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i PAVEMENT DESIGN AND EVALUATION THE REQUIRED MATHEMATICS AND ITS APPLICATIONS F. Van Cauwelaert Editor: Marc Stet Federation of the Belgian Cement Industry B-1170 Brussels, Rue Volta 9.

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Page 1: Pavement design and evaluation

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PAVEMENT DESIGN AND EVALUATION

THE REQUIRED MATHEMATICSAND ITS APPLICATIONS

F. Van Cauwelaert

Editor: Marc Stet

Federation of the Belgian Cement IndustryB-1170 Brussels, Rue Volta 9.

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INTRODUCTION

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Pavement Design and Evaluation: The Required Mathematics and Its Applications

Keywords: textbook for pavement engineers, high order mathematical solutions for, rigid and flexiblepavement, mechanistic methods, practical applications in the field of pavement engineering

Preface

This book is intended for Civil Engineers and more specific for Pavement Engineers, who are interestedin the more advanced field of pavement engineering made available through the theory of highmathematics. In my extensive carrier as a Civil Engineer by profession, I noticed that some of mycolleagues feel uneasy when it comes to high-level theoretical and computational work. Perhaps this isbecause of the fact that as soon as they graduated, they are confronted with many and important practicalproblems where derivatives and integrals do not appear very useful at fist sight. However, at manyoccasions, conferences, seminars and other meetings, I realised that Civil Engineers remain excited aboutthe mathematical fundamentals of their art. Slow but sure grew the idea of writing a textbook on HighMathematics conceivable for practising Pavement Engineers. The primary objective of this book is toserve two purposes: first, to introduce the basic principles which must be known by people dealing withpavements; and secondly to present the theories and methods in pavement design and evaluation that maybe used by students, designers engineering consultants, highway and airport agencies, and researchers atuniversities. In addition, some of the new concepts developed in recent years to improve the methods ofpavement systems are explained. This book is written in a relatively simple way so that it may befollowed by people familiar with basic engineering courses in mathematics and pavement design.

This book consists of 26 chapters and is divided into two parts. The first part, which include chapters 1-9,covers the mathematics required by most of the problems related to pavement engineering. The assumedmathematical knowledge is that of high school level plus some basic elements of trigonometry andanalysis. The second part, which includes chapters 10-26, is concerned with practical solutions as facedby Pavement Engineers in the assessment of rigid and flexible pavements. The rigorous mathematicalsolutions are presented in the Appendix, explaining on complex variables.

Grateful acknowledgement is offered to the Fédération de l’Industrie Cimentière, Belgium (the Federationof the Belgian Cement Industry) for their intellectual and financial support in the process of therealisation of this book.

Thanks is due to my dear friend, Marc Stet, for proof-reading and editing the manuscript. His helpfulcomments to the mathematical and editorial content are highly appreciated.

Frans Van Cauwelaert

Brussels, December 2003.

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Part 1: The Required MathematicsChapter 1 covers the Laplace differential equation; the solution of for a great series of applications: itpresents the Bessel function of zero order, solution of the Laplace equation in axisymmetric co-ordinatesthat as used in great parts of this book. The Laplacian with coefficients different from 1 will be applied inproblems of anisotropic elasticity, the double Laplacian in most of the problems of multi-layer theory andthe extended Laplacian in the problems of rigid pavement on Winkler or Pasternak foundations. Chapter 2presents the gamma function, the factorial function for non-integers, required for the definition of Besselfunctions of non-integer order. Chapter 3 gives the general solutions of the different Bessel equations, theBer and Bei functions, and the modified form of the Bessel equation. Bessel functions are solutions of theLaplacian in polar or cylinder co-ordinates, used for applications with axial symmetry, e.g. in multi-layered structures. Trigonometric functions are solutions of the Laplacian in Cartesian co-ordinates,applied in cases of beams and rectangular slabs. Chapter 4 deals with the most important properties ofBessel functions:- derivatives, functions of half order,- asymptotic values required to express boundary conditions that must remain valid at infinite

distances,- indefinite integrals, equations between Bessel functions of different kind (required for integrations in

the complex plane).

Chapter 5 presents the beta function required for the resolution of definite integrals of Bessel functions.Chapter 6 gives the solutions for a series of important definite integrals of Bessel functions; among othersthe Poisson integral giving an integral representation of any Bessel function of the first kind. Chapter 7presents the hypergeometric function of Gauss required for the resolution of the infinite integrals ofBessel functions. Chapter 8 presents the most important infinite integrals of Bessel functions of directapplication in pavement analysis, especially in multi-layer theory. Chapter 9 presents the most importantinfinite integrals of Bessel functions resolved in the complex plane. They are essentially of application inslab theory.

Part 2: The ApplicationsThe second part focuses on the applications in the field of pavement engineering: rigid en flexiblepavements. Chapter 10 gives the basic solutions (equilibrium equations) for rigid pavements (theory ofstrength of materials) and flexible pavements (theory of elasticity) .This chapter gives the basic equationson continuum mechanics in different co-ordinate systems. Chapter 11 presents the integral transforms,Fourier’s expansion, Fourier’s integral, Hankel’s integral, required for the expression of discontinuousfunctions: the loads in pavement applications.

Chapters 12 to 19 concern mostly rigid pavements. Chapter 12 gives 3 simple applications on an elasticsubgrade: a beam subjected to a single load, a beam subjected to a distributed load, a slab subjected to asingle load. Chapter 13 gives the complete analytical solution for a beam on a Pasternak foundation (bothof infinite and finite extent). Chapter 14 gives the analytical solution for a circular slab on a Pasternakfoundation subjected to an axi-symmetric load (both of infinite and finite extent). Chapter 15 gives thecomplete analytical solution for a rectangular slab on a Pasternak foundation (both of infinite and finiteextent). Chapter 16 gives the analytical solution for a superposition of several slabs included the analysisof the adhesion between the slabs. Chapter 17 gives a back-calculation method for rigid pavement basedon Pasternak’s theory. Chapter 18 presents a solution for the computation of thermal stresses in rigidslabs on a Pasternak foundation. Chapter 19 presents two practical tests of interest with rigid slabs: thediametral test and a test for the determination of k and G in situ.

Chapters 20 to 26 concern mostly flexible pavements. Chapters 20 and 21 present the complete theory forsemi-infinite bodies subjected to all sorts of loads. Chapter 20 presents the Boussinesq problem: stresses

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INTRODUCTION

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and displacements in a semi-infinite body under a circular flexible plate uniformly loaded. It generalisesthe solution to different vertical loads (isolated, rigid, rectangular) and to orthotropic bodies. Chapter 21presents the solution a semi-infinite body subjected to shear stresses: radial and one-directional. Chapter22 gives the analytical solution for a multi-layered structure, included the problem of the adhesionbetween the layers, that of an eventual fixed bottom and that of an anisotropic subgrade. Chapter 23presents the numerical procedure required for the solution of a multi-layered structure. Chapter 24presents the theory at the base of the back-calculation methods for flexible pavements. Chapter 25presents the numerical procedure required for the back-calculation method for flexible pavements.Chapter 26 presents a practical test of interest with both types of road structures (rigid or flexible) multi-layered structures: the ovalisation test.

The Appendices gives the basic theory of complex numbers, especially the integration in the complexplane.

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.

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CONTENTS

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Table of Contents

Page

Preface............................................................................................................................................. iii

PART 1: THE REQUIRED MATHEMATICS................................................................................ 1

Chapter 1 The Laplace Equation.................................................................................................... 11.1 Introductory note................................................................................................................. 11.2 Derivation of the Laplace equation in polar co-ordinates from the Laplace equation in Cartesian

co-ordinates........................................................................................................................ 21.3 Equations related to the Laplace equation.............................................................................. 31.3.1 The Laplacian with coefficients different from 1. ............................................................... 31.3.2 The double Laplacian....................................................................................................... 31.3.3 The extended Laplacian ................................................................................................... 31.4 Resolution of the Laplace equation ....................................................................................... 41.4.1 Resolution by separation of the variables........................................................................... 41.4.2 Resolution by means of the characteristic equation (Spiegel, 1971) ..................................... 51.4.3 Resolution by means of indicial equations ......................................................................... 6

Chapter 2 The Gamma Function.................................................................................................... 92.1 Introductory note................................................................................................................. 92.2 Helpful relations.................................................................................................................. 92.3 Definition of the Gamma Function ......................................................................................102.4 Values of Γ(1/2) and Γ(-1/2)...............................................................................................11

Chapter 3 The General Solution of the Bessel Equation. .............................................................. 153.1 Introductory note................................................................................................................153.2 Helpful relations.................................................................................................................163.3 Resolution of the Bessel equation (Bessel functions of the first kind) .....................................173.4 Resolution of the Bessel equation for p an integer (Bessel functions of the second and third

kind)..................................................................................................................................193.4.1 For n integer, Jn = (-)n J-n .................................................................................................193.4.2 Bessel functions of the second kind..................................................................................193.4.3 Bessel functions of the third kind .....................................................................................213.5 The modified Bessel equation. ............................................................................................213.6 The ber and bei functions....................................................................................................233.7 The ker and kei functions....................................................................................................243.8 Resolution of the equation ∇2∇2w + w =0............................................................................253.9 The modified form of the Bessel equation ............................................................................25

Chapter 4 Properties of the Bessel Functions. .............................................................................. 274.1 Introductory note................................................................................................................274.2 Helpful relations.................................................................................................................274.3 Derivatives of Bessel functions ...........................................................................................294.3.1 Derivative of (rt)pJp(rt)....................................................................................................294.3.2 Derivative of (rt)-pJp(rt)..................................................................................................29

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4.3.3 Derivative of Jp(rt)..........................................................................................................294.4 Bessel functions of half order..............................................................................................304.4.1 Values of J1/2(rt), J-1/2(rt), J3/2(rt), J-3/2(rt)...........................................................................304.5 Asymptotic values..............................................................................................................314.5.1 Asymptotic values for Jp and J-p.....................................................................................314.5.2 Asymptotic values for Yp and Y-p ....................................................................................334.5.3 Asymptotic values for Hp

(1) and Hp(2)................................................................................33

4.5.4 Asymptotic values for Ip and I-p .......................................................................................334.5.5 Asymptotic value for Kn..................................................................................................354.5.6 Asymptotic values for ber and bei....................................................................................364.5.7 Asymptotic values for ker and kei....................................................................................364.6 Indefinite integrals of Bessel functions ................................................................................374.6.1 Fundamental relations .....................................................................................................374.6.2 The integral ∫rnJ0(rt)dr .....................................................................................................374.7 Relations between Bessel functions of different kind ............................................................384.7.1 Bessel functions with argument –rt ..................................................................................384.7.2 Relations between the three kinds of Bessel functions .......................................................384.7.3 Bessel functions of purely imaginary argument.................................................................39

Chapter 5 The Beta Function ....................................................................................................... 415.1 Introductory note................................................................................................................415.2 Helpful relations.................................................................................................................415.3 Definition of the beta function.............................................................................................415.4 Relation between beta and gamma functions ........................................................................425.5 The duplication formula for gamma functions ......................................................................42

Chapter 6 Definite integrals of Bessel functions ........................................................................... 456.1 Helpful relations.................................................................................................................456.2 Gegenbauer’s integral.........................................................................................................456.3 Sonine’s first finite integral.................................................................................................466.4 Sonine’s second finite integral.............................................................................................476.5 Poisson’s integral...............................................................................................................48

Chapter 7 The hypergeometric type of series ............................................................................... 497.1 Introductory note................................................................................................................497.2 Helpful relations.................................................................................................................497.3 Definition ..........................................................................................................................507.4 Properties of the multiple product (α)m ................................................................................517.4.1 A relation for (1 - b - m)m-n ..............................................................................................517.4.2 A relation for (a+b+1+m) m/22 m.........................................................................................517.4.3 A relation for Γ(a-n) .......................................................................................................527.4.4 The theorem of Vandermonde .........................................................................................527.4.5 The product of two Bessel functions with the same argument ............................................537.5 The hypergeometric series of Gauss 2F1[a,b;c;z]...................................................................547.5.1 Elementary properties .....................................................................................................547.5.2 Integral representation of the hypergeometric function ......................................................547.5.3 Value of F[a,b;c;z] for z = 1 ............................................................................................557.5.4 Convergence of the series F[a,b;c;z].................................................................................557.5.5 The product of two Bessel functions with different arguments ...........................................56

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Chapter 8 Infinite Integrals of Bessel Functions........................................................................... 578.1 Introductory note................................................................................................................578.2 Useful relations ..................................................................................................................578.3 The integral ∫e-atJν(bt)tµ-1dt ..................................................................................................608.3.1 Resolution of the integral................................................................................................608.3.2 Particular value...............................................................................................................618.3.3 The integral ∫ e-atcos(bt)dt................................................................................................638.3.4 The integral ∫ e-atsin(bt)dt.................................................................................................638.3.5 The integral ∫e-atcos(bt)sin(bt) dt ......................................................................................638.3.6 The integral ∫e-at sin(bt)/tdt...............................................................................................638.3.7 The integral ∫e-at sin(bt)tdt................................................................................................648.4 The integral ∫e-atJν(bt)Jν(ct)tµ-1dt ..........................................................................................648.4.1 Transformation of the integral..........................................................................................648.4.2 The integral ∫e-atsin(bt)sin(ct)t-1dt .....................................................................................648.4.3 The integral ∫e-atsin(bt)sin(ct)dt ........................................................................................658.4.4 The integral ∫∫e-amsin(bt)sin(cs)/(ts)dsdt ............................................................................658.4.5 The integral ∫∫me-amsin(bt)sin(cs)/(ts)dsdt .........................................................................678.5 The integral ∫e-atJµ(bt)Jν(ct)tλ-1dt ..........................................................................................688.6 The discontinuous integral ∫Jµ(at)Jν(bt)t-λdt...........................................................................688.6.1 Resolution of the integral................................................................................................688.6.2 The integral ∫Jµ(at)Jν(at)t-λdt.............................................................................................718.6.3 The integral ∫Jµ(at)Jµ-2k-1(bt)dt..........................................................................................728.6.4 Particular solutions of the integral ∫Jµ(at)Jν(bt)t-λdt ...........................................................728.6.5 Particular solution of the integral ∫Jν(at)Jν(bt)/tλdt..............................................................73

Chapter 9 Bessel functions in the complex plane .......................................................................... 779.1 Introductory note................................................................................................................779.2 Helpful relations.................................................................................................................779.3 Proof of Γ(z)Γ(1-z) = π/sin(πz) ...........................................................................................789.4 The Hankel’s contour integral for 1/Γ(z)..............................................................................809.5 The integral representation of Jν(z) ......................................................................................819.6 The integral representation of Iν(z) ......................................................................................839.7 The integral representation of Kν(z). ....................................................................................839.8 The integral representation of Kv(xz)....................................................................................849.9 Resolution of ∫xν+1 Jν(ax)/(x2+k2)dx .....................................................................................859.10 Resolution of ∫xρ -1 Jµ(bx)Jν(ax)/(x2+k2)dx.............................................................................899.11 Resolution of ∫sin(bx)cos(ax)/x/(x4 + k4)dx ..........................................................................90

PART 2: THE APPLICATIONS ................................................................................................... 93

Chapter 10 Laplace Equation in Pavement Engineering ............................................................ 9310.1 Equilibrium equation for beams in pure bending...................................................................9310.1.1 Sign conventions.........................................................................................................9310.1.2 Assumptions ...............................................................................................................9510.1.3 Bending moment and bending stress ............................................................................9510.1.4 The radius of curvature ...............................................................................................9610.1.5 Equilibrium ................................................................................................................96

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10.2 Equilibrium equation for bent plates....................................................................................9710.2.1 Bending moment and shear forces................................................................................9710.2.2 Equilibrium ................................................................................................................9910.3 Compatibility equation for a homogeneous, elastic, isotropic body submitted to forces applied

on its surface ...................................................................................................................10010.3.1 Principle of equilibr ium.............................................................................................10110.3.2 The principle of continuity.........................................................................................10110.3.3 The principle of elasticity ..........................................................................................10210.3.4 Stress potential..........................................................................................................10210.4 Compatibility equation for a homogeneous, elastic, anisotropic body submitted to forces applied

on its surface ...................................................................................................................10310.5 Basic equations of continuum mechanics in different co-ordinate systems ...........................10410.5.1 Plane polar co-ordinates ............................................................................................10410.5.2 Axi-symmetric Cylindrical Co-ordinates ....................................................................10510.5.3 Non symmetric cylindrical co-ordinates .....................................................................10610.5.4 Cartesian volume co-ordinates ...................................................................................10710.5.5 Axi-symmetric Cylindrical Co-ordinates for an orthotropic body..................................109

Chapter 11 The Integral Transforms........................................................................................ 11111.1 Introductory note..............................................................................................................11111.2 Helpful relations...............................................................................................................11211.3 The Fourier expansion (Spiegel, 1971)...............................................................................11211.3.1 Definitions................................................................................................................11211.3.2 Proof of the Fourier expansion. ..................................................................................11311.3.3 Example ...................................................................................................................11411.4 The Fourier integral..........................................................................................................11511.4.1 Definition .................................................................................................................11511.4.2 Proof of Fourier’s integral theorem............................................................................11511.4.3 Example ...................................................................................................................11611.5 The Hankel’s transform....................................................................................................11711.5.1 Definition .................................................................................................................11711.5.2 Example ...................................................................................................................11711.5.3 Application of the discontinuous integral of Weber and Schafheitlin ............................118

Chapter 12 Simple Applications of Beams and Slabs on an Elastic Subgrade ........................... 12112.1 The elastic subgrade .........................................................................................................12112.2 The beam on an elastic subgrade subjected to an isolated load: ∂4w/∂x4+Cw=0....................12312.3 The beam on an elastic subgrade subjected to a distributed load: ∂4w/∂x4+Cw=0..................12512.4 The infinite slab subjected to an isolated load: ∇2∇2w +Cw = 0 ..........................................125

Chapter 13 The Beam Subjected to a Distributed Load and Resting on a Pasternak Foundation............................................................................................................................... 129

13.1 The basic differential equations .........................................................................................12913.2 Case of a beam of infinite length .......................................................................................13013.2.1 Solution of the differential equation............................................................................13013.2.2 Application...............................................................................................................13313.3 Case of a beam of finite length with a free edge 13313.3.1 Solution #1...............................................................................................................13313.3.2 Application...............................................................................................................13513.3.3 Solution #2...............................................................................................................13713.3.4 Application...............................................................................................................137

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13.4 Case of a finite beam with a joint.......................................................................................13813.4.1 Solution #1...............................................................................................................13813.4.2 Application...............................................................................................................14013.4.3 Solution# 2...............................................................................................................14113.4.4 Application...............................................................................................................14213.4.5 Proof that, in de case of a Winkler foundation, at a joint Q = γ T..................................143

Chapter 14 The Circular Slab Subjected to a Distributed Load and Resting on a PasternakFoundation............................................................................................................. 145

14.1 The basic differential equations .........................................................................................14514.2 Case of a slab of infinite extent .........................................................................................14614.2.1 Solution of the differential equation............................................................................14614.2.2 Application 1............................................................................................................14714.2.3 Application 2............................................................................................................14814.3 Case of a slab of finite extent with a free edge ...................................................................14814.3.1 Solution #1...............................................................................................................14814.3.2 Solution #2...............................................................................................................15114.3.3 Application of solution #2 .........................................................................................15114.4 Case of a slab of finite extent with a joint...........................................................................15214.4.1 Solution #1...............................................................................................................15214.4.2 Solution #2. ..............................................................................................................15214.4.3 Application of solution #2 .........................................................................................153

Chapter 15 The Rectangular Slab Subjected to a Distributed Load and Resting on a PasternakFoundation............................................................................................................. 155

15.1 The basic differential equations .........................................................................................15515.2 Resolution of the deflection equation.................................................................................15515.3 Boundary conditions.........................................................................................................15815.4 Case of a slab of finite extent with free edge ......................................................................15815.4.1 Solution #1...............................................................................................................15815.4.2 Solution #2...............................................................................................................15915.5 Case of a slab of finite extent with a joint...........................................................................15915.5.1 Solution #1...............................................................................................................15915.5.2 Solution # 2..............................................................................................................16015.5.3 Application...............................................................................................................161

Chapter 16 The Multislab......................................................................................................... 16316.1 Theoretical justification....................................................................................................16316.2 General model..................................................................................................................16316.3 Full slip at each of the interfaces.......................................................................................16316.4 Full friction at the first interface, full slip at the second interface .........................................16416.5 Full friction at both interfaces ...........................................................................................16516.6 Partial friction ..................................................................................................................16616.6.1 Application...............................................................................................................168

Chapter 17 Back-calculation of Concrete Slabs ........................................................................ 17117.1 Back-calculation of moduli ...............................................................................................17117.2 Case where the load can be considered as a point load ........................................................17117.3 Computations...................................................................................................................17217.4 Case when the load is considered as distributed..................................................................17317.5 Comparison of the two methods ........................................................................................175

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17.6 Influence of the reference deflection..................................................................................17517.7 Analysis of field data........................................................................................................17617.8 Example of back-calculation.............................................................................................177

Chapter 18 Thermal Stresses in Concrete Slabs ....................................................................... 17918.1 Thermal stresses ..............................................................................................................17918.2 Slab of great length...........................................................................................................17918.2.1 Differential equilibrium equation ...............................................................................17918.2.2 Solution of the equilibrium equation for g < 1.............................................................18018.2.3 Solution of the equilibrium equation for g = 1.............................................................18018.2.4 Solution of the equilibrium equation for g > 1.............................................................18018.2.5 Boundary conditions .................................................................................................18018.2.6 Expression of the moment for g < 1............................................................................18018.2.7 Expression of the moment for g = 1............................................................................18218.2.8 Expression of the moment for g > 1............................................................................18218.2.9 Verification of the expression of the maximum moment for g = 1.................................18318.2.10 Equation of the thermal stress....................................................................................18318.2.11 Example ...................................................................................................................18418.3 Rectangular slab...............................................................................................................18418.3.1 Differential equation of equilibrium...........................................................................18418.3.2 Boundary conditions .................................................................................................18518.3.3 Examples .................................................................................................................18518.4 Circular slab ....................................................................................................................18718.4.1 Equilibrium equation.................................................................................................18718.4.2 Solution of the equilibrium equation...........................................................................18818.4.3 Boundary conditions .................................................................................................18818.4.4 Resulting moment .....................................................................................................18818.4.5 Comparison between the models for rectangular and circular slabs ...............................18918.5 Extension to a multi-slab system.......................................................................................189

Chapter 19 Determination of the Parameters of a Rigid Structure ........................................... 19319.1 Determination of the Young’s modulus of a concrete slab...................................................19319.1.1 Resolution of the compatibility equation.....................................................................19319.1.2 Equations for the stresses ..........................................................................................19319.1.3 Boundary conditions .................................................................................................19319.1.4 Stresses and displacements ........................................................................................19419.1.5 Tangential normal stress along the vertical diameter....................................................19519.2 Determination of the characteristics k and G of the subgrade ..............................................19519.2.1 Equilibrium equation for a Pasternak subgrade ...........................................................19519.2.2 Eccentrically loaded plate-bearing test........................................................................19619.2.3 Vertical load.............................................................................................................19819.2.4 Moment....................................................................................................................19919.2.5 Determination of k and G ..........................................................................................201

Chapter 20 The Semi-Infinite Body Subjected to a Vertical Load ............................................ 20320.1 Introductory note..............................................................................................................20320.2 The semi-infinite body subjected to a vertical uniform circular pressure ..............................20320.3 The semi-infinite body subjected to an isolated vertical load ...............................................20620.4 The semi-infinite body subjected to a circular vertical rigid load..........................................20720.5 The semi-infinite body subjected to a vertical uniform rectangular pressure.........................20920.6 Comparison between the vertical stresses. Principle of de Saint-Venant ..............................211

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20.7 The orthotropic body subjected to a vertical uniform circular pressure.................................21120.8 The orthotropic body subjected to a vertical uniform rectangular pressure ...........................215

Chapter 21 The Semi-Infinite Body Subjected to Shear Loads ................................................. 21721.1 The semi-infinite body subjected to radial shear stresses.....................................................21721.2 The semi-infinite body subjected to a one-directional asymmetric shear load .......................21821.3 The semi-infinite body subjected to a shear load symmetric to one of its axis’s ....................221

Chapter 22 The Multilayered Structure ................................................................................... 22522.1 The multilayered structure ................................................................................................22522.2 Solutions of the continuity equations .................................................................................22622.3 Boundary conditions.........................................................................................................22722.4 Determination of the boundary constants ...........................................................................22822.5 The fixed bottom condition ...............................................................................................23122.6 The orthotropic subgrade ..................................................................................................232

Chapter 23 The Resolution of a Multilayered Structure ........................................................... 23323.1 Choice of the integration formula ......................................................................................23323.2 Values at the origin...........................................................................................................23423.3 The geometrical scale of the structure................................................................................23423.4 Width of the integration steps............................................................................................23523.4.1 Influence of the moduli on the integration step............................................................23523.4.2 Influence of the radii of the loads on the integration step .............................................23523.4.3 Influence of the offset distance on the integration step.................................................23623.4.4 Modification of the step width ...................................................................................23623.5 Stresses and displacements at the surface...........................................................................23623.6 Stresses and displacements in the first layer .......................................................................237

Chapter 24 The Theory of the Back-Calculation of a Multilayered Structure .......................... 24124.1 The surface modulus.........................................................................................................24124.2 Equivalent layers..............................................................................................................24224.3 Equivalent semi-infinite body............................................................................................24224.4 Analysis of a deflection basin ............................................................................................24324.4.1 Analysis of a three-layer on a linear elastic subgrade ...................................................24324.4.2 Analysis of a three-layer with a very stiff base course .................................................24424.4.3 Analysis of a three layer with a very weak base course ................................................24524.4.4 Analysis of a two-layer on a subgrade with increasing stiffness with depth ...................24624.5 Algorithm of Al Bush III (1980) ........................................................................................247

Chapter 25 The Numerical Procedure of the Back-calculation of a Multilayered Structure ..... 24925.1 The analysis of a back-calculation program for a three-layered structure..............................24925.2 The sensitivity of the back-calculation procedure for a three layer structure .........................25125.2.1 The sensitivity to rounding off the values of the measured deflections ..........................25125.2.2 The sensitivity to the presence of a soft intermediate layer...........................................25125.2.3 The influence of fixing beforehand the value of one modulus.......................................25225.3 The sensitivity of the back-calculation procedure for a four layer structure ..........................25325.3.1 Value of the information given by the surface modulus................................................25425.3.2 The influence of fixing beforehand the value of one modulus.......................................25625.4 The influence of degree of anisotropy and Poisson’s ratio on the results of a back-calculation

procedure in the case of a semi-infinite subgrade ................................................................25725.4.1 Influence of the degree of anisotropy on the back-calculated moduli.............................257

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25.4.2 Influence of Poisson’s ratio of the subgrade on the deflections (n = 1) ..........................25825.5 The influence of Poisson’s ratio and degree of anisotropy on the results of a back-calculation

procedure in the case of a subgrade of finite thickness ........................................................25825.5.1 Influence of the degree of anisotropy on the back-calculated moduli.............................25825.5.2 Influence of Poisson’s ratio of the subgrade on the deflections (n = 1) ..........................259

Chapter 26 The Ovalisation Test .............................................................................................. 26126.1 Description of the ovalisation test......................................................................................26126.2 Interpretation of the results of the ovalisation test...............................................................26126.3 Slab with a cavity on an elastic subgrade subjected to a symmetrical load............................26226.3.1 Resolution in the case of a plain slab..........................................................................26226.3.2 Resolution in the case of a slab with a cavity...............................................................26326.4 Strains in the case of a non-symmetrical load.....................................................................26526.4.1 Stresses in a hollow cylinder subjected to a uniform external pressure..........................26526.4.2 Stresses in a hollow plate...........................................................................................26626.4.3 Application of the ovalisation test. .............................................................................271

References ............................................................................................................................... 273

Appendix Complex Functions ..................................................................................................... 275

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THE LAPLACE EQUATION

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PART 1: THE REQUIRED MATHEMATICS

Chapter 1 The Laplace Equation

1.1 Introductory note.

In mechanical engineering and thus also in the mechanics of civil engineering, one always starts with thefundamental requirement of equilibrium: equilibrium of the normal forces and equilibrium of themoments acting on the analysed body. Forces are resultants of stresses. In order to locate exactly thoseforces and determine, at least analytically, their amplitude, it is convenient to start from an infinitesimalsection of the body. On such an infinitesimal section, the stresses are necessarily infinitely small and,hence, can be considered as constant and uniformly distributed over the area of the infinitely smallsection. Hence the resulting force is simply the product of the constant stress by the area of the sectionand its point of application is the centre of gravity of the section, i.e. the midpoint of the section. All theresulting equations will then necessarily be differential equations and, nearly in almost all applications,the differential equation expressing equilibrium is a Laplace or an assimilated equation. The developmentof these equations are presented in chapter 10, the first chapter of Part 2 “Applications”. As in manyfields of engineering, also in Pavement engineering the differential equation of Laplace is the solution ofa great series of applications. Symbolically, Laplace equation is written as follows:

02 =∇ Φ (1.1)It is a homogeneous differential equation with second order partial derivatives.Function of the co-ordinate system, the equation is developed:

in plane Cartesian co-ordinates

0yx 2

2

2

2=+

Φ∂

Φ∂ (1.2)

in volume Cartesian co-ordinates

0zyx 2

2

2

2

2

2=++

Φ∂

Φ∂

Φ∂ (1.3)

in polar co-ordinates

0r

1rr

1

r 2

2

22

2=++

∂θ

Φ∂∂Φ∂

Φ∂ (1.4)

in cylindrical co-ordinates

0zr

1rr

1

r 2

2

2

2

22

2=+++

Φ∂

∂θ

Φ∂∂Φ∂

Φ∂ (1.5)

In case of axial symmetry, ∂Φ /∂θ = 0 and equations (1.4) and (1.5) simplify in:

0rr

1r 2

2

=+∂Φ∂

∂Φ∂

(1.6)

0zrr

1r 2

2

2

2

=++∂

Φ∂∂Φ∂

∂Φ∂

(1.7)

Equation (1.2) is applied in chapter 10.3.4.Equation (1.4) will be utilised in chapter’s 10.5.1 and 19.1.Equation (1.7) will be utilised in chapter’s 10.5.2, 10.6, 20.1, 20.2, 20.3 and 20.4.

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1.2 Derivation of the Laplace equation in polar co-ordinates from the Laplace equation inCartesian co-ordinates.

Consider the relations between Cartesian and polar co-ordinates:

θcosrx =θsinry =

Hence222 yxr +=

)x/y(tan 1−=θExpress the partial derivatives

( )θ

∂∂

cosrx

yx

xxr

2/122==

+=

θ∂∂

sinry

yr

==

rsin

x

y

x

y1

1x 2

2

∂∂θ

−=

+

−=

rcos

x1

x

y1

1y

2

∂∂θ

=

+

=

Then

xxr

rx ∂∂θ

∂θΦ∂

∂∂

∂Φ∂

∂Φ∂

+=

∂θΦ∂θ

∂Φ∂

θ∂Φ∂

rsin

rcos

x−=

xxxr

xrx2

2

∂∂θ

∂Φ∂

∂θ∂

∂∂

∂Φ∂

∂∂

Φ∂+=

∂θΦ∂

θθ∂θ∂Φ∂

θθ∂θ

Φ∂θ

∂Φ∂

θ∂

Φ∂θ

Φ∂cossin

r

2r

cossinr2

sinr

1r

sinr1

rcos

x 2

2

2

22

22

2

22

2

2+−++=

In the same way, obtain:

∂θΦ∂

θθ∂θ∂Φ∂

θθ∂θ

Φ∂θ

∂Φ∂

θ∂

Φ∂θ

Φ∂cossin

r

2r

cossinr2

cosr

1r

cosr1

rsin

y 2

2

2

22

22

2

22

2

2−+++=

Make the sum:

2

2

22

2

2

2

2

2

r

1rr

1

ryx ∂θ

Φ∂∂Φ∂

Φ∂

Φ∂

Φ∂++=+ (1.8)

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THE LAPLACE EQUATION

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1.3 Equations related to the Laplace equation.

Besides the properly so called Laplace differential equation, there exists a series of useful differentialequations closely related to the Laplace equation.

1.3.1 The Laplacian with coefficients different from 1.

For example in plane Cartesian co-ordinates

0yC

1

x 2

2

2

2=+

Φ∂

Φ∂ (1.9)

Equation (1.9) is utilised in chapter’s 10.3.5 and 20.6.

1.3.2 The double Laplacian

The double Laplacian, or the Laplace operator applied to a Laplace equation, is also a solution of animportant series of applications.It writes

022 =∇∇ Φ (1.10)Developed, for example in volume co-ordinates, the double Laplacian is

0zyxzyx 2

2

2

2

2

2

2

2

2

2

2

2=

++

++

Φ∂

Φ∂

Φ∂

∂ (1.11)

Equation (1.10) is applied in chapter’s 10.3, 10.5.1,10.5.2, 10.5.3, 10.5.4, 20.1, 20.2, 20.3 and 20.4.

1.3.3 The extended Laplacian

Often the Laplacian equation is completed by derivatives of an order lower than the second. Here in polarco-ordinates:

0kr

1rr

1

r 2

2

22

2=+++ Φ

∂θ

Φ∂∂Φ∂

Φ∂ (1.12)

or more generally

CBA 222 =+∇+∇∇ ΦΦΦ (1.13)

The extended Laplacian is used in chapters 10.1.5, 10.2.2 and in chapter’s 12 to 19.

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1.4 Resolution of the Laplace equation

Three very general methods are applied in this book.

1.4.1 Resolution by separation of the variables

Solution in volume Cartesian co-ordinates:

Consider equation (1.3) and assume a solution such as Φ = f1(x)f2(y)f3(z).Applying (1.3) yields

0z

)z(f)y(f)x(f)z(f

y

)y(f)x(f)z(f)y(f

x

)x(f2

32

21322

2

13221

2=++

and dividing by f1(x)f 2(y)f 3(z)

0)z(f

z

)z(f

)y(fy

)y(f

)x(fx

)x(f

3

23

2

2

22

2

1

21

2

=++ ∂

∂∂

Each of the 3 terms of the sum is a function of one single variable; this results in

)z(fCz

)z(f)y(fC

y

)y(f)x(fC

x

)x(f332

32

2222

2

1121

2===

0CCC 321 =++a system with a large series of solutions of the differential equation.For example: f1(x) = cos(x), f 2(y) = cos(y), f 3(z) = ez√2.

Solution in axi-symmetric cylinder co-ordinates ( Bowman,1958):

Consider equation (1.7) and apply a solution such as Φ = f(r)g(z).Applying (1.7) yields

0z

)z(g)r(f)z(g

r)r(f

r1

)z(gr

)r(f2

2

2

2=+

∂+

∂∂

and dividing by f(r)g(z)

0)z(g

z

)z(g

)r(fr

)r(fr1

)r(fr

)r(f2

2

2

2

=++ ∂

∂∂

Thus

)z(Cgz

)z(g)r(Cf

rf

r1

r

)r(f2

2

2

2=−=+

∂∂∂

Choose Cg(z) = ez.Hence f(r) must be a solution of

0)r(fr

)r(fr1

r

)r(f2

2=++

∂∂

∂ (1.14)

Equation (1.14) is called the Bessel equation of zero order.

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THE LAPLACE EQUATION

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The function known as Bessel’s function of the first kind and of n- th order and denoted Jn(r) is defined asfollows (Bowman, 1958):

...)!3n(!3

)2/r()!2n(!2

)2/r()!1n(!1

)2/r(!n!0)2/r(

)r(Jn6n4n2n

n ++

−+

++

−=+++

(1.15)

Hence

...!3!3)2/r(

!2!2)2/r(

!1!1)2/r(

1)r(J642

0 +−+−= (1.16)

and

...!4!3)2/r(

!3!2)2/r(

!2!1)2/r(

2r

)r(J753

1 +−+−= (1.17)

Differentiating the series for J0(r) and comparing the result with the series for J1(r) results in:

)r(Jdr

)r(dJ1

0 −= (1.18)

Also, after multiplying the series for J1(r) by r and differentiating:

( ) )r(rJ)r(rJdrd

01 = (1.19)

Using (1.18), (1.19) can be rewritten in the form:

)r(rJdr

)r(dJr

drd

00 =

0)r(Jdr

)r(dJr1

dr

)r(Jd0

02

02

=++ (1.20)

Hence J0(r) is a solution of (1.14) and Φ = J0(r)ez is a solution of (1.7).This particular solution was developed to introduce, from the first chapter on, the Bessel functions. Besselfunctions are frequently used and discussed throughout this book.

1.4.2 Resolution by means of the characteristic equation (Spiegel, 1971)

This solution applies to homogeneous linear differential equations with constant coefficients defined as

0adxd

a...dx

da

dx

da n1n1n

1n

1n

n

0 =++++ −−

−Φ

ΦΦΦ (1.21)

It is convenient to adopt the notations DΦ , D2Φ, ..., DnΦ to denote dΦ/dx, d2Φ/dx2, ...dnΦ/dxn, where D,D2, ..., Dn are called differential operators.Using this notation, (1.21) transforms

0aDa...DaDa n1n1n

1n

0 =

++++ −− Φ (1.22)

Let Φ = emx, m = constant, in (1.22) to obtain

0a...mama n1n

1n

0 =+++ − (1.23)that is called the characteristic equation. It can be factored into

0)mm)...(mm)(mm(a n210 =−−− (1.24)which roots are m1, m2, ...mn.

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PAVEMENT DESIGN AND EVALUATION

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One must consider three cases:

Case 1. Roots all real and distinct.

Then em1x, em2x, ... emnx are n linearly independent solutions so that the required solution is:xm

nxm

2xm

1 n21 eC...eCeC +++=Φ (1.25)Case 2. Some roots are complex.

If a0, a1, ..., an are real, then when a+bi is a root of (1.23) so also is a-bi (where a and b are real and i =√(-1) ) . Then a solution corresponding to the roots a+bi and a-bi is:

( )bxsinCbxcosCe 21ax +=Φ (1.26)

where use is made of Euler’s formula eiu = cos u +i sin u (see Appendix).

Case 3. Some roots are repeated.

If m1 is a root of multiplicity k , then a solution is given by:

( ) xm1kk

2321 1exC...xCxCC −++++=Φ (1.27)

Example

Consider the double Laplacian in plane Cartesian co-ordinates:

0yyx

2x 4

4

22

4

4

4=++

Φ∂

∂∂

Φ∂

Φ∂ (1.28)

Choose as solution Φ = f(x) ey . Hence (1.28) reduces in

0)x(fx

)x(f2

x

)x(f2

2

4

4=++

∂ (1.29)

Equation (1.29) is a homogeneous linear differential equation of order 4. The characteristic equationwrites:

01m2m 24 =++ (1.30)and can be factored in:

0)im()im( 22 =−+ (1.31)Based on (1.26) and (1.27), the solution of (1.29) becomes:

( )xsinDxcosCxxsinBxcosA)x(f +++=and the solution of (1.28):

( )[ ]xsinDxcosCxxsinBxcosAe)y,x( y +++=Φ (1.32)

1.4.3 Resolution by means of indicial equations

Consider the differential equation:

0wkdx

wd 22

2=+ (1.33)

Assume a solution in the form of a indicial series

L44

33

2210 xaxaxaxaa)x(f ++++=

Apply (1.33)

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THE LAPLACE EQUATION

7

0xakxakakxa.3.4xa.2.3a.1.2 22

21

20

22432 =+++++ LL

This equation must be equal zero for all values of x. Hence the sum of the coefficients of each exponentof x must be individually zero.

0aka.1.2 02

2 =+

0aka.2.3 12

3 =+

0aka.3.4 22

4 =+…

First let a0 = 1 and a1 = 0Hence

)!p2(

k)(a

!4k

a!2

ka

p2p

p2

4

4

2

2 −==−= (1.34)

0aaa 1p231 LL ==== + (1.35)

The successive terms of (1.34) are the terms of the cosine series. Hence, the first solution of (1.33) isf(x) = cos(kx). The second solution is obtained by setting a0 = 0 and a1 = k. Obviously, the secondsolution of (1.33) is f(x) = sin(kx).

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THE GAMMA FUNCTION

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Chapter 2 The Gamma Function

2.1 Introductory note

In chapter 1, we have defined the Bessel function of the first kind and of order zero as:

( )!k!k

2/r)()r(J

k2k

0 ∑ −=

This equation can be generalised, for n an integer, in:

( ))!nk(!k

2/r)()r(J

nk2k

n +−=

+

∑When p is not an integer, the Bessel function of order p writes:

( ))1pk(!k

2/r)()r(J

pk2k

p ++−=

+

∑ Γwhere Γ(k+p+1) is called the gamma function of k+p+1.

For our purpose, the gamma function is essentially required to express Bessel functions of non-integerorder: it is the factorial function for non-integers. Observe that with these definitions Γ(n+1) = n!

2.2 Helpful relations

∫∞

−−=o

x1p dxex)p(Γ

)p(p)1p( ΓΓ =+

1)1( =Γ

!n)1n( =+Γ

πΓ =)2/1(

±∞=− )n(Γ

−+++++−==

1n1

...31

21

1)n()n('dn

)n(dγΓΓ

Γ

5772157.0=γ (Euler’s constant)

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PAVEMENT DESIGN AND EVALUATION

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2.3 Definition of the Gamma Function

The gamma function is defined by:

∫∞

−−=o

x1p dxex)p(Γ (2.1)

which is convergent for p > 0.

Applying the definition

∫∞

−=+o

xp dxex)1p(Γ ∫∞

−−∞− −−−=o

1pxo

xp dx)px)(e(ex ∫∞

−−=o

x1p dxexp )p(pΓ=

one obtains the recurrence formula:)p(p)1p( ΓΓ =+ (2.2)

By taking (2.1) as the definition of Γ(p) for p > 0, we can generalise the gamma function to p < 0 by useof (2.2) in the form

p)1p(

)p(+

Γ (2.3)

This process is called analytic continuation.

Applying (2.1) we determine the value of Γ(1)

∫∞ ∞−− =−==o

oxx 1edxe)1(Γ (2.4)

Hence, applying (2.2)1)1(1)2( == ΓΓ

!212)2(2)3( =•== ΓΓ!3123)2(23)3(3)4( =••=•== ΓΓΓ

!n)1n( =+ΓFor n being a positive integer.

2.3. Values of gamma functions.

The values of Γ(p) for non-integer values of p must be computed numerically. One obtains for 1 ≤ p ≤ 2

p 1.0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0Γ(p) 1.00 .951 .918 .898 .887 .886 .894 .909 .931 .962 1.00

Table 1 Values of Γ(p) for non-integer values of p

The value of Γ(p) can be computed for any value of p using (2.2)Γ(4.3) = 3.3 Γ(3.3) = 3.3 * 2.3 * Γ(2.3) =

3.3 * 2.3 * 1.3 * Γ (1.3) = 3.3 * 2.3 * 1.3 * 0.898 = 8.861

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THE GAMMA FUNCTION

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2.4 Values of Γ(1/2) and Γ(-1/2).

Two particular values of the gamma function, Γ(1/2) and Γ(-1/2), appear in many applications. Bydefinition:

∫∞

−−=o

x2/1 dxex)2/1(Γ

Write x =u2 and dx = 2udu

∫∞

−=o

u due2)2/1(2

Γ

Then

[ ]

= ∫∫

∞−

∞− dve2due22/1(

o

v

o

u2 22

Γ ∫ ∫∞∞

+−=o o

)vu( dudve422

Write u = r cos θ, v = r sin θ, du dv = r dr dθ

[ ] ∫ ∫∞

−=2/

o o

r2 rdrde4)2/1(2

πθΓ ∫=

2/

o

d2π

θ π=

Hence

πΓ =)2/1( (2.5)

and by (2.3)

πΓ

Γ 22/1

)2/1()2/1( −=−=− (2.6)

2.5. Values of Γ(n) for n = 0, -1, -2, ...

By definition

∫∞ −

=o

xdx

xe

)0(Γ

that we write

dx...x!3

1

x!2

1edx...

x!3

1

x!2

1x1

e)0(o

32x

32o

x ∫∫∞

−∞

+−+

−+−=Γ

Derive

⋅⋅⋅+−−

⋅⋅⋅−−=

⋅⋅⋅− −−−

32x

32x

32x

x!3

1

x!2

1

x

1e

x!3

1

x!2

1e

x!3

1

x!2

1e

dxd

Hence

dx...x!3

1

x!2

1e...

x!3

1

x!2

1e)0(

32o

x32

x

+−−

+−−= ∫

∞−−Γ

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PAVEMENT DESIGN AND EVALUATION

12

dx...x!3

1

x!2

1e

32o

x

+−+ ∫

∞−

∞−

+−−=

o32

x ...x!3

1

x!2

1e)0(Γ

∞−

+−+−+−−=

o32

xx!1

11...

x!3

1

x!2

1x!1

11e)0(Γ

∞=

+−−=

∞−−

o

x/1xx!1

11ee)0(Γ (2.7)

−∞=−

=−1

)0()1(

ΓΓ (2.8)

±∞=− )n(Γ (2.9)The gamma function is undefined when the value of the argument is zero or a negative integer.

2.6. Derivative of Γ(n).We call Γ’(n) the derivative of Γ(n) with respect to n. For simplicity we consider only the case of n beingan integer, which besides is the only case required for our purpose.

By definition

∫∞

−−=o

x1n dxexdnd

)n('Γ ( )∫∞

−−=o

x1n dxexdnd

∫∞

−−=o

x1n dxexlogx

We first compute Γ’(1)

∫∞

−=o

xdxexlog)1('Γ

Integrating by parts

∫ ∫ −=−= xxlogxdxxx

xlogxxdxlog

∫ ∫ −=−=2x

xlog2

xdx

xx

21

2x

xdxlogx2222

……………

∫ ∫ −=−= −−2

nn1n

n1n

n

xxlog

nx

dxxn1

xlognx

dxxlogx

Further

( )∫ ∫ ∫ −−−− −+−= dxxedxexlogxexxlogxdxexlog xxxx

∫ ∫ ∫ −−−− −+

−= dxex

!2.21

dxexlogx!2

1e

!2.2x

xlog!2

xdxexlogx x2x2x

22x

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THE GAMMA FUNCTION

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∫ ∫ ∫ −−−− −+

−= dxex

!3.31

dxexlogx!31

e!3.3

xxlog

!3x

dxexlogx!2

1 x3x3x33

x2

...

∫ ∫ ∫ −−−−− −+

−=

−dxex

!n.n1

dxexlogx!n

1e

!n.nx

xlog!n

xdxexlogx

)!1n(1 xnxnx

nnx1n

Assembling:

∫ −−−

++++−

++++= x

n32x

n32x e

!n.nx

...!3.3

x!2.2

x!1.1

xexlog

!nx

...!3

x!2

x!1x

dxexlog

∫ ∫ ∫ ∫ −−−− −−−−− dxex!n.n

1...dxex

!3.31

dxex!2.2

1dxxe

!1.11 xnx3x2x

( )∫ −−−

++++−−= x

n32xxx e

!n.nx

...!3.3

x!2.2

x!1.1

xexlog)1edxexlog

∫ ∫ ∫ ∫ −−−− −−−−− dxex!n.n

1...dxex

!3.31

dxex!2.2

1dxxe

!1.11 xnx3x2x

Integrating between 0 and ∞ yields

∫∞ ∞−∞− −=o

ox

ox exlogxlogdxexlog

...!3.3)4(

!2.2)3(

!1.1)2(

e...!3.3

x!2.2

x!1.1

x

o

x32

−−−−

+++−

∞− ΓΓΓ

( ) ( ) ...31

21

100exloglimexloglimxloglimxloglimdxexlogo

x

0x

x

x0xx

x −−−−+−+−−=∫∞

∞→→∞→

( ) ⋅⋅⋅−−−−−+= −

∞→

−∫ 31

21

11exloglimxloglimdxxelog x

0x0 x

x

∫∞

∞→

− −=

−−−−−==

o M

xM1

...31

21

1Mloglimdxexlog)1(' γΓ (2.10)

where γ = 0.5772157 is called Euler’s constant.

By extent:

( ) ∫ ∫∫∞ ∞

−−∞−−∞

−+−=o o

xxo

xx

o

dxxedxexlogxexxlogxdxexlog

∫ ∫ ∫∞ ∞ ∞

−−− +=o o o

xxx dxxedxexlogdxexlogx

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PAVEMENT DESIGN AND EVALUATION

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∫∞

− +−==o

x )2(dxexlogx)2(' ΓγΓ

++−=

21

1)3(!2)3(' ΓγΓ

...

−+++++−−=

1n1

...31

21

1)n()!1n()n(' ΓγΓ

1n1

...31

21

1)n()n('

−+++++−= γ

ΓΓ

(2.11)

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THE GENERAL SOLUTION OF THE BESSEL EQUATION

15

Chapter 3 The General Solution of the Bessel Equation.

3.1 Introductory note.

In the way trigonometric functions are solutions of Laplace equation in Cartesian co-ordinates, Besselfunctions are solutions of Laplace equation in polar or cylindrical co-ordinates. One could say that Besselfunctions is a three dimensional function whereas trigonometric functions are two dimensional functions.The fundamental characteristics of both functions are identical:

- if f(x) = cos(x) )x(fdx

)x(fd2

2−=

- if f(r) = J0(r) )r(fdr

)r(df

r

1

dr

)r(fd2

2−=+

Bessel functions are essentially utilised in problems presenting an axial symmetry: in pavementengineering, the common case of a circular load on a layered structure, i.e. nearly all applications ofchapters 12 to 26.

Jp and J-p are the two solutions of the first kind of Bessel’s equation

0tr

pdr

d

r

1

dr

d 222

2=+−+ φ

φφφ

However, when p = n, Jn = (-)n J-n.In that case, Bessel’s equation requires another second solution, a solution of the second kind, Yp , whichwill not be utilised in this book.

Ip and I-p are the two modified solutions of the first kind of Bessel’s modified equation:

0tr

pdr

d

r

1

dr

d 222

2=−−+ φ

φφφ

These also, aren’t used in this book. However, they allow introducing the Ker and Kei functions,solutions of:

0itdr

d

r

1

dr

d 22

2=++ φ

φφ

which is applied in the problem of a slab subjected to an isolated load (chapter 12.4) and in theovalisation test (chapter 26.4).

The modified form of Bessel’s equation:

0FrFr

p

dr

dF

r

1)21(

dr

Fd )1(2222

222

2

2=+

−+−+ −γγβ

γαα

permits us to illustrate the existence of relations between Bessel functions of order ½ and trigonometricfunctions (chapter 4.3) Those relations allow us to establish asymptotic approximations for Besselfunctions (chapter 4.5), required for the numerical computations of functions of high arguments.

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PAVEMENT DESIGN AND EVALUATION

16

3.2 Helpful relations

( ))1kp(!k

2/rt)()rt(J

k2p

0

kp ++

−=+∞

∑ Γ

( ))1kp(!k

2/rt)()rt(J

k2p

0

kp ++−

−=+−∞

− ∑ Γ

)rt(J)()rt(J nn

n −=− for n integer

( )[ ] ∑− −−−

−+=1n

0

nk2

nn !k)2/rt()!1kn(1

)2/rtlog()rt(J2

)rt(Yπ

γπ

)!kn(!k)2/rt(

kn1

...31

21

11

k1

...31

21

11

)(1 nk2

0

k+

++++++++++−−

+∞

∑π

)rt(iY)rt(J)rt(H pp)1(

p +=

)rt(iY)rt(J)rt(H pp)2(

p −=

∑∞ +

−++

==0

k2p

pp

p )1pk(!k)2/rt(

)irt(Ji)rt(IΓ

∑−

−−−

−=1n

0k2n

kn

)2/rt(!k

)!1kn()(

21

)rt(K

++−+−

+−+ ∑

∞ ++ )1kn(

21

)1k(21

)2/rtlog()!kn(!k

)2/rt()(

0

k2n1n ΨΨ

k1

...31

21

11

)1k()1( +++++−=+−= γΨγΨ

)rt(ibei)rt(ber)irt(I)irti(J 00 ±=±=±

∑∑∞ +∞

++−=−=

0

2k4k

0

k4k

)!1k2()!1k2()2/rt(

)()rt(bei)!k2()!k2(

)2/rt()()rt(ber

)rt(kei)rtker()irt(K0 ±=±

[ ] ( )...

21

1!2!2

2/rt)rt(bei

42/rtlog()rt(ber)rtker(

4

+−++−=

πγ

[ ] ( ) ( )...

31

21

1!3!3

2/rt!1!12/rt

)rt(bei4

)2/rtlog()rt(bei)rt(kei62

++−+−+−=

πγ

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17

3.3 Resolution of the Bessel equation (Bessel functions of the first kind)

Recall the definition of the Laplacian in cylinder co-ordinates given by § 1.5:

0zr

1rr

1

r 2

2

2

2

22

2=+++

Φ∂

∂θ

Φ∂∂Φ∂

Φ∂ (3.1)

Consider next solution obtained by the method of separation of the variablestze)pcos()r(F θΦ =

and apply it to (3.1) that transforms into:

0FtFr

prF

r1

r

F 22

2

2

2=+−+

∂∂

∂ (3.2)

The general solution of equation (3.2) can be found by the method of the indicial equations (Wayland,1970). Therefore assume that the solution can be written as a series such as:

2n

0nraF ++

∑= α (3.3)

Hence

∑∞

+++=0

1nnra)n(

rF αα

∂∂

(3.4)

∑∞

+−++=0

n2

2r)1n)(n(

r

F ααα∂

∂ (3.5)

Replace the terms in (3.2) by (3,3), (3.4) and (3.5):

[ ]{ } 0ratrap)n()1n)(n(0

2nn

2nn

2 =+−++−++∑∞

+++ ααααα

Take out the first two terms of the first summation:

ra)1p)(1p(a)p(r 1022 +++−+

− αααα

0ratra)pn)(pn(0

2nn

2

2

nn =

+++−++ ∑∑∞

+∞

αα

If the terms in r0 and r1 are equal zero, the relation becomes homogeneous regarding the exponents of r.Therefore choose a0 = arbitrary, a1 = 0 and α = ± p.Rearrange the indices

[ ] 0rtaa)p22n)(2n(r0

2n2n2n =

+±++∑∞

++

α (3.6)

If (3.6) has to be identical zero whatever the value of m, each term of the summation has to be equal zero.Hence the recurrence formula

2n2n taa)p22n)(2n( −=±++ + (3.7)

and because a1 = 0, all the terms with an odd index are also equal zero.

Finally resulting in:

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18

20

20

20

2 2t

)p1(1a

)p1(22ta

)p22(2ta

a

±×−=

±×−=

±−=

40

22

4 2t

)p2)(p1(21a

)p24(4ta

a

±±×=

±−=

60

24

6 2t

)p3)(p2)(p1(321a

)p26(6ta

a

±±±××−=

±−=

..............s2

0s

s2 2t

)ps)...(p2)(p1(!sa)(

a

±±±−

=

If p is different from zero or not an integer, one obtains two linearly independent solutions

∑∑∞ +−∞ +

++−−+

++−=

0

k2pk

0

k2pk

)1kp(!k)rt(

)(B)1kp(!k

)2/rt()(AF

ΓΓ (3.8)

The corresponding series are called Bessel functions of the first kind and noted

∑∞ +

++−=

0

k2pk

p )1kp(!k)2/rt(

)()rt(JΓ

(3.9)

∑∞ +−

− ++−−=

0

k2pk

p )1kp(!k)2/rt(

)()rt(JΓ

(3.10)

The Bessel functions of the first kind and of order 0 and 1 are represented in Figure 3.1.

Figure 3.1 Bessel functions of the first kind and of order 0 and 1

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THE GENERAL SOLUTION OF THE BESSEL EQUATION

19

3.4 Resolution of the Bessel equation for p an integer (Bessel functions of the second and thirdkind)

3.4.1 For n integer, Jn = (-)n J-n

When p is an integer, let say n, solutions (3.9) and (3.10) are not linearly independent any more. Indeedwhen p = -n

∑ ∑∑∞ ∞ +−− +−+−

− +−−+

+−−=

+−−=

0 n

k2nk

1n

0

k2nk

k2nk

n )!kn(!k)2/rt(

)()!kn(!k

)2/rt()(

)!kn(!k)2/rt(

)()rt(J

When 0, 1, 2, 3, ..., k = n-1, (-n+k)! = ± ∞ and

0)!kn(!k

)2/rt( k2n=

+−

+−

Write k = n + j

∑ ∑∞ ∞ +

++−

− +−=

+−−=

n 0

j2njn

k2nk

n !j)!jn()2/rt(

)()!kn(!k

)2/rt()()rt(J

∑∞ +

− −=+

−−=0

nn

j2njn

n )rt(J)()!jn(!j

)2/rt()()()rt(J

Hence the two solutions are not linearly independent.

3.4.2 Bessel functions of the second kind

In order to find a second linearly independent solution, we define the function

)psin(

)rt(J)pcos()rt(J)rt(Y pp

p π

π −−= (3.11)

which is valid for p not an integer.Then we define for p = n

)psin(

)rt(J)pcos()rt(Jlim)rt(Y pp

npn π

π −

−=

We search the solution for n = 0.

