Solution to Problem 630 | Moment Diagrams by Parts

$\Sigma M_{R2} = 0$

$4R_1 + 200(2) = \frac{1}{2}(3)(400)(1)$

$R_1 = 50 \, \text{N}$
 

$\Sigma M_{R1} = 0$

$4R_2 = 200(6) + \frac{1}{2}(3)(400)(3)$

$R_2 = 750 \, \text{N}$
 

$(Area_{AB}) \, \bar{X}_A = \frac{1}{2}(4)(200)(\frac{8}{3}) - \frac{1}{4}(3)(600)(\frac{17}{5})$

$(Area_{AB}) \, \bar{X}_A = -463.33 \, \text{N}\cdot\text{m}^3$           answer
 

The value of (Area_AB) barred(X)_A is negative; therefore point A is below the tangent through B, thus the tangent through B slopes downward to the right. See the approximate elastic curve shown to the right and refer to the rules of sign for more information.