00:01
Hi, in this question we have the solid rod bc that has a diameter.
00:06
So diameter of bc is given as 30 mm, length is l and it is made of aluminium for which the allowable shearing stress is given as 25 mpa.
00:22
Now we have rod ab which is hollow have outer diameter.
00:28
So diameter of ab that is the outer diameter is given as 25 mm and this also has length l and it is made of a brass having shearing stress of 50 mpa.
00:46
So in part a we have to determine the largest inner diameter of rod ab.
00:52
So here the value of diameter of bc is 30, diameter of ab which is the outer diameter is 25.
01:02
Now we know that tau al that is shearing stress of aluminium is equals to 16 times divided by pi d cube.
01:21
So we can write the value of tau of brass divided by r0 of ab equals to t into 32 divided by pi into doab raised to power 4 minus denab raised to power 4.
01:48
So this is the outer diameter minus the inner diameter.
01:53
So we have tbc equals to tab equals to t.
01:58
So now from these two equations we can write pi into tbc cube into tau a divided by 16 as equals to tau of brass into pi into 25 4 minus din raised to power 4 divided by 32 into 12 .5.
02:25
So on solving for din we get the value of diameter of the inner side as 15 .18 mm.
02:35
So this is the largest inner diameter of the rod ab...