)psin(

)rt(J)pcos()rt(Jlim)rt(Y pp

0p0 π

π −

−=

Apply de L’Hospital’s rule

)pcos(p

)rt(J)psin()rt(J)pcos(

p

)rt(J

)rt(Y

pp

p

0 ππ∂

∂πππ

∂ −−−=

−=

p

)rt(J

p

)rt(J1)rt(Y

pp0 ∂

πCompute the derivatives of Jp(rt) and J-p(rt):

∑∞ +

++++

−++

−=0

k2pkp

)1kp()1kp('

)2/rtlog()1kp(!k

)2/rt()(

p

)rt(J

ΓΓ

Γ∂

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20

∑∞ +−−

++−++−

+−++−

−=0

k2pkp

)1kp()1kp('

)2/rtlog()1kp(!k

)2/rt()(

p

)rt(J

ΓΓ

Γ∂

and for p → 0

∑∞

++

−−=0

k2k

0 )1k()1k('

)2/rtlog(!k!k)2/rt(

)(2

)rt(YΓΓ

πBy (2.11)

k1

...31

21

11

)1k()1k('

+++++−=++

γΓΓ

Hence

[ ]

+++

+−++= ...

31

21

1!3!3)2/rt(

21

1!2!2)2/rt(

!1!1)2/rt(

)2/rtlog()rt(J2

)rt(Y642

00 γπ

(3.12)For n ≠ 0, (3.12) is extended to

[ ]{ } ∑− −−−

−+=1n

0

nk2

nn !k)2/rt()!1kn(1

)2/rtlog()rt(J2

)rt(Yπ

γπ

(3.13)

∑∞ +

+

++++++++++−−

0

nk2k

)!kn(!k)2/rt(

kn1

...31

21

11

k1

...31

21

11

)(1π

The functions Yp(rt) are called functions of the second kind.

The Bessel functions of the second kind and of order 0 and 1 are represented in Figure 3.2.

Figure 3.2 Bessel functions of the second kind and of order 0 and 1

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21

3.4.3 Bessel functions of the third kind

Hankel introduced next pair of conjugate complex functions, with i = √(-1):

)rt(iY)rt(J)rt(H pp)1(

p += (3.14)

)rt(iY)rt(J)rt(H pp)2(

p −= (3.15)

These functions are called Hankel functions or Bessel functions of the third kind.

3.5 The modified Bessel equation.

If in § 3.3 we chose:)tzcos()pcos()r(F θΦ =

as solution of the Laplace equation, equation (3.2) would modify into:

0FtFr

prF

r1

r

F 22

2

2

2=−−+

∂∂

∂ (3.16)

Equation (3.16) is called the modified Bessel equation.Its solution can immediately be deduced from the original solution (3.8)

)irt(BJ)irt(AJ)rt(F pp −+= (3.17)where

∑∞ ++

++−=

0

k2ppk2k

p )1kp(!k)2/rt(i

)()irt(JΓ

∑∞ +

++=

0

k2pp

p )1kp(!k)2/rt(

i)irt(JΓ

Hence we define the function Ip, modified Bessel function of the first kind, as a real function, solution ofthe modified Bessel equation.

∑∞ +

−++

==0

k2p

pp

p )1kp(!k)2/rt(

)irt(JiIΓ

(3.18)

pp BIAI)rt(F −+= (3.19)

It is easy to deduce from (3.18) that when p is an integer, let say p = n, In = I-n. In that case, the secondsolution is usually defined by

−−=

→ np

)rt(I)rt(I

2)(

lim)rt(Kppn

npn (3.20)

which is called the modified Bessel function of the second kind.

The function Kp(rt) is defined for unrestricted values of p by the equation

ππ

psin

)rt(I)rt(I

2)rt(K pp

p−

= − (3.21)

Applying de L’Hospital’s rule on (3.20) and (3.21), one verifies that)rt(Klim)rt(K p

npn

→=

In a similar way as given in § 3.4.2 one obtains:

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22

∑−

−−−

−=1n

0k2n

kn

)2/rt(!k

)!1kn()(

21

)rt(K

(3.22)

++−+−

+−+ ∑

∞ ++ )1kn(

21

)1k(21

)2/trln()!kn(!k

)2/rt()(

0

k2n1n ΨΨ

k1

...31

21

11

)1k()1( +++++−=+−= γΨγΨ

The modified Bessel functions of the first kind and of order 0 and 1 are represented in Figure 3.3.

Figure 3.3 Modified Bessel functions of the first kind of order 0 and 1

The modified Bessel functions of the second kind and of order 0 and 1 are given in Figure 3.4.

Figure 3.4 Modified Bessel functions of the second kind and of order 0 and 1

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23

3.6 The ber and bei functions.

Consider next modified Bessel equation:

0Ft)i(rF

r1

r

F 22

2=±−+

∂∂

∂ (3.23)

which solutions of the first kind are

...!4!4)2/rt(

!3!3)2/rt(i

!2!2)2/rt(

!1!1)2/rt(i

1)irt(I8642

0 +−−+=

...!4!4)2/rt(

!3!3)2/rt(i

!2!2)2/rt(

!1!1)2/rt(i

1)irt(I8642

0 ++−−=−

We define:

)rt(ibei)rt(ber)irt(I0 ±=± (3.24)where

...!6!6)2/rt(

!4!4)2/rt(

!2!2)2/rt(

1)rt(ber1284

+−+−= (3.25)

...!5!5)2/rt(

!3!3)2/rt(

!1!1)2/rt(

)rt(bei1062

++−= (3.26)

Notice that:

( ) ( ) ( ) ( ) ( ) ( )2

irtJirtJ2

irtiJirtiJ2

irtIirtI)rt(ber 000000 +−

=−+

=−+

= (3.27)

( ) ( ) ( ) ( ) ( ) ( )i2

irtJirtJi2

irtiJirtiJi2

irtIirtI)rt(bei 000000 −−

=−−

=−−

= (3.28)

The ber and bei functions can be depicted in Figure 3.5

Figure 3.5 The ber and bei functions

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24

3.7 The ker and kei functions.

Again consider equation (3.23), which solutions of the second kind are:

( ) ( ) ...31

21

1!3!3

2irt

21

1!2!2

2irt

!1!1

2irt

2irt

logirtIirtK

642

00

++

+

+

+

+

+−= γ

Recall that:

4i

2rt

loge21

2rt

logilog21

2rt

logilog2rt

log2

irtlog 2/i ππ +=+=+=+=

and applying (3.24)

( ) [ ] ...31

211

!3!32rt

i211

!2!22rt

!1!12rt

i4

i2rtlog)rt(ibei)rt(berirtK

642

0

++

+

+

+++−= πγ

( ) [ ] ...31

21

1!3!3

2rt

i21

1!2!2

2rt

!1!12rt

i4

i2rt

log)rt(ibei)rt(berirtK

642

0

++

+

+

−+−−=−

πγ

We define:

( ) ( )2

irtKirtK)rtker( 00 −+

= (3.29)

( ) ( )i2

irtKirtK)rt(kei 00 −−

= (3.30)

[ ] ( )...

21

1!2!2

2/rt)rt(bei

4)2/rtlog()rt(ber)rtker(

4

+−++−=

πγ (3.31)

[ ] ( ) ( )...

31

21

1!3!3

2/rt!1!12/rt

)rt(ber4

)2/rtlog()rt(bei)rt(kei62

++−+−+−=

πγ (3.32)

The ker and kei functions are given in figure 3.6.

Figure 3.6 The ker and kei functions

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THE GENERAL SOLUTION OF THE BESSEL EQUATION

25

3.8 Resolution of the equation ∇2∇2w + w =0

Consider the next equation:

0wdrdw

r1

dr

wddrd

r1

dr

d2

2

2

2=+

+

+ (3.33)

The solution can be obtained by splitting (3.33) into two simultaneous differential equations

zdrdw

r1

dr

wd2

2=+ (3.34)

wdrdz

r1

dr

zd2

2−=+ (3.35)

Both equations are verified together if iwz m= . Indeed, if, for example, z = iw

(3.34) becomes 0iwdrdw

r1

dr

wd2

2=−+ , and

(3.35) becomes 0wdrdw

r1

idr

wdi

2

2=++ or 0iw

drdw

r1

dr

wd2

2=−+

Hence the solution of (3.33) is given by the solution of

0w)i(drdw

r1

dr

wd2

2=±−+ (3.36)

The equations (3.25), (3.26), (3.31) and (3.32) give the solution of (3.36), for its two signs

)r(Dkei)rker(C)r(Bbei)r(Aberw +++= (3.37)

3.9 The modified form of the Bessel equation

Consider the next differential equation:

0FrFr

prF

r1

)21(r

F )1(2222

222

2

2=+

−+−+ −γγβ

γα∂∂

α∂

∂ (3.38)

which has for solutions, as can be verified by substitution,

)r(JBr)r(JAr)r(F ppγαγα ββ −+= (3.39)

or, if p is an integer,

)r(YBr)r(JAr)r(F ppγαγα ββ += (3.40)

An interesting application of (3.38) is the resolution of the Laplace equation in two- dimensionalCartesian co-ordinates

0yx 2

2

2

22 =+=∇

Φ∂

Φ∂Φ

Consider a solution such as Φ = F(x)ety

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PAVEMENT DESIGN AND EVALUATION

26

Hence

0e)x(Ftx

)x(F ty22

22 =

+=∇

∂Φ (3.41)

Comparing (3.41) with (3.38) yields:t12/1 === βγα

2/1p0p 222 ==− γαHence

)tx(JBx)tx(JAx)x(F 2/12/1

2/12/1

−+= (3.42)However

)txsin(B)txcos(A)x(F += (3.43)

is also a solution of (3.41). Therefore, one must conclude that there exists a relation between Besselfunctions of order ± 1/2 and trigonometric functions.

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PROPERTIES OF THE BESSEL FUNCTIONS

27

Chapter 4 Properties of the Bessel Functions.

4.1 Introductory note

This chapter deals with the most important properties of Bessel functions: derivatives, functions of halforder, asymptotic values, indefinite integrals and relations between functions of different kind.

4.2 Helpful relations

[ ] )rt(J)rt(t)rt(J)rt(drd

1pp

pp

−=

[ ] )rt(J)rt(t)rt(J)rt(drd

1pp

pp

+−− −=

)rt(Jrp

)rt(tJ)rt(Jdrd

p1pp −= −

)rt(Jrp

)rt(tJ)rt(Jdrd

p1pp +−= +

)rt(Jrtp2

)rt(J)rt(J p1p1p =+ +−

t)rt(rJ

dr)rt(rJ 10

=∫

∫ −=t

)rt(Jdr)rt(J 0

1

)rtsin(rt2

)rt(J 2/1 π=

)rtcos(rt2

)rt(J 2/1 π=−

[ ]rtrt2/1 ee

rt2

1)rt(I −−=

π

[ ]rtrt2/1 ee

rt2

1)rt(I −

− +=π

rt2/12/1 e

rt2KK −

− ==π

−−≅

2p

4rtcos

rt2

)rt(J pππ

π for high values of rt

++≅− 2

p4

rtsinrt2

)rt(J pππ

π for high values of rt

−−≅

2p

4rtsin

rt2

)rt(Y pππ

π for high values of rt

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PAVEMENT DESIGN AND EVALUATION

28

++≅− 2

p4

rtcosrt2

)rt(Y pππ

π for high values of rt

)4/2/prt(i)1(p e

rt2

H πππ

−−≅ for high values of rt

)4/2/prt(i)2(p e

rt2

H πππ

−−−≅ for high values of rt

( )i)2/1p(rtrtp ee

rt21

)rt(I π

π++−+≅ for high values of rt

−≅

82rt

cosrt2

e)rt(ber

2/rt ππ

for high values of rt

−≅

82rt

sinrt2

e)rt(bei

2/rt ππ

for high values of rt

+≅ −

82rt

cosert2

)rtker( 2/rt ππ for high values of rt

+−≅ −

82rt

sinert2

)rt(kei 2/rt ππ for high values of rt

( ) )z(JezeJ)z(J ppii

pp ±±± ==− ππ

( ) )z(JezeJ ppmiim

p ±± = ππ

2

)rt(H)rt(H)rt(J

)2(p

)1(p

p+

=

)psin(

)pcos()rt(Y)rt(Y)rt(J pp

p π

π−= −

)psin(i

e)rt(J)rt(J)rt(H

ippp)1(

p π

π−− −

=

)psin(i

e)rt(J)rt(J)rt(H

ippp)2(

p π

π

−= −

)rt(J)psin()mpsin(

e2)rt(He)rte(H pip)1(

pimpim)1(

p πππππ −− −=

)rt(J)psin()mpsin(

e2)rt(He)rte(H pip)2(

pimpim)2(

p πππππ += −

)()( )2/2/ ππ ip

pip rteJirtI −=

)rti(He2

i)rt(K )1(p

2/pip

ππ=

)rti(He2

i)rt(K )2(p

2/pip

ππ−=

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PROPERTIES OF THE BESSEL FUNCTIONS

29

4.3 Derivatives of Bessel functions

4.3.1 Derivative of (rt)pJp(rt)

By definition of the Bessel function

[ ]

++−= ∑

+

+

0k2p

k2p2k

pp

)1kp(!k2

)rt()(

drd

)rt(J)rt(drd

Γ

∑∞

+

−+

++

+−=

0k2p

1k2p2k

)1kp(!k2

t)rt)(kp(2)(

Γ

)kp(!k2

t)rt()( 1k2p

1k2p2

0

k

+−= −+

−+∞

∑Γ

[ ]∑∞

+−

+−

++−−=

0k2)1p(

k2)1p(kp

1k)1p(!k2

)rt()()rt(t

Γ

[ ] )rt(J)rt(t)rt(J)rt(drd

1pp

pp

−= (4.1)

4.3.2 Derivative of (rt)-pJp(rt)

One finds similarly

[ ] )rt(J)rt(t)rt(J)rt(drd

1pp

pp

+−− −= (4.2)

4.3.3 Derivative of Jp(rt)

Deriving the left member of (4.1) by parts yields

[ ] )rt(Jrp

)rt(tJ)rt(Jdrd

p1pp −= − (4.3)

deriving the left member of (4.2) by parts yields

[ ] )rt(Jrp

)rt(tJ)rt(Jdrd

p1pp +−= + (4.4)

Particularly

[ ] )rt(tJ)rt(tJ)rt(Jdrd

110 −== − (4.5)

Adding (4.3) and (4.4) yields

[ ] [ ])rt(J)rt(J2t

)rt(Jdrd

1p1pp +− −= (4.6)

and subtracting yields the recurrence formula for Bessel functions

)rt(Jrtp2

)rt(J)rt(J p1p1p =+ +− (4.7)

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PAVEMENT DESIGN AND EVALUATION

30

4.4 Bessel functions of half order

4.4.1 Values of J1/2(rt), J-1/2(rt), J3/2(rt), J-3/2(rt)

Expanding the Bessel functions in their series and recalling that Γ(1/2) = √π yields:

)rtsin(rt2

)rt(J 2/1 π= (4.8)

)rtcos(rt2

)rt(J 2/1 π=− (4.9)

−= )rtcos(

rt)rtsin(

rt2

)rt(J 2/3 π (4.10)

+−=− )rtsin(

rt)rtcos(

rt2

)rt(J 2/3 π (4.11)

4.3.2. Values of I1/2(rt), I-1/2(rt)

Expanding the Bessel functions in their series yields

[ ]rtrt2/1 ee

rt21

)rt(I −−=π

(4.12)

[ ]rtrt2/1 ee

rt21

)rt(I −− +=

π (4.13)

4.3.3. Values of K1/2(rt), K-1/2(rt)

By definition (3.21)

rt2/12/12/1 e

rt2)2/sin()rt(I)rt(I

2)rt(K −− =

−=

ππ

π (4.14)

rt2/12/12/1 e

rt2)2/sin()rt(I)rt(I

2)rt(K −−

− =−

ππ

(4.15)

Hence)rt(K)rt(K 2/12/1 −= (4.16)

4.4. Values of K0(rt) and Kn(rt)

By definition (3.21)

ππ

psin

)rt(I)rt(I

2lim)rt(K pp

0p0

−= −

→Applying de l’Hospital’s rule yields and omitting the lim sign

−=

−= −

dp

)rt(dI

dp

)rt(dI

21

pcosdp

)rt(dI

dp

)rt(dI

2)rt(K pp

pp

0 πππ

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PROPERTIES OF THE BESSEL FUNCTIONS

31

++−

+−= ∑∑

+−

)1pk(!k)2/rt(

dpd

)1pk(!k)2/rt(

dpd

21

)rt(Kpk2pk2

0 ΓΓ

+−+−+−++−−

=−−

∑ )1pk()1pk()1pk(')2/rt()1pk()2/rtlog()2/rt(

!k2)2/rt(

)rt(Kppk2

0 ΓΓΓΓ

++++++−++

−)1pk()1pk(

)1pk(')2/rt()1pk()2/rtlog()2/rt( pp

ΓΓΓΓ

Letting p → 0

++

+−= ∑ )1k()1k('

)2/rtlog(!k!k)2/rt(

)rt(Kk2

0 ΓΓ

Developing Γ’(k+1) as in § 3.3.2 yields

[ ]

++++−=

21

1!2!2)2/rt(

!1!1)2/rt(

)2/rtlog()rt(I)rt(K42

00 γ

⋅⋅⋅

+++

31

21

1!3!3)2/rt( 6

(4.17)

For n ≠ 0, (4.17) is relatively easily extended to

[ ] ∑ −+ −−−++−=

n

0

nk2k

n1n

n )2/rt(!k

)!1kn()(21

)2/rtlog()rt(I)()rt(K γ

[ ]∑∞ +

+++−+

0

nk2n

)!nk(!k)2/rt(

)kn()k()(21

ΦΦ (4.18)

4.5 Asymptotic values

4.5.1 Asymptotic values for Jp and J-p

Transform Bessel equation (3.2):

0FtFr

prF

r1

r

F 22

2

2

2=+−+

∂∂

by writing F(rt) = G(rt)/(rt)1/2

0Gr

4/1pt

r

G2

22

2

2=

−−+

∂ (4.19)

When p = ½, the equation reduces to

0Gtr

G 22

2=+

∂ (4.20)

whose general solution is given by)rtsin(B)rtcos(AG +=

Hence

)rtsin()rt(B)rtcos()rt(AF 2/12/1 −− += (4.21)

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PAVEMENT DESIGN AND EVALUATION

32

but also:)rt(DJ)rt(CJF 2/12/1 += − (4.22)

Equations (4.21) and (4.22) confirm, as we knew by (4.8) and (4.9), that there exists a relation betweenthe Bessel functions of half order and the trigonometric functions, and that this relation is valid for allvalues of the argument, thus in particular for high values of the argument. Hence the equations:

)rtsin(rt2

)rt(J 2/1 π=

)rtcos(rt2

)rt(J 2/1 π=−

can be considered as the asymptotic equations for the Bessel functions of half order.Thus (4.19) must have, for p not being an integer, two approximate values for high values of the argumentsuch as

)rtcos()rt(A)rt(J p2/1

pp α+≅ − (4.23)

)rtsin()rt(B)rt(J p2/1

pp β+≅ −− (4.24)

The coefficients αp and βp must be determined in such a way that equations (4.23) and (4.24) are linearlyindependent and compatible with the definitions of Jp and J-p.Derive, with respect to r, Jp in (4.23)

)rtsin(t)rt(A)rtcos()rt(A2t

)rt(J p2/1

pp2/3

p'p αα +−+−= −−

For high values of the argument the first term can be neglected against the second

)rtsin(t)rt(A)rt(J p2/1

p'p α+−≅ −

Further (4.3) and (4.4) simplify for high values of the argument

)rt(tJ)rt(J 1p'p −≅

)rt(tJ)rt(J 1p'p +−≅

Hence

)rtcos()rt(tA)rtsin(t)rt(A 1p2/1

1pp2/1

p −−

−− +≅+− αα (4.25)

)rtcos()rt(tA)rtsin(t)rt(A 1p2/1

1pp2/1

p +−

+− +−≅+− αα (4.26)

If we choose Ap = Ap-1 = Ap+1 = A and αp = k - pπ/2 in such a way that A and k are independent from p ,we notice that equations (4.25) and (4.26) are satisfied. Indeed

−+=

−+−

2)1p(

krtcosA2

pkrtsinA

ππ

Then

−+≅ −

2p

krtcos)rt(A)rt(J 2/1p

π

This equation must be satisfied for all values of p, thus also for p=1/2 for which

−−==

44rtcos

rt2

)rtsin(rt2

)rt(J 2/1ππ

ππHence A = (2π)1/2 and k = - π/4

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PROPERTIES OF THE BESSEL FUNCTIONS

33

Finally

−−≅

2p

4rtcos

rt2

)rt(J pππ

π (4.27)

Replacing p by –p in (4.27) yields

++≅− 2

p4

rtsinrt2

)rt(J pππ

π (4.28)

4.5.2 Asymptotic values for Yp and Y-p

Since Yp and Y-p are the paired second solutions with Jp and J-p we shall admit implicitly the followingasymptotic equations

−−≅

2p

4rtsin

rt2

)rt(Y pππ

π (4.29)

++≅− 2

p4

rtcosrt2

)rt(Y pππ

π (4.30)

4.5.3 Asymptotic values for Hp(1) and Hp

(2)

By application of the definitions (3.14) and (3.15), one immediately finds:

)2/p4/rt(i)1(p e

rt2

H πππ

−−= (4.31)

)2/p4/rt(i)2(p e

rt2

H πππ

−−−= (4.32)

4.5.4 Asymptotic values for Ip and I-p

Transform Bessel equation (3.16)

0FtFr

prF

r1

r

F 22

2

2

2=−−+

∂∂

by writing F(rt) = G(rt)/(rt)1/2

0Gr

4/1pt

r

G2

22

2

2=

−+−

∂ (4.33)

When p = ½, the equation reduces to

0Gtr

G 22

2=−

∂ (4.34)

whose general solution is given byrtrt BeAeG −+=

Hencert2/1rt2/1 e)rt(Be)rt(AF −−− += (4.35)

but also

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PAVEMENT DESIGN AND EVALUATION

34

)rt(DI)rt(CIF 2/12/1 −+= (4.36)Equation (4.35) is valid for all values of the argument, thus in particular for high values of the argument.

Hence the equations:

[ ]rtrt2/1 ee

rt21

)rt(I −−=π

[ ]rtrt2/1 ee

rt21

)rt(I −− +=

π

can be considered as the asymptotic equations for the modified Bessel functions of half order.Hence (4.33) must have, for p not an integer, two approximate values for high values of the argumentsuch as

+= ++−++− )2/1p(rt)2/1p(rt2/1pp ee)rt(A)rt(I βα (4.37)

+= +−+−+−+−−

)2/1p(rt)2/1p(rt2/1pp ee)rt(B)rt(I βα (4.38)

The coefficients α and β must be determined in such a way that equations (4.37) and (4.38) are linearlyindependent and compatible with the definitions of Ip and I-p.Derive, with respect to r, Ip in (4.37)

[ ])2/1p(rt)2/1p(rt2/3p

'p ee)rt(A

2t

)rt(I ++−++− +−= βα

−+ ++−++− )2/1p(rt)2/1p(rt2/1p eet)rt(A βα

For high values of the argument the first term can be neglected against the second

−≅ ++−++− )2/1p(rt)2/1p(rt2/1pp eet)rt(A)rt('I βα

Further simplify the derivatives of Ip(rt) for high values of the argument

)rt(tJ)rt(I 1p'p −≅

)rt(tJ)rt(I 1p'p +≅

Hence

− ++−++− )2/1p(rt)2/1p(rt2/1p eet)rt(A βα

+= −+−−+−−

)2/1p(rt)2/1p(rt2/11p eet)rt(A βα (4.39)

− ++−++− )2/1p(rt)2/1p(rt2/1p eet)rt(A βα

+= ++−++−+

)2/3p(rt)2/3p(rt2/11p eet)rt(A βα (4.40)

If we choose Ap = Ap-1 = Ap+1 = A and α = 0, β = πi in such a way that A and β are independent from p ,we notice that equations (4.39) and (4.40) are satisfied.

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PROPERTIES OF THE BESSEL FUNCTIONS

35

− ++−− i)2/1p(rtrt2/1 ee)rt(A π

+= −+−− i)2/1p(rtrt2/1 ee)rt(A π

Hence

+≅ ++−− i)2/1p(rtrt2/1p ee)rt(A)rt(I π

This equation must be satisfied for all values of p, thus also for p=1/2 for which

[ ]rtrt2/1 ee

rt21

)rt(I −−=π

Hence A = (2π)1/2

Finally

+≅ ++− i)2/1p(rtrtp ee

rt21

)rt(I π

π (4.41)

Replacing p by –p in (4.41) yields

+≅ +−+−−

i)2/1p(rtrtp ee

rt21

)rt(I π

π (4.42)

4.5.5 Asymptotic value for Kn

Consider definition (3.20):

np

)rt(I)rt(I

2)(

lim)rt(Kppn

npn −

−−=

→ (4.43)

For large values of the argument

npeeee

rt21

2)(

)rt(Ki)2/1p(rtrti)2/1p(rtrtn

n −−−+−

≅++−−−− ππ

π

np

eeie

rt21

2)(

)rt(K

ipiprtn

n −

−≅

−− ππ

π

Applying de l’Hospital’s rule yields

rtn

n e)ncos(rt2

22)(

)rt(K −−≅ ππ

π

rtn e

rt2)rt(K −≅

π (4.44)

In this form Kn(rt) may be generalised into

rtp e

rt2)rt(K −≅

π (4.45)

where p not an integer and also

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PAVEMENT DESIGN AND EVALUATION

36

rt0 e

rt2)rt(K −≅

π (4.46)

4.5.6 Asymptotic values for ber and bei

For high values of the argument (4.27) yields

( )

4irtcos

irt2

irtJ2/1

πWith (i)-1/4 = e-iπ/8 and √i = (1 + i)/√2

2

eeee

4irtcos

2/rt)4/2/rt(i2/rt)4/2/rt(i

+

=

−−−− πππ

2ee

4irtcos

2/rt)4/2/rt(i ππ −−≅

( ) )8/2rt(i2/rt

0 ert2

eirtJ π

π−−≅

( )

−−

−≅

82

rtsini

82

rtcos

rt2

eirtJ

2/rt

0ππ

π (4.47)

Similarly

( )

−+

−≅−

82

rtsini

82

rtcos

rt2

eirtJ

2/rt

0ππ

π (4.48)

Hence

( ) ( )

−=

+−≅

82rt

cosrt2

e2

irtJirtJ)rt(ber

2/rt00 π

π (4.49)

( ) ( )

−=

−−≅

82rt

sinrt2

ei2

irtJirtJ)rt(bei

2/rt00 π

π (4.50)

4.5.7 Asymptotic values for ker and kei

Recall equations (3.29) and (3.30)

( ) ( )2

irtKirtK)rtker( 00 −+

= (4.51)

( ) ( )i2

irtKirtK)rt(kei 00 −−

= (4.52)

Apply (4.46)

irt0 e

irt2)irt(K −=

π

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PROPERTIES OF THE BESSEL FUNCTIONS

37

2/)i1(rt8/i eert2

+−−= ππ

)8/2/rt(i2/rt eert2

ππ +−−=

+−

+= −

82rt

sini82

rtcose

rt2)irt(K 2/rt

0πππ

(4.53)

Similarly obtain

++

+=− −

82rt

sini82

rtcose

rt2)irt(K 2/rt

0πππ

(4.54)

Adding and subtracting (4.53) and (4.54) yields

+= −

82rt

cosert2

)rtker( 2/rt ππ (4.55)

+−= −

82rt

sinert2

)rt(kei 2/rt ππ (4.56)

4.6 Indefinite integrals of Bessel functions

4.6.1 Fundamental relations

In § 4.2.we have derived the following derivatives of Bessel functions:

[ ] )rt(J)rt(t)rt(rtJdrd

01 =

[ ] )rt(tJ)rt(Jdrd

10 −=

From those equations we easily deduce next fundamental integrals

∫ =t

)rt(rJdr)rt(rJ 1

0 (4.57)

∫ −=t

)rt(Jdr)rt(J 0

1 (4.58)

4.6.2 The integral ∫rnJ0(rt)dr

Integrating by parts solves the integral:

∫ ∫ −−−= dr)rt(Jr

t)1n(

t)rt(J

rdr)rt(Jr 11n1n

0n

∫ ∫ −−

− −+−= dr)rt(Jr

t)1n(

)rt(Jt

rdr)rt(Jr 0

2n0

1n

11n

Hence

∫ ∫ −− −−

−+= dr)rt(Jr

t

)1n()rt(Jr

t

)1n(t

)rt(Jrdr)rt(Jr 0

2n2

2

01n

21n

0n (4.59)

If n is odd, formula (4.59) leads to (4.57). If n is even, formula (4.59) leads to ∫J0(rt)dr, which istabulated.

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PAVEMENT DESIGN AND EVALUATION

38

4.7 Relations between Bessel functions of different kind

4.7.1 Bessel functions with argument –rt

Since Bessel’s equation is unaltered if rt is replaced by –rt, we must expect the functions J±p(-rt) to besolutions of the equations satisfied by J±p(rt). Considering the relation eπi = -1, we may write

)rte(J)rt(J ipp

π=−We can even consider the more general function Jp(emπirt) where m is an integer.

( )∑ ++−=

+

)1pk(!k2/rte

)()rte(Jk2pim

kimp Γ

ππ

Restricting the complex exponent to its principal value we get

( ) pim)mk2pm(ik2pim eee πππ == ++

Hence

( ) )rt(Je)1pk(!k

)2/rt()(erteJ p

pimk2p

kpimimp

πππΓ∑ =

++−=

+ (4.60)

4.7.2 Relations between the three kinds of Bessel functions

Consider (3.14) and (3.15)

)rt(iY)rt(J)rt(H pp)1(

p +=

)rt(iY)rt(J)rt(H pp)2(

p −=Hence by multiplying Yp(rt) by cos(pπ) and subtracting from Y-p(rt)

2

)rt(H)rt(H)rt(J

)2(p

)1(p

p+

= (4.61)

Consider (3.11)

)psin(

)rt(J)pcos()rt(J)rt(Y pp

p π

π −−=

Replace p by –p

)psin(

)rt(J)pcos()rt(J)rt(Y pp

p π

π +−= −

Hence by subtracting and dividing by 2

)psin(

)pcos()rt(Y)rt(Y)rt(J pp

p π

π−= − (4.62)

Consider (3.14) together with (3.11)

)rt(iY)rt(J)rt(H pp)1(

p +=

)psin(

)rt(J)pcos()rt(Ji)rt(J)rt(H pp

p)1(

p π

π −−+=

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PROPERTIES OF THE BESSEL FUNCTIONS

39

[ ])psin(i

)rt(J)psin(i)pcos()rt(J)rt(H pp)1(

p π

ππ −+−−=

)psin(i

e)rt(J)rt(J)rt(H

ippp)1(

p π

π−− −

= (4.63)

and similarly

)psin(i

e)rt(J)rt(J)rt(H

ippp)2(

p π

π

−= − (4.64)

Add equations (4.63) and (4.64) together

2

HH)rt(J

)2(p

)1(p

p+

= (4.65)

Multiply equation (4.63) by eipπ and (4.64) by e-ipπ and add together

2

e)rt(He)rt(H)rt(J

ip)2(p

ip)1(p

p

ππ −

−+

= (4.66)

In equation (4.63) replace rt by rt emπI

)psin(i

e)rte(J)rte(J)rte(H

ipimp

impim)1(

p π

ππππ

−− −

=

together with equation (4.60)

)sin(

)()()( ))1(

π

ππππ

pi

ertJertJerteH

ipp

impp

impim

p

−−

− −=

( ) ( ))psin(i

eee)rt(J

)psin(i

e)rt(J)rt(Je)rte(H

impimpipp

ippp

impim)1(

p ππ

ππππππ −

+−

=−−−

−−

applying equation (4.63)

)rt(J)psin()mpsin(

e2)rt(He)rte(H pip)1(

pimpim)1(

p πππππ −− −= (4.67)

and similarly

)rt(J)psin()mpsin(

e2)rt(He)rte(H pip)2(

pimpim)2(

p πππππ += − (4.68)

4.7.3 Bessel functions of purely imaginary argument

Consider (3.18)

∑∞ +

−++

==0

k2p

pp

p )1kp(!k)2/rt(

)irt(Ji)rt(IΓ

Recall that eπi/2 = i, e3πi/2= -i.Hence

)irt(Je)rt(I p2/ip

pπ−= (4.69)

Apply (3.21)

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PAVEMENT DESIGN AND EVALUATION

40

ππ

psin

)rt(I)rt(I

2)rt(K pp

p−

= −

ππ

ππ

psin

)irt(Je)irt(Je

2)rt(K p

2/ipp

2/ip

p

−− −

=

ππ

ππ

psini

)irt(Je)irt(Je

2i)rt(K p

ipp2/ip

p

−− −

=

Hence by (4.63)

)irt(He2

i)rt(K )1(p

2/ipp

ππ= (4.70)

and similarly

)irt(He2

i)rt(K )2(p

2/ipp

ππ−= (4.71)

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THE BETA FUNCTION

41

Chapter 5 The Beta Function

5.1 Introductory note

The gamma function, Γ(n), is a function with one variable. The beta function, B(m,n), is a function withtwo variables. The function can be defined as a product of Γ- functions. It is useful in the expression ofsolutions of definite integrals of Bessel functions, which, for particular values of their parameters, can bewritten in simple closed-forms.

5.2 Helpful relations

∫ −− −=1

0

1n1m dx)x1(x)n,m(B

∫ −−=2/

0

1n21m2 dsincos2)n,m(Bπ

ϑϑϑ

)m,n(B)n,m(B =

)nm()n()m(

)n,m(B+

ΓΓ

=

+−

21

)p2()p(21

p2 1p2 ΓΓΓΓ

5.3 Definition of the beta function

The beta function is defined by Euler’s integral of the first kind

∫ −− −=1

0

1n1m dx)x1(x)n,m(B (5.1)

which is convergent for m > 0 and n > 0.

Writing x = cos2θ and 1- x = sin 2θ, (5.1) transforms into

∫ −−=2/

0

1n21m2 dsincos2)n,m(Bπ

ϑϑϑ (5.2)

Replacing x by 1 - y in (5.1) yields)m,n(B)n,m(B = (5.3)

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PAVEMENT DESIGN AND EVALUATION

42

5.4 Relation between beta and gamma functions

By definition:

∫∞

−−=0

1mx dxxe)m(Γ

Hence, one may write

∫ ∫∞ ∞

−−−−=⋅0 0

1ny1mx dyyedxxe)n()m( ΓΓ

Write x = χ2, y = η2

∫ ∫∞ ∞

−−−−=⋅0 0

1n21m2 dede4)n()m(22

ηηχχΓΓ ηχ

( )∫ ∫∞∞

−−+−=⋅0 0

1n21m2 dde4)n()m(22

ηχηχΓΓ ηχ

Write χ = r cosϑ, η = r sinϑ

∫∞

−−−+−=⋅0

1n21m22n2m2r rdrdsincosre4)n()m(2

ϑϑϑΓΓ

∫ ∫∞

−−−+−=⋅0

2/

0

1n21m21n2m2r dsincosdrre4)n()m(2

πϑϑϑΓΓ

∫∞

−+−=⋅0

1n2m2r drre)n,m(B2)n()m(2

ΓΓ

Write r2 = ρ

)nm()n,m(Bde)n,m(B)n()m(0

1nm +⋅==⋅ ∫∞

−+− ΓρρΓΓ ρ

Hence

)nm()n()m(

)n,m(B+⋅

ΓΓ (5.4)

5.5 The duplication formula for gamma functions

Consider the function

( )∫=2/

0

p2 d2sinJπ

ϑϑ

Write u = 2ϑ

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43

)1p(2)2/1p()2/1(

uducosusinudusinudusin21

J2/

0

0p22/

0

p2

0

p2+

+==== ∫∫∫ Γ

ΓΓπππ

However, one also has

( )∫ ∫==2/

0

2/

0

p2p2 d)cossin2(d2sinJπ π

ϑϑϑϑϑ

∫ +++

=⋅= −−2/

0

1p2p2p21p2)1p2(

)2/1p()2/1p(2dsincos22J

π

ΓΓΓ

ϑϑϑ

)p2(p2)2/1p()2/1p(

2J 1p2Γ

ΓΓ ++= −

Hence

)p2()2/1()p()2/1p(2 1p2 ΓΓΓΓ =+− (5.5)

Equation (5.5), called “the duplication formula for gamma functions”, will often be used in furtherdevelopments.

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45

Chapter 6 Definite integrals of Bessel functions

6.1 Helpful relations

+=

π

νννννν

νν ΓνΓϑϑω

ω

0

2

)bt()at(

)bt(J)at(J21

21

2dsin)t(

)t(J

ϑω cosab2ba 222 −+=

∫ +++

+++

=2/

0

1211

1 dcossin)sinz(J)1(2

z)z(J

πνµ

µν

ν

νµ ϑϑϑϑνΓ

∫ ++++++

+

+=

2/

02/)1(22

2/122111

)yx(

)yx(Jyxdcossin)cosy(J)sinx(J

π

νµνµ

νµνµ

νµ ϑϑϑϑϑ

∫+=

2/

0

2 dcos)sinxcos()2/1()2/1(

)2/x(2)x(J

πν

ν

ν ϑϑϑΓνΓ

6.2 Gegenbauer’s integral

The important integral:

+=

π

νννννν

νν ΓνΓϑϑω

ω

0

2

)bt()at(

)bt(J)at(J21

21

2dsin)(J

(6.1)

ϑω cosab2ba 222 −+=which is valid for ν > -1/2, has been first proved by Gegenbauer (Watson, 1966), hence its name.

Develop the Bessel function in its series

( )( )∫ ∫∑

++−=

+π πν

ν

νν

νν ϑϑ

νΓω

ωϑϑ

ω

ω

0 0

2k2

k2 dsin)1k(!kt

2/t)(dsin

)t(

)t(J

∫∑ ++

−=π

νν ϑϑνΓ

ω

0

2k2k2

k dsin)1k(!k

)(2t

)(2

Integrate the series term by term.

( ) ( ) ( ) jkj22k

0j

k2k2 cosab2ba)!jk(!j

!k)( −

=−+

−== ∑ ϑωω

Consider the integral

∫π

ν ϑϑϑ0

2m dsincos

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PAVEMENT DESIGN AND EVALUATION

46

If m = 2n

( )∫ ∫ ++==π π

νν νϑϑϑϑϑϑ0

2/

0

2n22n2 2/1,2/1nBdsincos2dsincos

If m = 2n + 1

( )∫ ∫ =−=+π π

νν ϑϑϑϑϑϑ0 0

2n221n2 0sindsinsin1dsincos

Hence

( ))2()1(!1)2/1()2/1(

ba2t

)1()1()2/1()2/1(

2dsin)t(

)t(J 222

0

2++

++

+++

=∫ νΓνΓνΓΓ

νΓνΓνΓΓ

ϑϑω

ωπνν

νν

( )

++

++

+++

+

+

)3()2(!2)2/1()2/3(

ba4)2()1(!2)2/1()2/1(

ba2t 22222

4

νΓνΓνΓΓ

νΓνΓνΓΓ

( ) ( )

+

++

+++

+++

+

− ....)4()2(!3)2/1()2/3(

baba6)4()1(!3)2/1()2/1(

ba2t 2222322

6

νΓνΓνΓΓ

νΓνΓνΓΓ

rearranging somewhat the terms, results in:

++

+−

++=

πνν

νν

νΓνΓνΓνΓΓϑϑ

ω

ω

0

422 ...

)3(!2)2/at(

)2(!1)2/at(

)1(1

)2/1()2/1(2dsin)t(

)t(J

++

++

+× ...

)3(!2)2/bt(

)2(!1)2/bt(

)1(1 42

νΓνΓνΓHence

∫ +=π

νννννν

νν νΓΓϑϑω

ω

0

2

)bt()at(

)bt(J)at(J)2/1()2/1(2dsin

)t(

)t(J

6.3 Sonine’s first finite integral

The integral:

)z(Jdcossin)sinz(J)1(2

z1

2/

0

1211

++++

+=

+∫ νµ

πνµ

µν

νϑϑϑϑ

νΓ (6.2)

which is valid when both µ and ν exceed -1, expresses any Bessel function in terms of an integralinvolving a Bessel function of lower order. It was proved by Sonine (Watson, 1966) by expanding theintegrand in powers of z, and by simply applying the definition of the beta function (chapter 5).

=+

∫ +++ 2/

0

1211

dcossin)sinz(J)1(2

z πνµ

µν

νϑϑϑϑ

νΓ

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DEFINITE INTEGRALS OF BESSEL FUNCTIONS

47

∑∫

+++−=

++

++++++

)1()1k(!k2

dcossinz

)(k2

2/

0

121k221k2

k

νΓµΓ

ϑϑϑ

νµ

πνµνµ

∑++++++

+++−= +++

+++

)2k()1()1k(!k2

)1()1k(z)( 1k2

1k2k

νµΓνΓµΓ

νΓµΓνµ

νµ

)z(J 1++= νµ

6.4 Sonine’s second finite integral

The integral:

∫ ++++++

+

+=

2/

02/)1(22

2/122111

)yx(

)yx(Jyxdcossin)cosy(J)sinx(J

π

νµνµ

νµνµ

νµ ϑϑϑϑϑ

(6.3)which is valid when both µ and ν exceed -1, is also due to Sonine (Watson, 1966).

It is also proved by expanding the integrand in powers of x and y.

∫ ++2/

0

11 dcossin)cosy(J)sinx(Jπ

νµνµ ϑϑϑϑϑ

( ) ( )∫ ∑ ∑

=

=

+++++

++++−=

2/

0 0k 0j

11j2k2jk d

)1j()1k(!j!kcossin2/cosy2/sinx

)(π νµνµ

ϑνΓµΓ

ϑϑϑϑ

∑ ++++++++

−=++

+)1j()1k(!j!k

)1j,1k(B)2/y()2/x()(

21 j2k2

jkνΓµΓ

νµνµ

∑ ++++−=

+++

)2jk(!j!k)2/y()2/x(

)(21 j2k2

jkνµΓ

νµ

( )[ ] ( )[ ]∑ ++++−= +

)2jk(!j!k2/y(2/x

)()2/y()2/x(21

j2k2jk

νµΓνµ

Applying the binomial theorem yields

( ) ( )[ ] ( )[ ] ( )[ ]∑++ +

=+jk

0

j2k2jk22 2/y2/x!j!k)!jk(

2/y2/x

Hence, with m = k + j

∫ ++2/

0

11 dcossin)cosy(J)sinx(Jπ

νµνµ ϑϑϑϑϑ

( ) ( ) ( )[ ]∑ ++++

−=)2m(!m

2/y2/x)()2/y(2/x

21

m22m

νµΓνµ

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PAVEMENT DESIGN AND EVALUATION

48

( )( )

∑ +++

+

+

=

+++

++++ )1m(!m

4/yx)(

4/yx2

)y()x(

1m222

m1

221 νµΓ

νµ

νµνµ

νµ

( )( )

+

+

= ++++21

221

21

22

yxJ

yx

yxνµνµ

νµ

6.5 Poisson’s integral

Often one wants to verify from which value of the argument the asymptotic expansions, developed inchapter 4, can be considered as sufficiently accurate. Poisson’s integral (Watson, 1966) brings a fullytrustworthy answer to this question. Indeed, any Bessel function of the first kind can be expressed as

∫+=

2/

0

2 dcos)sinxcos()2/1()2/1(

)2/x(2)x(J

πν

ν

ν ϑϑϑΓνΓ

(6.4)

provided that ν > - 1/2. The bounds being finite, this integral can safely be numerically integrated.

Equation (6.4) can also be written as

∫+=

2/

0

2 dsin)cosxcos()2/1()2/1(

)2/x(2)x(J

πν

ν

ν ϑϑϑΓνΓ

(6.5)

To prove (6.5), expand cos(xcosθ) as a cos series

( )∫ ∑ −

+

2/

0

2k2k2k d

)!k2(sincosx

)()2/1()2/1(

2/x2 π ννϑ

ϑϑΓνΓ

Apply the beta function

( )∑ ++

++−

+=

)1k()!k2()2/1()2/1k(x

)()2/1()2/1(

2/x k2k

νΓνΓΓ

ΓνΓ

ν

Apply the duplication formula (5.5) for gamma functions

( ) ∑−

++−−

−=1k2k2

k21

)1k()!k2()!1k()2/1()!1k2(x

)()2/1(

2/xνΓ

ΓΓ

ν

( )∑ ++

−=+

)1k(!k2/x

)(k2

kνΓ

ν

which is the definition of Jν(x).

One realises that when (6.5) is proved, (6.4) is automatically proved too, because of the properties of thebeta integral.

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49

Chapter 7 The hypergeometric type of series

7.1 Introductory note

The hypergeometric function of Gauss is a solution of a differential equation similar to the Besselequation. It can be expressed as a product of multiple products, sort of truncated Γ functions. Thefunction is useful in the expression of solutions of infinite integrals of Bessel functions, which forparticular values of the parameters can then be written in simple closed-forms.

7.2 Helpful relations

( ) )1n)...(2)(1(n −+++= ααααα

( ) 10 =α

[ ] ( ) ( ) ( )( ) ( ) ( )∑

==

0n

n

nqn2n1

npn2n1q21p21qp z

...!n

...z;,...,;,...,F

ρρρ

αααρρρααα

[ ] ∑= n

n

nn12 z

)c(!n)b()a(

z;c;b,aF

mnn

mn )mb()1()nb1( −− +−=−−

m

mmm2

m)1ba(2

2ba2

1ba

2

)m1ba(++

++

++

=+++

n

n)1a(

)a()1()na(

+−−=+−

ΓΓ

n

n

0mmnmmn )ba()b()a(C +=∑

=−

[ ])ac()bc()abc()c(

1;c;b,aF12 −−−−

=ΓΓ

ΓΓ

[ ] a12 )z1(z;b;b,aF −−=

[ ] [ ] 1z,c;0,bFz;c;b,0F 1212 ==

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PAVEMENT DESIGN AND EVALUATION

50

[ ]( )∫

−−

=−−−1

0a

1bc1b

12 dxzx1

)x1(x)bc()b(

)c(z;c;b,aF

ΓΓΓ

[ ]

−−−

−=

z1z

;c;bc,aF)z1(

1z;c;b,aF 12a12

7.3 Definition

The series of hypergeometric type are solutions of differential equations that will not be handled here.However they allow to express a great number of relations in a convenient compact notation, invented byPochhammer (Watson, 1966). Further, their properties, especially those of the series 2F1, also called thehypergeometric series of Gauss, are very helpful in the resolution of infinite integrals of Bessel functions.

We write: ( ) 1)()1n)...(2)(1( 0n =−+++= αααααα (7.1)

and define the generalised hypergeometric function

[ ]!1z

...

...1z;,...,;,...,F

q21

p21q21p21qp ρρρ

αααρρρααα +=

...!2

z)1()...1()1(

)1()...1()1( 2

qq2211

pp2211+

+++

++++

ρρρρρρ

αααααα

[ ] ( ) ( ) ( )( ) ( ) ( )∑

==

0n

n

nqn2n1

npn2n1q21p21qp z

...!n

...z;,...,;,...,F

ρρρ

αααρρρααα (7.2)

As an example, it is evident that

( )( )

−+

+=

2

10 2z

;1F1

2/z)z(J ν

νΓ

ν

ν

If αI or ρI are neither zero neither negative integers, the series can be written as

[ ]( ) ( )

( ) ( )∑

∏∞

=−

=

=

++

++

+=1n

n

1n

0kq1

1n

0kp1

q21p21qp !nz

k...k

k...k

1z;,...,;,...,F

ρρ

αα

ρρρααα

( ) ( )( ) ( )

( )( ) ( )( )

( )( ) ( ) ( ) !nz

k...k

k...k

...

...1

n

1n1n

0k

1n

0kqq11

1n

0k

1n

0kpp11

p1

q1 ∑∏ ∏

∏ ∏∞

=−

=

=

=

=

++

++

+=

ρΓρΓρρΓ

ααΓααΓ

αΓαΓ

ρΓρΓ

[ ] ( ) ( )( ) ( )

( ) ( )( ) [ ] !n

zn...n

n...n

...

...z;,...,;,...,F

n

0n q1

p1

p1

q1q21p21qp ∑

= ++

++=

ρΓρΓ

αΓαΓ

αΓαΓ

ρΓρΓρρρααα (7.3)

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51

If one of the parameters αI is either zero either an integer, the series reduces in a finite polynomial.The most common series is the so called hypergeometric function of Gauss, 2F1(a,b;c;z).

7.4 Properties of the multiple product (α)m

7.4.1 A relation for (1 - b - m)m-n

We show that:

nmm

nm )nb()1()mb1( −− +−=−−Indeed

)1nmmb1)(2nmmb1)...(1mb1)(mb1()mb1( nm −−+−−−−+−−+−−−−=−− −)nb)(1nb)...(mb2)(mb1()mb1( nm −−−−−−−−−=−− −

)1mb)(2mb)...(1nb)(nb()1()mb1( nmnm −+−++++−=−− −

nmnm

nm )nb()1()mb1( −−

− +−=−− (7.4)When n = 0, (7.4) simplifies in

mm

m )b()1()mb1( −=−− (7.5)

7.4.2 A relation for (a+b+1+m)m/22m

We show that:

m

mmm2

m)1ba(

22ba

21ba

2

)m1ba(++

++

++

=+++

Develop the series

...2

)4ba(

2

)3ba(

2

)2ba(1

2

)m1ba(6

34

22

1m2

m +++

+++

+++

+=+++

642 2

)6ba)(5ba)(4ba(

2

)4ba)(3ba(

2

)2ba(1

+++++++

+++++

+++=

+++++

+

++

++

+

++

++

+++

++

++

+=)2ba)(1ba(

12

2ba2

2ba1

21ba

21ba

)1ba(2

2ba2

1ba

1

)3ba)(2ba)(1ba(

22

2ba1

22ba

22ba

22

1ba1

21ba

21ba

++++++

+

++

+

++

++

+

++

+

++

++

+

Hence

m

mmm2

m)1ba(

22ba

21ba

2

)m1ba(++

++

++

=+++

(7.6)

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52

7.4.3 A relation for Γ(a-n)

By definition of the gamma function, we have:)na()na)...(2a)(1a()a( −−−−= ΓΓ

)na()na)...(2a)(1a()1( n −+−+−+−−= Γ

)na()1a()1( nn −+−−= Γ

Hence

n

n)1a(

)a()1()na(

+−−=−

ΓΓ (7.7)

7.4.4 The theorem of Vandermonde

The theorem of Vandermonde (Watson, 1966) states that:

m

m

0nnmnnm )ba()b()a(C +=∑

=−

where mCn is the symbol for the combination of m elements in groups of n elements. We shall prove therelation for m = 3; however, the proof can easily be extended to any value of m.

∑=

− =3

0nn3nn3 )b()a(C

)1b)(b)(a(3)2b)(1b)(b( ++++=+++++ )2a)(1a)(a()b)(1a)(a(3

)1b)(b)(a()2b)(1b)(b( ++++)b)(1a)(a(2)1b)(b)(a(2 ++++=+++++ )2a)(1a)(a()b)(1a)(a(

[ ] =++++++ )1a(aab2)1b(b)2ba(

[ ]=+−−−−++++− )1ab)(ab()ab)(2b(2)2b)(1b(a

3)ba()ba)(1ba)(2ba( +=+++++and

∑=

− +=3

0n3n3nn3 )ba()b()a(C

Hence, using the same procedure, we prove that

m

m

0nnmnnm )ba()b()a(C +=∑

=− (7.8)

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53

7.4.5 The product of two Bessel functions with the same argument

Consider the product of two Bessel functions with the same argument (at)

∑ ∑∞

=

=

++

++−⋅

++−=⋅

0m 0n

n2n

m2m

)1n(!n)2/at(

)1()1m(!m

)2/at()1()at(J)at(J

νΓµΓ

νµ

νµ

The coefficient of (-)m(at/2)µ+ν+2m , obtained from the sum of next products,

∑=

+−+−

++−

+−+−−

m

0n

n2n

n2m2nm

)1n(!n)at(

)1()1nm()!nm(

)2/at()1(

νΓµΓ

νµ

is

∑= +++−+−

m

0n )1n(!n)1nm()!nm(1

νΓµΓ

∑= +++−+

++++⋅

−⋅

++++=

m

0n )1n()1nm()1m()1m(

!n)!nm(!m

)1m()1m(!m1

νΓµΓνΓµΓ

νΓµΓ

∗++++

= ∑= )1m()1m(!m

Cnmm

0n νΓµΓ

[ ] [ ])1n()1nm(

)1n()1n)....(m()1nm()1nm)...(m(+++−+

++++++−++−++∗

νΓµΓνΓννµΓµµ

[ ][ ])1nmm)...(m()1nm))...(m()1m()1m(!m

Cnmm

0n++−+++−++

++++= ∑

=ννµµ

νΓµΓ

[ ] [ ]∑=

++++−−+−−−−−−+−−−−−

=m

0n

nmn

nm )1m()1m(!m)1nmm)...(m()()1nm)...(m()(

CνΓµΓ

ννµµ

∑=

−++++

−−−−−=

m

0n

nmnm

nm )1m()1m(!m)m()m()(

CνΓµΓνµ

by (7.8)

)1m()1m(!m)m2()( m

m

++++−−−−

=νΓµΓ

νµ

by (7.5)

)1m()1m(!m)1m( m

+++++++

=νΓµΓ

νµ

Hence

∑∞ ++

+++++++

−=0

mm2

m)1m()1m(!m)1m()2/at(

)()at(J)at(JνΓµΓνµνµ

νµ (7.9)

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54

7.5 The hypergeometric series of Gauss 2F1[a,b;c;z]

The series 2F1[a,b;c;z], usually noted F[a,b;c;z], is a solution of the differential equation (Wayland, 1970)

[ ] 0abydzdy

z)1ba(cdz

yd)z1(z

2

2=−++−+−

as can be verified by substitution.

7.5.1 Elementary properties

If a, b or c are neither zero neither negative integers, the series can be written as (7.3)

[ ]!n

z)nc(

)nb()na()b()a(

)c(z;c;b,aF

n

0∑∞

+++

ΓΓΓΓ

Γ (7.10)

When a or b are equal zero or a negative integer, the series reduces to a polynomial.

When a = 0[ ] 1z;c;b,0F = (7.11)

When a = -1

[ ] zcb

1z;c;b,1F −=− (7.12)

Next equation is always true[ ] [ ]z;c;a,bFz;c;b,aF = (7.13)

Consider the expansion as a Taylor series around z = 0 of the function (1 - z)-a

...!3

z)2a)(1a(a

!2z

)1a(a!1z

a1)z1(32

a +++++++=− −

...z)2c)(1c(c!3

)2c)(1c(c)2a)(1a(az

)1c(c!2)1c(c)1a(a

zc!1ca

1 32 +++

+++++

+++

+⋅

+=

[ ]z;c;c,aF=Hence, whatever c

[ ] a)z1(z;c;c,aF −−= (7.14)

7.5.2 Integral representation of the hypergeometric function

We show here how the hypergeometric series can be represented as a definite integral. In several cases,the solution of this integral allows to express the infinite series as a closed form analytical expression. Westart with equation (7.10):

[ ]!n

z)nc(

)nb()na()b()a(

)c(z;c;b,aF

n

0∑∞

+++

ΓΓΓΓ

Γ

!n

z)bc()nc(

)bc()nb()na()b()a(

)c( n

0n∑∞

= −+−++

=ΓΓ

ΓΓΓΓΓ

Γ

∑∫∞

=

−−−+ +−−

=0n

1

0

n1bc1nb dx

!nz

)na()x1(x)bc()b()a(

)c(Γ

ΓΓΓΓ

Page 69: Pavement design and evaluation

THE HYPERGEOMETRIC TYPE OF SERIES

55

∫ ∑∞

=

−−− +−−

=1

0 0n

n1bc1b dx

!n)zx(

)na()x1(x)bc()b()a(

)c(Γ

ΓΓΓΓ

Expand the sum

∑∞

=+++++=+

0n

2n...

!2)zx(

)2a(!1

zx)1a()a(

!n)zx(

)na( ΓΓΓΓ

++++= ...

!2)zx(

)1a(a!1

zxa1)a(

[ ] a)zx1)(a(zx;c;c,aF)a( −−== ΓΓHence

[ ] ∫−

−−

=−−−1

0a

1bc1bdx

)zx1(

)x1(x)bc()b(

)c(z;c;b,aF

ΓΓΓ

(7.15)

7.5.3 Value of F[a,b;c;z] for z = 1

[ ] ∫−

−−

=−−−1

0a

1bc1bdx

)x1(

)x1(x)bc()b(

)c(1;c;b,aF

ΓΓΓ

∫ −−−− −−

=1

0

1abc1b dx)x1(x)bc()b(

)c(ΓΓ

Γ

)abc,b()bc()b(

)c(−−

−= Β

ΓΓΓ

[ ])ac()bc()abc()c(

1;c;b,aF−−−−

=ΓΓ

ΓΓ (7.16)

7.5.4 Convergence of the series F[a,b;c;z]

When z is positive, the series F[a,b;c;z] converges only when z is smaller than 1. When z is negative, theseries may converge for z> 1, but the result remains indefinite when strictly applying the definition:

[ ] ...z)2c)(1c(c!3

)2b)(1b(b)2a)(1a(az

)1c(c!2)1b(b)1a(a

z!1ba

1z;c;b,aF 32 +++

++++−

+++

+⋅

−=−

The problem can be solved by a transformation of the integral form.

