00:01
Hello students here we have the stepped steel shaft that has the diameter as d1 equal to 36 mm and d2 equal to 32 mm and here it is twisted by the torque t at the each end.
00:19
So here l1 equal to 0 .9 m and l2 equal to 0 .75 m.
00:30
So here the allowable shear stress is given as shear stress equal to 28 mpa and the maximum allowable twist is theta equal to 1 .8 degree.
00:53
So from this we have to determine the maximum permissible torque.
00:57
So first we can write the polar moment of inertia.
01:05
Polar moment of inertia is j equal to pi divided by 32 d power 4.
01:20
So for the first segment we can get the polar moment of inertia j1 equal to pi divided by 32 0 .036 the whole power 4 that is equal to 6 .803 into 10 power minus 7 m power 4 and we can get the theta 1 as angle twist is t1 l1 divided by g into j1.
01:55
Here g is the shear stress module so that can be calculated as tau allowable divided by gamma.
02:12
So here tau allowable is given as 28 mpa and from this we can get our g value as g equal to 28 into 10 power 86 mpa.
02:40
So we will get theta 1 equal to t1 into 0 .9 divided by 28 into 10 power 6 into 6 .803 into 10 power minus 7.
02:58
So we will get theta 1 equal to 0 .03141 radians.
03:14
Similarly we can get j2 equal to pi divided by 32 0 .03 to the whole power 4 that is equal to 3 .371 into 10 power minus 7 m power 4.
03:33
So we can calculate the theta 2 as from the total shift...