Mechanics of Materials, 7th Edition

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

Chapter 1 - Problems: 1.18

Answer

$$P_{MAX} = 67.86 \space kN$$

Work Step by Step

Solution: 1) Shear Force Distribution Areas: For Shearing areas: $A=\pi * D * t_{PLATE}$ $\bullet Steel Rod:$ $A_{STEEL} = \pi * 12 *10^{-3} * 10 * 10^{-3} = 3.770 *10 ^{-4} \space m^2$ $\bullet Aluminium Plate:$ $A_{ALUMINIUM} = \pi * 40 *10^{-3} * 8 * 10^{-3} = 1.005 *10 ^{-3} \space m^2$ 2) Max Allowable Force: Max force can be obtained from the shear stress equation as follows: $\tau_{MAX} = \frac{P_{MAX}}{A} \space \Rightarrow \space P_{MAX} = \tau_{MAX} * A$ Now Calculate the max force needed to shear aluminium and steel: $\bullet Steel Rod:$ $P_{STEEL} = 180*10^6 *3.770 *10 ^{-4} = 67860 \space N$ $P_{STEEL} = 67.860 \space kN$ $\bullet Aluminium Rod:$ $P_{ALUMINIUM} = 70*10^6 * 1.005 *10 ^{-3} = 70350 \space N$ $P_{ALUMINIUM} = 70.350 \space kN$ 3)Taking the smallest of the above values as the limiting one we get the final answer: $\boxed{P_{MAX} = 67.86 \space kN} \leftarrow \space\space ANS$
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