[ ] ∫−

−−

=−−−1

0a

1bc1bdx

)zx1(

)x1(x)bc()b(

)c(z;c;b,aF

ΓΓΓ

Write y = 1 -x, dy = - dx

[ ] [ ]a

aaaa yz1

z1)z1(zyz1)y1(z1)zx1(

−+−=+−=−−=−

Page 70: Pavement design and evaluation

PAVEMENT DESIGN AND EVALUATION

56

[ ] ∫

−+

−−=

−−−1

0a

1bc1b

ady

yz1

z1

y)y1(

)z1(

1)bc()b(

)c(z;c;b,aF

ΓΓΓ

−+

−−=

−−−1

0a

1b1bc

ady

yz1

z1

)y1(y

)z1(

1)bc()b(

)c(ΓΓ

Γ

[ ]

−−−

−=

z1z

;c;bc,aF)z1(

1z;c;b,aF a

(7.17)

When z < - 1, 0 < -z/(1-z) < 1, and the series converges.

7.5.5 The product of two Bessel functions with different arguments

Consider the product of two Bessel functions with different arguments

∑ ∑∞

=

=

++

++−⋅

++−=⋅

0m 0n

n2n

m2m

)1n(!n)2/bt(

)1()1m(!m

)2/at()1()bt(J)at(J

νΓµΓ

νµ

νµ

The coefficient of (-)maµbν(t/2)µ+ν+2m , obtained from the sum of next products,

∑=

+−+−

++−

+−+−−

m

0n

n2n

n2m2nm

)1n(!n)bt(

)1()1nm()!nm(

)2/at()1(

νΓµΓ

νµ

is

∑=

+++−+−

m

0n

n2n2m2

)1n(!n)1nm()!nm(ba

νΓµΓ

[ ][ ]∑

=

++−+++−+++−+−++−+−++

=m

0n

n2m2

ab

)1()1)...(1n)(n()1nm()1m(!n)!nm()1nm()1nm)...(1m)(m(

aνΓνννµΓµΓ

µΓµµµ

[ ] n2m

0n n

nm2

ab

)1)(1()1m(!n)!nm()1nm)...(1m)(m()1(

a

++++−−+−−+−−−−−

= ∑= ννΓµΓ

µµµ

[ ] n2m

0n n

nn

m2ab

)1)(1()1m(!n!m)nm)...(1m(m)m()1(

a

++++−−−−−

= ∑= ννΓµΓ

µ

∑=

++++−−−

=m

0n

n

2

2

n

nnm2

a

b)1(!n)1()1m(!m

)m()m(a

ννΓµΓµ

[ ])1()1(!

/;1;, 222

+Γ++Γ+−−−

=νµ

νµmm

abmmFa m

Hence

[ ]∑∞

=+ ++Γ

+−−−−⋅

+Γ=

0

222

)1(!/;1;,)2/(

)()1(2

)()()()(

m

mm

mmabmmFatbtat

btJatJµ

νµννµ

νµ

νµ (7.18)

Page 71: Pavement design and evaluation

INFINITE INTEGRALS OF BESSEL FUNCTIONS

57

Chapter 8 Infinite Integrals of Bessel Functions.

8.1 Introductory note.

This chapter presents the most important infinite integrals of Bessel functions of direct application inpavement analysis, especially in multilayer theory.

8.2 Useful relations

−+

+++

+

+=∫

+−−

2

2

0

1at

a

b;1;

21

,2

F)1(a

)()2/b(dtt)bt(Je ν

νµνµ

νΓ

νµΓνµ

νµ

ν

( )

++

+−+

++

+=∫

+−−

22

2

0 222

1at

ba

b;1;

21

,2

F

)1(ba

)()2/b(dtt)bt(Je ν

νµνµ

νΓ

νµΓνµ

νµ

ν

∫∞

=0

0 b1

dt)bt(J

∫∞

=0

1 1dtt

)bt(J

∫∞

=0

1 b1

dt)bt(J

( )∫∞

+=

02/122

0at

ba

1dt)bt(Je

( )∫∞

+=

02/322

0at

ba

atdt)bt(Je

( )∫∞

+

−=

02/522

222

0at

ba

ba2dtt)bt(Je

( )∫∞

+−=

02/122

1at

ba

a1

b1

dt)bt(Je

Page 72: Pavement design and evaluation

PAVEMENT DESIGN AND EVALUATION

58

( )∫∞

+=

02/322

1at

ba

btdt)bt(Je

( )∫∞

+=

02/522

21

at

ba

ab3dtt)bt(Je

∫∞

+=

022

at

ba

adt)btcos(e

∫∞

+=

022

at

ba

bdt)btsin(e

∫∞

+=

022

at

b4a

bdt)btsin()btcos(e

= −

∞−∫ a

btandt

t)btsin(

e 1

0

at

( )∫∞

+=

0222

at

ba

ab2tdt)btsin(e

)12(a

)2()bc(dtt)ct(J)bt(Je

20

1at

+

+=

+

∞−−∫

νΓπ

νµΓνµ

νµ

νν

−+−+

+++πν φφ

φν

νµνµ

0

22

22dsin

a

cosbc2cb;1;

212

,2

2F

∫∞

++

−+=

022

22at

)cb(a

)cb(alog

41

dtt

)ctsin()btsin(e

∫∞

++−

−+=

02222

at

)cb(a

1

)cb(a

12a

dt)ctsin()btsin(e

Page 73: Pavement design and evaluation

INFINITE INTEGRALS OF BESSEL FUNCTIONS

59

∫ ∫∞∞

0 0

ma dsdtets

tcsintbsin2/1222 )cba(a

bcarctan

2 ++=

π

∫ ∫∞∞

0 0

ma dsdtets

tcsintbsinm

2/12222222

222

))()(()2(

2 cbacabacbabc

++++++

∫∞

+++−−

+=

0

1at

a)1(2

cbdtt)ct(J)bt(Je

λνµνµ

νµλ

νµνΓ

m

2

2

2

2

0 a4

b

b

c;1;m,mF

)1m(!m)m2(

+−−−

+++++∑

∞νµ

µΓλνµΓ

∫∞

+−

++−

+

+−+

=0

1

21

)1(

21

2a

bdt

t

)bt(J)at(J

λνµΓνΓ

λνµΓ

λλν

ν

λνµ

+

+−−+−+2

2

a

b;1;

21

,2

1F ν

λµνλνµ if b < a

∫∞ −

++−

+++

++−

+−+

=0

1

21

21

21

)(2

1

2

adt

t

)at(J)at(J

λµνΓ

λνµΓ

λνµΓ

λΓλνµ

Γ

λ

λ

λνµ

∫∞

=0

dtt

)bt(J)at(Jλ

νν

( ) ( )

++

+−+−

++

+

+−

+− 222

22

212

22 ba

ba4;1;

432

,4

12F

ba)1(2

12

212

)ab(ν

λνλν

νΓλ

Γ

λνΓ

λνλ

ν

Page 74: Pavement design and evaluation

PAVEMENT DESIGN AND EVALUATION

60

8.3 The integral ∫e-atJν(bt)tµ -1dt

8.3.1 Resolution of the integral

Expanding the Bessel function in its series solves the integral

∑ ∫∫∞ ∞

−−+++∞

−−++

−=0 0

at1k2k2

k

0

1at dtet)1k(!k

)2/b()(dtt)bt(Je νµ

νµ

ν νΓ

Apply equation (2.1) of the gamma function

∑∫∞

++

+∞−−

++

++−=

0k2

k2k

0

1at

a)1k(!k

)k2()2/b()(dtt)bt(Je

νµ

νµ

ννΓ

νµΓ

Expand the series so as to obtain

∫∞

+−−

+

+++

−+

+=

02

21at

a

b)1(!1

21

21a)1(

)()2/b(dtt)bt(Je

ν

νµνµ

νΓ

νµΓνµ

νµ

ν

−++

+

++

++

+

+

+

+ ...a

b)2)(1(!2

12

12

11

224

4

νν

νµνµνµνµ

and by (7.11)

∫∞

+−−

−+

+++

+

+=

02

21at

a

b;1;

21

,2

Fa)1(

)()2/b(dtt)bt(Je ν

νµνµ

νΓ

νµΓνµ

νµ

ν (8.1)

Strictly spoken, the series converges for b < a only. Therefore transform the series by application of theintegral transform of the hypergeometric function (7.17):

( )∫∞

+−−

++

+−+

++

+=

022

2

222

1at

ba

b;1;

21

,2

F

)1(ba

)()2/b(dtt)bt(Je ν

νµνµ

νΓ

νµΓνµ

νµ

ν (8.2)

Convergence at the origin (t = 0) requires µ + ν > 0. Indeed, expand the integrand in series

dtt...)1(

)2/bt(...

!2ta

at1dtt)bt(Je0

1

0

221at∫ ∫

∞−

∞−−

+

−+−= µ

νµ

ν νΓ

Integration of the first term of the series leads to

∫∞ ∞

+−+++

=+

0 0

1 t)1()(

)2/b(dtt

)1()2/b( νµ

ννµ

ν

νΓνµνΓ

Convergence is secured for t = 0 if µ + ν > 0.

Page 75: Pavement design and evaluation

INFINITE INTEGRALS OF BESSEL FUNCTIONS

61

8.3.2 Particular value.

We have seen in § 7.5.1 that for particular values of its parameters, the hypergeometric function reducesto simple expressions.

The case where µ = ν + 1.

1ba

b;1;0,

2F 22

2

=

+

++

ννµ

Hence

( )∫∞

+−

++

+=

0 212

22

at

)1(ba

)12()2/b(dtt)bt(Je

νΓ

νΓν

νν

ν (8.3)

∫∞

+=

0220

at

ba

1dt)bt(Je (8.4)

( )∫∞

+=

02/322

1at

ba

btdt)bt(Je (8.5)

∫∞

=0

0 b1

dt)bt(J (8.6)

∫∞

=0

21b

1tdt)bt(J (8.7)

The case where (µ + ν)/2 = ν + 1.

2222

2

22

2

ba

aba

b;1;1,

21

Fba

b;1;

21

,1F+

=

+

++−=

+

+−+ νννν

Hence

( )∫∞

++−

++

+=

0 232

22

1at

)1(ba

)22()2/b(adtt)bt(Je

νΓ

νΓν

νν

ν (8.8)

( )∫∞

+=

02/322

0at

ba

atdt)bt(Je (8.9)

( )∫∞

+=

02/522

21

at

ba

ab3dtt)bt(Je (8.10)

Page 76: Pavement design and evaluation

PAVEMENT DESIGN AND EVALUATION

62

The case where µ = 3 and ν = 0.

(1 - µ + ν)/2 = -1, and ( )22

2

22

2

ba2b3

1ba

b;1;1,

23

F+

−=

+

− .

Hence

( ) 2/522

22

0

20

at

ba

ba2dtt)bt(Je

+

−=∫

∞− (8.11)

The case where the integral representation reduces to a simple integral.

µ = 1 and ν = 1

[ ]∫ −−=−

−=

=

1

02/1

20

z11z2

dx)zx1(

)x1(x)1()1(

)2(z;2;1,

21

Fz;2;21

,1FΓΓ

Γ

Hence

∫∞

++=

022

2

221at

ba

b;2;

21

,1Fba

2/bdt)bt(Je

( )∫∞

+−=

+−−

+

+=

022

2/1

22

2

2

22

221at

ba

a1

b1

ba

b11

b

ba2

ba

2/bdt)bt(Je (8.12)

∫∞

=0

1 b1

dt)bt(J (8.13)

µ = 0 and ν = 1

[ ]∫ −−=−

−=

1

02/1

20z11

z2

dx)zx1(

)x1(x)1()1(

)2(z;2;1,

21

FΓΓ

Γ

( )

−+=

++=∫

∞− aba

b1

ba

b;2;1,

21

F)2(ba

)2()2/b(dt

t)bt(J

e 22

022

2

2/122

1at

Γ

Γ (8.14)

∫∞

=0

1 1dtt

)bt(J (8.15)

µ = ν = 1/2

Apply (7.16)

=

+

=

−=

− −

−1

0

1

2

2

2/1

2

2

2

2

ab

tanba

dx

a

xb1

x)1()2/1(

)2/3(

a

b;

23

;21

,1Fa

b;

23

;1,21

FΓΓ

Γ

Page 77: Pavement design and evaluation

INFINITE INTEGRALS OF BESSEL FUNCTIONS

63

=

−= −

∞−−∫ a

btan

b2

a

b;

23

;1,21

F)2/3(a

)1()2/b(dtt)bt(Je 1

02

22/12/1

2/1at

πΓΓ

(8.16)

8.3.3 The integral ∫ e-atcos(bt)dt

This integral, together with the sin integral, occurs very often in physical problems. The solution is anapplication of (8.8). By (4.9), transform the cos function in the corresponding Bessel function

∫∫∞

−−

∞− =

0

2/12/1

at

0

at dtt)bt(Je2b

dt)btcos(eπ

Hence by (8.8) with ν = 1/2

( )∫∞ −

+=

+=

02222

2/1at

ba

a

)2/1(ba

)1()2/b(a2b

dt)btcos(eΓ

Γπ (8.17)

8.3.4 The integral ∫ e-atsin(bt)dt

By (4.8) transform the sin function in the corresponding Bessel function

∫∫∞

−∞

− =0

2/12/1

at

0

at dtt)bt(Je2b

dt)btsin(eπ

Hence by (8.3)

( )∫∞

+=

+=

02222

2/1at

ba

b

)2/3(ba

)2()2/b(2b

dt)btsin(eΓ

Γπ (8.18)

8.3.5 The integral ∫e-atcos(bt)sin(bt) dt

This integral can immediately be linked with (8.18)

∫∫∞

−∞

+==

022

at

0

at

b4a

bdt)bt2sin(e

21

dt)btsin()btcos(e (8.19)

8.3.6 The integral ∫e-at sin(bt)/tdt

By (4.8)

∫ ∫∞ ∞

−−− =0 0

2/12/1

atat dtt)bt(Je2b

dtt

)btsin(e

π

By (8.16)

=

= −

∞−−∫ a

btan

ab

tanb2

2b

dtt

)btsin(e 1

0

1atπ

π (8.20)

Page 78: Pavement design and evaluation

PAVEMENT DESIGN AND EVALUATION

64

8.3.7 The integral ∫e-at sin(bt)tdt

By equation (4.8)

∫ ∫∞ ∞

−− =0 0

2/32/1

atat dtt)bt(Je2b

tdt)btsin(eπ

By equation (8.2)

( )∫∞

+−

+=

022

2

2/322

2/12/3

2/1at

ba

b;

23

;21

,23

F)2/3(ba

)3()2/b(dtt)bt(Je

Γ

Γ

By equation (7.15)

( ) 2/12222

2

ba

a

ba

b;

23

;23

,21

F+

=

+−

Hence

( )∫∞

+=

0222

at

ba

ab2tdt)btsin(e (8.21)

8.4 The integral ∫e-atJν(bt)Jν(ct)tµ -1dt

8.4.1 Transformation of the integral

The integral can be transformed using Gegenbauer’s integral (6.1).

∫ ∫∫∞

−+−∞

−−

+=

0 0

21at

0

1at dtdsint)t(J

e)2/1()2/1(

)2/bc(dtt)ct(J)bt(Je

πννµ

νν

νµ

νν φφω

ωΓνΓ

−+

+++

+

+=

+

πν

νµ

νφφ

ων

νµνµ

νΓπ

νµΓ

0

22

2

2dsin

a;1;

212

,2

2F

)12(a

)2()bc( (8.22)

where φω cosbc2cb 222 −+= .The hypergeometric function reduces to an elementary function if µ = 1 or 2.

8.4.2 The integral ∫e-atsin(bt)sin(ct)t-1dt

Transform the integral by equation (5.8)

∫ ∫∞ ∞

−− =0 0

2/12/1atat dt)ct(J)bt(Je

2bcdt

t)ctsin()btsin(

Apply equation (8.22) with µ = 1 and ν = 1/2

∫ ∫∞

−=

0 02

2

2at dsin

a;

23

;23

,1Fa2

bcdt

t)ctsin()btsin(

e φφωπ

Apply equation (7.14)

Page 79: Pavement design and evaluation

INFINITE INTEGRALS OF BESSEL FUNCTIONS

65

∫ ∫∫−

∞−

++=

−++=

π

φφ

φφ

0

1

1222222

0

at

xbc2

cba

dx41

dcosbc2cba

dsin2bc

dtt

)ctsin()btsin(e

∫∞

−+

++=

022

22at

)cb(a

)cb(alog

41

dtt

)ctsin()btsin(e (8.23)

8.4.3 The integral ∫e-atsin(bt)sin(ct)dt

∫ ∫∞ ∞

−− =0 0

2/12/1atat tdt)ct(J)bt(Je

2bcdt)ctsin()btsin(e

π

Apply equation (8.22) with µ = 2 and ν = 1/2

∫ ∫∞

−=

0 02

2

3at dsin

a;

23

;2,23

Fa

bcdt)ctsin()btsin(e φφ

ωπ

Apply equation (7.14)

∫∫ −++=

∞−

π

φφ

φφ

02222

0

at d)cosbc2cba(

dsinabcdt)ctsin()btsin(e

( )∫∫−

∞−

−++=

1

122220

at

bcx2cba

dxabcdt)ctsin()btsin(e

∫∞

++−

−+=

02222

at

)cb(a

1

)cb(a

12a

dt)ctsin()btsin(e (8.24)

8.4.4 The integral ∫∫e-amsin(bt)sin(cs)/(ts)dsdt

Consider the double integral

∫ ∫∞∞

−=0 0

madsdtets

tcsintbsinI

where m2 = t2 + s2 Let θρ cost = and θρ sins = .

θρθθρ

θρθρπρ dde

sincos)sincsin()cosbsin(

dsdtets

tcsintbsin

0

2/

0 0

a

0

ma∫ ∫ ∫∫∞ ∞

−∞

− =

By (8.23)

θθθθθ

θθθθθθ

π

dsincosbc2sinccosba

sincosbc2sinccosbalog

sincos1

41

I2/

022222

22222

−++

+++=

which we rewrite

Page 80: Pavement design and evaluation

PAVEMENT DESIGN AND EVALUATION

66

θ

θθ

θθθθ

θθ

θθ

πd

sin)ca(cos)ba(

sincosbc21

sin)ca(cos)ba(

sincosbc21

logsincos

141

I

222222

2222222/

0

+++−

++++

= ∫

Expand the log function as a Taylor series, with z < 1

⋅⋅⋅++=

−+

5z

3z

z2z1z1

log53

( ) ( ) θθθ

θθθθ

π

d51

31

sin)ca(cos)ba(

sincosbc2sincos

241

I2/

0

53222222∫

⋅⋅⋅+++

+++=

The integral

[ ]θ

θθ

θθπ

dsin)ca(cos)ba(

)sin(cosI

2/

0n222222

1n

∫+++

=−

can be solved by letting u = tanθ.

[ ]du

u)ca()ba(

u)bc2(1k2

121

I0 0k

1k222222

k21k2

∫ ∑∞ ∞

=+

+

++++=

Using

( ) ( ) ( )∫ ∫∞ ∞ −

∞−

++

−+

+−=

+0 0k2222

2k2

20

k22222

1k2

1k2222

k2du

u

u)1k2(

k4

1

uk4

udu

u

u

βαββαββα

( ) ( )∫ ∫∞ ∞ −

++

−=

+0 0k2222

2k2

21k2222

k2du

u

u)1k2(

k4

1du

u

u

βαββα

( ) 21u

arctan1

u

u

00222

παβα

βαββα

==+

∞∞

[ ] [ ]

+

+++

++=

2/32222

33

2/12222 )ca)(ba(

cb3.2

1

)ca)(ba(

bc2

[ ]

⋅⋅⋅+

+++

2/52222

55

)ca)(ba(

cb5.4.2

3.1

Hence

++=

2/1222/122 )ca()ba(

bcarcsin

2I

π (8.25)

or

Page 81: Pavement design and evaluation

INFINITE INTEGRALS OF BESSEL FUNCTIONS

67

2/1222 )cba(a

bcarctan

2I

++=

π (8.26)

8.4.5 The integral ∫∫me-amsin(bt)sin(cs)/(ts)dsdt

Consider the double integral

∫ ∫∞∞

−=0 0

madsdtets

scsintbsinmI

where m2 = t2 + s2

Let θρ cost = and θρ sins =

θρθθ

θρθρπρ dde

sincos)sincsin()cosbsin(

dsdtets

scsintbsinm

0

2/

0 0

a

0

ma∫ ∫ ∫∫∞ ∞

−∞

− =

By (8.24)

θθθθθθθ

π

d)sinccosb(a

1

)sinccosb(a

1sincos

1a

21

I2222

2/

0

++−

−+= ∫

Let u = tanθ

[ ]∫∞

++−+

+=

022222

2du

u)ca(bcu2)ba(u

u12a

I

[ ]∫∞

++++

+−

022222

2du

u)ca(bcu2)ba(u

u12a

++++

++−+

+=

022222

22222

22 u)ca(bcu2ba

u)ca(bcu2balog

)ca(4

aI

++++

++−+

++

022222

22222

22 u)ca(bcu2ba

u)ca(bcu2balog

)ba(4

a

( )•

++

++

++

2/12222222cbaa

bc

ca

1

ba

12a

++

+++

++

−+

02/1222

22

2/1222

22

)cba(a

bcu)ca(arctan

)cba(a

bcu)ca(arctan

2/12222222

222

)cba)(ca)(ba(

)cba2(bc2

I++++

++=

π (8.27)

Page 82: Pavement design and evaluation

PAVEMENT DESIGN AND EVALUATION

68

8.5 The integral ∫e-atJµ (bt)Jν(ct)tλ-1dt

The integral is solved using equation (7.18) established for the product of two Bessel functions withdifferent arguments.

∫∞

−− =0

1at dtt)ct(J)bt(Je λνµ

[ ]∫∑∞

−∞ −+++

+ +++−−−−

+ 0

at

0

221m2m2mdte

)1m(!mb/c;1;m,mFt)2/b()(

)1(2

cbµΓ

νµ

νΓ

λνµ

νµ

νµ

Apply the definition (2.1) of the gamma function

∫∞

+++−− ×

+=

0

1at

a)1(2

cbdtt)ct(J)bt(Je λνµνµ

νµλ

νµνΓ

m

2

2

2

2

0 a4

b

b

c;1;m,mF

)1m(!m)m2(

+−−−

+++++

× ∑∞

νµµΓ

λνµΓ (8.28)

8.6 The discontinuous integral ∫Jµ(at)Jν(bt)t-λdt

This integral is of great importance in the resolution of physical problems because it allows expressingdiscontinuous boundary conditions. Many applications in pavement design will be based on the propertiesof this integral, the so-called discontinuous integral of Weber and Schafheitlin (Watson, 1966).

8.6.1 Resolution of the integral

The solution of the integral depends on the relative values of a and b, hence, the term discontinuous toqualify the integral. We assume that b < a, and solve the integral as the limiting function of anexponential function

∫∫∞

∞=

0

ct

0c0

dtt

)bt(J)at(Jelimdt

t

)bt(J)at(Jλ

νµλ

νµ

Expand the Bessel function with the smallest argument (here bt)

∫ ∑∫∞ −++

++−=

0

k2k2kct

0c0

dt)1k(!k

t)2/b()()at(Jelimdt

t

)bt(J)at(J

νΓ

λνν

µλνµ

∑ ∫∞

−+−+

→ ++−

=0

k2ctk2k

0cdtt)at(Je

)1k(!k)2/b()(

lim λνµ

ν

νΓApply equation (8.2) on the infinite integral

( )•

++

+−++•

++−

= ∑ +−++

+

→)1(ca

)1k2()2/a()1k(!k

)2/b()(lim

21k2

22

k2k

0cµΓ

λνµΓνΓ λνµ

µν

Page 83: Pavement design and evaluation

INFINITE INTEGRALS OF BESSEL FUNCTIONS

69

++

+−−+−++22

2

ca

a;1;

2k2

,2

1k2F µ

λνµλνµ

We now go over to limit by setting c = 0. For a proof that the limit of the hypergeometric series, when ctends to 0, is the same as the value of the series when c = 0, (see Watson, 1966). Hence

( )∑∫ +−++

+∞ +−++++

−=

21k2

2

k2k

0 a

)1k2()2/a()1k(!k

)2/b()(dt

t

)bt(J)at(Jλνµ

µν

λνµ λνµΓ

νΓ

+

+−−+−+++

1;1;2

k2,

21k2

F)1(

λνµλνµµΓ

Apply equation (7.16) on the term between accolades

∫∞

++−=

01 2a

bdt

t

)bt(J)at(Jνµλν

ν

λνµ

−++−

++−+++

++−+

−∑

2k21

2k22

)2/1()1k(!k

)k21(ab

21

)(k2k2

k

λνµΓ

λνµΓ

ΓνΓ

λνµΓ

Let p2/)k21( =++−+ λνµ

++−++

= ∑∫∞

+−k

21

)1k(!k

ab

)(

2a

bdt

t

)bt(J)at(J

k2k

01 λνµ

ΓνΓλλν

ν

λνµ

+− )2/1p(2

)2/1()p2(1p2 Γ

ΓΓ

Apply equation (5.5) on the term between accolades

∫ ∑∞

+−

++−++

+

+−+

=0

k2k

1k

21

)1k(!k

k2

1ab

)(

2a

bdt

t

)bt(J)at(J

λνµΓνΓ

λνµΓ

λλν

ν

λνµ

Apply equation (7.7) on

++−Γ k

21λνµ

∑∫

++−

++

+−−

+

+−+

= +−

21

)1k(!k

21

k2

1ab

2a

bdt

t

)bt(J)at(J k

k2

10

λνµΓνΓ

λµνλνµΓ

λλν

ν

λνµ

Apply equation (7.3) on

+

+−+k

21λνµ

Γ

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PAVEMENT DESIGN AND EVALUATION

70

++−

+

+−+

= +−

∫2

1)1(

21

2a

bdt

t

)bt(J)at(J1

0λνµ

ΓνΓ

λνµΓ

λλν

ν

λνµ

k2

k

kk

ab

)1(!k2

12

1

+

+−−

+−+

∑ ν

λµνλνµ

Hence

∫∞

+−

++−

+

+−+

=0

1

21

)1(

21

2a

bdt

t

)bt(J)at(J

λνµΓνΓ

λνµΓ

λλν

ν

λνµ

+

+−−+−+2

2

a

b;1;

21

,2

1F ν

λµνλνµ (8.29)

When a < b, one obviously obtains

∫∞

+−

++−

+

+−+

=0

1

21

)1(

21

2b

adt

t

)bt(J)at(J

λµνΓµΓ

λµνΓ

λλµ

µ

λνµ

+

+−−+−+2

2

b

a;1;

21

,2

1F µ

λνµλµν (8.30)

In order to establish the convergence conditions at the origin develop the Bessel functions in series andintegrate

∞+−+∞+

+−+=∫

0

1

0

.....1

t)(2

b)(2

adt

t

)bt(J)at(J

λνµµΓνΓ

λνµ

λµν

For t = 0, convergence requires µ + ν + 1 > λFor high values of the arguments apply equation (4.27)

∫ ∫∞ ∞

−−

−−

−−=

0 0

1 dt24

btcos24

atcostab

2dt

t

)bt(J)at(J νππµππ

πλ

λνµ

For t = ∞, convergence requires λ +1 > 0.

Hence the convergence conditions become:

µ + ν + 1 > λ > -1

Page 85: Pavement design and evaluation

INFINITE INTEGRALS OF BESSEL FUNCTIONS

71

8.6.2 The integral ∫Jµ(at)Jν(at)t-λdt.

The solution of the integral is given by equations (8.29) with a = b. Apply equation (7.16)

)ac()bc()abc()c(

)1;c;b,a(F−−−−

=ΓΓ

ΓΓ

=∫∞

0

)()(dt

t

atJatJλ

νµ

++−

Γ

+++

Γ

++−

Γ+Γ

Γ

+−+

Γ−

21

21

21

)1(

)(2

1

2

1

λµνλνµλνµν

λλνµ

λ

λa (8.31)

When λ = 0, equation (8.31) is undefined. Hence, when a = b, the requirements for convergence areµ + ν + 1 > λ > 0

However, when µ - ν = 2k + 1 the convergence conditions remainµ + ν + 1 > λ > -1

Indeed, by equation (8.31)

−+

=∫∞

→→ k2

)1k(

)(a1

limdtt

)at(J)at(Jlim

0 00 λΓΓ

λΓλλ

νµ

λ

By equation (7.7)

n

n

)1a()a()(

)na(+−

−=−

ΓΓ

+−−

=

−+ →→

2!k

12

))((

a1

limk

2)1k(

)(a1

lim k

k

00 λΓ

λλΓ

λΓΓ

λΓλλ

By equation (5.5)

( )2/1!k

12

22

1)(

a1

limdtt

)at(J)at(Jlim k

1k

0 00 Γ

λλΓ λ

λλνµ

λ

+−

+

=

−∞

→→ ∫

Hence, when µ - ν = 2k +1

∫∞ −−

−=0

21

a21

)(dt)at(J)at(Jνµ

νµ (8.32)

Page 86: Pavement design and evaluation

PAVEMENT DESIGN AND EVALUATION

72

8.6.3 The integral ∫Jµ(at)Jµ-2k-1(bt)dt

Applying (8.29), (8.30) and (8.32), one obtains

when b < a

∫∞

−−

−−

−−−

+−

−=

02

2

k2

1k2

1k2a

b;k2;k,kF

)1k()k2(a

)k(bdt)bt(J)at(J µµ

ΓµΓ

µΓµ

µ

µµ (8.33)

when b = a

∫∞

−− −=0

k1k2 a2

1)(dt)bt(J)at(J µµ (8.34)

when b > a

∫∞

+−− =

++−

−+

−=

02

2

11k2 0b

a;1;1k,kF

)k()1(b

)k(adt)bt(J)at(J µµ

ΓµΓ

µΓµ

µ

µµ (8.35)

8.6.4 Particular solutions of the integral ∫Jµ(at)Jν(bt)t-λdt

Value of the integral forIntegral

b < a b = a b > a

∫∞

0

dtt

)bt(J)at(J µµ µ

µ

ab

21

µ21 µ

µ

ba

21

∫∞

−0

1 dt)bt(J)at(J µµ µ

µ

a

b 1−

a21

0

∫∞

001 dt)bt(J)at(J

a1

a21 0

∫∞

0

dtt

)btcos()atsin(2π

0

dt)btsin()at(J0

0∫∞

0 ∞ 22 ab

1

∫∞

00 dt)btcos()at(J

22 ba

1

−∞ 0

Page 87: Pavement design and evaluation

INFINITE INTEGRALS OF BESSEL FUNCTIONS

73

8.6.5 Particular solution of the integral ∫Jν(at)Jν(bt)/tλdt.

In the special case in which the Bessel functions are of the same order, the discontinuous integral can beevaluated in a simple form.Apply equation (6.1)

φφω

ωΓνΓ

νπ

νν

λνν

λνν dsin

)t(

)t(J)2/1()2/1(

t2ab

t

)bt(J)at(J 2

0

2

∫+

=

where φω cosab2ba 222 −+=

( ) ( )∫ ∫ ∫∞ ∞ −

+

=0 0 0

2 dtdsint)t(J

2/12/12

ab

dtt

)bt(J)at(J πν

ν

λνν

ν

λνν φφ

ω

ωΓνΓ

Apply equation (8.2)

+

−+

+

−+=

−+−

∫ 1;1;2

,212

F)1(

)12(

2

1dtt

)t(J12

0

νλλν

νΓω

λνΓ

ω

ωλνν

λνν

ν

Apply equation (7.16)

+

−+

+=

+

−+

21

222

)2/1()1(1;1;

2,

212

Γλν

Γ

ΓνΓν

λλν

+

−+

−+=

−+−

∫2

1222

)2/1()12(

2

1dtt

)t(J12

Γλν

Γ

Γ

ω

λνΓ

ω

ωλνν

λνν

ν

∫ ∫∞

−+

+

−+

−+

+

=0 0

12

2d

sin

21

222

)12(

2

1

212

2ab

dtt

)bt(J)at(J π

λν

ν

ν

ν

λνν φ

ω

φλ

Γλν

Γ

λνΓν

Γ

Apply equation (5.5)

)2/1(212

222

2)12(

2

Γ

λνΓ

λνΓ

λνΓ

λν

−+

−+

=−+

( )∫ ∫∞

−+

+

−+

+

=0 0

12

2d

sin

21

)2/1(

212

212

2

abdt

t

)bt(J)at(J π

λν

ν

λ

ν

λνν φ

ω

φλ

ΓΓ

λνΓ

νΓ

Expand the integral in powers of cosφ, with α = 2ab/(a2+b2)

[ ] ⋅⋅⋅

+

+−

+−

+

+−

+=− +−

!2cos

12

122

12!1

cos2

121cos1

2212 φλνλν

αφλν

αφα λν

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PAVEMENT DESIGN AND EVALUATION

74

( )∫ +−

−+

π

λν

νφ

φ

φ

0 212

22

2d

cosab2ba

sin

( )φφφ

αλν

πν

λν ∫ ∑

+−

+

=+−

0

2k

k

k

212

22

dsincos!k

212

ba

1

Apply the definition (5.2) of the beta function, noticing that

when k = 2j + 1

∫ =+π

ν φφφ0

21j2 0dsincos

when k =2j

∫ ∫ ++==π π

νν νφφφφφφ0

2/

0

2j22j2 )2/1,2/1j(Bdsincos2dsincos

Hence

( )∫ +−

−+

π

λν

νφ

φ

φ

0 212

22

2d

cosab2ba

sin

( )∑ ++

+

+

+−

+

=+− )1j()!j2(

212

21j2

212

ba

1

j2

j2

212

22νΓ

νΓΓα

λν

λν

By (5.5)

)j(2

)j2()2/1()2/1j( 1j2 Γ

ΓΓΓ −=+

( )∑

++

+

+−

+

=+− )1j(2!j

212

21

212

ba

1j2

j2

j2

212

22 νΓ

νΓΓα

λν

λν

Take into account that

∑ ∑

+−

+−

=

+−

jjj2j2

432

412

2

212

λνλν

λν

Page 89: Pavement design and evaluation

INFINITE INTEGRALS OF BESSEL FUNCTIONS

75

( )∫ +−

−+

π

λν

νφ

φ

φ

0 212

22

2d

cosab2ba

sin

( )∑ +

+−

+−

++

+=

+− j

j2

jj

212

22)1(!j

232

212

)1(ba

)2/1()2/1(ν

αλνλν

νΓ

νΓΓλν

Hence

∫∞

=0

dtt

)bt(J)at(Jλ

νν

( ) ( )

++

+−+−

++

+

+−

+− 222

22

212

22 ba

ba4;1;

432

,4

12F

ba)1(2

12

212

)ab(ν

λνλν

νΓλ

Γ

λνΓ

λνλ

ν

(8.36)

Equation (8.36) is valid for a > b or a < b. In this form of the result, the discontinuity is masked.

Page 90: Pavement design and evaluation

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76

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BESSEL FUNCTIONS IN THE COMPLEX PLANE

77

Chapter 9 Bessel functions in the complex plane

9.1 Introductory note

This chapter presents the most important infinite integrals of Bessel functions resolved in the complexplane. They are essential for application in slab theory.

9.2 Helpful relations

zsin)z1()z(

ππ

ΓΓ =−

∫ −=C

zt dttei2

1)z(

1πΓ

∫∫∞

−−−−=0

tsinhzt

0

dteesin

d)sinzcos(1

)z(J νπ

ν πνπ

θθνθπ

∫∫∞

−−−−=0

tcoshzt

0

dteesin

d)coszcos(1

)z(I νπ

ν πνπ

θθνθπ

∫∞

−=0

tcoshz tdtcoshe)z(K νν

( )du

xu

xucos

21

x

)z2(21

)xz(K0 2

122

∫∞

++

+

=νν

ν

νΓ

νΓ

∫∞ +

=+0

22

1)ak(Kkdx

kx

)ax(Jxν

ννν

∫∞

=+0

0220 )ak(Kdx

kx

)ax(xJ

)ak(K)bk(Ik)(dxkx

)ax(J)bx(Jx 22/)(

022

1

νµρνµρνµ

ρ−−+

∞ −

−−=+

∫ when a > b

( ))ba(k)ba(k2

022 ee

k4dx

)kx(x

)axcos()bxsin( +−−−∞

−=+

∫π

when a > b

( )∫∞

−−+− −−=+0

)ab(k)ab(k222 ee2

k4dx

)kx(x

)axcos()bxsin( π when a < b

Page 92: Pavement design and evaluation

PAVEMENT DESIGN AND EVALUATION

78

+

−−

=+

+−

−−∞

∫ k2ba

cosek2ba

cosek4

dx)kx(x

)axcos()bxsin( k2

bak

2

ba

40

44π

when a < b

+

−−

−=+

+−

−−∞

∫ k2ab

cosek2ab

cose2k4

dx)kx(x

)axcos()bxsin( k2

abk

2

ab

40

44π

when a > b

9.3 Proof of Γ(z)Γ(1-z) = π/sin(πz)

Consider the infinite integral

∫∞ −

+0

1pdx

x1x

(9.1)

which we solve in two ways.To begin with, assume 0 < p < 1.Let x/(1+x) = y.

∫ ∫∞

−−

−=+

0

1

0

p1p1p

dy)y1(ydxx1

x (9.2)

By (5.1) and (5.4)

∫ −=−=−−1

0

p1p )p1()p()p1,p(Bdy)y1(y ΓΓ (9.3)

Consider now

∫ +

C

1pdz

z1z

(9.4)

contour integral in the complex plane.Since z =0 is a branch point (Appendix), choose the contour of Figure 9.1, where AB and GH are actuallycoincident with the x – axis but are shown separately for visual purposes. The integrand has the polez = -1, lying within C.

Figure 9.1 Contour integral in the complex plane.

A B

G

D

E

F

-1

H

Page 93: Pavement design and evaluation

BESSEL FUNCTIONS IN THE COMPLEX PLANE

79

The residue at z = -1 = eπi is

i)1p(1pi1p

1ze)e(

z1z

)1z(lim ππ −−−

−→==

++ (9.5)

Then

i)1p(C

1pie2dz

z1z ππ −

−=

+∫or, omitting the integrands,

i)1p(

HJAGHBDEFGAB

ie2 ππ −=+++ ∫∫∫∫

∫∫ ∫+

++

++

−− r

Ri2

)i2R

r

2

0i

i1pi1pdx

xe1

xe(d

Re1

Rei)(Redx

x1x

π

ππ

θ

θθθ

i)1p(0

2i

i1piie2d

re1

ire)re( π

πθ

θθπθ −

−=

++ ∫

where we have to use z = xe2πi for the integral along GH since the argument of z is increased by 2π ingoing around the circle BDEFG.Take the limit as r → 0 and R → ∞

0de

R1

i)e(

R

1limd

Re1

Rei)(Relim

2

0 i

pi

p1R

2

0i

i1pi

R=

+=

+∫∫ −∞→

∞→

π

θ

θπ

θ

θθθθ

0dre1

i)e(rlimd

re1

ire)re(lim

0

2i

pip

0r

0

2i

i1pi

0r=

+=

+∫∫ →

→ πθ

θ

πθ

θθθθ

with 0 < p < 1.Hence

∫ ∫∞

−−−

=+

++

0

0i)1p(

i2

1pi21pie2dx

xe1

)xe(dx

x1x π

π

ππ

[ ] i)1p(

0

1p)1p(i2 ie2dx

x1x

e1 ππ π −∞ −

− =+

− ∫

∫∞

−−

−−=

−=

−=

+0

ipip)1p(i2

i)1p(1p

sinee

i2

e1

ie2dx

x1x

ππππ

πππ

π (9.6)

Equalizing (9.3) and (9.6) proves

zsin)z1()z(

ππ

ΓΓ =− (9.7)

We have proven that (9.7) is valid for 0 < p <1.This can be extended along the real axis for non-integer z. If we make the substitutionz = x + 1 for 0 < x < 1, we find

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PAVEMENT DESIGN AND EVALUATION

80

[ ] =−

+−=−+=+−+=−

)x()1x()x(x

)x()1x()1x(1)1x()z1()z(ΓΓ

ΓΓΓΓΓΓ

ππ

ππ

ππ

ΓΓzsin)1xsin(xsin

)x1()x( =+

=−=−−=

9.4 The Hankel’s contour integral for 1/Γ(z).

Consider the integral ∫ −C

zt dtte around the path of Figure 9.2.

Figure 9.2 Hankel’s contour integral

Omitting the argument

∫ ∫∫∫ ++=+

C

R

R

zt dtteρ

π

π

ρ

Replace t by reiθ. For the first integral θ = -π, for the second r = ρ and for the third θ = π. Hence

( ) drereedtte iz

CR

irezt i πρ ππ −−

−−∫ ∫−

=

( ) θρρ θπ

πθρ θ

deiee iz

ie i −+

−∫+

( ) dreree iziR re i ππρ

π −∫+

Assume that z < 1 and take the limit for ρ → 0 and R → ∞. The second integral tends to zero and

∫ ∫∫ ∞

∞ −−−−−− −−=0

0

zizrzizr

C

zt dreredreredtte ππ

[ ]∫∞ −−− −=0

zizizr dreere ππ

∫∞ −−=0

zr drrezsini2 π

)z1(zsini2 −= Γπ

R

R

ρ

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BESSEL FUNCTIONS IN THE COMPLEX PLANE

81

and by equation (9.7)

∫ −=C

zt dttei21

)z(1

πΓ (9.8)

We have defined equation (9.8) for values of z < 1. However in this equation the integral defines ananalytic function for all z, hence, as for equation (9.7) defined in § 9.2, by the principle of analyticcontinuation equation (9.8) can be used for the reciprocal of the gamma function throughout the z plane,i.e. throughout the complex plane.

9.5 The integral representation of Jν(z)

We start with Hankel’s contour integral (9.8) wherein we replace z by (ν + m + 1) the argument of thegamma function in the successive denominators of the series expressing Jν(z):

( ) ∫ −−−=++ C

1mt dttei21

1m1 ν

πνΓHence

∑∞ +

++−

=0

2

12

)m(!m)/z()(

)z(Jmm

νΓ

ν

ν

∫∑∞

−−−=

C 0

t12m

dtet!m

)t4/z()(i2

)2/z()z(J ν

ν

ν π

∫ −−−=C

1t)t4/z( dtteei2

)2/z()z(J

2 νν π

Let t =zew/2 and dt = tdw

( ) ( )∫ −− −=C

www dw2/zeexp2/zeexpei21

)z(J νν π

∫ −=C

wsinhzw dweei21

)z(J νν π

(9.9)

Perform the integration around the path of Figure 9.3, consisting of three sides of a rectangle with verticesat ∞-πi, -πi, πi and ∞ + πi.

Figure 9.3 Integration around the contour

We write w = t ± iπ on the sides parallel to the real axis and w = 0 ± iθ on the lines joining 0 to ± iπ.Neglecting the integrands

-iπ

0

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PAVEMENT DESIGN AND EVALUATION

82

+++= ∫∫∫∫

∞ π

π

π

π

ν πi

i

i

i

i)z(J

0

0

21

∫∞

−=π

νπ

iwsinhzw dwee

iInt

21

1

∫∞

−−−=0

21

1 dteei

Int )itsinh(z)it( ππνπ

∫∞

−−=0

21

1 dteeei

Int tsinhzit πννπ

∫∞

−−−=0

21

1 dteeei

Int tsinhzit πννπ

Similarly

∫∞

−−−=0

21

4 dteeei

Int tsinhzit πννπ

With

[ ] [ ]πνπ

νπνπνπνπππ

πνπν sinsinicossinicos

iee

iii −=−−−=−−

21

21

dteesin

IntInt tsinhzt −∞

−∫−=+0

41 νπνπ

Next

∫−

−=0

21

νπ

i

wsinhzw dweei

Int

∫ −−=0

21

θθν θπ

deeInt )isinh(zi

∫ −=π

θθν θπ

021

2 deeInt sinizi

∫ −=π

θνθ θπ

021

2 deInt )sinz(i

Similarly

∫ −−=π

θνθ θπ

021

3 deInt )sinz(i

)sinz(cosee )sinz(i)sinz(i θνθθνθθνθ −=+ −−− 2Finally

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BESSEL FUNCTIONS IN THE COMPLEX PLANE

83

∫∫∞

−−−−=00

1dtee

sind)sinz(cos()z(J tsinhztν

π

ν πνπ

θθνθπ

(9.10)

9.6 The integral representation of Iν(z)

In analogy to the previous paragraph, one obtains

∫∫∞

−−−−=00

1dtee

sind)sinz(cos()z(I tcoshztν

π

ν πνπ

θθνθπ

(9.11)

9.7 The integral representation of Kν(z).

Consider (9.11)

∫ ∫∞

−−−−=π

νν π

νπθθνθ

π0 0

tcoshzt dteesin

d)coszcos(1

)z(I

Form

∫ ∫∞

−+− +−=

πν

ν πνπ

θθνθπ

0 0

tcoshzt dteesin

d)coszcos(1

)z(I

∫∞

−− =−

0

tcoshz tdtcoshesin

2)z(I)z(I νπνπ

νν

Apply equation (3.21)

νππ νν

ν sin)z(I)z(I

2)z(K

−= −

Hence

∫∞

−=0

tcoshz tdtcoshe)z(K νν (9.12)

Equation (9.12) can be written as follows

[ ] dtee21

dtee21

dteee21

)z(K t

0

tcoshzt

0

tcoshztt

0

tcoshz ννννν

−∞

−∞

−−∞

− ∫∫∫ +=+=

Transform the integral with the positive exponent eνt by setting t = -t and dt = -dt.Hence

dtee21

dtee21

)z(K t

0

tcoshzt

0

tcoshz ννν

−∞

−−−∞

− ∫∫ +−=

dtee21

dtee21

)z(K t

0

tcoshzt0

tcoshz ννν

−∞

−−

∞−

− ∫∫ +=

∫∞

∞−

−−= dte21

)z(K ttcoshz νν (9.13)

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84

Under this form the finite value of the integrand at infinite is evident.

9.8 The integral representation of Kv(xz)

A second integral representation, more appropriate for numerical integration, is gained from equation

(9.13). However we have to evaluate before the value of the integral ∫∞

0

2λβλαλ dcose (Spiegel, 1974).

Let

∫∞

−==0

2λβλβα αλ dcose),(II

Then

λβλλβ

αλ dsineI

∫∞

−=

∂∂

0

2

Idcosesine

∫∞

−∞−−=−=

00 222

22

αβ

λβλαβ

βλα

αλαλ

Thus

αβ

β 21

−=∂∂I

Ior

αβ

β 2−=

∂∂

Ilog

Integration with respect to β yields

1

2

4CIlog +−=

αβ

αββα 42 /Ce),(II −==C is a constant for all values of β, hence

∫∞

−==0

20 λα αλ de),(IC

Let x = αλ2

απ

α

Γ

α 21

2

21

2

1

0

21 === ∫∞

−− )/(dxexC x/

αβαλαπ

λβλ 4

0

22

21 /edcose −

∞− =∫ (9.14)

We now prove that (Watson, 1966)

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BESSEL FUNCTIONS IN THE COMPLEX PLANE

85

( )du

zu

)xucos(

)/(x

)z)(/()xz(K

/∫∞

++

+=

0212221

221νν

ν

νΓ

νΓ (9.15)

Therefore we shall show that equations (9.13) and (9.15) are identical

( )du

zu

)xucos(dte

)z(

x/

ttcoshxz ∫∫∞

+

∞−

−−

+

+=

021222

1

221

21

νν

ν

ννΓΓ (9.16)

If we write t = s(u2+z2), the expression on the right is equal to

( )( )[ ] xudsducoszusexpsdtdu

zu

xucoset //

t/

∫ ∫∫ ∫∞∞

−∞∞

+

−−+−=

+ 0 0

2221

0 02122

21ν

ν

ν

and if we change the order of the integration

( )[ ] [ ]dsszexpsxuducossuexpdtdu

zu

xucoset //

t/221

0 0

2

0 02122

21−−=

+

−∞∞∞∞

+

−−

∫ ∫∫ ∫ νν

ν

We apply equation (9.14)

[ ]∫∞

=−

0

2212

421

21

sx

expsxuducossuexp /Γ

and obtain

( )ds

sx

szexpszu

du)xucos(/∫ ∫

∞ ∞−

+

−−

=

+

+

0 0

221

2122 421

21

21 ν

νΓνΓ

Set s = ½ xe-t so that equation (9.15) is proven

( ) ∫∫∞

∞−

−−∞

+

=

+

+ dte

)z(

xdu

zu

)xucos( ttcoshxz/

νν

ν

νΓνΓ

221

21

21

02122

(9.17)

9.9 Resolution of ∫xν +1 Jν(ax)/(x2+k2)dx

This sort of integrals is usually solved by integrating in the complex plane the function where the Besselfunction of the first kind Jν is replaced by the Hankel function of the first kind Hν

(1), for reasons whichwill appear obvious at the end of the paragraph,

∫+

+

C22

)1(1dz

kz

)az(Hz νν

around a contour containing one or more poles of the integrand.The conditions of convergence are

a ≥ 0ν < 3/2

Consider the contour represented at figure 9.4 which includes the pole z = ik and avoids the branch pointz = 0. It is required that ℜ (k) > 0.The residue at z = ik is

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PAVEMENT DESIGN AND EVALUATION

86

)ikz)(ikz()az(Hz)ikz(

limi2)1(1

ikzC+−

−=

+

=∫ νν

π

ik2H)ik(

i2)aik)(1(1

C

νν

π+

=∫

)aik(Heik)aik(Hiik )1(2/i)1(

πννν

νν ππ ==∫

Figure 9.4 Integrating in the complex plane

and by equation (4.69)

)ak(Kk2C

νν=∫

Now we compute the integral around the contour, where neglecting the integrands

∫∫∫∫∫ +++=−

0r

R0

R

rC π

π

The first and the third integral are solved on the axis of the real values of the variables, thus the variable zmay be replaced by x.

∫+

=+R

r22

)1(1

1 dxkx

)ax(HxInt ν

ν

∫−

+

+=

r

R22

)1(1

3 dxkx

)ax(HxInt ν

ν

To solve this integral, replace x by –y.

∫+

−−=

+R

r22

)1(1

3 dyky

)ay(H)y(Int ν

ν

∫+

−=+R

r22

i)1(i1

3 dyky

)aye(H)e(yInt

πν

νπν

Replace now y by x

ik

-R R-r r

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87

∫+

−=+R

r22

i)1(i1

3 dxkx

)axe(H)e(xInt

πν

νπν

Add Int1 and Int3

[ ]∫

+

−=

+R

r22

i)1(i)1(1

13 dxkx

)axe(He)ax(HxInt

πν

νπν

ν

Hence by equation (4.66)

∫+

=+R

r22

1

3 dxkx

)ax(Jx2Int ν

ν

The solution of Int2 is difficult and requires a good knowledge of the properties of complex numbers(Appendix). Taking into account that we solve the integral for large values of R, we use the asymptoticnotation for Hν

(1)

θπ θπ

θ

ππθ

θνν θ

dRekeR

eRea

2eR

Int i

02i22

)4/2/vRea(ii

i)1(1

2

i

∫+

=

−−++

θπ θπ

θ

ππθθθ

θνν

dRekeR

eeRea

2eR

Int i

02i22

4/2/v)sini(cosiaRi

i)1(1

2 ∫+

−−+++

θππ

θ

νπθθθθνν

dkeR

eeea

2eeR

Int0

2i22

)12(4/isinaRcosiaR2/ii)1(2/12

2 ∫+

+−−−+−+

∫+

−+π θν

θπ

022

sinaR2/3

2 dkR

ea2

RInt

∫ −−≤2/

0

sinaR2/12 deR

a2

2Intπ

θν θπ

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88

Figure 9.5 Sinθ ≥ 2θ /π.

It can be deduced from Figure 9.5 that sinθ ≥ 2θ /π. Hence

∫ −−≤2/

0

2/aR2/12 deR

a2

2Intπ

θν θπ

( ) 2/34/aR2 Re1

a2

a4

Int −−−≤ νππ

Hence lim Int2 = 0 for R → ∞ if ν < 3/2.We finally show that lim Int4 = 0 for r → 0.

∫+

=+0

i22i2

i)1(1i

4 direker

)are(H)re(Int

π

θθ

θν

νθθ

We replace Hν(1) by its expression for small values of the argument

θθ

πθ

νν der

)a(frInt

0

2i22

4 ∫−+=

It follows immediately that lim Int4 = 0 for r → 0.And so we have proven that

∫∞ +

=+0

22

1)ak(Kkdx

kx

)ax(Jxν

ννν

(9.18)

if 0 ≤ ν < 3/2.In particular

∫∞

=+0

0220 )ak(Kdx

kx

)ax(xJ (9.19)

2θ /π

π/2 π

1

θ0

sinθ

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89

9.10 Resolution of ∫xρ -1 Jµ(bx)Jν(ax)/(x2+k2)dx

The solution of the integral

∫∞ −

+022

1

dxkx

)ax(J)bx(Jx νµρ

is obtained in a very similar way as the previous integral.The conditions of convergence are

a ≥ bν - µ < ρ < 4ρ + µ - ν = 2n

Consider the contour of figure 9.4.

dzkz

)az(H)bz(Jz22

11

∫+

−νµ

ρ

The value of the contour integral is

ik2

)aik(H)bik(J)ik(i2

)1(1νµ

ρ

π−

=∫Applying equations (4.68) and (4.69) yields

ii2

)ak(Ke)bk(Ieki2i2

2/i2/i21

ππ

ννπ

µµπρρ −−−

=∫)ak(I)bk(Iki2 2

νµρυµρ −−+−=∫

The value of the residue can only be real if ρ + µ - ν is an even integer. Hence the condition ρ + µ - ν = 2n

Hence

)ak(K)bk(Ik)1(2 2nνµ

ρ −−−=∫Now we compute the integral around the contour, where neglecting the integrands

∫∫∫∫∫ +++=−

0r

R0

R

r π

π

Applying the same method as in the previous case, one obtains

∫+

=−R

r22

1

13 dxkx

)ax(J)bx(Jx2Int

νµρ

Similarly it is shown that lim Int2 = 0 for R → ∞ if ρ < 4 and that lim Int4 = 0 for r → 0 ifν -µ < ρ.

So we have proved that

)ak(K)bk(Ik)(dxkx

)ax(J)bx(Jx 22/)(

022

1

νµρνµρνµ

ρ−−+

∞ −

−−=+

∫ (9.20)

For high values of the argument lim Iµ (bk) = C1 ebk and lim Kν (ak) = C2 e-ak.

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PAVEMENT DESIGN AND EVALUATION

90

Hence convergence of the result requires a > b.One obtains in particular applying equations (4.8), (4.9), (4.12) and (4.15), with a > b

( ))ba(k)ba(k2

022 ee

k4dx

)kx(x

)axcos()bxsin( +−−−∞

−=+

∫π

(9.21)

( ))ba(k)ba(k

022 ee

4dx

)kx(

)axcos()bxsin(x −−+−∞

−=+

∫π

(9.22)

Also for a < b

( ) abforee4

dx)kx(

)axcos()bxsin(x )ab(k)ab(k

022 >+=

+−−+−

∫π

(9.23)

However, integral (9.21) has no immediate solution for a < b. Indeed in that case ν - µ < ρ is not verified:µ = -1/2, ν = 1/2, ρ = 1.We then transform the integral considering that

+−=

+ 22222 kx

xx1

k

1

)kx(x

1

The integral becomes

∫ ∫∫∞ ∞∞

+−=

+0 0222

0222 dx

kx

)axcos()bxsin(x

k

1dx

x)axcos()bxsin(

k

1dx

kx(x

)axcos()bsin(

The solution of the first integral of the second member is given in § 8.5.4. Hence for a < b

( )∫∞

−−+− −−=+0

)ab(k)ab(k222 ee2

k4dx

)kx(x

)axcos()bxsin( π (9.24)

Note:When the condition ρ + µ - ν = 2n is not met, it can be shown (Watson, 1966) that the general equationtakes the form:

∫∞

−−

−=

−+

+−+

+0

2v22

1

k)ak(K)bk(I)ax(Y2

sin)ax(J2

coskx

)bx(Jx ρµνν

µρ

νµρνµρ

9.11 Resolution of ∫sin(bx)cos(ax)/x/(x4 + k4)dx

The solution of

dx)kx(x

)axcos()bxsin(

044∫

+can easily be reduced to the previous case by transforming the denominator

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91

+−

+=

+ 22

221

2244 zx

1

zx

1

ik2

1

kx

1

where z12 = -ik2 and z2

2 = ik2 (Pronk, 1991)Considering the same conditions for convergence

a ≥ bν - µ < ρ < 4ρ + µ - ν = 2n

one obtains applying (9.21)

+−

+∫ ∫∞ ∞

0 022

221

22 dx)zx(x

)axcos()bxsin(dx

)zx(x

)axcos()bxsin(

ik2

1

( ) ( )

−−−= +−−−+−−− )ba(z)ba(z

22

)ba(z)ba(z21

22221 ee

z4ee

z4ik2

1 ππ

( ) ( )[ ])ba(z)ba(z)ba(z)ba(z4

2221 eeeek8

1 +−−−+−−− −−−=

with z1 =k √(-i) =(1 –i)k/√2 and z2 =k √(i) =(1+i)k/√2

+−

+=

+−++−−−−−− k2baik

2baik

2bak

2baik

2baik

2ba

4 eeeeeek8

π

with eia + e-ia = cosa + i sina +cos a –i sina =2 cosa

+

−−

=+

+−

−−∞

∫ k2ba

cosek2ba

cosek4

dx)kx(x

)axcos()bxsin( k2

bak

2

ba

40

44π

(9.25)

When a < b, the boundary conditions are not met: ν - µ = ρ . The solution is obtained by splitting theintegral

∫ ∫∫∞ ∞∞

+−=

+ 0 022

2

20

22 dx)kx(x

)axcos()bxsin(xk1

dxx

)axcos()bxsin(dx

)kx(x)axcos()bxsin(

Hence the result is:

+−

−−=

+

+−

−−∞

∫ k2

abcosek

2

abcose2

k4dx

)kx(x)axcos()bxsin( k

2ab

k2ab

40

44

π (9.26)

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93

PART 2: THE APPLICATIONS

Chapter 10 Laplace Equation in Pavement Engineering

10.1 Equilibrium equation for beams in pure bending.

Consider the beam AB transversally loaded as shown in figure 10.1 (Timoshenko, 1953)

Figure 10.1 Transversally loaded beam

The middle portion CD of the beam is free from shear force; the bending moment. Mx = Pa is uniformbetween C and D. This condition is called ‘pure bending’. This problem will be solved under the“strength of materials approach” and not, such as in §10.3 and 10.4, under the “Theory of ElasticityApproach”. The assumptions will be more restrictive than those of the theory of elasticity, however thesolutions will be simpler and, in most cases, even correct.

