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: 2.63

Answer

$$\begin{aligned} \tan \theta &=0.39887 \end{aligned}$$

Work Step by Step

$\begin{aligned} A &=(0.150)(0.006)=0.9 \times 10^{-3} \mathrm{m}^{2} \\ \sigma_{x} &=\frac{P}{A}=\frac{200 \times 10^{3}}{0.9 \times 10^{-3}}=222.22 \times 10^{6} \mathrm{Pa} \\ \varepsilon_{x} &=\frac{\sigma_{x}}{E}=\frac{222.22 \times 10^{6}}{105 \times 10^{9}}=2.1164 \times 10^{-3} \\ \varepsilon_{y} &=-08_{x}=-(0.34)\left(2.1164 \times 10^{-3}\right) \\ &=-0.71958 \times 10^{-3} \end{aligned}$ $\begin{aligned} \tan \theta &=\frac{4\left(1+\varepsilon_{y}\right)}{10\left(1+\varepsilon_{x}\right)} \\ &=\frac{4\left(1-0.71958 \times 10^{-3}\right)}{10\left(1+2.1164 \times 10^{-3}\right)} \\ &=0.39887 \end{aligned}$
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