Solution to Problem 629 | Moment Diagrams by Parts

$10R_1 = 400 + \frac{1}{2}(6)(200)(2)$

$R_1 = 160 \, \text{lb}$
 

$\Sigma M_{R1} = 0$

$10R_2 + 400 = \frac{1}{2}(6)(200)(8)$

$R_2 = 440 \, \text{lb}$
 

$(Area_{AB})\,\bar{X}_A = \frac{1}{2}(10)(1600)(\frac{20}{3}) - 400(8)(6) - \frac{1}{4}(6)(1200)(\frac{44}{5})$

$(Area_{AB})\,\bar{X}_A = 18\,293.33 \, \text{lb}\cdot\text{ft}^3$           answer
 

$(Area_{AB})\,\bar{X}_B = \frac{1}{2}(10)(1600)(\frac{10}{3}) - 400(8)(4) - \frac{1}{4}(6)(1200)(\frac{6}{5})$

$(Area_{AB})\,\bar{X}_B = 11\,706.67 \, \text{lb}\cdot\text{ft}^3$           answer