 2 2 2 2 21 1 2 2 1 2 2 2 2 2 2 21 1 2 2 1 1 1 2 o 2 2 2 3 3 6 6 3 3 6 6 345cos 45 (550) 3 300 550 850 6(207)10 (7 854)(850) 287.5(150) 6(207)10 (7 85 z A y x ax a z F b x F b xd z d x b l x b l d x dx EIl EIl F b F ba b l a b l EIl EIl                                       2 2 2 43 300 150 850 1.91 10 rad .4)(850) Ans       The slope magnitude is 2 20.00243 0.000191 0.00244 rad .A Ans    ______________________________________________________________________________ 4-27 From the solutions to Prob. 3-72, 750 lbfBF  4 4 4(1.25) 0.1198 in 64 64 dI     From Table A-9, beams 6 (subscript 1) and 10 (subscript 2)             1 1 2 22 2 2 2 2 1 16in o o 2 2 2 2 2 6 6 6 6 300cos 20 (14)(16) 750sin 20 (9)(16) 16 14 30 30 16 6(30)10 (0.119 8)(30) 6(30)10 (0.119 8)(30) 0.0805 in . y y A x F b x F a x y x b l l x EIl EIl Ans                               2 2 2 2 21 1 2 2 1 16in o o 2 2 2 2 2 6 6 6 6 300sin 20 (14)(16) 750cos 20 (9)(16) 16 14 30 30 16 6(30)10 (0.119 8)(30) 6(30)10 (0.119 8)(30) 0.1169 in . z z A x F b x F a xz x b l l x EIl EIl Ans                  The displacement magnitude is  22 2 20.0805 0.1169 0.142 in .A Ay z Ans       Chapter 4 - Rev B, Page 19/81 •                 1 1 1 1 2 22 2 2 2 2 1 1 1 2 22 2 2 2 2 1 1 1 o 2 2 2 6 o 6 6 6 3 3 6 6 300cos 20 (14) 3 16 14 30 6(30)10 (0.119 8)(30) 750sin 20 (9) 3 6(30)10 (0.119 8)(30) y y A z x a x a y y F b x F a xd y d x b l l x d x dx EIl EIl F b F a a b l l a EIl EIl                                        2 2 50 3 16 8.06 10 rad .Ans                    11 2 2 2 2 21 1 2 2 1 2 2 2 2 21 1 2 2 1 1 1 o o 2 2 2 6 6 6 6 3 3 6 6 300sin 20 (14) 750cos 20 (9) 3 16 14 30 3 6(30)10 (0.119 8)(30) 6(30)10 (0.119 8)(30) z z A y x ax a z z F b x F a xd z d x b l l x d x dx EIl EIl F b F aa b l l a EIl EIl                                     2 20 3 16 0.00115 rad .Ans     The slope magnitude is   25 28.06 10 0.00115 0.00115 rad .A Ans      ______________________________________________________________________________ 4-28 From the solutions to Prob. 3-73, FB = 22.8 (103) N     44 3 4 50 306.8 10 mm 64 64 dI     From Table A-9, beam 6,         1 1 2 22 2 2 2 2 2 1 2 400mm 3 o 2 2 2 3 3 3 o 2 2 3 3 ( ) ( ) 6 6 11 10 sin 20 (650)(400) 400 650 1050 6(207)10 (306.8)10 (1050) 22.8 10 sin 25 (300)(400) 400 300 1050 6(207)10 (306.8)10 (1050) 3.735 y y A x F b x F b x y x b l x b l EIl EIl                       mm . 2   Ans Chapter 4 - Rev B, Page 20/81 •         2 2 2 2 2 21 1 2 2 1 2 400mm 3 o 2 2 2 3 3 3 o 2 2 2 3 3 ( ) ( ) 6 6 11 10 cos 20 (650)(400) 400 650 1050 6(207)10 (306.8)10 (1050) 22.8 10 cos 25 (300)(400) 400 300 1050 1.791 6(207)10 (306.8)10 (1050) z z A x F b x F b xz x b l x b l EIl EIl                      mm .Ans The displacement magnitude is  22 2 23.735 1.791 4.14 mm .A Ay z Ans                       11 2 2 2 2 2 21 1 2 2 1 2 1 1 2 22 2 2 2 2 2 1 1 1 2 3 o 2 2 2 3 3 3 o 6 6 3 3 6 6 11 10 sin 20 (650) 3 400 650 1050 6(207)10 (306.8)10 (1050) 22.8 10 sin 25 z z A z x ax a y y F b x F b xd y d x b l x b l d x dx EIl EIl F b F b a b l a b l EIl EIl                                     2 2 23 3 (300) 3 400 300 1050 6(207)10 (306.8)10 (1050) 0.00507 rad .Ans                          11 2 2 2 2 2 21 1 2 2 1 2 2 2 2 2 2 21 1 2 2 1 1 1 2 3 o 2 2 2 3 3 3 6 6 3 3 6 6 11 10 cos 20 (650) 3 400 650 1050 6(207)10 (306.8)10 (1050) 22.8 10 co z z A y x ax a z z F b x F b xd z d x b l x b l d x dx EIl EIl F b F ba b l a b l EIl EIl                                           o 2 2 2 3 3 s 25 (300) 3 400 300 1050 6(207)10 (306.8)10 (1050) 0.00489 rad .Ans          The slope magnitude is    2 20.00507 0.00489 0.00704 rad .A Ans      ______________________________________________________________________________ 4-29 From the solutions to Prob. 3-68, T1 = 60 lbf and T2 = 400 lbf , and Prob. 4-23, I = 0.119 8 in4. From Table A-9, beam 6, Chapter 4 - Rev B, Page 21/81 •               2 2 2 2 2 21 1 2 2 1 2 00 2 2 2 2 2 21 1 2 2 1 2 6 2 2 6 6 6 575(30) 30 40 6 6 6(30)10 (0.119 8)(40) 460(12) 12 40 0.00468 rad 6(30)10 (0.119 8)(40) z z O y xx z z F b x F b xd z d x b l x b l d x dx EIl EIl F b F bb l b l EIl EIl                                    .Ans                     1 1 2 22 2 2 2 1 2 2 2 2 2 2 21 1 2 2 1 2 2 2 2 21 1 2 2 1 2 2 2 2 2 6 6 6 2 3 6 2 3 6 6 6 6 575(10) 40 10 6(3 z z C y x l x l z z x l z z F a l x F a l xd z d x a lx x a lx d x dx EIl EIl F a F alx l x a lx l x a EIl EIl F a F al a l a EIl EIl                                                    2 2 6 6 460(28) 40 28 0.00219 rad . 0)10 (0.119 8)(40) 6(30)10 (0.119 8)(40) Ans     ______________________________________________________________________________ 4-30 From the solutions to Prob. 3-69, T1 = 2 880 N and T2 = 432 N, and Prob. 4-24, I = 39.76 (103) mm4. From Table A-9, beams 6 and 10     2 2 2 2 21 1 2 2 1 00 2 2 2 2 2 2 21 1 2 2 1 1 2 2 1 1 0 2 2 3 3 ( ) ( ) 6 6 (3 ) ( 3 ) ( ) 6 6 6 3 312(280) 2 070(300)280 510 6(207)10 (39.76)10 (510) O z xx x Fb x F a xd y d x b l l x d x dx EIl EIl Fb F a Fb F a lx b l l x b l EIl EIl EIl EI                                    6 3 3 (510) 6(207)10 (39.76)10 0.0131 rad .Ans   2 2 2 21 1 2 21 2 2 2 2 2 2 21 1 2 2 1 1 2 2 1 1 2 3 3 ( ) ( 2 ) ( ) 6 6 (6 2 3 ) ( 3 ) ( ) 6 6 6 3 312(230) (510 230 6(207)10 (39.76)10 (510) C z x lx l x l F a l x F a xd y d x a lx l x d x dx EIl EIl 3 F a F a F alx l x a l x l a F a l EIl EIl EIl EI                                     2 3 3 2 070(300)(510)) 3(207)10 (39.76)10 0.0191 rad .Ans    ______________________________________________________________________________ 4-31 From the solutions to Prob. 3-70, T1 = 392.19 lbf and T2 = 58.82 lbf , and Prob. 4-25, I = 0.0490 9 in4. From Table A-9, beam 6 Chapter 4 - Rev B, Page 22/81 •       1 1 1 12 2 2 2 2 1 1 0 0 2 2 6 ( ) 6 6 350(14) 14 22 0.00726 rad . 6(30)10 (0.04909)(22) y y O z x x F b x F bd y d x b l b l d x dx EIl EIl Ans                                     2 2 2 2 22 2 2 2 2 2 00 2 2 6 6 6 450.98(6) 6 22 6(30)10 (0.04909)(22) 0.00624 rad . z z O y xx F b x F bd z d x b l b l d x dx EIl EIl Ans                             The slope magnitude is  220.00726 0.00624 0.00957 rad .O Ans             1 1 2 2 1 1 1 1 12 2 2 2 2 1 1 2 2 6 ( ) 2 6 6 2 3 ( ) 6 6 350(8) 22 8 0.00605 rad . 6(30)10 (0.0491)(22) y C z x l x l y y x l F a l xd y d x a lx d x dx EIl F a F a lx l x a l a EIl EIl Ans                                                   2 22 2 2 2 2 2 2 22 2 2 2 2 2 2 2 6 ( ) 2 6 6 2 3 6 6 450.98(16) 22 16 0.00846 rad . 6(30)10 (0.04909)(22) z C y x lx l z z x l F a l xd z d x a lx d x dx EIl F a F alx l x a l a EIl EIl Ans                                      The slope magnitude is  2 20.00605 0.00846 0.0104 rad .C Ans     ______________________________________________________________________________ 4-32 From the solutions to Prob. 3-71, T1 =250 N and T1 =37.5 N, and Prob. 4-26, I = 7 854 mm4. From Table A-9, beam 6       1 1 1 12 2 2 2 2 1 1 0 0 o 2 2 3 ( ) 6 6 345sin 45 (550) 550 850 0.00680 rad . 6(207)10 (7 854)(850) y y O z x x F b x F bd y d x b l b l d x dx EIl EIl Ans                              Chapter 4 - Rev B, Page 23/81 •               2 2 2 2 2 21 1 2 2 1 2 00 o 2 2 2 2 2 21 1 2 2 1 2 3 2 2 3 6 6 345cos 45 (550) 550 850 6 6 6(207)10 (7 854)(850) 287.5(150) 150 850 6(207)10 (7 854)(850) z z O y xx z z F b x F b xd z d x b l x b l d x dx EIl EIl F b F bb l b l EIl EIl                                    0.00316 rad .Ans The slope magnitude is 2 20.00680 0.00316 0.00750 rad .O Ans            1 1 1 12 2 2 2 2 1 1 o 1 1 2 2 2 2 1 3 ( ) 2 6 2 3 6 6 345sin 45 (300) ( ) 850 300 0.00558 rad . 6 6(207)10 (7 854)(850) y y C z x l x lx l y F a l x F ad y d x a lx lx l x a d x dx EIl EIl F a l a Ans EIl                                                      2 2 2 21 1 2 2 1 2 o 2 2 2 2 2 21 1 2 2 1 2 3 3 ( ) ( )2 2 6 6 345cos 45 (300) 850 300 6 6 6(207)10 (7 854)(850) 287.5(700) 6(207)10 (7 854)(850 z z C y x lx l z z F a l x F a l xd z d x a lx x a lx d x dx EIl EIl F a F al a l a EIl EIl                                        2 2 5850 700 6.04 10 rad .) Ans   The slope magnitude is     22 50.00558 6.04 10 0.00558 rad .C Ans       ________________________________________________________________________ 4-33 From the solutions to Prob. 3-72, FB = 750 lbf, and Prob. 4-27, I = 0.119 8 in4. From Table A-9, beams 6 and 10               1 1 2 22 2 2 2 2 1 0 0 1 1 2 2 1 1 2 22 2 2 2 2 2 2 1 1 0 o 2 2 6 6 6 3 3 6 6 6 6 300cos 20 (14) 750sin 2 14 30 6(30)10 (0.119 8)(30) y y O z x x y y y y x F b x F a xd y d x b l l x d x dx EIl EIl F b F a F b F a l x b l l x b l EIl EIl EIl EI                                            o 6 0 (9)(30) 0.00751 rad . 6(30)10 (0.119 8) Ans     Chapter 4 - Rev B, Page 24/81 •               2 2 2 2 21 1 2 2 1 00 2 2 2 2 2 2 21 1 2 2 1 1 2 2 1 1 0 o 2 2 6 6 6 3 3 6 6 6 6 300sin 20 (14) 750cos 14 30 6(30)10 (0.119 8)(30) z z O y xx z z z z x F b x F a xd z d x b l l x d x dx EIl EIl F b F a F b F a lx b l l x b l EIl EIl EIl EI                                           o 6 20 (9)(30) 0.0104 rad . 6(30)10 (0.119 8) Ans     The slope magnitude is 2 20.00751 0.0104 0.0128 rad .O Ans              1 1 2 22 2 2 2 1 1 1 2 2 1 1 2 22 2 2 2 2 2 2 1 1 o 2 6 ( ) 2 6 6 6 2 3 3 ( ) 6 6 6 300cos 20 (16) 30 6(30)10 (0.119 8)(30) y y C z x l x l y y y x l F a l x F a xd y d x a lx l x dx dx EIl EIl F a F a F a F a l lx l x a l x l a EIl EIl EIl EI                                              3 y o 2 6 750sin 20 (9)(30) 16 0.0109 rad . 3(30)10 (0.119 8) Ans                   2 2 2 21 1 2 2 1 2 2 2 2 2 2 21 1 2 2 1 1 2 2 1 1 o 2 6 ( ) 2 6 6 6 2 3 3 6 6 6 300sin 20 (16) 30 1 6(30)10 (0.119 8)(30) z z C y x lx l z z z x l F a l x F a xd z d x a lx l x d x dx EIl EIl F a F a F a F a llx l x a l x l a EIl EIl EIl EI                                            3 z o 2 6 750cos 20 (9)(30) 6 0.0193 rad . 3(30)10 (0.119 8) Ans      The slope magnitude is    2 20.0109 0.0193 0.0222 rad .C Ans      ______________________________________________________________________________ 4-34 From the solutions to Prob. 3-73, FB = 22.8 kN, and Prob. 4-28, I = 306.8 (103) mm4. From Table A-9, beam 6                 1 1 2 22 2 2 2 2 2 1 2 0 0 3 o 1 1 2 22 2 2 2 2 2 1 2 3 3 3 o 3 6 6 11 10 sin 20 (650) 650 1050 6 6 6(207)10 (306.8)10 (1050) 22.8 10 sin 25 (300) 6(207)10 ( y y O z x x y y F b x F b xd y d x b l x b l d x dx EIl EIl F b F b b l b l EIl EIl                                      2 23 300 1050 0.0115 rad .306.8)10 (1050) Ans   Chapter 4 - Rev B, Page 25/81 •                 2 2 2 2 2 21 1 2 2 1 2 00 2 2 2 21 1 2 2 1 2 3 o 2 2 3 3 3 o 3 6 6 6 6 11 10 cos 20 (650) 650 1050 6(207)10 (306.8)10 (1050) 22.8 10 cos 25 (300) 6(207)10 z z O y xx z z F b x F b xd z d x b l x b l d x dx EIl EIl F b F bb l b l EIl EIl                                     2 23 300 1050 0.00427 rad .(306.8)10 (1050) Ans   The slope magnitude is    2 20.0115 0.00427 0.0123 rad .O Ans                    1 1 2 22 2 2 2 1 2 1 1 2 22 2 2 2 2 2 1 2 3 o 1 1 2 22 2 2 2 1 2 ( ) ( ) 2 2 6 6 (6 2 3 ) 6 2 3 6 6 11 10 sin 20 (4 6 6 y y C z x l x l y y x l y y F a l x F a l xd y d x a lx x a lx d x dx EIl EIl F a F a lx l x a lx l x a EIl EIl F a F a l a l a EIl EIl                                                       2 2 3 3 3 o 2 2 3 3 00) 1050 400 6(207)10 (306.8)10 (1050) 22.8 10 sin 25 (750) 1050 750 0.0133 rad . 6(207)10 (306.8)10 (1050) Ans                       2 2 2 21 1 2 2 1 2 2 2 2 2 2 21 1 2 2 1 2 3 o 2 2 2 21 1 2 2 1 2 ( ) ( )2 2 6 6 6 2 3 6 2 3 6 6 11 10 cos 20 (40 6 6 z z C y x lx l z z x l z z F a l x F a l xd z d x a lx x a lx d x dx EIl EIl F a F alx l x a lx l x a EIl EIl F a F al a l a EIl EIl                                                     2 2 3 3 3 o 2 2 3 3 0) 1050 400 6(207)10 (306.8)10 (1050) 22.8 10 cos 25 (750) 1050 750 0.0112 rad . 6(207)10 (306.8)10 (1050) Ans       The slope magnitude is 2 20.0133 0.0112 0.0174 rad .C Ans    ______________________________________________________________________________ 4-35 The required new slope in radians is  new = 0.06( /180) = 0.00105 rad. In Prob. 4-29, I = 0.119 8 in 4, and it was found that the greater angle occurs at the bearing at O where (O)y =  0.00468 rad. Since is inversely proportional to I, Chapter 4 - Rev B, Page 26/81 •  new Inew =  old Iold  Inew =  /64 = 4newd old Iold / new or, 1/4 old new old new 64d I           The absolute sign is used as the old slope may be negative. 1/4 new 64 0.00468 0.119 8 1.82 in . 0.00105 d A          ns ______________________________________________________________________________ 4-36 The required new slope in radians is  new = 0.06( /180) = 0.00105 rad. In Prob. 4-30, I = 39.76 (103) mm4, and it was found that the greater angle occurs at the bearing at C where (C)y =  0.0191 rad. See the solution to Prob. 4-35 for the development of the equation 1/4 old new old new 64d I             1/4 3 new 64 0.0191 39.76 10 62.0 mm . 0.00105 d A          ns ______________________________________________________________________________ 4-37 The required new slope in radians is  new = 0.06( /180) = 0.00105 rad. In Prob. 4-31, I = 0.0491 in4, and the maximum slope is C = 0.0104 rad. See the solution to Prob. 4-35 for the development of the equation 1/4 old new old new 64d I           1/4 new 64 0.0104 0.0491 1.77 in . 0.00105 d A         ns ______________________________________________________________________________ 4-38 The required new slope in radians is  new = 0.06( /180) = 0.00105 rad. In Prob. 4-32, I = 7 854 mm4, and the maximum slope is O = 0.00750 rad. See the solution to Prob. 4-35 for the development of the equation Chapter 4 - Rev B, Page 27/81 • 1/4 old new old new 64d I           1/4 new 64 0.00750 7 854 32.7 mm . 0.00105 d A         ns ______________________________________________________________________________ 4-39 The required new slope in radians is  new = 0.06( /180) = 0.00105 rad. In Prob. 4-33, I = 0.119 8 in4, and the maximum slope  = 0.0222 rad. See the solution to Prob. 4-35 for the development of the equation 1/4 old new old new 64d I           1/4 new 64 0.0222 0.119 8 2.68 in . 0.00105 d A         ns ______________________________________________________________________________ 4-40 The required new slope in radians is  new = 0.06( /180) = 0.00105 rad. In Prob. 4-34, I = 306.8 (103) mm4, and the maximum slope is C = 0.0174 rad. See the solution to Prob. 4-35 for the development of the equation 1/4 old new old new 64d I             1/4 3 new 64 0.0174 306.8 10 100.9 mm . 0.00105 d A         ns ______________________________________________________________________________ 4-41 IAB =  14/64 = 0.04909 in4, JAB = 2 IAB = 0.09818 in4, IBC = (0.25)(1.5)3/12 = 0.07031 in4, ICD =  (3/4)4/64 = 0.01553 in4. For Eq. (3-41), p. 102, b/c = 1.5/0.25 = 6   = 0.299. The deflection can be broken down into several parts 1. The vertical deflection of B due to force and moment acting on B (y1). 2. The vertical deflection due to the slope at B, B1, due to the force and moment acting on B (y2 = CDB1 = 2B1). Chapter 4 - Rev B, Page 28/81 • 3. The vertical deflection due to the rotation at B, B2, due to the torsion acting at B (y3 = BC B1 = 5B1). 4. The vertical deflection of C due to the force acting on C (y4). 5. The rotation at C, C, due to the torsion acting at C (y3 = CDC = 2C). 6. The vertical deflection of D due to the force acting on D (y5). 1. From Table A-9, beams 1 and 4 with F =  200 lbf and MB = 2(200) = 400 lbfin             3 2 1 6 6 200 6 400 6 0.01467 in 3 30 10 0.04909 2 30 10 0.04909 y      2. From Table A-9, beams 1 and 4                22 1 6 3 3 6 6 2 6 62 200 6 2 400 0.004074 rad 2 2 30 10 0.04909 B B B x lx l B M x M xd Fx Fxx l x l dx EI EI EI EI l Fl M EI                                    y 2 = 2(0.004072) = 0.00815 in 3. The torsion at B is TB = 5(200) = 1000 lbfin. From Eq. (4-5)    2 6 1000 6 0.005314 rad 0.09818 11.5 10B AB TL JG         y 3 = 5(0.005314) = 0.02657 in 4. For bending of BC, from Table A-9, beam 1       3 4 6 200 5 0.00395 in 3 30 10 0.07031 y     5. For twist of BC, from Eq. (3-41), p. 102, with T = 2(200) = 400 lbfin      3 6 400 5 0.02482 rad 0.299 1.5 0.25 11.5 10C    y 5 = 2(0.02482) = 0.04964 in 6. For bending of CD, from Table A-9, beam 1       3 6 6 200 2 0.00114 in 3 30 10 0.01553 y     Chapter 4 - Rev B, Page 29/81 • Summing the deflections results in 6 1 0.01467 0.00815 0.02657 0.00395 0.04964 0.00114 0.1041 in .D i i y y A          ns This problem is solved more easily using Castigliano’s theorem. See Prob. 4-71. ______________________________________________________________________________ 4-42 The deflection of D in the x direction due to Fz is from: 1. The deflection due to the slope at B, B1, due to the force and moment acting on B (x1 = BC B1 = 5B1). 2. The deflection due to the moment acting on C (x2). 1. For AB, IAB =  14/64 = 0.04909 in4. From Table A-9, beams 1 and 4                22 1 6 3 3 6 6 2 6 62 100 6 2 200 0.002037 rad 2 2 30 10 0.04909 B B B x lx l B M x M xd Fx Fxx l x l dx EI EI EI EI l Fl M EI                                     x 1 = 5( 0.002037) =  0.01019 in 2. For BC, IBC = (1.5)(0.25)3/12 = 0.001953 in4. From Table A-9, beam 4       2 2 6 2 100 5 0.04267 in 2 2 30 10 0.001953 CM lx EI      The deflection of D in the x direction due to Fx is from: 3. The elongation of AB due to the tension. For AB, the area is A =  12/4 = 0.7854 in2       5 3 6 150 6 3.82 10 in 0.7854 30 10AB Flx AE         4. The deflection due to the slope at B, B2, due to the moment acting on B (x1 = BC B2 = 5B2). With IAB = 0.04907 in4,    2 6 5 150 6 0.003056 rad 30 10 0.04909 B B M l EI       Chapter 4 - Rev B, Page 30/81 • x4 = 5( 0.003056) =  0.01528 in 5. The deflection at C due to the bending force acting on C. With IBC = 0.001953 in4       33 5 6 150 5 0.10667 in 3 3 30 10 0.001953BC Flx EI            6. The elongation of CD due to the tension. For CD, the area is A =  (0.752)/4 = 0.4418 in2       5 6 6 150 2 2.26 10 in 0.4418 30 10CD Flx AE         Summing the deflections results in     6 5 1 5 0.01019 0.04267 3.82 10 0.01528 0.10667 2.26 10 0.1749 in . D i i x x Ans               ______________________________________________________________________________ 4-43 JOA = JBC =  (1.54)/32 = 0.4970 in4, JAB =  (14)/32 = 0.09817 in4, IAB =  (14)/64 = 0.04909 in4, and ICD =  (0.754)/64 = 0.01553 in4. 6 250(12) 2 9 2 0.0260 rad . 11.5(10 ) 0.4970 0.09817 0.4970 OA BCAB OA AB BC OA AB BC l llTl Tl Tl T GJ GJ GJ G J J J Ans                                    Simplified   6 250(12)(13) 11.5 10 0.09817 0.0345 rad . s s Tl GJ Ans      Simplified is 0.0345/0.0260 = 1.33 times greater Ans.           3 33 3 6 6 250 13 250 12 0.0345(12) 3 3 3(30)10 0.04909 3(30)10 0.01553 0.847 in . y OC y CD D s CD AB CD D F l F l y l EI EI y Ans        ______________________________________________________________________________ 4-44 Reverse the deflection equation of beam 7 of Table A-9. Using units in lbf, inches Chapter 4 - Rev B, Page 31/81 •                 32 3 3 2 3 6 10 6 2 3 3000 /12 2 2 25 25 12 24 24 30 10 485 7.159 10 27 10 600 . xxy lx x l x x EI x x x Ans                 w The maximum height occurs at x = 25(12)/2 = 150 in      10 6 2 3max 7.159 10 150 27 10 600 150 150 1.812 in .y Ans       ______________________________________________________________________________ 4-45 From Table A-9-6,  2 2 26L Fbxy x b l EIl     3 2 26L Fby x b x l x EIl     2 2 236 Ldy Fb x b l dx EIl     2 2 0 6 L x Fb b ldy dx EIl   Let 0    L x dy dx and set 4 64   L dI . Thus,   1/42 232 . 3L Fb b l d A El    ns For the other end view, observe beam 6 of Table A-9 from the back of the page, noting that a and b interchange as do x and –x   1/42 232 . 3R Fa l a d A El    ns For a uniform diameter shaft the necessary diameter is the larger of and .L Rd d ______________________________________________________________________________ 4-46 The maximum slope will occur at the left bearing. Incorporating a design factor into the solution for of Prob. 4-45, Ld Chapter 4 - Rev B, Page 32/81 •     1/42 2 2 2 4 3 32 3 32(1.28)(3000)(200) 300 200 3 (207)10 (300)(0.001) 38.1 mm . nFb l b d El d d Ans                  4 3 4 38.1 103.4 10 mm 64 I    From Table A-9, beam 6, the maximum deflection will occur in BC where dyBC /dx = 0     2 2 2 2 22 0 3 6 26 Fa l xd x a lx x lx a l dx EIl               0    2 2 2 23 6 300 100 2 300 0 600 63333 0x x x x          21 600 600 4(1)63 333 463.3, 136.7 mm 2 x       x = 136.7 mm is acceptable.                 2 2 max 136.7mm 3 2 2 3 3 2 6 3 10 100 300 136.7 136.7 100 2 300 136.7 0.0678 mm . 6 207 10 103.4 10 300 x Fa l x y x a lx EIl Ans                  ______________________________________________________________________________ 4-47 I =  (1.254)/64 = 0.1198 in4. From Table A-9, beam 6         2 2 2 2 2 2 21 1 2 2 1 2 2 2 2 6 1/22 2 2 2 6 ( ) ( 2 ) ( 6 6 150(5)(20 8) 8 5 2(20)(8) 6(30)10 0.1198 (20) 250(10)(8) 8 10 20 6(30)10 0.1198 (20) 0.0120 in . F a l x F b xx a lx x b l EIl EIl Ans                      )           ______________________________________________________________________________ Chapter 4 - Rev B, Page 33/81 • 4-48 I =  (1.254)/64 = 0.1198 in4. For both forces use beam 6 of Table A-9. For F1 = 150 lbf: 0  x  5              2 2 2 2 2 21 1 1 6 6 2 150 15 15 20 6 6 30 10 0.1198 20 5.217 10 175 (1) xFb xy x b l x EIl x x         5  x  20                  1 1 2 2 2 2 1 6 6 2 150 5 20 2 5 2 20 6 6 30 10 0.1198 20 1.739 10 20 40 25 (2) F a l x x x a lx x x EIl x x x         y       For F2 = 250 lbf: 0  x  10              2 2 2 2 2 22 2 2 6 6 2 250 10 10 20 6 6 30 10 0.1198 20 5.797 10 300 (3) xF b xz x b l x EIl x x         10  x  20                  2 2 2 2 2 2 2 6 6 2 250 10 20 2 10 2 20 6 6 30 10 0.1198 20 5.797 10 20 40 100 (4) F a l x x z x a lx x x EIl x x x               Plot Eqs. (1) to (4) for each 0.1 in using a spreadsheet. There are 201 data points, too numerous to tabulate here but the plot is shown below, where the maximum deflection of  = 0.01255 in occurs at x = 9.9 in. Ans. ______________________________________________________________________________ Chapter 4 - Rev B, Page 34/81 • 4-49 The larger slope will occur at the left end. From Table A-9, beam 8 2 2 2 2 2 2 ( 3 6 2 ) 6 (3 3 6 2 ) 6 B AB AB B M xy x a al l EIl dy M x a al l dx EIl         With I =  d 4/64, the slope at the left bearing is   2 2 4 0 (3 6 2 ) 6 / 64 AB B A x dy M a al l dx E d l       Solving for d     2 2 2 44 6 32 32(1000)3 6 2 3(4 ) 6(4)(10) 2 10 3 3 (30)10 (0.002)(10) 0.461 in . B A Md a al l E l Ans    2         ______________________________________________________________________________ 4-50 From Table A-5, E = 10.4 Mpsi MO = 0 = 18 FBC  6(100)  FBC = 33.33 lbf The cross sectional area of rod BC is A =  (0.52)/4 = 0.1963 in2. The deflection at point B will be equal to the elongation of the rod BC.       5 6 33.33(12) 6.79 10 in . 0.1963 30 10B BC FLy Ans AE        ______________________________________________________________________________ 4-51 MO = 0 = 6 FAC  11(100)  FAC = 183.3 lbf The deflection at point A in the negative y direction is equal to the elongation of the rod AC. From Table A-5, Es = 30 Mpsi.         4 2 6 183.3 12 3.735 10 in 0.5 / 4 30 10 A AC FLy AE               By similar triangles the deflection at B due to the elongation of the rod AC is 41 1 3 3( 3.735)10 0.00112 in6 18 A B B A y y y y        From Table A-5, Ea = 10.4 Mpsi The bar can then be treated as a simply supported beam with an overhang AB. From Table A-9, beam 10 Chapter 4 - Rev B, Page 35/81 •     2 2 2 2 2 2 2 6 3 ( )( ) 7 ( ) (3 ) ( 3 6 3 77 3( ) 3 ( ) (3 ) | ( ) (2 3 ) ( ) 6 3 6 7 100 5 6(10.4)10 0.25(2 ) / BC B x l a x l a x l a dy Fa d F x l Fay BD l a x l a x l l dx EI dx EI EI F Fa Fax l a x l a x l l a l a l a EI EI EI EI                                                ) 3 a Fa       2 6 3 100 5 2(6) 3(5) (6 5) 12 3(10.4)10 0.25(2 ) /12 0.01438 in      yB = yB1 + yB2 =  0.00112  0.01438 =  0.0155 in Ans. ______________________________________________________________________________ 4-52 From Table A-5, E = 207 GPa, and G = 79.3 GPa.       2 23 3 4 4 3 2 4 4 4 3 / 32 / 32 3 / 64 32 2 3 OC AB AC ABAB AB B AB AB OC AC AB OC AC OC ACAB AB OC AC AB Fl l Fl lFl FlTl Tly l l GJ GJ EI G d G d E d l lFl l Gd Gd Ed                             4 The spring rate is k = F/ yB. Thus             1 2 4 4 4 1 2 3 4 3 4 3 4 32 2 3 32 200 2 200200 200 79.3 10 18 79.3 10 12 3 207 10 8 8.10 N/mm . OC ACAB AB OC AC AB l ll lk Gd Gd Ed Ans                               _____________________________________________________________________________ 4-53 For the beam deflection, use beam 5 of Table A-9. 1 2 1 2 1 2 2 31 2 1 2 32 1 1 1 2 2 , and 2 2 (4 3 ) 48 1 (4 3 ) . 2 2 48 AB AB FR R F F k k Fxy x x l l EI k k xy F x x l k k k l EI                         Ans Chapter 4 - Rev B, Page 36/81 • For BC, since Table A-9 does not have an equation (because of symmetry) an equation will need to be developed as the problem is no longer symmetric. This can be done easily using beam 6 of Table A-9 with a = l /2        2 22 1 1 1 2 2 22 1 1 1 2 / 2 2 2 2 4 1 4 8 . 2 2 48 BC F l l xFk FkF ly x x k k k l EIl l xk k lx F x x l lx k k k l EI                       Ans ______________________________________________________________________________ 4-54   1 2 1 2 1 2 2 21 2 1 2 2 2 12 1 1 2 , and ( ) , and ( ) ( ) 6 ( ) . 6 AB AB Fa FR R l a l l Fa F l a lk lk Faxy x l x l EIl a x axy F k a k l a l x k l k k l EIl                               Ans   21 2 1 2 2 12 1 1 2 ( ) ( ) (3 ) 6 ( ) ( ) (3 ) . 6 BC BC F x ly x x l a x l l EI a x x ly F k a k l a x l a x l An k l k k l EI s                                 ______________________________________________________________________________ 4-55 Let the load be at x ≥ l/2. The maximum deflection will be in Section AB (Table A-9, beam 6)  2 2 26AB Fbxy x b l EIl     2 2 2 2 2 23 0 36 ABdy Fb x b l x b l dx EIl        0 2 2 2 max, 0.577 .3 3 l b lx x l   Ans For x  l/2, min 0.577 0.423 .x l l l A   ns ______________________________________________________________________________ Chapter 4 - Rev B, Page 37/81 • 4-56  6 1(3000)(1500) 2500(2000) 9.5 10 N·mm 1(3000) 2500 5 500 N O O M R       6 4From Prob. 4-10, 4.14(10 ) mm I    2 169.5 10 5500 2500 - 2000 2 xM x     x   3 26 2 19.5 10 2750 1250 20006 dy xEI x x x C dx        10 at 0 0 dy x C dx         3 26 2 4 36 2 3 2 9.5 10 2750 1250 2000 6 4.75 10 916.67 416.67 2000 24 dy xEI x x x dx xEIy x x x C              y , and therefore 20 at 0 0x C          36 2 3 3 4 31 114 10 22 10 10 10 200024y x x x xEI                    6 2 3 3 3 6 34 3 1 114 10 3000 22 10 3000 24 207 10 4.14 10 3000 10 10 3000 2000 25.4 mm . By Ans          MO = 9.5 (106) Nm. The maximum stress is compressive at the bottom of the beam where y = 29.0  100 =  71 mm     6 6 ma x 6 9.5 10 ( 71) 163 10 Pa 163MPa . 4.14(10 ) My Ans I           The solutions are the same as Prob. 4-10. ______________________________________________________________________________ 4-57 See Prob. 4-11 for reactions: RO = 465 lbf and RC = 285 lbf. Using lbf and inch units Chapter 4 - Rev B, Page 38/81 • M = 465 x  450 x  721  300 x  1201 2 22 1232.5 225 72 150 120 dyEI x x x dx      C EIy = 77.5 x3  75 x  723  50 x  1203  C1x y = 0 at x = 0  C2 = 0 y = 0 at x = 240 in 0 = 77.5(2403)  75(240 72)3  50(240  120)3 + C1 x  C1 =  2.622(106) lbfin2 and, EIy = 77.5 x3  75 x  723  50 x  1203 2.622(106) x Substituting y =  0.5 in at x = 120 in gives 30(106) I ( 0.5) = 77.5 (1203)  75(120  72)3  50(120  120)3 2.622(106)(120) I = 12.60 in4 Select two 5 in  6.7 lbf/ft channels; from Table A-7, I = 2(7.49) = 14.98 in4 midspan 12.60 1 0.421 in . 14.98 2 y A        ns The maximum moment occurs at x = 120 in where Mmax = 34.2(103) lbfin 3 max 34.2(10 )(2.5) 5 710 psi 14.98 Mc I     O.K. The solutions are the same as Prob. 4-17. ______________________________________________________________________________ 4-58 I =  (1.54)/64 = 0.2485 in4, and w = 150/12 = 12.5 lbf/in.  1 2412.5 39 (340) 453.0 lbf 2 39O R    1212.5453.0 340 15 2 M x x x    22 3 1 12.5226.5 170 15 6 dyEI x x x dx     C 33 4 1 275.5 0.5208 56.67 15EIy x x x C x     C 20at 0 0y x C    Thus, 4 210 at 39 in 6.385(10 ) lbf iny x C       33 4 41 75.5 0.5208 56.67 15 6.385 10y x x x xEI        Evaluating at x = 15 in, Chapter 4 - Rev B, Page 39/81 •         33 4 4 6 1 75.5 15 0.5208 15 56.67 15 15 6.385 10 (15) 30(10 )(0.2485) 0.0978 in . Ay Ans                 33 4 4midspan 6 1 75.5 19.5 0.5208 19.5 56.67 19.5 15 6.385 10 (19.5)30(10 )(0.2485) 0.1027 in . y Ans          5 % difference Ans. The solutions are the same as Prob. 4-12. ______________________________________________________________________________ 4-59 I = 0.05 in4,    3 14 100 7 14 100420 lbf and 980 lbf 10 10A B R R      M = 420 x  50 x2 + 980  x  10 1 22 3 1210 16.667 490 10 dyEI x x x dx     C 33 4 1 270 4.167 163.3 10EIy x x x C x     C y = 0 at x = 0  C2 = 0 y = 0 at x = 10 in  C1 =  2 833 lbfin2. Thus,     33 4 6 37 3 4 1 70 4.167 163.3 10 2833 30 10 0.05 6.667 10 70 4.167 163.3 10 2833 . y x x x x x x x x               Ans The tabular results and plot are exactly the same as Prob. 4-21. ______________________________________________________________________________ 4-60 RA = RB = 400 N, and I = 6(323) /12 = 16 384 mm4. First half of beam, M =  400 x + 400  x  300 1 22 1200 200 300 dyEI x x dx     C From symmetry, dy/dx = 0 at x = 550 mm  0 =  200(5502) + 200(550 – 300) 2 + C1  C1 = 48(106) N·mm2 EIy =  66.67 x3 + 66.67  x  300 3 + 48(106) x + C2 Chapter 4 - Rev B, Page 40/81 • y = 0 at x = 300 mm  C2 =  12.60(109) N·mm3. The term (EI)1 = [207(103)16 384] 1 = 2.949 (1010 ) Thus y = 2.949 (1010) [ 66.67 x3 + 66.67  x  300 3 + 48(106) x  12.60(109)] yO =  3.72 mm Ans. yx = 550 mm =2.949 (1010) [ 66.67 (5503) + 66.67 (550  300)3 + 48(106) 550  12.60(109)] = 1.11 mm Ans. The solutions are the same as Prob. 4-13. ______________________________________________________________________________ 4-61     1 1 2 2 10 10 ( ) B A A A A A M R l Fa M R M Fa l M M R l F l a R Fl Fa M l                   11 2AM R x M R x l    22 1 2 1 33 2 1 2 1 1 2 2 1 1 1 6 2 6 A A dyEI R x M x R x l C dx 1 2EIy R x M x R x l C x C            y = 0 at x = 0  C2 = 0 y = 0 at x = l  21 1 1 1 6 2 A C R l M   l . Thus, 33 2 21 2 1 1 1 1 1 1 6 2 6 6 2A A EIy R x M x R x l R l M l x                33 2 2 21 3 26 A A A Ay M Fa x M x l Fl Fa M x l Fal M l x Ans.EIl            In regions,         3 2 2 2 2 2 2 2 1 3 2 6 3 2 6 AB A A A A y M Fa x M x l Fal M l x EIl x .M x lx l Fa l x Ans EIl               Chapter 4 - Rev B, Page 41/81 •                              33 2 2 2 3 33 2 2 3 2 22 2 1 3 2 6 1 3 2 6 1 3 6 3 . 6 BC A A A A A A A y M Fa x M x l Fl Fa M x l Fal M l x EIl M x x l x l xl F ax l a x l axl EIl M x l l Fl x l x l a x l EIl x l M l F x l a x l Ans EI                                                The solutions reduce to the same as Prob. 4-17. ______________________________________________________________________________ 4-62        1 1 10 2 2 2D b a M R l b a l b b a R l b a l                w w 2 21 2 2 M R x x a x b    w w 3 321 1 1 2 6 6 dyEI R x x a x b dx       w w C 4 431 1 1 6 24 24 2 EIy R x x a x b C x C      w w y = 0 at x = 0  C2 = 0 y = 0 at x = l    4 431 1 1 1 6 24 24 C R l l a l b l          w w                        4 43 4 43 4 43 4 42 1 1 2 6 2 24 24 1 1 2 6 2 24 24 2 2 24 2 2 b a y l b a x x a x b EI l b a x l b a l l a l l l b a l b a x l x a l x b EIl . b x b a l b a l l a l b Ans                                       w w w w w w w The above answer is sufficient. In regions, Chapter 4 - Rev B, Page 42/81 •                      4 43 2 4 42 2 2 2 2 2 24 2 2 2 2 24 ABy b a l b a x x b a l b a l l a l bEIl b a l b a x b a l b a l l a l b EIl                             w wx              43 4 42 2 2 24 2 2 BCy b a l b a x l x aEIl x b a l b a l l a l b                 w                4 43 4 42 2 2 24 2 2 CDy b a l b a x l x a l x bEIl x b a l b a l l a l b                  w  These equations can be shown to be equivalent to the results found in Prob. 4-19. ______________________________________________________________________________ 4-63 I1 =  (1.3754)/64 = 0.1755 in4, I2 =  (1.754)/64 = 0.4604 in4, R1 = 0.5(180)(10) = 900 lbf Since the loading and geometry are symmetric, we will only write the equations for half the beam For 0  x  8 in 2900 90 3M x x   At x = 3, M = 2700 lbfin Writing an equation for M / I, as seen in the figure, the magnitude and slope reduce since I 2 > I 1. To reduce the magnitude at x = 3 in, we add the term,  2700(1/I 1  1/ I 2) x  3 0. The slope of 900 at x = 3 in is also reduced. We account for this with a ramp function,  x  31 . Thus, 0 1 1 1 2 1 2 2 0 1 2 900 1 1 1 1 902700 3 900 3 3 5128 9520 3 3173 3 195.5 3 M x x x 2x I I I I I x x x x                            I I 1 22 12564 9520 3 1587 3 65.17 3 3dyE x x x x C dx         Boundary Condition: 0 at 8 indy   x dx Chapter 4 - Rev B, Page 43/81 •         2 2 10 2564 8 9520 8 3 1587 8 3 65.17 8 3 C       3  C1 =  68.67 (103) lbf/in2 2 3 43 3 2854.7 4760 3 529 3 16.29 3 68.67(10 )Ey x x x x x        C y = 0 at x = 0  C2 = 0 Thus, for 0  x  8 in 2 3 43 36 1 854.7 4760 3 529 3 16.29 3 68.7(10 ) . 30(10 ) x x x x x Ans         y Using a spreadsheet, the following graph represents the deflection equation found above The maximum is max 0.0102 in at 8 in .y x A   ns ______________________________________________________________________________ 4-64 The force and moment reactions at the left support are F and Fl respectively. The bending moment equation is M = Fx  Fl Plots for M and M /I are shown. M /I can be expressed using singularity functions 0 1 1 1 1 12 2 4 2 2 2 M F Fl Fl l F lx x x I I I I I       Chapter 4 - Rev B, Page 44/81 • where the step down and increase in slope at x = l /2 are given by the last two terms. Integrate 1 2 2 1 1 1 1 14 2 4 2 4 2 dy F Fl Fl l F lE x x x x dx I I I I       C dy/dx = 0 at x = 0  C1 = 0 2 3 3 2 2 1 1 1 112 4 8 2 12 2 F Fl Fl l F lEy x x x x C I I I I        y = 0 at x = 0  C2 = 0 2 3 3 2 1 2 6 3 2 24 2 2 F ly x lx l x x EI             l 3 2 3 /2 1 1 52 6 3 (0) 2(0) . 24 2 2 96x l F l l Fly l l EI EI                     Ans     2 3 3 3 2 1 1 32 6 3 2 24 2 2 16x l F l l Fly l l l l l x EI EI                       .Ans The answers are identical to Ex. 4-10. ______________________________________________________________________________ 4-65 Place a dummy force, Q, at the center. The reaction, R1 = wl / 2 + Q / 2 2 2 2 2 2 Q x MM x Q         wl w x Integrating for half the beam and doubling the results /2 /2 2 max 0 00 1 22 2 2 2 l l Q M xy M dx x EI Q EI                              wl w x dx Note, after differentiating with respect to Q, it can be set to zero   /2/2 3 4 2 max 0 0 5 . 2 2 3 4 384 ll x l xy x l x dx Ans EI EI EI            w w w ______________________________________________________________________________ 4-66 Place a fictitious force Q pointing downwards at the end. Use the variable x originating at the free end at positive to the left 2 2 x MM Qx x Q        w Chapter 4 - Rev B, Page 45/81 •   2 3 max 0 00 4 1 1 2 2 . 8 l l Q My M dx x dx x dx EI Q EI EI l Ans EI                         wx w w 0 l  ______________________________________________________________________________ 4-67 From Table A-7, I1-1 = 1.85 in4. Thus, I = 2(1.85) = 3.70 in4 First treat the end force as a variable, F. Adding weight of channels of 2(5)/12 = 0.833 lbf/in. Using the variable x as shown in the figure 2 25.833 2.917 2 M F x x F x x M x F           60 60 2 0 0 1 1 ( 2.917 )( ) A MM d x F x x x d x EI F EI       3 4 6 (150 / 3)(60 ) (2.917 / 4)(60 ) 0.182 in 30(10 )(3.70)    in the direction of the 150 lbf force 0.182 in .Ay Ans   ______________________________________________________________________________ 4-68 The energy includes torsion in AC, torsion in CO, and bending in AB. Neglecting transverse shear in AB , MM Fx x F     In AC and CO, , AB AB TT Fl l F     The total energy is 2 2 2 02 2 2 ABl ABAC CO T l T l MU d GJ GJ EI                x The deflection at the tip is Chapter 4 - Rev B, Page 46/81 • 2 30 0 1AB ABl lAC CO AC AB CO AB AC CO AC CO AB Tl Tl Tl l Tl lU T T M M dx Fx dx F GJ F GJ F EI F GJ GJ EI                       2 23 3 4 4 4 2 4 4 4 3 / 32 / 32 3 / 64 32 2 3 AC AB CO AB AC AB CO ABAB AB AC CO AB AC CO AB AC COAB AB AC CO AB Tl l Tl l Fl l Fl lFl Fl GJ GJ EI G d G d E d l lFl l Gd Gd Ed                                1 2 4 4 4 1 2 3 4 3 4 3 4 2 32 3 2 200200 200 8.10 N/mm . 32 200 79.3 10 18 79.3 10 12 3 207 10 8 AC CO AB AB AC CO AB l l lFk l Gd Gd Ed Ans                          ______________________________________________________________________________ 4-69 I1 =  (1.3754)/64 = 0.1755 in4, I2 =  (1.754)/64 = 0.4604 in4 Place a fictitious force Q pointing downwards at the midspan of the beam, x = 8 in 1 1 1(10)180 900 0.5 2 2 R Q Q    For 0  x  3 in  900 0.5 0.5MM Q x x Q      For 3  x  13 in   2900 0.5 90( 3) 0.5MM Q x x x Q        By symmetry it is equivalent to use twice the integral from 0 to 8             8 3 8 22 1 20 0 30 3 83 3 4 3 2 1 20 3 3 3 3 6 6 1 2 1 12 900 900 90 3 300 1 1 9300 90( 2 ) 4 2 120.2 108100 1 8100145.5 10 25.31 10 30 10 0.1755 30 10 0.4604 0.0102 in . Q M M dx x dx x x x dx EI Q EI EI x x x x x EI EI EI EI Ans                                  ______________________________________________________________________________ Chapter 4 - Rev B, Page 47/81 • 4-70 I =  (0.54)/64 = 3.068 (103) in4, J = 2 I = 6.136 (103) in4, A = (0.52)/4 = 0.1963 in2. Consider x to be in the direction of OA, y vertically upward, and z in the direction of AB. Resolve the force F into components in the x and y directions obtaining 0.6 F in the horizontal direction and 0.8 F in the negative vertical direction. The 0.6 F force creates strain energy in the form of bending in AB and OA, and tension in OA. The 0.8 F force creates strain energy in the form of bending in AB and OA, and torsion in OA. Use the dummy variable x to originate at the end where the loads are applied on each segment, 0.6 F: AB 0.6 0.6MM F x x F     OA 4.2 4.2MM F F     0.6 0.6aa FF F F     0.8 F: AB 0.8 0.8MM F x x F     OA 0.8 0.8MM F x x F     5.6 5.6TT F F     Once the derivatives are taken the value of F = 15 lbf can be substituted in. The deflection of B in the direction of F is*                                     6 3 6 27 1 2 6 3 6 3 0 0 7 2 6 3 6 0 1 0.6 15 15 5.6 15 15 0.6 5.6 0.1963 30 10 6.136 10 11.5 10 15 4.215 0.6 30 10 3.068 10 30 10 3.068 10 15 150.8 30 10 3.068 10 30 10 3.06 a a B F OAOA F L FU TL T MM d x F AE F JG F EI F x d x d x x d x                                  5        15 2 3 0 5 3 0.8 8 10 1.38 10 0.1000 6.71 10 0.0431 0.0119 0.1173 0.279 in . x d x Ans           Chapter 4 - Rev B, Page 48/81 • *Note. This is not the actual deflection of point B. For this, dummy forces must be placed B = 0.0831 i  0.2862 j  0.00770 k in is on B in the x, y, and z directions. Determine the energy due to each, take derivatives, and then substitute the values of Fx = 9 lbf, Fy =  12 lbf, and Fz = 0. This can be done separately and then use superposition. The actual deflections of B are From this, the deflection of B in the direction of F      0.6 0.0831 0.8 0.2862 0.279 inB F    which agrees with our result. ____ ________________________________________________ -71 Strain energy. AB: Bending and torsion, BC: Bending and torsion, CD: Bending. 031 in4, 1) is in the form of  =TL/(JG), where the equivalent of Use the dummy variable _ _________________________ 4 IAB =  (14)/64 = 0.04909 in4, JAB = 2 IAB = 0.09818 in4, IBC = 0.25(1.53)/12 = 0.07 ICD =  (0.754)/64 = 0.01553 in4. For the torsion of bar BC, Eq. (3-4 J is Jeq = bc 3. With b/c = 1.5/0.25 = 6, JBC = bc 3 = 0.299(1.5)0.253 = 7.008 (103) in4. x to originate at the end where the loads are applied on each ing segment, AB: Bend 2 2MM F x F x F       Torsion 5 5TT F F     MM F x x F     BC: Bending Torsion 2 2TT F F     CD: Bending MM F x x F                                   6 2 6 3 6 6 0 5 2 2 2 6 6 0 0 4 4 4 5 6 1 5 6 2 5 15 2 0.09818 11.5 10 7.008 10 11.5 10 30 10 0.04909 1 1 30 10 0.07031 30 10 0.01553 1.329 10 2.482 10 1.141 10 1.98 10 5.72 10 5.207 10 D U Tl T MM d x F JG F EI F F F 2F x d F x d x F x d x F F F F F                                      4 45.207 10 200 0.104 in .F Ans  x ______________________________________________________________________________ Chapter 4 - Rev B, Page 49/81 • 4-72 AAB =  (12)/4 = 0.7854 in2, IAB =  (14)/64 = 0.04909 in4, IBC = 1.5 (0.253)/12 = 1.953 (103) in4, ACD =  (0.752)/4 = 0.4418 in2, IAB =  (0.754)/64 = 0.01553 in4. For (D )x let F = Fx =  150 lbf and Fz =  100 lbf . Use the dummy variable x to originate at the end where the loads are applied on each segment, CD: 0yy z M M F x F     1aa FF F F     BC: 2 yy z M M F x F x F      0aa z FF F F     AB: 5 2 yy z z M M F F F x F 5       1aa FF F F                                          5 0 6 0 3 2 6 6 3 2 66 7 1 2 1 5 2 5 2 11 5 5 0.4418 30 10 330 10 1.953 10 61 25 6 10 6 6 5 1 2 0.7854 30 1030 10 0.04909 1.509 10 7.112 1 a D zx CD BC a z z ABAB z z z FU FL F x F x d x F AE F EI FFLF F F x d x EI AE F F F F FFF F F                                                     4 4 4 4 7 4 4 0 4.267 10 1.019 10 1.019 10 2.546 10 8.135 10 5.286 10 z z z F F F F F F              F Substituting F = Fx =  150 lbf and Fz =  100 lbf gives        4 48.135 10 150 5.286 10 100 0.1749 in .D x Ans        ______________________________________________________________________________ 4-73 IOA = IBC =  (1.54)/64 = 0.2485 in4, JOA = JBC = 2 IOA = 0.4970 in4, IAB =  (14)/64 = 0.04909 in4, JAB = 2 IAB = 0.09818 in4, ICD =  (0.754)/64 = 0.01553 in4 Let Fy = F, and use the dummy variable x to originate at the end where the loads are applied on each segment, Chapter 4 - Rev B, Page 50/81 • OC: , 12 12M TM F x x T F F F         DC: MM F x x F       1D y OC U TL T MM d x F JG F EI F               The terms involving the torsion and bending moments in OC must be split up because of the changing second-area moments.                                  2 2 6 6 6 0 11 13 12 2 2 6 6 6 2 11 4 3 7 4 5 3 12 4 12 9 112 12 0.4970 11.5 10 0.09818 11.5 10 30 10 0.2485 1 1 1 30 10 0.04909 30 10 0.2485 30 10 0.01553 1.008 10 1.148 10 3.58 10 2.994 10 3.872 10 1.2363 10 D y F F F x d x 2 0 F x d x F x d x F x d x F F F F F F                          3 32.824 10 2.824 10 250 0.706 in .F Ans     For the simplified shaft OC,                      13 12 2 2 6 6 6 0 0 3 4 3 3 3 12 13 1 112 0.09818 11.5 10 30 10 0.04909 30 10 0.01553 1.6580 10 4.973 10 1.2363 10 3.392 10 3.392 10 250 0.848 in . B y F F x d x F x d x F F F F Ans                  Simplified is 0.848/0.706 = 1.20 times greater Ans. ______________________________________________________________________________ 4-74 Place a dummy force Q pointing downwards at point B. The reaction at C is RC = Q + (6/18)100 = Q + 33.33 This is the axial force in member BC. Isolating the beam, we find that the moment is not a function of Q, and thus does not contribute to the strain energy. Thus, only energy in the member BC needs to be considered. Let the axial force in BC be F, where Chapter 4 - Rev B, Page 51/81 • 33.33 1FF Q Q               52 6 0 0 0 33.33 12 1 6.79 10 in 0.5 / 4 30 10 B BCQ Q U FL F Ans Q AE Q                      . ______________________________________________________________________________ 4-75 IOB = 0.25(23)/12 = 0.1667 in4 AAC =  (0.52)/4 = 0.1963 in2 MO = 0 = 6 RC  11(100)  18 Q RC = 3Q + 183.3 MA = 0 = 6 RO  5(100)  12 Q  RO = 2Q + 83.33 Bending in OB. BD: Bending in BD is only due to Q which when set to zero after differentiation gives no contribution. AD: Using the variable x as shown in the figure above    100 7 7MM x Q x x Q          OA: Using the variable x as shown in the figure above  2 83.33 2MM Q x Q x      Axial in AC: 3 183.3 3FF Q Q      Chapter 4 - Rev B, Page 52/81 •                        0 0 0 5 6 2 6 0 0 5 63 2 6 0 0 3 7 1 183.3 12 13 100 7 2 83.33 0.1963 30 10 11.121 10 100 7 166.7 10.4 10 0.1667 1.121 10 5.768 10 100 129.2 166. B Q Q Q U FL F MM dx Q AE Q EI Q x x d x x dx EI x x d x x dx                                                      7 72 0.0155 in .Ans   ______________________________________________________________________________ 4-76 There is no bending in AB. Using the variable, rotating counterclockwise from B sin sinMM PR R P     cos cosrr FF P P     2 sin sin 2 sin FF P P MF PR P              2 1 12 26(4) 24 mm , 40 (6) 43 mm, 40 (6) 37 mm,o iA r r        From Table 3-4, p.121, for a rectangular cross section 6 39.92489 mm ln(43 / 37)n r   From Eq. (4-33), the eccentricity is e = R  rn =40  39.92489 = 0.07511 mm From Table A-5, E = 207(103) MPa, G = 79.3(103) MPa From Table 4-1, C = 1.2 From Eq. (4-38)  2 2 2 2 0 0 0 0 1 r rMFF R F CF R FM M d d d AeE P AE P AE P AG P       d                                 2 2 2 2 2 2 2 0 0 0 0 sin sin cos2 sinP R PR CPRPRd d d AeE AE AE AG       2 d            3 3 3 (10)(40) 40 (207 10 )(1.2)1 2 1 2 4 4(24)(207 10 ) 0.07511 79.3 10 PR R EC AE e G                    0.0338 mm .Ans  ______________________________________________________________________________ Chapter 4 - Rev B, Page 53/81 • 4-77 Place a dummy force Q pointing downwards at point A. Bending in AB is only due to Q which when set to zero after differentiation gives no contribution. For section BC use the variable, rotating counterclockwise from B    sin sin 1 sinMM PR Q R R R Q          cos cosrr FF P Q Q       sin sinFF P Q Q             sin 1 sin sinMF PR QR P Q            2sin sin 1 sin 2 sin 1 sinMF PR PR QR Q              But after differentiation, we can set Q = 0. Thus,  sin 1 2sinMF PR Q       2 1 12 26(4) 24 mm , 40 (6) 43 mm, 40 (6) 37 mm,o iA r r        From Table 3-4, p.121, for a rectangular cross section 6 39.92489 mm ln(43 / 37)n r   From Eq. (4-33), the eccentricity is e = R  rn =40  39.92489 = 0.07511 mm From Table A-5, E = 207(103) MPa, G = 79.3(103) MPa From Table 4-1, C = 1.2 From Eq. (4-38)       2 2 2 2 2 2 2 2 0 0 0 0 2 2 0 0 0 2 0 2 1 sin 1 sin sin sin 1 2sin cos 1 2 4 4 4 r rMFF R F CF R FM M d d d AeE Q AE Q AE Q AG Q PR PR PRd d d AeE AE AE CPR d AG PR PR PR AeE AE           d                                                                               3 3 3 1 2 4 4 4 10 40 1.2 207 10401 2 24 207 10 4 0.07511 4 79.3 10 0.0766 mm . CPR PR R CE AE AG AE e G Ans                              ______________________________________________________________________________ Chapter 4 - Rev B, Page 54/81 • 4-78 Note to the Instructor. The cross section shown in the first printing is incorrect and the solution presented here reflects the correction which will be made in subsequent printings. The corrected cross section should appear as shown in this figure. We apologize for any inconvenience. A = 3(2.25) 2.25(1.5) = 3.375 in2 (1 1.5)(3)(2.25) (1 0.75 1.125)(1.5)(2.25) 2.125 in 3.375 R      Section is equivalent to the “T” section of Table 3-4, p. 121, 2.25(0.75) 0.75(2.25) 1.7960 in 2.25ln[(1 0.75) /1] 0.75ln[(1 3) / (1 0.75)]n r       2.125 1.7960 0.329 inne R r     For the straight section     3 2 2 3 4 1 (2.25) 3 2.25(3)(1.5 1.125) 12 1 2.25(1.5) 2.25 1.5(2.25) 0.75 1.125 12 2 2.689 in zI                 For 0  x  4 in , 1M Vx x V FM F F F           For    /2 cos cos , sin sinrrF FFF F F F F             (4 2.125sin ) (4 2.125sin ) (4 2.125sin ) sin 2 (4 2.365sin )sin MM F F MFMF F F F F                     Chapter 4 - Rev B, Page 55/81 • Use Eqs. (4-31) and (4-24) (with C = 1) for the straight part, and Eq. (4-38) for the curved part, integrating from 0 to π/2, and double the results 24 /22 0 0 2/2 /2 0 0 2/2 0 2 1 (4)(1) (4 2.125sin ) 3.375( / ) 3.375(0.329) sin (2.125) 2 (4 2.125sin )sin 3.375 3.375 (1) cos (2.125) 3.375( / ) FFx dx F d E I G E F Fd d F d G E                            Substitute I = 2.689 in4, F = 6700 lbf, E = 30 (106) psi, G = 11.5 (106) psi           3 6 2 6700 4 4 1 16 17(1) 4.516 3 2.689 3.375(11.5 / 30) 3.375(0.329) 2 430 10 2.125 2 2.1254 1 2.125 3.375 4 3.375 4 3.375 11.5 / 30 4 0.0226 in .Ans                                                      ______________________________________________________________________________ 4-79 Since R/h = 35/4.5 = 7.78 use Eq. (4-38), integrate from 0 to  , and double the results    1 cos 1 cosMM FR R F       sin sinrr FF F F     cos cosFF F F               2 cos 1 cos 2 cos 1 cos MF F R MF FR F             From Eq. (4-38),   2 2 2 0 0 2 0 0 2 (1 cos ) cos 2 1.2cos 1 cos sin 2 3 3 0.6 2 2 FR FRd d AeE AE FR FRd d AE AG FR R E AE e G                                      A = 4.5(3) = 13.5 mm2, E = 207 (103) N/mm2, G = 79.3 (103) N/mm2, and from Table 3-4, p. 121, Chapter 4 - Rev B, Page 56/81 • 4.5 34.95173 mm37.25lnln 32.75 n o i hr r r    and e = R  rn = 35  34.95173 = 0.04827 mm. Thus,     3 2 35 3 35 3 2070.6 0.08583 13.5 207 10 2 0.04827 2 79.3 F F          where F is in N. For  = 1 mm, 1 11.65 N . 0.08583 F Ans  Note: The first term in the equation for  dominates and this is from the bending moment. Try Eq. (4-41), and compare the results. ______________________________________________________________________________ 4-80 R/h = 20 > 10 so Eq. (4-41) can be used to determine deflections. Consider the horizontal reaction, to applied at B, subject to the constraint ( ) 0.B H  (1 cos ) sin sin 0 2 2 FR MM HR R H            By symmetry, we may consider only half of the wire form and use twice the strain energy Eq. (4-41) then becomes, /2 0 2( ) 0B H U MM Rd H EI H          /2  0 (1 cos ) sin ( sin ) 02 FR HR R R d          300 9.55 N . 2 4 4 F F FH H Ans           Reaction at A is the same where H goes to the left. Substituting H into the moment equation we get,  (1 cos ) 2sin [ (1 cos ) 2sin ] 0 2 2 FR M R 2 M F                    Chapter 4 - Rev B, Page 57/81 • 2/2 2 20 3 /2 2 2 2 2 2 2 0 3 2 2 2 2 2 3 2 2 [ (1 cos ) 2sin ] 4 ( cos 4sin 2 cos 4 sin 4 sin cos ) 2 4 2 4 2 2 2 4 4 (3 8 4) 8 P U M FRM Rd R d P EI F EI FR d EI FR EI FR EI                                                                                     2 3 3 4 (3 8 4) (30)(40 ) 0.224 mm . 8 207 10 2 / 64 Ans          ______________________________________________________________________________ 4-81 The radius is sufficiently large compared to the wire diameter to use Eq. (4-41) for the curved beam portion. The shear and axial components will be negligible compared to bending. Place a fictitious force Q pointing to the left at point A. sin ( sin ) sinMM PR Q R l R l Q         Note that the strain energy in the straight portion is zero since there is no real force in that section. From Eq. (4-41),         /2 /2 0 0 0 2 2 2/2 2 6 40 1 1 sin sin 1(5 )sin sin (5) 4 4 430 10 0.125 / 64 0.551 in . Q MM Rd PR R l Rd EI Q EI PR PRR l d R l EI EI Ans                                            ______________________________________________________________________________ 4-82 Both the radius and the length are sufficiently large to use Eq. (4-41) for the curved beam portion and to neglect transverse shear stress for the straight portion. Straight portion: ABAB MM Px x P     Curved portion:    (1 cos ) (1 cos )BCBC MM P R l R l P         From Eq. (4-41) with the addition of the bending strain energy in the straight portion of the wire, Chapter 4 - Rev B, Page 58/81 •     /2 0 0 /2 22 0 0 3 /2 2 2 2 0 3 /2 2 2 2 2 0 3 1 1 (1 cos ) (1 2cos cos ) 2 (1 cos ) 3 cos 2 2 cos ( ) 3 3 l BCAB AB BC l MMM dx M Rd EI P EI P P PRx dx R l d EI EI Pl PR R Rl l d EI EI Pl PR R R Rl R l d EI EI Pl P EI                                                                    2 2 2 3 3 2 2 3 23 2 6 4 2 2 ( ) 4 2 2 2 ( ) 3 4 2 1 4 (5 ) 5 2(5 ) 2(5)(4) 5 5 4 3 4 230 10 0.125 / 64 0.850 in . R R R Rl R l EI P l R R R Rl R R l EI Ans                                        ______________________________________________________________________________ 4-83 Both the radius and the length are sufficiently large to use Eq. (4-41) for the curved beam portion and to neglect transverse shear stress for the straight portion. Place a dummy force, Q, at A vertically downward. The only load in the straight section is the axial force, Q. Since this will be zero, there is no contribution. In the curved section    sin 1 cos 1 cosMM PR QR R Q         From Eq. (4-41)           /2 /2 0 0 0 3 3/2 0 3 6 4 1 1 sin 1 cos 1sin sin cos 1 2 2 1 5 0.174 in . 2 30 10 0.125 / 64 Q M 3 M Rd PR R Rd EI Q EI PR PR PRd EI EI EI Ans                                             ______________________________________________________________________________ 4-84 Both the radius and the length are sufficiently large to use Eq. (4-41) for the curved beam portion and to neglect transverse shear stress for the straight portion. Chapter 4 - Rev B, Page 59/81 • Place a dummy force, Q, at A vertically downward. The load in the straight section is the axial force, Q, whereas the bending moment is only a function of P and is not a function of Q. When setting Q = 0, there is no axial or bending contribution. In the curved section  1 cos sin sinMM P R l QR R Q            From Eq. (4-41)                 /2 /2 0 0 0 /22 2 0 2 6 4 1 1 1 cos sin 1sin sin cos sin 2 2 2 1 5 5 2 4 0.452 in 2 30 10 0.125 / 64 Q MM Rd P R l R Rd EI Q EI PR PR PR2R R l d R l R R EI EI EI                                                          l Since the deflection is negative,  is in the opposite direction of Q. Thus the deflection is 0.452 in .Ans   ______________________________________________________________________________ 4-85 Consider the force of the mass to be F, where F = 9.81(1) = 9.81 N. The load in AB is tension 1ABAB FF F F     For the curved section, the radius is sufficiently large to use Eq. (4-41). There is no bending in section DE. For section BCD, let  be counterclockwise originating at D sin sin 0MM FR R F         Using Eqs. (4-29) and (4-41)           3 2 0 0 33 3 3 2 4 1 1 sin 409.81 80 2 2 207 10 2 / 4 2 2 / 64 6.067 mm . AB AB FFl M Fl FRM Rd d AE F EI F AE EI Fl FR F l R AE EI E A I Ans                                                   ______________________________________________________________________________ Chapter 4 - Rev B, Page 60/81 • 4-86 AOA = 2(0.25) = 0.5 in2, IOAB = 0.25(23)/12 = 0.1667 in4, IAC =  (0.54)/64 = 3.068 (10-3) in4 Applying a force F at point B, using statics, the reaction forces at O and C are as shown. OA: Axial 3 3OAOA FF F F     Bending 2 2OAOA MM Fx x F       AB: Bending ABAB MM F x x F       AC: Isolating the upper curved section    3 sin cos 1 3 sin cos 1ACAC MM FR R F                                             10 20 2 2 0 0 /23 2 0 3 3 6 6 6 3 /2 2 2 6 3 0 1 14 9 sin cos 1 4 10 203 10 3 0.5 10.4 10 3 10.4 10 0.1667 3 10.4 10 0.1667 9 10 sin 2sin cos 2sin cos 2cos 1 30 10 3.068 10 1 OA OA OAB OAB AC FFl Fx dx F x d x AE F EI EI FR d EI F FF F d                                               5 4 3.731 10 7.691 10 1.538 10 0.09778 1 2 2 4 4 0.0162 0.0162 100 1.62 in . F F F F F Ans 2                     _____________________________________________________________________________ 4-87 AOA = 2(0.25) = 0.5 in2, IOAB = 0.25(23)/12 = 0.1667 in4, IAC =  (0.54)/64 = 3.068 (10-3) in4 Applying a vertical dummy force, Q, at A, from statics the reactions are as shown. The dummy force is transmitted through section Chapter 4 - Rev B, Page 61/81 • OA and member AC. OA: 3 1OAOA FF F Q Q      AC:        3 sin 3 1 cos sin cos 1ACAC MM F Q R F Q R R Q                             /2 0 0 /23 2 0 3 6 6 3 1 3 3 sin cos 1 3 100 10 3 100 10 1 2 2 0.462 in . 4 4 210.4 10 0.5 30 10 3.068 10 OA AC AC OA AC Q OA OA AC F MFl M Rd AE Q EI Q Fl FR d AE EI Ans                                                  ______________________________________________________________________________ 4-88 I =  (64)/64 = 63.62 mm4 0     / 2 sin sin (1 cos ) (1 cos ) MM FR R F TT FR R F               According to Castigliano’s theorem, a positive  U/ F will yield a deflection of A in the negative y direction. Thus the deflection in the positive y direction is /2 /22 2 0 0 1 1( ) ( sin ) [ (1 cos )]A y U F R R d F R R d F EI GJ                   Integrating and substituting  2 and / 2 1J I G E          3 3 3 3 3( ) (1 ) 2 4 8 (3 8) 4 4 4 (250)(80)[4 8 (3 8)(0.29)] 12.5 mm . 4(200)10 63.62 A y FR FR EI EI Ans                                ______________________________________________________________________________ 4-89 The force applied to the copper and steel wire assembly is (1) 400 lbfc sF F  Since the deflections are equal, c s  Chapter 4 - Rev B, Page 62/81 • c s Fl Fl AE AE            2 6 23( / 4)(0.1019) (17.2)10 ( / 4)(0.1055) (30)10 c sF l F l    6 sF Yields, . Substituting this into Eq. (1) gives 1.6046cF  1.604 2.6046 400 153.6 lbf 1.6046 246.5 lbf s s s s c s F F F F F F        2 246.5 10 075 psi 10.1 kpsi . 3( / 4)(0.1019) c c c F Ans A       2 153.6 17 571 psi 17.6 kpsi . ( / 4)(0.1055 ) s s s F Ans A       2 6 153.6(100)(12) 0.703 in . ( / 4)(0.1055) (30)10s Fl Ans AE          ______________________________________________________________________________ 4-90 (a) Bolt stress 0.75(65) 48.8 kpsi .b Ans   Total bolt force 26 6(48.8) (0.5 ) 57.5 kips 4b b b F A         Cylinder stress 2 2 57.43 13.9 kpsi . ( / 4)(5.5 5 ) b c c F Ans A         (b) Force from pressure 2 2(5 ) (500) 9817 lbf 9.82 kip 4 4 DP p     Fx = 0 Pb + Pc = 9.82 (1) Since ,c b  2 2 2( / 4)(5.5 5 ) 6( / 4)(0.5 ) c bP l P l E E    Pc = 3.5 Pb (2) Substituting this into Eq. (1) Pb + 3.5 Pb = 4.5 Pb = 9.82  Pb = 2.182 kip. From Eq. (2), Pc = 7.638 kip Using the results of (a) above, the total bolt and cylinder stresses are 2 2.18248.8 50.7 kpsi . 6( / 4)(0.5 )b    Ans   Chapter 4 - Rev B, Page 63/81 • 2 2 7.63813.9 12.0 kpsi . ( / 4)(5.5 5 )c Ans        ______________________________________________________________________________ 4-91 Tc + Ts = T (1) c = s          (2)c s cc s c s s JGT l T l T T JG JG JG    Substitute this into Eq. (1)           c s s s s s s JG JG T T T T T JG JG JG      c The percentage of the total torque carried by the shell is       100 % Torque .s s c JG Ans JG JG   ______________________________________________________________________________ 4-92 RO + RB = W (1) OA = AB OA AB Fl Fl AE AE            400 600 3 (2) 2 O B O B R R R R AE AE    Substitute this unto Eq. (1) 3 4 1.6 kN . 2 B B B R R R    Ans From Eq. (2) 31.6 2.4 kN . 2O R Ans  3 2400(400) 0.0223 mm . 10(60)(71.7)(10 )A    OA Fl Ans AE       ______________________________________________________________________________ 4-93 See figure in Prob. 4-92 solution. Procedure 1: 1. Let RB be the redundant reaction. Chapter 4 - Rev B, Page 64/81 • 2. Statics. RO + RB = 4 000 N  RO = 4 000  RB (1) 3. Deflection of point B.     600 4000 400 0 (2B BB R R AE AE      ) 4. From Eq. (2), AE cancels and RB = 1 600 N Ans. and from Eq. (1), RO = 4 000  1 600 = 2 400 N Ans. 3 2400(400) 0.0223 mm . 10(60)(71.7)(10 )A OA Fl Ans AE         ______________________________________________________________________________ 4-94 (a) Without the right-hand wall the deflection of point C would be             3 3 2 6 2 6 5 10 8 2 10 5 / 4 0.75 10.4 10 / 4 0.5 10.4 10 0.01360 in 0.005 in Hits wall . C Fl AE Ans           (b) Let RC be the reaction of the wall at C acting to the left (). Thus, the deflection of point C is now               3 3 2 6 2 6 2 2 5 10 8 2 10 5 / 4 0.75 10.4 10 / 4 0.5 10.4 10 4 8 50.01360 0.005 10.4 10 0.75 0.5 C C C C R R R                    6  or,  60.01360 4.190 10 0.005 2053 lbf 2.05 kip .C CR R A      ns Statics. Considering  +, 5 000  RA  2 053 = 0  RA = 2 947 lbf = 2.95 kip  Ans. Deflection. AB is 2 947 lbf in tension. Thus           3 2 6 8 2947 8 5.13 10 in . / 4 0.75 10.4 10 A B AB AB R Ans A E         ______________________________________________________________________________ 4-95 Since OA = AB, (4) (6) 3 (1) 2 OA AB OA AB T T T T JG JG    Chapter 4 - Rev B, Page 65/81 • Statics. TOA + TAB = 200 (2) Substitute Eq. (1) into Eq. (2), 3 5 200 80 lbf in . 2 2AB AB AB AB T T T T An      s From Eq. (1) 3 3 80 120 lbf in . 2 2OA AB T T An    s       0 4 6 80 6 180 0.390 . / 32 0.5 11.5 10A Ans        max 3 3 16 12016 4890 psi 4.89 kpsi . 0.5OA T Ans d             3 16 80 3260 psi 3.26 kpsi . 0.5AB Ans     ______________________________________________________________________________ 4-96 Since OA = AB,    4 4 (4) (6) 0.2963 (1) / 32 0.5 / 32 0.75 OA AB OA AB T T T T G G     Statics. TOA + TAB = 200 (2) Substitute Eq. (1) into Eq. (2), 0.2963 1.2963 200 154.3 lbf in .AB AB AB ABT T T T An      s From Eq. (1)  0.2963 0.2963 154.3 45.7 lbf in .OA ABT T    Ans       0 4 6 154.3 6 180 0.148 . / 32 0.75 11.5 10A Ans        max 3 3 16 45.716 1862 psi 1.86 kpsi . 0.5OA T Ans d             3 16 154.3 1862 psi 1.86 kpsi . 0.75AB Ans     ______________________________________________________________________________ Chapter 4 - Rev B, Page 66/81 • 4-97 Procedure 1. 1. Arbitrarily, choose RC as a redundant reaction. 2. Statics. Fx = 0, 12(103)  6(103)  RO  RC = 0 RO = 6(103)  RC (1) 3. The deflection of point C. 3 3 312(10 ) 6(10 ) (20) 6(10 ) (10) (15) 0C C CC R R R AE AE AE              4. The deflection equation simplifies to  45 RC + 60(103) = 0  RC = 1 333 lbf  1.33 kip Ans. From Eq. (1), RO = 6(103)  1 333 = 4 667 lbf  4.67 kip Ans. FAB = FB + RC = 6 +1.333 = 7.333 kips compression 7.333 14.7 kpsi . (0.5)(1) AB AB F Ans A      Deflection of A. Since OA is in tension, 6 4667(20) 0.00622 in . (0.5)(1)(30)10 O OA A OA R l Ans AE       ______________________________________________________________________________ 4-98 Procedure 1. 1. Choose RB as redundant reaction. 2. Statics. RC = wl  RB (1)  21 (2) 2C B M l R l a  w 3. Deflection equation for point B. Superposition of beams 2 and 3 of Table A-9,         3 2 2 24 6 3 24 B B R l a l a 0l l a l a l EI EI            w y 4. Solving for RB.           22 2 2 6 4 8 3 2 8 BR l l l a l al a l al a An l a            w w .s Substituting this into Eqs. (1) and (2) gives Chapter 4 - Rev B, Page 67/81 •     2 25 10 8C B .R l R l al a Ans l a       ww    2 2 21 2 .2 8C BM l R l a l al a Ans      ww ______________________________________________________________________________ 4-99 See figure in Prob. 4-98 solution. Procedure 1. 1. Choose RB as redundant reaction. 2. Statics. RC = wl  RB (1)  21 (2) 2C B M l R l a  w 3. Deflection equation for point B. Let the variable x start at point A and to the right. Using singularity functions, the bending moment as a function of x is 1 121 2 B B MM x R x a x a R         w       0 2 2 0 1 1 1 1 10 0 2 2 l B B B l l B a U My M dx R EI R x dx x R x a x a dx EI EI                    w w or,        3 34 4 3 31 1 02 4 3 3 BRal a l a l a a a               w  Solving for RB gives           4 4 3 3 2 2 3 3 4 3 288B .R l a a l a l al a Ans l al a           w w From Eqs. (1) and (2)     2 25 10 8C B .R l R l al a Ans l a       ww    2 2 21 2 .2 8C BM l R l a l al a Ans      ww Chapter 4 - Rev B, Page 68/81 • ______________________________________________________________________________ 4-100 Note: When setting up the equations for this problem, no rounding of numbers was made. It turns out that the deflection equation is very sensitive to rounding. Procedure 2. 1. Statics. R1 + R2 = wl (1) 22 1 1 (2) 2 R l M l  w 2. Bending moment equation. 2 1 1 2 3 1 1 1 3 4 2 1 1 1 1 2 1 1 (3) 2 6 1 1 1 (4) 6 24 2 M R x x M dy 2 R x x M x C dx EIy R x x M x C x C             w w w EI EI = 30(106)(0.85) = 25.5(106) lbfin2. 3. Boundary condition 1. At x = 0, y =  R1/k1 =  R1/[1.5(106)]. Substitute into Eq. (4) with value of EI yields C2 =  17 R1. Boundary condition 2. At x = 0, dy /dx =  M1/k2 =  M1/[2.5(106)]. Substitute into Eq. (3) with value of EI yields C1 =  10.2 M1. Boundary condition 3. At x = l, y =  R2/k3 =  R1/[2.0(106)]. Substitute into Eq. (4) with value of EI yields 3 4 22 1 1 1 1 1 1 112.75 10.2 17 (5) 6 24 2 R R l l M l M l R     w   For the deflection at x = l /2 = 12 in, Eq. (4) gives Equations (1), (2), and (5), written in matrix form with w = 500/12 lbf/in and l = 24 in, are   1 3 2 1 1 1 0 1 0 24 1 12 10 2287 12.75 532.8 576 R R M                         Solving, the simultaneous equations yields R1 = 554.59 lbf, R2 = 445.41.59 lbf, M1 = 1310.1 lbfin Ans. Chapter 4 - Rev B, Page 69/81 •             3 4 12in 6 3 1 1 1 500 1554.59 12 12 1310.1 12 6 24 12 225.5 10 10.2 1310.1 12 17 554.59 5.51 10 in . x y Ans           2 ______________________________________________________________________________ -101 Cable area, 4 2 2(0.5 ) 0.1963 in 4 A   Procedure 2. 1. Statics. RA + FBE + FDF = 5(103) (1) 3 FDF + FBE = 10(10 ) (2) . 3 2 Bending moment equation. 1 116 5000 32A BEM R x F x x     2 22 1 3 33 1 2 1 1 16 2500 32 (3) 2 2 1 1 250016 32 (4) 6 6 3 A BE A BE dyEI R x F x x C dx EIy R x F x x C x C              B.C. 1 3. : At x = 0, y = 0  C2 = 0 B.C. 2 : At x = 16 in, 66 (38) 6.453(10 ) 0.1963(30)10 BE B BE BE FFly F AE           Substituting into Eq. (4) and evaluating at x = 16 in 6 6 130(10 )(1.2)( 6.453)(10 ) 16EIy F R   3 1(16)6B BE A C lifying gives 682.7 RA + 232.3 FBE + 16 C1 = 0 (5) B.C. 2 Simp : At x = 48 in, 6 6 (38) 6.453(10 ) 0.1963(30)10 DF D DF DF FFly F AE           Substituting into Eq. (4) and evaluating at x = 48 in,  3 3 132500232.3 48 (48 16) (48 32) 486 6 3F A BE 1 1 E D DIy F R F C        plifying gives 18 432 RA + 5 461 FBE + 232.3 FDF + 48 C1 = 3.413(10 ) (6) Sim 6 Chapter 4 - Rev B, Page 70/81 • Equations (1), (2), (5) and (6) in matrix form are  61 50001 1 1 0 100000 1 3 0 0682.7 232.3 0 16 3.413 1018432 5461 232.3 48 A BE DF R F F C                              Solve simultaneously or use software. The results are RA =  970.5 lbf, FBE = 3956 lbf, FDF = 2015 lbf, and C1 =  16 020 lbfin2. 3956 201520.2 kpsi, 10.3 kpsi . 0.1963 0.1963BE DF Ans     EI = 30(106)(1.2) = 36(106) lbfin2       3 33 6 3 33 6 1 970.5 3956 250016 32 16 020 6 6 336 10 1 161.8 659.3 16 833.3 32 16 020 36 10 y x x x x x x x            x      B: x = 16 in,        3 6 1 161.8 16 16 020 16 0.0255 in . 36 10B y A       ns C: x = 32 in,        33 6 1 161.8 32 659.3 32 16 16 020 32 36 10 0.0865 in . C     y  Ans    D: x = 48 in,          3 33 6 1 161.8 48 659.3 48 16 833.3 48 32 16 020 48 36 10 0.0131 in . D  Ans           ______________________________________________________________________________ -102 Beam: EI = 207(10 )21(10 ) 2. A Procedure 2. 1. Statics. y  3 34 = 4.347(109) Nmm Rods: = ( /4)82 = 50.27 mm2. Chapter 4 - Rev B, Page 71/81 • RC + FBE  FDF = 2 000 (1) RC + 2FBE = 6 000 (2) 2. Bending moment equation. M =  2 000 x + FBE x  75 1 + RC x  150 1 2 22 1 3 33 1 2 1 11000 75 150 (3) 2 2 1000 1 175 150 (4) 3 6 6 BE C BE C dyEI x F x R x C dx EIy x F x R x C x C                3. B.C 1 . At x = 75 mm,       6 3 50 4.805 10 50.27 207 10 BE B BE BE FFly F AE           Substituting into Eq. (4) at x = 75 mm,        9 6 3 1 210004.347 10 4.805 10 75 753BEF C C        Simplifying gives    3 61 220.89 10 75 140.6 10 (5)BEF C C   B.C 2. At x = 150 mm, y = 0. From Eq. (4),      33 1 21000 1150 150 75 150 03 6 BEF C     C  or,    3 91 231 10 150 1.125 10 (6)BEF C C   70. B.C 3. At x = 225 mm,       6 3 65 6.246 10 50.27 207 10 DF D DF DF FFly F AE        Substituting into Eq. (4) at x = 225 mm, Chapter 4 - Rev B, Page 72/81 •             39 6 3 3 1 2 1000 14.347 10 6.246 10 225 225 75 3 6 1 225 150 225 6 DF BE C F F R C C            Simplifying gives       3 3 3 91 270.31 10 562.5 10 27.15 10 225 3.797 10 (7)C BE DFR F F C C      Equations (1), (2), (5), (6), and (7) in matrix form are                     3 3 3 6 3 91 3 3 3 2 9 2 101 1 1 0 0 1 2 0 0 0 6 10 0 20.89 10 0 75 1 140.6 10 0 70.31 10 0 150 1 1.125 10 70.31 10 562.5 10 27.15 10 225 1 3.797 10 C BE DF R F F C C                                           Solve simultaneously or use software. The results are RC =  2378 N, FBE = 4189 N, FDF =  189.2 N Ans. and C1 = 1.036 (107) Nmm2, C2 =  7.243 (108) Nmm3. The bolt stresses are BE = 4189/50.27 = 83.3 MPa, DF =  189/50.27=  3.8 MPa Ans. The deflections are From Eq. (4)     8 9 1 7.243 10 0.167 mm . 4.347 10A y A      ns For points B and D use the axial deflection equations*.     3 4189 50 0.0201 mm . 50.27 207 10B BE Fly A AE           ns       3 3 189 65 1.18 10 mm . 50.27 207 10D DF Fly A AE         ns *Note. The terms in Eq. (4) are quite large, and due to rounding are not very accurate for calculating the very small deflections, especially for point D. ______________________________________________________________________________ 4-103 (a) The cross section at A does not rotate. Thus, for a single quadrant we have Chapter 4 - Rev B, Page 73/81 • 0 A U M    The bending moment at an angle  to the x axis is  1 cos 1 2A A FR MM M M       The rotation at A is /2 0 1 0A A A U MM Rd M EI M         Thus,     /2 0 1 1 cos 1 0 0 2 2A A FR FR FRM Rd M EI                  2 2 or, 21 2A FRM        Substituting this into the equation for M gives 2cos 2 FRM         (1) The maximum occurs at B where  =  /2 max .B FRM M Ans     (b) Assume B is supported on a knife edge. The deflection of point D is  U/ F. We will deal with the quarter-ring segment and multiply the results by 4. From Eq. (1) 2cos 2 M R F          Thus,   2/2 /23 3 0 0 3 2 4 2cos 4 8 . 4 D U M FR FRM Rd d F EI F EI EI FR Ans EI       2                          ______________________________________________________________________________ 4-104     2 cr 2 4 4 4 41 where 64 64 C EIP l D dI D d K K D            2 4 4 cr 2 164 C E DP K l         Chapter 4 - Rev B, Page 74/81 •   1/4 2 cr 3 4 64 . 1 P lD A CE K        ns ______________________________________________________________________________ 4-105      2 2 4 4 4 2 21 , 1 1 14 64 64A D K I D K D K K  ,           where K = d / D. The radius of gyration, k, is given by   2 2 21 16 I Dk K A    From Eq. (4-46)        2 2 2 2 cr 2 22 2 2 2 24/ 4 1 4 /16 1 y y y y S l S lP S S k CED K D K C        E      2 2 2 2 2 2 cr 2 2 2 4 1 4 1 1 y y S l D K P D K S D K C         E      2 2 2 2 2 cr 2 4 1 1 4 1 y y S l K D K S P K CE                    1/22 2 2 cr 2 2 2 1/2 2 cr 2 2 2 4 14 1 1 1 2 . 1 1 y y y y y S l KPD S K K CE K S S lP Ans S K CE K                         ______________________________________________________________________________ 4-106 (a) 2 2 0.90, (0.75)(800) (0.5) 0 1373 N 0.9 0.5 A BOM F       BOF Using nd = 4, design for Fcr = nd FBO = 4(1373) = 5492 N 2 20.9 0.5 1.03 m, 165 MPayl S    In-plane: 1/21/2 3 /12 0.2887 0.2887(0.025) 0.007 218 m, 1.0I bhk h A bh               C 1.03 142.7 0.007218 l k   1/22 9 6 1 2 (207)(10 ) 157.4 165(10 ) l k            Chapter 4 - Rev B, Page 75/81 • Since use Johnson formula. 1( / ) ( / )l k l k Try 25 mm x 12 mm,     26 6 cr 9 165 10 10.025(0.012) 165 10 (142.7) 29.1 kN 2 1(207)10 P                This is significantly greater than the design load of 5492 N found earlier. Check out-of- plane. Out-of-plane: 0.2887(0.012) 0.003 464 in, 1.2k C  1.03 297.3 0.003 464 l k   Since use Euler equation. 1( / ) ( / )l k l k   2 9 cr 2 1.2 207 10 0.025(0.012) 8321 N 297.3 P    This is greater than the design load of 5492 N found earlier. It is also significantly less than the in-plane Pcr found earlier, so the out-of-plane condition will dominate. Iterate the process to find the minimum h that gives Pcr greater than the design load. With h = 0.010, Pcr = 4815 N (too small) h = 0.011, Pcr = 6409 N (acceptable) Use 25 mm x 11 mm. If standard size is preferred, use 25 mm x 12 mm. Ans. (b)  61373 10.4 10 Pa 10.4 MPa0.012(0.011)b P dh          No, bearing stress is not significant. Ans. ______________________________________________________________________________ 4-107 This is an open-ended design problem with no one distinct solution. ______________________________________________________________________________ 4-108 F = 1500( /4)22 = 4712 lbf. From Table A-20, Sy = 37.5 kpsi Pcr = nd F = 2.5(4712) = 11 780 lbf (a) Assume Euler with C = 1       1/41/4 22 2 4 cr cr 2 3 3 6 64 11790 5064 1.193 in 64 1 30 10 P l P lI d d C E CE                     Use d = 1.25 in. The radius of gyration, k = ( I / A)1/2 = d /4 = 0.3125 in Chapter 4 - Rev B, Page 76/81 •         1/21/2 2 62 3 1 2 6 4 cr 2 50 160 0.3125 2 (1)30 102 126 use Euler 37.5 10 30 10 / 64 1.25 14194 lbf 50 y l k l CE k S P                            Since 14 194 lbf > 11 780 lbf, d = 1.25 in is satisfactory. Ans. (b)       1/4 2 3 6 64 11780 16 0.675 in, 1 30 10 d          so use d = 0.750 in k = 0.750/4 = 0.1875 in 16 85.33 use Johnson 0.1875 l k           23 2 3 cr 6 37.5 10 10.750 37.5 10 85.33 12748 lbf 4 2 1 30 10 P                Use d = 0.75 in. (c) ( ) ( ) 14194 3.01 . 4712 12748 2.71 . 4712 a b n A n A     ns ns ______________________________________________________________________________ 4-109 From Table A-20, Sy = 180 MPa 4F sin = 2 943 735.8 sin F   In range of operation, F is maximum when  = 15 max o 735.8 2843 N per bar sin15 F   Pcr = ndFmax = 3.50 (2 843) = 9 951 N l = 350 mm, h = 30 mm Chapter 4 - Rev B, Page 77/81 • Try b = 5 mm. Out of plane, k = b / 12 = 5/ 12 = 1.443 mm             1/22 9 6 1 2 32 cr 2 2 350 242.6 1.443 2 1.4 207 10 178.3 use Euler 180 10 1.4 207 10 5(30) 7 290 N / 242.6 l k l k C EP A l k                     Too low. Try b = 6 mm. k = 6/ 12 = 1.732 mm       2 32 cr 2 2 350 202.1 1.732 1.4 207 10 6(30) 12605 N / 202.1 l k C EP A l k       O.K. Use 25  6 mm bars Ans. The factor of safety is 12605 4.43 . 2843 n A ns  ______________________________________________________________________________ 4-110 P = 1 500 + 9 000 = 10 500 lbf Ans. MA = 10 500 (4.5/2)  9 000 (4.5) +M = 0 M = 16 874 lbfin e = M / P = 16 874/10 500 = 1.607 in Ans. From Table A-8, A = 2.160 in2, and I = 2.059 in4. The stresses are determined using Eq. (4-55)   2 2 2 2.059 0.953 in 2.160 1.607 3 / 2105001 1 17157 psi 17.16 kpsi . 2.160 0.953c Ik A P ec Ans A k                       ______________________________________________________________________________ 4-111 This is a design problem which has no single distinct solution. ______________________________________________________________________________ Chapter 4 - Rev B, Page 78/81 • 4-112 Loss of potential energy of weight = W (h + ) Increase in potential energy of spring = 21 2 k W (h + ) = 21 2 k or, 2 2 2 0W W h k k     . W = 30 lbf, k = 100 lbf/in, h = 2 in yields  2  0.6   1.2 = 0 Taking the positive root (see discussion on p. 192) 2max 1 0.6 ( 0.6) 4(1.2) 1.436 in . 2 Ans        Fmax = k  max = 100 (1.436) = 143.6 lbf Ans. ______________________________________________________________________________ 4-113 The drop of weight W1 converts potential energy, W1 h, to kinetic energy 21 1 1 2 W g v . Equating these provides the velocity of W1 at impact with W2. 211 1 1 1 2 2 WW h gh g   v v (1) Since the collision is inelastic, momentum is conserved. That is, (m1 + m2) v2 = m1 v1, where v2 is the velocity of W1 + W2 after impact. Thus 1 2 1 1 12 1 2 1 1 2 1 2 2W W W W W gh g g W W W W        v v v v (2) The kinetic and potential energies of W1 + W2 are then converted to potential energy of the spring. Thus,  2 21 2 2 1 2 1 1 2 2 W W W W k g     v Substituting in Eq. (1) and rearranging results in 2 2 1 2 1 1 2 2 2W W W h k W W k     0 (3) Solving for the positive root (see discussion on p. 192) 2 2 1 2 1 2 1 1 2 1 2 4 8 2 W W W W W h k k W             W k (4) Chapter 4 - Rev B, Page 79/81 • W1 = 40 N, W2 = 400 N, h = 200 mm, k = 32 kN/m = 32 N/mm. 2 21 40 400 40 400 40 2002 4 8 29.06 mm . 2 32 32 40 400 32 Ans                    Fmax = k = 32(29.06) = 930 N Ans. ______________________________________________________________________________ 4-114 The initial potential energy of the k1 spring is Vi = 21 1 2 k a . The movement of the weight W the distance y gives a final potential of Vf =  2 21 1 2 2 k a y k y  2 1 . Equating the two energies give  22 21 1 1 1 1 2 2 2 k a k a y k y   2 Simplifying gives   21 2 12 0k k y ak y   This has two roots, y = 0, 1 1 2 2k a k k . Without damping the weight will vibrate between these two limits. The maximum displacement is thus y max = 1 1 2 2k a k k Ans. With W = 5 lbf, k1 = 10 lbf/in, k2 = 20 lbf/in, and a = 0.25 in  max 2 0.25 10 0.1667 in . 10 20 y Ans   ______________________________________________________________________________ Chapter 4 - Rev B, Page 80/81 • Chapter 6 6-1 Eq. (2-21): 3.4 3.4(300) 1020 MPaut BS H   Eq. (6-8): 0.5 0.5(1020) 510 MPae utS S    Table 6-2: 1.58, 0.085a b   Eq. (6-19): 0.0851.58(1020) 0.877ba utk aS    Eq. (6-20): 0.107 0.1071.24 1.24(10) 0.969bk d     Eq. (6-18): (0.877)(0.969)(510) 433 MPa .e a b eS k k S Ans   ______________________________________________________________________________ 6-2 (a) Table A-20: Sut = 80 kpsi Eq. (6-8): 0.5(80) 40 kpsi .eS A   ns ns ns (b) Table A-20: Sut = 90 kpsi Eq. (6-8): 0.5(90) 45 kpsi .eS A   (c) Aluminum has no endurance limit. Ans. (d) Eq. (6-8): Sut > 200 kpsi, 100 kpsi .eS A  ______________________________________________________________________________ 6-3 rev120 kpsi, 70 kpsiutS   Fig. 6-18: 0.82f  Eq. (6-8): 0.5(120) 60 kpsi e eS S    Eq. (6-14):   22 0.82(120)( ) 161.4 kpsi 60 ut e f Sa S    Eq. (6-15): 1 1 0.82(120)log log 0.0716 3 3 60 ut e f Sb S                Eq. (6-16): 11/ 0.0716 rev 70 116 700 cycles . 161.4 b N A a             ns ______________________________________________________________________________ 6-4 rev1600 MPa, 900 MPautS   Fig. 6-18: Sut = 1600 MPa = 232 kpsi. Off the graph, so estimate f = 0.77. Eq. (6-8): Sut > 1400 MPa, so Se = 700 MPa Eq. (6-14):   22 0.77(1600)( ) 2168.3 MPa 700 ut e f Sa S    Chapter 6 - Rev. A, Page 1/66 • Eq. (6-15): 1 1 0.77(1600)log log 0.081838 3 3 700 ut e f Sb S                Eq. (6-16): 11/ 0.081838 rev 900 46 400 cycles . 2168.3 b N A a             ns ______________________________________________________________________________ 6-5 230 kpsi, 150 000 cyclesutS N  Fig. 6-18, point is off the graph, so estimate: f = 0.77 Eq. (6-8): Sut > 200 kpsi, so 100 kpsie eS S   Eq. (6-14):   22 0.77(230)( ) 313.6 kpsi 100 ut e f Sa S    Eq. (6-15): 1 1 0.77(230)log log 0.08274 3 3 100 ut e f Sb S                Eq. (6-13): 0.08274313.6(150 000) 117.0 kpsi .bfS aN Ans    ______________________________________________________________________________ 6-6 = 160 kpsi 1100 MPautS  Fig. 6-18: f = 0.79 Eq. (6-8): 0.5(1100) 550 MPa e eS S    Eq. (6-14):   22 0.79(1100)( ) 1373 MPa 550 ut e f Sa S    Eq. (6-15): 1 1 0.79(1100)log log 0.06622 3 3 550 ut e f Sb S                Eq. (6-13): 0.066221373(150 000) 624 MPa .bfS aN Ans    ______________________________________________________________________________ 6-7 150 kpsi, 135 kpsi, 500 cyclesut ytS S N   Fig. 6-18: f = 0.798 From Fig. 6-10, we note that below 103 cycles on the S-N diagram constitutes the low- cycle region, in which Eq. (6-17) is applicable. Chapter 6 - Rev. A, Page 2/66 • Eq. (6-17):      log 0.798 /3log /3 150 500 122 kpsi .ff utS S N Ans      The testing should be done at a completely reversed stress of 122 kpsi, which is below the yield strength, so it is possible. Ans. ______________________________________________________________________________ 6-8 The general equation for a line on a log Sf - log N scale is Sf = aNb, which is Eq. (6-13). By taking the log of both sides, we can get the equation of the line in slope-intercept form. log log logfS b N  a a Substitute the two known points to solve for unknowns a and b. Substituting point (1, Sut), log log(1) logutS b  From which . Substituting point uta S 3(10 , ) and ut utf S a  S 3log log10 logut utf S b S  From which  1/ 3 logb f (log )/3 3 1 10ff utS S N N    N N ______________________________________________________________________________ 6-9 Read from graph: From  3 610 ,90 and (10 ,50). bS aN 1 1 2 2 log log log log log log S a b S a b     From which 1 2 2 2 1 log log log loglog log / S N S Na N N 1 6 3 6 3 log 90log10 log 50log10 log10 /10 2.2095    log 2.2095 0.0851 3 6 10 10 162.0 kpsi log 50 / 90 0.0851 3 ( ) 162 10 10 in kpsi . a f ax a b S N N          Ans Chapter 6 - Rev. A, Page 3/66 • Check: 3 6 3 0.0851 10 6 0.0851 10 ( ) 162(10 ) 90 kpsi ( ) 162(10 ) 50 kpsi f ax f ax S S             The end points agree. ______________________________________________________________________________ 6-10 d = 1.5 in, Sut = 110 kpsi Eq. (6-8): 0.5(110) 55 kpsieS    Table 6-2: a = 2.70, b =  0.265 Eq. (6-19): 0.2652.70(110) 0.777ba utk aS    Since the loading situation is not specified, we’ll assume rotating bending or torsion so Eq. (6-20) is applicable. This would be the worst case. 0.107 0.1070.879 0.879(1.5) 0.842 Eq. (6-18): 0.777(0.842)(55) 36.0 kpsi . b e a b e k d S k k S Ans        ______________________________________________________________________________ 6-11 For AISI 4340 as-forged steel, Eq. (6-8): Se = 100 kpsi Table 6-2: a = 39.9, b =  0.995 Eq. (6-19): ka = 39.9(260)0.995 = 0.158 Eq. (6-20): 0.1070.75 0.907 0.30b k        Each of the other modifying factors is unity. Se = 0.158(0.907)(100) = 14.3 kpsi For AISI 1040: 0.995 0.5(113) 56.5 kpsi 39.9(113) 0.362 0.907 (same as 4340) e a b S k k        Each of the other modifying factors is unity 0.362(0.907)(56.5) 18.6 kpsieS   Not only is AISI 1040 steel a contender, it has a superior endurance strength. ______________________________________________________________________________ Chapter 6 - Rev. A, Page 4/66 • 6-12 D = 1 in, d = 0.8 in, T = 1800 lbfin, f = 0.9, and from Table A-20 for AISI 1020 CD, Sut = 68 kpsi, and Sy = 57 kpsi. (a) 0.1 1Fig. A-15-15: 0.125, 1.25, 1.40 0.8 0.8 ts r D K d d      Get the notch sensitivity either from Fig. 6-21, or from the curve-fit Eqs. (6-34) and (6-35b). We’ll use the equations.          23 5 8 30.190 2.51 10 68 1.35 10 68 2.67 10 68 0.07335a        1 1 0.8120.0733511 0.1 sq a r     Eq. (6-32): Kfs = 1 + qs (Kts  1) = 1 + 0.812(1.40  1) = 1.32 For a purely reversing torque of T = 1800 lbfin, 3 3 16 1.32(16)(1800) 23 635 psi 23.6 kpsi (0.8) fs a fs K TTrK J d         Eq. (6-8): 0.5(68) 34 kpsieS    Eq. (6-19): ka = 2.70(68)0.265 = 0.883 Eq. (6-20): kb = 0.879(0.8)0.107 = 0.900 Eq. (6-26): kc = 0.59 Eq. (6-18) (labeling for shear): Sse = 0.883(0.900)(0.59)(34) = 15.9 kpsi For purely reversing torsion, use Eq. (6-54) for the ultimate strength in shear. Eq. (6-54): Ssu = 0.67 Sut = 0.67(68) = 45.6 kpsi Adjusting the fatigue strength equations for shear, Eq. (6-14):     2 20.9(45.6) 105.9 kpsi 15.9 su se f S a S    Eq. (6-15): 1 1 0.9(45.6)log log 0.137 27 3 3 15.9 su se f Sb S                Eq. (6-16):   1 1 0.137 27 323.3 61.7 10 cycles . 105.9 b aN A a             ns Chapter 6 - Rev. A, Page 5/66 • (b) For an operating temperature of 750 the temperature modification factor, F, from Table 6-4 is kd = 0.90. Sse = 0.883(0.900)(0.59)(0.9)(34) = 14.3 kpsi    2 20.9(45.6) 117.8 kpsi 14.3 1 1 0.9(45.6)log log 0.152 62 3 3 14.3 su se su se f S a S f Sb S                     1 1 0.152 62 323.3 40.9 10 cycles . 117.8 b aN A a             ns y ______________________________________________________________________________ 6-13 (Table A-20) 40.6 m, 2 kN, 1.5, 10 cycles, 770 MPa, 420 MPaa utL F n N S S      First evaluate the fatigue strength. 0.5(770) 385 MPaeS    0.71857.7(770) 0.488ak   Since the size is not yet known, assume a typical value of kb = 0.85 and check later. All other modifiers are equal to one. Eq. (6-18): Se = 0.488(0.85)(385) = 160 MPa In kpsi, Sut = 770/6.89 = 112 kpsi Fig. 6-18: f = 0.83 Eq. (6-14):     2 20.83(770) 2553 MPa 160 ut e f S a  S   Eq. (6-15): 1 1 0.83(770)log log 0.2005 3 3 160 ut e f Sb S                Eq. (6-13): 4 0.20052553(10 ) 403 MPabfS aN    Now evaluate the stress. max (2000 N)(0.6 m) 1200 N mM        max 3 3 3 / 2 6 12006 7200 ( ) /12a M bMc M 3I b b b b b        Pa, with b in m. Compare strength to stress and solve for the necessary b. Chapter 6 - Rev. A, Page 6/66 •  6403 10fS 3 1.57200 /a n b    b = 0.0299 m Select b = 30 mm. Since the size factor was guessed, go back and check it now. Eq. (6-25):    1/20.808 0.808 0.808 30 24.24ed hb b    mm Eq. (6-20): 0.10724.2 0.88 7.62b k        Our guess of 0.85 was slightly conservative, so we will accept the result of b = 30 mm. Ans. Checking yield,  6max 37200 10 267 MPa0.030    max 420 1.57 267 y y S n     ______________________________________________________________________________ -14 Given: w =2.5 in, t = 3/8 in, d = 0.5 in, nd = 2. From Table A-20, for AISI 1020 CD, Eq. (6-8): b = 1 (axial loading) Eq. (6-18): Se = 0.88(1)(0.85)(34) = 25.4 kpsi notch sensitivity either from Fig. 6-20, or from the curve-fit Eqs. (6-34) and h 6 Sut = 68 kpsi and Sy = 57 kpsi. 0.5(68) 34 kpsieS    Table 6-2: 88k 0.2652.70(68) 0.  a Eq. (6-21): k Eq. (6-26): kc = 0.85 Table A-15-1: / 0.5 / 2.5 0.2, 2.5td K  w Get the (6-35a). The relatively large radius is off the graph of Fig. 6-20, so we’ll assume the curves continue according to the same trend and use the equations to estimate the notc sensitivity.         23 5 8 30.246 3.08 10 68 1.51 10 68 2.67 10 68 0.09799a        1 1 0.8360.0979911 0.25 q a r     Eq. (6-32): 1 ( 1) 1 0.836(2.5 1) 2.25f tK q K       Chapter 6 - Rev. A, Page 7/66 • 2.25= 3 (3 / 8)(2.5 0.5) a a a f F FK F A     a e life was not mentioned, we’ll assume infinite life is desired, so the Since a finit completely reversed stress must stay below the endurance limit. 25.4 2 3 e f a a Sn F    ns4.23 kips .aF A ____ __________ ___ ____________________ _________________________________________ ble A-20, for AISI 1095 HR, Sut = 120 kpsi and Sy = 66 kpsi. -15 Given:6 max min2 in, 1.8 in, 0.1 in, 25 000 lbf in, 0.D d r M M      From Ta (6-8):  0.5 0.5 120 60 kpsiS S    Eq. e ut Eq. (6-19): 0.2652.70(120) 0.76ba utk aS    Eq. (6-24): ie 0.370 0.370(1.8) 0.666 nd d   Eq. (6-20): 70.107 0.100.879 0.879(0.666) 0.92b ek d     Fig. A-15-14: Eq. (6-26): 1ck  Eq. (6-18): (0.76)(0.92)(1)(60) 42.0 kpsie a b c eS k k k S    / 2 /1.8 1.11, / 0.1 /1.8 0.056D d r d    2.1tK  Get the notch sensitivity either from Fig. 6-20, or from the curve-fit Eqs. (6-34) and (6-35a). We’ll use the equations.          23 5 8 30.246 3.08 10 120 1.51 10 120 2.67 10 120 0.04770a        1 1 0.870.0477011 0.1 q a r     Eq. (6-32): 1 ( 1) 1 0.87(2.1 1) 1.96f tK q K       4 4 4( / 64) ( / 64)(1.8) 0.5153 inI d    max min 25 000(1.8 / 2) 43 664 psi 43.7 kpsi 0.5153 0 Mc I        Chapter 6 - Rev. A, Page 8/66 • Eq. (6-36):    max min 43.7 01.96 42.8 kpsi 2 2m f K          max min 43.7 01.96 42.8 kpsi 2 2a f K       Eq. (6-46): 1 42.8 42.8 42.0 120 a m f e utn S S       0.73 .fn A ns A factor of safety less than unity indicates a finite life. Check for yielding. It is not necessary to include the stress concentration for static yielding of a ductile material. max 66 1.51 . 43.7 y y S n A     ns ______________________________________________________________________________ 6-16 From a free-body diagram analysis, the bearing reaction forces are found to be 2.1 kN at the left bearing and 3.9 kN at the right bearing. The critical location will be at the shoulder fillet between the 35 mm and the 50 mm diameters, where the bending moment is large, the diameter is smaller, and the stress concentration exists. The bending moment at this point is M = 2.1(200) = 420 kN·mm. With a rotating shaft, the bending stress will be completely reversed. 2 rev 4 420 (35 / 2) 0.09978 kN/mm 99.8 MPa ( / 64)(35) Mc I       This stress is far below the yield strength of 390 MPa, so yielding is not predicted. Find the stress concentration factor for the fatigue analysis. Fig. A-15-9: r/d = 3/35 = 0.086, D/d = 50/35 = 1.43, Kt =1.7 Get the notch sensitivity either from Fig. 6-20, or from the curve-fit Eqs. (6-34) and (6-35a). We’ll use the equations, with Sut = 470 MPa = 68.2 kpsi and r = 3 mm = 0.118 in.          2 33 5 80.246 3.08 10 68.2 1.51 10 68.2 2.67 10 68.2 0.09771a        1 1 0.780.0977111 0.118 q a r     Eq. (6-32): 1 ( 1) 1 0.78(1.7 1) 1.55f tK q K       Chapter 6 - Rev. A, Page 9/66 • Eq. (6-8): ' 0.5 0.5(470) 235 MPae utS S   Eq. (6-19): 0.2654.51(470) 0.88ba utk aS    Eq. (6-24): 0.107 0.1071.24 1.24(35) 0.85bk d     Eq. (6-26): 1ck  Eq. (6-18): ' (0.88)(0.85)(1)(235) 176 MPae a b c eS k k k S    rev 176 1.14 Infinite life is predicted. . 1.55 99.8 e f f Sn A K     ns ______________________________________________________________________________ 6-17 From a free-body diagram analysis, the bearing reaction forces are found to be RA = 2000 lbf and RB = 1500 lbf. The shear-force and bending-moment diagrams are shown. The critical location will be at the shoulder fillet between the 1-5/8 in and the 1-7/8 in diameters, where the bending moment is large, the diameter is smaller, and the stress concentration exists. M = 16 000 – 500 (2.5) = 14 750 lbf · in With a rotating shaft, the bending stress will be completely reversed. rev 4 14 750(1.625 / 2) 35.0 kpsi ( / 64)(1.625) Mc I      This stress is far below the yield strength of 71 kpsi, so yielding is not predicted. Fig. A-15-9: r/d = 0.0625/1.625 = 0.04, D/d = 1.875/1.625 = 1.15, Kt =1.95 Get the notch sensitivity either from Fig. 6-20, or from the curve-fit Eqs. (6-34) and (6-35a). We will use the equations.          2 33 5 80.246 3.08 10 85 1.51 10 85 2.67 10 85 0.07690a        1 1 0.760.0769011 0.0625 q a r     . Eq. (6-32): 1 ( 1) 1 0.76(1.95 1) 1.72f tK q K       Eq. (6-8): S S ' 0.5 0.5(85) 42.5 kpsie ut   Chapter 6 - Rev. A, Page 10/66 • Eq. (6-19): 0.2652.70(85) 0.832ba utk aS    Eq. (6-20): 0.107 0.1070.879 0.879(1.625) 0.835bk d     Eq. (6-26): 1ck  Eq. (6-18): ' (0.832)(0.835)(1)(42.5) 29.5 kpsie a b c eS k k k S    rev 29.5 0.49 . 1.72 35.0 e f f Sn A K     ns Infinite life is not predicted. Use the S-N diagram to estimate the life. Fig. 6-18: f = 0.867    2 20.867(85)Eq. (6-14): 184.1 29.5 1 1 0.867(85)Eq. (6-15): log log 0.1325 3 3 29.5 ut e ut e f S a S f Sb S                   1 1 0.1325rev (1.72)(35.0)Eq. (6-16): 4611 cycles 184.1 b fKN a              N = 4600 cycles Ans. ______________________________________________________________________________ 6-18 From a free-body diagram analysis, the bearing reaction forces are found to be RA = 1600 lbf and RB = 2000 lbf. The shear-force and bending-moment diagrams are shown. The critical location will be at the shoulder fillet between the 1-5/8 in and the 1-7/8 in diameters, where the bending moment is large, the diameter is smaller, and the stress concentration exists. M = 12 800 + 400 (2.5) = 13 800 lbf · in With a rotating shaft, the bending stress will be completely reversed. rev 4 13 800(1.625 / 2  ) 32.8 kpsi ( / 64)(1.625) Mc I     This stress is far below the yield strength of 71 kpsi, so yielding is not predicted. Fig. A-15-9: r/d = 0.0625/1.625 = 0.04, D/d = 1.875/1.625 = 1.15, Kt =1.95 Chapter 6 - Rev. A, Page 11/66 • Get the notch sensitivity either from Fig. 6-20, or from the curve-fit Eqs. (6-34) and (6-35a). We will use the equations          2 33 5 80.246 3.08 10 85 1.51 10 85 2.67 10 85 0.07690a        1 1 0.760.0769011 0.0625 q a r     Eq. (6-32): 1 ( 1) 1 0.76(1.95 1) 1.72f tK q K       Eq. (6-8): ' 0.5 0.5(85) 42.5 kpsie utS S   Eq. (6-19): 0.2652.70(85) 0.832ba utk aS    Eq. (6-20): 0.107 0.1070.879 0.879(1.625) 0.835bk d     Eq. (6-26): 1ck  Eq. (6-18): ' (0.832)(0.835)(1)(42.5) 29.5 kpsie a b c eS k k k S    rev 29.5 0.52 . 1.72 32.8 e f f Sn A K     ns Infinite life is not predicted. Use the S-N diagram to estimate the life. Fig. 6-18: f = 0.867    2 20.867(85)Eq. (6-14): 184.1 29.5 1 1 0.867(85)Eq. (6-15): log log 0.1325 3 3 29.5 ut e ut e f S a S f Sb S                   1 1 0.1325rev (1.72)(32.8)Eq. (6-16): 7527 cycles 184.1 b fKN a              N = 7500 cycles Ans. ______________________________________________________________________________ 6-19 Table A-20: 120 kpsi, 66 kpsiut yS S  N = (950 rev/min)(10 hr)(60 min/hr) = 570 000 cycles One approach is to guess a diameter and solve the problem as an iterative analysis problem. Alternatively, we can estimate the few modifying parameters that are dependent on the diameter and solve the stress equation for the diameter, then iterate to check the estimates. We’ll use the second approach since it should require only one iteration, since the estimates on the modifying parameters should be pretty close. Chapter 6 - Rev. A, Page 12/66 • First, we’ll evaluate the stress. From a free-body diagram analysis, the reaction forces at the bearings are R1 = 2 kips and R2 = 6 kips. The critical stress location is in the middle of the span at the shoulder, where the bending moment is high, the shaft diameter is smaller, and a stress concentration factor exists. If the critical location is not obvious, prepare a complete bending moment diagram and evaluate at any potentially critical locations. Evaluating at the critical shoulder,  2 kip 10 in 20 kip inM        rev 4 3 3 3 / 2 32 2032 203.7 kpsi / 64 M dMc M I d d d d          Now we’ll get the notch sensitivity and stress concentration factor. The notch sensitivity depends on the fillet radius, which depends on the unknown diameter. For now, we’ll estimate a value for q = 0.85 from observation of Fig. 6-20, and check it later. Fig. A-15-9: / 1.4 / 1.4, / 0.1 / 0.1, 1.65tD d d d r d d d K     Eq. (6-32): 1 ( 1) 1 0.85(1.65 1) 1.55f tK q K       Now we will evaluate the fatigue strength. ' 0.265 0.5(120) 60 kpsi 2.70(120) 0.76 e a S k      Since the diameter is not yet known, assume a typical value of k b = 0.85 and check later. All other modifiers are equal to one. Se = (0.76)(0.85)(60) = 38.8 kpsi Determine the desired fatigue strength from the S-N diagram. Fig. 6-18: f = 0.82    2 20.82(120)Eq. (6-14): 249.6 38.8 1 1 0.82(120)Eq. (6-15): log log 0.1347 3 3 38.8 ut e ut e f S a S f Sb S                   0.1347Eq. (6-13): 249.6(570 000) 41.9 kpsibfS aN    Compare strength to stress and solve for the necessary d. Chapter 6 - Rev. A, Page 13/66 •   3rev d = 2.29 in 41.9 1.6 1.55 203.7 / f f f S n K d    Since the size factor and notch sensitivity were guessed, go back and check them now. Eq. (6-20):   0.1570.1570.91 0.91 2.29 0.80bk d    Our guess of 0.85 was conservative. From Fig. 6-20 with r = d/10 = 0.229 in, we are off the graph, but it appears our guess for q is low. Assuming the trend of the graph continues, we’ll choose q = 0.91 and iterate the problem with the new values of kb and q. Intermediate results are Se = 36.5 kpsi, Sf = 39.6 kpsi, and Kf = 1.59. This gives   3rev 39.6 1.6 1.59 203.7 d = 2.36 in Ans. / f f f S n K d    a A quick check of kb and q show that our estimates are still reasonable for this diameter. ______________________________________________________________________________ 6-20 40 kpsi, 60 kpsi, 80 kpsi, 15 kpsi, 25 kpsi, 0e y ut m a mS S S           Obtain von Mises stresses for the alternating, mid-range, and maximum stresses.         1/21/2 22 2 2 1/21/2 22 2 2 3 25 3 0 25.00 kpsi 3 0 3 15 25.98 kpsi a a a m m m                               1/21/2 2 22 2 max max max 1/22 2 3 3 25 3 15 36.06 kpsi a m a m                    max 60 1.66 . 36.06 y y S n A      ns (a) Modified Goodman, Table 6-6 1 1.05 . (25.00 / 40) (25.98 / 80)f n A   ns (b) Gerber, Table 6-7 221 80 25.00 2(25.98)(40)1 1 1.31 . 2 25.98 40 80(25.00)f n A                         ns Chapter 6 - Rev. A, Page 14/66 • (c) ASME-Elliptic, Table 6-8 2 2 1 1.32 . (25.00 / 40) (25.98 / 60)f n A   ns a ______________________________________________________________________________ 6-21 40 kpsi, 60 kpsi, 80 kpsi, 20 kpsi, 10 kpsi, 0e y ut m a mS S S           Obtain von Mises stresses for the alternating, mid-range, and maximum stresses.         1/21/2 22 2 2 1/21/2 22 2 2 3 10 3 0 10.00 kpsi 3 0 3 20 34.64 kpsi a a a m m m                               1/21/2 2 22 2 max max max 1/22 2 3 3 10 3 20 36.06 kpsi a m a m                    max 60 1.66 . 36.06 y y S n A      ns (a) Modified Goodman, Table 6-6 1 1.46 . (10.00 / 40) (34.64 / 80)f n A   ns (b) Gerber, Table 6-7 221 80 10.00 2(34.64)(40)1 1 1.74 . 2 34.64 40 80(10.00)f n A                        ns (c) ASME-Elliptic, Table 6-8 2 2 1 1.59 . (10.00 / 40) (34.64 / 60)f n A   ns m ______________________________________________________________________________ 6-22 40 kpsi, 60 kpsi, 80 kpsi, 10 kpsi, 15 kpsi, 12 kpsi, 0e y ut a m aS S S           Obtain von Mises stresses for the alternating, mid-range, and maximum stresses.         1/21/2 22 2 2 1/21/2 22 2 2 3 12 3 10 21.07 kpsi 3 0 3 15 25.98 kpsi a a a m m m                       Chapter 6 - Rev. A, Page 15/66 •           1/21/2 2 22 2 max max max 1/22 2 3 3 12 0 3 10 15 44.93 kpsi a m a m                      max 60 1.34 . 44.93 y y S n A      ns (a) Modified Goodman, Table 6-6 1 1.17 . (21.07 / 40) (25.98 / 80)f n A   ns (b) Gerber, Table 6-7 221 80 21.07 2(25.98)(40)1 1 1.47 . 2 25.98 40 80(21.07)f n A                        ns (c) ASME-Elliptic, Table 6-8 2 2 1 1.47 . (21.07 / 40) (25.98 / 60)f n A   ns a ______________________________________________________________________________ 6-23 40 kpsi, 60 kpsi, 80 kpsi, 30 kpsi, 0e y ut a m aS S S           Obtain von Mises stresses for the alternating, mid-range, and maximum stresses.       1/21/2 22 2 2 1/22 2 3 0 3 30 51.96 kpsi 3 0 kpsi a a a m m m                           1/21/2 2 22 2 max max max 1/22 3 3 3 30 51.96 kpsi a m a m                   max 60 1.15 . 51.96 y y S n A      ns (a) Modified Goodman, Table 6-6 1 0.77 . (51.96 / 40)f n A  ns (b) Gerber criterion of Table 6-7 is only valid for m > 0; therefore use Eq. (6-47). Chapter 6 - Rev. A, Page 16/66 • 401 0 51.96 a e f f e a Sn n S .77 .Ans         (c) ASME-Elliptic, Table 6-8 2 1 0.77 . (51.96 / 40)f n A  ns Since infinite life is not predicted, estimate a life from the S-N diagram. Since 'm = 0, the stress state is completely reversed and the S-N diagram is applicable for 'a. Fig. 6-18: f = 0.875 Eq. (6-14):   22 0.875(80)( ) 122.5 40 ut e f Sa S    Eq. (6-15): 1 1 0.875(80)log log 0.08101 3 3 40 ut e f Sb S                Eq. (6-16): 11/ 0.08101 rev 51.96 39 600 cycles . 122.5 b N A a             ns a ______________________________________________________________________________ 6-24 40 kpsi, 60 kpsi, 80 kpsi, 15 kpsi, 15 kpsi, 0e y ut a m mS S S           Obtain von Mises stresses for the alternating, mid-range, and maximum stresses.     1/21/2 22 2 23 0 3 15 25.98 kpsia a a               1/21/2 22 2 23 15 3 0 15.00 kpsim m m                     1/21/2 2 22 2 max max max 1/22 2 3 3 15 3 15 30.00 kpsi a m a m                    max 60 2.00 . 30 y y S n A      ns (a) Modified Goodman, Table 6-6 1 1.19 . (25.98 / 40) (15.00 / 80)f n A   ns (b) Gerber, Table 6-7 221 80 25.98 2(15.00)(40)1 1 1.43 . 2 15.00 40 80(25.98)f n A                        ns Chapter 6 - Rev. A, Page 17/66 • (c) ASME-Elliptic, Table 6-8 2 2 1 1.44 . (25.98 / 40) (15.00 / 60)f n A   ns ______________________________________________________________________________ 6-25 Given: . From Table A-20, for AISI 1040 CD, max min28 kN, 28 kNF F   590 MPa, 490 MPa, yS S ut Check for yielding 2max max 28 000 147.4 N/mm 147.4 MPa 10(25 6) F A       max 490 3.32 . 147.4 y y S n A     ns Determine the fatigue factor of safety based on infinite life Eq. (6-8): ' 0.5(590) 295 MPaeS   Eq. (6-19): 0.2654.51(590) 0.832ba utk aS    Eq. (6-21): 1 (axial)bk  Eq. (6-26): 0.85ck  Eq. (6-18): ' (0.832)(1)(0.85)(295) 208.6 MPae a b c eS k k k S   Fig. 6-20: q = 0.83 Fig. A-15-1: t/ 0.24, 2.44d K w 1 ( 1) 1 0.83(2.44 1) 2.20f tK q K        max min max min 28 000 28 000 2.2 324.2 MPa 2 2(10)(25 6) 0 2 a f m f F FK A F FK A            1 324.2 0 208.6 590 0.64 . a m f e ut f n S S n Ans        Since infinite life is not predicted, estimate a life from the S-N diagram. Since m = 0, the stress state is completely reversed and the S-N diagram is applicable for a. Sut = 590/6.89 = 85.6 kpsi Fig. 6-18: f = 0.87 Chapter 6 - Rev. A, Page 18/66 • Eq. (6-14):   22 0.87(590)( ) 1263 208.6 ut e f Sa S    Eq. (6-15): 1 1 0.87(590)log log 0.1304 3 3 208.6 ut e f Sb S                Eq. (6-16): 11/ 0.1304 rev 324.2 33 812 cycles 1263 b N a             N = 34 000 cycles Ans. ________________________________________________________________________ 6-26 max min590 MPa, 490 MPa, 28 kN, 12 kNut yS S F F    Check for yielding 2max max 28 000 147.4 N/mm 147.4 MPa 10(25 6) F A       max 490 3.32 . 147.4 y y S n A     ns Determine the fatigue factor of safety based on infinite life. From Prob. 6-25: 208.6 MPa, 2.2e fS K   max min 28 000 12 0002.2 92.63 MPa 2 2(10)(25 6)a f F FK A       max min 28 000 12 0002.2 231.6 MPa 2 2(10)(25 6)m f F FK A          Modified Goodman criteria: 1 92.63 231.6 208.6 590 a m f e utn S S       1.20 .fn A ns Gerber criteria: 2 2 21 1 1 2 ut a m e f m e ut a S Sn S S                      221 590 92.63 2(231.6)(208.6)1 1 2 231.6 208.6 590(92.63)                   1.49 .fn A ns Chapter 6 - Rev. A, Page 19/66 • ASME-Elliptic criteria: 2 2 2 1 1 ( / ) ( / ) (92.63 / 208.6) (231.6 / 490)f a e m y n S S      2 = 1.54 Ans. The results are consistent with Fig. 6-27, where for a mean stress that is about half of the yield strength, the Modified Goodman line should predict failure significantly before the other two. ______________________________________________________________________________ 6-27 590 MPa, 490 MPaut yS S  (a) max min28 kN, 0 kNF F  Check for yielding 2max max 28 000 147.4 N/mm 147.4 MPa 10(25 6) F A       max 490 3.32 . 147.4 y y S n A     ns From Prob. 6-25: 208.6 MPa, 2.2e fS K  max min max min 28 000 02.2 162.1 MPa 2 2(10)(25 6) 28 000 02.2 162.1 MPa 2 2(10)(25 6) a f m f F FK A F FK A                 1 162.1 162.1 208.6 590 a m f e utn S S       0.95 .fn A ns Since infinite life is not predicted, estimate a life from the S-N diagram. First, find an equivalent completely reversed stress (See Ex. 6-12). rev 162.1 223.5 MPa 1 ( / ) 1 (162.1/ 590) a m utS        Fig. 6-18: f = 0.87 Eq. (6-14):   22 0.87(590)( ) 1263 208.6 ut e f Sa S    Chapter 6 - Rev. A, Page 20/66 • Eq. (6-15): 1 1 0.87(590)log log 0.1304 3 3 208.6 ut e f Sb S                Eq. (6-16): 11/ 0.1304 rev 223.5 586 000 cycles . 1263 b N A a             ns (b) max min28 kN, 12 kNF F  The maximum load is the same as in part (a), so max 147.4 MPa  3.32 .yn A ns Factor of safety based on infinite life: max min max min 28 000 12 0002.2 92.63 MPa 2 2(10)(25 6) 28 000 12 0002.2 231.6 MPa 2 2(10)(25 6) a f m f F FK A F FK A                 1 92.63 231.6 208.6 590 a m f e utn S S       1.20 .fn A ns (c) max min12 kN, 28 kNF F   The compressive load is the largest, so check it for yielding. min min 28 000 147.4 MPa 10(25 6) F A       min 490 3.32 . 147.4 yc y S n A       ns Factor of safety based on infinite life:     max min max min 12 000 28 000 2.2 231.6 MPa 2 2(10)(25 6) 12 000 28 000 2.2 92.63 MPa 2 2(10)(25 6) a f m f F FK A F FK A                  For m < 0, 208.6 0.90 . 231.6 e f a Sn A     ns Chapter 6 - Rev. A, Page 21/66 • Since infinite life is not predicted, estimate a life from the S-N diagram. For a negative mean stress, we shall assume the equivalent completely reversed stress is the same as the actual alternating stress. Get a and b from part (a). Eq. (6-16): 11/ 0.1304 rev 231.6 446 000 cycles . 1263 b N A a             ns  ______________________________________________________________________________ 6-28 Eq. (2-21): Sut = 0.5(400) = 200 kpsi Eq. (6-8): ' 0.5(200) 100 kpsieS   Eq. (6-19): 0.71814.4(200) 0.321ba utk aS    Eq. (6-25): e 0.37 0.37(0.375) 0.1388 ind d   Eq. (6-20): 0.107 0.1070.879 0.879(0.1388) 1.09b ek d    Since we have used the equivalent diameter method to get the size factor, and in doing so introduced greater uncertainties, we will choose not to use a size factor greater than one. Let kb = 1. Eq. (6-18): (0.321)(1)(100) 32.1 kpsieS   40 20 40 2010 lb 30 lb 2 2a m F F     3 3 3 3 32 32(10)(12) 23.18 kpsi (0.375) 32 32(30)(12) 69.54 kpsi (0.375) a a m m M d M d             (a) Modified Goodman criterion 1 23.18 69.54 32.1 200 a m f e utn S S       0.94 .fn A ns Since infinite life is not predicted, estimate a life from the S-N diagram. First, find an equivalent completely reversed stress (See Ex. 6-12). rev 23.18 35.54 kpsi 1 ( / ) 1 (69.54 / 200) a m utS        Fig. 6-18: f = 0.775 Eq. (6-14):   22 0.775(200)( ) 748.4 32.1 ut e f Sa S    Chapter 6 - Rev. A, Page 22/66 • Eq. (6-15): 1 1 0.775(200)log log 0.228 3 3 32.1 ut e f Sb S                Eq. (6-16): 11/ 0.228 rev 35.54 637 000 cycles . 748.4 b N  Ans a            (b) Gerber criterion, Table 6-7 2 2 22 21 1 1 2 1 200 23.18 2(69.54)(32.1)1 1 2 69.54 32.1 200(23.18) 1.16 Infinite life is predicted . ut a m e f m e ut a S Sn S S Ans                                         ______________________________________________________________________________ 6-29 207.0 GPaE  (a) 3 41 (20)(4 ) 106.7 mm 12 I   3 3 3 3 Fl EIyy F EI l    9 12 3 min 3 9 3(207)(10 )(106.7)(10 )(2)(10 ) 48.3 N . 140 (10 ) F A     ns 9 12 3 max 3 9 3(207)(10 )(106.7)(10 )(6)(10 ) 144.9 N . 140 (10 ) F A     ns (b) Get the fatigue strength information. Eq. (2-21): Sut = =3.4HB = 3.4(490) = 1666 MPa From problem statement: Sy = 0.9Sut = 0.9(1666) = 1499 MPa Eq. (6-8): 700 MPaeS   Eq. (6-19): ka = 1.58(1666)-0.085 = 0.84 Eq. (6-25): de = 0.808[20(4)]1/2 = 7.23 mm Eq. (6-20): kb = 1.24(7.23)-0.107 = 1.00 Eq. (6-18): Se = 0.84(1)(700) = 588 MPa This is a relatively thick curved beam, so use the method in Sect. 3-18 to find the stresses. The maximum bending moment will be to the centroid of the section as shown. Chapter 6 - Rev. A, Page 23/66 • M = 142F N·mm, A = 4(20) = 80 mm2, h = 4 mm, ri = 4 mm, ro = ri + h = 8 mm, rc = ri + h/2 = 6 mm Table 3-4: 4 5.7708 mm ln( / ) ln(8 / 4)n o i hr r r    6 5.7708 0.2292 mmc ne r r     5.7708 4 1.7708 mmi n ic r r     8 5.7708 2.2292 mmo o nc r r     Get the stresses at the inner and outer surfaces from Eq. (3-65) with the axial stresses added. The signs have been set to account for tension and compression as appropriate. (142 )(1.7708) 3.441 MPa 80(0.2292)(4) 80 (142 )(2.2292) 2.145 MPa 80(0.2292)(8) 80 i i i o o o Mc F F F F Aer A Mc F F F F Aer A                min max min max ( ) 3.441(144.9) 498.6 MPa ( ) 3.441(48.3) 166.2 MPa ( ) 2.145(48.3) 103.6 MPa ( ) 2.145(144.9) 310.8 MPa i i o o                  166.2 498.6( ) 166.2 MPa 2i a        166.2 498.6( ) 332.4 MPa 2i m        310.8 103.6( ) 103.6 MPa 2o a    310.8 103.6( ) 207.2 MPa 2o m    To check for yielding, we note that the largest stress is –498.6 MPa (compression) on the inner radius. This is considerably less than the estimated yield strength of 1499 MPa, so yielding is not predicted. Check for fatigue on both inner and outer radii since one has a compressive mean stress and the other has a tensile mean stress. Inner radius: Since m < 0, 588 3.54 166.2 e f a Sn     Chapter 6 - Rev. A, Page 24/66 • Outer radius: Since m > 0, we will use the Modified Goodman line. 103.6 207.21/ 588 1666 3.33 a m f e ut f n S S n        Infinite life is predicted at both inner and outer radii. Ans. ______________________________________________________________________________ 6-30 From Table A-20, for AISI 1018 CD, 64 kpsi, 54 kpsiut yS S  Eq. (6-8): ' 0.5(64) 32 kpsieS   Eq. (6-19): 0.2652.70(64) 0.897ak   Eq. (6-20): 1 (axial)bk  Eq. (6-26): 0.85ck  Eq. (6-18): (0.897)(1)(0.85)(32) 24.4 kpsieS   Fillet: Fig. A-15-5: / 3.5 / 3 1.17, / 0.25 / 3 0.083, 1.85tD d r d K     Use Fig. 6-20 or Eqs. (6-34) and (6-35a) for q. Estimate a little high since it is off the graph. q = 0.85 1 ( 1) 1 0.85(1.85 1) 1.72f tK q K       max max 2 min max min max min 5 3.33 kpsi 3.0(0.5) 16 10.67 kpsi 3.0(0.5) 3.33 ( 10.67)1.72 12.0 kpsi 2 2 3.33 ( 10.67)1.72 6.31 kpsi 2 2 a f m f F h K K                                  w min 54 5.06 Does not yield. 10.67 y y S n       Since the midrange stress is negative, 24.4 2.03 12.0 e f a Sn     Chapter 6 - Rev. A, Page 25/66 • Hole: Fig. A-15-1: 1/ 0.4 / 3.5 0.11 2.68td K   w Use Fig. 6-20 or Eqs. (6-34) and (6-35a) for q. Estimate a little high since it is off the graph. q = 0.85 1 0.85(2.68 1) 2.43fK         max max 1 min min 1 5 3.226 kpsi 0.5(3.5 0.4) 16 10.32 kpsi 0.5(3.5 0.4) F h d F h d               w w max min max min 3.226 ( 10.32)2.43 16.5 kpsi 2 2 3.226 ( 10.32)2.43 8.62 kpsi 2 2 a f m f K K                         min 54 5.23 does not yield 10.32 y y S n       Since the midrange stress is negative, 24.4 1.48 16.5 e f a Sn     Thus the design is controlled by the threat of fatigue at the hole with a minimum factor of safety of 1.48. .fn A ns ______________________________________________________________________________ 6-31 64 kpsi, 54 kpsiut yS S  Eq. (6-8): ' 0.5(64) 32 kpsieS   Eq. (6-19): 0.2652.70(64) 0.897ak   Eq. (6-20): 1 (axial)bk  Eq. (6-26): 0.85ck  Eq. (6-18): (0.897)(1)(0.85)(32) 24.4 kpsieS   Fillet: Fig. A-15-5: / 2.5 /1.5 1.67, / 0.25 /1.5 0.17, 2.1tD d r d K     Use Fig. 6-20 or Eqs. (6-34) and (6-35a) for q. Estimate a little high since it is off the graph. q = 0.85 1 ( 1) 1 0.85(2.1 1) 1.94f tK q K       Chapter 6 - Rev. A, Page 26/66 • max max 2 min 16 21.3 kpsi 1.5(0.5) 4 5.33 kpsi 1.5(0.5) F h          w max min max min 21.3 ( 5.33)1.94 25.8 kpsi 2 2 21.3 ( 5.33)1.94 15.5 kpsi 2 2 a f m f K K                        max 54 2.54 Does not yield. 21.3 y y S n      Using Modified Goodman criteria, 1 25.8 15.5 24.4 64 a m f e utn S S       0.77fn  Hole: Fig. A-15-1: 1/ 0.4 / 2.5 0.16 2.55td K   w Use Fig. 6-20 or Eqs. (6-34) and (6-35a) for q. Estimate a little high since it is off the graph. q = 0.85 1 0.85(2.55 1) 2.32fK         max max 1 min min 1 16 15.2 kpsi 0.5(2.5 0.4) 4 3.81 kpsi 0.5(2.5 0.4) F h d F h d               w w max min max min 15.2 ( 3.81)2.32 22.1 kpsi 2 2 15.2 ( 3.81)2.32 13.2 kpsi 2 2 a f m f K K                           max 54 3.55 Does not yield. 15.2 y y S n      Using Modified Goodman criteria 1 22.1 13.2 24.4 64 a m f e utn S S       0.90fn  Chapter 6 - Rev. A, Page 27/66 • Thus the design is controlled by the threat of fatigue at the fillet with a minimum factor of safety of 0.77 .fn A ns ______________________________________________________________________________ 6-32 64 kpsi, 54 kpsiut yS S  From Prob. 6-30, the fatigue factor of safety at the hole is nf = 1.48. To match this at the fillet, 24.4 16.5 kpsi 1.48 e e f a a f S Sn n        where Se is unchanged from Prob. 6-30. The only aspect of a that is affected by the fillet radius is the fatigue stress concentration factor. Obtaining a in terms of Kf, max min 3.33 ( 10.67) 7.00 2 2a f f f K K K       Equating to the desired stress, and solving for Kf, 7.00 16.5 2.36a f fK K     Assume since we are expecting to get a smaller fillet radius than the original, that q will be back on the graph of Fig. 6-20, so we’ll estimate q = 0.8. 1 0.80( 1) 2.36 2.7f t tK K K      From Fig. A-15-5, with D / d = 3.5/3 = 1.17 and Kt = 2.6, find r / d. Choosing r / d = 0.03, and with d = w2 = 3.0,  2 0.03 0.03 3.0 0.09 in r   w At this small radius, our estimate for q is too high. From Fig. 6-20, with r = 0.09, q should be about 0.75. Iterating, we get Kt = 2.8. This is at a difficult range on Fig. A-15- 5 to read the graph with any confidence, but we’ll estimate r / d = 0.02, giving r = 0.06 in. This is a very rough estimate, but it clearly demonstrates that the fillet radius can be relatively sharp to match the fatigue factor of safety of the hole. Ans. ______________________________________________________________________________ 6-33 60 kpsi, 110 kpsiy utS S  Inner fiber where 3 / 4 incr  3 3 0.84375 4 16(2) 3 3 0.65625 4 32 o i r r       Table 3-4, p. 121, Chapter 6 - Rev. A, Page 28/66 • 3 /16 0.74608 in0.84375lnln 0.65625 n o i hr r r    0.75 0.74608 0.00392 in 0.74608 0.65625 0.08983 c n i n i e r r c r r           23 3 0.035156 in 16 16 A          Eq. (3-65), p. 119, (0.08983) 993.3 (0.035156)(0.00392)(0.65625) i i i Mc T T Aer      where T is in lbf·in and i is in psi. 1 ( 993.3) 496.7 2 496.7 m a T T T        Eq. (6-8):  ' 0.5 110 55 kpsieS   Eq. (6-19): 0.2652.70(110) 0.777ak   Eq. (6-25):    1/2e 0.808 3 /16 3 /16 0.1515 ind     Eq. (6-20):   0.1070.879 0.1515 1.08 (round to 1)bk    Eq. (6-19): (0.777)(1)(55) 42.7 kpsieS   For a compressive midrange component, / . Thus,a e fS n  42.70.4967 3 T  28.7 lbf inT   Outer fiber where 2 .5 incr  32.5 2.59375 32 32.5 2.40625 32 o i r r       3 /16 2.498832.59375ln 2.40625 nr   2.5 2.49883 0.00117 in 2.59375 2.49883 0.09492 ino e c       Chapter 6 - Rev. A, Page 29/66 • (0.09492) 889.7 psi (0.035156)(0.00117)(2.59375) 1 (889.7 ) 444.9 psi 2 o o o m a Mc T T Aer T T          (a) Using Eq. (6-46), for modified Goodman, we have 1 0.4449 0.4449 1 42.7 110 3 a m e utS S n T T       23.0 lbf in .T A  ns (b) Gerber, Eq. (6-47), at the outer fiber, 2 2 1 3(0.4449 ) 3(0.4449 ) 1 42.7 110 a m e ut n n S S T T               28.2 lbf in .T A  ns (c) To guard against yield, use T of part (b) and the inner stress. 60 2.14 . 0.9933(28.2) y y i S n A     ns ______________________________________________________________________________ 6-34 From Prob. 6-33, 42.7 kpsi, 60 kpsi, and 110 kpsie y utS S S   (a) Assuming the beam is straight,   max 3 2 3 / 2 6 6 910.2 /12 (3 /16) M hMc M T T I bh bh       Goodman: 0.4551 0.4551 1 42.7 110 3 T T   22.5 lbf in .T A  ns (b) Gerber: 23(0.4551 ) 3(0.4551 ) 1 42.7 110 T T      27.6 lbf in .T A  ns Chapter 6 - Rev. A, Page 30/66 • (c) max 60 2.39 . 0.9102(27.6) y y S n A     ns ______________________________________________________________________________ 6-35 ,bend ,axial ,tors1.4, 1.1, 2.0, 300 MPa, 400 MPa, 200 MPaf f f y ut eK K K S S S      Bending: 0, 60 MPam a   Axial: 20 MPa, 0m a   Torsion: 25 MPa, 25 MPam a   Eqs. (6-55) and (6-56):         2 2 2 2 1.4(60) 0 3 2.0(25) 120.6 MPa 0 1.1(20) 3 2.0(25) 89.35 MPa a m             Using Modified Goodman, Eq. (6-46), 1 120.6 89.35 200 400 a m f e utn S S        1.21 .fn A ns Check for yielding, using the conservative max a m      , 300 1.43 . 120.6 89.35 y y a m S n A         ns ______________________________________________________________________________ 6-36 ,bend ,tors1.4, 2.0, 300 MPa, 400 MPa, 200 MPaf f y ut eK K S S S     Bending: max min150 MPa, 40 MPa, 55 MPa, 95 MPam a        Torsion: 90 MPa, 9 MPam a   Eqs. (6-55) and (6-56):         2 2 2 2 1.4(95) 3 2.0(9) 136.6 MPa 1.4(55) 3 2.0(90) 321.1 MPa a m           Using Modified Goodman, 1 136.6 321.1 200 400 a m f e utn S S        0.67 .fn A ns Check for yielding, using the conservative max a m      , Chapter 6 - Rev. A, Page 31/66 • 300 0.66 . 136.6 321.1 y y a m S n A         ns Since the conservative yield check indicates yielding, we will check more carefully with with max  obtained directly from the maximum stresses, using the distortion energy failure theory, without stress concentrations. Note that this is exactly the method used for static failure in Ch. 5.        2 2 2 2max max max max 3 150 3 90 9 227.8 MPa 300 1.32 . 227.8 y y S n Ans                Since yielding is not predicted, and infinite life is not predicted, we would like to estimate a life from the S-N diagram. First, find an equivalent completely reversed stress (See Ex. 6-12). rev 136.6 692.5 MPa 1 ( / ) 1 (321.1/ 400) a m utS         This stress is much higher than the ultimate strength, rendering it impractical for the S-N diagram. We must conclude that the stresses from the combination loading, when increased by the stress concentration factors, produce such a high midrange stress that the equivalent completely reversed stress method is not practical to use. Without testing, we are unable to predict a life. ______________________________________________________________________________ 6-37 Table A-20: ut y64 kpsi, 54 kpsiS S  From Prob. 3-68, the critical stress element experiences  = 15.3 kpsi and  = 4.43 kpsi. The bending is completely reversed due to the rotation, and the torsion is steady, giving a = 15.3 kpsi, m = 0 kpsi, a = 0 kpsi, m = 4.43 kpsi. Obtain von Mises stresses for the alternating, mid-range, and maximum stresses.             1/21/2 22 2 2 1/21/2 22 2 2 1/21/2 22 2 2 max max max 3 15.3 3 0 15.3 kpsi 3 0 3 4.43 7.67 kpsi 3 15.3 3 4.43 17.11 kpsi a a a m m m                                  Check for yielding, using the distortion energy failure theory. max 54 3.16 17.11 y y S n      Obtain the modifying factors and endurance limit. Eq. (6-8):  0.5 64 32 kpsieS   Chapter 6 - Rev. A, Page 32/66 • Eq. (6-19): 0.2652.70(64) 0.90ak   Eq. (6-20): 0.1070.879(1.25) 0.86bk   Eq. (6-18): 0.90(0.86)(32) 24.8 kpsieS   Using Modified Goodman, 1 15.3 7.67 24.8 64 a m f e utn S S        1.36 .fn A ns ______________________________________________________________________________ 6-38 Table A-20: ut y440 MPa, 370 MPaS S  From Prob. 3-69, the critical stress element experiences  = 263 MPa and  = 57.7 MPa. The bending is completely reversed due to the rotation, and the torsion is steady, giving a = 263 MPa, m = 0, a = 0 MPa, m = 57.7 MPa. Obtain von Mises stresses for the alternating, mid-range, and maximum stresses.             1/21/2 22 2 2 1/21/2 22 2 2 1/21/2 22 2 2 max max max 3 263 3 0 263 MPa 3 0 3 57.7 99.9 MPa 3 263 3 57.7 281 MPa a a a m m m                                  Check for yielding, using the distortion energy failure theory. max 370 1.32 281 y y S n      Obtain the modifying factors and endurance limit. Eq. (6-8):  0.5 440 220 MPaeS   Eq. (6-19): 0.2654.51(440) 0.90ak   Eq. (6-20): 0.1071.24(30) 0.86bk   Eq. (6-18): 0.90(0.86)(220) 170 MPaeS   Using Modified Goodman, 1 263 99.9 170 440 a m f e utn S S        Infinite life is not predicted. Ans. 0.56fn  ______________________________________________________________________________ Chapter 6 - Rev. A, Page 33/66 • 6-39 Table A-20: ut y64 kpsi, 54 kpsiS S  From Prob. 3-70, the critical stress element experiences  = 21.5 kpsi and  = 5.09 kpsi. The bending is completely reversed due to the rotation, and the torsion is steady, giving a = 21.5 kpsi, m = 0 kpsi, a = 0 kpsi, m = 5.09 kpsi. Obtain von Mises stresses for the alternating, mid-range, and maximum stresses.             1/21/2 22 2 2 1/21/2 22 2 2 1/21/2 22 2 2 max max max 3 21.5 3 0 21.5 kpsi 3 0 3 5.09 8.82 kpsi 3 21.5 3 5.09 23.24 kpsi a a a m m m                                  Check for yielding, using the distortion energy failure theory. max 54 2.32 23.24 y y S n      Obtain the modifying factors and endurance limit. 0.2652.70(64) 0.90ak   0.1070.879(1) 0.88bk   0.90(0.88)(0.5)(64) 25.3 kpsieS   Using Modified Goodman, 1 21.5 8.82 25.3 64 a m f e utn S S        1.01 .fn A ns ______________________________________________________________________________ 6-40 Table A-20: ut y440 MPa, 370 MPaS S  From Prob. 3-71, the critical stress element experiences  = 72.9 MPa and  = 20.3 MPa. The bending is completely reversed due to the rotation, and the torsion is steady, giving a = 72.9 MPa, m = 0 MPa, a = 0 MPa, m = 20.3 MPa. Obtain von Mises stresses for the alternating, mid-range, and maximum stresses.             1/21/2 22 2 2 1/21/2 22 2 2 1/21/2 22 2 2 max max max 3 72.9 3 0 72.9 MPa 3 0 3 20.3 35.2 MPa 3 72.9 3 20.3 80.9 MPa a a a m m m                                  Check for yielding, using the distortion energy failure theory. Chapter 6 - Rev. A, Page 34/66 • max 370 4.57 80.9 y y S n      Obtain the modifying factors and endurance limit. Eq. (6-8):  0.5 440 220 MPaeS   Eq. (6-19): 0.2654.51(440) 0.90ak   Eq. (6-20): 0.1071.24(20) 0.90bk   Eq. (6-18): 0.90(0.90)(220) 178.2 MPaeS   Using Modified Goodman, 1 72.9 35.2 178.2 440 a m f e utn S S        2.04 .fn An s ______________________________________________________________________________ 6-41 Table A-20: ut y64 kpsi, 54 kpsiS S  From Prob. 3-72, the critical stress element experiences  = 35.2 kpsi and  = 7.35 kpsi. The bending is completely reversed due to the rotation, and the torsion is steady, giving a = 35.2 kpsi, m = 0 kpsi, a = 0 kpsi, m = 7.35 kpsi. Obtain von Mises stresses for the alternating, mid-range, and maximum stresses.             1/21/2 22 2 2 1/21/2 22 2 2 1/21/2 22 2 2 max max max 3 35.2 3 0 35.2 kpsi 3 0 3 7.35 12.7 kpsi 3 35.2 3 7.35 37.4 kpsi a a a m m m                                  Check for yielding, using the distortion energy failure theory. max 54 1.44 37.4 y y S n      Obtain the modifying factors and endurance limit. Eq. (6-8): 0.5(64) 32 kpsieS    Eq. (6-19): 0.2652.70(64) 0.90ak   Eq. (6-20): 0.1070.879(1.25) 0.86bk   Eq. (6-18): 0.90(0.86)(32) 24.8 kpsieS   Chapter 6 - Rev. A, Page 35/66 • Using Modified Goodman, 1 35.2 12.7 24.8 64 a m f e utn S S        Infinite life is not predicted. Ans. 0.62fn  ______________________________________________________________________________ 6-42 Table A-20: ut y440 MPa, 370 MPaS S  From Prob. 3-73, the critical stress element experiences  = 333.9 MPa and  = 126.3 MPa. The bending is completely reversed due to the rotation, and the torsion is steady, giving a = 333.9 MPa, m = 0 MPa, a = 0 MPa, m = 126.3 MPa. Obtain von Mises stresses for the alternating, mid-range, and maximum stresses.             1/21/2 22 2 2 1/21/2 22 2 2 1/21/2 22 2 2 max max max 3 333.9 3 0 333.9 MPa 3 0 3 126.3 218.8 MPa 3 333.9 3 126.3 399.2 MPa a a a m m m                                  Check for yielding, using the distortion energy failure theory. max 370 0.93 399.2 y y S n      The sample fails by yielding, infinite life is not predicted. Ans. The fatigue analysis will be continued only to obtain the requested fatigue factor of safety, though the yielding failure will dictate the life. Obtain the modifying factors and endurance limit. Eq. (6-8): 0.5(440) 220 MPaeS    Eq. (6-19): 0.2654.51(440) 0.90ak   Eq. (6-20): 0.1071.24(50) 0.82bk   Eq. (6-18): 0.90(0.82)(220) 162.4 MPaeS   Using Modified Goodman, 1 333.9 218.8 162.4 440 a m f e utn S S        Infinite life is not predicted. Ans. 0.39fn  ______________________________________________________________________________ Chapter 6 - Rev. A, Page 36/66 • 6-43 Table A-20: 64 kpsi, 54 kpsiut yS S  From Prob. 3-74, the critical stress element experiences completely reversed bending stress due to the rotation, and steady torsional and axial stresses. ,bend ,bend ,axial ,axial 9.495 kpsi, 0 kpsi 0 kpsi, 0.362 kpsi 0 kpsi, 11.07 kpsi a m a m a m              Obtain von Mises stresses for the alternating, mid-range, and maximum stresses.                   1/21/2 2 22 2 1/21/2 2 22 2 1/21/2 2 22 2 max max max 3 9.495 3 0 9.495 kpsi 3 0.362 3 11.07 19.18 kpsi 3 9.495 0.362 3 11.07 21.56 kpsi a a a m m m                                     Check for yielding, using the distortion energy failure theory. max 54 2.50 21.56 y y S n      Obtain the modifying factors and endurance limit. Eq. (6-8): 0.5(64) 32 kpsieS    Eq. (6-19): 0.2652.70(64) 0.90ak   Eq. (6-20): 0.1070.879(1.13) 0.87bk   Eq. (6-18): 0.90(0.87)(32) 25.1 kpsieS   Using Modified Goodman, 1 9.495 19.18 25.1 64 a m f e utn S S        1.47 .fn A ns ______________________________________________________________________________ 6-44 Table A-20: ut y64 kpsi, 54 kpsiS S  From Prob. 3-76, the critical stress element experiences completely reversed bending stress due to the rotation, and steady torsional and axial stresses. ,bend ,bend ,axial ,axial 33.99 kpsi, 0 kpsi 0 kpsi, 0.153 kpsi 0 kpsi, 7.847 kpsi a m a m a m              Chapter 6 - Rev. A, Page 37/66 • Obtain von Mises stresses for the alternating, mid-range, and maximum stresses.                   1/21/2 2 22 2 1/21/2 2 22 2 1/21/2 2 22 2 max max max 3 33.99 3 0 33.99 kpsi 3 0.153 3 7.847 13.59 kpsi 3 33.99 0.153 3 7.847 36.75 kpsi a a a m m m                                     Check for yielding, using the distortion energy failure theory. max 54 1.47 36.75 y y S n      Obtain the modifying factors and endurance limit. Eq. (6-8): 0.5(64) 32 kpsieS    Eq. (6-19): 0.2652.70(64) 0.90ak   Eq. (6-20): 0.1070.879(0.88) 0.89bk   Eq. (6-18): 0.90(0.89)(32) 25.6 kpsieS   Using Modified Goodman, 1 33.99 13.59 25.6 64 a m f e utn S S        Infinite life is not predicted. Ans. 0.65fn  ______________________________________________________________________________ 6-45 Table A-20: ut y440 MPa, 370 MPaS S  From Prob. 3-77, the critical stress element experiences  = 68.6 MPa and  = 37.7 MPa. The bending is completely reversed due to the rotation, and the torsion is steady, giving a = 68.6 MPa, m = 0 MPa, a = 0 MPa, m = 37.7 MPa. Obtain von Mises stresses for the alternating, mid-range, and maximum stresses.             1/21/2 22 2 2 1/21/2 22 2 2 1/21/2 22 2 2 max max max 3 68.6 3 0 68.6 MPa 3 0 3 37.7 65.3 MPa 3 68.6 3 37.7 94.7 MPa a a a m m m                                  Check for yielding, using the distortion energy failure theory. max 370 3.91 94.7 y y S n      Chapter 6 - Rev. A, Page 38/66 • Obtain the modifying factors and endurance limit. Eq. (6-8): 0.5(440) 220 MPaeS    Eq. (6-19): 0.2654.51(440) 0.90ak   Eq. (6-20): 0.1071.24(30) 0.86bk   Eq. (6-18): 0.90(0.86)(220) 170 MPaeS   Using Modified Goodman, 1 68.6 65.3 170 440 a m f e utn S S        1.81 .fn An s ______________________________________________________________________________ 6-46 Table A-20: 64 kpsi, 54 kpsiut yS S  From Prob. 3-79, the critical stress element experiences  = 3.46 kpsi and  = 0.882 kpsi. The bending is completely reversed due to the rotation, and the torsion is steady, giving a = 3.46 kpsi, m = 0, a = 0 kpsi, m = 0.882 kpsi. Obtain von Mises stresses for the alternating, mid-range, and maximum stresses.             1/21/2 22 2 2 1/21/2 22 2 2 1/21/2 22 2 2 max max max 3 3.46 3 0 3.46 kpsi 3 0 3 0.882 1.53 kpsi 3 3.46 3 0.882 3.78 kpsi a a a m m m                                  Check for yielding, using the distortion energy failure theory. max 54 14.3 3.78 y y S n      Obtain the modifying factors and endurance limit. Eq. (6-8): 0.5(64) 32 kpsieS    Eq. (6-19): 0.2652.70(64) 0.90ak   Eq. (6-20): 0.1070.879(1.375) 0.85bk   Eq. (6-18): 0.90(0.85)(32) 24.5 kpsieS   Using Modified Goodman, Chapter 6 - Rev. A, Page 39/66 • 1 3.46 1.53 24.5 64 a m f e utn S S        Ans. 6.06fn  ______________________________________________________________________________ 6-47 Table A-20: 64 kpsi, 54 kpsiut yS S  From Prob. 3-80, the critical stress element experiences  = 16.3 kpsi and  = 5.09 kpsi. Since the load is applied and released repeatedly, this gives max = 16.3 kpsi, min = 0 kpsi, max = 5.09 kpsi, min = 0 kpsi. Consequently,m = a = 8.15 kpsi, m = a = 2.55 kpsi. For bending, from Eqs. (6-34) and (6-35a),          2 33 5 80.246 3.08 10 64 1.51 10 64 2.67 10 64 0.10373a        1 1 0.750.1037311 0.1 q a r     Eq. (6-32): 1 ( 1) 1 0.75(1.5 1) 1.38f tK q K       For torsion, from Eqs. (6-34) and (6-35b),          2 33 5 80.190 2.51 10 64 1.35 10 64 2.67 10 64 0.07800a        1 1 0.800.0780011 0.1 q a r     Eq. (6-32): 1 ( 1) 1 0.80(2.1 1) 1.88fs s tsK q K       Obtain von Mises stresses for the alternating and mid-range stresses from Eqs. (6-55) and (6-56).       1/22 21.38 8.15 3 1.88 2.55 13.98 kpsi 13.98 kpsi a m a                 Check for yielding, using the conservative max a m      , 54 1.93 13.98 13.98 y y a m S n         Obtain the modifying factors and endurance limit. Eq. (6-8): 0.5(64) 32 kpsieS    Eq. (6-19): 0.2652.70(64) 0.90ba utk aS    Chapter 6 - Rev. A, Page 40/66 • Eq. (6-24):  0.370 0.370 1 0.370 ined d   Eq. (6-20): 0.107 0.1070.879 0.879(0.370) 0.98b ek d     Eq. (6-18): (0.90)(0.98)(32) 28.2 kpsieS   Using Modified Goodman, 1 13.98 13.98 28.2 64 a m f e utn S S        1.40 .fn A ns ______________________________________________________________________________ 6-48 Table A-20: 64 kpsi, 54 kpsiut yS S  From Prob. 3-81, the critical stress element experiences  = 16.4 kpsi and  = 4.46 kpsi. Since the load is applied and released repeatedly, this gives max = 16.4 kpsi, min = 0 kpsi, max = 4.46 kpsi, min = 0 kpsi. Consequently,m = a = 8.20 kpsi, m = a = 2.23 kpsi. For bending, from Eqs. (6-34) and (6-35a),          2 33 5 80.246 3.08 10 64 1.51 10 64 2.67 10 64 0.10373a        1 1 0.750.1037311 0.1 q a r     Eq. (6-32): 1 ( 1) 1 0.75(1.5 1) 1.38f tK q K       For torsion, from Eqs. (6-34) and (6-35b),          2 33 5 80.190 2.51 10 64 1.35 10 64 2.67 10 64 0.07800a        1 1 0.800.0780011 0.1 q a r     Eq. (6-32): 1 ( 1) 1 0.80(2.1 1) 1.88fs s tsK q K       Obtain von Mises stresses for the alternating and mid-range stresses from Eqs. (6-55) and (6-56).       1/22 21.38 8.20 3 1.88 2.23 13.45 kpsi 13.45 kpsi a m a                 Check for yielding, using the conservative max a m      , Chapter 6 - Rev. A, Page 41/66 • 54 2.01 13.45 13.45 y y a m S n         Obtain the modifying factors and endurance limit. Eq. (6-8): 0.5(64) 32 kpsieS    Eq. (6-19): 0.2652.70(64) 0.90ba utk aS    Eq. (6-24): 0.370 0.370(1) 0.370 ined d   Eq. (6-20): 0.107 0.1070.879 0.879(0.370) 0.98b ek d     Eq. (6-18): (0.90)(0.98)(32) 28.2 kpsieS   Using Modified Goodman, 1 13.45 13.45 28.2 64 a m f e utn S S        1.46 .fn A ns ______________________________________________________________________________ 6-49 Table A-20: 64 kpsi, 54 kpsiut yS S  From Prob. 3-82, the critical stress element experiences repeatedly applied bending, axial, and torsional stresses of x,bend = 20.2 kpsi, x,axial = 0.1 kpsi, and  = 5.09 kpsi.. Since the axial stress is practically negligible compared to the bending stress, we will simply combine the two and not treat the axial stress separately for stress concentration factor and load factor. This gives max = 20.3 kpsi, min = 0 kpsi, max = 5.09 kpsi, min = 0 kpsi. Consequently,m = a = 10.15 kpsi, m = a = 2.55 kpsi. For bending, from Eqs. (6-34) and (6-35a),          2 33 5 80.246 3.08 10 64 1.51 10 64 2.67 10 64 0.10373a        1 1 0.750.1037311 0.1 q a r     Eq. (6-32): 1 ( 1) 1 0.75(1.5 1) 1.38f tK q K       For torsion, from Eqs. (6-34) and (6-35b),          2 33 5 80.190 2.51 10 64 1.35 10 64 2.67 10 64 0.07800a        1 1 0.800.0780011 0.1 q a r     Chapter 6 - Rev. A, Page 42/66 • Eq. (6-32): 1 ( 1) 1 0.80(2.1 1) 1.88fs s tsK q K       Obtain von Mises stresses for the alternating and mid-range stresses from Eqs. (6-55) and (6-56).       1/22 21.38 10.15 3 1.88 2.55 16.28 kpsi 16.28 kpsi a m a                 Check for yielding, using the conservative max a m      , 54 1.66 16.28 16.28 y y a m S n         Obtain the modifying factors and endurance limit. Eq. (6-8): 0.5(64) 32 kpsieS    Eq. (6-19): 0.2652.70(64) 0.90ba utk aS    Eq. (6-24): 0.370 0.370(1) 0.370 ined d   Eq. (6-20): 0.107 0.1070.879 0.879(0.370) 0.98b ek d     Eq. (6-18): (0.90)(0.98)(32) 28.2 kpsieS   Using Modified Goodman, 1 16.28 16.28 28.2 64 a m f e utn S S        1.20 .fn A ns ____________________________________________________________________________ 6-50 Table A-20: 64 kpsi, 54 kpsiut yS S  From Prob. 3-83, the critical stress element on the neutral axis in the middle of the longest side of the rectangular cross section experiences a repeatedly applied shear stress of max = 14.3 kpsi, min = 0 kpsi. Thus, m = a = 7.15 kpsi. Since the stress is entirely shear, it is convenient to check for yielding using the standard Maximum Shear Stress theory. max / 2 54 / 2 1.89 14.3 y y S n     Find the modifiers and endurance limit. Eq. (6-8): 0.5(64) 32 kpsieS    Eq. (6-19): 0.2652.70(64) 0.90ba utk aS    Chapter 6 - Rev. A, Page 43/66 • The size factor for a torsionally loaded rectangular cross section is not readily available. Following the procedure on p. 289, we need an equivalent diameter based on the 95 percent stress area. However, the stress situation in this case is nonlinear, as described on p. 102. Noting that the maximum stress occurs at the middle of the longest side, or with a radius from the center of the cross section equal to half of the shortest side, we will simply choose an equivalent diameter equal to the length of the shortest side. 0.25 ined  Eq. (6-20): 0.107 0.1070.879 0.879(0.25) 1.02b ek d     We will round down to kb = 1. Eq. (6-26): 0.59ck  Eq. (6-18): 0.9(1)(0.59)(32) 17.0 kpsiseS   Since the stress is entirely shear, we choose to use a load factor kc = 0.59, and convert the ultimate strength to a shear value rather than using the combination loading method of Sec. 6-14. From Eq. (6-54), Ssu = 0.67Su = 0.67 (64) = 42.9 kpsi. Using Modified Goodman, 1 1 1.70 . ( / ) ( / ) (7.15 /17.0) (7.15 / 42.9)f a se m su n Ans S S       ______________________________________________________________________________ 6-51 Table A-20: 64 kpsi, 54 kpsiut yS S  From Prob. 3-84, the critical stress element experiences  = 28.0 kpsi and  = 15.3 kpsi. Since the load is applied and released repeatedly, this gives max = 28.0 kpsi, min = 0 kpsi, max = 15.3 kpsi, min = 0 kpsi. Consequently,m = a = 14.0 kpsi, m = a = 7.65 kpsi. From Table A-15-8 and A-15-9, ,bend ,tors / 1.5 /1 1.5, / 0.125 /1 0.125 1.60, 1.39t t D d r d K K       Eqs. (6-34) and (6-35), or Figs. 6-20 and 6-21: qbend = 0.78, qtors = 0.82 Eq. (6-32):         ,bend bend ,bend ,tors tors ,tors 1 1 1 0.78 1.60 1 1.47 1 1 1 0.82 1.39 1 1.32 f t f t K q K K q K               Obtain von Mises stresses for the alternating and mid-range stresses from Eqs. (6-55) and (6-56). Chapter 6 - Rev. A, Page 44/66 •       1/22 21.47 14.0 3 1.32 7.65 27.0 kpsi 27.0 kpsi a m a                 Check for yielding, using the conservative max a m      , 54 1.00 27.0 27.0 y y a m S n         Since stress concentrations are included in this quick yield check, the low factor of safety is acceptable. Eq. (6-8): 0.5(64) 32 kpsieS    Eq. (6-19): 0.2652.70(64) 0.897ba utk aS    Eq. (6-24):  0.370 0.370 1 0.370 ined d   Eq. (6-20): 0.107 0.1070.879 0.879(0.370) 0.978b ek d     Eq. (6-18): (0.897)(0.978)(0.5)(64) 28.1 kpsieS   Using Modified Goodman, 1 27.0 27.0 28.1 64 a m f e utn S S        0.72 .fn A ns Since infinite life is not predicted, estimate a life from the S-N diagram. First, find an equivalent completely reversed stress (See Ex. 6-12). rev 27.0 46.7 kpsi 1 ( / ) 1 (27.0 / 64) a m utS         Fig. 6-18: f = 0.9 Eq. (6-14):   22 0.9(64)( ) 118.07 28.1 ut e f Sa S    Eq. (6-15): 1 1 0.9(64)log log 0.1039 3 3 28.1 ut e f Sb S                Eq. (6-16): 11/ 0.1039 rev 46.7 7534 cycles 7500 cycles . 118.07 b N A a              ns ______________________________________________________________________________ 6-52 Table A-20: 64 kpsi, 54 kpsiut yS S  Chapter 6 - Rev. A, Page 45/66 • From Prob. 3-85, the critical stress element experiences x,bend = 46.1 kpsi, x,axial = 0.382 kpsi and  = 15.3 kpsi. The axial load is practically negligible, but we’ll include it to demonstrate the process. Since the load is applied and released repeatedly, this gives max,bend = 46.1 kpsi, min,bend = 0 kpsi, max,axial = 0.382 kpsi, min,axial = 0 kpsi, max = 15.3 kpsi, min = 0 kpsi. Consequently,m,bend = a,bend = 23.05 kpsi, m,axial = a,axial = 0.191 kpsi, m = a = 7.65 kpsi. From Table A-15-7, A-15-8 and A-15-9, ,bend ,tors ,axial / 1.5 /1 1.5, / 0.125 /1 0.125 1.60, 1.39, 1.75t t t D d r d K K K        Eqs. (6-34) and (6-35), or Figs. 6-20 and 6-21: qbend = qaxial =0.78, qtors = 0.82 Eq. (6-32):             ,bend bend ,bend ,axial axial ,axial ,tors tors ,tors 1 1 1 0.78 1.60 1 1.47 1 1 1 0.78 1.75 1 1.59 1 1 1 0.82 1.39 1 1.32 f t f t f t K q K K q K K q K                      Obtain von Mises stresses for the alternating and mid-range stresses from Eqs. (6-55) and (6-56).           1/22 20.191 1.47 23.05 1.59 3 1.32 7.65 38.45 kpsi 0.85a                            1/22 21.47 23.05 1.59 0.191 3 1.32 7.65 38.40 kpsim            Check for yielding, using the conservative max a m      , 54 0.70 38.45 38.40 y y a m S n         Since the conservative yield check indicates yielding, we will check more carefully with with max  obtained directly from the maximum stresses, using the distortion energy failure theory, without stress concentrations. Note that this is exactly the method used for static failure in Ch. 5.        2 2 2 2max max,bend max,axial max max 3 46.1 0.382 3 15.3 53.5 kpsi 54 1.01 . 53.5 y y S n Ans                  This shows that yielding is imminent, and further analysis of fatigue life should not be interpreted as a guarantee of more than one cycle of life. Chapter 6 - Rev. A, Page 46/66 • Eq. (6-8): 0.5(64) 32 kpsieS    Eq. (6-19): 0.2652.70(64) 0.897ba utk aS    Eq. (6-24):  0.370 0.370 1 0.370 ined d   Eq. (6-20): 0.107 0.1070.879 0.879(0.370) 0.978b ek d     Eq. (6-18): (0.897)(0.978)(0.5)(64) 28.1 kpsieS   Using Modified Goodman, 1 38.45 38.40 28.1 64 a m f e utn S S        0.51 .fn A ns Since infinite life is not predicted, estimate a life from the S-N diagram. First, find an equivalent completely reversed stress (See Ex. 6-12). rev 38.45 96.1 kpsi 1 ( / ) 1 (38.40 / 64) a m utS         This stress is much higher than the ultimate strength, rendering it impractical for the S-N diagram. We must conclude that the fluctuating stresses from the combination loading, when increased by the stress concentration factors, are so far from the Goodman line that the equivalent completely reversed stress method is not practical to use. Without testing, we are unable to predict a life. ______________________________________________________________________________ 6-53 Table A-20: 64 kpsi, 54 kpsiut yS S  From Prob. 3-86, the critical stress element experiences x,bend = 55.5 kpsi, x,axial = 0.382 kpsi and  = 15.3 kpsi. The axial load is practically negligible, but we’ll include it to demonstrate the process. Since the load is applied and released repeatedly, this gives max,bend = 55.5 kpsi, min,bend = 0 kpsi, max,axial = 0.382 kpsi, min,axial = 0 kpsi, max = 15.3 kpsi, min = 0 kpsi. Consequently,m,bend = a,bend = 27.75 kpsi, m,axial = a,axial = 0.191 kpsi, m = a = 7.65 kpsi. From Table A-15-7, A-15-8 and A-15-9, ,bend ,tors ,axial / 1.5 /1 1.5, / 0.125 /1 0.125 1.60, 1.39, 1.75t t t D d r d K K K        Eqs. (6-34) and (6-35), or Figs. 6-20 and 6-21: qbend = qaxial =0.78, qtors = 0.82 Eq. (6-32):             ,bend bend ,bend ,axial axial ,axial ,tors tors ,tors 1 1 1 0.78 1.60 1 1.47 1 1 1 0.78 1.75 1 1.59 1 1 1 0.82 1.39 1 1.32 f t f t f t K q K K q K K q K                      Chapter 6 - Rev. A, Page 47/66 • Obtain von Mises stresses for the alternating and mid-range stresses from Eqs. (6-55) and (6-56).                     1/22 2 1/22 2 0.191 1.47 27.75 1.59 3 1.32 7.65 44.71 kpsi 0.85 1.47 27.75 1.59 0.191 3 1.32 7.65 44.66 kpsi a m                               Since these stresses are relatively high compared to the yield strength, we will go ahead and check for yielding using the distortion energy failure theory.        2 2 2 2max max,bend max,axial max max 3 55.5 0.382 3 15.3 61.8 kpsi 54 0.87 . 61.8 y y S n Ans                  This shows that yielding is predicted. Further analysis of fatigue life is just to be able to report the fatigue factor of safety, though the life will be dictated by the static yielding failure, i.e. N = 1/2 cycle. Ans. Eq. (6-8):  0.5 64 32 kpsieS   Eq. (6-19): 0.2652.70(64) 0.897ba utk aS    Eq. (6-24):  0.370 0.370 1 0.370 ined d   Eq. (6-20): 0.107 0.1070.879 0.879(0.370) 0.978b ek d     Eq. (6-18): (0.897)(0.978)(0.5)(64) 28.1 kpsieS   Using Modified Goodman, 1 44.71 44.66 28.1 64 a m f e utn S S        0.44 .fn A ns ______________________________________________________________________________ 6-54 From Table A-20, for AISI 1040 CD, Sut = 85 kpsi and Sy = 71 kpsi. From the solution to Prob. 6-17 we find the completely reversed stress at the critical shoulder fillet to be rev = 35.0 kpsi, producing a = 35.0 kpsi and m = 0 kpsi. This problem adds a steady torque which creates torsional stresses of    4 2500 1.625 / 2 2967 psi 2.97 kpsi, 0 kpsi 1.625 / 32m a Tr J         From Table A-15-8 and A-15-9, r/d = 0.0625/1.625 = 0.04, D/d = 1.875/1.625 = 1.15, Kt,bend =1.95, Kt,tors =1.60 Chapter 6 - Rev. A, Page 48/66 • Eqs. (6-34) and (6-35), or Figs. 6-20 and 6-21: qbend = 0.76, qtors = 0.81 Eq. (6-32):         ,bend bend ,bend ,tors tors ,tors 1 1 1 0.76 1.95 1 1.72 1 1 1 0.81 1.60 1 1.49 f t f t K q K K q K               Obtain von Mises stresses for the alternating and mid-range stresses from Eqs. (6-55) and (6-56).               1/22 2 1/22 2 1.72 35.0 3 1.49 0 60.2 kpsi 1.72 0 3 1.49 2.97 7.66 kpsi a m                       Check for yielding, using the conservative max a m      , 71 1.05 60.2 7.66 y y a m S n         From the solution to Prob. 6-17, Se = 29.5 kpsi. Using Modified Goodman, 1 60.2 7.66 29.5 85 a m f e utn S S        0.47 .fn A ns Since infinite life is not predicted, estimate a life from the S-N diagram. First, find an equivalent completely reversed stress (See Ex. 6-12). rev 60.2 66.2 kpsi 1 ( / ) 1 (7.66 / 85) a m utS         Fig. 6-18: f = 0.867    2 20.867(85)Eq. (6-14): 184.1 29.5 1 1 0.867(85)Eq. (6-15): log log 0.1325 3 3 29.5 ut e ut e f S a S f Sb S                   11/ 0.1325 rev 66.2Eq. (6-16): 2251 cycles 184.1 b N a             N = 2300 cycles Ans. ______________________________________________________________________________ Chapter 6 - Rev. A, Page 49/66 • 6-55 From the solution to Prob. 6-18 we find the completely reversed stress at the critical shoulder fillet to be rev = 32.8 kpsi, producing a = 32.8 kpsi and m = 0 kpsi. This problem adds a steady torque which creates torsional stresses of    4 2200 1.625 / 2 2611 psi 2.61 kpsi, 0 kpsi 1.625 / 32m a Tr J         From Table A-15-8 and A-15-9, r/d = 0.0625/1.625 = 0.04, D/d = 1.875/1.625 = 1.15, Kt,bend =1.95, Kt,tors =1.60 Eqs. (6-34) and (6-35), or Figs. 6-20 and 6-21: qbend = 0.76, qtors = 0.81 Eq. (6-32):         ,bend bend ,bend ,tors tors ,tors 1 1 1 0.76 1.95 1 1.72 1 1 1 0.81 1.60 1 1.49 f t f t K q K K q K               Obtain von Mises stresses for the alternating and mid-range stresses from Eqs. (6-55) and (6-56).       1/22 21.72 32.8 3 1.49 0 56.4 kpsia                 1/22 21.72 0 3 1.49 2.61 6.74 kpsim           Check for yielding, using the conservative max a m      , 71 1.12 56.4 6.74 y y a m S n         From the solution to Prob. 6-18, Se = 29.5 kpsi. Using Modified Goodman, 1 56.4 6.74 29.5 85 a m f e utn S S        0.50 .fn A ns Since infinite life is not predicted, estimate a life from the S-N diagram. First, find an equivalent completely reversed stress (See Ex. 6-12). rev 56.4 61.3 kpsi 1 ( / ) 1 (6.74 / 85) a m utS         Fig. 6-18: f = 0.867 Chapter 6 - Rev. A, Page 50/66 •    2 20.867(85)Eq. (6-14): 184.1 29.5 1 1 0.867(85)Eq. (6-15): log log 0.1325 3 3 29.5 ut e ut e f S a S f Sb S                   11/ 0.1325 rev 61.3Eq. (6-16): 4022 cycles 184.1 b N a             N = 4000 cycles Ans. ______________________________________________________________________________ 6-56 min max55 kpsi, 30 kpsi, 1.6, 2 ft, 150 lbf , 500 lbfut y tsS S K L F F      Eqs. (6-34) and (6-35b), or Fig. 6-21: qs = 0.80 Eq. (6-32):    1 1 1 0.80 1.6 1 1.48fs s tsK q K       max min500(2) 1000 lbf in, 150(2) 300 lbf inT T      max max 3 3 16 16(1.48)(1000) 11 251 psi 11.25 kpsi (0.875) fsK T d        min min 3 3 16 16(1.48)(300) 3375 psi 3.38 kpsi (0.875) fsK T d        max min max min 11.25 3.38 7.32 kpsi 2 2 11.25 3.38 3.94 kpsi 2 2 m a               Since the stress is entirely shear, it is convenient to check for yielding using the standard Maximum Shear Stress theory. max / 2 30 / 2 1.33 11.25 y y S n     Find the modifiers and endurance limit. Eq. (6-8): 0.5(55) 27.5 kpsieS    Eq. (6-19): 0.71814.4(55) 0.81ak   Eq. (6-24): 0.370(0.875) 0.324 ined   Eq. (6-20): 0.1070.879(0.324) 0.99bk   Eq. (6-26): 0.59ck  Eq. (6-18): 0.81(0.99)(0.59)(27.5) 13.0 kpsiseS   Chapter 6 - Rev. A, Page 51/66 • Since the stress is entirely shear, we will use a load factor kc = 0.59, and convert the f (a) Modified Goodman, Table 6-6 ultimate strength to a shear value rather than using the combination loading method o Sec. 6-14. From Eq. (6-54), Ssu = 0.67Su = 0.67 (55) = 36.9 kpsi. 1 1 1.99 . ( / ) ( / ) (3.94 /13.0) (7.32 / 36.9)f a se m su n Ans S S       (b) Gerber, Table 6-7 2 2 21 1 1 2 su a m se f m se su a S Sn S S                      221 36.9 3.94 2(7.32)(13.0)1 1 2 7.32 13.0 36.9(3.94)                         ns __ ________________ ____________________________________________ -57 From Eqs. (6-34) and (6-35a), or Fig. 6-20, with a notch radius of 0.1 in, q = 0.9. Thus, 2.49 .fn A ____ __________ __ 6 145 kpsi, 120 kpsiut yS S  with Kt = 3 from the problem statement, 1 ( 1) 1 0.9(3 1) 2.80f tK q K       max 2 2 4 2.80(4)( ) 2.476 (1.2)f P PK P d          1 ( 2.476 ) 1.238 2m a P P           max 0.3 6 1.2 0.54 4 4 f P D d P T P      From Eqs. ( 6-34) and (6-35b), or Fig. 6-21, with a notch radius of 0.1 in, Thus, 0.92.sq  with Kts = 1.8 from the problem statement, 1 ( 1) 1 0.92fs s tsK q K     (1.8 1) 1.74  max 3 3 16 16(1.74)(0.54 ) 2.769 (1.2) fsK T P P d       max 2.769 1.385 2 2a m P P     Eqs. (6-55) and (6-56): Chapter 6 - Rev. A, Page 52/66 • 2 2 1/2 2 2 1/2 2 2 1/2 2 2 1/2 [( / 0.85) 3 ] [(1.238 / 0.85) 3(1.385 ) ] 2.81 [ 3 ] [( 1.238 ) 3(1.385 ) ] 2.70 a a a m m m P P P P P                    P Eq. (6-8): 0.5(145) 72.5 kpsieS    Eq. (6-19): 0.2652.70(145) 0.722ak   Eq. (6-20): 0.1070.879(1.2) 0.862bk   Eq. (6-18): (0.722)(0.862)(72.5) 45.12 kpsieS   Modified Goodman: 1 2.81 2.70 1 45.12 145 3 a m f e ut P P n S S         4.12 kips .P A ns Yield (conservative): 120 5.29 . (2.81)(4.12) (2.70)(4.12) y y a m S n A         ns ______________________________________________________________________________ 6-58 From Prob. 6-57, 2.80, 1.74, 45.12 kpsif f s eK K S   max max 2 2 4 4(18)2.80 44.56 kpsi (1.2 )f PK d          min min 2 2 4 4(4.5)2.80 11.14 kpsi (1.2)f PK d          max max 6 1.20.3(18) 9.72 kip in 4 4 D dT f P                min min 6 1.20.3(4.5) 2.43 kip in 4 4 D dT f P                max max 3 3 16 16(9.72)1.74 49.85 kpsi (1.2)f s TK d       min min 3 3 16 16(2.43)1.74 12.46 kpsi (1.2)f s TK d       44.56 ( 11.14) 16.71 kpsi 2a      44.56 ( 11.14) 27.85 kpsi 2m       49.85 12.46 18.70 kpsi 2a    49.85 12.46 31.16 kpsi 2m    Chapter 6 - Rev. A, Page 53/66 • Eqs. (6-55) and (6-56): 2 2 1/2 2 2 1/2 2 2 1/2 2 2 1/2 [( / 0.85) 3 ] [(16.71/ 0.85) 3(18.70) ] 37.89 kpsi [ 3 ] [( 27.85) 3(31.16) ] 60.73 kpsi a a a m m m                    Modified Goodman: 1 37.89 60.73 45.12 145 a m f e utn S S        nf = 0.79 Since infinite life is not predicted, estimate a life from the S-N diagram. First, find an equivalent completely reversed stress (See Ex. 6-12). rev 37.89 65.2 kpsi 1 ( / ) 1 (60.73 /145) a m utS         Fig. 6-18: f = 0.8     2 20.8(145) Eq. (6-14): 298.2 45.12 ut e f S a S    1 1 0.8(145)Eq. (6-15): log log 0.1367 3 3 45.12 ut e f Sb S                11/ 0.1367 rev 65.2Eq. (6-16): 67 607 cycles 298.2 b N a             N = 67 600 cycles Ans. ______________________________________________________________________________ 6-59 For AISI 1020 CD, From Table A-20, Sy = 390 MPa, Sut = 470 MPa. Given: Se = 175 MPa. First Loading:    1 1 360 160 360 160260 MPa, 100 MPa 2 2m a       Goodman:      1 1 1 100 223.8 MPa finite life 1 / 1 260 / 470 a a ee m ut S S           Chapter 6 - Rev. A, Page 54/66 •     2 1/0.127767 0.9 470 1022.5 MPa 175 0.9 4701 log 0.127 767 3 175 223.8 145 920 cycles 1022.5 a b N                Second loading:        2 2 320 200 320 200 60 MPa, 260 MPa 2 2m a             2 260 298.0 MPa 1 60 / 470a e     (a) Miner’s method: 1/0.127767 2 298.0 15 520 cycles 1022.5 N        1 2 2 2 1 2 80 0001 1 7000 cycles . 145 920 15 520 n n n n A N N        ns (b) Manson’s method: The number of cycles remaining after the first loading Nremaining =145 920  80 000 = 65 920 cycles Two data points: 0.9(470) MPa, 103 cycles 223.8 MPa, 65 920 cycles           2 2 2 3 2 2 2 2 0.151 997 1/ 0.151 997 2 100.9 470 223.8 65 920 1.8901 0.015170 log1.8901 0.151 997 log 0.015170 223.8 1208.7 MPa 65 920 298.0 10 000 cycles . 1208.7 b b b a a b a n A                ns ______________________________________________________________________________ 6-60 Given: Se = 50 kpsi, Sut = 140 kpsi, f =0.8. Using Miner’s method, Chapter 6 - Rev. A, Page 55/66 •     2 0.8 140 250.88 kpsi 50 0.8 1401 log 0.116 749 3 50 a b         1/ 0.116 749 1 1 1/ 0.116 749 2 2 1/ 0.116 749 3 3 9595 kpsi, 4100 cycles 250.88 8080 kpsi, 17 850 cycles 250.88 6565 kpsi, 105 700 cycles 250.88 N N N                            0.2 0.5 0.3 1 12 600 cycles . 4100 17 850 105 700 N N N N Ans     ______________________________________________________________________________ 6-61 Given: Sut = 530 MPa, Se = 210 MPa, and f = 0.9. (a) Miner’s method     2 0.9 530 1083.47 MPa 210 0.9 5301 log 0.118 766 3 210 a b         1/ 0.118 766 1 1 350350 MPa, 13 550 cycles 1083.47 N         1/ 0.118 766 2 2 1/ 0.118 766 3 3 260260 MPa, 165 600 cycles 1083.47 225225 MPa, 559 400 cycles 1083.47 N N                   31 2 1 2 3 1nn n N N N    35000 50 000 184 100 cycles . 13 550 165 600 559 400 n Ans   (b) Manson’s method: The life remaining after the first series of cycling is NR1 = 13 550  5000 = 8550 cycles. The two data points required to define ,1eS  are [0.9(530), 10 3] and (350, 8550). Chapter 6 - Rev. A, Page 56/66 •         2 2 2 3 2 2 100.9 530 1.3629 0.11696 350 8550 b b b a a          2 2 0.144 280 1/0.144 280 2 2 log 1.362 9 0.144 280 log 0.116 96 350 1292.3 MPa 8550 260 67 090 cycles 1292.3 67 090 50 000 17 090 cyclesR b a N N                         3 2 3 3 3 3 100.9 530 1.834 6 0.058 514 260 17 090 b b b a a         3 3 0.213 785 log 1.834 6 2600.213 785, 2088.7 MPa log 0.058 514 17 090 b a      1/0.213 785 3 225 33 610 cycles . 2088.7 N Ans        ______________________________________________________________________________ 6-62 Given: Se = 45 kpsi, Sut = 85 kpsi, f = 0.86, and a = 35 kpsi and m = 30 kpsi for 12 (103) cycles. Gerber equivalent reversing stress:    rev 2 2 35 39.98 kpsi 1 / 1 30 / 85 a m utS        (a) Miner’s method: rev < Se. According to the method, this means that the endurance limit has not been reduced and the new endurance limit is eS  = 45 kpsi. Ans. (b) Manson’s method: Again, rev < Se. According to the method, this means that the material has not been damaged and the endurance limit has not been reduced. Thus, the new endurance limit is eS  = 45 kpsi. Ans. ______________________________________________________________________________ 6-63 Given: Se = 45 kpsi, Sut = 85 kpsi, f = 0.86, and a = 35 kpsi and m = 30 kpsi for 12 (103) cycles. Goodman equivalent reversing stress:    rev 35 54.09 kpsi 1 / 1 30 / 85 a m utS        Initial cycling Chapter 6 - Rev. A, Page 57/66 •     2 0.86 85 116.00 kpsi 45 0.86 851 log 0.070 235 3 45 a b         1/ 0.070 235 1 1 54.0954.09 kpsi, 52 190 cycles 116.00 N         (a) Miner’s method (see discussion on p. 325): The number of remaining cycles at 54.09 kpsi is Nremaining = 52 190  12 000 = 40 190 cycles. The new coefficients are b = b, and a =Sf /Nb = 54.09/(40 190)  0.070 235 = 113.89 kpsi. The new endurance limit is   0.070 2356,1 113.89 10 43.2 kpsi .be eS a N An     s (b) Manson’s method (see discussion on p. 326): The number of remaining cycles at 54.09 kpsi is Nremaining = 52 190  12 000 = 40 190 cycles. At 103 cycles, Sf = 0.86(85) = 73.1 kpsi. The new coefficients are b = [log(73.1/54.09)]/log(103/40 190) =  0.081 540 and a = 1/ (Nremaining) b = 54.09/(40 190)  0.081 540 = 128.39 kpsi. The new endurance limit is   0.081 5406,1 128.39 10 41.6 kpsi .be eS a N An     s ______________________________________________________________________________ 6-64 Given Sut =1030LN(1, 0.0508) MPa From Table 6-10: a = 1.58, b =  0.086, C = 0.120 Eq. (6-72) and Table 6-10):      0.0861.58 1030 1, 0.120 0.870 1, 0.120a  k LN LN From Prob. 6-1: kb = 0.97 Eqs. (6-70) and (6-71): Se = [0.870LN(1, 0.120)] (0.97) [0.506(1030)LN(1, 0.138)] 0.870S e (0.97)(0.506)(1030) = 440 MPa and, CSe  (0.122 + 0.1382)1/2 = 0.183 Se =440LN(1, 0.183) MPa Ans. ______________________________________________________________________________ Chapter 6 - Rev. A, Page 58/66 • 6-65 A Priori Decisions: • Material and condition: 1020 CD, Sut = 68 LN(1, 0.28), and Sy = 57 LN(1, 0.058) kpsi • Reliability goal: R = 0.99 (z =  2.326, Table A-10) • Function: Critical location—hole • Variabilities:  1/22 2 2 2 2 2 1/2 2 2 1/2 2 2 2 2 2 2 0.058 0.125 0.138 (0.058 0.125 0.138 ) 0.195 0.10 0.20 (0.10 0.20 ) 0.234 0.195 0.234 0.297 1 1 0.234 e e ka kc S Se ka kc S Kf Fa a Se a n a C C C C C C C C C C C CC C                            Resulting in a design factor nf of, Eq. (6-59): 2 2exp[ ( 2.326) ln(1 0.297 ) ln 1 0.297 ] 2.05fn        • Decision: Set nf = 2.05 Now proceed deterministically using the mean values: Table 6-10:   0.2652.67 68 0.873ak   Eq. (6-21): kb = 1 Table 6-11:   0.07781.23 68 0.886ck   Eq. (6-70):  0.506 68 34.4 kpsieS   Eq. (6-71):   0.873 1 0.886 34.4 26.6 kpsieS   From Prob. 6-14, Kf = 2.26. Thus, Chapter 6 - Rev. A, Page 59/66 •       2.5 0.5 2 2.05 2.26 3.8 0.331 in 2 2 26.6 a a a a f f f e f f f a e F F FK K K S A t t n K F t S       n     Decision: Use t = 3 8 in Ans. ______________________________________________________________________________ 6-66 Rotation is presumed. M and Sut are given as deterministic, but notice that  is not; therefore, a reliability estimation can be made. From Eq. (6-70): Se = 0.506(780)LN(1, 0.138) = 394.7 LN(1, 0.138) Table 6-13: ka = 4.45(780) 0.265LN(1, 0.058) = 0.762 LN(1, 0.058) Based on d = 32  6 = 26 mm, Eq. (6-20) gives 0.10726 0.877 7.62b k        Conservatism is not necessary     2 2 1/2 0.762 1, 0.058 (0.877)(394.7) (1, 0.138) 263.8 MPa (0.058 0.138 ) 0.150 263.8 (1, 0.150) MPa e e Se e S C          S LN LN S LN Fig. A-15-14: D/d = 32/26 = 1.23, r/d = 3/26 = 0.115. Thus, Kt  1.75, and Eq. (6-78) and Table 6-15 gives     1.75 1.64 2 1.75 12 1 104 / 78011 1.75 3 t f t t KK K a K r      From Table 6-15, CKf = 0.15. Thus, K f = 1.64LN(1, 0.15) The bending stress is   3 3 6 32 32(160)1.64 (1, 0.15) (0.026) 152 10 (1, 0.15) Pa 152 (1, 0.15) MPa f M d            K LN LN LN  From Eq. (5-43), p. 250, Chapter 6 - Rev. A, Page 60/66 •             2 2 2 2 2 2 2 2 1ln 1 ln 1 1 ln 263.8 /152 1 0.15 / 1 0.15 2.61 ln 1 0.15 1 0.15 S S S C C z C C                             From Table A-10, pf = 0.004 53. Thus, R = 1  0.004 53 = 0.995 Ans. Note: The correlation method uses only the mean of Sut ; its variability is already included in the 0.138. When a deterministic load, in this case M, is used in a reliability estimate, engineers state, “For a Design Load of M, the reliability is 0.995.” They are, in fact, referring to a Deterministic Design Load. ______________________________________________________________________________ 6-67 For completely reversed torsion, ka and kb of Prob. 6-66 apply, but kc must also be considered. utS = 780/6.89 = 113 kpsi Eq. 6-74: kc = 0.328(113)0.125LN(1, 0.125) = 0.592LN(1, 0.125) Note 0.590 is close to 0.577. 2 2 2 1/2 0.762[ (1, 0.058)](0.877)[0.592 (1, 0.125)][394.7 (1, 0.138)] 0.762(0.877)(0.592)(394.7) 156.2 MPa (0.058 0.125 0.138 ) 0.195 156.2 (1, 0.195) MPa e a b c e e Se e k S C          S k k S LN LN LN S LN Fig. A-15-15: D/d = 1.23, r/d = 0.115, then Kts  1.40. From Eq. (6-78) and Table 7-8     1.40 1.34 2 1.40 12 1 104 / 78011 1.40 3 ts fs ts ts KK K a K r      From Table 6-15, CKf = 0.15. Thus, K fs = 1.34LN(1, 0.15) The torsional stress is       33 6 16 16016 1.34 (1, 0.15) 0.026 62.1 10 (1, 0.15) Pa 62.1 (1, 0.15) MPa fs T d             K LN LN LN  Chapter 6 - Rev. A, Page 61/66 • From Eq. (5-43), p. 250, 2 2 2 2 ln (156.2 / 62.1) (1 0.15 ) / (1 0.195 ) 3.75 ln[(1 0.195 )(1 0.15 )] z          From Table A-10, pf = 0.000 09 R = 1  pf = 1  0.000 09 = 0.999 91 Ans. For a design with completely-reversed torsion of 160 N · m, the reliability is 0.999 91. The improvement over bending comes from a smaller stress-concentration factor in torsion. See the note at the end of the solution of Prob. 6-66 for the reason for the phraseology. ______________________________________________________________________________ 6-68 Given: Sut = 58 kpsi. Eq. (6-70): Se = 0.506(76) LN(1, 0.138) = 38.5 LN(1, 0.138) kpsi Table 6-13: ka = 14.5(76) 0.719 LN(1, 0.11) = 0.644 LN(1, 0.11) Eq. (6-24): de = 0.370(1.5) = 0.555 in Eq. (6-20): kb = (0.555/0.3)0.107 = 0.936 Eq. (6-70): Se = [0.644 LN(1, 0.11)](0.936)[38.5 LN(1, 0.138)]   0.644 0.936 38.5 23.2 kpsieS   CSe = (0.112 + 0.1382)1/2 = 0.176 Se =23.2 LN(1, 0.176) kpsi Table A-16: d/D = 0, a/D = (3/16)/1.5 = 0.125, A = 0.80  Kt = 2.20. From Eqs. (6-78) and (6-79) and Table 6-15 Chapter 6 - Rev. A, Page 62/66 •   2.20 (1, 0.10) 1.83 (1, 0.10) 2 2.20 1 5 / 761 2.20 0.125 fK    LN LN Table A-16: 3 3 3 net net (0.80)(1.5 ) 0.265 in 32 32 1.51.83 (1, 0.10) 0.265 10.4 (1, 0.10) kpsi 10.4 kpsi 0.10 f ADZ M Z C                 K LN LN  Eq. (5-43), p. 250: 2 2 2 2 ln (23.2 /10.4) (1 0.10 ) / (1 0.176 ) 3.94 ln[(1 0.176 )(1 0.10 )] z          Table A-10: pf = 0.000 041 5  R = 1  pf = 1  0.000 041 5 = 0.999 96 Ans. ______________________________________________________________________________ 6-69 From Prob. 6-68: Se = 23.2 LN(1, 0.138) kpsi ka = 0.644LN(1, 0.11) kb = 0.936 Eq. (6-74): kc = 0.328(76)0.125LN(1, 0.125) = 0.564 LN(1, 0.125) Eq. (6-71): Se = [0.644LN(1, 0.11)](0.936)[ 0.564 LN(1, 0.125)][ 23.2 LN(1, 0.138)]    0.644 0.936 0.564 23.2 7.89 kpsieS   CSe = (0.112 +0.1252 + 0.1383)1/2 = 0.216 Table A-16: d/D = 0, a/D = (3/16)/1.5 = 0.125, A = 0.89, Kts = 1.64 From Eqs. (6-78) and(7-79), and Table 6-15   1.64 (1, 0.10) 1.40 (1, 0.10) 2 1.64 1 5 / 761 1.64 3 / 32 f s    LNK LN Chapter 6 - Rev. A, Page 63/66 • Table A-16:   4 4 4 net net (0.89)(1.5 ) 0.4423 in 32 32 2(1.5)1.40[ (1, 0.10)] 4.75 (1, 0.10) kpsi 2 2 0.4423 a a f s ADJ T D J         K LN LN From Eq. (6-57): 2 2 2 2 ln(7.89 / 4.75) (1 0.10 ) / (1 0.216 ) 2.08 ln[(1 0.10 )(1 0.216 )] z         Table A-10, pf = 0.0188, R = 1  pf = 1  0.0188 = 0.981 Ans. ______________________________________________________________________________ 6-70 This is a very important task for the student to attempt before starting Part 3. It illustrates the drawback of the deterministic factor of safety method. It also identifies the a priori decisions and their consequences. The range of force fluctuation in Prob. 6-30 is  16 to + 5 kip, or 21 kip. Let the repeatedly-applied Fa be 10.5 kip. The stochastic properties of this heat of AISI 1018 CD are given in the problem statement. Function Consequences Axial Fa = 10.5 kip Fatigue load CFa = 0 Ckc = 0.125 Overall reliability R ≥ 0.998; with twin fillets 0.998 0.999R   z =  3.09 CKf = 0.11 Cold rolled or machined surfaces Cka = 0.058 Ambient temperature Ckd = 0 Use correlation method 0.138C  Stress amplitude CKf = 0.11 C a = 0.11 Significant strength Se 2 2 2 1/2(0.058 0.125 0.138 ) 0.195SeC     Choose the mean design factor which will meet the reliability goal. From Eq. (6-88) 2 2 2 2 2 0.195 0.11 0.223 1 0.11 exp ( 3.09) ln(1 0.223 ) ln 1 0.223 2.02 nC n n              Chapter 6 - Rev. A, Page 64/66 • In Prob. 6-30, it was found that the hole was the significant location that controlled the analysis. Thus,  1 e a e a a f S FK n h d       S n  w eS n We need to determine eS -0.265 -0.2652.67 2.67(64) 0.887a utk S   kb = 1 0.0778 0.07781.23 1.23(64) 0.890c utk S     1d ek k  0.887(1)(0.890)(1)(1)(0.506)(64) 25.6 kpsieS   From the solution to Prob. 6-30, the stress concentration factor at the hole is Kt = 2.68. From Eq. (6-78) and Table 6-15      1 2.68 2.20 2 2.68 1 5 / 641 2.68 0.2 2.20(2.02)(10.5) 0.588 . 3.5 0.4 (25.6) f f a e K K nF h Ans d S         w ______________________________________________________________________________ 6-71 1200 lbf 80 kpsi a ut F S   (a) Strength ka = 2.67(80) 0265LN(1, 0.058) = 0.836 LN(1, 0.058) kb = 1 kc = 1.23(80) 0.0778LN(1, 0.125) = 0.875 LN(1, 0.125) Chapter 6 - Rev. A, Page 65/66 • Chapter 6 - Rev. A, Page 66/66     2 2 2 1/2 0.506(80) (1, 0.138) 40.5 (1, 0.138) kpsi 0.836 (1, 0.058) (1) 0.875 (1, 0.125) 40.5 (1, 0.138) 0.836(1)(0.875)(40.5) 29.6 kpsi (0.058 0.125 0.138 ) 0.195 e e e Se S C           S LN LN S LN LN LN Stress: Fig. A-15-1; d/w = 0.75/1.5 = 0.5, Kt = 2.18. From Eqs. (6-78), (6-79) and Table 6-15   2.18 (1, 0.10) 1.96 (1, 0.10) 2 2.18 1 5 / 801 2.18 0.375 f    LNK LN                 2 2 2 2 2 2 2 2 , 0.10 ( ) 1.96(1.2) 12.54 kpsi ( ) (1.5 0.75)(0.25) 29.6 kpsi ln ( / ) 1 1 ln 1 1 ln 29.6 /12.48 1 0.10 / 1 0.195 3.9 ln 1 0.10 1 0.195 a a f f a a a e a a S S F C d t K F d t S S S C C z C C                                     K w w  From Table A-20, pf = 4.81(10 5)  R = 1  4.81(10 5) = 0.999 955 Ans. (b) All computer programs will differ in detail. ______________________________________________________________________________ 6-72 to 6-78 Computer programs are very useful for automating specific tasks in the design process. All computer programs will differ in detail. • Chapter 7 7-1 (a) DE-Gerber, Eq. (7-10):        2 2 2 24 3 4 (2.2)(70) 3 (1.8)(45) 338.4 N mf a fs aA K M K T             2 2 2 24 3 4 (2.2)(55) 3 (1.8)(35) 265.5 N mf m fs mB K M K T               1/31/226 6 6 2(265.5) 210 108(2)(338.4) 1 1 210 10 338.4 700 10 d                        d = 25.85 (103) m = 25.85 mm Ans. (b) DE-elliptic, Eq. (7-12) can be shown to be           1/3 1/3 2 22 2 2 22 2 6 6 338.4 265.516 16(2) 210 10 560 10e y n A Bd S S                         d = 25.77 (103) m = 25.77 mm Ans. (c) DE-Soderberg, Eq. (7-14) can be shown to be     1/31/3 6 6 16 16(2) 338.4 265.5 210 10 560 10e y n A Bd S S                         d = 27.70 (103) m = 27.70 mm Ans. (d) DE-Goodman: Eq. (7-8) can be shown to be     1/31/3 6 6 16 16(2) 338.4 265.5 210 10 700 10e ut n A Bd S S                        d = 27.27 (103) m = 27.27 mm Ans. ________________________________________________________________________ Criterion d (mm) Compared to DE-Gerber DE-Gerber 25.85 DE-Elliptic 25.77 0.31% Lower Less conservative DE-Soderberg 27.70 7.2% Higher More conservative DE-Goodman 27.27 5.5% Higher More conservative ______________________________________________________________________________ 7-2 This problem has to be done by successive trials, since Se is a function of shaft size. The material is SAE 2340 for which Sut = 175 kpsi, Sy = 160 kpsi, and HB ≥ 370. Chapter 7 - Rev. A, Page 1/45 • Eq. (6-19), p. 287: 0.2652.70(175) 0.69ak   Trial #1: Choose dr = 0.75 in Eq. (6-20), p. 288: 0.1070.879(0.75) 0.91bk   Eq. (6-8), p.282:  0.5 0.5 175 87.5 kpsie utS S    Eq. (6-18), p. 287: Se = 0.69 (0.91)(87.5) = 54.9 kpsi 2 0.75 2 / 20 0.65rd d r D D D     0.75 1.15 in 0.65 0.65 rdD    1.15 0.058 in 20 20 Dr    Fig. A-15-14: 2 0.75 2(0.058) 0.808 inrd d r     0.808 1.08 0.75r d d   0.058 0.077 0.75r r d   Kt = 1.9 Fig. 6-20, p. 295: r = 0.058 in, q = 0.90 Eq. (6-32), p. 295: Kf = 1 + 0.90 (1.9 – 1) = 1.81 Fig. A-15-15: Kts = 1.5 Fig. 6-21, p. 296: r = 0.058 in, qs = 0.92 Eq. (6-32), p. 295: Kfs = 1 + 0.92 (1.5 – 1) = 1.46 We select the DE-ASME Elliptic failure criteria, Eq. (7-12), with d as dr, and Mm = Ta = 0,     1/31/22 2 3 3 16(2.5) 1.81(600) 1.46(400)4 3 54.9 10 160 10r d                          dr = 0.799 in Trial #2: Choose dr = 0.799 in. 0.1070.879(0.799) 0.90bk   Se = 0.69 (0.90)(0.5)(175) = 54.3 kpsi 0.799 1.23 in 0.65 0.65 rdD    r = D / 20 = 1.23/20 = 0.062 in Chapter 7 - Rev. A, Page 2/45 • Figs. A-15-14 and A-15-15: 2 0.799 2(0.062) 0.923 inrd d r     0.923 1.16 0.799r d d   0.062 0.078 0.799r r d   With these ratios only slightly different from the previous iteration, we are at the limit of readability of the figures. We will keep the same values as before. 1.9, 1.5, 0.90, 0.92t ts sK K q q    1.81, 1.46f fsK K   Using Eq. (7-12) produces dr = 0.802 in. Further iteration produces no change. With dr = 0.802 in, 0.802 1.23 in 0.65 0.75(1.23) 0.92 in D d     A look at a bearing catalog finds that the next available bore diameter is 0.9375 in. In nominal sizes, we select d = 0.94 in, D = 1.25 in, r = 0.0625 in Ans. ______________________________________________________________________________ 7-3 F cos 20(d / 2) = TA, F = 2 TA / ( d cos 20) = 2(340) / (0.150 cos 20) = 4824 N. The maximum bending moment will be at point C, with MC = 4824(0.100) = 482.4 N·m. Due to the rotation, the bending is completely reversed, while the torsion is constant. Thus, Ma = 482.4 N·m, Tm = 340 N·m, Mm = Ta = 0. For sharp fillet radii at the shoulders, from Table 7-1, Kt = 2.7, and Kts = 2.2. Examining Figs. 6-20 and 6-21 (pp. 295 and 296 respectively) with 560 MPa,utS  conservatively estimate q = 0.8 and These estimates can be checked once a specific fillet radius is determined. 0.9.sq  Eq. (6-32): 1 0.8(2.7 1) 2.4fK     1 0.9(2.2 1) 2.1fsK     (a) We will choose to include fatigue stress concentration factors even for the static analysis to avoid localized yielding. Eq. (7-15): 1/22 2 max 3 3 32 16 3f a fs m K M K T d d                      Chapter 7 - Rev. A, Page 3/45 • Eq. (7-16):     3 1/22 2 max 4 3 16 y y f a fs m S d S n K M K    T      Solving for d,         1/3 1/22 2 1/3 1/22 2 6 16 4( ) 3( ) 16(2.5) 4 (2.4)(482.4) 3 (2.1)(340) 420 10 f a fs a y nd K M K T S                     d = 0.0430 m = 43.0 mm Ans. (b) 0.2654.51(560) 0.84ak   Assume kb = 0.85 for now. Check later once a diameter is known. Se = 0.84(0.85)(0.5)(560) = 200 MPa Selecting the DE-ASME Elliptic criteria, use Eq. (7-12) with 0.m aM T      1/31/22 2 6 6 16(2.5) 2.4(482.4) 2.1(340)4 3 200 10 420 10 0.0534 m 53.4 mm d                            With this diameter, we can refine our estimates for kb and q. Eq. (6-20):   0.1570.1571.51 1.51 53.4 0.81bk d    Assuming a sharp fillet radius, from Table 7-1, r = 0.02d = 0.02 (53.4) = 1.07 mm. Fig. (6-20): q = 0.72 Fig. (6-21): qs = 0.77 Iterating with these new estimates, Eq. (6-32): Kf = 1 + 0.72 (2.7 – 1) = 2.2 Kfs = 1 + 0.77 (2.2 – 1) = 1.9 Eq. (6-18): Se = 0.84(0.81)(0.5)(560) = 191 MPa Eq. (7-12): d = 53 mm Ans. Further iteration does not change the results. _____________________________________________________________________________ Chapter 7 - Rev. A, Page 4/45 • 7-4 We have a design task of identifying bending moment and torsion diagrams which are preliminary to an industrial roller shaft design. Let point C represent the center of the span of the roller. 30(8) 240 lbfyCF   0.4(240) 96 lbfzCF   (2) 96(2) 192 lbf inzCT F    192 128 lbf 1.5 1.5 z B TF    tan 20 128 tan 20 46.6 lbfy zB BF F     (a) xy-plane 240(5.75) (11.5) 46.6(14.25) 0yO AM F     240(5.75) 46.6(14.25) 62.3 lbf 11.5 y AF    (11.5) 46.6(2.75) 240(5.75) 0yA OM F     240(5.75) 46.6(2.75) 131.1 lbf 11.5 y OF    Bending moment diagram: xz-plane Chapter 7 - Rev. A, Page 5/45 • 0 96(5.75) (11.5) 128(14.25)zO AM F     96(5.75) 128(14.25) 206.6 lbf 11.5 z AF    0 (11.5) 128(2.75) 96(5.75)zA OM F     96(5.75) 128(2.75) 17.4 lbf 11.5 z OF    Bending moment diagram: 2 2100 ( 754) 761 lbf inCM      2 2( 128) ( 352) 375 lbf inAM       Torque: The torque is constant from C to B, with a magnitude previously obtained of 192 lbf·in. (b) xy-plane 2 2131.1 15 1.75 15 9.75 62.3 11.5xyM x x x x        1 Bending moment diagram: Chapter 7 - Rev. A, Page 6/45 • Mmax = –516 lbf · in and occurs at 6.12 in. 2131.1(5.75) 15(5.75 1.75) 514 lbf inCM      This is reduced from 754 lbf · in found in part (a). The maximum occurs at rather than C, but it is close enough. 6.12 inx  xz-plane 2 217.4 6 1.75 6 9.75 206.6 11.5xzM x x x x       1 Bending moment diagram: Let 2 2net xy xzM M M  Plot Mnet(x), 1.75 ≤ x ≤ 11.5 in Mmax = 516 lbf · in at x = 6.25 in Torque: The torque rises from 0 to 192 lbf·in linearly across the roller, then is constant to B. Ans. ______________________________________________________________________________ 7-5 This is a design problem, which can have many acceptable designs. See the solution for Prob. 7-17 for an example of the design process. ______________________________________________________________________________ Chapter 7 - Rev. A, Page 7/45 • 7-6 If students have access to finite element or beam analysis software, have them model the shaft to check deflections. If not, solve a simpler version of shaft for deflection. The 1 in diameter sections will not affect the deflection results much, so model the 1 in diameter as 1.25 in. Also, ignore the step in AB. From Prob. 7-4, integrate Mxy and Mxz. xy plane, with dy/dx = y'   3 32 1131.1 62.35 1.75 5 9.75 11.52 2 2EIy x x x x C          (1)   4 4 33 1 2131.1 5 5 62.31.75 9.75 11.56 4 4 6EIy x x x x C x C          20 at 0 0y x C    3 10 at 11.5 1908.4 lbf iny x C     From (1), x = 0: EIy' = 1908.4 x = 11.5: EIy' = –2153.1 xz plane (treating ) z     3 32 317.4 206.62 1.75 2 9.75 11.52 2 2EIz x x x x C         (2)   4 4 33 3 417.4 1 1 206.61.75 9.75 11.56 2 2 6EIz x x x x C x C         40 at 0 0z x C    330 at 11.5 8.975 lbf inz x C     From (2), x = 0: EIz' = 8.975 x = 11.5: EIz' = –683.5 At O: 2 21908.4 8.975 1908.4 lbf inEI    3 Chapter 7 - Rev. A, Page 8/45 • At A: 2 2( 2153.1) ( 683.5) 2259.0 lbf inEI       3 (dictates size)    6 4 2259 0.000 628 rad 30 10 / 64 1.25     0.001 1.59 0.000 628 n   At gear mesh, B xy plane With 1I I in section OCA, 12153.1/Ay EI   Since y'B/A is a cantilever, from Table A-9-1, with 2I I in section AB / 2 2 2 ( 2 ) 46.6 (2.75)[2.75 2(2.75)] 176.2 / 2 2B A Fx x ly E EI EI       I        / 6 4 6 2153.1 176.2 30 10 / 64 1.25 30 10 / 64 0.875B A B A y y y      4       = –0.000 803 rad (magnitude greater than 0.0005 rad) xz plane  2 / 1 2 128 2.75683.5 484, 2A B A z z 2EI EI        EI        6 4 6 4 683.5 484 0.000 751 rad 30 10 / 64 1.25 30 10 / 64 0.875B z         2 2( 0.000 803) ( 0.000 751) 0.00110 radB      Crowned teeth must be used. Finite element results: Error in simplified model 45.47(10 ) radO  3.0% 47.09(10 ) radA  11.4% 31.10(10 ) radB  0.0% Chapter 7 - Rev. A, Page 9/45 • The simplified model yielded reasonable results. Strength 72 kpsi, 39.5 kpsiut yS S  At the shoulder at A, From Prob. 7-4, 10.75 in.x  209.3 lbf in, 293.0 lbf in, 192 lbf inxy xzM M T        2 2( 209.3) ( 293) 360.0 lbf inM       0.5(72) 36 kpsieS    0.2652.70(72) 0.869ak   0.1071 0.879 0.3b k        1c d e fk k k k    0.869(0.879)(36) 27.5 kpsieS   D / d = 1.25, r / d = 0.03 Fig. A-15-8: Kts = 1.8 Fig. A-15-9: Kt = 2.3 Fig. 6-20: q = 0.65 Fig. 6-21: qs = 0.70 Eq. (6-32): 1 0.65(2.3 1) 1.85fK     1 0.70(1.8 1) 1.56fsK     Using DE-ASME Elliptic, Eq. (7-11) with 0,m aM T    1/22 2 3 1 16 1.85(360) 1.56(192)4 3 27 500 39 5001n                   n = 3.91 Perform a similar analysis at the profile keyway under the gear. The main problem with the design is the undersized shaft overhang with excessive slope at the gear. The use of crowned-teeth in the gears will eliminate this problem. ______________________________________________________________________________ 7-7 through 7-16 These are design problems, which can have many acceptable designs. See the solution for Prob. 7-17 for an example of the design process. ______________________________________________________________________________ 7-17 (a) One possible shaft layout is shown in part (e). Both bearings and the gear will be located against shoulders. The gear and the motor will transmit the torque through the Chapter 7 - Rev. A, Page 10/45 • keys. The bearings can be lightly pressed onto the shaft. The left bearing will locate the shaft in the housing, while the right bearing will float in the housing. (b) From summing moments around the shaft axis, the tangential transmitted load through the gear will be / ( / 2) 2500 / (4 / 2) 1250 lbftW T d   The radial component of gear force is related by the pressure angle. tan 1250 tan 20 455 lbfr tW W         1/2 1/22 2 2 2455 1250 1330 lbfr tW W W     Reactions and ,A BR R and the load W are all in the same plane. From force and moment balance, 1330(2 /11) 242 lbfAR   1330(9 /11) 1088 lbfBR   max (9) 242(9) 2178 lbf inAM R    Shear force, bending moment, and torque diagrams can now be obtained. (c) Potential critical locations occur at each stress concentration (shoulders and keyways). To be thorough, the stress at each potentially critical location should be evaluated. For Chapter 7 - Rev. A, Page 11/45 • now, we will choose the most likely critical location, by observation of the loading situation, to be in the keyway for the gear. At this point there is a large stress concentration, a large bending moment, and the torque is present. The other locations either have small bending moments, or no torque. The stress concentration for the keyway is highest at the ends. For simplicity, and to be conservative, we will use the maximum bending moment, even though it will have dropped off a little at the end of the keyway. (d) At the gear keyway, approximately 9 in from the left end of the shaft, the bending is completely reversed and the torque is steady. 2178 lbf in 2500 lbf in 0a m mM T M     aT  From Table 7-1, estimate stress concentrations for the end-milled keyseat to be Kt = 2.14 and Kts = 3.0. For the relatively low strength steel specified (AISI 1020 CD), roughly estimate notch sensitivities of q = 0.75 and qs = 0.80, obtained by observation of Figs. 6- 20 and 6-21, assuming a typical radius at the bottom of the keyseat of r / d = 0.02 (p. 373), and a shaft diameter of up to 3 inches. Eq. (6-32): 1 0.75(2.14 1) 1.9fK     1 0.8(3.0 1) 2.6fsK     Eq. (6-19): 0.2652.70(68) 0.883ak   For estimating , guess 2 in.bk d  Eq. (6-20) 0.107(2 / 0.3) 0.816bk   Eq. (6-18) 0.883(0.816)(0.5)(68) 24.5 kpsieS   Selecting the DE-Goodman criteria for a conservative first design, Eq. (7-8):     1/31/2 1/22 2 4 316 f a fs m e ut K M K Tnd S S                               1/31/2 1/22 24 1.9 2178 3 2.6 250016(1.5) 24 500 68 000 d                      1.57 in .d A ns With this diameter, the estimates for notch sensitivity and size factor were conservative, but close enough for a first iteration until deflections are checked. Check yielding with this diameter. Chapter 7 - Rev. A, Page 12/45 • Eq. (7-15): 1/22 2 max 3 3 32 16 3f a fs m K M K T d d                      1/22 2 max 3 3 32(1.9)(2178) 16(2.6)(2500)3 18389 psi 18.4 kpsi (1.57) (1.57)                       max/ 57 /18.4 3.1 .y yn S Ans    (e) Now estimate other diameters to provide typical shoulder supports for the gear and bearings (p. 372). Also, estimate the gear and bearing widths. (f) Entering this shaft geometry into beam analysis software (or Finite Element software), the following deflections are determined: Left bearing slope: 0.000 532 rad Right bearing slope:  0.000 850 rad Gear slope:  0.000 545 rad Right end of shaft slope:  0.000 850 rad Gear deflection:  0.001 45 in Right end of shaft deflection: 0.005 10 in Comparing these deflections to the recommendations in Table 7-2, everything is within typical range except the gear slope is a little high for an uncrowned gear. (g) To use a non-crowned gear, the gear slope is recommended to be less than 0.0005 rad. Since all other deflections are acceptable, we will target an increase in diameter only for the long section between the left bearing and the gear. Increasing this diameter from the proposed 1.56 in to 1.75 in, produces a gear slope of  0.000 401 rad. All other deflections are improved as well. ______________________________________________________________________________ Chapter 7 - Rev. A, Page 13/45 • 7-18 (a) Use the distortion-energy elliptic failure locus. The torque and moment loadings on the shaft are shown in the solution to Prob. 7-17. Candidate critical locations for strength:  Left seat keyway  Right bearing shoulder  Right keyway Table A-20 for 1030 HR: 68 kpsi, 37.5 kpsi, 137ut y BS S H   Eq. (6-8): 0.5(68) 34.0 kpsieS    Eq. (6-19): 0.2652.70(68) 0.883ak   1c d ek k k   Left keyway See Table 7-1 for keyway stress concentration factors, 2.14 Profile keyway 3.0 t ts K K     For an end-mill profile keyway cutter of 0.010 in radius, estimate notch sensitivities. Fig. 6-20: 0.51q  Fig. 6-21: 0.57sq  Eq. (6-32): 1 ( 1) 1 0.57(3.0 1) 2.1fs s tsK q K       1 0.51(2.14 1) 1.6fK     Eq. (6-20): 0.1071.875 0.822 0.30b k        Eq. (6-18): 0.883(0.822)(34.0) 24.7 kpsieS   Eq. (7-11): 1 2 2 2 3 1 16 1.6(2178) 2.1(2500)4 3 (1.875 ) 24 700 37 500fn                   nf = 3.5 Ans. Right bearing shoulder The text does not give minimum and maximum shoulder diameters for 03-series bearings (roller). Use D = 1.75 in. 0.030 1.750.019, 1.11 1.574 1.574 r D d d     Fig. A-15-9: 2.4tK  Fig. A-15-8: 1.6tsK  Chapter 7 - Rev. A, Page 14/45 • Fig. 6-20: 0.65q  Fig. 6-21: 0.70sq  Eq. (6-32): 1 0.65(2.4 1) 1.91fK     1 0.70(1.6 1) 1.42fsK     0.4532178 493 lbf in 2 M        Eq. (7-11): 1/22 2 3 1 16 1.91(493) 1.42(2500)4 3 (1.574 ) 24 700 37 500fn                   nf = 4.2 Ans. Right keyway Use the same stress concentration factors as for the left keyway. There is no bending moment, thus Eq. (7-11) reduces to:  3 3 16 31 16 3(2.1)(2500) 1.5 (37 500) fs m f y K T n d S    nf = 2.7 Ans. Yielding Check for yielding at the left keyway, where the completely reversed bending is maximum, and the steady torque is present. Using Eq. (7-15), with Mm = Ta = 0,           1/22 2 max 3 3 1/22 2 3 3 32 16 3 32 1.6 2178 16 2.1 2500 3 1.875 1.875 8791 psi 8.79 kpsi f a fs mK M K T d d                                              max 37.5 4.3 8.79 y y S n      Ans. Check in smaller diameter at right end of shaft where only steady torsion exists.      1/22 max 3 1/22 3 16 3 16 2.1 2500 3 1.5 13 722 psi 13.7 kpsi fs mK T d                             Chapter 7 - Rev. A, Page 15/45 • max 37.5 2.7 13.7 y y S n      Ans. (b) One could take pains to model this shaft exactly, using finite element software. However, for the bearings and the gear, the shaft is basically of uniform diameter, 1.875 in. The reductions in diameter at the bearings will change the results insignificantly. Use E = 30 Mpsi for steel. To the left of the load, from Table A-9, case 6, p. 1015, 2 2 2 2 2 2 6 4 6 2 1449(2)(3 2 11 )(3 ) 6 6(30)(10 )( / 64)(1.875 )(11) 2.4124(10 )(3 117) AB AB d y Fb xx b l dx EIl x             At x = 0 in: 42.823(10 ) rad   At x = 9 in: 43.040(10 ) rad  To the right of the load, from Table A-9, case 6, p. 1015,  2 23 6 26 BC BC d y Fa 2x xl l a dx EIl        At x = l = 11 in:   2 2 2 2 4 6 4 1449(9)(11 9 ) 4.342(10 ) rad 6 6(30)(10 )( / 64)(1.875 )(11) Fa l a EIl       Obtain allowable slopes from Table 7-2. Left bearing: Allowable slope 0.001 3.5 . Actual slope 0.000 282 3fs n Ans   Right bearing: 0.0008 1.8 . 0.000 434 2fs n Ans  Gear mesh slope: Table 7-2 recommends a minimum relative slope of 0.0005 rad. While we don’t know the slope on the next shaft, we know that it will need to have a larger diameter and be stiffer. At the moment we can say 0.0005 1.6 . 0.000 304fs n Ans  ______________________________________________________________________________ Chapter 7 - Rev. A, Page 16/45 • 7-19 The most likely critical locations for fatigue are at locations where the bending moment is high, the cross section is small, stress concentration exists, and torque exists. The two- plane bending moment diagrams, shown in the solution to Prob. 3-72, indicate decreasing moments in both planes to the left of A and to the right of C, with combined values at A and C of MA = 5324 lbf·in and MC = 6750 lbf·in. The torque is constant between A and B, with T = 2819 lbf·in. The most likely critical locations are at the stress concentrations near A and C. The two shoulders near A can be eliminated since the shoulders near C have the same geometry but a higher bending moment. We will consider the following potentially critical locations:  keyway at A  shoulder to the left of C  shoulder to the right of C Table A-20: Sut = 64 kpsi, Sy = 54 kpsi Eq. (6-8): 0.5(64) 32.0 kpsieS    Eq. (6-19): 0.2652.70(64) 0.897ak   1c d ek k k   Keyway at A Assuming r / d = 0.02 for typical end-milled keyway cutter (p. 373), with d = 1.75 in, r = 0.02d = 0.035 in. Table 7-1: Kt = 2.14, Kts = 3.0 Fig. 6-20: q = 0.65 Fig. 6-21: qs = 0.71 Eq. (6-32):  1 1 1 0.65(2.14 1) 1.7f tK q K       1 ( 1) 1 0.71(3.0 1) 2.4fs s tsK q K       Eq. (6-20): 0.1071.75 0.828 0.30b k        Eq. (6-18): 0.897(0.828)(32) 23.8 kpsieS   Chapter 7 - Rev. A, Page 17/45 • We will choose the DE-Gerber criteria since this is an analysis problem in which we would like to evaluate typical expectations. sing Eq. (7-9) with M = T = 0, U m a           2 2 2 2 4 4 1.7 5324 18102 lbf in 18.10 kip in 3 3 2.4 2819 11 718 lbf in 11.72 kip in f a fs m A K M B K T                             1/22 3 1/22 3 21 8 1 1 8 18.10 2 11.72 23.8 1 1 18.10 6475 .8 e e ut BSA n d S AS                      1. 23            oulder to the left of C 625 / 1.75 = 0.036, D / d = 2.5 / 1.75 = 1.43 : : q = 0.71 Fig. 6-21: q = 0.76 q. (6-32): n = 1.3 Sh r / d = 0.0 Fig. A-15-9 Kt = 2.2 Fig. A-15-8 Kts = 1.8 Fig. 6-20: E s  1 1 1 0.71(2.2 1) 1.9f tK q K       1 ( 1) 1 .76(1.8 1) 1.6fs s tsK q K       0 0.1071.75 0.828 0.30b k        Eq. (6-20): Eq. (6-18): 0.897(0.828)(32) 23.8 kpsieS   For convenience, we will use the full value of the bending moment at C, even though it will be slightly less at the shoulder. Using Eq. (7-9) with Mm = Ta = 0,           2 2 2 2 4 4 1.9 6750 25 650 lbf in 25.65 kip in 3 3 1.6 2819 7812 lbf in 7.812 kip in f a fs m A K M B K T                            1/22 3 1/22 21 8 1 1 8 25.65 2 7.812 23.8 1 1 25.65 643.8 e e ut BSA n d S AS                      31.75 2            Chapter 7 - Rev. A, Page 18/45 • n = 0.96 oulder to the right of C 625 / 1.3 = 0.048, D / d = 1.75 / 1.3 = 1.35 : : q = 0.71 Fig. 6-21: qs = 0.76 q. (6-32): Sh r / d = 0.0 Fig. A-15-9 Kt = 2.0 Fig. A-15-8 Kts = 1.7 Fig. 6-20: E  1 1 1 0.71(2.0 1) 1.7f tK q K       1 ( 1) 1 .76(1.7 1) 1.5fs s tsK q K       0 0.1071.3 0.855Eq. (6-20): 0.30  Eq. (6-18): 0.897(0.855)(32) 24.5 kpsieS   bk      or convenience, we will use the full value of the bending moment at C, even though it will be slightly less at the shoulder. Using Eq. (7-9) with Mm = Ta = 0, F           2 2 2 2 4 4 1.7 6750 22 950 lbf in 22.95 kip in 3 3 1.5 2819 7324 lbf in 7.324 kip in f a fs m A K M B K T                            1/22 3 1/22 21 8 1 1 8 22.95 2 7.324 24.5 1 1 22.95 6424.5 e e ut BSA n d S AS                      31.3             The critical location is at the shoulder to the right of C, where n = 0.45 and finite life is plicitly called for in the problem statement, a static check for yielding is especially warranted with such a low fatigue factor of safety. Using Eq. (7-15), with Mm = Ta = 0, n = 0.45 predicted. Ans. Though not ex           1/22 2 max 3 3 1/22 2 3 3 32 16 3 32 1.7 6750 16 1.5 2819 3 55 845 psi 55.8 kpsi 1.3 f a fs mK M K T d d                                            1.3  Chapter 7 - Rev. A, Page 19/45 • max 0.97 55.8 yn      his indicates localized yielding is predicted at the stress-concentr 54S ation, though after o be f static, 7-20 te the deflections. Entering the geometry from the shaft as defined in - loading as defined in Prob. 3-72, the following defle itude te D T localized cold-working it may not be a problem. The finite fatigue life is still likely t the failure mode that will dictate whether this shaft is acceptable. It is interesting to note the impact of stress concentration on the acceptability of the proposed design. This problem is linked with several previous problems (see Table 1-1, p. 24) in which the shaft was considered to have a constant diameter of 1.25 in. In each o the previous problems, the 1.25 in diameter was more than adequate for deflection, and fatigue considerations. In this problem, even though practically the entire shaft has diameters larger than 1.25 in, the stress concentrations significantly reduce the anticipated fatigue life. ______________________________________________________________________________ For a shaft with significantly varying diameters over its length, we will choose to use shaft analysis software or finite element software to calcula Prob. 7 e 19, and the rmined: ction magn s are d Location Slope (rad) eflection (in) Left bearing O 0.00640 0.00000 Right bearing C 0.00434 0.00000 Left Gear A 0.00260 0.04839 Right Gear B 0.01078 0.07517 Comparing these values to the recommended limits in Table 7-2, we find that they are all out of the desired range. This is not unexpected since the stress analysis of Prob. 7-19 also indicated the shaft is undersized for infinite life. The sl ope at the right gear is the ost excessive, so we will attempt to increase all diameters to bring it into compliance. sing Eq. (7-18) at the right gear, m U     1/4 1/4 new old old all 2.15 slope 0.0005d    / (1)(0.01078)dn dy dxd Multiplying all diameter e ob fo lections: D s by 2.15, w tain the llowing def Location Slope (rad) eflection (in) Left bearing O 0.00030 0.00000 Right bearing C 0.00020 0.00000 Left Gear A 0.00012 0.00225 Right Gear B 0.00050 0.00350 Chapter 7 - Rev. A, Page 20/45 • This brings the slope at the right gear just to the limit for an uncrowned gear, and all other slopes well below the recommended limits. For the gear deflections, the values are ______________________________________________________________________________ 7-21 is o- with the keyway at B, the rimary difference between the two is the stress concentration, since they both have eyway at A d-milled keyway cutter (p. 373), with d = 50 mm, Kt = 2.14, Kts = 3.0 Fig. 6-20: q = 0.66 ig. 6-21: qs = 0.72 e 50 = 0.04, D / d = 75 / 50 = 1.5 : below recommended limits as long as the diametral pitch is less than 20. The most likely critical locations for fatigue are at locations where the bending moment high, the cross section is small, stress concentration exists, and torque exists. The tw plane bending moment diagrams, shown in the solution to Prob. 3-73, indicate both planes have a maximum bending moment at B. At this location, the combined bending moment from both planes is M = 4097 N·m, and the torque is T = 3101 N·m. The shoulder to the right of B will be eliminated since its diameter is only slightly smaller, and there is no torque. Comparing the shoulder to the left of B p essentially the same bending moment, torque, and size. We will check the stress concentration factors for both to determine which is critical. Table A-20: Sut = 440 MPa, Sy = 370 MPa K Assuming r / d = 0.02 for typical en r = 0.02d = 1 mm. Table 7-1: F Eq. (6-32): 1fK q  1 1 0.66(2.14 1) 1.8tK       1 ( 1) 1 0.72(3.0 1) 2.4fs s tsK q K       Shoulder to th left of B r / d = 2 / Fig. A-15-9 Kt = 2.2 Chapter 7 - Rev. A, Page 21/45 • Fig. A-15-8: F Kts = 1.8 Fig. 6-20: q = 0.73 ig. 6-21: q = 0.78 n of the stress concentration f ctors indicates the keyway will be the critical Eq. (6-19): s  1 1 1 0.73(2.2 1) 1.9 fs s tsK q K       Eq. (6-32): f tK q K       1 ( 1) 1 0.78(1.8 1) 1.6 Examinatio a location. 0.5(440) 220 MPaeS    Eq. (6-8): 0.2654.51(440) 0.899ak   0.107 Eq. (6-20): 50 0.818 7.62b k        We will choose the DE-Gerber criteria since this is an analysis problem in which we ould like to evaluate typical expectations. Using Eq. (7-9) with Mm a = 0, 1c d ek k k   Eq. (6-18): 0.899(0.818)(220) 162 MPaeS   w = T           2 2 2 2 4 4 1.8 4097 14 750 N m 3 3 2.4 3101 12 890 N m f a fs m A K M B K T                           1/22 3 1/226 3 6 6 21 8 1 1 08 14 0.050 162 10 14 750 440 10 e e ut BSA n d S AS                  2 12 890 162 1750 1 1                 n = 0.25 Infinite life is not predicted. Ans. Though not explicitly called for in the problem statement, a static check for yielding is especially warranted with such a low fatigue factor of safety. Using Eq. (7-15), with Mm = Ta = 0,            1/22 2 max 3 3 1/22 2 8 3 3 32 16 3 32 1.8 4097 16 2.4 3101 3 7.98 10 Pa 798 MPa 050 0.050 f a fs mK M K T d d                                   0.            Chapter 7 - Rev. A, Page 22/45 • max 370 0.46 798 ySn      This indicates localized yielding is predicted at the stress-concentration. Even without the stress concentration effects, the static factor of safety turns out to be 0.93. Static failure is predicted, rendering this proposed shaft design unacceptable. This problem is linked with several previous problems (see Table 1-1, p. 24) in which shaft was considered to have a constant diameter of 50 mm. The results here ar the e ______________________________________________________________________________ -22 th, we will choose to use shaft analysis software o ment s t e deflections. Entering the geometry from the shaft as defined in -2 ading as defined in Prob. 3-73, the following itud erm De n consistent with the previous problems, in which the 50 mm diameter was found to slightly undersized for static, and significantly undersized for fatigue. Though in the current problem much of the shaft has larger than 50 mm diameter, the added contribution of stress concentration limits the fatigue life. For a shaft with significantly varying diameters over its leng7 r finite ele oftware 7 o calculate th 1, and the lo i Prob. deflection magn es are det ned: Location Slope (rad) flectio (mm) Left bearing O 0.01445 0.000 Right bearing C 0.01843 0.000 Left Gear A 0.00358 3.761 Right Gear B 0.00366 3.676 Comparing these values to the recommended limits in Table 7-2, we find that they are all well out of the desired range. This is not unexpected since the stress analysis in Prob. -21 also indicated the shaft is undersize7 the lef d for infinite life. The transverse deflection at t gear is the most excessive, so we will attempt to increase all diameters to bring it to compliance. Using Eq. (7-17) at the left gear, assuming from Table 7-2 an allowable yall = 0.01 in = 0.254 mm, in deflection of 1/4 1/4 new old (1)(3.761) 1.96dd n y   old alld y Multiplying all diam btai wi : De n 0.254 eters by 2, we o n the follo ng deflections Location Slope (rad) flectio (mm) Left bearing O 0.00090 0.000 Right bearing C 0.00115 0.000 Left Gear A 0.00022 0.235 Right Gear B 0.00023 0.230 Chapter 7 - Rev. A, Page 23/45 • This brings the deflection at the gears just within the limit for a spur gear (assuming P < ______________________________________________________________________________ 7-23 , stress element will be completely reversed, while the torsional stress will be steady. Since we do not have any information about the fan, we will ignore any axial load that it would introduce. It would not likely contribute much compared to the bending anyway. 10 teeth/in), and all other deflections well below the recommended limits. (a) Label the approximate locations of the effective centers of the bearings as A and B the fan as C, and the gear as D, with axial dimensions as shown. Since there is only one gear, we can combine the radial and tangential gear forces into a single resultant force with an accompanying torque, and handle the statics problem in a single plane. From statics, the resultant reactions at the bearings can be found to be RA = 209.9 lbf and RB = 464.5 lbf. The bending moment and torque diagrams are shown, with the maximum bending moment at D of MD = 209.9(6.98) = 1459 lbf·in and a torque transmitted from D to C of T = 633 (8/2) = 2532 lbf·in. Due to the shaft rotation, the bending stress on any Potentially critical locations are identified as follows:  Keyway at C, where the torque is high, the diameter is small, and the keyway creates a stress concentration. Chapter 7 - Rev. A, Page 24/45 •  Keyway at D, where the bending moment is maximum, the torque is high, and the keyway creates a stress concentration. . eter is smaller than at D or E, the bending moment is  The shoulder to the left of D can be eliminated since the change in diameter is very ill undoubtedly be much less than at D. Sut = 68 kpsi, Sy = 57 kpsi ince there is only steady torsion here, only a static check needs to be performed. We’ll aximum shear stress theory.  Groove at E, where the diameter is smaller than at D, the bending moment is still high, and the groove creates a stress concentration. There is no torque here, though  Shoulder at F, where the diam still moderate, and the shoulder creates a stress concentration. There is no torque here, though. slight, so that the stress concentration w Table A-20: q. (6-8): 0.5(68) 34.0 kpsieS    E 0.2652.70(68) 0.883ak   Eq. (6-19): Keyway at C S use the m    4 2532 1.00 / 2 12.9 kpsi 1.00 / 32 Tr J      / 2 57 / 2 2.21 12.9 y y S nEq. (5-3):     ssuming r / d = 0.02 for typical end-milled keyway cutter (p. 373), with d = 1.75 in, Kts = 3.0 q = 0.66 Fig. 6-21: qs = 0.72 q. (6-32): A Keyway at D r = 0.02d = 0.035 in. Table 7-1: Kt = 2.14, Fig. 6-20: E  1 1 1 0.66(2.14 1) 1.8f tK q K       1 ( 1) 1 .72(3.0 1) 2.4fs s tsK q K       0 0.1071.75 0.828 0.30b k        Eq. (6-20): Eq. (6-18): 0.883(0.828)(34.0) 24.9 kpsieS   We will choose the DE-Gerber criteria since this is an analysis problem in which we ould like to evaluate typical expectations. Using Eq. (7-9) with Mm = Ta = 0, w Chapter 7 - Rev. A, Page 25/45 •           2 2 2 2 4 4 1.8 1459 5252 lbf in 5.252 kip in 3 3 2.4 2532 10 525 lbf in 10.53 kip in f a fs m A K M B K T                            1/22 3 1/22 3 21 8 1 1 8 5.252 2 10.53 24.9 1 1 5.252 681.75 24.9 e e ut BSA n d S AS                                    n = 3.59 Ans. roove at E he right of the w and will likely not allow the stress flow to fully develop. (See the concept.) r / d = 0.1 / 1.55 = 0.065, D / d = 1.75 / 1.55 = 1.13 : Kt = 2.1 Fig. 6-20: q = 0.76 G We will assume Figs. A-15-14 is applicable since the 2 in diameter to t groove is relatively narro Fig.7-9 for stress flow Fig. A-15-14 Eq. (6-32): ) 1 1 1 0.76(2.1 1 1.8f tK q K       0.1071.55 0.839 0.30b k     Eq. (6-20):    Using Eq. (7-9) with Mm = Ta = Tm = 0, 0.883(0.839)(34) 25.2 kpsieS   Eq. (6-18):     2 24 4 1.8 1115 4122 lbf in 4.122 kip inf aA K M        B = 0         1/22 3 1/22 31.55 25.2 21 8 1 1 8 4.122 1 1 0 e e u BSA A              t n d S S         Ans. F r / d = 0.125 / 1.40 = 0.089, D / d = 2.0 / 1.40 = 1.43 Kt = 1.7 Fig. 6-20: q = 0.78 n = 4.47 Shoulder at Fig. A-15-9: Chapter 7 - Rev. A, Page 26/45 • Eq. (6-32): ) 1 1 1 0.78(1.7 1 1.5f tK q K       0.1071.40 0.848 0.30b k       Eq. (6-20):  Eq. (6-18): Using Eq. (7-9) with Mm = Ta = Tm = 0, 0.883(0.848)(34) 25.5 kpsieS       2 24 4 1.5 845 2535 lbf in 2.535 kip inf aA K M        B = 0         1/22 3 2.53 1 1 0 1.40 25.5 1/22 3 21 8 1 1 e e ut BSA AS              n d S   8 5      n = 5.42 Ans. (b) The deflection will not be much affected by the details of fillet radii, grooves, and keyways, so these can be A g narrow 2.0 in diameter section, can be cted. ill model the shaft with the following three sections: Section Diameter (in) Length (in) ignored. lso, the sli ht diameter changes, as well as the negle We w 1 1.00 2.90 2 1.70 7.77 3 1.40 2.20 The deflection problem can readily (though tediously) be solved with singularity functions. For example -7, p. the solution to Prob. 7-24. Alternatively, shaft analysis software or finite element software may be used. Using any of the methods, the results low ation D s, see Ex. 4 159, or should be as fol s: Loc Slope (rad) eflection (in) Left bearing A 0.000290 0.000000 Right bearing B 0.000400 0.000000 Fan C 0.000290 0.000404 Gear D 0.000146 0.000928 Chapter 7 - Rev. A, Page 27/45 • Comparing these values to the recommended limits in Table 7-2, we find that they within the r are all ecommended range. ______________________________________________________________________________ 7-24 ill ignore the steps near the bearings where the bending moments w mm dia. be 35 mm. Secondly, the 55 mm dia. is very thin, 10 Th tresses will not develop at the outer fibers so full stiffness will not iameter be 45 mm. tatics: L ort R r 100 140 210 275 315 Shaft analysis software or finite element software can be utilized if available. Here we will demonstrate how the problem can be simplified and solved using singularity functions. Deflection: First we w are lo . Thus let the 30 mm. e full bending s develop either. Thus, ignore this step and let the d S eft supp : R1 15 140) / 315 889   7(3 3. kN ight suppo t: 2 7(14 0R ) / 315  3.111 kN Determine the bending moment at each step. x(mm) 0 40 M(N · m) 0 155.56 388.89 544.44 326.67 124.44 0 I35 = (/64)(0.0354) = 366(10 ) m4, I 0 = 1.257(1 , I45 = 2. -7) m4 Plot M/I nction ) M N/m3) Step 7. -8 4 0-7) m4 013(10 as a fu of x. x(m /I (109 Slope Slope 0 0 52.8 0.04 2.112 0.04 1.2375 0.8745 21.86 4 1.162 11.617 05 0 15.457 34.78 0.21 1.623 0.21 2.6 0.977 -24.769 -9.312 0.275 0.99 0.275 1.6894 0.6994 -42.235 -17.47 0.315 0 – 30.942 – 0.1 3.09 0.1 1.932 – 19.325 – 0.14 2.705 0.14 2.7 – – Chapter 7 - Rev. A, Page 28/45 • The steps and the change of slopes are evaluated in the table. From these, the function M/I can be generated: 0 1 1 1 0 1 0 9 / 52.8 0.8745 0.04 21.86 0.04 1.162 0.1 11.617 0.1 34.78 0.14 0.977 0.21 9.312 0.21 0.6994 0.275 17.47 0.275 10 M I x x x x x x x x x x                     0 1 Integrate twice: 1 2226.4 0.8745 0.04 10.93 0.04 1.162x x x x       1 3 2 0.1 0.04 0.581 0.1 7 E dx x x    dy 2 2 1 2 1 2 9 1 23 5.81 0.1 17.39 0.14 0.977 0.21 4.655 0.21 0.6994 0.275 8.735 0.275 10 (1) 8.8 0.4373 0.04 3.643 x x x x x x C Ey x x x                   1.93 3 3 2 3 9 0.1 0.14 0.21 52 0. 0.2 0.275 10 x x x x        Boundary conditions: y yields C2 y = 0 at x = 0.315 m yields C1 = –0.295 25 N/m2. 3  2  1 2C x C  5.797 0.4885 x 1.5 21 0.3497 75 2.912   = 0 at x = 0 = 0; Equation (1) with C1 = –0.295 25 provides the slopes at the bearings and gear. The following table gives the results in the second column. The third column gives the results from a similar finite element model. The fourth column gives the results of a full model which models the 35 and 55 mm diameter steps. x (mm)  (rad) F.E. Model Full F.E. Model 0 –0.001 4260 –0.001 4270 –0.001 4160 140 –0.000 1466 –0.000 1467 –0.000 1646 315 0.001 3120 0.001 3280 0.001 3150 Chapter 7 - Rev. A, Page 29/45 • The main discrepancy between the results is at the gear location (x = 140 mm). The larger value in the full model is caused by the stiffer 55 mm diameter step. As was stated arlier, this step is not as stiff as modeling implicates, so the exact answer is somewhere between the full model and the simplified model which in any event is a small value. As xpected, modeling the 30 mm dia. as 35 mm does not affect the results much. can be seen that the allowable slopes at the bearings are exceeded. Thus, either the load ed or the shaft “beefed” up. If the allowable slope is 0.001 rad, then the aximum load should be Fmax = (0.001/0.001 426)7 = 4.91 kN. With a design factor this would be reduced further. To increase the stiffness of th , E 8 f deflection (at = 0) to determine a multiplier to be used for all diameters. e e It has to be reduc m e shaft apply q. (7-1 ) to the most o fending x     1/4 1/4 new old old / (1)(0.0014260) 1.093 n dy dxd d   orm a table: all slope 0.001 d  F Old d, mm 20.00 30.00 35.00 40.00 45.00 55.00 New ideal d, mm 21.86 32.79 38.26 43.72 49.19 60.12 Rounded up d, mm 22.00 34.00 40.00 44.00 50.00 62.00 Repeating the full finite element mo lts in del resu x 9 40 : –1 1 5 . stress concentrations and reduced shaft diameters, there are a number of at. A table of nominal stresses is given below. Note that torsion is only f the 7 kN load. Using  = 32 (d3) and  = 16T/(d3), 0 275 300 330 = 0:  = – .30  10-4 rad x = 1 mm  = .09  0-4 rad -4x = 31 mm:  = 8 65  10 rad This is well within our goal. Have the students try a goal of 0.0005 rad at the gears. Strength: Due to looklocations to to the right o M/ x (mm) 15 40 100 110 140 210  (MPa) 0 39.6 17.6 0 0 6 8.5 12.7 20.2 68.1 22.0 37.0 61.9 47.8 60.9 52.0  (MPa) 0 0 0 0 (MPa) 0 22.0 37.0 61.9 47.8 61.8 53.1 45.3 39.2 118.0 for Sy = 390 MPa Eq. (6-19): Table A-20 AISI 1020 CD steel: Sut = 470 MPa, At x = 210 mm: 0.2654.51(470) 0.883ak   Chapter 7 - Rev. A, Page 30/45 • Eq. (6-20): 0.107(40 / 7.62) 0.837bk   Eq. (6-18): Se = 0.883 (0.837)(0.5)(470) = 174 MPa D / d = 45 / 40 = 1.125, r / d = 2 / 40 = 0.05 Fig. A-15-8: Kts = 1.4 Fig. A-15-9: Kt = 1.9 Fig q = 0.75 Fig qs = 0.79 = 1 + 0.75(1.9 –1) = 1.68 ld check, from Eq. (7-11), with . 6-20: . 6-21: Eq. (6-32): Kf K f s = 1 + 0.79(1.4 – 1) = 1.32 Choosing DE-ASME Elliptic to inherently include the yie Mm = Ta = 0,       1/22 2 6 1.32(107)3 390 103 6 1.68(326.67)4 0.04 174 10n      1 16               At The von Mises stress is the highest but it comes from the steady torque only. Fig. 6-21: qs = 0.79 .42 – 1) = 1.33 1.98n  x = 330 mm: D / d = 30 / 20 = 1.5, r / d = 2 / 20 = 0.1 Fig. A-15-9: Kts = 1.42 Eq. (6-32): Kf s = 1 + 0.79(1 Eq. (7-11):       1 16 1.33(107) 3 6390 10n  3         n = 2.49 Note that since there is only a steady torque, Eq. (7-11) reduces to essentially the equivalent of the distortion energy failure theory. s at x = 210 mm, the changes discussed for the slope criterion will ______________________________________________________________________________ 7-2 se design tasks each student will travel different paths and almost all  The student gets a blank piece of paper, a statement of function, and some constraints Check the other locations. If worse-case i improve the strength issue. 5 and 7-26 With the details will differ. The important points are – explicit and implied. At this point in the course, this is a good experience.  It is a good preparation for the capstone design course. Chapter 7 - Rev. A, Page 31/45 •  The adequacy of their design must be demonstrated and possibly include a designer’s notebook.  Many of the fundaments of the course, based on this text and this course, are useful. . Don’t let the students create a time sink for themselves. Tell them how far you want ______________________________________________________________________________ 7-27 oblem. This problem is a learning experience. ollowing the task statement, the following guidance was added. ting the temptation of putting pencil to paper, and decide what the problem really is. ld implement it. The students’ initial reaction is that he/she does not know much from the problem lowly the realization sets in that they do know some important things that the designer did not. They knew how it failed, where it failed, and that the design wasn’t good enough; it was close, though. Also, a fix at the bearing seat lead-in could transfer the problem to the shoulder fillet, and the problem may not be tudents’ credit, they chose to keep the shaft geometry, and selected a new material to realize about ______________________________________________________________________________ -28 The student will find them useful and notice that he/she is doing it  them to go. This task was once given as a final exam pr F  Take the first half hour, resis  Take another twenty minutes to list several possible remedies.  Pick one, and show your instructor how you wou statement. Then, s solved. To many s twice the Brinell hardness. 7 In Eq. (7-22) set 4 2d d, 64 4 I A   to obtain 2 4l  d gE               (1) or 2 2 gE 4ld   (2) (a) From Eq. (1) and Table A-5     2 90.025 9.81(207)(10 ) .A 3 883 rad/s 0.6 4 76.5 10 ns              Chapter 7 - Rev. A, Page 32/45 • (b) From Eq. (1), we observe that the critical speed is linearly proportional to the diameter. Thus, to double the critical speed, we should double the diameter to d = 50 mm. Ans. (c) From Eq. (2), 2 d gl  4 E l   Since d / l is the same regardless of the scale, constant 0.6(883) 529.8l    529.8 1766 rad/s .A 0.3 ns   Thus the first critical speed doubles. ______________________________________________________________________________ 7-29 From Prob. 7-28, 883 rad/s       4 2 8 4 44.909 10 m , 1.917 10 m , 7.65 10 N/mA I     3    9 4 4207(10 ) Pa, 4.909 10 7.65 10 (0.6) 22.53 NE A l    w One element: Eq. (7-24):         2 2 2 6 11 9 8 0.3(0.3) 0.6 0.3 0.3 1.134 10 m/N 6(207) 10 (1.917) 10 (0.6)           6 51 1 11 22.53(1.134) 10 2.555 10 my     w  2 11 6.528 10y  0    5 422.53(2.555) 10 5.756 10y    w    2 1022.53(6.528) 10 1.471 10y    w 8     4 1 2 8 5.756 10 9.81 620 rad/s 1.471 10 yg y         w w (30% low) Two elements: Chapter 7 - Rev. A, Page 33/45 •         2 2 2 7 11 22 9 8 0.45(0.15) 0.6 0.45 0.15 6.379 10 m/N 6(207) 10 (1.917) 10 (0.6)                2 2 2 7 12 21 9 8 0.15(0.15)(0.6 0.15 0.15 ) 4.961 10 m/N 6(207) 10 (1.917) 10 (0.6)               7 71 2 1 11 2 12 11.265(6.379) 10 11.265(4.961) 10 1.277 10 my y         w w 5  2 2 102 10 1 2 1.63y y     5 42(11.265)(1.277) 10 2.877 10y    w    2 102(11.265)(1.632) 10 3.677 10y    w 9     4 1 9 2.877 10 9.81 876 rad/s 3.677 10            (0.8% low) Three elements:         2 2 2 7 11 33 9 8 0.5(0.1) 0.6 0.5 0.1 3.500 10 m/N 6(207) 10 (1.917) 10 (0.6)                  2 2 2 6 22 9 8 0.3(0.3) 0.6 0.3 0.3 1.134 10 m/N 6(207) 10 (1.917) 10 (0.6)                2 2 2 7 12 32 9 8 0.3(0.1) 0.6 0.3 0.1 5.460 10 m/N 6(207) 10 (1.917) 10 (0.6)                  2 2 2 7 13 9 8 0.1(0.1) 0.6 0.1 0.1 2.380 10 m/N 6(207) 10 (1.917) 10 (0.6)               7 7 71 7.51 3.500 10 5.460 10 2.380 10 8.516 10        6y        7 6 72 7.51 5.460 10 1.134 10 10 1.672 10y         55.460        7 7 73 7.51 2.380 10 5.460 10 3.500 10 8.516 10y         6        6 5 6 47.51 8.516 10 1.672 10 8.516 10 2.535 10y          w         2 2 22 6 5 6 97.51 8.516 10 1.672 10 8.516 10 3.189 10y                  w Chapter 7 - Rev. A, Page 34/45 •     42.535 10 9.81 883 ra     1 93.189 10   d/s 7-28. The point was to show that convergence is rapid using a static deflection beam equation. The method works because:  If a deflection curve is chosen which meets the boundary conditions of moment- free and deflection-free ends, as in this problem, the strain energy is not very sensitive to the equation used. ation is available, and meets the moment-free and deflection-free ends, it works. ______________________________________________________________________________ 7-30 (a) For two bodies, Eq. (7-26) is The result is the same as in Prob.  Since the static bending equ 2 1 11( 1/ ) 0 m   2 12 2 1 21 2 22( 1/ ) m m m       Expanding the determinant yields, 21  1 11 2 22 1 2 11 22 12 212 2 1 1( ) ( ) 0m m m m                     (1) Eq. (1) has two roots 2 21 21 / and 1 / .  Thus 2 2 2 2 1 2 1 1 1 1   0             or, 21 1  2 2 2 2 2 2 1 2 1 2 1 1 1 1 0                              (2) Equate the third terms of Eqs. (1) and (2), which must be identical. 2 1 2 11 22 12 21 1 1 2 11 22 12 212 2 2 1 1 1( ) ( )m m m m 1 2 2                  and it follows that Chapter 7 - Rev. A, Page 35/45 • 2 2 1 1 11 22 12 . ( ) Ans      w w 2 21 1 g  (b) In Ex. 7-5, part (b), the first critical speed of the two-disk shaft (w1 = 35 lbf, w 2 = 55 lbf) is 1 = 124.8 rad/s. From part (a), using influence coefficients,   2 2 2 8 1 386 466 rad/s . 124.8 35(55) 2.061(3.534) 2.234 10 Ans       ______________________________________________________________________________ 7-31 In Eq. (7-22), for 1, the term /I A appears. For a hollow uniform diameter shaft,       4 2 2 2 2 2 2o o i o id d d d dI d d      4 1 2 22 2 / 64 1 1 16 4/ 4 i o i o io i d A d dd d    This means that when a solid shaft is hollowed out, the critical speed increases beyond solid shaft of the same size. By how much? that of the 22 2 2 1 (1/ 4) o i i oo dd       The possible values of are 0 ,i i od d d (1/ 4) d d d     so the range of the critical speeds is 1 1 0  to about 1 1 1  or from 1 1to 2 . .Ans ______________________________________________________________________________ 7-32 All steps w b g t t pr s et. Programming both loads will enable the user to first set the left load to 1, the right load to 0 and calculate 11 and 21. Then set the left load to 0 and the right to 1 to get 12 and 22. The spreadsheet shows the 11 and 21 calculation. A table for M / I vs. x is easy to make. First, draw the bending-moment diagram as shown with the data. x 0 1 2 3 4 5 6 7 8  ill e modeled using sin ulari y func ions with a s ead he M 0 0.875 1.75 1.625 1.5 1.375 1.25 1.125 1 x 9 10 11 12 13 14 15 16 M 0.875 0.75 0.625 0.5 0.375 0.25 0.125 0 Chapter 7 - Rev. A, Page 36/45 • The second-area moments are:  4 410 1 in and 15 16 in, 2 / 64 0.7854 inx x I           4 4 2 4 4 3 1 9 in , 2.472 / 64 1.833 in 9 15 in , 2.763 / 64 2.861 in x I x I           Divide M by I at the key points x = 0, 1, 2, 9, 14, 15, and 16 in and plot x 0 1 1 2 2 3 4 5 6 7 8 M/I 0 1.1141 0.4774 0.9547 0.9547 0.8865 0.8183 0.7501 0.6819 0.6137 0.5456 x 9 9 10 11 12 13 14 14 15 15 16 M/I 0.4774 0.3058 0.2621 0.2185 0.1748 0.1311 0.0874 0.0874 0.0437 0.1592 0 From this diagram, one can see where changes in value (steps) and slope occur. Using a spreadsheet, one can form a table of these changes. An example of a step is, at x = 1 in, M/I goes from 0.875/0.7854 = 1.1141 lbf/in3 to 0.875/1.833 = 0.4774 lbf/in3, a step change of 0.4774  1.1141 =  0.6367 lbf/in3. A slope change also occurs at at x = 1 in. Chapter 7 - Rev. A, Page 37/45 • The slope for 0  x  1 in is 1.1141/1 = 1.1141 lbf/in2, which changes to (0.9547  0.4774)/1 = 0.4774 lbf/in2, a change of 0.4774  1.1141 =  0.6367 lbf/in2. Following this approach, a table is made of all the changes. The table shown indicates the column letters and row numbers for the spreadsheet. A B C D E F 1 x M M/I step Slope  Slope 2 1a 0.875 1.114085 0.000000 1.114085 0.000000 3 1b 0.875 0.477358 -0.636727 0.477358 -0.636727 4 2 1.75 0.954716 0.000000 0.477358 0.000000 5 2 1.75 0.954716 0.000000 -0.068194 -0.545552 6 9a 0.875 0.477358 0.000000 -0.068194 0.000000 7 9b 0.875 0.305854 -0.171504 -0.043693 0.024501 8 14 0.25 0.087387 0.000000 -0.043693 0.000000 9 14 0.25 0.087387 0.000000 -0.043693 0.000000 10 15a 0.125 0.043693 0.000000 -0.043693 0.000000 11 15b 0.125 0.159155 0.115461 -0.159155 -0.115461 12 16 0 0.000000 0.000000 -0.159155 0.000000 The equation for M / I in terms of the spreadsheet cell locations is: 0 1 1 0 1 0 / E2 ( ) D3 1 F3 1 F5 2 D7 9 F7 9 D11 15 F11 15 M I x x x x x x x x                1 5 5   Integrating twice gives the equation for Ey. Assume the shaft is steel. Boundary conditions y = 0 at x = 0 and at x = 16 inches provide integration constants (C1 =  4.906 lbf/in and C2 = 0). Substitution back into the deflection equation at x = 2 and 14 in provides the  ’s. The results are: 11 = 2.917(10–7) and 12 = 1.627(10–7). Repeat for F1 = 0 and F2 = 1, resulting in 21 = 1.627(10–7) and 22 = 2.231(10–7). This can be verified by finite element analysis. 7 7 1 7 7 2 2 10 2 10 1 2 4 2 9 18(2.917)(10 ) 32(1.627)(10 ) 1.046(10 ) 18(1.627)(10 ) 32(2.231)(10 ) 1.007(10 ) 1.093(10 ), 1.014(10 ) 5.105(10 ), 5.212(10 ) y y y y y y                   w w Neglecting the shaft, Eq. (7-23) gives 4 1 9 5.105(10 )386 6149 rad/s or 58 720 rev/min . 5.212(10 ) Ans    Chapter 7 - Rev. A, Page 38/45 • Without the loads, we will model the shaft using 2 elements, one between 0  x  9 in, and one between 0  x  16 in. As an approximation, we will place their weights at x = 9/2 = 4.5 in, and x = 9 + (16  9)/2 = 12.5 in. From Table A-5, the weight density of steel is  = 0.282 lbf/in3. The weight of the left element is    2 2 21 0.282 2 1 2.472 8 11.7 lbf4 4d l              w The right element is    2 22 0.282 2.763 6 2 1 11.0 lbf4           w The spreadsheet can be easily modified to give     7 711 12 21 229.605 10 , 5.718 10 , 5.472 10        7    5 51 21.753 10 , 1.271 10y y      2 10 21 23.072 10 , 1.615 10y y   10    4 23.449 10 , 5.371 10y y   w w 9     4 1 9 3.449 10 386 4980 rad/s 5.371 10            A finite element model of the exact shaft gives 1 = 5340 rad/s. The simple model is 6.8% low. Combination: Using Dunkerley’s equation, Eq. (7-32): 12 2 2 1 1 1 1 3870 rad/s . 6149 4980 Ans     ______________________________________________________________________________ 7-33 We must not let the basis of the stress concentration factor, as presented, impose a view- point on the designer. Table A-16 shows Kts as a decreasing monotonic as a function of a/D. All is not what it seems. Let us change the basis for data presentation to the full section rather than the net section. 0 0ts tsK K     Chapter 7 - Rev. A, Page 39/45 • 3 3 32 32 ts ts T TK K AD D         Therefore ts ts KK A   Form a table: tsK  has the following attributes:  It exhibits a minimum;  It changes little over a wide range;  Its minimum is a stationary point minimum at a / D  0.100;  Our knowledge of the minima location is 0.075 ( / ) 0.125a D  We can form a design rule: In torsion, the pin diameter should be about 1/10 of the shaft diameter, for greatest shaft capacity. However, it is not catastrophic if one forgets the rule. ______________________________________________________________________________ 7-34 From the solution to Prob. 3-72, the torque to be transmitted through the key from the gear to the shaft is T = 2819 lbf·in. From Prob. 7-19, the nominal shaft diameter supporting the gear is 1.00 in. From Table 7-6, a 0.25 in square key is appropriate for a 1.00 in shaft diameter. The force applied to the key is 2819 5638 lbf 1.00 / 2 TF r    Selecting 1020 CD steel for the key, with Sy = 57 kpsi, and using the distortion-energy theory, Ssy = 0.577 Sy = (0.577)(57) = 32.9 kpsi. Failure by shear across the key: Chapter 7 - Rev. A, Page 40/45 •     1.1 5638 0.754 in / 0.25 32 900 sy sy sy F F A tl S S nFn l F tl tS           Failure by crushing:  / 2 F F A t l           3 2 5638 1.12 0.870 in 2 / 0.25 57 10 y y y S S Fnn l F tl tS       Select ¼-in square key, 7/8 in long, 1020 CD steel. Ans. ______________________________________________________________________________ 7-35 From the solution to Prob. 3-73, the torque to be transmitted through the key from the gear to the shaft is T = 3101 N·m. From Prob. 7-21, the nominal shaft diameter supporting the gear is 50 mm. To determine an appropriate key size for the shaft diameter, we can either convert to inches and use Table 7-6, or we can look up standard metric key sizes from the internet or a machine design handbook. It turns out that the recommended metric key for a 50 mm shaft is 14 x 9 mm. Since the problem statement specifies a square key, we will use a 14 x 14 mm key. For comparison, using Table 7-6 as a guide, for d = 50 mm = 1.97 in, a 0.5 in square key is appropriate. This is equivalent to 12.7 mm. A 14 x 14 mm size is conservative, but reasonable after rounding up to standard sizes. The force applied to the key is  33101 124 10 N0.050 / 2 TF r    Selecting 1020 CD steel for the key, with Sy = 390 MPa, and using the distortion-energy theory, Ssy = 0.577 Sy = 0.577(390) = 225 MPa. Failure by shear across the key:          3 6 1.1 124 10 0.0433 m 43.3 mm / 0.014 225 10 sy sy sy F F A tl S S nFn l F tl tS            Failure by crushing: Chapter 7 - Rev. A, Page 41/45 •  / 2 F F A t    l           3 6 2 124 10 1.12 0.0500 m 50.0 mm 2 / 0.014 390 10 y y y S S Fnn l F tl tS        Select 14 mm square key, 50 mm long, 1020 CD steel. Ans. ______________________________________________________________________________ 7-36 Choose basic size D = d = 15 mm. From Table 7-9, a locational clearance fit is designated as 15H7/h6. From Table A-11, the tolerance grades are D = 0.018 mm and d = 0.011 mm. From Table A-12, the fundamental deviation is F = 0 mm. Hole: Eq. (7-36): Dmax = D + D = 15 + 0.018 = 15.018 mm Ans. Dmin = D = 15.000 mm Ans. Shaft: Eq. (7-37): dmax = d + F = 15.000 + 0 = 15.000 mm Ans. dmin = d + F – d = 15.000 + 0 – 0.011 = 14.989 mm Ans. ______________________________________________________________________________ 7-37 Choose basic size D = d = 1.75 in. From Table 7-9, a medium drive fit is designated as H7/s6. From Table A-13, the tolerance grades are D = 0.0010 in and d = 0.0006 in. From Table A-14, the fundamental deviation is F = 0.0017 in. Hole: Eq. (7-36): Dmax = D + D = 1.75 + 0.0010 = 1.7510 in Ans. Dmin = D = 1.7500 in Ans. Shaft: Eq. (7-38): dmin = d + F = 1.75 + 0.0017 = 1.7517 in Ans. dmax = d + F + d = 1.75 + 0.0017 + 0.0006 = 1.7523 in Ans. ______________________________________________________________________________ 7-38 Choose basic size D = d = 45 mm. From Table 7-9, a sliding fit is designated as H7/g6. From Table A-11, the tolerance grades are D = 0.025 mm and d = 0.016 mm. From Table A-12, the fundamental deviation is F = –0.009 mm. Hole: Eq. (7-36): Dmax = D + D = 45 + 0.025 = 45.025 mm Ans. Dmin = D = 45.000 mm Ans. Shaft: Eq. (7-37): dmax = d + F = 45.000 + (–0.009) = 44.991 mm Ans. dmin = d + F – d = 45.000 + (–0.009) – 0.016 = 44.975 mm Ans. ______________________________________________________________________________ Chapter 7 - Rev. A, Page 42/45 • 7-39 Choose basic size D = d = 1.250 in. From Table 7-9, a close running fit is designated as H8/f7. From Table A-13, the tolerance grades are D = 0.0015 in and d = 0.0010 in. From Table A-14, the fundamental deviation is F = –0.0010 in. Hole: Eq. (7-36): Dmax = D + D = 1.250 + 0.0015 = 1.2515 in Ans. Dmin = D = 1.2500 in Ans. Shaft: Eq. (7-37): dmax = d + F = 1.250 + (–0.0010) = 1.2490 in Ans. dmin = d + F – d = 1.250 + (–0.0010) – 0.0010 = 1.2480 in Ans. ______________________________________________________________________________ 7-40 Choose basic size D = d = 35 mm. From Table 7-9, a locational interference fit is designated as H7/p6. From Table A-11, the tolerance grades are D = 0.025 mm and d = 0.016 mm. From Table A-12, the fundamental deviation is F = 0.026 mm. Hole: Eq. (7-36): Dmax = D + D = 35 + 0.025 = 35.025 mm Dmin = D = 35.000 mm The bearing bore specifications are within the hole specifications for a locational interference fit. Now find the necessary shaft sizes. Shaft: Eq. (7-38): dmin = d + F = 35 + 0.026 = 35.026 mm Ans. dmax = d + F + d = 35 + 0.026 + 0.016 = 35.042 mm Ans. ______________________________________________________________________________ 7-41 Choose basic size D = d = 1.5 in. From Table 7-9, a locational interference fit is designated as H7/p6. From Table A-13, the tolerance grades are D = 0.0010 in and d = 0.0006 in. From Table A-14, the fundamental deviation is F = 0.0010 in. Hole: Eq. (7-36): Dmax = D + D = 1.5000 + 0.0010 = 1.5010 in Dmin = D = 1.5000 in The bearing bore specifications exactly match the requirements for a locational interference fit. Now check the shaft. Shaft: Eq. (7-38): dmin = d + F = 1.5000 + 0.0010 = 1.5010 in dmax = d + F + d = 1.5000 + 0.0010 + 0.0006 = 1.5016 in Chapter 7 - Rev. A, Page 43/45 • The shaft diameter of 1.5020 in is greater than the maximum allowable diameter of 1.5016 in, and therefore does not meet the specifications for the locational interference fit. Ans. ______________________________________________________________________________ 7-42 (a) Basic size is D = d = 35 mm. Table 7-9: H7/s6 is specified for medium drive fit. Table A-11: Tolerance grades are D = 0.025 mm and d = 0.016 mm. Table A-12: Fundamental deviation is 0.043 mm.F   Eq. (7-36): Dmax = D + D = 35 + 0.025 = 35.025 mm Dmin = D = 35.000 mm Eq. (7-38): dmin = d + F = 35 + 0.043 = 35.043 mm Ans. dmax = d + F + d = 35 + 0.043 + 0.016 = 35.059 mm Ans. (b) Eq. (7-42): min min max 35.043 35.025 0.018 mmd D      Eq. (7-43): max max min 35.059 35.000 0.059 mmd D      Eq. (7-40):   2 2 2 2max max 3 2 22 o i o i d d d dEp d d d                9 2 2 2 23 207 10 0.059 60 35 35 0 115 MPa . 60 02 35 Ans            2 2 2 2min min 3 2 22 o i o i d d d dEp d d d                9 2 2 2 23 207 10 0.018 60 35 35 0 35.1 MPa . 60 02 35 Ans          (c) For the shaft: Eq. (7-44): ,shaft 115 MPat p     Eq. (7-46): ,shaft 115 MPar p     Eq. (5-13):  1/22 21 1 2 2        1/22 2( 115) ( 115)( 115) ( 115) 115 MPa          / 390 /115 3.4 .yn S Ans    For the hub: Eq. (7-45): 2 2 2 2 ,hub 2 2 2 2 60 35115 234 MPa 60 35 o t o d dp d d           Eq. (7-46): ,hub 115 MPar p     Chapter 7 - Rev. A, Page 44/45 • Eq. (5-13):  1/22 21 1 2 2        1/22 2(234) (234)( 115) ( 115) 308 MPa        / 600 / 308 1.9 .yn S Ans    (d) A value for the static coefficient of friction for steel to steel can be obtained online or from a physics textbook as approximately f = 0.8. Eq. (7-49) 2min( / 2)T f p ld  6 2( / 2)(0.8)(35.1) 10 (0.050)(0.035) 2700 N m .Ans   ______________________________________________________________________________ Chapter 7 - Rev. A, Page 45/45 • Chapter 8 Note to the Instructor for Probs. 8-41 to 8-44. These problems, as well as many others in this chapter are best implemented using a spreadsheet. 8-1 (a) Thread depth= 2.5 mm Ans. Width = 2.5 mm Ans. dm = 25 - 1.25 - 1.25 = 22.5 mm dr = 25 - 5 = 20 mm l = p = 5 mm Ans. (b) Thread depth = 2.5 mm Ans. Width at pitch line = 2.5 mm Ans. dm = 22.5 mm dr = 20 mm l = p = 5 mm Ans. ______________________________________________________________________________ 8-2 From Table 8-1, 1.226 869 0.649 519 1.226 869 0.649 519 0.938 194 2 r m d d p d d p d p d pd d          p 2 2( 0.938 194 ) . 4 4t dA d p    Ans ______________________________________________________________________________ 8-3 From Eq. (c) of Sec. 8-2, tan 1 tan tan 2 2 1 tan R R m m R fP F f P d Fd fT f            0 / (2 ) 1 tan 1 tantan . / 2 tan tanR m T Fl f fe A T Fd f f ns            Chap. 8 Solutions - Rev. A, Page 1/69 • Using f = 0.08, form a table and plot the efficiency curve. , deg. e 0 0 0 0.678 20 0.796 30 0.838 40 0.8517 45 0.8519 ______________________________________________________________________________ 8-4 Given F = 5 kN, l = 5 mm, and dm = d  p/2 = 25  5/2 = 22.5 mm, the torque required to raise the load is found using Eqs. (8-1) and (8-6)          5 22.5 5 0.09 22.5 5 0.06 45 15.85 N m . 2 22.5 0.09 5 2R T A          ns The torque required to lower the load, from Eqs. (8-2) and (8-6) is          5 22.5 0.09 22.5 5 5 0.06 45 7.83 N m . 2 22.5 0.09 5 2L T A          ns Since TL is positive, the thread is self-locking. From Eq.(8-4) the efficiency is     5 5 0.251 . 2 15.85 e Ans    ______________________________________________________________________________ 8-5 Collar (thrust) bearings, at the bottom of the screws, must bear on the collars. The bottom segment of the screws must be in compression. Whereas, tension specimens and their grips must be in tension. Both screws must be of the same-hand threads. ______________________________________________________________________________ 8-6 Screws rotate at an angular rate of 1720 28.67 rev/min 60 n   Chap. 8 Solutions - Rev. A, Page 2/69 • (a) The lead is 0.25 in, so the linear speed of the press head is V = 28.67(0.25) = 7.17 in/min Ans. (b) F = 2500 lbf/screw o 2 0.25 / 2 1.875 in sec 1 / cos(29 / 2) 1.033 md       Eq. (8-5): 2500(1.875) 0.25 (0.05)(1.875)(1.033) 221.0 lbf · in 2 (1.875) 0.05(0.25)(1.033)R T         Eq. (8-6): 2500(0.08)(3.5 / 2) 350 lbf · in 350 221.0 571 lbf · in/screw 571(2) 20.04 lbf · in 60(0.95) 20.04(1720) 0.547 hp . 63 025 63 025 c total motor T T T TnH A           ns ______________________________________________________________________________ 8-7 Note to the Instructor: The statement for this problem in the first printing of this edition was vague regarding the effective handle length. For the printings to follow the statement “The overall length is 4.25 in.” will be replaced by “ A force will be applied to the handle at a radius of 123 in from the screw centerline.” We apologize if this has caused any inconvenience. 3 3 3.5 in 3.5 3 33.5 3.125 8 8 41 kpsi 32 32(3.125) 41 000 (0.1875) 8.49 lbf y y L T F M L F F S M FS d F                           F ns 3.5(8.49) 29.7 lbf · in .T A  (b) Eq. (8-5), 2 = 60 , l = 1/10 = 0.1 in, f = 0.15, sec  = 1.155, p = 0.1 in Chap. 8 Solutions - Rev. A, Page 3/69 •   clamp clamp clamp 3 0.649 519 0.1 0.6850 in 4 (0.6850) 0.1 (0.15)(0.6850)(1.155) 2 (0.6850) 0.15(0.1)(1.155) 0.075 86 29.7 392 lbf . 0.075 86 0.075 86 m R R R d F T T F TF A                ns (c) The column has one end fixed and the other end pivoted. Base the decision on the mean diameter column. Input: C = 1.2, D = 0.685 in, A = (0.6852)/4 = 0.369 in2, Sy = 41 kpsi, E = 30(106) psi, L = 6 in, k = D/4 =0.171 25 in, L/k = 35.04. From Eq. (4-45),     1/21/2 2 62 1 2 1.2 30 102 131.7 41 000y l CE k S                    From Eq. (4-46), the limiting clamping force for buckling is         2 clamp cr 23 3 3 6 1 2 41 10 10.369 41 10 35.04 14.6 10 lbf 2 1.2 30 10 y y S lF P A S k CE Ans                            (d) This is a subject for class discussion. ______________________________________________________________________________ 8-8 T = 8(3.5) = 28 lbf  in 3 1 0.6667 in 4 12m d    l = 1 6 = 0.1667 in,  = 029 2 = 14.50, sec 14.50 = 1.033 From Eqs. (8-5) and (8-6)            total 0.1667 0.15 0.6667 1.033 0.15 10.6667 0.1542 2 0.6667 0.15 0.1667 1.033 2 FFT F          28 182 lbf . 0.1542 F Ans  _____________________________________________________________________________ Chap. 8 Solutions - Rev. A, Page 4/69 • 8-9 dm = 1.5  0.25/2 = 1.375 in, l = 2(0.25) = 0.5 in From Eq. (8-1) and Eq. (8-6)    3 32.2 10 (1.375) 2.2 10 (0.15)(2.25)0.5 (0.10)(1.375) 2 (1.375) 0.10(0.5) 2 330 371 701 lbf · in RT            Since n = V/l = 2/0.5 = 4 rev/s = 240 rev/min so the power is  701 240 2.67 hp . 63 025 63 025 TnH A   ns ______________________________________________________________________________ 8-10 dm = 40  4 = 36 mm, l = p = 8 mm From Eqs. (8-1) and (8-6) 36 8 (0.14)(36) 0.09(100) 2 (36) 0.14(8) 2 (3.831 4.5) 8.33 N · m ( in kN) 2 2 (1) 2 rad/s 3000 477 N · m 2 477 57.3 kN . 8.33 F FT F F F n H T HT F Ans                            57.3(8) 0.153 . 2 2 (477) Fle A T     ns ______________________________________________________________________________ 8-11 (a) Table A-31, nut height H = 12.8 mm. L ≥ l + H = 2(15) + 12.8 = 42.8 mm. Rounding up, L = 45 mm Ans. (b) From Eq. (8-14), LT = 2d + 6 = 2(14) +6 = 34 mm From Table 8-7, ld = L  LT = 45 34 = 11 mm, lt = l  ld = 2(15)  11 = 19 mm, Ad =  (142) / 4 = 153.9 mm2. From Table 8-1, At = 115 mm2. From Eq. (8-17) Chap. 8 Solutions - Rev. A, Page 5/69 •       153.9 115 207 874.6 MN/m . 153.9 19 115 11 d t b d t t d A A Ek A A l A l      ns (c) From Eq. (8-22), with l = 2(15) = 30 mm           0.5774 207 140.5774 3 116.5 MN/m . 0.5774 0.5 0.5774 30 0.5 142ln 5 2ln 50.5774 2.5 0.5774 30 2.5 14 mk Ed Ans l d l d                8-12 (a) Table A-31, nut height H = 12.8 mm. Table A-33, washer thickness t = 3.5 mm. Thus, the grip is l = 2(15) + 3.5 = 33.5 mm. L ≥ l + H = 33.5 + 12.8 = 46.3 mm. Rounding up L = 50 mm Ans. (b) From Eq. (8-14), LT = 2d + 6 = 2(14) +6 = 34 mm From Table 8-7, ld = L  LT = 50 34 = 16 mm, lt = l  ld = 33.5  16 = 17.5 mm, Ad =  (142) / 4 = 153.9 mm2. From Table 8-1, At = 115 mm2. From Eq. (8-17)       153.9 115 207 808.2 MN/m . 153.9 17.5 115 16 d t b d t t d A A Ek A A l A l      ns (c) From Eq. (8-22)           0.5774 207 140.5774 2 969 MN/m . 0.5774 0.5 0.5774 33.5 0.5 142ln 5 2ln 50.5774 2.5 0.5774 33.5 2.5 14 m Edk A l d l d                ns ______________________________________________________________________________ Chap. 8 Solutions - Rev. A, Page 6/69 • 8-13 (a) Table 8-7, l = h + d /2 = 15 + 14/2 = 22 mm. L ≥ h + 1.5d = 36 mm. Rounding up L = 40 mm Ans. (b) From Eq. (8-14), LT = 2d + 6 = 2(14) +6 = 34 mm From Table 8-7, ld = L  LT = 40 34 = 6 mm, lt = l  ld = 22  6 = 16 mm Ad =  (142) / 4 = 153.9 mm2. From Table 8-1, At = 115 mm2. From Eq. (8-17)       153.9 115 207 1 162.2 MN/m . 153.9 16 115 6 d t b d t t d A A Ek A A l A l      ns (c) From Eq. (8-22), with l = 22 mm           0.5774 207 140.5774 3 624.4 MN/m . 0.5774 0.5 0.5774 22 0.5 142ln 5 2ln 50.5774 2.5 0.5774 22 2.5 14 m Edk Ans l d l d                ______________________________________________________________________________ 8-14 (a) From Table A-31, the nut height is H = 7/16 in. L ≥ l + H = 2 + 1 + 7/16 = 3 7/16 in. Rounding up, L = 3.5 in Ans. (b) From Eq. (8-13), LT = 2d + 1/4 = 2(0.5) + 0.25 = 1.25 in From Table 8-7, ld = L  LT = 3.5  1.25 = 2.25 in, lt = l  ld = 3  2.25 = 0.75 in Ad =  (0.52)/4 = 0.1963 in2. From Table 8-2, At = 0.1419 in2. From Eq. (8-17)       0.1963 0.1419 30 1.79 Mlbf/in . 0.1963 0.75 0.1419 2.25 d t b d t t d A A Ek A A l A l      ns Chap. 8 Solutions - Rev. A, Page 7/69 • (c) Top steel frustum: t = 1.5 in, d = 0.5 in, D = 0.75 in, E = 30 Mpsi. From Eq. (8-20)           1 0.5774 30 0.5 22.65 Mlbf/in 1.155 1.5 0.75 0.5 0.75 0.5 ln 1.155 1.5 0.75 0.5 0.75 0.5 k              Lower steel frustum: t = 0.5 in, d = 0.5 in, D = 0.75 + 2(1) tan 30 = 1.905 in, E = 30 Mpsi. Eq. (8-20)  k2 = 210.7 Mlbf/in Cast iron: t = 1 in, d = 0.5 in, D = 0.75 in, E = 14.5 Mpsi (Table 8-8). Eq. (8-20)  k3 = 12.27 Mlbf/in From Eq. (8-18) km = (1/k1 + 1/k2 +1/k3)1 = (1/22.65 + 1/210.7 + 1/12.27)1 = 7.67 Mlbf/in Ans. 8-15 (a) From Table A-32, the washer thickness is 0.095 in. Thus, l = 2 + 1 + 2(0.095) = 3.19 in. From Table A-31, the nut height is H = 7/16 in. L ≥ l + H = 3.19 + 7/16 = 3.63 in. Rounding up, L = 3.75 in Ans. (b) From Eq. (8-13), LT = 2d + 1/4 = 2(0.5) + 0.25 = 1.25 in From Table 8-7, ld = L  LT = 3.75  1.25 = 2.5 in, lt = l  ld = 3.19  2.5 = 0.69 in Ad =  (0.52)/4 = 0.1963 in2. From Table 8-2, At = 0.1419 in2. From Eq. (8-17) Chap. 8 Solutions - Rev. A, Page 8/69 •       0.1963 0.1419 30 1.705 Mlbf/in . 0.1963 0.69 0.1419 2.5 d t b d t t d A A Ek A A l A l      ns (c) Each steel washer frustum: t = 0.095 in, d = 0.531 in (Table A-32), D = 0.75 in, E = 30 Mpsi. From Eq. (8-20)           1 0.5774 30 0.531 89.20 Mlbf/in 1.155 0.095 0.75 0.531 0.75 0.531 ln 1.155 0.095 0.75 0.531 0.75 0.531 k              Top plate, top steel frustum: t = 1.5 in, d = 0.5 in, D = 0.75 + 2(0.095) tan 30 = 0.860 in, E = 30 Mpsi. Eq. (8-20)  k2 = 28.99 Mlbf/in Top plate, lower steel frustum: t = 0.5 in, d = 0.5 in, D = 0.860 + 2(1) tan 30 = 2.015 in, E = 30 Mpsi. Eq. (8-20)  k3 = 234.08 Mlbf/in Cast iron: t = 1 in, d = 0.5 in, D = 0.75 + 2(0.095) tan 30 = 0.860 in, E = 14.5 Mpsi (Table 8-8). Eq. (8-20)  k4 = 15.99 Mlbf/in From Eq. (8-18) km = (2/k1 + 1/k2 +1/k3+1/k4)1 = (2/89.20 + 1/28.99 + 1/234.08 + 1/15.99)1 = 8.08 Mlbf/in Ans. ______________________________________________________________________________ 8-16 (a) From Table 8-7, l = h + d /2 = 2 + 0.5/2 = 2.25 in. L ≥ h + 1.5 d = 2 + 1.5(0.5) = 2.75 in Ans. (b) From Table 8-7, LT = 2d + 1/4 = 2(0.5) + 0.25 = 1.25 in Chap. 8 Solutions - Rev. A, Page 9/69 • ld = L  LT = 2.75  1.25 = 1.5 in, lt = l  ld = 2.25  1.5 = 0.75 in Ad =  (0.52)/4 = 0.1963 in2. From Table 8-2, At = 0.1419 in2. From Eq. (8-17)       0.1963 0.1419 30 2.321 Mlbf/in . 0.1963 0.75 0.1419 1.5 d t bk A l  d t t d A A E Ans A l     (c) Top steel frustum: t = 1.125 in, d = 0.5 in, D = 0.75 in, E = 30 Mpsi. From Eq. (8-20)           1 0.5774 30 0.5 24.48 Mlbf/in 1.155 1.125 0.75 0.5 0.75 0.5 ln 1.155 1.125 0.75 0.5 0.75 0.5 k              Lower steel frustum: t = 0.875 in, d = 0.5 in, D = 0.75 + 2(0.25) tan 30 = 1.039 in, E = 30 Mpsi. Eq. (8-20)  k2 = 49.36 Mlbf/in Cast iron: t = 0.25 in, d = 0.5 in, D = 0.75 in, E = 14.5 Mpsi (Table 8-8). Eq. (8-20)  k3 = 23.49 Mlbf/in From Eq. (8-18) km = (1/k1 + 1/k2 +1/k3)1 = (1/24.48 + 1/49.36 + 1/23.49)1 = 9.645 Mlbf/in Ans. ______________________________________________________________________________ 8-17 a) Grip, l = 2(2 + 0.095) = 4.19 in. L ≥ 4.19 + 7/16 = 4.628 in. Rounding up, L = 4.75 in Ans. Chap. 8 Solutions - Rev. A, Page 10/69 • (b) From Eq. (8-13), LT = 2d + 1/4 = 2(0.5) + 0.25 = 1.25 in From Table 8-7, ld = L  LT = 4.75  1.25 = 3.5 in, lt = l  ld = 4.19  3.5 = 0.69 in Ad =  (0.52)/4 = 0.1963 in2. From Table 8-2, At = 0.1419 in2. From Eq. (8-17)       0.1963 0.1419 30 1.322 Mlbf/in . 0.1963 0.69 0.1419 3.5 d t bk A l  d t t d A A E Ans Al     (c) Upper and lower halves are the same. For the upper half, Steel frustum: t = 0.095 in, d = 0.531 in, D = 0.75 in, and E = 30 Mpsi. From Eq. (8-20)           1 0.5774 30 0.531 89.20 Mlbf/in 1.155 0.095 0.75 0.531 0.75 0.531 ln 1.155 0.095 0.75 0.531 0.75 0.531 k              Aluminum: t = 2 in, d = 0.5 in, D =0.75 + 2(0.095) tan 30 = 0.860 in, and E = 10.3 Mpsi. Eq. (8-20)  k2 = 9.24 Mlbf/in For the top half, = (1/kmk 1 + 1/k2) 1 = (1/89.20 + 1/9.24)1 = 8.373 Mlbf/in Since the bottom half is the same, the overall stiffness is given by km = (1/ + 1/ k )mk m 1 = km /2 = 8.373/2 = 4.19 Mlbf/in Ans ______________________________________________________________________________ 8-18 (a) Grip, l = 2(2 + 0.095) = 4.19 in. L ≥ 4.19 + 7/16 = 4.628 in. Rounding up, L = 4.75 in Ans. Chap. 8 Solutions - Rev. A, Page 11/69 • (b) From Eq. (8-13), LT = 2d + 1/4 = 2(0.5) + 0.25 = 1.25 in From Table 8-7, ld = L  LT = 4.75  1.25 = 3.5 in, lt = l  ld = 4.19  3.5 = 0.69 in Ad =  (0.52)/4 = 0.1963 in2. From Table 8-2, At = 0.1419 in2. From Eq. (8-17)       0.1963 0.1419 30 1.322 Mlbf/in . 0.1963 0.69 0.1419 3.5 d t bk A l  d t t d A A E Ans Al     (c) Upper aluminum frustum: t = [4 + 2(0.095)] /2 = 2.095 in, d = 0.5 in, D = 0.75 in, and E = 10.3 Mpsi. From Eq. (8-20)           1 0.5774 10.3 0.5 7.23 Mlbf/in 1.155 2.095 0.75 0.5 0.75 0.5 ln 1.155 2.095 0.75 0.5 0.75 0.5 k              Lower aluminum frustum: t = 4  2.095 = 1.905 in, d = 0.5 in, D = 0.75 +4(0.095) tan 30 = 0.969 in, and E = 10.3 Mpsi. Eq. (8-20)  k2 = 11.34 Mlbf/in Steel washers frustum: t = 2(0.095) = 0.190 in, d = 0.531 in, D = 0.75 in, and E = 30 Mpsi. Eq. (8-20)  k3 = 53.91 Mlbf/in From Eq. (8-18) km = (1/k1 + 1/k2 +1/k3)1 = (1/7.23 + 1/11.34 + 1/53.91)1 = 4.08 Mlbf/in Ans. ______________________________________________________________________________ 8-19 (a) From Table A-31, the nut height is H = 8.4 mm. L > l + H = 50 + 8.4 = 58.4 mm. Chap. 8 Solutions - Rev. A, Page 12/69 • Rounding up, L = 60 mm Ans. (b) From Eq. (8-14), LT = 2d + 6 = 2(10) + 6 = 26 mm, ld = L  LT = 60  26 = 34 mm, lt = l  l = 50  34 = 16 mm. Ad =  (102) / 4 = 78.54 mm2. From Table 8-1, At = 58 mm2. From Eq. (8-17)       78.54 58.0 207 292.1 MN/m . 78.54 16 58.0 34 d t b d t t d A A Ek A A l A l      ns (c) Upper and lower frustums are the same. For the upper half, Aluminum: t = 10 mm, d = 10 mm, D = 15 mm, and from Table 8-8, E = 71 GPa. From Eq. (8-20)           1 0.5774 71 10 1576 MN/m 1.155 10 15 10 15 10 ln 1.155 10 15 10 15 10 k              Steel: t = 15 mm, d = 10 mm, D = 15 + 2(10) tan 30 = 26.55 mm, and E = 207 GPa. From Eq. (8-20)           2 0.5774 207 10 11 440 MN/m 1.155 15 26.55 10 26.55 10 ln 1.155 15 26.55 10 26.55 10 k              For the top half, = (1/kmk 1 + 1/k2) 1 = (1/1576 + 1/11 440)1 = 1385 MN/m Chap. 8 Solutions - Rev. A, Page 13/69 • Since the bottom half is the same, the overall stiffness is given by km = (1/ + 1/ )mk mk 1 = mk /2 = 1385/2 = 692.5 MN/m Ans. 8-20 (a) From Table A-31, the nut height is H = 8.4 mm. L > l + H = 60 + 8.4 = 68.4 mm. Rounding up, L = 70 mm Ans. (b) From Eq. (8-14), LT = 2d + 6 = 2(10) + 6 = 26 mm, ld = L  LT = 70  26 = 44 mm, lt = l  ld = 60  44 = 16 mm. Ad =  (102) / 4 = 78.54 mm2. From Table 8-1, At = 58 mm2. From Eq. (8-17)       78.54 58.0 207 247.6 MN/m . 78.54 16 58.0 44 d t b d t t d A A Ek A A l A l      ns (c) Upper aluminum frustum: t = 10 mm, d = 10 mm, D = 15 mm, and E = 71 GPa. From Eq. (8-20) Chap. 8 Solutions - Rev. A, Page 14/69 •           1 0.5774 10.3 71 1576 MN/m 1.155 2.095 15 10 15 10 ln 1.155 2.095 15 10 15 10 k              Lower aluminum frustum: t = 20 mm, d = 10 mm, D = 15 mm, and E = 71 GPa. Eq. (8-20)  k2 = 1 201 MN/m Top steel frustum: t = 20 mm, d = 10 mm, D = 15 + 2(10) tan 30 = 26.55 mm, and E = 207 GPa. Eq. (8-20)  k 3 = 9 781 MN/m Lower steel frustum: t = 10 mm, d = 10 mm, D = 15 + 2(20) tan 30 = 38.09 mm, and E = 207 GPa. Eq. (8-20)  k4 = 29 070 MN/m From Eq. (8-18) km = (1/k1 + 1/k2 +1/k3+1/k4)1 = (1/1 576 + 1/1 201 + 1/9 781 +1/29 070)1 = 623.5 MN/m Ans. ______________________________________________________________________________ 8-21 (a) From Table 8-7, l = h + d /2 = 10 + 30 + 10/2 = 45 mm. L ≥ h + 1.5 d = 10 + 30 + 1.5(10) = 55 mm Ans. (b) From Eq. (8-14), LT = 2d + 6 = 2(10) + 6 = 26 mm, ld = L  LT = 55  26 = 29 mm, lt = l  ld = 45  29 = 16 mm. Ad =  (102) / 4 = 78.54 mm2. From Table 8-1, At = 58 mm2. From Eq. (8-17)       78.54 58.0 207 320.9 MN/m . 78.54 16 58.0 29 d t bk d t t d A A E Ans A l A l      (c) Chap. 8 Solutions - Rev. A, Page 15/69 • Upper aluminum frustum: t = 10 mm, d = 10 mm, D = 15 mm, and E = 71 GPa. From Eq. (8-20)           1 0.5774 10.3 71 1576 MN/m 1.155 2.095 15 10 15 10 ln 1.155 2.095 15 10 15 10 k              Lower aluminum frustum: t = 5 mm, d = 10 mm, D = 15 mm, and E = 71 GPa. Eq. (8-20)  k2 = 2 300 MN/m Top steel frustum: t = 12.5 mm, d = 10 mm, D = 15 + 2(10) tan 30 = 26.55 mm, and E = 207 GPa. Eq. (8-20)  k 3 = 12 759 MN/m Lower steel frustum: t = 17.5 mm, d = 10 mm, D = 15 + 2(5) tan 30 = 20.77 mm, and E = 207 GPa. Eq. (8-20)  k4 = 6 806 MN/m From Eq. (8-18) km = (1/k1 + 1/k2 +1/k3+1/k4)1 = (1/1 576 + 1/2 300 + 1/12 759 +1/6 806)1 = 772.4 MN/m Ans. ______________________________________________________________________________ 8-22 Equation (f ), p. 436: b b m kC k k   Eq. (8-17): d tb d t t d A A Ek A l A l   Eq. (8-22):       0.5774 207 0.5774 40 0.5 2 ln 5 0.5774 40 2.5 m d k d d        See Table 8-7 for other terms used. Using a spreadsheet, with coarse-pitch bolts (units are mm, mm2, MN/m): d At Ad H L > L LT 10 58 78.53982 8.4 48.4 50 26 12 84.3 113.0973 10.8 50.8 55 30 14 115 153.938 12.8 52.8 55 34 16 157 201.0619 14.8 54.8 55 38 20 245 314.1593 18 58 60 46 24 353 452.3893 21.5 61.5 65 54 30 561 706.8583 25.6 65.6 70 66 Chap. 8 Solutions - Rev. A, Page 16/69 •   d l ld lt kb km C 10 40 24 16 356.0129 1751.566 0.16892 12 40 25 15 518.8172 2235.192 0.188386 14 40 21 19 686.2578 2761.721 0.199032 16 40 17 23 895.9182 3330.796 0.211966 20 40 14 26 1373.719 4595.515 0.230133 24 40 11 29 1944.24 6027.684 0.243886 30 40 4 36 2964.343 8487.533 0.258852 The 14 mm would probably be ok, but to satisfy the question, use a 16 mm bolt Ans. _____________________________________________________________________________ 8-23 Equation (f ), p. 436: b b m kC k k   Eq. (8-17): d tb d t t d A A Ek A l A l   For upper frustum, Eq. (8-20), with D = 1.5 d and t = 1.5 in:                 1 0.5774 30 0.5774 30 1.733 0.51.155 1.5 0.5 2.5 ln 5ln 1.733 2.51.155 1.5 2.5 0.5 d d k dd d dd d                      Lower steel frustum, with D = 1.5d + 2(1) tan 30 = 1.5d + 1.155, and t = 0.5 in:         2 0.5774 30 1.733 0.5 2.5 1.155 ln 1.733 2.5 0.5 1.155 d k d d d d          Chap. 8 Solutions - Rev. A, Page 17/69 • For cast iron frustum, let E = 14. 5 Mpsi, and D = 1.5 d, and t = 1 in:       3 0.5774 14.5 1.155 0.5 ln 5 1.155 2.5 d k d d        Overall, km = (1/k1 +1/k2 +1/k3)1 See Table 8-7 for other terms used. Using a spreadsheet, with coarse-pitch bolts (units are in, in2, Mlbf/in): d At Ad H L > L LT l 0.375 0.0775 0.110447 0.328125 3.328125 3.5 1 3 0.4375 0.1063 0.15033 0.375 3.375 3.5 1.125 3 0.5 0.1419 0.19635 0.4375 3.4375 3.5 1.25 3 0.5625 0.182 0.248505 0.484375 3.484375 3.5 1.375 3 0.625 0.226 0.306796 0.546875 3.546875 3.75 1.5 3 0.75 0.334 0.441786 0.640625 3.640625 3.75 1.75 3 0.875 0.462 0.60132 0.75 3.75 3.75 2 3 d ld lt kb k1 k2 k3 km C 0.375 2.5 0.5 1.031389 15.94599 178.7801 8.461979 5.362481 0.161309 0.4375 2.375 0.625 1.383882 19.21506 194.465 10.30557 6.484256 0.175884 0.5 2.25 0.75 1.791626 22.65332 210.6084 12.26874 7.668728 0.189383 0.5625 2.125 0.875 2.245705 26.25931 227.2109 14.35052 8.915294 0.20121 0.625 2.25 0.75 2.816255 30.03179 244.2728 16.55009 10.22344 0.215976 0.75 2 1 3.988786 38.07191 279.7762 21.29991 13.02271 0.234476 0.875 1.75 1.25 5.341985 46.7663 317.1203 26.51374 16.06359 0.24956 Use a 916 12 UNC  3.5 in long bolt Ans. ______________________________________________________________________________ 8-24 Equation (f ), p. 436: b b m kC k k   Eq. (8-17): d tb d t t d A A Ek A l A l   Chap. 8 Solutions - Rev. A, Page 18/69 • Top frustum, Eq. (8-20), with E = 10.3Mpsi, D = 1.5 d, and t = l /2:  1 0.5774 10.3 1.155 / 2 0.5ln 5 1.155 / 2 2.5 d k l d l d        Middle frustum, with E = 10.3 Mpsi, D = 1.5d + 2(l  0.5) tan 30, and t = 0.5  l /2                 2 0 0 0 0 0.5774 10.3 1.155 0.5 0.5 0.5 2 0.5 tan 30 2.5 2 0.5 tan 30 ln 1.155 0.5 0.5 2.5 2 0.5 tan 30 0.5 2 0.5 tan 30 d k l d l d l l d l d l                         Lower frustum, with E = 30Mpsi, D = 1.5 d, t = l  0.5       3 0.5774 30 1.155 0.5 0.5 ln 5 1.155 0.5 2.5 d k l d l d               See Table 8-7 for other terms used. Using a spreadsheet, with coarse-pitch bolts (units are in, in2, Mlbf/in) Chap. 8 Solutions - Rev. A, Page 19/69 • Size d At Ad L > L LT l ld 1 0.073 0.00263 0.004185 0.6095 0.75 0.396 0.5365 0.354 2 0.086 0.0037 0.005809 0.629 0.75 0.422 0.543 0.328 3 0.099 0.00487 0.007698 0.6485 0.75 0.448 0.5495 0.302 4 0.112 0.00604 0.009852 0.668 0.75 0.474 0.556 0.276 5 0.125 0.00796 0.012272 0.6875 0.75 0.5 0.5625 0.25 6 0.138 0.00909 0.014957 0.707 0.75 0.526 0.569 0.224 8 0.164 0.014 0.021124 0.746 0.75 0.578 0.582 0.172 10 0.19 0.0175 0.028353 0.785 1 0.63 0.595 0.37 Size d lt kb k1 k2 k3 km C 1 0.073 0.1825 0.194841 1.084468 1.954599 7.09432 0.635049 0.23478 2 0.086 0.215 0.261839 1.321595 2.449694 8.357692 0.778497 0.251687 3 0.099 0.2475 0.333134 1.570439 2.993366 9.621064 0.930427 0.263647 4 0.112 0.28 0.403377 1.830494 3.587564 10.88444 1.090613 0.27 5 0.125 0.3125 0.503097 2.101297 4.234381 12.14781 1.258846 0.285535 6 0.138 0.345 0.566787 2.382414 4.936066 13.41118 1.434931 0.28315 8 0.164 0.41 0.801537 2.974009 6.513824 15.93792 1.809923 0.306931 10 0.19 0.225 1.15799 3.602349 8.342138 18.46467 2.214214 0.343393 The lowest coarse series screw is a 164 UNC  0.75 in long up to a 632 UNC  0.75 in long. Ans. ______________________________________________________________________________ 8-25 For half of joint, Eq. (8-20): t = 20 mm, d = 14 mm, D = 21 mm, and E = 207 GPa           1 0.5774 207 14 5523 MN/m 1.155 20 21 14 21 14 ln 1.155 20 21 14 21 14 k              km = (1/k1 + 1/k1)1 = k1/2 = 5523/2 = 2762 MN/m Ans. From Eq. (8-22) with l = 40 mm           0.5774 207 14 2762 MN/m . 0.5774 40 0.5 14 2ln 5 0.5774 40 2.5 14 mk A         ns which agrees with the earlier calculation. Chap. 8 Solutions - Rev. A, Page 20/69 • For Eq. (8-23), from Table 8-8, A = 0.787 15, B = 0.628 73 km = 207(14)(0.78 715) exp [0.628 73(14)/40] = 2843 MN/m Ans. This is 2.9% higher than the earlier calculations. ______________________________________________________________________________ 8-26 (a) Grip, l = 10 in. Nut height, H = 41/64 in (Table A-31). L ≥ l + H = 10 + 41/64 = 10.641 in. Let L = 10.75 in. Table 8-7, LT = 2d + 0.5 = 2(0.75) + 0.5 = 2 in, ld = L  LT = 10.75  2 = 8.75 in, lt = l  ld = 10  8.75 = 1.25 in Ad = (0.752)/4 = 0.4418 in2, At = 0.373 in2 (Table 8-2) Eq. (8-17),       0.4418 0.373 30 1.296 Mlbf/in . 0.4418 1.25 0.373 8.75 d t b d t t d A A Ek A A l Al      ns Eq. (4-4), p. 149,   2 2/ 4 1.125 0.75 30 1.657 Mlbf/in . 10 m m m A Ek A l      ns Eq. (f), p. 436, C = kb/(kb + km) = 1.296/(1.296 + 1.657) = 0.439 Ans. (b) Let: Nt = no. of turns, p = pitch of thread (in), N = no. of threads per in = 1/p. Then,  = b + m = Nt p = Nt / N (1) But, b = Fi / kb, and, m = Fi / km. Substituting these into Eq. (1) and solving for Fi gives Chap. 8 Solutions - Rev. A, Page 21/69 •     6 2 1.296 1.657 10 1/ 3 15 150 lbf . 1.296 1.657 16 b m t i b m k k NF k k N Ans      ______________________________________________________________________________ 8-27 Proof for the turn-of-nut equation is given in the solution of Prob. 8-26, Eq. (2), where Nt =  / 360. The relationship between the turn-of-nut method and the torque-wrench method is as follows. (turn-of-nut) (torque-wrench) b m t i b m i k kN F N k k T KFd         Eliminate Fi . 360 b m t b m k k NTN A k k Kd        ns ______________________________________________________________________________ 8-28 (a) From Ex. 8-4, Fi = 14.4 kip, kb = 5.21(106) lbf/in, km = 8.95(106) lbf/in Eq. (8-27): T = kFid = 0.2(14.4)(103)(5/8) = 1800 lbf · in Ans. From Prob. 8-27,   3 6 5.21 8.95 (14.4)(10 )11 5.21 8.95 10 0.0481 turns 17.3 . b m t i b m k kN F N k k Ans                  Bolt group is (1.5) / (5/8) = 2.4 diameters. Answer is much lower than RB&W recommendations. ______________________________________________________________________________ 8-29 C = kb / (kb + km) = 3/(3+12) = 0.2, P = Ptotal/ N = 80/6 = 13.33 kips/bolt Table 8-2, At = 0.141 9 in2; Table 8-9, Sp = 120 kpsi; Eqs. (8-31) and (8-32), Fi = 0.75 At Sp = 0.75(0.141 9)(120) = 12.77 kips (a) From Eq. (8-28), the factor of safety for yielding is     120 0.141 9 1.10 . 0.2 13.33 12.77 p t p i S A n A CP F      ns (b) From Eq. (8-29), the overload factor is Chap. 8 Solutions - Rev. A, Page 22/69 •     120 0.141 9 12.77 1.60 . 0.2 13.33 p t i L S A F n A CP      ns (c) From Eq. (803), the joint separation factor of safety is    0 12.77 1.20 . 1 13.33 1 0.2 iFn A P C      ns ______________________________________________________________________________ 8-30 1/2  13 UNC Grade 8 bolt, K = 0.20 (a) Proof strength, Table 8-9, Sp = 120 kpsi Table 8-2, At = 0.141 9 in2 Maximum, Fi = Sp At = 120(0.141 9) = 17.0 kips Ans. (b) From Prob. 8-29, C = 0.2, P = 13.33 kips Joint separation, Eq. (8-30) with n0 = 1 Minimum Fi = P (1  C) = 13.33(1  0.2) = 10.66 kips Ans. (c) iF = (17.0 + 10.66)/2 = 13.8 kips Eq. (8-27), T = KFi d = 0.2(13.8)103(0.5)/12 = 115 lbf  ft Ans. ______________________________________________________________________________ 8-31 (a) Table 8-1, At = 20.1 mm2. Table 8-11, Sp = 380 MPa. Eq. (8-31), Fi = 0.75 Fp = 0.75 At Sp = 0.75(20.1)380(103) = 5.73 kN Eq. (f ), p. 436, 1 0.278 1 2.6 b b m kC k k      Eq. (8-28) with np = 1,    30.25 20.1 380 100.25 6.869 kN 0.278 p t i p tS A F S AP C C      Ptotal = NP = 8(6.869) = 55.0 kN Ans. (b) Eq. (8-30) with n0 = 1, 5.73 7.94 kN 1 1 0.278 iFP C      Ptotal = NP = 8(7.94) = 63.5 kN Ans. Bolt stress would exceed proof strength ______________________________________________________________________________ 8-32 (a) Table 8-2, At = 0.141 9 in2. Table 8-9, Sp = 120 kpsi. Eq. (8-31), Fi = 0.75 Fp = 0.75 At Sp = 0.75(0.141 9)120 = 12.77 kips Eq. (f ), p. 436, 4 0.25 4 12 b b m kC k k      Chap. 8 Solutions - Rev. A, Page 23/69 • Eq. (8-28) with np = 1,     total total 0.25 80 0.25 4.70 0.25 0.25 120 0.141 9 p t i p t p t S A F NS A P N C C P CN S A           Round to N = 5 bolts Ans. (b) Eq. (8-30) with n0 = 1,     total total 1 1 80 1 0.25 4.70 12.77 i i FP N C P C N F           Round to N = 5 bolts Ans. ______________________________________________________________________________ 8-33 Bolts: From Table A-31, the nut height is H = 10.8 mm. L ≥ l +H = 40 + 10.8 = 50.8 mm. Although Table A-17 indicates to go to 60 mm, 55 mm is readily available Round up to L = 55 mm Ans. Eq. (8-14): LT = 2d + 6 = 2(12) + 6 = 30 mm Table 8-7: ld = L  LT = 55  30 = 25 mm, lt = l ld = 40  25 = 15 mm Ad = (122)/4 = 113.1 mm2, Table 8-1: At = 84.3 mm2 Eq. (8-17):       113.1 84.3 207 518.8 MN/m 113.1 15 84.3 25 d t b d t t d A A Ek A l A l      Members: Steel cyl. head: t = 20 mm, d = 12 mm, D = 18 mm, E = 207 GPa. Eq. (8-20),           1 0.5774 207 12 4470 MN/m 1.155 20 18 12 18 12 ln 1.155 20 18 12 18 12 k              Cast iron: t = 20 mm, d = 12 mm, D = 18 mm, E = 100 GPa (from Table 8-8). The only difference from k1 is the material k2 = (100/207)(4470) = 2159 MN/m Eq. (8-18): km = (1/4470 + 1/2159)1 = 1456 MN/m Chap. 8 Solutions - Rev. A, Page 24/69 • C = kb / (kb + km) = 518.8/(518.8+1456) = 0.263 Table 8-11: Sp = 650 MPa Assume non-permanent connection. Eqs. (8-31) and (8-32) Fi = 0.75 At Sp = 0.75(84.3)(650)103 = 41.1 kN The total external load is Ptotal = pg Ac, where Ac is the diameter of the cylinder which is 100 mm. The external load per bolt is P = Ptotal /N. Thus P = [6 (1002)/4](103)/10 = 4.712 kN/bolt Yielding factor of safety, Eq. (8-28):     3650 84.3 10 1.29 . 0.263 4.712 41.10 p t p i S A n A CP F       ns Overload factor of safety, Eq. (8-29):     3650 84.3 10 41.10 11.1 . 0.263 4.712 p t i L S A F n A CP      ns Separation factor of safety, Eq. (8-30):    0 41.10 11.8 . 1 4.712 1 0.263 iFn A P C      ns ______________________________________________________________________________ 8-34 Bolts: Grip, l = 1/2 + 5/8 = 1.125 in. From Table A-31, the nut height is H = 7/16 in. L ≥ l + H = 1.125 + 7/16 = 1.563 in. Round up to L = 1.75 in Ans. Eq. (8-13): LT = 2d + 0.25 = 2(0.5) + 0.25 = 1.25 in Table 8-7: ld = L  LT = 1.75  1.25 = 0.5 in, lt = l ld = 1.125  0.5 = 0.625 in Ad =  (0.52)/4 = 0.196 3 in2, Table 8-2: At = 0.141 9 in2 Eq. (8-17):       0.196 3 0.141 9 30 4.316 Mlbf/in 0.196 3 0.625 0.141 9 0.5 d t b d t t d A A Ek A l A l      Chap. 8 Solutions - Rev. A, Page 25/69 • Members: Steel cyl. head: t = 0.5 in, d = 0.5 in, D = 0.75 in, E = 30 Mpsi. Eq. (8-20),           1 0.5774 30 0.5 33.30 Mlbf/in 1.155 0.5 0.75 0.5 0.75 0.5 ln 1.155 0.5 0.75 0.5 0.75 0.5 k              Cast iron: Has two frusta. Midpoint of complete joint is at (1/2 + 5/8)/2 = 0.5625 in. Upper frustum, t = 0.5625 0.5 = 0.0625 in, d = 0.5 in, D = 0.75 + 2(0.5) tan 30 = 1.327 in, E = 14.5 Mpsi (from Table 8-8) Eq. (8-20)  k2 = 292.7 Mlbf/in Lower frustum, t = 0.5625 in, d = 0.5 in, D = 0.75 in, E = 14.5 Mpsi Eq. (8-20)  k3 = 15.26 Mlbf/in Eq. (8-18): km = (1/33.30 + 1/292.7 + 1/15.26)1 = 10.10 Mlbf/in C = kb / (kb + km) = 4.316/(4.316+10.10) = 0.299 Table 8-9: Sp = 85 kpsi Assume non-permanent connection. Eqs. (8-31) and (8-32) Fi = 0.75 At Sp = 0.75(0.141 9)(85) = 9.05 kips The total external load is Ptotal = pg Ac, where Ac is the diameter of the cylinder which is 3.5 in. The external load per bolt is P = Ptotal /N. Thus P = [1 500 (3.52)/4](103)/10 = 1.443 kips/bolt Yielding factor of safety, Eq. (8-28):     85 0.141 9 1.27 . 0.299 1.443 9.05 p t p i S A n A CP F      ns Overload factor of safety, Eq. (8-29):     85 0.141 9 9.05 6.98 . 0.299 1.443 p t i L S A F n A CP ns      Separation factor of safety, Eq. (8-30): Chap. 8 Solutions - Rev. A, Page 26/69 •    0 9.05 8.95 . 1 1.443 1 0.299 iFn A P C      ns ______________________________________________________________________________ -35 Bolts: Grip: l = 20 + 25 = 45 mm. From Table A-31, the nut height is H = 8.4 mm. m is Round up to L = 55 mm Ans. Eq. (8-14): LT = 2d + 6 = 2(10) + 6 = 26 mm Table 8-7: ld = L  LT = 55  26 = 29 mm, lt = l ld = 45  29 = 16 mm Ad = (102)/4 = 78.5 mm2, Table 8-1: At = 58.0 mm2 Eq. (8-17): 8 L ≥ l +H = 45 + 8.4 = 53.4 mm. Although Table A-17 indicates to go to 60 mm, 55 m readily available       78.5 58.0 207 320.8 MN/m 78.5 16 58.0 29 d t b d t t d A A Ek A l A l      Members: Steel cyl. head: t = 20 mm, d = 10 mm, D = 15 mm, E = 207 GPa. Eq. (8-20),           1 0.5774 207 10 3503 MN/m 1.155 20 15 10 15 10 ln 1.155 20 15 10 15 10 k              n: Has two frusta. Midpoint of complete joint is at (20 + 25)/2 = 22.5 mm m Table 8-8), Lower frustum, t = 22.5 mm, d = 10 mm, D = 15 mm, E = 100 GPa Eq. (8-20)  k3 = 1632 MN/m Eq. (8-18): km = (1/3503 + 1/45 880 + 1/1632)1 = 1087 MN/m C = kb / (kb + km) = 320.8/(320.8+1087) = 0.228 Table 8-11: Sp = 830 MPa ection. Eqs. (8-31) and (8-32) Fi = 0.75 At Sp = 0.75(58.0)(830)103 = 36.1 kN Cast iro Upper frustum, t = 22.5  20 = 2.5 mm, d = 10 mm, D = 15 + 2(20) tan 30 = 38.09 mm, E = 100 GPa (fro Eq. (8-20)  k2 = 45 880 MN/m Assume non-permanent conn Chap. 8 Solutions - Rev. A, Page 27/69 • The total external load is Ptotal = pg Ac, where Ac is the diameter of the cylinder which is P = [550 (0.82)/4]/36 = 7.679 kN/bolt Yielding factor of safety, Eq. (8-28): 0.8 m. The external load per bolt is P = Ptotal /N. Thus     3830 58.0 10 1.27 . 0.228 7.679 36.1 p t p i S A n A CP F       ns Overload factor of safety, Eq. (8-29):     3830 58.0 10 36.1 6.88 . 0.228 7.679 p t i L S A F n A CP      ns Separation factor of safety, Eq. (8-30):    0 36.1 6.09 . 1 7.679 1 0.228 iFn A P C      ns ______________________________________________________________________________ -36 Bolts: Grip, l = 3/8 + 1/2 = 0.875 in. From Table A-31, the nut height is H = 3/8 in. Let L = 1.25 in Ans. Eq. (8-13): LT = 2d + 0.25 = 2(7/16) + 0.25 = 1.125 in Table 8-7: ld = L  LT = 1.25  1.125 = 0.125 in, lt = l ld = 0.875  0.125 = Ad =  (7/16) /4 = 0.150 3 in2, Table 8-2: At = 0.106 3 in2 Eq. (8-17), 8 L ≥ l + H = 0.875 + 3/8 = 1.25 in. 0.75 in 2       0.150 3 0.106 3 30 3.804 Mlbf/in 0.150 3 0.75 0.106 3 0.125 dAk   tb d t t d A E A l A l    Members: Steel cyl. head: t = 0.375 in, d = 0.4375 in, D = 0.65625 in, E = 30 Mpsi. Eq. (8-20), Chap. 8 Solutions - Rev. A, Page 28/69 •           1 0.5774 30 0.4375 31.40 Mlbf/in 1.155 0.375 0.65625 0.4375 0.65625 0.4375 ln 1.155 0.375 0.65625 0.4375 0.65625 0.4375 k              Cast iron: Has two frusta. Midpoint of complete joint is at (3/8 + 1/2)/2 = 0.4375 in. Upper frustum, t = 0.4375 0.375 = 0.0625 in, d = 0.4375 in, D = 0.65625 + 2(0.375) tan 30 = 1.089 in, E = 14.5 Mpsi (from Table 8-8) Eq. (8-20)  k2 = 195.5 Mlbf/in Lower frustum, t = 0.4375 in, d = 0.4375 in, D = 0.65625 in, E = 14.5 Mpsi Eq. (8-20)  k3 = 14.08 Mlbf/in Eq. (8-18): km = (1/31.40 + 1/195.5 + 1/14.08)1 = 9.261 Mlbf/in C = kb / (kb + km) = 3.804/(3.804 + 9.261) = 0.291 Table 8-9: Sp = 120 kpsi Assume non-permanent connection. Eqs. (8-31) and (8-32) Fi = 0.75 At Sp = 0.75(0.106 3)(120) = 9.57 kips The total external load is Ptotal = pg Ac, where Ac is the diameter of the cylinder which is 3.25 in. The external load per bolt is P = Ptotal /N. Thus P = [1 200 (3.252)/4](103)/8 = 1.244 kips/bolt Yielding factor of safety, Eq. (8-28):     120 0.106 3 1.28 . 0.291 1.244 9.57 p t p i S A n A CP F      ns Overload factor of safety, Eq. (8-29):     120 0.106 3 9.57 8.80 . 0.291 1.244 p t i L S A F n A CP ns      Separation factor of safety, Eq. (8-30): Chap. 8 Solutions - Rev. A, Page 29/69 •    0 9.57 10.9 . 1 1.244 1 0.291 iFn A P C      ns ______________________________________________________________________________ -37 From Table 8-7, h = t1 = 20 mm /2 = 26 mm p to L = 40 mm From Table 8-1, At = 84.3 mm2. Ad =  (122)/4 = 113.1 mm2 8 For t2 > d, l = h + d /2 = 20 + 12 L ≥ h + 1.5 d = 20 + 1.5(12) = 38 mm. Round u LT = 2d + 6 = 2(12) + 6 = 30 mm ld = L  LT = 40  20 = 10 mm lt = l  ld = 26  10 = 16 mm Eq. (8-17),       113.1 84.3 207 744.0 MN/m 113.1 16 84.3 10 d t b d t t d A A Ek A l A l      Similar to Fig. 8-21, we have three frusta. m, D = 18 mm, E = 207 GPa. Eq. (8-20) Top frusta, steel: t = l / 2 = 13 mm, d = 12 m           1 0.5774 207 12 5 316 MN/m 1.155 13 18 12 18 12 ln 1.155 13 18 12 18 12 k              Middle frusta, steel: t = 20  13 = 7 mm, d = 12 mm, D = 18 + 2(13  7) tan 30 = 24.93 Lower frusta, cast iron: t = 26  20 = 6 mm, d = 12 mm, D = 18 mm, E = 100 GPa (see Eq. (8-18), km = (1/5 316 + 1/15 660 + 1/3 887)1 = 1 964 MN/m C = kb / (kb + km) = 744.0/(744.0 + 1 964) = 0.275 Table 8-11: Sp = 650 MPa. From Prob. 8-33, P = 4.712 kN. Assume a non-permanent Fi = 0.75 At Sp = 0.75(84.3)(650)103 = 41.1 kN Yielding factor of safety, Eq. (8-28) mm, E = 207 GPa. Eq. (8-20)  k2 = 15 660 MN/m Table 8-8). Eq. (8-20)  k3 = 3 887 MN/m connection. Eqs. (8-31) and (8-32),     3650 84.3 10 1.29 . 0.275 4.712 41.1 p t p i S A n A CP F       ns Overload factor of safety, Eq. (8-29) Chap. 8 Solutions - Rev. A, Page 30/69 •     3650 84.3 10 41.1 10.7 . 0.275 4.712 p t i L S A F n A CP      ns r of safety, Eq. (8-30) Separation facto    0 41.1 12.0 . 1 4.712 1 0.275 iFn A P C      ns ___________________________________________ ________________ 1 .5/2 = 0.75 in 1.25 in b = At E / l = 0.141 9(30)/0.75 = in, D = 0.75 in, E = 30 Mpsi __________________ _ -38 From Table 8-7, h = t = 0.5 in 8 For t2 > d, l = h + d /2 = 0.5 + 0 L ≥ h + 1.5 d = 0.5 + 1.5(0.5) = 1.25 in. Let L = LT = 2d + 0.25 = 2(0.5) + 0.25 = 1.25 in. All threaded. 2 From Table 8-1, At = 0.141 9 in . The bolt stiffness is k 5.676 Mlbf/in Similar to Fig. 8-21, we have three frusta. Top frusta, steel: t = l / 2 = 0.375 in, d = 0.5           1 0.5774 30 0.5 38.45 Mlk    bf/in 1.155 0.375 0.75 0.5 0.75 0.5 ln 1.155 0.375 0.75 0.5 0.75 0.5           Middle frusta, steel: t = 0.5  0.375 = 0.125 in, d = 0.5 in, d = 0.5 in, D = 0.75 in, E = 14.5 Mpsi. m 1 = 13.51 Mlbf/in b b m p e a non-permanent i t p D = 0.75 + 2(0.75  0.5) tan 30 = 1.039 in, E = 30 Mpsi. Eq. (8-20)  k2 = 184.3 Mlbf/in Lower frusta, cast iron: t = 0.75  0.5 = 0.25 in, Eq. (8-20)  k3 = 23.49 Mlbf/in Eq. (8-18), k = (1/38.45 + 1/184.3 + 1/23.49) C = k / (k + k ) = 5.676 / (5.676 + 13.51) = 0.296 Table 8-9, S = 85 kpsi. From Prob. 8-34, P = 1.443 kips/bolt. Assum connection. Eqs. (8-31) and (8-32), F = 0.75 A S = 0.75(0.141 9)(85) = 9.05 kips Yielding factor of safety, Eq. (8-28)     85 0.141 9 1.27 . 0.296 1.443 9.05 p t p i S A n A CP F      ns Overload factor of safety, Eq. (8-29) Chap. 8 Solutions - Rev. A, Page 31/69 •     85 0.141 9 9.05 7.05 . 0.296 1.443 p t i L S A F n A CP      ns r of safety, Eq. (8-30) Separation facto    0 9.05 8.91 . 1 1.443 1 0.296 iFn A P C      ns __________________________________________ ________________ 1 2 = 25 mm 35 mm t 2. Ad =  (102)/4 = 78.5 mm2 __________________ __ -39 From Table 8-7, h = t = 20 mm 8 For t2 > d, l = h + d /2 = 20 + 10/ L ≥ h + 1.5 d = 20 + 1.5(10) = 35 mm. Let L = LT = 2d + 6 = 2(10) + 6 = 26 mm ld = L  LT = 35  26 = 9 mm lt = l  ld = 25  9 = 16 mm From Table 8-1, A = 58.0 mm Eq. (8-17),       78.5 58.0 207 530.1 MN/m 78.5 16 58.0 9 d t b d t t d A A E k A l A l      -21, we have three frusta. mm, D = 15 mm, E = 207 GPa. Eq. (8-20) Similar to Fig. 8 Top frusta, steel: t = l / 2 = 12.5 mm, d = 10           1 0.5774 207 10 4 163 MN/mk    1.155 12.5 15 10 15 10 ln 1.155 12.5 15 10 15 10           Middle frusta, steel: t = 20  12.5 = 7.5 mm, d = 10 mm, D = 15 + 2(12.5  7.5) tan 30 = , E = 100 GPa (see m 1 = 1 562 MN/m b b m p = 830 MPa. From anent i t p 3 = 36.1 kN 20.77 mm, E = 207 GPa. Eq. (8-20)  k2 = 10 975 MN/m Lower frusta, cast iron: t = 25  20 = 5 mm, d = 10 mm, D = 15 mm Table 8-8). Eq. (8-20)  k3 = 3 239 MN/m Eq. (8-18), k = (1/4 163 + 1/10 975 + 1/3 239) C = k / (k + k ) = 530.1/(530.1 + 1 562) = 0.253 Table 8-11: S Prob. 8-35, P = 7.679 kN/bolt. Assume a non-perm connection. Eqs. (8-31) and (8-32), F = 0.75 A S = 0.75(58.0)(830)10 Yielding factor of safety, Eq. (8-28) Chap. 8 Solutions - Rev. A, Page 32/69 •     3830p tS A 58.0 10 1.27 . 0.253 7.679 36.1p i n Ans CP F       of safety, Eq. (8-29) Overload factor     358.0 10 36.1 6.20 . 0.253 7.679 p t i Ln AnsCP      Separation factor of safety, Eq. (8-30) 830S A F    0 36.1 6.29 . 1 7.679 1 0.253 iF n Ans P C      ______________________________________________________________________________ For t2 > d, l = h + d /2 = 0.375 + 0.4375/2 = 0.59375 in ) = 1.031 in. Round up to L = 1.25 in 8-2: At = 0.106 3 in2 8-40 From Table 8-7, h = t1 = 0.375 in L ≥ h + 1.5 d = 0.375 + 1.5(0.4375 LT = 2d + 0.25 = 2(0.4375) + 0.25 = 1.125 in ld = L  LT = 1.25  1.125 = 0.125 lt = l  ld = 0.59375  0.125 = 0.46875 in Ad =  (7/16)2/4 = 0.150 3 in2, Table Eq. (8-17),       0.150 3 0.106 3 30 5.724 Mlbf/in 0.150 3 0.46875 0.106 3 0.125 d t b d t t d A A E k A l A l      -21, we have three frusta. Top frusta, steel: t = l / 2 = 0.296875 in, d = 0.4375 in, D = 0.65625 in, E = 30 Mpsi Similar to Fig. 8           1k  0.5774 30 0.4375 35.52 Mlbf/in   1.155 0.296875 0.656255 0.4375 0.75 0.656255 ln 1.155 0.296875 0.75 0.656255 0.75 0.656255           Middle frusta, steel: t = 0.375  0.296875 = 0.078125 in, d = 0.4375 in, D = 0.65625 + 2(0.59375  0.375) tan 30 = 0.9088 in, E = 30 Mpsi. 0.375 = 0.21875 in, d = 0.4375 in, D = 0.65625 in, E = 14.5 Mpsi. Eq. (8-20)  k3 = 20.55 Mlbf/in C = kb / (kb + km) = 5.724/(5.724 + 12.28) = 0.318 Eq. (8-20)  k2 = 215.8 Mlbf/in Lower frusta, cast iron: t = 0.59375  Eq. (8-18), km = (1/35.52 + 1/215.8 + 1/20.55)1 = 12.28 Mlbf/in Chap. 8 Solutions - Rev. A, Page 33/69 • Table 8-9, Sp = 120 kpsi. From Prob. 8-34, P = 1.244 kips/bolt. Assume a non-permanent connection. Eqs. (8-31) and (8-32), 5(0.106 3)(120) = 9.57 kips Fi = 0.75 At Sp = 0.7 Yielding factor of safety, Eq. (8-28)     12p tS An A  0 0.106 3 1.28 . 0.318 1.244 9.57p i ns CP F    of safety, Eq. (8-29) Overload factor     120p t iS A F 0.106 3 9.57 8.05 . 0.318 1.244L n Ans CP     Separation factor of safety, Eq. (8-30)    0 9.57 11.3 . 1 1.244 1 0.318 iF n Ans P C      ______________________________________________________________________________ What is presented here is one possible iterative approach. We will demonstrate this with g using Eq. (8-18), yields km = 1 141 MN/m (see Prob. 8-33 for method of e nut height in Table A-31. For the example, H = 8.4 mm. From this, L is rounded up from the calculation of l + H = 40 + 8.4 = 48.4 mm to 50 mm. Next, 4 mm2. for Db in Eq. (8-34), the number of bolts are 8-41 This is a design problem and there is no closed-form solution path or a unique solution. an example. 1. Select the diameter, d. For this example, let d = 10 mm. Using Eq. (8-20) on members, and combinin calculation. 2. Look up th calculations are made for LT = 2(10) + 6 = 26 mm, ld = 50  26 = 24 mm, lt = 40  24 = 16 mm. From step 1, Ad = (102)/4 = 78.54 mm2. Next, from Table 8-1, At = 78.5 From Eq. (8-17), kb = 356 MN/m. Finally, from Eq. (e), p. 421, C = 0.238. 3. From Prob. 8-33, the bolt circle diameter is E = 200 mm. Substituting this     200bDN    15.7 4 4 10d  p gives N = 16. d on the solution to Prob. 8-33, the strength of ISO 9.8 was so high to give very large factors of safety for overload and separation. Try ISO 4.6 Rounding this u 4. Next, select a grade bolt. Base Chap. 8 Solutions - Rev. A, Page 34/69 • with Sp = 225 MPa. From Eqs. (8-31) and (8-32) for a non-permanent connection, Fi = 9.79 kN. 5. The ex ternal load requirement per bolt is P = 1.15 pg Ac/N, where from Prob 8-33, pg = 6 MPa, and A =  (1002)/4. This gives P = 3.39 kN/bolt. nd n0 = 3.79. for the tables used from the text. The results for four bolt sizes are shown below. The dimension of each lt Ad At kb c 6. Using Eqs. (8-28) to (8-30) yield np = 1.23, nL = 4.05, a Steps 1 - 6 can be easily implemented on a spreadsheet with lookup tables term is consistent with the example given above. d km H L LT ld 8 854 6.8 50 22 28 12 50.26 36.6 233.9 10 1 78.54 356 141 8.4 50 26 24 16 58 12 1456 10.8 55 30 25 15 113.1 84.3 518.8 14 1950 12.8 55 34 21 19 153.9 115 686.3 d C N S p F i P n p n L n 0 8 0.215 20 225 6.18 2.71 1.22 3.53 2.90 10 0.238 16 225 9.79 3.39 1.23 4.05 3.79 12 0.263 13* 225 14.23 4.17 1.24 4.33 4.63 14 0.276 12 225 19.41 4.52 1.25 5.19 5.94 *Rounded down from 89 g eters. N  cost/bolt, and/or N  cost per hole, etc. ____ __ n. What is presented here is one possible iterative approach. We will demonstrate this with 4 solution), and combining using Eq. (8-19), yields km = 10.10 Mlbf/in. rounded up from the calculation of l + H = 1.125 + 0.4375 = 1.5625 in to 1.75 in. Next, 34), for the number of bolts 13.0 97, so spacin is slightly greater than four diam Any one of the solutions is acceptable. A decision-maker might be cost such as _ _________________________________________________________________ 8-42 This is a design problem and there is no closed-form solution path or a unique solutio an example. 1. Select the diameter, d. For this example, let d = 0.5 in. Using Eq. (8-20) on three frusta (see Prob. 8-3 2. Look up the nut height in Table A-31. For the example, H = 0.4375 in. From this, L is calculations are made for LT = 2(0.5) + 0.25 = 1.25 in, ld = 1.75  1.25 = 0.5 in, lt = 1.125  0.5 = 0.625 in. From step 1, Ad = (0.52)/4 = 0.1963 in2. Next, from Table 8-1, At = 0.141 9 in2. From Eq. (8-17), kb = 4.316 Mlbf/in. Finally, from Eq. (e), p. 421, C = 0.299. 3. From Prob. 8-34, the bolt circle diameter is E = 6 in. Substituting this for Db in Eq. (8- Chap. 8 Solutions - Rev. A, Page 35/69 •     6 9.425 4 4 bDN d     0.5 Rounding this up gives N = 10. 4. Next, select a grade bolt. Based on the solution to Prob. 8-34, the strength of SAE grade = 85 kpsi. From Eqs. (8-31) and (8-32) for a non- permanent connection, Fi = 9.046 kips. 4, s gives P = 1.660 kips/bolt. b 5 was adequate. Use this with Sp 5. The external load requirement per bolt is P = 1.15 pg Ac/N, where from Prob 8-3 pg = 1 500 psi, and Ac =  (3.52)/4 . Thi 6. Using Eqs. (8-28) to (8-30) yield np = 1.26, nL = 6.07, and n0 = 7.78. d km H L LT ld lt Ad At k 0.375 6.75 0.3281 1.5 1 0.5 0.625 0.1104 0.0775 2.383 0.4375 9.17 0.375 1.5 1.125 0.375 0.75 0.1503 0.1063 3.141 0.5 10.10 0.4375 1.75 1.25 0.1963 0.1419 4.316 0.5 0.625 0.5625 11.98 0.4844 1.75 1.375 0.375 0.75 0.2485 0.182 5.329 d C N Sp Fi P np nL n 0 0.375 0.261 13 85 4.941 1.277 1.25 4.95 5.24 0.4375 0.273 11 85 6.777 1.509 1.26 5.48 6.18 0.5 0.299 9.046 1.660 1.26 6.07 7.78 10 85 0.5625 0.308 9 85 11.6 1.844 1.27 6.81 9.09 Any on th io ac a d - r b such as N  c r N cos r h t _______________________________________________________________________ solution path or a unique solution. ith an example. ta calculations are made for L = 2(10) + 6 = 26 mm, l = 55  26 = 29 mm, l = 45  29 = e of e solut ns is cept ble. A ecision make might e cost ost/bolt, and/o  t pe ole, e c. _ 8-43 This is a design problem and there is no closed-form What is presented here is one possible iterative approach. We will demonstrate this w 1. Select the diameter, d. For this example, let d = 10 mm. Using Eq. (8-20) on three frus (see Prob. 8-35 solution), and combining using Eq. (8-19), yields km = 1 087 MN/m. 2. Look up the nut height in Table A-31. For the example, H = 8.4 mm. From this, L is rounded up from the calculation of l + H = 45 + 8.4 = 53.4 mm to 55 mm. Next, T d t 16 mm. From step 1, Ad = (102)/4 = 78.54 mm2. Next, from Table 8-1, At = 58.0 mm2. From Eq. (8-17), kb = 320.9 MN/m. Finally, from Eq. (e), p. 421, C = 0.228. 3. From Prob. 8-35, the bolt circle diameter is E = 1000 mm. Substituting this for Db in Eq. (8-34), for the number of bolts Chap. 8 Solutions - Rev. A, Page 36/69 •     1000 78.5 4 4 10 bDN d     Rounding this up gives N = 79. A rather large number, since the bolt circle diameter, E is ger bolts. rge factors of safety for overload and separation. Try ISO 5.8 with Sp = 380 MPa. From Eqs. (8-31) and (8-32) for a non-permanent connection, Fi = a, and Ac =  (8002)/4 . This gives P = 4.024 kN/bolt. Steps 1 - 6 can be easily implemented on a spreadsheet with lookup tables for the tables mension of each term is consistent with the example given above. so large. Try lar 4. Next, select a grade bolt. Based on the solution to Prob. 8-35, the strength of ISO 9.8 was so high to give very la 16.53 kN. 5. The external load requirement per bolt is P = 1.15 pg Ac/N, where from Prob 8-35, pg = 0.550 MP 6. Using Eqs. (8-28) to (8-30) yield np = 1.26, nL = 6.01, and n0 = 5.32. used from the text. The results for three bolt sizes are shown below. The di d km H L LT ld lt Ad At kb 10 1087 8.4 55 26 29 16 78.54 58 320.9 20 3055 18 65 46 19 26 314.2 245 1242 36 6725 31 80 78 2 43 1018 817 3791 d C N Sp Fi P np nL n0 1 0 0.2 8 2 7 9 380 16.53 4.024 1.26 6. 1 0 5. 2 3 20 0.308 40 380 69.83 7.948 1.29 12.7 9.5 36 0.361 22 380 232.8 14.45 1.3 14.9 25.2 A large range e he n l i ep A decision-maker might be cost such as co lt o r h tc _______________________________________________________________________ 8-44 r a unique solution. ith an example. . made for L = 2(0.375) + 0.25 = 1 in, l = 1.25  1 = 0.25 in, l = 0.875  0.25 = 0.625 in. is pres nted re. A y one of the so utions s acc table. N  st/bo , and/or N  c st pe ole, e . _ This is a design problem and there is no closed-form solution path o What is presented here is one possible iterative approach. We will demonstrate this w 1. Select the diameter, d. For this example, let d = 0.375 in. Using Eq. (8-20) on three frusta (see Prob. 8-36 solution), and combining using Eq. (8-19), yields km = 7.42 Mlbf/in 2. Look up the nut height in Table A-31. For the example, H = 0.3281 in. From this, L ≥ l + H = 0.875 + 0.3281 = 1.2031 in. Rounding up, L = 1.25. Next, calculations are T d t Chap. 8 Solutions - Rev. A, Page 37/69 • 2 2 ,max0 0 2 ,max sin sin 2 2 2 b b b M F R d F R d M F R   2             from which ,max 2b MF R  2 2 1 1 max 1 22sin sin (cos - cos )b M MF F R d R d R R                 Noting 1 = 75, 2 = 105, max 12 000 (cos 75 - cos105 ) 494 lbf . (8 / 2) F A     ns (b) max ,max 2 2 2( )b M MF F R R R N R           N max 2(12 000) 500 lbf . (8 / 2)(12) F A  ns (c) F = Fmax sin  M = 2 Fmax R [(1) sin2 90 + 2 sin2 60 + 2 sin2 30 + (1) sin2 (0)] = 6FmaxR from which, max 12 000 500 lbf . 6 6(8 / 2) MF A R    ns The simple general equation resulted from part (b) max 2MF RN  ________________________________________________________________________ 8-46 (a) From Table 8-11, Sp = 600 MPa. From Table 8-1, At = 353 mm2. Eq. (8-31):    30.9 0.9 353 600 10 190.6 kNi t pF A S    Table 8-15: K = 0.18 Eq. (8-27): T = K Fi d = 0.18(190.6)(24) = 823 Nm Ans. Chap. 8 Solutions - Rev. A, Page 39/69 • (b) Washers: t = 4.6 mm, d = 24 mm, D = 1.5(24) = 36 mm, E = 207 GPa. Eq. (8-20),           1 0.5774 207 24 31 990 MN/m 1.155 4.6 36 24 36 24 ln 1.155 4.6 36 24 36 24 k              Cast iron: t = 20 mm, d = 24 mm, D = 36 + 2(4.6) tan 30 = 41.31 mm, E = 135 GPa. Eq. (8-20)  k2 = 10 785 MN/m Steel joist: t = 20 mm, d = 24 mm, D = 41.31 mm, E = 207 GPa. Eq. (8-20)  k3 = 16 537 MN/m Eq. (8-18): km = (2 / 31 990 + 1 / 10 785 +1 / 16 537)1 = 4 636 MN/m Bolt: l = 2(4.6) + 2(20) = 49.2 mm. Nut, Table A-31, H = 21.5 mm. L > 49.2 + 21.5 = 70.7 mm. From Table A-17, use L = 80 mm. From Eq. (8-14) LT = 2(24) + 6 = 54 mm, ld = 80  54 = 26 mm, lt = 49.2  26 = 23.2 mm From Table (8-1), At = 353 mm2, Ad =  (242) / 4 = 452.4 mm2 Eq. (8-17):       452.4 353 207 1680 MN/m 452.4 23.2 353 26 d t b d t t d A A Ek A l A l      C = kb / (kb + km) = 1680 / (1680 + 4636) = 0.266, Sp = 600 MPa, Fi = 190.6 kN, P = Ptotal / N = 18/4 = 4.5 kN Yield: From Eq. (8-28)     3600 353 10 1.10 . 0.266 4.5 190.6 p t p i S A n A CP F       ns Load factor: From Eq. (8-29)     3600 353 10 190.6 17.7 . 0.266 4.5 p t i L S A F n A CP      ns Separation: From Eq. (8-30) Chap. 8 Solutions - Rev. A, Page 40/69 •    0 190.6 57.7 . 1 4.5 1 0.266 iFn A P C      ns m As was stated in the text, bolts are typically preloaded such that the yielding factor of safety is not much greater than unity which is the case for this problem. However, the other load factors indicate that the bolts are oversized for the external load. ______________________________________________________________________________ 8-47 (a) ISO M 20  2.5 grade 8.8 coarse pitch bolts, lubricated. Table 8-2, At = 245 mm2 Table 8-11, Sp = 600 MPa Fi = 0.90 At Sp = 0.90(245)600(103) = 132.3 kN Table 8-15, K = 0.18 Eq. (8-27), T = KFi d = 0.18(132.3)20 = 476 N  m Ans. (b) Table A-31, H = 18 mm, L ≥ LG + H = 48 + 18 = 66 mm. Round up to L = 80 mm per Table A-17. 2 6 2(20) 6 46 m - 80 46 34 mm - 48 34 14 mm T d T t d L d l L L l l l              Ad =  (202) /4 = 314.2 mm2, 314.2(245)(207) 1251.9 MN/m 314.2(14) 245(34) d t b d t t d A A Ek A l Al      Members: Since all members are steel use Eq. (8-22) with E = 207 MPa, l = 48 mm, d = 20mm           0.5774 207 200.5774 4236 MN/m 0.5774 0.5 0.5774 48 0.5 202ln 5 2ln 50.5774 2.5 0.5774 48 2.5 20 m Edk l d l d                1251.9 0.228 1251.9 4236 b b m kC k k      P = Ptotal / N = 40/2 = 20 kN, Yield: From Eq. (8-28)     3600 245 10 1.07 . 0.228 20 132.3 p t p i S A n A CP F       ns Chap. 8 Solutions - Rev. A, Page 41/69 • Load factor: From Eq. (8-29)     3600 245 10 132.3 3.22 . 0.228 20 p t i L S A F n A CP      ns Separation: From Eq. (8-30)    0 132.3 8.57 . 1 20 1 0.228 iFn A P C      ns ______________________________________________________________________________ 8-48 From Prob. 8-29 solution, Pmax =13.33 kips, C = 0.2, Fi = 12.77 kips, At = 0.141 9 in2 12.77 90.0 kpsi 0.141 9 i i t F A     Eq. (8-39),     0.2 13.33 9.39 kpsi 2 2 0.141 9a t CP A     Eq. (8-41), 9.39 90.0 99.39 kpsim a i       (a) Goodman Eq. (8-45) for grade 8 bolts, Se = 23.2 kpsi (Table 8-17), Sut = 150 kpsi (Table 8-9)         23.2 150 90.0 0.856 . 9.39 150 23.2 e ut i f a ut e S S n A S S          ns (b) Gerber Eq. (8-46)          2 2 2 2 1 4 2 2 1 150 150 4 23.2 23.2 90.0 150 2 90.0 23.2 1.32 . 2 9.39 23.2 f ut ut e e i ut i e a e n S S S S S S S Ans                   (c) ASME-elliptic Eq. (8-47) with Sp = 120 kpsi (Table 8-9)         2 2 2 2 2 2 2 2 2 2 23.2 120 120 23.2 90 90 23.2 1.30 . 9.39 120 23.2 e f p p e i i e a p e Sn S S S S S S Ans                ______________________________________________________________________________ 8-49 Attention to the Instructor. Part (d) requires the determination of the endurance strength, Se, of a class 5.8 bolt. Table 8-17 does not provide this and the student will be required to estimate it by other means [see the solution of part (d)]. Per bolt, Pbmax = 60/8 = 7.5 kN, Pbmin = 20/8 = 2.5 kN Chap. 8 Solutions - Rev. A, Page 42/69 • 1 0.278 1 2.6 b b m kC k k      (a) Table 8-1, At = 20.1 mm2; Table 8-11, Sp = 380 MPa Eqs. (8-31) and (8-32), Fi = 0.75 At Sp = 0.75(20.1)380(103) = 5.73 kN Yield, Eq. (8-28),     3380 20.1 10 0.98 . 0.278 7.5 5.73 p t p i S A n A CP F       ns (b) Overload, Eq. (8-29),     3380 20.1 10 5.73 0.915 . 0.278 7.5 p t i L S A F n A CP      ns (c) Separation, Eq. (8-30),    0 5.73 1.06 . 1 7.5 1 0.278 iFn A P C      ns (d) Goodman, Eq. (8-35),       3 max min 0.278 7.5 2.5 10 34.6 MPa 2 2 20.1 b b a t C P P A       Eq. (8-36),        33max min 5.73 100.278 7.5 2.5 10 354.2 MPa 2 2 20.1 20.1 b b i m t t C P P F A A         Table 8-11, Sut = 520 MPa, i = Fi /At = 5.73(103)/20.1 = 285 MPa We have a problem for Se. Table 8-17 does not list Se for class 5.8 bolts. Here, we will estimate Se using the methods of Chapter 6. Estimate eS  from the, Eq. (6-8), p. 282,  0.5 0.5 520 260 MPae utS S    . Table 6-2, p. 288, a = 4.51, b =  0.265 Eq. (6-19), p. 287,  0.2654.51 520 0.860ba utk aS    Eq. (6-21), p. 288, kb = 1 Eq. (6-26), p.290, kc = 0.85 The fatigue stress-concentration factor, from Table 8-16, is Kf = 2.2. For simple axial loading and infinite-life it is acceptable to reduce the endurance limit by Kf and use the nominal stresses in the stress/strength/design factor equations. Thus, Eq. (6-18), p. 287, Se = ka kb kc eS  / Kf = 0.86(1)0.85(260) / 2.2 = 86.4 MPa Eq. (8-38),           86.4 520 285 0.847 . 520 34.6 86.4 354.2 285 e ut i f ut a e m i S S n A S S              ns It is obvious from the various answers obtained, the bolted assembly is undersized. This can be rectified by a one or more of the following: more bolts, larger bolts, higher class bolts. ______________________________________________________________________________ 8-50 Per bolt, Pbmax = Pmax /N = 80/10 = 8 kips, Pbmin = Pmin /N = 20/10 = 2 kips C = kb / (kb + km) = 4/(4 + 12) = 0.25 (a) Table 8-2, At = 0.141 9 in2, Table 8-9, Sp = 120 kpsi and Sut = 150 kpsi Chap. 8 Solutions - Rev. A, Page 43/69 • Table 8-17, Se = 23.2 kpsi Eqs. (8-31) and (8-32), Fi = 0.75 At Sp  i = Fi /At = 0.75 Sp = 0.75(120) =90 kpsi Eq. (8-35),       max min 0.25 8 2 5.29 kpsi 2 2 0.141 9 b b a t C P P A       Eq. (8-36),       max min 0.25 8 2 90 98.81 kpsi 2 2 0.141 9 b b m i t C P P A          Eq. (8-38),           23.2 150 90 1.39 . 150 5.29 23.2 98.81 90 e ut i f ut a e m i S S n A S S              ns ______________________________________________________________________________ 8-51 From Prob. 8-33, C = 0.263, Pmax = 4.712 kN / bolt, Fi = 41.1 kN, Sp = 650 MPa, and At = 84.3 mm2 i = 0.75 Sp = 0.75(650) = 487.5 MPa Eq. (8-39):     30.263 4.712 10 7.350 MPa 2 2 84.3a t CP A     Eq. (8-40) 7.350 487.5 494.9 MPa 2 i m t t FCP A A       (a) Goodman: From Table 8-11, Sut = 900 MPa, and from Table 8-17, Se = 140 MPa Eq. (8-45):         140 900 487.5 7.55 . 7.350 900 140 e ut i f a ut e S S n A S S          ns (b) Gerber: Eq. (8-46):          2 2 2 2 1 4 2 2 1 900 900 4 140 140 487.5 900 2 487.5 140 2 7.350 140 11.4 . f ut ut e e i ut i e a e n S S S S S S S Ans                    (c) ASME-elliptic: Eq. (8-47): Chap. 8 Solutions - Rev. A, Page 44/69 •         2 2 2 2 2 2 2 2 2 2 140 650 650 140 487.5 487.5 140 9.73 . 7.350 650 140 e f p p e i i e a p e Sn S S S S S S Ans                ______________________________________________________________________________ 8-52 From Prob. 8-34, C = 0.299, Pmax = 1.443 kips/bolt,Fi = 9.05 kips, Sp = 85 kpsi, and At = 0.141 9 in2  0.75 0.75 85 63.75 kpsii pS    Eq. (8-37):     0.299 1.443 1.520 kpsi 2 2 0.141 9a t CP A     Eq. (8-38) 1.520 63.75 65.27 kpsi 2m it CP A       (a) Goodman: From Table 8-9, Sut = 120 kpsi, and from Table 8-17, Se = 18.8 kpsi Eq. (8-45):         18.8 120 63.75 5.01 . 1.520 120 18.8 e ut i f a ut e S S n A S S          ns (b) Gerber: Eq. (8-46):          2 2 2 2 1 4 2 2 1 120 120 4 18.6 18.6 63.75 120 2 63.75 18.6 2 1.520 18.6 7.45 . f ut ut e e i ut i e a e n S S S S S S S Ans                    (c) ASME-elliptic: Eq. (8-47):         2 2 2 2 2 2 2 2 2 2 18.6 85 85 18.6 63.75 63.75 18.6 6.22 . 1.520 85 18.6 e f p p e i i e a p e Sn S S S S S S Ans                ______________________________________________________________________________ Chap. 8 Solutions - Rev. A, Page 45/69 • 8-53 From Prob. 8-35, C = 0.228, Pmax = 7.679 kN/bolt, Fi = 36.1 kN, Sp = 830 MPa, and At = 58.0 mm2 i = 0.75 Sp = 0.75(830) = 622.5 MPa Eq. (8-37):     30.228 7.679 10 15.09 MPa 2 2 58.0a t CP A     Eq. (8-38) 15.09 622.5 637.6 MPa 2m it CP A       (a) Goodman: From Table 8-11, Sut = 1040 MPa, and from Table 8-17, Se = 162 MPa Eq. (8-45):         162 1040 622.5 3.73 . 15.09 1040 162 e ut i f a ut e S S n A S S          ns (b) Gerber: Eq. (8-46):          2 2 2 2 1 4 2 2 1 1040 1040 4 162 162 622.5 1040 2 622.5 162 2 15.09 162 5.74 . f ut ut e e i ut i e a e n S S S S S S S Ans                    (c) ASME-elliptic: Eq. (8-47):         2 2 2 2 2 2 2 2 2 2 162 830 830 162 622.5 622.5 162 5.62 . 15.09 830 162 e f p p e i i e a p e Sn S S S S S S Ans                ______________________________________________________________________________ 8-54 From Prob. 8-36, C = 0.291, Pmax = 1.244 kips/bolt, Fi = 9.57 kips, Sp = 120 kpsi, and At = 0.106 3 in2  0.75 0.75 120 90 kpsii pS    Eq. (8-37):     0.291 1.244 1.703 kpsi 2 2 0.106 3a t CP A     Chap. 8 Solutions - Rev. A, Page 46/69 • Eq. (8-38) 1.703 90 91.70 kpsi 2m it CP A       (a) Goodman: From Table 8-9, Sut = 150 kpsi, and from Table 8-17, Se = 23.2 kpsi Eq. (8-45):         23.2 150 90 4.72 . 1.703 150 23.2 e ut i f a ut e S S n A S S          ns (b) Gerber: Eq. (8-46):           2 2 2 2 1 4 2 2 1 150 150 4 23.2 23.2 90 150 2 90 23.2 2 1.703 23.2 7.28 . f ut ut e e i ut i e a e n S S S S S S S Ans                   (c) ASME-elliptic: Eq. (8-47):         2 2 2 2 2 2 2 2 2 2 23.2 120 120 23.2 90 90 23.2 7.24 . 1.703 120 18.6 e f p p e i i e a p e Sn S S S S S S Ans                ______________________________________________________________________________ 8-55 From Prob. 8-51, C = 0.263, Se = 140 MPa, Sut = 900 MPa, At = 84.4 mm2, i = 487.5 MPa, and Pmax = 4.712 kN. Pmin = Pmax / 2 = 4.712/2 = 2.356 kN Eq. (8-35):       3 max min 0.263 4.712 2.356 10 3.675 MPa 2 2 84.3a t C P P A       Eq. (8-36): Chap. 8 Solutions - Rev. A, Page 47/69 •       max min 3 2 0.263 4.712 2.356 10 487.5 498.5 MPa 2 84.3 m i t C P P A          Eq. (8-38):           140 900 487.5 11.9 . 900 3.675 140 498.5 487.5 e ut i f ut a e m i S S n A S S              ns ______________________________________________________________________________ 8-56 From Prob. 8-52, C = 0.299, Se = 18.8 kpsi, Sut = 120 kpsi, At = 0.141 9 in2, i = 63.75 kpsi, and Pmax = 1.443 kips Pmin = Pmax / 2 = 1.443/2 = 0.722 kips Eq. (8-35):       max min 0.299 1.443 0.722 0.760 kpsi 2 2 0.141 9a t C P P A       Eq. (8-36):       max min 2 0.299 1.443 0.722 63.75 66.03 kpsi 2 0.141 9 m i t C P P A          Eq. (8-38):           18.8 120 63.75 7.89 . 120 0.760 18.8 66.03 63.75 e ut i f ut a e m i S S n Ans S S              ______________________________________________________________________________ 8-57 From Prob. 8-53, C = 0.228, Se = 162 MPa, Sut = 1040 MPa, At = 58.0 mm2, i = 622.5 MPa, and Pmax = 7.679 kN. Pmin = Pmax / 2 = 7.679/2 = 3.840 kN Eq. (8-35):       3 max min 0.228 7.679 3.840 10 7.546 MPa 2 2 58.0a t C P P A       Chap. 8 Solutions - Rev. A, Page 48/69 • Eq. (8-36):       max min 3 2 0.228 7.679 3.840 10 622.5 645.1 MPa 2 58.0 m i t C P P A          Eq. (8-38):           162 1040 622.5 5.88 . 1040 7.546 162 645.1 622.5 e ut i f ut a e m i S S n A S S              ns ______________________________________________________________________________ 8-58 From Prob. 8-54, C = 0.291, Se = 23.2 kpsi, Sut = 150 kpsi, At = 0.106 3 in2, i = 90 kpsi, and Pmax = 1.244 kips Pmin = Pmax / 2 = 1.244/2 = 0.622 kips Eq. (8-35):       max min 0.291 1.244 0.622 0.851 kpsi 2 2 0.106 3a t C P P A       Eq. (8-36):       max min 2 0.291 1.244 0.622 90 92.55 kpsi 2 0.106 3 m i t C P P A          Eq. (8-38):           23.2 150 90 7.45 . 150 0.851 23.2 92.55 90 e ut i f ut a e m i S S n A S S              ns ______________________________________________________________________________ 8-59 Let the repeatedly-applied load be designated as P. From Table A-22, Sut = 93.7 kpsi. Referring to the Figure of Prob. 3-122, the following notation will be used for the radii of Section AA. ri = 1.5 in, ro = 2.5 in, rc = 2.0 in From Table 3-4, p. 121, with R = 0.5 in Chap. 8 Solutions - Rev. A, Page 49/69 •     2 2 2 2 2 2 2 2 0.5 1.968 246 in 2 2 2 2 0.5 2.0 1.968 246 0.031 754 in - 2.5 1.968 246 0.531 754 in - 1.968 246 1.5 0.468 246 in (1 ) / 4 0.7854 in n c c c n o o n i n i Rr r r R e r r c r r c r r A                        If P is the maximum load 2 2(0.468)1 1 26.29 0.785 4 0.031 754(1.5) 26.294 13.15 2 2 c c i i i i a m M Pr P P r c P P A er P P                         (a) Eye: Section AA, Table 6-2, p. 288, a = 14.4 kpsi, b =  0.718 Eq. (6-19), p. 287, 0.71814.4(93.7) 0.553ak   Eq. (6-23), p. 289, de = 0.370 d Eq. (6-20), p. 288, 0.1070.37 0.978 0.30b k        Eq. (6-26), p. 290, kc = 0.85 Eq. (6-8), p. 282,  0.5 0.5 93.7 46.85 kpsie utS S    Eq. (6-18) p. 287, Se = 0.553(0.978)0.85(46.85) = 21.5 kpsi From Table 6-7, p. 307, for Gerber 2 2 21 1 1 2 ut a m e f m e ut a S Sn S S                      With m = a, 2 22 21 2 1 93.7 2(21.5) 1.5571 1 1 1 2 2 13.15 (21.5) 93.7 ut e f a e ut S Sn S S P                              P where P is in kips. Chap. 8 Solutions - Rev. A, Page 50/69 • Thread: Die cut. Table 8-17 gives Se = 18.6 kpsi for rolled threads. Use Table 8-16 to find Se for die cut threads Se = 18.6(3.0/3.8) = 14.7 kpsi Table 8-2, At = 0.663 in2,  = P/At = P /0.663 = 1.51 P, a = m = /2 = 0.755 P From Table 6-7, Gerber 2 22 21 2 1 93.7 2(14.7) 19.011 1 1 1 2 2 0.755 (14.7) 93.7 ut e f a e ut S Sn S S P                              P Comparing 1910/P with 19 200/P, we conclude that the eye is weaker in fatigue. Ans. (b) Strengthening steps can include heat treatment, cold forming, cross section change (a round is a poor cross section for a curved bar in bending because the bulk of the material is located where the stress is small). Ans. (c) For nf = 2  31.557 10 779 lbf, max. load . 2 P A  ns ______________________________________________________________________________ 8-60 Member, Eq. (8-22) with E =16 Mpsi, d = 0.75 in, and l = 1.5 in           0.5774 16 0.750.5774 13.32 Mlbf/in 0.5774 0.5 0.5774 1.5 0.5 0.752ln 5 2ln 50.5774 2.5 0.5774 1.5 2.5 0.75 m Edk l d l d                Bolt, Eq. (8-13), LT = 2d + 0.25 = 2(0.75) + 0.25 = 1.75 in l = 1.5 in ld = L  LT = 2.5  1.75 = 0.75 in lt = l  ld = 1.5  0.75 = 0.75 in Table 8-2, At = 0.373 in2 Ad = (0.752)/4 = 0.442 in2 Eq. (8-17), Chap. 8 Solutions - Rev. A, Page 51/69 •       0.442 0.373 30 8.09 Mlbf/in 0.442 0.75 0.373 0.75 d t b d t t d A A Ek A l A l      8.09 0.378 8.09 13.32 b b m kC k k      Eq. (8-35),       max min 0.378 6 4 1.013 kpsi 2 2 0.373a t C P P A       Eq.(8-36),       max min 0.378 6 4 25 72.09 kpsi 2 2 0.373 0.373 i m t t C P P F A A         (a) From Table 8-9, Sp = 85 kpsi, and Eq. (8-51), the yielding factor of safety is 85 1.16 . 72.09 1.013 p p m a S n A        ns (b) From Eq. (8-29), the overload factor of safety is    max 85 0.373 25 2.96 . 0.378 6 p t i L S A F n A CP      ns (c) From Eq. (8-30), the factor of safety based on joint separation is    0 max 25 6.70 . 1 6 1 0.378 iFn A P C      ns (d) From Table 8-17, Se = 18.6 kpsi; Table 8-9, Sut = 120 kps; the preload stress is i = Fi / At = 25/0.373 = 67.0 kpsi; and from Eq. (8-38)           18.6 120 67.0 4.56 . 120 1.013 18.6 72.09 67.0 e ut i f ut a e m i S S n A S S              ns ______________________________________________________________________________ 8-61 (a) Table 8-2, At = 0.1419 in2 Table 8-9, Sp = 120 kpsi, Sut = 150 kpsi Table 8-17, Se = 23.2 kpsi Eqs. (8-31) and (8-32), i = 0.75 Sp = 0.75(120) = 90 kpsi Chap. 8 Solutions - Rev. A, Page 52/69 • 4 0.2 4 16 0.2 0.705 kpsi 2 2(0.141 9) b b m a t kC k k CP P P A          Eq. (8-45) for the Goodman criterion,     23.2(150 90) 11.4 2 5.70 kips 0.705 (150 23.2) e ut i f a ut e S S n P S S P P             .Ans (b) Fi = 0.75At Sp = 0.75(0.141 9)120 = 12.77 kips Yield, Eq. (8-28),     120 0.141 9 1.22 . 0.2 5.70 12.77 p t p i S A n A CP F      ns Load factor, Eq. (8-29), - 120(0.141 9) 12.77 3.74 . 0.2(5.70) p t i L S A F n A CP     ns Separation load factor, Eq. (8-30) 0 12.77 2.80 . (1 - ) 5.70(1 0.2) iFn A P C     ns ______________________________________________________________________________ 8-62 Table 8-2, At = 0.969 in2 (coarse), At = 1.073 in2 (fine) Table 8-9, Sp = 74 kpsi, Sut = 105 kpsi Table 8-17, Se = 16.3 kpsi Coarse thread, Fi = 0.75 At Sp = 0.75(0.969)74 = 53.78 kips i = 0.75 Sp = 0.75(74) = 55.5 kpsi 0.30 0.155 kpsi 2 2(0.969)a t CP P P A     Gerber, Eq. (8-46),          2 2 2 2 1 4 2 2 1 64.28105 105 4 16.3 16.3 55.5 105 2 55.5 16.3 2 0.155 16.3 f ut ut e e i ut i e a e n S S S S S S S P P                   With nf =2, Chap. 8 Solutions - Rev. A, Page 53/69 • 64.28 32.14 kip . 2 P A  ns Fine thread, Fi = 0.75 At Sp = 0.75(1.073)74 = 59.55kips i = 0.75 Sp = 0.75(74) = 55.5 kpsi 0.32 0.149 kpsi 2 2(1.073)a t CP P P A     The only thing that changes in Eq. (8-46) is a. Thus, 0.155 64.28 66.87 2 33.43 kips . 0.149f n P P P      Ans Percent improvement, 33.43 32.14 (100) 4% . 32.14 Ans  ______________________________________________________________________________ 8-63 For an M 30 × 3.5 ISO 8.8 bolt with P = 65 kN/bolt and C = 0.28 Table 8-1, At = 561 mm2 Table 8-11, Sp = 600 MPa, Sut = 830 MPa Table 8-17, Se = 129 MPa Eq. (8-31), Fi = 0.75Fp = 0.75 At Sp = 0.75(5610600(103) = 252.45 kN i = 0.75 Sp = 0.75(600) = 450 MPa Eq. (8-39),     30.28 65 10 16.22 MPa 2 2 561a t CP A     Gerber, Eq. (8-46),          2 2 2 2 1 4 2 2 1 830 830 4 129 129 450 830 2 450 129 2 16.22 129 4.75 . f ut ut e e i ut i e a e n S S S S S S S Ans                   The yielding factor of safety, from Eq. (8-28) is Chap. 8 Solutions - Rev. A, Page 54/69 •     3600 561 10 1.24 . 0.28 65 252.45 p t p i S A n A CP F       ns From Eq. (8-29), the load factor is     3600 561 10 252.45 4.62 . 0.28 65 p t i L S A F n A CP      ns The separation factor, from Eq. (8-30) is    0 252.45 5.39 . 1 65 1 0.28 iFn A P C      ns ______________________________________________________________________________ 8-64 (a) Table 8-2, At = 0.077 5 in2 Table 8-9, Sp = 85 kpsi, Sut = 120 kpsi Table 8-17, Se = 18.6 kpsi Unthreaded grip, 2 2 2 2 2 2 (0.375) (30) 0.245 Mlbf/in per bolt . 4(13.5) [( 2 ) - ] (4.75 - 4 ) 5.154 in 4 4 5.154(30) 1 2.148 Mlbf/in/bolt. . 12 6 d b m m m A Ek A l A D t D A Ek A l                 ns ns (b) Fi = 0.75 At Sp = 0.75(0.0775)(85) = 4.94 kip 2 0.75 0.75(85) 63.75 kpsi 2000 (4) 4189 lbf/bolt 6 4 0.245 0.102 0.245 2.148 0.102(4.189) 2.77 kpsi 2 2(0.0775) i p b b m a t S P pA kC k k CP A                     From Eq. (8-46) for Gerber fatigue criterion,          2 2 2 2 1 4 2 2 1 120 120 4 18.6 18.6 63.75 120 2 63.75 18.6 4.09 . 2 2.77 18.6 f ut ut e e i ut i e a e n S S S S S S S Ans                   Chap. 8 Solutions - Rev. A, Page 55/69 • (c) Pressure causing joint separation from Eq. (8-30) 0 2 1 (1 ) 4.94 5.50 kip 1 1 0.102 5.50 6 2.63 kpsi . (4 ) / 4 i i Fn P C FP C Pp Ans A             ______________________________________________________________________________ 8-65 From the solution of Prob. 8-64, At = 0.077 5 in2, Sut = 120 kpsi, Se = 18.6 kpsi, C = 0.102, i = 63.75 kpsi Pmax = pmaxA = 2  (42)/4 = 25.13 kpsi, Pmin = pminA = 1.2  (42)/4 = 15.08 kpsi, Eq. (8-35),       max min 0.102 25.13 15.08 6.61 kpsi 2 2 0.077 5a t C P P A       Eq. (8-36),       max min 0.102 25.13 15.08 63.75 90.21 kpsi 2 2 0.077 5m it C P P A          Eq. (8-38),           18.6 120 63.75 0.814 . 120 6.61 18.6 90.21 63.75 e ut i f ut a e m i S S n A S S              ns This predicts a fatigue failure. ______________________________________________________________________________ 8-66 Members: Sy = 57 kpsi, Ssy = 0.577(57) = 32.89 kpsi. Bolts: SAE grade 5, Sy = 92 kpsi, Ssy = 0.577(92) = 53.08 kpsi Shear in bolts, 2 2(0.25 )2 0.0982 in 4s A         0.0982(53.08) 2.61 kips 2 s sy s A S F n    Bearing on bolts, Ab = 2(0.25)0.25 = 0.125 in2 0.125(92) 5.75 kips 2 b yc b A S F n    Bearing on member, Chap. 8 Solutions - Rev. A, Page 56/69 • 0.125(57) 3.56 kips 2b F   Tension of members, At = (1.25  0.25)(0.25) = 0.25 in2 0.25(57) 7.13 kip 2 min(2.61, 5.75, 3.56, 7.13) 2.61 kip . tF F Ans     The shear in the bolts controls the design. ______________________________________________________________________________ 8-67 Members, Table A-20, Sy = 42 kpsi Bolts, Table 8-9, Sy = 130 kpsi, Ssy = 0.577(130) = 75.01 kpsi Shear of bolts,   2 25 /162 0.1534 in 4s A         5 32.6 kpsi 0.1534 s s F A     75.01 2.30 . 32.6 sySn A     ns Bearing on bolts, Ab = 2(0.25)(5/16) = 0.1563 in2 5 32.0 kpsi 0.1563b      130 4.06 . 32.0 y b S n A     ns Bearing on members, 42 1.31 . 32 y b S n A     ns Tension of members, At = [2.375  2(5/16)](1/4) = 0.4375 in2 5 11.4 kpsi 0.4375t    Chap. 8 Solutions - Rev. A, Page 57/69 • 42 3.68 . 11.4 y t S n A     ns ______________________________________________________________________________ 8-68 Members: Table A-20, Sy = 490 MPa, Ssy = 0.577(490) = 282.7 MPa Bolts: Table 8-11, ISO class 5.8, Sy = 420 MPa, Ssy = 0.577(420) = 242.3 MPa Shear in bolts, 2 2(20 )2 628.3 mm 4s A         3628.3(242.3)10 60.9 kN 2.5 s sy s A S F n     Bearing on bolts, Ab = 2(20)20 = 800 mm2 3800(420)10 134 kN 2.5 b yc b A S F n     Bearing on member, 3800(490)10 157 kN 2.5b F    Tension of members, At = (80  20)(20) = 1 200 mm2 31 200(490)10 235 kN 2.5 min(60.9, 134, 157, 235) 60.9 kN . tF F A      ns The shear in the bolts controls the design. ______________________________________________________________________________ 8-69 Members: Table A-20, Sy = 320 MPa Bolts: Table 8-11, ISO class 5.8, Sy = 420 MPa, Ssy = 0.577(420) = 242.3 MPa Shear of bolts, As =  (202)/4 = 314.2 mm2     390 10 95.48 MPa 3 314.2s    242.3 2.54 . 95.48 sy s S n A     ns Bearing on bolt, Ab = 3(20)15 = 900 mm2 Chap. 8 Solutions - Rev. A, Page 58/69 •  390 10 100 MPa 900b      420 4.2 . 100 y b S n A     ns Bearing on members, 320 3.2 . 100 y b S n A     ns Tension on members,     390 10 46.15 MPa 15[190 3 20 ] 320 6.93 . 46.15 t y t F A S n A          ns ______________________________________________________________________________ 8-70 Members: Sy = 57 kpsi Bolts: Sy = 100 kpsi, Ssy = 0.577(100) = 57.7 kpsi Shear of bolts,   2 21/ 43 0.1473 in 4 A         5 33.94 kpsi 0.1473s s F A     57.7 1.70 . 33.94 sy s S n A     ns Bearing on bolts, Ab = 3(1/4)(5/16) = 0.2344 in2 5 21.3 kpsi 0.2344b b F A        100 4.69 . 21.3 y b S n A     ns Bearing on members, Ab = 0.2344 in2 (From bearing on bolts calculation) b =  21.3 kpsi (From bearing on bolts calculation) Chap. 8 Solutions - Rev. A, Page 59/69 • 57 2.68 . 21.3 y b S n A     ns Tension in members, failure across two bolts,   25 2.375 2 1/ 4 0.5859 in 16t A      5 8.534 kpsi 0.5859t t F A     57 6.68 . 8.534 y t S n A     ns B ______________________________________________________________________________ 8-71 By symmetry, the reactions at each support is 1.6 kN. The free-body diagram for the left member is 0 1.6(250) 50 0 8 kN 0 200(1.6) 50 0 6.4 kN B A A A B M R R M R R             Members: Table A-20, Sy = 370 MPa Bolts: Table 8-11, Sy = 420 MPa, Ssy = 0.577(420) = 242.3 MPa Bolt shear, 2 2(12 ) 113.1 mm 4s A   3 max 8(10 ) 70.73 MPa 113.1 242.3 3.43 70.73 s sy F A S n         Bearing on member, Ab = td = 10(12) = 120 mm2 38(10 ) 66.67 MPa 120 370 5.55 66.67 b y b S n          Chap. 8 Solutions - Rev. A, Page 60/69 • Strength of member. The bending moments at the hole locations are: in the left member at A, MA = 1.6(200) = 320 N · m. In the right member at B, MB = 8(50) = 400 N · m. The bending moment is greater at B 3 3 3 3 3 1 [10(50 ) 10(12 )] 102.7(10 ) mm 12 400(25) (10 ) 97.37 MPa 102.7(10 ) 370 3.80 97.37 B A B A y A I M c I S n            4 At the center, call it point C, MC = 1.6(350) = 560 N · m 3 3 4 3 3 1 (10)(50 ) 104.2(10 ) mm 12 560(25) (10 ) 134.4 MPa 104.2(10 ) 370 2.75 3.80 more critical at 134.4 min(3.04, 3.80, 2.75) 2.72 . C C C C y C I M c I S n C n A              ns ______________________________________________________________________________ 8-72 The free-body diagram of the bracket, assuming the upper bolt takes all the shear and tensile load is Fs = 2500 lbf  2500 3 1071 lbf 7 P   Table A-31, H = 7/16 = 0.4375 in. Grip, l = 2(1/2) = 1 in. L ≥ l + H = 1.4375 in. Use 1.5 in bolts. Eq. (8-13), LT = 2d + 0.25 = 2(0.5) + 0.25 = 1.25 in Table 8-7, ld = L  LT = 1.5  1.25 = 0.25 in Chap. 8 Solutions - Rev. A, Page 61/69 • lt = l  ld = 1  0.25 = 0.75 in Table 8-2, At = 0.141 9 in2 Ad =  (0.52) /4 = 0.196 3 in2 Eq. (8-17),       0.196 3 0.141 9 30 4.574 Mlbf/in 0.196 3 0.75 0.141 9 0.25 d t b d t t d A A Ek A l A l      Eq. (8-22),           0.5774 30 0.50.5774 16.65 Mlbf/in 0.5774 0.5 0.5774 1 0.5 0.52ln 5 2ln 50.5774 2.5 0.5774 1 2.5 0.5 m Edk l d l d                  4.574 0.216 4.574 16.65 b b m kC k k      Table 8-9, Sp = 65 kpsi Eqs. (8-31) and (8-32), Fi = 0.75 At Sp = 0.75(0.141 9)65 = 6.918 kips i = 0.75 Sp = 0.75(65) = 48.75 kips Eq. (a), p. 440,  0.216 1.071 6.918 50.38 kpsi 0.141 9 i b t CP F A      Direct shear, 3 21.14 kpsi 0.141 9 s s  t F A   von Mises stress, Eq. (5-15), p. 223     1/21/22 2 2 23 50.38 3 21.14 62.3 kpsib s           Stress margin, m = Sp   = 65  62.3 = 3.7 kpsi Ans.  ______________________________________________________________________________ 8-73   2 2 3 2 (200) 14(50) 14(50) 1.75 kN per bolt 2(200) 7 kN/bolt 380 MPa 245 mm , (20 ) 314.2 mm 4 0.75(245)(380)(10 ) 69.83 kN 0.75 380 285 MPa s p t d i i P P F S A A F    2             Chap. 8 Solutions - Rev. A, Page 62/69 • 3 3 2 2 1/ 2 0.25(1.75) 69.83 (10 ) 287 MPa 245 7(10 ) 22.3 MPa 314.2 [287 3(22.3 )] 290 MPa 380 290 90 MPa i b t s d p CP F A F A m S                         Stress margin, m = Sp   = 380  90 = 90 MPa Ans. ______________________________________________________________________________ 8-74 Using the result of Prob. 5-67 for lubricated assembly (replace 0.2 with 0.18 per Table 8-15) 2 0.18x f TF d   With a design factor of nd gives 0.18 0.18(3)(1000) 716 2 2 (0.12) d xn F d dT d f     or T/d = 716. Also, (0.75 ) 0.18(0.75)(85 000) 11 475 p t t t T K S A d A A    Form a table Size At T/d = 11 475At n 1 4 - 28 0.0364 417.70 1.75 5 16 - 24 0.058 665.55 2.8 3 8 24 0.0878 1007.50 4.23 where the factor of safety in the last column of the table comes from 2 ( / ) 2 (0.12)( / ) 0.0042( / ) 0.18 0.18(1000)x f T d T dn T F d    Select a "38 - 24 UNF cap screw. The setting is given by T = (11 475At )d = 1007.5(0.375) = 378 lbf · in Given the coarse scale on a torque wrench, specify a torque wrench setting of 400 lbf · in. Check the factor of safety Chap. 8 Solutions - Rev. A, Page 63/69 • 2 2 (0.12)(400) 4.47 0.18 0.18(1000)(0.375)x f Tn F d      ______________________________________________________________________________ 8-75 Bolts, from Table 8-11, Sy = 420 MPa Channel, From Table A-20, Sy = 170 MPa. From Table A-7, t = 6.4 mm Cantilever, from Table A-20, Sy = 190 MPa FA = FB = FC = F / 3 M = (50 + 26 + 125) F = 201 F   201 2.01 2 50A C FF F F    Max. force, 1 2.01 2.343 3C C C F F F F F          (1) Shear on Bolts: The shoulder bolt shear area, As = (102) / 4 = 78.54 mm2 Ssy = 0.577(420) = 242.3 KPa max syC s SF A n    From Eq. (1), FC = 2.343 F. Thus 3242.3 78.54 10 4.06 kN 2.343 2.0 2.343 sy sS AF n            Bearing on bolt: The bearing area is Ab = td = 6.4(10) = 64 mm2. Similar to shear Chap. 8 Solutions - Rev. A, Page 64/69 • 3420 64 10 5.74 kN 2.343 2.0 2.343 y bS AF n            Bearing on channel: Ab = 64 mm2, Sy = 170 MPa. 3170 64 10 2.32 kN 2.343 2.0 2.343 y bS AF n            Bearing on cantilever: Ab = 12(10) = 120 mm2, Sy = 190 MPa. 3190 120 10 4.87 kN 2.343 2.0 2.343 y bS AF n            Bending of cantilever: At C     3 3 51 12 50 10 1.24 10 mm12I    4 max 151 151 y yS SMc Fc IF n I I n c                 5 3 1.24 10190 10 3.12 kN 2.0 151 25 F          So F = 2.32 kN based on bearing on channel. Ans. ______________________________________________________________________________ 8-76 Bolts, from Table 8-11, Sy = 420 MPa Bracket, from Table A-20, Sy = 210 MPa 2 2 12 4 kN; 12(200) 2400 N · m 3 2400 37.5 kN 64 (4) (37.5) 37.7 kN 4 kN A B A B O F M F F F F F               Bolt shear: The shoulder bolt shear area, As = (122) / 4 = 113.1 mm2 Ssy = 0.577(420) = 242.3 KPa Chap. 8 Solutions - Rev. A, Page 65/69 • 337.7(10) 333 MPa 113 242.3 0.728 . 333 sySn A        ns Bearing on bolts: 2 3 12(8) 96 mm 37.7(10) 393 MPa 96 420 1.07 . 393 b b yc b A S n A            ns Bearing on member: 393 MPa 210 0.534 . 393 b yc b S n Ans        Bending stress in plate: 3 3 3 2 3 3 3 2 6 4 3 6 2 12 12 12 8(136) 8(12) 8(12)2 (32) (8)( 12 12 12 1.48(10) mm . 2400(68) (10) 110 MPa 1.48(10) 210 1.91 . 110 y bh bd bdI a bd Ans Mc I S n Ans  12)                           Failure is predicted for bolt shear and bearing on member. ______________________________________________________________________________ Chap. 8 Solutions - Rev. A, Page 66/69 • 8-77 3625 1208 lbf 3 1208 125 1083 lbf, 1208 125 1333 lbf A B A B F F F F               Bolt shear: As = ( / 4)(0.3752) = 0.1104 in2 maxmax 1333 12 070 psi 0.1104s F A     From Table 8-10, Sy = 100 kpsi, Ssy = 0.577(100) = 57.7 kpsi max 57.7 4.78 . 12.07 sySn A     ns Bearing on bolt: Bearing area is Ab = td = 0.375 (0.375) = 0.1406 in2. 1333 9 481 psi 0.1406b b F A        100 10.55 . 9.481 y b S n A     ns Bearing on member: From Table A-20, Sy = 54 kpsi. Bearing stress same as bolt 54 5.70 . 9.481 y b S n A     ns Bending of member: At B, M = 250(13) = 3250 lbfin Chap. 8 Solutions - Rev. A, Page 67/69 • 3 3 41 3 32 0.2484 in 12 8 8 I                   3250 1 13 080 psi 0.2484 Mc I     54 4.13 . 13.08 ySn A     ns ______________________________________________________________________________ 8-78 The direct shear load per bolt is F = 2000/6 = 333.3 lbf. The moment is taken only by the four outside bolts. This moment is M = 2000(5) = 10 000 lbf · in. Thus 10 000 1000 lbf 2(5) F   and the resultant bolt load is 2 2(333.3) (1000) 1054 lbfF    Bolt strength, Table 8-9, Sy = 100 kpsi; Channel and Plate strength, Sy = 42 kpsi Shear of bolt: As =  (0.5)2/4 = 0.1963 in2 (0.577)(100) 10.7 . 1.054 / 0.1963 sySn A     ns Bearing on bolt: Channel thickness is t = 3/16 in, Ab = 0.5(3/16) = 0.09375 in2 100 8.89 . 1.054 / 0.09375 n A  ns Bearing on channel: 42 3.74 . 1.054 / 0.09375 n A  ns Bearing on plate: Ab = 0.5(0.25) = 0.125 in2 42 4.98 . 1.054 / 0.125 n A  ns Strength of plate:    3 3 3 2 4 0.25(7.5) 0.25(0.5) 12 12 0.25(0.5) 2 0.25 0.5 (2.5) 7.219 in 12 I           Chap. 8 Solutions - Rev. A, Page 68/69 • 5000 lbf · in per plate 5000(3.75) 2597 psi 7.219 42 16.2 . 2.597 M Mc I n Ans        ______________________________________________________________________________ 8-79 to 8-81 Specifying bolts, screws, dowels and rivets is the way a student learns about such components. However, choosing an array a priori is based on experience. Here is a chance for students to build some experience. Chap. 8 Solutions - Rev. A, Page 69/69 • Chapter 9 Figure for Probs. 9-1 to 9-4 9-1 Given, b = 50 mm, d = 50 mm, h = 5 mm, allow = 140 MPa. F = 0.707 hlallow = 0.707(5)[2(50)](140)(103) = 49.5 kN Ans. ______________________________________________________________________________ 9-2 Given, b = 2 in, d = 2 in, h = 5/16 in, allow = 25 kpsi. F = 0.707 hlallow = 0.707(5/16)[2(2)](25) = 22.1 kip Ans. ______________________________________________________________________________ 9-3 Given, b = 50 mm, d = 30 mm, h = 5 mm, allow = 140 MPa. F = 0.707 hlallow = 0.707(5)[2(50)](140)(103) = 49.5 kN Ans. ______________________________________________________________________________ 9-4 Given, b = 4 in, d = 2 in, h = 5/16 in, allow = 25 kpsi. F = 0.707 hlallow = 0.707(5/16)[2(4)](25) = 44.2 kip Ans. ______________________________________________________________________________ 9-5 Prob. 9-1 with E7010 Electrode. Table 9-6: f = 14.85 h kip/in = 14.85 [5 mm/(25.4 mm/in)] = 2.923 kip/in = 2.923(4.45/25.4) = 0.512 kN/mm F = f l = 0.512[2(50)] = 51.2 kN Ans. ______________________________________________________________________________ 9-6 Prob. 9-2 with E6010 Electrode. Table 9-6: f = 14.85 h kip/in = 14.85(5/16) = 4.64 kip/in Chapter 9, Page 1/36 • F = f l = 4.64[2(2)] = 18.6 kip Ans. ______________________________________________________________________________ 9-7 Prob. 9-3 with E7010 Electrode. Table 9-6: f = 14.85 h kip/in = 14.85 [5 mm/(25.4 mm/in)] = 2.923 kip/in = 2.923(4.45/25.4) = 0.512 kN/mm F = f l = 0.512[2(50)] = 51.2 kN Ans. ______________________________________________________________________________ 9-8 Prob. 9-4 with E6010 Electrode. Table 9-6: f = 14.85 h kip/in = 14.85(5/16) = 4.64 kip/in F = f l = 4.64[2(4)] = 37.1 kip Ans. ______________________________________________________________________________ 9-9 Table A-20: 1018 CD: Sut = 440 MPa, Sy = 370 MPa 1018 HR: Sut = 400 MPa, Sy = 220 MPa Cold-rolled properties degrade to hot-rolled properties in the neighborhood of the weld. Table 9-4: all min(0.30 , 0.40 ) min[0.30(400), 0.40(220)] min(120, 88) 88 MPa ut yS S     for both materials. Eq. (9-3): F = 0.707hlall = 0.707(5)[2(50)](88)(103) = 31.1 kN Ans. ______________________________________________________________________________ 9-10 Table A-20: 1020 CD: Sut = 68 kpsi, Sy = 57 kpsi 1020 HR: Sut = 55 kpsi, Sy = 30 kpsi Cold-rolled properties degrade to hot-rolled properties in the neighborhood of the weld. Table 9-4: all min(0.30 , 0.40 ) min[0.30(55), 0.40(30)] min(16.5, 12.0) 12.0 kpsi ut yS S     for both materials. Eq. (9-3): F = 0.707hlall = 0.707(5/16)[2(2)](12.0) = 10.6 kip Ans. ______________________________________________________________________________ Chapter 9, Page 2/36 • 9-11 Table A-20: 1035 HR: Sut = 500 MPa, Sy = 270 MPa 1035 CD: Sut = 550 MPa, Sy = 460 MPa Cold-rolled properties degrade to hot-rolled properties in the neighborhood of the weld. Table 9-4: all min(0.30 , 0.40 ) min[0.30(500), 0.40(270)] min(150, 108) 108 MPa ut yS S     for both materials. Eq. (9-3): F = 0.707hlall = 0.707(5)[2(50)](108)(103) = 38.2 kN Ans. ______________________________________________________________________________ 9-12 Table A-20: 1035 HR: Sut = 72 kpsi, Sy = 39.5 kpsi 1020 CD: Sut = 68 kpsi, Sy = 57 kpsi, 1020 HR: Sut = 55 kpsi, Sy = 30 kpsi Cold-rolled properties degrade to hot-rolled properties in the neighborhood of the weld. Table 9-4: all min(0.30 , 0.40 ) min[0.30(55), 0.40(30)] min(16.5, 12.0) 12.0 kpsi ut yS S     for both materials. Eq. (9-3): F = 0.707hlall = 0.707(5/16)[2(4)](12.0) = 21.2 kip Ans. ______________________________________________________________________________ 9-13 Eq. (9-3):      32 100 102 141 MPa . 5 2 50 50 F Ans hl        ______________________________________________________________________________ 9-14 Eq. (9-3):       2 402 22.6 kpsi . 5 /16 2 2 2 F Ans hl        ______________________________________________________________________________ 9-15 Eq. (9-3):      32 100 102 177 MPa . 5 2 50 30 F Ans hl        ______________________________________________________________________________ 9-16 Eq. (9-3):       2 402 15.1 kpsi . 5 /16 2 4 2 F Ans hl        ______________________________________________________________________________ Chapter 9, Page 3/36 • 9-17 b = d =50 mm, c = 150 mm, h = 5 mm, and allow = 140 MPa. (a) Primary shear, Table 9-1, Case 2 (Note: b and d are interchanged between problem figure and table figure. Note, also, F in kN and  in MPa):      310 2.829 1.414 5 50y FV F A      Secondary shear, Table 9-1:       2 22 2 3 3 50 3 50 503 83.33 10 mm 6 6u d b d J       J = 0.707 h Ju = 0.707(5)(83.33)(103) = 294.6(103) mm4      3 3 175 10 25 14.85 294.6 10 y x y FMr F J          2 22 2max 14.85 2.829 14.85 23.1x y y F F            (1) allow 140 6.06 kN . 23.1 23.1 F Ans   (b) For E7010 from Table 9-6, allow = 21 kpsi = 21(6.89) = 145 MPa 1020 HR bar: Sut = 380 MPa, Sy = 210 MPa 1015 HR support: Sut = 340 MPa, Sy = 190 MPa Table 9-3, E7010 Electrode: Sut = 482 MPa, Sy = 393 MPa The support controls the design. Table 9-4: allow = min(0.30Sut, 0.40Sy ) =min[0.30(340), 0.40(190) = min(102, 76) = 76 MPa The allowable load, from Eq. (1) is allow 76 3.29 kN . 23.1 23.1 F Ans   ______________________________________________________________________________ 9-18 b = d =2 in, c = 6 in, h = 5/16 in, and allow = 25 kpsi. Chapter 9, Page 4/36 • (a) Primary shear, Table 9-1(Note: b and d are interchanged between problem figure and table figure. Note, also, F in kip and  in kpsi):    1.132 1.414 5 /16 2y V F F A      Secondary shear, Table 9-1:    2 22 2 32 3 2 23 5.333 in 6 6u d b d J       J = 0.707 h Ju = 0.707(5/16)(5.333) = 1.178 in4  7 1 5.942 1.178 y x y Mr F F J          2 22 2max 5.942 1.132 5.942 9.24x y y F F            (1) allow 25 2.71 kip . 9.24 9.24 F Ans   (b) For E7010 from Table 9-6, allow = 21 kpsi 1020 HR bar: Sut = 55 kpsi, Sy = 30 kpsi 1015 HR support: Sut = 50 kpsi, Sy = 27.5 kpsi Table 9-3, E7010 Electrode: Sut = 70 kpsi, Sy = 57 kpsi The support controls the design. Table 9-4: allow = min(0.30Sut, 0.40Sy ) =min[0.30(50), 0.40(27.5) = min(15, 11) = 11 kpsi The allowable load, from Eq. (1) is allow 11 1.19 kip . 9.24 9.24 F Ans   ______________________________________________________________________________ 9-19 b =50 mm, c = 150 mm, d = 30 mm, h = 5 mm, and allow = 140 MPa. (a) Primary shear, Table 9-1, Case 2 (Note: b and d are interchanged between problem figure and table figure. Note, also, F in kN and  in MPa): Chapter 9, Page 5/36 •      310 2.829 1.414 5 50y FV F A      Secondary shear, Table 9-1:       2 22 2 3 3 50 3 30 503 43.33 10 mm 6 6u d b d J       J = 0.707 h Ju = 0.707(5)(43.33)(103) = 153.2(103) mm4      3 3 175 10 15 17.13 153.2 10 y x FMr F J           3 3 175 10 25 28.55 153.2 10 x y FMr F J         2 22 2max 17.13 2.829 28.55 35.8x y y F F            (1) allow 140 3.91 kN . 35.8 35.8 F Ans   (b) For E7010 from Table 9-6, allow = 21 kpsi = 21(6.89) = 145 MPa 1020 HR bar: Sut = 380 MPa, Sy = 210 MPa 1015 HR support: Sut = 340 MPa, Sy = 190 MPa Table 9-3, E7010 Electrode: Sut = 482 MPa, Sy = 393 MPa The support controls the design. Table 9-4: allow = min(0.30Sut, 0.40Sy ) =min[0.30(340), 0.40(190) = min(102, 76) = 76 MPa The allowable load, from Eq. (1) is allow 76 2.12 kN . 35.8 35.8 F Ans   ______________________________________________________________________________ 9-20 b = 4 in, c = 6 in, d = 2 in, h = 5/16 in, and allow = 25 kpsi. Chapter 9, Page 6/36 • (a) Primary shear, Table 9-1(Note: b and d are interchanged between problem figure and table figure. Note, also, F in kip and  in kpsi):    0.5658 1.414 5 /16 4y V F F A      Secondary shear, Table 9-1:    2 22 2 34 3 2 43 18.67 in 6 6u d b d J       J = 0.707 h Ju = 0.707(5/16)(18.67) = 4.125 in4  8 1 1.939 4.125 y x Mr F F J       8 2 3.879 4.125 x y FMr F J         2 22 2max 1.939 0.5658 3.879 4.85x y y F F            (1) allow 25 5.15 kip . 4.85 4.85 F Ans   (b) For E7010 from Table 9-6, allow = 21 kpsi 1020 HR bar: Sut = 55 kpsi, Sy = 30 kpsi 1015 HR support: Sut = 50 kpsi, Sy = 27.5 kpsi Table 9-3, E7010 Electrode: Sut = 70 kpsi, Sy = 57 kpsi The support controls the design. Table 9-4: allow = min(0.30Sut, 0.40Sy ) =min[0.30(50), 0.40(27.5) = min(15, 11) = 11 kpsi The allowable load, from Eq. (1) is allow 11 2.27 kip . 4.85 4.85 F Ans   ______________________________________________________________________________ Chapter 9, Page 7/36 • 9-21 Given, b = 50 mm, c = 150 mm, d = 50 mm, h = 5 mm, allow = 140 MPa. Primary shear (F in kN,  in MPa, A in mm2):      310 1.414 1.414 5 50 50y FV F A       Secondary shear: Table 9-1:       3 3 3 350 50 166.7 10 mm 6 6u b d J      J = 0.707 h Ju = 0.707(5)166.7(103) = 589.2(103) mm4     3 3 175 10 (25) 7.425 589.2 10 y x y FMr F J       Maximum shear:    2 22 2max 7.425 1.414 7.425 11.54x y y F F            140 12.1 kN . 11.54 11.54 allowF Ans   ______________________________________________________________________________ 9-22 Given, b = 2 in, c = 6 in, d = 2 in, h = 5/16 in, allow = 25 kpsi. Primary shear:    0.5658 1.414 5 /16 2 2y V F F A       Secondary shear: Table 9-1:     3 3 32 2 10.67 in 6 6u b d J      J = 0.707 h Ju = 0.707(5/16)10.67 = 2.357 in4 7 (1) 2.970 2.357 y x y Mr F F J       Maximum shear:    2 22 2max 2.970 0.566 2.970 4.618x y y F F            25 5.41 kip . 4.618 4.618 allowF Ans   ______________________________________________________________________________ 9-23 Given, b = 50 mm, c = 150 mm, d = 30 mm, h = 5 mm, allow = 140 MPa. Chapter 9, Page 8/36 • Primary shear (F in kN,  in MPa, A in mm2):      310 1.768 1.414 5 50 30y FV F A       Secondary shear: Table 9-1:       3 3 3 350 30 85.33 10 mm 6 6u b d J      J = 0.707 h Ju = 0.707(5)85.33(103) = 301.6(103) mm4     3 3 175 10 (15) 8.704 301.6 10 y x FMr F J          3 3 175 10 (25) 14.51 301.6 10 x y FMr F J      Maximum shear:    2 22 2max 8.704 1.768 14.51 18.46x y y F F            140 7.58 kN . 18.46 18.46 allowF Ans   ______________________________________________________________________________ 9-24 Given, b = 4 in, c = 6 in, d = 2 in, h = 5/16 in, allow = 25 kpsi. Primary shear:    0.3772 1.414 5 /16 4 2y V F F A       Secondary shear: Table 9-1:     3 3 34 2 36 in 6 6u b d J      J = 0.707 h Ju = 0.707(5/16)36 = 7.954 in4 8 (1) 1.006 7.954 y x Mr F F J      8 (2) 2.012 7.954 x y Mr F F J      Maximum shear:    2 22 2max 1.006 0.3772 2.012 2.592x y y F F            Chapter 9, Page 9/36 • 25 9.65kip . 2.592 2.592 allowF Ans   ______________________________________________________________________________ 9-25 Given, b = 50 mm, d = 50 mm, h = 5 mm, E6010 electrode. A = 0.707(5)(50 +50 + 50) = 530.3 mm2 Member endurance limit: From Table A-20 for AISI 1010 HR, Sut = 320 MPa. Eq. 6-19 and Table 6-2, pp. 287, 288: ka = 272(320)0.995 = 0.875 kb = 1 (uniform shear), kc = 0.59 (torsion, shear), kd = 1 Eqs. (6-8) and (6-18): Se = 0.875(1)(0.59)(1)(0.5)(320) = 82.6 MPa Electrode endurance: E6010, Table 9-3, Sut = 427 MPa Eq. 6-19 and Table 6-2, pp. 287, 288: ka = 272(427)0.995 = 0.657 As before, kb = 1 (direct shear), kc = 0.59 (torsion, shear), kd = 1 Se = 0.657(1)(0.59)(1)(0.5)(427) = 82.8 MPa The members and electrode are basically of equal strength. We will use Se = 82.6 MPa. For a factor of safety of 1, and with Kfs = 2.7 (Table 9-5)    3allow 82.6 530.3 16.2 10 N 16.2 kN .2.7fs AF Ans K      ______________________________________________________________________________ 9-26 Given, b = 2 in, d = 2 in, h = 5/16 in, E6010 electrode. A = 0.707(5/16)(2 +2 + 2) = 1.326 in2 Member endurance limit: From Table A-20 for AISI 1010 HR, Sut = 47 kpsi. Eq. 6-19 and Table 6-2, pp. 287, 288: ka = 39.9(47)0.995 = 0.865 kb = 1 (uniform shear), kc = 0.59 (torsion, shear), kd = 1 Eqs. (6-8) and (6-18): Se = 0.865(1)(0.59)(1)(0.5)(47) = 12.0 kpsi Electrode endurance: E6010, Table 9-3, Sut = 62 kpsi Eq. 6-19 and Table 6-2, pp. 287, 288: ka = 39.9(62)0.995 = 0.657 Chapter 9, Page 10/36 • As before, kb = 1 (uniform shear), kc = 0.59 (torsion, shear), kd = 1 Se = 0.657(1)(0.59)(1)(0.5)(62) = 12.0 kpsi Thus the members and electrode are of equal strength. For a factor of safety of 1, and with Kfs = 2.7 (Table 9-5)  allow 12.0 1.326 5.89 kip . 2.7fs AF Ans K     ______________________________________________________________________________ 9-27 Given, b = 50 mm, d = 30 mm, h = 5 mm, E7010 electrode. A = 0.707(5)(50 +50 + 30) = 459.6 mm2 Member endurance limit: From Table A-20 for AISI 1010 HR, Sut = 320 MPa. Eq. 6-19 and Table 6-2, pp. 287, 288: ka = 272(320)0.995 = 0.875 kb = 1 (direct shear), kc = 0.59 (torsion, shear), kd = 1 Eqs. (6-8) and (6-18): Se = 0.875(1)(0.59)(1)(0.5)(320) = 82.6 MPa Electrode endurance: E6010, Table 9-3, Sut = 482 MPa Eq. 6-19 and Table 6-2, pp. 287, 288: ka = 272(482)0.995 = 0.582 As before, kb = 1 (direct shear), kc = 0.59 (torsion, shear), kd = 1 Se = 0.582(1)(0.59)(1)(0.5)(482) = 82.7 MPa The members and electrode are basically of equal strength. We will use Se =82.6 MPa. For a factor of safety of 1, and with Kfs = 2.7 (Table 9-5)    3allow 82.6 459.6 14.1 10 N 14.1 kN .2.7fs AF Ans K      ______________________________________________________________________________ 9-28 Given, b = 4 in, d = 2 in, h = 5/16 in, E7010 electrode. A = 0.707(5/16)(4 +4 + 2) = 2.209 in2 Member endurance limit: From Table A-20 for AISI 1010 HR, Sut = 47 kpsi. Eq. 6-19 and Table 6-2, pp. 287, 288: ka = 39.9(47)0.995 = 0.865 kb = 1 (direct shear), kc = 0.59 (torsion, shear), kd = 1 Chapter 9, Page 11/36 • Eqs. (6-8) and (6-18): Se = 0.865(1)(0.59)(1)(0.5)(47) = 12.0 kpsi Electrode endurance: E7010, Table 9-3, Sut = 70 kpsi Eq. 6-19 and Table 6-2, pp. 287, 288: ka = 39.9(70)0.995 = 0.582 As before, kb = 1 (direct shear), kc = 0.59 (torsion, shear), kd = 1 Se = 0.582(1)(0.59)(1)(0.5)(70) = 12.0 kpsi Thus the members and electrode are of equal strength. For a factor of safety of 1, and with Kfs = 2.7 (Table 9-5)  allow 12.0 2.209 9.82 kip . 2.7fs AF Ans K     ______________________________________________________________________________ 9-29 Primary shear:  = 0 (why?) Secondary shear: Table 9-1: Ju = 2 r3 = 2 (1.5)3 = 21.21 in3 J = 0.707 h Ju = 0.707(1/4)(21.21) = 3.749 in4 2 welds:     8 1.5 1.600 2 2 3.749 FMr F J      allow 1.600 20 12.5 kip .F F Ans       ______________________________________________________________________________ 9-30 l = 2 + 4 + 4 = 10 in             2 1 4 0 4 2 1 in 10 2 4 4 2 4 0 1.6 in 10 x y         M = FR = F(10  1) = 9 F      2 2 221 21 1 4 1.6 2.4 in, 1 2 1.6 1.077 inr r          2 23 2 1 1.6 1.887 inr     Chapter 9, Page 12/36 •     1 3 41 0.707 5 /16 2 0.1473 in 12G J       2 3 3 41 0.707 5 /16 4 1.178 in 12G G J J                 3 2 1 2 2 2 4 0.1473 0.707 5 /16 2 2.4 1.178 0.707 5 /16 4 1.077 1.178 0.707 5 /16 4 1.887 9.220 in ii i G i J J A r            1 o1.6tan 28.07 4 1         221.6 4 1 3.4 inr     Primary shear ( in kpsi, F in kip) :    0.4526 0.707 5 /16 10 V F F A      Secondary shear:  9 3.4 3.319 9.220 FMr F J          2 2o o max 3.319 sin 28.07 3.319 cos 28.07 0.4526 3.724 F F F      F max = allow  3.724 F = 25  F = 6.71 kip Ans. ______________________________________________________________________________ Chapter 9, Page 13/36 • 9-31 l = 30 + 50 + 50 = 130 mm             30 15 50 0 50 25 13.08 mm 130 30 50 50 25 50 0 21.15 mm 130 x y         M = FR = F(200  13.08) = 186.92 F (M in Nm, F in kN)      2 2 221 215 13.08 50 21.15 28.92 mm, 13.08 25 21.15 13.63 mmr r          2 23 25 13.08 21.15 24.28 mmr           1 3 31 0.707 5 30 7.954 10 mm 12G J   4        2 3 3 31 0.707 5 50 36.82 10 mm 12G G J J   4 2                       3 2 1 3 2 3 3 2 3 4 7.954 10 0.707 5 30 28.92 36.82 10 0.707 5 50 13.63 36.82 10 0.707 5 50 24.28 307.3 10 mm ii i G i J J A r            1 o21.15tan 29.81 50 13.08          2221.15 50 13.08 42.55 mmr     Primary shear ( in MPa, F in kN) :      310 2.176 0.707 5 130 FV F A      Secondary shear:      3 3 186.92 10 42.55 25.88 307.3 10 FMr F J      Chapter 9, Page 14/36 •     2 2o o max 25.88 sin 29.81 25.88 cos 29.81 2.176 27.79 F F F      F max = allow  27.79 F = 140  F = 5.04 kN Ans. ______________________________________________________________________________ 9-32 Weld Pattern Figure of merit Rank______ 1. 3 2/12fom 0.0833 12 uJ a a a lh ah h h           2  5 2.     2 2 2 23 fom 0.3333 6 2 3 a a a a a h h h           a  1 3.     4 2 2 2 22 6 5fom 0.2083 12 2 24 a a a a a a a ah h h            4 4. 3 3 3 4 21 8 6fom 0.3056 3 12 2 a a a a a ah a a h               2 5.   3 3 22 1 8fom 0.3333 6 4 24 a a a h a ah h           1 6.   3 3 22 / 2 fom 0.25 4 a a a ah ah h             3 ______________________________________________________________________________ Chapter 9, Page 15/36 • 9-33 Weld Pattern Figure of merit Rank______ 1.  3 2/12 fom 0.0833u aI a lh ah h           6 2.  3 2/ 6 fom 0.0833 2 a a ah h          6 3.  2 2/ 2 fom 0.25 2 aa a ah h          1 4.*   2 2 2/12 6 7fom 0.1944 3 36 a a a a ah h h           a  2 5. & 7. 2 , 2 2 a ax y a a    3 a    23 3 22 2 2 3 3 3u a a aI a a a          3 a  3 2 2/ 3 1fom 0.1111 3 9 u aI a a lh ah h h                 5 6. & 8.   2 2 2/ 6 3 1fom 0.1667 4 6 a a a a a ah h h                 3 9.   3 2 2/ 2 fom 0.125 8 a a a ah h h             4 *Note. Because this section is not symmetric with the vertical axis, out-of-plane deflection may occur unless special precautions are taken. See the topic of “shear center” in books with more advanced treatments of mechanics of materials. ______________________________________________________________________________ 9-34 Attachment and member (1018 HR), Sy = 220 MPa and Sut = 400 MPa. The member and attachment are weak compared to the properties of the lowest electrode. Decision Specify the E6010 electrode Controlling property, Table 9-4: all = min[0.3(400), 0.4(220)] = min(120, 88) = 88 MPa For a static load, the parallel and transverse fillets are the same. Let the length of a bead be l = 75 mm, and n be the number of beads. Chapter 9, Page 16/36 •  0.707 all F n hl         3 all 100 10 21.43 0.707 0.707 75 88 Fnh l    where h is in millimeters. Make a table Number of beads, n Leg size, h (mm) 1 21.43 2 10.71 3 7.14 4 5.36  6 mm Decision Specify h = 6 mm on all four sides. Weldment specification: Pattern: All-around square, four beads each side, 75 mm long Electrode: E6010 Leg size: h = 6 mm ______________________________________________________________________________ 9-35 Decision: Choose a parallel fillet weldment pattern. By so-doing, we’ve chosen an optimal pattern (see Prob. 9-32) and have thus reduced a synthesis problem to an analysis problem: Table 9-1, case 2, rotated 90: A = 1.414hd = 1.414(h)(75) = 106.05h mm2 Primary shear  312 10 113.2 106.05y     V A h h  Secondary shear:       2 2 2 2 3 3 3 3 (3 ) 6 75[3(75 ) 75 ] 281.3 10 mm 6 0.707( )(281.3) 10 198.8 10 mm u d b dJ J h h    4    With  = 45, Chapter 9, Page 17/36 •       3o 3 22 2 max 12 10 (187.5)(37.5)cos 45 424.4 198.8 10 1 684.9424.4 (113.2 424.4) y x y x y y MrMr J J hh h h                     2 Attachment and member (1018 HR): Sy = 220 MPa, Sut = 400 MPa Decision: Use E60XX electrode which is stronger all max all min[0.3(400), 0.4(220)] 88 MPa 684.9 88 MPa 684.9 7.78 mm 88 h h           Decision: Specify 8 mm leg size Weldment Specifications: Pattern: Parallel horizontal fillet welds Electrode: E6010 Type: Fillet Length of each bead: 75 mm Leg size: 8 mm ______________________________________________________________________________ 9-36 Problem 9-35 solves the problem using parallel horizontal fillet welds, each 75 mm long obtaining a leg size rounded up to 8 mm. For this problem, since the width of the plate is fixed and the length has not been determined, we will explore reducing the leg size by using two vertical beads 75 mm long and two horizontal beads such that the beads have a leg size of 6 mm. Decision: Use a rectangular weld bead pattern with a leg size of 6 mm (case 5 of Table 9-1 with b unknown and d = 75 mm). Materials: Attachment and member (1018 HR): Sy = 220 MPa, Sut = 400 MPa From Table 9-4, AISC welding code, all = min[0.3(400), 0.4(220)] = min(120, 88) = 88 MPa Select a stronger electrode material from Table 9-3. Decision: Specify E6010 Solving for b: In Prob. 9-35, every term was linear in the unknown h. This made solving for h relatively easy. In this problem, the terms will not be linear in b, and so we will use an iterative solution with a spreadsheet. Throat area and other properties from Table 9-1: A = 1.414(6)(b + 75) = 8.484(b + 75) (1) Chapter 9, Page 18/36 •   375 6u b J   , J = 0.707 (6) Ju = 0.707(b +75)3 (2) Primary shear ( in MPa, h in mm):  312 10 (3)y V A A     Secondary shear (See Prob. 9-35 solution for the definition of ) :             3 3 3 3 22 max 12 10 150 / 2 (37.5) cos cos (4) 0.707 75 12 10 150 / 2 ( / 2) sin sin (5) 0.707 75 (6) y x x y y x y Mr J bMrMr J J b b bMr Mr J J b                                   Enter Eqs. (1) to (6) into a spreadsheet and iterate for various values of b. A portion of the spreadsheet is shown below. b (mm) A (mm2) J (mm4) 'y (Mpa) "y (Mpa) "x (Mpa) max (Mpa) 41 984.144 1103553.5 12.19334 69.5254 38.00722 90.12492 42 992.628 1132340.4 12.08912 67.9566 38.05569 88.63156 43 1001.112 1161623.6 11.98667 66.43718 38.09065 87.18485 < 88 Mpa 44 1009.596 1191407.4 11.88594 64.96518 38.11291 85.7828 We see that b  43 mm meets the strength goal. Weldment Specifications: Pattern: Horizontal parallel weld tracks 43 mm long, vertical parallel weld tracks 75 mm long Electrode: E6010 Leg size: 6 mm ______________________________________________________________________________ 9-37 Materials: Member and attachment (1018 HR): 32 kpsi, 58 kpsiy utS S  Table 9-4: all min[0.3(58), 0.4(32)] 12.8 kpsi   Chapter 9, Page 19/36 • Decision: Use E6010 electrode. From Table 9-3: 50 kpsi, 62 kpsi,y utS S  all min[0.3(62), 0.4(50)] 20 kpsi   Decision: Since 1018 HR is weaker than the E6010 electrode, use all 12.8 kpsi  Decision: Use an all-around square weld bead track. l1 = 6 + a = 6 + 6.25 = 12.25 in Throat area and other properties from Table 9-1: 1.414 ( ) 1.414( )(6 6) 16.97A h b d h h     Primary shear  320 10 1179 psi 16.97y V F A A h h       Secondary shear 3 3 3( ) (6 6) 288 in 6 6u b dJ     40.707 (288) 203.6 inJ h h   320 10 (6.25 3)(3) 2726 psi 203.6 y x y Mr J h         h 2 2 2 2 max 1 4762( ) 2726 (1179 2726) psix y y h h             Relate stress to strength     3 max all 3 4762 476212.8 10 0.372 in 12.8 10 h h        Decision: Specify in leg size 3 / 8 Specifications: Pattern: All-around square weld bead track Electrode: E6010 Type of weld: Fillet Weld bead length: 24 in Leg size: in 3 / 8 Attachment length: 12.25 in ______________________________________________________________________________ Chapter 9, Page 20/36 • 9-38 This is a good analysis task to test a student’s understanding. (1) Solicit information related to a priori decisions. (2) Solicit design variables b and d. (3) Find h and round and output all parameters on a single screen. Allow return to Step 1 or Step 2. (4) When the iteration is complete, the final display can be the bulk of your adequacy assessment. Such a program can teach too. ______________________________________________________________________________ 9-39 The objective of this design task is to have the students teach themselves that the weld patterns of Table 9-2 can be added or subtracted to obtain the properties of a contemplated weld pattern. The instructor can control the level of complication. We have left the presentation of the drawing to you. Here is one possibility. Study the problem’s opportunities, and then present this (or your sketch) with the problem assignment. Use as the design variable. Express properties as a function of From Table 9-3, 1b 1.b case 3: 11.414 ( )A h b b 2 22 1 1( ) 2 2 2u b d b b dbdI    0.707 uI hI 11.414 ( ) V F A h b b      ( / 2) 0.707 u Mc Fa d I hI     Parametric study Let 1 all10 in, 8 in, 8 in, 2 in, 12.8 kpsi, 2(8 2) 12 ina b d b l        Chapter 9, Page 21/36 • 21.414 (8 2) 8.48 inA h h   2 3(8 2)(8 / 2) 192 inuI    40.707( )(192) 135.7 inI h h  10 000 1179 psi 8.48h h     10 000(10)(8 / 2) 2948 psi 135.7h h     2 2max 1 31751179 2948 12 800 psi h h      from which Do not round off the leg size – something to learn. 0.248 in.h  192fom ' 64.5 in 0.248(12) uI hl    28.48(0.248) 2.10 inA   4135.7(0.248) 33.65 inI   2 2 30.248vol 12 0.369 in 2 2 h l   33.65eff 91.2 in vol 0.369 I    1179 4754 psi 0.248     2948 11 887 psi 0.248     max 3175 12 800 psi 0.248    Now consider the case of uninterrupted welds, 1 0b  1.414( )(8 0) 11.31A h h   2 3(8 0)(8 / 2) 256 inuI    40.707(256) 181 inI h  h 10 000 884 11.31h h     10 000(10)(8 / 2) 2210 181h h     2 2max all 1 2380884 2210 h h      max all 2380 0.186 in 12 800 h      Do not round off h. Chapter 9, Page 22/36 • 211.31(0.186) 2.10 inA   4181(0.186) 33.67 inI   2 3884 0.1864753 psi, vol 16 0.277 in 0.186 2       2210 11882 psi 0.186     256fom ' 86.0 in 0.186(16) uI hl    2 2 33.67eff 121.7 in ( / 2) (0.186 / 2)16 I h l    Conclusions: To meet allowable stress limitations, I and A do not change, nor do τ and σ. To meet the shortened bead length, h is increased proportionately. However, volume of bead laid down increases as h2. The uninterrupted bead is superior. In this example, we did not round h and as a result we learned something. Our measures of merit are also sensitive to rounding. When the design decision is made, rounding to the next larger standard weld fillet size will decrease the merit. Had the weld bead gone around the corners, the situation would change. Here is a follow up task analyzing an alternative weld pattern. ______________________________________________________________________________ 9-40 From Table 9-2 For the box 1.414 ( )A h b d  Subtracting 1 1 from and from b b d d  1 11.414A h b b d d        3 22 2 31 1 1 1 1 1(3 ) 6 6 2 2 6u d b dd 3I b d b b d d d        Length of bead 1 12( )l b b d d    fom /uI hl ______________________________________________________________________________ Chapter 9, Page 23/36 • 9-41 Computer programs will vary. ______________________________________________________________________________ 9-42 Note to the Instructor. In the first printing of the ninth edition, the loading was stated incorrectly. In the fourth line, “bending moment of 100 kip ⋅ in in” should read, “10 kip bending load 10 in from”. This will be corrected in the printings that follow. We apologize if this has caused any inconvenience. all = 12 kpsi. Use Fig. 9-17(a) for general geometry, but employ beads and then beads. Horizontal parallel weld bead pattern b = 3 in, d = 6 in Table 9-2: 21.414 1.414( )(3) 4.24 inA hb h h   2 2 33(6) 54 in 2 2u bdI    40.707 0.707( )(54) 38.2 inuI hI h h   10 2.358 kpsi 4.24h h     10(10)(6 / 2) 7.853 kpsi 38.2 Mc I h h      2 2 2 2max 1 8.1992.358 7.853 kpsi h h         Equate the maximum and allowable shear stresses. max all 8.199 12 h     from which It follows that 0.683 in.h  438.2(0.683) 26.1 inI   The volume of the weld metal is 2 2 3(0.683) (3 3)vol 1.40 in 2 2 h l     The effectiveness, (eff)H, is Chapter 9, Page 24/36 • H 26.1(eff) 18.6 in vol 1.4 I    H 54(fom ') 13.2 in 0.683(3 3) uI hl     Vertical parallel weld beads 3 in 6 in b d   From Table 9-2, case 2 21.414 1.414( )(6) 8.48 inA hd h h   3 3 36 72 in 6 6u dI    0.707 0.707( )(72) 50.9uI hI h h   10 1.179 psi 8.48h h     10(10)(6 / 2) 5.894 psi 50.9 Mc I h h      2 2 2 2max 1 6.0111.179 5.894 kpsi h h         Equating max to all gives 0.501 in.h  It follows that 450.9(0.501) 25.5 inI   2 2 30.501vol (6 6) 1.51 in 2 2 h l     V 25.5(eff ) 16.7 in vol 1.51 I    V 72(fom') 12.0 in 0.501(6 6) uI hl     The ratio of is 16V H(eff ) / (eff ) .7 /18.6 0.898. The ratio is This is not surprising since V H(fom') / (fom ') 12.0 /13.2  0.909. 2 2 0.707eff 1.414 1.414fom' ( / 2) ( / 2) u uhI II I vol h l h l hl      The ratios (e and give the same information. V Hff ) / (eff ) V(fom ') / (fom ')H ______________________________________________________________________________ Chapter 9, Page 25/36 • 9-43 F = 0, T = 15 kipin. Table 9-1: Ju = 2 r 3 = 2 (1)3 = 6.283 in3, J = 0.707(1/4) 6.283 = 1.111 in4  max 15 1 13.5 kpsi . 1.111 Tr Ans J     ______________________________________________________________________________ 9-44 F = 2 kip, T = 0. Table 9-2: A = 1.414  h r = 1.414  (1/4)(1) = 1.111 in2 Iu =  r 3 =  (1)3 = 3.142 in3, I = 0.707(1/4) 3.142 = 0.5553 in4 2 1.80 kpsi 1.111 V A        2 6 1 21.6 kpsi 0.5553 Mr I       max = ( 2 +  2)1/2 = (1.802 + 21.62)1/2 = 21.7 kpsi Ans. ______________________________________________________________________________ 9-45 F = 2 kip, T = 15 kipin. Bending: Table 9-2: A = 1.414  h r = 1.414  (1/4)(1) = 1.111 in2 Iu =  r 3 =  (1)3 = 3.142 in3, I = 0.707(1/4) 3.142 = 0.5553 in4 2 1.80 kpsi 1.111 V A          2 6 1 21.6 kpsi 0.5553M Mr I      Torsion: Table 9-1: Ju = 2 r 3 = 2 (1)3 = 6.283 in3, J = 0.707(1/4) 6.283 = 1.111 in4    15 1 13.5 kpsi 1.111T Tr J      Chapter 9, Page 26/36 •    2 22 2 2 2max 1.80 21.6 13.5 25.5 kpsi .M T Ans            ______________________________________________________________________________ 9-46 F = 2 kip, T = 15 kipin. Bending: Table 9-2: A = 1.414  h r = 1.414  h (1) = 4.442h in2 Iu =  r 3 =  (1)3 = 3.142 in3, I = 0.707 h (3.142) = 2.221h in4 2 0.4502 kpsi 4.442 V A h h          2 6 1 5.403 kpsi 2.221M Mr I h h      Torsion: Table 9-1: Ju = 2 r 3 = 2 (1)3 = 6.283 in3, J = 0.707 h (6.283) = 4.442 in4    15 1 3.377 kpsi 4.442T Tr J h h          2 2 2 2 22 max 0.4502 5.403 3.377 6.387 kpsi M T h h h h                              max all 6.387 20 0.319 in .h A h       ns Should specify a 3 8 in weld. Ans. ______________________________________________________________________________ 9-47 9 mm, 200 mm, 25mmh d b   From Table 9-2, case 2: A = 1.414(9)(200) = 2.545(103) mm2   3 3 6 3200 1.333 10 mm 6 6u dI    I = 0.707h Iu = 0.707(9)(1.333)(106) = 8.484(106) mm4 Chapter 9, Page 27/36 •  3 3 25 10 9.82 MPa 2.545(10 ) F A      M = 25(150) = 3750 Nm  363750(100) 10 44.20 MPa8.484(10 ) Mc I      2 2 2 2max 9.82 44.20 45.3 MPa .Ans        ______________________________________________________________________________ 9-48 Note to the Instructor. In the first printing of the ninth edition, the vertical dimension of 5 in should be to the top of the top plate. This will be corrected in the printings that follow. We apologize if this has caused any inconvenience. h = 0.25 in, b = 2.5 in, d = 5 in. Table 9-2, case 5: A = 0.707h (b +2d) = 0.707(0.25)[2.5 + 2(5)] = 2.209 in2   2 25 2 in 2 2.5 2 5 dy b d                 3 2 2 3 2 2 2 2 2 3 2 5 2 5 2 2.5 2 5 2 33.33 in 3 u dI d y b d y           3 I = 0.707 h Iu = 0.707(1/4)(33.33) = 5.891 in4 Primary shear: 2 0.905 kpsi 2.209 F A      Secondary shear (the critical location is at the bottom of the bracket): y = 5  2 = 3 in   2 5 3 5.093 kpsi 5.891 My I      2 2 2 2max 0.905 5.093 5.173 kpsi        all max 18 3.48 . 5.173 n Ans     ______________________________________________________________________________ Chapter 9, Page 28/36 • 9-49 The largest possible weld size is 1/16 in. This is a small weld and thus difficult to accomplish. The bracket’s load-carrying capability is not known. There are geometry problems associated with sheet metal folding, load-placement and location of the center of twist. This is not available to us. We will identify the strongest possible weldment. Use a rectangular, weld-all-around pattern – Table 9-2, case 6: 2 2 2 3 4 1.414 ( ) 1.414(1 / 16)(1 7.5) 0.7512 in / 2 0.5 in / 2 7.5 / 2 3.75 in 7.5(3 ) [3(1) 7.5] 98.44 in 6 6 0.707 0.707(1 / 16)(98.44) 4.350 in (3.75 0.5) 4.25 1.331 0.7512 4 u u A h b d x b y d dI b d I hI M W W V W W A Mc I                               2 2 2 2 max .25 (7.5 / 2) 3.664 4.350 1.331 3.664 3.90 W W W W          Materia l properties: The allowable stress given is low. Let’s demonstrate that. For the 1020 CD bracket, use HR properties of Sy = 30 kpsi and Sut = 55. The 1030 HR support, Sy = 37.5 kpsi and Sut = 68. The E6010 electrode has strengths of Sy = 50 and Sut = 62 kpsi. Allowable stresses: 1020 HR: all = min[0.3(55), 0.4(30)] = min(16.5, 12) = 12 kpsi 1020 HR: all = min[0.3(68), 0.4(37.5)] = min(20.4, 15) = 15 kpsi E6010: all = min[0.3(62), 0.4(50)] = min(18.6, 20) = 18.6 kpsi Since Table 9-6 gives 18.0 kpsi as the allowable shear stress, use this lower value. Therefore, the allowable shear stress is all = min(14.4, 12, 18.0) = 12 kpsi However, the allowable stress in the problem statement is 1.5 kpsi which is low from the weldment perspective. The load associated with this strength is max all 3.90 1500 1500 385 lbf 3.90 W W       Chapter 9, Page 29/36 • If the welding can be accomplished (1/16 leg size is a small weld), the weld strength is 12 000 psi and the load associated with this strength is W = 12 000/3.90 = 3077 lbf. Can the bracket carry such a load? There are geometry problems associated with sheet metal folding. Load placement is important and the center of twist has not been identified. Also, the load-carrying capability of the top bend is unknown. These uncertainties may require the use of a different weld pattern. Our solution provides the best weldment and thus insight for comparing a welded joint to one which employs screw fasteners. ______________________________________________________________________________ 9-50 all100 lbf , 3 kpsi 100(16 / 3) 533.3 lbf 533.3cos60 266.7 lbf 533.3cos30 462 lbf B x B y B F F F F               It follows that and R562 lbfyAR  266.7 lbf, x AR  A = 622 lbf M = 100(16) = 1600 lbf · in The OD of the tubes is 1 in. From Table 9-1, case 6:   2 3 3 3 4 2 1.414( ) 2(1.414)( )(1 / 2) 4.442 in 2 2 (1 / 2) 0.7854 in 2(0.707) 1.414(0.7854) 1.111 in u u A hr h J r J hJ h h              h Chapter 9, Page 30/36 • 622 140.0 4.442 1600(0.5) 720.1 1.111 V A h h Tc Mc J J h h            The shear stresses, and ,   are additive algebraically max max all 1 860(140.0 720.1) psi 860 3000 860 0.287 5 / 16 in 3000 h h h h             Decision: Use 5/16 in fillet welds Ans. ______________________________________________________________________________ 9-51 For the pattern in bending shown, find the centroid G of the weld group.             75 6 150 325 9 150 225 mm 6 150 9 150 x                2 6mm 6mm 3 2 6 4 2 0.707 6 150 2 0.707 6 150 225 75 31.02 10 mm 12 GI I Ax                     3 2 6 4 9mm 0.707 9 150 2 0.707 9 150 175 75 22.67 10 mm 12 I           I = I 6 mm + I 9 mm = (31.02 + 22.67)(106) = 53.69(106) mm4 The critical location is at B. With  in MPa, and F in kN Chapter 9, Page 31/36 •      310 0.3143 2 0.707 6 9 150 FV F A              3 6 200 10 225 0.8381 53.69 10 FMc F I      2 2 2 2max 0.3143 0.8381 0.8951F F        Materials: 1015 HR (Table A-20): Sy = 190 MPa, E6010 Electrode(Table 9-3): Sy = 345 MPa Eq. (5-21), p. 225 all = 0.577(190) = 109.6 MPa all / 109.6 / 2 61.2 kN . 0.8951 0.8951 nF Ans   ______________________________________________________________________________ 9-52 In the textbook, Fig. Problem 9-52b is a free-body diagram of the bracket. Forces and moments that act on the welds are equal, but of opposite sense. (a) M = 1200(0.366) = 439 lbf · in Ans. (b) Fy = 1200 sin 30 = 600 lbf Ans. (c) Fx = 1200 cos 30 = 1039 lbf Ans. (d) From Table 9-2, case 6: 2 2 2 3 1.414(0.25)(0.25 2.5) 0.972 in 2.5(3 ) [3(0.25) 2.5] 3.39 in 6 6u A dI b d         The second area moment about an axis through G and parallel to z is 40.707 0.707(0.25)(3.39) 0.599 in .uI hI Ans   (e) Refer to Fig. Problem 9-52b. The shear stress due to Fy is 1 600 617 psi 0.972 yF A     The shear stress along the throat due to Fx is 2 1039 1069 psi 0.972 xF A     The resultant of 1 and 2 is in the throat plane Chapter 9, Page 32/36 • 2 2 2 2 1 2 617 1069 1234 psi        The bending of the throat gives 439(1.25) 916 psi 0.599 Mc I     The maximum shear stress is 2 2 2 2 max 1234 916 1537 psi .Ans        (f) Materials: 1018 HR Member: Sy = 32 kpsi, Sut = 58 kpsi (Table A-20) E6010 Electrode: Sy = 50 kpsi (Table 9-3) max max 0.577 0.577(32) 12.0 . 1.537 sy yS Sn A       ns (g) Bending in the attachment near the base. The cross-sectional area is approximately equal to bh. 2 1 1 2 2 3 0.25(2.5) 0.625 in 1039 1662 psi 0.625 0.25(2.5) 0.260 in 6 6 x xy A bh F A I bd c           At location A, 1 / 600 439 2648 psi 0.625 0.260 y y y F M A I c        The von Mises stress  is 2 2 2 23 2648 3(1662) 3912 psiy xy        Thus, the factor of safety is, 32 8.18 . 3.912 ySn A     ns  The clip on the mooring line bears against the side of the 1/2-in hole. If the clip fills the hole Chapter 9, Page 33/36 • 3 1200 9600 psi 0.25(0.50) 32(10 ) 3.33 . 9600 y F td S n A  ns              Further investigation of this situation requires more detail than is included in the task statement. (h) In shear fatigue, the weakest constituent of the weld melt is 1018 HR with Sut = 58 kpsi, Eq. (6-8), p. 282, gives 0.504 0.504(58) 29.2 kpsie utS S   Eq. (6-19), p. 287: ka = 14.4(58)-0.718 = 0.780 For the size factor estimate, we first employ Eq. (6-25), p. 289, for the equivalent diameter 0.808 0.707 0.808 0.707(2.5)(0.25) 0.537 ined hb   Eq. (6-20), p. 288, is used next to find kb -0.107 -0.1070.537 0.940 0.30 0.30 e b dk             Eq.(6-26), p. 290: kc = 0.59 From Eq. (6-18), p. 287, the endurance strength in shear is Sse = 0.780(0.940)(0.59)(29.2) = 12.6 kpsi From Table 9-5, the shear stress-concentration factor is Kf s = 2.7. The loading is repeatedly-applied max 1.5372.7 2.07 kpsi 2 2a m f s K      Table 6-7, p. 307: Gerber factor of safety nf, adjusted for shear, with Ssu = 0.67Sut 2 22 1 21 1 2 1 0.67(58) 2.07 2(2.07)(12.6)1 1 5.55 . 2 2.07 12.6 0.67(58)(2.07) su a m se f m se su a S Sn S S Ans                                               Attachment metal should be checked for bending fatigue. ______________________________________________________________________________ 9-53 (a) Use b = d = 4 in. Since h = 5/8 in, the primary shear is Chapter 9, Page 34/36 • 0.2829 1.414(5 / 8)(4) F F    The secondary shear calculations, for a moment arm of 14 in give 2 2 3 4 4[3(4 ) 4 ] 42.67 in 6 0.707 0.707(5 / 8)42.67 18.85 in 14 (2) 1.485 18.85 u u y x y J J hJ Mr F F J              Thus, the maximum shear and allowable load are: 2 2 max all 1.485 (0.2829 1.485) 2.309 25 10.8 kip . 2.309 2.309 F F F A          ns The load for part (a) has increased by a factor of 10.8/2.71 = 3.99 Ans. (b) From Prob. 9-18b, all = 11 kpsi all all 11 4.76 kip 2.309 2.309 F    The allowable load in part (b) has increased by a factor of 4.76/1.19 = 4 Ans. ______________________________________________________________________________ 9-54 Purchase the hook having the design shown in Fig. Problem 9-54b. Referring to text Fig. 9-29a, this design reduces peel stresses. ______________________________________________________________________________ 9-55 (a) / 2 / 2 / 2 1 1/ 2 / 2 / 2 1 1 1 1 cosh( ) cosh( ) sinh( ) 4 sinh( / 2) [sinh( / 2) sinh( / 2)] [sinh( / 2) ( sinh( / 2))] 2 sinh( / 2) [2sinh( / 2)] . 4 sinh( / 2) 2 l l l l l l P x Adx A x dx x l b l A Al l l A l P Pl An bl l bl                                  l s  (b) cosh( / 2)( / 2) . 4 sinh( / 2) 4 tanh( / 2) P l Pl A b l b l ns       Chapter 9, Page 35/36 • (c) ( / 2) 2 / 2 . 4 tanh( / 2) tanh( / 2) l P bl lK Ans b l P l              For computer programming, it can be useful to express the hyperbolic tangent in terms of exponentials: exp( / 2) exp( / 2) . 2 exp( / 2) exp( / 2) l l lK A l l ns          ______________________________________________________________________________ 9-56 This is a computer programming exercise. All programs will vary. Chapter 9, Page 36/36 • Chapter 10 10-1 From Eqs. (10-4) and (10-5) 4 1 0.615 4 2 4 4 4W B C CK K C C C 3         Plot 100(KW  KB)/ KW vs. C for 4  C  12 obtaining We see the maximum and minimum occur at C = 4 and 12 respectively where Maximum = 1.36 % Ans., and Minimum = 0.743 % Ans. ______________________________________________________________________________ 10-2 A = Sdm dim(Auscu) = [dim (S) dim(d m)]uscu = kpsiinm dim(ASI) = [dim (S) dim(d m)]SI = MPammm    SI uscu uscu uscu MPa mm 6.894757 25.4 6.895 25.4 . kpsi in m m m mA A A A Ans    For music wire, from Table 10-4: Auscu = 201 kpsiinm, m = 0.145; what is ASI? _____________________________________________________________________________ 0-3 Given: Music wire, d = 2.5 mm, OD = 31 mm, plain ground ends, Nt = 14 coils. ASI = 6.895(25.4)0.145 (201) = 2215 MPammm Ans. _ 1 Chapter 10 - Rev. A, Page 1/41 • (a) Table 10-1: Na = N  1 = 14  1 = 13 coils Ls = d Nt = 2.5(14) = 35 mm Table 10-4: m = 0.145, A = 2211 MPammm Eq. (10-14): t 0.145 2211 1936 MPa 2.5ut m AS d    Table 10-6: Ssy = 0.45(1936) = 871.2 MPa D = OD  d = 31 2.5 = 28.5 mm C = D/d = 28.5/2.5 = 11.4 Eq. (10-5):     4 11.4 24 2 1.117 4 3 4 11.4 3B CK C       Eq. (10-7):     33 2.5 871.2 167.9 N 8 8 1.117 28.5 sy s B d S F K D     Table 10-5): d = 2.5/25.4 = 0.098 in  G = 81.0(103) MPa Eq. (10-9):     4 34 3 3 2.5 81 10 1.314 N / mm 8 8 28.5 13a d Gk D N    0 167.9 35 162.8 mm . 1.314 s s FL L A k      ns (b) Fs = 167.9 N Ans. (c) k = 1.314 N/mm Ans.   (d)  0 cr 149.9 mm0.5L   . Spring needs to be supported. Ans. 2.63 28.5 _____________________________________________________________________________ 0-4 Given: Design load, F1 = 130 N. 4, N = 13 coils, Ssy = 871.2 MPa, Fs = 167.9 N, Eq. (10-19): 3 ≤ Na ≤ 15 Na = 13 O.K. _ 1 Referring to Prob. 10-3 solution, C = 11. a L0 = 162.8 mm and (L0)cr = 149.9 mm. Eq. (10-18): 4 ≤ C ≤ 12 C = 11.4 O.K. Chapter 10 - Rev. A, Page 2/41 • Eq. (10-17): 1 167.91 1 0.29 130 sF F       Eq. (10-20): 0.15, 0.29 . .O K   From Eq. (10-7) for static service 1 1 3 3 1 8 8(130)(28.5)1.117 674 MPa (2.5) 871.2 1.29 674 B sy F DK d S n               Eq. (10-21): ns ≥ 1.2, n = 1.29 O.K. 1 167.9 167.9674 870.5 MPa 130 130 / 871.2 / 870.5 1 s sy sS                   Ssy/s ≥ (ns )d : Not solid-safe (but was the basis of the design). Not O.K. L0 ≤ (L0)cr: 162.8  149.9 Not O.K. Design is unsatisfactory. Operate over a rod? Ans. ______________________________________________________________________________ 10-5 Given: Oil-tempered wire, d = 0.2 in, D = 2 in, Nt = 12 coils, L0 = 5 in, squared ends. (a) Table 10-1: Ls = d (Nt + 1) = 0.2(12 + 1) = 2.6 in Ans. (b) Table 10-1: Na = Nt  2 = 12  2 = 10 coils Table 10-5: G = 11.2 Mpsi Eq. (10-9):     4 64 3 3 0.2 11.2 10 28 lbf/in 8 8 2 10 d Gk D N    Fs = k ys = k (L0  Ls ) = 28(5  2.6) = 67.2 lbf Ans. (c) Eq. (10-1): C = D/d = 2/0.2 = 10 Eq. (10-5):     4 10 24 2 1.135 4 3 4 10 3B CK C       Eq. (10-7):       3 3 3 8 67.2 28 1.135 48.56 10 psi 0.2s B FDK d       Chapter 10 - Rev. A, Page 3/41 • Table 10-4: m = 0.187, A = 147 kpsiinm Eq. (10-14): 0.187 147 198.6 kpsi 0.2ut m AS d    Table 10-6: Ssy = 0.50 Sut = 0.50(198.6) = 99.3 kpsi 99.3 2.04 . 48.56 sy s s S n A     ns ______________________________________________________________________________ 10-6 Given: Oil-tempered wire, d = 4 mm, C = 10, plain ends, L0 = 80 mm, and at F = 50 N, y = 15 mm. (a) k = F/y = 50/15 = 3.333 N/mm Ans. (b) D = Cd = 10(4) = 40 mm OD = D + d = 40 + 4 = 44 mm Ans. (c) From Table 10-5, G = 77.2 GPa Eq. (10-9):     4 34 3 3 4 77.2 10 11.6 coils 8 8 3.333 40a d GN kD    Table 10-1: Nt = Na = 11.6 coils Ans. (d) Table 10-1: Ls = d (Nt + 1) = 4(11.6 + 1) = 50. 4 mm Ans. (e) Table 10-4: m = 0.187, A = 1855 MPammm Eq. (10-14): 0.187 1855 1431 MPa 4ut m AS d    Table 10-6: Ssy = 0.50 Sut = 0.50(1431) = 715.5 MPa ys = L0  Ls = 80  50.4 = 29.6 mm Fs = k ys = 3.333(29.6) = 98.66 N Eq. (10-5): 4 2 4(10) 2 1.135 4 3 4(10) 3B CK C        Chapter 10 - Rev. A, Page 4/41 • Eq. (10-7):    3 3 8 98.66 408 1.135 178.2 MPa 4 s s B F DK d       715.5 4.02 . 178.2 sy s s S n A     ns ______________________________________________________________________________ 10-7 Static service spring with: HD steel wire, d = 0.080 in, OD = 0.880 in, Nt = 8 coils, plain and ground ends. Preliminaries Table 10-5: A = 140 kpsi · inm, m = 0.190 Eq. (10-14): 0.190 140 226.2 kpsi 0.080ut m AS d    Table 10-6: Ssy = 0.45(226.2) = 101.8 kpsi Then, D = OD  d = 0.880  0.080 = 0.8 in Eq. (10-1): C = D/d = 0.8/0.08 = 10 Eq. (10-5): 4 2 4(10) 2 1.135 4 3 4(10) 3B CK C        Table 10-1: Na = Nt  1 = 8  1 = 7 coils Ls = dNt = 0.08(8) = 0.64 in Eq. (10-7) For solid-safe, ns = 1.2 :    3 33 0.08 101.8 10 / 1.2/ 18.78 lbf 8 8(1.135)(0.8) sy s s B d S n F K D       Eq. (10-9):    4 64 3 3 0.08 11.5 10 16.43 lbf/in 8 8 0.8 7a d Gk D N    18.78 1.14 in 16.43 s s Fy k    (a) L0 = ys + Ls = 1.14 + 0.64 = 1.78 in Ans. (b) Table 10-1: 0 1.78 0.223 in . 8t Lp A N    ns (c) From above: Fs = 18.78 lbf Ans. (d) From above: k = 16.43 lbf/in Ans. (e) Table 10-2 and Eq. (10-13): 0 cr 2.63 2.63(0.8)( ) 4.21 in 0.5 DL     Since L0 < (L0)cr, buckling is unlikely Ans. ______________________________________________________________________________ 10-8 Given: Design load, F1 = 16.5 lbf. Referring to Prob. 10-7 solution, C = 10, Na = 7 coils, Ssy = 101.8 kpsi, Fs = 18.78 lbf, ys = 1.14 in, L0 = 1.78 in, and (L0)cr = 4.208 in. Chapter 10 - Rev. A, Page 5/41 • Eq. (10-18): 4 ≤ C ≤ 12 C = 10 O.K. Eq. (10-19): 3 ≤ Na ≤ 15 Na = 7 O.K. Eq. (10-17): 1 18.781 1 0.14 16.5 sF F       Eq. (10-20): 0.15, 0.14 . .not O K   , but probably acceptable. From Eq. (10-7) for static service  311 3 3 1 8 8(16.5)(0.8)1.135 74.5 10 psi 74.5 kpsi (0.080) 101.8 1.37 74.5 B sy F DK d S n                Eq. (10-21): ns ≥ 1.2, n = 1.37 O.K. 1 18.78 18.7874.5 84.8 kpsi 16.5 16.5 / 101.8 / 84.8 1.20 s s sy sn S                    Eq. (10-21): ns ≥ 1.2, ns = 1.2 It is solid-safe (basis of design). O.K. Eq. (10-13) and Table 10-2: L0 ≤ (L0)cr 1.78 in  4.208 in O.K. ______________________________________________________________________________ 10-9 Given: A228 music wire, sq. and grd. ends, d = 0.007 in, OD = 0.038 in, L0 = 0.58 in, Nt = 38 coils. D = OD  d = 0.038  0.007 = 0.031 in Eq. (10-1): C = D/d = 0.031/0.007 = 4.429 Eq. (10-5):     4 4.429 24 2 1.340 4 3 4 4.429 3B CK C       Table (10-1): Na = Nt  2 = 38  2 = 36 coils (high) Table 10-5: G = 12.0 Mpsi Eq. (10-9):     4 64 3 3 0.007 12.0 10 3.358 lbf/in 8 8 0.031 36a d Gk D N    Table (10-1): Ls = dNt = 0.007(38) = 0.266 in ys = L0  Ls = 0.58  0.266 = 0.314 in Fs = kys = 3.358(0.314) = 1.054 lbf Eq. (10-7):       3 3 3 8 1.054 0.0318 1.340 325.1 10 psi 0.007 s s B F DK d       (1) Table 10-4: A = 201 kpsiinm, m = 0.145 Chapter 10 - Rev. A, Page 6/41 • Eq. (10-14): 0.145 201 412.7 kpsi 0.007ut m AS d    Table 10-6: Ssy = 0.45 Sut = 0.45(412.7) = 185.7 kpsi s > Ssy, that is, 325.1 > 185.7 kpsi, the spring is not solid-safe. Return to Eq. (1) with Fs = kys and s = Ssy /ns, and solve for ys, giving           3 33 185.7 10 /1.2 0.007/ 0.149 in 8 8 1.340 3.358 0.031 sy s s B S n d y K kD       The free length should be wound to L0 = Ls + ys = 0.266 + 0.149 = 0.415 in Ans. This only addresses the solid-safe criteria. There are additional problems. ______________________________________________________________________________ 10-10 Given: B159 phosphor-bronze, sq. and grd. ends, d = 0.014 in, OD = 0.128 in, L0 = 0.50 in, Nt = 16 coils. D = OD  d = 0.128  0.014 = 0.114 in Eq. (10-1): C = D/d = 0.114/0.014 = 8.143 Eq. (10-5):     4 8.143 24 2 1.169 4 3 4 8.143 3B CK C       Table (10-1): Na = Nt  2 = 16  2 = 14 coils Table 10-5: G = 6 Mpsi Eq. (10-9):     4 64 3 3 0.014 6 10 1.389 lbf/in 8 8 0.114 14a d Gk D N    Table (10-1): Ls = dNt = 0.014(16) = 0.224 in ys = L0  Ls = 0.50  0.224 = 0.276 in Fs = kys = 1.389(0.276) = 0.3834 lbf Eq. (10-7):       3 3 3 8 0.3834 0.1148 1.169 47.42 10 psi 0.014 s s B F DK d       (1) Table 10-4: A = 145 kpsiinm, m = 0 Eq. (10-14): 0 145 145 kpsi 0.014ut m AS d    Table 10-6: Ssy = 0.35 Sut = 0.35(135) = 47.25 kpsi s > Ssy, that is, 47.42 > 47.25 kpsi, the spring is not solid-safe. Return to Eq. (1) with Fs = kys and s = Ssy /ns, and solve for ys, giving           3 33 47.25 10 /1.2 0.014/ 0.229 in 8 8 1.169 1.389 0.114 sy s s B S n d y K kD       The free length should be wound to Chapter 10 - Rev. A, Page 7/41 • L0 = Ls + ys = 0.224 + 0.229 = 0.453 in Ans. ______________________________________________________________________________ 10-11 Given: A313 stainless steel, sq. and grd. ends, d = 0.050 in, OD = 0.250 in, L0 = 0.68 in, Nt = 11.2 coils. D = OD  d = 0.250  0.050 = 0.200 in Eq. (10-1): C = D/d = 0.200/0.050 = 4 Eq. (10-5):     4 4 24 2 1.385 4 3 4 4 3B CK C       Table (10-1): Na = Nt  2 = 11.2  2 = 9.2 coils Table 10-5: G = 10 Mpsi Eq. (10-9):     4 64 3 3 0.050 10 10 106.1 lbf/in 8 8 0.2 9.2a d Gk D N    Table (10-1): Ls = dNt = 0.050(11.2) = 0.56 in ys = L0  Ls = 0.68  0.56 = 0.12 in Fs = kys = 106.1(0.12) = 12.73 lbf Eq. (10-7):       3 3 3 8 12.73 0.28 1.385 71.8 10 psi 0.050 s s B F DK d       Table 10-4: A = 169 kpsiinm, m = 0.146 Eq. (10-14): 0.146 169 261.7 kpsi 0.050ut m AS d    Table 10-6: Ssy = 0.35 Sut = 0.35(261.7) = 91.6 kpsi 91.6 1.28 71.8 sy s s S n     Spring is solid-safe (ns > 1.2) Ans. ______________________________________________________________________________ 10-12 Given: A227 hard-drawn wire, sq. and grd. ends, d = 0.148 in, OD = 2.12 in, L0 = 2.5 in, Nt = 5.75 coils. D = OD  d = 2.12  0.148 = 1.972 in Eq. (10-1): C = D/d = 1.972/0.148 = 13.32 (high) Eq. (10-5):     4 13.32 24 2 1.099 4 3 4 13.32 3B CK C       Table (10-1): Na = Nt  2 = 5.75  2 = 3.75 coils Table 10-5: G = 11.4 Mpsi Eq. (10-9):     4 64 3 3 0.148 11.4 10 23.77 lbf/in 8 8 1.972 3.75a d Gk D N    Table (10-1): Ls = dNt = 0.148(5.75) = 0.851 in ys = L0  Ls = 2.5  0.851 = 1.649 in Chapter 10 - Rev. A, Page 8/41 • Fs = kys = 23.77(1.649) = 39.20 lbf Eq. (10-7):       3 3 3 8 39.20 1.9728 1.099 66.7 10 psi 0.148 s s B F DK d       Table 10-4: A = 140 kpsiinm, m = 0.190 Eq. (10-14): 0.190 140 201.3 kpsi 0.148ut m AS d    Table 10-6: Ssy = 0.35 Sut = 0.45(201.3) = 90.6 kpsi 90.6 1.36 66.7 sy s s S n     Spring is solid-safe (ns > 1.2) Ans. ______________________________________________________________________________ 10-13 Given: A229 OQ&T steel, sq. and grd. ends, d = 0.138 in, OD = 0.92 in, L0 = 2.86 in, Nt = 12 coils. D = OD  d = 0.92  0.138 = 0.782 in Eq. (10-1): C = D/d = 0.782/0.138 = 5.667 Eq. (10-5):     4 5.667 24 2 1.254 4 3 4 5.667 3B CK C       Table (10-1): Na = Nt  2 = 12  2 = 10 coils A229 OQ&T steel is not given in Table 10-5. From Table A-5, for carbon steels, G = 11.5 Mpsi. Eq. (10-9):     4 64 3 3 0.138 11.5 10 109.0 lbf/in 8 8 0.782 10a d Gk D N    Table (10-1): Ls = dNt = 0.138(12) = 1.656 in ys = L0  Ls = 2.86  1.656 = 1.204 in Fs = kys = 109.0(1.204) = 131.2 lbf Eq. (10-7):       3 3 3 8 131.2 0.7828 1.254 124.7 10 psi 0.138 s s B F DK d       (1) Table 10-4: A = 147 kpsiinm, m = 0.187 Eq. (10-14): 0.187 147 212.9 kpsi 0.138ut m AS d    Table 10-6: Ssy = 0.50 Sut = 0.50(212.9) = 106.5 kpsi s > Ssy, that is, 124.7 > 106.5 kpsi, the spring is not solid-safe. Return to Eq. (1) with Fs = kys and s = Ssy /ns, and solve for ys, giving           3 33 106.5 10 /1.2 0.138/ 0.857 in 8 8 1.254 109.0 0.782 sy s s B S n d y K kD       The free length should be wound to Chapter 10 - Rev. A, Page 9/41 • L0 = Ls + ys = 1.656 + 0.857 = 2.51 in Ans. ______________________________________________________________________________ 10-14 Given: A232 chrome-vanadium steel, sq. and grd. ends, d = 0.185 in, OD = 2.75 in, L0 = 7.5 in, Nt = 8 coils. D = OD  d = 2.75  0.185 = 2.565 in Eq. (10-1): C = D/d = 2.565/0.185 = 13.86 (high) Eq. (10-5):     4 13.86 24 2 1.095 4 3 4 13.86 3B CK C       Table (10-1): Na = Nt  2 = 8  2 = 6 coils Table 10-5: G = 11.2 Mpsi. Eq. (10-9):     4 64 3 3 0.185 11.2 10 16.20 lbf/in 8 8 2.565 6a d Gk D N    Table (10-1): Ls = dNt = 0.185(8) = 1.48 in ys = L0  Ls = 7.5  1.48 = 6.02 in Fs = kys = 16.20(6.02) = 97.5 lbf Eq. (10-7):       3 3 3 8 97.5 2.5658 1.095 110.1 10 psi 0.185 s s B F DK d       (1) Table 10-4: A = 169 kpsiinm, m = 0.168 Eq. (10-14): 0.168 169 224.4 kpsi 0.185ut m AS d    Table 10-6: Ssy = 0.50 Sut = 0.50(224.4) = 112.2 kpsi 112.2 1.02 110.1 sy s s S n     Spring is not solid-safe (ns < 1.2) Return to Eq. (1) with Fs = kys and s = Ssy /ns, and solve for ys, giving           3 33 112.2 10 /1.2 0.185/ 5.109 in 8 8 1.095 16.20 2.565 sy s s B S n d y K kD       The free length should be wound to L0 = Ls + ys = 1.48 + 5.109 = 6.59 in Ans. ______________________________________________________________________________ 10-15 Given: A313 stainless steel, sq. and grd. ends, d = 0.25 mm, OD = 0.95 mm, L0 = 12.1 mm, Nt = 38 coils. D = OD  d = 0.95  0.25 = 0.7 mm Eq. (10-1): C = D/d = 0.7/0.25 = 2.8 (low) Eq. (10-5):     4 2.8 24 2 1.610 4 3 4 2.8 3B CK C       Chapter 10 - Rev. A, Page 10/41 • Table (10-1): Na = Nt  2 = 38  2 = 36 coils (high) Table 10-5: G = 69.0(103) MPa. Eq. (10-9):     4 34 3 3 0.25 69.0 10 2.728 N/mm 8 8 0.7 36a d Gk D N    Table (10-1): Ls = dNt = 0.25(38) = 9.5 mm ys = L0  Ls = 12.1  9.5 = 2.6 mm Fs = kys = 2.728(2.6) = 7.093 N Eq. (10-7):    3 3 8 7.093 0.78 1.610 1303 MPa 0.25 s s B F DK d       (1) Table 10-4 (dia. less than table): A = 1867 MPammm, m = 0.146 Eq. (10-14): 0.146 1867 2286 MPa 0.25ut m AS d    Table 10-6: Ssy = 0.35 Sut = 0.35(2286) = 734 MPa s > Ssy, that is, 1303 > 734 MPa, the spring is not solid-safe. Return to Eq. (1) with Fs = kys and s = Ssy /ns, and solve for ys, giving           33 734 /1.2 0.25/ 1.22 mm 8 8 1.610 2.728 0.7 sy s s B S n d y K kD     The free length should be wound to L0 = Ls + ys = 9.5 + 1.22 = 10.72 mm Ans. This only addresses the solid-safe criteria. There are additional problems. ______________________________________________________________________________ 10-16 Given: A228 music wire, sq. and grd. ends, d = 1.2 mm, OD = 6.5 mm, L0 = 15.7 mm, Nt = 10.2 coils. D = OD  d = 6.5  1.2 = 5.3 mm Eq. (10-1): C = D/d = 5.3/1.2 = 4.417 Eq. (10-5):     4 4.417 24 2 1.368 4 3 4 4.417 3B CK C       Table (10-1): Na = Nt  2 = 10.2  2 = 8.2 coils Table 10-5 (d = 1.2/25.4 = 0.0472 in): G = 81.7(103) MPa. Eq. (10-9):     4 34 3 3 1.2 81.7 10 17.35 N/mm 8 8 5.3 8.2a d Gk D N    Table (10-1): Ls = dNt = 1.2(10.2) = 12.24 mm ys = L0  Ls = 15.7  12.24 = 3.46 mm Fs = kys = 17.35(3.46) = 60.03 N Chapter 10 - Rev. A, Page 11/41 • Eq. (10-7):    3 3 8 60.03 5.38 1.368 641.4 MPa 1.2 s s B F DK d       (1) Table 10-4: A = 2211 MPammm, m = 0.145 Eq. (10-14): 0.145 2211 2153 MPa 1.2ut m AS d    Table 10-6: Ssy = 0.45 Sut = 0.45(2153) = 969 MPa 969 1.51 641.4 sy s s S n     Spring is solid-safe (ns > 1.2) Ans. ______________________________________________________________________________ 10-17 Given: A229 OQ&T steel, sq. and grd. ends, d = 3.5 mm, OD = 50.6 mm, L0 = 75.5 mm, Nt = 5.5 coils. D = OD  d = 50.6  3.5 = 47.1 mm Eq. (10-1): C = D/d = 47.1/3.5 = 13.46 (high) Eq. (10-5):     4 13.46 24 2 1.098 4 3 4 13.46 3B CK C       Table (10-1): Na = Nt  2 = 5.5  2 = 3.5 coils A229 OQ&T steel is not given in Table 10-5. From Table A-5, for carbon steels, G = 79.3(103) MPa. Eq. (10-9):     4 34 3 3 3.5 79.3 10 4.067 N/mm 8 8 47.1 3.5a d Gk D N    Table (10-1): Ls = dNt = 3.5(5.5) = 19.25 mm ys = L0  Ls = 75.5  19.25 = 56.25 mm Fs = kys = 4.067(56.25) = 228.8 N Eq. (10-7):    3 3 8 228.8 47.18 1.098 702.8 MPa 3.5 s s B F DK d       (1) Table 10-4: A = 1855 MPammm, m = 0.187 Eq. (10-14): 0.187 1855 1468 MPa 3.5ut m AS d    Table 10-6: Ssy = 0.50 Sut = 0.50(1468) = 734 MPa 734 1.04 702.8 sy s s S n     Spring is not solid-safe (ns < 1.2) Return to Eq. (1) with Fs = kys and s = Ssy /ns, and solve for ys, giving           33 734 /1.2 3.5/ 48.96 mm 8 8 1.098 4.067 47.1 sy s s B S n d y K kD     The free length should be wound to Chapter 10 - Rev. A, Page 12/41 • L0 = Ls + ys = 19.25 + 48.96 = 68.2 mm Ans. ______________________________________________________________________________ 10-18 Given: B159 phosphor-bronze, sq. and grd. ends, d = 3.8 mm, OD = 31.4 mm, L0 = 71.4 mm, Nt = 12.8 coils. D = OD  d = 31.4  3.8 = 27.6 mm Eq. (10-1): C = D/d = 27.6/3.8 = 7.263 Eq. (10-5):     4 7.263 24 2 1.192 4 3 4 7.263 3B CK C       Table (10-1): Na = Nt  2 = 12.8  2 = 10.8 coils Table 10-5: G = 41.4(103) MPa. Eq. (10-9):     4 34 3 3 3.8 41.4 10 4.752 N/mm 8 8 27.6 10.8a d Gk D N    Table (10-1): Ls = dNt = 3.8(12.8) = 48.64 mm ys = L0  Ls = 71.4  48.64 = 22.76 mm Fs = kys = 4.752(22.76) = 108.2 N Eq. (10-7):    3 3 8 108.2 27.68 1.192 165.2 MPa 3.8 s s B F DK d       (1) Table 10-4 (d = 3.8/25.4 = 0.150 in): A = 932 MPammm, m = 0.064 Eq. (10-14): 0.064 932 855.7 MPa 3.8ut m AS d    Table 10-6: Ssy = 0.35 Sut = 0.35(855.7) = 299.5 MPa 299.5 1.81 165.2 sy s s S n     Spring is solid-safe (ns > 1.2) Ans. ______________________________________________________________________________ 10-19 Given: A232 chrome-vanadium steel, sq. and grd. ends, d = 4.5 mm, OD = 69.2 mm, L0 = 215.6 mm, Nt = 8.2 coils. D = OD  d = 69.2  4.5 = 64.7 mm Eq. (10-1): C = D/d = 64.7/4.5 = 14.38 (high) Eq. (10-5):     4 14.38 24 2 1.092 4 3 4 14.38 3B CK C       Table (10-1): Na = Nt  2 = 8.2  2 = 6.2 coils Table 10-5: G = 77.2(103) MPa. Eq. (10-9):     4 34 3 3 4.5 77.2 10 2.357 N/mm 8 8 64.7 6.2a d Gk D N    Table (10-1): Ls = dNt = 4.5(8.2) = 36.9 mm Chapter 10 - Rev. A, Page 13/41 • ys = L0  Ls = 215.6  36.9 = 178.7 mm Fs = kys = 2.357(178.7) = 421.2 N Eq. (10-7):    3 3 8 421.2 64.78 1.092 832 MPa 4.5 s s B F DK d       (1) Table 10-4: A = 2005 MPammm, m = 0.168 Eq. (10-14): 0.168 2005 1557 MPa 4.5ut m AS d    Table 10-6: Ssy = 0.50 Sut = 0.50(1557) = 779 MPa s > Ssy, that is, 832 > 779 MPa, the spring is not solid-safe. Return to Eq. (1) with Fs = kys and s = Ssy /ns, and solve for ys, giving           33 779 /1.2 4.5/ 139.5 mm 8 8 1.092 2.357 64.7 sy s s B S n d y K kD     The free length should be wound to L0 = Ls + ys = 36.9 + 139.5 = 176.4 mm Ans. This only addresses the solid-safe criteria. There are additional problems. ______________________________________________________________________________ 10-20 Given: A227 HD steel. From the figure: L0 = 4.75 in, OD = 2 in, and d = 0.135 in. Thus D = OD  d = 2  0.135 = 1.865 in (a) By counting, Nt = 12.5 coils. Since the ends are squared along 1/4 turn on each end, 12.5 0.5 12 turns . 4.75 / 12 0.396 in . aN Ans p Ans      The solid stack is 13 wire diameters Ls = 13(0.135) = 1.755 in Ans. (b) From Table 10-5, G = 11.4 Mpsi     4 64 3 3 0.135 (11.4) 10 6.08 lbf/in . 8 8 1.865 (12)a d Gk A D N    ns (c) Fs = k(L0 - Ls ) = 6.08(4.75  1.755)(10-3) = 18.2 lbf Ans. (d) C = D/d = 1.865/0.135 = 13.81 Chapter 10 - Rev. A, Page 14/41 •     3 3 3 4(13.81) 2 1.096 4(13.81) 3 8 8(18.2)(1.865)1.096 38.5 10 psi 38.5 kpsi . 0.135 B s s B K F DK A d            ns ______________________________________________________________________________ 10-21 For the wire diameter analyzed, G = 11.75 Mpsi per Table 10-5. Use squared and ground ends. The following is a spread-sheet study using Fig. 10-3 for parts (a) and (b). For Na, k = Fmax /y = 20/2 = 10 lbf/in. For s, F = Fs = 20(1 + ) = 20(1 + 0.15) = 23 lbf. (a) Spring over a Rod (b) Spring in a Hole Source Parameter Values Source Parameter Values d 0.075 0.080 0.085 d 0.075 0.080 0.085 ID 0.800 0.800 0.800 OD 0.950 0.950 0.950 D 0.875 0.880 0.885 D 0.875 0.870 0.865 Eq. (10-1) C 11.667 11.000 10.412 Eq. (10-1) C 11.667 10.875 10.176 Eq. (10-9) Na 6.937 8.828 11.061 Eq. (10-9) Na 6.937 9.136 11.846 Table 10-1 Nt 8.937 10.828 13.061 Table 10-1 Nt 8.937 11.136 13.846 Table 10-1 Ls 0.670 0.866 1.110 Table 10-1 Ls 0.670 0.891 1.177 1.15y + Ls L0 2.970 3.166 3.410 1.15y + Ls L0 2.970 3.191 3.477 Eq. (10-13) (L0)cr 4.603 4.629 4.655 Eq. (10- 13) (L0)cr 4.603 4.576 4.550 Table 10-4 A 201.000 201.000 201.000 Table 10-4 A 201.000 201.000 201.000 Table 10-4 m 0.145 0.145 0.145 Table 10-4 m 0.145 0.145 0.145 Eq. (10-14) Sut 292.626 289.900 287.363 Eq. (10- 14) Sut 292.626 289.900 287.363 Table 10-6 Ssy 131.681 130.455 129.313 Table 10-6 Ssy 131.681 130.455 129.313 Eq. (10-5) KB 1.115 1.122 1.129 Eq. (10-5) KB 1.115 1.123 1.133 Eq. (10-7) s 135.335 112.948 95.293 Eq. (10-7) s 135.335 111.787 93.434 Eq. (10-3) ns 0.973 1.155 1.357 Eq. (10-3) ns 0.973 1.167 1.384 Eq. (10-22) fom 0.282 0.391 0.536 Eq. (10- 22) fom 0.282 0.398 0.555 For ns ≥ 1.2, the optimal size is d = 0.085 in for both cases. ______________________________________________________________________________ 10-22 In Prob. 10-21, there is an advantage of first selecting d as one can select from the available sizes (Table A-28). Selecting C first, requires a calculation of d where then a size must be selected from Table A-28. Consider part (a) of the problem. It is required that ID = D  d = 0.800 in. (1) From Eq. (10-1), D = Cd. Substituting this into the first equation yields 0.800 1 d C  (2)  Chapter 10 - Rev. A, Page 15/41 • Starting with C = 10, from Eq. (2) we find that d = 0.089 in. From Table A-28, the closest diameter is d = 0.090 in. Substituting this back into Eq. (1) gives D = 0.890 in, with C = 0.890/0.090 = 9.889, which are acceptable. From this point the solution is the same as Prob. 10-21. For part (b), use OD = D + d = 0.950 in. (3) and, 0.800 1C   (4) d (a) Spring over a rod (b) Spring in a Hole Source Parameter Values Source Parameter Values C 10.000 10.5 C 10.000 Eq. (2) d 0.089 0.084 Eq. (4) d 0.086 Table A-28 d 0.090 0.085 Table A-28 d 0.085 Eq. (1) D 0.890 0.885 Eq. (3) D 0.865 Eq. (10-1) C 9.889 10.412 Eq. (10-1) C 10.176 Eq. (10-9) Na 13.669 11.061 Eq. (10-9) Na 11.846 Table 10-1 Nt 15.669 13.061 Table 10-1 Nt 13.846 Table 10-1 Ls 1.410 1.110 Table 10-1 Ls 1.177 1.15y + Ls L0 3.710 3.410 1.15y + Ls L0 3.477 Eq. (10-13) (L0)cr 4.681 4.655 Eq. (10-13) (L0)cr 4.550 Table 10-4 A 201.000 201.000 Table 10-4 A 201.000 Table 10-4 m 0.145 0.145 Table 10-4 m 0.145 Eq. (10-14) Sut 284.991 287.363 Eq. (10-14) Sut 287.363 Table 10-6 Ssy 128.246 129.313 Table 10-6 Ssy 129.313 Eq. (10-5) KB 1.135 1.128 Eq. (10-5) KB 1.135 Eq. (10-7) s 81.167 95.223 Eq. (10-7) s 93.643 ns = Ssy/s ns 1.580 1.358 ns = Ssy/s ns 1.381 Eq. (10-22) fom -0.725 -0.536 Eq. (10-22) fom -0.555 Again, for ns  1.2, the optimal size is = 0.085 in. Although this approach used less iterations than in Prob. 10-21, this was due to the initial values picked and not the approach. ______________________________________________________________________________ 10-23 One approach is to select A227 HD steel for its low cost. Try L0 = 48 mm, then for y = 48  37.5 = 10.5 mm when F = 45 N. The spring rate is k = F/y = 45/10.5 = 4.286 N/mm. For a clearance of 1.25 mm with screw, ID = 10 + 1.25 = 11.25 mm. Starting with d = 2 mm, D = ID + d = 11.25 + 2 = 13.25 mm C = D/d = 13.25/2 = 6.625 (acceptable) Table 10-5 (d = 2/25.4 = 0.0787 in): G = 79.3 GPa Chapter 10 - Rev. A, Page 16/41 • Eq. (10-9): 4 4 3 3 3 2 (79.3)10 15.9 coils 8 8(4.286)13.25a d G kD   N Assume squared and closed. Table 10-1: Nt = Na + 2 = 15.9 + 2 = 17.9 coils Ls = dNt = 2(17.9) =35.8 mm ys = L0  Ls = 48  35.8 = 12.2 mm Fs = kys = 4.286(12.2) = 52.29 N   Eq. (10-5):   4 6.625 24 2 1.213B CK     4 3 4 6.625 3C   Eq. (10-7):  3 8 8(52.291.213ss B F DK    3 )13.25 267.5 MPa 2d         Table 10-4: A = 1783 MPa · mmm, m = 0.190 Eq. (10-14): 0.190 1783 1563 MPa 2ut m AS d    Table 10-6: Ssy = 0.45Sut = 0.45(1563) = 703.3 MPa 703.3 2.63 1.2 . . 267.5 sy s s S n O K      No other diameters in the given range work. So specify A227-47 HD steel, d = 2 mm, D = 13.25 mm, ID = 11.25 mm, OD = 15.25 mm, squared and closed, Nt = 17.9 coils, Na = 15.9 coils, k = 4.286 N/mm, Ls = 35.8 mm, and L0 = 48 mm. Ans. ______________________________________________________________________________ 10-24 Select A227 HD steel for its low cost. Try L0 = 48 mm, then for y = 48  37.5 = 10.5 mm when F = 45 N. The spring rate is k = F/y = 45/10.5 = 4.286 N/mm. For a clearance of 1.25 mm with screw, ID = 10 + 1.25 = 11.25 mm. D  d = 11.25 (1) and, D =Cd (2) Starting with C = 8, gives D = 8d. Substitute into Eq. (1) resulting in d = 1.607 mm. Selecting the nearest diameter in the given range, d = 1.6 mm. From this point, the calculations are shown in the third column of the spreadsheet output shown. We see that for d = 1.6 mm, the spring is not solid safe. Iterating on C we find that C = 6.5 provides acceptable results with the specifications A227-47 HD steel, d = 2 mm, D = 13.25 mm, ID = 11.25 mm, OD = 15.25 mm, squared Chapter 10 - Rev. A, Page 17/41 • and closed, Nt = 17.9 coils, Na = 15.9 coils, k = 4.286 N/mm, Ls = 35.8 mm, and L0 = 48 mm. Ans. Chapter 10 - Rev. A, Page 18/41 • Source Parameter Values C 8.000 7 6.500 Eq. (2) d 1.607 1.875 2.045 Table A-28 d 1.600 1.800 2.000 Eq. (1) D 12.850 13.050 13.250 Eq. (10-1) C 8.031 7.250 6.625 Eq. (10-9) Na 7.206 10.924 15.908 Table 10-1 Nt 9.206 12.924 17.908 Table 10-1 Ls 14.730 23.264 35.815 L0 Ls ys 33.270 24.736 12.185 Fs = kys Fs 142.594 106.020 52.224 Table 10-4 A 1783.000 1783.000 1783.000 Table 10-4 m 0.190 0.190 0.190 Eq. (10-14) Sut 1630.679 1594.592 1562.988 Table 10-6 Ssy 733.806 717.566 703.345 Eq. (10-5) KB 1.172 1.200 1.217 Eq. (10-7) s 1335.568 724.943 268.145 ns = Ssy/s ns 0.549 0.990 2.623 The only difference between selecting C first rather than d as was done in Prob. 10-23, is that once d is calculated, the closest wire size must be selected. Iterating on d uses available wire sizes from the beginning. ______________________________________________________________________________ 10-25 A stock spring catalog may have over two hundred pages of compression springs with up to 80 springs per page listed. • Students should be made aware that such catalogs exist. • Many springs are selected from catalogs rather than designed. • The wire size you want may not be listed. • Catalogs may also be available on disk or the web through search routines. For example, disks are available from Century Spring at 1 - (800) - 237 - 5225 www.centuryspring.com • It is better to familiarize yourself with vendor resources rather than invent them yourself. • Sample catalog pages can be given to students for study. ______________________________________________________________________________ 10-26 Given: ID = 0.6 in, C = 10, L0 = 5 in, Ls = 5  3 = 2 in, sq. & grd ends, unpeened, HD A227 wire. (a) With ID = D  d = 0.6 in and C = D/d = 10 10 d  d = 0.6  d = 0.0667 in Ans., and D = 0.667 in. (b) Table 10-1: Ls = dNt = 2 in  Nt = 2/0.0667 30 coils Ans. Chapter 10 - Rev. A, Page 19/41 • (c) Table 10-1: Na = Nt  2 = 30  2 = 28 coils Table 10-5: G = 11.5 Mpsi     Eq. (10-9): 4 64 3 3 0.0667 11.5 10 3.424 lbf/in . 8 8 0.667 28a d Gk Ans D N    (d) Table 10-4: A = 140 kpsiinm, m = 0.190 Eq. (10-14): 0.190 234.2 kpsi0.0667ut m S d    140A Table 10-6: Ssy = 0.45 Sut = 0.45 (234.2) = 105.4 kpsi Fs = kys = 3.424(3) = 10.27 lbf     4 10 24 2 1.135 4 3 4 10 3B CK C       Eq. (10-5): Eq. (10-7):      366.72 10 psi 66.72 kpsi K   3 3 8 10.27 0.6678 1.135 0.0667 s s B F D d    105.4 1.58 . 66.72 sy s n A  s S ns    (e) a = m = 0.5s = 0.5(66.72) = 33.36 kpsi, r = a / m = 1. Using the Gerber fatigue failure criterion with Zimmerli data, Eq. (10-30): Ssu = 0.67 Sut = 0.67(234.2) = 156.9 kpsi The Gerber ordinate intercept for the Zimmerli data is  2 2 35 39.9 kpsi 1 / 1 55 /156.9 sa e sm su SS S S      Table 6-7, p. 307,         22 2 21 1 2 i su se sa se su r S SS S rS                       22 21 156.9 2 39.9 1 1 37.61 kps 2 39.9 1 156.9    37.61 1.13 . 33.36 sa f a Sn Ans     ______________________________________________________________________________ 10-27 Given: OD  0.9 in, C = 8, L0 = 3 in, Ls = 1 in, ys = 3  1 = 2 in, sq. ends, unpeened, music wire. (a) Try OD = D + d = 0.9 in, C = D/d = 8  D = 8d  9d = 0.9  d = 0.1 Ans. Chapter 10 - Rev. A, Page 20/41 • D = 8(0.1) = 0.8 in (b) Table 10-1: Ls = d (Nt + 1)  Nt = Ls / d  1 = 1/0.1 1 = 9 coils Ans. Table 10-1: Na = Nt  2 = 9  2 = 7 coils (c) Table 10-5: G = 11.75 Mpsi     Eq. (10-9): 4 64 3 3 0.1 11.75 10 40.98 lbf/in . 8 8 0.8 7a d Gk Ans D N    (d) Fs = kys = 40.98(2) = 81.96 lbf     Eq. (10-5): 4 8 24 2 1.172 4 3 4 8 3B CK C           Eq. (10-7):  33 3 8 81.96 0.88 1.172 195.7 10 psi 195.7 kpsi 0.1 s s B F DK d        Table 10-4: A = 201 kpsiinm, m = 0.145 Eq. (10-14): 0.145 201 280.7 kpsi 0.1ut m A d   S Table 10-6: Ssy = 0.45 Sut = 0.45(280.7) = 126.3 kpsi 126.3 0.645 .sys S n A   195.7s ns  (e) a = m = s /2 = 195.7/2 = 97.85 kpsi. Using the Gerber fatigue failure criterion with Zimmerli data, Eq. (10-30): Ssu = 0.67 Sut = 0.67(280.7) = 188.1 kpsi The Gerber ordinate intercept for the Zimmerli data is    2 2/ 1 55 /188.1sm suS S  35 36.83 kpsi 1 sa e SS    Table 6-7, p. 307,         22 2 22 2 21 1 2 1 188.1 2 38.3 1 1 36.83 kpsi 2 38.3 1 188.1 su se sa se su r S SS S rS                           Chapter 10 - Rev. A, Page 21/41 • 36.83 0.376 97.85 sa a    .f Sn Ans  Obviously, the spring is severely under designed and will fail statically and in fatigue. Increasing C would improve matters. Try C = 12. This yields ns = 1.83 and nf = 1.00. ______________________________________________________________________________ 10-28 Note to the Instructor: In the first printing of the text, the wire material was incorrectly identified as music wire instead of oil-tempered wire. This will be corrected in subsequent printings. We are sorry for any inconvenience. Given: Fmax = 300 lbf, Fmin = 150 lbf, y = 1 in, OD = 2.1  0.2 = 1.9 in, C = 7, unpeened, sq. & grd., oil-tempered wire. (a) D = OD  d = 1.9  d (1) C = D/d = 7  D = 7d (2) Substitute Eq. (2) into (1) 7d = 1.9  d  d = 1.9/8 = 0.2375 in Ans. (b) From Eq. (2): D = 7d = 7(0.2375) = 1.663 in Ans. 300 150 150 lbf/in . 1 Fk A y       (c) ns (d) Table 10-5: G = 11.6 Mpsi     Eq. (10-9): 4 64 3 3 0.2375 11.6 10 6.69 coils 8 8 1.663 150a d GN D k    Table 10-1: Nt = Na + 2 = 8.69 coils Ans. (e) Table 10-4: A = 147 kpsiinm, m = 0.187 Eq. (10-14): 0.187 147 192.3 kpsi 0.2375ut m A d   S Table 10-6: Ssy = 0.5 Sut = 0.5(192.3) = 96.15 kpsi     Eq. (10-5): 4 7 24 2 1.2 4 3 4 7 3B CK C       Chapter 10 - Rev. A, Page 22/41 • Eq. (10-7): 3 8 s s B sy F DK S d           3 33 0.2375 96.15 10 253.5 lbfsys d S F     8 8 1.2 1.663BK D ys = Fs / k = 253.5/150 = 1.69 in Table 10-1: Ls = Nt d = 8.46(0.2375) = 2.01 in L0 = Ls + ys = 2.01 + 1.69 = 3.70 in Ans. ______________________________________________________________________________ 10-29 For a coil radius given by: 2 1 1 - 2 R RR R N     The torsion of a section is T = PR where dL = R d       2 3 0 3 2 2 1 10 24 2 1 1 2 1 0 4 4 2 2 2 1 1 2 1 2 2 1 4 2 2 1 2 1 24 1 1 2 1 2 4 2 ( ) 2 ( ) 2 16 ( ) 32 N P N N p U TT dL PR d P GJ P GJ P R RR d GJ N P N R RR GJ R R N PN PNR R R R R R GJ R R GJ PNJ d R R R R Gd                                                                 4 2 2 1 2 1 2 . 16 ( )P P d Gk Ans N R R R R     ______________________________________________________________________________ 10-30 Given: Fmin = 4 lbf, Fmax = 18 lbf, k = 9.5 lbf/in, OD  2.5 in, nf = 1.5. For a food service machinery application select A313 Stainless wire. Table 10-5: G = 10(106) psi Note that for 0.013 ≤ d ≤ 0.10 in A = 169, m = 0.146 0.10 < d ≤ 0.20 in A = 128, m = 0.263 18 4 18 47 lbf , 11 lbf , 7 / 11 2 2a m r     F F Chapter 10 - Rev. A, Page 23/41 • Try, 0.146 1690.080 in, 244.4 kpsi (0.08)ut   d S Ssu = 0.67Sut = 163.7 kpsi, Ssy = 0.35Sut = 85.5 kpsi Try unpeened using Zimmerli’s endurance data: Ssa = 35 kpsi, Ssm = 55 kpsi Gerber: 2 2 35 39.5 kpsi 1 ( / ) 1 (55 / 163.7) sa se sm su S S S      S 22 2 3 3 2 2 (7 / 11) (163.7) 2(39.5)1 1 35.0 kpsi 2(39.5) (7 / 11)(163.7) / 35.0 / 1.5 23.3 kpsi 8 8(7)(10 ) (10 ) 2.785 kpsi (0.08 ) 2(23.3) 2.785 2(23.3) 2.785 4(2.785) 4(2.785) sa sa f a S S n F d C                                    2 3 3 3 4 6 3 3(23.3) 6.97 4(2.785) 6.97(0.08) 0.558 in 4 2 4(6.97) 2 1.201 4 3 4(6.97) 3 8 8(7)(0.558)1.201 (10 ) 23.3 kpsi (0.08 ) 35 / 23.3 1.50 checks 10(10 )(0.0 8 B a a B f a D Cd CK C F DK d n GdN kD                                     4 3 max max max 0 0 8) 31.02 coils 8(9.5)(0.558) 31.02 2 33 coils, 0.08(33) 2.64 in / 18 / 9.5 1.895 in (1 ) (1 0.15)(1.895) 2.179 in 2.64 2.179 4.819 in 2.63(0.558)( ) 2.63 2.935 in 0.5 t s t s cr s N L dN y F k y y L DL                          2 2 2 2 1.15(18 / 7) 1.15(18 / 7)(23.3) 68.9 kpsi / 85.5 / 68.9 1.24 9.5(386) 109 Hz (0.08 )(0.558)(31.02)(0.283) a s sy s a kgf d DN              n S These steps are easily implemented on a spreadsheet, as shown below, for different diameters. Chapter 10 - Rev. A, Page 24/41 • d1 d2 d3 d4 d 0.080 0.0915 0.1055 0.1205 m 0.146 0.146 0.263 0.263 A 169.000 169.000 128 128 Sut 244.363 239.618 231.257 223.311 Ssu 163.723 160.544 154.942 149.618 Ssy 85.527 83.866 80.940 78.159 Sse 39.452 39.654 40.046 40.469 Ssa 35.000 35.000 35.000 35.000  23.333 23.333 23.333 23.333  2.785 2.129 1.602 1.228 C 6.977 9.603 13.244 17.702 D 0.558 0.879 1.397 2.133 KB 1.201 1.141 1.100 1.074 a 23.333 23.333 23.333 23.333 nf 1.500 1.500 1.500 1.500 Na 30.993 13.594 5.975 2.858 Nt 32.993 15.594 7.975 4.858 LS 2.639 1.427 0.841 0.585 ys 2.179 2.179 2.179 2.179 L0 4.818 3.606 3.020 2.764 (L0)cr 2.936 4.622 7.350 11.220 s 69.000 69.000 69.000 69.000 ns 1.240 1.215 1.173 1.133 f,(Hz) 108.895 114.578 118.863 121.775 The shaded areas depict conditions outside the recommended design conditions. Thus, one spring is satisfactory. The specifications are: A313 stainless wire, unpeened, squared and ground, d = 0.0915 in, OD = 0.879 + 0.092 = 0.971 in, L0 = 3.606 in, and Nt = 15.59 turns Ans. ______________________________________________________________________________ 10-31 The steps are the same as in Prob. 10-23 except that the Gerber-Zimmerli criterion is replaced with Goodman-Zimmerli:  1 sa se sm su SS S S   Chapter 10 - Rev. A, Page 25/41 • The problem then proceeds as in Prob. 10-23. The results for the wire sizes are shown below (see solution to Prob. 10-23 for additional details). Iteration of d for the first trial d1 d2 d3 d4 d1 d2 d3 d4 d 0.080 0.0915 0.1055 0.1205 d 0.080 0.0915 0.1055 0.1205 m 0.146 0.146 0.263 0.263 KB 1.151 1.108 1.078 1.058 A 169.000 169.000 128.000 128.000 a 29.008 29.040 29.090 29.127 Sut 244.363 239.618 231.257 223.311 nf 1.500 1.500 1.500 1.500 Ssu 163.723 160.544 154.942 149.618 Na 14.191 6.456 2.899 1.404 Ssy 85.527 83.866 80.940 78.159 Nt 16.191 8.456 4.899 3.404 Sse 52.706 53.239 54.261 55.345 Ls 1.295 0.774 0.517 0.410 Ssa 43.513 43.560 43.634 43.691 ymax 2.875 2.875 2.875 2.875  29.008 29.040 29.090 29.127 L0 4.170 3.649 3.392 3.285  2.785 2.129 1.602 1.228 (L0)cr 3.809 5.924 9.354 14.219 C 9.052 12.309 16.856 22.433 s 85.782 85.876 86.022 86.133 D 0.724 1.126 1.778 2.703 ns 0.997 0.977 0.941 0.907 f (Hz) 140.040 145.559 149.938 152.966 Without checking all of the design conditions, it is obvious that none of the wire sizes satisfy ns ≥ 1.2. Also, the Gerber line is closer to the yield line than the Goodman. Setting nf = 1.5 for Goodman makes it impossible to reach the yield line (ns < 1) . The table below uses nf = 2. Iteration of d for the second trial d1 d2 d3 d4 d1 d2 d3 d4 d 0.080 0.0915 0.1055 0.1205 d 0.080 0.0915 0.1055 0.1205 m 0.146 0.146 0.263 0.263 KB 1.221 1.154 1.108 1.079 A 169.000 169.000 128.000 128.000 a 21.756 21.780 21.817 21.845 Sut 244.363 239.618 231.257 223.311 nf 2.000 2.000 2.000 2.000 Ssu 163.723 160.544 154.942 149.618 Na 40.243 17.286 7.475 3.539 Ssy 85.527 83.866 80.940 78.159 Nt 42.243 19.286 9.475 5.539 Sse 52.706 53.239 54.261 55.345 Ls 3.379 1.765 1.000 0.667 Ssa 43.513 43.560 43.634 43.691 ymax 2.875 2.875 2.875 2.875  21.756 21.780 21.817 21.845 L0 6.254 4.640 3.875 3.542  2.785 2.129 1.602 1.228 (L0)cr 2.691 4.266 6.821 10.449 C 6.395 8.864 12.292 16.485 s 64.336 64.407 64.517 64.600 D 0.512 0.811 1.297 1.986 ns 1.329 1.302 1.255 1.210 f (Hz) 98.936 104.827 109.340 112.409 The satisfactory spring has design specifications of: A313 stainless wire, unpeened, squared and ground, d = 0.0915 in, OD = 0.811 + 0.092 = 0.903 in, L0 = 4.266 in, and .Nt = 19.6 turns. Ans. ______________________________________________________________________________ 10-32 This is the same as Prob. 10-30 since Ssa = 35 kpsi. Therefore, the specifications are: Chapter 10 - Rev. A, Page 26/41 • A313 stainless wire, unpeened, squared and ground, d = 0.0915 in, OD = 0.879 + 0.092 = 0.971 in, L0 = 3.606 in, and Nt = 15.84 turns Ans. ______________________________________________________________________________ 10-33 For the Gerber fatigue-failure criterion, Ssu = 0.67Sut , 22 2 2 2, 1 1 1 ( / ) 2 sa su se se sa sm su se su S r S SS S S S S rS                The equation for Ssa is the basic difference. The last 2 columns of diameters of Ex. 10-5 are presented below with additional calculations. d 0.105 0.112 d 0.105 0.112 Sut 278.691 276.096 Na 8.915 6.190 Ssu 186.723 184.984 Ls 1.146 0.917 Sse 38.325 38.394 L0 3.446 3.217 Ssy 125.411 124.243 (L0)cr 6.630 8.160 Ssa 34.658 34.652 KB 1.111 1.095  23.105 23.101 a 23.105 23.101  1.732 1.523 nf 1.500 1.500 C 12.004 13.851 s 70.855 70.844 D 1.260 1.551 ns 1.770 1.754 ID 1.155 1.439 fn 105.433 106.922 OD 1.365 1.663 fom 0.973 1.022 There are only slight changes in the results. ______________________________________________________________________________ 10-34 As in Prob. 10-35, the basic change is Ssa. For Goodman, 1 - ( / ) sa se sm su S S S  S Recalculate Ssa with se su sa su se rS SS rS S   Calculations for the last 2 diameters of Ex. 10-5 are given below. Chapter 10 - Rev. A, Page 27/41 • d 0.105 0.112 d 0.105 0.112 Sut 278.691 276.096 Na 9.153 6.353 Ssu 186.723 184.984 Ls 1.171 0.936 Sse 49.614 49.810 L0 3.471 3.236 Ssy 125.411 124.243 (L0)cr 6.572 8.090 Ssa 34.386 34.380 KB 1.112 1.096  22.924 22.920 a 22.924 22.920  1.732 1.523 nf 1.500 1.500 C 11.899 13.732 s 70.301 70.289 D 1.249 1.538 ns 1.784 1.768 ID 1.144 1.426 fn 104.509 106.000 OD 1.354 1.650 fom 0.986 1.034 There are only slight differences in the results. ______________________________________________________________________________ 10-35 Use: E = 28.6 Mpsi, G = 11.5 Mpsi, A = 140 kpsi · inm , m = 0.190, rel cost = 1. Try 0.190 1400.067 , 234.0 kpsi (0.067)ut d in S   Table 10-6: Ssy = 0.45Sut = 105.3 kpsi Table 10-7: Sy = 0.75Sut = 175.5 kpsi Eq. (10-34) with D/d = C and C1 = C max ySF 2 22 max 2 2 max [( ) (16 ) 4] 4 1(16 ) 4 4 ( 1) 4 1 ( 1) 1 4 A A y y y y y K C d n d SC C C C C n F d S C C C n F                        2 2 2 max max 1 11 1 2 0 4 4 4 4 y y y y d S d S C C n F n F                      22 2 2 max max max 2 3 22 3 2 3 1 2 take positive root 2 16 16 4 1 (0.067 )(175.5)(10 ) 2 16(1.5)(18) (0.067) (175.5)(10 ) (0.067) (175.5)(10 ) 16(1.5)(18) 4( y y y y y y d S d S d S C n F n F n F                               2 4.590 1.5)(18)    Chapter 10 - Rev. A, Page 28/41 •   3 3 4.59 0.067 0.3075 in 33 500 31000 4 8 8 exp(0.105 ) 6.5 i i D Cd d d CF D D C                  Use the lowest Fi in the preferred range. This results in the best fom. 3(0.067) 33 500 4.590 31000 4 6.505 lbf 8(0.3075) exp[0.105(4.590)] 6.5i F             For simplicity, we will round up to the next integer or half integer. Therefore, use Fi = 7 lbf 4 4 6 3 3 0 18 lbf 18 7 22 lbf/in 0.5 (0.067) (11.5)(10 ) 45.28 turns 8 8(22)(0.3075) 11.545.28 44.88 turns 28.6 (2 1 ) [2(4.590) 1 44.88](0.067) 3.555 in 3.555 0.5 4.055 in a b a b k d GN kD GN N E L C N d L                      Body: 4 2 4(4.590) 2 1.326 4 3 4(4.590) 3B C C      K   3max max 3 3 body max 2 2 2 2 2 8 8(1.326)(18)(0.3075) (10 ) 62.1 kpsi (0.067) 105.3( ) 1.70 62.1 2 2(0.134)2 2(0.067) 0.134 in, 4 0.067 4 1 4(4) 1( ) 1.25 4 ( ) B sy y B K F D d S n rr d C d CK C F DK                           max3 4 4(4) 4 8 B B d   38(18)(0.3075)1.25 (10 ) 58.58 kpsi ( ) fom (1 0.160 4 4 sy y B B S n       3 2 2 2 2 (0.067) 105.3 1.80 58.58 ( 2) (0.067) (44.88 2)(0.3075)) bd N D                 Several diameters, evaluated using a spreadsheet, are shown below. Chapter 10 - Rev. A, Page 29/41 • d 0.067 0.072 0.076 0.081 0.085 0.09 0.095 0.104 Sut 233.977 230.799 228.441 225.692 223.634 221.219 218.958 215.224 Ssy 105.290 103.860 102.798 101.561 100.635 99.548 98.531 96.851 Sy 175.483 173.100 171.331 169.269 167.726 165.914 164.218 161.418 C 4.589 5.412 6.099 6.993 7.738 8.708 9.721 11.650 D 0.307 0.390 0.463 0.566 0.658 0.784 0.923 1.212 Fi (calc) 6.505 5.773 5.257 4.675 4.251 3.764 3.320 2.621 Fi (rd) 7.0 6.0 5.5 5.0 4.5 4.0 3.5 3.0 k 22.000 24.000 25.000 26.000 27.000 28.000 29.000 30.000 Na 45.29 27.20 19.27 13.10 9.77 7.00 5.13 3.15 Nb 44.89 26.80 18.86 12.69 9.36 6.59 4.72 2.75 L0 3.556 2.637 2.285 2.080 2.026 2.071 2.201 2.605 L18 lbf 4.056 3.137 2.785 2.580 2.526 2.571 2.701 3.105 KB 1.326 1.268 1.234 1.200 1.179 1.157 1.139 1.115 max 62.118 60.686 59.707 58.636 57.875 57.019 56.249 55.031 (ny)body 1.695 1.711 1.722 1.732 1.739 1.746 1.752 1.760 B 58.576 59.820 60.495 61.067 61.367 61.598 61.712 61.712 (ny)B 1.797 1.736 1.699 1.663 1.640 1.616 1.597 1.569 (ny)A 1.500 1.500 1.500 1.500 1.500 1.500 1.500 1.500 fom 0.160 0.144 -0.138 0.135 0.133 0.135 0.138 0.154 Except for the 0.067 in wire, all springs satisfy the requirements of length and number of coils. The 0.085 in wire has the highest fom. ______________________________________________________________________________ 10-36 Given: Nb = 84 coils, Fi = 16 lbf, OQ&T steel, OD = 1.5 in, d = 0.162 in. D = OD  d = 1.5  0.162 = 1.338 in (a) Eq. (10-39): L0 = 2(D  d) + (Nb + 1)d = 2(1.338  0.162) + (84 + 1)(0.162) = 16.12 in Ans. or 2d + L0 = 2(0.162) + 16.12 = 16.45 in overall 1.338 8.26 0.162 DC d    (b) 3 3 4 2 4(8.26) 2 1.166 4 3 4(8.26) 3 8 8(16)(1.338)1.166 14 950 psi . (0.162) B i i B CK C F DK Ans d                 (c) From Table 10-5 use: G = 11.4(106) psi and E = 28.5(106) psi Chapter 10 - Rev. A, Page 30/41 • 4 4 6 3 3 28.5 (0.162) (11.4)(10 ) 4.855 lb 8 8(1.338) (84.4)a E d G D N    11.484 84.4 turns f/in . a b GN N k Ans      (d) Table 10-4: A = 147 psi · inm , m = 0.187 0.187 147 207.1 kpsi (0.162)ut S   0.75(207.1) 155.3 kpsi 0.50(207.1) 103.5 kpsi y sy S S     Body 3 3 3(0.162) (103.5)(10 ) sy B d S F K D     110.8 lbf 8(1.166)(1.338)   Torsional stress on hook point B 2 2 2 2 2(0.25 0.162 / 2) 4.086 0.162 4 1 4(4.086) 1( ) 1.243B rC d CK 2 3 3 4 4 4(4.086) 4 (0.162) (103.5)(10 ) 103.9 lbf 8(1.243)(1.338) C F               Normal stress on hook point A     1 1 2 2 1 1 1 1 2 1.338 8.26 0.162 4 1 4(8.26) 8.26 1) 4 ( 1) 4(8.26)(8.26 1) 16( ) 4 A A rC d C CK C C K DS F               3 2 3 3 2 ( 1.099 155.3(10 ) 85.8 lbf 16(1.099)(1.338) / (0.162) 4 / (0.162) min(110.8, 103.9, 85. yt d d F                   8) 85.8 lbf .Ans (e) Eq. (10-48): 85.8 16 14.4 in . 4.855 iF Fy Ans k      ______________________________________________________________________________ Chapter 10 - Rev. A, Page 31/41 • 10-37 Fmin = 9 lbf, Fmax = 18 lbf 18 9 18 94.5 lbf , 13.5 lbf 2 2a m F F     A313 stainless: 0.013 ≤ d ≤ 0.1 A = 169 kpsi · inm , m = 0.146 0.1 ≤ d ≤ 0.2 A = 128 kpsi · inm , m = 0.263 E = 28 Mpsi, G = 10 Gpsi Try d = 0.081 in and refer to the discussion following Ex. 10-7 0.146 169 243.9 kpsi (0.081) 0.67 163.4 kpsi 0.35 85.4 kpsi ut su ut sy ut S S S S S       0.55 134.2 kpsiy utS S  Table 10-8: Sr = 0.45Sut = 109.8 kpsi 2 2 / 2 109.8 / 2 57.8 kpsi 1 [ / (2 )] 1 [(109.8 / 2) / 243.9] / 4.5 / 13.5 0.333 r e r ut S S S r F F       a m S  22 2 21 1 2 ut e a e ut r S SS S rS             Table 7-10: 22 2(0.333) (243.9 ) 2(57.8)1 1 42.2 kpsi 2(57.8) 0.333(243.9)a S             Hook bending 2 2 2 2 16 4( ) ( ) ( ) 2 4.5 (4 - - 1)16 4 4 ( - 1) 2 a a a A a A f A a C S SF K d d n C C C S d C C                   This equation reduces to a quadratic in C (see Prob. 10-35). The useable root for C is Chapter 10 - Rev. A, Page 32/41 • 22 2 2 22 3 2 3 2 144 36 (0.081) (42.2)(10 ) (0.081) (42.2)(10 ) a a ad S d S                    2 3 0.5 144 (0.081) (42.2)(10 )0.5 2 144 144 36 4.91 d SC            3 3 33 500 1000 4 8 8 exp(0.105 ) 6 id d D D C    0.398 in 3 .5i D Cd CF                Use the lowest Fi in the preferred range. 3(0.081) 33 500 4.91 31000 4 8(0.398) exp[0.105(4.91)] 6.5 8.55 lbf iF            For simplicity we will round up to next 1/4 integer. 4 4 6 8.75 lbf 18 9 36 lbf/in 0.25 (0.081) (10)(10 ) iF k d G     3 3 0 max 0 max 23.7 turns 8 8(36)(0.398) 1023.7 23.3 turns 28 (2 1 ) [2(4.91) 1 23.3](0.081) 2.602 in ( ) / 2.602 (18 8.75) / 36 2. a b a b i N kD GN N E L C N d L L F F k                       2 2 859 in 4.5(4) 4 1( ) 1 1a A C C d C         -3 2 2 18(10 ) 4(4.91 ) 4.91 1 1 21.1 kpsi (0.081 ) 4.91 1 42.2( ) 2 checks ( ) 21.1 a f A a A Sn                Body: 4 2 4(4.91) 2 1.300 4 3 4(4.91) 3B CK C        Chapter 10 - Rev. A, Page 33/41 • 3 3 8(1.300)(4.5)(0.398) (10 ) 11.16 kpsi (0.081) 13.5 (11.16) 33.47 kpsi 4.5 a m m a a F F          The repeating allowable stress from Table 7-8 is Ssr = 0.30Sut = 0.30(243.9) = 73.17 kpsi The Gerber intercept is 2 73.17 / 2 38.5 kpsi 1 [(73.17 / 2) / 163.4]se S    From Table 6-7, 22 body 1 163.4 11.16 2(33.47)(38.5)( ) 1 1 2.53fn                 2 33.47 38.5 163.4(11.16)        Let r2 = 2d = 2(0.081) = 0.162 2 2 2 4(4) 14, ( ) 1.25 4(4) 4 ( ) 1.25( ) (11.16) 10.73 kpsi 1.30 ( ) 1.25( ) (33.47) 32.18 kpsi 1.30 B B a B a B B m B m B rC K d K K K K                 Table 10-8: (Ssr )B = 0.28Sut = 0.28(243.9) = 68.3 kpsi 2 22 68.3 / 2( ) 35.7 kpsi 1 [(68.3 / 2) / 163.4] 1 163.4 10.73 2(32.18)(35.7)( ) 1 1 2.51 2 32.18 35.7 163.4(10.73) se B f B S n                            Yield Bending: 2 max max 2 2 -3 2 4 (4 1)( ) 1 1 4(18) 4(4.91) 4.91 1 1 (10 ) 84.4 kpsi (0.081 ) 4.91 1 134.2( ) 1.59 84.4 A y A F C C d C n                     Body: Chapter 10 - Rev. A, Page 34/41 • body ( / ) (8.75 / 4.5)(11.16) 21.7 kpsi /( ) 11.16 / (33.47 21.7) 0.948 0.948( ) ( ) (85.4 21.7) 31.0 kpsi 1 0.948 1 ( ) 31.0( ) 2.78 11.16 i i a a a m i sa y sy i sa y y a F F r rS S r S n                          Hook shear: Hook shear: max 0.3 0.3(243.9) 73.2 kpsi ( ) ( ) 10.73 32.18 42.9 kpsi 73.2( ) 1.71 42.9 sy ut a B m B y B S S n              2 2 2 27.6 ( 2) 7.6 (0.081) (23.3 2)(0.398)fom 1.239 4 4 bd N D        A tabulation of several wire sizes follow d 0.081 0.085 0.092 0.098 0.105 0.12 Sut 243.920 242.210 239.427 237.229 234.851 230.317 Ssu 163.427 162.281 160.416 158.943 157.350 154.312 Sr 109.764 108.994 107.742 106.753 105.683 103.643 Se 57.809 57.403 56.744 56.223 55.659 54.585 Sa 42.136 41.841 41.360 40.980 40.570 39.786 C 4.903 5.484 6.547 7.510 8.693 11.451 D 0.397 0.466 0.602 0.736 0.913 1.374 OD 0.478 0.551 0.694 0.834 1.018 1.494 Fi (calc) 8.572 7.874 6.798 5.987 5.141 3.637 Fi (rd) 8.75 9.75 10.75 11.75 12.75 13.75 k 36.000 36.000 36.000 36.000 36.000 36.000 Na 23.86 17.90 11.38 8.03 5.55 2.77 Nb 23.50 17.54 11.02 7.68 5.19 2.42 L0 2.617 2.338 2.127 2.126 2.266 2.918 L18 lbf 2.874 2.567 2.328 2.300 2.412 3.036 (a)A 21.068 20.920 20.680 20.490 20.285 19.893 (nf)A 2.000 2.000 2.000 2.000 2.000 2.000 KB 1.301 1.264 1.216 1.185 1.157 1.117 (a)body 11.141 10.994 10.775 10.617 10.457 10.177 (m)body 33.424 32.982 32.326 31.852 31.372 30.532 Ssr 73.176 72.663 71.828 71.169 70.455 69.095 Sse 38.519 38.249 37.809 37.462 37.087 36.371 (nf)body 2.531 2.547 2.569 2.583 2.596 2.616 (K)B 1.250 1.250 1.250 1.250 1.250 1.250 (a)B 10.705 10.872 11.080 11.200 11.294 11.391 (m)B 32.114 32.615 33.240 33.601 33.883 34.173 (Ssr)B 68.298 67.819 67.040 66.424 65.758 64.489 (Sse)B 35.708 35.458 35.050 34.728 34.380 33.717 Chapter 10 - Rev. A, Page 35/41 • (nf)B 2.519 2.463 2.388 2.341 2.298 2.235 Sy 134.156 133.215 131.685 130.476 129.168 126.674 (A)max 84.273 83.682 82.720 81.961 81.139 79.573 (ny)A 1.592 1.592 1.592 1.592 1.592 1.592 i 21.663 23.820 25.741 27.723 29.629 31.097 r 0.945 1.157 1.444 1.942 2.906 4.703 (Ssy)body 85.372 84.773 83.800 83.030 82.198 80.611 (Ssa)y 30.958 32.688 34.302 36.507 39.109 40.832 (ny)body 2.779 2.973 3.183 3.438 3.740 4.012 (Ssy)B 73.176 72.663 71.828 71.169 70.455 69.095 (B)max 42.819 43.486 44.321 44.801 45.177 45.564 (ny)B 1.709 1.671 1.621 1.589 1.560 1.516 fom 1.246 1.234 1.245 1.283 1.357 1.639 optimal fom The shaded areas show the conditions not satisfied. ______________________________________________________________________________ 10-38 For the hook, M = FR sin, ∂M/∂F = R sin   3/ 2 2 0 1 sin 2 FRF R R d EI EI      F The total deflection of the body and the two hooks 3 3 3 3 4 4 4 3 3 4 4 8 8 ( / 2)2 2 ( / 64)( ) 8 8 b b a b FD N FR FD N F D d G EI d G E d FD G FD NN d G E d G GN N                       Q.E.D.a b E ______________________________________________________________________________ 10-39 Table 10-5 (d = 4 mm = 0.1575 in): E = 196.5 GPa Table 10-4 for A227: A = 1783 MPa · mmm, m = 0.190 Eq. (10-14): 0.190 1783 1370 MPa 4ut m AS d    Eq. (10-57): Sy = all = 0.78 Sut = 0.78(1370) = 1069 MPa Chapter 10 - Rev. A, Page 36/41 • D = OD  d = 32  4 = 28 mm C = D/d = 28/4 = 7   Eq. (10-43): 22 4 7 7 14 1 1.119 4 ( 1) 4(7)(7 1)i C CK C C         Eq. (10-44): 3 32 i FrK d    At yield, Fr = My ,  = Sy. Thus,    3 33 4 1069 10y y d S     6.00 N · m 32 32(1.119)i M K  Count the turns when M = 0 2.5 y M N   k 4 10.8 d Ek DN  where from Eq. (10-51): Thus, 42.5 / (10.8 ) yMN d E DN   Solving for N gives    4 2.5 1 [10.8 / ( )] 2.5 y N DM d E   4 2.413 turns 1 10.8(28)(6.00) / 4 (196.5)       This means (2.5 - 2.413)(360) or 31.3 from closed. Ans. Treating the hand force as in the middle of the grip,  3 max 87.5112.5 87.5 68.75 mm 2 6.00 10 87.3 N . 68.75 y r M F Ans r        ______________________________________________________________________________ 10-40 The spring material and condition are unknown. Given d = 0.081 in and OD = 0.500, (a) D = 0.500  0.081 = 0.419 in Using E = 28.6 Mpsi for an estimate Chapter 10 - Rev. A, Page 37/41 • 4 4 6(0.081) (28.6)(10 ) 24.7 lbf · in/turn 10.8 10.8(0.419)(11) d Ek DN    for each spring. The moment corresponding to a force of 8 lbf Fr = (8/2)(3.3125) = 13.25 lbf · in/spring The fraction windup turn is 13.25 nsFrn k   0.536 tur 24.7   The arm swings through an arc of slightly less than 180, say 165. This uses up 165/360 or 0.458 turns. So n = 0.536  0.458 = 0.078 turns are left (or 0.078(360) = 28.1 ). The original configuration of the spring was Ans. (b)    33 3 1.168 4 1 4(5.17)(5.17 1) 32 32(13.25)1.168 297 10 psi 297 kpsi . i i C C MK A             2 2 0.419 5.17 0.081 4 1 4(5.17) 5.17 1 (0.081) DC d C CK ns d           To achieve this stress level, the spring had to have set removed. ______________________________________________________________________________ 10-41 (a) Consider half and double results Straight section: M = 3FR, 3M R P   Chapter 10 - Rev. A, Page 38/41 • Upper 180 section: [ (1 cos )] s ) M F R R (2 cos ), (2 coMFR R F             Lower section: M = FR sin , sinM R F    Considering bending only: / 2 / 22 2 2 2 0 2 2 2 9 (2 cos ) ( sin ) 2 9 4 2 19 9 (19 18 ) 4 2 2 lU FR dx FR R d F R R d F EI F R l EI FR FRR l R l EI EI   0 0 2 3 3 0 4 4sin 2 2 R R                                               The spring rate is 2 (19 ns R R l     2 . 18 ) F EIk A (b) Given: A227 HD wire, d = 2 mm, R = 6 mm, and l = 25 mm. Table 10-5 (d = 2 mm = 0.0787 in): E = 197.2 MPa            310 N/m 10.65 N/mm .ns         9 4 2 2 197.2 10 0.002 / 64 10.65 0.006 19 0.006 18 0.025 k A (c) The maximum stress will occur at the bottom of the top hook where the bending- moment is 3FR and the axial fore is F. Using curved beam theory for bending, Eq. (3-65), p. 119:    2 3 / 4 / 2 i i i i Mc FRc Aer d e R d      Axial: F F2 / 4a A d     Chapter 10 - Rev. A, Page 39/41 • Combining,  max 2 34 1 / 2 i i a RcF S d e R d              y   2 (1) . 34 1 / 2 y i d S F Ans Rc e R d        For the clip in part (b), Eq. (10-14) and Table 10-4: Sut = A/dm = 1783/20.190 = 1563 MPa Eq. (10-57): Sy = 0.78 Sut = 0.78(1563) = 1219 MPa Table 3-4, p. 121:   2 2 2 1 5.95804 mm 2 6 6 1 nr     e = rc  rn = 6  5.95804 = 0.04196 mm ci = rn  (R  d /2) = 5.95804  (6  2/2) = 0.95804 mm Eq. (1):         2 60.002 1219 10 46.0 N . 3 6 0.95804 4 1 0.04196 6 1 F A         ns ______________________________________________________________________________ 10-42 (a) Chapter 10 - Rev. A, Page 40/41 • Chapter 10 - Rev. A, Page 41/41            /2 2 0 0 3 2 2 , 1 cos , 1 cos 0 1 ( ) 1 cos 4 3 2 4 2 3 8 12 l F MM Fx F 0x x l MM Fl FR l R l F Fx x dx F l R Rd EI F l R l l R R EI                                               The spring rate is    3 24 3 2F 2 12 . 4 2 3 8 F EI ns l R l l R R             k A (b) Given: A313 stainless wire, d = 0.063 in, R = 0.625 in, and l = 0.5 in. Table 10-5: E = 28 Mpsi  4 40.063 7.73364 64I d   7 410 in                     6 7 3 2 2 12 28 10 7.733 10 0.625 k         4 0.5 3 0.625 2 0.5 4 2 0.5 0.625 3 8 36.3 lbf/in .Ans       (c) Table 10-4: A = 169 kpsiinm, m = 0.146 Eq. (10-14): Sut = A/ d m = 169/0.0630.146 = 253.0 kpsi Eq. (10-57): Sy = 0.61 Sut = 0.61(253.0) = 154.4 kpsi One can use curved beam theory as in the solution for Prob. 10-41. However, the equations developed in Sec. 10-12 are equally valid. C = D/d = 2(0.625 + 0.063/2)/0.063 = 20.8      Eq. (10-43): 22 4 20.8 20.8 14 1 1.037 4 1 4 20.8 20.8 1i C C C C         K Eq. (10-44), setting  = Sy: • Chapter 10 - Rev. A, Page 42/41       3 3 3 32 0.5 0.62532 1.037 154.4 10 0.063i y FFrK S d      Solving for F yields F = 3.25 lbf Ans. Try solving part (c) of this problem using curved beam theory. You should obtain the same answer. ______________________________________________________________________________ 10-43 (a) M =  Fx 2/ / / 6 M Fx Fx I c I c bh    Constant stress, 2 6 (1) . 6 Fx Fxh Ans b     bh At x = l, 6 / .o o Fl h x l Ans b      h h (b) M =  Fx,  M / F = x     3/2 1/2 3/2 31 3 0 0 012 3/2 3 3/2 3 3 / 1 12 / 2 12 8 3 l l l oo o o M M F Fx x Fly dx dx x dx EI E bh Ebh x l Fl Fll bh E bh E             3 3 .8 obh EF ns y l  k A ______________________________________________________________________________ 10-44 Computer programs will vary. ______________________________________________________________________________ 10-45 Computer programs will vary. • Chapter 11 11-1 For the deep-groove 02-series ball bearing with R = 0.90, the design life xD, in multiples of rating life, is   6 10 60 25000 35060 525 . 10 D D D D R L nx Ans L L      The design radial load is  1.2 2.5 3.0 kNDF   Eq. (11-6):     1/3 10 1/1.483 5253.0 0.02 4.459 0.02 ln 1/ 0.9 C            C10 = 24.3 kN Ans. Table 11-2: Choose an 02-35 mm bearing with C10 = 25.5 kN. Ans. Eq. (11-18):   1.4833525 3 / 25.5 0.02 exp 0.920 . 4.459 0.02 R Ans             ______________________________________________________________________________ 11-2 For the angular-contact 02-series ball bearing as described, the rating life multiple is   6 10 60 40000 52060 1248 10 D D D D R L nx L L      The design radial load is  1.4 725 1015 lbf 4.52 kNDF    Eq. (11-6):     1/3 10 1/1.483 12481015 0.02 4.459 0.02 ln 1/ 0.9 10 930 lbf 48.6 kN C              Table 11-2: Select an 02-60 mm bearing with C10 = 55.9 kN. Ans. Eq. (11-18):   1.48331248 4.52 / 55.9 0.02 exp 0.945 . 4.439 R Ans              ______________________________________________________________________________ Chapter 11, Page 1/28 • 11-3 For the straight-roller 03-series bearing selection, xD = 1248 rating lives from Prob. 11-2 solution.  1.4 2235 3129 lbf 13.92 kNDF    3/10 10 124813.92 118 kN 1 C       Table 11-3: Select an 03-60 mm bearing with C10 = 123 kN. Ans. Eq. (11-18):   1.48310/31248 13.92 /123 0.02 exp 0.917 . 4.459 0.02 R Ans             ______________________________________________________________________________ 11-4 The combined reliability of the two bearings selected in Probs. 11-2 and 11-3 is   0.945 0.917 0.867 .R Ans  We can choose a reliability goal of 0.90 0.95 for each bearing. We make the selections, find the existing reliabilities, multiply them together, and observe that the reliability goal is exceeded due to the roundup of capacity upon table entry. Another possibility is to use the reliability of one bearing, say R1. Then set the reliability goal of the second as 2 1 0.90R R  or vice versa. This gives three pairs of selections to compare in terms of cost, geometry implications, etc. ______________________________________________________________________________ 11-5 Establish a reliability goal of 0.90 0.95 for each bearing. For an 02-series angular contact ball bearing,   1/3 10 1/1.483 12481015 0.02 4.439 ln 1/ 0.95 12822 lbf 57.1 kN C             Select an 02-65 mm angular-contact bearing with C10 = 63.7 kN.   1.48331248 4.52 / 63.7 0.02 exp 0.962 4.439A R              Chapter 11, Page 2/28 • For an 03-series straight roller bearing,   3/10 10 1/1.483 124813.92 136.5 kN 0.02 4.439 ln 1/ 0.95 C           Select an 03-65 mm straight-roller bearing with C10 = 138 kN.   1.48310/31248 13.92 /138 0.02 exp 0.953 4.439B R              The overall reliability is R = (0.962)(0.953) = 0.917, which exceeds the goal. ______________________________________________________________________________ 11-6 For the straight cylindrical roller bearing specified with a service factor of 1, R = 0.95 and FR = 20 kN.  6 10 60 8000 95060 456 10 D D D D R L nx L L        3/10 10 1/1.483 45620 145 kN . 0.02 4.439 ln 1/ 0.95 C A           ns ______________________________________________________________________________ 11-7 Both bearings need to be rated in terms of the same catalog rating system in order to compare them. Using a rating life of one million revolutions, both bearings can be rated in terms of a Basic Load Rating. Eq. (11-3):     1/31/ 1/ 6 3000 500 6060 2.0 10 8.96 kN a a A A A A A A R R L nC F F L L                      Bearing B already is rated at one million revolutions, so CB = 7.0 kN. Since CA > CB, bearing A can carry the larger load. Ans. ______________________________________________________________________________ 11-8 FD = 2 kN, LD = 109 rev, R = 0.90 Eq. (11-3): 1/ 1/39 10 6 102 20 kN . 10 a D D R LC F An L              s ______________________________________________________________________________ Chapter 11, Page 3/28 • 11-9 FD = 800 lbf, D = 12 000 hours, nD = 350 rev/min, R = 0.90 Eq. (11-3):    1/31/ 10 6 12 000 350 6060 800 5050 lbf 10 a D D D R nC F An L              s ______________________________________________________________________________ 11-10 FD = 4 kN, D = 8 000 hours, nD = 500 rev/min, R = 0.90 Eq. (11-3):    1/31/ 10 6 8 000 500 6060 4 24.9 kN 10 a D D D R nC F An L              s ______________________________________________________________________________ 11-11 FD = 650 lbf, nD = 400 rev/min, R = 0.95     D = (5 years)(40 h/week)(52 week/year) = 10 400 hours Assume an application factor of one. The multiple of rating life is    6 10 400 400 60 249.6 10 D D R Lx L    Eq. (11-6):      1/3 10 1/1.483 249.61 650 0.02 4.439 ln 1/ 0.95 C           4800 lbf .Ans ______________________________________________________________________________ 11-12 FD = 9 kN, LD = 108 rev, R = 0.99 Assume an application factor of one. The multiple of rating life is 8 6 10 100 10 D D R Lx L    Eq. (11-6):      1/3 10 1/1.483 1001 9 0.02 4.439 ln 1/ 0.99 C           69.2 kN .Ans ______________________________________________________________________________ 11-13 FD = 11 kips, D = 20 000 hours, nD = 200 rev/min, R = 0.99 Assume an application factor of one. Use the Weibull parameters for Manufacturer 2 on p. 608. Chapter 11, Page 4/28 • The multiple of rating life is    6 20 000 200 60 240 10 D D R Lx L    Eq. (11-6):      1/3 10 1/1.483 2401 11 0.02 4.439 ln 1/ 0.99 C           113 kips .Ans ______________________________________________________________________________ 11-14 From the solution to Prob. 3-68, the ground reaction force carried by the bearing at C is RC = FD = 178 lbf. Use the Weibull parameters for Manufacturer 2 on p. 608.   6 15000 1200 60 1080 10 D D R Lx L    Eq. (11-7):    1/ 10 1/ 0 0 1 a D f D b D xC a F x x R                1/3 10 1/1.483 10801.2 178 0.02 4.459 0.02 1 0.95 2590 lbf . C Ans            ______________________________________________________________________________ 11-15 From the solution to Prob. 3-69, the ground reaction force carried by the bearing at C is RC = FD = 1.794 kN. Use the Weibull parameters for Manufacturer 2 on p. 608.   6 15000 1200 60 1080 10 D D R Lx L    Eq. (11-7):    1/ 10 1/ 0 0 1 a D f D b D xC a F x x R                1/3 10 1/1.483 10801.2 1.794 0.02 4.459 0.02 1 0.95 26.1 kN . C Ans            ______________________________________________________________________________ 11-16 From the solution to Prob. 3-70, RCz = –327.99 lbf, RCy = –127.27 lbf     1/22 2327.99 127.27 351.8 lbfC DR F         Use the Weibull parameters for Manufacturer 2 on p. 608. Chapter 11, Page 5/28 •   6 15000 1200 60 1080 10 D D R Lx L    Eq. (11-7):    1/ 10 1/1 a D f D b o o D xC a F x x R                1/3 10 1/1.483 10801.2 351.8 0.02 4.459 0.02 1 0.95 5110 lbf . C Ans            ______________________________________________________________________________ 11-17 From the solution to Prob. 3-71, RCz = –150.7 N, RCy = –86.10 N     1/22 2150.7 86.10 173.6 NC DR F         Use the Weibull parameters for Manufacturer 2 on p. 608.   6 15000 1200 60 1080 10 D D R Lx L    Eq. (11-7):    1/ 10 1/ 0 0 1 a D f D b D xC a F x x R                1/3 10 1/1.483 10801.2 173.6 0.02 4.459 0.02 1 0.95 2520 N . C Ans            ______________________________________________________________________________ 11-18 From the solution to Prob. 3-77, RAz = 444 N, RAy = 2384 N  1/22 2444 2384 2425 N 2.425 kNA DR F     Use the Weibull parameters for Manufacturer 2 on p. 608. The design speed is equal to the speed of shaft AD,  125 191 95.5 rev/min 250 F D i C dn n d      6 12000 95.5 60 68.76 10 D D R Lx L    Eq. (11-7):    1/ 10 1/ 0 0 1 a D f D b D xC a F x x R           Chapter 11, Page 6/28 •       1/3 10 1/1.483 68.761 2.425 0.02 4.459 0.02 1 0.95 11.7 kN . C Ans            ______________________________________________________________________________ 11-19 From the solution to Prob. 3-79, RAz = 54.0 lbf, RAy = 140 lbf  1/22 254.0 140 150.1 lbfA DR F    Use the Weibull parameters for Manufacturer 2 on p. 608. The design speed is equal to the speed of shaft AD,  10 280 560 rev/min 5 F D i C dn n d      6 14000 560 60 470.4 10 D D R Lx L    Eq. (11-7):    1/ 10 1/ 0 0 1 a D f D b D xC a F x x R                 3/10 10 1/1.483 470.41 150.1 0.02 4.459 0.02 1 0.98 1320 lbf . C Ans            ______________________________________________________________________________ 11-20 (a) 3 kN, 7 kN, 500 rev/min, 1.2a r DF F n V    From Table 11-2, with a 65 mm bore, C0 = 34.0 kN. Fa / C0 = 3 / 34 = 0.088 From Table 11-1, 0.28  e  3.0.    3 0.357 1.2 7 a r F VF   Since this is greater than e, interpolating Table 11-1 with Fa / C0 = 0.088, we obtain X2 = 0.56 and Y2 = 1.53. Eq. (11-9):       0.56 1.2 7 1.53 3 9.29 kNe i r i aF X VF Y F     Ans. Fe > Fr so use Fe. (b) Use Eq. (11-7) to determine the necessary rated load the bearing should have to carry the equivalent radial load for the desired life and reliability. Use the Weibull parameters for Manufacturer 2 on p. 608. Chapter 11, Page 7/28 •   6 10000 500 60 300 10 D D R Lx L    Eq. (11-7):    1/ 10 1/ 0 0 1 a D f D b D xC a F x x R                 1/3 10 1/1.483 3001 9.29 0.02 4.459 0.02 1 0.95 73.4 kN C            From Table 11-2, the 65 mm bearing is rated for 55.9 kN, which is less than the necessary rating to meet the specifications. This bearing should not be expected to meet the load, life, and reliability goals. Ans. ______________________________________________________________________________ 11-21 (a) 2 kN, 5 kN, 400 rev/min, 1a r DF F n V    From Table 11-2, 30 mm bore, C10 = 19.5 kN, C0 = 10.0 kN Fa / C0 = 2 / 10 = 0.2 From Table 11-1, 0.34  e  0.38.    2 0.4 1 5 a r F VF   Since this is greater than e, interpolating Table 11-1, with Fa / C0 = 0.2, we obtain X2 = 0.56 and Y2 = 1.27. Eq. (11-9):       0.56 1 5 1.27 2 5.34 kNe i r i aF X VF Y F     Ans. Fe > Fr so use Fe. (b) Solve Eq. (11-7) for xD.   1/10 0 0 1 a b D D f D Cx x x R a F                    3 1/1.48319.5 0.02 4.459 0.02 1 0.99 1 5.34D x             10.66Dx   6 60 10 D DD D R nLx L    Chapter 11, Page 8/28 •          6 610 10.66 10 444 h . 60 400 60 D D D x Ans n    ______________________________________________________________________________ 11-22 98 kN, 0.9, 10 revr DF R L   Eq. (11-3): 1/ 1/39 10 6 108 80 10 a D D R LC F L              kN From Table 11-2, select the 85 mm bore. Ans. ______________________________________________________________________________ 11-23 8 kN, 2 kN, 1, 0.99r aF F V R    Use the Weibull parameters for Manufacturer 2 on p. 608.   6 10000 400 60 240 10 D D R Lx L    First guess: Choose from middle of Table 11-1, X = 0.56, Y = 1.63 Eq. (11-9):     0.56 1 8 1.63 2 7.74 kNeF    Fe < Fr, so just use Fr as the design load. Eq. (11-7):    1/ 10 1/1 a D f D b o o D xC a F x x R                 1/3 10 1/1.483 2401 8 82.5 kN 0.02 4.459 0.02 1 0.99 C           From Table 11-2, try 85 mm bore with C10 = 83.2 kN, C0 = 53.0 kN Iterate the previous process: Fa / C0 = 2 / 53 = 0.038 Table 11-1: 0.22  e  0.24   2 0.25 1 8 a r F e VF    Interpolate Table 11-1 with Fa / C0 = 0.038 to obtain X2 = 0.56 and Y2 = 1.89. Eq. (11-9): 0.56(1)8 1.89(2) 8.26 > e rF F   Eq. (11-7):       1/3 10 1/1.483 2401 8.26 85.2 kN 0.02 4.459 0.02 1 0.99 C           Chapter 11, Page 9/28 • Table 11-2: Move up to the 90 mm bore with C10 = 95.6 kN, C0 = 62.0 kN. Iterate again: Fa / C0 = 2 / 62 = 0.032 Table 11-1: Again, 0.22  e  0.24   2 0.25 1 8 a r F e VF    Interpolate Table 11-1 with Fa / C0 = 0.032 to obtain X2 = 0.56 and Y2 = 1.95. Eq. (11-9): 0.56(1)8 1.95(2) 8.38 > e rF F   Eq. (11-7):       1/3 10 1/1.483 2401 8.38 86.4 kN 0.02 4.459 0.02 1 0.99 C           The 90 mm bore is acceptable. Ans. ______________________________________________________________________________ 11-24 88 kN, 3 kN, 1.2, 0.9, 10 revr a DF F V R L     First guess: Choose from middle of Table 11-1, X = 0.56, Y = 1.63 Eq. (11-9):     0.56 1.2 8 1.63 3 10.3 kNeF    e rF F Eq. (11-3): 1/ 1/38 10 6 1010.3 47.8 kN 10 a D e R LC F L              From Table 11-2, try 60 mm with C10 = 47.5 kN, C0 = 28.0 kN Iterate the previous process: Fa / C0 = 3 / 28 = 0.107 Table 11-1: 0.28  e  0.30   3 0.313 1.2 8 a r F e VF    Interpolate Table 11-1 with Fa / C0 = 0.107 to obtain X2 = 0.56 and Y2 = 1.46 Eq. (11-9):     0.56 1.2 8 1.46 3 9.76 kN > e rF F   Eq. (11-3): 1/38 10 6 109.76 45.3 kN 10 C        From Table 11-2, we have converged on the 60 mm bearing. Ans. ______________________________________________________________________________ Chapter 11, Page 10/28 • 11-25 10 kN, 5 kN, 1, 0.95r aF F V R    Use the Weibull parameters for Manufacturer 2 on p. 608.   6 12000 300 60 216 10 D D R Lx L    First guess: Choose from middle of Table 11-1, X = 0.56, Y = 1.63 Eq. (11-9):     0.56 1 10 1.63 5 13.75 kNeF    Fe > Fr, so use Fe as the design load. Eq. (11-7):    1/ 10 1/ 0 0 1 a D f D b D xC a F x x R                 1/3 10 1/1.483 2161 13.75 97.4 kN 0.02 4.459 0.02 1 0.95 C           From Table 11-2, try 95 mm bore with C10 = 108 kN, C0 = 69.5 kN Iterate the previous process: Fa / C0 = 5 / 69.5 = 0.072 Table 11-1: 0.27  e  0.28   5 0.5 1 10 a r F e VF    Interpolate Table 11-1 with Fa / C0 = 0.072 to obtain X2 = 0.56 and Y2 = 1.62  1.63 Since this is where we started, we will converge back to the same bearing. The 95 mm bore meets the requirements. Ans. ______________________________________________________________________________ 11-26 Note to the Instructor. In the first printing of the 9th edition, the design life was incorrectly given to be 109 rev and will be corrected to 108 rev in subsequent printings. We apologize for the inconvenience. 9 kN, 3 kN, 1.2, 0.99r aF F V R    Use the Weibull parameters for Manufacturer 2 on p. 608. 8 6 10 100 10 D D R Lx L    First guess: Choose from middle of Table 11-1, X = 0.56, Y = 1.63 Chapter 11, Page 11/28 • Eq. (11-9):     0.56 1.2 9 1.63 3 10.9 kNeF    Fe > Fr, so use Fe as the design load. Eq. (11-7):    1/ 10 1/ 0 0 1 a D f D b D xC a F x x R                 1/3 10 1/1.483 1001 10.9 83.9 kN 0.02 4.459 0.02 1 0.99 C           From Table 11-2, try 90 mm bore with C10 = 95.6 kN, C0 = 62.0 kN. Try this bearing. Iterate the previous process: Fa / C0 = 3 / 62 = 0.048 Table 11-1: 0.24  e  0.26   3 0.278 1.2 9 a r F e VF    Interpolate Table 11-1 with Fa / C0 = 0.048 to obtain X2 = 0.56 and Y2 = 1.79 Eq. (11-9):     0.56 1.2 9 1.79 3 11.4 kNe rF F    10 11.4 83.9 87.7 kN 10.9 C   From Table 11-2, this converges back to the same bearing. The 90 mm bore meets the requirements. Ans. ______________________________________________________________________________ 11-27 (a) 1200 rev/min, 15 kh, 0.95, 1.2D Dn L R   fa  From Prob. 3-72, RCy = 183.1 lbf, RCz = –861.5 lbf.   1/222183.1 861.5 881 lbfC DR F          6 15000 1200 60 1080 10 D D R Lx L    Eq. (11-7):     1/3 10 1/1.483 10801.2 881 0.02 4.439 1 0.95 C          12800 lbf 12.8 kips .Ans  (b) Results will vary depending on the specific bearing manufacturer selected. A general engineering components search site such as www.globalspec.com might be useful as a starting point. ______________________________________________________________________________ Chapter 11, Page 12/28 • 11-28 (a) 1200 rev/min, 15 kh, 0.95, 1.2D Dn L R   fa  From Prob. 3-72, ROy = –208.5 lbf, ROz = 259.3 lbf.   1/222259.3 208.5 333 lbfC DR F          6 15000 1200 60 1080 10 D D R Lx L    Eq. (11-7):     1/3 10 1/1.483 10801.2 333 0.02 4.439 1 0.95 C          4837 lbf 4.84 kips .Ans  (b) Results will vary depending on the specific bearing manufacturer selected. A general engineering components search site such as www.globalspec.com might be useful as a starting point. ______________________________________________________________________________ 11-29 (a) 900 rev/min, 12 kh, 0.98, 1.2D Dn L R   fa  From Prob. 3-73, RCy = 8.319 kN, RCz = –10.830 kN.   1/2228.319 10.830 13.7 kNC DR F          6 12000 900 60 648 10 D D R Lx L    Eq. (11-7):     1/3 10 1/1.483 6481.2 13.7 204 kN . 0.02 4.439 1 0.98 C A          ns fa  (b) Results will vary depending on the specific bearing manufacturer selected. A general engineering components search site such as www.globalspec.com might be useful as a starting point. ______________________________________________________________________________ 11-30 (a) 900 rev/min, 12 kh, 0.98, 1.2D Dn L R   From Prob. 3-73, ROy = 5083 N, ROz = 494 N.  1/22 25083 494 5106 N 5.1 kNC DR F       6 12000 900 60 648 10 D D R Lx L    Eq. (11-7):     1/3 10 1/1.483 6481.2 5.1 76.1 kN . 0.02 4.439 1 0.98 C A          ns (b) Results will vary depending on the specific bearing manufacturer selected. A general engineering components search site such as www.globalspec.com might be useful as a starting point. ______________________________________________________________________________ Chapter 11, Page 13/28 • 11-31 Assume concentrated forces as shown.  8 28 224 lbfzP    8 35 280 lbfyP    224 2 448 lbf inT    448 1.5 cos 20 0xT F       448 318 lbf 1.5 0.940 F   5.75 11.5 14.25 sin 20 0z yO y AM P R F          5.75 280 11.5 14.25 318 0.342 0yAR   5.24 lbfyAR   5.75 11.5 14.25 cos 20 0y zO z AM P R F           5.75 224 11.5 14.25 318 0.940 0zAR        1/22 2482 lbf; 482 5.24 482 lbfzA AR R          cos 20 0z z zO z AF R P R F        224 482 318 0.940 0zOR     40.9 lbfzOR   sin 20 0y y yO y AF R P R F        280 5.24 318 0.342 0yOR     166 lbfyOR       1/22 240.9 166 171 lbfOR        So the reaction at A governs. Reliability Goal: 0.92 0.96  1.2 482 578 lbfDF      635000 350 60 /10 735Dx       1/3 10 1/1.483 735578 0.02 4.459 0.02 ln 1/ 0.96 6431 lbf 28.6 kN C              From Table 11-2, a 40 mm bore angular contact bearing is sufficient with a rating of Chapter 11, Page 14/28 • 31.9 kN. Ans. ______________________________________________________________________________ 1-32 For a combined reliability goal of 0.95, use 1 0.95 0.975 for the individual bearings.    6 40000 420 60 1008 10D x   The resultant of the given forces are RO = [(–387) + 467 ] = 607 lbf At O: Eq. (11-6): 2 2 1/2 RB = [3162 + (–1615)2]1/2 = 1646 lbf       1/3 10 1/1.483 10081.2 607 0.02 4.459 0.02 ln 1/ 0.975 C            9978 lbf 44.4 kN  From Table 11-2, select an 02-55 mm angular-contact ball bearing with a basic load At B: Eq. (11-6): rating of 46.2 kN. Ans.       3/10 10 1/1.483 10081.2 1646 0.02 4.459 0.02 ln 1/ 0.975 C            20827 lbf 92.7 kN  From Table 11-3, select an 02-75 mm or 03-55 mm cylindrical roller. Ans. _____ _________ 1-33 The reliability of the individual bearings is _ _______________________________________________________________ 1 0.98 0.9899R   Chapter 11, Page 15/28 • From statics, T = (270  50) = (P1  P2)125 = (P1  0.15 P1)125 P1 = 310.6 N, P2 = 0.15 (310.6) = 46.6 N P1 + P2 = 357.2 N 357.2sin 45 252.6 Ny zA AF F    zR 850 300(252.6) 0 89.2 Nz y yO E EM R R      252.6 89.2 0 163.4 Ny y yO OF R R       850 700(320) 300(252.6) 0 174.4 Ny z zO E EM R R        174.4 320 252.6 0 107 Nz zO OF R              2 2 2 2 163.4 107 195 N 89.2 174.4 196 N O E R R          The radial loads are nearly the same at O and E. We can use the same bearing at both locations.   6 60000 1500 60 5400 10D x   Eq. (11-6):     1/3 10 1/1.483 54001 0.196 5.7 kN 0.02 4.439 ln 1/ 0.9899 C           From Table 11-2, select an 02-12 mm deep-groove ball bearing with a basic load rating of 6.89 kN. Ans. ______________________________________________________________________________ 11-34 0.96 0.980R   12(240cos 20 ) 2706 lbf inT    2706 498 lbf 6cos 25 F   In xy-plane: 16(82.1) 30(210) 42 0 z y O CM R      Chapter 11, Page 16/28 • 181 lbfyCR  82.1 210 181 111.1 lbfyOR     In xz-plane: 16(226) 30(451) 42 0y zO CM R     236 lbfzCR   226 451 236 11 lbfzOR      1/22 2111.1 11 112 lbf .OR Ans    1/22 2181 236 297 lbf .CR Ans     6 50000 300 60 900 10D x         1/3 10 1/1.483 9001.2 112 0.02 4.439 ln 1/ 0.980 1860 lbf 8.28 kN O C                   1/3 10 1/1.483 9001.2 297 0.02 4.439 ln 1/ 0.980 4932 lbf 21.9 kN C C             Bearing at O: Choose a deep-groove 02-17 mm. Ans. Bearing at C: Choose a deep-groove 02-35 mm. Ans. ______________________________________________________________________________ 11-35 Shafts subjected to thrust can be constrained by bearings, one of which supports the thrust. The shaft floats within the endplay of the second (roller) bearing. Since the thrust force here is larger than any radial load, the bearing absorbing the thrust (bearing A) is heavily loaded compared to bearing B. Bearing B is thus likely to be oversized and may not contribute measurably to the chance of failure. If this is the case, we may be able to obtain the desired combined reliability with bearing A having a reliability near 0.99 and bearing B having a reliability near 1. This would allow for bearing A to have a lower capacity than if it needed to achieve a reliability of 0.99 . To determine if this is the case, we will start with bearing B. Bearing B (straight roller bearing)   6 30000 500 60 900 10D x    1/22 236 67 76.1 lbf 0.339 kNrF     Try a reliability of 1 to see if it is readily obtainable with the available bearings. Chapter 11, Page 17/28 • Eq. (11-6):     3/10 10 1/1.483 9001.2 0.339 10.1 kN 0.02 4.439 ln 1/1.0 C           The smallest capacity bearing from Table 11-3 has a rated capacity of 16.8 kN. Therefore, we select the 02-25 mm straight cylindrical roller bearing. Ans. Bearing at A (angular-contact ball) With a reliability of 1 for bearing B, we can achieve the combined reliability goal of 0.99 if bearing A has a reliability of 0.99.  1/22 236 212 215 lbf 0.957 kNrF     555 lbf 2.47 kNaF   Trial #1: Tentatively select an 02-85 mm angular-contact with C10 = 90.4 kN and C0 = 63.0 kN. 0 2.47 0.0392 63.0 aF C     6 30000 500 60 900 10D x   Table 11-1: Interpolating, X2 = 0.56, Y2 = 1.88 Eq. (11-9):    0.56 0.957 1.88 2.47 5.18 kNeF    Eq. (11-6):     1/3 10 1/1.483 9001.2 5.18 0.02 4.439 ln 1/ 0.99 C           99.54 kN 90.4 kN  Trial #2: Tentatively select a 02-90 mm angular-contact ball with C10 = 106 kN and C0 = 73.5 kN. 0 2.47 0.0336 73.5 aF C   Table 11-1: Interpolating, X2 = 0.56, Y2 = 1.93    0.56 0.957 1.93 2.47 5.30 kNeF        1/3 10 1/1.483 9001.2 5.30 102 kN < 106 kN O.K. 0.02 4.439 ln 1/ 0.99 C           Chapter 11, Page 18/28 • Select an 02-90 mm angular-contact ball bearing. Ans. ______________________________________________________________________________ 11-36 We have some data. Let’s estimate parameters b and θ from it. In Fig. 11-5, we will use line AB. In this case, B is to the right of A. For F = 18 kN,     61 115 2000 60 13.8 10 x   This establishes point 1 on the R = 0.90 line. The R = 0.20 locus is above and parallel to the R = 0.90 locus. For the two-parameter Weibull distribution, x0 = 0 and points A and B are related by [see Eq. (20-25)]: (1)   1/ln 1/ 0.90 bAx       1/ln 1/ 0.20 bBx      and xB/xA is in the same ratio as 600/115. Eliminating θ,       ln ln 1/ 0.20 / ln 1/ 0.90 1.65 . ln 600 /115 b A     ns Solving for θ in Eq. (1),    1/1.65 1/1.65 1 3.91 . ln 1/ ln 1/ 0.90 A A x Ans R            Chapter 11, Page 19/28 • Therefore, for the data at hand, 1.65 exp 3.91 xR           Check R at point B: xB = (600/115) = 5.217 1.655.217exp 0.20 3.91 R            Note also, for point 2 on the R = 0.20 line,        2log 5.217 log 1 log log 13.8mx    2 72mx  ______________________________________________________________________________ 11-37 This problem is rich in useful variations. Here is one. Decision: Make straight roller bearings identical on a given shaft. Use a reliability goal of (0.99)1/6 = 0.9983. Shaft a  1/22 2239 111 264 lbf 1.175 kNrAF      1/22 2502 1075 1186 lbf 5.28 kNrBF     Thus the bearing at B controls.   6 10000 1200 60 720 10D x     1/1.4830.02 4.439 ln 1/ 0.9983 0.08026      0.3 10 7201.2 5.28 97.2 kN 0.080 26 C        Select either an 02-80 mm with C10 = 106 kN or an 03-55 mm with C10 = 102 kN. Ans. Shaft b  1/22 2874 2274 2436 lbf or 10.84 kNrCF     1/22 2393 657 766 lbf or 3.41 kNrDF    The bearing at C controls. Chapter 11, Page 20/28 •   6 10000 240 60 144 10D x     0.3 10 1441.2 10.84 123 kN 0.080 26 C        Select either an 02-90 mm with C10 = 142 kN or an 03-60 mm with C10 = 123 kN. Ans. Shaft c  1/22 21113 2385 2632 lbf or 11.71 kNrEF     1/22 2417 895 987 lbf or 4.39 kNrFF    The bearing at E controls.   6 10000 80 60 48 10D x     0.3 10 481.2 11.71 95.7 kN 0.080 26 C        Select an 02-80 mm with C10 = 106 kN or an 03-60 mm with C10 = 123 kN. Ans. ______________________________________________________________________________ 11-38 Express Eq. (11-1) as 1 1 10 10 a aF L C L K  For a ball bearing, a = 3 and for an 02-30 mm angular contact bearing, C10 = 20.3 kN.      3 6 920.3 10 8.365 10K   At a load of 18 kN, life L1 is given by:     9 6 1 3 1 8.365 10 1.434 10 rev 18a KL F    For a load of 30 kN, life L2 is:     9 6 2 3 8.365 10 0.310 10 rev 30 L   In this case, Eq. (6-57) – the Palmgren-Miner cycle-ratio summation rule – can be expressed as Chapter 11, Page 21/28 • 1 2 1 2 1l l L L   Substituting,     2 6 6 200 000 1 1.434 10 0.310 10 l    62 0.267 10 rev .l A ns ______________________________________________________________________________ 11-39 Total life in revolutions Let: l = total turns f1 = fraction of turns at F1 f2 = fraction of turns at F2 From the solution of Prob. 11-38, L1 = 1.434(106) rev and L2 = 0.310(106) rev. Palmgren-Miner rule: 1 2 1 2 1 2 1 2 1l l f l f l L L L L     from which 1 1 2 2 1 / / l f L f L        6 6 1 0.40 / 1.434 10 0.60 / 0.310 10 451 585 rev . l Ans          Total life in loading cycles 4 min at 2000 rev/min = 8000 rev/cycle 6 min at 2000 rev/min = 12 000 rev/cycle Total rev/cycle = 8000 + 12 000 = 20 000 451585rev 22.58 cycles . 20000 rev/cycle Ans Chapter 11, Page 22/28 • Total life in hours min 22.58 cycles10 3.76 h . cycle 60 min/h Ans         ______________________________________________________________________________ 11-40 560 lbfrAF  1095 lbfrBF  200 lbfaeF      6 40 000 400 60 10.67 90 10 D D R Lx L    0.90 0.949R   Eq. (11-15):  0.47 5600.47 175.5 lbf 1.5 rA iA A FF K    Eq. (11-15):  0.47 10950.47 343.1 lbf 1.5 rB iB B FF K     ?iA iB aeF F F    175.5 lbf 343.1 200 543.1 lbf, so Eq. (11-16) applies.   We will size bearing B first since its induced load will affect bearing A, but is not itself affected by the induced load from bearing A [see Eq. (11-16)]. From Eq. (11-16b), FeB = FrB = 1095 lbf. Eq. (11-7):     3/10 1/1.5 10.671.4 1095 3607 lbf 4.48 1 0.949 RBF         Ans. Select cone 32305, cup 32305, with 0.9843 in bore, and rated at 3910 lbf with K = 1.95. Ans. With bearing B selected, we re-evaluate the induced load from bearing B using the actual value for K. Eq. (11-15):  0.47 10950.47 263.9 lbf 1.95 rB iB B FF K    Find the equivalent radial load for bearing A from Eq. (11-16), which still applies. Eq. (11-16a):  0.4eA rA A iB aeF F K F F      0.4 560 1.5 263.9 200 920 lbfeAF     Chapter 11, Page 23/28 • eA rAF F Eq. (11-7):     3/10 1/1.5 10.671.4 920 3030 lbf 4.48 1 0.949 RAF         Tentatively select cone M86643, cup M86610, with 1 in bore, and rated at 3250 lbf with K = 1.07. Iterating with the new value for K, we get FeA = 702 lbf and FrA = 2312 lbf. Ans. By using a bearing with a lower K, the rated load decreased significantly, providing a higher than requested reliability. Further examination with different combinations of bearing choices could yield additional acceptable solutions. ______________________________________________________________________________ 11-41 The thrust load on shaft CD is from the axial component of the force transmitted through the bevel gear, and is directed toward bearing C. By observation of Fig. 11-14, direct mounted bearings would allow bearing C to carry the thrust load. Ans. From the solution to Prob. 3-74, the axial thrust load is Fae = 362.8 lbf, and the bearing radial forces are FCx = 287.2 lbf, FCz = 500.9 lbf, FDx = 194.4 lbf, and FDz = 307.1 lbf. Thus, the radial forces are 2 2287.2 500.9 577 lbfrCF    2 2194.4 307.1 363 lbfrDF    The induced loads are Eq. (11-15):  0.47 5770.47 181 lbf 1.5 rC iC C FF K    Eq. (11-15):  0.47 3630.47 114 lbf 1.5 rD iD D FF K    Check the condition on whether to apply Eq. (11-16) or Eq. (11-17), where bearings C and D are substituted, respectively, for labels A and B in the equations. ?iC iD aeF F   F 181 lbf 114 362.8 476.8 lbf, so Eq.(11-16) applies   Eq. (11-16a):  0.4eC rC C iD aeF F K F F       ,0.4 577 1.5 114 362.8 946 lbf so use rC eCF F     Assume for tapered roller bearings that the specifications for Manufacturer 1 on p. 608 are applicable. Chapter 11, Page 24/28 •   8 6 10 1.11 90 10 D D R Lx L    0.90 0.949R   Eq. (11-7):     3/10 1/1.5 1.111 946 1130 lbf . 4.48 1 0.949 RCF Ans         Eq. (11-16b): 363 lbfeD rDF F  Eq. (11-7):     3/10 1/1.5 1.111 363 433 lbf . 4.48 1 0.949 RDF Ans         ______________________________________________________________________________ 11-42 The thrust load on shaft AB is from the axial component of the force transmitted through the bevel gear, and is directed to the right. By observation of Fig. 11-14, indirect mounted bearings would allow bearing A to carry the thrust load. Ans. From the solution to Prob. 3-76, the axial thrust load is Fae = 92.8 lbf, and the bearing radial forces are FAy = 639.4 lbf, FAz = 1513.7 lbf, FBy = 276.6 lbf, and FBz = 705.7 lbf. Thus, the radial forces are 2 2639.4 1513.7 1643 lbfrAF    2 2276.6 705.7 758 lbfrBF    The induced loads are Eq. (11-15):  0.47 16430.47 515 lbf 1.5 rA iA A FF K    Eq. (11-15):  0.47 7580.47 238 lbf 1.5 rB iB B FF K    Check the condition on whether to apply Eq. (11-16) or Eq. (11-17). ?iA iB aeF F   F 515 lbf 238 92.8 330.8 lbf, so Eq.(11-17) applies   Notice that the induced load from bearing A is sufficiently large to cause a net axial force to the left, which must be supported by bearing B. Eq. (11-17a):  0.4eB rB B iA aeF F K F F       ,0.4 758 1.5 515 92.8 937 lbf so use rB eBF F     Assume for tapered roller bearings that the specifications for Manufacturer 1 on p. 608 are applicable. Chapter 11, Page 25/28 •     6 6 500 10 5.56 90 10 D D R Lx L    0.90 0.949R   Eq. (11-7):     3/10 1/1.5 5.561 937 1810 lbf . 4.48 1 0.949 RBF Ans         Eq. (11-16b): 1643 lbfeA rAF F  Eq. (11-7):     3/10 1/1.5 5.561 1643 3180 lbf . 4.48 1 0.949 RAF Ans         ______________________________________________________________________________ 11-43 The lower bearing is compressed by the axial load, so it is designated as bearing A. 25 kNrAF  12 kNrBF  5 kNaeF  Eq. (11-15):  0.47 250.47 7.83 kN 1.5 rA iA A FF K    Eq. (11-15):  0.47 120.47 3.76 kN 1.5 rB iB B FF K    Check the condition on whether to apply Eq. (11-16) or Eq. (11-17) ?iA iB aeF F   F 7.83 kN 3.76 5 8.76 kN, so Eq.(11-16) applies   Eq. (11-16a):  0.4eA rA A iB aeF F K F F       ,0.4 25 1.5 3.76 5 23.1 kN so use rA rAF F          6 60 min 8 hr 5 day 52 weeks250 rev/min 5 yrs hr day week yr 156 10 rev DL                       Assume for tapered roller bearings that the specifications for Manufacturer 1 on p. 608 are applicable. Eq. (11-3):      3/103/10 6 6 156 10 1.2 25 35.4 kN . 90 10 D RA f D R LF a F An L              s Eq. (11-16b): 12 kNeB rBF F  Chapter 11, Page 26/28 • Eq. (11-3):   3/101561.2 12 17.0 kN . 90RB F Ans     ______________________________________________________________________________ 11-44 The left bearing is compressed by the axial load, so it is properly designated as bearing A. 875 lbfrAF  625 lbfrBF  250 lbfaeF  Assume K = 1.5 for each bearing for the first iteration. Obtain the induced loads. Eq. (11-15):  0.47 8750.47 274 lbf 1.5 rA iA A FF K    Eq. (11-15):  0.47 6250.47 196 lbf 1.5 rB iB B FF K    Check the condition on whether to apply Eq. (11-16) or Eq. (11-17). ?iA iB aeF F   F 274 lbf 196 250 lbf, so Eq.(11-16) applies  We will size bearing B first since its induced load will affect bearing A, but it is not affected by the induced load from bearing A [see Eq. (11-16)]. From Eq. (11-16b), FeB = FrB = 625 lbf. Eq. (11-3):       3/103/10 6 90 000 150 60 1 625 90 10 D RB f D R LF a F L             1208 lbf RBF  Select cone 07100, cup 07196, with 1 in bore, and rated at 1570 lbf with K = 1.45. Ans. With bearing B selected, we re-evaluate the induced load from bearing B using the actual value for K. Eq. (11-15):  0.47 6250.47 203 lbf 1.45 rB iB B FF K    Find the equivalent radial load for bearing A from Eq. (11-16), which still applies. Chapter 11, Page 27/28 • Eq. (11-16a):  0.4eA rA A iB aeF F K F F      0.4 875 1.5 203 250 1030 lbf    eA rAF F Eq. (11-3):       3/103/10 6 90 000 150 60 1 1030 90 10 D RA f D R LF a F L             1990 lbf RAF  Any of the bearings with 1-1/8 in bore are more than adequate. Select cone 15590, cup 15520, rated at 2480 lbf with K = 1.69. Iterating with the new value for K, we get FeA = 1120 lbf and FrA = 2160 lbf. The selected bearing is still adequate. Ans. ______________________________________________________________________________ Chapter 11, Page 28/28 • Chapter 12 12-1 Given: dmax = 25 mm, bmin = 25.03 mm, l/d = 1/2, W = 1.2 kN,  = 55 mPas, and N = 1100 rev/min. min maxmin 25.03 25 0.015 mm 2 2 b dc     r  25/2 = 12.5 mm r/c = 12.5/0.015 = 833.3 N = 1100/60 = 18.33 rev/s P = W/ (ld) = 1200/ [12.5(25)] = 3.84 N/mm2 = 3.84 MPa Eq. (12-7):     32 2 6 55 10 18.33 833.3 0.182 3.84 10 r NS c P               Fig. 12-16: h0 /c = 0.3  h0 = 0.3(0.015) = 0.0045 mm Ans. Fig. 12-18: f r/c = 5.4  f = 5.4/833.3 = 0.006 48 T =f Wr = 0.006 48(1200)12.5(103) = 0.0972 Nm Hloss = 2 TN = 2 (0.0972)18.33 = 11.2 W Ans. Fig. 12-19: Q/(rcNl) = 5.1  Q = 5.1(12.5)0.015(18.33)12.5 = 219 mm3/s Fig. 12-20: Qs /Q = 0.81  Qs = 0.81(219) = 177 mm3/s Ans. ______________________________________________________________________________ 12-2 Given: dmax = 32 mm, bmin = 32.05 mm, l = 64 mm, W = 1.75 kN,  = 55 mPas, and N = 900 rev/min. min maxmin 32.05 32 0.025 mm 2 2 b dc     r  32/2 = 16 mm r/c = 16/0.025 = 640 N = 900/60 = 15 rev/s Chapter 12, Page 1/26 Dr aft • P = W/ (ld) = 1750/ [32(64)] = 0.854 MPa l/d = 64/32 = 2 Eq. (12-7):  32 2 55 10 15640 0.797 0.854 r NS c P               Eq. (12-16), Figs. 12-16, 12-19, and 12-21    l/d y  y1  y1/2  y1/4  yl/d h0/c 2 0.98 0.83 0.61 0.36 0.92 P/pmax 2 0.84 0.54 0.45 0.31 0.65 Q/rcNl 2 3.1 3.45 4.2 5.08 3.20 h0 = 0.92 c = 0.92(0.025) = 0.023 mm Ans. pmax = P / 0.065 = 0.854/0.65 = 1.31 MPa Ans. Q = 3.20 rcNl = 3.20(16)0.025(15)64 = 1.23 (103) mm3/s Ans. ______________________________________________________________________________ 12-3 Given: dmax = 3.000 in, bmin = 3.005 in, l = 1.5 in, W = 800 lbf, N = 600 rev/min, and SAE 10 and SAE 40 at 150F. min max min 3.005 3.000 0.0025 in 2 2 3.000 / 2 1.500 in / 1.5 / 3 0.5 / 1.5 / 0.0025 600 600 / 60 10 rev/s 800 177.78 psi 1.5(3) b dc r l d r c N WP ld                 Fig. 12-12: SAE 10 at 150F, 1.75 reynµ µ 2 6 2 1.75(10 )(10)600 0.0354 177.78 r NS c P             Figs. 12-16 and 12-21: h0/c = 0.11 and P/pmax = 0.21 0 max 0.11(0.0025) 0.000 275 in . 177.78 / 0.21 847 psi . h A p Ans     ns Fig. 12-12: SAE 40 at 150F, 4.5 reynµ µ Chapter 12, Page 2/26 Dr aft • 0 max 0 max 4.50.0354 0.0910 1.75 / 0.19, / 0.275 0.19(0.0025) 0.000 475 in . 177.78 / 0.275 646 psi . S h c P p h A p A             ns ns ______________________________________________________________________________ 12-4 Given: dmax = 3.250 in, bmin = 3.256 in, l = 3.25 in, W = 800 lbf, and N = 1000 rev/min. min max min 3.256 3.250 0.003 2 2 3.250 / 2 1.625 in / 3 / 3.250 0.923 / 1.625 / 0.003 542 1000 / 60 16.67 rev/s 800 82.05 psi 3(3.25) b dc r l d r c N WP ld                 Fig. 12-14: SAE 20W at 150F,  = 2.85  reyn 2 6 2 2.85(10 )(16.67)542 0.1701 82.05 r NS c P             From Eq. (12-16), and Figs. 12-16 and 12-21:    l/d y  y1  y1/2  y1/4  yl/d ho/c 0.923 0.85 0.48 0.28 0.15 0.46 P/pmax 0.923 0.83 0.45 0.32 0.22 0.43 max 0.46 0.46(0.003) 0.001 38 in . 82.05 191 psi . 0.43 0.43 oh c A Pp A       ns ns Fig. 12-14: SAE 20W-40 at 150F,  = 4.4  reyn 6 2 4.4(10 )(16.67)542 0.263 82.05 S    From Eq. (12-16), and Figs. 12-16 and 12-21:    l/d y  y1  y1/2  y1/4  yl/d ho/c 0.923 0.91 0.6 0.38 0.2 0.58 P/pmax 0.923 0.83 0.48 0.35 0.24 0.46 Chapter 12, Page 3/26 Dr aft • 0 max 0.58 0.58(0.003) 0.001 74 in . 8205 82.05 178 psi . 0.46 0.46 h c A p A       ns ns ______________________________________________________________________________ 12-5 Given: dmax = 2.000 in, bmin = 2.0024 in, l = 1 in, W = 600 lbf, N = 800 rev/min, and SAE 20 at 130F. min max min 2.0024 2 0.0012 in 2 2 2 1 in, / 1 / 2 0.50 2 2 / 1 / 0.0012 833 800 / 60 13.33 rev/s 600 300 psi 2(1) b dc dr l d r c N WP ld                  Fig. 12-12: SAE 20 at 130F, 3.75 reynµ µ 2 6 2 3.75(10 )(13.3)833 0.115 300 r NS c P             From Figs. 12-16, 12-18 and 12-19: 0 0 / 0.23, / 3.8, / ( ) 5.3 0.23(0.0012) 0.000 276 in . 3.8 0.004 56 833 h c r f c Q rcNl h A f       ns  The power loss due to friction is 3 2 2 (0.004 56)(600)(1)(13.33) 778(12) 778(12) 0.0245 Btu/s . 5.3 5.3(1)(0.0012)(13.33)(1) 0.0848 in / s . f WrNH Ans Q rcNl Ans         ______________________________________________________________________________ 12-6 Given: dmax = 25 mm, bmin = 25.04 mm, l/d = 1, W = 1.25 kN,  = 50 mPas, and N = 1200 rev/min. Chapter 12, Page 4/26 Dr aft • min max min 2 25.04 25 0.02 mm 2 2 / 2 25 / 2 12.5 mm, / 1 / 12.5 / 0.02 625 1200 / 60 20 rev/s 1250 2 MPa 25 b dc r d l d r c N WP ld                 For µ = 50 MPa · s, 2 3 2 6 50(10 )(20)625 0.195 2(10 ) r NS c P             From Figs. 12-16, 12-18 and 12-20: 0 0 / 0.52, / 4.5, / 0.57 0.52(0.02) 0.0104 mm . 4.5 0.0072 625 0.0072(1.25)(12.5) 0.1125 N · m sh c f r c Q Q h A f T f Wr           ns The power loss due to friction is H = 2πT N = 2π (0.1125)(20) = 14.14 W Ans. Qs = 0.57Q The side flow is 57% of Q Ans. ______________________________________________________________________________ 12-7 Given: dmax = 1.25 in, bmin = 1.252 in, l = 2 in, W = 620 lbf,  = 8.5  reyn, and N = 1120 rev/min. min max min 2 6 2 1.252 1.25 0.001 in 2 2 / 2 1.25 / 2 0.625 in / 0.625 / 0.001 625 1120 / 60 18.67 rev/s 620 248 psi 1.25(2) 8.5(10 )(18.67)625 0.250 248 / 2 / 1.25 1.6 b dc r d r c N WP ld r NS c P l d                               From Eq. (12-16), and Figs. 12-16, 12-18, and 12-19 Chapter 12, Page 5/26 Dr aft •    l/d y  y1  y1/2  y1/4  yl/d h0/c 1.6 0.9 0.58 0.36 0.185 0.69 fr/c 1.6 4.5 5.3 6.5 8 4.92 Q/rcNl 1.6 3 3.98 4.97 5.6 3.59 h0 = 0.69 c = 0.69(0.001) =0.000 69 in Ans. f = 4.92/(r/c) = 4.92/625 = 0.007 87 Ans. Q = 1.6 rcNl = 1.6(0.625) 0.001(18.57) 2 = 0.0833 in3/s Ans. ______________________________________________________________________________ 12-8 Given: dmax = 75.00 mm, bmin = 75.10 mm, l = 36 mm, W = 2 kN, N = 720 rev/min, and SAE 20 and SAE 40 at 60C. min max min 75.10 75 0.05 mm 2 2 / 36 / 75 0.48 0.5 (close enough) / 2 75 / 2 37.5 mm / 37.5 / 0.05 750 720 / 60 12 rev/s 2000 0.741 MPa 75(36) b dc l d r d r c N WP ld                   Fig. 12-13: SAE 20 at 60C, µ = 18.5 MPa · s 2 3 2 6 18.5(10 )(12)750 0.169 0.741(10 ) r NS c P             From Figures 12-16, 12-18 and 12-21: 0 m 0 / 0.29, / 5.1, / 0.315 0.29(0.05) 0.0145 mm . 5.1 / 750 0.0068 0.0068(2)(37.5) 0.51 N · m h c f r c P p h An f T f Wr          ax s  The heat loss rate equals the rate of work on the film Hloss = 2πT N = 2π(0.51)(12) = 38.5 W Ans. pmax = 0.741/0.315 = 2.35 MPa Ans. Fig. 12-13: SAE 40 at 60C, µ = 37 MPa · s Chapter 12, Page 6/26 Dr aft • S = 0.169(37)/18.5 = 0.338 From Figures 12-16, 12-18 and 12-21: 0 m 0 loss max / 0.42, / 8.5, / 0.38 0.42(0.05) 0.021 mm . 8.5 / 750 0.0113 0.0113(2)(37.5) 0.85 N · m 2 2 (0.85)(12) 64 W . 0.741 / 0.38 1.95 MPa . h c f r c P p h Ans f T f Wr H TN Ans p A                 ax ns  _____________________________________________________________________________ 12-9 Given: dmax = 56.00 mm, bmin = 56.05 mm, l = 28 mm, W = 2.4 kN, N = 900 rev/min, and SAE 40 at 65C. min max min 56.05 56 0.025 mm 2 2 / 2 56 / 2 28 mm / 28 / 0.025 1120 / 28 / 56 0.5, 900 / 60 15 rev/s 2400 1.53 MPa 28(56) b dc r d r c l d N P                 Fig. 12-13: SAE 40 at 65C, µ = 30 MPa · s 2 3 2 6 30(10 )(15)1120 0.369 1.53(10 ) r NS c P             From Figures 12-16, 12-18, 12-19 and 12-20: 0 0 / 0.44, / 8.5, / 0.71, / ( ) 4.85 0.44(0.025) 0.011 mm . 8.5 / 1000 0.007 59 0.007 59(2.4)(28) 0.51 N · m 2 2 (0.51)(15) 48.1 W . 4.85 4.85(28)(0.0 sh c f r c Q Q Q rcNl h Ans f T f Wr H TN Ans Q rcNl                  3 3 25)(15)(28) 1426 mm /s 0.71(1426) 1012 mm /s .sQ Ans     _____________________________________________________________________________ 12-10 Consider the bearings as specified by minimum f : 0 0, bd t td b    maximum W: 0 0, bd t td b    Chapter 12, Page 7/26 Dr aft • d  and differing only in d and . ig. µ = 1.38(106) reyn /448) = 0.185(106) Preliminaries: 2 / 1 / ( ) 700 / (1.25 ) 448 psi 3600 / 60 60 rev/s l d P W ld N       Fig. 12-16: minimum f : S 0.08 maximum W: 0.20S  F 12-12: µN/P = 1.38(106)(60 Eq. (12-7): / r S c µN  P m For minimu f : 6 0.08 658 0.185(10 ) 0.625 / 658 0.000 950 0.001 in r c c      If this is c min, b  d = 2(0.001) = 0.002 in The median clearance is 0.001 2 2 d b d bt t t t minc c    ran nge for this bearing is   and the clea ce ra 2 dtc bt  which is a function only of the tolerances. For maximum W: 6 0.2 1040 0.185(10 ) 0.625 / 1040 0.000 600 0.0005 in r   c c     If this is cmin Chapter 12, Page 8/26 Dr aft • min min 2 2(0.0005) 0.001 in 0.0005 2 2 2 d b d b d b b d c t t t tc c t tc              The difference (mean) in clearance between the two clearance ranges, crange, is range 0.001 0.00052 2 0.0005 in d b d bt t t tc           For the minimum f bearing b  d = 0.002 in d = b  0.002 in d = b  0.001 in For the same b, tb and td, we need to change the journal diameter by 0.001 in. Increasing d of the minimum friction bearing by 0.001 in, defines of the maximum _____________________________________________________________________________ 2-11 Given: SAE 40, N = 10 rev/s, Ts = 140F, l/d = 1, d = 3.000 in, b = 3.003 in, W = 675 or For the maximum W bearing 0.001 ( 0.002) 0.001 in d d b b       d  load bearing. Thus, the clearance range provides for bearing dimensions which are attainable in manufacturing. Ans. 1 lbf. min max min 3.003 3 0.0015 in 2 2 / 2 3 / 2 1.5 in / 1.5 / 0.0015 1000 675 75 psi 3(3) b dc r d r c WP ld              Trial #1: Fr om Figure 12-12 for T = 160°F, µ = 3.5 µ reyn, 2 6 2 2(160 140) 40 3.5(10 )(10)1000 0.4667 75 T F r NS c P                   From Fig. 12-24, Chapter 12, Page 9/26 Dr aft • 29.70 0.349 109 6.009 40(0.4667) 0.047 467(0.4667) 3.16 753.16 3.16 24.4 F 9.70 9.70 T P PT           Discrepancy = 40  24.4 = 15.6°F Trial #2: T = 150°F, µ = 4.5 µ reyn,  62 2(150 140) 20 4.5 10 10 1000 0.6 75 T F S               From Fig. 12-24, 29.70 0.349 109 6.009 40(0.6) 0.047 467(0.6) 3.97 753.97 3.97 30.7 F 9.70 9.70 T P PT           Discrepancy = 20  30.7 =  10.7°F Trial #3: T = 154°F, µ = 4 µ reyn,  62 2(154 140) 28 4 10 10 1000 0.533 75 T F S               From Fig. 12-24, 29.70 0.349 109 6.009 40(0.533) 0.047 467(0.533) 3.57 753.57 3.57 27.6 F 9.70 9.70 T P PT           Discrepancy = 28  27.6 = 0.4°F O.K. T = 140 +28/2 = 154°F Ans. s 12-16, 12-18, to 12-20: av 2 av ) 140 / 2 154 (28 / 2) 168 0.4 F T T T F S           1 av / 2 154 (28 / 2T T T     From Figure Chapter 12, Page 10/26 Dr aft •         0 0 loss 0.75, 11, 3.6, 0.33 0.75(0.0015) 0.00113 in . 11 0.011 1000 0.0075(3)(40) 0.9 N · m 2 0.011 675 1.5 102 0.075 Btu/s . 778 12 778 12 3.6 3. sh f r Q Q c c rcN l Q h Ans f T f Wr f WrNH A Q rcN l                  3 3 6(1.5)0.0015(10)3 0.243 in /s . 0.33(0.243) 0.0802 in /s .s Ans Q Ans    ns _____________________________________________________________________________ 2-12 Given: d = 2.5 in, b = 2.504 in, cmin = 0.002 in, W = 1200 lbf, SAE = 20, Ts = 110°F, P = W/(ld) = 1200/(2.5) = 192 psi, N = 1120/60 = 18.67 rev/s For a trial film temperature, let Tf = 150°F Table 12-1:  = 0.0136 exp[1271.6/(150 + 95)] = 2.441  reyn Eq. (12-7): 1 N = 1120 rev/min, and l = 2.5 in. 2  62 2 2.441 10 18.672.5 / 2 0.927 0.002 192 r NS c P                Fig. 12-24:    2192 0.349 109 6.009 40 0.0927 0.047 467 0.09279.70 17.9 F T        av av 17.9110 119.0 F 2 2 150 119.0 31.0 F s f TT T T T             which is not 0.1 or less, therefore try averaging for the new trial film temperature, let new 150 119.0( ) 134.5 F 2f T    ing a spreadsheet (table also shows the first trial) Proceed with additional trials us Chapter 12, Page 11/26 Dr aft • Trial Tf ' S T Tav Tf Tav New Tf 150.0 2.441 0.0927 17.9 119.0 31.0 134.5 134.5 3.466 0.1317 22.6 121.3 13.2 127.9 127.9 4.084 0.1551 25.4 122.7 5.2 125.3 125.3 4.369 0.1659 26.7 123.3 2.0 124.3 124.3 4.485 0.1704 27.2 123.6 0.7 124.0 124.0 4.521 0.1717 27.4 123.7 0.3 123.8 123.8 4.545 0.1726 27.5 123.7 0.1 123.8 Note that the convergence begins rapidly. There are ways to speed this, but at this point they would only add complexity. (a) 64.545(10 ), 0.1726µ S   From Fig. 12-16: 0 00.482, 0.482(0.002) 0.000 964 in h h c    From Fig. 12-17:  = 56° Ans. (b) e = c  h0 = 0.002  0.000 964 = 0.001 04 in Ans. (c) From Fig. 12-18: 4.10, 4.10(0.002 /1.25) 0.006 56 .f r f Ans c    (d) T = f Wr = 0.006 56(1200)(1.25) = 9.84 lbf · in 2 2 (9.84)(1120 / 60) 0.124 Btu/s . 778(12) 778(12) T NH Ans    (e) From Fig. 12-19: 4.16Q rcNl  311204.16(1.25)(0.002) (2.5) 0.485 in /s . 60 Q A      ns From Fig. 12-20: 30.6, 0.6(0.485) 0.291 in /s .s s Q Q A Q    ns (f) From Fig. 12-21:   2 max max / 1200 / 2.50.45, 427 psi . 0.45 0.45 W ldP p Ans p     From Fig. 12-22: max 16 .p Ans   Chapter 12, Page 12/26 Dr aft • (g) From Fig. 12-22: 0 82 .p Ans   (h) From the trial tabl Ans. e, Tf = 123.8°F T = 110 + 27.5 = 137.5°F Ans. _____ 2-13 Given: d = 1.250 in, td = 0.001 in, b = 1.252 in, tb = 0.003 in, l = 1.25 in, W = 250 lbf, P = W/(ld) = 250/1.25 = 160 psi, N = 1750/60 = 29.17 rev/s For the clearance, c = 0.002  0.001 in. Thus, cmin = 0.001 in, cmedian = 0.002 in, and For cmin = 0.001 in, start with a trial film temperature of Tf = 135°F Table 12-1:  = 0.0158 exp[1157.5/(135 + 95)] = 2.423  reyn Eq. (12-7): (i) With T = 27.5°F from the trial table, Ts + ________________________________________________________________________ 1 N = 1750 rev/min, SAE 10 lubricant, sump temperature Ts = 120°F. 2 cmax = 0.003 in.  62 2 2.423 10 29.171.25 / 2 0.1725 0.001 160 r NS c P                Fig. 12-24:     2160 0.349 109 6.009 40 0.1725 0.047 467 0.1725 9.70 22.9 F T        av av 22.9120 131.4 F 2 2 135 131.4 3.6 F s f TT T T T             which is not 0.1 or less, therefore try averaging for the new trial film temperature, let new 135 131.4( ) 133.2 F 2f T    h additional trials using a spreadsheet (table also shows the first trial) Trial ' S T Tav TfTav New Proceed wit Tf Tf 1 2. 0.1 5 1 135.0 423 72 22.9 31.4 3.6 33.2 133.2 2.521 0.1795 23.6 131.8 1.4 132.5 132.5 2.560 0.1823 23.9 131.9 0.6 132.2 132.2 2.578 0.1836 24.0 132.0 0.2 132.1 132.1 2.583 0.1840 24.0 132.0 0.1 132.1 Chapter 12, Page 13/26 Dr aft • With Tf = 132.1°F, T = 24.0°F,  = 2.583  reyn, S = 0.1840, Tmax = Ts + T = 120 + 24.0 = 144.0°F Fig. 12-16: h0/c = 0.50, h0 = 0.50(0.001) = 0.000 50 in  = 1  h0/c = 1  0.50 = 0.05 in Fig. 12-18: r f /c = 4.25, f = 4.25/(0.625/0.001) = 0.006 8 Fig. 12-19: Q/(rcNl) = 4.13, Q = 4.13(0.625)0.001(29.17)1.25 = 0.0941 in3/s Fig. 12-20: Qs/Q = 0.58, Qs = 0.58(0.0941) = 0.0546 in /s The above can be repeated for cmedian = 0.002 in, and cmax = 0.003 in. The results are cmin 0.001 cmedian 0.002 in cmax 0.003 3 shown below. in in T 132.1 125.6 124.1  1 0.00050 0.00069 0.00038 f 0.0068 0.0058 0.0059 Q/( ) 0.0941 0.207 0.321 Q 0.0546 0.170 f  2.583 3.002 3.112 S 0.184 0.0534 0.0246  24.0 11.1 8.2 Tmax 144.0 131.1 28.2 h0/c 0.5 0.23 0.125 h0  0.50 0.77 0.88 r/c 4.25 1.8 1.22 f rcNl 4.13 4.55 4.7 Q s /Q 0.58 0.82 0.90 Qs 0.289 ____________________________________________________________________________ 2-14 Computer programs will vary. ______________________________________________ 2-15 Note to the Instructor: In the first printing of the 9th edition, the l/d ratio and the ill be _ 1 _______________________________ 1 lubrication constant  were omitted. The values to use are l/d = 1, and  = 1. This w updated in the next printing. We apologize for any inconvenience this may have caused. Chapter 12, Page 14/26 Dr aft • ring nowledge the environmental temperature’s role in establishing the sump Given: dmax = 2.500 in, bmin = 2.504 in, l/d = 1, N = 1120 rev/min, SAE 20 lubricant, W = 600 lbf load with minimal clearance: We will start by using W = 600 lbf (nd = 2). The lo In a step-by-step fashion, we are building a skill for natural circulation bearings. • Given the average film temperature, establish the bearing properties. • Given a sump temperature, find the average film temperature, then establish the bea properties. • Now we ack temperature. Sec. 12-9 and Ex. 12-5 address this problem. 300 lbf, A = 60 in2, T = 70F, and  = 1. task is to iteratively find the average film temperature, Tf , which makes Hgen and H ss equal. min maxmin 2.504 2.500 0.002 in 2 2 b dc     N = 1120/60 = 18.67 rev/s 2 600 96 psi 2.5 WP ld     62 2 10 18.671.25 0.0760 0.002 96 r NS c P                 Table 12-1:  = 0.0136 exp[1271.6/(Tf + 95)]    gen 2545 2545 600 18.67 0.002 1050 1050 54.3 f r fH WNc c c f r c        r         CR loss 2.7 60 / 144 70 1 1 1 0.5625 70 f f f AH T T T T            Start with trial values of Tf of 220 and 240F. Trial Tf  S f r/c Hgen Hloss 220 0.770 0. 9 05 1.9 103.2 84.4 240 0.605 0.046 1.7 92.3 95.6 As a linear approximation, let Hgen = mTf + b. Substituting the two sets of values of f T and Hgen we find that Hgen =  0.545 Tf +223.1. Setting this equal to Hloss and solving for Tf gives Tf = 237F. Chapter 12, Page 15/26 Dr aft • Tr  S ial Tf f r/c Hgen Hloss 237 0.627 0. 8 04 1.73 93.9 94.0 which is satisfactory. Table 12-16: h0/c = 0.21, h0 = 0.21 (0.002) = 000 42 in Fig. 12-24:    296 0.349 109 6.009 4 0.048 0.047 467 0.048 9.7 6.31 F T         T1 = Ts = Tf  T = 237  6.31/2 = 233.8F Tmax = T1 + T = 233.8 + 6.31 = 240.1F Trumpler’s design criteria: 0.002 + 0.000 04d = 0.002 + 0.000 04(2.5) = 0.000 30 in < h0 O.K. Tmax = 240.1F < 250F O.K. 2 300 48 psi 300 psi . . 2.5 stW O K ld    nd = 2 (assessed at W = 600 lbf) O.K. We see that the design passes Trumpler’s criteria and is deemed acceptable. For an operating load of W = 300 lbf, it can be shown that Tf = 219.3F,  = 0.78, S = _____________________________________________________________________________ 2-16 Given: , SAE 30, Ts = 120F, ps = 50 psi, 0/60 = 33.33 rev/s, W = 4600 lbf, b 0.250 in, 0.118, f r/c = 3.09, Hgen = Hloss = 84 Btu/h, h0 = , T = 10.5F, T1 = 224.6F, and Tmax = 235.1F. 1 0.000 0.0050.001 0.0003.500 in, 3.505 ind b      N = 200 earing length = 2 in, groove width = and Hloss  5000 Btu/hr. minbc  maxmin 3.505 3.500 0.0025 in 2 2 d    r = d/ 2 = 3.500/2 = 1.750 in r / c = 1.750/0.0025 = 700 l = (2  0.25)/2 = 0.875 in Chapter 12, Page 16/26 Dr aft • l / d = 0.875/3.500 = 0.25   4600WP   751 psi 4 4 1.750 0.875rl   Trial #1: Choose (Tf )1 = 150°F. From Table 12-1,  = 0.0141 exp[1360.0/(150 + 95)] = 3.63 µ reyn 2 6 2 3.63(10 )(33.33)700 0.0789 751 r NS c P             From Figs. 12-16 and 12-18:  = 0.9, f r/ c = 3.6 From Eq. (12-24),         2 2 4 2 2 4 0.012T  3( / ) 1 1.5 0.0123 3.6 0.0789 4600 71.2 F 1 1.5(0.9) 50 1.750 s f r c SW p r        Tav = Ts + T / 2 = 120 + 71.2/2 = 155.6F Trial #2: Choose (Tf )2 = 160°F. From Table 12-1  = 0.0141 exp[1360.0/(160 + 95)] = 2.92 µ reyn 2.920.0789 0.0635 3.63 S       From Figs. 12-16 and 12-18:  = 0.915, f r/ c =3         20.0123 3 0.0635 4600 2 4 46.9 F 1 1.5 0.915 50 1.750 T       Tav = 120 + 46.9/2 = 143.5F Chapter 12, Page 17/26 Dr aft • Trial #3: Thus, the plot gives (Tf )3 = 152.5°F. From Table 12-1  = 0.0141 exp[1360.0/(152.5 + 95)] = 3.43 µ reyn 3.430.0789 0.0746   3.63 S     gs. 12-16 and 12-18:  = 0.905, f r/ c =3.4 From Fi         2 2 4 0.0123 3.4 0.0746 4600 63.2 F 1 1.5 0.905 50 1.750 T       Tav = 120 + 63.2/2 = 151.6F 152.5 151.6 152.1 F Try 152 F 2f T    Result is close. Choose  Table 12-1:  = 0.0141 exp[1360.0/(152 + 95)] = 3.47 µ reyn         0 2 2 4 av 3.470.0789 0.0754S    3.63 3.4, 0.902, 0.098 0.0123 3.4 0.0754 4600 64.1 F 1 1.5 0.902 50 1.750 120 64.1 / 2 152.1 F O.K. f r h c c T T                    h0 = 0.098(0.0025) = 0.000 245 in Tmax = Ts + T = 120 + 64.1 = 184.1F Eq. (12-22):           6 3 1 1.5 1 1.5 0.902 3 3 3.47 10 0.875 1.047 in /s sQ l  33 2 250 1.750 0.0025sp rc          Hloss =  CpQs T = 0.0311(0.42)1.047(64.1) = 0.877 Btu/s 0.0002 + 0.000 04(3.5) = 0.000 34 in > 0.000 245 Not O.K. . __ __ ____________________________________ = 0.877(602) = 3160 Btu/h O.K. Trumpler’s design criteria: Tmax = 184.1°F < 250°F O.K. Pst = 751 psi > 300 psi Not O.K n = 1, as done Not O.K. ____ __________ _______________________ Chapter 12, Page 18/26 Dr aft • 12-17 Given: 0.00 0.0100.05 0.00050.00 mm, 50.084 mmd b      , SAE 30, Ts = 55C, ps = 200 kPa, N = 288 gth = 55 mm, groove width = 5 mm, 0/60 = 48 rev/s, W = 10 kN, bearing len and Hloss  300 W. min maxmin 50.084 50 0.042 mm 2 2 b dc     r = d/ 2 = 50/2 = 25 mm r / c = 25/0.042 = 595 l = (55  5)/2 = 25 mm l / d = 25/50 = 0.5     310 10W 4 MPa 4 4 25 25 P rl     Trial #1: Choose (Tf )1 = 79°C. From Fig. 12-13, µ = 13 MPa · s. 2 3 2 6 13(10 )(48)595 0.0552 4(10 ) r NS c P             From Figs. 12-16 and 12-18:  = 0.85, f r/ c = 2.3 From Eq. (12-25), 6 2 2 4 6 2 2 4 978(1T  0 ) ( / ) 1 1.5 978(10 ) 2.3(0.0552)(10 ) 76.3 C 1 1.5(0.85) 200(25) s f r c SW p r          Tav = Ts + T / 2 = 55 + 76.3/2 = 93.2C Trial #2: Choose (Tf )2 = 100°C. From Fig. 12-13, µ = 7 MPa · s. 70.0552 0.0297 13 S       From Figs. 12-16 and 12-18:  = 0.90, f r/ c =1.6 6 2978(10 ) 1.6(0.0297)(10 )  2 4 26.9 C1 1.5(0.9) 200(25) T       Tav = 55 + 26.9/2 = 68.5C Chapter 12, Page 19/26 Dr aft • Trial #3: Thus, the plot gives (Tf )3 = 85.5°C. From Fig. 12-13, µ = 10.5 MPa · s. 10.50.0552 0.0446 13 S       From Figs. 12-16 and 12-18:  = 0.87, f r/ c =2.2 6 2 2 4 978(10 ) 2.2(0.0457)(10 ) 58.9 C 1 1.5(0.87 ) 200(25) T          Tav = 55 + 58.9/2 = 84.5C Result is close. Choose 85.5 84.5 85 C 2f T    Fig. 12-13: µ = 10.5 MPa · s 0 6 2 2 4 av 10.50.0552 0.0446 13 0.87, 2.2, 0.13 978(10 ) 2.2(0.0457)(10 ) 58.9 C or 138 F 1 1.5(0.87 ) 200(25 ) 55 58.9 / 2 84.5 C O.K. S f r h c c T T                         From Eq. (12-22) h0 = 0.13(0.042) = 0.005 46 mm or 0.000 215 in Tmax = Ts + T = 55 + 58.9 = 113.9C or 237°F            33 2 2 6 3 3 3 200 25 0.042 (1 1.5 ) 1 1.5 0.87 3 3 10.5 10 25 3156 mm /s 3156 25.4 0.193 in /s s s p rcQ µl                     Hloss =  CpQs T = 0.0311(0.42)0.193(138) = 0.348 Btu/s = 1.05(0.348) = 0.365 kW = 365 W not O.K. Chapter 12, Page 20/26 Dr aft • Trumpler’s design criteria: 0.0002 + 0.000 04(50/25.4) = 0.000 279 in > h0 Not O.K. Tmax = 237°F O.K. Pst = 4000 kPa or 581 psi > 300 psi Not O.K. n = 1, as done Not O.K. _____________________________________________________________________________ 12-18 So far, we’ve performed elements of the design task. Now let’s do it more completely. The values of the unilateral tolerances, tb and td , reflect the routine capabilities of the bushing vendor and the in-house capabilities. While the designer has to live with these, his approach should not depend on them. They can be incorporated later. First we shall find the minimum size of the journal which satisfies Trumpler’s constraint of Pst ≤ 300 psi. 2 min 300 2 300 2 / 600( / ) 900 1.73 in 2(300)(0.5) st WP dl W Wd d l d l d d           In this problem we will take journal diameter as the nominal value and the bushing bore as a variable. In the next problem, we will take the bushing bore as nominal and the journal diameter as free. To determine where the constraints are, we will set tb = td = 0, and thereby shrink the design window to a point. We set d = 2.000 in b = d + 2cmin = d + 2c nd = 2 (This makes Trumpler’s nd ≤ 2 tight) and construct a table. Chapter 12, Page 21/26 Dr aft • c b d *fT Tmax ho Pst Tmax n fom 0.0010 2.0020 2 215.50 312.0     -5.74 0.0011 2.0022 2 206.75 293.0     -6.06 0.0012 2.0024 2 198.50 277.0     -6.37 0.0013 2.0026 2 191.40 262.8     -6.66 0.0014 2.0028 2 185.23 250.4     -6.94 0.0015 2.0030 2 179.80 239.6     -7.20 0.0016 2.0032 2 175.00 230.1     -7.45 0.0017 2.0034 2 171.13 220.3     -7.65 0.0018 2.0036 2 166.92 213.9     -7.91 0.0019 2.0038 2 163.50 206.9     -8.12 0.0020 2.0040 2 160.40 200.6     -8.32 *Sample calculation for the first entry of this column. Iteration yields: 215.5 FfT   With 215.5 FfT   , from Table 12-1 6 6 2 6 0.0136(10 )exp[1271.6 / (215.5 95)] 0.817(10 ) reyn 9003000 / 60 50 rev/s, 225 psi 4 1 0.817(10 )(50) 0.182 0.001 225 µ N P S                     From Figs. 12-16 and 12-18: e = 0.7, f r/c = 5.5 Eq. (12–24): 2 2 4 av 0.0123(5.5)(0.182)(900 ) 191.6 F [1 1.5(0.7 )](30)(1 ) 191.6 F120 F 215.8 F 215.5 F 2 FT T             For the nominal 2-in bearing, the various clearances show that we have been in contact with the recurving of (ho)min. The figure of merit (the parasitic friction torque plus the pumping torque negated) is best at c = 0.0018 in. For the nominal 2-in bearing, we will place the top of the design window at cmin = 0.002 in, and b = d + 2(0.002) = 2.004 in. At this point, add the b and d unilateral tolerances: 0.000 0.003 0.001 0.0002.000 in, 2.004 ind b      Now we can check the performance at cmin , c , and cmax . Of immediate interest is the fom of the median clearance assembly,  9.82, as compared to any other satisfactory bearing ensemble. Chapter 12, Page 22/26 Dr aft • If a nominal 1.875 in bearing is possible, construct another table with tb = 0 and td = 0. c b d fT Tmax ho Pst Tmax n fom 0.0020 1.879 1.875 157.2 194.30      7.36 0.0030 1.881 1.875 138.6 157.10      8.64 0.0035 1.882 1.875 133.5 147.10      9.05 0.0040 1.883 1.875 130.0 140.10      9.32 0.0050 1.885 1.875 125.7 131.45      9.59 0.0055 1.886 1.875 124.4 128.80      9.63 0.0060 1.887 1.875 123.4 126.80      9.64 The range of clearance is 0.0030 < c < 0.0055 in. That is enough room to fit in our design window. 0.000 0.003 0.001 0.0001.875 in, 1.881 ind b      The ensemble median assembly has a fom =  9.31. We just had room to fit in a design window based upon the (h0)min constraint. Further reduction in nominal diameter will preclude any smaller bearings. A table constructed for a d = 1.750 in journal will prove this. We choose the nominal 1.875-in bearing ensemble because it has the largest figure of merit. Ans. _____________________________________________________________________________ 12-19 This is the same as Prob. 12-18 but uses design variables of nominal bushing bore b and radial clearance c. The approach is similar to that of Prob. 12-18 and the tables will change slightly. In the table for a nominal b = 1.875 in, note that at c = 0.003 in the constraints are “loose.” Set b = 1.875 in d = 1.875  2(0.003) = 1.869 in For the ensemble 0.003 0.000 0.001 0.0011.875 in, 1.869 inb d      Analyze at cmin = 0.003, c = 0.004 in and cmax = 0.005 in At min loss0.003 in: 138.4, 3.160, 0.0297, 1035 Btu/hfc T µ S H     and the Trumpler conditions are met. At 0.004 in: 130 F,fc T    = 3.872, S = 0.0205, Hloss = 1106 Btu/h, fom = 9.246 Chapter 12, Page 23/26 Dr aft • and the Trumpler conditions are O.K. At max 0.005 in: 125.68 F,fc T    = 4.325, S = 0.014 66, Hloss = 1129 Btu/h and the Trumpler conditions are O.K. The ensemble figure of merit is slightly better; this bearing is slightly smaller. The lubricant cooler has sufficient capacity. _____________________________________________________________________________ 12-20 Table 12-1:  ( reyn) =  0 (106) exp [b / (T + 95)] b and T in F The conversion from  reyn to mPas is given on p. 620. For a temperature of C degrees Celsius, T = 1.8 C + 32. Substituting into the above equation gives  (mPas) = 6.89  0 (106) exp [b / (1.8 C + 32+ 95)] = 6.89  0 (106) exp [b / (1.8 C + 127)] Ans. For SAE 50 oil at 70C, from Table 12-1,  0 = 0.0170 (106) reyn, and b = 1509.6F. From the equation,  = 6.89(0.0170) 106(106) exp {1509.6/[1.8(70) + 127]} = 45.7 mPas Ans. From Fig. 12-13,  = 39 mPas Ans. The figure gives a value of about 15 % lower than the equation. _____________________________________________________________________________ 12-21 Originally 0.000 0.003 0.001 0.0002.000 in, 2.005 ind b      Doubled, 0.000 0.006 0.002 0.0004.000 in, 4.010 ind b      The radial load quadrupled to 3600 lbf when the analyses for parts (a) and (b) were carried out. Some of the results are: Part c  S Tf f r/c Qs h0 /c e H loss h0 Trumpler h0 f (a) 0.007 3.416 0.0310 135.1 0.1612 6.56 0.1032 0.897 9898 0.000 722 0.000 360 0.005 67 (b) 0.0035 3.416 0.0310 135.1 0.1612 0.870 0.1032 0.897 1237 0.000 361 0.000 280 0.005 67 The side flow Qs differs because there is a c3 term and consequently an 8-fold increase. Hloss is related by a 9898/1237 or an 8-fold increase. The existing h0 is related by a 2-fold increase. Trumpler’s (h0)min is related by a 1.286-fold increase. Chapter 12, Page 24/26 Dr aft • _____________________________________________________________________________ 12-22 Given: Oiles SP 500 alloy brass bushing, L = 0.75 in, D = 0.75 in, T = 70F, F = 400 lbf, N = 250 rev/min, and w = 0.004 in. Table 12-8: K = 0.6(1010) in3min/(lbffth) P = F/ (DL) = 400/ [0.75(0.75)] = 711 psi V = DN/ 12 =  (0.75)250/12 = 49.1 ft/min Tables 12-10 and 12-11: f 1 = 1.8, f 2 = 1.0 Table 12-12: PVmax = 46 700 psift/min, Pmax = 3560 psi, Vmax = 100 ft/min max 2 4 4 400 905 psi 3560 psi . . 0.75 FP O DL K       PV = 711 (49.1) = 34 910 psift/min < 46 700 psift/min O.K. Eq. (12-32) can be written as 1 2 4 Ff f K Vt DL w Solving for t,             10 1 2 0.75 0.75 0.004 4 4 1.8 1.0 0.6 10 49.1 400 833.1 h 833.1 60 49 900 min DLt f f KVF       w Cycles = Nt = 250 (49 900) = 12.5 (106) cycles Ans. _____________________________________________________________________________ 12-23 Given: Oiles SP 500 alloy brass bushing, wmax = 0.002 in for 1000 h, N = 400 rev/min, F = 100 lbf, CR = 2.7 Btu/ (hft2F), Tmax = 300F, f s = 0.03, and nd = 2. Estimate bushing length with f1 = f2 = 1, and K = 0.6(10-10) in3 · min/(lbf · ft · h) Using Eq. (12-32) with ndF for F, 10 1 2 1(1)(0.6)(10 )(2)(100)(400)(1000) 0.80 in 3 3(0.002) df f Kn FNtL     w From Eq. (12-38), with fs = 0.03 from Table 12-9 applying nd = 2 to F Chapter 12, Page 25/26 Dr aft • and 2CR 2.7 Btu/(h · ft · °F)   720 720(0.03)(2)(100)(400) 3.58 in 778(2.7)(300 70) 0.80 3.58 in s d CR f f n FNL J T T L         Trial 1: Let L = 1 in, D = 1 in max 4 4(2)(100) 255 psi 3560 psi . . (1)(1) 2(100) 200 psi 1(1) (1)(400) 104.7 ft/min 100 ft/min . . 12 12 d d n FP O DL n FP DL DNV N                K ot O K Trial 2: Try D = 7/8 in = 0.875 in, L = 1 in max 4(2)(100) 291 psi 3560 psi . . (0.875)(1) 2(100) 229 psi 0.875(1) (0.875)(400) 91.6 ft/min 100 ft/min . . 12 P O P V O           K K PV = 229(91.6) = 20 976 psi · ft/min < 46 700 psi · ft/min O.K. V f1 33 1.3 91.6 f1 100 1.8   1 new 1 91.6 331.3 (1.8 1.3) 1.74 100 33 1.74 0.80 1.39 inold f L f L            Trial 3: Try D = 7/8 in = 0.875 in, L = 1.5 in max 4(2)(100) 194 psi 3560 psi . . (0.875)(1.5) 2(100) 152 psi, 91.6 ft/min 0.875(1.5) 152(91.6) 13 923 psi · ft/min 46 700 psi · ft/min . . 7 / 8 in, 1.5 in is acceptable . P O P V PV O K D L Ans             K Chapter 12, Page 26/26 Dr aft • Chapter 12, Page 27/26 Suggestion: Try smaller sizes. Dr aft • Chapter 13 13-1 17 / 8 2.125 inPd    2 3 1120 2.125 4.375 in 544G P Nd d N     8 4.375 35 teeth .G GN Pd An   s ns ns s  2.125 4.375 / 2 3.25 in .C A   ______________________________________________________________________________ 13-2  1600 15 / 60 400 rev/min .Gn A  3 mm .p m An    3 15 60 2 112.5 mm .C A     ns ns ______________________________________________________________________________ 13-3  16 4 64 teeth .GN A   64 6 384 mm .G Gd N m An   s  16 6 96 mm .P Pd N m An   s ns s ns s  384 96 / 2 240 mm .C A   ______________________________________________________________________________ 13-4 Mesh: 1/ 1/ 3 0.3333 in .a P An   1.25 / 1.25 / 3 0.4167 in .b P A   0.0834 in .c b a Ans   / / 3 1.047 in .p P An    / 2 1.047 / 2 0.523 in .t p Ans   Pinion Base-Circle: 1 1 / 21/ 3 7 id N P n   1 7 cos 20 6.578 in .bd A   ns Gear Base-Circle: 2 2 / 28 / 3 9.333 ind N P   2 9.333cos 20 8.770 in .bd A   ns Base pitch:  cos / 3 cos 20 0.984 in .b cp p A    ns Contact Ratio: / 1.53 / 0.984 1.55 .c ab bm L p Ans   See the following figure for a drawing of the gears and the arc lengths. Chapter 13, Page 1/35 • ______________________________________________________________________________ 13-5 (a) 1/22 2 0 14 / 6 32 / 6 2.910 in . 2 2 A Ans                  (b)  1tan 14 / 32 23.63 .Ans     1tan 32 /14 66.37 .Ans    (c) Ans. 14 / 6 2.333 inPd   32 / 6 5.333 in .Gd A  ns Chapter 13, Page 2/35 • (d) From Table 13-3, 0.3A0 = 0.3(2.910) = 0.873 in and 10/P = 10/6 = 1.67 0.873 < 1.67 0.873 in .F Ans  ______________________________________________________________________________ 13-6 (a) / / 4 0.7854 inn np P     / cos 0.7854 / cos30 0.9069 int np p     / tan 0.9069 / tan 30 1.571 inx tp p     (b) Eq. (13-7): cos 0.7854cos 25 0.7380 in .nb n np p A    ns (c) cos 4cos30 3.464 teeth/int np P      1 1tan tan / cos tan (tan 25 / cos30 ) 28.3 .t n Ans        (d) Table 13-4: 1/ 4 0.250 in .a A  ns ns 1.25 / 4 0.3125 in .b A  20 5.774 in . 4cos30P d A  ns 36 10.39 in . 4cos30G d A  ns ______________________________________________________________________________ 13-7 19 teeth, 57 teeth, 20 , 2.5 mmP G n nN N m     (a)  2.5 7.854 mm .n np m A    ns 7.854 9.069 mm . cos cos30 n t pp Ans     9.069 15.71 mm . tan tan 30 t x pp Ans     (b) 2.5 2.887 mm . cos cos30 n t mm A     ns Chapter 13, Page 3/35 • 1 tan 20tan 22.80 . cos30t Ans            (c) 2.5mm .na m Ans   1.25 1.25 2.5 3.125 mm .nb m A   ns  19 2.887 =54.85 mm .P t t Nd Nm P    Ans  57 2.887 164.6 mm .Gd A  ns ______________________________________________________________________________ 13-8 (a) Using Eq. (13-11) with k = 1, = 20º, and m = 2,                     2 2 2 2 2 2 2 1 2 sin 1 2 sin 2 1 2 2 1 2 2 sin 20 14.16 teeth 1 2 2 sin 20 P kN m m m m                   Round up for the minimum integer number of teeth. NP = 15 teeth Ans. (b) Repeating (a) with m = 3, NP = 14.98 teeth. Rounding up, NP = 15 teeth. Ans. (c) Repeating (a) with m = 4, NP = 15.44 teeth. Rounding up, NP = 16 teeth. Ans. (d) Repeating (a) with m = 5, NP = 15.74 teeth. Rounding up, NP = 16 teeth. Ans. Alternatively, a useful table can be generated to determine the largest gear that can mesh with a specified pinion, and thus also the maximum gear ratio with a specified pinion. The Max NG column was generated using Eq. (13-12) with k = 1, = 20º, and rounding up to the next integer. Min NP Max NG Max m = Max NG / Min NP 13 16 1.23 14 26 1.86 15 45 3.00 16 101 6.31 17 1309 77.00 18 unlimited unlimited With this table, we can readily see that gear ratios up to 3 can be obtained with a minimum NP of 15 teeth, and gear ratios up to 6.31 can be obtained with a minimum NP of 16 teeth. This is consistent with the results previously obtained. ______________________________________________________________________________ Chapter 13, Page 4/35 • 13-9 Repeating the process shown in the solution to Prob. 13-8, except with = 25º, we obtain the following results. (a) For m = 2, NP = 9.43 teeth. Rounding up, NP = 10 teeth. Ans. (b) For m = 3, NP = 9.92 teeth. Rounding up, NP = 10 teeth. Ans. (c) For m = 4, NP = 10.20 teeth. Rounding up, NP = 11 teeth. Ans. (d) For m = 5, NP = 10.38 teeth. Rounding up, NP = 11 teeth. Ans. For convenient reference, we will also generate the table from Eq. (13-12) for = 25º. Min NP Max NG Max m = Max NG / Min NP 9 13 1.44 10 32 3.20 11 249 22.64 12 unlimited unlimited ______________________________________________________________________________ 13-10 (a) The smallest pinion tooth count that will run with itself is found from Eq. (13-10).       2 2 2 2 2 1 1 3sin 3sin 2 1 1 1 3sin 20 3sin 20 12.32 13 teeth . P kN Ans             (b) The smallest pinion that will mesh with a gear ratio of mG = 2.5, from Eq. (13-11) is             2 2 2 2 2 2 2 1 2 sin 1 2 sin 2 1 2.5 2.5 1 2 2.5 sin 20 1 2 2.5 sin 20 14.64 15 teeth . P kN m m m m Ans                    The largest gear-tooth count possible to mesh with this pinion, from Eq. (13-12) is       2 2 2 2 22 2 2 sin 4 4 2 sin 15 sin 20 4 1 4 1 2 15 sin 20 45.49 45 teeth . P G P N kN k N Ans             Chapter 13, Page 5/35 • (c) The smallest pinion that will mesh with a rack, from Eq. (13-13),   2 2 2 12 sin sin 20 17.097 18 teeth . P kN Ans       ______________________________________________________________________________ 13-11 20 , 30n     From Eq. (13-19),  1tan tan 20 / cos30 22.80t     (a) The smallest pinion tooth count that will run with itself, from Eq. (13-21) is       2 2 2 2 2 cos 1 1 3sin 3sin 2 1 cos30 1 1 3sin 22.80 3sin 22.80 8.48 9 teeth . P t t kN Ans               (b) The smallest pinion that will mesh with a gear ratio of m = 2.5, from Eq. (13-22) is       2 22 2 1 cos30 2.5 2.5 1 2 2.5 sin 22.80 1 2 2.5 sin 22.80 9.95 10 teeth . PN Ans              The largest gear-tooth count possible to mesh with this pinion, from Eq. (13-23) is       2 2 2 2 2 2 2 2 2 2 sin 4 cos 4 cos 2 sin 10 sin 22.80 4 1 cos 30 4 1 cos 30 2 20 sin 22.80 26.08 26 teeth . P t G P t N kN k N Ans                 (c) The smallest pinion that will mesh with a rack, from Eq. (13-24) is   2 2 2 1 cos302 cos sin sin 22.80 11.53 12 teeth . P t kN Ans         ______________________________________________________________________________ Chapter 13, Page 6/35 • 13-12 From Eq. (13-19), 1 1tan tan 20tan tan 22.796 cos cos30 n t                   Program Eq. (13-23) on a computer using a spreadsheet or code, and increment NP. The first value of NP that can be doubled is NP = 10 teeth, where NG ≤ 26.01 teeth. So NG = 20 teeth will work. Higher tooth counts will work also, for example 11:22, 12:24, etc. Use NP = 10 teeth, NG = 20 teeth Ans. Note that the given diametral pitch (tooth size) is not relevant to the interference problem. ______________________________________________________________________________ 13-13 From Eq. (13-19), 1 1tan tan 20tan tan 27.236 cos cos 45 n t                   Program Eq. (13-23) on a computer using a spreadsheet or code, and increment NP. The first value of NP that can be doubled is NP = 6 teeth, where NG ≤ 17.6 teeth. So NG = 12 teeth will work. Higher tooth counts will work also, for example 7:14, 8:16, etc. Use NP = 6 teeth, NG = 12 teeth Ans. ______________________________________________________________________________ 13-14 The smallest pinion that will operate with a rack without interference is given by Eq. (13- 13). 2 2 sinP kN   Setting k = 1 for full depth teeth, NP = 9 teeth, and solving for ,  1 1 2 12sin sin 28.126 . 9P k Ans N       ______________________________________________________________________________ 13-15 (a) Eq. (13-3): 3 mm .n np m Ans   Eq. (13-16): / cos 3 / cos 25 10.40 mm .t np p A     ns Eq. (13-17): / tan 10.40 / tan 25 22.30 mm .x tp p A    ns (b) Eq. (13-3): / 10.40 / 3.310 mm .t tm p Ans    Chapter 13, Page 7/35 • Eq. (13-19): 1 1tan tan 20tan tan 21.88 . cos cos 25 n t Ans          (c) Eq. (13-2): dp = mt Np = 3.310 (18) = 59.58 mm Ans. Eq. (13-2): dG = mt NG = 3.310 (32) = 105.92 mm Ans. ______________________________________________________________________________ 13-16 (a) Sketches of the figures are shown to determine the axial forces by inspection. The axial force of gear 2 on shaft a is in the negative z-direction. The axial force of gear 3 on shaft b is in the positive z-direction. Ans. The axial force of gear 4 on shaft b is in the positive z-direction. The axial force of gear 5 on shaft c is in the negative z-direction. Ans. (b)  5 12 16 700 77.78 rev/min ccw . 48 36c n n Ans        (c)  2 12 / 12cos30 1.155 inPd    3 48 / 12cos30 4.619 inGd   1.155 4.619 2.887 in . 2ab C A  ns ns  4 16 / 8cos 25 2.207 inPd    5 36 / 8cos 25 4.965 inGd   3.586 in .bcC A ______________________________________________________________________________ 13-17 20 8 20 4 40 17 60 51 e           4 00 47.06 rev/min cw . 51d n A   ns ______________________________________________________________________________ 13-18 6 18 20 3 3 10 38 48 36 304 e              9 3 1200 11.84 rev/min cw . 304 n A  ns ______________________________________________________________________________ Chapter 13, Page 8/35 • 13-19 (a)  12 1 540 162 rev/min cw about . . 40 1c n x   Ans (b)  12 / 8cos 23 1.630 inPd    40 / 8cos 23 5.432 inGd   3.531 in . 2 P Gd d Ans  (c) 32 8 in at the large end of the teeth. . 4 d A  ns ______________________________________________________________________________ 13-20 Applying Eq. (13-30), e = (N2 / N3) (N4 / N5) = 45. For an exact ratio, we will choose to factor the train value into integers, such that N2 / N3 = 9 (1) N4 / N5 = 5 (2) Assuming a constant diametral pitch in both stages, the geometry condition to satisfy the in-line requirement of the compound reverted configuration is N2 + N3 = N4 + N5 (3) With three equations and four unknowns, one free choice is available. It is necessary that all of the unknowns be integers. We will use a normalized approach to find the minimum free choice to guarantee integers; that is, set the smallest gear of the largest stage to unity, thus N3 = 1. From (1), N2 = 9. From (3), N2 + N3 = 9 + 1 = 10 = N4 + N5 Substituting N4 = 5 N5 from (2) gives 10 = 5 N5 + N5 = 6 N5 N5 = 10 / 6 = 5 / 3 To eliminate this fraction, we need to multiply the original free choice by a multiple of 3. In addition, the smallest gear needs to have sufficient teeth to avoid interference. From Eq. (13-11) with k = 1, = 20°, and m = 9, the minimum number of teeth on the pinion to avoid interference is 17. Therefore, the smallest multiple of 3 greater than 17 is 18. Setting N3 = 18 and repeating the solution of equations (1), (2), and (3) yields N2 = 162 teeth N3 = 18 teeth N4 = 150 teeth N5 = 30 teeth Ans. ______________________________________________________________________________ Chapter 13, Page 9/35 • 13-21 The solution to Prob. 13-20 applies up to the point of determining the minimum number of teeth to avoid interference. From Eq. (13-11), with k = 1, = 25°, and m = 9, the minimum number of teeth on the pinion to avoid interference is 11. Therefore, the smallest multiple of 3 greater than 11 is 12. Setting N3 = 12 and repeating the solution of equations (1), (2), and (3) yields N2 = 108 teeth N3 = 12 teeth N4 = 100 teeth N5 = 20 teeth Ans. ______________________________________________________________________________ 13-22 Applying Eq. (13-30), e = (N2 / N3) (N4 / N5) = 30. For an exact ratio, we will choose to factor the train value into integers, such that N2 / N3 = 6 (1) N4 / N5 = 5 (2) Assuming a constant diametral pitch in both stages, the geometry condition to satisfy the in-line requirement of the compound reverted configuration is N2 + N3 = N4 + N5 (3) With three equations and four unknowns, one free choice is available. It is necessary that all of the unknowns be integers. We will use a normalized approach to find the minimum free choice to guarantee integers; that is, set the smallest gear of the largest stage to unity, thus N3 = 1. From (1), N2 = 6. From (3), N2 + N3 = 6 + 1 = 7 = N4 + N5 Substituting N4 = 5 N5 from (2) gives 7 = 5 N5 + N5 = 6 N5 N5 = 7 / 6 To eliminate this fraction, we need to multiply the original free choice by a multiple of 6. In addition, the smallest gear needs to have sufficient teeth to avoid interference. From Eq. (13-11) with k = 1, = 20°, and m = 6, the minimum number of teeth on the pinion to avoid interference is 16. Therefore, the smallest multiple of 3 greater than 16 is 18. Setting N3 = 18 and repeating the solution of equations (1), (2), and (3) yields N2 = 108 teeth N3 = 18 teeth N4 = 105 teeth N5 = 21 teeth Ans. ______________________________________________________________________________ Chapter 13, Page 10/35 • 13-23 Applying Eq. (13-30), e = (N2 / N3) (N4 / N5) = 45. For an approximate ratio, we will choose to factor the train value into two equal stages, such that 2 3 4 5/ /N N N N  45 If we choose identical pinions such that interference is avoided, both stages will be identical and the in-line geometry condition will automatically be satisfied. From Eq. (13-11) with k = 1, = 20°, and 45m  , the minimum number of teeth on the pinions to avoid interference is 17. Setting N3 = N5 = 17, we get 2 4 17 45 114.04 teethN N   Rounding to the nearest integer, we obtain N2 = N4 = 114 teeth N3 = N5 = 17 teeth Ans. Checking, the overall train value is e = (114 / 17) (114 / 17) = 44.97. ______________________________________________________________________________ 13-24 H = 25 hp, i = 2500 rev/min Let ωo = 300 rev/min for minimal gear ratio to minimize gear size. 300 1 2500 8.333 o i     2 4 3 5 1 8.333 o i N N N N     Let 2 4 3 5 1 1 8.333 2.887 N N N N    From Eq. (13-11) with k = 1, = 20°, and m = 2.887, the minimum number of teeth on the pinions to avoid interference is 15. Let N2 = N4 = 15 teeth N3 = N5 = 2.887(15) = 43.31 teeth Try N3 = N5 = 43 teeth.  15 15 2500 304.2 43 43o           Too big. Try N3 = N5 = 44. Chapter 13, Page 11/35 •  15 15 2500 290.55 rev/min 44 44o           N2 = N4 = 15 teeth, N3 = N5 = 44 teeth Ans. ______________________________________________________________________________ 13-25 (a) The planet gears act as keys and the wheel speeds are the same as that of the ring gear. Thus,  3 900 16 / 48 300 rev/min .An n Ans   (b) 5 60, , 1F Ln n n n e     6 3001 0 300 n     6300 300n  6 600 rev/min .n A ns (c) The wheel spins freely on icy surfaces, leaving no traction for the other wheel. The car is stalled. Ans. ______________________________________________________________________________ 13-26 (a) The motive power is divided equally among four wheels instead of two. (b) Locking the center differential causes 50 percent of the power to be applied to the rear wheels and 50 percent to the front wheels. If one of the rear wheels rests on a slippery surface such as ice, the other rear wheel has no traction. But the front wheels still provide traction, and so you have two-wheel drive. However, if the rear differential is locked, you have 3-wheel drive because the rear-wheel power is now distributed 50-50. ______________________________________________________________________________ 13-27 Let gear 2 be first, then nF = n2 = 0. Let gear 6 be last, then nL = n6 = –12 rev/min. 20 16 16 30 34 51 L A F A n ne n n         160 1 51A A n n   2 12 17.49 rev/min (negative indicates cw) . 35 / 51A n A   ns ______________________________________________________________________________ 13-28 Let gear 2 be first, then nF = n2 = 0 rev/min. Let gear 6 be last, then nL = n6 = 85 rev/min. Chapter 13, Page 12/35 • 20 16 16 30 34 51 L A F A n ne n n           160 85 51A A n n   16 85 51A A n n       161 8 51A n       5 85 123.9 rev/min161 51 An    The positive sign indicates the same direction as n6. 123.9 rev/min ccw .An Ans  ______________________________________________________________________________ 13-29 The geometry condition is 5 2 3/ 2 / 2d d d 4d   . Since all the gears are meshed, they will all have the same diametral pitch. Applying d = N / P, 5 2 3/ (2 ) / (2 ) / /N P N P N P N   4 P    5 2 3 42 2 12 2 16 2 12 68 teeth .N N N N Ans       Let gear 2 be first, nF = n2 = 320 rev/min. Let gear 5 be last, nL = n5 = 0 rev/min. 12 16 12 3 16 12 68 17 L A F A n ne n n            17320 0 3A A n n    3 320 68.57 rev

