Rotation between A and B
$\theta_{AB} = \dfrac{1}{EI}(Area_{AB}) = 0$
$\frac{1}{2}L(R_AL) + M_AL = 0$
$\frac{1}{2}R_AL^2 + M_AL = 0$
$R_AL + 2M_A = 0$
$R_A = -\dfrac{2M_A}{L}$ → equation (1)
Deviation of B from tangent through A
$t_{B/A} = \dfrac{1}{EI}(Area_{AB}) \cdot \bar{X}_B = -\Delta$
$\frac{1}{2}L(R_AL)(\frac{1}{3}L) + M_AL(\frac{1}{2}L) = -EI\Delta$
$\frac{1}{6}R_AL^3 + \frac{1}{2}M_AL^2 = -EI\Delta$
$R_AL^3 + 3M_AL^2 = -6EI\Delta$
Substitute RA defined in equation (1)
$\left( -\dfrac{2M_A}{L} \right)L^3 + 3M_AL^2 = -6EI\Delta$
$-2M_AL^2 + 3M_AL^2 = -6EI\Delta$
$M_AL^2 = -6EI\Delta$
$M_A = -6EI\Delta / L^2$
Substitute MA to equation (1)
$R_A = -\dfrac{2(-6EI\Delta / L^2)}{L}$
$R_A = \dfrac{12EI\Delta}{L^3}$
Moment at B (See moment diagram by parts)
$M_B = M_A + R_AL$
$M_B = -6EI\Delta / L^2 + (12EI\Delta / L^3)L$
$M_B = -6EI\Delta / L^2 + 12EI\Delta / L^2$
$M_B = 6EI\Delta / L^2$
Thus,
$M_B = -M_A = 6EI\Delta / L^2$ okay!
