Q. An overhanging beam is shown in the following figure. Draw she force and bending moment diagrams.

SOLUTION:

Total load = 3 * 2 + 20 + 10 + 5 * 4 = 56kN
To find the reactions
Taking moments about A
Anticlockwise moments = clockwise moments
R{B} * 7 + (3 * 2) * 2/2 = 20 * 3 + 10 * 8.5 + (5 * 4)(4/2 + 3)
R{B} * 7 + 6 = 60 + 85 + 100 = 245
R{B} = (245 – 6)/7 = 239/7 = 34.14kN
R{A} = 56 – 34.14 = 21.86kN
(OR)
Taking moments about B
R{A} * 7 + 10 * 1.5 = 20 * 4 + (5 * 4) * 4/2 + (3 * 2)(2/2 + 7)
R{A} * 7 + 15 = 80 + 40 + 48 = 168
R{A} = (168 – 15)/7 = 153/7 = 21.86kN
SFD ( SHEAR FORCE DIAGRAM)
F{C} = F{F} = 0
F{A} = 15.86kN (same in AD)
F{D} = 15.86kN
F{B} = – 24.14kN (just left of B)
F{B} = 10kN (Just right of B)
F{E} = 10kN (same in BE)
Max + ve SF occurs at A = 15.86kN(Same in AD) and
Max -ve SF occurs at B = – 24.14kN
BMD ( BENDING MOMENT DIAGRAM)
M{E} = M{C} = M{F} = 0
M{A} = – (3 * 2) * 2/2 = – 6kNm
M{B} = – 10 * 1.5 = – 15kNm
M{D} = 21.86 * 3 – (3 * 2) (2/2+3)
= 65.58 – 24 = 41.58kNm
To find Point of contra flexure
Write the BM equation upto G and equate it to zero
To find a
21.86(a – 2) = (3 * 2)(2/2 + (a – 2)) = 6 + 6a – 12 21.86a – 43.72 = 6 + 6a – 12
21.86a – 6a = 6 – 12 + 43.72
15.86 * a = 37.72
a = 37.72/15.86 = 2.378m
To find b: 34.14b = 10(1.5 + b) + (5b) * b/2
= 15 + 10b + 2.5b²
= 2.5b² – 24.14b + 15 = 0
By solving quadratic equation b = 0.6675m from B
.. Point of contra flexure occurs at 0.378 m from A and 0.6675 m from B.