Mechanics of Materials, 7th Edition

Published by McGraw-Hill Education
ISBN 10: 0073398233
ISBN 13: 978-0-07339-823-5

Chapter 11 - Problems - Page 776: 11.1

Answer

(A)$u=43.1 \frac{in.lb}{in^{3}}$ (B)$u= 72.8 \frac{in.lb}{in^{3}}$ (C)$u= 172.4 \frac{in.lb}{in^{3}}$

Work Step by Step

We will just replace the units in the formula to determine the module of resilience. Modulus of elasticity : $E= 29. 10^{6} psi$ $σy$ : The uniaxiaxl tensile yield strength : $σy= 50 ksi$ u= The modules of resilence is defined as the maximum energy that can be absorbed per unit volume without creating permanent distortions: Modules of resilience $u= \frac{σ^{2}_{y}}{2E}$ $u = \frac{(50.10^{3})^{2}}{2.(29.10^{6})}$ $u=43.1 \frac{in.lb}{in^{3}}$ (B) Modulus of elasticity E=$29.10^{6} psi$ Yield strength : $σ_{y}=65 ksi$ Modulus of resilience : $u= \frac{σ^{2}_{y}}{2E}$ $u=\frac{(65.10^{3})^{2}}{2.(29.10^{6})}$ $u= 72.8 \frac{in.lb}{in^{3}}$ c) Modulus of elasticity : $E= 29.10^{6}$ Yield strength $σ_{y}= 100 ksi$ Modulus of resilience $u= \frac{σ^{2}_{y}}{2E}$ $u=\frac{(100.10^{3})^{2}}{2.(29.10^{6})}$ $u= 172.4 \frac{in.lb}{in^{3}}$
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