Models of Mechanics

Models of Mechanics

SOLID MECHANICS AND ITS APPLICATIONS Volume 138 Series Editor:

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

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

For a list of related mechanics titles, see final pages.

Models of Mechanics by

ANDERS KLARBRING Department of Mechanical Engineering, Linköping University, Linköping, Sweden

A C.I.P. Catalogue record for this book is available from the Library of Congress.

ISBN-10 ISBN-13 ISBN-10 ISBN-13

1-4020-4834-3 (HB) 978-1-4020-4834-0 (HB) 1-4020-4835-1 (e-book) 978-1-4020-4835-7 (e-book)

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Preface

This is a textbook on models and modelling in mechanics. The term model, or more precisely mathematical model, refers to a mathematical problem with an observational or experimental connection to the real physical world. By modelling is meant the intellectual process of obtaining a model. Mechanics in this context involves not only the mechanics of mass points and rigid bodies, but also continuum mechanics of solids and fluids as well as traditional engineering mechanics of beams, cables, pipe flow and wave propagation. Thus, the text may be conceived as having an unusually large scope on relatively few pages; this is made possible through a unified approach provided by the open scheme of classical mechanics. The open scheme consists of concepts of time and space together with universal laws of mass conservation, and linear and angular momenta. By this an almost algorithmic approach to modelling emerges, which forms the connecting thread of the text. This unified approach to engineering mechanics in a broad sense may be the most important novel feature of this book. The book is written at an intermediate level. An extensive and ideal background is basic courses in engineering mechanics, mechanics of materials, structural mechanics and fluid mechanics, as well as applied courses in the machine design or the civil engineering areas. A student having part of such a background will here get the knowledge that makes it possible to organize and structure more effectively what she has learnt before, as well as a platform for studying and deriving new models. Students with more of an applied math h matics background also benefit from the unified approach, and for them it provides a short cut to models and concepts not usually discussed in a mathematically rigorous setting. The book should not be confused with one of the many existing texts on continuum mechanics. These usually do not present a general enough setting for including the mechanics of mass points or the me anics of beams and other intrinsically one-dimensional bodies. Thus, even though this is sometimes claimed, three-dimensional continuum mechanics is not a wide enough frame to mathematically organize known facts of applied mechanics. On the other hand, a student having digested this text can go on

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Preface

in several directions. One could be advanced continuum mechanics of fluids and solids, and another mechatronics and control. Even though the text covers a large number of models, in a sense only a part of what may be considered the subject of models of mechanics is treated. On the one hand, the observations or experiments that shape any model of physical events are only sparingly discussed, and, on the other hand, numerical solution techniques, and the related theory of variational methods, are not considered. However, what is treated here is the basic understanding of what is a model, what is its foundation, its rigorous mathematical derivation from universal principles and its general properties. This seems to be the essential starting point also for entering into experimental investigations and numerical solution methods. Moreover, the emergence in recent years of equation-based finite element programs (FEMLAB, FlexPDE, Fastflo) has strengthened the need for better modelling skills. The book consists of three parts and appendices. Part I has two chapters and contains a general background. Chapter 1 discusses the concept of a model and its relation to the physical world. It also briefly gives the idea of an open scheme and indicates the connecting thread or modelling algorithm followed in later parts. Chapter 2 gives more details about the open scheme and discusses concepts of space, time, vectors and tensors. It presents a way of organizing the laws or assumptions that are the foundation of mechanics, and singles out three universal laws that must be true in any mechanical model. Part II, having five chapters, gives the definitions necessary to make the universal laws explicit. Chapter 3 introduces four geometric models of bodies, and in later chapters these models are discussed in detail. Chapter 4 treats the discrete model of mass points, and Newton’s second law and the law of action and reaction are derived as basic theorems from the universal laws. Chapter 5 discusses a geometrically one-dimensional model that eventually leads to theories of cables, pipe flow and beams. In fact, already Chapter 6 introduces the concepts and specializations necessary to arrive at fluid mechanics one-dimensional problems that represent pipe flow. Chapter 7 contains the three-dimensional model and here we come close to what is traditionally handled in books on continuum mechanics. Therefore, this treatment is somewhat less complete than the corresponding treatment of discrete and one-dimensional geometric models. Part III of the book introduces, in five chapters, those particular laws that are needed to arrive at models which are also well-posed mathematical problems. Such models are called complete in this text and have the ability to function as an operator that delivers an answer for a given set of data. The particular laws needed to obtain complete problems are categorized as kinematic constraints, constitutive laws and force laws. Chapter 8 discusses general principles for particular laws, such as coordinate, dimensional and observer invariance. Chapter 9 is devoted to the small displacement theories prevalent in classical mechanics of materials. Complete models of trusses, different beam theories and linear elasticity are derived. Chapter 10 continues the discussion on pipe flow, and complete problems of

Preface

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compressible and incompressible flow as well as flow in flexible pipes. Chapter 11 presents complete problems of fluid mechanics such as that based on Navier-Stokes’ equations. Chapter 12 gives complete problems based on the introduction of kinematic constraints. In particular, alternative derivations of beam theories, previously derived as intrinsically one-dimensional theories, are presented. There is also a derivation of the complete problem of rigid body mechanics. Finally, five appendices provide mathematical background material, a note on physical units and a list of mathematical symbols. Large parts of this text have been used for several years as material for a masters course (i.e., a course in the fourth year of the Swedish “civilingenjor” ¨ program). In addition to traditional lectures and classwork sessions, this course is based on a number of home assignments and a paper assignment. Some of the larger home assignments are extended versions of Exercises 8.2, 9.7 and 11.2, and they generally require use of MATLAB or FEMLAB.The paper assignment means the writing of a paper in an accepted scientific format, where all the modelling steps from universal laws to complete model, as well as numerical solutions and comparison with experimental facts, are included. One-dimensional wave propagation problems, such as those modelling blood flow in arteries, Section 10.3, and shallow water flow, Exercises 6.5 and 10.5, have proven ideal for such a paper assignment. The main focus of the course has been on discrete and one-dimensional models. The three-dimensional theory has been included but some parts, like Chapter 12, are better saved for a Ph.D. course. Many students and colleagues have influenced this text directly and indirectly. In particular, I would like to thank Lars-Erik Andersson, Ulf Edlund, Peter Ireman, Lars Johansson, Matts Karlsson, Jaroslav Mackerle, Magnus Sethson, Niclas Stromberg ¨ and Jonas St˚ ˚ alhand for valuable comments on early versions of the manuscript. A special thanks also goes to the editor of the SMIA series, Graham Gladwell, who’s comments considerably improved my “final” manuscript.

Anders Klarbring Link¨ o¨ping January 2006

Contents

Part I General Background 1

Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 1.1 The Concept of a Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 1.2 Models, Reality and The Conceptual World . . . . . . . . . . . . . . . . 6 1.3 The Idea of an Open Scheme . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

2

Open scheme . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1 The Physical Space , Geometric Vectors and Tensors . . . . . . . 2.2 Laws of Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2.1 Universal Laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2.2 Particular Laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

11 11 14 14 17 17

Part II Basic Models: Geometry and Universal Laws 3

Bodies and Their Placements in E . . . . . . . . . . . . . . . . . . . . . . . . . Body Model 3.1 Di 3.2 One-Dimensional Body Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2.1 Pipe Flow Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3 Two-Dimensional Body Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4 Three-Dimensional Body Model . . . . . . . . . . . . . . . . . . . . . . . . . i M d l f B di 3.5 R fi d G

21 21 23 25 25 27 29

4

Discrete Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . d C 4.1 M ................................ fM 4.1.1 C 4.2 Linear and Angular Momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . f h C fM 4.2.1 M

31 31 32 32 33

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4.3 System of Forces, and Force and Torque Resultants . . . . . . . . . 4.3.1 Change of Base Point for the Torque Resultant . . . . . . . 4.4 Explicit forms of Euler’s Laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.4.1 Change of Base Point for Euler’s Laws . . . . . . . . . . . . . . . fA dR 4.5 Th Th ’ L 4.6 N 4.7 A Complete Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.7.1 Example: An Inverse Kepler Problem. . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

33 35 36 36 37 38 38 39 41

One-Dimensional Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . d C 5.1 M ................................ 5.2 Linear and Angular Momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.3 System of Forces, and Force and Torque Resultants . . . . . . . . . 5.4 Explicit Forms of Euler’s Laws 5.5 Local Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.6 Equations of Motion in Natural Parameterization and Natural Frame . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . lP 5.6.1 N 5.6.2 Equations of Motion in Natural Base . . . . . . . . . . . . . . . . . 5.7 A Complete Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.7.1 Example: A Circular Beam. . . . . . . . . . . . . . . . . . . . . . . . . . 5.8 Another Complete Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.8.1 Example: Suspension Bridge . . . . . . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

45 45 46 46 47 48 49 49 53 55 56 57 58 60

6

Pipe Flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1 Kinematic Constraint, Velocity and Acceleration . . . . . . . . . . . . . 6.2 Spatial and Material Representations, and Time Derivatives. . . 6.3 Forces and Couple in a Pipe Bend . . . . . . . . . . . . . . . . . . . . . . . . . 6.4 Continuity Equation and Control Domain . . . . . . . . . . . . . . . . . . . 6.5 Volume of the Pipe and Isochoric Motion . . . . . . . . . . . . . . . . . . . 6.6 Equation of Motion in Terms of Pressure . . . . . . . . . . . . . . . . . . . 6.7 Conservative Forces and Incompressibility . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

63 63 64 66 67 69 71 73 75

7

Three-Dimensional Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.1 Spatial and Material Representations, and Time Derivatives . . dI h M 7.2 V l 7.3 Mass and the Continuity Equation . . . . . . . . . . . . . . . . . . . . . . . . . 7.4 Linear and Angular Momentum. . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.5 Stress Tensor, System of Forces, and Force and Torque Resultants l d Sh S 7.5.1 N 7.5.2 The Stress Tensor and the System of Forces . . . . . . . . . .

77 77 79 80 81 81 82 82

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7.6 Explicit Forms of Euler’s Laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84 7.7 Equations of Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85 Part III Complete Models by Adding Particular Laws 8

9

Particular Laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.1 Three Types of Particular Laws . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.2 General Principles for Particular Laws. . . . . . . . . . . . . . . . . . . . . . 8.3 Examples of Particular Laws for the Discrete Model . . . . . . . . . 8.4 Examples of Particular Laws for the One-Dimensional Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.5 Examples of Particular Laws for the Pipe Flow Model. . . . . . . . 8.6 Examples of Particular Laws for the Three-Dimensional Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

89 89 90 93 94 995 . 95 98

Small Displacement Models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99 9.1 Di M d l dG lS 99 9.1.1 Linearizing a Simple Discrete Model . . . . . . . . . . . . . . . . . 99 9.1.2 Truss Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103 9.1.3 Work Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105 9.1.4 Abstraction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107 9.1.5 Use of Abstract Work Equations in Modeling . . . . . . . . . 108 9.2 B M d l . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109 9.3 Linear Elasticity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .117 9.3.1 Elasticity Coefficients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .120 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124

10 Pipe Flow Models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .127 10.1 Incompressible Pipe Flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .127 10.2 Compressible . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129 10.3 Incompressible Pipe Flow in a Flexible Pipe . . . . . . . . . . . . . . . . . 134 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135 11 Models of Fluid Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .137 11.1 Inviscid Fluid . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 138 11.1.1 Elastic Fluid . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 138 11.1.2 Incompressible Fluid . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139 11.2 Viscous Fluid . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140 11.2.1 Navier-Stokes’ Equation for Plane Steady Laminar Flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141 11.2.2 Dimensionless Form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142 11.3 Statics of Fluids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143

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11.3.1 Archimedes’ Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .144 11.3.2 Density and Pressure of the Atmosphere . . . . . . . . . . . .146 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .146 12 Kinematic Constraints, Beams and Rigid Bodies . . . . . . . . . . . 149 12.1 Some Examples and the General Idea . . . . . . . . . . . . . . . . . . . . . . 149 12.1.1 Discrete Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .150 12.1.2 Incompressible Linear Elasticity . . . . . . . . . . . . . . . . . . . . . 151 12.1.3 Incompressible Viscous Fluid . . . . . . . . . . . . . . . . . . . . . . . .152 12.2 Beam Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153 12.2.1 Geometric Shape and Prerequisites . . . . . . . . . . . . . . . . . . 153 12.2.2 Equilibrium Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . .155 12.2.3 Displacement Fields and Constitutive Assumptions . . . . 157 12.3 Rigid Body Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163 12.3.1 Geometry of Rigid Body Motion . . . . . . . . . . . . . . . . . . . . .163 12.3.2 Euler’s Second Law for a Rigid Body . . . . . . . . . . . . . . . . .168 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .173 A

Sets and Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .175 A.1 The Notion of a Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .175 A.2 The Notion of a Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .176

B

Euclidean Point and Vector Spaces . . . . . . . . . . . . . . . . . . . . . . . . . 179 B.1 Euclidean Vector Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 180 B.2 Euclidean Point Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 185

C

Tensors and Some Mathematical Background . . . . . . . . . . . . . .187 C.1 Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .187 C.2 Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .190 C.2.1 Divergence Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .193 C.2.2 Localization Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . .194 C.2.3 Fundamental Theorems of Variational Calculus . . . . . . .194 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .195

D

Note on Physical Dimensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .197

E

Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .199

References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .201 Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .203

Part I

General Background

1 Introduction

This introductory chapter provides an overview of models and modelling in mechanics. When specific assumptions and developments enter in later chapters, it is essential to be able to relate these to an overall picture. Section 1.1 explains the concept of a complete model of mechanical phenomena. Such models are in a sense final outcomes of chains of modelling steps and they are frequently presented within frames further on in the text. The presented view on models is a refinement of what can be found broadly within engineering science literature, see, e.g., Ljung and Glad [13]. Section 1.2 describes how it is possible to view the relation between mathematical models and the real world. A more thorough discussion on these topics can be found in Hestenes [10]. The introductory chapter of Truesdell and Toupin [19] also makes good reading on the relation between mechanics and mathema tics. In Section 1.3 we discuss the particular and universal laws of mechanics and introduce the concept of an open scheme. Based on this concept a modelling method or algorithm emerges and this textbook is essentially based on applying such a method.

1.1 The Concept of a Model In a broad sense a model is a simplified representation of an object or system that we want to understand or investigate. Models can be real or physical, as utilized, for instance, by architects to understand the three-dimensional structure of buildings. Modern versions of such models are computer based virtual ones. Models can also be mental or verbal. However, in this text we are solely concerned with mathematical models. By this is meant a model which employs mathematical concepts or symbols. In fact, we will be even more re trictive and investigate only mathematical models which represent mechanical

3

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1 Introduction

systems within the realms of classical mechanics1 . Furthermore, it will be useful to talk about complete mathematical models. Such a model consists of a set of equations or other mathematical rules that, when provided with appropriate data, produces an output, i.e., an answer, see Figure 1.1. Thus, a complete model is a specification of the looser term mathematical model: a mathematical model may become a complete model on specifying what we regard as data and answer. The term mathematical model is frequently used synonymously with the terms theory, mathematical theory, physical theory or even mathematical-physical theory. Two examples of complete mathematical models of mechanics are given below to fix ideas:

Fig. 1.1. A complete mathematical model as an operator producing an answer from appropriate data.

1. Free flight of a mass point: The acceleration a of a mass point is the second time derivative of the position x and it is proportional to the force f applied to the point, the coefficient of proportionality being the mass m, i.e., d2 x f = ma, a = 2 . (1.1) dt Note that a, x and f are geometric vectors2. Equations (1.1) constitute a mathematical model that becomes complete on choosing what is the data and what is the answer. The data can be a force function f = f (t, x, ddtx ), the mass m and initial conditions. The answer is then the position vector as a function of time, x = x(t), i.e., the trajectory. An alternative, simpler, complete problem based on (1.1) is to let x = x(t) and m be the data, and f = f (t) the answer. 2. A stretched bar: An elastic bar is built in at one end and pulled by a force f at the other end, see Figure 1.2. A cross section of the bar is identified by a coordinate x that belongs to the closed interval [0, L]. 1

2

By classical mechanics we mean those parts of mechanics (the theory of motion and force) that involve a Euclidean version of physical space and an independent time. Thus, it includes solid and fluid mechanics as well as the theories of mass points and rigid bodies. Geometric vectors are introduced in Chapter 2 and more thoroughly discussed in Appendices B and C.

1.1 The Concept of a Model

5

Fig. 1.2. A bar stretched by a force f .

The bar is built in at x = 0 and the force is applied at x = L. The elastic property of a cross section is defined by the function EA(x) and the displacement field is denoted u = u(x). The following equations define the problem: d du (EA(x) ) = 0, x ∈ (0, L), (1.2) dx dx du(L) = f. (1.3) u(0) = 0, EA(L) dx Equation (1.2) is a differential equation and equations (1.3) are the corresponding boundary conditions. The data of the model may now be, for instance, the force f and the cross section elasticity function EA(x), while the displacement field u = u(x) is the answer. A complete mathematical model has a certain level of broadness: the box in Figure 1.1 should be operative for a whole class of data, not just for one particular item. For instance, any force function f = f (t, x, ddtx ) from a very large class of such functions can be put into Model 1 to obtain the corresponding position x = x(t). Thus, the idea of a complete model is very close to the basic mathematical notion of a mapping, function or operator (see Appendix A). A mapping consists of a domain, a co-domain and a rule that assigns a member of the co-domain to each member of the domain. The subset of the co-domain that is in this way connected to the domain by the function is called the range of the function. In Model 1 the domain could be all possible force functions and the co-domain all trajectories. However, there is a difference between the purely mathematical notion of a mapping and the concept of a complete model: the latter has an interpretation in terms of objects or phenomena in the real world. The nature of the interplay between models and the real world is discussed in the next subsection. Implicit in the above discussion is the question of well-posedness of mathematical problems. A mathematical problem is well-posed if it has one and only one solution, i.e., existence and uniqueness of solutions holds, and the solution depends continuously on the data. Thus, a well-posed mathematical problem

6

1 Introduction

with a physical interpretation defines a complete model, and the mapping represented by the complete model exists and is continuous. However, there are situations where it is useful to think that a mathematical problem defines a complete model even when it is not well-posed. For instance, in nonlinear elasticity the possibility of buckling phenomena introduces multiple solutions, but these solutions are generally distinct and give useful information. This text will not, in general, discuss questions of well-posedness. The mathematical sophistication needed for this is beyond the intended scope of the book. However, the property of uniqueness of a solution will be commented on in relation to statements of complete problems.

1.2 Models, Reality and the Conceptual World W Mechanics, as any branch of science, is based on theory, experiment and ob vation. By theory is meant a conceptual world that should be kept sharply distinct from the physical or real world it characterizes. Experiments and observations deal with the coupling of the conceptual world to the real world. The conceptual world describes regularities in the behavior of real objects or processes, but should not be confused with these. It is not inherent in nature and is very much a “free creation of the human mind”3 . In principle, different conceptual worlds may equally well describe the same regularities in the physical world. We may think of the conceptual world of mechanics as consisting of (A) The concepts of classical mechanics. These are classical concepts of Euclidean space-time geometry and concepts derived from them such as body, velocity and acceleration, as well as the implicitly defined concepts of mass and force. (B) Rules and specifications that connect and add structure to the concepts. (C) Complete models constructed on the basis of (A) and (B). The space of classical mechanics is Euclidean. In such a space we can talk of geometric vectors: force, velocity and acceleration turn out to be such vectors. In addition to space we need a measure of time which makes it possible to talk of change of position and thereby of velocity and acceleration. We say that a concept is implicitly defined4 , as in (A), when it is related to other concepts only by a set of rules. This is classical in geometry where concepts such as point, line and plane are defined by the properties expressed by axioms. In mechanics we start on the foundations given by Euclidean geometry and add the concept of a body consisting of material points, and velocity 3 4

Quote from Einstein. Sometimes the term undefined is used instead of implicitly defined. This is not used here because it gives the wrong impression: implicitly defined concepts are well-defined by the rules, rather than not defined at all.

1.2 Models, Reality and the Conceptual World

7

and acceleration of these points. Next, we add concepts of mass and force. Rules defining these implicit concepts are usually not known as axioms or postulates in mechanics. Instead, we talk about laws or assumptions, where the term laws is used for what seems to the more important rules. It is important for one’s understanding of mechanics to realize that the concept of force belongs to the conceptual world and is a creation of the human mind. It is related to conceptions of muscular intensity and external objects acting on our bodies, but this is not the main motivation for its use. Rather, the main justification for the concept of force is that valid models for motion and deformation can be created on its basis. Note that other concepts such as velocity and acceleration also clearly belong to the conceptual domain, but they seem to be less difficult to grasp than that of force, maybe since they are more directly related to physical observations. On the basis of implicitly defined concepts and rules, i.e., on the basis of (A) and (B), we build complete models. Such models, if they are of any use, should relate to relevant facts from the real world. This is achieved by a two–way activity as shown in Figure 1.3. When we construct the complete model we are inspired by features of the physical world that are important for the studied phenomena while irrelevant features are disregarded, i.e., an idealization is made. When a solution of a complete model is at hand, we want this solution to correspond to something in the real world. This is achieved by an interpretation of the solution. For both of these activities we need to choose a referent in the real world for a concept in the conceptual world. For instance, the mass point in Model 1 above may have the planet referent if a model of the solar system is considered, or in a case of molecular Tellus as dynamics a single molecule may be the referent.

Fig. 1.3. The connection between the real or physical world and the conceptual world of concepts and theory.

When a solution of a complete model is interpreted by use of referents we can compare this solution with known experimental or observational facts. If the comparison gives close agreement we say that we understand the physical

8

1 Introduction

observation or experiment. Sometimes we can also adjust the model by choosing parameters to fit the experiments. When a model fits experiments to a desired degree within a domain of applications, the model is validated. Once a model is validated it can be used, within the validated domain, for simulation of complex systems, the behavior of which are not known beforehand. Today, this is usually an exercise in computational mechanics involving numerical methods such as those of finite differences and finite elements. In this context we can also ask the question: why can the conceptual world of mechanics, by which models are built, be trusted in some experimental sense? A direct comment to this question is that it is not possible to validate the conceptual world directly. What one can do is to validate individual models and if this is possible in a large number of cases, the conceptual world is taken to be correct. One should also realize that even if a certain model poorly represents experimental data, the first conclusion is not that classical mechanics is not capable of representing the phenomena. Rather, as indicated in (A) and (B), when constructing models we make, besides the more fundamental rules, a number of specifications, which reflect a level of refinement of the model. For instance, we can use different geometrical shapes as models of bodies. A body may be discrete, i.e., consisting of a finite number of ma a rial points, or it could consist of a continuum of points in one, two or three dimensions. Furthermore, we make special assumptions such as saying that a body is rigid, elastic and so on. Thus, by making refinements within the realm of classical mechanics we can represent a very large domain of physical phenomena, and probably no other theory has a larger body of experimental and observational evidence to support it. Given the success of classical me h nics, a necessary requirement on theories which claim to have – in principle – a larger domain of applicability, such as quantum mechanics and relativity theory, is that they are consistent with classical mechanics. However, these refined theories are frequently less practical than classical mechanics, even to the extent that domains of applicability do not overlap at all in practice.

1.3 The Idea of an Open Scheme As should be clear from the preceding section, building a model involves a large number of rules and specifications and, clearly, these have different levels of generality. The most important rules are usually called laws, and among the laws there are three that are valid throughout all of classical mechanics. We call them universal laws or physical principles and they are: • • •

Conservation of mass. Euler’s law of linear momentum. Euler’s law of angular momentum.

As will be seen in upcoming chapters, for mechanics of particles, Euler’s laws are essentially equivalent to Newton’s laws, but these laws are also applicable to continuum models which is not true with Newton’s laws.

1.3 The Idea of an Open Scheme

9

Now, the universal laws are stated in terms of the concepts of momentum, force and torque, and these need to be made precise in order for the universal laws to be of any use. To that end we choose a geometric model of a body and specify what degrees of freedom a material point posses. We also need to specify what is the force system on the body. Here the idea of a cut principle, originating with Euler and relating to the familiar concept of a free body diagram, appears. Finally, we need assumptions defining how forces depend on the motion. These assumptions are called particular laws. It is by these that we characterize, e.g., if the mechanical body behaves like a fluid or a solid, or if it can be regarded as incompressible or rigid. It is natural to divide particular laws into three groups: • • •

Kinematic constraints. Force laws. Constitutive laws.

The logic behind this division is indicated in Chapter 2 and thoroughly treated in Part III of the text. Thus, there is a hierarchy of assumptions and specifications leading to a complete model. However, the upper level of this hierarchy is general and valid for all situations. We call it an open scheme. It is the skeleton of classical mechanics and may be summarized as follows: 1. Specification of the space in which mechanical objects move, and related mathematical concepts such as vectors and tensors. 2. Universal laws or physical principles. The open scheme is open since the universal laws contain notions which have to be made precise before their consequences in terms of, for instance, partial differential equations, can be derived and models can be obtained. This process of making precise is not algorithmic in the strict sense of the word. It certainly contains a creative ingredient, which has been identified as a pattern recognition skill. Nevertheless, we can offer a step-by-step approach to modelling in classical mechanics, using the skeleton provided by the open scheme. Modelling – step-by-step 1. The first step is to define what is a body and its elements, the material points. In this text, essentially four models of material bodies will be introduced: the discrete, one-dimensional, two-dimensional and threedimensional models. 2. Next one needs to define the ingredients of the universal laws. These are: mass, linear and angular momentum, and total force and torque. Such definitions are dependent on which body model is considered, and even within each model slightly different definitions are possible. After these definitions of basic notions have been introduced, the universal laws

10

1 Introduction

produce differential equations of continuity and motion. However, these equations still do not, in general, constitute a complete mathematical model: after it is decided what is the data, they are not sufficient to uniquely define an answer. 3. To close the modelling and obtain a solvable problem we need to introduce particular laws which are of three kinds: kinematic constraints, force laws and constitutive laws. Following this introduction, in Chapter 2 the details of the open scheme are given. Steps 1 and 2 in the modelling process are covered in Part II of the text. The first step is performed in Chapter 3. The second step, for different geometrical models, is performed in subsequent chapters of Part II. In Part III, step 3 of the modelling process is realized.

Exercises 1.1. For what do we use complete models? 1.2. Discuss the physical referents of the concept of force. 1.3. Classification of models. Make a list of complete models of physical phenomena that you have encountered during your previous education. Group these models into four categories: mechanical models, thermomechanical (or thermal) models, electromechanical (or electrical) models and others.

2 Open Scheme

This chapter presents the open scheme of mechanics, as discussed in Chapter 1. The space in which mechanical objects move is specified, and related math matical concepts such as geometrical vectors and tensors are defined. Next, the laws of mechanics are discussed. Discussions on the upper level of the conceptual world of mechanics, i.e., of the open scheme, are implicit in every textbook on mechanics. However, introductory texts usually start by mass point mechanics, citing Newton’s laws as basic postulates. Unfortunately, such a starting point is not possible for continuous bodies without making unnatural assumptions. Books on continuum mechanics therefore start from equations of momentum balance, i.e., from Euler’s laws. That such an approach is a general one valid for all of me anics, including mass point mechanics, is not always mentioned; exceptions are work by Truesdell and co-workers, [19, 18, 17], and a few textbooks (Fox [9] and Uhlhorn [20]) intended for introductory courses.

2.1 The Physical Space E, Geometric Vectors and Tensors Mechanics deals with motion of material bodies. To measure motion one needs the notion of time, and a way to specify place of identifiable parts of bodies. Such parts are known as material points. Time is given by having access to a clock and is represented mathematically by a scalar t. Position or place of a material point is measured with respect to some physical object which we regard as essentially undeformable. Depending on the application under consideration, different physical objects may play this role of such a reference frame. For instance, the walls of a laboratory based on the surface of the earth suffice in many cases. For astronomical considerations, the sun and the fixed stars taken as a complex is a usual reference frame. In any case, we think that it is possible to identify individual points, places or positions, and these items form a space which we denote by E and call the physical space. Together with a measure of time, E may be called a frame of reference or an observer.

11

12

2 Open Scheme

Fig. 2.1. Two different identifications of physical space: E 1 is defined by the sun and the fixed stars – a heliocentric reference frame; E 2 is defined by the walls of a laboratory placed on the surface of the earth.

The letter E is chosen since in classical mechanics we assume that the physical space is a Euclidean1 space and therefore geometric vectors may be thought of as directed line segments in E. Given a pair of points x and y in E, the geometric vector v from x to y is written y − x, i.e.,

v = y − x.

(2.1)

This text presumes no background knowledge in Euclidean geometry and geometric vectors other than that given in basic courses in mathematics and mechanics, but, for reference, precise definitions of Euclidean point spaces and corresponding vector spaces are given in Appendix B. However, the difference between a geometric vector and its rectangular scalar components is emphasized in the following. Let {e1 , e2 , e3 } be a right-handed orthonormal set of base vectors. Then every geometric vector v may be represented as v = v1 e1 + v2 e2 + v3 e3 , 1

Euclidean space, or more precisely Euclidean point space, is the standard space that one meets in elementary courses in vector algebra and mechanics. It is a space where a rectangular coordinate system can be used to represent points. A precise definition of the Euclidean point space is given in Appendix B.

2.1 The Physical Space E , Geometric Vectors and Tensors

13

where (v1 , v2 , v3 ) are the scalar components of v. Note now that these components depend on the choice of base vectors: a different set of base vectors gives different scalar components for the same geometric vector. Expressed differently, v is invariant with respect to base vectors while the components (v1 , v2 , v3 ) are not. Since equations expressing mechanical truth should not depend on the particular coordinate system chosen, it is frequently useful and illuminating to express these equations directly in the geometric vectors instead of components. Such equations can, when actual numerical computations are performed, be transformed to equations in the scalar components by specifying a particular coordinate system. A similar remark as that concerning vectors and their components, can be made concerning the representation of points in E. Let again {e1 , e2 , e3 } be a right-handed orthonormal set of base vectors and let o be a specific point in E called the origin. We then define the position vector x of the point x and express this vector in its components as follows: x = x − o = x1 e1 + x2 e2 + x3 e3 . Thus, given a right-handed orthonormal set of base vectors and an origin we may identify x with the components (x1 , x2 , x3 ). In the following we will, however, as much as possible, use an abstract point of view, and points will be seen as identical to their components only when this makes mathematical definitions or manipulations easier. Note that this notation is different from that usually found in elementary mechanics courses, where points are denoted A, B, C, . . . and the definition −− → of a vector using two points is written AB. The present notation has the advantage of suggesting the operation of adding a vector and a point, i.e., (2.1) is equivalently written as y = v + x. We say that v translates x. Geometric vectors formed from two points of E naturally have the physical dimension of length. However, vectors of velocity, acceleration and force also need to be measured with respect to the chosen frame of reference. Thus, by agreeing on predetermined scales, such vectors may also be seen as geometric vectors related to two points of E. This is utilized when we make drawings and include, for instance, both acceleration and force vectors in the same drawing. Even though real mechanics always happens in a three-dimensional space, frequently we simplify models by making them two-dimensional. Thus, occasionally in this text E can be thought of as a two-dimensional space. In fact, when solving a mechanics problem on paper, one draws figures, and E may then have the paper itself, which certainly has a two-dimensional character, as referent. As will be clear in the sequel, many mechanical models are more easily formulated if we define certain geometrical objects which in a sense extend the notion of a geometric vector. We call these objects tensors: a tensor2 T is a linear map that assigns to each geometric vector u a new geometric vector. 2

The object defined here would sometimes be called a second-order tensor.

14

2 Open Scheme

v = T u.

(2.2)

A tensor is related to a three by three matrix similarly to the way a vector is related to its components: given a right-handed orthonormal set of base vectors {e1 , e2 , e3 }, the nine components Tij of a tensor T are defined by Tij = ei · T ej ,

i, j = 1, 2, 3,

where a central dot indicates scalar product of geometric vectors. With this definition v = T u is equivalent to the matrix equation. ⎡ ⎤ ⎡ ⎤⎡ ⎤ v1 T11 T12 T13 u1 ⎣ v2 ⎦ = ⎣ T21 T22 T23 ⎦ ⎣ u2 ⎦. (2.3) v3 T31 T32 T33 u3 To indicate that a tensor T , and similarly for a vector, has a certain matrix representation we use the symbol ∼, i.e., ⎡ ⎤ T11 T12 T13 T ∼ ⎣ T21 T22 T23 ⎦. T31 T32 T33 Note, however, as is the case for vectors, the matrix or component repre sentation of a tensor depends on the set of base vectors, while the tensor itself is a purely geometrical object which makes sense even without introducing a particular base.

2.2 Laws of Mechanics 2.2.1 Universal Laws Mechanics consists of different theories which are applied to different areas of natural phenomena and engineering. Some examples of such theories are beam theory, rigid body theory, linear elasticity, truss models, viscous and inviscid fluid models, etc. Each theory is based on a set of assumptions or postulates. When such assumptions are felt to be of central importance they are frequently called laws. Some of these laws are particular to the specific theory, while others have counterparts in all mechanical theories. Over three hundred years of observations, experiments and striving for unification and order have shown that it is useful to single out three laws that, given the right interpretation, hold for all (classical) mechanical theories: the universal laws. We will state these laws in mathematical form, but it should be emphasized that these statements are, at the present stage of refinement, to be thought of as general ideas rather than as mathematical equations. The reason is that the ingredients of the laws – the notion of a body and its parts, mass, linear and angular momentum, total force and torque – can be made precise in different ways and this is the subject of upcoming chapters. The universal laws are as follows:

2.2 L

fM h

i

15

Conservation of mass The mass of any part of a body, M (P), does not change with time: ∂ M (P) = 0. (2.4) ∂t Euler’s law of linear momentum The change of linear momentum, p(P), equals the force resultant, f (P): ∂ p(P) = f (P). ∂t

(2.5)

Euler’s law of angular momentum The change of angular momentum with respect to a fixed point o, ho (P), equals the torque resultant with respect to the same point o, co (P): ∂ ho (P) = co (P). ∂t

(2.6)

The letter P stands for a part of a body. The whole body is denoted as B. The notations used in (2.4) through (2.6) implies that all concepts involved in the laws depend on which part of the body is considered. Thus, when building a specific mechanical theory, as a rule, we have to start by defining mathematically what is a body and what are its parts. Once this is done, we can go on to define mass, linear and angular momentum, total force and total torque. Mass and linear and angular momentum will have essentially the same structure in all specific theories: mass will always have an additive property; for a mass point with mass m and velocity v, linear momentum is mv, i.e., the product of mass and velocity; angular momentum is the moment (in the cross product sense) of this product about a point. Total force and torque are more difficult to define than the momentum quantities, and such definitions may look quite different from one specific model to another. However, they are, generally, based on the cut principle, i.e., on an appropriate free body diagram for the body part under consideration. The bold face notation used for linear and angular momentum, total force and total torque indicates that these quantities are geometric vectors. Euler’s laws, as they are stated here, have the structure of conservation laws: the momentum quantities are subject to conservation unless they are altered by inflow of momentum in terms of forces and torques. Other forms of these laws, which are obtained by using conservation of mass, have the structure of linear relations between force and acceleration, and between torque and angular acceleration. These forms will be derived for different models of bodies in the upcoming chapters. It is interesting to note that the conservation forms of Euler’s laws remain correct in special relativity. Note that the validity of (2.5) and (2.6), in the sense of being a correct description of physical observations, is not to be expected for all interpretations of E, relative to which vector quantities are measured. Experience has shown that if we study mechanics on earth, a physical space E defined by the surface

16

2 Open Scheme

of the earth works well, while a study of the solar system needs an E defined by the fixed stars. On the other hand, if E is the floor of a merry-go-round we cannot expect (2.5) and (2.6) to be in resonable agreement with observations. Thus, there is a certain class of interpretations of E for which (2.5) and (2.6) are valid, and these are known as inertial spaces. For a non-inertial space we can obtain the correct form of Euler’s laws by transforming from an inertial space. Pseudo-forces3 known as centrifugal and Coriolis forces then enter. Note that this text is not concerned with non-inertial spaces and pseudo-forces are not treated. The universal laws can certainly not be regarded as obvious truths, as is evident from their historical background. Euler’s laws generalize experience from special cases, and faith in these laws has gradually been built up from earlier more specific laws formulated by Huygens, Newton and others. They were stated by Euler in 1776. Euler’s law of linear momentum can be seen as a generalization to more general mechanical systems of Newton’s second law, which is a law that relates to mass points only. When Newton’s second law was first conceived, in the 17th century, experiments on impact seem to have played an important role. Consider two mass points with masses mA and mB and velocities v A (t) and v B (t) at time t. Experiments then show that the sum of linear momentum is conserved, i.e., mA v A (t1 ) + mB v B (t1 ) = mA v A (t2 ) + mB v B (t2 ),

(2.7)

where t1 is a time just before impact and time t2 is a time just after. Equatttion (2.7) means that momentum is transferred from one mass point to the other. The momentum-changing influence on point A from point B during the time interval from t1 to t2 is called an impulse and is denoted I A (t1 , t2 ), i.e., I A (t1 , t2 ) = mA v A (t2 ) − mA v A (t1 ).

(2.8)

By putting a general time t into this equation instead of t2 , taking the time derivative with respect to t, and identifying the time derivative of the impulse with the total force, we obtain a special case of (2.5). The need for Euler’s second law, on angular momentum, as a basic postulate besides the first law, may not be obvious. Especially since it seems to be standard practice in elementary books on particle mechanics to derive (2.6) as a consequence of (2.5) and certain assumptions concerning the nature of internal forces. However, for more general mechanical models, such a derivation is not possible without making highly unnatural assumptions. Moreover, the reader may also recall her training in elementary statics, where clearly both force and torque equilibrium are needed as basic axioms. The most basic example of this is probably the law of the lever. Finally, we mention that a 3

Real forces can always be considered as arising due to interaction between bodies. In the case of internal forces, both interacting bodies reside inside the system, while for external forces, one of the bodies is outside of the system. Pseudo-forces lack this interpretation of interaction.

2.2 Laws of Mechanics

17

deeper investigation of this question reveals that the existence of two vector laws is related to the fact that a rigid body in physical space has six degrees of freedom. 2.2.2 Particular Laws In the previous subsection we presented three laws which experience has shown to be universal, i.e., valid for all mechanical problems. However, experience has also shown that these laws are not sufficient in the sense that they do not in general give complete models from which we can find the motion or forces of interest. Therefore, the universal laws have to be complemented by force laws, constitutive laws and kinematic constraints. A typical force law is the law of gravitation. A constitutive law is a particular type of force law, for instance, Hooke’s law of elasticity, that relates to internal forces and somehow reflects the constitution of the specific material. Examples of kinematic constraints are that rigid obstacles in space cannot be penetrated or that the material is incompressible or rigid. As a rule, a kinematic constraint cannot be given without specifying the nature of the force maintaining it.

Exercises 2.1. Show that given a right-handed orthonormal set of base vectors {e1 , e2 , e3 }, (2.2) and (2.3) are equivalent. 2.2. Is conservation of mass an obvious truth? What if chemical or nuclear reactions take place? 2.3. Make a list of particular laws that you have used in previous studies of mechanics.

Part II

Basic Models: Geometry and n niversal Laws

3 Bodies and Their Placements in E

A central concept in mechanics is that of a body, or, more precisely, a material body. A body is an abstract or mathematical notion that represents anything that we intend to analyze by methods from mechanics. It can be, for instance, a vehicle such as a cars, airplane or boat, an astronomical object such as a star or planet as well as a gas or fluid. In the mathematical theory a body is a set which contains identifiable material points. Physically these points correspond to a mark or dot somewhere in or on the physical object and have (obviously?) nothing to do with atoms or subatomic particles. Mechanics certainly does not deny the existence of such particles, but simply does not need such ideas to be a useful theory. From a geometric point of view there are basically four alternative models of bodies and material points. The distribution of material points of each such model represents different geometric dimensions: zero (discrete), one, two and three. The three latter models will sometimes be called continuum models. Frequently, a three-dimensional body model is conceived to be more correct than lower dimensional models since E, in which the body resides, is threedimensional. However, from the point of view of consistency, lower dimensional models can be just as correct; the final determination of the validity of a model can be made only by comparison with observations or experiments. Even though each model is logically distinct from other models we, use the same notation B for a body in all cases. Similarly, a material point is denoted by X and the placement map, that takes material points in B and places them into E, is denoted by φt in every model.

3.1 Discrete Body Model The model of this section is the one studied in elementary courses of mechanics of particles. It concerns a finite number of material points. These material points are tracked by labelling them by, e.g., 1, 2, 3, etc. When studying a

21

22

3 Bodies and Their Placements in E

model consisting of, say, N material points, we introduce the set1 B = {1, . . . , N }, which from a mathematical point of view is the body. A part of the body, denoted P is simply a subset of {1, . . . , N }, i.e., P ⊂ {1, . . . , N }. A typical member of B is denoted by the capital letter X. At each time instant, each material point occupies a place in the physical space E. An element, member of, or place in, E is denoted by the lower case letter x. We say that every X ∈ B is mapped to a place x ∈ E. We denote this mapping by φt and write x = φt (X), (3.1)

which is illustrated in Figure 3.1. Since material points move in space, i.e., occupy different places at different times, the mapping φt is time-dependent, which we indicate by the subscript t. For a fixed time, we may call φt a placement of the body into E. For a fixed material point X, φt (X) is the motion of that point. We write Bt = φt (B) for the collective position of all material points in E at time t.

Fig. 3.1. The mapping φt for a discrete body model.

It is reasonable to assume that two or more material points cannot occupy the same place in E at the same time. If this is satisfied there is a one-to-one correspondence between B and Bt , and, therefore, there exists an inverse of φt at all times, i.e., there exists a mapping from Bt to B, denoted by φ−1 t , such that φt (φ−1 φ−1 x ∈ Bt , X ∈ B. t (x)) = x, t (φt (X)) = X 1

Appendix A gives a short discussion of the basic mathematical notions of sets and functions.

3.2 One-Dimensional Body Model

23

The velocity, V (X, t), of the material point X is the time derivative of φt , i.e., ∂ V (X, t) = φt (X). ∂t Similarly, the acceleration is defined as A(X, t) =

∂ ∂2 V (X, t) = 2 φt (X). ∂t ∂t

Note that velocity and acceleration are geometric vectors and, therefore, boldface notation is used. They become geometric vectors since the definition of a derivative contains the difference between two objects and these objects are points in this case. This is analogous to the definition of a tangent vector in the next subsection.

3.2 One-Dimensional Body Model In elementary mechanics there are several theories where properties are assumed to vary in one direction only. Examples include bars, ropes, strings and beams from solid mechanics, and flow in pipes from fluid mechanics. One way of making clear what is the basic character of these theories is to say that they are one-dimensional models of material points. As for the discrete model, the material points occupy places in E. However, we cannot use the natural numbers to define the body simply because in this case there are more material points than natural numbers. The appropriate mathematical structure that represents the body is an interval of the real line R, i.e.2 , B = [Xs , Xe ] ⊂ R, where Xs is the start material point and Xe is the end material point and we require that Xs < Xe , i.e., Xs  = Xe so that a body cannot be a single material point. A part of B is an interval included in [Xs , Xe ], which is not a single point, i.e., P = [X1 , X2 ] ⊂ [Xs , Xe ],

Xs ≤ X 1 < X 2 ≤ X e .

As for the discrete model, there is a map, again denoted by φt , that takes material points of B and places them into E. Equation (3.1) is valid also for the one-dimensional model, with the change of interpretation indicated. We call the mapping φt a placement and φt (X) is the motion of the material point X, with velocity and acceleration defined as in the previous section. The set Bt = φt (B) is the physical configuration of the body we are investigating. The set B is in some cases a pure thought construction, while in other 2

Square brackets indicate a closed interval of the real line. An open interval is indicated by standard brackets.

24

3 Bodies and Their Placements in E

Fig. 3.2. The mapping φt for the one-dimensional model.

situations it has a clear referent in reality. An example of the latter situation is when a flexible elastic string is considered. When all loads are removed from such a string it will change to a natural unstressed configuration and may be straightened and placed alongside a ruler to give each material point a number X. An example when it is not so natural to think of a physical counterpart for B is in fluid mechanics type problems such as flow in pipes. However, in principle it is always possible to assume that a set B exists and this turns out to be an extremely useful idea. The mapping φt is assumed to be differentiable as many times as necessary with respect to both X and t, and for a fixed t it should be one-to-one and invertible. In fact, it is natural to assume a condition which is slightly stronger than invertibility. To state this condition, define the derivative φt (X + ∆X) − φt (X) ∂φt (X) = lim . ∆X→0 ∂X ∆X

(3.2)

Clearly, this is a geometric vector since φt (X + ∆X) φt (X) is the difference between two points of E. A natural condition is now that    ∂φt (X)    (3.3)  ∂X  > 0, for all X ∈ B. If this is not satisfied, material points are not placed on a line in E, but rather collapsed into one place and φt is then not one-to-one. The absolute value in (3.3) may be imagined to characterize the relative change in length of the infinitesimal element dX as it is stretched by the mapping φt : if the vector (3.2) has less than unit length, a local shrinking occurs, and if it is longer than a unit vector then there is a local stretching. When (3.3) is satisfied, Bt = φt (B) represents a curve in E and (3.2) is a tangent vector to this curve, as seen in Figures 3.2 and 3.3.

