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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.

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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.

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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

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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.