Using conjugate beam method find the vertical deflection at D and slope at A for the S.S. beam loaded as shown in fig. in terms of EI.

(1) Support rxn calculation

$\sum M_A=0$

$50\times 4-V_B\times 5.5=0$

$V_B=36.36kN$

$\sum F_y=0$

$V_A-50+36.36=0$

$V_A=13.64kN$

(2)Draw BMD

$B.M_A=0$

$B.M_C=13.64\times 2=27.28kN.m$

$B.M_D=36.36\times 1.5=54.54 kN.m$

$B.M_B=0 ![enter image description here][2]$

(3)Draw $\frac{M}{EI}$ diagram

(4)Draw conjugate beam and consider $\frac{M}{EI}$ diagram as loading on the beam

$\sum M_A=0$

$(\frac{1}{2}\times 2\times \frac{27.28}{EI})(\frac{2}{3}\times 2)+(\frac{1}{2}\times 2\times \frac{13.6}{EI})(2+\frac{2}{3}\times 2)+(2\times \frac{13.64}{EI})(2+\frac{2}{2})+(\frac{1}{2}\times 1.5\times \frac{54.54}{EI})(4+\frac{1}{3}\times 1.5)-V_B'\times 5.5=0$

$5.5V_B'=\frac{1}{EI}[36.37]+\frac{1}{EI}[45.33]+\frac{1}{EI}[81.84]+\frac{1}{EI}[184.07]$

$5.5V_B'=\frac{1}{EI}[36.37+45.33+81.84+184.07]$

$=\frac{1}{EI}[347.61]$

$V_B'=\frac{347.61}{5.5EI}=\frac{63.20}{EI}$

$V_B'=\frac{63.20}{EI}$

$\sum F_y=0$

$V_A'=(\frac{1}{2}\times 2\times \frac{27.28}{EI})-(\frac{1}{2}\times 2\times \frac{13.6}{EI})-(2\times \frac{13.64}{EI})-(\frac{1}{2}\times 1.5\times \frac{54.54}{EI})+\frac{63.20}{EI}$

$V_A'\frac{1}{E}[27.28+13.6+27.28+40.90-63.20]$

$V_A'=\frac{45.86}{EI}$

(5)$Q_A$ in real beam = S.F. @ A in conjugate beam

$S.F._{AL}=0$

$S.F._{AR}=\frac{45.86}{EI}$

(6)$y_D$ in real beam=B.M @ D in conjugate beam

$B.M_D=\frac{63.20}{EI}\times 1.5-[\frac{1}{2}\times 1.5\times \frac{54.54}{EI}](\frac{1}{3}\times 1.5)$

$=\frac{1}{EI}[94.8-20.45]$

$B.M_D=\frac{74.35}{EI}$