10.1.1 Sign conventions

Figure 10.2 Sign conventions

P P

P P

A B

C D

x

a

x dx

σy

τxy

yA C

σy

dy

B

τyx

τxy

σxσx

D

τxy

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Consider the stresses acting on the different faces of ABCD: σx and σy are called normal stresses in theindicated directions and τxy and τyx are called shear stresses acting in a plane normal to the direction of thefirst index and in a direction given by the second index [Conventions of Timoshenko (1953)]. The normalstresses are taken positive when they produce tension and negative when they produce compression. Theshear stress on any face of an element will be considered positive when it has a clockwise moment withrespect to a centre inside the element (Figure 10.3a). If the moment is counter-clockwise with respect to acentre inside the element, the shear stress is negative (Figure 10.3b).

Figure 10.3 a and b Positive and negative shear stresses

The equality of complementary shear stresses, such as τxy and τyx on the faces of a rectangular elementcan be established from the equilibrium conditions of the element (Figure 10.4).

Figure 10.4 Equality of complementary stresses

We write the equilibrium of the moments due to the shear stresses

0dxdydydx xyyx =⋅+⋅ ττHence we write

yxxy ττ −= (10.1)

x

y dx

dyτxy

τyx

τxy

τyx

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10.1.2 Assumptions

Consider a prismatic beam with an axial plane of symmetry, which is taken as the xy-plane. When theapplied loads also act in such a plane of symmetry, bending will take place only in that plane. Assumethat the material is homogeneous and obeys Hooke’s law:

( )yxx Eµσσε −=

1 (10.2)

( )xyy Eµσσε −=

1 (10.3)

Furthermore assume that as the bending moment is uniform, the bending deformation will also beuniform. This implies that each cross-section, originally plane, is assumed to remain plane and normal tothe longitudinal fibres of the beam. Hence there is no deformation in the vertical direction and Hooke’slaw can be simplified:

Ex

ε = (10.4)

As a result of the loading shown in figure 10.1, fibres on the lower side of the beam are elongated slightlywhile those on the upper side are slightly shortened. Somewhere in between the top and the bottom of thebeam, there is a layer of fibres that remain unchanged in length. This is called the neutral surface. Theintersection of the neutral surface with the axial plane xy is called the neutral axis of the beam.

10.1.3 Bending moment and bending stress

Consider a section of the beam parallel to the xy – plane, with a segment ab orthogonal to the neutral axis(Figure 10.5). Call w, the deflection in the y – direction.

Figure 10.5 Bending moment and bending stress

Considering the assumptions of § 10.1.2., one can accept that the segment a-b remains orthogonal to theneutral axis after bending. In the plane xy, segment ab will rotate over an angle equal to – dw/dx. If u isthe displacement in the x – direction, one may write

dxdw

yu −=

The strain is defined as the relative displacement

x- dw/dx

x

a ax

yy

b

b

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2

2

xdx

wdy

dxdu

−==ε (10.5)

The stress is given by Hooke’s law

2

2

xxdx

wdyEE −== εσ (10.6)

The bending moment is obtained by integration of (10.2)

∫− −

−=−==2/h

2/h2

22/h

2/h

3

2

2

xdx

wdEI

3y

dx

wdEydyM σ (10.7)

The bending stress is obtained by (10.6) and (10.7)

IMy

x =σ (10.8)

10.1.4 The radius of curvature

It is known that the radius of curvature of an analytical function is given by the second derivative of thefunction, here d2w/dx2. We have defined the bending stress as positive when in tension. Thus when thetensile stress is located at the positive side of the vertical y-axis, the radius of curvature is oriented in theopposite direction. Hence:

2

2

dx

wdR1

−= (10.9)

REI

M = (10.10)

10.1.5 Equilibrium

Write now the equilibrium of the moments in a cross-section (Figure 10.6)

dxdx

dM2dx

T2 =

3

3

dx

wddx

dMEI1

EIT

−== (10.11)

and write the equilibrium of the vertical forces

dxdxdT

qdxpdx −=−

4

4

dx

wddxdT

EI1

EIqp

=−=−

(10.12)

EI)x(q)x(p

w22 −=∇∇ (10.13)

where p is the intensity of a load distributed over the beam. The double Laplacian of the deflection varieswith the value of the distributed load.

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Figure 10.6 Equilibrium of moments

10.2 Equilibrium equation for bent plates

Consider a transversally loaded slab as given in figure 10.7.

Figure 10.7 Equilibrium for bent plates

Postulate the hypothesis that the thickness of the slab can be considered to be thin against it's otherdimensions and that the deflections are small in comparison with the thickness. These basic assumptionsof the Theory of the Strength of Materials, allow to consider the mid plane as a neutral plane wherein thehorizontal displacements are admitted to be zero.

10.2.1 Bending moment and shear forces

Consider a section of the slab parallel to the xz – plane, with a segment ab orthogonal to the neutral midplane (Figure 10.8). Name w, the deflection in the z-direction.

y

P

Q

x

dx

q

T + (dT/dx)dx)

M M + (dM/dx)dx

T

p

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Figure 10.8 Bending moments and shear forces

Again, one may accept that the segment ab remains orthogonal to the neutral plane after bending. In theplane xz, segment ab will rotate over an angle equal to -∂w/∂x and in the plane yz, a similar segment willrotate over an angle -∂w/∂y. If u is the displacement in the x-direction, one can write:

xw

zu∂∂

−=

and if v is the displacement in the y-direction,

yw

zv∂∂

−=

The strains can be deduced from the displacements

2

2

xx

wz

xu

∂∂∂

ε −==

2

2

yy

wz

yv

∂∂∂

ε −==

yxw

z2xv

yu 2

xy ∂∂∂

∂∂

∂∂

γ −=+=

and the stresses by Hooke’s law

+

−−=

2

2

2

2

2xy

w

x

w

1

Ez

∂µ

µσ

+

−−=

2

2

2

2

2yx

w

y

w

1

Ez

∂µ

µσ (10.14)

yxw

1Ez 2

xy ∂∂∂

µτ

+−=

The bending and torsion moments are obtained by integration of (10.14).

( )∫−

+

−−==

2/h

2/h2

2

2

2

2

3

xxy

w

x

w

112

EhdzzM

∂µ

µσ

yx

y

-∂w/∂xx

b

b

z

a

z

a

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and with D, the stiffness of the plate,

∫−

+−==

2/h

2/h2

2

2

2

xxy

w

x

wDdzzM

∂µ

∂σ (10.15)

∫−

+−==

2/h

2/h2

2

2

2

yyx

w

y

wDdzzM

∂µ

∂σ (10.16)

∫−

−−==2/h

2/h

2

xyxy yxw

)1(DdzzM∂∂

∂µτ (10.17)

∫−

−=−==2/h

2/h

2

xyyxyx yxw

)1(DMdzzM∂∂

∂µτ (10.18)

The moments can also be expressed in function of the radii of curvature

+=

+=

yxy

yxx R

1R1

DMR1

R1

DM µµ (10.19)

10.2.2 Equilibrium

Consider an elementary parallelepiped of dimension dx.dy.h (Figure 10.9).

Figure 10.9 An elementary parallelepiped

Equilibrium of the moments with regard to the x-axis.

0Ty

M

x

My

yxy =+−∂

∂ (10.20)

Equilibrium of the moments with regard to the y-axis.

0Tx

My

Mx

xyx =+−−∂

∂∂

∂ (10.21)

dx

dy

Ty

Tx

My

Myx

Mxy

Mx

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Equilibrium of the vertical forces with regard to the z-axis

0qpy

T

xT yx =−++

∂∂

(10.22)

Replacing in (10.20) and (10.21) the moments by their values of (10.15) to (10.18), one obtains theequations for the shear forces

+−=

2

2

2

2

xy

w

x

wx

DT∂

∂∂∂

(10.23)

+−=

2

2

2

2

yy

w

x

wy

DT∂

∂∂∂

(10.24)

Replacing in (10.22) the shear forces by their values of (10.23) and (10.24), one obtains the equilibriumor continuity equation for a thin slab submitted to pure bending

Dqp

y

w

yx

w2

x

w4

4

22

4

4

4 −=++

∂∂

∂ (10.25)

10.3 Compatibility equation for a homogeneous, elastic, isotropic body submitted to forces appliedon its surface

The problem is solved in two dimensions. Consider the elementary rectangle ABCD, with unit thicknessas represented in Figure 10.10.

Figure 10.10 Elementary rectangle

Consider the stresses acting on the different faces of ABCD: σx and σy are called normal stresses in theindicated directions and τxy and τyx are called shear stresses acting in a plane normal to the direction of thefirst index and in a direction given by the second index [conventions of Timoshenko (1970)]. Thestresses, such as represented in Figure , are considered to be positive. The normal stresses are takenpositive when they produce tension and negative when they produce compression. The positive directionsof the components of shearing stress on any side are taken as the positive directions of the co-ordinateaxes if a tensile stress on the same side would have the positive direction of the corresponding axis. If the

τyx

dyx

τxy

σy

σxσx + (∂σx/∂x)dx

y A C

DBτyx + (∂τyx/∂x)dx

dy

σy + (∂σy/∂y)dy

τxy + (∂τxy/∂y)dy

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tensile stress has a direction opposite to the positive axis, the positive directions of the shear stressesshould be reversed.Assume that the rectangle ABCD is small enough to admit that the stresses are uniformly distributedalong its sides and therefore that the stress resultants pass through the centre of gravity of the rectangle.For simplicity, consider that the body forces (for example: weight) can be neglected.

The resolution is based on three principles: equilibrium, continuity and elasticity.

10.3.1 Principle of equilibrium

The body must remain in equilibrium under the applied forces. Therefore the sums of the horizontalforces, the vertical forces and the moments around the centre of gravity must be zero. From Figure wecan deduce that:

0yxxyx =+

∂τ

∂∂σ

0yxyyx =+

∂σ

∂τ

yxxy ττ =

The system of three equations reduces in a system of two equations:

0yxxyx =+

∂τ

∂∂σ

(10.26)

0yxyxy =+

∂σ

∂τ (10.27)

System (10.26) and (10.27) of two equations contains three unknowns. In order to solve the problem,more information is required.

10.3.2 The principle of continuity

Continuity means that the material of the body remains continuous, that the body remains uncracked afterloading. An equation between linear and angular strains can express this.If u is the displacement of agiven point in the x direction and v the displacement in the y direction, then define (Figure 10.11):

xv

yu

yv

xu

xyyx ∂∂

∂∂

γ∂∂

ε∂∂

ε +=== (10.28)

where εx and εy are the linear strains in the indicated direction, and γxy is the angular strain in the indicatedplane.If one assumes continuity of the material, one must assume continuity of the mathematical relationships,thus derivability of equation (10.28). Derive twice each of the three equations of (10.28) and add theresults so as to obtain the continuity equation

yxxy

xy2

2y

2

2x

2

∂∂

γ∂

ε∂

ε∂=+ (10.29)

The system now consists in three equations with six unknowns: more information is required to solve theproblem.

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102

Figure 10.11 Principle of continuity

10.3.3 The principle of elasticity

Let us assume that the body behaves elastically, that there is a linear relationship between strains andstresses. This was first observed by Hooke (1678) and therefore is referred to as Hooke’s law.

E

yxx

µσσε

−=

Exy

yµσσ

ε−

= (10.30)

E

)1(2

G1 xy

xyxyτµ

τγ+

==

where E is called Young’s modulus, µ Poisson’s ratio and G the shear modulus of the material. Equation(10.30) applies for the state of plane stress.

10.3.4 Stress potential

In order to solve the system of the six equations (10.26) to (10.30), assume that there exists a potentialfunction from which the stresses can be deduced (Airy, 1862).

Seek for such an equation that the equilibrium equations (10.26) and (10.27) are satisfied.

yxxy

2

xy2

2

y2

2

x ∂∂Φ∂

τ∂

Φ∂σ

Φ∂σ −=== (10.31)

Replace the stresses in (10.30) using equations (10.31) and replace the strains in the continuity equation(10.29) in order to obtain:

0yxyx

222

2

2

2

2

2

2

2=∇∇=

+

+ Φ

Φ∂

Φ∂

∂ (10.32)

u u + du

εx = ∂u/∂x

εy = ∂v/∂y

x,y x + dx, ydx

γ2

x, y + dy x + du, y + dy

γ1 = tg(∂v/∂x) ≈ ∂v/∂x

γ2 = tg(∂u/∂y) ≈ ∂u/∂y

γ = γ1 + γ2

γ = ∂u/∂y + ∂v/∂x

γ1

x + dx, y + dv

x x + u x + dx x + dx + u + du

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Equation (10.32) is the so-called continuity or compatibility equation: the double Laplacian of the stresspotential is zero.

10.4 Compatibility equation for a homogeneous, elastic, anisotropic body submitted to forcesapplied on its surface

Consider the same elementary rectangle as that represented in Figure 10.10, but assume that the Young’smoduli in horizontal and vertical directions differ from each other (Figure 10.12):

Figure 10.12 Elementary rectangle with horizontal and vertical Young;’s moduli

Write Ex = E/n and Ey = E. The equilibrium equations are similar to those presented in § 10.3.1

0yxxyx =+

∂τ

∂∂σ

(10.33)

0yxyxy =+

∂σ

∂τ (10.34)

Furthermore the continuity equation remains unchanged

∂ ε

∂ ε

∂ γ

∂ ∂

2

2

2

2

2x y xy

y x x y+ = (10.35)

However the elasticity conditions are modified due to the anisotropy. In matrix format, Hooke’s law iswritten as:

=

xy

y

x

xy

y

x

G1

00

0E1

E

0n/En/E

1

τσσ

µ

ν

τεε

(10.36)

Due to energetic principles, the matrix must be symmetrical (Lekhnitskii, 1963). Hence

n/EEνµ

= and ν = µ/n.

dyx

τxy

σy

σxσx + (∂σx/∂x)dx

y A C

DBτyx + (∂τyx/∂x)dx

dy

σy + (∂σy/∂y)dy

τxy + (∂τxy/∂y)dy

Ex

Ey

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It can further be shown (Barden, 1963, and, more generally, Van Cauwelaert, 1983) that if, as in isotropicelasticity, there is a relation between Young’s modulus, Poisson’s ratio and shear modulus, it isnecessary, though not sufficient, that this equation should be

E2n1

G1 µ++

= (10.37)

With n = 1, one gets the corresponding isotropic equation.

Hooke’s law then becomes

E

n yxx

µσσε

−=

Exy

yµσσ

ε−

= (10.38)

E

)2n1(

G1 xy

xyxyτµ

τγ++

==

The stress potentials, satisfying equilibrium, remain unaltered

yxxy

2

xy2

2

y2

2

x ∂∂Φ∂

τ∂

Φ∂σ

Φ∂σ −=== (10.39)

Replacing the stresses in (10.38) and the strains in (10.35), one obtains the compatibility equation for ananisotropic body in plane stress co-ordinates:

0y

nxyx 2

2

2

2

2

2

2

2=

+

+

Φ∂

Φ∂

∂ (10.40)

10.5 Basic equations of continuum mechanics in different co-ordinate systems

10.5.1 Plane polar co-ordinates

Using the same methodology as in § 10.3, one establishes next equations

Equilibrium:

0rr

1r

rrr =−

++ θθ σσ∂θ

∂τ∂

∂σ

(10.41)

0r

2rr

1 rr =++ θθθ τ∂

∂τ∂θ

∂σ

Strains and displacements:

rv

rv

ru

rv

ru

ru

rr −+=+==∂∂

∂θ∂

γ∂θ∂

ε∂∂

ε θθ (10.42)

Elasticity:

Er

rθµσσ

ε−

=

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105

Erµσσ

ε θθ

−= (10.43)

θθ τµ

γ rr E)1(2 +

=

Stress Potential:

−==+=

∂θΦ∂

∂∂

τ∂

Φ∂σ

∂θ

Φ∂∂Φ∂

σ θθ r1

rrr

1rr

1r2

2

2

2

2r (10.44)

Compatibility equation:

0r

1rr

1

rr

1rr

1

r 2

2

22

2

2

2

22

2=

++

++

∂θ

Φ∂∂Φ∂

Φ∂

∂θ

∂∂∂

∂ (10.45)

10.5.2 Axi-symmetric Cylindrical Co-ordinates

Equilibrium:

0rzr

rrzr =−

++ θσσ∂

∂τ∂

∂σ

(10.46)

0rzrrzzrz =++

τ∂

∂σ∂

∂τ

Strains and displacements:

rw

zu

zw

ru

ru

rzzr ∂∂

∂∂

γ∂∂

εε∂∂

ε θ +==== (10.47)

Elasticity:

E

( )zrr

σσµσε θ +−

=

E

( )zr σσµσε θ

θ+−

= (10.48)

E)( rz

zθσσµσ

ε+−

=

rzrz E)1(2τ

µγ

+=

Stress potential (Love, 1927):

−∇=

2

22

rrz ∂

Φ∂Φµ

∂∂

σ

−∇=

rr1

z2

∂Φ∂

Φµ∂∂

σθ

−∇−=

2

22

zz

)2(z ∂

Φ∂Φµ

∂∂

σ (10.49)

−∇−=

2

22

rzz

)1(r ∂

Φ∂Φµ

∂∂

τ

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106

−∇−

+=

2

22

z)1(2

E1

w∂

Φ∂Φµ

µ

zrE1

u2

∂∂∂+

−=Φµ

Compatibility equation:

0zrr

1

rzrr1

r 2

2

2

2

2

2

2

2=

++

++

Φ∂∂Φ∂

Φ∂

∂∂∂

∂ (10.50)

10.5.3 Non symmetric cylindrical co-ordinates

Equilibrium:

0rzr

1r

rrzrr =−

+++ θθ σσ∂

∂τ∂θ

∂τ∂

∂σ

0rzr

1r

rzzzrz =+++τ

∂∂σ

∂θ∂τ

∂∂τ θ (10.51)

0r

2zr

1r

rzr =+++ θθθθ τ∂

∂τ∂θ

∂σ∂

∂τ

Strains and displacements:

rw

zu

zw

rv

ru

ru

rzzr ∂∂

∂∂

γ∂∂

ε∂θ∂

ε∂∂

ε θ +==+== (10.52)

∂θ∂

∂∂

γ∂∂

∂θ∂

γ θθ rw

zv

rv

rv

ru

zr +=−+=

Elasticity:

E)( zr

rσσµσ

ε θ +−=

E)( zr σσµσ

ε θθ

+−= (10.53)

E)( rz

zθσσµσ

ε+−

=

θθ τµ

γ rr E)1(2 +

= rzrz E)1(2τ

µγ

+= zz E

)1(2θθ τ

µγ

+=

Stress potentials (Muki, 1960):

∂θΨ∂

∂θ∂Ψ∂

Φ∂Φµ

∂∂

σ2

2

2

22

rr

1rr

1

rz−+

−∇=

∂θΨ∂

∂θ∂Ψ∂

θ

Φ∂Φ∂

Φµ∂∂

σθ 2

2

2

2

22

r

1rr

1

r

1rr

1z

+−

∂−−∇=

−∇−=

2

22

zz

)2(z ∂

Φ∂Φµ

∂∂

σ (10.54)

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107

zr21

z)1(

r1 2

2

22

z ∂∂Ψ∂

Φ∂Φµ

∂θ∂

τϑ −

−∇−=

zr21

z)1(

r

2

2

22

rz ∂θ∂Ψ∂

Φ∂Φµ

∂∂

τ +

−∇−=

2

2

2

22

rz2

1

rrrzr1

Ψ∂

Ψ∂Φ∂Φ∂

∂θ∂∂

τ θ −−

+−=

−∇−

+=

2

22

z)1(2

E1

w∂

Φ∂Φµ

µ

∂∂

−∂∂

∂+−=

θΨΦµ

r1

zrE1

u2

∂∂

+∂∂

∂+−=

rzr1

E1

v2 Ψ

θΦµ

Compatibility equations:

0zr

1rr

1

rzr

1rr

1

r 2

2

2

2

22

2

2

2

2

2

22

2=

+++

+++

Φ∂

∂θ

Φ∂∂Φ∂

Φ∂

∂θ

∂∂∂

∂ (10.55)

0zr

1rr

1

r 2

2

2

2

22

2=+++

Ψ∂

∂θ

Ψ∂∂Ψ∂

Ψ∂

10.5.4 Cartesian volume co-ordinates

Equilibrium:

0zyxxzxyx =++

∂∂τ

∂τ

∂∂σ

0zyxzyyyx =++

∂τ

∂σ

∂τ (10.56)

0zyxzzyzx =++

∂∂σ

∂τ

∂∂τ

Strains and displacements:

zw

yv

xu

zyx ∂∂

ε∂∂

ε∂∂

ε === (10.57)

zu

xw

yw

zv

xv

yu

zxyzxy ∂∂

∂∂

γ∂∂

∂∂

γ∂∂

∂∂

γ +=+=+=

Elasticity:

E

)( zyxx

σσµσε

+−=

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108

E

)( xzyy

σσµσε

+−=

E

)( yxzz

σσµσε

+−= (10.58)

yzyz E)1(2τ

µγ

+=

zxzx E)1(2τ

µγ

+=

xyxy E)1(2τ

µγ

+=

Stress potentials (Van Cauwelaert, 1985):

yxxz

2

2

22

x ∂∂Ψ∂

Φ∂Φµ

∂∂

σ +

−∇=

yxyz

2

2

22

y ∂∂Ψ∂

Φ∂Φµ

∂∂

σ −

−∇=

−∇−=

2

22

zz

)2(z ∂

Φ∂Φµ

∂∂

σ (10.59)

zx21

z)1(

y

2

2

22

yz ∂∂Ψ∂

Φ∂Φµ

∂∂

τ −

−∇−=

zy21

z)1(

x

2

2

22

zx ∂∂Ψ∂

Φ∂Φµ

∂∂

τ +

−∇−=

−−−=

2

2

2

23

xyyx2

1zyx ∂

Ψ∂

Ψ∂∂∂∂

Φ∂τ

∂−∇−

+=

2

22

z)1(2

E1

Φµµ

∂∂

−∂∂

∂+−=

yzxE1

u2 ΨΦµ

∂∂

−∂∂

∂+−=

xzyE1

v2 ΨΦµ

Compatibility equations:

0zyxzyx 2

2

2

2

2

2

2

2

2

2

2

2=

++

++

Φ∂

Φ∂

Φ∂

∂ (10.60)

0zyx 2

2

2

2

2

2=++

Ψ∂

Ψ∂

Ψ∂

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10.5.5 Axi-symmetric Cylindrical Co-ordinates for an orthotropic body

An orthotropic body has a Young’s modulus in the horizontal plane Eh that is different of the modulus inthe vertical direction Ev. We call n = Ev/Eh the degree of anisotropy. As proposed in paragraph 10.4, for atwo-dimensional body, we accept equation (10.37)

E2n1

G1 µ++

=

Further we shall also accept the same equation between the Poisson’s ratios as between the moduli, asproposed by Eftimie (1973) and Van Cauwelaert (1983)

ν = (10.61)

where ν is Poisson’s ratio in the horizontal plane.

Equilibrium:

0rzr

rrzr =−

++ θσσ∂

∂τ∂

∂σ

(10.62)

0rzrrzzrz =++

τ∂

∂σ∂

∂τ

Strains and displacements:

rw

zu

zw

ru

ru

rzzr ∂∂

∂∂

γ∂∂

εε∂∂

ε θ +==== (10.63)

Elasticity:

E

(n )zrr

σσµσε θ +−

=

E

(n )zr σσµσε θ

θ+−

= (10.64)

E)( rz

zθσσµσ

ε+−

=

rzrz E2n1

τµ

γ++

=

Stress potential (Lekhnitskii, 1963,Van Cauwelaert, 1983):

+−∇+=

2

22

rr

)1(n)n(z ∂

Φ∂µΦµµ

∂∂

σ

+−∇+=

rr1

)1(n)n(z

2∂Φ∂

µΦµµ∂∂

σθ

+−∇−++=

2

2222

zz

)1(n)nnn(z ∂

Φ∂µΦµµ

∂∂

σ (10.65)

+−∇−=

2

222

rzz

)1(n)n(r ∂

Φ∂µΦµ

∂∂

τ

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110

+−∇++−=

2

2222

z)1(n)2n1)(n(

E1

w∂

Φ∂µΦµµ

zrE)n)(1(n

u2

∂∂∂++

−=Φµµ

Compatibility equation:

0zn

nrr

1

rzrr1

r 2

2

2

22

2

2

2

2

2

2=

−++

++

Φ∂

µ

µ∂Φ∂

Φ∂

∂∂∂

∂ (10.66)

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Chapter 11 The Integral Transforms.

11.1 Introductory note

We demonstrated in Chapter 10 that the functions expressing equilibrium or continuity must be solutionsof Laplace or assimilated equations. Consider Laplace equation in two-dimensional co-ordinates

0yx 2

2

2

2=

∂+

∂ φφ

Look for a solution obtained by separation of the variables. Indeed, such types of solutions are mucheasier in expressing boundary conditions. Assume solution:

yxecos=φSuppose now that, utilising this solution, we want to express that a vertical stress, σy, acting on ahorizontal boundary, located at y = 0, has next form (boundary condition)

- for –a < x < a: σy = f(x) = - p- for | x | > a: σy = f(x) = 0

Applying (10.31), we obtain the following expression for the vertical stressy

y xecos−=σand, for y = 0,

xcos)x(fy −==σAt first sight, the boundary condition cannot be satisfied with the chosen solution because of theoscillating character of the cos – function. With help of an integral transformation a cos- function canbecome discontinuous, but uniform. We first introduce in the original solution a “dummy” variable (avariable independent of the differential equation)

mye)mxcos(=φin such a way that φ is still solution of the Laplace equation. If we multiply now φ by a function F(m) ofthe dummy variable and integrate φ from 0 to ∝, we still have a solution satisfying the Laplace equation.

∫∞

=0

my dm)m(Fe)mxcos(φ

The vertical stress writes now, for y = 0,

∫∞

−==0

2y dm)m(F)mxcos(m)x(fσ

We have performed an integral transform on f(x), where F(m) is called the kernel of the transform.The purpose of this chapter is to determinate expressions for F(m) satisfying a series of boundaryconditions. In this particular case, we will find

3m

)masin(p2)m(F

π=

The transform is called a Fourier transform or a Fourier integral.

Generally, when the solutions of the Laplace equations are expressed in Cartesian co-ordinates, using cosand sin – functions, the transforms will be Fourier transforms; when the solutions are expressed in polaror cylinder co-ordinates, using Bessel functions, the transforms will be Hankel transforms.

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11.2 Helpful relations

∑∞

=

++=

1nnn

0L

xnsinb

Lxn

cosa2

a)x(f

ππ

∫−

=L

Ln dx

Lxn

cos)x(fL1

∫−

=L

Ln dx

Lxn

sin)x(fL1

[ ] ααααα dxsin)(Bxcos)(A)x(f0∫∞

+=

∫ ∫∞

∞−

∞−

== xdxsin)x(f1

)(Bxdxcos)x(f1

)(A απ

ααπ

α

∫∞

=0

0 dm)m(F)mr(mJ)r(F

∫∞

=0

0 dr)r(F)mr(rJ)m(F

11.3 The Fourier expansion (Spiegel, 1971)

11.3.1 Definitions

A function f(x) is meant to have a period T if for all x, f(x + T) = f(x), where T is a positive constant. Theleast value of T > 0 is called the least period or simply the period of f(x). Let f(x) be defined in theinterval (-L, L) and outside the interval by f(x + 2L) = f(x), i.e. assume that f(x) has a period 2L. TheFourier series or Fourier expansion corresponding to f(x) is given by

∑∞

=

++=

1nnn

0L

xnsinb

Lxn

cosa2

a)x(f

ππ (11.1)

where the Fourier coefficients an and bn are

∫−

=L

Ln dx

Lxn

cos)x(fL1

(11.2)

∫−

=L

Ln dx

Lxn

sin)x(fL1

(11.3)

A function f(x) is called odd if f(-x) = - f(x). A function f(x) is called even if f(-x) = f(x).A half range Fourier sine or cosine series is a series in which only sine terms or only cosine terms arepresent respectively. When a half range series corresponding to a given function is desired, the functionis generally defined in the interval (0, L) and then the function is specified as odd or even, so that it isclearly defined in the other half of the interval (-L, 0). In such a case we have for half range sine series

∫==L

0nn dx

Lxn

sin)x(fL2

b0aπ

(11.4)

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113

and for half range cosine series

∫==L

0nn dx

Lxn

cos)x(fL2

a0bπ

(11.5)

11.3.2 Proof of the Fourier expansion.

Assume that:- f(x) is defined and single-valued except possibly at a finite number of points in (- L, L)- f(x) is periodic outside (- L, L) with period 2L- f(x) and f’(x) are piecewise continuous in (- L, L).

then the series (11.1) with coefficients (11.2) and (11.3) converges to- f(x) is x is a point of continuity- [f(x + 0) + f(x – 0)]/2 if x is a point of discontinuity.

Consider the series

∑∞

=

++=

1nnn L

xnsinb

Lxn

cosaA)x(fππ

assuming that it converges uniformly to f(x) in (- L, L). Multiply by L

xmcos

π and integrate from – L to L

∫ ∑∫−

=−

+=L

L 1nn

L

L

dxL

xncos

Lxm

cosadxL

xmcosAdx

Lxm

cos)x(fππππ

dxL

xnsin

Lxm

cosb1n

nππ∑

=+

Applying the trigonometric functions product equations, one proves easily that

∫∫−−

==L

L

L

L

0dxL

xnsin

Lxm

sindxL

xncos

Lxm

cosππππ

when m ≠ n

∫∫−−

==L

L

L

L

LdxL

xnsin

Lxm

sindxL

xncos

Lxm

cosππππ

when m = n

∫−

=L

L

0dxL

xncos

Lxm

sinππ

0dxL

xksindx

Lxk

cosL

L

L

L∫∫

−−

==ππ

if k = 1,2, 3,

We obtain by integration

∫−

=L

LmLadx

Lxm

cos)x(fπ

if m ≠ 0

Thus

∫−

=L

Lm dx

Lxm

cos)x(fL1

if m = 1,2,3,… (11.6)

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Similarly multiplying by L

xmsin

π and integrating from – L to L proves

∫−

=L

Lm dx

Lxm

sin)x(fL1

if m = 1,2,3,… (11.7)

Integrating f(x) from – L to L yields

∫−

=L

L

AL2dx)x(f

∫−

=L

L

dx)x(fL21

A

Put m = 0 in (11.6)

∫−

=L

L0 dx)x(f

L1

a

Hence

2a

A 0= (11.8)

Which proves the Fourier expansion formula.

11.3.3 Example

Express the function f(x) represented in Figure 11.1 as a Fourier series.

Figure 11.1 Example of Fourier series

The function is an even function with period π. Hence

∑∞

=+=

1nn

0 )nx2cos(a2

a)x(f

dx)nx2cos(p2

dx)nx2cos(p2

dx)nx2cos(02

a2/

a

a

a

a

2/n ∫∫∫ ++=

π

ππππ

)na2sin(n

p2an π

=

πap4

a0 =

+= ∑

=1n n)na2sin()nx2cos(

ap2

)x(fπ

(11.9)

2a 2a

π

p

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11.4 The Fourier integral

11.4.1 Definition

Assume that- f(x) is defined and single-valued except possibly at a finite number of points in (- L, L)- f(x) is periodic outside (- L, L) with period 2L- f(x) and f’(x) are piecewise continuous in (- L, L).- Also assume that ∫ f(x)dx converges, i.e. is absolutely integrable in (- ∞, ∞).

The Fourier’s integral theorem states that

[ ] ααααα dxsin)(Bxcos)(A)x(f0∫∞

+=

(11.10)where

∫∞

∞−

= uducos)u(f1

)(A απ

α (11.11)

∫∞

∞−

= udusin)u(f1

)(B απ

α (11.12)

Fourier’s integral theorem can also be written in the form

∫ ∫∞

=

−∞=

−=0 u

dud)ux(cos)u(f1

)x(fα

ααπ

(11.13)

When f(x) is either an even or an odd function, Fourier’s integral theorem simplifies into

[ ]dxxcos)(A)x(f0∫∞

= αα (11.14)

[ ]dxxsin)(B)x(f0∫∞

= αα (11.15)

where

∫∞

=0

uducos)u(f2

)(A απ

α (11.16)

∫∞

=0

udusin)u(f2

)(B απ

α (11.17)

11.4.2 Proof of Fourier’s integral theorem

We demonstrate Fourier’s integral theorem by use of the Fourier expansion defined by (11.1).

∑∞

=

++=

1nnn

0L

xnsinb

Lxn

cosa2

a)x(f

ππ

where

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116

∫−

=L

Ln du

Lun

cos)u(fL1

∫−

=L

Ln du

Lun

sin)u(fL1

By substitution

∫ ∑ ∫−

=

∞−

−+=L

L 1ndu)xu(

Ln

cos)u(fL1

du)u(fL2

1)x(f

π (11.18)

If we assume that ∫f(u)du converges, the first term of the second member of (11.18) approaches zerowhen L → ∞. Hence

∑ ∫∞

=

∞−∞→−=

1nLdu)xu(

Ln

cos)u(fL1

lim)x(fπ

Calling ∆α = π/L, we can now write that

−= ∫∞

∞−→du)xu(cos)u(flim

1)x(f

0α∆α∆

π α∆

⋅⋅⋅−+−+ ∫ ∫

∞−

∞−

du)xu(3cos)u(fdu)xu(2cos)u(f α∆α∆

∫ ∑∞

∞−

=→−=

1n0du)u(f)xu(ncos

1lim)x(f α∆α∆

πα∆

We recognise the definition of the integral:

∑ ∫∞

=

→=

1n 00dacosncoslim αα∆α∆

α∆

Hence we may write

∫ ∫∫ ∑∞ ∞

∞−

∞−

=→−=−=

01n0du)ux(cos)u(fd

1du)u(f)xu(ncos

1lim)x(f αα

πα∆α∆

πα∆ (11.19)

which is Fourier’s integral formula (11.13).

11.4.3 Example

Express the function f(x) represented Figure 11.2. as a Fourier series

Figure 11.2 Example of Fourier integral

The function f(x) is an even function, hence

2ap

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∫∞

=0

d)xcos()(A)x(f ααα

απα

απ

α)asin(p2

du)ucos(p1

)(Aa

a

== ∫−

∫∞

=0

d)asin()xcos(p2

)x(f αα

ααπ

(11.20)

11.5 The Hankel’s transform

11.5.1 Definition

The Hankel’s transform, sometimes called the Fourier-Bessel Integral, is represented by next pair ofintegrals

∫∞

=0

dm)m(F)mr(mJ)r(f ν (11.21)

∫∞

=0

dr)r(f)mr(rJ)m(F ν (11.22)

as far as ν ≥ ½. Generally, the transform is applied on the Bessel function J0, often solution of differentialequations in polar or cylinder co-ordinates:

∫∞

=0

0 dm)m(F)mr(mJ)r(f (11.23)

∫∞

=0

0 dr)r(f)mr(rJ)m(F (11.24)

The proof of equations (11.21) and (11.22) is given in the specialised literature by the so-called method ofdual integrals. However this method is beyond the scope of this book.

11.5.2 Example

We start with a simple application of the transform, merely given as an example. As we will see in theapplications, the Bessel function of the first kind and of zero order, J0(mr), is solution of many differentialequations expressed in polar or in cylinder co-ordinates. Assume that we want to express a discontinuouscondition on the function f(r) such as:

- for r < a, f(r) = p- for r > a, f(r) = 0

We apply Hankel’s transform

∫∞

=0

0 dm)m(F)mr(mJ)r(f

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∫ ∫∞

+=a

0 a00 dr)mr(rJ0dr)mr(rJp)m(F

m)ma(J

pa)m(F 1=

∫∞

=0

10 dm)ma(J)mr(Jpa)r(f (11.25)

11.5.3 Application of the discontinuous integral of Weber and Schafheitlin

The properties of the discontinuous integral of Weber and Schafheitlin, developed in §8.6, will allow usto verify that equation (11.25) corresponds with the assumed conditions about f(r)

pdm)ma(J)mr(Jpa0

10 =∫∞

for r < a

∫∞

=0

10 0dm)ma(J)mr(Jpa for r >a

Hence, without having formally proven Hankel’s transform integral, we can nevertheless use theproperties of the transform applying the Weber – Schafheitlin integral. We now demonstrate how to applyit with the example of § 11.5.2. Assume that we want to express a discontinuous condition on the functionf(r) such as

- for r < a, f(r) = p- for r > a, f(r) = 0

using the properties of the Weber – Schafheitlin integral.

Assume that the solution can be expressed as follows

∫∫∞

==0

100

0 dmm

)ma(J)mr(JCdm

m

)ma(J)mr(mJC)r(f

λµ

λµ

(11.26)

Determine the values of C, µ and λ in order to meet the assumed conditions.

For r < a, the solution of equation (11.26) is

( )p

a

r;1;

21

,2

1F

21

1

21

2a

rC)r(f

2

2

1

0=

+−−+−

++

+−

=−

λµλµλµ

ΓΓ

λµΓ

λλ (11.27)

For r > a, the solution of equation (11.26) becomes

( )0

r

a;1;

21

,2

1F

21

1

21

2r

aC)r(f

2

2

1=

+

+−+−

++−

+

+−

=−+

µλµλµ

λµΓµΓ

λµΓ

λλµ

µ(11.28)

Equation (11.27) must be equal to p. Hence C = pa and the hypergeometric series must be equal to 1, thuslimited to its first term. This requires the second term between brackets to be zero: -µ - λ + 1 = 0.

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Equation (11.28) must be equal to zero. Hence the gamma function in the denominator left of thehypergeometric series must be infinite, or the argument of the gamma function must be zero: -µ +λ +1 =0. Hence, we become 2 equations for the determination of λ and µ, resulting in: λ = 0, µ = 1.

The integral transform of f(r) becomes

∫∞

=0

10 dm)ma(J)mr(Jpa)r(f (11.29)

which is exactly the same equation as (11.25) obtained by application of Hankel’s transform.

The reader will realise that many different boundary conditions can be expressed using the presentedmethod.

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Chapter 12 Simple Applications of Beams and Slabs on an Elastic Subgrade

12.1 The elastic subgrade

Consider a section of a beam represented in Figure 12.1.

Figure 12.1 Beam on an elastic subgrade

The beam is subjected to a distributed load p at its surface, a vertical soil reaction q and a shear τ at itsbottom. The equilibrium equation of the vertical forces is given by

EIqp

dx

wd4

4 −= (12.1)

However equation (12.1) contains two unknowns, w and q, hence, in order to be able to solve thedifferential equation, we must have more information on the behaviour of the soil. Winkler (1867)suggested that the subgrade reacts elastically as a layer of vertical springs. He proposed a linear relationbetween the vertical reaction q and the deflection w of the beam

kwq = (12.2)

where the constant k, expressed in N/mm3, is called the modulus of subgrade reaction.Equation (12.1) transforms into

EIp

EIkw

dx

wd4

4=+ (12.3)

Almost a century later, as recalled by Pronk (1993), Pasternak (1954) suggested that the subgrade couldalso react in the horizontal direction. He presented his subgrade model as a two-layer consisting of a shearlayer supported by the classical Winkler layer. The shear layer is characterised by its shear modulus G (inN/mm) and the Winkler layer by its modulus of reaction k (Figure 12.2). The contact pressure betweenbeam and shear layer is p1 and between the shear layer and the Winkler’s subgrade q.

E

p

h

k, Gq

dx

τ

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Figure 12.2 Plate-Pasternak model

By Winkler’s assumption kwq =

Vertical equilibrium in the beam is expressed by

14 ppwEI −=∇

The vertical equilibrium in the shear layer can be deduced from Figure 12.3.

Figure 12.3 Vertical equilibrium in Pasternak’s shear layer

0dxdxdT

qdxdxp1 =+−

dxdT

qp1 −=−

T is the shear force per unit of width in the shear layer. Pasternak (1954) assumed that the shear force isproportional with the variation of the deflection (figure 12.4).

dxdw

GT =

p

E

G

k

p1

q

dx

T + dT/dx

T

q

p1

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Figure 12.4 Shear force is proportional to deflection

Hence

2

2

dx

wdG

dxdT

=

and

2

2

1dx

wdGkwp −=

Finally, the equilibrium equation for a beam on a Pasternak foundation writes

EIp

wEIk

wEIG

w 24 =+∇−∇ (12.4)

12.2 The beam on an elastic subgrade subjected to an isolated load: ∂4w/∂x4+Cw=0

Consider the beam of infinite length as given in Figure 12.5.

Figure 12.5 Beam of infinite length on an elastic subgrade

The beam is submitted to a single load P at its surface and a soil reaction q.

For this simple application we adopt Winkler’s assumption. Note that p = 0, because the load is reducedto a single load. Hence equation (12.3) transforms into:

P

h E

kq

dx

T

T + dT/dx

dw

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124

0EIkw

x

w4

4=+

∂ (12.5)

We notice that the factor EI/k has the dimensions of a length to the fourth power, hence we set

41

kEI

l

=

and call l the elastic length of the structure. Finally, equation (12.5) transforms into

0l

w

x

w44

4=+

∂ (12.6)

We solve this differential equation by resolution of the characteristic equation (§ 1.4.2).

The characteristic equation, written m4 + 1/l4 can be factorised in:

0l

im

li

mli

mli

m =

−−

−+

+ (12.7)

where i = √(-1).

Squaring both sides of next equations allows verifying that

2i1

i2i1

i−

=−+

=

and equation (12.7) can be transformed in

−−

−+

+−

++

2li1

m2li1

m2li1

m2li1

m (12.8)

The solution of equation (12.8) is then

++

+=

2lx

sinD2l

xcosCe

2lx

sinB2l

xcosAew 2l

x

2l

x

(12.9)

In order to determine the integration constants A, B, C and D we write down the boundary conditions:- (1) for x = ± ∞, w = 0- (2) for x = 0, w is maximum, hence dw/dx = 0- (3) for x = 0, T = -P/2.

Condition (1) requires splitting equation (12.9) into two parts. The first part (equation 12.10) is valid onthe left side of the beam, for negative values of x

+=

2Lx

sinB2l

xcosAew 2l

x

L (12.10)

The second part (equation 12.11) is valid on the right side of the beam, for positive values of x

+=

2lx

sinD2l

xcosCew 2l

x

R (12.11)

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This requires a fourth boundary condition

- (4) for x = 0, wL = wR

The solution of the system is satisfied by

kl42P

EI42Pl

DCBA3

====−=

The equation valid on the right side of the load is

−=

2lx

sin2l

xcose

kl42P

w 2l/xL (12.12)

The equation valid on the left side of the load is

+= −

2lx

sin2l

xcose

kl42P

w 2l/xR (12.13)

The deflection in the middle of the beam (x = 0) is

kl42P

w 0x == (12.14)

12.3 The beam on an elastic subgrade subjected to a distributed load: ∂4w/∂x4+Cw=0

Consider the beam of Figure subjected to a distributed load p over a width 2a. The deflection in the axisof the load is obtained by integrating (12.12) over the width 0 – a and (12.13) over the width –a – 0.The solution is

2la

sinekl

pa2w 2l/a

20x == (12.15)

12.4 The infinite slab subjected to an isolated load: ∇2∇2w +Cw = 0

Westergaard solved this problem in 1926, mainly following the same method. Consider a slab of infiniteextend subjected to an isolated load P and a soil reaction q. Apply the solution developed in § 10.2.

Dq

y

w

xx

w2

x

w4

4

22

4

4

4−=

∂+

∂∂

∂+

∂ (12.16)

Here also, equation (12.16) contains only the negative term -q because there is no distributed load appliedon top of the beam. Because of the axial symmetry, transform (12.16) in polar co-ordinates

0Dkw

drdw

r1

dr

wddrd

r1

dr

d2

2

2

2=+

+

+ (12.17)

Because of the axial symmetry the terms in ∂/∂θ vanish. The solution is obtained by splitting (12.17) intotwo simultaneous differential equations

zdrdw

r1

dr

wd2

2=+

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126

wdrdz

r1

dr

zd2

2−=+

Both equations are together verified if z = µ iw. Hence the solution of (12.17) is given by the solution of(12.18)

0w)i(drdw

r1

dr

wd2

2=−+ µ (12.18)

and the solution of (12.18) is given in § 3.6.)l/r(dkei)l/rker(c)l/r(bbei)l/r(aberw +++= (12.19)

The boundary conditions are the same as in previous paragraph.- (1) for r = ∞, w = 0- (2) for r = 0, w must be finite- (3) for r = 0, w is maximum hence dw/dr = 0- (4) for r = 0, T = -P/(2πρ).

where ρ is the radius of a small circle around the load.

Boundary condition (1) requires a and b to be zero because the asymptotic values of ber and bei areinfinite (equations (4.49) and (4.50). Boundary condition (2) requires c to be zero because ker(0) isundefined. Hence the solution reduces in

)l/r(dkeiw = (12.20)Boundary condition (3) writes

[ ]

⋅⋅⋅−+

⋅⋅⋅−=

→ rl21

!1!1)l2/r(

)l2/rlog(l2!1!1

)l2/r(2dr

)l/r(dkeilim

2

0rγ

0l2!1!1

)l2/r(2l2!1!1)l2/r(4

4

3=⋅⋅⋅+

⋅⋅⋅−−

π (12.21)

Hence, boundary condition (3) is satisfied. The constant d is determined by the fourth boundary condition

πρ2P

drdw

r1

dr

wddrd

DT2

2−=

+−=

Limited to the first terms, we derived

[ ] )l/r(ber4

)l2/rlog()l/r(bei)l/r(keiπ

γ −+−=

!2!2)l2/r(

1)l/r(ber4

−=

!1!1)l2/r(

)l/r(bei2

=

Hence

22

2

rl

1drdw

r1

dr

wddrd

−=

+

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127

For r = ρ 2l

Dd

2P

Tρπρ

=−=

Dl

2P

d2

π−=

[ ]Dl

2P

)l/r(ber4

)l2/rlog()l/r(beiw2

ππ

γ

−+−−= (12.22)

In the axis of the load (r = 0)

D8Pl

w2

= (12.23)

The deflection in the axis of the load a slab subjected to a distributed load could be obtained byintegrating equation (12.21). However this is hard work. As will be shown in chapter 14, the solution canbe obtained in a simpler and more general way using Fourier’s transform developed in chapter 11.

Observe that only three boundary conditions were required to determine the boundary constants:conditions (1), (2), and (4). However the equilibrium equation (12.17) is a differential equation of thefourth order and thus requires 4 boundary conditions to be met. Interesting is that the fourth boundarycondition, condition (3), although not necessary for the determination of the unknown boundaryconstants, is essential for identifying that solution (12.19) is the appropriate solution of the problem.

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129

Chapter 13 The Beam Subjected to a Distributed Load and Resting on a PasternakFoundation

13.1 The basic differential equations

Consider the beam presented in Figure 13.1.

Figure 13.1 Beam on a Pasternak foundation

The equilibrium equation is given by (12.4)

EIp

EIkw

dx

wdEIG

dx

wd2

2

4

4=+− (13.1)

A particular solution of equation (13.1) allows to express the load as a pressure p uniformly distributedover a width 2a. From this point on we shall express the loads by means of the appropriate integraltransform. In the case of a beam, we will use Fourier’s integral (11.10) and more specific the integral(11.20). By using this method, differential equation (13.1) is the unique equation to solve in the case of abeam of infinite length. Indeed the distances c and d from the axis of the load to the edges of the beam arelarge enough so that the end effects can be neglected. For shorter beams, supplementary differentialequations are necessary in order to express the boundary conditions of each specific case. The differentialequation necessary to express the boundary conditions of the beam itself is the homogeneous equationcorresponding to (13.1)

0EIkw

dx

wdEIG

dx

wd2

2

4

4=+− (13.2)

Making use of complementary solutions of the homogeneous equation does not affect the state of loadingand do not alter the physics of the problem. The differential equations that are required to express theboundary conditions outside the beam are of two kinds. Either the beam has free edges at the end. Thedifferential equation then reflects a soil subjected to horizontal shear forces. Formula (13.1) can berewritten as follows

pkwdx

wdG

dx

wdEI

2

2

4

4=+− (13.3)

There is no vertical load, hence p = 0; there is no slab, hence EI = 0.

G

x

d

2a

c

p

E

y

h

k q

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130

0kwdx

wdG

2

2=+− (13.4)

In the second case, the beam has a joint separating it from the adjacent beam. The required differentialequation can again be deduced from equation (13.1). There is no load, hence p = 0, and the equation

0EIkw

dx

wdEIG

dx

wd2

2

4

4=+− (13.5)

is again the homogeneous equation: this time applicable to the unloaded neighbouring beams. We recallthat moments and shear forces can be deduced from the equations for the deflections by next equations:- moment in the beam

2

2

dx

wdEIM −= (13.6)

- shear force in the beam

3

3

dx

wdEIT −= (13.7)

- shear force in the subgrade

dxdw

GT = (13.8)

13.2 Case of a beam of infinite length

13.2.1 Solution of the differential equation

Let 1/l4 = k/EI and 2g/l2 = G/EI. Equation (13.1) now becomes

EIp

wl

1

dx

wd

l

g2

dx

wd42

2

24

4=+− (13.9)

We express the load by equation (11.20) developed in the example.

∫∞

<==0

axforpdtt

)l/atsin()l/xtcos(p2p

π (13.10)

∫∞

>==0

axfor0dtt

)l/atsin()l/xtcos(p2p

π (13.11)

In order to satisfy the equilibrium equation the deflection w must be express similarly

∫∞

=0

dtt

)l/atsin()l/xtcos()t(A

p2w

π (13.12)

where A(t) is a function of the integration variable. Introducing equation s(13.10) and (13.12) into (13.9)yields

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THE BEAM ON A PASTERNAK FOUNDATION

131

( )∫∞

++=

024

dt1gt2tt

)l/atsin()l/xtcos(kp2

(13.13)

The explicit solution of equation (13.13) depends on the value of g. It is obtained by a reduction of thedegree of the denominator from 4 to 2 followed by an application of the integrals in the complex planeequations (9.15) and (9.16).

- For g < 1

Integral (13.13) is split in the 2 integrals:

( ) ( )

−+−

++−= ∫∫

∞∞

022

0222

dt)ia(tt

)l/atsin()l/xtcos(dt

)i(tt

)l/atsin()l/xtcos(

g1

ikp

wββαπ

(13.14)

where 2

22

g1

g22

g12

g1

−=

−=

−=

+=

αββα

γβα

Applying equations (9.15) and (9.16) one obtains- For x < -a

+

−+

=+

l)ax(

sinl

)ax(cose

k2p

w l)ax(

βγ

βα

−−

−−

l)ax(

sinl

)ax(cose l

)ax(β

γβ

α

(13.15)

- For –a < x < a

+−

−=−−

l)xa(

sinl

)xa(cose2

k2p

w l)xa(

βγ

βα

+

++

−+−

l)xa(

sinl

)xa(cose l

)xa(β

γβ

α

(13.16)

- For x > a

+−

=−−

l)ax(

sinl

)ax(cose

k2p

w l)ax(

βγ

βα

+

++

−+−

l)xa(

sinl

)xa(cose l

)xa(β

γβ

α

(13.17)

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132

- For g > 1

( ) ( )

+−

+−= ∫∫

∞∞

022

0222

dtatt

)l/atsin()l/xtcos(dt

tt)l/atsin()l/xtcos(

1g

1k4p

(13.18)

where 1gg1gg 22 −−=−+= βα

Applying (915) and (9.16) one obtains

- For x < -a

−−

−=

−+−+

2

l)ax(

l)ax(

2

l)ax(

l)ax(

2

eeee

1gk4

pw

αβ

ααββ

(13.19)

- For –a < x < a

−−

−−−

−=

+−+−

−−

2

l)xa(

l)xa(

2

l)xa(

l)xa(

2

ee2ee2

1gk4

pw

αβ

ααββ

(13.20)

- For x > a

−−

−=

+−

−−

+−

−−

2

l)ax(

l)ax(

2

l)ax(

l)ax(

2

eeee

1gk4

pw

αβ

ααββ

(13.21)

- For g = 1

Applying the Rule of de l’Hospital for g → 1, one gives

- For x < -a

−−

+

−=−+

lax

2el

ax2e

k4p

w lax

lax

(13.22)

- For –a < x < a

+

+−

+−=+

−−

lxa

2el

xa2e4

k4p

w lxa

lxa

(13.23)

- For x > a

+

+−

+=+

−−

lxa

2el

ax2e

k4p

w lxa

lax

(13.24)

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133

13.2.2 Application

Consider the infinite beam given in Figure 13.2.

Figure 13.2 Infinite beam on a Pasternak foundation

For different values of G, deflections, moments and shear forces in the beam and in the shear layer arelisted in Table 13.1. One will note the positive influence of an increasing shear resistance of the soil.

G x = 0 x = 500 x = 1000

w 1.17 0.90 0.48M 37923 6848 -6959

T-beam 0 -47 -120

T-subgrade 0 0 0w 1.09 0.85 0.47M 35303 5781 -6271

T-beam 0 -42 -1010000

T-subgrade 0 -8 -7w 0.90 0.71 0.44M 28621 3230 -4425

T-beam 0 -30 -550000

T-subgrade 0 -28 -25

Table 13.1 Deflections, moments and shear forces in a beam of infinite length.

13.3 Case of a beam of finite length with a free edge

13.3.1 Solution #1

Consider the beam with one edge as presented in Figure 13.3. In order to reduce the amount ofmathematics, we will consider a beam with only one free edge and of infinite length in the other directionwhich solution requires only 2 or 3 boundary equations. The mathematics for a beam with two edges areexactly the same but the number of boundary equations is double.

With the first boundary condition we will demonstrate that the moment in the beam and at the edge mustbe zero; all authors even in the presence of dowels or other transfer devices accept this condition. Hence,M(x=c) = 0. If the edge is free, the shear force in the beam and at the edge should also be zero, which istrue in the case of a Winkler foundation. In the case of a Pasternak foundation the shear force in the beamshould be zero but all together the shear force in the shear layer should be continuous from the inside to

h = 200 E= 20000

2a = 200

yk = 0,1

x

G

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134

the outside of the beam. However, the differential equation of equilibrium (13.2) is of the fourth order:hence we can only express a total of 4 boundary conditions with two edges, and thus only a total of 2boundary conditions with only one edge. Hence we have only 1 equation left to express the shearconditions at the edge of the beam.

Figure 13.3 Half-infinite beam with free edge on a Pasternak foundation

The first manner is to express that the sum of the shear force in the beam plus the shear force in the shearlayer at the beam side of the edge is equal to the shear force in the shear layer at the outer side of theedge. Hence we write: (Tbeam + Tsoilin) (x=c) = Tsoilout (x=c) . This does not assure us that the shear force inthe beam will be zero at the edge, but practical applications show that this type of solution is close toreality. Mathematically we need 2 complementary solutions, wA and wB, of the homogeneous equilibriumequation

0wl

1

dx

wd

l

g2

dx

wd42

2

24

4=+− (13.25)

and one complementary solution, wc given in equation (13.4)

0kwdx

wdG

2

2=+− (13.26)

Again the explicit equations for the complementary solutions depend on the value of g. The parameters αand β are the same as in § 13.2.

- For g < 1

lx

cosew lx

α

=

To standardise to the general solution, we write the solution as follows

lx

cosek2p

w lx

α

= (13.27)

Indeed, multiplying the solution of a differential equation by a constant does not affect the result.

lx

sinek2p

w lx

α

= (13.28)

- For g= 1

h, E

G

k qy

x

c

2a

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THE BEAM ON A PASTERNAK FOUNDATION

135

lx

A ek2p

w = (13.29)

lx

B xek2p

w = (13.30)

- For g > 1

lx

A ek2p

= (13.31)

lx

B ek2p

= (13.32)

The reader will have noted that we now have a solution comprising of two terms with a positive exponent.The reason is that we have located the edge on the positive side of the x-axis. Hence the values of thefunctions wA and wB will be smaller for all the values of x < c so that in the numerical computationoverflow problems are automatically avoided. The solution of (13.26) is

g21

lx

Gk

xc eew

−−== (13.33)

For the complementary solution outside of the beam, we have chosen the negative exponent, because thefunction must vanish for x → ∞. The moment cancels by writing for x = c

[ ] 0BBwAAww2dx

2d=++ (13.34)

The boundary condition for the shear force becomes for x = c

[ ] [ ]cBA23

3w

dxd

CBwAwwdxd

l

g2

dx

d=++

− (13.35)

The system consists of two equations with 3 unknowns. To solve we need one more equation. Usuallyone chooses an equation expressing the deflections outside are a fraction of the deflections on the slabside

[ ] CBA CwBwAww =++δ (13.36)

where 0 ≤ δ ≤ 1. The solution of the problem consists now of a system of 3 equations, (13.34), (13.35)and (13.36), and 3 unknowns.

