Moment of Inertia
$I = \sum \dfrac{bd^3}{12} = \dfrac{190(190^3)}{12} - \dfrac{110(110^3)}{12}$
$I = 96\,400\,000 ~ \text{mm}^4$
$f_b = \dfrac{Mc}{I}$
$8.3 = \dfrac{M(190/2)(1000^2)}{96\,400\,000}$
$M = 8.422 ~ \text{kN}\cdot\text{m}$
$M = \dfrac{PL}{4}$
$8.422 = \dfrac{P(4)}{4}$
$P = 8.422 ~ \text{kN}$
$V_{max} = 0.5P = 4.211 ~ \text{kN}$
Spacing of Screws
$s = \dfrac{RI}{VQ}$
Part 1: Spacing of screws at A
$Q = 190(40)(75) = 570\,000 ~ \text{mm}^3$
$R = 2 \times 890 = 1,780 ~ \text{N}$
$s = \dfrac{1,780(96\,400\,000)}{4\,211(570\,000)} = 71.49 ~ \text{mm}$
Use s = 70 mm answer
Part 2: Spacing of screws at B
$Q = 110(40)(75) = 330\,000 ~ \text{mm}^3$
$R = 2 \times 890 = 1,780 ~ \text{N}$
$s = \dfrac{1,780(96\,400\,000)}{4\,211(330\,000)} = 123.48 ~ \text{mm}$
Use s = 123 mm answer
