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.