**Concept of Shear Force and Bending moment in beams: **

When the beam is loaded in some arbitrarily manner, the internal forces and moments are developed and the terms shear force and bending moments come into pictures which are helpful to analyze the beams further. Let us define these terms

Fig 1

Now let us consider the beam as shown in fig 1(a) which is supporting the loads P_{1}, P_{2}, P_{3} and is simply supported at two points creating the reactions R_{1} and R_{2} respectively. Now let us assume that the beam is to divided into or imagined to be cut into two portions at a section AA. Now let us assume that the resultant of loads and reactions to the left of AA is �F’ vertically upwards, and since the entire beam is to remain in equilibrium, thus the resultant of forces to the right of AA must also be F, acting downwards. This forces �F’ is as a shear force. The shearing force at any x-section of a beam represents the tendency for the portion of the beam to one side of the section to slide or shear laterally relative to the other portion.

Therefore, now we are in a position to define the shear force �F’ to as follows:

At any x-section of a beam, the shear force �F’ is the algebraic sum of all the lateral components of the forces acting on either side of the x-section.

**Sign Convention for Shear Force: **

The usual sign conventions to be followed for the shear forces have been illustrated in figures 2 and 3.

**Fig 2: Positive Shear Force**

**Fig 3: Negative Shear Force**

**Bending Moment: **

**Fig 4**

Let us again consider the beam which is simply supported at the two prints, carrying loads P_{1}, P_{2} and P_{3} and having the reactions R_{1} and R_{2} at the supports Fig 4. Now, let us imagine that the beam is cut into two potions at the x-section AA. In a similar manner, as done for the case of shear force, if we say that the resultant moment about the section AA of all the loads and reactions to the left of the x-section at AA is M in C.W direction, then moment of forces to the right of x-section AA must be �M’ in C.C.W. Then �M’ is called as the Bending moment and is abbreviated as B.M. Now one can define the bending moment to be simply as __the algebraic sum of the moments about an x-section of all the forces acting on either side of the section __

**Sign Conventions for the Bending Moment: **

For the bending moment, following sign conventions may be adopted as indicated in Fig 5 and Fig 6.

**Fig 5: Positive Bending Moment**

**Fig 6: Negative Bending Moment **

Some times, the terms �Sagging’ and Hogging are generally used for the positive and negative bending moments respectively.

**Bending Moment and Shear Force Diagrams: **

The diagrams which illustrate the variations in B.M and S.F values along the length of the beam for any fixed loading conditions would be helpful to analyze the beam further.

Thus, a shear force diagram is a graphical plot, which depicts how the internal shear force �F’ varies along the length of beam. If x dentotes the length of the beam, then F is function x i.e. F(x).

Similarly a bending moment diagram is a graphical plot which depicts how the internal bending moment �M’ varies along the length of the beam. Again M is a function x i.e. M(x).

**Basic Relationship Between The Rate of Loading, Shear Force and Bending Moment: **

The construction of the shear force diagram and bending moment diagrams is greatly simplified if the relationship among load, shear force and bending moment is established.

Let us consider a simply supported beam AB carrying a uniformly distributed load w/length. Let us imagine to cut a short slice of length dx cut out from this loaded beam at distance �x’ from the origin �0′.

Let us detach this portion of the beam and draw its free body diagram.

The forces acting on the free body diagram of the detached portion of this loaded beam are the following

• The shearing force F and F+ dF at the section x and x + dx respectively.

• The bending moment at the sections x and x + dx be M and M + dM respectively.

• Force due to external loading, if �w’ is the mean rate of loading per unit length then the total loading on this slice of length dx is w. dx, which is approximately acting through the centre �c’. If the loading is assumed to be uniformly distributed then it would pass exactly through the centre �c’.

This small element must be in equilibrium under the action of these forces and couples.

Now let us take the moments at the point �c’. Such that

**Conclusions:** From the above relations,the following important conclusions may be drawn

• From Equation (1), the area of the shear force diagram between any two points, from the basic calculus is the bending moment diagram

• The slope of bending moment diagram is the shear force,thus

Thus, if F=0; the slope of the bending moment diagram is zero and the bending moment is therefore constant.’

• The maximum or minimum Bending moment occurs where

The slope of the shear force diagram is equal to the magnitude of the intensity of the distributed loading at any position along the beam. The �ve sign is as a consequence of our particular choice of sign conventions

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