 Today, we're going to talk about beam analysis and work on creating shear and moment diagrams. So on our previous video discussing internal forces, we made the claim that internal forces can and do vary within a body, depending on their location within the body. So let's see if we can observe this by considering one of our simplifications, a simply supported beam. In other words, we think about having something like a plank of wood or a steel beam, and we're going to consider it to be simply supported. It is pinned on one end, so it's not allowed to slide side to side at that end, but there is a roller at the other end, so it can be considered statically determinant. And we are going to load it with some force in the middle, and that force is going to be equidistant from the two sides of the beam. If the beam has length L, then it's halfway between, so the distance between the sides and the middle load are each L over 2. And if we do a simple reaction, some are forces in the y direction, and some of the moments about a point, we'll find that the reaction here at point A is going to be, if we call this point A and this point B, the reaction at point A is going to be in the y direction equal to half of the force, and the reaction at point B is similar to going to be half of the force. So now we've determined the external forces on this beam. Well we'd like to know what's happening inside the beam, and the primary reason to know what's inside the beam is because ultimately if the beam is going to fail, if it's going to break, it's going to occur at some location inside the beam where the forces are too great for the material in the beam, and it either tears or crushes or otherwise changes and the beam will fail. So let's make a cut across the beam. There's actually two places to cut across the beam. We could cut here in the first half of the beam, or we could cut over on the right in the second half of the beam. Those are going to be slightly different, so let's consider each of them separately. Let's start with our first half cut. If we cut the beam to the left of that loading force and look at a free body diagram just for that part, we'll notice that we still have the reaction, the external reaction force here of F over 2, but we no longer have the loading F applied. And in order to balance that F over 2, if we're just concerned about the sum of forces in the y direction, we're going to need some sort of reaction. Well in this case the reaction is going to be shear. A common letter to be used for shear force is the letter V, and we'll notice in order for these to be balanced that that V has to be equal to the F over 2. Let's consider what happens on the other half. We still have the support on the right side, which is equal to F over 2. We have this force F that's acting on that right portion at that same distance L over 2. Now if we want to again consider ourselves as balancing, we're going to need a force at the cut. Here's the other half of our shear pair. Notice if you remember from our internal forces, we described a particular convention as being if we're in this direction with our shear, if we're counterclockwise with our shear in this direction, this is considered to be positive shear. So we have down here on the left, the other half of the shear is up on the right. And we'll notice that if we add V and the F over 2 over here, it has to be equal to the F that's there. In other words, V plus F over 2 has to be equal to F. The positive values have to be equal to the negative values, or the sum of all those forces have to be 0. So we notice that also over here that we get the shear force must be equal to F over 2. And so both of these balance. The shear acts to balance on both halves of this cut beam. Let's cut the beam a little further down. Let's do a second half cut. In other words, we'll cut it to the right side of the force. When we do so, we'll see that the left-hand side here still has the reaction at A. But now contains the reaction, I mean, the loading F. And the other side has the reaction on the far right side, the reaction at B of F over 2, but doesn't have the loading F because we've made the cut to the right of that. So this time, if we want to look at our shear, we'll notice that if we want to point the arrows the appropriate way, these are actually negative shear values. It's being defined in the clockwise direction here for our shear pair. So our negative shear, our shear opposite the normal positive convention, must be equal to. And if we look over here on the right side, that must be equal to F over 2, it must balance the F over 2 on the right side. And similarly, if we consider the sums over here, we can see that F over 2 moving up plus the negative V must be equal to the F on the other side. So the negative V must be equal to F over 2. Or the other way you could say that is that the shear itself would be equal to negative F over 2. So the shear is moving in our negative direction. Notice that our positive shear, if we look at our positive shear up above, even though the positive shear is sort of determined by a counterclockwise rotation, what the positive shear tends to do is creates a clockwise rotation on the parts. Is more in line with our normal definition of positive moment. Similarly, our negative shear creates a counterclockwise motion on each of the parts. Let's capture all this information in what we call a shear diagram. We'll sketch the entire beam, the length of the beam, although drawing the beam itself is not necessary here. And we'll notice, here we have the force F over 2. We have the force in the middle of F and the force on the right of F over 2 as part of our loading and reaction forces. This area here to the left has a shear of F over 2. This area to the right has a shear of negative F over 2. The region to the right is a region of negative shear. In other words, the shear has this convention. And the region to the left has positive shear. You can sort of think about that shear as resisting the spin that's being applied on the sides by going in the other direction. The other nice thing about this is the way the convention has it set up allows us to think about moving up and down. Here I have a force that's applied up. So my shear at the end, we can consider to be jumping immediately up that amount F over 2. There are no changes, but when it gets to the loading force in the middle, it jumps down an amount of F. So in this case, it goes to F over 2 plus another F over 2 down. And now we're at negative F over 2, at which point we continue at that same negative F over 2 value until we jump up again to the equilibrium 0 point at the other hinge. And there is our shear diagram for a simply loaded beam where the load is applied at the middle. So now let's consider