Macaulay's method provides a simplified way to analyze beam bending by writing one equation for the entire beam's bending moment rather than separate equations for each load section. This allows boundary conditions from any part of the beam to determine integration constants. Poisson's ratio is the ratio of lateral to axial strain in a loaded system, and standard equations are derived relating stress and strain in multiple dimensions.

1. Macaulay’s Method & Poisson’s Ratio
Author: Leicester College
Date created:
Date revised: 2009
Abstract:
This section examines a method of analysing the bending moment and shear force
distribution along a uniformly loaded, simply supported beam. The method used is
Macaulay’s and the stages involved in this method are described in detail. The method
also determines the position of the maximum bending moment, the slope and deflection
at this point.
Poisson’s ratio is the ratio between lateral and axial strain in a directly loaded system.
The derivation of standard equations and their further development is shown.
Contents
Macaulay’s Method & Poisson’s Ratio..............................................................................................1
Macaulay’s method for deriving bending moments..........................................................................2
Determine R1 and R2..........................................................................................................................4
Poisson’s Ratio...................................................................................................................................8
Credits...............................................................................................................................................10
These files support the Edexcel HN unit – Mechanical Principles (NQF L4)
File name Unit Key words
outcome
Macaulay’s Do not use this file.
beams
Balancing 4.1 Beams, balancing, rotation, mass, stress, shafts
Flywheels 4.3,4.4 Kinetic, battery, flywheel, inertia, energy storage, energy
Macaulay’s 2.1,2.2,2.3 Beams, stress, loading, UDL, slope, deflection, method
method
Poisson 1.1, 1.2 Poisson ratio, equations, axial, lateral, strain, stress
Selecting beams 2.2 Beams, columns, selection, slenderness, stress, section modulus
For further information regarding unit outcomes go to Edexcel.org.uk/ HN/ Engineering / Specifications
© Leicester College 2009 This work is licensed under a Creative Commons Attribution 2.0 License.

2. Macaulay’s Method
Macaulay’s method for deriving bending moments
For a beam with a number of concentrated loads, separate bending moment equations
have had to be written for each part of the beam between adjacent loads.
Integration of each expression then gives the slope and deflection relationships for each
part of the beam.
Each slope and deflection relationship includes constants of integration and these have to
be determined for each part of the beam. The constants of integration are then found by
equating slopes and deflections given by the expressions on each side of each load.
This can be very laborious if there are many loads. A much less laborious way of tackling
the problem is to use the method known as Macaulay’s method.
Macaulay’s method involves writing just one equation for the bending moment of the
entire beam and enables boundary conditions for any part of the beam to be used to
obtain the constants of integration.
Page 2 of 10

3. Macaulay’s Method
Lets look at a typical problem
A simply supported beam of length 3.0m has concentrated loads of 40 kN at 1.0m from
one end and 60kN at 2.0 m from the same end. (Shown below)
a. What is the deflection and slope of the beam under the 40kN load?
b. What is the maximum deflection and at what position along the beam does it
occur?
The beam has a value of EΙ = 20MNm 2
40kN 60kN X
R1 R2
1m
2m
x X
3M
Page 3 of 10

4. Macaulay’s Method
Determine R1 and R2
Taking moments about R1
CWM = ACWM
(40 x 1) +( 60 x 2) = R2 x 3
R2 53.3kN and R1 = 46.7kN
Considering the bending moment about section XX due to the forces to the left of the
section, then:
M = R1 x − 40{ x − 1} − 60{ x − 2} (40 and 60 are loads on the beam)
d2y M
=−
dx 2 EI
d2y 1
=− ( R1 x − 40{ x − 1} − 60{ x − 2} )
dx 2 EΙ
Integrating this expression gives:
dy 1 R1x 2 40{ x − 1} 2 60{ x − 2} 2
== − − − +A ( Beam slope)
dx EΙ 2 2 2
Integrating again gives:
40{ x − 1} 60{ x − 2}
3 3
1 R1x 3
y=− -− - Ax + b (Beam deflection)
EΙ 6 6 2
When x = 0 we have y = 0, B = 0. For this example the following figures are known
NOTE:
dy Slope of the y= Deflection
= graph
dx
Page 4 of 10

