 If you remember from Friday, we were looking at the results of these tensile tests that are taken on the sample solids, usually supplied by the manufacturer, threaded on each end. About a quarter of an inch in diameter is a fairly standard sample size, but that's more tradition than anything. There's also lots of examples of a sample that looks more like this being tested rather than this kind of thing. It just does take a different machine or at least different jaws of the machine. Typically, about two inches on either side of two little marks that are put in, and then that is the starting length, what we call the L0 of the piece, and then it's put in the jaws and can either be pulled in tension or pushed in compression, very often bolt, not both with the same P, a load, or some representation of the load on the y-axis, typically in terms of the normal strain, which is the applied load over the original area, and if you remember the original areas of concern, especially later in the test because there is necking that goes on, and then the response to that load in terms of the strain was on the x-axis. And remember that can typically be very, very small, especially in the early stages of the test. And we were looking at typical structural steels, like low-carbon steels, a general structural representative of what we call a ductile material. There's two typical responses of ductile materials. The steels tend to do something like this. As the load increases, a little bit of strain is resident, and it occurs in a very linear fashion for quite some time. There along here, if the load is removed, the response of the material is to return to its original size, which means it returns to zero strain. Any material who, once the load is removed, has zero residual strain in it is considered an elastic material, and this is the elastic region. And you can imagine just how steep this is, is pretty important to the different materials. And then we do tend to leave that linear region, go through a couple different parts to it as we went over. Those are not a particularly great concern to us because we will do most of our look at things in the elastic region. As you can imagine, that's highly desirable for structural materials that they can be loaded and unloaded, and they always return to a state of zero strain, generally quite desirable. So this might be an indication of the end of that proportional limit, but very, very close to that, we also have what we call the yield stress, and that's the stress that's in the back of your book. If you open up the back of your book, you don't have to do it now. We'll use this as we go along. It might be helpful to at least photocopy the back pages of the book where there's lots and lots of different types of materials, and one of the most important quantities in those tables is this yield stress. The highest you can go for the stress and still remain within that elastic limit for different materials. The peak there is typically the ultimate stress, and I believe that's also on the table. There's two tables in the back. One is in English humans, one in SIU. And I believe the ultimate limit is in most of those materials, or most of those tables. And then we have then also a stress at rupture. This is well into the region where that necking occurs, where the narrowing down of the material in the test section itself causes a drastic reduction in the area that's standing that stress, and the real curve actually goes something like this, if you use the actual cross-sectional area as opposed to using the original or nominal cross-sectional area there. So we looked at that in some detail yesterday, or Friday, there's just a refresher for it. However, this is a bit of a false picture that I've drawn, because the real stress strain diagram drawn more to scale. If you remember, we had some very, very small numbers down here, and I did not draw the x-axis to scale, not to either a logarithmic scale, which we couldn't do because we have zero in that scale, nor to a proper linear scale. And that's because there's very, very little strain in this elastic region for these low-carbon steels. They don't elongate very much. So the true diagram actually is more something like this, where it's very, very difficult even to see the slope in that early region because there's so little strain through that elastic region. So a couple of the problems that I want you to do are actually nothing more than an attempt to accurately portray this in some usable way. One way to do that, this is what the true diagram looks like. You can see it's very, very steep in here. It's almost imperceptible that there's any slope at all. And then it does something like what I've drawn. So there's a couple different ways you can handle this fact, that the most important region essentially disappears in a scale drawn. One thing you can do is you can break out the scale, the linear region. Notice that this left-hand diagram only goes up to .03, which is way, way back in here, almost completely disappearing. Break that out, draw it to one scale, get a good look at the linear region, and as you can guess, this slope is going to be very important