 Okay, so I want to talk about a special case, which is internal beam loading. So when we have beams that are under bending in other scenarios, what that means. So if I go ahead and say we have beams, which can be supported in a variety of ways, they can be what we call simply supported, which is they're pinned on one end and they're free to roll on the other end, cantilevered, which means they're fixed on one end and free floating on the other, and overhanging, which is like pinned, but that rolling or kind of bendable or movable position is slid in from the end of the beam. So great, we have these different types of beams. Now, if I want to know what's going on in a beam, I have let's say I have a cantilevered beam, I'll draw that over, and it has just an end load applied, let's call it P. Now, if I want to know what's going on, just like when I look at internal forces, normally I would make a cut somewhere through there, and then I take and separate out a section of that beam, and I'm going to pick the section of the cut to the right, which has the externally applied load P on it. So again, just like with internal forces in any other rigid body, I need to know what's happening internally, and I need to balance this force P. So I have a balancing force, which we call V, and P would tend to cause rotation clockwise, so I need a balancing counterclockwise moment to balance out both the force translation in the vertical direction and the rotational motion. So that gives me my internal forces, and I've left off my axial internal force because I have nothing applied to my beam that's axial. So these internal bending and shear forces are great, and we can model those and calculate them with free body diagrams and equilibrium equations. But the trick is that it's dependent on where I made my cut, so over here I just chose a location for the cut, but if I pick a different location, I'm going to have different resulting forces, right, particularly my bending moment, because the bending moment balances P based on how far away it is, based on that distance. So if that distance is different, then I'm going to have a different resulting bending moment. So that's something, a key point that we need to consider, and it's important for machine component design, because a lot of times the different things that we talk about, you know, that are machine components are mounted to shafts. There's a lot of rotational motion in machines. So things tend to be mounted on shafts, and shafts are pretty good examples of beams of this type, so it's nice to have an understanding of how to model those. So if I scroll down a little bit here, let's take another example. So suppose I have a beam, and I'm just going to do a simply supported beam, pinned, roller, and the reason we do pinned and roller is because of preventing the beam from being over constrained. So if you fix too many components of the beam, then it's over constrained and the solution procedure gets more difficult. So I'm going to apply a load P, and I'm going to say it's at a distance of half the length of the beam. So it's in the center of the beam equal from either side. Now, if I go ahead and say, well, what's a free body diagram of this look like? Again, my load P, and then I have to have reaction loads. Typically, we might say, well, let's call that RA, just say it's that position B, we're just, you know, making up labels. And then we have another one, RB. Now, solving for these reaction loads is a free body diagram, right? We can go ahead and do that, do a force balance and a moment balance. In this case, we'd only have two equations, some of forces in the Y, some of moments around some point. Now, this one's nice and easy because we said that this load P is centered. We know that it's going to be evenly distributed at these reactions. So we can go ahead and say that each reaction is equal to one half. Now, if I want to understand the internal forces in this beam, I need to do my internal force analysis, which is, again, making a cut. And I'm going to label this cut one. And I might need more than one cut, right? Because the situation is different if I'm talking about the left-hand side of the, in relation to the force P, or if I'm talking about the right-hand side, those are two different scenarios, right? And so we can analyze each separately. So, for cut one, if I again draw the two resulting free body diagrams that would come from this, I would see that I have my force P applied on this side, P over two applied over here. And then I have my resulting internal balancing forces. So I have a shear, shear, a moment, moment. And let's see if I label those V and M, M and V. And of course, those are going to be the same at either side of the cut, because really the cut is one location. I'm just looking at it from the left or from the right and analyzing it that way. And then, of course, I'm just missing my other reaction over here, which is P over two. Now, what I'm really interested then in is the difference between this and the other side. But before we get to that, let's go ahead and draw that other side. So let's say now we want to look at cut two, and I draw my free body diagram. It's going to have a force P. It's going to have a reaction P over two. It's going to have the same labeled shear and bending moment. Over here, it's got P over two, same labeled shear and