 I don't expect it to take too long, but you have the time. It actually had been three problems before, but I knocked it back to two for you tomorrow. So I think you'll have plenty of time, but you've got to let little extra in you since everybody can stay and all kicked out. Other than the fact that class comes in after that, but that's like an hour after, right? So it should be okay. All right, any questions or a go? Remember to open book, open notes. All right, we're working on a problem on Friday for some of you there should remain a problem. Everybody look at it over the weekend? I'll take bite of your snort. That means no. There's no football. You still hung over from last week? Sometimes those things can take a while. All right, we had a simple stand of some kind that when loaded must remain level. The trouble being that two legs were not the same length. Now, we know from what we've been looking at so far, which is new to us this term. It's nothing we had to look at last term. In fact, let's look at this as we would have looked at it last term. We don't know what this placement is, and so we would have set up our equations. We would have summed the forces, know that they summed to zero. So a free body diagram of the beam itself gives us that the two reaction forces must equal the load. Sum the moments, maybe about B, who told us what, that F, A, which we don't know yet, times its moment arm, oh, remember this was three meters, must equal the load times its moment arm, which is three minus X meters. No slap so far. You guys could have done that this much in your sleep. We have three unknowns, the two Fs and the X, so we need another equation. Last fall, we would have pulled in that other equation, being just the sum of the moments about some other point. A is as easy as any. Could do it about the load itself, but so the reaction at B times its moment arm has got to equal the load times its moment arm, just the X. And that would have done it for us last fall. Everybody did that, I think. If we just thought about it for a split second or two, we would have just put the load dead center. There's a trouble with that, that we didn't have, now that we're looking at real engineering materials and how they deform under loads. We have a problem with just using our statics analysis from last fall, which then makes FA and FB equal. So we'll say here, from last fall, that gives us a symmetric load. Well, it depends. If that's what you did, that's all you did, just took what we did from last fall, yeah, you get a symmetric load, and there's a trouble with that. Who did it this way, found the load to be symmetric, and placed the load dead center? Something like this, something to be in the arrow of both terms, X, the load, X is one. X is one? Yeah. Where we're here? Yeah. You got that from this? Yeah. Because with just this alone, that should have put the load right in the center. So this is what we would have seen last fall. Load right in the center, and two reactions out of that. And this would put the load, DJ, what'd you do? Put it right in the center. DJ, what'd you do? This is the one from last Friday we're taking out again. I was having a hard time solving this on the AISC, and they got a cancel out of the AISC. Okay, Pat? I didn't get that. Divide each over three, add them together, solve for X. Divide what? Bite three? Divide it, move this FAA and FB, take the three out for another side, about three meters. So you have everything in terms of X, and then put both of those in there. Yeah, and you get X. All right, so that's A divided by three, one minus X over three. That's this one divided by three, and put it ready there. Plus this one divided by the side of the three, which is 80X over three, equals 80. Right, that's these two put into there. These two divided by three, and put it in there. Is that okay? Let's see the 80 cancels. If we can multiply by three now, make it easier. We get one minus X over three, plus X equals three. What do we get? Two-thirds X equals... Okay, well this is algebra now, though, right? If I made a mistake, X equals three. X equals three. This doesn't work either. That puts it right there. So that must be some movement somewhere. Okay, well you get 1.5 just by inspection. So there must be an algebra error in there somewhere. Multiply by three. So that's just this. Oh, multiply by three, are in the truck. So this is one minus X over three. That's the... Oh, yeah. No, it's okay. That's the 80's canceled, and then multiplied by three. And this was 80 canceled, so that's one multiplied by three X three. Is that okay? Yeah, same thing I had. If you carry the one over, we get minus X over three, plus three X over three equals two. Now how do we get the three out of that? Two-thirds X equals two. Times the big three halves of both sides. Times the big X equals three again. Yeah, that's the same thing we had. Well, I don't make any sense either. Something's wrong somewhere. And then eight, is this distance... That's three minus X. That's 80 times three minus X. You have to do it really fast. You get 240 minus 80 X plus 80 X, all over three. What? You just distribute it really fast. You get 80 times three is 240 minus 80 X, and then plus 80 X. So 80 X is canceled, so it's 240 divided by three equals... Okay. 80. 