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.
410 Câble avec une charge centrée
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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
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 lecture, we will see how
a cable can carry a transverse load.
Until now, the loads that we have applied
on the cables acted in the axis of the cable.
Now, we have, as you can see
on the slide of the title, a rope
which is held by two men, on which acts a load in the middle.
In this lecture, we are going to study
the free-body of the load to obtain the internal forces in
both segments of cable on the left and on the right.
We will take an interest in the forces at the supports.
Then, we will make some variation studies about the influence of
the load which is applied and of the geometry of the cable.
In the case which interests us here, we
have a load, which acts in the middle of the cable.
And we are going to identify a free-body
which cuts the cable in two places and which includes the applied load.
I draw again this free-body nearby.
So, we have both segments of cable, the acting load Q
and the free-body cuts the segments of cable in two places.
This implies that, at this place here, I have to replace the cut segment of
cable by the effect which is inside
this segment of cable, or more exactly, the internal force.
Then, not knowing yet this internal force, I can already tell you, that
is not a secret for a cable, that it is a tensile internal force
which acts as until now, along the axis of the cable.
We have here two
internal forces, which I am going to call N1 on the left and N2 on the right.
Let's now look at this little video of a cable first without load on
which we add a load of 10 Newtons. To be able
to work on this structure, it is convenient to model it
replacing the real structure by what is called a structural sketch.
The structural sketch represents, to scale,
the structure with its good proportions,
the loads which are applies and the supports.
So, I copy both segments of cable in
the part on the right, with the correct inclination.
And at their intersection, acts
the load of 10 Newtons.
On the left and on the right, we have elements which are
fixed supports. They are supports which does not
enable any movements.
Here, likewise, in the
structural sketch, we have on the left and on the right a fixed support.
Sometimes the fixed support is also symbolized
in this way, like a small triangle, the support being located at the top of the triangle.
We draw again the structural sketch
here. That is quite easy because within
the framework of this construction, the distance
between the supports, which we
call the span, which is designated by the letter
l, is equal to 1.2 meters. And the
vertical distance between the supports
and the lower point of the cable, which we
call the rise,
f, is equal to 0.6 meters.
Thus, the angle is 45 degrees, which facilitated the drawing.
We have here, acting, a load of 10
Newtons. We first take an interest in the
free-body of the load and we draw it again
below. So, we have a segment
of cable parallel to the piece of cable on
the left, a segment parallel to the one of the cable on the right,
a load of 10 Newtons, and then both
internal forces on the left and on the right which are in
the extension of the cut segments of cable
and as I indicated before, we are going to have N1 and N2.
Let's now continue with the Cremona diagram in which we want to draw
the force of 10 Newtons which acts downwards.
I draw it with a length of 10 centimeters to make a drawing to scale.
So, I have 10 Newtons, equal to, we are going to say 100 millimeters.
For the free-body of the load to be in
equilibrium, it is necessary that the three forces which act
on this free-body then, the load of 10 Newtons, the internal force
N1 and the internal force N2, should be converging on only one point.
Well, that is good, we can see well that indeed,
they are converging on the point of hanging of the load.
Then, on the other hand, that their vectorial
sum should be zero, so we are going to construct the
vectorial sum of the load of 10 Newtons and of both internal forces N1 and N2.
So, I trace straight lines at 45 degrees
in my graph which correspond to the inclination
of the internal forces N1 and N2, thus
this segment here is parallel to this and also to this.
This segment here is parallel to the segment on
the left in the free-body or in the overall system.
I notice,
indeed, that the internal force N2 acts
in the direction I had supposed, that is to say that it pulls
on the free-body. An internal force which pulls on the free-body is
a tensile internal force. And the tension, for us, is red,
then I am going to draw this arrow again,
here, in red. And, since I have drawn to scale,
I can see that the internal force, here, is equal to 7.1 Newtons.
So, as I have drawn to scale, I directly obtain the numerical
value of the internal forces. Likewise,
for the internal force
N1 that I also measure at
7.1 Newtons. Seventy-one
millimeters give 7.1 Newtons. We can
now copy these internal forces in the real system.
That is to say that in the real system, we are going to replace
the basic color of each bar by
the conventional color for the tension, red.
And we will write, we will sketch the internal force drawing
a kind of small piece of cable with an element of tension and then,
here, we will write 7.1 Newtons.
Likewise,
for the segment on the left, also
7.1 Newtons. We now want to take an interest in
what happens near the supports. We are going to
study a free-body which cuts the cable and which
stops just before the support on the left. I draw again this
free-body. Once again, there will be a segment
of cable which is parallel to the segment on the right.
We are going to draw the beginning of the support. What is the internal force in
this cable ?
Well, we know it, it is the internal force N1 which is equal to 7.1 Newtons.
Of course, we could say that the internal force just near
the support is 7.1 Newtons which goes in the other direction.
Actually, in the field of
construction, it is often useful to know the
vertical and horizontal components of the forces at the
supports and that is what we are going to do here.
So, what would interest us, it would be to know the
vertical component here
and the horizontal one of these internal forces.
And this, we are going to be able to do it directly in the Cremona diagram.
In the Cremona diagram, we were taking an interest in the internal force N1.
