Solution to Problem 236 Statically Indeterminate

$\delta_{co} = \delta_{st}$

$\left( \dfrac{\sigma L}{E} \right)_{co} = \left( \dfrac{\sigma L}{E} \right)_{st}$

$\dfrac{\sigma_{co} (160)}{120\,000} = \dfrac{\sigma_{st} (240)}{200\,000}$

$10 \sigma_{co} = 9 \sigma_{st}$
 

When σ_st = 140 MPa
$\sigma_{co} = \frac{9}{10} (140)$

$\sigma_{co} = 126 \, \text{ MPa } > 70 \, \text{ MPa } \,$ (not okay!)
 

When σ_co = 70 MPa
$\sigma_{st} = \frac{10}{9} (70)$

$\sigma_{st} = 77.78 \, \text{ MPa } \lt 140 \, \text{ MPa } \,$ (okay!)
 

Use σ_co = 70 MPa and σ_st = 77.78 MPa.
 

$\Sigma F_V = 0$

$2P_{co} + P_{st} = W$

$2(\sigma_{co} \, A_{co}) + \sigma_{st} \, A_{st} = Mg$

$2 \, [ \, 70(900) \, ] + 77.78(1200) = M(9.81)$

$M = 22 358.4 \, \text{ kg}$           answer