Tensor analysis for engineers. Transformations. Mathematics. Applications 9781683929642

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Tensor Analysis for Engineers Third Edition

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Tensor Analysis for Engineers Transformations • Mathematics • Applications Third Edition

Mehrzad Tabatabaian, PhD, PEng

MERCURY LEARNING AND INFORMATION Dulles, Virginia Boston, Massachusetts New Delhi

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Copyright ©2023 by Mercury Learning And Information LLC. All rights reserved. An imprint of Walter De Gruyter GmbH This publication, portions of it, or any accompanying software may not be reproduced in any way, stored in a retrieval system of any type, or transmitted by any means, media, electronic display or mechanical display, including, but not limited to, photocopy, recording, Internet postings, or scanning, without prior permission in writing from the publisher. Publisher: David Pallai Mercury Learning and Information 22841 Quicksilver Drive Dulles, VA 20166 [email protected] www.merclearning.com (800) 232-0223 M. Tabatabaian. Tensor Analysis for Engineers: Transformations • Mathematics • Applications, Third Edition. ISBN: 978-1-68392-964-2 The publisher recognizes and respects all marks used by companies, manufacturers, and developers as a means to distinguish their products. All brand names and product names mentioned in this book are trademarks or service marks of their respective companies. Any omission or misuse (of any kind) of service marks or trademarks, etc. is not an attempt to infringe on the property of others. Library of Congress Control Number: 2023937708 232425321

This book is printed on acid-free paper in the United States of America.

Our titles are available for adoption, license, or bulk purchase by institutions, corporations, etc. For additional information, please contact the Customer Service Dept. at 800-232-0223(toll free). All of our titles are available in digital format at www.academiccourseware.com and other digital vendors. The sole obligation of Mercury Learning and Information to the purchaser is to replace the book, based on defective materials or faulty workmanship, but not based on the operation or functionality of the product.

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To my teachers and mentors for their invaluable transfer of knowledge and direction.

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Contents Prefacexiii About the Author CHAPTER 1:

xv INTRODUCTION 1.1 Index Notation—The Einstein Summation Convention

1

CHAPTER 2:

COORDINATE SYSTEMS DEFINITION

3

CHAPTER 3:

BASIS VECTORS AND SCALE FACTORS

5

CHAPTER 4: CONTRAVARIANT COMPONENTS AND TRANSFORMATIONS CHAPTER 5: COVARIANT COMPONENTS AND TRANSFORMATIONS

2

9 13

CHAPTER 6: PHYSICAL COMPONENTS AND TRANSFORMATIONS15 CHAPTER 7:

TENSORS—MIXED AND METRIC

CHAPTER 8: METRIC TENSOR OPERATION ON TENSOR INDICES 8.1 Example: Cylindrical Coordinate Systems 8.2 Example: Spherical Coordinate Systems

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viii • Contents CHAPTER 9:

DOT AND CROSS PRODUCTS OF TENSORS 9.1 Determinant of an N × N Matrix Using Permutation Symbols

29 34

CHAPTER 10: GRADIENT VECTOR OPERATOR—CHRISTOFFEL SYMBOLS35 10.1 Covariant Derivatives of Vectors—Christoffel Symbols of the 2nd Kind 35 10.2 Contravariant Derivatives of Vectors 39 10.3 Covariant Derivatives of a Mixed Tensor 40 st 10.4 Christoffel Symbol Relations and Properties—1 and 2nd Kinds 41 CHAPTER 11: DERIVATIVE FORMS—CURL, DIVERGENCE, LAPLACIAN51 11.1 Curl Operations on Tensors 51 11.2 Physical Components of the Curl of Tensors—3D Orthogonal Systems 54 11.3 Divergence Operation on Tensors 55 11.4 Laplacian Operations on Tensors 57 11.5 Biharmonic Operations on Tensors 58 11.6 Physical Components of the Laplacian of a Vector—3D Orthogonal Systems 59 CHAPTER 12: CARTESIAN TENSOR TRANSFORMATION— ROTATIONS65 12.1 Rotation Matrix 67 12.2 Equivalent Single Rotation: Eigenvalues and Eigenvectors 67 CHAPTER 13: COORDINATE INDEPENDENT GOVERNING EQUATIONS75 13.1 The Acceleration Vector—Contravariant Components 76 13.2 The Acceleration Vector—Physical Components 78 13.3 The Acceleration Vector in Orthogonal Systems—Physical Components 79 13.4 Substantial Time Derivatives of Tensors 82 13.5 Conservation Equations—Coordinate Independent forms 85

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Contents • ix

CHAPTER 14: COLLECTION OF RELATIONS FOR SELECTED COORDINATE SYSTEMS 89 14.1 Cartesian Coordinate System 89 14.2 Cylindrical Coordinate Systems 91 14.3 Spherical Coordinate Systems 93 14.4 Parabolic Coordinate Systems 96 14.5 Orthogonal Curvilinear Coordinate Systems98 CHAPTER 15: RIGID BODY ROTATION: EULER ANGLES, QUATERNIONS, AND ROTATION MATRIX 15.1 Active and Passive Rotations 15.2 Euler Angles 15.3 Categorizing Euler Angles 15.4 Gimbal Lock-Euler Angles Limitation 15.5 Quaternions-Applications for Rigid Body Rotation 15.6 From a Given Quaternion to Rotation Matrix 15.7 From a Given Rotation Matrix to Quaternion 15.8 From Euler Angles to a Quaternion 15.9 Putting it all Together

103 104 107 113 121 123 135 138 139 140

CHAPTER 16: MECHANICAL STRESS TRANSFORMATION: ANALYTICAL AND MOHR’S CIRCLE METHODS 143 16.1 Plane Stress Condition 144 16.2 Principal Stresses and Directions: Eigenvalues and Eigenvectors 152 16.3 Analysis of Transformed Stresses: Mohr’s Circle Graphical Method 153 16.4 3D Stress Transformation and Analysis 169 16.5 Principal Directions: Eigenvectors 177 16.6 Octahedral Stresses in Principal Coordinate System 183 16.7 Octahedral Stresses and Deviatoric Stresses 186 16.8 von Mises Yield Criterion vs Octahedral Shear Stress 188

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x • Contents CHAPTER 17: THE WORKED EXAMPLES 17.1 Example: Einstein Summation Conventions 17.2 Example: Conversion from Vector to Index Notations 17.3 Example: Oblique Rectilinear Coordinate Systems 17.4 Example: Quantities Related to Parabolic Coordinate System 17.5 Example: Quantities Related to Bi-Polar Coordinate Systems 17.6 Example: Application of Contravariant Metric Tensors 17.7 Example: Dot and Cross Products in Cylindrical and Spherical Coordinates 17.8 Example: Relation between Jacobian and Metric Tensor Determinants 17.9 Example: Determinant of Metric Tensors Using Displacement Vectors 17.10 Example: Determinant of a 4 × 4 Matrix Using Permutation Symbols 17.11 Example: Time Derivatives of the Jacobian 17.12 Example: Covariant Derivatives of a Constant Vector 17.13 Example: Covariant Derivatives of Physical Components of a Vector 17.14 Example: Continuity Equations in Several Coordinate Systems 17.15 Example: 4D Spherical Coordinate Systems 17.16 Example: Complex Double Dot-Cross Product Expressions 17.17 Example: Covariant Derivatives of Metric Tensors 17.18 Example: Active Rotation Using Single-Axis and Quaternions Methods

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191 191 192 193 197 200 203 203 205 206 207 207 208 209 209 210 213 214 214

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Contents • xi

17.19 Example: Passive Rotation Using Single-Axis and Quaternions Methods 17.20 Example: Successive Rotations Using Quaternions Method CHAPTER 18: EXERCISES

216 218 219

REFERENCES227 INDEX231

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Preface This is the third edition of the published textbook: Tensor Analysis for Engineers. In this edition, we have expanded the content on the application of mechanical stress transformation. We include topics like 2D and 3D stress tensor transformation, Mohr’s circle graphical method, and octahedral shear stress with applications in static failure theories for solid media. We added several worked-out numerical examples related to the above-mentioned topics. In the second edition, we expanded the content on the rigid body rotation and Cartesian tensors by including Euler angles and quaternions methods. In addition to the rotation matrix method, presented in the first edition and included in this edition, we collect all three methods in this volume of the textbook. In this edition, the quaternions and their algebraic calculation rules are presented. We also discuss the active and passive rotations and present several worked-out examples using the Euler angles and quaternions methods applications and their interrelations. The problem of gimbal lock is also analyzed and presented with detailed worked out examples. Additional references were added to the list of references for the second edition. In engineering and science physical quantities are often represented by mathematical functions, namely tensors. Examples include temperature, pressure, force, mechanical stress, electric/magnetic fields, velocity, enthalpy, entropy, etc. In turn, tensors are categorized based on their rank, i.e. rank zero, one, and so forth. The so-called scalar quantities (e.g. temperature) are tensors of rank zero. Likewise, velocity and force are tensors of rank one and mechanical stress and gradient of velocity are tensors of rank two. In Euclidean space, which could be of dimension , we can define several coordinate systems for our calculation and measurement of physical quantities. For example, in a 3D

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xiv • Preface space, we can have Cartesian, cylindrical, and spherical coordinate systems. In general, we prefer defining a coordinate system whose coordinate surfaces (where one of the coordinate variables is invariant or remains constant) match to the physical problem geometry at hand. This enables us to easily define the boundary conditions of the physical problem to the related governing equations, written in terms of the selected coordinate system. This action requires transformation of the tensor quantities and their related derivatives (e.g., gradient, curl, divergence) from Cartesian to the selected coordinate system or vice versa. The topic of tensor analysis (also referred to as “tensor calculus,” or “Ricci’s calculus,” since originally developed by Ricci (1835-1925), [1], [2]), is mainly engaged with the definition of tensor-like quantities and their transformation among coordinate systems and others. The topic provides a set of mathematical tools which enables users to perform transformation and calculations of tensors for any well-defined coordinate systems in a systematic way- it is a “machine.” The merit of tensor analysis is to provide a systematic mathematical formulation to derive the general form of the governing equations for arbitrary coordinate systems. In this book, we aim to provide engineers and applied scientists with the tools and techniques of tensor analysis for applications in practical problem solving and analysis activities. The geometry is limited to the Euclidean space/geometry, where the Pythagorean Theorem applies, with well-defined Cartesian coordinate systems as the reference. We discuss quantities defined in curvilinear coordinate systems, like cylindrical, spherical, parabolic, etc. and present several examples and coordinates sketches with related calculations. In addition, we listed several worked-out examples for helping the readers with mastering the topics provided in the prior sections. A list of exercises is provided for further practice for readers. Mehrzad Tabatabaian, PhD, PEng Vancouver, BC June 7, 2023

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About the Author Dr. Mehrzad Tabatabaian is a faculty member at the Mechanical Engineering Department, School of Energy at BCIT. He has several years of teaching and industry experience. Dr. Tabatabaian is currently Chair of the BCIT School of Energy Research Committee. He has published several papers in scientific journals and conferences and has written textbooks on multiphysics and turbulent flow modelling, advanced thermodynamics, tensor analysis, direct energy conversion, and Bond Graph modelling method. He holds several registered patents in the energy field resulting from years of research activities. Dr. Tabatabaian volunteered to help establish the Energy Efficiency and Renewable Energy Division (EERED), a new division at Engineers and Geoscientists British Columbia (EGBC). He was Chair of this division for several years. Mehrzad Tabatabaian received his BEng from Sharif University of Technology (former AUT) and advanced degrees from McGill University (MEng and PhD). He has been an active academic, professor, and engineer in leading alternative energy, oil, and gas industries. Mehrzad has also a Leadership Certificate from the University of Alberta and holds an EGBC P.Eng. License.

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CHAPTER

1

INTRODUCTION Physical quantities can be represented mathematically by t­ ensors. In further sections of this book we will define tensors more rigorously; however, for the introduction we will use this definition. An example of a tensor-like quantity is the temperature in a room (which could be a function of space and time) expressed as a scalar, a tensor of rank zero. Wind velocity is another example, which can be defined when we know both its magnitude and speed—a scalar ­quantity—and its direction. We define velocity as a vector or a t­ ensor of rank one. Scalars and vectors are familiar quantities to us and we encounter them in our daily life. However, there are quantities, or tensors of rank two, three, or higher that are normally dealt with in technical engineering computations. Examples include mechanical stress in a continuum, like the wall of a pressure vessel—a tensor of rank two—the modulus of elasticity or viscosity in a fluid—tensors of rank four—and so on. Engineers and scientists calculate and analyze tensor quantities, including their derivatives, using the laws of physics, mostly in the form of governing equations related to the phenomena. These laws must be expressed in an objective form as governing equations, and not subjected to the coordinate system considered. For example, the amount of internal stress in a continuum should not depend on what coordinate system is used for calculations. Sometimes tensors and their involved derivatives in a study must be transformed from one coordinate system to another. Therefore, to satisfy these ­technical/engineering needs, a mathematical “machine” is required to perform these operations accurately and systematically between arbitrary coordinate systems. Furthermore, the communication of technical computations requires precise definitions for tensors to guarantee a reliable level of standardization, i.e., identifying true ­tensor quantities from apparently

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2 • Tensor Analysis for Engineers, 3E tensor-like or non-tensor ones. This machine is called tensor analysis [2], [3], [4], [5], [6], [7]. The subject of tensor analysis has two major parts: a) definitions and properties of tensors including their calculus, and b) rules of transformation of tensor quantities among different coordinate systems. For example, consider again the wind velocity vector. By using tensor analysis, we can show that this quantity is a true tensor and transform it from a Cartesian coordinate system to a spherical one, for example. A major outcome of tensor analysis is having general relations for gradient-like operations in arbitrary coordinate systems, including gradient, curl, divergence, Laplacian, etc. which appear in many governing equations in engineering and science. Using tensor notation and definitions, we can write theses governing equations in explicit coordinate-­ independent forms.

1.1 INDEX NOTATION—THE EINSTEIN SUMMATION CONVENTION Writing expressions containing tensors could become cumbersome, especially when higher ranked tensors and higher dimensional space are involved. For  example, we usually use hatted arrow symbol for vectors (like A) and boldfont symbols for second rank tensors (like A). But this approach is very limited for expressing, for example, the modulus of elasticity, a 4th ranked tensor. Another limitation shows up when writing the components of a tensor in  N -­dimension space. For example, in 3D we write vector A in a Cartesian coor 3     dinate system as A = ∑Ai Ei = A1 E1 + A2 E2 + A3 E3 , where Ai is the compo 1 nent and Ei the unit vector. Taking this approach for N -dimension space and carrying the summation symbol (i.e., Σ) is cumbersome and seems unnecessary. The Einstein summation convention allows us to dispose of the summation symbol, if we carry summation operation for repeated indices in product-type expressions (unless otherwise specified). Using this approach, we       can write A = Ai Ei and a tensor of rank 4, for example, as T = Tijkl Ei E j Ek El , with all ­indices range for 1 to N. In practice, however, we even ignore ­writing the unit vectors and just use the component representing the original t­ ensor; hence this method is also referred to as component or index notation, or  ­ umber A ≡ Ai and T ≡ Tijkl . In addition, we represent a tensor’s rank by the n of free (i.e., not repeated) indices. We use these definitions and ­conventions ­throughout this book.

