Q.A Simply supported beam of span 4 m is carrying a udl of 10 kN/m on entire span. Draw SFD and BMD.

SOLUTION:
The load on the beam is symmetrical
RA = RB = (10 × 4)/2 = 20kN

SFD (SHEAR FORCE DIAGRAM)
F{A} = 20kN
F{C} = 0kN
F{B} = – 20 kN
Max + ve SF occurs at A = 20 kN and
Max – ve SF occurs at B = – 20 kN

BMD ( BENDING MOMENT DIAGRAM)
MA = MB = 0
MC = 20×2 -(10×2)2/2
= 40 – 20 = 20 KNm

Or
Mmax = MC = wl²/8 = (10 × 4²)/8 = 20 kNm