to us. And then have the rest of the diagram drawn to a much larger scale. So that's one way you can handle this fact, to have two graphs that represent the two of them on different scales side by side. Another way to do it is to use color, and you can either have one of the color diagrams at the top, or you can have them both at the bottom, you could have had the red one here for the red line, and then the blue one right above it for the blue line. This is much like how your book approaches it. For those of you that have had me for Physics 1 before, you know that I don't like this way to do it. Why not? It's relatively clear that the blue line goes at the blue scale, the red line goes at the red scale. The trouble is, if you photocopy this, you get none of that, because all the lines are then black in a photocopy. There might be a very slight difference in just how black the lines are, but if that is photocopied, if you wanted to photocopy that and have it in your notebook, that method of displaying the differences is useless. Another way that's typical, and this is commonly found in the scientific literature, is to put a little arrow that indicates to which scale the two axes pertain, and so instead of any color being used, just an arrow says read the bottom scale, this one says read the bow scale. If you do it this way, make sure that you don't have the top line referring to the bottom scale, and the bottom line referring to the top scale. It's just a chance of confusion for the reader, and anytime you're making graphs like this, you're trying to make them as clear as possible, I assume. So there's at least one problem in the homework where we give you some of this test data, allow you to graph it, that's all I did in these plots, myself I just took the data out of those problems, and practice, decide some way, not only figure out how to get the graphing software to do what you want, but also decide how you like it as a way to represent the data more accurately as part of the homework, and then you can also, I think you're also supposed to pick off some of the yield limits and the significant stresses in other places as well. Oh, I think I used an R for rupture stress, rather than a B for breaking stress. Different authors use different things. So those are the three possibilities. Feel free to even come up with your own possibility as to how you might want to represent these different ways to represent the very same thing in a way that's useful for the reader. Remember that's your concern, ultimately. All right, so let's take a look then at this elastic region and what it can mean to us as designers. Just on the elastic region, we can get a little bit more information than we might otherwise see. So in greatly expanded scale, something like that. Since this does go through the origin, then any spot on that will actually give you the slope of that linear region. So the slope is, let's see, it's just going to be the stress over the strain at that point. And the steeper that line is, the greater the stress the material can absorb for a given strain response. So that's pretty important, as you can imagine. This is then a material property. It comes right from the tests. The test comes right from the manufacturer. So it's a publicized value, generally publicized by the manufacturer of the material, the structural members themselves. Often, a certain customer might want to do their own tests as well. In that case, they'll order some of these samples pieces from the manufacturer and redo the test themselves just to make sure. This is known as the elastic modulus or modulus of elasticity, which is the meaning of the capital E. You may remember this from Physics 1 as Young's modulus, half of the y. It's not, it's fairly typical that there'll be different symbols for things in physics than there are for the very same quantity in engineering. Then to streamline things a little bit, of course, we can look at this ratio and we realize that we can cancel a couple of light terms top and bottom. So it's really S over A as the ratio, which is a little bit quicker, a little bit, a little bit. You don't think that's going to be on the test? S over A? When I taught this course seven years ago or something and I could do that joke, I've been waiting 20 years to do that joke from what I took it and my professor did it for me. I thought that's a good joke. I'm keeping that one, but I never taught it until 20 years later. So like I said, this is very much a property of the material. We will take it as a constant, which for the most part it is. There's very little that could change that. Maybe some extreme temperature limits might cause some differences in that, but they have to be pretty extreme. It actually might be such that the material itself is really changing in structure at higher temperatures that would change that. So we take that pretty much as a property, just like we would with density, and that's why it's in the book, in the back of the book, listed as a material quantity. Real quick, just to show you, aluminum alloys are also ductile materials, but they have a slightly different characteristic curve in that they tend