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CHAPTER

2

COORDINATE SYSTEMS DEFINITION For measuring and calculating physical quantities associated with geometrical points in space we require coordinate systems for reference. These systems (for example, in a 3D space) are composed of surfaces that mutually intersect to specify a geometrical point. For reference, an ideal system of coordinates called a Cartesian system is defined, in Euclidean space, such that it composes of three flat planes. These planes are considered by default to be mutually perpendicular to each other to form an orthogonal Cartesian system. In general, the orthogonality is not required to form a coordinate system—this kind of coordinate system is an oblique or slanted system. A Cartesian coordinate system, although ideal, is central to engineering and scientific calculations, since it is used as the reference compared to other coordinate systems. It also has properties, as will be shown in further sections, that enable us to calculate the values of tensors [3]. For practical purposes, sometimes it is more convenient to consider curved planes instead of flat ones in a coordinate system. For example, for cases when a cylindrical or a spherical water or an oil tank is involved, we prefer that all or some of coordinate surfaces match the geometry of the tank. For a cylindrical coordinate system, we consider the Cartesian system again but replace one of its planes with a cylinder, whereas for a spherical system we replace two flat planes with a sphere and a cone.

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4 • Tensor Analysis for Engineers, 3E

FIGURE 2.1  Sketches of Cartesian, Cylindrical, and Spherical coordinate systems.

Many other coordinate systems are defined/used in practice such as ­parabolic, bi-spherical, etc. Figure 2.1 shows some examples of common coordinate systems. For organizing the future usage of symbols for arbitrary coordinate systems we define coordinate variables with superscripted letters, for reasons that will be explained in further sections. For Cartesian systems we use yi and for other arbitrary systems we use x i —note that i is merely an index, not a raised power. For example, in a 3D space we have y1 ≡ X , y2 ≡ Y, and y3 ≡ Z where ( X , Y, Z ) are common notations for Cartesian coordinates. In general, for N dimensional systems we have

 yi ≡ ( y1 , y2 , y3 ,, yN ) Cartesian coordinates  i 2.1 1 2 3 N  x ≡ ( x , x , x ,, x ) arbitrary coordinates In this book, we limit the geometry to that known as Euclidean geometry [2], [4], [5]. Therefore, curved space geometry (i.e., Riemann geometry), space-time, and discussions of General Relativity are not included. However, curvilinear coordinate systems and oblique non-orthogonal systems are covered, where applicable, and defined with reference to Euclidean geometry/space.

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CHAPTER

3

BASIS VECTORS AND SCALE FACTORS For the measurement of quantities we need to define metrics or scales in whatever coordinate system we use. These scales are, usually, vectors defined at a point (such as the origin) and are tangent to the coordinate surfaces at that point. These vectors are called basis vectors. For Cartesian system yi we  define basis vector Ei (see Figure 3.1). Note that subscript i (i.e., i = 1,2,, N) is merely an index and the reason why we used it as a subscript for basis ­vectors willbe  explained in a further section. Now, considering an incremental vector ds , from point P to neighboring point P′ , we define the direc tion of the basis vector as moving from P to P′ or Ei = yi P → yi + dyi P′ .   For example, E1 = y1 P → y1 + dy1 P′. Now we define the magnitude of Ei such that the magnitude of distance from P to P′ is given as



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 ds ( i ) = dyi Ei 3.1 Also, we can get the distance = PP′ ds = ( i ) dyi in a Cartesian coordinate  system. Therefore, the basis vectors magnitude is unity, or Ei = 1 . In other words, the basis vectors in Cartesian system have unit lengths and    are ­well-known unit vectors (usually represented by i , j, k ). This is more than a trivial result and as shown in further sections, enables us to calculate/­ measure quantities in other coordinate systems  with reference to the Cartesian ­system. The directed distance vector ds is given as  N i   = ds ∑ = dy Ei dyi Ei 3.2 i =1

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6 • Tensor Analysis for Engineers, 3E



 The magnitude of this vector is the square root of the dot-product of ds with itself,      ds = ds ⋅ ds = dyi dy j Ei ⋅ E j 3.3

 FIGURE 3.1  A Cartesian system with unit vectors and an incremental vector ds.

Assuming the Pythagorean theorem holds in Euclidean space we have



 = ds

= dyi dyi

∑ ( dy )

i 2

3.4

i

From Equations 3.3 and 3.4, we can conclude that



  1 for i = j 3.5 Ei ⋅ E j =   0 for i ≠ j

Which also shows that the basis/unit vectors in Cartesian system yi are mutually perpendicular or the system is orthogonal, as defined.



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Now we assume an arbitrary system x i, which may be neither rectilinear nor orthogonal (see Figure 3.2). In this system the distance between two  points—say P and P′—is the same when considering the vector ds . In other words, the vector components in system x i compared to those in system yi could change along with the basis vector corresponding to system x i such that the vector itself remains the same, or as an invariant. This requirement is simply a statement of independence of physical quantities (and the laws of nature) regardless of the coordinate system we consider for calculation and analysis. Therefore, we have   i = ds dx = ei dyi Ei 3.6

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Basis Vectors and Scale Factors • 7

 where ei is the basis vector corresponding to system x i. In general the basis  vector ei can vary, both in magnitude and/or direction, from point to point in space. Also, the dx i may be dimensionless, like the angle coordinate in a polar coordinate system.

FIGURE 3.2  Sketches of a curvilinear coordinate system and basis vectors1.

 The magnitude of basis vector ei is the scale factor hi, or



 hi = ei 3.7



Note that hi is unity for a Cartesian coordinate system and may be d ­ ifferent from unity in general curvilinear systems. We will derive formulae for the calculation of hi in an arbitrary coordinate system in further sections.  The unit vector e ( i ) can be defined as  ei  e ( i) = ( no summation on i ) 3.8 hi Obviously, in a Cartesian system the unit vectors and the basis vectors are identical, since hi is unity.

  Common copyright (cc), https://commons.m.wikimedia.org/wiki/File:Vector_1-form. svg#mw-jump-to-license 1

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CHAPTER

4

CONTRAVARIANT COMPONENTS AND TRANSFORMATIONS For transformations between systems x i and Cartesian yi , we must have the functional relationships between their coordinate variables. For example,

x i = Fi ( y1 , y2 ,, yN )4.1 gives N number of functions transforming yi to x i system. Inversely [4], we can transform from x i to yi system using function Gi given as



yi = Gi ( x1 , x2 ,, x N ) 4.2

For example, for a 2D polar coordinate system, ( r ,q ) ≡ ( x1 , x2 ) with reference to Cartesian system ( X , Y ) ≡ ( y1 , y2 ) we have 1 1 2  X = r cosq  y = x cos ( x ) ≡ 2 . Or inversely,  1 2 Y = r sin q  y = x sin ( x ) 2 2  1 r x ( y1 ) + ( y2 ) = X2 + Y2 =    ≡ .   y2  −1  Y  q = tan    x2 = tan −1  1  X   y  

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10 • Tensor Analysis for Engineers, 3E

Now by taking partial derivatives of Gi (see Equation 4.2) we can relate i the differentials dyi , in a Cartesian system, to dx in an arbitrary coordinate ­system as dyi =



∂yi j dx 4.3 ∂x j

1 For example, after expanding Equation 4.3 for i = 1 we get, dy = 1 dy =

∂y1 1 ∂y1 2 ∂y1 N ∂yi  . The transformation coefficient can be dx + dx + + dx ∂x j ∂x1 ∂x2 ∂x N  ∂y1 ∂y1    1  ∂x ∂x N  ∂yi  written in matrix form as j =     . The determinant of the ∂x  N  ∂yN   ∂y  ∂x N   ∂x1 transformation coefficient matrix is defined as the Jacobian of the transformation, or

= 

∂ ( y1 , y2 ,, yN ) = ∂ ( x1 , x2 ,, x N )



∂y1 ∂x1  ∂yN ∂x1

∂y1 1 ∂y1 2 dx + 2 dx +  ∂x1 ∂x

∂y1 ∂x N   4.4





∂yN ∂x N

The Jacobian can be interpreted as the density of the space. In other words, let’s say that we have  = 5 for a given system x i . This means that we have packed 5 units of Cartesian space into a volume in x i space through transformation from Cartesian to the given system. The smaller the Jacobian, the smaller the space density would be and vice versa. Similarly, we can use function Fi (see Equation 4.1) to have

dx i =

∂x i dy j4.5 ∂y j

It can be shown [4], [2], that the determinant of transformation coefficient ∂x i is the inverse of  , or ∂y j

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Contravariant Components and Transformations • 11

∂ ( x1 , x2 ,, x N ) = =  ∂ ( y1 , y2 ,, yN ) −1



∂x1 ∂y1  ∂x ∂y1

N

∂x1 ∂yN   



4.6

∂x ∂yN  N

Equations 4.3 and 4.5 show a pattern for the transformation of differentials dyi and dx i, respectively. That is, the partial derivatives of the c­ orresponding coordinates appear in the numerator of the transformation coefficient. Nevertheless, one can ask: does this pattern maintain for general system-tosystem transformation? The short answer is “yes” [7]. Hence, for arbitrary systems x i and x′i we have

= dx i

∂x i ∂x′i j j i ′ ′ = dx and dx dx 4.7 ∂x′ j ∂x j

Any quantity, say A i, that transforms according to Equation 4.7 is defined as a contravariant type, with the standard notation of having the index i as a superscript. Therefore, transformation A i  A′i reads

= A′i

∂x′i j ∂x j i j = A and A A′ 4.8 ∂x j ∂x′i

Obviously, performing the related calculations requires the functional relations between the two systems, i.e., x i = func ( x′1 , x′2 ,, x′N ) or x′i = func ( x1 , x2 ,, x N ) . This in turn requires having the Cartesian system as a reference for calculating the values of A j or A′i, since x i and x′i are arbii trary. For example, transforming A directly from a spherical to a cylindrical system requires having the functional relations between these coordinates with reference to the Cartesian system as the main reference. The contravariant component of a vector has a geometrical meaning as well. To show this, we consider a rectilinear non-orthogonal system ( x1 , x2 ) in 2D, as shown in Figure 4.1. The components of vector A can be obtained by drawing parallel lines to the coordinates x1 and x2 to find contravariant components A1 and A2, respectively.

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12 • Tensor Analysis for Engineers, 3E

FIGURE 4.1  Contravariant components of a vector in an oblique coordinate system.

With reference to Figure 4.1, we can also find another set of components, A1 and A2 of the same vector A by drawing perpendicular lines to the ­coordinates x1 and x2 . This is shown in Figure 4.2. Obviously, components A1 and A2 are different in magnitude from the contravariant components. We define A1 and A2 as covariant components. In the next section we define the transformation rule for covariant quantities.

FIGURE 4.2  Covariant components of a vector in an oblique coordinate system.

We can conclude from these definitions that contravariant and covariant components of a vector in a Cartesian system are identical and there is no distinction between them.

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CHAPTER

5

COVARIANT COMPONENTS AND TRANSFORMATIONS



We use the standard notation of writing the index i as a subscript for  ­covariant quantities. We consider basis vector ei as a covariant quantity. We can rewrite Equation 3.6, considering two arbitrary systems x j and x′k  and the fact that ds is coordinate-system independent or invariant, as    ′k e′k dx j e j 5.1 = ds dx = ∂x j   After substituting for dx j , using Equation 4.7, we get dx′k e′k = k dx′k e j or ∂x′



 ∂x j   dx′k  e′k − k e j  = 0 5.2 ∂x′  

Since dx′k is arbitrary (i.e., the relation is valid for any choice of system = x′k and selections 0 , for all values of i = 1,2,, N ) of dx′k i ≠ 0 and dx′k ≠ i = therefore the expression in the bracket must be equal to zero, or we have



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∂x j   e′k = k e j 5.3 ∂x′

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14 • Tensor Analysis for Engineers, 3E

All quantities, say Ai , that transform according to Equation 5.3 defined as covariant type, with the standard notation of writing the index i as a ­subscript. Therefore, transformation Ai  A′j reads

= Ai

∂x′ j ∂x i ′ ′ = A and A Ai 5.4 j j ∂x i ∂x′ j

See Figure 4.2 for the geometrical interpretation of the covariant c­ omponent of a vector.

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CHAPTER

6

PHYSICAL COMPONENTS AND TRANSFORMATIONS





Having defined the contravariant and covariant quantities, like those of the components of a vector, one can ask this question: which one of these two types of components is the actual vector’s components in magnitude? The short answer is “none”! However, note that the combination of the contravariant or covariant components and their corresponding basis vectors (contravariant basis vectors will be defined in further section) gives the magnitude of the vector correctly. To find the actual magnitude of the vector, we should find the unit vectors and then the components of the vector corresponding to these unit vectors, or the physical components. In other words, the physical components of a vector (or in general, a tensor) are scalars whose physical dimensions and magnitudes are those of the components vectors tangent  to the coordinates’ lines. Therefore, for vector A , if we designate A ( i ) as  its physical component and the corresponding unit vector as e ( i ) we can write, then   i = A A= ei A ( i ) e ( i ) 6.1  Now, using Equation 3.8 and substituting for ei  i A e ( i ) hi = A ( i ) e ( i ) . Therefore,

we can write

A ( i ) = A i hi , ( no summation on i ) 6.2 Or A (1 ) = A1 h1 , A ( 2 ) = A2 h2 , and so forth. The scale factor hi is the parameter that turns the contravariant component into the physical component, when multiplied by it. Again, we can conclude that in a Cartesian system the physical and contravariant/covariant components are identical, since hi = 1 .

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16 • Tensor Analysis for Engineers, 3E Up to this point we have defined contravariant, covariant, and physical components of a vector along with covariant and physical/unit basis vectors. In principle, we can represent a vector by any combination of its components with the corresponding basis vectors. This requires having the contravariant basis vector definition. We will define this quantity and how it is transformed in a further section. First, we must have the definition of a tensor, as well as a specific tensor called a metric tensor.

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CHAPTER

7

TENSORS—MIXED AND METRIC In previous sections, we defined vectors or tensors of rank one. Higher order tensors could be defined based on similar definitions. For example, a tensor of rank two requires more than one free index, like mechanical stress or strain tensors. In general, we have the contravariant components written as A ij for tensor A . We avoid placing double-arrows over the symbol for tensors, for simplicity and generality. Since, in principle, we can have tensors of rank N, designating them with N number of hatted arrows is not a practical exercise. The invariant quantity is written as

 A = A ij ei e j 7.1

Since A is invariant, its value remains the same regardless of the coordinate system used. We transform A from an arbitrary coordinate system x′i  ij   to another system, say x i , or A′ e′i e′j = A km ek em . Now, using Equation 5.3,   we transform the covariant basis vectors ek and em to system x′i . Therefore, ∂x′i ∂x′ j    we have A km ek em = A km k m e′i e′j . Using these last two relations, we obtain ∂x ∂x j i ′ ′   ∂ x ∂ x  0 . Therefore, since basis vectors are non-zero, e′i e′j  A′ij − k m A km  = ∂x ∂x   we have

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A′ij =

∂x′i ∂x′ j km A 7.2 ∂x k ∂x m

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18 • Tensor Analysis for Engineers, 3E Quantities that transform according to Equation 7.2 are defined as doubly contravariant tensors, of rank two. Similarly, we can define mixed tensors, for example of rank two, as

A′ji =

∂x′i ∂x m k Am 7.3 ∂x k ∂x′ j

The rule for transformation of tensors is like those given for vectors, i.e., the contravariant component of the tensor transforms like a contravariant vector and the covariant component like a covariant vector. Also, we can expand the definition to a doubly covariant tensor, like

A′ij =

∂x k ∂x m Akm 7.4 ∂x′i ∂x′ j

Equations 7.2–7.4 are useful relations defining second rank tensors and their transformations. Expanding this rule, we can write the transformation of a N -rank tensor, for example contravariant components, as

A′ijN =

∂x′i ∂x′ j ∂x′N kmM  A 7.5 ∂x k ∂x m ∂x M

In an arbitrary curvilinear system, the Pythagorean Theorem applies but does not appear in the same form as it does in flat Cartesian systems. This is mainly due to the fact that basis vectors in an arbitrary system change in magnitude and/or direction, from point to point. Therefore, measuring the distance ds between two points on a curved surface, like a sphere, for example, requires scaling the coordinates by multiplying them with some scalar coefficients. These coefficients are the components of a tensor, called a metric tensor, associated with the coordinate system selected. Recalling Equation 3.6,  we can find the square of the magnitude of differential distance by dot-­ ds     product operation, or ds2 = ds ⋅ ds = ( dx i ei ) ⋅ dx j e j . The result is

