Hi and welcome to module 22 of applications in engineering mechanics. Today we're going to calculate the tensile load in a cable support system supporting a uniform load. Previous to this, in the earlier modules we did a cable system supporting concentrated loads. And we're also going to determine the required length of a cable. under a uniform load given the, the length of the span, and, and the sag. So, as I mentioned before, we've already done the cable systems under a concentrated load like an aerial tram or a, a traffic light type situation. this is a suspension bridge under a uniform load. So, what we're going to do is we're going to assume that the load and the bridge itself is, this, this bridge portion of this beam portion if far out weighs the rest of the elements of the structure which can be considered massless in comparison. the other application we talked about was self-weight. When you had like a rope swing or power lines. what we find is, in the condition where we have a uniform load, the cable forms a, a, parabolic shape. Now, under self weight you get a shape which is called a catenary curve. And I'll let you study about catenary curves on your own. I'd like to proceed with a span. Where we have a, a, a uniform load and again the rest of the bridge, we're going to consider massless. And I'm going to use symmetry to just look at the right half of the bridge, and we're going to want to find the tension in the cable. and to do that, we're going to first of all look at the load itself and, I can tell from my earlier course Introduction to Engineering Mechanics, what this load is, what the resultant force is because Q sub x or, is, is a constant. It's Q sub zero and it's expressed in units per, per units of force per length. And so for half the bridge the resultant force will be, that Q sub 0 times L over 2. And it will act since it's uniform and constant right at the center, or at L over 4 from the right hand side. And so, this x at L over 4 from point B. And so, once we know F sub R We can now sum forces in the y direction to find the vertical component of the tension in the cable up here at point B. So if I sum forces in the vertical direction, set it equal to 0, I find that T sub V equals Q sub 0, L over 2. Okay and so that takes care of the vertical component of the tension. Let's continue, on now. And here I've got my resulting force q sub zero times L over 2. It's equal and opposite the vertical component of the tension. let's go ahead and find the horizontal component of the tension here at the center of the bridge by summing moments about point B. And so, I've got sum of the moments power point B equals 0 [UNKNOWN]. I'll choose counter clockwise positive [COUGH] and so I get F sub h is going to cause a clockwise, tendency for a clockwise rotation. So that's negative in accordance with my sine convention. Its moment arm is the max sag, H. And then I've got the resultant force here acting down, so it's going to have a tendency to cause a counterclockwise rotation, so that's positive in this according to my sine convention. So that's going to be, q sub zero L over 2, times. It's moment arm, which is L over 4. All that equals zero, and now I can solve for the horizontal force component in the cable, down here at the center, f sub h equals q0L squared over 8 H. And we can look at our free-body diagram and see that Fh has to balance Th, the horizontal component of the tension here at the end of the bridge at point b. And so this is equal to T sub H. And so with that, we see that okay, I've now got the, I know what the cable's tension is at the center. I know what the resultant load is, and I know both the horizontal and vertical component of the tension at point B. And so my question to you is, where is the max tension in the cable going to occur? So, think about that and then come on back. So what you eh, should have found or what you should have concluded was that because the x-component is the same throughout, just like we saw with a concentrated load. We have the x component here, q sub 0 L squared over 8H and the horizontal component here the same. And so therefore, wherever you have your max vertical component of tension is where you're going to get the max tension in the cable itself. And so that's going to occur right here at point B. So let's go ahead and calculate that. I can use Pythagorean's theorem to add up the horizontal and vertical components at B. So I get T max equals the square root of T sub v squared. Plus T sub H squared, which is equal to the square root of, T sub V was q 0 times L over 2. All that squared. Plus, q0 L squared over 8 H, squared. And if you do that multiplication out and simplify, you're going to get T max equals Q sub 0L over 2 times the square root of 1. Plus L over 4H squared and so that's the max tension in the cable. Okay, so let's go ahead and have you do a problem on your own before we do that though, I did also want to talk about the total length. the total length, you can find the derivation of the total length for a suspension type cable system. it, in any standard statics textbook, the, the total length is given by this expression here. you will have to know what the max Sag is and you'll have to know what the span length is. So now you, you know how to find the tension in the cable. You know how to find the max length, and so you can work on this worksheet, which I have a solution for you in the module handouts. You got a 50 foot span. The max sag is 6 feet. And, what you need to do is, if you know that the cable can only support 5,000 pounds, what's the max load that can be handled by this bridge? And also find the total length of the cable. And once you've done that you should have a good basic understanding of suspension bridge systems.
