Question 3 of 8: Factor of Safety of a Stressed Plate

Question 3: Factor of Safety of a Stressed Plate (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. A thin plate in plane stress with $\sigma_x=180$ MPa, $\sigma_y=160$ MPa, $\tau_{xy}=-100$ MPa and yield strength $\sigma_Y=402$ MPa.

Find. The factor of safety against first yield by (a) von Mises and (b) maximum-shear (Tresca).

Approach. Reduce the plane-stress state to its in-plane principal stresses (the out-of-plane principal is zero), then divide the yield strength by each criterion’s equivalent stress.

1. Principal stresses. $\sigma_{avg}=\tfrac12(180+160)=170$, $R=\sqrt{(\tfrac{180-160}{2})^2+(-100)^2}=100.5$, so $\sigma_1=270.5$, $\sigma_2=69.5$ MPa; the plate is thin, so $\sigma_3=0$.
2. Von Mises equivalent stress. $\sigma_{vm}=\sqrt{\sigma_x^2-\sigma_x\sigma_y+\sigma_y^2+3\tau_{xy}^2}=\sqrt{180^2-180\cdot160+160^2+3(100)^2}=243.3$ MPa.
3. Von Mises FoS. $N_{vm}=\sigma_Y/\sigma_{vm}=402/243.3=\boxed{1.65}$.
4. Tresca FoS. With $\sigma_2 \gt 0$ the extreme principals are $\sigma_1$ and $\sigma_3=0$, so $N_T=\sigma_Y/(\sigma_1-\sigma_3)=402/270.5=\boxed{1.49}$.
5. Interpret. Tresca gives the lower (more conservative) factor because it ignores the intermediate principal $\sigma_2$, which von Mises credits with load-carrying capacity.