00:01
In this question we are given, this is a beam, this is the loading, the value of loading is 1 .65 pound per feet, all right? this is one support, this is another support.
00:16
The distance between support is 6 .6 feet and this overhang is 2 .9 feet.
00:26
All right? so if i name them, this is a and this is b.
00:33
So, sigma m .a would be zero.
00:38
That will give us what, rb, if this is re, rb and r.
00:42
So 175 into 6 .6 plus 2 .9 square by 2 minus rb into 6 .6.
00:55
That must be equal to 0.
00:57
All right.
00:58
So we will get our value of rb from here.
01:01
Rb would be, let me solve it.
01:05
175 into 6 .6 plus 2 .9 square by 2 all right divided by 6 .6.
01:15
So this is coming out to be 1196 .4 or we can say 5.
01:21
This much pounds is the value of r b and sigma f vertical is equal to 0.
01:27
This will give us what? r a so r a would be 175 multiplied by 6 .6 plus 2 .9 minus 1196 .5 so this is coming out to be 466 i guess yeah this much pounds this is re now if you try to do the free body diagram for the beam from a to b this is 466 this is 175 pound per feed and this is some distance x this is vx this is mx this is m x so vx would be equal to 466 minus 175 x all right and mx would be 466 x minus 175 x square by these are the sfd and bmd equations all right i will draw them in the end all right now if i put x is equal to 6 .6.
02:53
Then number two i get.
02:55
So vx would be 466 minus 175 into 6 .6.
03:03
So this would be minus 689 pounds.
03:08
And mx would be what? 466 into 6 .6 minus 175 by 2 into 6 .6 square.
03:25
So this is minus 735 .9 pounds...