Modular ratio

$n_{st} = \dfrac{E_{st}}{E_{wd}} = \dfrac{200}{10} = 20$

$n_{al} = \dfrac{E_{al}}{E_{wd}} = \dfrac{70}{10} = 7$

Area and centroid

Transformed steel

$A_1 = 1600(20) = 32\,000 ~ \text{mm}^2$
$y_1 = \frac{1}{2}(20) = 10 ~ \text{mm}$

Wood

$A_2 = 80(150) = 12\,000 ~ \text{mm}^2$

$y_2 = 20 + \frac{1}{2}(150) = 95 ~ \text{mm}$

Transformed aluminum

$A_3 = 560(50) = 28\,000 ~ \text{mm}^2$

$y_3 = 20 + 150 + \frac{1}{2}(50) = 195 ~ \text{mm}$

Total area

$A = A_1 + A_2 + A_3 = 32\,000 + 12\,000 + 28\,000$

$A = 72\,000 ~ \text{mm}^2$

Location of the Neutral Axis from the top of the beam

$A\bar{y} = \Sigma A_n y_n$
$72\,000\bar{y} = 32\,000(10) + 12\,000(95) + 28\,000(195)$

$\bar{y} = 96.11 ~ \text{mm}$

Moment of Inertia about NA

$I_{NA} = \dfrac{bd^3}{12} + Ad^2$

$I_{NA_1} = \dfrac{1600(20^3)}{12} + 32\,000(\bar{y} - 10)^2 = 238\,350\,617.3 ~ \text{mm}^4$

$I_{NA_2} = \dfrac{80(150^3)}{12} + 12\,000(\bar{y} - 95)^2 = 22\,514\,814.81 ~ \text{mm}^4$

$I_{NA_3} = \dfrac{560(50^3)}{12} + 28\,000(195 - \bar{y})^2 = 279\,645\,679 ~ \text{mm}^4$

$I_{NA} = 238\,350\,617.3 + 22\,514\,814.81 + 279\,645\,679$

$I_{NA} = 540\,511\,111.11 ~ \text{mm}^4$

Moment capacity of the beam

$f_b = \dfrac{Mc}{I}$

Based on steel

$\dfrac{120}{n_{st}} = \dfrac{M(\bar{y})(1000^2)}{I_{NA}}$

$M = 33.74 ~ \text{kN}\cdot\text{m}$

Based on Wood

$10 = \dfrac{M(\bar{y} - 20)(1000^2)}{I_{NA}}$

$M = 71.02 ~ \text{kN}\cdot\text{m}$

Based on Aluminum

$\dfrac{80}{n_{al}} = \dfrac{M(220 - \bar{y})(1000^2)}{I_{NA}}$

$M = 49.9 ~ \text{kN}\cdot\text{m}$

For safe value of M, use M = 33.74 kN·m *answer*