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.
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From the course by École Polytechnique Fédérale de Lausanne
L'art des structures 1 : Câbles et arcs
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From the lesson
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.
Meet the Instructors
Olivier L. BurdetDr
Laboratoire de Construction en Béton
Aurelio Muttoni
Hello.
In this video, we are going to study the effect of friction, and to consider
how a body can stay in equilibrium while it is on an inclined plane.
We are going to look at what is the effect
of the operating surfaces, what is the influence of the inclination.
We are going to look at, in which case there will be a sliding,
and in which case there will be equilibrium.
Obviously, it is the case which interests us
more, since our structures must be in equilibrium.
Let's start by considering our test bench.
This test bench will be characterized at each measure by
the maximal inclination, which will be tolerable for a surface condition.
That is to say for which angle
the block that we can see, will not slide on the surface.
We can note that there are two tracks on our
test bench, one Teflon track, on the left, white.
It is a synthetic material whose the
property is to offer very little sliding.
And, on the right, we simply have a
wood track that we will use to compare to.
And, you can see that
the block has several surfaces which
will enable us to study several surface conditions.
We start with the block, whose the lower face
is in Teflon, which slides on the Teflon track.
This is the condition where there will be the least friction.
Indeed, we only get 9 degrees of possible inclination.
When we make sliding the slab, the lower face
of the wood block, on the Teflon
track, we get an inclination of 11 degrees.
When we make sliding the block with the lower
face in rubber on Teflon, we reach 13 degrees.
Finally, if we put
the block whose the lower face is in sandpaper
on the Teflon track, we
reach a much larger angle of 34 degrees.
Let's look now,
quickly, at what happens when we make sliding on the wood track.
The Teflon block, with the lower part in Teflon on the
wood track, is able to stay in equilibrium with an angle of 15 degrees.
Wood on wood, we get 18 degrees.
Sandpaper on wood, we get 41 degrees.
Then, the best result is rubber on wood with 45 degrees.
Well, maybe you now understand why, when we make
hike in the mountains, we use
shoes with rubber soles.
It does not work very well on Teflon, but it
works very well when the surface is a little bit rougher.
We are going to summarize all the results in this table.
So, for the Teflon track,
we got an angle of 9 degrees with Teflon, of 11 degrees with
wood, of 13 degrees with rubber, and of 34 degrees with sandpaper.
For the wood track, we got 15 degrees with Teflon.
We can see that it is not the same value than with wood on Teflon.
Why ?
Because it is not exactly the same wood.
If you remember,
to have seen the wood block which was in a quite hard wood, while the track
was made by a wood a little bit softer. Wood on wood, we get 18 degrees,
sandpaper 41 degrees, and rubber 45 degrees.
We will come back, later, to this table to complete it.
We want, now, to look at the equilibrium of the block,
on an inclined plane.
So, it is subjected to gravity, to its weight G.
Then, there is a contact force which acts under the
structure and a friction force which prevents it or not to slide.
To solve this question of equilibrium, we simply
pass, a free-body under our block.
Thus, we include, all the weight, and
we are going, thanks to the free-body, to reveal
the contact force and the friction
force which act just under the block.
So, here, I draw again the free-body, on the right.
The gravity force which I had already drawn, it obviously acts in this way
we have G,
the weight of the block. The contact force just acts at
the interface, here, between the block and the inclined plane.
It is perpendicular to this inclined plane, and its magnitude
is equal to the projection of G on this perpendicular,
to the surface. Well, I am going to draw, here, G.
I represent it by a noncontinuous line, because G does not really act here.
When I am going to make its projection, on the perpendicular to the surface.
Well, we project it in this way, perpendicularly.
And I can read that this contact force C,
it has this magnitude.
So, we have here, again, a perpendicular.
And the third force which acts on this body, it acts, here.
It is the friction force.
Well, we can notice one thing, it is
that these two forces, the friction force
and the contact force, they intersect, here.
And there is no intersection with the weight.
Thus, we do not have what we had to get.
What we must get, to have the equilibrium,
is an intersection of the three forces on only one point.
This is a little mistake.
Note that the complete solving would not be extremely
complicated, it is simply not useful, at this stage.
What we are going to do, for the rest of this course, is that we are going to consider
a slab much thinner.
Thus, we are going to make a much smaller mistake.
In theory, we would need a slab with an infinitesimal
thickness, then we would not make any mistake anymore.
So, we have G which acts on the slab. We have the contact force.
Then, I am going to draw it on the center of gravity, as if it acted
on the center of gravity.
It is not totally right, but
the mistake is quite small, the friction force.
Now, let's look in the polygon of forces what it means.
We copy the force corresponding to G.
Then,
we add the force
corresponding to C, to the contact
force. So, the contact force, it
has a length which goes until the projection
of G on the vertical. You can see, that is approximately right.
If now, we trace the resultant
of these two forces. what we can notice is that this
resultant is here. It is parallel to the slope.
It is Rcg.
And it is a force which pushes our body downwards.
Well, that is this force which is going to make our body sliding.
If there was not any friction, our
body would slide, for example, like a skier on
a slope.
There is a very little friction, when we ski.
Thus we have, as result, a driving force.
What happens ?
This is the case which interests us, when the body does not slide.
Well, when the body does not slide, it means that there is
a friction force which acts in the upwards direction.
I draw it a little bit staggered,
simply for us to be able to see it.
Here, a friction force, which
opposes to the sliding, and which closes the polygon of forces.
Thus, our body is in equilibrium.
Let's now look at what are the reasons for
a body to slide, or for a body to be in equilibrium.
So, we will have
a body in equilibrium, if
the magnitude of the
resultant of C and G is less than or
equal to the magnitude of the maximal friction
force. Why the maximal friction force ?
Well, imagine precisely that we have the case of
the rubber block which can hold until 45 degrees.
If we put it on a slope of only
20 degrees, it does not mobilize all the force.
And it is not the friction force of 45 degrees which acts.
Because if this friction force would act,
then, our body would go up the slope thanks to the friction.
That is not right.
What is right, is that the friction force adapts to correspond
exactly to the force which is necessary not to have sliding.
This, until the maximum value that it can reach.
We have seen that for Teflon on Teflon, the force would not be large.
However, when we have rubber on wood,
well, the force is quite large.
We often write this value of the
maximal friction force, as (Greek) nu times
the magnitude of the contact force. We have, here,
a perpendicular. And, we can recognize,
here, the inclination angle alpha, that we find again,
here, in this construction. And we can notice that the
maximum friction
force will be
equal to the contact force
times the tangent of alpha max, which
is the maximum angle that we have been able to measure.
So, in other words, nu equal to the tangent
of alpha max. Let's come back to this table, to complete it.
We are going to add here, each time, alpha max.
Then, here, we are going to say nu equal to the tangent of alpha max.
And we have the values, here, zero point 158,
zero point 194, zero point 674,
zero point 231, zero point 268,
zero point 325, zero point 869,
and one point zero in this case, of 45 degrees.
In this video, we have seen the reasons for
which a body slides or not on an inclined plane.
That is that the friction. There is
no sliding if the maximum friction which can be mobilized
is larger than the driving force resulting
from the weight and from the contact force.
We have seen that the friction coefficient nu can
be deducted from a simple experiment taking the tangent of
the maximal possible angle of inclination.
We will not really come back on friction thereafter,
but it is important to understand well that
these friction forces do not have a nature
fundamentally different from the other
forces, with which we will be confronted.
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