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.
240 Dimensionnement
Loading...
Del curso dictado por École Polytechnique Fédérale de Lausanne
L'art des structures 1 : Câbles et arcs
24 calificaciones
De la lección
Matériaux et dimensionnement
Cette semaine vous présente les principales propriétés des matériaux utilisées pour la construction des structures. Les concepts de dimensionnement et de coefficient de sécurité sont introduits.
Conoce a los instructores
Olivier L. BurdetDr
Laboratoire de Construction en Béton
Aurelio Muttoni
Hello, in this video, we will deal with dimensioning which is
the operation of choosing the dimensions to give to an element
of structure, in such a way that it can resist the internal forces
to which it is subjected, and not to deform too much.
The principles of, dimensioning are,
then, one the one hand to ensure the structural safety,
that is to say that the element of structure should not break under a normal use.
On the other hand, that its use, also,
can be made in a comfortable way.
For example, without exaggerated deformations or vibrations.
And then, finally, we wish to build
structures which are economical, which
do not uselessly waste materials, and
which can also behave in a sustainable way.
We have two fundamental components
in the operations of dimensioning.
On the one hand, the service limit state.
You can see in the illustration, a roof which, which is not flat,
and for which there is no water drainage system which is foreseen.
Well, the result, it is a puddle of water
which stagnates, which is not aesthetically pleasing, in which, maybe,
mosquitoes will be able to develop, and which has
clearly negative effects on the quality of the behavior.
So, we want to limit the deformations.
And on the other hand, as I already said, it is not
something we will deal with within the framework of this course.
But, the service limit state is also linked to the limitation of vibrations.
For example, if we build a footbridge, and if people are scared
to step onto this footbridge because it vibrates
too much, well, we will have not built a
structure which is satisfactory. The ultimate limit
state ULS (in French ELU), while the service limit state is abbreviated SLS (in French ELS).
The ultimate limit state deals with the structural safety.
So, we absolutely have to avoid a failure of the structure.
In the picture you can see, below, there
were two children in the back of the car.
This is something that we cannot tolerate.
We do dangerous jobs, when
we deal with constructions.
And it is unacceptable that a structure which
we have designed behave in such a way.
So, we absolutely have to avoid failure, and it means
that we will have to think about failure scenarios.
It can be, for example, the impact of a truck on the column of a building.
To ensure that even in a such case, there will not be a failure and that
people who have not been involved are injured or worse,
in an accident. So, we want a very, very low
probability that a failure occur.
It is quite radically distinguishable, by the way, from the service limit
state, because, we talk about deformations.
Well, does it deform too much, or just a little
bit too much ?
This is something which is probably debatable.
However, failure is not debatable.
We want, in what follows, to take
the example of an elevator, because it is a simple
example, which corresponds to the knowledge, on
structures and on internal forces that we have learned up to now.
So, on an elevator, we have a cable
which links the elevator to the upper part of the building.
And this cable,
is used both to make the elevator go up and go down, but
also to support it, and thus, to insure the safety of its occupants.
Well, let's think for a while, what does it
mthe service limit state mean for an elevator ?
Well, it clearly means to be able to easily enter in
and exit from the elevator.
Let's say that if when the elevator is full and that it
arrives at a given floor, it is two or three centimeters too low, there is
a significant risk that the people are going to trip over
the ground at the exit, and this, it is
something which is not good for the usage.
So we want that the elevator arrive quite accurately at the level of the floor.
Conversely, we absolutely want to avoid, for the ultimate limit state,
we absolutely want to avoid failure.
That is to say, the fracture of the cable, and obviously,
with as consequence, the risk of fall of the elevator.
So, it is clear enough.
Now, we are going to see, how we can,
with the tools that we know, solve this problem.
Let's start by the service limit state. I remind you that it is abbreviated ELS.
We have, here, put some data about this elevator.
First, we have two elements in the elevator, the self-weight
G which corresponds to the self-weight of the cabin.
Its value is given as 10 kiloNewtons.
That is to say that the maker told us,
well, the elevator that we are going to deliver,
will weigh 10 kiloNewtons. And then, there is the number of persons
that we can put in the elevator, who are 10 persons of 80 kilograms, it
corresponds to 800 kilograms. That is to say 8 kiloNewtons of force.
