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.
430 Câble avec une charge excentrée
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Из курса от партнера École Polytechnique Fédérale de Lausanne
L'art des structures 1 : Câbles et arcs
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Câbles avec une charge
Les câbles sont le premier type de structure que nous abordons. Il est très important que vous compreniez parfaitement toutes les leçons en rapport avec les câbles, car les concepts et les constructions graphiques que nous dériverons seront réutilisés dans toutes les leçons suivantes du cours.
Nous commençons par le cas le plus simple, celui des câbles avec une seule force. Sur la base des conditions d'équilibre dans le plan, nous obtenons les efforts dans les divers tronçons du câble ainsi que les forces aux appuis. Il est ensuite possible de dimensionner le câble.
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Olivier L. BurdetDr
Laboratoire de Construction en Béton
Aurelio Muttoni
Hello.
In this video, we will see what are the
internal forces in a cable which is subjected to a
load, which acts transversally, but not in
the middle, but which acts in an eccentric way.
In a very similar way that what we
have done in the previous video, we will study
the free-body of the load, we will obtain the
internal forces in the segments of cable, then we will study
the free-bodies of the supports, to obtain the
forces at the supports, and we will
finish by a case a little bit particular
of a force which is not only eccentric, but also inclined.
In this video, you can see what happens, when on
a cable which did not have any loads, we apply an eccentric load.
Once again, as before, the cable takes a rectilinear shape,
between the supports and the applied load. As before, we
construct the structural sketch tracing parallels
to the different segments of cable in
an independent sketch.
So, we have a load of 10 Newtons which acts.
And then, for the record, this segment is parallel to this one.
This segment is parallel to this one. We start
by identifying the free-body including the load and
by drawing it again in an independent way.
We have a piece of cable on the left, another
piece of cable on the right, the load of 10 Newtons,
and the internal force in the left part
of the cable, N1, and in its right part,
N2. We want, in
the Cremona diagram, to draw
with a length of 10 centimeters,
the load of 10 Newtons, equal to 100 millimeters, there.
Then, we trace the parallels to both segments of cable, N2 first,
and N1 afterwards.
So, here, we have an internal force
N2 which is equal to 8.5
Newtons. Then, here, we have the internal force
N1 which is equal to 4.5 Newtons.
Then, we
can take an interest in
the free-body of the support
on the left which
includes a segment of
cable with an internal force of 4.5
Newtons. That is
this internal force N1 along which we travel in the other
direction, and for which we are going
to obtain a vertical force at the
support V1 and an horizontal one
H1. V1 is equal to
2.4 Newtons,
while H1 is equal to
3.7 Newtons.
We can do like before, that is to say
to identify the part of the Cremona diagram which corresponds
to the orange free-body. Likewise, we are going to take a
pink free-body for the support on the right.
We have a segment of cable
on which acts an internal force of 8.5
Newtons. So, we will have the forces
at the support V2 and H2. I should have drawn,
before, V1 and H1 for
the support on the left. In the Cremona diagram,
using the internal force N2 in the other direction, it becomes an arrow with two
heads, we obtain the force H2, whose we
can see that it is equal to H1, thus 3.7 Newtons,
and then, the vertical force on
the right which is equal to 7.6 Newtons,
the sum being equal once again to 10 Newtons,
which is the applied load. We are going
to copy again, I first indicate the free-bodies
used in the Cremona diagram.
So, the sky blue free-body, is the
free-body of the load to obtain N1 and N2.
Then, both free-bodies which have been
used to get the internal forces at the supports.
We copy the results to the structural sketch,
so both segments of cable are in tension.
Thus, we can draw them in red.
The internal force
N2 is equal to 8.5 Newtons in tension.
And the internal force N1 is equal to 4.5 Newtons in tension.
We can see that the method which we used to solve this case
is very similar to the one we have seen in the previous video.
Indeed, we do not have anymore
values of the internal forces in the segments of
cable which are equal, as well as
certain components of the forces at the supports are different.
Let's now look at,
the case a little bit more complicated of a
cable with a load which is neither centered, neither vertical.
So, we have an eccentric load which is a little bit inclined.
