Q. A cantilever beam of span 8 m carries an udl of 6 kN/m over entire span. In addition, it also carries three point loads of 10 kN, 20 kN and 15 kN at 2 m, 5 m and 7 m respectively from its fixed end. Draw S.F and B.M diagrams.

SOLUTION:

Total load = 10+20+15+ 6×8 = 93 kN

S.F.D ( SHEAR FORCE DIAGRAM)
FA = 93 kN ( 10 +20 + 15 + 6×8 = 93 kN)
FC =81 kN( 20 + 15 + 6 × 6 + 10 = 81 kN)
FD = 53 kN (20+15+ 6×3 = 53 kN)
FE = 21kN(15 + 6 × 1 = 21kN)
FB = 0
Max S.F occurs at fixed end A = FA =F max =93 kN

B.M.D ( BENDING MOMENT DIAGRAM)

M{B} = 0
M{E} = – (6 * 1) * 1/2 = – 3kNm
M{D} = – 15 * 2 – (6 * 3) * 3/2 = – 57 kNm
M{C} = – 20 * 3 – 15 * 5 – (6 * 6) * 6/2 = – 243 kNm
M{A } = – 10 * 2 – 20 * 5 – 15 * 7 -(6*8)* 8/2
= – 20 – 100 – 105 – 192 = – 417kNm
Max B.m occurs at fixed end A and its value
MA =M max = -417 kNm