00:01
So we're asked to find the directions at supports a and d for part a, as well as part b, finding the internal forces and bending moments in plain each duke.
00:14
So, for part a, part a, we need to treat the shaft as a simply supported beam with forces applied at specific points.
00:32
So we're going to let a and d be supports, and we're going to let b and c be the location of the applied forces.
00:55
Oops, b applied forces.
01:05
And the forces are force b is equal to 1 ,000.
01:16
That 400 newtons and force c is equal to 4 ,200 newtons.
01:33
All right now we're going to look at the distances, which we'll put over here.
01:41
Distance.
01:43
Oops, distances.
01:51
All right, so a and b is equal to 160 millimeters b and c is equal to 90 millimeters.
02:08
C and d is equal to 160 millimeters.
02:17
Okay? and now the beam is in static equilibrium.
02:23
So, all right, adjust our color.
02:28
So for one, the sum of forces, of vertical forces, sorry.
02:38
I'm going to set it equal to zero.
02:42
And that's going to be represented as r -a plus r -d minus f -b -b -minus f -c, it's equal to zero.
03:02
Okay, so now we're just going to input our values.
03:09
So we have, actually start, we're going to put this on the other side.
03:20
Forces on the other side.
03:23
So this will be rewritten as r -a plus r -d is equal to f -b plus f -c.
03:38
All right.
03:40
Now we input our values for f -b.
03:45
And fc which is 1 ,400 newtons plus 4 ,200 newtons.
03:57
And simplified out is going to give us 5 ,600 newtons.
04:07
All right, so now we need to find the sum of moments at a.
04:13
So, sum of moments, a moment a is equal to zero...