The moment at B due to R

_{A} is R

_{A}L and the moment at B due to triangular load is -1/6 w

_{o}L

^{2}
**Solution of R**_{A} by Moment-Area Method

$EI \, t_{A/B} = (Area_{AB}) \, \bar{X}_A = 0$

$\frac{1}{2}L(R_AL)(\frac{2}{3}L) – \frac{1}{4}L(\frac{1}{6}w_oL^2)(\frac{4}{5}L) = 0$

$\frac{1}{3}R_AL^3 – \frac{1}{30}w_oL^4 = 0$

$\frac{1}{3}R_AL^3 = \frac{1}{30}w_oL^4$

$R_A = \dfrac{w_oL}{10}$ *answer*

**Shear and Moment Diagrams**

See the shear and moment diagrams here: /reviewer/strength-materials/problem-705-solution-propped-beam