Q. A cantilever 5m long carries three point loads of 20 kN, 30 kN and 15 kN at 1 m, 2.5 m and 4 m respectively from free end. Draw SF and BM diagrama. Calculate S.F. and B.M at 4.5 m from free end.

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SOLUTION:

Total load = 15+ 30 + 20 = 65 kN

SHEAR FORCE DIAGRAM

FA = 65 kN (same between AC)

Fc =(just left of B) = 65 kN

Fc =(just right of B) 50 kN

FD= (just left of D) = 50 kN

FD =(just right of D) = 20 kN

FE =(just left of E) = 20 kN

FE= (just right of E) = 0 kN

FB = 0 kN

Max SF occurs at A = 65 kN (same between AC)

BENDING MOMENT DIAGRAM

M{E} = M{B} = 0
M{D} = – 20 * 1.5 = – 30kNm
M{C} = – 30 * 1.5 – 20 * 3 = – 105kNm
M{A} = – 15 * 1 – 30 * 2.5 – 20 * 4 = – 170kNm

Max. B.M occurs at support A

MA =M max =-170 kNm

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