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Hello.

In this video, we will see what are the

internal forces in a cable which is subjected to a

load, which acts transversally, but not in

the middle, but which acts in an eccentric way.

In a very similar way that what we

have done in the previous video, we will study

the free-body of the load, we will obtain the

internal forces in the segments of cable, then we will study

the free-bodies of the supports, to obtain the

forces at the supports, and we will

finish by a case a little bit particular

of a force which is not only eccentric, but also inclined.

In this video, you can see what happens, when on

a cable which did not have any loads, we apply an eccentric load.

Once again, as before, the cable takes a rectilinear shape,

between the supports and the applied load. As before, we

construct the structural sketch tracing parallels

to the different segments of cable in

3:07

and N1 afterwards.

So, here, we have an internal force

N2 which is equal to 8.5

Newtons. Then, here, we have the internal force

N1 which is equal to 4.5 Newtons.

Then, we

can take an interest in

the free-body of the support

on the left which

includes a segment of

cable with an internal force of 4.5

Newtons. That is

this internal force N1 along which we travel in the other

direction, and for which we are going

to obtain a vertical force at the

support V1 and an horizontal one

H1. V1 is equal to

2.4 Newtons,

while H1 is equal to

3.7 Newtons.

We can do like before, that is to say

to identify the part of the Cremona diagram which corresponds

to the orange free-body. Likewise, we are going to take a

pink free-body for the support on the right.

5:16

We have a segment of cable

on which acts an internal force of 8.5

Newtons. So, we will have the forces

at the support V2 and H2. I should have drawn,

before, V1 and H1 for

the support on the left. In the Cremona diagram,

using the internal force N2 in the other direction, it becomes an arrow with two

heads, we obtain the force H2, whose we

can see that it is equal to H1, thus 3.7 Newtons,

and then, the vertical force on

the right which is equal to 7.6 Newtons,

the sum being equal once again to 10 Newtons,

which is the applied load. We are going

to copy again, I first indicate the free-bodies

used in the Cremona diagram.

So, the sky blue free-body, is the

free-body of the load to obtain N1 and N2.

Then, both free-bodies which have been

used to get the internal forces at the supports.

We copy the results to the structural sketch,

so both segments of cable are in tension.

Thus, we can draw them in red.

The internal force

N2 is equal to 8.5 Newtons in tension.

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And the internal force N1 is equal to 4.5 Newtons in tension.

We can see that the method which we used to solve this case

is very similar to the one we have seen in the previous video.

Indeed, we do not have anymore

values of the internal forces in the segments of

cable which are equal, as well as

certain components of the forces at the supports are different.

Let's now look at,

the case a little bit more complicated of a

cable with a load which is neither centered, neither vertical.

So, we have an eccentric load which is a little bit inclined.

We introduce it in our relaxed cable, by means of a dynamometer.

The force which acts is exactly of 10 Newtons.

So, we exactly have the same applied load than before.

However, it does not act

vertically. Here, we have copied the structural

sketch of our configuration. So, we have, here, acting

on an inclined axis, a load of 10 Newtons.

And as before, we take an interest in the

free-body which includes the load.

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with the internal force of 8.3

Newtons in the

cable and the forces at the support V2 and

H2. We reuse

the vector

N2. Then, we introduce

H2, and we finish by the force

at the support V2. H2 is equal to

7 Newtons

and V2 is equal to 4.5

Newtons.

We can note that in this configuration, well,

the sum of the horizontal forces at the support

is not zero, since precisely, there is

a horizontal component in the load which is applied.

You maybe wondered, whether it be in this video, or in the

previous video, why, after having drawn the

load of 10 Newtons, I have systematically always

taken the internal force N2 which was on the right of this load, and afterwards the internal force N1.

And you maybe wondered, why

did not I do the reverse ?

Actually, I already applied, without officially saying it to you,

a convention which we are going to use for the rest of

this course, which is that, when we deal with the forces

which act on a free-body, we take them in a

systematic order which is the counterclockwise order.

So, we turn around the free-body, in the counterclockwise

order, starting by where the forces are known.

So, here, we know the load of 10 Newtons.

Afterwards, we will have the element two which is N2.

And finally, the element three which is N1.

And then, in the Cremona diagram,

this is the first element that we have traced.

Here, the second element.

Here, the third element.

We can, obviously, proceed with the opposite convention.

That is to say, to turn in the clockwise order.

Will the equilibrium be modified ?

No, since regarding the

treatment of each free-body, we insure the equilibrium,

and we can of course ensure the equilibrium by adding first N1, and afterwards N2.

Actually, that is what has been done in this figure, from the book The art of

structures, in which we have dealt with this free-body. And if we closely

look at this, indeed we have firstly dealt with the

force, then the internal force N1 and finally

the internal force N2. So, we have used the force which

is indicated by Q, then N1 and finally N2.

You can see that the diagram is similar

to the one we did before.

The results are obviously right, and are the same.

You can try to do this job by yourself.

To simplify the correction and the communication

within this course, we ask you,

for this course, for the exercises, for your

reasoning, also with the assistants or with your questions, to systematically

use the counterclockwise order, as I have

shown you in this lecture.

Within the framework of this video, we have generalized the approach

method to solve the equilibrium of a cable subjected to a transversal load.

We have first

dealt with the case of an eccentric load, then of an eccentric and inclined load.

We have seen that the order in which

we consider the internal forces and the loads which act

on a free-body, has an influence on the

shape and on the construction of the Cremona diagram.

For the rest of this course, we

will always use the counterclockwise

order for the construction of the Cremona diagrams.

We have seen how to

systematically obtain the internal forces in the segments

of cable, and how to copy them in

the structural sketch to communicate these internal forces.

Finally, we have seen how to obtain the forces

at the supports and their horizontal and vertical components.