Mechanics of Materials, 7th Edition

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

Chapter 2 - Problems - Page 109: 2.61

Answer

$$E=28.973 \times 10^{3} \mathrm{psi}$$ $$v=0.44444$$ $$\sigma=10.0291 \times 10^{3} \mathrm{psi}$$

Work Step by Step

$A=\frac{\pi}{4} d^{2}=\frac{\pi}{4}\left(\frac{5}{8}\right)^{2}=0.306796 \mathrm{in}^{2}$ $P=800 \mathrm{lb}$ $\sigma_{y}=\frac{P}{A}=\frac{800}{0.306796}=2.6076 \times 10^{3} \mathrm{psi}$ $\varepsilon_{y}=\frac{\delta_{y}}{L}=\frac{0.45}{5.0}=0.090$ $\varepsilon_{x}=\frac{\delta_{x}}{d}=\frac{-0.025}{0.625}=-0.040$ $E=\frac{\delta_{y}}{\varepsilon_{y}}=\frac{2.6076 \times 10^{3}}{0.090}=28.973 \times 10^{3} \mathrm{psi}$ $v=-\frac{\varepsilon_{x}}{\varepsilon_{y}}=\frac{-0.040}{0.090}=0.44444$ $\sigma=\frac{E}{2(1+v)}=\frac{28.973 \times 10^{3}}{(2)(1+0.44444)}=10.0291 \times 10^{3} \mathrm{psi}$
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