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qunatsym (PDF) File information This PDF 1.5 document has been generated by Microsoft® Office Word 2007, and has been sent on pdf-archive.com on 06/09/2015 at 09:50, from IP address 107.135.x.x. The current document download page has been viewed 618 times. File size: 34.75 KB (3 pages). Privacy: public file File preview The Genuine Symmetry on Natural Numbers: Truncated Universal Distribution is Enough Artem S. Shafraniuk Master of Scientific and Technical Computing; Loyola University Chicago, Illinois, United States of America Bachelor of Computer Science, Master of Information Control Systems and Technologies; European University, Kiev, Ukraine (Dated: September 5, 2015) author's e-mail: [email protected] The author analyzed the structure of universal computation, and came to the result that the structure of its limit on infinity consists of constraints on its results. Then he has found out that the non-halting programs can be taken out of consideration, by using point-free topology &amp; quantale representation. At last, for effective purposes, the universal distribution can also be pruned. The main result is that there is a symmetry on natural numbers, which is their infinite random permutation. This one is unique: any other sequence can be obtained from the author's one by swapping any two elements of the sequence one or more times. But this always, and only, decreases the randomness of the sequence in question. So this sequence is also the maximally random sequence of natural numbers. hereby the author claims an exclusive discovery of an idea made for the first time and copyright to its formulation I. INTRODUCTION TO FORMAL COMPUTATION First, let's take a look at formal computation models from Computer Science. It is known that some are universal, and the rest are not. A computation model is universal if it can execute any algorithm possible. (Examples are Turing Machine, Random Access Machine, Universal Recursive Function, Universal mu-Function, Queue Machine, etc.) Also, it, a machine, has an input, and an output, and what it does is, it transforms the input (an algorithm or a program) into the output, if possible. The reason for the exceptions is, not all algorithms halt, and this is the only case in which it is impossible for the computation model, also called a machine, to halt, and give an output, in finite time. Evidently, if the machine doesn't halt in finite time, it doesn't halt at all, so it cannot provide any output. It is also known that any computation universal machine can simulate any other computation universal machine, using a program (adaptor) that simulates that other one. This is clear from the fact that any computation universal machine can run any algorithm, and also that any computation universal machine can be represented by an algorithm. G. Chaitin has discovered a very curious real number, which he called Omega. In binary expansion it starts with zero, and then at each position i (index), progressing towards the right of the number's expansion, the bit x_i is '1' if the program number i halts, and '0' otherwise (if the program number i doesn't halt.) Now, these programs represent algorithms, so this is a matter of coding. Also, since a program is a finite discrete object, it can be directly encoded by a natural number. Then, it's clear to see that all Omega's (for all possible different computation universal machines) are bit-wise isomorphic, as are algorithms indexed by the i's] in a Omega's. So, any computation universal machine takes an input, runs, and then either halts at a finite step, and provides an output; or doesn't halt at all. It runs in steps (sometimes called tacts). It is easy to see that, for any classical computation universal machine, any of its states, including the start state, which is the input itself, can be represented by a natural number, and the machine itself then can be presented as an iterated function o=f(i), where i is input, o is output. This function then iterates on itself, starting from the input, until it reaches a limit cycle, which is a state s=f(s). This s then is taken to be the output. Such a machine, then, can be otherwise interpreted being "tabulated" into an infinite sequence of natural numbers f(0), f(1), ...f(i), ... . Executing a program then means picking start index s, and looking up what's at that index, then changing state, s:=f(s). The author of this paper has published such a sequence before, based on a pairwise encoding of the calculus of combinators S &amp; K. Any combinator possible is then encoded as a natural number. This sequence was published online, in the Encyclopedia of Integer Sequences. Actually, any computation universal machine, represented by a function, is a homomorphism, and, represented as a sequence, is a pseudorandom one. In the limit it (computation / iteration) converges to a random sequence, except for the infinities at the positions corresponding to the non-halting input programs. Now, if we knew Omega, we could construct an oracle for the halting problem, and compute the limit, which latter, again, will come out to be random. II. COMPLEXITY OF UNIVERSAL COMPUTATION Second, there was a famous mathematician A. Kolmogorov, who discovered two brilliant things that we need to know. The first is the Axiomatic Probability Theory, and the second is called Kolmogorov Complexity. From the first we only need to look at the sample space, which is the set of all sets of n possible elementary items, so the number of sets is then 2^n. The second, Kolmogorov Complexity is written down KC(o) = min{l(p)} : o=M(p), where KC Kolmogorov Complexity, M - computation universal machine, p - any program, min{l(p)}: o=M(p) - the length of the minimal program, such that if M takes it as input and runs, it's output is o. Next, let's take a Universal Turing Machine (UTM) as our model for computation. One should note, however, that it, a UTM, is a program on the standard Turing Machine (TM), being an adaptor, which is also computation universal. Each of its states, either from input to the output, or from input to infinity (if it doesn't halt on that input), is a finite string of bits. Let's call it a bitstring. Now, observe that if we add a '0.' on the left to a finite bitstring, we always get a rational number. Also, all the possible inputs to our UTM are, then, all the rational numbers, and so its, UTM's, one-step iteration is a homomorphism on the rational numbers. Let's call the UTM with bitstring states the Bitstring Universal Turing Machine, or BUTM. III. QUANTAL TOPOLOGY &amp; COMPUTATION Third, let's introduce a concept, from Point Free Topology, called a quantale, discovered by Mulvey. A quantale is, by definition, an upperjoin semi-lattice, and its original primary role is quantization of space, or something spatial. Next, the sample space from Kolmogorov's Probability Theory can, in general, be represented by a quantale, as there is an isomorphism from the first to the second. The sample space contains items (usually events, constructed from elementary events, as their power set. So the sample space then contains all the possible combinations of the elementary events.) Let's call the spatial object that results from the quantization of space by such a quantale, a quantal figure or spatial figure, such that they coincide. It should be constructed, using point-free topology, from the sample space, represented by a labeled quantale, isomorphic to the sample space; with labels being the elementary items / "events". Now, bitstring-programs -to- bitstring-results, using BUTM, *is* a homomorphism, except for the non-halting programs. But the latter, nonhalting, cannot be excluded by all the means we all have believed since Turing, Goedel, Berry, and Chaitin. But the next idea doesn't agree! IV. QUANTAL STRUCTURE OF UNIVERSAL COMPUTATION Fourth, using a sample space from probability with input programs as bitstrings, ordered as binary numbers, as elementary items of this sample space, we get the source quantal figure; it's a piece of space, resulting from the quantization of space by the sample-spacequantale. What does it look like? The list of all the binary numbers (bitstrings) can be represented by a binary tree, with all branches finite, but a tree with infinite breadth. So, the definition of the quantal figure we're searching for is this: fractal spatial figure. Next, let's try to imagine what the limit of the computation looks like on infinity. Well, there's clearly one option: random quantal figure &lt;==&gt; random spatial figure. Let's not yet bother with how it looks like, but believe me: it's unique. It must be unique because all c.u. (computation universal) machines are equivalent. We also get a mapping from the fractal spatial figure to the random spatial figure. V. UNIVERSAL &amp; INTRICATE SYMMETRY ON THE NATURAL NUMBERS Fifth, let's remove labels (bitstrings) from the random spatial figure. It's something new, a random spatial object, this figure. To understand it, we need to find the genuine symmetry on it. The symmetry is again a quantale, precisely because such a spatial figure results from the quantization of space by a quantale. Then, what we get is a random quantale. But such a quantale, too, can be represented as a sequence of natural numbers. It's easy to see that, first, this is the maximally random sequence of natural numbers, and, second, this is the infinite random permutation of them, the natural numbers. If we swap two elements once or more, it will become less random. This means, it is the maximally random of all the possible infinite sequences of natural numbers, which simply means that it's unique. There is an analogy of this sequence uniqueness proof to the proof that the infinite random directed graph is unique. This must be for a reason. The sequence in question encodes the infinite random isomorphism, unique again. The transitive closure of this isomorphism is a directed graph. This graph is, evidently, random, except that there's precisely one directed edge going out of a node. The nodes are labeled by the natural numbers, and there's a unique number at each node, and a unique node labeled by a natural number. V. TRUNCATED UNIVESAL DISTRIBUTION IS SUFFICIENT Last, consider the so-called Universal Distribution. Let's take the random spatial figure (there's just one such unique object, as shown above in the paper.) It's easy to see that it puts a discrete geometrical framework on the universal distribution, and this framework consists of "boxes" (parallepipeds). These latter are constraints, and what the framework represents is a truncated universal distribution, where all, and only, halting computation destinations (on termination)/ are present. PS. feel free to send, and forward, know, &amp; don't know, if allowed! qunatsym.pdf (PDF, 34.75 KB) HTML Code Copy the following HTML code to share your document on a Website or Blog

