Q. A simply supported beam of span 4 m carries a point load of 10 kN at a distance of 1 m from left hand support, Draw SFD. and B.MD.
SOLUTION:
First To find reactions
As the load is unsymmetrical the reactions are not equal. The load applied is nearer to support A and hence the load on support A is more compared to support B.
Total load= 10kN
Taking moments about A
RB × 4 = 10 × 1
RB = 10/4 = 2.5kN
RA = 10 – 2.5 = 7.5kN
(OR)
RA × 4 = 10 × 1
RB= 10/4 = 2.5kN
RA = 10 – 2.5 = 7.5kN
(OR)
RA × 4 = 10 × 3 (taking moments about B)
RA = 30/4 = 7.5kN
SFD ( SHEAR FORCE DIAGRAM)
Draw S.F.D from left support follow sign convention
F{A} = 7.5kN
F{B} = – 2.5kN
F{C} (just left of c)=+ 7.5kN
F{C} (just right of c)= – 2.5kN
Max S.F occurs at support A and is uniform in between A{C} = 7.5kN
