$I_{1-1} = I_w + 2nI_x = \dfrac{200(254^3)}{12} + 2(20)(32.7 \times 10^6)$

$I_{1-1} = 1\,581\,117\,733 ~ \text{mm}^4$

Measured from Section 1-1, the distance of extreme wood fiber and extreme steel fiber are equal. In this case, no need to investigate both materials.

$\dfrac{f_{bs}}{n} = \dfrac{120}{20} = 6 ~ \text{MPa}$

$f_{bw} = 8 ~ \text{MPa}$

Since *f*_{bs} / *n* < *f*_{bw}, steel is critical.

$\dfrac{f_{bs}}{n} = \dfrac{Mc}{I_{1-1}}$

$6 = \dfrac{M(127)}{1\,581\,117\,733}$

$M = 74\,698\,475.57 ~ \text{N}\cdot\text{mm}$

$M = 74.7 ~ \text{kN}\cdot\text{m}$ *answer*