Q. A simply supported beam of span 6 m. Three concentrated load of 10 kN, 20 kN and 30 kN are acting at 2 m, 3 m and 5 m respectively from left hand support. Draw SFD and BMD.

SOLUTION:
Total load = 10 + 20 + 30 = 60kN
To find the reactions
Taking moments about A
RB × 6 = 10 × 2 + 20 × 3 + 30 × 5
= 20 + 60 + 150 = 230
RB = 230/6 = 38.33kN
RA = 60 – 38.33 = 21.67kN

SFD (SHEAR FORCE DIAGRAM)
F{A} = 21.67kN (same in AC)
F{c} = 21.67kN
F{D} = 11.67kN
F{E} = – 38.33kN
F{B} = – 38.33kN
Max +ve SF occurs at A = 21.67kNand
Max – ve SF occurs at B = – 38.33kN

BMD (BENDING MOMENT DIAGRAM)
M{A} = M{B} = 0 ;
M{C} = 21.67 × 2 = 43.34kNm
M{D} = 21.67 × 3 – 10 × 1 = 55.01kNm
M{E} = 38.33 × 1 = 38.33kNm
.. Max BM occurs at D (where SF changes its sign) = Mmax = MD = 55.01kNm