Bending Moment and shear force diagram for Simple supported beam

Problem 5-2

Determine the values and draw the diagrams for shear force and bending moment due to the imposed load on simple supported beam shown in figure 5-2(a) Figure 5-2(a)

Solution:

The simple supported beam is a beam which has one end hinged and the other roller supported. Hinged support will have two reactions (vertical and horizontal) and roller will have one (vertical). In the case of beam shown in figure 5-2(a), there is no horizontal force and hence the hinge support will have only one reaction. 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:

Shear force at a section is the algebraic sum of the all the forces on either side of the section. Here we consider the forces on the left hand side of the section and take upward force as positive and downward force as negative. 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 as there will be a sudden change in the shear force vaslues at that point.

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;

To draw the shear force diagram we take the distances on x-axis and the shear force values on y-axis and join the points. Wherever there is point load the shear force diagram will be parallel to x-axis and in case of uniform load the shear force diagram is a sloping straight line as shown in shear force diagram (SFD) plotted in figure 5-2(b)

Bending moment values: The bending moment at a section is the algebraic sum of the moment of all the forces on either side of the section. Hogging bending moment is taken as negative and sagging bending moment as positive. Click here to get more about nature of bending moment diagram due to different loadings.

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;

To draw the bending moment diagram the distances are taken on x-axis and bending moment values on y-axis. Bending moment diagram will be a straight line when there is point load and it is parabolic when there is uniformly distributed load (udl). The values of bending moment are plotted in BMD in figure 5-2(b) 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;

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