Thermal Stresses -- Advanced Theory and Applications (Solid Mechanics and Its Applications)

The first paper ever published on thermal stresses and thermoelasticity: Duhamel, J.-M.-C., Second mémoire sur les phéno...

Author: Richard B. Hetnarski | M. Reza Eslami

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The first paper ever published on thermal stresses and thermoelasticity: Duhamel, J.-M.-C., Second mémoire sur les phénomènes thermo-mécaniques, J. de l’École Polytechnique, tome 15, cahier 25, 1837, pp. 1–57.

Thermal Stresses – Advanced Theory and Applications

SOLID MECHANICS AND ITS APPLICATIONS Volume 158 Series Editor:

G.M.L. GLADWELL Department of Civil Engineering University of Waterloo Waterloo, Ontario, Canada N2L 3GI

Aims and Scope of the Series The fundamental questions arising in mechanics are: Why?, How?, and How much? The aim of this series is to provide lucid accounts written by authoritative researchers giving vision and insight in answering these questions on the subject of mechanics as it relates to solids. The scope of the series covers the entire spectrum of solid mechanics. Thus it includes the foundation of mechanics; variational formulations; computational mechanics; statics, kinematics and dynamics of rigid and elastic bodies: vibrations of solids and structures; dynamical systems and chaos; the theories of elasticity, plasticity and viscoelasticity; composite materials; rods, beams, shells and membranes; structural control and stability; soils, rocks and geomechanics; fracture; tribology; experimental mechanics; biomechanics and machine design. The median level of presentation is the first year graduate student. Some texts are monographs defining the current state of the field; others are accessible to final year undergraduates; but essentially the emphasis is on readability and clarity.

For other titles published in this series, go to www.springer.com/series/6557

Richard B. Hetnarski

•

M. Reza Eslami

Thermal Stresses – Advanced Theory and Applications

123

Richard B. Hetnarski Department of Mechanical Engineering Rochester Institute of Technology Rochester, New York USA [email protected]

ISBN 978-1-4020-9246-6

M. Reza Eslami Department of Mechanical Engineering Amirkabir University of Technology (Tehran Polytechnic) Tehran Iran [email protected]

e-ISBN 978-1-4020-9247-3

Library of Congress Control Number: 2008936149 c Springer Science+Business Media, B.V. 2009 No part of this work may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, microfilming, recording or otherwise, without written permission from the Publisher, with the exception of any material supplied specifically for the purpose of being entered and executed on a computer system, for exclusive use by the purchaser of the work. Printed on acid-free paper springer.com

The authors dedicate this book to the memory of their Parents

Jan Hetnarski (1884 - 1966)

Mohammad Sadegh Eslami (1900 - 1980)

and

and

Izabela Hetnarska (1893 - 1980)

Zinat Shahrestani (1925 - 2006)

Love of their sons was overwhelming. Non Omnis Moriar.

Preface The authors are pleased to present Thermal Stresses – Advanced Theory and Applications. This book will serve a wide range of readers, in particular, graduate students, PhD candidates, professors, scientists, researchers in various industrial and government institutes, and engineers. Thus, the book should be considered not only as a graduate textbook, but also as a reference handbook to those working or interested in areas of Applied Mathematics, Continuum Mechanics, Stress Analysis, and Mechanical Design. In addition, the book provides extensive coverage of great many theoretical problems and numerous references to the literature. The field of Thermal Stresses lies at the crossroads of Stress Analysis, Theory of Elasticity, Thermodynamics, Heat Conduction Theory, and advanced methods of Applied Mathematics. Each of these areas is covered to the extend it is necessary. Therefore, the book is self-contained, so that the reader should not need to consult other sources while studying the topic. The book starts from basic concepts and principles, and these are developed to more advanced levels as the text progresses. Nevertheless, some basic preparation on the part of the reader in Classical Mechanics, Stress Analysis, and Mathematics, including Vector and Cartesian Tensor Analysis is expected. While selecting material for the book, the authors made every effort to present both classical topics and methods, and modern, or more recent, developments in the field. The book comprises ten chapters. Chapter 1 treats, among other topics, the basic laws of Thermoelasticity, with descriptions and mathematical formulations of stresses, deformations, constitutive laws, the equations of equilibrium and motion, the compatibility conditions, and an introduction to two-dimensional thermoelasticity. Chapter 2 is devoted to the necessary topics of thermodynamics. Detailed attention is given to the first and second laws of thermodynamics, Fourier’s law of heat conduction, and more advanced topics, namely, generalized thermoelasticity and second sound phenomenon, thermoelasticity without energy dissipation, variational formulation of mechanics, the reciprocity theorem, and the discussion of initial and boundary conditions. In the generalized theory of thermoelasticity, the Lord-Shulman, the Green-Lindsay, and the Green-Naghdi models are treated, and these models are then presented in a unified formulation for heterogeneous/anisotropic materials. vii

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Preface

Basic problems of Thermoelasticity are discussed in Chapter 3, where the analogy of thermal gradient and body forces is presented, and general solutions are derived in rectangular Cartesian, cylindrical, and spherical coordinate systems. Chapter 4 is devoted to problems of heat conduction, again treated in various coordinate systems. Steady state one-, two-, and three-dimensional problems are discussed, and necessary mathematical methods, like the use of Fourier series and Bessel functions, are introduced. Engineering applications are treated in Chapter 5. Various kinds of beams, including rectangular, bimetallic, and curved beams are discussed in detail, and more advanced or modern aspects, such as functionally graded beams, are treated. In Chapter 6, thermal stresses in disks, cylinders, and spheres are treated, including functionally graded cylinders and spheres. A short Chapter 7 presents an analysis of thermal expansion in piping systems, a unique introduction to this frequently encountered engineering application, a topic of importance, treated by advanced design codes. In Chapter 8, the theories of coupled and generalized thermoelasticity are presented. To the authors’ knowledge, such extensive treatment of these topics has never before been given in a textbook. Finite and boundary element methods are the topic of Chapter 9. The Galerkin finite element is introduced, and the methods of generalized thermoelasticity are applied to disks and spheres. Also, problems of functionally graded beams and layers are presented. The last chapter, Chapter 10, is devoted to the analysis of creep. First, general definitions and the theory are presented, and then the problems related to thermal effects are discussed. This chapter deals with useful and efficient numerical techniques to handle creep problems of structures subjected to thermal stresses. At the end of all chapters, except Chapters 9 and 10, there are a number of problems for the students to solve. In total, there are 47 problems. Also, at the end of each chapter, there is a list of literature. The book was prepared over a number of years, with each author working separately, except for a one-month period when both authors were Visiting Scholars at the Institute of Fundamental Technological Research of the Polish Academy of Sciences in Warsaw, where they worked jointly. The authors take this opportunity to thank the authorities of the Institute for the invitation and for providing needed space and services. The authors express their gratitude to Professor Theodore R. Tauchert of the University of Kentucky for reading Chapters 1–3 and 7, to Professor Józef Ignaczak of the Polish Academy of Sciences in Warsaw for reading Chapters 3, 6, and 8, to Dr. Xiangmin Zhou of the University of Minnesota for reading

Preface

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Chapter 9, and to Professor Jacek Skrzypek of Kraków University of Technology for reading Chapter 10. Their comments and corrections greatly improved the text and helped eliminate errors. Also, the help of Mr. A. Bagri, a graduate student of the second author, in preparation and checking of the manuscript is appreciated. The authors express thanks to Professor G.M.L. Gladwell, the Series Editor, for his support of the book’s publication and his advice regarding language problems, and to Mrs. Nathalie Jacobs, Publishing Editor, Mechanical Engineering, at Springer, for her kind offer to publish the book and her patience with shifting deadlines for delivering of the manuscript. Richard B. Hetnarski M. Reza Eslami September 2008

Contents Preface

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Notation – A Short List

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Historical Note. Beginnings of Thermal Stresses Analysis

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1 Basic Laws of Thermoelasticity 1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . 2 Stresses and Tractions . . . . . . . . . . . . . . . . . . 3 Equations of Motion . . . . . . . . . . . . . . . . . . . 4 Coordinate Transformation. Principal Axes . . . . . . . 5 Principal Stresses and Stress Invariants . . . . . . . . . 6 Displacement and Strain Tensor . . . . . . . . . . . . . 7 Compatibility Equations. Simply Connected Region . . 8 Compatibility Conditions. Multiply Connected Regions 9 Constitutive Laws of Linear Thermoelasticity . . . . . 10 Displacement Formulation of Thermoelasticity . . . . . 11 Stress Formulation of Thermoelasticity . . . . . . . . . 12 Two-Dimensional Thermoelasticity . . . . . . . . . . . 13 Michell Conditions . . . . . . . . . . . . . . . . . . . . 14 Problems . . . . . . . . . . . . . . . . . . . . . . . . . .

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1 1 2 4 6 7 10 14 17 21 23 24 27 35 39

2 Thermodynamics of Elastic Continuum 1 Introduction . . . . . . . . . . . . . . . . . . . 2 Thermodynamics Definitions . . . . . . . . . . 3 First Law of Thermodynamics . . . . . . . . . 4 Second Law of Thermodynamics . . . . . . . . 5 Variational Formulation of Thermodynamics . 6 Thermodynamics of Elastic Continuum . . . . 7 General Theory of Thermoelasticity . . . . . . 8 Free Energy Function of Hookean Materials . 9 Fourier’s Law and Heat Conduction Equation 10 Generalized Thermoelasticity, Second Sound . 11 Thermoelasticity without Energy Dissipation .

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43 43 44 45 46 48 50 54 57 60 61 70

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Contents 12 13 14 15 16 17

A Unified Generalized Thermoelasticity . Uniqueness Theorem . . . . . . . . . . . Variational Principle of Thermoelasticity Reciprocity Theorem . . . . . . . . . . . Initial and Boundary Conditions . . . . . Problems . . . . . . . . . . . . . . . . . .

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3 Basic Problems of Thermoelasticity 1 Introduction . . . . . . . . . . . . . . . . . . . . . 2 Temperature Distribution for Zero Thermal Stress 3 Analogy of Thermal Gradient with Body Forces . 4 General Solution of Thermoelastic Problems . . . 5 Solution of Two-Dimensional Navier Equations . . 6 General Solution in Cylindrical Coordinates . . . 7 Solution of Problems in Spherical Coordinates . . 8 Problems . . . . . . . . . . . . . . . . . . . . . . .

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105 . 105 . 106 . 110 . 113 . 117 . 120 . 123 . 126

4 Heat Conduction Problems 1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 Problems in Rectangular Cartesian Coordinates . . . . . . . . 2.1 Steady State One-Dimensional Problems . . . . . . . . 2.2 Steady Two-Dimensional Problems. Separation of Variables . . . . . . . . . . . . . . . . . . . . . . . . 2.3 Fourier Series . . . . . . . . . . . . . . . . . . . . . . . 2.4 Double Fourier Series . . . . . . . . . . . . . . . . . . . 2.5 Bessel Functions and Fourier-Bessel Series . . . . . . . 2.6 Nonhomogeneous Differential Equations and Boundary Conditions . . . . . . . . . . . . . . . . 2.7 Lumped Formulation . . . . . . . . . . . . . . . . . . . 2.8 Steady State Three-Dimensional Problems . . . . . . . 2.9 Transient Problems . . . . . . . . . . . . . . . . . . . . 3 Problems in Cylindrical Coordinates . . . . . . . . . . . . . . 3.1 Steady-State One-Dimensional Problems (Radial Flow) 3.2 Steady-State Two-Dimensional Problems . . . . . . . . 3.3 Steady-State Three-Dimensional Problems . . . . . . . 3.4 Transient Problems . . . . . . . . . . . . . . . . . . . . 4 Problems in Spherical Coordinates . . . . . . . . . . . . . . . 4.1 Steady-State One-Dimensional Problems . . . . . . . . 4.2 Steady-State Two- and Three-Dimensional Problems . 4.3 Transient Problems . . . . . . . . . . . . . . . . . . . . 5 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

75 79 88 91 95 98

131 . 131 . 132 . 132 . . . .

135 137 140 142

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153 158 163 166 175 175 179 191 194 199 200 201 211 216

Contents

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5 Thermal Stresses in Beams 1 Introduction . . . . . . . . . . . . . . . . . . 2 Thermal Stresses in Beams . . . . . . . . . . 3 Deflection Equation of Beams . . . . . . . . 4 Boundary Conditions . . . . . . . . . . . . . 5 Shear Stress in a Beam . . . . . . . . . . . . 6 Beams of Rectangular Cross Section . . . . . 7 Transient Stresses in Rectangular Beams . . 8 Beam with Internal Heat Generation . . . . 9 Bimetallic Beam . . . . . . . . . . . . . . . 10 Functionally Graded Beams . . . . . . . . . 11 Transient Stresses in FGM Beams . . . . . . 12 Thermal Stresses in Thin Curved Beams and 13 Deflection of Thin Curved Beams and Rings 14 Problems . . . . . . . . . . . . . . . . . . . .

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219 219 220 223 225 226 227 231 232 234 235 240 242 243 249

6 Disks, Cylinders, and Spheres 1 Introduction . . . . . . . . . . . . . . . . . . . 2 Cylinders with Radial Temperature Variation 3 Thermal Stresses in Disks . . . . . . . . . . . 4 Thick Spheres . . . . . . . . . . . . . . . . . . 5 Thermal Stresses in a Rotating Disk . . . . . 6 Non-axisymmetrically Heated Cylinders . . . . 7 Method of Complex Variables . . . . . . . . . 8 Functionally Graded Thick Cylinders . . . . . 9 Axisymmetric Stresses in FGM Cylinders . . . 10 Transient Thermal Stresses in Thick Spheres . 11 Functionally Graded Spheres . . . . . . . . . . 12 Problems . . . . . . . . . . . . . . . . . . . . .

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253 . 253 . 254 . 258 . 260 . 263 . 265 . 271 . 278 . 289 . 295 . 303 . 309

7 Thermal Expansion in Piping Systems 1 Introduction . . . . . . . . . . . . . . . 2 Definition of the Elastic Center . . . . 3 Piping Systems in Two Dimensions . . 4 Piping Systems in Three Dimensions . 5 Pipelines with Large Radius Elbows . . 6 Problems . . . . . . . . . . . . . . . . .

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351 . 351 . 352 . 354 . 357

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8 Coupled and Generalized Thermoelasticity 1 Introduction . . . . . . . . . . . . . . . . . . . . . 2 Governing Equations of Coupled Thermoelasticity 3 Coupled Thermoelasticity for Infinite Space . . . 4 Variable Heat Source . . . . . . . . . . . . . . . .

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317 317 318 322 329 335 346

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Contents 5 6 7 8 9 10 11 12

One-Dimensional Coupled Problem . . . . . . . . . . . Propagation of Discontinuities . . . . . . . . . . . . . . Half-Space Subjected to a Harmonic Temperature . . . Coupled Thermoelasticity of Thick Cylinders . . . . . . Green–Naghdi Model of a Layer . . . . . . . . . . . . . Generalized Thermoelasticity of Layers . . . . . . . . . Generalized Thermoelasticity in Spheres and Cylinders Problems . . . . . . . . . . . . . . . . . . . . . . . . . .

9 Finite and Boundary Element Methods 1 Introduction . . . . . . . . . . . . . . . . . . 2 Galerkin Finite Element . . . . . . . . . . . 3 Functionally Graded Layers . . . . . . . . . 4 Coupled Thermoelasticity of Thick Spheres . 5 Generalized Thermoelasticity of FG Spheres 6 Generalized Thermoelasticity of FG Disk . . 7 Higher Order Elements . . . . . . . . . . . . 8 Functionally Graded Beams . . . . . . . . . 9 Boundary Element Formulation . . . . . . .

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360 366 377 380 385 394 402 405

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413 . 413 . 414 . 422 . 428 . 440 . 451 . 463 . 466 . 476

10 Creep Analysis 1 Introduction . . . . . . . . . . . . . . . . . . . . 2 Creep of Metals . . . . . . . . . . . . . . . . . . 3 Constitutive Equation of Uniaxial Creep . . . . 4 Creep Relaxation, Linear Rheological Models . . 5 Three-Dimensional Governing Equations . . . . 6 Creep Potential, General Theory of Creep . . . 7 Stress Function for Creep Problems . . . . . . . 8 Creep Linearization . . . . . . . . . . . . . . . . 9 Creep Relaxation of Axisymmetric Stresses . . . 10 Creep Relaxation of Non-axisymmetric Stresses 11 Thermoelastic Creep Relaxation in Beams . . .

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Index

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499 499 503 510 512 514 518 522 525 531 536 540 553

Notation – A Short List The list contains basic notation. In particular chapters, additional local notation is used. The system of notation is similar to that used in other engineering textbooks. In particular, scalars appear as light face letters, and vectors and tensors as bold face straight letters. A B bi cy cσ c D Dij eijk eij E E F F gij G h H(.) i I I1 , I2 , I3 J J1 , J 2 , J3 k K L M

area Boussinesq’s function components of body force vector per unit mass specific heat on path y specific heat at constant stress specific heat at constant strain domain; Biot’s free energy function rate of deformation tensor components permutation symbol Green strain tensor components Young’s modulus strain tensor free energy function per unit volume force Euclidean metric tensor components shear modulus = μ; gravitational energy; Gibbs thermodynamic potential heat transfer coefficient by convection Heaviside function intrinsic energy per unit mass moment of inertia of an area; Intrinsic energy per unit volume invariants of stress tensor polar moment of inertia invariants of deviatoric stress tensor thermal conductivity kinetic energy; bulk modulus velocity gradient tensor bending moment; Michell’s function xv

xvi n p pi q q Q r R s S sij t t0 , t1 , t2 tn T T0 u, U U x, y, z r, φ, z r, φ, θ W X α δ(.) δij ij κ λ, μ ν ρ σij θ Φ ψ ωij

Notation – A Short List unit outer normal vector pressure entropy flux vector components heat flux vector specified heat flux heat heat produced per unit time per unit mass heat produced per unit time per unit volume entropy per unit mass; Laplace transform parameter entropy per unit volume second Piola-Kirchhoff stress tensor components time relaxation times traction vector absolute temperature initial temperature displacement vector internal energy; generalized strain energy function Cartesian coordinates cylindrical coordinates spherical coordinates work body force vector per unit volume coefficient of linear thermal expansion Dirac delta function Kronecker symbol strain tensor components diffusivity Lamé constants (μ = G) Poisson’s ratio density stress tensor components temperature change Airy stress function displacement potential rotation tensor components

Historical Note. Beginnings of Thermal Stresses Analysis Compared to the history of the theory of elasticity, which is traced to Robert Hooke and Edmé Mariotte in the seventeenth century or, even earlier, to Galileo Galilei in the 16th century, the history of thermoelasticity and thermal stresses is much younger. First paper on thermoelasticity, by J.M.C. Duhamel, was read before the French Academy of Sciences in Paris on February 23, 1835, ´ and published in the Journal de l’Ecole Polytechnique in 1837 [1]. Duhamel’s paper contained the formulation of boundary value problems and also the derivation of equations for the coupling of the temperature field and the body’s deformation. To keep this in perspective, the appearance of Duhamel’s paper occurred not very long after C.L.M.H. Navier’s seminal paper on the foundations of the theory of elasticity [2] was read to the Academy on May 14, 1821, and published in 1827, and J.B.J. Fourier’s treatise on the theory of heat [3] was published in 1822. At present, we are still in the second century of development of this field of Mechanics. The formulation of thermoelasticity equations is due to F. Neumann [4] in 1885, to E. Almansi [5] in 1897, to O. Tedone [6] in 1906, and to W. Voigt [7] in 1910. Most of early works were devoted to static problems. Integration of thermoelasticity equations was reduced to problems of action of body forces with the potential of which density is the temperature of the body. Thermoelasticity problems were reduced to elasticity problems, where a body is acted upon by volume and surface forces. The method was developed by creators of the theory of elasticity, B. de Saint-Venant, G. Lamé, and P.S. Laplace. Namely, problems of thermoelasticity were solved by finding solutions of Lamé displacement equations when a body is acted upon by arbitrary mass forces. Thus, many basic thermoelasticity problems were considered within classic theory of elasticity. Besides the development of the theory, a number of specific problems were solved. We should mention the work of C.W. Borchardt [8] of 1873 on a solution in integral representation for a sphere acted upon by an arbitrarily distributed temperature, and also papers by J. Hopkinson [9] of 1874 on thermal stresses in a sphere, by A. Leon [10] of 1905 on a hollow cylinder, and by S. Timoshenko [11] of 1925 on bi-metallic strips. xvii

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Further progress was made by M.A. Biot [12] who, in 1935, analyzed properties of two-dimensional distributions of thermal stresses, and by J.N. Goodier [13] who, in 1937, introduced the notion of the thermoelastic potential and considered the effect of non-continuous temperature fields. A paper by A. Signorini [14] of 1930 on finite thermoelastic deformations should also be mentioned. Of the papers published before the Second World War, papers by N. Muschelishvili [15] of 1923 and by P.F. Papkovich [16] of 1937 should be cited. We should also list the following three papers that appeared after the War: by J. Lighthill and J. Bradshaw [17] of 1949 on thermal stresses in turbine blades, by S.S. Manson [18] of 1947 on gas turbine disks, and by J. Aleck [19] of 1949 on thermal stresses in rectangular plates. During and after the Second World War, the requirements associated with new technologies contributed to a wave of research on thermal stresses. Gaining knowledge of the distribution of temperature in specific situations, finding thermal stresses in parts of complex mechanical systems, assessment of allowed stresses in various materials and in various loading conditions, matters of stability, problems of viscoelasticity, of fatigue, and thermal shock, became topics of active research both theoretical and experimental. Of the theoretical nature, a paper by W.M. Maysel [20] of 1941 on the generalization of the BettiMaxwell theorem to thermal stresses, as well as the work by N.O. Myklestadt [21] of 1942 on thermal stresses in a body with ellipsoidal inclusion should be mentioned. Steady-state problems of thermal stresses in a half-space were considered by D. Mindlin and D.H. Cheng [22] in 1950, and by B. Sen [23] who, in 1951, introduced the notion of nucleus of thermoelastic displacements, a fundamental development in the theory of steady-state problems. A dynamic counterpart to the static nucleus was proposed by J. Ignaczak [24]. An important development was the publication in 1953 of the first book on thermal stresses by E. Melan and H. Parkus [25]. The book provided a consistent description and analysis of most of the results received in steady-state thermoelasticity up to that time, and its publication provided a source from which teaching of the subject could be conducted. The book was subsequently complemented by the publication, in 1959, of the book by H. Parkus [26] on non-stationary thermal stresses. Within the next few years a number of other extensive texts appeared. Among them: 1. A very informative and carefully written monograph by B.A. Boley and J.H. Weiner [27], published in 1960. Besides its clear treatment of linear thermoelasticity, the book contained also some non-linear effects. 2. Somewhat earlier, in 1957, a practical description of thermal stresses applications for airplanes etc. by B.E. Gatewood [28] was published. The book contains also many theoretical analyses.


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3. The first book on thermoelasticity by W. Nowacki [29], in Polish, appeared in 1960. 4. The publication of the book by W. Nowacki in 1960 was followed two years later by the arrival of W. Nowacki’s monograph [30] in English, Thermoelasticity, an extensive and complete treatise on the subject. The second edition of the book, revised and enlarged, appeared in 1986. 5. Also, W. Nowacki [31], published a more specialized monograph on Dynamical Problems of Thermoelasticity, first in Polish in 1966, and then its translation appeared in English in 1975. 6. A large monograph on thermoelasticity was published by J.L. Nowi´ nski [32] in 1978. 7. It was only many years later that a book specifically written as a textbook on thermal stresses was prepared by N. Noda, R.B. Hetnarski, and Y. Tanigawa [33]. Returning to research papers related to a half-space, detailed analysis of steady-state problem for half-space was provided by E. Sternberg and E.L. McDowell [34] in 1957. Some results contained in that paper were independently received, also in 1957, by W. Nowacki [35] who considered a problem with discontinuous boundary conditions for temperature on the surface of the half-space. Three years later, in 1960, appeared a work by I.N. Sneddon and F.J. Lockett [36] in which steady-state problems for a half-space and a layer were considered, and the authors received close-form solutions for some types of surface heating. While considering papers related to a half-space, we should mention the publications on quasi-static problems. When changing of the temperature is a slow process, one can disregard inertia terms in the equations of motion. To a group of first publications on such problems belong papers by M.A. Sadovsky [37] of 1955, J.L. Bailey [38] of 1958, and the analysis of this type of problems by W. Nowacki [29] in his first book on thermoelasticity in Polish in 1960. A pioneering work on dynamic thermoelasticity was a paper in Russian by V.I. Danilovskaya [39] of 1950, where she solved a one-dimensional problem of stresses in a half-space due to a thermal shock applied to the bounding plane. Similar problems were subsequently considered by T. Mura [40] in 1952, and later by E. Sternberg and J.G. Chakravorty [41] who in 1959 analyzed the behavior of a stress wave as a result of a change of boundary condition for temperature. In the same category are papers by J. Ignaczak [42] of 1957, and by B.A. Boley and A.D. Barber [43] of 1957. A large amount of research on dynamic problems of thermoelasticity followed. A broad subject to which much attention was given in the 1950’s was plates and disks. We should mention papers by G. Horvay [44] of 1952 on perforated

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plates, and [45] of 1954, on rectangular strips. Also, in 1957 appeared a paper by H. Parkus [46] on a heated disk. Of other problems with practical applications are problems of thermal stresses in shells, like those treated by H. Parkus [47–49] in 1950–1951, and by W. Nowacki [50] in 1956. Other problems of that time were complex problems of thermal stresses in axisymmetric bodies, such as circular cylinders with stresses caused by a non-steady and discontinuous temperature applied to the cylinder’s surface. To this group belong papers by R. Trostel [51] and by T. Mura [52], both of 1956, and by J. Ignaczak [53] and by M. Sokolowski [54], both of 1958. Another area of research that was initiated in the 1950’s was problems of anisotropic bodies. A method of operator determinants allowed reduction of the problem of solving of complex differential equations to quasi-biharmonic equations and to the application of displacement functions similar to Galerkin’s displacement functions. We will mention works by W.H. Pell [55] of 1946, and a series of works by Polish authors: J. Nowi´ nski [56] of 1955, J. Nowi´ nski, W. Olszak, and W. Urbanowski [57] of 1956, J. Mossakowski [58] of 1957, and also W. Nowacki [59] of 1958, and from other works, by B. Sharma [60] of 1958. A number of researchers turned their attention to the investigation of inelastic effects, such as visco-thermoelasticity effects. A book by T. Alfrey [61] on polymers of 1948 should be mentioned. We list papers by H.S. Tsien [62] of 1950, W.T. Read [63] of 1950, A.M. Freudenthal [64] and [65] both of 1954, and W. Prager [66] of 1956, and [67] of 1958. Although J.M.C. Duhamel presented equations of thermoelasticity with coupling of field of deformation with field of temperature already in 1837, only papers published 120 years later by M.A. Biot [68] of 1956 and M. Lessen [69] of 1957 gave a new impulse to do research in this area. In classic thermoelasticity, a problem of temperature was solved first, and then stresses were received from Duhamel-Neumann equations. But both theoretical considerations and simple experiments show that a change of displacements in a body accompanies a change of temperature, and a change of temperature is accompanied by a change of displacements. Thus, treating a dynamic problem of thermoelasticity in stresses requires simultaneous solution of the stress equation of motion and the heat conduction equation in which appears the time derivative of first stress invariant. A number of important papers on coupled thermoelasticity were published in the next few years after Biot’s paper appeared in 1956. For example, J.H. Weiner [70] published in 1957 a proof on uniqueness of solutions of coupled equations of thermoelasticity. Analysis of wave propagation, including Rayleigh waves, in thermoelastic bodies were the subject of papers that appeared in 1958: by P. Chadwick and I.N. Sneddon [71], H. Deresiewicz [72], and F.J. Lockett [73] and, published in 1959 [74]. Of other papers published on the subject in 1959, we should mention papers by I.N. Sneddon [75], G. Eason and I.N. Sneddon [76], W. Nowacki [77], and G. Paria [78].


xxi

The theory of coupled thermoelasticity, as well as developed later theory of generalized thermoelasticity which was initiated by the paper by H.W. Lord and Y. Shulman [79] of 1967, are extensively treated in this book in Chapters 2, 8, and 9, and numerous references are provided. In this Note, we do not attempt to describe the development of the theory of thermal stresses for the time after 1960. If we did, the Note would take substantially more space. However, we direct interested readers to the following sources of information: (A) A review of papers on thermal stresses edited by one of the authors (RBH) contained in two special issues of Applied Mechanics Reviews: Vol. 44, issue No. 8–9, August–September 1991, with four review articles: T.R. Tauchert, Thermally induced flexure, buckling, and vibration of plates. D.H. Allen, Thermomechanical coupling in inelastic solids. J. Ignaczak, Domain of influence results in generalized thermoelasticity – a survey. N. Noda, Thermal stresses in materials with temperature-dependent properties. Vol. 50, issue No. 9, September 1997, with three review articles: E.A. Thornton, Aerospace thermal-structural testing technology. S.A. Dunn, Using nonlinearities for improved stress analysis by thermoelastic techniques. K.K. Tamma and R.R. Namburu, Computational approaches with applications to non-classical and classical thermomechanical problems. (B) In the years 1986–1999, five volumes of Thermal Stresses handbook, all edited by the first author (RBH), were published. First four volumes were published by North-Holland, Elsevier, in Amsterdam, and the fifth volume was published by Lastran in Rochester, N.Y. Each volume contains chapters with state-of-the-art coverage of specific areas of research and extensive bibliography. The lists of contents are as follows: Thermal Stresses I, North-Holland, Elsevier, Amsterdam, 1986, 547 pages. 1. R.B. Hetnarski, Basic Equations of the Theory of Thermal Stresses, pp. 1–21. 2. T.R. Tauchert, Thermal Stresses in Plates – Statical Problems, pp. 24–141. 3. J. Padovan, Anisotropic Thermal Stress Analysis, pp. 143–262.

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Historical Note. Beginnings of Thermal Stresses Analysis 4. D.P.H. Hasselman and J.P. Singh, Criteria for the Thermal Stress Failure of Brittle Structural Ceramics, pp. 263–298. 5. L. Karlsson, Thermal Stresses in Welding, pp. 299–389. 6. N. Noda, Thermal Stresses in Materials with Temperature-Dependent Properties, pp. 391–483. Thermal Stresses II, North-Holland, Elsevier, Amsterdam, 1987, 441 pages. 1. T.R. Tauchert, Thermal Stresses in Plates – Dynamical Problems, pp. 1–56. 2. H. Sekine, Thermal Stress Singularities, pp. 57–117. 3. F. Ziegler and H. Irschik, Thermal Stress Analysis Based on Maysel’s Formula, pp. 119–188. 4. R.A. Heller and S. Thangjitham, Probabilistic Methods in Thermal Stress Analysis, pp. 189–268. 5. R.S. Dhaliwal, Micropolar Thermoelasticity, pp. 269–328. 6. G.R. Halford, Low-Cycle Thermal Fatigue, pp. 329-428. Thermal Stresses III, North-Holland, Elsevier, Amsterdam, 1989, 573 pages. 1. J.R. Barber and M. Comninou, Thermoelastic Contact Problems, pp. 1–106. 2. F. Ziegler and F.G. Rammerstorfer, Thermoelastic Stability, pp. 107– 189. 3. T. Inoue, Inelastic Constitutive Relationships and Applications to Some Thermomechanical Processes Involving Phase Transformations, pp. 191–278. 4. J. Ignaczak, Generalized Thermoelasticity and its Applications, pp. 279–354. 5. S.A. L ukasiewicz, Thermal Stresses in Shells, pp. 355–553. Thermal Stresses IV, North-Holland, Elsevier, Amsterdam, 1996, 546 pages. 1. E.A. Thornton, Experimental Methods for High-Temperature Aerospace Structures, p. 1–89. 2. S.A. Dunn, Non-Linear Effects in Stress Measurement by Thermoelastic Techniques, pp. 91–154. 3. G.L. England and Chiu M. Tsang, Thermally Induced Problems in Civil Engineering Problems, pp. 155–275.


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4. K.K. Tamma, An Overview of Non-Classical/Classical Thermal structural Models and Computational Methods for Analysis of Engineering Structures, pp. 277–378. 5. L. Librescu and Weiqing Lin, Thermomechanical Postbuckling of Plates and Shells Incorporating Non-Classical Effects, pp. 379–452. 6. L.G. Hector, Jr., and R.B. Hetnarski, Thermal Stresses in Materials Due to Laser Heating, pp. 453–531. Thermal Stresses V, Lastran, Rochester, NY, 1999, 542 pages. 1. C.T. Herakovich and J. Aboudi, Thermal Effects in Composites, pp. 1–142. 2. K.K. Tamma and A.F. Avila, An Integrated Micro/Macro Modeling and Computational Methodology for High Temperature Composites, pp. 143–256. 3. R. Wojnar, S. Bytnar, and A. Galka, Effective Properties of Elastic Composites Subject to Thermal Fields, pp. 257–466. 4. N. Rajic, Material Characterization Using the Thermoplastic Effect, pp. 467–534. We should include in the list two review papers that were also used in the preparation of the Note: 1. J. Ignaczak, Development of Thermoelasticity in years 1945-1960 (in Polish), Rozprawy In˙zynierskie, 8, 3, 1960, pp. 581–600. 2. B.A. Boley, Thermal Stresses: A Survey, Thermal Stresses in Severe Environments, Proc. of Int. Conference, Virginia Polytechnic Institute and State University, Blacksburg, Virginia, published by Plenum Press, 1980, pp. 1– 11. Also, two special issues of the Journal of Thermal Stresses contain bibliographies on thermal stresses: 1. Bibliography on Thermal Stresses; Compiled by T.R. Tauchert and R.B. Hetnarski, Journal of Thermal Stresses, Vol. 9, Supplement, 1986, pp. (i)–(v), 1–128. The issue covers publications in JTS in Vol. 1–7 and partially Vol. 8 (years 1978–1985). 2. Bibliography on Thermal Stresses in Shells; Compiled by F.W. Keene and R.B. Hetnarski, Journal of Thermal Stresses, Vol. 13, No. 4, 1990, pp. 341– 545. The issue contains alphabetical listing with abstracts of some articles, listing of particular types of shells (thin, thick, composite, etc.), the authors index, list of journals, etc.

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Finishing this Note, it is worth mentioning that two developments in recent decades provided additional push to the growth of the field of Thermal Stresses; • In 1978, an international scientific journal, the Journal of Thermal Stresses, started to appear. The Journal, now a monthly publication, is at present in the 31st year of publication. It should be noted that the Journal publishes regularly lists of papers on thermal stresses that appeared in other journals (Publications on Thermal Stresses), which means that an interested reader may get a better view of what is being published in the area of thermal stresses in the world literature. • In 1995, the First International Congress on Thermal Stresses (then called a Symposium), was held at Shizuoka University, Hamamatsu, Japan. Since then, the Congresses have been held every two years, consecutively on three continents. The most recent one, the Seventh Congress, was held at the National Taiwan University of Science and Technology in Taipei, on June 4–7, 2007; it brought together scientists and engineers from 23 countries. International Congresses on Thermal Stresses (ICTS) are affiliated with the International Union of Theoretical and Applied Mechanics (IUTAM). For each of the Congresses extensive proceeding volumes have been published which present actual developments in the area of thermal stresses. Moreover, invited lectures presented at each Congress have been published in special issues of the Journal of Thermal Stresses. The list of Congresses is as follows: International Congresses on Thermal Stresses (ICTS) (first two were called Symposia) First International Symposium on Thermal Stresses and Related Topics Thermal Stresses ’95 Shizuoka University, Hamamatsu, Japan Principal Local Organizer: Naotake Noda June 5–7, 1995 Second International Symposium on Thermal Stresses and Related Topics Thermal Stresses ’97 Rochester Institute of Technology, Rochester, New York Principal Local Organizer: Richard B. Hetnarski June 8–11, 1997

Historical Note. Beginnings of Thermal Stresses Analysis Third International Congress on Thermal Stresses Thermal Stresses ’99 Cracow University of Technology, Cracow, Poland Principal Local Organizer: Jacek Skrzypek June 13–17, 1999 Fourth International Congress on Thermal Stresses Thermal Stresses 2001 Osaka Prefecture University in cooperation with Osaka Institute of Technology Principal Local Organizer: Yoshinobu Tanigawa June 8–11, 2001 Fifth International Congress on Thermal Stresses Thermal Stresses 2003 Virginia Polytechnic Institute and State University Blacksburg, Virginia Principal Local Organizer: Liviu Librescu June 8–11, 2003 Sixth International Congress on Thermal Stresses Thermal Stresses 2005 Technische Universität, Vienna, Austria Principal Local Organizers: Rudolf Heuer and Franz Ziegler May 26–29, 2005 Seventh International Congress on Thermal Stresses Thermal Stresses 2007 National Taiwan University of Science and Technology (NTUST) Taipei, Taiwan Principal Local Organizer: Ching-Kong Chao June 4–7, 2007 Eight International Congress on Thermal Stresses Thermal Stresses 2009 University of Illinois Urbana-Champaign Urbana-Champaign, Illinois Principal Local Organizer: Martin Ostoja-Starzewski June 1–4, 2009

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Bibliography [1] Duhamel, J.-M.-C., Second mémoire sur les phénomènes thermo´ mécaniques, J. de l’Ecole Polytechnique, tome 15, cahier 25, 1837, pp. 1-57. [2] Navier, C.L.M.H., Mémoire sur les lois de l’équilibre et du movement des corps solides élastiques, Mém. Acad. Sci., Paris, tome VII, pp. 375–393, 1827. [3] Fourier, J.B.J., Théorie analytique de la chaleur, Firmin Didot, Paris, 1822. [4] Neumann, F., Vorlesung u ¨ber die Theorie des Elasticit¨ at der festen K¨ orper und des Licht¨ athers, Teubner, Leipzig, 1885. [5] Almansi, E., Use of the Stress Function in Thermoelasticity, Mem. Reale Accad. Sci. Torino, Series 2, Vol. 47, 1897. [6] Tedone, O., Allgemeine Theoreme der matematischen Elastizitätslehre (Integrationstheorie), Encyklopädie der matematischen Wissenschaften, Vol. 4, Part D, pp. 55–124 and pp. 125–214 (second article written with A. Timpe.), 1906. [7] Voigt, W., Lehrbuch der Kristallphysik, Teubner, Berlin, 1910. [8] Borchardt, C.W., Untersuchungen u ¨ber die Elasticität fester isotropen Körper unter Ber¨ ucksichtigung der Wärme, Mber Akad. der Wissenschaften, 9, Berlin, 1873. [9] Hopkinson, J., Thermal Stresses in a Sphere, Whose Temperature Is a Function of r Only, Mess. Math., Vol. 8, 1879, p. 168. [10] Leon, A., On Thermal Stresses, Der Bautechniker, Vol. 26, 1904, p. 968. [11] Timoshenko, S., Bending and Buckling of Bi-metallic Strips, J. Opt. Soc. of Am., Vol. 11, 1925, p. 233. [12] Biot, M.A., A General Property of Two-Dimensional Thermal Stress Distribution, Phil. Mag., VII, 1935. xxvii

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[13] Goodier, J.N., On the Integration of the Thermo-Elastic Equations, Phil. Mag., VII, 23, 1937, p. 1017. [14] Signorini, A., Sulle deformazioni termoelastiche finite, Proc. 3rd Int. Congr. Appl. Mech., Vol. 2, Stockholm, 1930, pp. 80–89. [15] Muschelishvili, N., Sur l’équilibre des corps élastiques soumis l’action de la chaleur, Bull. de l’Université Tiflis, No. 3, 1923. [16] Papkovich, P.F., On the General Solution of Thermal Stresses (in Russian), Prikl. Mat. Mech., New Series, Vol. 1, No. 2, 1937. [17] Lighthill, J. and Bradshaw, J., Thermal Stresses in Turbine Blades, Phil. Mag., Vol. 40, 1949, p. 770. [18] Manson, S.S., Determination of Elastic Stresses in Gas Turbine Discs, NACA-Rep., Vol. 871, 1947. [19] Aleck, J., Thermal Stresses in a Rectangular Plate Clamped Along an Edge, J. Appl. Mech., Vol. 16, No. 118, 1949. [20] Maysel, V.M., A Generalization of Betti-Maxwell Theorem to the Case of the State of Thermal Stresses and Some Applications (in Russian), Dokl. Acad. Nauk SSSR, Vol. 30, 1941, pp. 115–118. [21] Myklestadt, N.O., Two Problems of Thermal Stress in the Infinite Solid, J. Appl. Mech., Vol. 9, No. 136, 1942. [22] Mindlin, R.D. and Cheng, D.H., Thermoelastic Stress in the Semi-Infinite Solid, J. Appl. Phys., Vol. 21, 1950, p. 931. [23] Sen, B., Note on the Stresses Produced by Nuclei of Thermoelastic Strain in a Semi-infinite Solid, Quart. Appl. Math., Vol. 8, 1951, p. 365. [24] Ignaczak, J., A Dynamic Nucleus of Thermoelastic Strains in an Elastic Infinite Space and Semi-Space, Bull. Acad. Polon., Ser. Techn. Vol. 5, 1959, p. 305. [25] Melan, E. and Parkus, H., W¨ armespannungen infolge station¨ arer Temperaturfelder, Springer, Wien, 1953. [26] Parkus, H., Instationäre Wärmespannungen, Springer, Wien, 1959. [27] Boley, B.A. and Weiner, J.H., Theory of Thermal Stresses, Wiley, New York, 1960. [28] Gatewood, B.E., Thermal Stresses, McGraw-Hill, New York, 1957.

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[29] Nowacki, W., Problems of Thermoelasticity (in Polish), PWN-Polish Scientific Publishers, Warszawa, 1960. [30] Nowacki, W., Thermoelasticity, Pergamon Press, Oxford, and PWNPolish Scientific Publishers, Warszawa, 1962; 2nd edition, PWN-Polish Scientific Publishers, Warszaawa, and Pergamon Press, Oxford, 1986. [31] Nowacki, W., Dynamic Problems of Thermoelasticity (in Polish), PWNPolish Scientific Publishers, Warszawa, 1966. The English translation: Noordhoff, Leyden, 1975. [32] Nowi´ nski, J.L., Theory of Thermoelasticity with Applications, Sijthoff and Noordhoff, Alpen aan den Rijn, 1978. [33] Noda, N., Hetnarski, R.B., and Tanigawa, Y., Thermal Stresses, Lastran, Rochester, 2000; 2nd edition, Taylor and Francis, New York, 2003. [34] Sternberg, E. and McDowell, E.L., On the Steady-State Thermoelastic Problem for the Half-Space, Quart. Appl. Math., Vol. 14, 1957, p. 381. [35] Nowacki, W., A Three-Dimensional Thermoelastic Problem with Discontinuous Boundary Conditions, Archiwum Mechaniki Stosowanej, Vol. 9, 1957, p. 319. [36] Sneddon, I.N. and Lockett, F.J., On the Steady-State Thermoelastic Problem for the Half-Space and the Thick Plate, Quart. J. Appl. Math., Vol. 18, 1960. [37] Sadowsky, M.A., Thermal Shock on a Circular Surface of Exposure of an Elastic Half-Space, J. Appl. Mech., Vol. 22, 1955, p. 177. [38] Bailey, J.L., A Thermoelastic Problem in the Half-Space, Dissertation, Michigan State Univ., 1958. [39] Danilovskaya, V.I., Thermal Stresses in an Elastic Half-Space Arising After a Sudden Heating of Its Boundary (in Russian), Prikl. Mat. Mech., Vol. 14, No. 3, 1950. [40] Mura, T., Thermal Strains and Stresses in Transient State, Proc. 2nd Japan Nat. Congr. Appl. Mech., Vol. 19, 1952, p. 9. [41] Sternberg, E. and Chakravorty, J.G., On Inertia Effects in a Transient Thermoelastic Problem, J. Appl. Mech., Vol. 26, No. 4, 1959. [42] Ignaczak, J., Thermal Displacements in an Elastic Semi-Space Due to Sudden Heating of the Boundary Plane, Archiwum Mechaniki Stosowanej, Vol. 9, 1957, p. 395.

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[43] Boley, B.A. and Barber, A.D., Dynamic Response of Beams and Plates to Rapid Heating, J. Appl. Mech., Vol. 24, 1957, p. 413. [44] Horvay, G., Thermal Stresses in Perforated Plates, Proc. 1st U.S. Nat. Congr. Appl. Mech., 1952, p. 247. [45] Horvay, G., Thermal Stresses in Rectangular Strips, I, II, Proc. 2nd U.S. Nat. Congr. Appl. Mech., 1954, p. 313. [46] Parkus, H., Stress in a Centrally Heated Disc, Proc. 2nd U.S. Nat. Congr. Appl. Mech., 1954, p. 307. [47] Parkus, H., Die Grundgleichungen der allgemeinen Zylinderschale, ¨ Osterr. Ing. Arch., Vol. 6, 1951, p. 30. [48] Parkus, H., Die Grundgleichungen der Schalentheorie in allgemeinen Ko¨ ordinaten, Osterr. Ing. Arch., Vol. 4, 1950, p. 160. [49] Parkus, H., Wärmespannungen in Rotationschalen bei drehsymmet¨ rischer Temperaturverteilung, Sitsungsber. Osterr. Akad. Wiss., Abt. IIa, 1951, p. 160. [50] Nowacki, W., Thermal Stresses in Cylindrical Disks (in Polish), Archiwum Mechaniki Stosowanej, Vol. 8, No. 1, 1956. [51] Trostel, R., Instationäre Wärmespannungen in Hohlzylindern mit Kreisringquerschnitt, Ing. Arch., Vol. 24, No. 1, 1956. [52] Mura, T., Dynamical Thermal Stresses Due to Thermal Shocks, Res. Rep., Faculty of Eng. Meiji Univ., 1956, p. 8. [53] Ignaczak, J., Thermal Stresses in a Long Cylinder Heated in a Discontinuous Manner over the Lateral Surface, Archiwum Mechaniki Stosowanej, 1958, p. 25. [54] Sokolowski, M., Axially-Symmetrical Problems of Thermo-Elasticity for a Cylinder of Unlimited Length, Bull. Acad. Pol. Sci., Serie Tech. Vol. 6, 1958, p. 207. [55] Pell, W.H., Thermal Deflection of Anisotropic Thin Plates, Quart. Appl. Math., 4, 1946, p. 27. [56] Nowi´ nski, J., Thermal Stresses in a Thick-Walled Spherical Vessel of Transversally Isotropic Material (in Polish), Archiwum Mechaniki Stosowanej, Vol. 7, 1955, pp. 363–374.

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[57] Nowi´ nski, J., Olszak, W., and Urbanowski W., On the Thermoelastic Problem in the Case of Bodies of any Type of Curvilinear Orthotropy, Archiwum Mechaniki Stosowanej, Vol. 7, No. 2, 1955, pp. 247–265. [58] Mossakowski, J., The State of Stress and Displacement in a Thin Anisotropic Plate Due to a Concentrated Source of Heat, Archiwum Mechaniki Stosowanej, Vol. 9, No. 5, 1957, pp. 563–577. [59] Nowacki, W., Thermal Stresses in Anisotropic Bodies (in Polish), Archiwum Mechanicki Stosowanej, Vol. 6, 1954, p. 481. [60] Sharma, B., Thermal Stresses in Transversally Isotropic Semi-Infinite Solids, J. Appl. Mech., Vol. 25, 1958. [61] Alfrey, T., Mechanical Behavior of High Polymers, Interscience Publ., New York, 1948. [62] Tsien, H.S., A Generalization of Alfrey’s Theorem for Viscoelastic Media, Quart. Appl. Math., Vol. 8, 1950, pp. 104–106. [63] Read, W.T., Stress Analysis for Compressible Viscoelastic Materials, J. Appl. Phys., Vol. 21, 1950, pp. 671–674. [64] Freudenthal,, A.M., On Inelastic Thermal Stresses in Flight Structures, J. Aero. Sci., Vol. 21, 1954, pp. 772–778. [65] Freudenthal,, A.M., Effect of Rheological Behavior on Thermal Stresses, J. Appl. Phys., Vol. 25, 1954. [66] Prager, W., Thermal Stresses in Viscoelastic Structures, ZAMP, Vol. 7, 1956. [67] Prager, W., Non-Isothermal Plastic Deformation, Proc. Kong. Nederl. Akad. van Wetenschappen, Series B, Vol. 61, No. 3, 1958, pp. 176–182. [68] Biot, M.A., Thermoelasticity and Irreversible Thermodynamics, J. Appl. Phys., Vol. 27, No. 3, 1956, pp. 240–253. [69] Lessen, M., Thermoelasticty and Thermal Shock, J. Mech. Phys. Solids, Vol. 5, 1956, p. 57. [70] Weiner, J.H., A Uniqueness Problem for Coupled Thermoelastic Problems, Quart. Appl. Math., Vol. 15, No. 1, 1957, pp. 102–105. [71] Chadwick, P. and Sneddon, I.N., Plane Waves in an Elastic Solid Conducting Heat, J. Mech. Phys. Solids, Vol. 6, 1958, p. 223.

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[72] Deresiewicz, H., Solutions of the Equations of Thermoelasticity, Proc. 3rd U.S. Nat. Congr. Appl. Mech., 1958. [73] Lockett, F.J., Effect of Thermal Properties of a Solid on the Velocity of Rayleigh Waves, J. Mech. Phys. Solids, Vol. 7, 1958, p. 71. [74] Lockett, F.J., Longitudinal Elastic Waves in Cylinders and Tubes – Including Thermoelastic Effects, Proc. Edinburgh Math. Soc., Vol. 11, No. 3, 1959. [75] Sneddon, I.N., The Propagation of Thermal Stresses in Thin Metallic Rods, Proc. Royal Soc. Edinburgh, Section A, Vol. 65, 1959. [76] Eason, G. and Sneddon, I.N., The Dynamic Stresses Produced in Elastic Bodies by Uneven Heating, Proc. Royal Soc. Edinburgh, Section A, Vol. 65, 1959. [77] Nowacki, W., Some Dynamic Problems of Thermoelasticity, Archiwum Mechaniki Stosowanej, Vol. 11, No. 2, 1959. [78] Paria, G., Coupling of Elastic and Thermal Deformations, Applied Sci. Res. (A), Vol. 7, 1958, p. 6. [79] Lord, H.W. and Shulman, Y., A Generalized Dynamical Theory of Thermoelasticity, J. Mech. Phys. Solids, Vol. 15, No. 5, 1967, pp. 299–309.

Chapter 1 Basic Laws of Thermoelasticity

The basic laws of thermoelasticity, similar to those of the theory of elasticity, include the equations of motion, the compatibility equations, and the constitutive law. This chapter begins with the derivation of the basic laws of linear thermoelasticity, where the linear strain-displacement relations are obtained, following the general discussion of Green and Almansi nonlinear strain tensors. The necessity and sufficiency of the compatibility conditions for the simply and multiply connected regions are presented for three-dimensional conditions. These conditions are then reduced to the two-dimensional case called Michell conditions. The classical general and simple thermoelastic plane strain and plane stress formulations are presented.

1

Introduction

In general, the variation of the temperature field within an elastic continuum results in thermal stresses. The influence of the temperature field in the governing equations of thermoelasticity is through the constitutive law. The theory of linear thermoelasticity is based on linear addition of thermal strains to mechanical strains. While the equilibrium and compatibility equations remain the same as for elasticity problems, the main difference rests in the constitutive law. On this basis, many techniques that have been developed to solve the elasticity problems are applicable as well to the thermoelasticity problems. There are, however, special classes of thermoelasticity problems, such as coupled thermoelasticity problems, which require entirely different mathematical approaches and means of analysis. Even for these classes of problems, some of the basic equations remain the same. It is, therefore, necessary to define the basic laws of thermoelasticity and to derive the governing equations. In this chapter, the thermoelastic material is assumed to be homogeneous and isotropic, with constant material properties. R.B. Hetnarski, M.R. Eslami, Thermal Stresses – Advanced Theory and Applications, Solid Mechanics and its Applications 158, c Springer Science + Business Media B.V. 2009

1

2

2

Chapter 1. Basic Laws of Thermoelasticity

Stresses and Tractions

Consider a continuum under the action of external and body forces occupying a given configuration V and surface boundaries A at a time t. The continuum reacts to the external forces to reach the equilibrium state. Considering a small element V with a part of its surface A associated with its unit outer normal vector n, the reaction of the element under equilibrium condition to its surroundings is a force F acting on A, as shown in Fig. 1.2-1. The force F is a function of location of V in V and the orientation of A. As A varies through V , both magnitude and direction of F change. When A approaches zero, the ratio of F/A approaches a finite value dF/dA. This vector is called the traction vector and is denoted by tn . Therefore, dF tn = (1.2-1) dA The superscript n indicates the surface for which n is a unit outer normal vector. The traction vector tn is related to the normal and shearing stresses at the point of consideration. Consider an infinitesimal tetrahedron at point P formed by three surfaces parallel to the coordinate planes and one inclined. The inclined plane is characterized by its unit outer normal vector n. The traction vector on the plane is tn with three components tn1 , tn2 , and tn3 along the axes of the coordinate system. The normal and shear stresses are defined on the planes parallel to the coordinate planes, as shown in Fig. 1.2-2, where sign conventions are observed. Calling the area of the inclined surface by A, the projections of this surface on the coordinate planes are A1 = n1 A A2 = n2 A A3 = n3 A

(1.2-2)

where n1 , n2 , and n3 are the cosine directions of n. When the height of the tetrahedron h approaches zero, A → dA, and Ai → dAi (i = 1, 2, 3).

Figure 1.2-1: Stress vector at a point.

2. Stresses and Tractions

3

Figure 1.2-2: Traction vector on a inclined plane. Assuming the variations of stresses inside the tetrahedron element to be infinitesimal quantities of higher orders, the equilibrium of elements requires the balance of the forces acting on the element along the coordinate axes. Along the axis x1 this reduces to 1 1 u1 hdA −σ11 n1 dA − σ21 n2 dA − σ31 n3 dA + tn1 dA + hX1 dA = ρ¨ 3 3

(1.2-3)

where h is the height of vertex P from the base dA, X1 is the component of u1 is the inertia force in x1 direction. the body force in x1 direction, and ρ¨ As the volume of the element approaches zero, h → 0 and Eq. (1.2-3) reduces to (1.2-4) tn1 = σ11 n1 + σ21 n2 + σ31 n3 This is Cauchy’s formula relating the traction vector to the stresses. Generalizing Cauchy’s formula leads to the following tni = σji nj

(1.2-5)

This relation insures that the given stress tensor at a point fully defines the components of the traction vector on any given orientation about the point. While the stress tensor at a point in a given coordinate system of a continuum under equilibrium state is fixed, the magnitude and the orientation of the traction vector change on the surface element around the point. It has been proved that the loci of the end of the traction vector at a point move on an ellipsoid called Lamé ellipsoid. The orientation and dimensions of Lamé ellipsoid differ from point to point in a continuum, but if the external forces are kept constant, Lamé ellipsoid at a fixed point is fixed in the stress space.

4

3


Equations of Motion

Consider an elastic body acted upon by an arbitrary traction surface force tn and a body force X per unit volume. We assume that the body occupies the volume V and is bounded by an exterior surface A at a time t. The resulting force acting on the body is

Fi =

A

tni dA +

Xi dV

V

(1.3-1)

Using Cauchy’s formula tni = σji nj

(1.3-2)

Gauss theorem can be used to transform the surface integral of the traction forces to the volume integral as A

tni dA =

σji nj dA =

A

V

σji,j dV

(1.3-3)

Thus, the total force acting on the body is, in components

Fi =

V

(σji,j + Xi )dV

(1.3-4)

Defining the linear momentum by Pi =

V

ρu˙ i dV

(1.3-5)

Newton’s law of motion requires that Fi = P˙ i

(1.3-6)

Upon substitution from Eqs. (1.3-4) and (1.3-5) into Eq. (1.3-6), we obtain

V

(σji,j + Xi )dV =

V

ρ¨ ui dV

(1.3-7)

Since the volume V is arbitrary, Eq. (1.3-7) reduces to the following equation of motion ui (1.3-8) σji,j + Xi = ρ¨ Solution of the equation of motion requires a specification of boundary conditions for the evaluation of the constants of integration. The boundary conditions in terms of the known tractions on the boundary directly follow from Cauchy’s formula. Once the traction forces on the boundary are known, they are related to the stresses on the boundary through Eq. (1.2-5), which in expanded form becomes ˜11 n1 + σ ˜21 n2 + σ ˜31 n3 tñ1 = σ n t˜2 = σ ˜12 n1 + σ ˜22 n2 + σ ˜32 n3 n t˜3 = σ ˜13 n1 + σ ˜23 n2 + σ ˜33 n3

(1.3-9)

3. Equations of Motion

5

where tñ1 , tñ2 , and tñ3 are the known components of the traction force on the boundary, n1 , n2 , and n3 are the cosine directions of the vector normal to the boundary, and σ ˜ij are the stresses at the boundary. The equilibrium condition further requires that the resultant moment of the surface and volume forces vanishes with respect to an arbitrary point. If xi is a component of the position vector at the point of application of forces, then the moment of momentum of the body with respect to the origin of the coordinate system is Hi =

V

eijk xj ρu˙ k dV

(1.3-10)

where eijk is the permutation symbol and is 1 when (i, j, k) permute like (123), is −1 when (ijk) permute like (321), and 0 when any two indices are equal. The moment of the traction forces tni and body forces Xi with respect to the origin is Mi =

A

eijk xj tnk dA +

V

eijk xj Xk dV

(1.3-11)

Introducing Cauchy’s formula into the surface integral of Eq. (1.3-11) and using Gauss theorem, the surface integral representing the moment of traction forces is transformed into a volume integral as A

eijk xj tnk dA =

V

= V

(eijk xj σlk ),l dV

eijk σjk dV +

V

eijk xj σlk,l dV

(1.3-12)

In order to obtain Eq. (1.3-12) we used xj,l = δjl , where δij is Kronecker delta, being 1 when i = j and 0 when i = j. Substituting Eq. (1.3-12) into Eq. (1.311) gives Mi =

V

eijk (σjk + xj σlk,l + xj Xk )dV

(1.3-13)

The balance of moments is satisfied through Euler equation as Mi = H˙ i

(1.3-14)

which happens to be V

eijk [σjk + xj (σlk,l + Xk − ρ¨ uk )]dV = 0

(1.3-15)

In view of the equation of motion, the terms inside the parentheses vanish and Eq. (1.3-15) yields V

eijk σjk dV = 0

(1.3-16)

which implies that σjk = σkj , or the stress tensor is symmetric. Thus, if the stress tensor is symmetric, Euler equation is identically satisfied.

6

4


Coordinate Transformation. Principal Axes

For a fixed and stationary state of loading, the components of the stress tensor at a point of a continuum vary in different directions around the point. Once a coordinate system is fixed at the point, the stress tensor can be written in that system provided that the directions of the coordinate system are fixed in the space xi . If the coordinate system rotates about the point, the stress components change, even though the external state of loading is unchanged. Let us consider an arbitrary Cartesian coordinate system xi of which the origin is fixed at a point P in the continuum. External mechanical and thermal loading as well as the body forces are assumed to be constant. The stress tensor associated with the xi system will be called σij . Now we let the xi system rotate in space about point P to reach a new orientation defined by the coordinate system xi . The stress tensor in the xi system is called σ ij . Since stress is a tensor of rank two, the law of tensor transformation is observed and the two stress tensors σij and σ ij are related according to that law. Assume that the linear relation between two coordinate systems is known as xi = xi (x1 , x2 , x3 )

(1.4-1)

Thus the transformation law is written as ∂xi ∂xj σ ¯mn = σij ∂xm ∂xn

(1.4-2)

In a two-dimensional space, the law of coordinate transformation between two Cartesian coordinates systems (x, y) and (x,y) rotated through an angle φ, as shown in Fig. 1.4-1, is x = x cos φ − y sin φ y = x sin φ + y cos φ

(1.4-3)

From Eq. (1.4-2) the components of stresses in the two different reference systems become y

y syy

syx sxy

sxx x

f

x

Figure 1.4-1: Coordinate transformation.

5. Principal Stresses and Stress Invariants ∂x 2 ∂x ∂y ∂y ) σxx + 2( )σxy + ( )2 σyy ∂x ∂x ∂x ∂x ∂x 2 ∂x ∂y ∂y 2 = ( ) σxx + 2( )σxy + ( ) σyy ∂y ∂y ∂y ∂y ∂x ∂x ∂x ∂y ∂y ∂x ∂y ∂y = σxx + σxy + σxy + σyy ∂x ∂y ∂x ∂y ∂x ∂y ∂x ∂y

7

σ xx = ( σ yy σ xy

(1.4-4)

Using the relations (1.4-3), the above equations reduce to σ xx = σxx cos2 φ + 2σxy sin φ cos φ + σyy sin2 φ σ yy = σxx sin2 φ − 2σxy sin φ cos φ + σyy cos2 φ σ xy = (−σxx + σyy ) sin φ cos φ + σxy (cos2 φ − sin2 φ)

(1.4-5)

or, since sin2 φ = 12 (1 − cos 2φ) and cos2 φ = 12 (1 + cos 2φ) σxx + σyy σxx − σyy + cos 2φ + σxy sin 2φ 2 2 σxx + σyy σxx − σyy = − cos 2φ − σxy sin 2φ 2 2 σxx − σyy =− sin 2φ + σxy cos 2φ 2

σ xx = σ yy σ xy

(1.4-6)

These equations relate the stresses in (x, y) system to the stresses in any other arbitrary system (x,y). It follows from Eqs. (1.4-6) that there exists an angle φ for which the shear stress is zero. The coordinate axes associated with this condition are called the principal axes and the corresponding stresses are called the principal stresses.

5

Principal Stresses and Stress Invariants

The magnitude and direction of the traction vector tn at a point of a stressed body depends upon the direction of the surface passing through the point. In general, the direction of the traction vector does not coincide with the unit outer normal vector n. In the particular case when the direction of the traction vector and the normal vector n coincide, the plane is called the principal plane. Since the traction vector has no projection on this plane, the shear stresses vanish. The normal axis to this plane is called a principal axis and the associated normal stress is called a principal stress. Denoting the principal stress by σ, it follows that tni = σni . Substituting this into Cauchy’s formula (1.3-2) gives (1.5-1) σni = σij nj Replacing the left-hand side by σnj δij and rearranging the equation, we obtain (σij − δij σ)nj = 0

(1.5-2)

8


This equation has a set of nontrivial solutions for nj if the determinant of the coefficients of nj vanishes, i.e., |σij − δij σ| = 0

(1.5-3)

Expanding the determinant for σ gives −σ 3 + I1 σ 2 − I2 σ + I3 = 0

(1.5-4)

where I1 , I2 , and I3 are the first, the second, and the third invariant of the stress tensor, namely I1 = σ11 + σ22 + σ33 σ σ σ σ I2 = 22 23 + 11 13 σ32 σ33 σ31 σ33 σ 11 I3 = σ21 σ31

σ12 σ13 σ22 σ23 σ32 σ33

σ 11 + σ21

σ12 σ22

(1.5-5)

It may be shown that Eq. (1.5-4) has always three real roots along the principal axes. For each value of the principal stress there corresponds a unit vector n representing the direction cosine of the unit outer normal vector acting on the principal plane. Denoting the principal stresses in three orthogonal directions by σ1 , σ2 , and σ3 , the associated unit outer normal vectors acting on the corresponding principal planes are n1 , n2 , n3 , respectively. That is, σ1 acts along the unit vector n1 , σ2 acts along n2 , and σ3 acts along n3 . The directions of the vectors n1 , n2 , and n3 define a system of coordinates consisting of the principal axes. Since on the principal planes shear stresses vanish, the invariants of the stress tensor in terms of the principal stresses reduce to I1 = σ1 + σ2 + σ3 I2 = σ2 σ3 + σ3 σ1 + σ1 σ2 I3 = σ1 σ2 σ3

(1.5-6)

It will be proved now that for a state of stress where the stress tensor is symmetric, the directions of the principal axes are mutually orthogonal [1]. Writing Eq. (1.5-1) for the first and the second principal stresses, we get σ1 n1i = σij n1j

(1.5-7)

σ2 n2i

(1.5-8)

=

σij n2j

where n1i and n2i represent the vectors n1 and n2 written in components. The inner products of Eq. (1.5-7) and n2i , and Eq. (1.5-8) and n1i become σ1 n1i n2i = σij n1j n2i σ2 n1i n2i

=

σij n2j n1i

(1.5-9) (1.5-10)

5. Principal Stresses and Stress Invariants

9

Since the stress tensor is symmetric, the right-hand side of Eq. (1.5-10) can be written as σij n2j n1i = σji n2i n1j = σij n2i n1j (1.5-11) Now, substituting Eq. (1.5-11) into Eq. (1.5-10) and then subtracting it from Eq. (1.5-9) gives (σ1 − σ2 )n1i n2i = 0 (1.5-12) Since the principal stresses are in general different, it follows that n1i n2i = 0

(1.5-13)

which means that the unit vectors n1 and n2 are orthogonal. By the same reasoning it may be proved that n3 is also mutually orthogonal to n1 and n2 . In the plasticity and creep analysis of structures, another form of the stress tensor is important. We define the deviatoric stress tensor Sij by Sij = σij − σm δij

(1.5-14)

where σm is the mean stress given by σm =

1 (σ1 + σ2 + σ3 ) = I1 3 3

(1.5-15)

and δij is Kronecker delta. The principal deviatoric stress is obtained from the following determinant equation |Sij − Sδij | = 0

(1.5-16)

Expanding this determinant yields S 3 + J2 S − J3 = 0

(1.5-17)

where J1 = S 1 + S2 + S 3 = 0 1 J2 = Sij Sij 2 1 J3 = Sij Sjk Ski 3

(1.5-18)

are the three invariants of the stress deviatoric tensor. The second invariant of the stress deviatoric tensor, J2 , is important in the plasticity and creep analysis of structures as it is proportional to the distortion strain energy function and is related to Huber-von Mises-Hencky yield criterion.

10

6


Displacement and Strain Tensor

Consider a body in its original undeformed configuration described in the coordinate system (a1 , a2 , a3 ) fixed to the body. A point P of the body has the coordinates ai (i = 1, 2, 3) in this system. At a later time the body is deformed (and moved) to a new position Q. Along with the body, the original coordinate system (a1 , a2 , a3 ) is transformed into the deformed configuration (x1 , x2 , x3 ). The coordinates of the point Q in the deformed coordinates are xi , (i = 1, 2, 3). It is assumed that the change and deformation of the body is continuous and the point transformation is one-to-one. Let us confine ourselves to the rectangular Cartesian coordinates, and assume that the law of coordinate transformation between the original and the deformed coordinates is known and given as [1] (1.6-1) xi = xî (a1 , a2 , a3 ) with the unique inverse transformation law î (x1 , x2 , x3 ) ai = a

(1.6-2)

The differentials of a position vector of a point in the original and deformed coordinates are dr0 = ai dai dr = gi dxi

(1.6-3)

where ai and gi are the unit vectors in the original and deformed coordinates, respectively. The differentials of line elements in the original and the deformed coordinates are dS02 = dr0 .dr0 = ai .aj dai daj = δij dai daj dS 2 = dr.dr = gi .gj dxi dxj = δij dxi dxj (1.6-4) From Eqs. (1.6-1) and (1.6-2) ∂xi dal ∂al ∂ai dai = dxl ∂xl dxi =

(1.6-5)

Substituting into Eqs. (1.6-4) yields ∂ai ∂aj dxl dxm ∂xl ∂xm ∂xi ∂xj dS 2 = δij dxi dxj = δij dal dam ∂al ∂am dS02 = δij dai daj = δij

(1.6-6)

6. Displacement and Strain Tensor

11

The difference of the line elements may be written in terms of the variables in the original coordinates (a1 , a2 , a3 ), or the deformed coordinates (x1 , x2 , x3 ) as follows ∂xα ∂xβ − δij )dai daj ∂ai ∂aj ∂aα ∂aβ dS 2 − dS02 = (δij − δαβ )dxi dxj ∂xi ∂xj dS 2 − dS02 = (δαβ

(1.6-7)

Denoting Green strain tensor in terms of the strains in the original coordinates as ∂xα ∂xβ 1 eij = (δαβ − δij ) (1.6-8) 2 ∂ai ∂aj and Almansi strain tensor in terms of the strains in the deformed coordinates as Eij =

1 ∂aα ∂aβ ) (δij − δαβ 2 ∂xi ∂xj

(1.6-9)

the difference of squares of length of the line elements in terms of the variables in either coordinates becomes dS 2 − dS02 = 2eij dai daj dS 2 − dS02 = 2Eij dxi dxj

(1.6-10)

The displacement vector u is now defined. From Fig. 1.6-1 r = r0 + u

(1.6-11)

u = r − r0

(1.6-12)

or

Figure 1.6-1: Displacement vector.

12


In components form ui = x i − a i

(1.6-13)

since ∂uα ∂xα = + δαi ∂ai ∂ai ∂uα ∂aα = δαi − ∂xi ∂xi

(1.6-14)

substituting into the expression for eij yields 1 ∂xα ∂xβ − δij ] [δαβ 2 ∂ai ∂aj ∂uα ∂uβ 1 + δαi )( + δβj ) − δij ] = [δαβ ( 2 ∂ai ∂aj 1 ∂uj ∂ui ∂uα ∂uα eij = [ + + ] 2 ∂ai ∂aj ∂ai ∂aj

eij =

(1.6-15)

The relation for Eij is 1 2 1 = 2 1 Eij = 2

Eij =

∂aα ∂aβ ] ∂xi ∂xj ∂uα ∂uβ [δij − δαβ (− + δαi )(− + δβj )] ∂xi ∂xj ∂ui ∂uα ∂uα ∂uj + − ] [ ∂xi ∂xj ∂xi ∂xj [δij − δαβ

(1.6-16)

For the infinitesimal theory of elasticity the displacement gradient uα,i is small, so that the quadratic terms in Eqs. (1.6-15) and (1.6-16) may be neglected and strain tensors eij and Eij are both reduced to the linear form as 1 (uj,i + ui,j ) (1.6-17) 2 For an infinitesimal strain, the difference between Eq. (1.6-15) and Eq. (1.616) are removed, and the partial derivatives are referred to the rectangular Cartesian coordinates. Here ij is a symmetric second order tensor called the strain tensor. The derivative of ui is a second order tensor, and it is ij =

1 1 ui,j = (ui,j + uj,i ) + (ui,j − uj,i ) 2 2

(1.6-18)

ui,j = ij − ωij

(1.6-19)

or

6. Displacement and Strain Tensor

13

where ωij is a skew symmetric tensor called the rotation tensor, ωji =

1 (ui,j − uj,i ) = −ωij 2

(1.6-20)

The rotation tensor ωij in the i, j plane can be represented by its corresponding vector ωk which is perpendicular to the i, j plane. These are related by 1 ekij ωij 2 ωij = eijk ωk

ωk =

(1.6-21)

In terms of the conventional coordinate system, the six strain-displacement relations reduce to ∂u ∂v ∂w yy = zz = ∂x ∂y ∂z 1 ∂u ∂v 1 ∂v ∂w = ( + ) yz = ( + ) 2 ∂y ∂x 2 ∂z ∂y

xx = xy

1 ∂w ∂u zx = ( + ) 2 ∂x ∂z (1.6-22)

We also introduce the following definitions γxy =

∂u ∂v + ∂y ∂x

γyz =

∂v ∂w + ∂z ∂y

γzx =

∂w ∂u + ∂x ∂z

(1.6-23)

The symmetric strain tensor ij has many properties similar to the stress tensor. For a strain tensor at a point of a body, there are three orthogonal principal axes and three principal planes in which the shear strains are zero on the principal planes, and the normal strains along the principal axes are called the principal strains. Mohr circle for strain is drawn in a similar way to Mohr circle for stress. The first invariant of strain tensor has a simple geometric meaning in the case of the infinitesimal strain. The first strain invariant is I1 = 11 + 22 + 33

(1.6-24)

Consider a cubic element of original size l1 , l2 , and l3 in an unstrained state. Let the sides of the element be parallel to the principal axes, and 11 , 22 , and 33 be the principal strains after deformation. In the strained state the length of edges of the element are l1 (1 + 11 ), l2 (1 + 22 ), and l3 (1 + 33 ), and remain orthogonal to each other. The volume change of the element due to straining is ΔV = l1 l2 l3 (1 + 11 )(1 + 22 )(1 + 33 ) − l1 l2 l3

(1.6-25)

Neglecting the infinitesimal terms of order two and three, ΔV = l1 l2 l3 (11 + 22 + 33 )

(1.6-26)

14


Thus

ΔV (1.6-27) V Therefore, the first invariant of the strain tensor represents the change in volume per unit volume. The quantity I1 is called the dilatation. The infinitesimal theory of elasticity and thermoelasticity is limited to the linear strain-displacement relations. For large deflection analysis, such as problems of buckling, the theory of finite elasticity based on nonlinear straindisplacement relations is used. I1 = 11 + 22 + 33 =

7

Compatibility Equations. Simply Connected Region

The compatibility conditions for a simply connected region are obtained by elimination of the displacement components between the strain-displacement relations. These conditions insure the continuity and single-valuedness of all three components of displacements in the continuum. The strain-displacement relations of the small deformation theory of elasticity are ij =

1 (ui,j + uj,i ) 2

(1.7-1)

Differentiating twice gives ij,kl =

1 (ui,jkl + uj,ikl ) 2

Interchanging the subscripts results in the following 1 (uk,lij + ul,kij ) 2 1 ik,jl = (ui,kjl + uk,ijl ) 2 1 jl,ik = (uj,lik + ul,jik ) 2 kl,ij =

From these relations, the compatibility equation follows as ij,kl + kl,ij − ik,jl − jl,ik = 0

(1.7-2)

This is the tensorial form of the compatibility equations in the rectangular Cartesian coordinates. From 81 equations represented by Eq. (1.7-2) only six equations are independent, and the remaining ones are either repetitions or are identical symmetries of ij and kl .

7. Compatibility Equations. Simply Connected Region

15

The six components of the compatibility equation in the rectangular Cartesian coordinates in conventional form are ∂ 2 xx ∂ 2 yy ∂ 2 xy + = 2 ∂y 2 ∂x2 ∂x∂y 2 ∂ 2 yz ∂ yy ∂ 2 zz + =2 ∂z 2 ∂y 2 ∂y∂z 2 2 ∂ zz ∂ xx ∂ 2 xz + = 2 ∂x2 ∂z 2 ∂x∂z 2 ∂ ∂xz ∂xy ∂yz ∂ xx = ( + − ) ∂y∂z ∂x ∂y ∂z ∂x ∂ ∂xy ∂yz ∂xz ∂ 2 yy = ( + − ) ∂x∂z ∂y ∂z ∂x ∂y ∂ 2 zz ∂ ∂xz ∂yz ∂xy = ( + − ) ∂x∂y ∂z ∂y ∂x ∂z

(1.7-3)

We will now prove that satisfying the compatibility equations is the necessary and sufficient condition for the continuity and single-valuedness of the displacement field in a simply connected region. A region is called simply connected if every simple closed curve within it can be continuously contracted into a point without having to leave the region at any stage. A simple closed curve is a closed, sectionally smooth curve which does not cross itself. A region which is not simply connected is called multiply connected. This definition of a simply connected region is valid for both two- and three-dimensional regions. Thus, a three-dimensional region contained between two concentric spheres is a simply connected region, while a two-dimensional region between two concentric circles is a multiply connected region. Consider a continuum of a simply connected region in the xi system. At a point P0 with coordinates x0i , the displacements u0i and the rotations ωij0 are specified. At an arbitrary neighboring point P¯ with coordinates x¯i the displacement ui can be obtained with a line integral joining point P0 with point P¯ along a single-valued curve C as [1] ui (¯ xl ) = u0i (x0l ) +

P¯ P0

dui = u0i (x0l ) +

P¯ ∂ui P0

∂xj

dxj

(1.7-4)

Since ui,j = ij − ωij , Eq. (1.7-4) can be written as ui (¯ xl ) = u0i (x0l ) +

P¯ P0

ij dxj −

P¯ P0

ωij dxj

(1.7-5)

The latter integral should be replaced by an integral in terms of the strain tensor ij . To do this, it is recalled that since P¯ is fixed with respect to the

16


integral, d¯ xj = 0, and adding it to the latter integral of Eq. (1.7-5) has no effect. With such consideration, the integration by parts yields P¯ P0

ωij (dxj − d¯ xj ) = (¯ xj − x0j )ωij0 +

P¯ P0

(¯ xj − xj )

∂ωij dxk ∂xk

(1.7-6)

But it is verified that ∂ωij ∂jk ∂ik = − ∂xk ∂xi ∂xj

(1.7-7)

Substituting Eq. (1.7-7) into Eq. (1.7-6) and finally into Eq. (1.7-5) gives ui (¯ xl ) = u0i (x0l ) − (¯ xj − x0j )ωij0 + where xj − xj )( Uik = ik − (¯

P¯ P0

Uik dxk

∂jk ∂ik − ) ∂xi ∂xj

(1.7-8)

(1.7-9)

Equation (1.7-8) relates the displacement components at point P¯ to those at point P0 . If the displacement components are to be unique and single-valued at an arbitrary point P¯ , the latter integral of Eq. (1.7-8) must be independent of the path. This requires that the integral around any closed curve C must vanish, that is C

Uik dxk = 0

(1.7-10)

The three integrals given by Eq. (1.7-10) are called Cesàro integrals. The path independency of function Uik is that it must be an exact differential. The condition for Uik to be an exact differential, using the Stoke’s theorem, is ∂Uik ∂Uim − =0 ∂xm ∂xk

(1.7-11)

This condition of single-valuedness is valid for a simply connected region. By substituting Eq. (1.7-9) into Eq. (1.7-11) it is readily verified that some terms in this equation cancel out, and the remaining terms, those containing (¯ xj −xj ), form exactly the compatibility equation (1.7-2) which is identically zero. Therefore, it is proved that the necessary and sufficient condition for the continuity and single-valuedness of displacement field in a simply connected region is that the compatibility equation be satisfied in region D and over its boundary C0 , see Fig. 1.8-1. For a multiply connected region, however, the compatibility equation is a necessary condition for the continuity and single-valuedness of the displacement field, but it is not sufficient. The sufficient condition for single-valuedness requires additional conditions which will be discussed in the next section.

8. Compatibility Conditions. Multiply Connected Regions

17

Figure 1.8-1: Simply and multiply connected plane regions.

8

Compatibility Conditions. Multiply Connected Regions

The continuity and single-valuedness of displacements for a simply connected region is assured as long as the compatibility equation is satisfied. However, satisfying the compatibility equation is a necessary condition for multiply connected region but this condition is not sufficient. Figure 1.8-1 shows two twodimensional regions, of which one is simply connected (the one on the left) and the other is multiply connected. If a function ui is to be single-valued, the integral in Eq. (1.7-8) between points P0 and P¯ must be independent of the path joining the points. To derive the proper rule, we confine ourselves temporarily to two-dimensional problems and consider the functions P (x, y) and Q(x, y) which are defined and continuous in a domain D. The line integral (P dx + Qdy) is said to be independent of the path in D if for every pair of end points P0 and P¯ in D the value of the line integral P¯

(P dx + Qdy)

(1.8-1)

P0

is the same for all paths C from P0 to P¯ . Then the value of the line integral depends upon the end points P0 and P¯ , but not the path joining P0 and P¯ . Theorem I If the line integral (P dx + Qdy) is independent of the path in D, then

(P dx + Qdy) = 0

(1.8-2)

C

on every simple closed path C in D. Theorem II If Eq. (1.8-2) holds for every simple closed path in D, then (P dx + Qdy) is independent of the path in D. Theorem III If P (x, y) and Q(x, y) have continuous partial derivatives in D and (P dx + Qdy) is independent of the path in D, then

18

Chapter 1. Basic Laws of Thermoelasticity ∂P ∂Q = ∂y ∂x

(1.8-3)

Theorem IV Let P (x, y) and Q(x, y) have continuous derivatives in D and let D be simply connected. If ∂Q ∂P = ∂y ∂x

(1.8-4)

in D, then (P dx + Qdy) is independent of the path in D. The proofs of the above theorems are given in any textbook of calculus, and will be omitted here. Now we only recall Green’s theorem for a simply connected region D in three dimensions. Consider a three-dimensional domain D with surface S bounded with the closed curve C0 . According to Green’s theorem, the integral over the curve C0 can be transformed into the integral in the domain D as

∂R ∂Q − ) cos (n, x) ∂z C0 D ∂y ∂R ∂Q ∂P ∂P − ) cos (n, y) + ( − ) cos (n, z)]ds +( ∂z ∂x ∂x ∂y (P dx + Qdy + Rdz) =

[(

(1.8-5)

The term cos (n, x) is the cosine of the angle between the normal vector n to the surface S and the x axis. The proof of the latter theorem follows from Green’s theorem. Now, the results may be extended to a multiply connected region. For simplicity, we confine ourselves to two-dimensional problems and consider the ∂Q ∂P or closed curve C of Green’s theorem which encloses a point at which ∂y ∂x fails to exist, and thus the theorem cannot be applied. The following theorem provides proper explanation for the line integral in this case. Theorem V Let P (x, y) and Q(x, y) be continuous and have continuous derivatives in a region D. Let R be a closed region in D whose boundary consists of N + 1 distinct simple closed curves C0 , C1 , C2 ,. . . , CN , where C0 includes C1 , C2 ,. . . , CN in its interior. Then

(P dx + Qdy) = C0

N

(P dx + Qdy)

(1.8-6)

i=1 Ci

The proof of this theorem is received by passing from a multiply connected region to a simply connected region by means of N cuts between the curves C0 , C1 , C2 ,. . . , CN . Applying Theorem I, the path’s independency of line integrals causes the line integrals cancel out each other and the remaining integrals are like in Eq. (1.8-6). When the region is multiply connected, the displacement components obtained from Eq. (1.7-8) are not single-valued. While satisfying the compatibility

8. Compatibility Conditions. Multiply Connected Regions

19

Figure 1.8-2: Multiply connected region with N cuts. equation is a necessary condition for continuity of the displacement field, it is not sufficient. The sufficiency condition of the single-valuedness of function ui in a multiply connected region will be now established. Consider a closed region D where its boundary consists of N + 1 distinct simple closed curves C0 , C1 , C2 ,. . . , CN . The boundary curve C0 includes C1 , C2 ,. . . , CN in its interior. This region is changed into a simply connected by N cuts, as indicated in Fig. 1.8-2. Cesàro integrals for such a region become [1] C

Uik dxk =

C0

Uik dxk −

N s=1 Cs

Uik dxk

(1.8-7)

Since the region is made simply connected, Eq. (1.7-10) or (1.8-6) are used to give

C0

Uik dxk =

N

s=1 Cs

Uik dxk

(1.8-8)

We set Cesàro integral equal to zero for each closed path in D, as

Cs

Uik dxk =

Cs

[ik − (¯ xj − xj )(

∂jk ∂ik − )]dxk = 0 ∂xi ∂xj s = 1, 2, ..., N

(1.8-9)

or Cs

Uik dxk =

Cs

[il + ejik eknm (¯ xj − xj )ln,m ]dxl = 0 s = 1, 2, ..., N

(1.8-10)

Once Eq. (1.8-10) is satisfied over each closed interior contour Cs of the region D, Cesàro integral for the path C0 , which encloses all the interior closed paths Cs , must necessarily also vanish.

20


There are similar conditions for the single-valuedness of the rotation function ωk . From Eqs. (1.6.19) and (1.6-21) [2] ui,j = ij + ejik ωk

(1.8-11)

The conditions for integration of Eq. (1.8-11) require that epmj (ij + ejik ωk ),m = epmj ij,m + epmj ejik ωk,m = 0

(1.8-12)

But recalling the following relationship between the permutation symbol and Kronecker symbol (1.8-13) eijk eklm = δil δjm − δim δjl Eq. (1.8-12) can be written as epmj ejik ωk,m = (δpi δmk − δpk δmi )ωk,m = δpi ωk,k − ωp,i

(1.8-14)

Substituting Eq. (1.8-14) into Eq. (1.8-12) and recalling that ωk,k = 0, gives −ωp,i + epmj ij,m = 0

(1.8-15)

Integrating this equation results in the following relationship for ωk at a point P¯ in terms of its value ωk0 at point P0 ωk = ωk0 +

P¯ P0

ekij lj,i dxl

(1.8-16)

For ωk to be single-valued the integral in Eq. (1.8-16) should vanish along the closed interior contours of the region Cs

ekij lj,i dxl = 0

s = 1, 2, . . . , N

(1.8-17)

Satisfaction of Eqs. (1.8-10) and (1.8-17) in a multiply connected region constitutes the sufficient conditions for single-valuedness of displacements and rotations of the multiply connected region. Equation (1.8-10) can be further simplified by noting that the quantities

x¯j

Cs

ejik ekmn ln,m dxl

in Eq. (1.8-10) should vanish when ωk is to be single-valued. Thus, Eq. (1.8-10) is simplified to the form Cs

(il − ejik ekmn xj ln,m )dxl = 0

s = 1, 2, ..., N

(1.8-18)

Equations (1.8-17) and (1.8-18) are called the three-dimensional Cesàro conditions.

9. Constitutive Laws of Linear Thermoelasticity

9

21

Constitutive Laws of Linear Thermoelasticity

In the classical theory of linear thermoelasticity the components of the strain tensor are linear functions of the components of the stress tensor and the components of the strain tensor due to the temperature change; that is [3,4] ij = eij + Tij

(1.9-1)

where eij denotes the elastic strain and Tij stands for the thermal strain. Consider a cubic element whose temperature is raised from the reference temperature T0 at which strains and stresses are zero, to the temperature T . The sides of the element are free from tractions. The thermal strain of the element due to the temperature change is Tij = α(T − T0 )δij

(1.9-2)

where T − T0 is the temperature change, and α is the coefficient of linear thermal expansion. The relation (1.9-2) represents a property of an isotropic body, in which a temperature change T − T0 results in no change of shear angles, the only result being a change of volume of the element. The elastic strain tensor is linearly proportional to the stress tensor as eij =

1 ν (σij − σkk δij ) 2G 1+ν

(1.9-3)

where G is the shear modulus and ν is Poisson’s ratio. Equation (1.9-3) is known as the constitutive law of linear elasticity, or Hooke’s law. From Eqs. (1.9-2) and (1.9-3) the total strain tensor is ij =

1 ν (σij − σkk δij ) + α(T − T0 )δij 2G 1+ν

(1.9-4)

Equation (1.9-4) is called the constitutive law of linear thermoelasticity. Solving this equation for the stress tensor σij gives σij = 2G[ij +

ν 1+ν (kk − α(T − T0 ))δij ] 1 − 2ν ν

(1.9-5)

It is sometimes useful to write the stress-strain relations in terms of Lamé constants λ and μ, where μ is the same as shear modulus G. In terms of Lamé constants, λ = 2Gν/(1 − 2ν) and μ = G, the strain tensor is related to the stress tensor by ij =

1 λ (σij − σkk δij ) + α(T − T0 )δij 2μ 3λ + 2μ

(1.9-6)

22


Solving for the stress tensor σij , gives σij = 2μij + [λkk − α(3λ + 2μ)(T − T0 )]δij

(1.9-7)

From Eq. (1.9-4), denoting the first invariant of the strain tensor by e = I1 = xx + yy + zz and the first invariant of the stress tensor by I1 = σxx + σyy + σzz , we obtain e − 3α(T − T0 ) =

1 1 − 2ν I1 2G 1 + ν

(1.9-8)

For an isotropic elastic material in a state of uniform temperature, Eq. (1.9-8) reduces to 1+ν e = 3Ke (1.9-9) I1 = 2G 1 − 2ν The constant K is called the bulk modulus. The quantity e, for the infinitesimal displacement field, is the change of volume per unit volume of the material, see Eq. (1.6-27). Thus, Eq. (1.9-9) relates the first invariant of the stress tensor to the first invariant of the strain tensor. For the special case of hydrostatic compression σxx = σyy = σzz = −p,

σxy = σyz = σzx = 0

(1.9-10)

the first invariant of the stress tensor is I1 = −3p. Substituting for e from Eq. (1.6-27) into Eq. (1.9-9), we obtain p ΔV =− (1.9-11) V K where V is the volume and ΔV is the change of volume. Equation (1.9-11) states that a material under hydrostatic compression is compressible, provided that the bulk modulus of the material is of a finite value. For incompressible materials, the bulk modulus K approaches infinity. For this special case, from Eqs. (1.9-11) and (1.9-9), ν = 1/2 and since G = E/2(1 + ν), G = E/3. The value of ν = 1/2 is an upper limit of Poisson’s ratio. The stress-strain relations in terms of Young’s modulus and Poisson’s ratio are frequently used, and they are xx = yy = zz = xy = yz = zx =

1 [σxx − ν(σyy + σzz )] + α(T − T0 ) E 1 [σyy − ν(σzz + σxx )] + α(T − T0 ) E 1 [σzz − ν(σxx + σyy )] + α(T − T0 ) E σxy 2G σyz 2G σzx 2G

(1.9-12)

10. Displacement Formulation of Thermoelasticity

23

The relationship between Lamé constants λ and μ, Young’s modulus E, Poisson’s ratio ν, and the bulk modulus K are [1] (see also the complete table in Hetnarski and Ignaczak [5], p. 144) Eν 2Gν G(E − 2G) 3Kν 3K(3K − E) = = = = (1 + ν)(1 − 2ν) 1 − 2ν 3G − E 1+ν 9K − E 2G =K− 3 E λ(1 − 2ν) 3 3KE 3K(1 − 2ν) μ=G= = = (K − λ) = = 2(1 + ν) 2ν 2 9K − E 2(1 + ν) λ(1 + ν)(1 − 2ν) G(3λ + 2G) E = 2G(1 + ν) = = = 3K(1 − 2ν) ν λ+G 9K(K − λ) 9KG = = 3K + G 3K − λ E λ 3K − E λ 3K − 2G ν= −1= = = = 2G 2(λ + G) 6K 3K − λ 2(3K + G) E 2G 2G(1 + ν) GE λ(1 + ν) K= =λ+ = = = (1.9-13) 3(1 − 2ν) 3 3(1 − 2ν) 3(3G − E) 3ν λ=

10

Displacement Formulation of Thermoelasticity

The equation of motion in terms of stresses was derived in Section 1.3, see Eq. (1.3-8) ui (1.10-1) σij,j + Xi = ρ¨ The stresses can be expressed in terms of strains and then in terms of displacements. Substituting for the strain tensor ij in terms of the displacement ui , Eq. (1.9-7) gives σij = μ(ui,j + uj,i ) + [λuk,k − α(3λ + 2μ)(T − T0 )]δij

(1.10-2)

Taking the partial derivative of Eq. (1.10-2) and substituting into the equilibrium equation (1.10-1) yields μui,kk + (λ + μ)uk,ki − (3λ + 2μ)αT,i + Xi = ρ¨ ui

(1.10-3)

Equation (1.10-3) is called Navier equation. It is expressed in terms of the displacement components along the three coordinate axes. The boundary conditions must be satisfied on the surface boundary of the body. If the traction components on the boundary are given as tni , Cauchy’s formula, see Eq. (1.3-2), tni = σij nj (1.10-4)

24


states the boundary conditions. Since, however, the problem formulation is in terms of displacements, the prescribed traction on the boundary can be related to the displacement components by Eqs. (1.10-4) and (1.10-2) as tni = {μ(ui,j + uj,i ) + [λuk,k − α(3λ + 2μ)(T − T0 )]δij }nj

(1.10-5)

Writing Eq. (1.10-5) in components yields ∂u ∂u ∂u ∂u ∂v ∂w + ny + nz ) + G(nx + ny + nz ) ∂x ∂y ∂z ∂x ∂x ∂x Eα(T − T0 ) nx − 1 − 2ν ∂v ∂v ∂v ∂u ∂v ∂w tny = λeny + G(nx + ny + nz ) + G(nx + ny + nz ) ∂x ∂y ∂z ∂y ∂y ∂y Eα(T − T0 ) ny − 1 − 2ν ∂w ∂w ∂w ∂u ∂v ∂w tnz = λenz + G(nx + ny + nz ) + G(nx + ny + nz ) ∂x ∂y ∂z ∂z ∂z ∂z Eα(T − T0 ) nz − (1.10-6) 1 − 2ν tnx = λenx + G(nx

where nx , ny , and nz are the cosine directions of the unit outer normal vector to the boundary, and e = xx + yy + zz is the first invariant of the strain tensor. The traction boundary conditions along with the thermal boundary conditions will fully define the displacement and temperature fields. It should be noted that the solution of Navier equation (1.10-3), simultaneously satisfies compatibility condition and the constitutive law and, therefore, is an acceptable solution of a problem of thermoelasticity. It is important to notice that this formulation is justified for simply connected bodies, where the compatibility condition is necessary and sufficient for single-valuedness of the solution. For a multiply connected domain, the solution of Eq. (1.10-3) must be obtained along with Cesàro condition of Section 1.8, see Eqs. (1.8-17) and (1.8-18).

11

Stress Formulation of Thermoelasticity

In many problems the boundary conditions are defined in terms of stresses. In this case it is more convenient to write the governing equations of thermoelasticity in terms of the stress tensor. The equations given in the following are derived for the dynamic problems of thermoelasticity [4].

11. Stress Formulation of Thermoelasticity

25

The governing equations of thermoelasticity in terms of displacement components and in the absence of body forces are, see Eq. (1.10-3) ui μui,kk + (λ + μ)uk,ki − (3λ + 2μ)αT,i = ρ¨

(1.11-1)

Using the strain-displacement relation ij =

1 (ui,j + uj,i ) 2

(1.11-2)

changing the index i in Eq. (1.11-1) into j and then differentiating it with respect to index i, and adding the results to Eq. (1.11-1) differentiated with respect to the index j, results in the following equation expressed in terms of the strain tensor μij,kk + (λ + μ)kk,ij − (3λ + 2μ)αT,ij = ρïj

(1.11-3)

Substituting for the strain from the stress-strain relations (1.9-6) ij =

1 λ (σij − σkk δij ) + α(T − T0 )δij 2μ 3λ + 2μ

(1.11-4)

into Eq. (1.11-3), yields 2(λ + μ) λ σkk,ij − σmm,kk δij + 2μα(T,ij + T,kk δij ) 3λ + 2μ 3λ + 2μ 1 λ 2μα σ ¨kk δij ) + 2 T¨δij = 2 (¨ σij − (1.11-5) c2 3λ + 2μ c2

σij,kk +

where c22 = μ/ρ. Contraction of Eq. (1.11-5) gives 221 σkk +

α(3λ + 2μ) (4μT,kk − 3ρT¨) = 0 λ + 2μ

where 221 = ∇2 −

1 ∂2 c21 ∂t2

c21 =

λ + 2μ ρ

(1.11-6)

(1.11-7)

Using Eq. (1.11-6), we transform Eq. (1.11-5) into the following final form of the stress compatibility equation of Beltrami-Michell type, where the body forces are absent [4,5] 2(λ + μ) λ 1 1 ¨kk δij σkk,ij + ( − )σ 3λ + 2μ 3λ + 2μ c22 c21 3λ + 2μ 5λ + 4μ +2μα[T,ij + T,kk δij ] − αρT¨δij = 0 λ + 2μ λ + 2μ 222 σij +

where 222 = ∇2 −

1 ∂2 c22 ∂t2

(1.11-8)

(1.11-9)

26


In terms of the operators 221 and 222 , Eq. (1.11-8) is simplified to the form 222 {221 σij −

α(3λ + 2μ) [2μ(T,ij − T,mm δij ) + ρT¨δij ]} = 0 λ + 2μ

(1.11-10)

For the thermoelastic static condition, this equation reduces to [σij,mm −

2μ(3λ + 2μ)α (T,ij − T,mm δij )],kk = 0 λ + 2μ

(1.11-11)

The stress-temperature field equation of the dynamic theory of thermoelasticity may be obtained by eliminating the displacement and strain tensors from the governing equations. When a body force exists, the derivative of the equation of motion (1.10-1), utilizing Eq. (1.11-2), leads to 1 (1.11-12) [σik,kj + σjk,ki + Xi,j + Xj,i ] = ρïj 2 Substituting for the components of the strain tensor from Eq. (1.11-4), Eq. (1.11-12) becomes ρ λ 1 1 [σik,kj + σjk,ki ] − [¨ σij − σ ¨kk δij ] = − [Xi,j + Xj,i ] + ραT¨δij 2 2μ 3λ + 2μ 2 (1.11-13) It may be shown that the stress tensor field of the dynamic theory of thermoelasticity, in addition to Eq. (1.11-13), should satisfy the stress compatibility equation of Beltrami-Michell type [5] 2(λ + μ) λ 1 1 ¨kk δij σkk,ij + ( 2 − 2) σ 3λ + 2μ 3λ + 2μ c2 c1 3λ + 2μ 5λ + 4μ + T,kk δij ] − αρT¨δij + Fij = 0 λ + 2μ λ + 2μ

222 σij + +2μα[T,ij where

Fij = Xi,j + Xj,i +

λ Xm,m δij λ + 2μ

(1.11-14)

(1.11-15)

For quasi-static problems, where the variation of temperature field with the time is slow, and in the absence of body forces, Eq. (1.11-5) reduces to σij,kk +

2(λ + μ) λ σkk,ij − σmm,kk δij + 2μα(T,ij + T,kk δij ) = 0 (1.11-16) 3λ + 2μ 3λ + 2μ

The contraction of Eq. (1.11-16) with respect to indices i and j, i.e., i = j, yields 4μ(3λ + 2μ)α (1.11-17) T,kk σmm,kk = − λ + 2μ

12. Two-Dimensional Thermoelasticity

27

Substituting σmm,kk from Eq. (1.11-17) into Eq. (1.11-16) yields 2(λ + μ) 3λ + 2μ σkk,ij + 2μα(T,ij + T,kk δij ) = 0 3λ + 2μ λ + 2μ

σij,kk +

(1.11-18)

The Eqs. (1.11-5) or (1.11-14) in components constitute a system of six equations governing the six components of stress. For multiply connected regions, these equations offer the necessary condition for single-valuedness of stress fields but they are not sufficient. For sufficiency, Cesàro integrals along the interior boundaries must be satisfied. These conditions were derived in Section 1.8 for single-valuedness of the displacement and rotation fields, respectively, in the form Cs

(il − ejik ekmn xj ln,m )dxl = 0

s = 1, 2, ..., N

(1.11-19)

Cs

ekij lj,i dxl = 0

s = 1, 2, ..., N

(1.11-20)

Using the stress-strain relation, the strain tensor is replaced by the stress tensor and the general Cesàro integral equations in terms of stresses take the form [6] 1+ν ν (σil − ejik ekmn xj σln,m )dxl − (σpp δil − ejik ekml xj σpp,m )dxl E Cs E Cs +α Cs

[(T − T0 )δil − ejik ekml xj T,m ]dxl = 0

s = 1, 2, ..., N (1.11-21)

and 1+ν ν ekmn σln,m dxl − ekml σpp,m dxl E E Cs Cs +α Cs

ekml T,m dxl = 0

s = 1, 2, ..., N

(1.11-22)

For a multiply connected domain the simultaneous consideration of Eqs. (1.1121) and (1.11-22) and the governing equations in terms of the stresses should provide the complete solution which assures satisfaction of the continuity conditions.

12

Two-Dimensional Thermoelasticity

In the classical theory of elasticity there are situations where a problem may be well approximated by the two-dimensional theory. These formulations are classified as either plane stress or plane strain states. Each class of problems is itself divided into simple or generalized plane stress, and simple or generalized

28


plane strain, respectively. The following discussion is devoted to the detailed understanding of each condition and its assumptions [7,8]. Simple plane stress state According to this formulation, the two-dimensional stresses are functions of the variables x and y, and the transverse stresses are zero, i.e., σyy = σyy (x, y) σxx = σxx (x, y) σzz = σxz = σyz = 0

σxy = σxy (x, y)

The stress-strain relations from Hooke’s law reduce to 1 xx = (σxx − νσyy ) + α(T − T0 ) E 1 yy = (σyy − νσxx ) + α(T − T0 ) E 1 xy = σxy 2G Solving for the stresses we receive E σxx = [(xx + νyy ) − α(1 + ν)(T − T0 )] 1 − ν2 E [(yy + νxx ) − α(1 + ν)(T − T0 )] σyy = 1 − ν2 E xy σxy = 1+ν The equations of motion for two-dimensional problems are ∂σxx ∂σxy ∂ 2u + +X =ρ 2 ∂x ∂y ∂t ∂2v ∂σxy ∂σyy + +Y =ρ 2 ∂x ∂y ∂t

(1.12-1)

(1.12-2)

(1.12-3)

(1.12-4)

The compatibility equations (1.7-3) are all identically satisfied except the equation ∂ 2 xy ∂ 2 xx ∂ 2 yy (1.12-5) + = 2 ∂y 2 ∂x2 ∂x∂y Substituting the strain-displacement relations (1.6-22) into Eqs. (1.12-3) and finally into Eq. (1.12-4) results in the following equations of motion in terms of displacements for the plane stress state 1 + ν ∂ ∂u ∂v ∂ 2u ∂ 2u Eα ∂T ∂2u + ) + G ( + ) − + X = ρ ∂x2 ∂y 2 1 − ν ∂x ∂x ∂y 1 − ν ∂x ∂t2 ∂2v ∂2v 1 + ν ∂ ∂u ∂v Eα ∂T ∂2v G( 2 + 2 ) + G ( + )− +Y =ρ 2 ∂x ∂y 1 − ν ∂y ∂x ∂y 1 − ν ∂y ∂t (1.12-6) G(


29

Substituting the strains from Eq. (1.12-2) into the compatibility equation (1.12-5), gives ∂2 ∂2 ∂ 2 σxy 2 (σ − νσ ) + (σ − νσ ) + Eα∇ T = 2(1 + ν) xx yy yy xx ∂y 2 ∂x2 ∂x∂y

(1.12-7)

Differentiating the first of Eqs. (1.12-4) with respect to x and the second of Eqs. (1.12-4) with respect to y and adding gives ∂ 2 σxx ∂ 2 σyy ∂X ∂Y ∂ u¨ ∂¨ v ∂ 2 σxy + + + − ρ( + ) = −2 ∂x2 ∂y 2 ∂x ∂y ∂x ∂y ∂x∂y

(1.12-8)

Eliminating σxy between Eqs. (1.12-7) and (1.12-8) gives (

∂2 ∂X ∂Y ∂2 ∂ u¨ ∂¨ v + )(σxx +σyy )+Eα∇2 T = −(1+ν)[ + −ρ( + )] (1.12-9) ∂x2 ∂y 2 ∂x ∂y ∂x ∂y

Either Eqs. (1.12-6) or (1.12-9) are the governing equations of two-dimensional simple plane stress problem. The solution for the stresses must satisfy the following boundary conditions tnx = nx σxx + ny σyx tny = nx σxy + ny σyy tnz = 0

(1.12-10)

where n is the unit outer normal vector to the boundary and nx and ny are the cosine directions of n. Generalized plane stress state Consider a thin flat plate of thickness 2h. The origin of the coordinate axes is fixed to the middle plane of the plate such that the axes x and y are laying in the middle plane, and the z-axis is perpendicular to the middle plane. All the loadings are applied in planes parallel to both x and y axes, so the faces of the plate are free from traction. The loadings and the body forces X and Y are assumed to be symmetrically distributed with respect to the middle plane. The component of the body force along the z-axis is assumed to be zero. As the result of such loadings, the displacement along the z-axis of points on the middle plane, w, is zero, and at all other points is very small so that it may be ignored compared to the other two components of displacement u and v along the x and y axes. In addition, the variations of u and v across the thickness are assumed to be small. Consequently, the u and v components of displacement may be assumed to be the average displacements through the thickness as [9] u¯(x, y) =

1 h u(x, y, z)dz 2h −h

30

Chapter 1. Basic Laws of Thermoelasticity 1 h v(x, y, z)dz 2h −h 1 h w(x, y, z)dz = 0 w(x, ¯ y) = 2h −h v¯(x, y) =

(1.12-11)

where the bar quantities indicate the average values through the thickness. The last expression for w¯ is zero due to the symmetry assumption of w across the thickness. Since the faces of the plate are traction free, the transverse stresses must vanish at ±h, σzz (x, y, ±h) = σxz (x, y, ±h) = σyz (x, y, ±h) = 0

(1.12-12)

From this assumption, it follows that ∂σxz (x, y, ±h) ∂σyz (x, y, ±h) = =0 ∂x ∂y

(1.12-13)

From the third of equations of motion, it follows that ∂σzz (x, y, ±h) =0 ∂z

(1.12-14)

Since σzz and its derivative with respect to z are zero on the faces of the plate, and the plate is thin, it is well justified to assume that σzz is zero everywhere across the thickness of the plate. The average values of the other stress components across the thickness are 1 h σxx dz σ ¯xx = 2h −h h 1 σ ¯yy = σyy dz 2h −h h 1 σxy dz σ ¯xy = 2h −h

(1.12-15)

Also, the mean values of the body forces across the thickness are h ¯= 1 X Xdz 2h −h 1 h Y¯ = Y dz 2h −h

(1.12-16)

and Z¯ = 0. The equations of equilibrium in terms of the mean values are obtained by integration over the thickness, and they are ¯yx ∂σ ¯xx ∂ σ ¯ =0 + +X ∂x ∂y ∂σ ¯xy ∂ σ ¯yy + + Y¯ = 0 ∂x ∂y

(1.12-17)


31

The mean values of the stress-strain relations from Hooke’s law are 1 ¯yy ) + α(T¯ − T0 ) (¯ σxx − ν σ E 1 ¯yy = ¯xx ) + α(T¯ − T0 ) (¯ σyy − ν σ E 1 ¯xy = σ ¯xy 2G

¯xx =

where

1 h T¯(x, y) = T (x, y, z)dz 2h −h

(1.12-18)

(1.12-19)

Solving Eqs. (1.12-18) for the stresses, we obtain E [(¯xx + ν¯yy ) − α(1 + ν)(T¯ − T0 )] 1 − ν2 E = [(¯yy + ν¯xx ) − α(1 + ν)(T¯ − T0 )] 1 − ν2 E = ¯xy 1+ν

σ ¯xx = σ ¯yy σ ¯xy

(1.12-20)

These equations are similar to Eqs. (1.12-3), but for the average values. The equations of equilibrium and compatibility equations for the average values are likewise similar to those presented earlier for the simple plane stress. Simple plane strain state In the case of simple plane strain condition the displacement components u and v are assumed to be independent of the z-coordinate, and w = 0 u = u(x, y)

v = v(x, y)

w=0

(1.12-21)

The strain-displacement relations are ∂u ∂x ∂v yy = yy (x, y) = ∂y 1 ∂v ∂u + ) xy = xy (x, y) = ( 2 ∂x ∂y zz = xz = yz = 0 xx = xx (x, y) =

(1.12-22)

From Hooke’s law, for zz = 0, we have σzz = ν(σxx + σyy ) − Eα(T − T0 )

(1.12-23)

32


Using Eq. (1.12-23), the stress-strain relations reduce to 1 − ν2 ν [σxx − σyy ] + α(1 + ν)(T − T0 ) E 1−ν 1 − ν2 ν = [σyy − σxx ] + α(1 + ν)(T − T0 ) E 1−ν 1 = σxy 2G

xx = yy xy

(1.12-24)

The equations of motion and the compatibility equations remain the same as given by Eqs. (1.12-4) and (1.12-5). The equations of motion in terms of displacements are obtained using a similar procedure, and for the plane strain state they are 1 ∂ ∂u ∂v Eα ∂T ∂2u ∂ 2u ∂ 2u + ) + G ( + ) − + X = ρ ∂x2 ∂y 2 1 − 2ν ∂x ∂x ∂y 1 − 2ν ∂x ∂t2 ∂ ∂u ∂v Eα ∂T ∂2v ∂ 2v ∂ 2v 1 ( + )− +Y =ρ 2 G( 2 + 2 ) + G ∂x ∂y 1 − 2ν ∂y ∂x ∂y 1 − 2ν ∂y ∂t (1.12-25) G(

The compatibility equation in terms of the stresses for the plane strain problem is obtained to be (

∂2 ∂2 Eα 2 1 ∂X ∂Y ∂ u¨ ∂¨ v + )(σxx + σyy ) + ∇ T =− [ + − ρ( + )] ∂x2 ∂y 2 1−ν 1 − ν ∂x ∂y ∂x ∂y (1.12-26)

Generalized plane strain state The problem of axially unrestrained plane strain deformation, such that the axial force is zero and the axial strain is constant, is called the general plane strain problem [10,11]. Let us consider a problem in which the z-axis coincides with the axial direction. From Hooke’s law, the stress-strain relation along the z-axis is 1 zz = (1.12-27) [σzz − ν(σxx + σyy )] + α(T − T0 ) E Due to the assumption of zero axial load, the integration of axial stress σzz over the area should vanish, i.e.,

Fz =

A

σzz dxdy = 0

(1.12-28)

Integrating Eq. (1.12-27) over the area, with the consideration of Eq. (1.12-28), yields σxx + σ ¯yy ) + EαT¯ (1.12-29) Ezz = −ν(¯


33

where σ ¯xx =

1 σxx dxdy A A

σ ¯yy =

1 σyy dxdy A A

1 T¯ = (T − T0 )dxdy A A (1.12-30)

¯yy = 0, and Eq. (1.12-29) When external mechanical loads are zero, σ ¯xx = σ results in (1.12-31) Ezz = EαT¯ Substituting into Eq. (1.12-27), the axial stress is σzz = ν(σxx + σyy ) + Eα(T¯ − T + T0 )

(1.12-32)

For an axisymmetric plane strain condition, such as a thick-walled cylinder, we have, relative to the cylindrical coordinates (r, φ, z), [6,7,10] σrr + σφφ =

Eα ¯ (T − T + T0 ) 1−ν

(1.12-33)

and when there is no axial load σzz =

Eα ¯ (T − T + T0 ) 1−ν

(1.12-34)

Thus, for this special case, the generalized plane strain state gives σzz = σrr + σφφ

(1.12-35)

When the body forces and inertia terms are absent, the governing equations may be reduced to a more simple and convenient form by introducing Airy stress function. After introducing Airy stress function Φ defined as [1] σxx =

∂ 2Φ ∂y 2

σyy =

∂ 2Φ ∂x2

σxy = −

∂2Φ ∂x∂y

(1.12-36)

the equations of motion (1.12-4) are identically satisfied and Eq. (1.12-9) for plane stress condition reduces to (

∂2 ∂2 + )(σxx + σyy ) + Eα∇2 T = 0 ∂x2 ∂y 2

(1.12-37)

Substituting for stresses from Eq. (1.12-36), the governing equation in terms of Airy stress function for the plane stress condition reduces to ∇2 ∇2 Φ + Eα∇2 T = 0

(1.12-38)

Similarly, when the body forces and inertia terms are zero, the compatibility equation (1.12-26) for the plane strain condition reduces to (

∂2 Eα 2 ∂2 + )(σxx + σyy ) + ∇ T =0 2 2 ∂x ∂y 1−ν

(1.12-39)

34


Figure 1.12-1: Two-dimensional domain. and by introducing Airy stress function, this equation is further simplified for the plane strain state to the form ∇2 ∇2 Φ +

Eα 2 ∇ T =0 1−ν

(1.12-40)

The stress function Φ is obtained through the integration of Eqs. (1.12-38) or (1.12-40), knowing the distribution of the temperature. The constants of integration in the expression for the stress function are obtained by application of the known physical traction or kinematical boundary conditions of the problem. Consider a two-dimensional domain D with the boundary S as shown in Fig. 1.12-1. An element of the boundary is ds and the unit outer normal vector is n. The angle between n and the x-axis is φ. The cosine directions of n are nx = cos (n, x) = cos φ = dy/ds and ny = cos (n, y) = sin φ = −dx/ds. For pure thermal stresses, the traction forces on the boundary are considered to be zero, and from Eqs. (1.12-10) and (1.12-36) ∂ 2Φ ∂ 2 Φ ∂y + ∂y 2 ∂s ∂x∂y ∂ 2 Φ ∂y ∂ 2 Φ + ∂x∂y ∂s ∂x2

∂x ∂ ∂Φ = ( )=0 ∂s ∂s ∂y ∂x ∂ ∂Φ = ( )=0 ∂s ∂s ∂x

(1.12-41)

The integration yields ∂Φ =A ∂x

∂Φ =B ∂y

where A and B are constants. Integrating the stress function Φ over the boundary S yields

Φ=

S ∂Φ 0

∂s

S

(A

ds = 0

∂x ∂y + B )ds = Ax + By + C ∂s ∂s

(1.12-42)

13. Michell Conditions

35

where C is a constant. The partial derivative of Φ with respect to the normal direction to the boundary is ∂x ∂y ∂Φ =A +B (1.12-43) ∂n ∂n ∂n Since the stresses are related to the second derivative of the stress function, the constants A, B, and C may be assumed to be zero. Thus, from Eq. (1.12-42) and (1.12-43), the boundary condition for Φ is ∂Φ =0 (1.12-44) ∂n This boundary condition is valid for any problem solved by means of Airy stress function. It is recalled that the governing equations and the conditions obtained in this section are restricted to simply connected regions. The general and particular solutions to Eq. (1.12-38) or (1.12-40) were suggested by P.F. Papkovich, and they will be discussed in Chapter 3. Φ=

13

Michell Conditions

The three-dimensional Cesàro conditions expressing the sufficiency and singlevaluedness of displacements and rotations, ui and ωi , can be reduced to twodimensional problems of a multiply connected region. J.H. Michell, however, derived similar conditions for two-dimensional problems from an entirely different approach and without recourse to the reduction from the three-dimensional theory [12]. It was proved that the compatibility condition in terms of Airy stress function for the plane stress condition is [c.f., Eq. (1.12-38)] ∇2 ∇2 Φ + Eα∇2 T = 0

(1.13-1)

This equation is the necessary and sufficient condition for the displacement field to be continuous and single-valued within a simply connected region. For a multiply connected region, where the body is made of N distinct boundaries, the above equation is a necessary but not a sufficient condition. To obtain the additional required conditions, a region of main boundary C0 and several other boundaries within the region, such as Ci , i = 1, 2, . . . , N in Fig. 1.13-1, may be considered. In this region, with external and body forces applied to the stresses to be unique and single-valued, the strain components must be single-valued. In twodimensional problems the strains are related to the displacement components by the formulas xx =

∂u ∂x

yy =

∂v ∂y

1 ∂u ∂v xy = ( + ) 2 ∂y ∂x

(1.13-2)

36


Figure 1.13-1: Two-dimensional multiply connected region. Strain components, being continuous and single-valued, do not necessarily imply continuity and single-valuedness of the displacement field. The additional conditions required, besides the compatibility equation, to insure the singlevaluedness of the displacement field, are the single-valuedness of the rotation ωz , and the u-component and the v-component of the displacement. The rotation vector is defined as 1 ∂v ∂u ωz = ( − ) (1.13-3) 2 ∂x ∂y or ∂u ∂v ωz = − (1.13-4) + xy = − xy ∂y ∂x ∂u ∂v or must be continuous. Since xy is continuous, for ωz to be continuous ∂y ∂x ∂u requires that According to Cauchy’s integral theorem, continuity of ∂y

d( Ci

∂u )=0 ∂y

i = 1, 2, . . . , N

(1.13-5)

But

∂u ∂ ∂u ∂ ∂u )= ( )dx + ( )dy ∂y ∂x ∂y ∂y ∂y Using definitions of the strains and Eq. (1.13-4) d(

(1.13-6)

∂xx ∂xy ∂yy ∂u )= dx + (2 − )dy ∂y ∂y ∂y ∂x

(1.13-7)

d(

and the stress-strain relations 1 xx = (σxx − νσyy ) + α(T − T0 ) E 1 yy = (σyy − νσxx ) + α(T − T0 ) E 1+ν xy = σxy E

(1.13-8)

13. Michell Conditions

37

Equation (1.13-7) becomes ∂u 1 d( ) = ∂y E

∂σxx ∂σyy ∂σyy ∂σyy + − −ν ]dx ∂y ∂y ∂y ∂y ∂σxy ∂σyy ∂σxx ∂σxy + 2ν − +ν ]dy +[2 ∂y ∂y ∂x ∂x ∂T ∂T +Eα( dx − dy) ∂y ∂x [

Using the equations of motion, Eqs. (1.12-4), and the definition of Airy stress function, Eq. (1.12-36), this equation becomes

1 ∂ 2 ∂u ∂ 2 ∂ ∂2Φ ∇ Φdy − ∇ Φdx + (1 + ν)( dx d( ) = − ∂y E ∂x ∂y ∂x ∂x∂y ∂ ∂2Φ ∂T ∂T + (1.13-9) dy) + Eα( dy − dx) ∂y ∂x∂y ∂x ∂y but since ∂ ∂ dx ∂ dy = + ∂s ∂x ds ∂y ds ∂ ∂ dx ∂ dy ∂ dy ∂ dx = + = − ∂n ∂x dn ∂y dn ∂x ds ∂y ds

(1.13-10)

it follows that −d(

1 ∂ ∂T ∂2Φ ∂u ) = [ ∇2 Φds + (1 + ν)d ] + α ds ∂y E ∂n ∂x∂y ∂n

(1.13-11)

and, finally, single-valuedness of the rotation requires that 1 ∂ 2 ∂T 1+ν ∂2Φ − ∇ Φds − )−α ds = 0 d( E Ci ∂n E ∂x∂y Ci Li ∂n

(1.13-12)

∂2Φ is single-valued, and the second ∂x∂y integral automatically vanishes. The remaining condition reduces to Since stresses are single-valued, σxy = −

∂T 1 ∂ 2 ∇ Φds + α ds = 0 E Ci ∂n Ci ∂n

(1.13-13)

The second condition is the single-valuedness of u which requires that

du = Ci

( Ci

∂u ∂u dx + dy) = 0 ∂x ∂y

(1.13-14)

Integrating by parts along a closed curve Ci we obtain

du = Ci

∂u P0 ∂u P0 ∂u ∂u x + y − [xd( ) + yd( )] = 0 ∂x P0 ∂y P0 Ci ∂x ∂y

(1.13-15)

38


The first two terms are zero, and the last term, using the definition for strains, becomes

(x Ci

∂xx ∂xx ∂xx ∂xy ∂yy +y )dx + [x + y(2 − )]dy = 0 ∂x ∂y ∂y ∂y ∂x

(1.13-16)

Substituting for strains from Hooke’s law, Eq. (1.13-8), and using Airy stress function ∂∇2 Φ ∂T ∂∇2 Φ ∂T 1 (x −y )ds − α (x −y )ds E Ci ∂s ∂n ∂s ∂n Ci ∂2Φ ∂ 2Φ 1+ν [xd( 2 ) + yd( )] = 0 (1.13-17) 2+ E ∂x ∂x∂y Ci

−

The integration by parts of the third term yields

∂ 2Φ ∂2Φ ∂ 2 Φ P0 ∂ 2 Φ ∂ 2Φ ∂ 2 Φ P0 ) + yd( + y − [ dx + )] = x dy] ∂x2 ∂x∂y ∂x2 P0 ∂x∂y P0 ∂x∂y Ci Ci ∂x2 dx dy = − (σyy dx − σxy dy) = − (σyy − σxy )ds ds ds Ci Ci [xd(

= Ci

(σyy ny + σxy nx )ds =

Ci

tny ds

Substitution of the last term into Eq. (1.13-17) gives the second condition in the form ∂∇2 Φ ∂T ∂∇2 Φ ∂T 1 (x −y )ds + α (x −y )ds E Ci ∂s ∂n ∂s ∂n Ci 1+ν n − ty ds = 0 i = 1, 2, . . . , N E Ci

(1.13-18)

where tny is the component of the surface force acting on the boundary in the y-direction. The third condition comes from the single-valuedness of the v-component of displacement, that is dv = 0 (1.13-19) Ci

which, by a similar treatment as that of the condition (1.13-14), results in the following condition 1 ∂∇2 Φ ∂T ∂∇2 Φ ∂T 1+ν n (y tx ds = 0 +x )ds + α (y + x )ds + E Ci ∂s ∂n ∂s ∂n E Ci Ci i = 1, 2, . . . , N (1.13-20)

where tnx is the x-component of the force acting on the boundary. The three equations (1.13-13), (1.13-18), and (1.13-20) are known as Michell continuity

14. Problems

39

conditions and together with the compatibility equations constitute all the necessary and sufficient conditions for the displacement field to be continuous and single-valued for a multiply connected region. Michell conditions must be satisfied along each individual inner boundary curve of the region. The same procedure may be followed for the plane strain condition, provided that E, ν, and α are replaced by E/(1 − ν 2 ), ν/(1 − ν), and α(1 + ν), respectively.

14

Problems

1. Show by direct transformation from the coordinate system (x, y, z) to another coordinate system (m, n, l) that the mean stress and shear stress intensity, as defined bellow, are invariants. σxx + σyy + σzz = σmm + σnn + σll 2 2 2 (σxx − σyy ) + (σyy − σzz )2 + (σzz − σxx )2 + 6(σxy + σyz + σzx ) 2

2 2 2 + σnl + σlm ) = (σmm − σnn )2 + (σnn − σll )2 + (σll − σmm )2 + 6(σmn

2. Find the radial displacement in the case of the strain with spherical symmetry. 3. Find the radial displacement in the case of small axisymmetric strains, where the displacement uz in the direction of z-axis is assumed to be zero. How do the results change if uz is constant? 4. Derive Eqs. (1.11-5), (1.11-10), (1.11-11), (1.11-13), and (1.11-15). 5. Derive Eqs. (1.11-20) and (1.11-21). 6. Derive Michell conditions in polar coordinates.

Bibliography [1] Fung, Y.C., Foundations of Solid Mechanics, Prentice Hall, Englewood Cliffs, New Jersey, 1965. [2] Kovalenko, A.D., Thermoelasticity, Basic Theory and Applications, Wolters-Noordhoff, Groningen, The Netherlands, 1969. [3] Carlson, D.E., Linear Thermoelasticity, Encyclopedia of Physics, Vol. VIa/2, C. Truesdell, ed., Springer, Berlin, 1972. [4] Nowacki, W., Thermoelasticity, 2nd edition, PWN-Polish Scientific Publishers, Warsaw, and Pergamon Press, Oxford, 1986. [5] Hetnarski, R.B. and Ignaczak, J., Mathematical Theory of Elasticity, Taylor and Francis, New York, 2004. [6] Boley, B.A. and Weiner, J.H., Theory of Thermal Stresses, Wiley, New York, 1960. [7] Noda, N., Hetnarski, R.B., and Tanigawa, Y., Thermal Stresses, 2nd edition, Taylor and Francis, New York, 2003. [8] Nowinski, J.L., Theory of Thermoelasticity with Applications, Sijthoff and Noordhoff, Alpen aan den Rijn, 1978. [9] Saada, A.S., Elasticity, Theory and Applications, Pergamon Press, New York, 1974. [10] Fridman, Y.B., Strength and Deformation in Nonuniform Temperature Fields, Consultants Bureau, New York, 1964. [11] Burgreen, D., Elements of Thermal Stress Analysis, C.P. Press, New York, 1971. [12] Michell, J.H., On the Direct Determination of Stress in an Elastic Solid, with Application to the Theory of Plates, Proc. London Math. Soc., Vol. 31, pp. 100–124, 1899.

41

Chapter 2 Thermodynamics of Elastic Continuum A new presentation of the thermodynamic principles for solid elastic continuum is given. The first and the second laws of thermodynamics in variational form are stated, and the variational principle of thermodynamics in terms of entropy follows. The principle of thermoelasticity linearization is discussed and the classical coupled as well as the generalized (with second sound effect) theories are derived using the linearization technique. A unique generalized formulation, considering Lord–Shulman, Green–Lindsay, and Green–Naghdi models, for the heterogeneous anisotropic material is presented, where the formulation is properly reduced to those of isotropic material. The uniqueness theorem and the variational form of the generalized thermoelasticity are derived, and the exposition of the Maxwell reciprocity theorem concludes the chapter.

1

Introduction

The basic laws of thermodynamics are discussed in this chapter and their extension to deformable bodies is presented. Restricting ourselves to the deformations of elastic bodies in the range of small deformation theory of elasticity, the deformation process falls into the discussion of conservative systems and the reversible thermodynamic processes, and these will constitute the main topics of this chapter. The first and second laws of thermodynamics are discussed and the general statement of variational law of thermodynamics in terms of entropy of reversible and irreversible systems are analyzed. The thermodynamics of elastic continuum and the related uniqueness theorem are then presented. The general laws governing coupled thermoelasticity problems are given next and the discussion is extended to the coupled thermoelasticity with the second sound effect. R.B. Hetnarski, M.R. Eslami, Thermal Stresses – Advanced Theory and Applications, Solid Mechanics and its Applications 158, c Springer Science + Business Media B.V. 2009

43

44

2

Chapter 2. Thermodynamics of Elastic Continuum

Thermodynamics Definitions

To describe the general laws of thermodynamics, the following terms are used in the text. A continuum is a medium with a geometric configuration in the Euclidean space which exhibits continuous macroscopic properties. The neighboring molecules remain neighboring under the action of any loading conditions and no geometric discontinuity may occur. A part of a continuum which is separated from the rest of the continuum is called a system. A closed system does not exchange matter with its surroundings. If a system does not interact with the surroundings, it is called an isolated system. The control volume is a system which has a defined boundary. This boundary may be fixed or deformable, and the continuum may cross the boundaries. The state of a system is defined when all the information needed to characterize the system is known. This information is expressed in terms of a property. The properties are the macroscopic characteristics of the system defined by means of statistical averaging procedures. The properties are sometimes called the state variables, and for a system they are the density, specific volume, specific pressure, temperature, velocity, enthalpy, entropy, and internal energy. If a property depends upon mass, such as volume and energies of all kinds, it is called an extensive property. On the other hand, if it is independent of mass, such as temperature, pressure, velocity, etc., it is called an intensive property. The specific values of extensive properties, the values per unit mass, are called intensive properties. The process is the change of any state of the system or control volume. For a thermodynamic process to proceed from a given state to a final state, the functional relationship between the properties must be known. When a property of the system is expressed by a single-valued mathematical relationship to a set of other system properties, the mathematical relation is called the equation of state. The equation of state helps to define the state variables of a system or control volume during a thermodynamic process. It is also used to define all the other state variables of a system or control volume at the equilibrium condition. The proper choice of state variables in the equation of state is arbitrary. In mechanics of a deformable body, the equation of state is called the constitutive law, and it relates the state of stress to the state of strain. In any process of solid deformable materials, in the sense of deformation under loading conditions, the constitutive laws are the essential tool in defining the state of equilibrium [1,2]. The thermodynamic equilibrium of a system or control volume occurs when the values of all the state variables are independent of the time. In contrast, when the state variables vary with the time the system is undergoing a process. A cycle is a process whose initial and final states are identical. The number of properties needed to describe a process is normally larger than the number of properties required to define the system or control volume at equilibrium. For example, in plasticity problems the state of stress and total strain fully define

3. First Law of Thermodynamics

45

the state of equilibrium, while during the plastic loading, other properties, such as temperature and internal variables, are required. This information must be included in the constitutive law of the material. Work is a form of energy which is transferred by the boundaries of a system or control volume and is defined as positive if the system or control volume is capable of doing work on the surroundings. On the other hand, work done on a system or control volume is defined as negative. Heat is a form of energy which is transferred through the boundaries of a system or control volume during a process due to the difference of the temperature with surroundings. The heat transferred to a system or control volume is considered positive, and heat transferred from a system or control volume is considered negative. Work and heat are not properties. Thermal insulation is a method that prevents the transfer of heat to or from a system or control volume. For a system with thermal insulation any process that takes place is called adiabatic. At this point it is worthwhile to define the concept of temperature which is an intensive property of the system. When several systems are in thermal equilibrium, a single-valued function which is identical between the systems must describe the thermal equilibrium and identify the quality of the equilibrium. This function may be called temperature and it is identical among systems in thermal equilibrium with each other. The choice of a temperature scale is arbitrary. The temperature whose definition is so postulated and which is measured on a scale defined on an arbitrary thermometer is called the empirical temperature.