3.3 Two-Dimensional Body Model

25

t (X) . It is produced in the limit process given in Fig. 3.3. The tangent vector ∂φ∂X (3.2), where Xn = X + ∆X approaches X.

3.2.1 Pipe Flow Model There is generally a difference between solid mechanics problems like those representing beams and ropes, and fluid mechanics problems like flow in pipes. In the former case, for instance when a bar is vibrating, the curve represented by the particles in the physical space looks different for every time instance. In the latter case the material particles will for every time occupy a subset of a fixed line in the physical space. This line, or pipe, is denoted by C. Thus, in fluid mechanics problems φt (B) ⊂ C, as seen in Figure 3.4. Since every material point X will always be mapped to a point along the line or pipe C, we can keep track of the location of material points in E by using a path coordinate s. The mapping φt may then be seen as a mapping from B to C and we may write3 s = φt (X).

(3.4)

3.3 Two-Dimensional Body Model Two dimensional models in the field of strengths-of-materials are called sheets, plates and shells. A two-dimensional theory in fluid mechanics is Reynold’s equation in hydrodynamic lubrication. From a geometric point of view, the characteristic of such theories is that we may label material points by pairs of real numbers. Let R2 denote the space of pairs of real numbers and let Ω stand for a subset of R2 that is sufficiently nice to allow the mathematical 3

The mapping defined by (3.4) is denoted s(X) in Chapter 6.

26

3 Bodies and Their Placements in E

Fig. 3.4. Pipe flow model.

operations involved in the theory to be permissible. Typically this means that Ω has a piecewise smooth boundary, and is the closure of a connected open set so that it has a well-defined interior. The set Ω represents the body, i.e., B = Ω ⊂ R2 . Parts of bodies, P, are simply nice subsets of Ω, i.e., P ⊂ Ω. As in the one-dimensional case, (3.1) is valid given the appropriate interpretation and a restriction similar to (3.3) holds. Note that Bt= φt (B) is generally a curved two-dimensional surface imbedded in the three-dimensional space E. Figure 3.5 illustrates the elements involved in the two-dimensional model.

Fig. 3.5. The mapping φt for a two-dimensional body model.

3.4 Three-Dimensional Body Model

27

3.4 Three-Dimensional Body Model Since physical space is three-dimensional, and since it is very natural to imag ine that material points fill three-dimensional regions of physical space, threedimensional theories have a special status and are frequently conceived as being more real than corresponding lower-dimensional counterparts. Furthermore, by making assumptions on displacement, velocity and/or stress fields, such as assuming that they vary only in one or two space dimensions, it is possible to derive lower-dimensional theories from the corresponding threedimensional ones. Examples of this will be given in Chapter 12. Also for the three-dimensional body model, (3.1) is valid if given the right interpretation. Following the one and two-dimensional case, the body could be taken as a nice subset of R3 , the space of triples of real numbers. However, it is also convenient, and used in this text, to imagine that the body B is a subset of E, see Figure 3.6. We call such a subset a reference configuration. A part of a body is defined as a nice subset of B, i.e., P ⊂ B ⊂ E. A useful possibility is to let B coincide with Bt = φt (B) at time t = 0, i.e., φt for t = 0 is the identity map.

Fig. 3.6. In the three-dimensional body model we take B to be a subset of E .

In the one-dimensional model a local measure of shrinking and elongation is obtained by the limit in (3.2). We want to obtain a similar measure in the three-dimensional case. However, now ∆X is the difference between two points in E and therefore a vector. An object that generalizes the tangent vector (3.2) to the three-dimensional model is the tensor F (X, t) defined by the expression

28

3 Bodies and Their Placements in E

φt (X + ∆X) − φt (X) = F (X, t)∆X + o(∆X), where o(∆X) is any vector valued function that approaches zero faster than ∆X, usually denoted “small o of ∆X”. This definition suggests the expression dx = F dX which may be interpreted as if F maps the (infinitesimal) line element dX in B into the line element dx in Bt . The tensor F (X, t) is known as the deformation gradient. Given an orthonormal base {e1 , e2 , e3 } and an origin o we may write φt (X) − o = φt1 (X1 , X2 , X3 )e1 + φt2 (X1 , X2 , X3 )e2 + φt3 (X1 , X2 , X3 )e3 where Xi and φti are the coordinates of X and φt , respectively. Omitting arguments, the following matrix representation of F (X, t) is then valid: ⎡ ∂φ ∂φ ∂φ ⎤ t1

t1

t1

⎢ ∂X1 ⎢ ⎢ ∂φt2 F (X, t) ∼ ⎢ ⎢ ∂X 1 ⎢ ⎣ ∂φ

∂X2 ∂φt2 ∂X2 ∂φt3 ∂X2

∂X3 ∂φt2 ∂X3 ∂φt3 ∂X3

t3

∂X1

⎥ ⎥ ⎥ ⎥. ⎥ ⎥ ⎦

The determinant of F = F (X, t), denoted det F , plays an important role in the further development of three-dimensional models. We can calculate det F in the standard way from the matrix representation of F , and it can be shown that the value will be independent of the particular base used. In fact, it holds that F u · (F v × F w) det F = u · (v × w) for any set of geometric vectors {u, v, w} such that the denominator is nonzero. Here × is the cross product (see Appendix B.1). Now, u · (v × w) represents the volume of a parallelepiped, see Figure 3.7, with positive sign if {u, v, w} is right-handed and negative sign otherwise. Similarly, F u · (F v × F w) represents the volume (with sign) of the parallelepiped after mapping with F .Thus, det F is the relative change of volume under mapping by F . A negative value of det F means that F inverts volumes. We do not accept this from a physically reasonable mapping φt and therefore we require det F > 0.

(3.5)

This is a condition analogous to (3.3) for the one-dimensional model. The determinant of the deformation gradient, det F , represents relativity of volumes also in the infinitesimal sense. This may be seen from the changeof-variables formula, familiar from basic calculus, which, under condition (3.5), implies that   dV Vx = det F dV VX , (3.6) Pt

P

3.5 R fi

dG

i M d l

f B di

29

Fig. 3.7. The determinant of F , det F , can be represented as the quotient of volumes of parallelepipedes.

where dV VX and dV Vx represent volume elements in B and Bt , respectively. Equation (3.6) means that the volume of the subbody P under the mapping φt , i.e., Pt = φt (P), can be calculated by integrating over P and compensating by the relativity of volumes expressed by det F .

3.5 Refined Geometric Models of Bodies The concept of the material point used in previous subsections does not include any notion of rotation or deformation of the point itself. A possible extension in this direction, usually referred to as a Cosserat theory, is to associate with each material point X and each time t a set of vectors di (X, t), i = 1, . . . , n, called directors. For instance, the rotation of a material point can be characterized by a set of orthonormal vectors {d1 (X, t), d2 (X, t), d3 (X, t)} and the motion of a body is then given by φt (X, t), d1 (X, t), d2 (X, t), d3 (X, t) for X ∈ B. In plane (two-dimensional) problems this description can be simplified, and rotation can be represented by a scalar value ϕt (X), specifying a rotation with respect to a reference line. The motion is then given by φt (X, t), ϕt (X) for X ∈ B.

30

3 Bodies and Their Placements in E

In this work we will use the latter possibility when developing beam theory in Chapter 9. Apart from that we will generally suppose that a material point has a placement but no rotation or deformation. However, when developing mechanical models for material with microstructure, e.g., composites or granular material, such possibilities may be considered.

4 Discrete Model

In this chapter the discrete body model will be given additional structure, resulting in the theory of mass point mechanics studied in basic courses. The additional structure is precisely what is needed to apply the universal laws introduced in Section 2.2.1. That is, we need to define mass, linear and angular momentum as well as force and torque resultants. As a first simple consequence of the universal laws and these definitions we derive an expression for the acceleration of the center of mass, showing that a system of material points behaves like a single material point if motion of the center of mass is under study. Next, we specify the force system, giving force and torque resultants. As a consequence of Euler’s laws we derive the law of action and reaction. In elementary courses on the mechanics of particles this law is usually taken as a basic postulate. In this text, on the contrary, it is derived as a theorem. This, perhaps unusual, approach has the advantage that it can be generalized to continuum models.

4.1 Mass and its Conservation Each material point X ∈ B = {1, . . . , N } is given a positive scalar mX , called the mass of the point. The mass M (P) of a part P of the body is the sum of masses of points included in the part: M (P) = mX . X∈P

Since a part of a body may be just one particle, a direct consequence of the law of conservation of mass is ∂ mX = 0 for all X ∈ B. ∂t That is, each material point has a constant mass.

31

(4.1)

32

4 Discrete Model

4.1.1 Center of Mass For a body B, the center of mass xc , for a given placement φt , is defined as follows: (φt (X) − o)mX , (4.2) (xc − o)M (B) = X∈B

where xc − o is the position vector of the center of mass and φt (X) − o is the position vector of the point φt (X) which is occupied by the material point X at time t. The center of mass may be seen as the point of application of the resultant of the system of forces produced by the weights of the material points in B. The velocity and the acceleration of the center of mass are defined as ∂ ∂ v c = xc , ac = v c . ∂t ∂t By using (4.1) and definition (4.2) we find V (X, t)mX , ac M (B) = A(X, t)mX . (4.3) v c M (B) = X∈B

X∈B

In the next subsection we will see that these equations may be used to draw conclusions on the motion of the center of mass.

4.2 Linear and Angular Momentum The linear momentum for a part P of a discrete body is the sum of the momentum of all its material points. The momentum of a material point is in turn the product of mass and velocity and we thus define linear momentum as p(P) = V (X, t)mX . X∈P

A position vector of the place φ t(X), occupied by the material point X, is written r(X, t) = φt (X) − o, where o is the origin. The angular momentum with respect to o for a part P may then be defined, as for the linear momentum, as ho (P) = r(X, t) × V (X, t)mX . X∈P

We calculate the time derivatives of these vectors, since Euler’s laws require such derivatives. Using the law of conservation of mass, in the form (4.1), one finds that ∂ p(P) = A(X, t)mX . (4.4) ∂t X∈P

Also by (4.1) we have

4.3 System of Forces, and Force and Torque Resultants

33

∂ ∂ ho (P) = [r(X, t) × V (X, t)] mX ∂t ∂t X∈P

and from ∂ [r(X, t) × V (X, t)] = V (X, t) × V (X, t) + r(X, t) × A(X, t) ∂t = r(X, t) × A(X, t) we find

∂ ho (P) = r(X, t) × A(X, t)mX . ∂t

(4.5)

X∈P

4.2.1 Motion of the Center of Mass Euler’s law of linear momentum, the expression given in (4.3) for acceleration of the center of mass and the above formula for the time derivative of p, (4.4), give f (B) = ac M (B). (4.6) That is, force = acceleration × mass, if by force is meant the resultant force (yet to be precisely defined), by mass is meant the total mass and by acceleration is meant the acceleration of the center of mass. Thus, in a sense, a system of material points can be considered as one single material point if the motion of the center of mass is under consideration.

4.3 System of Forces, and Force and Torque Resultants What is left to define for discrete systems, before Euler’s laws are specified, are the force and torque resultants. As a preparatory for this we need to define what is meant by forces, or, rather, what is the system of forces considered. Such a system is defined by the following two items which hold for all times under consideration: (i) For every material point X ∈ B there is a geometric vector f e (X), called an external force vector. (ii) For every ordered pair1 (Xk , X ) of material points of B, there is a geometric vector f i (Xk , X ), called the internal force with which Xk acts on X .2

1 2

An ordered pair is simply two numbers in order where the two possible orderings of the numbers define two different ordered pairs. The vector f i (Xk , X ) when Xk = X does not enter the analysis, so its value is irrelevant.

34

4 Discrete Model

We also denote by P −1 the complement of P in B = {1, . . . , N }. The total force and torque for a discrete model are now defined as f (P) = f (X, P), (4.7) co (P) =



X∈P

r(X, t) × f (X, P),

(4.8)

X∈P

where f (X, P) = f e (X) +



f i (Xk , X)

Xk ∈P −1

is the part of the total force related to the material point X when part P of the body B is considered. This part of the total force originates from sources external to B and from particles in the complement P −1 : only the internal forces f i (Xk , X ) such that Xk ∈ P −1 , are included in the total force and the total torque. This implies that “actions between” material points both included in P do not contribute but cancel out. An intuitive explanation of this fact is that “one cannot lift oneself by the bootstraps”. See Figures 4.1 and 4.2 for a geometric illustration. There may be a force system defined for each time instant under consideration. Therefore, a more complete notational convention should indicate that force-like variables introduced above are functions of time, as is the case for velocities, accelerations and position vectors. However, for simplicity, such a notational scheme is not used in this text. A convention not to show time dependency explicitly will be used for all force-like variables in the sequel since confusion due to this is not likely.

Fig. 4.1. System of forces for a discrete model of three material points.

In Figure 4.2 a system of three material points is depicted. The part P = {2, 3}, with P −1 = {1}, is considered. Equations (4.7) and (4.8) read in this

4.3 System of Forces, and Force and Torque Resultants

35

case f ({2, 3}) = f (2, {2, 3}) + f (3, {2, 3}), co ({2, 3}) = (φt (2) − o) × f (2, {2, 3}) + (φt (3) − o) × f (3, {2, 3}), where f (2, {2, 3}) = f e (2) + f i (1, 2) and f (3, {2, 3}) = f e (3) + f i (1, 3). The definition of forces on P can be thought of as if we have cut out P from

Fig. 4.2. The individual forces contributing to f ({2, 3}) and co ({2, 3}) for the three point system of Figure 4.1, are shown in solid lines. Dashed forces are forces interior to {2, 3} or forces acting on the part {1}. We imagine a cut that isolates points 2 and 3.

the surrounding and represented the action of the surrounding on P by means of forces. This is the first appearance in this text of the very important cut principle which will be used for other body models in the sequel. The reader should compare this idea with the free body diagram of elementary courses. 4.3.1 Change of Base Point for the Torque Resultant The torque resultant depends on, or is calculated with respect to, the origin, a point in E which is, at least in theory, completely arbitrarily chosen. Therefore, we naturally ask ourselves: what is the relation between torque resultants, calculated for two different points in E, but for the same force system? Say that we are interested in a point x and the origin o. A direct calculation gives

36

4 Discrete Model

cx (P) =



(φt (X) − x) × f (X, P)

X∈P

=



(r(X, t) + (o − x)) × f (X, P)

X∈P

=



r(X, t) × f (X, P) + (o − x) ×

X∈P



f (X, P)

X∈P

= co (P) + (o − x) × f (P), where we have used definitions of total force and torque given above in (4.7) and (4.8). The result is the following connection formula for two torque resultants of the same force system with respect to different points cx (P) = co (P) + (o − x) × f (P).

(4.9)

Below we will use this formula to conclude that the law of angular momentum is form invariant with respect to the base point.

4.4 Explicit Forms of Euler’s Laws We have in previous subsections given the definitions of concepts involved in Euler’s laws (linear and angular momentum, and total force and torque) and derived some preliminary consequences. It is useful at this stage to summarize how Euler’s laws appear, given these definitions and by using, as done above, the law of conservation of mass. From (4.4), (4.5), (4.7) and (4.8) we find the following: Euler’s law of linear momentum f (X, P) = A(X, t)mX . X∈P

Euler’s law of angular momentum r(X, t) × f (X, P) = r(X, t) × A(X, t)mX . X∈P

(4.10)

X∈P

(4.11)

X∈P

Note that since f (X, P) depends on P, in general f (X, P) =  A(X, t)mX even though a superficial look at (4.10) might suggest otherwise. The similar conclusion holds for (4.11). Note also that these forms of Euler’s laws do not have the character of conservation laws. However, they are the most useful forms for our purposes. 4.4.1 Change of Base Point for Euler’s Laws Euler’s law of angular momentum, in the form of (4.11), holds for every choice of base point, not only for the origin, as used above. This follows from the formula (4.9) and Euler’s laws:

4.5

h

h

fA

dR

37

cx (P) = co (P) + (o − x) × f (P) r(X, t) × A(X, t)mX + (o − x) × A(X, t)mX = X∈P

=



X∈P

(φt (X) − x) × A(X, t)mX .

X∈P

The form (2.6) of Euler’s law of angular momentum is also valid for any point of E, not only the origin o, provided that this point is not a function of time, i.e., a moving point. On the other hand, the form derived here is valid even if the base point x is a function of time. The reason for this is simply that no time derivative of x appears in the above calculations.

4.5 The Theorem of Action and Reaction Newton’s third law or the law of action and reaction is given as a basic postulate in elementary books on point mass mechanics. Here we will derive it as a theorem from Euler’s laws. Consider a discrete body B consisting of at least two material points denoted A and B. We write Euler’s laws of linear momentum for the sub-bodies consisting of each of these points as well as for the sub-body consisting of both points: f e (A) + f i (Xk , A) + f i (B, A) = A(A, t)mA , Xk ∈B\{A,B}

f e (B) +



f i (Xk , B) + f i (A, B) = A(B, t)mB ,

Xk ∈B\{A,B}

f e (A) + f e (B) +

Xk ∈B\{A,B}

f i (Xk , A) +



f i (Xk , B)

Xk ∈B\{A,B}

= A(A, t)mA + A(B, t)mB , where the notation A\B means the set of elements which are in A but not in B. Adding the first two of these equations and comparing the result with the last equation gives f i (A, B) = −f i (B, A).

(4.12)

A similar calculation using Euler’s law of angular momentum (4.11) gives r(A, t) × f i (B, A) + r(B, t) × f i (A, B) = 0.

(4.13)

Equations (4.12) and (4.13) mean that the internal forces are of equal magnitude, have opposite directions and have a common line of action. That is, they satisfy Newton’s third law or the law of action and reaction. Figure 4.3 shows two situations: one that satisfies (4.12) but not (4.13) and another one that satisfies both equations.

38

4 Discrete Model

Fig. 4.3. The forces in frame I satisfy (4.13), while those in II do not.

4.6 Newton’s Laws In most elementary books on particle mechanics, the basic postulates are the action and reaction principle, i.e., (4.12) and (4.13), together with Euler’s law for one material point, X, usually called Newton’s (second) law, i.e., f e (X) + f i (Xk , X) = A(X, t)mX . (4.14) Xk ∈B\{X}

From these postulates one derives Euler’s laws, (4.10) and (4.11). Since we have shown that, conversely, (4.12), (4.13) and (4.14) follow from Euler’s laws, one concludes that, for particle systems, there is an equivalence between the approach used here and that given in most elementary text books. However, our goal is to proceed beyond systems consisting of a finite number of mate rial points, to continuum systems. Then, in general, it is not possible to start from Newton’s law and the law of action and reaction. The approach of this text, where Euler’s laws are taken as basic postulates, is, on the other hand, universal and can be applied to all body models.

4.7 A Complete Model Newtons’s law (4.14) and the laws of action and reaction (4.12) and (4.13), or equivalently Euler’s laws, do not, since the forces must be regarded as unknowns, constitute a system of equations from which one can solve the motion of a set of particles. To obtain such a system one needs to add constitutive

4.7 A Complete Model

39

equations for internal forces and force laws for external forces. This will be discussed in Part III. However, if one is interested in the inverse problem, where the forces are obtained from a given motion, such a problem can, in some cases, be formulated at the present stage. A discussion of this follows: For a given motion, the right hand side of (4.14) is known, and (4.12), (4.13) and (4.14) constitute a linear equation system for the forces at each time instant. It is a question whether this equation system determines the forces. We can get an idea of the nature of the answer by counting the number of equations and the number of unknowns. To that end, (4.12) reduces the number of unknown internal forces to one for each pair of points, and using this information in (4.13) we obtain the direction of the force. Thus, (4.12) and (4.13) leave only the magnitude of one internal force for each pair of material points as an unknown3 . Equation (4.14) represents three scalar equations for each point and adds the unknown vector f e (X). Thus, in conclusion, the equation system represented by (4.12), (4.13) and (4.14) leaves one scalar undetermined for each pair of points, and only for the case of a single point, for which there are no internal forces, can one obtain a uniquely solvable system. However, in many specific applications we have additional information on the nature of the forces; for instance, we may know that certain forces have a particular direction or may be zero, and such information may result in a solvable system. Logically, these conditions must be regarded as particular laws and should belong to Part III. Nevertheless, for didactic reasons their existence is acknowledged at this stage. Without being more explicit about these force conditions, they are called simple particular laws and are defined simply as conditions needed to obtain a solvable system. Such a solvable system is an analogy to what in statics is known as a statically determinate problem. Having obtained a solvable problem a complete model is at hand, which we formulate as follows: Force problem of particle mechanics: Given a set of material points B, masses mX , accelerations A(X, t) and appropriate simple particular laws, find internal forces f i (Xk , X ) and external forces f e (X) such that (4.12), (4.13) and (4.14) are satisfied. For linear particular laws, the mathematical form of this model is a linear equation system, and its mathematical well-posedness follows when the system matric is non-singular. 4.7.1 Example: An Inverse Kepler Problem As a very simple example of this force problem we will consider the case of a single material point with mass m in uniform circular motion. Since there 3

Provided that the placements of points do not coincide, i.e., material points do not occupy the same place in E .

40

4 Discrete Model

is only one point involved, no internal forces are present and the problem consists in obtaining f e (X) from (4.14). Circular motion takes place in a plane of E. We use unit vectors e1 and e2 to span this plane and place the origin at the center of the circle. The position vector r(t) = r(X, t) is then given by r(t) = r cos(ωt)e1 + r sin(ωt)e2 = re(t), where r is the magnitude of the position vector and also the radius of the circle, ω is the angular velocity and e(t) is a unit vector in the direction of r(t). This motion is illustrated in Figure 4.4.

Fig. 4.4. The geometry of circular motion.

To obtain the right hand side of (4.14) we calculate the acceleration: A(t) =

∂2 r(t) = −rω 2 cos(ωt)e1 − rω 2 sin(ωt)e2 = −rω 2 e(t) ∂t2

and from (4.14), letting f = f e (X) and m = mX , we obtain f = −mrω 2 e(t). Thus, we find that the force is directed from the particle to the center of the circle at all times. A force that is directed along lines that for all times goes through a single common point, the centre of force, is known as a central force4 . Since we can imagine that forces exist simultaneously at all points in space, regardless of the presence of a particle or not, we talk of a field of forces and therefore the term central force field is often used. 4

The term central forces is used also for internal forces that satisfy the law of action and reaction, (4.12) and (4.13).

4.7 A Complete Model

41

Note also that for a given radius of the circle and for a given mass, the magnitude of f is not known unless we know the angular velocity, i.e., the speed of the rotation. An important case is when this speed follows Kepler’s 3rd law. Let T be the period of revolution, i.e., ωT = 2π, then Kepler’s 3rd law says that r3 = C, T2 where C is a constant. Thus, we find f = −m

4π 2 C e(t), r2

which is, basically, the famous inverse square law of gravitation. Note that the first two laws of Kepler were included in the assumption of a uniform circular motion, but, of course, these laws also apply to the more general cases of elliptic, parabolic and hyperbolic motion.

Exercises 4.1. Consider a discrete system in a state of pure translatory acceleration, meaning that there is a function Atr (t) of time only such that A(X, t) = Atr (t) for all X ∈ B, i.e., all material points have the same acceleration. Use the definition of the center of mass, xc , to show that for this situation Euler’s laws result in f (B) = Atr (t)M (B) and co (B) = (xc − o) × Atr (t)M (B). Thus, co (B) = (xc − o) × f (B). Comparing with (4.9) shows that the force system, when the body is in a state of pure translatory acceleration, must be such that the torque calculated with respect to xc is the zero vector. We say that the force system has a central axis which goes through xc . From facts in later chapters it will be clear that these conclusions hold for both one-dimensional and three-dimensional continuum bodies as well as for the discrete body since they do not depend on the precise definition of the force system. 4.2. A body is said to be in equilibrium if it is in a state of constant translatation, meaning that there is a vector V const such that V (X, t) = V const ,

(4.15)

for all X and t, i.e., the velocity is constant for all material points and time. Clearly, if (4.15) holds, from Euler’s laws one finds that co (B) = 0,

f (B) = 0.

(4.16)

4 Discrete Model

42

A body is said to be in quasi-equilibrium if it satisfies (4.16). Are equilibrium and quasi-equilibrium the same thing? What holds in this respect for a single particle, a rigid body or a general body? Give examples. Show that if a body is in quasi-equilibrium, its center of mass has a constant velocity. 4.3. The angular momentum with respect to the center of mass of a body B is defined as hxc (B) = (φt (X) − xc ) × V (X, t)mX . X∈B

The spin angular momentum is defined as hspin xc (B) =



(φt (X) − xc ) ×

X∈B

(a) Show that

∂ [φt (X) − xc ]mX . ∂t

hxc (B) = hxspin (B). c

(b) Show that ∂ ∂ ∂ ho (B) = hspin p(B). xc (B) + (xc − o) × ∂t ∂t ∂t (c) Use the result of (b), the formula (4.9) and Euler’s laws to show the spin equation of motion: cxc (B) =

∂ spin h (B). ∂t xc

(4.17)

This form of Euler’s law of angular momentum is useful in rigid body mechanics as will be commented on in Chapter 12. 4.4. Assume that the system of forces in the discrete model is extended by external and internal couple vectors, ce (X) and ci (Xk , X ), respectively, such that the term c(X, P), X∈P

where c(X, P) = ce (X) +



ci (Xk , X)

Xk ∈P −1

is added to the total torque co (P) in (4.8). Show that (4.13) must then be changed to r(A, t) × f i (B, A) + r(B, t) × f i (A, B) + ci (B, A) + ci (A, B) = 0, so we can no longer assume that f i (B, A) and f i (A, B) have a common line of action.

4.7 A Complete Model

43

4.5. For an extended force system, such as in Exercise 4.4, show that ce (X) + ci (Xk , X) = 0 (4.18) Xk ∈B\{X}

for all X ∈ B. Compare this result with (4.14). Should we extend the theory so as to get something different from the zero vector on the right hand side of (4.18)? 4.6. Prove equation (4.13). 4.7. Prove that Euler’s laws follow from (4.12), (4.13) and (4.14). 4.8. A material point is in circular motion under the influence of a force which is always directed towards a fixed point of the circumference of the circle, a center of force. Show that the magnitude of the force is proportional to ρ−5 where ρ is the distance between the point and the center of force. 4.9. A material point moves along a trajectory which in circular coordinates (r, ϕ) is given by r = r(ϕ) = exp(−ϕ). Assume a central force directed towards the origin and show that the magnitude of the force is proportional to r−3 .

5 One-Dimensional Model

In this chapter we add structure to the one-dimensional body model. We follow the same strategy as for the discrete model. That is, we define mass, and give consequences of mass conservation, introduce linear and angular momentum and a force system. This gives explicit forms of Euler’s laws from which an action and reaction principle and equations of motion are derived. These equations of motion are to be logically compared to Newton’s law (4.14), but are in this case differential equations defined on the interval B. Now, for several types of problems, such as those of fluid mechanics, it is not so obvious how to define B in practice, even though, in principle, such an interval exists. Therefore, it is useful to have equations expressed in a natural coordinate for the curve Bt = φt (B) in E. This is developed for the plane two-dimensional case, and equations of motion are given in a natural normal-tangential frame.

5.1 Mass and its Conservation Each material point X ∈ B = [Xs , Xe ] is given a positive scalar ρ0 (X). Thus, we define a scalar field1 on B. It is called the referential mass distribution of B and has the physical dimension of mass per length. The mass of the part P = [X1 , X2 ] of the body, denoted M (P), is as in all models the “sum” of the masses of its points. However, since a continuum of particles is considered, we need to use the integral operation and write  X2 M (P) = ρ0 (X) dX. X1

We intend to evaluate the consequence of the law of conservation of mass for this mass function. Taking the time derivative of the above expression, and using that M(P) is constant in time one concludes that 1

A definition of the term field is given in Appendix C

45

46

5 One-Dimensional Model



X2 X1

∂ ρ0 (X) dX = 0, ∂t

for X1 and X2 such that Xs ≤ X1 < X2 ≤ Xe . Since this should hold for all possible P = [X1 , X2 ] it seems reasonable to assume that the integrand must be zero for the whole interval [Xs , Xe ]. In fact, there is a mathematical result known as the localization theorem that has exactly this implication. A precise statement of this theorem is given in Appendix C where it is labeled Theorem 2. Thus, the law of conservation of mass states that, for all X in the interval [Xs , Xe ], ∂ ρ0 (X) = 0, ∂t which we interpret by saying that the referential mass distribution is constant in time.

5.2 Linear and Angular Momentum Natural generalizations of expressions for linear and angular momentum from the discrete model to the one-dimensional model are obtained by changing sums to integrals. For a part P = [X1 , X2 ] of a one-dimensional body B we define  X2  X2 p(P) = V (X, t)ρ0 (X) dX, ho (P) = r(X, t) × V (X, t)ρ0 (X) dX. X1

X1

We calculate the time derivatives of these vectors following a formally identical scheme to the discrete case: using the law of conservation of mass one obtains  X2 ∂ p(P) = A(X, t)ρ0 (X) dX, (5.1) ∂t X1 ∂ ho (P) = ∂t



X2

r(X, t) × A(X, t)ρ0 (X) dX.

(5.2)

X1

5.3 System of Forces, and Force and Torque Resultants The system of forces has two ingredients, similarly to the discrete model: a definition of external forces and a definition of internal forces based on a cut principle: (i) For every X ∈ [Xs , Xe ] there is a geometric vector, q 0 (X), called the external force vector per unit reference length.2 2

It would also be possible to define an external couple vector for every X, see Exercise 5.1.

5.4 Explicit Forms of Euler’s Laws

47

(ii) For every X ∈ [Xs , Xe ] there exist vectors, f + (X), f − (X), m+ (X) and m− (X). We call f + (X) and f − (X) positive and negative cut forces, and m+ (X) and m− (X) positive and negative cut couples.

Fig. 5.1. The force system for a part P of a one-dimensional body.

With reference to Figure 5.1, the resultant force and torque for a part P of a one-dimensional body is now defined as  X2 f (P) = q 0 (X) dX + f − (X1 ) + f + (X2 ), (5.3) X1



X2

co (P) = X1

r(X, t) × q 0 (X) dX + rt (X1 , t) × f − (X1 ) + r t (X2 , t) × f + (X2 ) + m− (X1 ) + m+ (X2 ).

(5.4)

Note that we follow our convention not to explicitly indicate that forcelike variables may be functions of time.

5.4 Explicit Forms of Euler’s Laws From (5.1) through (5.4) we obtain explicit forms of Euler’s laws: Euler’s law of linear momentum  X2  q 0 (X) dX + f − (X1 ) + f + (X2 ) = X1

X2

A(X, t)ρ0 (X) dX.

(5.5)

X1

Euler’s law of angular momentum 

X2

X1

r(X, t) × q 0 (X) dX + r(X1 , t) × f − (X1 ) + r(X2 , t) × f + (X2 ) + m− (X1 ) + m+ (X2 )  X2 r(X, t) × A(X, t)ρ0 (X) dX. = X1

(5.6)

48

5 One-Dimensional Model

5.5 Local Equations The counterpart of the law of action and reaction in the one-dimensional case is the following result: For any X in the interval [X1 , X2 ], f + (X) = −f − (X),

m+ (X) = −m− (X).

(5.7)

The proof of this fact can be obtained as (4.12) and (4.13) were obtained. That is, we state (5.5) and (5.6) for three different intervals: P1 = [X1 , X2 ], P2 = [X2 , X3 ] and P3 = [X1 , X3 ]. These are such that P3 = P1 ∪ P2 . An alternative proof is to let X1 and X2 in (5.5) and (5.6) approach the common point X. If the integrands are bounded, the integrals in (5.5) disappear in this limit process and the first equation of (5.7) is obtained. Similarly, the second equation in (5.7) follows from (5.6). Due to the facts expressed by (5.7) we use the notation f (X) = f + (X) = −f − (X),

m(X) = m+ (X) = −m− (X),

for all X in the interval [X1 , X2 ]. Newton’s law (4.14) was obtained by writing (4.10) (or (4.11)) for P including only one point X. The corresponding equations in the one-dimensional model are the equations of motion which read as follows: For any X in the interval [X1 , X2 ], ∂f (X) + q 0 (X) = ρ0 (X)A(X, t), (5.8) ∂X ∂m(X) ∂φt (X) × f (X) + = 0. (5.9) ∂X ∂X There are at least two simple proofs of these results. The first is based on using Theorem 2 of Appendix C. The second is based on taking the derivative of Euler’s equations with respect to X2 and letting X2 = X. The differential equations (5.8) and (5.9) are vector equations: they are not related to any particular coordinate system, or system of base vectors. However, they can easily be used as a basis for deriving equations of motion in any particular coordinate system. For instance, assume a fixed right-handed ortonormal set of base vectors {e1 , e2 , e3 } and an origin o, which defines a cartesian coordinate system. Any vector involved in (5.8) and (5.9) can be represented in this base: f (X) = f1 (X)e1 + f2 (X)e2 + f3 (X)e3 , m(X) = m1 (X)e1 + m2 (X)e2 + m3 (X)e3 , A(X, t) = A1 (X, t)e1 + A2 (X, t)e2 + A3 (X, t)e3 , q 0 (X) = q01 (X)e1 + q02 (X)e2 + q03 (X)e3 , φt (X) − o = φt1 (X)e1 + φt2 (X)e2 + φt3 (X)e3 .

5.6 Equations of Motion in Natural Parameterization and Natural Frame

49

Substituting these representations into (5.8) and (5.9) one obtains six equations. The three equations related to (5.8) read as follows: ⎧ ∂f (X) ⎪ ⎪ 1 + q01 (X) = ρ0 (X)A1 (X, t) ⎪ ⎪ ⎪ ⎪ ∂X ⎨ ∂ff2 (X) (5.10) + q02 (X) = ρ0 (X)A2 (X, t) ⎪ ⎪ ∂X ⎪ ⎪ ⎪ ⎪ ⎩ ∂ff3 (X) + q03 (X) = ρ0 (X)A3 (X, t). ∂X In the next section we will obtain the equations of motion related to a set of base vectors that are normal and tangential to the curve B t, a socalled natural base.

5.6 Equations of Motion in Natural Parameterization and Natural Frame The main result of this section is a reformulation of the equations of motion (5.8) and (5.9). A natural parameterization, denoted s, of the curve Bt is introduced, and (5.8) and (5.9) are transformed from equations with X as an independent variable to equations with s as the variable. Next, a natural tangential-normal orthonormal base is introduced and the equations of motion are expressed in this base. The curvature of Bt enters these equations. 5.6.1 Natural Parameterization The local equations of motion (5.8) and (5.9) are differential equations where the independent variables are time t and the coordinate X. The latter parameterizes material points. Practically, the parameterization X may be achieved by a procedure based on the knowledge of a particular reference configuration for the body. For instance, for an elastic perfectly flexible string, the body returns to a natural length when all forces are removed. By putting such a relaxed straightened body alongside a ruler we may give each point a number which we use as the coordinate X. However, for many types of problems it is not so natural, even if in principle possible, to allocate a coordinate value to each material point. For instance, for a fluid there is no natural stress free configuration to which the body returns when forces are removed. Because of this, we like to define a coordinate which labels points along the current configuration Bt , and gives a theory which is not based on tracking material points. This will be done in the following. We start with some general facts concerning curves: A curve in physical space E is a mapping from the real numbers R to E, which we denote for instance ψ, i.e., x = ψ(q), q ∈ R, x ∈ E,

50

5 One-Dimensional Model

where q is the parameter of the curve. It is required that the tangent vector ∂ψ(q)/∂q never be zero. Obviously, a special example of a curve is given by the placement φt , where X is the parameter, cf. Figures 3.2 and 3.3 where a tangent vector is shown. A natural parameterization is a parameterization where the tangent vector ∂ψ(q)/∂q is a unit vector. We may then think that q measures the length of the curve. In general, the placement φt is not a curve with a natural parameterization and X does not measure length along the deformed body. However, for any curve it is always possible to redefine the parameterization so that it becomes natural, cf. Figure 5.2. This is done by defining an additional curve ψt , with natural parameterization s, that has a domain that includes Bt . The parameter s can then act as parameterization for the original curve as well. We can relate

Fig. 5.2. The placement φt and the natural parameterization ψt .

s and X since both φt and ψt are one-to-one: each X picks out an s through the function s(X) defined by s = (ψt )−1 (φt (X)) = s(X), or, equivalently, s and X are related by ψt (s) = φt (X).

(5.11)

Inversely, for each s which is such that ψt (s) is included in the domain of φt , we may define a function X(s):

X = (φt )−1 (ψt (s)) = X(s).

5.6 Equations of Motion in Natural Parameterization and Natural Frame

51

We are now ready to rewrite the local equations of motion (5.8) and (5.9) into a form where the natural parameterization s is the independent parameter. We simply premultiply these equations by ∂X(s)/∂s, insert X = X(s) and use the chain rule. Starting with (5.8) one obtains in this way ∂f (X(s)) ∂X(s) ∂X(s) + q 0 (X(s)) = ρ0 (X(s)) A(X(s), t). ∂s ∂s ∂s Here appear, quite naturally, the functions ρ (s, t) = ρ0 (X(s))

(5.12)

∂X(s) ∂s

and

∂X(s) . ∂s The dependence on time t for the first function comes from the fact that the function X(s) is time dependent even though this is not notationally explicit. The function q(s) also clearly depends on time but here we follow our convention of not explicitly indicating this for force-like variables. These functions have important physical interpretations. Remember that the referential mass distribution ρ0 could be integrated (summed) to give the total mass for a part of a body. The same holds for the function ρ (s, t), i.e., from the change of variables formula it follows that  X2  s2 M (P) = ρ0 (X) dX = ρ (s, t) ds, q(s) = q 0 (X(s))

X1

s1

where s1 = s(X1 ) and s2 = s(X2 ). The quantity ρ (s, t) may be called the mass distribution function corresponding to the placement φt . Both functions ρ0 and ρ have the physical dimension of mass per length. The function q(s) has an interpretation similar to ρ (s, t): it is an external force vector which measures force per unit length of Bt . Similarly, as for the mass, the total body force for a part of a body may be calculated either by integrating over B or over Bt , i.e.,  X2  s2 q 0 (X) dX = q(s) ds. X1

s1

Turning now to equation (5.9), we premultiply by ∂X(s)/∂s, insert X = X(s) and use the chain rule to obtain ∂φt (X(s)) ∂m(X(s)) × f (X(s)) + = 0. (5.13) ∂s ∂s Using (5.11) we find that a unit tangent vector, which we denote by et (s)3, appears in (5.13), i.e., 3

This is the first occurrence in this text of the subscript t in the meaning of tangential. Previously such a subscript has referred to time. This double meaning should not be too confusing since the reference to time occurs only for curves and placements of bodies.

52

5 One-Dimensional Model

et (s) =

∂ψt (s) ∂φt (X(s)) = . ∂s ∂s

(5.14)

In the next subsection the unit tangent vector et (s) will be supplemented by two other unit vectors to form a local orthonormal set of base vectors depending on s. The final reformulation in this section, of (5.8) and (5.9), is achieved by viewing accelerations and forces as functions of s. We use the lower case letter to indicate the acceleration when it is seen as a function of s, i.e., a(s, t) = A(X(s), t). We also write f (s) = f (X(s)) and m(s) = m(X(s)), even though this may be a slight abuse of notation since clearly two different functions are given the same notation. With definitions of ρ (s, t) and q(s) introduced above one now obtains from (5.12) and (5.13) the equations of motion expressed in the variable s: ∂f (s) + q(s) = ρ (s, t)a(s, t), ∂s

(5.15)

∂m(s) = 0. (5.16) ∂s There is an alternative derivation of these equations that will be used later. It is based on stating Euler’s equations in terms of the natural parameter s, from which (5.15) and (5.16) follow by localization on using Theorem 2 of Appendix C. We briefly recollect these steps. By change of variables in (5.1) and (5.2) we obtain  s2 ∂ p(P) = a(s, t)ρ (s, t) ds, (5.17) ∂t s1  s2 ∂ ho (P) = (ψt (s) − o) × a(s, t)ρ (s, t) ds, (5.18) ∂t s1 et (s) × f (s) +

and the force and torque resultants become  s2 f (P) = q(s) ds + f − (s1 ) + f + (s2 ),

(5.19)

s1



s2

co (P) = s1

(ψt (s) − o) × q(s) ds + (ψt (s1 ) − o) × f − (s1 ) + (ψt (s2 ) − o) × f + (s2 ) + m− (s1 ) + m+ (s2 ), (5.20)

where s1 = φt (X1 ) and s2 = φt (X2 ). Inserting (5.17) through (5.20) into Euler’s equations gives an explicit form of these laws from which (5.15) and (5.16) follow by localization. Note the fundamental difference between the two forms of Euler’s laws: the integration in (5.17) through (5.20) is done for integration domains that depend on time, while in the form in terms of the variable X this is not the case.

5.6 Equations of Motion in Natural Parameterization and Natural Frame

53

5.6.2 Equations of Motion in Natural Base In this subsection we will express the equations of motion (5.15) and (5.16) in an orthonormal basis where one base vector is the tangent vector et (s), introduced in the previous subsection. Such a base is usually referred to as a natural base or frame. Note that since the tangent vector et (s) is a function of s - the location along Bt - it is clear that the base will also vary along Bt . For simplicity, we will assume that the curve Bt belongs to a flat twodimensional subspace of E, i.e., to a plane. The unit vector eb is taken to be a normal vector to the plane containing Bt . We have, thus, introduced two orthogonal unit vectors, et (s) and eb . A third vector, called the normal vector, is introduced by using the vector product: en (s) = eb × et (s). The vector triple {et (s), en (s), eb } is then a right-handed orthonormal base of E for each s. Since the equations of motion (5.15) and (5.16) contain derivatives with respect to s, we need to calculate the derivatives of et (s) and en (s). It turns out that such a calculation introduces the idea of curvature of the curve Bt . To that end we conclude that et (s) · et (s) = 1 ⇒ et (s) · et (s) · eb = 0 ⇒ eb ·

∂et (s) = 0, ∂s

∂et (s) = 0. ∂s

t (s) , implying that Thus, et (s) and eb are orthogonal to ∂ e∂s

∂et (s) = κ(s)en (s) ∂s

(5.21)

for some scalar κ(s), which has the physical dimension [length]−1 . The sign of κ(s) is defined by the predetermined direction of s. The absolute value of κ(s) is called the curvature. This is explained by the formula    ∂et (s)   |κ(s)| =  ∂s  which shows that |κ(s)| measures the change of tangential direction. Furthermore 1/|κ(s)| is called the radius of curvature. The origin of this notation comes from the fact that three non collinear points on the curve Bt uniquely determine a circle. In the limit, when these three points approach each other, the circle becomes what is called the circle of curvature and its radius becomes the radius of curvature, see Figure 5.3. Having determined the derivative of the tangent vector, we consider the derivative of the normal vector. An expression for this derivative follows from

54

5 One-Dimensional Model

Fig. 5.3. The circle of curvature, with radius that equals the radius of curvature 1/|κ(s)|, is obtained in the limit when three defining points approach each other.

calculating its components in the natural basis. The component in front of et (s) is obtained from et (s) · en (s) = 0 ⇒ et (s) ·

∂en (s) ∂et (s) + · en (s) = 0 ∂s ∂s

from which we obtain, by using (5.21), that et (s) ·

∂en (s) = −κ(s). ∂s

The component in front of en (s) is zero since en (s) · en (s) = 1 ⇒ en (s) ·

∂en (s) = 0. ∂s

Finally, we note that the component in front of eb is also zero since eb · en (s) = 0 ⇒ eb ·

∂en (s) = 0. ∂s

Thus, it can be concluded that ∂en (s) = −κ(s)et (s). ∂s

(5.22)

We have now constructed a base, derived the properties (5.21) and (5.22) for this base and are, thus, prepared to investigate how the equations of motion, (5.15) and (5.16), appear in the new base. Following the ideas at the end

5.7 A Complete Model

55

of Section 5.5, where cartesian coordinates were investigated, we should then represent the vectors in the base. For simplicity it will be assumed that the vectors have the following reduced form: f (s) = ft (s)et (s) + fn (s)en (s), q(s) = qt (s)et (s) + qn (s)en (s),

m(s) = mb (s)eb ,

a(s, t) = at (s, t)et (s) + an (s, t)en (s).

