Calculation of support reaction:

Let $V_a$ and $V_c$ be the support reaction at B and C

$\sum F_y=0$

$V_b+V_c+2.5+2.5=6*3$

$V_b+V_c=13$ (1)

$\sum M_c=0$

$2.5*(0.5+3)+V_b*3-6*3*\frac{3}{2}-2.5*0.5=0$

$V_b=6.5KN$

$V_c=13-V_b=13-6.5=6.5KN$

SF analysis:

SF @ A=2.5KN

SF @ B but just on LHS of B=2.5KN

SF @ B but just on RHS of B=2.5+6.5=9KN

SF @ D=-2.5KN

SF @ C but just on RHS of C=-2.5KN

SF @ C but just on LHS of C=-2.5-6.5=-9KN

Consider any section in BC from A at a distance x

Section of shear zero

Equating the general equation for shear to zero

$2.5+6.5-6(x-0.5)=0$

x=2m from A

BM analysis:

BM @ A=0, BM @ D=0

BM @ B=2.5*0.5=1.25KNm

BM @ C=2.5*0.5=1.25 KNm

$BM_{max} at x=2m$

$BM_{max} at x=2m$

$BM_{max}$ in the region BC distant x from A,

$BM_{max}=2.5*x+6.5(x-0.5)-6(x-0.5)\frac{(x-0.5)^2}{2}$

x=2m

$BM_{max}=8KNm$