$M_1L_1 + 2M_2(L_1 + L_2) + M_3L_2 + \dfrac{6A_1\bar{a}_1}{L_1} + \dfrac{6A_2\bar{b}_2}{L_2} = 0$

Where,

$L_1 = L$
$L_2 = 0$

$M_1 = -\frac{1}{4}PL$

$\dfrac{6A_1\bar{a}_1}{L_1} = \frac{3}{8}PL^2$

$\dfrac{6A_2\bar{b}_2}{L_2} = 0$

Thus,

$-\frac{1}{4}PL(L) + 2M_2(L + 0) + 0 + \frac{3}{8}PL^2 + 0 = 0$

$2L\,M_2 = -\frac{1}{8}PL^2$

$M_2 = -\frac{1}{16}PL$

$M_2 = \Sigma M_{\text{to the left of }M_2}$

$-\frac{1}{16}PL = RL - P(\frac{5}{4}L) - P(\frac{1}{2}L)$

$-\frac{1}{16}PL = RL - \frac{5}{4}P) - \frac{1}{2}PL$

$RL = \frac{27}{16}PL$

$R = \frac{27}{16}P$ *answer*

**Check using area-moment method**

$EI \, t_{B/C} = 0$

$(Area_{BC}) \cdot \bar{X}_B = 0$

$\frac{1}{2}L(RL)(\frac{2}{3}L) - \frac{1}{2}(\frac{1}{2}L)(\frac{1}{2}PL)[ \, \frac{1}{2}L + \frac{2}{3}(\frac{1}{2}L) \, ]$

$\frac{1}{2}L(\frac{1}{4}PL)(\frac{1}{3}L) - \frac{1}{2}L(\frac{5}{4}PL)(\frac{2}{3}L) = 0$

$\frac{1}{3}RL^3 - \frac{5}{48}PL^3 - \frac{1}{24}PL^3 - \frac{5}{12}PL^3 = 0$

$\frac{1}{3}RL^3 = \frac{9}{16}PL^3$

$R = \frac{27}{16}P$ *okay*