No acceleration component in the eb -direction means that the curve is assumed to belong to a fixed two-dimensional subspace of B, having eb as normal vector, for all times. For f (s) and q(s) there is no component out of the plane and the moment m(s) has a bending action directed perpendicularly to the plane. Substituting into the equations of motion, (5.15) and (5.16), and using (5.21) and (5.22) one finally obtains ∂fft (s) − fn (s)κ(s) + qt (s) = ρ (s, t)at (s, t), ∂s ∂ffn (s) + ft (s)κ(s) + qn (s) = ρ (s, t)an (s, t), ∂s ∂mb (s) fn (s) + = 0. ∂s

(5.23) (5.24) (5.25)

5.7 A Complete Model As in the discrete model, the equations of motion (5.8) and (5.9), or any of its representations in a particular coordinate system or base, do not represent a model from which we can solve the motion of a one-dimensional body. To obtain such a model we need to add constitutive equations and force laws. However, inverse problems, where we seek the forces given the motion, can be formulated by adding only limited assumptions on the behavior of internal and external forces. For instance, these assumptions can take the form of boundary conditions where cut forces and couples are prescribed at an end of a body, or they can be statements saying that the external forces have prescribed values. That is, a complete model is formulated by adding conditions, which we may call simple particular laws, to the equilibrium equations. As a special case of this procedure we take a static problem, i.e., a(s, t) = 0, with a known time independent placement φt (X), prescribed external forces and appropriate boundary conditions, where we want to solve for the cut forces and couples:

56

5 One-Dimensional Model

One-dimensional force problem: Given a distribution of external forces qt (s) and qn (s), a curvature κ(s), s ∈ (0, L), and boundary conditions f¯n = fn (0),

f¯t = ft (0),

m ¯ b = mn (0),

find fn = fn (s), ft = ft (s) and mb = mb (s) such that ∂fft − fn κ(s) + qt (s) = 0, ∂s

(5.26)

∂ffn + ft κ(s) + qn (s) = 0, ∂s ∂mb fn + = 0. ∂s

(5.27) (5.28)

The well-posedness of this model follows from the standard theory of first order differential equations, sometimes known as dynamic systems.

5.7.1 Example: A Circular Beam Consider a one-dimensional body in the form of a cut open circle with radius R. At the cut, two forces, in opposite tangential directions, with magnitude F , are applied to the two cut surfaces. The situation is depicted in Figure 5.4. We want to solve for the distributions of cut forces and couples. This problem

Fig. 5.4. A curved beam in the form of a cut open circle.

can be treated as a special case of the one-dimensional force problem stated above. There are no distributed external forces, and the curvature is 1/R, so the governing equations are fn ∂fft − = 0, ∂s R

∂ffn ft + = 0, ∂s R

fn +

∂mb = 0, ∂s

5.8 Another

omplete

odel

57

and the boundary conditions of the complete problem are −F = ft (0),

0 = fn (0),

0 = mb (0).

It is straightforward to verify that the solution of this initial boundaryvalue problem is   s s s fn (s) = F sin , ft (s) = −F cos , mb (s) = F R cos − 1 . R R R Noting that L = 2πR one finds that the boundary conditions at s = L are automatically satisfied. In fact, the problem would be unsolvable if external forces were not in equilibrium, since the governing partial differential equations assume such a situation. Note also that it is the deformed configuration of the beam that is a circle. Apparently we do not need any information on the form of the undeformed body to obtain the cut forces and couples. This is typical for a statically determinate problem. If we make an assumption of small displacements, see Section 9, there is no distinction between deformed and undeformed geometry and this is frequently how problems of this type are viewed in elementary courses.

5.8 Another Complete Model We will consider a complete model that contains, as special cases, the classical problems of the suspension bridge and the catenary cable. A basic simple particular law that is behind these problems is to take the one-dimensional body as being completely flexible, i.e., there is no non-zero cut couple, which we express as m(s) = 0 for all s. (5.29) Equation (5.16) immediately reveals that this condition implies that f (s) is collinear with et (s), i.e., f (s) = ft (s)et (s). (5.30) In the following this result will be substituted into (5.15) and the resulting equation expressed in a cartesian coordinate system; a restriction to two space dimensions is then used for simplicity. The position of a material point X = X(s) is written as φt (X(s)) = x(s)e1 + y(s)e2 , where x(s) and y(s) are functions specifying the current position in space of the one-dimensional body, and {e1 , e2 } is an orthonormal set of base vectors. This implies that we may write the tangent vector as et (s) =

∂x(s) ∂φt (X(s)) ∂y(x) = e1 + e2 . ∂s ∂s ∂s

(5.31)

58

5 One-Dimensional Model

Furthermore, the external force vector is written as q(s) = q1 e1 (s) + q2 e2 (s), which when used in (5.15), together with (5.30) and (5.31), and taking the acceleration to be zero, leads to   ∂ ∂x(s) ft (s) + q1 (s) = 0, (5.32) ∂s ∂s   ∂ ∂y(s) ft (s) + q2 (s) = 0. (5.33) ∂s ∂s If we consider a problem where the external forces q1(s) and q 2(s) are known, then (5.32) and (5.33) constitute a system of two equations for three unknown functions x(s), y(s) and ft (s). Thus, such a problem is under-determined, and we need to add additional specific conditions. One simple such condition is to assume that the one-dimensional body is inextensible, i.e., we think of a chain or a stiff cable. This means that we can regard the length of the deformed cable as known and equal to, say, L. Furthermore, if ∂x(s)/∂s > 0, then the y-coordinate may be considered as a function of the x-coordinate4 , i.e. y = y(x) and we obtain   2     right dy(x) 2 2 ds = dx + dy = 1+ dx = L, (5.34) dx Bt Bt left where right and left are extreme values of the x-coordinate. Note also that we use straight d in the derivative expression since functions are of one variable only. We may now formulate the following complete model: Inextensible cable problem: Given external forces q1 (s) and q2 (s), s ∈ (0, L), and appropriate boundary conditions, find the functions x(s), y(s) and ft (s) such that (5.32), (5.33) and (5.34) are satisfied. It should be noted that the equations of this model are non-linear, and that we can, therefore, not expect a solution to be unique in general. In fact, the suspension bridge example in the next subsection will show non-uniqueness. However, solutions are generally distinct, and, as in the example, physical insight may be used to single out which solution is the more interesting. Therefore, we regard also models of this type as complete models. 5.8.1 Example: Suspension Bridge A cable is fixed at both ends. A cartesian coordinate system is introduced by means of the base {e1 , e2 }, defining x and y directions. This base is oriented

5.8 Another

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59

Fig. 5.5. A cable is loaded by a uniform loading w in the y-direction. This configuration may represent a suspension bridge.

so that the loading is zero in the x-direction, see Figure 5.5. The loading in the the y-direction is a constant force per unit length in x-coordinate, positive in the negative y-direction and denoted w. Since q2 (s) is a force per unit length in the s-coordinate, we have in infinitesimal terms, w dx = −q2 (s) ds, i.e., q2 (s) = −w

dx(s) . ds

(5.35)

This type of loading occurs for a suspension bridge, where w represents the weight of the deck that is carried by a cable through a dense distribution of vertical wires. We introduce this loading into (5.32) and (5.33) and obtain   d dy(s) dx(s) dy(s) ft (s) = w = =⇒ ft (s) = wx(s) + A, (5.36) ds ds ds ds   d dx(s) dx(s) ft (s) = 0 = =⇒ ft (s) = B, (5.37) ds ds ds where A and B are integration constants. We now assume that dx(s)/ds > 0, so we may view y as a function of x, and also that ft (s) = 0. Then, (5.36) and (5.37) result in B

dy(x) = wx + A dx

= =⇒

By(x) =

w 2 x + Ax + C, 2

where C is a third integration constant. Thus, we conclude that the cable takes the form of a parabola. Introducing new constants b = A/B and c = C/B we write 4

This conclusion follows from the implicit function theorem.

60

5 One-Dimensional Model

w 2 x + bx + c. 2B The constants b and c can be eliminated by placing the coordinate system, see Figure 5.5, so that y(x) = dy(x)/dx = 0 when x = 0. Then, b = c = 0. The constant B is determined by using the inextensibility condition (5.34), from which we find  right   wx 2 1+ dx = L. (5.38) B left y(x) =

This is an equation having two roots, one the negative of the other. Thus, the solution is non-unique. The two solutions correspond to the tangent inner force being a tension or a compression force. For a compression force the bridge takes the shape of a parabola with a maximum. This shape is from a mechanical point of view highly unstable. The tension force, on the other hand, gives the shape shown in Figure 5.5, and this is the solution that in most engineering situations is the interesting one. We now like to obtain the tangent force ft (s). From (5.37) we get    2  wx 2 dy(x) ds =B 1+ ft (s) = B =B 1+ . dx dx B From this we can see that the constant B > 0 represents the lowest value of ft (s), denoted ftmin , i.e.,  ft (s) = (fftmin )2 + w2 x2 .

Exercises 5.1. Extend the derivation of the results in Section 5.5 by (a) adding a couple per unit reference length, l(X), for every X ∈ [Xs , Xe ]. This extension is needed in Section 9.2. (b) adding concentrated forces and couples, K(X) and H(X), respectively, at a finite number of points in the interval B. 5.2. The cartesian form of equation (5.8) was derived in (5.10). Derive the cartesian form of (5.9). 5.3. Show that the function s(X), introduced in Section 5.6, is given by   X  ∂φt (X)    s(X) = ss +  ∂X  dX, Xs where ss = s(Xs ), if we assume that ∂s/∂X > 0.

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5.4. Extend the derivation of (5.23) through (5.25) to the case when no re triction on the form of vectors is imposed. Retain, however, the assumption of a plane problem (this assumption is removed in the next exercise). Try to interpret the results. 5.5. The natural frame {et (s), en (s), eb } was introduced only for the plane case, i.e., eb was independent of s. Extend the derivation of (5.23) through (5.25) to the case of a general curve Bt . You need to start by deriving the so called Frenet’s formulas, which you need to find in the literature. The plane form of these are (5.21) and (5.22). 5.6. Consider the complete model of Section 5.8. Show that if the external forces q1 (s) and q2 (s) are zero and the boundary conditions are such that ft (s) = 0 for all s, then the cable is straight in its deformed configuration. By the way: is it reasonable to assume that ft (s) < 0? 5.7. This is the classical problem of the catenary cable. It concerns the complete model of Section 5.8. The problem is very similar to the suspension bridge problem treated in that section, the difference is that the load w now has a constant given value per unit length of the cable, not per unit of the x-coordinate as was the case for the bridge. Derive the following representation of the geometry of the hanging cable: y(x) =

 T  wx cosh −1 , w T

and the following expression for the force: ft (x) = T cosh

 wx  , T

where x and y refer to a cartesian coordinate system with the origin at the extreme point of the curve and the y-direction coinciding with the load direction as in Section 5.8, and T is the horizontal component of the tensile force in the cable. 5.8. Extend the one-dimensional model to a two-dimensional one, i.e., derive equations of motion for a plate or a shell. (This is an extensive exercise.)

6 Pipe Flow

The pipe flow model is essentially a special case of the one-dimensional model studied in the previous chapter. An additional assumption is introduced in that the mapping ψt is regarded as time independent and, consequently, we write it without the subscript t in the sequel. The curve defined by ψ and the parameterization s is then interpreted as the pipe C through which the material points flow, cf. Section 3.2 and Figure 3.4. Pipe flow raises the need for certain geometrical concepts, mostly related to fluid mechanics problems. These are as follows: spatial (Eulerian) and ma rial (Lagrangian) fields, material time derivatives, isochoric motion and the continuity equation. In the first part of this chapter we will go through these concepts. The presentation is important as such, but this one-dimensional setting is also quite useful as a preparation for similar developments for the three-dimensional case, treated in the next chapter. We will also introduce the concepts of volume and pressure that are related to a three-dimensional view of bodies, but which, nevertheless, can be discussed in the present onedimensional setting.

6.1 Kinematic Constraint, Velocity and Acceleration The requirement that the pipe C is a fixed curve in E, i.e., that ψ is time independent, introduces a constraint on velocities. This is obtained by taking the time derivative of the counterpart of (5.11), i.e., φt (X) = ψ(s(X)). We then use the definition of velocity, i.e., V (X, t) =

∂ψ(s(X)) ∂ψ(s(X)) ∂s(X) = = V (X, t)et (s), ∂t ∂s ∂t

where s = s(X),

63

64

6 Pipe Flow

∂s(X) ∂t is the signed speed of the material point X and

(6.1)

V (X, t) =

et (s) =

∂ψ(s) ∂s

is a unit tangent vector since ψ is a natural parameterization, see Chapter 5. Using this representation of the velocity, (5.21) and (6.1), we may calculate the acceleration of a material point X in pipe flow: A(X, t) =

∂V (X, t) ∂ V (X, t) = et (s) + V (X, t)2 κ(s)en (s) ∂t ∂t = At (X, t)et (s) + An (X, t)en (s),

where s = s(X) and At (X, t) =

∂V (X, t) , ∂t

An (X, t) = V (X, t)2 κ(s).

(6.2)

This decomposition of the acceleration vector was, in fact, used already at the end of Section 5.6. However, there lower case letters were used for accele ration since it was seen as a function of s. In this respect see the next section.

6.2 Spatial and Material Representations, and Time Derivatives It has been seen in the preceding sections that the acceleration of a mate r ial point may sometimes be regarded as a function of the material point X and sometimes as a function of the position along the pipe represented by s. Velocity and speed may similarly be given two representations. Generally, we use lower case letter for velocity, speed or acceleration when viewed as functions of s and upper case letters when viewed as functions of X. We have the relations v(s, t) = V (X(s), t), v(s, t) = V (X(s), t),

a(s, t) = A(X(s), t),

at (s, t) = At (X(s), t),

an (s, t) = An (X(s), t).

We will now generalize these observations and then conclude that every mechanical(or thermomechanical) field has a material (function of time and X) as well as spatial (function of time and s) representation. Every material field can be made spatial and vice versa. A spatial representation fs of the material field f is given by fs (s, t) = f (X(s), t). Inversely, the material representation gm of the spatial field g is given by

6.2 Spatial and Material Representations, and Time Derivatives

65

gm (X, t) = g(s(X), t). Obviously, (gm )s = g,

(ffs )m = f.

We frequently study partial derivatives of fields with respect to time, as when defining velocity and acceleration. The following notations are intro duced for such derivatives. The time derivative of a spatial field is denoted by a prime, i.e., ∂g(s, t) g  (s, t) = . ∂t The partial time derivative of a material field is denoted by a superposed dot, i.e., ∂f (X, t) f˙(X, t) = . ∂t Clearly, the time derivative of a material field represents the change of a quantity when keeping fixed or following a material point. To be able to study the same kind of change when a spatial representation is at hand, we introduce the material time derivative of a spatial field, which we again denote by a superposed dot. It is defined as follows: g˙ = ((g)˙ m )s

⇔

g(s, ˙ t) = g˙ m (X, t) for s = s(X).

(6.3)

That is, one makes a material function out of the spatial function, then performs the time derivative and, finally, transforms the result back to the spatial form. By letting gm be f we also find from (6.3) that f˙(X, t) = f˙s (s, t),

(6.4)

for s = s(X), which is the reason for using the dot notation for what are seemingly two different operations. A relation between the material and spatial time derivative of a spatial field follows easily from the chain rule. Assuming for simplicity that g is a scalar valued function we have g(s, ˙ t) =

∂ ∂g(s, t) g(s(X), t) |X=X(s) = v(s, t) + g  (s, t). ∂t ∂s

(6.5)

An explanation of equation (6.5) in less abstract terms is useful. To that end, consider an observer that records the temperature T of the water in a river by traveling in a boat that moves with the stream. Such an observer will experience two types of changes of the temperature. The first one is connected to a change in time, and for a small time interval ∆t it is approximately given by ∂T ∆t. ∂t The second type of change comes from the movement of the boat. If the location of the boat is given by a coordinate s this change is approximately

66

6 Pipe Flow

∂T ∆s, ∂s where ∆s is a small change in location. Since for a small time interval, it approximately holds that ∆s = v∆t, where v is the velocity of the stream, we find that the total change in temperature recorded by the observer is ∆T ≈

∂T ∂T ∆t + v∆t. ∂t ∂s

It now follows that

∆T ∂T ∂T T˙ = lim = + v, ∆t→0 ∆t ∂t ∂s which is equation (6.5) for g taken as the temperature T . An important example of the use of material time derivative follows from the first equation of (6.2), which with the introduced dot notation reads At (X, t) = V˙ (X, t). We have at (s, t) = At (X, t) for s = s(X) and (6.4) gives V˙ (X, t) = v(s, ˙ t), which with (6.5) leads to at (s, t) = v  (s, t) + v(s, t)

∂v(s, t) . ∂s

(6.6)

That is, the tangential acceleration in a spatial representation is the sum of two terms, the second one being the convection term.

6.3 Forces and Couple in a Pipe Bend At this stage we may formulate a slight extension of the complete force problem of Section 5.7, which was obtained by assuming zero acceleration. We consider a solid static pipe in which a fluid substance flows. Both the solid and the fluid substances are treated simultaneously as a one-dimensional body, and equations (5.23) through (5.25) are valid if ρ is taken to be the mass per unit length of the moving fluid substance. Since this problem is only a small variation of the complete problem of Section 5.7 we refrain from formulating explicitly a new complete problem, but move on to an example. Consider a pipe in the shape of a quarter circle with radius R, as seen in Figure 6.1. Inside the pipe a fluid substance with constant mass distribution ρ flows with constant speed v. There are no external forces, so equations (5.23) through (5.25) become in this case ∂fft fn − = 0, ∂s R ∂ffn v2 ft + = ρ , ∂s R R ∂mb fn + = 0, ∂s

(6.7) (6.8) (6.9)

6.4 Continuity Equation and Control Domain

67

Fig. 6.1. A pipe shaped as a quarter circle .

where we have used that the acceleration is given by (6.2). As boundary conditions we can use the conditions at the free end s = πR/2. They are π ft = fn = mb = 0 at s = R . 2 One concludes by substitution that the differential equations (6.7) through (6.9) together with these boundary conditions are satisfied by    s s s ft = ρ v 2 1 − sin , fn = −ρ v 2 cos , mb = ρ v 2 R sin − 1 . R R R The forces and couple at s = 0 become ft = ρ v 2 , fn = −ρ v 2 and mb = −ρ v 2 R. These may be called reaction forces and couple since they occur where the pipe is fixed to the surrounding. They may also be obtained by using Euler’s laws written for a control volume, see the next section and Exercises 6.2 and 6.3. One then concludes that these forces and couples are independent of the particular form of the pipe: only the direction and location of in- and out-flow matters. Note that the cut forces and couple obtained here do not relate in particular to either the solid pipe substance or the fluid substance. The theory is not detailed enough to make such a distinction.

6.4 Continuity Equation and Control Domain Conservation of mass implies an important differential equation called the continuity equation that will be derived in this subsection. In Section 5.6 we introduced a mass distribution function by the formula ρ (s, t) = ρ0 (X(s))

∂X(s) . ∂s

(6.10)

68

6 Pipe Flow

This formula is constructed so that both the referential mass distribution ρ 0 and the mass distribution ρ can be used to express total mass for a part of a body P as follows:  X2  s2 M (P) = ρ0 (X) dX = ρ (s, t) ds, X1

s1

where s1 = s(X1 ) and s2 = s(X2 ). Equation (6.10) is a spatial equation. It can be expressed in material form by substituting X = X(s) to obtain ρ (s(X), t)

∂s(X) = ρ0 (X). ∂X

∂ From the conclusion ∂t ρ0 (X) = 0, derived in section 5.1, and the definition of the speed V (X, t), we have   ∂ ∂s(X) ∂s(X) ∂V (X(s), t) + ρ (s, t) = 0, ρ (s(X), t) = ρ˙  (s, t) ∂t ∂X ∂X ∂X ∂ where we used the identity ∂t ρ (s(X), t) = ρ˙  (s, t), which follows from (6.4) or the first equality of (6.5). Using the chain rule and the definition of spatial speed, v(s, t) = V (X(s), t), we then find

ρ˙  (s, t) + ρ (s, t)

∂v(s, t) = 0, ∂s

or, alternatively, on using (6.5), ρ (s, t) +

∂ [ρ (s, t)v(s, t)] = 0. ∂s

(6.11)

This is a partial differential equation in the spatial variable s and time t called the continuity equation. For an alternative derivation of this equation, using Reynolds’ transport equation, see Exercise 6.1. A useful version of the continuity equation can be obtained by introducing the concept of a control domain. A control domain is a fixed subset R of the pipe C, which is such that R ⊂ φt (B) for all times of interest. ψ(s) ∈ R when s belongs to the interval [¯1 , s¯2 ]. Note the difference between R and Pt defined by the interval [s1 , s2 ] = [s(X1 ), s(X2 )]: the latter depends on time and always contains the same material points, while R is fixed in space and may be occupied by different parts of the body at each time. Integrating (6.11) we find  s¯2  s¯2 ∂ ∂ ρ (s, t) ds = − (ρ (s, t)v(s, t)) ds. s1 ∂t s1 ∂s Since ¯1 and ¯2 are independent of time, the time derivative can be moved outside the integral on the left hand side. The right hand side can be integrated by parts and we find

6.5 Volume of the Pipe and Isochoric Motion

∂ ∂t



s¯2

ρ (s, t) ds = ρ (¯1 , t)v(¯1 , t) − ρ (¯2 , t)v(¯2 , t).

69

(6.12)

s1

This equation has a very clear interpretation: The integral is the mass contained in R at time t and ρ (s, t)v(s, t) is the mass flux at s, i.e., the mass per time measure that passes the cross section at s. Thus, (6.12) says that the change of mass inside R equals the inflow minus the outflow of mass. Control domain versions of Euler’s laws are given in Exercise 6.2.

6.5 Volume of the Pipe and Isochoric Motion Even though we are developing a one-dimensional theory where material points are placed along a line, the physical objects we intend to simulate do have three-dimensional shape when looked at in the magnifying glass. Therefore, it is convenient to introduce the three-dimensional concept of cross section area: for each s along the pipe and for each time t we imagine a perpendicular cut that has an area α(s, t) > 0, see Figure 6.2. The volume

Fig. 6.2. The concept of cross section area along the pipe C.

for a placement φt of a part P = [X1 , X2 ], i.e., the volume of Pt = φt (P), is defined as  s2 Vol(P Pt ) = α(s, t) ds, s1

where s1 = s(X1 ) and s2 = s(X2 ). We are interested in how the volume changes with time, and to that end rewrite Vol(P Pt ) as an integral over P instead of an integral over Pt by means of the change of variables formula. One then obtains 

X2

Vol(P Pt ) =

α(s(X), t) X1

and the time derivative becomes

∂s(X) dX, ∂X

70

6 Pipe Flow

∂ Vol(P Pt ) = ∂t



X2



∂s(X) ∂α(s(X), t) ∂V (X, t) + α(s(X), t) ∂X ∂t ∂X X   s12  ∂v(s, t) = α(s, ˙ t) + α(s, t) ds, ∂s s1

 dX (6.13)

where in the second integral we have changed back to integrating over Pt. A motion is said to be isochoric if ∂ Vol(P Pt ) = 0 ∂t for every Pt and t. By Theorem 2 in Appendix C, the localization theorem, we find from (6.13) that for isochoric motions α(s, ˙ t) + α(s, t)

∂ ∂ ∂v(s, t) = α(s, t) + (α(s, t)v(s, t)) = 0, ∂s ∂t ∂s

(6.14)

where we have used (6.5). Assuming that α(s, t) = α(s) is independent of time1 we find that α(s)v(s, t) = function of t only. The product α(s)v(s, t) is the volume flux, that is, the volume of material per time measure that passes the cross section at s. Thus, we have found that for an isochoric motion, the volume flux is constant along the pipe. It is also seen that a small area implies a high velocity and conversely. Next, we investigate what an assumption of isochoric motion means for a one-dimensional density function ρ(s, t), defined by the formula ρ (s, t) = α(s, t)ρ(s, t).

(6.15)

Clearly, the function ρ(s, t) has the physical dimension mass per volume. Substituting (6.15) into the continuity equation (6.11) we find α (s, t)ρ(s, t) + α(s, t)ρ (s, t) + ρ(s, t)

∂(α(s, t)v(s, t)) ∂ρ(s, t) + α(s, t)v(s, t) = 0. ∂s ∂s

For isochoric motion, according to (6.14), the first and third terms in this expression add to zero, and the remaining two terms can be simplified by using the material time derivative of ρ(s, t). One finds, since α(s, t) > 0, that α(s, t)ρ(s, ˙ t) = 0

= =⇒

ρ(s, ˙ t) = 0.

That is, for isochoric motion the one-dimensional density is constant when following a material point. 1

This is an assumption that is not valid for, e.g., blood flow in arteries and veins since these are elastic, and pressure will change the cross-section area.

6.6 Equation of Motion in Terms of Pressure

71

6.6 Equation of motion in terms of pressure In the previous section we introduced a quasi three-dimensional view of pipe flow: we conceived a pipe with cross-section area and, thus, a threedimensional shape, while still maintaining that variables are functions of the pipe coordinate s only. Here we will expand on this view in a way which re ates to the physical behavior of fluids at rest or having negligible frictional resistance. The forces acting on any surface of such a fluid will be directed normally to the surface and can be represented by a pressure p, supposed positive when pressing onto the surface. For a pipe, this pressure is taken to be a function of the pipe coordinate s only (apart from a dependence on time which we do not express explicitly), i.e., p = p(s). We will derive equations of motion where the momentum is based on a strictly one-dimensional view, but the force system is based on the threedimensional notion of pressure. The time derivatives of linear and angular momentum are given by (5.17) and (5.18), while the total force for a part Pt = [s1 , s2 ] is given by   f (P) = q V (s) dV Vx + (−p(s))n dAx , (6.16) ˆt P

ˆt ∂P

where Pˆt is the three-dimensional subset of E formed from cross section areas as s goes from s1 to s2 , and ∂ Pˆt is the boundary of this domain, cf. Figure 6.3. Moreover, q V (s) is a force per unit volume measure and n is the outward unit normal vector of ∂ Pˆt . The divergence theorem (C.21) implies that   p(s)n dAx = ∇p dV Vx , ˆt ∂P

ˆt P

where the vector ∇p is the gradient of p. From now on we will assume that the pipe is straight, i.e., et (s) is independent of s and consequently written et . In such a case, and since p depends on s only, the gradient can be written ∇p =

∂p(s) et . ∂s

ˆ t ,but varying For any field f (s, t) defined on the three-dimensional domain P in the et -direction only, it holds that   s2 f (s, t) dV Vx = α(s, t)f (s, t) ds. ˆt P

s1

Therefore, we may now rewrite (6.16) as  s2  s2 ∂p(s) ˜ (s) ds − f (P) = α(s, t)et ds, q ∂s s1 s1

(6.17)

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6 Pipe Flow

where ˜ (s) = q V (s)α(s, t) q is a force per unit length. From (5.17) and (6.17) we obtain the following explicit form of Euler’s law of linear momentum:   s2  ∂p(s) q˜ (s) − α(s, t)et − a(s, t)ρ (s, t) ds = 0. (6.18) ∂s s1 We now write ˜n, q˜ (s) = ˜t (s)et + q ˜ n is the component of q˜(s) that is orthogonal to et . Furthermore, since where q the pipe is straight a(s, t) = v(s, ˙ t)et . From (6.18), by using localization, we now get q˜t (s) −

∂p(s) α(s, t) = v(s, ˙ t)ρ (s, t) ∂s

(6.19)

˜ n = 0. The later condition is a constraint on the loading. Roughly as well as q speaking one may say that it comes from the assumption that p depends on s only. Equation (6.19) may be compared to the general equation of motion (5.15). One then finds f (s) = −p(s)α(s, t)et ,

˜ (s) + p(s) q(s) = q

∂α(s, t) et . ∂s

That p(s) contributes to q(s) may be somewhat surprising, but it comes from the fact that the change in cross section area along the pipe implies that the pressure has a component in the et -direction, as seen in Figure 6.3. We illustrate this for a circular cross section. Viewing the radius r of a cross

Fig. 6.3. Pressure acting on a part of a pipe.

6.7 Conservative Forces and Incompressibility

73

section as a function of s, i.e., r = r(s), we see that the pressure will act in the et -direction with a magnitude of p(s)∂r(s)/∂s, and the total force from the circumferential pressure, on the section shown in Figure 6.3, in the s-direction, is  s2  s2 ∂r(s) ∂α(s, t) p(s) π2r(s) ds = p(s) ds. ∂s ∂s s1 s1 Thus, the force per unit length q(s) may be written as above. If we write down the total couple, similarly to the total force in (6.16), and use (5.18), we may write Euler’s law of angular momentum in an explicit form. A similar, but somewhat more cumbersome analysis, will show that we need to imagine that the curve ψ defines the locus of center points of each cross section. If this is not the case, one has to introduce a moment per unit reference length, as in Exercise 5.1. One should also note that (6.19) seems to be used also for curved pipes ˜ n = 0. A closer examination of this fact shows that and for situations when q it may be a good approximation if the curvature of the pipe is small compared to the size of cross section areas. In the following we will also take this view. Finally, we rewrite (6.19) in a form that will be used later. With ρ (s, t) = ρ(s, t)α(s, t) one finds −

1 ∂p(s) q˜t (s) + = v(s, ˙ t). ρ(s, t) ∂s ρ(s, t)α(s, t)

(6.20)

Equation (6.20) will be the basic equation of Chapter 10, where we will continue our discussion on pipe flow problems. However, already in the next section we will show some special solutions of this equation that are based on assumptions of incompressibility, conservative forces and constant crosssection.

6.7 Conservative Forces and Incompressibility In this section, the concept of conservative forces is introduced through an example where ˜t (s) is due to gravity. We also state simple particular laws leading to an incompressible fluid. Note that we assume that (6.20) is valid for a curved pipe in the following. A particular case of a conservative force occurs when ˜t (s) is due to gravity. The direction of gravity is given by the unit vector eg , and the gravitational acceleration is, as usual, a constant denoted by g. The gravitational force per unit length of the pipe is then gρ(s, t)α(s, t)eg , and we may write q˜t (s) = gρ(s, t)α(s, t)eg · et (s).

(6.21)

For simplicity let the pipe be located in a flat two-dimensional subspace of E, spanned by eg and a unit vector ex , together forming a two-dimensional

74

6 Pipe Flow

orthonormal base. For an origin o, the position of a material point X = X(s) is given by φt (X(s)) − o = −h(s)eg + x(s)ex , where h(s) is conceived as the height above a reference plane including the origin, and x(s) is the coordinate in a direction perpendicular to the direction of gravity. Thus, using (5.14) we find et (s) = −

∂h(s) ∂x(s) eg + ex . ∂s ∂s

(6.22)

Putting (6.22) into (6.21) one finds q˜t (s) ∂h(s) = −g . ρ(s, t)α(s, t) ∂s Clearly, we can introduce a function U (s) = gh(s) which is a potential function in the sense that q˜t (s) ∂U (s) =− . ρ(s, t)α(s, t) ∂s

(6.23)

Any force, not just a force due to gravity, with the property that it can be derived from the derivative of a potential is called a conservative force. Next, we introduce the simple particular law relating to an incompressible fluid. For isochoric motion we have found, in Section 6.5, that the material derivative of ρ(s, t) is zero. That is, the density stays the same for every material point. We will use this property, but first we define the related property of homogeneity. A motion is said to be homogeneous if ρ(s, t) is independent of s. A material is now said to be an incompressible fluid if every possible motion is homogeneous and isochoric and p(s) can take any value. From the relation between material and spatial derivative, ρ(s, ˙ t) = ρ (s, t) + v(s, t)

∂ρ(s, t) , ∂s

we find that ρ(s, t) = ρin = constant

(6.24)

for all possible motions of an incompressible fluid. We may now substitute (6.23) into (6.20), using (6.24) and (6.5), to obtain −

1 ∂p(s) ∂U (s) ∂v(s, t) ∂v(s, t) − = + v(s, t) . ρin ∂s ∂s ∂t ∂s

(6.25)

For the special case when the cross-section area is time independent, i.e., α(s, t) = α(s), we may use the conclusion from Section 6.5 that for isochoric motion and for a cross-section that is time independent, the volume flux q(t) = α(s)v(s, t),

(6.26)

6.7 Conservative Forces and Incompressibility

75

is a function of time only. Substituting (6.26) into (6.25) we find    ∂ 1 ∂q(t) q(t)2 ∂ 1 − (p + ρin U (s)) = ρin + , ∂s α(s) ∂t α(s) ∂s α(s) which may be integrated from 0 to s to give − p(s) − ρin U (s) + p(0) + ρin U (0) ∂q(t) ρin q(t)2 = ρin I(s) + ∂t 2 

where



 1 1 − , (α(s))2 (a(0))2

(6.27)

s

ds . 0 α(s) If we have boundary conditions of known pressure at s = 0 and s = L, where we can think of L as the total length of the pipe, then the left hand side of (6.27) is known, and what is at hand is an ordinary differential equation for q(t) which, together with the initial condition, is uniquely solvable. By substituting this solution into (6.27) we get the pressure function p(s). It is interesting to note that the solution depends on α(0) and α(L), but not on cross-section dimensions between the end points of the pipe. In the case of steady state, when q(t) is time independent, implying that v(s, t) = v(s) is also time independent, (6.27) gives I(s) =

−

p(s) p(0) v(s)2 v(0)2 + − U (s) + U (0) = − , ρin ρin 2 2

or expressed differently, p(s) v(s)2 + U (s) + = constant. ρin 2

(6.28)

This equation is Bernoulli’s equation for incompressible steady flow and it can clearly be obtained directly from (6.25).

Exercises 6.1. Let g(s, t) be a spatial field defined on the pipe C and let P = [X1 , X2 ] be a part of a one-dimensional body which at time t occupies Pt = φt (P) ⊂ C. Show that  s2  s2 ∂ g(s, t) dc = g  (s, t) dc + g(s2 , t)v(s2 , t) − g(s1 )v(s1 , t), (6.29) ∂t s1 s1 where s1 = s(X1 ) and s2 = s(X2 ). This equation is the one-dimensional version of Reynolds’ transport equation. Identify g with the mass distribution function ρ and use (6.29) to make an alternative derivation of the continuity equation (6.11).

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6 Pipe Flow

6.2. Euler’s laws can be written for a control domain R, defined by the interval [¯1 , s¯2 ], instead of writing them for a part P of a body. Use the equations of motion (5.15) and (5.16) to derive the following control domain versions of Euler’s laws: ∂ f (R) = ∂t



s¯2

ρ (s, t)v(s, t) ds s1

+ ρ (¯2 , t)v 2 (¯2 , t)et (¯2 ) − ρ (¯1 , t)v 2 (¯1 , t)et (¯1 ),

co (R) =

∂ ∂t



s ¯2

(ψ(s) − o) × ρ v ds

s1

+ ρ (¯2 , t)v 2 (¯2 , t)(ψ(¯2 ) − o) × et (¯2 ) − ρ (¯1 , t)v 2 (¯1 , t)(ψ(¯1 ) − o) × et (¯1 ), where f (R) and co (R) are defined similarly to f (P) and co (P). What are the interpretations of these equations? 6.3. Use Euler’s laws of Exercise 6.2 to calculate the reaction forces and couple at s = 0 for the pipe bend in Figure 6.1. Conclude that these forces and couple are independent of the particular shape of the pipe. 6.4. Explain why (6.11) and (6.14) have the same mathematical form. Use this insight to give an alternative derivation of (6.14). 6.5. The ideas in Section 6.6 can be used to establish the equation of motion for shallow water flow. Let the density be constant, denoted by ρin , and let the pressure and volume force be given by p(s, y) = p∗ + ρin g(h(s, t) − y), q V = −ρin gen , ∗

where p is a fixed atmospheric pressure at the water surface, g is the gravitational acceleration, en is the upward unit vector and y is the corresponding coordinate, and h(s, t) is the water level above some horizontal reference. Show that the equation of motion becomes ∂v(s, t) ∂v(s, t) ∂h(s, t) + v(s, t) +g = 0. ∂t ∂s ∂s

(6.30)

6.6. For a given left hand side and taking s equal to a fixed length, (6.27) can be considered as a non-linear ordinary differential equation for the function q(t). Give an analytical solution for this differential equation.

7 Three-Dimensional Model

In this chapter the three-dimensional geometric model introduced in Chapter 3 is filled out with details. Similar developments are found in standard text-books on continuum mechanics and omitted details may be easily found elsewhere. Therefore, the presentation in this chapter is less complete than in previous chapters where other geometrical models were treated. If the reader were to pick one particular book on continuum mechanics in order to go deeper into the material of this chapter, the closest in notations and spirit would probably be Gurtin [5]. In Chapter 6 we found that there are two different types of formulations of equations of motions available: the first one in terms of the variable X, which labels material points, and the second one in terms of a pipe coordinate s, which identifies points along the pipe. In the three-dimensional case we have a similar situation: there is a material coordinate X ∈ B and a spatial coordinate x ∈ E. Since there is a one-to-one correspondence between X ∈ B and x ∈ Bt , we may use a description in terms of any of these two variables when writing Euler’s laws. The traditional approach is to state Euler’s laws using x as variable and deriving equations of motion valid on the domain Bt . Equations defined on B may then be obtained by a change of variable. We will follow this traditional path in this chapter. However, before stating Euler’s laws in explicit form for the three-dimensional model, we need some preliminaries such as definition of mass, density, derivation of continuity equation, etc.

7.1 Spatial and Material Representations, and Time Derivatives Spatial representations of velocity and acceleration are obtained from the material representations V (X, t) and A(X, t) and the placement φt as follows: v(x, t) = V ((φt )−1 (x), t),

a(x, t) = A((φt )−1 (x), t).

77

78

7 Three-Dimensional Model

As in the case of pipe flow we can speak generally about material fields (function of time and X) and spatial fields (function of time and x): every material field can be made spatial and vice versa. A spatial representation fs of the material field f is given by fs (x, t) = f ((φt )−1 (x), t). Inversely, the material representation gm of the spatial field g is given by gm (X, t) = g(φt (X), t). Again, as for pipe flow, the following notations are introduced for derivatives with respect to time. The time derivative of a spatial field is denoted by a prime, i.e. ∂g(x, t) g  (x, t) = . ∂t The partial time derivative of a material field is denoted by a superposed dot, i.e. ∂f (X, t) f˙(X, t) = . ∂t Concerning the material time derivative of a spatial field, the definition given in (6.3) holds also in the three-dimensional case, as well as the conclusions in (6.4). However, the generalization of (6.5) will depend on whether the field is a scalar field or a vector field. Proofs of the following two results can be found in Gurtin [5]. (i) Let ϕ(x, t) be a scalar field and v(x, t) the spatial velocity. Then ϕ(x, ˙ t) = ϕ (x, t) + v(x, t) · ∇ϕ(x, t),

(7.1)

where ∇ϕ(x, t), the gradient of ϕ(x, t), is a geometric vector, see Appendix C. (ii) Let ξ(x, t) be a vector field and v(x, t) the spatial velocity. Then ˙ ξ(x, t) = ξ  (x, t) + ∇ξ(x, t)v(x, t),

(7.2)

where ∇ξ(x, t), the gradient of ξ(x, t), is a tensor field. This object is also defined in Appendix C. The gradient of a vector field can be used to define the velocity gradient as follows: L(x, t) = ∇v(x, t), where v is the spatial velocity. From the chain rule of differentiation we have a formula for the time derivative of the deformation gradient: ∂ F (X, t) = L(x, t)F (X, t), ∂t

x = φt (X).

7.2 V l

dI

h

M

79

Another formula which will be useful later, but which is somewhat non-trivial to establish (see, e.g., Gurtin [5]) is ∂ det F (X, t) = det F (X, t) div v(x, t), ∂t

x = φt (X),

(7.3)

where div v(x, t) = tr∇v(x, t) = trL(x, t) and tr denotes the trace of a tensor, see Appendix C.

7.2 Volume and Isochoric Motion The volume of a part P of a body B at time t is given by   Vol(P Pt ) = dV Vx = det F (X, t) dV VX , Pt

P

where a change of coordinates provides the equality between the integrals. We will study how the volume changes with time. Using (7.3) and a change of coordinates, we can conclude that  ∂ ∂ Vol(P Pt ) = det F (X, t) dV VX ∂t ∂t  P  = det F (X, t) div v(φt (X), t) dV VX = P

Pt

div v(x, t) dV Vx . (7.4)

A motion is isochoric if

∂ Vol(P Pt ) = 0 ∂t for every Pt and t. By the three-dimensional localization theorem, Theorem 3 of Appendix C, we find from (7.4) that for isochoric motions div v(x, t) = 0, for all x ∈ Bt and all times t. Another interesting characterization of isochoric motion follows by applying the divergence theorem, (C.22), to (7.4). One then finds  v(x, t) · n(x, t) dAx = 0, ∂ Pt

where ∂P Pt is the boundary of Pt and n(x, t) is its outward unit normal. This states that there is no net flux across the boundary ∂P Pt .

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7 Three-Dimensional Model

7.3 Mass and the Continuity Equation As with the one-dimensional model, we define for each material point X ∈ B a scalar variable ρ0 (X), called the reference density, with physical dimension mass per volume. The mass of a part of the body P is defined as  M (P) = ρ0 (X) dV VX . P

Using the universal law of mass conservation and the three-dimensional version of the localization theorem, Theorem 3 of Appendix C, we find ∂ ρ0 (X) = 0, ∂t

(7.5)

for all X ∈ B. The existence of a one-to-one mapping φt between B and Bt makes it possible to express the mass of a part of a body as an integral over Pt = φt (P). Define the density ρ(x, t) at a point x ∈ Bt by ρ(x, t) det F (X, t) = ρ0 (X),

(7.6)

for x = φt (X). Changing variables in the definition of mass then gives  M (P) = ρ(x, t) dV Vx . Pt

Note that the density ρ(x, t) is clearly not required to be independent of time as is the case for the reference density ρ(X). From the conclusion (7.5) and from (7.6), we find by using (7.3) that ∂ det F (X, t) ∂t = ρ(x, ˙ t) det F (X, t) + ρ(x, t) det F (X, t) div v(x, t) = 0,

ρ(x, ˙ t) det F (X, t) + ρ(x, t)

and since det F (X, t) > 0 we have ρ(x, ˙ t) + ρ(x, t) div v(x, t) = 0.

(7.7)

Using (7.1) and (C.16) we find from (7.7) that ρ (x, t) + div (ρ(x, t)v(x, t)) = 0.

(7.8)

Equations (7.7) and (7.8) are two versions of the continuity equation which are direct consequences of conservation of mass. One concludes from (7.7) that for isochoric motion the material time derivative of the density is zero.

7.5 Stress Tensor, System of Forces, and Force and Torque Resultants

81

7.4 Linear and Angular Momentum As for the mass, linear and angular momentum may be expressed either as functions over P or as functions over Pt = φt (P). Both of these versions are given below. The linear momentum is defined as   p(P) = V (X, t)ρ0 (X) dV VX = v(x, t)ρ(x, t) dV Vx P

Pt

and the angular momentum is   ho (P) = r(X, t) × V (X, t)ρ0 (X) dV VX = (x − o) × v(x, t)ρ(x, t) dV Vx . P

Pt

The time derivatives of these vectors become, by using conservation of mass,   ∂ p(P) = A(X, t)ρ0 (X) dV VX = a(x, t)ρ(x, t) dV Vx , (7.9) ∂t P Pt ∂ ho (P) = ∂t

 P

r(X, t) × A(X, t)ρ0 (X) dV VX  = (x − o) × a(x, t)ρ(x, t) dV Vx . (7.10) Pt

7.5 Stress Tensor, System of Forces, and Force and Torque Resultants As in the one-dimensional case, we distinguish between external forces which have their cause outside the body, and internal forces which are caused by the interaction between the parts of the body. In the one-dimensional case the definition of forces was based on a cut principle: when part of a body is removed to form a sub-body, the action of the removed part is represented by cut forces and couples. These forces and couples are located at the cut, or the contact, between the parts and are therefore frequently referred to as contact forces and couples. The principle of representing the action of the removed part by contact forces is different from how internal forces are conceived in the discrete model where, in principle, a material point is acted on by all other points. This is called action at a distance. The cut principle in the one-dimensional case was introduced by Euler, but for the three-dimensional case it was given in full generality by Cauchy some 75-100 years later, in 1823. Cauchy assumed the existence, for each time t, of a force per unit area s(n, x) for every unit vector n and point x ∈ Bt , called the traction vector. For a cut surface S through Bt , s(n, x) is the force per unit area which the material on the positive side of S exerts upon that on

82

7 Three-Dimensional Model

the negative side. Note that any cut through a point x with the the same normal vector n gives the same vector s(n, x). Thus, the traction vector is independent of, for instance, the curvature of S. This cut principle of Cauchy is illustrated in Figure 7.1. In the following subsections we will present the main consequences of the cut principle when joined with Euler’s laws. The cut principle gives a system of forces which in turn gives explicit form to Euler’s laws, from which equations of motion and a form of action and reaction follow as mathematical theorems.

Fig. 7.1. Cauchy’s cut principle.

