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22-Mec-A7 Advanced Strength of Materials · May 2015

Question 3 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 2015 — 07-Mec-A7 Advanced Strength of Materials. Open-book, 3 hours. Eight problems of equal value; any five constitute a complete paper. All eight are solved here.

Reference texts. Ugural & Fenster, Advanced Strength and Applied Elasticity (4th ed.); Boresi & Schmidt, Advanced Mechanics of Materials (6th ed.); Hibbeler, Mechanics of Materials (10th ed.); Timoshenko & Goodier, Theory of Elasticity.

Question 3: 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. A closed-end thick cylinder loaded by internal and external pressure in the ratio 6:1.

Given data
Internal radius, ri0.11 m
External radius, ro0.18 m
Elastic limit, σY330 MPa
Poisson’s ratio, v0.31
Pressure ratiopi = 6 pe

Find. The allowable internal pressure pi by (a) Tresca and (b) von Mises.

p_i = 6 p_er_i = 0.11 mr_o = 0.18 mp_e
Cross-section: internal pressure pi pushing outward on the bore, external pressure pe = pi/6 on the outer wall.

Approach. Use the Lamé solution; the critical point is the bore. Take the closed-end axial stress $\sigma_z=\tfrac12(\sigma_r+\sigma_\theta)$ as the intermediate principal stress, then apply each yield criterion and scale linearly in pi.

  1. Lamé stresses at the bore. With $\sigma_{r,\theta}=A\mp B/r^2$ and $p_e=p_i/6$, at $r=r_i$ the radial stress is $\sigma_r=-p_i$ and $$\sigma_\theta-\sigma_r=\frac{2B}{r_i^2}=\frac{2(p_i-p_e)r_o^2}{r_o^2-r_i^2}=\frac{\tfrac53 p_i\,(0.0324)}{0.0203}=2.660\,p_i.$$
  2. (a) Maximum-shear-stress (Tresca) criterion. Yield occurs when $\sigma_\theta-\sigma_r=\sigma_Y$, so $$2.660\,p_i=330\ \text{MPa}\ \Rightarrow\ \boxed{p_i=124\ \text{MPa}\ (\text{Tresca}).}$$ At this pressure $\sigma_r=-124$, $\sigma_\theta=+206$, $\sigma_z=+41$ MPa.
  3. (b) Von Mises criterion. Using all three principal stresses, $$\sigma_{vm}=\sqrt{\tfrac12\big[(\sigma_r-\sigma_\theta)^2+(\sigma_\theta-\sigma_z)^2+(\sigma_z-\sigma_r)^2\big]}=2.304\,p_i.$$ Setting $\sigma_{vm}=\sigma_Y=330$ MPa, $$\boxed{p_i=143\ \text{MPa}\ (\text{von Mises}).}$$ Von Mises permits about 15% more pressure than Tresca, because it credits the intermediate axial stress with relieving the peak shear.
Final results — Question 3
CriterionAllowable pi
Maximum shear stress (Tresca)124 MPa
Von Mises (distortion energy)143 MPa