bending moment. Okay, so I've got my two free body diagrams. Now, typically what we would do here then is draw something that we call a shear and bending moment diagram. And really all I'm trying to do is represent the shear, V, and the bending moment M as a function of X. So if I say, I'm going to go back up here to my original drawing, if I say that X is some coordinate position along my beam, I can go ahead and draw what I think the shear would be. Now, if I look at my first cut location one and look at my free body diagram here on the left, I can say, well, by summation of forces in the Y, V has to be equal to P over two, right? And that's going to be the same no matter where I cut up until I hit point P, right? In which case I jump to the other side, which is down here at the cut free body diagram labeled two. Now I have V down, P over two up and P over two down. So V in this case has to be negative because of the direction of the arrow I've drawn, which is down on the left hand side here. And that's by convention, down and then up is like my standard positive convention for drawing shear. So what that tells me is that this actually is pointing the opposite direction, which means it's negative. So if I come up here to my drawing, I can say, well, my shear is P over two, and then I hit that center line where P is, and then it becomes negative P over two, and it goes back to zero. So the shaded area under here represents that plot with P over two and negative P over two for my shear. Now if I look at bending moment, I can do the same sort of thing. Starting with number one free body diagram number one, I can see that, okay, my moment has to balance this P over two, and it's going to be, of course, P over two times however far away I am. So as I move position with increasing position X, I can see that my moment is going to get larger with X because it's, you know, force times distance effectively. So increasing, increasing, increasing, increasing, and then I hit P, right? What happens after P is now I'm in this region two, and I have a contribution to the moment now in the opposite direction from P at an increasing, increasing as I move to the right with increasing distance from P. So I can go ahead and say, well, now it's going down, and we could work out the equations to this, but I'm just kind of walking through it slowly here. And the shaded area under here is a plot of my moment. Now, what is this maximum moment? Because usually when we're, you know, doing stress, this is really what we're interested in. Is this maximum value? So I'm going to scroll down here a little bit to get this on the screen better. So I have this maximum moment, and what's it going to be equal to? Well, it's right in the middle, where the moment is at the location of P, which means P doesn't contribute, but P over two does. So P over two at a distance of L over two, which means that my maximum moment in this case is P L over four, and it occurs where L is one, where we're at a position of L over two. So I get a plot that looks like that. Now this one was relatively straightforward in that we only had this one point load, it was in the center and all of that. Now one thing you might notice in looking at looking at these plots is that there's a relationship between moment and shear, right? And that relationship is that it's an integration. So if I integrate my shear from left to right, and remember that integration is accumulation of area under a curve. So as I integrate from left to right, I'm increasing, increasing, increasing, just like my plot down here, then it drops negative. Negative means I'm decreasing, so decreasing, decreasing, decreasing, until I hit zero. So moment is the integration of shear, and it's a nice, you know, mathematical way to think about what's happening when we plot it like this. But we can also just do these cut methods and figure out what's going on once we do free body diagrams and balance these out. And of course the same thing can exist in torque. So I'll show that really quickly, just a simple drawing. So if I have, for example, a shaft with a pulley on it, this is kind of hard for me to draw, but do my best. Shaft with a pulley on it, and the shaft continues, and then there's another pulley. And suppose, you know, this pulley has some belt pulling on it that way, and this pulley has a belt pulling on it this way, and that's how they're, you know, transmitting force between the two. And then, you know, we usually assume that we have bearings at either end, so there's no load being carried there. So if I have bearings, that means this is free to rotate, which means there can't be any load in it. But between these two bearings, I'm going to have the resulting torque based on whatever the, you know, radius of these pulleys is. So if I were to draw a torque diagram along the position X of this shaft, I'd see that I have nothing, nothing, nothing. Then I get to that first pulley, torque, second pulley, nothing after that. So my torque would be whatever that, you know, force times radius would end up being, and it would only exist between those two pulleys. So I can do a torque diagram just like I can do a shear and bending moment diagram, depending on, you know, what's all going on in my system. Okay, well that's all I have for beam loading. We'll talk about that more as we do other example problems.