Mechanics of Materials, 7th Edition

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

Chapter 1 - Problems: 1.2

Answer

$$ d_{AB} \geq 22.57 \space mm$$ $$d_{BC} \geq 40.16 \space mm$$

Work Step by Step

1) Internal Forces: $\bullet AB:$ Taking a cut between points A and B and applying the equilibrium equations: $\sum {F_x} = 0; \space\space -60\space kN + N_{AB} = 0$ $N_{AB} = 60 \space kN \space (Tension)$ $\bullet BC:$ Taking a cut between points B and C and applying the equilibrium equations: $\sum {F_x} = 0; \space\space -60\space kN + 2*125\space kN + N_{BC} = 0$ $N_{BC} = -190 \space kN \space (Compression)$ 2)Minimum Cross-sectional Area/Diameter From equation of the average normal stress: $\sigma_{avg} = \frac{N_{int}}{A} \space\space \Rightarrow A = \frac{N_{int}}{\sigma_{avg}}$ Since $A = \frac{1}{4} \pi d^2 \Rightarrow \space\space d=\sqrt{\frac{4A}{\pi}}$ $\bullet AB:$ $A_{AB} \geq \frac{60*10^3\space kN}{150 * 10^6 \space Pa} = 4*10^{-4} \space m^2$ $d_{AB} \geq \sqrt{\frac{4A}{\pi}} = \boxed {22.57 \space mm} \space\space\space\leftarrow ANS1$ $\bullet BC:$ $A_{min, BC} \geq \frac{190*10^3\space kN}{150 * 10^6 \space Pa} = 1.267*10^{-3} \space m^2$ $d_{AB} \geq \sqrt{\frac{4A}{\pi}} = \boxed {40.16 \space mm} \space\space\space\leftarrow ANS2$
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