Q. Q.A beam of length 11 m is simply supported on a span of 8 m with an over hang of 3 m in the right hand portion. It carries a udl of 10 kN/m on a length of 3 m from LHS and a point load are 3kN,6kN and 5kN at the distance 5m, 7m and 8m from LHS. Draw SF and BM diagrams and state the position of point of contra flexure.
SFD ( SHEAR FORCE DIAGRAM)
F{A} = 24.375kN
F{c} = – 5.625kN (same in CD)
FD =-8.625 kN(same inCE )
F{B} = – 14.625kN (same in EB)
FB =(just right ofB )=5kN (same in BF)
F{F} = 5kN
Max +veSF occurs at A = 24.375kN and and Max ve SF occurs at B = – 14.625 (same in EB)
BMD( BENDING MOMENT DIAGRAM)
M{A} = M{F} = 0
M{B} = – 5 * 3 = – 15kNm
M{F} = 19.625 * 1 – 5 * 4 = – 0.375 kNm
M{E} = 24.375 * 7 – 3 * 2 – (10 * 3)(3/2 + 4)
= 170.625 – 6 – 165 = – 0.375 kNm
M{D} = 24.375 * 5 – (10 * 3)(3/2 + 2)
= 121.875 – 105 = 16.875kNm
M{C} = 24.375 * 3 – (10 * 3) * 3/2
= 73.125 – 45 = 28.125kNm
x/24.375 = (3 – x)/5.625 ; 5.625 = 73.125 – 24.375x
30x = 73.125 ; x = 73.125/30 = 2.4375m
M max =M 2.4375
= 24.375 * 2.4375 – (10 * 2.4375) * 2.4375/2
= 29.707 kNm
To find point of contra flexure
Equate the BM upto H is equal to zero
19.625(1 + a) – 6a – 5(4 + a) = 0
19.625 + 19.625a – 6a – 20 – 5a = 0
8.625a = 0.375; a = 0.375/8.625 = 0.04347m
Point of contra flexure occurs at 0.04347 m from E
