Principles of continuum mechanics: an introduction for engineers [2 ed.] 9781107199200, 9781108183161

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Principles of

Continuum Mechanics A n I n t r o d u c t i o n f o r E n g i n ee r s

J. N. Reddy Second Edition

Principles of Continuum Mechanics Conservation and Balance Laws with Applications Second Edition Continuum mechanics deals with the stress, deformation, and mechanical behaviour of matter as a continuum rather than a collection of discrete particles. The subject is interdisciplinary in nature, and it is gaining increased attention in recent times primarily because of a need to understand a variety of phenomena at different spatial scales. The second edition of Principles of Continuum Mechanics provides a concise yet rigorous treatment of the subject of continuum mechanics and elasticity at the senior undergraduate and first-year graduate levels. It prepares engineer-scientists for advanced courses in traditional as well as emerging fields, such as biotechnology, nanotechnology, energy systems, and computational mechanics. The large number of examples and exercise problems contained in the book systematically advance the understanding of vector and tensor analysis, basic kinematics, balance laws, field equations, constitutive equations, and applications. A solutions manual is available for the book. J. N. Reddy is Distinguished Professor, Regents Professor, and the Holder of Oscar S. Wyatt Endowed Chair in the Department of Mechanical Engineering at Texas A & M University. He is internationally-recognized for his research and education in applied and computational mechanics. The shear deformation plate and shell theories that he developed bear his name (the Reddy third-order shear deformation theory and the Reddy layerwise theory) in the literature. The finite element formulations and models he developed have been implemented into commercial software like ABAQUS, NISA, and HyperXtrude. He is the author of nearly 600 journal papers and 20 textbooks, some of them with multiple editions.

Principles of Continuum Mechanics Conservation and Balance Laws with Applications J . N . R E D DY Texas A & M University

Second Edition

University Printing House, Cambridge CB2 8BS, United Kingdom One Liberty Plaza, 20th Floor, New York, NY 10006, USA 477 Williamstown Road, Port Melbourne, VIC 3207, Australia 4843/24, 2nd Floor, Ansari Road, Daryaganj, Delhi – 110002, India 79 Anson Road, #06–04/06, Singapore 079906 Cambridge University Press is part of the University of Cambridge. It furthers the University’s mission by disseminating knowledge in the pursuit of education, learning, and research at the highest international levels of excellence. www.cambridge.org Information on this title: www.cambridge.org/9781107199200 DOI: 10.1017/9781108183161 c J. N. Reddy 2018  This publication is in copyright. Subject to statutory exception and to the provisions of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press. First Edition published 2010 Printed in the United States of America by Sheridan Books, Inc. A catalogue record for this publication is available from the British Library. Library of Congress Cataloging-in-Publication Data ISBN 978-1-107-19920-0 Hardback Additional resources for this publication at www.cambridge.org/Reddy Cambridge University Press has no responsibility for the persistence or accuracy of URLs for external or third-party Internet websites referred to in this publication and does not guarantee that any content on such websites is, or will remain, accurate or appropriate.

“Let us together be protected and nourished by God’s blessings. Let us together join our mental forces in strength for the benefit of humanity. Let our efforts at learning be luminous, filled with joy, and endowed with the force of purpose. Let us never be poisoned with the seeds of hatred for anyone. Let there be peace and serenity in all the three universes.” – Invocation from Taittiriya Upanishad

That which is not given is lost.

Contents

Preface to the Second Edition Preface to the First Edition Symbols used in the Book

page xi xiii xvii

1

INTRODUCTION 1.1 Continuum Mechanics 1.2 Examples of Engineering Systems 1.3 Objective of the Study 1.4 Summary Problems References for Additional Reading

1 1 5 7 8 10 11

2

VECTORS AND TENSORS 2.1 Motivation 2.2 Definition of a Vector 2.3 Vector Algebra 2.3.1 Unit Vector 2.3.2 Zero Vector 2.3.3 Vector Addition 2.3.4 Multiplication of Vector by Scalar 2.3.5 Scalar Product of Vectors 2.3.6 Vector Product 2.3.7 Triple Products of Vectors 2.3.8 Plane Area as a Vector 2.3.9 Components of a Vector 2.4 Index Notation and Summation Convention 2.4.1 Summation Convention 2.4.2 Dummy Index 2.4.3 Free Index 2.4.4 Kronecker Delta and Permutation Symbols 2.4.5 Transformation Law for Different Bases 2.5 Theory of Matrices 2.5.1 Definition of a Matrix

13 13 13 14 15 15 15 16 18 19 22 24 26 29 29 29 30 30 33 35 35

Contents

viii

2.5.2 Matrix Addition and Multiplication of a Matrix by a Scalar 2.5.3 Symmetric and Skew Symmetric Matrices 2.5.4 Matrix Multiplication 2.5.5 Inverse and Determinant of a Matrix 2.5.6 Positive-Definite and Orthogonal Matrices 2.5.7 Eigenvalues and Eigenvectors 2.6 Vector Calculus 2.6.1 The Del Operator 2.6.2 Divergence and Curl of a Vector 2.6.3 Cylindrical and Spherical Coordinate Systems 2.6.4 Gradient, Divergence, and Curl Theorems 2.7 Tensors 2.7.1 Dyads 2.7.2 Matrix Form of a Second-Order Tensor 2.7.3 Transformation of Components of a Dyad 2.7.4 Tensor Calculus 2.8 Summary Problems References for Additional Reading 3

KINEMATICS OF A CONTINUUM 3.1 Deformation and Configuration 3.2 Engineering Strains 3.2.1 Normal Strain 3.2.2 Shear Strain 3.3 General Kinematics of a Continuum 3.3.1 Configurations of a Continuous Medium 3.3.2 Material and Spatial Descriptions 3.3.3 Displacement Field 3.4 Analysis of Deformation 3.4.1 Deformation Gradient 3.4.2 Various Types of Deformation 3.4.3 Green Strain Tensor 3.4.4 Infinitesimal Strain Tensor 3.4.5 Principal Values and Principal Planes of Strain 3.5 Rate of Deformation and Vorticity Tensors 3.5.1 Velocity Gradient Tensor 3.5.2 Rate of Deformation Tensor 3.5.3 Vorticity Tensor and Vorticity Vector 3.6 Compatibility Equations 3.7 Summary Problems References for Additional Reading

37 37 39 40 44 44 50 50 51 53 55 57 57 58 60 60 61 62 66 69 69 70 70 71 75 75 76 80 82 82 84 87 91 94 95 95 96 97 99 101 102 109

Contents

ix

4

STRESS VECTOR AND STRESS TENSOR 4.1 Introduction 4.2 Stress Vector, Stress Tensor, and Cauchy’s Formula 4.3 Transformations of Stress Components and Principal Stresses 4.3.1 Transformation of Stress Components 4.3.2 Principal Stresses and Principal Planes 4.4 First and Second Piola–Kirchhoff Stress Tensors 4.4.1 Introduction 4.4.2 First Piola–Kirchhoff Stress Tensor 4.4.3 Second Piola–Kirchhoff Stress Tensor 4.5 Summary Problems References for Additional Reading

111 111 112 120 120 123 125 125 126 127 128 129 132

5

CONSERVATION OF MASS AND BALANCE OF MOMENTA AND ENERGY 5.1 Introduction 5.1.1 Preliminary Comments 5.1.2 Velocity and Acceleration 5.2 Conservation of Mass 5.2.1 Conservation of Mass in Spatial Description 5.2.2 Conservation of Mass in Material Description 5.2.3 Reynolds Transport Theorem 5.3 Balance of Momenta 5.3.1 Principle of Balance of Linear Momentum 5.3.2 Principle of Balance of Angular Momentum 5.4 Thermodynamic Principles 5.4.1 Introduction 5.4.2 Balance of Energy for One-Dimensional Flows 5.4.3 Energy Equation for a Three-Dimensional Continuum 5.5 Summary Problems References for Additional Reading

133 133 133 134 135 135 139 141 141 141 153 154 154 155 158 161 161 166

CONSTITUTIVE EQUATIONS 6.1 Introduction 6.2 Elastic Solids 6.2.1 Introduction 6.2.2 Generalized Hooke’s Law for Orthotropic Materials 6.2.3 Generalized Hooke’s Law for Isotropic Materials 6.3 Constitutive Equations for Fluids 6.3.1 Introduction 6.3.2 Ideal Fluids

167 167 169 169 169 172 177 177 177

6

Contents

x

6.3.3 Viscous Incompressible Fluids Heat Transfer 6.4.1 General Introduction 6.4.2 Fourier’s Heat Conduction Law 6.4.3 Newton’s Law of Cooling 6.4.4 Stefan–Boltzmann Law 6.5 Summary Problems References for Additional Reading

178 178 178 179 179 180 180 181 183

APPLICATIONS IN HEAT TRANSFER, FLUID MECHANICS, AND SOLID MECHANICS 7.1 Introduction 7.2 Heat Transfer 7.2.1 Governing Equations 7.2.2 Analytical Solutions of One-Dimensional Heat Transfer 7.2.3 Axisymmetric Heat Conduction in a Cylinder 7.2.4 Two-Dimensional Heat Transfer 7.3 Fluid Mechanics 7.3.1 Preliminary Comments 7.3.2 Summary of Equations 7.3.3 Inviscid Fluid Statics 7.3.4 Parallel Flow (Navier–Stokes Equations) 7.3.5 Diffusion Processes 7.4 Solid Mechanics 7.4.1 Governing Equations 7.4.2 Analysis of Bars 7.4.3 Analysis of Beams 7.4.4 Principle of Superposition 7.4.5 Analysis of Plane Elasticity Problems 7.5 Summary Problems References for Additional Reading

185 185 185 185 189 192 193 195 195 196 197 198 202 205 205 207 211 216 221 229 230 240

Answers to Selected Problems

241

Index

247

6.4

7

Preface to the Second Edition This second edition of Principles of Continuum Mechanics has the same objective as the first, namely, to present the concepts of continuum mechanics in a simple yet mathematically rigorous manner and illustrate the concepts through simple applications from heat transfer, fluid mechanics, and solid mechanics. The subject of continuum mechanics deals with the stress, deformation, and mechanical behavior of matter as a continuum rather than a collection of discrete particles. The subject is interdisciplinary in nature, and it is gaining increased attention primarily because of a need to understand a variety of phenomena at different spatial scales. Formulations of the mathematical models of any phenomena in nature rely heavily on the knowledge of continuum mechanics. Mathematical models and their numerical evaluation are aids to design and manufacturing. The most critical step in arriving at a design of a system (or a component thereof) that is both functional and cost-effective is the construction of a physics-based mathematical model. It is in this connection that a course on continuum mechanics is most helpful. Principles of Continuum Mechanics provides a concise yet rigorous treatment of the subject of continuum mechanics and elasticity at the senior undergraduate and first-year graduate levels. In all of the chapters of the second edition, additional explanations, examples, and exercise problems have been added. No attempt has been made to enlarge the scope or increase the number of topics covered. The large number of examples and exercise problems contained in the book systematically advance the understanding of vector and tensor analysis, basic kinematics, balance laws, field equations, and constitutive equations. The book may be used as a textbook for a first course on continuum mechanics as well as elasticity. A solutions manual has also been prepared for the book. The solutions manual is available from the publisher only to instructors who adopt the book as a textbook for a course. Since the publication of the first edition, several users of the book communicated their comments and compliments as well as errors they found, for which the author thanks them. All of the errors known to the author have been corrected in the current edition. Drafts of the manuscript of this book prior to its publication were read by the author’s doctoral students, who have made suggestions for improvements. In particular, the author wishes to thank Archana Arbind, Parisa Khodabakhshi, Jinseok Kim, and Namhee Kim for their help. The author is grateful to the following professional colleagues for their friendship, encouragement, and constructive comments on the book:

xii

Preface to the Second Edition

Shailendra Joshi, National University of Singapore C. W. Lim, City University of Hong Kong A. Rajagopal, Indian Institute of Technology, Hyderabad Jani Romanoff, Aalto University, Finland E. C. N. Silva, University of S˜ao Paulo, Brazil Arun Srinivasa, Texas A&M University, College Station Karan Surana, University of Kansas Vinu Unnikrishnan, University of Alabama, Tuscaloosa C. M. Wang, University of Queensland, Australia The author also expresses his sincere thanks to Mr. Steven Elliot, Senior Editor of Mechanical Engineering at Cambridge University Press, for his continued encouragement, friendship, and support in producing this book. The author requests readers to send their comments and corrections to [email protected]. J. N. Reddy College Station

Preface to the First Edition This book is a simplified version of the author’s book, An Introduction to Continuum Mechanics with Applications, 2nd ed., published by Cambridge University Press (New York, 2008), intended for use as an undergraduate text book. As most modern technologies are no longer discipline-specific but involve multidisciplinary approaches, undergraduate engineering students should be educated to think and work in such environments. Therefore, it is necessary to introduce the subject of principles of continuum mechanics (i.e., laws of physics applied to science and engineering systems) to undergraduate students so that they have a strong background in the basic principles common to all disciplines and are able to work at the interface of science and engineering disciplines. A first course on principles of mechanics provides an introduction to the basic concepts of stress and strain and conservation principles, and prepares engineer-scientists for advanced courses in traditional as well as emerging fields such as biotechnology, nanotechnology, energy systems, and computational mechanics. Undergraduate students with such backgrounds may seek advanced degrees in traditional (e.g., aerospace, civil, electrical, mechanical, physics, applied mathematics) as well as interdisciplinary degree programs (e.g., bioengineering, engineering physics, nanoscience and engineering, biomolecular engineering, and so on). There are not many books on principles of mechanics that are written keeping undergraduate engineering or science students in mind. A vast majority of books on the subject are written for graduate students of engineering and tend to be more mathematical and too advanced to be of use for third year or senior undergraduate students. This book presents the subjects of mechanics of materials, fluid mechanics, and heat transfer in unified form using the conservation principles of mechanics. It is hoped that the book, which is simple, facilitates understanding of the main concepts of the previous three courses under a unified framework. With a brief discussion of the concept of a continuum in Chapter 1, a review of vectors and tensors is presented in Chapter 2. Since the analytical language of applied sciences and engineering is mathematics, it is necessary for all students of this course to familiarize themselves with the notation and operations of vectors, matrices, and tensors that are used in the mathematical description of physical phenomena. Readers who are familiar with the topics of this chapter may refresh or skip and go to the next chapter. The subject of kinematics, which deals with geometric changes without regard to the forces causing the deformation, is discussed in Chapter 3. Measures of engineering normal and shear strains and definitions of mathematical strains are introduced here. Both simple onedimensional systems as well as two-dimensional continua are used to illustrate the strain and strain–rate measures introduced. In Chapter 4, the concepts of

xiv

Preface to the First Edition

stress vector and stress tensor are introduced. It is here that readers are presented with entities that require two directions – namely, the plane on which they are measured and the direction in which they act – to specify them. Transformation equations among components of stress tensors referred to two different orthogonal coordinate systems are derived, and principal values and principal planes (i.e., eigenvalue problems associated with the stress tensor) are also discussed. Chapter 5 is dedicated to the derivation of the governing equations of mechanics using the conservation principles of continuum mechanics (or laws of physics). The principles of conservation of mass, linear momentum, angular momentum, and energy are presented using one-dimensional systems as well as general threedimensional systems. The derivations are presented in invariant (i.e., independent of a coordinate system) as well as in component form. The equations resulting from these principles are those governing stress and deformation of solid bodies, stress and rate of deformation of fluid elements, and transfer of heat through solid media. Thus, this chapter forms the heart of the course. Constitutive relations that connect the kinematic variables (e.g., density, temperature, deformation) to the kinetic variables (e.g., internal energy, heat flux, and stresses) are discussed in Chapter 6 for elastic materials, viscous fluids, and heat transfer in solids. Chapter 7 is devoted to the application of the field equations derived in Chapter 5 and constitutive models presented in Chapter 6 to problems of heat conduction in solids, fluid mechanics (inviscid flows as well as viscous incompressible flows), diffusion, and solid mechanics (e.g., bars, beams, and plane elasticity). Simple boundary-value problems are formulated and their solutions are discussed. The material presented in this chapter illustrates how physical problems are analytically formulated with the aid of the equations resulting from the conservation principles. As stated previously, the present book is an undergraduate version of the author’s book An Introduction to Continuum Mechanics (Cambridge University Press, New York, 2008). The presentation herein is limited in scope when compared to the author’s graduate level textbook. The major benefit of a course based on this book is to present the governing equations of diverse physical phenomena from a unified point of view, namely, from the conservation principles (or laws of physics) so that students of applied science and engineering see the physical principles as well as the mathematical structure common to diverse fields. Readers interested in advanced topics may consult the author’s continuum mechanics book cited above or other titles listed in references therein. The author is pleased to acknowledge the fact that the manuscript was tested with the undergraduate students in the College of Engineering at Texas A&M University as well as in the Engineering Science Program at the National University of Singapore. The students, in general, have liked the contents and the simplicity with which the concepts are introduced and explained. They also expressed the feeling that the subject is more challenging than most at the undergraduate level but a useful prerequisite to graduate courses in engineering.

Preface to the First Edition

xv

The book contains so many mathematical expressions that it is hardly possible not to have typographical and other kinds of errors. The author wishes to thank in advance those who are willing to draw the author’s attention to typos and errors, using the e-mail address [email protected]. J. N. Reddy College Station

The one exclusive sign of thorough knowledge is the power of teaching. Aristotle It is the supreme art of the teacher to awaken joy in creative expression and knowledge. Albert Einstein A good teacher can inspire hope, ignite the imagination, and instill a love of learning. Brad Henry Teaching is a very noble profession that shapes the character, caliber, and future of an individual. If the people remember me as a good teacher, that will be the biggest honour for me. A. P. J. Abdul Kalam

Symbols used in the Book The symbols that are used throughout the book for various important quantities are defined in the following list. In some cases, the same symbol has different meaning in different parts of the book; it should be clear from the context.

Symbol

Meaning

a aij cv , cp C C0 Cijkl d da dA df dF ds dS dv dV dx dX D, Di D

Acceleration vector, Dv Dt Coefficients of matrix [A] Specific heat at constant volume and pressure, respectively Concentration; deformed configuration Undeformed configuration Elastic stiffness coefficients Diameter Area element (vector) in spatial description Area element (vector) in material description Force on a small elemental area ∆a in C Pre-image of df in C0 Surface element in current configuration (= dΓ) Surface element in reference configuration (= dΓ0 ) Volume element in current configuration (= dΩ) Volume element in reference configuration (= dΩ0 ) Line element (vector) in current configuration Line element (vector) in reference configuration Diffusion coefficients T Symmetric part of the velocity  gradient tensor, L = (∇v) ; 1 T that is, D = 2 (∇v) + ∇v Rectangular Cartesian components of D D ∂ Material time derivative, Dt = ∂t +v·∇ Internal energy per unit mass Unit vector Unit basis vector in the direction of vector A Basis vector in the xi -direction Basis vectors in the (r, θ, z) system Basis vectors in the (R, φ, θ) system Basis vectors in the (x, y, z) system Basis vectors in the (x1 , x2 , x3 ) system Alternating symbol

Dij D/Dt e ˆ e ˆA e ei ˆθ , e ˆz ) (ˆ er , e ˆφ , e ˆθ ) (ˆ eR , e ˆy , e ˆz ) (ˆ ex , e ˆ2 , e ˆ3 ) (ˆ e1 , e eijk

xviii

Symbols used in the Book

Symbol

Meaning

E E, E1 , E2 E ˆi E


Green–Lagrange strain tensor, E = 12 FT · F − I Young’s moduli (modulus of elasticity) Internal heat generation per unit mass Unit base vector along the Xi material coordinate direction (i = 1, 2, 3) Components of the Green–Lagrange strain tensor in the (x1 , x2 , x3 ) system (i, j = 1, 2, 3) Components of the Green strain tensor E in the cylindrical coordinate system (r, θ, z) Components of the Green strain tensor E in the rectangular coordinate system (x, y, z) Load per unit length of a bar Body force vector Body force components in the x, y, and z directions T Deformation gradient, F = (∇0 x) Acceleration due to gravity; internal heat generation per unit volume Shear modulus (modulus of rigidity) Height of the beam; thickness; heat transfer coefficient Heat input to the system Second moment of area of a beam; current density Unit second-order tensor Invariants of a second-order tensor Determinant of F, J = |F| (Jacobian) Diffusion flux Spring constant; thermal conductivity Thermal conductivity tensor Electrical conductivity Kinetic energy Direction cosines Length Velocity gradient tensor, L = (∇v)T Bending moment in beam problems Unit normal vector in the current configuration ˆ ith component of the unit normal vector n ˆ Components of the unit normal vector n Axial force in beam problems Unit normal vector in the reference configuration ˆ Ith component of the unit normal vector N

Eij Err , Eθθ , Erθ , · · · Exx , Eyy , Exy , · · · f f fx , fy , fz F g G h H I I I1 , I2 , I3 J J k k ke K `ij L L M ˆ n ni (nx , ny , nz ) N ˆ N NI

Symbols used in the Book

xix

Symbol

Meaning

p

Internal pressure of a pressure vessel; thermodynamic or hydrostatic pressure Hydrostatic pressure; point load in beams; perimeter First Piola–Kirchhoff stress tensor Distributed transverse load on a beam Intensity of the distributed transverse load in beams ˆ Heat flux normal to the boundary, qn = ∇ · n Heat flux vector in the current configuration Mass flow rate; volume rate of flow Rate of heat due to current density Radial coordinate in the cylindrical polar system; r = |r| Position vector in cylindrical coordinates, x Cylindrical coordinate system Radial coordinate in the spherical coordinate system; R = |R|; universal gas constant; radius of curvature Position vector in the spherical coordinate system Spherical coordinate system Second Piola–Kirchhoff stress tensor Components of the second Piola–Kirchhoff stress tensor in the rectangular coordinate system (x1 , x2 , x3 ) Time Stress vector; traction vector ˆj Stress vector on xi -plane, ti = σij e Temperature Displacement vector Displacements in the (x, y, z) coordinate system Displacements in the (x1 , x2 , x3 ) coordinate system Displacements in the (r, θ, z) coordinate system Internal (or strain) energy Velocity, v = |v| ˆ , vn = v · n ˆ Projection of v onto n Components of velocity vector v in (x1 , x2 , x3 ) system Components of velocity vector v in (r, θ, z) system Velocity vector, v = Dx Dt ˆ) Velocity vector normal to the plane (whose normal is n Shear force in beam problems; scalar potential of body forces Power input Position vector in the current configuration Rectangular Cartesian coordinates Rectangular Cartesian coordinates Position vector in the reference configuration

