Problem 721 | Propped beam with decreasing load by moment-area method

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