The angle between tangents through A and B is zero
$EI\,\theta_{AB} = (\text{Area}_{AB}) = 0$
$\frac{1}{2}L(R_AL) + MAL - \frac{1}{3}(\frac{1}{2}L)(\frac{1}{8}w_oL^2) = 0$
$\frac{1}{2}R_AL^2 + M_AL - \frac{1}{48}w_oL^3 = 0$
$M_A = \frac{1}{48}w_oL^2 - \frac{1}{2}R_AL$ → equation (1)
The deviation of B from a tangent line through A is zero
$EI\,t_{B/A} = (\text{Area}_{AB}) \cdot \bar{X}_B = 0$
$\frac{1}{2}L(R_A L)(\frac{1}{3}L) + M_A L(\frac{1}{2}L) - \frac{1}{3}(\frac{1}{2}L)(\frac{1}{8}w_o L^2)[ \, \frac{1}{4}(\frac{1}{2}L) \, ] = 0$
$\frac{1}{6}R_A L^3 + \frac{1}{2} M_A L^2 - \frac{1}{384}w_oL^4 = 0$
$64R_AL + 192M_A = w_oL^2$ → equation (2)
Substitute M_{A} of equation (1) to equation (2) above
$64R_AL + 192(\frac{1}{48}w_oL^2 - \frac{1}{2}R_AL) = w_oL^2$
$64R_AL + 4w_oL^2 - 96R_AL = w_oL^2$
$32R_AL = 3w_oL^2$
$R_A = \frac{3}{32}w_oL$
Substitute R_{A} to equation (1)
$M_A = \frac{1}{48}w_oL^2 - \frac{1}{2}(\frac{3}{32}w_oL)L$
$M_A = \frac{1}{48}w_oL^2 - \frac{3}{64}w_oL^2$
$M_A = -\frac{5}{192}w_oL^2$ answer
Sum up moments at right support B
$M_B = M_A + R_AL - \frac{1}{8}w_oL^2$ → see the right end of moment diagram by parts
$M_B = -\frac{5}{192}w_oL^2 + (\frac{3}{32}w_oL)L - \frac{1}{8}w_oL^2$
$M_B = -\frac{5}{192}w_oL^2 + \frac{3}{32}w_oL^2 - \frac{1}{8}w_oL^2$
$M_B = -\frac{11}{192}w_oL^2$ answer