7.5.1 Normal and Shear Stress For simplicity, write s = s(n, x). For each normal vector n and point x ∈ Bt we may define sn = sn n, sn = n · s, st = s − sn , where sn and sn are the normal stress and the normal stress vector, respectively, and st is the shear stress vector, see Figure 7.2. Given any unit vector t such that t · n = 0, the scalar st = t · st = t · s is the shear stress in the direction t. 7.5.2 The Stress Tensor and the System of Forces Stress components are defined relative to an orthonormal base {e1 , e2 , e3 }. The stress components at a point x ∈ Bt are defined as follows:

7.5 Stress Tensor, System of Forces, and Force and Torque Resultants

83

Fig. 7.2. Normal and shear stress vectors.

σij (x) = ei · s(ej , x),

i, j = 1, 2, 3.

There are 9 stress components. The component σii is the normal stress1 on the cut surface with normal vector ei and σij , i = j, is the shear stress in the direction of ei on the cut surface with normal vector ej , see Figure 7.3.

Fig. 7.3. Illustration of components of the stress tensor.

If the stress vector is a linear function of the vector n we write s = T n,

(7.11)

where T = T (x) is a tensor called the stress tensor. Comparing with (C.3) in Appendix C one finds that the components of this tensor in the {e1 , e2 , e3 }system are σij = σij (x), i.e., ⎡ ⎤ σ11 σ12 σ13 T ∼ ⎣ σ21 σ22 σ23 ⎦. σ31 σ32 σ33 1

The summation convention is not used in this work, so σii represents σ11 , σ22 and σ33 and not σ11 + σ22 + σ33 .

84

7 Three-Dimensional Model

It can be proved from Euler’s laws that the linearity assumed in writing (7.11) is indeed true and that internal forces may thus be represented by a stress tensor. The proof will be partly outlined below after we have formally introduced the system of forces: (i) For every x ∈ Bt there is a geometric vector b(x) called the external force vector per unit volume. (ii) For every x ∈ Bt and unit vector n there is a geometric vector s(n, x) called the traction vector. The force and torque resultants for a part P of B at time t are now defined as   b(x) dV Vx + s(n, x) dAx , (7.12) f (P) =  c(P) =

Pt

Pt

(x − o) × b(x) dV Vx +

∂ Pt

 ∂ Pt

(x − o) × s(n, x) dAx ,

(7.13)

where n is the normal vector to the boundary ∂P Pt .

7.6 Explicit Forms of Euler’s Laws Putting together the expressions for the derivatives of linear and angular momentum, (7.9) and (7.10), with the expressions for total force and torques, (7.12) and (7.13), we obtain Euler’s law of linear momentum    b(x) dV Vx + s(n, x) dAx = Pt

∂ Pt

Pt

a(x, t)ρ(x, t) dV Vx .

(7.14)

Euler’s law of angular momentum   (x − o) × b(x) dV Vx + (x − o) × s(n, x) dAx Pt ∂ Pt  = (x − o) × a(x, t)ρ(x, t) dV Vx . (7.15) Pt

7.7 Equations of Motion It can be proved that a necessary and sufficient condition for Euler’s laws (7.14) and (7.15) to be satisfied is that there exists a tensor field T = T (x), x ∈ Bt such that (i) equation (7.11) holds, i.e., s = T n; (ii) T is symmetric, i.e., T = T T , where a superscript T denotes the transpose of a matrix, see Appendix C;

7.7 Equations of Motion

85

(iii) T satisfies the equation of motion div T (x) + b(x) = ρ(x, t)a(x, t).

(7.16)

For a proof we refer to, e.g., Gurtin [5]. Essentially, (7.11) follows by applying (7.14) to a tetrahedron with unit normals −e1 , −e2 , −e3 and n. The symmetry of T follows from (7.15) and can be considered as a local form of this equation. Similarly, the equation of motion (7.16) is a local form of (7.14). Note that (7.11) contains a form of the law of action and reaction: by taking n to be positive and negative one finds s(n, x) = −s(−n, x). The symmetry of T implies that σij = σji and therefore there are only 6 independent components of the stress tensor.

Exercises 7.1. Verify that spatial and material definitions of p(P) and ho (P) are equivalent. Verify equations (7.9) and (7.10) 7.2. The mean stress of a part P of a three-dimensional body at time t is denoted by T¯ (P Pt ) and is defined by  Vol(P Pt )T¯ (P Pt ) = T dV Vx . Pt

Show that Pt )T¯ (P Pt ) = Vol(P





Pt

(b − ρa) ⊗ (x − o) dV Vx +

∂ Pt

s ⊗ (x − o) dV Vx ,

where ⊗ is the dyadic product, see Appendix C. Note that in the absence of body forces and accelerations, the mean stress is completely given by boundary values of the traction vector. 7.3. Suppose in the result of the previous exercise that the body-part P is in equilibrium and that there are no body forces. Suppose also that ∂P Pt consists of two surfaces, S0 and S1 , such that S1 is enclosing S0 and that s = −π0 n on S0 , s = −π1 n on S1 . 

Show that T¯ (P Pt ) = −

π1 V1 − π0 V0 V 1 − V0

 I,

where V1 and V0 are the volumes enclosed by S1 and S0 , respectively.

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7 Three-Dimensional Model

7.4. A planet-like spherical body with radius R and uniform density ρ has an angular velocity ω. Show that the mean stress in the planet is a hydrostatic pressure (1/5)ρgR plus a biaxial tension (1/5)ρR2 ω 2 perpendicular to the polar axis, where g is the acceleration on the planet’s surface due to its own gravitation. 7.5. Consider a three-dimensional body that moves inside a fixed boundary in a stationary motion, i.e., for every x ∈ Bt , v(x, t) = v(x),

ρ(x, t) = ρ(x).

Conclude that such a body is in quasi-equilibrium, cf. Exercise 4.2.

Part III

Complete Models by Adding Particular Laws

8 Particular Laws

In Part II we derived general conclusions that follow from the universal laws of mechanics. It should be clear that the universal laws are generally not sufficient to study a particular natural phenomenon or engineering problem. A first obvious reason for this fact is that there is nothing in these laws that distinguish between, e.g., the behavior of a solid substance or a fluid substance, which from common sense have very different mechanical behaviors. Thus, to obtain complete problems we generally need to introduce assumptions which characterize such behaviors. This chapter contains an introduction to these assumptions, which we call particular laws. In Section 8.1 we give a classification of constitutive laws, and in Section 8.2 we go through some general principles that are guides in formulating such laws. In the remaining sections of this chapter we present some examples of particular laws. Some of these will be used to formulate complete models in upcoming chapters, and some are presented as illustrations of general ideas.

8.1 Three Types of Particular Laws Generally, the extra conditions needed to obtain complete models can be divided into kinematic constraints, constitutive laws and force laws. Kinematic Constraints These are conditions that restrict the motion of maaterial points. Typical such conditions are incompressibility and the condition giving the theory of rigid bodies. Others are, for instance, contact conditions such as the restriction that two bodies cannot penetrate each other, i.e., two material points cannot occupy the same place in physical space. A kinematical constraint was in fact introduced in our discussion of pipe flow, where the placement of material points was restricted to belong to the pipe. Boundary conditions are often of kinematic nature, where we impose that the boundary of a configuration Bt has a prescribed placement. Generally, a kinematic constraint is followed by a condition on

89

90

8 Particular Laws

the force maintaining the restriction. For instance, if we have frictionless contact between two bodies, the contact force, maintaining the impenetrability contact condition, is prescribed to be compressive and having a direction that is normal to the contact surface. Forces related to kinematic constraints are usually called reaction forces. Constitutive Laws These are laws that govern internal forces. Typically we build these laws on some experimental observations where we have somehow isolated the law in question: the linear behavior of elastic solids at small strain is found from simple uniaxial tests, and the linear behavior of viscous fluids is found by using a Couette viscometer. Having found that a law is valid in a situation where it is isolated, we make the hypothesis that it holds also in more general situations. Force Laws These are laws that govern external forces. The most well known such law is probably that of gravitation. The force of gravitation is isolated when we let a body fall freely. We make the hypothesis that the same force also acts in situations when other forces are present. A special simple case of a force law that we frequently use is when a force is set to a fixed prescribed value for the simple reason that we know the force from calculations or considerations relating to events outside the present system. Note that kinematic constraints can be seen as extreme cases of force laws and constitutive laws, but it is, anyway, conceptually useful to to keep them as a separate class. In any case, constitutive laws (and some kinematic constraints) are relations between motion φt and an internal force, i.e., f i , f and m, and T , depending on the dimension of the model, and force laws (and some kinematic constraints) are relations between motion and appropriate external forces. Moreover, since the distinction between internal and external forces depends on how the system boundary is set, it may, in some cases, be somewhat arbitrary whether a particular law should be regarded as a constitutive law or a force law, but in this case also a classification is useful.

8.2 General Principles for Particular Laws A general comment on particular laws is that it is clearly naive to think that such laws are solely due to well planned physical experiments, even though facts from these should be of primary importance. We are, in fact, usually guided in the search for particular laws by previous experience with related or less general theories, and the reason for choosing a particular law may then be more a matter of mathematical convenience than experimental evidence. Moreover, there are general principles that any particular law should satisfy. Below we give some invariance principles that are of utmost importance. Further discussion on these issues can be found in many sources, but the most comprehensive are probably those in Truesdell and Toupin [19] and Truesdell and Noll [18].

8.2 General Principles for Particular Laws

91

Coordinate Invariance Any relation with a physical content should be invariant under the choice of a particular coordinate system. In this text, and widely used elsewhere, see, e.g., Gurtin [5], we achieve this by using a coordinate–free representation in terms of geometric vectors and tensors that makes equations independent of any coordinates. Dimensional Invariance In addition to coordinate invariance we can conclude that any relation describing physical behavior should also be invariant with respect to the physical units used, e.g., the meaning of an equation should not change if we use meters instead of yards as a length unit. Such invariance implies that it is possible to rewrite equations in a dimensionless form. This sometimes gives striking insight into a problem; an example is the dimensionless form of Navier-Stokes’ equations, see Section 11, from which we identify the Reynolds number as a dimensionless combination of quantities that is of relevance for the physical behavior of a viscous fluid. Observer Invariance An intuitive view of an observer is that it is someone or something which maps physical events into the Euclidean point space E. As discussed in Chapter 2, this is done on the basis of physical objects such as the walls of a laboratory, or heavenly bodies such as the sun and the stars. Based on this view we understand that distance is invariant to a change of the observer and therefore the images that two observers produce in E will differ by a rotation and a translation but not in shape or dimension. In mathematical terms we say that two motions φt (X) and φ∗t (X) are related by a change in observer when φ∗t (X) = q(t) + Q(t)(φt (X) − o)

(8.1)

for every material point X and time t. Here q(t) is a point in E, representing translation, and Q(t) is a proper1 orthogonal tensor, representing rotation. The particular mathematical form of equation (8.1) is justified in Section 12.3, where we treat rigid body motion. Furthermore, forces and couples – e.g., f i and f e of the discrete model, f , m and q 0 of the one-dimensional model and s and b of the threedimensional model – act on configuration Bt = φt (B). Thus, two observers which interpret motion as φt (X) and φ∗t (X) will also interpret forces differently. If the first observer sees a force as a vector s then the second observer will see it rotated by Q(t), i.e., as s∗ = Q(t)s.

(8.2)

Now, a particular law which is such that it relates a motion φ t(X) to a force, say s, is said to be independent of the observer if it holds that φ∗t (X) and s∗ , given as in (8.1) and (8.2) are also related by the same law. 1

An orthogonal tensor Q satisfies det Q = ±1. The proper orthogonal tensors are those for which the plus sign holds. These tensors represent rotations.

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8 Particular Laws

A situation when we more of less intuitively rely on independency of the observer is when force measurements need to be made in rotating reference frames. If these measurements are based on a constitutive law such as a Hooke-type law of a spring, we are likely to established this law in a static frame, but assume that it is also valid in the rotating frame. For instance, let a mass be kept in circular motion by a spring attached to the mass and the center of the circle. In attempting to measure the centrifugal force on the mass, we may measure the elongation of the spring. Coordinate invariance and dimensional invariance should be valid for all equa tions of mechanical theory, also for those obtained from Euler’s laws. However,Euler’s laws are not independent of the observer. The class of observer transformations for which Euler’s laws are invariant are those that connect inertial frames. In non-inertial frames, pseudo-forces, such as centrifugal forces, have to be added. However, we still take as a physical principle that the constitutive laws and the force laws should obey independency of the observer; in the example of the rotating spring, this principle states that the only effect from rotation is the centrifugal force. The three invariance principles above are general principles, in the sense that no reasonable particular law should violate them2 . There are, however, other principles which are valid for large classes of particular laws, but which can be violated. One such principle is that of local action for constitutive laws in continuum theories, which states that the stress (or cut forces and couples) at a material point X should not depend on the motion outside an arbitrary small neighborhood of X. Since the motion close to X is characterized by its gradient, a possible implementation of local action in the three-dimensional theory is to say that the stress depends on the deformation gradient F . Ideal materials, i.e., constitutive laws, where the force (or stress) depends on the deformation gradient, but not on higher order gradients, are called simple materials. It seems that only very exotic behaviors need constitutive laws which are not simple, and they will not be discussed further in this text. Another principle for constitutive laws of continuum bodies, which is certainly not generally applicable, but which is sometimes very useful, is that of isotropy. An ideal material is said to be isotropic if it possesses no preferred directions, i.e., its mechanical response is the same no matter what is the orientation of the material. Certainly, there are real materials which are not isotropic, examples being wood or composite materials. However, in cases when the principle is applicable it grately simplifies equations. Finally, we mention the of is related to the dynamical behavior of a mechanical system: if the system returns to its initial state after an arbitrarily small perturbation is applied and removed, then the 2

Exceptions to this are based on the small displacement as sumption. The cancellation of higher order terms involved in these theories destroys the possibility of satisfying independency of the observer.

8.3 Examples of Particular Laws for the Discrete Model

93

system is said to be stable in that state; if an arbitrarily small perturbation sends the system off to a new state, the initial state was unstable. However, such a definition of stability is frequently mathematically difficult to apply, and some more intuitive definitions of stability have appeared, e.g., in Section 9.3 an elastic constitutive law is said to be stable if elongation is accompanied by tension. Stability turns out to be strongly related to uniqueness of solutions of complete problems.

8.3 Examples of Particular Laws for the Discrete Model A kinematic restriction for the discrete model of Chapter 4, frequently introduced and discussed in the literature, is that the distance between material points is constant, i.e., considering the material points Xk and X one assumes that |φt (Xk ) − φt (X X )| is independent of time. (8.3) When such restrictions are stated for all pairs of points of the body B, we obtain a model which is usually referred to as a rigid body model. However, the physical counterpart of a rigid body is probably a bulky body with continuous distribution of matter. The point-wise distribution of mass of the discrete model may therefore not be appropriate. A more natural rigid body model follows by introducing (8.3) for all material points of the three-dimensional geometric model. This is treated in Chapter 12. For pairs of points not restricted by (8.3), a constitutive law may be introduced for the internal forces related to the pair: considering again points Xk and X we assume f i (Xk , X ) = −f i (X X , Xk ) = f (d) n(φt (Xk ), φt (X X )), where n(φt (Xk ), φt (X X )) =

(8.4)

φt (Xk ) − φt (X X ) , |φt (Xk ) − φt (X X )|

is the unit vector pointing from φ(X X ) to φ(Xk ), and f (d) is a possibly nonlinear function of the distance between the pair of points, the distance being given by d = |φt (Xk ) − φt (X X )|. It is a simple matter to show that the constitutive law defined by (8.4) is independent of the observer (Exercise 8.1). Note also that it is consistent with the law of action and reaction. Probably the most well-known example of f (d) is Newton’s law of gravitation: mk m f (d) = G 2 , d where G is the universal constant of gravitation, and mk and m are masses of the two points under consideration. If f (d) is aimed at representing the

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8 Particular Laws

springy behavior of a solid (massless) body (simply, a spring) connecting the two points, we expect that there is a distance d0 such that f (d0 ) = 0 and a stable behavior of the spring corresponds to ∂f (d)/∂d > 0. The examples of gravitation and response of a spring, represent elastic constitutive behaviors: a certain value of d, at a particular time, corresponds to a unique internal force and the previous time history of d has no influence. On the other hand, if the force depends on the time derivative of d, or more generally, on its whole previous history, then the constitutive law is usually referred to as plastic or viscous.

8.4 Examples of Particular Laws for the One-Dimensional Model As indicated in Section 8.2, in continuum models (i.e., one-, two- and threedimensional geometric models) a general constitutive idea is the principle of local action, which states that the internal forces and couples at a material point X depend on the motion only in an arbitrarily small neighborhood of X. For a Cosserat theory, with the motion specified by φt (X) and directors di (X, t), i = 1, . . . , n, see Section 3.5, an implementation of the idea of local action is to say that internal forces and moments at X depend on φt (X) and di (X, t), i = 1, . . . , n and the derivatives ∂φt (X) ∂di (X, t) and , i = 1, . . . , n. ∂X ∂X Dependence on higher order derivatives could, in principle, also be included, but is useful only in very special cases. Thus, we may assume the existence of ˆ such that constitutive functions fˆ and m ∂φt (X) ∂di (X, t) f (X) = fˆ (φt (X), di (X, t), , (8.5) , X), ∂X ∂X ∂φt (X) ∂di (X, t) ˆ t (X), di (X, t), m(X) = m(φ , , X). (8.6) ∂X ∂X The principle of independency of the observer can be shown to require, see ˆ should not be present. Antman [1], that the first two arguments in fˆ and m That is to say that while, e.g., f (X) may depend on the derivative of the placement it cannot depend on the placement itself. Intuitively this makes sense since the response should depend on relative positions of material points (expressed by the derivative) and not on their absolute locations. Equations (8.5) and (8.6), with restrictions introduced by the independency of the observer principle, are general elastic constitutive equations valid for large displacement problems. For the details of such quite complicated problems we refer to Antman [1]. In this text we will treat elastic one-dimensional problems, i.e., beam problems, only in the small displacement case, and for such problems constitutive equations are introduced in an alternative way in Section 9.3.

8.6 Examples of Particular Laws for the Three-Dimensional Model

95

8.5 Examples of Particular Laws for the Pipe Flow Model In the discussion of constitutive equations for the one-dimensional model in the previous section we put down relations valid for a particular material point X. This is a material or Lagrangean point of view which is appropriate for a solid substance. The pipe flow model is intended primarily to model the behavior of a fluid substance. In such a case it is more natural to search for a constitutive law which is valid for a particular point along the pipe, i.e. for a particular coordinate s. This is a spatial or Eulerian point of view. Based on the principle of local action we may assume that internal forces, represented by the pressure p, depend on ∂s(X)/∂X. This means, from equations (6.10) and (6.15) and by assuming a constant cross-section, that the pressure depends on the density function ρ(s, t). Thus, we may suggest the following general constitutive law of elastic type: p(s) = pˆ(ρ(s, t)),

(8.7)

where pˆ is the constitutive function. A kinematic constraint has been introduced in the pipe flow model in that the material points have to follow the pipe. This introduces additional reaction forces which maintain the constraint.

8.6 Examples of Particular Laws for the Three-Dimensional Model An implementation of the principle of local action in the three-dimensional case is to assume that the stress tensor T (X) is dependent on the derivative of the deformation, i.e., on the deformation gradient F (X, t). The following general constitutive equation can then be assumed T (X) = Tˆ (F (X, t), X),

(8.8)

where Tˆ is the constitutive function. The constitutive equation (8.8) represents an elastic behavior: a given deformation, represented by F (X, t), completely specifies the stress, independently of the time derivative of F (X, t) or, more generally, of its previous history. Elasticity is usually associated with solid bodies. However, (8.8) also includes, as special cases, the ideal and elastic fluids. Such fluids are defined by the property that for any cut surface S through the body, the traction vector is normal to the cut and, thus, the shear stress is zero. In mathematical terms this reads s(n, x) = −p(n, x)n, (8.9)

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8 Particular Laws

where p(n, x) is the so-called hydrostatic pressure. By comparing (8.9) with (7.11) we see that this pressure is actually the same for all directions n at a point x. Thus, p(n, x) = p(x) and T (x) = −p(x)I,

(8.10)

where I is the identity tensor. An elastic fluid is now defined by (8.10) and a constitutive function pˆ such that p(x) = pˆ(ρ(x, t)), (8.11) where ρ(x, t) is the density. Note that formally the same law was given for pipe flow in (8.7). However, these two laws relate to different geometric models and, in particular, equation (8.10) was not needed for the pipe flow case.

Fig. 8.1. The function pˆ for an elastic fluid. For the ideal fluid, ρ(x, t) is a constant, ρin (subscript in stands for incompressible), and p(x) can take any value.

The ideal fluid is an extreme case of (8.11), where the slope of the function pˆ goes to infinity and we set ρ(x) to a constant, see Figure 8.1. This means that the material derivative of ρ(x, t) is zero, which according to the continuity equation (7.7) means that div v(x, t) = 0 and, thus, the motion of an ideal fluid is always isochoric. Constitutive assumptions which allow only isochoric motion are said to define an incompressible material. A fluid differs from a solid in that it cannot support shear stresses without motion. However, the elastic and ideal fluids defined above do not include any possibility of developing shear stresses at all. Such constitutive equations work well for studies of, e.g., flow of air at low speeds, but are not appropriate for studies of, e.g., flowing water. An extension of (8.10) that takes shear forces due to shearing motion of the fluid into account should include the velocity gradient tensor L(x, t) = ∇v(x, t),

8.6 Examples of Particular Laws for the Three-Dimensional Model

97

which measures the relative velocity of material points. A linear extension of (8.10) that takes shear forces into account is ˜ T (x) = (−p(x) + λtrD(x, t))I + 2˜ µD(x, t),

(8.12)

where

1 (L(x, t) + L(x, t)T ) 2 is the rate of deformation tensor, which is the symmetric part of L(x, t), ˜ and µ and λ ˜ are viscosity coefficients. Since for isochoric motion D(x, t) =

div v(x, t) = tr L(x, t) = tr D(x, t) = 0, the constitutive law (8.12) reduces to T (x) = −p(x)I + 2˜ µD(x, t)

(8.13)

for such cases. This constitutive law together with div v(x, t) = 0 defines an incompressible fluid. The constitutive law (8.13) is a special case of a so called Newtonian fluid which is defined by saying that the stress T (x) is a linear function of the velocity gradient tensor L(x, t). However, it can be shown that a necessary and sufficient condition for such incompressible fluids to be independent of the observer is that it is of the form (8.13). Thus, (8.13) defines the most general Newtonian fluid we can think of. It is remarkable that it is characterized by one constant only. Stability or uniqueness of solutions will require that this constant µ ˜ > 0. Furthermore, (8.13) contains no reference to the orientation of the material and therefore defines an isotropic material. Thus, one concludes that a (Newtonian) fluid is always isotropic. Concerning constitutive laws that define solid material, this text will deal with them in the limited context of small displacements only. Small displacement theories are well known within strength-of-materials and are termed beam (one-dimensional), and plate (two-dimensional) theories, and linear elasticity (three-dimensional). Since an approximation is involved in these theories, constitutive laws will not satisfy independency of the observer. However, we can talk of isotropic material and the most general isotropic linear elastic constitutive law is then T (x) = λ(tr E(x, t))I + 2µE(x, t),

(8.14)

where E(x, t) is the small displacement strain tensor, and λ and ν are called Lame’s ´ elasticity coefficients. Note the remarkable resemblance in form between (8.12) and (8.14). However, the physical meanings of these equations are quite different since D(x, t) appears in the first one and E(x, t) in the other. Stability or uniqueness of solutions for boundary value problems based on (8.14) follows if µ > 0 and 2µ + 3λ > 0, see Section 9.3.

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Exercises 8.1. Show that the constitutive law (8.4) is independent of the observer. 8.2. Consider a discrete body consisting of two material points. Let one of the points be fixed at o and assume that the internal forces between the points are governed by (8.4) with k f (d) = α , d where k and α are constants such that k > 0 and −1 ≤ α ≤ 2. The classical Newton’s law of gravitation is recovered as a special case when α = 2 and k = Gm1 m2 , where G is the universal constant of gravity and m1 and m2 are the masses of the two points. When α = −1, a linear spring connects the two points. Introduce this constitutive law into Newton’s equation, use a suitable coordinate system and solve the resulting system of ordinary differential equations numerically (using, e.g., MATLAB) for different values of α. What happens when −1 < α < 2? Consult the literature on the two-body central force problem and Bertrand’s theorem, e.g., Jose´ and Saletan [11].

9 Small Displacement Models

Well known models from the field of strength-of-materials, such as those of beams, plates and linear elasticity, are based on a small displacement assumption. Elementary presentations of such theories usually introduce the small displacement assumption on physical grounds. However, at least in the elastic case, a mathematically clear procedure for deriving small displacement theories is to put down equations valid for the full large displacement case and then linearize these equations at a configuration having zero internal forces. An alternative, perhaps less strict approach, is to work by induction. That is, we study in detail a simple small example, identify what seems to be general consequences of a linearization and use these consequences as assumptions when deriving more refined theories. This is basically the route used in the present text. By considering a simple example, the following consequences, which are next generalized and used as assumptions when deriving small displacement theories of one- and three-dimensional problems, are identified: (i) Equations of motion are stated in a reference configuration where internal forces are zero, without any reference to the deformed or loaded state. In a three-dimensional model this has the consequence that we do not have to distinguish between material coordinates X and spatial coordinates x. (ii) Internal forces are linear functions of the displacements. The second item is usually interpreted through the introduction of the concept of strain, so that we have a geometric relation where strain is a linear function of the displacement, and a constitutive equation where internal forces are linear functions of strain.

9.1 Discrete Models and General Structure 9.1.1 Linearizing a Simple Discrete Model Consider a discrete system consisting of four material points, see Figure 9.1. These points are associated with reference positions

99

100

9 Small Displacement Models

Fig. 9.1. A simple model with four material points.

φ0 (X),

X = 1, 2, 3, 4

in physical space E. We may think of these reference positions as initial positions of the material points. By the influence of forces the material points generally take other positions in space at times larger than the initial time. However, in this simple example we assume that only X = 4 is free to move while the other material points are fixed at their reference positions for all times. We write, for the position of the free material point at an arbitrary time, φt (4) = φ0 (4) + u, where u is called the displacement vector. Furthermore, we treat the problem as two-dimensional, i.e., either we think that E is two-dimensional or we consider all positions and vectors as belonging to a two-dimensional subspace of a three-dimensional E. We write Newton’s law for the free point. For simplicity, only a static situation is studied, so the acceleration of the free point is taken to be zero. Newton’s law then becomes an equilibrium equation which we write as fe +

3

f iA = 0,

(9.1)

A=1

where, in the notation of Chapter 3, f e = f e (4), f iA = f i (A, 4),

A = 1, 2, 3.

The conditions of action and reaction put constraints on f iA , A = 1, 2, 3. General expressions that satisfy these constraints are (see Section 8.3)

9.1 Di

M d l

dG

lS

f iA = −ffA (dA )nA (u), where nA (u) =

101

(9.2)

φ0 (4) − φ0 (A) + u φt (4) − φ0 (A) = |φt (4) − φ0 (A)| |φ0 (4) − φ0 (A) + u|

is the unit vector directed from the position of the point X = A to the point X = 4, and fA (dA ) is a scalar valued function of the distance dA = dA (u) = |φt (4) − φ0 (A)| = |φ0 (4) − φ0 (A) + u|. A non-linear complete problem consists of finding, for a given external force f e , the displacement u such that (9.1) and (9.2) are satisfied. We are interested in a linearized version of this problem. To that end, we assume that the functions fA are such that fA (dA (0)) = 0.

(9.3)

We note that (9.1) is already a linear function so we need only consider (9.2). Note that fA (dA (u)) = fA (dA (0)) +

∂ffA (dA (0)) ∇dA (0) · u + o(u), ∂dA

nA (u) = nA (0) + o(1), where the “residual” function o(·) is defined in Appendix C.2 and nA (0) = ∇dA (0), so equation (9.2) becomes, with (9.3),   ∂ffA (dA (0)) f iA = −nA (0) nA (0) · u + o(u). ∂dA We introduce this equation, for A = 1, 2, 3, into (9.1) and neglect higher order terms to obtain   3 ∂ffA (dA (0)) e nA (0) nA (0) · u . (9.4) f = ∂dA A=1

A linear complete problem now consists of solving u from (9.4) for given f e . This is clearly a small displacement problem since we have neglected higher order terms of u. Equation (9.4) has a form which, as we will see later, is shared by all small displacement theories. As a first step in revealing this structure we introduce the notation eA = nA (0) · u. (9.5) Since

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9 Small Displacement Models

dA (u) − dA (0) = eA + o(u), we may interpret eA as giving the first order change of distance between material points X = A and X = 4. We call eA the (generalized) strain.1 The term within large brackets in (9.4) is the first order change of fA . It is denoted by σA and called the (generalized) stress2 . With this notation (9.4) can be written as 3 fe = nA (0)σA (9.6) A=1

and by introducing the notation EA =

∂ffA (dA (0)) ∂dA

we may write σA = EA eA .

(9.7)

Equation (9.6) can be interpreted as an equilibrium equation where the geometric information used to write it down is that of the reference positions, or reference configuration, i.e., nA (0) and not nA (u) is used. Equation (9.7) may be considered as a constitutive equation related to the first order quantities, and EA is called an elasticity coefficient. Equations (9.5), (9.6) and (9.7) contain the consequences of linearization as indicated in the introduction of the chapter: (i) Equilibrium (a special case of equation of motion) is stated in relation to the reference configuration which is assumed to correspond to zero internal forces (a stress free configuration3 ). (ii) Internal forces are linear functions of displacements, and these linear functions are naturally written as a composition between two other linear functions: one geometrical and one constitutive. Generalizations of these consequences may be used as direct assumptions when writing down small displacement theories for other geometrical models than the discrete one. This will be utilized for beams and for linear elasticity in upcoming subsections. However, equations (9.5), (9.6) and (9.7), in fact, contain further structure, not yet mentioned, that can be used in such a generalization. To see this, it is useful to introduce a matrix notation. Let {e1 , e2 } be a 1

2 3

This is the first time the concept of strain appears in this text. Strain is here regarded as any measure of change of shape and size (not orientation) of a mechanical system. When such a general notion of strain is used we usually talk of generalized strain to indicate a broader meaning than that of elementary strengthof-materials texts. Generalized stress is related to generalized strain through the concept of work as discussed below in Section 9.1.3. Linearizing around a non-stress free configuration leads to the linear theory of elastic stability.

9.1 Discrete Models and General Structure

103

set of orthonormal vectors that form a base of the two-dimensional subspace of E that is under consideration. Vectors can be expressed in this base, i.e., fe =

2

fie ei ,

nA (0) =

i=1

fe = as well as matrices

f1e f2e

nAi ei ,

u=

i=1

and column vectors can be introduced 

2

 ,



 u1 u= , u2

⎤ n11 n12 D = ⎣ n21 n22 ⎦, n31 n32 ⎡

2

u i ei ,

i=1

⎡

⎤ e1 e = ⎣ e2 ⎦, e3

⎡

⎤ σ1 σ = ⎣ σ2 ⎦, σ3

⎤ E1 0 0 E = ⎣ 0 E2 0 ⎦. 0 0 E3 ⎡

With these notations, (9.5), (9.6) and (9.7) can be written as e = Du,

f e = DT σ,

σ = Ee,

(9.8)

where T indicates transpose of a matrix or a vector. The three equations in (9.8) can be put together to form f e = Ku,

(9.9)

where K = DT ED is the stiffness matrix. Equation (9.9) is the matrix form of (9.4). It is a remarkable fact that the transpose of the geometrically defined matrix D appears in the equilibrium equation f e = DT σ. This has to do with how generalized stress and strain are defined, i.e., how equation (9.4) was broken up into (9.5), (9.6) and (9.7). As we will see below, it is also related to the concept of mechanical work. However, before we take a look at that, the mechanical model of a truss will be briefly considered and we will see that such a model has a structure that is completely analogous to the structure of the present simple problem. 9.1.2 Truss Model A geometric view of a truss is that it consists of a number of joints or nodes distributed in the physical space E together with bars or segments that connect the nodes, see Figure 9.2. Each bar connects two nodes and when pulled or pushed by opposite forces at these nodes it displays a linear elastic behaviour, i.e., it may be regarded as an elementary mechanical spring element. Moreover, bars are joined so that they are free to rotate relative to each other without any resistance. Now, if we consider a static situation and

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9 Small Displacement Models

Fig. 9.2. A typical two-dimensional truss structure.

small displacements, it turns out that an extended version (many material points) of the linearized problem of the last section is a good model for a truss. In a sense, a truss is a referent of the model of the last section, where reference positions of material points correspond to nodes, and the bars are the interaction lines between these points. The nodes of the are decomposed into fixed fi nodes and free nodes. The fixed nodes are fixed in space for all times and have zero displacements. The free nodes are free to be displaced by the application of external forces. Displacement vectors of free nodes are denoted ui and external force vectors are denoted f e(X i), see Figure 9.2. We will formulate a complete problem which consists in finding these displacement vectors given the external forces. As generalized strains we take elongations of bars, which we collect into a column vector e. All displacement vectors of free nodes, i.e., ui , form a column vector u and to the first order (a small displacement assumption) we may write e = Du, (9.10) for a matrix D that is built up by the elements of direction vectors of the bars. This is simply the same equation as (9.5), but written for many bars simultaneously. The generalized stresses are the forces transmitted by the bars and they are collected in a column vector σ. These forces are at each free node in equilibrium with the external forces f e (Xi ), that are collected in a column vector f e . This equilibrium can be expressed, simultaneously for all free nodes, as DT σ = f e, (9.11) which is equation (9.6), but now for many free nodes. As is apparent from (9.5) and (9.6), D contains normal vectors of bars as rows and DT contains the same vectors as columns. Finally, we may introduce a matrix E with

9.1 Discrete Models and General Structure

105

nonzero elements on the diagonal only. These nonzero elements are the elasticity coefficients of the bars, and as a third equation we have σ = Ee.

(9.12)

Altogether we may define the following complete model for a truss: Truss problem: Given a vector of external forces f e , find the displacement vector u such that Ku = f e ,

where K = DT ED.

(9.13)

The matrix K is called the stiffness matrix and has been, for the special case, introduced already in the previous subsection. If all elasticity coefficients are greater than zero, K can be shown to be positive semi-definite. When bars and fixed nodes are arranged in such a way that rigid body displacements of the structure are prevented, K is non-singular and the above problem has a unique solution. Equations (9.10), (9.11), (9.12) and (9.13) may be represented as in Figure 9.3.

Fig. 9.3. The structure of the truss equations.

9.1.3 Work Equations We give a proposition showing that the appearance of DT in the equilibrium equation is equivalent to the existence of a work equivalence. Introduce the set of kinematically admissible displacements and strains: K = {(u, e) | e = Du}. This is the set of pairs of vectors that satisfy the geometric equation of the model. We may now state the claimed equivalence:

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9 Small Displacement Models

Proposition 1 All (σ, f e ) which satisfy DT σ = f e

(9.14)

f eT u = σ T e for all (u, e) ∈ K,

(9.15)

are such that and conversely. Proof : To prove that (9.14) implies (9.15) we take the scalar product (in the sense aT b for two column vectors a and b) with any u: (DT σ)T u = f eT u. Since (DT σ)T u = σ T Du = σ T e the result follows. Conversely, (9.15) may be written as f eT u = σ T Du for all u, i.e., (f e − DT σ)T u = 0 for all u. If u is a column vector of length n we may take n different u’s, each with one entrance being 1 and all others being zero, and in that way we retrieve all the rows of the matrix equation (9.14). A reversed version of Proposition 1 can also be given. Introduce the set of statically admissible forces and stresses: S = {(f e , σ) | f e = DT σ}. The proof of the following Proposition is completely analogous to that of Proposition 1. Proposition 2 All (u, e) which satisfy Du = e

(9.16)

f eT u = σ T e for all (f e , σ) ∈ S,

(9.17)

are such that and conversely. Equation (9.15) is usually called the principle of virtual displacements and equation (9.17) the principle of virtual forces. An explanation for the term virtual follows below. The scalar f eT u may be interpreted as the mechanical work of the external force f e . The equation f eT u = σ T e says that this work is equal to σ T e. Since σ represents internal forces, we call σ T e the internal work, while f eT u is known as the external work. Thus, (9.15) and (9.17) say that the internal work equals the external work.

9.1 Discrete Models and General Structure

107

We can now return to the concepts of generalized stresses and strains. A generalized strain is any measure of deviation from rigid body displacement, i.e., change of shape and size. The generalized stress σ corresponding to a generalized strain e is the quantity that can be multiplied, in a scalar product sense, with the strain to produce the internal work such that the equivalence f eT u = σ T e is valid. Generalized stresses and strains that represent internal work in this way are said to be work conjugate. Kinematically admissible displacements and strains are frequently refereed to as virtual displacements and strains, since such quantities are not constrained by equilibrium or constitutive equations. In this terminology the left and right hand sides of (9.15) would be called external and internal virtual work, respectively. Similarly, one refers to statically admissible forces and stresses as virtual forces and stresses, respectively. It is not clear if anything is gained from this terminology, but it has a firm tradition in many subfields of applied mechanics. 9.1.4 Abstraction A very important point is now that the principles expressed by the above propositions have a much broader context than that of a truss or a discrete model. The mathematical structure involved can be generalized to an abstract setting, see Tonti [16], and this abstract setting can then be used to obtain small displacement theories for all one-, two- and three-dimensional geometric models. Below we give the abstraction that in subsequent sections is applied. The abstract setting involves the following elements: 1. a linear space4 U of configuration variables. In the special examples of a truss this set was a set of column vectors u of displacements. 2. a linear space F of force-like variables. In the special examples of a truss this set was that of column vectors f e of nodal forces. 3. a bilinear functional defined for elements of U and F that will be denoted

f, uE for f ∈ F and u ∈ U. It should be such that for every u ∈ U different from zero, there exists at least on f ∈ F such that f, uE = 0 and an analogous requirement holds for every f ∈ F. The spaces U and F are then said to be put in separating duality by the bilinear form. For the truss, the operation aT b for two column vectors a and b plays the role of bilinear functional. 4. a linear space E of generalized strains. In the special examples of a truss this set was a set of column vectors e. 5. a linear space S of generalized stresses. In the special examples of a truss this set was a set of column vectors σ. 4

A linear space is a set of objects for which the operations of addition and multiplication by a scalar are defined, and the addition and scalar multiple axioms hold, see Appendix B.1.

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9 Small Displacement Models

6. a bilinear functional defined for elements of E and S that will be denoted

σ, eI for σ ∈ S and e ∈ E. It should define a separating duality. 7. a linear operator D mapping from U to E. 8. the adjoint operator D∗ of D which maps from S to F and which is defined by

D∗ σ, uE = σ, DuI for all u ∈ U and σ ∈ S. The adjoint operator corresponds to the transpose of D for the truss.

Generalizations of (9.15) and (9.17) to the abstract structure given in the above items should be obvious. The work equivalence becomes

f, uE = σ, eI , where indices E and I stand for external and internal. Figure 9.4 gives a view

Fig. 9.4. The general structure of the abstract scheme.

of the abstract structure or scheme. When comparing with Figure 9.3 we can see that what is missing in Figure 9.4 is a connection between the right and the left side of the figure. This connection is achieved by introducing special assumptions of constitutive nature between elements of E and S as well as declaring the forces as known. The constitutive assumption should be a linear mapping if we intend to continue our generalization of the truss equations. 9.1.5 Use of Abstract Work Equations in Modeling Generalized versions of Propositions 1 and 2 can be used in the following ways: (i) Assume that we know how strain is related to the configuration variables, i.e., assume that we know the equation e = Du. Assume also that we can write down what is the external work f, uE and the internal work

σ, eI . Then, by means of the principle of virtual displacements,

f, uE = σ, eI for all (u, e) such that e = Du, we can obtain the equilibrium equation f = D∗ σ.

(9.18)

9.2 B

M d l

109

(ii) Assume that we know the equilibrium equation f = D∗ σ and expressions for external and internal work. Then, by means of the principle of virtual forces,

f, uE = σ, eI for all (f, σ) such that f = D∗ σ,

(9.19)

we can obtain obtain the geometric equation e = Du. The application under (i) is the most well-known use of work equivalences.This is probably since it has an extension to large displacement and dynamic situations, and since geometric ideas are closer to intuition than those relating to forces. In fact, in the extended form, (9.18) may be used as an alternative to Euler’s laws, i.e., as an alternative to the universal equations of mechanics. However, at this moment, in this text, we are in the situation that general equations of motion, which may be specialized to equilibrium equations, were derived in Chapters 4, 5 and 7, while geometric equations for one- and threedimensional small displacement theories are not yet at hand. Therefore, in subsequent sections we will use (ii) to derive such equations.

9.2 Beam Models In this section we will use item (ii) of Subsection 9.1.5 to derive geometric equations for a beam. The corresponding equilibrium equations are essentially known from Chapter 5, and these equations are now supplemented by constitutive equations, that are based on the geometric equations, so that complete problems are obtained. In order not to treat too many ideas at the same time we will consider a somewhat special case. Extensions are indicated in the exercises. The beam is taken to be straight of length L; it is clamped at its end points and a plane (two-dimensional) case is considered. A natural coordinate along the beam is s. It is loaded by forces per unit length qn and qt and there is a couple per unit length, lb . The configuration is shown in Figure 9.5. The equilibrium equations can be inferred from (5.23), (5.24) and (5.25) and become (for simplicity, here and in the sequel, the argument s is omitted) ∂ffn + qn = 0, ∂s

(9.20)

∂fft + qt = 0, (9.21) ∂s ∂mb + lb = 0. fn + (9.22) ∂s Note that we are considering a straight beam so the curvature κ is zero. Furthermore, the couple per unit length lb was not present in the derivation of (5.24) but such a generalization is easily obtained, see Exercise 5.1 and

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9 Small Displacement Models

Fig. 9.5. Forces and couples acting on a clamped, straight and plane beam.

Chapter 12. Equations (9.20), (9.21) and (9.22) represent f = D∗ σ in the abstract setting. To apply (ii) of Subsection 9.1.5 we need expressions for internal and external work. The abstract external forces are represented by forces per unit length qn and qt and a couple per unit length lb . We get the external work by multiplying these by displacements un and ut and a rotation ϕ, defined along the length of the beam, and integrating. That is,  L (qn un + qt ut + lb ϕ) ds. (9.23)

f, uE = 0

The generalized stress σ is represented by ft, fn and mb. The corresponding representations of generalized strains are denoted εt , εn and ω and we may write  L

σ, eI =

0

(ffn εn + ft εt + mb ω) ds.

(9.24)

The geometric meanings of εt , εn and ω are not known at this stage. In fact, that is what we intend to obtain by using the principle of virtual forces (9.19). By inserting the equilibrium equations (9.20), (9.21) and (9.22) into (9.23), integrating by parts and utilizing un (0) = ut (0) = ϕ(0) = un (L) = ut (L) = ϕ(L) = 0, i.e., the clamping conditions, one obtains   L ∂un ∂ut ∂ϕ

f, uE = fn ( − ϕ) + ft + mb ds. ∂s ∂s ∂s 0 The principle of virtual forces now reads   L ∂un ∂ut ∂ϕ fn ( − ϕ − εn ) + ft ( − ε t ) + mb ( − ω) ds = 0, ∂s ∂s ∂s 0

9.2 Beam Models

111

for all ft , fn and mb . By use of a fundamental theorem of variational calculus, which is given as Theorem 4 in Appendix C, one obtains the following geometric equations, representing e = Du in the abstract setting: εn =

∂un − ϕ (shearing), ∂s

(9.25)

∂ut (elongation), (9.26) ∂s ∂ϕ ω= (bending). (9.27) ∂s We have thus obtained in what way the generalized strains εn , εt and ω are functions of the displacement variables un and ut and the rotation ϕ. Since the configuration of the beam is defined by associating not only displacement but also rotation to each material point we have the situation of refined geometric models, discussed in Section 3.5. We say that the rotation is a director and it is illustrated by drawing short lines along the beam, see Figure 9.6. Vertical lines are taken to represent ϕ = 0. The requirement that generalized strains should be zero for rigid body displacements is satisfied by saying that directors deform rigidly when they remain normal to the direction of the beam, i.e., for ϕ = ∂un /∂s. The difference between the slope of the beam and the director, i.e. εn , can be interpreted as a shearing of the cross-section. These items are illustrated in Figure 9.6. The generalized strain εt represents local elongation εt =

Fig. 9.6. The kinematics of shearing of a beam. The line segments indicate the directors, i.e., rotations of material points. When the segment is normal to the beam then the shearing is zero and, in general, the shearing is the difference between the slope ∂un /∂s and the director angle ϕ.

of the beam, and ω is a quantity representing bending. Simple constitutive laws are now, as for the truss case, linear relations between generalized stresses and strains. We write such relations as fn = kεn ,

(9.28)

ft = (EA)εt ,

(9.29)

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9 Small Displacement Models

mb = (EI)ω,

(9.30)

where k = k(s), (EA) = (EA)(s) and (EI) n= (EI)(s) are parameters that are regarded as given and that may vary alo g the length of the beam: k is the shear coefficient; (EA) is the elongation stiffness; and (EI) is the bending stiffness. The rather strange looking notations (EA) and (EI) are used because, as we will see in Chapter 12, beam theory can also be derived as a constrained three-dimensional theory. It will then be clear that (EA) can be split into a geometric part, the area A, and a material part, the elasticity coefficient E, and similarly for (EI).

