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Hello, this video

deals with the case of a cable subjected to two symmetric loads which act on it.

We are going to solve this case identifying

and solving the equilibrium of each load

in an individual way using an appropriate

free-body, then we will combine these results together.

We will get

the internal forces in each segment of cable and we will also obtain

the forces at the supports at the ends of the cable.

Here, you have a cable without any loads on which

we simultaneously hang up two loads at predetermined positions.

As you can see, the cable take a trapezoidal

shape with the supports on the upper

part of the trapezium and both loads on the lower part.

We have, on the left, a structural sketch representing

our structure and on the right, we are going to construct

a Cremona diagram combining

the various internal forces in a single graphical construction.

We firstly identify a free-body around the force on the left.

We can draw again this free-body just

below, with an horizontal segment,

an inclined segment, the load of 10

Newtons and, acting on the

ends, on the edge of the free-body, the internal forces in the

cables which we wish to determine.

In the Cremona diagram, we can draw this force of 10 Newtons.

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In this Cremona diagram, we are going to actually start with

the internal force of 7.4 Newtons acting horizontally.

Then, we will have a load of 10 Newtons and we are going to close the polygon of forces

corresponding to the pink free-body, here, by this internal force.

Then, if we measure it, we are going to see that it has the same magnitude,

it also has a value of 12.4 Newtons. That is quite logical since our

structure is symmetric. I identify the contribution to the

Cremona diagram of the pink free-body. We now want to

take an interest in the free-body concerning the supports.

Well, I draw again this free-body

independently. We have a piece of cable, the end

of the cable acting on this free-body, there is of course

the internal force of 12.4

Newtons which pulls this support

downwards. In the Cremona diagram, we now reuse

the internal force of 12.4 Newtons pulling downwards then,

we finish

the construction with the vertical component on the left

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with a piece of cable and

the end. The internal force is equal

to 12.4 Newtons. I did not copy it

here, also 12.4 Newtons and I did not copy it

in the free-body either. We have a tensile internal force

of 12.4 Newtons. On the yellow

free-body acts the internal force of 12.4

Newtons, going down towards the left. Then,

the horizontal internal force on the right

which is equal to 7.4 Newtons.

Then, the vertical internal force on the right

which is equal to 10 Newtons. I also

identify the contribution to

the Cremona diagram, of the yellow

free-body. If we apply a second

load on our system, actually it does not change its shape, but now, as

it was the case for a system with only one load, the internal forces are doubled.

Thus, we would just have to take again the Cremona diagram which we have previously

made and to read it with a

doubled scale to directly get the internal forces.

Or, of course, we could integrally reconstruct it.

Let's now deal with a practical exercise of dimensioning of the cable.

It is now a cable on which acts not

anymore 10 Newtons, but a permanent load of 12 kiloNewtons,

12000 Newtons so it is a quite significant

load on which is added a variable load of 8 kiloNewtons.

Let's see how to proceed to obtain the diameter of the cable

knowing that the strength of the cable fsd is 320 Newtons per square millimeter.

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We first have to get the design load, Qd, which acts on this

cable. Qd is equal to 1.35

times G plus 1.5

times Q or for us, 1.35

times 12 plus 1.5 times 8,

that is to say 28.2 kiloNewtons.

We are now going to have a design load,

I draw it here in pink, a little bit staggered, Qd.

By the way, I should not have moved it, we are going

to draw it over, here, Qd equal to 28.2 kiloNewtons.

Naturally, we could made a new Cremona diagram

for the load of 28.2 kiloNewtons but it is not necessary since we

have already the solution for 10 Newtons, and therefore we now want

to multiply everything by the ratio between 28200 Newtons

and 10 Newtons to directly obtain our internal forces.

The larges internal force, we have obtained it before in both inclined

parts of the cable, it was 12.4 Newtons

for an applied load of twice 10 Newtons.

Now, we want to see what is the

result for 28.2

kiloNewtons and actually, because we do not want to have thousands of Newtons, we

would be interested to have this answer in kiloNewtons, at least for the moment.

This answer is simply going to be 12.4

divided by 10, this is a proportion, times 28.2.

And this, it gives us

35 kiloNewtons. Thus the internal force in these

two segments of cable, for the dimensioning,

we have Nd is equal to 35

kiloNewtons. We now want

to do the dimensioning of the cable. The dimensioning is

the choice of the dimensions, actually we

want to know the diameter of the cable. We

have then Nd is equal to

35 kiloNewtons. That is to say

35000 Newtons and we have a material which

resists to 320 Newtons per square millimeter thus the minimal required

area is 35000 divided

by 320, that is to say

109 square millimeter, so

the diameter is still the square root of 4 times

A divided by pi : it is equal

to square root of 4 times 109 divided by

pi, that is to say 11.8 millimeters,

thus we choose a cable of 12

millimeters diameter which will be adequate

to resist to the design internal forces.

We also could obviously do the verification

of the transitional part of the cable, you

know, the one which had 7.4 Newtons for an applied load of twice 10 Newtons,

but actually, this cable would have a smaller diameter, that is why

we only did the dimensioning for the part with the largest internal force.

In this video, we have seen how

to solve the equilibrium of a cable subjected to two symmetric

loads, we have proceeded expressing in the Cremona

diagram, the equilibrium of each free-body around

the load on the left and around the load on the right.

We have seen how to obtain the internal forces in each segment of cable,

then we have proceeded to the dimensioning performing a

load combination to get the design load.

Thereafter, a cross-multiplication, a

proportion to obtain the design internal force

in the cable and finally we have been able to determine the diameter of the cable.

During the construction, we have also determined

the forces at the supports in a very

similar way to what we did before

for the case of the cable with only one force.

A question which maybe

came to your mind is, but, what happens

if I do not know the geometry of the cable ?

Because that is true that until now, we have had photos of the geometry

of the cable, well, we are going to take an interest

in this question in the following lectures.