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Hello.

In this video, I am going to introduce you to arched structures, to discuss about

their shapes and about a very important problem

which concerns them which is their stability.

We will start with an arch subjected to two

loads, we will solve its equilibrium and we will

see that it is very similar to what we have done until now with cables.

We will make some comparisons, then we will see that the stability,

respectively the instability of the arches is what

enormously differentiates them from cables, and

we will see how to choose the proper shape to give to an arch.

In this video, you can see a very simplified

arch which is constituted of three rectilinear wood elements,

on which I apply two symmetric loads of 10

Newtons which you recognize and this structure is

in equilibrium.

We are now going to solve this structure and in a symmetric way, I am also

going to solve what we have already done a long time ago, the case

of a cable structure which is symmetric to this arched structure.

That is quite instructive.

Let's start by the cable structure that we know well.

So, we have a load of 10 Newtons on the top and a

load of 10 Newtons on the bottom. If we

look at a free-body on the

left, we

have a horizontal rectilinear element, an inclined element,

the load of 10 Newtons, and the equilibrium

of this element will be obtained

by having a horizontal line which corresponds to the

segment in the construction and then an inclined

line which corresponds to the segment which goes up towards the left

support. So, we know well this

construction. We thus obtain the internal force

in the horizontal segment and the internal force

in the inclined segment, what gives us tensile internal

forces because these internal forces pull on the free-body.

This is the contribution of the sky blue free-body.

If we proceed in the same way for an orange free-body on

the right, well, its equilibrium will be obtained using

the internal force in the horizontal part of the cable in the other direction, the

force of 10 Newtons, and finally the inclined internal force.

I quickly draw the free-body nearby, a horizontal part,

an inclined part, the load of 10 Newtons and both internal forces which

are parallel to the internal forces in the Cremona diagram

which have the same orientation.

They are two tensile internal forces, thus all our structure is in tension.

Then, if we look at the equilibrium in the other direction for each

support, on the left, on the right, I am not going to draw the corresponding free-bodies,

you can do it by yourself if you wish but we have

already done it in the past, we obtain

a vertical component which is equal to 10

Newtons and a horizontal component which has a certain value.

And if we copy them in the real system, we then have here on the right a vertical

force at the support and a horizontal force

at the support, and likewise, on the left.

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is going to give us a force which

pushes towards the left and a force which pushes upwards.

If I now draw again the corresponding free-body, I have

a horizontal part, an inclined part and a load of 10 Newtons.

I now have

an internal force which pushes on the free-body going up towards the right, that is an

internal force which pushes on the free-body thus it is a compressive internal force and, likewise

in the horizontal part, I have a horizontal internal force thus I can draw again,

in blue, these internal forces in the

free-body. I can identify the sky blue

free-body here.

We can see that, by construction, the

internal forces are exactly the same as before but

they have changed sign, they have become negative, compressive internal forces.

If we look at the free-body on the right,

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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

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Arches are fundamentally unstable.

The next videos in this

lecture will show how we can

choose the correct shape of an arch to ensure

that precisely, we do not have this problem of stability.

I even so reassure you about stability.

We have ancient constructions, which date back

from the Roman time for example, or from the Middle-Ages,

which use arches and which did not collapse, therefore the fact that an

arch is potentially unstable does not mean that it

is going to collapse, but we have to be careful.

This video shows how we can keep the symmetry, the mirror

effect between the cable and the arch if we add a load.

I am going first to add a load on the cable on the blackboard.

It automatically changes shape, then I have added a load on

the arch on the foreground, and making go up the point where I have added the

second load, and giving it a symmetric shape

in relation to the one of the cable, then, I have obtained a stable solution.

This is nice but generally, we do not have someone who is kindly going to change

the shape of the arch therefore we cannot rely on this method.

So, what is the ideal shape that an arch should have ?

We have seen that it was the mirror shape of the one of a cable.

Indeed, here, you look at the opposite shape of the

Golden Gate bridge which we have already seen before.

And, we compare it here, on the right, to a bridge

by the Swiss engineer, Robert Maillart, and we can see a very similar shape.

I use the applet i-Cremona to demonstrate the ideal shape of an arch.

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We can see that there is a support here, another here.

I apply loads where there are the pillars which go down from the deck to

the arch and here, we do not see them very well, I guess where they should be.

I introduce these loads, this, it is okay.

Then I activate the funicular polygon and I give it

the proper shape and we can immediately see that,

indeed, that is the same shape that we would have typically obtained

for a suspension bridge, which is correct for an arch with lots of loads.

Thus, it is a parabola in the present case, a second order parabola.

In general, we will use the shape of the funicular polygon

under the permanent loads to define the shape of the arch.

Once again, we have an arch designed by Christian

Menn, in canton Graubünden, in Switzerland. Here it is also quite clear where

the loads act. And the initial shape of

the arch is defined by the permanent loads, the weight

of the structure, essentially the weight of the deck because the arch is not very

heavy, but, obviously, when we make an accurate

calculation, we take into account all of these weights.

To avoid the problem of instability, it is necessary that the

funicular polygon should be inside the matter all the time.

I am going to show you an example of this later.

And, for that, we will define the thickness of the arch

under the influence of the variable loads. Thus, the variable loads

are used to determine the thickness of the arch.

This applet shows the bridge over the Salginatobel by Robert Maillart.

You can see that this bridge does not have a very

classic shape, its thickness is variable.

Under the effect of the permanent loads, we can

give it this shape.

Actually, the internal forces are always going to pass by this point here on the top then I

am going to redo the solution defining that the

internal forces must always pass by this red point.

You are going to see the influence that it is going to have.

I now add a quite significant variable load.

For example, it can simulate a truck. Then, I make it move

on to the bridge.

Look at what happens with the blue funicular polygon,

as well in the left part as in the right part.

It tends to go up under the load.

Remember, I have said that the arch had to go up where

the load is applied and to go down on the other side.

But, in any case, along all the length of the bridge, thanks

to the shape which has been given to it, the polygon of forces stays

all the time inside the concrete.

This means that our structure is stable.

In this video, I have introduced the shape and the stability of arches.

I have shown that there is a big

similarity at the level of the solution with graphical statics

with what what we have done with cables

thus it is relatively simple since we are already experienced.

However, because of the phenomenon

of instability, the shape to give to the arches

is very important, as well as its thickness.

It is important to consider the permanent loads to define the main

shape of the arch and the variable loads to define the thickness.