Mechanics of Materials, 7th Edition

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

Chapter 2 - Problems - Page 73: 2.2


$$\delta = 0.0303 \space in$$ $$\sigma_{AVG} = 15.28 \space ksi$$

Work Step by Step

1. Average Tensile Stress: $\sigma_{AVG} = \frac{P}{A}$ For a constant, circular cross section: $A = \frac{\pi D^4}{4}$ Substituting all known values into the tensile stress equation, obtain average tensile stress: $\sigma_{AVG} = \frac{P}{A} = \frac{4 * 750 \space lb}{\pi * (1/4)^2 \space in^2} = 15278\space psi = \boxed{15.28\space ksi}$ $\leftarrow ANS (b)$ 2. Young's Law (strain calculation) $\sigma = E\epsilon$ Substituting all known values and solving for the strain: $\epsilon = \frac{\sigma}{E} = \frac{15.28 \space ksi}{29000 \space ksi} = 5.269*10^-4 \space in/in$ Finally the actual elongation of the rope is calculated as follows: $\delta = \epsilon * L = 5.269*10^-4 \space ft/ft * 4.8 ft = 2.529*10^-3 \space ft = \boxed{0.0303 \space in}$ $\leftarrow ANS (a)$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.