(

)



  = ds2 dx i dx j ei ⋅ e j 7.6



The term in the bracket, on the R.H.S of Equation 7.6, is a scalar (i.e., a t­ ensor of rank zero) resulting from a dot-product operation on the basis ­vectors. Hence, we can write it as     = ei ⋅ e j ei e= hi h j cos a j cos a 7.7

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(

)

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Tensors—Mixed and Metric • 19

where a is the angle between the tangents to x i and x j axes of the c­ oordinate system at the selected point in space. Geometrically speaking, Equation 7.7 is the multiplication of the magnitude of a basis vector with the projection of the other one along the direction of the original one. Therefore, this ­quantity could be used to identify whether a system (or at least the basis vectors selected) is orthogonal or not. Also, it could be a metric for ­measuring the distance over a curved surface. It has more properties and applications, which we will encounter in future sections; hence, it is designated by a s­ ymbol and a name, i.e., metric tensor gij   gij= ei ⋅ e j 7.8



It is easily seen from Equation 7.8 that a metric tensor is symmetric since order in a dot-product is irrelevant (i.e., commutative). Also, if gij = 0 for i ≠ j then the coordinates of the system x i are orthogonal. For example, the metric tensor in a 3D orthogonal system can be represented by a 3 × 3 matrix containing null off-diagonal elements,



 g11  gij =  0  0

0   0  , orthogonal system 7.9 g33 

0 g22 0

It would be useful to apply metric tensor definitions and expand Equation 7.6 using Equations 7.7 and 7.9 for an orthogonal 3D system, which gives ds2 = h12 ( dx1 ) + h22 ( dx2 ) + h32 ( dx3 ) 7.10 2



2

2

For example, in a Cartesian system we recover the familiar form of 2 2 2 2 the Pythagorean   theorem (i.e., ds = ( dX ) + ( dY ) + ( dZ ) ), since g11 = g22 = g33 = Ei ⋅ Ei = 1. Similarly, for a spherical coordinate system ( r ,j ,q )  no sum on i

2 2 2 g= 1 , h= g= r 2 , and h= g= r 2 sin 2 j (see Example 8.2 we have h= 1 11 2 22 3 33 2 2 for scale factors). Therefore, we obtain ds = ( dr ) + r 2 ( dj )2 + r 2 sin 2 j ( dq )2 .

So far, we have not shown that gij is a tensor. To do this, we transform the metric tensor to another arbitrary coordinate system, say x′i . We then can   ∂x k ∂x m   ∂x k ∂x m ′ ⋅ e e g = gkm ; write g′ij = e′i ⋅ e′j = i . Therefore, we have ( ) m ij k  ∂x′ ∂x′ j  ∂x′i ∂x′ j = gkm

that is, when compared to Equation 7.4 we can conclude that gkm is a doubly covariant tensor.

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20 • Tensor Analysis for Engineers, 3E As mentioned in the previous sections, to calculate or measure tensorlike quantities we use a Cartesian system as the reference. To evaluate gij  we use Equation 7.8 and transform the covariant basis vector ei to the k m     ∂y ∂y Cartesian system, yi . Therefore, gij = ei ⋅ e j = i Ek ⋅ Em . The quanj ∂x ∂x   tity Ek ⋅ Em , or the dot-product of the basis/unit vectors in the Cartesian ­system, is either one or zero, due to orthogonality of the coordinates. Hence,   1, when k = m . Using this property, we can write Ek ⋅ Em =  0, when k ≠ m

(



gij =

)

∂yk ∂yk 7.11 ∂x i ∂x j

Using Equation 7.11, we can readily conclude that the metric   tensor in a Cartesian system is the familiar Kronecker delta, d = Ei ⋅ E j . For 3D ij 1 0 0    Cartesian coordinates, we receive a diagonal unity matrix d ij = 0 1 0  . 0 0 1  It can be shown ([3], [4]) that the determinant of a metric tensor is equal to the square of the Jacobian of the transformation matrix, or

= g g=  2 7.12 ij

Therefore, in an orthogonal system xi , having gij = 0 for i ≠ j , we receive

= g g= h12 h12  hN2 , orthogonal system 7.13 11 g22  g NN   gii e= hi2 (no sum on i ) (see Equation 7.7). From Note that= i ei Equations 7.12 and 7.13 we can conclude that the Jacobian for an orthogonal system is equal to the products of the scale factors, or



 = h1 h2  hN , orthogonal system 7.14 For example, in a spherical coordinate system the Jacobian is = h= r 2 sin j (see Example 8.2 for scale factors). spherical r hj hq

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CHAPTER

8

METRIC TENSOR OPERATION ON TENSOR INDICES An important and useful application of metric tensor is that it can be used to lower and raise indices of a tensor. Therefore, we can change a contravariant component of a tensor to a covariant one, or vice versa, by multiplying the appropriate metric tensor to the tensor at hand. For example, let’s define the covariant component of a vector (i.e., a tensor j of rank one) as Ai = A gij , where A j is the given contravariant compoj nent. Now, we show that the quantity A gij is transformed like a covariant quantity, according to Equation 5.4. Considering systems x i and x′i ,  ∂x′ j k   ∂x m ∂x n   ∂x′ j ∂x n  ∂x m k we can write A′ j g′ij = = A g A gmn .   k  i j mn  j  k i   ∂x ∂x′  ∂x′  ∂x   ∂x′ ∂x′ 1 , n = k ∂x′ j ∂x n ∂x n = = d kn , where d kn =  is j k k ∂x ∂x′ ∂x  0, n ≠ k a mixed second rank tensor (a Kronecker delta). Therefore, we have  ∂x′ j ∂x n  ∂x m k ∂x m n k ∂x m n = A g = d A g ( k ) mn ∂x′i ( A gmn ) . The expression in  k mn j  i ∂x′i   ∂x ∂x′  ∂x′ n

Note that the expression

=A

the last bracket is actually Am, according to our definition. Finally, we have ∂x m ′ j g′ij = A′i A= Am , which clearly shows that the quantity A′ j g′ij is trans∂x′i formed like a covariant component (see Equation 5.4). Therefore, we have, also using the symmetry property of a metric tensor or gij = g ji ,

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j = Ai A= gij A j g ji 8.1

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22 • Tensor Analysis for Engineers, 3E



Using Equation 8.1, we can write the expression for a contravariant basis vector, or   ei = e j gij 8.2



which shows that the metric tensor lowers the contravariant index, whilethe dummy index is dropped out. This will enable us to write the vector A in terms of covariant or contravariant basis vectors, as     i = A A= ei A i gij= ej A i gij = e j A j e j 8.3



Note that by using physical components and corresponding unit vectors, we can write  j  i = A A= ei A= A ( k ) e ( k ) 8.4 je

(

) (

)

So far, we have defined the doubly covariant metric tensor gij . Similarly, the   doubly contravariant metric tensor is defined g ij = e i ⋅ e j . Therefore, we can write   ij  i = A A= ei A j g= ei A j e j Also, the combination of the two types of metric tensor gives the mixed one, as g ij g jk = d ki 8.5



ij Equation 8.5 can be derived as follow: we have= A i g= A j g ij g jk A k . In this 

(

=Aj

)

0 , but expression, we write A = d A , therefor we have A d − g g jk = ij k since A is arbitrary the expression in the bracket is zero. Hence g g jk = d ki. i

i k

k

k

i k

ij

With reference to Figure  3.2, Figure 4.1, Figure 4.2, and Equation 8.3 we can sketch the vector A in terms of its covariant/contravariant components with their corresponding basis vectors, as shown in Figure 8.1 (see Example 15.3). Users may also want to watch a related video at https://www.youtube.com/watch?v=CliW7kSxxWU.

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Metric Tensor Operation on Tensor Indices • 23

FIGURE 8.1  Sketch for covariant and contravariant components of a vector and basis vectors in a non-orthogonal coordinate system xi and alternate system x i .

In the following subsections, we present two examples demonstrating related calculations for cylindrical and spherical coordinate systems.

8.1  EXAMPLE: CYLINDRICAL COORDINATE SYSTEMS Consider a cylindrical polar coordinate system ( x1 , x2 , x3 ) ≡ ( r ,q , z ) where r is the radial distance, q the azimuth angle, and z the elevation from X − Y plane, w.r.t. the Cartesian coordinates ( y1 , y2 , y3 ) ≡ ( X , Y, Z ) , as shown in Figure 8.2. The functional relations corresponding to cylindrical and Cartesian systems are:

FIGURE 8.2  Cylindrical coordinate system.

= cosq  y1 x1 sin x2  X r=  2  = r sin q ≡  y= x1 sin x2. Find the inverse relations (i.e., x i = Fi ( y1 , y2 ,, yN ) Y  Z z=  y3 x 3 =     x i = Fi ( y1 , y2 ,, yN ) , the basis vectors ei and e i for the cylindrical coordinate system in terms of the Cartesian unit vectors. Also find the scale factors,  unit vectors, and metric tensors gij and g ij and line element magnitude ds .

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24 • Tensor Analysis for Engineers, 3E Solution: The functions Fi can be obtained by solving for X , Y, Z , using the functional relation, given

 1 1 2 2 2 x y y 2 2 = + ( ) ( )   r X +Y =    y1    2 −1  Y  . To find the covariant basis ­vectors, x tan −1  2  = q tan   ≡ = X  y   z = Z  x3 = Y 3   j 1 2 3    ∂y  ∂y  ∂y   ∂y  we use ei = i E j . Therefore, we receive e1 = 1 E1 + 1 E2 + 1 E3 = cosq E1 + sin q E2 ∂x ∂x ∂x ∂x 1  2  3  1  2  3      ∂y ∂y ∂y ∂y ∂y  ∂y . Similarly, , e = E + E + E = − r sin q E + r cos q E E + E + E = cos q E + sin q E 2 1 2 3 1 2 1 2 3 1 2 2 2 2 ∂x ∂x ∂x ∂x1 ∂x1 ∂x1 1 2 3   ∂y  ∂y  ∂y  and e3 = 3 E1 + 3 E2 + 3 E3 = E3 . In terms of coordinates notation, we ∂x ∂x ∂x have     er cosq E1 + sin q E2 =     − r sin q E1 + r cosq E2  eq =    ez = E3  ∂x i  For calculating contravariant basis vectors, we receive e i = j E j. After ∂y ­expansion and using the functional relations, we get     e r cosq E1 + sin q E2 =  sin q  cosq   q − E1 + E2 e = r r   z  e = E3 The scale factors are the magnitudes of the covariant basis vectors. Hence,     h= h= er ⋅ er = 1 , h= h= eq ⋅ e= r , and h= h= 1 . The unit vectors 1 r 2 q q 3 z  e  e ( i ) = i , are hi     e ( r ) cosq E1 + sin q E2 =     − sin q E1 + cosq E2  e (q ) =    e ( z ) = E3

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Metric Tensor Operation on Tensor Indices • 25

  Note that unit vectors e ( r ) and e (q ) in cylindrical coordinate system change with location and are not constant vectors.   2 2 2 h= 1 , gqq= h= r 2 , g= h= 1 , and The metric tensor is gij = ei ⋅ e j , or g= rr r q zz z the off-diagonal elements of the metric tensor are null, hence the coordinate 1 0 0    system is orthogonal, or gij = 0 r 2 0  . The Jacobian= is  h= r. r hq hz 0 0 1    The contravariant metric tensor can be calculated using g ij = e i ⋅ e j , or 1 0 0    ij g = 0 r −2 0  , which is equal to the inverse of covariant metric tensor. 0 0 1 

Another way for calculating the contravariant basis vectors is by using the    2    ij  11  33  22  e 1 g= e1 e1 = e 3 g= e3 e3 . relation e i = g e j , or= , e g= e2 r −2 e2 , and=  2 2 2 2 For the line element we have, ds = ( hi dx i ) = ( dr ) + ( rdq ) + ( dz ) .

8.2  EXAMPLE: SPHERICAL COORDINATE SYSTEMS Consider a spherical polar coordinate system ( x1 , x2 , x3 ) ≡ ( r ,j ,q ) where r is the radial distance, j the polar/meridian angle, and q the azimuthal angle w.r.t. the Cartesian coordinates ( y1 , y2 , y3 ) ≡ ( X , Y, Z ) , as shown in Figure 8.3. The functional relations corresponding to spherical and Cartesian systems are:

FIGURE 8.3  Spherical coordinate system.

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26 • Tensor Analysis for Engineers, 3E

= sin j cosq  y1 x1 sin x2 cos x3  X r=   = ≡  y2 x1 sin x2 sin x3 . Find the inverse relations (i.e., Y r sin j sin q = 1 2   3 = cos j  Z r=  y x cos x   x i = Fi ( y1 , y2 ,, yN ) , the basis vectors ei and e i for the spherical coordinate system in terms of the Cartesian unit vectors. Also find the scale factorsand  unit vectors and metric tensors gij and g ij and line element magnitude ds . Solution: The functions Fi can be obtained by solving for X , Y, Z , using corresponding func­  2 2 2  x1 =  y1 ) + ( y2 ) + ( y3 ) (  2 2 2  r= X + Y + Z       y3 Z  2 −1 −1  = = ≡  x cos  tional relations, or j cos   2 2 2 2 2 2  X +Y +Z     ( y1 ) + ( y2 ) + ( y3 )    Y q = tan −1     y1  X  x3 = tan −1  2    y 

  .  

 ∂y  To find the covariant basis vectors, we use ei = i E j. Therefore, we get ∂x 1 2 3     ∂y  ∂y  ∂y  e1 = 1 E1 + 1 E2 + 1 E3 = sin x2 cos x3 E1 + sin x3 sin x2 E2 + cos x2 E3 Similarly, ∂x ∂x ∂x 1 2 3     ∂y  ∂y  ∂y  e2 = 2 E1 + 2 E2 + 2 E3 = x1 cos x2 cos x3 E1 + x1 sin x3 cos x2 E2 − x1 sin x2 E3 , and ∂x ∂x ∂x 1  2    ∂y ∂y3   ∂y e3 = E + E + E3 = − x1 sin x2 sin x3 E1 + x1 cos x3 sin x2 E2 . In terms of coor1 2 3 3 3 ∂x ∂x ∂x dinate variables, we have      er = sin j cosq E1 + sin j sin q E2 + cos j E3       ej = r cos j cosq E1 + r cos j sin q E2 − r sin j E3    − r sin j sin q E1 + r sin j cosq E2  eq = j

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Metric Tensor Operation on Tensor Indices • 27

The scale factors are the magnitudes of the basis vectors. Hence       h= h= er ⋅ er = 1 , h= h= eq ⋅ e= r sin j . h= ej ⋅ e= r , and h= 1 r 3 q q 2 j j  e  The unit vectors e ( i ) = i , are hi      e ( r ) = sin j cosq E1 + sin j sin q E2 + cos j E3       e (j ) = cos j cosq E1 + cos j sin q E2 − sin j E3    − sin q E1 + cosq E2  e (q ) = Note that unit vectors in spherical coordinate system change with l­ocation   2 h= 1, and are not constant vectors. The metric tensor is gij = ei ⋅ e j , or g= rr r 2 2 2 2 2 gjj= h= r , gqq= h= r sin j , and the off-diagonal elements of j q the metric tensor are null, hence the coordinate system is orthogonal, 1 0  0   = r 2 sin j . The g ii = hi−2 , is  h= gij = 0 r 2 0  . The Jacobian r hj hq  2 0 0 ( r sin j )  1 0  0   or g ij = 0 r −2 0  , then we can calculate the contravariant basis  −2  0 0 ( r sin j )    1     1   vectors, as e j = g ij ei , or e 1 = e1 , e 2 = 2 e2 , and e 3 = 2 2 e3 which yields r r sin j

     e r = sin j cosq E1 + sin j sin q E2 + cos j E3   j cos j cosq  cos j sin q  sin j  E1 + E2 − E3 e = r r r  sin q  cosq   q e E E2 = − + 1  r sin j r sin j   For the line element we have, ds =

TAE3E.CH08_1pp.indd 27

( h dx )

i 2

i

=

( dr )2 + ( rdj )2 + ( r sin j dq )2 .

