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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,
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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.
Olivier L. BurdetDr
Aurelio Muttoni
