$\Sigma M_{hinge\,\,support} = 0$
$5P_{br} - 3P_{co} = 0$
$5\sigma_{br} \, A_{br} - 3\sigma_{co} \, A_{co} = 0$
$5(90)(1200) - 3\sigma_{co}(1500) = 0$
$\sigma_{co} = 120 \, \text{MPa}$
$\delta = \dfrac{\sigma L}{E}$
$\delta_{br} = \dfrac{90(2000)}{100\,000} = 1.8 \, \text{mm}$
$\delta_{co} = \dfrac{120(3000)}{120\,000} = 3 \, \text{mm}$
By ratio and proportion
$\dfrac{\delta_{T(co)} - \delta_{co}}{3} = \dfrac{\delta_{br} - \delta_{T(br)}}{5}$
$5\delta_{T(co)} - 5\delta_{co} = 3\delta_{br} - 3\delta_{T(br)}$
$5(16.8 \times 10^{-6})(3000) \, \Delta T - 5(3) = 3(1.8) - 3(18.7 \times 10^{-6})(2000) \, \Delta T$
$0.3642 \, \Delta T = 20.4$
$\Delta T = 56.01^\circ \text{C}$ drop in temperature answer
