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Hello. In this lecture, we are going to solve

our first truss, a truss with four nodes.

As we have seen it in the previous lecture,

it is an arch-cable to which we have added an additional cable.

In the first course, we have seen

that this is a method to stabilize the arch.

It is indeed efficient but we did not calculate the internal force

which acts in this cable.

Now, we are going to see how to solve this structure

proceeding node by node.

We will ask the question : "By which node should we start ?"

And then, we will obtain the internal forces in each of the bars of our truss.

We have here a truss with four nodes and we can wonder :

"By which node can we start ?"

If I look at the node at the bottom on the left,

I can see that it is not possible to solve it because I have here

one, two, three and four unknowns,

because the support force has two components, we have here five unknowns.

However, the methods that we have seen do not enable us to solve a system

with more than two unknowns.

If I look at the right support, the situation is a bit better

since I have one, two, and the vertical force, three unknowns.

Unfortunately, I cannot solve a system with three unknowns,

only a system with two.

If I look at this node, I have a known force but

one, two, three unknowns which are the internal forces in the bars.

It is not going to be possible.

However, if I look at this node here,

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Around this subsystem, we are going to turn

counterclockwise,

starting by what we know, that is to say the force of 20N.

After this force, we are going to meet the internal force in the bar A-B.

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We will be able to read these values to scale in the Cremona diagram.

I am not going to do it now but you can do it

with an accurate drawing at home.

You will obtain the actual internal force in Newtons in each bar.

The contribution of node B to the Cremona diagram is symbolized by this triangle.

4:06

We can now move to node C because we had three unknowns

but we just got one of those,

the internal force in bar B-C.

This node is subjected to a force of 10N, the internal force in B-C,

the internal force in A-C

and the internal force in C-D.

I introduce the force of 10N.

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which is subjected to the vertical support reaction RD,

because it is a mobile support.

Turning counterclockwise,

we are going to meet the internal force in C-D which we already know in the other direction,

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which is subjected to the support reaction RA.

I am not going to draw it because I do not know in which direction it is going to act

but we are going to find it out soon.

What are the internal forces acting on node A ?

We are going to start by the internal force in A-D

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My drawing is quite complicated, that is why I made available for you

the complete solution of the structure on the next slide,

starting by node B

with its contribution indicated by this orange triangle,

then moving to node C with these two orange triangles,

then to node D, this upper triangle

and finally node A, the lower quadrilateral that we just made.

You can also find the accurate internal forces in each of the elements of the structure.

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It is also possible to create a truss with four nodes by adding a timber diagonal,

for example, instead of a chain,

so an element which can at once resist

to tension and compression.

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Let's see how we are going to solve this element.

It is very similar to what we have done the last time.

I am then going to do it quite quickly.

We are going to start by node C because it is a node where there are only two unknowns.

We know the force of 10N

and this node is also subjected to

the horizontal internal force in B-C which we are going to meet first,

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Coming back to the free-body,

we can see that the internal force in B-C pushes on the free-body,

as well as the internal force in C-D.

So both internal forces, here, are in compression :

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on which a vertical support reaction RD acts

but we do not know it for the moment, as well as two internal forces, C-D and B-D which we already know

and the internal force in the lower chain, in all likelihood in tension, AD.

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here, vertical, which is still equal to 12.5N.

I remind you - we have seen that on the lecture about the obtaining of the forces at the supports -

that whatever the arch-cable, the support reactions will be identical for a given combination of loads.

It is not surprising that we get the same result for these two configurations.

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On this graph, I have copied the results of the calculations

of the three configurations which we have seen in these past two lectures.

First, here, you have the arch-cable.

Here, you have the truss with a diagonal made by a cable,

a diagonal in tension.

And then here, the truss with a diagonal in compression.

Let's have a look at how to compare these various configurations.

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In the three configurations, we can first recognize an arch.

In the case of the arch-cable, it takes a funicular shape.

In the case of both trusses, it is a fixed shape,

the one we have given to our truss.

Note that the three configurations that we are comparing have the same height

since the left bar always has the same inclination.

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If we get a closer look, the internal forces in the left bar

20N, 23N, 20N in compression are very similar.

The internal force in the upper bar, 11.5N in the middle, 10.4N, is also similar.

Here, it is only 5.8N but we can see that there is another bar in compression

and both together reaches roughly 10N.

If we look at the tensile internal forces, we first always have tension on the bottom of the structure.

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The internal forces, 10.1N, are identical in the configurations of the arc-cable and of the

truss with a diagonal in compression.

Here, it is 7.2, it is lower but we also have a diagonal with 5N.

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This is an important observation, it means that the system does not change

in a dramatic way if we change its setup a bit.

It also means that if you want to quickly estimate the internal forces

in a future truss, a quick calculation simply using an arch-cable

with the correct loads will already give you

a very good approximation.

I here have models of five configurations of trusses,

always with the same distance between the supports and the same total height.

We can recognize in all the cases an upper shape of arch

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If we compare this configuration with this one,

we can see that the diagonal, here, has been given to us as being in tension

so this diagonal is going to be in compression.

This diagonal, we have already calculated it before

so we know that it is in tension.

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We can thus note a big variety in the possibilities of design of trusses.

We can also wonder : "Can we calculate them all ?"

If we look at this truss, here on the top, I cannot calculate it.

I have two, three, four, five unknowns here.

Here, I have one, two, three unknowns in addition to the force at the support.

Even if I independently determined the force at the support, what I can do,

I would not succeed to solve the structure because I have three bars.

Here, likewise, I know one force and I have three bars.

I know one force and I have three bars.

Actually, we are not going to be able to calculate this type of trusses

with the methods we have studied in this course.

But how could we know if we can calculate or not a truss ?

This will be the topic of the next video.

In this lecture about our first truss with four nodes,

we have seen that starting from a chain which was stabilizing the arch-cable,

we have obtained a new structure which is called a truss.

We have seen that the method of solving it

to obtain the internal forces is similar to what we have seen till now,

we have not introduced any new methods of calculation.

Simply, we must start by a node where there are only

two unknowns, for example two bars for which we do not know the internal force.

We have seen that the diagonals in a truss

can be in tension or in compression

and that the order of magnitude of the internal forces is similar to the arch-cable.

Finally, as we just saw, certain structures are not

directly computable using the methods of this course.

And the next video will enable you to understand

the ones that we are able to calculate.