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SHEAR FORCE AND BENDING MOMENT DIAGRAM FOR A CANTILEVER BEAM.

SFD &  BMD for a cantilever with a point Load at it's free end: 


Consider a cantilever beam of span "L" fixed at end A and free at end B carrying a point load at end B. 
                  Consider a section X-X at a distance x from free end B & consider the right portion of the section. 
The shear force at this section will be equal to the resultant of forces acting on right/left portion of the assumed section as per the definition of shear force. But the resultant force acting on right portion of X-X is "W" acting in downward direction. Since W lies in right portion of X-X,  it will be considered positive as per sign conventions of SFD.
Hence shear force at X-X is positive.

Let Fx  = shear force at X-X. 
Fx  = +W _______________ (1)

The shear force will be constant at all the sections between A and B since there is no other load present between A and B. 

The SFD is shown below, 
Let,  Mx  = Bending moment at X-X
The B.M at X-X will be given by, 
Mx  =  - W×X ___________(2)

The B.M will be negative as the moment due to the load "W"  will be clockwise for right portion of X-X (as per the sign conventions of B.M) and also the beam will bend in a manner that the convexity will be at the top of the beam. 
                    From eq(2), it is evident that the Bending moment at any section between A &  B is directly proportional to the distance of section X-X from free end.

At x=0, i. e at B,  B. M = 0
At x=L, i.e at A,  B.M = - WL
Hence B.M follows straight line rule. 

The BMD is shown below, 
Problem:
Draw SFD &  BMD for the cantilever beam shown below carrying point loads at different locations. 
Solution:

Shear force diagram :

Shear force at D = 200N

S.F just to the right of C = 200N
(S.F will remain constant between D&C)

S.F just to the left of C = 200+100
 = 300N(Due to point load 100N at C) 

S.F just to the right of B = 300N
(S.F will remain constant between C&B) 

S.F just to the left of B =300+50=350N
(Due to point Load 50N at B) 

S.F at A = 350N

Bending Moment diagram:

B.M at D = 0

The B.M at any section between C & D at a distance x from D is given by,
Mx = -200(x),  which follows a straight line law

At C,  x = 4
B.M at C = - 200X4 = - 800Nm

The B.M at any section between B &  C at a distance x from D is given by,  
Mx  = - 200(x)-100(x-4), which follows a straight line law.

At B,  x = 8
B.M at B = - (200X8) +(-100X4) = -2000Nm 

The B.M at any section between A & B at a distance of x from D is given by,
Mx  = 200(x)-100(x-4)-50(x-8),  which follows a straight line law. 

At A, x = 10
B.M at A = - 200(10)-100(10-4)-50(10-8)
                = - 2700Nm

MD = 0Nm
Mc  = - 800Nm
MB  = - 2000Nm
MA  = - 2700Nm

BMD is shown below,

SFD &  BMD for a cantilever  carrying Uniformly distributed load (UDL):



Consider a cantilever beam of span "L" fixed at end A and free at end B carrying a UDL of w per unit length over it's entire span.

Consider a section X-X at a distance x from free end B & consider the right portion of the section. 

The shear force at this section will be equal to the resultant of forces acting on right/left portion of the assumed section as per the definition of shear force. But the resultant force acting on right portion of X-X is 
W(length of right portion) = Wx acting in downward direction. Since Wx lies in right portion of X-X,  it will be considered positive as per sign conventions of SFD.

Hence shear force at X-X is positive.
Let Fx  = shear force at X-X. 
Fx = +WX ___________(1)

The above eq shows that S.F forms a straight line law.
At,  x = 0
S.F = 0
At,  x = L
S.F = WxL = WL

SFD is shown below, 
Let,  Mx  = Bending moment at X-X

The B.M will be negative as the moment due to the load "WX" will be clockwise for right portion of X-X (as per the sign conventions of B.M) and also the beam will bend in a manner that the convexity will be at the top of the beam. 

UDL of a section is converted into a point Load acting at the C.G of the section.  

The B.M at X-X will be given by, 
Mx  = - (Total load on right portion) x  
       (distance of C.G of right portion from X)

Mx  =  - W×X(X/2) = - WX/2 ________(2)

From the above eq it is evident that the Bending moment at a section is proportional to the square of the distance of the section from free end, which follows a parabolic law. 

At B,  x = 0
B.M at B = 0

At A,  x = L
B.M at A = - (WL/2)

The BMD is shown below, 
Problem:
Draw SFD &  BMD for the cantilever beam shown below carrying point load 2KN at a distance of 2m from free end and a UDL of 10KN/m run over a length of 3m from the free end. 
Solution:
Shear force diagram :
S.F at D = 0

S.F just to the right of point C = 10x2 = 20KN

S.F just to the left of point C  = (10x2)+2 = 22KN

S.F just to the right of the point B = 10(3+2)+2 = 52KN

S.F just to the left of point B = 10(5)+2 = 52KN

Since there is no load between A & B, the shear force will remain constant between A & B

SFD is shown below, 
Bending Moment diagram :

B.M at D = 0

B.M at C = - 10(2)(2/2) = - 20KNm

B.M at B = - 10(5)(5/2)-2(3) = - 131KNm

B.M at A = - 10x5(5/2 + 2)-2x5
 = - 235KNm  

The B.M between D & C and C & B varies by a parabolic law. But the B.M between A & B varies by a straight line law. 

The bending moment diagram is shown below, 



























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