So now I wanted to look at a suspension bridge and I kind of want to compare this to the cable state bridge which is why I use the same span and I use the same distributed load. So the cable state bridge tends to have more cables. It has these vertical cables we often refer to as suspenders or suspender cables. Their job is really to take the load up to this main cable and that upper cable is the main cable that's draping over. That main cable gets connected back to abutments and goes up and over the towers. So the suspenders, they're important but their main job is really just to take this load from the deck and take it up to the main cables. So analyzing them is fairly straightforward. It's like what we've done previously, summing forces in the wide direction. We don't even have an angle that we have to deal with. So online we've posted all the steps to go through that. The bottom line for those suspenders is that we have a certain diameter that's required. Now use the same allowable stress that I used for the cable state bridge but the diameter of the suspenders ends up being 3.03 millimeters and that's kind of the story I wanted to tell with that. The diameter is quite a bit smaller so the diameter just for comparison of the cable state bridge, we used the same size cable for all those cables. So for the cable state cables that was 6.4 millimeters. So it was significantly less for the suspenders but that makes sense. There were only eight cables for the cable state bridge. Now there's 18 cables in this main span for the suspenders so we can use a smaller diameter. Again I was using the same strength. But now the main thing I want to look at is this main cable. So I want to calculate, I need to calculate the tension force in the main cable. That will be my first goal. And then I also want to calculate the diameter of a steel cable that I would use for that main cable. So to do that I'm going to look at a free body diagram of focusing on that main cable. So I'll put that down here. Here's my main cable. I am going to look at half of the span. So let's let that 9 meters equal L, that's the full span. And so this would be L over 2 or 4.5 meters. This, since this is a distributed load, this will still carry that distributed load. And that distributed load is W, 4.48 kilo Newton per meter, distributed over that length. And then I'm going to look at the forces. So I have a force, I'm going to call that T mid. So that's a tension force at the middle. And then I'll have a tension force at the support. I can resolve that into a TX and a TY. And my goal is to calculate the tension force in the support. There will be suspenders in here carrying that force up to that main cable. But I'm really focusing on this main cable. And these forces T support and T mid are the tension forces in that cable. So to figure out what those values are, we can still use equilibrium. We have horizontal and vertical equilibrium. But because we're looking at a piece of this bridge, we also have rotational equilibrium. And that is where I'm going to actually start. So I'm going to start with rotation. And I want no rotation, if I want to make this work. I'm going to use the support as my rotation point. So I will do something called summing the moments. We'll talk about moments more when we look at arches and even more when we look at beams. Again, it's a fancy way of saying I don't want any rotation. So I'll assume clockwise positive. So I'm going to look at these forces and see what's happening. This distributed force is going to try to pull clockwise around my support point. I'm going to look at a distributed load. I'm going to make the total load. So the total load will be w times l over 2. Because this is a rectangular load, it's evenly distributed. And it's acting at a point here that will be l over 4. So if this is half the span, it'll act right in the middle. And that will be my total force. The reason I needed to do that, or it's helpful to do that, is causing this rotation will be the total load and the distance makes a difference. So rotation is going to be affected by how far away it is. So when I sum my moments, I will have w l over 2. That's my total force acting at a distance that is l over 4. Okay, again, we'll talk more about moments later in the course. That's balanced out by this T-mid, which is this attention force at the middle. I left off the distance, but that distance is a sag and is 2 meters. And that's important because that's the distance that this force, this attention force, is acting from my support. So that will be 2 meters, which I often refer to as d, which is the sag in my suspension bridge. And that's something we'd play with as a designer, how far should the span be and how much sag should we use. Sending that equal to 0, I can solve for this attention force. So this attention force, and it ended up being w l squared over 8 d. And that's going to allow me to solve for this attention force at the mid. We know all these values, so we know the distributed load, 4.48 kilo Newton's per meter. We know the length. Again, I'm using 9 meters as the length, and that squared over 8. And then this d is the sag, and that is 2 meters. And I can calculate this T attention force at the middle. That attention force at the middle ends up being 22.7 kilo Newton's. So that's the attention force maximum at the middle. It turns out that the attention force at the support is higher, but if we use horizontal and vertical equilibrium, we can solve for that. I'm not going to write the equations. We've done horizontal and vertical equilibrium quite a bit. But if we look at horizontally T mid and T x are the only two things happening. So T x is going to equal T mid if we don't want this to move horizontally. And that is also then 22.7 kilo Newton's. And vertically, we have just T y and the total load acting downward, which is w l over 2. That value ends up being 20.2 kilo Newton's. That's the vertical and horizontal component of the attention force at the support. If I want the actual tension force at the support, I would take T x squared plus T y squared square root. And that value ends up being 30.4 kilo Newton's. This is the maximum tension force in the main cable. So that's what I would design for. I designed for the 30.4. If I wanted to calculate the diameter, I would again use my equation. I'll use a different color so we can see it in the middle. I can calculate the stress in the cable. We've done this one before, but it is force over area and keep that less than the allowable stress. I use the same allowable stress that I use for the cable state bridge. You can solve for the cross sectional area, putting in this 30.4 for T and my allowable stress was 310,850 kilo Newton's per meter squared. You can solve for area and then I can solve for diameter. The diameter of these main cables, if you do that, the main cables of the suspension bridge ends up being 11.2 millimeters. So quite a bit higher than the cables of the cable state bridge, but the suspenders are quite a bit lighter. In the end, these are the values for the suspension bridge. The total amount of steel for a suspension bridge tends to be greater. I'm sure that would be the scenario here than a cable state bridge. Because we're using all these suspenders and larger main cables, we tend to be able to make our suspension bridges span a lot further. So there's pros and cons to using a cable state bridge versus a suspension bridge. But this was to give you a little bit of background on how you would analyze a suspension bridge.