pavement design and evaluation
DESCRIPTION
just about pavementTRANSCRIPT
i
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.
PAVEMENT DESIGN AND EVALUATION
ii
INTRODUCTION
iii
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.
PAVEMENT DESIGN AND EVALUATION
iv
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
INTRODUCTION
v
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.
PAVEMENT DESIGN AND EVALUATION
vi
.
CONTENTS
vii
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
PAVEMENT DESIGN AND EVALUATION
viii
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
CONTENTS
ix
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
PAVEMENT DESIGN AND EVALUATION
x
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
CONTENTS
xi
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
PAVEMENT DESIGN AND EVALUATION
xii
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
CONTENTS
xiii
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
PAVEMENT DESIGN AND EVALUATION
xiv
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
THE LAPLACE EQUATION
1
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.
PAVEMENT DESIGN AND EVALUATION
2
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
2θ
∂∂θ
−=
+
−=
rcos
x1
x
y1
1y
2
2θ
∂∂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)
THE LAPLACE EQUATION
3
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.
PAVEMENT DESIGN AND EVALUATION
4
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.
THE LAPLACE EQUATION
5
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.
PAVEMENT DESIGN AND EVALUATION
6
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)
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).
PAVEMENT DESIGN AND EVALUATION
8
THE GAMMA FUNCTION
9
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)
PAVEMENT DESIGN AND EVALUATION
10
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
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
THE GAMMA FUNCTION
11
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
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
THE GAMMA FUNCTION
13
∫ ∫ ∫ −−−− −+
−= 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
PAVEMENT DESIGN AND EVALUATION
14
∫∞
− +−==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)
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.
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
THE GENERAL SOLUTION OF THE BESSEL EQUATION
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:
PAVEMENT DESIGN AND EVALUATION
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
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∂
∂
PAVEMENT DESIGN AND EVALUATION
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
THE GENERAL SOLUTION OF THE BESSEL EQUATION
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:
PAVEMENT DESIGN AND EVALUATION
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
THE GENERAL SOLUTION OF THE BESSEL EQUATION
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
PAVEMENT DESIGN AND EVALUATION
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
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
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.
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
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−=
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)
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 −−
− =−
=π
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
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)
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
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
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.
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
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
0π
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
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.
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 −−+=
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)
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)
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+
=Γ
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)
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+⋅
=Γ
ΓΓ (5.4)
5.5 The duplication formula for gamma functions
Consider the function
( )∫=2/
0
p2 d2sinJπ
ϑϑ
Write u = 2ϑ
DEFINITE INTEGRALS OF BESSEL FUNCTIONS
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.
PAVEMENT DESIGN AND EVALUATION
44
DEFINITE INTEGRALS OF BESSEL FUNCTIONS
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
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
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
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.
THE HYPERGEOMETRIC TYPE OF SERIES
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 ==
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)
THE HYPERGEOMETRIC TYPE OF SERIES
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)
PAVEMENT DESIGN AND EVALUATION
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)
THE HYPERGEOMETRIC TYPE OF SERIES
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)
PAVEMENT DESIGN AND EVALUATION
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∑∞
+++
=Γ
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∑∞
+++
=Γ
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
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(
2Γ
[ ] 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(
−+−=+−=−−=−
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)
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
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
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
++++++
=π
∫∞
+++−−
+=
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
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.
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)
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
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)
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(
eπ
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)
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
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
Iπ
[ ]
⋅⋅⋅+
+++
2/52222
55
)ca)(ba(
cb5.4.2
3.1
Hence
++=
2/1222/122 )ca()ba(
bcarcsin
2I
π (8.25)
or
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)
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
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(
1µ
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
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
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)
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
4π
0
dt)btsin()at(J0
0∫∞
0 ∞ 22 ab
1
−
∫∞
00 dt)btcos()at(J
22 ba
1
−∞ 0
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
Fλ
Γλν
Γ
ΓνΓν
λλν
+
−+
−+=
−+−
∞
∫2
1222
)2/1()12(
2
1dtt
)t(J12
0λ
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
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
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.
