 All right, darn it, so wish that had gone in. I love it when students throw cans in the classroom. All right, today we're gonna continue our look at the fact that the materials in real life, and certainly the way we study them in this class, deflect the fact that all these beams we've been looking at, however they might be supported, however they might be loaded, all kinds of things we've looked at as possibilities in this class. All of these type of things can cause beams to deflect in some way. You know, this kind of load. Let's see, with those three supports, certainly the beam can't go down at those points, but in between it can do all kinds of things. So it'd probably do something like this, I don't know what it would do under that kind of load. Of course, greatly exaggerated, and we can have other types of loads of various kinds, and there'll be other deflections caused by these other types of beams. Certain things we just know. For example, of course, the beam can't go lower than this, and it can't go lower than that, and because it's embedded in the wall, it's gonna have no angle to it at the wall. So maybe this one will deflect something like that. We figure, again, greatly exaggerated. But that's just sort of our common sense take on this, and that's not good enough. So what we need is some kind of way to actually calculate what this deflection is. So we'll use the little symbol V. I'm not sure exactly why, but that was good as anything. Deflection from some neutral, unloaded state, which, of course, is just simply straight. We want to figure out just what deflection curves are as a function of x. Certainly it has to do with x, because as we go along the axial direction of the beam, there's the load changes with x, there's supports at x, all of these type of things we have to figure out. So as we go through this, we're gonna come up with five equations. So we're gonna keep track of our equations as we accumulate the five of them. And we may need all five for certain problems. We may need just, well, we might need just one of them, because here's our first deflection equation. We might just know what the deflection curve is, sometimes in certain classes, certain types of problems like these. You just already know what you've got somehow. I don't know exactly how that would take form, but I guess if we had a beam that was supporting everywhere along it, that we had a beam that was laying on the curve, no matter how we load that beam, it's not gonna deflect, then we'd know the deflection curve for that problem. So I don't know, something like that I guess we could do. But it's a possibility, so we have to have it up there. Plus, as we're gonna find, these five equations are very closely related to each other. And the end product that we want anyway is that we do want to get back to this load curve, other than this deflection curve. That's our ultimate goal. So that's our first equation there. All right, so let's start to put this together. We have some, well, let's see, let me draw it in a more regular way. It doesn't really matter what we've got going here. We've got some position x, some deflection at that point where we're gonna zoom in and look at some piece of the beam that's deflected. So this is nothing more than a picture of us taking a zoomed in look here someplace, one of these spots. And we just have to narrow in on it to really figure out what's going on. So that's the kind of thing we've got right now. So there'll be, of course, along here neutral axis. And with respect to that neutral axis, the beam has some curvature at that point. And we, before it called that radius a curvature row, so we'll continue with that for now. And at that point, we'll know something about the slope of the beam, that kind of thing being, let's see, the slope. Well, that's the, if the neutral axis is taking the shape of this load curve and the slope is the derivative of that with respect to x. And of course, that's also the tangent of whatever angle that the beam is happening to make there. It's curved, it's loaded, it's got some slope to it there. So that's all we're looking at is that little piece right there. Except for the fact that these are very small deflections. We're talking about small, actually not small. We're talking about the angle here. So from very small angles to that slope. Remember, all these are absurdly exaggerated drawings. What we're really talking about is not even millimeters of deflection, maybe tenths of millimeters of deflection over the span of a 20 foot beam or something. Very, very small deflections. So when that's the case, then tangent theta is approximately equal to theta for very, very small deflections. So we'll take that and then that gives us our second load curve. Which is that now we have a way to relate the first derivative of the load curve. So we've taken a step one beyond. So now our second equation is that one. We're already making pretty good progress. We're looking for that load curve. If we know how the slope changes with x, then we could just integrate that and we have our load curve. It would be all set and we could start continuing whatever the design is that we're trying to do for that load curve. So we're already making some progress. How many equations I say we're going to come up with? Five. We've got two of them done. We're going to be out of here by about 24 paths. We're just going to be flying. Dewey's going to be sorry she missed this class because we're just working so well. Hi Dewey. Everybody? So we have that one. Colin. Senator Nesterkart. All right. So we're doing, we're making great progress. Let's