3

First Law of Thermodynamics

A system can interact with its surroundings and may transfer energy to the surrounding medium. This energy may be in the form of heat or work. Denoting the first variation of a function in terms of its variables by δ, the first law of thermodynamics states that the first variation of the heat absorbed by a system Q is equal to the increase of the differential of internal energy change U minus the first variation of the work done on the system W , all referred to unit volume, i.e., δQ = dU − δW (2.3-1) We assume the mathematical meaning of δ-operation as path dependent, with the energy interactions between two end states depending upon the end states as well as the path of variations. On the other hand, d-operation in mathematical sense is the variation of the function in terms of all the variables involved in the function, including the time. Note that δ-operation does not apply to the time domain, while d-operation does.

46


For an adiabatic system where the transferred heat through the boundaries of the system is zero, the first law of thermodynamics reduces to the conservation of mechanical energy dU − δW = 0

(2.3-2)

For a system which completes a cyclic process, the first law reduces to δQ + δW = 0

(2.3-3)

that is, the sum of the first variation of the heat absorbed by a system and the first variation of the work done on the system is zero. The net change of internal energy is zero, because according to the definition of a cyclic process the initial and final state variables are identical. Equation (2.3-1) is a statement of conservation of energy and, therefore, it is a principal law of mechanics. The total internal energy per unit volume U is related to several other types of energies. The most conventional form is to relate it to the kinetic energy K, the gravitational energy G, and the internal energy E, all per unit volume, as [2] U =K +G+E

(2.3-4)

This concept may be extended to a system on a macroscopic or microscopic scale.

4

Second Law of Thermodynamics

The first law of thermodynamics as stated by Eqs. (2.3-1) to (2.3-3) does not show the direction of the process. This law is quantitive rather than directional. To fully define the process of energy transfer, a law defining the direction of energy transfer must be stated. The second law of thermodynamics defines the direction of the process. Consider a system under a thermodynamic process. Two distinct processes are considered, reversible and irreversible. When a system can reach its initial state from the final state through the same path of intermediate states which were passed from the initial state to the final state, the process is called reversible. Otherwise, the process is called irreversible. According to the second law of thermodynamics, when a system completes a cycle the following inequality holds

δQ ≤0 (2.4-1) T where T is the absolute temperature. This is called Clausius inequality. The equality sign refers to the reversible process. The important conclusion of

4. Second Law of Thermodynamics

47

this inequality is the direction of a thermodynamical process. An irreversible process of a system along a closed path takes place in such a way that the quantity presented on the left-hand side of Eq. (2.4-1) is always negative. For a reversible process δQrev =0 (2.4-2) T The concept of entropy as a thermodynamic property which was introduced by Clausius [3] and in earlier work by Rankine [4,5], directly follows from Eq. (2.4-2). The entropy change of a reversible cycle is dS =

δQrev T

(2.4-3)

where S is a thermodynamic property. The entropy change of a system may be observed along a closed cycle. Consider a system which transfers from state (1) to state (2) through an arbitrary process while the return path from state (2) to state (1) takes place as a reversible process. From Eq. (2.4-2) 2 δQ 1

From Eq. (2.4-3)

T

1 δQrev

+

1 δQrev 2

T

≤0

(2.4-4)

1

dS = (S1 − S2 ) (2.4-5) T 2 The term (S1 − S2 ) is called the entropy change, and it is a thermodynamic property. Substituting into Eq. (2.4-4) =

2

2 δQ 1

T

≤ S2 − S1

(2.4-6)

The left-hand side is the entropy transfer from state (1) to state (2). Equation (2.4-6) states that the entropy transfer of a thermodynamic process can never exceed its entropy change. Equation (2.4-4) with the aid of Eq. (2.4-6) can be stated as follows Sgen = (S2 − S1 ) −

2 δQ 1

T

≥0

(2.4-7)

where Sgen is the generated entropy in the defined closed cycle. This inequality states that in an arbitrary closed cycle, the thermodynamic process take places in such a way that the total entropy generation is never negative. Equation (2.4-7) can be modified for a thermodynamic differential path. The total entropy change of a differential path of a system dS is divided into two parts: dSe – the part of entropy exchange of the system with environment, and dSi – the entropy change inside the system [2] dS = dSe + dSi

(2.4-8)

48


The entropy exchange of the system with environment is related to the absorbed heat by δQ (2.4-9) dSe = T For the reversible process dSi is zero, while for the irreversible process dSi is always positive, dSi = 0 dSi > 0

for reversible process for irreversible process

(2.4-10)

and dSi < 0 never occurs in nature. For any thermodynamic state, two independent properties will define the system. The temperature and the entropy, called the fundamental properties, can fully define the thermodynamic state. While the temperature is an intensive property of the system, the entropy is an attribute of the matter and thus is an extensive property.

5

Variational Formulation of Thermodynamics

Consider a thermodynamic system at a static equilibrium. According to definition, the static thermodynamic equilibrium is a situation in which none of the state variables vary with the time. In addition, in the state of thermodynamic equilibrium no changes occur in boundary conditions of the system and thus no spontaneous state associated with the change of boundary conditions may occur. The variational formulation of the state of equilibrium will be based on the second law of thermodynamics and the direction of possible entropy change. The mathematical treatment is presented in the sense of variational calculus. We denote the system at equilibrium by A and the virtual state by B [2]. The thermodynamic properties of the system A are U , S, σij , ij , and T while the same properties of the virtual state B are U + δU , S + δS, σij + δσij , ij + δij , T + δT . From the first law of thermodynamics any virtual state should satisfy the equation

Since δW = σij δij System A at equilibrium U, S, σij , ij , T

δQ = dU − δW

(2.5-1)

δQ = dU − σij δij

(2.5-2)

Virtual system B neighboring to A (U + dU ), (S + δS), (σij + δσij ) (ij + δij ), (T + δT )

5. Variational Formulation of Thermodynamics

49

We will first focus on a system with rigid boundaries under a virtual state at constant internal energy. Thus δij = 0 and dU = 0. From the first law it follows that δQ = 0. Now the virtual system B has the entropy S + δS compared with system A with the entropy S at thermodynamic equilibrium. If we assume that δS > 0 it follows that the entropy of virtual system S + δS is larger than that of the system at equilibrium. Now, if any spontaneous changes can occur in system B to bring it to the equilibrium state A, they should occur in the direction of decreasing entropy. No such change is permitted by the second law of thermodynamics. Only spontaneous virtual processes with increasing entropy can bring the system B to the equilibrium state A. This means that the entropy of the system A at equilibrium is maximum with respect to all the virtual states B. Therefore, the variational formulation at thermodynamic equilibrium can be stated S = max

(U = const,

ij = const)

(2.5-3)

In terms of the first variation of entropy, the necessary and sufficient condition for a thermodynamic system to be at equilibrium is δSU ≤ 0

(2.5-4)

which states that for an isolated system to be in thermodynamic equilibrium, the entropy variation for all possible paths is either zero or negative provided that the internal energy is kept constant. An alternative statement of thermodynamic equilibrium of an isolated system is based on energy. An isolated system is in thermodynamic equilibrium if for all possible variations its energy is either zero or positive, provided that the entropy remains constant: (2.5-5) δUS ≥ 0 That is, a thermodynamic isolated system is in the state of equilibrium if its energy is at relative minimum with respect to all neighboring states, where the entropy content of the system is kept constant, or U = min

(S = const)

(2.5-6)

The laws stated in Eqs. (2.5-3) and (2.5-6) are the most general statements of the rules of mechanics for a system in equilibrium. This system may be referred to the state of dynamic equilibrium or static equilibrium of a continuum, and it can range from a compressible and an incompressible continuum at motion to a deformable solid continuum under general deformation process. These rules constitute the most important laws of nature and show the direction of events in the present cycle of nature.

50

6


Thermodynamics of Elastic Continuum

Consider a continuous elastic solid under thermal and mechanical loads. The foundations of thermoelasticity of elastic solids are based on the general laws of classical thermodynamics with the consideration of the principle of conservation of energy. For an elastic body, the six components of strain tensor ij and the temperature T , two state properties, constitute a complete set of state variables which define the state of the system. Once these two state properties are defined at each point of the elastic continuum, the system is completely defined and the other state variables become known. We recall that once a certain state variable is expressed as a single-valued function of a set of other state variables, then this functional relationship is called the equation of state. Equation (2.3-4) suggests that the kinetic energy of the system can be separated from the rest of the internal energy. Thus, it is assumed that [6,7] U =I +K

(2.6-1)

where I is called the intrinsic energy per unit volume. Substituting into the first law of thermodynamics, yields δQ + δW = dI + dK

(2.6-2)

Since heat and work are not state variables and are path dependent, δQ and δW denote small increments in a short time t. From the second law of thermodynamics for a reversible process δQ = T dS

(2.6-3)

The total work done on the system by the external surface traction forces tni and the body forces Xi is = δW

A

(tni dui )dA +

V

(Xi dui )dV

(2.6-4)

where ui are the displacement components. Using Green’s integral theorem and Cauchy’s formula, the surface integration is replaced by integration over the volume, and thus = δW

V

(σij dij + ρ¨ ui dui )dV

(2.6-5)

where σij,j +Xi is replaced by ρ¨ ui . The work done by the forces per unit volume is thus ui dui δW = σij dij + ρ¨

(2.6-6)

The kinetic energy per unit volume of the body under the action of surface and body forces is 1 (2.6-7) K = ρu˙ i u˙ i 2

6. Thermodynamics of Elastic Continuum

51

It is easily verified that dK = ρ¨ ui dui

(2.6-8)

δW − dK = σij dij

(2.6-9)

From Eqs. (2.6-6) and (2.6-8)

Substituting into the first law gives σij dij = dI − δQ

(2.6-10)

dI − T dS = σij dij

(2.6-11)

S = S(ij , T ) I = I(ij , T )

(2.6-12)

Using Eq. (2.6-3) yields Now, let us select

as the fundamental equations of the strained system, where S and I are the entropy and the intrinsic energies per unit volume, respectively. Once the system is at elastic equilibrium, two independent state variables, such as strain tensor and temperature, will fully define the state of the system and, therefore, Eqs. (2.6-12) represent a set of single-valued functions throughout the continuum. Differentiating Eqs. (2.6-12) yields dS = S,ij dij + S,T dT dI = I,ij dij + I,T dT

(2.6-13)

Note that dI − T dS = d(I − T S) + SdT

(2.6-14)

Thus, we obtain dI − T dS = (I − T S),ij dij + (I − T S),T dT + SdT

(2.6-15)

Let us introduce Helmholtz free energy per unit volume F F = I − TS

(2.6-16)

It follows that the free energy, as a state variable, is a function of the strain tensor and the temperature, and thus from Eqs. (2.6-11) and (2.6-14) F,ij dij + F,T dT = σij dij − SdT

(2.6-17)

From this equation the following results are obtained ∂F ∂ij ∂F S=− ∂T

σij =

(2.6-18)

52


Thus, for an isothermal process there exists a scalar function F whose partial derivative with respect to strain tensor gives the corresponding stress tensor. Now, introducing Gibbs thermodynamic potential G G = F − σij ij = I − T S − σij ij

(2.6-19)

it follows that ∂G ∂σij ∂G S=− ∂T ij = −

(2.6-20)

That is, if Gibbs thermodynamic potential is given, the strain tensor can be obtained by partial differentiation of this function with respect to the stress tensor. It further follows that once the intrinsic internal energy is known, Eq. (2.6-11) yields ∂I ∂ij ∂I T = ∂S σij =

(2.6-21)

It is worth comparing the foregoing conclusions with the case of isothermal linear elasticity, where the state of stress is fully defined by the given state of strain or vise versa. In the theory of linear elasticity the differential of strain energy function per unit volume of the material is defined as 1 (2.6-22) dU0 = σij dij 2 Geometrically, this relation represents the area under the stress-strain curve in the stress-strain space. The stress tensor is obtained from the partial differentiation of the strain energy function with respect to the strain tensor, that is, σij =

∂U0 ∂ij

(2.6-23)

The area above this curve is called the complementary strain energy function per unit volume and is shown as γ0 , see Fig. 2.6-1. Once the function γ0 is known, the strain tensor is obtained from the partial differentiation of γ0 with respect to the stress tensor as ij =

∂γ0 ∂σij

(2.6-24)

At this point we may generalize the above conclusion to the reversible adiabatic process of elastic bodies. The free energy function F for a reversible

6. Thermodynamics of Elastic Continuum

53

σ

γ0 U0

ε

Figure 2.6-1: Strain energy and the complementary strain energy. adiabatic process is the proper generalization of strain energy function U0 of isothermal deformation of linear elastic continuum. Similarly, Gibbs thermodynamic potential function G for a reversible adiabatic deformation of elastic continuum replaces the complementary strain energy function γ0 of isothermal deformation. Our attention is now focused on the definition of specific heat of elastic continuum. Let the path y represent a defined thermodynamic process. According to definition, the specific heat cy on the path y is defined as

cy =

1 δQ ρ dT

= y

T dS ρ dT

(2.6-25) y

or

T cy = ρ

∂S dij ∂S + ∂ij dT ∂T

(2.6-26) y

For the particular case of constant strain tensor process, the path y coincides with dij = 0 and the specific heat cy is called c , which is known as the specific heat at constant strain. From Eqs. (2.6-26) and (2.6-18) it follows that c =

T ∂2F T ∂S =− ρ ∂T ρ ∂T 2

(2.6-27)

From Eq. (2.6-18) we have ∂S ∂ 2F ∂σij =− =− ∂ij ∂ij ∂T ∂T

(2.6-28)

Substituting Eqs. (2.6-27) and (2.6-28) into Eq. (2.6-26) yields cy = −

T ∂σij dij + c ρdT ∂T

(2.6-29)

54


On the other hand, since ij = ij (σij , T ), dij at constant stress is ∂ij dT (2.6-30) ∂T Substituting into Eq. (2.6-29), the specific heat at constant stress is found to be T ∂σij ∂ij (2.6-31) + c cσ = − ρ ∂T ∂T This equation relates the specific heat at constant stress and the strain for the general solid elastic continuum. dij =

7

General Theory of Thermoelasticity

In the derivations of thermoelastic governing equations discussed in the preceding sections, it was assumed that the deformation and temperature change, compared to undeformed original dimensions of the body and the reference temperature, were small. These assumptions made it possible to assume linearity of the governing equations, and thus the principle of superposition holds true. The linearity assumption precludes applications to cases of large deformations occurring in buckling and post-buckling of thin structures, large deformations of composite materials, and flutter problems. The linearity assumption in temperature changes is violated whenever variation in temperature exceeds 30o C to 40o C in a structure under non-isothermal conditions, see Parkus [8]. These conditions frequently occur in structural design problems. The nonlinear governing equation for large deformation of a structure is of the kinematical type and takes into account the full nonlinear Green or Almansi strain tensors. These strain tensors were defined in Chapter 1, see Eq.(1.6-15), where Green strain tensor was the strain measured with respect to the original undeformed configuration of the body as

1 eij = 2

∂uj ∂ui ∂uα ∂uα + + ∂ai ∂aj ∂ai ∂aj

(2.7-1)

Green strain tensor is associated with the second Piola-Kirchhoff stress tensor sij . The related nonlinear equation of motion in terms of Piola-Kirchhoff stress tensor is, see Parkus [8] [sij (δkj + uk,j )],i + ρ0 bk = ρ0 u¨k

(2.7-2)

where bi is the body force component per unit mass, and ρ0 is the mass density in the undeformed state, related to the mass density ρ in the deformed state by

ρdV = V

V0

ρ0 dV0

(2.7-3)

7. General Theory of Thermoelasticity

55

The relation between the volume element in deformed and undeformed conditions are obtained using the general rules of the coordinate transformation between general curvilinear coordinates. Since the discussion of the tensors in general curvilinear coordinates is excluded from this book, we will just give a brief discussion in this section and refer to Fung [2] for more detailed information. The volume element in original undeformed coordinates is dV0 = da1 da2 da3 . In the deformed coordinates system dV = dS1 ·(dS2 ×dS3 ) = g1 .(g2 × g3 )da1 da2 da3 , where gi are the base vectors defined in the general coordinates. The final result for the volume element between two coordinate systems is √ (2.7-4) dV = gdV0 where g = |gij | = |δij + 2eij |

(2.7-5)

Here gij is the Euclidean metric tensor, a principal tensor of rank two defined in coordinate transformation between two general coordinate systems. Substituting Eq. (2.7-4) into Eq. (2.7-3) gives ρ0 ρ= √ g

(2.7-6)

Returning again to the energy equation (2.6-2), the rate of the first law of thermodynamics in integral form and in terms of the quantities per unit mass may be written as [8–10] d d 1 u˙ j u˙ j dm + idm = bj u˙ j dm + tnj u˙ j dA dt m dt m m A 2

+ m

rdm −

A

Qj nj dA

(2.7-7)

where i is the intrinsic energy per unit mass, bj is the body force per unit mass, tnj is the traction force, r is the heat produced per unit time and unit mass, and Qj is the heat flux through the surface of the body being positive outward in positive direction of the unit outward normal vector nj . The terms on the lefthand side of Eq. (2.7-7) are the rate of change of kinetic and intrinsic energies. The first two terms in the right-hand side of this equation are the rates of the work done by all the body and external traction forces, the third term is the heat produced per unit time inside the body, and the fourth term is the heat transported into the body from an external source. According to the principle of rate of work, the rate of change of kinetic energy is equal to the rate of all forces including the external and internal forces, that is, see Parkus [8], d 1 bj u˙ j dm + tnj u˙ j dA − sij e˙ ij dV0 u˙ j u˙ j dm = dt m 2 m A V0

(2.7-8)

56


where V0 refers to the volume of the body at its initial configuration. Substituting Eq. (2.7-8) into Eq. (2.7-7) yields V0

sij e˙ ij dV0 −

A

Qi ni dA =

m

(i˙ − r)dm

(2.7-9)

Using Gauss theorem, as indicated below, for an arbitrary function fi A

fi ni dA =

V

1 √ √ (fi g),i dV g

(2.7-10)

and with the aid of Eq. (2.7-4) and dm = ρ0 dV0 , Eq. (2.7-9) is finally reduced to (2.7-11) sij e˙ ij − qi,i = ρ0 (i˙ − r) √ where qi = Qi g is the heat flux vector per unit area of the undeformed body within the body. To complete the system of the governing equations, the second law of thermodynamics demanding positive production of entropy is added. In terms of Clausius inequality, this law states that [11] r Qi ni d sdm ≥ dm − dA dt m m T A T

(2.7-12)

where s is the entropy per unit mass. From Gauss theorem (2.7-10) and Eq. (2.7-4) qi ρ0 (T s˙ − r) + qi,i − ( ) T,i ≥ 0 (2.7-13) T Equations (2.7-1), (2.7-2), (2.7-5), (2.7-6), and (2.7-11) and the inequality (2.713) constitute the complete system of equations governing the general nonlinear theory of thermoelasticity. Introducing Helmholtz’s free energy function for unit mass f = i − Ts

(2.7-14)

and assuming that f, s, qi , and sij are functions of eij , T , and T,i , the constitutive laws governing the material behavior may be obtained. From Eq. (2.7-11) and the inequality (2.7-13), the following relations are derived [8]: ∂f ∂eij ∂f s=− ∂T ∂f =0 ∂T,i ˙ qi,i = ρ0 (r − T s) qi T,i ≤ 0 sij = ρ0

(2.7-15) (2.7-16) (2.7-17) (2.7-18) (2.7-19)

8. Free Energy Function of Hookean Materials

57

Equations (2.7-15) and (2.7-16) are the constitutive laws for stress and entropy, respectively, and Eq. (2.7-18) is the energy equation. Since it was assumed that qi = qi (eij , T, T,i )

(2.7-20)

we may refer to this equation as the general constitutive law for the heat flux vector and, as noted, it does not depend upon the free energy function and by itself is an independent equation. The linearized theory follows from the assumption that the field variables are infinitesimal, so that their second and higher order terms may be neglected in the governing equations. We begin to linearize starting from Eqs.(2.7-1) and (2.7-2). The linearized strain-displacement relations and the equation of motion for the case of infinitesimal deformations are, respectively, 1 (ui,j + uj,i ) 2 + Xi = ρuï

ij =

(2.7-21)

σij,j

(2.7-22)

where Xi = ρbi is the body force component per unit volume, and sij and eij are reduced to Cauchy stress tensor σij and the linear strain tensor ij , respectively. The energy equation (2.7-18) reduces to qi,i = ρ(r − sT ˙ 0)

(2.7-23)

where the distinction between ρ0 and ρ in the original and deformed coordinates is eliminated, and T0 is the initial uniform temperature. Expanding f and qi in Maclaurin’s series and substituting the results in Eqs. (2.7-15) and (2.7-16), with the consideration of Eq. (2.7-17), results in the following linear equations: σij = Cijkl kl − βij (T − T0 ) ρc (T − T0 ) + βij ij ρs = S = T0 qi = −kij T,j

(2.7-24) (2.7-25) (2.7-26)

These equations are the basic constitutive laws used in the linear theory of thermoelasticity.

8

Free Energy Function of Hookean Materials

In this section we will derive the free energy function of a general Hookean material. It is recalled that the free energy function is the proper generalization

58


of the strain energy function of isothermal deformation and thus, similarly to the derivation of the strain energy, the free energy will be derived based on the known stress-strain relations [6,7]. Consider a general Hookean material with the stress-strain relations ij = Dijkl σkl + αij (T − T0 )

(2.8-1)

σij = Cijkl kl − βij (T − T0 )

(2.8-2)

or where Dijkl , Cijkl , αij , and βij are the general anisotropic material constants. The free energy function is found from the integration of Eq. (2.6-18) with the known stress-strain relations. From Eqs. (2.6-18) and (2.8-2) ∂F = C11kl kl − β11 (T − T0 ) ∂11 Integrating with respect to 11 yields 1 F = C1111 211 + 11 [C1122 22 + C1133 33 2 +C1112 12 + C1121 21 + C1113 13 + C1131 31 +C1123 23 + C1132 32 − β11 (T − T0 )] +f1 (22 , 33 , 12 , 21 , 13 , 31 , 23 , 32 , T ) σ11 =

(2.8-3)

(2.8-4)

where f1 (., T ) is a function independent of 11 . From Eqs. (2.6-18), (2.8-2), and (2.8-4) we obtain ∂F ∂f1 = σ22 = C22kl kl − β22 (T − T0 ) = C1122 11 + ∂22 ∂22 Equation (2.8-5) is satisfied provided that C2211 = C1122 and

(2.8-5)

∂f1 = C2222 22 + C2233 33 + C2212 12 + C2221 21 ∂22 +C2213 13 + C2231 31 + C2223 23 + C2232 32 − β22 (T − T0 )(2.8-6) Integrating Eq. (2.8-6) with respect to 22 gives 1 f1 = C2222 222 + 22 [C2233 33 + C2212 12 + C2221 21 2 +C2213 13 + C2231 31 + C2223 23 + C2232 32 − β22 (T − T0 )] +f2 (33 , 12 , 21 , 13 , 31 , 23 , 32 , T ) (2.8-7) where f2 (., T ) is a function independent of 11 and 22 . Substituting Eq. (2.8-7) into Eq. (2.8-4), yields 1 1 F = C1111 211 + C2222 222 − β11 (T − T0 )11 − β22 (T − T0 )22 2 2 +11 (C1122 22 + C1133 33 + · · · + C1132 32 ) +22 (C2233 33 + C2212 12 + · · · + C2232 32 ) +f2 (33 , 12 , 21 , 13 , 31 , 23 , 32 , T ) (2.8-8)

8. Free Energy Function of Hookean Materials

59

If this procedure is repeated for all the strain components, the free energy function results in 1 (2.8-9) F = Cijkl ij kl − βij (T − T0 )ij + f9 (T ) 2 where f9 (T ) is a function of temperature alone and once f9 is obtained, the free energy is fully determined. From Eqs. (2.6-18) and (2.8-2) we obtain ∂2F ∂S ∂σij =− = −βij = ∂T ∂T ∂ij ∂ij

(2.8-10)

Substituting Eqs. (2.8-10) and (2.6-27) into the first of Eqs. (2.6-13) gives ρc dS = βij dij + dT (2.8-11) T Integrating Eq. (2.8-11) from the reference state at T0 , where S = 0, F = 0, σij = 0, and ij = 0, to the state at T , gives S = βij ij + ρc ln (

T ) T0

(2.8-12)

On the other hand, from Eqs. (2.6-18) and (2.8-9) S = βij ij −

∂f9 ∂T

(2.8-13)

∂f9 Comparing Eqs. (2.8-12) and (2.8-13), is obtained, which upon integration ∂T yields T (2.8-14) f9 = −ρc [T ln ( ) − (T − T0 )] T0 Substituting into Eq. (2.8-9), the expression for free energy function becomes T 1 F = Cijkl ij kl − βij (T − T0 )ij − ρc [T ln ( ) − (T − T0 )] 2 T0

(2.8-15)

Gibbs thermodynamic potential and the intrinsic energy are obtained from Eqs. (2.6-19) and (2.6-16) as T 1 G = − Cijkl ij kl − ρc [T ln ( ) − (T − T0 )] 2 T0 1 I = Cijkl ij kl + T0 βij ij + ρc (T − T0 ) (2.8-16) 2 For the isotropic elastic continuum the constants βij , αij , and Cijkl are reduced to the expressions β11 = β22 = β33 = α(3λ + 2μ) = β α11 = α22 = α33 = α Cijkl = λδij δkl + (δik δjl + δil δjk )μ

(2.8-17)

60


and the free energy function becomes 1 T F = λii kk + μij ij − β(T − T0 )ii − ρc [T ln ( ) − (T − T0 )] 2 T0

9

(2.8-18)

Fourier’s Law and Heat Conduction Equation

The heat balance for an element of a body relating the components of the heat flux vector q to the rate of specific heat influx Q is [1] Q˙ = −qi,i

(2.9-1)

According to Fourier’s law of heat conduction, the heat flux across a unit surface in unit time into a continuum is related to the temperature distribution through the following empirical law qi = −kij T,j

(2.9-2)

where kij is the coefficient of thermal conduction for a general anisotropic material. From the second law of thermodynamics δQ = T dS

(2.9-3)

Using the rate form of Eq. (2.9-3) and from Eqs. (2.6-13), (2.6-27), and (2.810), the following equation is obtained dS Q˙ = T S˙ = T dt ∂S dT ∂S dij + ) = T( ∂ij dt ∂T dt dT dij = T βij + c ρ dt dt

(2.9-4)

From Eqs. (2.9-1) and (2.9-2) the above equation yields θ c ρT˙ = (kij T,j ),i − T0 (1 + )βij ˙ij T0

(2.9-5)

where θ = T − T0 is the temperature change. This equation is the most general form of the first law of thermodynamics. It describes the law of thermal balance in a heterogeneous anisotropic solid continuum. For a homogeneous isotropic solid this equation reduces to θ c ρT˙ = kT,ii − T0 (1 + )β ˙ii T0

(2.9-6)

10. Generalized Thermoelasticity, Second Sound

61

If the temperature change θ is small compared to the reference temperature T0 , Eq. (2.9-6) may be approximately written in the simpler form c ρT˙ = kT,ii − T0 β ˙ii

(2.9-7)

In most of the practical cases of coupled thermoelasticity, Eq. (2.9-7) is precise enough to be used in conjunction with the thermoelasticity equations for the solution of coupled temperature and stress fields. If in addition to the heat flux through the boundary of a solid continuum there is also the heat generated within the body at a rate of R per unit volume and unit time, Eq. (2.9-7) is modified to give kT,ii = c ρT˙ − R + T0 β ˙ii

(2.9-8)

In Eqs. (2.9-7) or (2.9-8) the field of temperature is coupled with the field of strain and, therefore, any attempt to define the temperature distribution within the body should be with simultaneous consideration of thermoelasticity equations. This situation occurs when the rate of application of a thermomechanical load is rapid enough to produce thermal stress waves. However, in most practical engineering problems the thermomechanical load is of the steady-state nature or vary sufficiently slowly with the time so as not to excite inertia effects. Such problems are called quasi-static and quasi-stationary. Neglecting the strain rate in Eq. (2.9-8), the first law of thermodynamics for heat conduction in a solid continuum reduces to kT,ii = cρT˙ − R

(2.9-9)

where c = c is the specific heat of the material. For a steady-state condition the heat conduction Eq. (2.9-9) is further reduced to kT,ii = −R

(2.9-10)

This equation along with the imposed boundary conditions fully defines the field of temperature distribution in the solid body.

10

Generalized Thermoelasticity, Second Sound

When the time rate of change of thermal boundary conditions on a structure, or the time rate of change of thermal sources in a structure, is comparable with the structural vibration characteristics, thermal stress waves are produced. Under these circumstances the solution of the problem for the fields of stresses and temperature must be obtained through the coupled equations of

62


thermoelasticity. In terms of the displacement components and the temperature and for an isotropic solid, the governing equations are, see Eqs. (1.10-3) and (2.9-8) ui μui,kk + (λ + μ)uk,ki + Xi − (3λ + 2μ)αT,i = ρ¨ ˙ kT,ii − ρcT = T0 β ˙ii − R

(2.10-1) (2.10-2)

The thermoelasticity equations (2.10-1) and (2.10-2) are of a mixed hyperbolicparabolic type. To obtain the solution for the temperature and the displacements, and finally the stresses, Eqs. (2.10-1) and (2.10-2) must be solved simultaneously. Due to the mathematical properties of the mixed equations, a part of the effects produced by an external is felt immediately at all points of the body. In other words, if an isotropic and homogeneous elastic body is exposed to a thermal shock, the effect of shock reaches instantaneously to all points from the point of application. For coupled thermoelastic problems this means that the effect of thermal shock influences the pattern and magnitude of stresses and temperature instantaneously everywhere in the solution domain. Physically, it means that the speed of propagation of a thermoelastic disturbance is infinite. Since this behavior is physically unacceptable and inconsistent with the heat transfer mechanisms, the theory based on Eqs. (2.10-1) and (2.10-2) must be modified with an improved mathematical model. On this basis, improved models with the finite speed of wave propagation have been proposed. The improved theories make use of a modified Fourier’s law of heat conduction or modified constitutive law and entropy expression. The theories based on a modified Fourier’s law lead to the governing equations of the hyperbolic type which admit finite speeds of thermoelastic disturbances. According to such theories, the heat propagation is viewed as a wave phenomenon rather than a diffusion process. The wave type heat propagation is referred to as second sound, while the first sound is the usual sound (wave). The theories involving the second sound effect are motivated by experiments and exhibit the actual occurrence of second sound at low temperatures and for small time intervals, especially at the beginning of the application of a thermal shock. The effects of second sound in the generalized thermoelastic solids are in general small and short lived. To overcome the mathematical discrepancy associated with the infinite speed of propagation of thermoelastic disturbances in the classical coupled thermoelasticity, the improved theories which admit finite speed for thermoelastic signals have been formulated either by incorporating a heat fluxrate term into Fourier’s law or by including the temperature-rate among the constitutive variables. The resulting theories are referred to as the coupled thermoelasticity with second sound effect, or the generalized theory of coupled thermoelasticity.


63

Lord–Shulman (LS) model The basis of the model proposed by Lord and Shulman in 1967 [12] is to modify Fourier’s law of heat conduction by introducing the concept of a relaxation time. The rate form of the first law of thermodynamics for a thermally conducting deformable body in the state of small strains and temperature change for unit mass is σij ˙ij + ρT s˙ = ρi˙ (2.10-3) where s and i are the entropy and intrinsic energy densities, respectively, and qi,i = −ρT s˙

(2.10-4)

According to the principle of positive entropy production in a deformable elastic solid, the heat flux qi and the temperature T cannot be in an arbitrary functional relationship. According to Fourier’s law of heat conduction, this relationship for a general anisotropic solid is qi = bT,i + Bij T,j

(2.10-5)

where b and Bij are material constants. For an isotropic elastic solid this relation reduces to k>0 (2.10-6) qi = −kT,i According to Lord and Shulman theory, the most general, tensorially valid, linear relation between qi and T,i , that takes into account the rate of the heat flow is [12] (2.10-7) qi + aq˙i + Aij q˙j = bT,i + Bij T,j where a and Aij are positive material constants. For an isotropic material this relation reduces to (2.10-8) qi + t0 q˙i = −kT,i where t0 is the relaxation time, representing the time-lag needed to establish steady state heat conduction in a volume element when a temperature gradient is suddenly imposed on the element [12]. Helmholtz’s free energy function f (ij , T ) from Eq. (2.7-14) for unit mass is (2.10-9) f (ij , T ) = i(ij , T ) − T s(ij , T ) The derivative of f with respect to time is df ∂f ˙ ∂f T ˙ij + = dt ∂ij ∂T

(2.10-10)

64


From Eqs. (2.6-18) the entropy is related to the free energy function f by ∂f ∂ij ∂f s=− ∂T

σij = ρ

(2.10-11)

Substituting the second of Eq. (2.10-11) into Eq. (2.10-4) gives qi,i = −ρT s˙ = ρT

∂2f d ∂f ∂ 2f ( ) = ρT ( 2 T˙ + ˙ij ) dt ∂T ∂T ∂ij ∂T

(2.10-12)

From Eqs. (2.10-8) and (2.10-12) the energy equation in terms of the free energy function per unit mass is defined as

2 ∂2f ˙ ¨) + ∂ f (˙ij + t0 ïj ) T ( T + t kT,ii = −ρT 0 ∂T 2 ∂ij ∂T 2 3 ∂ 2f ∂3f ˙ )( ∂ f + 2T ∂ f ) T −ρt0 (T˙ )2 ( 2 + T ) + ( ˙ ij ∂T ∂T 3 ∂ij ∂T ∂ij ∂T 2

3 ∂ f +(˙ij )2 (T 2 (2.10-13) ) ∂ij ∂T

The second bracketed term on the right-hand side of Eq. (2.10-13) represents nonlinear terms of temperature and strain, and may be neglected in the framework of the linear thermoelasticity theory. Hence, the energy equation in terms of the free energy function per unit mass reduces to kT,ii = −ρT [

∂2f ∂2f ˙ ¨ T ) + ( T + t (˙ij + t0 ïj )] 0 ∂T 2 ∂ij ∂T

(2.10-14)

Defining the dimensionless temperature change as (T − T0 ) θ¯ = T0

(2.10-15)

and according to the definition of the specific heat at constant strain from Eq. (2.6-27) ∂2f (2.10-16) c = −T ∂T 2 where f is the free energy per unit mass. The free energy function is expanded into a power series in terms of the three strain invariants and the dimensionless ¯ according to Boley and Weiner [13], and the stresstemperature change θ, strain-temperature relation from the first of Eqs. (2.10-11) becomes σij = a1 δij + a2 (2I1 δij ) + a3 (I1 δij − ij ) ¯ ij ) + higher order terms +a7 (θδ

(2.10-17)


65

where I1 and I2 are the first and second invariants of the strain tensor and ai ’s are the coefficients in the series for f . The energy equation (2.10-14) becomes T {2a13 + 2a17 I1 + 6a25 θ¯ + · · ·}(T˙ + t0 T¨) T02 T − {a7 δij + a11 (δij I1 − ij ) + 2a14 I1 δij T0 ¯ ij + a19 [I (δij I − ij ) + I δij ] +2a17 θδ 1 1 2 +a21 (ik jk − ij I1 + δij I2 ) + 3a24 I12 δij + · · ·}(˙ij + t0 ïj )

kT,ii = −

(2.10-18)

The specific heat is c =

−T (2a13 + 2a17 I1 + 6a25 θ¯ + · · ·) ρT02

(2.10-19)

For the linear theory of thermoelasticity the linear terms in the strain ij and the temperature θ¯ are retained in Eqs. (2.10-17) and (2.10-18), and the nonlinear terms are ignored. The constant coefficients a1 to a7 in Eq. (2.10-17) are thus obtained in terms of Lamé constants as a1 = 0 a2 = (λ + 2μ)/2 a3 = −2μ a7 = −(3λ + 2μ)αT0

(2.10-20)

Substituting into Eq. (2.10-17) results in the known linear constitutive relation σij = λkk δij + 2μij − (3λ + 2μ)α(T − T0 )δij

(2.10-21)

¯ the energy equation is reduced to For constant c and small θ, kT,ii = ρc (T˙ + t0 T¨) + (3λ + 2μ)αT0 (˙kk + t0 ¨kk )

(2.10-22)

Equations (2.10-1) and (2.10-22) constitute the governing hyperbolic displacement-temperature equations of generalized thermoelasticity with one relaxation time t0 in an isotropic solid due to Lord and Shulman. For a general anisotropic homogeneous solid the governing equations of LS types read [14] kij T,ji = (1 + t0

∂ )[ρc T˙ + T0 βij u˙ i,j ] ∂t

(2.10-23)

and Cijkl uk,lj − βij T,j + Xi = ρ¨ ui

(2.10−23a)

66


Green–Lindsay (GL) model Green and Lindsay introduced two different lag times t1 and t2 in the stressstrain relations and the entropy expression. This model is based on the work of M¨ uller [15] who developed a more direct approach to include the second sound effect. He obtained the results by considering restrictions on a class of constitutive equations for the linear theory of thermoelasticity with the help of the proposed entropy inequality involving an entropy flux vector. In the theory that M¨ uller developed, he discussed the restrictions which arise when the volume heat supply and the body force are zero. On the other hand, Green and Lindsay took the problem one step further by discussing restrictions on the same class of constitutive equations, with the help of an entropy production inequality proposed by Green and Laws [16] which allows both the volume heat supply and the body force to be non-zero. The inequality (2.7-13) for entropy production may be replaced by a more general one qi (2.10-24) ρ(T˜s˙ − r) + qi,i − T˜,i ≥ 0 T˜ where T˜ is a constitutive scalar function postulated to be positive. The function T˜ is called the thermodynamic temperature [17] and its reciprocal (T˜)−1 is called the coldness function [15]. The inequality (2.10-24) is reduced to (2.7-13) for the classical thermoelasticity when T˜ = T . Introducing an energy function f˜ for unit mass as f˜ = i − T˜s

(2.10-25)

it is postulated that f˜ and T˜ are functions of T, T˙ , T,i and eij . Here, f˜ reduces to Helmholtz free energy function when T˜ = T . If we assume that sij is independent of e˙ ij , qi is independent of T˙,i , and s is independent of T¨, then the energy equation (2.7-11), the inequality (2.10-24), and Eq. (2.10-25) result in the following equations as the counterparts of Eqs. (2.7-15) to (2.7-18), and the inequality (2.7-19), as proposed by Green and Lindsay [18] ∂ f˜ ∂eij ˜ ∂f ∂ T˜ +s =0 ∂ T˙ ∂ T˙ ∂ f˜ qi ∂ T˜ + =0 ∂T,i ρ0 T˜ ∂ T˙ ∂ f˜ 1 ∂ T˜ ˙ ∂ f˜ ˙ ( T,i + s˙ T˜ + qi,i = 0 +s )T+ ∂T ∂T ∂T,i ρ0 ∂ f˜ ∂ T˜ ˙ qi ∂ T˜ ( +s )T+ T,i ≤ 0 ∂T ∂T ρT˜ ∂T

sij = ρ0

(2.10-26) (2.10-27) (2.10-28) (2.10-29) (2.10-30)


67

It is further postulated that T˜ = T˜(T, T˙ )

(2.10-31)

Equations (2.10-26), (2.10-27), (2.10-28), and (2.10-31), respectively, are the constitutive equations for the stress, entropy, heat flux, and temperature. Equation (2.10-29) is the energy equation and the inequality (2.10-30) is the law of entropy production. These equations along with Eqs. (2.7-1), (2.72), and (2.7-5) constitute the general governing equations of the nonlinear temperature-rate dependent thermoelasticity. It is important to realize that qi , as well as sij and s, are now derivable from the energy function through Eq. (2.10-28), rather than by the hypothesis proposed by Fourier’s law. It is interesting to note that by setting T˜ = T , Eq. (2.10-28) becomes independent of qi and thus a separate relation is needed to describe qi and the temperature relationship, which is the case of classical coupled thermoelasticity theory and Fourier’s law. The general theory presented in this section may now be linearized through Maclaurin series expansion about the reference state. Let us expand T˜ and f˜ in Maclaurin’s series with the assumption that T˜(T, 0) = T , (2.10-32) T˜ = T + t1 [T˙ + α1 (T − T0 )T˙ + α2 T˙ 2 ] 1 ρf˜ = Cijkl ij kl − βij ij [(T − T0 ) + t1 T˙ ] 2 t1 1 1 −( ) [{ρc [(T − T0 ) + t2 T˙ ] − Ci T,i }T˙ − kij T,i T,j ] T0 2 2 (2.10-33) The expanded expressions for T˜ and f˜ are substituted into Eqs. (2.10-26) to (2.10-29) and the inequality (2.10-30) to arrive at σij = Cijkl kl − βij [(T − T0 ) + t1 T˙ ] (2.10-34) Ci ρc ) [(T − T0 ) + t2 T˙ ] − ( ) T,i + βij ij T0 T0 qi = −(Ci T˙ + kij T,j ) qi,i = ρ(r − sT ˙ 0) ρc (t1 − t2 )y02 + 2Ci y0 yi + kij yi yj ≥ 0 ρs = S = (

(2.10-35) (2.10-36) (2.10-37) (2.10-38)

where t1 , t2 , α1 , α2 , and Ci are some new material constants not previously defined in the classical coupled thermoelasticity or Lord and Shulman model. Also, y0 and yi are some arbitrary real variables. Eliminating s and qi from Eqs. (2.10-35) to (2.10-37) and substituting for σij from Eq. (2.10-34) into the linearized equation of motion (1.3-8) results in the following displacementtemperature field equations of generalized thermoelasticity due to Green and Lindsay for anisotropic homogeneous materials (2.10-39) kij T,ji + ρr = ρc (T˙ + t2 T¨) − 2Ci T˙,i + T0 βij u˙ i,j ˙ Cijkl uk,lj − βij (T + t1 T ),j + Xi = ρ¨ ui (2.10-40)

68


For isotropic materials, these equations reduce to the following system of equations (2.10-41) kT,ii + ρr = ρc T˙ + ρc t2 T¨ + (3λ + 2μ)αT0 u˙ i,i μui,jj + (λ + μ)uj,ji − (3λ + 2μ)αT,i − t1 (3λ + 2μ)αT˙,i +Xi = ρ¨ ui (2.10-42) It is noted that both the equation of motion and the energy equation are more general than those of the classical coupled thermoelasticity theory, while in LS theory the equation of motion is identical with the classical theory and only the energy equation is changed. It is to be recalled that Fourier’s law of heat conduction remains unchanged in GL theory. Ignaczak [19] suggests the system of governing equations for an anisotropic homogeneous body which contains the two foregoing models as ui Cijkl uk,lj − βij T,j − t1 βij T˙,j + Xi = ρ¨ ¨ kij T,ji + ρ(r + t0 r) ˙ = ρc (t0 + t2 )T + ρc T˙ +t0 T0 βij uï,j + T0 βij u˙ i,j − 2Ci T˙,i

(2.10-43) (2.10-44)

It is simple to note that when t1 = t2 = t0 = Ci = 0 the above system reduces to that of the classical coupled thermoelasticity. When t0 = 0 and t1 = t2 = Ci = 0 the system reduces to that of LS theory and for Ci , t1 and t2 = 0 and t0 = 0 GL theory is obtained. The parameter r in Eq. (2.10-43) represents the heat source. Comparing Lord–Shulman (LS) and Green–Lindsay (GL) models While both LS and GL theories entail hyperbolic energy equations and admit second sound, there exist fundamental differences between the two theories. The differences are as follows [20]: 1. While LS theory modifies the energy equation, GL theory modifies the constitutive laws. 2. Due to the ad hoc approach admitted by LS theory, it does not provide any new material constants other than the thermal relaxation time t0 . On the other hand, GL theory is based on thermodynamic principles and offers five new material constants that are not present in the classical theory, namely t1 , t2 , and Ci . The Ci constants appear in Fourier’s law as qi = −(Ci T˙ + kij T,j )

(2.10-45)

The conductivity tensor kij is observed to be symmetric. 3. In the derivations of GL theory, it is observed that t1 ≥ t2

(2.10-46)


69

Thus if t2 ≥ 0 then t1 ≥ 0, and if the theory is to admit the second sound, S and σij must depend on T˙ . It is postulated that t2 ≥ 0

(2.10-47)

Then, in view of inequality (2.10-46), t2 = 0 whenever t1 = 0. This means that if σij is independent of T˙ , then S is also independent of T˙ and the heat transfer equation becomes parabolic, thus showing that the theory is no longer admitting the second sound. This case is not observed in LS theory where both σij and S are independent of T˙ , but still the theory contains a hyperbolic type heat transfer equation, thus admitting the second sound. 4. The heat conduction law of GL theory, Eq. (2.10-45), does not contain a heat flux-rate term. For a material having a center of symmetry at every point, this law reduces to the classical Fourier’s law. Accordingly, GL theory admits the second sound without violating the classical Fourier’s law. 5. The symmetry condition of conductivity tensor (kij = kji ) is an integral part of GL theory, while this is not the case for LS theory. The two theories are thus structurally different from each other, and one cannot be obtained from the other, or as a particular case of the other. While the heat flux rate term incorporates the second sound phenomenon in LS theory, the temperature rate terms in the constitutive laws do the same in GL theory. Dropping q˙i from the constitutive law (2.10-8) reduces LS theory to the classical theory, and dropping T˙ from the constitutive laws (2.10-34) and (2.10-35) reduces GL theory to the classical theory. At this point it is appropriate to present some information regarding the physical values of the relaxation time for different materials. Different researchers have reported their experimental results for the value of the relaxation time associated with Lord and Shulman model. In a review article, Chandrasekharaiah [14,20] reports that different researchers have found the value of t0 to range from 10−10 s for gases to 10−14 s for metals. The value of the relaxation time for liquids falls in between. Studies made by Vedavarz et al. [21] indicate that the value of t0 for aluminum, being a function of temperature, ranges from 10−11 –10−6 s at cryogenic temperatures to 10−14 –10−11 s at room temperatures and smaller than 10−14 s at high temperatures. Therefore, for such materials the importance of Lord and Shulman model becomes relevant when a structure is exposed to a thermal shock load of very small time interval and/or very large heat flux. Recent studies show that the value of relaxation time for some materials may be much larger than those reported. Experimental studies made by Kaminski [22] show that the value of t0 may be as large as 11 s for glass ballotini and 21 s for sand at laboratory temperatures. Studies made by Vedavarz et al. [21] indicate that the value of t0 may be as high as 10 to 1,000 s for the organic materials and tissues and about 16 s for processed meat.

70

11


Thermoelasticity without Energy Dissipation

Green and Naghdi [23] in 1993 postulated a new concept in generalized thermoelasticity which is called the thermoelasticity without energy dissipation. The principal feature of this theory is that in contrast to the classical thermoelasticity associated with Fourier’s law of heat conduction, the heat flow does not involve energy dissipation. Also, the same potential function which is defined to derive the stress tensor is used to determine the constitutive equation for the entropy flux vector. In addition, the theory permits the transmission of heat as thermal waves at finite speeds. The general discussion is derived from the nonlinear theory, and the linear theory is then presented. The general idea is postulated in a paper by Green and Naghdi [24] in making use of the general entropy balance. While the basic developments in [24] are general, the particular application is confined to the flow of heat in a stationary rigid solid transmitted by conduction and by the heat pulse propagated as thermal waves at finite speed. Three types of constitutive response functions are suggested. Type I, after linearization of the theory, is the same as the classical heat conduction theory (based on Fourier’s law), while the types II and III permit propagation of thermoelastic disturbances with a finite speed. A finite body B with material points X fixed in the original reference system Ko is considered. The position vector of the material point (particle) in the original coordinates system is given by (X1 , X2 , X3 ). In the deformed configuration the body occupies a region R. The deformed coordinate system is called K where the position vector of the material point X is (x1 , x2 , x3 ). Now, the thermal variables are defined as follows [23,24]: (a) Thermal displacement: ϑ = ϑ(X1 , X2 , X3 , t) (b) Empirical temperature: θ = ϑ˙ (c) Temperature: T which depends on θ and the properties of the material such that ∂T >0 (2.11-1) T > 0 and ∂θ ∂ϑ (d) Thermal displacement gradient: Υi = ∂Xi ∂θ (e) Temperature gradient: Γi = which is related to Υj as ∂xi ˙ j = Fji Γi Υ (f) External rate of supply of entropy per unit mass: η (g) External rate of supply of heat per unit mass: r = T η (h) Internal rate of production of entropy per unit mass: ξ (i) Internal rate of production of heat per unit mass: T ξ

(2.11-2)

11. Thermoelasticity without Energy Dissipation

71

(j) Entropy density per unit mass: s (k) heat density per unit mass: T s (l) Internal flux of entropy per unit mass: −ζ (m) Internal flux of heat per unit mass: −h = −ζθ In the above definitions Fij is the deformation gradient tensor defined as Fij =

∂Xi ∂Xj

(2.11-3)

where Xi is the coordinate transformation law defined by xi = Xi (X1 , X2 , X3 , t)

vi = x˙ i

(2.11-4)

A superposed dot indicates the material time differentiation holding Xi fixed. Furthermore, the velocity gradient tensor Lij is defined as Lij =

∂vi = Dij + Wij ∂xj

(2.11-5)

where Dij is the rate of deformation tensor and Wij is the vorticity tensor. The law of mass conservation and the equation of motion are ρ˙ + ρ vi,i = 0 σij,j + ρbi = ρv˙ i

(2.11-6) (2.11-7)

where ρ is the mass density in the deformed coordinates, bi is the body force per unit mass, σij is Cauchy stress tensor (tni = σij nj ), tni is the traction force acting on a surface where its unit outer normal vector is n, all referred to the deformed coordinate system. Representing the components of entropy flux vector and the heat flux vector by pi and qi , respectively, the local field equation for the balance of entropy is ρs˙ = ρ (η + ξ) − pi,i

(2.11-8)

where ζ = p i ni

qi = T pi

(2.11-9)

Eliminating the external body force bi and the external supply of entropy η from the local field equation for the energy balance results in the following reduced energy equation σij Lij − pi gi − ρ(ψ˙ + sT˙ ) − ρT ξ = 0

(2.11-10)

where ψ is the specific Helmholtz free energy function (per unit mass) and gi is the temperature gradient vector in the deformed coordinate system, i.e., gi = T,i =

∂T ∂xi

(2.11-11)

72


Thermoelasticity without energy dissipation (type II) Consider a homogeneous material and assume that the constitutive relations for ψ, T, σij , pi , s, and ξ are functions of the variables θ, Υi , Fij , and Xi . For simplicity, we may define θ, Υi , Fij as the independent variables and introduce the constitutive equations in explicit dependence of Xi . The constitutive laws for the specific free energy, the temperature, the stress, the entropy flux, the internal rate of production of entropy, and the entropy density are assumed to be functions of the independent variables as Υi , Fij ) ψ = ψ(θ, σij = σij (θ, Υi , Fij ) ξ = ξ(θ, Υi , Fij )

; ; ;

T = T (θ, Υi , Fij ) pi = pi (θ, Υi , Fij ) s = s(θ, Υi , Fij )

(2.11-12)

Substituting Eqs. (2.11-12) into the reduced energy equation (2.11-10) yields ∂ ψ ∂ ψ ∂ T ˙ ∂ T + s ) θ + [ρ( + s ) Fji Γi ρT ξ + ρ( ∂θ ∂θ ∂Υj ∂Υj ∂ ψ ∂ T ∂ T + s ) Fkj ] Lij ] + [−σij + ρ( ∂θ ∂Fik ∂Fik ∂ T ∂Υi ∂ T ∂Fij +( pj ) +( pk ) =0 ∂Υi ∂xj ∂Fij ∂xk +pi Γi

(2.11-13)

where symbols with a hat show the values of the functions. In the third term, relation (2.11-2) is used. Comparing this equation with a general polynomial equation of the form a + a 1 y 1 + a 2 y 2 + a 3 y 3 + a4 y 4 + a 5 y 5 = 0

(2.11-14)

we get a = ρ T ξ ∂ ψ ∂ T + s ) a1 = ρ( ∂θ ∂θ ∂ T ∂ ψ ∂ T a2 = ρ( + s ) Fji + pi ∂Υj ∂Υj ∂θ a3 = −σij + ρ( ∂ T pj ∂Υi ∂ T a5 = pk ∂Fij

∂ ψ ∂ T + s ) Fkj ∂Fik ∂Fik

a4 =

(2.11-15)

11. Thermoelasticity without Energy Dissipation

73

with the variables ˙ y1 = θ,

y2 = Γi ,

y3 = Lij ,

y4 =

∂Υi , ∂xj

y5 =

∂Fij ∂xk

(2.11-16)

Since the reduced energy equation (2.11-13) must be identically satisfied for all processes and under the restrictions imposed by the functional dependence of the constitutive equations (2.11-12), we may choose all the variables of Eqs. (2.11-16) to be equal to zero and a = 0 in the first of Eqs. (2.11-15). Now the reduced energy equation takes the form 5

an y n = 0

(2.11-17)

n=1

The left side of this equation must be identically zero. Thus, when all vari∂Fij ables except the arbitrary variable y5 = are chosen to be zero, since the ∂xk components of entropy flux vector pk (or components of heat flux vector qk ) ∂ T are necessarily nonzero from the physical point of view, then = 0. Next, ∂Fij we may choose that all variables defined in Eq. (2.11-16) to be zero except ∂Υi y4 = . In this case, since the components of entropy flux vector pj are ∂xj ∂ T nonzero, it follows that = 0. Thus ∂Υi ∂ T =0 ∂Fij ∂ T =0 ∂Υi

(2.11-18)

Recalling Eqs. (2.11-12), the above results suggest that T is only a function of θ, i.e., (2.11-19) T = T (θ) Thus, we may choose the empirical temperature θ = T − T0 , where T0 is a ∂ T constant. Then = 1. ∂θ Υi , Fij ), or ψ = ψ(T, Υi , Fij ). From Eqs. (2.11-12) it follows that ψ = ψ(θ, By setting a1 = a2 = a3 = 0 in Eq. (2.11-15) and with the help of Eqs. (2.11-18) and (2.11-19) we obtain s=−

∂ψ ∂θ

pi = −ρFji

; σij = ρ ∂ψ ∂Υj

∂ψ Fkj ∂Fik

; ξ=0

(2.11-20)

74


Equations (2.11-18) and (2.11-20) are the necessary and sufficient conditions for the reduced energy equation to be satisfied under the assumptions (2.11-12). The developments presented by Green and Naghdi and discussed in this section are based on the assumption that the internal rate of entropy production ξ is zero. Physically, this means that the thermoelastic analysis is based on the assumption that there is no energy dissipation. It is further noted from Eqs. (2.11-20) that the entropy flux vector is determined by the partial derivative of the specific free energy function with respect to the thermal displacement gradient vector Υj . Linearized theory The linearized theory follows from the assumption that the temperature change θ, the thermal displacement ϑ, and the displacement components ui = Xi − Xi are small compared to their reference values and thus their higher order terms may be neglected. It is further assumed that the time and space derivatives of θ, ϑ, and ui are small and therefore their higher order terms may be neglected. To present the linearized theory, Green and Naghdi [23] postulated a specific Helmholtz free energy function as a quadratic function of infinitesimal temperature, strain, and thermal displacement gradient. For the isotropic material this expression is 1 ρ0 c θ 2 E α∗ κ∗ − Υi Υi θ ii + ρ0 ψ = λ(ii )2 + μ ij ji − 2 2T0 3(1 − 2ν) 2T0

(2.11-21)

where κ∗ is a constant, ij is the component of strain tensor, and α∗ is the coefficient of volume thermal expansion. From Eqs. (2.11-9) and (2.11-20) the components of stress tensor σij , the entropy density per unit mass s, the components of entropy flux vector pi , and the components of heat flux vector qi are σij = λkk δij + 2μ ij − ρ0 s = ρ 0 c

E α∗ θ δij 3(1 − 2ν)

θ E α∗ ii + T0 3(1 − 2ν)

κ∗ Υi T0 qi = −κ∗ Υi pi = −

(2.11-22)

where δij is the Kronecker delta. The expression for the specific internal energy ε is ρ0 ε = ρ0 (ψ + s θ) ρ0 c θ 2 κ∗ 1 + Υi Υi = λ(ii )2 + μ ij ji + 2 2T0 2T0

(2.11-23)

12. A Unified Generalized Thermoelasticity

75

Substituting for σij , s, and pi from Eqs. (2.11-22) into the equation of motion (2.11-7) and Eq. (2.11-8), eliminating the distinction between ρ and ρ0 , yields (λ + μ) uj,ji + μ ui,jj −

E α∗ θ,i + ρ0 bi = ρ0 uï 3(1 − 2ν)

E α∗ T0 ρ0 c θ¨ + uï,i = ρ0 r + κ∗ θ,ii 3(1 − 2ν)

(2.11-24) (2.11-25)

These equations constitute the displacement-temperature equations of coupled thermoelasticity without energy dissipation for an isotropic material.

12

A Unified Generalized Thermoelasticity

The conventional theory of thermoelasticity is based on Fourier’s heat conduction law. Due to the parabolic nature of the energy equation of this theory, infinite propagation speeds for the thermal disturbances are predicted. The concept of the hyperbolic nature involving finite speeds of thermal disturbances is reported by Maxwell [25] for the first time, known as the second sound. Chester [26] provides some justification to the fact that the so-called second sound must exist in any solid. Most of the approaches that came out to overcome the unacceptable prediction of the classical theory are based on the general notion of relaxing the heat flux in the classical Fourier’s heat conduction equation, thereby introducing a non-Fourier effect. Three distinct generalized thermoelasticity theories admitting the second sound effect, as discussed earlier, are Lord–Shulman (LS) [12], Green–Lindsay (GL) [18], and Green– Naghdi (GN) [23] theories. Ignaczak [19] suggested a combined system of coupled equations for LS and GL theories. Also, the same author reported a survey of the domain of influence for the results of LS and GL theories [27]. Francis [28], Ignaczak [19], and Chandrasekharaiah [14,20] have reported brief reviews of these theories. In this section, a new unified formulation for the generalized coupled thermoelasticity theories based on LS, GL, and GN models is proposed. The unifier parameters are introduced to consolidate the equations of LS, GL, and GN theories in a single system of equations for the anisotropic and heterogeneous materials. The equations are also simplified for the isotropic and homogeneous materials [29,30]. The fundamental equations of Lord–Shulman, Green–Lindsay, and Green– Naghdi theories in a unified form are presented here introducing the terms η and t3 as unifier terms. These equations in general form are The equations of motion: ∇. σ + ρb = ρ¨ u

(2.12-1)

76


The linear strain-displacement relations: 1 E = (∇u + (∇u) ) 2

(2.12-2)

Hooke’s law for the linear thermoelastic materials: σ = C E − β(T − T0 + t1 T˙ )

(2.12-3)

The energy balance equation: ∇. q = R − T0 S˙

(2.12-4)

The entropy relationship: S=(

ρc 1 )(T + t2 T˙ − T0 ) + β : E − C . ∇T T0 T0

(2.12-5)

The heat conduction equation: T˙ η q + η τ q˙ + t3 q˙ = −η K ∇T − t3 K ∇T˙ − t3 K∗ ∇T − C

(2.12-6)

where ρ is the mass density, σ is Cauchy’s stress tensor, u is the displacement vector, b is the body force vector per unit mass, q is the heat flux vector, T0 is the reference temperature, T is the absolute temperature, S is the entropy per unit volume, R is the internal heat source per unit volume per unit time, E is the strain tensor, β is the second order tensor of stress-temperature moduli, K is the second order tensor of thermal conductivity, C is the forth order tensor of elastic moduli, and c is the specific heat. Also, τ is the second order tensor is a of relaxation times in LS model, t1 and t2 are the relaxation times, C vector of new material constants proposed by Green and Lindsay, and K∗ is the second order tensor of new material constants associated with GN theory. Also, η and t3 are terms introduced to consolidate all theories into a unified system of equations. In Eqs. (2.12-1) to (2.12-6) the superscript dot (.) denotes the differentiation with respect to time. Meanwhile, ∇ is the del operator and indicates the gradient of a function, (∇.) denotes the divergence operator, and the superscript prime ( ) indicates the transpose of the matrix. In Eqs. (2.12-1) to (2.12-6) the double dot product of two second order tensors means the trace of their product. For example, β : E in index form is βij Eij . The product of C E is a second order tensor and in the index notation is Cijkl Ekl , the product . ∇T is a of τ q is a first order tensor and in the index notation is τij qj , and C scaler which in the index notation is Ci T,i . Equations (2.12-1) to (2.12-6) include the governing coupled system of = 0, equations for the classical theory of thermoelasticity when τ = 0, C t1 = t2 = t3 = 0 and η = 1. The given system of equations yields the equa = 0 and t = t = t = 0. The governing tions of LS theory when η = 1, C 1 2 3

12. A Unified Generalized Thermoelasticity

77

equations for GL theory are obtained when τ = 0, η = 1 and t3 = 0. To = 0, obtain the governing equations for GN theory type III, we set τ = 0, C η = t1 = t2 = 0, and t3 = 1. Two special cases of GN theory, namely type II and I, may be obtained from the equations of GN theory type III by setting K → 0 and K∗ → 0, respectively. To obtain GN theory type II, from the equations of GN theory type III, we set K → 0. When K∗ → 0 the equations of GN theory type III reduce to GN theory type I, which is identical with the classical theory of thermoelasticity. To obtain the system of equations in terms of displacement and temperature, terms σ, E, q and S may be eliminated with the help of Eqs. (2.12-1) to (2.12-6). To this end, we apply del operator with dot sense, (∇ . ), to Eq. (2.12-6) to achieve ˙ + t3 ∇ . q˙ η ∇ . q + η ∇ . (τ q) T˙ ) (2.12-7) = −η ∇ . (K ∇T ) − t3 ∇ . (K ∇T˙ ) − t3 ∇ . (K∗ ∇T ) − ∇ . (C Substituting the first and third terms on the left-hand side of Eq. (2.12-7) using Eq. (2.12-4) gives ˙ + η ∇ . (τ q) ¨ ˙ + t3 (R˙ − T0 S) η (R − T0 S) T˙ ) (2.12-8) = −η ∇ . (K ∇T ) − t3 ∇ . (K ∇T˙ ) − t3 ∇ . (K∗ ∇T ) − ∇ . (C Also, the second term, which only contributes to LS theory, should be substituted by using Eq. (2.12-6). Thus, for simplicity, we only use the terms which = 0, η = 1, appear in LS theory. Equation (2.12-6) for LS theory, the case (C t3 = 0), may be solved to find the heat flux in terms of the temperature as −1

q = −B

t −∞

τ −1 K∇ T dt B

(2.12-9)

where B = exp(t τ −1 ) = = exp(t τ −1 ) = B

∞ n t n=0 n! ∞ n

(τ −1 )n

t (τ −1 )n n! n=0

(2.12-10)

Here, (τ −1 )0 = I, and I is the identity tensor. Differentiation of Eq. (2.12-9) with respect to the time variable, considering ∂(B−1 ) ∂B −1 that = − B−1 B = − τ −1 B−1 , and multiplying the result by τ ∂t ∂t leads to τ q˙ = B−1

t

−∞

τ −1 K∇ T dt − K∇ T B

(2.12-11)

78


Substituting Eq. (2.12-11) in Eq. (2.12-8) gives

˙ + η ∇ . B−1 η (R − T0 S)

t −∞

τ −1 K∇ T dt + t (R ˙ − T0 S) ¨ B 3

T˙ ) = −t3 ∇ . (K ∇T˙ ) − t3 ∇ . (K∗ ∇T ) − ∇ . (C

(2.12-12)

Now, substituting S from Eq. (2.12-5) into Eq. (2.12-12) and utilizing Eq. (2.12-2), considering the values for the unifier terms η and t3 and the parameters t0 , t1 , t2 , and noting that β and E are symmetric tensors, the energy balance equation in terms of the displacement vector and temperature may be obtained. Also, eliminating the stress and strain tensors with the use of Eqs. (2.12-1) to (2.12-3) leads to the equations of motion in terms of the displacement vector and the temperature. This system of equations is t3 ∇. (K∇T˙ ) + t3 ∇. (K∗ ∇T ) + η∇.

−1

B

t −∞

−1 B τ K(∇ T ) dt

˙ − t3 T0 β : (∇¨ −ρ c (t2 + t3 )T¨ − η ρ c T˙ − η T0 β : (∇u) u) (2.12-13) +2 C . (∇T˙ ) + (∇. C)T˙ + ηR + t3 R˙ = 0 ∇. (C (∇ u)) − (∇. β) T − T0 + t1 T˙ − β ∇ T + t1 T˙ + ρb = ρ¨ u (2.12-14) To obtain the differential form of LS theory, Eq. (2.12-13) is differentiated with respect to the time variable t. Reconsideration of Eqs. (2.12-13) and (2.12-14) for GN theory type II reveals that no damping term appeared in the system of equations and, therefore, GN theory type II is known as the thermoelasticity without energy dissipation. For the isotropic materials, the material properties are C = λI⊗I + μ(I⊗I + I⊗I) β = β I ; τ = t0 I K = k I ; K∗ = κ∗ I = exp(t/t 0) I B = exp(t/t0 ) I ; B =0 C

(2.12-15)

or in index notation Cijkl = λδij δkl + μ(δik δjl + δil δjk ) βij = βδij ; τij = t0 δij Kij = kδij ; Kij∗ = κδij Bij = exp(t/t0 ) δij Ci = 0

;

0 ) δij B ij = exp(t/t

(2.12-16)

where λ and μ are Lamé constants, k is the thermal conductivity, t0 is the relaxation time proposed by Lord and Shulman, and ⊗ indicates the tensor

13. Uniqueness Theorem

79

product. Equations (2.12-13) and (2.12-14) for the isotropic heterogeneous medium appear in the form

k t exp(−(t − t )/t0 ) (∇ T )dt t0 −∞ ˙ − t3 T0 β(∇.¨ −ρc(t2 + t3 )T¨ − ηρcT˙ − ηT0 β(∇.u) u) + ηR + t3 R˙ = 0 (2.12-17)

t3 ∇. (k∇T˙ ) + t3 ∇. (κ∗ ∇T ) + η∇.

λ∇(∇. u) + (∇λ)(∇. u) + μ∇. (∇u) + (∇u)

+ (∇u) + (∇u) . (∇μ) − (∇β) T − T0 + t1 T˙ − β∇ T + t1 T˙

+ρb = ρ¨ u

(2.12-18)

This coupled system of equations may be reduced to obtain the equations for the isotropic homogeneous body as t

k t3 k∇. (∇T˙ ) + t3 κ∗ ∇. (∇T ) + η exp(−(t − t )/t0 )∇. (∇ T )dt t0 −∞ ˙ − t3 T0 β(∇.¨ −ρc(t2 + t3 )T¨ − ηρcT˙ − ηT0 β(∇.u) u) + ηR + t3 R˙ = 0 (2.12-19) λ∇(∇. u) + μ∇. (∇u) + (∇u) − β∇ T + t1 T˙ + ρb = ρ¨ u (2.12-20) Also, these equations may be written as t

k t3 κ∗ ∇2 T + t3 k∇2 T˙ + η exp(−(t − t )/t0 )(∇2 T )dt − ρc(t2 + t3 )T¨ t0 −∞ ˙ − t3 T0 β(∇. u ¨ ) + ηR + t3 R˙ = 0 −ηρcT˙ − ηT0 β(∇. u) (2.12-21) 2 ˙ μ∇ u + (λ + μ)∇(∇. u) − β∇ T + t1 T + ρb = ρ¨ u (2.12-22) where ∇2 is Laplace operator. Meanwhile, the stress-displacement relations and the heat conduction equation for the isotropic materials are

σ = λ(∇. u)I + μ (∇u) + (∇u) − β(T − T0 + t1 T˙ )I (2.12-23) (2.12-24) ηq + η t0 q˙ + t3 q˙ = −η k∇T − t3 k∇T˙ − t3 κ∗ ∇T

13

Uniqueness Theorem

The uniqueness theorems of linear classical and generalized coupled thermoelasticity have been presented by a number of researchers. Among the first who contributed to the subject was Nowacki [31] who discussed the displacementtemperature uniqueness theorem of the linear classical thermoelasticity. The stress-temperature uniqueness theorem for the generalized thermoelasticity based on Green and Lindsay model was given by Ignaczak [32]. The theorem covers an unconventional initial-boundary-value problem in which the initial

80


stress and stress rates are prescribed. The same author presented the stressheat flux uniqueness theorem for the generalized thermoelasticity with single relaxation time (Lord and Shulman model) [33]. The existence theorems in a functional space of Sobolev type for both LS and GL models were proved by Bem [34,35]. The other references in this subject are Ignaczak [36], Sherief and Dhaliwal [37], Bem [38], Iesan [39], and Wang and Dhaliwal [40]. The more general discussion on the uniqueness theorems may be found in the works by Chirita for micropolar structures [41], by Iesan [42] for nonsimple materials, and by Dhaliwal and Sherief [43] for anisotropic materials. Chandrasekharaiah [44] proved a uniqueness theorem for the case where heatflux is included among the independent constitutive variables. The same author presented a uniqueness theorem in the theory of thermoelasticity without energy dissipation [45]. Wang and Dhaliwal [46] proved that an initial-boundary value problem of the linear generalized thermoelasticity due to Lord–Shulman for anisotropic unbound domain has a unique solution. In this section, the uniqueness theorems of linear generalized thermoelasticity of homogeneous and isotropic material based on Lord and Shulman and Green and Lindsay models are presented. Lord–Shulman model An isotropic and homogeneous linear elastic continuum occupying the volume V with boundary surface A is considered. The volume region V is assumed to be regular and simply connected. The field equations referred to the Cartesian coordinates are [36] 1 in V for t ≥ 0 ij = (ui,j + uj,i ) 2 σij,j + Xi = ρ¨ ui in V for t ≥ 0 ˙ −qi,i + R = ρc θ + (3λ + 2μ)αT0 ˙ii σij = 2μij + λkk δij − (3λ + 2μ)αθδij Lqi = −kθ,i

(2.13-1) (2.13-2) in V for t ≥ 0 (2.13-3) in V for t ≥ 0 (2.13-4) (2.13-5)

where θ = T − T0 , T0 being the reference temperature, and the operator L is defined as ∂ (2.13-6) L = 1 + t0 ∂t The relaxation time t0 and the material constants are all positive quantities as ρ>0 μ>0

k>0 c > 0 (3λ + 2μ) > 0

α>0 t0 > 0

(2.13-7)


81

With the governing equations given by Eqs. (2.13-1) to (2.13-5), the boundary conditions are assumed of the form ui (xm , t) = u¯i (xm , t) m = 1, 2, 3 on A × [0, ∞) ¯ m , t) m = 1, 2, 3 on A × [0, ∞) θ(xm , t) = θ(x

(2.13-8)

where A is the boundary of the body. The boundary conditions for the displacement components ui and the temperature are specified by the known functions ¯ respectively. It is assumed that the prescribed initial conditions are u¯i and θ, ui (xm , 0) = u0i (xm ) θ(xm , 0) = θ0 (xm )

u˙ i (xm , 0) = vi (xm ) ˙ m , 0) = θ1 (xm ) θ(x

m = 1, 2, 3 m = 1, 2, 3

in V in V (2.13-9)

Theorem Consider a regular region of linear thermoelastic material occupying the volume V with boundary surface A. There exists at most one set of single-valued functions for the stress tensor σ(xm , t), strain tensor (xm , t), and heat flux qi (xm , t) of class C (1) , and displacement ui (xm , t) and temperature T (xm , t) of class C (2) in V + A for t ≥ 0 so that Eqs. (2.13-1) to (2.13-9) are satisfied. Proof To prove the uniqueness of the solution, it is sufficient to show that the linear set of governing equations (2.13-1) to (2.13-5) subjected to the homogeneous initial and boundary conditions in V ui (xm , 0) = u˙ i (xm , 0) = 0 m = 1, 2, 3 ˙ θ(xm , 0) = θ(xm , 0) = 0 m = 1, 2, 3 in V ui (xm , t) = 0 θ(xm , t) = 0 m = 1, 2, 3 on

(2.13-10) A × [0, ∞) (2.13-11)

results in the solution ui (xm , t) = 0

and

θ(xm , t) = 0

in

V × [0, ∞)

(2.13-12)

The proof may follow by introducing the notations [36] Lui = Ui Lθ = Θ

Lij = Υij Lqi = Qi

Lσij = Σij (2.13-13)

where the operator L is as defined by Eq. (2.13-6). Applying the operator L to Eqs. (2.13-1) to (2.13-5) and using the definitions given by Eqs. (2.13-13) yield

82

Chapter 2. Thermodynamics of Elastic Continuum 1 (Ui,j + Uj,i ) 2 Σij,j = ρUï ˙ + (3λ + 2μ)αT0 Υ ˙ ii −Qi,i = ρc Θ Σij = 2μΥij + λΥkk δij − (3λ + 2μ)αΘδij Qi = −kθ,i Υij =

(2.13-14)

The application of the operator L to the initial conditions (2.13-10) gives in V Ui (xm , 0) = U˙ i (xm , 0) = 0 m = 1, 2, 3 ˙ Θ(xm , 0) = Θ(xm , 0) = 0 m = 1, 2, 3 in V (2.13-15) and applying it to the boundary conditions (2.13-11) yields Ui (xm , t) = 0

on A × [0, ∞)

Θ(xm , t) = 0

(2.13-16)

Multiplying the time rate form of the first of Eqs. (2.13-14) by Σij and integrating it over the volume, using the divergence theorem, results in

V

˙ ij dV = Σij Υ

V

= A

Σij U˙ i,j dV =

U˙ i Σij nj dA −

V V

(Σij U˙ i ),j − U˙ i Σij,j dV

U˙ i Σij,j dV

(2.13-17)

where nj is the component of the unit outer normal vector to the boundary. According to the first part of Eq. (2.13-16), Ui (xm , t) = 0 on the boundary for the time [0, ∞). Based on this assumption, U˙ i (xm , t) = 0 on A × [0, ∞), resulting in vanishing of the surface integral in the right-hand side of Eq. (2.1317), and yielding ˙ ij dV = − U˙ i Σij,j dV Σij Υ (2.13-18) V

V

Multiplying the second of Eqs. (2.13-14) by U˙ i and integrating over the volume, we have ∂ 1 ˙ ˙ Ui Ui dV ρU˙ i Uï dV = (2.13-19) U˙ i Σij,j dV = ∂t V 2 V V Subtraction of Eq. (2.13-19) from Eq. (2.13-19) gives V

˙ ij + ∂ ( 1 U˙ i U˙ i ) dV = 0 Σij Υ ∂t 2

(2.13-20)

˙ ij and integrated over Now, the fourth of Eqs. (2.13-14) is multiplied by Υ the volume to give

V

˙ ij dV = Σij Υ

V

˙ ij + λΥkk Υ ˙ mm − (3λ + 2μ)αΘΥ ˙ kk dV 2μΥij Υ

∂ 1 1 = μ(Υij − Υmm δij )(Υij − Υnn δij ) ∂t 3 3 V

(3λ + 2μ) 2 ˙ + (Υkk ) − (3λ + 2μ)αΘΥkk dV 6

(2.13-21)


83

Substituting Eq. (2.13-21) into Eq. (2.13-20), we obtain ∂ 1 1 (3λ + 2μ) (Υkk )2 μ(Υij − Υmm δij )(Υij − Υnn δij ) + ∂t V 3 3 6 1 ˙ kk dV = 0 + ρU˙ i U˙ i dV − (3λ + 2μ)α ΘΥ (2.13-22) 2 V

Eliminating Qi from the third and the fifth of Eqs. (2.13-14), we arrive at ˙ + (3λ + 2μ)αT0 Υ ˙ kk kθ,ii = ρc Θ

(2.13-23)

Multiplying Eq. (2.13-23) by Θ and integrating over V yields

ρc ∂ Θ2 dV k (Θθ,i ),i dV − k θ,i Θ,i dV = 2 ∂t V V V ˙ kk dV +(3λ + 2μ)αT0 ΘΥ

(2.13-24)

V

The first integral on the left-hand side of Eq. (2.13-24) vanishes due to the divergence theorem and the second part of Eq. (2.13-16). By the definition of Θ from Eq. (2.13-13), Eq. (2.13-24) reduces to

k ∂ ρc 2 kt0 Θ + θ,i θ,i dV + θ,i θ,i dV ∂t V 2T0 2T0 T0 V

+(3λ + 2μ)α V

˙ kk dV = 0 ΘΥ

(2.13-25)

Adding Eqs. (2.13-22) and (2.13-25) gives ∂ 1 1 (3λ + 2μ) μ(Υij − Υmm δij )(Υij − Υnn δij ) + (Υkk )2 ∂t V 3 3 6 1 ρc 2 kt0 k + ρU˙ i U˙ i + Θ + θ,i θ,i dV + θ,i θ,i dV = 0 (2.13-26) 2 2T0 2T0 T0 V

Using the homogeneous initial condition given by the second of Eqs. (2.13-10) and Eqs. (2.13-15), and the strain-displacement relations given by the first of Eqs. (2.13-14) at t = 0, and integrating Eq. (2.13-26) with respect to the time yields V

1 1 (3λ + 2μ) Υmm δij )(Υij − Υnn δij ) + (Υkk )2 3 3 6 1 ρc 2 kt0 k t + ρU˙ i U˙ i + Θ + θ,i θ,i dV + dτ θ,i θ,i dV = 0 2 2T0 2T0 T0 0 V (2.13-27) μ(Υij −

84


Considering the positive values of the material constants given by Eqs. (2.13-7), all components of the integrand in Eq. (2.13-27) are non-negative. Therefore, Eq. (2.13-27) implies that U˙ i = 0

Θ=0

in V × [0, ∞)

(2.13-28)

Equation (2.13-28) and the first of Eqs. (2.13-15) suggest that Ui = 0

Θ=0

in V × [0, ∞)

(2.13-29)

or, considering the definition of the operator L from Eq. (2.13-6), ui + t0 u˙ i = 0

θ + t0 θ˙ = 0

in V × [0, ∞)

(2.13-30)

Integration of Eq. (2.13-30) yields t ui (xm , t) = ui (xm , 0) exp (− ) t0 t θ(xm , t) = θ(xm , 0) exp (− ) t0

(xm , t) ∈ V × [0, ∞) (xm , t) ∈ V × [0, ∞) m = 1, 2, 3

(2.13-31)

Due to the assumption of the homogeneous initial conditions given by Eqs. (2.13-10), we arrive at Eq. (2.13-12), which concludes the proof. Green–Lindsay model The fundamental system of field equations governing the homogeneous isotropic linear thermoelasticity proposed by Green and Lindsay is [18]: the strain-displacement relations: ij =

1 (ui,j + uj,i ) 2

(2.13-32)

the equation of motion σij,j + Xi = ρuï

(2.13-33)

the energy equation −qi,i + R = ρc θ˙ + ρc t2 θ¨ + (3λ + 2μ)αT0 ˙kk

(2.13-34)

the stress-strain relation ˙ ij σij = 2μij + λkk δij − (3λ + 2μ)α(θ + t1 θ)δ

(2.13-35)

and the heat conduction law qi = −kθ,i

(2.13-36)


85

where θ = T − T0 is the temperature change above the reference temperature, and R = ρr is the heat source function. To derive the uniqueness theorem for the stress-temperature formulation, ui , ij , and qi are eliminated from Eqs. (2.13-32) to (2.13-36) and the governing equations in terms of the stress tensor σij and the temperature change θ reduce to the following [32,47] ρ−1 σ(ik,kj) + ρ−1 X(i,j) −

... 1 λ (¨ σij − σ ¨kk δij ) − α(θ¨ + t1 θ)δij = 0 2μ 3λ + 2μ (2.13-37)

1 τ θ¨ + θ˙ = − (−kθ,ii + αT0 σ˙ kk − R) (2.13-38) cs where 1 σ(ik,kj) = (σik,kj + σjk,ki ) 2 1 X(i,j) = (Xi,j + Xj,i ) 2 cs = ρc + 3(3λ + 2μ)T0 α2 ρc ρc τ = (1 − ) t1 + t2 (2.13-39) cs cs Here, cs is the specific heat per unit volume at constant stress. From Eqs. (2.10-45), (2.10-46), (2.13-39) and the definition of the material constants it follows τ ρc ≤ ≤1 (2.13-40) 0 0 are given for the displacement, traction, temperature, and heat flux. Consider a traction boundary condition on A1 , a given kinematical boundary condition on A2 , a specified temperature on A3 , and a specified heat flux on A4 : x ∈ A1 t > 0 σji nj = tni (x, t) ui = Ui (x, t) x ∈ A2 t > 0 θ = Θ(x, t) x ∈ A3 t > 0 qi ni = Q(x, t) x ∈ A4 t > 0

(2.15-5) (2.15-6)

where tni (x, t), Ui (x, t), Θ(x, t), and Q(x, t) are, respectively, the known traction, the displacement, the temperature, and the heat flux, in their respective

15. Reciprocity Theorem

93

domains, and that A = A1 + A2 = A3 + A4 . The heat flux is related to the temperature through the following general law [18] qi = −(Ci θ˙ + kij θ,j )

(2.15-7)

Once the distribution for the body force Xi and the heat generation R in x ∈ V and t ≥ 0 and the functions tni , Ui , Θ, and Q are known, a solution for the displacement ui , the stress σij , and the temperature θ for x ∈ V and t > 0 are uniquely determined. We associate this problem with the system {(Xi , R, tni , Ui , Θ, Q); (ui , σij , θ)}. Now, the reciprocal theorem is stated as follows: Reciprocal theorem Consider two problems associated with the following systems: (1)

n(1)

, Ui , Θ(1) , Q(1) ); (ui , σij , θ(1) )]

(2)

n(2)

, Ui , Θ(2) , Q(2) ); (ui , σij , θ(2) )]

[(Xi , R(1) , ti [(Xi , R(2) , ti

(1)

(1)

(1)

(2)

(2)

(2)

(2.15-8)

where each system satisfies the governing equations (2.15-1) to (2.15-3) and Eq. (2.15-7) and the initial and boundary conditions (2.15-4) and (2.15-5) to (2.15-6). Laplace transforms of Eqs. (2.15-1) to (2.15-3) and the boundary conditions (2.15-5) to (2.15-6), and considering the initial conditions (2.15-4), yield ∗(m)

σij

= Cijrs u∗(m) − βij (1 + t1 s)θ∗(m) rs

∗(m)

∗(m)

σji,j + Xi R∗(m) + ∗(m)

qi

(2.15-9)

∗(m)

= ρ 0 s 2 ui

∗(m) kij θ,ij

(2.15-10)

− ρcsθ∗(m) − ρct2 s2 θ∗(m) − ∗(m)

= −(Ci sθ∗(m) + kij θ,j

∗(m) T0 βij sui,j

)

+

∗(m) 2Ci sθ,i

=0 (2.15-11) (2.15-12)

with the boundary conditions ∗(m)

σji

∗(m) ui ∗(m)

n∗(m)

nj = ti

∗(m) Ui ∗(m)

= θ =Θ ∗(m) qi ni = Q∗(m)

on A1 on A2 on A3 on A4

(2.15-13)

where the superscript (m) is 1 and 2 and (∗ ) denotes Laplace transforms of the function. Equation (2.15-10) is written for m = 1 and m = 2, the first ∗(2) ∗(1) equation is multiplied by ui and the second equation by (−ui ) and the resulting equations are added to arrive at

94

Chapter 2. Thermodynamics of Elastic Continuum V

∗(1) ∗(2) ui

∗(2) ∗(1) ui )dV

− Xi

(Xi

∗(1) ∗(2)

+ V

∗(2) ∗(1)

=

(σij ui

V

∗(1) ∗(2)

− σij ui

),j dV

∗(2) ∗(1)

(σij ui,j − σij ui,j )dV

(2.15-14)

Applying the divergence theorem, the first term on the right-hand side of Eq. (2.15-14) is transformed into the surface integral with the use of Eq. (2.15-13). The second integral on the right-hand side of Eq. (2.15-14) is also replaced with the aid of Eq. (2.15-9). Thus, Eq. (2.15-14) is written as

V

∗(1) ∗(2) (Xi ui

+ A2

−

∗(2) ∗(1) Xi ui )dV

∗(1) ∗(2) (σij Ui

−

+ A1

n∗(1) ∗(2) ui

(ti

∗(2) ∗(1) σij Ui )nj dA

n∗(2) ∗(1) ui )dA

− ti

∗(2)

= V

−βij (1 + t1 s)(θ∗(1) ui,j

∗(1)

−θ∗(2) ui,j )dV

(2.15-15)

When the body has a center of symmetry, while being anisotropic, Ci = 0 and from Eq. (2.15-11), using the divergence theorem, and with the help of Eq. (2.15-12) and boundary conditions (2.15-13), we obtain

−βij T0 s

∗(1)

∗(2)

(ui,j θ∗(2) − ui,j θ∗(1) )dV +

V

∗(2)

= V

= V

V

(R∗(1) θ∗(2) − R∗(2) θ∗(1) )dV

∗(1)

kij (θ∗(1) θ,ij − θ∗(2) θ,ij )dV ∗(2)

kij (θ∗(1) θ,j

= A3

∗(1)

− θ∗(2) θ,j ),i dV

∗(2) kij (Θ∗(1) θ,j

−

∗(1) Θ∗(2) θ,j )ni dA

+ A4

(Q∗(1) θ∗(2) − Q∗(2) θ∗(1) )dA (2.15-16)

Solving Eq. (2.15-15) for

∗(2)

V

∗(1)

−βij (θ∗(1) ui,j − θ∗(2) ui,j )dV

and substituting into Eq. (2.15-16) gives

T0 s[

V

∗(1) ∗(2) ui

(Xi

∗(1)

+ A2

∗(2)

(σij Ui

= (1 + st1 )[

V

∗(2) ∗(1) ui )dV

− Xi

∗(2)

∗(1)

− σij Ui

+ A1

n∗(1) ∗(2) ui

(ti

)nj dA]

(R∗(1) θ∗(2) − R∗(2) θ∗(1) )dV +

∗(2)

−Θ∗(1) θ,j )ni dA +

A4

n∗(2) ∗(1) ui )dA

− ti

A3

∗(1)

kij (Θ∗(2) θ,j

(θ∗(1) Q∗(2) − θ∗(2) Q∗(1) )dA]

(2.15-17)

Using the convolution theorem, the inverse Laplace transforms of Eq. (2.15-17) results in the desired reciprocal theorem in the form

T0

V

(1)

[Xi

(2)

(2)

∗ ui − Xi

(1)

∗ ui ]dV +

A1

n(1)

[ti

(2)

n(2)

∗ ui − ti

(1)

∗ ui ]dA

16. Initial and Boundary Conditions

+ A2

+t1

(2) [Ui

(R

∗

(1)

V

(1) σij

−

(1) Ui

∗

95

(2) σij ]nj dA

∗ θ(2) − R(2) ∗ θ(1) )dV +

+t1 (Θ(2) ∗ θ,j − Θ(1) ∗ θ,j )]ni dA + (1)

(2)

=

A3

A4

V

(R(1) θ(2) − R(2) θ(1) )dV (1)

(2)

kij [(Θ(2) ∗ θ,j − Θ(1) ∗ θ,j ) [(Q(2) ∗ θ(1) − Q(1) ∗ θ(2) )

+t1 (Q(2) ∗ θ(1) − Q(1) ∗ θ(2) )]dA

(2.15-18)

where the following notations are used: f ∗g =

t 0

t

f (x, t − τ )g(x, τ )dτ

∂g (x, τ )dτ (2.15-19) ∂τ 0 It is noted that while Green and Lindsay theory involves two relaxation times t1 and t2 , only t1 appears in the reciprocal theorem. By setting t1 = 0 in Eq. (2.1518), the reciprocal theorem of the classical theory of coupled thermoelasticity is obtained. f ∗ g =

16

f (x, t − τ )

Initial and Boundary Conditions

The differential equation of heat conduction derived in the previous section should be solved along with the initial and boundary conditions to yield the temperature distribution. In general, the number of initial and boundary conditions in the direction of each independent variable must be equal to the order of the highest derivative of the governing differential equation. With this in mind and noting that the condition on the time is called the initial condition and the condition on the space is called the boundary condition, the total number of initial and boundary conditions for any differential equation can be readily determined. For the heat conduction problems the solution of Eq. (2.9-9) of the previous section, regardless of the mathematical method employed, are required with one initial and six boundary conditions, two in each of the x, y, and z directions. In this section, a detailed discussion of possible initial and boundary conditions for the heat conduction problems is presented [58]. Initial condition In the differential equation of heat conduction only the first derivative of the temperature with respect to time is involved. Therefore, only one initial condition is essential in order to evaluate the constant of integration obtained in the solution of the equation, and this condition may be generally stated as limt→0 T (r, t) = T0 (r)

(2.16-1)

96


where T0 (r) is a known function of the space coordinates x, y, and z. The functional relationship of the temperature and space coordinates, therefore, must be given at zero time. Boundary conditions The boundary conditions most frequently used in the heat conduction in solids can be classified as follows: 1. Prescribed temperature The temperature at the surface of a boundary is known to be a constant or a function of time and/or space T (r, t)|r=a = T¯(a, t)

(2.16-2)

2. Prescribed heat flux The heat flux across the surface of a boundary is known to be a constant or a function of time and/or space. The mathematical description of this condition is ∂T (2.16-3) ) = ±q ” ∂n where the left-hand side of Eq. (2.16-3) is the amount of heat conducted through the surface of the body, and the right-hand side is the amount of heat flux radiated into the boundary surface by an outside source. The plus or minus signs on both sides of the equation depend on the direction of the heat transferred to the surface. It is recalled that the left side of Eq. (2.16-3) is the ∂T ∂ denotes the differentiation statement of Fourier’s law, qn = −k( ), and ∂n ∂n along the normal to the boundary. Practical examples of such a boundary condition are the exposure of a solid to the radiation of sun, and the exposure of the tubes of a water tube boiler to the radiation of hot gases produced by burning of fuel inside the chamber. ±k(

3. Insulation (no heat flux) This is a special form of the previous boundary condition, where the radiated heat flux to or from the boundary is zero. Mathematically, this condition is described using Eq. (2.16-3) as ∂T ∂n

=0

(2.16-4)

b

This condition has many applications in practical problems, where the solid is protected against heat loss by the insulation materials. An example is the insulation of pipes containing hot flowing liquids. 4. Heat transfer to the ambient by convection When a boundary of a solid is exposed to the ambient air, in the absence of external radiation the heat flux transferred to or from the ambient through the solid body is

16. Initial and Boundary Conditions qc = h(Tb − T∞ )

97 (2.16-5)

where h is the heat transfer coefficient by convection, Tb the temperature of the solid’s boundary, and T∞ is the temperature of the ambient at a distance far from the boundary. Considering the principle of the thermal balance on the boundary, the heat transferred by conduction to the boundary must be equal to the heat transferred to the ambient by convection, or ∂T ±k ∂n

= h(Tb − T∞ )

(2.16-6)

b

∂ where denotes the differentiation along the normal to the boundary. The ∂n direction of the heat flux from or to the boundary will define the positive or negative sign in Eq. (2.16-6). This is shown in Figs. 2.16-1 and 2.16-2.

Figure 2.16-1: −k(

∂T )b − h(Tb − T∞ ) = 0 ∂n

Figure 2.16-2: +k(

∂T )b − h(Tb − T∞ ) = 0 ∂n

98


The values of the heat transfer coefficient h for typical materials mostly encountered in conduction problems are given in the engineering tables and charts.

17

Problems

1. Verify that if all thermal quantities are required to be indifferent, the energy equation and the entropy inequalities are all invariant with change of reference frame. 2. The governing equations of thermoelasticity including the classical coupled, LS, and GL models (in the absence of body forces) are ˙ ,j − Cijkl uk,l,j = 0 ρ¨ ui + βij T0 (θ + t1 θ) ρc(t0 + t2 )θ¨ + ρcθ˙ + βij (u˙ i,j + t0 uï,j ) − 2Ci θ˙,i − kij θ,ji = 0 (a) Simplify these equations for one-dimensional problems. (b) Does orthotropy of the material alter the form of equations ? 3. Solve the equations of Problem 2 for a half-space orthotropic material under the following initial and boundary conditions: – at t = 0 the stress is zero everywhere in the half-space; – at t = 0, θ = 0 everywhere; – at the plane x = 0, the Heaviside step function condition for temperature is given θ(t) = 0 for t < 0 θ(t) = T0 for t > 0 4. Check the solutions of Problem 3 for elastic and temperature waves. 5. Simplify the solutions of Problem 3 for the following three theories and compare the results. (a) Classical theory, t0 = t1 = t2 = 0 (b) GL theory, t0 = 0 (c) LS theory, t1 = t2 = 0. 6. Based on Green and Lindsay assumptions, Eqs. (2.10-34) and (2.10-35), obtain the generalized thermoelastic equations (2.10-39) and (2.10-40). Simplify for the isotropic materials. 7. Using Lord and Shulman assumption, Eq. (2.10-8), obtain the energy Eq. (2.10-23).

Bibliography [1] Obert, E.F., Concepts of Thermodynamics, McGraw-Hill, New York, 1960. [2] Fung, Y.C., Foundations of Solid Mechanics, Prentice Hall, Englewood Cliffs, New Jersey, 1965. [3] Clausius, R., On Different Forms of the Fundamental Equations of the Mechanical Theory of Heat and Their Convenience for Application, Phy. (Leipzig), Ser. 2, Vol. 125, pp. 313, 1865. [4] Rankine, W.J.M., On the Hypothesis of Molecular Vortices, or Centrifugal Theory of Elasticity, and its Connexion with the Theory of Heat, Phil. Mag., Ser. 4, No. 67, November 1855, pp. 354–363 and No. 68, pp. 411– 420, December 1855. [5] Rankine, W.J.M., On the Thermal Energy of Molecular Vortices, Trans. Royal Soc. Edinburgh, Vol. 25, pp. 557–566, 1869. [6] Takeuti, Y., Foundations for Coupled Thermoelasticity, J. Therm. Stresses, Vol. 2, pp. 323–339, 1979. [7] Takeuti, Y., A Treatise on the Theory of Thermal Stresses, Vol. 1–2, University of Osaka Prefecture, 1978. [8] Parkus, H., Thermoelasticity, 2nd edition, Springer, New York, 1976. [9] Melan, E. and Parkus, H., W¨ armespannungen infolge stationärer Temperaturfelder, Springer, Vienna, 1953. [10] Parkus, H., Instationäre Wärmespannungen, Springer, Vienna, 1959. [11] Truesdell, C. and Noll, W., The Non-Linear Field Theories of Mechanics, Encyclopedia of Physics, Vol. III/3, Springer, Berlin, 1965. [12] Lord, H.W. and Shulman, Y., A Generalized Dynamical Theory of Thermoelasticity, J. Mech. Phys. Solids, Vol. 15, pp. 299–309, 1967. 99

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[13] Boley, B.A. and Weiner, J.H., Theory of Thermal Stresses, Wiley, New York, 1960. [14] Chandrasekharaiah, D.C., Hyperbolic Thermoelasticity: A Review of Recent Literature, Appl. Mech. Rev., Vol. 51, No. 12, Part 1, pp. 705–729, December 1998. [15] M¨ uller, I., The Coldness, A Universal Function in Thermoelastic Bodies, Arch. Rat. Mech. Anal., Vol. 41, No. 5, pp. 319–332, 1971. [16] Green, A.E. and Laws, N., On the Entropy Production Inequality, Arch. Rat. Mech. Anal., Vol. 45, No. 1, pp. 47–53, 1972. [17] Suhubi, E.S., Thermoelastic Solids, Chapter 2 in: Continuum Physics II, A.C. Eringen (Ed.), Academic Press, New York, 1975. [18] Green, A.E. and Lindsay, K.A., Thermoelasticity, J. Elasticity, Vol. 2, No. 1, pp. 1–7, March 1972. [19] Ignaczak, J., Linear Dynamic Thermoelasticity: A Survey, Shock Vib. Dig., Vol. 13, No. 9, pp. 3–8, 1981. [20] Chandrasekharaiah, D.S., Thermoelasticity with Second Sound: A Review, Appl. Mech. Rev., Vol. 39, No. 3, pp. 355–376, March 1986. [21] Vedavarz, A., Kumar, S., and Moallemi, K., Significance on Non-Fourier Heat Waves in Microscale Conduction, DSC, Vol. 32, Micromechanical Sensors, Actuators, and Systems, ASME, p. 109, 1991. [22] Kaminski, W., Hyperbolic Heat Conduction Equation for Materials with a Nonhomogeneous Inner Structure, J. Heat Transfer, Vol. 112, pp. 555– 560, 1990. [23] Green, A.E. and Naghdi, P.M., Thermoelasticity Without Energy Dissipation, J. Elasticity, 31, pp. 189–208, 1993. [24] Green, A.E. and Naghdi, P.M., A Re-examination of the Basic Postulate of Thermomechanics, Proc. Royal Soc. New York, Vol. A432, pp. 171–194, 1991. [25] Maxwell, J.C., On the Dynamical Theory of Gases, Phil. Trans. Roy. Soc. New York, Vol. 157, pp. 49–88, 1867. [26] Chester, M., Second Sound in Solids, Phys. Rev., Vol. 131, pp. 2013–2015, 1963. [27] Ignaczak, J., Domain of Influence Results in Generalized Thermoelasticity, A Survey, App. Mech. Rev., Vol. 44, No. 9, pp. 375–382 , 1991.

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[28] Francis, P.H., Thermomechanical Effects in Elastic Wave Propagation: A Survey, J. Sound Vib., Vol. 21, pp. 181-192, 1972. [29] Bagri, A. and Eslami, M.R., A Unified Generalized Thermoelasticity Formulation; Application to Thick Functionally Graded Cylinders, J. Therm. Stresses, Vol. 30, No. 9 and 10, pp. 911-930, 2007. [30] Bagri, A. and Eslami, M.R., A Unified Generalized Thermoelasticity: Solution for Cylinders and Spheres, Int. J. Mechanical Sci., Vol. 49, pp. 1325-1335, 2007. [31] Nowacki, W., Dynamic Problems of Thermoelasticity, English Edition, Ed. by Francis, P.H., Hetnarski, R.B., Noordhoff Int. Pub., Leyden, The Netherlands, 1975. [32] Ignaczak, J., A Uniqueness Theorem for Stress-Temperature Equations of Dynamic Thermoelasticity, J. Therm. Stresses, Vol. 1, No. 2, pp. 163–170, 1978. [33] Ignaczak, J., Uniqueness Theorem in Generalized Thermoelasticity, J. Therm. Stresses, Vol. 2, No. 2, pp. 171–176, 1979. [34] Bem, Z., Existance of a Generalized Solution in Thermoelasticity with One Relaxation Time, J. Therm. Stresses, Vol. 5, No. 2, pp. 195–206, 1982. [35] Bem, Z., Existence of a Generalized Solution in Thermoelasticity with Two Relaxation Times, Part II, J. Therm. Stresses, Vol. 6, No. 2–4, pp. 281–299, 1983. [36] Ignaczak, J., A Note on Uniqueness in Thermoelasticity with One Relaxation Time, J. Therm. Stresses, Vol. 5, No. 3–4, pp. 257–264, 1982. [37] Sherief, H.H. and Dhaliwal, R.S., A Uniqueness Theorem and Variational Principle for Generalized Thermoelasticity, J. Therm. Stresses, Vol. 3, No. 4, pp. 223–230, 1980. [38] Bem, Z., Existence of a Generalized Solution in Thermoelasticity with Two Relaxation Times, Part I, J. Therm. Stresses, Vol. 5, No. 3–4, pp. 395–404, 1982. [39] Iesan, D., Reciprocity, Uniqueness, and Minimum Principles in Dynamic Theory of Thermoelasticity, J. Therm. Stresses, Vol. 12, No. 4, pp. 465– 482, 1989. [40] Wang, J. and Dhaliwal, R.S., Uniqueness in Generalized Thermoelasticity in Unbounded Domains, J. Therm. Stresses, Vol. 16, No. 1, pp. 1–10, 1993.

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Chapter 3 Basic Problems of Thermoelasticity Not all types of temperature changes in a solid continuum result in creation of thermal stresses. The chapter begins with the discussion of the condition on what type of temperature distribution causes thermal stresses. The analogy of temperature gradient with body forces is stated. Then the theoretical analysis of thermal stress problems is presented in three main classical coordinate systems, that is, the rectangular Cartesian coordinates, the cylindrical coordinates, and the spherical coordinates. In discussing the analytical methods of solution, Airy stress function, Boussinesq function, the displacement potential, Michell function, and Papkovich functions are defined and the general solution in three coordinate systems are given in terms of these functions.

1

Introduction

In this chapter, some basic problems of the theory of thermoelasticity are discussed. The answers will be given to the following important questions: 1. Do all kinds of distribution of the temperature in a solid body create thermal stresses? 2. What are the relations between the temperature distribution, the boundary conditions, and the thermal stresses? 3. Can we relate the thermal stresses to the body forces? In addition, will be presented some general methods of solutions of problems of thermoelasticity in two and three dimensions. Navier equations will be derived R.B. Hetnarski, M.R. Eslami, Thermal Stresses – Advanced Theory and Applications, Solid Mechanics and its Applications 158, c Springer Science + Business Media B.V. 2009

105

106

Chapter 3. Basic Problems of Thermoelasticity

in three coordinate systems, namely, in the rectangular Cartesian coordinates, the cylindrical coordinates, and the spherical coordinates. The general solution of the set of governing equations in each of these coordinate systems will be given. Although the final solutions depend on specific boundary conditions, the general solutions in this chapter will be given without the consideration of the effect of the boundary conditions. However, in the following chapters, some proposed methods of solution will be applied to physical problems with boundary conditions stated.

2

Temperature Distribution for Zero Thermal Stress

Generally, when a body is exposed to a thermal gradient, it is expected that thermal stresses will be developed within the body. The question arises whether any type of thermal gradient results in thermal stresses. Consider a freely supported body, so that no constraint prevents its thermal expansion. We further assume that the boundary traction t¯ni and the body force Xi are zero. Setting all the stress components equal to zero, σij = 0, the surface boundary condition t¯ni = nj σ ¯ij (3.2-1) is identically satisfied, and the governing equation in terms of the stresses, Eq. (1.11-18), reduces to T,ij +

3λ + 2μ T,kk δij = 0 λ + 2μ

(3.2-2)

In the expanded form, this equation reads ∂ 2T 1+ν 2 =0 )∇ T + 1−ν ∂x2 1+ν 2 ∂ 2T ( =0 )∇ T + 1−ν ∂y 2 1+ν 2 ∂ 2T ( )∇ T + 2 = 0 1−ν ∂z ∂ 2T ∂ 2T = 0, = 0, ∂x∂y ∂y∂z (

∂2T =0 ∂z∂x

(3.2-3)

Adding up the first three equations results in ∇2 T = 0. This means that the only possible temperature distribution which produces zero thermal stresses in a body of simply connected region is when ∇2 T = 0 ∂2T ∂2T ∂2T = = =0 ∂x∂y ∂y∂z ∂z∂x

(3.2-4)

2. Temperature Distribution for Zero Thermal Stress

107

The unique solution for the temperature distribution satisfying Eq. (3.2-4) is T − T0 = a + bx + cy + dz = B0 + Bi xi

(3.2-5)

where a, b, c, and d are some arbitrary constants of integration and B0 = a, B1 = b, B2 = c, and B3 = d. A temperature distribution of this form will not produce any thermal stresses in a body of simply connected region provided that the body has not been constrained by its boundary in any direction. For multiply connected region, in addition to Eq. (1.11-18), Cesàro integral equations (1.11-21) and (1.11-22) must be satisfied for zero thermal stresses. Equation (1.11-21) for σij = 0 reduces to [1] Cs

[(T − T0 )δil − ejik ekml xj T,m ]dxl = 0

s = 1, 2, ..., N

(3.2-6)

Substituting for temperature from Eq. (3.2-5) and using the identity on permutation symbol, Eq. (1.8-13), this equation yields

B0

Cs

dxl +

Cs

Bi xi dxl −

Cs

Bk xk dxl +

Cs

Bl xm dxm = 0

(3.2-7)

This equation is identically satisfied as the second and third terms cancel and the first and last integrals on the closed curve Cs are zero. The second Cesàro integral equation for σij = 0, from Eq. (1.11-22), yields Cs

ekml T,m dxl = 0

s = 1, 2, ..., N

(3.2-8)

s = 1, 2, ..., N

(3.2-9)

Substituting T from Eq. (3.2-5) gives Cs

ekml Bm dxl = 0

This equation is clearly satisfied. It is, therefore, concluded that a linear distribution of temperature in a body, of either simply or multiply connected region, results in zero thermal stresses provided that the boundaries are free of traction. The displacement components, not including the rigid body motion, corresponding to the temperature distribution (3.2-5) are obtained from the straindisplacement relations. Since the stresses are zero, then ∂u = α(T − T0 ) = α(a + bx + cy + dz) ∂x ∂v = α(T − T0 ) = α(a + bx + cy + dz) yy = ∂y ∂w zz = = α(T − T0 ) = α(a + bx + cy + dz) ∂z ∂u ∂v ∂v ∂w ∂w ∂u + = + = + =0 ∂y ∂x ∂z ∂y ∂x ∂z xx =

(3.2-10)

108


Integration of these partial differential equations, following a procedure described by Love [2] (page 127), and by Timoshenko and Goodier [3] (page 252), yields b u = α[(a + bx + cy + dz)x − (x2 + y 2 + z 2 )] 2 c v = α[(a + bx + cy + dz)y − (x2 + y 2 + z 2 )] 2 d 2 w = α[(a + bx + cy + dz)z − (x + y 2 + z 2 )] 2

(3.2-11)

The condition on temperature distribution for a two-dimensional plane stress or plane strain problem to have zero thermal stresses is obtained by setting the stress components equal to zero in the proper governing equations. We assume that in either case, the temperature distribution is independent of z, and is a function of x and y. Furthermore, we will first establish the condition for simply connected regions. Plane stress problems For plane stress problems the condition is obtained by setting in Eq. (1.12- 9) σxx = σyy = σxy = 0

(3.2-12)

Since the body force and the inertia terms are also excluded, Eq. (1.12-9) for zero thermal stress condition reduces to ∇2 T =

∂ 2T ∂2T + =0 ∂x2 ∂y 2

(3.2-13)

In this case the strains are xx = yy = zz = α(T − T0 ) xy = yz = zx = 0

(3.2-14)

Plane strain problems For simple plane strain problems, from Eq. (1.12-26), the condition for zero in-plane stresses, σxx = σyy = σxy = 0, is ∇2 T =

∂ 2T ∂2T + =0 ∂x2 ∂y 2

(3.2-15)

The stress in z-direction is σzz = ν(σxx + σyy ) − Eα(T − T0 ) = −Eα(T − T0 )

(3.2-16)

2. Temperature Distribution for Zero Thermal Stress

109

That is, the harmonic temperature distribution satisfying Eq. (3.2-15) results in zero in-plane stresses, but non-zero σzz . The strains for this case, from Eq. (1.12-24), are zz = 0 xx = yy = (1 + ν)α(T − T0 ) xy = yz = zx = 0

(3.2-17)

For multiply connected regions, the temperature distribution causing zero thermal stresses should satisfy Michell conditions in addition to the harmonic equation (3.2-15). Setting the stress components equal to zero, Eq. (1.13-12) for plane the stress condition gives Cs

∂T ds = 0 ∂n

(3.2-18)

In addition, Eqs. (1.13-18) and (1.13-20) yield

(x Cs

∂T ∂T ∂T ∂T −y )ds = + x )ds = 0 (y ∂s ∂n ∂s ∂n Cs

(3.2-19)

Equation (3.2-18) means that the heat flux through each of the interior holes must be zero. The same conditions are obtained for the plane strain condition. Therefore, the temperature distribution causing zero thermal stresses in the plane stress and plane strain conditions of a multiply connected region must satisfy Eqs. (3.2-15), (3.2-18), and (3.2-19). The temperature distribution for the three-dimensional problems in which the thermal displacements are zero is easily obtained by considering u=v=w=0

(3.2-20)

This yields xx = yy = zz = xy = yz = zx = σxy = σyz = σzx = 0

(3.2-21)

From the stress-strain relations, Eq. (1.9-5), it follows that the corresponding stresses are Eα(T − T0 ) σxx = σyy = σzz = − (3.2-22) 1 − 2ν This result is equivalent to a state of hydrostatic stresses for an incompressible material where the normal components of the stresses in any direction are equal. This situation occurs when a body is heated and is prevented to expand in any direction.

110

3


Analogy of Thermal Gradient with Body Forces

The problems of thermal stresses can be formulated in such a way that the effects of thermal gradient are considered as the body forces. This method reduces the problem to those of elasticity problems in the presence of body forces. For this purpose, we may consider a pure thermal stress problem, where body and surface forces are assumed to be zero. The equations of motion (1.103) for the static condition and in the absence of body forces are ∂e + μ∇2 u − ∂x ∂e (λ + μ) + μ∇2 v − ∂y ∂e (λ + μ) + μ∇2 w − ∂z (λ + μ)

Eα ∂T =0 1 − 2ν ∂x Eα ∂T =0 1 − 2ν ∂y Eα ∂T =0 1 − 2ν ∂z

(3.3-1)

where e = xx + yy + zz . The boundary conditions, from Eq. (1.10-6) with tni = 0 are ∂u ∂u ∂u ∂u ∂v ∂w + ny + nz ) + G(nx + ny + nz ) ∂x ∂y ∂z ∂x ∂x ∂x Eα(T − T0 ) nx = 0 − 1 − 2ν ∂v ∂v ∂v ∂u ∂v ∂w λeny + G(nx + ny + nz ) + G(nx + ny + nz ) ∂x ∂y ∂z ∂y ∂y ∂y Eα(T − T0 ) − ny = 0 1 − 2ν ∂w ∂w ∂w ∂u ∂v ∂w λenz + G(nx + ny + nz ) + G(nz + ny + nz ) ∂x ∂y ∂z ∂z ∂z ∂z Eα(T − T0 ) − nz = 0 1 − 2ν (3.3-2)

λenx + G(nx

where nx , ny , and nz are the cosine directors of the unit outer normal vector to the boundary. Comparison of these equations with those of elasticity equations reveals that a thermal stress problem may be treated as an elasticity (nonthermal) problem in the presence of body forces if an equivalent body force with components Eα ∂T 1 − 2ν ∂x

Eα ∂T 1 − 2ν ∂y

Eα ∂T 1 − 2ν ∂z

in x, y, and z-directions, respectively, is considered. It is similarly concluded that the term Eα(T − T0 )/(1 − 2ν) in Eqs. (3.3-2) is equivalent to the surface

3. Analogy of Thermal Gradient with Body Forces

111

force acting on the boundary in an elasticity (non-thermal) problem. In other words, thermal stress problems are similar to elasticity problems where a body is in the state of uniform constant temperature, but placed in a body force field equivalent to a hydrostatic state of stress p=−

Eα (T − T0 ) 1 − 2ν

(3.3-3)

The stresses obtained by this equivalent body force should be superimposed on the stresses obtained by stress-strain relations without thermal expansion. Equation (3.3-3) has another significant property in interpreting the thermal effects in solid bodies. Consider a small cube of an elastic solid under a uniform temperature change (T − T0 ). If we allow the cube to expand freely, the thermal strains are xx = yy = zz = α(T − T0 ) xy = yz = zx = 0 In order to bring back the sides of the cube to its original length, a hydrostatic pressure must act on all sides of the cube. It is easily verified that the pressure given in Eq. (3.3-3) is the hydrostatic pressure which forces the cube to its original size and, therefore, brings all the thermal expansions to zero. It should be noticed, however, that the dimensions of the hydrostatic pressure are not the same as that of body force, the former one being force per unit area, and the latter being force per unit volume. Let us now consider a heated body with zero body forces and free of surface traction, but subjected to some arbitrary thermal gradient. The thermal stresses can be calculated from Eqs. (3.3-1) once the displacement components are obtained, using the boundary condition (3.3-2). This problem may be treated as an isothermal elasticity problem provided we make a transformation of the stresses Eα(T − T0 ) 1 − 2ν Eα(T − T0 ) σyy = σyy − 1 − 2ν Eα(T − T0 ) σzz = σzz − 1 − 2ν σyz = σyz σxy = σxy − σxx = σxx

σzx = σzx

(3.3-4)

where non-primed stresses are the real existing stresses in a heated body, and primed stresses are the stresses in an unheated body which is subjected to the equivalent body force. The temperature distribution in Eqs. (3.3-4) may be as general as T = T (x, y, z). The relations between the strains and stresses

112


resulting from the equivalent body forces are 1 + σzz )] [σ − ν(σyy E xx 1 yy = [σyy − ν(σxx + σzz )] E 1 zz = [σzz − ν(σxx + σyy )] E 1 1 xy = yz = σxy σ 2G 2G yz

xx =

zx =

1 σ 2G zx

(3.3-5)

As expected, relations between ij and σij exclude the term α(T − T0 ) which exists in thermal stress problems, because the effect of temperature is now included in the body force. Substitution of Eqs. (3.3-4) in the equation of motion (1.3-8), with X = Y = Z = 0 and ρ¨ ui = 0 gives ∂σ ∂σxx ∂σ Eα ∂T + xy + xz − =0 ∂x ∂y ∂z 1 − 2ν ∂x ∂σ ∂σ ∂σxy Eα ∂T + yy + yz − =0 ∂x ∂y ∂z 1 − 2ν ∂y ∂σ ∂σ Eα ∂T ∂σxz + yz + zz − =0 ∂x ∂y ∂z 1 − 2ν ∂z

(3.3-6)

The boundary conditions in terms of new stresses become Eα(T − T0 ) nx 1 − 2ν Eα(T − T0 ) σyx nx + σyy ny + σyz nz = ny 1 − 2ν Eα(T − T0 ) σzx nx + σzy ny + σzz nz = nz 1 − 2ν nx + σxy ny + σxz nz = σxx

(3.3-7)

A review of Eqs. (3.3-5) to (3.3-7) shows that a thermal stress problem can be solved as an isothermal elasticity problem provided that we take a body force field with components in x, y, and z-directions given by Eα ∂T 1 − 2ν ∂x Eα ∂T Y = 1 − 2ν ∂y Eα ∂T Z = 1 − 2ν ∂z X =

and a surface traction with components Eα(T − T0 ) nx X¯ = 1 − 2ν

(3.3-8)

4. General Solution of Thermoelastic Problems Eα(T − T0 ) Y¯ = ny 1 − 2ν Eα(T − T0 ) Z¯ = nz 1 − 2ν

113

(3.3-9)

After the set of primed stresses, σij , are obtained, the true stresses are calculated using Eqs. (3.3-4).

4

General Solution of Thermoelastic Problems

Equilibrium equations in terms of displacement components were obtained to be, see Eq. (3.3-1), Eα ∂T ∂e + μ∇2 u = ∂x 1 − 2ν ∂x ∂e Eα ∂T 2 (λ + μ) + μ∇ v = ∂y 1 − 2ν ∂y ∂e Eα ∂T (λ + μ) + μ∇2 w = ∂z 1 − 2ν ∂z (λ + μ)

(3.4-1)

where the temperature distribution T is assumed to be known from heat transfer equations. The complete solution of these equations may be represented as ui = ugi + uTi

(i = 1, 2, 3)

(3.4-2)

where ugi is the general solution of Eqs. (3.4-1) with the right-hand terms equal zero, and uTi is a particular solution. The general solution of Eqs. (3.4-1) with the right-hand terms equal zero was proposed by Papkovich [4,5] in the form ugi =

∂ (xk Φk ) − 4(1 − ν)Φi ∂xi

(3.4-3)

where Φi is a solution of the harmonic equation ∇2 Φi = 0

(3.4-4)

A particular solution of Eqs. (3.4-1) was suggested by Papkovich [5] and Goodier [6] in the form ∂ψ (3.4-5) uTi = ∂xi where ψ is a scalar function satisfying the equation ∇2 ψ =

1+ν α(T − T0 ) 1−ν

(3.4-6)

114


Combining Eq. (3.4-3) with Eq. (3.4-5), the total solution of Eqs. (3.4-1) reduces to ∂ ui = (ψ + xk Φk ) − 4(1 − ν)Φi (3.4-7) ∂xi This method of solution satisfies all the conditions for both a simply-connected region as well as a multiply-connected region. The function ψ is called the displacement potential, and represents a particular solution in thermoelasticity problems. In solving quasi-static thermoelasticity problems by the displacement potential method, we should obtain a solution for the displacement potential ψ. This step of solution is performed after the temperature function has been determined from the heat transfer equations and the thermal boundary conditions. The displacements and stresses obtained in this step will generally not satisfy the boundary conditions. To complement the solution, the general solution for displacement corresponding to isothermal elasticity, ugi , must be added to the displacement components obtained from the displacement potential. The constants of integration involved in the general solution are to be determined in such a manner that the total solution, the sum of ugi and uTi , satisfies all the boundary conditions. Plane stress Let us now apply the results obtained above to the simple case of plane stress problems. Equations (3.4-1) in this case reduce to 1−ν 2 ∂ ∂u ∂v ∂T ∇ u+ ( + ) = 2α 1+ν ∂x ∂x ∂y ∂x ∂ ∂u ∂v ∂T 1−ν 2 ∇ v+ ( + ) = 2α 1+ν ∂y ∂x ∂y ∂y

(3.4-8)

By introducing the displacement potential ψ through the relations u=

∂ψ ∂x

v=

∂ψ ∂y

(3.4-9)

Eqs. (3.4-8) become ∂ 2 [ ∇2 ψ − 2α(T − T0 )] = 0 ∂x 1 + ν ∂ 2 [ ∇2 ψ − 2α(T − T0 )] = 0 ∂y 1 + ν

(3.4-10)

Equations (3.4-10) are both satisfied if ∇2 ψ = (1 + ν)α(T − T0 )

(3.4-11)

4. General Solution of Thermoelastic Problems

115

The stress-displacement relations for the plane stress condition, see Eq. (1.122), reduce to E ∂u ∂v [( + ν ) − (1 + ν)α(T − T0 )] 2 1 − ν ∂x ∂y E ∂v ∂u = [( + ν ) − (1 + ν)α(T − T0 )] 1 − ν 2 ∂y ∂x E ∂u ∂v = ( + ) 2(1 + ν) ∂y ∂x

σxx = σyy σxy

(3.4-12)

which, upon substitution of Eqs. (3.4-9), and using Eq. (3.4-11), become ∂2ψ ∂y 2 ∂2ψ = −2G 2 ∂x ∂ 2ψ =G ∂x∂y

σxx = −2G σyy σxy

(3.4-13)

The stresses obtained through Eqs. (3.4-13) will clearly not satisfy the boundary conditions. To complete the formulation, the general solution corresponding to isothermal elasticity must be added to the stresses in Eqs. (3.4-13). To incorporate the general solution, we may use either Eqs. (3.4-3) and (3.4-4), or, instead, the complementary solution given by Airy stress function. In the latter case, the biharmonic equation representing the general solution of an isothermal elasticity problem in terms of Airy stress function is ∇4 Φ = 0

(3.4-14)

where Φ is Airy stress function. The relations between the stresses and Airy stress function in the plane stress condition are ∂2Φ ∂y 2 ∂ 2Φ = ∂x2 ∂2Φ =− ∂x∂y

σxx = σyy σxy

(3.4-15)

Therefore, the complete solution for a thermoelasticity problem in terms of Airy stress function Φ and the displacement potential ψ for the plane stress condition becomes σxx =

∂2 (Φ − 2Gψ) ∂y 2

116

Chapter 3. Basic Problems of Thermoelasticity ∂2 (Φ − 2Gψ) ∂x2 ∂2 =− (Φ − 2Gψ) ∂x∂y

σyy = σxy

(3.4-16)

Plane strain For the plane strain condition, when zz = 0, the equilibrium equations in terms of displacement components reduce to ∂ ∂u ∂v ∂T ( + ) = 2(1 + ν)α ∂x ∂x ∂y ∂x ∂ ∂u ∂v ∂T (1 − 2ν)∇2 v + ( + ) = 2(1 + ν)α ∂y ∂x ∂y ∂y

(1 − 2ν)∇2 u +

(3.4-17)

which upon substitution of Eqs. (3.4-9) yields ∇2 ψ =

1+ν α(T − T0 ) 1−ν

(3.4-18)

To find the relations between the stresses and the displacement potential, we take the stress-strain relations for the plane strain conditions, ∂u ∂v E [(1 − ν) +ν − (1 + ν)α(T − T0 )] (1 + ν)(1 − 2ν) ∂x ∂y E ∂v ∂u = [(1 − ν) +ν − (1 + ν)α(T − T0 )] (1 + ν)(1 − 2ν) ∂y ∂x ∂u ∂v = G( + ) (3.4-19) ∂y ∂x

σxx = σyy σxy

Substituting Eqs. (3.4-9) and Eq. (3.4-18) into Eq. (3.4-19), we get ∂2ψ ∂y 2 ∂2ψ = −2G 2 ∂x ∂2ψ = 2G ∂x∂y

σxx = −2G σyy σxy

(3.4-20)

The complete solution in terms of the displacement potential ψ and the stress function Φ is obtained from Eqs. (3.4-16). The axial stress in the plane strain condition, when zz = 0, is obtained by noting that (3.4-21) σzz = ν(σxx + σyy ) − Eα(T − T0 ) which in terms of the displacement potential ψ is σzz = −2G∇2 ψ

(3.4-22)

5. Solution of Two-Dimensional Navier Equations

117

Equations (3.4-16) in polar coordinates are 1 ∂ 1 ∂2 + 2 2 )(Φ − 2Gψ) r ∂r r ∂φ ∂2 σφφ = 2 (Φ − 2Gψ) ∂r ∂ 1 ∂ σrφ = − [ (Φ − 2Gψ)] ∂r r ∂φ

σrr = (

(3.4-23)

These formulas may be used for both plane stress and plane strain problems.

