Solution to Problem 631 | Moment Diagrams by Parts

$\Sigma M_A = 0$

$4R_2 + M = 100(4)(2)$

$R_2 = 200 - \frac{1}{4}M$
 

$\Sigma M_B = 0$

$4R_1 = 100(4)(2) + M$

$R_1 = 200 + \frac{1}{4}M$
 

$(Area_{AB}) \, \bar{X}_A = 0$

$\frac{1}{2}(4)(800 - M)(\frac{4}{3}) - \frac{1}{3}(4)(800)(1) = 0$

$\frac{8}{3}(800 - M) = \frac{3200}{3}$

$M = 400 \, \text{lb}\cdot\text{ft}$           answer
 

The uniform load over span AB will cause segment AB to deflect downward. The moment load equal to 400 lb·ft applied at the free end will cause the slope through B to be horizontal making the deviation of A from the tangent through B equal to zero. The downward deflection therefore due to uniform load will be countered by the moment load.