Fig. 9.7. The structure of the beam equations. Equation numbers refer to those in the text.

The structure of the equations for a straight plane beam, just derived, is illustrated in Figure 9.7. By eliminating fn , ft , mb , εn , εt and ω we obtain the following equations, which provide the direct coupling between un , ut and ϕ, on one hand, and qn , qt and lb , on the other, as shown in Figure 9.7:   ∂ut ∂ (9.31) (EA) + qt = 0, ∂s ∂s    ∂ ∂un − ϕ + qn = 0, k (9.32) ∂s ∂s     ∂un ∂ ∂ϕ k −ϕ + (EI) + lb = 0. (9.33) ∂s ∂s ∂s

9.2 Beam Models

113

These differential equations can be solved when appropriate boundary conditions are specified. We have already considered the clamped boundary conditions un (0) = ut (0) = ϕ(0) = un (L) = ut (L) = ϕ(L) = 0. A generalization of these conditions is when, on the boundary, values of functions are prescribed differently from zero. Boundary conditions of this type, i.e., when the dependent variables of the differential equations are prescribed, are called essential or Dirichlet boundary conditions. A different type of boundary conditions is when forces or couples are prescribed, i.e., given fn∗ , ft∗ and m∗b we require that fn (ξ) = fn∗ , ft (ξ) = ft∗ and mb (ξ) = m∗b for ξ = 0 or ξ = L. When using the constitutive laws (9.28), (9.29) and (9.30) and the geometric conditions (9.25), (9.26) and (9.27), then such boundary conditions become ft∗ = (EA)(ξ) fn∗ = k(ξ)



∂ut (ξ) , ∂s

 ∂un (ξ) − ϕ(ξ) , ∂s

(9.34) (9.35)

∂ϕ(ξ) , (9.36) ∂s for ξ = 0 or ξ = L. Boundary conditions of this type are known as natural or Neumann boundary conditions. Such conditions can be fitted into the structure of Figure 9.7 where they are seen as additional equilibrium conditions, see Exercise 9.3. This fact is actually the reason why they are called natural conditions. Our problem now consists of the differential equations (9.31), (9.32) and (9.33), together with appropriate boundary conditions. However, a quick look at these equations shows that equations involving the tangential displacement ut , directed along the beam, are not coupled to equations involving the displacement in the normal direction of the beam, un , and the rotation of directors ϕ. Thus, the problem decouples into two problems. These problems are formally stated as follows: m∗b = (EI)(ξ)

Bar problem: Given a distribution of force per unit length qt = qt (s), s ∈ (0, L), and at each end of the beam natural or essential boundary conditions, find the displacement field ut = ut (s) such that (9.31) is satisfied for all s ∈ (0, L). Timoshenko beam problem: Given distributions of force per unit length qn = qn (s), couple per unit length lb = lb (s), s ∈ (0, L), and natural or essential boundary conditions at each end of the beam, find the displacement field un = un (s) and the rotation ϕ = ϕ(s) such that (9.32) and (9.33) are satisfied for all s ∈ (0, L). The bar problem should be known from elementary strength-of-materials textbooks. However, in the setting used there it is usually isolated from the

114

9 Small Displacement Models

problems of bending and shearing. We have seen that this is not essential: in a small displacement situation, for a straight beam with simple decoupled constitutive laws, the problem of elongation due to a force directed along the beam, i.e., the bar problem, is unaffected by shearing and bending, and conversely. Note that a one-dimensional straight solid loaded axially is usually called a bar (see Section 9.1.2) and the term beam is saved for such a solid that is loaded normally to its length direction. However, here we have found it convenient to use the term beam for both cases, except for the name of the actual complete problem. If the bar problem is familiar from elementary treatments, this is probably not so for the Timoshenko beam problem. This problem is usually presented as a refinement of what is known as the Euler-Bernoulli beam theory. The refinement is that deformations due to shear are included in the former but not in the latter. Here we take the opposite route. Having derived the Timoshenko beam problem we introduce the idea of rigidity in shearing and obtain the classical Euler-Bernoulli beam theory. A model expressing rigidity in shearing is obtained by formally letting the shear stiffness k approach infinity. Note that for a well-behaved qt , equation (9.21) implies that fn is bounded. Therefore, (9.28) implies that εn → 0 when k → ∞, and by (9.25) we obtain ϕ=

∂un . ∂s

(9.37)

This condition is now used to eliminate ϕ from all relevant equations. Taking the derivative of (9.33) with respect to s and inserting the result into (9.32) we obtain the Euler-Bernoulli differential equation   ∂2 ∂ 2 un ∂lb + qn − = 0. (9.38) − 2 (EI) ∂s ∂s2 ∂s The essential boundary conditions associated with this equation are fixed values of un and ∂un /∂s at the ends of the beam. The natural boundary conditions are, firstly, the condition of fixed couple, obtained from (9.36) by introducing (9.37), ∂ 2 un (ξ) m∗b = (EI)(ξ) (9.39) ∂s2 and, secondly, the condition of fixed force, obtained by using (9.22), (9.30) and (9.37),   ∂ ∂ 2 un (ξ) ∗ fn = − (EI)(ξ) − lb (ξ), (9.40) ∂s ∂s2 where ξ = 0 or ξ = L. We can now formulate the following model:

9.2 Beam Models

115

Euler-Bernoulli beam problem: Given distributions of force per unit length qn = qn (s) and couple per unit length lb = lb (s), s ∈ (0, L), and at each end of the beam natural or essential boundary conditions5 , find the displacement field un = un (s) such that (9.38) is satisfied for all s ∈ (0, L). As indicated above, the Euler-Bernoulli beam is rigid in shear, but can bend, i.e., (EI) is finite. A reversed type of beam behavior is when the bending stiffness (EI) goes to infinity while the shear stiffness k remains finite. From the equilibrium equations we conclude that mb is finite for a well-behaved loading and, thus, ω = ∂ϕ/∂s → 0 as (EI) → ∞, meaning that ϕ is constant. We put this constant to zero, reducing (9.32) to   ∂ ∂un k + qn = 0. (9.41) ∂s ∂s Essential boundary conditions are fixed values for un and natural boundary conditions are ∂un (ξ) fn∗ = k(ξ) , ∂s for ξ = 0 or ξ = L. We now have the following model: Shear beam problem: Given a distribution of force per unit length qn = qn (s), s ∈ (0, L), and at each end of the beam a natural or essential boundary condition, find the displacement field un = un (s) such that (9.41) is satisfied for all s ∈ (0, L). From a mathematical point of view all four models of this section are linear elliptic boundary value problems, for which unique solutions exist given some regularity conditions of the data. Therefore they certainly work as the operator illustrated in Figure 1.1 and can be regarded as complete models. Example: A Cantilever Beam Problem We wish to determine the displacement un of the plane cantilever beam with constant shear coefficient and bending stiffness as shown in Figure 9.8. The beam is considered to be a Timoshenko beam, so this is an example of the complete problem entitled the Timoshenko beam problem. However, in the analytical treatment we apply here, it is convenient not to perform the elimination of generalized stresses and strains leading to the differential equations (9.32) and (9.33). Rather, with the loading shown in Figure 9.8 we solve the equilibrium equations 5

To obtain a well-posed (complete) problem for the fourth order Euler-Bernoulli equation (9.38), natural and essential boundary conditions cannot be mixed arbitrarily. For instance, it is conflicting to simultaneously prescribe both the displacement and the force.

116

9 Small Displacement Models

Fig. 9.8. Cantilever beam.

∂ffn = 0, ∂s

fn +

and the constitutive equations   ∂un fn = k −ϕ , ∂s

∂mb =0 ∂s

mb = (EI)

(9.42)

∂ϕ ∂s

(9.43)

together with the boundary conditions un = 0,

ϕ=0

(9.44)

mb = 0

(9.45)

at s = 0 and fn = −P,

at s = L. The first equilibrium equation of (9.42), together with the first boundary condition of (9.45), lead to fn = −P for all s ∈ (0, L), i.e., the shear force is constant along the beam. Inserting this shear force into the second equilibrium equation and using the second boundary condition of (9.45), we find mb = P (s − L). These known internal force and couple distributions can now be included in the constitutive equations (9.43) to obtain the displacement un and the rotation ϕ. Starting with the second constitutive equation we find ∂ϕ P (s − L) = (EI) , ∂s which when integrated, and taking into account the second boundary condition of (9.44), leads to   P 1 2 ϕ= s − Ls . (9.46) (EI) 2 The first constitutive equation gives   ∂un −P = k −ϕ , ∂s

9.3 Linear Elasticity

117

into which we insert (9.46), integrate and use the first boundary condition of (9.44). We then obtain the final result as   P 1 3 1 2 P un = s − Ls − s. (9.47) (EI) 6 2 k The first term of (9.47) is due to bending flexibility and the second term is due to shear flexibility. In fact, usage of Euler-Bernoulli’s beam theory leads to a displacement given by the first term in (9.47), while the shear beam theory gives the second term.

9.3 Linear Elasticity The equations of linear elasticity will be formulated as a boundary value problem posed on a domain B ⊂ E. In the small displacement setting we do not distinguish between Bt = φt (B) and B0 = φ0 (B), where the latter can without loss of generality be thought of as coinciding with B. The equilibrium equation follows from (7.16) by putting a = 0. Thus, there exists a symmetric tensor field T over B such that div T + b = 0

in B,

(9.48)

where b is the external body force per unit volume. Omitting arguments and writing “in B” means that the independent variable x (or X) belongs to B. In addition to (9.48), Cauchy’s equation on the boundary of B, s = Tn

on ∂B,

(9.49)

should also be regarded as an equilibrium equation. In the following we will use item (ii) in Subsection 9.1.5 to derive a geometric equation showing how to define a strain that is work conjugate to T , as a linear function of the displacement vector u defined on B. To that end, we need expressions for internal and external work. The external work is defined as  

f, uE = b · u dV + s · u dA. B

∂B

To express the internal virtual work we introduce a symmetric strain tensor E, which has a matrix representation as follows: ⎤ ⎡ ε11 ε12 ε13 E ∼ ⎣ ε21 ε22 ε23 ⎦. ε31 ε32 ε33 The internal work is expressed as 

σ, eI =

B

T : E dV,

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9 Small Displacement Models

where the double dot indicates inner product of tensors. Expressed in the components of the tensors, this inner product can be written as T :E=

3 3

σij εij .

i=1 j=1

Note that, since T is symmetric, it makes no sense to let E be nonsymmetric. This is because any tensor A can be uniquely decomposed as A = Asy + Aan , where Asy is symmetric and Aan is anti-symmetric, and, as seen in Exercise C.1 in Appendix C, S : A = S : Asy for a symmetric tensor S. Substituting the equilibrium equations (9.48) and (9.49) into f, uE and using the divergence theorem, equation (C.24) of Appendix C, one finds 

f, uE = T : ∇u dV. (9.50) B

The next step is to compare this expression with σ, e e I, i.e., to form the work equivalence, and then use (9.19) to obtain the geometric equation. However, we then need to use a fundamental theorem of variational calculus in the form of Theorem 5 in Appendix C, and since this theorem refers to symmetric tensors, we need to rewrite (9.50) somewhat. Indeed, we find that, see Exercise C.1, T : ∇u = T :

 1 ∇u + (∇u)T . 2

Using this result in (9.50) and equalizing internal and external work, we find that the principle of virtual forces gives     1 T : E− ∇u + (∇u)T dV = 0, 2 B for all symmetric tensors T . Theorem 5 of Appendix C then gives E=

 1 ∇u + (∇u)T , 2

(9.51)

which is the sought geometric equation. What remains to be specified in the theory of linear elasticity is a constitutive law connecting T and E. For simplicity, we will consider only the case of an isotropic material, i.e., a model that has the same response in all directions. It can be shown, see e.g. the Appendix of Gurtin [5], that the most general linear and isotropic constitutive law is the following:

9.3 Linear Elasticity

119

Fig. 9.9. The structure of the equations of linear elasticity. Equation numbers refer to those in the text.

T = λ(tr E)I + 2µE,

(9.52)

where λ and µ are Lame’s ´ elasticity coefficients. The structure of the equations of linear elasticity are now shown in Figure 9.9. The boundary conditions related to linear elasticity are natural boundary conditions, where the traction vector s is a prescribed given value s∗ , i.e., s∗ = T n, and essential boundary conditions where the displacement is a prescribed given value u∗ , i.e., u∗ = u. In summary, we have the following complete model: Linear elasticity: Given a volume force on B and essential or natural boundary conditions on ∂B, find fields of displacements, stresses and strains such that (9.48), (9.51) and (9.52) are satisfied on B. An alternative statement of the complete problem of linear elasticity is obtained by eliminating stresses and strains and, thus, using only displacements as dependent variables. The resulting equation is known as Navier’s equation, see Exercise 9.9. From a mathematical point of view, the problem of linear elasticity is an elliptic boundary value problem. After having introduced, in the next subsection, some physically reasonable bounds for the elasticity coefficients, we will show that it has a unique solution.

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9 Small Displacement Models

9.3.1 Elasticity Coefficients The constitutive law (9.52) is frequently expressed by using coefficients different from the Lam´´e coefficients. In this subsection we will introduce such coefficients by means of examples and relate them to the Lam´ ´e coefficients. We will also discuss the bounds on elasticity coefficients that are induced by a requirement of a monotone stress-strain behavior. As a first step we invert the constitutive equation (9.52). Taking its trace we find tr T = λ(tr E)tr I + 2µ tr E = (3λ + 2µ)tr E, where in the last step we used tr I = 3. Thus, we immediately find   1 λ E= (tr T )I . T− 2µ 3λ + 2µ

(9.53)

We are now ready to look at three different examples representing pure tension, simple shearing, and uniform compression or expansion: Example 1: Pure Tension Consider a solid (e.g., a bar) in pure tension. Referred to a orthonormal base {e1 , e2 , e3 }, the loading is in the direction of e1 . A stress tensor that satisfies the equilibrium conditions is then ⎡ ⎤ σ00 T = σe1 ⊗ e1 ∼ ⎣ 0 0 0 ⎦, 000 where σ is the value of the applied stress. Inserting such a stress into (9.53) we find ⎤ ⎡ ⎤ ⎡ λ+µ 0 0 ε11 0 0 µ(3λ+2µ) σ ⎥ ⎢ λ 0 − 2µ(3λ+2µ) σ 0 E ∼ ⎣ 0 ε22 0 ⎦ = ⎣ ⎦. λ 0 0 ε33 0 0 − σ 2µ(3λ+2µ)

The usual definitions of Young’s modulus E and Poisson’s ratio ν are that these material constants satisfy σ = Eε11 ,

ε22 = ε33 = νε11 ,

for a situation of pure tension. Obviously this happens for E=

µ(3λ + 2µ) , λ+µ

and inverting these relations gives

ν=

λ , 2(λ + µ)

9.3 Linear Elasticity

λ=

Eν , (1 + ν)(1 − 2ν)

121

E . 2(1 + ν)

µ=

Concerning displacement components in pure tension, the geometric equation (9.51) can be integrated to give u1 = ε11 X1 ,

u2 = ε22 X2 ,

u3 = ε33 X3 ,

up to integration constants that represent rigid body motion. Example 2: Simple Shearing Consider a simple shear deformation defined by u1 = γX2 ,

u2 = 0,

u3 = 0,

for which the only non-zero strain component is ε12 =

1 γ. 2

Inserting this into (9.52) we find that the only non-zero stress component is σ12 = µγ. This result explains why the Lam´ ´e coefficient µ is frequently called the shear modulus. Example 3: Uniform Compression or Expansion For a body of arbitrary shape with a pressure boundary condition, i.e., a boundary condition of the form s∗ = −pn, where n is the outward unit normal vector and p is a constant, one concludes that the spherical stress tensor T = −pI satisfies the equilibrium equations. We can think of the body as submerged in a fluid with the hydrostatic pressure p. By substituting this stress tensor into the constitutive equation (9.53) we find E = εI = − where ε=−

p I, 3λ + 2µ

p 3λ + 2µ

is a uniform strain in any direction of the body.

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9 Small Displacement Models

For a body submitted to this type of uniform pressure boundary condition one may ask what is its resulting volume change. Relative volume change is described by det F and from x = φt (X) = X + u one finds det F = (1 + ε)3 = 1 + 3ε + o(ε) = 1 + tr E + o(ε). Thus, to the first order, tr E = 3ε, represents the change in volume and one concludes that p tr E = − , κ where κ = 23 µ + λ is called the bulk or compression modulus. Up to integration constants, that represent rigid body displacement, the geometric equation (9.51) now gives p u 1 = − X1 , κ

p u 2 = − X2 , κ

p u 3 = − X3 . κ

Bounds on Elasticity Coefficients We generally expect the physical behavior of an elastic material to be such that when the strain increases, the stress should also increase, i.e., the relation between stress and strain is monotonic. This will induce constraints on what values can be taken by the elasticity coefficients λ, µ, κ, E and ν. The monotonic behavior is mathematically expressed by saying that T :E=

3 3

σij εij ≥ 0,

(9.54)

i=1 j=1

for all T and E that are related by the constitutive law (9.52). To evaluate this inequality we introduce the deviatoric part of the strain tensor: 1 dev E = E − (tr E)I, 3 where I is the identity tensor. The operators dev and tr represent orthogonal projections in the sense that, first, for any tensor S, 1 tr (dev S) = tr S − (tr S)tr I = 0, 3 dev ((tr S)I) = 0, where we have used tr I = 3, and, secondly, any tensor S can be represented as 1 S = (tr S)I + dev S. 3

9.3 Linear Elasticity

123

Fig. 9.10. Schematic illustration of the decomposition of a second order tensor S into orthogonal parts.

These properties are illustrated in Figure 9.10. We learned earlier that tr E represents a volume change, i.e., dilatation, so dev E represents the remaining part of the deformation, namely the shape change. By using this decomposition, we can write the constitutive law (9.52) as T = κ(tr E)I + 2µ dev E, (9.55) where κ = 23 µ + λ is the compression modulus. Thus, the constitutive law can be decomposed into two independent terms, one dealing with deviation and the other with dilatation. From this expression for the stress we find T : E = κ(tr E)2 + 2µ dev E : dev E, from which it is concluded that the monotonicity requirement (9.54) implies that κ ≥ 0, µ ≥ 0. (9.56) Expressions for κ and µ in terms of E and ν are κ=

E , 3(1 − 2ν)

µ=

E . 2(1 + ν)

Thus, (9.56) holds if and only if 1 . (9.57) 2 The existence of negative values for Poisson’s ratio may come as somewhat of a surprise, and one even sees comments in the literature claiming that no such materials exist. However, these comments are based on a limited view on the relation between the conceptual and the real world. Certainly, a material with microstructure can be a referent of linear elasticity and it is not difficult to construct a microstructure that displays the behavior of negative Poisson’s ratio as it is seen in, e.g., a research monograph on topology optimization [2]. E ≥ 0,

−1 ≤ ν ≤

124

9 Small Displacement Models

Uniqueness Theorem In the following we will show that the linear elasticity model can indeed be regarded as a complete model: for a monotonic constitutive law satisfying (9.54), and boundary conditions that prevent rigid body motion, equations (9.48), (9.51) and (9.52) uniquely define fields of displacements, stresses and strains. The proof is classical and is built on assuming the contrary, i.e., one assumes that (u1 , E 1 , T 1 ) and (u2 , E 2 , T 2 ) are two different solutions for the same data. Then the differences u = u1 − u2 ,

E = E1 − E2,

T = T 1 − T 2,

must be a solution of a model with zero data, i.e., a model defined by b = s∗ = u∗ = 0. The work equation for such a model is simply  T : E dV = 0.

σ, eI = B

When (9.54) holds with κ > 0 and µ > 0, this means that T = E = 0 everywhere on B. A zero small strain field on a domain, i.e., E = 0, implies that the corresponding displacement field must have the same mathematical form as that of a rigid velocity field, see Section 12.3. For such a displacement field, if at least three points of ∂B, that are not on a line, are prevented from moving by the condition u∗ = 0, then the whole displacement field must be zero, i.e., u = 0. The final conclusion is that the assumption of two different solutions was wrong and a contradiction that implies uniqueness is reached.

Exercises 9.1. (Characterization of trusses) A truss is said to be unstable if it can move without deforming its bars, i.e., Du = 0 has a nontrivial solution u = 0. If Du = 0 has only the trivial solution u = 0, then the truss is said to be stable. (a) Verify that a truss is stable if and only if D has linearly independent columns. For stable trusses we distinguish between statically determinate trusses, for which it is possible to solve (9.11) uniquely for the bar forces σ, and statically indeterminate trusses, for which we need to include constitutive equations and form the stiffness matrix K to obtain such bar forces. (b) Conclude that for a statically determinate truss, D is a square matrix and for a statically indeterminate truss, D has more rows than columns. For unstable trusses we distinguish between rigid motion trusses and mechanisms. A truss is a rigid body truss if all solutions of Du = 0 represent rigid body displacements of the whole truss. For a mechanism the truss can deform without stretching or compressing its bar members.

9.3 Linear Elasticity

125

(c) Give simple examples of the four types of trusses defined above. 9.2. (Unstable truss) For an unstable truss the equilibrium equation D T σ = f e has a solution only for certain forces f e . (a) Verify that the solvability condition for the equilibrium equation is f eT u = 0 for all u such that Du = 0. (b) Consider a rigid motion truss. Show that the condition stated in part (a) corresponds to force and torque equilibrium for the whole truss. 9.3. Redo the derivation of the geometric equations (9.25), (9.26) and (9.27) in the case when one end of the beam, s = L, is free and subject to external forces and couples fn∗ , ft∗ and m∗b . The definition of external virtual work is then  L

f, uE = fn∗ un (L) + ft∗ ut (L) + m∗b ϕ(L) + (qn un + qt ut + lb ϕ) ds. 0

and the equilibrium equations (9.20), (9.21) and (9.22) have to be supplemented by end equilibrium conditions fn (L) = fn∗ , ft (L) = ft∗ and mb (L) = m∗b . 9.4. Redo the derivation of the geometric equations (9.25), (9.26) and (9.27) for a beam with curvature κ. Conclude that for non-zero curvature the problem does not decouple into a bar and a beam problem. 9.5. Extend the one-dimensional model (the beam model) to a two-dimensional one, i.e., derive equations for a plate or a shell. (This is an extensive exercise . 9.6. Show that when (EI) and k are independent of s, one can eliminate ϕ from equations (9.32) and (9.33) to obtain ∂ 4 un (EI) ∂ 2 qn ∂lb + + (9.58) − qn = 0. 4 2 ∂s k ∂s ∂s 9.7. Consider the clamped beam with uniform normal force distribution as in Figure 9.11. Use the three beam theories, i.e, Timoshenko’s, Euler-Bernoulli’s and the shear beam theories, to calculate the transverse displacement un . Compare the obtained results. (EI)

Fig. 9.11. Clamped beam subjected to uniform force.

126

9 Small Displacement Models

9.8. Make a detailed derivation of (9.50). 9.9. Derive Navier’s equation for the case without body forces: µ∇2 u + (λ + µ)(∇(div u)) = 0, where the Laplace operator is defined by ∇2 u = div (∇u). In cartesian components this equation reads µ

3 ∂ 2 ui j=1

3 ∂ 2 uj + (λ + µ) = 0, 2 ∂xj ∂xj ∂xi j=1

i = 1, 2, 3.

10 Pipe Flow Models

In this chapter we will study problems based on equations which were derived in Chapter 6. When representing cut forces by a pressure p = p(s), Euler’s equations were shown to lead to equation (6.20), which is the basic equation of motion in pipe flow. Rewriting the material time derivative on the right hand side of (6.20) by means of (6.5), we obtain −

q˜t (s) ∂v(s, t) ∂v(s, t) 1 ∂p(s) + = + v(s, t) . ρ(s, t) ∂s ρ(s, t)α(s, t) ∂t ∂s

(10.1)

This is the first basic equation used in this chapter. In Section 10.1 it will be combined with assumptions of incompressibility and constant cross-section area. In Section 10.2 compressible pipe flow and constant cross-section area is considered. Finally, in Section 10.3 we consider again an incompressible fluid, but this time contained in a flexible pipe.

10.1 Incompressible Pipe Flow In this section we will study the equation of motion (10.1) for an incompressible fluid. The reader should compare this section with Section 6.7, where a problem for incompressible pipe flow with conservative forces was studied. That treatment contained the possibility of a varying cross section area. Here we consider only a constant cross section area, but, on the other hand, treat both a conservative and a non-conservative case. We first give a reminder of the concept of incompressibility in pipe flow. A fluid is incompressible if the following three items hold: •

The motion is isochoric. It was shown in Section 6.5 that this implies ∂ ∂ α(s, t) + (α(s, t)v(s, t)) = 0. ∂t ∂s

127

(10.2)

128

•

10 Pipe Flow Models

The fluid is homogeneous, i.e., ρ(s, t) = ρ(t)

•

= =⇒

∂ρ(s, t) = 0. ∂s

(10.3)

The pressure p(s) can take any value.

As was shown in Section 6.5, the continuity equation, i.e., conservation of mass, implies that for isochoric motion, ρ(s, ˙ t) =

∂ ∂ ρ(s, t) + v(s, t) ρ(s, t) = 0. ∂t ∂s

(10.4)

Thus, from this and (10.3) we find that for an incompressible fluid, the density can be taken as a constant in time and space, which we denote by ρin . In this section and the next we will consider the special case where the cross section area is constant in time and space, i.e., we look at a pipe with rigid walls without tapering. We write α(s, t) = α0 = constant. Equation (10.2) then implies that the velocity is constant for all positions along the pipe, i.e., v(s, t) = v(t). Thus, all points along the pipe have the same velocity and, at least for a straight pipe, we may view the motion as the motion of a rigid body. Furthermore, a constant cross section area implies qt (s) = ˜t (s) (see Section 6.6 for these notations) and we can rewrite equation (10.1) as −

∂p(s) qt (s) ∂v(t) + = ρin . ∂s α0 ∂t

(10.5)

We may now formulate the following complete model: Incompressible pipe flow: Given a force function qˆt such that qt (s) = ˆt (v(t), t), an interval (0, L) representing the pipe, a time interval (0, T ), initial conditions and boundary conditions, find functions v(s, t) and p(s) for s ∈ (0, L) and t ∈ (0, T ) such that (10.5) is satisfied. Now we proceed to give explicit solutions of this problem for two force functions. Make an integration of (10.5) from s = 0 to s = L: p(0) − p(L) +

1 α0

 0

L

qt (s) ds = ρin L

∂v(t) . ∂t

(10.6)

10.2 Compressible Pipe Flow

129

This equation parallels (6.27) in Section 6.7, which was derived under the condition of conservative forces, but for a cross section that may depend on s. As a special case of (10.6) we consider qt (s) = 0 and assume that p(0) and p(L) are time independent. For an initial velocity v0 one then finds from (10.6) that p(0) − p(L) v(t) = t + v0 . (10.7) ρin L Clearly, this equation represents a solution of a special case of the problem in Section 6.7. This case of qt (s) = 0, or, more generally, q t (s ( ) being conservative, represents a pipe flow analogy of what in the three-dimensional case would be called an inviscid fluid. Such a fluid has no shear resistance to flow. The simplest type of fluid having such a resistance is a viscous Newtonian fluid. The pipe flow version of a Newtonian fluid would correspond to the assumption qt (s) = −Cv(t), (10.8) α0 for some constant C > 0, i.e., a linear relation between force and velocity. Substituting (10.8) into (10.6) gives p(0) − p(L) = ρin L

∂v(t) + CLv(t). ∂t

(10.9)

For a given pressure difference this is an ordinary differential equation. When the pressure difference is constant its general solution can be written as   C p(0) − p(L) v(t) = A exp − t + , (10.10) ρin CL where A is a constant, which is determined by the initial condition v(0) = v0 . One finds, with this condition, that      p(0) − p(L) C C v(t) = (10.11) 1 − exp − t + v0 exp − t . CL ρin ρin A familiar real life situation where this equation applies is the sudden opening of a water tap. We can then assume that a fixed pressure difference is established, and the velocity of the water changes from the initial value v0 to the new steady state value (p(0) − p(L))/CL.

10.2 Compressible Pipe Flow In this section we will continue to look at the case of a constant cross section area, but omit the assumption of incompressibility. We will also assume that qt (s) = 0. Equation (10.1) becomes in this case:

130

10 Pipe Flow Models

−

∂p(s) = ρ(s, t) ∂s



 ∂v(s, t) ∂v(s, t) + v(s, t) . ∂t ∂s

(10.12)

The continuity equation (6.11) with (6.15), for constant cross section area, takes the form ∂ ∂ ρ(s, t) + [ρ(s, t)v(s, t)] = 0. (10.13) ∂t ∂s As indicated several times before in the text, these two equations - based on universal laws - are not sufficient for obtaining a complete mathematical problem. In the previous section the assumption of incompressibility was introduced to obtain a complete problem. Such an assumption may be reasonable for modelling of some fluid substances and physical phenomena, but it is not sufficient for other phenomena such as wave propagation, responsible for, say, noise in water pipes. A constitutive assumption that may be more appropriate in such cases is that the pressure is some non-linear function of the density at each point along the pipe, i.e., p(s) = pˆ(ρ(s, t)),

(10.14)

where pˆ is the non-linear function. Fluids with such constitutive behavior and no viscous effects are known as elastic fluids. Our problem consists now of equations (10.12), (10.13) and (10.14). With initial and boundary conditions, these equations can be made into a complete mathematical problem. However, before we explicitly formulate such a problem, a preliminary step, where the pressure is eliminated, will be performed. This step is facilitated if we introduce the function c(ρ), defined by ∂ pˆ(ρ) = c2 (ρ). ∂ρ For reasons that will soon become clear, c(ρ) will be called the speed of sound. Clearly, c(ρ) is real valued only if the derivative of pˆ(ρ) is non-negative, which is a physically reasonable assumption since the pressure is likely to rise due to compression. Substituting (10.14) into (10.12) and using the definition of the speed of sound one finds c2 (ρ) ∂ρ ∂v ∂ρ + +v = 0, ρ ∂s ∂t ∂t

(10.15)

where we have omitted arguments. Together with the continuity equation (10.13), (10.15) constitutes a system which we write       ∂ ρ v ρ ∂ ρ =− 2 . (10.16) c (ρ)/ρ v ∂s v ∂t v Compressible pipe flow: Given an interval (0, L) representing the pipe, a time interval (0, T ), initial conditions and boundary conditions, find functions v(s, t) and ρ(s, t) for s ∈ (0, L) and t ∈ (0, T ) such that (10.16) is satisfied.

10.2 Compressible Pipe Flow

131

Introducing the notation x = (ρ, v), we may write equation (10.16) as ∂x ∂x = −A(x) , ∂t ∂s 

where A(x) =

(10.17)

 v ρ . c2 (ρ)/ρ v

This is an example of a quasi-linear partial differential equation: it is linear in derivatives of x but depends non-linearly with respect to x itself. If A(x) has real eigenvalues, a system of this type is called hyperbolic. For the system (10.16), the eigenvalues are v ± c(ρ), which are real if c(ρ) 2 is positive, i.e., if the relation between p and ρ has a non-negative derivative. The hyperbolic quasi-linear partial differential equation (10.16) describes interesting and important physical phenomena such as sonic booms and shocks. However, in the following we will be content with investigating a linear version of (10.16): we will consider small perturbations of the solution around the constant density ρ∗ and the velocity v = 0. This means that we wish to solve the following system of linear partial differential equations, obtained by taking A(x) to be the constant matrix A(x∗ ) for x∗ = (ρ∗ , 0): ∂v ∂ρ + ρ∗ = 0, ∂t ∂s

(10.18)

∂v c2 (ρ∗ ) ∂ρ = 0. + ∂t ρ∗ ∂s

(10.19)

The function ρ should here be regarded as a perturbation of ρ ∗, i.e., the density is ρ + ρ∗ . The system (10.18) and (10.19) can be rewritten as a second order wave equation. Take the derivative of (10.18) with respect to t and the derivative of (10.19) with respect to s: ∂2ρ ∂2v + ρ∗ = 0, 2 ∂t ∂s∂t

∂2v c2 (ρ∗ ) ∂ 2 ρ + = 0. ∂t∂s ρ∗ ∂s2

Substituting the second equation into the first one gives 2 ∂2ρ 2 ∗ ∂ ρ − c (ρ ) = 0. ∂t2 ∂s2

(10.20)

This is the classical wave equation for ρ. By taking the derivative of (10.18) with respect to s and the derivative of (10.19) with respect to t, and substituting one result into the other, the same wave equation is shown to be valid also for v. From the general theory for wave equations one knows that the solution of (10.20) can be written as ρ(s, t) = f (s − ct) + g(s + ct),

(10.21)

132

10 Pipe Flow Models

where f and g are general functions that are determined from initial and boundary conditions and c = c(ρ∗ ). In the following we prove (10.21) starting directly from the system (10.18) and (10.19). This has the advantage of naturally connecting solutions for ρ and v. Introduce the change of variables ξ = s + ct,

η = s − ct.

Regarding v and ρ as functions of ξ and η, and using the chain rule, we find that (10.18) and (10.19) take the form   ∂ρ ∂ρ ∂v ∂v c −c + ρ∗ + = 0, ∂ξ ∂η ∂ξ ∂η   ∂v ∂v c2 ∂ρ ∂ρ −c + ∗ + c = 0. ∂ξ ∂η ρ ∂ξ ∂η Dividing the first equation by c, multiplying the second by ρ ∗/c 2 and adding and subtracting equations, one obtains ∂ρ ρ∗ ∂v + = 0, ∂ξ c ∂ξ ∂ρ ρ∗ ∂v − = 0. ∂η c ∂η ∗

∗

From these results it is concluded that the combinations ρ+ ρc v and ρ − ρc v are functions only of η and ξ, respectively. Denoting such functions 2f and 2g we may write ρ∗ ρ(s, t) + v(s, t) = 2f (s − ct), (10.22) c ρ∗ (10.23) ρ(s, t) − v(s, t) = 2g(s + ct). c Solving (10.22) and (10.23) for ρ and v one finds ρ(s, t) = f (s − ct) + g(s + ct), v(s, t) =

c (f (s − ct) − g(s + ct)). ρ∗

This is a solution of the system of linear partial differential equations (10.18) and (10.19). The functions f and g are determined from initial and boundary values. For the understanding of the behavior of this solution, we study the situation when g = 0 so that ρ(s, t) = f (s − ct). It is then clear that ρ(s, 0) = ρ(s + s1 , s1 /c).

10.2 Compressible Pipe Flow

133

Thus, at time t = 1s /c, ρ has the same value at s + 1 s as it had at s at time t = 0. Clearly, this means a propagation of an initial shape of the function ρ forward in the space variable s. The situation is illustrated in Figure 10.1, from which it is also clear why we called c the speed of sound: sound is a propagation of a perturbation of the density of a medium. Note that in this linear case the shape of the solution function is not changed, it is only translated in space. This is not exactly true in the initial non-linear problem defined by (10.16). However, for a small non-linearity we can still expect that solutions propagate similarly to the linear case, but a gradual perturbation of the shape occurs. Finally, it should be clear that the function g represents propagation to the left, and the full solution is the conjunction of these two propagations.

Fig. 10.1. The propagation of a solution of (10.18) and (10.19) when g = 0. To the left is a solution at time t = 0 and to the right is the solution at t = t1 .

Equations (10.22) and (10.23) give a means of coupling the state at one end of a pipe to that at the other end. Let L be the length of the pipe and let T be a fixed time value chosen such that cT = L. One then finds from (10.22) that ρ(0, t) +

ρ∗ v(0, t) = 2f (ct) c = 2f (L − c(T + t)) = ρ(L, t + T ) +

ρ∗ v(L, t + T ). c

A reversed version of this equation is established by using (10.23). In summary we have the following two equations ρ(0, t) +

ρ∗ ρ∗ v(0, t) = ρ(L, t + T ) + v(L, t + T ). c c

ρ∗ ρ∗ v(L, t) = ρ(0, t + T ) − v(0, t + T ). c c In hydraulics one makes frequent use of equations of this type where pressure and volume flux at the ends of the pipe are related. Such equations are known as transmission line equations. ρ(L, t) −

134

10 Pipe Flow Models

10.3 Incompressible Pipe Flow in a Flexible Pipe In this section the cross-section area will be considered as an unknown function alongside the pressure p and the velocity v. The fluid is again considered as incompressible and we will treat the case ˜t (s) = 0. The conclusion that the density is constant for an incompressible fluid holds also for a non-constant cross-section area and (10.1) may therefore be written as ∂v(s, t) 1 ∂p(s) ∂v(s, t) + v(s, t) + = 0. ∂t ∂s ρin ∂s

(10.24)

To this equation of motion we may add the equation (10.2) characterizing isochoric flow. These two equations do not constitute a complete problem since we have three unknown functions in the two equations. We may close the system by assuming an elastic constitutive law of the form α(s, t) = α ˆ (p(s)),

(10.25)

for some nonlinear function α. ˆ Using (10.25) to eliminate α from (10.2) we get ∂p(s) ∂p(s) ∂v(s, t) + v(s, t) + β(p(s)) = 0, (10.26) ∂t ∂s ∂s where ∂α ˆ (p(s)) β(p(s)) = α ˆ (p(s))/ . ∂s Equations (10.24) and (10.26) may be collected into the following system of equations, which should be compared to (10.16):       v β(p) ∂ p ∂ p =− 1 v . (10.27) ∂t v ∂s v ρin The following complete model can now be formulated: Incompressible pipe flow in a flexible pipe: Given an interval (0, L) representing the pipe, a time interval (0, T ), initial conditions and boundary conditions, find functions v(s, t) and p(s) for s ∈ (0, L) and t ∈ (0, T ) such that (10.27) is satisfied. Equation (10.27) is again a  system of the type (10.17). The eigenvalues of the system matrix are v ± β(p)/ρin , which are real, giving a hyperbolic ˆ and its derivative are positive. system, if both α(p) As a particular example of the elastic relation (10.25) we may use what could be obtained from the theory of pressure vessels, see Exercise 10.3. Assuming circular cross-section with radius r one can establish that r = r0 +

r02 p, Eh

(10.28)

10.3 Incompressible Pipe Flow in a Flexible Pipe

135

where r0 is a referential radius, E is Young’s modulus and h is the wall thickness. Substituting (10.28) into β(p) gives β(p) =

Eh p + . 2r0 2

(10.29)

Linearizing (10.27), using (10.29), at v = p = 0 gives ∂p(s) Eh ∂v(s, t) + = 0, ∂t 2r0 ∂s

(10.30)

∂v(s, t) 1 ∂p(s) + = 0. ∂t ρin ∂s

(10.31)

A comparison of these equations with (10.18) and (10.19) in the previous section shows that (10.30) and (10.31) have a general solution in terms of two functions f and g such that p(s) = f (s − ct) + g(s + ct), v(s, t) =

1 ρin c

(f (s − ct) − g(s + ct)), 

where c=

Eh 2ρin r0

(10.32)

is the wave propagation velocity or speed of sound. Formula (10.32) is known as the Moens-Korteweg equation. It is used in biomechanics for estimating pulse motion in the arterial system, i.e., the propagation of a heart beat. However, it should be said that it may be a crude model, since soft tissues are not well modelled by the small displacement linear elasticity theory that is the base for equations (10.28) and (10.29).

Exercises 10.1. Solve the compressible pipe flow problem when q t(s) is proportional to v 2 (s, t) instead of v(s, t). 10.2. Assume that (10.14) is invertible and establish transmission line equations in terms of pressure and volume f lux. 10.3. Establish equations (10.28) and (10.29). 10.4. A very useful dimensionless form of the system (10.27) may be derived. Introduce the same dimensionless variables and fields as in Section 11.2.2 and show that if we take υ = c as given in (10.32), then (10.27) becomes

136

10 Pipe Flow Models

∂ ∂ tˆ where

      ˆ p) ∂ pˆ pˆ vˆ β(ˆ =− , vˆ 1 vˆ ∂ˆ s vˆ ˆ p) = 1 + pˆ . β(ˆ 2

10.5. In Exercise 6.5 we established the equation of motion for shallow water flow, (6.30), where h(s, t) represented the water level above some reference plan. Let h0 (s) be the distance between this reference plan and the bottom of the sea. Then the isochoric equation (10.2) can be written as ∂h(s, t) ∂ + [h(s, t) + h0 (s)v(s, t)] = 0. ∂t ∂s This equation and (6.30) constitute a hyperbolic system. Linearize this system around v(s, t) = h(s, t) = 0 and show that √ the wave propagation velocity of the linearized system for constant h0 is gh0. This formula explains why tsunamis propagate with such large speeds!

11 Models of Fluid Mechanics

In this chapter we will derive complete three-dimensional models representing fluid flow. All such models are, in principle, based on an equation of motion which has been derived as a consequence of Euler’s laws, and an equation of continuity, which has been derived as a consequence of conservation of mass. Both of these derivations can be found in Chapter 7. A slight rewriting of the equation of motion (7.16), relevant for this chapter, follows from a(x, t) = ˙ v(x, t) and (7.2), and reads div T (x) + b(x) = ρ(x, t) [v  (x, t) + (∇v(x, t))v(x, t)].

(11.1)

The equation of continuity was derived as equations (7.7) or (7.8) and both versions are repeated here: ρ(x, ˙ t) + ρ(x, t) div v(x, t) = 0,

ρ (x, t) + div(ρ(x, t)v(x, t)) = 0.

(11.2)

Equations (11.1) and (11.2) represent four scalar equations. If we regard b(x) as known and T (x), v(x, t) and ρ(x, t) as fields to be determined, i.e., unknowns, we seem to lack six equations (assuming T (x) to be symmetric). Below we will study three types of constitutive equations that close the mathematical system and lead to complete problems that model important physical processes. Equations (11.1) and (11.2), as well as the constitutive equations, make no reference to a domain B consisting of material points X, i.e., the formulation in this chapter is strictly spatial in terms of x ∈ E as an independent variable. Therefore, we may think that the fluid domain is time independent; it is denoted by R, and is in essence a control domain. Several complete models of fluid flow is derived in this chapter. In what sense these models are based on well-posed problems is not proved or even stated in details since it is well beyond the scope of this book. For statements and an overview of such results we refer to Pironneau [15].

137

138

11 Models of Fluid Mechanics

11.1 Inviscid Fluid Constitutive equations of elastic and ideal fluids were introduced in Section 8.6. A characteristic of both these types of is that the stress tensor can be expressed as T (x) = −p(x)I, (11.3) where I is the identity tensor and p(x) is the hydrostatic pressure. Thus, there are no shear stresses, and such fluids are called inviscid fluids. 11.1.1 Elastic Fluid The constitutive equation of an elastic fluid is defined by (11.3) and a constitutive function pˆ such that p(x) = pˆ(ρ(x, t)).

(11.4)

The function pˆ may in most applications be assumed to have a positive derivative. By inserting the constitutive equations (11.3) and (11.4) into the equation of motion (11.1) we obtain −∇ˆ p(ρ(x, t)) + b(x) = ρ(x, t) [v  (x, t) + (∇v(x, t))v(x, t)] .

(11.5)

This equation is usually known as the Euler equation of an inviscid fluid and together with (11.2) constitutes four scalar equations for the four scalar unknowns represented by v(x, t) and ρ(x, t). This system of partial differential equations is posed on a domain R ⊂ E, and need to be augmented by conditions on the boundary of R. To that end it is usually assumed that, on a part of the boundary, we know the contact pressure and/or the velocity. Knowing the pressure means, through (11.4), that the density is also known. For an inviscid fluid it only makes sense to prescribe the velocity in the normal direction of the boundary of R: since no shear forces exist in such a fluid a tangential motion of the boundary cannot influence the medium in the interior of the domain. In summary, we have the following complete model: Elastic inviscid fluid flow: Given a volume force on R and boundary conditions on ∂R, find fields of velocity and density such that (11.2) and (11.5) are satisfied on R. Note that the problem contained in this model is, in general, not well-posed: if discontinuities (shocks) develop a so-called entropy condition needs to be added to have a unique solution.

