Solution to Problem 266 Thermal Stress

$\delta_T = \alpha \, L \, \Delta T$
 

 

$\delta_{T(st)} = (6.5 \times 10^{-6})(15)(100)$

$\delta_{T(st)} = 0.00975$
 

$\delta_{T(al)} = (12.8 \times 10^{-6})(10)(100)$

$\delta_{T(al)} = 0.0128 \, \text{in}$
 

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

$\left( \dfrac{PL}{AE} \right)_{st} + \left( \dfrac{PL}{AE} \right)_{al} = 0.00975 + 0.0128$
 

Where P_st = P_al = P. Thus,
$\dfrac{P(15)}{1.5(29 \times 10^6)} + \dfrac{P(10)}{2(10 \times 10^6)} = 0.02255$

$P = 26\,691.84 \, \text{lb}$
 

$\sigma = \dfrac{P}{A}$

$\sigma_{st} = \dfrac{26\,691.84}{1.5} = 17\,794.56 \, \text{ psi}$          answer

$\sigma_{al} = \dfrac{26\,691.84}{2.0} = 13\,345.92 \, \text{ psi}$          answer