$10R_1 + 3P = 6(100)$
$R_1 = 60 - 0.30P$
$EI \, t_{A/C} = 0$
$(Area_{AC}) \, \bar{X}_A = 0$
$\frac{1}{2}(10)(600 - 3P)(\frac{20}{3}) - \frac{1}{2}(6)(600)(8) = 0$
$P = 56 \, \text{ lb}$ answer
Thus,
$240 - 1.2P = 172.8 \, \text{ lb}$
$600 - 3P = 432 \, \text{ lb}$
Under the 100-lb load:
$EI \, t_{B/C} = (Area_{BC}) \, \bar{X}_B$
$EI \, t_{B/C} = \frac{1}{2}(6)(172.8)(2) + \frac{1}{2}(6)(432)(4) - \frac{1}{2}(6)(600)(4)$
$EI \, t_{B/C} = -979.2 \, \text{ lb}\cdot\text{ft}^3$
The negative sign indicates that the elastic curve is below the reference tangent.
Therefore,
$EI \, \delta_B = 979.2 \, \text{ lb}\cdot\text{ft}^3$ downward answer
