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.

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.

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.

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

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.