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Hello.

In this video, we will talk about the thrust of arches and about the pressure line.

We will see how thrust can be carried by adjacent arches.

We will see that the position of the pressure

line can vary inside the structure.

We will see the stress dsitribution in the cross-section which can be elastic

or plastic.

And, it will lead us to the limit eccentricity that the

pressure line can reach inside a given structure.

And we will see that this eccentricity can

be exceeded and that the pressure line

can get out of the matter if the

material has a tensile strength.

In this applet, we have modelled an arch,

we have already introduced its two supports and the uniform loads which act on it.

Then we introduce the funicular polygon.

And we add a point of passage.

It is thus an arch with three hinges, a hinge at each support, and

a hinge approximately in the middle of the arch, in this configuration.

What we can see is that the blue line remains

in the center of the matter, not exactly

since it is an uniformly distributed load

and the arch has been designed for a load shaped like a catenary.

But we are going to keep this shape. What we have seen in the applet, I quickly

draw it again. So, we have a

pressure line, roughly

in the middle of the structure. We call

this, the pressure line. The load

which acts on our arch is a uniformly distributed

load, g. And our arch has a

span of l between both supports,

and its rise is the

maximum height of the arch

according to the line which links the supports.

We know that the forces at the supports of

the arch are, on the left and on the right, vertically, a force V,

which is equal to g times l over 2,

and horizontally, a force

H, and this force H, I do not think we have done it for the arches but we had

done it in a symmetric way for the cables, its value

is g times l squared over 8 times f. We can see that it is necessary to have

a force at the support which pushes the arch inwards.

It means that the arch pushes on its supports

horizontally

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outwards.

That is what we call the thrust of the arch.

In our configuration, we

have supports which are able to support this thrust.

In reality, it will always be necessary

to be able to withstand it otherwise the arch will collapse.

How can we withstand this thrust ?

Always a load g.

And an arch which tends to push on its

supports because it has an inclined compressive internal

force which arrives to the supports. Well, a solution, which we have seen, already a

certain time ago, it is to use flying buttresses.

A solution which is efficient,

the horizontal component of the thrust is

carried by the forces which are slightly inclined in the columns

and a little bit more inclined in the flying buttresses.

And the whole thrust H, on the left and on the right, is

carried by the pillars and by the flying buttresses.

But it is a solution which is relatively cumbersome and that is

a little bit a problem.

On the other hand, you have maybe note in certain cathedrals, for example,

there are intermediate pillars and they have no flying buttresses.

How does it work ?

This is an interesting property of the adjacent arches.

Let's take, for example, this bridge, which we have already seen in the heading of this lecture.

It has a certain length and it is a bridge which carries trains, so

locomotives, which are relatively heavy loads.

And yet, this bridge does not seem to have extremely thick pillars.

The ones I draw are rather thicker than they are in reality.

What do we have here ? We have, on the top of each of these pillars, we

have an arch with

arch stones, and then a thrust.

What I draw is a little bit twisted but you get the message!

There is just at the end, we reach here,

what we call the abutment, it is the end of the bridge.

Then, here, we are going to find ourselves with a wall,

you have already seen that,

there we go, here we have our valley. What happens?

Let's look at a free-body, which is located here.

Let's draw again our free-body, bigger.

On the left, we have a part of arch, on the right, a part of arch.

And here, a part of pillar. We have a pressure which comes

from the left, a pressure which comes from the right.

Let's make a polygon of forces, what do we have ?

We have the internal force Pd, then the internal force Pg and we

can notice that both horizontal components

cancel each other out in such a way that the internal force

in the column, here, on the bottom, is vertical.

That is very interesting. So, we can notice

that the thrust is canceled in adjacent arches.

So, it supposes that the arches are exactly identical, if they are

not exactly identical, the

thrust is only partially canceled but for the most part.

So, this is an interesting property

of adjacent arches which enables to

only carry the thrust on the abutment, because on the

abutment, here, we are going to have a thrust

which cannot be offset.

However, the advantage we have is that we are close

to the ground, we are simply going to make a big

wall and this big wall is going to carry this thrust,

and to transmit it to the ground which does not move.

This is what

we have already seen in the applet. When we have an arch with three

hinges, well, the shape of the funicular polygon is

defined, there is only one solution. So, here we have the span, the

rise

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to temperature differences and

to support settlements.

Thus, if the supports move, which can happen since since we are going to apply

big forces on them, further to the arch construction, well,

this arch is not going to change its shape. And that is not going to pose an essential

problem for the structure.

Then, it is an advantage, and it explains why

arches with three hinges have been quite popular, at the

beginning of the industrial era, during the nineteenth

century and at the beginning of the twentieth

century because they were structures which were easy to understand.

