By symmetry
$C_V = B_V = 400 ~ \text{lb}$
$\Sigma M_A = 0$
$8B_H = 6B_V + 8(500)$
$8B_H = 6(400) + 8(500)$
$B_H = 800 ~ \text{lb}$
Moment at B
$M_B = 500(2) = 1000 ~ \text{lb}\cdot\text{ft}$
Axial Force at B
$P_a = 400 + B_V \cos \alpha + B_H \cos \theta$
$P_a = 400 + 400(4/5) + 800(3/5)$
$P_a = 1200 ~ \text{lb compression}$
Maximum Compressive Stress Will Occur at Point B
$\sigma_c = \dfrac{P_a}{bd} + \dfrac{6M_B}{bd^2}$
$\sigma_c = \dfrac{1200}{4(4)} + \dfrac{6(1000)(12)}{4(4^2)}$
$\sigma_c = 1200 ~ \text{psi}$ answer
Axial, Shear, and Moment Diagrams
Not actually necessary but just in case you need it.
$P_1 = 800 \cos \theta = 800(3/5) = 480 ~ \text{lb}$
$P_2 = 900 \sin \theta = 900(4/5) = 720 ~ \text{lb}$
$P_3 = 800 \cos \theta = 800(3/5) = 480 ~ \text{lb}$
$P_4 = 400 \sin \theta = 400(4/5) = 320 ~ \text{lb}$
$P_5 = 500 \sin \theta = 500(4/5) = 400 ~ \text{lb}$
$V_1 = 800 \sin \theta = 800(4/5) = 640 ~ \text{lb}$
$V_2 = 900 \cos \theta = 900(3/5) = 540 ~ \text{lb}$
$V_3 = 800 \sin \theta = 800(4/5) = 640 ~ \text{lb}$
$V_4 = 400 \cos \theta = 400(3/5) = 240 ~ \text{lb}$
$V_5 = 500 \cos \theta = 500(3/5) = 300 ~ \text{lb}$
