Mechanics of Materials, 7th Edition

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

Chapter 2 - Problems: 2.3

Answer

$$P = 9.817 \space kN$$ $$\sigma = 500 \space MPa$$

Work Step by Step

1. Lateral Elongation formula: $\delta = \epsilon * L = \frac{\sigma}{E} * L$ Solving the above formula for sigma: $\delta = \frac{\sigma}{E} * L \Rightarrow \sigma = \frac {\delta * E}{L} = =\frac{45*10^{-3}\space m * 200 * 10^9\space N/m^2}{18\space m} = \boxed{500\space MPa} $ $\leftarrow ANS (b)$ 2. Axial Stress Formula $\sigma = \frac{P}{A}$ for a constant circular section of cable: $A = \frac{\pi * D^2}{4} = 1.963 * 10^{-5}\space m^2$ $500*10^6 \space N/m^2 = \frac {P}{1.963 * 10^{-5}\space m^2} \Rightarrow P = 9817\space N = \boxed{9.82 \space kN}$ $\leftarrow ANS (a)$
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