00:01
So, here in this question, first we calculate the polar moment of inertia for solid circular shaft.
00:06
The polar moment of inertia for a solid circular shaft.
00:23
The formula for polar moment for inertia of the solid circular shaft is given by j j is equals to pi into d raised to the power 4 upon 32, where d is the diameter of the shaft.
00:36
And plugging in the values, j is equals to pi into 0 .018 raised to the power 4 upon 32.
00:45
Remember that we need to convert the diameter from millimeters to meters, that is from 18 millimeters meters to 0 .018 meters.
00:54
So, this comes out to be 1 .15 into 10 raised to the power minus 7 meter raised to the power 4.
01:00
Then, we calculate the maximum shear stress in the shaft.
01:06
Now, the maximum shear stress that is tau max is equals to t into d by 2 into j where t is the torque applied on the shaft given that torque is 200 newton meters tau max is equals to 200 into 0 .018 upon 2 into 1 .15 into 10 to the power minus 7 this comes out to be 156 .52 megapascals then since the the maximum shearing stress is greater than yield strength since tau max, tau max is greater than yield strength, yield strength that is 145, that is 145 megapascals...