00:01
So here it is given that our rod is taken under this trace with the pressure of p is equals to 30 kilo newton which can be written as 13 to 10 to the power 3 newtons the diameter of the rod was 20 millimeter.
00:21
Now after applying this kind of tensile stress in the first part we have to find the elongation of rod.
00:31
So for that let's have a look to the diagram first.
00:35
So let's suppose this is nothing but what the rod in which for this 150 mm of part have been elongated while this diameter of the rod is 20 mm okay and at this both the end the pressure is applied in opposite direction.
01:03
So to find the elongation first of all let us calculate the cross sectional area which would be equals to the pi by 4 into d square.
01:16
So this is equals to pi divided by 4 into 20 square will be 400 will comes out 100 pi.
01:26
Now obviously when we are applying the pressure at both the ends there will be a stress produced that is sigma which is given by the pressure applied per unit area and it will be equals to.
01:44
Now pressure was given that is 30 into 10 to the power 3 newton divided by area we have calculated 100 pi.
01:53
So after dividing it will comes out 95 .492 mega pascal.
02:01
Similarly let us calculate sigma epsilon that is epsilon equals to sigma divided by e which would be equals to 95 .492 divided by e which was given to us 70 into 10 to the power 3.
02:25
So after dividing it will comes out 1 .3641 into 10 to the power minus 3.
02:32
While if you remember strain is given by delta l by l.
02:40
So from here delta l would be equals to 1 .364 into 10 to the power minus 3 multiplied by the initial length that is l which was 150...