Chapter 13, Problem 84A

To determine

Temperature atwhich rivet must be cooled to fit into the rivet hole.

Answer to Problem 84A

Rivet must be cooled at $-1.2\times {10}^2°\text{C}$ .

Explanation of Solution

Given:

Diameter of rivet, $L_1\text{= 1}\text{.2250 cm}$

Diameter of hole, $L_2\text{= 1}\text{.2230 cm}$

Temperature of rive hole, $T\text{= 20}°\text{C}$

Linear coefficient of expansion, $\alpha {\text{ = 12×10}}^{\text{-6}}\text{per}°\text{C}$

Formula used:

Linear thermal expansion:

  $\Delta L=a{L}_1\Delta T$

Where, $\Delta L$ is the change in length  $L_1$ , $\alpha$ is a coefficient of linear expansion, $\Delta T$ is the change in temperature and $L_1$ is the length at temperature $T$

Calculation:

  $\left(L_2-L_1\right)=\alpha L_1\left(T_2-T_1\right)$

Where, $L_1$ is the length at temperature $T_1$ and $L_2$ is the length at temperature $T_2$ .

Now, substitute all the given values and solve:

  $L_2-L_1=12\times {10}^{-6}\text{per}°\text{C}\times 1.2250\text{cm}\times \left(T_2-T_1\right)$

  $1.2230-1.2250=12\times {10}^{-6}\times 1.2250\times \left(T_2-20\right)$

  $\frac{-0.0020}{14.7\times {10}^{-6}}=T_2-20$

  $T_2=\frac{-20\times {10}^2}{14.7}+20$

  $T_2=-1.36\times {10}^2+20$

  $T_2=-1.2\times {10}^2°\text{C}$

Conclusion:

At $-1.2\times {10}^2°\text{C}$ rivet must be cooled to fit into rivet hole.