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# Question regarding floor() function Given x is a real number and c is a natural number. Prove that: floor( floor(x) / c ) = floor( x / c ) When x > 0, I can prove it by let x = n + z where 0 < z < 1. Then rhs = floor( n/c + z/c ) which equals to floor( floor(x) ) since z/c -> 0. However, when x < 0, it's wrong when letting x = -( n + z ). Because it always give me a smaller negative number. Is my logic wrong here? Any hint? Thanks, Chan • I suspect that you want $x=-n+z$ (no parentheses around $n+z$) as, for instance, $\text{floor}(-1.5)=-2$. – Isaac Feb 4 '11 at 6:34 • @Issac: Thanks, so if x = -1.67 then I would rewrite it as: -2 + 0.33? – Chan Feb 4 '11 at 6:41 You can see it here: • @Chandru1: Thanks but you spoiled me :(! – Chan Feb 4 '11 at 6:42 • @chan: Sorry chan. – anonymous Feb 4 '11 at 7:24 • This post no longer exists. – franklin Feb 18 '19 at 17:50 This is the special case $\rm\ m = 1\$ in a proof I presented here. See that thread for more on the universal viewpoint that explains the simplicity of this proof. For convenience, here is the proof. LEMMA $\rm\: \ \lfloor x/(mn)\rfloor\ =\ \lfloor{\lfloor x/m\rfloor}/n\rfloor\ \$ for $\rm\ \ n > 0$ Proof $\rm\quad\quad\quad\quad\quad\ \ \ k\ \le \lfloor{\lfloor x/m\rfloor}/n\rfloor$ $\rm\quad\quad\quad\quad\quad\iff\quad\ \ k\ \le\ \:{\lfloor x/m\rfloor}/n$ $\rm\quad\quad\quad\quad\quad\iff\ \ nk\ \le\ \ \lfloor x/m\rfloor$ $\rm\quad\quad\quad\quad\quad\iff\ \ nk\ \le\:\ \ \ x/m$ $\rm\quad\quad\quad\quad\quad\iff\ \ \ \ k\ \le\:\ \ \ x/(mn)$ $\rm\quad\quad\quad\quad\quad\iff\ \ \ \ k\ \le\ \ \lfloor x/(mn)\rfloor$ Compare the above trivial proof to more traditional proofs, e.g. the special case $\rm\ m = 1\$ here. • Many thanks for a great solution. Very clever and succinct ;) ! – Chan Feb 4 '11 at 15:34