They are also easier to calculate but

nowadays, it is not a problem anymore.

If we have an arch with two hinges,

well, the pressure line

is maybe in the middle as I have drawn it before, but it

can eventually go down until the lowermost part or go up until the uppermost part.

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f min and f max

The

maximal

thrust, H max, is equal to g

l square over 8 f min.

So, it is the lower

configuration, for which the thrust is bigger

since the rise is smaller. The minimal

thrust, H min, is

equal to g l square over

8 f max.

Well, I forgot to put it but g, obviously,

it was the load which was applied on our structure.

And if we have an arch without hinge, always with the

load g, the pressure line is even more free to change.

It can start here, on the lowermost part of the arch,

on the extreme external part of the arch, pass by the lowermost part

in the middle of the span, and go until the extreme external part on the right.

Or on the contrary, it can start on the extreme internal

part on the left, go until the uppermost part,

and go down on the extreme internal part, on the right.

These configurations does not have the same span anymore, we have a minimal

span and a maximal span,

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While the minimal

thrust H

min is equal to

g l min square over

8 f max. It should be noted that it is a similar

solution to the one of the arch with two hinges, but which is even more

sensitive to the temperature

differences and to the support settlements. Then

it is a solution which

is absolutely adequate but we will avoid it, for example, if we have grounds which

are a little bit soft because with this solution we will have a risk of movement of the arch.

However, if we are, for example, on a mountain, we can cross a valley,

it is a solution which is absolutely commendable.

Let's come back to the case of the arch with

three hinges subjected to a uniformly distributed

load g and we want to look at the stresses which act in the cross-section.

First, we have to know

the dimensions of the section. Here we can see that the thickness of the arch has a

certain value, I am going to call it t.

And then, this arch has a

height h which is also identified here. We

are first going to look at the elastic stress

distribution which corresponds

to what we have seen until now in our course.

That is to say, when we have a compressive internal force N

which acts on a surface,

stresses act.

For everything to be in equilibrium, of stresses,

sigma is equal to N over A,

that is to say N over t times h. Thus, if we know

t and h and if we know the internal force N, we can determine the elastic stresses.

However, you

will see that a plastic stresses

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distribution will be useful to see how the pressure

line can indeed move inside a section.

This, without making a too complicated reasoning.

In a plastic stresses distribution, either the compressive

stress is equal to zero, either the compressive stress is equal to the

compressive strength fc.

So, we cannot have smaller stresses, they must

be equal to zero, or equal to fc.

How are we going to do?

We still have the compressive internal force N.

What we can see is that here, clearly, the compressive stress

which we had on the left with the elastic distribution was low.

Here, what we are going to do is that we are going to have a zone with this stress fc.

And this stress fc, here, it strongly

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acts. And obviously, it

means that we do not need to use all the width of the section h.

We are just going to use a small part of the section.

Well, that is not really like

that that it behaves in reality, but this way

to think, you will see, will help us later.

What we can see is that we have N is equal to fc times b times t.

What interests us is b since that is what we do not know.

So, b is equal to N divided by fc times t. If we reach

the maximum internal force which can be taken by this section, b becomes equal to

h and both approaches on the left and on the right are identical.

That will not happen very often with the arches, but it is theoretically possible.

In this arch with three hinges, to simplify the things, I

am going to place the hinge in the middle, in

such a way that the structure is symmetrical.

Obviously, it does not correspond exactly to the reality,

in what we have constructed or in what

we have observed in our model, but,

for this model, we can accept it.

What interests us is to see what is going to happen to

the pressure line if the loads which are applied on the arch change.

Well, how are they going to change ?

We are going to modify them. We are going to modify the forces which act here

on the left. We select them, the forces which act on the left

part, and I increase their value.

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What we can notice is that when I increase the load on the left part,

the arch goes up until approaching the extrados here, the external

surface and then in this right part, we can see that on the contrary,

the pressure line approaches the lower surface of the arch.

Now, I have selected the loads

which act on the right part of the arch. Likewise, I increase their value.

And what we can is is that, in the

same way, the arch goes up in the right part until

potentially touching the upper surface, and it goes down in

the left part until potentially touching the lower surface.

I copy in this picture what we have observed,

then, with an hinge on the

top. When the loads are

symmetrical, the pressure line is centered, approximately centered.

When the loads are asymmetrical, the pressure line goes up

in the left part and goes down in the right part.

And the opposite occurs when

we increase the internal forces in the right part of

the arch. We can observe that the pressure line

moves

inside the structure

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does not vary a lot. It means that maybe

we could take off matter in theses places, if we are interested.

We can notice this, we have already seen it

in certain examples of arches, particularly with two hinges.

