$10R_1 + 3P = 6(100)$

$R_1 = 60 - 0.30P$

$EI \, t_{A/C} = 0$

$(Area_{AC}) \, \bar{X}_A = 0$

$\frac{1}{2}(10)(600 - 3P)(\frac{20}{3}) - \frac{1}{2}(6)(600)(8) = 0$

$P = 56 \, \text{ lb}$ *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*