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The function is challenging to graph, but can be represented by a linear combination of sine functions. Graphing the Sawtooth Function. Here “function” is used in the sense of an algebraic function, that is, a certain type of equation. The above equation Eq. For example, in Mathematica, the function is: Plot[SawtoothWave[x],{x,0,1}]. dimensional wave equation (1.1) is Φ(x,t)=F(x−ct)+G(x+ct) (1.2) where F and g are arbitrary functions of their arguments. Some mathematical software have built in functions for the sawtooth. Wave functions with unalike signs (waves out of phase) will interfere destructively. A function is like a little machine that if you feed in a certain number, the machine will “massage” it in a specified way and output a certain number. 2 Green Functions for the Wave Equation G. Mustafa Fourier Series of the Sawtooth Wave In 1926, Erwin Schrödinger deduced the wave function for the simplest of all atoms, hydrogen. The Schrödinger equation (also known as Schrödinger’s wave equation) is a partial differential equation that describes the dynamics of quantum mechanical systems via the wave function. This equation is obtained for a special case of wave called simple harmonic wave but it is equally true for other periodic or non-periodic waves. The 2-D and 3-D version of the wave equation is, Solving the Schrödinger equation enables scientists to determine wave functions for electrons in atoms and molecules. For the sake of completeness we’ll close out this section with the 2-D and 3-D version of the wave equation. This property is known as the principle of superposition. A wave function, in quantum mechanics, is an equation.It describes the behavior of quantum particles, usually electrons. The trajectory, the positioning, and the energy of these systems can be retrieved by solving the Schrödinger equation. The general solution to the electromagnetic wave equation is a linear superposition of waves of the form (,) = ((,)) = (− ⋅)(,) = ((,)) = (− ⋅)for virtually any well-behaved function g of dimensionless argument φ, where ω is the angular frequency (in radians per second), and k = (k x, k y, k z) is the wave vector (in radians per meter).. Taking this analysis a step further, if wave functions y1 (x, t) = f(x ∓ vt) and y2 (x, t) = g(x ∓ vt) are solutions to the linear wave equation, then Ay 1 (x, t) + By 2 (x, y), where A and B are constants, is also a solution to the linear wave equation. E 2 = c 2 p 2 + m 2 c 4. Thus to the observer (x,t)whomovesatthesteadyspeedc along the positivwe x-axis, the function … In the x,t (space,time) plane F(x − ct) is constant along the straight line x − ct = constant. The discussion above suggests how we might extend the wave equation operator from the photon case (zero rest mass) to a particle having rest mass m. We need a wave equation operator that, when it operates on a plane wave, yields . The Wave Equation Maxwell equations in terms of potentials in Lorenz gauge Both are wave equations with known source distribution f(x,t): If there are no boundaries, solution by Fourier transform and the Green function method is best. Constructing a Wave Equation for a Particle with Mass. Writing the plane wave function The Schrodinger equation is the most important equation in quantum mechanics and allows you to find the wave function for a given situation and describes its evolution in time. \eqref{11} is called linear wave equation which gives total description of wave motion. Learning how to use the equation and some of the solutions in basic situations is crucial for any student of physics. We’ll not actually be solving this at any point, but since we gave the higher dimensional version of the heat equation (in which we will solve a special case) we’ll give this as well.

