Q.A beam of length 9 m is simply supported on a span of 5 m with an over hang of 4 m in the right hand portion. It carries a udl of 3 kN/m on a length of 5 m from LHS and a point load of 4 kN at 9 m from LHS. Draw SF and BM diagrams and state the position of point of contra flexure.
Total load 3 * 5 + 4 = 19 kN
To find reactions
Taking moments about A
R{B} * 5 = 4 * 9 + (3 * 5) * 5/2 = 73.5
R{B} = 73.5/5 = 14.7kN
RA =Total load – R{B} = 19 – 14.7 = 4.3kN
SFD SHEAR FORCE DIAGRAM
F{A} = 4.3kN
FB =(just left ofB )=-10.7 kN
FB (just right of B) = 4 kN (same in BC)
F{C} = 4kN
Max +veSF occurs at A = 4.3kN and
Max – ve SF occurs at B = 10.7kN BMD ( BENDING MOMENT DIAGRAM)
M{A} = M{C} = 0
M{B} = – 4 * 4 = – 16kNm
x/4.3 = (5 – x)/10.7
10.7x = 4.3 * 5 – 4.3x
10.7x + 4.3x = 21.5
15x = 21.5
x = 21.5/15 = 1.43m
M max =M1.43 =4.3*1.43 – (3*1.43)* 1.43/ 2
= 3.08165 kNm
Max + ve BM occurs at 1.43 m from A i.e., 3.08165 kNm and
Max-ve BM occurs at B = – 16 kNm
To find point of contra flexure
Equate the BM upto E is equal to zero
4.3a – (3×a) * a/2 = 0
4.3a – 1.5a² = 0
a(4 3 – 1.5a² ) 0
4.3 = 1.5a
a = 4.3/1.5 = 2.87m
The point of contra flexure occurs at 2.87 m from left support A