11.1 I

i id Fl id

139

11.1.2 Incompressible Fluid As for pipe flow, discussed in Chapter 10, an incompressible fluid is a fluid for which all motion is isochoric, the fluid is homogenous and the pressure can take any value. Isochoric motion means div v(x, t) = 0,

(11.6)

which when combined with an assumption of homogeneity and the equation of continuity (11.2), implies that the density is constant in time and space: a constant which we denote by ρin . Equation (11.5) is now replaced by −∇p(x) + b(x) = ρin [v  (x, t) + (∇v(x, t))v(x, t)] .

(11.7)

Equations (11.6) and (11.7) constitute a system of four equations for the four scalar unknowns represented by v(x, t) and p(x) and we may state the following complete model: Incompressible inviscid fluid flow: Given a volume force on R and boundary conditions on ∂R, find fields of velocity and pressure such that (11.6) and (11.7) are satisfied on R. Even though the difficulty related to discontinuities (shocks), indicated for the elastic inviscid fluid flow model, is removed by the isochoric condition, the well-posedness of this problem has only been shown for two-dimensional f low. The state of the situation is basically the same as for the Navier-Stokes’ viscous problem, derived and discussed in the next section. Bernoulli’s Theorem After introducing the concept of conservative body forces and an identity from vector calculus we derive Bernoulli’s theorem for an incompressible inviscid fluid flow. Conservative forces were introduced in Chapter 6 for pipe flow. The similar idea for incompressible three-dimensional fluid flow is that there exists a potential function U (x), x ∈ R, such that b(x)/ρin = −∇U (x).

(11.8)

A typical example of a conservative force acting on an incompressible fluid is the force due to gravity. In this case U (x) = g(x − o) · e 3 , where g is the gravitational acceleration, o the origin and e3 is an upward unit vector. It is a straightforward exercise in vector calculus (Exercise C.3) to prove the identity 1 (∇v)v = ∇(v · v) + (curl v) × v, (11.9) 2 for any vector field v on E. The curl of a vector field v, denoted curl v, is defined in Appendix C.2.

140

11 Models of Fluid Mechanics

Assume now steady state, i.e., v  (x, t) = 0, so that we can regard the velocity as a function of x only, as well as an incompressible fluid. It then follows from (11.7), (11.8) and (11.9) that   1 p(x) v(x) · ∇ + U (x) + v(x) · v(x) = 0. ρin 2 Thus, the quantity B(x) =

p(x) 1 + U (x) + v(x) · v(x) ρin 2

does not change in the direction of flow, which one could express by saying that B(x) is constant along stream lines, and this is the meaning of Bernoulli’s theorem. For an irrotational flow, i.e., curlv(x) = 0, B(x) is constant everywhere in the fluid domain. Furthermore, Bernoulli’s theorem has a natural extension to the compressible case, see Gurtin [5]. If we want to extend it further, to the viscous case, we need to include loss of energy due to internal friction.

11.2 Viscous Fluid We will consider a linear incompressible viscous fluid. The constitutive equation for such a fluid is given by (8.13) and is repeated here: T (x) = −p(x)I + 2˜ µD(x, t),

(11.10)

where µ ˜ is a viscosity coefficient and D(x, t) =

 1 ∇v(x, t) + (∇v(x, t))T 2

is the rate of deformation tensor. We like to substitute (11.10) into the equation of motion (11.1). To that end we need the vector identity (Exercise C.3) 2 div D(x, t) = div (∇v(x, t)) + ∇(div v(x, t)),

(11.11)

where the last term on the right hand side vanishes due to incompressibility, i.e., due to equation (11.6). From (11.10) and (11.1), with ρ = ρin , we now obtain ˜∇2 v(x, t) + b(x) = ρin [v(x, t) + (∇v(x, t))v(x, t)] , −∇p(x) + µ

(11.12)

where we use the notation ∇2 v(x, t) = div (∇v(x, t)). Equation (11.12) is known as Navier-Stokes’ equation of fluid mechanics. This equation together with the isochoric constraint (11.6) will constitute a complete problem when appropriate boundary conditions are prescribed. The boundary conditions are, as for the inviscid case, conditions on velocity and pressure. However,

11.2 Viscous Fluid

141

for the viscous case the whole velocity vector on the boundary can be prescribed, not only its normal part as for an inviscid fluid. In summary we have: Navier-Stokes fluid flow: Given a volume force on R and boundary conditions on ∂R, find fields of velocity and pressure such that (11.6) and (11.12) are satisfied on R. For two-dimensional flow it has been proved in what sense this problem possess a unique solution. However, it is an open problem to show well-posedness for three-dimensional flow. In fact, to prove or disprove the existence of a smooth solution is one of the Clay Institute $1 million Prize Problems, see Fefferman [8]. 11.2.1 Navier-Stokes’ Equation for Plane Steady Laminar Flow We consider a situation where flow occurs in one direction only, which is taken to be defined by the unit vector e1 , i.e., v(x, t) = v1 (x1 , x2 )e1 .

(11.13)

Here the velocity component v 1(x1 , x2) depends on coordinates x1 and x2 of the point x in the coordinate system defined by the orthonormal base {e1 , e2 , e3 }. The fact that velocity does not depend on the x3 -coordinate means that we consider plane flow. Having a velocity with the same direction over the whole flow domain can be seen as a case of laminar flow. That v1 (x1 , x2 ) does not depend on time means that we look at a steady state. With the flow (11.13), the condition (11.6) for isochoric motion becomes div v(x, t) =

∂v1 (x1 , x2 ) = 0. ∂x1

Thus, the velocity actually cannot depend on x1 and we write v 1(x 1, x 2) = v1 (x2 ). Due to this ⎤ ⎡ ⎤ ⎡ v1 0 ∂v1 /∂x2 0 0 0 ⎦, v(x, t) ∼ ⎣ 0 ⎦, ∇v(x, t) ∼ ⎣ 0 0 0 0 0 and we conclude that (∇v(x, t))v(x, t) = 0. Moreover, ∇2 v(x, t) =

∂ 2 v1 (x2 ) e1 . ∂x22

In summary, one concludes that for plane steady laminar flow, without body forces, Navier-Stokes’ equation is reduced to µ ˜

∂ 2 v1 (x2 ) ∂p(x1 , x2 , x3 ) = , 2 ∂x2 ∂x1

∂p(x1 , x2 , x3 ) ∂p(x1 , x2 , x3 ) = = 0. (11.14) ∂x2 ∂x3

142

11 Models of Fluid Mechanics

Since the left hand side of the first equation of (11.14) depends on x2 only, and since, due to the second part of (11.14), p(x1 , x2 , x3 ) depends on x1 only, it follows that ∂ 2 v1 (x2 ) µ ˜ = −∆, (11.15) ∂x22 where ∆ is a constant representing the slope of the linear pressure loss in the e1 -direction. Equation (11.15) is integrated for particular boundary conditions in Exercise 11.1. 11.2.2 Dimensionless Form As indicated in Section 8.2, dimensional invariance, which should hold for all equations describing physical phenomena, implies that it is possible to rewrite these equations in dimensionless form. Navier-Stokes’ equation is an equation for which considerable insight can be gained by obtaining its dimensionless version. Therefore we proceed to derive such a version. To that end, let  be a typical length (e.g., the diameter of a pipe) and υ a typical speed (e.g., an average velocity). Then we may introduce a dimensionless position vector x ˆ − o by x−o x ˆ−o= , (11.16)  where o is the origin, and a dimensionless time by tυ tˆ = . 

(11.17)

ˆ The dependent fields v and p are transformed to dimensionless versions v p ˆ by making the following scalings and 1 v(x, t), υ

(11.18)

1 p(x, t), υ 2 ρin

(11.19)

ˆ (ˆ v x, tˆ) = pˆ(ˆ x, tˆ) =

where x, x ˆ, t and tˆ are related as in (11.16) and (11.17), and where we have temporarily abandoned our convention that time is not shown explicitly as an argument in force-like variables. Since ∇p(x, t) =

υ 2 ρin ∇xˆ pˆ(ˆ x, tˆ), 

∇v(x, t) =

υ ˆ (ˆ ∇xˆ v x, tˆ), 

υ2 ∂ υ 2 ∂ ˆ ˆ ˆ (ˆ ∇ v(x, t) = v (ˆ x , t ), v x, tˆ), 2 xˆ ∂t  ∂ tˆ where ∇xˆ means the gradient with respect to the dimensionless position and similarly for ∇2xˆ , we find by substitution into (11.12) that ∇2 v(x, t) =

11.3 S

1 2 ˆ (ˆ ∇ v x, tˆ) = −∇xˆ pˆ(ˆ x, tˆ) + Re xˆ



f Fl d

 ∂ ˆ ˆ ˆ ˆ ˆ v (ˆ x, t) + (∇xˆ v (ˆ x, t))ˆ v (ˆ x, t) , ∂ tˆ

where Re =

143

(11.20)

υρin µ ˜

is the Reynolds number. Equation (11.20) indicates that all Navier-Stokes problems with the same Reynolds number and matching boundary conditions have the same dimensionless solution. This means, among other things, that we can figure out how to use physical models in experimental studies: if we are interested in in a medium with a certain density and viscosity, then we can actually use another medium if  and υ are adjusted to give the same Reynolds number as in the original case. Furthermore, it is known that flows satisfying Navier-Stokes equation change character abruptly when the average velocity is changed at a certain value, i.e., the flow changes from laminar to turbulent. This derivation shows that to characterize such a change we need only keep track of the Reynolds number since the overall character of flow depends on this number only.

11.3 Statics of Fluids Consider a fluid at rest, i.e., v(x, t) = 0 for all x ∈ R and all times t. The constitutive equation then reads T (x) = −p(x)I, irrespectively of the presence of viscosity. Moreover, the equation of motion (11.1) becomes, both for compressible and incompressible fluids, ∇p(x) = b(x).

(11.21)

A direct consequence of this equation is that a fluid cannot be at rest unless the volume forces can be derived as the gradient of a potential function, i.e., they must be conservative. We expand on this fact for a special case below, and the general case is treated in Exercise 11.3. In the following we will use (11.21) to prove Archimedes’ principle and derive equations for density and pressure in the atmosphere. To that end, let e3 be a unit vector in the upward vertical direction and let b be the force due to gravity, i.e., (11.22) b(x) = −ρ − (x)ge3 , where g is the gravitational acceleration and ρ(x) ρ( ) is the which in this static situation is not dependent on time but which could depend on position. Inserting (11.22) into (11.21) and writing the resulting equation in an orthonormal base {e1 , e2 , e3 } gives

144

11 Models of Fluid Mechanics

∂p(x1 , x2 , x3 ) ∂p(x1 , x2 , x3 ) = = 0, ∂x1 ∂x2 ∂p(x1 , x2 , x3 ) = −ρ(x1 , x2 , x3 )g. (11.23) ∂x3 From the first two equations of (11.23) we see that the pressure can depend on x3 only, i.e., p(x 1, x 2, x 3) = p(x 3). Inserting this into the last equation implies that the density also depends on x 3 only, i.e., ρ(x 1, x 2, x 3) = ρ(x 3). For a complete problem, where we want to solve for p(x) given the force, this later condition is a solvability condition on the data of the problem, i.e., on ρ(x), stemming from the fact that the load must be conservative. Such solvability conditions for the case of a general load are discussed in Exercise 11.3. Equation (11.23) can now be simplified to ∂p(x3 ) = −ρ(x3 )g. ∂x3

(11.24)

Introducing the potential function  Ψ (x3 ) = g

x3

ρ(ζ)dζ, 0

the solution of (11.24) becomes p(x3 ) = p(0) − Ψ (x3 ),

(11.25)

where p(0) is the pressure at x3 = 0. Equation (11.25) means that the pressure in a at rest, acted on by gravitational forces, depends only on the coordinate in the direction of gravitation. Moreover, the pressure is uniquely specified given a density distribution and the pressure at a certain depth, e.g., at a surface of the fluid. 11.3.1 Archimedes’ Principle Let a body be immersed or partly immersed into a fluid, see Figure 11.1. The domain of the immersed part is Ω and the outward unit normal vector on its boundary ∂Ω is denoted n. The total force exerted by the fluid on the body in Ω, the buoyancy force, is  FB = − p(x)n dSx , ∂Ω

where we have assumed that the pressure at the surface of the f is zero so that we can extend the integral over the whole of ∂Ω and not just the part in contact with the . We insert (11.25) into this expression for the force, and use the divergence theorem (C.21) and that the gradient of Ψ (x3 ) can be written as

11.3 Statics of Fluids

145

Fig. 11.1. Archimedes’ principle: The integral of fluid pressure on an object immersed into a fluid, giving a buoyancy force, equals the weight of a fictitious fluid occupying Ω with an opposite sign.

∇Ψ (x3 ) =

∂Ψ (x3 ) e3 = ρ(x3 )ge3 , ∂x3

(11.26)



to obtain F B = ge3

ρ(x3 ) dV Vx .

(11.27)

Ω

The right hand side of this equation is the opposite of the gravity force that would act on a volume of fluid identical in shape and position to that of the immersed part Ω of the body, and the equation says that this force equals the buoyancy force due to pressure from the on the immersed object. This is the first part of Archimedes’ principle. The second part is the fact that the line of action of the force goes through the center of gravity of the fictitious fluid . We go on to prove this. The torque from the pressure on the immersed fluid with respect to the point o is  M Bo = − (x − o) × p(x)n dSx . ∂Ω

Substituting (11.25), using the divergence theorem (C.23) and the fact that curl (x − o) = 0, we find that  Vx , (11.28) M Bo = − (x − o) × ∇Ψ (x3 ) dV Ω

where we have also used the vector analysis formula (C.18). Inserting (11.26) into (11.28) and using   (xc − o) ρ(x) dV = (x − o)ρ(x) dV, Ω

Ω

which defines the center of gravity xc of the fictitious fluid, we end up with

146

11 Models of Fluid Mechanics

M Bo = F B × (xc − o). This equation represents our claim that the buoyancy force acts through the center of gravity xc . Note that Archimedes’ principle holds both for compressible and incompressible fluids. In particular, it explains both why objects float in water and why a balloon can sail through the air. 11.3.2 Density and Pressure of the Atmosphere So far in this section we have considered the force b(x) to be data of our problem. When the force is given by (11.22) this means that the density ρ(x) must be considered as given. This is no difficulty for an incompressible fluid, where ρ(x) = ρin is constant and (11.25) can be directly integrated to give p(x3 ) = p(0) − ρin gx3 . However, for a compressible fluid, say the air of the atmosphere, ρ(x) is usually not known and to make (11.24) into a complete problem we need a constitutive equation in the form of (11.4). By inserting this equation into (11.24) we get ∂ρ(x3 ) ρ(x3 )g =− , ∂x3 ∂ pˆ(ρ(x3 ))/∂ρ

(11.29)

where we have assumed that ∂ pˆ(ρ(x3 ))/∂ρ = 0. Consider for simplicity a linear version of (11.4), i.e., p(x) = kρ(x), where k is a constant. Equations (11.29) can then be integrated to give gx3 gx , p(x3 ) = p(0)e− 3 , (11.30) k k where ρ(0) and p(0) may be considered as density and pressure at ground level. Note that the pressure is no longer a linear function of hight x3 as it is in the incompressible case. ρ(x3 ) = ρ(0)e−

Exercises 11.1. Solve equation (11.15), for a given pressure drop ∆, when x 2 ∈ [0, h] and the boundary conditions are v1 (0) = 0 and v1 (h) = ν, where ν is a given velocity. Calculate also the stresses in the fluid. Make a drawing of the situation to understand the physical origin of the boundary conditions and the pressure drop. 11.2. Consider steady laminar Navier-Stokes flow without body forces in a cylindrical pipe of length L and constant area A, where the pressures are ps

11.3 Statics of Fluids

147

and pe at the left and right ends of the pipe, respectively. Then the volume flux q satisfies A2 (ps − pe ), q= 8πµL ˜ where µ ˜ is the viscosity constant. Make a derivation of this equation which is known as Hagen-Poiseuille’s equation. 11.3. Conclude that (11.21) has a solution p(x) for given b(x) only if in a cartesian frame ∂bi (x1 , x2 , x3 ) ∂bj (x1 , x2 , x3 ) = , ∂xj ∂xi where i, j = 1, 2, 3. This is a necessary condition for b(x) to be derivable as the gradient of a potential function. Conclude also that for a force in the form of (11.22) this condition implies that the density depends on x3 only. 11.4. Use (11.27) and (11.30) to prove that the equilibrium height above ground of a balloon is approximately (for a balloon that is small compared to this height) proportional to ln(ρ(0)/ρb ), where ρb is the average density of the balloon.

12 Kinematic Constraints, Beams and Rigid Bodies

Kinematic constraints, or restrictions, were discussed in Chapter 8. Such constraints are always associated with forces needed to maintain them. Therefore they may be seen as extreme cases of force laws or constitutive laws depending on whether the maintaining force is internal or external. In this chapter we will mainly consider some kinematic constraints related to internal forces, which may thus be seen as extreme cases of constitutive laws. Kinematic constraints are widely discussed in the continuum mechanics literature; a more extensive presentation which is in line with the present one can be found in Podio-Guidugli [14]. We start the discussion in Section 12.1 by briefly considering constraints in a discrete system, as well as in incompressible linear elasticity and in the incompressible Newtonian fluid. These examples point to an approach to constraints where parameters of certain constitutive laws tend to infinity to produce the required equations. In Section 12.2 it will be seen how Euler-Bernoulli and Timoshenko beam theories, derived as small displacement one-dimensional theories in Section 9.2, can be seen as special cases of a small displacement three-dimensional theory when kinematic constraints are introduced. Finally, in Section 12.3 the theory of rigid bodies is discussed. A rigid body is seen as a three-dimensional continuum in which the rigidity is a kinematic constraint.

12.1 Some Examples and the General Idea In this section we consider particular cases of kinematic constraints. The discussion is aimed at setting the stage for the remaining two sections of this chapter. The presentation is somewhat informal and we have generally omitted arguments of time and place for quantities such as stress and strain.

149

150

12 Kinematic Constraints, Beams and Rigid Bodies

12.1.1 Discrete Problem Consider a discrete model consisting of one material point acted on by four forces, f e and f i , i = 1, 2, 3. A two-dimensional version of this problem is shown in Figure 12.1. The three forces f i are directed in three orthogonal directions e1 , e2 and e3 , and they are decoupled and linearly dependent on the displacement u of the material point in the following way: f i = −ki ui ei ,

i = 1, 2, 3,

where ui = u · ei and ki > 0 are constants. If we view these forces as produced by springs, we may think of ki as spring constants. The force f e may be seen as a given external force having a general direction. Equilibrium requires fe +

3

fi = 0

⇔

fe = −

i=1

3

ki ui ei .

(12.1)

i=1

Fig. 12.1. A two-dimensional version of the problem treated in this subsection. As k1 approaches infinity a problem with a kinematic constraint emerges.

We will now consider the thought experiment of letting one of the spring constants, say k1 , grow to infinity while the corresponding force stays bounded. This means that a kinematic constraint u 1 = e1 · u = 0

(12.2)

emerges and the corresponding force can be written f 1 = −λe1 ,

(12.3)

for some scalar λ. Equation (12.1) then becomes f e = λe1 +

3 i=2

ki ui ei ,

(12.4)

12.1 Some Examples and the General Idea

151

and (12.1) has been replaced by (12.2) and (12.4). The force f 1 can be called a reaction force which arises due to the kinematic constraint (12.2) and it can take any value necessary for equilibrium to hold. We can obtain a reaction free equilibrium equation, i.e., an equilibrium equation that does not contain the reaction force, by projecting onto the plane having e1 as a normal vector. The operator that achieves this projection is P = I − e1 ⊗ e1 = e2 ⊗ e2 + e3 ⊗ e3 , where I is the identity tensor. Letting P act on (12.4) we get (f e · e2 )e2 + (f e · e3 )e3 =

3

ki ui ei .

(12.5)

i=2

Here, f e ·e2 and f e ·e3 are the components of f e in the e2 – and e3 – directions. Similarly, we can obtain a pure reaction problem by projecting onto the e1 –direction. This is achieved by acting on (12.4) by the tensor e1 ⊗ e1 , and we then obtain (f e · e1 )e1 = λe1 . Note that the reaction force f 1 = −(f e · e1 )e1 is workless in the sense that f 1 · u = 0, for all u that satisfy (12.2). 12.1.2 Incompressible Linear Elasticity In this subsection we use an approach similar to that of the previous subsection to derive equations for incompressible linear elasticity. We also comment on other types of constraints in linear elasticity. In Section 9.3 we saw that the constitutive law of isotropic linear elasticity could be written as in T = κ(tr E)I + 2µ dev E,

(12.6)

where κ is the compression modulus and µ is the shear modulus. By using the operators tr and dev we can decompose (12.6) into two orthogonal parts, i.e., tr T = 3κ tr E,

dev T = 2µ dev E.

The constants κ and µ are to be compared to the h k :s of the previous subsection. By letting one of these or both approach infinity we can obtain constitutive equations for materials with certain kinematic constraints. If we let κ tend to infinity while tr T stays bounded we get T = λI + 2µ dev E,

(12.7)

12 Kinematic Constraints, Beams and Rigid Bodies

152

and tr E = 0,

(12.8) R

where λ is a scalar representing the reaction stress T = λI. It was concluded in Section 9.3 that trE represents, within the usual small displacement approximation of linear elasticity, the relative volume change, meaning that (12.7) and (12.8) are constitutive equations for an incompressible linear elastic material. If we let µ tend to infinity instead of κ we get a shape preserving material, and if both constants are made infinite, then we get a rigid material operating in small displacements. Substituting (12.7) into the equilibrium equation (9.48) we get ∇X λ + div (2µ dev E) + b = 0.

(12.9)

This equation is to be compared to (12.4) and, as in the previous subsection, we can operate on (12.9) by a projection operator to obtain a reaction free equilibrium problem. Since curl (∇X ϕ) = 0 for any scalar field ϕ, we conclude that curl is the needed projection operator. The reaction free equilibrium problem is thus curl (div (2µ dev E) + b) = 0. (12.10) The reaction stress T R = λI is workless in the sense that T R : E = 0, for all E that satisfy (12.8). We have seen that the isotropic constitutive law (12.6) can produce three types of kinematic constraints by letting coefficients tend to infinity: incompressibility, shape preservation and rigidity. If we are interested in other types of constraints we need to start from anisotropic linear elastic constitutive laws, which contain more material constants. For instance, a constraint of inextensibility in a certain direction follows from a transversely isotropic material. Such a material has one preferred direction: rotations of the material that keep this direction fixed cannot be experimentally detected. The kinematic constraint of inextensibility in a particular direction follows by letting the Young’s modulus associated with the preferred direction approach infinity. In Section 12.2 below we will also start from a transversely isotropic material and show that the kinematic constraint associated with Euler-Bernoulli beam theory follows by letting all material constants except the Young’s modulus of the preferred direction approach infinity. 12.1.3 Incompressible Viscous Fluid We concluded in Chapters 8 and 11 that the constitutive behavior of an incompressible linear viscous fluid (i.e., a Newtonian fluid) reads T = −pI + 2˜ µD,

div v = 0.

(12.11)

12.2 Beam Theory

153

We can compare these equations with (12.7) and (12.8). Inserting the first equation of (12.11) into the equation of motion, (7.16), we obtain NavierStokes’ equation (11.7). By operating on this equation with the curl operator we obtain an equation that is free of the reaction stress T R = −pI. This equation is known as the vorticity equation and can be shown to read ∂ω + curl (ω × v) − µ ˜∇2 ω = curl b, ∂t where ω = curl v is the vorticity. The reaction stress T R = −pI is workless in the sense that T R : D = 0, for all D that satisfy div v = tr D = 0.

12.2 Beam Theory In Section 9.2 a beam model has been introduced as an intrinsically onedimensional theory. This is usually referred to as a direct approach. It has the advantage of a certain simplicity. However, since we tend to look at beams as three-dimensional although slender bodies, certain aspects are missing in the direct approach. For instance, there is no reference to the three-dimensional Cauchy stress and therefore no possibility of deriving constitutive equations for cut forces and couples from known three-dimensional constitutive equations. The derivation of beam theory from a parent three-dimensional theory is known as a deductive approach. There are at least two different such approaches: (1) In the asymptotic approach we get the beam equations as a limit of three-dimensional equations when the beam tends to become infinitesimally thin. This is an approximate although systematic approach. (2) In the exact approach, to be used below, we regard the beam as a three-dimensional structure, but consisting of a material which due to kinematic constraints can deform only in a beam-like sense. A derivation of Timoshenko’s beam theory which follows this latter approach and which has inspired the present section is given in Lembo and Podio-Guidugli [12]. We start the derivation of beam equations by concluding that the beam equilibrium equations (9.20) through (9.22) follow from the three-dimensional equilibrium equations (9.48) and (9.49). Next, we introduce a kinematic constraint and constitutive laws compatible with this constraint. In this way we arrive at the same equations as were derived by a strictly one-dimensional reasoning in Section 9.2. 12.2.1 Geometric Shape and Prerequisites We consider a beam-like domain B, shown in Figure 12.2. The straight axis of B is defined by the origin o and the direction given by the unit vector es , i.e., a point P (s) on the axis is given by

12 Kinematic Constraints, Beams and Rigid Bodies

154

Fig. 12.2. The geometry of a beam-like domain B. It shows the origin o, a point P (s) along its axis, the cross-section A(s) and the orthonormal base {e1 , e2 , es }.

s ∈ (0, L),

P (s) = o + ses ,

(12.12)

when the beam has length L. An orthonormal base {e1 , e2 , es } is introduced and any point X in B can be written as X = P (s) + X1 e1 + X2 e2 ,

(12.13)

for some coordinates X1 , X2 and s. For each coordinate s, (X1 , X2 ) belongs to a subset A(s) of R2 , which we can regard as the cross-section of B at s. By using (12.12) and (12.13), any field f (X) on B can be seen as a function of (X1 , X2 , s), i.e., f (X) = f (X1 , X2 , s). Let P be a subset of B formed by restricting s to the interval (a, b), where 0 ≤ a < b ≤ L. A volume integral of f (X) over P can be written     b

P

f (X) dV VX = a

A(s)

f (X1 , X2 , s) dX1 dX2 ds,

(12.14)

and, similarly, a surface integral of f (X) over ∂P, the boundary of P, reads 

 ∂P

f (x) dAX =  + A(a)

a

b

 ∂ A(s)

 f (X1 , X2 , s) dr ds

f (X1 , X2 , a) dX1 dX2 +

 A(b)

f (X1 , X2 , b) dX1 dX2 , (12.15)

where dr is the integration element of the curve ∂A(s) which is the boundary of A(s). In Section 12.2.3 a further specialization of the geometry is used: there it is assumed that for each s the point (0, 0) is the centroid of the cross section A(s) and that its principal axes are the coordinate axes. In mathematical terms this reads  Xi dX1 dX2 = 0, i = 1, 2, (12.16) A(s)

12.2 Beam Theory

155

 A(s)

X1 X2 dX1 dX2 = 0.

(12.17)

Moreover, we have the following notations for the area of the cross section A(s) and its area principal moments of inertia:   A(s) = dX1 dX2 , Ii (s) = Xi2 dX1 dX2 , i = 1, 2. A(s)

A(s)

12.2.2 Equilibrium Equations As is clear from the discussion in Chapter 7, the equilibrium equations (9.48) and (9.49), and the symmetry of the stress tensor T (X) are equivalent to the vanishing of the total force and torque on any subset P of B, i.e.,   b(X) dV VX + s(X) dAX = 0, (12.18) P

 P



(X − o) × b(X) dV VX +

∂P

∂P

(X − o) × s(X) dAX = 0,

(12.19)

where b(X) is a force per unit volume and s(X) is a stress vector on the boundary of P. These equations may be viewed as derived from (7.14) and (7.15) when accelerations are set to zero and x is changed for X due to the small displacement assumption. We will now rewrite (12.18) in terms of the force per unit length q(s) and the cut force vector f (s). These are defined as   q(s) = b(X1 , X2 , s) dX1 dX2 + s(X1 , X2 , s) dr, A(s)

∂ A(s)

 f (s) =

A(s)

T (X1 , X2 , s)es dX1 dX2 .

Integrals in (12.18) will be rewritten by using (12.14) and (12.15). We P then use that, s(X) = T (X)n, where n is the outward unit vector on ∂ P. When integrating over A(a) and A(b) it holds that n = −es and n = es , respectively. Thus, it follows from (12.18) that 

b

q(s) ds − f (a) + f (b) = 0. a

Then, from



b

f (b) − f (a) = a

∂f (s) ds, ∂s

and the localization theorem, Theorem 2 of Appendix C.2, it finally follows that

12 Kinematic Constraints, Beams and Rigid Bodies

156

∂f (s) = 0. (12.20) ∂s Next, we will rewrite (12.19) in terms of the cut force vector f (s), the couple per unit length l(s) and the cut couple vector m(s). The first vector was defined above and the latter two are defined as  l(s) = (X1 e1 + X2 e2 ) × b(X1 , X2 , s) dX1 dX2 A(s)  + (X1 e1 + X2 e2 ) × s(X1 , X2 , s) dr, q(s) +

∂ A(s)

 m(s) =

A(s)

(X1 e1 + X2 e2 ) × T (X1 , X2 , s)es dX1 dX2 .

Using these definitions and X − o = X1 e1 + X2 e2 + ses , it follows from (12.19), (12.14) and (12.15) that 



b

ses × q(s) ds + a

b

l(s) ds a

− aes × f (a) + bes × f (b) − m(a) + m(b) = 0. We rewrite this equation by using  m(b) − m(a) = a

 bes × f (b) − aes × f (a) =

b

b

∂m(s) ds, ∂s

∂(sf (s)) ds ∂s  b  b ∂f (s) = ses × es × f (s) ds. ds + ∂s a a

es × a

It then follows by the use of (12.20) and localization that l(s) + es × f (s) +

∂m(s) = 0. ∂s

(12.21)

Equations (12.20) and (12.21) can be compared with (5.8) and (5.9) or (5.15) and (5.16), which are material and spatial versions of equations of motion for a one-dimensional theory. Here we consider small displacements, and equilibrium is set up disregarding deformations as discussed in Section 9. Equations (9.20) through (9.22) are two-dimensional equations. We arrive at these by assuming that the external forces and couples are such that the forces act in the es -e1 –plane and the external couple acts in the e2 –direction. We also take reduced forms of the cut force and couple. Using the same notations for components as in Chapters 5 and 9 this means that

12.2 Beam Theory

q(s) = qt (s)es + qn (s)e1 , f (s) = ft (s)es + fn (s)e1 ,

157

l(s) = lb (s)e2 ,

(12.22)

m(s) = mb (s)e2 .

(12.23)

By inserting these expressions into (12.20) and (12.21) we get ∂ffn (s) + qn (s) = 0, ∂s

(12.24)

∂fft (s) + qt (s) = 0, (12.25) ∂s ∂mb (s) fn (s) + + lb (s) = 0. (12.26) ∂s These three equations are the same as the equilibrium equations in the section on beam theory in Chapter 9 and constitute one third of the diagram in Figure 9.7. They are used directly in the Timoshenko beam problem, but in the Euler -Bernoulli problem we use an equation where fn (s) is eliminated. This equation is obtained by taking the derivative with respect to s of equation (12.26) and inserting in (12.24). We then find −

∂ 2 mb (s) ∂lb (s) − + qn (s) = 0. ∂s2 ∂s

(12.27)

It should be emphasized that these equilibrium equations are exact consequences of the original three-dimensional equilibrium equations, but they are not equivalent to these: we cannot derive the three-dimensional equations from the one-dimensional ones. In fact, the one-dimensional equilibrium equations may be seen as projections of the three-dimensional equations in the same sense as (12.5) and (12.10) have been seen as projected equilibrium equations in Section 12.1. 12.2.3 Displacement Fields and Constitutive Assumptions As discussed in Chapter 8, a kinematic constraint is a particular law and in the developments in this chapter it may be seen as an extreme case of a constitutive law: the material behavior is such that internal reactions develop to the extent necessary to keep the kinematic constraint satisfied. In the following we will develop such constitutive laws that are implicit in Euler-Bernoulli and Timoshenko beam theories. To match the restriction of the equilibrium equations to the plane case, i.e., the specifications of (12.20) and (12.21) to (12.24) through (12.27), and also to be compatible with the development in Section 9.2, we will use an assumption of plane deformation in the following. Plane deformation means that no displacement occurs in the e2 –direction and the remaining two displacement components vary only in the es -e1 –plane, i.e., if the displacement field u is seen as a function of (X1 , X2 , s) we have

158

12 Kinematic Constraints, Beams and Rigid Bodies

u(X1 , X2 , s) = u1 (X1 , s)e1 + us (X1 , s)es ,

(12.28)

where u1 (X1 , s) and us (X1 , s) are displacement components; the component in the e2 –direction, u2 , is zero. This displacement field implies that the following components of the strain tensor E are zero ε22 = ε2s

1 = 2

ε12

1 = 2

 

∂u2 = 0, ∂X2

∂us (X1 , s) ∂u2 + ∂s ∂X2

(12.29) 

∂u2 ∂u1 (X1 , s) + ∂X1 ∂X2

= 0,

(12.30)

= 0.

(12.31)



Euler-Bernoulli Displacement Field In the plane strain case the displacement field of the Euler-Bernoulli beam theory is the following: u(X1 , X2 , s) = u0 (s) − where

∂un (s) X 1 es , ∂s

(12.32)

u0 (s) = ut (s)es + un (s)e1 ,

is the displacement vector for points along the axis defined by (12.12). This displacement is illustrated in Figure 12.3. The displacement field (12.32) implies,

Fig. 12.3. The displacement field as given by (12.32). Note that this is a small displacement problem and the rotations and translations in the figure are grossly exaggerated.

in addition to (12.29) through (12.31), that the following strain components are zero:

12.2 Beam Theory

ε11 = ε1s =

1 2



∂u1 (X1 , s) = 0, ∂X1

∂u1 (X1 , s) ∂us (X1 , s) + ∂s ∂X1

159

(12.33)  = 0.

(12.34)

Conversely, under the plain strain assumption, one can show that (12.33) and (12.34) imply the Euler-Bernoulli displacement field (12.32). We proceed to do so: Equation (12.33) gives that u1 (X1 , s) is a function of s only, i.e., u1 (X1 , s) = un (s) for some function un (s). Inserting this into (12.34) and integrating gives ∂un (s) + ut (s), us (X1 , s) = −X1 ∂s for some function ut (s). By inserting into (12.28) we obtain (12.32). We have thus seen that the only strain component that is non-zero for the displacement field (12.32) is εss =

∂us (X1 , s) ∂ut (s) ∂ 2 un (s) = − X1 . ∂s ∂s ∂s2

The constitutive law that reflects this and the kinematic constraints (12.29) through (12.31), (12.33) and (12.34) should, in accordance with Section 12.1, be obtained by letting certain material constants approach infinity. As has been seen in Section 12.1, linear elastic isotropic material can in this way only produce three types of constraints, none of which is what we need here. Thus, we should start from a more general case, and the linear elastic transversely isotropic material, where the preferred direction coincides with the es –direction, is a possibility. Such materials have five independent material constants, and if we let four of these tend to infinity while requiring the stress to be bounded, the kinematic constraints (12.29) through (12.31), (12.33) and (12.34) emerge. The remaining material constant is a tension or Young-type constant which is denoted E. The constitutive law for the remaining nonreactive part of the stress tensor is   ∂ut (s) ∂ 2 un (s) − σss = Eεss = E X (12.35) 1 . ∂s ∂s2 In general, the value of E may vary from point to point in B, but in the following it is assumed that it varies only with respect to the s coordinate, i.e., E = E(s). Next, we will introduce the constitutive law, expressed by (12.35) and the fact that other stress components are reaction stresses, into the definitions of the cut force f (s) and the cut couple m(s). To that end we note that T (X1 , X2 , s)es = σ1s e1 + σ2s e2 + σss es .

(12.36)

Clearly, the stress components σij as well as the strain component εss depend on position in B. However, we have chosen not to make this notationally

160

12 Kinematic Constraints, Beams and Rigid Bodies

explicit. By introducing (12.36) and (12.35) into the definition of f (s), using that (0, 0) is the centroid of A(s), i.e., equation (12.16), and the assumption that E depends only on the s coordinate, we find   2 ∂ut (s) f (s) = σis dX1 dX2 ei + EA(s) es , ∂s A(s) i=1 where we have used the notation A(s) for the area of A(s), as introduced in Section 12.2.1. Comparing this expression for f (s) with that given in (12.23) we find  fn (s) = σ1s dX1 dX2 , (12.37) 

A(s)

0= A(s)

σ2s dX1 dX2 ,

(12.38)

∂ut (s) . (12.39) ∂s Equations (12.37) and (12.38) are conditions on reactive parts of the stress tensor. Equation (12.37) may be substituted into (12.24) to give a condition on the shear stress σ1s in terms of the external force bn (s). Equation (12.39) is the constitutive equation for the bar problem of Section 9.2. It can be divided into equation (9.29), relating generalized stress and strain, and equation (9.26), which is a purely geometric equation. The bar problem of Section 9.2 has now been completely derived from threedimensional theory. What has been gained compared to the direct onedimensional approach is that the constant of the constitutive law, EA(s), can now be seen as composed of a geometric contribution in A(s) and a material contribution E. To derive the constitutive equation of the Euler-Bernoulli beam problem we should introduce (12.36) and (12.35) into the definition of m(s). Using (12.16) and (12.17) we then find   ∂ 2 un (s) m(s) = (X1 σ2s − X2 σ1s )dX1 dX2 es + EI1 (s) e2 , ∂s2 A(s) ft (s) = EA(s)

where the definition of the area principle moment of inertia I1 , introduced in Section 12.2.1, has been used. Comparing this expression with (12.23) we find  0= (X1 σ2s − X2 σ1s )dX1 dX2 , (12.40) A(s)

∂ 2 un (s) . (12.41) ∂s2 Equation (12.40) is a constraint on reaction stresses, while (12.41) together with the equilibrium equation (12.27) define the Euler-Bernoulli beam problem, previously stated in Section 9.2. As with the beam problem, (12.41) can mb (s) = EI1 (s)

12.2 Beam Theory

161

be divided into a constitutive equation relating generalized stress and strain, and into a purely geometric equation. The advantage of the method used in this chapter as compared to the direct method of Section 9.2 is that we get some information on Cauchy stresses, and the constant EI1 (s) can be derived from geometric information and a knowledge of the Young’s modulus E. Timoshenko Displacement Field We will now derive the Timoshenko beam equations as we derived the EulerBernoulli theory above. We then replace (12.32) by the Timoshenko displacement field: u(X1 , X2 , s) = u0 (s) − ϕ(s)X1 es , where, as above,

(12.42)

u0 = ut (s)es + un (s)e1

is the displacement vector along the axis defined by (12.12). What is new, compared to (12.32), is the function ϕ(s). It represents rotation in the es -e1 – plane of material lines oriented in the e1 –direction. Note that all such lines are straight lines also after deformation, as illustrated in Figure 12.4.

Fig. 12.4. The displacement field as given by (12.42). Rotations and translations are grossly exaggerated.

In the corresponding development of Euler-Bernoulli’s beam theory one assumes, in addition to (12.42), that the material lines in the e1 –direction are orthogonal to the line defined by the deformed axis of the beam. This means that all material directions in the es -e1 –plane have the same rotation and no shearing occurs. In mathematical terms the extra assumption reads ϕ(s) = ∂un (s)/∂s, as should be familiar from Section 9.2. Euler-Bernoulli’s displacement field (12.32) has been shown to be equivalent to the vanishing of the two strain components ε11 and ε1s , as expressed in

162

12 Kinematic Constraints, Beams and Rigid Bodies

(12.33) and (12.34). The Timoshenko displacement field (12.42) can similarly be shown to be equivalent to (12.33) and, instead of (12.34), the condition   1 ∂ ∂ ∂u1 (X1 , s) ∂us (X1 , s) ε1s = + = 0. (12.43) ∂X1 2 ∂X1 ∂s ∂X1 Thus, we no longer have a constraint where the shear strain ε1s is zero. Rather, we have two non-zero strain components which are ∂us (X1 , s) ∂ut (s) ∂ϕ(s) = − X1 , ∂s ∂s ∂s     1 ∂u1 (X1 , s) ∂us (X1 , s) 1 ∂un (s) = = + − ϕ(s) . 2 ∂s ∂X1 2 ∂s εss =

ε1s

A constitutive law that reflects that all but these two strain components are zero can be found by starting, as in the section on Euler-Bernoulli beams, from a transversely isotropic material where the preferred axis coincides with the es –direction. We let three of the five independent material constants approach infinity. The remaining two are a Young’s modulus E and a shear modulus G such that ∂ut (s) ∂ϕ(s) σss = Eεss = E − X1 , (12.44) ∂s ∂s   ∂un (s) σ1s = G2ε1s = G − ϕ(s) . (12.45) ∂s As with E, the shear modulus G may depend on the position along the beam, i.e., on s, while no dependence with respect to other coordinates is assumed. Inserting (12.36), (12.44) and (12.45) into the definition of f (s), and taking (12.16) and the definition of the cross-section area A(s) into account, we find  f (s) = GA(s)

 ∂un (s) − ϕ(s) e1 ∂s  + A(s)

 σ2s dX1 dX2 e2 + EA(s)

Comparing this expression with that given in (12.23) we find   ∂un (s) fn (s) = GA(s) − ϕ(s) , ∂s

∂ut (s) es . ∂s

(12.46)

as well as equations (12.38) and (12.39). This means that we have found the bar problem of Section 9.2 also when starting from the Timoshenko displacement field. Moreover, we have derived, in (12.46), an equation which contains (9.25) and (9.28). Next, we insert (12.36), (12.44) and (12.45) into the definition of m(s). By using (12.16) and (12.17) we then find

12.3 Rigid Body Theory

 m(s) =

A(s)

X1 σ2s dX1 dX2 es + EI1 (s)

Comparing this expression with (12.23) we find  0= X1 σ2s dX1 dX2 ,

163

∂ϕ(s) e2 . ∂s

(12.47)

A(s)

∂ϕ(s) . (12.48) ∂s Equation (12.48) contains equations (9.27) and (9.30) and, thus, we have derived all equations of the Timoshenko beam theory of Section 9.2. It may seem as if the admissible stress is more restricted in the Timoshenko theory than in the Euler-Bernoulli theory. For instance, the shear stress σ1s becomes constant over each cross-section in the Timoshenko theory, while in the Euler-Bernoulli theory it is a reaction stress which need only satisfy some integral constraints. However, intuitively one may feel that the constraint (12.43) should result in some reactive stresses, not discussed here. It is shown in Lembo and Podio-Guidugli [12] that this is indeed the case, but since (12.43) is a higher order constraint, these reaction stresses are also of higher order and of non-symmetric nature. This topic is, however, well beyond what can be discussed in a text book at this level. mb (s) = EI1 (s)

12.3 Rigid Body Theory In this section we will show how classical rigid body theory can be seen as a three-dimensional continuum theory where the constitutive behavior of the inner forces is completely specified by a kinematic constraint. We start by looking at the geometry of rigid body motion, after which we look at what Euler’s second law implies for a rigid body. 12.3.1 Geometry of Rigid Body Motion A body B is said to be rigid if for all pairs of material points Xk , X ∈ B it holds that |φt (Xk ) − φt (X X )| is independent of time. (12.49) That is, the distance between all points is invariable. Condition (12.49) is our only particular law in what follows. It leads to the classical theory of rigid bodies.

164

12 Kinematic Constraints, Beams and Rigid Bodies

The Placement of a Rigid Body In this paragraph we will show that the placement mapping φt takes a particular form when the body is rigid. In fact, the condition (12.49) holds if and only if φt can be written as x = φt (X) = p(t) + Q(t)(X − P ),

(12.50)

where P ∈ B is an arbitrary reference point, p(t) = φt (P ) ∈ Bt and Q(t) is a proper orthogonal tensor1 , and where for convenience we have chosen the reference configuration B ⊂ E such that for some time t it holds that φt (X) = X for all X ∈ B. We proceed to prove this equivalence and show also how to represent Q(t) by use of two orthonormal bases. Next, we derive the classical result of Euler that any rigid body placement can be seen as the sum of a translation and a rotation. As the first step we prove that (12.50) implies (12.49). From (12.50) one concludes that for any two points Xk , X ∈ B it holds that φt (Xk ) − φt (X X ) = Q(t)(Xk − X ), and that therefore |φt (Xk ) − φt (X X )|2 = [Q(t)(Xk − X )] · [Q(t)(Xk − X )] = (Xk − X ) · [Q(t)T Q(t)(Xk − X )] = |Xk − X |2 , (12.51) where we have used definitions of the transpose of a tensor (C.9) and of orthogonal tensors (C.13). The last term of (12.51) is obviously independent of time. The proof of the converse, that (12.49) implies (12.50), follows Gurtin [5]. An alternative proof is contained in Exercises 12.3 through 12.5 and follows Curnier [7]. Let X and Y be arbitrary points of B. As stated above, at a certain time t it holds that φt (X) = X for all X ∈ B, so (12.49) implies that [φt (X) − φt (Y )] · [φt (X) − φt (Y )] = (X − Y ) · (X − Y ).