to have a linear region that doesn't end in such as nicely a defined way. They do have this necking response to rupture, a bit of a different looking curve. It's a little harder to define where the linear region ends. So one thing that's commonly done is a particular strain is arbitrarily chosen, something like, make sure to get the number right, 0.2%. And then parallel to the curve up from that, then they can pick that off, and that's often given then as the yield stress. One thing that's interesting about these type of curves and partly why that is chosen is that the test generally goes up the curve. If it gets to this point and the load is relieved, it tends to come down this line. Once you've gone past the true linear region up into the curve somewhere, if you relieve it, the load at that time will come down this line to a point where the load is completely removed, but some residual strain remains. That there is a permanent lengthening. Remember, that's what the strain is. It's the physical response to material to this strain, to the stress. There's some permanent stretch, if you will, that's left in the material now, even though the load's been completely removed. If the test is then resumed from that place, they go up this 2% offset curve and then, for the most part, back onto the regular curve. Sometimes with a little bit of change, maybe a little bit of a lowering of this, not of great concern to us, again, because we're going to spend almost all, if not all, of the rest of the term in this elastic region as we look at materials that behave in very predictable ways for us so that we can spend some good attention to them. An interesting curve in a very common structural solid is the test curve of concrete. And it does highlight the difference in concrete in terms of tension tests and compression tests. If this test is run on concrete in tension, it does something like this and then ruptures there. There's very little, if any, necking of concrete, so it doesn't have this curl over part that came more as an arithmetic fact of using the original area, even though the cross sectional area is now changing. Remember, that's what caused the curve to curl over a bit there. So we'll have some yield stress for the concrete in tension, but concrete is actually a much better compressive material. And so the compression test, which would look like that because we take compressive forces to be negative, but it's also a shortening of the material. It also would have a negative strain as well. These slopes are constant either side of the unloaded condition, but the concrete goes much farther in compression than it does in tension. We still label both the similar line, or would you label them both just similar lines? There may be some other little symbol on them that it's very rare you'd have a piece that needs to go into tension and compression in its lifetime. There are certain materials, you can imagine, certain engineering designs where material has to cycle in and out of compression and tension. Concrete is not real good for that, partly because it is so weak in tension, this can be a factor of almost somewhere around 8, and it's true for wood as well. The difference between these two is almost a factor of 8 in some concrete mixture. We're going to look at ways to handle, well, you've heard about ways to handle this by putting into the concrete reinforcing bars, or what's known as rebar. Give me a second to reset the tape. So we'll actually look at some design ways we can design concrete that it can do better in tension. Because even though it's not good in tension, as you'll see in a couple weeks, it's vitally important and unwaitable that concrete has to take some tension because of the way we're going to look at things. Actually, it's no big secret. We're going to spend a lot of time looking at loaded beams that have transverse loads, and their response to that, of course, is to deform something like that, exaggerated, of course, with some kind of load on it. We'll actually look at that bending in a couple weeks. As it bends like this, the top of the beam is going into compression, the bottom of the beam is going into tension. If you've got a concrete beam like this, it's unavoidable that the bottom of the beam is now in tension. So we're going to look at how to help that by putting in reinforcing bar, rebar, across the bottom of the beam, and the rebar then will absorb the tension, and the concrete will not. If you go look at some of the older concrete bridges around here, and New York is famous for having old bridges now, you'll see a lot of the concrete structure of the bridge is actually quite broken away at the bottom, maybe even exposing the rebar. In fact, there's a couple places out in front of the building that's done that, where you can see the exposed rebar. Because the concrete in tension is actually broken away, and just the rebar remains. But when we design it as we look at this technique, we completely take the concrete in tension out of the calculation anyway, and assume that it's all on the rebar. Anyway, that's for the future, a little bit of a taste of what's to come. The last material of any importance