how the moment varies across this beam. We'll consider the simply supported beam again, same beam, same loadings, same force in the same location, which means we also have the same reaction forces. And once again, we'll make our little slices, actually vertical slices, across the beam. And see if we can apply an appropriate moment to achieve static equilibrium. So if we look at the first half cut, I notice that I have this force F over 2 applied on the left side. And I've cut some distance x along the beam. In the previous case, that didn't matter. But in this case, it will. We notice that we already have, for the sum of forces in the y direction, we've already dealt with the shear force to balance that. But what we hadn't taken care of was the sum of moments. So let's go ahead and consider that now. And we're going to take the sum of moments around this point here where we made the cut. Doing so will allow us to ignore the shear force in our moment consideration. So in order to balance moments, I'm going to apply some sort of internal moment. Am I an internal bending moment at that point? Notice if I apply that internal bending moment at that point, I need to consider doing the same thing, equal and opposite, to the other portion of the beam. So let's draw the other portion of the beam here with our loading and our reaction. And record our distances here. This is l over 2. And this distance here, along the entire length here, is going to be equal to l minus x, where x is that distance that we've chosen so far. So whatever moment, internal moment we apply to the left side of the cut, we're going to have to apply that same moment in an opposite direction to the right side of the cut. And if you remember, our positive convention for the moment implies that we're going to be pulling on the top, tension on the top. So I've applied a positive moment in both of these directions. Although one of them is rotating clockwise, and one of them is rotating counterclockwise. So let's see if we can balance moments on this left-hand side. So I'm going to add up all the moments rotating in one direction. Let's go ahead and add up all the moments that are going clockwise to start with. If we're going around this center point, the moments that are making us go clockwise are f over 2 applied to a moment arm of x. So that's my clockwise moment. And that's going to have to be balanced with my counter clockwise moments, which is the internal moment that I'm applying. So now I already have a relationship between, or I already have a value for the internal moment. Let's see if that value still applies on the other side. Let's start by applying the clockwise moments. Here mi is a clockwise moment. We also have this moment that's being applied along this moment arm by the force f. And that's being applied in a clockwise direction. So I'll put plus f. And what length is that being applied to? Well, it looks like it's being applied to a length of l over 2 minus x. Notice it's easier to think about that as being we've taken x away from that half portion. So it's l over 2 minus x. If we want to balance that moment diagram, on the other side, we have a counterclockwise moment with a magnitude of f over 2 times a moment arm of l minus x. If we do a little bit of algebra there, we can see that the f times l over 2 and the f times l over 2 cancel in both of these places. We end up with mi, the internal moment mi, is equal to fx minus f over 2x or f over 2x, which not surprisingly balances what we applied on the other side. The key point here is we notice that this depends upon x. It depends on how far away you make the cut from the left side. Let's now consider a second half cut. In other words, a cut that's past the point of the force application will still consider that to be at position x. But this time, x is going to be greater than l over 2, whereas before, x had to be less than l over 2. Well, if we make that second half cut, we go ahead and draw in the reaction force as well as the applied load. On the other side, the shorter side, we simply have the reaction force. And we're going to go ahead and consider our internal moments applied at each side. Again, it's a pair of moments that oppose each other equal and opposite. And we're using the convention that this internal moment pair is positive when it creates tension on the top. So let's balance the moment equations for each of these pieces. For this piece here, we recognize our clockwise direction. Force f over 2 applied with a moment arm of x. Our counterclockwise direction is going to be f applied to a moment arm of, let's see here, that was moment arm x. This must be x minus l over 2. x minus l over 2. And also in that counterclockwise direction is our moment. If we solve that, we find out that the moment must be equal to fl over 2 minus fx over 2. Or if you prefer f over 2 times the difference l minus x. If I look at the other side, that one's much simpler. We notice this distance is simply l minus x. And so our moment balancing equation says the moment in the clockwise direction is equal to the moment in the counterclockwise direction, which is the force times the moment arm l minus x. But you'll notice is the same equation as the other side. f over 2l minus f over 2x. Let's gather this information in a moment diagram. I'll draw the beam here. And I'll recognize on the left-hand side of the beam that we have a formula for the moment. Mi is equal to f over 2 times x. And on the right side, we have a formula. Mi is equal to f over 2l minus f over 2x. If I plot those two things together, you'll notice I see on this right hand that I have a slope of f over 2. So that means I have a proportional relationship. So I will draw a line with a slope of f over 2. And when I get to the point where x equals l over 2, I'll find that Mi is equal to f over 2 times l over 2 or fl over 4. So this value here at the peak is equal to fl over 4. Notice if I plug that same thing into the other side, if I notice that x is equal to l over 2, I have that Mi is equal to f over 2l minus f over 2l over 2. Or I have fl over 2 minus fl over 4, which is again fl over 4. So as expected, we have the moments being the same at the point where they meet. In this case, my slope now is negative f over 2. This entire beam has a positive moment, which means it has a positive moment. And you'll notice that the slope here of f over 2 and the slope here of negative f over 2 are the same as the shear. In fact, that is a relationship between moment and shear. That the shear along a beam is equal to the slope of the moment, which makes problems like these particularly prone to solutions using calculus, which establishes the rate of change and the relationships between rates of change, derivatives, integrals, et cetera.