( E=200GPa, v=0.29 ) is subjected to a state of plane strain ( e zz = e xz = e y z =0) when the design loads are applied. At the critical point in the member, three of the stress components are s xx =60 Mpa , s yy = 240 MPa , and s xy =-80 MPa . The material has a yield stress s 0 =490 MPa . Based on the maximum shear-stress criterion, determine the factor of safety used in the design. I understand how to get two principle stress values of 270.42MPa and 29.58MPa. My lecturer used v(sigma_x + sigma_y) to find the third principle stress. Where did that formula come from? Solution We know that longitudinal strain e1 = (1/2*E) (sigma1 - v(sigma2+sigma3) We have been given the plane stress condition,,, so eZ = 0 0 = (1/2*E) (sigmaZ - v(sigmaX+sigmaY) therefore sigmaZ = v(sigmaX+sigmaY) .

1. ( E=200GPa, v=0.29 )
is subjected to a state of plane strain ( e zz =
e xz = e y z =0) when
the design loads are applied. At the critical point in the member,
three of the stress components are
s xx =60 Mpa , s yy = 240 MPa , and s xy =-80 MPa .
The material has a yield stress s 0 =490
MPa . Based on the
maximum shear-stress criterion, determine the factor of safety used
in the design.
I understand how to get two principle stress values of 270.42MPa and 29.58MPa. My
lecturer used
v(sigma_x + sigma_y)
to find the third principle stress. Where did that formula come from?
Solution

2. We know that longitudinal strain
e1 = (1/2*E) (sigma1 - v(sigma2+sigma3)
We have been given the plane stress condition,,, so eZ = 0
0 = (1/2*E) (sigmaZ - v(sigmaX+sigmaY)
therefore
sigmaZ = v(sigmaX+sigmaY)