Q. A simply supported beam of span 6 m carries three point loads of 60 kN, 40 kN and 70 kN at a distance of 2 m, 4 m and 6 m from left support Draw SFD and BMD.

SOLUTION :

Total load= 60 + 40 + 70 = 170kN
Taking moments about A
RB × 6 = 60 × 2 + 40 × 4 + 70 × 6
= 120 + 160 + 420 = 700
RB = 700/6 = 116.67kN
RA = 170 – 116.67 = 53.33kN

(OR)
SFD (SHEAR FORCE DIAGRAM)
F{A} = 53.33kN (same in AC)
F{c} = – 6.67kN (same in CD)
F{D} = – 46.67kN (same in DB)
Max +ve SF occurs at A = 53.33kN
Max – ve SF occurs at B = 46.67kN

BMD (BENDING MOMENT DIAGRAM)
M{A} = M{B} = 0
M{C} = 53.33 × 2 = 106.66kNm
M{D} = 116.67 × 2 – 70 × 2 = 93.34kNm
max + ve BM occurs at C = (where SF changes its sign)
MC =M max =106.66 kNm