13.3.2 Application

Consider the finite beam as presented in Figure 13.4.

Table 13.2 gives the response of the deflections, moments and shear forces in the beam and in the shearlayer are given for different values of G and a deflection ratio of 0 %. Table 13.3 for a deflection ratio of50 % and Table 13.4 for a deflection ratio of 100 %.

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136

Figure 13.4 Example of a half-infinite beam with free edge on a Pasternak foundation

G x = 0 x = 500 x = 1000

w 1.23 1.01 0.57M 37767 9005 0

T-beam 3 -40 00

T-subgrade 0 0 0w 1.17 1.01 0.69M 33848 6159 0

T-beam 1 -35 710000

T-subgrade 1 -7 -7w 1.00 0.91 0.76M 26304 2115 0

T-beam -1 -29 850000

T-subgrade 8 -15 -8Table 13.2 Deflections, moments and shear forces in a beam of finite length; Moment in beam is zeroat the edge; Shear forces in beam and shear layer inside are equal with the shear forces in shear layer

outside. Deflection ratio is 0 %.

G x = 0 x = 500 x = 1000

w 1.23 1.01 0.57M 37767 9005 0

T-beam 3 -40 00

T-subgrade 0 0 0w 1.17 0.99 0.65M 34162 6501 0

T-beam 1 -36 610000

T-subgrade 1 -7 -8w 0.95 0.80 0.56M 27728 3722 0

T-beam 0 -28 450000

T-subgrade 4 -21 -18Table 13.3 Deflections, moments and shear forces in a beam of finite length; Moment in beam is zeroat the edge; Shear forces in beam and shear layer inside are equal with the shear forces in shear layer

outside. Deflection ratio is 50 %.

k = 0.1

h = 200 E = 20000

G

p

y

x

200

1000

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137

G x = 0 x = 500 x = 1000

w 1.23 1.01 0.57M 37767 9005 0

T-beam 3 -40 00

T-subgrade 0 0 0w 1.16 0.97 0.62M 34446 6811 0

T-beam 2 -35 510000

T-subgrade 1 -7 -8w 0.92 0.74 0.44M 28562 4663 0

T-beam 1 -28 250000

T-subgrade 2 -25 -26

Table 13.4 Deflections, moments and shear forces in a beam of finite length; Moment in beam is zeroat the edge; Shear forces in beam and shear layer inside are equal with the shear forces in shear layer

outside. Deflection ratio is 100 %.

Note that the variation in the results due to the variation of the G-values is significantly more dominantthan the variation due to the deflection ratios. Also see that, however different from zero, the shear forcesin the beam at the edges of the beam are very low.

13.3.3 Solution #2

Several authors are convinced that it is very difficult, if not impossible, to measure the deflection of thesubgrade outside the beam, and thus a fortiori the ratio between the subgrade deflection outside the beamand the deflection on top of the beam at the edge. We have further shown that a variation of the deflectionratio had no great influence on the results. Hence a solution has been searched for in order to avoid theuse equation of (13.36). An option is to assume that the sum of the shear force at the edge of the beamand the shear force in the shear layer is equal to the shear force in the shear layer under an infinite beam.In that case equation (13.35) writes

[ ] [ ]wdxd

l

g2BwAww

dxd

l

g2

dx

d2BA23

3−=++

− (13.37)

Hence the system is reduced to the 2 equations (13.34) and (13.37) with 2 unknowns. In fact the systemignores what happens outside of the beam and refers to what it can support under the influence of aninfinite beam

13.3.4 Application

Again, we consider Figure 13.2, but this time we apply the solutions (13.34) and (13.37). Deflections,moments and shear forces in the beam and in the shear layer are given for different values of G are listedin Table 13.5.

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138

G x = 0 x = 500 x = 1000

w 1.23 1.01 0.57M 37767 9005 0

T-beam 3 -40 00

T-subgrade 0 0 0w 1.15 0.93 0.52M 35403 7853 0

T-beam 2 -36 110000

T-subgrade 1 -7 -8w 0.90 0.71 0.39M 28940 5090 0

T-beam 1 -28 150000

T-subgrade 1 -27 -26

Table 13.5 Deflections, moments and shear forces in a beam of finite length; Moment in beam is zeroat the edge; Shear forces in beam and shear layer inside are equal with the shear forces in shear layer

for a beam of infinite length.

Note that the results do not differ very much with those of the previous tables. Hence we advocate the useof the method presented in this paragraph, simply because it is more easier. The results are strictly thesame in the four tables for G = 0. This is in fact the Winkler case where no shear transfer is possiblethrough the subgrade and where the deflection is by hypothesis zero. It is evident that the results must bethe same

13.4 Case of a finite beam with a joint

13.4.1 Solution #1

Consider Figure 13.5 where we have represented a semi-infinite beam with a joint at a distance c from theaxis of the load.

Figure 13.5 Half-infinite beam on a Pasternak foundation

We prefer this configuration in order to simplify the mathematical expressions. The solution for theloaded beam is given by of equation (13.13), which expresses the load and the two solutions wA and wB

with positive exponents in x of the homogeneous differential equation (13.2).

Eh

k q

G

p

2a

c

y

x

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139

The solution for the unloaded beam is given by solutions wC and wD with negative exponents in x of thehomogeneous differential equation (13.2). Those solutions depend again on the value of g.

- For g < 1

lx

cosek2p

w lx

α−

= (13.38)

lx

sinek2p

w lx

α−

= (13.39)

- For g = 1

lx

C ek2p

w−

= (13.40)

lx

D xek2p

w−

= (13.41)

- For g > 1

lx

C ek2p

−= (13.42)

lx

D ek2p

−= (13.43)

The boundary conditions are for x = c.- the moment at the edge of the loaded beam is zero;- the moment at the edge of the unloaded beam is zero;- the shear force beam + soil) at the loaded side of the edge is equal with the shear force (beam +

soil) at the unloaded side of the edge.

We need one additional boundary condition. As for the case of the beam with an edge, we propose asfourth condition a relation between the deflections at both sides of the joint: the deflection at the unloadedside is a fraction of the deflection at the loaded side

loadedunloaded ww δ= (13.44)

where 0 ≤ δ ≤ 1. The equations for the boundary conditions are

[ ] 0BBwAAww2dx

2d=++ (13.45)

[ ] 0DDwCCw2dx

2d=+ (13.46)

[ ] [ ]DC23

3

BA23

3DwCw

dxd

l

g2

dx

dBwAww

dxd

l

g2

dx

d+

−=++

− (13.47)

[ ] DCBA DwCwBwAww +=++δ (13.48)The solution now consists of a solvable system of four equations (13.45), (13.46), (13.47) and (13.48) andfour unknowns.

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13.4.2 Application

Consider the same application as in Figure , but this time with a joint instead of an edge. Deflections,moments and shear forces in the beam and in the shear layer are given for different values of G in Table13.6, Table 13.7 and Table 13.8 respectively for deflection ratios of 0, 50 % and 100 %.

G x = 0 x = 500 x = 1000

W 1.23 1.01 0.57M 37767 9005 0

T-beam 3 -40 00

T-subgrade 0 0 0W 1.17 1.01 0.69M 33848 6159 0

T-beam 1 -35 710000

T-subgrade 1 -7 -7W 1.00 0.91 0.76M 26304 2115 0

T-beam -1 -29 850000

T-subgrade 8 -15 -8

Table 13.6 Deflections, moments and shear forces in a beam of finite length; Moments in beams arezero at the edge; Shear forces in beam and shear layer are equal in both beams; Deflection ratio is 0%.

G x = 0 x = 500 x = 1000

W 1.21 0.92 0.38M 39758 11149 0

T-beam 4 -41 -80

T-subgrade 0 0 0W 1.13 0.88 0.43M 36277 8807 0

T-beam 3 .37 .310000

T-subgrade 1 -7 -8W 0.93 0.77 0.50M 28181 4233 0

T-beam 0 -28 350000

T-subgrade 3 .24 -21

Table 13.7 Deflections, moments and shear forces in a beam of finite length; Moments in beams arezero at the edge; Shear forces in beam and shear layer are equal in both beams; Deflection ratio is

50%.

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141

G x = 0 x = 500 x = 1000

W 1.20 0.87 0.29M 40754 12221 0

T-beam 5 -42 -120

T-subgrade 0 0 0W 1.11 0.82 0.31M 37384 10013 0

T-beam 4 -38 -710000

T-subgrade 1 -7 -8W 0.90 0.70 0.37M 29095 5266 0

T-beam 1 -28 150000

T-subgrade 1 -28 -27

Table 13.8 Deflections, moments and shear forces in a beam of finite length; Moments in beams arezero at the edge; Shear forces in beam and shear layer are equal in both beams; Deflection ratio is

100%.

It is evident that for the beam with a joint, the deflection ratio is meaning full especially near the joint.Also, the ratio between the deflections on top on two beams can easily be measured with a good accuracy.However, as in the previous case, a simpler method is also given.

13.4.3 Solution# 2

Similar to the case of a beam with a free edge, an easier method can be derived. It is clear that the ratiobetween the deflections depends on the amount of shear force transferred through the joint (for exampleby means of the dowels or aggregate interlock). In § 13.4.5 we demonstrate for a slab on a Winklerfoundation there is an equation between the amount of transferred shear Q and the ratio between thedeflections:

TT12

Q γδ

δ=

+= (13.49)

where T is the shear force in the infinite beam.Thus if we knew the amount of transferred shear, two boundary conditions would be sufficient:

- the moment at the edge of the loaded beam is zero;- the shear force at the edge of the loaded beam is Q as determined by equation (13.49)

However, a similar equation cannot been established for a beam on a Pasternak foundation but practiceshows that it can be utilised without great difficulties. We therefore think that a method similar to the oneproposed for the free edge can also been applied in the case of the jointed edge. We utilise equation(13.34) and slightly transform equation (13.37) as follows

- for x = c

[ ] 0BBwAAww2dx

2d=++ (13.50)

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142

[ ] .dxdw

l

g2

dx

wdBwAww

dxd

l

g2

dx

d23

3

BA23

3−=++

− γ (13.51)

When γ = 0, we have a free edge behaviour, with only shear transfer in the shear layer. When γ = 1, wehave the case of f full shear transfer.

13.4.4 Application

Consider Figure13.4, but this time with a joint instead of an edge. Deflections, moments and shear forcesin the beam and in the shear layer are listed for different values of G in Table 13.9 to Table 13.11. Theshear transfer is varied between 0, 67 and 100%. The shear transfer of 67 % corresponds with a deflectionratio of 50 %.

G x = 0 x = 500 x = 1000

w 1.23 1.01 0.57M 37767 9005 0T-beam 3 -40 00

T-subgrade 0 0 0w 1.17 1.01 0.69M 33848 6159 0T-beam 1 -35 7

10000

T-subgrade 1 -7 -7w 1.00 0.91 0.76M 26304 2115 0T-beam -1 -29 850000

T-subgrade 8 -15 -8

Table 13.9 Deflections, moments and shear forces in a beam of finite length; Moments in beams arezero at the joint; Shear transfer at the joint is 0%.

G x = 0 x = 500 x = 1000

w 1.21 0.92 0.38M 39768 11160 0T-beam 4 -41 -8

0

T-subgrade 0 0 0w 1.12 0.85 0.37M 36882 9466 0T-beam 3 .37 -510000

T-subgrade 1 -7 -8w 0.89 0.68 0.34M 29283 5478 0T-beam 1 -27 350000

T-subgrade 0 -29 -29

Table 13.10 Deflections, moments and shear forces in a beam of finite length; Moments in beams arezero at the joint; Shear transfer at the joint is 67%.

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143

G x = 0 x = 500 x = 1000

w 1.20 0.87 0.29M 40754 12221 0T-beam 5 -42 -120

T-subgrade 0 0 0w 1.11 0.81 0.29M 37611 10261 0T-beam 4 -38 -8

10000

T-subgrade 1 -7 -9w 0.88 0.67 0.32M 29453 5699 0T-beam 2 -27 050000

T-subgrade 0 -30 -30

13.11 Deflections, moments and shear forces in a beam of finite length; Moments in beams are zero atthe joint; Shear transfer is 100%

13.4.5 Proof that, in de case of a Winkler foundation, at a joint Q = γ T

For x = c, let d - a =(c - a) /(l√2). Consider equation (13.16) and its successive derivatives.

)adcos(e)adcos(ew )ad()ad( +−−= +−−−

)adsin(e)adsin(eM )ad()ad( +−−= +−−−

[ ] [ ])adsin()adcos(e)adsin()adcos(eT )ad()ad( +−+−−−−= +−−−

Ignore the stiffness EI and the powers of l√2 in the denominators of M and T, which, anyhow, simplify inthe expressions of the boundary conditions. Hence, we may write T = w – M. Let

dddd DeDeCBeBAeA −− ====The boundary equations write

dsinDdcosC)dsinBdcosAw( +=++δ0dcosBdsinAM =+−

0dcosDdsinC =−)dsind(cosD)dsind(cosC)dsind(cosB)dsind(cosAT ++−=−++−

The solution of the system is

dcos1

wMTdsinMA

δδ

+−−

+=

dsin1

wMTdcosMB

δδ

+−−

+−=

Hence)dsind(cosB)dsind(cosATQ −++−=

δδ

+−−

−−=1

wMTMTQ

TT12

Q γδ

δ=

+=

and the relationship has been proven (Van Cauwelaert and Stet, 1998).

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145

Chapter 14 The Circular Slab Subjected to a Distributed Load and Resting on aPasternak Foundation

14.1 The basic differential equations

Figure 14.1 Circular slab on a Pasternak foundation

Naturally circular slabs are ‘rare’ in pavements. In fact only the case of a slab of infinite extent is ofinterest to us because it can be solved in polar co-ordinates, thus with only one variable; the radialdistance r. To conserve this advantage in the case of limited slabs, we will only consider in this chapterthe case of slabs with axial symmetry: the load is applied in the centre of the slab. The equilibrium'sequation for an axial-symmetrical load case on a Pasternak foundation is given by

D p

=Dkw

rw

r1

+ r

wDG

rw

r1

+ r

w

r

r1

+ r 2

2

2

2

2

2+

∂∂

∂∂−

∂∂

∂∂

∂∂

∂∂ (14.1)

where p is the pressure uniformly distributed over a circular area with radius a, r is the distance to the originof the cylindrical co-ordinates, D = Eh3/12(1-µ2) is the stiffness of the slab, E is Young's modulus, µ isPoisson's ratio, G is Pasternak’s shear modulus, k is the subgrade reaction modulus and h is the thickness ofthe slab. In the case of a circular slab, we will use Hankel’s integral and in particular the integral (11.23)developed in § 11.5. For finite slabs, supplementary differential equations will be necessary in order toexpress the boundary conditions of each specific case. The differential equation necessary to express theboundary conditions of the slab itself is the homogeneous equation corresponding to (14.1)

0=Dkw

rw

r1

+ r

wDG

rw

r1

+ r

w

r

r1

+ r 2

2

2

2

2

2+

∂∂

∂∂−

∂∂

∂∂

∂∂

∂∂ (14.2)

The differential equations that are required to express the boundary conditions outside the slab are of twotypes. The first type considers a slab terminated by a free edge. The differential equation then reflects asoil subjected to horizontal shear forces. It can be deduced from (14.1)

0=kw rw

r1

+ r

wG

2

2+

∂∂

∂∂− (14.3)

q

p2a

R

E

k

rh

G

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146

The second type is where the slab is separated by a joint with a circumcircular slab. The requireddifferential equation again can be deduced from equation (14.1).

0=Dkw

rw

r1

+ r

wDG

rw

r1

+ r

w

r

r1

+ r 2

2

2

2

2

2+

∂∂

∂∂−

∂∂

∂∂

∂∂

∂∂ (14.4)

which is again the homogeneous equation this time applicable to the unloaded adjacent slab. We recallthat moments and shear forces can be deduced from the equations for the deflections by next equations- radial moment in the slab

+−=

drdw

r1

dr

wdDM

2

2

r µ (14.5)

- tangential moment in the slab

+−=

drdw

r1

dr

wdDM

2

2

t µ (14.6)

- radial shear force in the slab

+−=

drdw

r1

dr

wddrd

DT2

2

r (14.7)

- tangential shear force in the slab

0drdw

r1

dr

wddd

r1

DT2

2

t =

+−=

θ (14.8)

- radial shear force in the subgrade

drdw

GT = (14.9)

- tangential shear force in the subgrade

0ddw

r1

GT ==θ

(14.10)

14.2 Case of a slab of infinite extent

14.2.1 Solution of the differential equation

Let 1/l4 = k/EI and 2g/l2 = G/EI. Equation (14.1) becomes

D p

=l

w

rw

r1

+ r

w

l

g2rw

r1

+ r

w

r

r1

+ r 42

2

22

2

2

2+

∂∂

∂∂−

∂∂

∂∂

∂∂

∂∂ (14.11)

We express the load by (11.23) developed in the example.

∫∞

<==0

10 arforpdt)l/ta(J)l/tr(Jl

pap (14.12)

∫∞

>==0

0 arfor0dt)l/ta(1J)l/tr(Jl

pap (14.13)

In order to satisfy the equilibrium equation the deflection w must be expressed in a similar manner.

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147

∫∞

=0

10 dt)l/ta(J)l/tr(J)t(Ap2

(14.14)

where A(t) is a function of the integration variable. Introducing (14.12) and (14.14) into (14.11) derives

( )∫∞

++=

02410 dt

1gt2t

)l/ta(J)l/tr(Jlkpa

w (14.15)

The explicit solution of equation (14.15) could be obtained through the methods developed in Chapter 9.However for values of g < 1, the solution contains complex variables which are extremely difficult toexpress as real expressions. Hence we will compute (14.15) numerically.

14.2.2 Application 1

Figure 14.2 Infinite slab on a Winkler foundation

One application can be solved analytically. Figure 14.2 presents a single load P on an infinite slab restingon a Winkler foundation. In order to express the load, we transform formula (14.12):

tl2

Pl2

taal1

limP)l/ta(Jl

palim

20a1

0a ππ==

→→

Formula (14.15) for the deflection transforms into

dt1t

)l/rt(tJ

kl2

Pw

04

02 ∫

+=

πIn the axis of the load J0(rt/l) = 1. Hence

∫∞

+=

042

dt1t

t

kl2

Pw

πLet t2 = tgθ

( ) kl8

Pd

cos1tg

1

kl4

Pw

20

222=

+= ∫

∞ϑ

θθπ (14.16)

Formula (14.16) is identical to equation (12.22).

P

E

k

h r

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148

14.2.3 Application 2

We consider the infinite slab given in Figure 14.3. The response expressed in deflections, radial momentsand stresses for different values of G are given in Table 14.1.

Figure 14.3 Example of a circular slab on a Pasternak foundation.

G r = 0 r = 250 r = 500

w 0.0920 0.0836 0.0683M 9335 3570 1269σ – radial 1.40 0.54 0.190

σ – tangential 1.40 0.70 0.35w 0.0856 0.0776 0.0631M 9017 3313 1111σ – radial 1.35 0.50 0.1710000

σ – tangential 1.35 0.66 0.33w 0.0679 0.0609 0.0491M 8078 2582 691σ – radial 1.21 0.39 0.1050000

σ – tangential 1.21 0.54 0.24

Table 14.1 Deflections, moments and stresses in a slab of infinite extent

14.3 Case of a slab of finite extent with a free edge

14.3.1 Solution #1

Consider Figure 14.4.By the first boundary condition we state that the radial moment in the slab and at the edge must be zero.Hence, Mr(r=R) = 0. As in the previous case of the beam with a free edge we will first express that thesum of the shear force in the slab plus the shear force in the shear layer at the slab side of the edge isequal to the shear force in the shear layer at the outer side of the edge. Thus we write:(Tslab + Tsoilin) (r=R) = Tsoilout (r=R) ,

p = 12a = 200

E = 20000

k = 0.1

h = 200r

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149

Figure 14.4 Finite circular slab on a Pasternak foundation

Mathematically we need 2 complementary solutions, wA and wB, of the homogeneous equilibrium equation

0=l

w

rw

r1

+ r

w

l

g2rw

r1

+ r

w

r

r1

+ r 42

2

22

2

2

2+

∂∂

∂∂−

∂∂

∂∂

∂∂

∂∂ (14.17)

and 1 complementary solution, wc of equation (14.3)

0kwdx

wdG

2

2=+− (14.18)

We assume that the solution of (14.17) is of the same type as the solution of equation (12.17) for a slabsubjected to an isolated load: a modified Bessel function of the first kind. Hence we consider that theLaplacian in polar co-ordinates is an operator with next property

)tr(ft)tr(fdrd

r1

dr

d 22

2=

+

Hence we can consider equation (4.17) as a characteristic equation (§1.4.2) which roots will give us theargument of the modified Bessel functions.

0l

1D

l

g2D

42

24 =+−

The solutions depend on the value of g.- For g < 1

2g1

i2

g1g1igD 2

1−

++

=−+= (14.19)

2g1

i2

g1g1igD 2

2−

−+

=−+= (14.20)

The complementary solutions are then

+

+

=+l

i(rBI

li(r

AIlkpa

BwAw 00BAβαβα

(14.21)

that we transform in

q

E

k

h

G

r

2ap

R

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150

+

+

+

+

li(r

Il

i(rI

i2B

lkpa

li(r

Il

i(rI

2A

lkpa

0000βαβαβαβα

and further in

( )+

−∑∑

=

−∞ k

0n

n2n2k2k2

n2

n

0

k2C)1(

!k!k)l2/(r

Alkpa

βα

( )

−∑∑

=

++−

+

∞ 1k

1n

1n2)1n2(k2k2

1n2

n

0

k2C)1(

!k!k)l2/(r

Blkpa

βα (14.22)

where 2

g12

g1 −=

+= βα

- For g = 11D 2,1 = (14.23)

+=+ )l/r(I

lr

B)l/r(AIlkpa

BwAw 10BA (14.24)

- For g >1

1ggD 21 −+= (14.25)

1ggD 22 −−= (14.26)

[ ]l/r(BI)l/r(AIlkpa

BwAw 00BA βα +=+ (14.27)

where 1gg1gg 22 −−=−+= βα

Equation (14.17) has 4 solutions. We have chosen the 2 Bessel functions of the first kind because theirvalues at the origin are finite: I0(0) = 1, I1(0) = 0. For the third boundary equation (14.18), which has 2solutions, we shall take the Bessel function of the second kind because its asymptotic value tends to zero.

=

g2lr

Kw 0C (14.28)

We write the boundary conditions. The deflection outside is a fraction of the deflection inside:[ ] CBA CwBwAww =++δ (14.29)

The radial moment at the slab side of the edge is zero:

[ ] 0BwAwwdrdw

r1

udr

wdBA2

2=++

+ (14.30)

The sum of the shear force in the slab and the shear force in the subgrade at the slab side is equal to theshear force in the subgrade outside:

[ ]dr

dwC

l

g2BwAww

drd

l

g2drd

r1

dr

ddrd C

2BA22

2−=++

+ (14.31)

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151

14.3.2 Solution #2

As explained in 0, we consider in the second solution that the sum of the shear force in the slab and theshear force in the subgrade is equal with the shear force in the subgrade in the case of a slab of infiniteextent. The boundary conditions are

[ ] 0BwAwwdrdw

r1

udr

wdBA2

2=++

+ (14.32)

[ ]drdw

l

g2BwAww

drd

l

g2drd

r1

dr

ddrd

2BA22

2−=++

+ (14.33)

14.3.3 Application of solution #2

Figure 14.5 Example of a slab of finite extent on a Pasternak foundation

Deflections, radial moments and stresses are given for different values of G in Table 14.2.

G r = 0 r = 250 r = 500w 0.4072 0.4022 0.3958M 6323 990 0σ – radial 0.95 0.15 00

σ – tangential 0.95 0.30 0.09w 0.3830 0.3780 0.3715M 6358 1035 0σ – radial 0.95 0.16 010000

σ – tangential 0.95 0.30 0.09w 0.3110 0.3060 0.2993M 6354 1106 0σ – radial 0.95 0.17 050000

σ – tangential 0.95 0.31 0.10

Table 14.2 Deflections, moments and stresses in a slab of finite extent; Moment in slab is zero at theedge; Shear force in slab and shear force in subgrade are equal with shear force; in subgrade for a

slab of infinite extent

k = 0.1

G

h = 200r

E = 20000

p = 12 a = 200

R = 500

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14.4 Case of a slab of finite extent with a joint

14.4.1 Solution #1

The solution for the loaded slab is given by equation (14.14), which expresses the load and the twosolutions wA and wB of the homogeneous differential equation (14.17). The solutions for the unloaded slabare the two other solutions of (14.17), wC and wD, which are the expressions for wA and wB wherein theBessel functions I of the first kind are replaced by Bessel functions K of the second kind. We write theboundary conditions:

The deflection outside is a fraction of the deflection inside[ ] DCBA DwCwBwAww +=++δ (14.34)

The radial moment at the loaded slab side of the joint is zero

[ ] 0BwAwwdrdw

r1

udr

wdBA2

2=++

+ (14.35)

The radial moment at the unloaded slab side of the joint is zero

[ ] 0DwCwdrdw

r1

udr

wdDC2

2=+

+ (14.36)

The sum of the shear force in the slab and the shear force in the subgrade at the loaded slab side is equalwith the sum of the shear force in the slab and the shear force in the subgrade at the unloaded slab side

[ ] [ ]DC22

2

BA22

2DwCw

drd

l

g2drd

r1

dr

ddrd

BwAwwdrd

l

g2drd

r1

dr

ddrd

+

+=++

+

(14.37)

14.4.2 Solution #2.

To reduce the number of boundary equations to two, we suggest to express the shear force in the loadedslab as a fraction of the shear force in the slab of infinite extent. The resulting boundary conditions are:

The radial moment at the loaded slab side of the joint is zero

[ ] 0BwAwwdrdw

r1

udr

wdBA2

2=++

+ (14.38)

The shear force in the loaded slab is a fraction of the shear force in the slab of infinite extent

[ ] wdrd

l

g2drd

r1

dr

ddrd

BwAwwdrd

l

g2drd

r1

dr

ddrd

22

2

BA22

2

+=++

+ γ (14.39)

where 0 ≤ γ ≤ 1.

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153

14.4.3 Application of solution #2

Figure 14.6 Example of a slab of finite extent on a Pasternak foundation

Deflections, radial moments and stresses are given for different values of G and for different percentagesof shear transfer are listed in Table 14.3 to Table 14.5.

G r = 0 r = 250 r = 500

w 0.4072 0.4022 0.3958M 6323 990 0σ – radial 0.95 0.15 00

σ – tangential 0.95 0.30 0.09w 0.3830 0.3780 0.3715M 6358 1035 0σ – radial 0.95 0.16 010000

σ – tangential 0.95 0.30 0.09w 0.3110 0.3060 0.2993M 6354 1106 0σ – radial 0.95 0.17 050000

σ – tangential 0.95 0.31 0.10

Table 14.3 Deflections, moments and stresses in a slab of finite extent; Moment in slab is zero at theedge. Shear transfer at the joint is 0 %

E = 20000

G

p = 1

k = 0.1

h = 200

2a = 200

r

R = 500

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154

G r = 0 r = 250 r = 500

w 0.1940 0.1878 0.1783M 7496 1871 0s – radial 1.12 0.28 00

s – tangential 1.12 0.44 0.14w 0.1821 0.1759 0.1666M 7429 1842 0σ – radial 1.11 0.28 010000

σ – tangential 1.11 0.44 0.14w 0.1472 0.1413 0.1326M 7130 1695 0σ – radial 1.07 0.25 050000

σ – tangential 1.07 0.41 0.13

Table 14.4 Deflections, moments and stresses in a slab of finite extent; Moment in slab is zero at theedge. Shear transfer at the joint is 67 %

G r = 0 r = 250 r = 500

w 0.0890 0.0821 0.0712M 8073 2305 0σ – radial 1.21 0.35 00

σ – tangential 1.21 0.51 0.16w 0.0831 0.0763 0.0656M 7957 2239 0σ – radial 1.19 0.34 010000

σ – tangential 1.19 0.50 0.16w 0.0665 0.0602 0.0505M 7512 1986 0σ – radial 1.13 0.30 050000

σ – tangential 1.13 0.46 0.15

Table 14.5 Deflections, moments and stresses in a slab of finite extent; Moment in slab is zero at theedge. Shear transfer at the joint is 100 %

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155

Chapter 15 The Rectangular Slab Subjected to a Distributed Load and Resting ona Pasternak Foundation

15.1 The basic differential equations

Figure 15.1 Slab on a Pasternak foundation

The equilibrium equation in cartesian co-ordinates for a slab on a Pasternak foundation is given by

Dp

= Dkw

+ y

w+

x

w

DG

- y

w+

x

w

y+

x 2

2

2

2

2

2

2

2

2

2

2

2

∂∂∂

∂∂

∂∂ (15.1)

where G/D = 2g/l2 and k/D = 1/l4.

The load is expressed as a double Fourier’s integral

∫ ∫∞∞

=0 0

2dsdt

ts)l/sbsin()l/sycos()l/atsin()l/xtcos(p4

(15.2)

where a and b are respectively the length and the width of a rectangular load with a uniformly distributedpressure p. The centre of gravity of the load is located in x = 0 and y = 0. The elastic length is l = [Eh3/12(1-µ2)k]1/4. A similar equation must express the deflection

dsdtts

(bs/l)sin(ys/l)cos(at/l)sin(xt/l)cost)C(s, = w

oo∫∫∞∞

(15.3)

Letting (15.2) and (15.3) in (15.1) yields

1+s+s2g+gt2+st2+t

l D

4p = t)C(s,

422224

4

2π (15.4)

15.2 Resolution of the deflection equation

In order to be able to express the boundary conditions at an edge or at a joint, we shall show that it is requiredto partially integrate the deflection equation with respect to the variable orthogonal to the concerned edge orjoint. To partially integrate (15.3) with respect to x, the denominator of (15.4) must be split into two factors:

z

p2a

Eh

G

k q

y

d

x

c

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156

)z)(tz+(t = 1+s+gs2+gt2+s2t+t 22

221

2422224 + (15.5)where- with g > 1

1-g-g+s = z

1-g+g+s = z

222

221

(15.6)

z+t

1 -

z+t

1

1-g

1

k2p

= t)C(s,21

222

222π (15.7)

- with g < 1

βαβα

i - = z

i + = z

2

1 (15.8)

g)+s( + g-1+)g+s(

21

= 22222α

g)+s( - g-1+)g+s(

21

= 22222β

z+t

1 -

z+t

1

g-1

i

k

2p = t)C(s,

22

221

222π (15.9)

The partial integration is based on next equations- with x > a

(xz/l)K(az/l)I zl)(ax

2= dt

)z+tt((ta/l)sin(tx/l)cos

1/2-1/2

1/2

22o

π∫∞

(15.10)

- with x < aand making use of the trigonometric identitycos(tx/l)sin(ta/l) = sin[t(x+a)/l] - sin(tx/l)cos(ta/l)

(az/2l)K(az/2l)Izla

2 = dt

)z+tt((at/l)sin

1/21/222o

π∫∞

(15.11)

(az/l)K(xz/l)Izl)(ax

2 = dt

)z+tt((ta/l)cos(tx/l)sin

1/2-1/2

1/2

22o

π∫∞

(15.12)

Knowing that

[ ]

e2z

= (z)K = (z)K

e - ez2

1 = (z)I

z-1/2-1/2

z-z1/2

π

π

the solutions are- with g > 1

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157

- with x > a

dss

(sb/l)sin(sy/l)cos

ze - e -

ze - e

1-g

1k2

p = w

21

/lza)+(x-/lza)-(x-

22

/lza)+(x-/lza)-(x-

o2

11

22

∫∞

π (15.13)

- with < a

dss

(sb/l)sin(sy/l)cos

ze-e-2

-

ze-e-2

1-g

1k2

p = w

21

/lzx)+(a-/lzx)-(a-

22

/lzx)+(a-/lzx)-(a-

o2

11

22

∫∞

π (15.14)

- with g < 1 and making use of the complex equation e(α±iß)z = cos(αz) ± i sin (ßz)

- with x > a

ds/l]a)+[(xsing)+(s+/l]a)+[(xcosg-1 e -

/l]a)-[(xsing)+(s+/l]a)-[(xcosg-1 e

1)+2gs+s(s

(sb/l)sin(sy/l)cos

g-1

1kp

= w

22/la)+(x-

22/la)-(x-

24o

2

∫∞

ββ

ββ

π

α

α (15.15)

- with x < a

ds/l]a)+[(xsing)+(s+/l]a)+[(xcosg-1 g-1

e -

/l]x)-[(asing)+(s+/l]x)-[(acosg-1 g-1

e - 2

1)+g2s+s(s

(sb/l)sin(sy/l)coskp

= w

222

/la)+(x-

222

/lx)-(a-

24o

∫∞

ββ

ββ

π

α

α (15.16)

- with g = 1 and making use of de l’Hospital’s rule for lim g = 1

- with x > a

ds)1+s(s

(sb/l)sin(sy/l)cosea)/l+(xs+1+2 -

ea)/l-(xs+1+2 k2

p = w

22a)z/l+(x-2

a)z/l-(x-2

o

π (15.17)

- with x < a

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158

ds)1+s(s

(sb/l)sin(sy/l)cosea)/l+(xs+1+2 -

ex)/l-(as+1+2 - 4 k2

p = w

22a)z/l+(x-2

x)z/l-(a-2

o

π (15.18)

where z = (s2 + 1)1/2.

15.3 Boundary conditions

As in the previous chapters we formulate the boundary conditions for a slab limited by a free edge or by ajoint. However in the case of a slab we must consider four boundary conditions at the edge:- the values of the moment;- the shear force;- the moment of torsion in the slab- the value of the shear force in the subgrade

The latter condition is an additional boundary condition compared to the case of a beam or a circular slab.Fortunately, it can be shown (Timoshenko, 1951) that at an edge, a unique equation may replace theconditions regarding the shear force and the moment of torsion

2

3

3

3xyxx

yx

w)2(

x

wy

MTV

∂∂

∂−+

∂=

∂−= µ (15.19)

yx

w)2(

y

wx

MTV

2

3

3

3yxyy

∂∂

∂−+

∂=

∂+= µ (15.20)

15.4 Case of a slab of finite extent with free edge

15.4.1 Solution #1

First we consider the case where the sum of the shear forces in slab and subgrade is equal to the shearforce in the subgrade outside the slab. We also limit the problem to that of a slab with only one edge.Mathematically we need 2 complementary solutions, wA and wB, of the homogeneous equilibriumequation

0 = Dkw

+ y

w+

x

w

DG

- y

w+

x

w

y+

x 2

2

2

2

2

2

2

2

2

2

2

2

∂∂

∂∂

∂∂

∂∂ (15.21)

- with g > 1

[ ] dss

)l/sbsin()l/sycos(e)s(Be)s(A

1g

1k2

pw

0

l/xzl/xz2B,A 21∫

∞+

−=

π (15.22)

- with g = 1

dss

)l/sbsin()l/sycos(e)s(B

lx

e)s(Ak2

pw

0

l/xzl/xzB,A ∫

+=

π (15.23)

- with g < 1

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159

[ ] dss

)l/sbsin()l/sycos(e)l/xsin()s(B)l/xcos()s(A

g1

1kp

w0

l/x2B,A ∫

∞+

−= αββ

π(15.24)

We also need the solution of the equilibrium equation of the subgrade outside the slab

0 = kw + y

w+

x

wG -

2

2

2

2

∂∂

∂∂ (15.25)

Considering that the edge is located on the positive side of the x – axis, we take the solution with thenegative exponent

dss

)l/sbsin()l/sycos(e)s(C

k2p

w0

)g2/(1slx

C

2

∫∞ +−

(15.26)

The boundary conditions are:- ratio between the deflections on both sides of the edge

[ ] CBA CwBwAww =++δ (15.27)- cancelling of the moment at the edge of the slab

[ ] 0BwAwwyx

BA2

2

2

2=++

∂+

∂µ (15.28)

- equality of the shear forces

[ ] C2BA22

3

3

3Cw

xl

g2BwAww

xl

g2

yx)2(

x ∂∂

−=++

∂∂

−∂∂

∂−+

∂µ (15.29)

15.4.2 Solution #2

We assume that the sum of the shear forces in slab and subgrade is equal to the shear force in thesubgrade under a slab of infinite extent. :The boundary conditions are- cancelling of the moment at the edge of the slab

[ ] 0BwAwwyx

BA2

2

2

2=++

∂+

∂µ (15.30)

- equality of the shear forces

[ ]xw

l

g2BwAww

xl

g2

yx)2(

x 2BA22

3

3

3

∂∂

−=++

∂∂

−∂∂

∂−+

∂µ (15.31)

15.5 Case of a slab of finite extent with a joint

15.5.1 Solution #1

We consider that the shear forces left from the joint (loaded slab) are equal to the shear forces right fromthe joint (unloaded slab). We need 2 supplementary solutions of the homogeneous equation (15.12)applying to the unloaded slab. We assume that the joint is located at the positive side of the x –axis,hence we shall only consider the functions with a negative exponent. The solutions are

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160

- with g > 1

[ ] dss

)l/sbsin()l/sycos(e)s(De)s(C

1g

1k2

pw

0

l/xzl/xz2D,C 21∫

∞−− +

−=

π (15.32)

- with g = 1

dss

)l/sbsin()l/sycos(e)s(D

lx

e)s(Ck2

pw

0

l/xzl/xzD,C ∫

∞−−

+=

π (15.33)

- with g < 1

[ ] dss

)l/sbsin()l/sycos(e)l/xsin()s(D)l/xcos()s(C

g1

1kp

w0

l/x2D,C ∫

∞−+

−= αββ

π

(15.34)The boundary conditions are:- ratio between the deflections on both sides of the joint

[ ] DCBA DwCwBwAww +=++δ (15.35)- cancelling of the moment at the edge of the loaded beam

[ ] 0BwAwwyx

BA2

2

2

2=++

∂+

∂µ (15.36)

- cancelling of the moment at the edge of the unloaded beam

[ ] 0DwCwwyx

DC2

2

2

2=++

∂+

∂µ (15.37)

- equality of the shear forces

[ ] =++

∂∂

−∂∂

∂−+

∂BA22

3

3

3BwAww

xl

g2

yx)2(

[ ]DC22

3

3

3DwCw

xl

g2

yx)2(

x+

∂∂

−∂∂

∂−+

∂µ (15.38)

15.5.2 Solution # 2

Let us consider that the shear force at the edge of the loaded beam is partially transferred and is thus equalto a fraction of the shear force in an infinite slab; further we consider that the shear force in the subgradeat the joint is equal with the shear force in the subgrade under an infinite slab. This can be expressed by 2boundary conditions applied at the edge of the loaded slab. The boundary conditions are:

- cancelling of the moment at the edge of the loaded beam

[ ] 0BwAwwyx

BA2

2

2

2=++

∂+

∂µ (15.39)

- equality of the shear forces

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161

[ ] =++

∂∂

−∂∂

∂−+

∂BA22

3

3

3BwAww

xl

g2

yx)2(

xw

l

g2

yx

w)2(

x

w22

3

3

3

∂∂

∂∂

∂−+

∂µγ (15.40)

Two equations are enough to express the boundary conditions. Note that if we adopt the second solutionfor both cases (free edge and joint), equations (15.39) and (15.40) solves the whole problem. Indeed forthe case of a free edge, just let γ = 0 in order to obtain boundary condition (15.31).

15.5.3 Application

Figure 15.2 Example of a slab on a Pasternak foundation.

Bending stresses at the bottom of the slab and deflections are given for different values of G in Table 15.1without shear transfer at the joint and in Table 15.2 with 50 % of shear transfer.

G x = 0 x = 500 x = 1000

σx 0.90 0.06 0σy 0.93 0.33 0.180w 0.098 0.078 0.053σx 0.87 0.05 0σy 0.89 0.30 0.1510000w 0.090 0.070 0.045σx 0.77 0.01 0σy 0.78 0.21 0.0950000w 0.068 0.050 0.030

Table 15.1 Bending stresses and deflections in a slab with a joint; The shear transfer is zero

2a = 200p = 0.7

E = 20000

G

k = 0.1

h = 200

x

xy

zq

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162

G x = 0 x = 500 x = 1000

σx 0.92 0.08 0σy 0.93 0.31 0.140w 0.096 0.072 0.040σx 0.88 0.06 0σy 0.89 0.28 0.1210000w 0.088 0.065 0.036σx 0.78 0.02 0σy 0.78 0.21 0.0850000w 0.067 0.048 0.026

Table 15.2 Bending stresses and deflections in a slab with a joint; The shear transfer is 50 %

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THE MULTISLAB

163

Chapter 16 The Multislab

16.1 Theoretical justification

Consider a pavement consisting of several superposed concrete layers (of different quality). It can bedesigned using the models developed for a single slab as far as the total thickness of the layers can still beconsidered as small against the whole structure. Mathematically one must assume that the radius ofcurvature (ρ = D/M) remains important, in other words, that the deflection at the surface of the upperlayer may be considered to be equal with the deflection at the base of the lower layer.

16.2 General model

Figure 16.1 The Multislab

The mechanical characteristics of the equivalent structure depend of the mechanical characteristics of thedifferent layers and the adhesion conditions at their interfaces. The three layered structure can be replacedby an equivalent single layered structure.

16.3 Full slip at each of the interfaces

The equilibrium equations for each layer writes

3

23

4

2

212

4

1

11

4

Dqp

wD

qpw

Dqp

w−

=∇−

=∇−

=∇ (16.1)

By hypothesis w = w1 = w2 = w3, hence q = p – (D1 + D2 + D3)∇4w. The equilibrium equation of theequivalent slab writes:

p1

g1

p2

q2

p3

q3G, k

h1E1

2a

E2

E3

h2

h3

p

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164

Dqp

DDDqp

w321

4 −=

++−

=∇ (16.2)

where D = Σ Di is the equivalent stiffness of the structure.

Consider a Pasternak’s subgrade. Hence

Dp

Dkw

wDG

w 24 =+∇−∇ (16.3)

The moment in x is given by

+−= 2

2

2

2

dywd

dxwd

DM µ (16.4)

that we may write as follows

( )

+++−=++= 2

2

2

2

321321 dywd

dxwd

DDDMMMM µ (16.5)

The bending stresses at the base of each of the layers are

DM

h

D6

dy

wd

dx

wd

h

D62i

i2

2

2

2

2i

ii =

+−= µσ (16.6)

16.4 Full friction at the first interface, full slip at the second interface

We replace the two upper layers by an equivalent layer which modulus is equal with the modulus of theupper layer. The transverse section of the equivalent layer is a T – section, as given in Figure 16.2. Thewidth of the vertical bar of the T – section is equal with the ratio of the moduli: b = E2/E1. Hence themoment of inertia of the transformed section remains equal with the moment of inertia of the initialsection.

Figure 16.2 Multislab with friction at the first interface; slip at the second

E2

E3

h2

E1 E1

E1

h3

E3

h1c

b

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165

The position of the neutral axis is given by:

( )( )21

2221

21

bhh2hhh2bh

c+

++= (16.7)

The moment of inertia of the T – section is

3hh)ch(3)ch(h3

b3

hch3ch3I

32

221

212

31

21

21

12

+−+−+

+−= (16.8)

The stiffness of the equivalent section is

)1(IE

D 21

12112 µ−

= (16.9)

The equilibrium equations are

3

24

12

24

Dqp

wD

qpw

−=∇

−=∇ (16.10)

Therefore we can write

Dqp

DDqp

w312

4 −=

+−

=∇ (16.11)

Dp

Dkw

wDG

w 24 =+∇−∇ (16.12)

The bending stresses are:

- at the base of the upper layer

DD

Ich

M 12

12

1xx

−=σ (16.13)

- at the base of the mid layer

1

212

12

21xx E

ED

DI

chhM

−+=σ (16.14)

- at the base of the lower layer

DD

h

M6 323

xx =σ (16.15)

16.5 Full friction at both interfaces

We replace the three layers with an equivalent layer which modulus is equal with the modulus of theupper layer. The equivalent transverse is a section with decreasing widths, as presented in Figure 16.3.The widths of the subsections are equal with the ratio’s of the moduli: b1 = E2/E1, b2 = E3/E1. Theposition of the neutral axis is given by

( ) ( )( )32211

2332312

22211

21

hbhbh2hhh2hh2bhhh2bh

c++

+++++= (16.16)

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166

The moment of inertia is given by

( ) ( )( )3/hhchchhb3/hchchI 322

21

2121

31

21

21123 +−+−++−=

( ) ( )( )3/hhchhchhhb 33

2312

21232 +−++−++ (16.17)

Figure 16.3 Multislab with friction at both interfaces

The bending stresses are:

- at the base of the upper layer

123

1xx I

chM

−=σ (16.18)

- at the base of the mid layer

1

2

123

21xx E

EI

chhM

−+=σ (16.19)

- at the base of the lower layer

1

3

123

321xx E

EI

chhhM

−++=σ (16.20)

16.6 Partial friction

Consider a structure of two layers with partial friction at their interface as presented in Figure 16.4. Weassume that a fraction F of the interface is in full friction and thus a fraction (1 – F) in full slip. Theequilibrium equations write:

E1

E1

h3

h2

h1 c

b2

b1

E3

E2

E1

E1

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THE MULTISLAB

167

- for the fraction in full slip

21

4

DDqp

w+−

=∇ (16.21)

- for the fraction in full friction

12

4

Dqp

w−

=∇ (16.22)

Figure 16.4 Partial friction at the interface

Multiply both terms of (16.21) by (1 – F) and both terms of (16.22) by F

( ) ( ) ( )( )qpF1DDwF1 214 −−=+∇−

( )qpFwDF 124 −=∇

Adding both equations, we obtain the equilibrium equation for a structure in partial friction.

( )( ) Dqp

FDDDF1qp

w1221

4 −=

++−−

=∇ (16.23)

The elastic length is( )( )

kD

kFDDDF1

l 12214 =++−

= (16.24)

The general differential equation writes

Dqp

Dkw

wDG

w 24 −=+∇−∇ (16.25)

Moments and stresses are obtained as in the previous paragraphs.

The question rises if the soil reaction q is the same under the interface with slip as under the interfacewith friction. Since slip and friction can be considered as randomly distributed, the deflection at thesurface will not discriminate the different situations. Hence w being continuous, by Winkler’s hypothesisq will also be continuous. Its value will be a weighted average between the subgrade reactions in bothcases.

F(1 - F)E1

E2b

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168

16.6.1 Application

Figure 16.5 Example of a two-layered multislab

The bending stresses are given in Table 16.1 at different depths and for different friction conditions. Thecomputations refer to the middle of the slab. The stresses can also be depicted from Figure 16.6.

Adhesion σx1(z = 0) σx1 (z = 200) σx2 (z = 200) σx2 (z = 500)

Full slip - 0.47 0.47 - 0.23 0.2350 % slip - 0.38 0.27 - 0.13 0.20

Table 16.1 Stresses and deflections function of the interface condition

p = 0.7

k = 0.1 G = 200000

E2 = 10000

E1 = 30000

h2 = 300

h1 = 200

2a = 200

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169

Figure 16.6 Stresses in a two-layered concrete structure function of the friction conditions at theinterface condition

0.01

-0.47- 0.38- 0.25

0.47 0.27 - 0.23- 0.13

0.23

0.20 0.15

0.02

Full friction

50 % friction

Full slip

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171

Chapter 17 Back-Calculation of Concrete Slabs

17.1 Backcalculation of moduli

By the word back-calculation we mean a procedure consisting in estimating the moduli of the differentlayers of a pavement based on the values of a series of deflections measured on the pavement. Theprocedure is intended to estimate a residual life of the pavement in order to, if required, design anstrengthening overlay over the existing pavement. The mathematical models are all based on theoreticalrelations existing between the deflections and the moduli of the layers. In the case of a concrete pavementwe mean by moduli the mechanical characteristics related to the pavement: the Young modulus E of theconcrete, the subgrade modulus k of Westergaard and the shear modulus G of Pasternak. Thebackcalculation procedure is based on Pasternak’s model.

17.2 Case where the load can be considered as a point load

The deflection under an isolated load P is given by

∫∞

++=

024

02 dm

1gm2m)l/mr(mJ

kl2P

)r(wπ

(17.1)

When r = 0, the integral can be solved analytically

- for g < 1

o2222 wkl4P

g1

1

g1

garctg

2kl4P

)0(wπ

ππ

=−

−−= (17.2)

- for g = 1

o22 wkl4P

kl4P

)0(wππ

== (17.3)

- for g > 1

o22

2

22 wkl4P

1gg

1ggln

1g

1kl8P

)0(wππ

=−−

−+

−= (17.4)

The value of k can be deduced from those results2

o

)0(ww

D4P

k

=

π (17.5)

where

kD

l)1(12

EhD 4

2

3

=−

Using equation (17.5) for the computation of k, we have an iterative backcalculation method in order toestimate the values of E and G giving computed deflections as close as possible to measured deflections.

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172

17.3 Computations

The backcalculation procedure is presented in Figure 17.1.

Figure 17.1 Model of backcalculation based on E

gdif1 < gdif2

Seed modulus E = 1000

g = 0

computation of k and the deflections

Edif2 = Σ wcomp – wmeas

Edif1 = Edif2

E = E + ∆E

g = 0

computation of k and the deflections

gdif1 = gdif2

gdif2 = Σ wcomp – wmeas

g = g + ∆g

computation of k and the deflections

gdif2 = Σ wcomp – wmeas

No

Yes

Edif1 = gdif1

Edif1 < Edif2 No

Yes

Results

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173

The flow chart of Figure 17.1 can easily be understood. The computation starts with the input of a seedvalue for the modulus E of the concrete. This value must differ from zero, to avoid divisions by zero. Theinitial value of g is set equal to zero. The corresponding value of k is computed by equation (17.5). Thecorresponding values of the deflections are computed for the same load and at the same distances as themeasured values of the deflections. The sum of the absolute differences between computed and measureddeflections, called Edif2, is the begin value of the E – loop, i.e. the loop for the estimation of the E – valuegiving the best fit. For each E – value, a second loop, the g - loop with begin value gdif2, estimates the g– value giving the best fit for the given E – value. When both loops have allowed to determine the E – andg– values, and by equation (17.5) the k – values, giving the best fit of all calculation, the best results havebeen derived.

17.4 Case when the load is considered as distributed

The deflection under a uniformly distributed circular load of pressure p and radius a is given by

∫∞

++=

024

10 dm1gm2m

)l/ma(J)l/mr(mJklpa

)r(w (17.6)

This integral can be solved analytically. So for g > 1

−=

βββ

ααα )l/a(K)l/r(I)l/a(K)l/r(I

1g2

1klpa

)r(w 10102

(17.7)

where

1gg1gg 22 −+=−−= βα

However, even when r = 0, the value of k cannot be deduced from previous equation. Hence the methodillustrated in Figure 17.1 cannot be used. However, if we consider the elastic length l instead of themodulus E, the problem can be solved (Stet and Van Cauwelaert, 2004) Indeed, when knowing l and g,the value of k can be obtained from equation (17.6) The backcalculation procedure is identical to that ofthe previous paragraph.

Further, in this case the value of k, corresponding with a given value of g, can be deduced from all thedeflection values and not only from the centre deflection. It seems more realistic to choose a mean valueof k obtained from all the deflection values.

Finally, in using the elastic length, the model can be used for backcalculation of a multilayered slab. Ofcourse the determination of each individual modulus will require some information, for example, theratio’s between the moduli or the knowledge of the value of one of the moduli.

The computations are presented in Figure 17.2.

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174

Figure 17.2 Model of backcalculation based on elastic length

Seed elastic length l = 1000

g = 0

computation of k and the deflections

ldif2 = Σ wcomp – wmeas

ldif1 = ldif2

l = l + ∆l

g = 0

computation of k and the deflections

gdif1 = gdif2

gdif2 = Σ wcomp – wmeas

g = g + ∆g

computation of k and the deflections

gdif1 < gdif2

gdif2 = Σ wcomp – wmeas

No

Yes

ldif1 = gdif1

ldif1 < ldif2 No

Yes

Results

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17.5 Comparison of the two methods

The second method, which considers the load as uniformly distributed, requires more computer time thanthe first, which considers the load as isolated. The question then rises if the assumption of an isolatedload, which does not exist in reality, has an influence on the results.Consider a slab with a thickness of 200 mm, a modulus of 30000 N/mm², resting on a subgrade with asubgrade modulus of 0.1 N/mm³ and a shear modulus of 20000 N/mm. The slab is subjected to a classicalfalling weight load of 100000 N uniformly spread over a circular loading plate with a radius of 150 mm.In Table 17.1 the deflections in µm computed by Pasternak’s model are given for the usual distances for aload distributed over a circle with a radius of 150 mm and for an isolated load of same magnitude.

Distances 0 300 600 900 1200 1500 1800 2100 2400Distributed 233 206 160 115 78 50 30 17 8Isolated 241 210 162 116 79 51 30 17 8

Table 17.1 Deflections under a distributed and an isolated load

The differences between the deflections are small. In order to verify if those differences can be neglected,we list in Table 17.2 the values of the mechanical characteristics backcalculated with the deflections ofTable in the assumption of the classical falling weight load.

Characteristics E k Gw (r = 0) = 233 30340 0.101 18443w (r = 0) = 241 26505 0.097 25353

Table 17.2 Backcalculated slab characteristics

The results differ more than 10 %. Hence we conclude that the method with the assumption of an isolatedload should not be applied.

17.6 Influence of the reference deflection

In the method developed for an isolated load, we took the centre deflection as reference deflection for thecomputation of k by (17.5). In the method developed for the distributed load, we are free to choose thereference deflection. In table 17.3 we compare the values of the mechanical characteristics backcalculatedwith centre deflection, the second deflection and the mean value of the 9 deflections as referencedeflection for the computation of k by (17.6). The fit is equal to the sum of the absolute differencesbetween the computed deflections and the measured deflections, sum divided by the number ofdeflections.

Reference deflection E k G Fit

w (r = 0) 30340 0.101 18443 0.199W (r = 300) 30414 0.101 18487 0.205w- mean 30587 0.101 18483 0.264

Table 17.3 Backcalculated values of the mechanical characteristics in function of the referencedeflection

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The differences are not significant. However, this is perhaps due to the fact that the analysed deflectionbasin is a theoretical one.

17.7 Analysis of field data

We analyse the deflection basin, presented in Table 17.4, obtained on a concrete slab of 265.3 mmthickness built on a granular sub-layer (CROW, 1999). The load was the classical falling weight load of100000 N. The deflections are expressed in µm.

Distances 0 300 600 1000 1500 2000 2500Deflections 142 135 120 98 75 55 43

Table 17.4 Deflections measured on a concrete pavement

In Table 17.5 we present the values of the mechanical characteristics obtained by a backcalculation basedon the three different reference deflections.

Reference deflection E k G Fitw (r = 0) 45112 0.028 153175 0.570w (r = 300) 42709 0.027 156730 0.544w- mean 46994 0.030 142354 0.581

Table 17.5 Backcalculated values of the mechanical characteristics of a concrete pavement

Again the differences do not seem to be very significant, although more important than those of Table17.3. The modulus of the concrete of the slab was dynamically determined on a core taken out of the slaband found to be equal to 46100 N/mm², hence the backcalculated values seem realistic. It seems that inthis case the method based on the mean value of all the deflections is the most reliable method.

The values of the subgrade modulus k seem very low. Indeed a value of 307 N/mm² was obtained for thesubgrade by a backcalculation performed with the methods developed for flexible pavements (Chapter24). This value corresponds roughly with a CBR- value of 30 % and thus, based on the publishedcorrelation’s between CBR and k, on a value for the subgrade modulus of about 0.1 N/mm³. For themoment this remains an open question. However, we know that the correlation’s between CBR and khave been established without taking into account the shear resistance of the soil. Therefore we present inTable 17.6 different pairs of k and G values resulting in a center deflection of 142 µm on a slab with amodulus of 45112 N/mm2 and a thickness of 265.3 mm.

E k G w (r = 0) Fit45112 0.103 0 142 13.85745112 0.070 50000 142 9.28645112 0.046 100000 142 5.00045112 0.028 153175 142 0.57145112 0.017 200000 142 5.714

Table 17.6 Pairs of k and G values for identical deflections

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Table 17.6 is not intended to proof anything, more data need to be analysed, but it clearly shows theinfluence of the value of G on the value of k. In this way, when G = 0, the backcalculated value of kcorresponds fairly well with a modulus of 300 N/mm².

The last column of Table 17.6 presents the values of the fit when comparing the deflection basincalculated with the assumed characteristics with the observed deflection basin (Table 17.4). It seemsevident that the best fit was obtained when taking in account the real characteristics of the subgrade. Itillustrates the fact that Pasternak’s model is certainly closer to reality than Winkler’s model.

17.8 Example of backcalculation

Table 17.7 gives a series of deflection basins obtained on an airfield with a concrete pavement. Thedeflections are expressed in µm, the distances and thickness in mm.

Slab 0 300 600 1000 1500 2000 2500 Thick-ness

1 142 135 120 98 75 55 43 265.32 139 131 113 92 74 54 42 280.23 104 98 82 67 59 44 36 275.04 122 112 94 73 59 45 37 263.25 120 109 97 75 59 42 33 277.46 165 152 128 106 78 55 41 276.27 143 133 113 93 74 55 41 272.98 181 168 145 120 88 60 42 272.39 150 136 115 96 76 58 45 289.7

10 158 144 123 104 82 60 46 271.5

Table 17.7 Deflection basins obtained in the field

Table 17.8 presents the results of the mechanical characteristics back-calculated with a full automaticprogram based on the flow sheet of Figure 17.2.

