$\sum M_A=0\ (\circlearrowright +ve)$

$-V_D\times5+(\frac{1}{2}\times6\times5)(\frac{2}{3}\times6)+10\times1.5$

$+(2\times2)(3+1)+(\frac{1}{2}\times2\times2)\times(3+\frac{2}{3}\times2)=0$

$\boxed{V_D=19.934\ kN}\ (\uparrow)$

$\sum F_Y=0(\uparrow+ve)$

$V_A+19.934-10-(2\times2)-\frac{1}{2}\times2\times2=0$

$\boxed{V_A=-3.934\ kN}\ (\downarrow)$

$\sum F_X=0(\rightarrow+ve)$

$-H_A+\frac{1}{2}\times6\times5=0$

$\boxed{H_A=15\ kN}\ (\leftarrow)$

1. Consider part AB:

$BM_A=0\\BM_B=15\times6-\frac{1}{2}\times6\times5\times\frac{1}{3}\times6\\ BM_B=\underline{60\ kNm}$

2. Consider part BC:

$\textbf{1. Shear force calculation: }(\uparrow|\downarrow+ve)$

$SF_{BL}=0$

$SF_{BR}=-3.434\ kN$

$SF_{EL}=-3.434\ kN$

$SF_{ER}=-3.934-10=13.934\ kN$

$SF_F=13.934\ kN$

$SF_C=-19.934\ kN$

$\textbf{2. Bending moment calculation: }(\circlearrowright|\circlearrowleft+ve)$

$BM_{B}=\underline{60\ kNm}\ \text{\{obtained already\}}$

$BM_{E}\ \text{\{considering part EBA\}}$

$=15\times6-\frac{1}{2}\times5\times6\times\frac{1}{3}\times6-3.934\times1.5$

$\phantom{BM_{E}\ \text{\{considering part EBA\}}}=\underline{54.01\ kNm}$

$BM_F\ \text{\{considering part FEBA\}}$

$=15\times6-3.934\times3-\frac{1}{2}\times5\times6\times\frac{1}{3}\times6-10\times1.5$

$\phantom{BM_{E}\ \text{\{considering part FEBA\}}}=\underline{33.198\ kNm}$

$BM_{F}\ \text{\{considering part CD\}}=0$

$\therefore \text{part CD does not bend at all.}$

$\underline{BM_C=0}$

3. Consider part CD: