0:17 → 0:20
In this problem we have this small structure here,
0:20 → 0:22
in which we have three elements.
0:22 → 0:25
This rod with circular section,
0:26 → 0:30
this bar with a square section,
0:31 → 0:34
and these two elements are joined together through this pin,
0:35 → 0:38
with diameter equal to 25 millimeters.
0:38 → 0:42
Then we need to find what is the average normal stress in the rod,
0:43 → 0:49
what is the average normal stress in the bar and what is the average shear stress in the pin.
0:50 → 0:56
Then we can start with the average stress in the rod.
0:56 → 1:00
So then we can cut here for example,
1:02 → 1:06
and we have here this applied load of a hundred kilo Newtons.
1:07 → 1:14
Then if we apply the equations of equilibrium, we know, we know that the sum of forces in the X direction is equal to zero,
1:15 → 1:21
then the internal reaction force is equal to 100 kilo Newtons.
1:22 → 1:32
Then from here we know that the average stress in the rod is equal to the force divided by the area,
1:32 → 1:38
so we know that this force is equal to 100 kilo Newtons,
1:40 → 1:49
and the area is equal to 1000 millimeters squared.
1:49 → 1:54
So this is equal to 100 mega Pascal's.
1:56 → 1:58
We can do the same for the bar,
2:01 → 2:05
we can cut here, and we have this.
2:08 → 2:13
We have here this point load of 100 kilo Newtons,
2:14 → 2:22
and of course then the internal reaction force is equal to 100 kilo Newton as well,
2:22 → 2:31
then we know that the average normal stress in the bar is equal to the force divided by the area,
2:31 → 2:32
so this is equal to...
2:36 → 2:39
then this is equal to fifty mega Pascal's,
2:42 → 2:49
and finally we can calculate the shear stress in the pin.
2:52 → 2:55
we can consider first the forces acting on the pin,
2:56 → 3:04
so this is the piece and we have here 50 kilo Newtons.
3:06 → 3:08
We have here 50 kilo Newtons as well,
3:09 → 3:14
we have here this point load of hundred kilo Newtons,
3:15 → 3:23
then for example if we cut this structure here this plane and here at this plane,
3:24 → 3:26
we have here something like this.
3:27 → 3:35
So 100 kilonewtons acting in this direction and we have here our shear stress,
3:37 → 3:39
and in the surface as well,
3:40 → 3:45
then we know that the average shear stress is equal to the force.
3:46 → 3:52
100 kilo Newtons divided by these areas,
3:52 → 3:56
this one and of course this one,
3:56 → 4:00
so then this is two times the area of the pin.
4:04 → 4:10
Then this is equal to 102 mega Pascal's.