see what's the next one. Oh, this one, this one you'll have to pull out of your calculus book. Which in other words means there's no possible way we could remember this even though it might look vaguely familiar when we get to it. So this one has to do with the curvature of some curve itself. It looks something like this. See if it doesn't look familiar. The second derivative of this load curve is related to something like this. I'm sure that most of you have already written it down because it just pops from the memory. Look familiar? Okay. Oh, no. But it does certainly look like the type of thing you would have done for a week or two in calculus shaking your head and going, why don't we do this kind of stuff? Well, here's why we did it. Here's one of the things it leads to. But the deal for us here that we need is remember, let's see, that's the thing we just came up with. That's the angle the curve is making at any point, which is very small. So if this is very small, what's the square of very small? Very smaller. So this is for our purposes zero, which means the bottom is one to the three halves, which is one. So we have that this is essentially the second derivative of our load curve, which you might think is our third equation because we've gone from the curve to the first derivative to the second derivative, but not quite because remember, this curvature business itself is not an easy thing for us to know. It's also not an easy thing for us to use. So we've got to do something else with that that will be a little bit more useful because that's just not all that useful. But we have dealt with that kind of thing before. Let's see if we can relate it to this drawing some. Actually, I think I'll just clean it up because we don't need the angle on here right now. So I'm going to just clean up that very same drawing. We've got this segment of the beam with some heavily exaggerated curvature to it, some neutral axis along there, and the radius of curvature down with respect to the neutral axis because there's a slightly different curve at the top here and a slightly different curve at the bottom. So we relate it to the neutral axis because remember, that's our point of no stress. All right, so we've got, let's see, we've got this whole business sometimes some angle delta theta which is very small. And remember, we measure y from the neutral axis. So let's see. A distance, let's see, down to some point here. That's y up from the neutral axis from the center of curvature that itself is a distance rho minus y then. If this portion of the beam had a length of delta x, that line will have a length we'll call delta x prime. Because it's, remember, above the neutral axis for this type of bending, we have compression below the neutral axis. We have tension. And I think the only other piece we need, now we don't really need it, so I'll lose it out. All right, we actually, we've done this type of thing. This is a familiar step from, from, jeez, I don't know, maybe the third week or something we did this. Because we can now write down what the strain is, which remember is the, a deflected length, a amount of deflection, amount of contraction or extension divided by the original length. So the amount of relative compression we have here is delta x minus delta x prime. Do you want them in that order? No, delta x prime, yeah. We want the new length minus the fold length. Yeah, that'll give us a negative over the original length. Remember, this is just from our definition of strain from a long time ago. Right, that was right, about third week or something, wasn't it? Remember how young you were back then, how fresh-faced and eager you were back then? Now you're, you're jaundiced and biased and bitter and tired. We have done this very thing. Let's see. That delta x prime, well that's actually the arc length at the radius r minus y. So the radius times the angle r minus y del theta. That's delta x prime. Right, that's just, that's just a way to do arc length. Delta x is the same kind of thing only the radius down there is rho, the radius of curvature. So this rho delta theta and then that's the same thing on the bottom. Which is nice because the del theta's canceled. We need a rho on the bottom there. Rho del theta. Off we have rho minus rho. So we're left with just minus y over rho. If you remember exactly the same thing we got three weeks ago. But we're almost setting out to tie this all together and we'll have our third equation momentarily. We've got this, this, this strain now. Remember too, the definition of Young's modulus. Remember that? What is it? Yeah, but it cancels top, it's s over 8. Because it cancels. So that means that that strain we can put in there we get our normal stress which is Young's modulus, a characteristic of the material times whatever strain we're seeing here. But we know that strain will now be minus. Except in the last few weeks we've come up with something else for the normal strain. Remember we got to all this business all we were talking about was simple axial loading that caused our normal stresses. Now we're talking about bending that causes our normal stresses. We developed that about a month ago. So we know that this also equals minus m y over i. So that means of course then these two things are equal to each other. So let's see if we can put that all together now. We can do something more with this top one. Minus e y minus e y times 1 over rho but 1 over rho is the second derivative of the load curve. Put those all together. So let's see. Let's write it up here so we don't make a mistake. The two minus signs cancel. So I'm done with those. E