Properties of W200 × 100

*A* = 12 700 mm

^{2}*t*_{f} = 23.7 mm

*d* = 229 mm

*I*_{xx} = 113 × 10

^{6} mm

^{4}
Maximum moment

$M_{max} = \dfrac{w_o L^2}{8} = \dfrac{30(10^2)}{8}$

$M_{max} = 375 \, \text{ kN}\cdot\text{m}$

Maximum shear

$V_{max} = \dfrac{w_o L}{2} = \dfrac{30(10)}{2}$

$V_{max} = 150 \, \text{ kN}$

By transfer formula for moment of inertia

$I_{NA} = 2 \, [ \, (113 \times 10^6) + (12700)(229/2)^2 \, ]$

$I_{NA} = 559\,000\,350 \, \text{ mm}^4$

Maximum flexural stress

$(\,f_b\,)_{max} = \dfrac{Mc}{I} = \dfrac{375(1000^2)(229)}{559\,000\,350}$

$(\,f_b\,)_{max} = 153.62 \, \text{ MPa}$ *answer*

Bolt pitch

$s = \dfrac{RI}{VQ_{NA}}$

$s = \dfrac{(30 \times 2)(559\,000\,350)}{150 \, [ \, 12\,700(229 /2) \, ]}$

$s = 153.77 \, \text{ mm}$ *answer*