P P q q0 qn q Q Qe r r (r, θ, z) R R (r, φ, θ) S Sij t t ti T u (u, v, w) (u1 , u2 , u3 ) (ur , uθ , uz ) U v vn (v1 , v2 , v3 ) (vr , vθ , vz ) v vn V W x (x, y, z) (x1 , x2 , x3 ) X

xx

Symbols used in the Book

Greek symbols Symbol

Meaning

α

Angle; coefficient of thermal expansion; kinetic energy coefficient Deformation mapping Dirac delta Components of the unit tensor, I (Kronecker delta)  Infinitesimal strain tensor, ε = 12 (∇0 u)T + ∇0 u Total energy stored per unit mass Rectangular components of the infinitesimal strain tensor Components of the infinitesimal strain tensor in the cylindrical coordinate system (r, θ, z) A typical scalar function; angular coordinate in the spherical coordinate system Viscous dissipation, Φ = τ : D; Airy stress function Shear strain in one-dimensional problems Internal entropy production; total boundary vorticity vector, ζ = 2ω Entropy density per unit mass; dashpot constant Extension ratio; Lam´e constant; eigenvalue Eigenvalues of a 3 × 3 matrix Lam´e constant; viscosity Poisson’s ratio; νij Poisson’s ratios Angular coordinate in the cylindrical and spherical coordinate systems; angle; absolute temperature Angles corresponding to maximum normal stress and maximum shear stress, respectively Mass density in the current configuration Mass density in the reference configuration Boltzman constant Cauchy stress tensor Components of the stress tensor, σ, in the rectangular coordinate system (x1 , x2 , x3 ) ˆ Normal and shear stresses on a plane with normal n Principal normal and shear stresses Components of the stress tensor σ in the cylindrical coordinate system (r, θ, z) Shear stress Viscous stress tensor Domain of a problem

χ δ δij ε  εij εrr , εθθ , εrθ , · · · φ Φ γ Γ ζ η λ λ1 , λ2 , λ3 µ ν θ θn , θs ρ ρ0 σ σ σij σn , σs σpi , σsi σrr , σθθ , σrθ , · · · τ τ Ω

Symbols used in the Book

Symbol

Meaning

Ω

Spin tensor or skew symmetric part of the velocity  gradient tensor, (∇v)T ; that is Ω = 12 (∇v)T − ∇v Components of spin tensor Ω in the rectangular coordinate system (x1 , x2 , x3 ) Components of the spin tensor Ω in the cylindrical coordinate system (, r, θ, z) Components of the spin tensor tensor Ω in the rectangular coordinate system (x, y, z) Angular velocity Axial (vorticity) vector of Ω, ω = 12 ∇ × u Components of vorticity vector ω in the rectangular coordinate system (x1 , x2 , x3 ) Components of vorticity vector ω in the rectangular coordinate system (x, y, z) Warping function; stream function Helmholtz free energy density; Prandtl stress function Gradient operator with respect to x Gradient operator with respect to X Laplace operator, ∇2 = ∇ · ∇ Biharmonic operator, ∇4 = ∇2 ∇2 Matrix of components of the enclosed tensor Column of components of the enclosed vector Symbol used for the dot product or scalar product Symbol used for the cross product or vector product

Ωij Ωrθ , Ωrz , Ωθz Ωxy , Ωxz , Ωyz ω ω ωi ωx , ωy , ωz ψ Ψ ∇ ∇0 ∇2 ∇4 [] {} · ×

xxi

Ignorance more frequently begets confidence than does knowledge: it is those who know little, and not those who know much, who so positively assert that this or that problem will never be solved by science. Charles Darwin Science can purify religion from error and superstition. Religion can purify science from idolatry and false absolutes. Pope John Paul II A yogi seated in a Himalayan cave allows his mind to wander on unwanted things. A cobbler, in a corner at the crossing of several busy roads of a city, is absorbed in mending a shoe as an act of service. Of these two, the latter is a better yogi than the former. Swami Vivekananda

xxii

Symbols used in the Book

Table 1 Conversion factors

Quantity

US customary unit

SI equivalent

Time Mass Length

s lb (mass) in ft lb/in3 lb/ft3 lb (force) kip (103 lb) lb/in2 (psi) ksi (103 psi) Msi (106 psi) lb in lb ft ft lb/s hp (550 ft lb/s) ◦ F ◦ ◦ F = 95 C + 32◦ F

s 0.4536 kg 25.4 mm 0.3048 m 27.68 × 103 kg/m3 16.02 kg/m3 4.448 N 4.448 kN 6.895 kN/m2 6.895 MN/m2 6895 MN/m2 0.1130 Nm 1.356 Nm 1.356 W 745.7 W 0.5556◦ C

Density Force Pressure or stress

Moment or torque Power Temperature Conversion formula

s = second; lb = pound; in = inch; ft = foot; hp = horse power; kg = kilogram (= 103 grams); m = meter; mm = millimeter (10−3 m); N = Newton; W = Watt; Pa = Pascal = N/m2 ; kN = 103 N; MN = 106 N; MPa = 106 Pa; GPa = 109 Pa

Note: Historical notes at the end of each chapter were based on information found at https://www.wikipedia.org/. Quotes by various people included in this book were found at different web sites. For example, visit: http://naturalscience.com/dsqhome.html, http://thinkexist.com/quotes/david hilbert/, http://www.yalescientific.org/2010/10/from-the-editor-imagination-in-science/, https://www.brainyquote.com/quotes/. The author is motivated to include the quotes at various places in his book for their wit and wisdom, although he cannot vouch for their accuracy.

1

INTRODUCTION One thing I have learned in a long life: that all our science, measured against reality, is primitive and childlike – and yet it is the most precious thing we have. Albert Einstein What we need is not the will to believe but the will to find out. Bertrand Russell Those who know how to think need no teachers. Mahatma Gandhi

1.1

Continuum Mechanics Matter is composed of discrete molecules, which in turn are made up of atoms. An atom consists of electrons, positively charged protons, and neutrons. Electrons form chemical bonds. An example of mechanical (that is, has no living cells) matter is a carbon nanotube (CNT), which consists of carbon molecules in a certain geometric pattern in equilibrium with each other, as shown in Fig. 1.1. Another example of matter is a biological cell, which is a fundamental unit of any living organism. There are two types of cells: prokaryotic and eukaryotic cells. Eukaryotic cells are generally found in multicellular organs and they have a true nucleus, distinct from a prokaryotic cell. Structurally, cells are composed of a large number of macromolecules (or large molecules). These macromolecules consist of large numbers of atoms and form specific structures, like chromosome and plasma membranes in a cell. Macromolecules occur under four major types: carbohydrates, proteins, lipids, and nucleic acids. To highlight the hierarchical nature of the structures formed by the macromolecule in a cell, let us analyze a chromosome. Chromosomes, which are carriers of hereditary traits in an individual, are found inside the nucleus of all eukaryotes. Each chromosome consists of a single nucleic acid macromolecule called deoxyribonucleic acid (DNA) (2.2–2.4 nanometers wide). These nucleic acids are in turn formed from the specific arrangement of monomers called mono-nucleotides (0.3–0.33 nanometers). The fundamental units of nucleotides are formed again by a combination of a specific arrangement of a phosphate radical, nitrogenous base, and a carbohydrate sugar. The hierarchical nature of the chromosome is as shown in Fig. 1.2(a). Similarly to the chromosomes, all the structures in a cell are formed from a combination of the macromolecules.

2

INTRODUCTION

(a)

(b)

Strip of a graphene sheet rolled into a tube

(n, 0) / zigzag

(m, m) / arm chair (m, n )chiral

Fig. 1.1 Carbon nanotubes (CNTs) with different chiralities

At the macroscopic scale, eukaryotic cells can be divided into three distinct regions: nucleus, plasma membrane, and cytoplasm having a host of other structures, as shown in Fig. 1.2(b). The nucleus consists of chromosomes and other protein structures and is the control center of the cell determining how the cell functions. The plasma membrane encloses the cell and separates the material outside the cell from inside. It is responsible for maintaining the integrity of the cell and also acts as channels for the transport of molecules to and from the cell. Cell membrane is made up of a double layer of phospholipid molecules (macromolecule), having embedded transmembrane proteins. The region between the cell membrane and the nucleus is the cytoplasm which consists of a gel-like fluid called cytosol, the cytoskeleton, and other macromolecules. Cytoskeleton forms the biomechanical framework of the cell and consists of three primary protein macromolecule structures of actin filaments, intermediate filaments, and microtubules. Cell growth, expansion, and replication are all carried out in the cytoplasm. The interactions between the different components of the cell are responsible for maintaining the structural integrity of the cell. The analysis of these interactions to obtain the response of the cell when subjected to an external stimulus (mechanical, electrical, chemical) is studied systematically under cell mechanics. The structural framework of primary macromolecular structures in a cell is shown in Fig. 1.2(c).

Continuum Mechanics

Figure 1.1.2 Nitrogen Base Phosphate

DNA (2.2-2.4 nm)

(a)

Nucleotides (0.3-0.33 nm) Nucleus

Cell membrane

Organelles

Cytoplasm (interior contents of the cell) Actin filaments Intermediate filaments

Network of actin filaments Microtubule Cytosol- heterogeneous fluid component Cytoskeleton – Filament network permeating the cell’s interior

(b)

(c) Cell membrane Animal cell

Microfilaments

Microtubule

Fig. 1.2 (a) Hierarchical nature of chromosome (b) Structure of a generalized cell (c) Macromolecular structure in a cell

3

4

INTRODUCTION

The study of matter at molecular or atomistic levels is very useful for understanding a variety of phenomena; but studies at these scales are not useful to solve common engineering problems. The understanding gained at the molecular level needs to be taken to the macrosopic scale (that is, scale that a human eye can see) to be able to study its behavior. Central to this study is the assumption that the discrete nature of matter can be overlooked, provided the length scales of interest are large compared to the length scales of a discrete molecular structure. Thus, matter at sufficiently large length scales can be treated as a continuum in which all physical quantities of interest, including density, are continuously differentiable. The subject of mechanics deals with the study of motion and forces in solids, liquids, and gases and the deformation or flow of these materials. In such a study, we make the simplifying assumption, for analysis purposes, that the matter is distributed continuously, without gaps or empty spaces (i.e., we disregard the molecular structure of matter). Such a hypothetical continuous matter is termed a continuum. In essence, in a continuum all quantities such as the density, displacements, velocities, stresses, and so on vary continuously so that their spatial derivatives exist and are continuous. The continuum assumption allows us to shrink an arbitrary volume of material to a point, in much the same way as we take the limit in defining a derivative, so that we can define quantities of interest at a point. For example, density (mass per unit volume) of a material at a point is defined as the ratio of the mass ∆m of the material to a small volume ∆V surrounding the point in the limit that ∆V becomes a value 3 , where  is small compared with the mean distance between molecules: ρ = lim ∆V

→3

∆m . ∆V

(1.1.1)

In fact, we take the limit  → 0. A mathematical study of mechanics of such an idealized continuum is called continuum mechanics. Engineers and scientists undertake the study of continuous systems to understand their behavior under “working conditions,” so that the systems can be designed to function properly and produced economically. For example, if we were to repair or replace a damaged artery in a human body, we must understand the function of the original artery and the conditions that led to its damage. An artery carries blood from the heart to different parts of the body. Conditions like high blood pressure and increase in cholesterol content in the blood may lead to deposition of particles in the arterial wall, as shown in Fig. 1.3. With time, accumulation of these particles in the arterial wall hardens and constricts the passage, leading to cardiovascular diseases. A possible remedy for such diseases is to repair or replace the damaged portion of the artery. This in turn requires an understanding of the deformation and stresses caused in the arterial wall by the flow of blood. The understanding is then used to design the vascular prosthesis (that is, artificial artery).

Examples of Engineering Systems

5

Fig. 1.3 Progressive damage of (plaque formation in) artery due to deposition of particles in the arterial wall

The primary objectives of this book are: (1) to study the conservation and balance principles in mechanics of continua and formulate the equations that describe the motion and mechanical behavior of materials; and (2) to present the applications of these equations to simple problems associated with flows of fluids, conduction of heat, and deformation of solid bodies. While the first of these objectives is an important topic, the reason for the formulation of the equations is to gain a quantitative understanding of the behavior of an engineering system. This quantitative understanding is useful in the design and manufacture of better products.

1.2

Examples of Engineering Systems Typical examples of engineering problems, which are sufficiently simple to cover in this course, are described below. At this stage of discussion, it is sufficient to rely on the reader’s intuitive understanding of concepts. Example 1.2.1 (A mechanical structure) We wish to design a diving board which must enable the swimmer to gain enough momentum for the swimming exercise. The diving board is fixed at one end and free at the other end (see Fig. 1.4). The board is initially straight and horizontal, and of length L and uniform cross-section A = bh. The design process consists of selecting the material with Young’s modulus E and cross-sectional dimensions b and h such that the board carries the weight W of the swimmer. The design criteria are that the stresses developed do not exceed the allowable stress and the deflection of the free end does not exceed a pre-specified value δ. A preliminary design of such systems is often based on mechanics of materials equations. The final design involves the use of more sophisticated equations, such as the threedimensional elasticity equations. The equations of elementary beam theory may be used to find a relation between the deflection δ of the free end in terms of the length L, cross-sectional dimensions b and h, Young’s modulus E, and weight W :

6

INTRODUCTION

δ=

4W L3 . Ebh3

(1.2.1)

Given δ (allowable deflection) and load W (maximum possible weight of a swimmer), one can select the material (Young’s modulus, E) and dimensions L, b, and h (which must be restricted to the standard sizes fabricated by a manufacturer). In addition to the deflection criterion, one must also check if the board develops stresses that exceed Figure 1.1.4 the allowable stresses of the material selected. Analysis of pertinent equations provides the designer with alternatives to select the material and dimensions of the board so as to have a cost-effective but functionally reliable structure.

L

b h

Fig. 1.4 A diving board fixed at left end and free at right end

Example 1.2.2 (Fluid flow) We wish to measure the viscosity µ of a lubricating oil used in rotating machinery to prevent damage of the parts in contact. Viscosity, like Young’s modulus of solid materials, is a material property that is useful in the calculation of shear stresses developed between a fluid and a solid body. A capillary tube is used to determine the viscosity of a fluid via the formula µ=

πd4 P1 − P2 , 128L Q

(1.2.2)

where d is the internal diameter, L is the length of the capillary tube, P1 and P2 are Figure the pressures1.1.5 at the two ends of the tube (oil flows from one end to the other, as shown in Fig. 1.5), and Q is the volume rate of flow at which the oil is discharged from the tube.

r

vx (r ) x

Internal diameter, d

P1

P2 L

Fig. 1.5 Measurement of viscosity of a fluid using capillary tube

Objective of the Study

7

Example 1.2.3 (Heat flow in solids) We wish to determine the heat loss through the wall of a furnace. The wall typically consists of layers of brick, cement mortar, and cinder block (see Fig. 1.6). Each of these materials provides a varying degree of thermal resistance. The Fourier heat conduction law dT q = −k (1.2.3) dx provides a relation between the heat flux q (heat flow per unit area) and gradient of temperature T . Here k denotes thermal conductivity (1/k is the thermal resistance) of the material. The negative sign in Eq. (1.2.3) indicates that heat flows from a highFigure 1.1.6 temperature region to a low-temperature region. Using the continuum mechanics equations, one can determine the heat loss when the temperatures inside and outside of the building are known. A building designer can select the materials as well as thicknesses of various components of the wall to reduce the heat loss (while ensuring necessary structural strength – a structural analysis aspect).

Cross-section of the wall

Furnace

x

Fig. 1.6 Heat transfer through a composite wall of a furnace

The previous three examples provide some indication of the need for studying the response of materials under the influence of external loads. The response of a material is consistent with the laws of physics and the constitutive behavior of the material. This book has the objective of describing the physical principles and deriving the equations governing the stress and deformation of continuous materials, and then solving some simple problems from various branches of engineering to illustrate the applications of the principles discussed and equations derived.

1.3

Objective of the Study The primary objective of this book, as already stated, is twofold: (1) use the physical principles to derive the equations that govern the motion and thermomechanical response of materials and systems; and (2) application of these equations for the solution of specific problems of engineering and applied science (e.g., linearized elasticity, heat transfer, and fluid mechanics). The governing

8

INTRODUCTION

equations for the study of deformation and stress of a continuous material are nothing but an analytical representation of the global laws of conservation of mass and balance of momenta and energy, and the constitutive response of the continuum. They are applicable to all materials that are treated as a continuum. Tailoring these equations to particular problems and solving them constitute a bulk of engineering analysis and design. The study of motion and deformation of a continuum (or a “body” consisting of continuously distributed material) can be broadly classified into five basic categories: (1) (2) (3) (4) (5)

Kinematics Conservation of mass Kinetics (balance of linear and angular momentum) Thermodynamics (first and second laws of thermodynamics) Constitutive equations.

Kinematics is a study of the geometric changes or deformation in a continuum, without the consideration of forces causing the deformation. The principle of conservation of mass ensures that the mass of the deforming medium is conserved when no mass is created or destroyed. Kinetics is the study of the static or dynamic equilibrium of forces and moments acting on a continuum, using the principles of balance of linear and angular momentum. This study leads to equations of motion as well as the symmetry of stress tensor in the absence of body couples. Thermodynamic principles are concerned with the balance of energy and relations among heat, mechanical work, and thermodynamic properties of the continuum. Constitutive equations describe thermomechanical behavior of the material of the continuum, and they relate the dependent variables introduced in the kinetic description to those introduced in the kinematic and thermodynamic descriptions. Table 1.1 provides a brief summary of the relationship between physical principles and governing equations, and physical entities involved in the equations. To the equations derived from physical principles, one must add boundary conditions of the system (and initial conditions if the phenomena are time-dependent) to complete the analytical description.

1.4

Summary In this chapter, the concept of a continuous medium is discussed and the major objectives of the present book, namely, (a) use of the principles of mechanics to derive the equations governing a continuous medium, and (b) applications of these equations in the solution of specific problems arising in engineering, are presented. Mathematical formulation of the governing equations of a continuous medium necessarily requires the use of vectors, matrices, and tensors – mathematical tools that facilitate analytical formulation of the natural laws. Therefore,

Summary

9

Table 1.1 Major topics of the present study, principles of mechanics used, resulting governing equations, and dependent variables involved

Topic of study

Physical principle

Resulting equations

Variables involved

1. Kinematics

Based on geometric changes

Strain–displacement relations Strain rate– velocity relations

Displacements and strains Velocities and strain rates

2. Mass

Conservation of mass

Continuity equation

Density and velocities

3. Kinetics

Balance of linear momentum Balance of angular momentum

Equations of motion Symmetry of stress tensor

Stresses, and velocities Stresses

4. Thermodynamics

First law

Energy equation

Second law

Clausius–Duhem inequality

Temperature, heat flux, stresses, and velocities Temperature, heat flux, and entropy

Constitutive axioms

Hooke’s law

5. Constitutive equations (not all relations are listed)

Newtonian fluids Fourier’s law Equations of state

6. Boundary conditions

All of the above principles and axioms

Relations between kinematic and kinetic variables

Stresses, strains, heat flux, and temperature Stresses, pressure, velocities Heat flux, and temperature Density, pressure, temperature All of the above variables

it is useful to gain certain operational knowledge of vectors, matrices, and tensors first. Chapter 2 is dedicated to this purpose. The study of principles of mechanics is broadly divided into topics outlined in Table 1.1. The first four topics constitute the subjects of Chapters 3 through 6, respectively. For the convenience of analysis, a continuum may be treated either as a solid or a fluid (liquids and gases), and equations derived in Chapters 3 through 6 are specialized in Chapter 7 to study, through some simple problems, the behavior of solids and fluids. Many of the concepts presented herein are the same as those that were most likely introduced in undergraduate courses on mechanics of materials, heat transfer, fluid mechanics, and material science. The present course brings together

10

INTRODUCTION

these courses under a common mathematical framework and, thus, may require mathematical tools as well as concepts not seen before. Readers must motivate and challenge themselves to learn the new mathematical concepts introduced here, because the language of engineers is mathematics. This subject also serves as a prelude to many graduate courses in engineering and applied sciences. While this book is self-contained for an introduction to continuum mechanics, there are several books that may provide an advanced treatment of the subject. The graduate level text book by the author, An Introduction to Continuum Mechanics with Applications, 2nd ed. (Cambridge University Press, New York, 2013), provides additional and advanced material. Interested readers may consult other titles listed in References for Additional Reading at the end of each chapter.

Problems 1.1

The end deflection (in the direction of the force) of a cantilever beam subjected to end load W (N) is given by δ=

W L3 , 3EIy

where L (m) is the length, E (N/m)2 is Young’s modulus, and Iy (m4 ) is the moment of inertia of the beam about the y-axis. If the beam is of length L = 3 m, and has a cross-section of channel shape with width b = 300 mm and height h = 80 mm, as shown Fig. P1.1, determine the maximum tensile and compressive stresses in the beam, if FigureinP1.1 the stress is σ(x, z) = M (x)z/Iy , where z is the transverse coordinate measured from the geometric centroid of the cross-section and M is the bending moment, which is a function of position x along the length of the beam. The bending moment is taken as positive clockwise at any section x along the beam.

z

W

c1 A1

L

x

b

A2

h

z t = 12 mm

c2

y

h = 80 mm b = 300 mm

Fig. P1.1 1.2

The velocity field of fully developed flow through a circular pipe of diameter d and length L is given by (see Fig. 1.5)   d2 dP r2 vz (r) = − 1−4 2 , 16µ dx d where x is the coordinate along the pipe, r is the radial coordinate, µ is the fluid viscosity, and dP/dx is the pressure gradient in the x-direction. Show that the viscosity µ is given by µ=−

πd4 dP , 128Q dx

References for Additional Reading

1.3

11

where Q is the total volume rate flow at any section of the pipe. The pressure gradient dP/dx can be taken as (P2 − P1 )/L to obtain the result in Eq. (1.2.2). Consider a furnace with a composite wall of three materials (see Fig. 1.6) with constant individual conductivities ki and thicknesses hi (i = 1, 2, 3). Assuming that the flow of heat is one-dimensional across the wall, determine the relation between the heat flow q and temperature T1 at the inside of the inner wall and the temperature T4 outside of the outer wall.

References for Additional Reading 1. 2.

Boal, D., Mechanics of the Cell, Cambridge University Press, New York, NY (2002). Reddy, J. N., An Introduction to Continuum Mechanics, 2nd ed., Cambridge University Press, New York, NY (2013).

Historical notes • Leonardo di ser Piero da Vinci (15 April 1452–2 May 1519), more commonly known as Leonardo da Vinci, was an Italian Renaissance polymath whose areas of interest included invention, painting, sculpting, architecture, science, music, mathematics, engineering, literature, anatomy, geology, astronomy, botany, writing, history, and cartography. He has been called the father of palaeontology, ichnology, and architecture, and is widely considered one of the greatest painters of all time. Sometimes credited with the inventions of the parachute, helicopter and tank, he epitomised the Renaissance humanist ideal.

