 So there are two typical beam supports. There are simple supports where we have the beam and it is supported by a pin on one side and a roller on the other side, which corresponds to two reaction forces in the x and y direction on the pin and one reaction force in the y direction on the roller. Another very regular type of beam support is the cantilever beam. In this case, we presume that the beam is unsupported, completely unsupported on one side and supported by what we call a fixed support on the other side. And in this case, the fixed support actually maintains three reactions, a reaction in the x direction and a reaction in the y direction. And then also, it must maintain some sort of moment in about the point where the beam is affixed to the wall. So we have three reactions in both cases. We have three unknowns. In this case, there are three forces, two in the vertical direction and one in the horizontal direction that are unknown. And in the second case, we have three unknowns, but two of them are reactions, x and y, and one of them is a moment. So in a previous video, we had an opportunity to consider what happened when we loaded a simply supported beam. Let's consider what's going to happen to our internal forces when we load a cantilever beam. So let me put a little bit more data on this cantilever beam. Let's call this point in the far left point A. We'll call the point on the far right, well, the point, the application point, we'll call point B. And the point in the far right, we'll call point C. We'll assume that the application point is some distance d away from the wall and that the beam itself as an entire length represented by small l. So we have two distances that are important here. The distance that's the length of the entire beam and then the distance of the application of the force. And we'll go ahead and give this force a magnitude F. Well, when we're working with something like this, let's consider our equilibrium equations. Equilibrium equations and something like this are fairly straightforward. If this is the only loading, load F. We can, first of all, do our sum of forces in the x direction. And we'll notice that the reaction at x is the only thing in the x direction. And so that must be 0. It must not have any magnitude. If we do our sum of forces in the y direction, we can see that the reaction at point A, reaction y at point A, maybe I should call it Ay, officially. This would be Ax, although it's unimportant. And the applied force in the downward direction must sum to 0. And that simplifies algebraically that we know that the reaction force must be equal in magnitude to the applied force. So that solution is relatively simple. And finally, we can take the sum of moments around some point. And this is most simply done by taking them around point A, the point at which the reaction moment is being applied. And we know the sum of those forces is going to be equal to the moment that we have applied here, which appears to be in a clockwise direction. But it must be opposed. Well, in this case, it says opposed. But it must be opposed by FD, the force that's being applied and the moment arm of that force. But notice that's in the same direction, the same clockwise direction. So if we sum those up and set them equal to 0 and then solve for the moment, we find that the moment must be equal to negative FD. In other words, it must be going in a counterclockwise direction, although we defined the initial arrow and drew the initial arrow in a clockwise direction. Well, now that we've solved for our various reactions, let's take a look at what the internal forces look like on this cantilever beam. So let's consider a slice through the beam. Our first slice will be to the left of the applied force. Left of applied force. And if we look at that, here's our slice here. First of all, let's consider what happens with the reactions. Well, let's apply the reactions. We have, first of all, this moment that we know is negative FD. So it's actually going in the other direction, but we'll keep the sign conventions we adopted at the beginning. And we also notice that we have a reaction force y that's equal to F. Well, if we consider the reactions, the internal reactions that must be happening on that face, we see we must have a vertical point here, a shear value that has to be equal to R of y, or actually has to be equal to F in this particular case. Because the reaction force, we know that R y minus v has to be equal to 0, so those two values must be equal. So we know what our shear has to be at that point. Now let's ask about our bending moment. Well, to consider our bending moment, first of all, we need to think about what our convention is for bending moments. And if you remember, the convention for bending moment said that a bending moment in this direction would be considered to be positive bending moment. If you remember, we were thinking about if we have an internal moment of that direction, that will create a tension on the bottom, which is the positive value for bending moment. Well, now we need to consider how to balance the moments in this case. We have our reaction moment here, or our bending moment. Let's call that mb for the time being. And we need to balance our moments. Well, now we need to select a point around which to balance the moments. I'm going to go ahead and select the point to be here at this point m, or at this point a. And we see that, first of all, we have a clockwise but negative moment. We have a clockwise m. We have a counterclockwise mb. And then we also have our shear being applied over whatever distance here. And this distance ends up being some distance x, because we don't know where we've made our cut, minus x. And we know the sum of all those moments has to be equal to 0. Well, let's plug in what we know. We know that the moment here is negative fd. We defined it as positive. We kept it as clockwise here, but we know that it has a negative value, that it's actually acting in a counterclockwise fashion. And if we solve this, if we replace this m with negative fd and we solve, we find out that our internal moment must be equal to, let's move that over to this side of the equation, negative fd minus v times x. Let's make sure we actually have that in the correct direction. Nope, we do not have that in the correct direction. Notice that the v, as we defined, was actually in our positive direction, in our clockwise direction. And so that ends up being positive in this location. So now that we've calculated the reaction forces, let's go ahead and consider what our shear and our moment diagram look like along the length of the beam, or at least along this left-hand side of the beam. Let's begin with the shear diagram. I'll draw the length of the beam. And if I remember, we know that there was a reaction force applied here at the far left side. And there was an equal applied force out at distance d. And we know that across that entire length, we had a shear that stayed the same. It did not have a dependency upon our position x. And it has a value of f, equivalent to the magnitude of the applied force. So in this case, our shear across this entire segment is positive f. And notice that corresponds with the application on the left-hand side of a force f, maintaining that same shear all the way through, and then dropping down with the application of the applied load back down to a value of 0. Well, while we're here, let's think about what actually happens on the rest of the beam to the right. If you think about a little segment cut out of the rest of the beam, that little segment, if we slice it off and we apply our internal forces, we notice that those internal forces balance with nothing. There are no applied loads on this little end section. And therefore, the values of all the internal reactions must be equal to 0. So our shear for the remainder of the beam stays at a value of 0. And our shear diagram jumps up to the value of f, stays constant, jumps down to a value of 0, and stays constant there. Now let's look at our moment diagram. In the case of the moment diagram, we have one place where we know the moment, what the moment must be. At that very far left, there is an applied reaction moment. And that applied reaction moment has a particular value, negative fd. So in order for us to, so the value at that far point is a moment of negative fd. Well, if I look at the moment equation we created earlier, let's bring that information down here. We recognize that the moment at any point x, the bending moment, is equal to negative fd, which we see we have here, plus v. And we know that it's v is equal to the force fx. So notice when x is equal to 0, then the value is indeed negative fd, which is what we've plotted here. How does that value change? Well, if we look, when x takes the value of d, when we get to the point where the load is being applied, if x equals d, then we get a moment of negative fd plus positive fd, or a moment value of 0. So our internal moment at the place where the force is applied is a value of 0. So now we have two points on our moment diagram. And we notice that this equation, the moment depends on x in a linear value. The slope of the line is f. So I can create the rest of the curve by drawing a direct line between these two points, recognizing that everywhere here is below the 0 point. Usually we establish 0 where the beam is. And so we have a negative moment that gets decreasingly large until it becomes 0 at the point where the load is applied. And in the same way that the shear is 0, there is also no internal moment for the remaining part of the beam. So now we've created the entire moment diagram. Notice our convention for negative moment is going to be tension on the top of the beam, which makes sense if you think about this beam perhaps as something like a diving board or if we're hanging something from it, it's going to tend to stretch the top and compress the bottom. And that's what a negative internal moment indicates. One other thing to recognize is, again, that the slope of the moment diagram is equal to the magnitude of the shear. Notice that is even true for this end here, where the slope is 0 corresponding to a shear of 0.