80. But what that sounds like, what you're reading that to me, blah, blah, blah, blah, blah, blah, blah. I'm just figuring out my mistake. Well, what's our mistake then? I don't know where it is on there. But it's 1.5? I don't have the answer now. Fv times three, 80 times X. Well, anyway, in here, A plus Fv... Just distribute the 80s versus the 90s, then by three. And that's how it equals to the other one. And 240 divided by three is 80. So that's all for today. It's just canceled. Yeah, it's not that bad. Anyway, so... You have to get 5.8, that's 3.2, and you get like 4.6, 4.8, something like that. Yeah. And those, then you just double that, and you divide your 80 by that, and you'll work with mine. A lot of that. Now I can see why math is not taught over the telephone. All right, so this is okay. There was just some algebra error, and there was given us the X over three. How do you get that by inspection now? How can you just assume that it's 1.5? Well, the symmetry of it, if this is moved over, then... Well, I guess we could solve it, but I mean, that's certainly the simplest of the solutions. The trouble is that this ignores the deformation that we've been studying, and remember the requirement was that the beam remain level. And we haven't taken that into account. So there's some kind of deformation here, del A and some there, del B, that in fact must be equal, which means, well, those are equal. That's really just a single unknown. We now have four unknowns, not three. So we need a fourth equation. Now that we're looking at other things, now that we're looking at the deformation of the materials, there's got to be other unknowns. What's our fourth equation then to find this unknown? There's four unknowns, and now these, they're equal, so it's really only one unknown. The left stream. All right, how do I write it? So that, okay, now there's a fifth unknown. We need to set this requirement sometimes, since it was a restriction on the problem. And you're right on the edge of what you need anyway. Set the two deformations equal. That's then what? Strain and A. Well, let's see if the easier way to do it is I'll make sure I've got the algebra right this time. That is del of A. The other equation has bala in it anyway, I think. That doesn't happen often enough. And then there's the same equation for Fv, Lv over. A lot of those things cancel out. The area of the two is the same. Remember, they both have the same diameter. They were also the same material, huh? I don't have it up there, but it was the same for both of them. Oh, that's 22. Now with that then, we have enough to solve for everything. And it doesn't change things a lot, but it's still something we needed to add that we hadn't had before. There's a slight shift of that load slightly past the halfway point, which makes sense because post A is longer, which means it'll take less force to make it deform as much as post B is. It's got to be a little bit less. And Fv is, the deformation in A and the platform remains level. From this, you can also find the deformation in the two of them. That's what I got when I solved this, isn't it in that equation? When you have a system set of equations, every unknown doesn't necessarily appear in every equation. Now I double-checked this one last night where my algebra went wrong. And then there was one other little piece that required to find your member final diameter after loading. Where would that come from? Once loaded, its natural response is to sort of bulge out like that. Lausanne's ratio, which we know because we know what the material was. I think it was .35? Yeah. Stress and strain in the radial direction over that in the axial direction. In the axial stress, we have that number. It's the, which we can get from here. That's full eight nulls here. And Frank, you did that. You thought that this needed to be in there and got a number, something like that? Yeah. All right, so the deformation of these materials brings in other unknowns. But they do respond in predictable ways. So it allows us to engineer for these things, which we couldn't have done last fall. All right, one last thing we need to bring up. Now let me double-check now that everybody's here. Everybody can stay afterwards tomorrow if need be. I don't think it will be actually short of the test from last year. But I don't want some to stay if someone can. This is not fair. So everybody can stay if need to. All right, so it's optional. If you want to stay and work a little bit more. If nothing else, just help take the pressure off a little bit. It's easy enough just to make simple algebra errors when you're under the gun. All right. Well, we've got two things in particular we've looked at as far as material properties. The Young's modulus. Let's see. That's the load over the area and the deformation over the length. Both the area and the length were original values, remember? Unloaded values. So these two things here are just constants of the geometry of the piece we're looking at which makes this essentially then the load we apply over the material that the structural members response to that. Young's modulus is essentially the material's response to a given load in a very predictable way that we can use. It might make sense to think, well, what if we have the same kind of thing not for the normal loading that we do but the shear loading we do? What if we have a shear stress divided by the shear response? Let's see if we remember what that is. Imagine we've got a piece of something like this. We've got some kind of shearing load on it like that. The real response to that is something like that. Remember we looked at that? The shear strain. What did we call this? This response to that? That's the shear strain, but what was the symbol? Yeah. So if we have the same type of thing here where we had the normal stress over the normal strain, now we have the shear stress over the shear strain. It might make sense that we have the same type of thing, some characteristic of the material that's load over response, which is just what this would be. The load over the response. Well, we can't really write that in the same type of way that we could here. This is indeed a material characteristic that we call G. It's called the modulus of rigidity that these two moduli are actually related function given there. We'll see that later. I don't know that we'll actually cover that. It's derived in the book. It's a simple derivation, but we can use it now if we wish. Is that a Poisson-Zerchian book? Yeah, that's Poisson-Zerchian. So we have a new material