Q. A simply supported beam of span 6 m, carries a udl of 4 kN/m over its right half of the span. A point load 6 kN is acting at 2 m from left hand support. Draw SFD and B.MD.

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SOLUTION:

Total load = 6 + 4 * 3 = 18kN
To find reactions
Taking moments about B
R{A} * 6 = 6 * 4 + (4 * 3) * 3/2 = 24 + 18 = 42kN.                                  R{A} = 42/6 = 7kN
R{B} = 18 – 7 = 11kN
SFD ( SHEAR FORCE DIAGRAM)
FA =7 kN (same from A to C)
FC =(just left of C )=7 kN
FC =(just right of C )=1 kN
FD = 1kN
F{B} = – 11kN
Max SF occurs at A = 7kN
BMD ( BENDING MOMENT DIAGRAM)
MA = MB = 0
M{C} = 7 * 2 = 14kNm
M{D} = 7 * 3 – 6 * 1 = 15kNm
By similar triangles
x/1 = (3 – x)/11
11x = 3 – x
12x = 3 ;
x = 3/12 = 0.25m ;
M max =11 * 2.75 – (4 * 2.75) 2.75 / 2
= 15.125kNm

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