that we need to look at before we really start to look at this stressed over straining business, brittle materials like glass or brick or ceramics typically have curves like this. Very little identifiable linear region, no necking at all, and a fairly low yield stress in tension especially. So brick is very good as a compressive structural material like concrete is. But since brick typically comes in individual small pieces, it's not really commercially viable to put in a rebar in it. OK, so let's look at this elastic modulus. Modulus of elasticity. And take another peak at it, see some of the things we can learn from it. Let's see, that's P over A. I'm not going to put the A0 on there. You have to understand that's the original area. Same thing for the length, save a little bit of trouble. So if we put those things together, we get something like that. If we move things around a little bit and do a little bit of just quick algebra, we'll see something that is very familiar, you'll see is very familiar to what you've seen before. And that's that. Yeah, that's what I want. Let's bring the P over all by itself and put everything else on the other side. So that's going to be E, A, del over L. Yeah, right? Yeah, that looks good. One thing we'll do on a lot of problems is we'll actually put this E and this A together. There's a single number for a particular problem because those are essentially constants. In fact, all three of those numbers are really nothing more than constants as far as a problem is concerned. These three are independent of any load, tension, compression, or no load at all. These three, this is a material constant. These are properties of the structure of the piece itself. So this is really some constant times a elongation. This should look quite familiar. It looks very much like hook's law for springs. And indeed, that's exactly what it is. These materials, elastically, very much like a spring does, just happens to be that for our common experience, these deformations are very, very small. And they also function out of tension. Springs are more off of torsion because the water twists as the spring expands. When you put the spring in tension, the material is actually twisting more than simply along it. Yeah, you're right. But the overall response to the spring and the pieces we're looking at in these cases are very, very much the same. We'll look at a simple problem just to start getting used to the numbers. So imagine we have a structural member in a nuclear facility of zirconium alamoid. Regular steels in a nuclear atmosphere, if you will, where there's a lot of bombardment due to different radioactive decay particles. Structural steel changes drastically under that kind of bombardment, which means that the modules of elasticity changes drastically. So typically, you have to use a little bit more exotic structural members in a nuclear power plant that are much, much sturdier against the nuclear bombardment. There's a thing called hydrogen embrittlement where the materials are constantly bombarded and they become much more brittle under that. Zirconium does not tend to do that. So let's imagine this is 3 foot in length. And I want you to find the radius given a couple things. The yield stress is 57.5 ksi. Remember, that's thousands of pounds per square inch. So this is 57,000 pounds per square inch. The modulus of elasticity is 14 times 10 to the third ksi. Now remember, the modulus of elasticity is units of pressure, like ksi, over units of inches per inch or meters per meter. So the ultimate unit, it's on the modulus of elasticity, are either Pascals or Psi or their brothers and sisters, ksi and kilopascals in the length. And then one other thing I'd like you to do is apply a factor of safety of 3. Find the radius of that piece given those bits of information. One more. We're going to need to load it, put it under a four kip compression load. Now that factor of safety, there's different ways you can apply it. The end result is going to be the same. One of the ways to apply it is you can reduce the yield stress by 3 and then work with a much smaller allowable yield stress. Or you can reduce that. Or you can calculate the final radius and then multiply it by 3. That's a little bit different because that changes the area by 9, not by 3. OK, I think you've got all the pieces there you need. All right, so given the modules, elasticity, given the load, we've got all the parts we need. We can then find the area and then from the area you find the radius. The stress is P over A, but you've got to get a factor of safety in there. And one way to do that is to make a ratio that we now have an allowable stress that's 1 third of the yield stress because the factor of safety is 3. So that's a fairly typical way to do it, reduce the properties of the material by the factor of safety. And then design for that stress, which is a much smaller stress. So you've got to do that first. Yeah, that's a more typical way to do it. There might be different standards in different industries or different companies, but you have to understand what it is. That's sigma allowable. Yeah, so you design for this much lower stress than the yield stress of the material proper just to start getting used to these numbers. So design for a yield stress of something like 19.2. And that factor of safety is semi-arbitrary. It's through experience that they decided 3 is about right. 