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Welcome back to module 18 of Applications in Engineering Mechanics.

Today we're going to look at the next major topic in

the course, which is structures supported by cable support systems.

And so we're going to give some examples of that.

I'll explain why cable support systems might be used.

And we're going to list the assumption that

we're going to use in analyzing cable support systems.

So here are some examples, one might be

cables under their own weight. here's a rope swing, or power lines

that are systems that are, are held up

by their own, on, on either side but they support their own weight, self-weight.

we may have applications where we have concentrated loads

like this concentrated load shown on this cable system.

Examples of real world structures might include a traffic light, or

perhaps a ski lift or an aerial tramp.

And finally, you'll also see cable support systems in, in suspension type bridges.

And so, those are all examples when we talk about a cable,

when I talk about a cable, I'm going to talk about a generic.

So it might be a rope, it might be a power line, a cable-like structure.

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So, first of all, I want you to do a little research

on your own, and find out why cable support systems might be

used, and then come on back and we'll do it together.

' Kay, the things that you should have found out that

are advantages for cable support systems that are, that they are strong.

they're lightweight and they are flexible engineering structures.

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And so, we make some assumptions based on those observations of cable systems.

We're going to

look at the concentrated load.

Cable systems first, the examples like the traffic

light or the arrow tram or ski lift.

And for these cables, we're going to assume that they're pin connected.

there's going to be no external moments, they're

perfectly flexible, no internal moments, and they're inextensible.

2:01

And so, let's look at the cable here, a typical cable.

This is actually a rope but analogous to a cable system.

There are no external moments, and so we can assume that they're pin connected.

They're perfectly flexible, there's no internal

moments within, within the, the cable itself.

And we are going to assume that its inextensible, that if

I put loads on it its not going to length it.

Now that opposed to a beam where you would have, you could have external moments.

You could have, you are going to have internal

moments as well as we saw when we drew

the shear moment, excuse me, the bending force moment diagrams.

And these are also inextensible, but that's the the, the comparison.

So if you go back with those assumptions, that's going to lead to the

fact that these cable systems, since they're

pin connected, they have no internal moments.

They can only be in tension, they can't be in corre, compression.

We can treat them as two force members.

And the

other structure we've already studied that's composed

of two force members were truss structures.

And so we're going to be able to analyze cable systems just like truss structures,

with the method of joints or the method of sections, or we could use geometry.

And you'll see how we do that in later problems.

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So here's our, our first cable system, concentrated load cable system.

Got two concentrated loads. We're going to do this in parts.

The first thing I want to do is find the reactions

at points A and B.

And so, my question to you is how you, how would you start that?

3:30

Okay?

What you should have said was to do the free body diagram.

Let's do that together, we have pin connections on both ends.

We've got B sub y and a B sub x force reaction, two

orthogonal directions on the right and the same thing over here on the left.

A sub y and A

sub x.

I've drawn A sub x to the left, I could've drawn

it to the right but I can see just intuitively that.

You're going to have A's of x and b's of A's of x to

the left and b's of x to the right because of the loads.

4:20

And so I'm going to get, you'll notice Ax, Ay and Bx, those three unknowns have their

line of actions going through A, so they don't have a tendency to cause a moment.

I'm going to be able to solve for By directly.

And so I've got the 500 pound force,

which is causing, tend to cause a clockwise rotation,

which is negative in a sort accordance with my sign convention.

The moment arm is the perpendicular distance between the line of action

of the 500 pound force, and point A, which is eight feet.

Do the same thing for the 1900 pound force times its moment arm, which is 24 feet.

and then the By is going to tend to cause

a counter clockwise rotation, so that's going to be positive,

times its momental arm which is 30 equals 0.

And if I solve for By, I'll find out

that By equals 1653.3 pounds, and that will be up.

And so I've got one of my external reactions.

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I'd like you guys to, you folks to do and

an eq, equilibrium equation now to find A sub y,

and when you have that come on back.

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Okay, and finding A sub y, what you should have done with

some forces in the y direction, substitute in your value of By.

Which we just found and you found IA sub y is 746.7 pounds up.

the only re, remaining reactions we need to solve for Ax and Bx.

To do that we'll

sum forces in the x direction, set it equal to 0.

So I've got minus Ax plus Bx equals 0, or Ax equals Bx.

So I, I have a relationship between the two.

I haven't solved for them explicitly.

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The way we're going to do that is either to do a, a joint cut or a section cut.

There's many different options you can use.

You can can go ahead and try some others, but I'm

going to take a section cut over here on the right hand side.

And for that section cut, I want you to now draw

the free body diagram, and we'll analyse that to find Bx.

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Okay, here is the, here is the free body diagram for that section cut.

I have my By at 1654.3, unknown Bx. I have the 1900 pound force down.

I, I have an unknown tension in the section from p1 to p2.

We don't even know the direction, because on the left, right hand side

we know this sag is four feet, but this sag over here is unknown.

So we're just going to say that's an unknown direction.

how would I go ahead and solve for Bx?

'Kay, what you should have said was you were going to sum moments about p2.

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by summing moments about p2, I'll be able

to solve for Bx directly, so let's do that.

I've got the 1653.3 pound force times its momental

arm, which is 6. Minus Bx times it's momental

arm which is 4 equals 0. And so I can saw for Bx which

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is 2480 pounds and it will be positive,

so it's to the right. And so there is another.

One of my external reactions. I know on my overall free

body diagram I had picked Ax to the left, and Bx to the right.

Since Ax equals Bx, Ax in this case will be 2480 pounds to the left.

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And so I have all the external reactions. at points A and B.

There is one other thing I want to know we'll use in, in later problems.

We see that the magnitude of Ax and the magnitude

of Bx are the same all the way along the cable.

And that's the same as any x component for a cut along

the cable, because there are no other loads in the x direction.

So anywhere I might cut the cable.

I'll know that the x direction or

the x component of the force will have to be equal to that 2480 pounds.

And that's going to be a very useful fact.

And we'll use that as we continue on and solve

the other two parts of the problems beginning with next module.