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CHAPTER

9

DOT AND CROSS PRODUCTS OF TENSORS We often must multiply tensors by each other. We consider two vectors for discussion here, without losing generality. When we multiply two scalars (i.e., tensors of rank zero) we just deal with arithmetic multiplication. But a vector has a direction, in addition to its magnitude, that should be taken into consideration for multiplication operations. In principle we can have two vectors  just multiply by each other like AB, which is a tensor of rank two, called a dyadic product. Or, multiply the vectors’ magnitudes and form a new vector with the resulting magnitude directed perpendicular to the plane containing   the original vectors, the cross-product A × B, which is a tensor of rank one. Or, multiply the vectors’ magnitudes and reduce the rank  of  the product by projecting one vector on the other one, the dot-product A ⋅ B , which is a tensor of rank zero. Obviously, for higher than rank-one tensors we would have a greater number of combinations as the result of their products. In general, compared to the rank of original tensors, dyadic product increases the rank, cross-product keeps the rank, and dot-product decreases the rank, i.e., the result is a tensor quantity of higher, the same, or lower rank (by one level/ rank) w.r.t to the original quantities, respectively.   We start with dot-product operations. Let’s consider A and B ;   two vectors   ⋅ B ( A i ei ) ⋅ B j e j , using the dot-product of these two vectors is given by A = the contravariant components of the vectors. We expand this expression, with Equations 7.7 and 7.8, to receive

(



TAE3E.CH09_1pp.indd 29

)

  j A= ⋅ B A i B= gij A i B j hi h j cos a 9.1

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30 • Tensor Analysis for Engineers, 3E



where a is the angle between the two vectors. Further, using Equation 8.1 we receive   A⋅B = A i B=i A j B j 9.2 Similarly, we could start with using the covariant components of the ­vector, but the conclusive results would be the same as the expression given by Equation 9.2. Readers may want to do this as an exercise. From Equation 9.2 we can conclude that the order of multiplication does not change the result of the dot-product, or A i Bi = Bi A i . In  an  orthogonal Cartesian system, either Equation 9.1 or 9.2 would result in A ⋅ B= A i B jd ij= A1 B1 + A2 B2 +  + A N BN= A1 B1 + A2 B2 +  + AN BN . The second equality results from the fact that in orthogonal Cartesian systems there is no distinction between contravariant and covariant components. An example of the dot-product quantity of two vectors is the mechanical work— the dot product of force and distance vectors. The cross-product operation can be established with a more general ­formulation using the permutation symbol (also referred to as the LeviCivita symbol) defined as +1 even permutation  ei1 i2 iN = −1 odd permutation 9.3 0 repeated index 



For example, for N = 6 we receive ei1 i2 i6. The permutation is considered even if an even number of interchanges of indices put the indices in arithmetic order (i.e., 123456 ), and is considered odd if an odd number of interchanges of indices put the indices in order. For any case with a repeated index the permutation symbol is zero. As an example, let’s consider e342165 . Examining the indices, we find out that 4 interchanges (i.e. ( 3 ↔1 )

(4↔2)

(4↔3)

(6 ↔ 5)

342165 → 142365 → 124365 → 123465 → 123456 ) put them in order; note that the sequence of interchanges is irrelevant. Therefore e342165 = 1 . Similarly, e362145 = −1 , since 5 interchanges put the indices in order. If any of the indices is repeated then it is zero, for example e= e= 0, etc. We 344165 142165 purposefully called the permutation symbol a symbol and not a tensor, for reasons that will be explained later in this section. Another way of identifying the even or odd number of permutations is to write down the given indices as a set and connect them with the equivalent number set but in arithmetic

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Dot and Cross Products of Tensors • 31

order. Then the number of points at the cross-section of the lines connecting the same numbers is the order of permutation. This method is shown in the following figure for e342165. The connecting lines intersect at six points; hence the permutation is even. 3

4

2

1

6

5

1

2

3

4

5

6

  Now considering two vectors A and B in a Cartesian system the cross product of these two vectors is another vector C , given as (we propose this expression, for now at least)



C i = eijk A j Bk = − eijk B j A k , Cartesian 9.4       A× B

B× A

 for the i -component of C . This relation clearly shows that the order of multiplication matters for cross-product operation. For example, in a ­   3D Cartesian system we have A = ( A1 , A2 , A3 ) and B = ( B1 , B2 , B3 ) and    we find their cross-product using Equation 9.4, C= ( C1 , C2 , C3 )= A × B as, = C1 A2 B3 − A3 B2 = C3 A1 B2 − A2 B1 . , C2 A3 B1 − A1 B3 , and=

Note that in Equation 9.4, we have written the relation for the contravari ant component of the resulting vector C. As we mentioned previously, in a Cartesian system contravariant or covariant components need not be considered; hence we usually place all indices as subscripts. But in an arbitrary system we should make sure that the resulting C i is transformed like a contravariant component/quantity. To show this, we set guidelines to form a tensor expression in an arbitrary coordinate system, x i as follows: a. Identify the rank of the expression representing the desired quantity in terms of a tensor; i.e., zero is a scalar, one is a vector, two or more a ­tensor of the corresponding rank. b. Identify and decide if the expression should be in contravariant or ­covariant form and place the indices appropriately (i.e., as subscript for covariant and superscript for contravariant). c. Form the expression based on guidelines (a) and (b).

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32 • Tensor Analysis for Engineers, 3E

d. Transform the expression from arbitrary system x i to another system, ∂x i ∂x′i say x′i (or vice versa) using the linear operators j or according to ∂x′ j ∂x the combination rule (see Sections 4 and 5). e. Show that the resulting expression recovers the related familiar form in Cartesian coordinates. Usually, starting from an expression form in a Cartesian system (as listed in part (e), above) gives a reasonable and usually the correct form to start with. Let’s apply these guidelines to the expression given by Equation 9.4. By observing the expression, we conclude that the contravariant components A j and Bk are correct forms since they transform like contravariant quantities. However, for the permutation symbol we need to make sure that it is a tensor quantity, i.e., it transforms like a tensor. It can be shown that the permutation symbol does not transform like a tensor unless a necessary adjustment is implemented [2], [4], [7] by multiplying it by the Jacobian of the transformation (see Equation 4.4). Now, we define the covariant and contravariant permutation tensors by multiplying Jacobian and inverse of Jacobian to the permutation symbol, respectively, given as



i1 i2 iN = ei1 i2 iN 9.5  i1 i2 iN =  −1 e i1 i2  iN   Therefore, the general expression for the vector C, for example, is Ci = eijk A j Bk and C i =  −1 eijk A j Bk . Using permutation tensor, we will get j k Ci = eijk A j Bk covariant component = ijk A B 9.6  i ijk A j Bk  −1 eijk A j Bk contravariant component = C =



Note that in Equation  9.6, for the covariant (contravariant) component of the resulting vector C , we used the combination of covariant (contravariant) permutation tensor and (covariant) components of the two  contravariant  contributing vectors A and B. j To show that, for example, ijk A Bk transforms like a covariant quantity, we write down its transformation between two arbitrary systems, or l  ∂x l ∂x m ∂x n   p ∂x′ j   q ∂x′k  p q ∂x ′ A′ j B′k  lmn i = A = B = A B d md n ijk  lmn j p  q  k  i p q ′ ′ ′ ′ ∂x ∂x ∂x   ∂x   ∂x  ∂x 

∂x l . Similarly, for the contravariant form (i.e., the second ∂x′i ­expression in Equation 9.6) it can be shown that it does indeed transform like a contravariant quantity. Readers may want to do this as an exercise.

(

lmn

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A m Bn )

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Dot and Cross Products of Tensors • 33

The cross-product of two vectors can be written in terms of their physical components as well. This can be done using, for example, the contravariant component when multiplied by the corresponding scale factor. Therefore, we can write for an orthogonal system (using Equations 9.5, 9.6, and 11.8) hi eijk hi eijk  h j hk A ( j ) B ( k )  , or hiC i = hi ijk A j Bk= , or C ( i ) = A j Bk      C( i )

eijk

C ( i) =





hi h j hk  A ( j ) B ( k )  , orthogonal 9.7

For orthogonal coordinate systems, using  = h1 h2 h3 , we receive 



( A × B) (1) =C (1) = A ( 2 ) B ( 3 ) − A ( 3 ) B ( 2 )   ( A × B) ( 2 ) =C ( 2 ) = A ( 3 ) B (1) − A (1) B ( 3 ) 9.8   ( A × B) ( 3 ) =C ( 3 ) = A (1) B ( 2 ) − A ( 2 ) B (1)

     



Expanding on these results, in a N -dimensional space, the generalized crossproduct of N − 1 vectors reads, using Equation 9.6,



 Ci1 = i1 i2 iN A i2 Bi3  Z iN 9.9  i1 i i i i i i C =  1 2 N A 2 B 3  Z N

 where now  vector C is perpendicular to the hyperplane formed by the N − 1 vectors A through Z .

Immediately, using Equation 9.9 we can conclude that in an arbitrary system the unit volume d is, [2], d  = dx1 dx2  dx N 9.10



For example, in the spherical system ( r , j , q ) we have d  = r 2 sin j drdj dq . Another useful extension is for the orthogonal systems (i.e., gi ≠ j = 0), for     which we get= gij ∏ = gii g11 g22  gNN . But g11 = e1 ⋅ e1 = h12 , g22 = e2 ⋅ e2 = h22 and so forth. Hence, gij = h12 h22  hN2 . Using the relationship gij =  2 , we finally reach at



TAE3E.CH09_1pp.indd 33

= 

h1 h2  hN orthogonal system 9.11

i

i

And hence = d

= h ∏

= h dx ( h dx )( h dx )( h ∏ i

i

i

1

1

2

2

N

dx N ) orthogonal system 9.12

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34 • Tensor Analysis for Engineers, 3E

9.1 DETERMINANT OF AN PERMUTATION SYMBOLS

MATRIX USING

Using the permutation symbol eijk (see Equation 9.3), we can write the eijk M i1 M j 2 Mk 3, ­determinant of a 3 × 3 matrix= [ M ] , as M e= ijk M1 i M2 j M3 k given by expansion based on row or column, respectively, and M ij represent the elements of the matrix [2], [4]. To expand on these results, we consider [ M ] to be an N × N matrix, (N ≥ 3). Therefore, we can write its determinant as = M e= ei1 i2 iN M i1 1 M i2 2  M iN N 9.13 i1 i2 iN M1 i1 M2 i2  M NiN See the application example in Section 17.10.

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CHAPTER

10

GRADIENT VECTOR OPERATOR—CHRISTOFFEL SYMBOLS In engineering and science, most if not all governing equations (e.g., equilibrium, Navier-Stokes, Maxwell’s equations) contain terms that involve derivatives of tensor quantities. These equations are mathematical models of related quantity transports, like momentum, mass, energy, etc. Different forms of derivatives result from the gradient operator, which is a vector that takes the derivative of and performs dyadic, dot-product, cross-product, etc. operations on the involved tensors. Depending on the rank of the tensors under the gradient operation, the result could be complex expressions. In the previous sections, we have defined and derived necessary formulae to tackle the derivation of gradient vector transformation and consequently find the general forms of related derivatives appearing in the governing equations.

10.1 COVARIANT DERIVATIVES OF VECTORS— CHRISTOFFEL SYMBOLS OF THE 2ND KIND  ∂ The covariant component of the gradient vector ∇ is defined as ∇ i = i , ∂x   ∂  or we can write the full vector as ∇ = e i i = e i∇ i . To show that ∇ i trans∂x forms like a covariant component between two arbitrary systems, we can ∂ ∂x j ∂ ∂x j ′i write ∇= = = ∇ j . This expression clearly shows that ∇′i is a ∂x′i ∂x′i ∂x j ∂x′i covariant quantity. The transformation from x′i to x i is obvious.

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36 • Tensor Analysis for Engineers, 3E When a gradient operates on a scalar, or tensor, of rank zero like temperature, pressure, or concentration, we can write it as ∇ iy and the ­corresponding ∂x j transformation is ∇′iy ′ = ∇′iy = i ∇ jy , using the fact that scalar y is ∂x′ ­invariant (or y = y ′ ) and independent of the coordinate systems selected.  Now, assuming that the gradient operates on a vector, say , which is also A  an invariant quantity, the  result is ∇A, or a tensor of rank two. The covariant component is then ∇ i A, which is a vector by itself. Expanding, we get

    ∇i A = ∇i A je j = e j ∇ i A j + A j ∇ i e j 10.1

(

)

(

)

(

)

The first term in the R.H.S of Equation 10.1 is the straight-forward derivative of the component A j . But the second term involves a gradient of the  basis vector e j , which is not necessarily zero in an arbitrary system since the basis vector’s magnitude and/or direction may change from point to point.  Note that in a Cartesian system, ∇ i E j = 0 , i.e., the second term in Equation  10.1, vanishes. Therefore, we can interpret ∇ i e j as a measure of the curvature of the arbitrary coordinate surface x i = constant (e.g., in a spherical coordinate system X 2 + Y 2 + Z 2 = r 2 is the surface of the sphere). The physi  cal meaning of ∇ i e j is this that it represents the rate of change of e j with  i respect to coordinate variable x in the direction of basis vector ek . This k quantity is represented by Γ kij (some authors use   instead) and is the  ij  Christoffel symbol of the second kind [8]. It may be useful to pause at this point and make sure that the physical/ geometrical meaning of the Christoffel symbol is well understood. To help with this understanding, we consider a simple 2D polar coordinate system, ( r,q ) ≡ ( x1 , x2 ) as shown in Figure 10.1.

FIGURE 10.1  2D Polar coordinate system.

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Gradient Vector Operator—Christoffel Symbols • 37







TAE3E.CH10_1pp.indd 37

  As shown, the covariant basis vectors er and eq change from point to point in space, e.g., points A and A’ , in general. Let’s consider, for example  ∂e  ∇q eq =q which represents the change for basis vector eq with respect to ∂q coordinate variable q . This derivative is a vector and has two components in the ( r,q ) coordinate system. Therefore, we can write it as a linear combina  tion of the basis vectors in polar coordinate system er and eq , or   ∂eq   ∇q eq = = a er + b eq 10.2 ∂q

Similar expressions can be written down for other possible derivatives (i.e.,    ∇q er , ∇ r eq , ∇ r er ). In our example (see Equation 10.2), the coefficients a and b are the corresponding Christoffel symbols, also  referred to as connec∂eq  tion coefficients. a is the component of the vector in the direction of er , ∂ q  ∂eq  r or Γqq and b is the component of the vector in the direction of eq , or ∂q Γqqq . Using these definitions, we can write all possible derivatives with their corresponding Christoffel symbols as   ∂eq r  q   ∂q = Γqq er + Γqq eq    ∂er = Γ r e + Γq e rq r rq q  ∂q 10.3    ∂eq = Γqr r er + Γqq r eq  ∂r  ∂e    r = Γ rrr er + Γqrr eq  ∂r The values of Christoffel symbols can be calculated using the functional relations between the polar and Cartesian systems. For this purpose, we need the relations for the basis vectors. Using Equation 5.3, we have   ∂X  ∂Y   ∂X  ∂Y  = = er E1 + E2 and eq E1 + E2 , where Ei are unit vectors in ∂r ∂r ∂q ∂q Cartesian system. Performing the calculations, using the functional relation between the polar and Cartesian systems, as X = r cosq and Y = r sin q , we receive     er cosq E1 + sin q E2 =   10.4  − r sin q E1 + r cosq E2  eq =

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38 • Tensor Analysis for Engineers, 3E    ∂eq  Using Equations 10.4 we receive = − r cosq E1 − r sin q E2 = − rer ,    ∂q    ∂er ∂e 1  ∂eq 1 eq , = − sin q E1 + cosq E2 = = − sin q E1 + cosq E2 = eq , and r = 0 . r ∂q ∂r r ∂r Finally, comparing these results with Equation 10.3, we find the Christoffel 1 r symbols values for polar coordinate system as Γqq = − r , Γqrq = Γqq r = , and r Γqqq =Γqrr =Γqr r =Γ rrq =Γ rrr =0 . Table 10.1 gives the summary of these results and the matrix form of these symbols for polar coordinate systems. TABLE 10.1  Christoffel symbols of the 2nd kind for a 2D polar coordinate system ( r ,q ) .

