So I wanted to look at one can lever beam and this I'm using the Eiffel Tower is my model or my example. It's fairly common. This is Eiffel Tower with a wind load. Engineers often will model at least for our first pass the overall building or tower in this case with lateral loads on it. In this case a wind load as a can lever beam. So the model of a can lever beam, maybe the more common one would be just a tall beam like that with the wind load on it. So the Eiffel Tower has a slightly different shape which is one of the things I want to look at. But I'm really doing the same thing I would do to model this can lever beam like this. So in a can lever beam we still need our three reactions as a minimum. The reactions that we get in this can lever beam are going to be the same as we get in the Eiffel Tower. So to counteract that wind force I need a horizontal reaction. Let's call that AX. I don't for the wind load have a vertical reaction but obviously for the weight of the Eiffel Tower I would have a vertical reaction for a different load case and I'll have a moment. So I need this moment to actually resist the wind load, resist rotation. So we can go through and we can solve for these reactions unless we do that. So the first thing we'll actually we can look at this. Moving force is horizontally. We have the wind load acting this way in the A-sub-AX acting opposite direction. The total wind load that we have acting is going to be 350 kilonewton per meter acting over the whole height which is 300 meters. That quantity is a large number 105,000 kilonewton and that's going to need to be reacted along the base. So we don't want it to slide across the ground. We need some connections there to resist that. So we would do that by summing force as horizontally equals zero and we would find that AX is going to counteract that applied wind load. We'd also want to look at the moment. So summing the moments, summing the moments about the base of the Eiffel Tower. So we have the wind load trying to rotate it clockwise in this case and that is that force 105,000 kilonewton. It's acting since this is evenly distributed or to sum it's acting in the middle. We are at 150 meters. That's counteracted. The only thing counteracting it is this moment. That's why we need a moment if we need that. If we need it to not rotate, moment equals zero. I can solve for that moment. The large number, 75,1,575,000 kilonewton. So when we model buildings as beams, we end up with large numbers. One thing with that moment, so if we look at the base, when we're generating that moment, we don't even have a constant base along the whole base. What we have are these two legs, actually four legs, and looking at just a front half. So we look at both halves. And in the two-dimensional case, I have these two legs and I want to generate a moment. That moment actually gets generated by an upward force and a downward force. So two forces acting equal and opposite at a certain distance. The distance or the base of the Eiffel Tower is 100 meters. That's going to generate the moment that I need to resist the wind force. So I use this arrow often, but generally what's happening is a force up and down. So in a standard beam, that would happen with nails or bolts on the edges of the beam. But it would be able to provide that rotational support. In the Eiffel Tower, its forces in those two legs and connections that will, will make that happen. So you could calculate that force that's required. This moment is going to have to be, this moment, this big moment is going to have to be that force times 100 meters. You could solve for the force. The force is going to be that big moment over 100 meters. Am I doing that right? But then I could solve for that, that actual force. So it's just another moment balance. So I have moments generating this way and these two, this couple, going the opposite direction. The one last thing I want to look at is how the shear and moment varies along the height. I'm not going to draw a free body diagram, but we'll kind of superimpose one on this beam. So if we look at this, what we're going to be trying to find, we want to find the force across the beam and the moment. As we vary from the top to the bottom, or from the bottom to the top, we can use the shear and moment diagrams to generate that. But if we look at this, we're at the very top, there's no load. So we'll have a shear of zero and a moment of zero. As we start moving down, we'll get more and more load and we'll get more and more shear and moment. So let's actually draw the diagrams, which you can produce with the beam simulator. We'll draw a shear diagram and a moment diagram. They're both starting at zero. I start at zero, since there's no load. As I move my way down, I get a little bit more load. As I move one meter down, I'd get a certain load. And I'd have a linear variation as I move down. And I get down to the base where I have my maximum. That maximum would actually equal this 105,000 kilonewtons. But that would be my variation of shear along the height of the Eiffel Tower. Moment, moment is instead of varying linearly. What happens is as I cut this little cross-section, I have a net force and I have a net distance. And they're both kind of a function of the distance. So we end up with a parabolic variation as opposed to a linear variation. So I start at zero and I very part parabolicly. The max moment at the base is this value. So it varies from zero down to the maximum at the base with a parabolic shape, which happens to be the same shape as the Eiffel Tower. And it was done that way on purpose. It gives it in a very efficient system. Again, that moment's going to be counteracted by these forces, acting in the legs that very also as we go up the height.