Question 1 of 8: Thick-Walled Cylinder – Allowable Internal Pressure

Question 1: Thick-Walled Cylinder – Allowable Internal Pressure (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 loaded by internal pressure ten times the external pressure, sized by the wall radii and the yield strength.

Find. The largest internal pressure p_i the wall can carry at first yield, evaluated separately with the von Mises and Tresca criteria.

Approach. Evaluate the Lamé stresses at the bore (where they are largest), take the closed-end axial stress as the mean of radial and hoop, then set each yield criterion equal to σ_Y and solve for p_i.

1. Bore stresses from Lamé. With $p_i=10p$ and $p_e=p$, at $r=r_i$ the radial stress is $\sigma_r=-p_i=-10p$ and the hoop stress is $$\sigma_\theta=\frac{p_i(r_i^2+r_o^2)-2p_e r_o^2}{r_o^2-r_i^2}=\frac{10p(0.01+0.0256)-2p(0.0256)}{0.0156}=19.538\,p.$$
2. Closed-end axial stress. For a capped cylinder the axial stress is the mean of the in-plane stresses, $\sigma_z=\tfrac12(\sigma_r+\sigma_\theta)=\tfrac12(-10p+19.538p)=4.769\,p$, so the ordered principals are $\sigma_\theta \gt \sigma_z \gt \sigma_r$.
3. Tresca criterion. Yield when $\sigma_\theta-\sigma_r=\sigma_Y$: $(19.538+10)p=29.538\,p=315\Rightarrow p=10.66$, hence $\boxed{p_i=10p=106.6\ \text{MPa}}$.
4. Von Mises criterion. $\sigma_{vm}=\sqrt{\tfrac12\big[(\sigma_\theta-\sigma_z)^2+(\sigma_z-\sigma_r)^2+(\sigma_r-\sigma_\theta)^2\big]}=25.58\,p=315\Rightarrow p=12.31$, hence $\boxed{p_i=123.1\ \text{MPa}}$.
5. Compare. Because $\sigma_z$ is exactly the mean of $\sigma_\theta$ and $\sigma_r$, von Mises gives $2/\sqrt3=1.155$ times the Tresca pressure—Tresca is the conservative design value.