$R_C = 80(8) = 640 \, \text{ lb}$

$M_C = 80(8)(4) = 2560 \, \text{lb}\cdot\text{ft}$

$t_{B/C} = \dfrac{1}{EI}(Area_{BC})\bar{X}_B$

$t_{B/C} = \dfrac{1}{EI}[ \, \frac{1}{2}(6)(3840)(2) - 6(2560)(3) - \frac{1}{3}(6)(1440)(1.5) \, ] \, (12^3)$

$t_{B/C} = \dfrac{1}{EI}[ \, 27\,360 \, ] \, (12^3)$

$t_{B/C} = \dfrac{1}{(1.5 \times 10^6)(40)}[ \, 27\,360 \, ] \, (12^3)$

$t_{B/C} = -0.787968 \, \text{ in}$

Thus, δ_{B} = | t_{B/C} | = 0.787968 in *answer*

Why is it that we are considering the reaction moment and reaction force to find deflection in this case but not in the case of previous problem. I know that we are trying to find the deflection at the free end in the previous problem and at a point on the beam in this problem but why are the considerations different? Please answer. I'm unable to comprehend this logic and I need help

One word: convenience. If we take the support as our moment center, we will have this moment diagram:

The blue area is not easy to calculate, much worse in finding its centroid. In the previous problem, the moment diagram by parts are just triangular, making it very convenient.