SOLUTION:

Total load = 2 * 2 = 4kN

S.F.D (SHEAR FORCE DIAGRAM)

F{A} = 4kN (same between A and C)
F{C} = 4kN
F{B} = 0kN
Max S.F occurs at A = 4kN

BMD( BENDING MOMENT DIAGRAM)

M{B} = 0
M{C} = – (2 * 2) * 2/2 = – 4kNm
M max =M A
= – (2 * 2) * (2/2 + 1) = – 8kNm
Max B.M occurs at Fixed end A
M max = -8 kNm

 

 

Q. A cantilever beam of span 3 m long carries a udl of 10 kN/m over a length of 2 m from free end. Draw S.F.D and B .M.D. Find the max. S.F. and Max B.M and also find S.F and B.M at 1m from the free end.

 

SOLUTION:

S.F.D(SHEAR FORCE DIAGRAM)
FA = 20 kN (10 x 2 = 20 kN)
FD = 20 kN (10 × 2 = 20 kN)
FC = 10 kN (10×1) = 10 kN)
FB = 0

Max S.F occurs at fixed end A
FA = Fmax = 20 kN
S.F at 1m from free end
FC= 10 kN

B.M.D( BENDING MOMENT DIAGRAM)

MB = 0
MC = – ( 10×1)1/2 = -5 kNm
MD = -(10×2)x 2/2 = -20 kNm
MA = -(10×2) (2/2+1) = -40 kNm.
Max BM occurs at fixed end A i.e..

MA = Mmax = – 40 kNm
=40 kNm (hogging)