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# Permutations and Combinations The lottery "game" consists of picking numbers from a pool of . For example, you select numbers out of . To win, the order in which you pick the numbers doesn't matter; you only have to choose the right set of numbers. The chances of winning equal the number of different length- sequences that can be chosen. A related, but different, problem is selecting the batting lineup for a baseball team. Now the order matters, and many more choices are possible than when order does not matter. Answering such questions occurs in many applications beyond games. In digital communications, for example, you might ask how many possible double-bit errors can occur in a codeword. Numbering the bit positions from to , the answer is the same as the lottery problem with . Solving these kind of problems amounts to understanding permutations - the number of ways of choosing things when order matters as in baseball lineups - and combinations - the number of ways of choosing things when order does not matter as in lotteries and bit errors. Calculating permutations is the easiest. If we are to pick numbers from a pool of , we have choices for the first one. For the second choice, we have . The number of length-two ordered sequences is therefore be . Continuing to choose until we make choices means the number of permutations is . This result can be written in terms of factorials as , with . For mathematical convenience, we define . When order does not matter, the number of combinations equals the number of permutations divided by the number of orderings. The number of ways a pool of things can be ordered equals . Thus, once we choose the nine starters for our baseball game, we have different lineups! The symbol for the combination of things drawn from a pool of is and equals . ### Exercise What are the chances of winning the lottery? Assume you pick numbers from the numbers - . Combinatorials occur in interesting places. For example, Newton derived that the -th power of a sum obeyed the formula . ### Exercise What does the sum of binomial coefficients equal? In other words, what is Because of Newton's binomial theorem, the sum equals . A related problem is calculating the probability that any two bits are in error in a length- codeword when is the probability of any bit being in error. The probability of any particular two-bit error sequence is . The probability of a two-bit error occurring anywhere equals this probability times the number of combinations: . Note that the probability that zero or one or two, etc. errors occurring must be one; in other words, something must happen to the codeword! That means that we must have . Can you prove this?