5. Macaulay’s Method
When x = 3m we have y = 0 . Substitute for x = 3
3
1 46.7 x 3 40 x 2 3 60x1
0=− - = 3A
EΙ 6 6 6
40 x 2 3
The 2 comes from x − 2 in this case x = 3 therefore 3 − 1 = 2
6
Hence A = −48.9. This becomes: A is a constant, which
once calculated is
carried through the
rest of the
calculations
And,
1 R1x 3 40{ x − 1} 3 60{ x − 2} 3
y=− - - − 48.9 x
E1 6 6 6
The deflection at x = 1.0m is:
10 3 46.7 x1.0 3
y= − 0 − 0 − 48.9 x1.0
20 x10 6 6
Note; The 10 3 is included because all the forces have until now been in kN..
The { x − 2} term is zero because it has a negative value when x = 1.0m. The { x − 1}
term is also zero. y = 2.06mm. .
The slope of the beam at x = 1.0m is given in a previous equation as:
dy 10 3 46.7 x1.0 2
=− − 0 − 0 − 48.9
dx 20 x10 6 2
Thus the slope is 1.28 x10 −3 rad.
Page 5 of 10

6. Macaulay’s Method
Maximum deflection occurs when dy dx = 0. Using the Integrated expression:
dy 1 46.7 x 2 40{ x − 1} 2 60{ x − 2}
2
=0=− − − 48.9
dx EΙ 2 2 2
By inspection of the problem it looks likely that the maximum deflection will occur between
the two load positions. Hence the solution of the above equation will be for x having a
value between 1 and 2m. With x having such a value the { x − 2} bracket will be zero and
the { x − 1} bracket can be written as ( x − 1) .
46.7 x 2 − 40( x − 1) − 0 − 2 x 48.9 = 0 .
2
Every term has been multiplied by 2, to remove the fractions. Therefore 48.9 x 2 to keep
the equation balanced.
Solve the brackets first:
( x − 1) ( x − 1)
− 40 x 2 + 40 x + 40 x − 40
46.7 x 2 − 40 x 2 + 80 x − 40 − 0 − 2 x 48.9 = 0
Re-introducing
6.7 x + 80 x − 40 − 97.8 = 0
2 the rest of the
equation
6.7 x 2 + 80 x − 137.8 = 0
The roots of a quadratic equation of the form ax 2 + bx + C = 0 are given.
x=
− 80 + (80 2
− 46.7 x − 137.8)
2 x6.7
− 80 + ( 6400 + 3693.04)
=
13.4
− 80 + (10093.04)
=
13.4 −
− 80 + ( 80 + 100.46)
=
13.4
Page 6 of 10

7. Macaulay’s Method
180.46
=− We ignore the negative value
13.4
20.46
= =1.527m
13.4
Now put the value of x back into the Macaulay expressions for the beam. We get;
x = 1.527m (only the positive value has any significance).
The deflection at this value of x is obtained using equation:
R1x 3 40{ x − 1} 60{ x − 2}
3 3
1
y=− − − − 48.9 x
EΙ 6 6 6
R11.527 3 40{1.527 − 1} 60{1.527 − 2}
3 3
1
− − − − 48.9 x
E1 6 6 6
10 3 46.7 X 1.527 3 40 X 0.527 3 60 X 0.4733
=− − − − 48.9 X 1.527
20 X 10 6 6 6 6
= -5 −5 ( 27.71 − 0.9757 − 1.0582 − 74.67 )
= - 5 x10 −5 X − 48.9939
= 2.449695 X 10 −3
y = 2.449695mm
Therefore
Page 7 of 10

8. Macaulay’s Method
Poisson’s Ratio
The ratio between LATERAL and AXIAL STRAIN
AXIAL STRAIN (X-X)
LATERAL
STRAIN
( Y – Y)
Given the greek letter - ν
Can be found by the ratio lateral strain / axial strain – it is dimensionless
When a member is in tension -
• The lateral strain represents a decrease in width ( -ve lateral strain)
• The axial strain represents an elongation (+ve axial strain)
The reverse is true when the member is in compression
Strain due to stress acting along X-X direction = σx / E
Strain due to stress acting along the Y-Y direction = - νσy / E
Page 8 of 10

9. Macaulay’s Method
Combining strain in the X-X direction we have;
1 εx = σx - νσy
E E
Combining strain in the X-X direction we have;
2 εy = σy - νσx
E E
WE can go further by using simultaneous equations and by substitution as
shown below;
3 σx = E (εx + νεy )
(1 - ν
2
)
AND
4 σy = E (εy + νεx )
(1 - ν
2
)
Equations 1 to 4 can be used to solve problems involving 3 dimensional
stress and strain
Page 9 of 10

10. Macaulay’s Method
Credits
This resource was created Leicester College and released as an open educational
resource through the Open Engineering Resources project of the Higher Education
Academy Engineering Subject Centre. The Open Engineering Resources project was
funded by HEFCE and part of the JISC/HE Academy UKOER programme.
© 2009 Leicester College
This work is licensed under a Creative Commons Attribution 2.0 License.
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The HEA logo is owned by the Higher Education Academy Limited may be freely distributed and copied for educational purposes only,
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The Leicester College name and logo is owned by the College and should not be produced without the express permission of the
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Page 10 of 10