$\delta_{steel} = \delta_{timber}$
$\left( \dfrac{\sigma L}{E} \right)_{steel} = \left( \dfrac{\sigma L}{E} \right)_{timber}$
$\dfrac{\sigma_{steel} L}{29 \times 10^6} = \dfrac{\sigma_{bronze} L}{1.5 \times 10^6}$
$1.5 \sigma_{steel} = 29 \sigma_{timber}$
When σtimber = 1200 psi
$1.5 \sigma_{steel} = 29(1200)$
$\sigma_{steel} = 23\,200 \, \text{ psi } = 23.2 \, \text{ ksi } \gt 20 \, \text{ ksi }$ (not ok!)
When σsteel = 20 ksi
$1.5(20 \times 1000) = 29 \sigma_{timber}$
$\sigma_{timber} = 1034.48 \, \text{ psi} \lt 1200 \, \text{ psi}$ (ok!)
Use σsteel = 20 ksi and σtimber = 1.03 ksi
$\Sigma F_V = 0$
$F_{steel} + F_{timber} = 300$
$(\sigma \, A)_{steel} + (\sigma \, A)_{timber} = 300$
$20 \,[ \, 4(8t) \, ] + 1.03(8^2) = 300$
$t = 0.365 \, \text{ in}$ answer
