Q.Simply supported beam of span 8m carries three point loads of 10KN, 20 kN and 10 kN are placed at a distance of 2m, 4m and 6 m from the LHS. Draw SFD and BMD.

by

Q.Simply supported beam of span 8m carries three point loads of 10KN, 20 kN and 10 kN are placed at a distance of 2m, 4m and 6 m from the LHS. Draw SFD and BMD.

SOLUTION:
Total load = 10 + 20 + 10 = 40 KN
As the given loading is symmetrical
RA=RB= 40/2 = 20kN

SFD (SHEAR FORCE DIAGRAM)
F{A}= 20kN
F{B}= 20kN
F{D}= 10kN
F{E}= – 10kN
F{B}= – 20kN
Max + ve SF occurs at A = 20 kN (same in AC) Max – ve SF occurs at B = -20 kN (same in EB)

BMD (BENDING MOMENT DIAGRAM)
MA = MB = 0
MC = ME = 20 × 2 = 40 kNm
MD = 20 × 4 – 10 × 2 = 60 KNm
Max BM occurs at D (where SF changes its sign)
= Mmax = 60 kNm

Leave a Comment