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.
$$\frac{F}{A}=\alpha E \Delta T=(25\times10^{-6})(70\times10^9\;N/m^2)(35^{\circ}-12^{\circ})$$
$$=4.0\times10^7\;N/m^2$$