Mechanics of Materials, 7th Edition

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

Chapter 1 - Problems - Page 21: 1.5

Answer

$$d_{IN} = 14.93 \space mm$$

Work Step by Step

1) Internal Forces: Taking a cut through point C and applying equilibrium equations: $\sum{F_y} = 0; \space\space N_{int} - 1200 \space N = 0 \space \Rightarrow N_{int} = 1200 \space N$ 2) Cross sectional Area: Cross sectional area of an annular cylinder: $A_{SXN} = \frac{\pi}{4}(d_{OUT}^2 - d_{IN}^2)$ 3) Average Normal Stress: $\sigma_{AVG} = \frac{N_{INT}}{A_{SXN}}$ Solving for $A_{SXN}$: $A_{SXN} = \frac{\pi}{4}(d_{OUT}^2 - d_{IN}^2) = \frac{N_{INT}}{\sigma_{AVG}}$ Solving the above equation for $d_{IN}$ $d_{IN} = \sqrt{d_{OUT}^2 - \frac{N_{INT}}{\sigma_{AVG}} * \frac{4}{\pi}} = \sqrt{(25*10^{-3} \space m)^2 - \frac{1200 \space N}{3.80 *10^6 \space Pa} * \frac{4}{\pi}}$ $d_{IN} = \boxed{14.93 \space mm} \space\space \leftarrow ANS$
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