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.
320 Equilibre de trois forces
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來自 École Polytechnique Fédérale de Lausanne 的課程
L'art des structures 1 : Câbles et arcs
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從本節課中
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.
與講師見面
Olivier L. BurdetDr
Laboratoire de Construction en Béton
Aurelio Muttoni
Hello. In this video, we will see how
to solve the equilibrium of a body, of a free-body subjected to three forces.
Until now, you remember that we have
solved the equilibrium of a body submitted to two forces.
For example, its weight and the ground reaction.
But we have not, it is the first time that we see
a body, if you look at the picture on the screen,
a body which is subjected to three forces :
its weight, the ground reaction, and then the effect
of a rope of which the person takes hold.
We are going to proceed in the following way, in this lecture.
Since we know how to obtain the resultant
of two forces, we are going to solve the equilibrium
of a system of three forces, obtaining first, the resultant of two of the forces.
Then, we will apply
the principles of the equilibrium of two forces, that we have seen previously.
Here, we have a video which shows a man which keeps
himself in equilibrium on the ground, taking hold of a rope.
It is clear that if we removed the rope, the person would fall down because he
leaned back too much, compared to what is possible.
This, we have already seen it in a previous video.
Let's look,
now, at the free-body which includes this person.
This free-body cuts the rope, passes under
the feet of the person, entirely surrounds the
person, thus, its weight. And the forces we have,
which acts on this free-body,
are the weight of the person G which is equal to 800 Newtons.
An internal force, inside the rope, I think that we can immediately
admit that it is a tensile internal force, which we denote by T.
And then, the effect of the ground that I am going to denote by S.
But, we do not really know
in which direction the effect of the ground acts.
Well, it is precisely what we want to see today.
We are going to start on the right, in the polygon
of forces with the weight of the person.
The person has a known weight of 800 Newtons.
And then, we also know that acts on this free-body, the internal force
T in the cable.
We can, thus, very easily obtain,
in the polygon of forces, the resultant of these two forces,
which we are going to name
R GT.
It is the resultant of the weight of the person, and of the tension.
And, we get, not only the, the
magnitude of the resultant, but also its orientation.
What we know from the previous video, it is that these, this resultant must
necessarily pass by the intersection of the line of action of the weight, and of the
line of action of the force in the rope, thus, by this
place, here. I copy, now,
the resultant, with its direction in the
real system. This,
it is the line
of action of the resultant. And then the resultant itself, well,
acts, here. We have named it, then, RGT.
Now, in our free-body, we still only have two forces, now.
The resultant RGT, and the effect of the ground S.
Are these two forces in equilibrium ?
Well, let's have a look.
What do we know, for two forces to be in equilibrium ?
Well, they have to be equal and opposite.
So, we can calculate the magnitude of the RGT which is
a little bit greater than 800 Newtons, let's say roughly 900 Newtons.
But now, we have to know by where it passes.
Well, what we can see, it is that for two
forces to be in equilibrium, it is then necessary, that these two
magnitudes should be equal. But also that, the line of action should be the
same one. So, that will only be possible if the force
F has the same line of action. I am going to erase the old forces.
It is now necessary that this force S should have the same line of action than the resultant,
what we can obtain in this this way.
And, at that time, we can see in the polygon of forces, that
this force has the same magnitude than the resultant
RGT. And for the equilibrium to be reached,
it is necessary that the polygon of forces
should be closed.
And at that time, well,
the resultant, the vectorial sum is also zero.
What we can notice, moreover, looking again at the drawing, on the right,
it is that these three forces, at that time, the force G, the force T, and the
force S are converging on one point.
We have, thus, two conditions for the equilibrium,
that the vectorial sum is zero, that the
polygon of forces is closed, and that these
three forces are converging on only one point.
Before going further, we want to look at what
happens, if the man, keeping his
feet at the same place, and without lowering his
hands, lowers its body bending the knees.
Thus, he makes his center of gravity go down.
What does it change to the equilibrium ?
Well, you have seen that in the video, I have
nevertheless moved a little bit my arms.
Well, that is not a big deal.
We can imagine that I did not move them.
And then, to simplify the reasoning, we are also
going to admit that the rope was perfectly horizontal.
That was not totally the case, but that was not that far from it either.