5

Solution of Two-Dimensional Navier Equations

Navier equations in the two-dimensional thermoelasticity for the steady-state temperature distribution may be solved by a series expansion. The governing steady-state thermoelastic equations for the plane strain condition are ∂ 2u ∂ 2v ∂θ ∂ 2u + (1 − 2ν) + = 2α(1 + ν) 2 2 ∂x ∂y ∂x∂y ∂x ∂θ ∂2v ∂2u ∂ 2v = 2α(1 + ν) 2(1 − ν) 2 + (1 − 2ν) 2 + ∂y ∂x ∂x∂y ∂y 2 2 ∂ θ ∂ θ + =0 ∂x2 ∂y 2

2(1 − ν)

(3.5-1)

where θ = T −T0 is the temperature change. Let us assume that the set of Eqs. (3.5-1) model a problem which exhibits some periodic nature in x direction. In this case the dependent functions u, v, and θ are expanded into a Fourier sine and cosine series as follows from [7]. Introducing the Fourier sine and cosine transformation, respectively, by ∞

2 1 f (x, y) sin mxdx f¯(x, y) = ( ) 2 π 0 2 1∞ f (x, y) cos mxdx fˆ(x, y) = ( ) 2 π 0

(3.5-2)

Eqs. (3.5-1) are transformed to the following form u − (K 2 − 1)mDˆ v = −βmθˆ (D2 − K 2 m2 )¯ (K 2 D2 − m2 )ˆ v + (K 2 − 1)mD¯ u = βmθˆ (D2 − m2 )θˆ = 0

(3.5-3)

118


where D=

d dy

β=

2(1 + ν) α 1 − 2ν

K2 =

2(1 − ν) 1 − 2ν

(3.5-4)

Equations (3.5-3) represent a system of ordinary differential equations which ˆ The solution happens may be solved for the transformed functions u¯, vˆ, and θ. to be u¯ = (A1 + myA2 ) cosh my + (B1 + myB2 ) sinh my vˆ = (A3 + myA4 ) cosh my + (B3 + myB4 ) sinh my θˆ = A cosh my + B sinh my

(3.5-5)

where the coefficients A, B, A1 to A4 , and B1 to B4 are the constants of integration and are functions of m and depend on the boundary conditions. These constants, however, are not independent of each other. Substituting the solution (3.5-5) into the first of Eqs. (3.5-3) and equating the coefficients of cosh my, sinh my, my cosh my, and my sinh my yields A2 = −B4 =

1 m(k 2

[m(K 2 − 1)(A3 + B1 ) − βB]

+ 1) −1 [m(K 2 − 1)(A1 + B3 ) − βA] A4 = −B2 = m(k 2 + 1)

(3.5-6)

The constants A2 , A4 , B2 , and B4 are eliminated from Eqs. (3.5-5) by substitution from Eqs. (3.5-6). Recalling the stress-strain relations for the plane strain condition 2G ∂u ∂v [(1 − ν) +ν − (1 − ν)αθ] 1 − 2ν ∂x ∂y 2G ∂v ∂u = [(1 − ν) +ν − (1 − ν)αθ] 1 − 2ν ∂y ∂x ∂u ∂v = G( + ) ∂y ∂x

σxx = σyy σxy

(3.5-7)

we receive the general solution for stresses in terms of the constants of integration σ ˆxx (x, y) = G{[K 2 (A1 + myA2 + A4 )m − 2A4 m +(k 2 − 2)(B3 + myB4 )m − βA] cosh my + [K 2 m(B1 + myB2 + B4 ) −2B4 m + m(K 2 − 2)(A3 + myA4 ) − βB] sinh my} σ ˆyy (x, y) = G{[m(K 2 − 2)(A1 + myA2 ) + K 2 m(A4 +B3 + myB4 ) − βA] cosh my + [m(K 2 − 2)(B1 + myB2 ) +mK 2 (B4 + A3 + myA4 ) − βB] sinh my}

5. Solution of Two-Dimensional Navier Equations σ ¯xy (x, y) = Gm[(A1 + 2myA2 + B2 − B3 ) sinh my +(B1 + 2myB2 + A2 − A3 ) cosh my]

119

(3.5-8)

The constants of integration are now obtained using the given kinematical and forced boundary conditions, as well as the thermal boundary conditions. The given kinematical boundary conditions must satisfy the first two of Eqs. (3.5-5). The thermal boundary conditions should be applied to the third of Eqs. (3.5-5). The stress boundary conditions should be applied to Eqs. (3.5-8). Once the constants of integration are obtained, the inverse Fourier transformation is applied and the functions are transformed to the original domain. As an example, a plane strain problem of an infinite, homogeneous, isotropic elastic layer occupying the region 0 ≤ y ≤ h is considered. The mechanical boundary conditions may be considered as in [7] σyy = σxy = 0 v = σxy = 0

on on

y=0 y=h

(3.5-9)

The thermal boundary conditions are a ≤ |x| ≤ b

θ = f (x) ∂θ =0 ∂y ∂θ =0 ∂y

0 ≤ |x| ≤ a 0 ≤ |x| < ∞

y=0 |x| > b y = 0 y=h

(3.5-10)

For the given mechanical and thermal boundary conditions, the constants of integration are obtained to be A2 = 0

A4 = 0 B2 = 0 B4 = 0 1 βm−1 A(m) tanh mh A3 = B1 = − 2 K2 − 1 −1 1 βm A(m) A1 = B3 = 2 K2 − 1

(3.5-11)

The constant A(m) is to be found from the second condition of Eqs. (3.5-10). Substituting the constants of integration from Eq. (3.5-11) into Eq. (3.5-5) yields βA(m) cosh m(y − h) 2m(K 2 − 1) cosh mh βA(m) sinh m(y − h) vˆ = 2m(K 2 − 1) cosh mh A(m) cosh m(y − h) θˆ = cosh mh u¯ =

(3.5-12)

120


The inverse transformation of Eqs. (3.5-12) provides the expressions for physical quantities of the displacements and the temperature in the original domain. They are ∞ A(m) cosh m(h − y) β sin mxdm u(x, y) = √ 2 m cosh mh 2π(K − 1) 0 ∞ β A(m) sinh m(h − y) cos mxdm v(x, y) = √ 2 m cosh mh 0 2π(K − 1) cosh m(h − y) 2 1∞ A(m) θ(x, y) = ( ) 2 cos mxdm (3.5-13) π cosh mh 0

6

General Solution in Cylindrical Coordinates

Consider the asymmetric three-dimensional thermoelastic problems in cylindrical coordinates (r, φ, z). The equation of motion (1.3-8) for the static condition in three-dimensional cylindrical coordinates in the absence of body forces are [8–12] σrr,r + r−1 σφr,φ + σzr,z + r−1 (σrr − σφφ ) = 0 σrφ,r + r−1 σφφ,φ + σzφ,z + 2r−1 σrφ = 0 σrz,r + r−1 σφz,φ + σzz,z + r−1 σrz = 0

(3.6-1)

The stress-strain relations are σrr = (2μ + λ)rr + λ(φφ + zz ) − β(T − T0 ) σφφ = (2μ + λ)φφ + λ(rr + zz ) − β(T − T0 ) σzz = (2μ + λ)zz + λ(rr + φφ ) − β(T − T0 ) σrz = 2μrz σφz = 2μφz σrφ = 2μrφ

(3.6-2)

where β = (3λ + 2μ)α. The strain-displacement relations in cylindrical coordinates are rr = u,r φφ = r−1 (u + v,φ ) zz = w,z 1 rφ = (r−1 u,φ + v,r − r−1 v) 2 1 φz = (v,z + r−1 w,φ ) 2 1 rz = (u,z + w,r ) 2

(3.6-3)

6. General Solution in Cylindrical Coordinates

121

Substituting Eqs. (3.6-3) into (3.6-2) and finally into (3.6-1) results in the equilibrium equations in terms of displacement components (2μ + λ)(u,rr + r−1 u,r − r−2 u) + μr−2 u,φφ + μu,zz + (λ + μ)r−1 v,rφ 3μ + λ − v,φ + (λ + μ)w,rz = βT,r r2 μ(v,rr + r−1 v,r − r−2 v) + (2μ + λ)r−2 v,φφ + μv,zz 3μ + λ +(λ + μ)r−1 u,rφ + u,φ + (μ + λ)r−1 w,zφ = βr−1 T,φ r2 μ(w,rr + r−1 w,r + r−2 w,φφ ) + (2μ + λ)w,zz + (λ + μ)(r−1 v,φz u,rz + r−1 u,z ) = βT,z (3.6-4) Based on the displacement formulations, the solution of the governing equations reduces to the following scalar equations for the thermoelastic displacement potential ψ, Michell’s function M , and the so called Boussinesq’s function B, 1+ν α(T − T0 ) 1−ν ∇4 M = 0 ∇2 B = 0 ∇2 ψ =

(3.6-5) (3.6-6) (3.6-7)

The displacement components are related to these functions. They are u = ψ,r − M,rz +

2 B,φ r

1 1 ψ,φ − M,zφ − 2B,r r r w = ψ,z + 2(1 − ν)∇2 M − M,zz v=

where ∇2 =

∂2 1 ∂ ∂2 1 ∂2 + + + ∂r2 r ∂r r2 ∂φ2 ∂z 2

(3.6-8)

(3.6-9)

Substituting the displacement components of Eqs. (3.6-8) into the straindisplacement relations (3.6-3) and the stress-strain relations (3.6-2), results in the following equations for stresses 2 2 σrr = 2G[ψ,rr − ∇2 ψ + (ν∇2 M − M,rr ),z + B,rφ − 2 B,φ ] r r 1 1 1 σφφ = 2G[ ψ,r + 2 ψ,φφ − ∇2 ψ + (ν∇2 M − M,r r r r 1 2 2 − 2 M,φφ ),z − B,rφ + 2 B,φ ] r r r

122


σzz = 2G{ψ,zz − ∇2 ψ + [(2 − ν)∇2 M − M,zz ],z } 1 1 1 M σrφ = 2G[ ψ,rφ − 2 ψ,φ + ( r r r r 1 1 −M,r ),φz − B,rr + B,r + 2 B,φφ ] r r 1 2 σrz = 2G{ψ,rz + [(1 − ν)∇ M − M,zz ],r + B,φz } r 1 1 σzφ = 2G{ ψ,zφ + [(1 − ν)∇2 M − M,zz ],φ − B,rz } (3.6-10) r r where G = μ is the shear modulus. Once the displacement potential ψ, Michell’s function M , and Boussinesq’s functions B are found, the displacements and the stresses are obtained from Eqs. (3.6-8) and (3.6-10), respectively. The general solution of the harmonic and biharmonic equations (3.6-5) to (3.6-7) may be obtained using the method of separation of variables. A very general solution of a harmonic equation may be written as shown in [11]. The solution of a harmonic equation in general cylindrical coordinates in terms of the variables r, φ, and z is a combination of each group of solutions, as given bellow. Selection of each group of the solution depends upon the boundary conditions.

Jn (βr) Yn (βr)

In (βr) Kn (βr) rn r−n

cos nφ sin nφ

cos nφ sin nφ

cos nφ sin nφ

cosh βz sinh βz

z 1

cos βz sin βz

ln r 1

J0 (βr) Y0 (βr)

I0 (βr) K0 (βr) φ 1

φ 1

z 1

φ 1

cosh βz sinh βz

cos βz sin βz

(3.6-11) As an example, a solution of harmonic partial differential equation in cylindrical coordinates with homogeneous boundary conditions along the z-axis is T (r, φ, z) =

[An In (βn r) + Bn Kn (βn r)][Cn cos nφ + Dn sin nφ]

n

×[En cos βn z + Fn sin βn z] If the boundary conditions along the z-axis are nonhomogeneous, the third bracket may be replaced with [En cosh βn z + Fn sinh βn z] Similarly, a very general solution of a biharmonic function may be written as the above plus the functions

7. Solution of Problems in Spherical Coordinates

βrJn+1 (βr) βrYn+1 (βr) βrJ1 (βr) βrY1 (βr)

rn+2 r−n+2

βrIn+1 (βr) βrKn+1 (βr) βrI1 (βr) βrK1 (βr)

φ 1

cos nφ sin nφ

r2 (ln r − 1) r2

cos nφ sin nφ

cosh βz sinh βz

cos nφ sin nφ

φ 1

φ 1

z 1

cos βz sin βz

z 1

rn r−n

cosh βz sinh βz

ln r 1

Jn (βr) Yn (βr)

J0 (βr) Y0 (βr)

cos βz sin βz

In (βr) Kn (βr)

I0 (βr) K0 (βr) cos nφ sin nφ

φ 1

φ 1

φ 1 z3 z2

123

z3 z2

cos nφ sin nφ

r 0

βz sin βz βz cos βz

βz sin βz βz cos βz r3 r ln r

βz sinh βz βz cosh βz cos nφ sin nφ

βz sinh βz βz cosh βz

cos φ sin φ

φ cos φ φ sin φ

z 1 z 1

(3.6-12) Solution of any harmonic and biharmonic equation can be obtained by proper combination of the functions listed in Eqs. (3.6-11) and (3.6-12).

7

Solution of Problems in Spherical Coordinates

Consider the equilibrium equations in spherical coordinate system (r, θ, φ), as shown in Fig. 3.7-1. The equations in dimensionless form and in terms of the displacement components are [13,14] 1 2(1 − ν) ¯ ur − 1 (ρ2 u¯r,ρ ),ρ − e¯,ρ + Δ¯ [(ρ sin θ¯ uθ ),θ 2 1 − 2ν ρ ρ sin θ 2(1 + ν) ¯ T,ρ = 0 +(ρ¯ uφ ),φ ],ρ − 1 − 2ν 1 2(1 − ν) 2(1 + ν) ¯ 1 T ],θ + 2 (ρ2 u¯θ,ρ ),ρ [ e¯ − u¯r,ρ − ρ 1 − 2ν 1 − 2ν ρ 1 uθ,φ − (sin θ¯ + 2 2 [¯ uφ ),θ ],φ = 0 ρ sin θ 2(1 + ν) ¯ 1 2(1 − ν) 1 T ],φ + 2 (ρ2 u¯φ,ρ ),ρ [ e¯ − u¯r,ρ − ρ sin θ 1 − 2ν 1 − 2ν ρ 1 1 − 2 { uφ ),θ ]},θ = 0 [¯ uθ,φ − (sin θ¯ ρ sin θ

(3.7-1)

(3.7-2)

(3.7-3)

124


Figure 3.7-1: Spherical coordinates. where the dimensionless quantities are defined as ρ=

r a

e¯ =

e αT0

T − T0 T¯ = T0

u¯i =

ui aαT0

i = r, θ, φ

(3.7-4)

where a is some characteristic length and T0 is the reference temperature. The ¯ are defined as dimensionless dilatation and Laplace operator Δ 1 1 2 (ρ u¯r ),ρ + [(sin θ¯ uθ ),θ + u¯φ,φ ] ρ2 ρ sin θ 1 ∂ ∂ 1 ∂2 ¯ = 1 ∂ (ρ2 ∂ ) + Δ ] [ (sin θ ) + ρ2 ∂ρ ∂ρ ρ2 sin θ ∂θ ∂θ sin θ ∂φ2

e¯ =

(3.7-5) (3.7-6)

Now, ∂/∂φ is applied to Eq. (3.7-2), then Eq. (3.7-3) is multiplied by sin θ and ∂/∂θ is applied to it. Adding the two resulting equations, e¯, u¯r , and T¯ are eliminated and the result is ¯ 1 [¯ uθ,φ − (sin θ¯ uφ ),θ ]} = 0 Δ{ sin θ

(3.7-7)

Introducing the thermoelastic displacement potential ψ by u¯θ =

1 ψ,θ ρ

u¯φ =

1 ψ,φ ρ sin θ

(3.7-8)

Eq. (3.7-7) is automatically satisfied by the definition of the thermoelastic displacement potential ψ given by Eq. (3.7-8). To obtain the relation between u¯r and ψ, Eq. (3.7-8) is substituted into Eq. (3.7-1) 2(1 − ν) 1 1+ν ¯ T ),ρ + 2 (¯ e− {[sin θ(¯ ur − ψ,ρ ),θ ],θ 1 − 2ν 1−ν ρ sin θ 1 + (¯ ur − ψ,ρ ),φφ } = 0 sin θ

(3.7-9)

7. Solution of Problems in Spherical Coordinates

125

Dividing Eq. (3.7-9) by ρ, gives 2(1 − ν) 1 1+ν ¯ 1 1 1 (¯ e− T ),ρ + 2 {[sin θ( u¯r − ψ,ρ ),θ ],θ 1 − 2ν ρ 1−ν ρ sin θ ρ ρ 1 1 1 + (3.7-10) ( u¯r − ψ,ρ ),φφ } = 0 sin θ ρ ρ The first term of Eq. (3.7-10) is related to Boussinesq’s function B by e¯ −

1+ν ¯ 1+ν T = (ρB),ρ 1−ν 1−ν

(3.7-11)

It is easily checked that the following relation is identically satisfied for any function F 1 1 (3.7-12) (ρF ),ρρ = 2 (ρ2 F,ρ ),ρ ρ ρ On this basis, and from the definition of Boussinesq’s function B, Eq. (3.7-10) is rewritten as 1 1 2(1 + ν) 1 2 1 (ρ B,ρ ),ρ + 2 {[sin θ( u¯r − ψ,ρ ),θ ],θ 2 1 − 2ν ρ ρ sin θ ρ ρ 1 1 1 + ( u¯r − ψ,ρ ),φφ } = 0 sin θ ρ ρ

(3.7-13)

Assuming the following relation for the functions B, u¯r , and ψ 2(1 + ν) 1 B = (¯ ur − ψ,ρ ) 1 − 2ν ρ

(3.7-14)

¯ =0 ΔB

(3.7-15)

then Eq. (3.7-13) gives From Eq. (3.7-14) the expression for u¯r is u¯r = ψ,ρ +

2(1 + ν) ρB 1 − 2ν

(3.7-16)

Substituting Eqs. (3.7-8) and (3.7-16) into Eq. (3.7-5), the expression for dilatation becomes ¯ + 2(1 + ν) 1 (ρ3 B),ρ (3.7-17) e¯ = Δψ 1 − 2ν ρ2 Substituting for e¯ in Eq. (3.7-11), the condition which the thermoelastic displacement potential ψ has to satisfy is obtained to be 4(1 − ν) ¯ = 1 + ν [T¯ − 1 (ρB),ρ − B] Δψ 1−ν 1 − 2ν 1 − 2ν

(3.7-18)

126


The displacement components are all related to the functions ψ and B. Once the solution of Eqs. (3.7-15) and (3.7-18) are obtained for B and ψ, the displacement components are calculated. It may be verified that the displacement components in terms of the functions ψ and B automatically satisfy the equilibrium equations (3.7-1) to (3.7-3). From stress-strain and strain-displacement relations it follows that the stress components are related to the functions ψ and B and they are 2−ν 1 ¯ T (ρB),ρ − (1 − ν)(1 − 2ν) 1−ν 1 1 − μ2 μ σ ¯θθ [ ψ,ρ + ψ,μμ − 2 ψ,μ ] 2 ρ ρ ρ ν 2−ν 1 ¯ + ρB,ρ + B− T (1 − ν)(1 − 2ν) (1 − ν)(1 − 2ν) 1−ν 1 μ 1 1 σ ¯φφ = [ ψ,ρ − 2 ψ,μ + 2 ψ,φφ ] 1+ν ρ ρ ρ (1 − μ2 ) ν 2−ν 1 ¯ T + ρB,ρ + B− (1 − ν)(1 − 2ν) (1 − ν)(1 − 2ν) 1−ν 1 1 1 ψ σ ¯rθ = −(1 − μ2 ) 2 [ ( ),ρρ + B,μ ] 1−ν ρ 1 − 2ν 1 μ 1 σ ¯θφ = − [ψ,μ + ψ] 2 1+ν ρ 1 − μ2 1 1 1 ψ σ ¯φr = ( ),ρ + B] (3.7-19) 1 [ 2 1 − 2ν (1 − μ ) 2 1 + ν ρ 1 1+ν 1 = 1+ν

σ ¯rr =

ψ,ρρ +

where the dimensionless stress is σ ¯ij = σij /EαT0 and μ = cos θ.

8

Problems

∂ (xk Φk ) − 4(1 − ν)Φi is the general so∂xi lution and uTi = ψ,i is the particular solution of the three-dimensional Navier equations.

1. Prove that in xyz-system ugi =

2. Prove Eqs. (3.6-4) using the stress-strain and strain-displacement relations in cylindrical coordinates. 3. Derive equivalent relations to Eqs. (3.6-4) for the plane stress and plane strain conditions. 4. Derive relations for the stresses, the stress function, and the displacement potential in cylindrical coordinates for the plane stress and the plane strain conditions.

8. Problems

127

5. Verify Eqs. (3.6-5) to (3.6-7). 6. Verify the solution of Eqs. (3.5-3) given by Eqs. (3.5-5). 7. Derive the equilibrium equations in terms of the displacement components in spherical coordinates, Eqs. (3.7-1) to (3.7-3). 8. Verify Eqs. (3.7-19). 9. Prove Eq. (3.7-15).

Bibliography [1] Boley, B.A. and Weiner, J.H., Theory of Thermal Stresses, Wiley, New York, 1960. [2] Love, A.E.H., A Treatise on the Mathematical Theory of Elasticity, Dover, New York, 1963. [3] Timoshenko, S. and Goodier, J.N., Theory of Elasticity, 2nd edition, McGraw-Hill, New York, 1951. [4] Papkovich, P.F., Expression for the General Integral of the Basic Equation of the Theory of Elasticity in Terms of Harmonic Functions [in Russian], Izvestiya AN USSR, Ser. math. i estestv. nauk, Vol. 10, 1932. [5] Papkovich, P.F., On the General Integral for Thermal Stresses [in Russian], PMM Vol. 1, No. 2, 1937. [6] Goodier, J.N., On the Integration of the Thermoelastic Equations, Phil. Mag., Vol. 7, 23, 1937. [7] Singh, B.M. and Dhaliwal, R.S., Mixed Boundary-Value Problems of Steady State Thermoelasticity and Electrostatics, J. Therm. Stresses, Vol. 1, No. 1, pp. 1–11, July 1978. [8] Takeuti, Y. and Noda, N., A Three-Dimensional Treatment of Transient Thermal Stresses in a Circular Cylinder Due to an Arbitrary Heat Supply, J. Appl. Mech., Vol. 45, No. 4, pp. 817–821, 1978. [9] Takeuti, Y., Zaime, S., and Noda, N., Thermal Stress Problems in Industry, 1: on Thermoelastic Distribution in Machining Metals, J. Therm. Stresses, Vol. 1, No. 2, pp. 199–210, 1978. [10] Takeuti, Y. and Noda, N., A General Treatise on the Three-Dimensional Thermoelasticity of Curvilinear Aelotropic Solids, J. Therm. Stresses, Vol. 1, No. 1, pp. 25–40, 1978. [11] Takeuti, Y. and Noda, N., Three-Dimensional Transient Thermal Stresses in a Finite Circular Cylinder under Nonaxisymmetric Temperature Distribution, J. Therm. Stresses, Vol. 3, No. 2, pp. 159–183, 1980. 129

130

Bibliography

[12] Ootao, T., Akai, T., and Tanigawa, Y., Three-Dimensional Transient Thermal Stress Analysis of a Nonhomogeneous Hollow Circular Cylinder Due to Moving Heat Source in Axial Direction, J. Therm. Stresses, Vol. 18, No. 5, pp. 497–512, 1995. [13] Kovalenko, A.D., Thermoelasticity, Basic Theory and Applications, Walters-Noordhoff, Groningen, The Netherlands, 1969. [14] Takeuti, Y. and Tanigawa, Y., Transient Thermal Stresses of a Hollow Sphere Due to Rotating Heat Source, J. Therm. Stresses, Vol. 5, No. 3–4, pp. 283–298, 1982.

Chapter 4 Heat Conduction Problems The ability of obtaining the temperature distribution in an elastic continuum through the solutions of the heat conduction equation is an essential tool for the analysis of thermal stress problems. The analytical methods of solution of heat conduction problems may be classified as the differential, the lumped, and the integral formulations. This chapter presents the analytical methods of solution of heat conduction problems based on the differential and lumped formulations. Similarly to the analysis given in Chapter 3, this chapter presents the method of analysis of heat conduction in three main classical coordinate systems, namely, the rectangular Cartesian coordinates, the cylindrical coordinates, and the spherical coordinates. The temperature distributions in each coordinate system for one-, two-, and three-dimensional problems for steady state and transient conditions are obtained. The temperature distributions in a number of structures of functionally graded materials are also presented. The basic mathematical techniques of solution are discussed. The extend of the heat conduction analysis given in this chapter should be adequate for further applications to various thermal stress problems.

1

Introduction

The differential equation of heat conduction was derived in Chapter 2 in conjunction with the initial and boundary conditions. We will now consider the methods of solution of this equation in order to receive the temperature distribution within the body. There are various analytical techniques to solve the heat conduction equation. In this text, the classical methods of solution of differential and partial differential equations using the separation of variables and series solution are employed. The discussion is presented in three classical systems of coordinates, namely, the rectangular Cartesian coordinates, the cylindrical coordinates, and the spherical coordinates. In each coordinate system, the steady state and transient problems will be discussed. For the R.B. Hetnarski, M.R. Eslami, Thermal Stresses – Advanced Theory and Applications, Solid Mechanics and its Applications 158, c Springer Science + Business Media B.V. 2009

131

132

Chapter 4. Heat Conduction Problems

method of solution of the heat transfer problems in this section one may refer to Arpaci [1], Carslaw and Jaeger [2], Holman [3], and Kreith [4].

2

Problems in Rectangular Cartesian Coordinates

The general form of the governing equation of heat conduction in solids in rectangular Cartesian coordinates was derived in Section 9 of Chapter 2. The anisotropic form of Eq. (2.9-9) is ∂T ∂ ∂T ∂ ∂T ∂T ∂ (kx )+ (ky ) + (kz ) = −R + ρc ∂x ∂x ∂y ∂y ∂z ∂z ∂t

(4.2-1)

where T is the absolute temperature and kx , ky , and kz are the coefficients of thermal conduction along the coordinate axes x, y, and z, respectively. We will consider the analytical methods of solution of this partial differential equation and we will discuss some examples.

2.1

Steady State One-Dimensional Problems

When the heat flow is in x direction only and kx = k is constant, the heat conduction equation (4.2-1) reduces to R d2 T =− 2 dx k

(4.2-2)

Integrating twice with respect to x yields the temperature distribution in the form R (4.2-3) dx dx + C1 x + C2 T =− k where C1 and C2 are the constants of integration. Now the following examples may be considered. Example 1 A flat plate of thickness L separates two liquids of different temperatures T1 and T2 , see Fig. 4.2-1. Find the temperature distribution across the thickness of the plate. Solution Since the internal energy generation is zero, Eq. (4.2-1) reduces to d2 T =0 dx2

(a)

2. Problems in Rectangular Cartesian Coordinates

133

Figure 4.2-1: Flat thick plate. The boundary conditions are T (0) = T1 ,

T (L) = T2

(b)

Integrating Eq. (a) twice gives T = C1 x + C2

(c)

Using the boundary conditions leads to the determination of constants C1 and C2 T2 − T1 C2 = T1 , C1 = (d) L Thus T2 − T1 T (x) = (e) x + T1 L Equation (e) is linear in x and, as expected, is independent of the thermal conductivity of the material. Example 2 Consider a plate of thickness 2L1 insulated from both sides by cladding material, see Fig. 4.2-2. The coefficients of heat conduction for the plate and for the cladding material are k1 and k2 , respectively. Assume a rate of internal heat generation per unit volume R produced in the plate. Compute the temperature distribution in the plate and cladding material if the convection heat transfer coefficient is h and the ambient temperature is T∞ . Solution There are two regions in which we are to solve the governing equation of heat conduction, the plate and the cladding material. The equation for the plate R d2 T1 + =0 (a) 2 dx k1

134


Figure 4.2-2: Flat thick plate with cladding material. with the boundary conditions dT1 (0) =0 dx

(b)

T1 (L1 ) = T2 (L1 )

(c)

dT1 (L1 ) dT2 (L1 ) = k2 dx dx The equation for the cladding material k1

d2 T2 =0 dx2

(d)

(e)

with the boundary conditions consisting of Eqs. (c) to (d), and −k2

dT2 (L2 ) = h[T2 (L2 ) − T∞ ] dx

(f )

The solutions of Eqs. (a) and (f ) are T1 = −

Rx2 + Ax + B 2k1

T2 = Cx + D

(g) (h)

Four constants A, B, C, and D are evaluated from the four boundary conditions (b), (c), (d), and (f ) A=0 (i) −R

L21 + B = CL1 + D 2k1 −RL1 = k2 C

(j) (k)


135

−k2 C = h(CL2 + D − T∞ )

(l)

Solving for the constants and substituting into Eqs. (g) and (h) yields the temperature distribution in the plate and cladding material, respectively

and

T1 − T∞ x k1 L2 k1 k2 = 1 − ( )2 − 2( ) − 2( )(1 + ) 2 RL1 /2k1 L1 k2 L1 k2 hL2

(m)

x L2 k2 T2 − T∞ = −( ) + ( )(1 + ) RL21 /2k2 L1 L1 hL2

(n)

Note: Many one-dimensional problems of heat conduction completely solved may be found in Chapter 3 of the book by Naotake Noda, Richard B. Hetnarski, and Yoshinbo Tanigawa, Thermal Stresses, Second edition, Taylor and Francis, New York, 2003, see [5].

2.2

Steady Two-Dimensional Problems. Separation of Variables

The general form of the governing partial differential equation for a steady two-dimensional problem in x and y directions is ∂ ∂T ∂T ∂ (kx )+ (ky ) = −R ∂x ∂x ∂y ∂y

(4.2-4)

where the thermal conductivities in x and y directions are assumed to be variable. The solution of the partial differential equation (4.2-4), when kx is a function of x and ky is a function of y, is obtained by the method of separation of variables, which is the most common method of solution of this partial differential equation. When the boundary conditions of a problem are specified, then according to this method, the solution is sought as the product of functions of each coordinate separately. This allows the constants of integration in each separated function to be found directly from the homogeneous boundary conditions, and the non-homogeneous boundary conditions be treated by using the concept of expansion into a series. To show the method let us assume a general form of Eq. (4.2-4) a1 (x)

∂ 2T ∂T ∂ 2T ∂T + a (x) (x)T + b (y) + b2 (y) + a + b3 (y)T = 0 (4.2-5) 2 3 1 2 2 ∂x ∂x ∂y ∂y

The solution of this equation may be taken in the product form as T (x, y) = X(x)Y (y)

(4.2-6)

136


where X(x) is a function of x alone, and Y (y) is a function of y alone. Upon substitution of Eq. (4.2-6) into Eq. (4.2-5) and after dividing of the whole equation by XY , we get [a1 (x)

d2 X dX 1 d2 Y dY 1 + a (x) (x)X] (y) + b2 (y) + a = −[b + b3 (y)Y ] 2 3 1 2 2 dx dx X dy dy Y (4.2-7)

The left-hand side of Eq. (4.2-7) is a function of the variable x only and the right-hand side is a function of y only. Therefore, we conclude that the only way that the above equation can hold is when both sides are equal to a constant, say, ±λ2 . This constant is called the separation constant. Considering this, the equation reduces to the following ordinary differential equations: d2 X dX + a2 (x) + [a3 (x) ∓ λ2 ]X = 0 2 dx dx

(4.2-8)

d2 Y dY + b2 (y) + [b3 (y) ± λ2 ]Y = 0 2 dy dy

(4.2-9)

a1 (x) b1 (y)

These equations may be solved by the techniques of the ordinary differential equations as two independent equations, and the constants of integration then may be found using the boundary conditions. In solving a partial differential equation by the method of separation of variables two questions may be raised: (1) Is it always possible to find the constants of integration using the given boundary conditions? (2) What sign should be considered for the separation constant? The answer to question (1) is positive, provided that the problem’s geometry is classical (rectangular or circular domains). The answer to question (2) depends on the problem’s geometry and the thermal boundary conditions. The nature of the solution must be compatible with the given boundary conditions. A pair of the homogeneous boundary conditions in a given direction requires harmonic solution in that direction. Therefore, the sign of the separation constant is selected in such a way that the solution in the direction of a pair of homogeneous boundary conditions leads to a harmonic solution. Referring to Eqs. (4.2-8) and (4.2-9), two boundary conditions are required in each x and y direction. If the boundary conditions in x direction, as an example, are homogeneous, the sign of the separation constant must be selected so that the solution in x direction leads to a harmonic function. The method described may be readily extended to the three-dimensional and transient problems as they will be discussed subsequently. Before we show some examples of this method, we may study the expansion into a Fourier series, as it will be needed in the treatment of such problems.


2.3

137

Fourier Series

Consider the following series ∞ ao g(x) = [an cos nx + bn sin nx] + 2 n=1

(4.2-10)

where ao , an , and bn , (n = 1, 2, ...) are constants. If this series converges for −π ≤ x ≤ +π, then the function g(x) is a periodic function with a period 2π, since sin (nx) and cos (nx) are periodic functions of period 2π, that is g(x) = g(x + 2π)

(4.2-11)

Now, we need to know whether it is possible to find a series solution of the above form for a given function f (x) so that upon expansion into a series of sine and cosine functions it converges to the given function. Consider an arbitrary function f (x). Let us assume that this function may be represented by a trigonometric series ∞ a0 [an cos nx + bn sin nx] (4.2-12) + f (x) = 2 n=1 It will be assumed that the above series converges to the value of the function f (x). With this assumption we may integrate both sides of Eq. (4.2-12) from −π to π π π π ∞ π a0 f (x)dx = ( (an cos nxdx + bn sin nxdx) (4.2-13) dx + −π −π 2 −π n=1 −π Evaluating each term gives π a0 dx = πa0 −π 2 π

π

an sin nx π |−π = 0 n −π −π π π bn cos nx π bn sin nxdx = bn sin nxdx = − |−π = 0 n −π −π an cos(nx)dx = an

Therefore,

cos nxdx =

(4.2-14)

π −π

or

f (x)dx = πa0

(4.2-15)

1π a0 = f (x)dx (4.2-16) π −π To calculate an and bn we need to know certain definite integrals. If n and k are integer numbers, then the following relations for n = k hold π

−π π −π

cos nx cos kxdx = 0 cos nx sin kxdx = 0

π

−π

sin nx sin kxdx = 0

(4.2-17)

138


and for n = k

π −π π −π

π

−π

(n = 0)

cos2 nxdx = π cos nx sin nxdx = 0 sin2 nxdx = π

(n = 0)

(4.2-18)

With the aid of the relations (4.2-17) and (4.2-18) we can compute the coefficients an and bn of the series (4.2-12). To find the coefficients an for n = 0, multiplying both sides of Eq. (4.2-12) by cos kx and integrating from −π to π yields π −π

f (x) cos kxdx =

π ∞ a0 π cos kxdx + (an cos nx cos kxdx 2 −π −π n=1 π

+bn

−π

sin nx cos kxdx)

(4.2-19)

From relations (4.2-17) and (4.2-18), all the integrals on the right-hand side of Eq. (4.2-19) equal to zero, except the integral with the coefficient ak , that is π −π

π

f (x) cos kxdx = ak

−π

cos2 kxdx = ak π

(4.2-20)

or

1π f (x) cos kxdx (4.2-21) π −π To obtain bn , we multiply both sides of Eq. (4.2-12) by sin kx and integrate from −π to π ak =

π

−π

or

π

f (x) sin kxdx = bk

−π

sin2 kxdx = bk π

(4.2-22)

1π f (x) sin kxdx (4.2-23) bk = π −π The coefficients a0 , ak , and bk determined by Eqs. (4.2-16), (4.2-21), and (4.223) are called Fourier coefficients of the function f (x), and the trigonometric series (4.2-12) is called Fourier series of the function f (x). One may ask what properties must the function f (x) possess in order to be possible to expand it into a Fourier series. The following theorem deals with this matter [6]. Theorem. If a periodic function f (x) with period 2π is piecewise monotonic and bounded in the interval [−π, π], then Fourier series constructed for this function converges at all points where f (x) is continuous, and it converges to the average of the right- and left-hand limits of f (x) at each point where f (x) is discontinuous.


139

Figure 4.2-3: The step function H(x) in interval [−π, π]. As an example we may consider the function f (x) = −1 for −π < x < 0 and f (x) = 1 for 0 ≤ x ≤ π, as shown in Fig. 4.2-3. This function is piecewise monotonic and bounded in the interval −π ≤ x ≤ π, and thus can be expanded into a Fourier series. Fourier coefficients are π 1π 1 0 f (x)dx = [ (−1)dx + (1)dx] = 0 ao = π −π π −π 0 π 1 0 1 sin kx 0 sin kx π ak = [ (−1) cos kxdx + cos kxdx] = [− |−π + | ]=0 π −π π k k 0 0 π 1 0 bk = [ (−1) sin kxdx + sin kxdx] π −π 0 cos kx π 2 1 cos kx 0 |−π − |0 ] = [1 − cos πk] = [ π k k πk 0 for k = 2, 4, 6, ... bk = 4 for k = 1, 3, 5, ... πk Therefore, 4 1 1 [sin x + sin 3x + · · · + sin (2n − 1)x] π 3 2n − 1 It is noticed that the expansion of the above function is in terms of sine functions which is an odd function of x, that is, sin x = − sin (−x), see Fig. 4.2-4, left part. This property of the function is apparent, as the function f (x) defined above is an odd function itself, i.e., f (x) = −f (−x). Unlike the odd functions, the even functions are such that f (x) = f (−x), see Fig. 4.2-4, right part. It may be further noticed that an odd function is symmetric with respect to the origin of the coordinate axes, whereas an even function is symmetrical with respect to the y-axis. As an example of an even function consider the periodic function f (x) = x2 , −π ≤ x ≤ π. We notice that f (x) = f (−x), and thus f (x) is an even function. Fourier coefficients are 1π 2 1 x3 π 2π 2 x dx = |−π = a0 = π −π π 3 3 f (x) =

140


Figure 4.2-4: Odd and even functions. 1π 2 1 x2 sin kx π 2π x cos kxdx = [ x sin kxdx] |−π − π −π π k k −π 1π 4 −2 x cos kx π [− |−π + cos kxdx] = [π cos kπ] = πk k k −π πk 2 4 for k = 2, 4, 6, ... = k2 4 for k = 1, 3, 5, ... − k2

ak =

1π 2 1 x2 cos kx π 2π x sin kxdx = [− x cos kxdx] |−π + π −π π k k −π 1π 2 x sin kx π sin kxdx] = 0 [ |−π − = πk k k −π

bk =

Therefore, Fourier expansion of the function f (x) becomes f (x) = x2 =

π2 1 1 − 4(cos x − 2 cos 2x + 2 cos 3x − · · ·) 3 2 3

We again notice that the even function f (x) = x2 is expanded into cosine functions which are even, that is, cos x = cos (−x). The function f (x) is piecewise monotonic, bounded, and continuous and, therefore, may be expanded into a convergent series.

2.4

Double Fourier Series

We have discussed the expansion of a function of a single variable into sine and cosine Fourier series. In many practical problems where a function is defined in terms of two independent variables, the solutions may be sought in a form of a series of products of sine and cosine functions. In this situation, the concept of expansion of a given function of two variables into a double series will be an essential tool to handle the non-homogeneous boundary conditions involved in the solution. Expansion of a function of two independent variables into a double Fourier series is discussed in this section. The method will be applied to the solutions of problems of heat conduction, thermal stresses, and deflection of plates.


141

To expand a given function of two variables into a double Fourier series, we follow precisely the same procedure as for single variable functions. Consider a given function f (x, y) defined in the rectangular region −a < x < a, −b < y < b. It is easily verified that for a set of two orthogonal functions, sin mπx/a, m = 1, 2, 3, ..., and cos nπy/b, n = 0, 1, 2, 3, ..., the product of any two functions obeys the following rules [6,7]: For m = i or n = j a b

mπx nπy iπx jπy sin )(sin sin )dxdy = 0 a b a b −a −b a b mπx nπy iπx jπy (sin cos )(sin cos )dxdy = 0 a b a b −a −b a b mπx nπy iπx jπy (cos sin )(cos sin )dxdy = 0 a b a b −a −b a b nπy iπx jπy mπx cos )(cos cos )dxdy = 0 (cos a b a b −a −b (sin

(4.2-24)

The only products of such functions whose integrals do not vanish over the rectangular region are those for which m = i and n = j a b

a b mπx mπx nπy 2 nπy 2 (sin (sin sin ) dxdy = cos ) dxdy a b a b −a −b −a −b a b a b mπx mπx nπy 2 nπy 2 = (cos (cos sin ) dxdy = cos ) dxdy a b a b −a −b −a −b = ab m = n = 1, 2, 3, ... (4.2-25)

If n = 0 in the second, m = 0 in the third, or either m = 0 or n = 0 in the last of the above integrals, the results of the integrations is doubled. If m = n = 0 in the last integral, its value is 4ab. With the above relations, and following the procedure given for Fourier expansion of a function of a single variable, we may expand a given function f (x, y) into a double Fourier series as ∞ ∞

∞ ∞ mπx mπx nπy nπy f (x, y) = Amn sin Bmn sin sin + cos a b a b m=1 n=1 m=1 n=0 ∞ ∞ ∞ ∞ mπx mπx nπy nπy + Cmn cos Dmn cos sin + cos a b a b m=0 n=1 m=0 n=0 (4.2-26)

To find the coefficients of the series, both sides of the above equation are multiplied by one of the functions (sin mπx/a)(sin nπy/b), (sin mπx/a)(cos nπy/b), (cos mπx/a)(sin nπy/b), or (cos mπx/a)(cos nπy/b), and integrated from −a to a with respect to x, and from −b to b with respect to y. Using the results

142


of Eqs. (4.2-24) and (4.2-25), we find nπy 1 ab mπx sin dxdy f (x, y) sin ab −a −b a b 1 ab mπx nπy Bmn = f (x, y) sin cos dxdy ab −a −b a b 1 ab mπx nπy f (x, y) cos sin dxdy Cmn = ab −a −b a b 1 ab mπx nπy f (x, y) cos Dmn = cos dxdy (4.2-27) ab −a −b a b The values of Bm0 , C0n , D0n , Dm0 are one half, and D00 is one quarter of the above values. If the function f (x, y) is an odd function of both x and y, then Bmn = Cmn = Dmn = 0 and it follows that Amn =

f (x, y) =

∞ ∞

Amn sin

m=1 n=1

mπx nπy sin a b

(4.2-28)

where Amn is obtained from the first of Eqs. (4.2-27).

2.5

Bessel Functions and Fourier-Bessel Series

Bessel functions Consider a differential equation of the form [8,9] d2 y dy + x + (x2 − p2 )y = 0 (4.2-29) 2 dx dx where p is any real, imaginary, or complex constant. The solution of this differential equation is sought in the form of a series of products of some power of x, that is x2

y = xr

∞

ak x k

(4.2-30)

k=0

The coefficient a0 is a nonzero constant. Equation (4.2-30) may be rewritten as ∞ y=

ak xr+k

(4.2-31)

k=0

A solution presented by this equation is complete if the coefficients ak are computed in such a way that the series (4.2-31) satisfies the differential equation (4.2-29). Therefore, taking the derivatives y =

y =

∞

(r + k)ak xr+k−1

k=0 ∞ k=0

(r + k)(r + k − 1)ak xr+k−2


143

and substituting in Eq. (4.2-29) yields x2

∞

(r + k)(r + k − 1)ak xr+k−2 + x

k=0

∞

(r + k)ak xr+k−1

k=0

+(x2 − p2 )

∞

ak xr+k = 0

(4.2-32)

k=0

To satisfy Eq. (4.2-32) for all the values of x, the coefficients of x to the power r, r +1, r +2, ..., r +k must be equal to zero, which yields a system of equations [(r − 1)r + r − p2 ]a0 = 0 [(r + 1)r + (r + 1) − p2 ]a1 = 0 [(r + 2)(r + 1) + (r + 2) − p2 ]a2 + a0 = 0 ................................................. [(r + k)(r + k − 1) + (r + k) − p2 ]ak + ak−2 = 0 These equations can be written in the form [r2 − p2 ]a0 = 0 [(r + 1)2 − p2 ]a1 = 0 [(r + 2)2 − p2 ]a2 + a0 = 0 .................................... [(r + k)2 − p2 ]ak + ak−2 = 0

(4.2-33)

The last of Eqs. (4.2-33) can be presented as [(r + k − p)(r + k + p)]ak + ak−2 = 0 Since a0 cannot be zero, therefore, from the first of Eqs. (4.2-33) r 2 − p2 = 0 The roots are r1 = p, r2 = −p. From the second of Eqs. (4.2-33), a1 = 0. First, a solution for r1 = p > 0 is considered. From the system of Eqs. (4.233) all the coefficients a1 , a2 , ... are computed in succession in terms of a0 . For instance, if we put a0 = 1, then ak = −

ak−2 k(2p + k)

(4.2-34)

which for different values of k yields a1 = 0,

a3 = 0,

and in general a2 = −

1 2(2p + 2)

a2ν+1 = 0

144


1 2 × 4(2p + 2)(2p + 4) 1 a2ν = (−1)ν (4.2-35) 2 × 4 × 6 × · · · × 2ν(2p + 2)(2p + 4) × · · · × (2p + 2ν) where ν is a natural number. Upon substitution of the coefficients ak from Eq. (4.2-35) into Eq. (4.2-30), we receive a4 =

x2 x4 + 2(2p + 2) 2 × 4(2p + 2)(2p + 4) x6 + · · ·] − 2 × 4 × 6(2p + 2)(2p + 4)(2p + 6)

y1 = xp [1 −

(4.2-36)

All the coefficients a2ν are determined, as for every k the coefficient of ak in Eqs. (4.2-33), (r + k)2 − p2 , is different from zero. The function y1 from Eq. (4.2-36) is a particular solution of Eq. (4.2-29). Now, we will establish a condition under which all the coefficients ak will be determined for the second root r2 = −p. This occurs if for any even natural number k the following inequalities are satisfied (r2 + k)2 − p2 = 0

(4.2-37)

or ±(r2 + k) = p But p = r1 , hence ±(r2 + k) = r1 Therefore, condition (4.2-37) is in this case equivalent to ±(r1 − r2 ) = k where k is an even natural number. But r1 = p,

r2 = −p

Thus r1 − r2 = 2p Therefore, if p is not equal to an integer, it is possible to write a second particular solution that is obtained from expression (4.2-36) by substituting −p for p x2 x4 + 2(−2p + 2) 2 × 4(−2p + 2)(−2p + 4) x6 + · · ·] − 2 × 4 × 6(−2p + 2)(−2p + 4)(−2p + 6)

y2 = x−p [1 −

(4.2-38)


145

It is easily verified that both series from expressions (4.2-36) and (4.2-38) converge for all the values of x. The series for y1 , in Eq. (4.2-36), multiplied by a certain constant, is called Bessel function of the first kind of order p and is designated by Jp . The series y2 in Eq. (4.2-38) is then J−p . The general solution to Eq. (4.2-29) for p not equal to an integer is y = C1 Jp (x) + C2 J−p (x)

(4.2-39)

where C1 and C2 are arbitrary constants and Jp (x) and J−p (x) are defined as Jp (x) =

∞

(−1)k

( x2 )p+2k k!Γ(p + k + 1)

(4.2-40)

(−1)k

( x2 )−p+2k k!Γ(−p + k + 1)

(4.2-41)

k=0

and J−p (x) =

∞

k=0

The function Γ(.) appearing in these equations is called the Gamma function and is defined for integer numbers as Γ(p + 1) = pΓ(p) = p! Γ(1) = 0! = 1

(4.2-42)

When p is a real number, then Γ(.) is defined as Γ(p)Γ(p − 1) =

π sin πp

(4.2-43)

The solution (4.2-39) is valid as long as p is not an integer number. When p = n, a natural number, Jp (x) and J−p (x) are not two independent solutions of Eq. (4.2-29) and it is easily verified that for p = n the following relation exists (4.2-44) J−n (x) = (−1)n Jn (x) Thus, the functions given in Eq. (4.2-39) are a constant multiple of the other and the solution is not complete, and therefore we must look for another independent solution to be combined with Jn (x) to give the complete solution. To obtain a second linearly independent solution, the function Yp (x) is defined as [9,10] Jp (x) cos pπ − J−p (x) (4.2-45) Yp (x) = sin pπ This combination of Jp (x) and J−p (x) is obviously a solution of Bessel equation for p not being an integer, as cos πp and sin πp are constant numbers. Note that Yp (x) is linearly independent of Jp (x). When p = n is an integer, Yp (x) assumes the indeterminate form 0/0. A rather tedious mathematical manipulation indicates that the limit when p → n exists and Yp (x) in this case is a solution of Bessel equation.

146


We, therefore, conclude that Eqs. (4.2-40) and (4.2-45) are two linearly independent solutions of Bessel equation (4.2-29) when p is any real or imaginary number, and the general solution of Eq. (4.2-29) can be written as y(x) = C1 Jp (x) + C2 Yp (x)

(4.2-46)

The function Yp (x) is known as Bessel function of the second kind of order p. In the differential equation (4.2-29) replacing x by ±ix we arrive at x2

dy d2 y + x − (x2 + p2 )y = 0 2 dx dx

(4.2-47)

This equation is called the modified Bessel equation and its solution is readily obtained by replacing x by ix in Eq. (4.2-46), thus y(x) = C1 Jp (ix) + C2 Yp (ix)

(4.2-48)

According to the definition, from Eq. (4.2-40) it follows that Jp (ix) =

∞ (−1)k ( 12 ix)p+2k k=0 k!Γ(p + k + 1) ∞ ( 12 x)p+2k p

=i

k=0

k!Γ(p + k + 1)

Defining the modified Bessel function of the first kind of order p as Ip (x) =

∞

( 12 x)p+2k k=0 k!Γ(p + k + 1)

(4.2-49)

we find that Jp (ix) = ip Ip (x)

(4.2-50)

When p = n is a natural number In (x) = i−n Jn (ix) and by interchanging n by −n we receive I−n (x) = in J−n (ix) = (−1)n in Jn (ix) = i−n Jn (ix) The comparison shows that In (x) = I−n (x)

(4.2-51)

Also, since k is an even natural number, from Eq. (4.2-49) we find that In (−x) = (−1)n In (x)

(4.2-52)


147

Equation (4.2-52) indicates that In (x) is odd or even, depending upon the value of n. When n is an even integer In (x) is an even function of x, and when n is odd integer In (x) is an odd function of x. The second solution of Eq. (4.2-47), which is linearly independent of Ip (x), is I−p (x) if p is not an integer. But, since I−p (x) is not in general independent of Ip (x) and for integer values of p it is a constant multiple of Ip (x), we introduce the function Kp (x) as 1 [I−p (x) − Ip (x)] Kp (x) = π 2 sin pπ

(4.2-53)

The function Kp (x) is called the modified Bessel function of the second kind of order p, and it is a solution linearly independent of Ip (x). The complete solution of Eq. (4.2-47) is thus shown to be y(x) = C1 Ip (x) + C2 Kp (x)

(4.2-54)

The graphical representations of Bessel functions and the modified Bessel functions of the first and second kinds, respectively, are presented in Figs. 4.2-5 and 4.2-6. More mathematical details are left out and the reader interested in Bessel functions may consult specialized books on the subject. Fourier-Bessel series Let us consider the piecewise continuous function f (x) defined in the interval 0 < x < a. Fourier-Bessel series is defined as f (x) =

∞

Aj Jn (ξj x)

(4.2-55)

j=1

where Aj =

a 1 xf (x)Jn (ξj x)dx Jn (ξj x) 2 0

j = 1, 2, ...

(4.2-56)

and Jn (ξj x) 2 is the square of the norm of Jn (ξj x) on the interval 0 < x < a, with weight function x, and is defined by Jn (ξj x) 2 = =

a

x[Jn (ξj x)]2 dx 0 ξj2 a2 [Jn (ξj a)]2 + (ξj2 a2 2ξj2

− n2 )[Jn (ξj a)]2

(4.2-57)

148

Chapter 4. Heat Conduction Problems 1

J0

J1

J2

Bessel functions

.5

J3

0 −.5

Y0

Y1

−1

Y2

−1.5 −2

0

2

4

x

6

8

10

Figure 4.2-5: Bessel functions of the first and second kind.

modified Bessel functions

4

3

I0

2

K 0 K1

I1

K2

I2 I3

1

0

0

1

2 x

3

4

Figure 4.2-6: Modified Bessel functions of the first and second kind. Now the coefficient Aj in Eq. (4.2-55) is obtained for three special cases. (1) If ξj , (j = 1, 2, ...) are the positive roots of the equation Jn (ξa) = 0

(4.2-58)

then the coefficient Aj is given by Aj =

a 2 xf (x)Jn (ξj x)dx a2 [Jn+1 (ξj a)]2 0

j = 1, 2, ...

(4.2-59)


149

(2) If ξj , (j = 1, 2, ...) are the positive roots of the equation bJn (ξa) + ξaJn (ξa) = 0

(b ≥ 0, b + n > 0)

(4.2-60)

which can also be written in the form (b + n)Jn (ξa) − ξaJn+1 (ξa) = 0

(4.2-61)

then the coefficient Aj is given as a 2ξj2 Aj = 2 2 xf (x)Jn (ξj x)dx (ξj a − n2 + b2 )[Jn (ξj a)]2 0

j = 1, 2, ... (4.2-62)

(3) If n = 0 in Eq. (4.2-55) and ξj , (j = 1, 2, ...) are the positive roots of the equation J0 (ξa) = 0

(4.2-63)

which can also be written in the form J1 (ξa) = 0

(4.2-64)

then the coefficient Aj is given as Aj =

a 2 xf (x)J0 (ξj x)dx a2 [J0 (ξj a)]2 0

j = 1, 2, ...

(4.2-65)

In the latter case, since ξ1 = 0 and J0 (0) = 1, it is more convenient to write Eq. (4.2-55) in the form f (x) = A1 +

∞

Aj J0 (ξj x)

(4.2-66)

j=2

where 2 a A1 = 2 xf (x)dx a 0 a 2 Aj = 2 xf (x)J0 (ξj x)dx a [J0 (ξj a)]2 0

j = 2, 3, ...

(4.2-67)

With the general discussion presented for the method of separation of variables, Fourier expansion of a function, Bessel function, and Fourier-Bessel expansion, we may now consider the solution to some problems.

150


Example 3 Consider a rectangular plate of isotropic homogeneous material subjected to the boundary conditions shown in Fig. 4.2-7. Three sides of the plate are maintained at constant zero temperature, and the fourth side is exposed to a variable temperature. Find the temperature distribution within the plate when the heat generation in the plate is absent. Solution The two-dimensional heat conduction equation is ∂2T ∂2T + =0 ∂x2 ∂y 2 and the boundary conditions are T (0, y) = 0 T (x, 0) = 0,

(a)

T (a, y) = 0 T (x, b) = f (x)

(b)

By the method of separation of variables, the solution is assumed in the form T (x, y) = X(x)Y (y)

(c)

Substituting Eq. (c) in (a) and dividing by XY yields −

1 d2 X 1 d2 Y = X dx2 Y dy 2

(d)

Since each side of Eq. (d) is a function of a different independent variable, the only way that both sides may be equal is that they must be each equal to a separation constant. Taking the separation constant as λ2 , the governing equation is −

1 d2 Y 1 d2 X = = λ2 X dx2 Y dy 2

Figure 4.2-7: Rectangular plate.

(e)


151

or

d2 X + λ2 X = 0 dx2 d2 Y − λ2 Y = 0 (f ) dy 2 The value of the separation constant is determined from the boundary conditions, and its sign depends upon the nature of the boundary conditions. Three cases are distinguished: (a) Case 1, λ2 = 0 For this case the solution of the differential equations (f ) leads to a simple integration Y = A3 + A 4 y (g) X = A1 + A2 x, where A1 through A4 are the constants of integration. From Eq. (c) T = (A1 + A2 x)(A3 + A4 y)

(h)

It is easily verified that the above solution cannot satisfy the given boundary conditions. Thus, this case should be excluded from the consideration. (b) Case 2, λ2 < 0 Solution of Eqs. (f ) is X = B1 e−λx + B2 eλx Y = B3 cos λy + B4 sin λy

(i)

where B1 through B4 are constants, and T = (B1 e−λx + B2 eλx )(B3 cos λy + B4 sin λy)

(j)

We may verify that this solution cannot satisfy the nonhomogeneous boundary condition along the side y = b, and thus it is not acceptable. (c) Case 3, λ2 > 0 Integration of Eqs. (f ) yields the following solution X = C1 cos λx + C2 sin λx Y = C3 e−λy + C4 eλy

(k)

where again C1 through C4 are the integration constants. To find these constants, the boundary conditions (b) must be used. Since three of the boundary conditions are homogeneous, they are directly applied to Eqs. (k), to yield X(0) = 0,

X(a) = 0,

Y (0) = 0

(l)

152


and the nonhomogeneous boundary condition is T (x, b) = f (x)

(m)

The first two of Eqs. (l) yield C1 = 0 sin λa = 0 or λn =

nπ a

n = 1, 2, 3, ...

(n)

The third of Eqs. (l) gives C3 = −C4 Substituting this into Eqs. (k) and then into (c) we find that a solution to the plate problem takes the form T =

∞

Cn sin

n=1

nπy nπx sinh a a

(o)

where Cn are arbitrary constants. To obtain the constants Cn , the nonhomogeneous boundary condition is used. In the following we let f (x) = T1 , where T1 is constant. Then Eq. (m) yields T1 =

∞

Cn sin

n=1

nπx nπb sinh a a

(p)

Expanding f (x) = T1 at y = b into a Fourier sine series gives T1 =

∞ 2T1 (−1)n+1 + 1 nπx sin π n=1 n a

(q)

Comparing Eqs. (p) and (q) results in Cn =

2T1 (−1)n+1 + 1 nπ sinh nπb a

(r)

and finally the solution for the temperature problem is T =

∞ (−1)n+1 + 1 nπx nπy 2T1 sinh nπb sin π n=1 n sinh a a a

This concludes the solution.

(s)


2.6

153

Nonhomogeneous Differential Equations and Boundary Conditions

In the previous section we studied a heat conduction problem in twodimensional rectangular coordinates governed by a homogeneous differential equation, where all boundary conditions except one were homogeneous, and the application of the nonhomogeneous boundary condition resulted in obtaining the final solution to the problem. Now the question arises on how we should deal with problems in which: (a) Two or more boundary conditions are nonhomogeneous. (b) The governing differential equation is nonhomogeneous. The first type of problems is easily handled by the principle of superposition. Since the governing differential equation for heat conduction in solids is linear, therefore, we may use the concept of linear superposition of the auxiliary problems. To describe the method we consider the following example. Example 4 Find the steady-state temperature distribution in the rectangular plate shown in Fig. 4.2-8. Solution By the transformation θ = T − T∞ , the governing equation for the temperature distribution (a) ∇2 T = 0 and the boundary conditions T (0, y) = f (y)

(b)

T (a, y) = T0

(c)

Figure 4.2-8: Rectangular plate.

154

Chapter 4. Heat Conduction Problems k −k

∂T (x, 0) =q ∂y

∂T (x, b) = h(T − T∞ ) ∂y

(d) (e)

are transformed to those shown in Fig. 4.2-9. Clearly, the boundary conditions shown in Fig. 4.2-8 are equivalent to those of Fig. 4.2-9. Since all the four boundary conditions are nonhomogeneous, the problem may be divided into four problems, each one having three homogeneous and one nonhomogeneous boundary conditions. By the simple transformation of θ = T − T∞ , the last condition becomes homogeneous and the problem reduces to the one shown in Fig. 4.2-9. Also, a solution of the problem may be divided into three problems as shown in Fig. 4.2-10, such that θ = θ1 + θ2 + θ3 where θ1 , θ2 , and θ3 are solutions to the appropriate problems. The choice of the origin of the coordinate axis is important when handling the boundary conditions. As noted in Fig. 4.2-10 the origin is selected at the corner where both sides are subjected to homogeneous boundary conditions.

Figure 4.2-9: The problem in terms of θ.

Figure 4.2-10: Decomposition of the main problem into three auxiliary problems with one nonhomogeneous boundary condition each.


155

It is obvious that the total solution is obtained when the three solutions based on a common coordinate system are added. The second kind of problems in which a nonhomogeneous differential equation is to be satisfied is also treated by a superposition method. For two-dimensional problems, the following example will illustrate a method of solution to a problem involving nonhomogeneous boundary conditions. Example 5 Consider an infinitely long bar of square cross section floating in a fluid of constant temperature T0 . If the heat transfer coefficient between the bar and the fluid is large compared with that of the bar and ambient, find a steadystate distribution of the temperature in the cross section of the bar. Solution It is convenient to take the coordinate axes along the sides of the cross section of the bar as shown in Fig. 4.2-11. In terms of θ = T − T∞ and the selected coordinate system, the governing differential equation of heat conduction becomes ∂ 2θ ∂ 2θ + =0 ∂x2 ∂y 2

(a)

subjected to the following boundary conditions θ(0, y) = θ0 −k

∂θ(L, y) = hθ(L, y) ∂x θ(x, 0) = θ0

−k

∂θ(x, L) = hθ(x, L) ∂y

(b)

where θ0 = T0 − T∞ . We assume a solution of the form θ(x, y) = θ1 (x, y) + θ2 (x, y)

Figure 4.2-11: Long bar of square cross section.

(c)

156


where θ1 and θ2 satisfy the following differential equations ∂ 2 θ1 ∂ 2 θ1 + =0 ∂x2 ∂y 2

∂ 2 θ2 ∂ 2 θ2 + =0 ∂x2 ∂y 2

(d)

and the boundary conditions θ1 (0, y) = 0 θ2 (0, y) = θ0 ∂θ1 (L, y) ∂θ2 (L, y) −k −k = hθ1 (L, y) = hθ2 (L, y) ∂x ∂x θ2 (x, 0) = 0 θ1 (x, 0) = θ0 ∂θ1 (x, L) ∂θ2 (x, L) −k −k (e) = hθ1 (x, L) = hθ2 (x, L) ∂y ∂y Since the set of the boundary conditions (b) involves two nonhomogeneous boundary conditions, we have divided the problem into two problems, each containing only one nonhomogeneous boundary condition. Now, each of the two problems is readily solvable by the separation of variables, and a complete solution is the sum of the form (c). Example 6 Consider a rectangular element of Example 4. We assume that in the element a constant amount of heat R per unit volume per unit time is generated and that the heat convection coefficient is large. The element may represent a currentconducting electric wire of rectangular cross section, as shown in Fig. 4.2-12. Find the temperature distribution in the cross section of the wire. Solution The governing equation is ∂2T R ∂2T + =− 2 2 ∂x ∂y k

Figure 4.2-12: Electric wire of rectangular cross section.

(a)


157

and the boundary conditions are T (L, y) = T∞

T (x, l) = T∞

∂T (0, y) =0 ∂x

∂T (x, 0) =0 ∂y

(b)

The last two boundary conditions are due to symmetry of the temperature distribution about x and y axes. By a simple transformation θ = T − T∞ we arrive at the following equation R ∂2θ ∂2θ + 2 =− 2 ∂x ∂y k

(c)

and the boundary conditions θ(L, y) = 0 ∂θ(0, y) =0 ∂x

θ(x, l) = 0 ∂θ(x, 0) =0 ∂y

(d)

Since the differential equation is nonhomogeneous, we assume a solution in either of the following forms θ(x, y) = ψ(x, y) + φ(x) θ(x, y) = ψ(x, y) + φ(y)

(e)

where ψ is a solution to the homogeneous equation and φ is a solution to the nonhomogeneous equation. Since the cause of nonhomogeneity in the differential equation, R/k, is assumed to be constant in this case, either one of the above solutions may be selected. On the other hand, if the nonhomogeneous term were a function of x, the first of equations (e) should be selected, and if it were a function of y, the second of equations (e) should be selected. Upon substitution of Eq. (e) into Eqs. (c) and (d) we find ∂ 2ψ ∂ 2ψ + 2 =0 ∂x2 ∂y

(f )

with the boundary conditions ∂ψ(0, y) =0 ∂x ∂ψ(x, 0) ψ(x, l) + φ(x) = 0 =0 ∂y ψ(L, y) = 0

and

d2 φ R + =0 dx2 k

(g)

(h)

158


with the boundary conditions dφ(0) =0 φ(L) = 0 (i) dx It is noted that the nonseparable form of Eq. (c) is reduced to the separable form for ψ, Eq. (f ), and to an ordinary differential equation for φ, Eq. (h), and these equations are easily solved. The solution for θ is ∞ (−1)n cosh λn y θ(x, y) 1 x 2 ] − 2 = [1 − ( ) cos λn x 3 RL2 /k 2 L n=0 (λn L) cosh λn l

(j)

where (2n + 1)π n = 0, 1, 2, ... (k) 2 This method can be applied to many partial differential equations, including problems in cylindrical coordinates, and unsteady problems, as will be shown later in this chapter. λn L =

2.7

Lumped Formulation

In many practical cases the differential formulation of heat conduction requires the solution in complex geometries where the boundary conditions are complicated. Furthermore, the nature of the geometry is such that the detailed analysis in particular direction does not provide valuable information. In this case, we may ignore the differential formulation in that direction and by averaging the temperature distribution simplify the solution while satisfying the boundary conditions. A lumped formulation of a problem in a specific direction means that it is independent of space variable in that direction. For this reason, proper consideration of boundary conditions in the lumped direction must be observed to insure the correctness of the solution. Depending on the problem’s geometry, we are allowed to lump in one or more space directions. The result of lumped formulation of a problem in any direction should, however, result in simplification of the solution while the boundary conditions and the required accuracy are maintained. In general, when a problem is lumped in one or more than one direction, the general form of distributed law of heat conduction is no longer valid and the governing equation is obtained by consideration of the heat balance of the lumped element. The following examples illustrate the lumped formulation for a triangular fin and a turbine blade. Example 7 Consider a triangular fin as shown in Fig. 4.2-13. The base temperature is assumed to be T0 . The coefficients of convective heat transfer to the ambient

2. Problems in Rectangular Cartesian Coordinates T0

159

h1 (T – T ) dx d (qx Ax ) dx qx Ax + dx

b

h1

dx

T x

L qx Ax

h2 (T – T ) dx

h2 (a)

(b)

Figure 4.2-13: Triangular fin. from the top and bottom surfaces are h1 and h2 , respectively, and it is assumed that b/L 1 and /L 1. The end surfaces in -direction are assumed to be insulated. We are to find the temperature distribution in the fin [1]. Solution From the assumptions of the problem, the solution domain is lumped in all directions except the x-direction. The heat balance of an element in x-direction is

qx Ax − [qx Ax +

d(qx Ax ) dx] − h1 (T − T∞ )ldx − h2 (T − T∞ )dx = 0 dx

(a)

Considering dθ bx Ax = (b) dx L the expression (a) reduces to the following differential equation for the heat balance in the fin dθ (h1 + h2 ) d2 θ − Lθ = 0 (c) x 2+ dx dx kb The boundary conditions are θ = T − T∞

qx = −k

θ(0) = finite,

θ(L) = T0 − T∞ = θ0

(d)

The solution of differential equation (c) is θ(x) = A1 I0 (x∗ ) + A2 K0 (x∗ )

(e)

√ where x∗ = 2m x, m = (h1 + h2 )L/kb. Using the boundary conditions, the constants of integration A1 and A2 are

A1 =

θ0 √ , I0 (2m L)

A2 = 0

(f )

160


Substituting into Eq. (e)

√ I0 (2m x) √ θ(x) = θ0 I0 (2m L)

(g)

Equation (g) represents the temperature distribution in the fin. Example 8 Consider a portion of a gas turbine blade shown in Fig. 4.2-14a. An element of the blade is shown in Fig. 4.2-14b and an idealized configuration of the cross section for the calculation purpose is shown in Fig. 4.2-14c. The blade is cooled over its base and receives heat over its other surfaces in convective and radiative form as a result of the flow of hot gases. The boundary conditions are shown in Fig. 4.2-14a. We may assume that the tip of the blade is insulated. We are to find a temperature field within the blade. Solution We take the x and y coordinates as shown in Fig. 4.2-14a. The equation of the thermal energy balance for an element shown in Fig. 4.2-14b is −

∂(qx Ax ) ∂(qy Ay ) dx − dy − 2h∞ (T − T∞ )dxdy + 2q2 dxdy = 0 ∂x ∂y

(a)

∂θ Taking θ = T − T∞ and noting that Ax = b( Lx )2 dy, Ay = b( Lx )2 dx, qx = −k ∂x ∂θ and qy = −k we obtain ∂y x

q⬙2

insulated

q y Ay

h T qx Ax

Tip q⬙2

h T

l

dx

q⬙2

∂ (qx Ax) dx qx Ax + ∂x

dy q⬙

qy Ay +

2

h ∞ T∞

h T ∂ (qy Ay) dy ¶y b L

q⬙1 h T h0,T0 base

y

b b( x ) 2 L

a

c

Figure 4.2-14: Turbine blade.

2. Problems in Rectangular Cartesian Coordinates ∂ 2 ∂θ ∂2θ (x ) + x2 2 − m2 θ = −n ∂x ∂x ∂y

161 (b)

where m2 = 2h∞ L2 /kb and n = 2q2 L2 /kb. The boundary conditions are θ(0, y) = finite −k

∂θ(L, y) = −q1 + h∞ θ(L, y) ∂x ∂θ(x, 0) =0 ∂y

−k

∂θ(x, l) = h0 [θ(x, l) − θ0 ] ∂y

(c)

where θ0 = T0 − T∞ . We assume the solution of Eq. (b) in the form of a sum of functions ψ(x, y) and φ(x) θ(x, y) = ψ(x, y) + φ(x)

(d)

where we require ψ(x, y) to satisfy the homogeneous equation ∂ 2ψ ∂ 2 ∂ψ (x ) + x2 2 − m 2 ψ = 0 ∂x ∂x ∂y

(e)

with the boundary conditions ψ(0, y) = finite −k

−k

∂ψ(L, y) = h∞ ψ(L, y) ∂x ∂ψ(x, 0) =0 ∂y

∂ψ(x, l) = h0 [ψ(x, l) + φ(x) − θ0 ] ∂y

(f )

and φ(x) is to satisfy the nonhomogeneous equation d 2 dφ (x ) − m2 φ = −n dx dx

(g)

with the boundary conditions φ(0) = finite ,

−k

dφ(L) = −q1 + h∞ φ(L) dx

(h)

The solution of Eq. (e) is obtained by the separation of variables method as ψ(x, y) = X(x)Y (y)

(i)

162


Substituting Eq. (i) in Eq. (e) yields dX d2 X + 2x + (λ2 x2 − m2 )X = 0 2 dx dx d2 Y − λ2 Y = 0 dy 2

x2

The solution for Y is obtained as Y (y) = C1 exp (λy) + C2 exp (−λy)

(j)

Introducing a transformation X = X ∗ /x 2 , the equation in x direction reduces to 1

x2

dX ∗ 1 d2 X ∗ + x + [λ2 x2 − (m2 + )]X ∗ = 0 2 dx dx 4

which is a Bessel equation, and its solution, using the transformation X=

1 1

x2

[C3 J(m2 + 1 ) 12 (λx) + C4 K(m2 + 1 ) 12 (λx)] 4

(k)

4

1

where J is Bessel function of the first kind and order (m2 + 1/4) 2 , and K is 1 the modified Bessel function of the second kind and order (m2 + 1/4) 2 , see Section 2.5 of this chapter. Equations (j) and (k) can be substituted in Eq. (i) and the constants of integration C1 , C2 , C3 , and C4 , and λ are evaluated using the boundary conditions (f ) and (h). A general solution φ(x) for Eq. (g) is obtained by presenting the solution in a power form of x as (l) φG (x) = xr where r is a constant to be found. Substituting this into the homogeneous form of Eq. (g), the following characteristic equation is obtained r 2 + r − m2 = 0

(m)

the solution of which is r1,2

1 =− ± 2

1 + m2 4

(n)

Then the general solution to Eq. (g) becomes φG (x) = Axr1 + Bxr2 =

1 x

1 2

1 2+ 1 ) 2 4

[Ax(m

1 2+ 1 ) 2 4

+ Bx−(m

]

(o)


163

Since the right-hand side of Eq. (g) is a constant, the particular solution is easily verified to be n φp = 2 (p) m and thus the complete solution to Eq. (g) becomes φ(x) =

1 x

1 2

1 2+ 1 ) 2 4

[Ax(m

1 2+ 1 ) 2 4

+ Bx−(m

]+

n m2

(q)

Therefore, a unique solution to the temperature problem for the blade is obtained if there are constants λ; C1 , C2 , C3 , C4 ; A and B such that the boundary conditions (f ) and (h) are satisfied.

2.8

Steady State Three-Dimensional Problems

When a solid body is exposed to different boundary conditions in three dimensions, the temperature gradient is produced in all three directions and the temperature distribution is a function of all three space coordinates. The heat conduction equation in three dimensions for a steady state temperature distribution is ∂ 2T ∂2T ∂2T + + =0 (4.2-68) ∂x2 ∂y 2 ∂z 2 A solution of Eq. (4.2-53) is easily obtained by the separation of variables method. The following examples illustrate the method. Example 9 Consider a solid body in the form of a rectangular parallelepiped as shown in Fig. 4.2-15. The surface temperatures are assumed to be T1 = constant on the surface x = 0, and T2 = constant on the surface x = a, and zero on the other surfaces. We are to find a temperature distribution within the body.

Figure 4.2-15: Rectangular parallelepiped.

164


Solution The differential equation for temperature distribution is ∂2T ∂ 2T ∂2T + + =0 ∂x2 ∂y 2 ∂z 2

(a)

The boundary conditions are T = T1 at x = 0 T = T2 at x = a T = 0 at y = 0, y = b, z = 0, z = c

(b)

The solution for the temperature distribution, using the separation of variables method, is T =

∞ ∞ m=1 n=1

Amn

T1 sinh l(a − x) + T2 sinh lx mπy nπz sin sin sinh la b c

where

2

l =

(c)

m 2 n2 + 2 π2 b2 c

(d)

The constant coefficients Amn are found from the first of Eqs. (b) as ∞ ∞

Amn sin

m=1 n=1

nπz mπy sin =1 b c

(e)

From above, it is verified that Amn is zero unless m and n are both odd numbers. Thus, Amn is obtained from the first of Eqs. (4.2-17) to be 4/π 2 mn. Finally, after substitution and taking odd numbers for m = 2p + 1 and n = 2q + 1, we obtain T =

∞ ∞ T1 sinh l(a − x) + T2 sinh lx (2p + 1)πy (2q + 1)πz 4 sin sin 2 π p=0 q=0 (2p + 1)(2q + 1) sinh la b c

(f ) where

(2p + 1)2 π 2 (2q + 1)2 π 2 l= + b2 c2 p = 0, 1, 2, ...

1 2

(g)

q = 0, 1, 2, ...

Example 10 Consider a rectangular body as shown in Fig. 4.2-16. It is required to obtain a steady-state temperature distribution within the body, where the surfaces


165

Figure 4.2-16: Rectangular parallelepiped. x = 0 and x = a are maintained at constant temperatures T1 and T2 , respectively, and the other surfaces are exposed to free convection to the ambient. For simplicity, take the ambient temperature as T∞ = 0. Solution The governing equation is ∂2T ∂ 2T ∂2T + + =0 ∂x2 ∂y 2 ∂z 2

(a)

and the boundary conditions are T (0, y, z) = T1 ∂T (x, b, z) k = −hT (x, b, z) ∂y ∂T (x, y, c) = −hT (x, y, c) k ∂z

T (a, y, z) = T2 ∂T (x, −b, z) k = hT (x, −b, z) ∂y ∂T (x, y, −c) k = hT (x, y, −c) ∂z

(b)

A solution satisfying the boundary conditions (b), is T =

∞ ∞ r=1 s=1

Ars

T1 sinh l(a − x) + T2 sinh lx cos αr y cos βs z sinh la

(c)

where αr = mπ/b, βs = nπ/c, and l = αr2 + βs2 . From the first of Eqs. (b) the constant Ars satisfies the following relation ∞ ∞

Ars cos (αr y) cos (βs z) = 1

(d)

r=1 s=1

The final solution, using the nonhomogeneous boundary conditions, is T =

∞ ∞ 4h2 [T1 sinh l(a − x) + T2 sinh lx] r=1 s=1

[(αr2 + h2 )b + h][(βs2 + h2 )c + h]

×

cos (αr y) cos (βs z) cos αr b cos βs c sinh la

(e)

166

2.9


Transient Problems

When the thermal boundary conditions vary with the time, the temperature distribution is also a function of the time. To formulate a transient heat conduction problem, both the initial conditions and the boundary conditions are needed. The governing heat conduction equation is

∂ 2T ∂2T ∂2T + + k ∂x2 ∂y 2 ∂z 2

= ρc

∂T −R ∂t

(4.2-69)

Example 11 Consider a plate of thickness 2L initially at uniform temperature T0 . The plate is suddenly immersed into a bath of constant temperature T∞ , see Fig. 4.2-17. Assuming the heat convection coefficient to be large, it is required to find a temperature distribution when the rate of heat generation within the plate R = const. Solution Taking θ = T − T∞ , the governing equation is κ

∂θ ∂2θ R = − 2 ∂x ∂t ρc

(a)

where κ = k/(ρc) is the diffusivity. The initial and boundary conditions are θ(x, 0) = θ0 = T0 − T∞ ∂θ(0, t) = 0, ∂x

θ(L, t) = 0

Figure 4.2-17: Thick plate.

(b)


167

The differential equation (a) is nonhomogeneous, and we let θ(x, t) = ψ(x, t) + φ(x)

(c)

where φ(x) satisfies the nonhomogeneous differential equation, i.e., d2 φ R + =0 dx2 k

(d)

subjected to the boundary conditions φ(L) = 0,

dφ(0) =0 dx

(e)

and ψ(x, t) satisfies the equation ∂2ψ ∂ψ = 2 ∂x ∂t

(f )

ψ(x, 0) = −φ(x) + θ0

(g)

κ subjected to the initial condition

and the boundary conditions ∂ψ(0, t) = 0, ∂x

ψ(L, t) = 0

(h)

A solution of Eq. (d) satisfying the boundary conditions (e) is φ(x) =

x RL2 [1 − ( )2 ] 2k L

(i)

To find ψ(x, t) the method of separation of variables is used and a solution takes the form ψ(x, t) = X(x)τ (t) (j) Substituting ψ(x, t) from Eq. (j) into Eq. (f ) we obtain d2 X + λ2 X = 0 dx2

(k)

with the boundary conditions dX(0) = 0, dx and

X(L) = 0

dτ + κλ2 τ = 0 dt

(l)

(m)

168


The initial condition is later applied to Eq. (m) using Fourier series. A solution of Eq. (k) is (n) Xn (x) = An cos λn x + Bn sin λn x where An and Bn are constants. Substituting (n) into the boundary conditions (l) gives Bn = 0 and λn =

(2n + 1)π , 2L

n = 0, 1, 2, 3, ...

(o)

A solution of Eq. (m) is also obtained by a simple integration as τn (t) = Cn exp(−κλ2n t)

(p)

and thus Eq. (j) implies ψ in the form ψ(x, t) =

∞

an exp(−κλ2n t) cos λn x

(q)

n=0

where an is also constant. To find the constant an , the initial condition ψ(x, 0) = −φ(x) + θ0 is used where φ(x) is substituted from Eq. (i). For θ0 = 0, using expansion into Fourier series we obtain an = −(−1)n

RL2 2 k (λn L)3

(r)

and finally ∞ θ(x, t) (−1)n 1 x 2 ] − 2 exp (−κλ2n t) cos λn x = [1 − ( ) 3 RL2 /k 2 L (λ L) n n=0

(s)

Example 12 Consider a homogeneous and isotropic beam with rectangular cross section of height a and width b, occupying the region 0 ≤ y ≤ a and −b/2 ≤ z ≤ b/2. The initial temperature of the beam is T0 at t = 0. The upper surface of the beam at y = a is exposed to a sudden constant heat flux q , while the other three surfaces are kept thermally insulated. Find the transient temperature distribution in the cross section of the beam. Solution The equation of one-dimensional transient heat conduction in the cross section of the beam is k

∂T ∂2T = ρc 2 ∂y ∂t

0 0, while the other surfaces are assumed to be


171

insulated. The material properties of the beam across the height are assumed to obey the exponential relations given by k = k0 exp(ξk y/a)

ρ = ρ0 exp(ξρ y/a)

c = c0 exp(ξc y/a)

(a)

where k0 , ρ0 , and c0 are the thermal conductivity, the mass density, and the specific heat of the base constituent material, respectively, and ξk , ξρ and ξc are some constant multipliers. Find a temperature field in the beam. Solution The equation of one-dimensional transient heat conduction in the cross section of the beam and the boundary and initial conditions, assuming θ = T − T1 , are

∂θ ∂θ ∂ k = ρc ∂y ∂y ∂t ∂θ(0, y) =0 θ(a, t) = 0 ∂y θ(y, 0) = θ0 = T0 − T1

(b)

Using the method of separation of variables, we let θ = f (y)τ (t). Substituting this solution and the material properties (a) in the first of Eqs. (b) gives e−ξy/a f

ξk df d2 f + 2 dy a dy

1 dτ = −λ2 κ0 τ dt

=

(c)

where ξ = ξρ + ξc − ξk and κ0 = k0 /ρ0 c0 is the thermal diffusivity of the base material, and λ is a constant. The solution is obtained for two different cases; when the thermal diffusivity is a position dependent function, i.e., ξ = 0, and when it is assumed to be a constant, say, ξ = 0. Case (a) Position-dependent thermal diffusivity, ξ = 0 : When this assumption is considered, the separation of variables for the time function τ (t) leads to dτ + κ0 λ2 τ = 0 dt The solution to this equation is τ = De−κ0 λ

2t

(d)

where D is a constant. The differential equation and the boundary conditions for the function f (y) become d2 f ξk df + + λ2 eξy/a f = 0 2 dy a dy df (0) =0 dy

f (a) = 0

(e)

172


Introducing the new variables e(ξ y/2a) = u, f = f/uξk /ξ , and 2λ a u/ξ = v, a solution for f (y) is obtained as

f =e

−(ξk y/2a)

2λa (ξy/2a) 2λa (ξy/2a) A J(ξk /ξ) ( ) + B Y(ξk /ξ) ( ) e e ξ ξ

(f )

where A and B are constants. Applying the boundary conditions (e) to Eq. (f ), the ratio of the constants B/A and the characteristic equation which can be solved for λ are obtained as J( ξk +1) ( ξ

2λa 2λa ξ/2 2λa ξ/2 2λa )Y( ξk ) ( e ) − J ( ξk ) ( e )Y( ξk +1) ( )=0 ξ ξ ξ ξ ξ ξ ξ J( ξk ) ( 2λξn a ) eξ/2 J( ξk +1) ( 2λξn a ) Bn ξ =− =− ξ An Y( ξk ) ( 2λξn a eξ/2 ) Y( ξk +1) ( 2λξn a ) ξ

(g)

ξ

Substituting the results in Eq. (f ), then using Eqs. (f ) and (d), a solution for the temperature distribution θ(y, t) becomes θ=

∞

⎡ −(κ0 λ2n t+ξk y/2a)

Cn e

⎣J(ξ /ξ) (ψ) − k

n=1

J(ξk /ξ) ( 2λξn a eξ/2 ) Y(ξk /ξ) ( 2λξn a eξ/2 )

⎤

Y(ξk /ξ) (ψ)⎦

(h)

where Cn is constant, ψ = 2λξn a eξy/2a . To obtain the constant of integration Cn the initial condition is used to give θ0 e(ξk y/2a) =

∞

⎡

J( ξk ) ( 2λξn a eξ/2 )

⎤

2λn a ξy/2a 2λn a ξy/2a ⎥ Cn ⎢ ) − ξ 2λn a ξ/2 Y( ξk ) ( )⎦ e e ⎣J( ξk ) ( ξ ξ ξ Y ( ξk ) ( ξ e ) ξ n=1 ξ

(i) To find the constant coefficient Cn , consider the function gn (y) defined as J( ξk ) ( 2λξn a eξ/2 ) 2λn a ξy/2a 2λn a ξy/2a gn (y) = J( ξk ) ( ) − ξ 2λn a ξ/2 Y( ξk ) ( ) e e ξ ξ ξ Y ( ξk ) ( ξ e ) ξ ξ

which with the aid of the ratio of B/A can also be written in the form J( ξk +1) ( 2λξn a ) 2λn a ξy/2a 2λn a ξy/2a Y ( ξk ) ( gn (y) = J( ξk ) ( )− ξ ) e e 2λ a n ξ ξ ξ Y( ξk +1) ( ξ ) ξ ξ

This function is a solution of the following equation d2 gn u2 2 du

dgn ξk 4λ2n a2 2 +u u − ( )2 gn = 0 + 2 du ξ ξ

2. Problems in Rectangular Cartesian Coordinates where u = eξy/2a . For (i = j) we have d2 gi u2 2 du

173

dgi ξk 4λ2i a2 2 +u u − ( )2 gi = 0 + du ξ2 ξ

4λ2j a2 2 dgj ξk +u u − ( )2 gj = 0 + 2 du ξ ξ Multiplying the first of these equations by gj /u and the second by gi /u and subtracting the results gives d2 gj u2 2 du

u(gj

4a2 2 d2 gi d2 gj dgi dgj 2 − g ) + (g λ − λ − g ) + i j i i j ugi gj = 0 du2 du2 du du ξ2

which can also be written in the form

d 4a2 dgi dgj u(gj − gi ) + 2 λ2i − λ2j ugi gj = 0 du du du ξ Integration of the above equation with respect to u from u = 1 to u = eξ/2 results in eξ/2

dgi dgj u(gj − gi ) du du 1

+

eξ/2 4a2 2 2 λ − λ ugi gj du = 0 i j ξ2 1

By using the relations for derivatives of Bessel functions given by d(J( ξk ) ( 2λξn a u)) ξ

du

=

− J( ξk +1) ( ξ

d(Y( ξk ) ( 2λξn a u))

ξk 2λn a 2λn a 2λn a u) + J ξk ( u) ξ 2 λn a u ( ξ ) ξ ξ

2λn a 2λn a ξk 2λn a u) + Y ( ξk ) ( u) ξ du ξ 2 λn a u ξ ξ ξ ξ/2 and noting that g(u) = 0 when u = e , it can be easily shown that the preceding equation reduces to the form ξ

=

− Y( ξk +1) (

eξ/2 1

a

or

0

ugi gj du = 0 eξy/a gi gj dy = 0

Thus, gi and gj (i = j) are the orthogonal set of functions with the weight eξy/a . We may use this property of gn to find the coefficient Cn . Multiplying Eq. (i) by eξy/a gi (y) and integrating from y = 0 to y = a gives θ0

⎡

a

J( ξk ) ( 2λξn a eξ/2 )

⎤

2λn a ξy/2a ⎦ ξ ⎣J ξk ( 2λn a eξy/2a ) − e ) dy Y ξk ( ( ξ ) ξ Y( ξk ) ( 2λξn a eξ/2 ) ( ξ ) ξ ξ ⎡ ⎤2 a J( ξk ) ( 2λξn a eξ/2 ) a a 2λ 2λ n n ξ eξy/2a ) − eξy/2a )⎦ dy e(ξy/a) ⎣J( ξk ) ( Y ξk ( ξ ξ Y ξk ( 2λξn a eξ/2 ) ( ξ ) ξ 0 ξ (ξ+ 2k

e

)y/2a

0

Cn =

(

ξ

)

(j)

174


Case (b) Constant thermal diffusivity, ξ = 0 : For this case, the solution for τ (t) is obtained from the ordinary differential equation dτ + κ0 λ2 τ = 0 (k) dt The solution of Eq. (k) is 2 τ = De−κ0 λ t (l) The differential equation and the boundary conditions for f (y) are d2 f ξk df + + λ2 f = 0 dy 2 a dy df (0) =0 dy

f (a) = 0

(m)

and a solution is f (y) = e−(ξk y/2a) [A cos (μ y) + B sin (μ y)] where

λ2 − (

μ=

(n)

ξk 2 ) 2a

Applying the boundary conditions (m) to Eq. (n), gives B ξk = A 2a μn tan (a μn ) = − Thus θ(y, t) =

∞

2 a μn ξk

n = 1, 2, . . .

Cn e−(κ0 λn t+ξk y/2a) 2

n=1

sin [μn (y − a)] cos (μn a)

(o)

From the initial condition ∞

θ0 e(ξk y/2a) =

n=1

Cn

sin [μn (y − a)] cos (μn a)

Multiplying this equation by sin [μm (y − a)] and integrating from y = 0 to y = a, considering the orthogonality of sin [μn (y − a)], we arrive at a

θ0 cos(μn a) Cn =

a 0

0

eξk y/2a sin [μn (y − a)]dy

sin2 [μn (y − a)]dy

(p)

3. Problems in Cylindrical Coordinates

3

175

Problems in Cylindrical Coordinates

A general form of the governing equation of heat conduction in cylindrical coordinates is k(

∂ 2T 1 ∂T ∂ 2T ∂T 1 ∂2T + + ) + R = ρc + ∂r2 r ∂r r2 ∂φ2 ∂z 2 ∂t

(4.3-1)

In the following, we will discuss one-, two-, and three-dimensional problems, both steady-state and transient.

3.1

Steady-State One-Dimensional Problems (Radial Flow)

When the temperature distribution in the cylindrical geometry is a function of radius only, Eq. (4.3-1) reduces to 1 dT R d2 T + + =0 2 dr r dr k

(4.3-2)

This case is often encountered in practical problems and the solution is obtained by a simple integration. Example 14 Consider a hollow cylinder of inner and outer radius a and b, respectively. It is required to obtain a temperature distribution in the cylinder when the rate of heat generation is zero, and the temperature at the inner radius is different from that at the outer radius. Solution The differential equation of heat conduction (4.3-2) becomes for this case d dT (r ) = 0 (a) dr dr The boundary conditions are T (a) = Ta ,

T (b) = Tb

(b)

Integrating Eq. (a) yields T = A + B ln r

(c)

where A and B are constants. Upon finding A and B from the boundary conditions, we receive T =

1 b r (Ta ln + Tb ln ) ln b/a r a

(d)

176


Now, if we assume that the inside surface is kept at Ta = constant and the outside surface is exposed to convection at ambient temperature T∞ , the boundary conditions become T (a) = Ta ,

k

dT (b) + h[T (b) − T∞ ] = 0 dr

(e)

Using these boundary conditions we obtain T =

b r 1 {Ta [1 + mb ln ] + mbT∞ ln } 1 + mb ln b/a r a

(f )

where m = h/k. Example 15 Consider the hollow cylinder of Example 12, but with a constant rate of heat generation per unit volume R. Find the temperature distribution in the cylinder. Solution The governing equation becomes R 1 d dT (r ) + = 0 r dr dr k Assuming the boundary conditions T (a) = Ta ,

T (b) = Tb

(a)

(b)

a solution is Rr2 4k Then, substituting Eq. (c) into the boundary conditions (b) we get T = A + B ln r −

Ta = A + B ln a −

Ra2 4k

Rb2 4k Solving Eq. (d) for A and B, the constants of integration become Tb = A + B ln b −

A = Ta −

(c)

(d)

ln a R 2 Ra2 (b − a2 )] + b [Tb − Ta + 4k 4k ln a

B=

1 R 2 (b − a2 )] b [Tb − Ta + 4k ln a

(e)


177

Upon substitution of A and B from Eq. (e) in Eq. (b), the temperature distribution is completely determined. Example 16 Consider a circular fin of constant thickness δ, as shown in Fig. 4.3-1. The temperature is specified at r = a and r = b, and the top and bottom surfaces are exposed to a free convection to the ambient at T∞ . We are to find a temperature field in the fin. Solution Considering an element as shown, the amount of the heat transferred to and from the element can be defined. The energy balance requires that the net thermal inflow and outflow to be zero, that is d (qr A)dr] − 2qc pdr = 0 (a) dr where p = 2πr is the inner perimeter of the circular element, A = 2πrδ and the flux qr according to Fourier law of conduction is qr A − [qr A +

qr = −k

dT dr

(b)

while qc = h(T − T∞ )

(c)

Substituting Eqs. (b) and (c) in Eq. (a) and taking θ = T − T∞ , we find that a steady state temperature must satisfy the differential equation d2 θ 1 dθ 2h + − θ=0 dr2 r dr kδ The boundary conditions are θ(a) = θa = Ta − T∞ ,

(d)

θ(b) = θb = Tb − T∞

(e)

Equation (d) represents the so called modified Bessel equation, and its general solution in the form of C1 I0 (mr) + C2 K0 (mr), employing the boundary conditions, reduces to qc pdr

dr

h,T Tb

qr A

r

qc pdr

qr A+

d (q A)dr dr r

h,T

Figure 4.3-1: Circular fin.

d

a b

178

Chapter 4. Heat Conduction Problems θ = θa

I0 (mr)K0 (mb) − I0 (mb)K0 (mr) I0 (ma)K0 (mb) − I0 (mb)K0 (ma)

−θb where m2 =

I0 (mr)K0 (ma) − I0 (ma)K0 (mr) I0 (ma)K0 (mb) − I0 (mb)K0 (ma)

(f )

2h . kδ

Example 17 Consider a turbine disk subjected to the flow of hot gases from its edges and cooled from its sides, as shown in Fig. 4.3-2. Find the temperature distribution within the disk. Solution The temperature distribution can be approximated by taking it constant along the thickness and varying symmetrically in radial direction. Considering an annular element around the center of the disk and writing the equation of thermal equilibrium, and taking θ = T − T∞ , gives d2 θ 1 dθ 2h + − θ=0 dr2 r dr kb This is the modified Bessel equation. The boundary conditions are

(a)

dθ(ro ) (b) =q dr The solution of the modified Bessel equation (a) subjected to the first boundary condition θ(0) = finite is 2h θ = CI0 (r ) (c) kb where C is a constant of integration, and I0 is the modified Bessel function of the first kind of zero order. The constant C is evaluated knowing the boundary conditions at r = ro . This yields θ(0) = finite,

C= k

k

q 2h kb

I1 (ro

2h ) kb

Figure 4.3-2: Turbine disk.

(d)


179

where the first derivative of I0 is equal to I1 . Therefore, the final expression for the temperature distribution becomes q

I0 (r

θ(r) = k 2h I1 ( ro kb

3.2

2h ) kb 2h ) kb

(e)

Steady-State Two-Dimensional Problems

In this section a solution of problems in cylindrical coordinates are presented. The problems include the flow of heat in regions bounded by cylindrical surfaces such as finite solid and hollow cylinders, and semi-cylinders. The problems depend on r and φ, or r and z and the method of separation of variables leads to a pair of separated equations, one being Bessel equation in r-direction. Indeed, general method of solution presented in this section is the method of separation of variables, although problems of this nature can also be treated by the method of conformal mapping and Laplace transforms. Example 18 Consider a solid cylinder of radius a and length L, Fig. 4.3-3. It is required to obtain a temperature distribution for various boundary conditions. (a) Assume that the surface z = 0 is kept at a prescribed temperature T0 (r), the surface z = L is kept at zero temperature, and the surface r = a is exposed to a free convection to the ambient at T∞ = 0. Solution The differential equation for the temperature distribution for the case of axial symmetry is ∂2T 1 ∂T ∂2T + =0 + ∂r2 r ∂r ∂z 2

Figure 4.3-3: Solid cylinder.

(a)

180


The boundary conditions are T (r, 0) = T0 (r)

T (r, L) = 0,

∂T (a, z) + h1 T = 0 T (0, z) = finite (b) ∂r where h1 = h/k. A solution is obtained using the method of separation of variables and the boundary conditions, and is given as T (r, z) =

∞

An

n=1

J0 (rαn ) sinh (L − z)αn sinh Lαn

(c)

where αn are the positive roots of the equation αJ0 (αa) + h1 J0 (αa) = −αJ1 (αa) + h1 J0 (αa) = 0

(d)

where An is constant. From the second of Eqs. (b), we have T0 (r) =

∞

An J0 (rαn )

(e)

n=1

which is Fourier-Bessel expansion of T0 (r). From Eq. (4.2-62) the constant An is obtained as a 2αn2 rT0 (r)J0 (rαn )dr a2 (h21 + αn2 )J02 (aαn ) 0

An =

(f )

If T0 (r) = T0 is constant, since 0x sn Jn−1 (s)ds = xn Jn (x), using Eq. (d) the temperature distribution reduces to T (r, z) = T0

∞ 2h1 J0 (rαn ) sinh (L − z)αn n=1

a(h21 + αn2 )J0 (aαn ) sinh Lαn

(g)

(b) Consider the same cylinder with surfaces z = 0 kept at a prescribed temperature T = T0 (r) and z = L exposed to free convection to ambient temperature at T∞ = 0. The boundary conditions are T (r, 0) = T0 (r)

∂T (r, L) + h1 T = 0 ∂z

∂T (a, z) + h1 T = 0 T (0, z) = finite ∂r The temperature distribution, similar to part (a), is found to be T (r, z) = T0

∞ n=1

An J0 (rαn )

αn cosh (L − z)αn + h1 sinh (L − z)αn αn cosh Lαn + h1 sinh Lαn

with the same definition for αn and An as before.

(h)

(i)


181

(c) We assume again the same cylinder with the surface r = a kept at the prescribed temperature T = T0 (z) and the other two surfaces at z = 0 and z = L exposed to a free convection to ambient temperature at T∞ = 0. The boundary conditions are ∂T (r, 0) − h1 T = 0 ∂z

∂T (r, L) + h1 T = 0 ∂z

T (a, z) = T0 (z)

T (0, z) = finite

(j)

Using the method of separation of variables, the temperature distribution for this case is T (r, z) =

∞

An I0 (rαn )

n=1

(αn cos αn z + h1 sin αn z) h1

(k)

where αn are the positive roots of the equation tan αL =

2αh1 α2 − h21

(l)

Using the third boundary condition T0 (z) =

∞

An I0 (aαn )

n=1

(αn cos αn z + h1 sin αn z) h1

(m)

the coefficient An is obtained by multiplying Eq. (m) by (αm cos αm z + h1 sin αm z) and integrating with respect to z from z = 0 to z = L, when m = n. Considering Eq. (l), we have An =

2h1 + h21 )L + 2h1 ]

I0 (aαn )[(αn2

×

L 0

T0 (z)[αn cos αn z + h1 sin αn z]dz

(n)

The constant An is substituted in Eq. (k) to give the final expression for the temperature distribution as T (r, z) = 2

∞ I0 (rαn )[αn cos αn z + h1 sin αn z] n=1

×

I0 (aαn )[(αn2 + h21 )L + 2h1 ]

L 0

T0 (z)[αn cos αn z + h1 sin αn z]dz

(o)

182


(d) The temperature distribution for a hollow cylinder with inner radius a and outer radius b, when the prescribed temperature on the surfaces at r = b, z = 0, and z = L is zero, but the surface at r = a is kept at a temperature T0 (z), can be similarly treated. The boundary conditions are T (r, 0) = 0 T (a, z) = T0 (z)

T (r, L) = 0 T (b, z) = 0

(p)

Using the method of separation of variables and the boundary conditions, the temperature distribution is T (r, z) =

∞

An

n=1

K0 (αn r)I0 (αn b) − I0 (mr)K0 (αn b) sin αn z I0 (αn b)

(q)

where αn = nπ/L. Applying the boundary condition at r = a, T0 (z) =

∞ n=1

An

K0 (αn a)I0 (αn b) − I0 (αn a)K0 (αn b) sin αn z I0 (αn b)

(r)

Multiplying Eq. (r) by sin αm z and integrating with respect to z from z = 0 to z = L, when m = n, we obtain An =

L I0 (αn b) 2 T0 (z) sin αn z dz L K0 (αn a)I0 (αn b) − I0 (αn a)K0 (αn b) 0

(s)

Substituting for An , the final expression for the temperature distribution is L ∞ I0 (αn r)K0 (αn b) − K0 (αn r)I0 (αn b) 2 T (r, z) = T0 (z) sin αn z dz sin αn z L n=1 I0 (αn a)K0 (αn b) − K0 (αn a)I0 (αn b) 0 (t)

Example 19 Consider a solid cylinder of outside radius b, or a hollow cylinder of inside radius a and outside radius b. The temperature is assumed to be constant along the axis of the cylinder, the z-axis, but varies in r and φ directions. It is also assumed that the temperature reaches its steady-state condition and that there is no heat generation in the cylinder [11]. We are to find a temperature distribution in the cylinder. Solution The heat conduction equation to be solved in this case is 1 ∂T 1 ∂ 2T ∂2T + =0 + ∂r2 r ∂r r2 ∂φ2

(a)


183

The boundary conditions are T (b, φ) = F (φ) = known T (0, φ) = finite for a solid cylinder T (a, φ) = Ti = known constant for a hollow cylinder T (r, φ) = T (r, φ + 2π) 1 ∂T (r, φ) 1 ∂T (r, φ + 2π) = r ∂φ r ∂φ

(b)

We take a solution in the product form T (r, φ) = R(r) Φ(φ)

(c)

Upon substitution in Eq. (a), and knowing that the solution in φ direction must be periodic, the selection of a proper separation constant results in the equation d2 Φ + λ2 Φ = 0 (d) dφ2 subjected to the boundary conditions Φ(φ) = Φ(φ + 2π),

dΦ(φ) dΦ(φ + 2π) = dφ dφ

(e)

and the separated equation in r-direction is d2 R 1 dR R + − λ2 2 = 0 2 dr r dr r

(f )

The boundary conditions on R will depend on whether the cylinder is solid or hollow. We will first consider a solid cylinder which in this case gives R(0) = finite

(g)

The solution of Eq. (d) is obtained as Φ = C cos λφ + D sin λφ

(h)

Applying the first condition of (e) defines the value of the separation constant λ = n,

n = 0, 1, 2, 3, ...

The second condition of Eq. (e) is automatically satisfied and thus a solution for Φ reduces to Φ(φ) = C0 + Cn cos nφ + Dn sin nφ

184


Equation (f ) is the Euler differential equation which is easily integrated to give R(r) = En rn + Gn r−n (i) Applying the boundary condition (g) for a solid cylinder gives Gn = 0 and thus a temperature distribution after substitution in Eq. (c), and with introduction of new constants, becomes T (r, φ) = A0 +

∞

rn (An cos nφ + Bn sin nφ)

(j)

n=1

To obtain the constants A0 , An , and Bn , we use the nonhomogeneous boundary condition on the outer boundary of the cylinder. From Eq. (j) and the first of Eqs. (b) the temperature distribution at r = b is F (φ) = A0 +

∞

bn (An cos nφ + Bn sin nφ)

n=1

Expanding F (φ) into a Fourier sine and cosine series with the period from 0 to 2π and equating the proper terms yields 1 2π F (φ)dφ 2π 0 2π 1 An bn = F (φ) cos nφdφ π 0 1 2π Bn bn = F (φ) sin nφdφ π 0 A0 =

and upon substitution into Eq. (j), the temperature distribution for a solid cylinder exposed to a nonaxisymmetric thermal boundary condition becomes fully determined. For a hollow cylinder of inner radius a and outer radius b subjected to a constant inside temperature Ti and a variable outside temperature, the same procedure is followed. The general solution of Eq. (a) is obtained by substituting Eqs. (h) and (i) into Eq. (c), and with the introduction of new constants, and with taking θ(r, φ) = T (r, φ) − Ti it becomes θ(r, φ) = A0 + B0 ln r +

∞ λ=1

r r [Aλ ( )λ + Bλ ( )−λ ] cos λφ a a

r r + [Cλ ( )λ + Dλ ( )−λ ] sin λφ a a

(k)


185

The constants of integration A0 , B0 , Aλ , Bλ , Cλ , Dλ and the separation constant λ must be obtained using the given boundary conditions. For simplicity, we assume that the given boundary condition F (φ) at r = b, has a line of symmetry passing through the center of the cylinder, thus the temperature distribution can be obtained for half of the cylindrical region. Measuring the angle φ from the line of symmetry the following boundary conditions apply ∂θ(r, 0) =0 ∂φ

∂θ(r, π) =0 ∂φ

θ(a, φ) = 0

θ(b, φ) = F (φ) − Ti (l)

From Eq. (k) it follows that

∞ ∂θ r r −λ[Aλ ( )λ + Bλ ( )−λ ] sin λφ = ∂φ λ=1 a a

r r + λ[Cλ ( )λ + Dλ ( )−λ ] cos λφ a a Applying the first of Eqs. (l) gives

∞ ∂θ r r |φ=0 = λ[Cλ ( )λ + Dλ ( )−λ ] = 0 ∂φ a a λ=1

which yields Cλ = Dλ = 0. The second condition of Eqs. (l) gives λ=n

with

n = 1, 2, 3, ...

Thus Eq. (k) becomes θ(r, φ) = A0 + B0 ln r +

∞

r r [An ( )n + Bn ( )−n ] cos nφ a a n=1

(m)

Applying the third of Eqs. (l) gives A0 + B0 ln a +

∞

(An + Bn ) cos nφ = 0

n=1

This may be written as A0 + B0 ln a = 0 ∞

(An + Bn ) cos nφ = 0

n=1

which gives A0 = −B0 ln a

Bn = −An

Substituting A0 and Bn into Eq. (m) results in θ(r, φ) = B0 ln

∞ r r r An [( )n − ( )−n ] cos nφ + a n=1 a a

(n)

186


The constants B0 and An are to be found from the last of Eqs. (l). The general method to compute these constants is again the expansion of F (φ) into a Fourier series and then equating it with Eq. (n) evaluated at r = b. A comparison of corresponding coefficients will result in obtaining of the constants B0 and An . As an example, we may assume that the temperature distribution at the outer boundary of the cylinder, F (φ), to have the following form

θ(b, φ) =

T1 − Ti T2 − Ti

0≤φ≤β β≤φ≤π

for for

where β is a known arbitrary angle between zero and π. Evaluating Eq. (n) at r = b and rearranging it in a dimensionless form gives ¯ φ) = θ(b, φ) − (T1 − Ti ) θ(b, T2 − T1 =

∞ B0 ln ab − (T1 − Ti ) b b An + [( )n − ( )−n ] cos nφ T2 − T1 a n=1 T2 − T1 a

(o)

Thus

¯ φ) = θ(b,

0 1

at at

0≤φ≤β β≤φ≤π

¯ φ) is Fourier expansion of the function θ(b, ∞ (−1)n ¯ φ) = π − β + 2 θ(b, sin n(π − β) cos nφ π π n=1 n

(p)

A comparison of Eqs. (p) and (o) yields π(T2 − Ti ) − β(T2 − T1 ) 1 ] b π ln a n 2(−1) sin n(π − β) T2 − T1 An = nπ ( ab )n − ( ab )−n B0 = [

Substituting the values of constants B0 and An into Eq. (n), the temperature distribution in a thick-walled cylinder subjected to the boundary conditions (l) becomes π(T2 − Ti ) − β(T2 − T1 ) r ln (q) T (r, φ) = Ti + a π ln ab

∞ 2(T2 − T1 ) (−1)n sin n(π − β) ( ar )n − ( ar )−n + cos nφ π n ( ab )n − ( ab )−n n=1


187

In this problem we discussed the temperature distribution in a hollow cylinder under a non-axisymmetric temperature distribution caused by non-axisymmetric thermal boundary conditions. The thermal boundary conditions were, however, of the same type on each single boundary, i.e., either of different specified temperatures or of different heat fluxes. We may have non-axisymmetric temperature distribution in a hollow cylinder caused by non-axisymmetric thermal boundary conditions, but of different types on each single boundary. This is like to having a cylinder exposed to a given heat flux on a part of its outer boundary while on the other part a specified temperature is applied. The solutions to problems with this type of boundary conditions are not easy to obtain and are classified as mixed boundary value problems. The solutions to such mixed problems are usually obtained by means of a system of infinite number of linear algebraic equations for an infinite number of unknown constants. The concept and formulations of the mixed boundary value problems are discussed by Kantorovich and Krylov [12] and by Hetnarski and Ignaczak [13]. Example 20 Consider a thick hollow cylinder of a functionally graded material (FGM) with inside radius a and outside radius b. The cylinder’s material is graded through the radial direction according to a power law distribution, and thus the coefficient of heat conduction is a function of r. The temperature distribution of the cylinder is to be calculated for two types of thermal boundary conditions; (A) when the boundary conditions result in a non-axisymmetric temperature distribution T = T (r, φ), and (B) when the boundary conditions result in an axisymmetric temperature distribution T = T (r, z). (A) Temperature distribution for non-axisymmetric condition Solution The heat conduction equation for a steady-state condition in the two-dimensional problem in polar coordinates and the thermal boundary conditions for FGM hollow cylinder are given, respectively, as T,rr + (

k (r) 1 1 + )T,r + 2 T,φφ = 0 k(r) r r

a≤r≤b

− π ≤ φ ≤ +π

(a)

C11 T (a, φ) + C12 T,r (a, φ) = f1 (φ) C21 T (b, φ) + C22 T,r (b, φ) = f2 (φ)

(b)

where k = k(r) is the thermal conduction coefficient, (the symbol ( ) denotes derivative with respect to r), a and b are the inner and outer radii of the hollow cylinder, respectively, and Cij are the constant thermal parameters related to

188


the conduction and convection coefficients. The functions f1 (φ) and f2 (φ) are known temperature distributions on the inner and outer radii, respectively. We assume that the nonhomogeneous heat conduction coefficient k(r) is a power function of r as r (c) k(r) = k0 ( )m3 l where k0 and m3 are material parameters and l is a characteristic length. Using Eqs. (a) and (c), the heat conduction equation becomes 1 1 T,rr + (m3 + 1) T,r + 2 T,φφ = 0 r r

(d)

Since T (r, φ) is a periodic function of φ, it may be written in the form of complex Fourier series as T (r, φ) =

∞

Tn (r)einφ

(e)

n=−∞

where Tn (r) is the coefficient of finite complex Fourier series of T (r, φ) and is Tn (r) =

1 π T (r, φ)e−inφ dφ 2π −π

(f )

Substituting Eq. (e) into Eq. (d), the following equation is obtained 1 n2 Tn (r) + (m3 + 1) Tn (r) − 2 Tn (r) = 0 r r

(g)

Equation (g) is the Euler equation and has solutions in the form of Tn (r) = An rβ

(h)

Substituting Eq. (h) into Eq. (g), the following characteristic equation is obtained β 2 + m 3 β − n2 = 0 The roots of this equation are βn1,2 =

−m3 m23 ¯ +( + n2 )1/2 2 4

Thus Tn (r) = An1 rβn1 + An2 rβn2

(i)

Substituting Eq. (i) into Eq. (e), gives T (r, φ) =

∞

(An1 rβn1 + An2 rβn2 )einφ

n=−∞

(j)


189

Using the boundary conditions (b) to determine the constants An1 and An2 , yields ∞

[(C11 aβn1 + C12 βn1 aβn1 −1 )An1 + (C11 aβn2 + C12 βn2 aβn2 −1 )An2 ]einφ

n=−∞

= f1 (φ) ∞

[(C21 bβn1 + C22 βn1 bβn1 −1 )An1 + (C21 bβn2 + C22 βn2 bβn2 −1 )An2 ]einφ

n=−∞

= f2 (φ) In these equations the right-hand sides are the coefficients of a finite complex Fourier series of the left-hand sides as (C11 aβn1 + C12 βn1 aβn1 −1 )An1 + (C11 aβn2 + C12 βn2 aβn2 −1 )An2 1 π = f1 (φ)e−inφ dφ 2π −π (C21 bβn1 + C22 βn1 bβn1 −1 )An1 + (C21 bβn2 + C22 βn2 bβn2 −1 )An2 1 π = f2 (φ)e−inφ dφ 2π −π These are a system of algebraic equations for the constant coefficients An1 and An2 . A solution to the system by Cramer’s method is 1 π [(C21 bβn2 + C22 βn2 bβn2 −1 )f1 (φ) − 2π −π (C11 aβn2 + C12 βn2 aβn2 −1 )f2 (φ)]e−inφ dφ/(Cˆ1 − Cˆ2 ) 1 π An2 = [(C11 aβn1 + C12 βn1 aβn1 −1 )f2 (φ) − 2π −π (C21 bβn1 + C22 βn1 aβn1 −1 )f1 (φ)]e−inφ dφ/(Cˆ1 − Cˆ2 )

An1 =

where Cˆ1 = (C11 aβn1 + C12 βn1 aβn1 −1 )(C21 bβn2 + C22 βn2 bβn2 −1 ) Cˆ2 = (C11 aβn2 + C12 βn2 aβn2 −1 )(C21 bβn1 + C22 βn1 bβn1 −1 )

(B) Temperature distribution for axisymmetric condition Solution An axially symmetric heat conduction equation in a steady-state condition for a functionally graded cylinder is

∂2T k (r) 1 + + ∂r2 k(r) r

∂T ∂2T + 2 = 0, ∂r ∂z

a < r < b, 0 < z < L

(k)

190


where k(r) is the thermal conduction coefficient, and the symbol ( ) denotes a derivative with respect to r. The thermal boundary conditions are assumed as T (r, 0) = 0,

T (r, L) = 0

(l)

C11 T (a, z) + C12 T,r (a, z) = f1 (z) C21 T (b, z) + C22 T,r (b, z) = f2 (z)

(m)

where Cij are the constant thermal parameters related to the conduction and convection coefficients. The functions f1 (z) and f2 (z) are known on the inner and outer radii, respectively. We assume that the nonhomogeneous thermal conduction coefficient k(r) is a power function of r as k(r) = k0 ( rl )m3 where k0 and m3 are the material parameters and l is a characteristic length. Using the definition of k = k(r), the heat conduction equation becomes 1 ∂T ∂ 2T ∂2T + (m + 1) =0 + 3 ∂r2 r ∂r ∂z 2

(n)

The solution of the heat conduction equation, satisfying the thermal boundary conditions given by Eqs. (l) is written in the form of Fourier series as T (r, z) =

∞ n=1

Tn (r) sin(

nπ z) L

(o)

where Tn (r) is the coefficient of sine Fourier series of T (r, z), and is Tn (r) = where ζn =

nπ . L

2L T (r, z) sin(ζn z) dz L 0

(p)

Substituting Eq. (o) into Eq. (n), gives d2 Tn (r) 1 dTn (r) + (m3 + 1) − ζn2 Tn (r) = 0 2 dr r dr

(q)

Equation (q) is the generalized Bessel equation, and has a general solution as Tn (r) = r−β [an Iβ (ζn r) + bn I−β (ζn r)]

(r)

where β = m23 , an , and bn are unknowns to be found using the thermal boundary conditions, and Iβ is the modified Bessel function of the first kind of the βth order. If β is a noninteger, Iβ and I−β are independent functions. In the case when β is an integer, the second kind of modified Bessel function Kβ should be used, instead of I−β . Substituting Eq. (r) into Eq. (o), yields T (r, z) =

∞ n=1

r−β [an Iβ (ζn r) + bn I−β (ζn r)] sin(ζn z)

(s)


191

Using the boundary conditions (m) to determine the constants an and bn , and using Iβ (ζn r), the derivative of Iβ (ζn r) with respect to r as Iβ (ζn r) = ζn Iβ−1 (ζn r) −

β Iβ (ζn r) r

yield an =

e1 e6 − e2 e4 , e3 e6 − e4 e5

bn =

e2 e3 − e1 e5 e3 e6 − e4 e5

where constants e1 to e6 are given as 2L 2L f1 (z) sin(ζn z) dz, e2 = f2 (z) sin(ζn z) dz e1 = L 0 L 0 e3 = C11 a−β Iβ (ζn a) − C12 βa−β−1 Iβ (ζn a) + C12 a−β Iβ (ζn a) (ζn a) e4 = C11 a−β I−β (ζn a) − C12 βa−β−1 I−β (ζn a) + C12 a−β I−β

e5 = C21 b−β Iβ (ζn b) − C22 βb−β−1 Iβ (ζn b) + C22 b−β Iβ (ζn b) (ζn b) e6 = C21 b−β I−β (ζn b) − C22 βb−β−1 I−β (ζn b) + C22 b−β I−β

The solution (s) makes sense if e3 e6 − e4 e5 = 0.

3.3

Steady-State Three-Dimensional Problems

The problems of this nature are in general not easy to solve unless the geometry and boundary conditions are simple. A mathematical tool to analyze these types of problems is the method of separation of variables. An example will show the procedure. Example 21 Consider a half of a solid cylinder 0 ≤ φ ≤ π of radius a and height h in which the surface temperature of all of its surfaces is zero except the semi-circular plane at the end of the cylinder, z = h, which is exposed to a known temperature distribution f (r, φ), Fig. 4.3-4. It is required to obtain a distribution of the steady-state temperature in the semi-cylinder. Solution The steady state temperature distribution should satisfy Laplace equation in cylindrical coordinates 1 ∂T ∂2T 1 ∂2T ∂ 2T + + =0 + ∂r2 r ∂r r2 ∂φ2 ∂z 2

(a)

192


Figure 4.3-4: Half of a solid cylinder. The boundary conditions are T (0, φ, z) = finite T (a, φ, z) = 0 T (r, 0, z) = 0 T (r, π, z) = 0 T (r, φ, 0) = 0 T (r, φ, h) = f (r, φ) = known

(b)

We assume the solution in the product form T (r, φ, z) = R(r)Φ(φ)Z(z)

(c)

Substituting Eq. (c) into Eq. (a) gives

2R

R Φ Z +r + + r2 =0 r R R Φ Z or

R R Z Φ (d) = −r2 −r − r2 = −α2 Φ R R Z where α2 is the separation constant. The negative sign for α is chosen because the solution in φ-direction must be periodic, T (r, φ, z) = T (r, φ + 2π, z). This choice yields Φ = A cos αφ + B sin αφ

From the third and fourth of boundary conditions (b) it follows that A=0 α = n = 1, 2, 3, ...


193

Thus, a solution in φ-direction becomes Φ = Bn sin nφ

(e)

From Eq. (d) the separation of the functions R and Z yields

n2 R 1R Z =− 2 + + = −λ2 − Z r R rR

(f )

where λ2 is a new separation constant and the choice for the negative sign has been made because the positive sign, upon using the boundary conditions, results in a trivial solution. From Eq. (f )

R +

1 n2 R + (λ2 − 2 )R = 0 r r

This is a standard form of Bessel equation which is readily solved to give R = CJn (λr) + DYn (λr) Applying the boundary conditions, the first of Eqs. (b) requires D = 0, as Yn approaches infinity when r → 0. The second of Eqs. (b) defines the characteristic value λ as Jn (aλ) = 0 The roots of this equation are λm a = ρmn , where ρmn are the roots of Jn (ρmn ) = 0. Thus, ρmn λm = (g) a and the solution in r-direction reduces to R = CJn (λm r)

(h)

Now, the differential equation in z-direction from Eq. (f ) becomes d2 Z − λ2m Z = 0 dz 2

(i)

Integration of this equation yields Z = E cosh λm z + F sinh λm z From the fifth of Eqs. (b) it follows that E = 0 and thus Z = F sinh λm z

(j)

194


Substituting Eqs. (e), (h), and (j) into Eq. (c) gives T (r, φ, z) =

∞ ∞

Amn sin nφJn (λm r) sinh λm z

(k)

m=1 n=1

The constant Amn may now be determined from the nonhomogeneous boundary condition at z = h T (r, φ, h) = f (r, φ) =

∞ ∞

Amn sin nφJn (λm r) sinh λm h

(l)

m=1 n=1

The expansion of the function f (r, φ) into Fourier-Bessel series, using Eq. (4.259), yields the value of Amn as Amn

2 = π

a 0

r[

π 0

f (r, φ) sin nφdφ]Jn (λm r)dr 2 sinh λm hJn+1 (λm a)

a2 2

(m)

While the temperature distribution is a function of all three coordinates, the solution was obtained in a simple manner as only one nonhomogeneous boundary condition was incurred. If more than one nonhomogeneous boundary condition is involved, then a decomposition problem technique described in Section 2.6 of this chapter should be employed to handle the nonhomogeneous boundary conditions.

3.4

Transient Problems

A general form of the governing heat conduction equation in cylindrical coordinates is obtained to be 1 ∂ ∂T 1 ∂ 2T ∂2T ρc ∂T (r )+ 2 2 + 2 = r ∂r ∂r r ∂φ ∂z k ∂t

(4.3-3)

In general, an analytical solution to this equation when the temperature is a function of space variables as well as the time is rather difficult to obtain, and for such problems numerical methods are recommended, especially, if the boundary conditions are complicated. In the following, analytical solutions for some typical simple problems are presented. Example 22 Consider an infinite solid cylinder of radius b subjected to an initial temperature T (r, 0) = g(r), where g(r) is a known function of r. If we keep the outer surface r = b at a constant temperature T0 , find the temperature distribution in the cylinder.


195

Solution The temperature distribution in this case is a function of the radius and the time, and thus the heat conduction equation, taking θ(r, t) = T (r, t) − T0 , becomes ∂θ ∂ 2 θ 1 ∂θ = κ( 2 + ) (a) ∂t ∂r r ∂r k where κ = ρc is the diffusivity. The boundary and initial conditions are θ(b, t) = 0 θ(r, 0) = g(r) − T0 = f (r) θ(0, t) = finite

(b)

A solution to Eq. (a) may be obtained by the separation of variables, but we may proceed in a different way by taking the solution in the form θ(r, t) = e−κλ t R(r) 2

(c)

where R is a function of the radius only. Upon substitution of Eq. (c) into Eq. (a) we obtain for R Bessel equation of order zero d2 R 1 dR + + λ2 R = 0 dr2 r dr

(d)

where λ2 is a constant which will be found from the boundary conditions. A solution to this equation, recalling that Bessel function of the second kind has an infinite value at r = 0, is R(r) = AJ0 (λr)

(e)

Substituting Eq. (e) into Eq. (c) gives θ(r, t) = Ae−κλ t J0 (λr) 2

(f )

The first boundary condition gives the characteristic values λ as J0 (λn b) = 0

(g)

This equation has an infinite number of real roots λ1 , λ2 , ..., λn , and thus Eq. (f ) becomes θ(r, t) =

∞

An e−κλn t J0 (λn r) 2

(h)

n=1

From the initial condition we can find the constant coefficients An f (r) =

∞ n=1

An J0 (λn r)

(i)

196


Equation (i) is Fourier-Bessel expansion of the function f (r). Multiplying both sides of Eq. (i) by rJ0 (λm r) then integrating from 0 to b and setting n = m results in the proper value for the constants An . For g(r) = T1 , where T1 is constant, θ1 = T1 − T0 and Fourier-Bessel expansion of θ1 , using Eq. (4.2-59), yields the expression for the constants An An =

2θ1 (λn b)J1 (λn b)

(j)

The final solution for the unsteady state temperature in the solid cylinder is ∞ T (r, t) − T0 e−κλn t J0 (λn r) =2 T1 − T0 n=1 (λn b)J1 (λn b) 2

(k)

Example 23 Consider a finite solid circular cylinder of radius b and length 2L subjected to a general prescribed heat flux on its boundary at r = b in the form of Qf (φ)g(z)H(t). Here Q is assumed to be the strength of the source of heat per unit area per unit time and H(t) is the Heaviside step function. The ends at z = ±L are exposed to free convection (h, T∞ ). It is required to find the transient three-dimensional temperature distribution in the cylinder, see Takeuti and Noda [14]. Solution The transient heat conduction equation is ∂ 2 θ 1 ∂θ 1 ∂2θ ∂2θ ∂θ = κ( 2 + + 2 2 + 2) ∂t ∂r r ∂r r ∂φ ∂z

(a)

where θ = T − T∞ . The boundary conditions are k

∂θ + hθ = Qf (φ)g(z)H(t) ∂r

∂θ ± hθ = 0 ∂z and the initial condition is

at

k

θ=0

at

at

r=b

z = ±L

t=0

(b)

(c)

Introducing the following dimensionless quantities ρ = r/b

ξ = z/b

ξ1 = L/b

¯ = hb/k h

τ = κt/b2

θ¯ = kθ/(Qb)

(d)


197

the governing equation and the boundary and initial conditions become ∂ θ¯ ∂ 2 θ¯ 1 ∂ θ¯ 1 ∂ 2 θ¯ ∂ 2 θ¯ = 2+ + 2 2+ 2 ∂τ ∂ρ ρ ∂ρ ρ ∂φ ∂ξ ∂ θ¯ ¯ ¯ + hθ = f (φ)g(ξ)H(τ ) ∂ρ ∂ θ¯ ¯ ¯ ± hθ = 0 ∂ξ θ¯ = 0

at at

at

(e)

ρ=1

ξ = ±ξ1 τ =0

(f )

Assuming g(ξ) to be symmetric with respect to ξ = 0, Laplace transforms of Eqs. (e) and (f ) are Δ1 θ¯∗ = sθ¯∗ (g) ∂ θ¯∗ ¯ ¯∗ 1 + hθ = f (φ)g(ξ) ∂ρ s ∂ θ¯∗ ¯ ¯∗ ± hθ = 0 ∂ξ

at

at

ρ=1

ξ = ±ξ1

(h)

where Laplace transforms are indicated by an asterisk and s is the Laplace transform parameter, and Δ1 =

1 ∂2 ∂2 1 ∂ ∂2 + + + ∂ρ2 ρ ∂ρ ρ2 ∂φ2 ∂ξ 2

A general solution of Eq. (g) subjected to the boundary conditions (h) is θ¯∗ =

∞ ∞ (An cos nφ + Bn sin nφ)Jn (μρ)Ck cos βk ξ k=1 n=0

¯ n (μ) − μJn+1 (μ)] s[(n + h)J

(i)

where Jn is Bessel function of the first kind and n-th order, μ2 = −(s + β 2 ), and 1 π f (φ)dφ 2π −π 1π f (φ)(cos nφ, sin nφ) dφ (An , Bn ) = π −π ξ1 ¯2 βk2 + h Ck = ¯ ¯ 2 ) −ξ1 g(ξ) cos βk ξdξ h + ξ1 (βk2 + h A0 =

and βk are the roots of the equation ¯ cos βξ1 = 0 β sin βξ1 − h

(j)

198


The inverse Laplace transform of Eq. (i) gives the temperature distribution in the cylinder θ¯ =

∞ k=1

Ck cos βk ξ

∞ n=0

(An cos nφ + Bn sin nφ)

In (βk ρ) ¯ n (βk ) + βk In+1 (βk ) (n + h)I

∞

exp[−(μ2nm + βk2 )τ ] 2μnm Jn (μnm ρ) 2 + ¯ ¯ (μnm + βk2 ){[n(n + h)/μ nm − μnm ]Jn (μnm ) − hJn+1 (μnm )} m=1

(k)

where In is the modified Bessel function of the first kind and n-th order, and μnm is the m-th positive root of the equation (n + h)Jn (μnm ) − μnm Jn+1 (μnm ) = 0 Example 24 Consider a circular brake disk of radius a and thickness 2L, as shown in Fig. 4.3-5, see Takeuti and Noda [15]. The disk is considered to have an annular contact on both sides when the brake operates, so that due to the mechanical contact between the brake and the disk a heat input of intensity Q0 is generated on both sides of the disk. This causes the temperature to vary axisymmetrically about the axis of the disk. The disk is assumed to be exposed to the convective heat transfer to the ambient T∞ from the sides z = ±L and r = a. The coefficient of thermal convection from the sides z = ±L is assumed to be h2 and from the side r = a to be h1 . We are to find a temperature field in the brake disk. Solution The governing heat conduction equation is ∂θ ∂ 2 θ 1 ∂θ ∂ 2 θ + 2 = (a) + 2 ∂ρ ρ ∂ρ ∂ξ ∂τ where θ = T − T∞ and the dimensionless quantities are defined as follows: r z κt ¯ i = hi L h ρ= ξ= τ= 2 (b) L L L k

Figure 4.3-5: Circular brake disk.

4. Problems in Spherical Coordinates

199

The initial and boundary conditions are θ=0

at

∂θ ¯ + h1 θ = 0 ∂ρ

τ =0 at

ρ = ρ0

∂θ ¯ at ξ = ±1 (c) ± h2 θ = Qf (ρ) ∂ξ where ρ0 = κ/L, and Q = Q0 L/κ, Q0 being the heat generated per unit time and per unit volume, and f (ρ) represents the distribution of Q over the surfaces z = ±L. Using the method of separation of variables for the space variables and the method of Laplace transforms for the time variable, the solution becomes ∞

cosh gn ξ δ0h¯ 2 e−gn τ fn J0 (gn ρ) − θ=Q ¯ 2 cosh gn ) gn2 2(gn sinh gn + h n=1 2

∞

2βm cosh βm ξe−(gn +βm )τ − 2 2 ¯ m=1 [(h2 + 1) sin βm + βm cos βm ](gn + βm ) 2

2

(d)

In this equation the distribution of heat flux on the surfaces ξ = ±1 has been expanded into Bessel function series, and fn is the coefficient of expansion f (ρ) =

∞

fn J0 (gn ρ)

(e)

n=1

where J0 (gn ρ) is Bessel function of the first kind and zero order, and gn are positive roots of the equation ¯ 1 J0 (gρ0 ) = 0 gJ0 (gρ0 ) − h with βm being positive roots of the equation ¯2 β tanh β = h ¯ 2 = 0, and δ0h¯ = 0 when h ¯2 = where δ0h¯ 2 = 1 when h 0. 2 The problem of temperature distribution in a long cylinder when a line source of heat rotates on its surface is given by Hetnarski [16].

4

Problems in Spherical Coordinates

The governing heat conduction equation in spherical coordinates has the form 1 ∂2 ∂2T 1 R ∂ ∂T 1 ρc ∂T (rT ) + + = (sin θ ) + (4.4-1) 2 2 2 2 2 r ∂r r sin θ ∂θ ∂θ k k ∂t r sin θ ∂φ where the variables are in accordance with Fig. 4.4-1. In its very general form, the analytical solution of this equation may be obtained by the use of separation of variables. The one- and two-dimensional cases are discussed in the subsections which follow.

200


Figure 4.4-1: Spherical coordinates.

4.1

Steady-State One-Dimensional Problems

The differential equation of heat conduction for radial temperature distribution in spherical coordinates reduces to R 1 d2 (rT ) + = 0 (4.4-2) 2 r dr k If R = const, a solution to this equation, including the particular solution, is T (r) = −

R 2 C1 r + + C2 6k r

(4.4-3)

where C1 and C2 are the constants of integration. Now we may apply this result to some problems which are frequently needed in design. Example 25 Consider a hollow thick sphere of inside radius a and outside radius b. The temperature distribution is a function of radius alone. Solution Consider the following cases: (a) The rate of heat generation in the material of the sphere is zero but the temperature at the inside surface is Ta and at the outside surface is Tb . The boundary conditions are T (a) = Ta ,

T (b) = Tb

(a)

Substituting these conditions in Eq. (4.4-3) and taking R = 0 yields T = Ta − (Ta − Tb )

(1 − a/r) (1 − a/b)

(b)


201

(b) The inside temperature is Ta and the outside surface is exposed to free convection to the ambient at T∞ . We assume R = 0. The boundary conditions are ∂T (b) (c) + h1 [T (b) − T∞ ] = 0 T (a) = Ta , ∂r where h1 = h/k. Substituting these conditions in Eq. (4.4-3) and evaluating the constants yields T (r) =

h1 b2 T∞ (r − a) + aTa [r(1 − h1 b) + h1 b2 ] r[a(1 − h1 b) + h1 b2 ]

(d)

(c) The sphere is solid and its surface temperature at r = b is Tb , and the rate of heat generated per unit volume is a constant R. The boundary conditions are T (0) = finite,

T (b) = Tb

(e)

The temperature distribution satisfying the boundary conditions is T =

4.2

R 2 (b − r2 ) + Tb 6k

(f )

Steady-State Two- and Three-Dimensional Problems

In spherical coordinates a method of separation of variables results in Legendre differential equation. For this reason, before proceeding further, we discuss Legendre equation and Legendre series. The differential equation (1 − x2 )

d2 y dy − 2x + n(n + 1)y = 0 dx2 dx

(4.4-4)

where n is a constant, is called Legendre differential equation. A general solution to this equation is obtained in a similar manner to that of Bessel equation, and it is, see Korn and Korn [10] y(x) = APn (x) + BQn (x)

(4.4-5)

where Pn (x) is Legendre polynomial of the first kind and degree n, and Qn (x) is Legendre polynomial of the second kind and degree n, and both polynomials have some definite relations with the variable x, as will be shown. The method at which we arrive at Eq. (4.4-5) is by assuming a series solution of Eq. (4.4-4) as y(x) =

∞

k=0

ak x k

(4.4-6)

202


which upon substitution in Eq. (4.4-4) results in a recurrence relation for ak ak+2 = −ak

(n − k)(n + k + 1) (k + 1)(k + 2)

k = 0, 1, 2, ...

