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22-Mec-A7 Advanced Strength of Materials · December 2016

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

National Exams — December 2016 · 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 as a study resource.

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

Check: the figure for Question 1 labels the applied load P = 7000 N, whereas the printed text reads “7800 N”. The drawn value on the exam figure governs, so 7000 N is used throughout.

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 thick cylinder loaded by internal pressure that is eight times the external pressure, yielding governed by the two classical criteria.

Given data
Inner radiusri = 0.055 m
Outer radiusro = 0.080 m
Pressure ratiopi = 8 pe
Yield strengthσY = 300 MPa
Poisson’s ratiov = 0.29

Find. The allowable pi from (a) Tresca and (b) von Mises, evaluated at the most-stressed point (the bore).

Approach. Use the Lamé solution to get the radial, hoop and axial stresses at the bore in terms of $p_i$ (with $p_e=p_i/8$), then set each yield criterion equal to $\sigma_Y$ and solve.

  1. Lamé stresses at the bore. With $p_e=p_i/8$, at $r=r_i$: $\sigma_r=-p_i$, $\ \sigma_\theta=\dfrac{p_i(r_i^2+r_o^2)-2p_e r_o^2}{r_o^2-r_i^2}$, and for closed ends $\sigma_z=\tfrac12(\sigma_r+\sigma_\theta)$. Numerically $\sigma_\theta=2.222\,p_i$, so $\sigma_\theta-\sigma_r=3.222\,p_i$ and $\sigma_z=0.611\,p_i$.
  2. (a) Maximum-shear (Tresca). The extreme principals at the bore are $\sigma_\theta$ and $\sigma_r$, so $\sigma_\theta-\sigma_r=\sigma_Y$: $$3.222\,p_i=300\ \text{MPa}\ \Rightarrow\ \boxed{p_i=90.4\ \text{MPa.}}$$
  3. (b) Von Mises. Using all three principals, $\sigma_{vm}=\sqrt{\tfrac12[(\sigma_\theta-\sigma_r)^2+(\sigma_\theta-\sigma_z)^2+(\sigma_z-\sigma_r)^2]}=2.874\,p_i$. Setting $\sigma_{vm}=\sigma_Y$: $$\boxed{p_i=104.4\ \text{MPa.}}$$
  4. Compare. Von Mises allows about 15.5 % more pressure than Tresca, because accounting for the intermediate axial stress $\sigma_z$ lowers the effective stress — Tresca is the conservative design criterion.
Final results — Q3
CriterionAllowable pi
Maximum shear stress (Tresca)90.4 MPa
Von Mises (distortion energy)104.4 MPa