Question 4 of 8: Thick-Walled Cylinder — Tresca and Von Mises

Question 4: 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 thick-walled cylinder with $p_i=5p_e$.

Find. Allowable internal pressure $p_i$ by (a) maximum-shear (Tresca), (b) Von Mises.

Approach. The worst point is the inner wall, where the Lamé hoop stress peaks and $\sigma_r=-p_i$. For a closed cylinder the axial stress equals the mean of $\sigma_r$ and $\sigma_\theta$, so it is the intermediate principal stress; both criteria then reduce to a multiple of $(\sigma_\theta-\sigma_r)$.

1. Lamé stresses at the inner wall. With $p_e=p_i/5$, $$\sigma_{\theta,i}=\frac{p_i(r_i^2+r_o^2)-2p_e r_o^2}{r_o^2-r_i^2}=p_i\frac{r_i^2+0.6\,r_o^2}{r_o^2-r_i^2}=1.5\,p_i,\qquad \sigma_{r,i}=-p_i.$$
2. Stress difference and axial stress. $\sigma_\theta-\sigma_r=1.5p_i+p_i=2.5\,p_i$; closed ends give $\sigma_z=\tfrac12(\sigma_r+\sigma_\theta)$, the intermediate principal stress.
3. Maximum-shear criterion (part a). Tresca sets $\sigma_1-\sigma_3=\sigma_\theta-\sigma_r=\sigma_Y$: $$2.5\,p_i=330\ \Rightarrow\ \boxed{p_i=132\ \text{MPa}}\quad(p_e=26.4\ \text{MPa}).$$
4. Von Mises criterion (part b). Because $\sigma_z$ is exactly the mean, $\sigma_{VM}=\tfrac{\sqrt3}{2}(\sigma_\theta-\sigma_r)=\sigma_Y$, i.e. the allowable difference grows by $2/\sqrt3$: $$p_i=\frac{2}{\sqrt3}\,(132)=\boxed{152\ \text{MPa}}.$$
5. Comparison. Von Mises permits $2/\sqrt3=1.155$ times the Tresca pressure — the familiar 15.5 % margin when the third principal stress sits midway between the other two.