00:01
A beam ab that is 5 meters long is supported by two pins at point c and d that are 2 meters apart.
00:15
The beam carries a uniformly distributed load of 0 .6 kilometers per meter, and the beam also carries two concentrated loads at point a and b at the ends of the beam that respectively have 9itudes of 3 kilnutons and 5 kilonutons.
00:32
We want to determine the location of the two supports so that both reactions are equal.
00:44
So first things first, let's try to determine the reaction forces c and d, knowing that they are equal.
00:56
Let's look at the sums of our forces along y, which must be equal to zero, tib equilibrium.
01:05
I am drawing our coordinate system as usual.
01:09
And so what we have is that rc, which is equal to r, so let's just label the reaction force r.
01:25
So two times the reaction force corresponding to rc plus rd must be equal to the sum of the concentrated loads plus the distributed loads.
01:42
And our distributed load has a total force of 0 .6 kilonuton, times 5 meters, corresponding to 3 kilonutons.
02:01
So 2r must be equal to 3 kilnoughtons plus 3 kilnoughtons plus 5 kilnoutons.
02:06
That is 11 kilonutons, which means that our reaction force is equal to 5 .5 kilnutons.
02:24
Now let's look at our other clearing condition.
02:26
We want to have that the sums of our torques are equal to 0, so we can use any points.
02:30
Let's use point a.
02:35
So our reaction forces will generate counterclockwise torques, whereas our distributed load and contrary load will generate clockwise torque.
02:55
So these will be in opposition.
02:58
Let's calculate each of their moments.
03:04
So our first reaction force will have a moment of r times x...