For concentrated load
P at midspan of a simply supported beam of span
L = 12 ft.
$V_{max} = \frac{1}{2}P$
$M_{max} = \frac{1}{4}PL = \frac{1}{4}P(12) = 3P$
From the cross section shown:
$I =\dfrac{8(8^3)}{12} - \dfrac{4(4^3)}{12} = 320 \, \text{ in}^4$
$Q_{NA} = 4(2)(2) + 6(2)(3) + 2(2)(1) = 56 \, \text{ in}^3$
From bending stress
$f_b = \dfrac{Mc}{I}$
$1400 = \dfrac{3P(12)(4)}{320}$
$P = 3111.11 \, \text{ lb}$
Maximum shear stress
$f_v = \dfrac{VQ_{NA}}{Ib} = \dfrac{\frac{1}{2}(3111.11)(56)}{320(4)}$
$f_v = 68.06 \, \text{ psi}$ answer
Spacing (or pitch) of screws, s
From Strength of Screws
$s = \dfrac{RI}{VQ_\text{screws}}$
For Horizontal Bolts
$Q_h = 4(2)(3) = 24 ~ \text{in.}^3$
$s_h = \dfrac{2(200)(320)}{(3111.11/2)(24)} = 3.42 ~ \text{in.}$ answer
For Vertical Bolts
$Q_v = 8(2)(3) = 48 ~ \text{in.}^3$
$s_v = \dfrac{2(200)(320)}{(3111.11/2)(48)} = 1.734 ~ \text{in.}$ answer
