Let $V_a$ and $V_b$ be the reactions at the left support A and right support B.

Total load on beam=$ 2*9+\frac{1}{2}*2*9=27kN$

OR for trapezoidal load on beam =$ \frac{1}{2}*9*(2+4)=27kN$

Now, distance of resultant load from left end of the beam i.e at C.

$27*x=18*4.5+(9*6)$

$x=5m$

Distance of resultant load from $A=5-1=4m$

Calculating support reactions:

$\sum M_A=0$ i.e taking moment about A

$V_b*6=27*4 i.e V_b=18kN$

$V_a$=total load-$V_b=27-18=9kN$

Now, consider any section x at a distance x from the left end.

a) Load intensity at the section x be 'w'.

$w'=2+\frac{4-2}{9}*x=2+\frac{2x}{9}$

b) Load on the beam b/w the left end C and section X

$V= - \int_0^x {w' dx}= \int_0^x 2+\frac{2x}{9} dx = -(2x +\frac{2.x^2}{2*9})$

V=$- (2x+\frac{x^2}{9})$

(Integration of the loading intensity gives us the shear force)

c) Moment of the above load about X

$M= - \int_0^x {V dx}= - \int_0^x 2x+\frac{x_2}{9} dx = -(\frac{2x^2}{2} +\frac{x^3}{3*9})$

M=$-(x^2+\frac{x^3}{27})$

Shear force analysis:

At any section in CA distant x from C, shear force

$\delta x=-(2x+\frac{x^2}{9})$

At, x=0, $\delta x=0$;

At z=1m,$\delta x=-(2*1+\frac{1^2}{9}= -2.11kN$;

At any section in AB distant x from C, shear force

$\delta x=V_a-(2x+\frac{x^2}{9})=9-(2x+\frac{x^2}{9})$

At x=1m, $\delta x=9-(2*1+\frac{1^2}{9}=6.89kN$

At x=7m, $\delta x=9-(2*7+\frac{7^2}{9}=-10.44kN$

At any section in BD distant x from C, shear force

$\delta x=(V_a+V_b)-(2x+\frac{x^2}{9})=27-(2x+\frac{x^2}{9})$

At x=7m, $\delta x=27-(2*7+\frac{7^2}{9}=7.56kN$

At x=9m, $\delta x=27-(2*9+\frac{9^2}{9}=0kN$

Section of zero shear in AB,

Equating the SF to zero i.e $9-(2x+\frac{x^2}{9}=0$

$x^2+18x-81=0$

On solving, x=3.728m

B.M analysis:

BM at A=$M_A=-(x^2+\frac{x^3}{27})=-(1^2+\frac{1^3}{27})=-1.037kNm$

BM at B=$M_B=9*6-(x^2+\frac{x^3}{27})=54-(7^2+\frac{7^3}{27})=-7.704kNm$

Maximum BM at x=3.728m

BM at E=$M_E=9*(2.728-1)-(x^2+frac{x^3}{27})$

$=24.552-(3.728^2+frac{3.728^3}{27})=8.735kNm$

Point of contraflexure: Equating the general expression for BM to zero,

$9(x-1)-(x^2+\frac{x^3}{27})=0$

$9x-9-x^2-\frac{x^3}{27}=0$

$243x-243-27x^2-x^3=0$ (multiplied by 27)

On solving, we get x=1.15m and x=6.14m