Solution to Problem 239 Statically Indeterminate

$\delta_{st} = \delta_{al} + 0.10$

$\left( \dfrac{\sigma \, L}{E} \right)_{st} = \left( \dfrac{\sigma \, L}{E} \right)_{al} + 0.10$

$\dfrac{\sigma_{st} (250)}{200\,000} = \dfrac{\sigma_{al} (249.90)}{70\,000} + 0.10$

$0.00125 \sigma_{st} = 0.00357 \sigma_{al} + 0.10$

$\sigma_{st} = 2.856 \sigma_{al} + 80$
 

$\Sigma F_V = 0$

$2P_{st} + P_{al} = 400\,000$

$2\sigma_{st} \, A_{st} + \sigma_{al} \, A_{al} = 400\,000$

$2(2.856 \sigma_{al} + 80)1200 + \sigma_{al} (2400) = 400\,000$

$9254.4 \sigma_{al} + 192\,000 = 400\,000$

$\sigma_{al} = 22.48 \, \text{ MPa}$           answer