a horizontal segment, an inclined segment and a load of

10 Newtons, we are going to reuse the

horizontal segment, to introduce the second load of

10 Newtons, and here I

directly draw it in blue but we are going to check it, we

have well a compressive internal force in the part which

goes up towards the support, and this internal force clearly goes in this direction, it pushes

on the free-body so it is in compression, likewise for the internal force here,

and we can identify the contribution of this

free-body to the Cremona diagram.

We are now going to look a little bit more in detail

at the contribution of what happens at the supports.

We have an element and the beginning of a support.

We have a compressive internal force which acts here.

And for this to be in equilibrium, we are going

to have a vertical force at the support like that, and a

horizontal force at the support like that. We are going to check it in

the Cremona diagram for the left support.

So, we have the inclined force in the other direction, then

the vertical force at the support,

and the horizontal force at the support which pushes towards the right.

So, we have here a vertical force and here

a horizontal force which pushes the structure inwards,

while previously, the horizontal force pulled the

structure outwards. Likewise, we can analyze the right

support with a compression which

we will use in the other direction in the Cremona diagram.

Then, a horizontal force at the support which pushes towards the left

and a vertical force at the support which pushes upwards.

So, we have a vertical force at the support upwards and a horizontal force

at the support towards the right which we can copy in the

diagram of the real structure. We can identify the contribution

of the various free-bodies to the Cremona diagram.

There we are, that's all.

Thus, if you know how to solve a cable, you know how to solve an arch.

I am not going to do a very long lecture.

We can stop now.

Not really, otherwise I may even

have started with arches. There is a little detail.

Let's watch this video on the arch which we have seen before.

I have added a force on the right. And what happens ?

Catastrophe ! The arch collapses.

That is the first time we have a collapse.

Until now, when we added a load on a cable,

the cable changed a little bit its shape and it was very happy.

But we have never had something which broke.

Here, the structure really

collapsed, and it is potentially dangerous.

Thus here, we have something which is not

similar to what we had until now with the cables.

Let's look at this in more detail.

What happens when we add a force on a cable which is subjected to

twice 10 Newtons and then on an arch which is also subjected

to twice 10 Newtons ? I am going to add a load of

10 Newtons on the right in both cases. What happens in the cable ?

We already know that, we have already seen it. What happens is that the cable is going

to go down a little bit on the right, and to go up a little bit on the left, then to change its shape.

So, this, it was one of the problems we had identified

with the cables, that is to say that they move a lot.

And that is all.

Now, we have well noticed that the construction

of the arch, it is the mirror shape of the construction of the cable.

Then let's make a mirror. What it means is

that the shape of the arch, when we add a

force, should be like that. This, it the shape of the funicular

polygon of the loads. We could calculate it but it is like that.

However, when we add a load on the right around

the point here on the right, what does it want to do ?

It wants to go down, it certainly does

not want to go up. But, it would be necessary that it should go up for the arch

to change its shape. This, it is a fundamental difference which

makes us enter in the true problematic of arches.

If we add a force on a cable, the funicular polygon,

that is to say the cable,

changes its shape but follows the applied

loads.

Thus, the structure is always stable.

Actually, we never talked about stability

because we did not have this problem.

Except if we put so many loads that we

mangage to break the cable but otherwise there

is no problem.

If we add a load on an arch, the funicular polygon