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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.

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- 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. The JISC logo is licensed under the terms of the Creative Commons Attribution-Non-Commercial-No Derivative Works 2.0 UK: England & Wales Licence. All reproductions must comply with the terms of that licence. The HEA logo is owned by the Higher Education Academy Limited may be freely distributed and copied for educational purposes only, provided that appropriate acknowledgement is given to the Higher Education Academy as the copyright holder and original publisher. The Leicester College name and logo is owned by the College and should not be produced without the express permission of the College. Page 10 of 10

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