480 Câble avec deux charges inégales

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Из курса от партнера École Polytechnique Fédérale de Lausanne

L'art des structures 1 : Câbles et arcs

24 оценки

École Polytechnique Fédérale de Lausanne

L'art des structures 1 : Câbles et arcs

24 оценки

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.

Из урока

Câbles avec plusieurs charges

Le cas suivant est celui de câbles avec deux charges, dont la résolution est un peu plus complexe que les câbles avec une seule charge. Une fois les efforts et forces aux appuis déterminées, le reste est comme la semaine précédente. Finalement, au travers de la méthode du câble auxiliaire, vous verrez comment généraliser au cas de câbles avec plusieurs charges.

470 Câble avec deux charges symétriques14:48

480 Câble avec deux charges inégales7:21

485 Méthode du câble auxiliaire20:27

Познакомьтесь с преподавателями

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

Hello.

In this video, we are going to solve the case

of a cable, which is subjected to two unequal loads.

We will see that the application of the method which

we have seen for a cable subjected to

two equal loads, is absolutely possible.

So, we will study a free-body for each load.

We will obtain the internal forces in each segment of cable,

and finally, we will obtain the corresponding forces at the supports.

So, when I place

a second load on the right of the cable, we can note that the shape of the cable

changes, then reaches the equilibrium and stabilizes.

We have here represented a structural sketch of this cable

with a load of 10 Newtons on the left and of

20 Newtons on the right. We first isolate

a free-body corresponding to the force on the left, which we immediately

draw again. This free-body

includes the segment of cable on the right, the

segment of cable on the left, and a load of 10 Newtons.

In the Cremona diagram, we introduce

this load of 10 Newtons,

a parallel

component to the transitional part of the cable,

and a parallel component to the left part of the cable.

We can directly read the value of

these internal forces, approximately 10 Newtons here, and

15 Newtons.

Copying these internal forces

on the free-body, we can see that this internal force of 15 Newtons here,

pulls on the free-body, it is thus a tensile internal force, thus this part of the

cable is in tension and for the right part,

this internal forces of 10 Newtons here, also pulls on the free-body, and

thus we also have a tensile internal force in the cable.

Well, it is not surprising to have a cable in tension,

we have here 10 Newtons, and here 15 Newtons.

We now take an interest

in the free-body on the right, including the load on the right, a part of the intermediate

segment of cable, and

a part of the segment of cable which goes up towards

the support on the right. So, a horizontal part of cable,

or almost horizontal, a part of cable which goes up, and

a load of 20 Newtons.

We copy this load in the Cremona diagram.

We are going to use the internal force of

10 Newtons, which we already

know, in the transitional part of the cable, the

load of 20 Newtons, and finally, to

go up towards the start of our polygon of forces,

we will have an internal force whose the value is roughly 22 Newtons.

We can copy this internal force in the free-body,

here 22 Newtons, we can note that it pulls on the free-body, thus this

element is in tension, and the last segment of cable is

thus subjected to an internal force of 22 Newtons.

I am going to do very quickly,

obtaining of the forces at the supports, because there is nothing

different from what we have seen previously.

We obtain

a vertical component on the left and a horizontal

component on the left traveling again along

the internal force of 15 Newtons, in the other direction.

The horizontal internal force on the left is equal to

10 Newtons. The vertical internal

force on the left is equal to 11 Newtons.

And on the right, we travel again along

the internal force of 22 Newtons, in the other direction,

then, the internal force on the right, we

can see that it is equal to the horizontal internal force on the left and has thus a value of 10 Newtons.

And finally, the vertical internal force on the right, which necessarily must,

but we can see it, have a value of 19 Newtons for the sum

to correspond to the applied load of 30 Newtons.

We are going to copy these elements.

So, we have 11 Newtons here, 10 Newtons horizontal,

19 Newtons vertical, and once again 10 Newtons horizontal.

We have seen in this solution that, on the

basis of a photograph or respectively of the structural system which

has been given, we can easily obtain the internal forces in

the segments of cable, as well as the forces at the supports.

In the next lecture,

we will see how to determine, without necessarily knowing in

advance the shape of the cable, the shape

or a possible shape for the cable without having

necessarily a photo or a model available before.

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