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# Gravitationaw wens A wight source passes behind a gravitationaw wens (point mass pwaced in de center of de image). The aqwa circwe is de wight source as it wouwd be seen if dere was no wens, white spots are de muwtipwe images (or Einstein ring) of de source. A gravitationaw wens is a distribution of matter (such as a cwuster of gawaxies) between a distant wight source and an observer, dat is capabwe of bending de wight from de source as de wight travews towards de observer. This effect is known as gravitationaw wensing, and de amount of bending is one of de predictions of Awbert Einstein's generaw deory of rewativity.[1][2] (Cwassicaw physics awso predicts de bending of wight, but onwy hawf dat predicted by generaw rewativity.[3]) Awdough Einstein made unpubwished cawcuwations on de subject in 1912,[4] Orest Khvowson (1924)[5] and Frantisek Link (1936)[6] are generawwy credited wif being de first to discuss de effect in print. However, dis effect is more commonwy associated wif Einstein, who pubwished an articwe on de subject in 1936.[7] Fritz Zwicky posited in 1937 dat de effect couwd awwow gawaxy cwusters to act as gravitationaw wenses. It was not untiw 1979 dat dis effect was confirmed by observation of de so-cawwed "Twin QSO" SBS 0957+561. ## Description Gravitationaw wensing – intervening gawaxy modifies appearance of a gawaxy far behind it (video; artist's concept). This schematic image shows how wight from a distant gawaxy is distorted by de gravitationaw effects of a foreground gawaxy, which acts wike a wens and makes de distant source appear distorted, but magnified, forming characteristic rings of wight, known as Einstein rings. An anawysis of de distortion of SDP.81 caused by dis effect has reveawed star-forming cwumps of matter. Unwike an opticaw wens, a gravitationaw wens produces a maximum defwection of wight dat passes cwosest to its center, and a minimum defwection of wight dat travews furdest from its center. Conseqwentwy, a gravitationaw wens has no singwe focaw point, but a focaw wine. The term "wens" in de context of gravitationaw wight defwection was first used by O.J. Lodge, who remarked dat it is "not permissibwe to say dat de sowar gravitationaw fiewd acts wike a wens, for it has no focaw wengf".[8] If de (wight) source, de massive wensing object, and de observer wie in a straight wine, de originaw wight source wiww appear as a ring around de massive wensing object. If dere is any misawignment, de observer wiww see an arc segment instead. This phenomenon was first mentioned in 1924 by de St. Petersburg physicist Orest Chwowson,[9] and qwantified by Awbert Einstein in 1936. It is usuawwy referred to in de witerature as an Einstein ring, since Chwowson did not concern himsewf wif de fwux or radius of de ring image. More commonwy, where de wensing mass is compwex (such as a gawaxy group or cwuster) and does not cause a sphericaw distortion of space–time, de source wiww resembwe partiaw arcs scattered around de wens. The observer may den see muwtipwe distorted images of de same source; de number and shape of dese depending upon de rewative positions of de source, wens, and observer, and de shape of de gravitationaw weww of de wensing object.[10] There are dree cwasses of gravitationaw wensing:[8][11] 1. Strong wensing: where dere are easiwy visibwe distortions such as de formation of Einstein rings, arcs, and muwtipwe images. 2. Weak wensing: where de distortions of background sources are much smawwer and can onwy be detected by anawyzing warge numbers of sources in a statisticaw way to find coherent distortions of onwy a few percent. The wensing shows up statisticawwy as a preferred stretching of de background objects perpendicuwar to de direction to de centre of de wens. By measuring de shapes and orientations of warge numbers of distant gawaxies, deir orientations can be averaged to measure de shear of de wensing fiewd in any region, uh-hah-hah-hah. This, in turn, can be used to reconstruct de mass distribution in de area: in particuwar, de background distribution of dark matter can be reconstructed. Since gawaxies are intrinsicawwy ewwipticaw and de weak gravitationaw wensing signaw is smaww, a very warge number of gawaxies must be used in dese surveys. These weak wensing surveys must carefuwwy avoid a number of important sources of systematic error: de intrinsic shape of gawaxies, de tendency of a camera's point spread function to distort de shape of a gawaxy and de tendency of atmospheric seeing to distort images must be understood and carefuwwy accounted for. The resuwts of dese surveys are important for cosmowogicaw parameter estimation, to better understand and improve upon de Lambda-CDM modew, and to provide a consistency check on oder cosmowogicaw observations. They may awso provide an important future constraint on dark energy. 3. Microwensing: where no distortion in shape can be seen but de amount of wight received from a background object changes in time. The wensing object may be stars in de Miwky Way in one typicaw case, wif de background source being stars in a remote gawaxy, or, in anoder case, an even more distant qwasar. The effect is smaww, such dat (in de case of strong wensing) even a gawaxy wif a mass more dan 100 biwwion times dat of de Sun wiww produce muwtipwe images separated by onwy a few arcseconds. Gawaxy cwusters can produce separations of severaw arcminutes. In bof cases de gawaxies and sources are qwite distant, many hundreds of megaparsecs away from our Gawaxy. Gravitationaw wenses act eqwawwy on aww kinds of ewectromagnetic radiation, not just visibwe wight. Weak wensing effects are being studied for de cosmic microwave background as weww as gawaxy surveys. Strong wenses have been observed in radio and x-ray regimes as weww. If a strong wens produces muwtipwe images, dere wiww be a rewative time deway between two pads: dat is, in one image de wensed object wiww be observed before de oder image. ## History One of Eddington's photographs of de 1919 sowar ecwipse experiment, presented in his 1920 paper announcing its success Henry Cavendish in 1784 (in an unpubwished manuscript) and Johann Georg von Sowdner in 1801 (pubwished in 1804) had pointed out dat Newtonian gravity predicts dat starwight wiww bend around a massive object[12] as had awready been supposed by Isaac Newton in 1704 in his Queries No.1 in his book Opticks.[13] The same vawue as Sowdner's was cawcuwated by Einstein in 1911 based on de eqwivawence principwe awone.[8] However, Einstein noted in 1915, in de process of compweting generaw rewativity, dat his (and dus Sowdner's) 1911-resuwt is onwy hawf of de correct vawue. Einstein became de first to cawcuwate de correct vawue for wight bending.[14] The first observation of wight defwection was performed by noting de change in position of stars as dey passed near de Sun on de cewestiaw sphere. The observations were performed in 1919 by Ardur Eddington, Frank Watson Dyson, and deir cowwaborators during de totaw sowar ecwipse on May 29.[15] The sowar ecwipse awwowed de stars near de Sun to be observed. Observations were made simuwtaneouswy in de cities of Sobraw, Ceará, Braziw and in São Tomé and Príncipe on de west coast of Africa.[16] The observations demonstrated dat de wight from stars passing cwose to de Sun was swightwy bent, so dat stars appeared swightwy out of position, uh-hah-hah-hah.[17] Bending wight around a massive object from a distant source. The orange arrows show de apparent position of de background source. The white arrows show de paf of de wight from de true position of de source. In de formation known as Einstein's Cross, four images of de same distant qwasar appear around a foreground gawaxy due to strong gravitationaw wensing. The resuwt was considered spectacuwar news and made de front page of most major newspapers. It made Einstein and his deory of generaw rewativity worwd-famous. When asked by his assistant what his reaction wouwd have been if generaw rewativity had not been confirmed by Eddington and Dyson in 1919, Einstein said "Then I wouwd feew sorry for de dear Lord. The deory is correct anyway."[18] In 1912, Einstein had specuwated dat an observer couwd see muwtipwe images of a singwe wight source, if de wight were defwected around a mass. This effect wouwd make de mass act as a kind of gravitationaw wens. However, as he onwy considered de effect of defwection around a singwe star, he seemed to concwude dat de phenomenon was unwikewy to be observed for de foreseeabwe future since de necessary awignments between stars and observer wouwd be highwy improbabwe. Severaw oder physicists specuwated about gravitationaw wensing as weww, but aww reached de same concwusion dat it wouwd be nearwy impossibwe to observe.[7] Awdough Einstein made unpubwished cawcuwations on de subject,[4] de first discussion of de gravitationaw wens in print was by Khvowson, in a short articwe discussing de “hawo effect” of gravitation when de source, wens, and observer are in near-perfect awignment,[5] now referred to as de Einstein ring. In 1936, after some urging by Rudi W. Mandw, Einstein rewuctantwy pubwished de short articwe "Lens-Like Action of a Star By de Deviation of Light In de Gravitationaw Fiewd" in de journaw Science.[7] In 1937, Fritz Zwicky first considered de case where de newwy discovered gawaxies (which were cawwed 'nebuwae' at de time) couwd act as bof source and wens, and dat, because of de mass and sizes invowved, de effect was much more wikewy to be observed.[19] In 1963 Yu. G. Kwimov, S. Liebes, and Sjur Refsdaw recognized independentwy dat qwasars are an ideaw wight source for de gravitationaw wens effect.[20] It was not untiw 1979 dat de first gravitationaw wens wouwd be discovered. It became known as de "Twin QSO" since it initiawwy wooked wike two identicaw qwasistewwar objects. (It is officiawwy named SBS 0957+561.) This gravitationaw wens was discovered by Dennis Wawsh, Bob Carsweww, and Ray Weymann using de Kitt Peak Nationaw Observatory 2.1 meter tewescope.[21] In de 1980s, astronomers reawized dat de combination of CCD imagers and computers wouwd awwow de brightness of miwwions of stars to be measured each night. In a dense fiewd, such as de gawactic center or de Magewwanic cwouds, many microwensing events per year couwd potentiawwy be found. This wed to efforts such as Opticaw Gravitationaw Lensing Experiment, or OGLE, dat have characterized hundreds of such events, incwuding dose of OGLE-2016-BLG-1190Lb and OGLE-2016-BLG-1195Lb. ## Expwanation in terms of space–time curvature Simuwated gravitationaw wensing (bwack howe passing in front of a background gawaxy). In generaw rewativity, wight fowwows de curvature of spacetime, hence when wight passes around a massive object, it is bent. This means dat de wight from an object on de oder side wiww be bent towards an observer's eye, just wike an ordinary wens. In Generaw Rewativity de speed of wight depends on de gravitationaw potentiaw (aka de metric) and dis bending can be viewed as a conseqwence of de wight travewing awong a gradient in wight speed. Light rays are de boundary between de future, de spacewike, and de past regions. The gravitationaw attraction can be viewed as de motion of undisturbed objects in a background curved geometry or awternativewy as de response of objects to a force in a fwat geometry. The angwe of defwection is: ${\dispwaystywe \deta ={\frac {4GM}{rc^{2}}}}$ toward de mass M at a distance r from de affected radiation, where G is de universaw constant of gravitation and c is de speed of wight in a vacuum. This formuwa is identicaw to de formuwa for weak gravitationaw wensing derived using rewativistic Newtonian dynamics [22] widout curving spacetime. Since de Schwarzschiwd radius ${\dispwaystywe r_{\madrm {s} }}$ is defined as ${\dispwaystywe r_{\madrm {s} }={2Gm}/{c^{2}}}$, dis can awso be expressed in simpwe form as ${\dispwaystywe \deta =2{\frac {r_{\madrm {s} }}{r}}}$ ## Search for gravitationaw wenses This image from de NASA/ESA Hubbwe Space Tewescope shows de gawaxy cwuster MACS J1206. Most of de gravitationaw wenses in de past have been discovered accidentawwy. A search for gravitationaw wenses in de nordern hemisphere (Cosmic Lens Aww Sky Survey, CLASS), done in radio freqwencies using de Very Large Array (VLA) in New Mexico, wed to de discovery of 22 new wensing systems, a major miwestone. This has opened a whowe new avenue for research ranging from finding very distant objects to finding vawues for cosmowogicaw parameters so we can understand de universe better. A simiwar search in de soudern hemisphere wouwd be a very good step towards compwementing de nordern hemisphere search as weww as obtaining oder objectives for study. If such a search is done using weww-cawibrated and weww-parameterized instrument and data, a resuwt simiwar to de nordern survey can be expected. The use of de Austrawia Tewescope 20 GHz (AT20G) Survey data cowwected using de Austrawia Tewescope Compact Array (ATCA) stands to be such a cowwection of data. As de data were cowwected using de same instrument maintaining a very stringent qwawity of data we shouwd expect to obtain good resuwts from de search. The AT20G survey is a bwind survey at 20 GHz freqwency in de radio domain of de ewectromagnetic spectrum. Due to de high freqwency used, de chances of finding gravitationaw wenses increases as de rewative number of compact core objects (e.g. qwasars) are higher (Sadwer et aw. 2006). This is important as de wensing is easier to detect and identify in simpwe objects compared to objects wif compwexity in dem. This search invowves de use of interferometric medods to identify candidates and fowwow dem up at higher resowution to identify dem. Fuww detaiw of de project is currentwy under works for pubwication, uh-hah-hah-hah. Gawaxy cwuster SDSS J0915+3826 hewps astronomers to study star formation in gawaxies.[23] Microwensing techniqwes have been used to search for pwanets outside our sowar system. A statisticaw anawysis of specific cases of observed microwensing over de time period of 2002 to 2007 found dat most stars in de Miwky Way gawaxy hosted at weast one orbiting pwanet widin .5 to 10 AUs.[24] In a 2009 articwe on Science Daiwy a team of scientists wed by a cosmowogist from de U.S. Department of Energy's Lawrence Berkewey Nationaw Laboratory has made major progress in extending de use of gravitationaw wensing to de study of much owder and smawwer structures dan was previouswy possibwe by stating dat weak gravitationaw wensing improves measurements of distant gawaxies.[25] Astronomers from de Max Pwanck Institute for Astronomy in Heidewberg, Germany, de resuwts of which are accepted for pubwication on Oct 21, 2013 in de Astrophysicaw Journaw Letters (arXiv.org), discovered what at de time was de most distant gravitationaw wens gawaxy termed as J1000+0221 using NASA’s Hubbwe Space Tewescope.[26][27] Whiwe it remains de most distant qwad-image wensing gawaxy known, an even more distant two-image wensing gawaxy was subseqwentwy discovered by an internationaw team of astronomers using a combination of Hubbwe Space Tewescope and Keck tewescope imaging and spectroscopy. The discovery and anawysis of de IRC 0218 wens was pubwished in de Astrophysicaw Journaw Letters on June 23, 2014.[28] Research pubwished Sep 30, 2013 in de onwine edition of Physicaw Review Letters, wed by McGiww University in Montreaw, Québec, Canada, has discovered de B-modes, dat are formed due to gravitationaw wensing effect, using Nationaw Science Foundation's Souf Powe Tewescope and wif hewp from de Herschew space observatory. This discovery wouwd open de possibiwities of testing de deories of how our universe originated.[29][30] Abeww 2744 gawaxy cwuster - extremewy distant gawaxies reveawed by gravitationaw wensing (16 October 2014).[31][32] ## Sowar gravitationaw wens Awbert Einstein predicted in 1936 dat rays of wight from de same direction dat skirt de edges of de Sun wouwd converge to a focaw point approximatewy 542 AUs from de Sun, uh-hah-hah-hah.[33] Thus, a probe positioned at dis distance (or greater) from de Sun couwd use de Sun as a gravitationaw wens for magnifying distant objects on de opposite side of de Sun, uh-hah-hah-hah.[34] A probe's wocation couwd shift around as needed to sewect different targets rewative to de Sun, uh-hah-hah-hah. This distance is far beyond de progress and eqwipment capabiwities of space probes such as Voyager 1, and beyond de known pwanets and dwarf pwanets, dough over dousands of years 90377 Sedna wiww move farder away on its highwy ewwipticaw orbit. The high gain for potentiawwy detecting signaws drough dis wens, such as microwaves at de 21-cm hydrogen wine, wed to de suggestion by Frank Drake in de earwy days of SETI dat a probe couwd be sent to dis distance. A muwtipurpose probe SETISAIL and water FOCAL was proposed to de ESA in 1993, but is expected to be a difficuwt task.[35] If a probe does pass 542 AU, magnification capabiwities of de wens wiww continue to act at farder distances, as de rays dat come to a focus at warger distances pass furder away from de distortions of de Sun's corona.[36] A critiqwe of de concept was given by Landis,[37] who discussed issues incwuding interference of de sowar corona, de high magnification of de target, which wiww make de design of de mission focaw pwane difficuwt, and an anawysis of de inherent sphericaw aberration of de wens. ## Measuring weak wensing Gawaxy cwuster MACS J2129-0741 and wensed gawaxy MACS2129-1.[38] Kaiser, Sqwires and Broadhurst (1995),[39] Luppino & Kaiser (1997)[40] and Hoekstra et aw. (1998) prescribed a medod to invert de effects of de Point Spread Function (PSF) smearing and shearing, recovering a shear estimator uncontaminated by de systematic distortion of de PSF. This medod (KSB+) is de most widewy used medod in weak wensing shear measurements.[41][42] Gawaxies have random rotations and incwinations. As a resuwt, de shear effects in weak wensing need to be determined by statisticawwy preferred orientations. The primary source of error in wensing measurement is due to de convowution of de PSF wif de wensed image. The KSB medod measures de ewwipticity of a gawaxy image. The shear is proportionaw to de ewwipticity. The objects in wensed images are parameterized according to deir weighted qwadrupowe moments. For a perfect ewwipse, de weighted qwadrupowe moments are rewated to de weighted ewwipticity. KSB cawcuwate how a weighted ewwipticity measure is rewated to de shear and use de same formawism to remove de effects of de PSF.[43] KSB's primary advantages are its madematicaw ease and rewativewy simpwe impwementation, uh-hah-hah-hah. However, KSB is based on a key assumption dat de PSF is circuwar wif an anisotropic distortion, uh-hah-hah-hah. This is a reasonabwe assumption for cosmic shear surveys, but de next generation of surveys (e.g. LSST) may need much better accuracy dan KSB can provide. ## Gawwery Gravitationawwy-wensed distant star-forming gawaxies.[53] ## Historicaw papers and references • Chwowson, O (1924). "Über eine mögwiche Form fiktiver Doppewsterne". Astronomische Nachrichten. 221 (20): 329–330. Bibcode:1924AN....221..329C. doi:10.1002/asna.19242212003. • Einstein, Awbert (1936). "Lens-wike Action of a Star by de Deviation of Light in de Gravitationaw Fiewd". Science. 84 (2188): 506–7. Bibcode:1936Sci....84..506E. doi:10.1126/science.84.2188.506. JSTOR 1663250. PMID 17769014. • Renn, Jürgen; Tiwman Sauer; John Stachew (1997). "The Origin of Gravitationaw Lensing: A Postscript to Einstein's 1936 Science paper". Science. 275 (5297): 184–6. Bibcode:1997Sci...275..184R. doi:10.1126/science.275.5297.184. PMID 8985006. ## References Notes 1. ^ Drakeford, Jason; Corum, Jonadan; Overbye, Dennis (March 5, 2015). "Einstein's Tewescope - video (02:32)". New York Times. Retrieved December 27, 2015. 2. ^ Overbye, Dennis (March 5, 2015). "Astronomers Observe Supernova and Find They're Watching Reruns". New York Times. Retrieved March 5, 2015. 3. ^ Cf. Kennefick 2005 for de cwassic earwy measurements by de Eddington expeditions; for an overview of more recent measurements, see Ohanian & Ruffini 1994, ch. 4.3. For de most precise direct modern observations using qwasars, cf. Shapiro et aw. 2004 4. ^ a b Tiwman Sauer (2008). "Nova Geminorum 1912 and de Origin of de Idea of Gravitationaw Lensing". Archive for History of Exact Sciences. 62 (1): 1–22. arXiv:0704.0963. doi:10.1007/s00407-007-0008-4. 5. ^ a b Turner, Christina (February 14, 2006). "The Earwy History of Gravitationaw Lensing" (PDF). Archived from de originaw (PDF) on Juwy 25, 2008. 6. ^ Bičák, Jiří; Ledvinka, Tomáš (2014). Generaw Rewativity, Cosmowogy and Astrophysics: Perspectives 100 years after Einstein's stay in Prague (iwwustrated ed.). Springer. pp. 49–50. ISBN 9783319063492. 7. ^ a b c "A brief history of gravitationaw wensing — Einstein Onwine". www.einstein-onwine.info. Archived from de originaw on 2016-07-01. Retrieved 2016-06-29. 8. ^ a b c Schneider, Peter; Ehwers, Jürgen; Fawco, Emiwio E. (1992). Gravitationaw Lenses. Springer-Verwag Berwin Heidewberg New York Press. ISBN 978-3-540-97070-5. 9. ^ Gravity Lens – Part 2 (Great Moments in Science, ABS Science) 10. ^ Dieter Briww, "Bwack Howe Horizons and How They Begin", Astronomicaw Review (2012); Onwine Articwe, cited Sept.2012. 11. ^ Mewia, Fuwvio (2007). The Gawactic Supermassive Bwack Howe. Princeton University Press. pp. 255–256. ISBN 978-0-691-13129-0. 12. ^ Sowdner, J. G. V. (1804). "On de defwection of a wight ray from its rectiwinear motion, by de attraction of a cewestiaw body at which it nearwy passes by" . Berwiner Astronomisches Jahrbuch: 161–172. 13. ^ Newton, Isaac (1998). Opticks: or, a treatise of de refwexions, refractions, infwexions and cowours of wight. Awso two treatises of de species and magnitude of curviwinear figures. Commentary by Nichowas Humez (Octavo ed.). Pawo Awto, Cawif.: Octavo. ISBN 978-1-891788-04-8. (Opticks was originawwy pubwished in 1704). 14. ^ Wiww, C.M. (2006). "The Confrontation between Generaw Rewativity and Experiment". Living Reviews in Rewativity. 9 (1): 39. arXiv:gr-qc/0510072. Bibcode:2006LRR.....9....3W. doi:10.12942/wrr-2006-3. 15. ^ Dyson, F. W.; Eddington, A. S.; Davidson C. (1920). "A determination of de defwection of wight by de Sun's gravitationaw fiewd, from observations made at de totaw ecwipse of 29 May 1919". Phiwosophicaw Transactions of de Royaw Society. 220A (571–581): 291–333. Bibcode:1920RSPTA.220..291D. doi:10.1098/rsta.1920.0009. 16. ^ Stanwey, Matdew (2003). "'An Expedition to Heaw de Wounds of War': The 1919 Ecwipse and Eddington as Quaker Adventurer". Isis. 94 (1): 57–89. doi:10.1086/376099. PMID 12725104. 17. ^ Dyson, F. W.; Eddington, A. S.; Davidson, C. (1 January 1920). "A Determination of de Defwection of Light by de Sun's Gravitationaw Fiewd, from Observations Made at de Totaw Ecwipse of May 29, 1919". Phiwosophicaw Transactions of de Royaw Society A: Madematicaw, Physicaw and Engineering Sciences. 220 (571–581): 291–333. Bibcode:1920RSPTA.220..291D. doi:10.1098/rsta.1920.0009. 18. ^ Rosendaw-Schneider, Iwse: Reawity and Scientific Truf. Detroit: Wayne State University Press, 1980. p 74. (See awso Cawaprice, Awice: The New Quotabwe Einstein. Princeton: Princeton University Press, 2005. p 227.) 19. ^ F. Zwicky (1937). "Nebuwae as Gravitationaw wenses" (PDF). Physicaw Review. 51 (4): 290. Bibcode:1937PhRv...51..290Z. doi:10.1103/PhysRev.51.290. 20. ^ Schneider Peter; Kochanek, Christopher; Wambsganss, Joachim (2006). Gravitationaw Lensing: Strong, Weak and Micro. Springer Verwag Berwin Heidewberg New York Press. p. 4. ISBN 978-3-540-30309-1. 21. ^ Wawsh, D.; Carsweww, R. F.; Weymann, R. J. (31 May 1979). "0957 + 561 A, B: twin qwasistewwar objects or gravitationaw wens?". Nature. 279 (5712): 381–384. Bibcode:1979Natur.279..381W. doi:10.1038/279381a0. PMID 16068158. 22. ^ Friedman, Y.; Steiner, J. M. (2017). "Gravitationaw Defwection in Rewativistic Newtonian Dynamics". Europhysics Letters. 117 (5): 59001. arXiv:1705.06967. Bibcode:2017EL....11759001F. doi:10.1209/0295-5075/117/59001. 23. ^ "Hewping Hubbwe". www.spacetewescope.org. Retrieved 29 October 2018. 24. ^ Cassan, A.; Kubas, D.; Beauwieu, J.-P.; Dominik, M.; Horne, K.; Greenhiww, J.; Wambsganss, J.; Menzies, J.; Wiwwiams, A. (2012). "One or more bound pwanets per Miwky Way star from microwensing observations". Nature. 481 (7380): 167–169. arXiv:1202.0903. Bibcode:2012Natur.481..167C. doi:10.1038/nature10684. 25. ^ Cosmowogy: Weak gravitationaw wensing improves measurements of distant gawaxies 26. ^ Sci-News.com (21 Oct 2013). "Most Distant Gravitationaw Lens Discovered". Sci-News.com. Retrieved 22 October 2013. 27. ^ van der Wew, A.; et aw. (2013). "Discovery of a Quadrupwe Lens in CANDELS wif a Record Lens Redshift". Astrophysicaw Journaw Letters. 777 (1): L17. arXiv:1309.2826. Bibcode:2013ApJ...777L..17V. doi:10.1088/2041-8205/777/1/L17. 28. ^ Wong, K.; et aw. (2014). "Discovery of a Strong Lensing Gawaxy Embedded in a Cwuster at z = 1.62". Astrophysicaw Journaw Letters. 789 (2): L31. arXiv:1405.3661. Bibcode:2014ApJ...789L..31W. doi:10.1088/2041-8205/789/2/L31. 29. ^ NASA/Jet Propuwsion Laboratory (October 22, 2013). "Long-sought pattern of ancient wight detected". ScienceDaiwy. Retrieved October 23, 2013. 30. ^ Hanson, D.; et aw. (Sep 30, 2013). "Detection of B-Mode Powarization in de Cosmic Microwave Background wif Data from de Souf Powe Tewescope". Physicaw Review Letters. 14. 111 (14): 141301. arXiv:1307.5830. Bibcode:2013PhRvL.111n1301H. doi:10.1103/PhysRevLett.111.141301. 31. ^ Cwavin, Whitney; Jenkins, Ann; Viwward, Ray (7 January 2014). "NASA's Hubbwe and Spitzer Team up to Probe Faraway Gawaxies". NASA. Retrieved 8 January 2014. 32. ^ Chou, Fewecia; Weaver, Donna (16 October 2014). "RELEASE 14-283 - NASA's Hubbwe Finds Extremewy Distant Gawaxy drough Cosmic Magnifying Gwass". NASA. Retrieved 17 October 2014. 33. ^ Einstein, Awbert (1936). "Lens-Like Action of a Star by de Deviation of Light in de Gravitationaw Fiewd". Science. 84 (2188): 506–507. Bibcode:1936Sci....84..506E. doi:10.1126/science.84.2188.506. PMID 17769014. 34. ^ Eshweman, Von R. (1979). "Gravitationaw wens of de sun: its potentiaw for observations and communications over interstewwar distances," Science, 205 (4411): 1133–1135. 35. ^ Geoffrey A. Landis, "Mission to de Gravitationaw Focus of de Sun: A Criticaw Anawysis," ArXiv, paper 1604.06351, Corneww University, 21 Apr 2016 (downwoaded 30 Apriw 2016) 36. ^ Cwaudio Maccone (2009). Deep Space Fwight and Communications: Expwoiting de Sun as a Gravitationaw Lens. Springer. ISBN 9783540729433. 37. ^ Landis, Geoffrey A., “Mission to de Gravitationaw Focus of de Sun: A Criticaw Anawysis,” paper AIAA-2017-1679, AIAA Science and Technowogy Forum and Exposition 2017, Grapevine TX, January 9–13, 2017. Preprint at arXiv.org (accessed 24 December 2016). 38. ^ "Gawaxy cwuster MACS J2129-0741 and wensed gawaxy MACS2129-1". www.spacetewescope.org. Retrieved 23 June 2017. 39. ^ Kaiser, Nick; Sqwires, Gordon; Broadhurst, Tom (August 1995). "A Medod for Weak Lensing Observations". The Astrophysicaw Journaw. 449: 460–475. arXiv:astro-ph/9411005. Bibcode:1995ApJ...449..460K. doi:10.1086/176071. 40. ^ Luppino, G. A.; Kaiser, Nick (20 January 1997). "Detection of Weak Lensing by a Cwuster of Gawaxies at z = 0.83". The Astrophysicaw Journaw. 475 (1): 20–28. arXiv:astro-ph/9601194. Bibcode:1997ApJ...475...20L. doi:10.1086/303508. 41. ^ Babu, Gutti Jogesh; Feigewson, Eric D. (2007). Statisticaw Chawwenges in Modern Astronomy IV: Proceedings of a Conference Hewd at Pennsywvania State University, University Park, Pennsywvania, USA, 12–15 June 2006, Vowume 371 (iwwustrated ed.). Astronomicaw Society of de Pacific. p. 66. ISBN 978-1-58381-240-2. 42. ^ Pwionis, Manowis; López-Cruz, O.; Hughes, D. (2008). A Pan-Chromatic View of Cwusters of Gawaxies and de Large-Scawe Structure (iwwustrated ed.). Springer Science & Business Media. p. 233. ISBN 978-1-4020-6940-6. 43. ^ Frederic Courbin, Dante Minniti, Frederic Courbin, Dante Minniti (2008). Gravitationaw Lensing: An Astrophysicaw Toow (iwwustrated ed.). Springer. p. 69. ISBN 978-3-540-45857-9. 44. ^ "Hubbwe sees de brightest qwasar in de earwy Universe". www.spacetewescope.org. Retrieved 10 January 2019. 45. ^ "On de hunt for newborn stars". www.spacetewescope.org. Retrieved 15 October 2018. 46. ^ "Warped and distorted". www.spacetewescope.org. Retrieved 24 September 2018. 47. ^ "Stretched out image of distant gawaxy". www.spacetewescope.org. Retrieved 16 January 2018. 48. ^ "Cosmic snake pregnant wif stars". www.spacetewescope.org. Retrieved 20 November 2017. 49. ^ "Hubbwe captures gawwery of uwtra-bright gawaxies". www.spacetewescope.org. Retrieved 8 June 2017. 50. ^ "Detaiwed wook at a gravitationawwy wensed supernova". www.spacetewescope.org. Retrieved 21 Apriw 2017. 51. ^ Loff, Sarah; Dunbar, Brian (February 10, 2015). "Hubbwe Sees A Smiwing Lens". NASA. Retrieved February 10, 2015. 52. ^ "Most distant gravitationaw wens hewps weigh gawaxies". ESA/Hubbwe Press Rewease. Retrieved 18 October 2013. 53. ^ "ALMA Rewrites History of Universe's Stewwar Baby Boom". ESO. Retrieved 2 Apriw 2013. Bibwiography

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Company: IBM India Pvt Limited Hello, IBM came here at Hyderabad on 22th June 06. I was one among the selected students. Selection process consisted of a written test, a technical interview and an HR interview. Written Test was of 4 sections consisting of 55 questions in 1 hr. No negative marking. Section 1(Verbal Ability): Consisted of 10 questions. 5 questions were fill in the blanks. They were quite easy n were mainly based on prepositions n use of articles. 1) He was out of work ______ six months before he found a new job. (from/during/for/in). 2)The Chairperson usually presides _____ the meeting. (at/over/during/on). 3) The credibility of the game has been destroyed by the match-fixing scandals in cricket.(choose the word nearest in meaning). 4)They cost \$10 _____ kilo(the/a/an). 5)Black Monday was ____ most dramatic in a huge list of horrible days (a/an/the). Next Reading Passage was given and 5 questions were given based on the passage. Section 2(Analytical Ability): Consisted of 25 questions. Duration:30 min. On a particular day, students who visited a canteen ordered either a burger, a pastry or a sandwich. 19 students had a burger, 25 had a pastry, and 27 had a sandwich. 7 students had a burger and a pastry but not a sandwich, 9 had a pastry and sandwich but not a burger, 5 had a burger and sandwich but not a pastry. 3 students had a burger, a pastry and a sandwich. How many students had only a burger? How many students had only a pastry? How many students had only a sandwich? Introducing a man, a woman said, “He is the only son of my mother’s mother”. How is the woman related to the man? (ans: Niece). Some were on data sufficiency. They were also quite easy ones Some questions were on time and work. For this refer R.S.Agarwal Some questions were on the following pattern: In a binary system 1 is written as \$ and 0 is written as *. In this system as 1 moves to the left the value of the number doubles itself. (Clearly this was the problem of binary conversions. You just needed to put \$ n *s in place of 1s n 0s after doing proper conversions). Section 3(Attention to detail): 10 questions Very easy section. questions were: If + is replaced by *, – is replaced by /, * is replaced by +, / is replaced by * then calculate 2/65*3+65-23 and so on. (Just hold on patiently and calculate to check which of the given options is correct) A set of questions were based as: There were conditions given as to how a marketing executive would be selected and in the following questions individuals with their qualifications were given. Based on the previous conditions given, u have to decide whether each one can qualify or not. (Easy set again) Refer Verbal Reasoning of RS Agarwal Section 4(Technical): 10 ques Sum questions on C, database, unix etc. * A directed graph is also called _______. * In a queue the end at which insertions are performed is called the ____ end and the end from where deletions are done is called the ______ end. * The operator used to get value at address stored in a pointer variable is____. * char a[]=”have a nice day!” ; then p+=7 points to ____ * what permission would Chmod 755 yield on a file ? * The state of the file system is contained in a ____ block. Remaining questions I don’t remember. Technical Interview: Questions based on C, Database, Unix, DS, CN what is trigger in DBMS. what happens when we open a file in r+ mode like whether file will be created if it s not there can we write to file? which topology takes minimum wiring options: star,bus,ring and complete etc Ur favourite subject. I choosed JAVA. Oops concepts. They go in detail in the subject. Also prepare well ur core subjects, Programs. They ask u everything whatever u mention in ur resume. HR Interview: They ask u all general questions. Which websites do you browse frequently n y? N they ask u questions based on it. What is your latest movie? They ask us to give the gist of the story? What did u like in the movie? Etc.. Questions on ur strengths. How could you add value to IBM if you got this job? What do you know about IBM? Why would you want to join IBM?etc., How do you feel if I reject you??etc., (Paper Submitted By: Shailaja)

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Artificial Intelligence Questions and Answers – Local Search Problems and Optimization Problems This set of Artificial Intelligence Questions and Answers for Campus interviews focuses on “Local Search Problems and Optimization Problems”. 1. In many problems the path to goal is irrelevant, this class of problems can be solved using ____________ a) Informed Search Techniques b) Uninformed Search Techniques c) Local Search Techniques d) Informed & Uninformed Search Techniques Explanation: If the path to the goal does not matter, we might consider a different class of algorithms, ones that do not worry about paths at all. Local search algorithms operate using a single current state (rather than multiple paths) and generally move only to neighbors of that state. 2. Though local search algorithms are not systematic, key advantages would include __________ a) Less memory b) More time c) Finds a solution in large infinite space d) Less memory & Finds a solution in large infinite space Explanation: Two advantages: (1) they use very little memory-usually a constant amount; and (2) they can often find reasonable solutions in large or infinite (continuous) state spaces for which systematic algorithms are unsuitable. 3. A complete, local search algorithm always finds goal if one exists, an optimal algorithm always finds a global minimum/maximum. a) True b) False Explanation: An algorithm is complete if it finds a solution if exists and optimal if finds optimal goal (minimum or maximum). 4. _______________ Is an algorithm, a loop that continually moves in the direction of increasing value – that is uphill. a) Up-Hill Search b) Hill-Climbing c) Hill algorithm d) Reverse-Down-Hill search Explanation: Refer the definition of Hill-Climbing approach. 5. When will Hill-Climbing algorithm terminate? a) Stopping criterion met b) Global Min/Max is achieved c) No neighbor has higher value d) All of the mentioned Explanation: When no neighbor is having higher value, algorithm terminates fetching local min/max. Sanfoundry Certification Contest of the Month is Live. 100+ Subjects. Participate Now! 6. What are the main cons of hill-climbing search? a) Terminates at local optimum & Does not find optimum solution b) Terminates at global optimum & Does not find optimum solution c) Does not find optimum solution & Fail to find a solution d) Fail to find a solution Explanation: Algorithm terminates at local optimum values, hence fails to find optimum solution. 7. Stochastic hill climbing chooses at random from among the uphill moves; the probability of selection can vary with the steepness of the uphil1 move. a) True b) False Explanation: Refer to the definition of variants of hill-climbing search. 8. Hill climbing sometimes called ____________ because it grabs a good neighbor state without thinking ahead about where to go next. a) Needy local search b) Heuristic local search c) Greedy local search d) Optimal local search Explanation: None. 9. Hill-Climbing approach stuck for which of the following reasons? a) Local maxima b) Ridges c) Plateaux d) All of the mentioned Explanation: Local maxima: a local maximum is a peak that is higher than each of its neighboring states, but lower than the global maximum. Ridges: Ridges result in a sequence of local maxima that is very difficult for greedy algorithms to navigate. Plateaux: a plateau is an area of the state space landscape where the evaluation function is flat. 10. ___________ algorithm keeps track of k states rather than just one. a) Hill-Climbing search b) Local Beam search c) Stochastic hill-climbing search d) Random restart hill-climbing search Explanation: Refer to the definition of Local Beam Search algorithm. 11. A genetic algorithm (or GA) is a variant of stochastic beam search in which successor states are generated by combining two parent states, rather than by modifying a single state. a) True b) False Explanation: Stochastic beam search, analogous to stochastic hill climbing, helps to alleviate this problem. Instead of choosing the best k from the pool of candidate successors, stochastic beam search chooses k successors at random, with the probability of choosing a given successor being an increasing function of its value. 12. What are the two main features of Genetic Algorithm? a) Fitness function & Crossover techniques b) Crossover techniques & Random mutation c) Individuals among the population & Random mutation d) Random mutation & Fitness function Explanation: Fitness function helps choosing individuals from the population and Crossover techniques defines the offspring generated. 13. Searching using query on Internet is, use of ___________ type of agent. a) Offline agent b) Online agent c) Both Offline & Online agent d) Goal Based & Online agent Explanation: Refer to the definitions of both the type of agent. Sanfoundry Global Education & Learning Series – Artificial Intelligence. To practice all areas of Artificial Intelligence for Campus Interviews, here is complete set of 1000+ Multiple Choice Questions and Answers on Artificial Intelligence. If you find a mistake in question / option / answer, kindly take a screenshot and email to [email protected]

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## Is it possible to map value to color/color code using a bilinear color gradient for visualization? Path Finder (Disclaimer: May be a little nuts. Meaning you do not necessarily need to be sane to enjoy this question/answer) Many custom visualizations can work with a colorcode in the event data to further enrich a chart. Is it possible to map a numerical value to a color / colorcode? E.g. 42.4 -> #FFA500 Furthermore, is it possible to map a value range to a color range? E.g. 0.0-1.0 -> green-red? Labels (2) • ### other 1 Solution Path Finder A nice usage scenario for chart annotation (e.g. heatmap) is to merge two values (e.g. local riskscore, global riskscore) into a single color. Of course the same could be achieved in coming up with an overall score first and mapping this into a colorcode. But since splunk can do, let's take the more colorful approach of "mapping two values into a 2D color space" ("Bilinear Interpolate Color Gradient"). As referenced in slide 26 of conf19's: SEC1374 - Augment Your Security Monitoring Use Cases with Splunk's Machine Learning Toolkit (@time 41:04) Colorize Graph using two riskscores `bilinearInterpolateColorGradient(inputVal1, inputVal2, colorspaceCol00, colorspaceColX0, colorspaceCol0Y, colorspaceColXY, "outcolor")` Goal: - Input two values in the range of 0..1 - Map into a 2D colorspace defined by the four colorcodes in the corners - Output a single colorcode in the format "#AABBCC" Result: The two input values will get mapped into a 2D colorspace like this one: 2DColorspace at HarmonicCode: harmoniccode.blogspot.com Bilinear color interpolation at HarmonicCode: harmoniccode.blogspot.com Code: (4 Macros) ```````comment(" Maps two values into a color space with four colors. Outputs a final color, smoothly interpolated between these four color, on basis of the two input values. https://harmoniccode.blogspot.com/2011/04/bilinear-color-interpolation.html x: Input value 1 (x axis in color space. left->right) y: Input value 2 (y axis in color space. bottom->top) col00: color in the bottom left corner of the color space. colX0: color in the bottom right corner of the color space. col0Y: color in the top left corner of the color space. colXY: color in the top right corner of the color space. outcolorfieldname: name of field in which the final color is written. ")` | eval _x = \$x\$ | eval _y = \$y\$ | eval _col00 = \$col00\$ | eval _colX0 = \$colX0\$ | eval _col0Y = \$col0Y\$ | eval _colXY = \$colXY\$ `interpolateColor(_col00, _colX0, _x, "_outcolorB")` `interpolateColor(_col0Y, _colXY, _x, "_outcolorT")` `interpolateColor(_outcolorB, _outcolorT, _y, "_outcolor")` | eval \$outcolorfieldname\$ = _outcolor `````` interpolateColor(4) ``````| eval _col1 = \$col1\$ | eval _col2 = \$col2\$ | eval _lerpt = \$lerpt\$ | eval _outcolor = "#ZZZZZZ" | eval _lerpt = if(_lerpt > 1, 1.0, _lerpt) | eval _lerpt = if(_lerpt < 0.0, 0.0, _lerpt) `splitColorIntoComponents(_col1, "_col1r", "_col1g", "_col1b")` `splitColorIntoComponents(_col2, "_col2r", "_col2g", "_col2b")` | eval _deltaRed = _col2r - _col1r, _deltaGreen = _col2g - _col1g, _deltaBlue = _col2b - _col1b | eval _outRed = _col1r + _lerpt * _deltaRed, _outGreen = _col1g + _lerpt * _deltaGreen, _outBlue = _col1b + _lerpt * _deltaBlue `genColorFromComponents(_outRed, _outGreen, _outBlue, "_outcolor")` | eval \$outcolorfieldname\$ = _outcolor `````` splitColorIntoComponents(4) ``````| eval _color = \$color\$ | rex field=_color "^#(?<_red>[0-9a-zA-Z]{2,2})(?<_green>[0-9a-zA-Z]{2,2})(?<_blue>[0-9a-zA-Z]{2,2})\$" | eval _red = tonumber(_red, 16), _green = tonumber(_green, 16), _blue = tonumber(_blue, 16) | eval \$oufieldname_red\$ = _red | eval \$oufieldname_green\$ = _green | eval \$oufieldname_blue\$ = _blue `````` genColorFromComponents(4) ``````| eval _red = \$red\$ | eval _green = \$green\$ | eval _blue = \$blue\$ | eval _outcolor = "#" + upper(printf("%02x", _red)) + upper(printf("%02x", _green)) + upper(printf("%02x", _blue)) | eval \$outcolorfieldname\$ = _outcolor `````` Testbed / Usage example ``````| makeresults count=484 | eval count = 22 | streamstats count as id | eval id = id - 1 | eval x = floor(id / count) / count, y = (id % count) / count, z = 0 | eval col00 = "#00AA00" | eval colX0 = "#FFA500" | eval col0Y = "#FFFF00" | eval colXY = "#FF0000" `bilinearInterpolateColorGradient(x, y, col00, colX0, col0Y, colXY, "outcolor")` | eval clusterId=outcolor, clusterColor=outcolor, z=0 | table clusterId, x, y, z, clusterColor **-> Render the generated events into MLTK's 3D Scatterplot** `````` Path Finder A nice usage scenario for chart annotation (e.g. heatmap) is to merge two values (e.g. local riskscore, global riskscore) into a single color. Of course the same could be achieved in coming up with an overall score first and mapping this into a colorcode. But since splunk can do, let's take the more colorful approach of "mapping two values into a 2D color space" ("Bilinear Interpolate Color Gradient"). As referenced in slide 26 of conf19's: SEC1374 - Augment Your Security Monitoring Use Cases with Splunk's Machine Learning Toolkit (@time 41:04) Colorize Graph using two riskscores `bilinearInterpolateColorGradient(inputVal1, inputVal2, colorspaceCol00, colorspaceColX0, colorspaceCol0Y, colorspaceColXY, "outcolor")` Goal: - Input two values in the range of 0..1 - Map into a 2D colorspace defined by the four colorcodes in the corners - Output a single colorcode in the format "#AABBCC" Result: The two input values will get mapped into a 2D colorspace like this one: 2DColorspace at HarmonicCode: harmoniccode.blogspot.com Bilinear color interpolation at HarmonicCode: harmoniccode.blogspot.com Code: (4 Macros) ```````comment(" Maps two values into a color space with four colors. Outputs a final color, smoothly interpolated between these four color, on basis of the two input values. https://harmoniccode.blogspot.com/2011/04/bilinear-color-interpolation.html x: Input value 1 (x axis in color space. left->right) y: Input value 2 (y axis in color space. bottom->top) col00: color in the bottom left corner of the color space. colX0: color in the bottom right corner of the color space. col0Y: color in the top left corner of the color space. colXY: color in the top right corner of the color space. outcolorfieldname: name of field in which the final color is written. ")` | eval _x = \$x\$ | eval _y = \$y\$ | eval _col00 = \$col00\$ | eval _colX0 = \$colX0\$ | eval _col0Y = \$col0Y\$ | eval _colXY = \$colXY\$ `interpolateColor(_col00, _colX0, _x, "_outcolorB")` `interpolateColor(_col0Y, _colXY, _x, "_outcolorT")` `interpolateColor(_outcolorB, _outcolorT, _y, "_outcolor")` | eval \$outcolorfieldname\$ = _outcolor `````` interpolateColor(4) ``````| eval _col1 = \$col1\$ | eval _col2 = \$col2\$ | eval _lerpt = \$lerpt\$ | eval _outcolor = "#ZZZZZZ" | eval _lerpt = if(_lerpt > 1, 1.0, _lerpt) | eval _lerpt = if(_lerpt < 0.0, 0.0, _lerpt) `splitColorIntoComponents(_col1, "_col1r", "_col1g", "_col1b")` `splitColorIntoComponents(_col2, "_col2r", "_col2g", "_col2b")` | eval _deltaRed = _col2r - _col1r, _deltaGreen = _col2g - _col1g, _deltaBlue = _col2b - _col1b | eval _outRed = _col1r + _lerpt * _deltaRed, _outGreen = _col1g + _lerpt * _deltaGreen, _outBlue = _col1b + _lerpt * _deltaBlue `genColorFromComponents(_outRed, _outGreen, _outBlue, "_outcolor")` | eval \$outcolorfieldname\$ = _outcolor `````` splitColorIntoComponents(4) ``````| eval _color = \$color\$ | rex field=_color "^#(?<_red>[0-9a-zA-Z]{2,2})(?<_green>[0-9a-zA-Z]{2,2})(?<_blue>[0-9a-zA-Z]{2,2})\$" | eval _red = tonumber(_red, 16), _green = tonumber(_green, 16), _blue = tonumber(_blue, 16) | eval \$oufieldname_red\$ = _red | eval \$oufieldname_green\$ = _green | eval \$oufieldname_blue\$ = _blue `````` genColorFromComponents(4) ``````| eval _red = \$red\$ | eval _green = \$green\$ | eval _blue = \$blue\$ | eval _outcolor = "#" + upper(printf("%02x", _red)) + upper(printf("%02x", _green)) + upper(printf("%02x", _blue)) | eval \$outcolorfieldname\$ = _outcolor `````` Testbed / Usage example ``````| makeresults count=484 | eval count = 22 | streamstats count as id | eval id = id - 1 | eval x = floor(id / count) / count, y = (id % count) / count, z = 0 | eval col00 = "#00AA00" | eval colX0 = "#FFA500" | eval col0Y = "#FFFF00" | eval colXY = "#FF0000" `bilinearInterpolateColorGradient(x, y, col00, colX0, col0Y, colXY, "outcolor")` | eval clusterId=outcolor, clusterColor=outcolor, z=0 | table clusterId, x, y, z, clusterColor **-> Render the generated events into MLTK's 3D Scatterplot** `````` Get Updates on the Splunk Community! #### The Splunk Success Framework: Your Guide to Successful Splunk Implementations Splunk Lantern is a customer success center that provides advice from Splunk experts on valuable data ... #### Splunk Training for All: Meet Aspiring Cybersecurity Analyst, Marc Alicea Splunk Education believes in the value of training and certification in today’s rapidly-changing data-driven ... #### Investigate Security and Threat Detection with VirusTotal and Splunk Integration As security threats and their complexities surge, security analysts deal with increased challenges and ...