(4.4-7)

Specifically, we define n(n + 1) a0 2! (n − 1)(n + 2) n(n + 1)(n − 2)(n + 3) a3 = − a4 = a1 a0 3! 4! (n − 1)(n + 2)(n − 3)(n + 4) a5 = a1 5!

a0 = a0

a1 = a1

a2 = −

We see that all ai are expressible in terms of a0 and a1 and, therefore, a solution being the sum of two independent solutions to Eq. (4.4-4) is obtained as n(n + 1) 2 n(n + 1)(n − 2)(n + 3) 4 x + x − · · · · · · · ·] 2! 4! (n − 1)(n + 2) 2 (n − 1)(n + 2)(n − 3)(n + 4) 4 + a1 x[1 − x + x − · · · · · · · ·] 3! 5! (4.4-8) y1 (x) = a0 [1 −

This solution is convergent as long as |x| < 1. It can be easily verified that when n is a positive even integer, the first expression in Eq. (4.4-8) has a finite number of terms, but the second expression remains to have infinite number of terms. In this case, when n is a positive even integer, the first expression reduces to a polynomial called Legendre polynomial, but the second expression remains an infinite series. When n is a positive odd integer the situation is opposite, that is, the second expression in Eq. (4.4-8) has a finite number of terms, called a Legendre polynomial, and the first expression is an infinite series. To obtain a standard form for Legendre polynomial in any case, when n is even or odd, it is customary to multiply the finite sum occurring in Eq. (4.4-8) by one of the following factors (−1)n/2 n! 2n [( n2 )!]

when n is even integer

(−1)(n−1)/2 (n + 1)! 2n ( n−1 )!( n+1 )! 2 2

when n is odd integer

This leads to the following general relations for the series in Eq. (4.4-8), which we call here Pn (x) and Qn (x) as indicated in Eq. (4.4-5): (a) When n is a positive odd integer greater than or equal to 3, then Pn (x) = (−1)(n−1)/2

1 × 3 × 5 × · · · · ·n 2 × 4 × 6 × · · · · ·(n − 1)


203

(n − 1)(n + 2) 3 (n − 1)(n + 2)(n − 3)(n + 4) 5 x + x + · · · · · ·] 3! 5! 2 × 4 × 6 × · · · · ·(n − 1) Qn (x) = (−1)(n+1)/2 1 × 3 × 5 × · · · · ·n n(n + 1) 2 n(n + 1)(n − 2)(n + 3) 4 ×[1 − x + x + · · · · · ·] 2! 4! (4.4-9)

×[x −

(b) When n is a positive even integer greater than or equal to 2, then 1 × 3 × 5 × · · · · ·(n − 1) 2 × 4 × 6 × · · · · ·n n(n + 1) 2 n(n + 1)(n − 2)(n + 3) 4 ×[1 − x + x + · · · · ·] 2! 4! 2 × 4 × 6 × · · · · ·n Qn (x) = (−1)n/2 1 × 3 × 5 × · · · · ·(n − 1) (n − 1)(n + 2) 3 (n − 1)(n + 2)(n − 3)(n + 4) 5 ×[x − x + x + · · · · ·] 3! 5! (4.4-10) Pn (x) = (−1)n/2

It is easily verified that the solution to Legendre differential equation as given by Eq. (4.4-5) is in terms of Legendre polynomials Pn (x), which has a finite number of terms depending on the value of n, and Legendre function Qn (x), which has an infinite number of terms, as defined in Eqs. (4.4-9) and (4.4-10). The first seven Legendre polynomials Pn (x) are P0 (x) = 1 P1 (x) = x 1 P2 (x) = (3x2 − 1) 2 1 P3 (x) = (5x3 − 3x) 2 1 P4 (x) = (35x4 − 30x2 + 3) 8 1 P5 (x) = (63x5 − 70x3 + 15x) 8 1 P6 (x) = (231x6 − 315x4 + 105x2 − 5) 16

(4.4-11)

Legendre polynomial Pn (x) is also expressible in the following form Pn (x) =

1 dn (x2 − 1)n 2n n! dxn

(4.4-12)

which is called Rodrigues formula. We also notice that Legendre polynomials have the following properties for all values of n

204

Chapter 4. Heat Conduction Problems Pn (1) = 1 Pn (−1) = (−1)n Pn (−x) = (−1)n Pn (x)

(4.4-13)

Differentiating Legendre differential equation (4.4-4) m times using Leibnitz formula we obtain the associated Legendre differential equation as (1 − x2 )

d2 y dy m2 − 2x ]y = 0 + [n(n + 1) − dx2 dx 1 − x2

(4.4-14)

The standard solution to this differential equation is y(x) = APnm (x) + BQm n (x)

(4.4-15)

where

dm Pn (x) (4.4-16) dxm is called the associated Legendre polynomial of the first kind of degree n and order m and m 2 m/2 d Qn (x) (x) = (1 − x ) (4.4-17) Qm n dxm is called the associated Legendre polynomial of the second kind of degree n and order m. The form of Pnm (x) for some values of m and n, when x = cos θ, is Pnm (x) = (1 − x2 )m/2

P11 (x) = (1 − x2 )1/2 = sin θ P21 (x) = 3x(1 − x2 )1/2 = 3 cos θ sin θ P22 (x) = 3(1 − x2 ) = 3 sin2 θ 3 3 P31 (x) = (5x2 − 1)(1 − x2 )1/2 = (5 cos2 θ − 1) sin θ 2 2 2 2 P3 (x) = 15x(1 − x ) = 15 cos θ sin2 θ P33 (x) = 15(1 − x2 )3/2 = 15 sin3 θ 5 5 P41 (x) = (7x3 − 3x)(1 − x2 )1/2 = (7 cos3 θ − 3 cos θ) sin θ 2 2 15 15 2 2 2 P4 (x) = (7x − 1)(1 − x ) = (7 cos2 θ − 1) sin2 θ 2 2 3 2 3/2 P4 (x) = 105x(1 − x ) = 105 cos θ sin3 θ P44 (x) = 105x(1 − x2 )2 = 105 sin4 θ (4.4-18) In discussing Legendre differential equation it was indicated that to have a solution, the condition |x| < 1 must be satisfied and, furthermore, the limit of convergence for the solution in Legendre polynomial was set to be −1 < x < +1. This suggests that we may take the variable in the form of x = cos θ. With such a change of variable, Eq. (4.4-4) reduces to dy 1 d (sin θ ) + n(n + 1)y = 0 sin θ dθ dθ

(4.4-19)


205

This is Legendre differential equation in terms of the variable θ, and thus the solution follows from Eq. (4.4-5): y = APn (cos θ) + BQn (cos θ)

(4.4-20)

Equation (4.4-19) is usually obtained when Laplace equation in spherical coordinates is solved by the method of separation of variables and the variable θ is considered as the co-latitude coordinate. Similarly to the trigonometric functions cos nx and sin nx, Legendre polynomials Pn (x) are orthogonal functions with respect to a weighting function w(x) = 1, over the interval −1 to +1, that is +1 −1

if m = n

(4.4-21)

n = 0, 1, 2, ....

(4.4-22)

Pm (x)Pn (x)dx = 0

and when m = n +1 −1

[Pn (x)]2 dx =

2 2n + 1

This property is used to expand any arbitrary function in terms of Legendre polynomials. Suppose f (x) is a continuous function and its derivative is continuous in the interval (−1, 1). Then the function f (x) can be expanded in a series of the form ∞ f (x) =

an Pn (x)

(4.4-23)

n=0

The coefficients an are obtained by multiplying both sides of Eq. (4.4-23) by Pm (x) and integrating over the interval (−1, 1). This gives +1 −1

f (x)Pm (x)dx =

∞

+1

an

n=0

−1

Pm (x)Pn (x)dx

Using Eqs. (4.4-21) and (4.4-22) yields +1 −1

+1

f (x)Pn (x)dx = an

and thus an =

−1

[Pn (x)]2 dx =

2an 2n + 1

2n + 1 +1 f (x)Pn (x)dx 2 −1

(4.4-24)

The integral in Eq. (4.4-24) is obtained as long as we know f (x). Similarly to Pn (x), we notice that Pnm (x) are also orthogonal in the same interval of (−1, 1) with respect to the weighting function w(x) = 1, that is +1 −1

Pnm (x)Pkm (x)dx = 0

for n = k

(4.4-25)

206


and

+1 −1

[Pnm (x)]2 dx =

(n + m)! 2 (n − m)! (2n + 1)

(4.4-26)

This property suggests that, like in the case of Legendre polynomial Pn (x), we can expand a given function F (x) in terms of Pnm (x) as F (x) =

∞

Cn Pnm (x)

n=0

Multiplying both sides of the above equation by Pkn (x) and integrating from −1 to 1 and using Eqs. (4.4-25) and (4.4-26) gives Cn =

(n − m)! (2n + 1) +1 F (x)Pnm (x)dx . (n + m)! 2 −1

Often, while solving Laplace equation in spherical coordinates, we have to deal with a series expansion involving Pnm (x) of the following form F (θ, φ) =

∞

(amn cos mφ + bmn sin mφ)Pnm (cos θ)

n=0

In this case we need to find the coefficients amn and bmn in terms of the known function F (θ, φ). To find amn we note that 2π

F (θ, φ) cos kφdφ = 0

∞ ∞

amn πPnm (cos θ)δmk

n=0 m=0

where δmk is the Kronecker symbol. This gives 2π

F (θ, φ) cos mφdφ = 0

∞

amn πPnm (cos θ)

n=0

Using the orthogonality condition for Pnm (x), multiplying both sides by sin θPsk (cos θ), integrating from 0 to π, and interchanging k with m in the final result, yields amn

2π (n − m)! (2n + 1) π m = sin θPn (cos θ)dθ F (θ, φ) cos mφdφ (n + m)! 2π 0 0

In a similar manner bmn is found. In Figs. 4.4-2 to 4.4-5 the plots of Legendre polynomials with respect to θ and x are shown [17]. We now present some useful relations for Legendre polynomials and the associated Legendre functions, as Pnm (−x) = (−1)n+m Pnm (x) Pnm (±1) = 0 for m > 0

(4.4-27)

4. Problems in Spherical Coordinates Pn (Cos θ)

207

1

0.8 0.6

P1

0.4

P2 0.2 0

20

40

θ 80

60

P3

−0.2 −0.4

Figure 4.4-2: Legendre polynomial Pn (cos θ) versus θ. Pn (x)

16 14 12

P3

10 8 6 4

P2

2

P1 1

1.2

1.4

1.6

1.8

x

2

Figure 4.4-3: Legendre polynomial Pn (x) versus x. The recurrence relations for Pn (x) and Qn (x) are Pn (x) = Qn (x) =

n/2

(−1)k (2n − 2)! xn−2k n k!(n − 2k)!(n − k)! 2 k=0 ∞ 2n (n + k)!(n + 2k)! −(n+2k+1) x k=0

k!(2n + 2k + 1)!

(2n + 1)xPn (x) = (n + 1)Pn+1 (x) + nPn−1 (x)

n = 1, 2, 3, ...

Pn+1 (x) − Pn−1 (x) = (2n + 1)Pn (x)

n = 1, 2, 3, ...

208


Qn (x)

2.5 2 1.5 1

Q0 0.5 0 −0.5 −1

Q3 0.2

0.4

0.6

x

0.8

Q2 Q1

Figure 4.4-4: Legendre polynomial Qn (x) versus x. Qn (x) 4

3

2

Q0

1

Q3 Q2 0

1

1.1

Q1 1.2

1.3

1.4

x

1.5

Figure 4.4-5: Legendre polynomial Qn (x) versus x.

(n + 1)Qn+1 (x) − (2n + 1)xQn (x) + nQn−1 (x) = 0 xQn (x) − Qn−1 (x) − nQn (x) = 0

(4.4-28)

Knowing the properties of Legendre polynomials, we return to our discussion on a general solution of the heat conduction equation in spherical coordinates. We may consider heat conduction in a spherical, or partly spherical, solid


209

body which is exposed to some kind of steady thermal fields. The steady-state temperature will then satisfy Laplace equation ∇2 T = 0

(4.4-29)

∂2T ∂T ∂T ∂2T 1 ∂ 2T + 2r sin θ + cos θ =0 + sin θ + ∂r2 ∂r ∂θ2 ∂θ sin θ ∂φ2

(4.4-30)

which in spherical coordinates is r2 sin θ

Any solution of this equation is called a spherical harmonic. This equation will be solved by the method of separation of variables by taking T (r, θ, φ) = R(r)F (θ, φ)

(4.4-31)

Substituting this in Eq. (4.4-30) yields r2 sin θ

d2 R dR ∂F ∂ 2F R ∂2F F + 2r sin θ + cos θR =0 F + sin θR + dr2 dr ∂θ2 ∂θ sin θ ∂φ2

Dividing by RF sin θ, rearranging and equating the separated function to a constant such as λ, gives r2 d2 R 2r dR + R dr2 R dr 1 ∂2F cos θ ∂F 1 ∂ 2F = −( + )=λ + F ∂θ2 F sin θ ∂θ F sin2 θ ∂φ2

(4.4-32)

It will be seen later that taking the separation constant λ = n(n + 1) is more convenient. Doing so, results in dR d2 R + 2r − n(n + 1)R = 0 2 dr dr ∂2F cos θ ∂F 1 ∂2F + + n(n + 1)F = 0 + ∂θ2 sin θ ∂θ sin2 θ ∂φ2

r2

(4.4-33) (4.4-34)

Equation (4.4-33) is the Euler differential equation and is readily solved to give A2 (4.4-35) rn+1 where A1 and A2 are constants of integration. Equation (4.4-34) still has to be separated, and thus by taking R(r) = A1 rn +

F (θ, φ) = Θ(θ)Φ(φ) and substituting in Eq. (4.4-34) we find cos θ dΘ d2 Φ 1 d2 Θ Φ + + n(n + 1)ΘΦ = 0 Θ Φ + dθ2 sin θ dθ sin2 θ dφ2

(4.4-36)

210


Dividing this equation by ΘΦ/ sin2 θ, and calling the separation constant m2 , gives 1 d2 Φ sin2 θ d2 Θ sin θ cos θ dΘ 2 + n(n + 1) sin + θ = − = m2 Θ dθ2 Θ dθ Φ dφ2 This yields sin2 θ

d2 Θ dΘ + sin θ cos θ + [n(n + 1) sin2 θ − m2 ]Θ = 0 2 dθ dθ

d2 Φ + m2 Φ = 0 2 dφ

(4.4-37) (4.4-38)

Solution to Eq. (4.4-38) is Φ = A3 cos mφ + A4 sin mφ

(4.4-39)

while the solution to Eq. (4.4-37) is obtained in terms of Legendre functions. In fact, Eq. (4.4-37) is the associated Legendre equation which by taking x = cos θ and knowing that dΘ dx dΘ dΘ = = − sin θ dθ dx dθ dx 2 d2 Θ dΘ dΘ d 2 d Θ (− sin θ ) = − cos θ + sin = θ dθ2 dθ dx dx dx2 it is modified to (1 − x2 )

d2 Θ dΘ m2 − 2x ]Θ = 0 + [n(n + 1) − dx2 dx 1 − x2

(4.4-40)

Equation (4.4-40) is the same as Eq. (4.4-14) and, therefore, its solution, after substitution of x = cos θ, is Θ = A5 Pnm (cos θ) + A6 Qm n (cos θ)

(4.4-41)

Substituting Eqs. (4.4-35), (4.4-39), and (4.4-41) into Eqs. (4.4-36) and (4.431), the general solution to Laplace equation in spherical coordinates becomes A2 )(A3 cos mφ + A4 sin mφ)[A5 Pnm (cos θ) + A6 Qm n (cos θ)] n+1 r (4.4-42) There are six constants of integration, A1 through A6 , which can be evaluated by using the thermal boundary conditions. We may notice that if the problem is designed in such a way that the temperature distribution is independent of the meridional angle φ, then the T (r, θ, φ) = (A1 rn +


211

temperature is only a function of r and θ and the equation in θ direction, since m = 0, is reduced to (1 − x2 )

d2 Θ dΘ − 2x + n(n + 1)Θ = 0 dx2 dx

(4.4-43)

which is Legendre differential equation and thus, from Eq. (4.4-5), Θ = A5 Pn (cos θ) + A6 Qn (cos θ)

(4.4-44)

Equation (4.4-44) is a solution in θ-direction for m = 0.

4.3

Transient Problems

The solutions to transient heat conduction problems in spherical coordinates are similarly obtained as those to steady-state problems. In the examples that follow some general problems of this nature are discussed. Example 26 Consider a hot solid sphere of radius b, initially at constant temperature T0 . The surface is exposed to convective cooling as the time increases. It is required to obtain a transient temperature in the sphere. Solution Taking θ(r, t) = T (r, t) − T∞ , the heat conduction equation is ∂θ ∂ 2 θ 2 ∂θ = κ( 2 + ) ∂t ∂r r ∂r

(a)

The boundary and initial conditions are −k

∂θ(b, t) = hθ(b, t) ∂r

∂θ(0, t) =0 ∂r θ(r, 0) = T0 − T∞ = θ0

(b) 1 2

˜ and employA solution to the problem, by interchanging the variable θ = θ/r ing the method of separation of variables along with the boundary conditions, is ∞

θ = r− 2 1

An J 1 (λn r) exp (−κλ2n t) 2

n=1

(c)

which can also be written in the form θ=

∞ n=1

2 sin λn r An π r

exp (−κλ2n t)

(d)

212


where λn is the n-th root of the characteristic equation bλ −1 k

tan bλ = − hb

(e)

Using the initial condition we have ∞

θ0 =

n=1

2 sin λn r An π r

(f )

Multiplying Eq. (f ) by r sin (λm r) and integrating with respect to r from r = 0 to r = b, when m = n, yields "

An =

π 2θ0 (sin bλn − bλn cos bλn ) 2 λn (bλn − sin bλn cos bλn )

Substituting for An gives θ = 2θ0

∞ (sin bλn − bλn cos bλn ) sin λn r n=1

bλn − sin bλn cos bλn

λn r

exp (−κλ2n t)

(g)

Example 27 Consider a solid sphere of radius b with its surface kept at zero temperature. The initial temperature of the sphere is a known function f (r, φ, θ) and we are to find a temperature T = T (r, φ, θ, t) inside the sphere. Solution The transient temperature distribution should satisfy the heat conduction equation 1 1 ∂2 ∂ ∂T 1 ∂2T ∂T (rT ) + ] = κ[ (sin θ ) + ∂t r ∂r2 r2 sin θ ∂θ ∂θ r2 sin2 θ ∂φ2

(a)

The boundary and initial conditions are T (b, φ, θ, t) = 0,

T (r, φ, θ, 0) = f (r, φ, θ) = known

(b)

Take a solution in the form T = e−κλ t F (r, φ, θ) 2

(c)

Substituting Eq. (c) in Eq. (a) and separating the function F (r, φ, θ), the following solution is obtained T (r, φ, θ, t) = (λr)−1/2 Jn+ 1 (λr)Pnm (cos θ)[A cos mφ + B sin mφ]e−κλ 2

2t

(d)


213

where m and n are positive integers. The first boundary condition at the outside surface, T = 0, is satisfied if λ is the root of Jn+ 1 (λb) = 0

(e)

2

In order to find the constants A and B, the known function f (r, φ, θ) is expanded in the form f (r, φ, θ) =

∞ ∞ ∞

(λr)− 2 Jn+ 1 (λr) 1

2

n=0 λ=0 m=0

×Pnm (cos θ)[Amnλ cos mφ + Bmnλ sin mφ]

(f )

where the summation over λ is over the roots of Eq. (e). Multiplying both sides of Eq. (f ) by cos mφ and integrating from 0 to 2π, gives 2π

f (r, φ, θ) cos mφdφ = π 0

∞ ∞

(λr)− 2 Jn+ 1 (λr)Pnm (cos θ)Amnλ 1

2

n=0 λ=0

Since +1 −1

[Pnm (cos θ)]2 d(cos θ) =

2(n + m)! (2n + 1)(n − m)!

and +1 −1

[Pnm (cos θ)][Pnk (cos θ)]d(cos θ) = 0

for m = k

thus +1 −1

=

Pnm (cos θ)d(cos θ)

2π

f (r, φ, θ) cos mφdφ 0 ∞

1 2(n + m)!π Amnλ (λr)− 2 Jn+ 1 (λr) 2 (2n + 1)(n − m)! λ=0

Again, multiplying both sides of this equation by r3/2 Jn+ 1 (λr) and integrating 2 from 0 to b yields b 0

r3/2 Jn+ 1 (λr)dr 2

+1 −1

Pnm (cos θ)d(cos θ)

2π

f (r, φ, θ) cos mφdφ 0

b ∞ 1 2(n + m)!π r3/2 Jn+ 1 (λr) Amnλ (λr)− 2 Jn+ 1 (λr)dr = 2 2 (2n + 1)(n − m)! 0 λ=0

From this equation, Amnλ is found. It is noticed that the coefficient π must be replaced by 2π when m = 0.

214


In a similar manner Bmnλ is obtained, and after substitution, a solution to the problem is T (r, φ, θ, t) =

∞ ∞ ∞

(λr)− 2 Jn+ 1 (λr) 1

2

n=0 λ=0 m=0

×Pnm (cos θ)[Amnλ cos mφ + Bmnλ sin mφ]e(−κλ

2 t)

(g)

Example 28 Consider an elastic solid sphere of radius r0 initially at zero temperature. At time t ≥ 0, part of its surface 0 ≤ θ ≤ θ0 at r = r0 is exposed to a heat flux q (θ), while the rest of the surface is insulated. Find the temperature distribution in the sphere. Solution The equation of nonsteady heat conduction in the spherical coordinates is ∂T 1 ∂ 1 ∂T 1 ∂ 2 ∂T (r )+ 2 [(1 − μ2 ) ] = (a) r2 ∂r ∂r r ∂μ ∂μ κ ∂t where μ = cos θ. The initial and boundary conditions are T (r, θ, 0) = 0

at

t≤0

∂T = q (μ) at r = r0 μ0 ≤ μ ≤ 1 t ≥ 0 ∂r ∂T = 0 at r = r0 − 1 ≤ μ ≤ μ0 t ≥ 0 k ∂r where μ0 = cos θ0 . We take the solution in the form [18] k

T (r, μ, t) = Ω(t) + Ts (r, μ) + T1 (r, μ, t)

(b)

(c)

where Ts is a steady state solution that satisfies the following equation and boundary conditions 1 ∂Ω (d) ∇2 Ts = κ ∂t ∂Ts k = q (μ) at r = r0 μ0 ≤ μ ≤ 1 ∂r ∂Ts k = 0 at r = r0 − 1 ≤ μ ≤ μ0 ∂r 1 r0 −1

0

Ts r2 drdμ = 0

and T1 satisfies the following system ∇2 T1 =

1 ∂T1 κ ∂t

(e)

4. Problems in Spherical Coordinates k

∂T1 =0 ∂r

at

215

r = r0

−1≤μ≤1

T1 (r, μ, 0) = −Ts (r, μ)

at

t≥0

t≤0

(f )

and for Ω(t) is [19] t 1 κ 2 Ω(t) = 2πr0 [ q (μ)dμ]dt kV 0 μ0

(g)

where V = 4πr03 /3 is the volume of the sphere. For a known function q (μ), the function Ω(t) is obtained by proper integration of Eq. (g). Once the function Ω is known from Eq. (g), it is substituted into Eq. (d) and the solution for Ts happens to be Ts (r, μ) = −

∞ c 2 c 2 2n + 1 r n r0 + r + r0 ( ) 10 6 2n r0 n=1

1 q (μ)

×{

−1

[

k

−

c r0 ]Pn dμ}Pn (μ) 3

(h)

where c = (1/κ)(dΩ/dt) and Pn (μ) is Legendre polynomial of the first kind and degree n. A solution T1 (r, μ, t), for the known function of Ts (r, μ) from Eq. (h), is obtained as T1 (r, μ, t) = −

∞ ∞ r0 2n + 1 r 1 ( )2 2 k n=0 m=1 βnm − n(n + 1) r0

Jn+ 1 (βnm r/r0 ) 1 2 2 2 [ Pn (μ)q (μ)dμ]Pn (μ)e−κβnm t/r0 × Jn+ 1 (βnm ) μ0 2

(i) where βnm are roots of Jn− 1 (βnm ) = 2

n+1 J 1 (βnm ) βnm n+ 2

(j)

and Jn+ 1 is Bessel function of the first kind and order (n + 12 ). 2 A complete solution for the temperature distribution of the sphere is the sum of Ω(t), Ts (r, μ), and T1 (r, μ, t) from Eqs. (g), (h), and (i), respectively. The problem of temperature distribution in a thick hollow sphere with a rotating heat source on its outer surface is given by Takeuti and Tanigawa [20]. They have used the method of Laplace transform to arrive at the solution for the temperature distribution.

216

5


Problems

1. Obtain a solution of Example 4 using the given boundary conditions. 2. Obtain a solution of Example 5 using the given boundary conditions. 3. Obtain the constants of integration of Example 8, using the given boundary conditions. 4. Consider an isolated pipe containing flowing fluid. The mean velocity and the temperature of the fluid at upstream are v and T0 , respectively. The inner and outer radii of the pipe are ri and ro , respectively. If a constant internal heat R is uniformly generated in the downstream half of the pipe, a steady state temperature distribution of the outer surface of the pipe is to be found. Neglect the axial conduction of the fluid and pipe. 5. Consider a hollow thick cylinder of inside radius a and outside radius b. The initial temperature of the cylinder is T = 0 at time t = 0. At time t > 0 a constant heat flux q is radiated to the side of the cylinder. The side z = 0 is insulated and z = L is at ambient temperature T∞ . The inside surface is kept at constant temperature T0 . Find the transient temperature distribution in the cylinder. 6. Consider a thick sphere of inside and outside radii a and b, respectively. At t = 0, the sphere is at uniform temperature T0 . The sphere is suddenly exposed to a constant heat flux q from one side. Find the transient temperature distribution in the sphere if the inside surface is kept at constant temperature T0 . The sphere is cooled by convection from the outer surface to the ambient at (h, T∞ ).

Bibliography [1] Arpaci, V.S., Conduction Heat Transfer, Addison-Wesley, Reading, Massachusets, 1966. [2] Carslaw, H.S. and Jaeger, J.C., Conduction of Heat in Solids, Clarendon, Oxford, 1959. [3] Holman, J.P., Heat Transfer, McGraw-Hill, New York, 1963. [4] Kreith, F., Principles of Heat Transfer, Intext Educational, New York, 1973. [5] Noda, N., Hetnarski, R.B., and Tanigawa, Y., Thermal Stresses, 2nd edition, Taylor and Francis, New York, 2003. [6] Pipes, L.A. and Harvill, L.R., Applied Mathematics for Engineers and Physicists, McGraw-Hill, New York, 1970. [7] Bowman, F., Introduction to Bessel Functions, Dover, New York, 1958. [8] McLachlan, N.W., Bessel Functions for Engineers, Clarendon, Oxford, 1955. [9] Johnson, D.G. and Johnson, J.R., Mathematical Methods in Engineering and Physics, Ronald, New York, 1965. [10] Korn, G.A. and Korn, T.M., Mathematical Handbook for Scientists and Engineers, McGraw-Hill, New York, 1968. [11] Sabbaghian, M. and Eslami, M.R., Creep Relaxation of Nonaxisymmetric Thermal Stresses in Thick Walled Cylinders, AIAA J., Vol. 12, No. 12, pp. 1652–1658, 1974. [12] Kantorovich, L.V. and Krylov, V.I., Approximate Methods of Higher Analysis, Interscience, New York, P. Noordhoff, Groningen, The Netherlands, 1964. [13] Hetnarski, R.B. and Ignaczak, J., Mathematical Theory of Elasticity, Taylor and Francis, New York, 2004. 217

218

Bibliography

[14] Takeuti, Y. and Noda, N., Three-Dimensional Transient Thermal Stresses in a Finite Circular Cylinder under Nonaxisymmetric Temperature Distribution, J. Therm. Stresses, Vol. 3, No. 2, pp. 159–183, 1980. [15] Takeuti, Y. and Noda, N., Thermal Stress Problems in Industry, 2: Transient Thermal Stresses in Disk Brake, J. Therm. Stresses, Vol. 2, No. 1, pp. 61–72, 1979. [16] Hetnarski, R.B., Stresses in Long Cylinder Due to Rotating Line Source of Heat, AIAA J., Vol. 7, No. 3, pp. 419–423, 1969. [17] Abramowitz, M. and Stegun, J., Handbook of Mathematical Functions, Dover, New York, 1965. [18] Cheung, J.B., Chen, T.S., and Thirumalai, K., Transient Thermal Stresses in a Sphere by Local Heating, J. Appl. Mech., Vol. 41, No. 4, pp. 930–934, 1974. [19] Olecer, N.Y., On the Theory of Conductive Heat Transfer in Finite Regions with Boundary Conditions of the Second Kind, Int. J. Heat Mass Tran., Vol. 8, pp. 529–556, 1965. [20] Takeuti, Y. and Tanigawa, Y., Transient Thermal Stresses of a Hollow Sphere Due to Rotating Heat Source, J. Therm. Stresses, Vol. 5, No. 3–4, pp. 283–298, 1982.

Chapter 5 Thermal Stresses in Beams As an application of the theory of thermoelasticity, thermal stress analysis of beams based on the elementary beam theory is the objective of this chapter. It begins with the derivation of formulas for axial thermal stresses and thermal lateral deflections in beams, and the associated boundary conditions are stated. The discussion on transient thermal stresses is presented, and the analysis of beams with internal heat generation follows. The formulas for thermal stresses in a bimetallic beam are discussed. The analysis of beams of functionally graded materials under steady state and transient temperature distributions is presented, and analysis of thermal stresses in curved beams concludes the chapter.

1

Introduction

Beams belong to the most frequently used elements in practical design problems. In this chapter, derivations of the beam’s governing equations based on elementary beam theory under mechanical and thermal loads are presented. Basically, Euler-Bernoulli assumption that the plane section remains plane under lateral deformation is used. The transverse shear stress is consequently ignored. Derivation of the governing equations for the thermal stresses and deflection equations of the beams are presented in Sections 2 through 5. In Section 6 the deflection and thermal stresses of rectangular beams are obtained. In Sections 7 and 8 the transient thermal stresses in beams under transient thermal loads and internal heat generation are given. Then, thermal stresses in nonhomogeneous beams, including beams with functionally graded materials, are discussed. Finally, thermal stresses in curved beams and rings are presented, where Euler-Bernoulli assumption is used to derive the thermal deflection and stresses.

R.B. Hetnarski, M.R. Eslami, Thermal Stresses – Advanced Theory and Applications, Solid Mechanics and its Applications 158, c Springer Science + Business Media B.V. 2009

219

220

2

Chapter 5. Thermal Stresses in Beams

Thermal Stresses in Beams

In accordance with Euler-Bernoulli hypothesis, a beam deflects in such a way that its plane sections remain plane after deformation and perpendicular to the beam’s neutral axis. Now, consider a beam under axial and lateral loads in x-y plane, as shown in Fig. 5.2-1. Consider two line elements of the beam, EF and GH, which are straight and along the axial direction with equal lengths before the load is applied. Element EF lies on the neutral axis, while the element GH is at a distance y from the neutral axis. The beam is assumed to be under the bending and axial loads so that it deflects in lateral direction. Considering Euler-Bernoulli hypothesis, the elongation of EF and GH elements may be written as F = (1 + )EF E# 0 # G H = (1 + )GH H ry + y G# = # ry EF

(5.2-1)

where and 0 are strains of GH and EF elements, respectively, and ry is the radius of curvature of the beam axis at y = 0 in the xy-plane. Dividing the second of Eqs. (5.2-1) by the first equation and using the last of Eqs. (5.2-1) gives 1+

(1 + ) y = ry (1 + 0 )

(5.2-2)

Then, using the small deformation theory, yields = 0 +

y y y + 0 ∼ = 0 + ry ry ry

(5.2-3)

Now, consider a beam with thermal gradients along the y and z-directions. In accordance with Euler-Bernoulli assumption, the axial displacement is a O

ry A E G C

y

F H D

B⬘

A⬘

B

x

E⬘ G⬘ C⬘

y

F⬘ H⬘ D⬘

Figure 5.2-1: Deflection of an element of beam.

2. Thermal Stresses in Beams

221

linear function of the coordinates y and z in the plane of cross section of the beam. Thus (5.2-4) u = C1 (x) + C2 (x)y + C3 (x)z where C1 , C2 , and C3 are coefficients, which are functions of x, the beam axis. Assuming thermal loading only, these coefficients may be obtained using the boundary conditions. Since the beam is in static equilibrium, the axial force and bending moments in y and z directions must vanish. These conditions in terms of the axial stress in the beam yield the following relations: A

σxx dA = 0,

A

σxx ydA = 0,

A

σxx zdA = 0

(5.2-5)

where dA = dydz. Equations (5.2-5) are sometimes called the equilibrium equations of the beam. In order to find C1 , C2 , and C3 , the axial strain is written from Eq. (5.2-4) as xx =

du y z dC1 dC2 dC3 + = + y+ z = 0 + dx dx dx dx ry rz

(5.2-6)

where ry and rz are the radii of curvatures of the beam axis in xy and xz planes, respectively, and 0 is the axial strain of the beam on the x-axis. The stress, according to Hooke’s law is σxx = E(xx − αθ)

(5.2-7)

where θ = T − T0 . Thus σxx = E[0 +

y z + − αθ] ry rz

(5.2-8)

Substituting Eq. (5.2-8) in Eqs. (5.2-5) and noting that 0 , ry , and rz are functions of x only, they may be taken outside of the integral over the area dydz, thus

1 1 ydA + zdA = αθdA ry A rz A A A 1 1 0 ydA + y 2 dA + yzdA = αθydA ry A rz A A A 1 1 0 zdA + yzdA + z 2 dA = αθzdA ry A rz A A A

0

dA +

(5.2-9)

From the above system of equations 0 , ry , and rz are calculated and substituted into Eq. (5.2-8) to obtain the axial stress. If we select the y and z axes as the centroid axes of the cross section, then

ydA = A

zdA = 0 A

(5.2-10)

222


and since, by definition, the moments of inertia and the product of inertia of the cross section are A

A A

y 2 dA = Iz z 2 dA = Iy yzdA = Iyz

(5.2-11)

then the system of Eqs. (5.2-9) may be solved for 0 , ry , and rz to give PT EA 2 E(Iy Iz − Iyz ) ry = Iy MT z − Iyz MT y 2 ) E(Iy Iz − Iyz rz = Iz MT y − Iyz MT z 0 =

(5.2-12)

where

PT =

EαθdA A

MT y = MT z =

EαθzdA A

EαθydA

(5.2-13)

A

Upon substitution of 0 , ry , and rz into Eq. (5.2-8) we obtain σxx = −Eαθ +

PT Iz MT y − Iyz MT z Iy MT z − Iyz MT y )y + ( )z +( 2 2 A Iy Iz − Iyz Iy Iz − Iyz

(5.2-14)

Equation (5.2-14) can be further simplified by taking the y and z axes in the principal directions of the cross sectional area of the beam. In this case Iyz = 0, and Eq. (5.2-14) simplifies to the form σxx = −Eαθ +

PT M T z y MT y z + + A Iz Iy

(5.2-15)

Equation (5.2-14) gives the axial stress in a beam subjected to thermal loading when the temperature distribution is a function of y and z. To find the strains, radii of curvature and thermal stresses due to the combined mechanical and thermal loads, the thermal moments must be replaced by the total moments acting on the beam in Eqs. (5.2-12), (5.2-14) and (5.2-15). Also, the term PT must be replaced by PT + PM , where PM is the axial load due to the external forces applied on the beam and the reaction forces at the boundary. Thus, in

3. Deflection Equation of Beams

223

general, when both mechanical and thermal moments are present, the relations for the axial strain, radii of curvature, and axial stress will be written in the form P EA 2 E(Iy Iz − Iyz ) ry = Iy Mz − Iyz My 0 =

rz =

2 ) E(Iy Iz − Iyz Iz My − Iyz Mz

(5.2-16)

and σxx = −Eαθ +

P Iz My − Iyz Mz Iy Mz − Iyz My )y + ( )z +( 2 2 A Iy Iz − Iyz Iy Iz − Iyz

(5.2-17)

in which the total moments and the axial load are My = M T y + M M y Mz = MT z + M M z P = PT + PM

(5.2-18)

where MT is the moment due to the thermal gradient, and MM is the mechanical moment due to action of the external forces and the reaction forces at the boundary of the beam.

3

Deflection Equation of Beams

Consider a beam of arbitrary cross section as shown in Fig. 5.3-1. The dimensions of the beam in y and z directions are assumed to be small enough so that σyy and σzz are negligible. Denoting deflection in y and z directions by v Z My

My

x

Mz

Mz y

Figure 5.3-1: Positive directions for the moments applied on the beam.

224


and w, respectively, the differential equation for the beam deflection is derived as follows. From the elementary beam theory and for a small deflection, the radii of curvature are related to the deflections of the beam as d2 v 1 d2 v dx2 ∼ = −⎡ − = ⎤ 3/2

2 ry dx2 dv ⎦ ⎣1 + dx d2 w dx2

1 d2 w ∼ = −⎡ − = ⎤

2 3/2 rz dx2 dw ⎣1 + ⎦ dx

(5.3-1)

where ry and rz are the radii of curvature of the beam axis in xy and xz planes, respectively. Substituting the relations for radii of curvature given by Eqs. (5.2-16) into Eqs. (5.3-1) leads to the deflection equations of the beam d2 v Iy Mz − Iyz My =− 2 2 ) dx E(Iy Iz − Iyz d2 w Iz My − Iyz Mz =− 2 2 ) dx E(Iy Iz − Iyz

(5.3-2)

When both the mechanical and the thermal moments act, the total moments are My = M T y + M M y Mz = MT z + M M z

(5.3-3)

where MT is the moment due to the thermal gradient, and MM is the mechanical moment due to the action of the external forces and the reaction forces at the boundary of the beam. When the principal directions of the cross section of the beam are selected as the y and z axes, Iyz = 0 and Eqs. (5.3-2) simplify to the form d2 v Mz =− 2 dx EIz My d2 w =− 2 dx EIy

(5.3-4)

In addition to the lateral deflections of the beam, the axial displacement of the beam may be of interest. The integration of Eq. (5.2-6) with respect to x, using the relations for the axial strain and radii of curvature given by Eqs. (5.2-16), results in an expression for the axial displacement as u = u0 +

x P 0

Iy Mz − Iyz My Iz My − Iyz Mz y+ z dx (5.3-5) + 2 ) 2 ) EA E(Iy Iz − Iyz E(Iy Iz − Iyz

4. Boundary Conditions

225

where u0 is the axial displacement at x = 0. Sometimes the average axial displacement for the cross section of the beam is of interest. The average of the axial displacement for the cross section is obtained from Eq. (5.3-5) as uav

1 = u dA A A x x

1 P Iy Mz − Iyz My = dx + u0 dA + ( y dA) 2 ) A A E(Iy Iz − Iyz A 0 EA 0

Iz My − Iyz Mz + ( z dA) dx 2 ) E(Iy Iz − Iyz A

(5.3-6)

Since the relations for the axial strain and radii of curvature given by Eqs. (5.2-16) are obtained when the y and z axes are the centroid axes of the cross section, Eq. (5.3-6) reduces to uav =

x 1 P u0 dA + dx A A 0 EA

(5.3-7)

It may be found from this equation that when a cantilever beam of constant cross section and length L is exposed to a thermal gradient and axial mechanical loads, the total average displacement of the free end of the beam is uav =

PL EA

(5.3-8)

and when the temperature rise, θ, is uniformly distributed through the cross section of the cantilever beam, Eq. (5.3-8) reduces to uav = αθL +

PM L EA

(5.3-9)

which is identical to the elongation of a bar under a uniform temperature rise θ and the axial mechanical load PM .

4

Boundary Conditions

Consider a beam of arbitrary cross section subjected to both mechanical and thermal loads. Take the x-axis along the axis of the beam. The following boundary conditions may exist at any end of the beam: 1. Simply supported end Let assume that the end x = L of the beam is simply supported in y direction. Thus, the deflection and the moment at this end must be zero, that is

w|x=L = 0

MM y |x=L

d2 w = −EIy 2 − MT y dx

=0 x=L

(5.4-1)

226


2. Built-in end At the built-in end the deflection and the slope of the beam must be zero. Thus, if the end x = L is assumed to be built-in, it follows that dw (5.4-2) |x=L = 0 dx 3. Free end At the free end the moment and the shear force must be zero, thus, if the end x = L is assumed to be free, it follows that [1,2] w|x=L = 0

d2 w =0 MM y |x=L = −EIy 2 − MT y dx x=L dMM y d dMT y d2 w (Qz )x=L = ( )x=L = − (EIy 2 )x=L − ( )x=L = 0 dx dx dx dx (5.4-3)

5

Shear Stress in a Beam

Consider an element of a beam of arbitrary cross section of the length dx, as shown in Fig. 5.5-1. The width of the beam, b, is assumed to be a function of y. If the distribution of the temperature change is assumed to be θ = θ(x, y), the thermal bending takes place in the x-y plane. The equilibrium equation for an element of the cross section of the beam is b2 a 1

b2 −b1

(σxy dx)dz +

−b1

y

b2 ∂σxx (σxx + dx)dy dz − ∂x −b1

a1 y

σxx dy dz = 0 (5.5-1)

Rearranging the terms yields σxy = −

a1 ∂σxx

∂x

y

dy

(5.5-2)

Substituting for σxx from Eq. (5.2-14), when the y and z axes are selected in principal directions and the beam is under thermal gradient load in y-direction, yields y

b2

∫ −b1(s xy dx) dz

b1 b2

b2

a

∫ −b1( ∫ y2(s

xx +

∂sxx dx( dy ( dz ∂x x

a1 y b2

a

∫ −b1( ∫ y2s xx dy ( dz

z dx

Figure 5.5-1: Transverse shear stress in a beam.

6. Beams of Rectangular Cross Section σxy = −

a1 ∂ y

∂x

227

[−Eαθ +

PT MT z ) y] dy +( A Iz

(5.5-3)

It is to be noted that Eq. (5.5-3) represents the shear stress due to the thermal gradient. If mechanical loads are present, the resulting transverse shear stress must be considered and added.

6

Beams of Rectangular Cross Section

Consider a beam of rectangular cross section of width b and height a subjected to the boundary conditions, shown in Fig. 5.6-1. The initial temperature of the beam is assumed to be zero [3,4]. The temperature of the upper and lower surfaces of the beam are then raised and maintained at constant temperatures T1 and T2 , respectively. A constant rate of internal energy generation R per unit volume per unit time is produced in the beam. The one-dimensional heat conduction equation is d2 T (5.6-1) k 2 +R=0 dy and the boundary conditions are T (a/2) = T1 ,

T (−a/2) = T2

(5.6-2)

Solving the heat conduction equation (5.6-2) subjected to the above boundary conditions leads to the temperature distribution in the beam in the form T1 − T2 T1 + T2 R 2 R (5.6-3) )y+( + a) T = (− ) y 2 + ( 2k a 2 8k Using the temperature distribution, the thermal force and thermal moment are a/2 1 T1 + T2 Ra2 + Eα(T − 0)bdy = EαA PT = EαθdA = 12k 2 A −a/2 a/2 Eα(T1 − T2 )Iz MT z = EαθydA = Eα(T − 0)ybdy = a A −a/2 T1

x

y

R a

O

z

T2 L

b

Figure 5.6-1: Rectangular beam.

228


Figure 5.6-2: Simply supported beam. Thus d2 v MT z + MM z =− 2 dx EIz in which MM z = M0z + Q0 x, and M0z and Q0 are the mechanical moment and shear (reaction) force at the left side boundary of the beam. Integrating twice yields v=−

M0z 2 Q0 3 α(T1 − T2 ) 2 x − x + C1 x + C2 x − 2a 2EIz 3EIz

(5.6-4)

For simply supported beams, as shown in Fig. 5.6-2, the boundary conditions are v(0) = v(L) = 0,

d2 v(0) d2 v(L) MT z = =− 2 2 dx dx EIz

Applying these boundary conditions to Eq. (5.6-4) gives M0z = Q0 = 0, C1 = α(T1 − T2 )L/(2a) and C2 = 0. In this case, the deflection is obtained as v=

α(T1 − T2 ) (xL − x2 ) 2a

(5.6-5)

The normal stress σxx from Eq. (5.2-17) is σxx = Eα

Ra2 2y 2 1 [( ) − ] 8k a 3

(5.6-6)

It may be found from this type of loading that when thermal gradient is independent of x, the reaction forces at the boundary of the simply supported beam are zero. For a cantilever beam, as shown in Fig. 5.6-3, the boundary conditions are v(0) =

MT z dMM z (L) d2 v(L) dv(0) =− , Qy (L) = = 0, =0 2 dx dx EIz dx

Applying the boundary conditions to Eq. (5.6-4) and calculating M0z , Q0 , and the constants C1 and C2 gives v=−

α(T1 − T2 ) 2 x 2a

(5.6-7)

6. Beams of Rectangular Cross Section

229

Figure 5.6-3: Cantilever beam. L

v

x

y (a) M0z

MMz Q0 x (b)

Figure 5.6-4: Fixed-simply supported beam. and the normal stress σxx is σxx = Eα

Ra2 2y 2 1 [( ) − ] 8k a 3

(5.6-8)

The shear stress for both cases may be easily shown to be zero. For a fixed-pinned condition, as shown in Fig. 5.6-4, the boundary conditions are dv(0) d2 v(L) MT z v(0) = v(L) = = 0, =− (5.6-9) 2 dx dx EIz Using the boundary conditions, yields v=

α(T1 − T2 ) 2 x3 (x − ) 4a L

(5.6-10)

where M0z = −3MT z /2 and Q0 = 3MT z /2L. In this case, the normal stress σxx is

σxx = Eα

R 2 3(T1 − T2 ) 3(T1 − T2 ) R 2 y + yx − y− a 2k 2aL 2a 24k

(5.6-11)

230


Figure 5.6-5: Fixed-fixed beam. The shear stress due to the thermal gradient from Eq. (5.5-2) is σxy = −

a/2 ∂σxx

∂x

y

dy =

3Eα(T1 − T2 )a 4y 2 (1 − 2 ) 16L a

(5.6-12)

For a fixed-fixed beam, as shown in Fig. 5.6-5a, the boundary conditions are v(0) = v(L) =

dv(0) dv(L) = =0 dx dx

Applying the boundary conditions to Eq. (5.6-4) gives M0z = −MT z , Q0 = 0 and results in zero values for the constants of integration C1 and C2 and thus v=0 The normal stress, when the beam is free to expand in the axial direction, is

σxx

R 2 T1 − T2 R 2 y − y− a = Eα 2k a 24k

(5.6-13)

When the axial expansion is prevented, as shown in Fig. 5.6-5b, a mechanical axial load PM develops. When the axial resulting load is less than the buckling load, the axial stress is obtained from the condition xx =

σxx + α (T − 0) = 0 E

(5.6-14)

that gives

σxx = Eα

R 2 T1 − T2 R 2 T1 + T2 y − y− a − 2k a 8k 2

(5.6-15)

7. Transient Stresses in Rectangular Beams

231

Substituting for the axial thermal stress from Eq. (5.6-15) into Eq. (5.2-17), the axial force is obtained as PM = −EAα [

R 2 T1 + T2 a + ] 12k 2

(5.6-16)

The shear stress for both cases may be easily shown to be zero. For the fixedfixed boundary conditions, the axial mechanical load may also be obtained from Eq. (5.3-7), where the total average axial displacement from x = 0 to x = L, uav , and the axial displacement at x = 0, u0 , are zero. Thus, the mechanical axial load is obtained as PM = −PT . From the last two types of boundary conditions it is found that there are special cases where no deformation occurs while thermal stresses exist. Also, there are some special cases of loading where thermal stress is zero but thermal deformation exist. For example, see Chapter 3 and the cases where the temperature is linearly distributed through the geometry of the body. For more basic problems related to thermal stresses in beams, the reader is recommended to refer to Chapter 2 of reference [4].

7

Transient Stresses in Rectangular Beams

Consider a simply supported rectangular cross section beam of height a and width b initially at reference temperature T0 . The side y = a/2 is suddenly exposed to a uniform heat flux q while other faces are insulated. The temperature distribution for this problem is obtained by solving the one-dimensional transient heat conduction. The origin of the coordinate system is considered at the centroid and along the principal direction of the cross section of the beam. The solution of the one-dimensional transient heat conduction equation for the temperature distribution is (see Example 12 of Chapter 4)

T (y, t) = T0 + −

aq 24κt 12y 2 12y + 2 + [ −1 24k a2 a a

∞ 48 nπ(2y + a) (−1)n (−n2 π2 κt/a2 ) e cos ] 2 2 π n=1 n 2a

(5.7-1)

where κ is the thermal diffusivity, k is the thermal conductivity, ρ is the mass density, and c stands for the specific heat of the material. Using the temperature distribution, the thermal force and the thermal moment are

PT = Eα

MT z

⎛

q Aκt a ⎞

∞ q Iz ⎝ 96 1 (−n2 π2 κt)/a2 ) ⎠ = Eα 1− 4 e 2k π n=1,3,5,... n4

(5.7-2)

232


It may be shown that the reaction forces and moments at the boundaries are zero and, therefore, the axial stress in the beam from Eq. (5.2-17) is

σxx

aq = Eα 2k

y2 1 − 2 12 a

+

96 y − 4( ) π a

∞ 4 nπ(2y + a) (−1)n (−n2 π2 κt)/a2 ) e cos 2 2 π n=1 n 2a ∞

⎤

1 (−n2 π2 κt/a2 ) ⎦ e 4 n=1,3,5,... n

(5.7-3)

Now consider the same thin rectangular cross section beam of height a and width b initially at reference temperature T0 . The side y = a/2 is suddenly exposed to a constant temperature T1 while the other faces are insulated. The solution of the one-dimensional heat conduction equation along the y-direction, considering the given boundary conditions, results in the temperature distribution ∞ 4(T1 − T0 ) (−1)n 2 (5.7-4) cos λn (y + a/2) e−λn κt T (y, t) = T1 + π n=1 (2n − 1) in which the characteristic value λn = (2n − 1)π/2a. The axial thermal force and the thermal bending moment from Eqs. (5.2-13) are calculated using the equation for temperature distribution from Eq. (5.7-4) as

∞ 1 8 2 e−λn κt PT = EαA(T1 − T0 ) 1 − 2 2 π n=1 (2n − 1)

MT z = −EαIz

∞ 48(T1 − T0 ) 1 2 e−λn κt [(2n − 1)π + 4(−1)n ] 3 3 π a n=1 (2n − 1) (5.7-5)

Since the reaction forces at the boundary of a simply supported beam under thermal gradient load are zero, the axial thermal stress in the beam from Eq. (5.2-14) is σxx = −Eα(T1 − T0 )

∞ n=1

−λ2n κt

e

8 4 (−1)n 1 cos λn (y + a/2) + 2 π (2n − 1) π (2n − 1)2

1 48 y + 3( ) [(2n − 1)π + 4(−1)n ] π a (2n − 1)3

8

(5.7-6)

Beam with Internal Heat Generation

Consider a simply supported beam of height a, length L, and width b. At t = 0, the temperature of the beam is uniform and equal to the ambient temperature T∞ . For t > 0, a heat is generated within the beam material at the rate R per unit time and unit volume. The sides y = ±b/2 are thermally insulated,

8. Beam with Internal Heat Generation

233

and the sides z = ±a/2 are exposed to free convection to ambient at (h, T∞ ), where h is the convective coefficient. The heat generation R is assumed in the form [5] (5.8-1) R = q0 e−μa/2 cosh (μz) where q0 and μ are assumed to be constants. The heat conduction equation takes the form R 1 ∂T ∂2T =− + 2 ∂z k κ ∂t

(5.8-2)

where κ = k/(ρc) is the thermal diffusivity. The initial and boundary conditions are T (z, 0) = T∞ ∂T (a/2, t) = h[T (a/2, t) − T∞ ] −k ∂z ∂T (−a/2, t) −k = −h[T (−a/2, t) − T∞ ] ∂z

(5.8-3)

Substituting the expression for the heat generation R from Eq. (5.8-1) into Eq. (5.8-2), and solving the transient partial differential equation (5.8-2) yields T (z, t) = T∞ +

∞

Bn (t) cos λn z

(5.8-4)

n=0

where λn are the roots of the characteristic equation λn tan (λn a/2) =

h k

and Bn (t) =

Gn 2 (1 − e−λn κt ) 2 κλn

where κC μ cos (λn a/2)(eμa/2 − e−μa/2 ) + λn sin (λn a/2)(eμa/2 + e−μa/2 ) Nn k μ2 + λ2n

1 sin (λn a) Nn = a+ 2 λn Gn =

Here, C = q0 e−μa/2 . The thermal stresses are obtained using the equations for thermal stresses in beams, see Eq. (5.2-15), as σxx = Eα [−(T − T∞ ) +

1 a/2 12z a/2 (T − T∞ )dz + 3 (T − T∞ )zdz] (5.8-5) a −a/2 a −a/2

234


Upon substitution from Eq. (5.8-4) into Eq. (5.8-5), the thermal stress distribution is σxx = −Eα

∞

Bn (t) cos λn z +

n=0

9

∞ 2Eα Bn (t) sin (λn a/2) a n=0 λn

(5.8-6)

Bimetallic Beam

Consider a simply supported beam made of two bonded layers of different materials, initially at reference temperature. The temperature is slowly raised to a constant uniform temperature, where the temperature change is θ. The origin (y = 0) is taken at the bonding surface of the bimetallic beam as shown in Fig. 5.9-1. The axial strains of each layer are expressed by [4] xxi =

σxxi y + αi θ = 0 + Ei ry

(i = 1, 2)

(5.9-1)

where 0 and ry denote the strain and the radius of curvature at the bonding surface y = 0, respectively. Since external forces do not act on the beam, then 0

a2

−a1 0 −a1

σxx1 bdy +

σxx2 bdy = 0

0

a2

σxx1 ybdy +

0

σxx2 ybdy = 0

(5.9-2)

Substituting stress σxxi from Eq. (5.9-1) into Eqs. (5.9-2), we get 2(E2 a2 + E1 a1 )b0 + (E2 a22 − E1 a21 ) 0

=2

−a1

a2

E1 α1 θbdy + 2

b ry

E2 α2 θbdy

0

3(E2 a22 − E1 a21 )b0 + 2(E2 a32 + E1 a31 ) 0

=6

−a1

a2

E1 α1 θybdy + 6

0

b ry

E2 α2 θybdy

(5.9-3)

L E1 ,α1 ,θ E2 ,α2 ,θ

a1 a2

x b

y

Figure 5.9-1: Bimetallic beam.

10. Functionally Graded Beams

235

From Eqs. (5.9-3) the strain 0 and the curvature 1/ry at the bonding surface (y = 0) are obtained as 2b2 θ [2(E1 α1 a1 + E2 α2 a2 )(E1 a31 D 3 +E2 a32 ) − (E2 α2 a22 − E1 α1 a21 )(E2 a22 − E1 a21 )] 2 6b2 θ = [(E2 α2 a22 − E1 α1 a21 )(E1 a1 + E2 a2 ) D −(E1 α1 a1 + E2 α2 a2 )(E2 a22 − E1 a21 )]

0 =

1 ry

(5.9-4)

(5.9-5)

where D = b2 [(E2 a22 − E1 a21 )2 + 4E1 E2a1 a2 (a1 + a2 )2 ] Substituting Eqs. (5.9-4) and (5.9-5) in Eq. (5.9-1) for thermal stresses gives 2Ei b2 θ [2(E1 α1 a1 + E2 α2 a2 )(E1 a31 + E2 a32 ) D 3 − (E2 α2 a22 − E1 α1 a21 )(E2 a22 − E1 a21 )] 2 6Ei b2 θ + [(E2 α2 a22 − E1 α1 a21 )(E1 a1 + E2 h2 ) D −(E1 α1 h1 + E2 α2 a2 )(E2 h22 − E1 a21 )] − Ei αi θ

σxxi =

(5.9-6)

Beams with more than two different layers are treated as the composites. For some basic problems of composite beams one may refer to reference [4]. Also, references [6–9] discuss the static, transient, and dynamic behavior of composite beams under thermal loads.

10

Functionally Graded Beams

Consider a beam made of a nonhomogeneous material, where its material properties vary continuously from one constituent material to another one (a beam made of a functionally graded material, or FGM beam or, simply, functionally graded beam), as shown in Fig. 5.10-1. The beam is initially at reference temperature T0 . The temperature of the beam is slowly raised to T = T0 +θ, where θ is the temperature change. Consider the system of coordinates (x1 , y1 , z1 ) in which the material properties of the functionally graded beam are introduced. We define another coordinate system (x, y, z) used to obtain the equations of thermal stress and deflection of the functionally graded beam. The coordinate transformation law between the coordinate systems (x, y, z) and (x1 , y1 , z1 ) is x = x1 ,

y = y1 − y ∗ ,

z = z1 − z ∗

(5.10-1)

236


Figure 5.10-1: FGM beam. where y ∗ and z ∗ are the distances along the y and z directions between the coordinate systems (x1 , y1 , z1 ) and (x, y, z), respectively. When Euler-Bernoulli assumption is valid, the strain at a position (x, y, z) is related to the strain and the radii of curvature of the beam on the x-axis as y z + (5.10-2) xx = 0 + ry rz where 0 , ry , and rz are the axial strain and the radii of curvature of the beam in y and z directions at y = z = 0, respectively. Also, from Hooke’s law for the isotropic linear thermoelastic materials σxx = E[xx − α(T − T0 )] y z + − αθ] = E[0 + ry rz

(5.10-3)

where E = E(x1 , y1 , z1 ) and α = α(x1 , y1 , z1 ). Since the beam is at static equilibrium A

σxx dA = 0

A

σxx ydA = 0

A

σxx zdA = 0

(5.10-4)

Substituting Eq. (5.10-3) in the equilibrium conditions (5.10-4), gives

1 1 EydA + EzdA = EαθdA ry A rz A A A 1 1 0 EydA + Ey 2 dA + EyzdA = EαθydA ry A rz A A A 1 1 0 EzdA + EyzdA + Ez 2 dA = EαθzdA (5.10-5) r r A A y A z A

0

EdA +

We now seek for a system of coordinates in which the thermal stresses and the deflection appear in a simple form. To obtain such a coordinate system,


237

consider the coordinate system (x, y, z) where the following conditions are satisfied EydA = EzdA = 0 (5.10-6) A

A

Solving the system of Eqs. (5.10-5) with respect to 0 , ry , and rz 0 =

PT , A∗ E0

ry =

∗2 ) E0 (Iy∗ Iz∗ − Iyz , ∗ ∗ MTz Iy − MTy Iyz

rz =

∗2 ) E0 (Iy∗ Iz∗ − Iyz ∗ ∗ MTy Iz − MTz Iyz

(5.10-7) we obtain

E dA A E0 E 2 Iy∗ = z dA, A E0 A∗ =

PT =

EαθdA, A

Iz∗ =

E 2 y dA, A E0

MTy =

EαθzdA, A

∗ Iyz =

A

MTz =

E yzdA E0 EαθydA A

(5.10-8) Here, E0 is a reference Young’s modulus, and it may be taken equal to Young’s modulus of one of the constituent materials of the functionally graded beam. Substituting Eqs. (5.10-7) into Eq. (5.10-3) results in the axial thermal stress σxx in the beam as σxx =

∗ ∗ MTz Iy∗ − MTy Iyz MTy Iz∗ − MTz Iyz E PT [ ∗ +( )y + ( )z] − Eαθ(5.10-9) ∗2 ∗2 E0 A Iy∗ Iz∗ − Iyz Iy∗ Iz∗ − Iyz

It may be easily checked that Eqs. (5.10-7) to (5.10-9) reduce to the equations for the homogeneous beams, Eqs. (5.2-11) to (5.2-14), when the material properties of the beam are constant. To find the position of origin of the coordinate system (x, y, z) in the cross section of the beam, Eq. (5.10-6) should be solved using the coordinate transformation law (5.10-1), which gives

y ∗ = A

E y1 dA E dA

A

z ∗ = A

E z1 dA (5.10-10) E dA

A

Once the functional relationship of Young’s modulus of the functionally graded beam is known, Eqs. (5.10-10) are solved to give y ∗ and z ∗ . This defines the transformation law between the coordinate systems (x1 , y1 , z1 ) and (x, y, z).

238


When the combined mechanical and thermal loads are considered, the thermal moments and thermal axial force must be replaced by the total moments and the total axial force acting on the beam as M y = M T y + MM y Mz = M T z + M M z P = PT + PM

(5.10-11)

where My and Mz are the total moments due to the thermal and external mechanical loads, MT y and MT z are the thermal moments, and MM y and MM z are the mechanical moments resulting from the action of the external forces and the reaction forces at the boundaries of the beam in y and z directions, respectively. Also, P is the total axial force resulting from the thermal gradient force PT and the mechanical axial force PM due to the external axial loads and reaction forces at the boundaries of the beam. To find the lateral deflection of the functionally graded beam under thermal loads, expressions for the radii of curvature from Eqs. (5.10-7) are substituted in Eqs. (5.3-1). Also, the axial displacement of the beam is obtained by substituting Eqs. (5.10-7) in Eqs. (5.2-6) for the axial strain xx and by the integration with respect to the axial variable, x. Now consider a beam of functionally graded materials (FGM) made of two constituent materials, metal and ceramic. Two types of graded profiles may be assumed across the beam’s cross section for the constituent mixture, namely the power law and the exponential law [10,11]. Using the variation profile, the axial thermal stresses are obtained for each FGM profile. Here, we give an example of a functionally graded beam with the exponential variation profile under thermal stresses. The exponential FGM Consider an FGM beam of the height a and width b with exponentially varying properties across the beam height as P (y) = Pl exp(ξ y1 /a),

0 ≤ y1 ≤ a

(5.10-12)

where Pl is the corresponding property of the lower surface of the beam, and ξ is a constant multiplier. The coordinate system in which the material properties of the beam are defined is such that 0 ≤ y1 ≤ a and −b/2 ≤ z1 ≤ b/2, see Fig. 5.10-2. The material properties vary exponentially from the lower surface of the beam to the upper surface across the beam height. The material properties, such as Young’s modulus E(y1 ), the coefficient of thermal expansion α(y1 ), and the thermal conductivity k(y1 ) are expressed by Eq. (5.10-12) as E(y) = El exp(ξE y1 /a)


239 x x1

y, y1

a o

y*

o1

z z1

b 2

b 2

b

Figure 5.10-2: Exponentially FGM beam.

α(y) = αl exp(ξα y1 /a) k(y) = kl exp(ξk y1 /a)

(5.10-13)

Poisson’s ratio ν is considered to be constant across the thickness. Using Eqs. (5.10-10), the distances of the origin of the coordinate system (x, y, z) from the origin of the material coordinate system (x1 , y1 , z1 ) are obtained to be

y∗ =

a ξE eξE − eξE + 1

z∗ = 0

ξE (−1 + eξE ) (5.10-14)

Substituting Eqs. (5.10-13) and (5.10-14) into Eqs. (5.10-8), and taking E0 = El , gives A∗ =

ab ξE (e − 1) ξE

3 ξ 1 b a −1 + e E ∗ Iy = 12 ξE ab Iz∗ = 3 −2y ∗ ξE a − y ∗ 2 ξE2 − 2a2 − 2eξE a2 ξE + 2eξE a2 ξE

+eξE a2 ξE2 − 2eξE y ∗ aξE2 + 2eξE y ∗ aξE + eξE y ∗2 ξE2

∗ Iyz =0

a

PT = MTz =

(El exp(ξE y1 /a))(αl exp(ξα y1 /a)) θ b dy1

0

a 0

MTy = 0

(El exp(ξE y1 /a))(αl exp(ξα y1 /a)) θ (y1 − y ∗ ) b dy1 (5.10-15)

240


Using Eqs. (5.10-9) and (5.10-15), the axial thermal stress in an exponential FGM beam is obtained. As an example, consider a functionally graded beam under the reference uniform temperature T0 . For the given boundary conditions, the uniform temperature is to be raised to Tf , where the temperature difference is θ = Tf − T0

(5.10-16)

Substituting this temperature difference in Eqs. (5.10-15) for the axial force and bending moment and integrating yields

−1 + eξE +ξα PT = El αl θ a b ξE + ξα El αl θ a b ∗ MTz = y ξα + a + y ∗ ξE + eξE +ξα ξE a + eξE +ξα a ξα (ξE + ξα )2 − aeξE +ξα + eξE +ξα y ∗ ξE + eξE +ξα y ∗ ξα

(5.10-17)

Substituting Eqs. (5.10-15) and (5.10-17) in Eq. (5.10-9), the axial thermal stress in the exponential FGM beam under uniform temperature rise is obtained.

11

Transient Stresses in FGM Beams

Consider a beam of rectangular cross section of the height a and width b made of functionally graded material under a transient temperature distribution. Two coordinate systems are defined in the cross section of the beam. The coordinate system (x1 , y1 , z1 ) in which the material properties of the functionally graded beam are introduced, and the coordinate system (x, y, z) which is used to obtain the equations of thermal stress and deflection of the functionally graded beam. The coordinate transformation law between the coordinate systems (x, y, z) and (x1 , y1 , z1 ) are given by Eq. (5.10-1). The beam is assumed to have an FGM profile described by the exponential functions. The material properties, referred to the (x1 , y1 , z1 ) coordinate system, are assumed in the form E(y1 ) = El exp(ξE y1 /a), 0 ≤ y1 ≤ a α(y1 ) = αl exp(ξα y1 /a), 0 ≤ y1 ≤ a k(y1 ) = kl exp(ξk y1 /a), 0 ≤ y1 ≤ a ρ(y1 ) = ρl exp(ξρ y1 /a), 0 ≤ y1 ≤ a c(y1 ) = cl exp(ξc y1 /a), 0 ≤ y1 ≤ a

(5.11-1)

The beam is initially assumed to be at a reference temperature T0 . Then the upper surface of the beam is raised to the temperature T1 , while the other

11. Transient Stresses in FGM Beams

241

surfaces of the beam are insulated. Assume that the thermal diffusivity of the beam is constant through the cross section of the beam, i.e., ξk = ξρ + ξc . The one-dimensional heat conduction equation for the transient condition is solved to obtain the temperature distribution. The temperature distribution in the cross section of the beam for the given boundary conditions is (see Example 13 of Chapter 4) ∞

T = T1 +

Cn e−[κl λn t+(ξk y1 /2a)] sin [μn (y1 − a)] 2

(5.11-2)

n=1

where (T0 − T1 )

a ξ y /2a e k 1 sin [μn (y1 − a)]

a0

Cn =

0

sin2 [μn (y1 − a)]

μ2n + (ξk /2a)2

λn =

(5.11-3)

In Eq. (5.11-2) κl = kl /ρl cl is the thermal diffusivity of the beam, and μn is the root of the characteristic equation 2 a μn ξk

tan (a μn ) = −

(5.11-4)

The distances between the coordinate systems (x1 , y1 , z1 ) and (x, y, z) are given by Eqs. (5.10-14) and the other terms required to find the transient axial thermal stress are given by Eqs. (5.10-15). The thermal bending moment and the axial thermal force are obtained by substituting the temperature distribution (5.11-2) into Eqs. (5.10-15)

PT = El αl a b +2

(T1 − T0 ) −1 + eξE +ξα

ξE + ξα

2 Cn e−κl λn t 2 μn a cos (μn a) 2 2 2 2 n=0 4(ξE + ξα − ξk ) + 4 ξk (ξE + ξα ) − 3 ξk + 4 μn a

∞

+2 sin (μn a) ξE + 2 sin (μn a) ξα − sin (μn a) ξk + 2 μn ae

η

MTz = a bEl αl

(T1 − T0 ) ∗ ξE +ξα ξE a + eξE +ξα ξα a + y ∗ ξα 2 y ξE + e (ξE + ξα )

−e

ξE +ξα

a−e ∞

ξE +ξα ∗

ξE +ξα ∗

y ξE − e

y ξα + a

Cn e−κl λn t − μn a cos (μn a) y ∗ η 2 − sin (μn a) μ2n a3 2 + μ2 a2 )2 (η n=1 n +y ∗ μ3n a3 cos (μn a) + 2 μn a2 cos (μn a) η + sin (μn a) η y ∗ μ2n a2 2

242

Chapter 5. Thermal Stresses in Beams + sin (μn a) η 3 y ∗ + sin (μn a) aη 2 − aeη μn y ∗ η 2 − a3 eη μn 3 y ∗

+a e μn − 2 a e μn η + a e μn η 4 η

3

2 η

2 η

2

(5.11-5)

where η = ξE + ξα − ξk /2

(5.11-6)

Substituting Eqs. (5.11-5) into Eq. (5.10-9), the transient axial thermal stress in the exponential FGM beam under the assumed transient temperature distribution is obtained.

12

Thermal Stresses in Thin Curved Beams and Rings

Consider a thin circular ring element, symmetrical about the plane of its centroid, as shown in Fig. 5.12-1. It is assumed that the ring is experiencing a temperature rise distribution which varies across the height of its cross section, θ = θ(z). It is further assumed that the end conditions of the ring are defined by the concentrated bending moment M and the force F , where the lateral deflection, tangential displacement, and rotation of the ring are w, v, and γ, respectively. The assumed thermal and mechanical loadings result in the deformations of the ring which take place in the plane of its original curvature. According to Euler-Bernoulli assumption, the strain of an element at a distance z from the centroidal axis due to bending and stretching of the ring is φφ = φ0 −

z γ r

(5.12-1)

g

v M+dM

M w,z

F+dF R

f df

Figure 5.12-1: An element of the ring.

F

13. Deflection of Thin Curved Beams and Rings

243

where φ0 is the strain of the centerline axis, and γ is the relative rotation of the deformed element of the ring. The sign ( ) indicates the derivative with respect to the variable φ. The thrust and the bending moment at the cross section of the ring, using Hooke’s law, are

E(φφ − αθ)dA = EA φ0 − PT

F = A

M= A

EI γ − MT r

E(φφ − αθ)zdA = −

(5.12-2)

where A is the area of the cross section of the ring, and I is the moment of inertia of the cross section. The thermal stress resultants are defined as

PT =

EαθdA

MT =

A

EαθzdA

(5.12-3)

A

The solution of Eqs. (5.12-2) for φ0 and γ results in F + PT AE (M + MT )r γ = EI

φ0 =

(5.12-4)

Substituting from Eqs. (5.12-4) into Eq. (5.12-1) and using Hooke’s law results in the expression for the thermal stress in thin curved beams as σφφ = −Eαθ +

13

F + PT (M + MT ) + z A I

(5.12-5)

Deflection of Thin Curved Beams and Rings

Consider a thin circular ring element, symmetrical about the plane of its centroid, as shown in Fig. 5.13-1. It is assumed that the ring is experiencing a temperature rise distribution which varies across the height of its cross section, θ = θ(z). The ring carries the distributed load p, q, and m in the directions where w, v, and γ are defined. It is further assumed that the end conditions of the ring are defined by the concentrated bending moment M and the forces F and N acting in the directions where γ, v, and w are defined, respectively. The assumed thermal and mechanical loadings result in the deformations of the ring which takes place in the plane of its original curvature. According to Euler-Bernoulli assumption, the strain of an element at a distance z from the centroidal axis due to bending and stretching of the ring is φφ = φ0 −

z γ r

(5.13-1)

244

Chapter 5. Thermal Stresses in Beams m(f)

N+dN

p(f) q(f) g

v

N M

M+dM w,z

F+dF

F

R f df

Figure 5.13-1: A curved beam. where φ0 is the strain of the centerline axis, and γ is the relative rotation of the deformed element of the ring. The sign ( ) indicates the derivative with respect to the variable φ. The strain of the centerline axis and the total rotation of the deformed element are v − w w + v φ0 = χ= (5.13-2) r r where the total rotation is the sum of the relative rotation γ and the angular deformation due to the shear at the natural axis, ψ, of the cross section of the ring w + v χ= =γ+ψ (5.13-3) r The equations of motion of the thin ring may be obtained by minimizing its associated functional. According to Hamilton’s principle, the ring under the described loadings should satisfy the following variational rule [12] δ

t1 φ2 ρAr

(

t0

φ1

{

2

(w˙ 2 + v˙ 2 +

B1 I γ) ˙ −[ (v − w)2 A 2

B2 r2 2 B3 + (γ ) + (w + v − rγ)2 − PT (v − w) + MT γ ] 2 2 +(pw + qv + mγ)r}dφ − [M γ − F v − N w]φφ21 )dt = 0

(5.13-4)

where the first term is the kinetic energy, the second term is the strain energy stored in the ring, the third term is the potential energy of external distributed forces, and the fourth term is the potential energy of the end forces acting on the ring, where M , F , and N are the given concentrated bending moment and the forces acting on the ends, respectively. The constants B1 , B2 , and B3 are


245

defined as

EA EI KAG B3 = B2 = 3 (5.13-5) r r r where A is the area of the cross section of the ring, I is the moment of inertia of the cross section, E is the modulus of elasticity, G is the shear modulus, and K is the form factor for shear. Using the method of calculus of variation, the stationary value of the functional of Eq. (5.13-4) with respect to w, v, and γ yields the coupled differential equations of motion B1 =

B1 (v − w) + B3 (w + v − rγ ) − ρArw¨ = PT − pr B1 (v − w ) − B3 (w + v − rγ) − ρAr¨ v = PT − qr B2 r2 γ + B3 r(w + v − rγ) − ρIr¨ γ = −MT − mr

(5.13-6)

and the boundary conditions at the end of the ring B3 (w + v − rγ) = N B1 (v − w) − PT = F −B2 r2 γ − MT = M

(5.13-7)

Static solution Considering a ring under static mechanical and thermal loads, the equations of motion (5.13-6) reduce to (B1 + B3 )Dv + (B3 D2 − B1 )w − B3 rDγ = PT − pr (B1 D2 − B3 )v − (B1 + B3 )Dw + B3 rγ = PT − qr B3 rv + B3 rDw + r2 (B2 D2 − B3 )γ = −MT − mr where D = d/dφ. Decoupling Eqs. (5.13-8) for w, v, and γ yields [12] D2 B1 + B2 PT − )(p − ) B3 B1 B2 r B2 (B1 + B3 ) 2 P D[ D − 1](q − T ) B1 B3 r M D(D2 + 1)(m + T ) r

D2 (D4 + 2D2 + 1)w = −rD2 ( r B2 1 − B2 +

B1 + B3 2 1 PT ) D − )(p − B1 B3 B2 r D4 D2 1 P −r( − + )(q − T ) B1 B3 B2 r 1 M − (D2 + 1)(m + T ) B2 r

D2 (D4 + 2D2 + 1)v = −rD(

(5.13-8)

246


1 P P [D(p − T ) − q + T ] B2 r r M 1 (D2 + 1)(m + T ) − rB2 r

D2 (D2 + 1)γ =

(5.13-9)

Subsequently, integrating twice yields D2 B1 + B2 PT − )(p − ) B3 B1 B2 r B2 (B1 + B3 ) P P [ D(q − T ) − (q − T )dφ] B1 B3 r r MT MT [D(m + ) + (m + )dφ] + C5 φ + C6 r r

(D2 + 1)2 w = −r( r B2 1 − B2

+

B1 + B3 PT PT 1 D(p − (p − )− )dφ] B1 B3 r B2 r 1 1 P P 1 −r[( D2 − )(q − T ) + { (q − T )dφ}dφ] B1 B3 r B2 r 1 MT MT − {m + + [ (m + )dφ]dφ} + F5 φ + F6 B2 r r

(D2 + 1)2 v = −r[

1 PT P )dφ − [ (q − T )dφ]dφ} { (p − B2 r r 1 M M − {m + T + [ (m + T )dφ]dφ} + H3 φ + H4 B2 r r r

(D2 + 1)γ =

(5.13-10)

where C5 , C6 , F5 , F6 , H3 , and H4 are constants of integration. A complete solution of Eqs. (5.13-10) is φ ξ

w=

[ 0

0

Φ(η) sin (ξ − η)dη] sin (φ − ξ)dξ

+C1 sin φ + C2 cos φ + C3 φ sin φ + C4 φ cos φ φ ξ

[

v= 0

0

Ω(η) sin (ξ − η)dη] sin (φ − ξ)dξ

+F1 sin φ + F2 cos φ + F3 φ sin φ + F4 φ cos φ φ

γ= 0

Ψ(ξ) sin (φ − ξ)dξ + H1 sin φ + H2 cos φ

(5.13-11)

where the functions Φ(φ), Ω(φ), and Ψ(φ) are the right-hand side functions of Eqs. (5.13-10), and C’s, F ’s, and H’s are the constants of integration. When the lateral loads p, q, and m are uniformly distributed along the ring and are constant and the temperature distribution is a function of the coordinate z, Eqs. (5.13-11) may be solved to give


247

w = C1 sin φ + C2 cos φ + C3 φ sin φ + C4 φ cos φ + C5 φ + C6 PT B1 + B2 PT MT m r +r(p − + − (q + )φ ) − r B1 B2 B2 B2 r B2 r v = F1 sin φ + F2 cos φ + F3 φ sin φ + F4 φ cos φ + F5 φ + F6 rpφ rq rq(φ2 − 4) m(2 − φ2 ) MT φ + + − + − B2 B3 2B2 2B2 B2 r γ = H1 sin φ + H2 cos φ + H3 φ + H4

1 MT φ q 2 mφ2 + pφ − 2 − (φ − 2) − B2 r 2 2r

(5.13-12)

The 16 constants of integration in Eqs. (5.13-12) are not entirely independent. The conditions ensuring that Euler equations (5.13-8) are identically satisfied by Eqs. (5.13-12) yield u, v, and γ in terms of six independent constants of integration, as follows w = C1 sin φ + C2 cos φ + C3 φ sin φ + C4 φ cos φ + C6 rp(B1 + B2 ) PT MT m r + − − (q + )φ − B1 B2 B1 B2 r B2 r v = −C1 cos φ + C2 sin φ + C3 (−φ cos φ +

l sin φ) n

l cos φ) + C6 φ + rH4 n MT φ rpφ rq(φ2 − 4 − 2B2 /B3 ) m(φ2 − 2) − − − + B2 r B2 2B2 2B2 +C4 (φ sin φ +

2B1 B3 C6 φ (C3 sin φ + C4 cos φ) + + H4 rn r 1 MT φ q mφ2 ] + [pφ − 2 − (φ2 − 2) − B2 r 2 2r

γ=

(5.13-13)

where l = B1 B2 + B1 B3 − B2 B3

n = B1 B2 + B1 B3 + B2 B3

The stress resultants defined by Eqs. (5.12-2) are 2B1 B2 B3 (C3 cos φ − C4 sin φ) − rp n 2B1 B2 B3 r M =− (C3 cos φ − C4 sin φ) − B2 rC6 − r2 (p − qφ) + rmφ n (5.13-14)

F =−

248


The shear stress resultant given by Timoshenko [13] is N = KAGψ =

KAG (w + v − rγ) r

(5.13-15)

where K is the form factor for shear, depending on the cross section geometry. Upon substitution of w, v, and γ 2B1 B2 B3 (C3 sin φ + C4 cos φ) + rp n

N=

(5.13-16)

As an example, one may consider a curved cantilever beam subjected to a uniformly distributed temperature change θ(z) =

θo + θi θo − θ i − z 2 a

(5.13-17)

where a is the thickness of the beam, θi is its inner face temperature change, and θo is its outer face temperature change. The thermal force and moment resultants are PT =

1 αEA(θo + θi ) 2

MT = − αEI

(θo − θi ) a

(5.13-18)

Now we will consider the curved beam of angular span ϕ which, in addition to thermal loading, is subjected to a constant lateral load p, directed inward, as shown in Fig. 5.13-2. The boundary conditions at the fixed end along with the natural boundary conditions at the free end are w=v=γ=0 w + v − rγ = 0

at φ = 0 v − w = PT

2

− B2 r γ = MT

Figure 5.13-2: A curved beam.

at

(5.13-19) φ=ϕ

14. Problems

249

After evaluating the six constants of integration, we obtain pr4 [(β3 − 1 − β1 ) sin ϕ sin φ + (β3 + 1 + β1 )φ sin (ϕ − φ) 2EI

PT r MT r2 +2(1 + β1 )(1 − cos φ)] − (1 − cos φ) + EA EI

w =

pr4 ) {(β3 + 1 − β1 ) sin(ϕ − φ) + (β3 + 1 + β1 )φ cos (ϕ − φ) 2EI −(β3 − 1 − β1 ) sin ϕ cos φ + 2[φ − (1 + β1 ) sin φ − sin ϕ]} MT r3 [β1 PT sin φ + (sin ϕ − φ)] + EI r

v =(

γ =

pr3 MT rφ [sin (ϕ − φ) − sin ϕ − φ] − EI EI

(5.13-20)

where β1 = B2 /B1 = I/Ar2 is the dimensionless constant of Winkler theory of curved beams, and β3 = B2 /B3 = EI/KAGr2 is the effect of shear deformation.

14

Problems

1. Consider a rectangular cross section beam (b × a) of length L. The temperature distribution across the beam thickness is given by Eq. (a) of Section 6. Calculate the transverse shear stress distribution in the beam’s cross section if two ends are simply supported. 2. A beam of rectangular cross section (b × a) is exposed to a constant heat flux q on the top surface (y = a/2) and to convective heat transfer at the bottom surface (y = −a/2) by (h, T∞ ). The heat is generated in the beam material at the rate of R per unit volume. The ends of the beam are simply supported. Obtain: (a) Temperature distribution (b) Thermal stresses (c) Thermal deflection (d) The transverse shear stress across the thickness The thermal boundary conditions make the sides of the beam insulated so that the temperature variation is only in y-direction. 3. The beam of Problem 2 is reconsidered. The initial temperature at t = 0 is T0 . At this instant of time, the heat is generated at the rate of R per unit of volume and unit of time. The sides at z = ±b/2 are thermally insulated while the top and bottom surfaces at y = ±a/2 are exposed to ambient at (h, T∞ ). Both ends of the beam are clamped. Find:

250


(a) Temperature distribution (b) Thermal stresses (c) Thermal deflection (d) The transverse shear stress across the thickness 4. Obtain the general expression for the axial thermal stresses of a beam, Eq. (5.2-7), if the cross section is circular. 5. Obtain the general expression for the transverse shear stress, Eq. (5.5-3), for a circular cross section beam. 6. A beam of the height 2a and width b is considered. The initial temperature of the beam is T0 . The beam is suddenly exposed to a rate of heat flux q at the top surface, while the bottom surface is in equilibrium with the ambient through the convective heat transfer at (h, T∞ ). The side surfaces of the beam are thermally insulated. Obtain the expression for thermal stresses. 7. A beam of the height 2a and width b is generating heat through its body at the rate given by R(z, t) = A(t) sin (ηz) where η is constant. The beam is initially at a reference temperature T0 . Thermal boundary conditions are convective heat transfer from the top and bottom surfaces, where the side surfaces are thermally insulated. Find the thermal stresses in the beam.

Bibliography [1] Boley, B.A. and Weiner, J.H., Theory of Thermal Stresses, Wiley, New York, 1960. [2] Timoshenko, S. and Woinowsky-Krieger, S., Theory of Plates and Shells, McGraw-Hill, New York, 1959. [3] Burgreen, D., Elements of Thermal Stress Analysis, C.P. Press, New York, 1971. [4] Noda, N., Hetnarski, R.B., and Tanigawa, Y., Thermal Stresses, 2nd edition, Taylor and Francis, New York, 2003. [5] Singh, J.R., Thomas, J.R., Jr., and Hasselman, D.P.H., Thermal Stresses in a Partially Absorbing Flat Plate Symmetrically Heated by Thermal Radiation and Cooled by Convection, J. Therm. Stresses, Vol. 3, No. 3, pp. 341–349, 1980. [6] Tanigawa, Y., Murakami, H., and Ootao, Y., Transient Thermal Stress Analysis of Laminated Composite Beam, J. Therm. Stresses, Vol. 12, pp. 25–39, 1989. [7] Chen, D., Cheng, S., and Gerhart, T.D., Thermal Stresses in Laminated Beams, J. Therm. Stresses, Vol. 5, pp. 67–74, 1982. [8] Eslami, M.R., Naghdi, A.Y., and Shiari, B., Static Analysis of Thermal Stresses in Beams Based on Layer-Wise Theory, Proc. 1st National Aerospace and Aeronautics Conf., Amirkabir Univ. of Tech., Iran, 1996. [9] Eslami, M.R., Naghdi, A.Y., and Shiari, B., Dynamic Analysis of Thermal Stresses in Beams Based on Layer-Wise Theory, Proc. ISME National Conf., Tabriz University, Iran, 1997. [10] Suresh, S. and Mortensen, A., Fundamentals of Functionally Graded Materials, IOM Communications, New York, 2003. [11] Praveen, G.N. and Reddy, J.N., Nonlinear Transient Thermoelastic Analysis of Functionally Graded Ceramic-Metal Plates, Int. J. Solids Struct., Vol. 35, pp. 4457–4476, 1998. 251

252

Bibliography

[12] Fettahlioglu, O.A. and Steele, T.K., Thermal Deformations and Stresses in Circularly Curved Thin Beams and Rings, J. Therm. Stresses, Vol. 11, No. 3, pp. 233–256, 1988. [13] Timoshenko, S., Strength of Materials, Vol. II, 3rd edition, Van Nostrand, Princeton, New Jersey, 1956.

Chapter 6 Disks, Cylinders, and Spheres Thick cylinders, spheres, and disks are components of many structural systems. Due to their capacity to withstand high pressures, radial loads, and radial temperature gradients, the problem of thermal stress calculations is an important design issue. This chapter presents the method to calculate thermal stresses in such structural members which are made either of homogeneous/isotropic materials or of functionally graded materials. The latter ones, classified as new materials, are mainly designed to withstand high temperatures and high temperature gradients, and they may be designed in such a way that the applied loads, mechanical or thermal, produce a uniform stress distribution across their radial direction. Functionally graded materials exhibit the unique design features, where by selection of proper grading profiles, stress distribution within the element may be optimized.

1

Introduction

Disks, cylinders, and spheres are among the basic elements which are frequently used in engineering design problems and, therefore, their design to withstand the expected loads is important. In this chapter the analysis of these elements when subjected to thermal loads is discussed. Of particular interest is the thermal stress analysis of thick hollow cylinders subjected to a non-axisymmetric temperature distribution. We recall that in Section 7 of Chapter 1 the compatibility conditions for simply connected regions were discussed. Moreover, it was observed that for multiply connected regions, in addition to the compatibility conditions as necessary conditions, Michell conditions had to be checked as sufficiency conditions. We note that very seldom in the classical theory of elasticity Michell conditions are used to check the sufficiency conditions of a multiply connected region. Thick hollow cylinders or spheres under axisymmetric mechanical or thermal loads are solved by means of the compatibility equations without reference to sufficiency R.B. Hetnarski, M.R. Eslami, Thermal Stresses – Advanced Theory and Applications, Solid Mechanics and its Applications 158, c Springer Science + Business Media B.V. 2009

253

254

Chapter 6. Disks, Cylinders, and Spheres

conditions for a multiply connected region. The reason is that Michell conditions are automatically satisfied for symmetric loadings. For non-symmetric loadings Michell conditions are not identically satisfied and they should be checked. The non-axisymmetrically heated thick cylinders are a classic example of the application of Michell conditions. It is noted that of all possible nonaxisymmetric temperature distributions in a thick cylinder, only those distributions that do not make Michell conditions to be identically satisfied, will contribute to thermal stresses. The application of complex variable method in thermoelasticity is also introduced in this chapter. The application is given for the non-axisymmetrically heated cylinder. This should help in giving a better understanding of Michell conditions.

2

Cylinders with Radial Temperature Variation

Consider a solid circular cylinder with axis oz of radius b subjected to the radial temperature variation, θ(r) = T (r) − T0 with T0 being the reference temperature, assuming plane strain condition zz = rz = φz = 0. The stressstrain relations are 1 [σrr − ν(σφφ + σzz )] + αθ E 1 = [σφφ − ν(σrr + σzz )] + αθ E = ν(σrr + σφφ ) − Eαθ

rr = φφ σzz

(6.2-1)

Solving for stresses in terms of strains yields E [(1 − ν)rr + νφφ − (1 + ν)αθ] (1 + ν)(1 − 2ν) E = [(1 − ν)φφ + νrr − (1 + ν)αθ] (1 + ν)(1 − 2ν)

σrr = σφφ

(6.2-2)

The equilibrium equation for axial symmetry is dσrr σrr − σφφ + =0 dr r

(6.2-3)

and the strain-displacement relations, with u being the radial displacement, are du u rr = φφ = (6.2-4) dr r

2. Cylinders with Radial Temperature Variation

255

Substituting Eq. (6.2-4) into Eq. (6.2-2) gives E du u [(1 − ν) + ν − (1 + ν)αθ] (1 + ν)(1 − 2ν) dr r E u du = [(1 − ν) + ν − (1 + ν)αθ] (1 + ν)(1 − 2ν) r dr

σrr = σφφ

(6.2-5)

Using Eqs. (6.2-5) in equilibrium equation (6.2-3) and simplifying, results in the equilibrium equation in terms of displacement u as 1 + ν dθ d 1 d(ur) [ ]= α (6.2-6) dr r dr 1 − ν dr The integration of Eq. (6.2-6) yields C2 1+ν α r θrdr + C1 r + (6.2-7) u= 1−ν r 0 r where C1 and C2 are the constants of integration. Since the displacement must be finite at r = 0, it follows that C2 must be zero. The components of strain from Eq. (6.2-4) are 1+ν α r 1+ν θrdr + C1 + rr = − αθ 2 1−ν r 0 1−ν r 1+ν α θrdr + C1 (6.2-8) φφ = 1 − ν r2 0 and the stresses from Eq. (6.2-2) are αr E E θrdr + C1 σrr = − 1 − ν r2 0 (1 + ν)(1 − 2ν) E Eαθ αr E σφφ = θrdr − + C1 2 1−ν r 0 1−ν (1 + ν)(1 − 2ν) (6.2-9) The constant C1 is found using the boundary condition σrr = 0

at

r=b

(6.2-10)

which yields α(1 + ν)(1 − 2ν) b C1 = θrdr (1 − ν)b2 0

(6.2-11)

Upon substitution into Eqs. (6.2-7) and (6.2-9), we receive 1+ν α r r2 b θrdr] [ θrdr + (1 − 2ν) 2 1−ν r 0 b 0 Eα 1 b 1 r σrr = θrdr − 2 θrdr] [ 2 1−ν b 0 r 0 Eα 1 b 1 r σφφ = θrdr + 2 θrdr − θ] [ 2 1−ν b 0 r 0 u=

(6.2-12)

256


The stress in axial direction, σzz , is obtained from the last of Eqs. (6.2-1) σzz =

Eα 2ν b θrdr − θ] [ 1 − ν b2 0

(6.2-13)

For hollow cylinders with inside radius a and outside radius b the same governing equations hold but the integration for displacement is carried from a, the inside radius, to r. Thus, from Eq. (6.2-7) u=

1+ν α r C2 θrdr + C1 r + 1−ν r a r

(6.2-14)

Substituting u from this equation into Eq. (6.2-4) and then into Eqs. (6.2-2), the radial stress becomes σrr = E[−

r C2 α C1 − θrdr + ] 2 (1 − ν)r a (1 + ν)(1 − 2ν) (1 + ν)r2

(6.2-15)

Applying the boundary conditions σrr = 0 σrr = 0

at at

r=a r=b

(6.2-16)

yields b (1 + ν)(1 − 2ν) α θrdr (1 − ν) (b2 − a2 ) a (1 + ν) αa2 b C2 = θrdr (1 − ν) (b2 − a2 ) a

C1 =

Substituting C1 and C2 into Eq. (6.2-14), the radial displacement and the stresses become r 1 + ν α (1 − 2ν)r2 + a2 b [ θrdr + θrdr] 1−ν r b 2 − a2 a a 1 Eα a2 b 1 r [ σrr = (1 − 2 ) θrdr − 2 θrdr] 1 − ν b 2 − a2 r r a a b Eα a2 1 r 1 (1 + ) θrdr + θrdr − θ] [ 2 σφφ = 1 − ν b − a2 r2 a r2 a

u=

(6.2-17) The axial stress from the last of Eqs. (6.2-1) is σzz =

Eα 2ν b θrdr − θ] [ 2 1 − ν b − a2 a

(6.2-18)

and the axial force Fz for this case, i.e., when zz = 0, is b

Fz =

a

2πrσzz dr

(6.2-19)

2. Cylinders with Radial Temperature Variation

257

If we assume that the inside temperature of the cylinder is Ta and the outside temperature is Tb , the temperature distribution becomes (see Eq. (d) of Example 14, Chapter 4) T =

b Td ) + Tb b ( ln r ln a

(6.2-20)

where Td = Ta − Tb . Substituting the temperature distribution T in Eqs. (6.217), the stresses for a hollow cylinder with fixed ends become EαTd b b2 b a2 [ln (1 − ) ln ] + b 2 2 2 r b −a r a 2(1 − ν) ln a 2 2 EαTd b b b a (1 + 2 ) ln ] − 2 σφφ = b [1 − ln 2 r b −a r a 2(1 − ν) ln a 2 νEαTd 2a b 2 b ln − ln ] − Eα(Tb − T0 ) σzz = b [1 − 2 2 b −a a ν r 2(1 − ν) ln a σrr = −

(6.2-21)

Now, consider the case of the generalized plane strain condition for a hollow thick cylinder. From Eq. (1.12-28), the axial load is zero b

Fz =

a

2πrσrr dr = 0

(6.2-22)

Rewriting Eq. (1.12-32) in the cylindrical coordinates, yields σzz = r(σrr + σφφ ) + Eα(θ¯ − θ)

(6.2-23)

where θ = T − T0 and b b 1 2π 2 θ¯ = θdA = θrdr = θrdr A A π(b2 − a2 ) a b 2 − a2 a

(6.2-24)

Substituting for σrr and σφφ from Eqs. (6.2-17) in Eq. (6.2-23) gives σzz =

Eα ¯ (θ − θ) 1−ν

(6.2-25)

On the other hand, from Eqs. (6.2-17) the sum of the radial and tangential stresses are σrr + σφφ =

b 2 Eα Eα ¯ [ 2 (θ − θ) θrdr − θ] = 2 1−ν b −a a 1−ν

(6.2-26)

Thus, for this special case σrr + σφφ = σzz

(6.2-27)

258


When a thick-walled cylinder is also subjected to an inside and outside pressure, pa and pb , respectively, the total stresses in the cylinder become the sum of the thermal stresses and mechanical stresses. The mechanical stresses are obtained from the same governing equations except that the term T vanishes. Thus, the simultaneous solution of these equations for the radial displacement u is given by Eq. (6.2-7) except the term involving the integral of the temperature vanishes, and the radial displacement becomes C2 (6.2-28) u = C1 r + r Substituting this expression into Eqs. (6.2-4) and then in Eqs. (6.2-2), and using the boundary conditions σrr = −pa σrr = −pb

at at

r=a r=b

(6.2-29)

gives the equations for radial and tangential stresses due to mechanical load only p a a2 b2 p b b2 a2 (1 − ) − (1 − ) b 2 − a2 r2 b 2 − a2 r2 p a a2 b2 p b b2 a2 σφφ = 2 (1 + ) − (1 + ) (6.2-30) b − a2 r2 b 2 − a2 r2 Therefore, when a thick cylinder is subjected to both thermal and mechanical stresses, the resulting stresses are the sum of Eqs. (6.2-21) and (6.2-30). σrr =

3

Thermal Stresses in Disks

Consider a thin circular disk subjected to a radial temperature variation, θ(r) = T (r) − T0 with T0 being the reference temperature. The stress and displacement components are functions of the radius, and assuming the plane stress condition, the equilibrium equation is [1] dσrr σrr − σφφ + =0 (6.3-1) dr r The strain displacement relations are du u rr = φφ = (6.3-2) dr r and the strain-stress relations in polar coordinates for the plane stress condition become 1 (σrr − νσφφ ) + αθ rr = E 1 φφ = (6.3-3) (σφφ − νσrr ) + αθ E

3. Thermal Stresses in Disks

259

Solving Eqs. (6.3-3) for stresses gives E [rr + νφφ − (1 + ν)αθ] 1 − ν2 E = [φφ + νrr − (1 + ν)αθ] 1 − ν2

σrr = σφφ

(6.3-4)

Upon introduction of Eqs. (6.3-2) into Eq. (6.3-4) and then into Eq. (6.3-1) the equilibrium equation in terms of displacements becomes dθ d 1 d(ur) [ ] = (1 + ν)α dr r dr dr

(6.3-5)

Integrating this equation, the radial displacement for a hollow disk of inside radius a and outside radius b becomes C2 α r θrdr + C1 r + u = (1 + ν) r a r

(6.3-6)

and the stresses from Eqs. (6.3-2) and (6.3-4) are Eα r EC1 EC2 σrr = − 2 θrdr + − r 1 − ν (1 + ν)r2 a r Eα EC1 EC2 σφφ = 2 θrdr − Eαθ + + r 1 − ν (1 + ν)r2 a (6.3-7) Applying the boundary conditions σrr = 0 at r = a and r = b, the constants C1 and C2 are found to be (1 − ν)α b θrdr b 2 − a2 a (1 + ν)αa2 b C2 = θrdr b 2 − a2 a

C1 =

Therefore, the radial displacement and the stresses for a hollow disk which is free of traction on the boundaries are b (1 + ν)α r (1 + ν)a2 + (1 − ν)r2 θrdr + θrdr α r (b2 − a2 )r a a a2 b Eα r Eα (1 − σrr = − 2 θrdr + 2 ) θrdr r (b − a2 ) r2 a a Eα r Eα a2 b σφφ = 2 θrdr − Eαθ + 2 ) θrdr (1 + r (b − a2 ) r2 a a

u=

(6.3-8)

260


For a solid disk free of traction on the outer boundary we take a → 0 and receive (1 + ν)α r (1 − ν)αr b u= θrdr + θrdr r b2 0 0 b r Eα b2 θrdr] σrr = 2 [ θrdr − 2 b r 0 0 b r 1 1 σφφ = Eα[ 2 θrdr + 2 θrdr − θ] b r 0 0 (6.3-9) At the center of the disk, for continuous temperature distribution, 1 r θrdr = 0 r→0 r 0 θ(0) 1 r lim 2 θrdr = r→0 r 2 0 lim

which results in zero radial displacement and equal value for σrr and σφφ σrr |r=0 = σφφ |r=0 = Eα [

4

1 b θ(0) θrdr − ] b2 0 2

(6.3-10)

Thick Spheres

The equilibrium equation for a thick spherical vessel of inside radius a and outside radius b subjected to radial temperature change is dσrr 2(σrr − σφφ ) + =0 dr r

(6.4-1)

and the strain-displacement relations are rr =

du dr

φφ =

u r

(6.4-2)

The strain-stress relations for spherical symmetry, i.e., when σθθ = σφφ , are 1 [σrr − 2νσφφ ] + αθ E 1 = [(1 − ν)σφφ − νσrr ] + αθ E

rr = φφ

(6.4-3) Substituting Eqs. (6.4-2) into Eq. (6.4-3) and then into (6.4-1), the equilibrium equation in terms of radial displacement u reduces to d 1 d(ur2 ) 1 + ν dθ [ 2 ]=α dr r dr 1 − ν dr

(6.4-4)

4. Thick Spheres

261

The integration of Eq. (6.4-4) yields u=

1+ν α r 2 C2 θr dr + C1 r + 2 2 1−ν r a r

(6.4-5)

Using Eq. (6.4-5) in Eq. (6.4-2) and then Eq. (6.4-3) and solving for the stresses yields E EC1 2α r 2 2EC2 θr dr + − 3 1−ν r a (1 − 2ν) (1 + ν)r3 r E Eαθ α EC1 EC2 = θr2 dr − + + 3 1−ν r a 1 − ν (1 − 2ν) (1 + ν)r3

σrr = − σφφ

Assuming the inside and outside pressures to be zero, σrr = 0 σrr = 0

at at

r=a r=b

gives the constants C1 and C2 which, upon substitution, yields 1+ν a3 b 2 b3 r 2 α [ θr dr + θr dr 1 − ν b 3 − a3 r 2 r r2 a 2(1 − 2ν) b 2 θr dr] r + 1+ν a b 2E a3 b 2 b3 r 2 α σrr = − [ θr dr + θr dr − θr2 dr] 1 − ν b 3 − a3 r 3 r r3 a a b E a3 b 2 b3 r 2 α σφφ = [ θr dr+ θr dr + 2 θr2 dr − (b3 − a3 )θ] 1 − ν b 3 − a3 r 3 r r3 a a (6.4-6) u=

For constant temperature change θ all the components of stresses vanish, and the radial displacement becomes proportional to the radius of the sphere u = αθr

(6.4-7)

For a solid sphere Eqs. (6.4-6) are used by letting the inside radius approach zero, thus 1+ν 1−ν 2Eα σrr = 1−ν Eα σφφ = 1−ν u=α

1 r 2 2(1 − 2ν) r b 2 θr dr + θr dr] r2 0 1 + ν b3 0 1 r 2 1 b 2 θr dr − 3 θr dr] [ 3 b 0 r 0 2 b 2 1 r 2 θr dr + 3 θr dr − θ] [ 3 r 0 b 0 [

(6.4-8)

262


At r = 0 the value of the displacement and the stresses are indeterminate, but considering that the following limits hold 1 r 2 lim θr dr = 0 r→0 r 2 0 r θ(0) 1 lim θr2 dr = r→0 r 3 3 0 then at r = 0 the radial displacement is zero and the stresses are σrr = σφφ =

2Eα 1 b 2 θ(0) [ 3 ] θr dr − 1−ν b 0 3

(6.4-9)

(6.4-10)

For an infinite body subjected to a radial temperature change, the results of the sphere can be applied by letting b → ∞ and a → 0. We then receive 1+ν α r 2 u= θr dr 1 − ν r2 0 2Eα 1 r 2 θr dr σrr = − 1 − ν r3 0 Eα 1 r 2 [ θr dr − r3 θ] σφφ = 1 − ν r3 0

(6.4-11)

When a thick sphere is subjected to an inside and outside pressure pa and pb , respectively, in addition to thermal loads, then the resulting stresses are the sum of mechanical and thermal stresses. The mechanical stresses are obtained in the same manner as the thermal stresses and thus, from Eq. (6.4-5), the radial displacement of such a sphere, subjected to an inside and outside pressure only, is u = C1 r +

C2 r2

Applying the boundary conditions σrr = −pa σrr = −pb

at at

r=a r=b

and using Eqs. (6.4-2) and (6.4-3) gives the following equations for the radial and tangential stresses for a thick sphere loaded by an inside and outside pressures, pa and pb , respectively, 1 b3 a3 3 3 [p a (1 − ) − p b (1 − )] a b b 3 − a3 r3 r3 1 b3 a3 3 3 = 3 [p a (1 + ) − p b (1 + )] a b b − a3 r3 r3

σrr = σφφ

(6.4-12)

Therefore, when a thick sphere is subjected to the inside and outside pressures in addition to the thermal gradient, the total resulting stresses are given by the sum of Eqs. (6.4-6) and (6.4-12).

5. Thermal Stresses in a Rotating Disk

5

263

Thermal Stresses in a Rotating Disk

Consider a thin circular disk of constant thickness rotating at an angular velocity ω. The temperature distribution is assumed to be axisymmetric and a function of radius r. The body force per unit volume due to centrifugal force is ρrω 2 , where ρ is the mass density of the disk material. Since the forces are functions of the radius and are axisymmetric, the shear stress is zero and the radial and tangential stresses are functions of the radius. Since the disk is thin, the plane stress condition is assumed. The equilibrium equation for axisymmetric stresses in a rotating disk is dσrr σrr − σφφ + + ρrω 2 = 0 dr r

(6.5-1)

The strain-displacement relations and the stress-strain relations for the plane stress condition are du u φφ = dr r du 1 = (σrr − νσφφ ) + αθ rr = dr E u 1 φφ = = (σφφ − νσrr ) + αθ r E rr =

(6.5-2)

(6.5-3)

We introduce the stress function Φ by the relations σrr =

Φ r

σφφ =

dΦ + ρr2 ω 2 dr

(6.5-4)

The equilibrium equation (6.5-1) is satisfied. Eliminating u from Eqs. (6.5-2) gives dφφ (6.5-5) + φφ − rr = 0 r dr The compatibility equation (6.5-5) in terms of the stress function, using Eqs. (6.5-3) and (6.5-2), becomes dθ Φ d2 Φ 1 dΦ + − 2 + (3 + ν)ρrω 2 = −Eα 2 dr r dr r dr

(6.5-6)

This differential equation is rewritten as dθ d 1 d [ (rΦ)] = −(3 + ν)ρrω 2 − Eα dr r dr dr

(6.5-7)

Integrating with respect to r gives Φ=−

r C2 3 + ν 3 2 Eα r θrdr + C1 + ρr ω − 8 r 2 r a

(6.5-8)

264


where a is the inside radius of the disk. For a solid disk a = 0. Here C1 and C2 are the constants of integration. The corresponding stresses from Eq. (6.5-4) are Φ C1 C2 3 + ν 2 2 Eα r θrdr + =− ρr ω − 2 + 2 r 8 r 2 r a dΦ = + ρr2 ω 2 dr 1 + 3ν 2 2 1 r C1 C2 =− θrdr] + ρr ω + Eα[−θ + 2 − 2 8 r a 2 r

σrr = σφφ

(6.5-9) For a solid disk (a = 0) with no external force at the outside radius b the boundary condition is σrr = 0 at r = b. Since the disk is solid, then C2 = 0, otherwise at r = 0 the stress becomes infinite. The constant C1 is 3 + ν 2 2 2Eα b θrdr ρb ω + 2 C1 = 4 b 0

(6.5-10)

Thus the stresses become 3+ν 1 b 1 r θrdr − 2 θrdr) ρω 2 (b2 − r2 ) + Eα ( 2 8 b 0 r 0 1 σφφ = ρω 2 [(3 + ν)b2 − (1 + 3ν)r2 ] 8 1 b 1 r +Eα [−θ + 2 θrdr + 2 θrdr] (6.5-11) b 0 r 0

σrr =

For a hollow disk of the inside radius a and outside radius b, assuming traction-free boundary conditions σrr = 0 at r = a and r = b, we receive C1 Eα b 3+ν θrdr = ρω 2 (b2 + a2 ) + 2 2 8 b − a2 a 3+ν Eαa2 b 2 2 2 C2 = − θrdr ρω a b − 2 8 b − a2 a

(6.5-12)

Substituting the constants C1 and C2 into Eq. (6.5-9), the stresses become 3+ν a2 b 2 1 r 2 2 2 2 θrdr σrr = ρω (b + a − 2 − r ) + Eα [− 2 8 r r a b b 1 a2 + 2 θrdr − θrdr] b − a2 a r2 (b2 − a2 ) a 3+ν a2 b2 1 + 3ν 2 σφφ = ρω 2 (b2 + a2 + 2 − r ) 8 r 3+ν b 1 r 1 b a2 +Eα[−θ + 2 θrdr + 2 θrdr − θrdr] r a b − a2 a r2 (b2 − a2 ) a (6.5-13)

6. Non-axisymmetrically Heated Cylinders

265

When thermal stresses are zero, the maximum stress due to centrifugal forces occurs at the inner radius and it is σφφ =

3+ν 2 2 1 − ν a2 ) ρb ω (1 + 4 3 + ν b2

(6.5-14)

For a solid disk and in the absence of thermal stresses the maximum stress occurs at the center of the disk and it is σrr = σφφ =

6

3+ν 2 2 ρb ω 8

(6.5-15)

Non-axisymmetrically Heated Cylinders

Consider a thick cylinder subjected to steady non-axisymmetric temperature distribution T = T (r, φ). For a cylinder with inside and outside radii a and b, respectively, the general temperature distribution is assumed in the form T (r, φ) = A0 + B0 ln

∞ ∞ r + Fn (r) cos nφ + Gn (r) sin nφ a n=0 n=1

(6.6-1)

where the arbitrary constants A0 and B0 and the functions Fn (r) and Gn (r) may be determined for various boundary conditions, as discussed in problems of Chapter 4. The first term on the right-hand side of Eq. (6.6-1), A0 , produces a uniform axial stress for a thick cylinder in the plane strain condition. The second term, being a function of the radius, results in an axisymmetric stress in the cylinder. The terms under the summation signs produce non-axisymmetric thermal stresses. The total stresses in the cylinder can be obtained by superposing the results of the axisymmetric and non-axisymmetric solutions. An axisymmetric solution has been obtained by Eq. (6.2-14). In the following we calculate thermal stresses corresponding to the asymmetric temperature T (r, φ) =

∞

Fn (r) cos nφ +

n=0

∞

Gn (r) sin nφ

(6.6-2)

n=1

To this end we use Michell conditions in cylindrical coordinates [2–6] 2π

2π ∂∇2 Φ x ∂∇2 Φ (y − )rdφ = −E1 α1 ∂r r ∂φ 0 0 2π 2π ∂∇2 Φ y ∂∇2 Φ (x (x + )rdφ = −E1 α1 ∂r r ∂φ 0 0

(y

2π ∂∇2 Φ 0

∂r

rdφ = −E1 α1

2π ∂T 0

∂r

∂T x ∂T − )rdφ ∂r r ∂φ ∂T y ∂T + )rdφ ∂r r ∂φ (6.6-3)

rdφ

at r = a

(6.6-4)

266


where x = r cos φ and y = r sin φ, and Φ is Airy stress function satisfying the following differential equation ∇4 Φ + E1 α1 ∇2 T = 0

(6.6-5)

For the plane strain condition α1 = α(1 + ν) and E1 = E/(1 − ν 2 ). The boundary conditions for the stress function are (see Eqs. (1.12-42) to (1.1244)) ∂Φ =0 r=b ∂r r=a Φ = a1 x + a 2 y + b 0 ∂Φ r=a = a1 cos φ + a2 sin φ ∂r

(6.6-6)

Φ=

(6.6-7)

where a1 , a2 , and b0 are arbitrary constants. If the temperature distribution of Eq. (6.6-2) is to be plane harmonic, i.e., a solution of ∇2 T = 0, the functions Fn (r) and Gn (r) must satisfy the relations 1 r 1 r

d dFn n2 (r ) − 2 Fn = 0 dr dr r d dGn n2 (r ) − 2 Gn = 0 dr dr r

n = 0, 1, 2, ... n = 1, 2, ...

(6.6-8)

The stress function related to a temperature distribution of Eq. (6.6-2) may be assumed in the form Φ(r, φ) =

∞ n=0

fn (r) cos nφ +

∞

gn (r) sin nφ

(6.6-9)

n=1

Substituting Eqs. (6.6-2) and (6.6-9) into Eq. (6.6-5), considering Eqs. (6.6-8), and equating the coefficients of cos nφ, yields d4 fn 2 d3 fn 1 + 2n2 d2 fn + − dr4 r dr3 r2 dr2 2 2 2 1 + 2n dfn n (n − 4) + fn = 0 + r3 dr r4

n = 0, 1, 2, ...

(6.6-10)

Also, substituting Eqs. (6.6-2) and (6.6-9) into the boundary conditions (6.6-6) and (6.6-7) and equating the coefficients of cos nφ, we obtain dfn (b) =0 n = 0, 1, 2, ... dr f0 (a) = b0 f1 (a) = a1 a fn (a) = 0 df0 (a) df1 (a) dfn (a) =0 = a1 =0 dr dr dr fn (b) =

for n ≥ 2 for n ≥ 2

(6.6-11)


267

Finally, substituting Eqs. (6.6-2) and (6.6-9) into Michell conditions (6.6-3) and (6.6-4) and equating the coefficients of cos nφ, we get 1 1 d f1 1 d df1 d ( − ) [− 2 + (r )] = −E1 α1 ( − )F1 (r) r dr r r dr dr r dr d 1 d dF0 df0 [ (r )] = −E1 α1 dr r dr dr dr

(6.6-12)

Similarly, for gn we obtain d4 gn 2 d3 gn 1 + 2n2 d2 gn + − dr4 r dr3 r2 dr2 2 2 2 1 + 2n dgn n (n − 4) + gn = 0 + r3 dr r4

n = 1, 2, ...

(6.6-13)

The boundary conditions for gn are dgn (b) =0 n = 1, 2, ... dr g1 (a) = a2 a gn (a) = 0 for n ≥ 2 dg1 (a) dgn (a) = a2 = 0 for n ≥ 2 dr dr gn (b) =

(6.6-14)

Michell conditions take the form 1 1 d g1 1 d dg1 d ( − ) [− 2 + (r )] = −E1 α1 ( − )G1 (r) r dr r r dr dr r dr

(6.6-15)

In the derivation of Michell conditions (6.6-12) and (6.6-15) the orthogonality condition of the trigonometric functions was used, i.e., 2π

sin nφ sin mφdφ = 0

2π

cos nφ cos mφdφ = 0

0 π

n = m (6.6-16) n = m = 0

Using the boundary conditions (6.6-11) and (6.6-14), the solution of Eqs. (6.610) and (6.6-13) becomes fn = gn = 0

for

n≥2

(6.6-17)

It is thus verified that Michell conditions are automatically satisfied for the temperature terms of n ≥ 2. That is, among all infinite number of terms of the temperature distribution of Eq. (6.6-2), all terms automatically satisfy Michell condition and thus they do not contribute to the thermal stresses, except for the terms associated with n = 0 and n = 1. This conclusion is important, since if Michell conditions were disregarded, all terms for n ≥ 2 of the series (6.6-2) could have been erroneously considered to contribute to thermal stresses, and thermal stresses would have been calculated with many terms which were not

268


actually existing. Since, however, for the plane strain condition zz = 0, we have σzz = ν(σrr + σφφ ) − Eα(T − T0 ), all terms of the temperature distribution (6.6-2) will appear in the expression for axial stress. This case is a classic example of the necessity and sufficiency of the compatibility and Michell conditions for uniqueness and single-valuedness of displacement field in a multiply-connected region discussed in Sections 8 and 13 of Chapter 1. Among all terms in Eq. (6.6-2), only terms related to n = 0 and n = 1 contribute to thermal stresses σrr , σφφ , and σrφ . In the following we let T (r, φ) = F0 (r) + F1 (r) cos φ + G1 (r) sin φ

(6.6-18) (0)

(0) and the thermal stresses related to F0 (r), are denoted by σrr , and σφφ , and (1) (1) (1) those related to F1 (r) cos φ + G1 (r) sin φ, are denoted by σrr , σφφ , and σrφ . The radial temperature distribution

T 0 (r) = F0 (r)

(6.6-19)

is associated with a stress function f0 (r) obtained from the solution of Eq. (6.6-10) for n = 0 as f0 (r) = C1 + C2 ln

r r r r + C3 ( )2 + C4 ( )2 ln a a a a

(6.6-20)

The corresponding stresses are 1 df0 r C2 2C3 C4 = 2 + 2 + 2 (1 + 2 ln ) r dr r a a a d2 f0 C2 2C3 C4 r = = − 2 + 2 + 2 (3 + 2 ln ) 2 dr r a a a

(0) = σrr (0)

σφφ

(6.6-21)

The constants C2 to C4 and b0 should be obtained using Eqs. (6.6-11) and the second of Eqs. (6.6-12). Taking the temperature distribution as T 0 (r) = F0 (r) = Td ln

r + Ta a

(6.6-22)

and substituting F0 (r) in the second of Eqs. (6.6-12), with the use of boundary (0) (0) (a) = σrr (b) = 0, results in the following values for the constants conditions σrr α1 E1 Td a2 b2 b ln 2 2 2(b − a ) a 2 α1 E1 Td a b C3 = [(1 + 2 ln )b2 − a2 ] 2 2 8(b − a ) a 2 α1 E1 Td a C4 = − 4

C2 = −

(6.6-23)


269

The constants C2 to C4 are substituted into Eqs. (6.6-21), and the final expression for radial and tangential stresses related to the axisymmetric temperature change become EαTd a2 b r b2 (1 − ) ln ] [− ln + 2 2 2 2(1 − ν) a b −a r a 2 2 EαTd a b r b = (1 + 2 ) ln ] [−1 − ln + 2 2 2(1 − ν) a b −a r a

(0) = σrr (0)

σφφ

(6.6-24)

To obtain the non-axisymmetric stresses, the temperature distribution T (1) (r, φ) must be considered in the form T (1) (r, φ) = F1 (r) cos φ + G1 (r) sin φ

(6.6-25)

The functions F1 (r) and G1 (r) are the solutions of Eqs. (6.6-8) for n = 1, and they are A0 + A1 r r B0 G1 (r) = + B1 r r

F1 (r) =

(6.6-26)

Thus

A0 B0 + A1 r) cos φ + ( + B1 r) sin φ (6.6-27) r r Since x = r cos φ, y = r sin φ, the terms related to A1 and B1 in the expression of temperature do not produce any thermal stress due to linear variation of temperature in x and y directions. Thus the remaining term of the temperature change which produces the non-axisymmetric thermal stress is T (1) (r, φ) = (

A0 B0 cos φ + sin φ (6.6-28) r r For n = 1, the stress functions f1 (r) and g1 (r) are the solutions of Eq. (6.6-10) and (6.6-13), and are identical and equal to r a r r r (6.6-29) f1 (r) = g1 (r) = C1 + C2 + C3 ( )3 + C4 ln a r a a a The stresses are related to the stress function as 1 ∂Φ 1 df1 f1 1 ∂2Φ (1) σrr = =( + 2 − 2 ) cos φ 2 r ∂r r ∂φ r dr r 2a 2r 1 = [− 3 C2 + 3 C3 + C4 ] cos φ r a ar ∂2Φ d2 f1 2a 6r 1 (1) = cos φ = [ 3 C2 + 3 C3 + C4 ] cos φ σφφ = 2 2 ∂r dr r a ar ∂ 1 ∂Φ d f1 (1) σrφ = − ( )= ( ) sin φ ∂r r ∂φ dr r 2a 2r 1 = [− 3 C2 + 3 C3 + (6.6-30) C4 ] sin φ r a ar T (1) (r, φ) =

270


The constants C1 to C4 and a1 are obtained from Eqs. (6.6-11) and the first of Eqs. (6.6-12). For the A0 term we receive E1 α1 A0 ab2 4(a2 + b2 ) E1 α1 A0 a3 C3 = 4(a2 + b2 ) E1 α1 A0 a C4 = − 2

C2 = −

(6.6-31)

Identical values are obtained for the constants of B0 term, which upon substitution in Eqs. (6.6-30) gives Eαr b2 a2 ) (1 − ) (A0 cos φ + B0 sin φ) (1 − 2(1 − ν)(a2 + b2 ) r2 r2 Eαr a 2 + b 2 a2 b 2 = − 4 ) (A0 cos φ + B0 sin φ) (3 − 2(1 − ν)(a2 + b2 ) r2 r 2 2 Eαr b a = (1 − 2 ) (1 − 2 ) (A0 sin φ − B0 cos φ) 2 2 2(1 − ν)(a + b ) r r (6.6-32)

(1) = σrr (1)

σφφ

(1)

σrφ

The complete solution for stresses due to the temperature distribution given by Eq. (6.6-18) is obtained by the superposition of the axisymmetric and nonaxisymmetric stresses. Adding Eqs. (6.6-24) to (6.6-32) yields EαTd a2 b r b2 (1 − ) ln ] [− ln + 2 2 2 2(1 − ν) a b −a r a 2 2 a b Eαr (1 − 2 ) (1 − 2 ) (A0 cos φ + B0 sin φ) + 2(1 − ν)(a2 + b2 ) r r EαTd a2 b r b2 (0) (1) = σφφ + σφφ = (1 + ) ln ] [−1 − ln + 2 2 2 2(1 − ν) a b −a r a 2 2 2 2 ab a +b Eαr − 4 ) (A0 cos φ + B0 sin φ) (3 − + 2 2 2 2(1 − ν)(a + b ) r r 2 Eαr b2 a (1) = σrφ = ) (1 − ) (A0 sin φ − B0 cos φ) (1 − 2(1 − ν)(a2 + b2 ) r2 r2 (6.6-33)

(0) (1) + σrr = σrr = σrr

σφφ

σrφ

For axial stress σzz we obtain σzz = ν(σrr + σφφ ) − Eα(T − T0 )

(6.6-34)

For the case of a generalized plane strain condition see pages 390–391 of [1].

7. Method of Complex Variables

7

271

Method of Complex Variables

Another technique to obtain the non-axisymmetric thermal stresses in thick cylinders is the method of complex variables. The complex variable method is extensively applied in plane problems of elasticity, such as in the treatments presented by Muskhelishvili [7] and Wang [8]. The method has also considerable advantages in the analysis of two-dimensional thermoelasticity as discussed by Nowacki [4] and Gatewood [2]. The treatment presented here follows that of Gatewood [2]. On the complex plane we define the variable z by two real variables x and y so that z = x + iy (6.7-1) √ where i = −1 is called the imaginary number. A function of the complex variable z has a real part ξ(x, y) and an imaginary part η(x, y) f (z) = f (x + iy) = ξ(x, y) + iη(x, y)

(6.7-2)

where ξ and η are functions of x and y. The function f (z) is called analytical or regular in a region R if it possesses a unique derivative at every point of R. Points at which the function f (z) ceases to have a derivative are called singular points of the analytical function. It is proved in the theory of analytic functions that the necessary and sufficient conditions that a function f (z) is analytic are ∂η ∂ξ ∂η ∂ξ = =− (6.7-3) ∂x ∂y ∂y ∂x These equations are called Cauchy-Riemann conditions. It is proved that if f (z) is analytic in region R, then the first order derivatives as well as the higher order derivatives exist in R. Once a function is analytic in R the following theorem holds: Cauchy-Goursat theorem. If a function f (z) is single-valued and analytic inside and on a closed curve C, then

f (z)dz = 0

(6.7-4)

C

This is the necessary and sufficient condition for the single-valuedness of the analytic function f (z) to be continuous in a simply connected region. For a multiply connected region with outer boundary curve C and inner boundary curves C1 , C2 , ..., Cn , the necessary and sufficient condition for an analytic function f (z) to be continuous in region R is

f (z)dz = C

f (z)dz + C1

C2

f (z)dz + · · ·

f (z)dz Cn

(6.7-5)

272


That is, when the integration over all the contours is performed in the same direction, the integral f (z) over the exterior curve C is equal to the sum of the integrals over the interior contours. These properties of the complex variables are sufficient for the description of the non-axisymmetric thermal stresses in a thick circular cylinder of the inside radius a and the outside radius b. To define the single-valued condition of the displacement field in the multiply connected region of the thick cylinder, one has to define the continuity condition of displacements. For the plane strain condition, where zz = 0, and for the zero thermal stress condition xx = yy = = (1 + ν)αθ Since

∂v ∂y

(6.7-7)

∂u ∂v = = (1 + ν)αθ ∂x ∂y

(6.7-8)

xx =

∂u ∂x

(6.7-6)

yy =

thus, for zero stress condition

On the other hand, if the shear stress is also zero, the shear strain should be zero and thus xy =

1 ∂v ∂u ( + )=0 2 ∂x ∂y

or

∂v ∂u =− ∂x ∂y

It is thus verified that for this condition the following relations must hold ∂u ∂v = ∂x ∂y

∂v ∂u =− ∂x ∂y

(6.7-9)

These are Cauchy-Riemann conditions of an analytic function on the complex plane for a function with real part u and imaginary part v. This analytic function may also be written in a different form. According to the definition, the rotation in x-y plane is ω=

∂v ∂u 1 ∂v ∂u ( − )= =− 2 ∂x ∂y ∂x ∂y

(6.7-10)

From Eqs. (6.7-6), (6.7-9), and (6.7-10) ∂ω ∂ = ∂x ∂y

∂ ∂ω =− ∂y ∂x

(6.7-11)

Thus Cauchy-Riemann conditions are valid for a function with a real part and imaginary part ω, f (z) = + iω (6.7-12)


273

It should be noted that this function is consistent with the case of zero thermal stresses for the plane strain condition. The exact differentials of the displacement components are ∂u ∂u dx + dy = dx − ωdy ∂x ∂y ∂v ∂v dx + dy = ωdx + dy dv = ∂x ∂y

du =

The change of total displacement from a point 1 in the domain to another point 2 in the domain in the complex plane is 2

(Δu)12 + i(Δv)12 =

2

(du + idv) =

2

( + iω)(dx + idy) =

1

1

f (z)dz 1

(6.7-13) Therefore, the condition for the single-valuedness of displacement field is that the integration of function f (z), which is introduced in the case of zero thermal stresses, be independent of path. This means

f (z)dz = 0

(6.7-14)

The change of rotation from a point 1 to a point 2 is (Δω)12 =

2 ∂ω

(

1

∂x

dx +

∂ω dy) ∂y

(6.7-15)

From Eqs. (6.7-8) and (6.7-10), considering Eqs. (6.7-6) and (6.7-11), we have ∂θ ∂ω = −(1 + ν)α ∂x ∂y ∂ω ∂θ = (1 + ν)α ∂y ∂x

(6.7-16)

Thus 2

(Δω)12 = (1 + ν)α

(− 1

∂θ ∂θ dx + dy) ∂y ∂x

(6.7-17)

From Fig. 6.7-1 the direction cosine of the unit outer normal vector to the boundary has components dy dx = ds dn dx dy = ny = − ds dn nx =

(6.7-18)

274


Figure 6.7-1: Unit normal vector to a two-dimensional boundary. Thus the integration for rotation becomes 2

∂θ ∂x ∂θ ∂y + )ds ∂y ∂s ∂x ∂s 1 2 ∂θ ∂x ∂θ ∂y = (1 + ν)α ( + )ds ∂y ∂n 1 ∂x ∂n

(Δω)12 = (1 + ν)α

(−

or

(Δω)12 = (1 + ν)α

2 ∂θ 1

∂n

ds

(6.7-19)

Thus, the condition for the single-valuedness of rotation for the case of zero thermal stresses reduces to ∂θ ds = 0 (6.7-20) ∂n To obtain an expression for the function f (z), strain and rotation should be substituted into Eq. (6.7-12). From Eq. (6.7-19) we have

ω = ω(x0 ) + (1 + ν)α

∂θ ds ∂n

(6.7-21)

Substituting for and ω from Eqs. (6.7-6) and (6.7-21) into Eq. (6.7-12) yields

f (z) = (1 + ν)α[θ + i

r

∂θ dφ] + i ω(x0 ) ∂r

(6.7-22)

Since the integration of the function f (z) over a close path should be zero and the integration of the constant terms over a close path is equal to zero, therefore, the constants can be omitted from the above expression for f (z). Thus ∂θ dφ] (6.7-23) f (z) = (1 + ν)α[θ + i r ∂r where r and φ are the polar coordinates.