Let's now think about the case of the cabin.
It is very easy to adjust the cabin when it
is empty, for it to stop, exactly at the level of the floors,
isn't it ?
What we wish, is now that, if this, if in
this empty cabin, we add 10 persons, then, 8 kiloNewtons.
Well, this cubicle, goes down to not more, to not more than delta l equal to 10 millimeters.
We consider than 10 millimeters, it is acceptable.
People are going to raise a little bit their foot.
And if the cabin is too high, or too low of 10 millimeters,
then there will not be any problems for the use of the elevator.
Now, we need another data.
It is the length of the cable, and for this
exercise, we are going to take a length of 50 meters.
Then, when the cabin is in the
stationary position and it is empty, it is at position 0.
And then, it is going to go down of delta l,
when we are going to place the loads Q inside.
I put only one man.
Obviously, there would be ten persons.
And delta l, we know, has to be less than or equal to 10 millimeters.
How can we solve this ?
Well, using the formulae which we have seen previously.
We know
that epsilon is equal to delta l over l.
Then, delta l, we know it, not more than 10 millimeters.
l, 50 meters.
Thus, we can calculate epsilon.
What we know, on the other hand, it is that E the modulus of elasticity, our cabin,
I forgot to mention it, is made by steel.
Then the modulus of elasticity
is equal to 205 000 Newtons per square millimeter.
As we know, it is equal to sigma over epsilon.
Then epsilon, we know it, we just have seen it before.
But, we do not know sigma. Then, we can get sigma
equals to E times epsilon equals
to E times delta l over l.
Sigma being the stress in the cable.
How can we get it ? The stress in the cable is, is equal
to the internal force in the cable, divided by the section of the cable.
And it is thus equal
to E times delta l over l. What is the value of N ?
In our case, N is equal to the load of the occupants.
That is to say the 8 kiloNewtons that we
are going to add from the stationary position.
The weight of the cubicle does not play a role in this, in this case.
Since we know N, what, what are we looking for ?
We are looking for A, we get
that A is equal to N over E
times L over delta l. Here, I believe that we know everything.
Thus, we can substitute. N is equal to 8,000 Newtons.
We have to be careful, with the units. I am going to do
everything, in Newtons and in millimeters. The length is thus equal to 50,000 millimeters
divided by the modulus of elasticity of steel which is equal to
205,000 times the limit of 10 millimeters.
And all this, if I calculate it,
it is equal to 195 square millimeters
Then, how to get
the diameter of a cable whose we know the area ?
So, everyone knows the area of a cable, as pi r squared.
For us, it is much easier, for the
rest of this course, to use a very
similar formula which says that the area of a circle
is pi times the diameter squared, divided by 4.
If you try it mathematically, you will see that
it is equal to pi r squared, we do not use the radius, because we cannot
really measure the radius easily. So, what does it mean ?
It means that D is equal to the square root of 4 times A over pi.
And this number is roughly equal to 16 millimeters.
Thus, for the service limit state, it is necessary
that the diameter of the cable of the elevator should be 16 millimeters.
We can now deal with the ultimate limit state.
You remember of the idealized stress-strain
diagram, with, here, the value fs for steel.
But it is clear that when we buy steel,
we do not know if it is going to be exactly as
good as it should, as it should be.
On the other hand, it is very true within the framework of the case of the elevator.
Maybe it is going to wear out, with time.
Then, what we are going to do, it is that the
dimensioning is going to take place at a lower level.
We are going to consider a material which has the same modulus,
but whose the strength fsd. The design
strength fsd, is thus equal to the strength of the material,
divided by a certain gamma m, which is a partial safety
factor,
which is given by the standards, with
values which vary according to the material. For the case of the example we
are now dealing with, we are going to use
the value fsd of 80 Newtons
per square millimeter. This value is quite low, compared
to the steel strength.
We will see the steel strength a little bit later.
But it is totally justified, because it
is going to be used for an elevator, where
the wear and tear, the repeated loads, the impacts, etc, need to be taken into account.
Thus, we have to beware of the materials, but who has never tried to
sidle as the ninth person in
an elevator, which can only contain eight persons?
Or even after the ninth,
to try to be the tenth one, or the eleventh one?
We know that human beings, but also loads in general
do not tend to respect what is written in the standards.