Slab E k G Fit1 46994 0.030 142354 0.5812 31190 0.025 191129 0.7983 34344 0.020 350504 0.9744 21525 0.021 319031 0.7435 27541 0.031 246377 1.0926 25004 0.030 141047 1.2697 26655 0.023 210052 1.0718 34800 0.036 75353 1.3169 18612 0.020 221911 1.46110 30487 0.023 168312 1.714

Table 17.8 Back calculated characteristics based on the deflections of Table 17.7

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Chapter 18 Thermal Stresses in Concrete Slabs

18.1 Thermal stresses

Under the influence of a thermal gradient τ an elastic beam or a slab of uniform thickness and great lengthtakes the shape of a cylindrical surface. A circular elastic slab takes the shape of a spherical cap. Arectangular elastic slab takes the shape of a clastic surface (uplifted at the four corners). Let the thicknessof a beam being h, the difference of temperature between both sides of the beam being t and thecoefficient of thermal dilatation being α0 . The thermal strain is εth = α0t/2. Hence by (10.5) and (10.7)α0t/2 =h/(2R) or α0τ =1 /R, where R is the radius of curvature. This curvature can be suppressed byapplying on the ends of the beam such moments that Mc = EI/R =Eα0τ and at the boundaries of a slab ofgreat length such moments that Mc = D/R = Dα0. Considering that the thermal dilatation is the same onthe boundaries of a rectangular slab, the required moments Mx and My will be equal. Hence by (10.19)M = D(1 + µ)/R = Dα0τ (1 + µ) where D is the stiffness of the slab and µ its Poisson’s ratio. The thermaldeformation of a slab subjected to a thermal gradient is partially refrained by the stiffness of the subgrade.If the subgrade had an infinite stiffness, the thermal stress soliciting a slab with thickness h along itsboundaries would be equal to σ0 = Dα0τh/2I in the case of a slab of great length and equal toσ0 = D(1 + µ)α0τh/2I in the case of circular or rectangular slabs. The subgrade being deformable, thevalue of σ0 must be multiplied by a factor which value depends on the dimensions and the elastic lengthsof the slabs.

18.2 Slab of great length

18.2.1 Differential equilibrium equation

For convenience we take the x – axis parallel with traffic; the y – axis is perpendicular to traffic. Weadmit that the curvature of the slab takes the shape of a cylindrical surface deformed perpendicularly totraffic. It is the case of continuously reinforced concrete and probably also that of a lean concrete basecourse. Hence the problem is reduced to the case of a slab of great length in the x direction and of smallwidth in the y direction, with a stiffness D and subjected to a thermal gradient in the direction of thewidth. Due to the great length of the slab, all derivatives in x are zero : dnw/dxn = 0.

The equilibrium equation of an unloaded slab on a Pasternak foundation is

0Dkw

dywd

DG

dywd

2

2

4

4

=+− (18.1)

where w is the deflection for all y,G is Pasternak’s shear modulusk is Westergaard’s subgrade modulus.

Let G/D = 2g/l2, k/D = 1/l4.Hence

0l

w

dy

wd

l

g2

dy

wd42

2

24

4=+− (18.2)

The solution of (18.2) differs for g < 1, g = 1 or g > 1.

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18.2.2 Solution of the equilibrium equation for g < 1

The solution of equation (18.2) is

[ ] [ ]l/ysin(D)l/ycos(Cel/ysin(B)l/ycos(Aew l/yl/y ββββ αα +++= − (18.3)where

2g1

2g1 −

=+

= βα

18.2.3 Solution of the equilibrium equation for g = 1

The solution of equation (18.2) isl/yl/yl/yl/y DyeCyeBeAew −− +++= (18.4)

18.2.4 Solution of the equilibrium equation for g > 1

The solution of equation (18.2) is

l/yl/yl/yl/y DeCBeAew βαβα −− +++= (18.5)

where

1gg1gg 22 −−=−+= βα

18.2.5 Boundary conditions

The values of the constants A, B, C and D are determined by the boundary conditions. The width of theslab is L. The slab is subjected to a moment Mc on each of its boundaries which compensates the momentresulting from the curvature due to the thermal gradient. The case is symmetric. The boundary conditionsare:

- for all y, w(y) = w(-y)- for y = 0, the deflection presents a maximum: dw/dy = 0- for y = L/2, the moment is equal to Mc: -D d2w/dy2 = Mc

- for y = L/2, the shear force in the slab is zero: -D d³w/dy³ + G dw/dy = G dw/dy.

18.2.6 Expression of the moment for g < 1

Let z = L/(2l). The boundary conditions write:The deflections are symmetrical:

[ ] [ ] =+++ − l/ysin(D)l/ycos(Ce)l/ysin(B)l/ycos(Ae l/yl/y ββββ αα

(18.6)

[ ] [ ]l/ysin(D)l/ycos(Ce)l/ysin(B)l/ycos(Ae l/yl/y ββββ αα −+−−

For y = 0, dw/dy = 00=β+α−β+α DCBA (18.7)

HenceDBCA −== and

[ ] [ ]{ }l/ysin(B)l/ycos(Ae)l/ysin(B)l/ycos(Ae2w l/yl/y ββββ αα −++= −

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For y = L/2, d²w/dy² = -Mc/D

+

−++

−− )zcos(g1)zsin(gBe)zsin(g1)zcos(gAe 2z2z ββββ αα

(18.8)

D2lM

)zcos(g1)zsin(gBe)zsin(g1)zcos(gAe2

c2z2z −=

−−−

−+ −− ββββ αα

For y = L/2, d³w/dy³ = 0

( ) ( )[ ]++−−− )zsin(g21)zcos(g21Ae z βββαα

( ) ( )[ ]+++−− )zcos(g21)zsin(g21Be z βββαα

(18.9)

( ) ( )[ ]−+−−− )zsin(g21)zcos(g21Ae z βββαα

( ) ( )[ ] 0)zcos(g21)zsin(g21Be z =++−− βββαα

The solutions of the system of equations (18.8) and (18.9) are

( ) ( )[ ])zsinh()zcos(g21)zcosh()zsin(g21D

lM2A

2c αββαβα ++−−

∇−= (18.10)

( ) ( )[ ])zcosh()zsin(g21)zsinh()zcos(g21D

lM2B

2c αββαβα +−−−

∇= (18.11)

)z2sinh()z2sin( αββα +=∇

The resulting moment is MR = Mt – Mc = Mt - Dd2w /dy2

Introduce the values of A and B in the expression of the moment Mc; the resulting moment becomes

−−+= )l/ysinh()l/ysin(g1)l/ycosh()l/ycos(g

l

DADM 2

20y,R αβαβτα

−++ )l/ycosh()l/ycos(g1)l/ysinh()l/ysin(g

l

DB 22

αβαβ (18.12)

that we write

y0y,R CDM τα= (18.13)

For y = 0, the moment is maximum and equation (18.12) transforms into

++

−=)z2sinh()z2sin(

)zsinh()zcos()zcosh()zsin(21DM 0y,R αββα

αββαβατα (18.14)

For g = 0, one finds the equation established by Westergaard (1927)

++

−=)2/z2sinh()2/z2sin(

)2/zsinh()2/zcos()2/zcosh()2/zsin(21DM 0R τα (18.15)

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18.2.7 Expression of the moment for g = 1

Let z = L/(2l). The boundary conditions write:

The deflections are symmetrical:l/yl/yl/yl/yl/yl/yl/yl/y DyeCyeBeAeDyeCyeBeAe −−+=+++ −−−− (18.16)

For y = 0, dw/dy = 0

0DClB

lA

=++− ( 18.17)

HenceDCBA −==

For y = L/2, d²w/dy² = - Mc/D

( ) ( ) ( )D

lMeeClzeeCl2eeA

2czzzzzz −=−++++ −−− (18.18)

For y = L/2, d³w/dy³ = 0

( ) ( ) ( ) 0eeClzeeCl3eeA zzzzzz =++−+− −−− (18.19)

The solutions of the system of equations (18.18) and (18.19) are

[ ])zcosh(z)zsinh(3DlM

A2

c +∇

−= (18.20)

)zsinh(DlM

Cl2

c

∇= (18.21)

z2)z2sinh( +=∇The resulting moment is MR = Mt – Mc. Introduce the values of A and B in the expression of the momentMc; the resulting moment becomes

[ ])l/ysinh(Cly)l/ycosh(Cl2)l/ycosh(Al

D2DM

20y,R +++= τα (18.22)

that we write

y0y,R CDM τα= (18.23)

For y = 0, the moment is maximum and equation (18.22) writes

+

+−=

z2)z2sinh()zcosh(z)zsinh(

21DM 0y,R τα (18.24)

18.2.8 Expression of the moment for g > 1

Let z = L/(2l). The boundary conditions write:

The deflections are symmetrical:l/yl/yl/yl/yl/yl/yl/yl/y DeCeBeAeDeCeBeAe βαβαβαβα +++=+++ −−−−

(18.25)For y = 0, dw/dy = 0

0DCBA =−−+ βαβα (18.26)

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HenceDBCA ==

For y = L/2, d²w/dy² = - Mc/D

[ ] [ ]D

lMeeBeeA

2czz2zz2 −=+++ −− ββαα βα (18.27)

For y = L/2, d³w/dy³ = 0

[ ] [ ] 0eeBeeA zz3zz3 =−+− −− ββαα βα (18.28)

The solutions of the system of equations (18.27) and (18.28) are

)zsinh(DlM

A 32

c ββ∇

−= (18.29)

)zsinh(DlM

B 32

c αα∇

= (18.30)

[ ])zsinh()zcosh()zsinh()zcosh(2 αβαβαβ −=∇

The resulting moment is

++= )l/ycosh(B)l/ycosh(A

l

D2DM 22

20y,R ββαατα (18.31)

that we write

y0y,R CDM τα= (18.32)

The maximum moment in y = 0 is

−−

−=)zsinh()zcosh()zsinh()zcosh(

)zsinh()zsinh(1DM 0y,R αβαβαβ

ααββτα (18.33)

The moment in y = L/2 is

++= )zcosh(B)zcosh(A

l

D2DM 22

20y,R ββαατα

Replacing A and B by their values, one verifies that MRy = 0.

18.2.9 Verification of the expression of the maximum moment for g = 1

Passing to the limit for g → 1 one obtains by equations (18.15) or by (18.33) the value for the resultingmoment for g = 1

+

+−=

)z2sinh(z2)zcosh(z)zsinh(

21DM 0y,R τα (18.34)

This equation is identical to equation (18.24) which therefore is verified together with the basic equations(18.15) and (18.33).

18.2.10 Equation of the thermal stress

The thermal stress is given by the classical equation σ = MRh /(2I).

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18.2.11 Example

Consider a slab of great length, a width of 5000 mm and a thickness of 200 mm. Let E = 30000 N/mm²,µ = 0.20 and k = 0.1 N/mm³. The thermal gradient is 0.08 °C/mm and the coefficient of dilatation0.00001. The thermal stresses in function of the distance to the middle of the slab and the value ofPasternak’s modulus are given in Table 18.1.

G 0 500 1000 1500 2000 25000 2.63 2.49 2.06 1.34 0.49 020000 2.43 2.31 1.92 1.27 0.47 050000 2.18 2.08 1.74 1.69 0.45 0

Table 18.1 Thermal stresses in a slab of great length

18.3 Rectangular slab

18.3.1 Differential equation of equilibrium

We admit that the curvature of the slab takes the shape of two perpendicular cylindrical surfaces whichleads to a clastic surface (uplifted at the four corners). It is the case of a pavement built of concrete slabs.The problem has been considered by Bradbury (1938) as that of a slab with stiffness D built up of twoperpendicular beams subjected to a thermal gradient.

The equilibrium equation of an unloaded slab on a Pasternak foundation is

0Dkw

wDG

w 24 =+∇−∇ (18.35)

where w is the deflection for any x,yG is Pasternak’s shear modulusk is Westergaard’s subgrade modulus.

The thermal stress along the boundaries of the slab is uniform. Thus the slab is not subjected to torsion.As a result of this the moments of torsion are zero.

0yx

w)1(DMM

2

yxxy =∂∂

∂−=−= µ (18.36)

Hence Bradbury suggests approaching the solution of equation (18.35) by next equation)y(w)x(w)y,x(w += (18.37)

where the variables are separated.

The equilibrium equation simplifies in two equations

0EIkw

dxwd

EIG

dxwd

2

2

4

4

=+− (18.38)

0EIkw

dywd

EIG

dywd

2

2

4

4

=+− (18.39)

The solutions of equations (18.38) and (18.39) are given in § 18.2.

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18.3.2 Boundary conditions

Conditions related to the values of the moments along the boundaries of the slab

c02

2

2

2

x M)1()1(Dy

w

x

wDM µταµµ +=+=

∂+

∂−=

(18.40)

c02

2

2

2

y M)1()1(Dy

w

x

wDM µταµµ +=+=

∂+

∂−=

Mc in this equation has the same signification and value as in equation (18.8). Given equation (18.37), thecondition given in equation (18.40) can be split in two separated conditions

c02

2

2

2MD

y

wD

x

wD ==

∂−=

∂− τα (18.41)

c02

2

2

2MD

y

wD

x

wD µτµαµµ ==

∂−=

∂− (18.42)

In analogy the conditions related to the shear forces are simplified by

0xw

D3

3

=

∂∂

− (18.43)

0yw

D3

3

=

∂∂

− (18.44)

The conditions in equations (18.41) to (18.44) lead to same results as the conditions of the previousparagraph.

The moments in the slab are

+= yx0x,R CCDM µτα (18.45)

+= yx0y,R CCDM µτα (18.46)

When the slab is square the value of the moment along the boundaries of the slab appears in the equationfor the resulting moment along the diagonals.

+= = yx0R C)1(DM µτα (18.47)

18.3.3 Examples

Consider a square slab with a side of 5000 mm and a thickness of 200 mm. Let E = 30000 N/mm²,µ = 0.20 and k = 0.1 N/mm³. The thermal conditions remain the same. The thermal stresses along adiagonal are given in function of the distance to the centre and the value of Pasternak’s modulus inTable 18.2.

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x 0 500 1000 1500 2000 2500G y 0 500 1000 1500 2000 2500

σx 3.16 2.99 2.47 1.61 0.59 00

σy 3.16 2.99 2.47 1.61 0.59 0σx 2.92 2.77 2.30 1.52 0.56 0

20000σy 2.92 2.77 2.30 1.52 0.56 0σx 2.62 2.49 2.09 1.40 0.54 0

50000σy 2.62 2.49 2.09 1.40 0.54 0

Table 18.2 Thermal stresses in a slab of great length

The thermal stresses along a side of the slab are given in function of the distance to the centre and thevalue of Pasternak’s modulus on Table 18.3.

x 0 500 1000 1500 2000 2500G

y 2500 2500 2500 2500 2500 2500σx 2.63 2.49 2.06 1.34 0.49 0

0σy 0.53 0.50 0.41 0.27 0.10 0σx 2.43 2.31 1.92 1.27 0.47 0

20000σy 0.49 0.46 0.38 0.25 0.09 0σx 2.18 2.08 1.74 1.17 0.45 0

50000σy 0.44 0.42 0.35 0.23 0.09 0

Table 18.3 Thermal stresses along a side of a square slab

The results concerning the σx stresses are exactly the same as those obtained along the boundary of a slabof great length (Table 18.1). The results concerning the σy stresses do not respect the boundary conditionswhich stated that those stresses should be zero. The results are imputable to the application of theequation for a moment in a slab

∂∂

+∂∂

−=2

2

2

2

y,Ryw

xw

DM µ (18.48)

According to Bradbury, the derivative ∂2w/∂y2 cancels in y = 2500 but not the derivative ∂2w/∂x2. As aresult the stress σy cannot be zero in y = 2500. In order to avoid this Bradbury suggests to consider that atthe border the slab behaves as a beam. He applies a moment Mt = EI/R = EIα0τ.. In that case theresulting moment is given by

x0x,R CEIM τα= (18.49)

Equation (18.49) can be compared with equation (18.13) .The results become those of Table 18.4.

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x 0 500 1000 1500 2000 2500G

y 2500 2500 2500 2500 2500 2500σx 2.52 2.39 1.98 1.29 0.47 0

0σy 0 0 0 0 0 0σx 2.33 2.22 1.84 1.22 0.45 0

20000σy 0 0 0 0 0 0σx 2.09 2.00 1.67 1.12 0.43 0

50000σy 0 0 0 0 0 0

Table 18.4 Thermal stresses along the side of a square slab acotding to Bradbury

When he represents the boundary of a slab by a thin beam, Bradbury resolves the problem in a state ofplane stress. Nevertheless it seems evident that the considered case is one of plain strain (Timoshenko,1951, § 1). In that case the model of a slab of great length has to be adopted. The results are given inTable 18.5.

x 0 500 1000 1500 2000 2500G

y 2500 2500 2500 2500 2500 2500σx 2.63 2.49 2.06 1.34 0.49 0

0σy 0 0 0 0 0 0σx 2.43 2.31 1.92 1.27 0.47 0

20000σy 0 0 0 0 0 0σx 2.18 2.08 1.74 1.17 0.45 0

50000σy 0 0 0 0 0 0

Table 18.5 Thermal stresses along the side of a square slab following Timoshenko

The question rises: how far from the boarder have we to apply a model of a slab of great length (σy = 0)and from which distance on have we to apply the model of a rectangular slab (σy ≠ 0). Since de values ofσx given in Table 18.1 and in Table 18.5 are the same, we suggest to systematically adopt the model of arectangular slab and to accept that along the boundaries the stresses perpendicular to the boarders arezero. However the solution is only an approximate one. In the next section we will compare the solutionfor a rectangular slab with that for a circular slab, which can rigorously been established. Circular slabsare not built in reality but the comparison of the results will enable us to appreciate the effectiveness ofthe solution adopted for rectangular slabs.

18.4 Circular slab

For simplicity we shall limit us to the model of a circular slab on a Winkler foundation. Consider acircular slab with radius R.

18.4.1 Equilibrium equation

The equilibrium equation writes in polar co-ordinates

0Dkw

rw

r1

rw

rr1

r 2

2

2

2

=+

∂∂

+∂∂

∂∂

+∂∂

(18.50)

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18.4.2 Solution of the equilibrium equation

The solution of equation (18.50) is)l/r(Bbei)l/r(Aberw += (18.51)

where

....6!6

)l2/r(!4!4)l2/r(

!2!2)l2/r(

1)l/r(ber1284

−+−= (18.52)

....!5!5)l2/r(

!3!3)l2/r(

!1!1)l2/r(

)l/r(bei1062

+−= (18.53)

18.4.3 Boundary conditions

Let

∂∂

+∂

∂∂∂

=∂

∂=

∂∂

=r

berr1

rber

r3ber

rber

2berr

berr1

1ber2

2

2

2

∂∂

+∂

∂∂∂

=∂

∂=

∂∂

=r

beir1

rbei

r3bei

rbei

2beir

beir1

1bei2

2

2

2

The boundary conditions are with z = R/l

[ ] [ ]D

lM)z(1bei)z(2beiB)z(1ber)z(2berA

2c−=+++ µµ (18.54)

[ ] [ ] 0)z(3beiB)z(3berA =+ (18.55)where Mc = Dα0τ(1 + µ) (Timoshenko, 1951, § 14).

The solutions of the system of equations (18.54) and (18.55) are

)z(3beiDlM

A2

c

∇−= (18.56)

)z(3berDlM

B2

c∇

= (18.57)

[ ] [ ] )z(3ber)z(1bei)z(2bei)z(3bei)z(1ber)z(2ber ×+−×+=∇ µµ

18.4.4 Resulting moment

The resulting moment is

( ) ( ))l/r(1bei)l/r(2beil

DB)l/r(1ber)l/r(2ber

l

DA)1(DM

220R µµµτα +++++= (18.58)

that we write

r0R C)1(DM µτα += (18.59)

18.4.5 Comparison between the models for rectangular and circular slabs

Although very similar, equations (18.45) and (18.46) for a rectangular slab and (18.59) for a circular slabdiffer fundamentally. Equation (18.59) is directly based on a slab model. Equations (18.45) and (18.46)

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are based on a somewhat artificial addition of two models of slabs of a great length (rather beam models).It is therefore necessary to compare the results of both models in order to verify their reliability.

Consider a slab with a modulus of 30000 N/mm², a thickness of 200 mm, a Poisson ratio equal to 0.20.The subgrade modulus is equal to 0.05 N/mm³. The thermal conditions remain the same. In Table 18.6 wecompare the stresses in the centre of a circular slab with the stresses in the centre of inner and outersquare slabs for different radius values.

Inner square slab Circular slab Outer square slab

Side stress Radius stress Side stress14142 2.99 10000 3.00 20000 3.007071 3.26 5000 3.21 10000 3.094255 2.35 3000 2.97 6000 3.172828 0.92 2000 1.45 4000 2.131414 0.074 1000 0.13 2000 0.28707 0.005 500 0.009 1000 0.019

Table 18.6 Thermal stresses in circular and square slabs

In Table 18.7 we compare the stresses along the radius of a circular slab with the stresses along thediagonals of inner and outer square slabs.

DistancesSlab Dimensions 0 100 200 500 1000 1414 2000Square 4000 2.125 2.115 2.086 1.887 1.242 0.441 0Circular 2000 1.441 1.441 1.419 1.267 0.788 0.228 0Square 2828 0.915 0.906 0.879 0.704 0.233 0

Table 18.7 Thermal stresses in circular and square slabs

The results seem to be satisfactorily. Hence we conclude that the developed model for the computation ofthermal stresses in concrete slabs and base courses can be used.

18.5 Extension to a multi-slab system

The extension of the method to a multi-slab system can be done in two different ways.

If one may admit that the radii of curvature due to the thermal gradient are the same for the differentslabs, thus that the slabs remain in contact as well in the vertical direction as in the horizontal plane, oneshall consider the system as a multi-layered structure with friction at the interfaces.In that case, the moment Mc required to suppress the curvature is Mc = Dα0τ at the border of a long slaband Mc = Dα0τ (1 + µ1) at the border of a circular or rectangular slab. In those expressionsD = E1/(1 - µ1

2).Ieq where Ieq is the moment of inertia of the equivalent section with constant modulus. Theequivalent section is represented at Figure 18.1 for a bi-slab structure. The width of the lower slab isequal to E2/E1.(1 - µ1)2/(1 - µ2)2.

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Figure 18.1 Equivalent bi-slab structure

However this hypothesis seems rarely expressing the reality. If the radii of curvature are different(variable gradient with depth, different thermal dilatations) we admit that due to the weight of the appliedloads the slabs will remain in contact in the vertical direction (this hypothesis was admitted byWestergaard and Bradbury regarding the contact between slab and subgrade). In that case one shallconsider the system as a multi-layered structure with slip at the interfaces.

In the case of a tri-slab structure, the total moment acting on the border of the slab is given by nextequation

3c2c1cctotal MMMM ++=

303320221011ctotal DDDM τατατα ++=

The equation of equilibrium is deduced from Figure 18.2.

Figure 18.2 Equilibrium equation

The equilibrium equation of the upper slab is

1

11

4Dq

w −=∇

That of the intermediate slab

2

212

4D

qpw

−=∇

E1

E2 E1

E1

p1

q1

p2

q2

p3

q3

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That of the lower slab

3

23

4D

qpw

−=∇

The slabs remain vertically in contact. Hencewwww 321 ===

Adding the three equilibrium equations yields

( ) qqpqpqwDDD 22114

321 −=−+−+−=∇++

which is the equilibrium equation of the structure.Let D = D1 + D2 + D3 and q = kw - G∇2w.Then

0Dk

wDG

w 24 =+∇−∇

with k/D = 1/l4.

CommentIt seems that the pavements in continuously reinforced concrete and their eventual base-courses in leanconcrete should be treated by the model of the slab of great length. Also the base-courses in lean concreteof asphalt pavements. However the pavements built with concrete slabs and their eventual base-courses inlean concrete should both be treated by the model of the rectangular slab. It is indeed not possible toapply two different models on the same structure which, of course, would be the ideal solution.

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Chapter 19 Determination of the Parameters of a Rigid Structure

19.1 Determination of the Young’s modulus of a concrete slab

The purpose of the developed method is to determine the Young’s modulus of a concrete layer through adiametral test performed on a core taken out of the layer. The theory developed, in this regard, byMitchell (Timoshenko, 1948) is not really appropriate. Therefore we will briefly present a more generalmodel developed by Van Cauwelaert (1993).

19.1.1 Resolution of the compatibility equation

The problem is solved in polar co-ordinates. Consider a disc with diameter D (2R) subjected todiametrically opposite radial pressures p uniformly distributed over arcs 2αR. The compatibility equationis (§ 10.5.1)

0r1

rr1

rr1

rr1

r 2

2

22

2

2

2

22

2

=

∂∂

+∂∂

+∂∂

∂∂

+∂∂

+∂∂

θΦΦΦ

θ (19.1)

The solution of (19.1) is obtained by separation of the variables

[ ])m2()m2(mm DrCrBrAr)mcos( −+− +++= θΦThe stresses must be finite at the origin; hence the constants B and D vanish.

[ ])m2(m CrAr)mcos( ++= θΦ (19.2)

19.1.2 Equations for the stresses

Applying equation (10.44) one obtains

[ ]m2mr Cr)m2)(m1(Ar)m1(m)mcos( −++−= −θσ (19.3)

[ ]m2mr Cr)m2)(m1(Ar)m1(m)mcos( −++−= −θσ (19.4)

[ ]m2mr Cr)m1(Ar)m1()msin(m +−−−= −θτ θ (19.5)

19.1.3 Boundary conditions

For r = R, the boundary conditions areσr = - p for - α < θ < α, π - α < θ < π + α (19.6)

σr = 0, elsewhere (19.7)τrθ = 0, everywhere. (19.8)

We shall develop equation (19.3) as a Fourier series in order to express the discontinuous loadingconditions. Hence the boundary equations become

1CR)m2)(m1(AR)m1(m m2m =−++− −

0CR)m1(AR)m1( m2m =+−− −

The solutions of the system are

2R

A)m1(2m−

=−

2R

C)m1(m

=+

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194

and

−+

=

− m2m

r Rr

)m2(Rr

m2

)mcos( θσ (19.9)

+−

−=

− m2m

Rr

)m2(Rr

m2

)mcos( θσθ (19.10)

−=

− m2m

r Rr

mRr

m2

)msin( θτ θ (19.11)

In order to express conditions (19.6) and (19.7) expand (19.9) as a Fourier series (§ 11.3).

∑∞

=

+

+=

1nnn

0

Ln

sinbL

ncosa

2a

)(fπθπθ

θ

With 2L = π, the Fourier coefficients become

)n2sin(n

p2an α

π−=

παp4

a0 −=

0bn =Hence

)n2sin()n2cos(n

p2p2)(f

1n

αθππ

αθ ∑

=

−−= (19.12)

19.1.4 Stresses and displacements

The normal stresses are obtained by transforming (19.9) and (19.10) by (19.12)

−+

−−=

−∞

=∑

n2)1n(2

1nr R

r)n1(

Rr

n)n2sin()n2cos(n

p2p2αθ

ππα

σ (19.13)

+−

+−=

−∞

=∑

n2)1n(2

1n Rr

)n1(Rr

n)n2sin()n2cos(n

p2p2αθ

ππα

σθ (19.14)

The radial strain is given by

Er

rθµσσ

ε−

=

and the radial displacement by ∫= dru rε .

This results in

−−−= ∑∞

=1n

)n2sin()n2cos(Rr

)1(E

pDu αθµα

π

++−−

+

−+ +− 1n21n2

Rr

)1n2(n)n1(n1

Rr

1n21 µµ

(19.15)

Assuming the value of µ, the value of E can be deduced from (19.15) by a direct determination of thedisplacement u, along, for example, the horizontal diameter (Dac Chi Nguyen, 1991).

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19.1.5 Tangential normal stress along the vertical diameter

Assume that the disc is subjected to two opposite point loads, P = 2pαR, located at both ends of itsvertical diameter. The kernel of the Fourier series is then

DP4

n)n2sin(p2

lim0

=→

αα

Along the vertical diameter (θ = 0) the tangential stress is given by

−⋅⋅⋅

+

+

++−=

642642

Rr

4Rr

3Rr

2Rr

4Rr

3Rr

21DP4

DP2

ππσθ

DP2

πσθ = (19.16)

Equation (19.16) is the well-known formula for the tensile stress of rupture along the vertical diameter ofa loaded disc in the so-called indirect split-tensile or Brazilian tensile test.

19.2 Determination of the characteristics k and G of the subgrade

Pasternak (1954) himself proposed the method of the determination of the parameters k and G through aneccentrically loaded plate.

19.2.1 Equilibrium equation for a Pasternak subgrade

Consider a semi-infinite body subjected to a uniform pressure p. Isolate an elementary prism dxdy.

Figure 19.1 Stress distribution in a Pasternak subgradeThe vertical stress q expresses the soil reaction and Tx and Ty the lateral shear forces (Figure 19.1).Vertical equilibrium yields

dxdydy

dTTdydx

dxdT

TdxTdyTqdxdypdxdy yy

xxyx

+−

+−++=

dy

dT

dxdT

qp yx −−=

Assume

Ty + (dTy/dy)dy

Tx + (dTx/dx)dxTx

dy

dx

Ty

q

p

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kwq = (Winkler)

dydw

GTdxdw

GT yx == (Pasternak)

Hence

+−=

2

2

2

2

dy

wd

dx

wdGkwp (19.17)

and with G/k = l2

+−=

2

2

2

22

dy

wd

dx

wdlw

kp

(19.18)

or in polar co-ordinates

++−=

2

2

22

22

d

wd

r

1drdw

r1

dr

wdlw

kp

ϑ (19.19)

19.2.2 Eccentrically loaded plate-bearing test

Consider the eccentrically loaded circular plate presented in Figure 19.2.

Figure 19.2 Eccentrically loaded plate

Ne

WCWR

WL

2a

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Split the load in a normal centre load N and a moment M = eN (Figure 19.3).

Figure 19.3 Vertical resultant and moment

and, as can be depicted from Figure 19.4:

Figure 19.4 Eccentrically loaded plateOne has:- w0 = (wL + wR)/2- w1 = (wR – wL)/2 = atanα0 ≅ aαO

W0

+

N M

a0

WL

NM

r r

+

N

2a

N

y

x

e

M

θ

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198

19.2.3 Vertical load

Split N = NP + NL.:- NP is the resultant of the subgrade reaction underneath the plate (Figure 19.5).- NL is the resultant of the subgrade reaction outside the plate (Figure 19.6).

Figure 19.5 Deflection underneath the plate

The deflection under the rigid plate is constant. Hence the stress under the plate is also a constant.

0y kw=σand

20p akwN π=

Consider now the subgrade reaction outside the plate (Figure 19.6). Therefore assume that the distributionof the deflection outside the plate is the same as the deflection under an isolated load, with the boundarycondition that for r = a, w = w0

Figure 19.6 Deflection outside the plate

Recall equation (19.19)

++−=

2

2

22

22

d

wd

r

1drdw

r1

dr

wdlw

kp

ϑ

W0

2a

NL

2a

W0

NP

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Due to axial symmetry d/dθ = 0, hence

+−=

drdw

r1

dr

wdlw

kp

2

22

The equilibrium equation for an isolated load is given by the homogeneous equation

0drdw

r1

dr

wdlw

2

22 =

+−

which appropriate solution is (§ 3.5))l/r(AKw 0=

where K0 is the modified Bessel function of second kind and of order zero. For r = a, w = w0. Hence

)l/a(K)l/r(K

ww0

00=

and

)l/a(laK)l/a(K

2kwrdrd)l/r(K

)l/a(K1

kwrdrdN 10

0a

2

0 a

2

00

00zL

πθθσ

π π

∫ ∫ ∫ ∫∞ ∞

===

The total load is then

+=+=

)l/a(K)l/a(K

al

21akwNNN0

120LP π (19.20)

19.2.4 Moment

Figure 19.7 Deflection due to the moment

Consider Figure 19.7. Split the moment in M = MP + ML. The moment of the stresses underneath theplate is

04

4

0P a4k

a1

4a

kavI

M αππ

ασ ===

The moment of the stresses outside the plate is again obtained considering that the distribution of thedeflection is that of the deflection under an isolated load.

2a

α0

w1

M = e N

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The equilibrium equation becomes in this case

0d

wd

r

1drdw

r1

dr

wdlw

2

2

22

22 =

++−

ϑ (19.20)

The stresses along the circumference of the plate vary from 0 for θ = 0 to a maximum value for θ = π/2.Separating the variables of (19.20), we therefore use a sine function.

θsinvw =Hence (19.20) transforms in

0r

v

l

v

dr

dv

r

1

dr

vd222

2=−−+

which appropriate solution is)l/r(AKv 1=

andθsin)l/r(AKw 1=

For r = a and θ = π/2, w = aα0. Hence

)l/a(Ka

A1

0α= θα sin

)l/a(K)l/r(K

aw1

10=

Figure 19.8 Stress distribution due to the moment

The moment of the stresses outside the plate is deduced from Figure 19.8:

∫ ∫∞

θθσ0 a

zL rdrdsinr2M

∫ ∫∞

θθα0 a

221

10L drdsinr)l/r(K

)l/a(K1

ka2M

)l/a(Ka)l/a(K

lakM 2

2

10L απ=

The total moment becomes

rsinθ

r

N

θ

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201

+=

)l/a(K)l/a(K

al

41

kaM1

20

4 πα (19.21)

19.2.5 Determination of k and G

Let w0 = (wR + wL)/2 and w1 =aα0 = (wR - wL)/2. Hence

+

+=

)l/a(K)l/a(K

al

21a2

wwkN

0

12LR π (19.22)

+

−=

)l/a(K)l/a(K

al

41

a2

wwkNe

1

23LR π (19.23)

Divide (19.23) by (19.22):

( )( )

)l/a(lK/a)l/a(K

21

)l/a(lK/a)l/a(K

41

wwawwe

0

1

1

2

LR

LR

+

+=

−+

(19.24)

Knowing e, a, wR and wL, determine a/l in an iterative manner from (19.24) and k from (19.22).

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Chapter 20 The Semi-Infinite Body Subjected to a Vertical Load

20.1 Introductory note

Boussinesq solved the problem of a semi-infinite body subjected to an isolated load in 1885, which iscommonly referred to as the Boussinesq solution. At that time the stress potentials expressed in cylinderco-ordinates (Love, 1927) were not yet discovered and the use of integral transforms in civil engineeringmechanics, in particular the Hankel transform, were not very common. Nevertheless, Boussinesq solvedthe problem in a memoir of about 80 pages. All civil engineers know about this memoir but seldom arethose who have studied or at least read it.Fortunately, the mathematical methods available today offer simple solutions for a great series of loadingcases. To begin with we will look into the problem of a semi-infinite body subjected to a uniformlydistributed load, followed by the methodology first proposed by Burmister in 1943, and then derive theBoussinesq solution for an isolated load.

20.2 The semi-infinite body subjected to a vertical uniform circular pressure

In this application, we assume axial symmetry. Thus the problem will be solved in axi-symmetriccylindrical co-ordinates (§ 10.5.2). Consider a semi-infinite body subjected to a vertical pressure puniformly distributed over a circular area with radius a, given in Figure 20.2.

Figure 20.1 Semi-infinite body subjected to a distributed vertical pressure

The stress potential is a solution of the compatibility equation (10.50).

0zrr

1rzrr

1r 2

2

2

2

2

2

2

2

=

∂∂

+∂∂

+∂∂

∂∂

+∂∂

+∂∂ ΦΦΦ

(20.1)

The Bessel function J0(mr) is a solution of equation

2a

r

z

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204

0mrr

1

r2

2

2=+

∂+

∂Φ

ΦΦ (20.2)

Hence, if we assume a solution by separation of the variables )z(f)mr(J0=Φ , equation (20.1) can betransformed into

0)mr(J)z(fmz

)z(fm2

z

)z(f0

42

22

4

4=

+

∂−

∂ (20.3)

Using the resolution method by means of the characteristic equation (§ 1.4.2), we obtain the solution of(20.1)

−+−= −− mzmzmzmz

0 zDezCeBeAe)mr(JΦ (20.4)

Applying equation (10.49), we obtain the expressions for the stresses and the displacements

+−−−+−= − mzmz2mz2

0z e)mz21(CmeBmeAm)mr(mJ µσ

+−+ −mze)mz21(Dm µ (20.5)

−++++= − mzmz2mz2

0r e)mz21(CmeBmeAm)mr(mJ µσ

−+− −mze)mz21(Dm µ

−+++− − mzmz2mz21 e)mz1(CmeBmeAm

mr)mr(mJ

−− −mze)mz1(Dm (20.6)

µσθ 2DmeCme)mr(mJ mzmz0

−= −

−++++ − mzmz2mz21 e)mz1(CmeBmeAm

mr)mr(mJ

−− −mze)mz1(Dm (20.7)

+++−= − mzmz2mz2

1rz e)mz2(CmeBmeAm)mr(mJ µτ

−+ −mze)mz2(Dm µ (20.8)

−−−−−

+−= − mzmz2mz20 e)mz42(CmeBmeAm

m)mr(mJ

E1

w µµ

+−− −mze)mz42(Dm µ (20.9)

−+++

+= − mzmz2mz21 e)mz1(CmeBmeAm

m)mr(mJ

E1

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205

−− −mze)mz1(Dm (20.10)

The constants A, B, C and D are to be determined by the boundary conditions.The stresses must vanish ad infinite depth. Hence the constants related to positive exponents must bezero: A = C = 0. The boundary conditions at the surface (z = 0) express that

arforpz <−=σarfor0z >=σrwhatever0rz =τ

We apply a Hankel’s transform (§ 11.5) with kernel F(m) = paJ1(ma)/m, example (§ 11.4.3), in order toexpress the discontinuous character of the vertical stress at the surface

[ ] pdm)21(DmBm)ma(J)mr(Jpa0

210z −=−+−= ∫

∞µσ (20.11)

This requires the first condition

1)21(DmBm2 =−+ µWe further express that the load is strictly vertical, thus that the shear stress at the surface must be zero

[ ] 0dm2DmBm)ma(J)mr(Jpa0

211rz =+−= ∫

∞µτ (20.12)

This requires the second condition

02DmBm2 =+− µ

The system results in 1Dm2Bm2 == µHence the stress function becomes

( ) dmemzm

maJmrJpa mz−

+−=Φ ∫ µ2)()(

03

10 (20.13)

and the equations for the stresses are

( )∫∞

−+−=0

mz10z dmemz1)ma(J)mr(Jpaσ (20.14)

( ) −−= ∫∞

0

mz10r dmemz1)ma(J)mr(Jpaσ

( )∫∞

−−−−0

mz11 dmemz21mr

)ma(J)mr(Jpa µ (20.15)

+= ∫∞

0

mz10 dme2)ma(J)mr(Jpa µσθ

( )∫∞

−−−+0

mz11 dmemz21mr

)ma(J)mr(Jpa µ (20.16)

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∫∞

−−=0

mz11rz dmmze)ma(J)mr(Jpaτ (20.17)

( )∫∞

−+−+

=0

mz10 dmemz22m

)ma(J)mr(Jpa

E1

w µµ

(20.18)

( )∫∞

−−−+

−=0

mz11 dmemz21m

)ma(J)mr(Jpa

E1

u µµ

(20.19)

Equations (20.14) to (20.19) can be integrated in the axis of the load (r = 0, J0(mr) = 1). By (8.12) and(8.5)

( )

+−−=

2/322

3

zza

z1pσ (20.20)

( ) ( )

+−

+−−−=

+=

2/322

2

2/122

rr

za

az

za

z)21(21

2p

2µµ

σσσ θ (20.21)

0rz =τ (20.22)By (8.14) and (8.12)

( ) ( )( )

+−

+−

−+

−=

2/122

2/1222

za

z1pz

E1

zzapE

12w

µµ (20.23)

0u = (20.24)

At the surface, we find Boussineq’s well-known equation for the deflection

( )pa

E12

w2µ−

= (20.25)

20.3 The semi-infinite body subjected to an isolated vertical load

The solution is derived from the previous case by a simple transformation of the kernel of the Hankel’stransform. The function f(z) remains unaltered. The kernel for the distributed load is

m)ma(J

paK 1dis = (20.26)

The kernel for the isolated load, P = pπa2 is obtained by going over to the limit

ππ 2P

m)ma(J

aP

limm

)ma(JpalimK 1

0a1

0aiso ===

→→ (20.27)

The stress functions becomes

∫∞

−+−=0

mz2

0 dme)mz2(m

)mr(J2P

µπ

Φ (20.28)

The stresses and displacements are obtained replacing in equations (20.14) to (20.19) the kernel (20.26)by the kernel (20.27).In particular the vertical stress

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207

( )∫∞

−+−=0

mz0z dmemz1)mr(mJ

2Pπ

σ (20.29)

Applying (8.9) and (8.11), (20.29) can be integrated

( ) 2/522

3

zrz

z32P

+−=

πσ (20.30)

In the axis of the load we obtain the well-known formula of Boussinesq

2zz2

P3

πσ −= (20.31)

20.4 The semi-infinite body subjected to a circular vertical rigid load

Consider a semi-infinite body subjected to a vertical load transmitted through a rigid plate, such as, forexample, in the case of a plate-bearing test. This problem is solved in the specialised literature by the so-called method of dual integral. This method, as already mentioned, is beyond the scope of this book.However, utilising the properties of the discontinuous Weber-Schafheitlin integrals, we shall be able toobtain the appropriate solution.

Consider the stress function (20.4) that we have developed for the semi-infinite body subjected to adistributed load.

−−= −− mzmz

0 zDeBe)mr(JΦ (20.32)

Here also the function f(z) will remain unaltered, hence Bm2 = 2µ and Dm = 1. The surface boundaryconditions are

arfor0z >=σarforttanconsw <=

rwhatever0rz =τ

We will meet the boundary conditions by applying a Hankel’s transform which kernel we determine bythe properties of the discontinuous Weber-Schafheitlin integrals (§ 8.6):

dmm

)ma(J)mr(JK

0

0z ∫

∞−=

λνσ (20.33)

∫∞

+−

=0

10

2dm

m

)ma(J)mr(JE

)1(2Kw

λνµ

(20.34)

For r > a

+

+−+−

++−

+

+−

−=+− 2

2

1z

r

a;1;

21

,2

1F

21

)1(2r

21

aK ν

λνλνλν

ΓνΓ

λνΓ

σλλν

ν

The first condition writes: 01for0z =++−= λνσ

For r < a

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−−−

+

−=

+− 2

2

1

02

a

r;1;

2,

2F

2)1(2a

2r

E)1(2

Kwλνλν

λνΓΓ

λνΓ

µ

λλ

The second condition writes: 0forttanconsw =−−= λν The system results in2/12/1 =−= νλ .

∫∞

−=0

2/102/1

z dm)ma(J)mr(JmKσ

∫∞

−=0

0z dm)masin()mr(Ja2

σ

By § 8.6.4:

22zra

1a2

K−

−=π

σ

In order to determine the value of K, we integrate the value of σz over the circular area with radius a.

∫ ∫−

−=−π θ

π

2

0

a

022 ra

rdrda2

KP

a2a2

KP ππ

−=−

Hence

a22P

= (20.35)

Finally, for z = 0

∫∞

−=0

0z dm)masin()mr(Ja2

σ (20.36)

∫∞−

=0

02

dmm

)masin()mr(Ja

PE

1w

πµ

(20.37)

In the axis of the load (r = 0)

a2P

E1

dmm

)masin(a

PE

1w

2

0

2 µπ

µ −=

−= ∫

∞ (20.38)

Comparing equation (20.38) with the result obtained for a uniformly distributed load, equation (20.25)

aP

E)1(2

w2

πµ−

=

we obtain

4w

w

flexible

rigid π= (20.39)

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Hence, when trying to determine the modulus of the subgrade by a plate-bearing test, do not overestimateit by using the formula for a flexible load.

For z ≥ 0

∫∞

−+−=0

mz0z dme)mz1)(masin()mr(J

a2Pπ

σ (20.40)

which becomes for r = 0, by (8.19) and (8.21)

( )

+

+−=

222

22

zza

z3a2Pπ

σ (20.41)

20.5 The semi-infinite body subjected to a vertical uniform rectangular pressure

Together with the problem of the semi-infinite body subjected to shear loads (breaking forces) at itssurface, which is discussed in Chapter 21, this solution will allow to treat the problems of oblique loads.This problem is solved in cartesian co-ordinates (§ 10.5.4). Consider a semi-infinite body subjected to avertical pressure p distributed over a rectangular area with sides 2a and 2b.

The stress potential is a solution of the compatibility equation 10.60.

0zyxzyx 2

2

2

2

2

2

2

2

2

2

2

2=

++

++

Φ∂

Φ∂

Φ∂

∂ (20.42)

In the case of a vertical load the equation in Ψ is useless. If we assume a solution by separation of thevariables

)z(f)sycos()txcos(=Φequation (20.42) can be transformed into

( ) ( ) 0)sycos()txcos(stz

)z(fst2

z

)z(f 2222

222

4

4=

++

∂+−

∂ (20.43)

Letting m2 = t2 + s2, the solution for f(z) is identical to the solution for a semi-infinite body subjected to avertical pressure distributed over a circular area. Hence

mzmzmzmz zDezCeBeAe)z(f −− −+−= (20.44)The constants A, B, C and D are to be determined by the boundary conditions.The stresses must vanish at infinite depth. Hence the constants related to positive exponents must be zero:A = C = 0. The boundary conditions at the surface (z = 0) express that

bybandaxaforpz <<−<<−−=σbyandaxfor0z >>=σ

y,xwhatever0yzxz == ττIn order to express the discontinuous character of the vertical stress at the surface, we apply a doubleFourier’s integral (§ 11.4) with kernel

ts)sbsin()tasin(p4

)s,t(F2π

=

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[ ] pdsdt)21(DmBmts

)sbsin()sycos()tasin()txcos(p4

0 0

232z −=−+−= ∫ ∫

∞ ∞

µπ

σ (20.45)

This requires the first condition

1)21(DmBm 23 =−+ µWe further express that the load is strictly vertical, thus that the shear stresses at the surface must be zero

[ ] 0dsdt2DmssBmts

)sbsin()sycos()tasin()txsin(p4

0 0

22xz =+−= ∫ ∫

∞ ∞

µπ

τ (20.46)

[ ] 0dsdt2DmttBmts

)sbsin()sycos()tasin()txsin(p4

0 0

22yz =+−= ∫ ∫

∞ ∞

µπ

τ (20.47)

This requires the second condition

02DmBm2 =+− µThe system results in

1Dm2Bm 23 == µwhich are the same values as those obtained in the case of a distributed load over a circular area. Hencethe equations for the stresses and the displacements can directly been obtained from the results of § 20.2.

We are particularly interested in the equation for the vertical stress, which can be deduced from equation(20.14)

( )∫∞

−+−=0

mz10z dmemz1)ma(J)mr(Jpaσ

∫ ∫∞ ∞

−+−=0 0

mz2z dsdte)mz1(

ts)sbsin()sycos()tasin()txcos(p4

πσ (20.48)

This equation cannot be integrated analytically. However, in the axis of the load (x = 0, y = 0) (20.48)simplifies into an equation that can be integrated.

∫ ∫∞ ∞

−+−=0 0

mz2z dsdte)mz1(

ts)sbsin()tasin(p4

πσ (20.49)

Split equation (20.49) into two factors

∫ ∫∞

−∞

−=0

mz

021,z dsdte

ts)sbsin()tasin(p4

πσ

∫ ∫∞

−∞

−=0

mz

022,z dsdtmze

ts)sbsin()tasin(p4

πσ

By equation (8.26)

2/12221,z )baz(zab

arctanp2

++−=

πσ (20.50)

By equation (8.27)

2/12222222

222

2,z )baz)(bz)(az()baz2(zabp2++++

++−=

πσ (20.51)

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20.6 Comparison between the vertical stresses. Principle of de Saint-Venant

We have established the equations giving the vertical stress in the axis of vertical loads of different shapeswith different pressure distributions. The mean pressure is p = -1. In Table 20.1 we give the values of thevertical stresses in function of the relative depth, z/a, for identical loads (loads with same resultant andsame moment regarding the vertical axis). The utilised equations are (20.20) for the circular flexible load,(20.31) for the isolated load, (20.41) for the circular rigid load, (20.51) and (20.52) for the square flexibleload. The radius of the circular loads is a, the length of the side of the square load is πa .

Depth z/a (20.20) (20.31) (20.41) (20.51,52)

0.1 -1 -150 -0.505 -10.5 -0.910 -6.0 -0.560 -0.9061 -0.646 -1.50 -0.500 -0.6392 -0.284 -0.375 -0.260 -0.2825 -0.057 -0.060 -0.056 -0.057

10 -0.0148 -0.0150 -0.0148 -0.014820 -0.00374 -0.00375 -0.00373 -0.00374

Table 20.1 Vertical stresses for different loads

We conclude from Table 20.1 that from a certain depth on the stresses become identical whatever theshape of the load or the distribution of the pressure at the surface of the semi-infinite body. Those resultsare a perfect illustration of the principle of de Saint-Venant, which states that at a distance great enoughfrom an applied load the values of the stresses and the displacements are independent from the way theload is applied as far as resultant and moment remain identical.

20.7 The orthotropic body subjected to a vertical uniform circular pressure

This problem is explained in axi-symmetric cylindrical co-ordinates (§ 10.5.5). Consider the orthotropicsemi-infinite body subjected to a vertical pressure p uniformly distributed over a circular area with radiusa, as given in Figure 20.2. The stress potential is a solution of the compatibility equation (10.66).

0zn

nrr

1

rzrr1

r 2

2

2

22

2

2

2

2

2

2=

−++

++

Φ∂

µ

µ∂Φ∂

Φ∂

∂∂∂

∂ (20.52)

The Bessel function J0(mr) is a solution of equation

0mrr

1

r2

2

2=+

∂+

∂Φ

ΦΦ (20.53)

Hence, if we assume a solution by separation of the variables )z(f)mr(J0=Φ , equation (20.1) can betransformed into

0)mr(J)z(fmz

)z(f)2nn(m

z

)z(f)n( 0

42

2222

4

422 =

+

∂−+−

∂− µµ (20.54)

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Figure 20.2 Orthotropic body subjected to a distributed vertical pressure

Using the resolution method by means of the characteristic equation (§ 1.4.2), we obtain the solution of(20.1)

−+−= −− smzsmzmzmz

0 DeCeBeAe)mr(JΦ (20.55)

where

22

22

n

ns

µ

µ

−= (20.56)

We use next notations2Am)1(nA µ+=2Bm)1(nB µ+=2Csm)n(nC µ+=2Dsm)n(nD µ+=

Applying equation (10.65), we obtain the expressions for the stresses and the displacements

+++−= −− smzsmzmzmz

0z DeCeBeAe)mr(mJσ (20.57)

+++= −− smz2smz2mzmz

0r DesCesBeAe)mr(mJσ

++

+++

++− −− smzsmzmzmz1 Den1

Cen1

BeAemr

)mr(mJµµ

µµ

(20.58)

22smzsmz

0n

)n1(DeCe)mr(mJ

µ

µσθ

+= −

++

+++

+++ −− smzsmzmzmz1 Den1

Cen1

BeAemr

)mr(mJµµ

µµ

(20.59)

2a

r

z

Eh

Ev

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−+−= −− smzsmzmzmz

1rz sDseBeAe)mr(mJτ (20.60)

++

−++

+−+

−= −− smzsmzmzmz0 sDe1n

sCe1n

BeAem

)mr(mJE

1w

µµ

µµµ

(20.61)

+++++++= −− smzsmzmzmz1 De)1(Ce)1(Be)n(Ae)n(

mE)mr(mJ

u µµµµ (20.62)

The stresses must vanish ad infinite depth. Hence the constants related to positive exponents must bezero: A = C = 0. The boundary conditions at the surface (z = 0) express that

arforpz <−=σarfor2/pz =−=σarfor0z >=σrwhatever0rz =τ

We apply a Hankel’s transform (§ 11.5) with kernel F(m) = paJ1(ma)/m, example (§ 11.5.2), in order toexpress the discontinuous character of the vertical stress at the surface

[ ] pdmDB)ma(J)mr(Jpa0

10z −=+−= ∫∞

σ (20.63)

This requires the first condition1DB =+

We further express that the load is strictly vertical, thus that the shear stress at the surface must be zero

[ ] 0dmDsB)ma(J)mr(Jpa0

11rz =+−= ∫∞

τ (20.64)

This requires the second condition0DsB =+

The system results in s1

1D

s1s

B−

=−

−=

All the equations for the stresses can now be established. Again we are particularly interested in thevertical stress, which equation writes

dmes1

1e

s1s

)ma(J)mr(Jpa0

mszmz10z ∫

∞−−

−+

−−−=σ (20.65)

In the axis of the load, we have

[ ]dmese)ma(Js1

pa

0

mszmz1z ∫

∞−− +−

−−=σ (20.66)

By (8.13)

+−

+−−=

2/1222/1222z)za(

1

)zsa(

1sz1pσ (20.67)

In the case of an isolated load equation (20.56) transforms into

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[ ]dmese)mr(mJ)s1(2

P

0

mszmz0z ∫

∞−− +−

−−=

πσ (20.68)

By (8.9)

++

+−

−−=

2222/322zrsz(

sz

)rz(

sz)s1(2

σ (20.69)

and in the axis of the load (r = 0)

2

2

2zs

ss1

z2

P ++−=

πσ (20.70)

If we compare equation (20.70) with the corresponding stress in an isotropic body of equation (20.31)

2zz2

P3

πσ −=

we notice that if s < 1, thus n > 1)isotropic()canisotropi( zz σσ >

thus stress concentration in the axis of the load for the anisotropic body; and if s > 1, thus n < 1)isotropic()canisotropi( zz σσ <

thus stress dispersion in the axis of the load for the anisotropic body.

Fröhlich (1934) had already observed this phenomenon in the early thirties. To take into account thedifferences between observed values and computed values of the vertical stress, he modified Boussinesq’sformula (20.31) as follows

2zz2

P

π

νσ −= (20.71)

The parameter ν was called the stress concentration factor of Fröhlich. Indeed for most of the soils ν > 3.Only for super consolidated clays, Fröhlich observed values ν < 3. In fact the range of values admitted byFröhlich was 2 < ν < 6. Similar observations were made regarding the values of the degree of anisotropyof orthotropic soils. In most of the cases n > 1, except again for super consolidated clays. Hence onecould imagine a relation between the stress concentration factor of Fröhlich and the degree of anisotropyof an orthotropic soil, based, for example, on the equation for the vertical stress in the axis of an isolatedload:

ν=++2

2

s

ss1 (20.72)

Assuming a Poisson’s ratio of 0.50, as usual for subgrades, equation (20.72) allows to fix a range of nbased on the range of ν values. One obtains

3n6.0 <<

This presents a particular interest. Indeed, Fröhlich’s relation was obtained by transforming Boussinesq’srelation “afterwards” i.e. without a rigorous mechanical analysis. It can be shown that Fröhlich’s relationdoes not respect the continuity principle. However, Veverka (1973) has proved that compatibility can berestored if the modulus of the semi-infinite body is a function of the bulk stress [p0 = (σx + σy + σz)/3]

k00 pEE = (20.73)

where

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215

23

k−−

=νν

(20.74)

Nevertheless, Fröhlich’s model does not allow the computation of all the stresses and displacements inthe case of semi-infinite bodies subjected to distributed loads of all sorts. However, all this can easily bedone by anisotropic elasticity, by applying relations (20.72) and (20.74).

2

2

ss1

s2s1k

−+

−+= (20.75)

Hence both theories are complementary. Anisotropic elasticity provides the mathematical tools whileFröhlich provides the values of the physical parameters. Further, equation (20.75) could create a linkbetween the, from a computational viewpoint, relatively simple anisotropic theory and the morecomplicated non-linear models developed by Ullidtz (1998).

In this paragraph, we have merely considered the degree of anisotropy n as a mechanical concept: theratio between the values of two Young’s moduli. It is the usual approach in pavement engineering.However, one can also consider anisotropy as a geotechnical concept. Next relation has been established(Van Cauwelaert and Cerisier, 1982) based on Prandtl’s bearing capacity’s model (Prandtl, 1921).

−+

−−=

)2/(sin1)2/(sin1

sin11

pc4

Ks 2

2

a ϕϕ

ϕ

whereKa is the active earth pressure coefficientc is the cohesion of the soilϕ is the angle of internal friction of the soilp is the uniform limit pressure in Prandtl’s model.

20.8 The orthotropic body subjected to a vertical uniform rectangular pressure

This problem is essentially of interest in geotechnical engineering. The stresses and displacements candirectly be obtained from equations (20.47) to (20.52) transformed with the appropriate kernel of § 20.4.The vertical stress, for example, becomes

[ ]dsdteses1

1ts

)sbsin()sycos()tasin()txcos(p4 smzmz

0 02z

−−∞∞

+−−

−= ∫ ∫π

σ

This equation can be integrated in the axis of the load

+

+++

++−

−−=

2/122222/1222z)zsba(sz

abarctan

)zba(z

abarctans

s11p2

πσ

The reader interested in more results (rigid plate) and the detailed mathematics will find the requiredinformation in Van Cauwelaert (1985).