275

Based on this formulation in the complex plane, a thick cylinder of inside radius a and outside radius b under a non-axisymmetric temperature distribution is considered. The temperature change distribution is represented by an infinite series given in the form θ1 (r, φ) =

∞

[An rn + Bn r−n ] cos nφ

(6.7-24)

n=1

where θ1 (r, φ) is the non-axisymmetric temperature distribution in the thick cylinder. To define the terms contributing to the thermal stresses, we should check Cauchy-Riemann conditions. Substituting Eq. (6.7-24) for temperature in Eq. (6.7-23) and integrating gives ∞

f (z) = (1 + ν)α

[(An rn + Bn r−n ) cos nφ + i(An rn − Bn r−n ) sin nφ] (6.7-25)

n=1

√

where i = −1, is the imaginary number. This equation may be further simplified by noting Euler formulas sin nφ =

einφ − e−inφ 2i

cos nφ =

einφ + e−inφ 2

Substituting these relations into Eq. (6.7-25) and simplifying yields f (z) = (1 + ν)α

∞

(An rn einφ + Bn r−n e−inφ )

(6.7-26)

n=1

Making use of the complex variable z, which in polar coordinates may be written as z = reiφ we get dz = ireiφ dφ and the following relations can be easily verified

n

z dz =

0 if n = −1 2πi if n = −1

(6.7-27)

Substituting from Eq. (6.7-26) into (6.7-14) and evaluating the integral, and considering Eq. (6.7-27), indicates that the only term in the temperature equation that produces a dislocation and, therefore, the non-axisymmetric thermal stress is B1 cos φ (6.7-28) θ11 = r The thermal stresses are obtained using the displacement potential ψ and Airy stress function Φ. Substituting for θ11 from Eq. (6.7-28) in the expression for displacement potential yields ∇2 ψ =

∂ 2 ψ 1 ∂ψ 1 + ν B1 1 ∂2ψ + =α + cos φ 2 2 2 ∂r r ∂r r ∂φ 1−ν r

(6.7-29)

276


The solution can be assumed to have the following form ψ = ψ0 (r) cos φ

(6.7-30)

Substituting Eq. (6.7-30) into Eq. (6.7-29) gives d 1 d 1 + ν B1 [ (rψ0 )] = α dr r dr 1−ν r

(6.7-31)

the solution of which is ψ0 (r) = α

1+ν 1 1 B1 r [ ln r − ] 1−ν 2 4

(6.7-32)

Therefore, 1+ν 1 1 B1 r [ ln r − ] cos φ 1−ν 2 4 Airy stress function is taken in the form ψ(r, φ) = α

Φ(r, φ) = (Ar3 +

B ) cos φ r

(6.7-33)

(6.7-34)

where A and B are constants to be determined from the boundary conditions. Having the solution for ψ and Φ, the stresses may be calculated from Eqs. (3.4-23) as 1 ∂ 1 ∂2 )(Φ − 2Gψ) ( + r ∂r r ∂φ2 ∂2 σφφ = 2 (Φ − 2Gψ) ∂r ∂ 1 ∂ σrφ = − [ (Φ − 2Gψ)] ∂r r ∂φ σrr =

(6.7-35)

Substituting Eqs. (6.7-33) and (6.7-34) into Eqs. (6.7-35) and performing the differentiations, yields 2B 1 + ν GαB1 − ] cos φ r3 1−ν r 2B 1 + ν GαB1 ] cos φ σφφ = [6Ar + 3 − r 1−ν r 2B 1 + ν GαB1 ] sin φ σrφ = [2Ar − 3 − r 1−ν r

σrr = [2Ar −

(6.7-36)

The constants A and B may be calculated using the boundary conditions. Considering a thick hollow cylinder of inside radius a and outside radius b, the boundary conditions are σrr = 0 σrr = 0

r=a r=b

(6.7-37)


277

This yields 1 + ν a2 b 2 GαB1 1 − ν a2 + b 2 1+ν 1 2A = GαB1 2 1 − ν a + b2

2B = −

(6.7-38)

where B1 is the constant in the temperature equation (6.7-28). Substituting in Eqs. (6.7-36) and simplifying, yields Eα a2 b2 r (1 − )(1 − )B1 cos φ 2(1 − ν) a2 + b2 r2 r2 Eα r a2 + b 2 a 2 b 2 σφφ = (3 − − 4 )B1 cos φ 2(1 − ν) a2 + b2 r2 r 2 2 Eα a b r σrφ = (1 − 2 )(1 − 2 )B1 sin φ 2(1 − ν) a2 + b2 r r

σrr =

(6.7-39)

The total stresses due to the non-axisymmetric temperature distribution of Eq. (6.7-24) and the radial part A0 +B0 ln r/a from Eq. (6.6-22) can be obtained by superposing the stresses due to the axisymmetric and non-axisymmetric parts of the temperature. Adding Eqs. (6.7-39) to the axisymmetric stresses, Eqs. (6.6-24), yields the expressions for the complete set of stresses in the cylinder subjected to the traction free boundary conditions of Eqs. (6.7-37). The results are EαB0 a2 b b2 r (1 − ) ln − ln ] [ 2 2 2 2(1 − ν) b − a r a a 2 2 a b r Eα (1 − 2 )(1 − 2 )B1 cos φ + 2 2 2(1 − ν) a + b r r b2 r EαB0 a2 b [ 2 = (1 + 2 ) ln − ln − 1] 2 2(1 − ν) b − a r a a 2 2 2 2 Eα a +b ab r + (3 − − 4 )B1 cos φ 2 2 2 2(1 − ν) a + b r r 2 2 r Eα a b = (1 − 2 )(1 − 2 )B1 sin φ 2(1 − ν) a2 + b2 r r

σrr =

σφφ

σrφ

(6.7-40)

The axial stress σzz may be calculated for the plane strain condition for the constrained cylinders, where zz = 0. For free cylinders where zz = const, the generalized plane strain condition, the axial stress is given by Eq. (1.12-35) as σzz = σrr + σφφ

(6.7-41)

Substituting the first and the second of Eqs. (6.7-40) into Eq. (6.7-41), the axial stress for a thick cylinder in general plane strain condition becomes

278

Chapter 6. Disks, Cylinders, and Spheres σzz =

8

EαB0 b b2 r 1 ln − ln − ] [ 2 2 (1 − ν) b − a a a 2 2 EαB1 a + b2 r + (2 − ) cos φ (1 − ν) a2 + b2 r2

(6.7-42)

Functionally Graded Thick Cylinders

Functionally graded materials (FGMs) are new advanced heat resisting materials used in modern technologies. In addition to superb heat properties and their potential to carry ultra high thermal stresses, they are corrosion and erosion resistant and have high fracture stiffness. The basic concept is to mix the ceramic and metal such that the material properties continuously vary from one constituent material to the other. In effect, coefficients of the governing equations for the temperature and stress distributions are coordinate dependent, as the material properties are functions of position. There are some analytical thermal and stress calculations for FGMs in the one-dimensional case for thick cylinders [9]. The authors have considered the nonhomogeneous material properties as linear functions of r. The thermal and stress analysis of these types of structures are sometimes carried out using the theory of laminated composites [10]. The material properties are, however, continuous functions of position and therefore there are some objections to the analogy of FGMs with composites. Obata and Noda [11,12] used the perturbation technique to derive the thermal stress equations of the thick hollow spheres and plates made of functionally graded materials under different temperature distributions. Obata et al. [13] presented the solution for thermal stresses of a thick hollow cylinder under two-dimensional transient temperature distribution made of functionally graded material. In the study of thermal stresses of FGM plates, spheres, cylinders, and semiinfinite bodies in two and three-dimensional cases, the traditional potential function method is used. This method of solution has limitations in choosing the boundary conditions for stresses and displacements. Using the power law distribution for the material constants, Jabbari et al. [14,15] presented an analytical solution for thick functionally graded cylinder under mechanical and thermal stresses. They presented a direct method of solution for Navier equations. In this section, the direct method of solution of Navier equations is presented which does not have the limitations of the potential function method as to handle the general types of mechanical and thermal boundary conditions. Consider a thick hollow cylinder of the inner radius a and outer radius b made of a functionally graded material [14]. The cylinder’s material is graded through the r-direction, thus the material properties are functions of r. Let u and v be the displacement components in the radial and circumferential directions, respectively. Then the strain-displacement relations are

8. Functionally Graded Thick Cylinders

279

rr = u,r v,φ u φφ = + r r 1 u,φ v rφ = ( + v,r − ) (6.8-1) 2 r r where (,) denotes partial derivative. The stress-strain relations for the planestrain condition are σrr = (λ + 2μ)rr + λφφ − (3λ + 2μ)αθ(r, φ) σφφ = (λ + 2μ)φφ + λrr − (3λ + 2μ)αθ(r, φ) σrφ = 2μrφ

(6.8-2)

where σij and ij (i, j = r, φ) are the stress and strain tensors, θ(r, φ) is the temperature change distribution determined from the heat conduction equation, α is the coefficient of linear thermal expansion, and λ and μ are Lamé coefficients. The equilibrium equations in the radial and circumferential directions, disregarding the body forces and the inertia terms, are 1 1 σrr,r + σrφ ,φ + (σrr − σφφ ) = 0 r r 1 2 σrφ ,r + σφφ ,φ + σrφ = 0 (6.8-3) r r To obtain the equilibrium equations in terms of the displacement components for the FGM cylinder, the functional relationship of the material properties must be known. Since the cylinder’s material is assumed to be graded along the r-direction, the modulus of elasticity and the coefficient of thermal expansion are assumed to be described with power laws as r E(r) = E0 ( )m1 l r m2 α(r) = α0 ( ) (6.8-4) l where E0 and α0 are the material constants, m1 and m2 are the power law indices of the material, and l is a characteristic length. We may further assume that Poisson’s ratio is constant. Using the relations (6.8-1) to (6.8-4), Navier equations in term of the displacements are 1 νm1 1 − 2ν 1 1 1 1 − 1) 2 u + ( ) 2 u,φφ + ( ) v,rφ u,rr + (m1 + 1) u,r + ( r 1−ν r 2 − 2ν r 2 − 2ν r (4 + 2m1 )ν − 3 1 (1 + ν)α0 +[ [(m1 + m2 )rm2 −1 θ + rm2 θ,r ] ] 2 v,φ = 2 − 2ν r (1 − ν)lm2 1 1 1 2 − 2ν 1 1 ) 2 v,φφ + ( ) u,rφ v,rr + (m1 + 1) v,r − (m1 + 1) 2 v + ( r r 1 − 2ν r 1 − 2ν r 3 − 4ν 1 2 + 2ν α0 rm2 −1 + m1 ) 2 u,φ = ( )( m2 )θ,φ +( (6.8-5) 1 − 2ν r 1 − 2ν l

280


To obtain the displacement components u and v, the temperature change distribution must be known. Using the thermal boundary conditions of Example 20 of Chapter 4, the temperature change distribution is θ(r, φ) =

+∞

(An1 rβn1 + An2 rβn2 )einφ

(6.8-6)

n=−∞

where βn1 , βn2 and the constants of integration An1 and An2 are determined in Example 20 of Chapter 4. With the given temperature distribution, Navier equations may be solved for the displacement components u(r, φ) and v(r, φ). The displacement components may be expanded into the complex Fourier series as u(r, φ) = v(r, φ) =

∞ n=−∞ ∞

un (r)einφ vn (r)einφ

(6.8-7)

n=−∞

where un (r) and vn (r) are the coefficients of the complex Fourier series of u(r, φ) and v(r, φ), respectively, and are 1 π u(r, φ)e−inφ dφ 2π −π 1 π v(r, φ)e−inφ dφ vn (r) = 2π −π

un (r) =

(6.8-8)

Substituting Eq. (6.8-6) and Eq. (6.7-7) into Eqs. (6.8-5), yields 1 νm1 in 1 (1 − 2ν)n2 1 un + (m1 + 1) un + [ −1− ] 2 un + ( ) v r 1−ν 2 − 2ν r 2 − 2ν r n (4 + 2m1 )ν − 3 1 (1 + ν)α0 +in[ [(m1 + m2 + βn1 )An1 rβn1 +m2 −1 ] 2 vn = 2 − 2ν r (1 − ν)lm2 +(m1 + m2 + βn2 )An2 rβn2 +m2 −1 ] (6.8-9) 1 (2 − 2ν)n2 1 in 1 vn + (m1 + 1) vn − [m1 + 1 + ] 2 vn + ( ) u r 1 − 2ν r 1 − 2ν r n 3 − 4ν 1 in(2 + 2ν) +in( α0 [An1 rβn1 +m2 −1 + An2 rβn2 +m2 −1 ] + m 1 ) 2 un = 1 − 2ν r (1 − 2ν)lm2 (6.8-10) Equations (6.8-9) and (6.8-10) are a system of ordinary differential equations having general and particular solutions. The general solutions are assumed as ugn (r) = Brη vng (r) = Crη

(6.8-11)


281

Substituting Eqs. (6.8-11) into Eqs. (6.8-9) and (6.8-10), yields νm1 (1 − 2ν)n2 −1− ]B 1−ν 2 − 2ν η (4 + 2m1 )ν − 3 +i[ + ]nC = 0 2 − 2ν 2 − 2ν 3 − 4ν η + + m1 ]nB + [η(η − 1) + (m1 + 1)η i[ 1 − 2ν 1 − 2ν (2 − 2ν)n2 ]C = 0 −m1 − 1 − 1 − 2ν

[η(η − 1) + (m1 + 1)η +

(6.8-12)

A nontrivial solution (B, C) of Eqs. (6.8-12) is obtained provided η satisfies the equation (1 − 2ν)n2 νm1 −1− ][η(η − 1) 1−ν 2 − 2ν (2 − 2ν)n2 η +(m1 + 1)η − m1 − 1 − ] + n2 [ 1 − 2ν 2 − 2ν (4 + 2m1 )ν − 3 η 3 − 4ν + (6.8-13) ][ + + m1 ] = 0 2 − 2ν 1 − 2ν 1 − 2ν

[η(η − 1) + (m1 + 1)η +

Equation (6.8-13) has four roots ηn1 to ηn4 . Thus a general solution to Eqs. (6.8-9) and (6.8-10) takes the form ugn (r) =

4

Bnj rηnj

j=1

vng (r) =

4

Nnj Bnj rηnj

(6.8-14)

j=1

where Nnj = Cnj /Bnj is obtained from the first of Eqs. (6.8-12) as Nnj =

i[ηj (ηj − 1) + (m1 + 1)ηj + ηj n[ 2−2ν +

νm1 − 1−ν (4+2m1 )ν−3 ] 2−2ν

1−

(1−2ν)n2 ] 2−2ν

j = 1, 2, 3, 4 (6.8-15)

For isotropic materials (m1 = 0) and for n = 1, Eq. (6.8-13) has multiple roots and hence a solution of the form of ln (r/r0 ), r0 > 0, must be considered for un (r) and vn (r). The particular solutions upn (r) and vnp (r) are assumed as upn (r) = Dn1 rβn1 +m2 +1 + Dn2 rβn2 +m2 +1 vnp (r) = Dn3 rβn1 +m2 +1 + Dn4 rβn2 +m2 +1

(6.8-16)

282


Substituting Eqs. (6.8-16) into Eqs. (6.8-9) and (6.8-10), yields d1 Dn1 rβn1 +m2 −1 + d2 Dn2 rβn2 +m2 −1 + d3 Dn3 rβn1 +m2 −1 +d4 Dn4 rβn2 +m2 −1 = d5 rβn1 +m2 −1 + d6 rβn2 +m2 −1

(6.8-17)

d7 Dn3 rβn1 +m2 −1 + d8 Dn4 rβn2 +m2 −1 + d9 Dn1 rβn1 +m2 −1 +d10 Dn2 rβn2 +m2 −1 = d11 rβn1 +m2 −1 + d12 rβn2 +m2 −1

(6.8-18)

where constants d1 to d12 are d1 = (βn1 + m2 + 1)(βn1 + m2 ) + (m1 + 1)(βn1 + m2 + 1) + −1 −

(1 − 2ν)n2 2 − 2ν

d2 = (βn2 + m2 + 1)(βn2 + m2 ) + (m1 + 1)(βn2 + m2 + 1) +

νm1 1−ν νm1 1−ν

(1 − 2ν)n2 2 − 2ν βn1 + m2 + 1 (4 + 2m1 )ν − 3 in( + ) 2 − 2ν 2 − 2ν βn2 + m2 + 1 (4 + 2m1 )ν − 3 + ) in( 2 − 2ν 2 − 2ν (1 + ν)(m1 + m2 + βn1 )α0 An1 (1 − ν)lm2 (1 + ν)(m1 + m2 + βn2 )α0 An2 (1 − ν)lm2 (βn1 + m2 + 1)(βn1 + m2 ) + (m1 + 1)(βn1 + m2 + 1) − m1 (2 − 2ν)n2 −1 − 1 − 2ν (βn2 + m2 + 1)(βn2 + m2 ) + (m1 + 1)(βn2 + m2 + 1) − m1 (2 − 2ν)n2 −1 − 1 − 2ν βn1 + m2 + 1 3 − 4ν + + m1 ) in( 1 − 2ν 1 − 2ν βn2 + m2 + 1 3 − 4ν in(2 + 2ν)α0 An1 in( + + m1 ) d11 = 1 − 2ν 1 − 2ν (1 − 2ν)lm2 in(2 + 2ν)α0 An2 d12 = (1 − 2ν)lm2 −1 −

d3 = d4 = d5 = d6 = d7 =

d8 =

d9 = d10 =

Equating the coefficients of the identical powers yields d1 Dn1 + d3 Dn3 = d5 d9 Dn1 + d7 Dn3 = d11

(6.8-19)


283

d2 Dn2 + d4 Dn4 = d6 d10 Dn2 + d8 Dn4 = d12

(6.8-20)

Equations (6.8-19) and (6.8-20) are a system of algebraic equations, where the solution is given by Cramer’s method as d5 d7 − d3 d11 d1 d7 − d3 d9 d1 d11 − d5 d9 = d1 d7 − d3 d9

Dn1 = Dn3

d6 d8 − d4 d12 d2 d8 − d4 d10 d2 d12 − d6 d10 = d2 d8 − d4 d10

Dn2 = Dn4

(6.8-21)

provided d1 d7 − d3 d9 = 0 and d2 d8 − d4 d10 = 0. The complete solutions for un (r) and vn (r) are the sum of the general and particular solutions, and are un (r) = ugn (r) + upn (r) vn (r) = vng (r) + vnp (r)

(6.8-22)

Thus un (r) =

4

Bnj rηnj + Dn1 rβn1 +m2 +1 + Dn2 rβn2 +m2 +1

j=1

vn (r) =

4

Nnj Bnj rηj + Dn3 rβn1 +m2 +1 + Dn4 rβn2 +m2 +1

(6.8-23)

j=1

For n = 0 the coefficient Nnj in Eq. (6.8-15) is equal to zero because the system of Eqs. (6.8-9) and (6.8-10) for n = 0 is reduced to the two decoupled ordinary differential equations 1 νm1 (1 + ν)α0 1 [(m1 + m2 + β01 ) u0 + (m1 + 1) u0 + ( − 1) 2 u0 = r 1−ν r (1 − ν)lm2 (6.8-24) × A01 rβ01 +m2 −1 + (m1 + m2 + β02 )A02 rβ02 +m2 −1 ] 1 1 v0 + (m1 + 1) v0 − (m1 + 1) 2 v0 = 0 (6.8-25) r r The solutions of Eqs. (6.8-24) and (6.8-25) are u0 (r) =

2

(B0j rη0j + D0j rβ0j +m2 +1 )

j=1

v0 (r) =

4

B0j rη0j

j=3

where η01,2 =

m2 νm1 −m1 ±( 1 −( − 1))1/2 2 4 1−ν

(6.8-26)

284


η03 = 1 η04 = −(m1 + 1) D0j =

l−m2 (1 + ν)(β0j + m1 + m2 )α0 A0j (1 − ν)[(β0j + m2 + 1)(β0j + m2 ) + (β0j + m2 + 1)(m1 + 1) + j = 1, 2

νm1 1−ν

− 1]

(6.8-27) Substituting Eqs. (6.8-23) and (6.8-26) into Eq. (6.8-7), give u(r, φ) =

2 j=1

4

[

4

Bnj rηnj

n=−∞,n=0 j=1

+Dn1 r v(r, φ) =

∞

(B0j rη0j + D0j rβ01 +m2 +1 ) + βn1 +m2 +1

∞

B0j rη0j +

j=3

+ Dn2 r [

βn2 +m2 +1

4

]einφ

Nnj Bnj rηj + Dn3 rβn1 +m2 +1

n=−∞,n=0 j=1

+Dn4 r

βn2 +m2 +1

]einφ

(6.8-28)

Substituting Eqs. (6.8-28) into Eqs. (6.8-1) and (6.8-2), the strains and stresses are obtained, where the stress distributions are σrr

E0 = (1 + ν)(1 − 2ν)lm1 −

2

[(1 − ν)η0j + ν]B0j rη0j +m1 −1 + [νβ0j + νm2 + 1

j=1

∞ 4 (1 + ν)α0 β0j +m1 +m2 ]D r + { [(1 − ν)ηnj + ν(inNnj + 1)] 0j lm2 n=−∞,n=0 j=1

×Bnj rηnj +m1 −1 + [(1 − ν)(βn1 + m2 + 1)Dn1 + ν(inDn3 + Dn1 ) (1 + ν)α0 An1 ]rβn1 +m1 +m2 + [(1 − ν)(βn2 + m2 + 1)Dn2 + ν(inDn4 + Dn2 ) − lm2 (1 + ν)α0 βn2 +m1 +m2 An2 ]r } einφ − lm2 σφφ

E0 = (1 + ν)(1 − 2ν)lm1

2

[(1 − ν)η0j + ν]B0j rη0j +m1 −1

j=1

+[(1 − ν)β0j + m2 + 1 − +

∞

{

4

(1 + ν)α0 ]D0j rβ0j +m1 +m2 lm2

[νηnj + (1 − ν)(inNnj + 1)]Bnj rηnj +m1 −1

n=−∞,n=0 j=1

(1 + ν)α0 An1 ] lm2 + [ν(βn2 + m2 + 1)Dn2 + (1 − ν)(inDn4 + Dn2 )

+[ν(βn1 + m2 + 1)Dn1 + (1 − ν)(inDn3 + Dn1 ) − ×rβn1 +m1 +m2

(1 + ν)α0 − An2 ]rβn2 +m1 +m2 } einφ lm2


285

σrφ

∞ 4 E0 η04 +m1 −1 = (η − 1)B r + { [in + (ηnj − 1)Nnj ] 04 04 m (1 + ν)l 1 n=−∞,n=0 j=1

×Bnj rηnj +m1 −1 + [inDn1 + (βn1 + m2 )Dn3 ]rβn1 +m1 +m2 + [inDn2

+(βn2 + m2 )Dn4 ]r

βn2 +m1 +m2

} einφ

(6.8-29)

To determine the constants Bnj , we may consider any general from of boundary conditions for displacements or stresses such as u(a, φ) = g1 (φ) u(b, φ) = g2 (φ) v(a, φ) = g3 (φ) v(b, φ) = g4 (φ)

(6.8-30)

σrr (a, φ) = g5 (φ) σrr (b, φ) = g6 (φ) σrφ (a, φ) = g7 (φ) σrφ (b, φ) = g8 (φ)

(6.8-31)

or

It is recalled that Eqs. (6.8-24) to (6.8-31) contain four unknowns Bn1 , Bn2 , Bn3 , and Bn4 . Therefore, four boundary conditions are required to evaluate the four unknowns. These boundary conditions may be selected from the list of conditions given in Eqs. (6.8-30) and (6.8-31). Assume that the four boundary conditions are specified from the list of Eqs. (6.8-30) and (6.8-31). The boundary conditions may be either the given displacements or stresses, or combination of both. We expand the given boundary conditions in the finite complex Fourier series gj (φ) =

∞

Gj (n)einφ

j = 1, ..., 4

(6.8-32)

j = 1, ..., 4

(6.8-33)

n=−∞

where 1 π Gj (n) = gj (φ)e−inφ dφ 2π −π

Using the four boundary conditions and Eqs. (6.8-28) and (6.8-29), the constants of integration Bnj are calculated. As an example, consider a thick hollow cylinder of inner radius a = 1 m and outer radius b = 1.2 m and set l = 1 m. Poisson’s ratio is assumed 0.3 and the modulus of elasticity and the thermal coefficient of expansion at the inner

286


radius are Ei = 200 Gpa and αi = 1.2 × 10−6 /◦ C, respectively. For simplicity of analysis we consider the power law of material properties be the same as m1 = m2 = m3 = m. A thick hollow cylinder with inside boundary traction free and a given temperature distribution T (a, φ) = 60 cos 3φ◦ C is assumed. The outside boundary is assumed to be radially fixed with zero temperature. Therefore, the assumed boundary conditions yield σrr (a, φ) = 0, σrφ (a, φ) = 0, u(b, φ) = 0 and v(b, φ) = 0. To obtain the temperature distribution (6.8-6) the constants of integration An1 and An2 are obtained from the method described in Example 20 of Chapter 4. The displacement and stresses are then obtained from the boundary conditions and Eqs. (6.8-28) and (6.8-29). Figure 6.8-1 shows the temperature distribution in the cross section of the cylinder, along the radius, and in the circumferential direction. Figure 6.8-2

Figure 6.8-1: Temperature distribution in the cross section of the cylinder.

Figure 6.8-2: Radial displacement in the cross section of the cylinder.


287

shows the resulting thermoelastic radial displacement due to the given temperature variations. The resulting circumferential displacement v is shown in Fig. 6.8-3. It is noted that, due to the assumed boundary conditions, the u and v displacements are zero at r = b, and follow the pattern of the temperature distribution at the inside surface at r/a = 1. Figures 6.8-4 to 6.8-6 show the distribution of the radial, circumferential, and shear thermal stresses in the cross section of the cylinder. It is interesting to see that all components of stresses follow a harmonic pattern on the outside surface. The radial and shear stresses are zero at the internal surface, due to the assumed boundary conditions. The effect of the power law index on the distribution of the radial thermal stress is shown in Fig. 6.8-7. This figure is the plot of σrr versus r/a at φ = π/3. It is shown that as m increases, the radial thermal stress increases.

Figure 6.8-3: Circumferential displacement in the cross section of the cylinder.

Figure 6.8-4: Radial thermal stress in the cross section of the cylinder.

288


Figure 6.8-5: Hoop thermal stress in the cross section of the cylinder.

Figure 6.8-6: Shear thermal stress in the cross section of the cylinder.

Figure 6.8-7: Radial distribution of radial thermal stress σrr at φ = π2 .

9. Axisymmetric Stresses in FGM Cylinders

9

289

Axisymmetric Stresses in FGM Cylinders

Axisymmetric thermal stress problems in thick cylinders of isotropic, anisotropic, and composite materials may be found in a number of references [16–19]. Thick cylinders of functionally graded materials (FGMs) under axisymmetric temperature distribution are discussed in references [20–25]. Using the power law functions for the FGM material coefficients and the direct method of solution, Navier equations are solved to obtain the mechanical and thermal stresses in thick cylinders. A solution for the mechanical and thermal stresses in the FGM thick cylinders of finite length under axisymmetric loading is given in [26,27]. More general solution of FGM thick cylinders under threedimensional temperature distribution is found in [28]. The analysis is based on the direct method of solution of Navier equations. The problem of thermal stresses in a functionally graded thick cylinder, when temperature variation is along the radial and axial directions is discussed in this section. Consider a hollow circular cylinder of finite length L, inner radius a, and outer radius b, made of functionally graded material. The cylinder’s material is graded through the r-direction, thus the material properties are functions of r. The axisymmetric cylindrical coordinates (r, z) are considered along the radial and axial directions, respectively. Let u and w be the displacement components in the radial and axial directions, respectively. Then the strain-displacement relations are rr =

∂u , ∂r

φφ =

u , r

zz =

∂w , ∂z

1 ∂u ∂w rz = ( + ) 2 ∂z ∂r (6.9-1)

The stress-strain relations of a functionally graded cylinder for axisymmetric condition are E(r) E(r)α(r) [(1 − ν)rr + ν(φφ + zz )] − θ(r, z) (1 + ν)(1 − 2ν) 1 − 2ν E(r) E(r)α(r) σφφ = [(1 − ν)φφ + ν(rr + zz )] − θ(r, z) (1 + ν)(1 − 2ν) 1 − 2ν E(r) E(r)α(r) σzz = [(1 − ν)zz + ν(rr + φφ )] − θ(r, z) (1 + ν)(1 − 2ν) 1 − 2ν E(r) σrz = (6.9-2) rz 1+ν σrr =

where θ(r, z) is the temperature change distribution determined from the first law of thermodynamics, σij and ij (i, j = r, z) are the stress and strain tensors, respectively, α(r) is the thermal expansion coefficient, E(r) is Young’s modulus, and ν is Poisson’s ratio.

290


The equilibrium equations in the axial and radial directions, disregarding the body forces and inertia terms, are 1 σzz,z + σrz,r + σrz = 0 r σrr − σφφ + σrz,z = 0 σrr,r + r

(6.9-3)

A comma denotes partial differentiation with respect to the space variables. To obtain the equilibrium equations in terms of the displacement components for the functionally graded cylinder, the functional relationship of the material properties must be known. Since the cylinder’s material is assumed to be graded along the r-direction, the modulus of elasticity and the coefficient of thermal expansion are assumed to be described with the power laws as E(r) = Eo ( rl )m1 and α(r) = αo ( rl )m2 , where E0 and α0 are the material constants, m1 and m2 are the power-law indices of the material, and l is a characteristic length. It is assumed that Poisson’s ratio ν is constant across the cylinder thickness. Using the relations (6.9-1) to (6.9-3), Navier equations in terms of the displacement components are 1 1 u,rz + [1 + m1 (1 − 2ν)] u,z + (1 − 2ν)w,rr + (1 − 2ν)(m1 + 1) w,r r r (1 + ν)αo m2 +2(1 − ν)w,zz = 2 r θ,z (6.9-4) lm2 1 1 1 − 2ν 1 u,zz + w,rz (1 − ν)u,rr + (m1 + 1)(1 − ν) u,r + [ν(m1 + 1) − 1] 2 u + r r 2 2 1 (m1 + m2 )(1 + ν)αo m2 −1 (1 + ν)αo m2 +νm1 w,z = r θ+ r θ,r (6.9-5) r lm2 lm2

Navier equations (6.9-4) and (6.9-5) are a non-homogeneous system of partial differential equations with non-constant coefficients. To solve these equations, the axisymmetric temperature distribution in the cylinder must be known. Let us consider Example 20 of Chapter 4 with axisymmetric temperature change distribution in the functionally graded cylinder given by Eq. (s) as θ(r, z) =

∞

r−β [an Iβ (ζn r) + bn I−β (ζn r)] sin(ζn z)

(6.9-6)

n=1

where ζn = nπ/L, β = m3 /2, m3 is the power law index of the conduction coefficient, and an and bn are obtained using the given thermal boundary conditions, as discussed in Example 20 of Chapter 4. Let us consider a thick FGM cylinder with ends free to move in the axial direction. To solve Navier equations for simply supported end conditions, the displacement components u(r, z) and w(r, z) are expanded in the form of Fourier series as u(r, z) =

∞ n=1

un (r) sin(ζn z)

w(r, z) =

∞ n=0

wn (r) cos(ζn z)

(6.9-7)


291

Substituting Eqs. (6.9-6) and (6.9-7) into Eqs. (6.9-4) and (6.9-5), yields 1 1 ζn un + ζn [1 + m1 (1 − 2ν)] un + (1 − 2ν)wn + (1 − 2ν)(m1 + 1) wn r r (1 + ν)α 0 m2 −β −2ζn2 (1 − ν)wn = 2ζn r [an Iβ (ζn r) + bn I−β (ζn r)] (6.9-8) lm2 1 1 2ν − 1 2 (1 − ν)un + (m1 + 1)(1 − ν) un + [ν(m1 + 1) − 1] 2 un + ζn un r r 2 1 1 (1 + ν)α0 m2 −β − ζn wn − νm1 ζn wn = r a n Iβ (ζn r) + bn I−β (ζn r) m 2 r l 2 (1 + ν)(m1 + m2 − β)α0 m2 −β−1 + r [an Iβ (ζn r) + bn I−β (ζn r)] (6.9-9) lm2 Equations (6.9-8) and (6.9-9) are a system of ordinary differential equations with non-constant coefficients, having the general and particular solutions. Since these equations are in the form of a generalized Bessel equation [29], therefore the general solutions are assumed as ugn (r) = rq [Cn1 Ip (ζn r) + Cn2 I−p (ζn r)] wng (r) = rq+1 [Cn3 Ip (ζn r) + Cn4 I−p (ζn r)]

(6.9-10)

Substituting Eqs. (6.9-10) into homogeneous parts of Eqs. (6.9-8) and (6.9-9) and making use of the relationship between the derivatives of the modified Bessel function, the final form for the general solution of the displacement components becomes [27] ugn (r) = rq1 [Cn1 Ip1 (ζn r) + Cn2 I−p1 (ζn r)] e7 e9 wng (r) = −rq1 +1 [ Cn1 Ip1 (ζn r) + Cn2 I−p1 (ζn r)] e8 e10 +rq2 +1 [Cn3 Ip2 (ζn r) + Cn4 I−p2 (ζn r)]

(6.9-11)

where Cn1 to Cn4 are the constants of integration to be obtained using the proper boundary conditions, and the constants e7 to e10 are e7 = ζn [q + 1 + m1 (1 − 2ν) + p] e8 = (1 − 2ν)(q + 1)(q + m1 + 1) + 2(1 − ν)p2 +p[(1 − 2ν)(2q + m1 + 3) + 2ν − p − 1] e9 = ζn [q + 1 + m1 (1 − 2ν) − p] e10 = (1 − 2ν)(q + 1)(q + m1 + 1) +2(1 − ν)p2 − p[(1 − 2ν)(2q + m1 + 3) + 2ν + p − 1] 1 (ν − 1)[4m1 + (ν − 1)(m1 + 2)2 ] p1 = 2(ν − 1) m1 m1 m1 p2 = , q1 = − , q2 = − −1 (6.9-12) 2 2 2

292


The particular solutions upn (r) and wnp (r) are assumed as upn (r) = rm2 −β+1 [Dn1 Iβ (ζn r) + Dn2 I−β (ζn r)] wnp (r) = rm2 −β+2 [Dn3 Iβ (ζn r) + Dn4 I−β (ζn r)]

(6.9-13)

Substituting Eqs. (6.9-13) into Eqs. (6.9-8) and (6.9-9) and making use of the relationship between the derivatives of Bessel function, the constant coefficients Dn1 to Dn4 are obtained and given in [27]. For n = 0 Eq. (6.9-8) becomes an ordinary differential equation w0 + (m1 + 1)w0 /r = 0 with the solution w0 =

2

C0kj rλj λ1 = −m1 λ2 = 0

(6.9-14)

j=1

The complete solution is, thus, the sum of the general and particular solutions for the displacement components u(r, z) and w(r, z) and are u(r, z) =

∞

{rm2 −β+1 [Dn1 Iβ (ζn r) + Dn2 I−β (ζn r)]

n=1

+r− w(r, z) =

∞

m1 2

{rm2 −β+2 [Dn3 Iβ (ζn r) + Dn4 I−β (ζn r)]

n=1 2

+

j=1

+

[Cn1 Ip1 (ζn r) + Cn2 I−p1 (ζn r)]} sin(ζn z)

C0kj rλj − r1−

m1 2

[

e7 Cn1 Ip1 (ζn r) e8

m1 e9 Cn2 I−p1 (ζn r)] + r− 2 [Cn3 Ip2 (ζn r) + Cn4 I−p2 (ζn r)]} cos(ζn z) e10 (6.9-15)

Substituting Eqs. (6.9-15) into Eqs. (6.9-1) and (6.9-2), the stresses are obtained as σrr =

∞ Eo r m 1 {[(m2 − β + 1)(1 − ν)Dn1 + νDn1 − ζn νr2 Dn3 ]rm2 −β Iβ (ζn r) (1 + ν)(1 − 2ν)lm1 n=1

+[(m2 − β + 1)(1 − ν)Dn2 + νDn2 − ζn νr2 Dn4 ]rm2 −β I−β (ζn r) m1 m1 (1 − ν) + ν]r− 2 −1 +(1 − ν)rm2 −β+1 [Dn1 Iβ (ζn r) + Dn2 I−β (ζn r)] + [− 2 m1 ×[Cn1 Ip1 (ζn r) + Cn2 I−p1 (ζn r)] + (1 − ν)r− 2 [Cn1 Ip 1 (ζn r) + Cn2 I−p (ζn r)] 1 m1 m1 e e9 7 Cn2 I−p1 (ζn r)] −ζn νr 2 [Cn3 Ip2 (ζn r) + Cn4 I−p2 (ζn r)] + ζn νr1− 2 [ Cn11 Ip1 (ζn r) + e8 e10 2 αo −ζn C0j rλj − m2 rm2 −β (1 + ν) [an Iβ (ζn r) + bn I−β (ζn r)]} sin(ζn z) (6.9-16) l j=1 σφφ =

∞ Eo r m 1 {[(m2 − β + 1)νDn1 + (1 − ν)Dn1 − ζn νr2 Dn3 ]rm2 −β Iβ (ζn r) m (1 + ν)(1 − 2ν)l 1 n=1


293

+[(m2 − β + 1)νDn2 + (1 − ν)Dn2 − ζn νr2 Dn4 ]rm2 −β I−β (ζn r) m1 m1 ν + (1 − ν)]r− 2 −1 +νrm2 −β+1 [Dn1 Iβ (ζn r) + Dn2 I−β (ζn r)] + [− 2 m1 ×[Cn1 Ip1 (ζn r) + Cn2 I−p1 (ζn r)] + νr− 2 [Cn1 Ip 1 (ζn r) + Cn2 I−p (ζn r)] 1 m1 m1 e e9 7 1− 2 [ Cn1 Ip1 (ζn r) + Cn2 I−p1 (ζn r)] −ζn νr 2 [Cn3 Ip2 (ζn r) + Cn4 I−p2 (ζn r)] + ζn νr e8 e10 2 αo C0j rλj − m2 rm2 −β (1 + ν) [an Iβ (ζn r) + bn I−β (ζn r)] sin(ζn z) (6.9-17) −ζn l j=1 σzz =

∞ Eo r m 1 {[(m2 − β + 2)νDn1 − ζn (1 − ν)r2 Dn3 ]rm2 −β Iβ (ζn r) (1 + ν)(1 − 2ν)lm1 n=1

+[(m2 − β + 2)νDn2 − ζn (1 − ν)r2 Dn4 ]rm2 −β I−β (ζn r) m1 m1 ν + ν]r− 2 −1 +νrm2 −β+1 [Dn1 Iβ (ζn r) + Dn2 I−β (ζn r)] + [− 2 m1 ×[Cn1 Ip1 (ζn r) + Cn2 I−p1 (ζn r)] + νr− 2 [Cn1 Ip 1 (ζn r) + Cn2 I−p (ζn r)] 1 −ζn (1 − ν)r

m1 2

+ζn (1 − ν)r

m 1− 21

−ζn

2 j=1

σrz =

[Cn3 Ip2 (ζn r) + Cn4 I−p2 (ζn r)] e7 e9 [ Cn1 Ip1 (ζn r) + Cn2 I−p1 (ζn r)] e8 e10

C0j rλj −

αo m2 −β r (1 + ν) [an Iβ (ζn r) + bn I−β (ζn r)] sin(ζn z) l m2

(6.9-18)

∞ E0 r m 1 {rm2 −β+1 {[(m2 − 2β + 2)Dn3 + ζn Dn1 ]Iβ (ζn r) 2(1 + ν)lm1 n=1

+[(m2 + 2)Dn4 + ζn Dn2 ]I−β (ζn ) + ζn rDn3 Iβ−1 (ζn r) + ζn rDn4 I−β−1 (ζn r)} m1 m1 e7 m1 e9 − 1 + p1 ) + ζn ]Cn1 Ip1 (ζn r) + [( − 1 − p1 ) +r− 2 {[( + ζn ]Cn2 I−p1 (ζn r) 2 e8 2 e10 e7 e9 Cn2 I−p1 −1 (ζn r)} −ζn r Cn1 Ip1 −1 (ζn r) − ζn r 8 e10 m1 m1 m1 + p2 )Cn3 Ip2 (ζn r) − ( − p2 )Cn4 I−p2 (ζn r) +r− 2 −1 {−( 2 2 2 +ζn rCn3 Ip2 −1 (ζn r) + ζn rCn4 I−p2 −1 (ζn r) + C0j λj rλj −1 } cos(ζn z) (6.9-19) j=1

To determine the displacements and stresses, four boundary conditions are required to evaluate the four unknowns Cn11 , Cn21 , Cn32 , and Cn42 . These four boundary conditions may be selected from the general boundary conditions for displacements and stresses. For example, the unknown constants may be obtained from the displacement boundary conditions u(a, z) = g1 (z), w(a, z) = g3 (z),

u(b, z) = g2 (z) w(b, z) = g4 (z)

(6.9-20)

σrr (b, z) = g6 (z) σrz (b, z) = g8 (z)

(6.9-21)

or from the stress boundary conditions σrr (a, z) = g5 (z), σrz (a, z) = g7 (z),

294


or from the mixed boundary conditions. This procedure will be used by expanding the given boundary conditions into Fourier series. Consider a hollow functionally graded cylinder with simply supported ends of the length L = 120 mm, with the inner radius a = 20 mm, outer radius b = 24 mm, and let us set l = 100 mm. Poisson’s ratio is assumed 0.3. The modulus of elasticity, the thermal coefficient of expansion, and the thermal conduction coefficient at the inner and outer radius are Ein = 66.2 GPa, αin = 10.3 × 10−6 /◦ C, kin = 18.1 W/mK, Eout = 117 GPa, αout = 7.11 × 10−6 /◦ C, and kout = 2.036 W/mK, respectively. The inside boundary is traction free with zero temperature. The outside boundary is assumed to be axially fixed with given temperature change distribution θ(b, z) = 30 sin(25πz)◦ C. Therefore, the assumed boundary conditions yield u(b, z) = 0, w(b, z) = 0, σrr (a, z) = 0, and σrz (a, z) = 0. Figure 6.9-1 shows the axial displacement along the axial and radial directions. The displacement is zero at the outer surface of the cylinder, according to the assumed boundary condition. Figures 6.9-2 and 6.9-3 show the distribution of the radial and axial thermal stresses of the cylinder. It is interesting to see that all components of stresses follow a harmonic pattern on the outside surface. The radial stress is zero at the inside surface, due to the assumed boundary conditions. The radial and axial stresses are zero at the ends in the cylinder, due to the assumed boundary conditions. The method of solution used in this section directly solves Navier equations by the power series method. In contrast to the potential function method, the advantage of this method is its ability to handle more general types of mechanical and thermal boundary conditions [25–28]. The proposed method may be extended to other types of mechanical and thermo-mechanical problems.

x 10−6 8 6 4 W (m)

2 0 −2 −4 −6 −8 0.024 0.023 0.022 0.021 r (m)

0.02 0

0.02

0.04

0.06

z (m)

Figure 6.9-1: Axial displacement.

0.08

0.1

0.12

10. Transient Thermal Stresses in Thick Spheres

295

x 107 1.5 1

srr (Pa)

0.5 0 −0.5 −1 −1.5 0.02 0.021

0.12 0.1

0.022

0.08 0.06

0.023

0.04 0.024

r (m)

0.02 0

z (m)

Figure 6.9-2: Radial thermal stress.

x 107 5

szz (Pa)

2.5

0

−2.5

−5 0.02 0.12

0.021

0.1 0.08

0.022

0.06 0.04

0.023 r (m)

0.02 0.024

0

z (m)

Figure 6.9-3: Axial thermal stress.

10

Transient Thermal Stresses in Thick Spheres

Transient thermal stresses in spheres are discussed by a number of authors. Cheung et al. [30] have analyzed the transient thermal stresses in a solid sphere. Their assumptions include the isotropic homogeneous material and a local

296


surface heating which results in a non-uniform temperature distribution. Takeuti and Tanigawa [31] have proposed a solution for thick hollow spheres with the consideration of coupling term in the energy equation, under a transient temperature distribution caused by a rotating heat source. The final solution that they have presented, however, is for an uncoupled energy equation with the temperature distribution as a function of (r, θ, t), where (r, φ, θ) are the spherical coordinates 0 ≤ r1 ≤ r ≤ r2 , 0 ≤ φ ≤ 2π, 0 ≤ θ ≤ π; and t is the time. The material presented in this section is from Cheung et al. [30]. Consider a solid sphere of radius ro and at the initial zero temperature T0 = 0. The known heat flux q (θ) is applied to the surface of the sphere at t ≥ 0 resulting in local heating of the sphere in the region 0 ≤ θ ≤ θ0 . The temperature distribution for this problem was obtained in Example 28 of Chapter 4, as (6.10-1) T (r, μ, t) = Ω(t) + Ts (r, μ) + T1 (r, μ, t) where μ = cos θ, and Ω(t), Ts (r, μ), and T1 (r, μ, t) were, respectively, obtained as t 1 κ Ω(t) = 2πro2 [ q (μ)dμ]dt kV 0 μ0 ∞ c 2 c 2 2n + 1 r n Ts (r, μ) = − ro + r + ro ( ) 10 6 2n ro n=1 1 q (μ)

cro ]Pn dμ}Pn (μ) 3 −1 ∞ ∞ 2n + 1 ro r 1 T1 (r, μ, t) = − ( )2 2 k n=0 m=1 βnm − n(n + 1) ro 1 Jn+ 1 (βnm r/ro ) 2 × × [ Pn (μ)q (μ)dμ] Jn+ 1 (βnm ) μ0 ×{

[

k

−

2

×Pn (μ)e−κβnm t/ro 2

2

(6.10-2)

where Pn (μ) is Legendre polynomial of the first kind and order n, V = 4πro3 /3 is the volume of the sphere, c = (1/κ)dΩ/dt, and βnm are the positive roots of the following equation: n+1 J 1 (βnm ) (6.10-3) Jn− 1 (βnm ) = 2 βnm n+ 2 Thermal stresses are the solution of Navier equation for the solid sphere with traction free surface. Navier equation in the rectangular Cartesian coordinate system is (6.10-4) uk,ki + (1 − 2ν)ui,kk = 2α(1 + ν)T,i and the stresses are related to the displacement components through Hooke’s law as ν 2(1 + ν) uk,k δij − α(T − T0 )δij ] σij = G[ui,j + uj,i + (6.10-5) 1 − 2ν 1 − 2ν


297

The boundary conditions for pure thermal stresses are for the traction free boundaries tni = σij nj = 0. The solution for thermal stresses in the solid sphere subjected to the described temperature distribution is the sum of the particular solution σijp and the general solution σijg . The general solution σijg is obtained from the homogeneous system of Eqs. (6.10-4) and (6.10-5) without the temperature terms, and it is associated with the surface traction induced by σijp . The particular solution σijp will be obtained from the system of Eqs. (6.10-4) and (6.10-5) with the temperature terms included in both equations. To obtain the particular solution, the thermoelastic displacement potential ψ is used. The relation between the displacement components and the function ψ is (6.10-6) ui = ψ,i Substituting ui from Eq. (6.10-6) into Eq. (6.10-4) yields ∇2 ψ = lα(T − T0 )

(6.10-7)

where l = (1 + ν)/(1 − ν). The solution of Eq. (6.10-7) is of the form ψ = ψ1 + ψ2 where

(6.10-8)

t

ψ1 = lακ

0

T1 (r, μ, t)dt

(6.10-9)

Substituting Eq. (6.10-8) into Eq. (6.10-7) yields the relation for ψ2 ∇2 ψ2 = lα(T − T0 ) − ∇2 ψ1

(6.10-10)

The expression for T1 from the third of Eqs. (6.10-2) is substituted into Eq. (6.10-9) and integrated to give the relation for ψ1 . With known ψ1 and T , Eq. (6.10-10) is solved for ψ2 . The results for ψ1 and ψ2 are added according to Eq. (6.10-8) to give the complete solution for the thermoelastic displacement potential ψ. The stresses associated with this function are the particular solution for the thermal stresses in the sphere. The particular solutions, called σijp , in spherical coordinates are related to the displacement potential ψ as ∂ 2 ψ¯ − T¯) ∂R2 1 1 ∂ ψ¯ ∂ 2 ψ¯ p + 2 (1 − μ2 ) 2 σ ¯θθ = 2[ R ∂R R ∂μ ¯ 1 ∂ψ − 2 μ − T¯] R ∂μ 1 ∂ ψ¯ 1 ∂ ψ¯ ¯ p = 2( − 2 μ − T) σ ¯φφ R ∂R R ∂μ

p σ ¯RR = 2(

298

Chapter 6. Disks, Cylinders, and Spheres 1 ∂ ψ¯ 1 ∂ 2 ψ¯ − ) R2 ∂μ R ∂R∂μ =σ ¯θφ =0

p = 2¯ μ( σ ¯Rθ p σ ¯Rφ

(6.10-11)

1

where μ ¯ = (1 − μ2 ) 2 , and the dimensionless terms are R=

r r0

ψ¯ =

τ=

T¯ =

κt r02

ψ lq0 r03 α/k

σ ¯ij =

(T −T0 ) q0 r0 /k

σij Glq0 r0 α/k

(6.10-12)

with q0 being a constant heat flux coefficient of the general expression for q (θ). The general solution of Eqs. (6.10-4) and (6.10-5) is obtained by means of the spherical harmonic Airy stress function. The stress function may be assumed in the form (6.10-13) Φ(r, μ) = rn Pn (μ) where Pn (μ) is Legendre polynomial of the first kind and order n. Using this form, the general solution is introduced by Sternberg et al. [32] as σ ¯ijg =

∞

cn σ ¯ij∗n +

n=0

∞

∗∗(n+1)

dn σ ¯ij

(6.10-14)

n=0

where the stresses related to σij∗n are ∗ σ ¯RR = n(n − 1)Rn−2 Pn (μ) ∗ σ ¯θθ = Rn−2 [μPn (μ) − n2 Pn (μ)] ∗ = Rn−2 [−μPn (μ) + nPn (μ)] σ ¯φφ ∗ σ ¯Rθ = −(n − 1)¯ μRn−2 Pn (μ) ∗ ∗ σ ¯θφ =σ ¯Rφ =0 ∗∗(n)

and the stresses associated with σ ¯ij

(6.10-15)

are

∗∗ σ ¯RR = [n2 (n − 3) − 2νn]Rn−1 Pn−1 (μ) ∗∗ = Rn−1 [(n + 4 − 4ν)μPn−1 (μ) − n(n2 + 2n − 1 + 2ν)Pn−1 (μ)] σ ¯θθ ∗∗ σ ¯φφ = Rn−1 [−(n + 4 − 4ν)μPn−1 (μ) + n(n − 3 − 4νn + 2ν)Pn−1 (μ)] ∗∗ σ ¯Rθ = (2 − n2 − 2ν)¯ μRn−1 Pn−1 (μ) ∗∗ ∗∗ σ ¯θφ = σ ¯Rφ = 0

(6.10-16)

where Pn (μ) = dPn (μ)/dμ. The coefficients cn and dn are evaluated using the boundary conditions. Due to the traction free condition, the boundary conditions are (6.10-17) σij nj = 0 at r = r0


299

This condition for the sphere under consideration reduces to σ ¯RR = σ ¯Rθ = 0 at R = 1. This yields ηn [3n + 2 + 2ν − n3 + 2νn] + ξn [2 − (n + 1)2 − 2ν] 2(n − 1)[n2 + n + 1 + ν(2n + 1)] n = 2, 3, 4, ... (6.10-18) ξn + nηn dn = n = 0, 1, 2, 3, ... (6.10-19) 2[n2 + n + 1 + ν(2n + 1)] cn =

p p where ξn and ηn are evaluated from σ ¯RR and σ ¯Rθ at R = 1 as

p σ ¯RR (1, μ, τ ) =

∞

ξn (τ )Pn (μ)

n=0 ∞

p σ ¯Rθ (1, μ, τ ) = μ ¯

ηn (τ )Pn (μ)

(6.10-20)

n=0

With the known temperature distribution, the particular solution for stresses p p and σ ¯Rθ are is obtained from Eq. (6.10-11), and at R = 1 the stresses σ ¯RR evaluated and substituted into Eq. (6.10-20). The expressions obtained for p p and σ ¯Rθ at R = 1 are expanded into Legendre polynomials and, therefore, σ ¯RR the functions ξn (τ ) and ηn (τ ) are obtained from Eq. (6.10-21). With the known ξn (τ ) and ηn (τ ), the coefficients cn and dn are known, and finally the complete general solution σ ¯ijg is obtained from Eq. (6.10-14). As an example, consider the case when q (θ) = q0 for μ0 ≤ μ ≤ 1. The solution for the temperature distribution reduces to T¯ = 3Aτ −

∞ ∞

Bnm Dn

n=0 m=1

Jn+ 1 (βnm R) 2

R

1 2

× Pn (μ)[e−βnm τ − 1] (6.10-21) 2

where A=

1 (1 − μ0 ) 2

Bnm =

1

Dn =

μ0

Pn (μ)dμ

2n + 1 2 − n(n + 1)]J [βnm n+ 1 (βnm )

(6.10-22)

2

The complete solution for the thermoelastic displacement potential from Eq. (6.10-7), using Eqs. (6.10-9) and (6.10-10), is ∞ ∞ 1 Bnm Dn Jn+ 12 (βnm R) 2 ψ = Aτ R + 1 2 2 βnm R2 n=0 m=1

×Pn (μ)[e−βnm τ − 1] 2

(6.10-23)

300


The expressions for ξn (τ ) and ηn (τ ), evaluated by using Eq. (6.10-20), are ξn (τ ) = 2 ηn (τ ) = 2

∞ Bnm Dn 2 βnm

m=1 ∞

n(n + 1)Jn+ 1 (βnm )[e−βnm τ − 1] 2

2

Bnm Dn 2 Jn+ 1 (βnm )[e−βnm τ − 1] 2 2 βnm m=1

(6.10-24)

The final solution for the stresses, which is the sum of the particular and general solution, is ∞ ∞

−2 Jn− 21 (βnm R) 1 βnm R R2 n=0 m=1 (n + 1)(n + 2) Jn+ 21 (βnm R) + ]Pn (μ) 1 2 R2 βnm R2

σ ¯RR = 2

βnm Dn [

×[e−βnm τ − 1] + 2

∞

n(n − 1)Rn−2 cn Pn (μ)

n=2

+

∞

(n + 1)[(n + 1)(n − 2) − 2ν]Rn dn Pn (μ)

(6.10-25)

n=1 ∞ ∞

σ ¯θθ = 2

βnm Dn

n=0 m=1

1

Jn− 1 (βnm R)

βnm R

R2

2

1

Pn (μ)

n+1 μ 1 − μ2 )P (μ) − P (μ) + P (μ)] n n 2 R2 2 R2 2 R2 n βnm βnm βnm ∞ Jn+ 1 (βnm R) 2 2 [e−βnm τ − 1] + Rn−2 cn [μPn (μ) × 1 R2 n=2

+[(1 −

−n2 Pn (μ)] +

∞

(

Rn dn (n + 5 − 4ν)μPn (μ)

n=1

−(n + 1)[(n + 1)2 + 2(n + 1) − 1 + 2ν]Pn (μ)

σ ¯φφ = 2

∞ ∞

βnm Dn

n=0 m=1

1

Jn− 1 (βnm R)

βnm R

R2

2

1

2

+

∞

(

∞

(6.10-26)

Pn (μ)

Jn+ 1 (βnm R) μ n+1 2 +[(1 − 2 2 )Pn (μ) − 2 2 Pn (μ)] 1 βnm R βnm R R2 ×[e−βnm τ − 1] +

)

Rn−2 cn [nPn (μ) − μPn (μ)]

n=2

Rn−2 dn −(n + 5 − 4ν)μPn (μ) + (n + 1)[n − 2 − 2(2n + 1)ν]

n=1

Pn (μ)

)

(6.10-27)

10. Transient Thermal Stresses in Thick Spheres ∞ ∞

301

−1 Jn− 12 (βnm R) 1 βnm R R2 n=0 m=1 n + 2 Jn+ 21 (βnm R) 2 + 2 2 ] Pn (μ) × [e−βnm τ − 1] 1 βnm R R2

σ ¯Rθ = 2¯ μ

−¯ μ

∞

βnm Dn [

(n − 1)Rn−2 cn Pn (μ) + μ ¯

n=2

∞

[2 − (n + 1)2 − 2ν]Rn dn Pn (μ)

n=1

(6.10-28) σ ¯Rφ = σ ¯θφ = 0

(6.10-29)

At the center of the sphere, where R → 0, the stresses must be independent of θ. This condition excludes all the terms related to n = 2 in Eqs. (6.1025) to (6.10-28). The temperature and stresses at the center of the sphere are obtained as limiting values for R → 0. This yields 2 ∞ 1 e−β0m τ − 1 3 ¯ √ ] T (0, μ, τ ) = 4A[ τ − 3/2 4 2π m=1 β0m J 1 (β0m ) 2

σ ¯RR (0, μ, τ ) = σ ¯θθ (0, μ, τ ) = σ ¯φφ (0, μ, τ ) 8 = 3

∞ 2 e−β0m τ − 1 A 3/2 11 m=1 β0m J 1 (β0m ) 2

2

σ ¯Rθ (0, μ, τ ) = 0

(6.10-30)

The plots of temperature, and tangential and radial stresses are shown in Figs. 6.10-1 to 6.10-3. The calculations are performed for θ0 = 45◦ and a

Figure 6.10-1: Temperature distribution in the sphere.

302


Figure 6.10-2: Tangential stress distribution in the sphere.

Figure 6.10-3: Radial stress distribution in the sphere.

typical brittle material with ν = 0.25. The calculations are also performed for a uniform heat flux when the entire surface of the sphere is heated by a constant heat flux q0 [30]. The figures show the distribution of the temperature and the stresses for dimensionless heating time at τ = 0.03 and 0.05 at angle θ = 2◦ . The results show a tensile stress concentration within the heated zone in the interior of the sphere. It is further noted that the magnitude of stresses due to uniform heating exceeds the magnitude of stresses induced by nonuniform heating.

11. Functionally Graded Spheres

11

303

Functionally Graded Spheres

An analytical solution for the stresses in spheres made of functionally graded materials is given by Lutz and Zimmerman [33]. They considered thick spheres under radial thermal loads, where radially graded materials with linear composition of the constituent materials were considered. Obata and Noda [34,35] studied the one-dimensional steady-state and transient thermal stresses in a functionally graded circular hollow cylinder and sphere using the perturbation method. Thermal stresses in materials with temperature dependent properties are discussed by Noda [36]. Using the power law functions for the material properties of the constituent materials, Eslami et al. [37] presented an analytical solution based on the direct solution of Navier equation for a thick FGM spherical vessel. Consider a thick hollow sphere of inside radius a and outside radius b made of FGM. The sphere’s material is graded through the radial r-direction. Thus, the material properties are functions of r. Let u be the displacement component along the radial direction. The strain-displacement relations are rr = u ,

φφ =

u r

(6.11-1)

where ( ) denotes the differentiation with respect to r. The stress-strain relations are σrr = λe + 2μrr − (3λ + 2μ)αθ(r) σφφ = λe + 2μφφ − (3λ + 2μ)αθ(r)

(6.11-2)

where σij and ij (i, j = r, φ) are the stress and strain tensors, θ(r) is the temperature change distribution determined from the heat conduction equation, α is the coefficient of thermal expansion, and λ and μ are Lamé constants related to the modulus of elasticity E and Poisson’s ratio ν as λ=

νE , (1 + ν)(1 − 2ν)

μ=

E 2(1 + ν)

(6.11-3)

The equilibrium equation in the radial direction, disregarding the body force and the inertia term, is 2 + (σrr − σφφ ) = 0 σrr r

(6.11-4)

To obtain the equilibrium equation in terms of the displacement component for the FGM sphere, the functional relationship of the material properties must be known. The sphere’s material is assumed to be described with a power law function of the radial direction as r r α(r) = m2 ( )n2 (6.11-5) E(r) = m1 ( )n1 , a a

304


where m1 and m2 are material constants and n1 and n2 are the power law indices of the material. We may further assume that Poisson’s ratio is constant. Using relations (6.11-1)–(6.11-5), Navier equation in term of the displacement is Ar2 u + Bru + Cu =

f (r) rn1 −2

(6.11-6)

where m1 1 − ν 1 + ν 1 − 2ν m1 1 − ν (n1 + 2) B= 1 + ν 1 − 2ν 2m1 1 − ν ν ( n1 − 1) C= 1 + ν 1 − 2ν 1 − ν A=

f (r) =

an1 [E αθ + Eα θ + Eαθ ] (1 − 2ν)

(6.11-7) (6.11-8)

The heat conduction equation in the steady-state condition for the onedimensional problem in spherical coordinates and the thermal boundary conditions for an FGM hollow sphere are given, respectively, as 1 2 (r k(r)θ (r)) = 0 r2 C11 θ (a) + C12 θ(a) = f1 C21 θ (b) + C22 θ(b) = f2

(6.11-9)

where k = k(r) is the thermal conductivity and Cij are either thermal conductivity k, or convection coefficient h, depending on the type of thermal boundary conditions. The terms f1 and f2 are known constants at the inside and outside radii. It is assumed that the thermal conduction coefficient k(r) is a power function of r as r (6.11-10) k(r) = m3 ( )n3 a where m3 and n3 are material parameters. Using Eq. (6.11-10), the heat conduction equation becomes 1 n3 +2 [r θ (r)] = 0 r2

(6.11-11)

Integrating Eq. (6.11-11) twice yields θ(r) = C1 r−(n3 +1) + C2

(6.11-12)

Using the boundary conditions, the second and third of Eq. (6.11-9), the constants C1 and C2 are obtained to be C1 =

C22 f1 − C12 f2 C12 [(n3 + 1)C21 b−(n3 +2) − C22 b−(n3 +1) ] − C22 [(n3 + 1)C11 a−(n3 +2) − C12 a−(n3 +1) ]

11. Functionally Graded Spheres C2 =

305

f1 [(n3 + 1)C21 b−(n3 +2) − C22 b−(n3 +1) ] − f2 [(n3 + 1)C11 a−(n3 +2) − C12 a−(n3 +1) ] C12 [(n3 + 1)C21 b−(n3 +2) − C22 b−(n3 +1) ] − C22 [(n3 + 1)C11 a−(n3 +2) − C12 a−(n3 +1) ]

(6.11-13) The proof that the denominator does not vanish is not given. We show that this formula is valid in a particular example given on the pages to follow. Navier equation for the radial displacement u is given by Eq. (6.11-6). Once the function f (r) is known, the equation may be solved analytically. Substituting f (r) from Eq. (6.11-8) into Eq. (6.11-6), and using Eqs. (6.11-5) and (6.11-12) we obtain Ar2 u + Bru + Cu = γrn2 −n3 + ωrn2 +1

(6.11-14)

where C1 m1 m2 (n1 + n2 − n3 − 1) 1 − 2ν C2 m1 m2 (n1 + n2 ) ω= 1 − 2ν γ=

(6.11-15)

Equation (6.11-14) is a nonhomogeneous Euler differential equation with the general and particular solutions. The general solution, ug , is obtained by assuming [37] ug (r) = Qrp

(6.11-16)

Substituting Eq. (6.11-16) into the homogeneous form of Eq. (6.11-14) yields Ap2 + (B − A)p + C = 0

(6.11-17)

where A, B, and C are obtained from Eq. (6.11-7). Equation (6.11-19) has two real roots p1 and p2 as p1,2 =

A−B±

(B − A)2 − 4AC 2A

(6.11-18)

The requirement on real values of p1,2 is satisfied in a particular example. Thus, the general solution is ug (r) = Q1 rp1 + Q2 rp2

(6.11-19)

The particular solution up (r) is assumed to have the form [37] up (r) = βrn2 −n3 + ζrn2 +1

(6.11-20)

Substituting Eq. (6.11-20) in Eq. (6.11-14) yields [(A(n2 − n3 − 1) + B)(n2 − n3 ) + C]βrn2 −n3 +[(A(n2 ) + B)(n2 + 1) + C]ζrn2 +1 = γrn2 −n3 + ωrn2 +1 (6.11-21)

306


Equating the coefficients of the identical powers yields γ [A(n2 − n3 − 1) + B](n2 − n3 ) + C ω ζ= (An2 + B)(n2 + 1) + C

β=

(6.11-22)

where A, B, and C are obtained from Eq. (6.11-7). The complete solution for u(r) is the sum of the general and particular solutions as u(r) = ug (r) + up (r)

(6.11-23)

u(r) = Q1 rp1 + Q2 rp2 + βrn2 −n3 + ζrn2 +1

(6.11-24)

Thus

Substituting Eq. (6.11-24) into Eqs. (6.11-1) and (6.11-2), the strains and stresses are obtained as rr = p1 Q1 rp1 −1 + p2 Q2 rp2 −1 + (n2 − n3 )βrn2 −n3 −1 + (n2 + 1)ζrn2 φφ = Q1 rp1 −1 + Q2 rp2 −1 + βrn2 −n3 −1 + ζrn2 (6.11-25) ( m1 Q1 [(ν − 1)p1 − 2ν]rn1 +p1 σrr = n1 a (1 + ν)(2ν − 1) +Q2 [(ν − 1)p2 − 2ν]rn1 +p2 C2 m2 (ν + 1) C1 m2 (ν + 1) +{ + ζ[n2 (ν − 1) − (ν + 1)]rn3 +1 + an2 an2 ) n1 +n2 −n3 +[(n2 − n3 )(ν − 1) − 2ν]β}r ( m1 σφφ = n1 −Q1 (νp1 + 1)rn1 +p1 − Q2 (νp2 + 1)rn1 +p2 a (1 + ν)(2ν − 1) C2 m2 (ν + 1) C1 m2 (ν + 1) +{[ − ζ(n2 ν + ν + 1)]rn3 +1 + n 2 a an2 ) n1 +n2 −n3 −(n2 ν − n3 ν + 1)β}r (6.11-26) To determine the constants Q1 and Q2 , the boundary conditions for stresses must be used. Consider the mechanical boundary conditions at the inside and outside radii as σrr (a) = −Pa ,

σrr (b) = −Pb

(6.11-27)

Substituting the first of Eqs. (6.11-26) into Eq. (6.11-27), the constants of integration are obtained Q1 =

ϕ22 X − ϕ12 Y , ϕ11 ϕ22 − ϕ12 ϕ21

Q2 =

ϕ11 Y − ϕ21 X ϕ11 ϕ22 − ϕ12 ϕ21

(6.11-28)

11. Functionally Graded Spheres

307

where ϕ11 = η[(ν − 1)p1 − 2ν]an1 +p1 ϕ12 = η[(ν − 1)p2 − 2ν]an1 +p2 ϕ21 = η[(ν − 1)p1 − 2ν]bn1 +p1 ϕ22 = η[(ν − 1)p2 − 2ν]bn1 +p2 X = −(Pa + τ (a)) Y = −(Pb + τ (b))

(6.11-29)

The proof that the denominators in Eqs. (6.11-28) do not vanish is not given. We show that this formula is valid in a particular example. In the above equations η and τ (r) are η=

an1 (1

m1 + ν)(2ν − 1)

C1 m2 (ν + 1) C2 m2 (ν + 1) + ζ(n2 (ν − 1) − (ν + 1)))rn3 +1 + n 2 a an2 n1 +n2 −n3 +((n2 − n3 )(ν − 1) − 2ν)β]r (6.11-30)

τ (r) = η[(

Now consider a thick hollow sphere of inner radius a = 1 m and outer radius b = 1.2 m. Poisson’s ratio is assumed to be constant and is taken to be 0.3, and the modulus of elasticity and the thermal coefficient of expansion at the inner radius are m1 = 200 GPa and m2 = 1.2 × 10−6 /◦ C, respectively. The boundary conditions for the temperature are taken as θ(a) = 10◦ C and θ(b) = 0◦ C. The hollow sphere is assumed to be under internal pressure of 50 MPa and zero external pressure (i.e., σrr (a) = −50 MPa and σrr (b) = 0 MPa). For different values of n, the temperature profile, the radial displacement, the radial stresses, and the hoop stresses along the radial direction are plotted in Figs. 6.11-1 to 6.11-4. The power index for the modulus of elasticity, coefficient of thermal expansion, and heat conduction coefficient are assumed to be identical (n1 = n2 = n3 = n) in these graphs. Figure 6.11-1 shows that as n in the power law increases, the temperature decreases. Figure 6.11-2 shows that for higher values for n, the radial displacement decreases. Figure 6.11-3 represents the radial stress along the radial direction, and it decreases as the power law index increases. The circumferential stress versus the radial direction is shown in Fig. 6.11-4. It is seen that for n < 1 the circumferential stress decreases along the radial direction. When n > 1, the situation is reversed and the circumferential stress increases along the radial direction. The curve associated with n = 1 shows that the variation of circumferential stress along the radial direction is minor, and it is almost uniform across the radius. To investigate the pattern of the stress distribution along the sphere radius, the effective stress σ ∗ = | σrr − σφφ | is plotted along the radial direction for

308

Chapter 6. Disks, Cylinders, and Spheres 1.2 1.0

n=−2 n=−1 n=0 n=1 n=2 n=3

T /T(a)

0.8 0.6 0.4 0.2 0 −0.2

1

1.02

1.04

1.06

1.08

1.1

1.12

1.14

1.16

1.18

1.2

r/a

Figure 6.11-1: Radial temperature distribution.

7

× 10−4

6.5 6

u/a

5.5 n=−2 n=−1 n=0 n=1 n=2 n=3

5 4.5 4 3.5 3 1

1.02 1.04 1.06 1.08

1.1

1.12 1.14 1.16 1.18

1.2

r/a

Figure 6.11-2: Radial displacement distribution.

different values of b/a and the power law index n. Figures 6.11-5 to 6.11-7 are plotted for b/a = 1.2, 2, and 3, respectively. It is interesting to note from Fig. 6.11-5 that for n = 3 the effective stress is almost uniform along the radius of the sphere. For larger b/a ratios, the curves related to n = 3 produce almost uniform distribution for the effective stress, see Figs. 6.11-6 and 6.11-7.

12. Problems

309 0.2 0

n=−2 n=−1 n=0 n=1 n=2 n=3

srr /Pi

−0.2 −0.4 −0.6 −0.8 −1 −1.2

1

1.02 1.04 1.06 1.08

1.1

1.12 1.14 1.16 1.18

1.2 r/a

Figure 6.11-3: Radial stress distribution. 3.2 3

n=3 n=2 n=1 n=0 n=−1 n=−2

2.8

σff /Pi

2.6 2.4 2.2 2 1.8 1.6 1

1.02 1.04 1.06 1.08

1.1

1.12 1.14 1.16 1.18 1.2 r/a

Figure 6.11-4: Circumferential stress distribution.

12

Problems

1. Obtain Michell conditions in cylindrical coordinates. For a hollow thick cylinder under radial temperature distribution, T = T (r), check the existence of thermal stresses through Michell conditions. 2. Consider a temperature change distribution in a thick hollow cylinder of the form

310

Chapter 6. Disks, Cylinders, and Spheres 0.3

0.26

n= 1

0.24

n= 2 n= 3

(GPa)

n= −2 n= −1 n= 0

0.22

√2 s *

0.28

0.2 0.18 0.16 0.14 0.12

1

1.02 1.04 1.06 1.08

1.1 r/a

1.12 1.14 1.16 1.18

1.2

Figure 6.11-5: Effective stress distribution for b/a = 1.2. 0.18 n= −2 n= −1 n= 0 n= 1 n= 2

0.16

√2 s *

(GPa)

0.14 0.12

n= 3 0.1

n= 3.5

0.08 0.06 0.04 0.02 0

1

1.1

1.2

1.3

1.4

1.5 r/a

1.6

1.7

1.8

1.9

2

Figure 6.11-6: Effective stress distribution for b/a = 2. θ(r, θ) =

∞

(An rn + Bn r−n ) cos nθ

n=1

where An and Bn are prescribed. The related stress function satisfies the equation ∇2 Φ =

∞

(C1m rm + C2m r−m )(C3m sin mθ + C4m cos mθ)

n=1

Check all the three Michell conditions to identify the terms of temperature which do not satisfy the conditions. Then use Navier equations in polar coordinates and with the use of Papkovich function obtain the stresses.

12. Problems

311 0.16 n= −2 n= −1 n= 0 n= 1 n= 2 n= 3 n= 4

0.14 0.12

√2 s *

(GPa)

0.1 0.08 0.06 0.04 0.02 0

1

1.2

1.4

1.6

1.8

2 r/a

2.2

2.4

2.6

2.8

3

Figure 6.11-7: Effective stress distribution for b/a = 3.

3. Consider a finite solid circular cylinder of the length 2l and radius r2 . (a) Use Papkovich function to obtain the expression for displacement components. (b) For the given temperature distribution T (r, z) = T0 + a0 (r) +

∞

an (r). cos kn z

n=1

where an (r) are prescribed and kn are constant values, obtain the associated thermal stresses. 4. Find the thermal stresses in a thick hollow sphere of inside and outside radii a and b, respectively, subjected to the general temperature variation of the form T = T (r, φ) and with the stress-free boundary conditions. 5. A rotating disk with variable thickness is shown in Fig. 6.12-1. The temperature variation across the thickness, z, is neglected and it is assumed to vary along the radius r. The disk profile is assumed to be the second order polynomial of the following form f (r) = (L − l)(

D−r 2 ) +l D−d

312


Figure 6.12-1: The disk profile, Problem 5. The mass density of the disk is ρ. The disk is rotating at constant speed ω. At outer radius of the disk turbine blades are installed so that their weight per unit length is w. Determine the mechanical and thermal stresses in the disk.

Bibliography [1] Hetnarski, R.B. and Ignaczak, J., Mathematical Theory of Elasticity, Taylor and Francis, New York, 2004. [2] Gatewood, B.E., Thermal Stresses, McGraw-Hill, New York, 1957. [3] Gatewood, B.E., Thermal Stresses in Long Cylindrical Bodies, Phil. Mag., Ser. 7, Vol. 32, pp. 282–301, 1941. [4] Nowacki, W., Thermoelasticity, 2nd edition, PWN-Polish Scientific Publishers, Warsaw, and Pergamon Press, Oxford, 1986. [5] Boley, B.A. and Weiner, J.H., Theory of Thermal Stresses, Wiley, New York, 1962. [6] Sabbaghian, M. and Eslami, M.R., Creep Relaxation of Non Axisymmetric Thermal Stresses in Thick Walled Cylinders, AIAA J. Vol. 12, No. 12, pp. 1652–1658, 1974. [7] Muskhelishvili, N.I., Some Basic Problems of the Mathematical Theory of Elasticity, Noordhoff, Groningen, Holland, 1953. [8] Wang, C.T., Applied Elasticity, McGraw-Hill, New York, 1953. [9] Zimmerman, R.W. and Lutz, M.P., Thermal Stress and Thermal Expansion in a Uniformly Heated Functionally Graded Cylinder, J. Therm. Stresses, Vol. 22, pp. 177–188, 1999. [10] Han, X., Liu, G.R., and Lam, K.Y., A Quadratic Layer Element for Analyzing Stress Waves in FGMs and Its Applications in Material Characterization, J. Sound Vib., Vol. 236, No. 2, pp. 307–321, 2000. [11] Obata, Y. and Noda, N., Transient Thermal Stresses in a Hollow Sphere of Functionally Gradient Material, Proceedings of Thermal Stresses Symposium, Shizuoka University, Hamamatsu, pp. 335–338, 1995. 313

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[12] Obata, Y. and Noda, N., Two-Dimensional Unsteady Thermal Stresses in a Partially Heated Plate Made of Functionally Graded Material, Proceedings of Thermal Stresses Symposium, Rochester Institute of Technology, Rochester, pp. 735–738, 1997. [13] Obata, Y., Kanayama, K., Ohji, T., and Noda, N., Two-Dimensional Unsteady Thermal Stresses in a Partially Heated Circular Cylinder Made of Functionally Graded Material, Proceedings of the Third Congress on Thermal Stresses, Kraków University of Tech., Kraków, Poland, pp. 595– 598, 1999. [14] Jabbari, M., Sohrabpour, S., and Eslami, M.R., Mechanical and Thermal Stresses in a Functionally Graded Hollow Cylinder due to Radially Symmetric Loads, Int. J. Pres. Ves. Pip., Vol. 79, pp. 493–497, 2002. [15] Jabbari, M., Sohrabpour, S., and Eslami, M.R., General Solution for Mechanical and Thermal Stresses in a Functionally Graded Hollow Cylinder due to Nonaxisymmetric Steady-State Loads, J. Appl. Mech., Vol. 70, pp. 111–118, 2003. [16] Katsuo, M., Sawa, T., Kawaguchi, K., and Kawamura, H., Axisymmetrical Thermal Stress Analysis of Laminated Composite Finite Hollow Cylinders Restricted at Both Ends in Steady State, Proceedings of the 1996 ASME International Mechanical Engineering Congress and Exposition, pp. 17–22, 1996. [17] Okumura, I.A. and Noda, N., Thermoelastic Potential Functions in Transversely Isotropic Solids and Their Applications. J. Therm. Stresses, Vol. 14, No. 3, pp. 309–331, 1991. [18] Misra, J.C. and Achari, R.M., On Axisymmetric Thermal Stresses in an Anisotropic Hollow Cylinder, J. Therm. Stresses, Vol. 3, No. 4, pp. 509– 520, 1980. [19] Chen, P.Y.P., Axisymmetric Thermal Stresses in an Anisotropic Finite Hollow Cylinder, J. Therm. Stresses, Vol. 6, No. 2–4, pp. 197–205, 1980. [20] Lu, Y., Xiao, J., and Zhang, K., Steady-State Temperature Distribution and Thermal Stress of Functionally Gradient Material Cylinder, Wuhan Jiaotong Keji Daxue Xuebao/Journal of Wuhan Transportation University, Vol. 21, No. 2, pp. 158–163 (in Chinese), 1997. [21] Horgan, C.O. and Chan, A.M., The Pressurized Hollow Cylinder or Disk Problem for Functionally Graded Isotropic Linearly Elastic Materials, J. Elasticity, Vol. 55, pp. 4359, 1999.

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[22] Tutuncu, N. and Ozturk, M., The Exact Solution for Stresses in Functionally Graded Pressure Vessels, Composites, Part B, Vol. 32, pp. 683–686, 2001. [23] Liew, K.M., Kitipornchai, S., Zhang, X.Z., and Lim, C.W., Analysis of the Thermal Stress Behaviour of Functionally Graded Hollow Circular Cylinders, Int. J. Solids Struct., Vol. 40, pp. 2355–2380, 2003. [24] Zhang, X.D., Liu, D.Q., and Ge. C., Thermal Stress Analysis of Axial Symmetry Functionally Gradient Materials under Steady Temperature Field, J. Funct. Grad. Mater., Vol. 25, pp. 452–455, 1994. [25] Obata, Y. and Noda, N., Steady Thermal Stresses in a Hollow Circular Cylinder and a Hollow Sphere of a Functionally Gradient Material, J. Therm. Stresses, Vol. 17, No. 3, pp. 471–487, 1994. [26] Jabbari, M., Bahtui, A., and Eslami, M.R., Axisymmetric Mechanical and Thermal Stresses in Thick Long FGM Cylinders, J. Therm. Stresses, Vol. 29, No. 7, pp. 643–663, 2006. [27] Bahtui, A., Jabbari, M., and Eslami, M.R., Mechanical Stresses in Thick FGM Pressure Vessels, Proceedings, Int. Congress and Exhibition on Pressure Vessel and Piping, OPE 2006, Channai, India, 7–9, Feb. 2006. [28] Jabbari, M., Mohazzab, A.H., Bahtui, A., and Eslami, M.R., Analytical Solution for Three-Dimensional Stresses in a Short Length FGM Hollow Cylinder, ZAMM, Vol. 87, No. 6, pp. 413–429, 2007. [29] Rice, R.G. and Do, D.D., Applied Mathematics and Modeling for Chemical Engineering, Wiley, New York, pp. 131–132, 1995. [30] Cheung, J.B., Chen, T.S., and Thirumalai, K., Transient Thermal Stresses in a Sphere by Local Heating, ASME J. Appl. Mech., Vol. 41, No. 4, pp. 930–934, 1974. [31] Takeuti, Y. and Tanigawa, Y., Transient Thermal Stresses of a Hollow Sphere due to Rotating Heat Source, J. Therm. Stresses, Vol. 5, No. 3–4, pp. 283-298, 1982. [32] Sternberg, E., Eubanks, E.A., and Sadowsky, M.A., On the Axisymmetric Problem of Elasticity Theory for a Region Bounded by Two Concentric Spheres, Proc. First US National Congress of Appl. Mech., ASME, New York, pp. 209–215, 1952. [33] Lutz, M.P. and Zimmerman, R.W., Thermal Stresses and Effective Thermal Expansion Coefficient of a Functionally Graded Sphere, J. Therm. Stresses, Vol. 19, pp. 39–54, 1996.

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[34] Obata, Y. and Noda, N., Steady Thermal Stress in a Hollow Circular Cylinder and a Hollow Sphere of a Functionaly Gradient Material, J. Therm. Stresses, Vol. 14, pp. 471–487, 1994. [35] Obata, Y. and Noda, N., Transient Thermal Stresses in a Hollow Sphere of Functionally Gradient Material, Proceedings of Thermal Stresses Symposium, Shizuoka University, Hamamatsu, pp. 335–338, 1995. [36] Noda, N., Thermal Stresses in Materials with Temperature-Dependent Properties, Chap. 6 in Thermal Stresses I, R.B. Hetnarski, ed., Elsevier Science, Amsterdam, 1986. [37] Eslami, M.R., Babaei, M.H., and Poultangari, R., Thermal and Mechanical Stresses in a Functionally Graded Thick Sphere, Int. J. Pres. Ves. Pip., Vol. 82, pp. 522–527, 2005.

Chapter 7 Thermal Expansion in Piping Systems Piping systems, such as those installed in refineries, are transmitters of high pressure fluids or gases at high temperatures, and failure of such systems may cause catastrophic damage. Advanced design codes have been developed to provide safety instructions for designers. Piping systems are initially installed at reference temperature. At working conditions, under elevated temperatures, the end expansion and contraction forces and bending moments are created. The resulting thermal expansions, due to the imposed constraints, may produce large stresses which, if not properly taken care of, may cause failure. This chapter presents a simple method of analysis of piping systems. The main scope of the chapter is to provide an analytical method to calculate the end expansion and contraction forces and bending moments. In a number of examples it is shown how the end forces and moments may be made smaller.

1

Introduction

Piping systems are essential components in many industries such as refineries, power plants, and chemical plants, where their prime purpose is the transport of fluid from one piece of equipment to another. Normally, the content fluid of the pipe is hot, and since the piping system is initially designed at reference temperature, the temperature change causes thermal expansion. If the ends of the piping system are restricted, which is usually the case, forces and moments are produced at the supports of the pipes causing thermal stresses in the system. The art of piping flexibility analysis is to give enough flexibility to the piping system so that the resulting stresses at all points of the system remain under a safe limit. Usually, this flexibility is designed with a loop or flexible joint at the ends. Therefore, the design procedure of a piping system is to consider the isometric of the piping system at the reference temperature. R.B. Hetnarski, M.R. Eslami, Thermal Stresses – Advanced Theory and Applications, Solid Mechanics and its Applications 158, c Springer Science + Business Media B.V. 2009

317

318

Chapter 7. Thermal Expansion in Piping Systems

Then, by means of analytical or numerical methods the restrained forces and moments at the support of the piping system are calculated and, finally, by sketching the free body diagram of each piping member using the appropriate codes and standards, the stresses are computed and compared with the safe limit. If the calculated stresses are above the allowable limit, a loop for the piping system at a proper location may be considered and the calculation procedure is repeated. There are many methods for the calculation of the restrained forces and moments of a piping system under thermal expansion [1–3]. In this chapter we discuss an analytical technique based on the elastic center method. The stiffness approach or the finite element method may also be employed for pipeline analysis. The stiffness approach is essentially derived from the structural analysis under mechanical loads. Since thermal loads behave similarly to mechanical loads, they may be included in the stiffness method of analysis of structures. In the following sections the elastic center method is described.

2

Definition of the Elastic Center

Consider a piping system in two-dimensions with clamped ends A and B. The isometric lines of the piping system are shown in Fig. 7.2-1. The piping system is assumed to be under a uniform reference temperature T0 . If we ignore the weight of the pipes, the reaction forces and moments at ends A and B at the reference temperature T0 are zero. Now, a hot fluid is passed through the piping system and the temperature of the piping system is raised to T . It is again assumed that the temperature T is constant through the length of the piping system. If the end B is considered free, due to thermal expansion of the piping system, it travels to point B , as seen from Fig. 7.2-1. To bring point B to B, considering the clamped condition, forces Fx , Fy , and a bending moment MB must be applied at point B, see Fig. 7.2-2. The application of these forces and moment at point B produce opposite and equal reaction forces Fx and Fy

Figure 7.2-1: Pipe at reference temperature and at raised temperature, without constraint at B.

2. Definition of the Elastic Center

319

Fx

X

B MB

Fy

Y

A

Fx

Fy

MA

Figure 7.2-2: Pipe at elevated temperature.

at A, as well as a bending moment MA which in general is not equal to MB . The pipeline analysis entails computation of the reaction forces and moments at the ends of the piping system, based upon free-body diagrams of each pipe element. The stress analysis of the pipe is then carried out in conjunction with an acceptable engineering code. To obtain the reaction forces Fx and Fy and the reaction moments MA and MB , the method of elastic center may be used. According to this method, the elastic center of the piping system is found and a coordinate system is fixed to it. In the coordinate system fixed to the elastic center, the bending moments do not appear and, therefore, the problem is reduced to two equations for two unknown forces Fx and Fy . To find the elastic center and the associated coordinate system, the Maxwell reciprocity theorem is used. A general treatment of this theorem is given in Section 15 of Chapter 2. The general reciprocity theorem, which is described in Chapter 2, is reduced to a simple law for a piping system in two-dimensions. The Maxwell reciprocity theorem The work done by the loads of the first state on the corresponding deformations of the second state is equal to the work done by the loads of the second state on the corresponding deformations of the first state. To apply the reciprocity theorem to a piping system in two-dimensions, consider Figs. 7.2-3a and b with the first and second states as defined below: The first state: – The load is the bending moment MB acting at point B. – The deformations are the horizontal and vertical displacements δxo and δyo , respectively, and the rotation φo at an arbitrary point O (which is elastically connected to the piping system).

320

Chapter 7. Thermal Expansion in Piping Systems fB X B

MB

0

δxB

δyB

X

B

b δyo

a

B⬘

Fy Fx

x

y

δxo Y

x

0

a

0⬘

b

Y

y A

A

(a)

(b)

Figure 7.2-3: (a) The first and (b) second states.

The second state: – The loads are Fx and Fy applied at point O. – The deformations are δxB , δyB , and the rotation φB , produced at point B. According to Maxwell theorem MB φB = Fx δxo + Fy δyo

(7.2-1)

Now, if the point O is selected so that the application of forces Fx and Fy at O do not produce any rotation in the piping system (and, consequently, no bending moment is produced), then φB = 0. From Eq. (7.2-1), since φB = 0, the left-hand side is zero and thus Fx δxo + Fy δyo = 0

(7.2-2)

Since Fx and Fy are, in general, different from zero, it follows that δxo = 0 δyo = 0

(7.2-3)

Since δxo and δyo are the displacements produced by MB acting at point B, Eqs. (7.2-3) provide a method of calculation of the location of the point O. This point is called the elastic center, and it is a fictitious point in the piping system. Its property is that if the forces Fx and Fy act at that point, no rotation and thus the moments appear at any point on the piping system. But, since Fx and Fy are needed at points other than O, moments MA and MB will appear in the system.

2. Definition of the Elastic Center

321

Figure 7.2-4: Position of the elastic center.

Figure 7.2-5: Displacement of the end point B due to the bending moment MB . Now consider a two-dimensional piping system with fixed ends A and B under elevated temperature, as shown in Fig. 7.2-4. The coordinate axes are fixed to the end B. The position of the elastic center is shown at point O with the coordinates a and b from the end B. We let the end B remain free to move. Due to the action of the bending moment MB at point B, this point is displaced to point B , as shown in Fig. 7.2-5, with horizontal and vertical displacements uB and vB and the end line rotation φB . The location of the elastic center O is found using Eqs. (7.2-3). From the elementary theory of strength of materials, the slope of a pipe element ds due to the constant bending moment MB is dφ =

MB ds EI

or

φB = MB

L

ds EI

(7.2-4)

322


where the integration is carried over the length of the pipe L. The horizontal displacement of the end B due to the rotation of an element ds at the vertical distance y from B is

uB =

L

y dφ = MB

L

y ds EI

(7.2-5)

Similarly, the vertical displacement of B due to the rotation of an element ds at the horizontal distance x from B is

vB =

L

x dφ = MB

L

xds EI

(7.2-6)

The displacements uB and vB and the rotation φB at the end B cause the point O to be displaced. We may assume that a rigid bar connects points O and B. The total horizontal and vertical displacements of point O due to the displacements uB and vB and the rotation φB at point B are, see Fig. 7.2-3b δxo = uB − bφB δyo = aφB − vB

(7.2-7)

Therefore, when the material and the pipe properties remain constant through the length of the piping system, from Eqs. (7.2-3) and upon the substitution from Eq. (7.2-4), the coordinates a and b of the elastic center O are

vB xds a= = L = x¯ φB L ds yds uB = L = y¯ b= φB L ds

(7.2-8)

It is noticed that the elastic center O of a two-dimensional piping system coincides with the center of mass of the isometric of the piping system.

3

Piping Systems in Two Dimensions

Consider a piping system in two-dimensions, as shown in Fig. 7.3-1. A hot fluid is passed through the piping system and its temperature is raised from T0 to T . It is assumed that T0 and T are uniform along the piping system. The global coordinate axes are fixed at one of the ends of the piping system. The rule is that the considered fixed end is arbitrarily made free, while the other end is fixed, and the global coordinate axes are chosen in opposite direction of the thermal expansion of the free end. Let, arbitrarily, the end B be free, while the end A remain fixed. The global coordinate system (x, y) is fixed to the point B and the axes are pointed in opposite directions to their thermal expansion. The location of the elastic center O is calculated and the global coordinate system is transferred to the point O.

3. Piping Systems in Two Dimensions

323

Figure 7.3-1: A piping system in two dimensions. The deformation of the piping system in the coordinates xy fixed to the elastic center O, and in terms of the thermal deflections, are Δxx + Δyx = Δx Δxy + Δyy = Δy

(7.3-1)

where Δxx = deflection of the piping system due to force Fx in x direction. Δyx = deflection of the piping system due to force Fy in x direction. Δx = thermal expansion of the piping system in x direction due to the temperature change (T − T0 ). Symbols Δxy , Δyy , and Δy are similarly defined. To obtain the thermal deformation, Castigliano theorem may be used. According to Castigliano theorem, the deflection of a piping system in a specific direction is the partial derivative of the strain energy with respect to the force in that specific direction. From the elementary theory of strength of materials, the strain energy of a pipe segment of length L under the axial force P and bending moment M is U=

L 2 P ds 0

2AE

+

L M 2 ds 0

2EI

(7.3-2)

where E is the modulus of elasticity, A is the cross sectional area of the pipe, and I is the moment of inertia of the pipe cross section. Neglecting the axial strain energy as small compared to the bending strain energy, we have U=

L M 2 ds 0

2EI

(7.3-3)

Consider an element ds of the piping system in the local coordinate system xy fixed to the elastic center. To obtain Δxx , we apply the force Fx to point O, Fig. 7.3-2. Since the deflection in x direction is required, an auxiliary force Fxa is applied at O, where the deflection is Δxx =

∂U |(F =0) ∂Fxa xa

(7.3-4)

324


From Fig. 7.3-2, the moment about the element ds is M = (Fx + Fxa )y where the moment in the clockwise direction is considered positive. Thus ∂U M ∂M = ds ∂Fxa EI ∂Fxa

Therefore, substituting into Eq. (7.3-4) and using M = (Fx + Fxa )y, yields Δxx =

Fx Ixx EI

(7.3-5)

where Ixx = y 2 ds is the line moment of inertia of the piping system about the x-axis. Deformation Δyx is calculated by applying the force Fy at point O along the y direction and the auxiliary force Fxa along the x-axis, as shown in Fig. 7.3-3. The deformation Δyx is Δyx

M ∂M ∂U = |(Fxa =0) = ds ∂Fxa EI ∂Fxa

The bending moment of forces acting at O about the element ds is M = Fxa y − Fy x Fxa

Fx

o

x

x y ds

y

Figure 7.3-2: Calculation of Δxx .

Figure 7.3-3: Calculation of Δyx .

(7.3-6)


325

Substituting into Eq. (7.3-6) gives Fy Ixy (7.3-7) EI where Ixy = xyds is the line product of inertia of the piping system about the xy-axes. To calculate Δxy , we apply the force Fx at point O and the auxiliary force Fya at O along the y-axis, as shown in Fig. 7.3-4. The deformation Δxy is Δyx = −

Δxy =

∂U M ∂M |(Fya =0) = ds ∂Fya EI ∂Fya

(7.3-8)

The bending moment about ds is M = Fx y − Fya x and the deformation Δxy is Fx Ixy (7.3-9) EI Finally, the deformation Δyy is obtained by applying the force Fy and the auxiliary force Fya at point O along the y-axis, as shown in Fig. 7.3-5. The deformation is ∂U M ∂M Δyy = |(Fya =0) = ds (7.3-10) ∂Fya EI ∂Fya Δxy = −

Figure 7.3-4: Calculation of Δxy .

Figure 7.3-5: Calculation of Δyy .

326


The bending moment of forces about the element ds is M = −(Fy + Fya )x Substituting into Eq. (7.3-10) gives Δyy =

Fy Iyy EI

(7.3-11)

where Iyy = x2 ds is the line moment of inertia of the piping system about the y-axis. Substituting Eqs. (7.3-5), (7.3-7), (7.3-9), and (7.3-11) into Eq. (7.3-1) yields the system of equilibrium equations of the piping system in terms of the forces in the coordinate system fixed to the elastic center as Fx Ixx − Fy Ixy = EIΔx −Fx Ixy + Fy Iyy = EIΔy

(7.3-12)

This is a system of two equations for two unknowns Fx and Fy . Once the configuration of the piping system in xy-coordinates is known, the line moments of inertia are calculated. Given the pipe material and the pipe cross section properties, the constants E and I are known. The values of Δx and Δy are thermal expansions of the piping system in x and y directions, which are calculated knowing the coefficient of thermal expansion, temperature change, and the projection of the piping system between the end points. These known values are substituted into Eqs. (7.3-12) and the system of equations is solved for Fx and Fy . The calculated forces are in the coordinate system fixed to the elastic center. The calculated forces with their moments are transferred to the end where the global coordinate is fixed. The forces Fx and Fy with their moments are transferred to the other end, changing their directions. The following example illustrates the technique discussed in this section. As a note from statics, the line moments of inertia of a pipe of length L in m-n plane, see Fig. 7.3-6, is

Figure 7.3-6: Moment of inertia of a straight pipe.


327

Imm = L¯ n2 3 L Inn = + Lm ¯2 12 Imn = Lm¯ ¯n

(7.3-13)

Equations (7.3-13) show that Imm and Inn are always positive, but Imn may be positive or negative, depending on the coordinates of its elastic center. The algebraic sign of m ¯ and n ¯ must be considered in calculating Imn . Example 1 Consider a two-dimensional piping system in xy-coordinates, as shown in Fig. 7.3-7. The lengths of the pipes are AB = 700 cm and BC = 900 cm. The modulus of elasticity of the pipe material is E = 20600 kN/cm2 , the outside diameter of the pipe is Do = 27.305 cm, and the wall thickness is t = 0.927 cm. The data is for linepipe No. 10, schedule 40, where the moment of inertia of the pipe cross section I = 6690 cm4 . The working temperature of the pipeline system is 385◦ C. Calculate the forces and moments at the end points of the piping system. The coefficient of thermal expansion of the pipe material is obtained from Table 7.3-1. Solution Point A is made free and the global coordinate system is fixed to point A with axes pointing in opposite directions to its thermal expansion directions, as shown in Fig. 7.3-7. The location of the elastic center is x¯ =

700 × 350 + 900 × 0 l1 x¯1 + l2 x¯2 = = 153.125 cm l1 + l2 700 + 900

y¯ =

l1 y¯1 + l2 y¯2 700 × 0 + 900 × 450 = = 253.125 cm l1 + l2 700 + 900 l1 = 700 cm

Fx

X

B A

MAz

253.125 cm

Fy

0 l2 = 900 cm

153.125 cm

C Y

Fx

y

Fy

MCz

Figure 7.3-7: Pipeline of Example 1.

x

328


Table 7.3-1: The coefficient of thermal expansion of a typical line pipe material. Reference temperature is assumed 25 ◦ C. Working temperature Linear expansion ◦ C cm/10 m 25 0 150 1.5305 200 2.1913 250 2.8787 300 3.6047 350 4.0555 400 4.9153 450 5.9718 500 6.7963 550 7.6192 600 8.4817 The line moments of inertia are Ixx = 700 × 253.1252 +

9003 + 900 × (450 − 253.125)2 = 140.48 × 106 cm3 12

7003 + 700 × (350 − 153.125)2 + 900 × 153.1252 = 76.82 × 106 cm3 12 = 700 × (−253.125)(−350 + 153.125) + 900 × (153.125)(450 − 253.125)

Iyy = Ixy

= 62.02 × 106 cm3 From Table 7.3-1 the thermal expansion for the given working temperature of the piping system is 4.7 cm/10 m. Thus, the thermal expansions in x and y directions are Δx = 700 × 4.7/1000 = 3.29 cm Δy = 900 × 4.7/1000 = 4.23 cm Substituting the line moments of inertia and the thermal expansions in Eqs. (7.3-12), yields 140.48 × 106 Fx − 62.02 × 106 Fy = 20600 × 3.29 × 6690 −62.02 × 106 Fx + 76.82 × 106 Fy = 20600 × 4.23 × 6690 Solving for Fx and Fy , gives Fx = 10.22 kN

Fy = 15.84 kN

The calculated forces Fx and Fy are placed at the elastic center O along the positive x and y directions. These forces are transferred to the global coordinates at the end A. The bending moment due to these forces, positive in the

4. Piping Systems in Three Dimensions

329

clockwise direction, is calculated and transferred to the end A. The directions of the forces Fx and Fy are reversed and transferred to the end C. The bending moment of the forces with changed directions is calculated and transferred to the end C. The moments at points A and C are MzA = −10.22 × 253.125 + 15.84 × (700 − 153.125) = 6259.8 kN cm MzC = 10.22 × (900 − 253.125) − 15.84 × 153.125 = 4185.56 kN cm

4

Piping Systems in Three Dimensions

The equilibrium equation of a three-dimensional piping system under thermal expansion in the coordinate system fixed to the elastic center can be established by means of the equilibrium of thermal deflection. In each direction, the total deflection of the piping system due to the reaction forces must be equal to the thermal expansion. For a piping system in three-dimensions the deflection equations are Δxx + Δyx + Δzx = Δx Δyx + Δyy + Δyz = Δy Δzx + Δzy + Δzz = Δz

(7.4-1)

where Δij is the deflection of the piping system in j-direction due to a force in i-direction and Δx , Δy , and Δz are the total thermal expansions of the piping system in x, y, and z direction, respectively. To evaluate the deflection Δij , Castigliano theorem may be used. The strain energy of a pipe segment of length L under the action of an axial force P , bending moment Mb , and torsional moment Mt , is U=

L 2 P ds 0

2AE

+

L Mb2 ds 0

2EI

+

L Mt2 ds 0

2GJ

(7.4-2)

where I and J are the moment of inertia and the polar moment of inertia of the pipe cross section, respectively. In the evaluation of Δij , the strain energy of the axial force may be ignored since it is small compared to the other terms. Since G = E/2(1 + ν) and for the pipe cross section J = 2I, therefore, GJ = EI/(1 + ν). The strain energy of a pipe under the bending moment Mb and the torque Mt is thus U=

L Mb2 ds 0

2EI

+

L Mt2 (1 + ν)ds 0

2EI

(7.4-3)

which suggests that the torsional strain energy of a pipe may be calculated similarly to that of the bending strain energy, but with an equivalent length

330


Figure 7.4-1: Applied forces at elastic center in x-y plane. of (1 + ν)ds. Now, consider a pipe element ds in x-y projection of the piping system, as shown in Fig. 7.4-1. To obtain Δyx , for instance, the real force Fy and the auxiliary force Fxa are applied at the elastic center O and the strain energy of the pipe element ds under the action of Fy and Fxa is calculated. From Castigliano theorem the deformation is ∂U Mb ∂Mb |Fxa =0 = ds for in-plane pipe member (7.4-4) ∂Fxa EI ∂Fxa ∂U Mt ∂Mt = |Fxa =0 = (1 + ν) ds for out-of-plane pipe member ∂Fxa EI ∂Fxa (7.4-5)

Δyx = Δyx

Forces Fy and Fxa produce bending moment Mb = Fxa y − Fy x about element ds, if ds lies in x-y plane. Otherwise, that is, if ds is an element of out-of-plane pipe, forces Fy and Fxa produce a torque Mt = Fxa y − Fy x. Upon substitution of the moment Mb , or the torque Mt , in Eqs. (7.4-4), or Eq. (7.4-5), and integrating over the length of the pipe, we obtain Δyx = −Fy

Ixy EI

(7.4-6)

where Ixy = xyds is the product of inertia of the isometric line of the pipe system in x-y plane. In a similar manner, the remaining deformations in the x-y plane, Δxx , Δyy , and Δxy , may be calculated. Projection of the piping system in the y-z plane provides four deformations Δyy (not to be confused with Δyy in the x-y plane), Δyz , Δzy , and Δzz , and projection in the x-z plane produce another set of four displacements Δxx , Δxz , Δzx , and Δzz , which in general, are Iii Δii = Fi i not summed (7.4-7) EI and Iij Δji = −Fj i = j j not summed (7.4-8) EI


331

Upon substitution into Eqs. (7.4-1), the equilibrium equations of the piping systems in three-dimensions under thermal expansions are obtained as (Ixx + Ixx )Fx − Ixy Fy − Ixz Fz = EIΔx −Ixy Fx + (Iyy + Iyy )Fy − Iyz Fz = EIΔy −Ixz Fx − Iyz Fy + (Izz + Izz )Fz = EIΔz

(7.4-9)

Here, Ixx is the moment of inertia in the x-y plane and Ixx is the moment of inertia in the x-z plane. Similarly, Iyy and Iyy are the line moments of inertia are the line moments of inertia in in the x-y and y-z planes, and Izz and Izz the y-z and x-z planes, respectively. In these equations the forces Fx , Fy , and Fz are unknowns. The quantity I is the moment of inertia of the pipe’s cross section, E is the modulus of elasticity at the design temperature, Δx , Δy , and Δz are the total thermal expansions of the piping system in x, y, and z directions, respectively, which are obtained knowing the coefficient of thermal expansion of the pipe’s material.

Example 2 Consider a piping system in three-dimensions, as shown in Fig. 7.4-2. The lengths of the pipes are AB = l1 = 700 cm, BC = l2 = 900 cm, and CD = l3 = 500 cm. The modulus of elasticity of the pipe material is E = 20600 kN/cm2 , Poisson’s ratio ν = 0.3, the outside diameter of the pipe is Do = 27.305 cm, and the wall thickness is t = 0.927 cm (the moment of inertia of the pipe cross section I = 6690 cm4 ). The working temperature of the pipeline system is 385◦ C. Calculate the forces and moments at the end points of the piping system. Solution Point A is made free and the global coordinate system is fixed to point A with axes pointing in opposite directions to its thermal expansion directions, as shown in Fig. 7.4-2. The projection of the pipeline in three planes, x-y, y-z, and z-x, are considered and the line moments of inertia in each projected plane are calculated. Z

700 cm

X B

A

D

900 cm

C

500 cm

Y

Figure 7.4-2: Pipeline isometric of Example 2.

332

Chapter 7. Thermal Expansion in Piping Systems 700 cm

A

B 440 cm

x

0 900 cm

108.89 cm

C, D y

Figure 7.4-3: Projection of the pipeline in x-y plane.

(A) In the x-y plane Figure 7.4-3 represents the projection of the pipeline in the x-y plane. Pipe CD is perpendicular to this plane and is called the out-of-plane pipe. Its equivalent length is (1 + ν) = 1.3 times its physical length. The location of the elastic center is 700 × 350 + 900 × 0 + 1.3 × 500 × 0 l1 x¯1 + l2 x¯2 + l3 x¯3 = x¯ = = 108.89 cm l1 + l2 + l3 700 + 900 + 1.3 × 500 l1 y¯1 + l2 y¯2 + l3 y¯3 700 × 0 + 900 × 450 + 1.3 × 500 × 900 = 440 cm = l1 + l2 + l3 700 + 900 + 1.3 × 500 The local coordinate system is drawn parallel to the global coordinate system from the elastic center O, and the line moments of inertia in the x-y plane are calculated as y¯ =

9003 + 900 × (450 − 440)2 + 1.3 × 500(900 − 440)2 12 = 333.9 × 106 cm3 7003 = + 700(350 − 108.89)2 + 900 × 108.892 + 1.3 × 500 × 108.892 12 = 87.66 × 106 cm3 = 700(−350 + 108.89)(−440) + 900(450 − 440)(108.89) +1.3 × 500(900 − 440)(108.89) = 107.8 × 106 cm3

Ixx = 700 × 4402 +

Iyy Ixy

(B) In the y-z plane Figure 7.4-4 is the projection of the pipeline in the y-z plane. Pipe AB is perpendicular to this plane and is called the out-of-plane pipe. Its equivalent length is 1.3 times its physical length. The location of the elastic center is


333

A, B

0 900 cm

z

54.113 cm

529.87 cm

D

C

y 500 cm

Figure 7.4-4: Projection of the pipeline in y-z plane. y¯ =

1.3 × 700 × 900 + 900 × 450 + 500 × 0 = 529.87 cm 1.3 × 700 + 900 + 500

z¯ =

1.3 × 700 × 0 + 900 × 0 + 500 × 250 = 54.113 cm 1.3 × 700 + 900 + 500

The local coordinate system is drawn parallel to the global coordinate system from the elastic center O and the line moments of inertia in the y-z plane are calculated as = 1.3 × 700 × 54.1132 + 900 × 54.1132 + Iyy

5003 + 500(250 − 54.113)2 12

= 34.902 × 106 cm3 9003 + 900(529.87 − 450)2 12 +500 × 529.872 = 351.54 × 106 cm3 = 1.3 × 700(−54.113)(−900 + 529.87) + 900(529.87 − 450)(−54.113) +500(250 − 54.113)(529.87) = 66.23 × 106 cm3

Izz = 1.3 × 700(900 − 529.87)2 + Iyz

(C) In the x-z plane Figure 7.4-5 is the projection of the pipeline in the x-z plane. Pipe BC is perpendicular to this plane and is called the out-of-plane pipe and its equivalent length is 1.3 times its physical length. The location of the elastic center is x¯ =

700 × 350 + 1.3 × 900 × 0 + 500 × 0 = 103.375 cm 700 + 1.3 × 900 + 500

z¯ =

700 × 0 + 1.3 × 900 × 0 + 500 × 250 = 52.743 cm 700 + 1.3 × 900 + 500

The local coordinate system is fixed to the elastic center O, parallel to the global coordinate system, and the line moments of inertia are calculated as

334

Chapter 7. Thermal Expansion in Piping Systems D z 103.375 cm

500 cm

x

0 A

B, C

52.743 cm

700 cm

Figure 7.4-5: Projection of the pipeline in x-z plane. Ixx = 700 × 52.7432 + 1.3 × 900 × 52.7432 +

Izz

Ixz

5003 + 500(250 − 52.743)2 12

= 35.07 × 106 cm3 7003 = +700(350 − 103.375)2 + 1.3 × 900 × 103.3752 + 500 × 103.3752 12 = 89 × 106 cm3 = 700(−350 + 103.375)(−52.743) + 1.3 × 900(103.375)(−52.743) +500(250 − 52.743)(103.375) = 12.92 × 106 cm3

Thermal expansions in three-dimensions, using Table 7.3-1, are Δx =

700 4.7 = 3.29 cm 1000

Δy =

900 4.7 = 4.23 cm 1000

Δz =

500 4.7 = 2.35 cm 1000

Substituting the line moments of inertia and the thermal expansions in the deflection equations gives [(333.9 + 35.07)Fx − 107.8Fy − 12.92Fz ] × 106 = 20600 × 3.29 × 6690 [−107.8Fx + (87.66 + 34.902)Fy − 66.23Fz ] × 106 = 20600 × 4.23 × 6690 [−12.92Fx − 66.23Fy + (351.54 + 89)Fz ] × 106 = 20600 × 2.35 × 6690 Solving for the forces Fx , Fy , and Fz yields Fx = 4.155 kN

Fy = 9.724 kN

Fz = 2.43 kN

5. Pipelines with Large Radius Elbows

335

These forces are transferred to the global coordinates at the end A, and they are transferred with opposite directions to the other end at D. The moments of the forces acting at points A and D are MxA = (900 − 529.87) × 2.43 − 54.113 × 9.724 = 373.221 kN cm MxD = −529.87 × 2.43 + (500 − 59.113) × 9.724 = 3048.22 kN cm MyA = 52.743 × 4.155 − (700 − 103.375) × 2.43 = −1230.65 kN cm MyD = −(500 − 52.743) × 4.155 + 103.375 × 2.43 = −1607.15 kN cm MzA = −440 × 4.155 + (700 − 108.89) × 9.724 = 3919.75 kN cm MzD = −108.89 × 9.724 + (900 − 440) × 4.155 = 852.45 kN cm The magnitude of forces of Example 2 may be compared with those of Example 1, to show that while the numerical data of two examples are identical, the pipeline in three-dimensions develops end forces of smaller magnitudes and thus it is more flexible. The magnitude of force Fx is reduced by 54.3% and the magnitude of force Fy is reduced by 38.6%.

5

Pipelines with Large Radius Elbows

Large radius elbows may be used in the design of a piping system. A twodimensional piping system with a large radius elbow connection is shown in Fig. 7.5-1. The calculation procedure and the deflection equations are essentially the same as discussed, except that the moments of inertia of the elbow must be computed. Consider an elbow of mid-radius R in the m-n plane, as shown in Fig. 7.5-2. The coordinates of the elastic center of the elbow are calculated: π/2 0

(¯ n − n)Rdθ = 0

Figure 7.5-1: Pipeline with large radius elbow.

(7.5-1)

336


Figure 7.5-2: Position of the elastic center (EC) of the elbow.

Figure 7.5-3: Product of inertia of the elbow. Substituting n = R sin θ and computing the integral, the coordinates of the elastic center are obtained as 2R π

m ¯ =n ¯=

(7.5-2)

To calculate the moment of inertia of the elbow about the elastic center we may use the parallel axis law. First, observe that the line moment of inertia of the elbow with respect to the m-axis is π/2

Im =

n2 Rdθ =

0

π/2

R3 sin2 θdθ =

0

πR3 4

(7.5-3)

From the parallel axis law Im = IEC +

πR 2 (¯ n) 2

(7.5-4)

Solving for IEC gives IEC = R3 (

2 π − ) = 0.1488R3 4 π

(7.5-5)

To compute the product of inertia Imn , consider an elbow in the m-n plane, as shown in Fig. 7.5-3. In this case we have π/2

Imn =

(a + R cos θ)(b + R sin θ)Rdθ 0

(7.5-6)


337

In evaluation of Eq. (7.5-6) a proper algebraic sign in the coordinate system must be observed. The moments of inertia and the product of inertia of an elbow are computed from Eqs. (7.5-3) and (7.5-6). Since, however, large radius elbows are flexible in bending, a correction factor must be considered. von Kármán bending rigidity factor K may be used as the adjustment factor. It is [2] K=

12λ2 + 10 12λ2 + 1

(7.5-7)

where

tR (7.5-8) r2 with t being the wall thickness of the pipe and r being the mean radius of the pipe’s cross section. There are other types of correction factors, with details provided in the references at the end of this chapter. The elastic center method described in this section has some limitations in applications. Some of these limitations may be removed and some are inherent in the method. In the derivation of Eqs. (7.4-9) the following assumptions are made: λ=

1. The weight of the pipeline is excluded. The reaction forces due to the weight of the pipeline can be calculated and superposed on the thermal forces at the supports. 2. The temperature of the pipeline is assumed to be uniform and constant. 3. Only one branch of the pipeline is considered. 4. The members of the system are assumed to be parallel to one of the coordinate axes. 5. The pipes are assumed to be clamped at the supports. The effect of elastic supports may be included in the deflection equations (7.4-9) provided that they are placed along the pipe member at the support. The deflection of the elastic support is then proportional to its spring constant and the reaction force at the support. For example, if a support exhibits elastic response in the x direction and its spring constant in x direction is kx , the proportional displacement of support in x direction is δx = Fx /kx . This displacement must be incorporated in the deflection equations (7.4-9) in x direction. The unknown forces Fx and, correspondingly, Fy and Fz , which are obtained by solution of Eqs. (7.4-9), are thus eliminated. As a general rule, the reaction forces and moments of a pipeline system increase if pipes of larger wall thickness are used, and the reaction forces and moments decrease if the length of the pipeline is increased. The reason

338


is simple, as in the first case the moment of inertia of a pipe cross section increases for larger wall thickness and this increases the numerical value of the right-hand side of Eqs. (7.4-9). On the other hand, if the length of the pipeline is increased, the length moment of inertia increases causing the left-hand side of Eqs. (7.4-9) to increase and, therefore, decrease the forces. The reason to design a loop of pipe to increase the total length in a pipeline system in order to decrease the reaction forces and moments is thus clear. A loop of pipe is made of a number of pipe large radius elbows. Loops, in addition to increase the total length of a pipeline, due to the large flexibility coefficient of elbows, efficiently increase the line moment of inertia of the pipeline system. Example 3 Reconsider Example 1, but with a large radius elbow of mean radius R = 100 cm connecting pipe elements AB = 600 cm and CD = 800 cm, as shown in Fig. 7.5-4. The other numerical data are identical with those of Example 1. Find the forces and moments at the ends A and D. Solution The fixed end A, arbitrarily chosen, is made free and the global coordinate axes are selected in opposite direction of movement of point A, as shown in Fig. 7.5-4. To apply the von Kármán coefficient for the equivalent length of elbows we find 0.927 × 100 tR = = 0.533 2 r [(27.305 − 0.927)/2]2 12λ2 + 10 K= = 3.05 12λ2 + 1

λ=

600 cm

B

X R=100 cm

A

C 222.133 cm

0

x 136.98 cm

Y

800 cm

y

D

Figure 7.5-4: Pipeline isometric of Example 3.


339

The line moment of inertia of the elbow with respect to its elastic center is IEC = R3 (π/4 − 2/π) = 148778.39 cm3 The equivalent length of the elbow is πR π × 100 = 3.05 × = 479.1 cm 2 2 The elastic center of the elbow is located at 2R/π = 0.6366R with respect to the vertical line drawn from point B. The coordinates of the elastic center of the pipeline are K×

600 × 400 + 479.1 × (1 − 0.6366) × 100 + 800 × 0 = 136.98 cm 800 + 600 + 479.1 600 × 0 + 479.1 × (1 − 0.6366) × 100 + 800 × 500 y¯ = = 222.133 cm 800 + 600 + 479.1 The local coordinate axes are transferred to the elastic center and the line moments of inertia of the pipeline about the axes x and y are x¯ =

Ixx = 600×222.1332 +148778.39×3.05+479.1×[222.133−(1−0.6366)×100]2 8003 + 800(500 − 222.133)2 = 150.78 × 106 cm3 12 6003 Iyy = + 600 × (400 − 136.98)2 + 148778.39 × 3.05 12 +479.1 × [136.98 − 100 × (1 − 0.6366)]2 + 800 × 136.982 = 79.752 × 106 cm3 +

The position of the elbow with respect to the local coordinates axes fixed to the elastic center is shown in Fig. 7.5-5. The line product of inertia of the elbow is π/2

Ixy =

0

(R cos θ + a)(−b − R sin θ)Rdθ

= −[

abRπ R3 + R2 (a + b) + ] 2 2

Figure 7.5-5: Position of the elbow with respect to the local coordinates.

340


Note that the elastic center of the elbow is located at the fourth quarter, where its x-coordinate is positive and its y-coordinate is negative. The line product of inertia of the pipeline, with the values R = 100 cm, a = 36.98 cm, and b = 122.133 cm, is Ixy = 600 × (−222.133)(−400 + 136.98) + 800 × (136.98)(500 − 222.133) 1003 36.98 × 122.133 × π × 100 −3.05 × [ + 1002 (36.98 + 122.133) + ] 2 2 = 63.407 × 106 cm3 The thermal expansions in the x and y directions are Δx =

700 4.7 = 3.29 cm 1000

900 4.7 = 4.23 cm 1000 Substituting into the deflection equations (7.3-12) yields Δy =

(150.78Fx − 63.407Fy ) × 106 = 20600 × 6690 × 3.29 (−63.407Fx + 79.752Fy ) × 106 = 20600 × 6690 × 4.23 Solving for Fx and Fy gives Fx = 9.135 kN

Fy = 14.57 kN

The bending moments at ends A and D are MzA = −4.135 × (222.133) + 14.57 × (700 − 136.98) = 6174 kN cm MzD = 4.135 × (900 − 222.133) − 14.57 × 136.98 = 4196.5 kN cm The effect of the large radius elbow is to reduce the reaction forces. The reduction in this example is 10.62% in the x direction and 8.02% in the y direction. Example 4 Consider a three-dimensional piping system with large radius elbows, as shown in Fig. 7.5-6. The pipe is No. 10 schedule 40 with modulus of elasticity of E = 20600 kN/cm2 , Poisson’s ratio ν = 0.3, and the operating temperature of 350◦ C. The outside diameter of the pipe is 27.305 cm and the pipe thickness is 0.927 cm. The moment of inertia of the cross section is I = 6690 cm4 . The coefficient of thermal expansion of the pipe material at 350◦ C is assumed to be α = 4.37 cm/10 m. The reaction forces and moments at fixed points A and F are required.


341

Figure 7.5-6: Pipe isometric in three-dimensions. Solution The fixed end A, arbitrarily chosen, is made free and the global coordinate axes are selected in opposite direction of movement at point A, as shown in Fig. 7.5-6. To apply the von Kármán coefficient for the equivalent length of the elbows we find 0.927 × 100 tR = = 0.533 r2 [(27.305 − 0.927)/2]2 12λ2 + 10 K= = 3.05 12λ2 + 1

λ=

Three projections of the pipeline isometric are sketched in the x-y, x-z, and y-z planes and the calculations are made in each plane as follows: (A) In the x-y plane The length of the elbows = πR/2 = 100π/2 = 157 cm, AB = 600 cm. The equivalent length of elbows BC = KπR/2 = 3.05 × 157 = 479.1 cm, CD = 800 cm. The equivalent length of the out-of-plane elbow is DE = 1.15 × 157 = 180.65 cm and the equivalent length of the out-ofplane pipe is EF = 1.3 × 400 = 520 cm. The elastic center of the elbow BC = t¯ = s¯ = 2R/π = 0.6366R. IEC = R3 (

π 2 − ) = 0.1488R3 4 π

Thus the location of the elastic center in the x-y plane is 600 × 400 + 479.1 × (1 − 0.6366) × 100 600 + 479.1 + 800 + 180.65 + 520 +800 × 0 + 180.65 × 0 + 500 × 0 = 99.78 cm ” 600 × 0 + 479.1 × (1 − 0.6366) × 100 + 800 × 500 y¯ = 600 + 479.1 + 800 + 180.65 + 520 +180.65 × (0.6366 × 100 + 900) + 520 × 1000 = 430.86 cm ”

x¯ =

342


where the equivalent length of the out-of-plane elbow is calculated as 1.15 × πR/2 = 180.65 cm. The moments of inertia are Ixx = 600 × 430.862 + 0.1488 × 1003 × 3.05 8003 12 +800 × (500 − 430.86)2 + 180.65 × (0.6366 × 100 + 900 − 430.86)2 +0.1488 × 1003 × 1.15 + 520 × (1000 − 430.86)2 = 452, 790, 862 cm3 6003 = + 600 × (400 − 99.78)2 + 0.1488 × 1003 × 3.05 12 +479.1×[99.78 − (1 − 0.6366) × 100]2 + 800 × 99.782 + 180.65 × 99.782 +520 × 99.78 = 89, 401, 815 cm3 +479.1 × [430.86 − (1 − 0.6366) × 100]2 +

Iyy

The product of inertia is π/2

Ixy = =(

0

(−a + R cos θ)(−b − R sin θ)Rdθ

πR R3 )ab − (b − a)R2 − 2 2

Thus we have a = 100 − 99.78 = 0.22 cm,

b = 430.86 − 100 = 330.86 cm

and Ixy = 600 × (−430.86)(−400 + 99.78) − 3.05 × [−(

100π ) × 0.22 × 330.86 2

1003 ] + 800(500 − 430.86)(99.78) 12 +180.65 × (0.6366 × 100 + 900 − 430.86)(99.78) +520(1000 − 430.86)(99.78) = 110, 690, 085 cm3 +(330.86 − 0.22) × 1002 +

(B) In the x-z plane AB = 600 cm, BC = 1.15 × πR/2 = 180.65 cm, CD = 1.3 × 800 = 1, 040 cm, DE = 1.15 × πR/2 = 180.65 cm, EF = 400 cm. The location of the elastic center in x-z plane is 600 × 400 + 180.65 × (1 − 0.6366) × 100 600 + 180.65 + 1040 + 180.65 + 400 +1040 × 0 + 180.65 × 0 + 400 × 0 = 102.68 cm ” 600 × 0 + 180.65 × 0 + 1040 × 0 y¯ = 600 + 180.65 + 1040 + 180.65 + 400 +180.65 × (1 − 0.6366) × 100 + 400 × 300 = 52.71 cm ”

x¯ =


343

The moments of inertia are Ixx = 600×52.712 + 180.65×52.712 + 1040×52.712 + 0.1488×1003 × 1.15 +180.65 × [52.71 − (1 − 0.6366) × 100]2 + 400 × (300 − 52.71)2 4003 + = 35, 072, 193 cm3 12 6003 Izz = + 600 × (400 − 102.68) + 0.1488 × 1003 × 1.15 12 +180.65 × [102.68 − (1 − 0.6366) × 100]2 + 1040 × 102.682 +180.65 × 102.682 + 400 × 102.682 = 89, 092, 478 cm3 Ixz = 180.65 × (−52.71)[102.68 − (1 − 0.6366) × 100] +600 × (−52.71)(−400 + 102.68) + 1040 × (−52.71)(102.68) +180.65 × [−52.71 + (1 − 0.6366) × 100] × (102.68) +400 × (300 − 52.71)(102.68) = 12, 995, 641 cm3

(C) In the y-z plane AB = 1.3 × 600 = 780 cm, BC = 1.15πR/2 = 180.65 cm, CD = 800 cm, λ = 0.533, K = 3.05, DE = πR/2 = 479.1 cm, EF = 400 cm. The location of the elastic center in the y-z plane is 780 × 1000 + 180.65 × [1000 − (1 − 0.6366) × 100] 780 + 180.65 + 800 + 479.1 + 400 +800 × 500 + 479.1 × (1 − 0.6366) × 100 + 400 × 0 = 519.54 cm ” 780 × 0 + 180.65 × 0 + 800 × 0 + 479.1 × (1 − 0.6366) × 100 z¯ = 780 + 180.65 + 800 + 479.1 + 400 +400 × 300 = 52.05 cm ” The moments of inertia are y¯ =

= 780×52.052 + 180.65×52.052 + 800 × 52.052 + 0.1488 × 1003 × 3.05 Iyy

Izz

+479.1 × [52.05 − (1 − 0.6366) × 100]2 + 400 × (300 − 52.05)2 4003 + = 35, 267, 056 cm3 12 = 780 × (1000 − 519.54)2 + 0.1488 × 1003 × 1.15 8003 +180.65 × [1000 − (1 − 0.6366) × 100 − 519.54]2 + 12 +800 × (500 − 519.54)2 + 0.1448 × 1003 × 3.05 +479.1 × [519.54 − (1 − 0.6366) × 100]2 + 400 × 519.542 = 479, 115, 626 cm3

344


The product of inertia is π/2

Iyz =

0

Iyz = (

(a − R cos θ)(b + R sin θ)Rdθ

πR R3 )ab − (b − a)R2 − 2 2

Thus a = 100 − 52.05 = 47.95 cm, b = 519.54 − 100 = 419.54 cm, and Iyz = 780 × (−52.05)(−1000 + 519.54) + 180.65 × [−1000 + (1 − 0.6366) ×100 + 519.54] × (−52.05) + 800 × (−52.05)(−500 + 519.54) π × 100 +3.05 × [ × 47.95 × 419.54 − (419.54 − 47.95) × 1002 2 1003 − ] + 400 × (300 − 52.05) × 519.54 = 71, 175, 902 cm3 2 At 350◦ C the value of thermal expansion is 4.37 cm/10 m. Therefore, the total thermal deformations in each direction are 700 = 3.06 cm 1000 1000 ΔY = 4.37 = 4.37 cm 1000 500 Δz = 4.37 = 2.19 cm 1000 Δx = 4.37

The piping system deflection equations in three-dimensions are )Fx − Iyx Fy − Izx Fz = EIΔx (Ixx + Ixx −Ixy Fx + (Iyy + Iyy )Fy − Izy Fz = EIΔy −Ixz Fx − Iyz Fy + (Izz + Izz )Fz = EIΔz

Substituting for the moments of inertia Iij , the modulus of elasticity, the moment of inertia of the cross sections, and the thermal expansions Δi in the deflection equations yields 487, 863, 055Fx − 110, 690, 085Fy − 12, 995, 641Fz = 421, 710, 840 −110, 690, 085Fx + 124, 668, 871Fy − 71, 175, 902Fz = 602, 247, 180 −12, 995, 641Fx − 71, 175, 902Fy + 568, 208, 104Fz = 301, 123, 590 Upon solution, the three forces in global coordinates at point A become Fx = 2.70 kN Fy = 7.94 kN Fz = 1.25 kN


345

Figure 7.5-7: Pipe isometric in x-y plane.

Figure 7.5-8: Pipe isometric in x-z plane. At point F opposite and equal forces are applied. From Fig. 7.5-7 the moments about the z-axis at points F and A are Mz |F = 2.7 ∗ (1000 − 430.86) − 7.94 × 99.78 = 744.4 kN cm Mz |A = −2.7 ∗ 430.86 + 7.94 × (700 − 99.78) = 3, 602.4 kN cm From Fig. 7.5-8 the moments about the y-axis are My |F = −2.7 × (500 − 52.71) + 1.25 × 102.68 = −1, 079.3 kN cm My |A = 2.7 × 52.71 − 1.25 × (700 − 102.68) = −604.3 kN cm From Fig. 7.5-9 the moments about the x-axis are Mx |F = −7.94 × (500 − 52.05) + 1.25 × 519.54 = −2, 907.3 kN cm Mx |A = 7.94 × 52.05 − 1.25 × (1000 − 519.54) = −187.3 kN cm The results show that for a piping system with fixed ends, the forces and moments may be large if 1. The pipe elements are short, and thus the whole piping system is short. 2. The moment of inertia of the pipe’s cross section is large. This means that the piping system is made of pipes of large outside radius, or the wall thickness of the pipe is large. 3. The working temperature of the piping system is high.

346


Figure 7.5-9: Pipe isometric in y-z plane.

There are many methods in practice to reduce the induced thermal forces and moments in a piping system. The proper design of flexible joints at the end connections of the piping system will reduce the induced forces. The proper design of loops at the proper locations along the piping system has similar effect. Also, of common practice is the pre-defined end displacements at cold condition of the piping system, so that at working temperature the pre-designed end displacements help to reduce the induced thermal forces and moments.

