Draw the shear force diagram and bending moment diagram for the beam loaded as shown in fig.

And locate the point contra flexure ( if any).

Let $V_A$ and $V_B$ be the support reaction at A and B respectively

$\sum F_y=0 (+ve)$

$V_A+V_B=8*4+10=42$ (1)

$\sum M_B=0 (+ve)$

$V_A*(4+5)-8*4*(\frac{4}{2}+5)+10*3=0$

$V_A=21.55KN$

$V_B=42-21.55=20.45KN$

Shear force analysis:

SF @ A=21.55KN

SF @ C=21.55-8*4=-10.45KN

SF @ B but just on LHS of B=-10.45KN

SF @ B but just on RHS of B=-10.45+20.45=10KN

SF @ D=10-10=0KN

Section of zero shear:

Consider ant section x in AC at distance x from A and equating the shear force equation to zero.

$21.55-8x=0$

x=2.69m

Bending moment analysis:

$BM @ A=0, BM @ D=0$

$BM @ B=-10*3=-30KNm$

$BM @ C=21.55*4-8*4*\frac{4}{2}=22.2 KNm$

$BM at x=2.69$ in region AC is,

$BM_x=21.55*2.69-8*\frac{2.69^2}{2}=29.02KNm$

Point of contraflexure: Between C and D

Let the BM be zero at distance x from A

$21.55x-\frac{8x^2}{2}=0$

On solving we get x=5.38 m from A