5 Engineering Example 3

5.1 Shear force and bending moment of a beam

Introduction

The beam is a fundamental part of most structures we see around us. It may be used in many ways depending as how its ends are fixed. One end may be rigidly fixed and the other free (called cantilevered ) or both ends may be resting on supports (called simply supported ). Other combinations are possible. There are three basic quantities of interest in the deformation of beams, the deflection, the shear force and the bending moment.

For a beam which is supporting a load of $w$ (measured in $\text{N}{\text{m}}^{-1}$ and which may represent the self-weight of the beam or may be an external load), the shear force is denoted by $S$ and measured in $\text{N}{\text{m}}^{-1}$ and the bending moment is denoted by $M$ and measured in $\text{N}{\text{m}}^{-1}$ .

The quantities $M$ , $S$ and $w$ are related by

$\frac{dM}{dz}=S$ (1)

and

$\frac{dS}{dz}=-w$ (2)

where $z$ measures the position along the beam. If one of the quantities is known, the others can be calculated from the Equations (1) and (2). In words, the shear force is the negative of the derivative (with respect to position) of the bending moment and the load is the derivative of the shear force. Alternatively, the shear force is the negative of the integral (with respect to position) of the load and the bending moment is the integral of the shear force. The negative sign in Equation (2) reflects the fact that the load is normally measured positively in the downward direction while a positive shear force refers to an upward force.

Problem posed in words

A beam is fixed rigidly at one end and free to move at the other end (like a diving board). It only has to support its own weight. Find the shear force and the bending moment along its length.

Mathematical statement of problem

A uniform beam of length $L$ , supports its own weight $w_o$ (a constant). At one end $\left(z=0\right)$ , the beam is fixed rigidly while the other end $\left(z=L\right)$ is free to move. Find the shear force $S$ and the bending moment $M$ as functions of $z.$

Mathematical analysis

As $w$ is a constant, Equation (2) gives

$S=-\int \; <!--nolimits\; -->wdz=-\int \; <!--nolimits\; -->w_{o}dz=-w_{o}z+C$ .

At the free end $\left(z=L\right),$ the shear force $S=0$ so $C=w_{o}L$ giving

$S=w_o\left(L-z\right)$

This expression can be substituted into Equation (1) to give

$M=\int \; <!--nolimits\; -->Sdz=\int \; <!--nolimits\; -->w_o\left(L-z\right)dz=\int \; <!--nolimits\; -->\left(w_{o}L-w_{o}z\right)dz=w_{o}Lz-\frac{w_o}{2}{z}^2+K$ .

Once again, $M=0$ at the free end $z=L$ so $K$ is given by $K=-\frac{w_o}{2}{L}^2.$ Thus

$M=w_{o}Lz-\frac{w_o}{2}{z}^2-\frac{w_o}{2}{L}^2$

The diagrams in Figure 8 show the load $w$ (Figure 8a), the shear force $S$ (Figure 8b) and the bending moment $M$ (Figure 8c) as functions of position $z.$

Figure 8 :

Interpretation

The beam deforms (as we might have expected) with the shear force and bending moments having maximum values at the fixed end and minimum (zero) values at the free end. You can easily experience this for yourselves: simply hold a wooden plank (not too heavy) at one end with both hands so that it is horizontal. As you try this with planks of increasing length (and hence weight) you will find it increasingly difficult to support the weight of the plank (this is the shear force) and increasingly difficult to keep the plank horizontal (this is the bending moment).

This mathematical model is an excellent description of real beams.