Q. A simply supported beam of span 6m carries a central point load of 8 kN. Draw SFD and BMD

SOLUTION:

As the beam is symmetrically loaded,
hence RA =R B = Total load /2 = 8/2 = 4 kN

SFD (SHEAR FORCE DIAGRAM)

FA =4 kN(from A to C it is 4 kN)
FC =(just left ofC )=4kN
FC =(just right ofC )=-4 kN
FB = – 4kN (same from C to B)
max S.F occurs at support A = 4kN

BMD (BENDING MOMENT DIAGRAM)
M{A} = M_{B} = 0 ( why because It’s simply supports)
M{C} = 4 × 3 = 12kNm

Max BM occurs at C where SF changes it sign
=M C =M max =12 kNm