Q. Draw SFD and BMD for the double over hanging beam as shown in figure

SOLUTION:
Total load = 5 + 20 + 15 + 10 = 50kN
To find the reactions
Taking moments about A
R{B} * 7 = 20 * 2 + 15 * 5 + 10 * 8.5 – 5 * 1
= 40 + 75 + 85 – 5 = 195
R{B} = 195/7 = 27.857kN
R{A} = 50 – 27.857 = 22.143kN
SFD ( SHEAR FORCE DIAGRAM)
FC =-5 kN(same ln(CA) )
F{A} = 17.143kN (same in AD)
F{D} = – 2.857kN (same in DE)
F{E} = – 17.857kN (same in EB)
F{B} = 10kN (sarne in BF)
F{F} = 10kN
Max +ve SF occurs in between A& D = 17.143kN and
Max – ve SF occurs in between EB = – 17.859kN
BMD ( BENDING MOMENT DIAGRAM)
M{C} = 0
M{F} = 0
M{R} = – 10 * 1.5 = – 15kNm
M{A} = – 5 * 1 = – 5kNm
M{D} = 22.143 * 2 – 5 * 3 = 29.286kNm
M{E} = 22.143 * 5 – 20 * 3 – 5 * 6 = 20.715kNm Max +veBM occurs at D = 29.286kNm and
max -ve BM occurs at B = – 15kNm
Points of Contra Flexure:
1. BM changes its sign at (1) near to A; let it be X1 from left side free end C equate BM upto (1) to zero and find X1
5X1 – 22.143(X1 – 1) = 0
5X1 – 22.143X1 + 22.143 = 0
– 17.143X1 = – 22.143 ;
X1= – 22.143/- 17.143 = 1.2916m FROM C
SIMILARLY:
2. BM changes its sign at (2) near B, let it be X2 from right side free end F.
Equate BM upto (2) to zero and find X2
-10* X2 +27.857* X2 – 41.7855 = 0
17.85 – X{2} = 41.7855 ;
X{2} = 41.7855/17.85 = 2.34m from F