Q. A simply supported beam of span 3 m carries two point loads of each 10 kN acting at a distances of 1 m and 2 m respectively from left hand support. Draw S.F.D and B.M.D.
Q. A simply supported beam of span 3 m carries two point loads of each 10 kN acting at a distances of 1 m and 2 m respectively from left hand support. Draw S.F.D and B.M.D.
Total load = 10 + 10 = 20kN
As the load is symmetrical hence
The reactions are equal
RA = RB = 20/2 = 10kN
SFD (SHEAR FORCE DIAGRAM )
F{A} = 10kN
F{B} = – 10kN
F{C} = (just left of C)= 10kN
F{C} (just right of C) = 0 kN
F{D} = (just left of D) = 0kN
F{D} = (just right of D) = – 10 kN
Max S.F occurs at support A
FA =F max =10 kN
BMD( BENDING MOMENT DIAGRAM )
M{A} = M{B} = 0
M{C} = 10 × 1 = 10kNm (taking moments from A)
Or
M{C} = 10 × 2 – 10 × 1 = 10 kN (taking moments from B)
M{D} = 10 × 1 = 10kNm
Max. BM occurs in between C and D and its value = 10kNm
