Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 13 - Temperature and Kinetic Theory - Problems - Page 386: 19

Answer

$4.0\times10^7\;N/m^2$.

Work Step by Step

The theory is discussed on page 367. The applied stress must counteract the thermal stress. Calculate the required stress. Here, E is Young’s modulus for aluminum, given in Table 9-1. $$\frac{F}{A}=\alpha E \Delta T=(25\times10^{-6})(70\times10^9\;N/m^2)(35^{\circ}-12^{\circ})$$ $$=4.0\times10^7\;N/m^2$$
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