00:01
The non -uniform varying loading in section cb can be treated as a combination of can be treated as a combination of uniformly distributed, uniformly distributed, distributed load of intensity, load of intensity 20 kn upon m and uniformly varying load and uniformly vary load which varies from 0 to 20 kn per meter from c to b.
01:38
So according to point ra, a point, p is equal to 22 .0 kn and 20 kn per meter.
01:46
Point c, rb and point b, 20 kn upon meter.
01:51
And to find out reaction forces at a and b, so sigma ma is equal to 0.
01:58
Substituting the values minus 22 .22 into 1 .5 plus 20 into 3 into 4 .5 plus 0 .5 into 3 into 20 into 5 minus as continuing so, minus rb into 6 is equal to 0.
02:28
So, rb is equal to 64 .5 kilo newton.
02:34
And after that, sigma fy is equal to 0.
02:38
And after that, substituting all the values.
02:40
So, ra plus 22 minus 20 into 3 minus 0 .5 into 3 into 20 plus rb is equal to 0.
02:52
So, ra plus 22 minus 20 into 3 minus 0 .5 into 3 into 20 plus 64 .5 is equal to 0.
03:08
So, ra is equal to 3 .5 kn.
03:15
And after that, to find out the shear force equation at different section of the beam taking section from the end a, for 0 is less than x is less than 1 .5.
03:28
So, b and x is equal to ra is equal to 3 .5 kilo newton...