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22-Mec-A7 Advanced Strength of Materials · May 2014

Question 4 of 8: Allowable Pressure in a Thick-Walled Cylinder

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format: National Exams, May 2014 — 07-Mec-A7 Advanced Strength of Materials. Open-book, 3 hours; any five of the eight problems constitute a complete paper and all problems are of equal value. All eight problems are solved.

Reference texts: A. C. Ugural & S. K. Fenster, Advanced Strength and Applied Elasticity, 4th ed.; A. P. Boresi & R. J. Schmidt, Advanced Mechanics of Materials, 6th ed.; R. C. Hibbeler, Mechanics of Materials, 10th ed.; S. P. Timoshenko & J. M. Gere, Theory of Elastic Stability, 2nd ed.

Question 4: Allowable Pressure in a Thick-Walled Cylinder (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. $r_i=50$ mm, $r_o=80$ mm; elastic limit $\sigma_Y=280$ MPa; $v=0.28$; $p_i=5\,p_e$.

Find. The allowable internal pressure $p_i$ by (a) von Mises and (b) maximum-shear (Tresca).

Approach. The worst point is the inner wall. Use Lamé’s equations to express the three principal stresses ($\sigma_r,\sigma_\theta,\sigma_z$) there in terms of $p_e$, then equate the von Mises and Tresca equivalent stresses to $\sigma_Y$ and back out $p_i=5p_e$.

  1. Lamé stresses at the inner wall. With $p_i=5p_e$, $p_e=p_e$, evaluating $\sigma=A\pm B/r^2$ at $r=r_i$ gives (per unit $p_e$) $$\sigma_r=-5\,p_e,\qquad \sigma_\theta=\frac{p_i(r_i^2+r_o^2)-2p_e r_o^2}{r_o^2-r_i^2}=8.128\,p_e,\qquad \sigma_z=\tfrac12(\sigma_r+\sigma_\theta)=1.564\,p_e,$$ where $\sigma_z$ is the closed-end axial stress (the intermediate principal stress).
  2. von Mises criterion. $\sigma_{\text{vM}}=\sqrt{\tfrac12\big[(\sigma_\theta-\sigma_z)^2+(\sigma_z-\sigma_r)^2+(\sigma_r-\sigma_\theta)^2\big]}=11.37\,p_e$. Setting this to $\sigma_Y$: $$p_e=\frac{280}{11.37}=24.6\ \text{MPa}\;\Rightarrow\;\boxed{p_i=5p_e=123.1\ \text{MPa}}.$$
  3. Maximum-shear (Tresca) criterion. The extreme principals are $\sigma_\theta$ and $\sigma_r$, so $\sigma_\theta-\sigma_r=13.128\,p_e=\sigma_Y$: $$p_e=\frac{280}{13.128}=21.3\ \text{MPa}\;\Rightarrow\;\boxed{p_i=5p_e=106.6\ \text{MPa}}.$$
  4. Compare. Von Mises permits $15\%$ more pressure than Tresca, because Tresca ignores the intermediate stress $\sigma_z$ and is therefore the more conservative (safer) criterion.

Check / assumption. The ends are taken as closed, so an axial stress $\sigma_z=\tfrac12(\sigma_r+\sigma_\theta)$ acts. For open ends ($\sigma_z=0$) the von Mises result would differ slightly; the closed-end reading is standard for a pressurised cylinder and is stated here explicitly.

CriterionAllowable $p_e$Allowable $p_i=5p_e$
von Mises$24.6$ MPa$123.1$ MPa
Maximum shear (Tresca)$21.3$ MPa$106.6$ MPa