PAVEMENT DESIGN AND EVALUATION
76
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
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
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
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
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
-iπ
0
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
2π
νπ
i
wsinhzw dweei
Int
∫ −−=0
21
2π
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
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)
PAVEMENT DESIGN AND EVALUATION
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)
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
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(
Cν
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
BESSEL FUNCTIONS IN THE COMPLEX PLANE
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
PAVEMENT DESIGN AND EVALUATION
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
BESSEL FUNCTIONS IN THE COMPLEX PLANE
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.
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
BESSEL FUNCTIONS IN THE COMPLEX PLANE
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)
PAVEMENT DESIGN AND EVALUATION
92
LAPLACE EQUATION IN PAVEMENT ENGINEERING
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
PAVEMENT DESIGN AND EVALUATION
94
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
LAPLACE EQUATION IN PAVEMENT ENGINEERING
95
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
xσ
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
PAVEMENT DESIGN AND EVALUATION
96
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.
LAPLACE EQUATION IN PAVEMENT ENGINEERING
97
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
PAVEMENT DESIGN AND EVALUATION
98
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
LAPLACE EQUATION IN PAVEMENT ENGINEERING
99
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
PAVEMENT DESIGN AND EVALUATION
100
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
LAPLACE EQUATION IN PAVEMENT ENGINEERING
101
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.
PAVEMENT DESIGN AND EVALUATION
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
LAPLACE EQUATION IN PAVEMENT ENGINEERING
103
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
PAVEMENT DESIGN AND EVALUATION
104
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−
=
LAPLACE EQUATION IN PAVEMENT ENGINEERING
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
PAVEMENT DESIGN AND EVALUATION
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)
LAPLACE EQUATION IN PAVEMENT ENGINEERING
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
+−=
PAVEMENT DESIGN AND EVALUATION
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
wΦ
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∂
LAPLACE EQUATION IN PAVEMENT ENGINEERING
109
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)
nµ
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
PAVEMENT DESIGN AND EVALUATION
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)
SIMPLE APPLICATIONS OF BEAMS AND SLABS ON AN ELASTIC SUBGRADE
111
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.
PAVEMENT DESIGN AND EVALUATION
112
11.2 Helpful relations
∑∞
=
++=
1nnn
0L
xnsinb
Lxn
cosa2
a)x(f
ππ
∫−
=L
Ln dx
Lxn
cos)x(fL1
aπ
∫−
=L
Ln dx
Lxn
sin)x(fL1
bπ
[ ] 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
aπ
(11.2)
∫−
=L
Ln dx
Lxn
sin)x(fL1
bπ
(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)
SIMPLE APPLICATIONS OF BEAMS AND SLABS ON AN ELASTIC SUBGRADE
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
aπ
if m = 1,2,3,… (11.6)
PAVEMENT DESIGN AND EVALUATION
114
Similarly multiplying by L
xmsin
π and integrating from – L to L proves
∫−
=L
Lm dx
Lxm
sin)x(fL1
bπ
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
SIMPLE APPLICATIONS OF BEAMS AND SLABS ON AN ELASTIC SUBGRADE
115
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
PAVEMENT DESIGN AND EVALUATION
116
∫−
=L
Ln du
Lun
cos)u(fL1
aπ
∫−
=L
Ln du
Lun
sin)u(fL1
bπ
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
SIMPLE APPLICATIONS OF BEAMS AND SLABS ON AN ELASTIC SUBGRADE
117
∫∞
=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
PAVEMENT DESIGN AND EVALUATION
118
∫ ∫∞
+=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.
SIMPLE APPLICATIONS OF BEAMS AND SLABS ON AN ELASTIC SUBGRADE
119
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.
PAVEMENT DESIGN AND EVALUATION
120
SIMPLE APPLICATIONS OF BEAMS AND SLABS ON AN ELASTIC SUBGRADE
121
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
PAVEMENT DESIGN AND EVALUATION
122
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
SIMPLE APPLICATIONS OF BEAMS AND SLABS ON AN ELASTIC SUBGRADE
123
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
PAVEMENT DESIGN AND EVALUATION
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)
SIMPLE APPLICATIONS OF BEAMS AND SLABS ON AN ELASTIC SUBGRADE
125
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=+
PAVEMENT DESIGN AND EVALUATION
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
−=
+
SIMPLE APPLICATIONS OF BEAMS AND SLABS ON AN ELASTIC SUBGRADE
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.
