$\Sigma M_{R2} = 0$

$12R_1 = 8P $

$R_1 = \frac{2}{3}P$

$\Sigma M_{R1} = 0$

$12R_2 = 4P$

$R_2 = \frac{1}{3} P$

$M_{max} = 4(\frac{2}{3}P) = \frac{8}{3}P \, \text{ lb}\cdot\text{ft}$

Based on allowable shearing force of beam

$f_v = \dfrac{VQ_{NA}}{Ib}$

Where:

*V* = 2/3 *P*

*Q*_{NA} = 6(1)(4.5) + 2 [ 5(1)(2.5) ] = 52 in^{3}

*I* = 8(10^{3})/12 - 6(8^{3})/12 = 410.67 in^{4}

*b* = 2 in

*f*_{v} = 120 psi

Thus,

$120 = \dfrac{\frac{2}{3}P(52)}{410.67(2)}$

$P = 2843.1 \, \text{ lb}$

Based on allowable shearing force of the screws

$s = \dfrac{RI}{VQ_{screw}}$

Where:

*R* = 2(300) = 600 lb

*V* = 2/3 *P*

*Q*_{screw} = 6(1)(4.5) = 27 in^{3}

*I* = 410.67 in^{4}

*s* = 5 in

Thus,

$5 = \dfrac{600(410.67)}{\frac{2}{3}P(27)}$

$P = 2737.8 \, \text{ lb}$

For safe value of *P*, use *P* = 2737.8 lb. *answer*

Bending stress:

$f_b = \dfrac{Mc}{I} = \dfrac{\frac{8}{3}(2737.8)(12)(5)}{410.67}$

$f_b = 1066.67 \, \text{ psi}$ *answer*