Plane-Stress Yield Check and Stress Rotation: 22-Mec-A7 Advanced Strength of Materials, May 2015, Q6

Question 6: Plane-Stress Yield Check and Stress Rotation (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Plane stress $\sigma_x=180$, $\sigma_y=80$, $\tau_{xy}=-60$ MPa; yield stress 275 MPa.

Find. (a) whether von Mises predicts yielding; (b) the normal and shear stresses on a plane rotated 60° clockwise.

Approach. Compute the von Mises equivalent stress and compare with the yield stress, then apply the plane-stress transformation equations at $\theta=-60^\circ$.

(a) Von Mises equivalent stress. For plane stress, $$\sigma_{vm}=\sqrt{\sigma_x^2-\sigma_x\sigma_y+\sigma_y^2+3\tau_{xy}^2}=\sqrt{180^2-180(80)+80^2+3(60)^2}=188\ \text{MPa}.$$ Since 188 MPa is well below the 275 MPa yield stress, $\boxed{\text{no yielding occurs}}$ (factor of safety $\approx 1.47$).
(b) Rotate 60° clockwise ($\theta=-60^\circ$). With $\tfrac{\sigma_x+\sigma_y}{2}=130$ and $\tfrac{\sigma_x-\sigma_y}{2}=50$ MPa, $$\sigma_{x'}=130+50\cos(-120^\circ)-60\sin(-120^\circ)=157\ \text{MPa},$$ $$\tau_{x'y'}=-50\sin(-120^\circ)-60\cos(-120^\circ)=73.3\ \text{MPa}.$$ The complementary face carries $\sigma_{y'}=103$ MPa, so the normal stresses still sum to 260 MPa as required by invariance of the trace.

Final results — Question 6
Quantity	Value
Von Mises stress	188 MPa (no yield; below 275 MPa)
σ_x′ at −60°	157 MPa
τ_x′y′ at −60°	73.3 MPa

Topic: Plane-stress transformation and the von Mises yield criterion.

Key relations: $\sigma_{x'}=\tfrac{\sigma_x+\sigma_y}{2}+\tfrac{\sigma_x-\sigma_y}{2}\cos2\theta+\tau_{xy}\sin2\theta$; $\tau_{x'y'}=-\tfrac{\sigma_x-\sigma_y}{2}\sin2\theta+\tau_{xy}\cos2\theta$; $\sigma_{vm}=\sqrt{\sigma_x^2-\sigma_x\sigma_y+\sigma_y^2+3\tau_{xy}^2}$.

Why this works: The von Mises stress collapses a multiaxial state into a single equivalent uniaxial stress that drives distortion; comparing it with the tensile yield stress gives a direct yes/no on yielding. The transformation equations then rotate the stress tensor to any plane—a clockwise rotation is simply a negative $\theta$.

Common pitfall: using $+60^\circ$ for a clockwise rotation (it must be $-60^\circ$), and dropping the sign of $\tau_{xy}$, which flips the transformed shear.

Source: Hibbeler, Mechanics of Materials, Ch. 9–10 (stress transformation, yield criteria).

Question 6 of 8: Plane-Stress Yield Check and Stress Rotation

Question 6: Plane-Stress Yield Check and Stress Rotation (20 marks)