6

Problems

1. Consider a three-dimensional pipeline configuration, where instead of fixed ends, the ends are supported by flexible joints. The flexible joints are assumed to work only along the pipeline axes. Modify the three-dimensional equilibrium equations of the pipeline system to include this type of boundary conditions. 2. Consider the pipeline isometric, as shown in Fig. 7.6-1, under temperature change ΔT = 400◦ C. The pipe is No. 8, schedule 30, with the modulus of elasticity E = 20600 kN/cm2 . The radius of elbows is R = 150 cm. Find the pipeline reaction forces and moments at points A and H. 3. Reconsider Problem 2, where both ends are supported by axial flexible joints with axial spring constant ka = 600 N/m. Calculate the pipeline reaction forces and moments. Discuss the results.

6. Problems

347 D

z

y

B

x

500cm

C

700cm

A

E

m

0c

60

F

G

400cm

H

Figure 7.6-1: Pipeline isometric, Problem 2. 4. Reconsider Problem 2 and find the reaction forces and moments when point I (the middle point of pipe CD) is fixed. Discuss the results.

Bibliography [1] Kellogg, M.W., Design of Piping Systems, 2nd edition, Wiley, New York, 1956. [2] Crocker, S. and King, R.C., (Eds), Piping Handbook, 5th edition, McGrawHill, New York, 1967. [3] Spotts, M.F., Mechanical Design and Analysis, Prentice Hall, Englewood Cliffs, New Jersey, 1965.

349

Chapter 8 Coupled and Generalized Thermoelasticity

A structure under thermal shock load, when the period of shock is of the same order of magnitude as the lowest natural frequency of the structure, should be analyzed through the coupled form of the energy and thermoelasticity equations. Analytical solutions of this class of problems are mathematically complex and are limited to those of an infinite body or a half-space, where the boundary conditions are simple. This chapter begins with the analytical solutions of a number of classical problems of coupled thermoelasticity for an infinite body and a half-space. Coupled thermoelasticity problem for a thick cylinder is discussed when the inertia terms are ignored. The generalized thermoelasticity problems for a layer, based on Green–Naghdi, Green–Lindsay, and Lord–Shulman models are discussed, when the analytical solution in the space domain is found. The solution in the time domain is obtained by numerical inversion of Laplace transforms. The generalized thermoelasticity of thick cylinders and spheres, in a unified form, is discussed, and problems are solved analytically in the space domain, while the inversion of Laplace transforms are carried out by numerical methods.

1

Introduction

Most problems in thermoelasticity are based on the assumption that the temperature field is governed by the first law of thermodynamics for a rigid heat conductor. Coupled problems of thermoelasticity take into account the time rate of change of the first invariant of the strain tensor in the first law of thermodynamics causing the dependence between the temperature and strain fields, and thus creating the coupling between elastic and thermal fields. This R.B. Hetnarski, M.R. Eslami, Thermal Stresses – Advanced Theory and Applications, Solid Mechanics and its Applications 158, c Springer Science + Business Media B.V. 2009

351

352

Chapter 8. Coupled and Generalized

situation occurs when the time rate of change of thermal boundary conditions, or other imposed thermal loads, is large enough to excite the thermal stress wave propagation. The mathematical treatment of coupled thermoelasticity problems by analytical methods is rather complicated, so that only very basic problems have been treated in the literature. Until recently, only problems of the infinite space, half-space, and the layer were analytically solved. Numerical methods, such as the finite element and boundary element techniques, have also been applied to solve this class of problems. A number of analytical solutions of onedimensional coupled thermoelasticity problems in rectangular and cylindrical geometries using Laplace transform have been published. In this chapter, the coupled thermoelasticity problems in the infinite space, the half-space, and the layer are treated by analytical methods. Moreover, the theoretical considerations are extended to some basic engineering problems, such as thick cylinders.

2

Governing Equations of Coupled Thermoelasticity

In Chapter 2, an energy equation for coupled thermoelasticity problems was derived. The general governing equations of coupled thermoelasticity are: The equation of motion ui (8.2-1) σij,j = ρ¨ The linear strain-displacement relations 1 ij = (ui,j + uj,i ) 2

(8.2-2)

Hooke’s law for a linear homogeneous isotropic thermoelastic material σij = 2μij + λkk δij − (3λ + 2μ)α(T − T0 )δij

(8.2-3)

The energy equation kT,ii − ρcT˙ − αT0 (3λ + 2μ)˙ii = −R

(8.2-4)

The first three equations can be combined to give the equation of motion in terms of displacement components as μui,kk + (λ + μ)uk,ki − (3λ + 2μ)αT,i = ρ¨ ui

(8.2-5)

Equations (8.2-4) and (8.2-5) are the displacement-temperature equations of coupled thermoelasticity for a solid elastic continuum. The initial and boundary conditions for thermal and mechanical loads must be stated along with the energy equation and the equation of motion to fully describe the problem.

2. Governing Equations

353

The thermal boundary conditions are satisfied through either of the following equations T = Ts T,n + aT = b

on ∂D for t > t0 on ∂D for t > t0

(8.2-6) (8.2-7)

where a and b are either constants or given functions on the boundary. These conditions are related to the specified temperature, convection, or radiation on the boundary. The mechanical boundary conditions may be specified through the traction vector on the boundary. The traction components are related to the stress tensor through Cauchy’s formula tni = σij nj

on ∂D for t > t0

(8.2-8)

where tni are the prescribed traction components on the boundary whose outer normal vector is n. This boundary condition relates the traction vector on the boundary to the stress components. On the other hand, in terms of the displacement components, Eq. (8.2-8) may be used to relate the traction vector on the boundary to the displacements. Using the constitutive law for linear thermoelasticity along with the strain displacement relations, the traction components are presented in terms of the displacement components as tni = μ(ui,j + uj,i )nj + λuk,k ni − (3λ + 2μ)α(T − T0 )ni

(8.2-9)

Either of Eqs. (8.2-8) and (8.2-9) relates the traction boundary conditions to the stress or displacement fields. Another form of the boundary condition is the prescribed displacement on the boundary as on ∂D (8.2-10) ui = u¯i where u¯i is the prescribed displacement on the boundary. The general governing equations of coupled thermoelasticity may also be written in terms of the stress tensor and temperature. In the following, we are to obtain the dynamic stress-temperature compatibility equations of BeltramiMichell type that together with the stress-temperature energy equation are necessary (but not sufficient) to formulate an initial-boundary problem of coupled thermoelasticity. A complete (necessary and sufficient) stress-temperature formulation of coupled thermoelasticity may be obtained as given by Ignaczak [1] (see page 301 in [1] when t1 = t0 = 0). Using Eq. (8-2.2), we can write Eq. (8-2.5) as [2] μij,kk + (λ + μ)kk,ij − (3λ + 2μ)αT,ij = ρïj

(8.2-11)

Substituting for the strain tensor from Eq. (8.2-3) 2μij = σij −

λ δij σkk + 2μα(T − T0 )δij 3λ + 2μ

(8.2-12)

354


The general governing equation is then obtained as 2(λ + μ) λ σkk,ij − δij σss,kk − 2βT,ij 3λ + 2μ 3λ + 2μ 1 λ +[2μαT,kk δij + 6(λ + μ)αT,ij ] = 2 (¨ σij − ¨kk ) δij σ c2 3λ + 2μ 1 +2μα 2 T¨δij (8.2-13) c2 σij,kk +

where c22 = μ/ρ. Contraction of Eq. (8.2-13) gives ¯ 21 σkk + 2

(3λ + 2μ)α (4μT,kk − 3ρT¨) = 0 λ + 2μ

where ¯ 21 = ∇2 − 2

1 ∂2 c21 ∂t2

c21 =

λ + 2μ ρ

(8.2-14)

(8.2-15)

Equations (8.2-13) and (8.2-14) can be combined to give the final form of the dynamic stress-temperature equations of Beltrami-Michell type 2(λ + μ) 1 1 λ ¨kk σkk,ij + ( 2 − 2 ) δij σ 3λ + 2μ c2 c1 3λ + 2μ 3λ + 2μ 5λ + 4μ +2μα(T,ij + T,kk δij ) − αρδij T¨ = 0 λ + 2μ λ + 2μ ¯ 22 σij + 2

where ¯ 22 = ∇2 − 2

1 ∂2 c22 ∂t2

(8.2-16)

(8.2-17)

The energy equation (8.2-4), which is in terms of the strain rate, is rewritten for the stress rate using Eq. (8.2-3) as kT,ii − (ρc + 3γα)T˙ −

γ σ˙ ii = −R (3λ + 2μ)

(8.2-18)

where γ = (3λ + 2μ)αT0 . Equations (8.2-16) and (8.2-18) constitute the stresstemperature equations of coupled thermoelasticity. These, in general, are not sufficient to formulate a coupled initial-boundary value problem. A complete stress-temperature formulation of the coupled thermoelasticity, within a generalized thermoelasticity with two relaxation times, is given by Ignaczak [1].

3

Coupled Thermoelasticity for Infinite Space

Returning to displacement formulations, and introducing the displacement vector U = ui + vj + wk, the vectorial form of Eqs. (8.2-4) and (8.2-5) is ˙ = −R k∇2 T − ρcT˙ − αT0 (3λ + 2μ) div U

(8.3-1)

3. Coupled Thermoelasticity for Infinite Space

355

and ¨ μ∇2 U + (λ + μ) grad divU − (3λ + 2μ)α grad T = ρU

(8.3-2)

The displacement can now be written as the sum of an irrotational and a potential part as given by Nowacki [2] U = grad ψ + curl Ω

(8.3-3)

where ψ is a scalar potential, and Ω is a vector potential. We may substitute Eq. (8.3-3) into Eqs. (8.3-1) and (8.3-2) to arrive at 1 ¨ (3λ + 2μ) ψ= α(T − T0 ) c21 λ + 2μ 1¨ ∇2 Ωi − 2 Ω i = 1, 2, 3 i = 0 c2 k∇2 T − ρcT˙ − αT0 (3λ + 2μ)∇2 ψ˙ = −R

∇2 ψ −

(8.3-4)

where c1 and c2 are the speed of propagation of the elastic longitudinal wave and the speed of the shear wave, respectively. Elimination of T between the first and the last of Eqs. (8.3-4) results in a single equation for ψ, namely (∇2 −

m1 R 1 ∂ 1 ∂2 β 2 T0 )(∇2 − 2 2 )ψ − ∇2 ψ˙ = − κ ∂t c1 ∂t (λ + 2μ)k k

(8.3-5)

with

β k β = (3λ + 2μ)α κ= λ + 2μ ρc and the equation for the components of vector Ω remains as m1 =

∇2 Ωi −

1¨ Ωi = 0 c22

i = 1, 2, 3

(8.3-6)

(8.3-7)

For quasi-steady problems, when the variation of temperature with respect to time is slow and the inertia effects are neglected, the system of equations reduces to αT0 (3λ + 2μ) 2 ˙ R 1 ∇2 T − T˙ − ∇ ψ=− κ k k α(3λ + 2μ) 2 ∇ψ= (T − T0 ) λ + 2μ

(8.3-8)

The function ψ can be eliminated from Eqs. (8.3-8) and the equation for heat conduction takes the form R ∇2 T − mT˙ = − k

(8.3-9)

356


where m=

1 α2 T0 (3λ + 2μ)2 + κ k(λ + 2μ)

(8.3-10)

When the temperature distribution is stationary, the dynamic effect of temperature field vanishes, and the coupling term in the governing equations vanishes. In this case, two independent equations for temperature distribution and displacement potential are R k α(3λ + 2μ) ∇2 ψ = (T − T0 ) (λ + 2μ)

∇2 T = −

(8.3-11)

The first equation is the well known differential equation of heat conduction in solids and the second equation involves the displacement potential for thermoelastic problems. A general solution of Eqs. (8.3-4) corresponding to Ω = 0 and R = 0 can be obtained in the following way. We take ψ in the form, see Kovalenko [3], ¯ y, z)ept ψ = ψ(x,

(8.3-12)

Substituting Eq. (8.3-12) into Eq. (8.3-5) we obtain (∇2 + δ12 )(∇2 + δ22 )ψ¯ = 0

(8.3-13)

2(1 − C) 1 + C 2 1/2 −p2 1+C ) ± [1 − +( )] } {(1 + 2 2c1 η η η

(8.3-14)

where δ12 , δ22 = and C=

T0 (3λ + 2μ)2 α2 ρ2 c21 c

η=

κp c21

(8.3-15)

Therefore, ψ takes the form ψ=

2

ψ¯j ept

(8.3-16)

j=1

where ψ¯j satisfy the equation (∇2 + δj2 )ψ¯j = 0

j = 1, 2

(8.3-17)

Once the solution for ψ¯j is obtained, the displacement vector from Eq. (8.3-3) is U=

2 j=1

grad ψ¯j ept

(8.3-18)

4. Variable Heat Source

357

The temperature distribution is obtained from the first of Eqs. (8.3-4) T − T0 = −

2 ρc21 p2 ( 2 + δj2 )ψ¯j ept α(3λ + 2μ) j=1 c1

(8.3-19)

The strain components are related to the displacement components kk = div U =

2

∇2 ψ¯j ept = −

j=1

2

δj2 ψ¯j ept

(8.3-20)

j=1

and finally from the stress-strain relations the stresses become σkl = 2μ[kl + δkl

2

(δj2 +

j=1

4

ρp2 ¯ pt )ψj e ] 2μ

(8.3-21)

Variable Heat Source

Let us consider a body of finite dimensions subjected to a rate of internal heat generation R(x, t) per unit volume and unit time as given by [4,5] R(x, t) = Q(t) cos

x L

(8.4-1)

where L is a characteristic length. It is required to find the field of displacement and temperature if the body is initially at rest and at uniform temperature. It is further assumed that the constraint on the displacement field is such that the displacement in x-direction is allowed while displacements in y and z directions are prevented, that is u = u(x, t)

v=w=0

(8.4-2)

The governing equations reduce to ∂T ∂2u ∂2T + R(x, t) = ρc + (3λ + 2μ)αT 0 ∂x2 ∂t ∂x∂t ∂σxx ∂2u =ρ 2 ∂x ∂t σxx = (λ + 2μ)xx − (3λ + 2μ)α(T − T0 ) σyy = σzz = λxx − (3λ + 2μ)α(T − T0 ) ∂u xx = ∂x k

(8.4-3) (8.4-4) (8.4-5) (8.4-6)

All other stresses and strains are zero. Eliminating xx from Eqs. (8.4-5) and (8.4-6) and substituting σxx in Eq. (8.4-4) gives the equation between displacement component u and the temperature. This equation along with the energy equation (8.4-3) become

358


∂ 2u ∂ 2u ∂T − ρ − (3λ + 2μ)α =0 (8.4-7) 2 2 ∂x ∂t ∂x ∂2T ∂T ∂2u k 2 − ρc − (3λ + 2μ)αT0 = −R (8.4-8) ∂x ∂t ∂x∂t Equations (8.4-7) and (8.4-8) are coupled and a general analytical solution may not be easily obtainable. However, in this case, where the functional relationship for R is given as a harmonic function of x, we may take the solution in nondimensional form as ρc x u = F (τ ) sin (8.4-9) L (3λ + 2μ)α L T − T0 x = G(τ ) cos (8.4-10) T0 L where τ = κt/L2 is the dimensionless time variable. Substituting these equations in Eqs. (8.4-7) and (8.4-8) gives (λ + 2μ)

d2 F −C G=0 dτ 2 dG dF Q(τ )L2 G+ + = dτ dτ T0 k F + D2

(8.4-11)

where D = k/(Lρc (λ + 2μ)/ρ) and C = [(3λ + 2μ)2 α2 T0 ]/[(λ + 2μ)ρc]. The initial conditions are dF (0) = G(0) = 0 (8.4-12) F (0) = dt The solutions of Eqs. (8.4-11) are composed of a general and a particular solutions. The general solution, taking Q as zero, is Fg (τ ) = A1 eη1 τ + A2 eη2 τ + A3 eη3 τ Gg (τ ) = B1 eη1 τ + B2 eη2 τ + B3 eη3 τ

(8.4-13)

where the constants A1 , A2 , . . . , B3 are the nontrivial roots of the following equations

η1 A1 + (1 + η1 )B1 = 0 (1 + η12 D2 )A1 − CB1 = 0 η2 A2 + (1 + η2 )B2 = 0 (1 + η22 D2 )A2 − CB2 = 0 η3 A3 + (1 + η3 )B3 = 0 (1 + η32 D2 )A3 − CB3 = 0

(8.4-14)

The eigenvalues ηi are the roots of the determinant of the coefficients, which reduces to (1 + η1 )(1 + η12 D2 ) + Cη1 = 0 (1 + η2 )(1 + η22 D2 ) + Cη2 = 0 (1 + η3 )(1 + η32 D2 ) + Cη3 = 0

(8.4-15)

4. Variable Heat Source

359

To obtain the particular solution, we assume Q(t) in Eq. (8.4-1) to have the following special form [5] Q = Q0 (1 − e−t/t0 ) = Q0 (1 − e−τ /τ0 )

(8.4-16)

where τ0 = κt0 /L2 . The particular solution based on Eq. (8.4-16) is then τ 3 e−τ /τ0 Q0 L2 C [1 + 0 ] T0 k m1 Q0 L2 τ0 (D2 + τ0 )e−τ /τ0 ] Gp = [1 + T0 k m1

Fp =

(8.4-17)

where m1 = (D2 + τ02 )(1 − τ0 ) + Cτ02 . The sum of Eqs. (8.4-13) and (8.4-17) represents the complete solution of Eqs. (8.4-7) and (8.4-8). The detailed analysis is given in [5]. Now, let us consider a heat source located at the origin of the coordinate system and varying harmonically with the time as [4] R = Q0 eiωt

(8.4-18)

where ω is the frequency of the fluctuation of the heat source and Q0 is a constant. The assumed heat generation produces the longitudinal waves. From Eqs. (8.3-4) 1 ¨ α(3λ + 2μ) ψ= (T − T0 ) c21 λ + 2μ 1 αT0 (3λ + 2μ) 2 ˙ R ∇2 T − T˙ − ∇ ψ=− κ k k

∇2 ψ −

(8.4-19)

The solution is taken in the form ¯ y, z)eiωt ψ = ψ(x, ¯ y, z)eiωt θ = T − T0 = θ(x,

(8.4-20)

which upon substitution in Eqs. (8.4-19) gives (∇2 + m22 )ψ¯ = m1 θ¯ Q0 (∇2 − m3 )θ¯ − m3 η2 ∇2 ψ¯ = − k

(8.4-21)

where α(3λ + 2μ) , λ + 2μ ω2 iω m22 = 2 , m3 = c1 κ m1 =

η1 =

αT0 (3λ + 2μ) k η2 = η1 κ

(8.4-22)

360


¯ From Eqs. (8.4-21) the function ψ¯ is eliminated to arrive at an equation for θ, ¯ which results in and then eliminating θ¯ to arrive at an equation for ψ, 1 (∇2 − m3 )(∇2 + m22 )θ¯ − m3 n∇2 θ¯ = − (∇2 + m22 )Q0 k m1 2 2 2 ¯ 2¯ (∇ − m3 )(∇ + m2 )ψ − m3 n∇ ψ = − Q0 k

(8.4-23)

where n = η1 κm1 . Integrating the second of Eqs. (8.4-23) in cylindrical coordinates yields Q0 m1 ∞ ∞ ξJ0 (ξr) cos γr ψ¯ = − 2 dξdγ (8.4-24) 2π k 0 0 F (ξ, γ) where F (ξ, γ) = (ξ 2 + γ 2 + k12 )(ξ 2 + γ 2 + k22 ) k12 + k22 = m3 (1 + n) − m22 k12 k22 = −m3 m22

(8.4-25)

which after integration gives ψ¯ =

−Q0 m1 (e−k1 r − e−k2 r ) 4πkr(k12 − k22 )

(8.4-26)

where r = (x21 + x22 + x23 )1/2 k1 = a1 + ib1 , k2 = a2 + ib2

a1 and a2 > 0 (8.4-27)

Integration of the first of Eqs. (8.4-23) yields θ¯ =

Q0 [(m22 + k12 )e−k1 r − (m22 + k22 )e−k2 r ] 4πkr(k12 − k22 )

(8.4-28)

Substituting ψ¯ and θ¯ from Eqs. (8.4-26) and (8.4-28) into Eqs. (8.4-20), the solutions of Eqs. (8.4-19) are obtained.

5

One-Dimensional Coupled Problem

Bahar and Hetnarski, in three papers [6], [7], and [8] developed and applied to the problems of coupled thermoelasticity a method called the state space approach to thermoelasticity, and then applied the method to a layered media [9]. In this approach, the method of the matrix exponential, based on CayleyHamilton theorem, is applied to equations of coupled thermoelasticity. The method allows to avoid the use of thermoelastic potential in solving a broad class of problems. As examples of subsequently published papers in which this

5. One-Dimensional Coupled Problem

361

method was utilized, the references [10–13] should be consulted. The concept follows the motivation led by Uflyand [14] and Lebedev et al. [15] to formulate Fourier transform solution of elastostatics in terms of the field variables that are inherent to the nature of the problem, instead of using Fourier transform of the Airy stress function developed by Sneddon [16]. Solutions to thermoelastic problems that do not make use of the thermoelastic potential are scarce in thermoelasticity. An example is the work of Dillon [17]. The technique involves Laplace transformations of the governing equations of one-dimensional coupled thermoelasticity along with the zero initial conditions. It will be shown that the elimination of Laplace transform of either the temperature or the displacement from the governing equations produces an ordinary differential equation in the transformed temperature or displacement, whose differential operator is identical to the one governing the transformed thermoelastic potential. The thermoelastic problem is, therefore, formulated in terms of the natural physical variables. Moreover, the characteristic equation governing the transformed temperature or displacement is the same as the one governing the transformed thermoelastic potential. Therefore, the usual values of the characteristic roots may be used to solve a given boundary value problem. Application of the method to half-space and layer problems are presented herein. The one-dimensional classical coupled thermoelastic equations are obtained from Eqs. (8.2-4) and (8.2-5) in which we let R = 0. By introducing a suitable set of dimensionless dependent and independent variables, and retaining the same symbols for the dimensionless quantities, the following dimensionless field equations are obtained ∂ ∂ 2 u¯(¯ x, t¯) ∂2 x, t¯) = η (8.5-1) ( 2 − )T¯(¯ ¯ ∂ x¯ ∂t ∂ x¯∂ t¯ ∂2 ∂ T¯(¯ x, t¯) ∂2 (8.5-2) x, t¯) = γ ( 2 − 2 )u(¯ ∂ x¯ ∂t ∂ x¯ where η = (3λ + 2μ)α/ρc, γ = (3λ + 2μ)αT0 /(λ + 2μ) and the nondimensional terms appeared in the preceding equations are related to the dimensional terms as c1 c1 x¯ = x; u¯ = u κ κ 2 c λ + 2μ t¯ = 1 t; T¯ = (T − T0 )/T0 ; c21 = κ ρ Taking Laplace transform of Eqs. (8.5-1) and (8.5-2), and making use of the zero initial conditions, leads to d2 x, s) du∗ (¯ ¯∗ (¯ − s) T x , s) = ηs 2 d¯ x d¯ x d2 dT¯∗ (¯ x, s) 2 ∗ ( 2 − s )¯ u (¯ x, s) = γ d¯ x d¯ x

(

(8.5-3) (8.5-4)

362


where ∗ indicates Laplace transform and s is the transform parameter. Calling the coupling parameter C = γη and eliminating T¯∗ (¯ x, s) between Eqs. (8.5-3) and (8.5-4) yields [D4 − s(s + 1 + C)D2 + s3 ]¯ u∗ (¯ x, s) = 0

(8.5-5)

Elimination of u¯∗ (¯ x, s) between Eqs. (8.5-3) and (8.5-4) gives [D4 − s(s + 1 + C)D2 + s3 ]T¯∗ (¯ x, s) = 0

(8.5-6)

where D = d/d¯ x. It is noticed that the same mathematical operator governs both the displacement and the temperature in the transformed domain. This differential operator is also the same for the transformed thermoelastic potential in Boley and Hetnarski [18].

Application to a half-space A half-space occupying the region x ≥ 0 subjected to the boundary conditions T¯(0, t¯) = T¯a H(t¯) and σ ¯xx (0, t¯) = 0, where H(t¯) is the Heaviside unit step function, is considered. The first of these boundary conditions describes a sudden thermal shock in the form of a suddenly applied temperature T¯a on the traction free boundary plane. The characteristic equation corresponding to either of Eqs. (8.5-5) or Eq. (8.5-6) is k 4 − s(s + 1 + C)k 2 + s3 = 0

(8.5-7)

Denoting the roots of Eqs. (8.5-7) by ±k1 and ±k2 , with positive real parts of k1 and k2 , the solutions of Eqs. (8.5-5) and (8.5-6) bounded at infinity in the transformed domain are obtained x, s) = A1 exp (−k1 x¯) + B1 exp (−k2 x¯) T¯∗ (¯ ∗ u¯ (¯ x, s) = A2 exp (−k1 x¯) + B2 exp (−k2 x¯)

(8.5-8) (8.5-9)

where A1 , A2 , B1 , and B2 are constants. Substituting Eqs. (8.5-8) and (8.5-9) into Eq. (8.5-4) yields [(k12 − s2 )A2 + γk1 A1 ] exp (−k1 x¯) + [(k22 − s2 )B2 + γk2 B1 ] exp (−k2 x¯) = 0 (8.5-10) Since Eq. (8.5-10) holds true for all values of x¯, the coefficients of the exponential terms must identically vanish, yielding A1 = −(k12 − s2 )

A2 γk1

B1 = −(k22 − s2 )

B2 γk2

(8.5-11)

Substituting A1 and B1 into Eq. (8.5-8) gives 1 k 2 − s2 k 2 − s2 T¯∗ (¯ x, s) = − [ 1 A2 exp (−k1 x¯) + 2 B2 exp (−k2 x¯)] γ k1 k2

(8.5-12)


363

The constants of integration A2 and B2 in Eqs. (8.5-9) are obtained using the boundary conditions T¯a T¯∗ (0, s) = (8.5-13) s γ T¯a d¯ u∗ (0, s) = (8.5-14) d¯ x s Using Eqs. (8.5-12) and (8.5-9) and inserting them into Eqs. (8.5-13) and (8.5-14) yield k 2 − s2 −γ T¯a k12 − s2 A2 + 2 B2 = (8.5-15) k1 k2 s γ T¯a (8.5-16) k1 A2 + k2 B2 = − s Solving Eqs. (8.5-15) and (8.5-16) for A2 and B2 and substituting their values into Eqs. (8.5-9) and (8.5-12) gives the expressions for temperature and displacement in the transformed domain in the form T¯a T¯∗ (¯ x, s) = [(k12 − s2 ) exp (−k1 x¯) − (k22 − s2 ) exp (−k2 x¯)] s(k12 − k22 ) (8.5-17) ¯a −γ T u¯∗ (¯ x, s) = (8.5-18) [k1 exp (−k1 x¯) − k2 exp (−k2 x¯)] s(k12 − k22 )

Application to a layer problem A layer defined in the region 0 ≤ x ≤ h with traction free surface at x¯ = 0 and ¯ is considered. The surface x¯ = 0 is rigidly attached to a foundation at x¯ = h ¯ suddenly exposed to a temperature of T (0, t¯) = T¯a H(t¯) while the other surface ¯ is exposed to a heat flux given by Q( ¯ t¯) = Q0 H(t¯), ¯ h, of the layer at x¯ = h ¯ is related ¯ where Ta and Q0 are constant values. non-dimensional heat flux Q ¯ to its dimensional counterpart Q by means of relation Q = (κ/kc1 T0 )Q. Laplace transform of the boundary conditions at the surfaces x = 0 and ¯ are x¯ = h T¯a d¯ u∗ (0, s) γ T¯a T¯∗ (0, s) = and = (8.5-19) s d¯ x s ¯ s) Q0 dT¯∗ (h, ¯ s) = 0 (8.5-20) = and u¯∗ (h, d¯ x s The solution of Eqs. (8.5-5) and (8.5-6) is T¯∗ (¯ x, s) = A1 exp (−k1 x¯) + A2 exp (k1 x¯) + A3 exp (−k2 x¯) + A4 exp (k2 x¯) (8.5-21) u¯ (¯ x, s) = B1 exp (−k1 x¯) + B2 exp (k1 x¯) + B3 exp (−k2 x¯) + B4 exp (k2 x¯) (8.5-22) ∗

364


Substituting Eqs. (8.5-21) and (8.5-22) into Eq. (8.5-4) yields [(k12 − s2 )B1 + γk1 A1 ] exp (−k1 x¯) + [(k12 − s2 )B2 − γk1 A2 ] exp (k1 x¯) +[(k22 − s2 )B3 + γk2 A3 ] exp (−k2 x¯) + [(k22 − s2 )B4 − γk2 A4 ] exp (k2 x¯) = 0 (8.5-23) Since Eq. (8.5-23) should hold true for all values of x¯, its coefficients should vanish identically, producing the relations between the constant coefficients (k12 − s2 )B1 γk1 2 (k − s2 )B3 A3 = − 2 γk2

(k12 − s2 )B2 γk1 2 (k − s2 )B4 A4 = 2 γk2

A1 = −

A2 =

(8.5-24)

Using the relations (8.5-24), Eq. (8.5-21) becomes 1 k 2 − s2 T¯∗ (¯ x, s) = − { 1 [B1 exp (−k1 x¯) − B2 exp (k1 x¯)] γ k1 k 2 − s2 + 2 [B3 exp (−k2 x¯) − B4 exp (k2 x¯)]} k2

(8.5-25)

The four constants of integration B1 to B4 in Eqs. (8.5-22) and (8.5-25) will now be obtained. Using the transformed boundary conditions (8.5-19) and (8.5-20), Eqs. (8.5-22) and (8.5-25) give k1 (B1 − B2 ) + k2 (B3 − B4 ) = −

γ T¯a s

(8.5-26)

k 2 − s2 γ T¯a k12 − s2 (B1 − B2 ) + 2 (B3 − B4 ) = − (8.5-27) k1 k2 s ¯ + B2 exp (k1 h) ¯ + B3 exp (−k2 h) ¯ + B4 exp (k2 h) ¯ =0 B1 exp (−k1 h) (8.5-28) 2 2 ¯ ¯ (k1 − s )[B1 exp (−k1 h) + B2 exp (k1 h)] ¯ + B4 exp (k2 h)] ¯ = Q0 γ (8.5-29) +(k22 − s2 )[B3 exp (−k2 h) s Solving Eqs. (8.5-26) to (8.5-29) gives ¯ γ[Q0 − T¯a k1 exp (k1 h)] ¯ 2s(k12 − k22 ) cosh k1 h ¯ γ[Q0 + T¯a k1 exp (−k1 h)] B2 = − 2 2 ¯ 2s(k1 − k2 ) cosh k1 h ¯ − Q0 ] γ[T¯a k2 exp (k2 h) B3 = 2 2 ¯ 2s(k1 − k2 ) cosh k2 h ¯ γ[Q0 + T¯a k2 exp (−k2 h)] B4 = − ¯ 2s(k12 − k22 ) cosh k2 h B1 =

(8.5-30) (8.5-31) (8.5-32) (8.5-33)


365

Substituting the constants of integration, Eqs. (8.5-30) to (8.5-33), into Eqs. (8.5-22) and (8.5-25) gives the expression of the displacement and the temperature in the transformed domain. An approximate small time inverse Laplace transforms of Eqs. (8.5-22) and (8.5-25) provide a solution in the real time domain [19]. A more general treatment of the problems of coupled thermoelasticity using the state-space approach is given by Sherief and Anwar [20]. The authors have considered the Lord and Shulman model of generalized thermoelasticity and have extended their previous work to include the heat source in the energy equation. The problem is then formulated for a two-dimensional domain. Laplace and Fourier integral transforms are employed to transform the governing equations. The resulting equations (two thermoelastic equations in x and y-directions and the energy equation) are finally being presented in terms of the variable y, and the formal analytical solutions are then obtained. Fourier transforms are inverted analytically, but the inverse Laplace transform is obtained using a numerical method. The technique is used to describe the behavior of a thick plate subjected to heating which varies exponentially with time on parts of its upper and lower surfaces. The plots for the temperature, the displacements u and v, and the stress σxx are presented for different locations in the plate at different times. Another technique to analytically approach solution to the coupled thermoelasticity problems is D’Alembert’s method. D’Alembert’s method, used for solving problems in mechanics, was generalized by Hetnarski [21] to treat problems of coupled thermoelasticity. It should be noted that D’Alembert’s method for a hyperbolic (wave) equations and its generalizations, like classical thermoelasticity, generalized thermoelasticity, etc., is equivalent to the method of Laplace transforms and inverse transforms for such equations. Returning to reference [21], a system of coupled equations was considered for one-dimensional problem in an infinite region, in which at time t = 0 three initial functions, that is, the initial displacement, the initial time derivative of displacement, and the initial temperature, as functions of space variable were given as arbitrary functions. The solution was assumed in the form of a displacement function and a temperature function , both of two variables, space and time. Each of these two functions consisted of a sum of three terms, and each term constituted the product of an unknown operator depending of both space variable and time, and one of the three assumed initial functions. The problem was reduced to finding the six unknown operators. For this purpose, Hetnarski employed the theory of initial functions given by Agaryev in [22]. D’Alembert’s method for solution to simplest problems of Green and Naghdi hyperbolic theory of thermoelasticity without energy dissipation was considered by Chandrasekharaiah [23]. Most recently, D’Alembert’s method was used by Rossikhin and Shitikova [24,25] in the problem of an impact of a thermoelastic rod against a heated rigid barrier, for both uncoupled and coupled strain and temperature fields.

366

6


Propagation of Discontinuities

According to the classical isothermal theory of elasticity, a strain wave applied to the end of a rod (or to the free surface of a half-space) is transmitted unaltered in shape along the rod, the portion of the rod ahead of the pulse remains undisturbed. In coupled thermoelasticity theory, however, two significant new phenomena arise; namely, a slow exponential decay with the time of the discontinuity in strain at the front of the pulse, and a precursor disturbance which precedes the pulse. A knowledge of the motion and magnitude of discontinuities in coupled thermoelastic problems is important in several respects. First, it provides an immediate estimate of the importance of thermoelastic damping as well as the distortion of an input pulse as it travels along the medium. Second, it is useful in assessing the accuracy of approximate theories. Finally, it describes the behavior of discontinuities which can provide considerable insight in the role of thermoelastic coupling. The method used in this work to calculate the magnitude and location of discontinuities is based on the work of Boley and Hetnarski [18], and the properties of Laplace transform provided in [26]. Consider a half-space confined to the region x ≥ 0. The general governing equations of the classical coupled thermoelasticity are θ˙ ˙ =0 − (β/k)T0 div U κ ¨ μ∇2 U + (λ + μ) grad div U − β grad θ = ρU

∇2 θ −

(8.6-1)

where β = (3λ + 2μ)α. We introduce a nondimensional spacial coordinate x¯, a nondimensional displacement component u¯ in x-direction, and a nondimensional time t¯, by referring the lengths and the time, respectively, to the quantities (κ/c1 ) and (κ/c21 ), where c1 is the velocity of propagation of elastic longitudinal wave. The stress σxx and the temperature change θ maintain their physical dimensions. Calling βT0 λ + 2μ

(8.6-2)

βa1 (3λ + 2μ)2 α2 T0 = (λ + 2μ)ρc ρc

(8.6-3)

a1 = the coupling parameter is then C=

The governing equations (8.6-1) for the one-dimensional problem of the halfspace reduce to ∂2 ∂ θ β − ) = 2 ¯ ∂ x¯ ∂ t T0 ρc ∂2 ∂2 a1 ( 2 − 2 ) u¯ = ∂ x¯ ∂ t¯ T0

(

∂ 2 u¯ ∂ x¯∂ t¯ ∂θ ∂ x¯

(8.6-4) (8.6-5)

6. Propagation of Discontinuities

367

Introducing the thermoelastic potential ψ which is related to the displacement and the temperature by u¯ =

θ 1 ∂2 ∂2 = ( 2 − 2) ψ T0 a1 ∂ x¯ ∂ t¯

∂ψ ∂ x¯

(8.6-6)

equation (8.6-5) is identically satisfied and the normal strain and stress in x-direction are xx =

∂2ψ ∂ x¯2

σxx = (λ + 2μ)xx − βθ = (λ + 2μ)

∂2ψ ∂ t¯2

(8.6-7)

Equation (8.6-4) in terms of the displacement potential becomes {

∂4 ∂3 ∂ ∂3 − }ψ = 0 [ + (1 + C)] + ∂ x¯4 ∂ x¯2 ∂ t¯ ∂ t¯ ∂ t¯3

(8.6-8)

The initial conditions at t = 0 are assumed to be ψ=

∂ψ ∂ 2ψ = 2 =0 ∂t ∂t

at

t=0

(8.6-9)

These conditions state that the body is undisturbed and at rest and at zero temperature at t = 0. Laplace transform of Eq. (8.6-8) is [

d4 d2 − (s + 1 + C)s + s3 ]ψ ∗ = 0 d¯ x4 d¯ x2

(8.6-10)

The general solution of Eq. (8.6-10) vanishing at x¯ → ∞ is ψ ∗ = Ae−k1 x¯ + Be−k2 x¯ where k1 and k2 are 1 s 2 = {s + 1 + C ± [s2 − 2(1 − C)s + (1 + C)2 ] 2 } k1,2 2

(8.6-11)

(8.6-12)

and A and B are the constants of integration to be found from the mechanical and thermal boundary conditions at x¯ = 0. The strain, the temperature change, and the stress in the transformed domain are d2 ψ ∗ d¯ x2 T0 d2 θ∗ = ( 2 − s2 )ψ ∗ a1 d¯ x ∗ σxx = (λ + 2μ)s2 ψ ∗ ∗xx =

(8.6-13)

Now, we may consider the solution in Laplace transform domain. There are basically 16 different fundamental cases associated with different combinations

368


of mechanical and thermal boundary conditions applied at the surface x¯ = 0. Based on these fundamental solutions, other cases with arbitrary timedependent boundary conditions may be constructed by means of Duhamel or convolution theorem without major difficulty. We use the notation H(t) to denote the Heaviside unit step function

H(t) =

0 1

for for

t0

(8.6-14)

The symbol δ(t) is used to denote the Dirac delta function dH(t) (8.6-15) dt The 16 cases considered are conventionally divided into four sets of problems each. In the first set (cases 1–4), combinations of step variations of strain (or stress) and temperature are prescribed; while in the second set (cases 5–8), the specification of the heat input takes the place of that of the temperature. Set 3 (cases 9–12) and set 4 (cases 13–16) differ from sets 1 and 2, respectively, in the fact the Dirac delta function replaces the Heaviside unit step function. Thus, it follows from Eq. (8.6-15) that δ(t) =

∗ = sψ ∗ ψi+8

for

i = 1, 2, ..., 8

(8.6-16)

The boundary conditions specified for each of the 16 problems are listed in Table 8.6-1. The solutions in Laplace transform domain are ψ1∗ = ψ3∗

a a (λ + 2μ) = 3 2 σa s (k1 − k22 )

∗ ×[(s2 − k22 )e−k1 x¯ − (s2 − k12 )e−k2 x¯ ] = ψ11

ψ∗ a (λ + 2μ) = 9 sσa s

∗ a1 θa k12 e−k1 x¯ − k22 e−k2 x¯ ψ10 × = T0 s3 (k12 − k22 ) s −k1 x ¯ −k2 x ¯ ∗ a1 θa e −e ψ = × = 12 2 2 T0 s(k1 − k2 ) s a s2 − k22 −k1 x¯ = e ( s[k1 (s2 − k22 ) − k2 (s2 − k12 )] k1 s2 − k12 −k2 x¯ ψ∗ − e ) = 13 k2 s ∗ ψ14 a1 θa (k2 /k1 )e−k1 x¯ − (k1 /k2 )e−k2 x¯ = = −( ) T0 s[k2 (s2 − k12 ) − k1 (s2 − k22 )] s

2 2 −k1 x ¯ 2 2 −k2 x ∗ σa k2 (s − k2 )e − k1 (s − k1 )e ¯ ψ15 = = λ + 2μ s3 [k2 (s2 − k22 ) − k1 (s2 − k12 )] s −k1 x ¯ −k2 x ¯ ∗ a1 θa e −e ψ =( ) = 16 2 2 2 2 T0 s[k2 (s − k2 ) − k1 (s − k1 )] s

ψ2∗ = ψ4∗ ψ5∗

ψ6∗ ψ7∗ ψ8∗

(8.6-17)


369

Table 8.6-1:: Different boundary conditions. Case 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

Boundary conditions xx = a H(t) θ=0 xx = 0 θ = θa H(t) σxx = σa H(t) θ=0 σxx = 0 θ = θa H(t) xx = a H(t) ∂θ/∂x = 0 xx = 0 ∂θ/∂x = −θa H(t) σxx = σa H(t) ∂θ/∂x = 0 σxx = 0 ∂θ/∂x = −θa H(t) xx = a δ(t) θ=0 xx = 0 θ = θa δ(t) σxx = σa δ(t) θ=0 σxx = 0 θ = θa δ(t) xx = a δ(t) ∂θ/∂x = 0 xx = 0 ∂θ/∂x = −θa δ(t) σxx = σa δ(t) ∂θ/∂x = 0 σxx = 0 ∂θ/∂x = −θa δ(t)

The location of discontinuities may be obtained using Theorem IX of reference [26] which is presented here for convenience. Let two functions F (t) and f (s) be related by the following Laplace inverse formula c+iy 1 f (s)eg(s,t) ds (8.6-18) lim F (t) = 2πi y→∞ c−iy where g = g(s, t) is a prescribed function. If there exists a constant K such that K m lim s [f (s) − ] = 0 m > 1, n > 0 (8.6-19) s→∞ sn and if 0 < n ≤ 1, there also exists a function ξ(t) such that lim sr [g(s, t) − sξ(t)] = 0

r >1−n

(8.6-20)

for ξ = 0 if n > 1 for ξ = 0 δF (t) = ⎪ K if n = 1 for ξ = 0 ⎪ ⎪ ⎩ ∞ if n < 1 for ξ = 0

(8.6-21)

δF (t) = F (ξ + 0) − F (ξ − 0)

(8.6-22)

s→∞

then

⎧ 0 ⎪ ⎪ ⎪ ⎨ 0

where represents the discontinuity in the function F .

370


In the particular case when s is large and f (s) =

K 1 [1 + O( p )] n s s

n > 0, p > (1 − n)

(8.6-23)

then Eq. (8.6-19) is satisfied (Theorem X of reference [26]). The solution of the present case for the strain, stress, and temperature distribution in the transformed domain, as given by Eq. (8.6-17), contains two terms, exp (−k1 x¯) and exp (−k2 x¯). The inversion to the time domain may have the general form as (xx , σxx , θ) =

1 c+i∞ ¯ ¯ [f,σ,θ (s)e(st−k1 x¯) + g,σ,θ (s)e(st−k2 x¯) ]ds 2πi c−i∞

(8.6-24)

For large values of s C(1 − C) 1 + O( 2 ) s s C(1 − C) 1 k22 = s − C − + O( 2 ) s s k12 = s2 + Cs + C +

(8.6-25)

or C C(4 − C) 1 + + O( 2 ) 2 8s s √ C 1 C(4 − 3C) k2 = s − √ − + O( 5/2 ) 2 s 8s3/2 s k1 = s +

(8.6-26)

We first consider the second term of Eq. (8.6-24) which contains the term exp (−k2 x¯). By √ inspection of Eq. (8.6-26) it appears that, due to the presence of the term s, it is impossible to satisfy condition (8.6-20), required when 0 < n ≤ 1, and, therefore, that the previously stated theorem is not applicable in such cases. It is thus verified that excepting the functions gT for cases 2 and 4, and gσ for case 4, condition (8.6-19) can always be satisfied with n > 1 and in these cases the method is applicable and reveals that no discontinuities arise. The consequence is that the terms containing the factor exp (−k2 x¯) always yield, after transformation, continuous functions in space and time domains and in fact give terms containing the error function, its integrals, and exponential integrals [26]. To study the discontinuities, we may consider the first term on the right-hand side of Eq. (8.6-24). Substituting k1 from Eq. (8.6-26) into the term exp (st − k1 x¯), with the consideration of Eq. (8.6-18), gives est−k1 x¯ = e¯x/2 eg(s,t,¯x)

(8.6-27)

C(4 − C) 1 g(s, t¯, x¯) = st¯ − [s + + O( 2 )] x¯ 8s s

(8.6-28)

¯

where

¯


371

It is noticed that the condition (8.6-20) can be satisfied for any value of n in the range 0 < n ≤ 1 (that is, whenever a non-zero discontinuity exists) by setting ξ = t¯ − x¯ and by choosing r < 1. This means that discontinuities can only occur at x¯ = t¯ which means, in terms of the dimensional variables, that all discontinuities travel with the velocity c1 , the velocity of propagation of the elastic longitudinal wave. The definition of discontinuities, as given by Eq. (8.6-22), may now be restated in terms of the physical variables t¯ and x¯, rather than the single variable ξ. It then takes the form δF (t¯, x¯) = F (t¯ + 0, x) − F (t¯ − 0, x¯) = −[F (t¯, x¯ + 0) − F (t¯, x¯ − 0)] at t¯ = x¯ (8.6-29) The classification of discontinuities and the examination of the functions f,σ,θ are given by Boley and Hetnarski [18]. In addition, the magnitude of discontinuity of each function is given by the authors. The problem of discontinuities produced by an instantaneous point source of heat, or by a suddenly applied Dirac delta function in an infinite thermoelastic solid is discussed by Wagner [27]. He has approached the problem using the fundamental matrix of the system of partial differential equations governing the classical coupled thermoelastic field. The basic concept is to obtain the solution of the fundamental matrix by simple definite integrals and power series. For the elements of this matrix, the constant and the linear term in the Taylor series about 0 with respect to the coupling constant are expressed explicitly by the error functions, the elementary functions, and the delta functions. We will try to discuss in this section the key equations and the solutions and to present some curves for the distribution of the displacement and the temperature showing the discontinuities, without the detailed analysis contained in reference [27]. Let us consider the system of partial differential equations of the classical coupled thermoelasticity, Eqs. (8.2-4) and (8.2-5). Denoting by A(∂) the matrix of the differential operators for the vector representation of the governing equations, the equation of motion and the energy equation, gives

A(∂)

U T

=

(ρ∂t2 − μΔ)I − (λ + μ)∇ · ∇T η∂t ∇T

β∇ ∂t − κΔ

U T

=

F R/ρc (8.6-30)

where U(x, t) is the displacement vector, T (x, t) is the temperature, ∂t = ∂/∂t, I is the 3 × 3 identity matrix, Δ is Laplace operator, and ∇ is the gradient operator. The symbol ∇T is the transpose of ∇. The constants η = T0 β/ρc, β = (3λ + 2μ)α, and κ = k/ρc. The body force vector is shown by F, and R is the heat supply per unit volume per unit time. The determinant of the matrix A(∂) is detA(∂) = (ρ∂t2 − μΔ)2 [(ρ∂t2 − (λ + 2μ)Δ)(∂t − κΔ) − βη∂t Δ]

(8.6-31)

372


The operator A(∂) is quasi-hyperbolic, due to the operator of the equation of motion, and hence it has a unique fundamental matrix EA given by [28]

EA =

IEW + H(∂)∇ · ∇T EP ∗ EW −η∂t ∇T EP

−β∇EP 2 (ρ∂t − (λ + 2μ)Δ)EP

(8.6-32)

where EP is the fundamental solution of P (∂) = [ρ∂t2 − (λ + 2μ)Δ](∂t − κΔ) − βη∂t Δ

(8.6-33)

and EW is the fundamental solution of W (∂) = ρ∂t2 − μΔ

(8.6-34)

In Eq. (8.6-32) the sign ∗ indicates the convolution operator in four dimensions, and (8.6-35) H(∂) = (λ + μ + ηβ)∂t − (λ + μ)κΔ is a heat operator. Introducing the dimensionless variables μ ηβ κρ C= a= λ + 2μ λ + 2μ λ + 2μ 3 5/2 ρ λ + 2μ κρ b=κ d= ξ= 3/2 λ + 2μ (λ + 2μ) β

γ=

(8.6-36)

the fundamental matrix EA in terms of the thermo-mechanical coupling parameter C becomes −1 ˆ . t / aξ (I Eγ + ∇ · ∇T Fγ,C ) −b−2 d−1 ∇EC a (8.6-37) EA = x −ab−1 ξ −1 Cd∂t ∇T EC b−3 (∂t2 − Δ)EC b In Eq. (8.6-37) EC is the fundamental solution of the mathematical operator PC = (∂t − Δ)(∂t2 − Δ) − C∂t Δ, and Eˆγ is the fundamental solution of the mathematical operator Wγ (∂) = ∂t2 − γΔ. Also Fγ,C = [(1 + C − γ)∂t − (1 − γ)Δ]EC ∗ Eˆγ

(8.6-38)

By selecting proper scaling parameters, we let a = b = ξ = d = 1 in Eq. (8.6-37). Therefore, the matrix EA may be transformed to a dimensionless form given by

Eγ,C =

I Eˆγ + ∇ · ∇T Fγ,C −C∂t ∇T EC

−∇EC (∂t2 − ∇)EC

(8.6-39)

Also, the displacement vector and the temperature caused by an instantaneous point source of heat at the origin are related to EC , by [27]

UC TC

=

−∇EC 2 (∂t − Δ)EC

(8.6-40)


373

Once the expression for EC is obtained, the solutions for the displacements and the temperature are obtained. Consider an infinite space with an instantaneous heat source produced by a delta function at the origin. Fourier-Laplace transforms of EC with a suitable integration path of the inverse Laplace transform yields the expression for EC , namely [28] 2 H(t) ∞ v2 + 1 + C 2 v +1+C sin (|x|v ) exp (−tv ) EC (t, x) = 2 2π |x| 0 v2 + 1 v2 + 1 dv s2 − 1 − C H(t)H(|x| − t) × 2 + Ress=1 {exp [s (ts − |x|)] v(v + 1 + C) 2π|x| s2 − 1 1 } C = 0 (8.6-41) × 2 s(s − 1 − C)

Expanding the expression (8.6-41) into a power series with respect to C, the thermo-mechanical coupling parameter, and ignoring the higher order terms (C 2 and higher), gives √ H(t) 2C 2|x| −|x|2 /4t |x| EC (t, x) = + 2(1 − C) erf ( √ ) { √ e 8π|x| π 2 t √ |x| |x| +[1 − C − C(t + |x|)(t + − 1)]et+|x| erfc ( t + √ ) 2 2 t √ |x| |x| −[1 − C − C(t − |x|)(t − − 1)]et−|x| erfc ( t − √ )} 2 2 t H(t)H(|x| − t) |x| + {C − 1 + [1 − C − C(t − |x|)(t − − 1)]et−|x| } 4π|x| 2 +O(C 2 ) (8.6-42) where erf(x) is the error function of the variable (x) and erfc(x) = 1−erf(x). The temperature distribution is obtained from the second of Eq. (8.6-40) with TC = (∂t2 − Δ)EC , and it is 2 H(t) ∞ v2 + 1 + C 2 v +1+C sin (|x|v ) exp (−tv ) TC (t, x) = 2 2π |x| o v2 + 1 v2 + 1 H(t)H(|x| − t) s2 − 1 − C (v 2 + 1 + C)v dv+ Res (ts − |x|)] {exp [s × s=1 v2 + 1 2π|x| s2 − 1 s(s2 − 1 − C) × } (8.6-43) s2 − 1

Expanding this expression into the power series with respect to C and keeping only the first two terms yields

374

Chapter 8. Coupled and Generalized H(t) (1 + Ct)|x| −|x|2 /4t − C(t + |x| + 1) et+|x| { √ 3/2 e 8π|x| πt √ √ |x| |x| × erfc ( t + √ ) + C(t − |x| + 1)et−|x| erfc ( t √ )} 2 t 2 t H(t)H(|x| − t) + (8.6-44) (|x| − t − 1) et−|x| + O(C 2 ) 4π|x|

TC (t, x) =

When the thermo-mechanical coupling parameter is zero (C = 0), Eq. (8.6-44) reduces to the well known fundamental solution of the heat operator (∂t − Δ)T as H(t) −|x|2 /4t e (8.6-45) TC=0 (t, x) = (4πt)3/2 Equation (8.6-44) for the temperature distribution of coupled thermoelasticity for C = 0.1 and fixed times t = 1, 1.2, 1.5, and 2 versus the modulus of x, |x|, is plotted in Fig. 8.6-1. Equation (8.6-45) for the temperature distribution of uncoupled thermoelasticity (C = 0) and the same fixed times t = 1, 1.2, 1.5, and 2 is plotted in Fig. 8.6-2, see [27]. The two entirely different kinds of the behavior of temperature distributions are clearly observed in Figs. 8.6-1 and 8.6-2. The coupled solution shows the wave front which travels along x-axis. It is noted that the coupled solution predicts a slightly higher temperature. The displacement uC produced by the Heaviside heat source is obtained from Eq. (8.6-40) and Eq. (8.6-42) and is x uC (t, x) = −∇EC = (8.6-46) fC (t, |x|) |x|

Figure 8.6-1: Temperature distribution produced by an instantaneous heat source, coupled case with C = 0.1.


375

Figure 8.6-2: Temperature distribution produced by an instantaneous heat source, uncoupled case with C = 0. where the function fC (t, |x|) is found to be [27] √ H(t) 2C t(1 − t)r −r2 /4t r √ fC (t, |x|) = { + 2(1 − C) erf ( √ ) e 2 8πr π 2 t 2 r −[r − 1 + C((1 − r)(t2 − t + 1) + (1 − 3t − r))]et+r 2 √ r × erfc ( t + √ ) − [r + 1 − C((1 + r)(t2 − t + 1) 2 t 2 √ r r + (1 − 3t + r))]et−r erfc ( t − √ )} 2 2 t H(t)H(r − t) + {C − 1 + [r + 1 − C((1 + r)(t2 − t + 1) 4πr2 r2 + (1 − 3t + r))]et−r } + O(C 2 ) for C → 0 (8.6-47) 2 Equation (8.6-47) has a discontinuity at t = r. The magnitude of this discontinuity is [28] lim [fC (t, t + δ) − fC (t, t − δ)]

δ→0

1 s2 − 1 − C Ress=1 {exp ts(s − 1) } 2πt s2 − 1 e−Ct/2 = 4πt =

(8.6-48)

Figures 8.6-3 and 8.6-4 show the distribution of the radial displacement versus the modulus of x [27]. It is noted that the displacements associated

376


Figure 8.6-3: Displacement produced by an instantaneous heat source, coupled case with C = 0.1.

Figure 8.6-4: Displacement produced by an instantaneous heat source, uncoupled case with C = 0. with the coupled solution are slightly smaller than in the uncoupled case. The curves are plotted for t = 1, 1.2, 1.5, and 2. It is to be noted that the two alternative solutions of coupled thermoelasticity, corresponding to a concentrated source of heat in an infinite space, are also available in the literature: (i) the solution in the form of a series in powers of C obtained by Hetnarski [29], and (ii) the solution in the form of a sum of two series in powers of the convolution coefficients associated with a decomposition theorem of coupled

7. Half-Space Subjected to a Harmonic Temperature

377

thermoelasticity, obtained by Jakubowska [30]. The results given by Wagner [27] are complimentary to those in [29] and [30]. The extension of the method presented in reference [29] is given by Hetnarski [31], and an effective method of inverting Laplace transform of exponential type that appear in the solution in these two papers appear in [32]. In addition, we may also refer to the treatment of one-dimensional coupled thermoelasticity problem for short times presented by Hetnarski [33].

7

Half-Space Subjected to a Harmonic Temperature

Analytical solutions of coupled thermoelasticity problems are, in general, restricted to the infinite space and the half-space. The problems are restricted to one-dimensional cases where the effect of boundary conditions, except at the bounding plane of the half-space, are mathematically simple to apply. The method of Laplace transform is, in general, appropriate and is usually used to solve the coupled equations (see e.g., Bahar and Hetnarski [6–8]). The application of Laplace transform to solve the coupled thermoelasticity is shown in the next section. However, different mathematical tools such as other integral transforms may be used for the analysis. The following example illustrates the technique. Consider a half-space in cylindrical coordinates where on its boundary at iωt ¯ . It is assumed that z = 0 the temperature varies harmonically as θ = θ(r)e the boundary z = 0 is stress free [3,4]. The given boundary condition causes an axisymmetric temperature distribution which, in turn, produces axisymmetric thermal stresses. The solution may be obtained by summing up the solutions of two problems: for an infiiωt ¯ is satisfied on the plane nite space where the boundary condition θ = θ(r)e z = 0, and for a half-space where the remaining boundary conditions are satisfied. The solution for the first problem is obtained by assuming the following form for the functions θ and ψ: iωt ¯ θ = θ(r)e iωt ¯ ψ = ψ(r)e

(8.7-1)

Substituting into Eqs. (8.4-19), where T − T0 = θ, gives (∇2 + m22 )ψ¯ = m1 θ¯ (∇2 − m3 )θ¯ − m3 η2 ∇2 ψ¯ = 0

(8.7-2)

where m22 = ω 2 /c21 , m1 = α(3λ+2μ)/(λ+2μ), m3 = iω/k, and η2 = καT0 (3λ+ 2μ)/k. Applying Fourier-Hankel integral transform, the system of equations reduces to

378

Chapter 8. Coupled and Generalized ˆ¯ = m ˆθ¯ −(ξ 2 + γ 2 − m22 )ψ 1 2 2 ˆ ˆ¯ + γ θ¯ˆ = 0 ¯ −(ξ + γ + m3 )θ + m3 η2 (ξ 2 + γ 2 )ψ 0

where

ˆθ(ξ, ¯ γ) =

(8.7-3)

2 ∞ ∞ ¯ θ(r, z)rJ0 (ξr) sin γzdrdz π 0 0

2 ∞ ∞ ¯ ψ(r, z)rJ0 (ξr) sin γzdrdz π 0 0 ∞ ˆθ¯ (ξ) = rJ0 (ξr)θ¯0 (r)dr 0 ˆ¯ γ) = ψ(ξ,

(8.7-4)

0

The solution of the system of Eqs. (8.7-3) is ˆ¯ ˆ¯ = − γm1 θ0 ψ F (ξ, γ) ˆ¯ ˆθ¯ = γ θ0 (ξ 2 + γ 2 − m2 ) 2 F (ξ, γ)

(8.7-5)

where F (ξ, γ) = (ξ 2 + γ 2 + k12 )(ξ 2 + γ 2 + k22 ) k12 + k22 = m3 (1 + η2 m1 ) − m22 k12 k22 = −m3 m22 ˆ¯ and ˆθ¯ from Eqs. (8.7-5) in (8.7-4) and inverting the transSubstituting for ψ formation gives

¯ z) = m1 ψ(r, or

2 ∞ ∞ ˆθ¯0 (ξ)γξJ0 (ξr) sin γz dξdγ π 0 0 F (ξ, γ)

2 m1 ∞ ˆ¯ ¯ θ0 (ξ)ξ(e−k1 z − e−k2 z )J0 (ξr)dξ ψ(r, z) = − 2 2 π k1 − k2 0

(8.7-6)

The stresses are then calculated as 1 ∂ ψ¯ ∂ 2 ψ¯ ¯ + 2 ) + ρω 2 ψ] r ∂r ∂z ∂ 2 ψ¯ ∂ 2 ψ¯ ¯ = −eiωt [2μ( 2 + 2 ) + ρω 2 ψ] ∂r ∂z 1 ∂ ψ¯ ∂ 2 ψ¯ ¯ = −eiωt [2μ( + 2 ) + ρω 2 ψ] r ∂r ∂r ∂ 2 ψ¯ = eiωt 2μ ∂r∂z

(1) σrr = −eiωt [2μ( (1)

σφφ

(1) σzz (1) σrz

(8.7-7)

7. Half-Space Subjected to a Harmonic Temperature

379

(1) (1) We check that on the plane z = 0 σzz = 0 and σrz = 0, where

(1) (r, 0, t) σrz

=

2 m1 eiωt ∞ 3 ˆ¯ α θ0 (α)J0 (αr)dα π k1 + k2 0

(8.7-8)

(2)

The solution for the second problem, σij , is obtained in such a way that on the plane z = 0 the following boundary conditions are satisfied: (2) σzz (r, 0, t) = 0 (1) (2) σrz (r, 0, t) + σrz (r, 0, t) = 0 (2) θ (r, 0, t) = 0

(8.7-9)

Both longitudinal and shear waves are present in the elastic half-space. Therefore, the following system of equations is considered for the second problem (∇2 − m3 )(∇2 + m22 )ψ¯(2) − m3 n∇2 ψ¯(2) = 0 ¯ =0 (∇2 + τ 2 )Ω

(8.7-10) (8.7-11)

¯ iωt . The displacements where τ 2 = ω 2 /c22 , n = m1 αT0 (3λ + 2μ)/ρc, and Ω = Ωe 2) ¯ ¯ are related to the functions ψ and Ω through the following relations ¯ ∂ ψ¯(2) ∂2Ω + ∂r ∂r∂z ¯ ∂ ψ¯(2) ∂ 2 Ω ¯ = + 2 + τ 2Ω ∂z ∂z

u¯(2) r = u¯(2) z

(8.7-12)

¯ are The stresses in terms of the displacements, or functions ψ¯(2) and Ω, (2) ¯ = 2μ(∂z2 − ∇2 )ψ¯(2) − ρω 2 ψ¯(2) + 2μ(∂z2 + τ 2 ),z Ω σ ¯zz ∂ 2 ψ¯(2) (2) ¯ σ ¯rz = 2μ + μ(2∂z2 + τ 2 ),r Ω ∂r∂z

(8.7-13)

where ∂z is the derivative with respect to z. The solution of Eqs. (8.7-10) can be represented by Hankel integral ψ¯(2) =

∞ 0

(Ae−λ1 z + Be−λ2 z )J0 (ξr)dξ

(8.7-14)

where λ1 and λ2 are the roots of the equation λ4 + [m22 − m3 (1 + n) − 2ξ 2 ]λ2 + ξ 4 − ξ 2 [m22 − m3 (1 + n)] − m3 m22 = 0 (8.7-15) of which the real parts are positive. The solution of Eq. (8.7-11) is assumed to have the form ¯= Ω

∞ 0

D(ξ)e−νz J0 (ξr)dξ

(8.7-16)

380


where

ν=

ξ2 − τ 2

The amplitude of the temperature is given by the relation 1 θ¯(2) = (∇2 + m22 )ψ¯(2) m1

(8.7-17)

Substituting Eq. (8.7-14) into Eq. (8.7-17) yields 1 ∞ [A(λ21 + m22 − ξ 2 )e−λ1 z θ¯(2) = m1 0 +B(λ22 + m22 − ξ 2 )e−λ2 z ]J0 (ξr)dξ

(8.7-18)

The constants A, B, and D are determined from the boundary conditions (8.7-9). Hence √ m1 μξ 3 ˆθ¯0 (ξ) νn2 A = − 2π k1 + k2 Δ1 n1 2μξ 2 − ρω 2 λ22 − λ21 B = −A D= A n2 2μνξ 2 n2

(8.7-19)

where Δ1 = (λ1 n2 − λ2 n1 )4μνξ 2 − (2μξ 2 − ρω 2 )(ν 2 + ξ 2 )(λ22 − λ21 ) n1,2 = λ21,2 + m22 − ξ 2 ¯ one can determine the displacements u¯(2) Knowing the functions ψ¯(2) and Ω, r , (2) (1) (2) (2) u¯z , and finally the stresses σ ¯ij . Adding the stresses σij to σij results in the final stresses σij . The second part of the solution is identical with that of the Lamb problem for the elastic half-space. More works on the inertia effect in the elastic half-space due to dynamic thermal stresses have been reported by other investigators. Among them we may refer to the works done by Danilovskaya [34], Mura [35], and Sternberg and Chakravorty [36].

8

Coupled Thermoelasticity of Thick Cylinders

The method of Laplace transform may be employed to study the coupled problems with finite boundaries. However, in carrying out the inversion by analytical methods, mathematical difficulties are incurred. The following problem is a classical example. When the effect of inertia term in thermoelastic equation in

8. Coupled Thermoelasticity of Thick Cylinders

381

radially symmetric cylindrical coordinates is considered, the analytical derivation of inverse Laplace transform is very complicated. The following problem is solved by means of Laplace transform method when the effect of inertia term is excluded. A thick cylinder of infinite length and the inside and outside radius a and b, respectively, is considered. For a radially symmetric loading condition, the thermoelasticity equations reduce to the following dimensionless form (Gosn and Sabbaghian [37]) ∂ T¯ ∂ 1 ∂ [ (¯ ru¯)] = ∂¯ r r¯ ∂¯ r ∂¯ r 1 ∂ ∂ ∂ ¯ 1 ∂ [ (¯ r )− ]T =C (¯ ru¯˙ ) r¯ ∂¯ r ∂¯ r ∂ t¯ r¯ ∂¯ r

(8.8-1) (8.8-2)

where u¯ = u¯(¯ r, t¯) and T¯ = T¯(¯ r, t¯) and C=

T0 β 2 ηβκ (3λ + 2μ)2 α2 T0 = 2 2 = 2 ρ(λ + 2μ)c ρ c1 c ρc1

and β = (3λ + 2μ)α

η=

βT0 k

κ=

(8.8-3) k ρc

(8.8-4)

The dimensionless quantities r¯, t¯, u¯, T¯, and σ ¯ij are related to the dimensional quantities r, t, u, T¯ and σij through the relations r¯ =

c1 r κ

c2 t¯ = 1 t κ

u¯ =

ρc31 u βκT0

T − T0 T¯ = T0

σ ¯ij =

σij βT0

(8.8-5)

The system of Eqs. (8.8-1) and (8.8-2) are solved using Laplace transform method. Applying Laplace transform to Eqs. (8.8-1) and (8.8-2) under quiescent initial conditions yields dT¯∗ d 1 d [ (¯ ru¯∗ )] = d¯ r r¯ d¯ r dr 1 d d 1 d (¯ r ) − s]T¯ = Cs (¯ ru¯∗ ) [ r¯ d¯ r d¯ r r¯ d¯ r

(8.8-6) (8.8-7)

r, s) and T¯∗ (¯ r, s) are the transformed displacement and the temperwhere u¯∗ (¯ ature, respectively. To solve the coupled equations (8.8-6) and (8.8-7) in the transformed domain, they are decoupled by eliminating T¯∗ from Eq. (8.8-6) using Eq. (8.8-7) and, similarly, eliminating u¯∗ from Eq. (8.8-7) using Eq. (8.8-6). Upon elimination, the decoupled equations are d 1 d d [ (¯ r ) − p2 ]T¯∗ = 0 d¯ r r¯ d¯ r d¯ r d 1 d d 1 d [ ( ru¯∗ ) = 0 )][¯ r ( ) − p2 ](¯ d¯ r r¯ d¯ r d¯ r r¯ d¯ r

(8.8-8) (8.8-9)

382


where p2 = s(1 + C). The solution of Eqs. (8.8-8) and (8.8-9) follows to be A1 T¯∗ (¯ r, s) = − 2 + A2 I0 (p¯ r) + A3 K0 (p¯ r) p B1 r¯ B2 u¯∗ (¯ r, s) = − 2 − 2 2 + B3 I1 (p¯ r) + B4 K1 (p¯ r) 2p p r¯

(8.8-10) (8.8-11)

where A1 , A2 , A3 , and B1 to B4 are the constants of integration and I0 , I1 , K0 , and K1 are the modified Bessel functions of orders zero and one and of first and second kind, respectively. The solutions given by Eqs. (8.8-10) and (8.8-11) must satisfy the coupled equations (8.8-6) and (8.8-7). Substitution of Eqs. (8.8-10) and (8.8-11) into the coupled equations (8.8-6) and (8.8-7), leads to A2 = pB3 A3 = −pB4 (8.8-12) A1 = −CB1 Considering these relations, Eqs. (8.8-10) and (8.8-11) become B1 T¯∗ (¯ r, s) = C 2 + pB3 I0 (p¯ r) − pB4 K0 (p¯ r) p B1 r¯ B2 r, s) = − 2 − 2 2 + B3 I1 (p¯ r) + B4 K1 (p¯ r) u¯∗ (¯ 2p p r¯

(8.8-13) (8.8-14)

The four constants of integrations B1 to B4 are obtained using the boundary conditions. For solid cylinders, Eqs. (8.8-13) and (8.8-14) reduce to C T¯∗ (¯ r, s) = 2 B1 + B3 pI0 (p¯ r) p B1 r¯ u¯∗ (¯ r, s) = − 2 + B3 I1 (p¯ r) 2p

(8.8-15) (8.8-16)

We may consider the lateral surface of the cylinder to be traction free, but subjected to a temperature change F (t). The boundary conditions in the physical space are then σrr = 0 θ = F (t)

for t ≥ 0 at r = b for t ≥ 0 at r = b

(8.8-17) (8.8-18)

To apply these conditions in the transform domain, the dimensionless stressstrain relations from Hooke’s law are considered as

where γ2 =

∂ u¯ u¯ + (1 − 2γ 2 ) − T¯ ∂¯ r r¯ ¯ u¯ 2 ∂u = (1 − 2γ ) + − T¯ ∂¯ r r¯

σ ¯rr =

(8.8-19)

σ ¯φφ

(8.8-20)

μ c2 = 22 λ + 2μ c1

c22 =

μ ρ

c21 =

λ + 2μ ρ

(8.8-21)

8. Coupled Thermoelasticity of Thick Cylinders

383

Laplace transforms of Eqs. (8.8-19) and (8.8-20) are d¯ u∗ u¯∗ + (1 − 2γ 2 ) − T¯∗ d¯ r r¯ u∗ u¯∗ ∗ 2 d¯ + − T¯∗ r, s) = (1 − 2γ ) σ ¯φφ (¯ d¯ r r¯

∗ σ ¯rr (¯ r, s) =

(8.8-22) (8.8-23)

The boundary conditions (8.8-17) and (8.8-18) in the transform domain take the form ∗ =0 σ ¯rr ∗ T¯ = f¯∗ (s)

at r¯ = ¯b at r¯ = ¯b

(8.8-24) (8.8-25)

where ¯b is the dimensionless radius of the cylinder (¯b = c1 b/κ), and f¯∗ (s) is Laplace transform of F (t)/T0 . Applying the boundary conditions (8.8-24) and (8.8-25) to Eqs.(8.8-15) and (8.8-16) and using Eq. (8.8-22) gives B1 =

2γ 2 p2 I1 (p¯b) ¯∗ f (s) L(p¯b)

B3 =

¯b(1 + C − γ 2 ) f¯∗ (s) L(p¯b)

(8.8-26)

with L(pb) = (1 + C − γ 2 )p¯bI0 (p¯b) − 2Cγ 2 I1 (p¯b)

(8.8-27)

Substituting the constants B1 and B3 in Eqs. (8.8-15) and (8.8-16) results in the solution in the transform domain as pbI0 (p¯ I1 (p¯b) ¯∗ r) ¯∗ f (s) + (1 + C − γ 2 ) f (s) (8.8-28) T¯∗ (¯ r, s) = 2Cγ 2 ¯ ¯ L(pb) L(pb) I1 (p¯b) ¯∗ r) ¯∗ 2 I1 (p¯ f f (s) u¯∗ (¯ r, s) = −¯ rγ 2 (s) + (1 + C − γ )b (8.8-29) L(p¯b) L(p¯b) The inverse Laplace transforms of these equations are obtained by means of the convolution integral, and they are T¯(¯ r, t) = 2Cγ 2 θ1 (¯ r, t) + (1 + C − γ 2 )θ2 (¯ r, t) 2 2 ¯ u¯(¯ r, t) = −¯ rγ u1 (¯ r, t) + b(1 + C − γ )u2 (¯ r, t)

(8.8-30) (8.8-31)

where u¯1 (¯ r, t¯) = θ1 (¯ r, t¯) = r, t¯) = θ¯2 (¯ r, t¯) = u¯2 (¯

t¯ 0

0

f (t¯ − ξ)g1∗ (¯ r, ξ)dξ

f (t¯ − ξ)g2∗ (¯ r, ξ)dξ

t¯ 0

t¯

f (t¯ − ξ)g3∗ (¯ r, ξ)dξ

(8.8-32) (8.8-33) (8.8-34)

384


and

1 λ0 +i∞ I1 (p¯b) g1∗ (¯ r, t¯) = [ ]exp(st¯)ds 2πi λ0 −i∞ L(p¯b) 1 λ0 +i∞ p¯bI0 (p¯ r) ∗ ¯ r , t) = [ g2 (¯ ]exp(st¯)ds 2πi λ0 −i∞ L(p¯b) r) 1 λ0 +i∞ I1 (p¯ r, t¯) = [ ] exp(st¯)ds g3∗ (¯ ¯ 2πi λ0 −i∞ L(pb)

(8.8-35) (8.8-36) (8.8-37)

The integrals of Eqs. (8.8-35) to (8.8-37) are evaluated according to the standard method of inverse Laplace transform. A closed curve C0 made up of the part of the circle C1 and the straight line L1 is considered. The radius of circle C1 is s = R0 and lies on the left of the line L1 defined by ξ = λ0 , where λ0 is a sufficiently large positive real number, so that the integrand singularities lie to the left of it. It is noted that the integrands in Eqs. (8.8-35) to (8.8-37) are single-valued and analytical functions of s on and within the closed contour C0 , except at the singular points which coincide with the roots of their common denominator L(pb). These roots are obtained from the equation (8.8-38) L(p¯b) = (1 + C − γ 2 )p¯bI0 (p¯b) − 2Cγ 2 I1 (p¯b) = 0 The roots of Eq. (8.8-38) are all imaginary and, in general, distinct. It is further shown that

LimR0 →∞

C0

|Ij∗ | exp(st¯)ds → 0

for j = 1, 2, 3

(8.8-39)

where I1∗ , I2∗ , and I3∗ are the integrands of Eqs. (8.8-35) to (8.8-37), respectively. From Cauchy’s residue theorem it follows that ∞ ηn J0 [ηn (¯ r/¯b)] T¯(¯ r, t¯) = {2Cγ 2 + (1 + C − γ 2 ) } J1 (ηn ) n=1 ×ρ1 Pn

t¯ 0

u¯(¯ r, t¯) =

f (t¯ − ξ)exp(−ρ1 ηn2 ξ)dξ

∞

{−¯ rγ 2 + ¯b(1 + C − γ 2 )

n=1

×ρ1 Pn

t¯ 0

f (t¯ − ξ)exp(−ρ1 ηn2 ξ)dξ

(8.8-40) r/¯b)] J1 [ηn (¯ } J1 (ηn ) (8.8-41)

where iηn , n = 0, 1, 2, ..., are the roots of Eq. (8.8-38) and ρ1 and Pn are defined as 1 ρ1 = ¯2 (8.8-42) b (1 + C) −2ϑηn2 (8.8-43) Pn = 2Cγ 2 (ϑ − 2Cγ 2 ) − ϑ(ηn2 ϑ − 2Cγ 2 ) ϑ = 1 + C − γ2

9. Green–Naghdi Model of a Layer

385

The solution given by Eqs. (8.8-40) and (8.8-41) depends on the integration of the given thermal boundary condition F (t). Once the function F (t) is known, the integrals are evaluated and the solution is known. As an example, let us consider the case of zero traction on the boundary, which is exposed to a temperature change varying harmonically in the time F (t) = T1 sin ωt

(8.8-44)

The dimensionless temperature function is f¯(t) = T1 sin ωt/T0 , where T1 is known. Substituting into Eqs. (8.8-40) and (8.8-41) and carrying the integrations yields ∞ r/¯b)] ηn J0 [ηn (¯ T1 [Pn [ϑ (8.8-45) − 2Cγ 2 ] T¯(¯ r, t¯) = T0 n=1 J1 (ηn ) α2 sin ω t¯ − ¯b2 (1 + C)ω cos ω t¯ + ω¯b2 (1 + C)exp[−ηn2 t¯/¯b2 (1 + C)] × n ηn4 + ¯b4 (1 + C)2 ω 2 ∞ ¯bJ1 [ηn (¯ T1 r/¯b)] u¯(¯ r, t¯) = [Pn [ϑ (8.8-46) + r¯γ 2 ] T0 n=1 J1 (ηn ) η 2 sin ω t¯ − ¯b2 (1 + C) cos ω t¯ + ω¯b2 (1 + C) exp[−ηn2 t¯/¯b2 (1 + C)] × n η 4 + ¯b4 (1 + C)2 ω 2 n

These are the dimensionless temperature distribution and the distribution of the displacement along the radial direction of the thick cylinder subjected to the boundary conditions (8.8-17) and (8.8-18) in which F = F (t) is given by Eq. (8.8-44).

9

Green–Naghdi Model of a Layer

Green and Naghdi have formulated three models of thermoelasticity for homogeneous and isotropic materials [38], which are labelled as models I, II, and III. The nature of these models is such that when the respective theories are linearized, model I is reduced to the classical thermoelasticity (based on Fourier’s law). The linearized versions of models II and III permit propagation of thermal waves at a finite speed. The model II, in particular, exhibits a feature that is not present in the other (known) thermoelastic models, as it does not sustain dissipation of thermoelastic energy [39]. In this model, the constitutive equations are derived by starting with reduced energy equation [23,40] and by including the thermaldisplacement gradient among the constitutive variables. Reference [39] includes the derivation of a complete set of governing equations of linearized version of the theory for homogeneous and isotropic materials in terms of displacement and temperature fields and a proof of the uniqueness of the solution for the corresponding initial boundary value problem. The uniqueness of the

386


solution of governing equations formulated in terms of stress and energy-flux are established in [41]. Chandrasekharaiah [42] studied the one-dimensional thermal wave propagation in a half-space based on Green–Naghdi (GN) model due to a sudden application of temperature to the boundary, using Laplace transform method. He obtained the analytical solution, in a closed form, for displacement, temperature, and stress, and analyzed his results in light of their counterparts in this work. Chandrasekharaiah and Srinath [43,44] have studied cylindrical/spherical waves caused by (i) a load applied to the boundary of the cylindrical/spherical cavity in an unbounded body, and (ii) a line/point heat source in an unbounded body, where in both cases GN type II model was used. Chandrasekharaiah [45] presented complete solutions of the governing field equations of GN theory. Sharma and Chauhan [46] investigated the disturbances produced in a half-space by the application of a mechanical point load and thermal source acting on the boundary of the half-space. The material is assumed to be homogeneous and isotropic. Laplace and Hankel transforms are used and different theories of generalized thermoelasticity are employed to provide a basis for the comparison of the results. Green–Naghdi third model [23,40] admits dissipation of energy. In this model, the constitutive equations are derived starting with a reduced energy equation, where the thermal-displacement gradient, in addition to the temperature gradient, are among the constitutive variables. Li and Dhaliwal [47] used the generalized thermoelasticity to solve a boundary value problem of an isotropic space with its plane boundary either held rigidly fixed or stress free and subjected to a sudden temperature increase to obtain the approximate small time solution for displacement, temperature, and stress fields by employing Laplace transform technique. The linear theory of thermoelasticity of Green–Naghdi (GN) type II (without energy dissipation) and type III (with energy dissipation) for homogeneous and isotropic materials are employed to study the thermal and mechanical waves in a layer. The disturbances are generated by a sudden application of the temperature to the boundary. The dimensionless form of the governing equations is solved utilizing Laplace transform method. Closed form solutions are obtained for a layer in Laplace domain, and a numerical inversion of Laplace transform method is used to obtain the temperature, displacement, and stress fields in the physical time domain. The thermo-mechanical wave propagation and reflection from the layer boundary are investigated and the influence of the damping parameter in the temperature, displacement, and stress fields in Green–Naghdi type III is discussed. A homogeneous isotropic thermoelastic solid is considered. In the absence of body force and heat generation, the governing field equations for the dynamic coupled generalized thermoelasticity in the time domain based on Green and Naghdi theory (type III) are written as [48] ui μui,jj + (λ + μ)uj,ij − βθ,i = ρ¨ ∗ ˙ ¨ ρcθ + βθ0 uï,i = κ θ,ii + k θ,ii

(8.9-1) (8.9-2)


387

Here, ui is the displacement tensor, θ is the temperature change above the uniform reference temperature T0 , ρ is the mass density, c is the specific heat, λ and μ are Lamé constants, β = (3λ+2μ)α, α is the coefficient of linear thermal expansion, κ∗ is a material constant characteristic, and k is the coefficient of heat conduction. Equations (8.9-1) and (8.9-2) reduce to Green and Naghdi type II theory by setting k = 0. The strain and stress tensors ij and σij associated with ui and θ are given by the following geometrical and constitutive relations, respectively, as 1 ij = (ui,j + uj,i ) 2 σij = 2μij + [λkk − α(3λ + 2μ)θ]δij

(8.9-3) (8.9-4)

It is convenient to introduce the dimensionless variables as 1 xi = xi , t = θ θ = , ij = T0

v 1 (λ + 2μ) ui t, ui = βT0 (λ + 2μ) 1 ij , σij = σij βT0 βT0

(8.9-5)

where is a standard length and v is a standard speed. Equations (8.9-1)(8.9-4) in the dimensionless form, dropping the hat for convenience, take the form c21 − c¯22 )uj,ij − c¯21 θ,i = uï c¯22 ui,jj + (¯ θ¨ + C u¨,ii = c¯2T θ,ii + c¯2k θ˙,ii 1 ij = (ui,j + uj,i ) 2 c¯2 c¯2 σij = (1 − 2 22 )ui,i δij + 22 (ui,j + uj,i ) − θδij c¯1 c¯1

(8.9-6) (8.9-7) (8.9-8) (8.9-9)

where (λ + 2μ) μ κ∗ 2 2 , c ¯ = , c ¯ = , 2 T ρv 2 ρv 2 ρcv 2 k β 2 T0 c¯2K = , C= ρcv ρc(λ + 2μ) c¯21 =

(8.9-10)

Here, c¯1 , c¯2 , and c¯T represent the nondimensional speeds of the purely elastic dilatational waves, shear waves, and thermal waves, respectively, and c¯K is the damping coefficient in the third GN model. Also, C is the usual thermoelastic coupling parameter. It is to be pointed out that the expression for the nondimensional speed cT is determined principally by the material constant κ∗ . Assume that the layer is initially at rest, thus ˙ 0) = 0 ui (x, 0) = u˙ i (x, 0) = θ(x, 0) = θ(x,

(8.9-11)

388


where x = (x1 , x2 , x3 ). Transforming Eqs. (8.9-6) and (8.9-7) into Laplace transform domain, yields c¯22 u∗i,jj + (¯ c21 − c¯22 )u∗j,ij − c¯21 θ,i∗ = s2 u∗i

(8.9-12)

∗ c2T + s¯ c2K )θ,ii s2 (θ∗ + Cu∗i,i ) = (¯

(8.9-13)

where u∗i and θ∗ are Laplace transforms of ui and θ, respectively. Also, s is the transform parameter.

One-dimensional waves in a layer Consider a layer with one-dimensional disturbances propagating along the xdirection in the interval (0 ≤ x ≤ 1). In terms of the nondimensional variables, Eqs. (8.9-12) and (8.9-13) reduce to the following system of governing field equations for u∗x and θ∗ d2 u∗x dθ∗ − ) = s2 u∗x 2 dx dx du∗x d2 θ∗ 2 ∗ s (θ + C c2K ) 2 ) = (¯ c2T + s¯ dx dx

c¯21 (

(8.9-14) (8.9-15)

and in Eqs. (8.9-8) and (8.9-9), du∗x − θ∗ dx du∗ = x dx

∗ σxx =

(8.9-16)

∗xx

(8.9-17)

where u∗x is the displacement component along the x-direction. The values of ∗ and ∗xx are the normal stress u∗x and θ∗ are functions of x and s. Also, σxx and strain along the x-direction, respectively. Using the solution proposed by Chandrasekharaiah [41], we find −(

x

x

−(

x

x

u∗x = A1 e V1 + A2 e V1 + A3 e V2 + A4 e V2 (8.9-18) 2 2 s V − c¯1 −( x )s ( x )s )(A1 e V1 − A2 e V1 ) θ∗ = 2 [( 1 c¯1 V1 2 V − c¯21 −( x )s ( x )s +( 2 )(A3 e V2 − A4 e V2 )] (8.9-19) V2 s −( x )s ( x )s −( x )s ( x )s ∗ σxx = − 2 [V1 (A1 e V1 − A2 e V1 ) + V2 (A3 e V2 − A4 e V2 )] c¯1 (8.9-20) 1 1 −( x )s ( x )s −( x )s ( x )s ∗xx = −s[( )(A1 e V1 − A2 e V1 ) + ( )(A3 e V2 − A4 e V2 )] V1 V2 (8.9-21) )s

(

)s

)s

(

)s


389

where 1 2 Vi = √ [¯ cT + s¯ c2K + (1 + C)¯ c21 + (−1)i+1 ]1/2 , i = 1, 2 2 = [{¯ c2T + s¯ c2K − (1 + C)¯ c21 }2 + 4C¯ c21 c¯2T ]1/2 = V12 − V22

(8.9-22) (8.9-23)

Equations (8.9-18) to (8.9-21), when u∗x → 0 as x → ∞, reduce to the general solution for the half-space. In this case A2 = A4 = 0

(8.9-24)

Consider a layer of unit dimensionless thickness exposed to a sudden application of the temperature to its boundary at x = 0 of the form θ(0, t) = Θ(t)H(t)

(8.9-25)

where H(t) is the Heaviside unit step function. In addition, the boundary conditions may be assumed as σxx (0, t) = 0,

dθ(1, t) = 0, dx

ux (1, t) = 0

(8.9-26)

where Θ(t) is the prescribed continuous function of t for t ≥ 0. Using the given boundary conditions, the coefficients appearing in Eqs. (8.9-18) to (8.9-21) become 1 V1 V22 e2s/V1 , s2 (V12 − V22 ) (1 + e2s/V1 ) 1 1 V1 V 2 , A2 = − 2 2 2 2 s (V1 − V2 ) (1 + e2s/V1 ) e2s/V2 1 V2 V 2 , A3 = 2 2 1 2 s (V2 − V1 ) (1 + e2s/V2 ) 1 V2 V 2 1 A4 = − 2 2 1 2 s (V2 − V1 ) (1 + e2s/V2 ) A1 =

(8.9-27)

where Ai (i = 1, 2, 3, 4) are in Laplace domain. Substituting the values of Ai from Eqs. (8.9-27) in Eqs. (8.9-18) to (8.9-21), the following expressions are obtained u∗x =

V1 V2 1 V1 V2 ( 2−x )s ( x )s ( 2−x )s [ (e V1 − e V1 ) − (e V2 2 2 2 2s/V 2s/V 1 2 s (V1 − V2 ) (1 + e ) (1 + e ) (

x

)s

−e V2 )] (8.9-28) 2 2 2 2 1 V1 V2 V2 V1 − c¯1 ( 2−x V1 V2 − c¯1 ( 2−x )s ( x )s )s [ (e V1 + e V1 ) − (e V2 θ∗ = 2 2 2 2s/V 2s/V 1 2 s¯ c1 V1 − V2 V1 1 + e V2 1 + e ( Vx )s

+e

2

)]

(8.9-29)

390

Chapter 8. Coupled and Generalized ∗ σxx =−

1 (V1 V2 )2 1 1 ( 2−x )s ( Vx )s ( 2−x )s V1 V2 1 [ (e + e ) − (e 2 2 2 2s/V 2s/V s¯ c1 V1 − V2 1 + e 1 1+e 2

( Vx )s

+e

2

)]

∗xx = −

s(V12 x ( V )s

+e

2

(8.9-30) 1 [ − V22 ) 1

V22 ( 2−x )s V1 (e 2s/V +e 1

( Vx )s

+e

1

)−

1

V12 ( 2−x )s V2 (e 2s/V +e 2

)]

(8.9-31)

The inverse Laplace transforms of these expressions, using the numerical method suggested by Durbin [49], provide the values of the functions ux , θ, σxx , and xx in physical domain in terms of x and t. The set of expressions (8.9-28) to (8.9-31) was obtained using the boundary conditions (8.9-25) and (8.9-26). Similar expressions may be obtained for different types of boundary conditions. To compare the accuracy of the formulations presented in this section, the formulations are reduced to those of a half-space. Thus, the thermal and stress boundary conditions for the half-space are assumed as similar to reference [42] as (8.9-32) θ(0, t) = H(t), σxx (0, t) = 0 The material parameters are assumed as [36] c¯1 = 1,

c¯T = 0.5,

C = 0.02,

c¯K = 0

(8.9-33)

Now, the results of this section are compared with the analytical results obtained by Chandrasekharaiah [42]. Figures 8.9-1 and 8.9-2 show the temperature and stress distributions versus the time and x. It is seen that the present results are in close agreement with the analytical solutions given in [42]. At the point of wave fronts, however, there are some minor differences due to the errors encountered in the numerical inverse Laplace transform.

Figure 8.9-1: Comparison of dimensionless temperature versus dimensionless x at time = 1.


391

Figure 8.9-2: Comparison of dimensionless stress versus dimensionless time at x = 1.

X=0

X

X=1.0

Figure 8.9-3:: Model of a layer. As another example, consider an elastic layer of finite thickness along the x-direction, as shown in Fig. 8.9-3. The boundary conditions are assumed as θ(0, t) = H(t),

σxx (0, t) = 0,

θ(1, t) = 0,

σxx (1, t) = 0

(8.9-34)

The traction forces along the edges x = 0 and x = 1 are zero, where the temperature along the edge x = 1 is assumed zero. The input temperature along the edge x = 0 is assumed to follow the Heaviside unit step function. Using the given boundary conditions, the constants of integration of Eqs. (8.9-18) to (8.9-21) are obtained. The results for the temperature and stress distributions versus x are shown in Figs. 8.9-4 and 8.9-5. The boundary conditions are shown to be satisfied in the figures. Figure 8.9-4 shows the variation of the dimensionless temperature against the dimensionless x at the values of the time 0.2, 0.6, 1.6, 2, and 2.2. Wave front at the dimensionless time t = 2.0 is at the location x = 1, the end of the layer. This indicates that the speed of propagation of the temperature wave is c¯T =

L 1 = = 0.5 t 2.0

(8.9-35)

392


Figure 8.9-4: Variation of dimensionless temperature versus dimensionless x for different values of the time.

Nondimensional stress

0.4 0.3 0.2 0.1 0 –0.1

0

0.2

0.4

0.6

0.8

1

–0.2 –0.3 –0.4 Nondimensional x

t=0.2 t=1 t=1.4

t=0.6 t=1.2

Figure 8.9-5: Variation of dimensionless stress versus dimensionless x for different values of the time. which was previously selected as the input data. Figure 8.9-5 shows the variation of the dimensionless stress against the dimensionless x at values of the time 0.2, 0.6, 1.0, 1.2 and 1.4. The stress wave front at t = 1.0 is located in x = 1, the end of the layer. From the figure 1 L = = 1.0 (8.9-36) t 1.0 which was previously given as the input data. In this diagram, the dimensionless values of the time t = 0.2, t = 0.6, and t = 1.0 belong to the transmitted stress waves, which are compressive. At t = 1.2 and t = 1.4, there is the reflection of the stress wave from the edge x = 1. As may be seen from the figure, the sign of the reflecting stress is changed when the stress wave front from the free boundary of the layer is reached to the thermal wave front. In other words, the stress changes from compressive to tensile values when the stress wave front from a traction free boundary hits the temperature wave front. This phenomenon takes place at a time between t = 1.2 and t = 1.4, more exactly, at t = 1.333. c¯1 =


393

Now consider the same layer, shown in Fig. 8.9-3, with the given boundary conditions as θ(0, t) = H(t),

σxx (0, t) = 0,

dθ (1, t) = 0, dx

ux (1, t) = 0 (8.9-37)

The edge at x = 0 is assumed to be traction free, but exposed to a temperature input in the form of a Heaviside unit step function. The edge at x = 1 is thermally insulated with a fixed displacement u = 0. The results for the temperature, displacement and stress versus x are obtained from Eqs. (8.9-28) to (8.9-30). The temperature and stress distributions are shown in Figs. 8.9-6 and 8.9-7, respectively. The figures show that the boundary conditions are satisfied. Figure 8.9-6 shows the variation of the dimensionless temperature versus the dimensionless x at t = 0.2, 0.6, 1.6, 2.0, and 2.2. The wave front at t = 2.0 reaches the boundary of the layer at x = 1. Thus, the speed of the thermal wave propagation is c¯T = 0.5. Figure 8.9-7 shows the variation of the dimensionless stress versus the dimensionless x at t = 0.2, 0.6, 1.0, and 1.2. The wave front at t = 1.0 hits the boundary of the layer at x = 1. Thus, the speed of the propagation of the mechanical disturbances is c¯1 = 1.0, as

Figure 8.9-6: Variation of dimensionless temperature versus dimensionless x for different value of the time. Nondimensional stress

0.1 0 –0.1 0

0.2

0.4

0.6

0.8

1

–0.2 –0.3 –0.4 –0.5 –0.6 –0.7

t=0.2 t=0.6 t=1 t=1.2

–0.8 Nondimensional x

Figure 8.9-7: Variation of dimensionless stress versus dimensionless x for different value of the time.

394


assumed previously. The stress waves at t = 0.2, 0.6, and 1, during the propagation, are compressive. At t = 1.2, during the reflection of the wave from the boundary at x = 1, the stress wave is again compressive, but doubled in magnitude due to the assumed fixed boundary conditions and the speeds of thermal and elastic waves.

10

Generalized Thermoelasticity of Layers

According to Lord–Shulman theory, Fourier’s law of heat conduction is modified by introducing the relaxation time [50]. In Green–Lindsay theory two relaxation times are introduced by modifying the stress-strain relation and the entropy density [51]. Green and Naghdi have formulated three models of thermoelasticity for homogeneous and isotropic materials [39], which are labeled as models I, II, and III. The nature of these models is such that when the respective theories are linearized, model I is reduced to the classical thermoelasticity and the linearized versions of models II and III permit propagation of thermoelastic waves with a finite speed. The model II, in particular, exhibits a feature that is not present in the other thermoelastic models, as it does not sustain dissipation of thermal energy [39]. In this model, the constitutive equations are derived starting with the reduced energy equation [38,40] and include the thermal-displacement gradient among the constitutive variables. Several investigators have employed these models to solve a variety of thermoelastic problems. Chen and Dargush [52] used the boundary element method to analyze the transient and dynamic problems in generalized thermo-elasticity of a half-space using Laplace transform method. Chen and Lin [53] proposed a hybrid numerical method based on Laplace transform and control volume method to analyze the transient coupled thermoelastic problems with relaxation times, with the exposition to a nonlinear radiation boundary condition. Hosseini Tehrani and Eslami [54] considered the boundary element formulation for the analysis of coupled thermoelastic problems in a finite domain and studied the coupling coefficient and the relaxation times effects on thermal and elastic waves propagation. Bagri and Eslami [55] used the transfinite element method to study the thermoelasticity of a disk under thermal shock loads using LS model and investigated the coupling coefficient and the relaxation time effects on the temperature and the stress wave propagation. Consider a homogeneous isotropic thermoelastic solid. In the absence of body force and heat generation, the governing field equations for the generalized coupled thermoelasticity in the time domain based on LS, GL, and GN theories may be written in a unified form as [56] ui (8.10-1) μui,jj + (λ + μ)uj,ij − β(θ,i + t1 θ˙,i ) = ρ¨ ˙ ¨ ηρcθ + ρc(t0 + t2 + t3 )θ + βT0 [(t0 + t3 )¨ ui,i + η u˙ i,i ] ∗ ˙ = (kη + κ t3 )θ,ii + kt3 θ,ii (8.10-2)

10. Generalized Thermoelasticity of Layers

395

where ui is the displacement tensor, θ is the temperature change with respect to the uniform reference temperature T0 , ρ is the mass density, c is the specific heat, λ and μ are the Lamé constants, β = (3λ + 2μ)α, α is the coefficient of linear thermal expansion, κ∗ is a material constant characteristic associated with GN theory, k is the coefficient of thermal conduction, t0 is the relaxation time proposed by Lord and Shulman, and t1 , t2 are the relaxation times proposed by Green and Lindsay. Also, η and t3 are terms introduced to consolidate all theories into a unified system of equations. The governing equations for LS theory may be obtained when η = 1, and t1 = t2 = t3 = 0. To obtain the governing equations for GL theory, one should set η = 1, t0 = t3 = 0. Equations (8.10-1) and (8.10-2) include the governing coupled system of equations for GN theory type III with η = t0 = t1 = t2 = 0, and t3 = 1. Two special cases of GN theory, namely the type II and type I, may be obtained from the equations of GN theory type III by setting k = 0 and κ∗ = 0, respectively. In other words, to obtain GN theory type II from the equations of GN theory type III we put k = 0, while when κ∗ = 0, the equations of GN theory type III reduce to those of GN theory type I, which is identical with the classical theory of thermoelasticity. Reconsideration of the equations of GN theory type II reveals that no damping term appeared in the system of equations and, therefore, GN theory type II is known as the thermoelasticity without energy dissipation. The stress tensor σij associated with ui and θ is given by the following geometrical and constitutive relationship ˙ ij σij = λui,i δij + μ(ui,j + uj,i ) − β(θ + t1 θ)δ

(8.10-3)

It is convenient to introduce the dimensionless variables as 1 v 1 (λ + 2μ) x¯i = xi , t¯ = t, u¯i = ui βT0 v v v v t¯0 = t0 , t¯1 = t1 , t¯2 = t2 , t¯3 = t3 1 θ θ¯ = , σ ¯ij = σij T0 βT0

(8.10-4)

where is a characteristic length and v is a characteristic speed. Equations (8.10-1) to (8.10-3) in terms of the dimensionless parameters, removing the bar for convenience, take the form c21 − c¯22 )uj,ij − c¯21 (θ,i + t1 θ˙,i ) = uï c¯22 ui,jj + (¯ (t0 + t2 + t3 )θ¨ + η θ˙ + C [(t0 + t3 )¨ ui,i + η u˙ i ] 2 2 2 ˙ = (η¯ cK + t3 c¯T )θ,ii + t3 c¯K θ,ii c¯2 c¯2 ˙ ij σij = (1 − 2 22 )(ui,i )δij + 22 (ui,j + uj,i ) − (θ + t1 θ)δ c¯1 c¯1

(8.10-5) (8.10-6)

(8.10-7)

396


Here (λ + 2μ) μ κ∗ 2 2 , c ¯ = , c ¯ = , 2 T ρv 2 ρv 2 ρcv 2 k β 2 θ0 c¯2K = , C= ρcv ρc(λ + 2μ)

c¯21 =

(8.10-8)

In Eqs. (8.10-8), c¯1 and c¯2 represent the nondimensional speeds of purely elastic dilatational and shear waves, respectively. Furthermore, c¯T and c¯K are, respectively, the speed of thermal wave disturbances and the damping coefficient in √ GN model, while c¯K / t0 + t2 is the speed of thermal wave disturbances in LS and GL models, and C is the usual thermoelastic coupling parameter. For the initial conditions we assume that the body is initially at rest and the initial values of temperature change and the rate of temperature are zero, thus ˙ 0) = 0 (8.10-9) ui (x, 0) = u˙i (x, 0) = θ(x, 0) = θ(x, where x = (x1 , x2 , x3 ). Laplace transform of Eqs. (8.10-5) to (8.10-7) by virtue of Eq. (8.10-9) yields c¯22 u∗i,jj + (¯ c21 − c¯22 )u∗j,ij − c¯21 (1 + st1 )θ,i∗ = s2 u∗i

(8.10-10)

s2 (t0 + t2 + t3 )θ∗ + C(t0 + t3 )u∗i,i + sη(θ∗ + Cu∗i,i ) ∗ = (η¯ c2K + t3 c¯2T + st3 c¯2K )θ,ii

σij∗ = (1 − 2

(8.10-11)

c¯22 ∗ c¯2 )(ui,i )δij + 22 (u∗i,j + u∗j,i ) − (1 + t1 s)θ∗ δij 2 c¯1 c¯1 (8.10-12)

where u∗i , θ∗ , and σij∗ are Laplace transforms of ui , θ and σij , respectively, and s is Laplace transform parameter. Now, consider a layer in the domain (0 ≤ x ≤ 1) subjected to onedimensional disturbances in the x-direction. Therefore, the characteristic length in Eqs. (8.10-4) is the layer thickness. In terms of the nondimensional variables, Eqs. (8.10-10) to (8.10-12) reduce to the following system of governing field equations

dθ∗ d2 u∗x − (1 + st ) = s2 u∗x (8.10-13) 1 dx2 dx

du∗x du∗x 2 ∗ ∗ s (t0 + t2 + t3 )θ + C(t0 + t3 ) + sη θ + C dx dx 2 ∗ dθ = [(η + st3 )¯ c2K + t3 c¯2T ] 2 (8.10-14) dx du∗ ∗ σxx = x − (1 + st1 )θ∗ (8.10-15) dx

c¯21


397

∗ where u∗x and σxx in the above equations are the transformed displacement component and the normal stress along the x-direction, respectively. Eliminating θ∗ from Eqs. (8.10-13) and (8.10-14) results in

D1

d4 u∗x d2 u∗x − D + D3 u∗x = 0 2 dx4 dx2

(8.10-16)

where D1 , D2 , and D3 are defined as

D1 = c¯21 (η + st3 )¯ c2K + t3 c¯2T

(8.10-17)

D2 = s2 (η + st3 )¯ c2K + t3 c¯2T + c¯21 s [s(t0 + t2 + t3 ) + η]

+¯ c21 (1 + st1 ) s2 C(t0 + t3 ) + sηC

D3 = s4 (t0 + t2 + t3 ) + s3 η

(8.10-18) (8.10-19)

The solution of Eq. (8.10-16) is readily known u∗x = A1 e−k1 x + A2 ek1 x + A3 e−k2 x + A4 ek2 x

(8.10-20)

where ki = √

1 [D2 + (−1)i+1 D22 − 4D1 D3 ]1/2 , i = 1, 2 2D1

(8.10-21)

Substituting the solution for ux ∗ into Eq. (8.10-13), the transformed temperature field may be obtained as θ∗ = A5 e−k1 x + A6 ek1 x + A7 e−k2 x + A8 ek2 x

(8.10-22)

where the unknown constants in Eq. (8.10-22) may be written in terms of constants in Eq. (8.10-20) s2 − c¯21 k12 A1 , c¯21 k1 (1 + st1 ) s2 − c¯21 k22 A7 = 2 A3 , c¯1 k2 (1 + st1 ) A5 =

c¯21 k12 − s2 A2 c¯21 k1 (1 + st1 ) c¯2 k 2 − s2 A8 = 2 1 2 A4 c¯1 k2 (1 + st1 ) A6 =

(8.10-23)

Substituting these results in Eq. (8.10-15), the transformed stress solution is obtained as ∗ σxx =−

s2 s2 [A1 e−k1 x − A2 ek1 x ] − [A3 e−k2 x − A4 ek2 x ] 2 k1 c¯1 k2 c¯21

(8.10-24)

The unknown constants A1 , A2 , A3 , and A4 are obtained by applying the boundary conditions. In the particular case of a half-plane x → ∞ and A2 = A4 = 0

(8.10-25)

398


As an example we consider a layer exposed to a sudden application of temperature shock to its boundary at x = 0 with the form θ∗ (0, t) = H(t)

(8.10-26)

where H(t) is the Heaviside unit step function. The remaining boundary conditions are as follows ∗ σxx (0, t) = 0,

dθ∗ (1, t) = 0, dx

u∗x (1, t) = 0

(8.10-27)

Therefore, the layer is stress free at x = 0, whereas it is fixed and thermally insulated at x = 1. The application of the above boundary conditions to Eqs. (8.10-20), (8.10-22), and (8.10-24) leads to

k1 1 k2 sinh(k1 (1 − x)) − sinh(k2 (1 − x)) s(B1 k1 − B2 k2 ) cosh(k1 ) cosh(k2 ) (8.10-28)

B1 k1 1 B2 k2 ∗ θ = cosh(k1 (1 − x)) − cosh(k2 (1 − x)) s(B1 k1 − B2 k2 ) cosh(k1 ) cosh(k2 ) (8.10-29)

s cosh(k1 (1 − x)) cosh(k2 (1 − x)) ∗ σxx = 2 − (8.10-30) + c¯1 (B1 k1 − B2 k2 ) cosh(k1 ) cosh(k2 ) u∗x =

where s2 − c¯21 k12 c¯21 k1 (1 + st1 ) s2 − c¯21 k22 B2 = 2 c¯1 k2 (1 + st1 ) B1 =

(8.10-31)

and ki (i = 1, 2) are given in Eq. (8.10-21). Similar expressions may be obtained for the layers with different types of boundary conditions. The inverse Laplace transforms of these expressions, using the numerical method suggested by Honig and Hirdes [57], provide the displacement, temperature and stress fields. To carry out the numerical analysis, the following material constants and parameters are considered c¯1 = 1,

c¯K = 1,

C = 0.02

(8.10-32)

For LS model, the characteristic speed v in Eq. (8.10-4) is chosen so that t0 = 1. With the assumed numerical values for the relaxation times, c¯K and c¯1 , the speeds of propagation of the elastic and thermal disturbances from Eq. (8.10-8) for LS theory become unity. Using the assumed initial and boundary conditions, the temperature and the stress are plotted and shown in the

10. Generalized Thermoelasticity of Layers t=0.25

1.4 Dimensionless temperature

399

t=0.5

1.2

t=0.75

1

t=1.25

0.8 0.6 0.4 0.2 0 –0.2

0

0.2

0.4

0.6

0.8

1

Dimensionless x

Figure 8.10-1: Variation of dimensionless temperature versus dimensionless x for different values of the time considering LS theory. 1 Dimensionless stress

0.5 0 –0.5 0

0.2

0.4

0.6

0.8

1

–1 –1.5 –2 –2.5

t=0.25 t=0.5

–3

t=0.75

–3.5

t=1.25

–4 Dimensionless x

Figure 8.10-2: Variation of dimensionless stress versus dimensionless x for different values of the time considering LS theory. figures. Figures 8.10-1 and 8.10-2 show the temperature and stress distributions versus x for several values of the time, using LS theory. In Figure 8.10-1 the temperature wave propagation is shown at several values of the time. At times t = 0.25, 0.5, 0.75 it shows the propagation of the temperature wave, and at t = 1.25 it shows the reflection of the wave from the end boundary of the layer at x = 1. The figure shows that the heat wave propagates with the speed of unity. Figure 8.10-2 shows the stress distributions across the layer at different values of the time. Curves related to t = 0.25, 0.5, 0.75 show the wave propagation, while at t = 1.25, it shows the reflection of the same waves. It is seen from the figure that the speed of the elastic wave takes the value of unity, as expected from the assumed parameters. For Green–Lindsay model, the relaxation times t1 and t2 are assumed to be equal and the characteristic speed v is chosen so that t1 = t2 = 1. Figures 8.10-3 and 8.10-4 show the distributions of the temperature and the stress through the thickness of the layer resulting from GL theory. The curves related to t = 0.25, 0.5, and 0.75 show the wave propagation, while at t = 1.25 they

400

Chapter 8. Coupled and Generalized t=0.25

Dimensionless temperature

1.4

t=0.5

1.2

t=0.75

1

t=1.25

0.8 0.6 0.4 0.2 0 –0.2 0

0.2

0.4

0.6

0.8

1

Dimensionless x

Figure 8.10-3: Variation of dimensionless temperature versus dimensionless x for different values of the time considering GL theory. 500 Dimensionless stress

400 300 200 100 0 –100 0

0.2

0.4

0.6

0.8

1

–200 –300 –400 –500

t=0.25 t=0.5 t=0.75 t=1.25

Dimensionless x

Figure 8.10-4: Variation of dimensionless stress versus dimensionless x for different values of the time considering GL theory. show the reflection of the same waves from the end boundary. It is seen from the stress wave distribution that when the layer is exposed to a temperature shock, the elastic and temperature waves are produced and propagate through the layer thickness. Due to the assumed numerical values for the relaxation times, c¯K and c¯1 , the speeds of propagation of these waves across the layer, are equal. For GN theory type II (¯ cK = 0), the characteristic speed is such that t3 = 1 and c¯T = 1. For this model, the temperature, displacement, and stress fields versus x are obtained from Eqs. (8.10-28) to (8.10-30). The temperature and stress distributions are shown in Figures 8.10-5 and 8.10-6. The figures show that the boundary conditions are satisfied. Figure 8.10-5 shows the variation of the dimensionless temperature versus the dimensionless x at t = 0.25, 0.5, 0.75, and 1.25. The wave front at t = 0.25 reaches the boundary of the layer at x = 0.25. Thus, the speed of thermal propagation is c¯T = 1. Figure 8.10-6 shows the variation of the dimensionless stress versus the


401


2.5 t=0.25 2

t=0.5 t=0.75

1.5

t=1.25

1 0.5 0 0

0.2

0.4

0.6

0.8

1

–0.5 Dimensionless x

Dimensionless stress

Figure 8.10-5: Variation of dimensionless temperature versus dimensionless x for different values of the time considering GN theory. 1 0.5 0 –0.5 0 –1 –1.5 –2 –2.5 –3 –3.5 –4 –4.5

0.2

0.4

0.6

0.8

1

t=0.25 t=0.5 t=0.75 t=1.25

Dimensionless x

Figure 8.10-6: Variation of dimensionless stress versus dimensionless x for different values of the time considering GN theory.

dimensionless x at t = 0.25, 0.5, 0.75, and 1.25. The wave front at t = 0.25 hits the boundary of the layer at x = 0.25. Thus, the speed of propagation of the mechanical disturbances is c¯1 = 1.0, as assumed previously. The stress waves at times t = 0.25, 0.5 and 0.75, during propagation, are compressive. At t = 1.25, during the reflection of the wave from the boundary at x = 1, the stress waves are again compressive, due to the assumed fixed boundary conditions. Comparison between these generalized coupled thermoelasticity theories shows that under thermal shock loading, GL theory predicts larger stresses compared to the other theories. Also, it can be seen from the figures that LS and GN theories predict similar stress wave distributions versus thickness at different times. Comparing the temperature wave fronts reveals that LS and GL models predict close results, which are damped through the layer as time is increased. On the other hand, GN model type II shows the temperature wave distribution without damping. This wave is reflected from the insulated boundary of the layer with the double value of its magnitude.

402


When mechanical shock loads are applied (the case is not shown here), LS model leads to larger value of the temperature at the wave front compared to GL model. This result is reported in reference [58]. In general, under thermal shock loads, GL theory predicts larger stresses compared to the other theories. LS and GN theories predict the smallest value of the stress at the wave front. LS and GL models predict almost identical temperature wave fronts. GN model type II shows the temperature wave distribution without damping, as expected.

11

Generalized Thermoelasticity in Spheres and Cylinders

Consider a sphere or a circular long cylinder, under plane strain condition exposed to a radially symmetric thermal shock load. Equations (8.10-10) to (8.10-12) may be written in the spherical or polar coordinates and in dimensionless form. Laplace transforms of the resulting equations become dθ∗ u∗ s2 d du∗ [ + m ] − (1 + t1 s) = 2 u∗ dr dr r dr c¯1 2 d m d (η¯ c2K + t3 c¯2T + st3 c¯2K )[ 2 + ] θ∗ − (t0 + t2 + t3 )s2 + ηs θ∗ dr r dr du∗ u∗ = C (t0 + t3 )s2 + ηs ( +m ) dr r ⎡ ⎤ du∗

∗ ⎥ u∗ c¯22 ⎢ c¯22 du∗ σrr 1 ∗ dr ⎢ ⎥ + m ) − (1 + t1 s)θ = 2 2 ⎣ u∗ ⎦ + (1 − 2 2 )( ∗ σφφ 1 c¯1 c¯1 dr r r

(8.11-1)

(8.11-2) (8.11-3)

where m is a parameter introduced to obtain the equations in cylindrical and spherical coordinates. For m = 1 the equations reduce to the cylindrical coordinates, and for m = 2 the equations in spherical coordinates are obtained. Equations (8.11-1) and (8.11-2) may also be written in the form dθ∗ d 1 d(rm u∗ ) [ m ] − ω12 u∗ = γ dr r dr dr 1 d(rm u∗ ) 1 d m dθ∗ 2 ∗ θ = ζ [ (r ) − ω ] 2 rm dr dr rm dr

(8.11-4) (8.11-5)

where s2 (t0 + t2 + t3 )s2 + ηs 2 ; ω = 2 c¯21 (η¯ c2K + t3 c¯2T + st3 c¯2K ) C [(t0 + t3 )s2 + ηs] γ ; γ = 1 + t1 s; ξ2 = (η¯ c2K + t3 c¯2T + st3 c¯2K ) ω12 =

ζ=

ξ2 γ

(8.11-6)

11. Generalized Thermoelasticity

403

Equation (8.11-5) is rewritten as [∇2 − ω22 ]θ∗ = ζ

1 d(rm u∗ ) rm dr

(8.11-7)

where the operator ∇2 is given as ∇2 =

1 d md m d d2 + (r ) = rm dr dr dr2 r dr

(8.11-8)

and the operator ∇1 is defined as ∇1 =

1 d(rm ) d m [ ]= + m r dr dr r

(8.11-9)

Applying the operator ∇1 to Eq. (8.11-4) yields 1 d m dθ∗ 1 d m d 1 d(rm u∗ ) 2 1 d m ∗ u ) = γ (r [ ]) − ω (r (r ) (8.11-10) 1 rm dr dr rm dr rm dr rm dr dr Eliminating u∗ with the use of Eqs. (8.11-7) and (8.11-10) results in 1 d md 2 (r [∇ − ω22 ]θ∗ ) − ω12 [∇2 − ω22 ]θ∗ = ξ 2 ∇2 θ∗ rm dr dr

(8.11-11)

Equation (8.11-11) is rewritten in the form ∇2 [∇2 − ω22 ]θ∗ − ω12 [∇2 − ω22 ]θ∗ = ξ 2 ∇2 θ∗

(8.11-12)

(∇2 − ω12 )(∇2 − ω22 )θ∗ = ξ 2 ∇2 θ∗

(8.11-13)

and then

To eliminate the term θ∗ using Eqs. (8.11-4) and (8.11-5), we may differentiate Eq. (8.11-5) with respect to r to give dθ∗ d 1 d m dθ∗ d 1 d(rm u∗ ) [ m (r )] − ω22 =ζ [ m ] dr r dr dr dr dr r dr

(8.11-14)

Eliminating the term θ∗ with the help of Eqs. (8.11-4) and (8.11-14) leads to

d 1 d d 1 d(rm u∗ ) m r [ m ] − ω12 u∗ m dr r dr dr r dr m ∗ d 1 d(r u ) d 1 d(rm u∗ ) 2 2 ∗ −ω2 [ m ] − ω1 u = ξ 2 [ m ] dr r dr dr r dr

(8.11-15)

With the definition of the operator 2 as 2 =

d 1 d m d2 m d m [ m (r )] = 2 + − dr r dr dr r dr r2

(8.11-16)

404


Equation (8.11-15) appears in the form (2 − ω12 )(2 − ω22 )u∗ = ξ 2 2 u∗

(8.11-17)

Also, Eqs. (8.11-13) and (8.11-17) may be written in the form (∇2 − D12 )(∇2 − D22 )θ∗ = 0 (2 − D12 )(2 − D22 )u∗ = 0

(8.11-18) (8.11-19)

where Di2 (i = 1, 2) are the roots of the characteristic equation D4 − (ω12 + ω22 + ξ 2 )D2 + ω12 ω22 = 0

(8.11-20)

Solutions of Eqs. (8.11-18) and (8.11-19) are obtained by changing the dependent functions as θ∗ = θ˜∗ /r(m−1)/2 and u∗ = u˜∗ /r(m−1)/2 with a change of variable r to r = r˜/Di [59,60]. The resulting equations are solved for the displacement and the temperature in Laplace transform domain as u∗ = r − θ∗ = r−

m−1 2

m−1 2

2 (

)

i=1 2 (

)

Ai I(m+1)/2 (Di r) + Bi K(m+1)/2 (Di r)

Gi I(m−1)/2 (Di r) + Hi K(m−1)/2 (Di r)

(8.11-21) (8.11-22)

i=1

where I(m+1)/2 and K(m+1)/2 are the modified Bessel functions of the first kind and second kind and order (m + 1)/2, respectively. Also, I(m−1)/2 and K(m−1)/2 are the modified Bessel functions of the first kind and second kind and order (m − 1)/2, respectively. In Eqs. (8.11-21) and (8.11-22), Ai , Bi , Gi , and Hi are the constants of integration that must be obtained using the given boundary conditions. Now, consider the following relations for the derivatives of the modified Bessel functions as ν Iν (Di r) r ν Iν (Di r) r ν d[Kν (Di r)] = −Di Kν−1 (Di r) − Kν (Di r) dr r ν = −Di Kν+1 (Di r) + Kν (Di r) r d[Iν (Di r)] = Di Iν−1 (Di r) − dr = Di Iν+1 (Di r) +

(8.11-23)

Substitution of Eqs. (8.11-21) and (8.11-22) into Eq. (8.11-4), considering Eqs. (8.11-23), results in Ai (Di2 − ω12 ) γDi Bi (Di2 − ω12 ) Hi = − γDi

Gi =

(8.11-24) (8.11-25)

12. Problems

405

Thus u∗ = r − θ∗ = r−

2 (

m−1 2

)

Ai I(m+1)/2 (Di r) + Bi K(m+1)/2 (Di r)

(8.11-26)

i=1 2

Di2 − ω12 Ai I(m−1)/2 (Di r) − Bi K(m−1)/2 (Di r) γDi i=1

m−1 2

(8.11-27)

Substituting Eqs. (8.11-26) and (8.11-27) into Eq. (8.11-3), considering Eqs. (8.11-23) for the derivatives of the modified Bessel functions, the radial and hoop stresses are obtained as ∗ σrr

=r

− m−1 2

∗ σφφ =r

2

ω12 Ai I(m−1)/2 (Di r) − Bi K(m−1)/2 (Di r) Di

i=1 c¯2 1 −2m 22 c¯1 r 2 − m−1 2

2

i=1 c¯2 −(2Di 22 − c¯1

Ai I(m+1)/2 (Di r) + Bi K(m+1)/2 (Di r)

(8.11-28)

c¯22 1 A I (D r) + B K (D r) i i i i (m+1)/2 (m+1)/2 c¯21 r

ω12 ) Ai I(m−1)/2 (Di r) − Bi K(m−1)/2 (Di r) Di

(8.11-29) Equations (8.11-26) to (8.11-29) are in Laplace transform domain. The inverse Laplace transforms should be used to bring the equations into the physical time domain. The inversion of Laplace transforms may be carried out numerically using a numerical method, such as the method given in [57].

12

Problems

1. Solve Eq. (8.3-5) for a solid spherical domain with the radial thermal flow when R = 0. 2. Consider a rod of length L thermally insulated along its length. The initial temperature at x = L is suddenly raised by T (L, t) = T0 e(−t/t0 ) Find the temperature and displacement distribution for the fixed boundary conditions at x = 0 and at the free end x = L. 3. A rod of length L is considered. The heat described by Q(x, t) = Q1 (t) cos (x/L) is generated along the rod. The end x = 0 is fixed, but the end x = L is free. Find the temperature and displacement distribution along the rod.

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[49] Durbin, F., Numerical Inversion of Laplace Transforms: An Efficient Improvement to Dubner and Abate’s Method, Computer J., Vol. 17, pp. 371– 376, 1974. [50] Lord, H.W. and Shulman, Y., A Generalized Dynamical Theory of Thermoelasticity, J. Mech. Phys. Solids, Vol. 15, pp. 299–309, 1967. [51] Green, A.E. and Lindsay, K.E., Thermoelasticity, J. Elasticity, Vol. 2, pp. 1–7, 1972. [52] Chen, J. and Dargush, G.F., Boundary Element Method for Dynamic Poroelastic and Thermoelastic Analysis, Int. J. Solids Struc., Vol. 32, No. 15, pp. 2257–2278, 1995. [53] Chen, H. and Lin, H., Study of Transient Coupled Thermoelastic Problems with Relaxation Times, Trans. ASME, J. Appl. Mech., Vol. 62, pp. 208–215, 1995. [54] Hosseini Tehrani, P. and Eslami, M.R., Boundary Element Analysis of Coupled Thermoelasticity with Relaxation Time in Finite Domain, J. AIAA, Vol. 38, No. 3, pp. 534–541, 2000. [55] Bagri, A. and Eslami, M.R., Generalized Coupled Thermoelasticity of Disks Based on the Lord–Shulman Model, J. Therm. Stresses, Vol. 27, No. 8, pp. 691–704, 2004. [56] Bagri, A., Taheri, H., Eslami, M.R., and Fariborz, S., Generalized Coupled Thermoelasticity of a Layer, J. Therm. Stresses, Vol. 29, No. 4. pp. 359– 370, 2006. [57] Honig, G. and Hirdes, U., A Method for the Numerical Inversion of Laplace Transforms, J. Comp. App. Math., Vol. 10, pp. 113–132, 1984. [58] Hosseini Tehrani, P. and Eslami, M.R., Boundary Element Analysis of Finite Domains Under Thermal and Mechanical Shock with the Lord– Shulman Theory, J. Strain Anal., Vol. 38, No.1, pp. 53–64, 2003. [59] Bagri, A. and Eslami, M.R., A Unified Generalized Thermoelasticity Formulation: Application to Thick Functionally Graded Cylinders, J. Therm. Stresses, special issue devoted to the 70th Birthday of Józef Ignaczak, Vol. 30, No. 9 and 10, pp. 911–930, 2007. [60] Bagri, A. and Eslami, M.R., A Unified Generalized Thermoelasticity: Solution for Cylinders and Spheres, Int. J. Mech. Sci., Vol. 49, pp. 1325–1335, 2007.

Chapter 9 Finite and Boundary Element Methods Because the analytical solutions to coupled and generalized thermoelasticity problems are mathematically complicated, the numerical methods, such as the finite and the boundary element methods, have become powerful means of analysis. This chapter presents a new treatment of the finite and the boundary element methods for this class of problems. The finite element method based on Galerkin technique is employed in order to model the general form of the coupled equations, and the application is then expanded to the two- and onedimensional cases. The generalized thermoelasticity problems for a functionally graded layer, a thick sphere, a disk, and a beam are discussed using Galerkin finite element technique. To show the strong rate of convergence of Galerkinbased finite element, a problem for a radially symmetric loaded disk with three types of shape functions, linear, quadratic, and cubic, is solved. It is shown that the linear solution rapidly converges to that of the cubic solution. The chapter concludes with the boundary element formulation for the generalized thermoelasticity. A unique principal solution satisfying both the thermoelasticity and the coupled energy equations is employed to obtain the boundary element formulation.

1

Introduction

Due to the mathematical complexities encountered in analytical treatment of coupled thermoelasticity problems, the finite element method is often preferred. The finite element method itself is based on two entirely different approaches, the variational approach based on Ritz method, and the weighted residual methods. The variational approach, which for elastic continuum is based on the extremum of the total potential and kinetic energies has deficiencies in handling the coupled thermoelasticity problems due to the controversial functional relation of the first law of thermodynamics. On the other hand, the R.B. Hetnarski, M.R. Eslami, Thermal Stresses – Advanced Theory and Applications, Solid Mechanics and its Applications 158, c Springer Science + Business Media B.V. 2009

413

414

Chapter 9. Finite and Boundary Element Methods

weighted residual method based on Galerkin technique, which is directly applied to the governing equations, is quite efficient and has a very high rate of convergence [1–5].