Symbol Meaning  r eq change w.r.t q Γqq  eq change w.r.t r Γqr r  er change w.r.t r Γ rrr  Γqqq eq change w.r.t q  er change w.r.t q Γqrq Γqrr

 in er direction  in er direction  in er direction  in eq direction  in eq direction

  er change w.r.t r in eq direction

Value −r

Symmetry −

0

= Γ

0

−

0

−

1/r 0

r rq

Matrix form  Γr Γ r = rrr Γq r

Γ rrq  r  Γqq 

 Γ rr Γq =  q

Γqrq  q  Γqq 



q

q qr

= Γ

Γq r

−

The number of independent Christoffel symbols can be obtained using N 2 ( N + 1) , for an N -dimensional system. For example, in a 2D polar coor2 4×3 dinate system ( r,q ) with N = 2 , we receive = 6 symbols, among them 2 only two are non-zero. Note that Christoffel symbols are symmetric with respect to the lower indices (see the next section). Now we continue with our general formulation and discussion, after establishing the physical/geometrical meaning of Christoffel symbols of ­ 2nd kind with this example.  Rewriting ∇ i e j using the definition of Christoffel symbol (see Equation 10.3), we have

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  ∇ie j = Γ kij ek 10.5

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Gradient Vector Operator—Christoffel Symbols • 39

   Plugging back into Equation 10.1, we receive ∇ i A = e j ∇ i A j + A j Γ kij ek . In this relation, for the last term k indices, since they are  we interchange j and   dummy indices. Then, ∇ i A = e j ∇ i A j + A k Γ ikj e=j  e j ∇ i A j + Γ ikj A k . The  expression in the bracket is a tensor, since both ∇ i A and e j are tensors. We j define the expression in the bracket as A, i where the comma in the subscript indicates differentiation and index i represents the covariant component of gradient vector. Hence, ∂A j = A, ji + Γ ikj A k 10.6 i ∂x

(





( (

)

)

)

Equation  10.6 is the covariant derivative of the contravariant component of vector A . Or   ∇i A = A, ji e j 10.7

10.2  CONTRAVARIANT DERIVATIVES OF VECTORS At this point in our discussion, it seems logical to seek the contravariant ­component of gradient vector, ∇ i . We can write this operator as ∇ i = g ij ∇ j using a metric tensor (see Section 8). Therefore, we can raise the derivative index by multiplying the contravariant metric tensor to the expression, or A,k ⋅i = g ij A ,kj 10.8







Equation 10.8  is the contravariant derivative of the contravariant component of vector A . Or   ∇i A = A,k ⋅i ek 10.9  From the start we could use the covariant component of vector A , to find the corresponding covariant and contravariant gradient components as well (see Equation 10.1). To this end, we can write the covariant derivative of the covariant component of vector A , as     ∇i A = ∇i A je j = e j ∇ i A j + A j ∇ i e j 10.10

(

)

(

)

(

)

The first term on the R.H.S of Equation 10.10 is straight-forward, the ­derivative of the component A j . The second term involves the gradient of

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40 • Tensor Analysis for Engineers, 3E



 the contravariant basis vector e j , for which we requirea new expression. To ∂e j   e, ji + Γ ikj e k . Due to the find this quantity, we use Equation 10.6 to write = i ∂x  fact that e, ji is a tensor (this can be shown by transformation of the terms involved between arbitrary systems, like A, ji ) and its value in a Cartesian system is null, hence  it is equal to zero in any arbitrary coordinate system.  j ∂e j  Therefore, = e, i + Γ ikj = e k 0 , or ∂x i   j ∂e j  ∇ie = = −Γ ikj e k 10.11 i ∂x j Substituting for ∇ i e from Equation 10.11, into Equation 10.10, we receive    ∇ i A j e j = e j ∇ i A j − A j Γ ikj e k . In this relation for the last term we interchange j and k indices, since they are dummy indices. Then, after factoring    out e j , we get ∇ i A j e j = e j ∇ i A j − Ak Γ kij . We define the expression in the bracket as A j,i where the comma in subscript indicates differentiation and i represents the covariant component of gradient vector. Hence

(

)

(

)

(





(

)

)

(

A= j, i

)

∂A j

− Γ kij Ak 10.12 ∂x i Equation  10.12 is the covariant derivative of the covariant component of vector A . Or   ∇i A = A j, i e j 10.13

10.3  COVARIANT DERIVATIVES OF A MIXED TENSOR Comparing Equations 10.6 and 10.12, we observe that for each index a term involving a Christoffel symbol is added to the expression on the R.H.S and when a covariant component of a vector is involved in a derivative operation, a minus sign is multiplied to the corresponding Christoffel symbol, contrary to a plus sign for when the contravariant component is involved. This rule can be used to extend the derivative calculation of a tensor of higher ranks. To show this, we consider a mixed tensor of third rank, for example Akij with   the invariant A = Akij ei e j e k . The covariant derivative can be written as

TAE3E.CH10_1pp.indd 40

∂Akij j i m + Γ nm Akjm + Γ nm Akim − Γ nk Amij 10.14 ∂x n   Therefore, ∇ n A = Akij, n ei e j e k which is the component n of a tensor of rank    four (i.e., ∇A = Akij, n ei e j e k e n ). ij A= k, n

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Gradient Vector Operator—Christoffel Symbols • 41

10.4 CHRISTOFFEL SYMBOL RELATIONS AND PROPERTIES—1ST AND 2ND KINDS The definition of a Christoffel symbol of the first kind, properties, and some relations for calculation of Christoffel symbols are discussed in this section. The properties are: 1. Symmetry of the lower indices: The Christoffel symbol of the second kind is symmetric w.r.t lower indices, or Γ kij =Γ kji 10.15



To show this, we use Equation 10.5, after changing index k → n , and  ­perform a dot product operation on it by contravariant basis vector e k. Or     e k ⋅ ∇ i e j = Γ nij e k ⋅ en = Γ kij. Therefore, we can write  d nk



  Γ kij = e k ⋅ ∇ i e j 10.16

(

)

We can use Equation 10.16 to calculate Christoffel symbol Γ kij in Cartesian coordinate system variables yi. After expanding the R.H.S ­ expression by appropriate transformation and using chain rule, we will receive  ∂x k   ∂yn ∂  ∂yp   ∂x k ∂yn ∂ ∂yp  m    e k= Ep  (E ⋅ Ep ) = ⋅ ∇ i e j  m Em  ⋅ i n  = j j m i n    ∂y  ∂x ∂y  ∂x  ∂y ∂x ∂y ∂x  dm

(

)

p

p ∂x k  ∂yn ∂x q  ∂ 2 y ∂x q . Where, in the last expression we inserted ( = 1,   ∂yp  ∂x i ∂yn  ∂x q ∂x j ∂x q  q

di

or using chain rule). Finally, we receive

p ∂x k ∂ 2 y Γ kij = p i j 10.17 ∂y ∂x ∂x

Equation 10.17 clearly shows the symmetry of Γ kij for its lower indices, since ∂ 2 yp ∂ 2 yp the order of differentiation is irrelevant, i.e., i j = j i . ∂x ∂x ∂x ∂x 2. Christoffel symbol is not a tensor: Because its value in a Cartesian ­system is zero; hence, if it transforms as a tensor it should be zero in any arbitrary system as well. But we know that this is not the case (see Equations 10.10 and 10.17).

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42 • Tensor Analysis for Engineers, 3E 3. Covariant derivative of the metric tensor: This quantity leads to a useful formula for calculating the value of a Christoffel symbol. First, we find the covariant derivative of the metric tensor, or       ∇ i g jk =∇ i e j ⋅ ek =e j ⋅ ( ∇ i ek ) + ∇ i e j ⋅ ek . Now, using Equation     10.5, we can write ∇ i ek = Γ nik en and ∇ i e j = Γ nij en . Therefore,   n  n  ∇ i g jk = Γ ik (e j ⋅ en ) + Γ ij (en ⋅ ek ) and finally we receive  

(

)

(

gnk

g jn



)

∇ i g jk =

∂g jk

= Γ nik g jn + Γ nij gnk 10.18 ∂x i Note that by comparing Equation 10.18 with the rule of covariant derivatives of tensors (see Equation 10.14) we can conclude that g jk, i = 0, a reasonable result considering that a metric tensor in a Cartesian system is a constant/ unity and hence its derivative is equal to zero in an arbitrary system. 4. Christoffel symbol of the first kind: So far, we have defined the Christoffel symbol of the second kind. We now derive formulas that can be used for calculating its value as well as defining a Christoffel symbol of the first kind. To do this, we manipulate Equation 10.18 by permuting the indices (i.e., i → j, j → k, and k → i ), twice in sequence. Hence we ∂g get two alternative but equivalent relations as kij = Γ nji g jn + Γ njk gni and ∂x ∂gij n n = Γ kj gin + Γ ki gnj . Now we subtract Equation 10.18 from the sum of ∂x k the latter two relations, using along the way the symmetry of both metric tensor and Christoffel symbols. The result reads



Γ nij gnk =

1  ∂g jk ∂gik ∂gij  + j − k  10.19  ∂x ∂x  2  ∂x i

In Equation 10.19, the quantity Γ nij gnk is the Christoffel symbol of the first kind, also written as Γ ij, k ≡ ij, k. We can use Γ nij gnk to find a relation for ­calculation of the Christoffel symbol of the second kind in terms of ­metric tensor. To do this we multiply it by g mk to get Γ nij gnk g mk =d nm Γ nij =Γ mij . Therefore, after multiplying Equation 10.19 by g mk, we get

= Γ mij

g mk  ∂g jk ∂gik ∂gij  + j − k  10.20  ∂x ∂x  2  ∂x i

Equation 10.20 states that the Christoffel symbol of the second kind is equal to the first kind multiplied by the doubly contravariant metric tensor related to the coordinate system in use. Observing Equation 10.19, a symmetry shows up. That is, the derivative is w.r.t. the coordinates indices and for each term in the bracket the remaining indices are used for the metric tensor.

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Gradient Vector Operator—Christoffel Symbols • 43

5. For an orthogonal coordinate system we can simplify Equation 10.20, using the property of gij and g ij being diagonal tensors or matrices ij (see Equation 7.9). In other words, we can write g= g= 0 for i ≠ j . ij Therefore, examining the terms on the R.H.S of Equation 10.20, we can ∂gij ∂g conclude that k = 0 and ikj ≠ 0 only when i = k . Also g mk ≠ 0 when ∂x ∂x ∂g jk m = k and ≠ 0 , since j cannot be equal to k . By implementing ∂x i these relations among the indices (i.e., i= k= m and i ≠ j ) and ­rewriting Equation 10.20, we receive g ii ∂gii Γ iij = , 2 ∂x j



no summmation on i 10.21

Equation 10.20, for i= j ≠ k = m reduces to



Γ kii =−

g kk ∂gii , 2 ∂x k

no summmation on i and k 10.22

For orthogonal coordinate systems, using Equation 7.13, we substitute for 2 hi−2 ∂ ( hi ) i 2 ii −2 , or gii = hi and g = hi in Equation 10.21 to obtain Γ ij = 2 ∂x j 1 ∂hi Γ iij = , no summmation on i, ( orthogonal ) 10.23 hi ∂x j Simply, by letting i = j in Equation 10.23, we receive



1 ∂hi Γ iii = , hi ∂x i

no summmation on i, ( orthogonal ) 10.24

2 hk−2 ∂ ( hi ) Similarly, Equation 10.22 yields Γ =− , or 2 ∂x k k ii



Γ kii =−

hi ∂hi , hk2 ∂x k

no summmation on i and k, ( orthogonal ) 10.25

6. Another useful relation can be derived for the value of Γ iin, which   is ei change w.r.t x n in the direction of ei. We use the fact that the derivative of the permutation tensor, or ,nijk is zero, since its value in the Cartesian coordinates is null. Using Equation 10.14, we can

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44 • Tensor Analysis for Engineers, 3E

∂ ijk j ijm i + Γ nm  mjk + Γ nm  imk + Γ knm = 0 . To simplify the ∂x n ­derivation, without losing generality, we select=i 1,=j 2,= k 3. Hence, ∂ 123 m 23 1 2 1m3 3 12 m we get = − ( Γ nm  + Γ nm  + Γ nm  ) . Collecting non∂x n 123 ∂ zero terms, we have = − ( Γ1n1  123 + Γ 2n2  123 + Γ 3n3  123 ) = − 123 Γ nii , n ∂x ­summation applies on i . Recalling that  123 = 1 /  and Γ nii = Γ ini , we ∂ 1 1 get = − Γ iin (recall  is the Jacobian). Performing the  n   ∂x    ­differentiation and rearranging terms, we receive write = ,nijk

1 ∂ Γ iin = n 10.26  ∂x



Note that Equation 10.26 works for arbitrary coordinate systems and it recovers Equation 10.23 when the system is orthogonal. 10.4.1. Example: Christoffel symbols for cylindrical and spherical coordinate systems In this example, we calculate and derive the relations for Christoffel symbols of the second kind, for cylindrical and spherical coordinate systems. The results are also expressed in terms of unit vectors (i.e., the physical components) for each coordinate system. Readers should note that the results depend on the order of coordinate axes, defined. In other words, attention should be given when reading comparable results from other references in relation to their defined corresponding coordinates axes. Cylindrical coordinates From the results of Example 8.1, for a cylindrical coordinate system ( r ,q , z ) (note the order of the coordinates defined here, see Figure 8.2) we can 2 2 write the non-zero terms of the metric tensor, as g= h= 1 , gqq= h= r 2, rr r q 2 rr and g= h= 1 . Therefore, the contravariant metric tensor reads g = 1, zz z gqq = r −2 , and g zz = 1 . Using Equations 10.20 and 10.25 and considering that indices i, j, k, m are various permutations of cylindrical variables r,q , z we receive the Christoffel symbol of the second kind as

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Gradient Vector Operator—Christoffel Symbols • 45



r Γqq

  Γ rrr  r  r Γ = Γq r   Γ rzr     Γq  q  qrr Γ = Γq r   Γqzr    Γz  z  zrr Γ = Γq r   Γ zzr  

