Problem: The state of plane stress shown occurs in an aluminum member made of an alloy with a tensile yield strength of 35 ksi.  Determine the factor of safety with respect to yield, using the

a) maximum-shearing-stress criterion
b) maximum-distortion-energy criterion

Solution: Some comments before starting...

From page 28, the theoretical F.S. is . But we are given . If we consider the approximate plot of stress vs strain, the yield stress and the ultimate stress are the same:
Here the maximum stress is that determined using Mohr's Circle. Since §1.6 showed that the yield stress for shear is one-half the normal yield stress, the F.S. becomes:

Before using either of the criteria, I use Mohr's Circle to determine the principal values and the maximum shearing stress:

If we do a rotation of the axes from xy to x'y' to eliminate , we get the principal values:

It is implied that is zero. Considering a rotation from y'z to y'z', we get the smaller (shaded) circle shown. It's clear that the third possible circle is also inside the circle first drawn, and the maximum shearing stress is R = 11.18 ksi.

a) Using the maximum-shearing-stress criterion,

b) Using the maximum-distortion-energy criterion,

¹ This is problem 6.82 from Beer & Johnston's Mechanics of Materials, 2nd Edition.