When a distinguished but elderly scientist states that something is possible, he is almost certainly right. When he states that something is impossible, he is very probably wrong. Arthur C. Clarke I can live with doubt and uncertainty and not knowing. I think it is much more interesting to live not knowing than to have answers that might be wrong. Richard Feynman

2

VECTORS AND TENSORS Mathematics is the language in which God has written the universe. Galileo Galilei Mathematics has a three-fold purpose. It must provide an instrument for the study of nature. But this is not all: it has a philosophical purpose, and, I daresay, an aesthetic purpose. Henri Poincar´e

2.1

Motivation In the mathematical description of equations governing a continuous medium, we derive relations between various quantities that are used to predict the response of the continuum by means of the laws of nature, such as Newton’s laws, and geometric representations of the medium. As a means of expressing a natural law or a relationship between quantities, a coordinate system in a chosen frame of reference is adopted. The mathematical form of the law or relationship thus depends upon the chosen coordinate system and may appear different in another type of coordinate system. The laws of nature, however, should be independent of the choice of a coordinate system, and we may seek to represent the law in a manner independent of a particular coordinate system (we always return to a particular coordinate system of our choice to solve the equations resulting from the physical law). A way of doing this is provided by objects called vectors and tensors. When vector and tensor notation is used, a particular coordinate system need not be introduced. Consequently, the use of vector and tensor notation in formulating natural laws leaves them invariant. A study of physical phenomena by means of vectors and tensors may lead to a deeper understanding of the problem in addition to bringing simplicity and versatility into the analysis. This chapter is dedicated to the study of algebra and calculus of physical vectors and tensors, as needed in the subsequent study.

2.2

Definition of a Vector The quantities encountered in analytical description of physical phenomena may be classified into the following two groups according to the information needed to specify them completely: scalars and nonscalars (see Table 2.1). The scalars are given by a single number. Nonscalars have not only a magnitude specified, but also additional information, such as direction. Nonscalars that obey certain rules

14

VECTORS AND TENSORS

Table 2.1 Classification of mathematical quantities

Scalars

Nonscalars

Mass Temperature Time Volume Length

Force Moment Stress Acceleration Displacement

(such as the parallelogram law of addition) are called vectors. Not all nonscalar quantities are vectors, unless they obey certain rules as discussed next. A physical vector is often shown as a directed line segment with an arrowhead at the end of the line, as shown in Fig. 2.1. The length of the line represents the magnitude of the vector and the arrow indicates the direction. In written or typed material, it is customary to place an arrow over the letter denoting the ~ In printed material, the letter used for the vector is commonly vector, such as A. denoted by a boldface letter, A, such as used in this study. The magnitude of the vector A is denoted by |A|, kAk, or A. Magnitude of a vector is a scalar. Actual computation of the magnitude of a general vector requires the notion of a “norm” of a vector. The “dot product” of a physical vector with itself gives the square of the length, as will be discussed in Section 2.3.5. Thus, mathematically we can only find the square of the magnitude of a vector.

• Magnitude (length) of the vector, A

A = Aeˆ A

•

eˆ A Origin of the vector

Fig. 2.1 Geometric representation of a physical vector

2.3

Vector Algebra In this section, we discuss various operations with vectors and physically interpret them. First, we introduce the notion of unit and zero vectors.

Vector Algebra

2.3.1

15

Unit Vector A vector of unit length is called a unit vector. The unit vector along A may be defined as follows: A ˆA = e . (2.3.1) A We may now write ˆA . A = Ae

(2.3.2)

Thus any vector may be represented as a product of its magnitude and a unit vector. A unit vector is used to designate direction. It does not have any physical ˆ. dimensions. We denote a unit vector by a “hat” above the boldface letter, e

2.3.2

Zero Vector A vector of zero magnitude is called a zero vector or a null vector. All null vectors are considered equal to each other without consideration as to direction. Note that a light face zero, 0, is a scalar and boldface zero, 0, is the zero vector.

2.3.3

Vector Addition Let A, B, and C be any vectors. Then there exists a vector A + B, called sum of A and B, such that the following rules hold: (1) A + B = B + A (commutative). (2) (A + B) + C = A + (B + C) (associative). (3) there exists a unique vector, 0, independent of A such that A + 0 = A (existence of zero vector).

(2.3.3)

(4) to every vector A there exists a unique vector −A (that depends on A) such that A + (−A) = 0 (existence of negative vector). The addition of two vectors is shown in Fig. 2.2(a). Note that the commutative property is essential for a nonscalar to qualify as a vector. The combination of the two diagrams in Fig. 2.2(a) gives the parallelogram shown in Fig. 2.2(b), and it characterizes the commutativity. Thus, we say that vectors add according to the parallelogram law of addition. The negative vector −A has the same magnitude as A, but has the opposite sense. Subtraction of vectors is carried out along the same lines. To form the difference A − B, we write A − B = A + (−B) and subtraction reduces to the operation of addition.

(2.3.4)

16

VECTORS AND TENSORS

B A A+B

B A

A

=

+

= B+A

B

B+A A B

A B (a)

(b)

Fig. 2.2 (a) Addition of vectors. (b) Parallelogram law of addition

As an example of a nonscalar that has magnitude and direction but does not obey the commutativity, consider finite rotation. Finite rotation has a magnitude θ and preferred direction as that in which a right-handed screw would advance when turned in the direction of rotation, as indicated in Fig. 2.3(a). Now consider two different rotations of a rectangular block, in certain order. The first rotation is about the z-axis by θz = +90◦ , followed by rotation θy = −90◦ about the y-axis. This sequence of rotations results in the final position indicated in Fig. 2.3(b). We may represent this pair of rotations as R1 + R2 . Now reversing the order of rotations, i.e., θy first and θz next, we obtain R2 + R1 [see Fig. 2.3(c)], which is not the same as the position achieved by R1 + R2 . Thus, a finite rotation is not a vector, although it has a direction and magnitude.

2.3.4

Multiplication of Vector by Scalar Let A and B be vectors and α and β be real numbers (scalars). To every vector A and every real number α, there corresponds a unique vector αA such that the following conditions hold: (1) α(βA) = (αβ)A (associative). (2) (α + β)A = αA + βA (distributive scalar addition). (3) α(A + B) = αA + αB (distributive vector addition).

(2.3.5)

(4) 1 · A = A · 1 = A, 0 · A = 0. A vector A and its scalar multiples αA for α > 1 and α < 1 are shown in Fig. 2.4.

Fig. 2.3 Vector Algebra

17

Defined direction (thumb) (with double arrow) Sense of rotation =z

(a)

b

b

b a

a z

a

y

x

a

a b

a b b Rotation 1 +90 deg about the z-axis

(b)

b a

a

a b

b

a

Rotation 1 90 deg about the y-axis

b

b

Final position

b

a

a

b

Rotation 2 90 deg about the y-axis a Final position a b a b b a Rotation 2 +90 deg about the z-axis

(c)

Fig. 2.3 (a) Preferred sense of rotation. (b) Rotation θz followed by rotation θy . (c) Rotation θy followed by rotation θz

A

α A (α > 1)

α A (α < 1)

Fig. 2.4 A typical vector A and its scalar multiple

Two vectors A and B are equal if their magnitudes are equal, |A| = |B|, and if their directions are equal. Consequently, a vector is not changed if it is moved parallel to itself. This means that the position of a vector in space, that is, the point from which the line segment is drawn (or the end without arrowhead), may be chosen arbitrarily. In certain applications, however, the actual point of location of a vector may be important, for instance, a moment or a force acting on a body. A vector associated with a given point is known as a localized or bound vector.

18

VECTORS AND TENSORS

Two vectors A and B are said to be linearly dependent if they are a scalar multiple of each other, that is, c1 A + c2 B = 0 for some nonzero scalars c1 and c2 . If two vectors are linearly dependent, then they are collinear (i.e., parallel). If three vectors A, B, and C are linearly dependent, then they are coplanar (i.e., in the same plane). A set of n vectors, {A1 , A2 , . . . , An }, is said to be linearly dependent if a set of n numbers c1 , c2 , . . . , cn can be found such that c1 A1 + c2 A2 + · · · + cn An = 0, where c1 , c2 , . . . , cn cannot all be zero. If this expression cannot be satisfied (i.e., all ci are zero), the set of vectors {A1 , A2 , . . . , An } is said to be linearly independent.

2.3.5

Scalar Product of Vectors When a force F acts on a mass point and moves through a displacement vector d [see Fig. 2.5(a)], the work done by the force vector is defined by the projection of the force in the direction of the displacement, as shown in Fig. 2.5(b), times the magnitude of the displacement. Such an operation may be defined for any two vectors. Since the result of the product is a scalar, it is called the scalar product. We denote this product as F · d ≡ (F, d) and it is defined as follows: F · d ≡ (F, d) = F d cos θ,

0 ≤ θ ≤ π,

(2.3.6)

where θ is the angle (measured counterclockwise) between vectors F and d. The scalar product is also known as the dot product or inner product. F = F eˆ F

F = F eˆ F

θ d = d eˆ d

(a) eˆ F ⋅ eˆ d = cos q, eˆ d ⋅ eˆ d = 1, eˆ F ⋅ eˆ F = 1

F

θ 

Projection of vector F on to vector d: F cos q eˆ d

   d=d eˆ d

(b)

Fig. 2.5 (a) Representation of work done. (b) Projection of a vector

A few simple results follow from the definition in Eq. (2.3.6), and they are listed next:

Vector Algebra

(1) (2)

(3)

Since A · B = B · A, the scalar product is commutative. If the vectors A and B are perpendicular to each other, then A · B = AB cos(π/2) = 0. Conversely, if A · B = 0, then either A or B is zero or A is perpendicular, or orthogonal, to B. If two vectors A and B are parallel and in the same direction, then A · B = AB cos 0 = AB, since cos 0 = 1. Thus the scalar product of a vector with itself is equal to the square of its magnitude: A · A = AA = A2 .

(4) (5)

(2.3.7)

ˆ is given by The orthogonal projection of a vector A in any direction e ˆ)ˆ (A · e e. The scalar product follows the distributive law also: A·(B + C) = (A · B) + (A · C).

2.3.6

19

(2.3.8)

Vector Product Consider the concept of moment due to a force. Let us describe the moment about a point O of a force F acting at a point P, such as shown in Fig. 2.6(a). By definition, the magnitude of the moment is given by M = F `,

F = |F|,

(2.3.9)

where ` is the perpendicular distance from the point O to the force F (called lever arm). If r denotes the vector OP and θ the angle between r and F as shown in Fig. 2.6(a) such that 0 ≤ θ ≤ π, we have ` = r sin θ and, thus M = F r sin θ.

(2.3.10)

A direction can now be assigned to the moment. Drawing the vectors F and r from the common origin O, we note that the rotation due to F tends to bring r into F, as can be seen from Fig. 2.6(b). We now set up an axis of rotation perpendicular to the plane formed by F and r. Along this axis of rotation we set up a preferred direction as that in which a right-handed screw would advance when turned in the direction of rotation due to the moment, as can be seen from ˆM and agree that it Fig. 2.7. Along this axis of rotation we draw a unit vector e represents the direction of the moment M. Thus, we have ˆM = F r sin θ e ˆM M = Me = r × F.

(2.3.11) (2.3.12)

According to this expression, M may be looked upon as a vector resulting from a special operation between the two vectors F and r. It is thus the basis for defining a product between any two vectors. Since the result of such a product is a vector, it is called the vector product.

20

VECTORS AND TENSORS

r×F

F

+

r

O

F θ

θ

P



O

r

(a)

(b)

Fig. 2.6 (a) Representation of a moment. (b) Direction of rotation

F M

êM

θ

r Fig. 2.7 Axis of rotation

The product of two vectors A and B is a vector C whose magnitude is equal to the product of the magnitudes of vectors A and B times the sine of the angle measured from A to B such that 0 ≤ θ ≤ π. The direction of vector C is specified by the condition that C be perpendicular to the plane of the vectors A and B and points in the direction in which a right-handed screw advances when turned so as to bring A into B, as shown in Fig. 2.8. The vector product is usually denoted by

Fig. 2.8

ˆ = AB sin θ e ˆ, C = A × B = AB sin(A, B) e

(2.3.13)

B sin θ sine of the angle between vectors A and B. This where sin(A,B) denotes the A×B

C = A ´B

B

ê

C = A ´B

B sinθ

θ

ê

A sinθ

A×B

B A A× sin A×B=−B Aθ

θ

B

A

θ B

θ −B×A

A

A

A×B=−B×A B×A=−A×B Fig. 2.8 Representation of the vector product

Vector Algebra

21

ˆA and product is called the cross product or vector product. When A = A e ˆB are the vectors representing the sides of a parallelogram, with A and B = Be B denoting the lengths of the sides, then the magnitude of the vector product ˆ= A × B represents the area of the parallelogram, AB sin θ. The unit vector e ˆA × e ˆB denotes the normal to the plane area. Thus, an area can be represented e as a vector (see Section 2.3.8 for additional discussion). The description of the velocity of a point of a rotating rigid body is an important example of geometrical and physical applications of vectors. Suppose a rigid body is rotating with an angular velocity ω about an axis, and we wish to describe the velocity of some point P of the body, as shown in Fig. 2.9(a). Let v denote the velocity at the point. Each point of the body describes a circle that lies in a plane perpendicular to the axis with its center on the axis. The radius of the circle, a, is the perpendicular distance from the axis to the point of interest. The magnitude of the velocity is equal to ωa. The direction of v is perpendicular to a and to the axis of rotation. We denote the direction of the velocity by the unit vector ˆ e. Thus we can write ˆ. v = ωa e

(2.3.14)

Let O be a reference point on the axis of revolution, and let OP = r. We then have a = r sinθ, so that ˆ. v = ω r sin θ e

(2.3.15)

The angular velocity is a vector since it has an assigned direction and magnitude, and obeys the parallelogram law of addition. We denote it by ω and represent its direction in the sense of a right-handed screw, as shown in Fig. 2.9(b). If we ˆr be a unit vector in the direction of r, we see that further let e

Fig. 2.9

ˆω × e ˆr = e ˆ sin θ. e

ω

(2.3.16)

eˆ w

ω

v

●

O●

θ

r

ω

a ●

P eˆ (a)

v

ω r (b)

Fig. 2.9 (a) Velocity at a point in a rotating rigid body. (b) Angular velocity as a vector

22

VECTORS AND TENSORS

With these relations we have v = ω × r.

(2.3.17)

Thus the velocity of a point of a rigid body rotating about an axis is given by the vector product of ω and a position vector r drawn from any reference point on the axis of revolution. From the definition of vector product the following few simple results follow: (1)

The products A × B and B × A are not equal. In fact, we have A × B ≡ −B × A.

(2)

(2.3.18)

Thus the vector product does not commute. We must therefore preserve the order of the vectors when vector products are involved. If two vectors A and B are parallel to each other, then θ = π or 0 and sin θ = 0. Thus A × B = 0.

(3)

Conversely, if A × B = 0, then either A or B is zero, or they are parallel vectors. It follows that the vector product of a vector with itself is zero; that is, A × A = 0. The distributive law still holds, but the order of the factors must be maintained: (A + B) × C = (A × C) + (B × C).

2.3.7

(2.3.19)

Triple Products of Vectors Now consider the various products of three vectors: A(B · C),

A · (B × C),

A × (B × C).

(2.3.20)

The product A(B · C) is merely a multiplication of the vector A by the scalar B · C. On the other hand, the product A · (B × C) is a scalar and it is termed the scalar triple product. It can be seen that the product A · (B × C), except for the algebraic sign, is the volume of the parallelepiped formed by the vectors A, B, and C, as shown in Fig. 2.10. We also note the following properties: (1)

The dot and cross can be interchanged without changing the value: A · B × C = A × B · C ≡ [ABC].

(2)

(2.3.21)

A cyclical permutation of the order of the vectors leaves the result unchanged: A · B × C = C · A × B = B · C × A ≡ [ABC].

(2.3.22)

Vector Algebra

(3)

If the cyclic order is changed, the sign changes: A · B × C = −A · C × B = −C · B × A = −B · A × C.

(4)

23

(2.3.23)

A necessary and sufficient condition for any three vectors, A, B, C to be coplanar is that A · (B × C) = 0. Note also that the scalar triple product is zero when any two vectors are the same. B×C A C B Fig. 2.10 Scalar triple product A · (B × C) as the volume of a parallelepiped

The vector triple product A × (B × C) is a vector normal to the plane formed by A and (B × C). The vector (B × C), however, is perpendicular to the plane formed by B and C. This means that A × (B × C) lies in the plane formed by B and C and is perpendicular to A, as shown in Fig. 2.11. Thus A × (B × C) can be expressed as a linear combination of B and C: A × (B × C) = c1 B + c2 C.

(2.3.24)

Likewise, we would find that (A × B) × C = d1 A + d2 B.

(2.3.25)

Thus, the parentheses cannot be interchanged or removed. It can be shown that c1 = A · C, and hence that Fig. 2.11

B×C

c2 = −A · B,

A × (B × C) = (A · C)B − (A · B)C.

A C

c2C c1B

B A × (B × C), perpendicular to both A and B × C Fig. 2.11 The vector triple product

(2.3.26)

24

VECTORS AND TENSORS

Example 2.3.1 Let A and B be any two vectors in space. Express vector A in terms of its components along (i.e., parallel) and perpendicular to vector B. Solution: The component of A along B is given by (A · ˆ eB ), where ˆ eB = B/B is the unit vector in the direction of B. The component of A perpendicular to B and in the plane of A and B is given by the vector triple product ˆ eB × (A × ˆ eB ). Thus, A = (A · ˆ eB )ˆ eB + ˆ eB × (A × ˆ eB ). Alternatively, using Eq. (2.3.26) with A = C = ˆ eB and B = A, we obtain ˆ eB × (A × ˆ eB ) = A − (ˆ eB · A)ˆ eB or A = (A · ˆ eB )ˆ eB + ˆ eB × (A × ˆ eB ).

(2.3.27)

Example 2.3.2 Find the equation of a plane connecting the terminal points of vectors A, B, and C. Assume that all three vectors are referred to a common origin. Solution: Let r denote the position vector. The vectors connecting the terminal points of vectors A, B, C, and r should be in the plane. Thus, for example, the scalar triple product of the vectors B − A, C − A, and r − A should be zero in order that they are co-planar, as shown in Fig. 2.12:

Fig. 2.12

(C − A) × (B − A) · (r − A) = 0.

ˆ1 , B = e ˆ2 , and C = e ˆ3 , then the equation of the plane is For example, if A = e −x − y − z + 1 = 0 or x + y + z = 1.

B−A

z

•

A •

y

•

B r C

r−A C−A

•

x Fig. 2.12 Equation of a plane connecting the terminal points of vectors A, B, and C

2.3.8

Plane Area as a Vector As indicated previously, the magnitude of the vector C = A × B is equal to the area of the parallelogram formed by the vectors A and B, as shown in Fig. 2.13(a). In fact, the vector C may be considered to represent both the magnitude and the direction of the product A and B. Thus, a plane area may be

Vector Algebra

25

looked upon as possessing a direction in addition to a magnitude, the directional character arising out of the need to specify an orientation of the plane in space. It is customary to denote the direction of a plane area by means of a unit vector drawn normal to that plane. To fix the direction of the normal, we assign Fig. 2.13 a sense of travel along the contour of the boundary of the plane area in question. The direction of the normal is taken by convention as that in which a righthanded screw advances as it is rotated according to the sense of travel along the boundary curve or contour, as shown in Fig. 2.13(b). Let the unit normal vector ˆ . Then the area can be denoted by S = S n ˆ. be given by n S = Snˆ

B

C = A×B eˆ

nˆ

θ

S

A (a)

(b)

Fig. 2.13 (a) Plane area as a vector. (b) Unit normal vector and sense of travel

Representation of a plane as a vector has many uses. The vector can be used to determine the area of an inclined plane in terms of its projected area, as illustrated in the next example. Example 2.3.3 (a) (b)

Determine the plane area of the surface obtained by cutting a cylinder of ˆ , as shown in cross-sectional area S0 with an inclined plane whose normal is n Fig. 2.14(a). Express the areas of the sides of the tetrahedron obtained from a cube (or a ˆ , as shown in Fig. 2.14(b), in prism) cut by an inclined plane whose normal is n terms of the area S of the inclined surface. x3

nˆ nˆ

ˆ0 n θ

S

ˆ0 n

S2 S3

nˆ 3

(a)

x2 S

S1

nˆ 2 S0

nˆ

nˆ 1

x1

(b)

Fig. 2.14 Vector representation of an inclined plane area

26

VECTORS AND TENSORS

Solution: (a) Let the plane area of the inclined surface be S, as shown in Fig. 2.14(a). First, we express the areas as vectors S0 = S0 n ˆ0

and

S=Sn ˆ.

ˆ 0 (if the angle between n ˆ and n ˆ 0 is acute; Since S0 is the projection of S along n otherwise the negative of it), ˆ0 = Sn ˆ ·n ˆ0. S0 = S · n

(2.3.28)

ˆ ·n ˆ 0 is the cosine of the angle between the two unit normal vectors. The scalar product n (b) For reference purposes we label the sides of the cube by 1, 2, and 3 and the normals and surface areas by (ˆ n1 , S1 ), (ˆ n2 , S2 ), and(ˆ n3 , S3 ), respectively (i.e., Si is the surface area of the plane perpendicular to the ith line or n ˆ i vector), as shown in Fig. 2.14(b). Then we have ˆ 3 = −ˆ ˆ 1 = −ˆ ˆ 2 = −ˆ n e1 , n e2 , n e3 , (2.3.29) ˆ ·e ˆ1 = Sn1 , S1 = S n

ˆ ·e ˆ2 = Sn2 , S2 = S n

ˆ ·e ˆ3 = Sn3 , S3 = S n

(2.3.30)

ˆ. where ni are the direction cosines of the unit normal n

2.3.9

Components of a Vector So far we have considered a geometrical description of a vector as a directed line segment. We now embark on an analytical description of a vector and some of the operations associated with this description. The analytical description of vectors is useful in expressing, for example, the laws of physics in analytical form. The analytical description of a vector is based on the notion of its components. In a three-dimensional space, a set of no more than three linearly independent vectors can be found. Let us choose any set and denote it as follows: e1 , e2 , e3 .

