Solving for reactions
$\Sigma M_{R2} = 0$
$6R_1 = 80(4)(4)$
$R_1 = 213.33 \, \text{ lb}$
$\Sigma M_{R1} = 0$
$6R_2 = 80(4)(2)$
$R_2 = 106.67 \, \text{ lb}$
From the conjugate beam
$\Sigma M_A = 0$
$6F_2 + \frac{1}{3}(4)(640) [\frac{3}{4}(4)] = \frac{1}{2}(4)(853.33) [\frac{2}{3}(4)] + \frac{1}{2}(2)(213.33) [4 + \frac{1}{3}(2)]$
$F_2 = 497.77 \, \text{ lb}\cdot\text{ft}^2$
$M_B = \frac{1}{2}(2)(213.33 [\frac{1}{3}(2)] - 2F_2$
$M_B = \frac{1}{2}(2)(213.33 [\frac{1}{3}(2)] - 2(497.77)$
$M_B = -853.32 \, \text{ lb}\cdot\text{ft}^3$
Thus, the deflection at B is
$EI \, \delta_B = M_B$
$EI \, \delta_B = -853.32 \, \text{ lb}\cdot\text{ft}^3$
$EI \, \delta_B = 853.32 \, \text{ lb}\cdot\text{ft}^3$ downward answer
