Contraction of steel rod, assuming complete freedom:
$\delta_{T(st)} = \alpha \, L \, \Delta T$
$\delta_{T(st)} = (11.7 \times 10^{-6})(900)(40)$
$\delta_{T(st)} = 0.4212 \, \text{mm}$
The steel rod cannot freely contract because of the resistance of aluminum rod. The movement of A (referred to as δA), therefore, is less than 0.4212 mm. In terms of aluminum, this movement is (by ratio and proportion):
$\dfrac{\delta_A}{0.6} = \dfrac{\delta_{al}}{1.2}$
$\delta_A = 0.5 \delta_{al}$
$\delta_{T(st)} - \delta_{st} = 0.5 \delta_{al}$
$0.4212 - \left( \dfrac{PL}{AE} \right)_{st} = 0.5 \left( \dfrac{PL}{AE} \right)_{al}$
$0.4212 - \dfrac{P_{st} (900)}{300(200\,000)} = 0.5 \left[ \dfrac{P_{al} (1200)}{1\,200(70\,000)} \right]$
$28080 - P_{st} = 0.4762P_{al}$ → Equation (1)
$\Sigma M_B = 0$
$0.6P_{st} = 1.2P_{al}$
$P_{st} = 2P_{al}$ → Equation (2)
Equations (1) and (2)
$28\,080 - 2P_{al} = 0.4762P_{al}$
$P_{al} = 11\,340 \, \text{N}$
$\sigma_{al} = \dfrac{P_{al}}{A_{al}} = \dfrac{11\,340}{1200}$
$\sigma_{al} = 9.45 \, \text{ MPa}$ answer
