Assuming complete freedom:
$\delta_T = \alpha \, L \, \Delta T$
$\delta_{T(co)} = (16.8 \times 10^{-6})(750)(95 - 10)$
$\delta_{T(co)} = 1.071 \, \text{mm}$
$\delta_{T(al)} = (23.1 \times 10^{-6})(750 - 0.18)(95 - 10)$
$\delta_{T(al)} = 1.472 \, \text{mm}$
From the figure:
$\delta_{T(al)} - \delta_{al} - \Delta = \delta_{T(co)} + \delta_{co}$
$1.472 - \left( \dfrac{PL}{AE} \right)_{al} - 0.18 = 1.071 + \left( \dfrac{PL}{AE} \right)_{co}$
$1.472 - \dfrac{2F(750 - 0.18)}{400(70\,000)} - 0.18 = 1.071 + \dfrac{F(750)}{500(120\,000)}$
$0.221 = (6.606 \times 10^{-5}) \, F$
$F = 3345.44 \, \text{N}$
$P_{co} = F = 3345.44 \, \text{N}$
$P_{al} = 2F = 6690.89 \, \text{N}$
$\sigma = P/A$
$\sigma_{co} = \dfrac{3345.44}{500} = 6.69 ~ \text{MPa}$ answer
$\sigma_{al} = \dfrac{6690.89}{400} = 16.73 ~ \text{MPa}$ answer
