$\Sigma F_H = 0$
$F_H = 2000 \cos 15^\circ ~ \text{lb}$
$\Sigma M_E = 0$
$6F_V = \frac{1}{4}(2000 \cos 15^\circ) + 2(2000 \sin 15^\circ)$
$F_V = \frac{250}{3}\cos 15^\circ + \frac{2000}{3}\sin 15^\circ ~ \text{lb}$
$F_V = \frac{250}{3}(\cos 15^\circ + 8 \sin 15^\circ) ~ \text{lb}$
At section through C
$M_C = \Sigma M_\text{to the right of C}$
$M_C = 3F_V$
$M_C = 250(\cos 15^\circ + 8 \sin 15^\circ) ~ \text{lb}\cdot\text{ft}$
$\sigma_a = \dfrac{P}{A} = \dfrac{2000 \cos 15^\circ}{2(6)}$
$\sigma_a = 160.99 ~ \text{psi}$
$\sigma_f = \dfrac{6M}{bd^2} = \dfrac{6[ \, 250(\cos 15^\circ + 8\sin 15^\circ) \, ] (12)}{2(6^2)}$
$\sigma_f = 759.11 ~ \text{psi}$
$\sigma_A = -\sigma_a - \sigma_f = -160.99 - 759.11$
$\sigma_A = -920.10 ~ \text{psi}$ answer
$\sigma_B = -\sigma_a + \sigma_f = -160.99 + 759.11$
$\sigma_B = 598.12 ~ \text{psi}$ answer
At section through D
By inspection, the maximum moment is at D, hence, the maximum normal stress will occur at the extreme fibers of section through D.
$M_D = \Sigma M_\text{to the right of D}$
$M_D = 4F_V$
$M_D = \frac{1000}{3}(\cos 15^\circ + 8 \sin 15^\circ) ~ \text{lb}\cdot\text{ft}$
$\sigma_a = \dfrac{P}{A} = \dfrac{2000 \cos 15^\circ}{2(6)}$
$\sigma_a = 160.99 ~ \text{psi}$
$\sigma_f = \dfrac{6M}{bd^2} = \dfrac{6[ \, \frac{1000}{3}(\cos 15^\circ + 8\sin 15^\circ) \, ] (12)}{2(6^2)}$
$\sigma_f = 1012.16 ~ \text{psi}$
$\sigma_\text{top} = -\sigma_a - \sigma_f = -160.99 - 1012.16$
$\sigma_\text{top} = -1173.15 ~ \text{psi}$ answer
$\sigma_\text{bottom} = -\sigma_a + \sigma_f = -160.99 + 1012.16$
$\sigma_\text{bottom} = 851.17 ~ \text{psi}$ answer
