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Hello, in this video, we are going to look at

the case of the cables subjected to many loads and subjected to distributed loads.

When loads are distributed, we will see that there is another unit of measure to

quantify these loads, we will see how to solve

the problems related to uniform loads,

we will see the shape that cables take under these kinds

of loads, the maximum internal forces, as well as the forces at the supports.

Let's start here with a cable subjected to eight loads.

Well, it is a quite large number of loads, but it is something

that we can solve by the methods which we have seen until now.

Let's first observe that the spacing between these

loads is constant according to

the horizontal axis. I am going to trace an horizontal straight line and

I copy the position of these loads, and we can note

that this spacing is regular. To my scale,

the spacing is roughly 35 millimeters

on the sheet, indeed, in reality, the spacing in

the model was 150 millimeters for each weight.

Let's look at how to solve this problem using the applet.

To change a little bit, I have already

introduced in the applet the eight loads, but

I introduced them on top of each other,

at the position of the resultant. Since all

these loads are symmetric, it is clear that the resultant

is equal to eight times Q and acts in the middle of the span.

The problem which I am solving here, corresponds actually to the one

that I would have, if I had taken a big free-body which includes all the forces,

and which cuts the first and the last segment of cable.

So, I take

each of these forces and I place them where they

must be placed, that is to say at the level of

each of the weights which act on the chain;

and what we can note, it is, in a non surprising way,

that the inclination which has initially been set

stays the same, since now, if I

draw once again a free-body, well, I include

all the forces, I have the resultant here, in

the middle, which did not change, and I cut the first and the last segment, thus the

system did not change, and it is logical that the inclination should still be the same.

What we can

also note, if we take a ruler, it is that the

distance between the line which links the supports here, and the

cable, is the same than the distance between the cable and the point

which corresponded to the hanging point of the eight forces before.

This s a very important geometrical

property, the ones among you who are very

good at mathematics may remember that it is

a property which is linked to second order parabolas.

Indeed, cables which are subjected to

uniform loads tend to take a parabolic shape.

Obviously, we can also solve this problem with the classic

graphical statics on the paper. So, we have here,

the angle which was specified, for the first

segment, we have already on the right, the

eight forces in the Cremona diagram, thus we

can simply copy this orientation, in

the Cremona diagram. The orientation of

the final part of the cable is also given by the position of the resultant

and of the last support, we can also

copy this inclination in the Cremona diagram.

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There is no problem to do this, except

that it is, obviously, more tedious that what we did before.

So, we know that this part here will correspond to the

first segment of cable, we can copy the inclination of the

second segment, in the structural sketch, to obtain

the inclination of this second segment, and so forth.

Getting ready, I realized that I did not succeed to do a job accurate

enough on the tablet, that is why I directly prepared the right

solution electronically, but for you there should be no problem, on a sheet of paper, it is

not something difficult if we make the effort to be a little accurate.

We thus have here, this segment which corresponds to this first segment of cable,

the second one, the third one and so forth,

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along all the cable.

If for example we make this drawing with

a electronic drawing program, as I

did myself, there is no problem

since the precision which we can reach is excellent.

We can observe certain things in

this diagram, first, in this configuration,

the largest internal force in the cable is observed

here, and in a similar way there, it means that is is,

each time, the segment which is closest to the support which has the largest internal force.

It is not universally valid, but it is very often the case for the

cables which are subjected to

uniform or regular loads, as it is

the case here.

Conversely, the segment in the middle of the cable,

is the one which has the smallest internal forces.

What does it mean ?

Well, it means that if you had to

build a cable in two segments, well, the

best would be to connect these two segments in

the middle, rather than do this near the supports.

The Golden Gate bridge is a typical

example of a cable subjected to lots of loads.

Let's first look at how this bridge works :

there are two main load-bearing cables,

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which bear between both pylons and which

extend beyond them. We have a certain number of vertical

cables, which link

the lower part of the bridge to these load-bearing cables, we

call these cables hangers,

because they hang up the load.

The load itself, is located on the deck, which is the useful part of the bridge,

the one where the vehicles run, in this case, cars.

Then, we can see on the lower part of the hangers, a very big

element of structure, with horizontal elements, and diagonal elements,

as well as vertical elements by the way.

This structure is called a truss, this is a type of structure which we

will see in the second part of the course, which will not be in this MOOC.

The function of the truss is really to take the

loads of the deck, and to bring them into the hangers.

We want to talk

about the x axis, talking about the longitudinal

axis of the bridge, the axis that the vehicles follow.

If we look at this photo, it is very clear that there is much

more material in the deck and in the trusses than in the cables.

Admittedly, the cables are quite significant, they have a diameter of approximately one meter,

essentially in steel, but the deck itself

is much larger and heavier.

What does it mean ? It means that the weight according

to x, is approximately the weight of the

deck, there is, that is true, the contribution of the

cable, but essentially, it is a constant.

When a cable is subjected to a constant load, it takes

the shape of a second order parabola. This structural sketch represents the

central span of the Golden Gate bridge, we can identify

the span, which is the distance between both supports, and we will

name it l, as before.

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And this length is typically expressed in meters.

Then, the load which acts, q, we denote it with a lower case, and

it is not expressed in kiloNewtons anymore, but in kiloNewtons per meter.

But if we look at the whole structure, well,

the resultant which acts in the middle, since the loads are

clearly distributed in a symmetric way, this

resultant has a magnitude of q times l, which clearly

has kiloNewtons as unit of measure, thus the resultant

stays a value of the same kind than the ones we are used to

work with. Let's now take this resultant

q times l, in the Cremona diagram,

we have this property of the parabola of the second order which says that if we double

the rise at mid-span, it directly

gives us the orientation of the tangent in the

left part, and in the right part of our cable.

It corresponds, as I showed you before in the applet,

to take a free-body which would cut very

close to the support, which would include all the forces,

and now, we can obtain the internal forces near the supports,

simply by copying these inclinations in the Cremona diagram.

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Thus, they are similar triangles and then using the properties

of the similar triangles, we can say that,

here we have f and f, and here we have l over 2, we

can thus say that 2 f divided by l over

2 is like what we have

here on the right, q l over 2, divided by H.

Well, we know q l over 2, we

know everything, except H,

so H is equal to q l square

divided by 8 f.

If we want to know the magnitude of the maximum internal force in the cable, N

max, we can observe that this triangle is a right-angled triangle, then, N

max, will be equal to the square root of V

squared, we already know V, q l over 2, plus H

squared that we just determined before. In this video,

we have seen how to work with cables

with numerous loads, or with uniform loads.

We have seen how, working with the resultant of all the

loads and using the geometric properties related to the shape of the cables

which take the shape of a second order parabola, we can

obtain the internal forces, as well as the forces at the supports, on a cable,

with numerous loads or uniform loads.

We have seen that the units of measure are different, the unit of measure of

the distributed loads is the kiloNewton per meter, or an unit of force

by unit of length, and we denote them with a lower case,

q, or g if it is a self-weight, rather than with a

capital Q or capital G, as we

did it for the concentrated loads until now.