Hello, in this video, we are going to look at

the case of the cables subjected to many loads and subjected to distributed loads.

When loads are distributed, we will see that there is another unit of measure to

quantify these loads, we will see how to solve

the problems related to uniform loads,

we will see the shape that cables take under these kinds

of loads, the maximum internal forces, as well as the forces at the supports.

Let's start here with a cable subjected to eight loads.

Well, it is a quite large number of loads, but it is something

that we can solve by the methods which we have seen until now.

Let's first observe that the spacing between these

loads is constant according to

the horizontal axis. I am going to trace an horizontal straight line and

I copy the position of these loads, and we can note

that this spacing is regular. To my scale,

the spacing is roughly 35 millimeters

on the sheet, indeed, in reality, the spacing in

the model was 150 millimeters for each weight.

Let's look at how to solve this problem using the applet.

To change a little bit, I have already

introduced in the applet the eight loads, but

I introduced them on top of each other,

at the position of the resultant. Since all

these loads are symmetric, it is clear that the resultant

is equal to eight times Q and acts in the middle of the span.

The problem which I am solving here, corresponds actually to the one

that I would have, if I had taken a big free-body which includes all the forces,

and which cuts the first and the last segment of cable.

So, I take

each of these forces and I place them where they

must be placed, that is to say at the level of

each of the weights which act on the chain;

and what we can note, it is, in a non surprising way,

that the inclination which has initially been set

stays the same, since now, if I

draw once again a free-body, well, I include

all the forces, I have the resultant here, in

the middle, which did not change, and I cut the first and the last segment, thus the

system did not change, and it is logical that the inclination should still be the same.

What we can

also note, if we take a ruler, it is that the

distance between the line which links the supports here, and the

cable, is the same than the distance between the cable and the point

which corresponded to the hanging point of the eight forces before.

This s a very important geometrical

property, the ones among you who are very

good at mathematics may remember that it is

a property which is linked to second order parabolas.

Indeed, cables which are subjected to

uniform loads tend to take a parabolic shape.

Obviously, we can also solve this problem with the classic

graphical statics on the paper. So, we have here,

the angle which was specified, for the first

segment, we have already on the right, the

eight forces in the Cremona diagram, thus we

can simply copy this orientation, in

the Cremona diagram. The orientation of

the final part of the cable is also given by the position of the resultant

and of the last support, we can also

copy this inclination in the Cremona diagram.