PAVEMENT DESIGN AND EVALUATION
128
THE BEAM ON A PASTERNAK FOUNDATION
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
PAVEMENT DESIGN AND EVALUATION
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
THE BEAM ON A PASTERNAK FOUNDATION
131
( )∫∞
++=
024
dt1gt2tt
)l/atsin()l/xtcos(kp2
wπ
(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)
PAVEMENT DESIGN AND EVALUATION
132
- For g > 1
( ) ( )
+−
+−= ∫∫
∞∞
022
0222
dtatt
)l/atsin()l/xtcos(dt
tt)l/atsin()l/xtcos(
1g
1k4p
wβ
(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)
THE BEAM ON A PASTERNAK FOUNDATION
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
PAVEMENT DESIGN AND EVALUATION
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
Aβ
α
=
To standardise to the general solution, we write the solution as follows
lx
cosek2p
w lx
Aβ
α
= (13.27)
Indeed, multiplying the solution of a differential equation by a constant does not affect the result.
lx
sinek2p
w lx
Bβ
α
= (13.28)
- For g= 1
h, E
G
k qy
x
c
2a
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
wα
= (13.31)
lx
B ek2p
wβ
= (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 %.
PAVEMENT DESIGN AND EVALUATION
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
THE BEAM ON A PASTERNAK FOUNDATION
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.
PAVEMENT DESIGN AND EVALUATION
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
THE BEAM ON A PASTERNAK FOUNDATION
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
Cβ
α−
= (13.38)
lx
sinek2p
w lx
Dβ
α−
= (13.39)
- For g = 1
lx
C ek2p
w−
= (13.40)
lx
D xek2p
w−
= (13.41)
- For g > 1
lx
C ek2p
wα
−= (13.42)
lx
D ek2p
wβ
−= (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.
PAVEMENT DESIGN AND EVALUATION
140
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%.
THE BEAM ON A PASTERNAK FOUNDATION
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)
PAVEMENT DESIGN AND EVALUATION
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%.
THE BEAM ON A PASTERNAK FOUNDATION
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).
PAVEMENT DESIGN AND EVALUATION
144
THE CIRCULAR SLAB ON A PASTERNAK FOUNDATION
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
PAVEMENT DESIGN AND EVALUATION
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.
THE CIRCULAR SLAB ON A PASTERNAK FOUNDATION
147
∫∞
=0
10 dt)l/ta(J)l/tr(J)t(Ap2
wπ
(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
PAVEMENT DESIGN AND EVALUATION
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
THE CIRCULAR SLAB ON A PASTERNAK FOUNDATION
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
PAVEMENT DESIGN AND EVALUATION
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)
THE CIRCULAR SLAB ON A PASTERNAK FOUNDATION
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
PAVEMENT DESIGN AND EVALUATION
152
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.
THE CIRCULAR SLAB ON A PASTERNAK FOUNDATION
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
PAVEMENT DESIGN AND EVALUATION
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 %
THE RECTANGULAR SLAB ON A PASTERNAK FOUNDATION
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
pπ
(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
PAVEMENT DESIGN AND EVALUATION
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
THE RECTANGULAR SLAB ON A PASTERNAK FOUNDATION
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
PAVEMENT DESIGN AND EVALUATION
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
THE RECTANGULAR SLAB ON A PASTERNAK FOUNDATION
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
∫∞ +−
=π
(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
PAVEMENT DESIGN AND EVALUATION
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(
xµ
[ ]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
THE RECTANGULAR SLAB ON A PASTERNAK FOUNDATION
161
[ ] =++
∂∂
−∂∂
∂−+
∂
∂BA22
3
3
3BwAww
xl
g2
yx)2(
xµ
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
PAVEMENT DESIGN AND EVALUATION
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 %
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
PAVEMENT DESIGN AND EVALUATION
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
THE MULTISLAB
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)
PAVEMENT DESIGN AND EVALUATION
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
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
PAVEMENT DESIGN AND EVALUATION
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
THE MULTISLAB
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
PAVEMENT DESIGN AND EVALUATION
170
BACK-CALCULATION OF CONCRETE SLABS
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
=−
=µ
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.
