Q. A beam 8 m long is simply supported at left end and at 6 m from left end. It carries an udl of 5 kNim over the over hanging portion and a print load of 10 kN at 3 m from left hand support. Draw SF and BM diagrams and locate the point of contra flexure.
SFD ( SHEAR FORCE DIAGRAM)
F{A} = 3.33kN same in AD)
F{D} = – 6.67kN same in DB)
F{c} = 0
F{B} = 10kN
Max + ve SF occurs at A = 3.33kN and at
Max + ve SF occurs at B = 10kN
Max – ve SF occurs at D (and in DB)= – 6.67kN. BMD:(BENDING MOMENT DIAGRAM)
M{A} = 0 ; M{C} = 0
M{B} = – (5 * 2) * 2/2 = (- 10)kNm
M_{D} = 3.33 * 3 = 9.99kNm
max + ve BM occurs at D = 9.99kNm and
max – ve BM occurs at B = – 10kNm
Point of contra flexure:
In BMD at (1) the BM is zero
Write the BM equation upto (1) and equate it to zero to find x1
16.67(X1 – 2) – (5 * 2)(X1 – 2/2) = 0
16.67X1 – 33.34 – 10X1 + 10 = 0
6.67X1 = 23.34
X1 = 23.34/6.67 = 3.5m
… Point of contra flexure occurs at 3.50 m from free end C. or 1.5 m from support B or 4.5 m from support A.
