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Shear Force and Bending Moment  for simple supported beam- Solved example

Problem 5-2

Calculate the values and draw the diagrams for shearing force and bending moment for the following beam shown in figure 5-2(a).

Figure 5-2(a)

Solution:

The given beam has two unknown reaction components which are calculated in Prob 4-1 as  Ay = 66 kN and By = 24 kN;

Shearing force values:

Wherever there is a point load we have to calculate the values of shearing force on both the sides (left and right) of that section.

FA-left = 0

FA-right = 66 kN;

FC = 66 – 20 x 4 = -14 kN;

FD-left = 66 – 20 x 4 = -14 kN;

FD-right = 66 – 20 x 4 – 10 = -24 kN;

FB-left = -24 kN;

FB-right = 0;

The values of shear force are plotted in SFD in figure 5-2(b)

Bending moment values:

MA = 0;

MC = 66 x 4 – 20 x 4 x 2 = 264-160 = 104 kNm;

MD = 66 x 8 – 20 x 4 x 6 = 528 – 480 = 48 kNm;

MB = 0;

The values of bending moment are plotted in BMD.in figure 5-2(b)

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Figure 5-2(b)

Throughout the span the bending moment is sagging in nature. To determine the maximum bending moment for the beam we use the relationship between shearing force and bending moment which says that the rate of change of bending moment (dM/dx) is equal to the shearing force (Fx) at the section. For maximum value of any function its first derivative should be equal to zero. Therefore at the point of maximum bending moment the shearing force must be equal to zero. From the shear force diagram it is clear that the shear force is zero between A and C. The point of zero shearing force can be easily determined by using the property of similar triangles.

Hence,  66/x = 14/(4-x) ;

Which yields, x =  3.3m;

Now calculate the bending moment at x = 3.3m which will be the maximum bending moment for this beam. Therefore,

Mmax = 66 x 3.3 -20 x (3.3) x (3.3/2) = 108.9 kNm;

More examples on shear and moment diagrams

Example 5-1   Example 5-3  Example 5-4

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 Last updated on Saturday October 26, 2013