2

Galerkin Finite Element

The general governing equations of the classical coupled thermoelasticity are the equation of motion (1.3-8) and the first law of thermodynamics, Eqs. (2.9-1) and (2.9-8), as ui in V σij,j + Xi = ρ¨ ˙ qi,i + ρcθ + βT0 ˙ii = R

in V

(9.2-1) (9.2-2)

where β = α(3λ + 2μ). These equations must be simultaneously solved for the displacement components ui and temperature change θ = T − T0 , where T0 is the reference temperature. The thermal boundary conditions are satisfied by either of the equations θ = θs θ,n + aθ = b

on A on A

for t > t0 for t > t0

(9.2-3) (9.2-4)

where θ,n is the gradient of temperature change along the normal to the surface boundary A, and a and b are either constants or given functions of temperature on the boundary. The first condition is related to the specified temperature and the second condition describes the specified convection and heat flux on the boundary. The mechanical boundary conditions are specified through the traction vector on the boundary. The traction components are related to the stress tensor through Cauchy’s formula given by tni = σij nj

on A

for t > t0

(9.2-5)

where tni is the prescribed traction component on the boundary surface whose outer unit normal vector is n. For displacement formulations, using the constitutive laws of linear thermoelasticity along with the strain-displacement relations, the traction components can be related to the displacements as tni = μ(ui,j + uj,i )nj + λuk,k ni − (3λ + 2μ)αθni

(9.2-6)

where θ = T − T0 is the temperature change above the reference temperature T0 . It is further possible to have kinematical boundary conditions where the displacements are specified on the boundary as ui = u¯i (s)

on A

for t > t0

(9.2-7)

2. Galerkin Finite Element

415

The system of coupled equations (9.2-1) and (9.2-2) does not have a general analytical solution. A finite element formulation may be developed based on Galerkin method. The finite element model of the problem is obtained by discretizing the solution domain into a number of arbitrary elements. In each base element (e), the components of displacement and temperature change are approximated by the shape functions (e)

ui (x1 , x2 , x3 , t) = Uim (t)Nm (x1 , x2 , x3 ) θ(e) (x1 , x2 , x3 , t) = θm (t)Nm (x1 , x2 , x3 )

m = 1, 2, . . . , r

(9.2-8) (9.2-9)

where r is the total number of nodal points in the base element (e). The summation convention is used for the dummy index m. This is a Kantrovitch type of approximation, where the time and space functions are separated into distinct functions. Here Uim (t) is the component of displacement at each nodal point, and θm (t) is the temperature change at each nodal point, all being functions of time. The shape function Nm (x1 , x2 , x3 ) is function of space variables. Substituting Eqs. (9.2-8) and (9.2-9) into Eq. (9.2-1) and applying the weighted residual integral with respect to the weighting functions Nm (x1 , x2 , x3 ), the formal Galerkin approximation reduces to V (e)

(σij,j + Xi − ρ¨ ui )Nl dV = 0

l = 1, 2, . . . , r

(9.2-10)

Applying the weak formulation to the first term, yields V (e)

(σij,j )Nl dv =

A(e)

σij nj Nl dA −

V (e)

∂Nl σij dV ∂xj

(9.2-11)

where nj is the component of the unit outer normal vector to the boundary. Substituting Eq. (9.2-11) in Eq. (9.2-10) gives A(e)

σij nj Nl dA−

V (e)

∂Nl σij dV + Xi Nl dV − ρ¨ ui Nl dV = 0 (9.2-12) ∂xj V (e) V (e)

According to Cauchy’s formula, the traction force components acting on the boundary are related to the stress tensor as tni = σij nj

(9.2-13)

Thus, the first term of Eq. (9.2-12) is A(e)

σij nj Nl dA =

A(e)

tni Nl dA

(9.2-14)

From Hooke’s law, the stress tensor is related to the strain tensor, or the displacement components, and temperature change θ as σij = G(ui,j + uj,i ) + λuk,k δij − βθδij

(9.2-15)

416


Substituting for σij in the second term of Eq. (9.2-12) yields V (e)

∂Nl ∂Nl σij dV = [G(ui,j + uj,i ) + λuk,k δij − βθδij ]dV ∂xj V (e) ∂xj

(9.2-16)

Substituting this expression in Eq. (9.2-12) gives

V (e)

−

ρ¨ ui Nl dV +

βθ V (e)

V (e)

∂Nl [G(ui,j + uj,i ) + λuk,k δij ]dV ∂xj

∂Nl dV = Xi Nl dV + tni Nl dA ∂xi V (e) A(e)

(9.2-17)

Now, the base element (e) with r nodal points is considered and the displacement components and temperature change in the element (e) are approximated by Eqs. (9.2-8) and (9.2-9). Using these approximation, Eq. (9.2-17) becomes [5]

( V (e)

ρNl Nm dV )U¨mi + (

∂Nl ∂Nm dV )Umi ∂xj ∂xj ∂Nl ∂Nm +( λ dV )Umj V (e) ∂xi ∂xj

G V (e)

∂Nl ∂Nm +( G dV )Umj ∂xj ∂xi V (e) ∂Nl −( β Nm dV )θm = Xi Nl dV + tni Nl dA ∂xi V (e) V (e) A(e) l, m = 1, 2, . . . , r i, j = 1, 2, 3 (9.2-18) Equation (9.2-18) is the finite element approximation of the equation of motion. Galerkin approximation of the energy equation given by Eq. (9.2-1) becomes V (e)

(qi,i + ρc

∂θ + T0 β u˙ i,i − R)Nl dV = 0 ∂t

l = 1, 2, . . . , r

(9.2-19)

The weak formulation of the heat flux gradient qi,i gives

V (e)

qi,i Nl dV =

( V (e)

∂qx ∂qy ∂qz (qi ni )Nl dA + + )Nl dV = ∂x ∂y ∂z A(e)

∂Nl − qi dV ∂xi V (e)

(9.2-20)

where A(e) is the boundary surface of the base element (e). Substituting Eq. (9.2-20) in Eq. (9.2-19) and rearranging the terms gives

ρc V (e)

∂Nl ∂θ qi dV + T0 β u˙ i,i Nl dV Nl dV − ∂t ∂xi V (e) V (e)

= V (e)

RNl dV −

A(e)

(qi ni )Nl dA

l = 1, 2, . . . , r

(9.2-21)


417

Substituting for the displacement components ui and temperature change θ their approximate values in the base element (e) from Eqs. (9.2-8) and (9.29) give

∂Nm ∂Nl ∂Nm k dV )θm + ( T0 β Nl dV )U˙ mi ( ∂xi ∂xi ∂xi V (e) V (e) +( ρcNm Nl dV )θ˙m = RNl dV − (qi ni )Nl dA (9.2-22) V (e)

V (e)

A(e)

Equation (9.2-22) is the finite element approximation of the coupled energy equation. Equations (9.2-18) and (9.2-22) are assembled into a matrix form resulting in the general finite element coupled equation given by ¨ + [C]{Δ} ˙ + [K]{Δ} = {F } [M ]{Δ}

(9.2-23)

where [M ], [C], and [K] are the mass, the damping, and the stiffness matrices, respectively. Matrix {Δ}T = < Ui , θ > is the matrix of unknowns, and {F } is the known mechanical and thermal force matrix. For a two-dimensional problem, l and m take the values 1, 2, . . . , r. In this case, Eq. (9.2-18) reduces into two equations in x and y-directions, as

( V (e)

∂Nl ∂y V (e) ∂Nl + λ ∂x V (e) G

+

= V (e)

∂Nl ∂Nm dV ∂x ∂x V (e)

∂Nl ∂Nm ∂Nm G dV Um + dV ∂y ∂y ∂x V (e)

∂Nl ∂Nm βNm dV Vm − dV θm ∂y ∂x V (e)

ρNl Nm dV )U¨m +

(2G + λ)

X Nl dV +

A(e)

tnx Nl dA

( V (e)

∂Nl ∂x V (e) ∂Nl + λ ∂y V (e) G

= V (e)

∂Nl ∂Nm dV ∂y V (e) ∂y

∂Nl ∂Nm ∂Nm G dV Vm + dV ∂x ∂x ∂y V (e)

∂Nm ∂Nl dV Um − dV θm βNm ∂x ∂y V (e)

ρNl Nm dV )V¨m + (2G + λ)

+

(9.2-24)

Y Nl dV +

A(e)

tny Nl dA

(9.2-25)

The energy equation (9.2-22) for a two-dimensional problem becomes

( V (e)

T0 β

∂Nm ∂Nm T0 β Nl dV )U˙ m + ( Nl dV )V˙ m ∂x ∂y V (e)

418


+(

V (e)

ρcNm Nl dV )θ˙m + (

k V (e)

∂Nm ∂Nl dV ∂x ∂x

∂Nm ∂Nl dV )θm = k RNl dV − (qi ni )Nl dA + ∂y ∂y V (e) V (e) V (e)

(9.2-26) The elements of the mass, damping, stiffness, and the force matrices of the base element (e) are ⎡

[ ⎢ ⎢

⎢ [M ](e) = ⎢ ⎢

⎤

V (e)

ρNl Nm dV ]

0⎥

0

⎣

0 [ V (e)

0

ρNl Nm dV ] 0

⎥ ⎥ 0⎥ ⎥ ⎦

(9.2-27)

0

The damping matrix is ⎡

⎤

0 0 0 ⎢ ⎥ 0 0 0 ⎢ ⎥ ⎥ [C](e) = ⎢ ⎣ ⎦ ∂Nm ∂Nm T0 β T0 β ρcNm Nl dV ] [ Nl dV ] [ Nl dV ] [ ∂x ∂y v(e) v(e) v(e) (9.2-28) and the stiffness matrix is ⎡

[K](e)

⎤

[K11 ] [K12 ] [K13 ] ⎢ = ⎣ [K21 ] [K22 ] [K23 ] ⎥ ⎦ [K31 ] [K32 ] [K33 ]

(9.2-29)

where

∂Nl ∂Nm ∂Nl ∂Nm G dV + dV ] ∂x ∂x ∂y ∂y V (e) V (e) ∂Nl ∂Nm ∂Nl ∂Nm lm [K12 ]=[ G λ dV + dV ] ∂y ∂x ∂x ∂y V (e) V (e) ∂Nl lm ] = −[ βNm dV ] [K13 ∂x V (e) ∂Nl ∂Nm ∂Nl ∂Nm lm [K21 ]=[ G λ dV + dV ] ∂x ∂y ∂y ∂x V (e) V (e) ∂Nl ∂Nm ∂Nl ∂Nm lm dV + dV ] (2G + λ) G [K22 ] = [ ∂y ∂y ∂x ∂x V (e) V (e) ∂Nl lm [K23 ] = −[ βNm dV ] ∂y V (e) lm lm ] = [K32 ]=0 [K31 ∂Nm ∂Nl ∂Nm ∂Nl lm [K33 ]=[ k k dV + dV ] (9.2-30) ∂x ∂x ∂y ∂y V (e) V (e) lm [K11 ]=[

(2G + λ)


419

The force matrix is

(e) {f }l

⎧ ⎪ ⎪ ⎪ ⎪ ⎨

{

V (e)

XNl dV +

=⎪ { ⎪

⎫

A(e)

V (e) Y Nl dV + ⎪ ⎪ A(e) ⎩

{

V (e)

RNl dV −

tnx Nl dA} ⎪ ⎪ ⎪ ⎪ ⎬

tny Nl dA}

V (e) (qi

ni )Nl dA}

⎪ ⎪ ⎪ ⎪ ⎭

(9.2-31)

and the unknown matrix is ⎧ ⎫ ⎪ ⎨ {U } ⎪ ⎬

{δ}(e) = ⎪ {V } ⎪ ⎩ {θ} ⎭

(9.2-32)

The initial and the general form of the thermal boundary conditions is one, or the combinations, of the following: θ(x, y, z, 0) = 0(x, y, z) at t = 0 θ(x, y, z, t) = θs on A1 and t > 0 qx nx + qy ny + qz nz = −q on A2 and t > 0 qx nx + qy ny + qz nz = h(θ + T0 − T∞ ) on A3 and t > 0 qx nx + qy ny + qz nz = σ(θ + T0 )4 − αab qr on A4 and t > 0 (9.2-33) where T0 (x, y, z) is the known initial temperature, θs is the known specified temperature change on a part of the boundary surface A1 , q is the known heat flux on the boundary A2 , h and T∞ are the convection coefficient and ambient temperature specified on a part of the boundary surface A3 , respectively, σ is Stefan-Boltzmann constant, is the radiation coefficient of the boundary surface, αab is the boundary surface absorption coefficient, and qr is the rate of thermal flux reaching the boundary surface per unit area all specified on boundary surface A4 . The cosine directors of the unit outer normal vector to the boundary in x, y, and z-directions are shown by nx , ny , and nz , respectively. According to the boundary conditions given by Eqs. (9.2-33), the last surface integral of the energy equation (9.2-22) may be decomposed into four integrals over A1 through A4 as

A(e)

−

(qi ni )Nl dA = A4

A2

q Nl dA −

A3

(σ(θ + T0 )4 − αab qr )Nl dA

h(θ + T0 − T∞ )Nl dA l = 1, 2, . . . , r

(9.2-34)

Note that the signs of the integrals in Eq. (9.2-34) depend upon the direction of the heat input. The positive sign is defined when the heat is given to the body, and is negative when the heat is removed from the body. That is, q is defined positive in Eq. (9.2-34), since we have assumed that the heat flux is

420


given to the body. On the other hand, we have assumed negative convection on the surface area A3 , which means the heat is removed from A3 boundary by convection. Similarly, the boundary A4 is assumed to radiate to the ambient, as the sign of this integral is considered negative. In order to discuss the method in more detail a one-dimensional problem is considered [6–8]. The equation of motion in terms of displacement is ∂ 2u ∂θ ∂2u = ρ − β ∂x2 ∂x ∂t2 and the first law of thermodynamics reduces to

(9.2-35)

(λ + 2G)

∂θ ∂2θ ∂2u − ρc − βT =0 (9.2-36) 0 ∂x2 ∂t ∂x∂t Taking a line element of length L, the approximating function for axial displacement for the base element (e) is assumed to be linear in x as k

u(x, t)(e) = Ni Ui + Nj Uj = < N >(e) {U }(e)

(9.2-37)

where the piecewise linear shape function < N > is Ni = (L−η)/L, Nj = η/L, and η = x − xi . Similarly, the temperature change is approximated by θ(x, t)(e) = Ni θi + Nj θj = < N >(e) {θ}(e)

(9.2-38)

Employing the formal Galerkin method and applying the weak form to the first and second terms of Eq. (9.2-35) and first term of Eq. (9.2-36) results in the following system of equations

L 0

−

ρNl Nm dη U¨m +

L

L

0

L 0

∂Nl ∂Nm (2G + λ) dη Um ∂η ∂η

L j ∂Nl βNm XNl dη dη θm = tnx Nl + ∂η 0 i

(9.2-39)

L ∂Nm T0 β ρcNm Nl dη θ˙m Nl dη U˙ m + ∂η 0 0

j L L ∂Nm ∂Nl + k RNl dη dη θm = −(qx nx )Nl + ∂η ∂η 0 0 i

(9.2-40)

This system of equations may be written in matrix form as ¨ + [C]{Δ} ˙ + [K]{Δ} = {F } [M ]{Δ}

(9.2-41)

where the mass, damping, stiffness and force matrices for first order element are 4 × 4 matrices and are defined as ⎡

ρNi Ni

L⎢ 0 ⎢ (e) ⎢ [M ] = ⎣ ρN 0 j Ni

0

0 ρNi Nj 0 0 0 ρNj Nj 0 0

⎤

0 0⎥ ⎥ ⎥ dη 0⎦ 0

(9.2-42)

2. Galerkin Finite Element ⎡

[C](e)

0 ⎢ ∂Ni L ⎢ T0 β Ni ⎢ ∂η ⎢ = ⎢ 0 0 ⎢ ⎣ ∂Ni T0 β Nj ∂η [K](e)

where

421 ⎤

0 ρcNi Ni

0 0 ⎥ ∂Nj T0 β Ni ρcNi Nj ⎥ ⎥ ∂η ⎥ dη ⎥ 0 0 0 ⎥ ⎦ ∂Nj ρcNj Ni T0 β Nj ρcNj Nj ∂η ⎡ ⎤ K11 K12 K13 K14 ⎢ 0 K22 0 K24 ⎥ ⎥ ⎢ ⎥ =⎢ ⎣ K31 K32 K33 K34 ⎦ 0 K42 0 K44

L

K11 =

(2G + λ)( 0

K12 = −

L 0

βNi L

K13 = K31 = K14 = −

L

L

K32 K33 K34 K44

∂Ni dη ∂η

0

∂Ni ∂Nj dη ∂η ∂η

∂Ni dη ∂η

∂Ni 2 ) dη ∂η 0 L ∂Nj ∂Ni = K42 = k dη ∂η ∂η 0 L ∂Nj =− βNi dη ∂η 0 L ∂Nj 2 = (2G + λ)( ) dη ∂η 0 L ∂Nj =− βNj dη ∂η 0 L ∂Nj 2 = k( ) dη ∂η 0

K22 = K24

0

(9.2-44)

∂Ni 2 ) dη ∂η

(2G + λ)

βNj

(9.2-43)

k(

(9.2-45) ⎧ ⎫ L L ⎪ ⎪ n ⎪ ⎪ ⎪ tx Ni + XNi dη ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 0 0 ⎪ ⎪ ⎪ ⎪ L ⎪ ⎪ L ⎪ ⎪ ⎪ ⎪ ⎪ RNi dη ⎪ ⎨ −qx Ni + ⎬ (e) 0 0 L L {F } = ⎪ ⎪ ⎪ ⎪ ⎪ tnx Nj + XNj dη ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 0 ⎪ ⎪ 0 ⎪ ⎪ ⎪ ⎪ L L ⎪ ⎪ ⎪ ⎪ ⎪ −q N + ⎪ ⎩ ⎭ RN dη x j j 0

(9.2-46)

0

Upon substitution of the shape functions in the foregoing equations, the submatrices for the base element (e) are

422

Chapter 9. Finite and Boundary Element Methods ⎡ ρL 3 ⎢ 0 ⎢ (e) [M ] = ⎢ ρL ⎣ 6

0 ⎡

0

⎣

0

(9.2-47) ⎤

0

0

ρcL 3

T0 β 2

ρcL 6

0

0

0

− T02β

ρcL 6

T0 β 2

⎡ (2G+λ) L ⎢ ⎢ 0 ⎢ [K](e) = ⎢ ⎢ (2G+λ) ⎢− L ⎣

0

0 ρL 3

⎤

0 0⎥ ⎥ ⎥ 0⎦ 0

0

⎢ T0 β ⎢− 2

[C](e) = ⎢ ⎢

ρL 6

0 0 0 0

β 2 k L − β2 − Lk

⎥ ⎥ ⎥ 0 ⎥ ⎦

ρcL 3

− (2G+λ) L 0 (2G+λ) L

0

⎧ n ⎪ −tx + XL ⎪ ⎪ 2 ⎪ 0 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ RL ⎪ ⎪ ⎨ qx + 2 0 (e) {F } = ⎪ ⎪ ⎪ ⎪ tnx + XL ⎪ 2 ⎪ ⎪ L ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ −q + RL x 2

(9.2-48)

⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭

β

⎤

2

⎥ − Lk ⎥ ⎥ ⎥

⎥ −2 β ⎥ ⎦

(9.2-49)

k L

(9.2-50)

L

and the matrix of unknown nodal values is

{Δ}(e) =

⎧ Ui ⎪ ⎪ ⎪ ⎨θ i

⎫ ⎪ ⎪ ⎪ ⎬

⎪ Uj ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ⎭

(9.2-51)

θj

3

Functionally Graded Layers

The conventional heat conduction theory assumes that the thermal disturbances propagate at infinite speeds. However, arguments questioning the validity of such a phenomenon appear in literature for specialized applications involving very short transient durations, sudden high heat flux situations, and/or very low temperatures near the absolute zero. The concept of the so-called hyperbolic nature involving finite speeds of thermal disturbance dates as far back as the times of Maxwell [9]. Thermal disturbances of a hyperbolic nature have also been derived using various approaches (Landau [10], Peshkov [11], Cattaneo [12], Vernotte [13]). Most of these approaches are based on the general notion of relaxing the heat flux in the classical Fourier heat conduction

3. Functionally Graded Layers

423

equation, thereby introducing a non-Fourier effect. There is also some contradiction to these non-classical propositions in thermoelasticity, with arguments questioning the applicability of the finite speeds of propagation in gases to that occurring in solid continua. Chester [14] provides some justification to the fact that the so-called second sound must exist in any solid since all solid continua exhibit phonon-type excitations. In an idealized solid, for example, the thermal energy can be transported by quantized electronic excitations, which are called free electrons, and by the quanta of lattice vibrations, which are called phonons. These quanta undergo collisions of a dissipative nature, causing a thermal resistance in the medium. A relaxation time t0 is associated with the average communication time between these collisions for the commencement of resistive flow. Though the convergence time for the solutions of the hyperbolic model to that of the parabolic model is small, it may become important when extremely short times are involved. It is under such situations that the assumption of a parabolic heat conduction model may lead to inaccurate modeling of the transient thermal behavior. As a consequence, a hyperbolic heat conduction model that allows for both transient heat conduction and finite nature of thermal energy transport is argued as a reasonable substitute for evaluating the propagation of thermal and thermally induced stress disturbances. Among the various propositions, the applicability of these non-classical theories is well summarized by Ignaczak [15] where the theories due to Lord and Shulman [16] and Green and Lindsay [17] are included. Excellent references are also available due to Chandrasekharaiah [18], and Joseph and Preziosi [19,20]. Functionally Graded Materials (FGMs) are high-performance heat resistant materials able to withstand ultra high temperatures and extremely large thermal gradients used in the aerospace industries. FGMs are microscopically inhomogeneous in which the mechanical properties vary smoothly and continuously from one surface to the other [21]. Typically, these materials are made from a mixture of ceramic and metal. The coupled thermoelasticity of a layer with isotropic material is discussed in the literature. Bagri et al. [22] proposed a new system of coupled equations that contains Lord–Shulman (LS), Green–Lindasy (GL), and Green–Naghdi (GN) models [16,17,23,24] in a unified form. They employed the suggested formulation and then analytically solved the coupled system of equations for a layer using Laplace transform. The effect of the thermomechanical coupling coefficient in the problems of coupled thermoelasticity is pointed out in references [25] and [26]. Except for a few cases of coupled problems, the general closed form solution for the coupled thermoelasticity problems is not available in the literature and most of the relevant problems are solved numerically. Among the numerical procedures, the boundary and finite element method as well as finite difference method are most considered for these type of problems. Tamma and Namburu [27] have presented an overview of these numerical methods. Chen and Lin [28] proposed a hybrid numerical method based on Laplace transform and control

424


volume method to analyze the transient coupled thermoelastic problems with relaxation times involving a nonlinear radiation boundary condition. Hosseini Tehrani and Eslami [29,30] considered the boundary element formulation for the analysis of coupled thermoelastic problems in a two-dimensional finite domain and studied the coupling coefficient effects on thermal and elastic waves propagation. The response of functionally graded materials under dynamic thermal loads and using the coupled form equations of thermoelasticity theories are found in just a few articles. Zhang et al. [31] modeled an isotropic ceramic-metal laminated beam subjected to an abrupt heating condition, and demonstrated the influence of thermomechanical coupling on the thermal shock response. Praveen and Reddy [32] studied the static and dynamic response of the functionally graded plates and showed that the response of FG plates is not intermediate to the response of the ceramic and metal plates. Bagri et al. [33] considered the classical coupled thermoelasticity theory to study the behavior of an FG layer under thermal shock load. The effect of the material composition profile on the distribution of temperature, displacement, and stress through the thickness of the layer is studied. Considering an FGM layer under thermal shock, the analysis may need the solution obtained through the coupled thermoelasticity equations. For severe thermal shocks, the generalized models, such as LS, GL, or GN models, may seem to be proper. In this section the dynamic response of a layer made of FGMs based on LS theory is investigated (Bagri et al. [34]). The power law form function is assumed for the material properties distribution. A suitable transfinite element method via Laplace transform is employed to find the temperature and displacement field solution in the space domain. Finally, the temperature, displacement and stress in the physical time domain are obtained using a numerical inversion of Laplace transform proposed by Honig and Hirdes [35]. The temperature, displacement and stress waves propagation and reflection from the boundaries of a layer are studied. Also, the relaxation time and material volume fraction effects on temperature, displacement and stress variations are investigated. Consider a ceramic-metal FG layer with thickness of L and assume that the properties of FG layer obey a power law function as x (9.3-1) P = ( )n (Pm − Pc ) + Pc L where x is the position from the ceramic rich side of the layer, P is the effective property of FGM, n is the power law index that governs the distribution of the constituent materials through the thickness of the layer, and Pm and Pc are the properties of metal and ceramic, respectively. Meanwhile, the subscripts m and c indicate the metal and ceramic features, respectively. For LS theory, in the absence of body forces and heat supply, the governing equations of an FG layer are as follows [34]


425

Equation of motion ∂σxx ∂ 2u =ρ 2 ∂x ∂t

(9.3-2)

Strain-displacement relation xx =

∂u ∂x

(9.3-3)

Stress-strain relation σxx = (2μ + λ)xx − β(T − T0 )

(9.3-4)

Combining these equations and writing in terms of the displacement u, yields ∂ 2 u ∂(λ + 2μ) ∂u ∂[β(T − T0 )] ∂2u + (9.3-5) − = ρ ∂x2 ∂x ∂x ∂x ∂t2 The heat conduction equation for this one-dimensional case when the derivative of the relaxation time with respect to the position variable is neglected and k is a function of position, from Eq. (2.12-17), becomes (λ + 2μ)

∂2T ∂2 ∂ ∂T ∂ ∂u ∂T (k ) − ρc − ρct0 2 − βT0 (t0 2 + ) =0 (9.3-6) ∂x ∂x ∂t ∂t ∂t ∂t ∂x where t0 is the relaxation time proposed by Lord and Shulman. The preceding equations may be introduced in dimensionless form for convenience. The nondimensional parameters are defined as

T − T0 ; T¯ = km Td t(λ t m + 2μm )cm 0 (λm + 2μm )cm t¯ = ; t¯0 = km km qx σxx ; σ ¯xx = q¯x = βm Td cm Td ρm (λm + 2μm ) x¯ =

u¯ =

xcm ρm (λm + 2μm )

1/2 (λm + 2μm )3/2 ρm cm u km βm Td

(9.3-7)

where the subscript m denotes the metal properties and term Td is a characteristic temperature used for normalizing the temperature. Using the dimensionless parameters, the governing equations (9.3-5) and (9.3-6) appear in the form

∂(λ + 2μ) ∂ ρ ∂2 (λ + 2μ) ∂ 2 1 − + u¯ (λm + 2μm ) ∂ x¯2 (λm + 2μm ) ∂ x¯ ∂ x¯ ρm ∂ t¯2 ∂ 1 ∂β + β )T¯ = 0 − ( βm ∂ x¯ ∂ x¯

2 k ∂2 ρc ∂ ∂ 1 ∂k ∂ − + ( + t¯0 2 ) T¯ km ∂ x¯2 km ∂ x¯ ∂ x¯ ρm cm ∂ t¯ ∂ t¯ βm T0 ∂ ∂ u¯ ∂2 − β(t¯0 2 + ) =0 ¯ ρm cm (λm + 2μm ) ∂t ∂ t¯ ∂ x¯

(9.3-8)

(9.3-9)

426


Also, the dimensionless stress-displacement-temperature relation and the heat conduction equation for the functionally graded layer based on LS theory are (λ + 2μ) ∂ u¯ β ¯ T − (λm + 2μm ) ∂ x¯ βm k ∂ T¯ ∂ q¯x q¯x + t¯0 =− ∂ t¯ km ∂ x¯

σ ¯xx =

(9.3-10) (9.3-11)

The layer is occupied in the region 0 ≤ x¯ ≤ 1 and the corresponding dimensionless boundary conditions are ¯ q¯x = 1 − (1 + 100t¯)e−100t ; T¯ = 0;

σ ¯xx = 0 u¯ = 0

at x¯ = 0 at x¯ = 1

(9.3-12)

To solve the coupled system of equations, the transfinite element method may be employed. To this end, the equations may be transformed to the space domain using Laplace transformation. Assume that the layer is initially at rest and the initial displacement, velocity, temperature, and temperature rate are zero. Applying Laplace transformation to Eqs. (9.3-8) and (9.3-9) give

1 (λ + 2μ) ∂ 2 ∂(λ + 2μ) ∂ ρ 2 ∗ + s u¯ − 2 (λm + 2μm ) ∂ x¯ (λm + 2μm ) ∂ x¯ ∂ x¯ ρm 1 ∂β ∂ − ( (9.3-13) + β )T¯∗ = 0 βm ∂ x¯ ∂ x¯

1 ∂k ∂ k ∂2 ρc ¯ + s(1 + t0 s) T¯∗ − km ∂ x¯2 km ∂ x¯ ∂ x¯ ρm cm βm T0 ∂ u¯∗ − βs(1 + t¯0 s) =0 (9.3-14) ρm cm (λm + 2μm ) ∂ x¯ To find the solution of the equations using the transfinite element method, the geometry of the layer may be divided into a number of discretized elements through the thickness of the layer. In the base element, the Kantorovich approximation for the displacement u and temperature T with identical shape functions is assumed as u¯∗(e) =

Ni U¯i∗

T¯∗(e) =

i=1

Ni T¯i∗

(9.3-15)

i=1

where Ni is the shape function and terms U¯i∗ and T¯i∗ are the unknown nodal values of displacement and temperature, respectively. Substituting Eqs. (9.3-15) into Eqs. (9.3-13) and (9.3-14) and then employing Galerkin finite element method, the following system of equations, applying the weak form to the terms of second order of derivatives of the space variable, is obtained:

[K11 ] [K21 ]

[K12 ] [K22 ]

¯∗ U

T¯∗

=

F∗ Q∗

(9.3-16)


427

The submatrices [K11 ], [K12 ], [K21 ], [K22 ] for the base element (e), and the global force matrices F and Q are ij ] [K11

ij [K12 ]

=

x ¯f

∂Ni ∂Nj ρs2 Ni Nj d¯ x + (λm + 2μm ) ∂ x¯ ∂ x¯ ρm

1 x¯e ∂Nj ∂β = βNj x + Ni Nj d¯ βm x¯f ∂ x¯ ∂ x¯

ij [K21 ]=

ij ] [K22

x¯e (λ + 2μ)

=

(9.3-18)

x¯e sβm T0 ∂Nj β(1 + t¯0 s)Ni d¯ x ρm cm (λm + 2μm ) x¯f ∂ x¯ x¯e k ∂Ni ∂Nj x ¯f

km

⎧ βc ¯ ∗ ⎫ T ⎪ ⎪ βm 1 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 0 ⎨ ⎬

(9.3-17)

(9.3-19)

ρc + s(1 + t¯0 s)Ni Nj d¯ x ∂ x¯ ∂ x¯ ρm c m ⎧ ⎫ (1 + t¯0c s)¯ qx∗ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 0 ⎨ ⎬

(9.3-20)

(9.3-21) . ⎪ ; {Q} = ⎪ . ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ . ⎪ . ⎪ ⎪ ⎪ ⎪ ⎪ ⎭ ⎩ ⎭ 0 0 In these equations x¯f and x¯e are positions of the first and last nodes of the base element along the x axis, respectively. System of Eq. (9.3-16) is solved in the space domain. To transform the results from Laplace transform domain into the real time domain, the numerical inverse Laplace transform technique given in reference [35] is used. Consider an FG layer composed of aluminum and alumina as metal and ceramic constituents, respectively. The reference temperature is assumed to be T0 = 298 K. The relaxation time of aluminum and alumina is assumed to be t¯0m = 0.64, t¯0c = 1.5625, respectively. The linear Lagrangian polynomials are used for the shape functions in the base element. Figures 9.3-1 and 9.3-2 show the temperature and stress waves propagation and reflection from the boundaries of the layer for n = 1. In Fig. 9.3-1, the times t¯ = 0.2, 0.4, 0.6, 0.8, 1.7 show the temperature wave propagation through the thickness of the layer, while the reflection of the temperature wave occurred at time t¯ = 1.9. Figure 9.3-2 shows that the maximum of stress occurs at the temperature wave front. In Figs. 9.3-1 and 9.3-2 it is seen that a conversion between the mechanical and thermal energies occurs at the temperature wave front. It may be found from the figures that the propagation velocity of waves varies through the thickness of the layer. Variation of the power law index n provide different profiles of the constituent materials through the thickness of the layer. When n = 0, the FGM profile becomes that of pure metal. Larger values of n provides more ceramic rich FGM. The effect of power law index, n, on variation of the temperature and stress at a point located at the middle point of the layer thickness is shown in Figs. 9.3-3 and 9.3-4. It is seen from Fig. 9.3-3 that when n increases, {F } = ⎪ ⎪ ⎪ ⎪ ⎪ ⎩

428


Figure 9.3-1: Distribution of the nondimensional temperature through the thickness of the layer for n = 1.

Figure 9.3-2: Distribution of the nondimensional stress through the thickness of the layer for n = 1. the speed of temperature wave decreases. In Fig. 9.3-4 it is shown that the amplitude of stress variation is decreased with the increase of n. The relaxation time effect on variation of the temperature and stress at the middle point of the thickness of the layer is investigated and is shown in Figs. 9.3-5 and 9.3-6. The value of n = 1 is considered for the power law index. It is seen that for the classical theory of thermoelasticity, the case when t0 = 0, smaller values for amplitude of temperature and resulting stress variations are obtained. Since with the increase of relaxation time the propagation velocity of temperature wave decreases, these maximum values of variations occur at the later times.

4

Coupled Thermoelasticity of Thick Spheres

The finite element analysis of coupled thermoelasticity of thick spheres and cylinders was studied by Li et al. [36], by Ghoneim [37], and by Eslami and

4. Coupled Thermoelasticity of Thick Spheres

429

Nondimensional temperature

3 2.5 2 1.5 1 0.5 0 −0.5

0

2

4

6

8

10 n=1

−1

n=2

Nondimensional time

n=5


Figure 9.3-3: Variations of the nondimensional temperature at the middle point of the thickness of the layer for different values of power index. 6 5 4 3 2 1 0 −1 0 −2 −3 −4 −5

2

4

6

8

10

n=1 n=2 n=5 Nondimensional time

Figure 9.3-4: Variations of the nondimensional stress at the middle point of the thickness of the layer for different values of power index. 2.5 Nondimensional temperature

Classical theory 2

LS theory

1.5 1 0.5 0 −0.5

0

2

4

6

8

10

12

14

−1 Nondimensional time

Figure 9.3-5: Variations of the nondimensional temperature at the middle point of the thickness of the layer for the classical and LS theories.



430 6 5 4 3 2 1 0 −1 0 −2 −3 −4 −5

2

4

6

8

10

12

14

Classical theory LS theory

Nondimensional time

Figure 9.3-6: Variations of the nondimensional stress at the middle point of the thickness of the layer for the classical and LS theories. Vahedi [38,39]. Galerkin method is basically used to obtain the finite element formulations. The analysis is also based on displacement formulation. Consider a thick-walled sphere of inside and outside radii rin and rout , respectively. For symmetric loading condition σθθ = σφφ

θθ = φφ

(9.4-1)

the equation of motion is ∂σrr σrr − σφφ +2 = ρ¨ u ∂r r and the strain-displacement relations are rr =

∂u ∂r

θθ =

u = φφ r

(9.4-2)

(9.4-3)

From Hooke’s law E [(1 − ν)rr + 2νφφ − (1 + ν)α(T − T0 )] (1 + ν)(1 − 2ν) E = [φφ + νrr − (1 + ν)α(T − T0 )] (1 + ν)(1 − 2ν)

σrr = σφφ

(9.4-4) The first law of thermodynamics for coupled condition is ∂2T 2 ∂T ρc + − T˙ = γ1 (˙rr + 2˙φφ ) − R ∂r2 r ∂r k

(9.4-5)

where γ1 = (3λ+2μ)αT0 /k. Elimination of the stresses from Eqs. (9.4-2), (9.4-3), and (9.4-4) results in the equation of motion in terms of the displacement (1 + ν) ∂T (1 + ν)(1 − 2ν)ρ ∂ 2 u 2 ∂u 2u + − 2 −α = u¨ 2 ∂r r ∂r r (1 − ν) ∂r (1 − ν)E

(9.4-6)


431

The energy equation, after substitution for strains, becomes ∂ 2T ρc γ1 ∂ 2 ∂T − T˙ = 2 (r2 u) + ˙ −R 2 ∂r r ∂r k r ∂r

(9.4-7)

The boundary conditions are in general given as −k

σrr nr = tnr

∂T = qr ∂r

(9.4-8)

where nr is the unit vector in radial direction. In terms of displacement, the boundary conditions at inside and outside radii are At r = rin E(1 − ν) ∂u ν 2u [ + − (1 + ν)(1 − 2ν) ∂r (1 − ν) r ρ c c21 t 2 ρ c c21 T (t) = T0 2 + ( ) − k k

(1 + ν) α(T − T0 )] = −Pa (t) (1 − ν)

ρ c c2 t t − k1 (9.4-9) −1 e

At r = rout u=0 ∂T k = −ho [T − T0 ] ∂r

(9.4-10)

where ho is the convection coefficient at inside and outside surfaces of the sphere, respectively, Ti (t) is the inside surface temperature which is assumed to vary in time and is applied as a thermal shock to the inside surface, T0 is the constant outside ambient temperature, Pa (t) is the applied pressure shock at the inside surface which may be considered zero. The governing equations are changed into dimensionless form through the following formulas (T − T0 ) T¯ = T0 1 − ν ρ c c1 u u¯ = ( ) 1 + ν k α T0 (1 − 2ν)σrr σ ¯rr = EαT0 ρ c c1 r r¯ = k ρ c c21 t t¯ = 3k 4 4

c1 = 5

(1 − ν)E (1 + ν)(1 − 2ν)ρ

(9.4-11)

432


Using these quantities, and in the absence of heat generation, the governing equations are expressed in dimensionless form (bar is dropped for convenience) ∂T ∂ ∂u 2u ( + )− = u¨ ∂r ∂r r ∂r 2 ∂T 1 ∂ ∂ ∂T ˙ ( )+ − T˙ = C 2 (r2 u) ∂r ∂r r ∂r r ∂r

(9.4-12) (9.4-13)

The boundary conditions are u 1 − 2ν ∂u + γ2 − T = − Pa (t) ∂r r EαT0 T = 1 + (t2 − t − 1 ) e−t

at r = a

(9.4-14)

and u=0 ∂T = −ηo T ∂r

at r = b

(9.4-15)

where a and b are the dimensionless inner and outer radii of the sphere, respectively, and the parameters used in these equations are defined as C=

2ν ho T0 (1 + ν)Eα2 ; γ2 = ; ηo = ρc(1 − ν)(1 − 2ν) (1 − ν) ρcc1

(9.4-16)

Due to the radial symmetry of loading conditions, the variations of the dependent functions are located along the radius of the sphere. Thus, the radius of the sphere is divided into a number of line elements (N E) with nodes i and j for the base element (e), as shown in Fig. 9.4-1. The displacement of element (e) is described by the linear shape function u(r, t) = α1 (t) + α2 (t)r

(9.4-17)

in terms of nodal unknown variables, the unknown coefficients α1 (t) and α2 (t) are found from Eq. (9.4-17) as Ui (t) = α1 + α2 ri Uj (t) = α1 + α2 rj

Figure 9.4-1:: The element (e) along the radius.

(9.4-18)


433

Solving for α1 and α2 and substituting in Eq. (9.4-17) yields u(η, t) =

L−η η Ui + Uj L L

(9.4-19)

where η is the variable in local coordinates, η = r − ri , and L is the element length L = rj − ri . Defining the linear shape functions as Ni =

L−η L

Nj =

η L

(9.4-20)

the displacement u is allowed to vary linearly in the base element (e) as (e)

u (η, t) = Ni Ui + Nj Uj =< Ni Nj >

Ui Uj

=< N >(e) {U }(e)

(9.4-21)

Similarly, the temperature variation in the element (e) is assumed to vary linearly T (e) (η, t) = Ni Ti + Nj Tj =< N >(e) {T }(e)

(9.4-22)

Using Eqs. (9.4-21) and (9.4-22) and applying the formal Galerkin method to the governing equations (9.4-12) and (9.4-13) for the base element (e) yields rj ∂

∂u 2u ∂T ∂2u + )− − 2 ]r2 Nm dr = 0 (9.4-23) r ∂r ∂t ri ∂r ∂r rj ∂ ∂T 2 ∂T ∂T 1 ∂ ∂u [ ( )+ − − C 2 (r2 )]r2 Nm dr = 0 r ∂r ∂t r ∂r ∂t ri ∂r ∂r m = i, j (9.4-24) [

(

Considering the change of variable η = r−ri and applying the weak formulation to the terms of second order derivatives gives L ∂[(η + ri )2 Nm ] ∂u

L

L ∂[(η + ri )2 Nm ] ∂T

L

2 ∂u 2u (η + ri ) Nm dη dη − − ∂η ∂η (η + ri ) ∂η (η + ri )2 0 0 L L ∂T ∂u L (η + ri )2 Nm (η + ri )2 Nm u¨dη = (η + ri )2 Nm (9.4-25) + dη + ∂η ∂η 0 0 0 2

L ∂T dη − dη + 2(η + ri )Nm (η + ri )2 Nm T˙ dη ∂η ∂η ∂η 0 0 0 L ∂ ∂T L +C Nm [(η + ri )2 u]dη ˙ = (η + ri )2 Nm ∂η ∂η 0 0 m = i, j (9.4-26)

Substituting the shape functions for u and T from Eqs. (9.4-21) and (9.4-22) yields

434


L

d[(η + ri )2 Nm ]

dN dN < > −2Nm [(η + ri ) < > − < N >] dη{U } dη dη dη 0 L L dN (η + ri )2 Nm < (η + ri )2 Nm < N > dη{U¨ } > dη{T } + + dη 0 0 ∂u L (9.4-27) = (η + ri )2 Nm ∂η 0

L

d[(η + ri )2 Nm ]

dN dN < > −2(η + ri )Nm < > dη{T } dη dη dη 0 L L d[(η + ri )2 N ] > dη{U˙ } (η + ri )2 Nm < N > dη{T˙ } + C Nm < + dη 0 0 dT L = (η + ri )2 Nm dη 0 m = i, j (9.4-28)

The terms on the right-hand side of Eqs. (9.4-27) and (9.4-28) are derived through the weak formulation and coincide with the natural boundary conditions. They cancel out each other between any two adjacent elements except the first node of the first element and the last node of the last element which coincide with the given boundary conditions on inside and outside surfaces of the sphere. These boundary conditions are −a

2 2 1 − 2ν Pa (t) = 2aγ2 U1 − a T1 + a

2 ∂u

∂η 1 UM = 0 T1 = 1 + (t2 − t − 1 ) e−t ∂T b2 = −b2 ηo TM ∂η M

EαT0

(9.4-29)

where the index 1 denotes the first note of the first element of the solution domain at r = a, and the index M denotes the last node of the last element of the solution domain at r = b. It is to be noted that due to the assumed boundary ∂T ∂u |1 and b2 |M vanish. Equations conditions at r = a and r = b, terms −a2 ∂η ∂η (9.4-27) and (9.4-28) are solved for the nodal unknown of the element (e) and are finally arranged in the form of the following matrix equations ¨ + [C]{Δ} ˙ + [K]{Δ} = {F } [M ]{Δ}

(9.4-30)

The definitions of the mass, damping, stiffness, and force matrices of Eqs. (9.4-30) for the base element (e) are

4. Coupled Thermoelasticity of Thick Spheres Mass matrix

435

⎡

⎤

m11 0 m13 ⎢ 0 0 0 [M ](e) = ⎢ ⎢ ⎣ m31 0 m33 0 0 0 where the components of the mass matrix are L

m11 =

0

L

m13 =

0

L

m31 =

0

L

m33 = Damping matrix

0

⎡

[C](e)

0 ⎢C ⎢ 21 =⎢ ⎣ 0 C41

0 0⎥ ⎥ ⎥ 0⎦ 0

(9.4-31)

(η + ri )2 Ni Ni dη (η + ri )2 Ni Nj dη (η + ri )2 Nj Ni dη (η + ri )2 Nj Nj dη

0 C22 0 C42

(9.4-32) ⎤

0 C23 0 C43

0 C24 ⎥ ⎥ ⎥ 0 ⎦ C44

(9.4-33)

where the components of the damping matrix are L

d[(η + ri )2 Ni ] dη dη 0 L d[(η + ri )2 Nj ] =C Ni dη dη 0 L d[(η + ri )2 Ni ] =C Nj dη dη 0 L d[(η + ri )2 Nj ] =C Nj dη dη 0

C21 = C C23 C41 C43

L

C22 =

0

L

C24 =

0

L

C42 =

0

L

C44 = Stiffness matrix

⎡

[K](e)

0

Ni

(η + ri )2 Ni Ni dη (η + ri )2 Ni Nj dη (η + ri )2 Nj Ni dη (η + ri )2 Nj Nj dη

K11 ⎢ 0 ⎢ =⎢ ⎣ K31 0

K12 K22 K32 K42

K13 0 K33 0

(9.4-34) ⎤

K14 K24 ⎥ ⎥ ⎥ dη K34 ⎦ K44

(9.4-35)

436


where the components of the stiffness matrix are K11 K13 K31 K33 K12 K14 K32 K34 K22 K24 K42 K44

L

d[(η + ri )2 Ni ] dNi

dNi = − 2(η + ri )Ni + 2Ni Ni ) dη dη dη dη 0 L

d[(η + ri )2 Ni ] dNj dNj = − 2(η + ri )Ni + 2Ni Nj ) dη dη dη dη 0 L

d[(η + ri )2 Nj ] dNi dNi = − 2(η + ri )Nj + 2Nj Ni ) dη dη dη dη 0

L d[(η + ri )2 Nj ] dNj dNj = − 2(η + ri )Nj + 2Nj Nj ) dη dη dη dη 0 L dNi = (η + ri )2 Ni dη dη 0 L dNj = (η + ri )2 Ni dη dη 0 L dNi = (η + ri )2 Nj dη dη 0 L dNj = (η + ri )2 Nj dη dη 0

L d[(η + ri )2 Ni ] dNi dNi − 2(η + ri )Ni = dη dη dη dη 0 L

d[(η + ri )2 Ni ] dNj dNj = dη − 2(η + ri )Ni dη dη dη 0 L

d[(η + ri )2 Nj ] dNi dNi = dη − 2(η + ri )Nj dη dη dη 0 L

d[(η + ri )2 Nj ] dNj dNj = dη − 2(η + ri )Nj dη dη dη 0 (9.4-36)

Also, the matrix {F } for this special case is related to the boundary conditions. With the inside temperature and pressure shocks and outside surface insulated, as given by Eqs. (9.4-29), the final assembled form of this matrix becomes ⎧ ⎫ 2 2 1−2ν ⎪ ⎪ 2aγ2 U1 − a T1 + a EαT0 Pa (t) ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 1 + (t2 − t − 1 ) e−t ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 0 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 0 ⎨ ⎬ (9.4-37) {F } = ⎪ . ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ . ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 0 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 0 ⎪ ⎪ ⎪ ⎪ ⎩ ⎭ 2 −b ηo TM


437

Moreover, the matrix of unknown nodal values is ⎧ Ui ⎪ ⎪ ⎪ ⎨

⎫ ⎪ ⎪ ⎪ ⎬

T {Δ}(e) = ⎪ i ⎪ Uj ⎪ ⎪ ⎪ ⎪ ⎩ ⎭ Tj

(9.4-38)

Using the linear shape functions (9.4-21) and (9.4-22) for {U } and {T } the component of the matrices are simplified as Components of mass matrix L(L2 + 5ri L + 10ri2 ) 30 L(3L2 + 10ri L + 10ri2 ) = m31 = 60 L(6L2 + 15ri L + 10ri2 ) = 30

m11 = m13 m33

(9.4-39)

Components of damping matrix (L2 + 4ri L − 6ri2 ) 12 (3L2 + 8ri L + 6ri2 ) =C 12 (L2 + 4ri L + 6ri2 ) = −C 12 (9L2 + 16ri L + 6ri2 ) =C 12 (L2 + 5ri L + 10ri2 ) =L 30 (3L2 + 10ri L + 10ri2 ) = C42 = L 60 (6L2 + 15ri L + 10ri2 ) =L 30

C21 = C C23 C41 C43 C22 C24 C44

(9.4-40)

Components of stiffness matrix L2 + ri L + ri2 L ri (L + ri ) = K31 = − L 1 2 1 1 = −K14 = − L − ri L − ri2 12 3 2 1 2 2 1 2 = −K34 = − L − ri L − ri 4 3 2 L2 + 3ri L + 3ri2 = K44 = −K24 = −K42 = 3L

K11 = K33 = K13 K12 K32 K22

(9.4-41)

438


The element matrices given by Eqs. (9.4-34) to (9.4-36) are generated within a loop to construct the general matrices of Eqs. (9.4-30), where after assembly of all the elements in the solution domain, they are solved using one of the numerical techniques of either time marching or modal analysis methods. As a numerical example, a thick sphere is considered with the following properties: E = 70 × 109 N/m2 , ν = 0.3, ρ = 2707 Kg/m3 , k = 204 W/m K, α = 23 × 10−6 1/K, c = 903 J/Kg K, T0 = 298 K. To study the effect of pure thermal shock, the pressure at the inner surface of the sphere is assumed to be zero (traction free condition) and the outer surface of the sphere is insulated (with ho = 0). The plot of internal thermal shock is shown in Fig. 9.4-2. The distribution of temperature, radial displacement, radial stress, and hoop stress at different times are plotted in Figs. 9.4-3 to 9.4-6. Figure 9.4-7 shows the time variation of the temperature, radial displacement, radial and hoop stresses at mid-point of the thickness of the sphere.

Nondimensional Temp.

1.4 1.2 1 0.8 0.6 0.4 0.2 0 0

10

20

30

40

50

Nondimensional time

Nondimensional Temperature

Figure 9.4-2: The temperature shock applied to the inner surface of the sphere. 0.7 0.6

t=0.4

0.5

t=0.8 t=1

0.4 0.3 0.2 0.1 0 1

1.2

1.4

1.6

1.8

2

Nondimensional radial position

Figure 9.4-3:: The temperature distribution at different values of the time.

Nondimensional displacement


439

0.04 0.02 0 −0.02 1

1.2

1.4

1.6

2

1.8

−0.04 −0.06 −0.08

t=0.4

−0.1

t=0.8 t=1

−0.12 −0.14 Nondimensional radial position

Figure 9.4-4:: The displacement distribution at different values of the time.

Nondimensional radial stress

0 1

1.2

1.4

1.6

1.8

2

−0.05 −0.1 −0.15 −0.2

t=0.4

−0.25

t=0.8 t=1

−0.3 Nondimensional radial position

Figure 9.4-5:: The radial stress distribution at different values of the time.

Nondimensional hoop stress

0 1

1.2

1.4

1.6

1.8

2

−0.1 −0.2 −0.3 −0.4

t=0.4

−0.5

t=0.8 t=1

−0.6 Nondimensional radial position

Figure 9.4-6:: The hoop stress distribution at different values of the time.

440

Chapter 9. Finite and Boundary Element Methods 1.5 Nondimensional temp. 1 Nondimensional Disp.

0.5

Nondimensional time 0 −0.5

0

2

4

6

8

10

12

14

−1 −1.5 Nondimensional radial stress


Figure 9.4-7: Time history of the temperature, displacement and stresses at mid-radius point of the thickness of the sphere.

5

Generalized Thermoelasticity of FG Spheres

The responses of functionally graded spheres under steady state conditions are reported in a number of references. Obata and Noda [40] studied onedimensional steady thermal stresses in a functionally graded circular hollow cylinder and a hollow sphere using the perturbation method. The analytical solution for the stresses in a linear functionally graded spheres under steady radial temperature distribution is reported in [41]. An analytical solution for the functionally graded thick spheres under combined steady mechanical and thermal loads is presented by Eslami et al. [42]. In their work, they reported the closed form solution for the case that an exponential change for the material composition is assumed. In this section a thick functionally graded sphere under thermal shock load is considered. For severe thermal shock load application, the second sound is excited and the generalized formulation is required to study the sphere behavior. We may consider the Green–Lindsay model and obtain the temperature and stress waves propagation in the FG sphere [43]. The material properties of the FG sphere are assumed to follow a power law form along the radial direction. The fundamental governing equations in GL theory are the equations of motion, linear strain-displacement relations, Hooke’s law for the linear thermoelastic materials, energy balance equation, entropy relationship, and the heat conduction equations as Equations of motion: ui (9.5-1) σij,j + ρbi = ρ¨ Linear strain-displacement relations: 1 ij = (ui,j + uj,i ) 2

(9.5-2)

5. Generalized Thermoelasticity of FG Spheres

441

Hooke’s law for linear thermoelastic materials: σij = Cijkl kl − βij (T + t1 T˙ − T0 )

(9.5-3)

Energy balance equation: qi,i = R − T0 S˙

(9.5-4)

Entropy relationship: S=(

Ci ρc )(T − T0 + t2 T˙ ) − ( )T,i + βij ij T0 T0

(9.5-5)

Heat conduction equation: qi = −(Ci T˙ + kij T,j )

(9.5-6)

where t1 , t2 , Ci are new material properties introduced by Green and Lindsay, and t1 , t2 are known as the relaxation times, see Section 10 of Chapter 2. Substituting for σij in Eqs. (9.5-1) using Eqs. (9.5-2) and (9.5-3) and eliminating S and qi between the Eqs. (9.5-4) to (9.5-6) results in a coupled system of equations in terms of displacement and temperature as

Cijkl uk,l − βij (T − T0 + t1 T˙ )

,j

+ ρbi = ρuï

(9.5-7)

(kij T,j ),i − ρcT˙ − ρct2 T¨ − (Ci T˙ ),i − Ci T˙,i − βij T0 u˙ i,j + R = 0 (9.5-8) For isotropic functionally graded materials, Ci = 0 and the other material properties are Cijkl = λδij δkl + μ(δik δjl + δil δjk ) βij = βδij kij = kδij

(9.5-9)

where the symbol δij is Kronecker delta. Substituting the preceding relations in Eqs. (9.5-7) and (9.5-8), and in the absence of body forces and internal heat generation, these equations appear in the form

λuk,k δij + μ(ui,j + uj,i ) − β(T − T0 + t1 T˙ )δij

(kT,i ),i − ρcT˙ − ρct2 T¨ − βT0 u˙ i,i = 0

,j

= ρuï

(9.5-10) (9.5-11)

Note that the material properties that appeared in the preceding equations for the FGMs are the effective material properties and are position dependent functions. Also, it may be seen that when t1 = t2 = 0, the preceding equations reduce to the equations for the classical theory of thermoelasticity that predicts an infinite speed for thermal wave propagation. The subscript comma (,) indicates the partial differentiation with respect to the space variables and the superscript dot (.) denotes the time differentiation.

442


The thermal stress-displacement and heat conduction relations for a functionally graded material are σij = λuk,k δij + μ(ui,j + uj,i ) − β(T − T0 + t1 T˙ )δij qi = kT,i

(9.5-12) (9.5-13)

The functional material properties of FGMs may be introduced by a relation that correlates the effective material properties of FGMs to the constituent material properties. For a ceramic-metal FGM, the composition of the metal and ceramic may be introduced by a simple law of mixture of material constituents. Thus, the effective material properties of FGMs may be defined as P = Vm Pm + Vc Pc = Vm (Pm − Pc ) + Pc

(9.5-14)

where P is the effective property of FGM, Vm and Vc are the volume fractions of metal and ceramic, and Pm and Pc are the properties of metal and ceramic, respectively. Meanwhile, the subscripts m and c indicate the metal and ceramic features, respectively. Now, consider a symmetrically loaded hollow sphere with inner and outer radii rin and rout , respectively. The sphere is assumed to be graded along the radial direction. Equations (9.5-10) to (9.5-13) in the spherical coordinates and for a symmetrically loaded hollow sphere become [34]

∂2 2 2 ∂λ ∂2 2 ∂ ∂(λ + 2μ) ∂ − + − ρ (λ + 2μ) + + u ∂r2 r ∂r r2 ∂r ∂r r ∂r ∂t2

∂ ∂T − β(T − T0 + t1 (9.5-15) ) =0 ∂r ∂t

2 ∂ ∂k ∂ ∂2 ∂ ∂2 k + + T − ρc − ρct 2 ∂r2 r ∂r ∂r ∂r ∂t ∂t2 ∂ ∂ 2 u=0 −βT0 ( ) + ∂t ∂r r

⎡

(9.5-16)

⎤

∂u ∂u 2u 1 ∂T 1 σrr = 2μ ⎣ ∂r ⎦ + λ( − β(T − T0 + t1 + ) ) σθθ 1 ∂r r ∂t 1 u/r

(9.5-17) ∂T (9.5-18) ∂r It is suitable to transform these equations into dimensionless form. For this reason the following dimensionless relations may be used qr = −k

r¯ =

t1 f2 ¯ t2 f2 r ¯ tf2 ; t= ; ; t¯1 = ; t2 = f1 f1 f1 f1

5. Generalized Thermoelasticity of FG Spheres σ ¯ij =

443

σij T − T0 f4 u qr f1 ; T¯ = ; u¯ = ; q¯r = f3 Td Td f1 f3 Td km Td

(9.5-19)

f1 = km /ρm cm f2 , f2 = (λm + 2μm )/ρm f3 = βm , f4 = (λm + 2μm ) where Td is a dimensionless characteristic temperature. Using the dimensionless relations, Eqs. (9.5-15) to (9.5-18) are written in the form

2 2 ∂ 2 ∂(λ + 2μ) ∂ 2 ∂λ ∂2 2 ∂ (λ + 2μ) + + ¯ − + − ρf 2 ¯2 u ∂¯ r2 r¯ ∂¯ r r¯2 ∂¯ r ∂¯ r r¯ ∂¯ r ∂t

f4 ∂ ∂ T¯ ¯ ¯ − β(T + t1 (9.5-20) ) =0 f3 ∂¯ r ∂ t¯

2 ∂2 2 ∂ ∂k ∂ ∂ ¯2 ∂ ) T¯ − f + t k + + f ρc( 1 2 ∂¯ r2 r¯ ∂¯ r ∂¯ r ∂¯ r ∂ t¯ ∂ t¯2 ∂ f1 f2 f3 T0 ∂ 2 β( ) u¯ = 0 − + ¯ f4 ∂ t ∂¯ r r¯

⎡

⎤

∂ u¯ ∂ T¯ 1 2μ ⎣ λ ∂ u¯ 2¯ β u 1 σ ¯rr ⎦+ = ( − (T¯ + t¯1 + ) ) ∂¯ r σ ¯θθ f4 u¯/¯ f4 ∂¯ r r¯ 1 f3 ∂ t¯ 1 r

q¯r = −

(9.5-21)

k ∂ T¯ km ∂¯ r

(9.5-22)

(9.5-23)

The boundary conditions at the inner and outer surfaces of the sphere must be defined. The inner surface may be assumed to be traction free but exposed to an abrupt temperature rise while the outer surface is fixed and thermally insulated. Thus, the mechanical and thermal boundary conditions at the inner and outer surfaces of the sphere are

T = T0 + Td ∂T = 0; ∂r

tf t f2 Td 2 t f2 Td − 2 1+ ( ) − − 1 e f1 f1 f1

u=0

; σrr = 0 at r = rin (9.5-24)

at r = rout

Using the dimensionless relations, the boundary conditions become ¯ T¯ = 1 + (t¯2 − t¯ − 1 ) e−t ; ∂ T¯ = 0; ∂¯ r

σ ¯rr = 0

at r¯ = a

u¯ = 0

at r¯ = b

(9.5-25)

Here, a and b are the dimensionless inner and outer radii, respectively. The variation of the given temperature boundary condition is shown in Fig. 9.5-1. The inner surface of the sphere is assumed to be fully ceramic and the outer surface to be fully metal. The outer surface is assumed to be fixed. The metal

444



1.4 1.2 1 0.8 0.6 0.4 0.2 0 0

10

20

30

40

50

Nondimensional time

Figure 9.5-1: Nondimensional temperature applied to the inner surface of the sphere versus nondimensional time.

Figure 9.5-2:: The power law index effect on metal volume fraction. volume fraction through the radial direction is assumed to vary according to a power law form as r¯ − a n (9.5-26) ) Vm = ( b−a where n is the power law index. The distribution of the material composition in the functionally graded sphere along the radial direction is shown in Fig. 9.5-2. As shown, when n = 0, the sphere is fully metal and when n → ∞ the sphere is fully ceramic. The power law index may be varied to obtain different profiles for the distribution of the constituent materials. Laplace transform method in conjunction with Galerkin finite element formulation may be used to obtain the solution of the coupled equations. Laplace transform is employed to transform the equations from the time domain into Laplace transform domain. It may be considered that the initial conditions for nondimensional displacement, velocity, temperature, and the rate of temperature are zero. Thus, applying Laplace transform to Eqs. (9.5-20) to (9.5-23) gives


445

∂2 2 ∂ ∂(λ + 2μ) ∂ 2 2 ∂λ (λ + 2μ) + − 2 + + − ρf22 s2 u¯∗ 2 ∂¯ r r¯ ∂¯ r r¯ ∂¯ r ∂¯ r r¯ ∂¯ r ¯1 f4 ∂ ∂β ∂ t T¯∗ = 0 − β(1 + st¯1 ) + β (9.5-27) + (1 + st¯1 ) f3 ∂¯ r ∂¯ r ∂¯ r

2 ∂ ∂k ∂ ∂2 k + + − f1 f2 ρcs(t¯2 s + 1) T¯∗ ∂¯ r2 r¯ ∂¯ r ∂¯ r ∂¯ r f1 f2 f3 T0 ∂ 2 ∗ − βs u¯ = 0 + f4 ∂¯ r r¯ ∗ σ ¯ rr ∗ σ ¯θθ

⎡

(9.5-28)

⎤

∂ u¯∗ 2μ ⎣ λ ∂ u¯∗ 2¯ β u∗ 1 ∗ 1 ¯ ⎦ ¯ = + ( − (1 + s t ) T + ) 1 ∂¯ r 1 1 f4 u¯∗ /¯ f4 ∂¯ r r¯ f3 r

(9.5-29) k ∂ T¯∗ (9.5-30) km ∂¯ r where s is Laplace transform parameter and the asterisk indicates Laplace transform of the terms. Also, Laplace transforms of the mechanical and thermal boundary conditions are 3s + 1 ∗ ; σ ¯rr = 0 at r¯ = a T¯∗ = (s + 1)3 ∂ T¯∗ (9.5-31) = 0; u¯∗ = 0 at r¯ = b ∂¯ r Now, the geometry of the sphere is divided into some discrete elements along the radius of the sphere, where the interpolation function Ni is used to estimate the temperature and displacement fields over the base element (e). Thus Ni U¯i∗ ; T¯∗(e) = Ni T¯i∗ (9.5-32) u¯∗(e) = q¯r∗ = −

i=1

i=1

where terms U¯i∗ and T¯i∗ are the nodal values of temperature and displacement, respectively, and is the number of nodes in the base element (e). Employing the approximated fields for the displacement and temperature in the base element (e), Eqs. (9.5-32), and applying Galerkin finite element method over the volume of the base element, V (e) , Eqs. (9.5-27) and (9.5-28) lead to

2 2 ∂λ ∂2 2 ∂ ∂(λ + 2μ) ∂ − 2 + + − ρf22 s2 u¯∗(e) (λ + 2μ) + 2 (e) ∂¯ r r ¯ ∂¯ r r ¯ ∂¯ r ∂¯ r r ¯ ∂¯ r V

¯1 f4 ∂ ∂β ∂ t ∗(e) T¯ − β(1 + st¯1 ) + β Ni dV = 0 (9.5-33) + (1 + st¯1 ) f3 ∂¯ r ∂¯ r ∂¯ r

V (e)

2 ∂ ∂k ∂ ∂2 k + + − f1 f2 ρcs(t¯2 s + 1) T¯∗(e) 2 ∂¯ r r¯ ∂¯ r ∂¯ r ∂¯ r

446


∂ f1 f2 f3 T0 2 ∗(e) − βs u¯ Ni dV = 0 + f4 ∂¯ r r¯

i = 1, . . . ,

(9.5-34)

Now, the weak formulation is applied to the terms with the second order derivatives with respect to the radial variable. Using the local coordinates η = r¯ − r¯f , where r¯f is the radius of the first node of the base element in radial direction, Eqs. (9.5-33) and (9.5-34) appear in the form L

(λ + 2μ) 2 ∂λ Ni (η + r¯f ) 2 − + ρf22 s2 u¯∗(e) 2 (η + r¯f ) (η + r¯f ) ∂η 0

f4 ∂ ∂β ¯∗(e) ∂ t¯1 ¯ ¯ T + β(1 + st1 ) +β + (1 + st1 ) f3 ∂η ∂η ∂η 2

L

∂ u¯∗(e) ∂ u¯∗(e) +(λ + 2μ)(η + r¯f ) dη = Ni (λ + 2μ)(η + r¯f )2 ∂η ∂η ∂η 0 (9.5-35) 2 ∂Ni

L

∂Ni ∂ ¯∗(e) T (η + r¯f ) f1 f2 ρcs(t¯2 s + 1)Ni + k ∂η ∂η 0 f1 f2 f3 T0 ∂ 2 2 ∗(e) + βsNi (η + r¯f ) u¯ dη + f4 ∂η η + r¯f ¯∗(e) L 2 ∂T = kNi (η + rf ) ∂η 0 2

(9.5-36)

where L is the length of the base element (e) along the radial direction. The right-hand side terms of Eqs. (9.5-35) and (9.5-36) cancel each other between any two adjacent elements, except the first node of the first element and the last node of the last element. These are nodes located on the inside and outside boundaries of the sphere and, therefore, may be related to the boundary conditions of the sphere. Thus

3s + 1 ∂ u¯∗(e) a2 f4 βc 2 ¯ ∗ ¯ U = ; −a ( λ + 2μ ) = 2aλ − (1 + t¯1c )T¯1∗ c c c 1 (s + 1)3 ∂η 1 f3 ∂ T¯∗(e) ∗ = 0; U¯M =0 (9.5-37) ∂η M

T¯1∗

Here, the subscripts 1 and M are referred to the nodes that are located on the inner and outer radius of the sphere, respectively. Substituting Eqs. (9.5-32) into Eqs. (9.5-35) and (9.5-36), the transfinite element formulation in matrix form is obtained as

[K11 ] [K21 ]

[K12 ] [K22 ]

¯ ∗(e) U

T¯∗(e)

=

{F ∗ } {Q∗ }

(9.5-38)


447

where the submatrices [K11 ], [K12 ], [K21 ], [K22 ] for the base element (e) and the global force matrices {F } and {Q} are ij ] [K11

L

= 0

(λ + 2μ)(η + r¯f )2

∂Ni ∂Nj + 2(λ + 2μ)Ni Nj ∂η ∂η

∂λ −2(η + r¯f ) Ni Nj + ρf22 s2 (η + r¯f )2 Ni Nj dη ∂η ij [K12 ]=

ij ] [K21

ij ] [K22

(9.5-39)

f4 L ∂ [β(1 + st¯1 )Nj ] (η + r¯f )2 Ni dη f3 0 ∂η

(9.5-40)

f1 f2 f3 T0 L ∂Nj = βs (η + r¯f )Ni + Ni Nj dη f4 ∂η 0

L

=

{F } =

0

2

(η + r¯f )

∂Ni ∂Nj f1 f2 ρc t¯2 s2 + s Ni Nj + k ∂η ∂η

⎧ ⎫ ¯ ∗ a 2 f 4 βc ¯ ∗ ⎪ ⎪ ⎪ 2aλc U1 − f3 T1 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 0 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩

. . . 0

⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭

;

{Q} =

⎧ ⎫ ⎪ ⎪0⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪0⎪ ⎪ ⎪ ⎪ ⎪ ⎨.⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩

.⎪ ⎪ ⎪ ⎪ ⎪ .⎪ ⎪ ⎪ ⎭ 0

(9.5-41)

dη

(9.5-42)

(9.5-43)

The results obtained here are for a ceramic-metal functionally graded sphere with aluminum and alumina as the metal and ceramic constituents. The sphere is considered to be initially at reference temperature T0 = 298 K and then it is exposed to the boundary conditions given by Eqs. (9.5-24). The transition in material composition in radial direction is assumed with a power law form function given by Eq. (9.5-26). The relaxation times in GL theory for the aluminum and alumina are assumed to be t¯1m = t¯2m = 0.64 and t¯1c = t¯2c = 1.5625, respectively. In order to study the wave propagation along the radius of the sphere, a linear material distribution is considered, i.e., n = 1, and the linear Lagrangian polynomials are used as interpolation functions to approximate the temperature and displacement fields over the elements. Figures 9.5-3 to 9.5-6 show the wave propagation through the radius of the sphere. As shown in Fig. 9.5-3, the temperature wave propagates with a finite speed along the radial direction. The speed of wave propagation changes with position owing to the material property varying in the sphere along the radial direction. The times t¯ = 0.4, 0.6, 0.8 indicate the wave propagation, while time t¯ = 2 shows the wave reflection from the outer surface of the sphere. A similar description may be given for the displacement and stresses wave propagation. In Figs. 9.5-4 to 9.5-6, the times t¯ = 0.4, 0.6 show the wave propagation from the inner surface of the sphere and the time t¯ = 0.8 represents the wave reflection from the outer surface. A remarkable point ob-

Chapter 9. Finite and Boundary Element Methods Nondimensional Temp.

448 1.2 1 0.8

t = 0.4

0.6

t = 0.6

0.4

t = 0.8

0.2

t=2

0 1.2

1

1.4

1.6

1.8

2

Nondimensional r

Figure 9.5-3: Distribution of the nondimensional temperature along the nondimensional radius of the disk for n = 1.

Nondimensional Disp.

0.03 t = 0.8

0.02 t = 0.6 t = 0.4

0.01 0 1

1.2

1.4

1.6

1.8

2

−0.01 −0.02 −0.03

Nondimensional r


Figure 9.5-4: Distribution of the nondimensional radial displacement along the nondimensional radius of the disk for n = 1. 0 −0.05 −0.1

1

1.2

1.4

1.6

1.8

2

t = 0.4

−0.15 −0.2 −0.25

t = 0.6 t = 0.8

−0.3 Nondimensional r

Figure 9.5-5: Distribution of the nondimensional radial stress along the nondimensional radius of the disk for n = 1. served from these figures is that the hoop stress wave is more influenced by the temperature than the displacement. Moreover, the figures reveal that the displacement and radial waves speeds are larger than the temperature wave speed and the maximum of displacement and corresponding radial stress occur at the temperature wave front.


5. Generalized Thermoelasticity of FG Spheres 0.2 0 1.2 −0.2 1 −0.4 −0.6 t = 0.4 −0.8 −1 t = 0.6 −1.2 −1.4 t = 0.8 −1.6 −1.8 −2

1.4

1.6

449

1.8

2

Nondimensional r


Figure 9.5-6: Distribution of the nondimensional hoop stress along the nondimensional radius of the disk for n = 1. 1.2 n=1 n=2 n=5

1 0.8 0.6 0.4 0.2 0 0

2

4

6 8 10 Nondimensional time

12

14

Figure 9.5-7: Variations of the nondimensional temperature at the middle radius of the sphere for different values of the power index. Different power law indices, n, provide different profiles for the material composition and, therefore, they influence the temperature, displacement, and stresses variations along the sphere radius. In order to study the effect of the power law index, the middle radius point of the sphere is considered. The plot of the temperature changes versus nondimensional time is illustrated in Fig. 9.5-7. The figure shows that with the increase of the power law index the temperature of the middle radius point raises at later time and therefore the thermal wave speed is reduced. The reason is that when power law index takes a greater value, the ceramic volume fraction increases. Also, the figure shows that with the increase of n, the temperature sustains for a period of time that may be observed in Fig. 9.5-7 for n = 5. It may be observed from Figs. 9.5-8 to 9.5-10 that a smaller displacement and stresses amplitude are obtained when n increases. Moreover, the amplitudes of displacement and stresses variations decrease with the increase of n. The reason may be found by considering Eq. (9.5-3), where the stresses are related to the temperature and the rate of temperature. Thus, with the increase of n the temperature and the rate of temperature decrease, resulting in the decrease of the corresponding displacement and stresses.

450


Nondimensional Disp.

0.15 0.1 0.05 0 −0.05

0

−0.1 −0.15

2

4

6

8

10

12

14



Figure 9.5-8: Variations of the nondimensional radial displacement at the middle radius of the sphere for different values of the power index.

0 −0.2 0

2

4

6

8

10

12

14

−0.4 −0.6 −0.8 −1

n=1 n=2 n=5

−1.2 −1.4 −1.6

Nondimensional time


Figure 9.5-9: Variations of the nondimensional radial stress at the middle radius of the sphere for different values of the power index.

0.5 0 −0.5

0

2

4

6

8

10

12

14

−1 −1.5 −2 −2.5


Figure 9.5-10: Variations of the nondimensional hoop stress at the middle radius of the sphere for different values of the power index.

6. Generalized Thermoelasticity of FG Disk

6

451

Generalized Thermoelasticity of FG Disk

The generalized coupled thermoelasticity of disks is reported in [44] and it is shown that for thermal shock problems the coupling coefficient has significant effect on the distribution of the thermal stresses, deformation, and temperature. Chen and Lin [28] proposed a hybrid numerical method based on Laplace transform and the control volume method to analyze the transient coupled thermoelastic problems with relaxation times involving a nonlinear radiation boundary condition. Since the solution of the coupled thermoelasticity equations of FGMs are difficult to achieve, the analytical solution of these problems is rare in the literature and the numerical solutions are most suitable for these types of problems. The dynamic coupled thermoelastic response of functionally graded cylinders and plates was studied by Reddy and Chin [45]. They compared the coupled and uncoupled problems results and the effect of volume fraction of constituent materials (ceramic and metal) using a power law distribution. The coupled thermoelasticity of FGM cylindrical shells based on the classical coupled thermoelastic assumption was studied by Bahtui and Eslami [46]. Bakhshi et al. [47] studied the coupled thermoelasticity of a functionally graded disk and showed the volume fraction effects on the distribution of temperature, displacement, and stresses. They assumed the classical coupled thermoelasticity theory for their analysis. In this section the behavior of a functionally graded disk under the generalized thermoelasticity assumption, where the second sound is excited, is studied. We may select Lord–Shulman (LS) theory for this study. The inner surface of a hollow disk is assumed to be under thermal shock load. The coupled system of equations for the functionally graded disk based on LS model is derived and employed to be solved for the temperature and stress distributions. To transform the governing equations from the time domain to the space domain, Laplace transformation is employed. The geometry of the disk is divided into a number of elements, where Galerkin finite element method is used to derive the base element stiffness and force matrices. Then, the system of equations is solved to obtain the unknown nodal temperature and displacement. Finally, the inverse Laplace transform proposed by Honig and Hirdes [35] is applied to obtain the results in the time domain. The general constitutive equations of LS model are the equations of motion, linear strain-displacement relations, Hooke’s law for the linear thermoelastic material, energy balance equation, entropy relationship, and the heat conduction equation. These equations may be described in terms of the displacement components and temperature as Equations of motion: ui σij,j + ρbi = ρ¨

(9.6-1)

452


Linear strain-displacement relations: 1 ij = (ui,j + uj,i ) 2 Hooke’s law for linear thermoelastic materials:

(9.6-2)

σij = Cijkl kl − βij (T − T0 )

(9.6-3)

qi,i = ρh − T0 S˙

(9.6-4)

Energy balance equation: Entropy relationship: S=(

ρc )(T − T0 ) + βij ij T0

(9.6-5)

Heat conduction equation: qi + t0 q˙i = −kij T,j

(9.6-6)

where ρ, σij , ui , bi , qi , T0 , T, S, h, ij , βij , kij , Cijkl , c, and t0 are the density, stress tensor, displacement component, body force component per unit mass, heat flux component, reference temperature, absolute temperature, entropy per unit volume, heat source per unit mass, strain tensor, tensor of stress-temperature moduli, tensor of thermal conductivity, tensor of elastic moduli, specific heat, and the relaxation time (proposed by Lord and Shulman), respectively. When t0 = 0 the preceding equations governed the equations of classical theory of thermoelasticity predicts an infinite speed for thermal wave propagation. Eliminating S, qi , ij , and σij using Eqs. (9.6-1) to (9.6-6), and neglecting the derivatives of relaxation time with respect to position variables, leads to the linear coupled field equations in terms of u and T as ui [Cijkl uk,l − βij (T − T0 )],j + ρbi = ρ¨ (kij T,j ),i − ρcT˙ − ρct0 T¨ − βij T0 (t0 uï,i + u˙ i,i ) + ρ(t0 h˙ + h) = 0

(9.6-7) (9.6-8)

The comma (,) is used to denote the partial differentiation with respect to the space variables and the superscript dot (.) denotes the time differentiation. Under the plane stress assumptions, for functionally graded materials, in the absence of heat source and body forces, Eqs. (9.6-7) and (9.6-8) appear in the form

¯ k,k δij + μ(uj,i + ui,j ) λu

,j

¯ − T0 )],i = ρ¨ − [β(T ui

¯ 0 (t0 uï,i + u˙ i,i ) = 0 (kT,i ),i − ρcT˙ − ρct0 T¨ − βT

(9.6-9) (9.6-10)

¯ and β¯ may be related to the Lamé constants and stress-temperature where λ moduli as ¯ = 2μ λ; β¯ = 2μ β λ λ + 2μ λ + 2μ


453

Here, k, β, ρ, c, λ, and μ are the effective heat conductivity, effective stresstemperature moduli, effective density, effective specific heat, and the effective Lamé constants of FGMs, respectively. For functionally graded materials, the thermal stress-displacement relation is ¯ k,k δij + μ(ui,j + uj,i ) − β(T ¯ − T0 )δij σij = λu

(9.6-11)

Here, the symbol δij is the Kronecker delta. Since FGMs consist of two constituent materials (mostly metal and ceramic), the effective properties of FGMs can be introduced as a function of the constituent material properties. This function relates the effective material properties of FGMs to the simple law of moisture of material constituents. Thus, for a ceramic-metal FGM, the effective properties may be defined as P = Vm Pm + Vc Pc = Vm (Pm − Pc ) + Pc

(9.6-12)

where P is the effective property of FGM, Vm and Vc are the volume fractions of metal and ceramic, and Pm and Pc are the properties of metal and ceramic, respectively. Consider an annular disk under axisymmetric thermal shock load applied into its inner surface. The coupled thermoelastic equations of the disk, with plane stress assumption and based on Lord–Shulman model of thermoelasticity theory, are obtained from Eqs. (9.6-9), (9.6-10), and (9.6-11) as

2 2 ¯ ¯ ¯ + 2μ) ∂ + 1 ∂ − 1 + ∂(λ + 2μ) ∂ + 1 ∂ λ − ρ ∂ (λ u ∂r2 r ∂r r2 ∂r ∂r r ∂r ∂t2 ∂ β¯ ∂ )(T − T0 ) = 0 −(β¯ + (9.6-13) ∂r ∂r

∂2 1 ∂ ∂k ∂ k + + 2 ∂r r ∂r ∂r ∂r 2 ¯ 0 (t0 ∂ + ∂ ) ∂ + −βT ∂t2 ∂t ∂r

σrr σθθ

⎡

∂2 ∂ − ρc − ρct0 2 T ∂t ∂t 1 u=0 r

(9.6-14)

⎤

¯ ∂u ¯ ∂u + u ) 1 − β(T − T0 ) = 2μ ⎣ ∂r ⎦ + λ( ¯ − T0 ) β(T ∂r r 1 u/r

(9.6-15)

Under the plane stress assumption, von Mises stress in an annular disk is 2 2 σe = (σrr + σθθ − σrr σθθ )1/2

(9.6-16)

These equations may be transformed into the dimensionless forms using the dimensionless parameters introduced as rˆ =

t0 A2 σij r ˆ tA2 ˆ ; t= ; t0 = ; σ îj = A1 A1 A1 A3 Td

454

Chapter 9. Finite and Boundary Element Methods T − T0 A4 ui q i A1 Tˆ = ; uî = ; qî = Td A1 A3 Td A5 Td ¯ m + 2μm )/ρm A1 = km /ρm cm A2 , A2 = (λ ¯ m + 2μm ), A5 = km A3 = β¯m , A4 = (λ

(9.6-17)

The subscript m denotes the metal property and Td is a characteristic temperature used to normalize the temperature. Using the introduced dimensionless parameters, Eqs. (9.6-13) to (9.6-16) are written as

2 2 ¯ ¯ ¯ + 2μ) ∂ + 1 ∂ − 1 + ∂(λ + 2μ) ∂ + 1 ∂ λ − ρA2 ∂ (λ ˆ 2 ˆ2 u ∂ˆ r2 rˆ ∂ˆ r rˆ2 ∂ˆ r ∂ˆ r rˆ ∂ˆ r ∂t ∂ β¯ ˆ A4 ∂ )T = 0 (9.6-18) − (β¯ + A3 ∂ˆ r ∂ˆ r

2 1 ∂ ∂k ∂ ∂ ∂2 ˆ0 ∂ ) Tˆ + t k + + A ρc( − A 1 2 ∂ˆ r2 rˆ ∂ˆ r ∂ˆ r ∂ˆ r ∂ tˆ ∂ tˆ2 A1 A2 A3 T0 ¯ ˆ ∂ 2 ∂ 1 ∂ − + uˆ = 0 β(t0 2 + ) ˆ ˆ A4 r rˆ ∂t ∂ t ∂ˆ

⎡

(9.6-19)

⎤

∂ uˆ ¯ ∂ uˆ uˆ 1 β¯Tˆ 1 λ 2μ ⎣ σ ˆrr ⎦ + ) = ( − ∂ˆ r + σ ˆθθ A4 uˆ/ˆ A4 ∂ˆ r rˆ 1 A3 1 r

(9.6-20)

2 2 σrr +σ ˆθθ −σ ˆrr σ ˆθθ )1/2 σ ê = (ˆ

(9.6-21)

In the uncoupled case, thermal field is not affected by displacement field and thus those terms related to displacement in Eq. (9.6-19) vanish. The uncoupled system of Eqs. (9.6-18) and (9.6-19) may be obtained when T0 = 0. In the preceding equations, it is to be noted that the FGM’s properties are position-dependent functions, and thus the elastic and thermal waves propagation speeds vary as a function of position in an FGM. The mechanical and thermal boundary conditions at the inside and outside surfaces of the disk must be specified. We assume that the outside surface of the disk is thermally insulated and the inside surface of the disk is subjected to a thermal shock such that its surface temperature is specified by the function

T = T0 + Td 1 − (1 + 100

t A2 −100 tAA2 1 )e A1

at r = r1

(9.6-22)

Also, the mechanical boundary condition on the inside surface and outside surface of the disk are traction free and fixed, respectively. Thus

T = T0 + Td ∂T = 0; ∂r

t A2 −100 tAA2 1 ; σ 1 − (1 + 100 )e rr = 0 A1

u=0

at r = r2

at r = r1 (9.6-23)


455

where r1 and r2 are the inner and outer radii of the disk, respectively. Using the dimensionless parameters, these equations become ˆ Tˆ = 1 − (1 + 100tˆ ) e−100t ; ∂ Tˆ = 0; ∂ˆ r

σ ˆrr = 0

at rˆ = a

uˆ = 0

at rˆ = b

(9.6-24)

Here, a and b are the dimensionless inner and outer radii, respectively. As discussed by Eq. (9.6-12), the FGM properties are position dependent functions. This functional relationship may be assumed to have a power law form [21,32]. For this reason, the metal volume fraction Vm for a functionally graded disk with inner radius a and outer radius b may be introduced as Vm = (

rˆ − a n ) b−a

(9.6-25)

where n is power law index that constitutes the metal constituent richness in functionally graded annular disk. Substituting this equation in Eq. (9.6-12) gives the position dependent material properties of FGMs in the geometry of the disk. The metal volume fraction change in an FGM is shown in Fig. 9.6-1. As shown, the fully metallic disk may be obtained when n = 0. The power law index may be varied to obtain different profiles for the distribution of materials. Laplace transformation may be employed to transform the equations from time domain to Laplace transform domain. Then, the problem may be solved using Galerkin finite element method. The geometry of the disk is divided into a number of elements through the radial direction. Then, the transfinite element procedure using Galerkin finite element method may be used to derive the stiffness and force matrices for the base element. Assume that the disk is initially at rest and the initial conditions for displacement, velocity, temperature, and the rate of temperature are zero. Thus

Figure 9.6-1:: The power law index effect on metal volume fraction.

456

Chapter 9. Finite and Boundary Element Methods u(r, 0) = 0; T (r, 0) = 0;

u(r, ˙ 0) = 0 ˙ T (r, 0) = 0

(9.6-26)

Applying Laplace transform to Eqs. (9.6-18) and (9.6-19) results in

2 ¯ ¯ ¯ + 2μ) ∂ + 1 ∂ − 1 + ∂(λ + 2μ) ∂ + 1 ∂ λ − ρA2 s2 uˆ∗ (λ 2 ∂ˆ r2 rˆ ∂ˆ r rˆ2 ∂ˆ r ∂ˆ r rˆ ∂ˆ r A4 ∂ ∂ β¯ ˆ∗ − (β¯ + (9.6-27) )T = 0 A3 ∂ˆ r ∂ˆ r

1 ∂ ∂k ∂ ∂2 k + + − A1 A2 ρcs(tˆ0 s + 1) Tˆ∗ ∂ˆ r2 rˆ ∂ˆ r ∂ˆ r ∂ˆ r A1 A2 A3 T0 ¯ ˆ ∂ 1 ∗ βs(t0 s + 1) − uˆ = 0 + A4 ∂ˆ r rˆ

(9.6-28)

where s is Laplace transform parameter and the asterisk denotes Laplace transform of the terms. Laplace transforms of the mechanical and thermal boundary conditions are Tˆ∗ =

10000 ; s(s + 100)2 ∂ Tˆ∗ = 0; ∂ˆ r

∗ σ ˆrr =0

at rˆ = a

uˆ∗ = 0

at rˆ = b

(9.6-29)

To solve the problem by the finite element method, the geometry of the disk is discretized into a number of radial elements along the radial direction and Galerkin finite element technique is considered to obtain the solution. The base element (e) is considered and the displacement uˆ∗ and temperature Tˆ∗ are approximated as uˆ∗(e) =

Ni Uî∗

Tˆ∗(e) =

i=1

Ni Tî∗

(9.6-30)

i=1

where Ni is the shape function approximating the displacement and temperature fields in the base element (e), and is the number of nodes in the base element. Terms Uî∗ and Tî∗ denote the nodal values of displacement and temperature, respectively. Using Eqs. (9.6-30) and applying Galerkin finite element method to the governing equations (9.6-27) and (9.6-28) for the base element (e), yields

2 ¯ ¯ ¯ + 2μ) ∂ + 1 ∂ − 1 + ∂(λ + 2μ) ∂ + 1 ∂ λ − ρA2 s2 uˆ∗(e) (λ 2 ∂ˆ r2 rˆ ∂ˆ r rˆ2 ∂ˆ r ∂ˆ r rˆ ∂ˆ r V (e)

A4 ∂ ∂ β¯ ˆ∗(e) − (β¯ + Ni dV = 0 i = 1, . . . , (9.6-31) )T A3 ∂ˆ r ∂ˆ r


457

∂2 1 ∂ ∂k ∂ k + + − A1 A2 ρcs(tˆ0 s + 1) Tˆ∗(e) 2 ∂ˆ r rˆ ∂ˆ r ∂ˆ r ∂ˆ r V (e)

A1 A2 A3 T0 ¯ ˆ ∂ 1 ∗(e) βs(t0 s + 1) − uˆ Ni dV = 0 i = 1, . . . , + A4 ∂ˆ r rˆ

(9.6-32)

Here, V (e) is the volume of the base element (e). With the aid of weak formulation and local coordinates η = rˆ − rˆf , where rˆf is the radius of the first node of the base element, Eqs. (9.6-31) and (9.6-32) may be presented in the form L

L

¯ + 2μ) ∂ ¯ + 2μ) ∂ ¯ + 2μ) ∂(λ (λ (λ + − 2 (η + rˆf ) ∂η (η + rˆf ) ∂η ∂η 0

¯ ¯ 1 A4 ¯ ∂ ∂λ ∂ β ˆ∗(e) 2 2 ∗(e) + − (β Ni (η + rˆf ) − ρA2 s uˆ + )T (η + rˆf ) ∂η A3 ∂η ∂η ⎫ L ¯ + 2μ)(η + rˆf ) ∂ uˆ∗(e) ⎬ ∂ Ni (λ ∂ uˆ∗(e) ¯ + dη = (λ + 2μ)Ni (η + rˆf ) ∂η ∂η ⎭ ∂η 0 −

(9.6-33)

k ∂ ∂k ∂ + − A1 A2 ρcs(tˆ0 s + 1) Tˆ∗(e) η + rˆf ∂η ∂η ∂η 0

A1 A2 A3 T0 ¯ ˆ ∂ 1 ∗(e) − uˆ Ni (η + rˆf ) βs(t0 s + 1) + A4 ∂η η + rˆf −

L ∂ [kNi (η + rˆf )] ∂ Tˆ∗(e) ∂ Tˆ∗(e) + dη = kNi (η + rf ) ∂η ∂η ∂η 0

(9.6-34)

where L is the length of element along the radius. The terms on the right-hand side of Eqs. (9.6-33) and (9.6-34) cancel each other between any two adjacent elements, except the nodes located on the boundaries of the disk. These boundary conditions in term of the nodal temperature and displacement are ¯ 10000 ∂ uˆ∗(e) ∗ ¯ ¯ c Uˆ ∗ − aA4 βc Tˆ∗ ˆ ; −a( λ + 2μ ) = λ T1 = c c 1 s(s + 100)2 ∂η 1 A3 1 ∂ Tˆ∗(e) ∗ = 0; UˆM =0 (9.6-35) ∂η M ∗ where Uˆ1∗ and UˆM are the radial displacements of the inner and outer boundaries of the disk, respectively. Also, the subscript 1 and M are referred to the first and last nodes of the solution domain that are located at the inner and outer radius of the disk, respectively. Due to the assumed boundary conditions,

ˆ∗(e) ∂u ˆ∗(e) = 0. Substituting Eqs. (9.6-30) into Eqs. (9.6-33) = 0 and ∂η 1 M

− ∂ T∂η

and (9.6-34), the transfinite element equations are expressed in matrix form as

[K11 ] [K21 ]

[K12 ] [K22 ]

ˆ ∗(e) U

Tˆ∗(e)

=

{F ∗ } {Q∗ }

(9.6-36)

458


The submatrices [K11 ], [K12 ], [K21 ], [K22 ] for the base element (e) and the global force matrices {F } and {Q} are ij [K11 ]

L

= 0

¯ + 2μ)(η + rˆf ) (λ

∂Ni ∂Nj ¯ + 2μ) Ni Nj + (λ ∂η ∂η η + rˆf

¯ ∂λ − Ni Nj + ρA22 s2 (η + rˆf )Ni Nj dη ∂η ij [K12 ]

ij ] [K21

A4 L ¯ ∂Nj ∂ β¯ β(η + rˆf )Ni = + (η + rˆf )Ni Nj dη A3 0 ∂η ∂η

(9.6-37) (9.6-38)

A1 A2 A3 T0 L ¯ ˆ ∂Nj βs(t0 s + 1) (η + rˆf )Ni = + Ni Nj dη A4 ∂η 0

ij [K22 ]=

(9.6-39)

L ∂ [kNi (η + rˆf )] ∂Nj

+ ρcA1 A2 s(tˆ0 s + 1)(η + rˆf )Ni Nj ∂η

∂k ∂Nj − dη (9.6-40) (η + rˆf ) + k Ni ∂η ∂η ∂η

0

⎧ ⎫ ¯ c Uˆ ∗ − aA4 β¯c Tˆ∗ ⎪ ⎪ λ ⎪ ⎪ 1 1 A3 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 0 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎬

{F } = ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩

. . . 0

⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭

;

⎧ ⎫ 0⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 0 ⎪ ⎪ ⎪ ⎪ ⎪ ⎨.⎪ ⎬

{Q} = ⎪ ⎪ .⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ .⎪ ⎪ ⎪ ⎪ ⎩ ⎪ ⎭ 0

(9.6-41)

The results obtained from the present formulation for a functionally graded annular disk are shown for several cases. These include investigating the wave propagation in FG disk and the power law index effects on variation of the temperature, displacement, radial stress, and hoop stress. Aluminum and alumina are considered as metal and ceramic constituents of the functionally graded annular disk. The relaxation time of aluminum and alumina are assumed to be t0m = 0.64, t0c = 1.5625, respectively. The reference temperature is T0 = 298 K. While the presented formulation is general, the linear Lagrangian polynomials are used for the shape functions in the base element, as shown in Fig. 9.6-2. The geometry of the disk is divided into 400 discretized elements along the radial direction. Now, the disk is assumed to be under the boundary conditions given by Eqs. (9.6-24). Figures 9.6-3 to 9.6-6 show the propagation of temperature, radial displacement, radial stress, and hoop stress waves, respectively, for the case that the material properties of the disk linearly change through the radius of the disk, i.e., when n = 1. As expected from LS theory it may be seen in Fig. 9.6-3 that the temperature waves propagate with finite speed of propagation (the second sound effect). Times tˆ = 0.2, 0.4, 0.6, 0.8, 1.7 show the wave propagation along the radius, while time tˆ = 1.9 represents


459

η node 2

node 1 ^r

f

N2(η) =

L

^r

N1(η) = 1 −

η

L

η

L

Linear element

Figure 9.6-2: The base element through the radius direction and the corresponding Lagrangian linear polynomial shape functions.

Figure 9.6-3: Distribution of the nondimensional temperature along the nondimensional radius of the disk for n = 1.

Figure 9.6-4: Distribution of the nondimensional radial displacement along the nondimensional radius of the disk for n = 1. the reflection of the temperature wave. These waves that propagate along the radial direction are in a circular form. It is to be noted that, using the presented formulation and nondimensional parameters, the dimensionless speed of thermal wave propagation in a pure ceramic disk is [35]

CˆT c =

kc ρm cm km ρc cc

1 = 0.16 ˆ t0c

(9.6-42)

460


Figure 9.6-5: Distribution of the nondimensional radial stress along the nondimensional radius of the disk for n = 1.

Figure 9.6-6: Distribution of the nondimensional hoop stress along the nondimensional radius of the disk for n = 1. The speed of thermal wave propagation for pure aluminum is obtained by replacing kc , ρc , cc with km , ρm , cm , the properties of aluminum, in Eq. (9.6-42). Thus, the thermal wave propagates in aluminum with the dimensionless speed of

CˆT m =

1 = 1.25 ˆ t0m

(9.6-43)

Therefore, the thermal waves in FGMs propagate with the speed of 0.16 < CˆT < 1.25, where CˆT is the nondimensional thermal wave speed in FGM. This phenomenon may be seen in Fig. 9.6-3, where the temperature wave reaches to the outer boundary of the disk in time interval (b − a)/1.25 < tˆ < (b − a)/0.16, i.e., 0.8 < tˆ < 6.25. This figure shows the disturbances in the vicinity of the thermal wave front, where a conversion between the thermal and mechanical energies may be observed. The distribution of the radial displacement along the radial direction of the disk is shown in Fig. 9.6-4. This figure shows that the displacement wave propagates along the radius of the disk at times tˆ = 0.2, 0.4, 0.6, and reflects from the outer boundary of the disk at the time


461

tˆ = 0.8. Comparison between Figs. 9.6-3 and 9.6-4 reveals that the maximum displacement occurs at the vicinity of the thermal wave front. The radial stress distribution is shown in Fig. 9.6-5. The similar interpretation may be cited in this figure for the wave propagation and reflection. Similar to the preceding notes, the nondimensional stress wave propagation speed in pure ceramic disk is 3 4 ¯ c + μc )ρm 4 (λ ˆ = 1.966 C1c = 5 ¯

(λm + μm )ρc

(9.6-44)

For pure aluminum, the thermal wave propagates with the speed of unity. On the other hand, the stress waves in FGMs propagate with the speed of 1 < Cˆ1 < 1.966, where Cˆ1 is the nondimensional stress wave speed. This occurrence may be seen in Fig. 9.6-5, where the stress wave reaches the outer boundary of the disk in time interval 0.5 < tˆ < 1. The hoop stress distribution along the radial direction may be observed in Fig. 9.6-6. It may be found that the hoop stress is more affected by the temperature wave than by the displacement wave. The effects of the power law index on the temperature, displacement, radial stress, and hoop stress at the middle point of the disk is shown in Figs. 9.6-7 to 9.6-10, respectively. In this case, the value of relaxation time and reference temperature are assumed as previously. It is seen that the maximum of the temperature, displacement, and stresses variations reduce when n is increased. Also, the amplitudes of oscillations of these waves are reduced when n has larger values. The reason is that the temperature wave propagates with a lower speed and, therefore, the amplitude of temperature, displacement, and stress waves become smaller. Thus, when a thermal shock is applied to the boundary of the disk, the functionally graded disk may be used as a thermal barrier to reduce the thermal shock effects.

Figure 9.6-7: Variations of the nondimensional temperature for points at the middle radius of the disk versus the nondimensional time for different values of the power index.

462


Figure 9.6-8: Variations of the nondimensional radial displacement for the points at the middle radius of the disk versus the nondimensional time for different values of the power index.


1 0.5 0 0

2

4

6

8

10

−0.5 −1 n=1

−1.5 −2

n=2 n=5 Nondimensional time

Figure 9.6-9: Variations of the nondimensional radial stress for the points at the middle radius of the disk versus the nondimensional time for different values of the power index.

Figure 9.6-10: Variations of the nondimensional hoop stress for the points at the middle radius of the disk versus the nondimensional time for different values of the power index.

7. Higher Order Elements

7

463

Higher Order Elements

Chen and Lin [28] proposed a hybrid numerical method based on Laplace transform and control volume method to analyze the transient coupled thermoelastic problems with relaxation times involving a nonlinear radiation boundary condition. Hosseini Tehrani and Eslami [29] considered the boundary element formulation for the analysis of coupled thermoelastic problems in a finite domain and studied the coupling coefficient and relaxation times effects on thermal and elastic wave propagations. In this section, a transfinite element method using Laplace transform is used to solve the coupled equations for an axisymmetrically loaded disk in the transformed domain. Elements of various orders are employed to investigate the effects of the number of nodes in an element. Finally, the temperature and displacement are inverted to obtain the actual physical quantities, using the numerical inversion of Laplace transform method proposed by Honig and Hirdes [35]. In the absence of the heat source and body forces and for isotropic materials, the nondimensionalized form of the generalized coupled thermoelastic equations of the axisymmetrically loaded circular disk based on Lord–Shulman theory in terms of the displacement and temperature may be written as [48]

1 ∂ ∂2 ∂T ∂2 1 + − u− − =0 2 2 2 ∂r r ∂r r ∂t ∂r

(9.7-1)

1 ∂ ∂ ∂2 ∂ + 1 + t0 T − 2 ∂r r ∂r ∂t ∂t

1 ∂2 ∂2 ∂3 1∂ −C t0 + + u=0 + ∂r∂t2 r ∂t2 ∂r∂t r ∂t

(9.7-2)

¯ + 2μ is the coupling coefficient. For the plane stress Here, C = T0 β¯2 / ρce λ 2μ 2μ ¯ = condition λ λ and β¯ = λ+2μ β. In the preceding equations ρ, u, T0 , T, λ+2μ ¯ ce , and t0 are the density, radial displacement, reference temperature, temβ, perature change, stress-temperature moduli, thermal conductivity, specific heat and the relaxation time (proposed by Lord and Shulman), respectively, while λ and μ are Lamé constants. The dimensionless thermal and mechanical boundary conditions are qin = −

∂T ; ∂r

T = 0;

u=0 σrr =

¯ u λ ∂u +¯ −T =0 ∂r λ + 2μ r

at r = a at r = b

(9.7-3)

where σrr and a and b are the radial stress and dimensionless inner and outer radii, respectively.

464


In order to derive the transfinite element formulation, Laplace transformation is used to transform the equations into Laplace transform domain. Applying Galerkin finite element method to the governing equations (9.7-1) and (9.7-2) for the base element (e), yields L 0

−

∂T 1 ∂ 1 − s2 u − Nm (η + ri ) − (η + ri ) ∂η (η + ri )2 ∂η L

∂ (Nm (η + ri )) ∂u ∂u + dη = Nm (η + ri ) ∂η ∂η ∂η 0 L 0

−

(9.7-4)

∂ 1 ∂ 1 u − s(1 + t0 s) T − C t0 s2 + s + (η + ri ) ∂η ∂η η + ri

∂ (Nm (η + ri )) ∂T ×Nm (η + ri ) + ∂η ∂η 6

L

∂T dη = Nm (η + ri ) ∂η 0

(9.7-5)

6

where u = nm=1 Nm Um and T = nm=1 Nm Tm . In the preceding equations, s, Nm , η = r − ri , ri , L, Um , and Tm are Laplace transform parameter, shape function, local coordinates, the radius of the i-th node of the base element, the length of element in the radial direction, nodal displacement, and the nodal temperature, respectively. The terms on the right-hand sides of Eqs. (9.7-4) and (9.7-5) cancel each other between any two adjacent elements, except the nodes located on the boundaries of the solution domain. These boundary conditions are

∂T −a = aqin ; ∂η 1

U1 = 0

TM = 0;

¯ λ = −¯ UM + bTM λ + 2μ

∂u b ∂η

M

(9.7-6)

The subscript 1 and M are referred to the first and last nodes of the solution domain, respectively. To investigate the accuracy of the method, a numerical example is considered. The material of the disk is assumed to be aluminum. The dimensionless inside and outside radii are a = 1 and b = 2. The dimensionless input heat flux is defined as the Heaviside unit step function. Since the applied boundary conditions are assumed to be axisymmetric, the radius of the disk is divided into 100 elements. These types of shape functions, linear, second order, and third order polynomials, as shown in Fig. 9.7-3, are used for the finite element model of the problem. Figures 9.7-1 and 9.7-2 show the wave propagation of the temperature and radial displacement along the radial direction [48]. The numerical values of

7. Higher Order Elements

465

0.9 Order1

0.8

Order2

Dimensionless T

0.7

Order3

0.6 t=1.2

0.5 0.4 0.3

t=0.6

0.2

t=0.2

0.1 0 −0.1 1

1.2

1.6

1.4

1.8

2

Dimensionless r

Figure 9.7-1: Distribution of the dimensionless temperature along the radius of the disk at three values of time for three types of elements.

0.18 order1

Dimensionless u

0.16

order2

0.14

t=1.2

order3

0.12 0.1 0.08 0.06

t=0.6

0.04 0.02

t=0.2

0 1

1.2

1.4

1.6

1.8

2

Dimensionless r

Figure 9.7-2: Distribution of the dimensionless displacement along the radius of the disk at three values of time for three types of elements.

the coupling parameter and the dimensionless relaxation time are assumed to be 0.01 and 0.64, respectively. The waves propagation are shown at several times. Two wave fronts for elastic and temperature waves are detected from the figures, as expected from LS model. It is seen from the figures that the results of the three types of shape functions for the assumed number of elements coincide. For smaller number of elements, the difference between the results obtained for different shape functions increases noticeably. For the assumed number of elements, the curves for radial displacement and temperature distribution are checked against the known data in the literature, where very close agreement is observed. Figure 9.7-1 clearly shows the temperature wave front (the second sound effect), which is propagating along the radius of the disk.

466

Chapter 9. Finite and Boundary Element Methods h node 1

node 2

N2 (h) =

L r

N1 (h) = 1 −

h L

h L

a) Linear element

node 1

node 2 L/2

node 3 L/2

L r

b) Quadratic element h node 1

h h )(1 − ) L L h h N2 (h) = 4 (1 − ) L L h (1 − 2 h ) N3 (h) = − L L N1 (h) = (1 − 2

h

node 2 L/3

node 3 L/3

L/3

L r c) Cubic element

node 4

h h h N1 (h) = (1 − 3 ) (1 − 3 )(1 − ) L 2 L L h h h 3 N2 (h) = 9 (1 − )(1 − ) L 2 L L h h h 9 N3 (h) = − (1 − 3 )(1 − ) L L 2L h h h 3 ) N4 (h) = (1 − 3 )(1 − 2L L L

Figure 9.7-3:: Elements with linear, second, and third order shape functions.

8

Functionally Graded Beams

The equations for a coupled thermoelastic beam, including the effects of shear deformation and rotatory inertia, are derived by Jones [49]. Coupled thermally induced vibrations of Euler-Bernoulli and Timoshenko beams with onedimensional heat conduction are investigated by Seibert and Rice [50]. Coupled thermoelasticity of beams made of homogeneous and isotropic material is discussed by Massalas and Kalpakidis [51,52]. The analytical solution of the coupled thermoelasticity of beams with Euler-Bernoulli assumption is given in reference [51], and that with Timoshenko assumption is given in reference [52]. In the treatment of these problems, a linear approximation for temperature variation across the thickness direction of the beam is considered. Finite element coupled thermoelastic analysis of composite Timoshenko beams is given by Maruthi Rao and Sinha [53], where the temperature variation across the thickness direction is neglected. Manoach and Ribeiro developed a numerical procedure to study the coupled large amplitude thermoelastic vibrations of Timoshenko beams subjected to the thermal and mechanical loads using the finite difference approximation and modal coordinate transformations [54]. Sankar solved the elastic problem of FGM beam and computed the thermal stresses in an FGM Euler-beam based on the uncoupled thermoelasticity assumption [55,56].

8. Functionally Graded Beams z

467

y h

FGM

x

b

l

Figure 9.8-1:: The beam and coordinates. This section presents the behavior of an FGM Euler-Bernoulli beam under lateral thermal shock with coupled thermoelastic assumption. The analysis is based on Galerkin finite element method, using a C 1 - continuous shape function [5,57]. Consider a beam of rectangular cross section with length l, height h, and width b, as shown in Figure 9.8-1. Using Euler-Bernoulli beam assumption with no deflection change in thickness direction, the axial displacement is u(x, z) = u0 − zw,x

(9.8-1)

where u is the axial displacement component, u0 is the displacement of a point on the reference plane, w is the lateral deflection, and z is measured across the thickness direction from the middle plane of the beam at x = 0. A comma in the subscript indicates partial differentiation. The FGM profile across the thickness direction of the beam, made of ceramic and metal constituent materials, may be assumed to follow a power law form as f (z) = fm + fcm (

2z + h n ) 2h

(9.8-2)

where f is any material property of the FGM, fm is metal property of FGM, fcm = fc −fm , fc denotes the ceramic property of FGM, and n is the power law index. The density, modulus of elasticity, coefficient of specific heat, coefficient of thermal expansion, and the conduction coefficient may be assumed to follow the power law form, indicated by Eq. (9.8-2). The location of the neutral axis of an FGM beam graded across the thickness is obtained as ρ(z)zdz (9.8-3) zN A = ρ(z)dz where ρ(z) is the mass density and N A is a symbol for the neutral axis where the integration is across the cross sectional area of the beam. For this formulation, z-coordinate is measured across the thickness direction from the FGM beam neutral axis. Assuming that the beam material is linearly elastic, the stress-strain relation for the FGM beam based on the assumed displacement components is σx = E(z)[x − α(z) θ]

(9.8-4)

468


where E is the modulus of elasticity, α is the coefficient of thermal expansion, θ = T − T0 is the temperature change, and T0 is the reference temperature. The bending moment resultant is expressed as follows

M= z

σx z dz

(9.8-5)

The temperature change across the thickness direction is assumed to be linear. This assumption is justified considering that the thickness of the beam is small with respect to its length [51,52]. Thus θ = θ1 (x, t) +

z θ2 (x, t) h

(9.8-6)

where θ1 and θ2 are unknowns to be found across the beam’s height and are coupled with the lateral deflection w of the beam. The equation of motion of a beam based on Euler-Bernoulli assumption is [50] (9.8-7) M,xx = I0 w,tt

where I0 = z ρ(z)dz is mass moment of inertia of the beam. Substituting Eq. (9.8-6) into Eq. (9.8-4), using Eq. (9.8-5), and neglecting u0 and the axial inertia effects, the equation of motion (9.8-7) becomes A1 w,xxxx + A2 θ1,xx + A3 θ2,xx − I0 w,tt = 0

(9.8-8)

where the A’ s are h/2

A1 =

−h/2

h/2

A3 =

−h/2

−E(z)z 2 dz −E(z)α(z)

h/2

A2 =

−h/2

−E(z)α(z)zdz

z2 dz h

Simply supported boundary conditions are considered for the beam and the beam is assumed to be initially at zero deflection w(0, t) = w(l, t) = 0, t>0 M (0, t) = M (l, t) = 0, t>0 w(x, 0) = 0, 0≤x≤l

(9.8-9)

The first law of thermodynamics for heat conduction in the coupled form is (kθ,i ),i − ρcv θ,t − α(3λ + 2μ)T0 (ii ),t = 0

i = 1, 2, 3

(9.8-10)

where k, cv , and ii are the thermal conductivity, specific heat, and normal strain, respectively, and λ and μ are Lamé constants. The energy equation for the beam under consideration reduces to k(z)θ,xx + (k(z)θ,z ),z − ρ(z)cv (z)θ,t − α(z)E(z)T0 zw,xxt = 0

(9.8-11)


469

The thermal boundary conditions may be assumed in the form of an applied heat flux q(t), convection hc , or specified temperature shock on the upper or lower surfaces of the beam. The energy equation is obtained assuming that the upper surface of the beam is exposed to a heat flux q(x, t) and the lower surface is under convection to the ambient with the coefficient hc . The beam is initially assumed to be at ambient temperature and the thermal boundary and initial conditions are assumed as θ(0, t) = θ(l, t) = 0, t>0 θ(x, 0) = 0, 0≤x≤l

(9.8-12)

Using Eq. (9.8-11), the residue of the energy equation may be made orthogonal with respect to dz and zdz to provide two independent equations for two independent functions θ1 and θ2 as [58] B1 θ1,xx + B2 θ2,xx + B3 θ1 + B4 θ2 + B5 θ1,t + B6 θ2,t +B7 w,xxt + q(x, t) = 0

(9.8-13)

C1 θ1,xx + C2 θ2,xx + C3 θ1 + C4 θ2 + C5 θ1,t + C6 θ2,t +C7 w,xxt + C8 q(x, t) = 0

(9.8-14)

where the B’ s and C’ s are defined as

B1 =

K(z)dz

B5 =

B2 =

−ρ(z)cv (z)dz

B6 =

C1 =

z K(z) dz h

z −ρ(z)cv (z) dz h

K(z)zdz

C2 =

B3 = −hc

K(z)

z2 dz h

−hc zlower h

B4 = B7 =

E(z)α(z)T0 zdz

C3 = −hc zlower

2 −hc zlower K(z) − dz C5 = −ρ(z)cv (z)zdz h h z2 C6 = −ρ(z)cv (z) dz C7 = E(z)α(z)T0 z 2 dz C8 = zupper h

C4 =

To solve the simultaneous governing equations, dimensionless values are defined as w¯ = λ1

kc w qavrg αc l2

x¯ = λ2

x l

κc t¯ = λ3 2 t h

θ θ¯ = λ4 T0

(9.8-15)

where qavrg and κc are the average heat flux at the top of the beam and thermal diffusivity, respectively. The bar values indicate dimensionless parameters. The parameters λi are dimensionless parameters introduced to enable the balance of the members of matrices in the FEM part of the solution. By changing λi more balanced FEM matrices including stiffness, capacitance, and mass matrices are obtained leading to convergence of the solution.

470


Using the dimensionless parameters, the three coupled governing equations are a1 w¯,¯xx¯x¯x¯ + a2 θ¯1,¯xx¯ + a3 θ¯2,¯xx¯ + a4 w¯,t¯t¯ = 0 x, t¯) = 0 b1 θ¯1,¯xx¯ + b2 θ¯2,¯xx¯ + b3 θ¯1 + b4 θ¯2 + b5 θ¯1,t¯ + b6 θ¯2,t¯ + b7 w¯,¯xx¯t¯ + b8 q(¯ ¯ ¯ ¯ ¯ ¯ ¯ x, t¯) = 0 c1 θ1,¯xx¯ + c2 θ2,¯xx¯ + c3 θ1 + c4 θ2 + c5 θ1,t¯ + c6 θ2,t¯ + c7 w¯,¯xx¯t¯ + c8 q(¯ (9.8-16) where the a’ s, b’ s, and c’ s are dimensionless constants of the coupled equations. Simultaneous solution of these equations provides the distribution of the lateral deflection of the beam w and the temperature parameters θ1 and θ2 . The system of coupled equations (9.8-16) consists of functions of the space variable x and time t. Traditional finite element solution for such problems is the time marching method. The solution presented in this section is obtained by transfinite element method, where time is eliminated using Laplace transform. Once the finite element solution in space domain is obtained, a numerical scheme is used for the inverse Laplace transform to find the final solution in real time domain. Applying Laplace transform to Eqs. (9.8-16), gives a1 W,¯xx¯x¯x¯ + a2 Θ1,¯xx¯ + a3 Θ2,¯xx¯ + a4 s2 W = 0 b1 Θ1,¯xx¯ + b2 Θ2,¯xx¯ + (b3 + b5 s)Θ1 + (b4 + b6 s)Θ2 + b7 sW,¯xx¯ + b8 Q = 0 c1 Θ1,¯xx¯ + c2 Θ2,¯xx¯ + (c3 + c5 s)Θ1 + (c4 + c6 s)Θ2 + c7 sW,¯xx¯ + c8 Q = 0 (9.8-17) where s is Laplace transform parameter and W = L[w] ¯ Θi = L[θ¯i ], i = 1, 2 Q = L[q]

(9.8-18)

The coupled system of Eqs. (9.8-17) may be solved by Galerkin finite element method. The beam is divided into a number of straight elements. The base element (e) along the length of the beam is considered. The unknown functions W , Θ1 , and Θ2 in (e) may be approximated with a third order interpolation function < N > as W = < Wp > {Np } Θ1 = < Θ1p > {Np } Θ2 = < Θ2p > {Np } p = 1, 2, 3, 4

(9.8-19)

Considering a C 1 -continuous shape function, the degrees of freedom for the element lateral deflection, for instance < Wp >, become < Wp > = < Wi W,x |i Wj W,x |j >

(9.8-20)


471

where W,x |i and W,x |j are the x-derivatives of W at node i and j. Similar approximation is considered for Θ1 and Θ2 . The third order C 1 -continuous shape functions < N > are x¯ x¯ N1 = 1 − 3( ¯ )2 + 2( ¯ )3 l l x¯ 2 N2 = x¯(1 − ¯ ) l x¯ 2 x¯ N3 = 3( ¯ ) − 2( ¯ )3 l l x ¯ x − ¯l)( ¯ )2 N4 = (¯ l

(9.8-21) (9.8-22) (9.8-23) (9.8-24)

where ¯l is the length of the base element. The assumed shape functions insure the continuity of the nodal degrees of freedom as well as their first derivative with respect to the variable x. Now, the formal Galerkin method may be applied to the system of Eqs. (9.8-17). This is done by making the residue of each of Eqs. (9.8-17) for the base element orthogonal with respect to the approximating functions (9.8-21). The final finite element equation of motion, after assembling the matrix equations of each individual element, is obtained as ([M ]s2 + [C]s + [K]){X} = {F }

(9.8-25)

where [M ], [C], [K], and {F } are mass, capacitance, stiffness, and force matrices, respectively. Matrix {X} is the matrix of unknowns containing six unknowns. The solution of Eq. (9.8-22) for the unknown matrix {X} is obtained in terms of Laplace transform parameter s. To obtain the solution in real time domain, the inverse Laplace transform must be carried out. This may be done numerically by the method proposed in [35]. To validate the formulations, the results of this paper are compared with the analytical solution of a homogeneous beam reported in reference [51]. An aluminum beam of length 0.25 m and height 0.0022 m with simply supported boundary conditions is assumed. The ends of the beam are assumed to be at ambient temperature T0 = 293◦ K. The upper surface of the beam is exposed to a step function heat flux of intensity q = 108 W/m2 , while the lower surface is assumed to be thermally insulated. Figure 9.8-2 shows the midpoint lateral deflection history of the heated beam for the coupled thermoelasticity assumptions reported by reference [51] and the present study. Figure 9.8-3 shows temperature change history between upper and lower surfaces at the midpoint of the heated beam for the coupled thermoelasticity assumptions reported by reference [51] and the present study. It is seen that the temperature history obtained by the coupled solution is diffused along the time and oscillated about a constant value with very small variations. Close agreements are observed between the two studies.

472

Chapter 9. Finite and Boundary Element Methods 0.09

Dimensionless lateral deflection

0.08 0.07 0.06 0.05 0.04 0.03 Massalas and Kalpakidis[4]

0.02

Present

0.01 0

0

0.2

0.4

0.6

0.8 1 1.2 Dimensionless time

1.4

1.6

1.8

2

Figure 9.8-2: Deflection history of an aluminum beam at midpoint with the coupled thermoelasticity assumption.

2


1.8 1.6 1.4 1.2 1 0.8 0.6

Massalas and Kalpakidis[4] Present

0.4 0.2 0

0

0.2

0.4

0.6


1.4

1.6

1.8

2

Figure 9.8-3: Temperature change history between upper and lower surfaces of an aluminum beam at midpoint with the coupled thermoelasticity assumption.

Consider an FGM beam with ceramic upper surface and metal lower surface. The material properties of metal and ceramic are given in Table 9.8-1. The mechanical boundary conditions at the ends of the beam are assumed for a beam to be simply supported. The thermal boundary conditions are assumed so that the ends of the beam are at ambient temperature at T0 = 293◦ K.


473

Table 9.8-1:: Material properties of metal and ceramic constituents. Metal: T i − 6Al − 4V Ceramic: ZrO2 Em = 66.2 (Gpa) Ec = 117.0 (Gpa) ν = 0.322 ν = 0.322 αm = 10.3 × 10−6 (1/K) αc = 7.11 × 10−6 (1/K) ρm = 4.41 × 103 (kg/m3 ) ρc = 5.6 × 103 (kg/m3 ) km = 18.1 (W/m K) kc = 2.036 (W/m K) cm = 808.3 (J/kg K) cc = 615.6 (J/kg K) 0.14


number of elements=4 number of elements=20,50,100

0.12 0.1 0.08 0.06 0.04 0.02 0

0

0.1

0.2

0.3

0.4 0.5 0.6 0.7 Dimensionless length

0.8

0.9

1

Figure 9.8-4: Lateral deflection versus length for n = 0 (full ceramic) and t¯ = 3.

The length and thickness of the beam is assumed to be 0.8 and 0.0025 m, respectively. While these numerical dimensions are used for an FGM beam, the effect of aspect ratio h/l for different values of thickness to length ratio is also examined. The upper side of the beam is subjected to a step function thermal shock with the strength q = 108 W/m2 while the lower side is subjected to convection to the surrounding ambient with coefficient of hc = 10000 mW 2K . To check the rate of convergence of Galerkin finite element method in relation to the total number of the elements in the solution domain, Fig. 9.8-4 is plotted for lateral deflection versus the length of the beam. The results are shown for 4, 20, 50, and 100 elements for full ceramic beam, where n = 0, and for time t¯ = 3. It is seen that the results rapidly converge for 20 elements and higher. This rapid convergence is the advantage of C 1 -continuous shape function and Galerkin finite element method. Figure 9.8-5 shows the lateral deflection of the mid-length of the beam with respect to the time for different values of the power law index n. The curve

474



0.14 0.12 0.1 0.08

Homogeneous(n=0) n=0.5 n=1

0.06

n=2

0.04 0.02 0

0

0.5

1


3

3.5

4

Figure 9.8-5: Lateral deflection history at the midpoint of the beam for different power law indices. 500


450 400 350 Homogeneous(n=0)

300

n=0.5

250

n=2 n=7

200

n=20

150 100 50 0 0

0.5

1


3

3.5

4

Figure 9.8-6: Temperature change history at the midpoint of the beam at the upper side for different power law indices. associated with n = 0 corresponds to pure ceramic beam. Due to the applied thermal shock, the beam vibrates. It is seen that with the increase of the power law index n in the range 0 to 2, the midpoint lateral deflection of the FGM beam is decreased. For 2 < n < 7 the lateral deflection is almost constant and minimum [57]. For larger values of power law index, the lateral deflection and oscillation frequency increase.


475

500 450 Dimensionless temperature

400 350 300

Homogeneous(n= 0) n= 0.5

250

n= 2 n= 7

200

n= 20

150 100 50 0 −0.5

−0.4

−0.3

−0.2

−0.1

0

0.1

0.2

0.3

0.4

0.5

Dimensionless height

Figure 9.8-7: Temperature change distribution at the midpoint of the beam across the thickness direction at t¯ = 3 for different power law indices.

Figure 9.8-6 shows the temperature history at the upper side and at midlength of the beam. Due to the applied step function thermal shock, the beam temperature peaks to a maximum value, and then is diffused with the time. The figure shows that for more metal rich FGM (larger values of n) the temperature distribution decreases in value due to higher conductivity of metal. Figure 9.8-7 shows the distribution of temperature changes across the thickness direction at the midpoint of the beam at t¯ = 3. As the value of n is increased, temperature distribution across the thickness of the FGM beam tends to be more uniform, but linear according to the assumption. Figure 9.8-8 shows the axial stress history. The axial stress is decreased as the value of n is increased. For larger values of n, the variation of the axial stress with respect to n is slow. Figure 9.8-9 shows the lateral deflection history at the midpoint of an FGM beam (n = 20). The figure shows the effect of the coupling term. The solid line shows the coupled thermoelastic solution, the thick dashed curve shows the solution for the uncoupled case, and the thin dashed curve depicts the coupled curve where the coupling coefficient is made five time larger. The actual dimensionless coupling term in this figure is assumed to be 3.7639 × 10−6 . The solution for actual value of the thermoelastic coupling term is not usually distinguishable from the uncoupled solution for different materials. The coupled solution shows that the coupling between strain and temperature fields slightly decreases the amplitude of vibration. Furthermore, it results in a decay of vibration amplitude with respect to time for long enough time. It is also observed that the frequency of vibration for the coupled case increases with time and is larger than that of the uncoupled case.

476

Chapter 9. Finite and Boundary Element Methods 0

−50

Dimensionless stress

−100 −150 −200

Homogeneous(n=0) n=0.5

−250

n=2

−300

n=7 n=20

−350 −400 −450 −500 0

0.5

1

1.5

2

2.5

3

3.5

4

Dimensionless time

Figure 9.8-8: Axial stress history at the midpoint of the beam at the upper side for different power law indices.


0.02 0.018 0.016 0.014 0.012 0.01 uncoupled

0.008

coupled magnified coupled

0.006 0.004 0.002 0

0

0.5

1

1.5

2

2.5

3

3.5

4

Dimensionless time

Figure 9.8-9: Lateral deflection history at the midpoint of the FGM beam for n = 20, the coupling effect.

9

Boundary Element Formulation

The effects of the thermal coupling on both plane harmonic thermoelastic waves in unbounded media and Rayliegh surface waves propagating along the free surface of a half-space, using the Lord–Shulman model, are reported by Nayfeh [59] and Nayfeh and Nemat-Nasser [60]. Puri [61], using the same theory, obtained exact solutions to the frequency equations and calculated exact values for the real and imaginary parts of the wave number. Agarwal [62]

9. Boundary Element Formulation

477

used the Green and Lindsay’s theory to determine the phase velocity, specific loss, attenuation coefficient, and amplitude ratio behavior for quasi-elastic and quasi-thermal modes by directly solving the frequency equations. Tamma and Zhou [63] have reviewed and demonstrated the time scale/space scale effect of the heat conduction models and subsequently the associated thermoelasticity models. A generalized thermoelasticity model is proposed by Zhou et al. [64] which utilizes the St. Venant-Kirchhoff hyperelasticity material model and the C- and F-processes heat conduction model. The resulting generalized thermoelasticity model includes the classical thermoelasticity model and the Lord– Shulman model as particular cases. Hosseini Tehrani and Eslami [65] showed the coupling effects in natural frequencies, temperature distribution and resonance amplitudes in a time harmonic problem by boundary element method. Prevost and Tao [66] demonstrated an implicit-explicit formulation adapted from a variation of the Newmark structural dynamic algorithm to solve a dynamic thermoelasticity problem based on the non-classical model. The above approach combines the algorithmic advantages of the explicit and the implicit methods into a single time integration procedure. Tamma and Railkar [67] demonstrated via tailored hybrid transfinite element formulations, the evaluation of a particular uncoupled class of non-Fourier stress wave disturbances. Their results showed the presence of significant thermal stresses due to nonFourier effects when the speeds of thermal waves and stress waves are equal under certain loading situations. For the case of unequal speeds of propagation, the relative magnitudes (Fourier versus non-Fourier) were shown to be almost comparable. More recently, Tamma [68,69] and Tamma and Namburu [27] studied various dynamic thermoelasticity problems of the non-classical type including effective approaches for stabilizing the oscillatory solution behavior. Chen and Dargush [70] used a boundary element method for transient and dynamic problems in generalized thermoelasticity in a half-space by using Laplace transform and infinite space fundamental solution. Chen and Lin [28] used Laplace transform and control volume method to study the transient coupled thermoelastic problem with relaxation times (Green and Lindsay model) in a half-space, and they found that the steep jumps occurring in the wave propagation of temperature, axial displacement, and axial stress depend on the values of the coupling coefficients and relaxation times. They also showed two waves propagating with different but finite speeds for problems with Green and Lindsay’s dynamic thermoelastic model. Eslami and Hosseini Tehrani [71] considered boundary element formulation for conventional coupled thermoelasticity in a two-dimensional finite domain and investigated coupling effect in temperature, displacement, and stress distribution. With the advent of laser technology, in which high heat flux conditions are encountered, its use for extremely short duration has found many applications [72–79]. Some typical applications include the annealing of semi-conductors, surface heating and melting of metals. Other applications include explosive

478


bonding and melting where high local heat fluxes are involved, nuclear boiling, etc. The basic nature of the thermal energy transport immediately after the application of the pulse and the resulting sustained temperature at the surface of the medium have been some topics of interest in such situations. In this section, a Laplace transform boundary element method is developed for the dynamic problem in generalized thermoelasticity with relaxation times involving a finite two-dimensional domain. The boundary element formulation is presented and a single heat excitation is used to drive the boundary element formulations. Aspects of numerical implementation are discussed. It is seen through various illustrative problems that this method has a good accuracy and efficiency in predicting the wave propagation of temperature, displacement, and stress. It should be noted that the distributions of temperature, displacement, and stress show discontinuities at their wave fronts. The thermo-mechanical wave propagation and reflection in a finite domain are investigated and the influence of coupling parameter and relaxation times on thermoelastic waves are also investigated. A homogeneous isotropic thermoelastic solid is considered. In the absence of body forces and heat flux, the governing equations for the dynamic generalized thermoelasticity in the time domain are written as ui = 0 (λ + μ)uj,ij + μui,jj − β(T,i + t1 T˙,i ) − ρ¨

(9.9-1)

kT,jj − ρce T˙ − ρce (t0 + t2 )T¨ − βT0 (t0 uï,i + u˙ i,i ) = 0

(9.9-2)

where λ, μ are Lamé constants, ui is the displacement vector, ρ is the mass density, T is the absolute temperature, T0 is the reference temperature, k is the coefficient of conduction, β is the stress-temperature modulus, ce is the specific heat, and t0 , t1 and t2 are the relaxation times proposed by Lord and Shulman [16] and Green and Lindsay [17], respectively. When t0 , t1 , and t2 vanish, Eqs. (9.9-1) and (9.9-2) reduce to the classical coupled theory. In LS theory (Lord and Shulman’s theory), t1 = t2 = 0 and Fourier’s law of heat conduction is modified by introducing the relaxation time t0 . In GL theory (Green and Lindsay’s theory), t0 = 0 and both the Duhamel-Neuman relations and entropy density function are modified by introducing two relaxation times t1 and t2 (Prevost and Tao [66]). It is convenient to introduce the usual dimensionless variables as x ¯ tc1 ¯ t0 c1 ¯ t1 c1 ¯ t2 c1 t0 = ; t= ; t1 = ; t2 = l l l l l σij (λ + 2μ)ui ¯ T − T0 ; u¯i = ; T = σ ¯ij = βT0 lβT0 T0 x¯ =

(9.9-3)

where l = k/ρce c1 is the dimensionless unit length, and c1 = (λ + 2μ)/ρ is the velocity of the longitudinal wave. Equations (9.9-1) and (9.9-2) take the form (dropping the bar for convenience)


479

λ+μ μ ui,jj + uj,ij − (T,i + t1 T˙i ) − uï = 0 λ + 2μ λ + 2μ T,jj − T˙ − (t0 + t2 )T¨ −

T0 β 2 (u˙ j,j + t0 u¨j,j ) = 0 ρce (λ + 2μ)

(9.9-4) (9.9-5)

Using the dimensionless equations, the stress wave speed is one and the speed of the temperature wave may be computed as (Prevost and Tao [66])

ct =

1 t0 + t2

(9.9-6)

Transferring Eqs. (9.9-4) and (9.9-5) to Laplace transform domain yields μ λ+μ ui,jj + uj,ij − (T,i + t1 sT,i ) − s2 ui = 0 λ + 2μ λ + 2μ T0 β 2 T,ii − sT − (t0 + t2 )s2 T − (suj,j + t0 s2 uj,j ) = 0 ρce (λ + 2μ)

(9.9-7) (9.9-8)

Equations (9.9-7) and (9.9-8) are rewritten in the matrix form as Lij Uj = 0

(9.9-9)

For the two-dimensional domain the operator Lij reduces to ⎡ ⎢

Lij = ⎢ ⎣

μ λ+μ Δ + λ+2μ D12 − s2 λ+2μ λ+μ DD λ+2μ 1 2 T0 β 2 − ρce (λ+2μ) s(1 + t0 s)D1

λ+μ DD λ+2μ 1 2 μ λ+μ Δ + λ+2μ D22 − s2 λ+2μ 2 0β − ρceT(λ+2μ) s(1 + t0 s)D2

Ui = [ u v

⎤

−D1 (1 + t1 s) ⎥ ⎥ −D2 (1 + t1 s) ⎦ Δ − s(1 + st0 + st2 )

T]

∂ (i = 1, 2) and Δ denotes the Laplacian. The boundary ∂xi conditions are assumed to be where Di =

ui = uî τi = σ îj nj T = Tˆ q = qî ni

on on

Γu Γτ

on on

ΓT Γq

(9.9-10)

îj , Tˆ, and qî are the known values of the displacement, stress, where uî , σ temperature, and the heat flux on the boundaries Γu , Γτ , ΓT , and Γq , respectively. Either of these boundary conditions or their combinations may be defined on the boundary.

