Q. A simply supported beam of span 10 m is loaded with concentrated loads of 20 kN, 40 kN and 20 kN at 2 m, 5 m and 8 m from left support respectively. Draw S.F.D and BMD and mark the values at salient points.

SOLUTION:
Total load = 20 + 40 + 20 = 80kN
As loading on the beam is symmetrical Hence
RA = RB = 80/2 = 40kN

SFD (SHEAR FORCE DIAGRAM)
F_{A} = 40kN (A to C constant)
F{C} = 20 kN (C to D constant)
F{D} = – 20 kN (D to E constant)
F{E} = – 40kN (E to B constant)
Max + ve SF occurs at A = 40kN and
Max – ve SF occurs at B = 40kN

BMD (BENDING MOMENT DIAGRAM)

MA = MB = 0
MC = ME = 40×2 = 80 kNm
MD = 40 × 5 – 20 × 3 = 140 kNm
Max B.M occurs at D = MD = Mmax = 140 KNm