Q. A simply supported beam AB of span 6m long carries a udl of 10 kN/m over a length of 2 m from the right support. In addition to that it carries two point loads of 12 kN and 15 kN at a distances of 2 m and 3 m respectively from left support. Draw SF and BM diagrams and indicating values at salient points.

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SOLUTION:


Total load = 12 + 15 + (10 * 2) = 47kN
To find the reactions
Taking moments about B
R{A} * 6 = (10 * 2) * 2/2 + 15 * 3 + 12 * 4
= 20 + 45 + 48 = 113
R{A} = 113/6 = 18.83kN
R{B} = 47 – 18.83 = 28.17kN
SFD ( SHEAR FORCE DIAGRAM)
Max + ve SF = F A = 18.83
F{C} = 18.83kN
F{D} = 6.83
F{E} = – 8.17
Max ve SF = F{B} = – 28.17kN
BMD (BENDING MOMENT DIAGRAM)
M{A} = M_{B} = 0
M{C} = 18.83 * 2 = 37.66 kNm
M{D} = 18.83 * 3 – 12 * 1 = 44.49kNm
M{E} = 28.17 * 2 – (10 * 2) 2/2 = 36.34
Max BM occurs at D = Mmax = 44.49 kNm

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