Properties of W200 × 100
A = 12 700 mm
2tf = 23.7 mm
d = 229 mm
Ixx = 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
