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 RA 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
