Mechanics of Materials, 7th Edition

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

Chapter 1 - Problems: 1.4

Answer

$$P = 6.752 \space kip$$

Work Step by Step

1) Setup: Cross sectional Areas: $A_{AB} = \frac{1}{4} \pi d_{AB}^2 = \frac{1}{4} \pi * 1.25^2 = 1.227 \space in^2$ $A_{BC} = \frac{1}{4} \pi d_{BC}^2 = \frac{1}{4} \pi *0.75^2 = 0.4418 \space in^2$ 2)Internal Forces $\bullet AB:$ Taking a cut between points A and B and applying the equilibrium equations: $\sum {F_y} = 0; \space \space N_{AB} -12 - P = 0; \space\space N_{AB} = 12+P$ $\bullet BC:$ Taking a cut between points B and C and applying the equilibrium equations: $\sum {F_y} = 0; \space \space N_{BC} - P = 0; \space\space N_{BC} =P$ 3)Average Axial Stress Average axial stress is defined as follows: $\sigma_{AVG} = \frac{N_{INT}}{A}$ The stated requirement in the problem is that $\sigma_{AB} = \sigma_{BC}$ Substituting the average stress equation into the above condition, we get: $$\frac{N_{AB}}{A_{AB}} = \frac{N_{BC}}{A_{BC}}$$ $$\frac{12+P}{A_{AB}} = \frac{P}{A_{BC}}$$ Solving for P: $$P = \frac{12*A_{BC}}{A_{AB}-A_{BC}} = \boxed{6.752 \space kip} \space\space \leftarrow ANS$$
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