Solution to Problem 632 | Moment Diagrams by Parts

$\Sigma M_B = 0$

$3R_1 + 200(1) = 800(2)(2)$

$R_1 = 1000 \, \text{N}$
 

$\Sigma M_A = 0$

$3R_2 = 200(4) + 800(2)(1)$

$R_1 = 800 \, \text{N}$
 

 

$(Area_{AB}) \, \bar{X}_A = \frac{1}{2}(2)(2000)(\frac{4}{3}) + \frac{1}{2}(1)(800)(\frac{7}{3}) - \frac{1}{3}(2)(1600)(\frac{3}{2}) - \frac{1}{2}(1)(400)(\frac{7}{3}) - \frac{1}{2}(1)(200)(\frac{8}{3})$

$(Area_{AB}) \, \bar{X}_A = 1\,266.67 \, \text{N}\cdot\text{m}^3$           answer
 

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