Solution to Problem 271 Thermal Stress

Stress in bronze when σ_st = 55 MPa
$\Sigma M_A = 0$

$4P_{br} + P_{st} = 2.5(80\,000)$

$4\sigma_{br}(1300) + 55(320) = 2.5(80\,000)$

$\sigma_{br} = 35.08 \, \text{MPa}$
 

 

By ratio and proportion:
$\dfrac{\delta_{T(st)} + \delta_{st}}{1} = \dfrac{\delta_{T(br)} + \delta_{br}}{4}$

$\delta_{T(st)} + \delta_{st} = \dfrac{\delta_{T(br)}}{4} + \dfrac{\delta_{br}}{4}$

$(\alpha \, L \, \Delta T)_{st} + \left( \dfrac{\sigma L}{E} \right)_{st} = \dfrac{(\alpha \, L \, \Delta T)_{br}}{4} + \left( \dfrac{\sigma L}{4E} \right)_{br}$

$(11.7 \times 10^{-6})(1500) \Delta T + \dfrac{55(1500)}{2000} = \dfrac{(18.9 \times 10^{-6})(3000) \Delta T}{4} + \dfrac{35.08(3000)}{4(83\,000)}$

$0.017\,55 \Delta T + 0.4125 = 0.014\,175 \Delta T + 0.317$

$\Delta T = -28.3^\circ C$
 

A temperature drop of 28.3°C is needed to stress the steel to 55 MPa.           answer