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.

overhanging beam with combined loading

Figure 5-4(a)


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

Calculation of support reactions

In the case of beam shown in figure 5-4(a), there are 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)

shear froce and bending moment diagram

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.

You can visit the following links of solved examples on Bending moment and shear force calculations and plotting of diagrams

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