Solution to Problem 599 | Spacing of Rivets or Bolts in Built-Up Beams

Properties of W200 × 100
A = 12 700 mm²
t_f = 23.7 mm
d = 229 mm
I_xx = 113 × 10⁶ mm⁴
 

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