Shigley's Mechanical Engineering Design 10th Edition

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

Chapter 1 - Problems - Page 36: 1-9

Answer

$n_{d}$ = 1.43 Ans

Work Step by Step

Problem 1-9 Given: Max. load = 1.10 P Min. area = $(0.95)^{2}$ A Min. strength = 0.85 S Analysis: We will use the design factor from eq (1-1) $n_{d}$=$\frac{loss-of-function-parameter}{maximum -allowable-parameter}$ $n_{d}$=$\frac{1.10}{(0.85)(0.95^{2})}$ $n_{d}$ = 1.43 Ans
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