(2.3.31)

This set is called a basis (or a base system). A basis is called orthonormal if the basis vectors are mutually orthogonal and have unit magnitudes. To distinguish the basis (e1 , e2 , e3 ) that is not orthonormal from one that is orthonormal, we ˆ2 , e ˆ3 ), where denote the orthonormal basis by (ˆ e1 , e ˆ1 · e ˆ2 = 0, e ˆ2 · e ˆ3 = 0, e ˆ3 · e ˆ1 = 0, e ˆ1 · e ˆ1 = 1, e ˆ2 · e ˆ2 = 1, e ˆ3 · e ˆ3 = 1. e

(2.3.32)

ˆ or (ˆ ˆy , e ˆz ) is used in place of (ˆ ˆ2 , e ˆ3 ). In some books the notation (ˆi, ˆj, k) ex , e e1 , e In view of the previous discussion of the cross product of vectors, we note the following relations resulting from the cross products of the basis vectors: ˆ1 × e ˆ1 = 0, e

ˆ1 × e ˆ2 = e ˆ3 , e

ˆ2 × e ˆ1 = −ˆ ˆ2 × e ˆ2 = 0, e e3 , e ˆ3 × e ˆ1 = e ˆ2 , e

ˆ1 × e ˆ3 = −ˆ e e2 , ˆ2 × e ˆ3 = e ˆ1 , e

ˆ3 × e ˆ2 = −ˆ ˆ3 × e ˆ3 = 0. e e1 , e

(2.3.33)

Vector Algebra

27

It is clear from the concept of linear dependence that we can represent any vector in three-dimensional space as a linear combination of the basis vectors: ˆ x + Ay e ˆ y + Az e ˆ z = A1 e ˆ 1 + A2 e ˆ 2 + A3 e ˆ3 . A = Ax e

(2.3.34)

ˆ 1 , A2 e ˆ2 , and A3 e ˆ3 are called the vector components of A, and The vectors A1 e A1 , A2 , and A3 are called scalar components of A associated with the basis ˆ2 , e ˆ3 ), as indicated in Fig. 2.15. (ˆ e1 , e A 3 e3 e3 A A 1e

1

e2

e1

A3eˆ 3

A 2 e2

eˆ 3 A1eˆ 1

A

eˆ 2

eˆ 1

(a)

A2eˆ 2

(b)

Fig. 2.15 Components of a vector in (a) a general coordinate system that is oblique, and (b) the rectangular Cartesian system

When the basis is orthonormal, A1 , A2 , and A3 are the physical components of the vector A; that is, the components have the same physical dimensions or units as the vector. A scalar multiple of a vector is the same as the vector whose components are the scalar multiples: αA = (αA1 )ˆ e1 + (αA2 )ˆ e2 + (αA3 )ˆ e3 .

(2.3.35)

Two vectors are equal if and only if their respective components are equal. That is, A=B implies that A1 = B1 , A2 = B2 , and A3 = B3 . The operations of vector addition, scalar product, and vector product of vectors can be expressed in terms of components, as discussed next. Addition of Vectors. The sum of vectors A and B is the vector C whose components are the sum of the respective components of vectors A and B:

A + B = A1 ˆ e1 + A2ˆ e2 + A3ˆ e3 + B1ˆ e1 + B2ˆ e2 + B3ˆ e3 = (A1 + B1 )ˆ e1 + (A2 + B2 )ˆ e2 + (A3 + B3 )ˆ e3 ≡ C1 ˆ e1 + C2ˆ e2 + C3ˆ e3 = C,

(2.3.36)

with C1 = A1 + B1 , C2 = A2 + B2 , and C3 = A3 + B3 . Scalar Product of Vectors. The scalar product of vectors A and B is the scalar

A · B = A1 ˆ e1 + A2ˆ e2 + A3ˆ e3 · B1ˆ e1 + B2ˆ e2 + B3ˆ e3 = A1 B 1 + A2 B 2 + A3 B 3 ,

(2.3.37)

where the orthonormal property (2.3.32) of the basis vectors is used in arriving at the last expression.

28

VECTORS AND TENSORS

Vector Product of Vectors. The vector product of vectors A and B is the vector

A × B = A1 ˆ e1 + A2ˆ e2 + A3ˆ e3 × B1ˆ e1 + B2ˆ e2 + B3ˆ e3 = (A2 B3 − A3 B2 )ˆ e1 + (A3 B1 − A1 B3 )ˆ e2 + (A1 B2 − A2 B1 )ˆ e3 , (2.3.38) where the relations in Eq. (2.3.33) are used in arriving at the final expression. Example 2.3.4 The velocity at a point in a flow field is v = 2ˆi + 3ˆj (m/s). Determine: (a) (b) (c) (d)

the the the the the

ˆ at the point, velocity vector vn normal to the plane n = 3ˆi − 4k angle between v and vn , tangential velocity vector vt on the plane, and mass flow rate across the plane through area A = 0.15 m2 if the density of fluid (water) is ρ = 103 kg/m3 and the flow is uniform.

Solution: (a) The unit vector normal to the plane is given by
ˆ . ˆ = 15 3ˆi − 4k n Then the velocity vector normal to the plane, vn , is equal to the projection of the velocity vector, v, along the normal to the plane [see Fig. 2.16(a)],
ˆ m/s. ˆ )ˆ ˆ = 0.24 3ˆi − 4k vn = (v · n n = 65 n (b) The angle between the vectors v and vn is given by       v · vn vn 1.2 θ = cos−1 = cos−1 = cos−1 √ = 70.6◦ . |v||vn | v 13 (c) The tangential velocity vector on the plane is given by
ˆ = 1.28ˆi + 3ˆj + 0.96k. ˆ vt = v − vn = 2ˆi + 3ˆj − 0.24 3ˆi − 4k ˆ) (d) The mass flow rate is given by (vn ≡ v · n Q = ρvn A = 103 × 1.2 × 0.15 = 180 (kg/s). Various vectors are depicted in Fig. 2.16(b).

A = 0.15 m 2

ˆ ˆ v = 2i+3j

vt

y v ˆj kˆ

z

θ iˆ

x

n = 3iˆ - 4kˆ

(a)

Fig. 2.16 Flow across a plane

vn

(b)

Index Notation and Summation Convention

2.4

Index Notation and Summation Convention

2.4.1

Summation Convention

29

Use of index notation facilitates writing long expressions in succinct form. For example, consider the component form of vector A A = A1 e1 + A2 e2 + A3 e3 ,

(2.4.1)

which can be abbreviated as A=

3 X

Ai ei or A =

i=1

3 X

Aj ej .

j=1

If we had chosen the notation A = Ax ex + Ay ey + Az ez , where (Ax , Ay , Az ) are the same as (A1 , A2 , A3 ) and the basis (ex , ey , ez ) is the same as (e1 , e2 , e3 ), it would not have been possible to write it with the summation convention. The summation index i or j is arbitrary as long as the ˆ. The expression can be further shortened by same index is used for both A and e omitting the summation sign and having the understanding that a repeated index means summation over all values of that index. Thus, the three-term expression A1 e1 + A2 e2 + A3 e3 can be simply written as A = Ai ei .

(2.4.2)

This notation is called the summation convention. The summation convention allows us to write several expressions or equations in a single statement. As we shall see shortly, the six relations in Eq. (2.3.32) can be written as a single equation with the help of index notation and certain symbols that we are about to introduce in Section 2.4.4.

2.4.2

Dummy Index The repeated index of any expression is called a dummy index because it can be replaced by any other symbol that has not already been used in that expression. Thus, the expression in Eq. (2.4.2) can also be written as A = Ai ei = Aj ej = Am em ,

(2.4.3)

and so on. As a rule, no index must appear more than twice in an expression. For example, Ai Bi Ci is not a valid expression because the index i appears more than twice. Other examples of dummy indices are Fi = Ai Bj Cj , Gk = Hk (2 − 3Ai Bi ) + Pj Qj Fk .

30

VECTORS AND TENSORS

Each of the above equations expresses three equations when the range of i and j is 1 to 3. The first equation, for example, is equal to the following three equations: F1 = A1 (B1 C1 + B2 C2 + B3 C3 ), F2 = A2 (B1 C1 + B2 C2 + B3 C3 ), F3 = A3 (B1 C1 + B2 C2 + B3 C3 ). This amply illustrates the usefulness of the summation convention in shortening long and multiple expressions into a single expression.

2.4.3

Free Index A free index is one that appears in every expression of an equation, except for expressions that contain real numbers (scalars) only. Index i in the equation Fi = Ai Bj Cj and k in the equation Gk = Hk (2 − 3Ai Bi ) + Pj Qj Fk above are free indices. Another example is Ai = 2 + Bi + Ci + Di + (Fj Gj − Hj Pj )Ei . The above expression contains three equations (i = 1, 2, 3). The expressions Ai = Bj Ck , Ai = Bj , and Fk = Ai Bj Ck do not make sense and should not arise, because the indices on the two sides of the equal sign do not match.

2.4.4

Kronecker Delta and Permutation Symbols It is convenient to introduce the Kronecker delta δij and alternating symbol eijk because they allow easy representation of the scalar product and vector product, respectively, of orthonormal vectors in a right-handed basis system. We define the dot product ˆ ei · ˆ ej as ˆ ei · ˆ ej = δij ,

(2.4.4)

where  δij =

1, if i = j 0, if i 6= j.

(2.4.5)

Thus, the single expression in Eq. (2.4.4) is the same as the six relations in Eq. (2.3.32). The Kronecker delta δij , due to its definition, modifies (or contracts) the subscripts in the coefficients of an expression in which it appears Ai δij = Aj ,

Ai Bj δij = Ai Bi = Aj Bj ,

δij δik = δjk .

As we shall see shortly, δij denote the Cartesian components of a second-order ˆi e ˆj = e ˆi e ˆi . unit tensor, I = δij e We define the cross product ˆ ei × ˆ ej as ˆ ei × ˆ ej ≡ eijk ˆ ek or eijk = ˆ ei × ˆ ej · ˆ ek = ˆ ei · ˆ ej × ˆ ek ,

(2.4.6)

Index Notation and Summation Convention

31

where

eijk

     

1,

if i, j, k are in a natural order and not repeated (i 6= j 6= k), = −1, if i, j, k are not in a natural order    and not repeated (i 6= j 6= k),   0, if any of i, j, k are repeated.

(2.4.7)

The symbol eijk is called the alternating symbol or permutation symbol. The natural order of i, j, and k is the order in which they appear alphabetically. A natural cyclic order means going from i to j and k, from j to k and i, or from k to i and j, as shown in Fig. 2.17(a). Going opposite to a natural cyclic order is shown in Fig. 2.17(b). By definition, the subscripts of the permutation symbol can be permuted in a natural cyclic order, without changing its value. An interchange of any two subscripts will change the sign (hence, interchange of two subscripts twice keeps the value unchanged): eijk = ekij = ejki ,

eijk = −ejik = ejki = −ekji ,

e123 = e312 = e231 = 1,

e213 = e321 = e132 = −1.

A formula that gives the correct value of eijk for any fixed values of i, j, and k is eijk = 12 (i − j)(j − k)(k − i) for any i, j, k = 1, 2, 3. ˆ2 , e ˆ3 ), the scalar and vector products can be In an orthonormal basis (ˆ e1 , e expressed in the index notation using the Kronecker delta and the alternating symbols: A · B = (Aiˆ ei ) · (Bj ˆ ej ) = Ai Bj δij = Ai Bi ,

(2.4.8)

A × B = (Aiˆ ei ) × (Bj ˆ ej ) = Ai Bj eijk ˆ ek .

Note that the components of a vector in an orthonormal coordinate system can be expressed as ˆi , Ai = A · e

j

k

i•

+

i•

•j (a)

+

j

k

•

•

(2.4.9)

•

•

•k

i•

-

i•

•j

-

•k

(b)

Fig. 2.17 (a) A natural cyclic order is going from any index to the next in the order it appears alphabetically (going from k to i makes it cyclic). (b) Opposite to a natural cyclic order is going in a direction opposite to that of a natural cyclic order

32

VECTORS AND TENSORS

and therefore we can express vector A as ˆi = (A · e ˆi )ˆ A = Ai e ei .

(2.4.10)

Further, the Kronecker delta and the permutation symbol are related by the identity, known as the e-δ identity (see Example 2.4.2), eijk eimn = δjm δkn − δjn δkm .

(2.4.11)

The permutation symbol and the Kronecker delta prove to be very useful in proving vector identities. Since a vector form of any identity is invariant (i.e., valid in any coordinate system), it suffices to prove it in one coordinate system. In particular, an orthonormal system is very convenient because we can use the index notation, permutation symbol and the Kronecker delta. The following examples contain several cases of incorrect and correct use of index notation and illustrate some of the uses of δij and eijk . Example 2.4.1 Discuss the validity of the following expressions: (1) am bs = cm (dr − fr ), (2) am bs = cm (ds − fs ), (3) ai = bj ci di , (4) xi xi = r2 , (5) ai = 3. Solution: (1) (2) (3)

(4) (5)

Not a valid expression because the free indices r and s do not match. Valid; both m and s are free indices. There are nine equations (m, s = 1, 2, 3). Not a valid expression because the free index j is not matched on both sides of the equality, and index i is a dummy index in one expression and a free index in the other; i cannot be used both as a free and dummy index in the same equation. The equation would have been valid if i on the left side of the equation is replaced with j; then there will be three equations). A valid expression, containing one equation: x21 + x22 + x23 = r2 . It is a valid expression in some branches of mathematics but it is not a valid expression in continuum mechanics because it violates form-invariance (material frame indifference) under a basis transformation (every component of a vector cannot be the same in all bases).

Example 2.4.2 Simplify the following expressions: (1) δij δjk δkp δpi , (2) εmjk εnjk , (3) (A × B) · (C × D). Solution: (1)

Successive contraction of subscripts yields the result: δij δjk δkp δpi = δij δjk δki = δij δji = δii = 3.

Index Notation and Summation Convention

(2)

33

Expand using the e-δ identity εmjk εnjk = δmn δjj − δmj δnj = 3δmn − δmn = 2δmn .

(3)

In particular, the expression εijk εijk is equal to 2δii = 6. Expanding the expression using the index notation, we obtain (A × B) · (C × D) = (Ai Bj eijk ˆ ek ) · (Cm Dn emnp ˆ ep ) = Ai Bj Cm Dn eijk emnp δkp = Ai Bj Cm Dn eijk emnk = Ai Bj Cm Dn (δim δjn − δin δjm ) = Ai Bj Cm Dn δim δjn − Ai Bj Cm Dn δin δjm = Ai Bj Ci Dj − Ai Bj Cj Di = Ai Ci Bj Dj − Ai Di Bj Cj = (A · C)(B · D) − (A · D)(B · C), where we have used the e-δ identity (2.4.11). Although the previous vector identity is established in an orthonormal coordinate system, it holds in a general coordinate system. That is, the above vector identity is invariant.

Example 2.4.3 ˆj in vector form. Rewrite the expression emni Ai Bj Cm Dn e ˆj = B. Examining the indices in the permutation symbol Solution: We note that Bj e and the remaining coefficients, it is clear that vectors C and D must have a cross product between them and the resulting vector must have a dot product with vector A. Thus we have ˆj = [(C × D) · A]B = (C × D · A) B. emni Ai Bj Cm Dn e

2.4.5

Transformation Law for Different Bases When the basis vectors are constant, that is, with fixed lengths (with the same units) and directions, the basis is called Cartesian. The general Cartesian system is oblique. When the basis vectors are unit and orthogonal (orthonormal), the basis system is called rectangular Cartesian, or simply Cartesian. In much of our study, we shall deal with Cartesian bases. Let us denote an orthonormal Cartesian basis by ˆy , e ˆz } {ˆ ex , e

or

ˆ2 , e ˆ3 }. {ˆ e1 , e

The Cartesian coordinates are denoted by (x, y, z) or (x1 , x2 , x3 ). The familiar rectangular Cartesian coordinate system is shown in Fig. 2.18(a). We shall always use right-handed coordinate systems. A position vector to an arbitrary point (x, y, z) or (x1 , x2 , x3 ), measured from the origin, is given by ˆ 1 + x2 e ˆ 2 + x3 e ˆ3 , r = xˆ ex + yˆ ey + zˆ ez = x1 e

(2.4.12)

34

VECTORS AND TENSORS

or, in summation notation, by ˆj , r = xj e

r · r = r2 = xi xi .

(2.4.13)

We shall also use the symbol x for the position vector r = x. The square of the length of a line element dr = dx is given by dr · dr = (ds)2 = dxj dxj = (dx1 )2 + (dx2 )2 + (dx3 )2 = (dx)2 + (dy)2 + (dz)2 .

(2.4.14)

Next we discuss the relationship between the components of two different orthonormal coordinate systems (i.e., coordinate transformations). Consider the ˆ¯1 , e ˆ¯2 , e ˆ¯3 }. ˆ2 , e ˆ3 } and the second coordinate basis {e first coordinate basis {ˆ e1 , e We can express the same vector in the coordinate system without bars (referred to as “unbarred”) and also in the coordinate system with bars (referred to as “barred”): ˆi = (A · e ˆi )ˆ A = Ai e ei ¯ ˆ ˆ ˆ¯i . ¯ ¯ = Ai ei = (A · ei )e

(2.4.15)

ˆ ˆ¯j ) ≡ `ji Ai , ¯j = Ai (ˆ A¯j = A · e ei · e

(2.4.16)

From Eq. (2.4.10) we have

where `ij are called the direction cosines and they are defined as ˆ¯i · e ˆj . `ij = e

(2.4.17) Equation (2.4.16) gives the relationship between the components (A¯1 , A¯2 , A¯3 ) and (A1 , A2 , A3 ), and it is called the transformation rule between the barred and unbarred components in the two coordinate systems. The coefficients `ij Fig. 2.18 are the direction cosines of the barred coordinate system with respect to the unbarred coordinate system [see Fig. 2.18(b)]: ˆ¯i and e ˆj . `ij = cosine of the angle between e z = x3

x3

.

(x, y,z ) = (x1 ,x 2 ,x3 ) ˆ z =e ˆ3 x = r e ˆ x =e ˆ1 e

ˆy =e ˆ2 e

x2

x3

(a)

x2 b

x3

x1

a

y = x2 x1

x = x1

(2.4.18)

l11 = cos q l31 = cos a l23 = cos b

q

x1 (b)

Fig. 2.18 Rectangular Cartesian coordinates

x2

Theory of Matrices

35

Note that the first subscript of `ij comes from the barred coordinate system and the second subscript from the unbarred system. Obviously, `ij is not symmetric (i.e., `ij 6= `ji ). The rectangular array of these components is called a matrix, which is the topic of the next section. Example 2.4.4 illustrates the computation of direction cosines, `ij . Example 2.4.4 ˆi (i = 1, 2, 3) be a set of orthonormal base vectors, and define a new right-handed Let e ˆ ˆ ¯1 · e ¯2 = 0) coordinate basis by (note that e ˆ ¯1 = 13 (2ˆ ˆ3 ) , e e1 + 2ˆ e2 + e ˆ ¯3 = ˆ e ¯ e1 × ˆ ¯ e2 =

1 √ 3 2

√1 2

ˆ ¯2 = e

ˆ2 ) , (ˆ e1 − e

(ˆ e1 + ˆ e2 − 4ˆ e3 ) .

Determine the direction cosines `ij of the transformation. ˆ ˆ ˆ ¯1 , e ¯2 , e ¯3 ) is a linearly independent set. We find that Solution: It can be verified that (e the linear relation ˆ ˆ ˆ ¯1 + α2 e ¯1 + α3 e ¯1 = 0 α1 e ˆi in the previous is trivial because the relations (obtained by setting the coefficients of e relation to zero), 2 α 3 1

+

√1 α2 2

+

1 √ α 3 2 3

= 0,

2 α 3 1

−

1 √ α 2 2

+

3

1 √

2

α3 = 0,

1 α 3 1

−

4 √ α 3 2 3

= 0,

yield α1 = α2 = α3 = 0. From Eq. (2.4.17) we have the following values of the direction cosines: ˆ ¯1 · e ˆ1 = 23 , `11 = e ˆ ¯2 · e ˆ1 = √12 , `21 = e ˆ ¯3 · e ˆ1 = `31 = e

ˆ ¯1 · e ˆ2 = 23 , `12 = e ˆ ¯2 · e ˆ2 = − √12 , `22 = e

1 √ , 3 2

2.5

Theory of Matrices

2.5.1

Definition of a Matrix

ˆ ¯3 · e ˆ2 = `32 = e

3

1 √

2

,

ˆ ¯1 · e ˆ3 = 13 , `13 = e ˆ ¯2 · e ˆ3 = 0, `23 = e 4 ˆ ¯3 · e ˆ 3 = − 3√ `33 = e . 2

In the preceding sections we studied the algebra of ordinary vectors and the transformation of vector components from one coordinate system to another. For example, the transformation equation (2.4.16) relates the components (A¯1 , A¯2 , A¯3 ) in the barred coordinate system (¯ x1 , x ¯2 , x ¯3 ) to components (A1 , A2 , A3 ) in the unbarred coordinate system (x1 , x2 , x3 ) of a vector A. Writing Eq. (2.4.16) in an expanded form, A¯1 = `11 A1 + `12 A2 + `13 A3 A¯2 = `21 A1 + `22 A2 + `23 A3 A¯3 = `31 A1 + `32 A2 + `33 A3 ,

(2.5.1)

36

VECTORS AND TENSORS

we see that there are nine coefficients relating the components Ai to A¯i . The form of these linear equations suggests writing down the scalars of `ij (coefficient in the jth column and the ith row) in the rectangular array   `11 `12 `13 [L] =  `21 `22 `23  . `31 `32 `33 This rectangular array [L] of scalars `ij is called a matrix, and the quantities `ij are called the elements of the matrix [L]. Thus, matrices can be viewed as transformations of one set of variables to another set. The matrix representations of linear equations and components of vectors and tensors are convenient and facilitate computations, as will be seen later. Often boldface letters are used to denote matrices, that is L ≡ [L]. The word “matrix” was first used in 1850 by James Sylvester (1814–1897), an English algebraist. However, Arthur Caley (1821–1895), professor of mathematics at Cambridge, was the first one to explore properties of matrices. Significant contributions in the early years were made by Charles Hermite, Georg Frobenius, and Camille Jordan, among others. If a matrix has m rows and n columns, we will say that it is an m by n (m × n) matrix, the number of rows always being listed first. The element in the ith row and jth column of a matrix [A] is generally denoted by aij (or Aij ), and we will designate a matrix by A = [A] = [aij ]. A square matrix is one that has the same number of rows as columns. An n × n matrix is said to be of order n. The elements of a square matrix for which the row number and the column number are the same (i.e., aii for any fixed i) are called diagonal elements or simply the diagonal. A square matrix is said to be a diagonal matrix if all of the off-diagonal elements are zero. The sum of the diagonal elements of any square matrix is called the trace. An identity (or unit) matrix, denoted by I = [I] = [δij ], is a diagonal matrix whose elements are all 1s. Examples of a diagonal and an identity matrix are given below:     5 0 0 0 1 0 0 0  0 −2 0 0   0 1 0 0   ,  I=  0   0 0 1 0 . 0 1 0 0 0 0 3 0 0 0 1 If the matrix has only one row or one column, we will normally use only a single subscript to designate its elements. For example,    x1  X= , Y = {y1 y2 y3 }, x  2  x3 denote a column matrix and a row matrix, respectively. Row and column matrices can be used to denote the components of a vector. One may use a row or column to display the components of a vector, dictated by the convenience.