PAVEMENT DESIGN AND EVALUATION
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
BACK-CALCULATION OF CONCRETE SLABS
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.
PAVEMENT DESIGN AND EVALUATION
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
BACK-CALCULATION OF CONCRETE SLABS
175
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
PAVEMENT DESIGN AND EVALUATION
176
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
BACK-CALCULATION OF CONCRETE SLABS
177
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
PAVEMENT DESIGN AND EVALUATION
178
THERMAL STRESSES IN CONCRETE SLABS
179
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.
PAVEMENT DESIGN AND EVALUATION
180
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 −++= −
THERMAL STRESSES IN CONCRETE SLABS
181
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)
PAVEMENT DESIGN AND EVALUATION
182
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)
THERMAL STRESSES IN CONCRETE SLABS
183
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).
PAVEMENT DESIGN AND EVALUATION
184
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.
THERMAL STRESSES IN CONCRETE SLABS
185
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.
PAVEMENT DESIGN AND EVALUATION
186
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.
THERMAL STRESSES IN CONCRETE SLABS
187
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)
PAVEMENT DESIGN AND EVALUATION
188
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)
THERMAL STRESSES IN CONCRETE SLABS
189
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.
PAVEMENT DESIGN AND EVALUATION
190
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
THERMAL STRESSES IN CONCRETE SLABS
191
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.
PAVEMENT DESIGN AND EVALUATION
192
DETERMINATION OF THE PARAMETERS OF A RIGID STRUCTURE
193
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
=+
PAVEMENT DESIGN AND EVALUATION
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).
DETERMINATION OF THE PARAMETERS OF A RIGID STRUCTURE
195
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
PAVEMENT DESIGN AND EVALUATION
196
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
DETERMINATION OF THE PARAMETERS OF A RIGID STRUCTURE
197
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
dθ
r r
+
N
2a
N
y
x
e
M
θ
PAVEMENT DESIGN AND EVALUATION
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
DETERMINATION OF THE PARAMETERS OF A RIGID STRUCTURE
199
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
PAVEMENT DESIGN AND EVALUATION
200
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:
∫ ∫∞
=π
θθσ0 a
zL rdrdsinr2M
∫ ∫∞
=π
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
DETERMINATION OF THE PARAMETERS OF A RIGID STRUCTURE
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).
PAVEMENT DESIGN AND EVALUATION
202
THE SEMI-INFINITE BODY SUBJECTED TO A VERTICAL LOAD
203
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
PAVEMENT DESIGN AND EVALUATION
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
uµ
THE SEMI-INFINITE BODY SUBJECTED TO A VERTICAL LOAD
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)
PAVEMENT DESIGN AND EVALUATION
206
∫∞
−−=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
THE SEMI-INFINITE BODY SUBJECTED TO A VERTICAL LOAD
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
PAVEMENT DESIGN AND EVALUATION
208
−−−
+
−
−=
+− 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
Kπ
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
Kπ
= (20.35)
Finally, for z = 0
∫∞
−=0
0z dm)masin()mr(Ja2
Pπ
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)
THE SEMI-INFINITE BODY SUBJECTED TO A VERTICAL LOAD
209
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
=
PAVEMENT DESIGN AND EVALUATION
210
[ ] 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)
THE SEMI-INFINITE BODY SUBJECTED TO A VERTICAL LOAD
211
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)
PAVEMENT DESIGN AND EVALUATION
212
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
THE SEMI-INFINITE BODY SUBJECTED TO A VERTICAL LOAD
213
−+−= −− 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
PAVEMENT DESIGN AND EVALUATION
214
[ ]dmese)mr(mJ)s1(2
P
0
mszmz0z ∫
∞−− +−
−−=
πσ (20.68)
By (8.9)
++
+−
−−=
2222/322zrsz(
sz
)rz(
sz)s1(2
Pπ
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
THE SEMI-INFINITE BODY SUBJECTED TO A VERTICAL LOAD
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).
PAVEMENT DESIGN AND EVALUATION
216
THE SEMI-INFINITE BODY SUBJECTED TO A SHEAR LOAD
217
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
PAVEMENT DESIGN AND EVALUATION
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
Kπ
=
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.