In addition to the internal force
in the rope, we have, as applied force
to this free-body, the weight of the person.
So, the person, here, went down.
And then, of course, we have the effect of the ground on the person.
How do we solve this problem ? Well, we just did it.
We have to find, first, the intersection of the line of action of
the internal force in the rope, and of the line of action of the
weight of the person.
So, I trace the line of action of the weight of the person in both free-bodies.
And then, I trace the line of action of the internal force in the string, then, horizontal.
And we obtain two intersections.
Well, since the person has carefully gone down,
keeping his center of gravity on the same line of action,
the lines of action, on the left and on the right,
are identical, and there is no problem.
This is the same point of intersection that we have there.
What's about the feet ? Well, likewise.
The feet having not
moved, the internal force under the feet S, passes by the same place than before.
Let's now look, in the Cremona diagram what, what happens.
So, we have, here, the weight of the person
of 800 Newtons. We have
the line of action of the internal force in
the rope, and then, a parallel, of course, and then
the line of action,
of the action of the ground on the person. Well, we have solved
our problem. Thus, we can, now, trace the
force of the ground S, and the tensile internal force
T that we get. We can see that the person
was less leaned in this
configuration, than he was before.
Thus, the tensile internal force T is smaller than before.
What we can also see, it is that each of these forces, in the polygon of forces,
is parallel to both other forces in the real system.
What does it mean ?
It means that only one polygon of forces
represents the equilibrium of these two configurations.
Thus, the person has lowered his centre of
gravity, but all the rest having remained similar,
well, the internal forces under the feet, in the
rope, and so forth, remained the same.
We can generalize this observation, saying that if we move
a force, not only a
load, on its line of action,
the equilibrium
does not change. Thus, the equilibrium,
and the internal forces in the elements of the structure does not change.
So, that is very convenient, because
that is true that maybe before, you
have asked yourself the question, particularly, I know
that it often happens, when we use the applet.
Is is correct to place the force
with the tip of the vector which is on the center of gravity ?
Should I put the tail of the vector on the center of gravity ?
Should I place the force anywhere else ?
Actually, as we now see, we can freely place this force, higher,
lower, even really lower, or really higher, as long as
the line of action passes by the center of
gravity, as the line of action does not change.
We can make the vector slide as we want to.
We have, here, a very similar configuration, where the man, instead
of pulling on a rope, leans on a column, in front of him.
It is clear that, one more time, without the column, the man would fall down,
since the resultant of his weight is considerably in front
of the position of the action of the ground on the person,
Here, in the corresponding free-body, then, we pass just under the hands,
then under the feet.
We have, acting on this free-body,
the weight G of 800 Newtons.
So, we have seen that we have not really used the solving, but
we can use it to graphic scale, to obtain the various internal forces.
Maybe we are going to do it later.
We are going, consequently, to have a compressive internal force
which acts from the column on the hands, which
I call C. And then, always, the effect of the
ground. So, we do not exactly know where it is.
Actually, we know it, since what we saw before is that it is necessary that the
line of action of the weight, which is,
here, the line of action of the compression in the arms, and the
line of action of the internal force under the feet
converge in only one point. And then, we can see, I had
quite well drawn my force, thus, the, the internal force S.
I can take off
the question mark.
Let's construct the entire polygon
of forces for this person, to scale.
So, we put 800 Newtons.
And we have a compressive internal force. Well, we do not know until where it
goes, but we can trace the line of action. Then,
we will have the internal force under the feet. We know the line of action, and then
now, that we, for the polygon to be closed, we can
accurately trace it. So, this is the internal force S.
And now, we can know the magnitude of the internal force C.
So, we are going to read them. To scale, there is no problem.
Here, we have roughly 38 millimeters, that is to say 380 Newtons.
And then, here, we have 900
Newtons for the internal force in the legs of the persons.
In this video, we have seen how
to solve the equilibrium of a free-body on which act
three forces.
For the free-body to be in equilibrium, we have seen
that it is necessary that the vectorial sum of the three forces should be zero.
In addition, we have seen, and that is another
way to express the fact that it is equal to zero, that the polygon of forces must be closed.
And then, we have seen that for the equilibrium to be possible, it is necessary that
the three forces which act on this
free-body converge in only one point.
