$\delta_{co} = \delta_{st}$
$\left( \dfrac{\sigma L}{E} \right)_{co} = \left( \dfrac{\sigma L}{E} \right)_{st}$
$\dfrac{\sigma_{co} (160)}{120\,000} = \dfrac{\sigma_{st} (240)}{200\,000}$
$10 \sigma_{co} = 9 \sigma_{st}$
When σst = 140 MPa
$\sigma_{co} = \frac{9}{10} (140)$
$\sigma_{co} = 126 \, \text{ MPa } > 70 \, \text{ MPa } \,$ (not okay!)
When σco = 70 MPa
$\sigma_{st} = \frac{10}{9} (70)$
$\sigma_{st} = 77.78 \, \text{ MPa } \lt 140 \, \text{ MPa } \,$ (okay!)
Use σco = 70 MPa and σst = 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
