Mechanics of Materials, 7th Edition

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

Chapter 1 - Problems - Page 24: 1.17


$$\tau_{FAIL} = 888.9 \space psi $$

Work Step by Step

Solution: 1) Area of Shear force Distribution: From Diagram in textbook, area of shear force distribution is: $A = 3\space in * 0.6\space in = 1.8 \space in^2$ 2)Failure Stress: $\tau_{FAIL} = \frac{F_{FAIL}}{A} = \frac{1600 \space lb}{1.8 \space in^2} = 888.9 \space psi$ $\boxed{\tau_{FAIL} = 888.9 \space psi} \leftarrow \space\space ANS$
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