00:01
Hi, in the first part of the question we have to determine the maximum shear stress it is given by root of m square plus t square where m is the bending moment, t is the torsional moment.
00:11
So, which one substitution it is root of 10 ,000 square plus 30 ,000 square which is equal to so the torsional stress will be equal to 31 .63 into 10 power 6 newton meter.
00:25
So, newton millimeter which is equal to pi by 60 into tau into d cube where tau is equal to so tau is equal to it is 0 .58 into sy which is the yield strength into so s divided by factor of safety so which is equal to 0 .58 into 700 by 2 which is equal to 203 mega pascal.
00:54
Then so we have 31 .63 into 10 power 6 which is equal to pi by 6 into 203 into d cube.
01:04
So from this we have d is equal to so d is nearly equal to so the diameter is equal to it is 92 .58 millimeter or it is nearly equal to 100 millimeter.
01:20
So this is the first part.
01:21
In the second part of the question we have so we are given with a value of t1 and t2.
01:27
Now t1 and t2 sigma by n factor of safety so we have the torsional stress on a is equal to tb which is equal to 50 newton meter.
01:42
So 50 newton meter then we have the net force acting along y is equal to 0 that is t1 plus t2 is equal to ra plus rb which on substitution we have 1200 plus 1200 plus 800 is equal to ra plus rb...