THE SEMI-INFINITE BODY SUBJECTED TO A SHEAR LOAD
219
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)
PAVEMENT DESIGN AND EVALUATION
220
+−−−= −−− 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)
THE SEMI-INFINITE BODY SUBJECTED TO A SHEAR LOAD
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 ∫ ∫
∞ ∞
=λ
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∫∞
=λ
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
PAVEMENT DESIGN AND EVALUATION
222
- 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
THE SEMI-INFINITE BODY SUBJECTED TO A SHEAR LOAD
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)
PAVEMENT DESIGN AND EVALUATION
224
THE MULTILAYERED STRUCTURE
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
PAVEMENT DESIGN AND EVALUATION
226
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
THE MULTILAYERED STRUCTURE
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.
PAVEMENT DESIGN AND EVALUATION
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
THE MULTILAYERED STRUCTURE
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−+⋅=
PAVEMENT DESIGN AND EVALUATION
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
−−
−
++
+=
THE MULTILAYERED STRUCTURE
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.
PAVEMENT DESIGN AND EVALUATION
232
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
sµ
µ−−
= (§20.7)
THE RESOLUTION OF A MULTILAYERED STRUCTURE
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.
PAVEMENT DESIGN AND EVALUATION
234
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
wµ
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.
THE RESOLUTION OF A MULTILAYERED STRUCTURE
235
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
PAVEMENT DESIGN AND EVALUATION
236
[ ]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
THE RESOLUTION OF A MULTILAYERED STRUCTURE
237
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
PAVEMENT DESIGN AND EVALUATION
238
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
THE RESOLUTION OF A MULTILAYERED STRUCTURE
239
( ) ( )( ) ( )∫
+
− +−++−+
+−−−+
a
a22222/1222
2222/122
zrx2rxzrx2ra
dxrx2raz2xaa3
2π
The integrals with Bessel functions of the same order can be integrated by the equations of § 8.4.
PAVEMENT DESIGN AND EVALUATION
240
THE THEORY OF BACKCALCULATION OF A MULTILAYER
241
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
PAVEMENT DESIGN AND EVALUATION
242
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
rµ
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
0µ
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.
THE THEORY OF BACKCALCULATION OF A MULTILAYER
243
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
PAVEMENT DESIGN AND EVALUATION
244
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.
THE THEORY OF BACKCALCULATION OF A MULTILAYER
245
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.
PAVEMENT DESIGN AND EVALUATION
246
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
THE THEORY OF BACKCALCULATION OF A MULTILAYER
247
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.
PAVEMENT DESIGN AND EVALUATION
248
NUMERICAL BACKCALCULATION OF A MULTILAYER
249
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−−
=
PAVEMENT DESIGN AND EVALUATION
250
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.
NUMERICAL BACKCALCULATION OF A MULTILAYER
251
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.
PAVEMENT DESIGN AND EVALUATION
252
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
NUMERICAL BACKCALCULATION OF A MULTILAYER
253
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.
PAVEMENT DESIGN AND EVALUATION
254
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
NUMERICAL BACKCALCULATION OF A MULTILAYER
255
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.
PAVEMENT DESIGN AND EVALUATION
256
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
NUMERICAL BACKCALCULATION OF A MULTILAYER
257
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
PAVEMENT DESIGN AND EVALUATION
258
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.
NUMERICAL BACKCALCULATION OF A MULTILAYER
259
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.
PAVEMENT DESIGN AND EVALUATION
260
THE OVALISATION TEST
261
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
PAVEMENT DESIGN AND EVALUATION
262
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
THE OVALISATION TEST
263
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)
PAVEMENT DESIGN AND EVALUATION
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.
THE OVALISATION TEST
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
PAVEMENT DESIGN AND EVALUATION
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
THE OVALISATION TEST
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 =+−−
PAVEMENT DESIGN AND EVALUATION
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)
THE OVALISATION TEST
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)
PAVEMENT DESIGN AND EVALUATION
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
THE OVALISATION TEST
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.
PAVEMENT DESIGN AND EVALUATION
272
REFERENCES
273
References
Airy (1862): “The Airy stress function”. Brit. Assoc. Advan. Sci. Rept.Barden (1963): “Stresses and Displacements in a Cross-Anisotropic Soil”. Geotechnique n° 13. London.Boussinesq (1885): “Application des Potentiels à l’Etude de l’Equilibre et du Mouvement des Solides élastiques”.