y d squared v dx squared. The second derivative of our load curve which is kind of what you might expect. We went from the curve to the first derivative and now we're at the second derivative. Equals m y over i. Y cancels. That's good. So we're going to be able to find out the beam deflection without worrying about just where we are in the beam. It's going to be the same for the entire beam. That kind of makes sense. It all goes together. And we'll rearrange things a little bit just to make them in a way that's useful. We get that. So here's our, which is nice, since the moment itself, remember, is a function of x. We started figuring that business out what the moment is with position on the beam for since back in sacks we did that. So the moment is a related second derivative of the beam. I don't know if that's supposed to be in yellow. I don't know why it turned to pink. That's terrible. Can you go down and paint the racing, please, and put it in yellow? We have no trouble figuring out the moment as a function of x. We've been doing that for months since last fall we've been doing that. So if we can figure out the moment curve, we know it's related to the second derivative of the curvature, the deflection of the beam, the beam curve, and we can integrate back and get the beam curve now in colors. That's awful, pretty bored. If you guys want to take a second and take a picture with your cell phones or something. Hey, Pat, you got your coffee mug there. Did you know there's free coffee at Starbucks if you're bringing your own mug for a root day? Yeah. Wish you'd known that unbranded. Man, we paid for that, didn't you, sucker? All you have to do is be able to handle the stomach going into Starbucks. All right, third equation there. This is going great. Everybody okay? Bobby, it's a matter. You wish you'd gotten free coffee. You didn't know that. Three curves there. All of them leading back to the shape of the beam itself when loaded. Now, the leftover part is very easy. The next two equations, if you think about it for a second, you come up with them. The next two equations are, remember that moment is related to shear, the slope of the, see when I've got the diagrams there. It gives us the slope of the moment curve. Well, we've got M right there. So this is D. I put EI over there. There's two reasons to put EI over there. One is that has nothing to do with the load. That's E, remember, is the material characteristic. I is the cross-section characteristic. Has nothing to do with the load. Also has to do with the fact that when we've done these problems, with the moment function, the moment is a function of X all by itself, we didn't come up with a moment divided by EI. So it's just a little more straightforward there. So then this equals EI, which of course is constant, and we get that the shear is proportional to the third derivative of the beam curve, which is good because we know how to come up with the shear as a function of X as well. That might even be a little bit easier than coming up with a moment as a function of X. So if we have the shear as a function of X, we've got the third derivative of the beam curve. We can integrate back to the beam curve itself. One more equation, five equations. And we have a feeling where it's coming from. Well, yeah, that's a good guess. That's going to be in there. We'll certainly follow, but where do we get the equation to do that? No, it's Q, third derivative. But remember we did our shear moment diagrams. We had moment was related to the shear, the shear was related to shear moment diagrams. We had one that related the slope of the shear curve, eight yourselves, and you say, oh man, I knew that. The load curve. Load curve happens to be often that it's zero because we are using point loads. Sometimes the distributed load curve is constant. Sometimes it's linear. We can handle anything, but the load curve itself is a function of X. And we get then the fourth derivative. And there we go. We've got the five curves, all of these things for us. And so we're ready to apply it. One of the easiest examples we've got, it's right in the book anyway, but this way you don't have to sit and go through that one. You don't have to take notes for this because it's in the book. You can just sit and pay attention to it because it's nothing more really than several steps of integration of simple polynomials. So we'll do that one together. We'll take the case of a simply supported cantilever beam with a point load at the end. About as easy a one as we can get so we can take our first cut through this. Simplicity's sake, we'll take X from the end. We don't have to. We can do X from the wall. But if we take X from the load, then the moment itself becomes a little bit easier. The moment is then just minus pX at any place, which makes sense. Remember, there's no moment at the end since it's a free end. And then the moment increases with distance because of that load at some point p. So we know that the shear must be equal to p there. Then we have a moment, a couple there that we need to counteract. And that couple must be the same size as the, the moment must be the same size as the couple, which is p times X. But that's a minus direction, so it's minus pX. So about as simple as we can get for a moment function there. If we take X measured from the free end, it just becomes simpler. But it doesn't matter. We could have done it from the other way. We just had a real fond of hard work if we can avoid it. All right. So that puts us in right here at the third equation. We now know that the moment is equal to minus pX is equal to EI, that's to