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: 1.3

Answer

$$\sigma_{AB} = 17.93 \space ksi \space (Tension)$$ $$\sigma_{BC} = 22.63 \space ksi \space (Tension)$$

Work Step by Step

1) Setup: $A_{AB} = \frac{1}{4} \pi d_{AB}^2 =1.227 \space in^2$ $A_{BC} = \frac{1}{4} \pi d_{BC}^2 =0.4418 \space in^2$ 2) Internal Forces: $\bullet AB:$ Taking a cut between points A and B and applying the equilibrium equations: $\sum{F_y} = 0; \space\space N_{AB} -10 -12 = 0 \Rightarrow N_{AB} = 22 kip \space\space(Tension)$ $\bullet BC:$ Taking a cut between points B and C and applying the equilibrium equations: $\sum{F_y} = 0; \space\space N_{BC} -10 = 0 \Rightarrow N_{BC} = 10 kip \space\space(Tension)$ 3) Average Normal Stress: From the formula for average normal Stress: $\sigma_{avg} = \frac{N_{int}}{A}$ $\bullet AB:$ $\sigma_{AB} = \frac{N_{AB}}{A_{AB}} = \frac{22 \space kip}{1.227 \space in^2} = \boxed{17.93 \space ksi\space(Tension)}$ $\space\space\leftarrow ANS1$ $\bullet BC:$ $\sigma_{BC} = \frac{N_{BC}}{A_{BC}} = \frac{10 \space kip}{0.4418\space in^2} = \boxed{22.63 \space ksi\space(Tension)}$ $\space\space\leftarrow ANS1$
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