Q. Draw SFD and BMD for an double over hanging beam as given below,

SOLUTION:
Total load = 2 * 2 + 4 + 2 * 2 = 12N
To find the reactions
The loading on the beam is symmetrical hence
R{A} = R{B} = 12/2 = 6N
SFD ( SHEAR FORCE DIAGRAM)
F{E} = 0
F{C} = 0
F A =(just left ofA )= – 4 N
FA =(just right ofA )=+2N (same in AD)
F{D} = 2N
F{B} = 4N
Max +ve SF occurs at B = 4N and
Max-ve SF occurs at A = – 4N
BMD ( BENDING MOMENT DIAGRAM)
M{E} = M{C} = 0
M{A} = – (2 * 2) * 2/2 = – 4Nm
M{B} = – (2 * 2) * 2/2 = – 4Nm
M{D} = 6 * 1 – (2 * 2)(2/2 + 1)
= 6 – 8 = – 2Nm