SOLUTION:

Total load = 10×2 = 20 kN

S.F.D (SHEAR FORCE DIAGRAM)
FB = Fc = 0
FA = 10×2 = 20 kN
max S.F occurs at support A = 20 kN

B.M.D ( BENDING MOMENT DIAGRAM)
MB = MC = 0
MA = -10 x 2 x 2/2 = -20 kNm
(or)
MA = -WL²/2 =[ -(10×2²)/2] = -20 kN

Max B.M occurs at fixed end at at A = 20 kNm (hogging)

Q. A cantilever beam of span 4m carries a udl of 10 kN/m from 1 m 3 m from fixed end. Draw S.F.D and B.M.D.

SOLUTION:

Total load = 10 ×2 = 20kN

S.F.D (SHEAR FORCE DIAGRAM)
FA 20 kN (same in AC)
Fc = 20 kN
F{D}=F{B}=0
Max SF occurs at A =F{max}=20kN

B.M.D ( BENDING MOMENT DIAGRAM)
MB=0
MD= 0
MC= – ( 10 × 2) 2/2 = – 20 KNm
MA = – ( 10 × 2) (2/2 +1) = – 40KNm
Max B.M occurs at support A
M{max}= – 40kNm