(12.52)

Taking the gradient of this expression with respect to Y Y, using (C.17) and the fact that the gradient of φt at Y is the deformation gradient F (Y, t), we find F (Y, t)T [φt (X) − φt (Y )] = X − Y.

(12.53)

Next, we take the gradient of (12.53) with respect to X. By using (C.20) we find F (Y, t)T F (X, t) = I. (12.54) By letting X = Y in (12.54) and comparing with (C.13) we see that F (X, t) is orthogonal at each X. Then its transpose coincides with its inverse and Y, implying that the (12.54) gives that F (X, t) = F (Y, t) for all X and Y deformation gradient is constant. It is clear that a placement that satisfies 1

Proper orthogonal tensors are defined in Appendix C.

12.3 Rigid Body Theory

165

φt (X) = φt (Y ) + F (t)(X − P ) has a constant deformation gradient F (t). Gurtin [5] shows that if B is a connected set, the converse also holds and thereby (12.50) follows. That Q(t) is proper follows from (3.5). Any orthogonal tensor can be represented as Q(t) =

3

ei (t) ⊗ E i ,

(12.55)

i=1

where {E 1 , E 2 , E 3 } is an orthogonal basis and ei (t) = Q(t)E i (i = 1, 2, 3) are vectors of another such a basis. This follows by noting that for any vector V,   3 3 [Q(t)E i ] ⊗ E i V = Q(t)V − Q(t) (E i · V )E i = 0, Q(t) − i=1

i=1

where we have used (B.4), (C.8) and the definition of the tensor product. An important interpretation of this result follows by inserting (12.55) into (12.50). We then find that φt (X) − φt (P ) =

3

[(X − P ) · E i ] ei (t),

(12.56)

i=1

which means that the components of the vector X−P in the basis {E 1 , E 2 , E 3 } are the same as the components of the vector φt (X) − φt (P ) in the basis {e1 (t), e2 (t), e3 (t)}. An essential property of proper orthogonal tensors is that they can be seen as effecting rotations of vectors. To show this, we first conclude that if Q(t) is a proper orthogonal tensor, then there is a unit vector N such that Q(t)N = N . The proof follows by taking the determinant of Q(t)T (Q(t) −I) = −(Q(t) − I)T , which is a direct consequence of Q(t)T Q(t) = I. One then finds that det (Q(t) − I) = 0, so Q(t) has an eigenvalue that is 1 and a corresponding eigenvector N such that (Q(t) − I)N = 0. Now, let E 2 and E 3 be such that {N , E 2 , E 3 } is an orthogonal base. Then (12.55) can be written as Q(t) = N ⊗ N +

3

ei (t) ⊗ E i .

i=2

The transformation of a vector V by Q(t) to v, i.e., v = Q(t)V, is illustrated in Figure 12.5. Since V · E i = v · ei (t) (i = 2, 3), it can be concluded that the vector is rotated around the axis defined by N .

166

12 Kinematic Constraints, Beams and Rigid Bodies

Fig. 12.5. The rotation of a vector V to v = Q(t)V as seen in a plane orthogonal to the axis of rotation.

The first term in (12.50) is clearly a translation: if Q(t) = I, then each material point X ∈ B is translated by the same vector p(t) − P from its ref rence position. Since we have just concluded that the second term in (12.50) represents a rotation we have the theorem that any rigid body placement can be seen as the sum of a translation and a rotation around a certain axis. This was first proved by Euler in 1776. The development in this paragraph explains the particular mathematical form of the observer transformation (8.1). Two observers are assumed to record motions that differ by a rigid body motion. This, in turn, reveals that there are dual interpretations of the basis {E 1 , E 2 , E 3 } and any vector that is represented in this base. In one interpretation, we may see X − P and φt (X) − φt (P ) = Q(t)(X − P ) as two recordings of the same physical referent. The first one as seen from the body fixed frame, and the second one as seen from the space fixed frame. This is in contrast to our view so far where these two vectors are different objects in the same frame. The Velocity of a Rigid Body In this paragraph we will show that the velocity of a rigid body can be written as V (X, t) = V (P, t) + ω(t) × (φt (X) − φt (P )), (12.57) where P ∈ B is an arbitrary reference point and ω(t) is the angular ve city vector of the rigid body at time t. In terms of the orthogonal basis

12.3 Rigid Body Theory

167

{e1 (t), e2 (t), e3 (t)} introduced in the previous paragraph, 3

ω(t) =

1 ∂ei (t) ei (t) × . 2 i=1 ∂t

The proof of (12.57) follows by direct differentiation of (12.50): V (X, t) = V (P, t) +

∂Q(t) (X − P ). ∂t

By using X − P = Q(t)Tt (φ (X) − t φ (P )) we obtain V (X, t) = V (P, t) +

∂Q(t) Q(t)T (φt (X) − φt (P )). ∂t

(12.58)

The tensor (∂Q(t)/∂t)Q(t)T appearing in this expression is anti-symmetric, which we conclude from  T ∂Q(t) ∂Q(t)T ∂Q(t) T T T Q(t) = −Q(t) =− Q(t) . QQ = I ⇒ ∂t ∂t ∂t Thus, by using (12.55) we find   ∂Q(t) 1 ∂Q(t) ∂Q(t)T T T Q(t) = Q(t) − Q ∂t 2 ∂t ∂t   3 ∂ei (t) 1 ∂ei (t) ⊗ ei (t) − ei (t) ⊗ = . (12.59) 2 i=1 ∂t ∂t By introducing this equation into (12.58) we get 3

1 V (X, t) = V (P, t) + 2 i=1



∂ei (t) [ei (t) · (φt (X) − φt (P ))] ∂t   ∂ei (t) − ei (t) · (φt (X) − φt (P )) . ∂t

Using now the vector identity (B.15) we arrive at (12.57). To interpret (12.57) three facts can be noted. (a) The second term in (12.57) vanishes when φt (X) belongs to the axis A = φt (P ) + αω, where α is some real number. (b) The absolute value of the relative velocity V (X, t) − V (P, t) is proportional to the perpendicular distance between A and φt (X). (c) V (X, t) − V (P, t) is directed perpendicularly to both φt (X) − φt (P ) and ω(t). These three facts are illustrated in Figure 12.6 and mean that the velocity of a rigid body can be seen as the sum of a velocity V (P, t) and the velocity produced by an instantaneous rotation around the axis A. It should be carefully noted that the directions of ω(t) and the vector N defined in the previous paragraph certainly do not coincide in general.

168

12 Kinematic Constraints, Beams and Rigid Bodies

Fig. 12.6. The relative velocity field V (X, t) − V (P, t) as seen in the plane orthogonal to ω(t).

12.3.2 Euler’s Second Law for a Rigid Body In rigid body mechanics we can obtain a complete problem that gives the velocity field from given forces without considering inner forces (represented by a stress tensor in the three-dimensional case). These forces, which are reaction forces related to the rigid body constraint (12.49), are removed by using Euler’s laws for the complete body B only. This may be thought of as using a projected equation in the same way as (12.5) and (12.10) were such projections. If in a particular case we are interested in inner forces, these can, of course, be somewhat characterized by anyway writing Euler’s laws for parts of bodies, but this will not give us a unique determination. Since we will use Euler’s second law only for the complete body, we will subsequently use the notations co (t) = co (B) and ho (t) = ho (B). This law now reads ∂ co (t) = ho (t), (12.60) ∂t where the angular momentum with respect to o was defined in Chapter 7 for the three-dimensional base model as:  ho (t) = (φt (X, t) − o) × V (X, t)ρ0 (X) dV VX . B

We need to substitute the velocity field of equation (12.57) into this expression. For simplicity, however, it will be assumed that the motion is such that the body is fixed at the origin o, i.e., p(t) = o for all times t, meaning that V (P, t) = 0 and therefore that V (X, t) = ω(t) × (φt (X) − o),

12.3 Rigid Body Theory

169

which when substituted into the expression for the angular momentum gives  ho (t) = (φt (X, t) − o) × [ω(t) × (φt (X) − o)]ρ0 (X) dV VX . (12.61) B

By using (B.15) we find that (12.61) can be written as ho (t) = J o (t)ω(t), where J o (t) =

 B

(12.62)

! |φt (X, t) − o|2 I − (φt (X, t) − o) ⊗ (φt (X, t) − o) ρ0 (X) dV VX

is the inertia tensor relative to the origin o. Substituting (12.62) into (12.60) gives the equation of motion for the rigid body. However, since J o (t) depends on time through the placement φt (X) it is somewhat inconvenient to use. We are led to a more useful form by substituting φt (X) − o = Q(t)(X − o) into the expression for the inertia tensor. This gives J o (t) = Q(t)J o (B)Q(t)T , (12.63) 

where J o (B) =

B

! |X − o|2 I − (X − o) ⊗ (X − o) ρ0 (X) dV VX

is the body inertia tensor, which is time independent. Equation (12.63) was derived by using (C.8) and (C.12). Equation (12.62) can now be written as H o (t) = J o (B)Ω(t),

(12.64)

where H o (t) = Q(t)T ho (t) is the body angular momentum and Ω(t) = Q(t)T ω(t) is the body angular velocity. Depending on how we view the tensor Q(t) – as a mapping between observers or configurations – as discussed in Section 12.3.1, these vectors can be thought of in different ways. Either as vectors rotated back to the reference configuration B or as the same objects as the angular momentum and the angular velocity, but viewed by an observer regarding the body as its fixed reference frame. Similarly, the body inertia tensor J o (B) can be seen as the inertia tensor as it is conceived by such a body fixed observer. We would like to write the Euler’s law (12.60) in body tensors and vectors. To that end, we rewrite its right hand side by using (12.63) and (12.64): ∂ ∂ ho (t) = [Q(t)J o (B)Ω(t)] ∂t ∂t ∂Q(t) ∂Ω(t) = Q(t)T Q(t)J o (B)Ω(t) + Q(t)J o (B) . (12.65) ∂t ∂t The first term of the right hand side of this expression can be rewritten by using (12.59) and (B.15), i.e.,

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12 Kinematic Constraints, Beams and Rigid Bodies

∂Q(t) Q(t)T Q(t)J o (B)Ω(t) ∂t = ω(t) × Q(t)J o (B)Ω(t) = Q(t)[Ω(t) × J o (B)Ω(t)],

(12.66)

where we have also used that for any orthogonal tensor Q and any vectors a and b it holds that2 Q(a × b) = (Qa) × (Qb). By using (12.65) and (12.66) in (12.60) we finally find ∂Ω(t) + Ω(t) × J o (B)Ω(t), (12.67) ∂t where we have defined the body torque with respect to o as C o (t) = Q(t)T co (t). We can now state the following complete model: C o (t) = J o (B)

Rigid body motion for a fixed point: Given a body torque C o (t) as a function of time and initial conditions for the body angular velocity, i.e., ¯ for a given vector Ω, ¯ find the time history of Ω(t) such that Ω(0) = Ω (12.67) is satisfied. That the mathematical problem of this model is well-posed follows from general theory for ordinary differential equations. Note that the usefulness of the above derivation is somewhat larger than may appear at first. By using the spin equation of motion (4.17) we can obtain an equation identical to (12.67) but stated for the center of mass xc instead of a fixed origin o. Such an equation is valid for a body that is free in space. We will subsequently study this problem for the case when the body torque is zero, but first we study some properties of the body inertia tensor. Using (B.6) and (B.15) we find that for any vectors a and b,  a · J o (B)b = [a × (X − o)] · [b × (X − o)]ρ0 (X) dV VX . (12.68) B

From this expression and the definition of the transpose of a tensor, (C.9), we find that J o (B) = J o (B)T , i.e., the body inertia tensor is symmetric. Moreover, since a scalar product of two identical vectors is positive, it follows that a · J o (B)a ≥ 0 for all vectors a. This is to say that J o (B) is positive semidefinite. In fact, an even sharper result follows from the conclusion that a·J o (B)a is zero for a = 0 only if a × (X − o) is zero for all X ∈ B, which is impossible unless B is one-dimensional and this situation is excluded. Thus, J o (B) is also positive definite. A symmetric tensor possesses three eigenvalues, and if it is positive definite these eigenvalues are positive. The eigenvalues of J o (B) are called principal moments of inertia and the corresponding orthogonal eigenvectors define the principal axes in the body frame. Denoting the principal moments of inertia by J1 , J2 and J3 we, thus, have that J o (B) =

3

Ji E i ⊗ E i ,

i=1 2

This is to say that an orthogonal tensor is its own adjugate.

(12.69)

12.3 Rigid Body Theory

171

if {E 1 , E 2 , E 3 } is oriented with the principle axes. From (12.69) we conclude that det J o (B) = J1 J2 J3 > 0 so J o (B) is invertible, which holds for any positive definite tensor. This means that there is an alternative statement of equation (12.67) in terms of the body angular momentum, i.e. from (12.64) and the invertibility of J o (B) we conclude that (12.67) can be written as C o (t) =

∂H o (t) + J o (B)−1 H o (t) × H o (t). ∂t

(12.70)

This leads to an alternative statement of the problem of rigid body motion for a fixed point, in terms of H o (t) as an unknown vector function. The Torque Free Problem We now look at the problem of free motion of a rigid body with a fixed point, i.e., we will set C o (t) = 0. For this case, solutions of (12.67) or alternatively (12.70) can be characterized by means of two conservation properties. The first one follows directly from the conclusion that if C o (t) = 0, then co (t) is also zero, and (12.60) gives that ho (t) is fixed vector in space. Therefore, from the orthogonality of Q(t) we have that h2 = |ho (t)|2 = |H o (t)|2 is constant for all times.

(12.71)

Thus, solutions of (12.70), when C (to) = 0, are constrained to a sphere in the vector space of body angular momentum vectors. Next, we show that another quantity that is conserved is the kinetic energy, which is defined as 1 1 ho (t) · ω(t) = H o (t) · Ω(t) 2 2 1 1 1 = ω(t) · J o (t)ω(t) = Ω(t) · J o (B)Ω(t) = H o (t) · J o (B)−1 H o (t), 2 2 2

K(t) =

where equivalence of different expressions follows from (12.62), (12.64) and the orthogonality of Q(t). The following chain of identities follows from (12.64), (12.67), (12.70), (C.9), the symmetry of J o (B)−1 and the orthogonality property of vector products, i.e., from Axiom 14 of Appendix B: 2

∂K(t) ∂[H o (t) · Ω(t)] = ∂t ∂t ∂H o (t) ∂Ω(t) = · Ω(t) + H o (t) · = [C o (t) − Ω(t) × H o (t)] · Ω(t) ∂t ∂t −1 + H o (t) · J o (B) [C o (t) − Ω(t) × H o (t)] = 2C o (t) · Ω(t).

Thus, for the torque free case the kinetic energy K = K(t) is time independent and, in addition to (12.71), we have that the body angular momentum vector H o (t) must be such that

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12 Kinematic Constraints, Beams and Rigid Bodies

2K = H o (t) · J o (B)−1 H o (t) is constant for all times.

(12.72)

Since J o (B)−1 is positive definite, (12.72) defines an ellipsoidal surface in the space of body angular momentum vectors. Thus, solutions of (12.70) must belong to the intersection between the sphere defined by (12.71) and the ellipsoid defined by (12.72). enables us to visualize solutions of the zero torque rigid body motion problem. To that end, note that the two constants K and h cannot be completely independent. Let the eigenvectors of J o (B) be arranged so that J3 ≤ J2 ≤ J1 . Since J o (B)−1 has eigenvalues Ji−1 (i = 1, 2, 3) and is positive definite, J1−1 |H o (t)|2 ≤ H o (t) · J o (B)−1 H o (t) ≤ J3−1 |H o (t)|2 . From these inequalities, (12.71) and (12.72), it follows that   2KJ J3 ≤ h ≤ 2KJ J1 .

(12.73)

We now think of K, i.e., the ellipsoid, as fixed and vary the value of h, i.e., the radius of the sphere, within the limits of (12.73). We then assume that √ the eigenvalues are strictly separated, i.e., J3 < J2 < J1 . If h = 2KJ J3 , the sphere lies inside the ellipsoid, and the intersection consists of two points at the √ ends of the√smallest semiaxis of the ellipsoid. If h is increased so that 2KJ J3 < h < 2KJ J2 , the intersection forms√two closed curves around the J1 , the sphere lies outside end of the smallest semiaxis. Similarly, if h = 2KJ the ellipsoid, with intersection at the two points forming the ends of the √ √ J2 < h < √2KJ J1 the intersection forms two largest semiaxis, and for 2KJ closed curves around these points. If h = 2KJ J2 the intersection consists of two circles that, in turn, intersect at the end of the middle semiaxis. In Figure 12.7 the fixed ellipsoid with intersecting curves is shown. To further understand the motion of the torque free rigid body problem we may investigate which stationary solutions are possible, i.e., solutions of (12.67) or (12.70) that are time independent. Such solutions satisfy J o (B)−1 H o (t) × H o (t) = 0, meaning that J o (B)−1 H o (t) and H o (t) should be parallel. If the principal moments of inertia are separated, i.e., J3 < J2 < J1 , this holds if and only if H o (t) = Hoi E i ,

i = 1, 2, 3,

(12.74)

for some values of Hoi . Clearly (12.74) implies that Ω(t) = Ωi E i (i=1,2,3) for some values Ωi and also that ω(t) is independent of time. From the geometric constructions of solutions as intersections between a sphere and an ellipsoid it is possible to understand that the solutions given in (12.74) are stable if i = 1, 3 and unstable for i = 2. The reason is that a small deviation from solutions at the end of the middle semiaxis (i = 2) will result in large perturbations,

12.3 Rigid Body Theory

173

Fig. 12.7. Intersections between the sphere, defined by (12.71), and the ellipsoid, defined by (12.72), shown as curves on the fixed ellipsoid.

while perturbations at the ends of the other semiaxis will only lead to small closed curves. Since the kinetic energy has different expressions, the condition (12.72) can be given in different forms. For instance, 2K = Ω · J o (B)Ω, defines an ellipsoid in the space of body angular velocity vectors Ω called the body ellipsoid of inertia, and 2K = ω · J o (t)ω, defines an ellipsoid in the space of angular velocity vectors ω called the space ellipsoid of inertia. Both of these objects may be seen as rigidly attached to the body. They describe the same inertia properties as seen from different observers. The space fixed observer views the space ellipsoid of inertia as moving in space with the body having one point fixed. A classical construction of Poinsot enables us to somehow characterize this motion, see Exercise 12.7.

Exercises 12.1. Verify the claim in Section 12.1.2 that a constraint of inextensibility can be derived from a transversely isotropic material.

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12 Kinematic Constraints, Beams and Rigid Bodies

12.2. Show that the Timoshenko displacement field (12.42) is equivalent to (12.33) and (12.43). 12.3. Prove that invariance of lengths of material line elements, i.e., (12.49), implies invariance of angles between such elements. (Hint: use the cosine rule.) Use this fact to show that [φt (X) − φt (P )] · [φt (Y ) − φt (P )] = (X − P ) · (Y − P )

(12.75)

for arbitrary but distinct points X, Y and P of B. This means that the scalar product of material line elements is preserved. 12.4. Define an operator r that takes vectors to vectors by the identity r(X) = φt (X) − φt (P ), where X = X − P . Use (12.75) to show that r(αX + βY ) · r(Z) = (αr(X) + βr(Y )) · r(Z)

(12.76)

for arbitrary constants α and β, and position vectors X = X − P , Y = Y − P and Z = Z − P . Since φt is one-to-one, (12.76) implies that r(αX + βY ) = αr(X) + βr(Y ), i.e., r is linear. 12.5. Since the operator r in the previous exercise is linear it can be represented by a tensor Q such that r(X) = QX,

X = X − P.

Use (12.52) to show that Q is orthogonal. ˆ 1, E ˆ 2, E ˆ 3 }, are re12.6. Any two orthogonal bases, say {E 1 , E 2 , E 3 } and {E ˆ lated as E i = M E i , (i = 1, 2, 3), where M is an orthogonal tensor. Show that for the orthogonal tensor in (12.55), Q(t) =

3

ei (t) ⊗ E i =

i=1

3

ˆ i, ˆ i (t) ⊗ E e

i=1

ˆi (t) = Q(t)M Q(t)T ei (t). where e 12.7. Poinsot’s theorem reads: “The space ellipsoid of inertia rolls without slipping on the plane that has the constant vector ho (t) as normal”. Verify this!

A Sets and Functions

Every student of engineering frequently comes across the word function and as well as the words mapping and operator. Most likely he/she has seen the mathematical definition of a function in an introductory mathematics course and perhaps also been told that mapping and operator are basically synonymous terms. Nevertheless, since in applied mathematics, mechanics and engineering, the term function is mostly (or at least frequently) used in a slightly vague sense, connected to a traditional style of notation that may be confusing (see below), it seems as if the more precise concept of a mathematical function becomes diffuse to many students. While I do not object to this common use of the notion of a function – it may be economical – I do believe that it is very useful to constantly have the exact mathematical definition in the background. Therefore, we will give such a definition together with the prerequisite notion of a set.

A.1 The Notion of a Set A set is a collection of objects. In mathematics, an object is frequently a natural or real number, a vector or point (see Appendix B), or a function (yet to be defined). When used in a physical model these objects will usually have referents in the real world. The objects are said to be members or elements of the set. In general, a set is denoted by a letter, say A, and if an object x is a member of A we write x∈A and the symbol ∈ should be read as “is an element of” or “belongs to”. We can define a particular set by simply listing all of its elements. Usually curly brackets are used for this. Say that A is the set of all natural numbers from 1 to 4, then A = {1, 2, 3, 4}. (A.1)

175

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A Sets and Functions

Another way of defining a set is by giving rules that identify its elements. For instance, the set (A.1) could be written A = {x : x is an integer, 1 ≤ x ≤ 4}. The colon should be read as “such that” and the comma “and”. However, in this text we never use this type of definition for a set. Furthermore, there are some symbols that can be taken as predefined definitions of sets. For instance, the symbol R is the set of real numbers, and R2 is the set of pairs of real numbers. Similarly, Rn is the set of n-tuples of real numbers. If A and B are two sets, we say that A is a subset of B if each element of A is also in B. This is denoted by the symbol ⊂ and we write A ⊂ B. The explicitly defined sets that we are using in this text are (i) the physical Euclidean space E and subsets of this space, and (ii) bodies, denoted B, whose elements are material points; depending on what body model is considered, the material points are natural numbers, real numbers, pairs of real numbers or points in a subset of E. Clearly, other sets, for instance, sets (sometimes called classes) of continuous functions, are alluded to in this text, but I have not felt a necessity to introduce particular notations for these sets.

A.2 The Notion of a Function A function is a rule that associates each element in a set A to an element in a set B. We write f :A→B and think of f as the rule. If y ∈ B is the element that is associated to x ∈ A we also write y = f (x) and y is called the image of x under f or the value of f at x. The set A is the domain and B the co-domain of the function. The subset of B consisting of all the images under f is called the range of the function, denoted f (A). Figure A.1 gives an illustration of the concept of a function. Note that a function consists of three objects: the rule f , the domain A and the co-domain B. However, a common practice is to talk about f as the function and let A and B be understood from the context. It is also fairly common to call f (x) the function, which following the scheme introduced here is far from correct since f (x) is not even a rule but rather an element of the co-domain. A reason for this usage is that it gives a shorthand for indicating which symbol is used for elements of the domain. A further common usage,

The Notion of a Function

177

Fig. A.1. The concept of a function

which may be confusing, but is sometimes useful, is to use the same symbol for both the rule and the image, i.e., one writes y = y(x).

B Euclidean Point and Vector Spaces

Many facts of mechanics are formulated by means of three-dimensional vectors. Examples of such vectors are position vectors, velocity vectors, acceleration vectors and force vectors. We call these vectors geometric vectors and notationally they are distinguished from other mathematical objects, such as scalars and points, by use of bold face letters. In the overwhelming majority of textbooks in mechanics and algebra, geometric vectors are conceived (or even defined) as directed line segments. This is a very useful mental image of a vector that gives considerable insight and intuitive feeling to problems of mechanics and other scientific fields. However, it should be realized that such an image is just but one possible interpretation of the mathematical structure represented by vectors and the operations related to vectors. In terms of the view of physical theories presented in Chapter 1, the directed line segment (in the physical sense of the word) is a real world referent of the mathematically structure defined within the conceptual world. In the following we define an axiomatic structure that can stand by itself, irrespective of the interpretation or which physical referent is chosen. The presentation can be seen as a complement to the treatment of most beginners’ textbooks. In fact, the common intuitive definitions of geometric vectors and the space in which mechanics takes place, should be sufficient for the understanding most of this text. On the other hand, the insight on how models of mechanics (and in this case geometry) interplay with the real world should be strengthened by the following sections. Besides, it ties up some loose ends usually identified by serious student of mechanics. The presentation starts by defining Euclidean vector space and then using this structure to define Euclidean point space. Thus, vectors comes before points, which is in contrast to the elementary treatment where points and space are assumed at the outset. The presentation is inspired by Chadwick [6] and the appendix in Trusdell [17]. Other useful references are Bowen and Wang [3, 4] and Uhlhorn [20].

179

180

B Euclidean Point and Vector Spaces

B.1 Euclidean Vector Space From a mathematical point of view geometric vectors are elements of an oriented three-dimensional Euclidean vector space. We will introduce this mathematical structure in a step-wise fashion starting with axioms for addition, and scalar multiples, then adding the scalar product axioms and finally, introducing the vector or cross product which leads to the idea of orientation. A vector space V over the real numbers R (real numbers are typically denoted α and β) is a set of vectors (vectors are typically denoted u, v and w) and the following rules or axioms: Addition axioms To every pair of vectors u and v there corresponds a vector u + v, called the sum, with the properties: 1. u + v = v + u (commutative property). 2. u + (v + w) = (u + v) + w (associative property). 3. There exists a vector 0 such that u + 0 = u for all u ∈ V. 4. For every u ∈ V there exists −u ∈ V such that u + (−u) = 0. Scalar multiple axioms To every vector u and real number α there corresponds a vector αu, called the product of u and α, with the properties: 5. 1u = u. 6. α(βu) = (αβ)u (associative property). 7. (α + β)u = αu + βu (distributive property). 8. α(u + v) = αu + αv (distributive property). Note that the +’s in 7 represent different operations. Usually we write u − v for u + (−v) and it holds that −u = (−1)u. A set of n vectors {v 1 , v 2 , . . . , v n } is linearly independent if there is no linear combination which equals to zero except the trivial one, i.e., α1 v 1 + α2 v 2 + . . . + αn v n = 0 is true only for α1 = α2 = . . . = αn = 0. A set of vectors which is not linearly independent is called linearly dependent. If there exists at least one set of n linearly independent vectors in V, but no such set with n + 1 vectors, then V is said to be n-dimensional. In an n-dimensional vector space V, a set of n linearly independent vectors is called a basis. Let {b1 , b2 , . . . , bn } be such a basis. It can then be shown that any vector u ∈ V can be uniquely represented in the basis, i.e., there are unique real numbers {α1 , α2 , . . . , αn } such that u = α1 b1 + α2 b2 + . . . + αn bn =

n

αi bi .

i=1

There are many referents of the mathematical structure of a vector space in the real world. The reader may verify that straight directed line segments1 1

To be more precise, equivalence classes of directed line segments form a vector space.

B.1 Euclidean Vector Space

181

drawn on a paper are physical referents of two-dimensional vectors, given that the parallelogram law represents addition and the product operation is taken as a scaling (with direction) of the directed line segments. This interpretation is carried over to the three-dimensional case in an obvious way, perhaps by thinking of vectors as rigid thin rods. A vector space becomes an Euclidean vector space if, in addition to the operations of addition and scalar multiple introduced above, there exists a scalar product as follows: Scalar product axioms To every pair of vectors u and v there corresponds a real number u · v, called the scalar product of u and v, with the properties: 9. u · v = v · u (commutative property). 10. (αu + βv) · w = α(u · w) + β(v · w) (linearity). 11. u · u ≥ 0, and u · u = 0 if and only if u = 0 (positive-definiteness). Two vectors u and v are said to be orthogonal if u · v = 0; the magnitude (or absolute value) of a vector u, denoted |u|, is defined by √ |u| = u · u. A consequence of the scalar product axioms is Cauchy’s inequality |u|2 |v|2 ≥ (u · v)2 , where equality holds if and only if u and v are linearly dependent. This inequality allows us to introduce the notion of angle between vectors, denoted by φ, which is defined by the equation u · v = |u||v| cos φ,

0 ≤ φ ≤ π.

(B.1)

In a Euclidean vector space we distinguish a particular class of bases, namely the orthonormal bases. A basis {e1 , e2 , . . . , en } is orthonormal if ei · ej = δij ,

i, j = 1, . . . , n,

(B.2)

where δij is Kronecker’s delta, which takes the value 1 when i = j, and 0 when i = j. In the following we will deal with three-dimensional spaces only; a vector u can be written as u = u1 e1 + u2 e2 + u3 e3 =

3

u i ei ,

(B.3)

i=1

where (u1 , u2 , u3 ) are the components of u relative to the orthonormal basis {e1 , e2 , e3 }. These components can be expressed as u i = u · ei ,

i = 1, 2, 3,

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B Euclidean Point and Vector Spaces

which is found from (B.3) by using (B.2). Thus, we can write equation (B.3) as 3 u= (u · ei )ei . (B.4) i=1

The representation (B.3) can be used to calculate the scalar product between two vectors. Let u and v have the components (u1 , u2 , u3 ) and (v1 , v2 , v3 ), respectively. Then, by using (B.2) and Axiom 10, one finds that  u·v =

3

 ⎛ u i ei

·⎝

i=1

3

⎞ vj ej ⎠

j=1

=

3 3 i=1 j=1

ui vj ei · ej =

3 3 i=1 j=1

ui vj δij =

3

ui vi . (B.5)

i=1

This formula shows that the scalar product is uniquely defined by the axioms introduced thus far. We will now add the vector or cross product operation to our Euclidean vector space. Vector product axioms To every pair of vectors u and v there corresponds a vector u × v, called the vector product of u and v, with the properties: 12. u × v = −v × u (anti-commutative property). 13. (αu + βv) × w = α(u × w) + β(v × w) (linearity). 14. u · (u × v) = 0. 15. (u × v) · (u × v) = (u · u)(v · v) − (u · v)2 .

Axiom 14 expresses that the vector u × v is orthogonal to both u and v. By using (B.1) one finds that Axiom 15 can be rewritten as |u × v| = |u||v| sin φ. Thus, when thinking of vectors as oriented line segments, we see that the absolute value of the vector product u × v represents the area of the parallelogram spanned by u and v. The triple scalar product of three vectors u, v and w is defined as u · (v × w). A consequence of the vector product axioms is that the following holds: u · (v × w) = v · (w × u) = w · (u × v) = −u · (w × v) = −w · (v × u) = −v · (u × w),

(B.6)

which will be used to show the following important results concerning the relations between vectors in an orthonormal base {e1 , e2 , e3 }: e1 × e2 = ±e3 ,

e3 × e1 = ±e2 ,

e2 × e3 = ±e1 .

(B.7)

B.1 Euclidean Vector Space

183

As will be commented on below, the ± sign indicates an indeterminancy in the vector product as defined by the axioms. To prove (B.7) we first conclude from Axiom 15 that |e1 × e2 |2 = 1,

|e3 × e1 |2 = 1,

|e2 × e3 |2 = 1.

(B.8)

Furthermore, from equation (B.4), Axiom 14 and (B.6) one finds e2 × e3 =

3

((e2 × e3 ) · ei )ei = ((e2 × e3 ) · e1 )e1 = ((e1 × e2 ) · e3 )e1 , (B.9)

i=1

and, similarly, e1 × e2 = ((e1 × e2 ) · e3 )e3 ,

e1 × e2 = ((e1 × e2 ) · e3 )e3 .

(B.10)

Thus, comparing (B.9) and (B.10) with (B.8) one finds (e1 × e2 ) · e3 = ±1 and equations (B.7), which we attempted to prove, follow from (B.9) and (B.10). Using Axiom 12 and equations (B.7) one finds that the following holds true: 3 ei × ej = ± εijk ek , i, j = 1, 2, 3, (B.11) k=1

where εijk is the permutation symbol, which takes the value 1 when i, j and k is a cyclic permutation of 1, 2 and 3; the value -1 when i, j and k is a non-cyclic permutation of 1, 2 and 3; and zero otherwise. The representation (B.3) was used above in (B.5) to calculate the scalar product. It will be used also to calculate the vector product: from (B.3), (B.11) and Axioms 13 and 14 one finds ⎞  ⎛ 3  3 ui ei × ⎝ vj ej ⎠ u×v = i=1

j=1

=

3 3 i=1 j=1

ui vj ei × ej = ±

3 3 3

εijk ui vj ek . (B.12)

i=1 j=1 k=1

It also follows from this result and (B.5) that the triple scalar product can be written as    u1 u2 u3  3 3 3   (B.13) u · (v × w) = ± εijk ui vj wk = ± det  v1 v2 v3  .  w1 w2 w3  i=1 j=1 k=1 The ± sign in the last two equations indicates that the axioms do not uniquely determine the vector product, contrary to the case with the scalar product.

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B Euclidean Point and Vector Spaces

However, the indeterminacy is not dramatic: there are, in a sense, two vector products obeying the axioms, one being the negative of the other. We resolve the indeterminacy by dividing the set of orthonormal bases into two classes, the right-handed and the left-handed bases. We take a particular base, ¯2 , e ¯3 }, and require that for this base it holds that say {¯ e1 , e ¯j = ¯i × e e

3

¯k , εijk e

i, j = 1, 2, 3,

(B.14)

k=1

i.e., we take the plus sign in (B.11). All other bases can now be compared to this base by calculating its triple product. From (B.4) we have for any base {e1 , e2 , e3 }: 3 ¯j )¯ ei = (ei · e ej , i = 1, 2, 3. i=1

By using this equation, (B.14) and linearity in both scalar and vector products, i.e., Axioms 10 and 13, one finds that (e1 × e2 ) · e3 =

3 3 3

¯j )(e3 · e ¯k ). ¯i )(e2 · e εijk (e1 · e

i=1 j=1 k=1

As has been found above, this triple scalar product equals 1 or -1. If it is 1, we ¯2 , e ¯3 }, the righttake {e1 , e2 , e3 } to belong to the same class of bases as {¯ e1 , e handed bases, and we use the + sign in (B.12) when calculating the vector product. If the product is −1, {e1, e2, e 3} belongs to the class of left-handed bases and we use the − sign in (B.12). The definitions of right-handedness and left-handedness are extended to all linearly independent sets of three vectors by looking at the sign of its triple scalar product. The referents of left-handed and right-handed bases in the real world, when the directed line segment interpretation is chosen, should be well known from elementary courses: for a right-handed base e1 , e2 and e3 correspond to the thumb, the index finger and the middle finger of the right hand, respectively. A three-dimensional Euclidean vector space with the operation of vector product and definitions of right-handedness and left-handedness is an oriented three-dimensional Euclidean vector space. A further useful connection between the operations of scalar and vector products can be established. We may arrive at this by recalling the so-called ε-δ identity 3 εijk εist = δjs δkt − δjt δks , i=1

which can be verified by direct trial. By using (B.5) and (B.12) we find through this identity that any vectors u, v and w satisfy u × (v × w) = (u · w)v − (u · v)w.

(B.15)

B.2 Euclidean Point Space

185

Note that this identity does not need the definition of right-handedness since minus signs cancel out on the left hand side.

B.2 Euclidean Point Space Geometric vectors viewed as directed line segments are conceived as connecting points in space. Thus, we seem to think that it is obvious that there is a space of points in which mechanics goes on and this space is frequently called physical space. In the following we will define this space mathematically. We will then use the structure of Euclidean vector space defined in the previous subsection and the result will be a Euclidean point space. A Euclidean point space E is a set of points (points are typically denoted x, y and z) which is related to a Euclidean vector space V in the following way: Every pair of points (x, y) in E defines a vector v ∈ V, called the vector from x to y, which is denoted y − x, i.e. v = y − x. (ii) y − x = (y − z) + (z − x). (iii) Given an arbitrary point o ∈ E and vector r ∈ V there exists a unique point x ∈ E such that r = x − o. (i)

The point o and the vector r in Axiom (iii) may be called the origin and the position vector, respectively. The space V is known as the translation space of E. Axiom (ii) is the parallelogram law for addition of vectors. Axiom (iii) says that given an origin o there is a one-to-one correspondence between position vectors and points. Therefore, we may say that the dimension of the point space equals that of the corresponding vector space. The distance between points x and y is the magnitude of the corresponding vector, i.e. |x − y|. Furthermore, Axiom (iii) indicates that points and vectors may be added to give a point, i.e. we may write x = u + y. Note, however, that the addition of two points is not a meaningful concept. The following results are proved from the axioms: x − x = o,

x − y = −(y − x).

C Tensors and Some Mathematical Background

In the following we will introduce the concept of a tensor, and present those results from algebra and analysis that are used in this book. The aim is to make the presentation reasonably self-contained, but readers are recommended to consult a more complete mathematical textbook if they are encountering these notions for the first time. Note, however, that the mathematics presented here is mostly needed for the derivation and analysis of the three-dimensional model. The presentations of the discrete and the one-dimensional models can be followed by using elementary mathematics of geometric vectors.

C.1 Algebra In this text a tensor will be a linear mapping from a three-dimensional Euclidean vector space to the same space, i.e., a mapping that linearly assigns a vector v to any vector u. That a mapping f is linear means that f (α1 u1 + α2 u2 ) = α1 f (u1 ) + α2 f (u2 )

(C.1)

for all vectors u1 and u2 , and all real numbers α1 and α2 . We frequently write the action of a tensor T on a vector u without the bracket notation that is used for general functions, i.e., T assigns to each vector u a vector v = T u. (C.2) Note that since vectors are geometric objects that are independent of any particular coordinate system or base, it is clear from (C.2) that tensors also have this property. This is the reason why it is logical to represent tensors by bold face notation. However, given an orthonormal base {e1 , e2 , e3 }, tensor components can be derived and used to represent tensors. To show this, we start by representing the two vectors u and v in (C.2) in components:

187

188

C Tensors and Some Mathematical Background

v = v1 e1 + v2 e2 + v3 e3 ,

u = u1 e1 + u2 e2 + u3 e3 ,

and we find from (C.2) that v1 e1 + v2 e2 + v3 e3 = T (u1 e1 + u2 e2 + u3 e3 ). By using (B.2) and the linearity of T we find vi =

3

(ei · T ej )uj ,

i = 1, 2, 3.

j=1

Thus, by introducing Tij = ei · T ej ,

i, j = 1, 2, 3,

(C.3)

we may write vi =

3

Tij uj ,

i = 1, 2, 3,

j=1

which in matrix form reads ⎡ ⎤ ⎡ ⎤⎡ ⎤ v1 T11 T12 T13 u1 ⎣ v2 ⎦ = ⎣ T21 T22 T23 ⎦ ⎣ u2 ⎦. v3 T31 T32 T33 u3

(C.4)

Equation (C.3) defines the components of T , and given this definition, (C.2) and (C.4) are equivalent equations. We may say that the matrix in (C.4) is a representation of T and write ⎡ ⎤ T11 T12 T13 T ∼ ⎣ T21 T22 T23 ⎦. T31 T32 T33 A similar notation for vectors would be ⎡ ⎤ u1 u ∼ ⎣ u2 ⎦. u3 Obviously, it is not strictly correct to use = instead of ∼ here, since matrix representations depend on the basis while the left hand sides of these relations do not. Since an orthonormal base is related to a cartesian coordinate system, the components Tij and ui are frequently called cartesian components. There are many definitions of tensors in the literature. The definition used here actually defines what is sometimes called a second-order tensor. In this terminology a vector would be a first-order tensor. A very common definition is based on satisfaction of rules for how tensor components transform under

C.1 Algebra

189

change of coordinate system. If these rules are satisfied, equations based on tensor components become form invariant to coordinate change. The present definition is automatically insensitive to coordinate change since vectors are coordinate independent. A member of a special class of tensors is formed from two vectors: the tensor (open or vector) product, u ⊗ v, of the two vectors u and v, is a tensor defined by the property that (u ⊗ v)w = (u · w)v for any vector w. From (C.3) it follows that the components of the tensor product u ⊗ v are ui vj (i, j = 1, 2, 3). It may be shown that any tensor T can be written as 3 3 T = Tij ei ⊗ ej . (C.5) i=1 j=1

This equation indicates that ei ⊗ ej (i, j = 1, 2, 3) may be seen as a basis for a space of tensors. As a particular case of (C.5) we have that the identity tensor I, defined by the property Iv = v for all vectors v, can be represented as I=

3

ei ⊗ ei .

(C.6)

i=1

The product ST of two tensors S and T is defined by (ST )v = S(T v) for all vectors v. In particular, for tensors a ⊗ b, c ⊗ d and T it holds that (a ⊗ b)(c ⊗ d) = (b · c)a ⊗ d,

(C.7)

T (a ⊗ b) = (T a) ⊗ b.

(C.8)

The trace of a tensor is denoted tr T and is a linear operator on tensors with the property tr (u ⊗ v) = u · v for all vectors u and v. By (C.5) and the linearity of tr we find tr T =

3

Tii .

i=1

That is, the trace of a tensor is the sum of its diagonal components in an orthonormal base. The transpose of a tensor T is denoted T T and is defined by Tu · v = u · TTv

(C.9)

190

C Tensors and Some Mathematical Background

for all vectors u and v. In terms of components we may write TT =

3 3

Tji ei ⊗ ej ,

i=1 j=1

and for the tensor product and arbitrary tensors S and T it holds that (u ⊗ v)T = v ⊗ u,

(C.10)

(ST )T = T T S T .

(C.11)

By using (C.8), (C.10) and (C.11) we find (u ⊗ c)T = u ⊗ (T T v).

(C.12)

A tensor Q is orthogonal if QQT = QT Q = I.

(C.13)

Thus, for an orthogonal tensor we get the inverse by forming the transpose. The determinant of a tensor is the determinant of any of its matrix representations, see Section 3.4. From (C.13) and standard properties of the determinant it follows that an orthogonal tensor has the determinant 1 or −1. The orthogonal tensors that have a determinant valued 1 are called proper orthogonal tensors. In Section 12.3 it was shown that proper orthogonal tensors represent rotations in the physical space.

C.2 Analysis In this section we consider fields over the Euclidean point space E. By a field we mean a mapping from (a subset of) E to a set of either scalars, points, vectors or tensors. Mechanics and thermomechanics contains many examples of such fields: (i)

The temperature inside a solid or fluid body (which represents a subset of E) varies from point to point and is a scalar field. (ii) The placement map φt for a three-dimensional body (see Chapter 3) maps a region B ⊂ E into E and is thus a point field. (iii) The velocity of a fluid or solid contained in a region of E is a typical vector field. (iv) Prime examples of tensor fields are stresses and strains inside a threedimensional body. We are generally interested in how fields vary over E. To measure this, the concept of a gradient is introduced. We are interested in gradients of scalar, point and vector fields. Let ϕ be a scalar field. The gradient of ϕ at x ∈ E is a vector field ∇ϕ(x) which satisfies

C.2 Analysis

191

ϕ(x + u) = ϕ(x) + ∇ϕ(x) · u + o(u), where o(u) is “small o of u” and is any function which approaches zero faster than its argument. The cartesian component representation of the gradient of a scalar field is ⎡ ∂ϕ ⎤ ⎢ ∂x1 ⎥ ⎥ ⎢ ⎢ ∂ϕ ⎥ ⎥. ⎢ ∇ϕ(x) ∼ ⎢ ⎥ ⎢ ∂x2 ⎥ ⎣ ∂ϕ ⎦ ∂x3 Next, let v be a vector field. The gradient of v at x ∈ E is a tensor field ∇v(x) which satisfies v(x + u) = v(x) + ∇v(x)u + o(u). The cartesian component representation of the gradient of a vector field is ⎡ ∂u ∂u ∂u ⎤ 1

1

1

∂x2 ∂u2 ∂x2 3 ∂u3 ∂x1 ∂x2

∂x3 ∂u2 ∂x3 ∂u3 ∂x3

⎢ ∂x1 ⎢ ⎢ ∂u2 ∇u(x) ∼ ⎢ ⎢ ∂x ⎢ 1 ⎣ ∂u

⎥ ⎥ ⎥ ⎥. ⎥ ⎥ ⎦

The gradient of a point field is defined as that of a vector field. A prime example is the deformation gradient F , see Chapter 3. Having defined the gradient of a vector field v, we can define the divergence of the same field. It is denoted div v, and is a scalar field defined by div v = tr ∇v. In cartesian components div v has a simple representation: div v =

∂v2 ∂v3 ∂v1 + + . ∂x1 ∂x2 ∂x3

We can also define the divergence of a tensor field T : div T is a vector field defined by (div T ) · v = div(T T v) for every constant vector v. In cartesian components it holds that ⎡ ∂T ∂T T12 ∂T T13 ⎤ T11 + + ⎢ ∂x1 ∂x2 ∂x3 ⎥ ⎥ ⎢ ⎢ ∂T T ∂T T ∂T T23 ⎥ 21 22 ⎥. ⎢ div T ∼ ⎢ + + ∂x2 ∂x3 ⎥ ⎥ ⎢ ∂x1 ⎣ ∂T T31 ∂T T32 ∂T T33 ⎦ + + ∂x1 ∂x2 ∂x3

192

C Tensors and Some Mathematical Background

The curl of a vector field v is denoted by curl v, and is a vector field with the property   ∇v − (∇v)T a = (curl v) × a (C.14) for every vector a. In cartesian components it holds that ⎡ ∂a

∂a2 ⎢ ∂x2 ∂x3 ⎢ ⎢ ∂a1 ∂a3 curl v ∼ ⎢ ⎢ ∂x3 − ∂x1 ⎢ ⎣ ∂a ∂a1 2 − ∂x1 ∂x2 3

−

⎤ ⎥ ⎥ ⎥ ⎥. ⎥ ⎥ ⎦

We also have the following useful expression for curl v in the orthonormal basis {e1 , e2 , e3 }: 3 3 3 ∂vk curl v = εijk ei . (C.15) ∂x j i=1 j=1 k=1

In the development of three-dimensional continuum models we sometimes need to take gradients, divergences and curls of products of scalar and vector fields. Therefore, we state some of the rules for such operations. Let ϕ be a scalar valued field and let u and v be two vector valued fields. Then div (ϕv) = ϕ div v + v · ∇ϕ,

(C.16)

∇(v · u) = (∇v)T u + (∇u)T v,

(C.17)

curl (ϕv) = ϕ curl v + ∇ϕ × v.

