00:01
In this question we are given that force in the x direction is 40 kilo newton.
00:07
So, sigma x would be 40 into 10 to the power 3 divided by what is the dimensions? dimension is 40 millimeter and 60 millimeters.
00:21
So divided by 40 into 60.
00:25
So this would be 1 ,000 or we can say 100 divided by 6 newton per millimeters.
00:32
This is sigma x.
00:34
Sigma y is unknown so as we know that strain in x direction would be equal to sigma x by e minus mu sigma y by is it so all right so assuming that compressive force is positive then f y is given to us that is tensile or we will see what comes so assuming compressive as positive whatever we will get will be our right so delta x is 0 .08 mm.
01:18
So strain is what? change in length over length.
01:23
So length in x was but 80 m m.
01:27
So 0 .08 divided by 80 is equal to sigma x is 100 by 6 minus mu with 0 .3.
01:43
So 0 .29 into sigma y is unknown.
01:50
And e is multiplied on the other side.
01:53
E is given to us 70 gpa, 70 into 10 to the power 9, alright? or we can say 70 into 10 to the power 3, newton per millimeter square.
02:07
Because sigma was in newton per millimeter square, so we will take accordingly...