characteristic. I believe it's in the back of the book. Anybody that you've got there's just real handy. Never mind, Jake. Double check. G is in there? In our tables? It's a material property just like the modulus of elasticity is. And we can use it in the same kind of way. So we'll look at it quick, see how we can do that. Two rigid plates fastened to some elastic material. This kind of thing is sometimes termed a shear block. I can actually use a sort of a shock absorber as a load is applied to the plates. The material will respond, can serve as a shock absorber for the material. Jake, how are you doing? Bob, you wouldn't have no trouble with these if you stuck with the technical free-hand sketching. Notice that the material actually responds a little bit different than we've been sketching in reality because of our assumption of perfect bonding of the material to those plates. But what we've always looked at before when we looked at this shear strain is we looked at the average and not necessarily worried about what the strain was individually at each plate. Specifically looking just what the slopes were at these two different places because the strain is different there than it is in the material and the interior of the material. But then we have to start looking at things like bonding strength and other considerations that are beyond the scope of this cluster right now. So let's put some numbers on there. Original dimensions, 50 millimeters back, 40 millimeters high, 160 long. Those are the original dimensions all in millimeters. Given that this material has a modulus of rigidity of 600 megapascals, find the load that would give a deformation we need to limit the deformation of this shear block to 0.8 millimeters. Maybe it'd be conflict with some other part of the machinery that it's a component of. So we need to find what load is possible. Be at that level or less. Let's see, that's going to come from, come from, let's see, well, if we knew the, or this is the shearing stress, we've got the area. Do we have this? Not, not specifically, but we certainly have enough information to find that, especially if we use the average values here. So we can get the load from the modulus of rigidity and the shear stress. Sorry, shear strain. So we've got G, the shear strain we can pull out. The area, you've got to make sure you've got the right area here. There's sort of several to choose from, as there are often in these three-dimensional problems more through the day for coming prepared. A little bit of algebra, mostly just getting used to the numbers, also making sure that you're using the proper area of that calculation. Millimeters and megapascals and things are getting, a lot of big things, a lot of small things, all mixed together. You've got to watch it and get it very carefully. The use of the same area is one of the easy places. It's the area over which that shear is being exerted. It's the 50 by the 160. What about gamma? What did you put in for gamma itself? This is the main process that I had too. Nothing yet? Still in conflict? You need to arm wrestle? That's the arc tangent of the 0.8 over the 40. That's where you got the 0.02? What is those two just divided straight? Which, if you remember, in radians the tangent is essentially equal to the angle. And we are talking about very small angles here. We have to exaggerate them greatly to the drawing. The 50 by 160 give us meters squared cans a week. We'll have newtons. Now we have radians. This only holds, it probably holds for zero degrees, but beyond that, that approximation is true only in radians. But another try, this is the problem I took off of tomorrow's test. We'll see if I should have left it on or not. It's very similar. Sure. Can I send it to the printer yet? And I haven't put it up on the board yet. In fact, good idea. Very similar problem. And shear block. There are classes that you can take that will worry about just how you do this bonding such that the failure is not there, assuming that I'm buying stuff from Earl. Add a little bit of free dimensionality to it. We're going to need that. We need the area which that plate's attached. That plate's loaded. Then with some load p causing the shear block to deform. We'll again now just look at the average response to form something like that. Since A piece there is B, the depth into the board is B, we need to find that deformation, that displacement there, the dimensions, the size of the load itself that can affect how far it's going to deform. But also then the material characteristic modulus of region. So it's not terribly unlike what we just did. They bungled it so. No, in terms of those, you'll have del as some fairly simple function of these five variables. On this plate, straight down. So how is it shifting when you're out? It's not shifting now. The whole piece just goes like that. It's a lot of impact. I told you, I don't want to do three dimensional things, but you guys make it. We can make this here. Get out of class question. The units, check out. You can check your units, because you know these are all going to be units of length. P is going to be in units of force. G is going to be in units of force per area. You can check your units. Once you've got del in there, this is all properties of the object, or the load, and the object itself. There's no... You're solving four. Del. You should have del as some function where the units work out in about very small angles. So, yeah, you've got that. Did you just take the tan part on there? Nope. Did the units work on that? Yeah, I guess they would. G is defined as the shear stress over the shear response. The shear stress is P over the area over which it acts, which is VH. Gamma now uses a small-angle approximation, which essentially, if you've got a tan, you just take it out. Yeah, being radiant, you can solve for P... Oh, no. I'll have to solve for del. That's the only way you've got for the alarm.