Mechanics of Materials, 7th Edition

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

Chapter 1 - Problems: 1.23

Answer

$$D_{PIN} = 1.030\space in$$ $$\sigma_{bearing}=38.83\space ksi$$

Work Step by Step

1) Internal Force in Link: Average Axial stress in link (calculated across the entire cross sectional area of the link, so $A_{link} = 2 * 0.25\space in^2$) $\sigma_{avg} = \frac{F}{A_{link}} = \frac{F}{0.5\space in^2} = -20 \space ksi$ (negative value for stress indicates compression) Solving for F: $F = |-20| \space ksi* 0.5\space in^2 = 10\space kip$ (directed in compression) 2) Diameter of Pin: Shear stress in pin: $\tau_{pin} = \frac{F}{A_{pin}} = \frac{10\space kip}{0.25 * \pi * D_{pin}^2} = 12.0 \space ksi$ Solving for unknown $D_{pin}$ : $D_{pin} = \sqrt{\frac{10}{0.25\pi * 12}} = \boxed{1.030\space in}\space\space\leftarrow ANS1$ 3)Bearing Stress (calculated across the width of the opening) $\sigma_{bearing} = \frac{F}{t * D_{pin}} = \frac{10\space kip}{0.25 * 1.030} = \boxed{38.83\space ksi}\space\space\leftarrow ANS2$
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