 Alright, on Friday we were working on a problem where we were looking at beam deflection. Did anybody skip Easter dinner and work through that problem? Do you have an excuse because it wouldn't load for you? So we had this problem of a uniformly distributed load, something like that. I think it was an 8-foot beam, no it was just an L-length beam, and the maximum load went up to some load, remember that's load per linear foot. So it got up to that. And we were looking for the deflection. Remember that business? We had four, well we had five equations, but they're all related to each other. They were all either differentials of each other or integrals of each other depending on which way you went through the list, but they're all very closely related. And from that we were getting what we're looking for, which is some function that describes the amount of deflection as a function of position on the beam. So that we could accurately predict just where that beam is going to be once it's loaded. So we started, we started that problem right on Friday. What was the, I think we had the first step set up, but just didn't quite get anywhere with this. What did we do first, remember? And what were we going to do with that? Yeah, that, the, remember we have this one relation where the load function gives us the negative slope of the shear curve. And then we, we knew that we could integrate or take the derivative of whatever, whichever direction we needed to go to get back to this load curve idea. So we had that little bit. We can now find the shear from that. We want it in the shear as we integrate through that. Okay, so we were, we were set up for those, those pieces there. Actually it won't be dealt with because we have an indefinite interval, so we're okay. Alright, what was the equation then for the load curve? Remember we, because of symmetry we only need to do it to halfway. So there's no sense going across the whole thing, especially since we can't get a single function for it. And so this actually needs to also be L over 2 because of that. What was the, the function that describes this? We had that, right? Zero here. If we take the intercept there, all we need then is the slope. The slope is the rise over the run, which is L over 2. So that function is that slope times X. So that's what we're going to integrate then. This is indefinite integrals because we'll, we'll use constant integration to establish those things. Integrate that real quick. Power by 1, bring down the power, have a constant of integration as well, right? How do we get rid of the constants of integration? Remember, it will be maybe all, I just remember you weren't here, so let me help what we're doing here. We're looking for this function of the deflection. The shear, or the derivative of that slope is the angle that the beam itself will take at any position for very small angles because the tangent is equal to the angle itself done in radians for very small angles. If we take the derivative of that, again, we get, if we take the derivative again, we get not quite the moment. Remember there's another term in there? There are actually two terms. By the way, that together is called the flexural rigidity, which is a pretty cool term if you think about it. It's kind of like jumbo shrimp. I mean, yeah, the load curve itself. And so that's where we did this indefinite integral, gave us a constant of integration that gives us the shear based on the load curve, but it's got this constant of integration in there. Remember how we get rid of those? With boundary conditions. Do we have any boundary condition for the shear that we could use? In fact, let's think what our boundary conditions could be. We're going to integrate, let's see, one, two, three, four times, so we suspect we'll have four constants of integration. We'll need four boundary conditions. Any places where we know something about the displacement, angle, moment, shear, or, well, we can't use the load because that's, that's, we don't have a constant of integration on the load curve itself, okay? Because of the pin support at x equals zero, the moment is zero. So we'll be able to use that on the next integration. Does the same equals the other as well? No, theta won't equals zero there because there's clearly some angular defect deflection right there at that point. Yeah, a cantilever support because we expect, once it's loaded, that its deflection will be something like that so that this angle, the slope there is zero, okay? That L over 2 with theta be 2? Because of the symmetry condition, we know that these two curves have to match each other and have a continuous slope. That's one thing that's definitely true of these beams. These beams cannot deflect such that there's sharp angles in, angle, the slope cannot be discontinuous because no beam's going to bend that way. These are elastic solids. They'll bend in nice continuous curves. So at the intercept, at the intersection of the two symmetry halves, we do know that the slope there will be zero. So we'll be able to use that on that integration. What else? There's two others. It's got to be one on the shear so we can get the constant integration at the shear level. And we haven't used this level yet either. So is there any boundary condition on the deflection or displacement itself? Constant. Is the shear constant? Remember the shear, the slope of the shear curve, the lower curve is constantly changing so the slope is constantly