$$\begin{gathered} 20-\mathrm{mm} \mathrm{diameter} \mathrm{steel} \mathrm{rod} \\ \mathrm{A}=314.16 {\mathrm{mm}}^2 \\ \mathrm{E}=200 \mathrm{GPa} \\ \mathrm{\alpha }=12.0\times {10}^{-6}{/}^{\circ }\mathrm{C} \end{gathered}$$

The force-temperature-deformation relationship for an axial member is given by the following equation.
(A tension force will be assumed.)

$\mathrm{\delta }=\frac{\mathrm{FL}}{\mathrm{AE}}+\mathrm{\alpha }∆\mathrm{TL}$