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

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

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

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

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

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

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

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

## Comments

## Why is it that we are

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

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.