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.
470 Câble avec deux charges symétriques
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En provenance du cours de École Polytechnique Fédérale de Lausanne
L'art des structures 1 : Câbles et arcs
24 notes
À partir de la leçon
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.
Rencontrer les enseignants
Olivier L. BurdetDr
Laboratoire de Construction en Béton
Aurelio Muttoni
Hello, this video
deals with the case of a cable subjected to two symmetric loads which act on it.
We are going to solve this case identifying
and solving the equilibrium of each load
in an individual way using an appropriate
free-body, then we will combine these results together.
We will get
the internal forces in each segment of cable and we will also obtain
the forces at the supports at the ends of the cable.
Here, you have a cable without any loads on which
we simultaneously hang up two loads at predetermined positions.
As you can see, the cable take a trapezoidal
shape with the supports on the upper
part of the trapezium and both loads on the lower part.
We have, on the left, a structural sketch representing
our structure and on the right, we are going to construct
a Cremona diagram combining
the various internal forces in a single graphical construction.
We firstly identify a free-body around the force on the left.
We can draw again this free-body just
below, with an horizontal segment,
an inclined segment, the load of 10
Newtons and, acting on the
ends, on the edge of the free-body, the internal forces in the
cables which we wish to determine.
In the Cremona diagram, we can draw this force of 10 Newtons.
Then, on this line of action,
we know that the horizontal load
will act, and finally, in parallel
with the segment of cable on the left, we can copy
the force which goes up towards the
left. well, we can
now finish the equilibrium
of the blue free-body on the left, directly
reading the value of the internal forces, here 7.4 Newtons in
the horizontal segment of cable. We have
tension here. 7.4
Newtons. Then, 12.4
Newtons in
the inclined segment on the left.
I identify the free-body
which has given us this construction, with
sky blue. We now want to study
the force,
the load on the right with this pink free-body which I also draw
again below,
with a horizontal segment, an inclined segment and the
load of 10 Newtons. Here, we have seen in
the blue free-body that this internal force had a magnitude of 7.4 Newtons.
Here, it will be the same in the other direction.
We already know it, here, we have 12.4 Newtons.
And here, we will have a tensile internal force whose we do not know the magnitude yet.
But we already know its inclination. Let's now consider the Cremona diagram, we
draw the second force of 10 Newtons.
In this Cremona diagram, we are going to actually start with
the internal force of 7.4 Newtons acting horizontally.
Then, we will have a load of 10 Newtons and we are going to close the polygon of forces
corresponding to the pink free-body, here, by this internal force.
Then, if we measure it, we are going to see that it has the same magnitude,
it also has a value of 12.4 Newtons. That is quite logical since our
structure is symmetric. I identify the contribution to the
Cremona diagram of the pink free-body. We now want to
take an interest in the free-body concerning the supports.
Well, I draw again this free-body
independently. We have a piece of cable, the end
of the cable acting on this free-body, there is of course
the internal force of 12.4
Newtons which pulls this support
downwards. In the Cremona diagram, we now reuse
the internal force of 12.4 Newtons pulling downwards then,
we finish
the construction with the vertical component on the left
and the horizontal component on
the left. We can see that this horizontal
component is also equal to 7.4 Newtons and the vertical
component is equal to 10 Newtons. We identify
the contribution of the free-body
to the global construction, and we now consider
the support on the right with a yellow free-body
with a piece of cable and
the end. The internal force is equal
to 12.4 Newtons. I did not copy it
here, also 12.4 Newtons and I did not copy it
in the free-body either. We have a tensile internal force
of 12.4 Newtons. On the yellow
free-body acts the internal force of 12.4
Newtons, going down towards the left. Then,
the horizontal internal force on the right
which is equal to 7.4 Newtons.
Then, the vertical internal force on the right
which is equal to 10 Newtons. I also
identify the contribution to
the Cremona diagram, of the yellow
free-body. If we apply a second
load on our system, actually it does not change its shape, but now, as
it was the case for a system with only one load, the internal forces are doubled.
Thus, we would just have to take again the Cremona diagram which we have previously
made and to read it with a
doubled scale to directly get the internal forces.
Or, of course, we could integrally reconstruct it.
Let's now deal with a practical exercise of dimensioning of the cable.
It is now a cable on which acts not
anymore 10 Newtons, but a permanent load of 12 kiloNewtons,
12000 Newtons so it is a quite significant
load on which is added a variable load of 8 kiloNewtons.
Let's see how to proceed to obtain the diameter of the cable
knowing that the strength of the cable fsd is 320 Newtons per square millimeter.
We first have to get the design load, Qd, which acts on this
cable. Qd is equal to 1.35
times G plus 1.5
times Q or for us, 1.35
times 12 plus 1.5 times 8,
that is to say 28.2 kiloNewtons.
We are now going to have a design load,
I draw it here in pink, a little bit staggered, Qd.
By the way, I should not have moved it, we are going
to draw it over, here, Qd equal to 28.2 kiloNewtons.
Naturally, we could made a new Cremona diagram
for the load of 28.2 kiloNewtons but it is not necessary since we
have already the solution for 10 Newtons, and therefore we now want
to multiply everything by the ratio between 28200 Newtons
and 10 Newtons to directly obtain our internal forces.
The larges internal force, we have obtained it before in both inclined
parts of the cable, it was 12.4 Newtons
for an applied load of twice 10 Newtons.
Now, we want to see what is the
result for 28.2
kiloNewtons and actually, because we do not want to have thousands of Newtons, we
would be interested to have this answer in kiloNewtons, at least for the moment.
This answer is simply going to be 12.4
divided by 10, this is a proportion, times 28.2.
And this, it gives us
35 kiloNewtons. Thus the internal force in these
two segments of cable, for the dimensioning,
we have Nd is equal to 35
kiloNewtons. We now want
to do the dimensioning of the cable. The dimensioning is
the choice of the dimensions, actually we
want to know the diameter of the cable. We
have then Nd is equal to
35 kiloNewtons. That is to say
35000 Newtons and we have a material which
resists to 320 Newtons per square millimeter thus the minimal required
area is 35000 divided
by 320, that is to say
109 square millimeter, so
the diameter is still the square root of 4 times
A divided by pi : it is equal
to square root of 4 times 109 divided by
pi, that is to say 11.8 millimeters,
thus we choose a cable of 12
millimeters diameter which will be adequate
to resist to the design internal forces.
We also could obviously do the verification
of the transitional part of the cable, you
know, the one which had 7.4 Newtons for an applied load of twice 10 Newtons,
but actually, this cable would have a smaller diameter, that is why
we only did the dimensioning for the part with the largest internal force.
In this video, we have seen how
to solve the equilibrium of a cable subjected to two symmetric
loads, we have proceeded expressing in the Cremona
diagram, the equilibrium of each free-body around
the load on the left and around the load on the right.
We have seen how to obtain the internal forces in each segment of cable,
then we have proceeded to the dimensioning performing a
load combination to get the design load.
Thereafter, a cross-multiplication, a
proportion to obtain the design internal force
in the cable and finally we have been able to determine the diameter of the cable.
During the construction, we have also determined
the forces at the supports in a very
similar way to what we did before
for the case of the cable with only one force.
A question which maybe
came to your mind is, but, what happens
if I do not know the geometry of the cable ?
Because that is true that until now, we have had photos of the geometry
of the cable, well, we are going to take an interest
in this question in the following lectures.
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