00:01
Hello students, here for the cylindrical shaft the axial force is given as axial force f equal to 300 lb and the e value is t equal to 4 lb feet and the shaft diameter d equal to 2 inch.
00:40
So, here first we have to calculate the principal stresses.
00:47
So, in order to calculate the principal stresses we have to calculate the three stresses that is first one is axial sigma axial.
01:01
So, we can write sigma axial equal to f by a.
01:08
So, here area of the system is a equal to pi into 3 by 2 the whole square that is equal to 3 .14 into 2 .5 divided by 2 the whole square that is equal to 4 .90625 inch the square.
01:32
So, from this we can get the sigma axial equal to 300 divided by 4 .90625.
01:44
So, we can write sigma axial equal to 61 .146 lb bar inch square.
01:59
So, in next section we have to calculate the bending stress.
02:08
So, bending stress is sigma bending.
02:16
So, the formula is sigma bending equal to mc divided by i.
02:24
This is our equation number 2.
02:27
So, we will get the i moment of inertia equal to pi r power 4 divided by 4 that is equal to 3 .14 into 1 .25 the whole power 4 divided by 4 that is equal to 1 .9165 inch the whole power 4 and c is half of the diameter.
03:07
So, we will get d by 2 equal to 1 .25 inch...