Transform the triangular load into a downward uniformly distributed load and upward increasing load. We do this so that we can easily draw the moment diagram by parts with moment center at the fixed support.

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

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

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

$\dfrac{w_oL^4}{30} + \dfrac{R_AL^3}{3} - \dfrac{w_oL^4}{8} = 0$

$\dfrac{R_AL^3}{3} = \dfrac{11w_oL^4}{120}$

$R_A = \dfrac{11w_oL}{40}$ *answer*

**Shear and Moment Diagrams**

See the shear and moment diagrams in this link: /reviewer/strength-materials/problem-706-solution-propped-beam-decreasing-load