Solution to Problem 263 Thermal Stress

Temperature at which δ_T = 3 mm:
$\delta_T = \alpha \, L \, (\Delta T)$

$\delta_T = \alpha \, L \, (T_f \, - \, T_i)$

$3 = (11.7 \times 10^{-6})(10\,000)(T_f \, - \, 15)$

$T_f = 40.64^\circ\text{C}$           answer
 

Required stress:
$\delta = \delta_T$

$\dfrac{\sigma \, L}{E} = \alpha \, L \, (\Delta T)$

$\sigma = \alpha \, E \, (T_f \, - \, T_i)$

$\sigma = (11.7 \times 10^{-6})(200\,000)(40.64 - 15)$

$\sigma = 60 \, \text{ MPa}$           answer