Q. A simply supported beam of span 6 m. A udl of 2 kN/m is acting on the right half of the span. A point load of 10 kN is at 2 m from the LHS. Draw SFD and BMD.

by

SOLUTION:

Total load = 10 + 2 * 3 = 16 kN
To find reactions
Taking moments about B
R{A} * 6 = 10 * 4 + (2 * 3) * 3/2 = 49
R{A} = 49/6 = 8.17kN
R{B} = 16 – 8.17 = 7.83kN
SFD( SHEAR FORCE DIAGRAM)
F{A} = 8.17kN
F{B} = – 7.83kN
F{C} = 8.17kN
F{D} = – 1.83kN
Max + ve SF occurs at A = 8.17kN (same in AC) and Max -veSF occurs at B = – 7.83kN
BMD ( BENDING MOMENT DIAGRAM)
M{A} = M{B} = 0
M{C} = 8.17 * 2 = 16.34kNm
M{D} = 8.17 * 3 – 10 * 1
= 24.51 – 10
= 14.51kNm ..
Max BM occurs at C (where SF changes its sign) M max =16.34 kNm

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