**Branch :**First Year-Engineering Syllabus

**Subject :**Elements of Mechanical Engineering

**Unit :**Beam and Trusses

## Shear force, axial force and bending moment

**Shear force, axial force and bending moment:**** **Consider a beam as shown in the figure.

Let us cut the beam at a distance of x mm from support into parts (1) and (2).

Part (2) is balanced under the action of applied forces and internal forces C. Internal forces on this part are these forces which are applied by part 2 on part 1. On the cut face at C following forces may act:

- A force tangential to surface
- A force normal to surface
- A moment

A shear force diagram shows the variation of shear force across the length. Shear force at a particular length can be found by taking a section at that point. For example, in order to find the shear force at a distance *x* in the cantilever beam, we cut the beam in two parts at a distance *x* from the end and consider the equilibrium of part (2). We could have considered part 1 as well, but then we have to find the support reactions first. Applying the vertical force balance to part (2), V_{2} =*P* .

Since V_{2} is downward on negative section, shear force is positive. Expression for shear force is independent of *x.*

Thus, the shear force is constant. At a slight distance away from point B towards right, the shear force is suddenly zero, as there is no force there. Actually, the point B is the point of singularity, where the shear force rises from 0 to *P*. Same applies to point A, where a slight distance away from point A towards left, there is no force. Thus shear force is zero there. Now, we can drawn the sl

For finding out the bending moment, consider the moment balance of part 2, taking the moment about C

*M *_{2} *P*(*L*-*x*) = 0

*M *_{2} = *-P* (*L*-*x*)

Since the negative moment is acting on the section whose area vector is negative, bending moment is positive.This is the equation of straight line. We can easily drawn the bending moment diagram, by finding out the bending moment at x = 0 and x = *L *and joining the points at x = 0 x = *L*, by straight line. At *x *= *L*, *M*_{2} = 0 and x = 0 *M*_{2} = - *PL* .

For finding out the axial force diagram, consider the horizontal force balance of section 2. Here we found

H_{2} = Q.Thus axial force is independent of *x *. The axial force diagram is drawn similar to shear force diagram.