say EI, EIL. And you can if you want to sing that out at any time. Jake, you're not in the mood? No? All right. Put this in right there. We're at d squared v dx squared. So we can integrate that. That will take us back one step to here. So we'll integrate that. Let's see the integral. We get minus pX. But then we integrate that once. We get minus pX squared over two. Is that right for integrating that moment? But this is an indefinite integral. So we have a constant of integration in there. This now is theta, a little piece there. Oh, we've got an EI in there. I'm sure I've got the pieces. EI, okay. We're all right. This is the part we're going to start with so we can take the next step in integration. However, to evaluate this constant of integration, remember how to get rid of those. Use boundary conditions. If you know some boundary condition that applies to the equation, then you can eliminate the constant of integration. We know that X equals L. The full length of the beam, theta equals zero. We know that because it's a cantilever support. It's an embedded support. So that's our boundary condition. Let's see. X equals L. Then we have minus P L squared over two plus C1 equals zero. Because theta is zero and X equals L. So C1 equals over two. Fair enough. With that one, that worked out okay. I think we got all the minus signs. Okay. So now we're at, let's see. Now we're at whatever C was, which is P L squared over two. And we can clean that up a little bit if we need to. But not too difficult. Just a polynomial. We'll integrate it once again, which will then give us our deflection curve. So minus P X. We're integrating this first term. That's cubed over, there's already a two there, so six plus, remember P L squared over two is a constant. So that integrates to X. And then another constant of integration. And that equals EI times the load curve. Yeah, because we integrated this side, integrated that side, got us down just to be as a function of X. So we iterated that constant of integration and what boundary condition can we use for if X equals L at the wall, the deflection is zero. Because that's a straight support. So X equals L, V equals zero. So we put that in. Minus P L cubed over six plus P L, and then another L. So this is P L cubed over two plus the constant we're trying to solve for equals zero. C two equals plus P L cubed over six minus P L cubed over two. Minus times all right. So now we can put everything back together. We get the load curve. No, it was P L squared and then X equals L for our boundary condition. So it becomes P L cubed. So of course then that cleans up to what? Minus P L cubed over three I think. And so now we can put all this back together into our load curve and we get then for the load curve and a final form. So we'll know just how much deflection with X. We know the beam is going to do something like this. But now we'll be able to draw it precisely. So put all the pieces back together. I'll clean it up a little bit for you. We get minus P over six EI. Remember that term EI is not load dependent. Minus X cubed plus three L minus two L cubed. And notice of course the units work out. All these terms, all three of these terms is meters cubed. And with everything else we've got over here then we land up with units of just meters. Take a step, we're done. Are they always going to be the same for every problem that we do? No, because we might not have a cantilever beam. If we have a simply supported beam like this under some kind of load that causes it to reflect we don't know what the angle is going to be at those. We do know that the deflection will be zero here but we don't know what the angle is going to be in any one of those. We might not know anything else. However, the boundary conditions are in table 12.1. 5.76 if you do have your books if you don't. Hang on a second and I'll put it up. A book is going to have these string of material books if you don't have it. These are the boundary conditions you can apply for anyone that has problems. A roller support or a simple pin support have the same boundary condition. No deflection at that point, that's what supports do, but they also can't support any moment. So if you have a problem where you start with the shear curve integrate back to the moment curve and get a constant integration there then you can use the boundary condition at a pin or a roller that there's no moment supported. That's a roller at the end. The third one is a roller in the middle. There could be moment there because there could be bending in the beam there but you know there's going to be no deflection. Same thing with a pin in the middle. We just use this cantilever beam. The two boundary conditions that come with it we just use both of those. A free end. We've been using those boundary conditions when we do the shear moment diagram. It's the fact that at a free end there's going to be no shear and no moment but there's certainly going to be some angle. Don't know what though. And then we don't see it very often but we could have two beams hinged in the middle. We know at that point the moment is zero because of the hinge. So those are available for you. Page 576, Table 121. Also, I probably should have left that up, also available to us since we're not lazy we're efficient for some simple loads all of these integrations are already done with applied boundary conditions for simple loads. Now we may not have that particular any one of these particular loads in any one of our problems so we've got to do it but if the load we're looking for is in here then we don't need to do the integration and you can see the