Theory of Matrices

2.5.2

37

Matrix Addition and Multiplication of a Matrix by a Scalar The sum of two matrices of the same size is defined to be a matrix of the same size obtained by simply adding the corresponding elements. If A is an m × n matrix and B is an m × n matrix, their sum is an m × n matrix, C, with cij = aij + bij

for all i, j.

(2.5.2)

A constant multiple of a matrix is equal to the matrix obtained by multiplying all of the elements by the constant. That is, the multiple of a matrix A by a scalar α, αA, is the matrix obtained by multiplying each of its elements with α:     a11 a12 . . . a1n αa11 αa12 . . . αa1n  a21 a22 . . . a2n   αa21 αa22 . . . αa2n      A= .. .. ..  , αA =  .. .. ..  .   . . ... .  . . ... .  am1

am2

...

amn

αam1

αam2

...

αamn

Matrix addition has the following properties: (1) (2) (3) (4) (5) (6)

Addition is commutative: A + B = B + A. Addition is associative: A + (B + C) = (A + B) + C. There exists a unique matrix 0, such that A + 0 = 0 + A = A. The matrix 0 is called zero matrix, all elements of which are zeros. For each matrix A, there exists a unique matrix −A such that A+(−A) = 0. Addition is distributive with respect to scalar multiplication: α(A + B) = αA + αB. Addition is distributive with respect to matrix multiplication (if the relevant products are defined; see Section 2.5.4): (A + B)C = AC + BC,

2.5.3

C(A + B) = CA + CB.

Symmetric and Skew Symmetric Matrices If A is an m×n matrix, then the n×m matrix obtained by interchanging its rows and columns is called the transpose of A and is denoted by AT . For example, consider the matrices     5 −2 1 3 −1 2 4  8  7 6  , B =  −6 A= (2.5.3) 3 5 7 .  2 4 3  9 6 −2 1 −1 9 0 The transposes of A and B are 

5 T  A = −2 1

8 7 6

  3 2 −1  −1 4 9  , BT =   2 3 0 4

 −6 9 3 6  . 5 −2  7 1

38

VECTORS AND TENSORS

The following two basic properties of a transpose should be noted: (1) (AT )T = A,

(2) (A + B)T = AT + BT .

(2.5.4)

A square matrix A of real numbers is said to be symmetric if AT = A. A square matrix is said to be unsymmetric or non-symmetric when it is not symmetric. A square matrix A is said to be skew-symmetric or antisymmetric if AT = −A. In terms of the elements of A, these definitions imply that A is symmetric if and only if aij = aji , and it is skew-symmetric if and only if aij = −aji . Note that the diagonal elements of a skew-symmetric matrix are always zero since aij = −aij implies aij = 0 for i = j. Examples of symmetric and skew-symmetric matrices, respectively, are     5 −2 12 21 0 −11 32 4  −2  11 2 16 −3  0 25 7   ,  .  12 16 13  −32 −25 8  0 15  21 −3

8

−4

19

−7

−15

0

Every square matrix A can be expressed as a sum of a symmetric matrix and a skew-symmetric matrix:

A = 12 A + AT + 12 A − AT or

(2.5.5) [A] = 12 [A] + [A]T + 12 [A] − [A]T . For example, consider the unsymmetric matrix   4 6 8  2 10 12  , 18 which can be expressed as    4 6 8 4  2 10 12  = 1   2  2 18 14 6 18  4 1 +  2 2 18  4 4 =  4 10 13 13

14

6 10 14 6 10 14

6

  8 12  +  6   8 12  −  6 

 13 0 13  +  −2 6 5

4 2 18

6 10 14

4 2 18

6 10 14 2 0 1

T  8  12   6 T  8  12   6 

−5 −1  . 0

A general square matrix of order n can have n2 independent elements, a skewsymmetric matrix can have only n(n − 1)/2 independent elements, and a symmetric matrix can have only n(n + 1)/2 independent elements. For example, when n = 3, the number of independent elements for general, skew-symmetric, and symmetric matrices become 9, 3, and 6, respectively.

Theory of Matrices

2.5.4

39

Matrix Multiplication ˆ1 +a2 e ˆ2 +a3 e ˆ3 in a Cartesian system. We can represent Consider a vector A = a1 e A as a product of a row matrix with a column matrix,     ˆ1   e  a1  ˆ2 e ˆ3 } ˆ1 + a2 e ˆ2 + a3 e ˆ3 . ˆ A = {a1 a2 a3 } = {ˆ e1 e = a1 e e a  2   2  ˆ3 e a3 Note that the vector A is obtained by multiplying the ith element in the row matrix with the ith element in the column matrix and adding them. This gives us a strong motivation for defining the product of two matrices. Let x and y be the column vectors (i.e., matrices with one column)     x1  y1           x2    y2   x= , y= . . . .  .       .    .       xm ym We define the product xT y      T x y = {x1 , x2 , . . . , xm }    

to be the scalar  y1    m X y2  = x y +x y +· · ·+x y = xi yi . (2.5.6) 1 1 2 2 m m ..  .  i=1   ym

It follows from Eq. (2.5.6) that xT y = yT x. More generally, let A = [aij ] be m × n and B = [bij ] be n × p matrices. The product AB is defined to be the m × p matrix C = [cij ] with   b1j         jth   b2j   cij = {i th row of [A]} = {ai1 , ai2 , . . . , ain } column . ..         of [B]   bnj n X = ai1 b1j + ai2 b2j + · · · + ain bnj = aik bkj . (2.5.7) k=1

The following comments are in order on the matrix multiplication, wherein A denotes an m × n matrix and B denotes a p × q matrix: (1)

(2)

The product AB is defined only if the number of columns n in A is equal to the number of rows p in B. Similarly, the product BA is defined only if q = m. If AB is defined, BA may or may not be defined. If both AB and BA are defined, it is not necessary that they be of the same size.

40

VECTORS AND TENSORS

(3) (4)

(5) (6) (7) (8) (9)

The products AB and BA are of the same size if and only if both A and B are square matrices of the same size. The products AB and BA are, in general, not equal AB 6= BA (even if they are of equal size); that is, the matrix multiplication is not commutative. For any real square matrix A, A is said to be normal if AAT = AT A; A is said to be orthogonal if AAT = AT A = I. If A is a square matrix, the powers of A are defined by A2 = AA, A3 = AA2 = A2 A, and so on. Matrix multiplication is associative: (AB)C = A(BC). The product of any square matrix with the identity matrix is the matrix itself. The transpose of the product is (AB)T = BT AT (note the order).

The next example verifies Property 9 above. Example 2.5.1 Verify the property 9 above using the matrices of matrix A and B is   5 −2 1  3 −1 2 7 6   8 −6 3 5 AB =  2 4 3  9 6 −2 −1 9 0

[A] and [B] in Eq. (2.5.3). The product   4  7 = 1

36 36 9 −57

−5 49 28 28

−2 39 18 43

 7 87  , 39  59

36 49 39 87

9 28 18 39

 −57 28  . 43  59

and 36  −5 T (AB) =  −2 7

36 49 39 87



Now compute the product   3 −6 9  5 3 6   −1 −2 BT AT =  2 5 −2  1 4 7 1

8 7 6

2 4 3

9 28 18 39

 −57 28  . 43  59

  36 −1  −5 9 = −2 0 7

Thus, (AB)T = BT AT is verified.

2.5.5

Inverse and Determinant of a Matrix If A is an n × n matrix and B is any n × n matrix such that AB = BA = I, then B is called an inverse of A. If it exists, the inverse of a matrix is unique (a consequence of the associative law). If both B and C are inverses for A, then by definition, AB = BA = AC = CA = I.

Theory of Matrices

41

Since matrix multiplication is associative, we have BAC = (BA)C = IC = C = B(AC) = BI = B. This shows that B = C, and the inverse is unique. The inverse of A is denoted by A−1 . A matrix is said to be singular if it does not have an inverse. If A is nonsingular, then the transpose of the inverse is equal to the inverse of the −1 T transpose: (A−1 ) = (AT ) . The actual computation of the inverse of a matrix requires evaluation of its determinant, which is discussed next. Let A = [aij ] be an n × n matrix. We wish to associate with A a scalar that in some sense measures the “size” of A and indicates whether or not A is nonsingular. The determinant of the matrix A = [aij ] is defined to be the scalar det A = |A| computed according to the rule detA = |aij | =

n X (−1)i+1 ai1 |Ai1 |,

(2.5.8)

i=1

where |Ai1 | is the determinant of the (n − 1) × (n − 1) matrix that remains on deleting out the ith row and the first column of A. For convenience we define the determinant of a zeroth-order matrix to be unity. For 1 × 1 matrices the determinant is defined according to |a11 | = a11 . For a 2 × 2 matrix A the determinant is defined by   a11 a12 a11 a12 = a11 a22 − a12 a21 . A= , |A| = a21 a22 a21 a22 In the above definition special attention is given to the first column of the matrix A. We call it the expansion of |A| according to the first column of A. One can expand |A| according to any column or row: |A| =

n X

(−1)i+j aij |Aij |,

(2.5.9)

i=1

where |Aij | is the determinant of the matrix obtained by deleting the ith row and jth column of matrix A. An numerical example of the calculation of a determinant is presented next. Example 2.5.2 Compute the determinant of the matrix  2 A= 1 2

5 4 −3

 −1 3 . 5

Solution: Using the definition in Eq. (2.5.9) and expanding by the first column, we have |A| = (−1)1+1 a11 |A11 | + (−1)2+1 a21 |A21 | + (−1)3+1 a31 |A31 |

42

VECTORS AND TENSORS

or 4 3 + (−1)3 a21 5 −1 + (−1)4 a31 5 −1 |A| = (−1)2 a11 −3 4 −3 5 5 3       = 2 (4)(5) − (3)(−3) + (−1) (5)(5) − (−1)(−3) + 2 (5)(3) − (−1)(4) = 2(20 + 9) − (25 − 3) + 2(15 + 4) = 74.

The cross product of two vectors the determinant, as follows: e ˆ2 e ˆ3 ˆ1 e A × B ≡ A1 A2 A3 B B B 1 2 3

A and B can be expressed as the value of



ˆ 1 + A2 e ˆ 2 + A3 e ˆ 3 × B1 e ˆ 1 + B2 e ˆ 2 + B3 e ˆ3 = A1 e = (A2 B3 − A3 B2 )ˆ e1 + (A3 B1 − A1 B3 )ˆ e2 + (A1 B2 − A2 B1 )ˆ e3 , and the scalar triple product C · (A × B) ≡

(2.5.10)

can be expressed as the value of a determinant: C1 C2 C3 A1 A2 A3 B1 B2 B3

= (A2 B3 − A3 B2 )C1 + (A3 B1 − A1 B3 )C2 + (A1 B2 − A2 B1 )C3 .

(2.5.11)

In general, the determinant of a 3 × 3 matrix A can be expressed in the form |A| = eijk a1i a2j a3k ,

(2.5.12)

where aij is the element occupying the ith row and the jth column of the matrix. ˆi , A2 = a2j e ˆj , The result in Eq. (2.5.12) can be established by taking A1 = a1i e ˆk and forming the scalar triple product and A3 = a3k e a11 a12 a13 A1 · A2 × A3 = eijk a1i a2j a3k = a21 a22 a23 . a a a 31

32

33

We note the following properties of determinants: (1) (2) (3) (4) (5)

det (AB) = det A · det B. det AT = det A. det (α A) = αn det A, where α is a scalar and n is the order of A. If A0 is a matrix obtained from A by multiplying a row or column of A by a scalar α, then det A0 = α detA. If A0 is the matrix obtained from A by interchanging any two rows or columns of A, then det A0 = −det A.

Theory of Matrices

(6) (7)

43

If A has two rows or columns one of which is a scalar multiple of another (that is, linearly dependent), det A = 0. If A0 is the matrix obtained from A by adding a multiple of one row or column to another, then det A0 = det A.

We define (in fact, the definition given earlier is an indirect definition) singular matrices in terms of their determinants. A matrix is said to be singular if and only if its determinant is zero. By property 6 above the determinant of a matrix is zero if it has linearly dependent rows (or columns). For an n × n matrix A the determinant of the (n − 1) × (n − 1) sub-matrix of A obtained by deleting row i and column j of A is called the minor of aij and is denoted by Mij (A). The quantity cof ij (A) ≡ (−1)i+j Mij (A) is called the cofactor of aij . The determinant of A can be cast in terms of the minor and cofactor of aij : detA =

n X

aij cof ij (A) for any value of j.

(2.5.13)

i=1

The adjunct (also called adjoint) of a matrix A is the transpose of the matrix obtained from A by replacing each element by its cofactor. The adjunct of A is denoted by AdjA. Now we have the essential tools to compute the inverse of a matrix. If A is nonsingular (that is, det A 6= 0), the inverse A−1 of A can be computed according to A−1 =

1 Adj A. det A

(2.5.14)

Example 2.5.3 Determine the inverse of the matrix of Example 2.5.2. Solution: We have, for example, 4 3 1 , M11 (A) = M (A) = 12 2 −3 5

3 , 5

1 M13 (A) = 2

4 , −3

cof 11 (A) = (−1)2 M11 (A) = 4 × 5 − (−3)3 = 29, cof 12 (A) = (−1)3 M12 (A) = −(1 × 5 − 3 × 2) = 1, cof 13 (A) = (−1)4 M13 (A) = 1 × (−3) − 2 × 4 = −11. The Adj(A) is given by  cof 11 (A) Adj(A) =  cof 21 (A) cof 31 (A)

cof 12 (A) cof 22 (A) cof 32 (A)

T  cof 13 (A) 29 cof 23 (A)  =  1 cof 33 (A) −11

The determinant is given by (expanding by the first row) |A| = 2(29) + 5(1) + (−1)(−11) = 74.

−22 12 16

 19 −7  . 3

44

VECTORS AND TENSORS

The inverse of A can now be computed using Eq. (2.5.14),   1  29 −22 19  −1 1 12 −7 . A = 74 −11 16 3 It can be easily verified that AA−1 = I.

2.5.6

Positive-Definite and Orthogonal Matrices A symmetric matrix A = [A] is said to be positive or positive-definite if {X}T [A]{X} = XT AX > 0.

(2.5.15)

holds for all nonzero vectors X = {X}. The expression {X}T [A]{X} represents a quadratic polynomial associated with matrix [A] with respect to vector {X}. A corollary to this definition is that a (symmetric) matrix [A] is positive if and only if there exists a nonsingular matrix [T ] such that [A] = [T ]T [T ] or A = TT T.

(2.5.16)

Then we have XT TT TX = YT Y,

Y ≡ TX,

(2.5.17)

which is always positive for all nonzero vectors {Y }. A nonsingular matrix Q = [Q] is said to be orthogonal if the following condition holds: QQT = QT Q = I

or Q−1 = QT .

(2.5.18)

From the foregoing definition it follows that the determinant of an orthogonal matrix is |Q| = ±1. If the determinant of [Q] is +1, then [Q] is called a rotation or a proper orthogonal matrix; otherwise, it is called an improper orthogonal matrix. A proper orthogonal matrix transforms a right-handed coordinate system into another right-handed coordinate system.

2.5.7

Eigenvalues and Eigenvectors It is conceptually useful to regard a matrix as an operator that changes, for example, a column vector into another column vector. In this regard it is of interest to inquire whether there are certain column vectors that have only their lengths, and not their directions, changed when operated on by a given matrix [S]. If such vectors {x} exist, they must satisfy the equation [S]{x} = λ{x}.

(2.5.19)

Such vectors {x} are called characteristic vectors, eigenvectors, or principal planes (defined by its normal vector) associated with the matrix [S] (which can

Theory of Matrices

45

be, for example, the underlying matrix of a tensor). The parameter λ is called a characteristic value, eigenvalue, or principal value, and it represents the change in length of the eigenvector {x} after it has been operated on by [S]. In view of the fact that x can be expressed as {x} = [I]{x}, Eq. (2.5.19) can also be written as ([S] − λ[I]) {x} = {0}.

(2.5.20)

Equation (2.5.20) represents a homogeneous set of linear equations for {x}. Therefore, a nontrivial solution, that is, a vector with at least one element of {x} nonzero, will not exist unless the determinant of the matrix [S] − λ[I] vanishes: |[S] − λ [I]| = 0.

(2.5.21)

The vanishing of this determinant yields an algebraic equation of degree n in λ, called the characteristic equation, when [S] is an n × n matrix. Second-order tensors, such as strain and stress tensors, are of interest in mechanics, where the eigenvalues and eigenvectors of the matrices associated with these tensors represent principal values and directions. Since the underlying matrix of second-order tensors is 3 × 3, the characteristic equation resulting from Eq. (2.5.21) is cubic in λ and yields three eigenvalues, say λ1 , λ2 , and λ3 . The character of these eigenvalues depends on the nature (i.e., real-valued, symmetric, positive-definite, and so on) of the matrix [S]. For a real-valued matrix [S], at least one of the eigenvalues will be real. The other two may be real and distinct, real and repeated, or complex conjugates. The vanishing of a determinant implies that the columns or rows of the matrix are linearly dependent and the three eigenvectors {x}(1) , {x}(2) , and {x}(3) , which denote three different orientations, are not unique. An infinite number of solutions exist, within a multiplicative constant, having at least n = 3 different orientations. A real-valued symmetric matrix [S] of order n has some desirable properties, as listed next: (1) (2) (3)

(4) (5)

All eigenvalues of [S] are real. If [S] is positive-definite, then the eigenvalues are strictly positive. Eigenvectors {x}(1) and {x}(2) associated with two distinct eigenvalues λ1 and λ2 are orthogonal: ({x}(1) )T {x}(2) = {0}. If all eigenvalues are distinct, then the associated eigenvectors are all orthogonal to each other. [S] always has n linearly independent eigenvectors, regardless of the algebraic multiplicities of the eigenvalues. For an eigenvalue of algebraic multiplicity m, it is possible to choose m eigenvectors that are mutually orthogonal. Hence, the set of n vectors can always be chosen to be linearly independent.

Returning to Eq. (2.5.20), let [S] be the matrix representation of a secondorder tensor S with respect to a rectangular Cartesian basis. Then the eigenvalue

46

VECTORS AND TENSORS

problem has the explicit form      s11 − λ s12 s13  x1   0   s21 = . s22 − λ s23  x 0  2    s31 s32 s33 − λ x3 0

(2.5.22)

For a nontrivial solution (i.e., at least one of the components x1 , x2 , and x3 is nonzero), we require that the determinant of the coefficient matrix be zero: s11 − λ s12 s13 s21 (2.5.23) s22 − λ s23 = 0. s s32 s33 − λ 31 The characteristic equation associated with Eq. (2.5.23) can be expressed in the form −λ3 + I1 λ2 − I2 λ + I3 = 0,

(2.5.24)

where I1 , I2 , and I3 are the invariants of S: I1 = sii ,

I2 =

1 2

(sii sjj − sij sij ) ,

I3 = |S|.

(2.5.25)

The invariants can also be expressed in terms of the eigenvalues (when known), I1 = λ1 + λ2 + λ3 ,

I2 = (λ1 λ2 + λ2 λ3 + λ3 λ1 ),

I3 = λ1 λ2 λ3 .

(2.5.26)

The eigenvector {x}(i) associated with any particular eigenvalue λi is calculated using Eq. (2.5.22), which gives only two independent relations among the (i) (i) (i) three components x1 , x2 , and x3 . Thus, two of the three components can be written in terms of the third, whose value is arbitrary (but nonzero). In other words, we can determine the eigenvectors only within a multiplicative constant. If the eigenvector is normalized such that it is a unit vector, then we use the following additional (i.e., third) condition to determine all three components: (i)

(i)

(i)

(x1 )2 + (x2 )2 + (x3 )2 = 1. (i)

(2.5.27)

(i)

(i)

For example, if x1 and x2 are expressed in terms of x3 [using Eq. (2.5.22)], (i) (i) (i) (i) say x1 = α x3 and x2 = β x3 , then Eq. (2.5.27) yields (i)

(x3 )2 = α2 + β 2 + 1


−1

1

(i)

or x3 = ± p

α2

+ β2 + 1

,

(2.5.28)

and the normalized eigenvector is ˆ (i) {X}

   α  = ±p . β α2 + β 2 + 1  1  1

(2.5.29)

The sign ± on the three eigenvectors should be selected such that we have a right-hand coordinate system (when the eigenvectors are orthonormal): ˆ (1) = x ˆ (2) × x ˆ (3) . x

(2.5.30)

Theory of Matrices

47

Example 2.5.4 Determine the eigenvalues and eigenvectors of the following matrix:   5 −1 [S] = . 3 1 Solution: The eigenvalue problem associated with the matrix [S] is to solve the equation [S]{X} − λ{X} = {0}:         5 −1 x1 1 0 x1 0 −λ = 3 1 x2 0 1 x2 0 

5−λ 3

−1 1−λ



x1 x2



 =

0 0

 .

(1)

The characteristic equation is obtained either from λ2 −I1 λ+I3 = 0 or from S −λ I = 0. Thus, λ2 − I1 λ + I3 = λ2 − 6 λ + 8 = 0; alternatively, 5−λ −1 = (5 − λ)(1 − λ) − (3)(−1) = 0 ⇒ λ2 − 6λ + 8 = 0. 3 1−λ The two roots of the quadratic equation are     p p λ1 = 12 6 − 62 − 4 × 8 = 2, λ2 = 12 6 + 62 − 4 × 8 = 4 To find the eigenvectors, we return to Eq. (1) and substitute for λ each of the (i) (i) eigenvalues λi and solve the resulting algebraic equations for (x1 , x2 ). For λ = λ1 = 2, we have   ( (1) )   5−2 −1 x1 0 = . (2) (1) 3 1−2 0 x2 (1)

(1)

Each row of the matrix equation in (2) yields the same condition 3x1 − x2 (1) (1) x2 = 3x1 . The eigenvector x(1) s given by   1 (1) (1) (1) {x} = x1 , x1 6= 0, arbitrary. 3

= 0 or

(1)

Usually, we take x1 = 1, as we are interested in the direction of the vector {x}(1) rather than in its magnitude. One may also normalize the eigenvector by using the condition (1)

(1)

(x1 )2 + (x2 )2 = 1. Then we obtain the following normalized eigenvector:     1 0.3162 {ˆ x}(1) = ± √110 =± . 3 0.9487

(3)

(4)

Using the same procedure, we can determine the eigenvector associated with λ2 = 4. Substituting for λ = λ2 = 4 into Eq. (1), we obtain   ( (2) )   5−4 −1 x1 0 = , (5) (2) 3 1−4 0 x2

48

VECTORS AND TENSORS

(2)

(2)

(2)

(2)

from which we obtain the condition x1 − x2 = 0 or x2 = x1 . The eigenvector x(2) is       1 1 0.7071 {x}(2) = or {ˆ x}(2) = ± √12 =± . (6) 1 1 0.7071 Because the matrix under consideration is not symmetric, we do not expect the eigenvectors to be orthogonal.

