**Bending Moment and shear force diagram for overhanging beam**

**Problem 5-4**

Determine the values and draw the diagrams for shear force and bending moment due to the imposed load on overhanging beam shown in figure 5-4(a) and find the position of point of contra-flexure, if any.

**Figure 5-4(a)**

**Solution:**

The overhanging beam is a beam which has unsupported length on one or both sides.

__Calculation of support reactions__

Applying the equations of static equilibrium we get;

ΣF_{x} = 0; **A**_{x} =
0;
(eq. 1)

ΣF_{y} = 0; **A**_{y} + **
C**_{y}** **– 10x4
– 20 ** = **0;

**A**_{y} + **C**_{y}**
**= 60 kN;
(eq. 2)

Considering z-axis passing through A, and
taking moment of all the forces about z-axis *(taking clockwise –ve
and anticlockwise +ve)*;

ΣM_{z} = 0; **C**_{y}
x 8 – 20x10 – 10x4x2
= 0; (eq. 3)

Solving eq. 3, we get **C**_{y}**
**=** **280/8 = **35 kN**;

Substituting the value of **C**_{y}**
**in eq. 2, we get **A**_{y}** **= **25 kN**;

**Shear force calculations**

In case of point load acting at a point, we should calculate shear force on both sides (left and right) of the point.

F_{A}-left = 0;

F_{A}-right = 25 kN;

F_{B} = 25 – 10x4
= -15 kN;

F_{C}-left = -15 kN;

F_{C}-right = 25 – 40 + 35 = 20 kN;

F_{D}-left = 20 kN;

F_{D}-right = 20 – 20 = 0;

The values of shear force are plotted in SFD in figure 5-4(b)

**Bending Moment Calculations**

M_{A} = 0;

M_{B} = 25x4
– 10x4x2
= 20 kNm;

M_{C} = 25x8
–10x4x6
= – 40 kNm;

The values of bending moment are plotted in BMD in figure 5-4(b)

**
Figure 5-4(b)**

Maximum +ve bending moment will occur at the point of zero shear force, which can be easily calculated by using the property of similar triangles of shear force diagram between A and B as given below;

25/x = 15/(4–x);

Which gives x = 2.5 m;

Therefore maximum +ve bending moment will occur at x = 2.5 m;

**M**_{max}** (+ve)** = 25x2.5 –
10x(2.5)x(2.5/2) = **31.25 kNm**

and **M**_{max}** (–ve)
**= **– 40 kNm**; at
point C.

It is evident from the bending moment diagram
that there is a point, at which the bending moment is equal to zero.
Such a point having different nature of bending moment on its two
sides (left and right) is known as point of __contra-flexure__.

**Point of contra-flexure** can be determined by
writing the equation of BM for part BC and put it equal to zero;

Mx= 25x – 10*4(x–2) = 0; x= 5.33 from A.

It can also be determined by considering the diagram between B and C and using the property of similar triangles;

40/a = 20/(4–a);

Which gives; a = 2.67m;

i.e., the point of contra-flexure is at a distance of 2.67 m from support C. This information will be helpful in providing rebars in case of reinforced concrete beams.

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