Mechanics of Materials, 7th Edition

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

Chapter 1 - Problems - Page 21: 1.6


(a) $a = 35.715$ m (b) Maximum normal stress $= 5.342$ MPa

Work Step by Step

First, write down the equations for the normal stress in members AB and BC respectively: $\sigma_{AB} = \frac{P_{o}}{\frac{\pi}{4}(15\times10^{-3})^{2}}$ $\sigma_{BC} = \frac{P_{1}}{\frac{\pi}{4}(10\times10^{-3})^{2}}$ Now, determine $P_{o}$ and $P_{1}$. The assembly has no external forces acting on it besides the force of gravity. Treat members AB and BC separately and calculate the force of gravity each is subjected to (note that the acceleration due to gravity is $9.81\frac{m}{s^{2}}$, and the density of brass, which is given in the problem, is $8470\frac{kg}{m^{3}}$): The weight of AB = $\frac{\pi}{4}(15\times10^{-3})^{2}(a)(8470)(9.81) = 14.683a$ Newtons The weight of BC = $\frac{\pi}{4}(10\times10^{-3})^{2}(b)(8470)(9.81) = 6.526b$ Newtons Summation of the forces in the y-direction in member AB: $\sum F_{y}=-14.683a-6.526b+P_{o}=0$ $P_{o} =14.683a+6.526b$ Summation of the forces in the y-direction in member BC: $\sum F_{y}=-6.526b+P_{1}=0$ $P_{1} =6.526b$ Equate variables $a$ and $b$ with a simple geometric relationship: $a + b = 100$ Using the above relationship, $P_{o}$ and $P_{1}$ can be equated in terms of either only $a$ or only $b$. Since the problem wants us to ultimately find the length of rod AB, let's equate $P_{o}$ and $P_{1}$ in terms of $a$. $P_{o} =14.683a+6.526(100-a)=8.157a+652.6$ $P_{1} =6.526(100-a)=652.6-6.526a$ We need to find the length of rod AB for which the maximum normal stress in ABC is a minimum. This occurs when $\sigma_{AB}$ is equal to $\sigma_{BC}$. $\frac{P_{o}}{\frac{\pi}{4}(15\times10^{-3})^{2}} = \frac{P_{1}}{\frac{\pi}{4}(10\times10^{-3})^{2}}$ $\frac{8.157a+652.6}{\frac{\pi}{4}(15\times10^{-3})^{2}} = \frac{652.6-6.526a}{\frac{\pi}{4}(10\times10^{-3})^{2}}$ ${(8.157a+652.6)}\times{\frac{\pi}{4}(10\times10^{-3})^{2}} = {(652.6-6.526a)}\times{\frac{\pi}{4}(15\times10^{-3})^{2}}$ $6.406\times10^{-4}a+0.051 = 0.115 - 1.153\times10^{-3}a$ $1.794\times10^{-3}a = 0.064$ $a = 35.715$ m Please note that each intermediate step contains rounded numbers, whereas the final value of $a$ was computed using raw data. Lastly, the question asks for the corresponding value of the maximum normal stress. Since $\sigma_{AB} = \sigma_{BC}$, use either to obtain the maximum normal stress: $\sigma_{AB} = \frac{{8.157a+652.6}}{\frac{\pi}{4}(15\times10^{-3})^{2}}$ $\sigma_{AB} = \frac{{8.157\times35.715+652.6}}{\frac{\pi}{4}(15\times10^{-3})^{2}}$ $\sigma_{AB} = 5341535.59$ Pa or $5.342$ MPa
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