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

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

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

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

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

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

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

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

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

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

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. …