2 probably would have been safe enough, but they went even to 3 for factor of safety. On the factor of safety, I guess, if you will. But that allows a lot of margin for miscalculations, assumptions, or maybe even experimental differences in what the yield stress really is. So typically, we don't design very, very important pieces like structural members and nuclear power plants right to the limit for the big factor of safety on. Then you've got the load. You can solve for the area. Then once you have the area, you can solve for the rays, which I hope you remember as Pirate Square. I remember just figuring it out. Something like 0.209 for the area. That's just the load of four kips and the allowable stress of 19.2. Then you should get a radius of 2.58 inches. Probably then, well, since this isn't a standard piece, you can make it any radius you want, since it's probably a one-off piece for whatever reactor. But if you're looking at some bolt or something, you want to make sure that you have a standard radius so you don't have to custom make every one of these parts to that exact radius. You might want to just make it 0.3 and buy stuff off the shelf that way. All right, one other quick look at the kind of thing. What would be the load find the allowable load for a deformation of 0.02 inches? And since we're talking about compression, that would actually be a decrease, it'd be a minus there, but the minus is our typical designation for a compressive load anyway. So given that the original length, you can find the strain. We've got the area from the first one. We know the elastic modulus, then you can solve for an allowable load to keep a strain, a deformation of no greater than that. So less than or equal to that strain. And you can solve for these. And if there's some deformation on there, you should be able to ask the pain that this is something like minus, and the minus being compression, minus 1.62 kips. Which was just an original given load, the one to know on this minimum allowable deformation whatever occurs with the design beyond that is a greater part of the problem than what we're looking at. What, I've got this, or well that, are you coming up with different numbers? Well then, stop. That means a strain of minus 0.02 inches, yeah that's 0.02, I read that right, over 36 inches, the original length was three feet. So that's an allowable strain of 0.0, what, overall? Did you use three feet? Yeah, well we'll just play with the tape back. So you guys are so used to me doing something wrong, you couldn't conceive of the possibility it wasn't me. Well I agree. There's Phil, and then you find that comforting. Okay, Phil did you find it comforting, and you agree with Chris? Usually. Yeah, okay. All right, so once you've found that in the strain, then you can use that in the elastic modulus to find the allowable load. Given the area we've just found, and the strain you just calculated, or should have calculated, get a 1.62. Remember though, big factor safety on that, so if it's 1.62, 1.63, it makes no difference because you're way, way under what the limits were by applying this factor safety. So that's a very typical way the factor safety is applied. It's right on the material properties themselves. Okay, any questions before we look at another problem? Question? That, from the area solved for the radius. That was from the allowable stress, which is the yield stress reduced by the factor of safety, which is then our given load, the four kips, divided by the area, which is unknown. So you had to find the area, then from the area to find the radius. Now, did it not work for you? Did you remember the square? Did anybody else get these numbers? So those are okay. You guys are making me nervous now. Yeah, thank you guys. It's just now a knee-jerk reaction to just automatically say my numbers are wrong. Once they weren't, catch off guard. All right, another problem. We have a simply pin beam. Supported by a cable there and holding a load of some kind. Maybe my mother-in-law, again, we haven't seen her in a couple weeks. Two feet and three feet along a five meter beam. And this is also three feet and that's four feet. So that's the basic setup. The deal is when this weight is applied, the beam, no, we're gonna treat this beam as a rigid beam, which remember means it does not undergo any bending itself. We're only looking at what happens to these cables once this load is applied and greatly exaggerated the beam turns down to there. The load drops a little bit and given that this point B drops by, let's say, 0.025 inches. Find, well, let's just make it the mass we'll have here. No, we'll find the load, W. Remember, W plus MG is all. Now, thanks, Tom. The reason it's doing that is because this cable is stretching due to the load. So I have a couple other pieces to give you. Software's quick today. All right, let me give you something about those cables themselves. There's two cables here. We'll, let's label the parts for reference. A, B, C, D, E. Call this down here C, this N, D, N, E. Both cables are of what's called A36 steel with a cross-sectional area cables B, E, and B, C. Nominal cross-sectional area of 0.002 square inches. Yeah, so given that, then we can find how much load we'll do that. A greater load will cause D, E to stretch even more and a lesser load will cause it to stretch less. So given this displacement of that point B, point B itself drops 0.025 inches, then find the how much load causes that kind of deformation in the cable D, E. Anybody happen to have their book here? Because you're going to need the modules to the elasticity of the A36 steel. So it's the very, very back, even past all pages still. I believe it's the very back cover, back inside cover. Okay, so take a chance to look at everybody's book, should be their A36 steel. And the top table is English units, the bottom one is SI units. So if you don't want to bring your book, you might want to video copy, sorry, photocopy those two pages and just keep them in your notebook. Tom, what does it have there for the modulus of elasticity for A36 steel? 29, what? Times zero. Five times 10 to the third. Times 10 to the third, KSI. Okay, so 10 to the sixth, PSI. So watch your units in all of these. Remember, we're working with some very, very large numbers and some very, very small numbers. All right, by geometry, you can find how much the elongates that will actually be its deformation. Then you can find out what load what must be in there to cause that. Make sure it gave you all the pieces. What you need is the force in the cable ED that will cause that kind of elongation. Once you have that force, you can then find W by a moment balance on the piece itself. So use the geometry to find how much the cable ED stretches. Use that and the material properties to find out the force that's doing that. Then that force should give you the weight by a moment balance on the structural member itself. The member put it a ABD. Scooch back in. What else had a booby trap? The things going off camera, off tablet. Remember these drawings, of course, are greatly exaggerated as are my qualifications. That doesn't make some sense. David, getting that? I just need to know what the problem was. Should get something like 0.0417 inches for the deformation of the cable. Is that about right? Don't happen to have a factor of safety on this one. Phil and Chris confirm that they're working as one of mine today. You didn't get that? Hope we can edit that now. Speaking French here in an American video. Phil, we're okay with that? Chris, you just not having a good day, is that what it is? Well, as far as I'm concerned, better you than me. All right, once you've got that, then you can get the strain. Once you've got the strain with the modules of elasticity, you can then find the load in that piece. And once you find that, you can then determine what W is. Just a simple proportion through the moments. What's not working? 0, 2, 5, 6, 1. None helps. Yeah, remember these are greatly exaggerated and not to scale. Nicely drawn, but not to scale. Okay, David, what is it? How much is it? Who are you supposed to find that? Okay, that's me. Because a certain mass will cause the deflection here I gave you. Will cause the deflection here you find, putting a certain strain into the cable. Once you find the strain, you can then find the force that caused it and then go back from there and find the weight. Check some numbers as we go along. So the strain is that 0, 4, 1, 7. Over the original length, remember the units must cancel on strain, it's unitless. If you get something like 0.116% can lead you to the stress in DE because we now have the strain in DE and we know the modulus of elasticity. A new piece of paper can't be used on the same time. So if we're looking for W, it'll be in pounds. You got something, Phil? Joe? What's the YB? Is that inches or anything? No, that's inches, yeah, sorry. I don't purposely mix units on these problems. No, I can't guarantee the book does or doesn't. What? That is the strain. You already have that one. All right, about the stress. I don't know if I have it separately. Yeah, I didn't calculate the stress separately because the load force is the, solves, solves. When you solve for the load, it's force over area and then the strain and the ratio of the stress and the strain of the marginal plasticism. So I don't separately have the stress but I do have this number, marginal units. Doesn't look right. Oh wait, yeah. So that's the one. It's on gas, the units work. Phil, you have this force that's in the cable. Do you have that separately? Like 67.18 pounds of force. Yeah. Or Tom, if you're looking for, the way you had it was 0.067 K. Then you can use that to find out what weight if you sum the moments about point A. That's expensive. I know. It's a nice paper though. What you need to do is contact somebody who will sponsor pads of this paper for us but they'll put their Cobra logo at the top and we get it for free. We can hang a little bit of advertising. We always can in this day and age. Joe, okay? Yeah, you got that. Now with the