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
Physics, Vol 16.Bush (1980): “Nondestructive Testing for Light Aircraft Pavements, Phase II”. FAA Report FAA-RD-80-9-II,
Department of Transportation, Federal Aviation Administration, Washington DC.Bradbury (1938): “Reinforced Concrete Pavements”. Wire Reinforcement Institute. Washington DC.CROW. (1998a): “Falling Weight Deflectometer Calibration Guide”, Record 18, Ede, the Netherlands.CROW. (1998b): “Deflection profile; not a Pitfall anymore”, Record 17, Ede, the Netherlands.De Jong, Peutz, Korswagen (1973): “Computer Program BISAR”. Royal Shell Laboratory, Amsterdam.Eckmann (1998): “The NOAH Software”. BCRA ’98, Trondheim, Norway.Eftimie (1973): “Starea de tensiure in terenurile anizotrofe de fundatie”. Buletinul Institutului Politechnic di Tasi.
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,
Denmark.Van Cauwelaert (1977): “Coefficients of Deformation of an Anisotropic Body”. ASCE Journal of the Engineering
Mechanics Division. New York.Van Cauwelaert, Cerisier (1982): “Bearing Capacity considered as a geotechnical concept”. BCRA’92. Trondheim.
Norway.Van Cauwelaert (1983): “L’élasticité anisotrope appliquée à la mécanique des milieux granulaires”. Thèse de
Doctorat. Ecole Polytechnique Fédérale de Lausanne. Suisse.Van Cauwelaert (1985): “Contraintes et Déplacements dans un massif semi-infini isotrope ou à anisotropie
transverse soumis à des charges rectangulaires souples et rigides en surface”. Revue Française de Géotechniquen° 28. Paris.
Van Cauwelaert, Lequeux, Delaunois (1986): “Computer Programs for the Determination of Stresses andDisplacements in Four layered Structures”. Waterways Experiment Station. Vicksburg. Mississippi.
PAVEMENT DESIGN AND EVALUATION
274
Van Cauwelaert, Alexander, White, Barker (1988): “Multilayer elastic program for Backcalculating Layer Moduli inPavement Evaluation”. Nondestructive Testing of Pavements and Backcalculation of Moduli. Bush III andBaladi Editors. STP 1026. ASTM, Philadelphia, PA.
Van Cauwelaert, Stet (1998): “Shear transfer and Deflection Ratio’s at a Joint of a Concrete Slab”. Proceedings ofthe 4th International Workshop on Theoretical Design on Concrete Pavements. CROW. Buçaco, Portugal.
Veverka (1973) : “ Modules, Contraintes et Déformations des massifs et couches granulaires”. Rapport deRecherche n° 162. Centre de Recherches Routières. Bruxelles.
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
York.Westergaard (1926): “Stresses in Concrete Pavements computed by Theoretical Analysis”. Proc. Highway Research
Board, Part 1.Westergaard (1927): “Analysis of Stresses in Concrete Pavements due to the Variation of Temperature”. Proc.
Highway Research Board.Westergaard (1948): “New Formulas for Stresses in Concrete Pavements of Airfields. Proceedings American
Society of Civil Engineers, Vol 113.Winkler (1867): “Die Lehre von der Elasticität und Festigkeit”. Prage Dominicus.
APPENDIX COMPLEX FUNCTIONS
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
PAVEMENT DESIGN AND EVALUATION
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 =
APPENDIX COMPLEX FUNCTIONS
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
PAVEMENT DESIGN AND EVALUATION
278
[ ]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
APPENDIX COMPLEX FUNCTIONS
279
∫ ==−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
PAVEMENT DESIGN AND EVALUATION
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
APPENDIX COMPLEX FUNCTIONS
281
∫−=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 =−
−
− −=
PAVEMENT DESIGN AND EVALUATION
282
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
APPENDIX COMPLEX FUNCTIONS
283
∫ ∫ ∫=−=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
∫∫ −−
=π
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
PAVEMENT DESIGN AND EVALUATION
284
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)
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.