$M_{max} = \frac{1}{4}PL = \frac{1}{4}P(12) = 3P \, \text{ lb}\cdot\text{ft}$
$V_{max} = \frac{1}{2}P$
$I = \frac{1}{12} (6)(12^3) = 864 \, \text{ in}^4$
From allowable flexural stress
$f_b = \dfrac{Mc}{I}$
$1200 = \dfrac{(3P \times 12)(6)}{864}$
$P = 4800 \, \text{ lb}$
Strength of bolt
$R = \dfrac{VQ_\text{1st plank}}{I}s$
$R = \dfrac{\frac{1}{2}(4800) \, [ \, 4(6)(4) \, ]}{864}(12)$
$R = 3200 \, \text{ lb}$
Normal force
$R = \mu N$
$3200 = 0.40N$
$N = 8000 \, \text{ lb}$
From tensile stress of bolt:
$\sigma = \dfrac{\text{Force}}{\text{Area}}$
$20\,000 = \dfrac{8000}{\frac{1}{4}\pi d^2}$
$d = 0.7136 \, \text{ in}$ answer
