00:01
Hello students, in this question we have a square and there is 150 mega pascal acting in this direction, 70 mega pascal acting in this direction and then we have a shear stress of 55 acting in this direction.
00:29
So what we can do is that we can now start calculating the principal stress.
00:34
So principal stress given by the equation sigma x or sigma 1 will be equal to sigma x plus sigma y by 2 plus square root of sigma x plus sigma y the whole square divided by 4 plus t x y the whole square.
00:58
T x y is the shear stress in the x -axis.
01:06
Similarly sigma 2 will be equal to sigma x plus sigma y by 2 plus square root of sigma x minus sigma y.
01:17
This is minus sign minus sigma y the whole square divided by 4 plus t x y the whole square right for two stresses.
01:39
So t max t max will be equal to square root of sigma x minus sigma y the whole square by 4 plus t x y square.
01:53
These are the definitions of all these stresses.
01:56
You can plug in the value and you get sigma 1 will be equal to sigma 2.
02:02
So same will be equal to 155 plus 150 divided by 2 plus square root of 55 minus 150 divided by 4 plus 70 the whole square shear stress right.
02:33
So now we can calculate that so sigma 1 is equal to sigma 2 will be equal to so here there is a square that will be equal to 108 .78 megapascal.
02:50
It is the sigma 1 sigma 2 value...