A beam 8.5 m long rest on the supports 5 m apart, the beam carries load as shown in fig. Draw the SFD and BMD showing all the important points.

$\sum F_Y = 0 (+ve)$

$V_A+V_B = 40+60+50*5 = 350$ (1)

$\sum M_B = 0 (+ve)$

$-40*(1.5+5)+V_A*5-50*5*\frac{5}{2}+60*2 = 0$

$V_A = 153 KN$ (2)

$V_B = 197KN$ from (1) and (2)

Shear force analysis:--

Shear force at $C = -40KN$

Shear force at any section CA, ie just before pt A

Shear force at A = -40KN and just after point A = $-40+V_A = -40+153 = 113KN$

Shear force at D = 60KN

Shear force just on RHS of $B = 60-197 = -137KN$

Finding the value of 'x' where bending moment is maximum, for that equating shear force equation to zero.

Let the section be 'x' meters from C

$-40+153-50x = 0$

x = 2.26m

Bending moment analysis:

BM at C = 0, BM at D = 0

BM at $A = -40*1.5 = 60KNm$

BM at $B = -60*2 = -120KNm$

$BM_x$ is maximum at distant x = 2.26m from C

$BM_x = 40*2.26+153*(2.26-1.5)-50\frac{(2.26-1.5)^2}{2} = 11.44KNm$