changing. What is zero? Is it? Let's sketch the shear curve. We know we're going to have symmetric supports there of how they, each support's got to hold half the load. The total load there is the area under, so each, each reaction is one half the total area or the area of just one side which is omega 0 L over 4, isn't it? So the shear is going to start there, omega 0 L over 4, and then do what? This is the slope of the shear curve starts at zero and gets increasingly greater to a maximum. So the shear curve is going to do, starts at zero and gets increasingly bigger. Shear curve is going to do something like that. That seem reasonable? Actually no, that's not going to work. It's going to reach a maximum and then, so it's going to go like that. No, that doesn't make sense either because we need to finish down here W0 L over 4 L. But what's going to happen with it? Well, at least we know what the shear is going to be at x equals zero. Then the slope gets less and less and less until it hits zero. But there's no discontinuity. Is this right? Starts at zero. Oh, there's a problem. And remember, that's the minus sign. At least we do know that the shear at zero is going to be W0 L over 4. I don't know what the rest of it's going to look like. There's our boundary condition. There's an extra bonus question there. What's the shear in the moment curve that will look like for that load? We'll go ahead with the problem at hand and figure that out later. We have time to think about it. x equals zero, that part is zero. So there's the first constant of integration. We integrate once again to get back to the moment. Zero x L plus C1, which I already know. C1 I know. I'm just leaving it in there for convenience. It's just easier to write it that way. And we know at x equals zero, the moment is zero. So that's zero, that's zero. And so C2 equals zero. That's good. So we're done with that one. We integrate again to get to the angle. And that's good because we have that zero at L over 2. So I'll give you that one. C3 is then, once you put x equals L over 2 and theta equals zero for the second boundary condition, you get C3 equals IW0 L cubed by 2. And remember each of these pieces here have to have the same unit. So you can watch how the L over 2's and everything all accumulate. We integrate it once more. That gets us then to the deflection curve that we were looking for. Minus W0 x now to the fifth over 60 L plus C1, which is W0 L over 4. There's a 2 down there already. So that's 8, 1 up. We're integrating it all. 8, 4. Is that right? This C1 becomes x cubed over 6. But C1 to W0 over 4 is over 24. Units of 8. OK, so they're working. C2 is missing. C3 is that. And we integrated once more so we'll have an x on it. Plus one last, one last 50 x, 3, 2, go around. I want you to 4. Yeah, 5 over 190. Yeah. Sorry? Well, we've got meters to the fifth over meters. That's meters to the fourth. Meeters to the fourth. Meeters. Yeah, that's L cubed. That one's L cubed. Why is that one, not L cubed. I'm sorry, not x cubed. Oh yeah, sorry. It's right there. L cubed. Yeah. So each of them have units of that part of meters to the fourth. All right. So what about this last constant of integration? We need something on the deflection itself. Sorry? We've got to plug in third condition. No, we already used that. We used that here. We can only use the bounding condition once. Because it's essentially, these are four unknowns. We need four equations. You can't use one of the equations twice again. Is the deflection 0x to 0? Of course. At x equals 0, the beam is fully supported. So the deflection is 0. So that's 0 when x, x, x equals 0. So C4 is itself 0. And so then we have the shear curve itself, which we can clean up a little bit if you want. If you're 0x, these polynomials are all calculated at any place along the length of the beam. That's the flexural rigidity. It's essentially the beam is already chosen for you when you have that number. Because you know, e from the material, remember that's the young's modulus. And i is the cross-sectional moment of inertia of the beam. So essentially what that's saying is you already have the beam chosen. You can know the material and the moment of inertia of the beam. If you don't have the beam chosen, you have to choose it such that those are still useful. If you have a minimum deflection you can withstand, you can choose EI, or you can calculate EI for any point along the beam. So that's the end of Friday's problem. So we've got to think of what that shear curve is going to look like. No, when we divide through by EI, it comes down here just like it is there. Okay, another method, remember this works for relatively simple curves, load curves that you've got them or fairly straightforward things. Remember a lot of these are in the table in the book. In fact, was that one? Remember we have this table, page, I think they're right after the page 808 in the eighth edition of our load curves. And we have, we don't have that one in there. Be careful. The 1, 2, 3, 4, 5, 6, 1 on the table looks kind of like it, but isn't quite. So the question becomes, is there some way to use the tables for those types of problems that aren't in the table? Normally distributed load, which could be the weight of the beam itself. If we were looking at the weight of the beam and the problem, it would look, it would be a load just like that. But then also imagine at the one quarter point, we also have a point load, maybe to