Mechanics of Materials, 7th Edition

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

Chapter 1 - Problems - Page 20: 1.1

Answer

$$ \sigma_{AB} = 84.88 \space\space MPa\space\space (Tension)$$ $$ \sigma_{BC} = -96.77 \space\space MPa\space\space (Compression)$$

Work Step by Step

Solution: 1)Set up: Area AB: $A_{AB} = \frac{1}{4}\pi d_1^2 = 7.0686 * 10^{-4} \space \space m^2$ Area BC: $A_{BC} = \frac{1}{4}\pi d_1^2 = 1.9635 * 10^{-3} \space\space m^2$ 2) Internal Forces: $\bullet AB:$ Taking a cut between points A and B and applying the equilibrium equation, we get: $\sum{F_x} = 0; \space\space -60 \space kN + N_{AB} = 0$ $N_{AB} = 60 \space kN \space (Tension)$ $\space\space\space\space\leftarrow N_{AB}$ $\bullet BC:$ Taking a cut between points B and C and applying the equilibrium equation, we get: $\sum{F_x} = 0; \space\space -60 \space kN + 2*125 \space kN + N_{BC} = 0$ $N_{BC} = -190 \space kN \space (Compression)$ $\space\space\space\space\leftarrow N_{BC}$ 3)Average Normal Stress Average Normal stress is equal to : $\sigma_{avg} = \frac{N}{A}$ $\bullet AB:$ $\sigma_{AB} = \frac{N_{AB}}{A_{AB}} = \frac {60*10^3 \space N}{7.0686*10^{-4} \space m^2} = 84.88*10^6 \space Pa = \boxed{84.88 \space MPa\space (Tension)}$ $\space\space\space\space\leftarrow ANS1$ $\bullet BC:$ $\sigma_{BC} = \frac{N_{BC}}{A_{BC}} = \frac {-190*10^3 \space N}{1.9635*10^{-3} \space m^2} = -96.77*10^6 \space Pa = \boxed{-96.77 \space MPa\space (Compression)}$ $\leftarrow ANS2$
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