Mechanics of Materials, 7th Edition

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

Chapter 2 - Problems: 2.1

Answer

$$D_{thread} = 0.564 \space mm$$ $$\sigma_{avg} = 36.3 \space MPa$$

Work Step by Step

Young's Law: $\sigma = E \epsilon$ 1. Substituting known values find tensile stress in string: $\sigma=3.3*10^9 N/m^2 * 0.011 \space m/m = 36.3*10^6 N/m^2 = \boxed{36.3 \space MPa}$ $\leftarrow ANS (b)$ 2. Average tensile Stress: $\sigma = \frac{P}{A}$ Assuming: $\bullet$ thread cross section is circular and constant $\bullet$ transverse contraction is negligible (ignoring Poisson's effect) $A = \frac{\pi D^2}{4}$ After substituting known values into tensile stress equation and solving for D: $\ 36.3*10^6 N/m^2= \frac{4 * 8.5 N}{\pi D^2} \Rightarrow D = 5.46*10^{-4} \space m = \boxed{0.546 \space mm}$ $\leftarrow ANS (a)$
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