Q. A simply supported beam of span 6 m carries udl of 20kN / m upto a distance of 3 m from left support and a point load of 50 kN is placed at a distance of 2 m from right support. Draw the SFD and BMD and indicate where max shear force and max B.M occurs.

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SOLUTION:


Total load = 20 * 3 + 50 = 110kN
To find the reactions
Taking moments about A
R{B} * 6 = 50 * 4 + 20 * 3 * 3/2 = 290
R{B} = 290/6 = 48.33kN
R_{A} = 110 – 48.33 = 61.67kN
SFD ( SHEAR FORCE DIAGRAM)
F_{A} = 61.67kN
F_{B} = – 48.33kN
F_{c} = 1.67kN
F_{D} = 1.67kN
Max +ve SF occurs at A = 61.67kN and
Max – ve SF occurs at B = – 48.33kN (same in DB)
B.M.D. ( BENDING MOMENT DIAGRAM)
M_{A} = M_{B} = 0
M_{C} = 61.67 * 3 – (20 * 3) 3/2 = 95.01kNm
M_{D} = 48.33 * 2 = 96.66kNm
Max BM occurs at D (where SF changes its sign)
M max =96.66 kNm.

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