## College Physics (4th Edition)

Published by McGraw-Hill Education

# Chapter 13 - Problems - Page 501: 95

#### Answer

The stress that develops in the girder is $4.80\times 10^7~N/m^2$

#### Work Step by Step

We can find the strain in the girder by considering the change in length $\Delta L$ that would occur if the girder was not constrained: $\frac{\Delta L}{L} = \alpha~\Delta T$ $\frac{\Delta L}{L} = (12\times 10^{-6}~K^{-1})(20~K)$ $\frac{\Delta L}{L} = 2.40\times 10^{-4}$ $Y = \frac{F/A}{\Delta L/L}$ $Y$ is Young's modulus $F$ is the force $A$ is the cross-sectional area $\Delta L$ is the change in length $L$ is the original length Note that $\frac{F}{A}$ is the stress and $\frac{\Delta L}{L}$ is the strain. We can find the stress $\frac{F}{A}$: $\frac{F}{A} = Y~\frac{\Delta L}{L}$ $\frac{F}{A} = (2.0\times 10^{11}~N/m^2)~(2.40\times 10^{-4})$ $\frac{F}{A} = 4.80\times 10^7~N/m^2$ The stress that develops in the girder is $4.80\times 10^7~N/m^2$

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.