Γ rrq r Γqq Γ rzq

Γ rrz  0 0 0     Γqr z  = 0 − r 0  Γ rzz  0 0 0 

Γqzq

Γqrz   0 1 / r 0     Γqq z  = 1 / r 0 0  10.27 Γqzz   0 0 0 

Γ rzq z Γqq Γ zzq

Γ rzz  0 0 0     Γqz z  = 0 0 0  Γ zzz  0 0 0 

Γqrq Γqqq

h ∂ ( hq ) 1 ∂ ( hq ) 1 ∂r ∂r 2 r 1 / r and Γqq For example, Γ12 = Γ 221 = Γqrq = = = = − q2 = −r = −r hq ∂r r ∂r ∂r hr ∂r h ∂ ( hq ) ∂r = − q2 = −r = − r . Recall that the symbols, given by Equation 10.27 are defined  hr ∂r ∂r ∂er   based on the covariant basis vectors, ei (see Equation 10.5), or = (1 / r ) eq  ∂q  Γqr q ∂eq  and = ( − r ) er . It is useful to write the Christoffel symbols in terms of ∂q  r Γqq

the ­physical components of the vectors involved. This is done by using   ∂ ( e (q ) ) ∂ ( re (q ) )     Equation 3.8 (i.e. ei = hi e ( i ) ), or = −e ( r ) , = − re ( r ) , hence ∂q ∂q r or the Christoffel symbol based on unit vector reads Γˆ qq = −1 (hatted to distinguish them from the relation given in Equation 10.27). Similarly, for    ∂ ( e ( r )) 1  ∂ ( e ( r ))  ∂er  = (1 / r ) eq we can write = ( re (q ) ) , hence = e (q ), or ∂q  ∂q r ∂q Γqr q 1. the Christoffel symbol based on unit vector reads Γˆ q = rq

r We could also calculate the Christoffel symbols Γˆ qrq and Γˆ qq , using the unit vectors or the physical components, as given in Example 8.1. Writing cosq  − sin q        these relations in vector form, we have e ( r ) =  sin q  , e (q ) =  cosq  ,  0   0     

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46 • Tensor Analysis for Engineers, 3E

0     ∂e ( r ) ∂e (q ) ∂e ( z )    and e ( z ) = 0  . Therefore, the derivatives are; = = = 0 , ∂r ∂r ∂r 1    − sin q  − cosq        ∂e ( r )  ∂e (q )  ∂e ( z )  ∂e ( r ) ∂e (q ) ∂e ( z )    = = cosq  e (q ) , = −e ( r ) , = 0, = = = 0  − sin q  = ∂q ∂q ∂q ∂z ∂z ∂z  0   0         ∂e ( r ) ∂e (q ) ∂e ( z ) = = = 0. Therefore, the non-zero Christoffel symbols calculated ∂z ∂z ∂z r based on the unit vectors are Γˆ qrq = 1 and Γˆ qq = −1. The final results can be summarized as



  Γˆ rrr   Γˆ r = Γˆ qr r  ˆr   Γ zr   Γˆ qrr  ˆq  ˆq Γ = Γq r  ˆq    Γ zr   Γˆ rrz  ˆz ˆz Γ = Γq r ˆz   Γ zr 

Γˆ rrq Γˆ r

qq

Γˆ rzq Γˆ qrq Γˆ q

qq

Γˆ qzq Γˆ rzq Γˆ z

qq

Γˆ zzq

Γˆ rrz  0   Γˆ qr z  = 0  Γˆ rzz  0 Γˆ qrz  0   Γˆ qq z  = 1  Γˆ qzz  0

0 0  −1 0  0 0  1 0  0 0  10.28 0 0 

Γˆ rzz  0 0 0     Γˆ qz z  = 0 0 0   Γˆ zzz  0 0 0 

Spherical coordinates From the results of Example 8.2, for a spherical system ( r ,j ,q ) , (note the order of the coordinates defined here, see Figure 8.3) we can write 2 2 the non-zero terms of the metric tensor as g= h= 1 , gjj= h= r 2 , and rr r j 2 2 2 gqq= h= r sin j . Therefore, the contravariant metric tensor reads g rr = 1 , q jj 2 g = 1 / r , and gqq = 1 / ( r 2 sin 2 j ) . Using Equations 10.20 and 10.25 and considering that indices i, j, k, m are various permutations of spherical ­variables r,j ,q we get the Christoffel symbol of the second kind as

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Gradient Vector Operator—Christoffel Symbols • 47



  Γ rrr  r  r Γ = Γj r   Γqr r      Γj  j  jrr Γ = Γj r   Γqj r     Γqrr   Γq = Γqj r   Γqq r  

Γ rrj r Γjj r Γqj

Γjrj Γjjj j Γqj Γqrj Γqjj Γqqj

j For example, Γ 233 = Γqq = −

Recall

that

the

Γ rrq  0 0 0    r  Γjq  = 0 − r 0  r  2   Γqq 0 0 − r sin j     0  Γjrq   0 1 / r    0 0  10.29 Γjjq  = 1 / r j   sin 2j  Γqq   0 0 −  2   Γqrq   0 0 1/r    q  Γjq  =  0 0 cot j  Γqqq  1 / r cot j 0  h3 ∂h3 r sin j ∂ ( r sin j ) sin 2j . = − 2 = − sin j cos j = − 2 2 ∂j h2 ∂x r 2

symbols are defined based on the covari ∂eq  sin 2j    ant basis vectors, ei (see Equation 10.5), or = −  ej . ∂q 2 We can write this  ∂ ( r sin j e (q ) ) as = ∂q

j Γqq

relation in terms of the physical components,  ∂ ( e (q ) )   sin 2j   = ( − cos j ) e (j ) . or −  re (j ) ,      ∂q 2   j Γˆ qq

Similarly, the other non-zero symbols can be written as; hj ∂hj ∂ ( r sin j ) h ∂h ∂r r r = − q2 q = − r sin j = − r sin 2 j , Γjj = − 2 = −r = − r , Γqq hr ∂r hr ∂r ∂r ∂r 1 ∂hj 1 ∂r 1 1 ∂hq 1 ∂ ( r sin j ) 1 Γqq r = = =, Γjj r = Γjrj = = = , Γqrq = hj ∂r r ∂r r hq ∂r r sin j r ∂r ∂ r sin j ( ) ∂ h 1 1 q and Γqjq = physical Γqqj = = = cot j . In terms of ­  hq ∂j r sin j ∂j ∂ej  components of the corresponding vectors, we have = ( − r ) er or  ∂j Γjr j   ∂ ( re(j ) ) ∂ e    r q ˆ = − re ( r ) = − e ( r ), hence Γjj = −1 . Also, = ( − r sin 2 j ) er ∂j ∂q  r Γqq

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48 • Tensor Analysis for Engineers, 3E  ∂ ( r sin j e(q ) )   r or = − sin j. = − r sin 2 j e ( r ) = − sin j ) e ( r ), hence Γˆ qq (   ∂q r ˆ Γqq   ∂e ( r ) ∂er    / r ) re (j ) e (j ) , hence Γˆ jrj = 1. Also, Also, or = (1 / r ) ej= (1=  ∂j ∂j Γjrj   ∂e ( r ) ∂er    == 1 / r ) eq or / r ) ( r sin j e (q ) ) sin j e (q ) , hence Γˆ qrq = sin j. ( (1=  ∂q ∂q q Γˆ qrq  Γrq ∂ ( re (j ) ) ∂ej    j ( r sin j e (q ) ) cos Also, = ( cot j )= eq or cot = j e (q ) , hence  ∂q  ∂ q Γˆ qjq Γqjq ˆΓq = cos j . jq The Christoffel symbols can be obtained directly using the unit vectors as well. From the results of Example 8.1, writing the physical components relasin j cosq  cos j cosq        tions in vector form, we have e ( r ) =  sin j sin q  , e (j ) =  cos j sin q  , and  cos j   − sin j      − sin q     ∂e ( r ) ∂e (j ) ∂e (q )    e (q ) =  cosq  . Therefore, the derivatives are; = = = 0 , ∂r ∂r ∂r  0    − sin j cosq  cos j cosq     ∂e (j )  ∂e (q ) ∂e ( r )      = −e ( r ) , = 0, = = cos j sin q  e (j ) ,  − sin j sin q  = ∂j ∂j ∂j  − cos j   − sin j      − sin j sin q  − cos j sin q    ∂e ( r )  ∂e (j )      sin j cosq  e (q ) sin j= , cos j cosq  e (q ) cos j , = = = ∂q ∂q     0 0     − cosq   ∂e (q )     = − e ( r ) sin j − e (j ) cos j . Therefore, the non-zero  − sin q  = ∂q    0  Christoffel symbols calculated based on the unit vectors are

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Gradient Vector Operator—Christoffel Symbols • 49

  Γˆ rrr Γˆ rrj Γˆ rrq  0 0 0    r    r r r ˆ ˆ ˆ ˆ = Γ Γj r Γjj Γ= 0 −1 0  jq    ˆr  ˆr ˆr     Γq r Γqj Γqq  0 0 − sin j    Γˆ jrr Γˆ jrj Γˆ jrq  0 1  0    ˆj ˆj  j j ˆ ˆ = Γj r Γjj Γjq  1 0 0  = Γ 10.30  ˆj  j j    ˆ ˆ − j 0 0 cos   Γq r Γqj Γqq      Γˆ qrr Γˆ qrj Γˆ qrq   0  0 sin j    ˆq  ˆq  q q ˆ ˆ 0 cos j  Γ = Γj r Γjj Γjq  =  0  ˆq  ˆq ˆq   0   Γq r Γqj Γqq  sin j cos j  Readers can use the Christoffel symbols of the 2nd kind calculated in this example and further calculate the Christoffel symbols of the 1st kind using Equation 10.19.

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CHAPTER

11

DERIVATIVE FORMS—CURL, DIVERGENCE, LAPLACIAN In the governing equations for physical phenomena we usually have terms which contain various forms of gradient operator, including the gradient ­vector itself—for example, when the gradient vector operates as cross product or dot product with another tensor quantity. The cross-product is called     the curl, ∇ × A and the dot-product is the divergence, ∇ ⋅ A. For example, in fluid mechanics the curl of the velocity vector is the vorticity vector and divergence of velocity vector vanishes, for incompressible fluids. As mentioned, one of the objectives of tensor analysis is to provide a tool to write down the governing equations in coordinate-independent forms while the covariancy or contravariancy of tensor quantities are correctly implemented in these equations. In the following sections, we derive the coordinate-independent relations for curl, divergence, Laplacian, and biharmonic operators of tensors for an arbitrary coordinate system.

11.1  CURL OPERATIONS ON TENSORS To form the curl operator expression, we first consider it for a vector in 3D coordinate space and then extend it to higher order dimension. In Section 9, we gave a list of guidelines for forming expressions for t­ ensors. Following these guidelines and using Equation 9.6,  we  can write the ­covariant component of the curl of a vector as Ci = ∇ × A = ijk ∇ j A k . We

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52 • Tensor Analysis for Engineers, 3E

know that ijk is a tensor (see Section 9). But the expression ∇ j A k does not transform like a tensor, due to extra terms appearing when transformed to another arbitrary coordinate system [4], [7]. However, if we use the results obtained in Section 10 and replace ∇ j A k with the tensor A ,k ⋅ j (i.e., the contravariant derivative of the contravariant component) then we get the proper expression as (see Equation 10.8) Ci = ijk A k, ⋅ j 11.1



Similarly, the contravariant component of the curl is written as, using Equations 9.6 and 10.12, C i =  ijk Ak, j 11.2



Both expressions given by Equations 11.1 and 11.2 are tensors, since terms are involved are tensors as well as both reduce to the Cartesian forms when ∂A written in the Cartesian coordinate system (i.e., Ci = eijk kj because the ∂y covariant and contravariant components are identical and Jacobian is unity). To have a more practical and explicit relation for the curl, we write Equation ∂A  ijk kj −  ijk Γ njk An examining the last term, C i  ijk A= 11.2 in detail or = k, j ∂x on the R.H.S, using the symmetry property of the Christoffel ­ symbol (i.e., Γ njk = Γ kjn ) and the anti-symmetry of the permutation symbol (i.e.,  ijk = − ikj) yields,  ijk Γ njk An = − ikj Γ nkj An . But after interchanging j ↔ k , dummy indices, only in the term on the R.H.S of the latter expression we receive − ikj Γ nkj An = − ijk Γ njk An. After comparing this result with the original expression (i.e., − ijk Γ njk An =  ijk Γ njk An) we can conclude that  ijk Γ njk An = 0. Therefore, Equation 11.2 can be written as

Ci =





(∇× A)

i

=

eijk  ∂Ak    11.3   ∂x j 

  The curl vector is then written as C = C i ei .

Similarly, we write Equation 11.1 in detail, also using Equation 10.8, or ∂A k k Ci = ijk A ,k ⋅ j = ijk g jn n + ijk g jnΓ nm A m . Examining the last term on the ∂x R.H.S, it turns out that it doesn’t vanish. Therefore, we have

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Derivative Forms—Curl, Divergence, Laplacian • 53



Ci =





(∇× A)

i

 ∂A k  k = g jn eijk  n + Γ nm A m  11.4  ∂x 

Note that the curl operation result  is a tensor of the  same rank as the original quantities. As in our example C is a vector, like ∇ and A . Now, we extend the discussion and find the relation for tensors of higher rank. For simplicity of writing the expressions, without losing generality, we consider a mixed tensor of second rank A ij . The invariant quantity is     A = A ij ei e j and the curl reads ∇ × A = ∇ × A ij ei e j . The gradient operator  just differentiates all the terms in front of it but the vector part of ∇ can   form a cross-product with either ei or e j, both are possible and legitimate operations. One can form, then two forms of the curl of the quantity A . In general, we can have N number of forms for the curl of a tensor of order N. For our example here, we have two forms:

(

)

1. We consider the case for which the cross-product operation occurring  with ei . Therefore, the contravariant component of A ij should be involved with the permutation tensor indices. write, using Equation 11.1,   We can  the component of curl of A as ∇ × ei A ij e j . Careful attention should be given to pick up the right indices to form the curl expression. Since we  picked the ei for cross-product operation with gradient, the covariant permutation tensor should be used, say kmn . Therefore, the index m corresponds to the contravariant differentiation and index n summed up with the contravariant component of the tensor A . The index k , is then the covariant component of the result or the curl tensor. The rank of the result should be the same as the original quantity A ij (i.e., rank two), hence the free covariant index j remains intact and we end up with a doubly-covariant tensor, or

(



)

 ∂A nj  = Ckj kmn = A nj ,m⋅ g mp ekmn  p + Γ npq A qj  11.5  ∂x    2. We consider the case for which the cross-product operation occurs with  e j . Therefore, the covariant component of A ij should be involved with the permutation tensor. We can write, using Equation 11.1, the compo   nent of curl of A as ∇ × e j A ij ei . Careful attentions should be given to pick up the right indices to form the curl expression. Since we picked  the e j for cross-product operation with gradient, the contravariant

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54 • Tensor Analysis for Engineers, 3E

permutation tensor should be used, say  kmn . Therefore, the index m corresponds to the covariant differentiation and index n is summed up with the covariant component of the tensor A . The index k , is then the contravariant component of the result or the curl. But the rank of the result should be the same as the original quantity A ij (i.e., rank two), hence the free contravariant index i remains intact and we end up with a doubly contravariant tensor, or

kmn i = C ki = An,m

ekmn  ∂Ani    11.6   ∂x m 

A similar operation can be used to write down the curl of a tensor of rank N, which has N possible outcomes.

