Q. A simply supported beam 8 m long carries point loads of 10 kN, 8 kN and 6 kN at distances of 2 m, 5 m and 6 m respectively from the left support. Draw SF and BM diagrams.

SOLUTION:
Total load = 10 + 8 + 6 = 24kN
To find reactions
Taking moments about A
RB × 8 = 10 × 2 + 8 × 5 + 6 × 6 = 96kN
RB = 96/8 = 12kN
RA= Total load – RB
= 24 – 12 = 12kN
(OR)
Taking moments about B
RA × 8 = 6 × 2 + 8 × 3 + 10 × 6 = 96
RA = 96/8 = 12kN
RA = Total load – RB
= 24 – 12 = 12kN

SFD (SHEAR FORCE DIAGRAM)
F{A} = 12 kN (same in AC )
F{C} = (just left of c)= 12kN
F{C} = (just right of c)= 2kN (same in CD)
F{D} =(just left ofD )=2 kN
F{D} (just right fD )= – 6 kN (same in DE)
F{E} = (just left of E )= – 6 kN
F{E} = (just right of E)= – 12kN (same in EB) F{B} = – 12kN
Max + ve SF occurs at A = 12KN
Max – ve SF occurs at B = 12 KN

BMD (BENDING MOMENT DIAGRAM)
MA = MB = 0
MC = 12×2 = 24KNm
MD = 12 × 5 – 10 × 3 =30kNm
ME = 12×2 = 24KN m
Max BM occurs at D = Mmax= 30KNm