Example 2.5.5 Determine the eigenvalues and eigenvectors  0 [S] =  1 1

of the following matrix:  1 1 0 1 . 1 0

Solution: The characteristic equation −λ3 + I1 λ2 − I2 λ + I3 = 0 in the present case is −λ3 − (−3)λ + 2 = 0, which can be expressed as (2 − λ)(1 + λ)2 = 0. Thus, the three roots are λ1 = 2, λ2 = −1, λ3 = −1. We note that λ = −1 is an eigenvalue with algebraic multiplicity of 2. The eigenvector components associated with λ = 2 are obtained from     (1)    x1    0  −2 1 1  (1)  1 −2 1  0 = , x2  x(1)    0  1 1 −2  3 which gives (1)

(1)

(1)

(1)

(1)

(1)

(1)

(1)

(1)

−2x1 + x2 + x3 = 0, x1 − 2x2 + x3 = 0, x1 + x2 − 2x3 = 0. (1)

(1)

(1)

Solution of these equations gives x1 = x2 = x3 . Thus the with λ1 = 2 is the vector     1   1 (1) 1 1 {x}(1) = x1 or {ˆ x}(1) = ± √13  1   1

eigenvector associated  

,



where {ˆ x}(1) denotes the normalized (unit) vector. The eigenvector components associated with λ = −1 are obtained from     (2)    x1    0  1 1 1  (2)  1 1 1  0 = . x2  x(2)    0  1 1 1  3 All three equations yield the same single relation: (2)

(2)

(2)

x1 + x2 + x3 = 0. (2)

(2)

(2)

Thus, values of two of the three components (x1 , x2 , x3 ) can be chosen arbitrarily.

Theory of Matrices

(2)

(2)

For the choice of x2 = 0 and x3 of it):   −1 0 {x}(2) =  1

49

= 1, we obtain the vector (or any nonzero multiples

 

(2)

x1

or {ˆ x}(2) = ± √12



   −1  0 .  1  (2)

(2)

A second independent vector can be found by choosing x2 = 1 and x3 obtain      −1   −1  (2) (2) (2) 1 1 {x} = x or {ˆ x} = ± √12 .  0  1  0 

= 0. We

Thus, in the present case, there exist two linearly independent eigenvectors associated with the double eigenvalue. The eigenvectors are not mutually orthogonal. We note that the matrix [S] considered here is symmetric (but not positive). Clearly, properties 1 and 3 concerning the eigenvalues and eigenvectors of a real-valued symmetric matrix are satisfied. As far as property 5 is concerned, it is possible to choose the values of two of the three components (x1 , x2 , x3 ) to have a set of linearly independent eigenvectors that are orthogonal. The second vector associated with λ = −1 could have (3) (3) been chosen by setting x1 = x3 = 1. We obtain      1   1  (2) (3) (3) 1 −2 −2 {x} = x or {ˆ x} = ± √6 .  1  1  1  Thus the three eigenvectors,    1  1 {ˆ x}(1) = √13 ,  1 

{ˆ x}(2) =

√1 2

   −1  0 ,  1 

are mutually orthogonal. Hence, we can write  √ √ √  2 √2 0 1 1  √2 − 3 0 3  1 0 6 1 1 1 −2 1    λ1 0 0 2 =  0 λ2 0  =  0 0 0 λ3 0

{ˆ x}(3) =

 √ 1 √2 1   √2 0 2 0 −1 0

√1 6

√ − 3 √0 3

 

 1  −2 ,  1 

 1 −2  1

 0 0 . −1

We close this section with a note that the discussion presented here on eigenvalue problems will be revisited in the later chapters in connection with the finding of the principal values and planes of strain and stress tensors. As will be shown in Section 2.7 on tensors, the underlying matrices of second-order tensors S are of order 3 × 3, denoted by [S]. Tensors were first conceived by Gregorio Ricci-Curbastro and his pupil Tullio Levi-Civita (both Italian mathematicians), who continued earlier works of Bernhard Riemann and Elwin Bruno Christoffel (both German mathematicians).

50

VECTORS AND TENSORS

2.6

Vector Calculus

2.6.1

The Del Operator The rate of change of a scalar field (such as the temperature of a continuous body) with distance is of importance. Let us denote a scalar field by φ = φ(x), x being the position vector in a rectangular Cartesian system (x1 , x2 , x3 ). Let us now denote a differential element with dx and its magnitude by ds ≡ |dx|. Then ˆ = dx/ds is a unit vector in the direction of dx, and we may express it as (by e the use of the chain rule of differentiation)     dφ dx ∂φ ∂φ ∂φ ∂φ ˆ· e ˆ1 ˆ2 ˆ3 = · =e +e +e . (2.6.1) ds eˆ ds ∂x ∂x1 ∂x2 ∂x3 The derivative (dφ/ds)eˆ is called the directional derivative of φ. We see that it is the rate of change of φ with respect to distance and that it depends on the ˆ in which the distance is taken, as shown in Fig. 2.19. The vector direction e ∂φ/∂x is called the gradient vector and is denoted by grad φ: ˆ1 grad φ ≡ e

∂φ ∂φ ∂φ ∂φ ∂φ ∂φ ˆ2 ˆ3 ˆx ˆy ˆz +e +e =e +e +e . ∂x1 ∂x2 ∂x3 ∂x ∂y ∂z

(2.6.2)

We interpret grad as some operator operating on function φ. This operator is denoted by grad φ ≡ ∇φ ∂ ∂ ∂ ∂ ∂ ∂ ˆ2 ˆ3 ˆx ˆy ˆz , +e +e =e +e +e (2.6.3) ∂x1 ∂x2 ∂x3 ∂x ∂y ∂z and is called the del operator. The gradient of a function defines both the direction magnitude of the maximum rate of increase of the function at any point. Fig.and 2.19 The del operator is a vector differential operator. It is important to note that whereas the del operator has some of the properties of a vector, it does not have them all, because it is an operator. For instance ∇ · A is a scalar, called the divergence of A (∂Ai /∂xi ), whereas A · ∇ is a scalar differential operator [Ai (∂/∂xi )]. Thus the del operator does not commute in this sense. ˆ1 ∇≡e

f (normal to the surface)

dx

eˆ =

•

∂x ∂s

•

Curve s

f( x )

x

x3

x + dx

x2 x1 Fig. 2.19 Directional derivative of a scalar function

Vector Calculus

51

When the scalar function φ(x) is set equal to a constant, φ(x) = c, a family of surfaces is generated for each value of c. If the direction in which the directional derivative is taken lies within a surface, then dφ/ds is zero because φ is a constant ˆ is tangent to a level surface. Then, it on a surface. In this case, the unit vector e ˆ follows from Eq. (2.6.1) that, if dφ/ds is zero, gradφ must be perpendicular to e and hence, perpendicular to the surface. Thus, if a surface is given by φ(x) = c, the unit normal to the surface is determined by grad φ . (2.6.4) |grad φ| ˆ may point in either The plus or minus sign appears because the direction of n direction away from the surface. If the surface is closed, the usual convention is ˆ pointing outward. to take n An important note is in order concerning the del operator. Two types of gradient are used in continuum mechanics: forward and backward gradients. The forward gradient is the usual gradient and backward gradient is the transpose of the forward gradient operator. To see the difference between the two types of gradient, consider a vector function A = Ai (x)ˆ ei . The forward and backward gradients of A are (both are second-order tensors; see Section 2.7)  ∂A → − ∂  i ˆj ˆi = ˆj e ˆi = Ai,j e ˆj e ˆi , ∇A = ∇A = e Ai e e (2.6.5) ∂xj ∂xj  T T ← − ∂Ai  ˆj e ˆi ˆi e ˆj , ∇A = ∇A = e = Ai,j e (2.6.6) ∂xj ← − where Ai,j = ∂Ai /∂xj . The backward gradient ∇, a more natural one, is often used in defining the deformation gradient, displacement gradient tensor, and velocity gradient tensor, which is introduced in Chapter 3. In the present book only one gradient operator (in bold), namely, the forward gradient operator ∇, a more common one, is used. To clarify, the transpose of the forward gradient is used to denote the backward gradient operator. ˆ=± n

2.6.2

Divergence and Curl of a Vector The dot product of a del operator with a vector is called the divergence of a vector and denoted by ∇ · A ≡ divA.

(2.6.7)

The divergence of a vector field represents the volume density of the outwardflux of the vector field. If we take the divergence of the gradient vector, we have div(grad φ) ≡ ∇ · ∇φ = (∇ · ∇)φ = ∇2 φ. 2

(2.6.8)

The notation ∇ = ∇ · ∇ is called the Laplacian operator. In Cartesian systems this reduces to the simple form

52

VECTORS AND TENSORS

∂2φ ∂2φ ∂2φ ∂2φ + + = . (2.6.9) ∂x2 ∂y 2 ∂z 2 ∂xi ∂xi The Laplacian of a scalar appears frequently in the partial differential equations governing physical phenomena. The curl of a vector is defined as the del operator operating on a vector by means of the cross product: ∇2 φ =

ˆi curl A = ∇ × A = eijk e

∂Ak . ∂xj

(2.6.10)

The curl of a vector function represents its rotation. If the vector field is the velocity of a fluid, curl of the velocity represents the rotation of the fluid at the point. ˆ denote the unit vector, taken positive outward, normal to the surface Let n Γ of a continuous medium occupying the region Ω, as shown in Fig. 2.20. In a Cartesian coordinate system, the unit normal vector can be expressed in terms of its components as ˆ = n1 e ˆ 1 + n2 e ˆ 2 + n3 e ˆ 3 = nx e ˆ x + ny e ˆ y + nz e ˆz . n

(2.6.11)

The components (n1 , n2 , n3 ) = (nx , ny , nz ) are called direction cosines because of the fact ˆ and the xi −axis. ni = cosine of the angle between n

(2.6.12)

ˆ · grad φ of a function φ is called the normal derivative of φ, The quantity n which represents the outward flux of φ normal to the boundary, and it is denoted by

Fig. 2.20

∂φ ˆ · grad φ = n ˆ · ∇φ. ≡n (2.6.13) ∂n In a rectangular Cartesian coordinate system (x, y, z), ∂φ/∂n takes the form ∂φ ∂φ ∂φ ∂φ = nx + ny + nz . ∂n ∂x ∂y ∂z n2 = ny

x2 = y

α

dy

Γ

dx

Ω

(2.6.14)

nˆ

n1 = nx n1 = nx = cos(nˆ , x ) = cosine of the angle between nˆ and x - axis n2 = ny = cos( nˆ , y ) = cosine of the angle between nˆ and y - axis n1 = nx = cos α , n2 = ny = sin α

x1 = x Fig. 2.20 A unit vector normal to a surface

Vector Calculus

53

Next, we present several examples to illustrate the use of index notation to prove certain identities involving vector calculus. Example 2.6.1 Establish the following identities using the index notation: (1) ∇(r) = rr , r = |r|.

(2) ∇(rp ) = p rp−2 r, p is an integer.

(3) ∇ × (∇F ) = 0.

Solution: (1) Note that we use the notation r ≡ x for a position vector, and r = |r| = √ √ r · r = xj xj . Consider ˆi ∇(r) = e

1 1 1 ∂r ∂ r x ˆi xi (xj xj )− 2 = = , ˆi ˆi 12 (xj xj ) 2 −1 2xi = e =e (xj xj ) 2 = e ∂xi ∂xi r r

from which we note the identity ∂r xi = . ∂xi r

(2.6.15)

(2) In this case we have ˆi ∇(rp ) = e

∂ ∂r ˆi ˆi = p rp−2 r. (rp ) = p rp−1 e = p rp−2 xi e ∂xi ∂xi

(3) Consider the expression  ∇ × (∇F ) = 2

ˆi e

∂ ∂xi



  ∂F ∂2F ˆ ˆk × ej = eijk e . ∂xj ∂xi ∂xj 2

F F We note that ∂x∂i ∂x is symmetric in i and j, ∂x∂i ∂x = j j Consider the kth component of the above vector:

eijk

∂2F , ∂xj ∂xi

whereas eijk = −ejik .

∂2F ∂2F ∂2F = −ejik = −eijk (renamed i as j and j as i). ∂xi ∂xj ∂xj ∂xi ∂xi ∂xj

Thus the expression, being equal to its own negative, is zero (i.e., the curl of the gradient of a function is zero). It also follows that eijk Sij = 0 whenever Sij is symmetric, Sij = Sji (here Sij stands for an expression with two free indices).

A list of vector operations in both vector notation and in Cartesian component form is presented in Table 2.2.

2.6.3

Cylindrical and Spherical Coordinate Systems Two commonly used orthogonal curvilinear coordinate systems are the cylindrical and the spherical coordinate systems (see Fig. 2.21). Table 2.3 contains a summary of the basic information for the two coordinate systems. It is clear from Table 2.3 that the matrices of direction cosines between the orthogonal rectangular Cartesian system (x, y, z) and the orthogonal curvilinear systems (r, θ, z) and (R, φ, θ) are as given in Eqs. (2.6.16) to (2.6.19).

54

VECTORS AND TENSORS

Table 2.2 Vector expressions and their Cartesian component forms [A, B, and C are ˆ2 , e ˆ3 ) are the Cartesian unit vectors] vector functions and U is a scalar function; (ˆ e1 , e

No.

Vector form

Component form

1. 2. 3. 4. 5.

A A·B A×B A · (B × C) A × (B × C) = B(A · C) − C(A · B)

ˆi Ai e Ai Bi ˆk eijk Ai Bj e eijk Ai Bj Ck ˆi eijk eklm Aj Bl Cm e

6.

∇U

7.

∇A

8.

∇·A

9.

∇×A

∂U ˆ e ∂xi i ∂Aj ˆ e ˆ e ∂xi i j ∂Ai ∂xi ∂A ˆk eijk ∂x j e i ∂ eijk ∂x (Aj Bk )

10.

∇ · (A × B) = B · (∇ × A) − A · (∇ × B)

11.

∇ · (U A) = U ∇ · A + ∇U · A

12.

∇ × (U A) = ∇U × A + U ∇ × A

13.

∇(U A) = ∇U A + U ∇A

14.

∇ × (A × B) = A(∇ · B) − B(∇ · A)

i

∂ (U Ai ) ∂xi ∂ eijk ∂x (U Ak )ˆ ei j ∂ ˆj ∂x (U Ak e ˆk ) e j ∂ eijk emkl ∂x (Ai Bj )ˆ el m

+ B · ∇A − A · ∇B 15. 16.

(∇ × A) × B = B · [∇A − (∇A)T ] ∇ · (∇U ) =

∂Aj ∂xi

ˆm e

Ak i ∂xj

ˆm e

eijk eklm Bl ∂2U ∂xi ∂xi

∇2 U

Fig. 2.21

∇ · (∇A) = ∇2 A

∂ 2 Aj ˆ e ∂xi ∂xi j

∇ × ∇ × A = ∇(∇ · A) − (∇ · ∇)A

∂ emil ejkl ∂x

19.

(A · ∇)B

20.

A(∇ · B)

iˆ Aj ∂B e ∂xj i ∂Bj ˆi ∂x Ai e j

17. 18.

ˆz e

z

z

(a)

R

y θ

ˆR e

ˆq e

r

x

2

ˆr e z

f

(b)

2

R = r2 + z 2 x

y

Line parallel ˆq to e

θ

y

R

z

ˆq e

ˆf e R= R

θ

y

x Line parallel to e ˆf

x

Fig. 2.21 (a) Cylindrical coordinate system. (b) Spherical coordinate system

Vector Calculus

Cylindrical coordinate system    ˆr  cos θ sin θ  e  ˆ = − sin θ cos θ e  θ  ˆz e 0 0    ˆx  cos θ − sin θ  e ˆy =  sin θ e cos θ   ˆ ez 0 0 Spherical coordinate system    ˆR  sin φ cos θ sin φ sin θ  e ˆφ =  cos φ cos θ cos φ sin θ e   ˆθ e − sin θ cos θ    ˆx  sin φ cos θ cos φ cos θ  e ˆy =  sin φ sin θ cos φ sin θ e   ˆz e cos φ − sin φ

2.6.4

 ˆx 0  e  ˆ 0 e  y ˆz e 1  ˆr 0  e ˆθ 0  e  ˆz 1 e

 

55

,

(2.6.16)

.

(2.6.17)

   

  ˆx  cos φ  e ˆ , − sin φ  e  y  ˆz 0 e   ˆR  − sin θ  e ˆ . cos θ  e  φ  ˆθ 0 e

(2.6.18)

(2.6.19)

Gradient, Divergence, and Curl Theorems Integral identities involving the gradient, divergence, and curl of a vector can be established from integral relations between volume integrals and surface integrals. These identities will be useful in later chapters when we derive the equations of a continuous medium. Let Ω denote a region in 0 the boundaries at x = 0 and x = L are kept at temperatures T0 and TL , respectively. Obtain the temperature distribution in the slab as a function of position x and time t. Obtain the steady state temperature distribution T (x, y) in a rectangular region, 0 ≤ x ≤ a, 0 ≤ y ≤ b for the boundary conditions qx (0, y) = 0, qy (x, b) = 0, qx (a, y) + hT (a, y) = 0, T (x, 0) = f (x).

7.6

7.7

Consider the steady flow of a fluid through a long, straight, horizontal circular pipe. The velocity field is given by   R2 dP r2 vr = 0, vθ = 0, vz (r) = − 1− 2 . 4µ dr R If the pipe is maintained at a temperature T0 on the surface, determine the steady-state temperature distribution in the fluid. Determine the temperature distribution in a long bar of rectangular cross-section with the boundary conditions shown in Fig. P7.7. You may use the principle of superposition.

y T(0, y) = T0

T(x,b) = f (x)

b

T(a, y) = T0 a T(x,0) = T0

Fig. P7.7

x

Problems

231

Fluid Mechanics 7.8

A fluid with density ρ and viscosity µ is placed between two vertical plates a distance 2a apart, as shown in Fig. P7.8. Suppose that the plate at x = a is maintained at a temperature T1 and the plate at x = −a is maintained at a temperature T2 , with T2 > T1 . Assuming that the plates are very long in the y-direction and hence that the temperature and velocity fields are only a function of x, determine the temperature T (x) and velocity vy (x). Assume that the volume rate of flow in the upward moving stream is the same as that in the downward moving stream and the pressure gradient is solely due to the weight of the fluid.

Temperature distribution, T ( x)

2a ● Cold plate

Hot plate T2 Velocity distribution,

T1

vy ( x )

y

● T 0

a

x

Fig. P7.8 7.9

7.10 7.11

7.12

An engineer is to design a sea lab 4 m high, 5 m width, and 10 m long to withstand submersion to 120 m, measured from the surface of the sea to the top of the sea lab. Determine (a) the pressure on the top and (b) the pressure variation on the side of the cubic structure. Assume the density of salt water to be ρ = 1, 020 kg/m3 . Compute the pressure and density at an elevation of 1,600 m for isothermal conditions. Assume P0 = 102 kPa, ρ0 = 1.24 kg/m3 at sea level. For the steady, two-dimensional flow between parallel plates of Problem 5.4, determine c such that the velocity field satisfies the principle of balance of linear momentum. Assume that the axial and shear stresses are related to the velocities by the relations   ∂vx ∂vy ∂vx σxy = µ + , σxx = 2µ − P, ∂y ∂x ∂x where µ, the viscosity of the fluid, is a constant and P is the pressure. Consider the steady flow of a viscous incompressible Newtonian fluid down an inclined surface of slope α under the action of gravity (see Fig. P7.12). The thickness of the fluid perpendicular to the plane is h and the pressure on the free surface is p0 , a constant. Use the semi-inverse method (that is, assume the form of the velocity field) to determine the pressure and velocity field.

y

vx

h x

Direction of gravity, ρ g

Fig. P7.12

α

232

APPLICATION TO PROBLEMS OF MECHANICS

7.13

Two immiscible fluids are flowing in the x-direction in a horizontal channel of length L and width 2b under the influence of a fixed pressure gradient. The fluid rates are adjusted such that the channel is half filled with Fluid I (denser phase) and half filled with Fluid II (less dense phase). Assuming that the weight of the fluid is negligible, determine the velocity field. Use the geometry and coordinate system shown in Fig. P7.13. Assume steady flow

Fixed wall

b

y

Less dense and less viscous fluid

μ2

Interface

x

b

μ1

Denser and more viscous fluid

Fixed wall

Fig. P7.13 7.14 7.15

Consider the flow of a viscous incompressible fluid through a circular pipe (see Section 7.3.4). Reformulate the problem when the weight of the fluid is taken into account. Consider a steady, isothermal, incompressible fluid flowing between two vertical concentric long cylinders with radii r1 = R and r2 = α R, as shown in Fig. P7.15.

Angular velocity of the outer cylinder, Ω

Stationary inner cylinder

+

αR

R

Fig. P7.15 If the outer one is rotating with an angular velocity Ω, show that the Navier–Stokes equations reduce to the following equations governing the circumferential velocity vθ = v(r) and pressure P :   v2 ∂P d 1 d ∂P ρ = , µ (rv) = 0, 0 = − + ρg. r ∂r dr r dr ∂z 7.16

Determine the velocity v and shear stress τrθ distributions. Consider the steady flow (along the axis of the cylinders) of a viscous, incompressible fluid in the annular region between two coaxial cylinders of radii R0 and αR0 , α < 1, as shown in Fig. P7.16. Take P¯ = P + ρgz. Determine the velocity and shear stress distributions in the annulus.

Problems

233

Velocity distribution

α R0 z

R0

r

Fig. P7.16

Solid Mechanics 7.17

An isotropic body (E = 210 GPa and ν = 0.3) with two-dimensional state of stress experiences the following displacement field (in mm): u1 = 3x21 − x31 x2 + 2x32 , u2 = x31 + 2x1 x2 ,

7.18

where xi are in meters. Determine the stresses and rotation of the body at point (x1 , x2 ) = (0.05, 0.02) m. Is the displacement field compatible? A two-dimensional state of stress exists in a body with the following components of stress: σ11 = c1 x32 + c2 x21 x2 − c3 x1 , σ22 = c4 x32 − c5 , σ12 = c6 x1 x22 + c7 x21 x2 − c8 , where ci are constants. Assuming that the body forces are zero, determine the conditions on the constants so that the stress field is in equilibrium and satisfies the compatibility equations.