original length which you didn't write down, three feet. Now you can find the strain in that piece. Once you find the strain, you can use the modus felasticity of that material which is given, material property. The load that will cause that strain is in here and the stress and so that's what I've done here and solved for it. It's just E epsilon A, all three of those are now known so you can find the load in there. And then from that find the extra load like that. Yeah, that looks like, that's right. Okay, with that though you can also find the strain in this piece and then determine where, actually where the load ends up because remember the load is going to draw due to the angle on the beam and the fact that its own cable is going to stretch a little bit. So there's a compounding effect of how far C will actually drop. The C will drop this delta YB plus its own elongation. It's a matter of grams for the load. Phil, what'd you get? For the load. For the mass I had 3.48 pounds of mass. All right, wait a minute. No, I didn't ask for the mass, that's for the weight. That's 111.97 pounds of weight. 112 pounds, notice that the weight C itself will drop by its only elongation in that cable plus the amount it's attachment point dropped. So there's two sets of numbers there that have this load dropping. And remember it could be that the material, none of them, there's no material failure but it could be that that weight dropped so far that whatever design purpose was is no longer viable. Okay, Joe, you all right? No, what's the matter? Are you just need more time or are you stuck somewhere? You got some of these numbers, did you get down to this stress? And then. I solved for, I have to equal some of that. And I solved for the stress. Right, so you can solve for the stress which itself is the force in EV over the area of the cable which I gave you. This strain is known and so you can solve for that force then and that should be the 67 pounds once you solve for this force because these things are all nominal. And then once you solve for that force you can do a moment balance about point A and find what weight is balancing that force. Can I help a little bit? Okay, look at it outside the class if you still have trouble come see me, we'll fix it up. All right, let's set up another problem. Imagine we have a structural member that we can approximate with this picture and it's loaded such that there's 30 kip at that free end. 45 at this interface between the two different pieces in that direction. And at midpoint there's a load like that of 75. Okay, so I've drawn those each at their point of application by whatever means. Some of the details, 12 inches, the 12 inches, and then 16 inches across there. An area here, 0.9 square inches and a third of that for that there. And again, A36 steel so it has the same modulus of elasticity. I want you to find the total deformation of that entire piece, 0.3 inches squared. 0.9 inches squared here or the bigger piece, 0.3 inches squared there on the smaller piece. Image now. All right, everybody got the general picture? What do we need to find that? What you need to realize is that each of the three sections will be formed by a certain amount because each of the three sections is under a different load, has different cross-sectional area. So we need to find how the individual pieces deform and then we can find the overall deformation by simply adding all those up. That's the picture. I'll get you started with, so let me label the points just for reference. C and D. So the section AB, to figure out how much it's going to deform, we need to find the load in that piece. Anything to do is to make an imaginary cut in that section to figure out what the load is in that section. We've got 30 kips there, 45 there, 75 there. These are all kips. I put an imaginary cut in the section AB so I can figure out what the load is there. And once I figure out what the load is, I can then use the modus elasticity, the cross-sectional area, and the original length to figure out the deformation in that section. We've got, what, 105 to the right, 45 to the left, so we need 60 kips in there to balance that. So now we know that in section AB it's deformation, which comes from however much load is in that piece, which we just found out is the 60. It's original length, which is the 12 inches. It's original area, the .9, and the given own modus elasticity. And after that, it's a matter of making sure the unit's all balanced. Just put the numbers in. Checking the units, kip, kip, inches squared, inches squared. We're left with just inches, which is what we'd expect, a deformation in inches. We'll carry an extra sig figure two through to the end. So we know that section AB is in tension. That's then a growth over just that region AB. Then you need to do it again between B and C, and then you need to do it again for C to D. You can do the same thing. So you'll get a change in length. Each one of the sections, and then the three of them added up will be the total deformation of the entire piece. All right, we're at the end. So you should be able to calculate one of those, and then we'll talk about that, wrap it up on Friday.