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Shear Force and Bending Moment Diagrams in simply supported overhanging Beams

A simple beam extending beyond its support on one end. is called as overhanging beam.
The overhanging may be one side or both sides of the supports.
Q1. Draw the shear force and bending moment diagrams in a simply supported overhanging beam carrying loads as shown in figure. locate point of contraflexure if any.
Solution : 
Step.1
Find the unknown support reactions at support A (Ax, Ay) and support C (Cy) at supports as shown in the figure.

Along direction x, there is no imposed force applied to the structure.
Thus the first equilibrium equation is: ∑ Fx = 0
Ax = 0
Along direction y, the imposed loads as well as support reactions.
Thus the second equilibrium equation  ∑ Fy = 0
Ay + Cy - 10x4 - 20 = 0
Ay + Cy = 60 --- (a)
Taking moments about A and apply the third equilibrium equation ∑ Mxy = 0
10x4x2 + 20x10 - Cy. 8 = 0 --- (b)
Cy = 280/8 = 35 kN
Substitute the Cy value in equation (a)
Ay = 60 - Cy = 60 - 35 = 25 kN
Step.2 : Segment AB
Take a section x-x in between A and B points at a distance x form support A
Consider left side of the section, the limits of x is valid in between A and just left side of B only.
ie. 0 < x < 4m
considering sign convention (Left side), the S.F and B.M. equations are
Vx = + Ay - 10x = 25 - 10.x ---(1)
Mx = + (Ay).x - 10x.x/2 = 25.x - 5.x.x ---(2)
substituting x values in the above equations,
When x= 0,
Vx = The shear force at A, VA = + 25 kN
Mx = The bending moment at A, MA = 0
When x= 4m,
Vb = The shear force at B,
Vb = 25- 10x4 = -15 kN
Mx = The bending moment at C,
Mc = 25x4 - 5x4x4 = 20 kN-m
Step.3 : Segment BC
Take a section x-x in between C and B points at a distance x form Free end D
Consider right side of the section, the limits of  x is valid in between C and just right side of B only.
 ie. 2m < x < 6m
considering sign convention (Right side), the S.F and B.M. equations are
Vx = - Cy + 20 = -35 + 20 = -15  kN             ---(3)
Mx = +(Cy).(x - 2) - 20.x = + 35.(x - 2) -20.x ---(4)
substituting x values in the above equations,
When x= 2m,
Vx =  The shear force at C, VC = -15 kN
Mx = The bending moment at A, 
Mc = 35(2-2) - 20.2 =  - 40 kN-m
When x= 6m,
Vb =  The shear force at B, 
Vb =  -15 kN
Mx = The bending moment at C, 
Mb = 35(6-2) - 20.6 = + 20 kN-m
Step.4 : Segment CD
Take a section x-x in between C and D points at a distance x form Free end D
Consider right side of the section, the limits of  x is valid in between D and just right side of C only.
 ie. 0 < x < 2m
considering sign convention (Right side), the S.F and B.M. equations are
Vx =  + 20 kN             ---(5)
Mx =  - 20.x =  - 20.x ---(6)
substituting x values in the above equations,
When x= 0
Vx =  The shear force at D, VD= +20 kN
Mx = The bending moment at C, 
Mc = - 20.2 = - 40 kN-m
When x= 6m,
Vb =  The shear force at B, 
Vb =  -15 kN
Plot the shear force and bending moment diagrams by using the above values.
(The negative values are plotted to scale below a horizontal reference line and positive values are plotted above the reference line.)
Shear Force and Bending Moment Diagrams

Point of Contraflexure:
By observing the B.M.Diagram, we can see a point at which the value of bending moment is zero and change its sign form +ve to -ve. This point is called point of contraflexure.
This lies in between B and C, ie. 2 < x < 4
Form equation (4),
Mx = + 35.(x - 2) - 20.x = 0
15.x - 70 = 0
x = 70/15 = 4.67m
( this as shown as 'a' in Fig.)
where a= x-2 = 4.67-2 = 2.67m
The Maximum Bending Moment:
The maximum B.M. will occurs at a point, where the shear force changes its sign convention.
By observing Shear Force Diagram, we may seen this at two points.
The points are at a distance of x from support A in AB segment, and at support C.
Therefore, the shear force equation (1) is equate to zero to identify the value of x
Vx = 25 - 10.x =0
x = 25/10 = 2.5 m
For the Maximum Bending Moment, substitute this x value in bending moment equation (2)
Mx = 25.x - 5.x.x
Mmax = 25x2.5 - 5(2.5)(2.5) = 31.25 kN-m
Assignment :
Q.1 Draw shear force and bending moment diagram of simply supported overhanging beam carrying point loads as shown in figure below. Where P= 10 kN, Q= 5 kN and L= 6m, a= 2m
Q.2. Draw shear force and bending moment diagram of simply supported overhanging beam carrying uniformly distributed load as shown in figure below.
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