Unit weight of wood

$\gamma = 8 \, \dfrac{\text{kN}}{\text{m}^3} \times \dfrac{1000 ~ \text{N}}{1 ~ \text{kN}} \times \left( \dfrac{1 ~ \text{m}}{1000 ~ \text{mm}} \right)^3$

$\gamma = 0.000\,008 ~ \text{N/mm}^3$

$w_o = \dfrac{3000}{400} + 0.000\,008(100d)$

$w_o = 7.5 + 0.0008d ~ \text{N/mm}$

**Part 1: Based on allowable bending stress (***F*_{b} = 24 MPa)

$M = \dfrac{w_o L^2}{8}$
$M = \dfrac{(7.5 + 0.0008d)(5000^2)}{8}$

$M = 3\,125\,000(7.5 + 0.0008d) ~ \text{N}\cdot\text{mm}$

$F_b = \dfrac{6M}{bd^2}$

$24 = \dfrac{6 [ \, 3\,125\,000(7.5 + 0.0008d) \, ]}{100d^2}$

$d = 245.2 ~ \text{mm}$ *answer*

**Part 2: Based on allowable shear stress (***F*_{v} = 1.24 MPa)

$V = R = \dfrac{w_o L}{2}$
$V = \dfrac{(7.5 + 0.0008d)(5000)}{2}$

$V = 2\,500(7.5 + 0.0008d)$

$F_v = \dfrac{3V}{2bd}$

$1.24 = \dfrac{3 \, [ \, 2\,500(7.5 + 0.0008d) \, ]}{2(100d)}$

$d = 232.44 ~ \text{mm}$ *answer*

**Part 3: Based on allowable deflection (δ = ***L*/240)

$\delta = \dfrac{L}{240} = \dfrac{5000}{240}$
$\delta = \dfrac{125}{6} \, \text{mm}$

$\delta = \dfrac{5w_o L^4}{384 EI}$

$\dfrac{125}{6} = \dfrac{5(7.5 + 0.0008d)(5000^4)}{384(18 \, 600) \left( \dfrac{100d^3}{12} \right)}$

$d = 268.9 ~ \text{mm}$ *answer*