Q. Draw S.F.D for the cantilever shown in Figure.

Q. Draw S.F.D for the cantilever shown in Figure.

SOLUTION:
Total load = 3×1 + 4 + 3×1 = 10 kN

SFD ( SHEAR FORCE DIAGRAM)
FE=10kN
FD=7kN (same in DC)
FC (just left of C) = 7 kN
FC (just right of C)=3 KN
FB =3kN
FA =0kN
MAX SHEAR FORCE occurs at A=10kN

BMD ( BENDING MOMENT DIAGRAM)
M{A}=0
M{B}= -(3 ×1) 1/2 = -1.5kNm
M{C}= -(3 × 1)(1/2+1) = – 4.5 kNm
M{D}= -4×1 – (3×1)(1/2+2) = – 11.5KNm
M{E}= -4×2 – (3×1) 1/2 – (3×1) ( 1/2+3 )
= -8 -1.5 – 10.5
= -20 KNm
Max BENDING MOMENT Occurs at E= ME=Mmax = -20 KNm