00:01
In this process we have been given the gauge length, gauge length is 10 inch, the diameter of the specimen is 0 .5 inch, when a new load of load of 9 kip is loaded or new diameter or change diameter is 0 .49935 inch.
00:33
Now let's calculate modulus of elasticity, modulus of elasticity of aluminium, so it will come as eal which will be equal to stress by strain, stress by strain.
00:58
Now if we see the stress strain curve then the value of stress is 70 and the value of strain is 0 .00614 correspondingly.
01:10
So, it will ultimately come as 11400 .65 ksi.
01:21
Now if we calculate the cross section area which is a which will be equal to pi by 4 d square, i am putting the value pi by 4 multiplied by 0 .5 inch square and it will ultimately come as 0 .19635 inch square.
01:45
Now when the specimen is loaded with, when the specimen is, when the specimen is loaded with 9 kip load, we know the change diameter is 0 .49935.
02:17
So, now if we calculate stress which is equal to p divided by a, p is here load which is equal to 9 kip.
02:29
So, our stress come as 9 divided by the area is we already calculated 0 .19635 and it will ultimately come as 45 .84 ksi...