480


Boundary integral equation In order to derive the boundary integral problem, we start with the following weak formulation of the differential equation set (9.9-9) for the fundamental solution tensor Vik∗ Ω

(Lij Uj )Vik∗ dΩ = 0

(9.9-11)

After integrating by parts over the domain and taking a limiting procedure approaching the internal source point to the boundary point, we obtain the following boundary integral equation

Ckj Uk (y, s) =

+ Γ

Γ

∗ τα (x, s)Vαj (x, y, s) − Uα (x, s)Σ∗αj (x, y, s)dΓ(x)

∗ ∗ T,n (x, s)V3j,n (x, y, s) − T (x, s)V3j,n (x, y, s)dΓ(x)

(9.9-12)

where Uα = uα (α = 1, 2) and U3 = T , and Ckj denotes the shape coefficient tensor. The kernel Σ∗αj in Eq. (9.9-12) is defined by Σ∗αj = [(

λ T0 β 2 (s + t0 s2 ) ∗ μ ∗ ∗ ∗ + + Vβj,α )]nβ Vkj,k V3j )δαβ + (Vαj,β λ + 2μ ρce (λ + 2μ) λ + 2μ (9.9-13)

Here, the fundamental solution tensor Vjk∗ must satisfy the differential equation lij Vjk∗ = −δik δ(x − y)

(9.9-14)

where lij is the adjoint operator of Lij in Eq. (9.9-9) and given by ⎡ ⎢

lij = ⎢ ⎣

μ λ+μ Δ + λ+2μ D12 λ+2μ λ+μ DD λ+2μ 1 2

− s2

D1 (1 + t1 s)

λ+μ DD λ+2μ 1 2 μ λ+μ Δ + λ+2μ D22 λ+2μ

+ s2

D2 (1 + t1 s)

T0 β 2 (s ρce (λ+2μ) T0 β 2 (s ρce (λ+2μ)

⎤

+ t0 s2 )D1 ⎥ + t0 s2 )D2 ⎥ ⎦ Δ − s(1 + t0 s + t2 s)

Fundamental solution In order to construct the fundamental solution we put the fundamental solution tensor Vij∗ of Eq. (9.9-14) in the following potential representation using the transposed co-factor operator μij of lij and the scalar function Φ∗ (Tosaka [80]) Vij∗ (x, y, s) = μij Φ∗ (x, y, s)

(9.9-15)

Substituting Eq. (9.9-15) into Eq. (9.9-14), we obtain the differential equation ΛΦ∗ = −δ(x − y) where Λ = det(lij ) =

μ (Δ − h21 )(Δ − h22 )(Δ − h23 ) λ + 2μ

(9.9-16) (9.9-17)


481

where h2i are a solution of h21 =

λ + 2μ 2 s μ

h22 + h23 = s2 + s(1 + t0 s + t2 s) + h22 h23 = s3 (1 + t0 s + t2 s)

T0 β 2 s(1 + t0 s + t1 s + t0 t1 s2 ) ρce (λ + 2μ) (9.9-18)

Here, h1 is the longitudinal wave velocity, h2 is the shear wave velocity, and h3 is the rotational wave velocity, and

λ + 2μ K0 (h1 r) K0 (h2 r) K0 (h3 r) Φ= + + 2πμ (h22 − h21 )(h23 − h21 ) (h23 − h22 )(h21 − h22 ) (h21 − h23 )(h22 − h23 ) (9.9-19) The fundamental solution tensor Vij∗ for two-dimensional domain is found as ∗ Vαβ = ∗ = V3α ∗ Vα3 =

V33∗ =

3

(ψk (r)δαβ − κk r,α r,β )

k=1 3 k=1 3 k=1 3

(α, β = 1, 2)

ξ´k (r)r,α ξk (r)r,α

ζk (r)

(9.9-20)

k=1

where Wk 2 λ+μ 2 λ + 2μ 2 [(hk − m2 )(h2k − m1 ) + ( )(hk − m1 − m3 m4 C )h ] 2π μ λ+μ k Wk (λ + μ) 2 λ + 2μ hk ×K0 (hk r) + [hk − m1 − m3 m4 C )] K1 (hk r) 2πμ λ+μ r Wk (λ + μ) 2 λ + 2μ [hk − m1 − m3 m4 C )]K2 (hk r) κk (r) = 2πμ λ+μ Wk ξ´k (r) = m4 (h2k − m2 )hk K1 (hk r) 2π Wk ξk (r) = Cm3 (h2k − m2 )hk K1 (hk r) 2π Wk 2 ζk (r) = (9.9-21) (h − m2 )(h2k − s2 )K0 (hk r) 2π k

ψk (r) =

and r = x − y; m1 = s(1 + t0 s + t2 s); m2 =

λ + 2μ 2 s μ

482

Chapter 9. Finite and Boundary Element Methods m3 = s(1 + t0 s); m4 = (1 + t1 s); C = Wi =

−1 (h2i − h2j )(h2k − h2i )

T0 β 2 ρce (λ + 2μ)

(i = 1, 2, 3 j = 2, 3, 1 k = 3, 2, 1) (9.9-22)

with K0 (hk r), K1 (hk r), and K2 (hk r) being the modified Bessel functions of the second kind and zero, first, and second order, respectively. In order to solve numerically the boundary element integral equation (9.912), the standard procedure is applied. When transformed numerical solutions are specified, transient solutions can be obtained by using an appropriate numerical inversion technique. In this section, a method presented by Durbin [81] is adopted for this numerical inversion. To compare the two-dimensional numerical results of this section with the analytical known solution for a half-space, a square plate subjected to heating at one edge with a step function is considered, as shown in Fig. 9.9-1 [29]. The plate is thermally insulated at three other edges. The plate material is considered stainless steel. Results are obtained along the axis of symmetry of the plate. The boundary element solution and the analytical solution of the coupled and uncoupled thermoelasticity of the half-space are given by Chen and Dargush [70] and Sternberg and Chakravorty [82], respectively. Both solutions are based on Laplace transform method. Figures 9.9-2 to 9.9-4 show a comparison of the dimensionless temperature, axial displacement, and axial stress at dimensionless length x = 1 (which is the location of the elastic wave front at the nondimensional time t = 1). The results are plotted for analytical solutions (Sternberg and Chakravorty [82]), the boundary element solution (Chen and Dargush [70]), and the present results for different coupling parameter. The case of C = 0 corresponds to the uncoupled solution. The coupled results are presented for C = 0.36 and C = 1 to match the results of Chen and y 6

θ(t) = θ0

-

6

-x

l

l

-

?

Figure 9.9-1:: A square plate subjected to thermal loading.


483

Figure 9.9-2:: Comparison of the dimensionless temperature at x = 1.

Figure 9.9-3:: Comparison of the dimensionless axial displacement at x = 1. Dargush. These values are unrealistically high for the material of interest. The data which are considered for the comparison purpose in Figs. 9.9-2 to 9.9-4 are related to the solution of coupled thermoelasticity in the half-space. In the half-space region, the solution domain reduces to a one-dimensional problem. It is seen that the displacement and uncoupled temperature results of the present analysis have good agreement with analytical and boundary element solutions for different coupling parameters. However, the coupled temperature and stress solution of the present work deviates from the half-space solution as time is advanced.

484

Chapter 9. Finite and Boundary Element Methods Nondimensional stress 0.4 0.2 0 −0.2 −0.4 −0.6

Analytical Present Sx C=0 Present Sx C=.36

−0.8

Present Sx C=1

−1 0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

Nondimensional time

Figure 9.9-4:: Comparison of the dimensionless axial stress at x = 1. For the uncoupled condition, the heat conduction equation is separately solved for the temperature distribution and is independent of the stress field. Since the data presented in Figs. 9.9-2 to 9.9-4 are related to the axis of is zero and the temperature distribution symmetry of the plate, thus qy = k ∂T ∂y is symmetric about the x-axis. Consequently, the distribution of temperature along the axis of symmetry of a two-dimensional domain coincides with that of the one-dimensional solution of the half-space. For the coupled solution, however, the influence of the stress field on the temperature distribution of a two-dimensional domain results in lower curve of temperature compared to the half-space solution. The displacement curves of the two-dimensional domain along the axis of symmetry of the domain almost coincide with those of the half-space. The reason is that due to symmetry, v = 0 along the x-axis and the only non-zero displacement is u. This condition coincides with the onedimensional half-space solution. The stress plot of two-dimensional domain along the axis of symmetry, shown in Fig. 9.9-4, is different from that of the half-space. The main reason for the difference is the existence of non-zero y in the two-dimensional domain. The effect of y causes that the peak of compressive stress occurs at a shorter time. The tensile stress produced by the application of thermal shock is larger in a finite domain compared to the half-space. This is shown in Fig. 9.9-4 for the range of time where the wave reflection is not produced yet. Now, consider a coated surface subjected to a laser beam. Since the effects of the coupled thermoelastic field is in general significant at a very short time at early stages of thermal shock application, the stress field is studied in the coated domain. When the length is too large compared to the thickness, as it is in the case of a coated surface, the problem may be modelled as shown


485

y 6 jjjjjjjj

θ(t) = θ0

6 -x l - l ? jjjjjjjj

Figure 9.9-5:: Model of a layer subjected to thermal loading. Temperature 1.2 Present 1

Chen&Lin T0=2

0.8 0.6 0.4 0.2 0 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 2.2 2.4 2.6 2.8 3 Nondimensional time

Figure 9.9-6: Comparison of the dimensionless temperature at the middle of the plate for LS theory. in Fig. 9.9-5. To compare the results with the solutions available for a halfspace, the square plate (l = 2 non-dimensional) of Fig. 9.9-5 is considered to experience a step function heating at one side and insulation at the other sides. The temperature and displacement wave propagations and the effect of the relaxation times are investigated along the axis of symmetry of the plate of unit dimensionless length and are compared with the half-space results, Figs. 9.9-6 to 9.9-8. The thermal load equation at x = 0 is considered as (see Fig. 9.9-9) θ(t) = t exp−5t

(9.9-23)

The thermoelastic stress wave front propagation along the axis of symmetry of the plate with different relaxation times for LS and GL theories are shown in Figs. 9.9-10 to 9.9-13.

486

Chapter 9. Finite and Boundary Element Methods Temperature 1 Present 0.8

Chen&Lin T1=T2=2.25

0.6

0.4

0.2

0 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 2.2 2.4 2.6 2.8 3 Nondimensional time

Figure 9.9-7: Comparison of the dimensionless temperature at the middle of the plate for GL theory. Axial displacement 2 Present

1.5

Chen T1=T2 = 2.25 Chen T0 = 2

1 0.5

Present & Chen

0 −0.5 −1 −1.5 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 2.2 2.4 2.6 2.8 3 Nondimensional time

Figure 9.9-8: Comparison of the dimensionless axial displacement at the middle of the plate for LS and GL theories. Figures 9.9-6 to 9.9-8 show the dimensionless temperature and displacement distribution versus dimensionless time for LS and GL theories with zero coupling parameter. The temperature distribution shows oscillation beyond the temperature wave front, which is due to the temperature wave reflection from the boundaries for both LS and GL theories in a finite domain. The distribution of axial displacement for LS model follows that of the half-space result. In GL model the temperature variation has more effects on displacement (t1 effect).


487

Nondimensional temperature 0.08

0.06

0.04

0.02

0 0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

Nondimensional time

Figure 9.9-9:: Pattern of the thermal loading, according to Eq. (9.9-23).

Nondimensional Sx 0.15

T0=0.64 Nondim. time=.5

0.1

Nondim. time=.7

0.05

Nondim. time=1

0 −0.05 −0.1 −0.15 −0.2 −0.25 0

0.2

0.4

0.6

0.8

1

1.2

1.4

X

Figure 9.9-10: Distribution of the dimensionless axial stress along the axis of symmetry of the domain shown in Fig. 9.9-5 for LS theory (t0 = 0.64). Beyond the temperature wave front in GL model, displacement distribution oscillates and deviates from the half-space result. Figures 9.9-10 to 9.9-13 show the stress distribution along the axis of symmetry of the domain shown in Fig. 9.9-5 under temperature loading of Eq. (9.9-23). The coupling parameter is 0.0168. The temperature distribution is nearly the same for the two theories [29] but the displacement and stress results for GL theory, especially when the stress wave speed is larger than that

488

Chapter 9. Finite and Boundary Element Methods Nondimensional Sx 0.15

T0=1.5625

0.1

Nondim. time=.5 Nondim. time=.7

0.05

Nondim. time=1

0 −0.05 −0.1 −0.15 −0.2 −0.25 0

0.2

0.4

0.8

0.6

1

1.2

1.4

X

Figure 9.9-11: Distribution of the dimensionless axial stress along the axis of symmetry of the domain shown in Fig. 9.9-5 for LS theory (t0 = 1.5625).

Figure 9.9-12: Distribution of the dimensionless axial stress along the axis of symmetry of the domain shown in Fig. 9.9-5 for GL theory (t1 = t2 = 0.64).

of the thermal waves, is many times as much as in LS theory. The peak values of the temperature decrease with the increase of x [29]. In LS and GL theories it is expected to have two waves propagation with different but finite speeds. The propagation speed of the elastic wave is larger than that of the thermal wave for t0 > 1 or t2 > 1. Figures 9.9-10 to 9.8-13 clearly show this phenomenon for the stress distribution. As the value of x is increased, the zone


489

Figure 9.9-13: Distribution of the dimensionless axial stress along the axis of symmetry of the domain shown in Fig. 9.9-5 for GL theory (t1 = t2 = 1.5625). between two steep jumps in the axial stress distribution grows wider. One of these jumps is located at the temperature wave front and the other at the stress wave front [29]. In this section, the hybrid application of Laplace transform and the boundary element methods is applied to analyze the generalized thermoelastic problems of two-dimensional finite domain and various loadings. The integral representation is derived directly from the governing differential equations in Laplace transform domain. By utilizing the infinite space adjoint fundamental solution, a boundary element formulation is obtained, thus eliminating the need for the volume discretization for the analysis of homogeneous media. Comparison with the available half-space solution is made and the applicability of the method is shown. It is found from this study that the magnitudes of steep jumps occurring in the wave propagations of temperature, axial displacement, and axial stress depend on the values of the coupling coefficients, relaxation times, and the chosen theory. In general, two wave fronts with different and finite speeds are observed in the distribution of temperature, displacement, and stress. It was shown that the elastic stress wave is faster than the thermal wave for t0 > 1 or t2 > 1. Further studies were carried out by the same authors to investigate the effects of the mechanical and thermal shocks on the stress wave fronts in the generalized thermoelasticity based on LS and GL models (Tehrani and Eslami [83–85]). Reviewing the paper based on LS theory [83], reveals an interesting conclusion in comparison of the results with the results obtained by GL model reported in Tehrani and Eslami [85]. In LS model, when ct > cs , due to the

490


lower temperature peak, the positive axial displacement produced by the compressive stress has larger peak compared to the case when cs > ct . That is, LS model does not sense the influence of heat (in contrast to GL model). However, due to the presence of the first and second rate of strain tensor with respect to the time in the energy equation, LS model predicts higher temperature distribution. Another conclusion follows from the comparison of the results of the coupled thermoelastic solution based on GL model [85] to that of the classical [84] or LS models [83]. It is observed from the results that the peak axial displacement predicted by GL model is up to about 5 times more than that of LS model at early stages of the shock application. The peak axial stress predicted by GL model is up to about 10 times higher than that predicted by LS model at early stages of the shock application. A justification for this point may be placed on the existence of the term t1 T˙i in Navier equations, which causes to absorb more thermal energy by the body deformation.

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[50] Seibert, A.G. and Rice, J.S., Coupled Thermally Induced Vibrations of Beams, AIAA J., Vol. 11, pp. 1033–1035, 1973. [51] Massalas, C.V. and Kalpakidis, V.K., Coupled Thermoelastic Vibration of a Simply Supported Beam, J. Sound Vib., Vol. 88, pp. 425–429, 1983. [52] Massalas, C.V. and Kalpakidis, V.K., Coupled Thermoelastic Vibration of a Timoshenko Beam, Lett. Appl. Eng. Sci., Vol. 22, pp. 459–465, 1984. [53] Maruthi Rao, D. and Sinha, P.K., Finite Element Coupled Thermostructural Analysis of Composite Beams, Comput. Struct., Vol. 63, pp. 539– 549, 1997. [54] Manoach, E. and Ribeiro, P., Coupled Thermoelastic Large Amplitude Vibrations of Timoshenko Beams, J. Mech. Sci., Vol. 46, pp. 1589–1606, 2004. [55] Sankar, B.V., An Elasticity for Functionally Graded Beams, J. Compos. Sci. Technol., Vol. 61, pp. 686–696, 2001. [56] Bhavani, V., Sankar, B.V., and Tzeng, J.T., Thermal Stresses in Functionally Graded Beams, AIAA J., Vol. 40, pp. 1228–1232, 2002. [57] Babai, M.H., Abbasi, M., and Eslami, M.R., Coupled Thermoelasticity of Functionally Graded Beams, J. Therm. Stresses, Vol. 31, pp. 680–697, 2008. [58] McQuillen, E.J. and Brull, M.A., Dynamic Thermoelastic Response of Cylindrical Shell, J. Appl. Mech., Vol. 37, pp. 661–670, 1970. [59] Nayfeh, A.H., Propagation of Thermoelastic Distribution in Non-Fourier Solids, AIAA J., Vol. 15, pp. 957–960, 1977. [60] Nayfeh, A.H. and Nemat-Nasser, S., Thermoelastic Waves in Solids with Thermal Relaxation, Acta Mech., Vol. 12, pp. 53–69, 1971. [61] Puri, P., Plane Waves in Generalized Thermoelasticity, Int. J. Eng. Sci., Vol. 11, pp. 735–744, 1973. [62] Agarwal, V.K., On Plane Waves in Generalized Thermoelasticity, Acta Mech., Vol. 31, pp. 185–198, 1979. [63] Tamma, K.K. and Zhou, X., Macroscale Thermal Transport and Thermo-Mechanical Interactions: Some Noteworthy Perspectives, J. Therm. Stresses, Vol. 21, pp. 405–450, 1998.

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Chapter 10 Creep Analysis

Under the combination of elevated temperature and mechanical loads, structural members tend to creep. Basic laws of creep are presented and the effect of temperature changes in the constitutive law of creep is discussed. The rheological models of two important engineering concepts of stress, namely, the loadand the deformation-controlled stresses, are presented. The chapter concludes with the description of numerical techniques of solutions to creep problems. The nature of pure thermal stresses as the deformation-controlled stresses, is presented in the example problems. It is shown that pure thermal stresses relax as the time advances.

1

Introduction

The creep phenomenon is an important design consideration in the analysis of structures. This importance is due to the increasingly large number of engineering applications of materials for the design of parts operating at high temperatures. Physically, when a solid is subjected to a force, the atomic lattice adjusts itself to oppose the applied force and maintain equilibrium. On a macroscopic scale the atomic adjustment is observed as a deformation when the lattice remains continuous. The response of this deformation to the applied stress varies with the magnitude and state of stress, the temperature, and the rate of deformation. The elastic deformation occurs when the strain appears and disappears simultaneously with the application and removal of the stress. The creep deformation results when the body undergoes continuous deformation under a constant load or stress. In general, the term creep is applied to a time-dependent stress-strain relation. The creep is observed in various types of materials such as metals, composites, plastics, concrete, rocks, and ice. The importance of creep deformation R.B. Hetnarski, M.R. Eslami, Thermal Stresses – Advanced Theory and Applications, Solid Mechanics and its Applications 158, c Springer Science + Business Media B.V. 2009

499

500

Chapter 10. Creep Analysis

in engineering problems reveals itself in particular at high temperature conditions. For example, the behavior of carbon steel at normal temperatures and moderate stresses is well described by the theory of elasticity, but it behaves quite differently at a temperature above 450◦ C. At this temperature the steel ceases to obey Hooke’s law, even at low stress, and its stress-strain curve is essentially dependent on the rate of deformation. For this reason the stressstrain curve which forms the basis of the theory of elasticity cannot be used in creep problems. In the last decades a considerable amount of research has been performed in the area of plasticity, the irreversible deformation of solids. It has been shown that the deformations of a solid body in the rate independent plasticity, to the first degree of approximation and at moderate temperatures, depend on the magnitude and the order of application of the acting forces. The time parameter should not be, however, completely ignored since, to some extent, the history of loading is of significance in the rate independent plasticity and this is why the concept of load path or loading history has been introduced in the theory of plasticity. The theory of creep was developed in close relation with the theory of plasticity, as both subjects have some common fundamental concepts. In other words, while there are important differences in the physical phenomena of creep and plasticity, the basic mathematical treatments of the problems involving creep in structures are similar to those in the theory of plasticity. Mathematically, the main difference is that in the theory of creep the stress-strain relation is time-dependent. Historically, the importance of creep was first noticed in the development and design of turbines, where the high temperature environment increases the rate of creep of the turbine disk and causes reduction in the space between the blades’ tips and the housing. Therefore, there was a direct need to assess the ability of materials to resist creep, and creep tests were developed for this purpose. Furthermore, the study of creep was divided into two branches, the metallurgical aspects of the creep of materials and the mathematical theory of creep problems. In metallurgy, there have been extensive developments in the determination of the stress-creep rate relations for different materials in different temperature ranges. In general, creep in materials is a function of the applied stress, temperature, time, and some structural characteristics. Furthermore, three characteristic periods of time have been observed during which the creep rate with time is sequentially decreasing, remaining essentially constant, and increasing. These are often called the periods of primary, secondary, and tertiary creep. The primary stage of creep, which begins immediately after application of the load, is usually very short in duration but the rate of creep is high. The amount of strain developed during the secondary creep period is large compared with that in the primary creep period. At this stage the creep curves are usually

1. Introduction

501 Rupture

Creep strain

T = Constant σ = Constant

Primary t1

Tertiary

Secondary Time

t2

tr

Figure 10.1-1: Creep stages at constant stress and temperature. approximated by a straight line. The tertiary stage of creep follows the secondary period and is characterized by a rather rapid increase in the creep rate which eventually causes rupture in the material. These three periods of creep are shown in Fig. 10.1-1. All of these stages of creep can be mathematically described by the constitutive equation of creep. The purpose of the constitutive equation is to provide a suitable means for calculating stresses and strains in a structure subjected to creep, where the intended lifetime of the structure is of particular importance. In the next section, the constitutive equations of creep are discussed in detail. Mathematically, the theory of creep has been basically derived from the theory of plasticity, and the creep problems are more or less treated as the plasticity problems. As an early general reference, we may mention the work of Norton [1], where the creep behavior of steel at high temperatures is discussed. In 1936, Nádai presented a few simple cases of the inelastic deformation in solids with a particular reference to the creep of metals in the strain-hardening range [2]. In this paper, only the constitutive equations of creep are discussed. In the same year, Bailey [3] discussed the design aspect of creep and proposed the expressions for creep under the general system of stresses. Soderberg [4] in 1936 published a treatise in which he presented a method of interpreting creep test data and he applied the method to several problems of machine design. The method involves a rational theory of plastic flow in polycrystalline materials to which empirical results from actual tests can be applied. The basic assumption in this proposed method is that the facts already established for the plastic flow at normal temperatures remain valid for high temperatures as well. Later, Wahl [5–7] analyzed the creep deformations in rotating disks based on Tresca criterion and the associated flow rule. The analytical results compared well with experimental data. In 1957, Weir [8] derived the equation of creep stresses in a thick-walled tube under internal pressure. A more

502


complete treatment of the creep theory can be attributed to Finnie and Heller [9], who discussed both metallurgical and mathematical aspects of creep. In 1959, Mendelson [10] proposed a general approach to the practical solution of creep problems. In the same year Rimrott [11] obtained the solution of creep in thick-walled tubes under internal pressure considering large strain. Later, in 1965 Taira [12] discussed both Huber-von Mises-Hencky and Tresca criteria and their associated flow rules to the creep of cylinders. Bhatnagar and Gupta [13], analyzed thick-walled orthotropic cylinder using the constitutive equations of anisotropic creep theory. As a general reference, Hult [14], Odqvist [15], Rabotnov [16], Smith and Nicolson [17], Penny and Marriot [18], Kachanov [19], and Fridman [20] should be mentioned. The theory of creep has been well established and discussed in these references, and many practical problems in structural mechanics are worked out. Before proceeding further, we will describe some of the basic terms which are frequently used in the literature of creep. Primary, secondary, and tertiary creep These three stages of creep are associated with constant stress and temperature conditions when the rate of creep under these conditions is decreasing, constant, and increasing, respectively. Figure 10.1-1 shows the stages of creep. Transient creep, relaxation Under some conditions when the load is constant, the stresses within the body vary with time. This stress redistribution, or stress relaxation, produces transient creep. Stationary creep When the stress redistribution with the time is completed, the body under the action of external forces creeps in a stationary state. This state of creep is analogous to the steady state condition in problems of mechanics. When such analysis is done for a specific problem, it means that either the state of stress does not vary with the time or the stress redistribution in the time is ignored. Recovery The term recovery is used when the strain hardening due to thermal softening is relieved.

2. Creep of Metals

2

503

Creep of Metals

A typical result from a long term creep test is shown in Fig. 10.2-1 for a tensile test of a bar at both constant temperature and stress. The time t is on the horizontal axis and the relative elongation = ΔL/L0 is on the vertical axis, where ΔL is the absolute elongation and L0 is the original length of the bar. When a bar experiences an initial strain 0 at t = 0, this strain is shown on -axis as the length OA. The magnitude depends on the load, and the initial strain may be all elastic or elastic and plastic. When the time advances, the length of the bar increases according to the AB portion of the curve and the rate of creep strain is decreasing with the time. This portion is called the primary creep. Its duration in most metals is short. At the time t1 , the straight portion BC of the curve begins where the rate of creep strain is constant with the time, and this portion is called the secondary creep. This stage is usually very long relative to the other stages of creep and that is why the designers often consider only this portion of the creep in their design calculations. The intersections of this straight line with the strain axis at the time t = 0 and t = t3 are denoted by 1 and 4 , respectively. This portion will eventually lead to the section CD, characterized by an increasing rate of creep strain and which finally leads to the rupture of the bar. This portion is called the tertiary creep. Depending on the temperature and stress conditions, rupture may appear in the form of brittle fracture or ductile deformation, where in the latter case it is usually accompanied by the formation of a neck [18]. The curve shown in Fig. 10.2-1 is a typical creep curve for a metal where the stress and the temperature during the test are kept constant. In general, creep of a metal is a function of the time as well as of the stress and the temperature. Therefore, the creep strain is expressed as a function c = f (t, σ, T )

(10.2-1) Rupture D

5

Creep strain

4

C

3

B 2

c

1 0 0

0

t1

= e + T+ p in general

Time

t2

tr

Figure 10.2-1: A typical creep curve.

504


where σ and T denote the stress and the temperature, respectively. It is generally accepted that the functional relationship between the creep strain and the time, stress, and temperature for the three phases of the creep may be more specifically defined than by Eq. (10.2-1), namely, by a product of three functions, each depending on one variable only [18] c = f1 (t) · f2 (σ) · f3 (T )

(10.2-2)

The separation of the time and stress functions, f1 (t) and f2 (σ), is implicit, while the separation of the temperature function, f3 (T ), is not generally suggested and sometimes is included in the time function [21]. In the following, each of the above functions is discussed separately and the functional relationship for the different phases of creep is given [22]. Time function for primary and secondary creep The temperature of the creeping material strongly affects the behavior of the body, therefore, in discussing the functional relationship between the creep strain and the time, stress, or temperature, the range of the testing or working temperature must be considered. The range of temperature for a specific metal or alloy is defined with respect to its melting point. Referring to the room temperature may not be appropriate, as it is too low to affect the creep properties of iron, while it is high enough temperature for lead. For this reason the temperature ratio T /Tm is introduced for the materials under the test where T is the material temperature and Tm is the melting temperature of the material, both measured in ◦ C. Under this assumption it is possible to obtain the same results for two different materials for a constant T /Tm . In Table 10.2-1 Table 10.2-1: Creep-time relaxation for metals and alloys after Garofalo [23]. T /Tm 0.05–0.3

0.2–0.7

Material Al, Ag, Au, Cd, Cu, Mg, Ti-Steel, Al – 10% Cu, and Cu – 3% Ag

Creep time relation = α ln t + c

Al, Ag, Brass, Cd, Cu, Iron and steel, Mg Mg – 2% Al, Ni, Pb, Pt, Sn, and Zn

= 0 + βtm and = 0 + βtm + kt

0.4–0.6

Ferritic and stainless steel

= 0 + t (1 − e−rt ) + ˙s t

0.96–0.99

Al, Au, Cu, and δ-Fe

= 0 + ˙s t

2. Creep of Metals

505

different creep-time relations for different T /Tm ranges are proposed for some pure metals and alloys [23]. When the temperature ratio T /Tm varies between 0.05 and 0.3, the logarithmic law for the time applies = α ln t + c

(10.2-3)

where α and c are constants. The experiments dealing with the logarithmic creep law have been conducted for short duration. Under these conditions the creep rate decreases as the time increases, as indicated by Eq. (10.2-3). Thus, Eq. (10.2-3) describes the primary creep. Equation (10.2-3), however, predicts an infinitely high initial creep rate at t = 0, which is physically not correct. The experimental results show a finite initial creep strain for some materials, even at low temperatures [23]. To include the initial creep strain in Eq. (10.2-3), we may consider c = α ln ν, where ν is some constant. Thus, Eq. (10.2-3) becomes = α ln νt This equation may be written in the form = α ln (1 + νt)

(10.2-4)

to predict an initial creep rate equal to ˙0 = αν. When the value of T /Tm is between 0.2 to 0.7, two relations are given for the time function in Table 10.2-1. The first relation describes the primary creep, and it is (10.2-5) = 0 + βtm where 0 is the initial strain due to loading, and β and m are constants which are independent of time. This relation well describes the creep in simple tension or shear. A simple addition of a linear term in t, such as kt, will also represent the secondary creep, thus (10.2-6) = 0 + βtm + kt describes the primary and secondary creep where 0 is the strain at t = 0, and β, m, and k are constants independent of time. For T /Tm between 0.4 and 0.6, the exponential law of creep is expressed by = 0 + t (1 − e−rt ) + ˙s t

(10.2-7)

where, again, 0 denotes the initial strain at t = 0, t is the limiting transient creep strain, r is the ratio of transient creep rate to the transient creep strain, and ˙s is the secondary creep rate [23]. When T /Tm is between 0.96 to 0.99, that is, when the material is near the melting point, the primary creep essentially disappears and the secondary creep is expressed by

506

Chapter 10. Creep Analysis = 0 + ˙s t

(10.2-8)

with the same definition for the terms as in Eq. (10.2-7). A general consequence of the above discussion is that the behavior of material in creep depends strongly upon its temperature. When the temperature is low relative to the material’s melting point, 0.05 < T /Tm < 0.3, the primary creep dominates and the logarithmic law, Eq. (10.2-4), applies. At moderate temperatures, 0.2 < T /Tm < 0.7, both the primary and secondary creeps are significant and must be accounted for in the design of materials. At relatively high temperatures, 0.96 < T /Tm < 0.99, the secondary creep is dominant and the primary creep is negligible. Stress function for primary and secondary creep The stress dependence on creep has been investigated by many researchers mainly for the secondary stage of the creep. It is assumed that the dependence of the primary creep on the applied stress is similar to that of the secondary creep’s dependence on the stress. For both metals and alloys in the annealed condition, the stress dependence on secondary creep rate, ˙s , at low stress level and constant temperature is given by the power relation (10.2-9) ˙s = Bσ n where B and n are stress independent material constants. For annealed metals and alloys n varies between 1 and 7 and does not seem to depend on crystalline structure [24–32]. The material constants B and to some extent n are temperature dependent. In most cases of creep at high-temperature n varies between 4 and 6 for pure metals and between 2 and 4 for alloys. There are a few exceptions, however, to these ranges. For example, copper exhibits low n values, and for aluminum at 920 K, n = 1.24. For γ-Fe, n = 1 at 1360◦ C (cf., Price et al. [33]). On the other hand, some iron-aluminum alloys exhibit high n values ranging between 4.6 and 6.8 (cf., Lawley et al. [28]). It is also found that n depends on the percentage of alloy’s elements (cf., Sellars and Quarrell [34]). At high stress levels the stress dependence on the secondary creep rate is given by (10.2-10) ˙s = B exp (βσ) where B and β are material constants independent of stress for the ranges of stress examined. This relation fits particularly well the crystals and polycrystals of annealed metals and alloys. The constants B and β are both temperature dependent as is reported by Kauzman [35], Cottrell and Aytekin [36], and Feltham [37,38,40], and Feltham and Mikean [39]. Equations (10.2-9) and (10.2-10) led the investigators to believe for some time that different mechanisms control the secondary creep at low and high

2. Creep of Metals

507

stress levels. It is very unlikely, however, that an abrupt change in mechanism occurs, and in fact it has been shown by Garofalo [23] that Eqs. (10.2-9) and (10.2-10) are both satisfied by a single stress function. This function is given by

˙ = B (sinh ασ)n

(10.2-11)

where B and α are constants at constant temperature. For values of α σ < 0.8, Eq. (10.2-11), expanding the function (sinh ασ)n into Taylor series and keeping only the first term, reduces to Eq. (10.2-9) and B αn = B. For values of ασ > 1.2, Eq. (10.2-11) reduces to Eq. (10.2-10), where B /2n = B and nα = β. Equation (10.2-11) fits the data for both metals and alloys, but to evaluate α, results for both high and low stresses must be available. Values of B , α, and n computed for various metals and alloys are given in Table 10.2-2. The parameters given in Table 10.2-2 are determined for each metal or alloy for a constant grain size. As shown in Table 10.2-2, n approaches unity for aluminum tested at 920 K, which is near its melting point. For this case, Eq. (10.2-11) shows the secondary creep rate to be proportional to sinh ασ. Such a relation has been proposed as a general stress-creep law by Kauzman [35] and Nádai and McVetty [41]. Experimental evidence shows that the hyperbolic-sine power function applies over a much wider range of stress and temperature and the hyperbolic-sine

Table 10.2-2: Computed values of B , α, and n.

Material

Test Temperature (◦ K)

B (sec−1 )

α (cm2 /kg)

n

Copper Copper Copper Copper Copper Copper Aluminum Aluminum Aluminum Aluminum3.1% Magnesium Austeniticstainless steel

673 723 773 823 903 973 477 533 920 531

6.80 × 10−6 7.80 × 10−6 6.10 × 10−6 7.20 × 10−6 4.70 × 10−6 8.30 × 10−6 2.78 × 10−6 1.94 × 10−5 2.67 × 10−8 4.17 × 10−6

1.57 × 10−3 2.00 × 10−3 2.79 × 10−3 3.26 × 10−3 5.71 × 10−3 9.35 × 10−3 5.65 × 10−3 7.28 × 10−3 1.78 × 10−3 2.96 × 10−3

3.57 3.36 3.39 3.38 2.66 2.26 5.00 4.55 1.24 2.26

977

1.47 × 10−8

1.11 × 10−3

3.64

1089

1.67 × 10−8

1.51 × 10−3

3.50

508


relation is a special case which may apply at temperatures near the melting point. It should be mentioned, however, that the mechanisms controlling creep near the melting point may differ in some cases from those controlling at lower temperatures. Also the creep behavior at these very high temperatures may be affected by oxidation.

Temperature function for creep Experimental evidence indicates that the creep of metals is a thermally activated process. In this process the creep rate dependence on temperature is defined through the factor exp (−ΔH/RT ), where ΔH is the activation energy for the process, R is the universal gas constant, and T is the absolute temperature. At low stress levels, the equation describing the stress-creep rate relation is given by Dorn [21] ˙ = S exp (−ΔH/RT )σ n

(10.2-12)

where S is a constant. For pure metals n and ΔH are found, according to Sherby and Dorn [42], to be independent of temperature. This is also true for annealed metals and alloys in the secondary creep range as reported by Weertman and Breen [43], Lawley et al. [28], and Garofalo et al. [44]. Exceptions from this rule, however, are aluminum and copper, for which n has been found to decrease as temperature increases. At high stress levels the stress-creep rate relation is given by Dorn [21] as

˙ = S exp (−ΔH/RT ) exp (βσ)

(10.2-13)

where S is a constant. For pure metals and for a limited temperature range β seems to be independent of temperature (Sherby et al. [45]). For annealed metals and alloys β has been reported by Feltham [46] and Garofalo [23] to increase in general with temperature. Creep-rupture, tertiary stage of creep The tertiary stage of creep, which follows after the secondary stage, results in rupture, and the material fails under the action of the applied stress at point D, where t3 = tr , see Fig. 10.2-1. There are many factors controlling the failure of the material due to the rupture. A number of these factors are the inter-crystalline fracture, re-crystallization, and the increasing stresses during necking. To obtain a relationship between the rupture time and the applied stress and temperature, we note that from Fig. 10.2-1 tr =

4 − 1 ˙s

(10.2-14)

2. Creep of Metals

509

To relate the rupture time tr to the total strain at the end of the primary and secondary stages of creep, Garofalo [23] suggested the following relations for the austenitic stainless steel tr = A(t2 − t1 )α

(10.2-15)

tr = B(t2 )β

(10.2-16)

and where he found parameters A and B to be temperature dependent, and α and β to be temperature independent. The latter are found to be very close to unity. The values of A and B for a type of stainless steel are given in Table 10.2-3. From Fig. 10.2-1, we find the following relations ˙s =

3 − 2 t2 − t1

(10.2-17)

or

3 − 1 (10.2-18) t2 Substituting Eqs. (10.2-17) and (10.2-18) into Eq. (10.2-14), and using Eqs. (10.2-15) and (10.2-16), gives ˙s =

tr =

A(3 − 2 ) B(3 − 1 ) = ˙s ˙s

(10.2-19)

For many materials (3 − 2 ) or (3 − 1 ) are temperature and stress dependent, and thus it is essential to use step-wise and incremental methods to calculate the rupture time. For some materials like copper, the rupture time is relatively short, and thus, it is approximately assumed that the values of (3 − 2 ) or (3 − 1 ) are independent of the applied stress and temperature and, therefore, Eq. (10.2-19) is simplified by assuming a constant value for the numerators of Eq. (10.2-19). That is, for low stresses, Eq. (10.2-12) suggested by Dorn [21] may be substituted in Eq. (10.2-19) to give tr = A0 exp (

ΔH −n )σ RT

(10.2-20)

Table 10.2-3: Parameters A and B for a type 316 austenitic stainless steel [22]. Test temperature A B ◦ C 593 1.81 1.48 704 2.24 2.09 816 2.53 2.30

510


where A0 = A(3 − 2 )/S = B(3 − 1 )/S . For high stresses, Eq. (10.2-13) may be used for tr ΔH tr = B0 exp ( ) exp (−βσ) (10.2-21) RT where B0 and β are temperature dependent.

3

Constitutive Equation of Uniaxial Creep

The relationship between the state of stress and the state of strain and their various time derivatives is known as the constitutive equation. In Section 2, the relationship between creep strain and time, stress, and temperature for the three stages of primary, secondary, and tertiary creep were discussed. Based on Eq. (10.2-2) and a proper choice of time, stress, and temperature functions, a constitutive equation of creep can be written for any stage of creep. Some researchers, however, have proposed a single equation for two or all three stages of creep. For example, Graham and Walles [47] proposed the following series equation for all three stages of creep c =

n

Ci σ αi tβi

(10.3-1)

i=1

which specifically is suggested to have the form c = C1 σ α1 t1/3 + C2 σ α2 t + (C3 σ α3 + C4 σ α4 )t3

(10.3-2)

and they claimed that any creep data can be accurately described with it. The three terms in Eq. (10.3-2) describe the primary, secondary, and tertiary creep, respectively. The effect of temperature is proposed to be included by a time-temperature parameter φ, φ = t(T − T )−γ

(10.3-3)

where γ and T are constants. The parameter φ may be used in Eq. (10.3-2) in place of t. The stress dependence of the creep strain for each stage is assumed to be of the form t

(c )i = βi

0

Ci (t − τ )βi −1 [σ(τ )]αi dτ

(10.3-4)

McVetty [48] has proposed a single equation for both the primary and the secondary creep as (10.3-5) c = Aσ α (1 − e−qt ) + Bσ β t where α, β, and q are constants and A and B are constants depending on temperature.

3. Constitutive Equation of Uniaxial Creep

511

Marin and Pao [49] suggested the assumption α = β in Eq. (10.3-5), which for the rate of creep reduces to dc = (Aqe−qt + B)σ α dt

(10.3-6)

Integrating Eq. (10.3-6) with respect to time yields t

c =

(Aqeq(t−τ ) + B)[σ(τ )]α dτ

(10.3-7)

0

Equations (10.3-4) and (10.3-7) are both of the general hereditary type t

c =

0

κ(t − τ )f (σ)dτ

(10.3-8)

where κ(t) is the proper time function of the constitutive equation of creep. Rabotnov [16] proposed the usual equation for the viscoelasticity for the strain-stress relation in general hereditary form t

ψ() = σ(t) + 0

κ(t − τ )σ(τ )dτ

(10.3-9)

Introducing the operation K ∗ (....) =

t 0

Eq. (10.3-9) becomes

κ(t − τ )(....)dτ

ψ() = (1 + K ∗ )σ(t)

(10.3-10)

(10.3-11)

For constant stress condition Eq. (10.3-9) reduces to an equation for a family of creep curves ψ() = σ(t)[1 + G(t)] (10.3-12) where G(t) = form

t 0

κ(t − τ )dτ . The time function G(t) is proposed to be of the G(t) = atβ

(10.3-13)

where a and β are some constants. It is proved that the proposed power law function for G(t) gives a satisfactory result. This suggests the following form for κ(t − τ ) κ(t − τ ) = aβ(t − τ )β−1 (10.3-14) The above equations are based on the reversible processes. Since the creep deformations are irreversible, the application of Eq. (10.3-11), which was proposed for reversible deformations, is unsuitable. Assume an irreversible creep deformation which reached to the point φ on the instantaneous plasticity curve, Fig. 10.3-1. Unloading at this point

512


Figure 10.3-1: Loading and unloading. follows a straight line according to elasticity rules with a slope equal to the initial slope. The equation of the unloading curve is φ = E( − ) + φ

(10.3-15)

Substituting this equation for φ into Eq. (10.3-11) gives E( − ) + φ = (1 + K ∗ )σ

(10.3-16)

This is the equation for the creep law in unloading. Here E is the modulus of elasticity of the material.

4

Creep Relaxation, Linear Rheological Models

Based on Newton’s law of viscosity, the stress and the rate of strain of a viscous material are related [16] by σ (10.4-1) ˙ = η where η is the coefficient of viscosity of the material. Hooke’s law expresses the stress-strain relations of an elastic body as =

σ E

(10.4-2)

Now, depending upon the initial and boundary conditions, there are two ways to define the state of stress or deformation in a viscoelastic material: (a)- Deformation controlled stress: If the material is subjected to a fixed initial deformation, Maxwell equation can be employed to determine the stress at any time t as

4. Creep Relaxation, Linear Rheological Models ˙ =

σ˙ σ + E η

513 (10.4-3)

For constant total strain condition, ˙ = 0. Integrating Eq. (10.4-3) in conjunction with the initial condition σ = σ0 at t = 0 and the boundary condition = const for t ≥ 0 yields t σ = σ0 exp (− ) (10.4-4) τ where τ = η/E. Equation (10.4-4) represents the stress relaxation for a Maxwell body. (b)- Load controlled stress: When the total load, and the resulting stress, is constant in time, Voigt viscoelastic model is obtained. The sum of Eqs. (10.4-1) and (10.4-2) for the stress becomes σ = E + η ˙

(10.4-5)

According to Voigt’s equation, when a body is under constant load, its deformation approaches the value σ0 /E. Thus, integrating Eq. (10.4-5), the strain as a function of the time is obtained as =

t σ0 [1 − exp (− )] E τ

(10.4-6)

Voigt viscoelastic model, as described by Eq. (10.4-6), does not relax under constant deformation condition. According to this model, stress remains constant as time advances. The parameter τ is called the delay time. Maxwell and Voigt viscoelastic models may be represented by systems of springs and dashpots, called the rheological models. Consider a spring of stiffness E and a dashpot of viscosity η. If a force corresponding to σ is applied and the resulting strain is , then Fig. 10.4-1a represents Maxwell body expressed by Eq. (10.4-4), and Fig. 10.4-1b represents Voigt body expressed by Eq. (10.4-6). An actual viscoelastic material will behave exactly neither as described by Maxwell model nor as described by Voigt model. It behaves as the combination of both, which then is called a general viscoelastic model. To describe the time dependent mechanical behavior of a material, it is possible to find a proper rheological model. For a given material, this model is different for different types of loading. For example, consider a steel bar fixed at point B, acted upon by a tensile force F which is kept constant, as shown in Fig. 10.4-2. The stress at any cross section is F/A, where A is the cross sectional area of the bar. Neglecting the decrease in A during the creep process, the acting stress is constant during loading while the creep strain increases permanently until rupture. Therefore, this mechanism is equivalent to Voigt rheological model. On the other hand, assume the same bar under constant total deformation. The ends A and B are fixed and do not move

514


Figure 10.4-1: Maxwell (a) and Voigt (b) rheological models.

Figure 10.4-2: A bar under constant load and constant deformation. in the direction of the bar axis. If the temperature of the bar becomes high enough to activate creep, the initial elastic stress in the bar relaxes toward lower state and as time increases, the initial elastic deformation decreases and creep deformation increases. This mechanism is equivalent to Maxwell viscoelastic body.

5

Three-Dimensional Governing Equations

The governing equations of the theory of creep and the method of treatment of the problems of creep are similar to the theory of plasticity where the same equations of equilibrium and compatibility as in the theory of elasticity are used in conjunction with the proper constitutive law and the stress-strain relations. Since the equilibrium and compatibility equations were discussed in Chapter 1, and the constitutive law was described in the previous section, the detailed discussion will now be devoted to the derivations of the stress-strain relations.

5. Three-Dimensional Governing Equations

515

Historically, the first approach to the stress-strain relations in elastic and plastic region was suggested by Saint-Venant in 1870 [50]. He proposed that the principal axes of stress and strain in a body under the action of external forces coincide. Later, in 1871, Lévy [51] proposed that the increments of total strain in a three-dimensionally loaded body are proportional to the deviatoric stresses. These relations were suggested in 1913 by von Mises [52] who worked independently and thus the relations are known as Lévy-Mises equations. They are (10.5-1) dij = Sij dλ where dij is the increment of total (elastic, plastic, and creep) strain tensor, Sij is the stress deviatoric tensor defined by Eq. (1.5-14), and dλ is a nonnegative quantity which may vary through the loading path and thus depends on the history of loading. For creep problems it is assumed that the material is incompressible during the creep process. Mathematically, this means cxx + cyy + czz = 0

(10.5-2)

where cxx , cyy , and czz are the creep strains in any orthogonal directions within the body. It is further assumed that the same relation exists between the increments of creep strains and the deviatoric stresses, that is, the principal directions of the stress tensor and the increments of the creep strain tensor coincide (10.5-3) dcij = Sij dλ Expanding Eq. (10.5-3) in terms of the actual stresses yields 2 1 dcxx = dλ[σxx − (σyy + σzz )] 3 2 2 1 c dyy = dλ[σyy − (σxx + σzz )] 3 2 2 1 c dzz = dλ[σzz − (σxx + σyy )] 3 2 dcxy = dλσxy dcyz = dλσyz dcxz = dλσxz

(10.5-4)

These creep strain-stress relations are not completely defined because the value of dλ is not known. In order to find dλ the definition of the effective stress and creep strain is used. Namely, it is easily verified that from Eqs. (10.5-4) the following expression is obtained (dcxx − dcyy )2 + (dcyy − dczz )2 + (dczz − dcxx )2 + 6[(dcxy )2 + (dcyz )2 + (dczx )2 ] 2 2 2 = (dλ)2 [(σxx − σyy )2 + (σyy − σzz )2 + (σzz − σxx )2 + 6(σxy + σyz + σzx )]

516


Recalling the definition for the octahedral stress and strain, the right-hand side of the above equation is proportional to the square of the octahedral shear stress, and the left-hand side is proportional to the square of the increment of the octahedral creep shear strain defined by 1 2 2 2 + σyz + σzx )] [(σxx − σyy )2 + (σyy − σzz )2 + (σzz − σxx )2 + 6(σxy 9 1 c 2 (dγoct ) = [(dcxx − dcyy )2 + (dcyy − dczz )2 + (dczz − dcxx )2 9 c 2 +6((dxy ) + (dcyz )2 + (dczx )2 )] (10.5-5)

(τoct )2 =

Thus the constant dλ in terms of the octahedral creep strain and stress becomes dλ =

c dγoct τoct

(10.5-6)

Substituting this expression in Eq. (10.5-3) yields dcij =

c dγoct Sij τoct

(10.5-7)

Equation (10.5-7) describes the relation between the increments of creep strains c c and the stresses for known relationship between dγoct and τoct . The relationship c c between dγoct and τoct is obtained through the experiment, where instead of the octahedral shear strain and stress the effective strain and stress are defined. The effective stress and the effective creep strain for the multi-axial stress and strain conditions, using Huber-von Mises-Hencky yield criterion, are defined, respectively, as 1 2 2 2 +σyz +σzx )]1/2 (10.5-8) σ ∗ = √ [(σxx −σyy )2 +(σyy −σzz )2 +(σzz −σxx )2 +6(σxy 2 √

2 [(dcxx − dcyy )2 + (dcyy − dczz )2 + (dczz − dcxx )2 3 +6((dcxy )2 + (dcyz )2 + (dczx )2 ) (10.5-9)

c∗

d =

For Tresca criterion, f (σij ) = σ1 − σ3 with σ1 > σ2 > σ3 , the effective strain is dc∗ = dc1

(10.5-10)

where the subscripts 1, 2, and 3 indicate the principal directions and σ1 , σ2 , and σ3 are the principal stresses. It is seen that when Huber-von Mises-Hencky yield criterion is used, the octahedral strain and stress are related to the effective strain and stress through a constant coefficient 3 σ ∗ = √ τoct 2 √ c∗ c d = 2dγoct

(10.5-11)

5. Three-Dimensional Governing Equations

517

It is noticed that for a uniaxial stress condition, such as in x-direction, the effective stress and increment of effective creep strain reduce to σ ∗ = σxx dc∗ = dcxx Now, the constant dλ in terms of the effective stress and the increment of effective creep strain becomes dλ =

3 dc∗ 2 σ∗

(10.5-12)

and the stress-creep strain relations become dcxx = dcyy = dczz = dcxy = dcyz = dcxz =

dc∗ 1 [σxx − (σyy + σzz )] ∗ σ 2 dc∗ 1 [σyy − (σxx + σzz )] σ∗ 2 dc∗ 1 [σzz − (σxx + σyy )] σ∗ 2 3 dc∗ σxy 2 σ∗ c∗ 3 d σyz 2 σ∗ 3 dc∗ σxz 2 σ∗

(10.5-13)

Equations (10.5-13) are the stress-strain relation in creep regime which are based on von Mises creep potential function and thus, when using the effective creep strain, one must be careful to use Eq. (10.5-9) for the calculations of the effective creep strain. In order to fully define the creep strain-stress relations, the relationship between the effective stress and the increment of effective creep strain must be known. This relationship is obtained from the experimental creep test. In Section 2, the creep behavior of materials at elevated temperatures for each stage of the primary, secondary, and tertiary creep were expressed by proper mathematical expressions. These experimentally based expressions were proposed for the uniaxial test conditions. Generalizing the uniaxial test result to the multi-axial stress condition, the uniaxial stress and creep strain relations are replaced by the effective stress and the increment of effective creep strain. The relation between the effective creep strain and the effective stress is considered as the constitutive law. Selecting a constitutive law, which best describes the creep behavior of the given material for each stage of the creep, and substituting it into Eqs. (10.5-13), fully determines the creep strain-stress relations.

518


Taking the general form of the constitutive law for creep as proposed by Eq. (10.2-2) and writing it in differential form gives dc = df1 (t) · f2 (σ) · f3 (T )

(10.5-14)

This relationship, which is experimentally established for a material, may be extended to the differential of effective creep strain and stress in the multi-state of stress condition. For three-dimensional stress-strain condition, Eq. (10.5-14) becomes (10.5-15) dc∗ = df1 (t) · f2 (σ ∗ ) · f3 (T ) which upon substitution into Eqs. (10.5-13) yields dcxx = dcyy = dczz = dcxy = dcyz = dcxz =

6

df1 (t) · f2 (σ ∗ ) · f3 (T ) 1 [σxx − (σyy + σzz )] σe 2 df1 (t) · f2 (σ ∗ ) · f3 (T ) 1 [σyy − (σxx + σzz )] σe 2 df1 (t) · f2 (σ ∗ ) · f3 (T ) 1 [σzz − (σxx + σyy )] σe 2 ∗ 3 df1 (t) · f2 (σ ) · f3 (T ) σxy 2 σe 3 df1 (t) · f2 (σ ∗ ) · f3 (T ) σyz 2 σe 3 df1 (t) · f2 (σ ∗ ) · f3 (T ) σxz 2 σe

(10.5-16)

Creep Potential, General Theory of Creep

In Section 5, stress-creep strain relations were obtained based on von Mises creep potential. According to von Mises assumption, the maximum shear stress and the maximum slide velocity are co-directional. A general theory, however, may be developed to define the creep strain-stress relations based on a creep potential function for a given material. Let us consider a system of rectangular coordinates, each axis representing a principal stress direction at a given point within a body under the action of different loads, as shown in Fig. 10.6-1. Every point in this system of coordinates represents the state of stress in the body. We plot the line ON in such a way that it makes equal angles with the three principal axes. Therefore, every point lying on this line has the coordinate σ1 = σ2 = σ3 = p, thus, representing a hydrostatic state of stress. The equation of any plane perpendicular to the line ON is √ σ1 + σ2 + σ3 = 3¯ r

6. Creep Potential, General Theory of Creep

519

s3

Q(s11,s22,s33)

␲-plane Q⬘

S

N

Q␲ p o

s2

s1

Figure 10.6-1: Stress space and yield surface. where r¯ is the distance of the plane from the origin along the line ON . The plane passing through the origin is called the π-plane and its equation is σ1 + σ2 + σ3 = 0. −→ The stress vector at any arbitrary point Q, OQ, representing the state of stress σ1 , σ2 , and σ3 can be decomposed into two components along the line ON and perpendicular to ON with the latter being parallel to the π-plane. The component along the line ON is √ 1 p = √ (σ1 + σ2 + σ3 ) = 3σm 3

(10.6-1)

where σm = 13 (σ1 + σ2 + σ3 ) is the mean stress. The component normal to ON is S 2 = OQ2 − p2 2 = σ12 + σ22 + σ32 − 3σm = (σ1 − σm )2 + (σ2 − σm )2 + (σ3 − σm )2 = S12 + S22 + S32 = 2J2

(10.6-2)

−→ From Eqs. (10.6-1) and (10.6-2) it follows that any vector OQ in the stress space can be decomposed into two components, one proportional to the hydrostatic stress and the other proportional to the deviatoric stresses. Equation (10.6-2) has three components on the stress axes equal to S1 , S2 , represents that S and S3 , which are the principal deviatoric stresses. The stress space, as shown in Fig. 10.6-1, is called Haigh-Westergaard stress space. The concept of stress space is useful to describe the creep potential function. It is evident from Fig. 10.6-1 that any other point Q , where the line QQ is parallel to ON ,

520


has the same deviatoric components as the point Q, but different hydrostatic component. The projection of the point Q on the π-plane, Qπ , has the same deviatoric stress components as the point Q, but its hydrostatic component is zero. Therefore, any point lying in the π-plane represents a state of stress in which its hydrostatic stress is zero. In the derivation of the creep-stress relations (10.5-16), which were based on Huber-von Mises-Hencky criterion, the assumption was made that the material is incompressible. This would physically mean that the hydrostatic stresses do not affect the creep process. In Haigh-Westergaard stress space this means that the creep potential function, which will be discussed shortly, makes a threedimensional surface, such that it never intersects the line ON . Indeed, later in this section it will be shown that Huber-von Mises-Hencky criterion is based on the energy dissipation of the shear stresses in the body. This criterion in Haigh-Westergaard stress space is expressed by a circular cylindrical surface revolving around the axis ON , and never intersects the ON axis. Now, let us consider a general condition where the creep potential function is given and the associated flow rule, the creep-stress relations, are desired. We define energy dissipation in creep as L = σij ij = L(σij )

(10.6-3)

where σij and ij are the stress and the creep strain tensors. The equation L(σij ) = const = k describes a surface of constant energy dissipation in HaighWestergaard stress space. A plot of the intersections of several of these surfaces for different values of energy dissipation with the π-plane is shown in Fig. 10.6-2. The contour of these lines in the π-plane cannot intersect each other, for otherwise the energy dissipation would not be uniquely expressed in terms of stress. Let us consider each of the contours shown in Fig. 10.6-2 to be s3

P2,s ij(2)

k2 k k1 P,s ij

o P1,s ij(1)

s2

s1

Figure 10.6-2: Surface of constant dissipation energy.

6. Creep Potential, General Theory of Creep

521

associated with a constant dissipation energy k1 , k, and k2 . Now suppose that (1) we are given a state of stress located on the surface L = k1 , σij = σij . At this point we assume that either an external agency applies an additional load or due to the stress redistribution in creep, the state of stress varies and reaches (2) to the surface L = k2 where the state of stresses is σij . This loading path intersects the surface L = k at a point P with the state of stress σij . During the loading, dL ≥ 0. This means that for each point P of the path P1 P (1)

(σij − σij )ij ≥ 0

(10.6-4)

If the points P1 and P lie on a single surface L = k, then the path P1 P lies entirely on this surface. The deformation rate vector, from Eq. (10.6-4), is then in the direction of the normal to the surface L(σij ) = k. Thus ij = h(σij )

∂L ∂σij

(10.6-5)

where h(σ)ij ) is a coefficient which is a function of the stress tensor. Equation (10.6-4) is a rephrasing of Drucker’s postulate for strain hardening definition [16] to the case of steady-state creep. According to this postulate, a stressed state σij∗ in the continuum is considered. An external agency slowly applies a load to the continuum to reach the state of stress σij . The load is then removed to bring the state of stress to σij∗ . The net work done on this cycle is either zero or positive. A function Υ(σij ) may be defined as h(σij )

∂L ∂Υ = ∂σij ∂σij

(10.6-6)

The function Υ is called the creep rate potential. The expressions (10.6-5) in terms of the creep rate potential is ij =

∂Υ ∂σij

(10.6-7)

The condition (10.6-7) is satisfied if h(σij ) = h(L). In this case L

Υ=

h(L)dL 0

Multiplying Eq. (10.6-5) by σij gives L = h(σij )

∂L σij ∂σij

(10.6-8)

Once the function L is known, the form of the function h(σij ) is fully determined from Eq. (10.6-8). Determining h from Eq. (10.6-8) and substituting

522


into Eq. (10.6-5) yields ij =

∂L L ∂L ∂σij σmn ∂σmn

(10.6-9)

Equation (10.6-9) suggests that once the function L is known, the strain tensor is derived and is proportional to the partial derivative of L with respect to the stress tensor. The converse formula exists in terms of a function U (ij ) which may be used to determine the stress tensor by the relation σij =

∂U ∂ij

(10.6-10)

It is easily verified that the function U is related to the stress and creep strain tensors as U = σij ij − Υ = L − Υ (10.6-11) The function U depends on ij . Differentiating both sides of the Eq. (10.6-11) with respect to ij gives ∂Υ ∂σkl ∂σkl ∂U = σij + kl − ∂ij ∂ij ∂σkl ∂ij

(10.6-12)

Using Eq. (10.6-7), the last two terms on the right-hand side of Eq. (10.6-12) cancel each other and we obtain Eq. (10.6-10). The function L(σij ) in the theory of creep is similar to the flow function in the theory of plasticity. The condition (10.6-4) states that the surface L = const is convex in the stress space. Similarly, the surface Υ(σij ) = const is convex in the stress space, and U (ij ) = const is convex in the rate of deformation space.

7

Stress Function for Creep Problems

Consider a two-dimensional plane strain creep problem. The governing equations are two equilibrium equations, one compatibility condition, and the creep strain-stress relations. Using the concept of stress function, the governing equations are reduced in terms of the stress function [1]. The equilibrium equations and the time-rate version of compatibility condition for two-dimensional problems are ∂σxx ∂σxy + =0 ∂x ∂y ∂σxy ∂σyy + =0 ∂x ∂y

(10.7-1)

7. Stress Function for Creep Problems ∂ 2 ˙xx ∂ 2 ˙yy ∂ 2 ˙xy + = 2 ∂y 2 ∂x2 ∂x∂y

523 (10.7-2)

The creep strain-stress relations in terms of the effective stress σ ∗ and the time-rate of the effective strain ˙∗ , see Eqs. (10.5-13), are ˙∗ 1 [σxx − (σyy + σzz )] ∗ σ 2 ˙∗ 1 ˙yy = ∗ [σyy − (σxx + σzz )] σ 2 ˙∗ 1 ˙zz = ∗ [σzz − (σxx + σyy )] σ 2 3 ˙∗ ˙xy = σxy 2 σ∗

˙xx =

(10.7-3)

For the plane strain condition ˙zz = 0, and thus σzz = 12 (σxx + σyy ) which upon substitution in the first two equations yields 3˙∗ (σxx − σyy ) 4σ ∗ 3˙∗ = −˙xx = ∗ (σyy − σxx ) 4σ 3˙∗ = ∗ σxy 2σ

˙xx = ˙yy ˙xy

(10.7-4)

Substituting Eqs. (10.7-4) into Eq. (10.7-2) gives the compatibility condition in the form ˙∗ ∂2 ∂2 3˙∗ ∂2 3 ( 2 − 2 ) [ ∗ (σxx − σyy ) ] = σxy ) ( ∂y ∂x 4σ ∂x∂y σ ∗

(10.7-5)

To write this condition in terms of stresses, the relationship between the effective creep strain and the effective stress must be considered. According to the definition of effective stress and strain, see Eqs. (10.5-8) and (10.5-9), for the plane strain condition 1 2 ]}1/2 σ ∗ = {3[ (σxx − σyy )2 + σxy 4 4 ˙∗ = [ (4˙2xx + ˙2xy )]1/2 3

(10.7-6)

Using the power law function for the constitutive equation of creep given by Eq. (10.2-9) ˙ = Bσ n the equation may be written to relate the rate of effective creep strain and the effective stress as [16] (10.7-7) ˙∗ = B(σ ∗ )n

524


It gives n−1 ˙∗ B(σ ∗ )n 1 2 = = B(σ ∗ )(n−1) = B{3 [ (σxx − σyy )2 + σxy ]} 2 ∗ ∗ σ σ 4

(10.7-8)

Substituting Eq. (10.7-8) in the compatibility condition (10.7-5) gives n−1 1 ∂2 ∂2 1 2 ( 2 − 2 ) [(σxx − σyy ){3 [ (σxx − σyy )2 + σxy ]} 2 ] 4 ∂y ∂x 4 2 n−1 ∂ 1 2 = ]} 2 ] (10.7-9) [σxy {3 [ (σxx − σyy )2 + σxy ∂x∂y 4

Introducing Airy stress function Φ defined by ∂2Φ ∂y 2 ∂2Φ σyy = ∂x2 ∂2Φ τxy = − ∂x∂y σxx =

(10.7-10)

the two equilibrium equations (10.7-1) are automatically satisfied, and the compatibility equation (10.7-8) in terms of the stress function becomes 1 ∂2 ∂2 ∂2Φ ∂2Φ ∂ 2Φ ∂ 2Φ 2 ∂ 2 Φ 2 n−1 ) [( − ) + 4( ( 2 − 2 ) {( 2 − )] 2 } 4 ∂y ∂x ∂y ∂x2 ∂y 2 ∂x2 ∂x∂y ∂2 ∂ 2 Φ 2 n−1 ∂ 2Φ ∂ 2Φ ∂ 2Φ 2 =− ) + 4( (10.7-11) { [( 2 − )] 2 } ∂x∂y ∂x∂y ∂y ∂x2 ∂x∂y In the derived Eq. (10.7-11), the equilibrium equations and the creep strainstress relations are satisfied. This equation with the given boundary conditions must be solved to obtain the creep stresses in the structure. The stress function and its normal gradient to the boundary should vanish on the problem boundary, as given by Eq. (1.12-44). For n = 1, Eq. (10.7-7) reduces to Hooke’s law and thus Eq. (10.7-11) reduces to the biharmonic equation for elasticity problems ∇4 Φ = 0

(10.7-12)

Analytical solution of Eq. (10.7-11) is, however, rather difficult. The numerical solution of this equation by finite difference is feasible. For special case of n = 3 the exponent (n − 1)/2 becomes unity and the equation becomes simple in form. Even in this simple form the analytical solution is not feasible.

8. Creep Linearization

8

525

Creep Linearization

In this section, a particular technique is used to linearize the nonlinear governing equations of creep. The technique is based on the assumption that Airy stress function and the stresses may be assumed in a series form, where each term of the stress function in the series is related to the corresponding term of the stresses. The stress distribution in the creep regime is, in general, governed by the nonlinear constitutive relation, such as Eq. (10.2-12), which results in a system of nonlinear partial differential equations for the governing equations. In the classical problems of creep the solution of these partial differential equations is obtained for some simple cases of one-dimensional problems where the governing equations are functions of only one space variable. The examples are the creep analysis of beams and radially symmetrically loaded disks, cylinders, and spheres. For more general cases of multi-axial state of creep no general analytical solution exists. For some particular cases, where the variations along one of the space variables are relatively smooth compared to other variables, the linearization of the stress-strain relation for creep is possible. Linearization leads to the linearized governing equations and, thus, an analytical solution is possible [53,54]. To show the technique, a non-axisymmetrically heated thick-walled cylinder is considered and its creep stresses are obtained using linearization of the governing equations. Consider a thick-walled cylinder of inner radius a and outer radius b, subjected to the inside pressure p. The temperature does not vary along its axis, but is in general a function of r and φ, with the assumption that the temperature variation in the φ direction is relatively small. Consider a temperature distribution of the following form θ(r, φ) = θa + θ∗ ln

∞ r +λ θn (r) cos nφ a n=1

(10.8-1)

where θa , θ∗ are constants, and θn (r) is a function of r. The parameter λ is assumed to be small relative to unity, so that its second, third, and higher order powers may be neglected. Using the incompressibility condition in the polar coordinates r, φ, and z, rr + φφ + zz = 0, and considering the plane strain condition zz = 0, the time rate version of the compatibility condition in a non-dimensional form for polar coordinates becomes 2 ∂ 2 ˙rr ∂ ˙rr ∂2 2 ∂ ˙rr − 3ρ = − ρ (r˙rφ ) ∂φ2 ∂ρ ∂ρ2 ∂ρ∂φ

where ρ = r/b.

(10.8-2)

526

Chapter 10. Creep Analysis 1 2

Assuming the plane strain condition, σzz = stress relations from Eqs. (10.5-13) become 3 4 3 ˙φφ = 4 3 ˙rφ = 2 ˙rr =

(σrr + σφφ ), the creep strain-

˙∗ (σrr − σφφ ) σ∗ ˙∗ (σφφ − σrr ) σ∗ ˙∗ σrφ σ∗

(10.8-3)

Here, σ ∗ and ˙∗ are the effective stress and the rate of effective strain, respectively. The stress distribution in the cylinder must satisfy the incompressibility and compatibility conditions, Eqs. (10.8-2) and (10.8-3), along with the equilibrium equations and the boundary conditions. The stresses will satisfy the equilibrium equations if a stress function Φ(r, φ) of the form [55] Φ(r, φ) = Φ0 (r) + λ

∞

Φn (r, φ)

(10.8-4)

n=1

is introduced, so that the following relations exist 0 = b2 σrr

1 ∂Φ0 ρ ∂ρ

n b2 σrr =

1 ∂ 2 Φn 1 ∂Φn + ρ2 ∂φ2 ρ ∂ρ

0 = b2 σφφ

∂ 2 Φ0 ∂ρ2 n = b2 σφφ

n b2 σrφ =−

(10.8-5) ∂ 2 Φn ∂ρ2

∂ 1 ∂Φn ( ) ∂ρ ρ ∂φ

(10.8-6)

where 0 σrr = σrr +λ 0 σφφ = σφφ +λ

∞

n σrr

n=1 ∞

n σφφ

n=1

σrφ = λ

∞

n σrφ

(10.8-7)

n=1

The superscript (0) indicates the axisymmetric solution and the superscript (n) indicates the additional stresses due to the non-axisymmetric temperature distribution. The problem is to find Φ such that Eqs. (10.8-2) and (10.8-3) along with the boundary conditions are satisfied. Let us assume that the constitutive law of creep is given as (10.8-8) ˙∗ = B0 ecθ σ ∗m


527

where B0 , c, and m are constants, and a superscript (∗) is referred to the effective stress or strain. Based on this equation, the relations between the rate of strains and stresses can be obtained. The effective stress from Eq. (10.8-5) for the stress components in the polar coordinates reduces to 2 1/2 ] σ ∗ = [(σrr − σφφ )2 + 4σrφ

(10.8-9)

Substituting Eqs. (10.8-9) and (10.8-8) into Eqs. (10.8-3), making use of Eq. (10.8-1), the creep stress-strain relations are defined. These relations may be linearized with a power series expansion in λ, assuming λ to be small relative to unity so that its second and higher order powers may be neglected. Expanding the creep coefficient B0 ecθ into a power series, using Taylor series expansion and substituting from Eq. (10.8-1), and neglecting the higher order terms of λ, yields B0 ecθ = B0 [1 + λc

∞

θn (r) cos nφ] exp [c(θa + θ∗ ln βρ)]

(10.8-10)

n=1

where θa and θ∗ are defined in Eq. (10.8-1) and are the coefficients of the symmetric part of the temperature distribution and β = b/a. Substituting Eqs. (10.8-7) in Eq. (10.8-9), expanding the right-hand side of Eq. (10.8-9) in the binomial series, and neglecting the higher order terms, we obtain σ ∗ = σ0∗ + λ

∞

σn∗

(10.8-11)

n=1 0 0 n n where σ0∗ = σφφ − σrr and σn∗ = σφφ − σrr .

Solution for axisymmetric stress 0 0 The axisymmetric solution for the stresses σrr and σφφ in the cylinder subjected to the axisymmetric temperature distribution θa + θ∗ ln βρ and the internal pressure P with the boundary conditions

σrr = −P 1 ρ= β

σrr = 0 ρ=1

(10.8-12)

is given by Eqs. (6.2-17) and (6.2-24). Carrying out the integrations for the given temperature distribution in Eqs. (6.2-17), the final form for the stresses becomes Ab−2K1 [1 − ρ−2K1 ] 2K1 Ab−2K1 0 σφφ = [(2K1 − 1)ρ−2K1 + 1] 2K1 0 0 − σrr = K1 K2 ρ−2K1 σ0∗ = σφφ 0 = σrr

(10.8-13)

528


where 2K1 P a−2K1 − b−2K1 1 cθ∗ K1 = (1 + ) m 2 2P K2 = 2K1 β −1

A=

β=

b a

(10.8-14)

Solution for non-axisymmetric stresses Substituting Eq. (10.8-13) in Eq. (10.8-11), and using Eqs. (10.8-10), the creep strain-stress relations, Eqs. (10.8-3), are expanded into a power series in λ which, keeping the linear terms, leads to ˙rr = ˙0rr + λ

∞

˙nrr

n=1 ∞

˙φφ = ˙0φφ + λ

˙nφφ

n=1

˙rφ = λ

∞

˙nrφ

(10.8-15)

n=1

We may consider that the function θn (r) in Eq. (10.8-1) has the general form θn (r) = Cn ρn + C−n ρ−n , with Cn and C−n being some constants. With this assumption for θn (r) in Eq. (10.8-10), using Eqs. (10.8-15) and (10.8-3), the strains are obtained to be 1 r 0rr = − B0 exp [c(θa + θ∗ ln )] σ0∗m 4 a nrr = −H1 [Cn bn ρn−2 + C−n b−n ρ−n−2 ] cos nφ 1 − H2 mρ2K1 −2 (σφn − σrn ) 4 n nrφ = H2 ρ2K1 −2 σrφ 1 H1 = cB0 ecθa β cθ∗ (K1 K2 )m ) 4 (10.8-16) H2 = B0 ecθa β cθ∗ (K1 K2 )m−1 where, due to the incompressibility condition, 0φφ = −0rr and nφφ = −nrr . Substituting the additional strains nrr , nφφ , and nrφ in the compatibility condition (10.8-2) and making use of the stress function, the compatibility condition for the additional stress functions, Φn , reduces to 3 2 4 ∂ 4 Φn ∂Φn 3 ∂ Φn 2 ∂ Φn 2 ∂ Φn + b ρ + b ρ − b ρ ρ + b 1 2 2 3 ∂ρ4 ∂ρ3 ∂ρ2 ∂ρ ∂ρ2 ∂φ2 3 2 4 ∂ Φn ∂ Φn ∂ Φn +b4 ρ + b5 + = Tn (ρ) cos nφ (10.8-17) 2 ∂ρ∂φ ∂φ2 ∂φ4

ρ4


529

where b1 = 4K1 − 2

b4 = 8

b2 = 4K12 − 8K1 + 3 b3 =

K1 20 + 6 − 4K1 − m m

b5 = −4K12 + 12K1 − 8 −

8K1 16 + m m

4 −2 m

(10.8-18)

and 8 [(2n − 2n2 )Cn bn+2 ρn+2−2K1 m −(2n + 2n2 )C−n b(−n+2) ρ−n+2−2K1 ] Tn (ρ) = cK1 K2

(10.8-19)

Equation (10.8-17) may be solved by assuming a stress function of the form Φn (ρ, φ) = Rn (ρ) cos nφ

(10.8-20)

Upon substitution of Eq. (10.8-20) in Eq. (10.8-17), an ordinary differential equation of Euler type is obtained as ρ4

3 2 d4 Rn 3 d Rn 2 2 d Rn + b ρ + (b − b n )ρ 1 2 3 dρ4 dρ3 dρ2 dRn −(b2 + n2 b4 )ρ + (n4 − b5 n2 )Rn = Tn (ρ) dρ

(10.8-21)

The general solution of this differential equation for R1 , R2 , ..., Rn is obtained as 1

2

3

4

RnG (ρ) = Cn1 ρSn + Cn2 ρSn + Cn3 ρSn + Cn4 ρSn

(10.8-22)

where Sni , i = 1, 2, 3, 4, are the solutions of the characteristic equation Sn (Sn − 1)(Sn − 2)(Sn − 3) + b1 Sn (Sn − 1)(Sn − 2) +(b2 − b3 n3 )Sn (Sn − 1) −(b2 + n2 b4 )Sn + (n4 − b5 n2 ) = 0

(10.8-23)

and Cn1 through Cn4 are the constants of integration to be found from the given boundary conditions. A particular solution is verified to be RnP (ρ) = Dn1 ρn+2−2K1 + Dn2 ρ−n+2−2K1

(10.8-24)

where Dn1 and Dn2 are obtained by substitution of RnP (ρ) in Eq. (10.8-21) and making use of Eq. (10.8-19). The solution of Eq. (10.8-21) is the sum of the general and particular solutions, Rn = RnG +RnP . The additional stress functions Φn are then obtained from Eq. (10.8-20) as 1

2

3

4

Φn (ρ, φ) = [Cn1 ρSn + Cn2 ρSn + Cn3 ρSn + Cn4 ρSn +Dn1 ρn+2−2K1 + Dn2 ρ−n+2−2K1 ] cos nφ

(10.8-25)

530


The additional stresses may now be calculated from Eqs. (10.8-6) to be cos nφ 1 Sn1 −2 1 {Cn ρ (Sn − n2 ) b2 2 3 +Cn2 ρSn −2 (Sn2 − n2 ) + Cn3 ρSn −2 (Sn3 − n2 )

n σrr =

+Cn4 ρSn −n (Sn4 − n2 ) + Dn1 ρL1 −2 (L1 − n2 ) +Dn2 ρL2 −2 (L2 − n2 )} n sin nφ 1 Sn1 −2 1 n σrφ = {Cn ρ (Sn − 1) b2 2 3 +Cn2 ρSn −2 (Sn2 − 1) + Cn3 ρSn −2 (Sn3 − 1) 4

2

+Cn4 ρSn −2 (Sn4 − 1) + Dn1 ρL1 −2 (L1 − 1) +Dn2 ρL2 −2 (L2 − 1)} cos nφ 1 Sn1 −2 1 1 n σφφ = {Cn ρ Sn (Sn − 1) b2 2 3 +Cn2 ρSn −2 Sn2 (Sn2 − 1) + Cn3 ρSn −2 Sn3 (Sn3 − 1) 4

+Cn4 ρSn −2 Sn4 (Sn4 − 1) + Dn1 ρL1 −2 L1 (L1 − 1) +Dn2 ρL2 −2 L2 (L2 − 1)} 4

(10.8-26)

where L1 = n + 2 − 2K1 and L2 = −n + 2 − 2K1 . Considering the assumed boundary conditions given by Eq. (10.8-12), the nonhomogeneous boundary condition, σrr = −P at the inner surface of the cylinder, is already satisfied by the axisymmetric solution. Therefore, the additional stresses should satisfy the homogeneous boundary conditions n 1 n n 1 n σrr ( , φ) = σrr (1, φ) = σrφ ( , φ) = σrφ (1, φ) = 0 β β

(10.8-27)

Using the conditions (10.8-27) and substituting into Eqs. (10.8-26), a set of four equations is obtained, which may be solved for the constants of integration Cn1 through Cn4 . The result in matrix form is ⎡ S 1 −2 1 α n (Sn − n2 ) ⎢ (Sn1 − n2 ) ⎢ ⎢ Sn1 −2 ⎣ α (Sn1 − 1)

(Sn1 − 2)

⎤

αSn −2 (Sn2 − n2 ) αSn −2 (Sn3 − n2 ) αSn −2 (Sn4 − n2 ) (Sn2 − n2 ) (Sn3 − n2 ) (Sn4 − n2 ) ⎥ ⎥ ⎥ 2 3 4 −2 Sn −2 2 Sn −2 3 Sn 4 α (Sn − 1) α (Sn − 1) α (Sn − 1) ⎦ (Sn2 − 2) (Sn3 − 2) (Sn4 − 2) 2

3

4

⎧ 1⎫ Cn ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ C2 ⎪ ⎬

⎧ 1 L −2 2 ⎫ Dn α 1 (n − L1 ) + Dn2 αL2 −2 (n2 − L2 ) ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎬ Dn1 (n2 − L1 ) + Dn2 (n2 − L2 ) n × ⎪ 3 ⎪ = ⎪ 1 L1 −2 2 L2 −2 C ⎪ ⎪ (1 − L1 ) + Dn α (1 − L2 ) ⎪ Dn α ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ n4 ⎪ ⎭ ⎩ ⎭ 1 2

Cn

Dn (1 − L1 ) + Dn (1 − L2 )

where α = 1/β. This matrix equation can be solved for Cn1 through Cn4 . It is clear that in solving the matrix, n has to vary from 1, 2, 3, .... to, theoretically, ∞.

9. Creep Relaxation of Axisymmetric Stresses

531

Therefore, at first, n is set equal to one and the matrix is solved for C11 through C14 . Then n is set equal to 2, and again the matrix is solved for C21 through C24 . The solution of the matrix for different values of n is repeated until the additional stresses from Eqs. (10.8-6) do not appreciatively change for higher values of n, and thus the series in Eqs. (10.8-6) converges. The theoretical derivation presented in this section represents a general method which can be applied to the problems of creep and plasticity or any other nonlinear stress analysis problem. The nature of the resulting differential equation after linearization depends on the constitutive relation between stress and strain, which is obtained using the experimental data. Any such constitutive law, although yielding a linear equation, may not result in an easily solved equation. However, since the differential equation is always linear, a solution can be obtained using classical procedures.

9

Creep Relaxation of Axisymmetric Stresses

A thick-walled cylindrical vessel of inside radius a and outside radius b is considered. Under different thermal boundary conditions at inside and outside radii, thermal stresses are developed. Due to the self-constraint nature of the thermal stresses in the special case where the stress limits are maintained in the elastic range, the thermal stresses relax toward lower stress condition. Thermal stresses in a thick cylinder due to an axisymmetric temperature distribution are given by Eqs. (6.2-17). Substituting the temperature change θ by (T − T0 ) we find Eα 1 − a2 /r2 b 1 r T rdr − T rdr] [ 2 1 − ν b − a2 r2 a a

(10.9-1)

Eα 1 + a2 /r2 b 1 r = T rdr + 2 T rdr − T ] [ 1 − ν b 2 − a2 a r a

(10.9-2)

σrr0 = σφφ0

Depending upon whether the cylinder is free to expand axially or is restricted, σzz can also be obtained. For the plane strain condition where the cylinder is restricted to expand in axial direction σzz0 = ν(σrr0 + σφφ0 ) − Eα(T − T0 )

(10.9-3)

and for the cylinder which is free to expand in the axial direction σzz0 = σrr0 + σφφ0

(10.9-4)

These stresses are the initial stresses at time zero and they are denoted by the subscript 0. The tangential stress given by Eq. (10.9-2) varies gradually from compressive at the inside to tensile at the outside of the cylinder if Ta > Tb , or vice versa if Ta < Tb .

532


The effective stress and strain rates are defined by Eqs. (10.5-8) and (10.5-9) as 1 (10.9-5) σ ∗ = √ [(σφφ − σrr )2 + (σrr − σzz )2 + (σzz − σφφ )2 ]1/2 2 √ 2 c ∗ [(˙φφ − ˙crr )2 + (˙crr − ˙czz )2 + (˙czz − ˙cφφ )2 ]1/2 ˙c = (10.9-6) 3 where ˙rr , ˙φφ , and ˙zz are the rate of creep strains in radial, tangential, and axial direction, respectively. Considering the secondary creep, the constitutive law of the secondary creep, Eq. (10.2-9), may be written between the effective stress and effective creep strain rate as discussed by Eq. (10.5-15). This yields ˙∗c = Bσ ∗n

(10.9-7)

where B and n are the material’s creep constants. The elastic stress-strain relations are 1 [σrr − ν(σφφ + σzz )] + α(T − T0 ) E 1 φφ = [σφφ − ν(σrr + σzz )] + α(T − T0 ) E 1 zz = [σzz − ν(σφφ + σrr )] + α(T − T0 ) E rr =

(10.9-8)

The equations for effective elastic stress and strain, from Eqs. (10.5-8) and (10.5-9), are 1 σ ∗ = √ [(σφφ − σrr )2 + (σrr − σzz )2 + (σzz − σφφ )2 ]1/2 2 √ 2 ∗ e = [(φφ − rr )2 + (rr − zz )2 + (zz − φφ )2 ]1/2 3

(10.9-9) (10.9-10)

Substituting Eqs. (10.9-8) into Eq. (10.9-10), using Eq. (10.9-4), results in the elastic effective elastic stress-strain relation as ∗e =

2 1+ν ∗ σ 3 E

(10.9-11)

The initial effective stress σ0∗ is obtained by substituting Eqs. (10.9-1), (10.9-2), and (10.9-4) or (10.9-3), into Eq. (10.9-9). Since the external mechanical forces are assumed to be absent, the total effective strain at any given point under pure thermal strain remains constant in time. Thus ∗ = ∗e + ∗c = constant

(10.9-12)

Differentiation of Eq. (10.9-12) with respect to time results in [55] d∗e d∗ =− c dt dt

(10.9-13)

9. Creep Relaxation of Axisymmetric Stresses

533

This equation suggests that the elastic strains are functions of time and redistribute due to creep as time advances. Combining Eqs. (10.9-7), (10.9-11), and (10.9-13) yields 3 BE dσ ∗ =− dt (10.9-14) σ ∗n 2 1+ν Integrating Eq. (10.9-14) in conjunction with the initial condition σ ∗ = σ0∗

at

t=0

(10.9-15)

gives

3 1−n (10.9-16) EBt]2/(1−n) 2 1+ν Substituting for the effective stress from Eq. (10.9-5) into Eq. (10.9-16) yields ∗(1−n)

σ ∗ = [σ0

−

∗(1−n)

(σφφ − σrr )2 + (σrr − σzz )2 + (σzz − σφφ )2 = 2[σ0

−

3 1−n EBt]4/(1−n) 2 1+ν (10.9-17)

For axially unconstrained cylinder the simultaneous solution of Eqs. (10.9-4) and (10.9-17) and the equilibrium equation dσrr σrr − σφφ + =0 dr r

(10.9-18)

provides the time dependent solution for the stresses σrr , σφφ , and σzz . A numerical solution for Eqs. (10.9-17) and (10.9-18) may be suggested as follows [55]. The numerical procedure used is basically an iterative method. Rearrange Eq. (10.9-17) to the form ∗(1−n)

t = [σ0

− σ ∗(1−n) ]

2 1+ν 3 (1 − n)EB

(10.9-19)

and divide the distance from a to b into m sections: 1. The initial elastic stresses σrr0 , σφφ0 , and σzz0 , and the effective stress σ0∗ are calculated from Eqs. (10.9-1), (10.9-2), (10.9-4), and (10.9-9) for the nodal points ri (i = 1, 2, ..., m + 1). 2. A creep reduction factor Ci is assumed. The initial value for the reduction factor is assumed to be equal for all nodal points as C1 = C2 = C3 = ... = Cm+1 = 0.1 3. All initial elastic tangential stresses σφφ0 calculated in step 1 are assumed to be reduced by creep by Ci percent to obtain σφφi = σφφ0 (1 − Ci )

i = 1, 2, ..., m + 1

(10.9-20)

534


Using σφφi , the equilibrium equation Δσrri = (σφφi − σrri )

Δr ri

(10.9-21)

and the boundary conditions (σrr )1 = (σrr )(m+1) = 0, the radial stress corresponding to the reduced σφφ is calculated for each nodal point (σrr )i+1 = (σrr )i + Δ(σrr )i

i = 1, 2, ..., m

(10.9-22)

Then, based on the reduced (σrr )i and (σφφ )i using Eq. (10.9-4), the reduced (σzz )i is calculated. The reduced stresses are substituted into Eq. (10.9-9) to obtain the reduced effective stress σi∗ for each nodal point. 4. The time ti , required for effective stress at each point to change from the ∗ to reduced value σi∗ is calculated from Eq. (10.9-19). initial value σ0i 5. The times ti ’s are compared with t1 and if their differences are greater than the convergence threshold δ, a new value of Ci is calculated for that point from (Ci )new = (Ci )old ×

t1 ti

i = 2, 3, ..., m + 1

(10.9-23)

after which steps 3 through 5 are repeated. 6. Once ti ’s converge, the process is iterated for the next increment of stress reduction. The radial and tangential stress relaxation for b/a = 1.5 and n = 4 are plotted in Figs. 10.9-1 and 10.9-2, respectively. A time parameter tp is used which relates to the actual time t by tp = tBE(αTd )n−1

(10.9-24)

Figure 10.9-1: Variation of the radial thermal stress with time for n = 4.

σff /EaTd

9. Creep Relaxation of Axisymmetric Stresses 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 −0.1 1 −0.2 −0.3 −0.4 −0.5 −0.6 −0.7 −0.8

535 tp=0 1.1 3.1 10

tp = t E B (E a Td)n-1 n=4 b/ a = 1.5

38 360

1.1

1.2

1.3

1.4

1.5 r/ a

Figure 10.9-2: Variation of the tangential thermal stress with time for n = 4.

σrr / EaTd

0.09 0.08

tp=104b1+3 B t

0.07

b1=2 b2=0.1

0.06

b/ a =1.5

tp=0 0.13 0.29

0.05 0.6

0.04 0.03

1.2

0.02

2.7

0.01 0 1

1.1

1.2

1.3

1.4

1.5

r/ a

Figure 10.9-3: Variation of the radial thermal stress with time for variable n. The creep constants n and B may be functions of temperature [15]. Assume a linear form for n as Tr − Tb n = β1 + β2 (10.9-25) Td where β1 is the value of n at a reference temperature Tb , β2 is a constant, Tr is the temperature at radius r, and Td = Ta − Tb . The results for b/a = 1.5 and the temperature dependent n, given by Eq. (10.9-27), are plotted in Figs. 10.9-3 and 10.9-4. In these figures the time parameter tp is defined as tp = 104β1 +3 Bt

(10.9-26)

Comparing Figs. 10.9-1 and 10.9-2 with Figs. 10.9-3 and 10.9-4 indicates that the stresses at the inner boundary and at points close to it decrease much faster when n is a function of temperature. Since Ta is assumed to be larger than Tb , n is larger at the inside radius, and thus faster creep relaxation is expected at and near the inner boundary.

536


σff /EaTd

0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 −0.1 1 −0.2 −0.3 −0.4 −0.5 −0.6 −0.7 −0.8

tp=104b1+3 B t

tp=0 0.13

b1 = 2 b2 = 0.1

0.29

b/ a =1.5

1.2

0.6 2.7

1.1

1.2

1.3

1.4

1.5 r/a

Figure 10.9-4: Variation of the tangential thermal stress with time for variable n.

10

Creep Relaxation of Non-axisymmetric Stresses

Consider a circular cylinder with inside and outside radii a and b, respectively. The temperature is assumed to be constant along the axis of the cylinder and in a steady-state condition. The temperature distribution is assumed to be circumferentially non-axisymmetric. The solution for temperature distribution for this type of boundary condition is given in Example 19 of Chapter 4. Let us assume that the temperature distribution has one axis of symmetry with respect to the circumferential angle φ. For this special case, Eq. (m) of Example 19 of Chapter 4 is introduced as T (r, φ) = C0 + D0 ln r +

∞

[Cn rn + Dn r−n ] cos nφ

(10.10-1)

n=1

We may consider, as an example, the case where the inside temperature is Ti , the outside surface of the thick cylinder has the temperature T1 in the interval −β < φ < +β, and the temperature T2 at the remaining part, as shown in Fig. 10.10-1. For the assumed boundary conditions, the constants of integration are found and the solution reduces to (T2 − Ti ) − (T2 − T1 )(β/π) ln r/a ln b/a ∞ 2(T2 − T1 ) (−1)n sin n(π − β) (r/a)n − (r/a)−n + cos nφ π n (b/a)n − (b/a)−n n=1 (10.10-2)

T (r, φ) = Ti +

Constants of integration of other types of thermal boundary conditions may be found by the techniques given in Chapter 4.

10. Creep Relaxation of Non-axisymmetric Stresses

537

Figure 10.10-1: Thick cylinder under non-axisymmetric temperature distribution. Initial thermal stresses in the cylinder The first two terms of the temperature distribution given by Eq. (10.10-1) produce axisymmetric thermal stresses and the third term produces nonaxisymmetric thermal stresses. Thermal stresses due to the axisymmetric part of the temperature distribution from Eqs. (6.2-17), where the temperature change θ is replaced by (T − T0 ), are Eα 1 − a2 /r2 b 1 r T rdr − T rdr] [ 2 1 − ν b − a2 a r2 a Eα 1 + a2 /r2 b 1 r = T rdr + T rdr − T ] [ 2 1 − ν b − a2 a r2 a

= σrr0 σφφ0

(10.10-3)

The axisymmetric part of the temperature distribution from Eq. (10.10-2) is used to evaluate the integrals of Eqs. (10.10-3) to arrive at σrr0 = σφφ0 =

2 EαB0 [ b (1 2(1−ν) b2 −a2

−

a2 ) ln ab r2

2 2 EαB0 [ b (1 − ar2 ) ln ab 2(1−ν) b2 −a2 σrφ0 =0

− ln ar ]

− ln ar − 1] (10.10-4)

Note that the boundary conditions are σrr = 0 at r = a and b. Thermal stresses due to the non-axisymmetric part of the temperature distribution are given by Eqs. (6.7-35) and are Eα a2 b2 r (1 − )(1 − )A1 cos φ 2(1 − ν) b2 + a2 r2 r2 Eα r b 2 + a 2 a2 b 2 σφφ0 = (3 − − 4 )A1 cos φ 2(1 − ν) b2 + a2 r2 r 2 2 Eα a b r σrφ0 = (1 − 2 )(1 − 2 )A1 sin φ 2 2 2(1 − ν) b + a r r

σrr0 =

(10.10-5)

538


The total initial stresses in the cylinder are, therefore, the sum of expressions given by Eqs. (10.10-4) and (10.10-5)

σrr0 = σrr0 + σrr0 σφφ0 = σφφ0 + σφφ0

σrφ0 = σrφ0 + σrφ0

(10.10-6)

The initial axial stress σzz0 for an axially unrestricted cylinder may be obtained from (10.10-7) σzz0 = σrr0 + σφφ0 In Eqs. (10.10-4) to (10.10-6) subscript “0” is used to indicate that these stresses are at t = 0. As time advances, the stresses change due to the creep relaxation. Thus, the initial stresses are changed and redistributed in the cylinder with time. Stress relaxation From Eq. (10.9-16) ∗(1−n)

σ ∗ = [σ0

−

3 1−n EBt]2/(1−n) 2 1+ν

(10.10-8)

Substituting for the effective creep stress in Eq. (10.10-8), gives [56] 2 (σφφ − σrr )2 + (σrr − σzz )2 + (σzz − σφφ )2 + 6σrφ 3 1−n ∗(1−n) = 2[σ0 − EBt]2/(1−n) 2 1+ν

(10.10-9)

Simultaneous solution of Eq. (10.10-9) and the equilibrium equations ∂σrr 1 + ∂r r ∂σrφ 1 + ∂r r

∂σrφ σrr − σφφ + =0 ∂φ r ∂σφφ 2σrφ + =0 ∂φ r

(10.10-10)

provides the time dependent stresses σrr , σφφ , σzz , and σrφ . The numerical solution discussed in the previous section is used to obtain the creep stress relaxations with the time and the dimensionless results are plotted for the thermal stresses. The results obtained for b/a = 1.5, β = π/3, Ti > T1 > T2 and n = 2 are plotted in Figs. 10.10-2 to 10.10-4. In theses figures, the actual time t is replaced by the time parameter tp defined as tp = (EαTd )n−1 EBt

(10.10-11)

where Td = T1 − T2 . The curves tp = 0 show the initial thermal stresses. The curves for the temperature dependent creep power n(T ) are given in [55].

10. Creep Relaxation of Non-axisymmetric Stresses

539

0.14

tp= t E B(E α Td)n-1

0.12

b/ a =1.5 n=2 r/ a =1.2 r/ a =1.4

σrr / EaTd

0.1 0.08

tp=0 0.18 0.6

0.06

tp=0 1.5 0.18

0.04

0.6 1.5

0.02 0 0

30

60

90

120

150

180

f⬚

Figure 10.10-2: Variation of radial thermal stress with time for n = 2. 1.5

tp= t E B(E α Td)n-1 b/ a =1.5 n=2

1

tp=0 0.18 0.6

σff /Ea Td

0.5

1.5

0 0

60

30

90

120

150

f⬚

180

1.5

−0.5

0.6

r/ a =1.0 r/ a =1.5

−1

0.18

tp=0 −1.5

Figure 10.10-3: Variation of tangential thermal stress with time for n = 2. 0.06

tp= t E B(E α Td)n-1 0.05

b/a=1.5 n=2 r/ a =1.2

σrf / Ea Td

0.04

r/ a =1.4 tp=0

0.03

0.18 0.02

0.6

tp=0

1.5

0.01

0.18 0.6

1.5 0 0

30

60

90

120

150

180

f⬚

Figure 10.10-4: Variation of shear thermal stress with time for n = 2.

540


11

Thermoelastic Creep Relaxation in Beams

Consider a beam of general cross section with an axis of symmetry about the z-axis, as shown in Fig. 10.11-1. The x-axis is assumed to be the neutral axis of the beam in bending. A bending moment M and an axial force P , both constant in time, act on the cross section. The moment M may be either mechanical or thermal, or both. The total strain at any instant of time is the sum of elastic strain e and creep strain c as [18,57] t = e + c (10.11-1) where the elastic strain from Hooke’s law is e =

σ E

(10.11-2)

From the assumption that the plane section remains plane after deformation, the total strain t at any point in the cross section at a distance z from the centroidal plane x-y is (10.11-3) t = Kz + ˆ where K is the curvature and ˆ is the strain of the centroidal plane of the beam. For the initial elastic condition at t = 0, the creep strain is zero and the elastic stress is obtained by multiplying Eq. (10.11-3) by E, using Eq. (10.11-2), as σ = EKz + Eˆ

(10.11-4)

Figure 10.11-1: A beam of cross section symmetric with respect to z-axis under bending moment M and axial force P .

11. Thermoelastic Creep Relaxation in Beams

541

When t > 0, creep strain occurs. Eliminating t between Eq. (10.11-1) and Eq. (10.11-3) and using Eq. (10.11-2) we obtain σ = EKz + E(ˆ − c )

(10.11-5)

The condition of static equilibrium of the beam under axial force P and bending moment M requires c2

P =

−c1

c2

σbdz

M=

−c1

σbzdz

(10.11-6)

Substituting stress from Eq. (10.11-5) into (10.11-6) yields P = EAˆ − M = EIK −

c2

Ec bdz

(10.11-7)

Ec bzdz

(10.11-8)

−c1

c2

−c1

where A and I are the area and the moment of the area of the cross section, respectively. Note that the integral of the term EKzbdz over the cross section is zero, as the coordinate axes are assumed to coincide at the centroid of the beam cross section. For constant loading condition in time, P˙ = M˙ = 0, the rate version of Eqs. (10.11-3), (10.11-5), (10.11-7), and (10.11-8) become ˙ + ˆ˙ ˙t = Kz ˙ + E(ˆ˙ − ˙c ) σ˙ = E Kz 1 ˙c bdz ˆ˙ = A 1 K˙ = ˙c bzdz I

t≥0

(10.11-9) (10.11-10) (10.11-11) (10.11-12)

Equations (10.11-9) to (10.11-12) are the governing equations controlling the creep relaxation in the beam. To obtain the stress as a function of time we integrate Eq. (10.11-10) with respect to time as ˙ + E(ˆ˙ − ˙c )]dt + C0 σ = [E Kz (10.11-13) where C0 is the constant of integration which is obtained using the initial condition t=0 (10.11-14) σ = σ0 where σ0 is the initial elastic stress on the cross section of the beam produced by the axial force P , bending moment M , and the thermal gradient. This yields C0 = σ0

(10.11-15)

542


Therefore,

t

σ = σ0 +

0

˙ + E(ˆ˙ − ˙c )]dt [E Kz

(10.11-16)

Upon substitutions from Eqs. (10.11-11) and (10.11-12) into Eq. (10.11-16) and assuming the constitutive creep law as given by Eq. (10.2-12) ˙c = Be

−ΔH RT

σn

(10.11-17)

Eq. (10.11-16) can be written as σ = σ0 + EB +

t z c2

(

0

I

−c1

e

−ΔH RT

σ n bzdz

−ΔH 1 c2 −ΔH n e RT σ bdz − e RT σ n )dt A −c1

(10.11-18)

Equation (10.11-18) is an integral equation which has to be solved for the stress σ as a function of initial stress σ0 and time. However, σ appears on both sides of the equation. To obtain σ, a numerical procedure based on the method of successive approximations may be used. The total strain at any time is obtained from Eqs. (10.11-10) and (10.11-9) ˙t = ˙c +

σ˙ E

(10.11-19)

Having ˙c from Eq. (10.11-17) and σ from Eq. (10.11-10), we may integrate Eq. (10.11-19) for t as a function of time. In order to calculate the stress from Eq. (10.11-18), the functional relationship between σ and z must be known at any time. Since stress appears on the left- and right-hand side of Eq. (10.11-18), the analytical solution is complicated. A simple and convenient method based on the successive approximations may be used to calculate the stress and strain. To use the method, Eqs. (10.11-18) and (10.11-19) are modified to the form z c2 −ΔH n dσ e RT σ bzdz = EB ( dt I −c1 −ΔH 1 c2 −ΔH n e RT σ bdz − e RT σ n ) + A −c1

(10.11-20)

1 c2 −ΔH n z c2 −ΔH n d e RT σ bzdz + e RT σ bdz) =B( dt I −c1 A −c1 (10.11-21) The numerical procedure 1. Use the analytical methods of elasticity and thermal stresses to obtain the initial elastic stress σ0 , and strain 0 . 2. Use the value of σ0 from step 1 to evaluate the integrals in the brackets on the right-hand side of Eqs. (10.11-20) and (10.11-21).


543

d dσ |t=0 and |t=0 from Eqs. (10.11-20) and (10.11-21). 3. Calculate dt dt 4. The stress and strain at the time t = Δt, a time increment away from t = 0, is obtained from dσ }t=0 dt d = 0 + Δtx{ }t=0 dt

σΔt = σ0 + Δtx{

(10.11-22)

Δt

(10.11-23)

5. To compute the stress and strain for the next time increment Δt, their values from Eqs. (10.11-22) and (10.11-23) are used as initial values in step 2 and the procedure is repeated for step 3. 6. The time may be advanced as far as desired and the stress and strain are found at any time (t + Δt) from the equations dσ }t=t dt d = t + Δtx{ }t=t dt

σt+Δt = σt + Δtx{

(10.11-24)

t+Δt

(10.11-25)

Consider a beam of rectangular cross section with the following assumptions: temperature at the top surface TA = 540◦ C, temperature at the bottom surface TB = 510◦ C, M = 45 N m, width b = 2.5 cm, height h = 7.6 cm, = 815◦ C. With (c1 = c2 = 3.8 cm), α = 9 × 10−6 cm/cm◦ C, E = 207 Gpa, ΔH RT these numerical values, Eq. (10.11-24) is evaluated according to the numerical procedure described to obtain the stress in the beam as a function of time. The results of stress relaxation for n = 2, and n = 3 are plotted in Figs. 10.11-2 to 10.11-4. The results are obtained in terms of a time parameter tp tp = tEB(EαTd )n

(10.11-26)

14 n=2

12

Z=1.5

σ/EaTd

10 Z=0

8 6 4 2

Z=−1.5

0 0

0.2

0.4

0.6 tp

0.8

1

1.2

Figure 10.11-2: Variation of stress with the time shown at top, middle, and lower surfaces for n = 2.

544


Figure 10.11-3: Variation of stress across the beam thickness at different times. 14 n=3

12

Z=1.5

10

σ/EaTd

8

Z=0

6 4 2 0 −2 −4

0

0.02

0.04

0.06

0.08

0.1 0.12 Z= −1.5

tp

Figure 10.11-4: Variation of stress the with time shown at top, middle, and lower surfaces for n = 3. where Td = |TA − TB |. The temperature distribution is assumed to have a parabolic form Td z (10.11-27) T = ( + 1)2 + TB 4 c For the materials subjected to thermal gradient, the creep strain-stress relations should be temperature dependent. This dependence, similar to the simple addition of the term αT to the normal strain components in the linear thermoelasticity problems, assures the effect of the thermal stresses in the creep process. From Figs. 10.11-1 to 10.11-3 it is seen that the point of zero stress shifts and, therefore, the neutral axis of the beam moves toward the lower tensile stress portion. As time is advanced, it sharply drops to a large compressive stress that for a sufficiently long time will exceed the allowable stress and cause failure of the beam. This is due to the nature of the applied stress


545

under mechanical moment M . Since the magnitude of the mechanical moment M is assumed to be constant, the resulting creep strain increases indefinitely resulting in increased mechanical stress. Thermoelastic and plastic creep relaxation of beams, thick cylinders, and spheres under mechanical and thermal stresses may be found in references [58–62]. The material under plastic deformation is assumed to obey the nonlinear strain hardening behavior and a numerical method is employed to obtain the creep relaxation curves. The problems of thick vessels under thermal and mechanical stresses are of practical interest, and the creep relaxation under elevated temperatures is an important design concern.

Bibliography [1] Norton, F.H., The Creep of Steel at High Temperatures, McGraw-Hill, New York, 1929. [2] Nádai, A. and Davis, E.A., The Creep of Metals II, Trans. ASME, J. Appl. Mech, Vol. 3, pp. A-7, March 1936. [3] Bailey, R.W., Design Aspects of Creep, J. Appl. Mech., Vol. 3, pp. A-1, March 1936. [4] Soderberg, C.R., The Interpolation of Creep Tests for Machine Design, Trans. ASME, Vol. 58, pp. 733–743, 1936. [5] Wahl, A.M., Creep Tests of Rotating Disks at Elevated Temperature and Comparison with Theory, J. Appl. Mech., pp. 225–235, Sept. 1954. [6] Wahl, A.M., Analysis of Creep in Rotating Disks Based on the Tresca Criterion and Associated Flow Rule, J. Appl. Mech., pp. 231–238, June 1956. [7] Wahl, A.M., Stress Distribution in Rotating Disks Subjected to Creep at Elevated Temperatures, J. Appl. Mech., pp. 299–305, 1957. [8] Weir, C.D., The Creep of Thick Tubes Under Internal Pressure, J. Appl. Mech., pp. 464–466, Sept. 1957. [9] Finnie, I. and Heller, W.R., Creep of Engineering Materials, McGraw-Hill, New York, 1959. [10] Mendelson, A., A General Approach to the Practical Solution of Creep Problems, Trans. ASME, Vol. 81, pp. 585–598, 1959. [11] Rimrott, F.P.J., Creep of Thick-Walled Tubes Under Internal Pressure Considering Large Strain, Trans. ASME, Vol. 81, pp. 271–275, 1959. [12] Taira, S., Creep of Thick-Walled Cylinders Under Internal Pressure at Elevated Temperature, The 8th Japan Congress on Testing Materials, Kyoto, Japan, pp. 53–60, 1965. 547

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[13] Bhatnagar, N.S. and Gupta, S.K., Analysis of Thick-Walled Orthotropic Cylinder in the Theory of Creep, J. Phys. Soc. Jpn., Vol. 27, No. 6, Dec. 1969. [14] Hult, J.A.H., Creep in Engineering Structures, Blaisdell, New York, 1966. [15] Odqvist, F.K.G., Mathematical Theory of Creep and Creep Rupture, Oxford University Press, Oxford, 1966. [16] Rabotnov, Y.N., Creep Problems in Structural Members, Wiley Interscience Div., Wiley, New York, 1969. [17] Smith, A.J. and Nicolson, A.M., Advances in Creep Design, Halsted Press Div., Wiley, New York, 1971. [18] Penny, R.K. and Marriot, D.L., Design for Creep, McGraw-Hill, New York, 1971. [19] Kachanov, L.M., The Theory of Creep, National Lending Library, New York, 1967. [20] Fridman, Y.B., Strength and Deformation in Nonuniform Temperature Field, Consultants Bureau, New York, 1964. [21] Dorn, J.E., Some Fundamental Experiments on High Temperature Creep, Mech. Phys. Solids, Vol. 3, pp. 85–116, 1954. [22] Skrzypek, J., Plasticity and Creep, Edited by Hetnarski, R.B., Begell House, New York, and CRC Press, Boca Raton, 1993. [23] Garofalo, F., Fundamentals of Creep and Creep Rupture in Metals, MacMillan Series in Material Science, MacMillan, New York, 1965. [24] Weertman, J., Theory of Steady State Creep Based on Dislocation Climb, J. Appl. Phys., Vol. 26, p. 1213, 1955. [25] Weertman, J., Creep of Aluminium Single Crystals, J. Appl. Phys., Vol. 27, p. 832, 1956. [26] Weertman, J., Creep of Indium, Lead and Some of Their Alloys with Various Metals, Trans. AIME, Vol. 218, p. 207, 1960. [27] Sherby, O.D., Creep of Polycrystalline Alpha and Beta Thallium, Trans. AIME, Vol. 212, p. 708, 1958. [28] Lawly, A., Coll, J.A., and Cahn, R.W., Influence of Crystallographic Order on Creep of Iron-Aluminium, Trans. AIME, Vol. 218, p. 166, 1960.

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[29] Tagart, W.J., Activation Energies for High Temperature Creep of Polycrystalline Magnesium, Acta Met., Vol. 9, p. 614, 1961. [30] McLean, D. and Hale, K.F., Structural Processes in Creep, London: Iron and Steel Inst., pp. 19, 1961. [31] Howard, E.M., Barmore, W.C., Mote, J.D., and Dorn, J.E., On the Thermally-Activated Mechanism of Prismatic Slip in the SilverAluminium Hexagonal-Intermediate Phase, Trans. AIME, Vol. 227, p. 1061, 1963. [32] Jones, R.B. and Harris, J.E., Proceedings of Joint International Conference on Creep, New York, The Institution of Mech. Eng., pp. 1–11, 1963. [33] Price, A.T., Holl, H.A., and Grennough, A.P., The Surface Energy and Self Diffusion Coefficient of Solid Iron Above 1350◦ C, Acta Met., Vol. 12, p. 49, 1964. [34] Sellars, C.M. and Quarrell, A.G., The High Temperature Creep of GoldNickel Alloys, Inst. Met., Vol. 90, p. 329, 1961–1962. [35] Kauzman, W., Flow of Solid Metals from the Standpoint of the Chamical Rate Theory, Trans. AIME, Vol. 143, p. 57, 1941. [36] Cottrell, A.H. and Aytekin, V., The Flow of Zinc Under Constant Stress, J. Inst. Met., Vol. 77, p. 389, 1950. [37] Feltham, P., The Plastic Flow of Iron and Plane Carbon Steels Above A3-Point, Proc. Phys. Soc., Vol. 66B, p. 865, 1953. [38] Feltham, P., On the Mechanism of High Temperature Creep in Metals with Special Reference to Polycrystalline Lead, Proc. Phys. Soc., Vol. 69B, p. 1173, 1956. [39] Feltham, P. and Mikean, J.D., Creep in Face-Centered Cubic Metals with Special Reference to Copper, Acta. Met., Vol. 7, p. 614, 1959. [40] Feltham, P., Stress Relaxation in Copper and Alpha-Brasses at Low Temperatures, J. Inst. Met., Vol. 89, p. 210, 1960–1961. [41] Nádai, A. and McVetty, P.G., Hyperbolic Sine Chart for Estimating Working Stresses of Alloys at Elevated Temperatures, Proc. ASTM, Vol. 43, p. 735, 1943. [42] Sherby, O.D. and Dorn, J.E., Some Observations on Correlations Between the Creep Behavior and the Resulting Structures in Alpha Solid Solutions, Trans. AIME, Vol. 197, p. 324, 1953.

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[43] Weertman, J. and Breen, J.E., Creep of Tin Single Crystals, J. Appl. Phys., Vol. 27, p. 1189, 1956. [44] Garofalo, F., Richmond, O., and Domis, W.F., Joint Int. Conf. on Creep, London: The Institution of Mech. Eng., pp. 1–31, 1963. [45] Sherby, O.D., Frenkel, R., Nadeau, J., and Dorn, J.E., Effect of Stress on the Creep Rates of Polycrystalline Aluminium Alloy Under Constant Structure, Trans. AIME, Vol. 200, p. 275, 1954. [46] Feltham, P., On the Activation Energy of High Temperature Creep in Metals, Phil. Mag., Vol. 2, p. 585, 1957. [47] Graham, A. and Walles, K.F.A., Relation Between Long and Short Time Properties of a Commercial Alloy, J. Iron Steel Inst., Vol. 179, 1955. [48] McVetty, P.G., Creep of Metals at Elevated Temperatures – The Hyperbolic Sine Relation Between Stress and Creep Rate, Trans. ASME, Vol. 65, 1943. [49] Marin, J. and Pao, Y.H., An Analytical Theory of the Creep Deformation of Materials, J. Appl. Mech., Vol. 20, 1953. [50] Saint-Venant, B., Mémoire sur l’établissement des équations différentialles des mouvements intérieurs opérés dans les corps solides ductiles au delà des limites o` u l’élasticité pourrait les ramener à leur premier état, Compt. Rend., Vol. 70, pp. 473–480, 1870. [51] Lévy, M., Mémoire sur les équations générales des mouvements intérieurs des corps solides ductile au dela limites o` u l’élasticité pourrait les ramener à leur premier état, Compt. Rend., Vol. 70, pp. 1323-1325, 1870. [52] von Mises, R., Mechanik der festen Körper in Plastisch deformablem Zustand, Göttinger Nachr. Math. Phys., Kl., pp. 582–592, 1913. [53] Rozenblyum, V.I., Axisymmetric Creep of Cylindrical Bodies with Temperature Variation Along the Axis, PMTF, No. 6, 1961. [54] Danyushevskii, A. and Listrinskii, G., Creep in a Nonuniformly Heated Thick-Walled Tube Subjected to Internal Pressure, Mech. Solids, Vol. 1, No. 2, pp. 72–73, 1966. [55] Eslami, M.R., Creep Linearization of Non-Axisymmetrically Heated Cylinders, AIAA J., Vol. 18, No.7, pp. 862–864, July 1980. [56] Sabbaghian, M. and Eslami, M.R., Creep Relaxation of Axisymmetric Thermal Stresses in Thick-Walled Cylindrical Vessels, ASME Paper No.74-PVP-9, 1974.

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[57] Sabbaghian, M. and Eslami, M.R., Creep Relaxation of Nonaxisymmetric Thermal Stresses in Thick-Walled Cylinders, AIAA J., Vol. 12, pp. 1652– 1658, Dec. 1974. [58] Eslami, M.R., Creep Relaxation of a Beam of General Cross Section Subjected to Mechanical and Thermal Loads, Trans. ASME, J. Mech. Design, Vol. 100, pp. 626–629, 1978. [59] Eslami, M.R. and Tafreshi, A., Thermoelastic-Plastic Analysis of Beams, Proc. Int. Conf. Constitutive Law in Eng. Mat., Chongqing, China, Aug. 10–13, 1989. [60] Eslami, M.R., Yazdan, M.R., and Salimi, E., Elastic-Plastic-Creep Analysis of Thick Cylinders and Spheres of Strain Hardening Materials, Proc. Int. Cong. Appl. Mech., Beijing, China, Aug. 21–25, 1989. [61] Eslami, M.R. and Loghman, A., Thermoelastic-Plastic-Creep Analysis of Thick Cylindrical Pressure Vessels of Strain Hardening Materials, Proc. ASME-PVP Conf., Hawaii, July 23–27, 1989. [62] Eslami, M.R. and Shariyat, M., A Technique to Distinguish the Primary and Secondary Stresses, Trans. ASME, J. Press. Vess. Tech., Vol. 117, pp. 1–7, 1995.

Index π-plane, 519 Activation energy of creep, 508 Airy stress function, 33, 113 Airy stress function, creep, 524 Airy stress function, plane strain, 33 Airy stress function, plane stress, 33 Analytic function, 271, 272 Anisotropic material, stress-strain relations, 58 Anisotropic solid, generalized energy equation, 65 Anisotropic solid, generalized thermoelasticity equation, 65 Axisymmetric temperature, 189 Base element, submatrices, 421, 427, 434, 447 Beam, bimetallic, 234 Beams, 540 Beams, boundary conditions, 225 Beams, coupled thermoelasticity, 467 Beams, curved, 242, 243 Beams, Euler-Bernoulli, 467 Beams, functionally graded, 467 Beams, Galerkin finite element, 470 Beams, shear stress, 226 Beams, thermal deflection, 227 Beams, thermal stresses, 222 Beams, transient thermal stress, 231 Beams, with internal heat generation, 233 Bessel equation, 142 Bessel equation, modified, 146 Bessel function, first kind, 145 Bessel function, modified first kind, 146

Bessel function, modified second kind, 147 Bessel function, second kind, 146 Betti-Maxwell reciprocity theorem, 91 Biharmonic solution, 122 Biot free energy function, 89 Body force analogy, 110 Boundary condition, convection, 96 Boundary condition, heat flux, 96 Boundary condition, insulation, 96 Boundary condition, temperature, 96 Boundary conditions, displacement, 24, 353 Boundary conditions, thermal, 353 Boundary conditions, traction, 4 Boundary element formulation, 480 Boundary integral equation, 480 Boussinesq function, 121, 125 Bulk modulus, 22 Castigliano theorem, 323, 329 Cauchy’s formula, 3 Cauchy-Goursat theorem, 271 Cauchy-Riemann condition, 271, 272 Cesàro condition, 20, 107 Cesàro integrals, 27 Circular brake disk, 198 Clausius inequality, 47, 56 Coefficient of thermal expansion, 21 Coldness function, 66 Compatibility equations, multiply connected, 19 Compatibility equations, simply connected, 14 Complex Fourier series, 280 553

554 Complex variables, 271 Complimentary strain energy function, 52 Constitutive equation of creep, 510 Constitutive law, 44 Constitutive law of entropy, 56 Constitutive law of heat flux, 57 Constitutive law of stress, 56 Constitutive law, linear thermoelasticity, 21 Control volume, 44 Convoloution operator, 372 Convolution theorem, 94 Coordinate transformation, 6 Coordinate transformation law, 10 Coordinates, deformed, 10, 54 Coordinates, original, 54 Coordinates, original undeformed, 10 Coupled energy equation, 60 Coupled equations, 62, 352, 417 Coupled thermoelasticity, 352, 381 Coupled thermoelasticity, FG layer, 423 Coupled thermoelasticity, half-space, 362, 377, 397 Coupled thermoelasticity, infinite space, 354 Coupled thermoelasticity, layer, 363, 386 Coupled thermoelasticity, one-dimensional, 361, 371, 420 Coupled thermoelasticity, thick cylinders, 381 Coupled thermoelasticity, thick spheres, 430 Coupled thermoelasticity, variable heat source, 357 Cramer’s rule, 283 Creep linearization, 525 Creep potential, 518 Creep rate potential, 521 Creep relaxation, 531, 536, 540

Index Creep relaxation, beam, 541 Creep relaxation, cylinder, 531, 536 Creep rupture, 508 Creep strain, 503 Creep, tertiary, 501 Creep, transient, 502 Creep, primary, 501 Creep, secondary, 501 Creep, stationary, 502 Creep, stress function, 506 Creep, temperature function, 508 Creep, time function, 505 Creep-stress relations, 515, 518 Curved beams, 242, 243 Cylinder, creep, 525, 536 Cylinder, non-axisymmetrically heated, 525, 536 Cylinders, axisymmetric stresses, 289 Cylinders, FGM, 278, 289 Cylinders, non-axisymmetric stresses, 265 Cylinders, non-axisymmetric temperature, 271, 278 Cylinders, radial temperature, 254 Cylindrical coordinates, 120 D’Alembert’s method, 365 Damping matrix, 418, 420, 435, 437 Deformation controlled stress, 512 Delay time, 513 Deviatoric stress, 515 Differential line element, 10 Dirac delta function, 368 Disk, radial temperature, 258 Disks, 463 Disks, rotating, 263 Displacement potential, 113, 121, 124, 125, 297 Displacement, gradient, 12 Displacement, vector, 11 Dissipation energy, 520

Index Double Fourier series, 140 Drucker’s postulate, 521 Effective creep strain, 516 Effective stress, 516 Elastic center, 319 Elastic constants, relations, 23 Elasticity theory, infinitesimal, 12 Elementary beam theory, 221 Energy equation, Green-Naghdi model, 75 Energy equation, Lord-Shulman model, 65 Energy principle, 49 Energy, internal, 45, 46 Entropy, 47 Entropy change, 47 Entropy expression, 59 Entropy expression, GL model, 66 Entropy expression, LS model, 64 Entropy generated, 47 Entropy law, 49 Entropy transfer, 47 Entropy, definition, 47 Equation of motion, nonlinear, 54 Equation of state, 44 Equations of motion, 4, 23 Equilibrium equations, cylindrical coordinates, 120 Euclidean metric tensor, 55 Euler differential equation, 209 Euler equation, 5 Euler formula, 275 Euler-Bernoulli assumption, 221, 242, 243 Exact differential, 16 FG layer, 423 FGM, Power law form, 442 FGM, power law form, 424, 444 Finite element, 417 Finite element, displacement formulation, 420 Finite element, FG layer, 423

555 Finite element, Galerkin, 415 Finite element, higher order, 463, 467 Finite element, one-dimensional, 420 Finite element, shape function, 445 Force matrix, 419, 420, 436 Fourier coefficients, 138 Fourier law, 60 Fourier law, general, 63 Fourier transformation, 117 Fourier-Bessel expansion, 196 Fourier-Hankel integral transform, 377 Fourier-Laplace transform, 373 Fouries series, 137 Functionally graded beam, 171 Functionally graded disk, 451 Functionally graded layer, 423 Functionally graded material, 187, 189, 278, 289, 303, 451 Functionally Graded materials, 423 Functionally graded sphere, 440 Fundamental solution, 480 Galerkin finite element, 415, 433, 445 Gauss theorem, 4, 5, 56 General solution, Cartesian coordinates, 113 General solution, plane strain, 116, 117 General solution, plane stress, 114 General solution, spherical coordinates, 124 Generalized plane strain, 257 Generalized theory, 54, 62, 70, 386 Generalized theory, comparison, 68 Generalized theory, GL model, 67 Generalized theory, layer, 396 Generalized theory, single model, 68 Generalized theory, unified form, 394 Generalized thermoelasticity, 385, 394, 451 Generalized thermoelasticity, unified theory, 75

556 Gibbs thermodynamics potential, 52, 59 Green theorem, 18 Green-Lindsay model, 66, 92, 400, 478 Green-Lindsay model, FG sphere, 440 Green-Naghdi model, 70, 386, 401 Haigh-Westergaard stress space, 520 Harmonic solution, 122 Harmonic temperature variation, 377 Harmonic varying heat source, 359 Heat balance, 60 Heat, definition, 45 Heaviside step function, 368 Helmholtz free energy, 51, 56, 59, 63, 71 Heterogeneous anisotropic material, 75 Hooke’s law, 21 Hookean material, 58 Hydrostatic stress, 519 Incompressibility condition of creep, 515 Initial condition, 95, 419 Intrinsic energy, 50, 55 Isoparametric element, 433, 437 Kantrovich approximation, 426 Kantrovitch approximation, 415 Lévy-Mises equation, 515 Lamé ellipsoid, 3 Laplace equation, spherical coordinates, 209 Laplace transform, 361 Laplace transforms, 197 Legendre equation, 201 Legendre equation, associated, 204 Legendre Polynomial, associated, 204 Legendre polynomials, 201, 206 Leibnitz formula, 204 Line moment of inertia, 326, 336

Index Linear momentum, 4 Linear temperature distribution, 107 Linear thermoelasticity, 65, 74 Linear thermoelasticity, GL model, 67 Linearize theory of thermoelasticity, 57 Load controlled stress, 513 Lord-Shulman model, 63, 399, 451, 463, 478 Lord-Shulman model, FG layer, 424 Lumped formulation, 158, 160 Maclaurin series, 57 Mass matrix, 418, 420, 435, 437 Maxwell body, 513 Maxwell equation, 512 Maxwell reciprocity theorem, 319 Method of complex variables, 271 Michell condition, 35 Michell conditions, 265 Mises yield criteria, 516 Mohr circle, 13 Moment of momentum, 5 Multiply connected region, 17, 39 Navier equations, 23, 113 Navier equations, FGM, 279, 290, 304 Navier equations, plane strain, 117 Newton’s law, 4 Newton’s law of viscosity, 512 Non-axisymmetric temperature, 187 Non-axisymmetric thermal stresses, 538 Nonaxisymmetric temperature, 182 Nonhomogeneous boundary conditions, 153 Nonhomogeneous differential equations, 153 Nonlinear theory of thermoelasticity, 56 Numerical method, creep relaxation, 533, 542

Index

557

Second sound, 62, 65 Secondary creep, 501, 505, 506, 510 Separation of variables, 135 Papkovich solution, 113 Shape function, 415 Path independent line integral, 17 Shape function, linear, 420 Permutation symbol, 5, 20 Shear modulus, 21 Piping system, large radius elbows, Simply connected region, 14 335 Single-valuedness of displacement Piping systems, 318 field, 15 Piping systems, equilibrium Single-valuedness, multiply equations, 326, 331 connected, 19 Piping systems, three-dimensions, 329 Single-valuedness, multiply connected Piping systems, two-dimensions, 322 region, 27 Plane strain, 108, 116, 254, 265, 523 Single-valuedness, rotation, 20 Plane strain, generalized, 32 Single-valuedness, simply connected, Plane strain, simple, 31 16 Plane stress, 108, 114, 258 Specific heat, 53 Plane stress, generalized, 29 Specific heat, constant strain, 53 Plane stress, simple, 28 Specific heat, constant stress, 54 Power law index effect, 428, 448 Speed of propagation, elastic wave, 396 Primary creep, 505, 506 Speed of propagation, thermal wave, Primary creep, 501, 510 396 Principal axes, 7, 13 Spheres, FGM, 303 Principal deviatoric stress, 9 Spheres, radial temperature, 260 Principal strains, 13 Spheres, thick, 296 Principal stresses, 7 Spherical coordinates, 123 Principle of rate of work, 55 State space approach, 361 Propagation of discontinuities, 366 Steffan-Boltzman constant, 419 Radial flow, 175 Stiffness matrix, 418, 420, 436, 437 Reciprocal theorem, 93 Strain energy function, 52 Reciprocity theorem, 91 Strain hardening, 521 Recovery, 502 Strain invariants, 13 Rectangular beams, 227 Strain tensor, 12 Relaxation, 502 Strain tensor, Almansi, 11, 54 Relaxation time, 63 Strain tensor, Green, 11, 54 Relaxation time effect, 428 Strain-displacement relations, 12, 13 Rheological model, 512 Strain-stress relations, 21 Rings, 242, 243 Stress formulation, 24 Rodrigues formula, 203 Stress function for creep, 522 Rotation tensor, 13 Stress invariant, 8 Rotation vector, 13 Stress invariants, deviatoric, 9 Rupture, 508 Stress space, 518 Octahedral shear stress, 516 Octaherdal creep strain, 516

558 Stress tensor, deviatoric, 9 Stress tensor, Piola-Kirchhoff, 54 Stress tensor, symmetry, 5 Stress tensor, transformation, 6 Stress traction, 2 Stress vector, 2 Successive approximation method, 542 System, 44 Temperature distribution, one-dimensional, 132, 133, 159, 175–178, 200 Temperature distribution, three-dimensional, 163, 165, 191 Temperature distribution, transient, 166, 168, 171, 194, 196, 198, 211 Temperature distribution, two-dimensional, 150, 153, 155, 156, 179, 182 Temperature wave speed, 479 Temperature, definition, 45 Tensor, Euclidean metric, 55 Tensor, skew symmetric, 13 Tertia ry creep, 501 Tertiary creep, 508 Thermal boundary conditions, 419 Thermal stress relaxation, 531, 536, 540 Thermodynamic definitions, 44 Thermodynamic properties, 45 Thermodynamic temperature, 66 Thermodynamic, adiabatic system, 45 Thermodynamic, extensive property, 44 Thermodynamic, intensive property, 44 Thermodynamic, thermal insulation, 45 Thermodynamics, 45

Index Thermodynamics, adiabatic system, 46 Thermodynamics, closed cycle, 47 Thermodynamics, cycle process, 46 Thermodynamics, first law, 45, 55, 60, 64 Thermodynamics, fundamental properties, 48 Thermodynamics, irreversible process, 47 Thermodynamics, reversible process, 47, 48 Thermodynamics, second law, 46, 56, 60 Thermodynamics, variational formulation, 49 Transfinite element, 426, 427, 446, 455, 463 transfinite element, 470 Transfinite element method, 464 Transient thermal stresses, 296 Tresca yield criteria, 516 Turbine blade, 160 Two-dimensional thermoelasticity, 28 Unified generalized theory, 75 Uniqueness theorem, Green-Lindsay model, 84 Uniqueness theorem, Lord-Shulman model, 80 Variational formulation, thermodynamics, 48 Variational principle, 89 Viscoelasticity, 511 Voigt body, 513 Volume change, 14 von Kármán bending rigidity factor, 337 Vorticity tensor, 71 Wave front, displacement, 377, 388 Wave front, stress, 388

Index Wave front, temperature, 374, 388 Weak formulation, 415, 426, 433, 446 Work, definition, 45 Yield surface, 518 Zero thermal strain, 109 Zero thermal stresses, displacements, 108

559 Zero thermal stresses, multiply connected, 107, 109 Zero thermal stresses, plane strain, 108 Zero thermal stresses, plane stress, 108 Zero thermal stresses, simply connected, 106

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