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Hello.

In this video, I am going to talk to you about continuous beams.

We have seen, in the previous lecture,

that Gerber beams do not only have advantages,

especially because of joints which can, at least in bridges,

or in particularly exposed situations, cause problems.

So, why not remove these joints?

We are going to see why and how we can remove them,

what are the consequences for the internal forces,

a resolution method, and finally, we will make some comments

on the shape to give to continuous beams.

Here, I show you the internal forces which we have determined

in a continuous beam.

A continuous beam with a central span which is equal to L, here,

and lateral spans which are approximately equal to 0.8 x L.

The purpose of these spans which are shorter on both ends

is to limit the internal forces.

You maybe remember, when we looked at

quite long Gerber beams, there was always

a longer distance between the arch and the cable

in the end part of the beam.

In the present case, with this configuration,

we can see that the arch and the cable roughly remain at the same distance

in all the spans, whether it be an edge span,

the first one or the last one, or else an intermediate span.

If that is possible, that is very advantageous

that the first and the last spans be shorter,

approximately 0.8 times the length of the other spans.

So here, when we have this configuration,

we will have 1/3 of the rise, f/3, here in the span

and then 2/3 of the rise, here, on the supports.

That means that 2/3 of the internal forces must be carried at the supports,

and 1/3 in the span.

Here we have the same beam which we have just seen,

with two spans L in the middle and two side spans 0.79 L

on the left and on the right.

We activate the resolution with the funicular polygon

and then we select the three support reactions.

Let's now push the "Control" button to have the three.

Then afterwards, I push the "Shift" button , releasing "Control"

to change their amplitude

and we can see that the shape of the funicular polygon gets closer

to what we want to have.

And here, we can see that the rise on the supports

is twice the rise at mid-span.

So here, with the applet, we have just determined the internal forces

as well as the support reactions which we can read here

for this configuration.

Let's now maybe look at something interesting

about this structure. We can see that there are here

two support reactions, one here, one here, one here, one here.

So we have six supports reactions plus, we have only one bar

and then, how many nodes do we have? Well, the nodes, actually,

we have two. We have a node here

and the other node at the end.

So we have 2x2, which is equal to 4.

We can thus see that this structure is statically indeterminate.

We have 6+1, 7 on the left, 4 on the right.

Our structure is thus three times statically indeterminate.

Yes actually, what we do not know are the three forces,

for example, the three support reactions, 1, 2 and 3.

However, if you remember, we just graphically determined them

through the use of the applet.

So, in this case here, the applet enabled us to solve

a statically indeterminate structure, of course, in a particular way

since we knew, we have set down that this value was the half of this one.

If we had not known it, we would not have been able to proceed with this resolution.

We can notice that, like in the Gerber beams,

the cable and the arch cross on various points and on these points,

there are no hinges anymore since we do not have a Gerber beam anymore

but a continuous beam.

But however, the arch and the cable cross on various points.

Here we have the example of the Mont-Blanc bridge in Geneva,

which has a shape which reminds a little bit the one we have just seen

for the case of a continuous beam.

And indeed, this bridge is constituted of a continuous beam.

The representation which we have done till now of the arches and the cables

for the beams is very good for continuity

compared to what we have done till now. However, it has a defect

which will become particularly apparent when we are going to start dealing

with the case of slabs. That is why I am going to introduce to you

an alternative representation.

This is, let's say, the usual representation of these arches and of these cables.

I propose you to represent

in the lower part of the drawing, a configuration - so, I copy here

all the points where the arch and the cable cross, you will see why

later. That is important.

And, in each part, I am going to represent an arch and a cable

so here, I indeed have an upper arch and a lower cable.

Then, here, I have an upper cable and a lower arch.

And then, to enable this arch to be correctly supported,

I must also have here an element which is in tension.

Likewise here, on the left, I am going to have again an arch and a cable

with a hanging element.

On the other side, that is going to be the same thing.

The interest of this representation is that, in this case,

the arches and the cables always remain inside the matter.

And this is quite interesting for the representation we will make soon

in the lecture about slabs.

I finish the drawing here, that is obviously repetitive since that is the same thing

for all the spans.

And what is interesting is that now we are going to compare

the arch-cables according to their respective length.

Here we have a nice example of a continuous beam bridge,

these are the Chillon viaducts which have been built

at the end of the 1960's on the shore of Lake Geneva.

And what configuration do we have?

We have a configuration where we have tension

in the entire upper part of the bridge.

This tension is carried by many prestressing cables.

And then in the lower part, we have compression.

That probably reminds you the configuration of the beam

with cantilevers of 0.5 times the length of the beam.

That is exactly this configuration which is used here

for a bridge which is continuous.

There are some places where there are joints but very rarely.

For the rest, it is simply a continuous beam.

In this lecture, we have seen that it is possible to design

and to build continuous beams without any joints.

The internal forces which act in these structures

are similar to the ones which act in a Gerber beam

with a regular crossing of the arch and the cable.

But these beams are statically indeterminate,

so, in general, we cannot solve them within the framework

of this course even though I have shown you

that it is possible to do it with the applet

with some additional hypotheses.

The shape which is given to a continous beam can be

either of constant depth as we have seen it, or of variable depth

which adapts according to the internal forces which act in the cross-section.