160 Efforts et contraintes

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From the course by École Polytechnique Fédérale de Lausanne

L'art des structures 1 : Câbles et arcs

24 ratings

École Polytechnique Fédérale de Lausanne

L'art des structures 1 : Câbles et arcs

24 ratings

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.

From the lesson

Introduction

Cette première semaine vous permettra de faire connaissance avec l'environnement du cours sur la plateforme i-structures ainsi de vous rappeler les principes de ce que sont les forces. Le sous-système est un élément décisif que nous utiliserons dans toute la suite du cours. Apprenez donc bien à l'utiliser pour résoudre les problèmes qui vous sont proposés. Prenez le temps de bien comprendre la terminologie que nous utilisons, faites les constructions graphiques en vous assurant que vous pouvez bien imprimer les feuilles et apprenez à utiliser l'applet i-cremona.

101 Présentation du cours12:14

120 Gravitation universelle7:45

140 Types de charges sur les structures9:43

150 Equilibre et sous-système9:03

160 Efforts et contraintes11:19

180 Applet i-Cremona15:08

Meet the Instructors

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

Hello, in this video, we will see the

concepts of internal force and of stress which

will enable us to understand what

happens inside the components of a structure.

On the basis of the example of a person,

standing on a stool, we will see how

the weight of this person is in

equilibrium with the gravitational force that this

person exerts on Earth, through the stool.

In this video, I stand up on a stool.

My weight of 800 Newtons is not directly borne by the ground anymore.

It is obvious that I still exert a gravitational

force on Earth of 800 Newtons too.

Then, these two forces must be in equilibrium.

But now, between us, there is the stool.

In order to sketch rather than to look at myself on the stool, we can

simply, sketch.

I am standing, on the stool which is directly standing, on Earth.

Acts on me, my weight, a gravitational force of 800 Newtons.

Acts on Earth, the gravitational force that I exert on

Earth, which is equal and opposite, of 800 Newtons.

And now, what interests us in this lecture, is what

happens, is what happens inside the stool ?

To answer this, we are going to isolate three free-bodies.

One for the man, one for Earth.

Well, it is not complete, but we include all Earth below.

And then, finally, one for the stool.

The forces that we

already know, are the weight of the person, 800 Newtons, and the gravitational

force that the man exerts on Earth, of 800 Newtons as well.

For that free-body "man" to be in equilibrium,

it is necessary that the stool exerts on him,

a force equal and opposite to his weight

of 800 Newtons, and on the same line of action.

Thus,

we have, here, 800 Newtons which are

the stool's effect on

the man. Likewise, below the stool,

we must have a force of 800 Newtons

which is the effect of the stool on the Earth.

Our two free-bodies of the man and of the Earth are now in equilibrium,

and now we have to deal with the free-body of the stool.

The stool is subjected to a

force of 800 Newtons which is the effect of the man on the stool.

And then, below,

also to a force of 800 Newtons which is the effect

of the Earth on the stool. If we disregard, what

is a reasonable hypothesis, the weight of the stool,

we can see that the stool, which has no weight, is nevertheless

subjected to forces which pass through it. These forces, inside

a structure, we call them internal forces.

This is the way I give you the following definition : an internal force is

a force inside

a structure. And, to be able to see

this force, it is necessary to use a free-body.

Then, we say that it is... this force is

revealed by the cut of a

free-body. Looking at an example,

hum, of a man who is now on the top of a column.

His weight is still 800 Newtons. And if we draw

a free-body which passes just under his feet,

it is obvious that, there must be an internal force in the column, of 800

Newtons, upwards, for the the man to be in equilibrium.

Thereafter, once again, we are going to... it is a little bit

less easy to admit, but we are going to do it anyway.

We are going to disregard the weight of the column.

If I take a free-body, in the middle of the column,

and there is no force inside the free-body since we disregard the weight

of the column. However, acts on the top of the column, an

internal force of 800 Newtons which is the effect of the man on the column.

And then, below our piece of column, we also

have an internal force of 800 Newtons which

is the effect of the ground on the column further to the person's weight.

The internal force, inside this component, which we

draw in blue, is a compressive internal force.

And conventionally, now, and for the rest of this course, the compression

will always be drawn in blue, and will always be negative.

It is a convention,

but it is important, and we will stick to it.

How to know, if an internal force which acts in a free-body is compressive ?

Well, we observe that, a compressive internal force

pushes

on the free-body. Let's look at

the element of column once again, on which

acts a compressive internal force of 800 Newtons,

which we denote by the letter N. We have here, N equal to minus 800 Newtons.

When I draw the force which acts on the subsystem, I do not need

to indicate a minus sign, since if the force pushes, it is compressive.

It is then, a minus sign, automatically.

This internal force acts

on the surface of the column, which we call its section.

The section, it is the surface of the

column, when we cut it perpendicularly to its axis.

Here, we denote this section by the letter A.

And it is expressed in units of area, typically, in square millimeters.

Of course, it could be square meters, or something else.

But, it is very often square millimeters.

We want,

now, to introduce the definition of the stress, as being

the internal force per unit of area. If we take a little element of

column, we recognize it, and if we say that it has an area

equal to 1, a unit surface. Then, we can define the

stress, as the internal force,

per unit of area.

We denote the stress by the greek letter sigma, and we write,

then, sigma equal to N, the internal force, divided by the area.

On this small element, we have a stress which pushes on the element.

This is a compressive stress.

And it is also negative per convention.

You probably wonder if it happens

to get positive internal forces and stresses.

And the answer is yes.

It happens when an element is in tension.

Here, I have a free-body of a window cleaner hung up to a cable.

We disregard the weight

of the cable which extends below the

window cleaner, which is, probably, very low.

And, we notice that for this man... let's say that his weight is still 800 Newtons.

For him to be in equilibrium, it is necessary that acts on the free-body,

an internal force N, equal and opposite, thus of 800 Newtons

upwards. If I take, now, a

free-body of the cable in the part, above the

window cleaner. We have, acting on this element,

we have, acting on this element of cable, an internal force N of

800 Newtons, upwards, and downwards

on his lower end. We draw, in

this section, the tension in

red. And, the tension is positive.

How do we know if an internal force is tensile ?

Well,

in a very similar way than what we have seen about the compression, an

internal force of

tension pulls

on the free-body.

Then,

the tensile stress

is also positive.

If we consider a little element of cable : sigma

equal to N over A. Since N is positive, sigma will be positive.

In this lecture, we have seen how to pass

from the loads to the internal forces, and from the internal forces to the stresses.

Internal forces are

forces which act inside elements of structure.

And stresses are internal forces per unit of area.

Compressive internal forces occur

when a force pushes on the free-body.

Tensile internal forces occur when a force pulls on the free-body.

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