Q. A cantilever beam 6 m long carries four point loads of 5 kN, 10 kN, 15 kN and 20 kN are acting at 2 m, 3 m, 4 m and 6 m respectively from fixed end. Draw SFD and B.M.D.

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SOLUTION;

Total load = 5 + 10 + 15 + 20 = 50 kN

SHEAR FORCE DIAGRAM

F A =F max = 50 kN (5 + 10 + 15 + 20) (same in between A &E)

F{B} = (just left of B)= 20kN

F{c} = (just left ofC )=35 kN

FC = (just right of C)= 20kN

F D =(just left ofD )=45 kN

F D =(just right ofD )=35 kN

F E =(just left ofE )=50 kN

F E =(just right ofE )=45 kN

BENDING MOMENT DIAGRAM

M{B} = 0

M{C} = – 20 * 2 = – 40kNm

M{D} = – 15 * 1 – 20 * 3 = – 75kNm

M{E} = – 10 * 1 – 15 * 2 – 20 * 4 = – 120 kNm

M max =MA = -5 * 2 -10 * 3 -15 * 4 – 20 * 6

= – 10 – 30 – 60 – 120

= – 220 kN

Max BM occurs at Fixed end A

M max =-220 kNm

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