one we just did is right in there. Tentaleever beam with a load right at the end. It's not an exact same form I have and they factor out some of the terms some of the common terms to get it down to a slightly different form but in general it's the very same solution we just made. So we didn't need to do it we could have just looked it up. Just be careful applying any of these make sure that your orientation is at the same place that some of the intermediate values you might need to use the deflection is in exactly the same direction or position or for example look at the second one if you can. I'll put it back up. A couple other concepts my favorite student is too Brandon does my favorite student for how many years I'm writing two. You've been a student here a lot longer than those two. The second one just as a precaution make sure that the A and the B that places the load is the same for your problem when you go to apply it as the A and the B that's used here than in the particular equations. Zero to A is the longer side so you have to make sure your beam looks like that you may need to imagine your beam being flipped over if you've got a P that's off center. So if we can match those great we're done we just look them up also we can check maximum deflections we know from our problem that's going to be at X equals zero and we can put that in solve the problem that makes those terms go away we get the maximum deflection as minus two over six is three minus P L Q over three E I think I did that right but that's in the book anyway that's what they've got there yeah minus V max right there or I mean V max right there we can also figure out what the maximum angle is by doing the same thing back to this equation to remember this right here itself is the angle so we can figure out the deflection at any point we can also figure out the maximum deflection the maximum angle all those types of things we need and then if we needed to we could go up and figure out the shear and the load well we've got the load curve but we could figure out the shear even though we could for this problem anyway alright let's start on another example one of the ones that's not in the book so we have to just make sure we know how to go through it because as you can imagine it's not often that the problems we have are going to be so nicely tabulated in the book there so nice simply supported beam a symmetrical triangle right there maximum of W O at the peak and a length of the beam L and then we want to find we want to find the load curve first thing you do of course is check the tables see if we got it there let's see we've got something that's kind of close at the bottom there we've got that which is kind of close looks like maybe this is just a doubling of what we have is that the case though this one has zero deflection at the light end and zero deflection at the heavy end we don't have zero deflection at the heavy end so we can't just take that one double it by flipping it over and joining the two solutions together it's not the same problem that we have there even though it kind of looks the same so we're going to have to do this one ourselves couple things let's see what can we use oh one thing we could do is we could recognize that this problem is symmetric so we don't need to do it across the whole beam we only need to do it half way then we can flip that solution over and join the two together all that we have to guarantee is that at the center the two solutions on either side meet in other words the deflection is the same for both sides I don't know how I write that it was to say same plus and minus mean on one side or the other of the center symmetry also that the angle must be not just the same on either side the angle must be zero because of the symmetry we know that it must have zero slope right there so that can help us a little bit we're doing a little bit smaller problem that way it's always nice to get a little bit smaller problem alright let's see you can take a second or two to come up with the equation for the load curve not too big a deal I hope omega zero over lx all that is is nx plus b for this line if we take x equals zero from any one of the ends then we get just nx plus b where that's the slope now when you're setting the course if we're take x equals zero at either end so we're okay so that puts us actually that puts us to the derivative of the shear which puts us in here at the fifth one we're all excited because we have lots of integrating to do now integration which will get us down to the shear a minus two omega zero x squared over two plus c one that look right boundary condition on the shear that can allow us to get rid of this constant of integration I think any of the boundary curves boundary conditions have to do with shear do they one does oh yeah the free end one does of course and we don't have a free end in this problem so we don't have a boundary condition a shear we're just going to have to carry that through as we do the next integration oh by the way this is e i valverine those tubes cancel oh lost the L I mean sorry so integrate it once more zero x cubed over L over three L plus c one x plus c two and we'll stop there yeah do we know any boundary conditions on the moment that we could use moment that either support is zero so if you're doing it from that direction then zero and even zero so we can come up with at least get rid of one of the one of the alright so we'll finish that real quick on Monday and then come up with another method of solving these which relates all this in the same way just can make some of the problems a little bit easier