$\delta_{T(al)} = (\alpha \, L \, \Delta T)_{al}$
$\delta_{T(al)} = (12.8 \times 10^{-6})(15)(120 - 60)$
$\delta_{T(al)} = 0.011\,52 \, \text{inch}$
$\delta_{T(st)} = (\alpha \, L \, \Delta T)_{st}$
$\delta_{T(st)} = (6.5 \times 10^{-6})(10)(120 - 60)$
$\delta_{T(st)} = 0.0039 \, \text{inch}$
$\delta_{T(al)} - \delta_{al} = \delta_{st} - \delta_{T(st)}$
$0.011\,52 - \left( \dfrac{PL}{AE} \right)_{al} = \left( \dfrac{PL}{AE} \right)_{st} - 0.0039$
$0.011\,52 - \dfrac{R(15)}{2(10 \times 10^6)} = \dfrac{(R + 50\,000)(10)}{3(29 \times 10^6)} - 0.0039$
$100\,224 - 6.525R = R + 50\,000 - 33\,930$
$84\,154 = 7.525R$
$R = 11\,183.25 \, \text{lbs}$
$P_{al} = R = 11\,183.25 \, \text{lbs}$
$P_{st} = R + 50\,000 = 61\,183.25 \, \text{lbs}$
$\sigma = \dfrac{P}{A}$
$\sigma_{al} = \dfrac{11\,183.25}{2} = 5\,591.62 \, \text{psi}$ answer
$\sigma_{st} = \dfrac{61\,183.25}{3} = 20\,394.42 \, \text{psi}$ answer