11.2 PHYSICAL COMPONENTS OF THE CURL OF TENSORS—3D ORTHOGONAL SYSTEMS In many applications, we consider orthogonal systems (i.e., gij = 0 , i ≠ j ). In this section, we derive expressions for the physical components of the curl of a vector A in 3D orthogonal systems. Recalling from previous sections  i   i A A= ei A= A ( i ) e ( i ) and hi = ei . (see Sections 3 and 8), we can write= ie    j d i j, for orthogonal systems we have ei ⋅ e i = Using ei ⋅ e = 1 which means that   the covariant basis vector ei and contravariant basis vector e i both point in   the same direction. For example, e 1 and e1 are along the same line and point in the same direction. Therefore,

 e i = 1 / hi 11.7

        Substituting ei = hi e ( i ) in A ( i ) e ( i ) = A i ei and e i = e ( i ) / hi in A ( i ) e ( i ) = Ai e i gives

i A= A i / hi ( i ) h= iA

no sum on i 11.8

1 2 3 Or A= A1 / h1 , A= A2 / h2 , A= A3 / h3 . (1) h= ( 2 ) h= ( 3 ) h= 1A 2A 3A

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Derivative Forms—Curl, Divergence, Laplacian • 55

 Now we would like to write the physical components of the curl of vector A . To   i eijk  ∂Ak  i do this we multiply Equation 11.3 by hi , or ∇ × A = hi h= hi iC  .   ∂x j  But hiC i = C ( i ) and Ak = hk A ( k ) (see Equation 11.8).  Therefore, we will get the physical component C ( i ) of the curl of vector A in terms of its physical component A ( k ) , as

(

C ( i ) = hi



eijk ∂ ( hk A ( k ) ) 

∂x j

)

no sum on i, orthogonal 11.9

Or, using  = h1 h2 h3 for orthogonal systems, we have           





(1 ) ( ∇ × A )=

(

1  ∂ ( h3 A ( 3 ) ) ∂ ( h2 A ( 2 ) )  −   h2 h3  ∂x2 ∂x3 

 ∂ ( h1 A (1 ) ) ∂ ( h3 A ( 3 ) )  −   11.10 ∂x3 ∂x1     1  ∂ ( h2 A ( 2 ) ) ∂ ( h1 A (1 ) )  ∇× A = − ( 3 ) C= (3)   h1 h2  ∂x1 ∂x2  



(∇× A= )(2)



C= (1 ) C= (2)

1 h1 h3

)

It can be concluded that for a Cartesian system we recover the familiar    ∂Az ∂Ay     ∂AX ∂AZ  expressions, or ∇ × A = C= , ∇ × A = C= − −    , 1 2 1 2 ∂X   ∂Z  ∂Y ∂Z     ∂AY ∂AX  h= h= 1. and ∇ × A = C= − 1 2 3 3   , since h= 3 ∂Y   ∂X

(

(

)

(

)

)

11.3 DIVERGENCE OPERATION ON TENSORS To form the divergence operator expression, we first consider it for a vector in 3D coordinate space and then extend the results to higher order dimension. In Section 9, we gave a list of guidelines for forming expressions. Following these guidelines and using Equation 9.2, we can write the divergence (using the covariant component of the gradient) as

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  ∇ ⋅ A = A ,ii 11.11

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56 • Tensor Analysis for Engineers, 3E which reduces to proper expression in Cartesian coordinates and shows that the divergence of a vector results in a scalar, or in general divergence operation reduces the rank of the tensor, on which it is operating, by one. To further simplify Equation 11.11, we write it explicitly (see Equation 10.6) as ∂A i i A = + Γ ini A n and plug in for Γ iin from Equation 10.26 (after ­interchanging ,i i ∂x ∂A i A i ∂ dummy indices i ↔ n), to get = + . After substituting for A,i i A ,ii ∂x i  ∂x i into Equation 11.11 and performing some manipulations, we receive   1 ∂ ( A i ) ∇⋅A = 11.12  ∂x i



In addition, Equation 11.11 can be written   inij terms of the covariant comij ponent of the vector using g , or ∇ ⋅ A = g A j , after expanding gives ,i   ij ij = ∇ ⋅ A g ij A j + A= g g A or after interchanging i ↔ j and using j j, i ,i , i

( )

(

( )

)

=0

the symmetry of metric tensor, we receive   ∇ ⋅ A = g ij Ai, j 11.13



Now we extend the discussion by considering a tensor of higher rank, for  example and without losing generality a tensor of second rank A = A ij ei e j.    The divergence of A is ∇ ⋅ A = ∇ ⋅ A ij ei e j in which we have choices   between ei or e j for the gradient vector performing a dot-product operation. Therefore, we have two cases for this example:       1. ∇ dot products with ei: In this case we get ∇ ⋅ A = ∇ ⋅ ei A ij e j . Note that the gradient differentiates all terms in front of it but as a vector  onlydotted with ei . In other words, the contravariant index i is involved in  the dot-product operation. Using Equation 11.11 we can write ∇ ⋅ A = A ij, i which is a tensor of rank one, as expected since divergence reduces the rank of A ij by one. But using Equation 11.12, we have ∂A ij i i + Γ im A= A mj − Γ mij Ami in which the first two terms on the R.H.S. j, i ∂x i i ∂A ij 1 ∂ A j i m gives , using Equation 11.12. Finally, we get + Γ im A j = ∂x i  ∂x i the result as

(

)

(

(



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)

)

(

)

i    1 ∂ A j i j ∇ ⋅ A = ∇ ⋅ ei A j e = − Γ mij Ami 11.14  ∂x i

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Derivative Forms—Curl, Divergence, Laplacian • 57

      2. ∇ dot products with e j : In this case we get ∇ ⋅ A = ∇ ⋅ e j A ij ei . Note that the gradient differentiates all terms in front of it but as a vector  only dotted with e j . In other words, the index j is involved  in the dot-­ product operation. Using Equation 11.11 we can write ∇ ⋅ A = g kj Aki , j which is a tensor of rank one, as expected since divergence reduces the rank of A ij by one. Therefore, we have

(



)

    ∇ ⋅ A = ∇ ⋅ e j A ij ei = Aki , . k 11.15

(

)

Using similar operations, we can write the expression for divergence of a ­tensor of rank N . Both relations given by Equations 11.14 and 11.15 reduce to familiar forms for divergence in a Cartesian coordinate system. For    ∂Ax ∂Ay ∂Az ­example, for divergence of vector A we get ∇ ⋅= . A A= + + i, i ∂x ∂y ∂z

11.4  LAPLACIAN OPERATIONS ON TENSORS A Laplacian operator is the result of a gradient operator forming a dot-product   with itself, or divergence of gradient, ∇ 2 = ∇ ⋅ ∇ . Performing Laplacian on   i a scalar y and using Equation 11.12, we can write ∇ 2y = ∇ ⋅ ∇y = ( ∇y ), i , i ∂y 1 ∂ ( ∇ y ) 2 g ij j . Therefore, or ∇ y = where ∇ iy = i  ∂x ∂x

(



)

(

)

1 ∂  ∂y  ∇ 2y = i  g ij j  11.16  ∂x  ∂x  Since ∇ 2 is a tensor of rank zero, then ∇ 2y is a scalar, as well. In general, the rank is determined by the quantity that Laplacian is operating on. We the discussion to find the Laplacian of a vector, for e­ xample  extend  A = A i ei . Since for Laplacian the dot-product operation is performed between the two gradients involved, we don’t have many choices, as was the case for the divergence operator, and hence the formulation is more definite in terms of the results. Using Equations 11.13 and 11.16, we can write the contravariant component of the gradient of vector A as g nm A,in and the components of the Laplacian as g nm A,in . Note that index m indicates the dot,m product operation between the contravariant component of the gradient of  A and the covariant derivative of the second gradient involved. Expanding nm i i nm the last expression, we receive g nm A = g,nm A,inm . But g,m = 0 , ,n m A, n + g

(

)

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58 • Tensor Analysis for Engineers, 3E since it is a tensor and its value is zero in the Cartesian coordinate system, hence it should be zero for an arbitrary system, as well. Therefore, we have 

A ) (g ( ∇= 2



i

nm

i A= g nm A,inm 11.17 , n ), m

The term A,inm which is the second covariant derivative is a new term, so ∂ ( A,in ) j i i i far. By expanding this term, we get= = + Γ mj A,nm ( A A,nj − Γ mn A,i j. , n ), m m ∂x ∂ ( Ai ) i i i k But = A,n A + Γ nk A and after substituting and rearranging similar ∂x n terms, we receive = A,inm

i k i ∂2 Ai k ∂Γ nk i ∂A i ∂A + A + Γ + Γ nk mj ∂x m ∂x n ∂x m ∂x m ∂x j 11.18 i j ∂A j j i k i k − Γ mn j + Γ mjΓ nk A − Γ mnΓ jk A ∂x

Similar expressions can be written for higher order tensors, using hints from Equation 11.18 for writing the terms involving Christoffel symbols  A with appropriate signs. The higher the rank of the more complicated the  2 related expression for ∇ A becomes. For an orthogonal coordinate system, Equation 11.17 simplifies, since g nm = 0 for m ≠ n , or (for N = 3 )  ∇ 2 A =i g11 A,i11 + g 22 A,i22 + g 33 A,i33 ( orthogonal )

(

)

11.5  BIHARMONIC OPERATIONS ON TENSORS A biharmonic operator is the result of a Laplacian operator operating on itself, or ∇ 4 =∇ 2 ( ∇ 2 ) . Performing the operation on a scalar y and using Equation 11.12, we can write



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   1 ∂  ∂  1 ∂  ij ∂y    ∇ 4y = k g kl l   g   11.19 ∂x   ∂x i  ∂x j     ∂x      ∇ 2y

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Derivative Forms—Curl, Divergence, Laplacian • 59

When a biharmonic operates on a vector,   thei result is a vector as well. For  example, for A = A i ei , we can write ∇ 4 A = B ei. Since the result is a vector we can use Equation 11.17 to write Bi = g lm g jk A,i jklm 11.20



The quantity A,i jklm is a lengthy expression and can be written in detail [2], [4], [7] using hints taken from Equation 11.18. In principle, one can extend the 2 2 2 ∇ ∇ discussion to find operators like ∇ 2 n =∇   . In practice we encounter n times

mostly up to the level of the biharmonic operator in governing equations.

11.6 PHYSICAL COMPONENTS OF THE LAPLACIAN OF A VECTOR—3D ORTHOGONAL SYSTEMS In this section, we derive expressions for physical components of the Laplacian of a vector. Relations for physical components are useful since they are the magnitude of the components using unit vectors as the scale of measurement (see Section 6).   We start with Equation 11.18 and write it for a vector, like A = A ( i ) e ( i )   1 ∂  ∂A  or ∇ 2 A = i  g ij j  . In an orthogonal system, we have g ij = 0 (when ∂x   ∂x  ii 2 i ≠ j ) and g = 1 / hi . Therefore, after substitution into the expression, we   1 ∂   ∂  A ( i ) e ( i )   2 will have ∇ A = i  2  . Performing the differentiations  ∂x  hi ∂x i  along with using related relations for Christoffel symbols and after some manipulations, we get the physical j -component as [7]  ∇2 A

( j)

 1 ∂h j ∂A ( i ) 1 ∂hi ∂A ( i )  = ∇2 A ( j ) + 2  2 i −  j h j hi2 ∂x j ∂x i   hi h j ∂x ∂x  A ( i ) ∂h j ∂h j A ( i ) ∂hi ∂hi   A ( j ) ∂h j ∂h j A ( k ) ∂hi ∂hi  + + − 2 2 i i +  11.21 j j  i i 3 3 h j hi ∂x ∂x   hi h j ∂x ∂x hk h j hi2 ∂x k ∂x j   hi h j ∂x ∂x  A ( i ) ∂   ∂h j  A ( i ) ∂   ∂h   i +   − j  j  i  i  2 2    ∂x  hi h j ∂x   ∂x  h j hi ∂x  

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60 • Tensor Analysis for Engineers, 3E

In Equation 11.21, summation is done on indices i and k but not on index j, since it is the free index. To show the application of Equation 11.21, we present two examples to calculate the components of Laplacian of a vector in cylindrical and spherical coordinate systems. Note that the Laplacian of a vector is a vector as well, which has three components. 11.6.1 Example: Physical components of the Laplacian of a vector—cylindrical systems

Consider a cylindrical coordinate system ( x1 , x2 , x3 ) ≡ ( r ,q , z ) where r is the radial distance, q the azimuthal angle, and z the vertical coordinate, w.r.t. the Cartesian coordinates ( y1 , y2 , y3 ) ≡ ( X , Y, Z ). The functional relations are  X = r cosq   Y = r sin q . Find the Laplacian expressions for a scalar and the ­physical Z = z  components of a vector in this system. Let A (1 ) = Ar , A ( 2 ) = Aq, and A ( 3 ) = Az . Solution:





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 ∂y  To find the covariant basis vectors, we use ei = i E j . Therefore, we get ∂x    ∂X  ∂Y  ∂Z   ∂X  ∂Y  e1 = E1 + E2 + E3 = cosq E1 + sin q E2. Similarly, e2 = E1 + E2 + ∂r ∂r ∂r ∂q ∂q    ∂Z   ∂X  ∂Y  ∂Z  E3 = − r sin q E1 + r cosq E2. And e3 = E1 + E2 + E3 = E3. Therefore, ∂q ∂z ∂z ∂z we have     er cosq E1 + sin q E2 =     − r sin q E1 + r cosq E2  eq =    ez = E3  The scale factors are the magnitudes of the basis vectors. Hence       h1 = hr = er ⋅ er = 1, h2 = hq = eq ⋅ eq = r , and h3 = hz = ez ⋅ ez = 1 . The e  unit vectors e ( i ) = i , are hi     e ( r ) cosq E1 + sin q E2 =     − sin q E1 + cosq E2  e (q ) =    e (j ) = E3  j

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Derivative Forms—Curl, Divergence, Laplacian • 61

0 0 1  2 The Jacobin = is  h= r and g = h , or g = 0 1 / r 0  . Now r hq hz 0 0 1  1 ∂  ∂y  using Equation 11.16 for orthogonal system i.e. ∇ 2y = i  g ii i  ,  ∂x  ∂x  ij since = g 0 for i ≠ j , and substituting for corresponding values we will get ii

−2 i

ij

1 ∂  ∂y  1 ∂ 2y ∂ 2y . In other words, the Laplacian operator + r + r ∂r  ∂r  r 2 ∂q 2 ∂z2 in cylindrical coordinate system reads

= ∇ 2y



= ∇2

1 ∂  ∂  1  ∂ 2  ∂ 2 11.22 r +  + r ∂r  ∂r  r 2  ∂q 2  ∂z2

Similarly, using Equation 11.21 we get the physical components of the Laplacian for the vector A, as



A 2 ∂A  2 ∇ 2 Ar − 2 q − 2r ∇ A r = r ∂q r   A ∂ A 2  2 r ∇ 2 Aq + 2 − 2q 11.23 ∇ A q = r ∂q r   2 2 ∇ A = ∇ Az z   Where the operator ∇ 2 is given  by Equation 11.22 and Ar, Aq , Az represent the physical components of A. 11.6.2 Example: Physical components of the Laplacian of a vector—spherical systems

Consider a spherical coordinate system ( x1 , x2 , x3 ) ≡ ( r , j , q ) where r is the radial distance, j the polar and q the azimuthal angle, w.r.t. the Cartesian coordinates ( y1 , y2 , y3 ) ≡ ( X , Y, Z ) . The functional relations are  X = r sin j cosq   Y = r sin j sin q . Find the Laplacian expressions for a scalar and the  Z = r cos j  ­physical components of a vector in this system. Let A (1 ) = Ar, A ( 2 ) = Aj, and A ( 3 ) = Aq.

