So this is a model I'm going to use for a simply supported beam. So I will build a truss bridge that was spanned between the two trees, which was 9 meters. So we're going to look at if we had used a beam instead of a truss, how would we design it? So this is the model I'm going to use. Looking at this, we could solve for this, these reactions, that's my first step, as I'm going to solve for the reactions. I could use rotational and vertical and horizontal equilibrium, but looking at the load, I have a total load acting downward in the middle that in everything is symmetric. The total downward load is going to be 5 kilo-neutens per meter times that 9 meters or 45, 45 kilo-neutens. Because everything is symmetric and it's evenly distributed, it really just get half of it to each side. So I'll have 22.5 and 22.5. And there's nothing acting horizontally. So my horizontal reaction will be zero and that will tell me my reactions. So that's my starting point, but I need to know internally what's happening. And we're going to draw diagrams here eventually. But let's look at one point in here. Okay, so I'm going to make a free body diagram. I'm going to cut this beam. Let's cut it at two meters over. And I can draw a diagram to represent that. So we can do the equilibrium. So let's look at this as two meters long. And I still have the applied load on it. Just five kilo-neutens per meter. And I have my reaction force. It's 22.5 kilo-neutens. And what I'm looking for is when I made this cut, I exposed the beam inside of the beam, I need to calculate the shear force v and the bending moment, bending moment m. Now I'm back to just doing equilibrium on this little piece. So let's do the shear force first. Summing forces in the y direction. So I have upward 22.5 kilo-neutens. I have acting downward. I'm acting downward a total of 10 kilo-neutens. Since it's 5 kilo-neutens per meter, but I'm acting over two meters. So that will be a downward 10 kilo-neutens and a downward v. And that all equals zero. So I can solve for that v which is shear force. And that should be 12.5 kilo-neutens. So I know what the shear force across this face is. And that's what I would use to design at that location. But I'd also want to calculate this bending moment. Bending moment tends to be more critical in beams. So let's go and look at the moments. Sum the moment. Let's sum the moment about this cut again. So that is at my two meters. So I'm looking at this point. So this is my point of rotation over here. So I'll have, let's go this way, positive 22.5 kilo-neutens. That's at a distance of two meters from my point. I could use this distributed load, but I can also just use the resultant load when I do the moment. So I have 10 total kilo-neutens pushing the opposite direction. So that is 10 kilo-neutens. And it's acting at a distance of one meter. And I always have the moment still acting in there minus m all equals zero. I should be able to solve for this m. See the other side? The moment is 35 kilo-neutens meters. So that is what I get for m. That's at one point. So for some reason I was designing right at that point. These would be the values I would use. But for this beam, I want to find the maximum values, the maximum of shear and moment. So I'm going to go and draw the diagrams, which is something you can get from the beam simulator. But let's look at what's happening along the whole length. So this will be my shear diagram. It will represent the shear at any location along the length. We know we have at the support 22.5 kilo-neutens. That's half of the load. So we get to the middle of this beam. We'll have another half of the load acting downward. So we're going to cross the zero point right at the middle and my shear. I'll try to make a straight line. So that's what my shear is going to look like. It'll come down to 22.5 here. My reaction support will bring it back up to zero. So this tells me the shear along the length. So if I'm worried about shear in my beam, which I would be for a very short wide beam, I'd be worried about the supports as the failure points in shear. And this would be my maximum value for that shear value. But now let's look also at the moment. For this beam, I expect the moment to be more critical if I were to look at both. Let's look at the moment along the length. Moment again always starts at zero. It turns out that the slope of the moment diagram equals the shear value. And the area under here equals the moment value. Again, you don't have to know how to do that. But that's what's going to help me draw this diagram. The diagram's going to start at zero. Have a certain slope. Come up to a maximum in the middle and then go back down to zero. That should be symmetric. It'll be even on both sides. But that tells me what's happening to the moment over the whole length. The maximum is right here at the middle is 50.6 kilo Newton meters. Now when we're designing beams, we look at many things. We look at the supports. We look at the moment. We look at the shear and we often look at the deflection as well. I'm just going to look at the moment today, which tends to be the more critical, most critical for most long slender beams. So we know the maximum moment in the beam. And we need to design the size for that maximum moment. When we design trusses and columns, we looked at stresses. So we're going to do the similar thing here. We're going to look at the stress in this beam. We're going to look at bending stress. So axial stress, which is tension or compression, you just took the force and divided by area. When we have bending, we use something called the flexure formula. We take the moment, we multiply by C and divide by I. We'll talk about these values, but the moment is just this maximum moment value that I'll put in. C is half the height of the cross section. Since bending happens about the center of the cross section, and moment of inertia is a quantity, we can look up the formula for it, but it is a resistance to bending. So that's my flexure formula that will tell me the stress in this beam, at least the maximum stress. And I want to keep that less than my allowable. We're going to assume that owl is going to use wood again. The allowable stress for the wood he's using is 30,000 kilonewton per meter squared. So this gives me a starting point. My goal is to design the cross section of the beam. So I need to make some assumptions, or I need to start playing with variables. I'm going to assume, with regard to assumed it's wood, I'm going to also assume that we're going to use a rectangular cross section. I'm going to assume actually that this width as well is 200 millimeters to give me a starting point. So that's the width of my wood. And then I'm going to determine what height I need. So that's my goal is to figure out that height. Looking at this equation, if I want to calculate the C value, C always goes from the middle to the outside edge. So it'll just be H over 2. Moment of inertia I can look up the value for a rectangular cross section is base times height cubed over 12. Again resistance to bending, but we can calculate it using that equation. So these values will get plugged into this equation. And that'll leave me with just one variable H. So I'll have 50.6 kilonewton meters times H over 2 over bH cubed b is, I'm going to change to meters, 0.2 meters times H cubed over 12. And that has to be less than 30,000 kilonewton per meter squared. The only unknown is H. So this will allow me to calculate the required height to make this beam work for this loading. And if we calculate it, we find that H is 225 millimeters. So if I will want to use a beam instead of a truss, she needs a cross section that is 200 millimeters wide by 225 millimeters high. If she wants to use rectangle cross section, we could also look at hollow sections eventually or eye beams. And that's something an engineer would play with, different dimensions, different shapes. But this is how you design a beam.