Q. A simply supported beam of span 6 m. A udl of 4 kN / m is acting over the entire span. Two point loads of 10 kN and 20 kN are acting at 2 m and 4 m respectively from the left end. Draw SFD and B.M.D.
Q. A simply supported beam of span 6 m. A udl of 4 kN / m is acting over the entire span. Two point loads of 10 kN and 20 kN are acting at 2 m and 4 m respectively from the left end. Draw SFD and B.M.D.
Taking moments about A
RB × 6 = 10 × 2 + 20 × 4 = 100
RB = 100/6 = 16.67kN
RA = 30 – 16.67 = 13.33kN
SFD (SHEAR FORCE DIAGRAM )
FA = 13.33 kN (same in AC)
FC = (just right of c)= 3.33kN (same in CD)
FD (just left of D)= 3.33 kN
FD = (just right of D) = – 16.67kN
FB = – 16.67kN
Max + ve SF occurs at A = 13.33 kN
and Max – ve SF occurs at B = 16.67
BMD (BENDING MOMENT DIAGRAM)
MA=MB=0
MC = 13.33 × 2 = 26.66KNm
MD = 16.67 × 2 = 33.34KNm
Max BMD occurs at D = 33.34 KNm
