Q. A beam of length 5 m, simply supported over a span of 4 m and is having 1 m overhang on the right hand support. It carries two point loads 2kN and 4 kN at the distances of 3 m and 5 m from LHS and a udi of 2 kN/m spread over 3 m from LHS. Draw SF and BM diagrama state the position and magnitude of max +ve BM and the position of point of contraflexure.

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SOLUTION:


Total load = (2 * 3) + 2 + 4 = 12kN
To find reactions
Taking moments about A
R{B} * 4 = 2 * 3 + 4 * 5 + (2 * 3) * 3/2
R{B} = 35/4 = 8.75kN
R{A} = 12 – 8.75 = 3.25kN
SFD ( SHEAR FORCE DIAGRAM)
F{A} = 3.25kN
F{B} = 4kN (same in BD)
F{D} = 4kN
F{c} = – 4.75kN(same in (B)
S{F} =cross at D and B
To find x use similar triangles
x/3.25 = (3 – x)/2.75
2.75x = 3.25 * 3 – 3.25x
6x = 9.75;
x = (9.75)/6 = 1.625m

BMD ( BENDING MOMENT DIAGRAM)
M{A} = M{D} = 0
M{B} = – 4 * 1 = – 4kNm
M{C} = 3.25 * 3 – (2 * 3) * 3/2
= 9.75 – 9 = 0.75kNm
M max =MD =M 1.625
=3.25 * 1.625 – 2 * 1.625 * 1.625/2
=2.640625 kNm

Max + ve BM occurs at 1.625 m from left support A = 2.640625kNm and
Max-ve Bm occurs at support B = – 4kNm

To find point of centre flexure: (where BM changes its sign)
Let us assume BM changes its sign at a distance of “a” from D
– 4 x a + 8.75(a – 1) = 0
– 4a + 8.75a – 8.75 = 0
4.75a = 8.75
a = 8.75/4.75 = 1.8421m
Point of contra flexure occurs at a distance of 1.8421 m from fre end D.

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