00:01
Now this question they want us to find the in -plane principal strength and the maximum in -plane shear strain and the average normal strength from the given value of x -y strength and shear strength.
00:16
So for the in -plane principal strength or epsilon 1 and epsilon 2, we can use equation 4 to find these values.
00:30
And if we solve equation 4 using the given values of xy strings and the shear strings, we will have that it will equal to 1808 10 minus 6 for absalom 1 and minus 128 and 128 10 minus 6 for epsilon 2.
00:54
And next we have to find the orientation of these two strengths using equation 5.
01:04
So if we solve equation 5, you have that ceta p will have value of minus 9 .217 degree and 80 .78 degree.
01:26
So next we have to find that which value of strength, epsilon 1, or epsilon 2 has which value of the degree.
01:38
So we can check by using the strength transformation equations, like equation 1.
01:46
If we find the x -datt plane strength using the cedar equal to minus 9 .2, we will get it's going to be 1806 which is equal to epsilon 1.
02:14
And that makes c dapy 1 is going to be minus 9 .217 degree and of course it makes the why that plane has value of 1 to 8 10 minus 6 of strength which is equal to the epsilon 2 and so cedar p 2 will be 80 .78 degree that's the first answers and for the second answers you can find the maximum in plane shear strain using equation now we will have that the max in plain strain, in plain shear strength is going to be 360 10 minus 6.
03:23
And for the average normal strain, we can find it using equation 7, which we will get the value of 30, 10 minus 6...