(C.18)

div (∇v)T = ∇div v,

(C.19)

We will also need that and that for any constant tensor A, ∇(Av) = A∇v.

(C.20)

The formulas (C.17) through (C.20) are easily established by writing them in equivalent cartesian component forms. For instance, in such components (C.17) reads 3 ∂(vi ui ) i=1

∂xj

=

3 3 ∂vi ∂ui ui + vi , ∂x ∂x j j i=1 i=1

j = 1, 2, 3,

which is obviously true from the elementary product rule for differentiation.

C.2 Analysis

193

C.2.1 Divergence theorems The physical meaning of the divergence operation is best understood from integral theorems (divergence theorems). Such theorems are given with reference to a region R ⊂ E with boundary ∂R. This region has to satisfy some regularity properties which are too technical to be stated here. There are different divergence theorems depending on whether we are considering scalar, vector or tensor fields and whether we are, e.g., considering scalar or vector products. We give four different versions in the following theorem. Theorem 1 Let ϕ, v and T be smooth scalar, vector and tensor fields on R, respectively. Then   ϕn dA = ∇ϕ dV, (C.21) ∂R

 

∂R

∂R



v · n dA =

R

div v dV,

(C.22)

curl v dV,

(C.23)



n × v dA =



∂R

R

 T n dA =

R

R

div T dV,

(C.24)

where n is the outward unit normal vector of ∂R. For the physical interpretation of (C.22) we may think of v as a velocity field. The scalar v · n dA is then the rate of volume of material entering R at a particular surface point, and the right hand side of (C.22) is the total rate of change of volume. Thus, the equality in (C.22) means that div v is a measure of this volume change, and since the region R can be made arbitrarily small, it is a local such measure. This fact can be appreciated by writing (C.22) (in a somewhat lose mathematical style) as  v · n dA = div v dV, for an infinitesimal volume element dV . A more precise indication of these facts are given in relation to the issue of isochoric (volume preserving) motions in Chapter 7. The divergence theorem (C.24) can be given a similar physical interpretation. As is clear in Chapter 7, if T is taken as the stress tensor, then T n represents the force per area on the surface element with unit outward normal n. By writing (C.24) for an infinitesimal element dV as  T n dA = div T dV, one concludes that divT is a local measure of the surface force on the element.

194

C Tensors and Some Mathematical Background

C.2.2 Localization theorems The universal principles of mechanics are global principles relating to a conglomerate of material points. However, complete models that we solve, analytically or numerically, are mostly of local character relating to each point. In the discrete model of mechanics, this transfer from global to local is trivial, but for continuum models more subtle mathematics is needed. Here we present two mathematical theorems that have the required function. The first is useful for the one-dimensional model while the second relates to the threedimensional one. Theorem 2 Let f (x) be a continuous function on an interval [c, d] of the real line. If  b f (x) dx = 0 a

for all a and b such that c ≤ a < b ≤ d, then, for all x in the interval [c, d], then f (x) = 0. Theorem 3 Let f (x) be a continuous function on a subset R of E. If  f (x) dV = 0 Ω

for every Ω ⊂ R, then f (x) = 0. The term continuous function should here be understood as a scalar, vector or tensor field. C.2.3 Fundamental theorems of variational calculus This text does not in general concern variational formulations of the problems encountered. Such formulations are seen as steps in the numerical modeling that naturally comes after the initial formulation of a complete problem and which we do not treat. However, an exception is Chapter 9 on small displacement theories, where work principles, of variational nature, are used and discussed. To derive consequences of work principles one generally needs the notion of separating duality which is classically related to the so-called fundamental theorem of variational calculus. Below we give two special versions of this theorem: the first one is useful for one-dimensional models and the second for three-dimensional ones. Theorem 4 Let f (x) be a continuous function on an interval [a, b] of the real line. If  b f (x)g(x) dx = 0 a

C.2 Analysis

195

for all continuous functions g(x), then, f (x) = 0. Theorem 5 Let S(x) be a continuous field of symmetric tensors on a subset R of E. If  S(x) : T (x) dV = 0 R

for all continuous fields of symmetric tensors T (x), then S(x) = 0.

Exercises C.1. Let S and T be tensors, where S is symmetric and T is arbitrary. Show that 1 S : T = S : (T + T T ). 2 C.2. Prove the identities (C.16) through (C.20). C.3. Prove the identities (11.9) and (11.11). Identity (11.9) follows from the definition of curl, i.e., (C.14), and (C.17). Identity (11.11) follows directly from (C.19).

D Note on Physical Dimensions

In Appendix B we defined the vector space V as a vector space over the real numbers. This means that distances in E are real numbers, which may seem somewhat in contrast to the convention that distance, as an experimentally measured quantity, has a physical dimension ([length]) and is not purely a numerical value (real number). However, if one regards the unit of length as fixed, it is clearly sufficient to give the numerical value of a distance to characterize it, and this is a motivation for the choice of structure of Appendix B. Similar arguments hold for other concepts such as force, velocity, acceleration and mass: it is assumed that units of length, time, mass and force are fixed, allowing velocity, acceleration and force to be considered as belonging to the same oriented three-dimensional Euclidean vector space, denoted by V in this text. This is also the reason that vector and scalar products between velocity vectors and force vectors are valid operations. Note also in this context that units of length, time, mass and force must be connected so as to make Euler’s laws dimensionally correct, i.e., [force] = [mass][acceleration] = [mass][length][time]−2 , where the square brackets indicate physical dimension.

197

E Notation

General conventions are that • • •

vectors and tensors are shown in bold face italic letters – examples: v, T ; points and scalars are shown in light face letters – examples: x, y; vector spaces, point spaces, bodies and parts of bodies are shown in script letters – examples: E, B, P, V.

Meaning of Main Symbols A a α B Bt b C c c curl D det div dV VX dV Vx E E E e ε F

material acceleration spatial acceleration pipe cross section area body subset of E occupied by body at time t force per unit volume subset of E representing a pipe torque speed of sound curl rate of deformation determinant divergence material integration element spatial integration element physical space or space of generalized strains strain tensor or unit vector Young’s modulus unit vector strain components deformation gradient

199

200

E Notation

f φt ϕ h J κ L M mX m n o ω Ω P p p ψ Q q R R r ρ s s σ T t tr u U V v V v Vol X x xc ∇ ⊗ × · φ˙ φ

force placement rotation of beam cross-section angular momentum inertia tensor curvature and compression modulus velocity gradient mass mass of discrete material point X couple normal vector origin angular velocity body angular velocity part of body linear momentum pressure naturally parametrized curve orthogonal tensor force per unit length the real line control domain position vector mass per unit length or volume (density) traction vector natural coordinate stress component stress tensor time trace displacement vector force potential material velocity spatial velocity material speed spatial speed volume material point belonging to B spatial point belonging to E center of mass gradient tensor product vector product scalar product material time derivative of φ spatial time derivative of φ

References

1. Antman, S.S. (1995) Nonlinear Problems of Elasticity, Springer, New York. 2. Bendsœ, M.P. and Sigmund, O. (2003) Topology Optimization: Theory, Methods and Applications, Springer, Berlin. 3. Bowen, R.M. and Wang, C.-C. (1976) Introduction to Vectors and Tensors, Volume 1: Linear and Multilinear Algebra, Plenum Press, New York. 4. Bowen, R.M. and Wang, C.-C. (1976) Introduction to Vectors and Tensors, Volume 2: Vector and Tensor Analysis, Plenum Press, New York. 5. Gurtin, M.E. (1981) An Introduction to Continuum Mechanics, Academic Press, Orlando. 6. Chadwick, P. (1976) Continuum Mechanics. Concise Theory and Problems, George Allen & Unwin Ltd, London. 7. Curnier, A. (2004) M´ Mecanique des solides d´ M ´eformables, Presses polytechniques et universitaires romandes, Lausanne. 8. Fefferman, C.L. (2000) Existence and smoothness of Navier-Stokes equation, http://www.claymath.org/millennium/. 9. Fox, E.A. (1967) Mechanics, Harper & Row, New York. 10. Hestenes, D. (1992) Modeling games in the newtonian world, Am. J. Phys. 60(8), pp. 732–748. 11. Jos´ e, J.V. and Saletan, E.J. (1998) Classical Dynamics, Canbridge University Press. 12. Lembo, M. and Podio-Guidugli, P. (2001) Internal constraints, reactive stresses, and the Timoshenko beam theory, Journal of Elasticity 65, pp. 131–148. 13. Ljung, L. and Glad, T. (1994) Modeling of Dynamic Systems, Prentice Hall, Engelwood Cliffs, NJ. 14. Podio-Guidugli, P. (2000) A Primer in Elasticity, Kluwer Academic, Dortrecht. Also in (2000) Journal of Elasticity 58, pp. 1–104. 15. Pironneau, O. (1989) Finite Element Methods for Fluids, John Wiley & Sons, Chichester. 16. Tonti, E. (1972) A mathematical model for physical theories, Atti della Academia nazionale dei Lincei. Classe di scienze fisiche, mat. e nat. Ser. 8. 52, pp. 175–181. 17. Truesdell, C. (1991) A First Course in Rational Continuum Mechanics, Academic Press, Boston.

201

202

References

18. Truesdell, C. and Noll, W. (1965) The Non-Linear Field Theories of Mechanics, Springer, Berlin. 19. Truesdell, C. and Toupin, R.A. (1960) The Classical Field Theories. Hanbuch der Physik (ed. S. Flugge), ¨ Springer, Berlin. Vol. III/1, pp. 226–858. 20. Uhlhorn, U. (1988) Teknisk mekanik 1F, Lunds Tekniska Hogskola. ¨ (in Swedish)

Index

acceleration, 23 in pipe flow, 64 action and reaction, 37, 85 angular momentum, 15, 32, 46, 52, 81 Archimedes’ principle, 143–146 bending of a beam, 111 bending stiffness, 112 Bernoulli’s equation, 75 Bernoulli’s theorem, 139–140 body, 15, 21–23, 27 part of, 15, 22, 23, 27 body angular momentum, 169 body angular velocity, 169 body fixed frame, 166 body inertia tensor, 169 boundary conditions Dirichlet, 113 essential, 113 natural, 113 Neumann, 113 bulk modulus, 122 cantilever beam, 115 catenary cable, 61 Cauch’s inequality, 181 complete mathematical model, see mathematical model, complete compression modulus, 122 conservation of mass, see universal laws, conservation of mass constitutive laws, see particular laws, constitutive laws continuity equation, 68, 80

continuum model, 21 control domain, 68, 137 Cosserat theory, 94 couples, 47 in pipe bend, 66 on beam, 109 cross section area, 69 curl, 192 curvature, 53 circle, 53 radius, 53 curve, 24, 49 cut principle, 15, 35, 46, 81 deformation gradient, 28 derivative of, 78 density of the atmosphere, 146 one-dimensional, 70 three-dimensional, 80 determinant, 28 dilatation, 123 directors, 29, 94, 111 divergence, 191 divergence theorems, 193 elastic fluid, 96, 130, 138 elasticity, 94 elasticity coefficients, 102, 120–123 ellipsoid of inertia, 173 elongation stiffness, 112 equations of motion in terms of pressure, 71 one-dimensional, 48–55

203

204

Index

three-dimensional, 84 equilibrium equation, 100, 117, 155 Euclidean point space, 185 dimension, 185 translation space, 185 Euclidean vector space, 180–185 axioms, 180–181 base, 180 orthonormal, 181 right- and left-handed, 184 cross product, 182 dimension, 180 oriented, 184 Euler’s equation of inviscid fluid, 138 Euler’s laws, see universal laws, Euler’s laws Euler-Bernoulli’s displacement field, 158–161 fields, 190 flexible cable, 57 flexible tube, 134 force laws, see particular laws, force laws forces central, 40 central axis, 41 conservative, 74, 139, 143 cut, 47 external, 33, 46 in pipe bend, 66 internal, 33 reaction, 151 resultant, 15, 33, 47, 52, 84 system of, 33–35, 46–47, 81–84 function, 176 linear, 187 geometric vector, see vector gradient, 190 gravitation, 93, 98 Hagen-Poiseuille’s equation, 147 homogeneous motion, 74, 128 ideal fluid, 96 implicitly defined concept, 6 incompressible material, 74, 96, 127, 139, 151–153

inertia tensor, 169 inertial space, 16 inextensibility, 152 inextensible cable, 58 inviscid fluid, 129, 138–140 irrotational flow, 140 isochoric motion, 70, 79, 127 isotropic material, 92, 118 Kepler’s problem, 39 kinematic admissibility, 105 kinematic constraints, see particular laws, kinematic constraints kinetic energy, 171 Kronecker’s delta, 181 laminar flow, 141 linear momentum, 15, 32, 46, 52, 81 local action, 92, 94, 95 localization theorems, 194 mass, 31, 45, 80 center of, 32 distribution, 51 flux, 69 referential distribution, 45 material point, 21 material representation, 64–66, 77–79 material time derivative, 65, 78 mathematical model, 3 beam, 109–117, 153–163 complete, 4 bar problem, 113 compressible pipe , 130 elastic inviscid fluid flow, 138 Euler-Bernoulli beam problem, 114 force problem of particle mechanics, 39 incompressible inviscid fluid , 139 incompressible pipe fl , 128 incompressible pipe in flexible pipe, 134 inextensible cable problem, 58 linear elasticity, 119 Navier-Stokes , 141 one-dimensional force problem, 55 rigid body motion, 170 shear beam problem, 115

Index Timoshenko beam problem, 113 truss problem, 105 dimensionless form, 135, 142 discrete, 21, 31–43, 99, 150–151 existence of solutions, 5 fluid mechanics, 137–147 hyperbolic, 131 idealization, 7 interpretation, 7 linear elasticity, 97, 117–124, 151–152 one-dimensional, 23, 45–61 pipe flow, 25, 63–76, 95, 127–136 referent, 7 refined geometry, 29, 111 refinement, 8 rigid body, 93, 152, 163–173 small displacements, 99–125 three-dimensional, 27, 77–86 truss, 103–105 two-dimensional, 25 uniquness of solutions, 5, 93 validation, 8 well-posed, 5 Moens-Korteweg’s equation, 135 motion, 22 natural base, 53–55 natural parameterization, 49–52 Navier’s equation, 125 Navier-Stokes’ equation, 140 plane steady laminar flow, 141 Newton’s law, 38 Newtonian f , 97, 129 observer invariance, 91 open scheme, 9, 11–17 origin, 13, 185 orthogonal tensor, 91, 164, 190 proper, 190 particular laws, 9, 17, 89–98 constitutive equations, 90 constitutive laws, 9, 17 force laws, 9, 17, 90 general principles, 90–93 coordinate invariance, 91 dimensional invariance, 91 observer invariance, 91

205

kinematic constraints, 9, 17, 89, 149–174 simple, 39, 55 permutation symbol, 183 physical space, 11 pipe, 25 placement, 22 plane deformation, 157 plane flow, 141 Poisson’s ratio, 120 position vector, 13, 185 pressure, 71, 128 of the atmosphere, 146 principle axes, 170 principle moments of inertia, 170 principle of virtual displacements, 106 principle of virtual forces, 106 quasi-equilibrium, 42 rate of deformation tensor, 97 reference configuration, 27 referent, see mathematical model, referent Reynolds number, 143 rotation, 166 separating duality, 107 set, 175 shallow water flow, 76, 136 shape preserving material, 152 shear coefficient, 112 shear modulus, 121 shearing of a beam, 111 simple material, 92 space fixed frame, 166 spatial representation, 64–66, 77–79 speed, 64 of sound, 130, 135 spin equation of motion, 42 stability, 92, 124 static admissibility, 106 statically determinate, 57, 124 steady state, 75, 140, 141 stiffness matrix, 103, 105 strain generalized, 102, 107, 110 tensor, 117

206

Index

stress generalized, 102, 107 mean, 85 normal, 82 shear, 82 tensor, 83 subset, 176 suspension bridge, 58 tangent vector, 24, 50 tensor, 13, 187 components, 14, 188 deviatoric part, 122 trace, 189 transpose, 189 tensor product, 189 Timoshenko’s displacement field, 161–163 torque body, 170 change of base point, 35 resultant, 15, 33, 47, 52, 84 traction vector, 81 translation, 166 transmission line equations, 133 transversely isotropic material, 159 uniqueness theorem, 124 universal laws, 8, 14–17 conservation of mass, 8, 15, 31, 45 Euler’s laws, 8, 15, 36, 47, 84 angular momentum law, 15, 36, 47, 84 change of base point, 36

control domain versions, 76 linear momentum law, 15, 36, 47, 84 variational calculus, 194 vector, 12, 179 angle, 181 angular velocity, 166 components, 181 displacement, 100 linearly dependent, 180 linearly independent, 180 magnitude, 181 orthogonal, 181 scalar components, 13 triple scalar product, 182 velocity, 23 in pipe flow, 63 velocity gradient, 78 viscous fluid, 140–143, 152–153 volume of a pipe, 69 of body part, 79 vorticity equation, 153 wave equation, 131 wave propagation, 129–135 work conjugate, 107 external, 106, 110, 117 internal, 106, 110, 117 virtual, 107 work equations, 105–109 workless constraints, 151 Young’s modulus, 120

Mechanics SOLID MECHANICS AND ITS APPLICATIONS Series Editor: G.M.L. Gladwell Aims and Scope of the Series The fundamental questions arising in mechanics are: Why?, How?, and How much? The aim of this series is to provide lucid accounts written by authoritative researchers giving vision and insight in answering these questions on the subject of mechanics as it relates to solids. The scope of the series covers the entire spectrum of solid mechanics. Thus it includes the foundation of mechanics; variational formulations; computational mechanics; statics, kinematics and dynamics of rigid and elastic bodies; vibrations of solids and structures; dynamical systems and chaos; the theories of elasticity, plasticity and viscoelasticity; composite materials; rods, beams, shells and membranes; structural control and stability; soils, rocks and geomechanics; fracture; tribology; experimental mechanics; biomechanics and machine design. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23.

R.T. Haftka, Z. G¨u¨ rdal and M.P. Kamat: Elements of Structural Optimization. 2nd rev.ed., 1990 ISBN 0-7923-0608-2 J.J. Kalker: Three-Dimensional Elastic Bodies in Rolling Contact. 1990 ISBN 0-7923-0712-7 P. Karasudhi: Foundations of Solid Mechanics. 1991 ISBN 0-7923-0772-0 Not published Not published. J.F. Doyle: Static and Dynamic Analysis of Structures. With an Emphasis on Mechanics and Computer Matrix Methods. 1991 ISBN 0-7923-1124-8; Pb 0-7923-1208-2 O.O. Ochoa and J.N. Reddy: Finite Element Analysis of Composite Laminates. ISBN 0-7923-1125-6 M.H. Aliabadi and D.P. Rooke: Numerical Fracture Mechanics. ISBN 0-7923-1175-2 J. Angeles and C.S. L´o´ pez-Caju´ n: Optimization of Cam Mechanisms. 1991 ISBN 0-7923-1355-0 D.E. Grierson, A. Franchi and P. Riva (eds.): Progress in Structural Engineering. 1991 ISBN 0-7923-1396-8 R.T. Haftka and Z. G¨u¨ rdal: Elements of Structural Optimization. 3rd rev. and exp. ed. 1992 ISBN 0-7923-1504-9; Pb 0-7923-1505-7 J.R. Barber: Elasticity. 1992 ISBN 0-7923-1609-6; Pb 0-7923-1610-X H.S. Tzou and G.L. Anderson (eds.): Intelligent Structural Systems. 1992 ISBN 0-7923-1920-6 E.E. Gdoutos: Fracture Mechanics. An Introduction. 1993 ISBN 0-7923-1932-X J.P. Ward: Solid Mechanics. An Introduction. 1992 ISBN 0-7923-1949-4 M. Farshad: Design and Analysis of Shell Structures. 1992 ISBN 0-7923-1950-8 H.S. Tzou and T. Fukuda (eds.): Precision Sensors, Actuators and Systems. 1992 ISBN 0-7923-2015-8 J.R. Vinson: The Behavior of Shells Composed of Isotropic and Composite Materials. 1993 ISBN 0-7923-2113-8 H.S. Tzou: Piezoelectric Shells. Distributed Sensing and Control of Continua. 1993 ISBN 0-7923-2186-3 W. Schiehlen (ed.): Advanced Multibody System Dynamics. Simulation and Software Tools. 1993 ISBN 0-7923-2192-8 C.-W. Lee: Vibration Analysis of Rotors. 1993 ISBN 0-7923-2300-9 D.R. Smith: An Introduction to Continuum Mechanics. 1993 ISBN 0-7923-2454-4 G.M.L. Gladwell: Inverse Problems in Scattering. An Introduction. 1993 ISBN 0-7923-2478-1

Mechanics SOLID MECHANICS AND ITS APPLICATIONS Series Editor: G.M.L. Gladwell 24. 25. 26. 27. 28. 29. 30. 31. 32.

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G. Prathap: The Finite Element Method in Structural Mechanics. 1993 ISBN 0-7923-2492-7 J. Herskovits (ed.): Advances in Structural Optimization. 1995 ISBN 0-7923-2510-9 M.A. Gonz´a´ lez-Palacios and J. Angeles: Cam Synthesis. 1993 ISBN 0-7923-2536-2 W.S. Hall: The Boundary Element Method. 1993 ISBN 0-7923-2580-X J. Angeles, G. Hommel and P. Kov´a´ cs (eds.): Computational Kinematics. 1993 ISBN 0-7923-2585-0 A. Curnier: Computational Methods in Solid Mechanics. 1994 ISBN 0-7923-2761-6 D.A. Hills and D. Nowell: Mechanics of Fretting Fatigue. 1994 ISBN 0-7923-2866-3 B. Tabarrok and F.P.J. Rimrott: Variational Methods and Complementary Formulations in Dynamics. 1994 ISBN 0-7923-2923-6 E.H. Dowell (ed.), E.F. Crawley, H.C. Curtiss Jr., D.A. Peters, R. H. Scanlan and F. Sisto: A Modern Course in Aeroelasticity. Third Revised and Enlarged Edition. 1995 ISBN 0-7923-2788-8; Pb: 0-7923-2789-6 A. Preumont: Random Vibration and Spectral Analysis. 1994 ISBN 0-7923-3036-6 J.N. Reddy (ed.): Mechanics of Composite Materials. Selected works of Nicholas J. Pagano. 1994 ISBN 0-7923-3041-2 A.P.S. Selvadurai (ed.): Mechanics of Poroelastic Media. 1996 ISBN 0-7923-3329-2 Z. Mr´o´ z, D. Weichert, S. Dorosz (eds.): Inelastic Behaviour of Structures under Variable Loads. 1995 ISBN 0-7923-3397-7 R. Pyrz (ed.): IUTAM Symposium on Microstructure-Property Interactions in Composite Materials. Proceedings of the IUTAM Symposium held in Aalborg, Denmark. 1995 ISBN 0-7923-3427-2 M.I. Friswell and J.E. Mottershead: Finite Element Model Updating in Structural Dynamics. 1995 ISBN 0-7923-3431-0 D.F. Parker and A.H. England (eds.): IUTAM Symposium on Anisotropy, Inhomogeneity and Nonlinearity in Solid Mechanics. Proceedings of the IUTAM Symposium held in Nottingham, U.K. 1995 ISBN 0-7923-3594-5 J.-P. Merlet and B. Ravani (eds.): Computational Kinematics ’95. 1995 ISBN 0-7923-3673-9 L.P. Lebedev, I.I. Vorovich and G.M.L. Gladwell: Functional Analysis. Applications in Mechanics and Inverse Problems. 1996 ISBN 0-7923-3849-9 ˇ Mechanics of Components with Treated or Coated Surfaces. 1996 J. Mencik: ISBN 0-7923-3700-X D. Bestle and W. Schiehlen (eds.): IUTAM Symposium on Optimization of Mechanical Systems. Proceedings of the IUTAM Symposium held in Stuttgart, Germany. 1996 ISBN 0-7923-3830-8 D.A. Hills, P.A. Kelly, D.N. Dai and A.M. Korsunsky: Solution of Crack Problems. The Distributed Dislocation Technique. 1996 ISBN 0-7923-3848-0 V.A. Squire, R.J. Hosking, A.D. Kerr and P.J. Langhorne: Moving Loads on Ice Plates. 1996 ISBN 0-7923-3953-3 A. Pineau and A. Zaoui (eds.): IUTAM Symposium on Micromechanics of Plasticity and Damage of Multiphase Materials. Proceedings of the IUTAM Symposium held in S`evres, Paris, France. 1996 ISBN 0-7923-4188-0 A. Naess and S. Krenk (eds.): IUTAM Symposium on Advances in Nonlinear Stochastic Mechanics. Proceedings of the IUTAM Symposium held in Trondheim, Norway. 1996 ISBN 0-7923-4193-7 D. Ie¸s¸an and A. Scalia: Thermoelastic Deformations. 1996 ISBN 0-7923-4230-5

Mechanics SOLID MECHANICS AND ITS APPLICATIONS Series Editor: G.M.L. Gladwell 49. 50. 51. 52.

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J.R. Willis (ed.): IUTAM Symposium on Nonlinear Analysis of Fracture. Proceedings of the IUTAM Symposium held in Cambridge, U.K. 1997 ISBN 0-7923-4378-6 A. Preumont: Vibration Control of Active Structures. An Introduction. 1997 ISBN 0-7923-4392-1 G.P. Cherepanov: Methods of Fracture Mechanics: Solid Matter Physics. 1997 ISBN 0-7923-4408-1 D.H. van Campen (ed.): IUTAM Symposium on Interaction between Dynamics and Control in Advanced Mechanical Systems. Proceedings of the IUTAM Symposium held in Eindhoven, The Netherlands. 1997 ISBN 0-7923-4429-4 N.A. Fleck and A.C.F. Cocks (eds.): IUTAM Symposium on Mechanics of Granular and Porous Materials. Proceedings of the IUTAM Symposium held in Cambridge, U.K. 1997 ISBN 0-7923-4553-3 J. Roorda and N.K. Srivastava (eds.): Trends in Structural Mechanics. Theory, Practice, Education. 1997 ISBN 0-7923-4603-3 Yu.A. Mitropolskii and N. Van Dao: Applied Asymptotic Methods in Nonlinear Oscillations. 1997 ISBN 0-7923-4605-X C. Guedes Soares (ed.): Probabilistic Methods for Structural Design. 1997 ISBN 0-7923-4670-X D. Fran¸c¸ ois, A. Pineau and A. Zaoui: Mechanical Behaviour of Materials. Volume I: Elasticity and Plasticity. 1998 ISBN 0-7923-4894-X D. Fran¸c¸ ois, A. Pineau and A. Zaoui: Mechanical Behaviour of Materials. Volume II: Viscoplasticity, Damage, Fracture and Contact Mechanics. 1998 ISBN 0-7923-4895-8 L.T. Tenek and J. Argyris: Finite Element Analysis for Composite Structures. 1998 ISBN 0-7923-4899-0 Y.A. Bahei-El-Din and G.J. Dvorak (eds.): IUTAM Symposium on Transformation Problems in Composite and Active Materials. Proceedings of the IUTAM Symposium held in Cairo, Egypt. 1998 ISBN 0-7923-5122-3 I.G. Goryacheva: Contact Mechanics in Tribology. 1998 ISBN 0-7923-5257-2 O.T. Bruhns and E. Stein (eds.): IUTAM Symposium on Micro- and Macrostructural Aspects of Thermoplasticity. Proceedings of the IUTAM Symposium held in Bochum, Germany. 1999 ISBN 0-7923-5265-3 F.C. Moon: IUTAM Symposium on New Applications of Nonlinear and Chaotic Dynamics in Mechanics. Proceedings of the IUTAM Symposium held in Ithaca, NY, USA. 1998 ISBN 0-7923-5276-9 R. Wang: IUTAM Symposium on Rheology of Bodies with Defects. Proceedings of the IUTAM Symposium held in Beijing, China. 1999 ISBN 0-7923-5297-1 Yu.I. Dimitrienko: Thermomechanics of Composites under High Temperatures. 1999 ISBN 0-7923-4899-0 P. Argoul, M. Fr´e´ mond and Q.S. Nguyen (eds.): IUTAM Symposium on Variations of Domains and Free-Boundary Problems in Solid Mechanics. Proceedings of the IUTAM Symposium held in Paris, France. 1999 ISBN 0-7923-5450-8 F.J. Fahy and W.G. Price (eds.): IUTAM Symposium on Statistical Energy Analysis. Proceedings of the IUTAM Symposium held in Southampton, U.K. 1999 ISBN 0-7923-5457-5 H.A. Mang and F.G. Rammerstorfer (eds.): IUTAM Symposium on Discretization Methods in Structural Mechanics. Proceedings of the IUTAM Symposium held in Vienna, Austria. 1999 ISBN 0-7923-5591-1

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P. Pedersen and M.P. Bendsøe (eds.): IUTAM Symposium on Synthesis in Bio Solid Mechanics. Proceedings of the IUTAM Symposium held in Copenhagen, Denmark. 1999 ISBN 0-7923-5615-2 S.K. Agrawal and B.C. Fabien: Optimization of Dynamic Systems. 1999 ISBN 0-7923-5681-0 A. Carpinteri: Nonlinear Crack Models for Nonmetallic Materials. 1999 ISBN 0-7923-5750-7 F. Pfeifer (ed.): IUTAM Symposium on Unilateral Multibody Contacts. Proceedings of the IUTAM Symposium held in Munich, Germany. 1999 ISBN 0-7923-6030-3 E. Lavendelis and M. Zakrzhevsky (eds.): IUTAM/IFToMM Symposium on Synthesis of Nonlinear Dynamical Systems. Proceedings of the IUTAM/IFToMM Symposium held in Riga, Latvia. 2000 ISBN 0-7923-6106-7 J.-P. Merlet: Parallel Robots. 2000 ISBN 0-7923-6308-6 J.T. Pindera: Techniques of Tomographic Isodyne Stress Analysis. 2000 ISBN 0-7923-6388-4 G.A. Maugin, R. Drouot and F. Sidoroff (eds.): Continuum Thermomechanics. The Art and Science of Modelling Material Behaviour. 2000 ISBN 0-7923-6407-4 N. Van Dao and E.J. Kreuzer (eds.): IUTAM Symposium on Recent Developments in Non-linear Oscillations of Mechanical Systems. 2000 ISBN 0-7923-6470-8 S.D. Akbarov and A.N. Guz: Mechanics of Curved Composites. 2000 ISBN 0-7923-6477-5 ISBN 0-7923-6489-9 M.B. Rubin: Cosserat Theories: Shells, Rods and Points. 2000 S. Pellegrino and S.D. Guest (eds.): IUTAM-IASS Symposium on Deployable Structures: Theory and Applications. Proceedings of the IUTAM-IASS Symposium held in Cambridge, U.K., 6–9 September 1998. 2000 ISBN 0-7923-6516-X A.D. Rosato and D.L. Blackmore (eds.): IUTAM Symposium on Segregation in Granular Flows. Proceedings of the IUTAM Symposium held in Cape May, NJ, U.S.A., June 5–10, 1999. 2000 ISBN 0-7923-6547-X A. Lagarde (ed.): IUTAM Symposium on Advanced Optical Methods and Applications in Solid Mechanics. Proceedings of the IUTAM Symposium held in Futuroscope, Poitiers, France, August 31–September 4, 1998. 2000 ISBN 0-7923-6604-2 D. Weichert and G. Maier (eds.): Inelastic Analysis of Structures under Variable Loads. Theory and Engineering Applications. 2000 ISBN 0-7923-6645-X T.-J. Chuang and J.W. Rudnicki (eds.): Multiscale Deformation and Fracture in Materials and Structures. The James R. Rice 60th Anniversary Volume. 2001 ISBN 0-7923-6718-9 S. Narayanan and R.N. Iyengar (eds.): IUTAM Symposium on Nonlinearity and Stochastic Structural Dynamics. Proceedings of the IUTAM Symposium held in Madras, Chennai, India, 4–8 January 1999 ISBN 0-7923-6733-2 S. Murakami and N. Ohno (eds.): IUTAM Symposium on Creep in Structures. Proceedings of the IUTAM Symposium held in Nagoya, Japan, 3-7 April 2000. 2001 ISBN 0-7923-6737-5 W. Ehlers (ed.): IUTAM Symposium on Theoretical and Numerical Methods in Continuum Mechanics of Porous Materials. Proceedings of the IUTAM Symposium held at the University of Stuttgart, Germany, September 5-10, 1999. 2001 ISBN 0-7923-6766-9 D. Durban, D. Givoli and J.G. Simmonds (eds.): Advances in the Mechanis of Plates and Shells The Avinoam Libai Anniversary Volume. 2001 ISBN 0-7923-6785-5 U. Gabbert and H.-S. Tzou (eds.): IUTAM Symposium on Smart Structures and Structonic Systems. Proceedings of the IUTAM Symposium held in Magdeburg, Germany, 26–29 September 2000. 2001 ISBN 0-7923-6968-8

Mechanics SOLID MECHANICS AND ITS APPLICATIONS Series Editor: G.M.L. Gladwell 90. 91.

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Y. Ivanov, V. Cheshkov and M. Natova: Polymer Composite Materials – Interface Phenomena & Processes. 2001 ISBN 0-7923-7008-2 R.C. McPhedran, L.C. Botten and N.A. Nicorovici (eds.): IUTAM Symposium on Mechanical and Electromagnetic Waves in Structured Media. Proceedings of the IUTAM Symposium held in Sydney, NSW, Australia, 18-22 Januari 1999. 2001 ISBN 0-7923-7038-4 D.A. Sotiropoulos (ed.): IUTAM Symposium on Mechanical Waves for Composite Structures Characterization. Proceedings of the IUTAM Symposium held in Chania, Crete, Greece, June 14-17, 2000. 2001 ISBN 0-7923-7164-X V.M. Alexandrov and D.A. Pozharskii: Three-Dimensional Contact Problems. 2001 ISBN 0-7923-7165-8 J.P. Dempsey and H.H. Shen (eds.): IUTAM Symposium on Scaling Laws in Ice Mechanics and Ice Dynamics. Proceedings of the IUTAM Symposium held in Fairbanks, Alaska, U.S.A., 13-16 June 2000. 2001 ISBN 1-4020-0171-1 U. Kirsch: Design-Oriented Analysis of Structures. A Unified Approach. 2002 ISBN 1-4020-0443-5 A. Preumont: Vibration Control of Active Structures. An Introduction (2nd Edition). 2002 ISBN 1-4020-0496-6 B.L. Karihaloo (ed.): IUTAM Symposium on Analytical and Computational Fracture Mechanics of Non-Homogeneous Materials. Proceedings of the IUTAM Symposium held in Cardiff, U.K., 18-22 June 2001. 2002 ISBN 1-4020-0510-5 S.M. Han and H. Benaroya: Nonlinear and Stochastic Dynamics of Compliant Offshore Structures. 2002 ISBN 1-4020-0573-3 A.M. Linkov: Boundary Integral Equations in Elasticity Theory. 2002 ISBN 1-4020-0574-1 L.P. Lebedev, I.I. Vorovich and G.M.L. Gladwell: Functional Analysis. Applications in Mechanics and Inverse Problems (2nd Edition). 2002 ISBN 1-4020-0667-5; Pb: 1-4020-0756-6 Q.P. Sun (ed.): IUTAM Symposium on Mechanics of Martensitic Phase Transformation in Solids. Proceedings of the IUTAM Symposium held in Hong Kong, China, 11-15 June 2001. 2002 ISBN 1-4020-0741-8 M.L. Munjal (ed.): IUTAM Symposium on Designing for Quietness. Proceedings of the IUTAM Symposium held in Bangkok, India, 12-14 December 2000. 2002 ISBN 1-4020-0765-5 J.A.C. Martins and M.D.P. Monteiro Marques (eds.): Contact Mechanics. Proceedings of the ˜ Peniche, Portugal, 3rd Contact Mechanics International Symposium, Praia da Consola¸c¸ao, 17-21 June 2001. 2002 ISBN 1-4020-0811-2 H.R. Drew and S. Pellegrino (eds.): New Approaches to Structural Mechanics, Shells and ISBN 1-4020-0862-7 Biological Structures. 2002 J.R. Vinson and R.L. Sierakowski: The Behavior of Structures Composed of Composite Materials. Second Edition. 2002 ISBN 1-4020-0904-6 Not yet published. J.R. Barber: Elasticity. Second Edition. 2002 ISBN Hb 1-4020-0964-X; Pb 1-4020-0966-6 C. Miehe (ed.): IUTAM Symposium on Computational Mechanics of Solid Materials at Large Strains. Proceedings of the IUTAM Symposium held in Stuttgart, Germany, 20-24 August 2001. 2003 ISBN 1-4020-1170-9

Mechanics SOLID MECHANICS AND ITS APPLICATIONS Series Editor: G.M.L. Gladwell 109. P. St˚a˚ hle and K.G. Sundin (eds.): IUTAM Symposium on Field Analyses for Determination of Material Parameters – Experimental and Numerical Aspects. Proceedings of the IUTAM Symposium held in Abisko National Park, Kiruna, Sweden, July 31 – August 4, 2000. 2003 ISBN 1-4020-1283-7 110. N. Sri Namachchivaya and Y.K. Lin (eds.): IUTAM Symposium on Nonlinear Stochastic Dynamics. Proceedings of the IUTAM Symposium held in Monticello, IL, USA, 26 – 30 August, 2000. 2003 ISBN 1-4020-1471-6 111. H. Sobieckzky (ed.): IUTAM Symposium Transsonicum IV. V Proceedings of the IUTAM Symposium held in G¨o¨ ttingen, Germany, 2–6 September 2002, 2003 ISBN 1-4020-1608-5 112. J.-C. Samin and P. Fisette: Symbolic Modeling of Multibody Systems. 2003 ISBN 1-4020-1629-8 113. A.B. Movchan (ed.): IUTAM Symposium on Asymptotics, Singularities and Homogenisation in Problems of Mechanics. Proceedings of the IUTAM Symposium held in Liverpool, United ISBN 1-4020-1780-4 Kingdom, 8-11 July 2002. 2003 114. S. Ahzi, M. Cherkaoui, M.A. Khaleel, H.M. Zbib, M.A. Zikry and B. LaMatina (eds.): IUTAM Symposium on Multiscale Modeling and Characterization of Elastic-Inelastic Behavior of Engineering Materials. Proceedings of the IUTAM Symposium held in Marrakech, Morocco, 20-25 October 2002. 2004 ISBN 1-4020-1861-4 115. H. Kitagawa and Y. Shibutani (eds.): IUTAM Symposium on Mesoscopic Dynamics of Fracture Process and Materials Strength. Proceedings of the IUTAM Symposium held in Osaka, Japan, 6-11 July 2003. Volume in celebration of Professor Kitagawa’s retirement. 2004 ISBN 1-4020-2037-6 116. E.H. Dowell, R.L. Clark, D. Cox, H.C. Curtiss, Jr., K.C. Hall, D.A. Peters, R.H. Scanlan, E. Simiu, F. Sisto and D. Tang: A Modern Course in Aeroelasticity. 4th Edition, 2004 ISBN 1-4020-2039-2 117. T. Burczy´n´ ski and A. Osyczka (eds.): IUTAM Symposium on Evolutionary Methods in Mechanics. Proceedings of the IUTAM Symposium held in Cracow, Poland, 24-27 September 2002. 2004 ISBN 1-4020-2266-2 118. D. Ie¸s¸an: Thermoelastic Models of Continua. 2004 ISBN 1-4020-2309-X 119. G.M.L. Gladwell: Inverse Problems in Vibration. Second Edition. 2004 ISBN 1-4020-2670-6 120. J.R. Vinson: Plate and Panel Structures of Isotropic, Composite and Piezoelectric Materials, Including Sandwich Construction. 2005 ISBN 1-4020-3110-6 121. Forthcoming 122. G. Rega and F. Vestroni (eds.): IUTAM Symposium on Chaotic Dynamics and Control of Systems and Processes in Mechanics. Proceedings of the IUTAM Symposium held in Rome, Italy, 8–13 June 2003. 2005 ISBN 1-4020-3267-6 123. E.E. Gdoutos: Fracture Mechanics. An Introduction. 2nd edition. 2005 ISBN 1-4020-3267-6 124. M.D. Gilchrist (ed.): IUTAM Symposium on Impact Biomechanics from Fundamental Insights to Applications. 2005 ISBN 1-4020-3795-3 125. J.M. Huyghe, P.A.C. Raats and S. C. Cowin (eds.): IUTAM Symposium on Physicochemical and Electromechanical Interactions in Porous Media. 2005 ISBN 1-4020-3864-X 126. H. Ding, W. Chen and L. Zhang: Elasticity of Transversely Isotropic Materials. 2005 ISBN 1-4020-4033-4 127. W. Yang (ed): IUTAM Symposium on Mechanics and Reliability of Actuating Materials. Proceedings of the IUTAM Symposium held in Beijing, China, 1–3 September 2004. 2005 ISBN 1-4020-4131-6

Mechanics SOLID MECHANICS AND ITS APPLICATIONS Series Editor: G.M.L. Gladwell 128. J.-P. Merlet: Parallel Robots. 2006 ISBN 1-4020-4132-2 129. G.E.A. Meier and K.R. Sreenivasan (eds.): IUTAM Symposium on One Hundred Years of Boundary Layer Research. Proceedings of the IUTAM Symposium held at DLR-G¨o¨ ttingen, Germany, August 12–14, 2004. 2006 ISBN 1-4020-4149-7 130. H. Ulbrich and W. G¨u¨ nthner (eds.): IUTAM Symposium on Vibration Control of Nonlinear Mechanisms and Structures. 2006 ISBN 1-4020-4160-8 131. L. Librescu and O. Song: Thin-Walled Composite Beams. Theory and Application. 2006 ISBN 1-4020-3457-1 132. G. Ben-Dor, A. Dubinsky and T. Elperin: Applied High-Speed Plate Penetration Dynamics. 2006 ISBN 1-4020-3452-0 133. X. Markenscoff and A. Gupta (eds.): Collected Works of J. D. Eshelby. Mechanics and Defects and Heterogeneities. 2006 ISBN 1-4020-4416-X 134. R.W. Snidle and H.P. Evans (eds.): IUTAM Symposium on Elastohydrodynamics and Microelastohydrodynamics. Proceedings of the IUTAM Symposium held in Cardiff, UK, 1–3 September, 2004. 2006 ISBN 1-4020-4532-8 135. T. Sadowski (ed.): IUTAM Symposium on Multiscale Modelling of Damage and Fracture Processes in Composite Materials. Proceedings of the IUTAM Symposium held in Kazimierz Dolny, Poland, 23–27 May 2005. 2006 ISBN 1-4020-4565-4 136. A. Preumont: Mechatronics. Dynamics of Electromechanical and Piezoelectric Systems. 2006 ISBN 1-4020-4695-2 137. M.P. Bendsøe, N. Olhoff and O. Sigmund (eds.): IUTAM Symposium on Topological Design Optimization of Structures, Machines and Materials. Status and Perspectives. 2006 ISBN 1-4020-4729-0 138. A. Klarbring: Models of Mechanics. 2006 ISBN 1-4020-4834-3

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