So, we are going to beware of the
loads, such as they are prescribed or indicated to us.
For example, in the case of the elevator, well, the maker told
us than the cabin was going to weigh 10 kiloNewtons.
It is maybe
right, but maybe he has made a
little mistake, that he is going to have to use a sheet metal
a little bit thicker, and that suddenly, it will weigh a little bit more.
Or else, during the life of the elevator, we are going to
have to add, aboard, a fire extinguisher which will also weigh a certain weight.
Or else, we will want to change the
decoration, to replace aluminum panels, maybe
by wood slabs, which can be thick, and then very heavy.
In short, we cannot trust, neither the loads
which are given to us, nor the users.
And the way in which we do this, it is
that we are going to work with a design load
which is obtained by punishing the loads
which are given to us. Then, the design load Qd
logically should be the sum of
the self-weight G and of the live load Q. But in
fact, we are going to punish G, by a factor
gamma G, and the load Q by a factor gamma Q, to take into account
of the fact that we are not totally sure of what is going to happen.
Gamma G is given by the standards,
and in a general way, in the European standards, it
is kept everywhere at 1 point 35 and likewise 1 point 5 for gamma Q.
Now, I would like nevertheless to point out that it is not,
because you know that, that you have the right
to be the fifteenth one to enter in an elevator for 10 persons.
Thus, at the ultimate limit state, we have the design
load Qd which is equal to 1.35 times the 10 kiloNewtons
of the weight of the cabin, plus 1.5 times
the 8 kiloNewtons of the weights, of the weight of the occupants
which is equal to 25.5 kiloNewtons. Thus, we have here, 1.5 Q,
1.35 G. And if we put everything together,
we get
a big load Qd. For our
free-body to be in equilibrium, it is necessary to have, here, a tensile internal
force Nd in the cable. And we
need that Nd be
necessarily equal to Qd, for the free-body to be in equilibrium.
Let's look, now, to the side of the strength.
The required strength, it is R which has to be
greater that the internal force Nd.
How do we get the strength ?
Well, the strength is equal to the area of the section of our cable
times fsd which is the design
strength, as I told you before.
For our case, we are going to take 80 Newtons per suqare millimeter.
Thus, we must have R which has to
be greater than or equal to Qd.
We are going to work with the equal part,
but obviously it will be necessary that the section, respectively,
the diameter of the cable which we will obtain,
should be at least the value that we are going to obtain.
We know fsd. Thus, we can write that A
required is equal to or greater than Qd over
fsd, where Qd is equal to 25,500
Newtons. And fsd is equal
to 80 Newtons per square millimeter
That is to say
319 square millimeters. Thus, the diameter
which interests us, it is 4 times 319 divided
by pi, square root of this number. It is roughly equal to 20 millimeters.
Then, now, we have two diameters
which we have to respect for our cable.
On the one hand, for the service limit state, 16 millimeters.
On the other hand, for the ultimate limit state, 20 millimeters.
So, it is always the biggest one of
the two diameters that we are going to take, of course.
Thus, in our case, we will need a cable with a diameter
equal to or greater than 20 millimeters, to be able to support our elevator.
I should clarify that it is more or less by
chance, that in our case, it is the ultimate limit
state with 20 millimeters which controls instead of
the service limit state with 16 millimeters.
These two calculations are independent.
They must be done each time, and it is only
once we have the result that we will be able to know or to decide
which one of the two, which one of both conditions controls the
dimensioning, in our case, once again, the ultimate limit state.
In this lecture on dimensioning, we
have begun by the criteria of dimensioning.
We have seen that there were criteria for the service limit state
ELS (SLS in English), and then other criteria for the ultimate limit state ELU (ULS in English).
We have seen that for this last limit state,
it is necessary to introduce partial safety factors.
Gamma m for the materials, or gamma G and gamma Q for the loads.
We are now going to proceed to exercises,
within the framework of this lecture.
Explora nuestro catálogo
Inscríbete de manera gratuita y obtén recomendaciones personalizadas, actualizaciones y ofertas.
Coursera brinda acceso universal a la mejor educación del mundo, al asociarse con las mejores universidades y organizaciones, para ofrecer cursos en línea.
© 2019 Coursera Inc. Todos los derechos reservados.