7.19–7.21: For the truss structures shown in Figs. P7.19–P7.21, determine the member stresses and strains. Assume linear elastic behavior, and let Ai be the area of cross-section and Ei be the modulus of the ith member. P h

C

O 1

1

h

A

2

3 b=4 4 D 5

a=3

For all members E, A (a and b in ft.) 2

a=3

P

Fig. P7.19

Fig. P7.20

2

L2

1

E = Young’s modulus Ai = cross-sectional area of the ith member A

L1 P

Fig. P7.21

B

234

APPLICATION TO PROBLEMS OF MECHANICS

7.22, 7.23: For the straight beam structures shown in Figs. P7.22 and P7.23, determine the transverse deflection as a function of position along the length of the beam. Assume linear elastic behavior with constant EI.

q0 z

F B

A

x

°°°

L

L

L/2

Fig. P7.22 7.24

7.25

Fig. P7.23

Consider a simply supported beam of length L under point loads F0 at x = L/4 and x = 3L/4 (the so-called four-point bending). Use the symmetry about x = L/2 to determine the transverse deflection w(x). Consider a semicircular curved beam of mean radius R. The beam is fixed at θ = π and subjected to a vertical upward load of P at θ = 0, as shown in Fig. P7.25. Determine the normal and shear forces (N, V ) and bending moment M at any section between θ = 0 and θ = π.

Fig. P7-25

. r

θ R

P

Fig. P7.25 7.26

A vertical 5 m high column used in the structure of a building is made from an I-section steel (E = 207 GPa) beam fixed to the foundations at its lower end. The I section is of depth 252 mm, breadth 203 mm, flange thickness 13.5 mm, and web thickness 8.0 mm. The maximum allowable compressive stress in the steel is 150 MN/m2 and the minimum safety Fig. factorP7-26 on axial force against failure by buckling is 2.0. Find the maximum weight the column can support (neglect the weight of the column itself) if (a) it is effectively free of constraint at its upper end, and (b) it is fixed at this end, with no lateral deflection allowed. N V

Fig. P7-25S

P P

.

·

Z

13.5Mmm 203 mm

r

R cos θ

θ

θ

8 mm L

x

(a)

R x

R sin θ

252 mmY P

(b)

225 mm Y

R 13.5 mm Z

Fig. P7.26

P

Problems

7.27

235

The plane pin-jointed structure shown in Figure P7.27 is used to support a vertical load F at a horizontal distance h from a vertical wall. The maximum allowable compressive force in the uniform strut BC, whose material has Young’s modulus of E and whose cross-section has a smallest moment of inertia of I, is given by the critical buckling load, Pcr . (a) Show that the ratio of the maximum allowable to the actual compressive force, P , in BC is given by Pcr π 2 EI = 2 sin2 θ cos θ, P h F where θ is the angle member BC makes with the vertical. (b) Find the optimum value of θ for a given value of I. With this optimum value of θ, find an expression for the lowest value of I which strut BC can have and just support the load F without buckling. Fig. P7.27 Neglect the weight of the member.

h

°

° B

A

F

θ

°C Fig. P7.27 7.28

Consider an isotropic, solid circular plate of radius a, constant thickness h, and spinning about the z-axis at an angular velocity of ω. Show that the governing equation is of the form i d h1 d 1 − ν2 2 − (ru) = ρω r, dr r dr E and its general solution is u(r) = C1 r +

7.29

C2 1 − ν 2 2 r3 − ρω , r E 8

where C1 and C2 are constants of integration. Use suitable boundary conditions to determine the constants of integration, Ch 1 and C2 , and evaluate the stresses σrr and B σθθ . A Derive the equilibrium equations governing the deformation of a homogeneous isotropic circular plate under the action of axisymmetric (about the z-axis) radial (f ) and transverse (q) forces. The free-body-diagram of an element of the plate with all relevant forces F is shown in Fig. P7.29. Note that the shear stresses are zero due to the symmetry. The stress resultants are defined by Z Nrr =

θ− h2

Z σrr dz, Nθθ =

h

C2

Z Mrr =

h 2

−h 2

h 2

−h 2

Z σrr z dz, Mθθ =

σθθ dz,

h 2

−h 2

σrr z dz.

236

APPLICATION TO PROBLEMS OF MECHANICS

a O

h 2

dr

r

r, θ = 0

θ

dθ z, w0 dθ

Qθ

Qr N rr

Nθθ

Mθθ

q(r)

M rr

h 2

Mθθ M rr +

Nθθ

Qθ

f(r)

∂Q r dr Qr + ∂r

∂M rr dr ∂r

N rr +

∂N rr dr ∂r

Fig. P7.29 7.30

Use the following displacement field for axisymmetric deformation of a circular plate and Hooke’s law to express the stress resultants of Problem 7.29 in terms of the displacements u and w: ur (r, z) = u(r) − z

7.31

dw , uθ (r, z) = 0 uz (r, z) = w(r). dr

Consider an isotropic, solid circular plate of radius a, constant thickness h, fixed at r = a, and subjected to uniformly distributed transverse load of intensity q0 . Show that the governing equation is of the form D

 d2 1 d  d2 w 1 dw  + + = q0 , 2 2 dr r dr dr r dr

where D = Eh3 /[12(1 − ν 2 )]. The general solution is w(r) = C1 + C2 log r + C3 r2 + C4 r2 log r +

7.32 7.33

q0 r4 . 64D

Use suitable boundary conditions to determine the constants of integration, Ci , i = 1, 2, 3, 4, and determine the deflection w(r) and bending moments Mrr (r) and Mθθ (r). For the plane elasticity problems shown in Figs. P7.32(a)–(d), write the known boundary conditions. You must know one element from each of the two pairs (ux , tx ) and (uy , ty ). An external hydrostatic pressure of magnitude p is applied to the surface of a spherical body of radius b with a concentric rigid spherical inclusion of radius a, as shown in Fig. P7.33. Determine the displacement and stress fields in the spherical body.

Problems

237

Figure P7-32 x2

τ

p

θ τ

(a)

τ

(b)

Spherical core, μ1 , λ1

b a

Spherical shell, μ 2 , λ2

x1

x2 Rigid core

τ0

τ0 (c)

(d)

b

Hollow cylindrical shaft μ , λ

b

a

τ0

τ0

x1

σ0

Figure P7-33

a

Fig. P7.32

p

Rigid inclusion

b a

Elastic sphere

Figure P7-34 7.34

Fig. P7.33

The lateral surface of a homogeneous, isotropic, solid cylinder of radius a, length L, and mass density ρ is bonded to a rigid surface. Assuming that the ends of the cylinder at z = 0 and z = L are traction-free (see Fig. P7.34), determine the displacement and stress fields in the cylinder due to its own weight. z

y = x2 r

θ

f = − ρ g eˆ z

f

L

x = x1

r

Fig. P7.34

Figure P7-34S z z

y = x2 r

APPLICATION TO PROBLEMS OF MECHANICS

238

Figure P7-35 A solid cylindrical body of radius a and height h is placed between two rigid plates, as

7.35

shown in Fig. P7.35. The plate at B is held stationary and the plate at A is subjected to a downward displacement of δ. Using a suitable coordinate system, write the boundary conditions for the following two cases: (a) when the cylindrical object is bonded to the plates at A and B; and (b) when the plates at A and B are frictionless. Rigid plate

A h

Lateral surface

z

σ zθ

σ zz

× σ zr

σ rz σ rθ

a

Cylinder

×

σ rr

×

B

Rigid plate

r

Fig. P7.35 7.36

Figure

Reconsider the concentric spheres of Problem 7.33. As opposed to the rigid core in Problem 7.33, suppose that the core is elastic and the outer shell is subjected to external pressure p (both are linearly elastic). Assuming Lam´ e constants of µ1 and λ1 P7-36 Figure for the core and µ2 and λ2 for the outerP7-37 shell (see Fig. P7.36), and that the interface is perfectly bonded at r = a, determine the displacements of the core as well as the shell. p

Spherical core μ1 , λ1

Rigid core

τ0

τ0

b

b

a

Hollow cylindrical shaft μ , λ

Spherical shell μ 2 , λ2

a

τ0

Fig. P7.36 7.37

7.38

Fig. P7.37

Consider a long hollow circular shaft with a rigid internal core (a cross-section of the shaft is shown in Fig. P7.37). Assuming that the inner surface of the shaft at r = a is perfectly bonded to the rigid core and the outer boundary at r = b is subjected to a uniform shearing traction of magnitude τ0 , find the displacement and stress fields in the problem. Interpret the following stress field obtained in case 3 of Example 7.4.7 using Fig. 7.25: σxx = 6c10 xy,

7.39

τ0

σyy = 0,

σxy = −3c10 y 2 .

Assume that c10 is a positive constant. Compute the stress field associated with the Airy stress function Φ(x, y) = Ax5 + Bx4 y + Cx3 y 2 + Dx2 y 3 + Exy 4 + F y 5 .

7.40

Interpret the stress field for the case in which A, B, and C are zero. Use Fig. 7.25 to sketch the stress field. Investigate what problem is solved by the Airy stress function (use Fig. 7.25 to discuss the results)   3A xy 3 B 2 Φ= xy − 2 + y . 4b 3b 4b

Problems

7.41

239

Show that the Airy stress function  
1
q0 Φ(x, y) = 3 x2 y 3 − 3b2 y + 2b3 − y 3 y 2 − 2b2 8b 5 Figure P7-41 satisfies the compatibility condition. Determine the stress field and find what problem it corresponds to when applied to the region −b ≤ y ≤ b and x = 0, a (see Fig. P7.41).

y b

x

b

a

Fig. P7.41

Fig.Determine P7-42 the Airy stress function for the stress field of the domain shown in Fig. P7.42 7.42 and evaluate the stress field. t

y

q (force per unit area)

b b

2b

x

a

Fig. P7.42 7.43

The thin cantilever beam shown in Fig. P7.43 is subjected to a uniform shearing traction of magnitude τ0 along its upper surface. Determine if the Airy stress function Fig. P7-42S   τ0 xy 2 xy 3 ay 2 ay 3 Φ(x, y) = xy − − 2 + + 2 4 b b b b

FigP7-43

y satisfies the compatibility condition and stress boundary conditions of the problem.

y

σ yy ( x , b) = q0 , σ xy ( x ,b) = 0 σ xx (0, y ) = 0, σ xy (0, y ) = 0

x

t0 b b

σ yy ( x , −b) = q0 , σ xy ( x , −b) = 0

V = q0ta

t

P =0 q ta 2 M= 0 2

x

2b

a Fig. P7.43 7.44

The curved beam shown in Fig. P7.44 is curved along a circular arc. The beam is fixed at the upper end and it is subjected at the lower end to a distribution of tractions statically ˆ1 . Assume that the beam is in a state equivalent to a force per unit thickness P = −P e of plane strain/stress. Show that an Airy stress function of the form   B Φ(r) = Ar3 + + C r log r sin θ r

240

APPLICATION TO PROBLEMS OF MECHANICS

provides an approximate solution to this problem and solve for the values of the constants A, B, and C.

Fig. P7-44

y = x2

r a

b

θ

x = x1

P

Fig. P7.44

Fig.References P7-44S for Additional Reading 1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

Bird, B., x2 Stewart, W. E., and Lightfoot, E. xN., Transport Phenomena, 2nd ed., John 2 Wiley & Sons, New York, NY (2002). Fenner, R. T. and Reddy, J. N., Mechanics of Solids σad Structures, 2nd ed., CRC Press, rθ = 0 σ rr = 0 Boca Raton, FL (2012). θ Holman, J. P., Heat Transfer,r7th ed., McGraw–Hill, New York, NYr (1990). Kreith, F., Manglik, R. M., and Bohn, M. S., Principles of Heat Transfer, 7th ed., σ rr = 0 a b Cengage Learning, Stamford, CT (2011). σ rθ = 0 Munson, B. R., Young, D. F., and Okiishi, T. H., Fundamentals of Fluid Mechanics, θ New York, NY (1990). θ John Wiley & Sons, x1 x1 rθ Reddy, J. N., AnPIntroduction to Continuum Mechanics, σ2nd ed., Cambridge University b b Press, New York, NY (2013). σ θθ dr = 0 σ rθ dr = P Reddy, J. N., Energy Principles and Variational Methods in Appliedσ θθ Mechanics, 3rd a a ed., John Wiley & Sons, New York, NY (2017). Sadd, M. H., Elasticity: Theory, Applications, and Numerics, 2nd ed., Academic Press, New York, NY (2009). Schlichting, H. and Gersten, K., Boundary Layer Theory, (translated from German by J. Kestin), 9th ed., McGraw–Hill, New York, NY (2017). Slaughter, W. S., The Linearized Theory of Elasticity, Birkh¨ aser, Boston, MA (2002).





Reason, observation, and experience; the holy trinity of science. Robert Green Ingersoll

Teachers are like candles, which consume themselves to brighten the lives of others. Author unknown

Though science can cause problems, it is not by ignorance that we will solve them. Isaac Asimov

Nothing is too wonderful to be true if it be consistent with the laws of nature. Michael Faraday

Answers to Selected Problems The only means of strengthening one’s intellect is to make up one’s mind about nothing – to let the mind be a thoroughfare for all thoughts. Not a select party. John Keats A man with a new idea is a crank until he succeeds. Mark Twain Full or partial answers to selected problems are included here for readers to check their solutions. Problems involving proofs are not included here (other than giving some hints).

Chapter 1: Introduction 2.1

Iy = 2.469 × 106 mm4 = 2.469 × 10−6 m4 , and σt = 2.245W kPa, σc = −7.475W kPa

2.2

πd dP Q == − 128µ dz   T4 −T1 q=− , and R

2.3

4

1 R

=



h3 k3

+

h2 k2

+

h1 k1



Chapter 2: Vectors and Tensors ˆB , r=A+C=A+βe

2.2 2.3 2.4 2.5

(C − A) × (B − A) · (r − A) = 0 |A| = 3, `11 = 23 , `12 = − 23 , `13 = − 13 The angle between vectors A and B is θ = 82.34◦ .
1 ˆ , Qm = 624.04 (kg/s) vn = √15 , θ = 39.42◦ , vt = 13 20ˆi + 26ˆj + 30k

2.7 2.8 2.9 2.11 2.12

2.14

2.17 2.18 2.19

and β is a real number.

13

The set is linearly independent. The set is linearly dependent: −5r1 + 2r2 + r3 = 0 The set is linearly independent. ˆj and e ˆn = `mn ¯ Use ¯ eˆi = `ij e eˆm .   0   √1  −1 √ √1 ˆ1  ˆ1   e  e 3 3 3  √2  3 1 0 √ √ ˆ ˆ2 (a) e = e 14 14 14    20   −4 ˆ3 ˆ3 e e √ √1 √5 42 42 42  1  1 √ √ 0 2 2  1 1 1 √ −2  ˆj ) L= 2 eˆi · e  (`ij = ¯ − 12

2.16

ˆB = e

B , |B|

2.1

2 √1 2

1 2

(a) |A| = −8. (b) |A| = −5    1 −1 −3 1 −2 (a) [A]−1 = − 18  4 4 −4 . (b) [A]−1 = − 15  −4 3 7 1 −5 3 −1 (a) [A] is not positive. (b) [A] is positive. (c) [A] is not positive. Q is nonsingular.

 −1 −1  −3

242

Answers to Selected Problems

2.21

√ √ ˆ (1) = ±(0, 0, 1). (c) λ1 = 4, λ2 = 2, (a) λ1 = 3.0, λ2 = 2(1 + 5), λ3 = 2(1 − 5); A 1 1 ˆ (1) = ± √ (0, 1, −1), A ˆ (2) = ± √ (0, 1, 1), A ˆ (3) = ±(1, 0, 0) λ3 = 1; A 2

2.22

2.23 2.24 2.28 2.29 2.30 2.31 2.37

2

ˆ (1) = ±(0.7300, 0.3372, 0.5945); (a) λ1 = 11.8242, λ2 = 1.2848, λ3 = −7.1090; A ˆ (2) = ±(0.4799, 0.8424, −0.5368); A ˆ (3) = ±(−0.6817, 0.4204, 0.5987). A ˆ (1) = ±(0.328, −0.737, 0.591), (b) λ1 = 3.24698, λ2 = 1.55496, λ3 = 0.19806; A ˆ (2) = ±(0.591, −0.328, −0.737), A ˆ (3) = ±(0.737, 0.591, 0.328) A ˆ (2) = 1 ± (−2, 2, 1) λ1 = 9, λ2 = −9, λ3 ; A 3

First show that grad(r) = rr and grad(rn ) = nrn−2 r. Use the gradient theorem. Use the divergence theorem. (b) 281 (c) 240 (d) (31, 25, 30) (e) (18, 15, 34) ˆi = δip e ˆp , e ˆj = δjq e ˆq , and e ˆk = δkr e ˆr and use their (a) See Problem 2.12 (e) Let e scalar triple product.   h  i rr ˆ rr ˆr = ∂E ˆr (b) ∇ · E = ∂E er + r1 Err − Eθθ e + r1 Err − Eθθ e ∂r ∂r

Chapter 3: Kinematics of a Continuum 3.1 3.2 3.3 3.4 3.5 3.6 3.8 3.9 3.10

3.11

3.12

3.13

3.14 3.15 3.16 3.17 3.19 3.20 3.22

εAB = 0.085 m/m εBE = 0.153 m/m εAC = 0.048 m/m The downward displacement of E is 2.29 mm (a) 0.0112◦ εAC = −0.00257, εBD = −0.00776
2 dvx 2 x = L 1 + 2L U0 dt (b) (X1 , X2 , X3 ) = (1, 2, 1)  2  k1 − 1 0 0 2   2[E] = 0 k2 − 1 0 0 0 k32 − 1  2  k1 − 1 e0 k1 k2 0 2 2   2[E] = e0 k1 k2 (1 + e0 )k2 − 1 0 0 0 k32 − 1   0 0 0   (c) 2[E] = 0 0 0 0 0 (2 + Bt)Bt   B2 A+B 0 2  (c) 2[E] = A+B A 0  0 0 0       cosh t sinh t 0 0 1 0 0 0 0      (c) [F ] = sinh t cosh t 0 , [D] = 1 0 0 , [W ] = 0 0 0  0 0 1 0 0 0 0 0 0 √ √ ˆ 1 = −0.7554 e ˆ1 +0.6552 e ˆ2 , n ˆ 2 = 0.6552 e ˆ1 +0.7554 e ˆ2 (b) λ1 = 7−5 2, λ2 = 7+5 2; n
2 E11 = 0, E12 = e2b0 , E22 = 12 eb0  2 E11 = 0, E12 = eb20 X2 , E22 = 12 2X2 eb20   E11 = −0.2+0.5 (−0.2)2 + (0.2 + 0.1X2 )2 , 2E12 = 0.5+(0.2+0.1X 2 )+(−0.2)(0.5)+   (0.2 + 0.1X2 )(−0.1 + 0.1X1 ), E22 = −0.1 + 0.1X1 + 0.5 (0.5)2 + (−0.1 + 0.1X1 )2

3 3 2 3 2 ε11 = 3X12 X2 + c1 2c32 + 3c22 X 2 − X2 , ε22 = − 2c2 + 3c2 X2 − X2 + 3c1 X1 X2 ,
 2ε12 = X1 X12 + c1 3c22 − 3X22 − 3c1 X1 X22   cos θ , εzz = 0 εrr = A, εrθ = 0, εrz = 0, εθθ = A, εzθ = 12 Br + C r

Answers to Selected Problems

3.24

3.25 3.26 3.27 3.28 3.29 3.32 3.34 3.36

The strain transformation equations are    cos2 θ sin2 θ 0 ε¯11       2θ 2θ    ε ¯ sin cos 0  22        ε¯33 0 0 1 =  2¯ ε23   0 0 0          2¯ ε13  0 0 0     2¯ ε12 − sin 2θ sin 2θ 0 EAC =

0 0 0 cos θ sin θ 0

1 2 − 12

0 0 0 − sin θ cos θ 0

sin 2θ sin 2θ 0 0 0 cos 2θ

               

243

ε11 ε22 ε33 2ε23 2ε13 2ε12

              

2 2 1 e1 +e2 +2(ae1 +be2 ) 2 a2 +b2

√ ˆ 2 = (2ˆ ˆ2 )/ 3 λ1 = 0, λ2 = 1, 000, n e1 + e 4c1 = c2 The strain field is not compatible. f (X2 , X3 ) = A + BX2 + CX3 , where A, B, and C are arbitrary constants. Dxx = 2x, Dyy = 2y, Dzz = 2z , Dxy = 1.5y, Dxz = −0.5z, Dyz = 0.5y + z, Ωzy = −0.5y − z, −Ωzx = Ωxz = 2.5z, −Ωxy = Ωyx = −0.5y. (b) Prove ω × v = Ω · v   ∂vi ∂xj ∂xk ∂xi ∂vj ∂xk ∂xi ∂xj ∂vk Hint: DJ = eijk ∂X + ∂X + ∂X Dt ∂X ∂X ∂X ∂X ∂X ∂X 1

2

3

1

2

3

1

2

3

Chapter 4: Stress Vector and Stress Tensor 4.3

4.4 4.5 4.7 4.8 4.9 4.10 4.11 4.12 4.13 4.14

√ ˆ = 2(ˆ ˆ| = ˆ3 ), |tn (a) tn e1 + 7ˆ e2 + e 204 MPa, θ = 120.89◦ , σn = −7.33 MPa, σs = 12.26 MPa. ˆ = 3.138ˆ ˆ | = 5.482 MPa, θ = 16.47◦ , σ = 5.257 MPa, (c) tn e1 − 3.138ˆ e2 + 3.219ˆ e3 , |tn n σs = 1.554 MPa ˆ | = 6, 608.08 psi, θ = 16.58◦ . (b) σ = 6, 333.33 psi, σ = 1, 885.62 psi. (a) |tn n s (2) (c) σ1 = 6, 656.85 psi, σ2 = 1, 000 psi, σ3 = −4, 656.85 psi (d) σs = −5, 656.4 psi σn = −2.833 MPa, σs = 8.67 MPa σn = −40.80 MPa, σs = −20.67 MPa σn = 3.84 MPa, σs = −17.99 MPa σn = 950 kPa, σs = −150 kPa σn = −76.60 MPa, σs = 32.68 MPa σ22 = 140 MPa, σs = 90 MPa σp1 = 972.015 kPa, σp2 = −72.015 kPa, θp1 = 36.65◦ , θp2 = 126.65◦ , (σs )max = 522.015 kPa. σp1 = 121.98 MPa, σp2 = −81.98 MPa, θp1 = 39.35◦ , θp2 = 129.35◦ , (σs )max = 101.98 MPa The transformation equations are σ ¯11 = σ11 cos2 θ + σ22 sin2 θ + (σ12 + σ21 ) cos θ sin θ σ ¯12 = (σ22 − σ11 ) cos θ sin θ + σ12 cos2 θ − σ21 sin2 θ σ ¯22 = σ11 sin2 θ + σ22 cos2 θ − (σ12 + σ21 ) cos θ sin θ.

