Q. A simply supported beam of span 8 m carries point loads of 10 kN, 30 kN and 20 kN at 2 m, 5 m and 7 m respectively from its LHS in addition to a udl of 7.5 kN/m over its entire length. Construct SF and BM diagrams and state the position and Magnitude of max. В.М.

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SOLUTION:


Load load = 10 + 30 + 20 + 7.5 * 8 = 120kN
To find reactions
R{B} * 8 = (7.5 * 8) 8/2 + 10 * 2 + 30 * 5 + 20 * 7
R{B} = 550/8 = 68.75kN
R{A} = 120 – 68.75 = 51.25kN
SFD ( SHEAR FORCE DIAGRAM)
F{A} = 51.25kN
F{c} = 36.25kN
F{D} = 3.75kN
F{E} = – 61.25kN
FB = – 68.75kN
Max +ve SF occurs at A = 51.25kN A and
max – veSF occurs at B = – 68.75 KN
BMD ( BENDING MOMENT DIAGRAM)
M{A} = M_{B} = 0
M{C} = 51.25 * 2 – (7.5 * 2) * 2/2 = 87.5kNm M{D} = 51.25 * 5 – 10 * 3 – (7.5 * 5) * 5/2 = 132.5kNm
M{E} = 68.75 * 1 – (7.5 * 1) * 1/2 = 65kNm
Max BM occurs at D (Where SF changes its sign) M max =M D =132.5 kNm

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