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# The reverse operation for the dot product? by cawthorne Tags: 3d engine, angles, dot product, field of view, rendering P: 7 Okay. First off. This isn't homework. I have been working on a personal project and if I could get an equation for this it would really help me out. I'll try and explain the problem the best I can. What I need: An equation/method where I can specify an angle theta and a 3D line and it returns the 3D coordinates of the points on the line which make the angle theta with the origin and the point (1, 0, 0). The problem: You have a 3D graph with origin O. The positional vector A with coordinates (1, 0, 0) and another position vector B. You could use the dot product function to work out the angle AOB, if you had B. But you don't. Dot Product function: cos(theta) = A.B/(|A|*|B|) You also know that B is somewhere on the 3D line r = (x, y, z) + lambda(a, b, c), where lambda is proportional to the displacement from (x, y, z) and (a, b, c) is the 3D gradient. This line can also be represented as a Cartesian line equation in the form x=y=z (but some other co-efficients etc.) e.g it could be 2x=5-y=(3z-2)/4. I want to specify an angle theta between BOA and work out position vector of A and the position vector of equation of the line B rests on. O will always have coordinates (0, 0, 0), but that doesn't mater as it's just a point of relevance. Also, it i worth mentioning that there may well be two solutions for B for a given angle theta if the line cuts through the angle twice, or one or no solutions. Actually come to think of it it could have infinite solutions if theta was 90 degrees (Point 90 degrees from the vector (1, 0, 0) are on the plane x = 0, because this is a flat plane, as a pose to a cone shape (the one value of theta where this happens), if the x coordinate of every point line r is also always equal to 0, then there will be infinite solutions). I like to think about the angle as a cone with the tip at the origin, (where every point on the surface of the cone is theta degrees away from the origin). Ramblings of a mad man: I have defined the function dotProduct (3DVector), which returns the angle AOB, when B is the parameter. The only things that I could work out are that for very large values of lambda in the line r the coordinate tends towards being a multiple of (a, b, c) (3D gradient), as the (x, y, z part gets infinitely small by comparison). And so the coordnate (a, b, c) can be treated as B if lambda is infinitely large. So we can do dotProduct((a, b, c)), which returns the same angle as though B where (a, b, c). Also I was thinking this: If you take the line r2 = (0, 0, 0) + lambda*(a, b, c). As in a line with the same gradient as r, but goes through the origin. Then the point on the line r, which is perpendicular to the line r2 through the origin point, is the closet to the origin the line r will get and if we treat that point as B then theta will be within 90 degrees +- the acute angle a point on the line r2 makes with A and O. And since r2 goes through the origin, there are only two angles any point on the line r2 can make with A and O (on acute one when the point is less than 90 degrees and one obtuse one it is over 90 degrees). The reason why I need this is because I have made a 3D engine, which takes into account perspective and that distortion effect you get when you take a panoramic photo with a large angle of view. To do the distortion effect I am just sampling the 3D line in 3D space very often and joining them with straight lines, but movement 3D space is only proportional to movement in the 2D rendering in very specific directions/positions relative to the camera. I have made it o it only samples as often as it need to (less when it is less distorted, more when it is more distorted etc.), but it would be a big help if I can figure out how to do this reverse dot product, because it will mean I can start rendering at the edge of a certain field of view, which will get rid of a lot of redundant computations. Thanks for any help guys. I'd really appreciate. Reagrds, Greg. P.S. - If this is the wrong section, please say. I will be happy to repost it elsewhere. << Mentor Note -- Use the Report button to ask the Mentors to move it >> Mentor P: 9,596 For each lambda, you have a known point r. You can calculate the angle α you get with that value of lambda, and solve the resulting equation (with the fixed angle) for lambda. You will get some special cases during the solution process, which correspond to "all lambdas are a solution" and similar things. P: 7 Quote by mfb For each lambda, you have a known point r. You can calculate the angle α you get with that value of lambda, and solve the resulting equation (with the fixed angle) for lambda. You will get some special cases during the solution process, which correspond to "all lambdas are a solution" and similar things. r = (x, y, z)+ lambda*(a, b, c) where x, y, z, a, b, c are know and lambda is known. let's say we want r when alpha = 80 degrees Okay so we have the equation alpha = acos(((x+ lambda*a, y + lambda*b, z + lambda*c).(1, 0, 0))/(((x+ lambda*a)^2 + (y+ lambda*b)^2 + (z + lambda*c)^2)^(0.5))*(1^2 + 0^2 + 0^2)^0.5)) (1^2 + 0^2 + 0^2)^0.5) = 1, so alpha = acos(((x+ lambda*a, y + lambda*b, z + lambda*c).(1, 0, 0))/(((x+ lambda*a)^2 + (y+ lambda*b)^2 + (z + lambda*c)^2)^(0.5))) since wee're using (1, 0 , 0), we only need to consider the x component of r for the dot product, so alpha = acos((x+ lambda*a)/(((x+ lambda*a)^2 + (y+ lambda*b)^2 + (z + lambda*c)^2)^(0.5))) Then I try to rearrange and get stuck cos^2(alpha) = (x^2 + 2*lambda*x*a + (lambda*a)^2)/((x+ lambda*a)^2 + (y+ lambda*b)^2 + (z + lambda*c)^2) If you can rearrange that for lambda I would be very grateful! My maths skills don't seem to be up to it :). P: 687 ## The reverse operation for the dot product? Hi, cawthorne, you are doing fine so far. You expanded the square in the numerator of the fraction, and eventually you will have to expand also the three squares in the denominator. This could be done now, or a bit later. The important next step is to notice that cos^2(alpha) is just a constant, because alpha is given. You can pass the whole denominator multiplying to the left side. Then distribute the constant cos^2(alpha), and then expand the squares as you did before. Finally you will be able to group the terms in lambda^2, in lambda and with no lambda, to find a solution of a quadratic equation. (You will see when you get there.) P: 7 Ahhh silly me :P. I did think it might have been a quadratic thing, but I got lost in all the extra terms (which I knew already!) :P. Thanks! P: 7 Quote by cawthorne Ahhh silly me :P. I did think it might have been a quadratic thing, but I got lost in all the extra terms (which I knew already!) :P. Thanks! By "I knew already" I meant that the extra terms where the input variables (constants for each given 3D vector), so I was silly to have been confused by them :P. Sometimes it's the easy parts which get you :P. Related Discussions Linear & Abstract Algebra 12 Calculus & Beyond Homework 2 Quantum Physics 2 Introductory Physics Homework 4 Calculus & Beyond Homework 1

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## Example of a conditional probability calculation Bill Harris has a fun little calculation of a conditional probability using three different data sources. Could be a good example for teaching intro probability or basic Bayesian inference. 1. Aleks says: I was quite impressed with Eliezer Yudkowsky's mini-course at http://yudkowsky.net/bayes/bayes.html 2. Andrew says: I'd prefer if Yudkowsky's page were called "Bayesian inference for binary inference." It's fine, but it doesn't capture most of what I see as applied Bayesian inference; see here. 3. Keith O'Rourke says: Posted something about this earlier re: Validation of Software by Cook et al. But if you “encode” the joint distribution from Yudkowsky's first example as a “data set” “R” code > datajoint 4. Bill says: Keith, Could you expand on that "Bayes theorem (a.k.a. Nearest Neighbors) " comment? That's a spin on Bayes theorem I've not seen before. Thanks Bill 5. Keith O'Rourke says: It’s just a rose by another name does not smell as sweet comment. Some would argue that Bayesian analysis simply involves a joint distribution of parameters θ (the unknowns) and observations x (random variables drawn from probability distribution with given unknown values of θ). With the joint distribution of (x, θ), a basic “axiom of inference” then says that probability statements about θ should be based on the conditional distribution of θ given the data observed xO, otherwise known as the posterior distribution of θ and here denoted by the posterior distribution π(θ | xO). This is a particular application of conditional probability in a two-stage system where we observe the outcome from the second stage (the data xO) and want to make statements about the concealed outcome from the first stage (the unknown parameters θ). This application is commonly referred to as Bayes theorem. This application of conditional probability could be viewed as the key step in what distinguishes a Bayesian analysis. For instance see Optimality and computations for relative surprise inferences. M. Evans, I.Guttman and T. Swartz, Canadian Journal of Statistics, Vol. 34, No. 1, 2006, 0 113-129. Now if the joint probability is specified exactly or approximately as a data set this conditioning step is the same as doing Nearest Neighbors. Now will this renaming make Bayes Theorem less mysterious for some? Would it be worthwhile to introduce students to Bayesian statistics with some minimalistic examples where joint distributions could be coded as data sets? (Now as Andrew pointed this is only one step of an applied Bayesian analysis) Keith

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# matlab display values on plot MATLAB or Matrix Laboratory is a programming language that was developed by MathWorks. Matlab supports plotting multiple lines on single 2D plane. Commented: Walter Roberson on 17 Apr 2020 Accepted Answer: Walter Roberson. How to display values on a plot: Vicky: 4/28/08 4:24 PM: hi, is there a way to display a value instead of circles for points on an x-y plot? How to display the minimum value from in plot?. If we want to put min(y) too (which means 2 values) what changes of this command we shall do? And best if it would display dynamically - after moving curosor the values would automatically show new pointed value. Show Hide all comments. Other MathWorks country sites are not optimized for visits from your location. Unable to complete the action because of changes made to the page. The column and row indices of Z are the x and y coordinates in the plane, respectively. When a programmer executes this program, MATLAB will display the plot as given above − We will show you one more example of the Matlab function plot y = x 2. Exsuse me, but in this example we want one value, the maxy. The axis equalcommand allows generating the plot with the same scale factors and the spaces on both axes. how can i plot them ? 11 answers. Learn more about error, output, matlab Asked 13th Dec, 2018; Usually RGB colors have values from 0 to 255. 0. To display it, you use. Here is one place you could read a bit about the practical aspects. At its simplest, an overlay can be implemented in terms of placing a non-visible axes above the visible one. However, if you want to increase the precision of the display in the text datatip, you can customize it. How to display the value of output on plot?. Unable to complete the action because of changes made to the page. hi, is there a way to display a value instead of circles for points on an x-y plot? https://www.mathworks.com/matlabcentral/answers/420177-display-value-on-a-plot#answer_337801, https://www.mathworks.com/matlabcentral/answers/420177-display-value-on-a-plot#answer_338117. Create a script file and type the following code − x = [0:5:100]; y = x; plot(x, y) When you run the file, MATLAB displays the following plot − Let us take one more example to plot the function y = x 2. You can modify any plot you create using either commands or the MATLAB GUI. An exploratory plot of your data enables you to identify discontinuities and potential outliers, as well as the regions of interest. ... 2/ I attempted to use "plot(ax,x,y)" ... but clearly in a bad way. Learn more about max, plot 1. 0 ⋮ Vote . I want to let the program calculate the x value when y=0.5 and indicate this point (x,0.5) on the plot with dashed lines connecting to the two axes. 2 Comments. Accelerating the pace of engineering and science. contourf(Z) creates a filled contour plot containing the isolines of matrix Z, where Z contains height values on the x-y plane.MATLAB ® automatically selects the contour lines to display. qqplot plots each data point in x using plus sign ('+') markers and draws two reference lines that represent the theoretical distribution. Learn more about app designer, bar plot In this post, I’ll examine the different overlays and how you might go about creating them in MATLAB. The value is stored as an on/off logical value of type matlab.lang.OnOffSwitchState. The height of the bar corresponds to the number of data points in the bin. Learn more about display value on a plot Representations of real-world objects (e.g., a motor vehicle or a section of the earth's topography) In the first case, it is generally desirable to select axis limits that provide good resolution in each axis direction and to fill the available space. Cari pekerjaan yang berkaitan dengan Display values plot matlab atau upah di pasaran bebas terbesar di dunia dengan pekerjaan 19 m +. Permalink. But also if i wana display the radius at the circle saying it is R=1.5 or any values how do i do it . Introduction to MATLAB Plot Function. Based on your location, we recommend that you select: . When you run the program, you will get a MATLAB graph along with the grid display. The grid oncommand allows you to put the grid lines on the graph. The generated plot gets assigned to a chart line object and its display gets customized by altering the attributes from the storing chart line object. MathWorks is the leading developer of mathematical computing software for engineers and scientists. You can also select a web site from the following list: Select the China site (in Chinese or English) for best site performance. I want to apply different data labels to each point in a scatter plot. How can I apply data labels to each point in a scatter plot in MATLAB 7.0.4 (R14SP2) ? Find the treasures in MATLAB Central and discover how the community can help you! In today's follow-up, I'll discuss how I computed the colors of the spectrum to display below the x-axis. Starting in R2018b, you can plot the variables of a table or timetable in a stacked plot. These are available to you in MATLAB Color Tools on the File Exchange and on … 4. This MATLAB function creates a 2-D line plot of the data in Y versus the corresponding values in X. 5. Starting in R2019b, you can display a tiling of plots using the tiledlayout and nexttile functions. This powerful language finds its utility in technical computing. Starting in R2019b, you can display a tiling of plots using the tiledlayout and nexttile functions. Also, you see the updates to the data properties of the plotted object update (such as XData). Find the treasures in MATLAB Central and discover how the community can help you! The procedure is as follows: Select a 2-D or 3-D axes from the New Subplots subpanel. There are pros and cons to this approach. (and the rest of the code appropriately), or add a separate. Follow 1,322 views (last 30 days) MathWorks Support Team on 27 Jun 2009. Basic Plot. You can use those numbers and divide the vector by 255 to use within MATLAB. Choose a web site to get translated content where available and see local events and offers. You may receive emails, depending on your. https://www.mathworks.com/matlabcentral/answers/157397-displaying-a-value-in-the-plot#answer_153941, https://www.mathworks.com/matlabcentral/answers/157397-displaying-a-value-in-the-plot#comment_753173, https://www.mathworks.com/matlabcentral/answers/157397-displaying-a-value-in-the-plot#comment_753182, https://www.mathworks.com/matlabcentral/answers/157397-displaying-a-value-in-the-plot#answer_153940. In a stacked plot, the variables are plotted in separate y-axes, but using a common x-axis. Other MathWorks country sites are not optimized for visits from your location. When we were rebuilding all of the graphics objects as MATLAB Objects, we found they they had several different variations of code that parsed this cell array case. However, the original workspace variables are not updated. Based on your location, we recommend that you select: . One can specify colors using a vector that gives the RGB triple where in MATLAB, each of the three values are numbers from 0 to 1. This example can draw two graphs of similar functions, but for the second time, it will decrease the increment’s value. y=(xˆ2+12x+24) for the value of x between 0 to 12. How to display values on a plot Showing 1-15 of 15 messages. Solution: The range of the x is 0 to 10. Could you help me ? Display complete axes of 3D plot in Matlab ? There are, however different scenarios for needing an overlay. And if you make a stacked plot from a timetable, the x-values are the row times. Ia percuma untuk mendaftar dan bida pada pekerjaan. Posted in group: comp.soft-sys.matlab: hi, is there a way to display a value instead of circles for points on an x-y plot? 3. num2str is fine (as would be sprintf) to use to format a value for display with additional text using the format string. Contents. Learn more about display value on a plot At some point, you’ll want to change the content of your plot in MATLAB. After you get the data looking just right, you might need to label certain items or perform other tasks to make the output look nicer. The MATLAB figure window displays plots. linehandle = plot( ___ ) returns a column vector of line series handles, using any of the arguments from previous syntaxes. Dear members. Display values in bar plot (app designer). as an simple example, suppose my data points are numbered: indic x y 1 1.2 0.66 Let us plot the simple function y = x for the range of values for x from 0 to 100, with an increment of 5. You can also use annotation object for an arrow pointing to the value and so on. In other words, just add 1 to your data before using loglog. Vote . Display value on a plot. Accelerating the pace of engineering and science. Toggle to show empty bins, specified as either 'off' or 'on', or as numeric or logical 1 (true) or 0 (false).A value of 'on' is equivalent to true, and 'off' is equivalent to false.Thus, you can use the value of this property as a logical value. I have program code in matlab with GUI. Choose a web site to get translated content where available and see local events and offers. After creating the axes, select it in the Plot Browser panel to enable the Add Data button at the bottom of the panel. Click the Add Data button to display the Add Data to Axes dialog. Display value on a plot. Display value on a plot. The column and row indices of Z are the x and y coordinates in the plane, respectively. I am not sure if I am the right track. Use the MATLAB ® Compiler SDK™ product to convert a MATLAB function (drawplot.m) to a method of a Java ® class (plotter) and wrap the class in a Java package (plotdemo).. Access the MATLAB function in a Java application (createplot.java) by instantiating the plotter class and using the MWArray class library to handle data conversion. Well, we don't know if you're "on the right track" or not, either since you didn't show what you tried nor indicate what the problem was, if any... ) to use to format a value for display with additional text using the format string. MATLAB: Workshop 15 - Linear Regression in MATLAB page 1 MATLAB Workshop 15 - Linear Regression in MATLAB Objectives: Learn how to obtain the coefficients of a “straight-line” fit to data, display the resulting equation as a line on the data plot, and display the equation and goodness-of … Note The Smith chart is commonly used to display the relationship between a reflection coefficient, typically given as S11 or S22, and a normalized impedance. We could also plot the above functions on different axes using the subplot() function in MATLAB. Numerical data displayed as line, mesh plots, or other specialized plot. MATLAB provides us with a convenient environment that can be used to integrate tasks like manipulations on matrix, plotting data and functions, implementing algorithms, creating user interfaces, etc. Plotting data is one of the most important tasks in Matlab programming. Learn more about plot, gui, matlab gui, guide, data Follow 1,274 views (last 30 days) MAHMOUD ALZIOUD on 18 Sep 2017. You can also select a web site from the following list: Select the China site (in Chinese or English) for best site performance. However, the high resolution MATLAB plot can be achieved through its inbuilt functions and resolution option (-r). The axis squarecommand generates a square plot. MathWorks is the leading developer of mathematical computing software for engineers and scientists. comp.soft-sys.matlab . You may receive emails, depending on your. Here is an example of a plot of a sine wave. Call the nexttile function to create the axes objects ax1 and ax2.Plot random data into each axes. To plot the variables of TT, use the stackedplot function. I will use and refer to several DIPUM3E functions. Now let’s take different examples to plot MATLAB graphs based on various mathematical functions. Problem 1: How to plot the MATLAB graph for the given equation in MATLAB? Please take note that as the value of increment decreases, the plotted graph will become smoother. The subplot() function is used to tell MATLAB how to split up the figure window and where to place the graph from each successive plot() command. Find the treasures in MATLAB Central and discover how the community can help you! It was a while ago now, but on April 27 I started explaining how I made this plot, which is from DIPUM3E (Digital Image Processing Using MATLAB, 3rd ed.):. This example can draw two graphs of similar functions, but for the second time, it will decrease the increment’s value. MATLAB: Displaying information about the data set by clicking on its plot. Problem 2: How to plot a Sin Function in MATLAB? 0. Posted by Doug Hull, May 30, 2012. A step size of h=1/16 is specified and Euler’s Method is used. Question. I have there some axes object called "axes1" - there my plot is displayed (simple plot of x, y values). The xlabel and ylabelcommands generate labels along x-axis and y-axis. James Tursa on 18 … We will approximate the solution to the D.E. Now I want to display this maximum value of x together with some text in a legend. Matlab extends its feature in 2D line plot to customize the plot presentation through the execution even after the plot is generated. labelpoints - matlab display values on plot Plotting volumetric data in MATLAB (1) I am working in Matlab and I have a 3d matrix with dimensions 384x512x160, which is made of 384x512 slices. Starting in R2019b, you can display a tiling of plots using the tiledlayout and nexttile functions. Example 1 . How to display values on a plot: Vicky: 4/28/08 4:24 PM: hi, is there a way to display a value instead of circles for points on an x-y plot? Call the tiledlayout function to create a 2-by-1 tiled chart layout. The function of y(x)= Sin(x) for 0 Next > How to label a series of points on a plot in MATLAB. display max ,min and a value of array in plot . display max ,min and a value of array in plot . The plotcommand can plot several sets of vectors. Matlab can generate multiple 2D line plots using the plot function within a loop. Then display grid lines in the bottom plot by passing ax2 to the grid function. plot( ___ ,Name,Value) provides additional options specified by one or more Name,Value pair arguments. The titlecommand allows you to put a title on the graph. qqplot(x) displays a quantile-quantile plot of the quantiles of the sample data x versus the theoretical quantile values from a normal distribution.If the distribution of x is normal, then the data plot appears linear. If you right click on a Data … The lines for data Y1, Y2,…,Yn with respect to their corresponding set of data X1, X2,.., Xn. I am plotting x vs y and I am taking out the maximum value of x with max(x). Then set the x-axis tick values for the lower plot by passing ax2 as the first input argument to the xticks function. y’= 1/y, y(0)=1. Solution: In the given equation, the range of the ‘x’ is 0 to 12. How to display values on a plot Showing 1-15 of 15 messages. Try to enter the following lines of code into your Matlab Command Window. I have a plot of several graphs organized by color. 2. The cell array of values is the same shape as the array of handles. The displayed plot shows your changes. thanks for that , i have fixed the axis and have define it is showing the line. how to plot a data points ? How to plot a figure with value on them as figure below. hello there, i have 15 (x) points named x1 to x15; and corresponding to this i have 15 (y) points named from y1 to y15. 0 ⋮ Vote. ', 'color',colors(i,:)) end. Answered: HAMZA NASSAR on 16 Sep 2020 Accepted Answer: MathWorks Support Team. Plotting Data Introduction. Call the tiledlayout function to create a 2-by-1 tiled chart layout. In general you can use plot(x1, y1, s1, x2, y2, s2, x3, y3, s3) where x1 and y1 are vectors of the same length and s1is an optional string. Vote. Call the tiledlayout function to create a 2-by-1 tiled chart layout. Create a plot. In this section, we will show you how to plot data, modify plots and save your work. Roberson on 17 Apr 2020 Accepted Answer: MathWorks Support Team on 27 Jun.! I 'll discuss how i computed the colors can be continuous or discrete by nature comment_753182... Have fixed the axis and have define it is Showing the line to reply ) Vicky 2008-04-28 UTC... There shows value of array in plot? time, it will decrease the ’! Mathematical computing software for engineers and scientists 2 values ) what changes of this command we shall?. Ax2.Plot random data into each axes too old to reply ) Vicky 2008-04-28 23:24:02.! Display below the x-axis tick values for the colors can be continuous or discrete by nature its feature 2D. On both axes y and i am the right track one of the panel the plotted object (! # answer_153940 add a separate and ax2.Plot data into each axes for from. Lower plot by passing ax2 as the array of values is the same scale and... Bit about the data is one of the ‘ x ’ is 0 to 12,. Difference between the range of the bar corresponds to the number of data type.. Practical aspects the lidar scan readings specified in scanObj x-values are the x is 0 to 255 can. The results plotted a way to display values on a plot of a table timetable... The lower plot by passing ax2 as the first input argument to the data set by clicking its. And resolution option ( -r ) or change how the plot command is used the! With some text in a stacked plot from a timetable, the maxy and. The x-values are the row times it would display dynamically - after moving curosor values... # answer_338117 that, i 'll discuss matlab display values on plot i computed the colors can be achieved through its inbuilt functions resolution! What changes of this command we shall do for needing an overlay a title on the graph button to it! //Www.Mathworks.Com/Matlabcentral/Answers/420177-Display-Value-On-A-Plot # answer_338117 column and row indices of Z are the row times plot data, modify and... The program, you can display a tiling of plots using the tiledlayout function to create axes. That point values how do i do it the original workspace variables are not updated 27 Jun 2009 for an. Of a table or timetable in a stacked plot, the plotted graph will become.... Updated with time the bin values ) what changes of this command we shall do increment matlab display values on plot the! Inbuilt functions and resolution option ( -r ) the execution even after the plot presentation through the execution even the... Allows you to put the grid display colors of the panel triples for the given equation, high... A value instead of circles for points on an x-y plot? this post, i have a Showing! Pointing to the xticks function of x with max ( x ) of output on plot? you could a! Plot function within a loop x together with some text in a stacked from... Plot of several graphs organized by color lidar scan readings specified in scanObj section, we recommend you... The practical aspects x ) between the range of the x and y coordinates in the.. Previous syntaxes 10 ) plot function within a loop a 2-by-1 tiled chart layout spaces on axes! And save your work on them as figure below achieved through its inbuilt functions and resolution option -r! Central and discover how the data set by clicking on its plot get translated content where and. So that we can compare the approximation with the same shape as the first input argument to grid... Similar functions, but for the colors of the ‘ x ’ is 0 to 10 ):! To use `` plot ( scanObj ) plots the lidar scan readings specified scanObj... Plots the lidar scan readings specified in scanObj vector of line series handles using! A programming matlab display values on plot that was developed by MathWorks values is the same shape the! Is generated the updates to the number of data points in the plane, respectively the. A scatter plot in MATLAB commands or the MATLAB RGB triples for lower. Datatip, you can also use annotation object for an arrow pointing to the page made to page. Select: above the visible one james Tursa on 18 … Histograms are plots that show the of! More about max, min and a value of type matlab.lang.OnOffSwitchState put min ( y ) ''... clearly! An example of a table or timetable in a scatter plot ax2.Plot random data into axes! Title on the graph circles for points on an x-y plot? placing a axes. To get translated content where available and see local events and offers axes dialog the xlabel and generate. Which are updated with time within a loop i attempted to use a log ( x+1 ) rather than log! The updates to the page axes, select it in the plane, respectively smithplot data! Can also use annotation object for an arrow pointing to the page this powerful language finds its utility technical! Display value on a plot ( too old to reply ) Vicky 2008-04-28 23:24:02 UTC potential outliers, well. Here is an example of a plot of your data before using matlab display values on plot sine wave solution is specified Euler. Display dynamically - after moving curosor the values would automatically show New pointed value from timetable! Through the execution even after the plot with the grid display feature in 2D plot. A Sin function in MATLAB programming as follows: select a 2-D or 3-D axes from New! Matlab command Window we recommend that you select: button to display it, you modify! Your data enables you to identify discontinuities and potential outliers, as well as the first input argument to value! Use text you can customize it high resolution MATLAB plot can be implemented in terms of placing non-visible! Not updated data on a plot a step size of h=1/16 is so... Bottom plot by passing ax2 as the value and so on knowing the MATLAB graph for the value x! On 18 … Histograms are plots that show the distribution of data points in the plot Browser provides the by. Increase the precision of the bar corresponds to the number of data generate labels along and... Can generate multiple 2D line plots using the tiledlayout function to create the objects. Scenarios for needing an overlay values in bar plot ( too old to reply ) Vicky 23:24:02... Old to reply ) Vicky 2008-04-28 23:24:02 UTC x ) for 0 < x < 10 to your before... Modify any plot you create using either commands or the MATLAB graph along with the true.... Based on your location, we recommend that you select: you use text you can a. - after moving curosor the values would automatically show New pointed value to. # answer_153940 in bar plot display max, plot When you run the,. Plot Browser panel to enable the add data to axes a way to display this maximum value x... Original workspace variables are plotted in separate y-axes, but using a common x-axis that point resolution option ( )! The program, you use text you can display a tiling of plots using the tiledlayout and nexttile.... Them in MATLAB argument to the page 0 < x < 10 in. The radius at the bottom of the code appropriately ), or add a legend or how... Just add 1 to your data enables you to identify discontinuities and potential outliers, well... # answer_338117 ( xˆ2+12x+24 ) for 0 < x < 10 function MATLAB. Value ) provides additional options specified by one or more Name, value pair arguments )! Is as follows: select a 2-D or 3-D axes from the New subpanel. Xdata ) designer, bar plot display max, min and a value x. 1/Y, y ) too ( which means 2 values ) what changes of this command we do! Are plots that show the distribution of data vs y and i am x. With max ( x ) transformation to display the data set by clicking on its plot designer bar... The spaces on both axes on the graph have a plot note that as the value stored! Rather than a log ( x ) the regions of interest this we... < 10 them as figure below multiple lines on single 2D plane there are, however scenarios... Sites are not updated the page the graph precision of the plotted will... Just add 1 to your data before using loglog values would automatically show New value! R=1.5 or any values how do i do it MATLAB Central and discover how the data one! … Histograms are plots that show the distribution of data will become smoother ALZIOUD 18!, plot When you run the program, you use text you can display a tiling plots. Labels along x-axis and y-axis MATLAB programming the first input argument to the page a. Plot? of TT, use the stackedplot function create a 2-by-1 chart... = plot ( ___, Name, value ) provides additional options specified by one or more Name value., or add a legend for an arrow pointing to the xticks function ) provides additional options specified by or... X vs y and i am the right track range of the spectrum to display this maximum value x! Also plot the MATLAB gui by clicking on its plot a title the! Display values in bar plot display max, plot When you run the program, you can display a of. Of y ( x ) a stacked plot, the x-values are the x and coordinates... The maxy of code into your MATLAB command Window and refer to several functions!

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# Conservation of energy, hypothetical situation Gold Member ## Main Question or Discussion Point I'm not sure the question belongs to classical physics, I apologize in case it's a no. Imagine a train or any object following a straight line motion. Suppose there's no friction between the ground and the train, such that its motion would go on forever. Now suppose that the train emits photons from its backside, parallel to the ground. So its "losing" energy and should go slower I believe. However I remember my professor said that even though photons are massless particles at rest, they carry a momentum. Hence... I'm tempted to say that the train should in fact go faster instead of going slower. I'm sure I'm confusing a lot of things. Can you explain me what would really happen in such an idealized situation? Related Classical Physics News on Phys.org Dale Mentor The train would go faster. This is the same principle as a rocket. Nabeshin Conservation of momentum. When the train emits a photon, it must recoil an equal amount. For conservation of energy, the energy that led to the creation of the photon was almost certainly not the kinetic energy of the train! Perhaps it was a battery connected to a laser or some such mechanism, but definitely not the train itself. So this line of argument is ineffective. Gold Member Ok thanks for the replies. Is there a way to be sure the photon wasn't emitted due to a loss of kinetic energy? What would happen in such a case? Is it even possible? I'm not even sure it makes sense to talk about a loss of KE since the train would gain some KE from the momentum of the photon... and hence it looks like possible for the train to keep a stable velocity and emitting photons, which is impossible. So it'd lose KE after all I guess, even though it emits photons, which also looks like an impossibility. Dale Mentor Actually, this depends on the frame of reference you are using. Let's say that we are generating photons by annihilating positrons and electrons, and that each time we are sending one photon straight forward and one straight behind. In the rest frame the momentum of the forward and backward photons are equal, so the train stays at rest and therefore the KE of the train is unchanged. Now, imagine from a reference frame in which the train is moving. The forward photon is blueshifted so it carries more momentum forward than the backward photon carries. So by conservation of momentum the trains momentum must decrease, but the train's velocity cannot change, so the momentum decreases due to the reduction in the mass of the train, therefore the KE is also reduced in this frame for the same reason. Gold Member Actually, this depends on the frame of reference you are using. Let's say that we are generating photons by annihilating positrons and electrons, and that each time we are sending one photon straight forward and one straight behind. In the rest frame the momentum of the forward and backward photons are equal, so the train stays at rest and therefore the KE of the train is unchanged. Now, imagine from a reference frame in which the train is moving. The forward photon is blueshifted so it carries more momentum forward than the backward photon carries. So by conservation of momentum the trains momentum must decrease, but the train's velocity cannot change, so the momentum decreases due to the reduction in the mass of the train, therefore the KE is also reduced in this frame for the same reason. Thanks a lot. Yeah during the day I had in mind that the mass of the train has to decrease... unfortunately my physics knowledge is too little to make any equation saying what I'm thinking regarding this problem. I get the idea, that's what matter. Problem solved!

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# Math posted by . Find all values of the missing digit that make the statement true 214,21_ is divisible by 11 • Math - A number divisible by 11 has the property that the sum of even digits minus the sum of odd digits is 0 or divisible by 11. 14641 is divisible by 11 because (4+4)-(1+6+1)=0 So if the missing digit is x, and if the number 21321x is divisible by 11, then (1+2+x)-(2+3+1)=0 solve for x. • Math - I am not following. The number you placed x in you used the number 21321x, but the number is 21421x. I am not following your steps to solve for x either • Math-corr. - Sorry, it was a typo. My apologies. For 21421X So if the missing digit is x, and if the number 21421x is divisible by 11, then (1+2+x)-(2+4+1)=0 solve for x. • Math - yo ## Similar Questions 2. ### Math- Find all values of the missing digit that make the statement true. 5,33_ is divisible by 11. A) 8 B) 6 C) 1 D) 5 3. ### math Find the value of the missing digit that makes the statement true. 214,21_ is divisible by 11 Determine whether the statement is true or false. Why? 4. ### math divisibility i need to write the missing digit to make each number divisible by 3 1,843,#89 i got the missing number as : 3 s/b 1,843,389 is that right? 5. ### math Find digits A and B in the number below so the folling condition are true. The 5-digit number must be divisible by 4. The 5-digit number must be divisible by 9. Digit A cannot be the same as Digit B. 12A3B Explain the steps you followed … 6. ### math Using the numbers 1 through 9 with no repeats, find a 9 number such that: the first digit is divisible by 1, the first two digits are divisible by 2, the first 3 digits are divisible by 3, and so on until we get to a 9 digit number … 7. ### Algebra ASAP so this is a fill in on a worksheet and I am having difficulty as the ones I inserted are incorrect can anybody help me how to do it all, sorry it's a long problem. Show that 3^2n − 1 is divisible by 8 for all natural numbers … 8. ### Math Looking to find the missing digit for each statement true. The number 874_ is divisible by 4. Please help how to solve 9. ### number theory The values A,B,C, and [(A/B)+C], are all integers which are divisible by 3. Then, which of the following statements must be true: A is divisible by 9 B is divisible by 9 C is divisible by 9 A and B are both divisible by 9 A,B,and C … 10. ### maths The values A,B,C, and [(A/B)+C], are all integers which are divisible by 3. Then, which of the following statements must be true: A is divisible by 9 B is divisible by 9 C is divisible by 9 A and B are both divisible by 9 A,B,and C … More Similar Questions

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### The Time-Temperature Graph We are going to heat a container that has 72.0 grams of ice (no liquid water yet!) in it. To make the illustration simple, please consider that 100% of the heat applied goes into the water. There is no loss of heat into heating the container and no heat is lost to the air. Let us suppose the ice starts at −10.0 °C and that the pressure is always one atmosphere. We will end the example with steam at 120.0 °C. There are five major steps to discuss in turn before this problem is completely solved. Here they are: 1) the ice rises in temperature from −10.0 to 0.00 °C. 2) the ice melts at 0.00 °C. 3) the liquid water then rises in temperature from zero to 100.0 °C. 4) the liquid water then boils at 100.0 °C. 5) the steam then rises in temperature from 100.0 to 120.0 °C Each one of these steps will have a calculation associated with it. WARNING: many homework and test questions can be written which use less than the five steps. For example, suppose the water in the problem above started at 10.0 °C. Then, only steps 3, 4, and 5 would be required for solution. To the right is the type of graph which is typically used to show this process over time. The ChemTeam hopes that you can figure out that the five numbered sections on the graph relate to the five numbered parts of the list just above the graph. Also, note that numbers 2 and 4 are phases changes: solid to liquid in #2 and liquid to gas in #4. Here are some symbols that will be used, A LOT!! 1) Δt = the change in temperature from start to finish in degrees Celsius (°C) 2) m = mass of substance in grams 3) Cp = the specific heat. Its unit is Joules per gram-degree Celsius (J / g °C is one way to write the unit; J g¯1 °C¯1 is another) 4) q = the amount of heat involved, measured in Joules or kilojoules (symbols = J and kJ) 5) mol = moles of substance. 6) ΔHfus is the symbol for the molar heat of fusion and ΔHvap is the symbol for the molar heat of vaporization. We will also require the molar mass of the substance. In this example it is water, so the molar mass is 18.0 g/mol. By the way, the p means the specific heat is measured at constant pressure; there is a related specific heat we will not discuss (yet) which is measured at constant volume. Not too suprisingly (I hope), it has the symbol Cv. Step One: solid ice rises in temperature As we apply heat, the ice will rise in temperature until it arrives at its normal melting point of zero Celsius. Once it arrives at zero, the Δt equals 10.0 °C. Here is an important point: THE ICE HAS NOT MELTED YET. At the end of this step we have SOLID ice at zero degrees. It has not melted yet. That's an important point. Each gram of water requires a constant amount of energy to go up each degree Celsius. This amount of energy is called specific heat and has the symbol Cp. Step Two: solid ice melts Now, we continue to add energy and the ice begins to melt. However, the temperature DOES NOT CHANGE. It remains at zero during the time the ice melts. Each mole of water will require a constant amount of energy to melt. That amount is named the molar heat of fusion and its symbol is ΔHfus. The molar heat of fusion is the energy required to melt one mole of a substance at its normal melting point. One mole of solid water, one mole of solid benzene, one mole of solid lead. It does not matter. Each substance has its own value. During this time, the energy is being used to overcome water molecules' attraction for each other, destroying the three-dimensional structure of the ice. The unit for this is kJ/mol. Sometimes you see older references that use kcal/mol. The conversion between calories and Joules is 4.184 J = 1.000 cal. Sometimes you also see this number expressed "per gram" rather than "per mole." For example, water's molar heat of fusion is 6.02 kJ/mol. Expressed per gram, it is 334.16 J/g. Notice how I shifted to Joules instead of kilojoules. This was done to keep the number within the ranges of ones to hundreds. Writing the value using kJ would require I write 0.33416. It is more understandable to write 334.16. Typically, the term "heat of fusion" is used with the "per gram" value. Step Three: liquid water rises in temperature Once the ice is totally melted, the temperature can now begin to rise again. It continues to go up until it reaches its normal boiling point of 100.0 °C. Since the temperature went from zero to 100, the Δt is 100. Here is an important point: THE LIQUID HAS NOT BOILED YET. At the end of this step we have liquid water at 100 degrees. It has not turned to steam yet. Each gram of water requires a constant amount of energy to go up each degree Celsius. This amount of energy is called specific heat and has the symbol Cp. There will be a different value needed, depending on the substance being in the solid, liquid or gas phase. Step Four: liquid water boils Now, we continue to add energy and the water begins to boil. However, the temperature DOES NOT CHANGE. It remains at 100 during the time the water boils. Each mole of water will require a constant amount of energy to boil. That amount is named the molar heat of vaporization and its symbol is ΔHvap. The molar heat of vaporization is the energy required to boil one mole of a substance at its normal boiling point. One mole of liquid water, one mole of liquid benzene, one mole of liquid lead. It does not matter. Each substance has its own value. During this time, the energy is being used to overcome water molecules' attraction for each other, allowing them to move from close together (liquid) to quite far apart (the gas state). The unit for this is kJ/mol. Sometimes you see older references that use kcal/mol. The conversion between calories and Joules is 4.184 J = 1.000 cal. Sometimes you also see this number expressed "per gram" rather than "per mole." For example, water's molar heat of vaporization is 40.7 kJ/mol. Expressed per gram, it is 2259 J/g or 2.259 kJ/g. By the way, this value can vary, depending on the molar heat value used and the molar mass of water used. For example: 40670 J/mol / 18.0 g/mol = 2257.56 J/g You have been warned! Typically, the term "heat of vaporization" is used with the "per gram" value. Step Five: steam rises in temperature Once the water is completely changed to steam, the temperature can now begin to rise again. It continues to go up until we stop adding energy. In this case, let the temperature rise to 120 °C. Since the temperature went from 100 to 120, the Δt is 20. Each gram of water requires a constant amount of energy to go up each degree Celsius. This amount of energy is called specific heat and has the symbol Cp. There will be a different value needed, depending on the substance being in the solid, liquid or gas phase.

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# Transposition/re-arrangement of equation Im having issues with my weak algebraic skills trying to correctly re-arrange the following equation for "d" 110=(24/(pi*d^2))+((3.5/(pi*d^4/64))*(d/2) i know that d=70 . I have no idea how to attack this transposition. Could someone please give me a few pointers or point me to a video that will help? tiny-tim Homework Helper hi russjai! (try using the X2 button just above the Reply box ) 110=(24/(π*d2))+((3.5/(π*d4/64))*(d/2) if i'm reading it right, that's of the form A = B/d2 + C/d3 you can multiply throughout by d3 to make it Ad3 - Bd - C = 0 there's no easy way to solve that, and the solution certainly won't be a whole number

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# Arithmo Rules - Missing ### The basics Arithmo–Missing is very similar to all the other Arithmo games in that it uses the basic math functions: addition, subtraction, multiplication, and division. Several people can play at the same time. The person to solve their puzzle first wins! In Arithmo–Missing you’re given a grid size, 2x2, 2x3, or 3x3, filled with numbers.. The object is to connect 3 adjacent numbers into equations to make a product from 13 to 24. We call these products Equa-13, Equa-14, Equa-15, and so on. Equa-15, for example, simply means "any three adjacent connected numbers on the grid that together can be made to equal 15." Remember that in the Arithmo world, "3 Connected Numbers" always means any 3 adjacent numbers on the grid that you can directly connect using a straight vertical, horizontal, or "L" shaped line. Once you connect 3 numbers, you add, subtract, multiply, or divide them together in any order or combination to make a product. Each grid in the Senior version of Missing! allows you to make connections and equations to come up with the various Equa-13 to Equa-24 products. However, you’ll find that in each Arithmo–Missing! puzzle, you won’t be able to make one or more of the Equa equations. Your mission is to figure out which equation you can’t make! ### Rules of the Game On your number filled grid, it is possible to make equations and connections of 3 adjacent numbers and make them equal a number from 13 to 24. We call these products Equa-13 to Equa-24. However, one or more of the Equas are missing. Your goal is to figure out which products you cannot make. Since you don’t know which ones are missing, you’ll have to make connections and find all of the Equas that do exist, until you can point out the ones that don’t. It’s simple: let’s use a basic example. You have a grid with numbers in it. You start forming connections of 3 adjacent numbers and make them equal a number from 13 to 24 using the four basic math operations however you want. But you’ll run into a problem: you’ll eventually realize that no matter what you do, you can’t make any of the connections and equations you come up with equal 15 (or any other number from 13 to 24). You might have found 13,14,16,17, and the others, but you can’t make an equation that equals 15! That means that Equa-15 is the missing equation, and you found it! Good work! Again, when we say "connections", "Equas", or "equations", we mean 3 adjacent numbers on the grid you can directly connect in a straight horizontal, vertical, or "L" shaped line and add, subtract, multiply, or divide together in any order or combination to equal from 13 to 24. ### Example Your Task: Find which Equas, from 13 to 24, are missing on the grid. 11 4 5 8 8 5 4 8 1 As you can see, we can make the following equations using the following connections: Equa-13 11+(8:4)=13, Equa-14 8+5+1=14, Equa-15 8+8-1=15, Equa-16 8x8:4=16, Equa-17 8+8+1=17, Equa-18 5+5+8=18, Equa-20 4+8+8=20, Equa-21 8+8+5=21, Equa-22 11x8:4=22, Equa-23 11+8+4=23, Equa-24 8x(8-5)=24 You can see that no matter how hard you try, one equation simply cannot be made no matter what math you do. Equa-19 can not be found on this grid. No matter which three adjacent numbers you connect, you will never be able to make them equal 19. The solution for this Arithmo–Missing game is: One Equa is missing, and that is Equa-19. That wasn’t too bad, right? Go find more missing Equas, and enjoy! ### Skill Levels Arithmo–Missing tasks are divided into 3 skill levels: Easy, Medium, Difficult.

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# 7 times a positive number n is decreased by 3, it is less than 25 Discussion in 'Calculator Requests' started by math_celebrity, May 31, 2021. Tags: 7 times a positive number n is decreased by 3, it is less than 25 7 times a positive number n: 7n Decreased by 3: 7n - 3 The phrase it is less than means an inequality. So we relate 7n - 3 less than 25 using the < sign to get our algebraic expression of: 7n - 3 < 25

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Blog # The Quest for 700: Weekly GMAT Challenge Each week Manhattan GMAT posts a GMAT question on our blog and follows up with the answer the next day. Are you up for the challenge? A function f(x) is defined as idempotent if, for every possible x, f(f(x)) = f(x). Which of the following functions is NOT idempotent? (A) f(x) = x (B) f(x) = –x (C) f(x) = |x| (D) f(x) = the greatest integer less than or equal to x (E) f(x) = the least integer greater than or equal to x ### Upcoming Events • Duke Fuqua (Round 3) • Ohio Fisher (Round 3) • Vanderbilt Owen (Round 3) • USC Marshall (Round 3) • Carnegie Mellon Tepper (Round 3) • Toronto Rotman (Round 3) • Cambridge Judge (Round 4) • UW Foster (Round 3) • Notre Dame Mendoza (Round 3) • Emory Goizueta (Round 3) • Oxford Saïd (Round 3) • IESE (Round 3) • Dartmouth Tuck (Round 3) • London Business School (Round 3) • Texas McCombs (Round 3) • Vanderbilt Owen (Round 4) • Berkeley Haas (Round 4)

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## Logical Operators [print-me] Logical operators typically involve boolean values and  yields boolean results. There are often use to create multiple or complex conditions. We have learned that a  boolean value can either be true or false. We can use the different logical operators available in C#. Read more ## Comparison Operators [print-me] C# has comparison operators which are used when comparing values or references of variables. These operators yield boolean results. Values yielded by comparison expressions are either true if the comparison is, in fact, true, and false if the comparison is false. These comparison operators are Read more ## Assignment Operators [print-me] Assignment Operators are shortcuts for modifying the value of a variable and immediately assigning the result to the variable itself. It is used to modify the value of the left operand depending on the right operand. The table shows assignment Read more ## C# Math C# Math Operators has available operators for use in mathematical calculations. By combining operators and operands, we have an expression which is how C# makes calculations. Here is a table of available C# Math Operators. C# Math Operators Operator Category Read more ## Expressions and Operators [print-me] C# operator :-Modern programming languages have a set C# operator which is one of the common elements of a programming language. C# offers mathematical operators, assignment operators, and comparison operators. There are different kinds of operators in C# and Read more ## Conversion Using the Convert Class [print-me] Convert Class – The .NET Framework has a static class available that can be used for converting values from one type to another. The System.Convertclass has some methods that can be used to convert to different target datatypes. You must Read more ## Explicit Conversion [print-me] Explicit conversion lets you force the program to convert a data into another data type if it doesn’t support implicit conversion. The explicit conversion has a tendency of losing or resulting to a modification of value so be cautious when Read more ## Implicit Conversion [print-me] An implicit conversion of variables is a kind of conversion that is automatically done by the compiler. A variable of one data type can be implicitly converted to another data type provided that its content can be handled by the type Read more ## Constant Variables [print-me] Constant variables, or simply constants, are variables whose value cannot be changed once initialized. Given that fact, constants can only do initialization since forgetting to initialize a constant will produce an error. Constants must be assigned with an initial Read more ## Using Variables [print-me] Variables in C# are containers inside the computer’s memory where you can place the data you need for your program. The term “variable” got its name from the fact that variables represent values that change. The data stored in these Read more ## C# Data Types [print-me] C# Data Types are predefined datatypes in C# used for storing the most basic type of data. This includes datatypes for storing numbers, characters, strings, and boolean values. They are called primitive data types in other languages because you Read more ## Variables [print-me] A variables is basically a storage in a computer’s memory where you can put a value that will be used by your program. Picture it as a container that holds your data. The contents of the container can be removed or Read more ## String Verbatim [print-me] Verbatim String allows you to ignore escape sequences and makes writing strings more natural and readable. When using escape sequences in string literals, you sometimes make a mistake of typing \\ for the “backslash” and type \ instead. This will produce an error Read more ## Escape Sequences [print-me] Escape sequences are character combinations starting with a backslash (\) and followed by a letter or digits inserted in a string literal to produce a modified output. For example, to produce a newline when outputting a string, we can use Read more ## C# Comments [print-me] When writing code, you might want to add some text that will serve as a reminder or a note for you or for anyone who will read your code. In C# (and most programming languages), you can do that Read more ## C# Errors [print-me] Most of the time, C# errors are encountered when creating a program. Almost all of the programs you see today suffered at least one error. Errors can be devastating if not fixed immediately. In any programming language, there are Read more ## IntelliSense [print-me] Perhaps one in all the foremost necessary options of Visual Studio is that the IntelliSense. IntelliSense is AN autocompletion tool additionally as a fast reference for categories, methods, and plenty of a lot of. I keep in mind victimisation Read more ## Simple Program in C# [print-me] Let’s produce our 1st program mistreatment the C# artificial language. The program can print a line of text during a console window. Note that we are going to be mistreatment Console Application because the project sort for the subsequent Read more ## Changing the Layout of Visual Studio [print-me] Visual Studio layout If you don’t like the position of the windows or the layout of the IDE, you have the ability to arrange them. By clicking and dragging the title bar of each window, you can make them Read more ## Exploring Visual Studio [print-me] overview of visual studio IDE Visual Studio has numerous windows and components, each providing functionality or displays certain information. Let’s further explore the different parts of Visual Studio by going deeper into the IDE. Create a new Windows Forms Read more ## Welcome to Visual Studio [print-me] We will now explore the important parts of the visual studio. After Visual Studio starts up, you will be welcomed by the following screenshot.   The numbered labels will be used to identify the different parts of the IDE. Read more ## Downloading and Installing Visual Studio [print-me] In this tutorial, I will show you how to download and install Visual Studio Community 2015. Downloading Visual Studio Community The Visual Studio Community 2015 is available as a free download. You can download the installer file in the Read more ## Visual Studio [print-me] Visual Studio is the all-in-one development environment for developing software using Microsoft Technologies. If you will be programming in C#, it is recommended to install and use Visual Studio. There are multiple versions of Visual Studio, both paid and free Read more ## .NET Framework [print-me] The .NET Framework is a platform created by Microsoft used for developing different kinds of applications for the Microsoft Windows platform. The .NET Framework can also be used for developing web applications, and even mobile applications. Several versions of the .NET Read more ## introduction to c# programming C# (pronounced as “c-sharp”) is an object-oriented general-purpose programming language created by Microsoft. C sharp can be used to build windows applications, web applications, web services, mobile applications, games, and many more. It is a language that even beginners and students Read more

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# Filling a scene with cubes I am trying to design a solution for simulating gaseous fluids in confined areas, I am using naiver-stokes equations for fluid dynamics but since it only covers incompressible fluids like liquid, i needed to find a way to simulate compression Recently, I came up with the solution which is, creating a field storing pressure data for each volumetric partition and I thought it would be a good idea to construct this field as a composition of cubes where each cube has homogeneous pressure value within itself. I tried to inflate a cube from a point and create other cubes from its surfaces recursively minding the collision with surfaces inside that scene, but this algorithm fails because there are lots of detailed cases involving complex holes and notches on the object. Is there a better way to construct this field? • Are you programming on paper or are you using specific tools / languages? By the way, this question belongs on SO (after a revision). – Job Dec 12 '11 at 19:16 • Paper oriented mostly, although i need to make a demo. Development environment is rather trivial. I just want some algorithms to solve the problem. SO is mostly about specific problems on specific prog. languages and tools. – Hgeg Dec 12 '11 at 20:47 • What's the paper you are reading and what's the name of the algorithm? – PhD Dec 13 '11 at 1:13 • My work is derived from the paper "Physically Based Modeling and Animation of Fire",Fedkiw et al, 2001. However there aren't any algorithms i took directly. That paper and most of the others use a fixed grids or voxels for pressure fields. However in my case these voxels need to be constructed according to the objects in the scene. – Hgeg Dec 13 '11 at 10:47 First off, the Navier-Stokes Equations don't assume incompressibility in general. You can simplify the equations if you assume incompressibility, but it's not necessary. Second, I have to ask: Do you have a really good reason not to be using off-the-shelf computational fluid dynamics software? Finite element analysis in general is not for the faint of heart. To address your actual question: Dividing the fluid volume into cubes should be straightforward. Personally, I would just divide the entire volume into cubes, then find the ones which are in the fluid. You can speed things up with a proper recursive fill (as suggested by Tydus), but you can also just check every single cube, if you don't mind trading a bit of CPU heat for some typing. Another aside, addressing your proposed solution: Assuming homogeneity within the cube (element) is an odd choice. It would be much more typical to use first order interpolation. If you're hoping to use a left-hand approximation to try to solve the differential equation for each element independently, beware: numerical instability looms, and your results may well be totally wrong. (In general, you should be thinking 'these results are probably wrong' whenever using numerical methods, but this probably is particularly probable). Typically, for this sort of problem, you need to assemble the full system of equations and solve simultaneously. (Like I said, not for the faint of heart). • The thing is, I am not trying to fit cubes into fluid volume. The system that contains the cube simulates the pressure field of the air, so that the gaseous fluid particles tend to move to the cube with the least pressure. Calculating the pressure value and particle velocity inside each cube, as you said, is straight forward. My question is to fill all the scene with cubes. As I said I couldn't find a solution that works for all(or at least most) of the cases. – Hgeg Dec 13 '11 at 10:54 • "I am not trying to fit cubes into fluid volume" and "My question is to fill all the scene with cubes." seem to be a direct contradiction. Please explain. "Calculating the pressure value and particle velocity inside each cube, as you said, is straight forward." Only if you have deep knowledge of the fundamentals of CFD - if you don't, be careful. – ipeet Dec 17 '11 at 3:21 A recursive "fill" should work for any shape no matter how complex. Unless a single cube is to big to fit in the hole. What are the cases where it fails? • I actually use some kind of recursive fill. The problem mostly caused by the fact that there is no specified minimum size of holes or shapes. For example, when inflation of a cube stops because of collision with a surface, next recursive calls are made on the each surface of the cube as you expected. However, the question is, how can I decide the initial position of the new cube on the surface of the old cube? I need to trace the holes by iterating over the surface, but the resolution is infinite, since there is no minimum limit for hole size. – Hgeg Dec 13 '11 at 11:01 • On collisions you can use smaller and smaller cubes. Filling in smaller and smaller holes. Like in calculus you get closer and closer to a limit, at smaller and smaller intervals. If the size of the cube is important in some way you could express the pressure relative to the cubes size to keep things proportional. It will not be easy if the shape is random and irregular because there will be all sorts of different curves. I might go for an imperfect solution where you just start off with "small" cubes and hope it gives a answer close to the truth. – Lord Tydus Dec 17 '11 at 4:00 I've seen an implementation of fluids using cellular automata, you could representate them with the minimum size that's meaningful to you. Cubes, spheres, whatever. The shape of the container shouldn't really matter with this method, just add the rules that generate new cells to "fill" complex shapes.

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Re: [newbie] from right to left • To: mathgroup at smc.vnet.net • Subject: [mg48424] Re: [newbie] from right to left • From: Mark Fisher <mark at markfisher.net> • Date: Sat, 29 May 2004 03:06:45 -0400 (EDT) • References: <c96i2u\$ika\[email protected]> • Sender: owner-wri-mathgroup at wolfram.com ```Here's one way: str = StringJoin @@ (ToString /@ {a, b, a, j, a, n, o, s, c, a, d, a, b, a, j, a, n, o, s, c, a, d}) First /@ (StringPosition @@ {str, "janos"}) First /@ ((StringLength[str] + 1) - ( StringPosition @@ (StringReverse /@ {str, "janos"}))) --Mark. János wrote: > Hi, > > Let's say I have a list of letter like symbols: > > lst = {a, b, a, j, a, n, o, s, c, a, d,a, b, a, j, a, n, o, s, c, a, d} > > and have a pattern > > pat = {j, a, n, o, s} > > If I want to match by going from left to right > > Position[Partition[lst, Length[pat], 1], pat] > > gives me positions {{4}, {15}}. So far so good. Now the genie came > out from the bottle and reversed the list, so now list reads like: > > revlist = Reverse[lst] > > revlist = {d, a, c, s, o, n, a, j, a, b, a, d, a, c, s, o, n, a, j, a, > b, a} > > and the genie say that now I have to read it from right to left and > find the pattern that way. I could now reverse the pattern > > revpat = Reverse[pat] > > revpat = {s, o, n, a, j} > > and match it up to the reversed list like: > > Position[Partition[revlst, Length[revpat], 1], revpat] > > {{4}, {15}} > > again, but that would not do me any good, because what I want is to > match it from right to left, so the position should be: > > {{8},{19}} > > that is, where "j" is found in revlist by reading from right to left. > Adding Length[pat] to the undesired result of course will give me the > "good" positions, but that looks somewhat "artificial". > > I can program it out procedurally, but that is not "nice" and slow. > > So, in summary, at odd occasions I read lst in western style and I can > find the pattern easily, but at even occasions, when my eyes are > reading the reversed lst from right to left, I have read it in arabic > style and I want to find the pattern that way too. > > Any good hint ? > > > János > ------------------------------------------------- > clear perl code is better than unclear awk code; but NOTHING comes > close to unclear perl code > http://www.faqs.org/faqs/computer-lang/awk/faq/ > ``` • Prev by Date: Re: reviving mathematica afte update to SuSE Linux 9.1

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It is currently 21 Oct 2017, 02:12 ### GMAT Club Daily Prep #### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email. Customized for You we will pick new questions that match your level based on your Timer History Track every week, we’ll send you an estimated GMAT score based on your performance Practice Pays we will pick new questions that match your level based on your Timer History # Events & Promotions ###### Events & Promotions in June Open Detailed Calendar Author Message Director Joined: 29 Nov 2012 Posts: 868 Kudos [?]: 1412 [0], given: 543 ### Show Tags 12 May 2013, 01:08 The pricing is quite different if you buy the book it would cost around 25 dollars and if you purchase the tests it costs around 49 dollars.. _________________ Click +1 Kudos if my post helped... Amazing Free video explanation for all Quant questions from OG 13 and much more http://www.gmatquantum.com/og13th/ GMAT Prep software What if scenarios http://gmatclub.com/forum/gmat-prep-software-analysis-and-what-if-scenarios-146146.html Kudos [?]: 1412 [0], given: 543 Intern Joined: 03 Oct 2013 Posts: 18 Kudos [?]: 7 [0], given: 38 Location: India Concentration: General Management, Strategy GMAT 1: 730 Q50 V39 ### Show Tags 03 Oct 2014, 12:11 fozzzy wrote: The pricing is quite different if you buy the book it would cost around 25 dollars and if you purchase the tests it costs around 49 dollars.. Can someone please help with this? I am facing a similar problem,if I buy a book it will cost me around 11\$ but buying access is 49\$.Is there any difference b/w the two sets of tests?? Kudos [?]: 7 [0], given: 38 Display posts from previous: Sort by Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.

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Extracting nonorthogonal sources in ICA/PCA/blind source seperation problem My problem is essentially a 'blind source separation' problem. I have 3 non-orthogonal sources (or basis functions) and N random linear combinations (mixes) of said sources. My problem is to obtain the sources from the mixes. Figure A shows the sources, B shows the mixes. Approaches taken: 1) PCA- I tried PCA on the mixes (fig.C), but the issue is that PCA will only give orthogonal bases, while my sources/bases are non-orthogonal. This issue with PCA is shown in figure D, where the data is clearly described by 2 non-orthogonal basis, but PCA (the solid lines) cant reconstruct them! 2) Factor Rotation - I tried applying some solutions form Factor Analysis (not my forte). The promax rotation (matlab: nw = rotatefactors(cov,'method','promax') ) is shown in fig. D with the dashed lines. As far as I can tell, factor rotation works with the principal components, not the original matrix and thus I have no idea how it could reconstruct the right basis. I think this only works with factor matrices, not with generic ones... 3) ICA - I tried overdetermined ICA with the fastICA algorithm (also not my forte, but I think I understand it better). I was hoping this would work since independent components (ICs) are non orthogonal. The solution is shown in fig.E. While the ICs are indeed nonorthogonal, they are NOT my original sources :-( Any other potential tips or leads or solutions would be greatly appreciated. Thanks!

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+ 1 What is the output of this code? print(int(1+2-3*4/5%6)) 17th Aug 2018, 8:30 AM + 2 Why 0? Why (3*4/5%6) is 3? 17th Aug 2018, 8:51 AM + 2 I think, it goes like this: 3 * 4 is 12; 12 / 5 is 2.4; 2.4 % 6 is 2.4; 1 + 2 - 2.4 is 0.6; int(0.6) is 0. 17th Aug 2018, 12:20 PM HonFu + 1 1 17th Aug 2018, 8:36 AM Dlite + 1 I don't understand. Why is 3*4/5%6=3? 17th Aug 2018, 9:35 AM

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IsabellLiebe 2016-09-26 07:37:21 If an object is propelled from ground level, the maximum height that it reaches is given by h = v2 sin2 θ , where θ is the angle between the ground 2g and the initial path of the object, v is the object’s initial velocity, and g is the acceleration due to gravity, 9.8 meters per second squared.

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# How to calculate class limits in statistics. Class Interval Arithmetic Mean Calculator, Definition, Formula & Calculation 2019-01-11 How to calculate class limits in statistics Rating: 5,7/10 293 reviews ## How to Calculate the Relative Frequency of a Class Class Width The difference between the upper or lower class limits of consecutive classes is the class width. This means that all data values must be included. The boundaries have one more decimal place than the raw data and therefore do not appear in the data. Mode: find the largest frequency - the corresponding value is the modal value or modal class. The class interval often called as units or classes or bins which are used to lump observations or data set in order to reduce the amount of data to make it easier to analyze or visualize Definition Class Interval Arithmetic Mean is the distribution represented by relative frequency counts or proportions of observations within different class intervals. Next ## statistics Sometimes you're locked into a certain number of classes because of the instructions. The researcher decides to choose 1 percent of the gas stations in New York and 1 percent of the gas stations in Connecticut for the sample. Class Interval Arithmetic Mean Calculator is an online statistics tool for data analysis programmed to represent a collection of variable data from a sample by lumping together into more manageable class intervals. Ungrouped Frequency Distribution A frequency distribution of numerical data. Histogram A graph which displays the data by using vertical bars of various heights to represent frequencies. This formula is applicable if we have to plot a frequency distribution, , relative frequency, ogive etc. Next ## How to Calculate the Relative Frequency of a Class This turns out to be 800 in New York and 200 in Connecticut. A simple calculation for the mean is to add all the integers and divide by the number of total integers in the sample. It includes examining both the process of obtaining data and understanding the data. Use the below online Class width calculator to calculate the Class Width Frequency Distribution. In this example, subtract the shortest player's height from the tallest player's height. Next ## statistics To find the lower, you subtract 0. By A frequency distribution shows the number of elements in a data set that belong to each class. All classes should have the same class width and it is equal to the difference between the lower limits of the first two classes. But it really depends on what you want to do. Use Discretion Kiran Gaunle is a freelancer based in New York. Then, let us find the class width of the given data using 5 classes. Next ## Class Interval Definition If you mean on a computer, go to the Start Menu, then to the Programs menu and then to the Accessories menu. For the supermarket example, the total number of observations is 200. A set with an upper bound is said to be bounded from above by that bound, a set with a lower bound is said to be bounded from below by that bound. Class Width The difference between the upper and lower boundaries of any class. However, this is not necessarily always true, and the above definition is sufficient. A statistic may not be meaningful and it can also mislead one to wrong conclusions. Next ## How to Calculate the Relative Frequency of a Class Use the below online Class width calculator to calculate the Class Width Frequency Distribution. The set of ideas which is intended to offer the way for making scientific implication from such resulting summarized data. Example of a class is a set of students who are alumni with same year of graduation. They are obtained by averaging the limits. The mean is the average in the data set. The vertical axis the cumulative frequency or relative cumulative frequency. Grouped data has been 'classified' and thus some level of data analysis has taken place, which means that the data is no longer raw. Next ## Class Width Calculator These are the indirect means by which statists promote andimplement their ideas. The grouping can be done differently with different class intervals also. It should of course be borne in mind that uncertainty does not imply ignorance but it refers to the incompleteness and the instability of data available. Age years Frequency Class Interval 0 - 9 15 10 10 - 19 18 10 20 - 29 17 10 30 - 49 35 20 50 - 79 20 30 Calculating Class Interval Given a set of raw or ungrouped data, how would you group that data into suitable classes that are easy to work with and at the same time meaningful? I have taken many course where we just did textbook projects. The set of data may also be termed a sample of a population. Median: calculate a running total of the frequencies - the first interval that is above half the total contains the median. Researchers transform information from qualitative studies into categories that researchers then count as frequencies. Next ## Calculate Frequency Distribution in Excel These classes can be divided based on different parameters. The raw data is not grouped. To learn more, see our. Class Limits Separate one class in a grouped frequency distribution from another. With this online Class Arithmetic Mean calculator you can effortlessly make your calculation for any data set Similar Resource. For example, it may be a frequency distribution of the heights of major league basketball players. Next ## Statistics: Frequency Distributions Graphs Construct the table after collecting heights for each member of the sample population i. As it embodies more of less all stages of the general process of learning, sometimes called scientific method , statistics is characterized as a science. Provide details and share your research! Class boundaries give the true class interval, and similar to class limits, are also divided into lower and upper class boundaries. But it is more informative to partition the range into quantiles check wikipedia. The first step is to determine how many classes you want to have. A statist knows that he cannot get away with-at least, notyet-openly declaring that your lif … e, money and property are notyours, so he advances these ideas by implicit means. Next ## Class Width Calculator Lengths have been measured to the nearest millimeter Find the mean of the data. Any data set accumulated for statistical purposes, such as the U. Diameter mm 35 — 39 40 — 44 45 — 49 50 — 54 55 — 60 Frequency 6 12 15 10 7 Solution: Step 1: Find the midpoint of each interval. Every time a poll is conducted or a study done and numbers reported, statistics are at work. It is not an object oriented language thus all data types are primitive types. In the table above, for the first class, 1 is the lower class limit while 3 is the upper class limit. Next

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# Function Intuitively, a function $$f$$ is a procedure (or machine) that takes an input and performs some operations to produce an output. For example, the function “+” takes a pair of numbers as input and produces their sum as output: on input (3, 6) it produces 9 as output, on input (2, 18) it produces 20 as output, and so on. Formally, in mathematics, a function $$f$$ is a relationship between a set $$X$$ of inputs and a set $$Y$$ of outputs, which relates each input to exactly one output. For example, $$-$$ is a function that relates the pair $$(4, 3)$$ to $$1,$$ and $$(19, 2)$$ to $$17,$$ and so on. In this case, the input set is all possible pairs of numbers, and the output set is numbers. We write $$f : X \to Y$$ (and say “$f$ has the type $$X$$ to $$Y$$”) to denote that $$f$$ is some function that relates inputs from the set $$X$$ to outputs from the set $$Y$$. For example, $$- : (\mathbb N \times \mathbb N) \to \mathbb N,$$ which is read “subtraction is a function from natural number-pairs to natural numbers.” $$X$$ is called the domain of $$f.$$ $$Y$$ is called the codomain of $$f$$. We can visualize a function as a mapping between domain and codomain that takes every element of the domain to exactly one element of the codomain, as in the image below. Talk about how they’re a pretty dang fundamental concept. Talk about how we can think of functions as mechanisms. Add pages on set theoretic and type theoretic formalizations. Talk about their relationships to programming. Talk about generalizations including partial functions, multifunctions, etc. Give some history, e.g. Church-Turing, Ackerman, recursion, etc. (Many of these todos should go into separate lenses; this definition here might be fine?) # Examples There is a function $$f : \mathbb{R} \to \mathbb{R}$$ from the real numbers to the real numbers which sends every real number to its square; symbolically, we can write $$f(x) = x^2$$. (TODO) Children: Parents: • Mathematics Mathematics is the study of numbers and other ideal objects that can be described by axioms. • Oog, I wrote a page about functions at /​p/​3vp because it didn’t occur to me that this page would be named differently… • Sorry :-p /​p/​3vp seemed too general a name for me back when I was first writing this, though I think /​p/​3vp is probably actually just fine. I suggest you merge the two and do what you want with them. (Right now I like my intro better than yours, but I might be biased :-p)

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# What does 0...0 and 1...1 mean I am very new to cryptography but can someone explain what exactly does 0...0 and 1...1 mean in general term. Is it that 1st and last bit can be 0,0 and 1,1 respectively and in-between bits can be anything $(0,1)$? In case it helps… I stumbled upon this while trying to solve the following multiple choice question: Let F be a block cipher with n-bit block length. Consider the message authentication code for 2n-bit messages defined by Mack(m1,m2)=Fk(m1⊕m2). Which of the following gives a valid attack on this scheme? • Obtain tag t on message m,0…0 (with m≠0…0), and then output the tag t on the message 0…0,0…0. • Obtain tag t on message m1,m2 (with m1≠m2), and then output the tag t on the message m2,m1. • Obtain tag t on message m,0…,0, and then output the tag t⊕(1…1) on the message m,1…1. • Obtain tag t on message m,m, and then output the tag 0…0 on the message 0…0,m. Please note that I do not want an answer to the multiple choice question. I merely want to know how to interpret the notation 0...0 and 1...1`. What exactly does $0...0$ and $1...1$ mean usually? This simply means a (more or less) long string of $0$s or $1$s or more clearly $000000...000000$ and $111111...111111$. Related notiational notes, you may have to use soon: Sometimes the notation $0^n$ and $1^n$ is also used for these strings with exactly $n$ zeroes and ones. Even more generally it's usually expected that people see the pattern, namely that nothing changes (or else there would be a note about this), for example $\lambda_0,...,\lambda_n$ is expected to be filled with $\lambda_i$ with $i\in \mathbb N_0, 0\leq i\leq n$ or relatedly $(\alpha,...,\alpha)\in \mathbb R^n, \alpha \in \mathbb R$ would be the same principle, that you'd have to fill the blanks with as many $\alpha$s as needed. One last note: If you'd want to indicate that you want a leading string $0$ and a trailing string $0$ and stuff you don't care about in the middle you'd usually notate it as $n\in\{0,1\}^*$ and $0||n||0$ which means "place an arbitrary string of arbitrary length between the zeroes". • Before anybody asks: No, there's no special reason why I used examples from standard math / linear algebra in the fourth paragraph. May 21, 2016 at 18:56

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Pacmath Lesson Plan • July 2022 • PDF This document was uploaded by user and they confirmed that they have the permission to share it. If you are author or own the copyright of this book, please report to us by using this DMCA report form. More details • Words: 1,079 • Pages: 4 PACMATH Lindsay Medley First Grade / Mathematics Common Core Standards:      Add and subtract two-digit numbers with and without regrouping. Represent commonly used fractions using words and physical models Model, represent and explain subtraction as comparison, take-away and part-to-whole. Represent data using objects, picture graphs and bar graphs Sort and classify objects by attributes, and organize data into categories in a simple table or chart. Lesson Summary: This lesson plan is to help student’s identity number by adding and subtracting while having fun using the arrows on the keyboard to play PacMath. Also to understand word problems. Estimated Duration: Estimated time would be about 25 minutes. I plan on making this lesson plan go on for at least 4 days. I would also like the students to take their time to write down their final score each day. Commentary: What is being challenged is the children have to answer the question correctly before being about be play PacMath. Students will get hooked on this game because they aren’t just playing PacMath the entire time. They have to answer the question first before moving on to the game. Instructional Procedures: The instructions for this lesson are answer the question, either a simple adding or subtracting question or a word problem, when the number of needed answers are answered correctly then you get to play a round of PacMath. The object of the game is to collect all the dots and try not to get hit by the floating ghost. If you collect the fruits while collecting the dots the ghost all turn blue and you can then go after them and get extra points. If a ghost gets you then you die and you go back to answering more questions. To move your PacMath man you must press the arrows up, down, left, or right on keyboard. Day 1: First 10 minutes: Give instructions and demonstrate the game. I will also give students a recording sheet to record their scores. Students will sit patiently and watch. 20 minutes: I will release students to a computer and walk around to see if anyone needs help. Students will play game quietly. Last 5 minutes: Students will stop playing game and write down their high score for the day. I will help students log off computers and make sure they write down scores. Day 2: First 10 minutes: I will give instructions of what the students will be doing today. 20 minutes: I will release students to a computer and walk around to see if anyone needs help. Students will play game quietly. Last 5 minutes: Students will stop playing game and write down their high score for the day. I will help students log off computers and make sure they write down scores. Day 3: First 10 minutes: I will give instructions of what the students will be doing today. 20 minutes: I will release students to a computer and walk around to see if anyone needs help. Students will play game quietly. Last 5 minutes: Students will stop playing game and write down their high score for the day. I will help students log off computers and make sure they write down scores. Day 4: First 10 minutes: I will give instructions of what the students will be doing today. 15 minutes: I will release students to a computer and walk around to see if anyone needs help. Students will play game quietly. Last 20 minutes: Students will stop playing game and write down their high score for the day. I will help students log off computers and make sure they write down scores. Then once all scores are entered I will collect the sheets and create a graph on the computer to see which day had the highest score all together. This graph will help me see how the students processed and grew within the past days. Use link http://nces.ed.gov/nceskids/graphing/classic/bar_pie_data.asp?ChartType=bar for bar graph. Pre-Assessment: A Pre-Assessment I would use a adding and subtracting worksheet pre-assessment. I would use it for a morning warm up before starting class just to get their brains working and awake. Scoring Guidelines: I would collect the pre-assessment, but I wouldn’t grade them, I would just look over them and see if they are understand the material being learned. If they did well I’d put a sticker on the sheet. Post-Assessment: For a post-assessment, I will give my students a quiz over all the material learned over the last few days. Scoring Guidelines: The guidelines for the quiz will be graded with a number score. It will be too see how well they listened during the week lesson. Differentiated Instructional Support How the instructions might be changed for gifted of accelerated students is maybe they can have headphones in and have the question read to them as the play the game. I could also have them buddy up so the kids that need extra help have their friends and also me. I might be diffacult for students because they will be using their fine motor skills to press the arrows on the keyboard to move PacMath man around. Discuss additional activities you could do to meet the needs of students who might be struggling with the material: Additional activities I could do for students with needs is, have all the questions that will be asked on the game written on paper and then they just answer on the paper and then create their own game to play once they have completed some questions. Extension http://www.abcya.com/first_grade_computers.htm This website has many different math games for first graders to help understand better. Homework Options and Home Connections There wouldn’t be any real homework assigned. I would just encourage parents to work with their kids with adding and subtracting and help them understand word problems better. Interdisciplinary Connections How this lesson can be integrated with other content areas is Language Arts and Reading. It can be integrated because with the word problems the students have to understand what is being asked in the problem therefore they have to read the question. Also students have to understand the purpose of the prompt and it has to make sense to them on what it is saying. Materials and Resources: For teachers Computer, Paper, Link to game, Link to graph, Projector Screen, pencil, Students scores, copy of questions being asked during game. For students Computer, recording sheet, Pencil, Link to game Key Vocabulary Word Problems, Graph, Recording Related Documents July 2022 0 Lesson Two Lesson Plan November 2019 177 Lesson Presentation Lesson Plan December 2019 145 June 2020 4 April 2020 37 Lesson Plan December 2019 102 More Documents from "Lindsay" October 2019 115 December 2019 76 November 2021 0 Reader S Digest All Time Favourite Songbook November 2019 229 January 2022 0 July 2022 0

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# about VCO output imdedance and Buffer - RF Cafe Forums RF Cafe Forums closed its virtual doors in 2010 mainly due to other social media platforms dominating public commenting venues. RF Cafe Forums began sometime around August of 2003 and was quite well-attended for many years. By 2010, Facebook and Twitter were overwhelmingly dominating online personal interaction, and RF Cafe Forums activity dropped off precipitously. Regardless, there are still lots of great posts in the archive that ware worth looking at. Below are the old forum threads, including responses to the original posts. NOTICE: The original RF Cafe Forum is available again for reading, and the new RF Cafe Blog is an active board. -- Antennas -- Systems Tommy Post subject: about VCO output imdedance and Buffer Posted: Mon Jul 30, 2007 5:25 am Lieutenant Joined: Sun Jul 29, 2007 10:30 pm Posts: 1 i have the question about VCO output impedance... 1. do not we really know VCO output impedance?? 2. in order to get enough output voltage swing, we use buffer in VCO output port. as i know, because we do not the output imdepance of VCO, the output swing should be check for the enough power for Mixer input power requirement by using buffer... 3. Actually...what is the exact usage of the buffer connected to VCO output port...moreover is there any method to check output impedance of VCO? 4. After buffer, the output impedance is usally 50 ohm..is that correct??? plz tell me anything wrong and things i asked... i hope i can get ur answer.. tommy Top IR Post subject: Posted: Mon Jul 30, 2007 9:51 am Joined: Mon Jun 27, 2005 2:02 pm Posts: 373 Location: Germany Hello Tommy, 1. The output impedance can be characterized by a parameter called pulling, which defines how much the frequency will change at a given load (Usually that gives -12dB return loss at all phases). I never saw that an output impedance of the VCO is measured directly. 2. The buffer is used for better isolation and to increase the output power of the VCO in order to be able to drive the LO port of the Mixer. 3. The buffer is an amplifier of which output impedance should be matched to 50 ohm. A good practice is to choose a buffer which is capable to an output power of at least 3dB more than the LO power and to insert a pad between the LO port and the buffer output to set the correct output power and to improve the matching between the LO port and the buffer output. 4. Yes, the buffer's impedance should be identical to the system characteristic impedance, which is usually 50 ohm. Top nubbage Post subject: Posted: Tue Jul 31, 2007 9:46 am General Joined: Fri Feb 17, 2006 12:07 pm Posts: 218 Location: London UK Hi Tommy To add a little to IR's point no.2: In Analyser-speak, the s-parameter S12 of a buffer amplifier is usually very low, and can be made even lower if negative feedback is used. Thus any reflected power from a load on the buffer does not find its way, out of phase, to the VCO and result in frequency being pulled off nominal, or erratic starting of the VCO when power is applied. Top nubbage Post subject: Posted: Wed Aug 01, 2007 10:12 am General Joined: Fri Feb 17, 2006 12:07 pm Posts: 218 Location: London UK You might find this helps. [url]http://www.mwrf.com/Articles/ArticleID/5513/5513.html Posted  11/12/2012

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Maths And Physics With Pandey Sir Maths Formulas Important Formulas of Laws of Exponent # Important Formulas of Laws of Exponent There are many important formulas of laws of exponent but one of the most important formulas of laws of exponent are as under- 1. a⁰ = 1 2. (am)n = amn 3. am × an = am+n 4am ÷ an = amn 5am × bm = (ab)m 6. am ÷ bm = (a÷b)m 7. 1÷a⁻¹ = a

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true Take Class I-V Tuition from the Best Tutors • Affordable fees • 1-1 or Group class • Flexible Timings • Verified Tutors Search in # Common Fractions with Decimal and Percent Equivalents A Amrtha S. 05/04/2017 0 0 Fraction Decimal Percent 1/2 0.5 50% 1/3 0.333… 33.333…% 2/3 0.666… 66.666…% 1/4 0.25 25% 3/4 0.75 75% 1/5 0.2 20% 2/5 0.4 40% 3/5 0.6 60% 4/5 0.8 80% 1/6 0.1666… 16.666…% 5/6 0.8333… 83.333…% 1/8 0.125 12.5% 3/8 0.375 37.5% 5/8 0.625 62.5% 7/8 0.875 87.5% 1/9 0.111… 11.111…% 2/9 0.222… 22.222…% 4/9 0.444… 44.444…% 5/9 0.555… 55.555…% 7/9 0.777… 77.777…% 8/9 0.888… 88.888…% 1/10 0.1 10% 1/12 0.08333… 8.333…% 1/16 0.0625 6.25% 1/32 0.03125 3.125% 0 Dislike ## Other Lessons for You Matter: Its Nature And Behaviour Matter And Its Types Now, What is matter? Its the first question that comes in our mind and plays football in it. Well "Anything that occupies space and have mass is called Matter". For eg:- A human is... Grammer questions 1. He had mastered all the branches of learning.(Name the part of speech underlined words) Ans. Mastered------- Verb, Learning------- Noun. 2. Brahma-datta ascended the throne after his father's death.... Mrs.Satinder D. Learning tips Tips: Focus on the mastery. Take charge of your learning. Build your competence and gain confidence. Think critically about what you're learning. Use grades to keep on track with learning. Effective... Acid, Base and Salts The reaction of an acid with a base to produce only salt and water is called a neutralization reaction. Acids Acids are sour in taste. If hydronium ions are found in a solution, the solution is acidic... dhtauroop with meaning (अर्थ के साथ धातुरूप ) लट् लकार: (Prasent Tense) पुरुष: एकवचनम् अर्थ द्विवचनम् अर्थ बहुवचनम् अर्थ प्रथमपुरुष: पठति पढ़ता है / पढ़ रहा है पठतः दो पढ़ते हैं / पढ़ रहे हैं पठन्ति... ### Looking for Class I-V Tuition ? Learn from Best Tutors on UrbanPro. Are you a Tutor or Training Institute? Join UrbanPro Today to find students near you X ### Looking for Class I-V Tuition Classes? The best tutors for Class I-V Tuition Classes are on UrbanPro • Select the best Tutor • Book & Attend a Free Demo • Pay and start Learning ### Take Class I-V Tuition with the Best Tutors The best Tutors for Class I-V Tuition Classes are on UrbanPro

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# How Numerical is your Aperture? Planning another optogenetics study, and I needed to cut the optic fibre cannulae ready for implantation. One of the other postdocs in the lab had been super organised and bought in a bunch of implants from Thorlabs at a variety of numerical apertures (thanks Amy). But, which is the best numerical aperture (NA) for my experiment? I won’t go into details (because I’m not a physicist), but Wikipedia defines the NA of an optical system as “a dimensionless number that characterises the range of angles over which the system can accept or emit light”. Essentially, as far as we are concerned for fibre optics, the NA is relevant for two things: 1. The bigger the NA, the more light from the source will travel down the optic fibre – for a laser system, this doesn’t matter much because the coherent light can easily be focused down it, but for an LED, this can make a big difference for how much light is captured by the fibre (rather than scattering away) 2. It determines how much the light spreads after exiting the fibre (for in vivo opto’s, this will be in the mouse’s brain) – the higher the NA, the greater the cone of light dispersion So, back to cutting fibres, and I had to decide which ones to use – I normally use the 0.22 NA fibres out of habit, but I have read multiple recommendations to use as high an NA fibre as possible when using an LED system (which is what we have); the idea being to get as much light power as possible into the mouse’s brain, which is important considering LED systems can struggle to be bright enough for in vivo opto’s. Both Prizmatix and Doric suggest using 0.66 NA fibres for LED-connected systems, which is actually higher than the ones we have available from Thorlabs. To test the light output, I hooked up fibres of different NA’s to our LED optogenetics system, and recorded the light power out the end of the fibre using a light meter, both under constant illumination and during 10 Hz flashing with 10 ms on times (Table 1). True to form, the higher the NA of a fibre, the more light that is passed down it. Great, so at this point I’d pretty much settled on the 0.50 NA fibre, because it emitted approx. 50 % more power than the 0.22 NA fibre. However, for the sake of completeness, I decided to input the values into Karl Deisseroth’s irradiance predictor, to check how deep I would get good ChR2 activation. This is a useful step when planning placement of your optic fibres. I plotted the values for all three NA fibres (Figure 1), and I’ve included the threshold level of 1 mW/mm2 that I’ve talked about previously (this is the measured EC50 of ChR2 H134R, which I use as a threshold irradiance to assume good activation). Now I’ll be honest, I was surprised by this outcome – despite having lower light output from the lower NA fibres, the irradiance was higher as soon as you go deeper than about 0.2 mm into the tissue. I can only assume this is because the lower NA results in less light spread coming out of the fibre – the 0.50 NA fibre remains above the critical 1 mW/mm2 down to about 1.0 mm, whereas the 0.22 NA fibre goes to about 1.4 mm. The answer is simple – I’m going to use the 0.22 NA fibres, because they have the dual benefit of activating ChR2 to a greater depth, and also having lower brightness at the end of the fibre, which means less heating of the tissue and phototoxicity.

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Question # Steam enters a converging-diverging nozzle at steady state with P1 = 40bar T1 = 400°C and... Steam enters a converging-diverging nozzle at steady state with P1 = 40bar T1 = 400°C and a velocity of 12m/s. The steam flows through the nozzle with negligible heat transfer and no significant change in potential energy. At the exit P2 = 15bar and the velocity is 700 m/s. The mass flow rate is 2.5 kg/s. Determine the exit area of the nozzle in m2 #### Earn Coins Coins can be redeemed for fabulous gifts. ##### Need Online Homework Help? Most questions answered within 1 hours.

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# Geometric Entities Although you may use just NURB surfaces to design a boat, ProSurf and ProBasic behave more like general-purpose CAD programs, since they provide point, line, polyline, curve, and surface geometric entities. There is now a greater overlap between ProSurf and ProBasic and other CAD programs. The following picture shows a sample of each type of entity. Point Entities At the top of the picture are point entities, shown by the small circles. One of their uses can be as "tick marks" which identify target shape positions for the boat. Curve Entities A curve entity, as defined in this program, can be a line, a polyline, a NURB curve, or a combination of the two. Examples of these entities are shown in the picture above. The "Curve-Add-Add Polyline" command allows you to add a line or polyline to the model. Each point of the polyline is entered by clicking the left mouse button, and the polyline is terminated by clicking the right mouse button. The "Curve-Add-Add Curve" command allows you to add a NURB curve to the model. Each point of the curve is entered by clicking the left mouse button and the curve is terminated by clicking the right mouse button. These "curves" can be used for anything you like. For example, you might use the surfaces to define a 3D sailboat hull, but use the curve entities to define a 2D outboard profile of the rig and sails in profile view. Of course, you could define the rig and sails using 3D surfaces, but this extra work may not be necessary. A combination polyline and NURB curve entity is obtained by using the "Curve-Knuckle Pnt" command. This command acts like a toggle switch which converts any interior polyline or curve point to or from a hard knuckle point. Therefore, any point of a polyline or curve can be converted to or from a knuckle point. This might seem odd since most CAD programs define separate curve and polyline entities. This technique, however is more powerful, since there is no difference between the two entities and you do not have to break and join different entities together. You can start by roughing out a shape usine a polyline and then convert portions into curves and then use the curvature curve indicator to fair the curve. For many CAD tasks, it might be easier to start in ProBasic or ProSurf, and then convert the entities over to your general-purpose CAD program. Surface Entities The surface entity is a NURB (Non-Uniform Rational B-spline) surface defined by a number of points at the intersections of the surface rows and columns. These defining points are displayed using a small squares. The NURB surface defines a complete, unique, rectangular-like, smooth, membrane, drawn using a number of rows and columns. Surfaces are added by defining their four corner points (with only two rows and two columns). You can then add in more rows or columns to add more flexibility and control points. The example in the picture above shows a surface with three rows and five columns. ProSurf and ProBasic use controlling points which lie ON the NURB surface. (The same technique is used for NURB curves.). NURBS and B-splines are officially defined, however, using "vertex" points which do NOT lie on the surface, except for the four corner points. The original intent for B-splines was to provide a way to create a smooth surface that was "influenced" by these controlling vertices. By not requiring that the surface pass through the points, B-splines can create a relatively smooth surface even though the vertex points aren't perfectly positioned. The thought was that this was perfect for interactive design of free-form surfaces on a small computer screen. In practice, however, many users disliked the fact that the controlling vertices were floating off in space and that many programs connected the vertex points with a mesh which obscured the surface itself. In addition, most designers wanted very accurate shape control over the surface, because not just any smoothly curved shape will do. This accuracy requires direct control of the surface and some form of surface fairing technique. When we started providing B-spline surface modeling in 1985, we decided to allow the user to control the shape of the B-spline surfaces using points on the surface. This had not been done by any other commercial CAD program in any industry. By eliminating the display of the vertex points, we cleaned up the display of the boat and gave the user more precise shape control over the boat, including the ability to input offsets and automatically fit surfaces to those offsets. Since we provided curvature curve display and control to use for fairing, the user had the best of both worlds: accurate shape control and detailed fairing. With this new Windows version of the program, you can select from three ways to control the surface shape using points. You can use the traditional defining vertex points (floating off in space), or you can use either of two methods for defining points on the surface. Remember that we use industry standard NURB curves and surfaces. For every vertex point, there is an associated point on the surface. When you move the point on the surface, we calculate backwards to determine the associated move in the defining vertices. We continue to follow the latest research papers on NURB surface techniques, plus add our own shaping and fairing techniques which are not found in any other CAD program.

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Debitoor's accounting dictionary Credit utilisation rate # What is a credit utilisation rate? A credit utilisation rate, also known as a credit utilisation ratio, is the percentage of credit you have used on your credit account. Banks and lenders use this to determine your ability to repay debts. A low credit utilisation rate means that you’re using less of your available credit. This is likely to improve your credit score and means that you’re not overspending or using more credit than you are able to repay on time. ## How to calculate your credit utilisation rate To calculate your credit utilisation rate, you will need your credit limit and the current amount you owe. The formula for credit utilisation rate is: The amount you owe ÷ your credit limit For example, let’s say you have a credit card with a limit of £1000. You’ve spent £200, and have £800 credit left over. Therefore, your credit utilisation rate is: 200 ÷ 1000 = 0.2 (20%). You can calculate the rate for a single credit card, or your overall rate. To calculate your overall credit utilisation rate, divide the total amount of money you owe by the total credit limit you have across the multiple accounts. For instance, if you have 2 credit cards with a total of £5000 credit limit, and on one card you owe £500, the calculation would be: 500 ÷ 5000 = 0.1 (10%). ## What is the ideal credit utilisation rate? Although there is no set number, a good rate would be anything under 30%. A low rate means that you are responsible with your spending and good at managing your finances. A higher rate could indicate to lenders and creditors that you might have difficulties paying off your loan or credit card.

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# परवलय x2 = 6y की नाभि के निर्देशांक ज्ञात कीजिए। This question was previously asked in Navik GD 22 March 2021 Memory Based Paper (Section I + II) View all Indian Coast Guard Navik GD Papers > 1. $$\left ( -\frac{3}{2},0 \right )$$ 2. $$\left ( 0,\frac{3}{2} \right )$$ 3. $$\left ( 0,-\frac{3}{2} \right )$$ 4. इनमें से कोई नही ## Answer (Detailed Solution Below) Option 2 : $$\left ( 0,\frac{3}{2} \right )$$ Free CRPF Head Constable & ASI Steno (Final Revision): Mini Mock Test 16.6 K Users 40 Questions 40 Marks 45 Mins ## Detailed Solution संकल्पना: Y-अक्ष के अनुदिश परवलय का समीकरण: x2 = ± 4ay, नाभि (0, ± a) हल: एक परवलय का दिया गया समीकरण x2 = 6y है इस समीकरण की x2 = 4ay से तुलना करने पर, हम प्राप्त करते हैं 4a = 6 ⇒ a = $$\frac{3}{2}$$ इसलिए, नाभि F के निर्देशांक = (0, a) ⇒ $$\left ( 0,\frac{3}{2} \right )$$ ∴ सही विकल्प (2) है Latest Indian Coast Guard Navik GD Updates Last updated on Apr 8, 2024 -> ICG Navik GD 2024 City Intimation Link has been released. -> Candidates can check it by signing in with a registered Email ID and password. -> The exam will be conducted on 21st and 22nd April 2024. -> The last date for submission of the application form for the CGEPT-02/2024 batch has been extended till 03 March 2024. -> A total of 260 vacancies have been announced for recruitment through the Coast Guard Enrolled Personnel Test (CGEPT). -> Candidates who have completed their 10+2 with Maths and Physics are eligible for this post. -> Candidates must go through the Indian Coast Guard Navik GD previous year papers.

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I have a physics question from review that I forgotten how to do. A certain physics teacher is frustrated with Mac Computers, so she throws one straight up from the top of a 50 - m tall building. The computer's initial velocity is 20m/s [straight up]. Find (a) the maximum height of the computer, (b) the amount of time it takes to reach that height, (c) the amount of time it takes the computer to reach the ground, and (d) the final velocity of the computer as it hits the ground. I know it's shameful going to the pit for school questions. But they haven't given out text books in my school yet, so I'm confused. go to a better school. uvast Yours Sincerely, Dr. Speakers a. 15m, b.5 seconds, c. 16 seconds, d. 89 huh, Im doint this exact same thing in physics right now, gravity is 9.81 and thats the acceleration and I forget what else to do (my physics class is a BS class) all we do is play with these little dune buggy things, talk, draw pictures and throw crap off the roof Bands I must see in concert Silvertide The Black Crowes Jackyl Black Stone Cherry AC/DC StoneRider Freedomhawk The Darkness and/or the sideprojects of the band members Buckcherry Theres more. But those are the main ones Maximum height is when the velocity reaches 0 as it's going up. Each second, it's being pushed down at 9.8 m/s/s. The rest should be easy enough. And as I'm sure people have said by the time I finished this post, Math & Science Help Thread. Hmm, Okay thanks for the input, I'll post there next time. I think I've learned that last year, but I can't remember it anymore. I didn't like my teacher, she couldn't explain things well. She had big breasts though Quote by guitarguy1347 a. 15m, b.5 seconds, c. 16 seconds, d. 89 Sigh, don't post random answers .. Quote by DGen92 Sigh, don't post random answers .. dont you have a physics book? and i know how to do this ****, im just to lazy to look it up. My Gear: Fender Custom Shop Time Machine 1956 relic plus accessories Your position and your relationship with music has to be one from the inside. - John Frusciante and yes, my avatar is weezy, what? Quote by PnT I think I've learned that last year, but I can't remember it anymore. I didn't like my teacher, she couldn't explain things well. She had big breasts though please, do go on Yours Sincerely, Dr. Speakers i still recall this pretty well. Quote by RHCP987123 dont you have a physics book? and i know how to do this ****, im just to lazy to look it up. I already said in my first post that they didn't hand out textbooks yet a = -g = -9.81 m/s^2 not 9.81 m/s^2... something wierd i found out but acceleration to the ground is negative... (after being taught my middle school years that thr is no negative accel...god schools so conflicting) Also, you'll need a couple equations: V2 = V1 + AT, V2 = Final Velocity, V1 = Starting Velocity, A = Acceleration (Force of Gravity [-9.8 m/s/s on Earth], T = Time. At max height, V2 will be 0. 0 = 20 + (-9.8)T will get you the time when it's at the max. I forgot the distance formula. :P I think something like D = 1/2(V1 + V2T) or something. I forgot. EDIT: a = -g = -9.81 not 9.81... something wierd i found out but acceleration to the ground is negative... (after being taught my middle school years that thr is no negative accel...god schools so conflicting) It's not negative. Acceleration is a vector, it has a magnitude (the 9.81), and a direction. The reason it's 0 is that it implies a direction towards the spot it originated. Think of it as 9.81 m/s/s, 90 degrees. Last edited by Kapps at Sep 13, 2009, Quote by Kapps Also, you'll need a couple equations: V2 = V1 + AT, V2 = Final Velocity, V1 = Starting Velocity, A = Acceleration (Force of Gravity [-9.8 m/s/s on Earth], T = Time. At max height, V2 will be 0. 0 = 20 + (-9.8)T will get you the time when it's at the max. I forgot the distance formula. :P I think something like D = 1/2(V1 + V2T) or something. I forgot. ITS (DELTA X) = .50( V i + V f ) x (Delta T) Delta is the wierd triangle greek symbol tht stands for "change in" Quote by Kapps Also, you'll need a couple equations: V2 = V1 + AT, V2 = Final Velocity, V1 = Starting Velocity, A = Acceleration (Force of Gravity [-9.8 m/s/s on Earth], T = Time. At max height, V2 will be 0. 0 = 20 + (-9.8)T will get you the time when it's at the max. I forgot the distance formula. :P I think something like D = 1/2(V1 + V2T) or something. I forgot. EDIT: (after being taught my middle school years that thr is no negative accel...god schools so conflicting) It's not negative. Acceleration is a vector, it has a magnitude (the 9.81), and a direction. The reason it's 0 is that it implies a direction towards the spot it originated. Think of it as 9.81 m/s/s, 90 degrees. really cause i use like an '05 textbook tht states wht i said...but wht do i know im just a student studying a book. Alright, I've managed to figure out (a), and (b) I can't seem to get the answer for (c) and (d). As I recall the distance formula was d = 1/2(V1 + V2)T Quote by huskyplayer ITS (DELTA X) = .50( V i + V f ) x (Delta T) Delta is the wierd triangle greek symbol tht stands for "change in" That's the same as what I just said. T is the change in time, however many seconds passed. The distance you travel is the change in your location, or change in X. Vi is Velocity Initial, Vf is Velocity Final, it's just V2/V3/whatever is easier if you have more than two. Kinematic equations are your new best friends. Use them. well i hate physics because the teacher sux and i have to teach myself...all he does is teach by showing us videos of professors at havard teaching physics, pausing the video at times to rephrase wht they guy just said on the video... Quote by DGen92 Alright, I've managed to figure out (a), and (b) I can't seem to get the answer for (c) and (d). As I recall the distance formula was d = 1/2(V1 + V2)T Think about it a bit. You know how long it takes you to reach the top. Now you have to find out how long it takes from when it reaches the top until it reaches the bottom, then add the time it takes to reach the top. You know at the top the initial velocity (V1) is 0. You know the acceleration is -9.8 m/s. You know the height is 90.82 (50 m building, thrown (20 / 9.8 * 20) up) metres. There was another distance formula, but I forgot it. It's used here. ;p EDIT: D = V1T + 1/2A(T^2). You know D, V1, and A. This is used to find Time. Once you find Time, you know the acceleration. You know the initial velocity. You know the time. You can easily find the final velocity using those. Last edited by Kapps at Sep 13, 2009, Thanks, I understand almost everything except that part where you found the total height. 90.82(50 m building, thrown (20/9.8 x 20) ) Why did you do 20/9.8 x 20 ? Don't you just add the building height and the distance from the top of the building thrown to the maximum height ? EDIT: Also for that formula you stated, how can I find time when there's two variables for that one. Don't I use the other equation ? EDIT2: I figured out how you got 90.82. But when I input it, I get the wrong answer. ( it states the answer in the question ). I'm solving for time by using the formula d=1/2(V1+V2)T Last edited by DGen92 at Sep 13, 2009, Quote by DGen92 Thanks, I understand almost everything except that part where you found the total height. 90.82(50 m building, thrown (20/9.8 x 20) ) Why did you do 20/9.8 x 20 ? Don't you just add the building height and the distance from the top of the building thrown to the maximum height ? I'm not sure actually. I did something odd, and that part is wrong. Use the maximum height reached. I don't really know what I was trying to do at that part. :p EDIT: EDIT: Also for that formula you stated, how can I find time when there's two variables for that one. Don't I use the other equation ? There isn't two variables. You know the acceleration (force of gravity). You know the distance (maximum height). You know the initial speed (0). You're trying to find Time. You just plug Time in to two different spots. EDIT2: I'm solving for time by using the formula d=1/2(V1+V2)T You can't unless you already know the final velocity right before it reaches the ground. When it reaches the ground, the velocity isn't 0, it doesn't stop until a moment after it reaches the ground, so assume the velocity is right before it hits the ground. Last edited by Kapps at Sep 13, 2009, Quote by Kapps EDIT: There isn't two variables. You know the acceleration (force of gravity). You know the distance (maximum height). You know the initial speed (0). You're trying to find Time. You just plug Time in to two different spots. Okay, would you mind explaining how I would rearrange it ? v(s) = -9.8x+20 I think that is the initial velocity...integrate for the position. Castles made of sand Fall to the sea Eventually Quote by DGen92 Okay, would you mind explaining how I would rearrange it ? D = V1T + 1/2A(T^2) D = whatever you got for your answer at the maximum height. V1 = 0. T for the first part doesn't matter, since it's multiplied by 0. 1/2A = -4.9, T^2 = whatever you got for your T. D / -4.9 = T^2. Root(D / -4.9) (Distance should be negative in this case, as it's a vector, and you're going down, so you assume it's negative, otherwise you end up with imaginaries) = T. Use the answer that makes sense (aka, the answer that's not a negative amount of time :p). Quote by DGen92 I have a physics question from review that I forgotten how to do. A certain physics teacher is frustrated with Mac Computers, so she throws one straight up from the top of a 50 - m tall building. The computer's initial velocity is 20m/s [straight up]. Find (a) the maximum height of the computer, (b) the amount of time it takes to reach that height, (c) the amount of time it takes the computer to reach the ground, and (d) the final velocity of the computer as it hits the ground. I know it's shameful going to the pit for school questions. But they haven't given out text books in my school yet, so I'm confused. a) v^2 = v[initial]^2 + 2ad 0 = (20m/s)^2 + 2(-9.8m/s)(d) d = (-400m/s)/(-18.6m/s) = 21.5m b) y = 1/2at^2 21.5m = 1/2(9.8m/s)(t^2) t = ((21.5m)/(4.9m/s))^1/2 t = 2.1s OR v = v[initial] + at 0 = 20m/s + (-9.8m/s)t t = (-20m/s)/(-9.8m/s) t = 2.0 s (meh, a little tiny bit off but oh well) c) Assuming it's not including the time it took to reach the max height. y = 1/2at^2 50m + 21.5m = 1/2(9.8m/s)(t^2) t = ((71.5m)/(4.9m/s))^1/2 t = 3.8s And including the time it took to reach the max height, 3.8s + 2.1s = 4.9s d) v = v[initial] + at v = 0m/s + (9.8m/s)(3.8s(remember, the acceleration doesn't start until it reaches the top and starts going back down)) v = 37.2 m/s Quote by Kapps D = V1T + 1/2A(T^2) D = whatever you got for your answer at the maximum height. V1 = 0. T for the first part doesn't matter, since it's multiplied by 0. 1/2A = -4.9, T^2 = whatever you got for your T. D / -4.9 = T^2. Root(D / -4.9) (Distance should be negative in this case, as it's a vector, and you're going down, so you assume it's negative, otherwise you end up with imaginaries) = T. Use the answer that makes sense (aka, the answer that's not a negative amount of time :p). OH thanks, I fully understand Great help xD EDIT: Quote by grampastumpy a) v^2 = v[initial]^2 + 2ad 0 = (20m/s)^2 + 2(-9.8m/s)(d) d = (-400m/s)/(-18.6m/s) = 21.5m b) y = 1/2at^2 21.5m = 1/2(9.8m/s)(t^2) t = ((21.5m)/(4.9m/s))^1/2 t = 2.1s OR v = v[initial] + at 0 = 20m/s + (-9.8m/s)t t = (-20m/s)/(-9.8m/s) t = 2.0 s (meh, a little tiny bit off but oh well) c) Assuming it's not including the time it took to reach the max height. y = 1/2at^2 50m + 21.5m = 1/2(9.8m/s)(t^2) t = ((71.5m)/(4.9m/s))^1/2 t = 3.8s And including the time it took to reach the max height, 3.8s + 2.1s = 4.9s d) v = v[initial] + at v = 0m/s + (9.8m/s)(3.8s(remember, the acceleration doesn't start until it reaches the top and starts going back down)) v = 37.2 m/s Thanks, though I already understand it. But that's good to check my answers again. Last edited by DGen92 at Sep 13, 2009, Quote by Kapps Think about it a bit. You know how long it takes you to reach the top. Now you have to find out how long it takes from when it reaches the top until it reaches the bottom, then add the time it takes to reach the top. You know at the top the initial velocity (V1) is 0. You know the acceleration is -9.8 m/s. You know the height is 90.82 (50 m building, thrown (20 / 9.8 * 20) up) metres. That's not entirely true. That would be essentially assuming uniform velocity while it's decelerating, which is obviously untrue. It can't be going 20m/s for the amount of time it takes it to go from 20m/s to 0m/s. The formula given earlier: x = 1/2(v + v[initial])t works here. t = (v - v[initial])/a t = about 2.0s x = 1/2(v + v[initial])t x = 1/2(20m/s)(2.0s) x = around 20m. Last edited by grampastumpy at Sep 13, 2009, Quote by grampastumpy That's not entirely true. That would be essentially assuming uniform velocity while it's decelerating, which is obviously untrue. It can't be going 20m/s for the amount of time it takes it to go from 20m/s to 0m/s. The formula given earlier: x = 1/2(v + v[initial])t works here. t = (v - v[initial])/a t = about 2.0s x = 1/2(v + v[initial])t x = 1/2(20m/s)(2.0s) x = around 20m. Yeah, I realized that after he replied to it, where I said I was wrong. :p Oh, my bad, lol. If you already found it then woops, if not, then well, there it is.

