Practice exam

[Resham Ramsay]

Are there solutions posted for the practice exam?

[Becca Wozniak]

The solutions are up. I do not understand problem 6. The shear stress
is 0 in the given configuration at an angle of 0 degrees so wouldn't
the principle angle be 0 degrees? And in the examples in class we
never found the maximum values of the normal and shear stresses in
order to find the principle angle or shear angle, we just used the
given values into the equations: (alpha)p = 1/2 tan-1(2*(sigma)xy/
((sigma)xx-(sigma)yy))...etc

[Mitul]

The idea of the problem was to show you that the shear and normal stresses vary with the orientation of the cube / object under force. 

I realized that the problem had a flaw and have corrected it. What you said Becca is right. But if you keep the problem the same as it was in homework 4. The solution posed will make more sense.

I apologize for the mix up.

[Kevan Chao]

Mitul,
For problem 6, where was the sigma-max and sigma-min given? If it wasn't given how were those values found, and why were they found? The rest of the calculations don't seem to use them

[Resham Ramsay]

I guess my question is the same as Becca's. I'm still confused as to
why we needed to find the max/min stresses. Shouldn't we just be able
to plug the given values for sigma xx/yy/xy into the angle equations?