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Chapter 21 The Semi-Infinite Body Subjected to Shear Loads

21.1 The semi-infinite body subjected to radial shear stresses

The problem is solved in a similar way as the body subjected to a vertical pressure (§ 20.2): the stressesare symmetrical regarding the axis of the load. Hence the stress-function, the solution of the compatibilityequations and the basic equations for the stresses are the same in both cases. Only the surface conditionsare different. One can assimilate them with the effect of a wheel that is pushed down on the surface of thesubgrade. It seems possible that a falling weight could induce a similar effect. Taking that into accountcould perhaps ameliorate the results of backcalculation of moduli based on deflections basins obtainedduring falling weight tests. However, therefore much more research and observation is required. Theboundary conditions express that

rwhatever0z =σarfor0rz <≠τ

arfor0rz >=τWe apply a Hankel’s transform (§ 11.5) with kernel F(m) =KJµ(ma)/mλ in order to express thediscontinuous character of the shear stress at the surface

[ ] )(fdm2DmBmm

)ma(J)mr(JmK

0

21rz τµτ λ

µ =+−= ∫∞

(21.1)

This requires the first condition

12DmBm2 =+− µWe further express that the load is strictly horizontal, thus that the vertical stress at the surface must bezero

[ ] 0dm)21(DmBmm

)ma(J)mr(JmK

0

20z =−+−= ∫

µσ λµ (21.2)

This requires the second condition

0)21(DmBm2 =−+ µ

The system results in 1Dm21Bm2 =+−= µTo determine the values of K, µ and λ we need to know the distribution of the shear stress at the interfaceload – subgrade. It is evident that the shear stress must be zero in the axis of the load. Hence we shallassume that at the surface the shear stress increases linearly from the center to the circumference of theload.

dmm

)ma(J)mr(JKdm

m

)ma(J)mr(mJK

01

1

0

1rz ∫∫

== λµ

λµτ (21.3)

For r < a

+−+−

+−−+

= −+−−

21)1(1

)2(

21)1(1

2ar

11)1(1rz λµΓΓ

λµΓ

τ λλ

+−−−+−−+2

2

ar

;2;2

1)1(1,

21)1(1

Fλµλµ

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218

In order to obtain a linear variation of the shear stress, the second term of the F – function must be zero.Hence the first condition: 03 =−− λµ .For r > a

( )F

21)1(1

)1(

21)1(1

2ra

11)1(rz

+−+−

+

+−−+

= −+−− λµΓµΓ

λµΓ

τ λλµ

µ

In order to satisfy the condition, the argument of the second Γ - function of the denominator must be zero.Hence the second condition: 01 =+− λµ . The system results in µ = 2, λ =1. The shear stress at thesurface writes

∫∞

==0

221 ar

Kdm)ma(J)mr(JKτ (21.4)

We determine K in function of the total shear load Q applied to the subgrade.

∫ ∫ ==π π

θ2

0

a

02

2

3a2K

drdar

KQ

Hence

a2Q3

=

and at the surface

∫∞

=0

21rz dm)ma(J)mr(Ja2

Q3π

τ (21.5)

We have written (21.5) with a positive sign. This means that, with the conventions of Timoshenko(chapter 10.3), the shear stresses are oriented towards the axis of the load. The other stresses anddisplacements are obtained in the usual manner. Hence, the equation for the vertical stress, for example,

∫∞

−−=0

mz20z dmmze)ma(J)mr(J

a2Q3π

σ (21.6)

21.2 The semi-infinite body subjected to a one-directional asymmetric shear load

Consider a semi-infinite body subjected to shear stresses acting in one direction, for example the x-direction in cartesian co-ordinates. This problem cannot be considered as a symmetric load case. Muki(1960) solved for the first time this problem in asymmetric cylinder co-ordinates. Thus he considered theshear load as circular, but acting in one specific direction. He discovered the stress potentials inasymmetric cylinder co-ordinates (§ 10.5.3) and solved the problem by expressing the stresses anddisplacements as a Fourier series of Bessel integrals. In other words he applied a Fourier-Besseltransform. In our opinion, his contribution was as important as those of Boussinesq (1885) and Love(1927). However, the involved mathematics are complicated.We got the idea that the problem could be easier solved in a more appropriate system of co-ordinates,namely the cartesian co-ordinates because the direction of the shear load could easily made parallel withone of the co-ordinate axis’s and as such determine the appropriate solutions of the compatibilityequations. Based on Muki’s work, we were able (Van Cauwelaert, 1985) to establish the stress potentialsin a system of cartesian co-ordinates both for an isotropic body as well as an orthotropic body (§ 10.5.4).It is the isotropic system that we will apply here to solve the problem of a semi-infinite body subjected toa one-directional shear load. We assume that the load is applied through a rectangular surface as in § 20.5.

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In a system of cartesian co-ordinates the solutions are of trigonometric type. Both cosine and sine havethe same properties and the choice of one of the other function has no fundamental mathematicalimplications. Indeed cosine and sine functions can be assimilated with the Bessel functions J-1/2 and J1/2 sothat a Hankel transform, or here, a Fourier transform can be applied on both functions. However in asystem of cylinder co-ordinates the two available solutions, J0(mr) and Y0(mr), have differentmathematical properties. A Hankel transform can only be applied on Bessel functions of the first kind.This explains the difficulty Muki has encountered in his solution.

The compatibility equations in Cartisian co-ordinates are

0zyxzyx 2

2

2

2

2

2

2

2

2

2

2

2

=

∂∂

+∂∂

+∂∂

∂∂

+∂∂

+∂∂ φφφ

(21.7)

0zyx 2

2

2

2

2

2

=

∂∂

+∂∂

+∂∂ ψψψ

(21.8)

The trigonometric solution will allow us to exactly define the required function.

Assume that the shear load is applied parallel to the x-axis. We assimilate the shear load with a breakingforce exercised through the wheel on the subgrade. Thus the corresponding shear stresses will be constantall over the surface of the load. This can be expressed by a double Fourier integral of the next form

∫ ∫∞ ∞

=0 0

2xz dsdtts

)tbsin()tycos()tasin()txcos(q4π

τ (21.9)

Hence we must choose the solutions of (21.7) and (21.8) that will allow us to express the surface shearstress in the x-direction as (21.9). The concerned stress potential is

zy21

z)1(

x

2

2

22

xz ∂∂∂

+

∂∂

−∇−∂∂

=ψφ

φµτ

Therefore we must have solutions in the form)z(f)sycos()txsin( 1=φ)z(f)sysin()txcos( 2=ψ

Remembering that the coefficients of positive exponents must be zero we finally obtain

−−= −− mzmz zDeBe)sycos()txsin(φ (21.10)

−= −mzFe)sysin()txcos(ψ (21.11)

where 222 stm += . The boundary conditions arebybandaxaforqxz <<−<<−=τ

byandaxfor0xz >>=τ

y,xwhatever0yzz == τσThe equations for the stresses are

[ ]mz2mz3z eDm)mz21(eBm)sycos()txsin( −− +−+−= µσ (21.12)

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+−−−= −−− mzmzmz2

xz mse2F

Dmte)mz2(teBm)sycos()txcos( µτ (21.13)

−−−−= −−− mzmzmz2

yz mte2F

Dmse)mz2(seBm)sysin()txsin( µτ (21.14)

The boundary equations become

12

22 −=+−= msF

DmttBmqxz µτ

0mt2F

Dms2sBm0 2yz =++−= µτ

0Dm)21(Bm0 3z =−+= µσ

The system results in

22

m

t)21(Bm µ−=

2m

tDm −=

2m

s2Fm =

The stresses becomemz

z ze)sycos()txsin(t −−=σ (21.15)

mz2

xz em

zt1)sycos()txcos( −

−=τ (21.16)

mzyz e

mzts

)sysin()txsin( −=τ (21.17)

We express the double Fourier integral

dsdtts

)sbsin()sycos()tasin()txcos(q4

0 02xz ∫ ∫

∞∞=

πτ (21.18)

By taking τxz as positive, we express that the shear stresses are oriented in the opposite side of the x-axis.The kernel of the Fourier transform is

ts)sb(tsin)tasin(q4

K2π

= (21.19)

Applying (21.19) to the equations (21.12), (21.13) and (21.14), we obtain

dsdtzes

)sbsin()sycos()tasin()txsin(q4 mz

0 02z

−∞∞

∫ ∫−=π

σ (21.20)

dsdtmt

z1ts

)sbsin()sycos()tasin()txcos(q4 2

0 02xz

−= ∫ ∫

∞∞

πτ (21.21)

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221

dsdtzem

)sbsin()sysin()tasin()txsin(q4 mz

0 02yz

−∞∞

∫ ∫=π

τ (21.22)

The other stresses and displacements can be obtained in the same way.

Equations (21.20) and (21.21) can be numerically integrated without difficulties. Indeed for s =0,sin(sb)/s = b and for t = 0, sin(ta)/t =a. However some problems could arise with (21.22) which,therefore is better transformed by letting

θρθρ sinscost ==Hence (21.22) becomes

ρθθρθρθρθρπ

τ ρπ

ddze)sinbsin()sinysin()cosasin()cosxsin(q4 z

2/

0 02yz

−∞

∫ ∫=

that now can also be numerically integrated without difficulties.

21.3 The semi-infinite body subjected to a shear load symmetric to one of its axis’s

Consider a semi-infinite body subjected to shear stresses acting in the x – direction but symmetrical to they – direction: τxz(x) = - τxz(-x). Then the expression at the surface for the shear stress must take a formsuch as

dsdtst

)sbsin()sycos()ta(J)txsin(K

0 0xz ∫ ∫

∞ ∞

µτ (21.23)

Hence the solutions of the compatibility equations become)z(f)sycos()txcos( 1=φ)z(f)sysin()txsin( 2=ψ

The values of µ and λ are determined in order to satisfy the boundary conditions in the x – direction.

- for –a < x < a τxz = f(x)- for | x | > a τxz = 0- for all x τxz (x)= - τxz (-x)

Knowing that 2

dss

)sbsin()sycos(

0

π=∫

, we only consider the integral

dtt

)ta(J)txsin(I

0∫∞

µ (21.24)

Transform equation (21.24)

dtt

)ta(J)tx(J

2x

I0

2/12/1

∫∞

−=

λµπ

Applying equations (8.25) and (8.26) one finds

- for | x | < a 02

2/32/1=

+−− λµ

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- for | x | > a 02

2/12/1=

++− λµ

Hence, µ = 3/2 and λ =1/2 and equation (21.23) transforms into

∫ ∫∞ ∞

=0 0

2/32/1xz dsdt

s)sbsin()sycos()ta(J)tx(J

Kτ (21.25)

For | x | < a and | y | < b

∫∞

=0

2/32/1xz dt)ta(J)tx(J22

xK

ππτ

2ax

a2Kxz

ππτ =

Let us admit that for x = a, τxz = -q then

a22Kq

ππ−=

ππ2a2

qK −=

ax

qxz −=τ

For x = - a, τxz = qFor x = 0, τxz = 0For x = a, τxz = - q

The final equation for τxz becomes

∫ ∫∞ ∞−

=0 0

2/32/1xz dsdt

s)sbsin()sycos()ta(J)tx(J

xaq2

πτ (21.26)

The concerned stress potential is

zy21

z)1(

x

2

2

22

xz ∂∂∂

+

∂∂

−∇−∂∂

=ψφ

φµτ

Remembering that the coefficients of positive exponents must be zero, the definite solutions of thecompatibility equations are

[ ]mzmz mzDeBe)sycos()txcos( −− −−=φ (21.27)

[ ]mzFe)sysin()txsin( −−=ψ (21.28)

where 222 stm += . The relations for the stresses become

[ ]mz2mz3z eDm)mz21(eBm)sycos()txcos( −− +−+−= µσ (21.29)

−−−= −−− mzmzmz2

xz mse2F

Dmte)mz2(teBm)sycos()txsin( µτ (21.30)

+−−−= −−− mzmzmz2

yz mte2F

Dmse)mz2(seBm)sysin()txcos( µτ (21.31)

The boundary equations write

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223

1ms2F

Dmt2tBmq 2xz =−−−= µτ

0mt2F

Dms2sBm0 2yz =+−= µτ

0Dm)21(Bm0 23z =−+= µσ

The system results in

22

m

t)21(Bm µ−=

2m

tDm −=

2m

s2Fm −=

The stresses become

dsdtezts

)sbsin()sysin()ta(J)txcos(a2q2 mz2/1

0 0

2/3z

−∞ ∞

∫ ∫=ππ

σ (21.32)

dsdtemt

z1st

)sbsin()sycos()ta(J)txsin(a2q2 mz2

0 02/1

2/3xz

−∞ ∞

−= ∫ ∫ππ

τ (21.33)

dsdteztm

)sbsin()sysin()ta(J)txcos(a2q2 mz2/1

0 0

2/3yz

−∞ ∞

∫ ∫=ππ

τ (21.34)

Other stress distributions can be obtained by setting, for | x | < a, the more general condition

n2

2/32/1−=

+−− λµ, where n is an integer.

We transform equation (21.26) into

∫ ∫∞ ∞

=0 0

2/12/3

xz dsdtst

)sbsin()sycos()ta(J)txsin(a2q2ππ

τ (21.35)

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225

Chapter 22 The Multilayered Structure

22.1 The multilayered structure

Consider the n-layered structure subjected to a load as presented in Figure 22.1.

Figure 22.1 Multilayered structure

Each layer is characterised by a thickness, a Young’s modulus, a Poisson’s ratio, and for the last layerconsidered as the subgrade eventually a degree of anisotropy.

The solutions for a two-layered structure and later for a three-layered structure subjected to verticaluniformly distributed loads were first published by Burmister (1943, 1944).

We present the solution for an n-layered structure subjected to the different loads analysed in Chapters 20and 22. The subgrade can either be isotropic or orthotropic, and can further be of infinite extent or limitedby a rigid bottom at a given depth.

The principal hypothesis will be that the different layers remain in contact. Hence, at the interfaces, thevertical stresses, the shear stresses and the vertical deflections are to be identical. A second hypothesis isthat there will be some horizontal friction between the layers. This will require an equation between thehorizontal displacements and the shear stresses.

h1

h2

h3

hn-2

hn-1

2a

z0

z1

z2

zn-2

zn-1

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22.2 Solutions of the continuity equations

Each layer will require a solution of the continuity equation. Depending on the load, this continuityequations and their solutions will be

- in the case of symmetrical circular loads

0zrr

1

rzrr

1

r 2

2

2

2

2

2

2

2=

∂+

∂+

∂+

∂+

∂ φφφ

- in the case of a uniformly distributed vertical pressure

- in the case of an isotropic layer

( )dmezDezCeBeAm

)ma(J)mr(Jpa

0

mzi

mzi

mzi

mzi

10∫∞

−− −+−=φ

- in the case of an orthotropic layer

( )dmeDeCeBeAm

maJmrJpa smz

ismz

imz

imz

i∫∞

−− −+−=0

10 )()(φ

- in the case of an isolated load

( )dmezDezCeBeA)mr(J2P

0

mzi

mzi

mzi

mzi0∫

∞−− −+−=

πφ

- in the case of a rigid load

( )dmezDezCeBeAm

)masin()mr(Ja2

P

0

mzi

mzi

mzi

mzi

0∫∞

−− −+−=π

φ

- in the case of a radial shear load

( )dmezDezCeBeAm

)ma(J)mr(Ja2

Q3

0

mzi

mzi

mzi

mzi

20∫∞

−− −+−=π

φ

where the index i is related to the concerned layer.

- in the case of rectangular loads

0zyxzyx 2

2

2

2

2

2

2

2

2

2

2

2=

∂+

∂+

∂+

∂+

∂ φφφ

0zyx 2

2

2

2

2

2=

∂+

∂+

∂ ψψψ

- in the case of a uniformly distributed vertical pressure

( )∫ ∫∞∞

−− −+−=0 0

mzi

mzi

mzi

mzi2

dsdtezDezCeBeAts

)btsin()ytcos()atsin()xtcos(

4

p

πφ

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227

- in the case of a uniformly distributed shear load in the x – direction

( )∫ ∫∞ ∞

−− −+−=0 0

mzi

mzi

mzi

mzi2 dsdtezDezCeBeA

ts)bssin()yscos()atsin()xtsin(

4qπ

φ

( )dsdteFeEts

)bssin()yssin()atsin()xtcos(4

q mzi

mzi

0 02

−∞ ∞

−= ∫ ∫πψ

- in the case of a shear load symmetric to one of its axis

( )∫ ∫∞ ∞

−− −+−−=0 0

mzi

mzi

mzi

mzi

2/32 dsdtezDezCeBeA

s)bssin()yssin(J)xtcos(

a8q

ππ

φ

( )dsdteFeEs

)bssin()yssin(J)xtcos(a8

q mzi

mzi

0 0

2/32

−∞ ∞

−−= ∫ ∫ππ

ψ

It must be noticed that in the case of a rigid load applied on a multilayered structure, the stress functionderived from the case of a rigid load on a semi-infinite body cannot ensure a deflection profile strictlycorresponding with a rigid load (deflection constant under the load). In the case of the semi-infinite bodythe f(z) functions at the surface for the vertical stress and the deflection are, excepted for a constant,identical. Hence the same boundary condition can be used for both stress and deflection. The secondboundary condition expresses that the shear stress is zero.In the case of the multilayer, the equations for the vertical stress and the deflection are no longer the samebecause of the presence of the terms in A1 and C1. However the difference of the multilayer deflectionprofile and the strictly rigid load profile remains small in such a way that the proposed stress function cansafely been used.

22.3 Boundary conditions

The values of the boundary “constants” (in fact they are not really constants, but functions of theintegration parameters m, t, s and of the geometrical and mechanical characteristics of the multilayeredstructure) Ai, Bi, Ci, Di, Ei and Fi are determined through the boundary equations. In the case of symmetricloads, the number of constants per layer is 4, excepted for the last layer for which the constants with apositive exponent must be zero. Hence we need 4n - 2 boundary equations. Two equations are required toexpress the loading conditions; four equations will express the interface conditions: 4(n – 1) + 2 = 4n – 2.If the load is a vertical load, the loading conditions will be

)P(fz =σ0rz =τ

If the load is a radial shear load, the loading conditions will be0z =σ

)Q(frz =τAt a depth zi, the vertical interface conditions will be

zjzi σσ =

rzjrzi ττ =

ji ww =where the index i refers to the bottom of layer i and the index j to the surface of the following layer j .The horizontal interface conditions can be of two sorts.

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228

If one assumes full friction between the layers, one shall write that the horizontal displacements are equal

ji uu =If one assumes full slip between the layers, one shall write that the shear stresses are zero

0rzjrzi == ττThis requires two sets of boundary conditions whether there is full friction or full slip between the layersand hence two different computer programs. In order to avoid this difficulty De Jong, Peutz andKorswagen, the authors of the BISAR program (De Jong et al, 1973) suggested next horizontal boundaryequation

( ) rziji )1(uu βταα −=−

- when α = 0, the case corresponds with full slip at the interface.- when α = 1, the case corresponds with full friction at the interface.

The parameter β is required to make the dimensions of the equality homogeneous. We recommend to takeβ = h1/Ei , thus the thickness of the first layer divided by the modulus of the upper layer of the interface.Notice that the characteristic β has the inverse dimensions of the characteristic k of Westergaard: it is alsoa spring constant.When 0 < α < 1, one could speak of partial friction at the interface. The interest of the equation is thatone can express both extreme horizontal conditions and any intermediate condition. However it seemsdifficult to appreciate the exact physical meaning of the parameter α. Therefore we suggested a secondequation, namely

ji uu α=- when α = 1, the case corresponds to full friction at the interface.- when α ≠ 1, one can speak of partial friction.

However, this equation cannot simulate a so-called ‘full slip’ interface condition. The method presents theadvantage that the factor α can be determined in situ by the ovalisation test (Chapter 26). But bothmethods present the disadvantage that one must assume that the friction parameters are constant all overthe horizontal interface.

In the case of asymmetric loads, the number of constants per layer is 6, excepted for the last layer forwhich the constants with a positive exponent must be zero. Hence we need 6n - 3 boundary equations.Three equations are required to express the loading conditions; six equations will express the interfaceconditions: 6(n – 1) + 3 = 6n – 3. The conditions are the same as in the previous case except that at thesurface there will be one condition regarding the vertical stress and two conditions regarding the shearstresses and at each interface one condion regarding the vertical stresses, one conditions regarding thedeflections, two conditions regarding the shear stresses and two conditions regarding the horizontaldisplacements.

22.4 Determination of the boundary constants

It will be necessary during the numerical integration procedure, to evaluate the values of the integrandsfor each value of the integration variable m or t and s. Therefore the values of the unknown parametersAi …, which vary with m or t and s, are to be computed for all values of the integration variables. Thosecomputations are performed by solving the system of boundary equations. Due to the presence ofexponentials, the system has to be handled with care. Indeed for increasing values of the integrationvariable, the positive exponentials tend to overflow and the negative, when in a denominator, lead todivisions by zero. The appropriate way to avoid numerical problems with the exponentials is to express

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229

the equations for the unknown parameters in such a way that their numerators contain only negativeexponents and their denominators at least one constant value and further only negative exponentials.

)e(fttanCons

)e(fA

mz2

mz1

i −

+=

When m tends to infinity, the value of the unknown parameter necessarily will tend to zero and thiswithout any numerical problem. However this is not so easy to achieve when keeping in mind that theequations for the unknown parameters can in no possible way be expressed in closed form when thenumber of layers exceeds three or four. We developed a matrix method to reach this target, which isillustrated for a symmetrical vertical load. The boundary conditions can be written in matrix form.

At the surface

( ) ( )TT111101 01DCBAM =

At the first interface

( ) ( )T222212T

111111 DCBAMDCBAM =At the i-th interface

( ) ( )Tjjjj2iT

iiii1i DCBAMDCBAM =At the last interface

( ) ( )Tnn2kT

kkkk1k DBMDCBAM =where for the simplicity of the writings j stands for i + 1 and k stands for n – 1.

Invert the matrices of the left side of the equations (the matrix inversion is left over as an exercise to thereader). Hence at the i-th interface

( ) ( )Tjjjj2i1

1iT

iiii DCBAMMDCBA −=

Matrix 11iM − can be split into two sub matrices in such a way that the exponents can be put aside of the

matrix brackets.ii mz

12imz

11i1

1i eMeMM += −−

11iM − is a (4,4) matrix (4 rows, 4 columns)

11iM is a (2,4) matrix built with the rows 1 and 3 of matrix 11iM −

12iM is a (2,4) matrix built with the rows 2 and 4 of matrix 11iM −

Also matrix 2iM can be split

ii mz22i

mz21i2i eMeMM −+=

2iM is a (4,4) matrix

21iM is a (4,2) matrix built with the columns 1 and 3 of matrix 2iM

22iM is a (4,2) matrix built with the columns 2 and 4 of matrix 2iM .

The boundary condition can now be split as follows

( ) ( ) ( )Tjjmz2

22i11iT

jj21i11iT

ii DBeMMCAMMCA i−+⋅=

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230

( ) ( ) ( )Tjjmz2

CiT

jjAiT

ii DBeMCAMCA i−+=

( ) ( ) ( )Tjj22i12iT

jjmz2

21i12iT

ii DBMMCAeMMDB i +⋅=

( ) ( ) ( )TjjDiT

jjmz2

BiT

ii DBMCAeMDB i +=

DiCiBiAi M,M,M,M are (2,2) matrices.

We start at the last interface condition. In the last layer, the semi-infinite body, A and C are zero. Henceapplying the equations established for the i – th interface, and we obtain

( ) ( )Tnnmz2

CkT

kk DBeMCA k−=

( ) ( )TnnDkT

kk DBMDB =Further

( ) ( ) ( )Tkkmz2

CjT

kkAjT

jj DBeMCAMCA j−+=

( ) ( ) ( )Tnnmz2

DkCjT

nnmz2

CkAjT

jj DBeMMDBeMMCA jk −− +=

( ) ( ) ( )

+= −−− T

nnDkCjT

nn)zz(m2

CkAjmz2T

jj DBMMDBeMMeCA jkj

Hence the equation for (Aj Cj)T contains at least the negative exponent jmz2e

−. We then simplify again

the writings ( ) ( )Tnnmz2

j,ACT

jj DBeMCA j−=

( ) ( ) ( )TkkDjT

kkmz2

BjT

jj DBMCAeMDB j +=

( ) ( ) ( )TnnDkDjT

nn)zz(m2

CkBjT

jj DBMMCAeMMDB jk += −−

Hence the equation for (Bj Dj)T contains at least no positive exponent so that we can write

( ) ( )Tnnj,BDT

jj DBMDB =

Continuing up to the first interface, we obtain

( ) ( )Tnnmz2

1,ACT

11 DBeMCA 1−=

( ) ( )Tnn1,BDT

11 DBMDB =

We split the matrix of the surface equation

( ) ( ) ( ) ( )TT110,BD

T110,AC

T111101 01DBMCAMDCBAM =+=

( )( ) ( )TTnn1,BD0,BD

mz21,AC0,AC 01DBMMeMM 1 =+−

The solution of this matrix equation takes next form

11

1

mz4mz2

mz221

ncebea

ebbB

−−

++

+=

11

1

mz4mz2

mz221

ncebea

eddD

−−

++

+=

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231

We can now formulate the equations for all the parameters. In order to respect the method that we havedefined in the beginning, we shall express each parameter associated with the less favourable exponent:

the parameters Ai and Ci with the highest possible positive exponent, i.e. with the exponent imzecorresponding with the depth of the bottom of the layer; the parameters Bi and Di with the lowest possible

negative exponent, i.e. with the exponent 1imze −− corresponding with the depth of the surface of the

layer.

( ) ( )Tnnmz

1,ACT

11mz DBeMCAe 11 −=

( ) ( )Tnn1,BDT

11 DBMDB =

( ) ( )Tnnmz

j,ACT

jjmz DBeMCAe jj −=

( ) ( )Tnnmz

j,BDT

jjmz DBeMDBe ii −− =

( )11

1n11n

mz4mz2

mzmz221

nmz

cebea

eebbBe

−−

−−−

++

+=

−−

( )11

1n11n

mz4mz2

mzmz221

nmz

cebea

eeddDe

−−

−−−

++

+=

−−

Hence all parameters, unless B1 and D1, can be numerically computed without any difficulty. The way B1

and D1 are computed is explained in Chapter 23.

22.5 The fixed bottom condition

In this case one can assume the presence of a rigid layer at a depth zn, one shall often speak of a “fixed”bottom at that depth in the subgrade. There are different possibilities of expressing this supplementarycondition. We prefer expressing that at the given depth both vertical and horizontal displacements arezero. In doing so we have two boundary conditions more that have to correspond with two unknownparameters more, namely An and Cn.

The matrix equation at the interface n – 1 now writes

( ) ( )Tnnnn2kT

kkkk1k DCBAMDCBAM =In order to modify as little as possible the solution developed in the previous paragraph, we shall expressAn and Cn in function of Bn and Dn in the equation. Therefore we write the displacement equations at thelevel of the fixed bottom as follows

( )Tnnmz

n,BCT

nnmz

n,AC DBeM)CA(eM nn −= ( )Tnnmz2

n,BCT

nnn,AC DBeM)CA(M n−=Then we express in the second member of the matrix equation

( ) ( )Tnnnn2kT

kkkk1k DCBAMDCBAM =

1nmzneA − and 1nmz

neC − in function of )zz2(mn

1nneB −−− and )zz2(mn

1nneD −−−

and the problem is solved.

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22.6 The orthotropic subgrade

Often the subgrade can be considered as orthotropic. The solution of this problem is fairly simple.Replace in the second member of the matrix equation at the last interface

( ) ( )Tnn2k

Tkkkk1k DBMDCBAM =

the matrix Mk2 with the isotropic equations for stresses and displacements by a modified matrix with thecorresponding orthotropic equations. The final result will then be expressed as follows

( )11

1n11n

mz4mz2

mzmz221

nmz

cebeaeebb

Be −−

−−−

+++

=−

( )11

1n11n

mz4mz2

smzmz221

nsmz

cebeaeedd

De −−

−−−

+++

=−

Where the ration of anisotropy 22

22

nn

µ−−

= (§20.7)

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233

Chapter 23 The Resolution of a Multilayered Structure

The resolution of a multilayered structure is obtained by a numerical integration of the stress functions,wherein, as shown in previous Chapter, the values of the boundary parameters are successively computedfor each value of the integration variable. This integration procedure requires a series of algorithms thatwe will develop now. Most of the text that follows is taken from our research for WES (Van Cauwelaertet al, 1986) and our contribution in the multilayer software called NOAH (Eckmann, 1998) and in thedevelopment of the rigid & flexible pavement design and evaluation program PAVERS® (Stet et al, 2001& 2004).

23.1 Choice of the integration formula

The accuracy of numerical integration depends on the integration formula and the length of theintegration interval at whose boundaries the function to be integrated has to be computed. Numericalintegration can be performed by two main types of integration formulas: the Newton-Coates closed-typeformulas or the Gauss integration formula. The latter has been applied by De Jong, Peutz and Korswagenin their BISAR program (1973). It provides the same accuracy with about half as many integration pointsas the former method, but needs a transformation of the integral so that its limits are (-1, 1). In all thischapter we will consider, as example, the case of a vertical uniformly distributed load. Then theexpressions to be integrated are of the following form

∫∞

=0

10 dm)mz(f)ma(J)mr(JpaInt

As can been seen, the limits integration are from zero to infinity. The integral includes a product ofoscillating Bessel functions (depending on the radii, a(i), of the loads and their distances, r(i), to theorigin of the co-ordinate system) multiplied by a function of exponentials (depending on thecharacteristics of the structure and the depth at which stresses are to be computed). Because of theoscillating character of the Bessel functions, the Gauss integration formula would be very adequate if theintegrations were performed within the limits of the successive zeros of the Bessel functions. However inthis case, radii of loads and distances have to be constant values; otherwise, a separate integration has tobe performed for each load. Consequently, as we were able to show (Van Cauwelaert et al, 1988), thetime saved in using this method will be more than lost in the successive integrations. Thus, in order toprovide for a method as general and as fast as possible, we use the Newton – Coates formulas.

The Newton – Coates formulas split the integration range over an even number of equally spaced abscisesand compute, by appropriate polynomials, the area between two abscises. The most widely used formulasof this type are Simpson’s rule and Weddles’rule. Simpson’ rule gives exact results for cubic functionsand is written as

[ ])n2(f...)3(f4)2(f2)1(f4)0(f3h

S +++++=

Weddle’s rule gives exact results for polynomials of degree five and is written as

[ ])n6(f...)7(f5)6(f2)5(f5)4(f)3(f6)2(f)1(f5)0(10

h3W +++++++++=

Since the function to be integrated is an exponential, the most appropriate rule is not known a priori. Inorder to be able to compare the two integration rules, a program was written for the simple case of a twolayer. This program was used to evaluate the number of intervals required to achieve the same accuracyusing both of the above rules. However, as a preliminary, an effort was made to determine the mostslowly converging response parameter (i.e. stress, displacement). Since there is little influence of theBessel functions (varying between – 1 and + 1) on convergence, computations were performed on theexponential function only.

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The results of the analysis show that the deflection at the first interface is the most sensitive responseparameter. Mathematically the reason for this is that only the vertical deflection shows a non-zero value atthe origin of the integration. Thus the analysis could be limited to this equation for comparing theappropriate integration rule. It revealed that Weddle’s rule in the particular case of a multilayeredstructure needs less intervals than Simpson’s rule to achieve an equal accuracy. On the basis of thisanalysis we adopt Weddle’s rule for numerical integrations.

23.2 Values at the origin

As can be seen from the basic equations, all the integrands are zero for m = 0 (at the origin on theintegration interval) except for the vertical deflection w; indeed

2a

m)ma(J)mr(J

lim 10

0m=

Hence, the values of the deflections in the different layers must be computed for m = 0, thus also thevalues of the unknown parameters Ai, Bi, Ci and Di. This cannot be done by the method developed inChapter 22, because of numerical redundancies. However, it can be done strictly analytically as we willillustrate in the case of an isotropic subgrade. For m = 0, all exponentials are equal to 1 and the boundaryconditions for the vertical stresses and the shear stresses simplify into

1D)21(BD)21(C)21(BA nnn111111 =−+=−+−−+ µµµ0D2BD2C2BA nnn111111 =+−=++− µµµ

Hence

nn 2B µ=1Dn =

Since ni21 w...w...ww ===== , we can write for all layers

[ ] ( )n

2n

nnn

ni E

12)42(DB

E1

µµ −

=−++

=

23.3 The geometrical scale of the structure

A general rule for the width of the integration steps, which we will establish further, must be independentof the geometry of the structure, i.e. for a two layered system, the required integration steps must beindependent of the thickness of the first layer and of the radii of the loads (initially expressed in lengthunits). In the case of only one load, the radius of this load is the most interesting scale factor because ofthe fact that the Bessel function J1(ma), expressing the influence of the area of the load, simplifies intoJ1(m). However, when dealing with different load radii, this simplification becomes useless. Further,considering the minimum influence, besides their sign, of the Bessel functions on the accuracy of theresults, it is much more efficient to choose a scale factor independent of the loads but that scales all thepossible structures, i.e. the thickness of the first layer.The choice of this scale factor is the more appropriate in that most of the problems related to accuracyarise in the neighborhood of the load at the surface and to a depth equal to the thickness of the first layer.For this reason, the thickness, h1, of the first layer is selected as the scale factor. In application, all lengthdimensions will be divided by the thickness of the first layer so that this thickness becomes equal to 1.After computations are completed, all the results related to lengths, in fact all of the displacements, willhave to be rescaled to their real values.

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Scale is not a problem for force. Stresses are linear functions of the unit pressure (total load on the surfacedivided by its area) so that stresses are obtained immediately in the same units as the input pressure.Moduli do not need to be scaled because only modular ratios are used. However, it is necessary to expressmoduli in the same units as stresses to avoid difficulties with the displacements. Pay attention to the factthat the friction parameter β (mm3/N) in the partial friction relationship suggested by De Jong et al (1973)must be scaled.

We proposed to define β by

i

1i E

h=β

Hence, β scaled writes

ii

11i E

1E

h/h==β

23.4 Width of the integration steps

23.4.1 Influence of the moduli on the integration step

The two-layer analysis shows that if one considers only the exponential part of the function, thedeflections decreases smoothly from the origin of integration (m = 0) on, and that 90 % of the deflectionat the first interface is obtained near the origin of the integration. Therefore, the total width of theintegration interval will depend on the number of integration steps required to ensure accuracy near theorigin. The analysis shows that the required step for a high modular ratio can be smaller than theintegration step for a small modular ratio. A regression analysis performed on a great series ofcomputations leads to next formula for the optimum step value:

1

2

EE

1.0step =

As may be noted, the equation goes through the origin because if the ratio between the moduli tends toinfinity the step width should tend to zero. For an n-layered structure, previous equation can begeneralised in

1

n

EE

1.0step =

When the modulus of the last layer is higher than the modulus of the first layer, or in the case of a fixedbottom, the value of the deflection at the origin rely on the ratio of both moduli, but, depending of thevalues of the intermediate moduli, values higher than the initial values can be obtained for increasingvalues of the integration variable. Therefore before starting the computation of stresses anddisplacements, it is necessary to search for the highest value of the deflection function, which, whendetermined, can be used to establish the step width.

23.4.2 Influence of the radii of the loads on the integration step

In the previous paragraph we considered a load radius equal to 1. Essentially this represents a radius ofthe loaded area equal to the thickness of the first layer after application of the scale factor. For smallvalues of the argument, the value of the Bessel function J1(ma) is equal to ma/2. Also the variation, as afunction of m, of the successive values of the total integrated function, will depend upon the value of theradius, a, of the load Consequently, the width of the integration step will necessarily depend upon thevalue of a. A regression analysis shows that next equation is appropriate

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[ ]1)a/hlog(1.0)ha(step)ha(step 111 +==≠When there are multiple loads of different radii, this equation has to be applied for the load with thelargest radius of loaded area.

23.4.3 Influence of the offset distance on the integration step

Previous evaluations were made in the assumption that the Bessel function J0(mr), expressing the distancer between the axis of the load and the vertical co-ordinate axis, has no influence on the accuracy of theresults. Indeed, when r = 0 and thus J0(mr) = 1, this function has no influence. When r is different fromzero, the absolute values of J0(mr) are smaller than one. The values of J0(mr) do not have any influenceon the accuracy of the results, but the oscillating nature of the function, between 1 and – 1, must be takeninto account. The successive roots of J0(x) are 2.4, 5.5, 8.7, 10.8, 14.9. The values of m in J0(mr), at whichthe above roots are met, depend on the values of r. It may be observed that for high values of r the rootsare very close to each other even for small values of m. The sign of the Bessel function changes after eachroot and thus also the sign of the function to be integrated. The change of the sign has little influence onthe deflection because most of the deflection is obtained near the origin. However, the change of sign hasa great influence on the magnitude of the stresses because of the exponential function, which still hassignificant values for large values of m. It is absolutely necessary to take those sign variations intoaccount to avoid errors in the values of the stresses.We have observed that a good accuracy is obtained when one compute at least 6 values of the function(let us call it a Weddle’s interval) between two roots. Therefore, when r is not zero, we will limit the stepwidth by a function of r, which insures at least 6 computations between the roots. The distance betweenthe roots is approximately equal to 3. Then the step limit can be expressed as follows

( )r5.0

6r

3mlim ==∆

This limitation increases,depending on the value of r, slightly the computer time.

23.4.4 Modification of the step width

The values of the integrands, without consideration of the Bessel functions, decrease more and moreslowly when the values of the integration variable increases. To take advantage of this fact, theintegration step can be increased for higher values of m and so the computation process can beaccelerated.We have shown that a same accuracy is achieved by multiplying the step by a factor of two when thedeflection function has decreased by one half. However, the step value cannot be increased indefinitelybecause of the oscillating nature of the Bessel functions. Hence, and for the same reasons as in previousparagraph, the step value has to be limited as follows

( )1h

a5.0

mlim =∆

23.5 Stresses and displacements at the surface

We have shown in chapter 22.4 that the parameters B1 and D1 contain a constant in their numerator whencomputed at the surface. Hence the numerical integration cannot be safely performed in that case. Weillustrate the solution with the example of an isotropic multilayer subjected to a symmetrical vertical load.

The boundary conditions at the surface are

1)21(mD)21(mCmBmA 11112

12

1 =−+−−+ µµ

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02mD2mCmBmA 11112

12

1 =++− µµWe know that the parameters A1 and C1 can be safely computed.Hence we express B1 and D1 in function of A1 and C1. We obtain

mC)42(2mA)41(2mB 1112

1112

1 µµµµ −+−+=

112

11 C)41(mA21mD µ−+−=Let us compute, for example, the vertical deflection

[ ]dm)21(D)21(mCmBmAm

)ma(J)mr(JE

1w 1111

21

21

0

10

1

1 µµµ

−+−−−+

−= ∫∞

Replacing B1 and D1 in function of A1 and C1

[ ]dm)21(mC2mA21m

)ma(J)mr(JE

)1(2w 11

21

0

10

1

21 µ

µ−+−

−= ∫

dmm

)ma(J)mr(JE

)1(2w

0

10

1

21 ∫

∞−=

µ

[ ]dm)21(mC2mA21m

)ma(J)mr(JE

)1(211

21

0

10

1

21 µ

µ−+−

−− ∫

The first integral can analytically be integrated (§ 8.6). The integral containing the terms A1 and C1 cansafely be numerically integrated. A similar algorithm can be applied for the other functions.

23.6 Stresses and displacements in the first layer

Original solution for integrals ∫e-atJµ(bt)Jν(ct)t-λdt.We know that parameters B1 and D1 contain a constant in their numerator when computed at the surface.

In the first layer, lower thus than the surface, they contain the exponential 1mze− . In expressing B1 andD1 in function of A1 and C1 we obtain integrals of the form

dmme)ma(J)mr(Jpa0

mz10

1∫∞

−− λ

that can be integrated analytically. For r = 0 the solution can be found in § 8.3. For r > 0 the methoddeveloped in § 8.5. could be used. However, when z1 is small, in particular after scaling, equation (8.28)converges very slowly in the case of displacements and even never in the case of stresses. To solve thisproblem, we have developed a method, based in fact on the theory of elasticity. We solve the problem foran isolated load, which can easily be done with the equations of § 8.3, and integrate then the result overthe area of the distributed load. The result is a finite integral that can numerically be solved withoutdifficulty. Let us illustrate this with the equation for the vertical stress in the case of a semi-infinite bodysubjected to an isolated load

( )∫∞

−+−=0

mz0z dmemz1)mr(mJ

2Pπ

σ

We analyse more in particular the integral

∫∞

−−=0

mz01,z dme)mr(mJ

2Pπ

σ

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By equation (8.9)

( ) 2/322i,z

rz

z2P

+−=

πσ

where the index i stands for “isolated”. We now integrate the equation for σz,i over the area of thedistributed load. Therefore we replace in the equation for the isolated load the distance r by the equation

( ) 2/122 cosr2r θρρ −+ . We thus can write that

∫ ∫=π

θρρσσ2

0

a

0i,zd,z dd

where d stands for “distributed”.

( )∫ ∫∫−++

=∞

−π

θρρ

θρρπ

2

0

a

02/32220

mz10

cosr2rz

ddz2p

dme)ma(J)mr(Jpa

More generally we write the complete series of integrals

( )∫ ∫∫−++

=∞

−π

θρρ

θρρπ

2

0

a

02/12220

mz10

cosr2rz

dda2

1dme

m)ma(J)mr(J

( )∫ ∫∫−++

=∞

−π

θρρ

θρρπ

2

0

a

02/32220

mz10

cosr2rz

ddza2

1dme)ma(J)mr(J

( )( )∫ ∫∫

−++

+−−=

∞−

π

θρρ

θρρθρρπ

2

0

a

02/5222

222

0

mz10

cosr2rz

ddcosr2rz2a2

1dmme)ma(J)mr(J

These integrals can partially be solved by letting x = ρcosθ, y =ρsinθ, dxdy = ρdρdθ and integrating with

respect to y within the limits ( ) 2/122 xa −+ and ( ) 2/122 xa −− . Hence

( )∫ ∫∫−

−+

−−

∞−

+−++=

a

a

xa

xa 22220

mz1022

22zxr2ryx

dxdya2

1dme

m)ma(J)mr(J

π

( ) ( )( ) ( )∫∫

+

∞−

−−+−+

−++−+=

a

a2/1222/1222

2/1222/1222

0

mz10 dxxazrx2ra

xazrx2ralog

a21

dmem

)ma(J)mr(Jπ

( )( ) ( )∫∫

+

∞−

+−++−+

−=

a

a2/12222/3222

2/122

0

mz10

zrx2razrx2rx

dxxaaz

dme)ma(J)mr(Jπ

( ) ( )( ) ( )∫∫

+

∞−

+−++−+

+−−−=

a

a 2222/3222

2222/122

0

mz10

zrx2rxzrx2ra

dxrx2raz2xaa3

1dmme)ma(J)mr(J

π

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( ) ( )( ) ( )∫

+

− +−++−+

+−−−+

a

a22222/1222

2222/122

zrx2rxzrx2ra

dxrx2raz2xaa3

The integrals with Bessel functions of the same order can be integrated by the equations of § 8.4.

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Chapter 24 The Theory of the Back-Calculation of a Multilayered Structure

The assessment of the structural condition using NDT technologies including H/FWD, radar, targetedcoring and material testing is relatively common practise today. One of the most useful applications ofNDT testing is to back-calculate the moduli of pavement components including the subgrade. The back-calculation of a multilayered structure consists in the estimation of the mechanical characteristics of thestructure based on the deflections measured in situ under the application of a load. The structuralcondition of in-service pavements is deduced from the H/FWD load response which involves theapplication of a simulated load to model the pavement structure. Based on the strain level of the actualtraffic and transfer functions for material fatigue, the pavement’s residual life can be estimated. In thischapter we examine in detail the theory involved and analyse the accuracy of the results.

24.1 The surface modulus

The theory of the surface modulus provides a good insight in the philosophy of back-calculation and thusis a good introduction to this problem. The surface modulus can directly be assessed from a deflectionmeasured at the surface of a multilayer at a given distance of the load. It corresponds to the value of themodulus of a semi-infinite body with the same deflection at the same distance.

Figure 24.1 The surface modulus

The deflection in a semi-infinite body subjected to an isolated load is given by (20.18)

( )∫∞

−+−+

=0

mz10 dmemz22m

)ma(J)mr(Jpa

E1

w µµ

(24.1)

on which is applied the kernel (20.27)

( )∫∞

−+−+

=0

mz0 dmemz22)mr(J

2P

E1

w µπ

µ (24.2)

2a

r

z

z1

r1

A

B

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In the axis of the load, by (8.4) and (8.9)

z1

)23(2P

E1

w µπ

µ−

+= (24.3)

At the surface, by (8.4)

r1

2P

E)1(2

w2

πµ−

= (24.4)

As a result, the deflections at the surface are equal to the deflections in the axis of the load when

)23(z)1(2

µ−−

= , or when 2z

r3z2

≥≥ , depending on the value of µ.

The conclusion is that the deflections at the surface of a semi-infinite body, and thus probably of amultilayer, give information about the deflections, and thus about the Young’s moduli, in the differentlayers of the structure. The surface modulus, E0(i), is then defined as the modulus corresponding with thedeflection w(i) measured at a distance r(i) of the load by next equation

( ) ( ))i(w)i(r

pa12)i(w)i(rP12

)i(E222

πµ −

≅−

= (24.5)

The surface modulus provides for no more than general information. Hence the deflection under adistributed load can be assimilated with the deflection under an isolated load.

24.2 Equivalent layers

We assume that two layers of a multilayer can be considered as equivalent when their stiffness’ are equal:

jjii DEDE = . The concept of stiffness comes from the theory of slabs on an elastic foundation as

defined in Chapter 12. Hence for the theory of Strength of Materials this assumption is correct if thereexists a full friction interface condition. Thus we assume that the thickness hi of a layer with modulus Ei isequivalent with the thickness hj of a layer with modulus Ej as far as next equation is satisfied.

3/1

j

iij E

Ehh

= (24.6)

24.3 Equivalent semi-infinite body

Consider Figure 24.2 representing a two-layer with moduli E1 and E2 on a semi-infinite body withmodulus E0. We replace the three-layer by an equivalent semi-infinite body. In the semi-infinite body, thedeflection in B at a depth z is the same as the deflection in A if the distance of A to the axis of the load isequal to 2z/3 (we assume the less favourable value µ = 0).

We conclude that in the case of a three layer, the values of the deflections measured at a distance from theload greater than 2(h1eq + h2eq)/3, allow for a fairly good estimation of the modulus of the subgrade.However the thickness’ h1eq and h2eq are not known. Hence, only the deflections measured at a greatdistance from the load can be used for an estimation of the modulus of the subgrade.

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Figure 24.2 Equivalent semi-infinite body

24.4 Analysis of a deflection basin

The evolution of the values of the surface moduli in function of the corresponding deflections givesuseful information about the respective stiffness’ of the successive layers of the multilayer. Most of theinformation hereafter is inspired from an excellent document edited by Van Gurp (CROW, 1998).

24.4.1 Analysis of a three-layer on a linear elastic subgrade

Consider the following classical pavement structure with layers of increasing stiffness placed over thesubgrade.

Layer E (N/mm²) µ H (mm) AdhesionSurface 10000 0.2 100 1Base 3000 0.3 300 1Subgrade 500 0.5 n/a n/a

Table 24.1 Normal three-layer structure on an elastic subgrade

Distances (mm) 0 300 600 900 1200 1500 1800 2100 2400Deflections (µm) 69.6 38.1 26.2 19.1 14.3 11.2 9 7.5 6.5E0 (N/mm²) 2155 656 477 436 443 446 463 476 481

Deflections and surface moduli with p = 1 N/mm², a = 100 mm

Table 24.2 Deflections and Surface moduli of normal three-layer structure

r r

E1h1

E2h2

E0

A A

E0h1,eq

E0h2,eq

E0B

z

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Back-calculated moduliE1 = 10036 E2 = 2993 E3 = 500 N/mm²

Equivalent depth of the subgradeheq = h1(E1/E3)1/3 + h2(E2/E3)1/3 = 271 + 545 = 816 mm; req = 2/3heq =544 mm.

0300

600900

12001500

180021002400

0 500 1000 1500 2000 2500

E0

Dis

tan

ces

Figure 24.3 Surface moduei plot of a normaol three layere structure

Although the back-calculation method has not yet been explained, we estimated it useful to present theresults in relation with this analysis on the evolution of the values of the surface modulus. Thecomputations presented no difficulties. Convergence was rapidly reached and the results are satisfactory.The value of the subgrade modulus can be estimated from the deflection located at 600 mm of the load(req = 544 mm); the value remains practically constant for the following deflections.

24.4.2 Analysis of a three-layer with a very stiff base course

Consider the following pavement structure comprising of three layers:

Layer E µ H AdhesionSurface 10000 0.2 100 1Base 25000 0.3 300 1Subgrade 500 0.5 n/a n/a

Table 24.3 Three-layer structure with stiff base on an elastic subgrade

Distances 0 300 600 900 1200 1500 1800 2100 2400Deflections 38.9 25.2 21.2 17.5 14.4 11.9 9.8 8.2 7E0 3856 992 590 476 434 420 425 436 446

Table 24.4 Deflections and Surface moduli of three-layer structure with stiff base

Back-calculated moduliE1 = 10071 E2 = 24716 E3 = 501 N/mm²Equivalent depth of the subgradeheq = h1*(E1/E0)1/3 + h2*(E2/E0)1/3 = 272 + 1101 = 1373; req = 914 mm.

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0

300

600

900

1200

1500

1800

2100

2400

0 500 1000 1500 2000 2500 3000 3500

E0D

ista

nce

s

Figure 24.4 Surface module plot of a structure with a stiff base layer

The back-calculation results are satisfactory. Convergence is reached a bit slower than in the previouscase. The presence of a rigid layer cannot be detected from the graph. The value of the subgrade moduluscan be estimated from the deflection located at 900 mm of the load (req = 914 mm); the value remainspractically constant for the following deflections.

24.4.3 Analysis of a three layer with a very weak base course

Consider a pavement system with a weak interlayer:

Layer E µ H AdhesionSurface 10000 0.2 100 1Base 50 0.3 300 1Subgrade 500 0.5 n/a n/a

Table 24.5 Three-layer structure with a soft interlayer on an elastic subgrade

Distances 0 300 600 900 1200 1500 1800 2100 2400Deflections 323.8 190.8 76.2 25.7 11.3 7.4 7.2 6.9 6.3E0 463 131 164 324 607 676 579 518 496

Table 24.6 Deflections and Surface moduli of three-layer structure with soft interlayer

Back-calculated moduliE1 = 10005 E2 = 50 E3 = 501 N/mm²

Equivalent depth of the subgradeheq = h1*(E1/E0)1/3 + h2*(E2/E0)1/3 = 271 + 139 = 410; req = 274 mm.

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0

300

600

900

1200

1500

1800

2100

2400

0 100 200 300 400 500 600 700

E0D

ista

nce

s

Figure 24.5 Surface module plot of three-layer structure with soft interlayer

The results are satisfactory. Convergence was reached very fast. The presence of the weak layer can verywell been detected from the graph: low values of the surface modulus between d(2) and d(4). The value ofthe subgrade modulus can be estimated from the deflection located at 2100 mm form the load (req = 914mm). It seems that the principle of equivalent layers cannot been applied in this case, probably because ofthe absence of friction at both interfaces of the base course.

24.4.4 Analysis of a two-layer on a subgrade with increasing stiffness with depth

Consider the multilayered pavement system of Table 24.7.

Layer E µ H AdhesionSurface 10000 0.2 100 1Base 3000 0.3 300 1Soil1 50 0.5 500 1Soil2 200 0.5 500 1Soil3 500 0.5 500 1Soil4 1000 0.5 n/a n/a

Table 24.7 Multi-layer structure on a subgrade with increasing stiffness with depth

Distances 0 300 600 900 1200 1500 1800 2100 2400Deflections 69.7 43.2 30.6 20.6 12.9 7.5 4 1.9 0.9E0 2152 579 408 405 484 667 1042 1880 3472

Table 24.8 Deflections and Surface moduli of multi-layer structure on a subgrade with increasingstiffness with depth

Backcalculated moduli:E1 = 11241 E2 = 348 E3 = 3455 N/mm²

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0

300

600

900

1200

1500

1800

2100

2400

0 500 1000 1500 2000 2500 3000 3500

E0

Dis

atan

ces

Figure 24.6 Surface moduli of multi-layer structure on a subgrade with increasing stiffness with depth

The results obtained with a three layer are entirely incorrect. Convergence was never obtained. Theincrease of the subgrade modulus can be detected on the graph from deflection d(4) on. Such an evolutionof the surface modulus indicates that the classical back-calculation procedure (with 3 or 4 layers) cannotbe applied.

24.5 Algorithm of Al Bush III (1980)

Many back-calculation programs are based on an algorithm developed by Al Bush of the WaterwaysExperiment Station. A broad regression analysis based on multilayer computations allowed him todiscover a sort of correlation between the deflections at the surface of a multi-layer and the logarithms ofthe values of the moduli of each layer. For example, in the case of a three layer

)i(dElog)i(cElog)i(bElog)i(a)i(w 321obs +++= (24.7)

where wobs(i) is the deflection observed at a distance r(i) of the load and a(i), b(i), c(i) and d(i) arecoefficients to be determined by a regression analysis based on the set of observed deflections. Equation(24.7) is the basic equation utilised in the regression analysis. However, the correlation subtending (24.7)is not of very high quality. Hence, the values of the back-calculated moduli giving computed deflectionsas close as possible to the observed deflections are not obtained by one application of (24.7). They haveto be computed in an iterative manner until a pre-required fit between observed and computed deflectionsis obtained. The programs differ from each other following the manner how the iterative computations areorganised.

In Chapter 25 we’ll analyse a program that we have developed for practical use.

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Chapter 25 The Numerical Procedure of the Backcalculation of a MultilayeredStructure

25.1 The analysis of a back-calculation program for a three-layered structure

We will explain the successive steps for the determination of the moduli of a three layered structure basedon a set of np (often 9) deflections.

1. Input of the measured deflections: np, zp(i), r(i)2. Input of the data related to the load: a, p3. Input of the seed moduli: E10, E20, E30.

The values of the seed moduli are required to start the procedure. It is advised to select seed values asclose as possible to the expected values.

4. Computation of the corresponding deflections: z0(i)5. Computation of the alternative moduli.

It is highly improbable that the values chosen as seed values would by chance be the expected valuesof the moduli. Therefore one has to assume new values for the moduli hopefully closer to theexpected values than the initially chosen values. In order to meet this assumption, we suggest tocompute the alternative moduli by a formula based on the deflection at the surface of a semi-infinite

body: paE

)1(2w

2µ−= .

10

1110

10

112

2

1011 ww

Eww

papa

)1(2

)1(2EE =

−=

µ

µ

In practice we assume that the first modulus E1 is proportional to the first deflection, zp(1), that thesecond modulus E2 is proportional to the second deflection, zp(2), and that the third modulus E3 isproportional to the last deflection, zp(np),

)1(zp)1(z

EE 01011 =

)2(zp)2(z

EE 02021 =

)1np(zp)np(z

EE 03031 =

6. Computation of the deflections with the alternative moduli.Compute 4 sets of deflections, z0(i), z1(i), z2(i) and z3(i), corresponding with the 4 sets of moduli.

)i(z,E,E,E 0302010)i(z,E,E,E 1302011)i(z,E,E,E 2302110)i(z,E,E,E 3312010

7. Estimation of the moduli by the method of the least squares.- Introduce the obtained results in Bush’s algorithm

)i(dElog)i(cElog)i(bElog)i(a)i(z 3020100 +++=)i(dElog)i(cElog)i(bElog)i(a)i(z 3020111 +++=)i(dElog)i(cElog)i(bElog)i(a)i(z 3021102 +++=)i(dElog)i(cElog)i(bElog)i(a)i(z 3120103 +++=

- Compute the values of the unknown parameters a(i), b(i), c(i) and d(i)

1011

01ElogElog)i(z)i(z

)i(a−−

=

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2021

02

ElogElog)i(z)i(z

)i(b−−

=

3031

03

ElogElog)i(z)i(z

)i(c−−

=

3020100 Elog)i(cElog)i(bElog)i(a)i(z)i(d −−−=Replace a(i), b(i), c(i) and d(i) by their values in order to obtain np equations for the deflectionsin function of the unknown moduli E1, E2, E3.

)i(dElog)i(cElog)i(bElog)i(a)i(z 3210 +++=Compute the sum of the squared differences between the computed deflections z0(i) and theobserved deflections zp(i).

[ ]∑ −np

1

20 )i(z)i(zp

Minimise the sum and deduce the values of the expected moduli E1, E2, E3.

0Elog 1

=∂

∂∑ 0Elog 2

=∂

∂∑ 0Elog 3

=∂

∂∑

8. Compute the deflections zres(i) resulting from the values of the expected moduli and compute the sum

of the absolute differences between the measured and the computed deflections ∑ −np

1res )i(z)i(zp .

If the difference is less than an a specified minimum limit, then the obtained values are the best fittingmoduli. If the difference is higher than the limit, start the whole process again from step 2 on, withthe computed values E1, E2, E3 as values for the new seed moduli.