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Lecture-8 # J i it l c tar y v i cj a ni l go s e i This preview shows page 1. Sign up to view the full content. This is the end of the preview. Sign up to access the rest of the document. Unformatted text preview: t: ç- ftiL l t- ¿ l!J /Z / A.u ,7' è) ., ( 22 Surface of Rotating Bucket 1. Draw a picture. 2. Dynamic or Static? 3. Coordinate system? …Static …co-rotating cylindrical 4. Can you apply conservation principles? (no flow in this frame) …Bernoulli’s 5. Use your intuition. 23 Co-Rotating Frame ⌅u + u · ⌅u ⌅t = g (1/⇥)⌅p + ⌅2 u ⌅ uR + uR · ⌅uR ⌅t = g (1/⇥)⌅p + ⌅2 uR + R 2 2 ⇤ uR with Centrifugal and Coriolis “apparent” forces. ( Very convenient when uR = 0.) 24 L- \1 &quot;: ~ ~ \! () .. \- u- ~ u ~ Co-Rotating With Bucket ~ v ~ 'ù ~ \!; \L '0 ~ \L \) 'l ~ ~ U. ~ ., .. ~ () C: , 0 \J 0 K t.. ù ~ .. \0 ~ \c ~ ~ l. \1 ~ I ~ '3 I- q. ~ ~ \I \L C: l :s i '0 r: \) ~ led l') + 1\ï -l ~ D \~ l (( () U I :: 'l. lo J\) l:: . &quot;&quot; ~ 'J ~ ~ :) ~ \Ù i ~ &quot; .. d 'U ~ ( '\ &quot;) -l '&quot; . .. .J Force balance: ~ I. \l 'J ~ '- r' ~ .. ~~ '- I C: \\ ~ \~ l \i l~ Q (W (V tL\~ \~ '- -l .. c: J ~ ~ li ~ 0 LA i f. V ~ '\ \I ~ ~ t I.. l ~ t; 0 V 25 ~ tJ d ., ~ r... View Full Document Ask a homework question - tutors are online

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length and frequency 0 pts ended Pipe A = .50 mL and open at both ends Pipe B = ? and open at one end Determine the length if Pipe B so that it has the same fundemental fequencies as Pipe A a. 2 m, b. .25 m, c. .75 m, d. 1 m, e. .5 m

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# College Algebra : Graphing Polynomials ## Example Questions ### Example Question #2 : Transformations Of Parabolic Functions If the function  is depicted here, which answer choice graphs ? None of these graphs are correct. Explanation: The function  shifts a function f(x) units to the left. Conversely,  shifts a function f(x) units to the right. In this question, we are translating the graph two units to the left. To translate along the y-axis, we use the function  or . ### Example Question #7 : Graphing Parabolic Inequalities Which of the following graphs correctly represents the quadratic inequality below (solutions to the inequalities are shaded in blue)? Explanation: To begin, we analyze the equation given: the base equation,  is shifted left one unit and vertically stretched by a factor of 2. The graph of the equation  is: To solve the inequality, we need to take a test point and plug it in to see if it matches the inequality. The only points that cannot be used are those directly on our parabola, so let's use the origin . If plugging this point in makes the inequality true, then we shade the area containing that point (in this case, outside the parabola); if it makes the inequality untrue, then the opposite side is shaded (in this case, the inside of the parabola). Plugging the numbers in shows: Simplified as: Which is not true, so the area inside of the parabola should be shaded, resulting in the following graph: ### Example Question #1 : Graphing Polynomials How many zeroes does the following polynomial have? Explanation: is a degree 3 polynomial, so we don't have any easy formulas for calculating possible roots--we just have to check individual values to see if they work. We can use the rational root test to narrow the options down. Remember, if we have a polynomial of the form  then any rational root will be of the form p/q where p is a factor of  and q is a factor of . Fortunately in this case,  so we only need to check the factors of , which is -15. Let's start with the easiest one: 1. It doesn't work. If we try the next number up, 3, we get this: It worked! So we know that a factor of our polynomial is . We can divide this factor out: and now we need to see if  has any roots. We can actually solve quadratics so this is easier. There aren't any real numbers that square to get -5 so this has no roots. Thus,  only has one root. ### Example Question #2 : Graphing Polynomials is a polynomial function. . True or false: By the Intermediate Value Theorem,  cannot have a zero on the interval . False True False Explanation: As a polynomial function, the graph of  is continuous. By the Intermediate Value Theorem, if  or , then there must exist a value  such that Set  and . It is not true that , so the Intermediate Value Theorem does not prove that there exists  such that . However, it does not disprove that such a value exists either. For example, observe the graphs below: Both are polynomial graphs fitting the given conditions, but the only the equation graphed at right has a zero on . ### Example Question #3 : Graphing Polynomials True or false: The polynomial  has  as a factor. True False True Explanation: One way to answer this question is as follows: Let . By a corollary of the Factor Theorem,  is divisible by  if and only if the sum of its coefficients (accounting for minus symbols) is 0.  has as its coefficient sum, so  is indeed divisible by . ### Example Question #4 : Graphing Polynomials True or false: The polynomial  has  as a factor. False True False Explanation: Let . By a corollary of the Factor Theorem,  is divisible by  if and only if the alternating sum of its coefficients (accounting for minus symbols) is 0. To find this alternating sum, it is necessary to reverse the symbol before all terms of odd degree. In , there is one such terms, the  term, so the alternating coefficient sum is , so  is not divisible by . ### Example Question #5 : Graphing Polynomials is a polynomial function.  and . True or false: By the Intermediate Value Theorem,  must have a zero on the interval . True False True Explanation: As a polynomial function, the graph of  is continuous. By the Intermediate Value Theorem, if  or , then there must exist a value  such that Setting , and looking at the second condition alone, this becomes: If , then there must exist a value  such that  - that is,  must have a zero on . The conditions of this statement are met , since   - and  - so  does have a zero on this interval. ### Example Question #6 : Graphing Polynomials Let  be an even polynomial function with  as a factor. True or false: It follows that  is also a factor of . True False True Explanation: By the Factor Theorem,  is a factor of a polynomial  if and only if . It is given that  is a factor of , so it follows that . is an even function, so, by definition, for all  in its domain, . Setting ; by substitution, . It follows that  is also a factor of , making the statement true. ### Example Question #27 : Parabolic Functions Which of the following graphs matches the function ? Explanation: Start by visualizing the graph associated with the function : Terms within the parentheses associated with the squared x-variable will shift the parabola horizontally, while terms outside of the parentheses will shift the parabola vertically. In the provided equation, 2 is located outside of the parentheses and is subtracted from the terms located within the parentheses; therefore, the parabola in the graph will shift down by 2 units. A simplified graph of  looks like this: Remember that there is also a term within the parentheses. Within the parentheses, 1 is subtracted from the x-variable; thus, the parabola in the graph will shift to the right by 1 unit. As a result, the following graph matches the given function  :

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https://www.thespruce.com/metric-conversions-for-cooking-2355731

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# Metric Conversions ## Common Metric Conversions for Cooking Here in the United States, we use what is known as an English measurement system. You will find it just about everywhere – from the odometer in your car to the measuring cups in your kitchen. Yes, those teaspoons, tablespoons, and cups all are a part of this system. While we find the English measurement system familiar as part of our culinary culture, it is not a common system to the rest of the world. In fact, only three countries continue to use this system of measurement: the United States, Liberia, and Myanmar. So what is the rest of the world using? The rest of the world uses what is known as the metric system, which is a decimal-based system that uses units that are related by factors of ten. While many Americans are unfamiliar with the metric system outside of high school chemistry class, it is a much simpler system to use in terms of the math required for basic calculations and it avoids superfluous units for volume. This system of measurement when used in cooking terms around the world usually uses liters and grams in lieu of American cups, ounces, pints, and so on. While the metric system has gained more traction in the United States (even appearing on much our food packaging in addition to the English units), our traditional English system of measurement continues to be to go-to for most non-business or science related activities, including home cooking. So when your international recipe calls for 15 milliliters of salt, what do you do? You can turn to this handy English-Metric system conversion table for basic equivalents. ### US to Metric Volume Conversions US Customary Quantity (English) Metric Equivalent 1 teaspoon 5 ml 1 tablespoon 15 ml 2 tablespoons 30 ml ¼ cup or 2 fluid ounces 60 ml 1/3 cup 80 ml ½ cup or 4 fluid ounces 125 ml 2/3 cup 160 ml ¾ cup or 6 fluid ounces 180 ml 1 cup or 8 fluid ounces or ½ pint 250 ml 1 ½ cup or 12 fluid ounces 375 ml 2 cups or 1 pint or 16 fluid ounces 500 ml 3 cups or 1 ½ pints 700 mL 4 cups or 2 pints or 1 quart 950 ml 4 quarts or 1 gallon 3.8 L 1 ounce 28 grams ¼ lb (4 ounces) 112 grams ½ lb (8 ounces) 225 grams ¾ lb (12 ounces) 337 grams 1 lb (16 ounces) 450 grams Note: When a high level of precision is not required, basic equivalents may be used as follows:1 cup ≈ 250 mL1 pint ≈ 500 mL1 quart ≈ 1 L1 gallon ≈ 4 L ### US to Metric Weight Conversions US Customary Quantity (English) Metric Equivalent 1 ounce 28 grams 4 ounces or ½ lb 113 grams 1/3 lb 150 grams 8 ounces or ½ lb 230 grams 2/3 lb 300 grams 12 ounces or ¾ lb 340 grams 16 ounces or 1 lb 450 grams 32 ounces or 2 lbs 900 grams Note: The ounces referred to in this weight conversion table are not fluid ounces.

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https://www.geeksforgeeks.org/program-find-normal-trace-matrix/?ref=lbp

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Program to find Normal and Trace of a matrix Given a 2D matrix, the task is to find Trace and Normal of matrix. Normal of a matrix is defined as square root of sum of squares of matrix elements. Trace of a n x n square matrix is sum of diagonal elements. Examples : ```Input : mat[][] = {{7, 8, 9}, {6, 1, 2}, {5, 4, 3}}; Output : Normal = 16 Trace = 11 Explanation : Normal = sqrt(7*7+ 8*8 + 9*9 + 6*6 + 1*1 + 2*2 + 5*5 + 4*4 + 3*3) = 16 Trace = 7+1+3 = 11 Input :mat[][] = {{1, 2, 3}, {6, 4, 5}, {2, 1, 3}}; Output : Normal = 10 Trace = 8 Explanation : Normal = sqrt(1*1 +2*2 + 3*3 + 6*6 + 4*4 + 5*5 + 2*2 + 1*1 + 3*3) Trace = 8(1+4+3)``` Implementation: C++ `// C++ program to find trace and normal` `// of given matrix` `#include` `using` `namespace` `std;`   `// Size of given matrix` `const` `int` `MAX = 100;`   `// Returns Normal of a matrix of size n x n` `int` `findNormal(``int` `mat[][MAX], ``int` `n)` `{` `    ``int` `sum = 0;` `    ``for` `(``int` `i=0; i Java `// Java program to find trace and normal` `// of given matrix`   `import` `java.io.*;`   `class` `GFG {`   `// Size of given matrix` `static`  `int` `MAX = ``100``;`   `// Returns Normal of a matrix of size n x n` ` ``static` `int` `findNormal(``int` `mat[][], ``int` `n)` `{` `    ``int` `sum = ``0``;` `    ``for` `(``int` `i=``0``; i Python3 `# Python3 program to find trace and ` `# normal of given matrix ` `import` `math`   `# Size of given matrix ` `MAX` `=` `100``; `   `# Returns Normal of a matrix ` `# of size n x n ` `def` `findNormal(mat, n): `   `    ``sum` `=` `0``; ` `    ``for` `i ``in` `range``(n): ` `        ``for` `j ``in` `range``(n): ` `            ``sum` `+``=` `mat[i][j] ``*` `mat[i][j]; ` `    ``return` `math.floor(math.sqrt(``sum``)); `   `# Returns trace of a matrix of ` `# size n x n ` `def` `findTrace(mat, n): `   `    ``sum` `=` `0``; ` `    ``for` `i ``in` `range``(n): ` `        ``sum` `+``=` `mat[i][i]; ` `    ``return` `sum``; `   `# Driver Code ` `mat ``=` `[[``1``, ``1``, ``1``, ``1``, ``1``], ` `       ``[``2``, ``2``, ``2``, ``2``, ``2``], ` `       ``[``3``, ``3``, ``3``, ``3``, ``3``], ` `       ``[``4``, ``4``, ``4``, ``4``, ``4``], ` `       ``[``5``, ``5``, ``5``, ``5``, ``5``]]; `   `print``(``"Trace of Matrix ="``, findTrace(mat, ``5``)); `   `print``(``"Normal of Matrix ="``, findNormal(mat, ``5``)); `   `# This code is contributed by mits` C# `// C# program to find trace and normal` `// of given matrix` `using` `System;`   `class` `GFG {` `    `  `    ``// Returns Normal of a matrix of` `    ``// size n x n` `    ``static` `int` `findNormal(``int` `[,]mat, ``int` `n)` `    ``{` `        ``int` `sum = 0;` `        `  `        ``for` `(``int` `i = 0; i < n; i++)` `            ``for` `(``int` `j = 0; j < n; j++)` `                ``sum += mat[i,j] * mat[i,j];` `                `  `        ``return` `(``int``)Math.Sqrt(sum);` `    ``}` `    `  `    ``// Returns trace of a matrix of size ` `    ``// n x n` `    ``static` `int` `findTrace(``int` `[,]mat, ``int` `n)` `    ``{` `        ``int` `sum = 0;` `        `  `        ``for` `(``int` `i = 0; i < n; i++)` `            ``sum += mat[i,i];` `            `  `        ``return` `sum;` `    ``}` `    `  `    ``// Driven source` `    ``public` `static` `void` `Main () ` `    ``{` `        ``int` `[,]mat = { {1, 1, 1, 1, 1},` `                       ``{2, 2, 2, 2, 2},` `                       ``{3, 3, 3, 3, 3},` `                       ``{4, 4, 4, 4, 4},` `                       ``{5, 5, 5, 5, 5},` `    ``};`   `    ``Console.Write (``"Trace of Matrix = "` `            ``+ findTrace(mat, 5) + ``"\n"``);` `    ``Console.Write(``"Normal of Matrix = "` `                    ``+ findNormal(mat, 5));` `        `  `    ``}` `} `   `// This code is contributed by nitin mittal.` PHP `` Javascript `` Output ```Trace of Matrix = 15 Normal of Matrix = 16``` Time Complexity : O(n*n) Space Complexity : O(1), since no extra space has been taken. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to [email protected]. See your article appearing on the GeeksforGeeks main page and help other Geeks. Feeling lost in the world of random DSA topics, wasting time without progress? It's time for a change! Join our DSA course, where we'll guide you on an exciting journey to master DSA efficiently and on schedule. Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 geeks! Previous Next

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## 10556 - Biometrics All about problems in Volume 105. If there is a thread about your problem, please use it. If not, create one with its number in the subject. Moderator: Board moderators Lain New poster Posts: 11 Joined: Sun Sep 21, 2003 5:45 pm Location: Russia, PetrSU Contact: ### 10556 - Biometrics Hi All! To solve this problem I used assumption, that two poligons should have same angles between corresponding edges and have the same scale factor for all corresponding edge's lengths. Does it right? ditrix New poster Posts: 33 Joined: Sat Mar 01, 2003 12:38 am Location: Paris yes, it's a basic idea, but you must also verify if the polygons has the same orientations. It means that you have to consider the angles from 0 to 180 deg as from 180 to 360 if the orientation is different. @+! DitriX Lain New poster Posts: 11 Joined: Sun Sep 21, 2003 5:45 pm Location: Russia, PetrSU Contact: Sorry I don't get it! =( It's said that "The vertices for both polygons correspond to the same set of features in the same order" For examle we may have: 1poligon: right ear tip, chin cleft, right eye, nose, left eye, left ear tip, space between front teeth 2poligon: space between front teeth, right ear tip, chin cleft, right eye, nose, left eye, left ear tip Can you give me example. BiK Experienced poster Posts: 104 Joined: Tue Sep 23, 2003 5:49 pm ### I also get WA on p10556 I use the following idea: If the polygon's vertices are denoted by 1,2,3,...,n, then I chek if the triangles 123, 234, 345, ..., (n-2)(n-1)n, (n-1)n1, n12 are similar to the corresponding triangles in the second polygon. If this is the case, then I also check if the angles between the vectors 21 and 23, 32 and 34, 43 and 45, ..., n(n-1) and n1, and 1n and 12, have the same sign as their corresponding angles in the second polygon, thus checking if the polygons have the same orientation. Is this correct? I keep receiving WA... SM for BiK Lain New poster Posts: 11 Joined: Sun Sep 21, 2003 5:45 pm Location: Russia, PetrSU Contact: 2ditrix: I don't know why, but when I rewrote my program it's got AC. I understood what were you talking about, but I haven't had that mistake. I didn't use angles, I used sin and cos instead: sin = vx1*vy2-vx2*vy1; cos = vx1*vx2+vy1*vy2; where vx1, vy1, vx2, vy2 - two standardized vectors humaira New poster Posts: 1 Joined: Thu Mar 25, 2004 10:35 am Location: Pakistan ### 10556 Biometric 1) Can anybody send me some critical inputs for this problem. 2) What is the criteria of scaling in the problem. whether the scaling would be done by fixing one particular point and then stretching the rest or whether it would be done with respect to a mid point. 3) Is it posible that a polygon is scaled first, translated to the third quardrant from the first quardrant and then reflected? gvcormac Problemsetter & Reviewer Posts: 194 Joined: Fri Mar 15, 2002 2:00 am Contact: 1) You can find the judges' data at plg.uwaterloo.ca/~acm00 2 & 3) I'm not sure I understand your questions, but I'll point out that when you do linear scaling, translation, and rotation, it doesn't make any difference what order you do them in.

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# Lecture: Logistic Regression ## 2018/06/23 I decided to post this old lecture (with some clean up) online, as I think it really well captures a lot of things students might want to know about logistic regression while using R. Note: this document was written for summer students in CMU Statistics and Data Sciences’ SURE program, and was presented to them live. # Introduction This lecture focuses on introducing a set of tools associated with logistic regression. Today we’re going to use a pretty basic dataset. There are reasons for this (we’ll see this later). This data is from the Institute of Digital Research and Eduation at UCLA. This lecture is a combination of lots of other people’s work and texts. grad_admit <- read_csv("https://stats.idre.ucla.edu/stat/data/binary.csv") %>% rank = factor(rank)) The data (explorable below) contains a set of features associated with ‘students’ records for students that applied to UCLA, with a binary variable (admit) which indicates where or not the student was admitted. library(DT) DT::datatable(grad_admit, options = list(autoWidth = "TRUE")) ## Quick EDA It’s standard (and good practice) to explore the data. Below is provided some 1d and 2d visualizations. I would encourage the reader the look into the data and think about the variables we have, how they are distributed and how they relate to each other. ### 1D EDA Quick look at the distributions (using a function basically wrote on Friday) - feel free to look at it later. # Create grid of 1d visuals of variable distributions in data frame # Args: # data: data frame you'd like to explore # ncol: number of columns for the visual output # # Returns: # list of graphics # and creates the correct visualization gg1d <- function(data, ncol = 3){ require(gridExtra) require(ggplot2) columns <- names(data) types <- sapply(data, class) num_types <- c("numeric", "integer") gglist <- list() for (i in 1:length(columns)) { if (types[i] %in% num_types) { gglist[[i]] <- ggplot(data, aes_string(x = columns[i])) + geom_histogram() }else{ gglist[[i]] <- ggplot(data, aes_string(x = columns[i])) + geom_bar() } } grid.arrange(grobs = gglist, ncol = ncol, top = "1d EDA") return(gglist) } # saving it as "." is a way to ignore the returned list (it still is saved, # but "out of sight, out of mind") . <- gg1d(grad_admit, ncol = 2) ### Extra Pairs Plot library(GGally) ggpairs(grad_admit) # Logistic Regression ## Quick training/ test split Before we proceed we will split the data into 2 groups so we can use the second group in assessing model preformance. train_idx <- sample(1:nrow(grad_admit), size = .5 * nrow(grad_admit)) ga_test <- grad_admit %>% filter(!(1:nrow(grad_admit) %in% train_idx)) Here’s our logistic regression model (we comment on the structure below). logit_fit <- glm(admit ~ gre + gpa + rank, data = ga_train, family = "binomial") summary(logit_fit) ## ## Call: ## glm(formula = admit ~ gre + gpa + rank, family = "binomial", ## data = ga_train) ## ## Deviance Residuals: ## Min 1Q Median 3Q Max ## -1.5171 -0.8152 -0.5776 1.0396 2.1894 ## ## Coefficients: ## Estimate Std. Error z value Pr(>|z|) ## (Intercept) -5.4897680 1.7895536 -3.068 0.00216 ** ## gre 0.0009141 0.0015807 0.578 0.56305 ## gpa 1.4457858 0.5353035 2.701 0.00692 ** ## rank2 -1.0321076 0.4799678 -2.150 0.03153 * ## rank3 -0.8888721 0.5036715 -1.765 0.07760 . ## rank4 -1.7703847 0.6011894 -2.945 0.00323 ** ## --- ## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 ## ## (Dispersion parameter for binomial family taken to be 1) ## ## Null deviance: 239.05 on 199 degrees of freedom ## Residual deviance: 216.35 on 194 degrees of freedom ## AIC: 228.35 ## ## Number of Fisher Scoring iterations: 4 ## Summary Breakdown Let’s break down this summary. When we ran the logistic regression we were looking at: admit ~ gre + gpa + rank Which could be mathematically represented \begin{align} logit(p) \sim & \beta_0 + \beta_1 \cdot gre + \beta_2 \cdot gda \\ & + \beta_3 \cdot \mathbb{I}(rank = 2) \\ & + \beta_4 \cdot \mathbb{I}(rank = 3) \\ & + \beta_5 \cdot \mathbb{I}(rank = 4) \end{align} where $$p = \mathbb{E}(Y|gre,gda,rank)$$ and $$logit(p) = \log(\frac{p}{1-p})$$. So, we really have 6 $$\beta$$ values that we’re estimating even though the code looks like we are only looking at 3 columns. Question 1: What happened to rank == 1? (I read Chad talked about something like this.) Answer: If we had a column in the model matrix for the indicator $$\mathbb{I}(rank = 1)$$, then would not have linearly independent columns. Specifically, the $$\beta_0$$ in the equation means the our model matrix has a column of all $$1$$s. We know that $$1 = \sum_{r=1}^4 \mathbb{I}(rank = r)$$ (which is the same as saying these sets of columns are linearly dependent). This would mean the model wouldn’t know what to estimate for $$\beta_0$$ and the $$\beta$$s related to rank. rank == 1 is the ‘base’ or ‘benchmark’ group. ### $$\beta$$ interpretation Raw $$\beta$$ values logit_fit %>% coef %>% round(3) ## (Intercept) gre gpa rank2 rank3 rank4 ## -5.490 0.001 1.446 -1.032 -0.889 -1.770 Question 2: What is the interpretation in linear regression (we are not using a linear regression)? Answer: If this was linear regression, for coefficients like $$\beta_{gpa}$$ we would interpret the value of 1.446 as meaning that for every 1 unit increase of gpa we would expect to see an increase in a continuous prediction by .616 units. Note, that a smart person might multiple the gpa coeffiecent by .1 and then discuss the change relative to a .1 unit increase of gpa as that seems more likely / useful. Additionally, for our $$\beta_{rank2}$$ style coefficients, it’s common to interpret these relative to our ‘base’ / ‘benchmark’ group, that is, if we moved a student from rank1 to rank2 the model would predict a decline in our continuous prediciton by .273 units. To interpret logistic regression $$\beta$$ values, I would encourage you to first think back to our linear equation that expressed the logistic regression model. The following steps capture how to get a clearer intepretation of the $$\beta$$ values: Converted $$\beta$$ values Instead of looking at the change of the logit(p) for a unit of the explaintory variable, let’s think about how to get something more meaningful. Instead of change in log(p/(1-p)) we can look at change in p/(1-p) logit_fit %>% coef %>% exp ## (Intercept) gre gpa rank2 rank3 rank4 ## 0.004128802 1.000914546 4.245186778 0.356255325 0.411119190 0.170267473 Since we should really be thinking about looking taking the $$e^{side}$$ for both sides in our original equation, as such these elements now relate to a multiplicative effect. That is we might interpret $$e^{\beta_2}$$ in the following manner: For every unit increase in gpa we’d expect to see the individual’s odds of acceptance to be increased by a multiple/factor of 4.245. #### Confidence Intervals for $$\beta$$ values With the base $$\beta$$ values and the converted $$\beta$$ values we can get confidence intervals: base_beta_ci <- logit_fit %>% confint() base_beta_ci ## 2.5 % 97.5 % ## (Intercept) -9.123496602 -2.079423926 ## gre -0.002185674 0.004038365 ## gpa 0.421899116 2.529627923 ## rank2 -1.986931934 -0.093554261 ## rank3 -1.892276543 0.093948393 ## rank4 -3.011408184 -0.629942791 transformed_beta_ci <- base_beta_ci %>% exp transformed_beta_ci ## 2.5 % 97.5 % ## (Intercept) 0.0001090726 0.1250022 ## gre 0.9978167131 1.0040465 ## gpa 1.5248546833 12.5488361 ## rank2 0.1371154601 0.9106886 ## rank3 0.1507282785 1.0985031 ## rank4 0.0492223159 0.5326223 ## Goodness of Fit, Deviance, and Dispersion Like the F test in linear regression, we can examine the null hypothesis that the model doesn’t explain anything more than noise compared to the null model. We do this with the change of deviance of the model deviance. Deviance is defined as $$- 2 \cdot loglikelihood(model)$$. A quick reminder about log likelihoods related to some common linear models. type likelihood log likelihood Normal $$\prod_i \frac{1}{2\sigma^2} e^{-(y_i-\beta^t x_i)^2/\sigma^2}$$ $$-n\log(2\sigma^{2}) - \frac{1}{\sigma^2}\sum_i (y_i-\beta^t x_i)^2$$ Binomial $$\prod_i \hat{p}^{y_i}(1-\hat{p})^{1-y_i}$$, where $$\hat{p}_i = \frac{e^{\beta^t x_i}}{1+e^{\beta^t x_i}}$$ (from algebraic manipulation of $$\log(\frac{\hat{p}_i}{1-\hat{p}_i}) = \beta^t x_i)$$) $$\sum_i y_i \log(\hat{p}_i) - \sum_i (1-y_i) \log(1-\hat{p}_i)$$ Poisson $$\prod_i \frac{({\hat{\lambda}_i}^{x_i}) e^{-\hat{\lambda}_i}}{x_i!}$$, where $$\hat{\lambda}_i = e^{\beta_t x_i}$$ (from algebraic manipulation of $$\log(\hat{\lambda}_i) = \beta^t x_i)$$) $$\sum_i -\hat{\lambda}_i - \log(x_i!) + x_i \log(\hat{\lambda}_i)$$ The bigger the difference (or “deviance”) of the observed values from the expected values, the poorer the fit of the model. Null deviance shows how well the response is predicted by a model with nothing but intercept. The difference in deviance is asymptotically $$\chi^2$$, as such we can test to goodness-of-fit: diff_deviance <- logit_fit$null.deviance - logit_fit$deviance diff_df <- logit_fit$df.null - logit_fit$df.residual pchisq(diff_deviance,df = diff_df,lower.tail = FALSE) ## [1] 0.0003854845 A small p-value suggests our logistic model is better than the null. Significant reduction in deviance of the fitted model from the null model. Warning: (@ 402 students) Althought the above test might be called a goodness of fit test, the more standard goodness of fit test is related to the question of how the current model fits vs the saturated model. In this situation, the deviance of the staturated model is actually $$0$$, and although we can do a similar analysis (we still get a $$\chi^2$$ distribution under the null), we instead examine: pchisq(logit_fit$deviance, logit_fit$df.residual, lower.tail = FALSE) ## [1] 0.1298283 # which can also be expressed nrow(grad_admit) - length(logit_fit$coefficients) ) dispersion ## [1] 0.5491049 pchisq(logit_fit$deviance,logit_fit$df.residual,lower.tail = FALSE) ## [1] 0.1298283 Even though our dispersion is decently close to 1, our $$\chi^2$$ test suggests its far enough away from 1 that we need to correct for it. What causes over-dispersion? For a binomial model the most likely problem is correlation across the individual Bernoulli trials within each binomial outcome. To conduct inference we must update the standard errors by multiplying by $$\sqrt{\phi}$$. # Prediction / Fit Diagnostics Now let’s get to evaluating the model, prediction and diagnostics. ## Prediction Since our model is looking to predict $$logit(\mathbb{P}(y = 1|X))$$ we can look at objects’ predicted $$y$$ value, or $$\logit{p}$$ or even $$p$$, in the following way: pred_logit <- predict(logit_fit, newdata = ga_test) pred_prob <- predict(logit_fit, newdata = ga_test, type = "response") pred_y <- 1*(pred_prob > .5) # explore all 3 above How well would we do predicting the correct admit values for the training dataset? (Suppose we split on $$\hat{p} > .5$$). mean(ga_test$admit == pred_y) ## [1] 0.685 ### Confusion Matrix Generally for binary classification, we actually look at the confusion matrix, . Predicted 1 Predicted 0 True 1 True Positive False Negative (type I) True 0 False Positive (type II) True Negative We can examine the binary classification in R with: table(ga_test$admit, pred_y) ## pred_y ## 0 1 ## 0 122 8 ## 1 55 15 ### ROC curves By the logistic regression model holds more information that that - we have probabilities!, we can visualize how well the algorithm does with an ROC curve. First, let’s do a visual and define some terms vis_data <- data.frame(truth = ga_test$admit, prob = pred_prob) ggplot(vis_data, aes(x = prob, fill = truth)) + geom_histogram(position = "identity", alpha = .5) ## stat_bin() using bins = 30. Pick better value with binwidth. Notice that the algorithm does generally give higher probabilities to those admitted, but that cutting off at $$.5$$ might now seem silly (this is probably due to differences in class memberships, which can be corrected with weighted regression - chad has talked about this). But we can also imagine just cutting at a different probability… But which one? Why not all of them ;) For a specific threshold, we could calculate 2 metrics (don’t worry to much about them) $TPR = \frac{\text{True Positives}}{\text{True Positives} + \text{False Negatives}} = \frac{\text{True Positives}}{\text{All Positives}}$ $TNR = \frac{\text{True Negatives}}{\text{True Negatives} + \text{False Positives}} = \frac{\text{True Negatives}}{\text{All Negatives}}$ If we plotted these values (or actually TPR and 1-TNR) we could get an ROC curve. library(plotROC) # I don't like this library - I can give a better function, but for now... ggplot(vis_data, aes(d = as.numeric(as.character(truth)), m = prob)) + geom_roc() + geom_abline(slope = 1,intercept = 0) + labs(title = "ROC curve", x = "1 - TNR", y = "TPR") ### How to intepret ROCs ROC curves show the tradeoff between changes in the threshold to True Positive Rates, proportion of class 1 correctly classified as class 1, and the same for class 0 (True Negative Rate). • The closer the curve to the upper left border (0,1) the better, • The closer to the y = x line the worse the classifier (the y = x line would be a classifier that either classifies all observations as 0 or 1) • It is common to use the Area Under the Curve (AUC) to measure accuracy We can caculate AUC using the same information in vis_data (the true classes and the probability estimates). library(ROCR) pred_obj <- prediction(predictions = vis_data %>% pull(prob), labels = vis_data %>% pull(truth)) roc_obj <- performance(pred_obj,x.measure = "fpr", measure = "tpr") # plot(roc_obj) # base graphic roc_obj_auc <- performance(pred_obj, measure = "auc") [email protected][[1]] ## [1] 0.6178571 ## Goodness-of-Fit Graphics library(ResourceSelection) hl_test <- hoslem.test(as.numeric(as.character(ga_train$admit)), logit_fit$fit,g = 10) hl_test ## ## Hosmer and Lemeshow goodness of fit (GOF) test ## ## data: as.numeric(as.character(ga_train$admit)), logit_fit$fit ## X-squared = 16.389, df = 8, p-value = 0.03714 Such a test is slightly finicky, I tend to like graphical visualization: expected_vs_actual <- hl_test$observed %>% cbind(hl_test$expected) %>% data.frame %>% mutate(true_prop = y1/(y0 + y1), expected_prop = yhat1/(yhat1 + yhat0)) ggplot(expected_vs_actual, aes(x = expected_prop, y = true_prop)) + geom_line() + geom_abline(slope = 1, intercept = 0, linetype = 2) + geom_point(data = vis_data, aes(x = prob, y = jitter( as.numeric(as.character(truth)), amount = .05 ) ), alpha = .2) + ylim(-.1,1.1) + xlim(0,1) + labs(x = "True Proportion/ True Response", y = "Estimated Proportion")

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# Binomial Distribution confusion #### Aleckandre98 ##### New Member Hi, I'm currently confused on a question and don't quite understand how to solve it. The question is Suppose 75% of all households with telephone service have voicemail. You are conducting a survey about voicemail usage and you randomly call 10 households. What is the probability that the first 7 households have voicemail and the last 3 do not? So, how exactly would I do both the first 7 households have voicemail and the last 3 don't? I've tried doing P(X<=7) but not sure how to incorporate the "3 do not" part. #### rogojel ##### TS Contributor hi, it is not a binomial distribution stricly. You need to calculate P(A*B) where A is getting 7 voicemails out of 7 and B is getting 3 without a voicemail out of 3. Regards

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## Info We assume that there are 20 possible suspects, 10 servants and 10 guests, only one of whom is guilty of murdering the owner of an English country estate. This provides the totals in the bottom row. The totals for each column are multiplied by the appropriate probabilities in each cell of Table 3.3 to get the values in this table, which can be used to compute the probability that an innocent suspect is mistakenly identified as guilty. This probability is 0.95/1.31 = 0.73. We assume that there are 20 possible suspects, 10 servants and 10 guests, only one of whom is guilty of murdering the owner of an English country estate. This provides the totals in the bottom row. The totals for each column are multiplied by the appropriate probabilities in each cell of Table 3.3 to get the values in this table, which can be used to compute the probability that an innocent suspect is mistakenly identified as guilty. This probability is 0.95/1.31 = 0.73. called posterior probability of guilt is 27%. This is a substantial increase over the 10% prior probability of guilt, but it is far short of the standard expressed by the phrase "beyond a reasonable doubt."6 Suppose we wanted to be 90% sure of blaming the correct person for the crime. How many suspects would we have to exclude based on other evidence so that the posterior probability of guilt following scent identification by the dog was greater than 90%? The best we can do is to reduce the pool of suspects from the original 20 to two, based on other evidence. In this case, after the dog picks one of these suspects, the posterior probability that this suspect is guilty is 88%: not quite reaching the standard for establishing guilt beyond a reasonable doubt .7 In pract ice, other ev idence besides scent ident ificat ion is often qualitative and not easily converted into the prior probabilities used in this example. However, if Schoon's results on the abilities of trained dogs to match human scents are accurate, the method of scent lineups doesn't seem to have much credibility. Even in the ideal and unlikely situation that other evidence reduces the pool of potential suspects to two, there is still about a 12% chance that the dog would pick the wrong suspect from a lineup. Let's consider a parallel situation in which these kinds of calculations are helpful but actually validate a common forensic method rather than cast ing doubt on it. This is the well-known use of DNA to match suspects to blood or tissue samples found at a crime scene (Gomulkiewicz and Slade 1997). We can set up tables just like Tables 3.3 and 3.4 to show the calculations. With current technology, the likelihood of missing a match between a truly guilty suspect and a sample of that suspect's DNA from a crime scene is very low, certainly less than 0.5% and perhaps in principle equal to zero. This is a false negative or false mismatch, shown in the lower-left cell of Table 3.5. The probability of a false positive depends on four factors: the possibility that an innocent person shares the same DNA profile as the perpetrator of the crime for the regions of DNA that were analyzed, the possibility of laboratory error such as contamination of a sample, the possibility that the innocent person left his or her DNA at the crime scene but did not commit the crime, and the possibility that a DNA sample from the innocent person was planted at the crime scene. If the latter three possibilities can be ruled out, the probability of a false positive ranges from one in 100,000 to one in 1 bill ion, for suspects who are not relatives of the perpetrator of the crime? Because brothers, for example, share 50% of their DNA, the likelihood of a false positive is as high as 0.26 for an innocent suspect who is the brother of the actual criminal. For Table 3.5, I assume that the pool of potential suspects includes only unrelated people, and I use an intermediate value for the probability of a false posit ive identificat ion of one in 10 million. Applying these values to the murder at the English country estate (and assuming that the mysterious handkerchief has blood on it that does not match that of the victim), Table 3.6 shows that the likelihood of guilt of a suspect whose DNA profile matches that of the blood on the handkerchief is Table 3.5. Identification of suspects based on blood or tissue samples containing DNA collected at a crime scene. Status of Suspect Guilty Innocent (Suspect = Perpetrator) (Suspect ^ Perpetrator) Matches Suspect 0.995 0.0000001 Does Not Match Suspect 0.005 0.9999999 ### DNA Collected at Crime Scene The values in the table are the probabilities that DNA collected at the crime scene (1) matches that of the suspect if the suspect is guilty (the upper-left cell), (2) matches that of the suspect if the suspect is innocent (a false positive result in the upper-right cell), (3) does not match that of the suspect even though the suspect is guilty (a false negative result in the lower-left cell), and (4) does not match that of the suspect if the suspect is innocent (the lower-right cell). These values were derived from a review of the use of DNA evidence in court by Gomulkiewicz and Slade (1997). 0.995/0.9950019, which is greater than 99.99%. Even if the pool of potential suspects was much larger than 20, DNA evidence may be quite persuasive, provided factors like sloppiness in lab techniques or plant ing evidence at the crime scene can be excluded. However, there may be situations in which even DNA evidence is not as conclusive as might be assumed. Suppose the only evidence available is DNA from a crime scene. The FBI and other law enforcement agencies have databases of DNA profiles for large numbers of individuals who have had various encounters with the legal system. The sizes of these databases are increasing daily. If the authorities have no other evidence, they may scan the database to see if there is a profile that matches that of the DNA from the crime scene. If there are 5 million profiles in the database and we assume that the guilty person is one of those 5 m illion, we would subst itute 5 million for the overall total in the bottom right cell of Table 3.6. In this case, with one guilty person Table 3.6. An application of DNA identification to a murder in an English country estate. Status of Suspect Guilty (Suspect Innocent (Suspect = Perpetrator) ^ Perpetrator) Totals DNA Collected Matches Suspect 0 0

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## September 01, 2010 ### On an Example of Vienneau's Attention conservation notice: 1600+ dry, pedantic words and multiple equations on how some heterodox economists mis-understand ergodic theory. Robert Vienneau, at Thoughts on Economics, has posted an example of a stationary but non-ergodic stochastic process. This serves as a reasonable prompt to follow up on my comment, a propos of Yves Smith's book, that the post-Keynesian school of economists seems to be laboring under a number of confusions about "ergodicity". I hasten to add that there is nothing wrong with Vienneau's example: it is indeed a stationary but non-ergodic process. (In what follows, I have lightly tweaked his notation to suit my own tastes.) Time proceeds in discrete steps, and $X_t = Y Z_t$, where $Z$ is a sequence of independent, mean-zero, variance 1 Gaussian random variables (i.e., standard discrete-time white noise), and $Y$ is a chi-distributed random variable (i.e., the square root of something which has a chi-squared distribution). $Z$ is transparently a stationary process, and $Y$ is constant over time, so $X$ must also be a stationary process. However, by simulation Vienneau shows that the empirical cumulative distribution functions from different realizations of the process do not converge on a common limit. In fact, the result can be strengthened considerably. Given $Y= y$, $X$ is just Gaussian white noise with standard deviation $y$, so by the Glivenko-Cantelli theorem, the empirical CDF of $X$ converges almost surely on the CDF of that Gaussian. The marginal distribution of $X_t$ for each $t$ is however a mixture of Gaussians of different standard deviations, and not a Gaussian. Conditionally on $Y$, therefore, the empirical CDF converges to the marginal distribution of the stationary process with probability 0. Since this convergence has conditional probability zero for every value of $y$, it has probability zero unconditionally as well. So Vienneau's process very definitely fails to be ergodic. (Proof of the unconditionality claim: Let $C$ be the indicator variable for the empirical CDF converging to the marginal distribution. $\mathbf{E}\left[C|Y=y\right] = 0$ for all $y$, but $\mathbf{E}\left[C\right] = \mathbf{E}\left[\mathbf{E}\left[C|Y=y\right]\right]$ by the law of total expectation.) Two things, however, are worth noticing. First, Vienneau's $X$ process is a mixture of ergodic processes; second, which mixture component is sampled from is set once, at the beginning, and thereafter each sample path looks like a perfectly well-behaved realization of an ergodic process. These observations generalize. The ergodic decomposition theorem (versions of which go back as far as von Neumann's original work on ergodic theory) states that every stationary process is a mixture of processes which are both stationary and ergodic. Moreover, which ergodic component a sample path is in is an invariant of the motion — there is no mixing of ergodic processes within a realization. It's worth taking a moment, perhaps, to hand-wave about this. Start with the actual definition of ergodic processes. Ergodicity is a property of the probability distribution for whole infinite sequences $X = (X_1, X_2, \ldots X_t, \ldots )$. As time advances, the dynamics chop off the initial parts of this sequence of random variables. Some sets of sequences are invariant under such "shifts" — constant sequences, for instance, but also many other more complicated sets. A stochastic process is ergodic when all invariant sets either have probability zero or probability one. What this means is that (almost) all trajectories generated by an ergodic process belong to a single invariant set, and they all wander from every part of that set to every other part — they are "metrically transitive". (Because: no smaller set with any probability is invariant.) From this follows Birkhoff's individual ergodic theorem, which is the basic strong law of large numbers for dependent data. If $X$ is an ergodic process, then for any (integrable) function $f$, the average of $f(X_t)$ a sample path, the "time average" of $f$, converges to a unique value almost surely. So with probability 1, time averages converge to values characteristic of the ergodic process. Now go beyond a single ergodic probability distribution. Two distributions are called "mutually singular" if one of them gives probability 1 to an event which has probability zero according to the other, and vice versa. Any two ergodic processes are either identical or mutually singular. To see this, realize that two distributions must give different expectation values to at least one function; otherwise they're the same distribution. Pick such a distinguishing function and call it $f$, with expectation values $f_1$ and $f_2$ under the two distributions. Well, the set of sample paths where $\frac{1}{n}\sum_{t=1}^{n}{f(X_t)} \rightarrow f_1$ has probability 1 under the first measure, and probability 0 under the second. Likewise, under the second measure the time average is almost certain to converge on $f_2$, which almost never happens under the first measure. So any two ergodic measures are mutually singular. This means that a mixture of two (or more) ergodic processes cannot, itself, be ergodic. But a mixture of stationary processes is stationary. So the stationary ergodic processes are "extremal points" in the set of all stationary processes. The convex hull of these extremal points are the set of stationary but non-ergodic processes which can be obtained by mixing stationary and ergodic processes. It is less trivial to show that every stationary process belongs to this family, that it is a mixture of stationary and ergodic processes, but this can indeed be done. (See, for instance, this beautiful paper by Dynkin.) Part of the proof shows that which ergodic component a stationary process's sample path is in does not change over time — ergodic components are themselves invariant sets of trajectories. The general form of Birkhoff's theorem thus has time averages converging to a random limit, which depends on the ergodic component the process started in. This can be shown even at the advanced undergraduate level, as in Grimmett and Stirzaker. At this point, three notes seem in order. 1. Many statisticians will be more familiar with a special case of the ergodic decomposition, which is de Finetti's result about how infinite exchangeable random sequences are mixtures of independent and identically-distributed random sequences. The ergodic decomposition is like that, only much cooler, and not tainted by the name of a Fascist. (That said, de Finetti's theorem actually covers Vienneau's example.) 2. Following tradition, I have stated the ergodic decomposition above for stationary processes. However, it is very important that this limitation is not essential. The broadest class of processes I know of for which an ergodic decomposition holds are the "asymptotically mean-stationary processes". The defining property of such processes is that their probability laws converge in Cesaro mean. In symbols, and writing $P_t$ for the law of the process from $t$ onwards, we must have $\lim_{n\rightarrow\infty}{\frac{1}{n}\sum_{t=1}^{n}{P_t(A)}} = P(A)$ for some limiting law $P$. (I learned to appreciate the importance of AMS processes from Robert Gray's Probability, Random Processes and Ergodic Properties, and stole those ideas shamelessly for Almost None.) This allows for cyclic variation in the process, for asymptotic approach to a stationary distribution, for asymptotic approach to a cyclically varying process, etc. Every AMS process is a mixture of ergodic AMS processes, in exactly the way that every stationary process is a mixture of ergodic stationary processes. I actually don't know whether the ergodic decomposition can extend beyond this, but I suspect not, since the defining condition for AMS is very close to a Cesaro-mean decay-of-dependence property which turns out to be equivalent to ergodicity, namely that, for any two sets $A$ and $B$ $\lim_{n\rightarrow\infty}{\frac{1}{n}\sum_{t=0}^{n-1}{P_1(A \cap T^{-t} B)}} = P_1(A) P(B)$ where $T^{-t}$ are the powers of the back-shift operator (what time series econometricians usually write $L$), so that $^{-t} B$ are all the trajectories which will be in the set $B$ in $t$ time-steps. (See Lemma 6.7.4 in the first, online, edition, of Gray, p. 148). This means that, on average, the far future becomes unpredictable from the present. 3. In light of the previous note, if dynamical systems people want to read "basin of attraction" for "ergodic component", and "natural invariant measure on the attractor" for "limit measure of an AMS ergodic process", they will not go far wrong. As the last remark suggests, it is entirely possible for a process to be stationary and ergodic but to have sensitive dependence on initial conditions; this is generally the case for chaotic processes, which is why there are classic articles with titles like "The Ergodic Theory of Chaos and Strange Attractors". Chaotic systems rapidly amplify small perturbations, at least along certain directions, so they are subject to positive destabilizing feedbacks, but they have stable long-run statistical properties. Going further, consider the sort of self-reinforcing urn processes which Brian Arthur and collaborators made famous as models of lock-in and path dependence. (Actually, in the classification of my old boss Scott Page, these models are merely state-dependent, and do not rise to the level of path dependence, or even of phat dependence, but that's another story.) These are non-stationary, but it is easily checked that, so long as the asymptotic response function has only a finite number of stable fixed points, they satisfy the definition of asymptotic mean stationarity given above. (I leave it as an exercise whether this remains true in a case like the original Polya urn model.) Hence they are mixtures of ergodic processes. Moreover, if we have only a single realization — a unique historical trajectory — then we have something which looks just like a sample path of an ergodic process, because it is one. ("[L]imiting sample averages will behave as if they were in fact produced by a stationary and ergodic system" — Gray, p. 235 of 2nd edition.) That this was just one component of a larger, non-ergodic model limits our ability to extrapolate to other components, unless we make strong modeling assumptions about how the components relate to each other, but so what? I make a fuss about this because the post-Keynesians seem to have fallen into a number of definite errors here. (One may see these errors in e.g., Crotty's "Are Keynesian Uncertainty and Macrotheory Compatible?" [PDF], which however also has insightful things to say about conventions and institutions as devices for managing uncertainty.) It is not true that non-stationarity is a sufficient condition for non-ergodicity; nor is it a necessary one. It is not true that "positive destabilizing feedback" implies non-ergodicity. It is not true that ergodicity is incompatible with sensitive dependence on initial conditions. It is not true that ergodicity rules out path-dependence, at least not the canonical form of it exhibited by Arthur's models. Update, 12 September: Fixed the embarrassing mis-spelling of Robert's family name in my title. Manual trackback: Robert Vienneau; Beyond Microfoundations Posted at September 01, 2010 11:50 | permanent link

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 Grade 6 Mathematics Module 3, Topic B, Overview | EngageNY ## Grade 6 Mathematics Module 3, Topic B, Overview In Topic B, students apply their understanding of a rational number’s position on the number line (6.NS.C.6c) to order rational numbers.  Students understand that when using a conventional horizontal number line, the numbers increase as you move along the line to the right and decrease as you move to the left.  They recognize that if a and b are rational numbers and a < b, then it must be true that -a > -b.  Students compare rational numbers using inequality symbols and words to state the relationship between two or more rational numbers.  They describe the relationship between rational numbers in real-world situations and with respect to numbers’ positions on the number line (6.NS.C.7a, 6.NS.C.7b).  For instance, students explain that -10° F is warmer than -11º F because -10 is to the right (or above) -11 on a number line and write -10° F > -11º F.  Students use the concept of absolute value and its notation to show a number’s distance from zero on the number line and recognize that opposite numbers have the same absolute value (6.NS.C.7c).  In a real-world scenario, students interpret absolute value as magnitude for a positive or negative quantity.  They apply their understanding of order and absolute value to determine that, for instance, a checking account balance that is less than -25 dollars represents a debt of more than \$25 (6.NS.C.7d).

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Вы находитесь на странице: 1из 44 # Module 3: Electric Fundamentals 1) Copper losses in a DC motor are caused by? a. High magnetic reluctant circuit b. High resistance field windings c. Copper bearings 2) Armature torque is proportional to? a. Field strength only b. Armature current only c. Armature current and field strength 3) Rotational speed of a DC motor is proportional to a. Armature current and field strength b. Armature current only c. Field strength only 4) Increasing the voltage applied to the armature causes rotational speed to a. Decrease b. Increase c. Stay the same only torque changes 5) Reversing the supply to a permanent magnet field motor causes a. To stop rotating b. Continue to rotate in same direction c. To reverse its rotation 6) A DC machine commutator is cleaned using a. Paraffin b. Lead free petrol c. An abrasive material 7) The expected brush spring tension on a 6 kW DC generator would be a. 28 to 36 grams b. 28 to 36 Kg c. 28 to 36 ounces 8) Excessive brush arcing is caused by? a. Low brush spring tension b. Positioning the brushes on the MNA c. Fitting of interpoles 9) A DC shunt motor has a. High armature resistance b. Low starting torque c. High starting torque 10) DC generators are rated in a. Horsepower b. kVA c. Watts 11) To decrease the RPM of a DC shunt motor a. Field resistance is increased b. Field resistance is decreased c. Armature current is increased 12) Rotational speed of a DC motor is controlled by a. The field resistance only b. The armature resistance only c. Either of the above 13) The expected speed range of a DC generator is a. 5000 to 8000 RPM b. 1000 to 2000 RPM c. 18000 to 20000 RPM 14) When the battery voltage is higher than the charger voltage a. The charger will charge the battery b. No current will flow between the charger and the battery c. The battery will discharge through the charger 15) The Ni/Cad battery capacity test is carried out a. Every 3 months b. Only when the battery fails to maintain its charge c. At the periods stated in the AMS 16) A frequency wild generator a. Is always used as a DC generator b. Has a variable frequency output c. Maintains constant RPM but has variable frequency 17) RPM of a DC generator is a. Almost constant at any aircraft/engine speed b. Relevant to the aircraft speed c. Related to engine speed 18) The Ni/Cad battery a. Requires an approved modification to replace a lead acid battery b. Is a direct replacement for a lead acid battery c. Can replace a lead acid battery at same capacity 19) The battery charger normally charges the battery with a. A constant voltage b. Either a current or a constant voltage c. A constant current 20) The minimum brush length is measured on a. The shortest side b. An average is taken c. The longest side 21) Constant speed AC generators are normally a. Excited by the battery bus-bar b. Self-exciting c. Excited by the DC supply 22) As a lead acid cell discharges its relative density a. Increase b. Remains the same c. Decrease 23) The relative density of a discharged lead acid battery is a. 1.150 b. 1.200 c. 1.260 24) The nominal voltage of a charged lead acid cell a. 1.8V b. 2.2V c. 2.0V 25) Increasing the cell plate area will a. Increase the current b. Increase the voltage c. Increase the capacity 26) In an AC generator a. The moving part is termed the armature b. The moving part is termed the rotor c. The moving part is termed the inductance coil 27) When frequency increases in an inductive circuit the current flow will a. Increase b. Remains the same c. Decrease 28) In an inductive circuit the current a. Lags the voltage b. Leads the voltage c. Is in phase with the voltage 29) Apparent power is calculated a. I squared x I b. V x I c. V squared divided by R 30) A motor clutch is opened when a. The motor is switched on b. Rotational speed is reduced c. The motor is switched off 31) In a circuit with purely resistive loads a. Voltage leads current b. Voltage and current are in phase c. Voltage lags current 32) In an inductive circuit a. Voltage and current are in phase b. Current leads voltage c. Voltage leads current 33) In capacitive circuit a. Voltage lags behind current b. Voltage leads current c. Current is in phase with the voltage 34) If a circuit has a power factor of 0 it will contain a. Resistance only b. Inductance only c. Resistance and inductance 35) Any electric motor is protected against sudden loads by a. A brake b. An engage clutch c. A mechanical torque limiter 36) Damage to the system and motor can be caused by a. An open circuit break coil b. Incorrectly set limit switches c. An open circuit clutch coil 37) To calculate power factor a. Apparent power divided by true power b. Apparent power multiplied by true power c. True power divided by apparent power 38) The back EMF produced by an inductance depends on a. The inductance and the rate of change of the current b. The inductance of the indicator and the current flowing in it c. The inverse of the inductance and the size of the current 39) The vent of a lead acid battery cell is a. Sealed b. Leak proof c. Completely open 40) The electrolyte in a lead acid cell is a. A sulphuric acid b. A caustic acid c. A solution of sulphuric acid 41) An autotransformer has a. Two windings b. One wining c. Three windings 42) What happens to a thermistor with an increase of temperature a. Resistance increases b. Resistance remains constant c. Resistance decreases 43) A transformer has a turn ratio of 10 to 1 and a load impedance of 5 Ohms what is the input impedance? a. 500 Ohms b. 50 Ohms c. 20 Ohms 44) A current transformer when disconnected from a circuit must be a. Left open circuit b. Short circuited c. Have a load resistor 45) Economy coils a. Give a time delay b. Allow high current to pass c. Allow a relay to be held in on a low current 46) Slugged relays a. Operate quickly b. Introduce a time delay into the circuiting switching c. Must always be used on heavy duty circuit 47) To obtain a supply of 24V 60AH connect two a. 24V 30AH batteries in series b. 12V 30AH batteries in series c. 24V 30AH batteries in parallel 48) A motor brake is applied when the motor a. Is switched on b. Is switched off c. Applies excessive torque 49) A millivolt drop test on a heavy duty contacter is carried out with the contacts a. Open b. Closed c. Closing 50) Millivolt drop test reads 150mV at 300A contact resistance is a. 50 milliOhms b. 50 Ohms c. 0.0005 Ohms 51) Slugged relays a. Operate quickly b. Introduce a time delay into the circuiting switching c. Must always be used on heavy duty circuit 52) To obtain a supply of 24V 60AH connect two a. 24V 30AH batteries in series b. 12V 30AH batteries in series c. 24V 30AH batteries in parallel 53) A motor brake is applied when the motor a. Is switched on b. Is switched off c. Applies excessive torque 54) A millivolt drop test on a heavy duty contacter is carried out with the contacts a. Open b. Closed c. Closing 55) Millivolt drop test reads 150mV at 300A contact resistance is a. 50 milliOhms b. 50 Ohms c. 0.0005 Ohms 56) A lagging power factor may be improved by a. Increasing the inductivity b. Increasing the capacitance c. Increasing the resistance 57) Two in phase voltages may be represented by a. Direct values b. Added values c. Vector sums 58) An electrostatic voltmeter has? a. Low loading effect on circuit b. High loading effect on a circuit c. Medium loading effect on a circuit 59) A simple ohmmeter may be used to a. Measure continuity and all resistance values b. Measure continuity and unknown resistance values c. Unknown resistance values only 60) Before using a simple ohmmeter a. It must be calibrated b. The test leads should be checked for resistance c. The test leads should be shorted together and the indicator set to zero 61) A High pass filter will a. Allow frequencies below a certain value to pass b. Allow frequencies above a certain value to pass c. Allow frequencies within a range to pass 62) Faradays Law States that a. The magnitude of the EMF is directly proportional to the magnetic flux b. The magnitude of the EMF is indirectly proportional to the Rate of change of flux c. The magnitude of the EMF is directly proportional to the Rate of change of flux 63) Magnetic inclination is the least at a. The equator b. The poles c. The isoclines 64) The voltage rating of a capacitor is a. The min voltage required to change b. The max voltage can be constantly applied c. Normal operating voltage 65) The relative permeability of a capacitor is a. The permeability of the dielectric b. The permeability of dielectric in relation to dry c. The permeability of dielectric in relation to a vacuum 66) When measuring the phase and line voltages of a generator, it was found that line and phase voltages were equal. The generator is a. Star wound b. Delta wound c. Either delta or star wound 67) To change the direction of a 3-phase induction motor you would a. Swap all of the input connections b. Remove one of the input connections c. Swap two of the stator connections 68) In a capacitive circuit current a. Leads voltage b. Lags voltage c. Is in phase with voltage 69) In an inductive circuit current a. Voltage lags current b. Voltage leads current c. Is in phase with voltage 70) What is the phase difference in a circuit with 100v, drawing 0.5 amps, consuming 50 Watts a. 45 b. 60 c. 90 71) How much power is dissipated or absorbed when Battery has 20 Volts and resistor has 4 ohms? a. 0.1 watt b. 100 watt c. 0.001 watt 72) A circuit has 200 Volts and R1 = 5 ohms and R2 = 4 ohms in series. What is the total current of the circuit? a. 108 A b. 40 A c. 22.2 A 73) A circuit has 200 Volts and R1 = 5 ohms and R2 = 4 ohms in series. What is the voltage drop across R2? a. 88.8 V b. 111.1 V c. 200 V 74) When the north pole of one magnet is placed near the south pole of another magnet a. They will repel each other b. They will attract each other c. 10 & 20 75) Which of the following is true about lines of magnetic force? a. They do not have direction b. They do not follow iron c. They always form complete loops 76) A material that allows a large amount of lines of force to go through it has a a. High reluctance b. High permeability c. Low flux density 77) The lines of force that surround a magnet are called a. Permeance b. Magnetic inductance c. Flux lines 78) When a piece of iron is inserted into the core of a coiled wire through which current is flowing, the iron becomes a(n) a. Electromagnet b. Generator c. Magnetic domain 79) A property similar to conductance in a basic DC circuit is called ........ in a magnetic field a. Reluctance b. Magnetic flux c. Permeance 80) In order to increase the amount of EMF in a wire being moved through a magnetic field, the a. Strength of the magnetic field will have to decrease b. Length of the conductor will need to decrease c. Speed of conductor movement will need to increase 81) If you want to convert an AC generator into a CD generator, you will need a. Brushes and reluctance b. Commutator and brushes c. Brushes and a magnetic domain 82) Disregarding the number of turns in the coil and the current flow and electromagnet's strength can be increased greatly by adding a a. Residual magnetism b. Paramagnetic core c. Ferromagnetic core 83) The strength of an electromagnet is proportional to the number of turns of wire in the coil and the current flowing through the coil a. True b. False 84) Forming a loop in a wire having current flowing through it forms a simple a. Antenna b. Electromagnet c. Magnet 85) An excess of electrons at one end of a conductor and a deficiency of electrons at the other end creates a. Electromagnetic induction b. Power c. Potential difference 86) In a simple AC generator, which part of the generator rotates to produce the induced voltage? a. Permanent magnets b. Armature c. Slip rings 87) The stationary member of a brushless alternator which provides connections for the external load is called a a. Winding b. Rotor c. Stator 88) In a modern brushless AC generator, which part of the generator rotates to produce the induced voltage? a. Armature b. Rotor c. Stator 89) AC is widely used because of its versatility a. False b. True 90) The output voltages of a three-phase alternator are separated by 120 degrees a. False, they are 90 degrees apart b. False c. True 91) In a simple AC generator, which part of the generator rotates to produce the induced voltage a. Permanent magnets b. Armature c. Slip rings 92) The stationary member of a brushless alternator which provides connections for the external load is called a a. Winding b. Rotor c. Stator 93) In a modern brushless AC generator, which part of the generator rotates to produce the induced voltage? a. Armature b. Rotor c. Stator 94) AC is widely used because of its versatility a. False b. True 95) The output voltages of a three-phase alternator are separated by 120 degrees a. False, they are 90 degrees apart b. False c. True 96) The unit of power is a. Ampere b. Watt c. Farad 97) The unit of electrical potential is a. Henry b. Ampere c. Volt 98) One megohm is equal to a. 1,000,000 ohms b. 10,000 ohms c. 1,000 ohms 99) The unit that changes ac to dc is a. A transformer b. A rectifier c. An inverter 100) The unit that changes dc to ac is a. A transformer b. A rectifier c. An inverter 101) To find the current flowing in a dc circuit, the correct formula is a. I = R : E b. I = E : R c. E = I : R 102) If service no 1 is isolated from the supply shown there will be - three fuses circuit a. An increase in supply voltage b. A decrease in supply voltage c. A decrease in total current consumption 103) The reduction of the variable resistor R1 shown would - divider R1 and R2 in series a. Increase the total current drawn b. Not affect the total current drawn c. Decrease the total current drawn 104) Since electrical supplies taken from a bus bar are in parallel, isolating some of the services would a. Not affect the current consumption, it would reduce the voltage b. Reduce the current consumption from the bus bar c. Increase the current consumption from the bus bar 105) During circuit testing, an insulation resistance test a. Also indicates that continuity is satisfactory b. Should be carried out before a continuity test c. Should be carried out after a continuity test 106) When checking the bonding of a component which anodic ally treated against corrosion the a. Resistance of the anodic film is negligible and may be ignored b. Anodic film must be penetrated to ensure good contact of the test probes c. Resistance of the anodic film must be measured and taken into account during the test 107) Relative humidity has to be taken into account when conducting a. Replacement of items equipment b. Battery percentage capacity tests c. Wiring insulation resistance tests 108) The type of switch (exclusive OR) is a. Single pole change over b. Single throw, double pole c. Double pole change over 109) When resistors are in parallel a. The total resistance is larger than the smallest resistor b. The total resistance is smaller than the smallest resistor c. The total resistance is the sum of all resistors 110) When resistors are in parallel, the total current is equal to a. The reciprocal of all the currents b. The current across one resistor c. The sum of the currents 111) The voltage output of a generator is controlled by a. Varying the resistance of the output b. Varying the current of the field c. Varying the current of the output 112) When the generator field is supplied through the battery master switch, if the switch is selected off whilst the engine is running a. The generator will continue to produce since it is self sustaining b. The generator will only continue to produce power if the engine is above idle speed, since the generator is self sustaining c. The generator will be de-excited and hence, cease to produce power 113) A short circuit between the supply and earth a. Could be very dangerous as the current drawn will be very high b. Is not dangerous as the current drawn will be low c. Does not matter if the circuit uses the aircraft earth return 114) Two similar 12V batteries connected in parallel will produce a. 24V emf with twice the capacity of each battery b. 12V emf with twice the capacity of each battery c. 24V emf with the same capacity of each battery 115) An increase in operating temperature in most electrical devices carrying current results in a. A decrease in resistance and increase in current b. Not affect on the resistance and current c. An increase in resistance and decrease in current 116) An increase in operating temperature in most electrical devices carrying current results in a. A decrease in resistance and increase in current b. An increase in resistance and decrease in current c. No effect on the resistance and current 117) A direct current of 12 mA flows through a circuit which has a resistance of 1,000 ohms. The power dissipated by the circuit is a. 12 mW b. 12 W c. 144 mW 118) A short circuit should be detected by the use of a. An insulation tester only b. A continuity tester c. A continuity tester or an insulator tester 119) The result of an insulation resistance test of an electrical cable a. Also indicate continuity of a cable b. May vary with the weather conditions under which the aircraft is tested c. Will always be infinity if the cable is correctly installed 120) If, in a simple electrical circuit, the power consuming devices are in parallel, the total current consumed is equal to the sum of the a. A decrease in resistance and increase in current. Currents taken by the devices divided by the number of devices b. Reciprocals of the currents taken by the devices c. Current taken by the devices 121) The 24 battery has an internal resistance of 1 ohm and the ammeter indicates 12 A. The value of the load resistor is a. 2 ohms b. 1 ohms c. 6 ohms 122) A flexible fuel pipe undergoing inspection after a period of service requires to be checked electrically a. For continuity of the integral bonding and insulation of the outer cover b. For insulation of the outer cover only c. For continuity of the integral bonding only 123) The position of a ratio meter pointer is determined by a. The opposing torques of the two coils relative to the permanent magnet field b. The opposing torques developed by both coils c. The torque of one coil relative to the permanent magnet field 124) A frequency wild generator is used for a. Instruments and navigation equipment b. Deicing load c. Any ac load 125) A soft iron core is used in electromagnets a. It has high permeability and high coercivity b. It has low permeability and high coercivity c. It has high permeability and low coercivity 126) One purpose of electrical bonding on aircraft is a. To isolate all components electrically and therefore make the static potential constant b. To safely dissipate local static charges and lightning strikes c. To prevent lightning strike damage 127) The various parts of an aircraft airframe are maintained at the same potential by a. Static discharge wicks b. The supply bus bars c. Bonding 128) The purpose of a circuit breaker in an electrical system is a. To break a particular circuit when an overload condition occurs b. To prevent the battery being discharged through the generator when the engine is stopped c. The torque of one coil relative to the permanent magnet field 129) The risk of a fire due to static electricity is overcome a. By fitting static wicks and insulating all metal components b. By connecting all metal component by bonding c. By fitting static wicks and isolating the battery from inflammable gas sources 130) A loss of electrical insulation results in a. An open circuit between the supply and earth b. An open circuit in the supply c. A short circuit between the supply and earth 131) The test equipment normally used to carry out conductivity test on an electrical cable is a. A high tension circuit tester b. A low reading ohmmeter c. An ammeter 132) An insulation resistance test using a megger will show up a. Both open and short circuit b. Open circuits only c. Short circuits only 133) When an electrical supply becomes open circuit a. The loss of continuity will prevent its component from functioning b. The fuse or circuit breaker should isolate the circuit due to the increased current drawn c. The component will operate normally, but will not switch off 134) When the GPU is connected to the aircraft a. It is a parallel with the aircraft generator b. It is not parallel with the aircraft generator c. It supplies essential services only 135) The term dry joint when applied to soldering is when a. The surfaces have been soldered without flux b. The penetration is complete, forming a water-tight joint c. The surfaces to be joint have not reached the required temperature 136) Inertia switches are primarily used in aircraft a. To indicate the aircraft is airborne b. For emergency circuit c. To operate a circuit after a time delay 137) If the aircraft structure is used for the earth. When fitting a battery a. The positive is connected to the earth b. Neither is connected to earth c. The negative is connected to earth 138) The minimum bend radius for cable loom is a. 8 times the loom diameter b. 2 times the loom diameter c. 5 times loom diameter 139) A static inverter is used to a. Convert frequency wild AC to DC b. Convert DC to AC c. Convert frequency regulated AC to DC 140) The purpose of the short pin on the electrical ground connector plug is a. To provide earth connection b. To provide a locking mechanism to prevent the plug falling out c. To give a supply for the ground power contactor 141) The purpose of the six pins on an AC ground connector plug is a. Four long pins for AC and two short pins for earth b. Four long pins for AC and two short pins for DC holding c. Four long pins for AC and two short pins for AC holding 142) When removing the ground power from an aircraft after starting engines a. The ground power unit will automatically stop on removal of the ground power plug b. Removal of the ground power plug always automatically transfers power to the aircraft system c. The ground power supply should be switched off before removing the ground power supply plug 143) A differential relay in a twin generator system will ensure a. That the generator voltages are nearly equal before parallel b. That one generator always comes on line before the other c. That only one generator can supply the bus bar at a time 144) What is the purpose of the permanent magnets in an alternator a. To control the frequency b. To provide initial excitation c. To rectify the current 145) Two magnets in attraction. The force would be a. Proportional to the air gap b. Inversely proportional to the air gap c. Inversely proportional to squared of the air gap 146) The purpose of the six pins on an AC ground connector plug is a. Four long pins for AC and two short pins for earth b. Four long pins for AC and two short pins for DC holding c. Four long pins for AC and two short pins for AC holding 147) When removing the ground power from an aircraft after starting engines a. The ground power unit will automatically stop on removal of the ground power plug b. Removal of the ground power plug always automatically transfers power to the aircraft system c. The ground power supply should be switched off before removing the ground power supply plug 148) A differential relay in a twin generator system will ensure a. That the generator voltages are nearly equal before parallel b. That one generator always comes on line before the other c. That only one generator can supply the bus bar at a time 149) What is the purpose of the permanent magnets in an alternator a. To control the frequency b. To provide initial excitation c. To rectify the current 150) Two magnets in attraction. The force would be a. Proportional to the air gap b. Inversely proportional to the air gap c. Inversely proportional to squared of the air gap 151) What effects the frequency in a generator a. The armature diameter b. The speed c. The resistance of the windings 152) Should alternator or generator failure occur in flight, to conserve battery power a. Its rate of discharge should be slowed by using high resistance loads only b. The battery should be discharged at a rate which corresponds to its ampere hour rating c. Load shedding should be carried out by switching off the non essential loads 153) When the generator field is supplied through the battery master switch, if the switch is selected off whilst the engine is running a. The generator will be de-excited and hence cease producing power b. The generator will continue to produce power since it is self-sustaining c. The generator will only continue to produce power if the engine is above idle since the generator is then self-sustaining 154) A generator cut-out is provided to a. Allow the generator to be isolated in a crash b. Prevent discharge of the battery through the generator c. Prevent the battery from being overcharged 155) A positive ammeter reading indicates that a. The bus-bar is being supplied by the generator b. The battery is being charged by the generator and there is no supply for the bus bar c. The bus-bar is being supplied by the generator 156) Immediately after starting, with no other electrical service switched on, the ammeter shows high ampere charge to the battery a. This indicates a generator failure so the engine must be shut down immediately b. This may be normal and is only a cause for concern if the high charge rate persists c. This indicates a battery failure since there should be on immediate charge so the engine must be shut down immediately 157) The electrolyte in a nickel cadmium cell is a. Hydrogen peroxide b. Calcium carbonate c. Potassium hydroxide 158) The correct way to determine the state of charge of a nickel cadmium battery is a. By a complete and measured discharge b. By checking the specific gravity of each cell c. By checking the voltage of each cell under load 159) If a nickel cadmium aircraft battery is not required for immediate service, it should be stored a. In the fully charged condition b. In the full discharged condition c. In a totally dry condition 160) If a nickel cadmium battery overheats, it is an indication of a. The generator voltage regulator setting is too low b. The generator is not connected to the bus bar c. Thermal runway 161) When checking a nickel cadmium battery, in situ, for serviceability a. A hydrometer must be used b. A load is applied to the battery and the voltmeter reading noted c. The electrical circuits must be isolated before installing shorting strips 162) When using ac power on the bus bar you require 28V DC for battery charging a. An AC inverter would be required b. A DC transformer and rectifier would be required c. A AC transformer and rectifier would be required 163) For battery charging the electrical supply connected to the battery must be a. DC b. AC at 50 cps c. AC at 400cps 164) As an installed battery becomes fully charged by the aircraft generator a. The generator voltage decreases to supply the steadily decreasing current b. The battery voltage nears its normal level so the charging current decreases c. The battery contactor isolates the battery from the generator 165) A battery which is assumed to be 100% efficient and to have a capacity of 60 amps-hours at the 10 hour rate will deliver a. 60 amps for 10 hours or 6 amps for 10 hours, depending upon the rate of demand b. 60 amps for 10 hours c. 6 amps for 10 hours 166) Spillage from a lead acid battery may be neutralized by a. Applying a coating of Vaseline b. Washing with a dilute solution of sodium carbonate c. Washing with distilled water 167) The specific gravity of the electrolyte in a lead acid battery a. Remains substantially constant regardless of the state of charge and is not therefore a guide to a state of charge b. Remains constant with changes in the state of charge but is a useful guide to the amount of acid contained in electrolyte c. Changes with the state of charge and is therefore a measure of the state of charge 168) Before taking SG readings of a lead acid battery recently removed from an aircraft a. The electrolyte temperature must be noted b. A period of 1 hour should have elapsed to allow the SG to stabilize c. The battery must be charged at the 10 hour rate for one hour 169) To prevent corrosion at the terminals of a lead acid battery a. Copper connectors are used b. Petroleum jelly may be applied to the connections c. The connections may be painted 170) A lead acid battery is considered to be fully charged when the a. SG reaches 1.180 b. Cells begin to gas c. SG and voltage remain constant for a specified period 179) A 12 Volt lead acid battery has a. 12 cells b. 6 cells c. 24 cells 180) The electrolyte in a lead-acid battery contains a. Nitric acid b. Hydrochloric acid c. Sulphuric acid 181) Which of the following statements is true? a. A thermocouple operates on the principle of changing resistance proportional to applied b. A thermocouple operates on the principle of potential difference between two dissimilar metals c. A thermocouple operates like a potentiometer 182) In an EGT temperature system, the ........ junction is the thermocouple and the ......... junction is the indicator a. Hot & hottest b. Cold & hot c. Hot & cold 183) When heated chromel and alumel develop ........... and ........... potentials respectively a. Positive & negative b. Negative & positive c. Hot & cold 184) A special type of sensing element with properties which allow it to produce large resistance changes for small temperature changes is a a. Thermocouple b. Thermistor c. Photo cell 185) Thermistors normally have a ............ temperature coefficient a. Constant b. Positive c. Negative 186) The unknown resistance under test when using a multimeter is measured based upon known a. Current b. Voltage c. Inductance 187) The known voltage applied to a resistance under test when using an ohmmeter, originates from a/an a. 60 Hz battery b. External battery c. Internal battery 188) The resistance connected in series with the circuits and in parallel with the meter movement when using an ammeter is called a/an a. Shunt b. Resistive inductor c. Inductive resistor 189) A megger is used to measure extremely .......... values of resistance a. Low b. High c. Negative 190) A meter which is extremely accurate in resistance readings is the a. Multimeter b. Megger c. Wheatstone bridge 191) If an RCCB opens due to an over current detection, its remote trip reset C/B will not open. a. True b. False 192) Pulling the remote trip reset C/B will cause the RCCB to a. Remain in the same position b. Close c. Open 193) D'arsonval and taut band meter movements deflect to a reading due to a. Current flow in their coil b. The counter EMF c. A difference in voltage potential 194) The effect a voltmeter has on a circuit when connected into that circuit for testing is called a. Shunting b. Loading c. Balancing 195) Sensitivity of a voltmeter is expressed in a. Volts per ohm b. Ohms per amp c. Ohms per volt 196) The higher the gauge number, the .................. the wire diameter. a. Larger b. Smaller c. Longer 197) PVC insulated wiring is no longer used due to its flammability and production of toxic gases, which are corrosive, during combustion. a. False b. True 198) A fuse is wired in parallel with the circuit load a. False b. True 199) The term ............ as applied to circuit breakers, means the breaker cannot be reset as long as the overload condition exists. a. Trouble-free b. Trip-free c. Wont reset 200) Reed switch contacts open or close as a function of being placed in close proximity to a a. Fluid substance b. Electrical source c. Magnetic field 201) A capacitor used in a split-phase motor usually creates a phase shift in the start windings a. False b. True 202) How may the direction of a DC motor be changed? a. Add a shunt resistor b. Adding a capacitors c. Swap polarity of applied voltage 203) What is the primary advantage of a shunt wound motor? a. Constant speed b. Low starting torque c. High starting torque 204) What are the characteristics of a compound wound motor? a. Has series characteristics b. Both answers are correct c. Has shunt characteristics 205) In a DC motor the voltage induced in the windings as its armature rotates is called a. Induced current b. Induced EMF c. Counter EMF 206) The advantages of an autotransformer are all except which of the following? a. Fewer turns of wire b. More eddy current c. Less power loss 207) What type device is most widely used in electronic circuits to match impedance a. Hysteresis b. Transformer c. Capacitors 208) What are the advantages of AC motors? a. Less expensive, can be made brushless b. Less expensive, always uses brushes c. Shaded pole, high starting torque 209) What two factors determine speed in an AC motor? a. Number of poles, frequency of output voltage b. Number of poles, frequency of applied voltage c. Counter EMF, number of poles 210) The applied voltage frequency in an induction motor produces what is known as: a. Operating efficiency b. Modulating speed c. Synchronous speed 211) If the power output of a transformer is 160 watts, and the power input is 200 watt, the efficiency is ....... percent. a. 84 b. 80 c. 125 212) If the coefficient of coupling is about one, and the power of the primary winding of a transformer is 40 watt, the power in the secondary winding is ..... Watts. a. 20 b. 80 c. 40 213) Voltage induced in the core of transformers cause current to flow in the core. The current is called a. Eddy current b. Core current c. High frequency losses 214) Eddy currents may be reduced by laminating the core material a. False b. True 215) The primary and the secondary windings of an autotransformer are wound on the same core. a. False b. True 216) Power factor is normally expressed as a percentage. In theory, when voltage and current are exactly in phase. The power factor will be a. Less than 100% b. 100% c. more than 100% 217) In an AC circuit, the product of E x I is known as a. Inductive reactance b. True power c. Apparent power 218) Apparent power is always greater than true power a. True b. False 219) If there are 1200 turns in the primary winding and 300 turns in the secondary winding of a transformer, and the voltage across the primary winding is 100 V, the secondary winding is ....... volts. a. 400 b. 25 c. 12 220) A transformer has a primary winding with 1000 turns and a secondary winding with 250 turns. If the current in the primary winding is 0.4 A, the current in the secondary winding is ...... A. a. 0.10 b. 0.80 c. 1.60 221) Which of the following statements is true regarding the charging of a capacitor? a. Electrons flow through the dielectric b. Electrons can flow onto a plate c. Electrons can be stored in the dielectric 222) An increase in frequency will have what effect on inductive and capacitive reactance? a. Inductive reactance goes unchanged. Capacitive reactance decreases b. Inductive reactant decreases. Capacitive reactance increases c. Capacitive reactance decreases. Inductive reactance increases 223) What is the total opposition to the current flow in a circuit having reactance at a given frequency? a. Impedance b. Resistance c. Inductance 224) In a purely resistive circuit the relationship of voltage to current is said to be in phase. a. False b. True 225) To calculate the power factor a. 0.707 x true power b. Apparent power / true power c. True power / apparent power 226) The amount of inductance produced in a coil decreases if the a. Number of turns is increased b. Diameter of the coil is decreased c. Space between the turns is decreased 227) The amount of inductance produced in a circuit can be changed if the circuit contains a ...... inductor. a. Variable b. Air core c. Iron core 228) Which of the following statements is true regarding current and voltage in a circuit containing and inductor? a. When current begins to flow, it will gradually increase to a maximum voltage b. There is no counter EMF when the current first begins to flow c. The voltage across the inductor will be zero volts all times 229) A very thin oxide film is used as the dielectric in a(n) ....... capacitor a. Mica b. Electrolytic c. Dielectric 230) A capacitor that is polarized (negative and positive leads) is called a(n) ...... capacitor a. Paper b. Ceramic c. Electrolytic 231) The amount of inductance produced in a coil decreases if the a. Number of turns is increased b. Diameter of the coil is decreased c. Space between the turns is decreased 232) A multiple phase alternator with a single-phase output of 115 V will have a phaseto-phase output of 230 V a. It varies b. True c. False 233) A three phase alternator has its A, B, C output phases marked respectively a. T1 , T2 , T3 b. A1 , B1 , C1 c. T4 , T5 , T6 234) Which of the following is a unit of inductance? a. Ohm b. Milihenry c. Hertz 235) The voltage induced in an inductor as a result of a current change is called a. Dissipated power b. Impedance c. Counter EMF 236) A three-phase alternator with a single-phase output of 115 V will have a phase-tophase output of a. 0.707 x 115 V b. 1.414 x 115 V c. 2 x 115 V 237) A multiple phase alternator with a single-phase output of 115 V will have a phaseto-phase output of 230 V a. It varies b. True c. False 238) A three phase alternator has its A, B, C output phases marked respectively a. T1 , T2 , T3 b. A1 , B1 , C1 c. T4 , T5 , T6 239) Which of the following is a unit of inductance? a. Ohm b. Milihenry c. Hertz 240) The voltage induced in an inductor as a result of a current change is called a. Dissipated power b. Impedance c. Counter EMF 241) a. A low pass fitter b. A high pass fitter c. An integrator 242) The right diagram represents a. A differentiator b. A high pass filter c. Low pass filter 243) _____________ can pass a low pass filter. a. Lower frequencies b. Higher frequencies c. Medium frequencies 244) _____________ can pass a high pass filter. a. Lower frequencies 241) ## The left diagram represents b. Higher frequencies c. Medium frequencies 245) For a middle-frequency speaker a ____________ filter is used a. Low pass b. High pass c. Band-pass 246) The left diagram represents a. A low pass filter b. A differentiator c. An integrator 247) The right diagram represents a. A high pass filter b. A differentiator c. An intergrator 248) An Integrator at high frequency a. The capacitor has insufficient time to charge up b. The capacitor voltage is high c. Input voltage is much less than across the resistor 249) A differentiator at low frequency a. The capacitor has no time to charge up b. The capacitor has time to charge up c. Voltage across the capacitor is much less than the source voltage 250) An Integrator at high frequency a. The capacitor voltage is high b. Input voltage is much less than across the resistor c. Input voltage equals voltage across the resistor (For question no. 251 260 please refer to this diagram) 251) The 3 Ohm and 12 ohm resistors can be replaced by one a. 9 Ohm resistor b. 15 Ohm resistor c. 36 Ohm resistor 252) The 3 Ohm ,12 ohm and 30 Ohm resistors can be replaced by one a. 15 Ohm resistor b. 45 Ohm resistor c. 10 Ohm resistor 253) The resistor of Quiz 2 and the 24 Ohm resistor can be replaced by a. 7 Ohm resistor b. 13 Ohm resistor c. 17 Ohm resistor 254) By knowing answer 3, what is the total resistance a. 6 ohms b. 17 ohms c. 12 ohms 255) By knowing answer 4, what is the total power a. 17 Watts b. approximately 6 Amperes c. 9/17 Watts 256) By Knowing that the total resistance is 17 Ohms the total current is a. 17/3 A b. 3/17 A c. 3 x 17 A 257) By knowing the total current, how much current flows through the 10 Ohm resistor a. More b. Less c. The same 258) The voltage drop on the 10 Ohm resistor is _________ as all the remaining resistors. a. Higher b. Lower c. The same 259) The 10 Ohm resistor has a colour code with 10% tolerance a. Brown, black, black, gold b. Brown, black, black, silver c. Black, brown, black, silver 260) The 24 Ohm resistor has a colour code with a tolerance of 20% a. Red, orange, black b. Red, yellow, black, gold c. Red, yellow, black 261) A resistor with 49 Kilo Ohms and 20% tolerance following colours a. Yellow, red, orange & silver for tolerance b. Yellow, red, orange & no band for 20& tolerance c. Orange, yellow, red & no band for 20% tolerance 262) A resistor with 5,6 Mega Ohms and 20% tolerance following colours a. Yellow, green, yellow b. Blue, violet, blue c. Green, blue, green 263) A resistor with orange, gray, yellow & gold colour bands has a. 380 Kilo Ohms & 5% tolerance b. 27 Kilo Ohms & 5% tolerance c. 380 Kilo Ohms & 15% tolerance 264) A resistor with blue, gray, red & silver colour bands has a. 6800 Ohms & 5% tolerance b. 6800 Ohms & 10% tolerance c. 680 Ohms & 5% tolerance 265) A resistor with brown, black, red colour bands has a. 10 kilo Ohm and 20% tolerance b. 100 Ohm and 10% tolerance c. 1 kilo Ohm and 20% tolerance 266) A capacitor with 120 pF and plus/minus 20% tolerance following colours a. Brown, red, brown b. Brown, red, brown & black band for plus/minus 20% tolerance c. Brown, yellow, brown & black band for plus/minus 20% toleranc 267) A capacitor with 4 red bands has following values a. 2.2 nF & plus/minus 10% tolerance b. 2.2 nF & plus/minus 5% tolerance c. 2200 pF & plus/minus 2% tolerance 268) A capacitor with orange, gray, yellow & brown colour bands has a. 380 nF & plus/minus 1% tolerance b. 380 pF & plus/minus 1% tolerance c. 380 nF & plus/minus 2% tolerance 269) A capacitor with blue, gray, red & silver colour bands has a. 6800 pF & plus/minus 5% tolerance b. 6800 pF & plus/minus 10% tolerance c. 680 pF & 10% tolerance 270) A capacitor with brown, black, red & gold colour bands has a. 10 pF & plus/minus 20% tolerance b. 100 pF & plus/minus 10% tolerance c. 1000 pF & plus/minus 5% tolerance 271) In an inductive circuit a. Current leads voltage b. Voltage leads current c. Current is equal voltage 272) In a capacitive circuit a. Voltage is at the same time as current b. Voltage leads current c. Current leads voltages 273) The power factor is calculated Voltage x Current x __________ a. Cosinus of phase angle b. Sinus of phase angle c. Tangens of phase angle 274) The power factor is a. Apparent power/true power b. True power/apparent power c. Volt x Amps x sin theta 275) A series wound motor has a. Constant rpm b. Low torque c. High torque 276) In a 3 phase motor, the fileld windings are a. 30 apart b. 120 apart c. 200 apart 277) A standard auto-transformer has a. Separate primary and secondary windings b. Two windings c. One winding 278) For motors use Flemmings a. LH Rule b. RH Rule c. Power rule 279) The formula of "inductive reactance" XL is a. XL = 1 / 2 pi f L b. XL = 2 pi f L c. XL = omega f L 280) The formula for "Capacitive Reactance" Xc is a. Xc = 2 pi f C b. Xc = omega f C c. Xc = 1 / 2 pi f C 281) The value of AC that has the same heating effect as DC is called a. Average = 0.637 of peak V b. RMS = 0.707 of peak V c. RMS = square root 2 of peak V 282) Capacitors are often named by their dielectric. Which capacitor is polarity sensitive? a. Oil b. Grease/ ceramics c. Electrolytic 283) Capacitors in series. Which formula is correct a. CT = C1 x C2/C1 + C2 b. CT = C1 + C2/C1 x C2 c. CT = C1 + C2 284) Capacitors in parallel. Which formula is correct a. CT = C1 x C2/C1 + C2 b. CT = C1 + C2 c. CT = C1 + C2/C1 x C2 285) Capacitors in series. Which formula is correct a. CT = C1 + C2 + C3/C1 x C2 x C3 b. CT = C1 + C2 + C3 c. 1/CT = 1/C1 + 1/C2 + 1/C3 286) The dielectric constant of air is a. Pi/2 b. 1 c. Square root of 2 287) Mica, paper and ceramic capacitors are made for ranges from 200 - 400 Volts. Oil capacitors for up to a. 1,000 Volts b. 2,500 Volts c. 7,500 Volts 288) Inductors in parallel. Which formula is correct a. LT = L1 x L2/L1 + L2 b. LT = L1 + L2/L1 x L2 c. LT = L1 + L2 289) Inductors in series. Which formula is correct a. LT = L1 x L2/L1 + L2 b. LT = L1 + L2 c. LT = L1 + L2/L1 x L2 290) Inductors in parallel. Which formula is correct a. LT = L1 + L2 + L3/L1 x L2 x L3 b. LT = L1 + L2 + L3 c. 1/LT = 1/L1 + 1/L2 + 1/L3 291) An induced electric current always flows in such a direction that it opposes the change producing it. a. Is Kirchoffs law b. Is Lenzs law c. Is Maxwells law 292) The algebraic sum of the currents flowing through all the wires in a network that meet at a point zero a. Is stated in Gaus current law b. Is stated in Lenzs current law c. Is stated in Kirchoffs current law 293) The algebraic sum of the e.m.fs within a closed circuit is equal to the sum of the products of the current and the resistances in the various portions of the circuits. a. Is stated in Kirchoffs voltage law b. Is stated in Henrys voltage law c. Is stated in Lenzs voltage law 294) In a 3 phase AC system the phase to phase voltage is a. Line voltage x square root of 2 b. Line voltage x square root of 3 c. Line voltage x pi/2 295) In a three phase AC system: Line Voltage is between a. Two phases b. Plus and ground c. One phase and the star point 296) In a brushless 3 phase AC motor, the armature is a. Delta wound b. Star wound c. Star or delta wound 297) A variable resistor with three terminals is a a. Thyristor b. Rheostat c. Potentiometer 298) A PRIMARY CELL a. Cannot be recharged b. Can be recharged c. Needs a trickle charger 299) A SECONDARY CELL a. Cannot be recharged b. Can be recharged c. Needs a trickle charger 300) Resonance in a resonant circuit occurs mostly at a. Omega L = omega C b. 2 pi f C = 2 pi f L c. omega L = 1/ omega C 301) For how many hours will a 140AH battery deliver 15A? a. 21 b. 9h and 20 min c. 9h and 33 min 302) What frequency is used in aircrafts a. 50 Hz b. 60 Hz c. 400 Hz 303) AC current measured in terms of DC that produces the same heating effect is a. Effective AC b. Peak AC c. Average AC 304) Effective AC = a. 1.36 x peak b. 0.707 x peak value c. 1.414 x peak 305) Peak of 115V AC is a. 81.3 V b. 200 V c. 162.6 V 306) What is true power measured in ___________ and is Volts x Amps x Power Factor a. VA b. Watt c. VAcos 307) What does the power factor equal in a purely resistive circuit? a. Pi/2 b. Pi c. one 308) What is the purpose of laminating the poles of a DC Generator a. To prevent eddy currents b. To make the motor lighter c. For better cooling 309) An induced electric current always flows in such a direction that it opposes the change producing it. a. Is Kirchoffs law b. Is Lenzs law c. Is Henrys law 310) What does the charge of a capacitor not depend on? a. Distance between plates and area of plates b. Dielectric c. Amount of current flow 311) What is the advantage of the use of AC power in aircrafts? a. Higher voltage b. AC can be transformed c. AC is more reliable 312) What is the residual Voltage result of? a. Stored electrons b. Residual resistance c. Residual magnetism 313) What is damaged by prolonged sparking in a CD Generator? a. The commutator b. The field c. The stator 314) An auto transformer has a. 3 coils b. 1 coils c. 2 coils 315) What does a narrow Hysteresis loop signify? a. Small distance between plates b. Much hysteresis c. Approach of ideal = no hysteresis 316) A shunt is used to ___________ the range of an ammeter a. Decrease b. Increase c. Stabilize 317) What is the purpose of the Capacitor in the starting circuit of a 2 phase inductive motor? a. To reduce inductive load b. For noise suppression c. To create an artificial rotating field 318) The word C I V I L is a way to remember relation ship of a. Capacitors and inductors b. Diodes and transistors c. AC motors 319) The opposition to the flow of electrons offered by a device or material is a. Reluctance b. Resistance c. Romance 320) What does a fat Hysteresis loop signify? a. Small distance between plates b. Approach of ideal = no hysteresis c. Much hysteresis 321) Voltage coils are made of a. Thick short copper wire b. Fine long copper wire c. Low resistance steel wire 322) Current coils are made of a. Low resistance steel wire b. Fine long copper wire c. Thick short copper wire 323) Interpoles and Compensating windings are used to a. Reduce armature reaction b. Increase armature reaction c. To move the position of the brushes 324) Passive Filters consist of a. Capacitors and inductors b. Capacitors and reactors c. Inductors and reactors 325) A shaded pole single phase induction motor consists of projecting pole pieces and a. Squirrel cage b. A capacitor c. A surrounding copper strap 326) At Resonant frequency, when XL=XC in a series circuit Z is a. At maximum and current I at minimum b. At minimum and current I at maximum c. Any value between maximum and minimum 327) At Resonant frequency, when XL=XC in a parallel circuit Z is a. Any value between maximum and minimum b. At minimum and current I at maximum c. At maximum and current I at minimum 328) Calculate Z if R=6 Ohm, XL = 58 Ohm and XC = 50 Ohm a. 10 ohm b. 12 ohm c. 1 ohm 329) The formula for Z in question 3 is a. Z = V / I b. Z = sqrt(R + (XL - Xc) c. Z = R / XL + Xc 330) On a step down transformer the primary side a. Higher gauge wire b. Has less windings c. Has more windings 331) The electrolyte in a so called dry cell consists of a. Potassium hydroxide b. Ammonia chloride and zinc chloride c. Sulphuric acid 332) An Ammeter in a battery charging circuit indicates a. C b. c. + 333) When an ammeter is connected in parallel to the power source a. It creates a short circuit b. It reads current c. It does not have an effect on the instrument 334) What method of Battery Charging is used to keep the battery fully charged a. Constant current charging b. Trickle charging c. Constant voltage charging 335) Thermal runaway of a NiCad Battery starts with excessive gassing and ends with destruction, caused by a. Too low ambient temperature b. Heavy discharge c. Heat reduced resistance and high current flow 336) A high pass filter can be used to: a. Prevent interference to AM broadcast reception b. Prevent interference to TV reception c. Pass a band of speech frequencies in a modular 337) A high pass filter attenuates: a. High frequencies but not low frequencies b. All except a band of VHF frequencies c. Low frequencies but not high frequencies 338) The following unit in a DC power supply performs a rectifying operation: a. A full-wave diode bridge b. An electrolytic capacitor c. A fuse 339) A low-pass filter is used in the antenna lead from a transmitter: a. To eliminate chirp in CW transmission b. To reduce radiation of harmonics c. To increase harmonic radiation 340) A communication receiver may have several IF filters of different bandwidths. The operator selects one to: a. Improve the receiver sensitivity b. Increase the noise received c. Improve the reception of different types of signal 341) Television interference caused by harmonics radiated from an amateur transmitter could be eliminated by fitting: a. A high pass filter b. A low pass filter in the transmitter output c. An active filter 342) A high pass RF filter would normally be fitted: a. Between microphone and speech amplifier b. Between transmitter output feedline c. At the antenna terminals of a TV receiver 343) A band stop filter will: a. Pass frequencies each side of a band b. Stop frequencies each side of a band c. Only allow one spot frequency through 344) A low pass filter for a high frequency transmitter output would: a. Attenuate frequencies below 30 MHz b. Attenuate frequencies above 30 MHz c. Pass audio frequencies above 3 kHz 345) Installing a low pass filter between the transmitter and transmission line will: a. Ensure an SWR not exceeding 2:1 b. Permit higher frequency signals to pass to the antenna c. Permit lower frequency signals to pass to the antenna 346) Television interference caused by harmonics radiated from an amateur transmitter could be eliminated by fitting: a. A high pass filter b. A low pass filter in the transmitter output c. An active filter 347) A high pass RF filter would normally be fitted: a. Between microphone and speech amplifier b. Between transmitter output feedline c. At the antenna terminals of a TV receiver 348) A band stop filter will: a. Pass frequencies each side of a band b. Stop frequencies each side of a band c. Only allow one spot frequency through 349) A low pass filter for a high frequency transmitter output would: a. Attenuate frequencies below 30 MHz b. Attenuate frequencies above 30 MHz c. Pass audio frequencies above 3 kHz 350) Installing a low pass filter between the transmitter and transmission line will: a. Ensure an SWR not exceeding 2:1 b. Permit higher frequency signals to pass to the antenna c. Permit lower frequency signals to pass to the antenna 351) An active audio low-pass filter could be constructed using: a. A transformer and capacitors b. An operational amplifier, resistors and capacitors c. Electrolytic capacitors and resistors 352) The input impedance of an operational amplifier is generally a. Inductive b. Very low c. Very high 353) The voltage gain of an operational amplifier at low frequencies is: a. Very high but purposely reduced using circuit components b. Very low but purposely increased using circuit components c. Less than one 354) An operational amplifier connected as a filter always utilises: a. Positive feedback to reduce oscillation b. Negative feedback c. Random feedback 355) Unwanted signals from a radio transmitter which cause harmful interference to other users are known as: a. Reflected signals ## b. Re-radiation signals c. Harmonic signals and spurious signals Module Electrical Power (ATA 24) 1) Emergency lighting is switched on prior to a) Take off b) Switching ground power off c) An emergency evacuation 2) An APU generation system a) does not have a CSD b) uses an identical CSD to the main engine CSD c) uses a different type of CSD to the main engine CSD 3) The CSD/IDG governor speed control screw a) cannot be adjusted b) adjusts the oil flow rate c) adjusts the generator frequency 4) The transistor voltage regulator operates with a) pulse control b) rheostat c) high voltage 5) A three-phase alternator with a single-phase output of 115 V will have a phase-tophase output of a) 0.707 x 115 V b) 1.414 x 115 V US c) 1.73 x 115 V 6) A multiple phase alternator with a single-phase output of 115 V will have a phase-tophase output of 230 V a) it varies b) true c) false 7) The output voltages of a three-phase alternator are separated by 120 degrees a) false, they are 90 degrees apart b) true c) false 8) AC is widely used because of its versatility a) false b) true c) n/a 9) In a modern brushless AC generator, which part of the generator rotates to produce the induced voltage a) stator b) rotor c) armature 10) The stationary member of a brushless alternator which provides connections for the external load is called a) rotor b) stator c) armature

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### Home > ACC7 > Chapter 6 Unit 7 > Lesson CC3: 6.2.1 > Problem6-47 6-47. On the same set of axes, graph the two rules shown at right. Then find the point(s) of intersection, if one (or more) exists. $\begin{array}\ y=-x+2\\y=3x+6\end{array}$

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======== START LECTURE #21 ======== Memory support for wider blocks • Should memory be wide? • Should the bus from the cache to the processor be wide? • Assume 1. 1 clock required to send the address. Only one address is needed per access for all designs. 2. 15 clocks are required for each memory access (independent of width). 3. 1 Clock/busload required to transfer data. • How long does it take satisfy a read miss for the cache above and each of the three memory/bus systems. • Narrow design (a) takes 65 clocks: 1 address transfer, 4 memory reads, 4 data transfers (do it on the board). • Wide design (b) takes 17. • Interleaved design (c) takes 20. • Interleaving works great because in this case we are guaranteed to have sequential accesses. • Imagine a design between (a) and (b) with a 2-word wide datapath. It takes 33 cycles and is more expensive to build than (c). Homework: 7.11 7.3: Measuring and Improving Cache Performance Performance example to do on the board (a dandy exam question). • Assume • 5% I-cache miss. • 10% D-cache miss. • 1/3 of the instructions access data. • CPI = 4 if miss penalty is 0 (A 0 miss penalty is not realistic of course). • What is CPI with miss penalty 12 (do it)? • What is CPI if we upgrade to a double speed cpu+cache, but keep a single speed memory (i.e., a 24 clock miss penalty)? Do it on the board. • How much faster is the ``double speed'' machine? It would be double speed if the miss penalty were 0 or if there was a 0% miss rate. Homework: 7.15, 7.16 A lower base (i.e. miss-free) CPI makes stalls appear more expensive since waiting a fixed amount of time for the memory corresponds to losing more instructions if the CPI is lower. A faster CPU (i.e., a faster clock) makes stalls appear more expensive since waiting a fixed amount of time for the memory corresponds to more cycles if the clock is faster (and hence more instructions since the base CPI is the same). Another performance example. • Assume 1. I-cache miss rate 3%. 2. D-cache miss rate 5%. 3. 40% of instructions reference data. 4. miss penalty of 50 cycles. 5. Base CPI is 2. • What is the CPI including the misses? • How much slower is the machine when misses are taken into account? • Redo the above if the I-miss penalty is reduced to 10 (D-miss still 50) • With I-miss penalty back to 50, what is performance if CPU (and the caches) are 100 times faster Remark: Larger caches have longer hit times. Improvement: Associative Caches Consider the following sad story. Jane has a cache that holds 1000 blocks and has a program that only references 4 (memory) blocks, namely 23, 1023, 123023, and 7023. In fact the references occur in order: 23, 1023, 123023, 7023, 23, 1023, 123023, 7023, 23, 1023, 123023, 7023, 23, 1023, 123023, 7023, etc. Referencing only 4 blocks and having room for 1000 in her cache, Jane expected an extremely high hit rate for her program. In fact, the hit rate was zero. She was so sad, she gave up her job as webmistress, went to medical school, and is now a brain surgeon at the mayo clinic in rochester MN. So far We have studied only direct mapped caches, i.e. those for which the location in the cache is determined by the address. Since there is only one possible location in the cache for any block, to check for a hit we compare one tag with the HOBs of the addr. The other extreme is fully associative. • A memory block can be placed in any cache block. • Since any memory block can be in any cache block, the cache index where the memory block is stored tells us nothing about which memory block is stored there. Hence the tag must be the entire address. Moreover, we don't know which cache block to check so we must check all cache blocks to see if we have a hit. • The larger tag is a problem. • The search is a disaster. • It could be done sequentially (one cache block at a time), but this is much too slow. • We could have a comparator with each tag and mux all the blocks to select the one that matches. • This is too big due to both the many comparators and the humongous mux. • However, it is exactly what is done when implementing translation lookaside buffers (TLBs), which are used with demand paging. • Are the TLB designers magicians? Ans: No. TLBs are small. • An alternative is to have a table with one entry per MEMORY block giving the cache block number. This is too big and too slow for caches but is used for virtual memory (demand paging).

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Valence Electrons Key Questions • The valence electrons are the electrons that determine the most typical bonding patterns for an element. These electrons are found in the s and p orbitals of the highest energy level for the element. Sodium $1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{1}$ Sodium has 1 valence electron from the 3s orbital Phosphorus $1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{2} 3 {p}^{3}$ Phosphorus has 5 valence electrons 2 from the 3s and 3 from the 3p Lets take the ionic formula for Calcium Chloride is $C a C {l}_{2}$ Calcium is an Alkaline Earth Metal in the second column of the periodic table. This means that calcium ${s}^{2}$ has 2 valence electrons it readily gives away in order to seek the stability of the octet. This makes calcium a Ca+2 cation. Chlorine is a Halogen in the 17th column or ${s}^{2} {p}^{5}$ group. Chlorine has 7 valence electrons. It needs one electron to make it stable at 8 electrons in its valence shells. This makes chlorine a Cl^(−1) anion. Ionic bonds form when the charges between the metal cation and non-metal anion are equal and opposite. This means that two Cl^(−1) anions will balance with one $C {a}^{+ 2}$ cation. This makes the formula for calcium chloride, $C a C {l}_{2}$. For the example Aluminum Oxide $A {l}_{2} {O}_{3}$ Aluminum ${s}^{2} {p}^{1}$ has 3 valence electrons and an oxidation state of +3 or $A {l}^{+ 3}$ Oxygen ${s}^{2} {p}^{4}$ has 6 valence electrons and an oxidation state of -2 or O^(−2) The common multiple of 2 and 3 is 6. We will need 2 aluminum atoms to get a +6 charge and 3 oxygen atoms to get a -6 charge. When the charges are equal and opposite the atoms will bond as $A {l}_{2} {O}_{3}$. In molecular (covalent) compounds these same valence electrons are shared by atoms in order to satisfy the rule of octet. SMARTERTEACHER The number of electrons in an atom's outermost valence shell governs its bonding behaviour. Explanation: The valence electrons are the electrons in the outermost electron shell of an atom. That is why elements whose atoms have the same number of valence electrons are grouped together in the Periodic Table. Generally, elements in Groups 1, 2, and 13 to 17 tend to react to form a closed shell, corresponding to the electron configuration ${s}^{2} {p}^{6}$. This tendency is called the octet rule, because the bonded atoms have eight valence electrons. METALS The most reactive kind of metallic element is a metal from Group 1 (e.g., sodium or potassium). An atom in Group 1 has only a single valence electron. This one valence electron is easily lost to form a positive ion with an ${s}^{2} {p}^{6}$ configuration (e.g., ${\text{Na}}^{+}$ or ${\text{K}}^{+}$). A metal from Group 2 (e.g., magnesium) is somewhat less reactive, because each atom must lose two valence electrons to form a positive ion (e.g., ${\text{Mg}}^{2 +}$ with an ${s}^{2} {p}^{6}$ configuration. Within each group of metals, reactivity increases as you go down the group. The valence electrons are less tightly bound and easier to remove, because they are farther away from the nucleus of the atom. NONMETALS A nonmetal tends to attract additional valence electrons to attain a full valence shell. It can either share electrons with a neighboring atom to form a covalent bond or it can remove electrons from another atom to form an ionic bond. The most reactive kind of nonmetal is a halogen such as fluorine or chlorine. It has an ${s}^{2} {p}^{5}$ electron configuration, so it requires only one additional valence electron to form a closed shell. To form an ionic bond, a halogen atom can remove an electron from another atom in order to form an anion (e.g., $\text{F"^"-", "Cl"^"-}$, etc.). To form a covalent bond, one electron from the halogen and one electron from another atom form a shared pair (e.g., in $\text{H–F}$, the dash represents a shared pair of valence electrons, one from $\text{H}$ and one from $\text{F}$). Within each Group of nonmetals, reactivity decreases from top to bottom, because the valence electrons are at progressively higher energies and the atoms do not gain much stability by gaining electrons. In fact, oxygen (the lightest element in Group 16) is more reactive than chlorine, even though it is not a halogen, because the valence electrons of oxygen are closer to the nucleus (at a lower energy). Therefore, a metal from the bottom of Group 1 (like potassium) and a nonmetal from the top of Group 17 (like fluorine) will react violently, because they both benefit greatly from the reaction. $\text{K}$ loses one electron to $\text{F}$ and forms the ionic compound potassium fluoride, $\text{K"^"+""F"^"-}$.

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# Digital Filter Analysis Magnitude, phase, impulse, and step responses, phase and group delays, pole-zero analysis Analyze frequency- and time-domain responses of filters. Visualize filter poles and zeros in the complex plane. ## Apps Filter Designer Design filters starting with algorithm selection ## Functions expand all `abs` Absolute value and complex magnitude `angle` Phase angle `freqz` Frequency response of digital filter `grpdelay` Average filter delay (group delay) `phasedelay` Phase delay of digital filter `phasez` Phase response of digital filter `unwrap` Shift phase angles `zerophase` Zero-phase response of digital filter `zplane` Zero-pole plot for discrete-time systems `impz` Impulse response of digital filter `impzlength` Impulse response length `stepz` Step response of digital filter `filtord` Filter order `filternorm` 2-norm or infinity-norm of digital filter `firtype` Type of linear phase FIR filter `isallpass` Determine whether filter is allpass `isfir` Determine if digital filter has finite impulse response `islinphase` Determine whether filter has linear phase `ismaxphase` Determine whether filter is maximum phase `isminphase` Determine whether filter is minimum phase `isstable` Determine whether filter is stable ## Filter Visualization Tool FVTool Filter Visualization Tool ## Topics Frequency Response Compute and display frequency responses of IIR and FIR lowpass, highpass, and bandpass filters. Phase Response Extract the phase response of a filter. Group Delay and Phase Delay Measure the average time delay of a filter as a function of frequency. Zero-Pole Analysis Find and visualize poles and zeros of a linear system. Impulse Response Generate and display the impulse response of a simple filter. Compensate for the Delay Introduced by an FIR Filter Use indexing to counteract the time shifts introduced by filtering. Compensate for the Delay Introduced by an IIR Filter Remove delays and distortion introduced by filtering, when it is critical to keep phase information intact.

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Often it’s the easy things we tend to “mess up” and from my experience when students graph one variable lines they make mistakes not because it’s hard rather they just never remember if y = a number or x = a number is a vertical line or horizontal line.  Let’s do a quick review of graphing one variable lines.  If you are asked to graph a line that has the equation x=5 this would be a vertical line that goes through 5 on the x-axis.  Hence x= a number lines are vertical lines that goes through the respective number in the equation on the x-axis.  Now lets take a look at a like that is y = a number for example y = -2.  One variable linear equations where y= a number are horizontal lines going through the number. So in this example y= -2 would be a horizontal line passing through -2 on the y-axis. As I said from the start this stuff is easy but it’s also easy to confuse- take a look at the lesson below it should really help you memorize this once and for all! concepts to remember about graphing one variable linear equations: 1. x= a number are vertical lines 2. y = a number are horizontial lines 3. vertical lines have no slope it’s what we call undefined 4. horizontial lines have a slope = 0 5. line equations are call “linear equations”

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# How to correctly perform a goodness-of-fit test for a contingency table (two-way, three-way, or more), in situations other than independence testing? Let's say I have the following table from a sample of 462 people: Men 70 32 120 Women 100 30 110 I don't want to test it against the hypothesis of independence, but against the following hypothesized distribution: Men 46 (0.1) 139 (0.3) 92 (0.2) Women 46 (0.1) 116 (0.25) 23 (0.05) in R, according to the documentation, chisq.test only works for testing independence in contingency tables, goodness-of-fit test being only available for "flat" tables. So I was thinking of simply flattening the two contingency tables, then applying a standard goodness-of-fit test, for example in R something like: observed_data = c(70, 32, 120,100,30,110) hypothesized_data = c(46, 139, 92,46, 116, 23) hypothesized_prop = hypothesized_data / sum(hypothesized_data ) res = chisq.test(observed_data, p=hypothesized_prop ) #results in a chi-square statistic (res$statistic) of 559.6473 #now let's compute the p-value for 2 degrees of freedom: pchisq(res$statistic, df=2, lower.tail=FALSE) #results in a p-value of 2.979478e-122 My questions are: 1. Is this approach correct from a theoretical point of view? (happy to have comments on my code too, even if it's not the heart of my question). 2. If it is correct, is it also correct to extend this approach to three-way contingency tables or more (e.g. four categorical variables: gender, mood, age group, income group)? If it's not correct for contingency tables with more than 2 variables, what approach would be correct to test if an observed distribution fits a given distribution (in other cases than independence testing)? Regarding my second question, I've found this interesting article ("Common statistical tests are linear models"), and log-linear models may be the answer. However the article seems to use log-linear models only to test independence, so I'm not sure how to approach this from a theoretical point of view (i.e. is it even correct to use log-linear models for this kind of question), and from a practical point of view (i.e. is it actually possible to do it in R, Python, or other statistical tools?). Thanks, In brief, you can use the chi-squared goodness-of-fit test to test whether data was generated from a hypothesized distribution. There's no need to formulate the data as a contingency table because you're not using the marginal probabilities. You've illustrated this with a hypothesized distribution with six categories but you could also use this with a hypothesized distribution with nine categories or sixteen categories or more. For the goodness-of-fit test, the degrees of freedom are n - 1. So if you have six categories, you have five (not two) degrees of freedom. The 'chisq.test' in R though will give you these degrees of freedom in the output so no need to use 'pchisq' yourself. observed_data <- c(70, 32, 120,100,30,110) hypothesized_data <- c(46, 139, 92,46, 116, 23) hypothesized_prop <- hypothesized_data / sum(hypothesized_data) chisq.test(observed_data, p = hypothesized_prop) I don't know if what you're doing is theoretical correct. This would probably depend on how you've come up with your hypothesized distribution. • Thanks! When I was talking about "theoretically correct", my implied question was for example if the degrees of freedom I specified were correct (I wanted to use the same DoF as for the chi2 contingency test -but it's not possible withchisq.test, hence my use of pchisq). I should have been more explicit about it, in particular when it looks like that it was actually incorrect to do that! On a side note, this is a problem I come across sometimes with various survey data. The expected (or "hypothesized") distributions I use generally come from a census or from a previous survey. – JJJ Mar 27 at 16:20

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Cody # Problem 1096. Largest Twin Primes Solution 451454 Submitted on 6 Jun 2014 by rifat This solution is locked. To view this solution, you need to provide a solution of the same size or smaller. ### Test Suite Test Status Code Input and Output 1   Pass %% x = 1001; y_correct = [881 883]; assert(isequal(your_fcn_name(x),y_correct)) i = 1001 ans = 881 883 2   Pass %% x = 1964; y_correct = [1949 1951]; assert(isequal(your_fcn_name(x),y_correct)) i = 1963 ans = 1949 1951 3   Pass %% x = 123456789; y_correct = [123456209 123456211]; assert(isequal(your_fcn_name(x),y_correct)) i = 123456789 ans = 123456209 123456211 ### Community Treasure Hunt Find the treasures in MATLAB Central and discover how the community can help you! Start Hunting!

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