Q. A simply supported beam of span 6 m, a udl of 2 kN/m is of length 4 m acting at middle of the span. Draw S.F.D and B.M.D

SOLUTION:
Total load = (2 × 4) = 8 kN
As the load on the beam is symmetrical
Hence RA =RB = Total load / 2 = 8/2 = 4kN

SFD (SHEAR FORCE DIAGRAM)
F{A} = 4kN (same from A to C)
F{c} = 4kN
F{D} = – 4kN
F{B} = – 4kN
F{E} = 0
Max S.F occurs @ support A = 4kN

BMD (BENDING MOMENT DIAGRAM)
M{A} = M{B} = 0
M{C} = M{D} = 4×1 = 4 kNm
Mmax =MF = 4 × 3 – (2 × 2)2/2
= 12 – 4 =