0:05

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.

1:19

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.

2:49

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.

9:49

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.

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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.

14:27

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.

16:13

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.

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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.