Now, in the orange free-body, we use the internal force N1
in the other direction, thus we place a head of
arrow at the other end and we take an interest in both
internal forces V and H which are
in equilibrium with this internal force. And it thus corresponds
to a vertical component V and an horizontal component H
of force at the support. We
can directly read their value in millimeters
and see that V is equal to 5 Newtons
and H is also equal to 5 Newtons.
We are going to do the same with the free-body of
the support on the right. I am going to be
very quick here. The internal force
is, of course, equal to 7.1
Newtons. And we have a vertical component
and a horizontal component. We
reuse the vector N2 but in the other
direction which corresponds to our pink free-body.
Then, we will have a
horizontal internal force
and a vertical internal force. We read once again
their value. 5 Newtons for
the vertical internal force and 5 Newtons
for the horizontal internal force. To understand well,
particularly if we want to come back to our solving later, I am going to indicate
in the Cremona diagram, at which
free-body corresponds which part of the construction.
So for the first part of the construction with the force of 10 Newtons and both
internal forces N1 and N2, it was the free-body of the load.
Then, we have had the orange
free-body with the internal force N1 and the
forces at the supports on the left and finally
the pink free-body with the internal force N2 and
the forces at the supports, H and V on the right.
You can note that I used the same symbol for H and V.
I knew beforehand that
these two quantities had the same value, otherwise
I would have used, and we are going to do it
later in other lectures, I would have taken H
left and H right, V left and V right.
We can note that the sum of both vertical forces, 2 times
5 is equal to 10 Newtons, thus equal to the applied load.
That is logical, this applied load must be transmitted to the supports.
What is interesting is to note that the sum
of both horizontal forces, we have an horizontal force on the top of the
Cremona diagram, which points towards the left, and an horizontal force, of
the same value 5 Newtons, which points
towards the right on the bottom. The sum of the horizontal forces is thus
zero, in this case. Last point
of glossary, the shape of the cable which we have thus obtained
is called the funicular polygon.
This is an important expression which will often come
back in the rest of this course.
Well, we have made this construction for a load of 10 Newtons.
Now, what happens, if as in the video
on the left, we add a second force of 10 Newtons ?
So, I propose you to think about it and why not to calculate the results.
What are the internal forces in the cable
under the effect of a load, not anymore of 10 Newtons, but of 20 Newtons ?
I meet you later to discuss about this solution.
Indeed, we have noted, when I added a second load,
on the system, that the shape of the system did not change.
Thus, the construction that we have made before
with 10 Newtons, we are going to be able to make it again.
I am going to make it again quickly.
Now, with a force
of 20 Newtons. Then,
we trace a straight line parallel to the first segment of cable.
A straight line parallel to the second segment
of cable. And, we can directly read the
internal forces, which not surprisingly, are simply doubled.
Of course we could have spared ourselves this
extra work, simply saying, well, the drawing
which I made before for
10 Newtons, I am going to use it for 20 Newtons.
If now I say that here, it is equal to
20 Newtons, it just changes the scale of the drawing and I can read
seventy-one millimeters which would correspond to 14.2 Newtons.
It would be, of course, much more reasonable.
If some of you said to themselves
that we could proceed in this way, well, congratulations, you have well understood how this works.
We now want to look at another
parameter which has a large influence.
So, I introduced to you already the concept of rise
which is the height between the supports and the lower point of the cable.
What happens if the rise changes ?
We have already solved the
basic configuration with a cable at 45 degrees which I draw in
red, and whose the Cremona construction is
already available on the right. Now, we want to look at
a variant with half the rise.
How am I going to proceed for the solving ?
Well, exactly in the same way than before.
So, I am going to make a free-body around the load of 10 Newtons.
I am going to cut a segment of cable on the left, a segment of cable on the right and
in the Cremona diagram on the right, I am going to copy parallels
to both segments of cable to close the polygon
of forces, which expresses the equilibrium. So, here I make the polygon of forces
which corresponds
to the pink configuration. And we can
note that the internal forces increase.
Here, it is equal to 11.2 Newtons.
For the record, in the red configuration, we had 7.1 Newtons.
We did not change the load, we just decreased the rise by half.
What happens if we are even more drastic and
that we divide the initial rise by four ?
Well, we obtain this red configuration where
the cable
is very horizontal.
Once again, we are going to make the same
construction, we trace the parallel to the orange segment.
And, likewise, the
left part.
And this time, the internal force reaches
20.6
Newtons. I am going to introduce now
the concept of slenderness, which is
equal to the ratio between the span and the rise,
it is something which is important, which we will see again later.
And what we can see is that if the ratio between
the span and the rise increases, that is to say either if the span
increases and the rise does not change, or, like in our case, if the span
does not change but the rise decreases, so, the internal forces increase.
Conversely, if the slenderness decreases, the internal forces decrease.
We will see that it is not a linear increase but
that it is an increase which can sometimes be very sensitive depending on
the type of structure we consider.
In this lecture, we have looked at the structural system which
applies for a cable subjected to a single load.
We have studied the free-body of the load,
we have obtained the internal forces in the various segments of cable.
We have seen that they were tensile internal forces.
We have also studied the internal forces which act
at the supports and the influence of the slenderness and we have
seen the relationships which exist between the load, if the load increases,
the internal forces increase, and the slenderness, if
slenderness increases, the internal forces also increase.
