S.F.D

B.M.D $\sum F_y=0$

$V_B+V_E=20*4+50+25*2$

$V_B+V_E+180$

and taking moment about E,

$\sum M_E = 0$

$-20*4*(\frac{4}{2}+5+5+2)+V_b*(5+5+2)-50*(5+2)-(25*2)*\frac{2}{2} = 0$

$V_B=126.67KN$

Now, $V_G = 180-V_B = 53.33KN$

B.M analysis:

BM in the section AB, distance * from A

$BM_x = -20*\frac{x^2}{2}$

At x = 0,

$BM_A = 0$

At x = 4,

$BM_B = -20*\frac{4^2}{2} = -160KNm$

$BM_c = -20*4(\frac{4}{2}+5)+126.67*5$

$BM_c = 73.30KNm $

$BM_E = 0$

BM in the section DE, distance x from E

$BM_x = 53.33*x-25\frac{x^2}{2}$

At x = 2

$BM_D = 53.33*2-25*\frac{2^2}{2} = 56.66$

${BM_C = 53.33*(5+2)-25*2*(\frac{2}{2}+5) }_{from- end-E} = 73.33KN$