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.