Solution to Problem 274 Thermal Stress

$\sigma_{al} = \sigma_{st}$

$\dfrac{R_1}{2} = \dfrac{(50\,000 - R_1)}{3}$

$3R_1 = 100\,000 - 2R_1$

$R_1 = 20\,000 \, \text{lbs}$
 

 

$\delta = \dfrac{PL}{AE}$

$\delta_{al} = \dfrac{20\,000(15)}{2(10 \times 10^6)} = 0.015 \, \text{inch}$

$\delta_{st} = \dfrac{(50\,000 - 20\,000)(10)}{3(29 \times 10^6)} = 0.003\,45 \, \text{inch}$
 

$\delta_{al} - \delta_{T(al)} = \delta_{st} + \delta_{T(st)}$

$0.015 - (12.8 \times 10^{-6})(15) \, \Delta T = 0.003\,45 + (6.5 \times 10^{-6})(10) \, \Delta T$

$0.011\,55 = 0.000\,257 \, \Delta T$

$\Delta T = 44.94^\circ F$
 

A drop of 44.94°F from the standard temperature will make the aluminum and steel segments equal in stress.       answer