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22-Mec-A7 Advanced Strength of Materials · May 2013

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 2013 — 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 as a study resource.

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. A thick cylinder with $p_i=3.5\,p_e$, yielding when the elastic limit is reached.

Given data
Inner radius / outer radiusa = 50 mm, b = 75 mm (a2 = 2500, b2 = 5625 mm2)
Elastic limitσY = 340 MPa
Poisson’s ratiov = 0.32
Pressure ratiopi = 3.5 pe

Find. The allowable internal pressure pi under (a) Tresca (max shear) and (b) von Mises yield criteria.

Approach. Use the Lamé equations to express the bore stresses in terms of pi (the critical location), then set each yield criterion equal to the elastic limit and solve for pi. Closed-end axial stress $\sigma_z=(\sigma_r+\sigma_\theta)/2$ is used.

  1. Lamé stresses at the bore (r = a). With $p_e=p_i/3.5$, the constants $A=0.2857p_i$ and $B/a^2=1.2857p_i$ give $$\sigma_r=-p_i,\qquad \sigma_\theta=1.5714\,p_i,\qquad \sigma_z=\tfrac12(\sigma_r+\sigma_\theta)=0.2857\,p_i.$$ These are the three principal stresses ($\sigma_\theta$ largest, $\sigma_r$ smallest).
  2. (a) Maximum shear stress (Tresca). Yield when $\sigma_1-\sigma_3=\sigma_Y$: $$\sigma_\theta-\sigma_r=(1.5714+1)\,p_i=2.5714\,p_i=340.$$ $$\boxed{p_i=132.2\ \text{MPa}\ \ (\text{Tresca})}$$
  3. (b) Von Mises. Yield when $$\tfrac12\big[(\sigma_\theta-\sigma_r)^2+(\sigma_r-\sigma_z)^2+(\sigma_z-\sigma_\theta)^2\big]=\sigma_Y^{2}.$$ Substituting the bore stresses gives an equivalent stress $2.227\,p_i=340$: $$\boxed{p_i=152.7\ \text{MPa}\ \ (\text{von Mises})}$$
  4. Comparison. Tresca is the more conservative criterion, permitting about 13% less pressure than von Mises — the expected ordering, since Tresca ignores the intermediate principal stress.
Question 4 — results
CriterionAllowable pi
Maximum shear stress (Tresca)132.2 MPa
Von Mises (distortion energy)152.7 MPa

Check: closed-end axial stress is assumed. Open ends (σz = 0) change the von Mises result only slightly, to 151.4 MPa; the Tresca value is unaffected because σz is the intermediate principal stress. The corresponding external pressure is pe = pi/3.5.