We introduce it in our relaxed cable, by means of a dynamometer.
The force which acts is exactly of 10 Newtons.
So, we exactly have the same applied load than before.
However, it does not act
vertically. Here, we have copied the structural
sketch of our configuration. So, we have, here, acting
on an inclined axis, a load of 10 Newtons.
And as before, we take an interest in the
free-body which includes the load.
This free-body cuts the cable on
the right and the cable on the left, and includes
the load of 10
Newtons. We are looking for the
internal forces N1 and N2 which act in
the cable.
We copy the load
of 10 Newtons to scale, that is to say 100
millimeters.
Then,
the parallels to the segment
N2,
then N1.
Which thus gives us, the internal forces N2 and
N1. N2
is equal to 8.3
Newtons and N1 is equal to
4.8 Newtons.
We copy these values in the structural sketch,
also indicating the type of internal force in the structure.
In this case, both elements are in tension.
We now take an interest in the support on the left.
I draw again the corresponding free-body,
with the internal force
of 4.8 Newtons.
And this free-body will be in equilibrium thanks to the
forces at the support V1 and
H1. In the Cremona diagram,
we had used N1 in this direction. We now use it
in the opposite direction, and we close the polygon of
forces thanks to its component
V1 and to the
force at the support H1. H1 is equal to
2.2 Newtons. V1
is equal to 4.2 Newtons.
Likewise, we are going to solve the free-body
corresponding to the support on the right;
with the internal force of 8.3
Newtons in the
cable and the forces at the support V2 and
H2. We reuse
the vector
N2. Then, we introduce
H2, and we finish by the force
at the support V2. H2 is equal to
7 Newtons
and V2 is equal to 4.5
Newtons.
We can note that in this configuration, well,
the sum of the horizontal forces at the support
is not zero, since precisely, there is
a horizontal component in the load which is applied.
You maybe wondered, whether it be in this video, or in the
previous video, why, after having drawn the
load of 10 Newtons, I have systematically always
taken the internal force N2 which was on the right of this load, and afterwards the internal force N1.
And you maybe wondered, why
did not I do the reverse ?
Actually, I already applied, without officially saying it to you,
a convention which we are going to use for the rest of
this course, which is that, when we deal with the forces
which act on a free-body, we take them in a
systematic order which is the counterclockwise order.
So, we turn around the free-body, in the counterclockwise
order, starting by where the forces are known.
So, here, we know the load of 10 Newtons.
Afterwards, we will have the element two which is N2.
And finally, the element three which is N1.
And then, in the Cremona diagram,
this is the first element that we have traced.
Here, the second element.
Here, the third element.
We can, obviously, proceed with the opposite convention.
That is to say, to turn in the clockwise order.
Will the equilibrium be modified ?
No, since regarding the
treatment of each free-body, we insure the equilibrium,
and we can of course ensure the equilibrium by adding first N1, and afterwards N2.
Actually, that is what has been done in this figure, from the book The art of
structures, in which we have dealt with this free-body. And if we closely
look at this, indeed we have firstly dealt with the
force, then the internal force N1 and finally
the internal force N2. So, we have used the force which
is indicated by Q, then N1 and finally N2.
You can see that the diagram is similar
to the one we did before.
The results are obviously right, and are the same.
You can try to do this job by yourself.
To simplify the correction and the communication
within this course, we ask you,
for this course, for the exercises, for your
reasoning, also with the assistants or with your questions, to systematically
use the counterclockwise order, as I have
shown you in this lecture.
Within the framework of this video, we have generalized the approach
method to solve the equilibrium of a cable subjected to a transversal load.
We have first
dealt with the case of an eccentric load, then of an eccentric and inclined load.
We have seen that the order in which
we consider the internal forces and the loads which act
on a free-body, has an influence on the
shape and on the construction of the Cremona diagram.
For the rest of this course, we
will always use the counterclockwise
order for the construction of the Cremona diagrams.
We have seen how to
systematically obtain the internal forces in the segments
of cable, and how to copy them in
the structural sketch to communicate these internal forces.
Finally, we have seen how to obtain the forces
at the supports and their horizontal and vertical components.
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