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Shear Force and Bending Moment - Solved example for overhanging beam

Problem 5-4

Calculate the values and draw the diagrams for shearing force and bending moment for the following overhanging beam shown in figure 5-4(a) and find the position of point of contra-flexure, if any.

Figure 5-4(a)

Solution:

Calculation of support reactions

The beam has three reaction components; Ax , Ay and Cy. The free-body-diagram is shown in figure 5-4(b).

Applying the equations of static equilibrium we get;

ΣFx = 0;  Ax = 0;                                (eq. 1)

ΣFy = 0; Ay + Cy –10x4 – 20  = 0;

Ay + Cy = 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);

ΣMz = 0;  Cy x 8 – 20x10 – 10x4x2 = 0;    (eq. 3)

Solving eq. 3, we get Cy = 280/8 = 35 kN;

Substituting the value of Cy in eq. 2, we get Ay = 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.

FA-left = 0;

FA-right = 25 kN;

FB = 25 – 10x4 = -15 kN;

FC-left = -15 kN;

FC-right = 25 – 40 + 35 = 20 kN;

FD-left = 20 kN;

FD-right = 20 – 20 = 0;

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

Bending Moment Calculations

MA = 0;

MB = 25x4 – 10x4x2 = 20 kNm;

MC = 25x8 –10x4x6 = – 40 kNm;

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

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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;

Mmax (+ve) = 25x2.5 – 10x(2.5)x(2.5/2) =  31.25 kNm

and Mmax (-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.

More examples on shear and moment diagrams

Example 5-1  Example 5-2  Example 5-3

You can also use our Free online calculators given below

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 Last updated on Saturday October 26, 2013