represent that there's a wall or some kind of low bearing partition right there. This particular problem is not in the tables. Hermeter, one of those two loads is comparable. Each has been at a significant contribution to the bending. And we'd expect it to be a little bit lopsided because of that asymmetric loading. So we figure it's going to be something like that and we need to find that curve. If you look in table appendix C, you see it's not there. That particular type of loading isn't there. So the two loads that are there. So we can break it into those two loads. That's the fourth one down in Appendix C. We'll add together the other load that is also in the table called superposition. Let's call this one C4. It's Appendix C sample 4. We'll call this one C2. It's the second one in the table. Notice the first one in the table is a symmetric point load. So that doesn't match. So be very careful of using the right load here. Another fairly serious precaution in the second. So for the uniformly distributed load, notice it gives you the load curve already integrated for you. This is the uniform load C4 origin. All the constants of integration figured out based on available boundary conditions. And so that's the deflection already given there. It's a matter of just filling in constants that we've got. We know that the load is 20 kilonewtons per meter. I'll give you an EI, 100 meganeuble meters squared for the B and A meters. What? That's all we need. That particular deflection then, this whole thing reduces to minus 8.33 times 10 to the minus 6. The minus means it's deflection down. They aren't always. There are certain bending moments that could cause a deflection up. X cubed minus 16 squared. All this is just putting in the known constants and simplifying a bit. Making sure the units work. This is in meganeuble meters. The W load is in kilonewtons, so you have to watch your units. And this will give a deflection then in meters. You'd expect, and if you graph this, you'll find, of course, symmetric as it should be. It's a symmetric load, so you get a symmetric deflection. That's the easy one. C2 is a little more difficult. Not more difficult. A little trickier. Minus P. That's whatever this particular load is. P. X over 48 Bi. Remember, this is just the integration already done for you with the constants of integration evaluated. 3L squared. Boy, sorry, I slipped one line. That's the line above. 6EIL. Let's see. What's B? B is the distance, no, write it down just like it is in the book. E squared minus x squared. But also is listed there. Only good. Other portion of the beam, you essentially just have to turn it around and do it from the other side, and the equations will match. Easy to do, if you're terrible with it. Easy to do on a spreadsheet. So, go ahead and evaluate from zero to A, and then do the back part on a spreadsheet. You can do it by adding the polynomials, too, but they just get a mess in the algebra. Just do it from the other side, and then that works for B up to L, zero to B on the opposite side. Furtet x go down zero to whatever A is, two meters. And then do the first equation. For the second equation, I had a column that did zero up to six meters, but then for graphing purposes, I took it from two down to eight. I used that column to calculate the deflection, but then I used this column to graph the deflection. But you've got to be careful with that restriction that L worked for, that this only works for up to the low. It gives you something like that, and then you have to graph to do it on a spreadsheet, because you can use a different column for x for the two, and then graph them together. And I'll show you that in a second. Just sprained it, big and cut it. They're pretty purple. I'll take it off. One becomes minus 187.5 when all the numbers are put together times 10 to the sixth, sorry, minus sixth, x, 28 minus x squared. And then the second one, you have to graph from the opposite side. Does anybody have the second one? Yep. Call this L and call this an R. Same as that? I didn't get it quite interesting. That's from the right. And if you graph those each independently, they do join up nicely as you'd want. They have to join up both by deflection, because we can't have a step jump in the deflection itself. We can't have a step jump in the slope either. So they have to match, I'm sure what the two of them do, but right there they have to have the same slope and the same deflection, so the two curves match. And when you do that, when you graph the two, the easy one, the symmetric one there is C4, and it's nice and symmetric as it looks. The point of the load is right in about here somewhere. Notice how the blue curve, C2, does match nicely in both slope and absolute deflection where the two curves come together. I graph the left hand curve on one side and then the right hand curve coming from the other. The two of them together give the total deflection added together. Of course, it's greatly exaggerated, but you see that the maximum deflection is going to be about, what, about 22 millimeters over an 8 meter beam. About a 24 foot beam, maybe about the size of this room, and you're going to have a deflection of almost an inch, which is certainly enough for other parts of the building to not fit if you don't account for that deflection, if you don't have some kind of flexibility built into the design itself. Okay, that's it. We'll do a couple more super position problems tomorrow.