Then, if we take an interest in the forces at the support,
using the force in the right leg and in the
left leg, in the other direction, we obtain the usual forces at the support,
that is to say an horizontal thrust and a vertical force, which is equal to 10 Newtons
of course, here. And then, we have a force,
V and a force H at each support. How is it with the arch-cable?
We are also going to load it with twice 10 Newtons.
Do not forget that there is a cable because, otherwise,
with the mobile support... You can see the symbol for the mobile support on the right, we will
come back to it in a little bit, it is a triangle which is on
1 or 2 casters, here, 2 casters, it symbolizes the fact that it can
move transversely, it is quite similar
to my little cart which you have seen before.
The Cremona diagram, with twice 10 Newtons,
is very similar to the one of the arch because, if we
make the equilibrium of the points of application of the
loads, we will obtain the same
thing. So, we are going to first take an interest
in the free-body which is just next to the support
on the right. What do we have in this free-body ?
Let's solve this free-body.
We obviously have an inclined leg of the
arch, we have the horizontal cable, and what
we know is that, we can only have a
vertical force at the support, which comes from the ground.
Well, let's come back to the Cremona diagram, it is this force,
here, in the leg along which we are going to travel in the other direction.
Then, we are going to add the tensile internal
force in the cable, and a vertical force at the support.
So, here, this is correct, we only have a vertical force
on this support.
Let's now look at the free-body close to
the left support, which I draw again.
What we have, acting on this free-body, is then an
inclined leg, which comes from the arch, the cable in
tension and, certain ground reactions which we will see a little bit later.
What we
know is that here, we have compression which acts and here,
we have tension, the force T which we have just determined before.
So, we have, acting on this free-body, the force T, well, I must draw it again
in the Cremona diagram, this time,
it is not possible to find a combination.
To this force, we add afterwards the diagonal internal
force in the leg and finally
we obtain that the vertical force at the support is the
only component which we have here, thus we only have a vertical component.
It is interesting to see that, just by introducing a single mobile support,
actually, we get the effect that on both supports, there are only vertical reactions.
Now, be careful, we got this effect because there
only are vertical loads acting on the system.
If we had an inclined load which acted
on the system, of course the support on the right
could only support a vertical component, however,
the support on the left could support an inclined component.
Let's see what happens if I now add an asymmetrical load
on my arch.
Well, as on a normal arch, we can see that
the arch tends to go down under the additional load and that actually
it is necessary to give it the opposite shape, that is to say to make the arch go up
here, where we apply the load for it to be stable.
What this means, is that arch-cables
behave, for the arch part, exactly as arches.
So, the stability problems are still here, and we
still need to stiffen them but we have seen before
that we know how to do it.
So, let's look at how to support the thrust
and let's see how it can be advantageous.
We have, here, a cathedral and if I indicate, here, the internal space
of the cathedral, we can see that it is actually quite limited.
There is a main nave and 2 secondary naves.
Everything we have around is essentially used for supporting the thrust.
On the right,
I have drawn the same cathedral, with still
the same volume which is useful for people
who are inside, and we have taken off everything which is used for supporting the thrust.
And, we can notice that it correspond to a lot of matter.
Of course, it makes the beauty and the
majesty of the cathedral, my purpose is not to take off
the flying buttresses, where they are, but if we
want to make modern and economical structures, it is interesting
to know that it is thrust.
If I simply introduce a cable, here, the
forces at the supports, under vertical loads, will only be
vertical and can thus directly go down in the
columns which we have on the left and on the right.
And all the flying buttresses and the other systems of buttress are not necessary anymore.
We do not need either to introduce a mobile support,
there are no casters because in this
case here, these columns, which are very long, very high, they
are flexible and thus they enable a slight transversal movement.
By the way, I think that you have a feeling, if we take off all
the matter of the flying buttresses, we feel a little bit that the arch will want to open.
Yes, it will want to open but when it will start
to open, the cable will be put in tension and
it will say : "I do not go further, there
will only have vertical internal forces from now on".
Let's now look at how to model an arch-cable with the applet i-Cremona.
I start by modelling an arch.
Well, you can see that it is really very similar.
So, I introduce a support on the left, a support on the right.
I activate the funicular polygon
and I give it the adequate shape.
There we go, it is probably the shape of the
funicular polygon of this vault.
Now, if I want to introduce a mobile support, I erase the second support.
If we introduce a mobile support, it will always be the second support.
I take it off, and I press the button "control"
on the keyboard or I click on the button "control" of the applet.
And I introduce, well, I am going to introduce it on the right, in this way you can
see well its symbol, a mobile support, which does not have the small
casters but we clearly see that the support is separated from the ground, it
can slide a little bit as on an air cushion, it is another symbol.
We can now place this support just at the place where it was before.
And, we can notice that, indeed, the thrust is
supported as for the other elements.
If I place my cursor in the middle of the cable, I obtain the internal force which acts
in this cable. Another example of arch-cable
structure, by necessity, it is a an arch, it has been used
for the shuttering of a big bridge, a quite imposing structure.
And because this bridge crosses a river, the builders of
that time have decided to create this arch and to transport it
making it float on 2 barges which we have on the left and on the right.
So, here, we have a barge, and, here, we have another barge.
Obviously, in this case, the only load which acts on this
arch is its self-weight but it is enough to create a thrust.
But, in this case, it is very, very
clear, the only things that the barges can offer,
they are ships, are vertical thrust, it is
the Archimedes's force, which is necessarily upwards.
So, it is absolutely impossible to support the thrust.
The solution which
has been adopted in this case, and we can see it in the water, with small marks,
small waves, is to introduce a series of cables, which link these
2 barges, in such a way that the thrust is compensated and that there is
only a vertical thrust, since obviously, it was impossible
to take an horizontal thrust with water.
Here is a picture
of the finished bridge, you can see the very ingenious aspect
of the solution which has been chosen, the
barge not only enables to move the formwork but
it also enables, once we have finished the arches,
lowering the formwork, to move it, to bring it
at the place where we want to build the second one, and so
forth, and thus to have a system which has
certainly been very economical for the construction of this bridge.
Until now, we do not have,
actually, seen yet how to really make mobile supports.
You have, on the left, the symbol for the mobile support.
So, we can enable, here, an horizontal movement, and we will only
have a vertical force, perpendicularly to the surface of the support.
In practice, we sometimes use supports shaped like a roller
but it is important to say that if the displacement
which we must have, must be significant, it is not
very easy and on the other hand, the surface
on which the rollers roll can easily becomes a
little bit dirty, if there are small stones which fall, if there is a little
bit of dirt, and thus these supports get fulled up and
do not necessarily work very well in the long run.
So, we have tried to find new solutions.
On the left, I show you a relatively old
solution, that is what we call a pendulum support.