Question 2 of 8: Thick-Walled Cylinder — Tresca and Von Mises

Question 2: Thick-Walled Cylinder — Tresca and Von Mises (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 closed-end thick cylinder carries an internal pressure that is seven times the external pressure; yielding is to just begin at the bore, where the stresses are largest.

Find. The allowable internal pressure $p_i$ from (a) Tresca and (b) von Mises, with first yield at the inner wall.

Approach. Evaluate the three Lamé principal stresses at the bore in terms of $p_i$ (using $p_e=p_i/7$), then set each yield criterion equal to $\sigma_Y$ and solve for $p_i$.

1. Lamé stresses at the bore. With $p_e=p_i/7$ and $r_o^2+r_i^2=8125\ \text{mm}^2$, $r_o^2-r_i^2=3125\ \text{mm}^2$, the hoop, axial (closed-end) and radial stresses become $$\sigma_\theta=\frac{p_i(r_o^2+r_i^2)-2p_e r_o^2}{r_o^2-r_i^2}=2.086\,p_i,\quad \sigma_z=\frac{p_i r_i^2-p_e r_o^2}{r_o^2-r_i^2}=0.5429\,p_i,\quad \sigma_r=-p_i.$$ Note $\sigma_z$ is exactly the mean of $\sigma_\theta$ and $\sigma_r$.
2. Maximum-shear (Tresca) criterion. The extreme principal stresses are $\sigma_\theta$ and $\sigma_r$, so $$\sigma_\theta-\sigma_r=(2.086+1.000)\,p_i=3.086\,p_i=\sigma_Y\ \Rightarrow\ \boxed{p_i=129.6\ \text{MPa}}$$
3. Von Mises criterion. $$\sigma_{vm}=\sqrt{\tfrac12\!\left[(\sigma_\theta-\sigma_z)^2+(\sigma_z-\sigma_r)^2+(\sigma_r-\sigma_\theta)^2\right]}=2.672\,p_i=\sigma_Y\ \Rightarrow\ \boxed{p_i=149.7\ \text{MPa}}$$
4. Consistency. Because the intermediate stress sits at the mean, the two predictions differ by the fixed factor $2/\sqrt3=1.155$: $149.7/129.6=1.155$. Von Mises is the less conservative (higher allowable), as expected.