Let's now look at what happens when the pressure line is

a little bit eccentric and we are going to look at that only in a plastic point of view.

We are going to have here, a pressure line a little bit

eccentric.

So, it was initially here, it

also represents the axis of our element. We have an

eccentricity e which has been created

outwards. In this case, we could have done it inwards.

And the stresses are shared in the following way.

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So, we have the stress fc which acts, we have

a limit width

which I call b l, the limit width and we have in the same

time here, the limit eccentricity.

So, what happens if we want to increase the eccentricity ?

Well, we still can go a little bit more on the left

with the internal force N, I should have written that this internal force was called N.

I think you had guessed it.

I could go further with the internal force N but

I cannot take a larger stress.

Right ?

Because I am at the level of the material strength.

And if I want to have something which is symmetrical...

I mean, I must have something which is symmetrical for the

resultant of all these stresses to be in equilibrium with the internal force N.

So, if I want to go further, I am going

to have a surface which is going to decrease, thus the force is going to decrease.

Actually, I could not go further than this eccentricity.

That is what we have a little bit in our case here.

We have an internal force which is supported on both

sides, which reaches a certain limit value.

If, instead of having rubber layer, we had a very fragile

material, as there is only a very small width

of maybe one centimeter in which the internal force can pass in

our timber arch, well, a failure could have occurred.

In reality, it is absolutely a case which is realistic.

So, the pressure line, well, it

cannot get out of the matter by all appearances.

However, in this video, I use a trick to enable

the pressure line to get out of the matter.

In a place where a crack largely opens, I

put a piece of adhesive tape which I seriously fix to the arch.

And afterwards, when I press on the arch, well, obviously in this place here, this

crack cannot open, and the arch

still works normally, but, in this place here

I can press stronger, without the crack opening.

What happens ?

Let's look at how it works in reality. So, in reality, in

this place here, the pressure line is outside the matter,

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well, in principle, it is not possible if we only have masonry,

we have seen it, we reach the limit eccentricity, we cannot go further.

We cannot get out of the matter.

However, we have seen that, thanks to my yellow

piece of adhesive tape here, I have been able to go further.

What happened?

Let's have a look. We have here our internal force,

we have our eccentricity, which is greater than

the limit eccentricity which we had before. Why?

Because we have introduced here this

element of adhesive tape which obviously cannot really

support compression but it is a place

where there is not a lot of compression.

We can see that there is a joint which tries to

open, thus we are going to have a tensile internal force here.

What does it mean ?

It means that our internal force will

now be supported by a compression part

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and this compression part will correspond to

the limit eccentricity, with

stresses here,

fc, and a certain width

and by the tensile internal

force T in the scotch. I propose you to look at how it

works by means of this applet. So, I apply here, a

force, which corresponds to the

compression, then it does not exactly act at the limit of the

section but not very far. And then, I make act a second force

which is the one of the adhesive tape, which acts in the other direction.

It is not very large, because this piece of adhesive tape is small,

it is just here, on the edge, these two forces are in principle parallel.

They do not need to be exactly parallel, but in principle, they are.

And now, I am going to use this button

"resultant" of the applet, and what do I see ?

You could have seen it if you had made the resultant by yourself,

by hand, it is absolutely possible.

It is part of the reasons for which we oblige you to

master well the resultants and what happens.

I think that if I had asked you,

what is the resultant of two forces which are

opposite, you would have say "yes, it is

not between both forces, it is outside".

Well, we take advantage of it. So, you can see that

adding a small red force here, we make move the

resultant outwards.

If we modify this force, you can see that we can make it move quite enough

outside but obviously, it will need

more adhesive tape, that is more matter.

In a concrete arch, it means reinforcement, in a steel arch,

actually it means that we must have a steel section which

is able to resist to this tension but actually, you remember

that among the arches we have seen, there were lots of arches which were

steel arches

which can support a certain tension. So, there are

the pure masonry arches, as the Pont du Gard,

the Roman arches, we are going to say the classic masonry arches.

But, there are also lots of arches

made by technological products which can have

a small eccentricity thanks to the fact that they

resist a little bit tension.

In this video, we have seen what is the pressure

line and the thrust of an arch, we have seen that it is necessary to support

this thrust, and that it can be supported

particularly, by opposite thrust, when we have adjacent arches.

We have seen what are the elastic and plastic stress

distributions in the arch, we have seen that there can be

an eccentricity of the pressure line inside

the structure according to, particularly, the loads which are applied,

we have seen that this eccentricity can reach a limit value but

that this limit value can also be exceeded, if we are in the presence

of a material which resists tension, or at least of a

material which has

reinforcements which can resist tension.