SOLUTION:

Total load = 4×3 + 2×3 = 18 kN

S.F.D ( SHEAR FORCE DIAGRAM)

FA=F{max}=18kN[4×3+2×3]

FB=0

Fc = 6 kN

FD = 6 kN

B.M.D ( BENDING MOMENT DIAGRAM)

MB=0

MC= -(2×3)3/2 = -9 kNm

MD= -(2×3)(3/2+1) = -15kNm

Mmax = MA = -(4×3)3/2-(2×3)(3/2+4)

= – 18 – 33

= – 51 KNm

Max B.M occurs at fixed end A

=Mmax= -51 kNm

 

 

Q. A cantilever beam of span 4 m long carries a udl of 2 kN/m for 1m from the fixed end and 4 kN/m for 1m from free end. Draw SFD & BMD.

SOLUTION:

Total load = 2×1 + 4×1=6kN

S.F.D ( SHEAR FORCE DIAGRAM)

FA = 6 kN

Fc = 4 kN

FD = 4 kN

FB = 0 kN

max SF occurs at A = 6 kN

B. M.D ( BENDING MOMENT DIAGRAM)

MB= 0

MD =-(4×1)1/2 = -2 kNm

MC = (4×1) (1/2+2) = -10 kNm

MA = – (4 × 1) (1/2 + 3) – (2 × 1) 1/2 = – 14 – 1 = – 15kNm

Max BM occurs at support A

MA =M max =-15 kNm