4.15 4.16 4.17

σ ¯xx = 15, 000psi, σ ¯xy = −5, 000psi, σ ¯yy = 7, 000 psi d = 0.7969 m σθθ = pD 4t

4.18 4.19 4.20 4.21

(b) tn = − 10 psi, ts = 8.576 psi (c) λ1 = 6.856, λ2 = −10.533, λ3 = −3.323 3 t = (50ˆ e1 − 150ˆ e2 − 50ˆ e3 )/3, tn = −16.67 MPa, ts = 52.7 MPa σ1 = 25 MPa, σ2 = 50 MPa, σ3 = 75 MPa, (σs )max = 25 MPa I1 = 150, I2 = 6, 875, I3 = 93, 750

244

Answers to Selected Problems

Chapter 5: Conservation of Mass and Balance of Momenta and Energy 5.2 5.4 5.5 5.7 5.8 5.9 5.14 5.15 5.16 5.17

∂ − ∂r (rρvr ) −

∂ − r ∂z (ρvz ) = r ∂ρ ∂t

∂ (ρvθ ) ∂θ

1 6

vavg = (3v0 − c) m/s (a) F = 24.12 N (b) F = 12.06 N (c) Fx = 45 N v2 = 9.9 m/s, Q = 19.45 liters/s (a) T = 0.15 N-m (b) ω0 = 50 rad/s = 477.5 rpm ˆ3 (kN-m) MA = 89 e
ρf1 = 0, ρf2 = a b2 + 2x1 x2 − x22 , ρf3 = −4abx3 c2 = −c6 = 3c4
3 P σxz = − 2I h2 − z 2 , σzz = 0 (I = 2bh ) 3 The equations are ∂σrr 1 ∂σrθ ∂σrz 1 + + + (σrr − σθθ ) + ρfr = 0 ∂r r ∂θ ∂z r ∂σrθ 1 ∂σθθ ∂σθz 2σrθ + + + + ρfθ = 0 ∂r r ∂θ ∂z r ∂σrz 1 ∂σθz ∂σzz σrz + + + + ρfz = 0 ∂r r ∂θ ∂z r

5.18

Satisfied identically for any A, B, and C.

Chapter 6: Constitutive Equations 6.2

6.3

6.4 6.5 6.6

E σkk = (1−2ν) εkk     σ 37.8  11            σ22    43.2              σ33 27.0 6 = 10 Pa σ23  21.6           σ13    0.0               σ12 5.4 I1 = 78.8 MPa, I2 = 1, 062.89 MPa2 , I3 = 17, 368.75 MPa3 J1 = 66.65 × 10−6 , J2 = 63, 883.2 × 10−12 , J3  = 244, 236 × 10−18 2

2

d d w N = EA du , M = −EI ddxw 2 , V = − dx EI dx2 dx

6.7

(a) T = 203.80◦ F. (b) T = 232.5◦ F

6.8

σ11 = 0, σ12 = 2

8µktx2 , (1+kt)2

σ22 =

(2µ+λ)k 1+kt b2 dP 2µ dx

6.10

− ∂P + µ ddyv2x = 0, ∂x

6.11

(a) ρ Dv = ρf − ∇p + µ∇2 v. Dt

6.14

(a) σ +

= kε + η dε dt

6.15

(a)

(b)

− ∂P = 0, − ∂P = 0, c = ∂y ∂z

(b) −∇p + ρf = ρ

η dσ = η dε . (b) σ = σ1 + σ2 k dt  dt  η dσ k2 + 1 + σ = k2 ε + η dε k1 dt k1 dt

1 σ η

+



∂v ∂t

1 dσ k2 dt

+ v · ∇v

=

k1 ε η



 + 1+

k1 k2



dε dt

Chapter 7: Applications in Heat Transfer, Fluid Mechanics, and Solid Mechanics 7.1

d − dr (rqr ) + rρQe = 0

Answers to Selected Problems

7.2 7.3 7.4 7.5 7.6 7.7 7.8 7.9 7.10

h
i 2 r 2 eR T (r) = T0 + ρQ4k 1− R h i
e e 0 c1 = k ρQ R02 − Ri2 + T0 − Ti log R , c2 = T0 + ρQ R02 + ck1 log R0 4k Ri 4k R P∞ 2 T −T0 2 L θ(x, t) = n=1 Bn sin λn x e−αλn t , Bn = L 0 f (x) sin λn x dx, θ = TL −T0 R P∞ cosh λn (b−y) T (x, y) = n=1 An cosh λ b cos λn x, An = a2 0a f (x) cos λn x dx n   3  µα2 R3 T (r) = T0 − 9k 0 1 − Rr 0

Let θ = T − T0 and θ is given by Eq. (7.2.49).  3   ρ β ga2 (T −T ) x vy (x) = r r 12µ 2 1 − x a a (a) P = γh = 1.2 MN/m2 . (b) P (z) = γ(120 + z) = 10(120 + z) kPa, P (4) = 1.24 MPa, where the coordinate z is measured from the top of the sea lab downward P = 82.314 kPa abs, ρ = 1.02 kg/m3 .

7.12

b2 ∂P 2µ ∂x 2 U (y) = ρgh2µsin α

7.13

U1 (y) =

7.11

c=

A= 7.14

245

 y 2h −

dP y 2 + µA y dx 2µ1 1   µ1 −µ2 b − dP dx 2 µ1 +µ2

y2 h2



+ B, U2 (y) =

dP y 2 dx 2µ2

+

A y µ2

+ B, B = − dP dx

R
4 r2 − R2 , Q = 2π 0R rvz (r) dr = − πR  8µ
r2 r2 1 vθ = r2 −r r Ω1 r12 − Ω2 r22 + 1r 2 (Ω1 − Ω2 ) 2 1 2 h

i ¯ R2 (1−α2 ) P r 2 r vz (r) = − ddz 1− R − log α log R , 4µ h  i
¯ R dvz dP r 1 2 τrz = −µ dr = − dz 4 2 R + (1 − α ) log α R r 1 ∂P 4µ ∂z

vz (r) =



b2 µ1 +µ2



,

∂P ∂z



7.15 7.16

7.17 7.18 7.19 7.20 7.21 7.22

σ11 = 96.88 MPa, σ22 = 64.597 MPa, σ33 = 48.443 MPa, σ12 = 4.02 MPa, σ13 = 0 MPa, σ23 = 0 c6 = −2c2 = −3c4 √ ε1 = E PA , ε2 = − E 2P A 1

1

2

2

P P ε1 = − 45 EA , ε4 = ε5 = 12 , ε3 = 25 EA F2 = P sin θ, F1 = −P sin θ cos θ

w1 (x) =

w2 (x) =

F L3 x2 x 1− L , 2 12EI L  F L3 x − 12EI 3 − 10 L +

q0 L4 x2 120EI L2



x 3 − 7L

P EA

0 ≤ x ≤ L,

 x2 x3 9L 2 − 2 L3 , L ≤ x ≤ 1.5L.  x2 x3 + 5L 2 − L3

7.23

w(x) =

7.24 7.25 7.26

11 F0 L w( L ) = 384 2 EI N = −P cos θ, V = −P sin θ, M = −P R(1 − cos θ) (a) 192 kN (b) 1092 kN

7.27

F (b) θ = 54.7◦ , I = 3 2π3h 2   E2 2 u(r) = r 3+ν − ar 2 1−ν ρω 2 a2 , 1+ν 8E     2 r2 σrr = 3+ν ρω 2 a2 1 − ar 2 , σθθ = 3+ν ρω 2 a2 1 − 1+3ν 8 8 3+ν a2   dMθθ d2 (rM ) − + r q(r) = 0 rr dr dr 2       2 1 dw Nrr = A du +νu , Nθθ = A ν du +u , Mrr = −D ddrw , 2 + ν r dr dr r dr r   2 1 dw Mθθ = −D ν ddrw , where A = Eh/(1 − ν 2 ) and D = Eh3 /[12(1 − ν 2 )]. 2 + r dr   2 2 2 0a w(r) = q64D 1 − ar 2

7.28

7.29 7.30

7.31

3

√

2

246

Answers to Selected Problems

7.33

  3 3 pr ur (r) = − 3Kbb3 +4µa 1 − ar3 , 3     1+2α(a3 /r 3 ) 1−α(a3 /r 3 ) σrr (r) = − p, σθθ (r) = σφφ = − p, 1+β 1+β

7.34 7.37 7.38 7.39 7.41 7.42

7.43

7.44

2µ , 3K

3

β = 2α ab3  2 uz (r) = − ρga 1− 4µ α=

r2 a2



, σθz = 0, σzr =

ρg r 2


2 2 r 0b uθ (r) = τ2µa − ar , σrθ = br2τ0 a c8 = c9 = c11 = c12 = 0
σxx = 2D 3x2 y − 2y 3 , σyy = 2Dy 3 , σxy = −6Dxy 2 ∇4 Φ = 0 issatisfied  3 2 2 y y x 0 + 5a − 53 yb3 , σxx = 3q 10 b 2b2 a2 b   3 σyy = q40 −2 − 3 yb + yb3 ,   2 σxy = 3q4b0 a x 1 − yb2 a   σxx = τ40 − 2x − 6xy + 2a + 6ay , b b b2 b2   2y 3y 2 τ0 σyy = 0, σxy = − 4 1 − b − b2 2 2

P a b A = − 2∆ , B = − P 2∆ , C=

P (a2 +b2 ) , ∆

∆ = b2 − a2 − (a2 + b2 ) log

b a

Index

A Absolute temperature, 177 Airy stress function, 227, 228, 239, 240 Analytical solution, 144, 189, 194, 209, 224 Angular: displacement, 81, 144 momentum, 8, 153 velocity, 21, 163, 232, 236 Anisotropic, 167, 175 Approximate solution, 224, 240 Axial vector, 97, 109 Axisymmetric: analysis, 108 boundary condition, 78 deformation, 93, 236 flow, 201 geometry, 93, 230 heat transfer, 188, 192 B Balance laws, 133 Balance of: angular momentum, 8, 153, 168, 204 energy, 8, 155–160, 168, 179, 187, 189, 230 forces, 120, 210 heat, 158 linear momentum, 8, 133, 142, 150 Bars, 74, 207–210 Barotropic, 177 Beams, 10, 74, 104, 146–148, 164, 206, 211–220, 229, 234–240 Beam theory, 5 Euler–Bernoulli, 74, 146, 212, 218 Bernoulli’s equations, 163 Biaxial state of strain, 108 Biharmonic: equation, 228, 229 operator, 228 Body: forces, 148, 151, 155, 159, 165, 198, 201, 227, 233 couples, 8, 154, 159 C Cantilever beam, 10, 165, 215, 239 Cartesian: basis, 33, 46, 128

components, 27, 30, 50–59, 117, 120, 150, 205 coordinate system, 27, 33, 39, 50–58, 62, 75, 85, 88, 92, 96, 107, 113, 186 tensor, 60, 117, 127 Cauchy: equations of motion, 182 formula, 111–116, 128, 149, 159 stress tensor, 116, 125, 128 stress vector, 116 Chain rule of differentiation, 50, 77, 82, 226 Characteristic: equation, 45, 63, 94 value, 45, 94 vectors, 45, 94 Clausius–Duhem inequality, 9, 161 Coefficient of themal expansion, 173 Cofactor, 43, 63 Collinear, 18 Compatibility: conditions, 99, 101, 108, 206, 224, 228, 239 equations, 99, 101, 233 strain, 99, 108 Composite: bar, 188, 209, 210 body, 188 cylinder, 188 rod, 181 wall, 7, 11, 188 Conduction: heat, 5, 7, 157, 159, 178, 182, 187, 192, 202 electrical, 202, 230 Conductivity: hydraulic, 203, 204 thermal, 7, 178, 186 electrical, 192, 230 Configuration, 8, 69, 75, 81 Conservation of mass, 8, 133, 135–140, 157, 168, 202, 206 Constitutive equations, 8, 167–180, 196 fluids, 177, 178, 197 heat transfer, 178 solids, 169–176, 224 viscoelastic, 182 Continuity conditions, 210 Continuity equation, 9, 136–142, 161, 162, 182, 196, 198, 200, 206 Continuum, 4, 8 Continuum mechanics, 4, 7, 10

248

Index

Control: surface, 135, 153, 155, 156 volume, 135, 136, 139, 143, 149, 153–157 Convection, 158, 178, 180, 187–191 Convective, 78, 134, 185, 197 heat transfer coeffcient, 188 Coordinates: cartesian, 33, 52, 58, 62, 75, 85, 107, 115, 182 cylindrical, 53–56, 64, 66, 93, 98, 107, 109, 161, 165, 200, 225 spherical, 53–56, 65, 66, 204 transformation of, 34, 62, 111 Coplanar, 18, 23, 62 Couette flow, 199 Creep, 169 Curl: of a vector, 51–53, 64, 97, 163 theorem, 55 Current density, 192, 230 Cylindrical coordinates: see Coordinates, cylindrical

Dummy index, 29, 32 Dyad, 57, 59, 60, 118 Dyadic, 57, 116, 128 E Eigenvalues, 44–60, 94, 108, 123, 128 Eigenvectors, 44–60, 94, 108, 123, 128 Elastic, 167, 169 Electromagnetic force, 115, 148 Elemental surface, 149 Energy equation, 196, 206 Engineering constants, 169, 172, 181 Engineering shear strains, 92 Engineering strains, 70, 90, 102 Equilibrium equations, 151, 165, 205, 222, 224, 236 Euler–Bernoulli beam theory, 74, 146, 212, 218 Euler–Bernoulli hypothesis, 212 Euler strain tensor, 102 Eulerian description, 76, 195 Exact solutions, 185, 197, 200

D F Darcy’s law, 204 Deformation, 8, 69, 76, 80, 92, 125, 169 analysis of, 82 gradient, 51, 70, 82, 83, 86, 105, 109, 139 homogeneous, 83, 84, 89, 90 infinitesimal, 92, 150, 152, 169, 206 isochoric, 85 mapping, 70, 75, 78, 83, 85, 91, 105, 126 nonhomogeneous, 86, 89 rate of, 70, 95, 96, 101, 105, 159, 178 uniform, 83, 89 Deformed configuration, 75, 81 Del operator, 50–52, 60 Density, 115, 135, 139, 142, 172, 177, 179, 186, 192, 202, 230, 238 Diagonal matrix, 36 Diffusion, 202 Diffusion coefficient, 202 Direction cosines, 35, 52, 60, 62, 223 Directional derivative, 50, 51 Displacement: angular, 81, 144 field, 80, 93, 99, 107, 206, 209, 233, 236–239 gradient, 51, 88, 91 rigid body, 70 vector, 18, 81, 83, 91, 93, 97, 105, 221, 224 Dissipation, 155, 160, 179, 182, 195 Divergence, 50, 55, 61, 137 Divergence theorem, 55, 137, 150 Dot product, 18, 30, 33, 51, 57 Double dot product, 59, 159

Fick’s law, 204 First law of thermodynamics, 9, 154–160 Fourier’s law, 9, 178, 179, 187, 194, 202, 204 Frame indifference, 32, 168 Free index, 30 G Generalized Hooke’s law, 169–173 Gradient, 7, 11, 50, 88 of temperature, 7, 187 operator (vector), 50, 51, 55, 60, 82 theorem, 46, 55, 115 Green (–Lagrange) strain tensor, 87–92, 101, 105 H Heat: conduction, 7, 157, 178, 182, 187, 192, 202 convection, 179 flow, 7 transfer, 7, 155–158, 178–180, 185–194 Heterogeneous, 167 Homogeneous: deformation, 83, 85 extension, 85 material, 93, 167, 236 motion, 83 stretch, 105 Hookean solids, 167, 169 Hooke’s law, 9, 169–173, 182, 236

Index

I Ideal fluid, 177 Incompressible fluid, 138, 162, 177, 178, 182, 195, 199, 204, 232 Incompressible materials, 137, 138, 160 Infinitesimal: deformation, 91, 150, 152, 169, 182, 206 strain, 74, 81, 92, 95, 99, 101, 108, 145 strain tensor, 91, 92 Inner product, 18 Internal energy, 155, 158, 160, 168, 187 Invariants, 46, 132 Inverse: mapping, 75, 78, 83 method, 224, 228 Inverse of a matrix, 40–44, 60, 84 Inviscid fluid, 177, 197 Irreversible process, 155, 230 Isochoric deformation, 85 Isothermal, 177, 205, 231 Isotropic material, 93, 167, 172–174, 177–179, 181, 186, 195, 203, 221, 225, 233, 236–238 J Jacobian, 83, 127 K Kinematics, 8, 69, 74, 82, 105, 109, 167, 170, 211 Kinematically infinitesimal, 130, 150, 152 Kinetic energy, 154, 158 Kronecker delta, 30, 32 L Lagrangian description, 76, 81, 91 Lam´ e constants, 173, 181 Laplacian operator, 51 Leibnitz rule, 134 Linear momentum, 133, 143, 150, 161, 231 Linearized elasticity, 7, 205 Linearly dependent, 18, 43 Linearly independent, 18, 26, 35, 45, 49, 62, 101 M Mapping, 70, 75, 78, 83, 85, 91, 105, 126 Material: coordinates, 75, 77, 81, 91, 175 coordinate system, 79 derivative, 76, 141 description, 76, 81, 104, 107, 134, 139, 169 frame indifference, 32, 146, 168 time derivative, 78, 80, 134, 137, 140

249

Matrices, 35–49 Matrix: addition, 37 determinant, 40–43 inverse, 40–43 minor of, 43 multiplication, 39 nonsingular, 44, 63, 83 singular, 43 skew-symmetric, 37, 38, 97, 102 Method of potentials, 225 N Navier–Stokes equations, 182, 185, 196–198, 233 Newtonian fluids, 9, 167, 177, 180, 182, 232 Newton’s: law of cooling, 179 second law, 115, 141, 145, 151, 161 third law, 112 Nonhomogeneous deformation, 86, 89 Nonion form, 58 Nonlinear elastic, 169 Non–Newtonian, 177 Normal derivative, 52 Normal stress, 113, 117, 122–124, 131, 155, 170, 173–178, 181 Null vector, 15 O Orthonormal, 26, 30–35, 47, 62, 63 Orthogonal, 19, 33, 40, 44, 48, 53, 60, 62, 117, 195 components, 117 coordinate system, 53, 55, 60 matrix, 44, 62 projection, 19, 62 tensor, 60 Orthotropic material, 169, 171, 174 Outflow, 55, 134, 141, 157 P Parallel flow, 198 Pendulum, 144 Perfect gas, 177 Permutation symbol, 30–33, 59 Plane strain, 221–224, 240 Plane stress, 131, 221–224 Poiseuille flow, 199, 200 Poisson’s ratio, 108, 170–173, 212 Potential energy, 154, 155, 157 Pressure: hydrostatic, 163, 178, 195, 197, 237 vessel, 119–122, 132, 175, 226 Principal:

250

Index

directions, 45, 95, 105, 130, 175 planes, 45, 49, 94, 123, 128 strains, 79, 94, 105 stresses, 111, 120, 123, 130 stretch, 85 values, 45, 49, 94, 111, 123, 128 Principle of superposition, 151, 174, 195 R Radiation, 159, 178, 180 Rate of deformation, 70, 95, 101, 159, 167, 177 Rigid-body motion, 70, 81, 84, 90 S Scalar product, 18, 26, 27, 30 Scalar triple product, 22–24, 42 Second law of thermodynamics, 8, 155, 161 Second-order tensor, 30, 45, 51, 57–61, 65, 84, 88, 94, 108, 111 Semi-inverse method, 225, 232 Shear: force, 147, 212–214, 219, 234 modulus, 169 strain, 70–74, 81, 89, 92, 104, 173, 175, 212 stress, 6, 113–125, 130, 155, 175, 195, 231, 233 Simple shear, 86, 89, 105 Singular, 41, 43 Skew-symmetric, 37, 38, 97, 102 Spatial coordinates, 76, 78, 91, 134 Spatial description, 76–78, 81, 105, 125, 134, 138 Spherical coordinates, 53–56, 65, 204 Spin tensor, 97, 109 Stefan–Boltzmann law, 180 Strain energy, 159, 172 Strain-displacement relations, 99, 205, 211, 220 Stress dyadic, 116 Stress-strain relations, 170, 173, 181, 183, 205, 219, 223 Stress vector, 57, 111–120, 126–129, 132, 149 Stretch, 74, 85, 105 Summation convention, 29–34 Superposition principle, 170, 197, 216–218, 231 Surface forces, 148 T Tensor components, 61, 89, 91, 93, 106, 132, 181 Tetrahedral element, 116, 118 Thermal conductivity, 7, 178, 179, 187

Thermal expansion coefficient, 169, 173, 206 Thermodynamic: form, 160 pressure, 178 principles, 8, 154–159 state, 177 Third-order tensor, 58, 59 Trace, 36, 59 Transformation of: basis, 32–35 components, 60, 128 coordinates, 34 strain components, 108, 245 stress components, 120 vector components, 35 Triple products of vectors, 22–24, 42 Tumor, 203, 204 U Undeformed body, 70–74, 76, 79, 82, 85, 88, 90, 126 Uniform deformation, 83, 84, 89 Unit: tensor, 30, 59 vector, 15, 19, 21, 24, 28, 46, 48, 50–53, 84, 112, 115 V Variational methods, 225 Vector, 13–28 addition, 15, 27, 57 calculus, 50–55 components, 27, 35, 48, 95 product, 19–22, 27, 30 Velocity: gradient tensor, 51, 80, 95, 102, 109, 159, 177, 180 vector, 28, 55, 78, 95, 150, 158, 177 Viscosity, 6, 10, 177, 197, 231 Viscous: dissipation, 160, 179, 195 fluid, 160, 167, 178–180, 182, 199, 230, 232 stress, 160, 178, 182 Volume change, 177 Vorticity: tensor, 95–98, 102, 105, 159 vector, 97, 109 Y Young’s modulus, 5, 10, 169, 206, 219, 235 Z Zero vector, 14, 44, 80 Zeroth-order tensor, 41, 59