Hello, in this video, I am going to talk to you

about the cable-stayed systems, the principle of their functioning, the way in which we

can solve them by means of the graphical statics and about the

diverse types of configuration which exist for the positioning of the cables.

In this video, you can see a timber bar

which is hung up to two inclined elements of chain.

When I add

two symmetric loads, this system is in equilibrium does not have any problem.

Let's see how to solve this system.

So we have, on the left and on the right, a

load of 10 Newtons which acts on our structure.

First, we look at the right part that I isolate with a free-body.

In our free-body, we have an horizontal element

of deck, an inclined element of cable and a load of 10 Newtons.

We are going to solve the equilibrium of this configuration.

So, we are going to have here a load of 10 Newtons.

Turning around the free-body, we first meet the chain.

As it is a chain, I know that it is

in tension, thus I directly draw it in tension.

And we have an element, a tensile internal force,

and we meet the horizontal which is in compression.

We have here a compression, here a tension and we can color our

cable-stayed system, well, I think we had already intuitively recognized it but now,

we have demonstrated it, it is a system in which the chains are in tension

and the element which is used as deck is in compression.

If we deal with the left part of our structure, we have

the same force here of 10 Newtons, the compressive internal

force in the other direction and the tension

which is equal to the tension in the other

cable. Here we have a more complex

structure with four times 10 Newtons hung up.

We can wonder if we can solve it.

At the least, if we identify this

free-body here, we can see that it is perfectly identical

to the one we had in the previous system, thus we can quietly

solve it. 10 Newtons.

I am going to number these tensions, I will call it

T1, this one T2, this one T3 and this one T4.

And I am going to take the advantage of the opportunity to number the compression 1, 2 and 3.

I come back to the tension. So I first have this element here which is in

tension. And I obtain here the internal force

T1. Then, the next element is in compression,

I obtain the compression C1.

Now, I am going to take an interest in this free-body

here which I draw nearby because it is interesting.

So we have the deck, we have an inclined cable and we have 10 Newtons.

What we also know is the compression C1 which comes from the right.

Then, turning in the counterclockwise direction

around our free-body, we are going to meet out force of 10 Newtons,

our load of 10 Newtons, then the compression C1 in the other direction, then the cable T2,

and finally the compression C2.

Obviously, we could continue towards the

left doing exactly the same construction.

I spare you this construction but I

do it myself to have the complete drawing.

So, here, it will be C3 with

both loads, the two last

loads of 10 Newtons and then the

tensions T3 and

T4.

Obviously, the compressions are used in both

directions for each of these construction. We can make a few

interesting observations on the basis of this solving.

What we can see is that the compression is not constant in the

deck, it is lower at

the end and becomes greater in the middle.

Likewise, the internal force in the cables T1 is greater than the internal force in the cables T2.

So, the cables which are more inclined support

more loads than the cables which are less inclined.

On the other hand, we can obviously see that the cables T2 are

distinctly shorter than the cables T1.

In this picture of the transporter bridge in Nantes, designed by

the french engineer Arnodin, we can see that the stay cables have the

same configuration than in our

model, that is to say a configuration with

a fan shape where all the stay cables are hung on the top of the mast.

This configuration has the advantage, we have seen it, that the

internal forces are lower in the stay cables which are less inclined,

however, the stay cables are quit long and

it is especially quite difficult to all hang them on the top of the mast.

Thus, it is a configuration which has a few problems.

Independently of this configuration, because the

central span is much longer than the

adjacent spans, it has been necessary to use

vertical cables to anchor the structure in the ground.

On the bridge of this illustration,

we have another configuration of the stay cables which we call the

harp shape, that is to say that the stay cables are parallel to each other.

Which is striking in this configuration, it is

that the length of the stay cables varies very, very quickly.

The external stay cables are very long, while the internal stay cables are very sort.

Let's look at what is about the internal forces.

I just draw the right part of a cable-stayed

bridge, such as we have calculated it before,

except that this time, the stay cables have a harp

configuration and to get something, I also load each stay cable

with 10 Newtons. Then I can draw here this load of

10 Newtons for the free-body on the right.

I trace a parallel to the stay cable.

This is the tension in the external stay cable and this, it is the

compression, so we are going to call them, T1, C1,

T2 and C2. Here I have T1 and C1.

If I look at the second stay cable, there is

also 10 Newtons and its orientation

is still the same. Well,

what is interesting is that T2 is equal to T1,

So the internal force in the stay cables is

the same. However, the compression C2 is

quite a bit greater, in this case

double, but it is really greater than C1.

So, we can imagine that when we reach

the end, when we have the shortest stay cables,

we have a huge compression. However, we have

the advantage to have internal forces in the stay cables which are constant.

Note that before, the internal forces in the stay cables

rather tended to decrease, which was favorable.

An advantage of the harp configuration is

that we have more space to place the

anchors of the stay cables which are not easy to

hang up on the top of the mast in the harp

configuration.

As you have probably guessed it, it is possible

to make transitional configurations, as we can see it here.

Here, a half-harp configuration.

The stay cables are not absolutely

parallel but they are anyway well distributed

on the height of the mast, which enables

to have a little bit the advantages of both solutions.

It is also possible to make cable-stayed bridges with multiple spans.

However, if we remember of the necessity to anchor in the ground, as we had seen it

for the bridge of Arnodin, it is actually

necessary that the pillars should be stiff enough

to help support the vertical

loads which are very significant in the exceptional cases.

But the bridge of Millau, which is in service since 2004, have shown that it was really

possible to make this kind of structure in an efficient and elegant way.

In this video about the cable-stayed systems,

we have seen how to solve them by means

of graphical statics.

We have seen that there are several possible configurations

for the stay cables, particularly the fan shape or

the harp shape and we have seen that there is a

large compression in the deck of this type of structure.

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