360 Equilibre de trois forces : résolution avec i-Cremona

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En provenance du cours de École Polytechnique Fédérale de Lausanne

L'art des structures 1 : Câbles et arcs

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École Polytechnique Fédérale de Lausanne

L'art des structures 1 : Câbles et arcs

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L'art des structures propose une découverte du fonctionnement des structures porteuses, telles que les bâtiments, les toitures ou les ponts. Ce cours présente les principes du dimensionnement et les structures en câbles et en arcs. Un deuxième cours présente les structures en treillis, en poutres et en cadres. Après avoir suivi ce cours, vous serez capable d'identifier les structures en câble, en arc et en arc et câble. Vous pourrez déterminer les efforts dans les éléments de structure principaux et procéder au choix de leurs dimensions en fonction du matériau de construction choisi. Votre maîtrise des constructions de statique graphique ainsi que de l'applet de calcul i-Cremona vous permettra de comparer diverses solutions et de choisir la forme la plus appropriée pour un type de structure donné. De manière générale, vous pourrez, en observant les structures autour de vous, reconnaître leur mode de fonctionnement, expliquer le choix de leurs dimensions et de leurs proportions.

À partir de la leçon

Equilibre dans le plan

Après avoir vu comment les structures unidimensionnelles (sur une seule ligne) sont en équilibre et peuvent être dimensionnées, nous passons aux structures bidimensionnelles, qui se situent dans un plan. Les conditions d'équilibre sont présentées.

310 Résultante de deux forces concourantes4:54

320 Equilibre de trois forces15:45

330 Frottement12:33

350 Tir à la corde11:30

360 Equilibre de trois forces : résolution avec i-Cremona4:26

Rencontrer les enseignants

Olivier L. Burdet
Dr
Laboratoire de Construction en Béton
Aurelio Muttoni

Hello.

In this video, we are going to see how to use the applet i-Cremona,

to solve of a system of three forces.

We will see how to introduce a force. I think that you remember this.

We will see how to introduce supports.

We will see how to activate and control the funicular polygon.

And finally, we will see how we can

obtain the internal forces which act within our structures.

So, we have, here, in our applet, the person,

whose known weight of 800 Newtons acts at the center of gravity.

Let's say here. Then, we are going to look at a

free-body, for example, here, which includes the person, which

passes just under his feet, which cuts the rope, and

which obviously includes the load which acts on the person.

We do not have, naturally, supports, but where the structure

has an intersection with the outside, where we are going to

have an internal force, we can also introduce a support, either

under the rope, either under the feet, I introduce first a support

under the feet.

I draw your attention on the fact that the first support

that I introduce has a little one inside the triangle,

this enables to know that it is the first support.

You will see that it is important later.

I introduce a second support, where the free-body cuts the rope.

Now, I can start the solving,

clicking on the button "funicular polygon".

This button "funicular

polygon", generates a funicular polygon, we are going to say, arbitrary.

But, we are going to be able to control it by

means of the following button, which is called initial tangent.

And thanks to this initial tangent, if we come closer to the support

one, there is a little line, a little black line which appears.

We click on this black line, and we keep the button of the mouse pushed down.

And it enables us to

give the correct shape to the funicular polygon.

That is to say that it follows the rope, here, and it passes, obviously, by the feet.

And if we look at the part on the right, we can

see that the construction of the funicular polygon has been done correctly.

If we place the cursor on the middle

of the segment, we can directly read the internal force.

Thus, the internal force in the legs is equal to 827 Newtons.

Then,

in the part on the right, this force has enlightened in yellow,

to indicate where it is located in the Cremona diagram.

Likewise, here, the internal force in the horizontal part,

in the rope, is equal to 225 red.

That is to say in tension.

And, they are going to appear.

That is almost horizontal, 178 degrees instead of 180.

If we place the cursor on the intersection

of both forces, well, or of the three forces, then we can see that

the equilibrium of these three forces is made by this polygon of forces.

That is useful for certain occasions, especially when we

will have more complicated structures.

I am now going to quickly redo the same

example, with the man who leans against a column.

So, again, a load of 800 Newtons.

Here, I do not draw the free-body, it is not necessary.

I place a support under the feet, a support against the column.

I activate the funicular polygon.

I solve it. Sometimes, it is a little bit difficult.

We can draw away as much as we want from the support.

Well, we can imagine the internal force in the arms, it is something like that.

So, we have, in the legs of the

person, a compressive internal force of 860 Newtons.

And in the arms, an internal force of roughly 250 Newtons.

In this lecture, we have seen that to solve a system of three

forces in the applet, it is possible

to introduce only one of the forces, and two supports.

These supports, when we use the resolution of the

funicular polygon, will enable us, once we will have

defined the proper geometry, to obtain the internal forces between each support and the force.

We will be able to read their value and their inclination in the applet

which will be used, for example, to later for a dimension the structural element.

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