Macaulay Method

A beam is a horizontal structural element that is capable of withstanding load primarily by resisting bending. The bending force induced into the material of the beam as a result of the external loads, own weight, span and external reactions to these loads is called a bending moment.

Describes the Macaulay Method for calculating the deflection of Beams

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Macaulay's Method

Using the Calculus to find expressions for the deflection of loaded beams (See Deflection of Beams Part 1) it is normally necessary to have a separate expression for the Bending Moment for each section of the beam between adjacent concentrated loads or reactions. Each section will produce its own equation with its own constants of integration. It will be appreciated that in all but the simplest cases the work involved will be laborious, the separate equations being linked together by equating slopes and deflections given by the expressions on either side of each "junction point.

However a method devised by Macaulay enables one continuous expression for bending moment to be obtained and provided that certain rules are followed the constants of integration will be the same for all sections of the beam.

Macaulay'\b{s method} (The double integration method) is a technique used in structural analysis to determine the deflection of Euler-Bernoulli beams. Use of Macaulay's technique is very convenient for cases of discontinuous and/or discrete loading.

It is advisable to deal with each different type of load separately.

Concentrated Loads

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Measuring $x$ from one end write down an expression for the Bending Moment in the last section of the beam enclosing all less than $x$ in square brackets, i.e.

$E\,I\,\frac{d^2y}{dx^2} = M = -\,W_1x + R[x - a] - W_2[x - b] - W_3[x - c]$

Subject to the condition that all terms for which the quantities in the square brackets are negative, are omitted ( i.e. given a value of zero), this equation may be said to represent the bending moment for all values of $x$ . If $x$ is less than $b$ then both the last two terms are omitted and so on.

The brackets are integrated as a whole. i.e.

$E\,I\,\frac{dy}{dx} = -\,W_1\frac{x^2}{2} + \frac{R}{2}[x - a]^2 - \frac{W_2}{2}[x - b]^2 - \frac{W_3}{2}[x - c]^2 + A$

And $E\,I\,y = -\,W_1\frac{x^3}{6} + \frac{R}{6}[x - a]^3 - \frac{W_2}{6}[x - b]^3 - \frac{W_3}{6}[x - c]^3\;+Ax + B$

By doing so it can be shown that the constants of integration are common to all sections of the beam, e.g. if $\displaystyle x = b - \Delta$

$E\,I\;\frac{dy}{dx} = -\left(\frac{W_1}{2} \right)(b - \Delta )^2 + \left(\frac{R}{2} \right)(b - \Delta - a)^2 + A$

And $E\,I\;y = -\left(\frac{W_1}{6} \right)(b - \Delta )^3 + \left(\frac{R}{6} \right)(b - \Delta - a)^3 + A(b - \Delta ) + B$

$E\,I\;\frac{dy}{dx} = -\left(\frac{W_1}{2} \right)(b + \Delta )^2 + \left(\frac{R}{2} \right)(b + \Delta - a)^2\;-\left(\frac{W_2}{2} \right)\Delta ^2+\;A'$

And $E\,I\;y = -\left(\frac{W_1}{6} \right)(b - \Delta )^3 + \left(\frac{R}{6} \right)(b + \Delta - a)^3\;-\left(\frac{W_2}{6} \right)\Delta ^3+\;A'(b + \Delta ) + B'$

Now as $\displaystyle \Delta \rightarrow 0$ the slope and deflection values must correspond (i.e.) at $x = b$ from which it can be seen that $A = A$ ' and $B = B$ '. The values of $A$ and $B$ are found as before ( Part 1).

Uniformly Distributed Loads.

Supposing that a uniformly distributed load is applied from a distance $a$ to a distance $b$ measured from one end $A$ . Then in order to obtain an expression for the Bending Moment at a distance $x$ from the end, which will apply for all values of $x$ , it is necessary to continue the loading up to the section at $x$ compensating this with an equal negative load from $b$ to $x$ (See diagram)

Hence $M = Rx - \displaystyle\frac{w}{2}[x - a]^2 + \displaystyle\frac{w}{2}[x - b]^2$

Each length of the loading acts at its centre of gravity. The square brackets are interpreted as before.

For $x>a$ but $<b$ , omit $[x - b]$ and:

Hence $M = Rx - \displaystyle\frac{w}{2}(x - a)^2$

This is clearly correct. The remaining steps of integration are the evaluation of the Constants now proceeds as before.

Concentrated Bending Moment

It is possible to write:

$E\,I\,\frac{d^2y}{dx^2} = M = -\,Rx + M_0[x - a]^0$

$E\,I\,\frac{dy}{dx} = -\,R\frac{x^2}{2} + M_0[x - a] + A\;\;\;\;\;\;etc.$

Example 1 [imperial]

Problem

A simply supported beam of length $L$ carries a load $W$ at a distance $a$ from one end and $b$ from the one $( a > b)$ .

Find the position and magnitude of the maximum deflection and show that the position is always approximately within $\displaystyle\frac{L}{13}$ of the centre.

Workings

The maximum deflection (i.e.) zero slope will occur on the length a since $a>b$

Taking the axes as shown in the diagram.

$E\,I\,\frac{d^2y}{dx^2} = M = \left(\frac{Wb}{L} \right)\,x - W[x - a]$

Integrating

$E\,I\,\frac{dy}{dx} = \left(\frac{Wb}{L} \right)\left( \frac{x^2}{2}\right) - \left(\frac{W}{2} \right)[x - a]^2 + A$

(1)

And Integrating again gives:-

$E\,I\,y = \left(\frac{Wb}{L} \right)\left( \frac{x^3}{6}\right) - \left(\frac{W}{6} \right)[x - a]^3 + Ax + B$

(2)

At $x = 0$ and $y = 0$ and therefore $B = 0$ .

At $x = 0$ , $y = L$

$\therefore\;\;\;\;\;A\,L\;= - \left(\frac{Wb}{L} \right)\left(\frac{L^3}{6} \right) + \left(\frac{W}{6} \right)b^3$

From which: $A\;= - \left(\displaystyle\frac{Wb}{6L} \right)\left(L^2 - b^2 \right)$

We need to find the value of $x$ when $\displaystyle \frac{dy}{dx}$ is zero. Using equation (#1) and omitting $[x - a]$ ( We can do this because at zero slope $x < a$ when $a > b$ )