Moment of inertia

$I = \dfrac{200(250^3)}{12} - \dfrac{150(200^3)}{12} = 160\,416\,666.7 ~ \text{mm}^4$

Maximum Shear

$V_{max} = \frac{2}{3}P$

Value of *P* based on the strength of nails

$R = 2(1300) = 2600 ~ \text{N}$
$Q = 150(25)(112.5) = 421\,875 ~ \text{mm}^3$

$s = \dfrac{RI}{VQ}$

$125 = \dfrac{2600(160\,416\,666.7)}{\frac{2}{3}P(421\,875)}$

$P = 11.86 ~ \text{kN}$

Value of *P* based on shear strength of wood

$Q = 200(125)(62.5) - 150(100)(50) = 812\,500 ~ \text{mm}^3$

$F_v = \dfrac{VQ}{Ib}$

$0.827 = \dfrac{\frac{2}{3}P(812\,500)}{160\,416\,666.7(50)}$

$P = 12.25 ~ \text{kN}$

For safe value of *P*, use *P* = 11.86 kN *answer for part (1)*

Maximum Flexural Stress

Maximum Moment

$M_{max} = 1.2(\frac{2}{3}P) = 0.8(11.86)$
$M_{max} = 9.488 ~ \text{kN}\cdot\text{m}$

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

$f_b = \dfrac{9.488(125)(1000^2)}{160\,416\,666.7}$

$f_b = 7.39 ~ \text{MPa}$ *answer for part (2)*