Q. A beam ABCDE of 8 m length with double over hang is loaded as shown in figure. Draw SFD and BMD. SOLUTION:
Total load= 10 + 40 + 20 + (6 * 6) = 106 kN
To find reactions
Taking moments about B
(Anticlockwise moment = clockwise moment)
R{D} * 5 + 10 * 1 + 6 * 1 * 1/2 = 40 * 3 + 20 * 7 + 6 * 5 * 5/2
RD * 5 + 10 + 3 = 120 + 140 + 75
R{D} = 322/5 = 64.4kN
RB= 106 – 64.4 = 41.6 kN
SFD ( SHEAR FORCE DIAGRAM)
Draw SFD as per the loading
F{A} = – 10kN
FD =(Just right ofB )=+25.6 kN
F B =(just left ofB )= -16 kN
F C =(just right ofC )= +7.6 kN
F C =(just left ofc )= -32.4 kN
F D =(iust left ofD )= -44.4 kN
F D =(just might ofD )= 20kN
F{E} = 20kN
Max +ve SF occurs at A = + 25.6 kN and
max -ve SF occurs at B = – 44.41
BMD( BENDING MOMENT DIAGRAM)
M{A} = M{E} = 0
M{B} = – 10 * 1 – 6 * 1 * 1/2 = – 13kNm
M{C} = 41.6 * 3 – 10 * 4 – 6 * 4 * 4/2 = 36.8kNm M{D} = – 20 * 2 = – 40kNm
Max + ve BM occurs at C = 36.8 kNm and
Max – ve BM occurs at D = – 40kNm
To find point of contra flexure
In double over hanging beams the point of contraflexure at two points in between supports a and b.
To find a
Find BM upto F and equate it to zero.
41.6(a – 1) – 10a – 6a * a/2 = 0
41.6a – 41.6 – 10a – 3a² = 0;
3a² – 31.6a +41.6 = 0
Solving quadratic equation a = 1.54m from A.
Find BM upto G and equate it to zero
64.4(b – 2) – 20b – 6(b – 2)((b – 2)/2) = 0
3b² – 56.4b + 116 = 0
By solving b = 2.35m
… Points of contra flexures occurs at 1.54 m from left end A and 2.35 m from right free end E.