Sometimes the effect of the formula of step 4 for the computation of the alternative moduli can be verydrastic in the sense that the value obtained for one of the moduli can be smaller than 1. The processbecomes then chaotic. Indeed, to compensate for the very low value of one of the moduli, the applicationof the least squares method will lead to a very high value for an another modulus and the results becomeworse after each loop.An appropriate test must be built in the program in order to stop the computations when this happens.Often the reason is as simple as an inappropriate choice of the seed moduli and a modification of some ofthem will solve the problem. Sometimes, but seldom, whatever the values of the seed moduli, the problemkeeps going on. This means that the input data are such that back-calculation, with the available tool, willbecome impossible.

In fact, back-calculation is an artificial procedure. The multilayer theory is intended to compute stressesand displacements of a multilayered structure subjected to a load at its surface and who’s mechanical andgeometrical characteristics are known. The solutions of the computations are unique and do not giveallowance for interpretation. In contrast, the computations in a backcalculation procedure areapproximates. First because they are based on approximate field data, and secondly because the resultsare approximate and vary in function of the chosen algorithm and in function of the before hand specifiedconvergence limit. To gain insight in the value of the obtained results, we analyse the influence of a seriesof factors inherent to the back-calculation procedure in the next paragraphs. The sensitivity is analysedfor a three and a four layer structure.

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25.2 The sensitivity of the back-calculation procedure for a three layer structure

We analyse the sensitivity to rounding off the values of the measured deflections and the sensitivity of theprocedure in the case of the presence of a weak layer. Furthermore we analyse the influence of fixing oneof the moduli of the structure.

25.2.1 The sensitivity to rounding off the values of the measured deflections

The quality of the results of back-calculation evidently depends on the number of measured deflectionsand the values of this deflections. Then the question arises how many deflections are required to define anadequate deflection basin and in how far the deflection values may be rounded off when taking, forexample, the mean value of three measurements. To analyse this, we consider the 3-layered structuredefined in table 25.1.

Layer Thickness (mm) E–modulus (N/mm²) Poisson’s ratio

Surface (AC) 100 10000 0.2Base (Stabilized sand) 300 3000 0.3Subgrade ∞ 500 0.5

Table 25.1: Three layered structure

In table 25.2, we compute the deflections under a falling weight (radius: 150 mm, pressure: 1,41 N/mm²)at the usual distances on a pavement (0, 300, 600, 900, 1200, 1500, 1800, 2100, 2400 mm) in the case offull friction at the interfaces. We call them the “reference” deflections, as if they were determined on site.

Distances (mm) 0 300 600 900 1200 1500 1800 2100 2400Deflections (µm) 192.3 120.9 83.2 60.5 45.5 35.5 28.7 23.9 20.5Surface modulus E0(i) 1650 656 477 437 436 447 461 474 474

Table 25.2: Deflections in µm due to inpact of a falling weight

Table 25.3 gives the values of the backcalculated moduli based on the deflection basin of reference.

Moduli E1 E2 E3 Fit LoopsWith 9 not rounded off deflections 9815 3070 497 0.19 8With 9 rounded off deflections 9940 3008 500 0.02 7With 6 not rounded off deflections 9837 3063 497 0.20 8With 6 rounded off deflections 9942 3007 500 0.01 8

Table 25.3: Back-calculated moduli

The fit is defined as the mean value of the absolute differences between the calculated and the referencedeflections; the loops are the number of iterations required to reach the specified convergence. Weconclude that, in the case of a three layer system rounding off or limiting the number of deflections to thefirst six does not seem to have a significant influence.

25.2.2 The sensitivity to the presence of a soft intermediate layer

We consider the 3-layered structure with a weak base defined in table 25.4.

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Layer Thickness (mm) E–modulus (N/mm²) Poisson’s ratio

Surface (AC) 100 10000 0.2Sandy material 200 200 0.5Subgrade ∞ 500 0.5

Table 25.4. Three-layered structure with a weak base

In table 25.5, we give the deflections for the structure of table 25.4 with the corresponding E0(i) values.

Distances 0 300 600 900 1200 1500 1800 2100 2400Deflections 106.6 72.6 25.6 14.2 11.6 9.8 8.3 7.1 6.2E0(i) 934 344 488 587 539 510 502 503 504

Table 25.5. Deflections in µm under falling weight on a structure with weak base

The back-calculation provides next results: E1 = 10021, E2 = 199, E3 = 500. The fit is 0.02 and thenumber of required loops is 5. Thus, the presence of a weak intermediate layer seems to have no influenceon the accuracy of the results in the case of a three layer structure.

25.2.3 The influence of fixing beforehand the value of one modulus

To improve the results of the back-calculation, setting one of the moduli to a fixed value can be ofinterest. Using the deflections of table 25.2, we investigate the influence of fixing on beforehand the valueof one of the moduli of the structure on the backcalculated values of the other moduli. In Tables 25.6 to25.8, we consecutively fix the values of moduli E1, E2 and E3. The “true” values of the fixed moduli areindicated between quotes.

E1 fixed E2 E3 Fit Loops12000 2576 503 0.15 511000 2726 502 0.08 5

“10000” 2907 500 0.03 69000 3130 499 0.09 68000 3418 497 0.19 6

Table 25.6: Back-calculation with E1 fixed

E1 E2 fixed E3 Fit Loops7984 3600 495 0.20 48863 3300 497 0.11 4

10006 “3000” 500 0.03 411511 2700 504 0.13 413515 2400 508 0.30 4

Table 25.7: Back-calculation with E2 fixed

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E1 E2 E3 fixed Fit Loops

18934 1551 600 1.78 714161 2098 550 0.96 510021 2988 “500” 0.02 46185 5276 450 1.19 73662 13360 400 2.67 7

Table 25.8 Back-calculation with E3 fixed

In table 25.9 we compare the relative influence of the different moduli.

Fixed modulus Variation E1 Variation E2 Variation E3E1 [- 20%, + 20%] [+ 14%, - 14%] [- 1%, + 1%]E2 [+ 27%, - 27%] [- 20%, + 20%] [+ 1%, - 1%]E3 [- 75%, + 75%] [+200 %, -200 %] [- 20%, + 20%]

Table 25.9 Relative influence of fixed values of moduli

We conclude that making an error of ± 20 %, when fixing the values of the moduli E1or E2 does not havea significant influence on the back-calculated values of the other moduli. However making the same error,when fixing the value of the modulus E3 of the subgrade yields considerable errors on the estimation ofthe values of the moduli E1 and E2. Thus, although often practised because the value of E3 is supposed tobe deductible from the value of the most outside deflection, this method should be avoided whenbackcalculating moduli.

25.3 The sensitivity of the back-calculation procedure for a four layer structure

The study into the sensitivity of the back-calculation procedure is now extended to a four layeredstructure. Indeed, experience thought that many problems can arise with such a pavement system. We willanalyse the sensitivity to rounding off the values of the measured deflections, the value of the informationgiven by the surface modulus, the presence weak interlayer and the influence of fixing beforehand thevalue of one modulus and of a fixed bottom..

Consider the 4-layered structure defined in Table 25.10.

Layer Thickness (mm) E–modulus (N/mm²) Poisson’s ratioSurface (CRCP) 200 35000 0.2Base (lean concrete) 200 10000 0.2Sub-base (sand) 300 300 0.5Subgrade ∞ 50 0.5

Table 25.10 Four layered structure

In Table 25.11, we compute the deflections under a standard falling weight (radius: 150 mm, pressure:1,41 N/mm²) at the usual distances for a concrete pavement (0, 300, 600, 900, 1200, 1500, 1800, 2100,2400 mm) in the case of full friction at the interfaces. Again we call them the “reference” deflections, asif they were determined on site.

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Distances (mm) 0 300 600 900 1200 1500 1800 2100 2400Deflections (µm) 399.4 380.9 359.6 336.4 312.4 288.6 265.6 243.8 223.5

Table 25.11 Reference deflections in µm due to inpact of a falling weight

In table 25.12 we give the values of the back-calculated moduli based on the deflection basin of reference.

Moduli E1 E2 E3 E4 Fit LoopsWith 9 not rounded off deflections 35144 10027 293 50 0.02 4With 9 rounded off deflections 40255 8103 415 50 0.21 5With 6 not rounded off deflections 35319 9891 306 50 0.01 4With 6 rounded off deflections 41932 7236 502 49 0.24 4

Table 25.12 Backcalculated moduli in N/mm2

We conclude that in the case of a four layer system, rounding off of averaged deflections should beavoided. On the contrary, limiting the number of deflections to the first six does not seem to have asignificant influence.

25.3.1 Value of the information given by the surface modulus

- Case of a classical structure with decreasing moduli

In Table 25.13 we computed de surface moduli for the deflections of Table 25.11.

Distances 0 300 600 900 1200 1500 1800 2100 2400Deflections 399.4 380.9 359.6 336.4 312.4 288.6 265.6 243.8 223.5E0(i) 794 208 110 79 63 55 50 46 44

Table 25.13 Surface moduli for a classical road structure

The surface moduli decrease regularly with the distance to the load. In the case of full friction between allthe layers, E0(n) gives a good indicative value, but not the exact value of the modulus of the subgrade.

- Case of the presence of a fixed bottom

Often a rocky layer is present at a certain depth and the vertical deflection vanishes at the correspondinglevel. In structural language one speaks of a so-called fixed bottom. This notion can also recover aphenomenon of a subgrade with increasing stiffness. Then the depth of the so-called fixed bottom is moredifficult to establish. In Table 25.14, we give the deflections for the structure of Table 25.10 with athickness of the subgrade (depth of the fixed bottom) of 5000 mm in the hypothesis of full friction at theinterfaces.

Distances 0 300 600 900 1200 1500 1800 2100 2400Deflections 248.8 230.4 209.6 187.3 164.4 142.1 121.0 101.3 83.4E0(i) 1275 344 189 141 121 112 109 112 119

Table 25.14 Deflections in µm under falling weight, bottom fixed at 5000 mm

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Increase of the E0(i) values from deflection d(8) reveals the presence of a fixed bottom. In Table 25.15,we calculate the depth of the fixed bottom giving the best fit for the backcalculated moduli.

Subgrade thickness E1 E2 E3 E4 Fit Loops5500 33465 11447 103 57 0.06 65000 35052 9918 309 50 0.02 54000 38968 7002 740 35 0.15 43000 43478 4790 1148 19 0.28 4

Table 25.15 Calculation of the depth of the fixed bottom

Fortunately, the depth giving the best fit is indeed 5000 mm. Nevertheless, the fit can be considered asvery satisfactory for all four thickness’. But the table shows that a wrong estimation of the depth of thefixed bottom can yield important erroneous estimations of the values of the moduli.

- Case of a weak intermediate layer.

We consider the 4-layered structure with a weak base defined in Table 25.16:

Layer Thickness (mm) E–modulus (N/mm²) Poisson’s ratioSurface (CRCP) 200 35000 0.2Sandy material 200 1000 0.5Stabilised sand 300 3000 0.3Subgrade 50 0.5

Table 25.16 Four-layered structure with a weak base

In Table 25.17, we present the deflections for the structure of Table 25.16 with the corresponding E0(i)values.

Distances 0 300 600 900 1200 1500 1800 2100 2400Deflections 106.7 90.0 74.3 60.2 48.3 38.7 31.3 25.7 21.0E0(i) 2973 881 534 439 411 410 422 441 461

Table 25.17 Deflections in µm under falling weight on a structure with weak base

Increase of the E0(i) values from deflection d(7) on reveals the presence of a weak intermediate layer.Table 25.18 lists the reference deflections and compare them to the back-calculated deflections.

Distances 0 300 600 900 1200 1500 1800 2100 2400Reference deflections 106.7 90.0 74.3 60.2 48.3 38.7 31.3 25.7 21.0Computed deflections 106.7 90.1 74.1 60.2 48.4 38.8 31.3 25.5 21.1

Table 25.18 Comparion of reference and back-calculated deflections

The fit is 0.09 which can be considered as excellent. However the back-calculated values of the moduliafter only 6 loops are: E1 = 38671, E2 = 8324, E3 = 194, E4 = 556 N/mm2.These values are completely wrong! Thus the presence of a weak intermediate layer has a profoundinfluence on the accuracy of the results.

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25.3.2 The influence of fixing beforehand the value of one modulus

Using the deflections of Table 25.11, we investigate the influence of fixing on beforehand one of themoduli of a four layer system. In Table 25.19 to Table 25.22, we consecutively fix the values of moduliE1, E2, E3 and E4. The “real” value of the fixed modulus is indicated between quotes.

E1 fixed E2 E3 E4 Fit Loops

42000 7730 410 50 0.19 438500 8804 354 50 0.10 5

“35000” 10085 290 50 0.01 532500 11183 238 50 0.07 528000 13765 126 50 0.24 6

Table 25.19 Back-calculation results with E1 fixed

E1 E2 fixed E3 E4 Fit Loops31511 12000 180 50 0.13 533224 11000 236 50 0.06 535201 “10000” 294 50 0.01 537498 9000 355 50 0.08 540178 8000 438 50 0.16 5

Table 25.20 Back-calculation results with E2 fixed

E1 E2 E3 fixed E4 Fit Loops36921 9079 360 50 0.08 536104 9499 330 50 0.05 535324 9925 “300” 50 0.02 534578 10357 270 50 0.03 533863 10796 240 50 0.06 5

Table 25.21 Back-calculation results with E3 fixed

E1 E2 E3 E4 fixed Fit Loops23527 18750 1 52 2.51 1228211 15572 3 51 0.44 735263 9955 298 “50” 0.01 650988 4163 760 49 0.64 780950 610 1112 48 1.09 10

Table 25.22 Back-calculation results with E4 fixed

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In Table 25.14 we examine the relative influence of the different moduli.

Fixed modulus Variation E1 Variation E2 Variation E3 Variation E4E1 [- 20%, + 20%] [+ 30%, - 30%] [- 49%, + 49%] ConstantE2 [+ 12%, - 12%] [- 20%, + 20%] [+ 40%, - 40%] ConstantE3 [- 4%, + 4%] [+ 9 %, - 9 %] [- 20%, + 20%] ConstantE4 [+ 82%, - 82%] [- 91%, + 91%] [+186%, -186%] [- 4%, + 4%]

Table 25.23 Relative influence of fixed values of moduli

We conclude that making an error of ± 20 %, when fixing the values of the moduli E1, E2 or E3, has notan importance of greater amplitude on the backcalculated values of the other moduli. However making anerror as small as ± 4 %, when fixing the value of the modulus E4 of the subgrade yields considerableerrors on the estimation of the values of the moduli E1, E2 and E3. Thus here also fixing the value of E4should be avoided when back-calculating moduli.

As a general conclusion, we note that in all tables of results, except of course in tables 25.8 and 25.23, thevalues of the fits remain very satisfactory. Nevertheless the under- or overestimates of the values of themoduli can be of major importance in residual life calculations. One must also realise that the precision ofa falling weight’s transducer can have a profound influence on the back-calculation result. Its accuracy is± 2% of the reading with a precision of ± 2 µm. It is important that H/FWD deflection results of alldevices must be reproducible. Calibration according to the COST approach or CROW Protocol ismandatory.

25.4 The influence of degree of anisotropy and Poisson’s ratio on the results of a back-calculationprocedure in the case of a semi-infinite subgrade

In most of the cases, the Poisson’s ratios of the layers and eventually the degree of anisotropy of thesubgrade are not known. Nevertheless their values are required as input values of the process and must beestimated. Important is to appreciate their influence. We examine the influence based on a three-layeredstructure.

E1 = 10000 N/mm² E2 = 3000 N/mm2 E3 = 500 N/mm²µ1 = 0.2 µ2 = 0.3 µ3 = 0.5 for n ≥1h1 = 100 mm h2 = 300 mm µ3 = n/2 for n < 1

The load is a uniform pressure of p = 1 N/mm² on a circular area with a = 100 mm

25.4.1 Influence of the degree of anisotropy on the back-calculated moduli

In Table 25.24 we give the “reference” deflections at the usual distances utilised in the case of flexiblepavements, for a degree of anisotropy n = 1.

Distances (mm) 0 300 600 900 1200 1500Deflections (µm) 70 38 26 19 14 11

Table 25.24 Reference deflections

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In Table 25.25 we present the backcalculated moduli for different values of the degree of anisotropy andthe corresponding computed deflections

n 0 300 600 900 1200 1500 Fit E1 E2 E3 loops1 70.0 38.0 26.0 18.8 14.2 10.0 0.06 9920 2924 506 51.5 70.0 38.0 25.9 18.8 14.2 10.2 0.14 10580 2732 595 52 70.0 38.0 25.8 18.8 14.3 10.3 0.16 10732 2693 656 55 70.0 38.0 25.8 18.8 14.2 10.2 0.15 10536 2753 900 50.5 70.0 38.1 25.6 18.8 14.4 10.5 0.26 11932 2386 448 40.2 70.0 38.2 25.2 18.7 14.6 10.9 0.47 14247 1949 377 5

Table 25.25 Back-calculated moduli in function of the degree of anisotropy

If the subgrade of reference (the observed subgrade) is isotropic, assuming a degree of anisotropy higherthat 1 has no influence on the results of the backcalculation. The value of modulus E3 increases slightlywhen n increases. This is normal since n = E3vert/E3hor when the table gives E3vert. However, assuminga degree of anisotropy smaller than 1 has a great influence on the values of the three moduli.

25.4.2 Influence of Poisson’s ratio of the subgrade on the deflections (n = 1)

In Table 25.26, we compute the surface deflections in function of Poisson’s ratio of the subgrade. Thedata are those of Table 25.24.

µ 0 300 600 900 1200 15000.5 70 38 26 19 14 110.4 70 39 27 20 15 120.3 70 39 27 20 16 120.2 69 38 26 20 15 120.1 67 36 25 19 15 12

Table 25.26 Deflections function of Poisson’s ratio of the subgrade

We conclude that the value assumed for the Poisson’s ratio of the subgrade has little influence on theresults.

25.5 The influence of Poisson’s ratio and degree of anisotropy on the results of a back-calculationprocedure in the case of a subgrade of finite thickness

Consider the three-layered structure given below:

E1 = 10000 N/mm² E2 = 2000 N/mm2 E3 = 100 N/mm²µ1 = 0.2 µ2 = 0.3 µ3 = 0.5 for n ≥1h1 = 100 mm h2 = 200 mm µ3 = n/2 for n < 1

h3 = 1000 mm

The load is a uniform pressure of p = 1 N/mm² on a circular area with a radius a = 100 mm

25.5.1 Influence of the degree of anisotropy on the back-calculated moduli

In Table 25.27 we give the “reference” deflections for a degree of anisotropy n = 1.

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Distances (mm) 0 300 600 900 1200Deflections (µm) 115 66 34 13 1

Table 25.27 “Reference” deflections for a degree of anisotropy n = 1.

In Table 25.28 we give the backcalculated moduli for different values of the degree of anisotropy and thecorresponding computed deflections.

n 0 300 600 900 1200 Fit E1 E2 E3 loops1 115.0 66.0 33.9 13.1 1.0 0.04 9550 2025 101 51.1 115.0 66.4 33.1 13.2 2.3 0.53 12519 1490 137 201.2 115.0 66.6 32.7 13.2 2.9 0.80 14191 1261 161 301.3 114.9 66.7 32.4 13.2 3.3 0.97 15276 1129 178 381.3 114.9 66.7 32.4 13.2 3.3 0.97 15276 1129 178 381.3 114.9 66.7 32.4 13.2 3.3 0.97 15276 1129 178 380.9 115.0 66.0 33.3 13.2 2.0 0.46 12189 1530 130 180.8 115.0 66.6 32.7 13.3 2.9 0.80 14579 1192 154 260.7 114.9 66.8 32.2 13.4 3.7 1.15 16887 923 174 59

Table 25.28 Influence of the degree of anisotropy on the backcalculated moduli

If the subgrade of reference (the observed subgrade) is isotropic, assuming a degree of anisotropydifferent from 1 has a great influence on the values of the results, which was not the case with a semi-infinite subgrade.

25.5.2 Influence of Poisson’s ratio of the subgrade on the deflections (n = 1)

In Table 25.29, we compute the surface deflections in function of Poisson’s ratio of the subgrade.

µ 0 300 600 900 12000.5 115 66 34 13 10.4 134 85 52 29 130.3 141 93 59 35 190.2 144 95 62 38 220.1 143 95 62 39 23

Table 25.29 Influence of Poisson’s ratio of the subgrade on the deflections

We conclude here that the value assumed for the Poisson’s ratio of the subgrade has a significantinfluence on the results. As a general conclusion we note that backcalculation of moduli is more delicatefor anisotropic subgrades and for subgrades with a finite thickness.

As a general conclusion we state that back-calculation of moduli, thus estimating the residual life of apavement, is a difficult, if not, a hazardous task.

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Chapter 26 The Ovalisation Test

26.1 Description of the ovalisation test

The ovalisation test developed by the French Laboratoire Central des Ponts et Chaussées (Kobisch,Peyronne, 1979) is a very interested field test that allows to measure, in all directions and ad all levels, thestrains in a hollow pavement subjected to a vertical load. The test can be of great help in the wholeevaluation process and allows a field verification of the back-calculation results. Furthermore, it givesdirect information about the adhesion at the interfaces of the layers. For our purpose it presents thespecific interest that the interpretation of the results requires both theories that we have utilised inpavement engineering. We briefly present the test and develop the involved mathematics in greater detail.The test consists in the determination, at different levels, of the variation of the diameter of a core holeunder the passage of a load, usually a 13 t axle (figure 26.1).

Figure 26.1 Variations of the diameter in a core hole

26.2 Interpretation of the results of the ovalisation test

The mathematical problem is to deduce the strain in the axis of the load of a plain pavement from strainsmeasured at the walls of a hollow pavement. In the case of a symmetrical load, we identify the problemwith that of a hollow plate subjected to a uniform circular vertical load and resting on a Winklerfoundation (a generalization of the Westergaard’s solution). The method allows us to determine the valueof a correction factor to apply to the measured strains in order to obtain the value of the strains in the axisof the load on a plain pavement. In the case of a non-symmetrical load, we shall establish the relationbetween the tangential strains at the boundary of a hollow plate and the strains in a plain plate subjectedto orthogonal horizontal forces acting in the plane of the plate and apply to the obtained results thecorrection factor determined for the case of a symmetrical load.

∆Φ < 0 shortening

∆Φ > 0 lengthening

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26.3 Slab with a cavity on an elastic subgrade subjected to a symmetrical load

Consider the slab presented in figure 26.2.

Figure 26.2 Slab with a cavity on an elastic subgrade

The slab with a cavity with radius b is subjected to a uniformly distributed pressure p over a circular areawith radius a. The problem is solved under the assumption of a Winkler foundation. It can also be solvedin the assumption of a Pasternak foundation. Hence, the solution is based on the theory of Strength ofMaterials.

26.3.1 Resolution in the case of a plain slab

This problem has previously been solved in Chapters 12 and 14. The load is expressed by

dm)l/ma(J)l/mr(Jl

pap 1

00∫

∞= (26.1)

and the deflection by

dm1m

)l/ma(J)l/mr(Jklpa

w0

41o∫

+= (26.2)

The stresses are given by

+−=

dr

dw

r

1

dr

wd

h

Dz12

2

2

3r µσ

)drdw

r1

drwd

(³h

Dz12

2

2

+−= µσθ

If r = 0, 2

rr

θθ

σσσσ

+== , hence

++−=

dr

dw

r

1

dr

wd)1(

h

Dz6

2

2

30r µσ

The tangential strain becomes

[ ])1(E1

)1(2E

1E r

rr0 µσµ

σσµσσε θθ

θ −=

+=

−=

2a

2b

p

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dm1m

)l/ma(Jm

kl

pa

2

z

04

12

30 ∫∞

+=θε (26.3)

If r > 0

dm1m

)l/ma(J)l/mr(mJ

rkl

zpa

04

112 ∫

+=θε (26.4)

26.3.2 Resolution in the case of a slab with a cavity

One will notice that is not necessary to modify the expression of the load and considering it as a ring. Theboundary conditions itself are sufficient to express the presence of the cavity. However, one must modifythe intensity of the pressure applied on a surface equal to π(a2 – b2) and not to πa2. Hence, we mustreplace in the expressions for the stresses and the deflection, the pressure p by a pressureq = a2 / (a2 – b2).

The general solution of the equilibrium equation writes

dm1m

)l/ma(J)l/mr(Jklqa

w0

41o∫

+= (26.5)

The boundary conditions are : in r = b, M = 0 and T = 0.To express them, we need two supplementary solutions of the homogeneous equation

0Dkw

r

w

r

1

r

w

rr

1

r 2

2

2

2=+

∂+

∂+

∂ (26.6)

Those solutions are given by equation (12.19))l/rker(AwA = (26.7)

)l/r(BkeiwB = (26.8)Out of the four solutions of equation (12.19), we have chosen the ker and kei functions, which tend tozero at infinity. We call the total solution

+++

= ∫∞

)l/r(Bkei)l/rker(Adm1m

)l/ma(J)l/mr(Jklqa

w0

41o

tot (26.9)

The boundary conditions become, for r = b,

0r

w

r

1

r

wDM tot

2tot

2

r =

∂+

∂−= µ (26.10)

0r

w

r

1

r

w

rDT tot

2tot

2

r =

∂+

∂−= (26.11)

The constants A and B are determined from the system equations (26.10) and (26.11).

The equations for stresses, strains and displacement in r = b write

0r

w

r

1

r

wI

Dz tot2

2tot

r =

∂+

∂−= µσ (26.12)

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264

∂+

∂−=

r

w

r

1

r

wI

Dz tot2

2totµσθ (26.13)

with σr = 0

2tot

22

rr

w)1(

EIDz

∂−−= µε (26.14)

drdw

r1

)1(EIDz tot2µεθ −−= (26.15)

drdw

)1(EIDz

ru tot2µεθ −−== (26.16)

For r = b

drdw

r1

)1(EIDz

bu tot2µ

φφ∆

εθ −−=== (26.17)

Then we define the correction parameter by

EE

R0r0r σ

εσ

φφ∆

θ== (26.18)

The value of R only depends on the geometrical parameters of the structure a/l and b/l. It can be shownthat the value of R is approximately independent of Poisson’s ratio.Apply Hooke’s law in r = 0

)1(E

0r00r µ

σεε θ −==

Hence, we obtain the radial strain in the axis of a plain slab by

)br(R

1)0r(0 =

−== θε

µε (26.19)

It appears, as can numerically be shown, that 0b

2Rlim→

= .

In table 26.1, a series of values of R for µ = 0.25 is presented.

2b/l\2a/l 0.1 0.15 0.2 0.25 0.3 0.35 0.4

0.05 1.872 1.914 1.937 1.950 1.959 1.966 1.9710.075 1.795 1.856 1.890 1.912 1.927 1.937 1.9460.1 - 1.797 1.842 1.871 1.891 1.906 1.918

0.15 - - 1.746 1.787 1.816 1.839 1.8560.2 - - - 1.705 1.741 1.770 1.7930.3 - - - - - 1.636 1.668

Table 26.1 Values of R vs 2a/l and 2b/l.

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265

26.4 Strains in the case of a non-symmetrical load

Often the wheel-load will be a dual wheel load with a non-symmetrical load distribution. The solution ofthe hollow plate subjected to horizontal forces will allow us to take into account the asymmetry of theproblem. Therefore we first present the solution for a hollow cylinder subjected to a uniform externalpressure, followed by the solution for the hollow plate. Both solutions are based on the theory ofelasticity.

26.4.1 Stresses in a hollow cylinder subjected to a uniform external pressure

Consider the hollow cylinder subjected to an external uniform tensile stress S given in figure 26.3.

Figure 26.3. The hollow cylinder subjected to a uniform tensile stress

Because of the axial symmetry, the problem will be solved in polar co-ordinates.The continuity equation writes

0rr

1

rrr

1

r 2

2

2

2=

∂+

∂+

∂ ΦΦ (26.20)

Develop equation (26.20)

0rr

1

rr

1

rr2

r 32

2

23

3

4

4=

∂∂

+∂

∂−

∂+

∂ ΦΦΦΦ

Make the substitution r = ez

0z

4z

4z 2

2

3

3

4

4=

∂+

∂−

∂ ΦΦΦ

The characteristic equation is

0G4G4G 234 =+−with roots G1,2 = 0, G3,4 = 2.Hence

DCeBzeAz)z(f z2z2 +++=and

DCrrlogBrrlogA 22 +++=Φ (26.21)The stresses are

S

2a

2b

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266

C2)rlog21(Br

A2r +++=σ (26.22)

C2)rlog23(Br

A2

+++−=θσ (26.23)

0r =θτ (26.24)

For reasons due to the radial displacements, which will not be developed here, B = 0. The boundaryconditions are

In r = b

0C2b

A2

=+

In r = a

SC2a

A2

=+

Hence

22

2

22

22

ba

aSC2

ba

baSA

−=

−−=

The stresses are

+

−=

−=

2

2

22

2

2

2

22

2

rr

b1

ba

Sa

r

b1

ba

Saθσσ (26.25)

26.4.2 Stresses in a hollow plate

Consider the plate given in figure 26.4.

Figure 26.4. Stresses in a hollow plate

Timoshenko (1948) presents the solution for a plate solicited in one direction (figure 26.4. left). Let’sgeneralize the solution for a plate solicited in two orthogonal directions (Figure26.4 right). Note b theradius of the hole in the plate and a the radius of a concentric circle, with a great in comparison with b.Transform the stresses expressed in Cartesian co-ordinates in polar co-ordinates.

ST

SLS

2b

2a

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267

θθτθσθσσ sincos2sincos xy2

y2

xr ++=

θθτθσθσσθ sincos2cossin xy2

y2

x −+=

)sin(cossincos)( 22xyxyr θθτθθσστ θ −+−=

This results in

θσ 2cos)STSL(21

)STSL(21

r −++= (26.26)

θσθ 2cos)STSL(21

)STSL(21

−−+= (26.27)

θτ θ 2sin)STSL(21

r −−= (26.28)

The stresses can be divided into two parts. A first part, constant and equal to (SL+ST)/2 are normalstresses, which can be determined by equation (26.25)

+=

2

2

22

2

rr

b1

ba

a2

STSLσ

+

+=

2

2

22

2

r

b1

ba

a2

STSLθσ (26.29)

The second part corresponds to an association between the normal stresses (SL - ST)cos2θ/2and the shear stresses (SL - ST)sin2θ/2.

The compatibility equation

0r

1

rr

1

rr

1

rr

1

r 2

2

22

2

2

2

22

2=

∂+

∂+

∂+

∂+

θ

ϕϕϕ

θ (26.30)

can be resolved by separation of the variables. It seems appropriated to express the θ-function as atrigonometric function and well as θθ 2cos)(f = so that the shear stresses can vanish when the normalstresses are maximal. Hence, we take

θϕ 2cos)r(f=so that equation (26.30) transforms into

02cosr

f4

dr

df

r

1

dr

fd

r

4

dr

d

r

1

dr

d222

2

22

2=

−+

−+ θ (26.31)

Develop equation (26.31)

0drdf

r

9

dr

fd

r

9

dr

fdr2

dr

fd32

2

23

3

4

4=+−+ (26.32)

Make the substitution zer =

0dz

df16

dz

fd4

dz

fd4

dz

fd2

2

3

3

4

4=+−−

The characteristic equation is

0G16G4G4G 234 =+−−

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268

0)4G)(4G(G)16G4G4G(G 223 =−−=+−−The roots are

2G1 = 2G2 −= 4G3 = 0G4 =Hence

Dr

1CBrArDBeCeAe)r(f

242z4z2z2 +++=+++= −

and

θϕ 2cos)Dr

1CBrAr(

242 +++= (26.33)

The stresses are

θθ

ϕϕσ 2cos

r

D4

r

C6A2

r

1

rr

1242

2

2r

++−=

∂+

∂= (26.34)

θϕ

σθ 2cosr

C6Br12A2

r 42

2

2

++=

∂= (26.35)

θθϕ

τ θ 2sinr

D2

r

C6Br6A2

r1

r 242

r

−−+=

∂∂

∂∂

−= (26.36)

The boundary conditions regarding σr and τrθ are

In r = a

)STSL(21

aD4

aC6

A2 24 −−=++

)STSL(21

aD2

aC6

Ba6A2 242 −−=−−+

In r = b

0bD4

bC6

A2 24 =++

0bD2

bC6

Bb6A2 242 =−−+

Let a tend to infinite

4)STSL(

A−−

= 0B = 4

b)STSL(C

4−−=

2b)STSL(

D2−

= (26.37)

Adding solutions (26.29), the final expressions for the stresses become

θσ 2cos)r

b4

r

b31(

2STSL

)r

b1(

2STSL

2

2

4

4

2

2

r −+−

+−+

= (26.38)

θσθ 2cos)r

b31(

2STSL

)r

b1(

2STSL

4

4

2

2+

−−+

+= (26.39)

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269

θτ θ 2sin)r

b2

r

b31(

2STSL

2

2

4

4

r +−−

−= (26.40)

The strains are

Er

rθµσσ

ε−

=E

rµσσε θ

θ−

=E

)1(2 rr

θθ

τµγ

+=

Hence

−+

−+

+= θε 2cos

r

b4

r

b31

2STSL

r

b1

2STSL

E1

2

2

4

4

2

2

r

+

−−

+

+− θµµ 2cos

r

b31

2STSL

r

b1

2STSL

4

4

2

2

−+

−+

+−−

+= θµµε 2cos

r

b4

r

b31

2STSL

r

b)1()1(

2STSL

E1

2

2

4

4

2

2

r

+−

− )2cos)rb3

1(2

STSL4

4

θµ (26.41)

The displacement is obtained by integration of equation (26.41)

∫= rdru ε

+−−

+

+−−

+= θµµ 2cos)

r

b4

r

br(

2STSL

r

a)1(r)1(

2STSL

E1

u3

42

−−

− θµ 2cos)r

b31(

2STSL

3

4 (26.42)

Let SLST

=λ , one obtains in r = b:

( ) +−

−++−−

+= θµµ 2cos)411(

2STSL

)1()1(2

STSLEb

u

−− )2cos)11(

2STSL

θµ

[ ]θ2cos)STSL(2STSLEb

u −++=

[ ]θλλ 2cos)1(21E

bSLu −++= (26.43)

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270

Consider figure 26.5:

Figure 26.5: Principal strains

Define the strain as the displacement divided by the initial length (this definition is not that of the theoryof elasticity where the strain is defined for infinitesimal lengths).

[ ]θλλφφ∆

2cos)1(21ESL

b2u2

−++== (26.44)

For 0=θ

[ ] )3(ESL

)1(21ESL

L

L λλλφφ∆

−=−++= (26.45)

For 2πθ=

[ ] )13(ESL

)1(21ESL

T

T −=−−+= λλλφφ∆

(26.46)

We define

λφφ∆

−== 3

ESL

R L

L

L (26.47)

13

EST

R T

T

T −== λφφ∆

(26.48)

Applying Hooke’s law one obtains for a plate without a hole

)1(ESL

)STSL(E1

L λµµε −=−= (26.49)

)(ESL

)SLST(E1

T µλµε −=−= (26.50)

Eliminating SL/E and λ out of equation (26.45) into (26.50) yields

ST

θ

SL

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271

+−+= )3(3

81

T

T

L

L

T

T

L

LL φ

φ∆φφ∆

µφφ∆

φφ∆

ε (26.51)

+−+= )3(3

81

T

T

L

L

T

T

L

LT φ

φ∆φφ∆

µφφ∆

φφ∆

ε (26.52)

26.4.3 Application of the ovalisation test.

Equations (26.51) and (26.52) give the relations between the strains in a plain plate and the strains at theboundaries of a hollow plate. However, these strains are the result of the application of unknown forces orstresses. These stresses are given by the solution developed in § 26.3 for the symmetrical case. The linkbetween the two solutions is given by the parameter R, for the first solution, and the parameters RL andRT, for the second solution. Those three parameters become equal to two when φ → 0 in the first model(symmetrical model, λ = 1) and when SL = ST in the second model (symmetrical case, λ = 1). Hence,both models provide identical results under the mentioned conditions. However, in reality φ ≠ 0 and, inthe assumption that SL > ST,

2

ESL

R L

L

L <=φφ∆

and 2

EST

R T

T

T <=φφ∆

Hence equations (26.51) and (26.52) underestimate the values of SL/E and ST/E and thus εL and εT by afactor R/2. The correct strains then become

L'L R

2εε = (26.53)

T'T R

2εε = (26.54)

In the symmetrical case (λ = 1), one finds that

)br(R

1)1(4

81

R2

)0r('T

'L =

−=

−==== θθ ε

µφφ∆

µεεε

which corresponds exactly with equation (26.19) established for the first solution.

Through all this chapter, we have assumed that LL / φφ∆ and TT / φφ∆ where the principal strains, hence,that L and T where the principal directions. In practice this is not necessarily always the case. Theprincipal strains are then obtained by application of the properties of Mohr’s circle on strains measured inthree directions.

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REFERENCES

273

References

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Gauthier-Villars, Paris.Bowman (1958) : “Introduction to Bessel Functions”. Dover Publications Inc. New York.Burmister (1943): “The Theory of Stresses and Displacements in Layered Systems and Applications to the Design

of Airport Runways”. Proc. of the Highway Research Board.Burmister (1944): “The General Theory of Stresses and Displacements in Layered Systems”. Journal of Applied

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Tome XIX, Fasc 1-2.Fröhlich (1934): “Druckverteilung in Baugrund”. Springer Verlag, Wien.Kobisch, Peyronne (1979): “L’ovalisation, une nouvelle méthode de mesure des déformations élastiques des

chausses”. Bulletin de Liaison Labo P. et Ch. n° 102. ParisLekhnitskii (1963): “Theory of Elasticity of an anisotropic Body”. Holden Day Inc. San Francisco.Love: “Mathematical Theory of Elasticity”. Cambridge University Press. New York.Muki: “Asymmetric Problems of the Theory of Elasticity”. North Holland Publishing Company. Amsterdam.Pasternak (1954): “On a new method of analysis of an elastic foundation by means of two foundation constants”.

Moscow: Gps. Izd. Lit. po Strait I Arkh. (In Russian).Prandtl (1921): “Uber die Eindringungsfestigkeit plastischer baustoffe und die Festigkeit von Schneiden”.

Zeitschrift fur Angewandte Mathematik und Mechanik. Vol. 1, N° 1. Pronk (1991): Personal communication.Pronk (1993): “The Pasternak foundation. An attractive alternative for the Winkler foundation”. Proc. of the 5th

Int’l. Conference on Concrete Pavement Design and Rehabilitation. Purdue University. West Lafayette. USA.Spiegel (1971): “Advanced Mathematics for Engineers and Scientists”. Schaum’s outline series. McGraw Hill Book

Company. New York.Stet, Thewessen and Van Cauwelaert. (2001), “Standard and Recommended Practices for FWD Evaluation and

Reporting Strength of Airfield Pavement; Standard submitted to NATO for Enquiry”, First European FWDUser’s Group Meeting, Delft University of Technology, February 2001.

Stet and Van Cauwelaert. (2004). “The Elastic Length: Key to the Analysis of Multi-layered Concrete Structures”.5th International CROW Workshop On Concrete Roads. Istanbul, Turkey.

Stet, Thewessen and Van Cauwelaert. (2004). „The Pavers® System. A Knowledge Sharing Concept in the Designand Assessment of Road, Airfield and Industrial Pavement“. 9th International Symposium On Concrete Roads.Istanbul, Turkey.

Timoshenko (1948): “Théorie de l’Elasticité”. Librairie Polytechnique Ch. Béranger. Paris et Liège.Timoshenko (1951): “Théorie des Plaques et des Coques”. Librairie Polytechnique Ch. Béranger. Paris et Liège.Timoshenko (1953): “Résistance des Matériaux”. Librairie Polytechnique Ch. Béranger. Paris et Liège.Ullidtz (1998): “Modeling Flexible Pavement Response and Performance”. Polyteknisk Forlag. Narayana Press,

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Watson (1966): “A Treatise on the Theory of Bessel Functions”. Cambridge University Press. Cambridge.Wayland (1970): “Complex variables applied in Science and Engineering”. Van Nostrand Reinhold Company. New

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275

Appendix Complex Functions

1 Complex variables. Basic definitions.

Given the complex number z = x + iy (A.1)

where x and y are real numbers, we call x the real part of z [x = ℜ( z)] and y the imaginary part of z[y = ℑ( z)] (although y itself is real).

Two complex numbers are equal if and only if their real and imaginary parts are equal, that is

222111 iyxzziyx +===+if and only if

2121 yyxx ==The sum of two complex numbers is defined by the following equation

)db(i)ca()idc()iba( +++=+++ (A.2)and the product, replacing i2 by -1,

)adbc(i)bdac()idc)(iba( ++−=++ (A.3)Since a complex number is completely determined by two parameters, its real and its complex parts, itcan readily be represented by a point in a Cartesian co-ordinate system: z = x + iy is represented by thepoint (x,y) (figure A.1)

Figure A.1 Point in a Cartesian co-ordinate system

A useful extension of the geometric interpretation is the expression of complex numbers in polar form.Using the usual relation between plane Cartesian and plane polar co-ordinates, we write

ϑϑ sinircosriyxz +=+=

( ) 2/122 yxr +=)x/ytan(a=ϑ

One speaks of the length of the radius vector, r = (x2 + y2)1/2, as the modulus or absolute value of thecomplex number z = x +iy. Manipulations of complex numbers in polar form are greatly facilitated by theintroduction of an identity between exponential and trigonometric functions:

ϑϑϑ sinicosei +=which is easily demonstrated by expanding each of the functions in its series.Hence we can write

x

z = x + iyy

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276

( ) ϑϑϑ iresinicosrz =+= (A.4)We conclude that the exponential function with imaginary exponent obeys the equation

)n2(ii ee πϑϑ +=Accepting the fact that the exponential function for complex and imaginary exponents obeys the samerules as it does for real functions, we may write

( )[ ]n321n321n321 ...iexpr...rrrz...zzz ϑϑϑϑ +++=

( ) ( )[ ]21212

1

2

1 sinicosrr

zz

ϑϑϑϑ −+−=

( )ϑϑ nsinincosrz nn += (A.5)

The extraction of roots requires us to take into consideration the ambiguity in representation of a complexnumber in polar form. Let us illustrate with the cubic root. We might be tempted to write

( ) ( )[ ]3/sini3/cosrerz 3/13/i3/13/1 ϑϑϑ +==We shall take r1/3 to mean the real cube root of the non-negative number r. The expression is a perfectlyrespectable cube root of z, but the theory of equations tells us that there should be three cube roots of anyreal number, and if we choose θ = 0 to make z real, we will have only one cube root. We can get aroundthis difficulty by representing z in the form

)n2(irez πϑ +=Then

)3/n23/(i3/13/1 erz πϑ += (A.6)Exploration of the right side of the equation shows that there are three (and only three) distinct values forz1/3:

3/i3/11 erz ϑ=

)3/23/(i3/12 erz πϑ +=

)3/43/(i3/13 erz πϑ +=

Interesting particular values are (with r = 1):

1)sin(i)cos(e i −=+= πππ (A.7)

i)2/sin(i)2/cos(e 2/i =+= πππ (A.8)

More generally (A.6) can be written

++

+=

nk2

nsini

nk2

ncosrz n/1n/1 πθπθ

(A.9)

This last equation is called ‘De Moivre’s theorem’

Following equations are easily deduced from the definition (A.1)

2121 zzzz +≤+

yxz +≤

2121 zzzz −≥−

2121 zzzz =

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277

2 Derivative of a complex function.

2.1. DefinitionsIf to each of a set of complex numbers with a variable z we may assume there corresponds one or morevalues of a variable w, then w is called a function of the variable z, written w = f z).A function is single-valued if for each value of z there corresponds only one value of w.If f(z) is single-valued in some region of the z plane, the derivative of z, denoted by f’(z), is defined as

z)z(f)zz(f

lim)z('f0z ∆

∆∆

−+=

→ (A.10)

provided the limit exists independent of the manner in which ∆z→ 0.If the limit exists for z = z0, then f(z) is called analytic at z0. If the limit exists for all z in a region ℜ, thenf(z) is called analytic in ℜ. The function f(z) is continuous at z0 if ).z(f)z(flim 0

zz 0

=→

2.2. The Cauchy –Riemann conditions.Let us establish the conditions required for the derivative to exist. Consider the complex variable

iyxz += (A.11)and the complex function

ψφ i)z(fw +== (A.12)We assume that the function is analytic thus that w = f(z) and its derivative dw/dz are both single-valuedand finite.

The derivative of w is given by (A.10). However, unlike real variables, ∆z = ∆x + i∆y is itself a complexvariable which can approach zero along infinitely many paths.

Figure A 2. Cauchy –Riemann conditions

Consider figure A 2. The point N = z + ∆z can approach the point M = z along any of the paths shown. Inparticular, let us consider two cases: (1) N tends to M parallel to the x-axis (along NO), thus ∆y = 0 and∆z = ∆x; (2) N tends to M parallel to the y-axis (along NQ), thus ∆x = 0 and ∆z = i∆y. In the general case,equation (A.10) becomes

[ ]yix

)y,x()yy,xx(i)y,x()yy,xx(lim

dzdw

0y0x ∆∆

ψ∆∆ψφ∆∆φ

∆∆ +

−+++−++=

→→

For case (1)

y

x

N = z + ∆z = x + ∆x +i(y + ∆y)

M = z = x + iy

O

Q

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[ ]x

)y,x()y,xx(i)y,x()y,xx(lim

dzdw

0x ∆ψ∆ψφ∆φ

−++−+=

→ (A.13)

and for case (2)[ ]yi

)y,x()yy,x(i)y,x()yy,x(lim

dzdw

0y ∆ψ∆ψφ∆φ

−++−+=

→ (A.14)

Recognising that the difference coefficients in (A.13) and (A.14) are partial derivatives with respect to xand y, we have respectively

yi

xdzdw

∂∂

+∂∂

=ψφ

(A.15)

xi

ydzdw

∂∂

−∂∂

=φψ

(A.16)

Now, if the derivative dw/dz is to be single-valued, it is necessary that (A.15) and (A.16) be equal.Hence

yxyx ∂∂

−=∂∂

∂∂

=∂∂ φψψφ

(A.17)

Equations (A.17) are called the Cauchy – Riemann equations. They are the necessary conditions thatw = f(z) be analytic.

3 Integration of a complex function

3.1 Introduction

The integration of a complex function is a difficult operation that requires the knowledge of the followingseries of preliminary steps :

- definition of a line integral:[ ]∫ +

Cdy)y,x(Qdx)y,x(P

- definition of a simple closed curve: a simple closed curve is a closed curve, which does notintersect itself anywhere.

- integration along a closed curve:

[ ]∫ ∫∫ℜ

∂∂

−∂∂

=+C

dxdyyP

xQ

QdyPdx

- Cauchy’s theorem:

∫ ∫ ==C C

0dz)z(fdz)z(f

- Pole of a function: if f(z) = Φ(z)/(z-a)n, Φ(a) ≠ 0, where Φ(z) is analytic in a region includingz = a, and if n is a positive integer, then z = a is called a pole of order n.

- Residue of a function: the residue of a function f(z) at a pole z = a, of order n, is defined bythe formula

[ ])z(f)az(dz

d)!1n(

1lima n

1n

1n

az1 −

−=

→−

- If f(z) is analytic in a region bounded by two closed curves C1 and C2 then

∫ ∫=1C 2C

dz)z(fdz)z(f

- If C is a simple closed curve having z = a as interior point then

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∫ ==−C n

1nifi2)az(

dzπ

∫ ==−C n

,.....4,3,2nif0)az(

dz

- Residue theorem: if f(z) is analytic in a region ℜ except for a pole of order n at z = a and ifC is any simple closed curve in ℜ containing z = a, then

∫ −=C 1ia2dz)z(f π (A.18)

3.2. Definition of a line integral.If f(z) is defined, single-valued and continuous in a region ℜ , we define the integral of f(z) along somepath C in ℜ from point z1 to point z2, where z1 = x1 + iy1, z2 = x2 + iy2, as

∫ ∫ ++=C

yx

yx

idydxivudzzf2,2

1,1

))(()(

∫∫ ++−=2y,1x

1y,1x

2y,1x

1y,1x

udyvdxivdyudx (A.19)

With this definition, the integral of a function of a complex variable can be made to depend on lineintegrals for real functions in the xy plane. Let C be a curve in the xy plane, which connects points A (a1,b1) and B(a2,b2) (figure A 3).

Figure A 3 Line Intergal

Let P(x,y) and Q(x,y) be single-valued functions defined at all points of C. Subdivide C into n parts bychoosing (n – 1) points on it given by (x1, y1), (x2, y2) …(xn-1, yn-1). Call ∆xk = xk – xk-1 and ∆yk = yk –yk-1, k= 1, 2, 3, …, n where (a1,b1) ≡ (x0,y0) and (a2,b2) ≡ (xn,,yn) and suppose that points (ξk,ηk) are chosen sothat they are situated on C between points (xk-1,yk-1) and (xk, yk). Form the sum

[ ]∑=

+n

1kkkkkkk y),(Qx),(P ∆ηξ∆ηξ

The limit of this sum as n → ∞ in such a way that all the quantities ∆xk and ∆yk approach zero, if suchlimit exists, is called a line integral along C and is denoted by

[ ]∫ +C

dy)y,x(Qdx)y,x(P (A.20)

or

A

B

a1 a2

b1

b2

y

x

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280

[ ]∫ +2b,2a

1b,1ady)y,x(Qdx)y,x(P (A.21)

3.3. Definition of a simple closed curve.We define a simple closed curve as a closed curve, which does not intersect itself anywhere.Mathematically, a curve in the xy plane is defined by the parametric equations x = φ(t) and y = ψ(t),where φ and ψ are single-valued and continuous in an interval t1 < t < t2. If φ(t1) = φ(t2) and ψ(t1) = ψ(t2)the curve is said to be closed. If φ(u) = φ(v) and ψ(u) =ψ(v) only when u = v (except in the special casewhere u = t1 and v = t2) the curve is closed and does not intersect itself and so is a simple closed curve. Ifwe look down upon a simple closed curve in the xy plane, the traversal of the curve in the counterclockwise direction is taken as positive while the traversal in the clockwise is taken as negative. Weassume that φ and ψ are piecewise differentiable in t1 < t < t2.

3.4. Integration along a simple closed curve.Let P, Q, ∂P/∂y, ∂Q/∂x be single-valued and continuous in a simply connected region ℜ bounded by asimple closed curve C. Consider figure A 4.

Figure A 4 Integration along a simple closed curve

Let the equations of the curves AEB and AFB be y =Y1(x) and y = Y2(x). If ℜ is the region bounded by C,we have

∫∫ ∫ ∫= =

∂∂

=∂∂

R

b

ax

)x(2Y

)x(1Yy

dxdyyP

dxdyyP

∫=

==b

ax

)x(2Y)x(1Yy dx)y,x(P

[ ]∫ −=b

a12 dxY,x(P)Y,x(P

∫ ∫−−=b

a

a

b21 dx)Y,x(Pdx)Y,x(P

y

xa b

e

f

A B

E

F

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∫−=C

Pdx

Similarly let the equations of curves EAF and EBF be x = X1(y) and x = X2(y). One gets

∫∫∫ =∂∂

CRQdydxdy

xQ

Adding both result yields

[ ]∫ ∫∫

∂−

∂∂

=+C R

dxdydyP

xQ

QdyPdx (A.22)

3.5. Cauchy’s theorem.Let C be a simple closed curve and f z) analytic within the region ℜ bounded by C as well as on C.Apply the definition (A.19) of an integral

∫ ∫ ++=C C

)idydx)(ivu(dz)z(f

. ∫ ∫∫ ++−=C CC

)udyvdx(i)vdyudx(dz)z(f

By (A.22)

[ ] ∫∫∫

∂∂

−∂∂

−=−RC y

uxv

vdyudx

[ ] ∫∫∫

∂∂

−∂∂

=+RC y

vxu

udyvdx

Since f(z) is analytic, ∂u/∂x = ∂v/∂y and ∂v/∂x =- ∂u/∂y.Hence the above integrals are zero and

∫ =C

0dz)z(f (A. 23)

and Cauchy’s theorem is proved.

3.6. Pole of a function.We define a singular point of a function f(z) as a value of z at which f(z) fails to be analytic. If f(z) isanalytic everywhere in some region except at an interior point z = a, we call z = a an isolated singularityof f(z). If f(z) = φ(z)/(z-a)n, φ(a) ≠ 0, where φ(z) is analytic in a region including z = a and if n is aninteger, then f(z) has an isolated singularity at z = a which is called a pole of order n.If n = 1, the pole is called a simple pole. A function can have other types of singularities besides poles.For example f(z) = √(z) has a branch point at z = 0: f(z) is not a single-valued function of z (cfr relation(A.6) giving the 3 roots of z1/3). The function f(z) =sin(z)/ z has a singularity at z = 0. However, due to thefact that z/)zsin(lim

0z → is finite, such a singularity is called a removable singularity.

3.7. Residue of a function.

Consider the function f1(z) which we expand in a Taylor series around z = a

⋅⋅⋅+−+−+= 22101 )az(b)az(bb)z(f

where

az11i

1i

1i )z(fdz

d)!1i(

1b =−

− −=

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Form the series

)az(a

)az(

a

)az(

a

)az(

)z(f)z(f 1

1n1n

nn

n1

−+⋅⋅⋅+

−+

−=

−= −

−+−−

⋅⋅⋅+−+−++ 2210 )az(a)az(aa (A.24)

which is called the Laurent’s series of f(z).Multiplying (A.24) by (z – a)n and taking the (n – 1)st derivative of both sides and letting z → a we find

[ ])z(f)az(dz

d)!1n(

1lima n

1n

1n

az1 −

−=

→− (A.25)

the residue of the function f(z) at the pole z = a ,where n is the order of the pole.For simple poles the calculation of the residue reduces to

)z(f)az(limaaz

1 −=→

− (A.26)

3.8. Region bound by two closed curves C1 and C2.

Consider figure A 5

Figure A 5 Region bound by two closed curves C1 and C2

The line AB (called a cross-cut) connects any point on C2 and a point on C1 .Hence by (A.23)

0dz)z(fAQPABRSBA

=∫Thus

∫ ∫ ∫ ∫ =+++AQPA AB BRSB BA

0dz)z(fdz)z(fdz)z(fdz)z(f

But

∫ ∫−=AB BA

dz)z(fdz)z(f

Hence

A

B

P

Q

R

S

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∫ ∫ ∫=−=AQPA BRSB BSRB

dz)z(fdz)z(fdz)z(f

and

∫ ∫=1C 2C

dz)z(fdz)z(f (A.27)

3.9. Resolution of the integral ∫ −C n)az(dz

Consider figure A 6.

Figure A 6 Resolution of the integral

Let C be a simple closed curve bounding a region having z = a as an interior point. Let C1 be a circle withradius ε having a centre at z = a. Since (z – a)n is analytic within and on the boundaries of the regionbounded by C and C1. we have by (A.27)

∫ ∫−

=−C 1C nn )az(

dz

)az(

dz

To evaluate this last integral, note that on C1, (z – a) = ε eiθ and dz = iε eiθ. The integral equals

∫∫ −−

θπ

θ

θθ

εθ

ε

ε 2

0

i)n1(1n

2

0inn

ide

id

e

ei

1nif0i)n1(

ei

)az(

dz2

0

i)n1(

1nC n≠=

−=

−∫πθ

ε (A.28)

If n = 1, the integral equals

i2di)az(

dzC

2

0n

πθπ

==−

∫ ∫ (A.29)

Notice that for n = 0, -1, -2, -3, …..the integrand is 1, (z – a), (z – a)2 …and is analytic everywhere insideC1, including z = a. Hence by (A.23) the integral is zero.

C

C1

ε

z = a

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3.10. The residue theorem

Let f(z) be analytic within and on the boundary of the region bounded by C except at a pole a within theregion, having a residue a-1.Expand f(z) in its Laurent’s series

)az(a

)az(

a

)az(

a)z(f 1

1n1n

nn

−+⋅⋅⋅+

−+

−= −

−+−−

⋅⋅⋅+−+−++ 2210 )az(a)az(aa

By integration we have

∫ ∫ ∫∫ −++

−+

−= −

−+−−

C C C1

1n1n

nn

Cdz

)az(a

.....dz)az(

adz

)az(

adz)z(f

dz.....)az(a)az(aaC

2210∫

+−+−++

Hence by (A.28) and (A.29)

∫ −=C 1ia2dz)z(f π (A.30)

which is the residue theorem.Because only the term involving a-1 remains, a-1 is called the residue of f(z) at the pole z = a.Further if f(z) is analytic within and on the boundary C of a region ℜ except at a finite number of polesa, b, c, …..within ℜ , having residues a-1, b-1, c-1 …respectively, then

....)cba(i2dz)z(f 111C+++= −−−∫ π (A.31)

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The author, Frans Van Cauwelaert, graduated in Civil Engineering at the University of Louvain(Belgium) and obtained a PhD in Technical Sciences at the Federal Polytechnic School of Lausanne(Switzerland).Frans can be considered as a nomad and a pioneer in the field of interest of pavement engineering, andis well known for his unique and rigorous mathematical solutions, and the ability to put advancedtheory into everyday practice. He is also a kind of geographical nomad, having worked in many partsall over the world: in Western Europe, Central Africa, Middle East and the United States of America.He must be considered a technical and scientific nomad too, having worked in the field as a youngengineer, in the laboratory, in design and engineering offices and in class rooms as well. He ended hisimpressive career as Head of the Department Promotion, Research and Development of the Federationof the Belgium Cement Industry (Febelcem).In the early seventies Frans became interested in the field of rational design of pavements anddeveloped several design programs: for the US Corps of Engineers, for universities, research centres,contractors and consultants. He is also one of the three musketeers, who started the Pavers Team.Retired, he finally found the time to reassembled 30 years of study, research and experience in hisbook.