Shigley's Mechanical Engineering Design 10th Edition

Published by McGraw-Hill Education
ISBN 10: 0073398209
ISBN 13: 978-0-07339-820-4

Chapter 9 - Problems - Page 499: 9-2

Answer

$F = 49.5$ kN

Work Step by Step

$F = 0.707 h l \tau_{allow} = 0.707(5)[2(50)](140)(10^{-3}) = 49.5$ kN
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