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62 • Tensor Analysis for Engineers, 3E Solution:

j  ∂y  To find the covariant basis vectors, we use ei = i E j . Therefore, we get ∂x     ∂X  ∂Y  ∂Z  e1 = E1 + E2 + E3 = sin j cosq E1 + sin j sin q E2 + cos j E3 . Similarly, ∂r ∂r ∂r      ∂Y ∂Z   ∂X e2 = E1 + E2 + E3 = r cos j cosq E1 + r cos j sin q E2 − r sin j E3 . And ∂j ∂j ∂j    ∂X  ∂Y  ∂Z  − r sin j sin q E1 + r sin j cosq E2 . Therefore, we e3 = E1 + E2 + E3 = ∂q ∂q ∂q have      er = sin j cosq E1 + sin j sin q E2 + cos j E3       ej = r cos j cosq E1 + r cos j sin q E2 − r sin j E3    e r j q E r j q E sin sin sin cos = − + 1 2 q  





The scale factors are the magnitudes of the basis vectors. Hence       h1 = hr = er ⋅ er = 1 , h2 = hj = ej ⋅ ej = r , and h3 = hq = eq ⋅ eq = r sin j .  e  The unit vectors e ( i ) = i , are hi      e ( r ) = sin j cosq E1 + sin j sin q E2 + cos j E3       e (j ) = cos j cosq E1 + cos j sin q E2 − sin j E3    − sin q E1 + cosq E2  e (j ) =    1 0 0   r 2 sin j and g ii = hi−2 , or g ij = 0 1 / r 2 The Jacobin= is  h= 0  r hj hq   2 2 1  0 1 / ( r sin j )  0 0 0   g ij = 0 1 / r 2 0  . Now using Equation 11.16 for orthogonal   2 2 0 1 / ( r sin j )  0 1 ∂  ∂y  system i.e. ∇ 2y = i  g ii i  , since = g ij 0 for i ≠ j, and substituting  ∂x  ∂x  ∂ 1 ∂  2 ∂y  1 for corresponding values we will = get ∇ 2y r + 2 2 r ∂r  ∂r  r sin j ∂j  ∂y  1 ∂ 2y . In other words, the Laplacian operator in  sin j + ∂j  r 2 sin 2 j ∂q 2  spherical coordinate system reads

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Derivative Forms—Curl, Divergence, Laplacian • 63



= ∇2

1 ∂  2 ∂  1 ∂  ∂  1 ∂2 11.24 r + 2  sin j + 2 2 2 r ∂r  ∂r  r sin j ∂j  ∂j  r sin j ∂q 2

Similarly, using Equation11.21 we receive the physical components of the Laplacian for the vector A, as



 2 2A 2 ∂(sin j Aj ) 2 ∂ Aq ∇ 2 Ar − 2 r − 2 − 2 ∇ A r = ∂j r r sin j r sin j ∂q   A  2 2 ∂A 2 cos j ∂Aq 11.25 ∇ 2 Aj − 2 j 2 + 2 r − 2 2 ∇ A j = ∂ ∂ r r r j j j q sin sin   2 A 2 ∂Ar 2 cos j ∂ Aj ∇ 2 Aq − 2 q 2 + 2 + 2 2 ∇ A q = r sin j r sin j ∂q r sin j ∂q  Where the operator ∇ 2 is given  by Equation 11.24 and Ar , Aj , Aq represent the physical components of A. In the next section, we focus on relations pertinent to tensor transformations between merely Cartesian coordinate systems.

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CHAPTER

12

CARTESIAN TENSOR TRANSFORMATION— ROTATIONS When transformation of tensors is performed from one Cartesian coordinate system to another one, relations derived in previous sections take simpler forms, called Cartesian tensors. As mentioned previously, a Cartesian system consists of three mutually perpendicular flat surfaces for which; the basis vectors are unit vectors, there is no distinction between covariant and contravariant components, and the components of a tensor are the physical components. Therefore, we use subscript indices for all types of components, regardless. We will consider two Cartesian systems, yi and y′i with a common origin— otherwise we can always redefine them to have a common origin with transforming, say, the origin of y′i to that of yi . Following the discussion from  Section 3, we can write the differential displacement ds in these systems, as

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   = ds dy = dy′j E′j 12.1 i Ei

  where Ei and E′j are the corresponding unit vectors. However, we ∂y have dyi = i dy′j , and after substituting in Equation 12.1 we get ∂y′j

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66 • Tensor Analysis for Engineers, 3E

 ∂yi    ∂yi   Ei = E′j or, 0 . Since dy′j is arbitrary, we conclude Ei − E′j  dy′j =    ∂y′j  ∂y′j  by i ↔ j , we have



 ∂y j  E′i = E j 12.2 ∂y′i  Now performing dot-product on both sides of Equation 12.2 with Ek gives,   ∂y j   ∂yk . Hence, Ek ⋅= E′i E j ⋅= Ek ∂y′i  ∂y′i d jk



  ∂y Ek ⋅ Ei′ =k 12.3 ∂y′i ∂y This means that the cosine of the angle between yk and y′i is equal to k , ∂y′i called the cosine direction.   ∂y′  Similarly, we can write Ek = i E′i , and after dot-product with E′j we receive ∂yk ′     ∂ y ′ ∂yi j ′j . Hence, interchanging i ↔ j, we receive E′j ⋅ E= E′i ⋅ E= k    ∂yk ∂yk d ij



  ∂y′ E′i ⋅ Ek =i 12.4 ∂yk ∂y′ This means that the cosine of the angle between yk and y′i is equal to i . ∂yk Therefore, comparing Equations 12.3 and 12.4 and considering the commutativity of the dot-product operation, we will have ∂y′i ∂yk 12.5 = ∂yk ∂y′i



 ∂y j  ∂y′j ∂yi ∂y′i  ∂y′j = dy′j and = E′i = Ej E j , but ∂y′j ∂yi ∂y′i ∂y j only in Cartesian coordinate systems. In other words, these relations prove that there is no distinction between covariant and contravariant components in a Cartesian system (see Equations 4.8 and 5.4, for comparison).

Equation 12.5 yields dyi =

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Cartesian Tensor Transformation—Rotations • 67

12.1  ROTATION MATRIX Now if we assume that coordinate system y′i is obtained by rotating coordi nates y j at an angle q about an axis parallel to a unit vector n = ( n1 , n2 , n3 ) , ∂yi′  = Rij is a function of q and n and it can be shown then the quantity that [9], it reads as

∂y j

(

)

Rij = ni n j + d ij − ni n j cosq + eijk nk sin q 12.6 Equation 12.6 is Rodrigues’ rotation formula, or the 3D rotation matrix. Readers should note that the Right-Hand-Rule convention applies to the rotation of coordinates with respect to the positive direction of unit vector   n when using Equation 12.6. For example, for n = ( 0,0,1 ) and q = p / 2 , Rij gives the transformation of x − y plane about the positive direction of z -axis for an angle of 90° . ∂y′i = dy′i = dy j Rij dy j which yields, y′i = Rij y j. Note that Now we integrate ∂y j the constant of integration is zero due to having a common origin for both  systems, and Rij is a constant for given n and q . Similarly, we can rotate  y′i at an angle (− q ) about the same unit vector n to recover the ­original −1 ′ R ji y′j, by having yi R= y j coordinates. Therefore, we can write= ij y j −1 Rij being equal to the inverse or transpose of Rij [4]. Further examining Equation 12.6, we can conclude that Rij for a given angle q is equal to R ji, or its transpose, calculated for angle ( −q ). This is the result of the property of the last term on the R.H.S of Equation 12.6 when i ↔ j to get R ji , − e jik nk sin q =− e jik nk sin( q ). The remaining terms are not or eijk nk sin q = affected by this index-change operation.

12.2 EQUIVALENT SINGLE ROTATION: EIGENVALUES AND EIGENVECTORS It would be useful to write the matrix form of the Cartesian coordinate rotations, as well. For example (in a 3D space with N = 3), y′i = Rij y j can be  y1′   R11 R12 R13   y1       written as  y2′  =  R21 R22 R23   y2 , and after using Equation 12.6, for  y′   R    3   31 R32 R33   y3   n = ( n1 , n2 , n3 ) and rotation angle q, we get

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68 • Tensor Analysis for Engineers, 3E

 y1′     y2′  =  y′   3  cosq + n1 n1 (1 − cosq )   n1 n2 (1 − cosq ) − n3 sin q  n1 n3 (1 − cosq ) + n2 sin q

n1 n2 (1 − cosq ) + n3 sin q cosq + n2 n2 (1 − cosq ) n2 n3 (1 − cosq ) − n1 sin q



n1 n3 (1 − cosq ) − n2 sin q   y1    n2 n3 (1 − cosq ) + n1 sin q   y2  cosq + n3 n3 (1 − cosq )   y3 

12.7

From Equation 12.7 (or 12.6) we can conclude that the trace of the rotation matrix (i.e., the sum of diagonal elements, Rii) is equal to 1 + 2 cosq, (note 2  that ni ni = n = n12 + n22 + n32 = 1, since n is a unit vector), or

 R −1 q = cos−1  ii  12.8  2 

Equation 12.8 is a useful relation for finding rotation angle q, once Rij is known/given. However, readers may ask: if Rij is known, how can one find the corresponding unit vector parallel to the axis of rotation? This q ­ uestion is all the more significant considering that we can have a sequence of ­rotations to arrive at a final desired orientation of the coordinate axes. For example, if we rotate the original system through a sequence of rotations and arrive at the final system orientation, i.e., yi → yit 1 →  → yitn → y′i ,   sequence

then we can apply Equation 12.7 for each sequence and calculate the final rotation matrix. That is {y′i } =  Rijtn   Rijtn−1    Rijt 1  {yi }. Note that for each   Rij   

rotation in the sequence, the corresponding rotation matrix is pre-multiplied to the previous ones, and the normal vector parallel to the rotation axis is defined based on the current coordinate system at hand in the sequence (see Example 12.2.1) Now, if we want to replace all intermediate rotations with just one equivalent rotation, Equation 12.8 gives the value of the angle for the desired equivalent single rotation. However, to find the axis of rotation parallel to the unit vector for the equivalent single rotation we need a mathematical procedure/ tool. This tool can be obtained by finding the real eigenvector of the given rotation matrix since, through all rotations, only the eigenvectors remain in the same direction [2], [4]. The mathematical procedure is as follows:

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Cartesian Tensor Transformation—Rotations • 69

We consider a normal unit vector ni perpendicular to a plane and write the components of the rotation tensor on this plane, or Rij n j . This quantity is a vector whose components are the elements of the rotation tensor on the plane considered. We identify this quantity as Rij n j = Ti . In general, the direction of the vector Ti does not necessarily coincide with the normal vector to the plane, as shown in Figure 12.1.

  FIGURE 12.1  Rotation matrix component T, about an arbitrary plane with normal n.

In other words, we could demand a specific plane and calculate its corresponding normal direction such that vector Ti ends up in the same direction as the normal to the plane considered. Therefore, we can write Ti ni = Rij n j ni. This quantity, which is the projection of Ti in the direction of ni, is a quadratic function of vector ni. To calculate its extremum value with the constraint that ni ni = 1 (which means that having the normal vector as a unit vector), we use the Lagrange multiplier method for extremizing the scalar quantity  = Rij n j ni − l ( ni ni − 1 ), where the constant l is the Lagrange ∂ multiplier, or the eigenvalue of Rij . This is achieved by letting = 0, ∂ni ∂ = 0 recovers the constraint, ni ni = 1) which leads to (note that the ∂l

(R

In



matrix

form,

ij

)

− ld ij n j = 0 12.9

Equation

can

be

written

as

R12 R13   n1   R11 − l    R22 − l e 23   n2  = 0. This system of equations has a non R21    R31 R32 R33 − l   n3  trivial (i.e., non-zero) solution, for ni, if the determinant of the matrix is zero, or R11 − l R12 R13 det Rij −= ld ij R21 R22 − l = e 23 0 12.10

(

)

R31

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12.9

R32

R33 − l

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70 • Tensor Analysis for Engineers, 3E

By evaluating the determinant, we get a cubic equation in terms of l (which has three roots, or eigenvalues) as  Rii R jj − Rij R ji  l 3 − ( Rii ) l 2 +  0 12.11  l − Rij =   2   1 3



2

Equation 12.11 is known as the characteristic equation for matrix Rij, where 1 is the first principal scalar invariant, which is equal to the trace of Rij; or 1 = Rii = l1 + l2 + l3 ,  2 is the second principal scalar invariant, which is  2 = l1 l2 + l2 l3 + l3 l1 , or the sum of the diagonal minors; and  3 is the third principal scalar invariant, which is equal to the determinant of Rij, or =  3 e= l1 l2 l3 . ijk R1 i R2 j R3 k Equation 12.11 contains several important properties and pieces of information about the system’s rotation matrix. Some examples related to the discussion here include: „„

„„ „„

„„

For symmetric tensors with real components there exist three real value answers for l (or l1 , l2 , and l3). This is not the case for Rij, since it is not symmetric, and we get only one real value l, [10]. The answers for l are the eigenvalues (or principal values) of Rij. The corresponding ni, calculated and normalized for each eigenvalue, are the eigenvectors (or principal directions) of Rij. A given eigenvector when multiplied by a constant real number (positive or negative) it gives a new vector, but all have the same direction parallel to the principal direction.

The method of finding equivalent rotation is very useful in practice for designing the motion of machine parts in robotic applications, such as a robot arm’s motion. 12.2.1 Example: Equivalent single rotation to sequential rotations of a Cartesian system Having a Cartesian system labeled as yi = ( y1 , y2 , y3 ) and its transformation to another system labeled as y′i = y′1 , y′2 , y′3 such that y′1 coincides with y2 , y2′ with y3 , and y3′ with y1 , as shown in Figure 12.2, find the overall equiva lent single rotation matrix for y′i = Rij y j , the axis of rotation direction n and the angle of rotation q.

(

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)

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Cartesian Tensor Transformation—Rotations • 71

FIGURE 12.2  Sequential rotations of a Cartesian coordinate system (left) and the equivalent single rotation (right).

Solution: We consider two sequential rotations: 1) rotation about y3 -axis with normal vector ( 0,0,1 ) with an angle of 90° (R.H.R. applies) to get the intermediate transformation ( y1 , y2 , y3 ) → ( y1′ , y2t , y3 ) , and 2) rotation about new y′1 with normal vector (1,0,0 ) with an angle of 90° (R.H.R. applies) to get the second transformation ( y1′ , y2t , y3 ) → ( y1′ , y′2 , y′3 ) (see Figure 12.2). Therefore, using Equation 12.7 and the data given for the first transformation with  y1′   0 1 0   y1        n = ( 0,0,1 ) we have  y2t  =  −1 0 0   y2 . Similarly, for the second y   0 0 1  y   3   3  y1′  1 0 0   y1′        rotation with n = (1,0,0 ) , we have  y2′  = 0 0 1   y2t . Note that  y′  0 −1 0   y   3   3 the second rotation is performed on the current system at hand, which is the result of the first rotation. Therefore, after substitution we can write  y1′  1 0 0   0 1 0   y1   y1′  0 1 0   y1               =  y2′  0 0 1   −1 0 0   y2 , or  y2′  = 0 0 1   y2 . The final  y′  0 −1 0   0 0 1   y   y′  1 0 0   y   3   3  3    3

0 1 0    rotation matrix is then Rij = 0 0 1 . Note that each rotation matrix 1 0 0 

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72 • Tensor Analysis for Engineers, 3E pre-multiplies the previous one in the sequence of rotations. Having the final rotation matrix, we can calculate the single rotation angle. Using Equation 12.8, 2p −1  R ii − 1  −1  0 − 1  we= get q cos= . The corresponding =     cos  2  3  2  rotation axis is the eigenvector associated with the real eigenvalue for 0  −l 1 0 1 0      Rij = 0 0 1  . We can write Rij − ld ij = 0 −l 1  =−l 3 + 1 =0 . 1 0 0   1 0 −l 

Hence, we get the characteristic equation as ( l − 1 ) ( l 2 + l + 1) = 0 with its only real root being l = 1 . Note that we could use Equation 12.11 instead for calculating the eigenvalues. For l = 1 , we get the eigenvector by solving 0 1 0   n1   n1   n2 = n1       0 0 1   n2  = (1 )  n2  , or  n3 = n2 . This system is not determinate but n = n n  1 0 0   n3  3  1  3  gives n= n= n3. However, the constraint for having n as a unit vector 1 2 3 gives n12 + n22 + n32 = 1, or 3 n12 = 1. Therefore, n= and finally n= n= 1 2 3 3   3 3 3 , , we have the equivalent rotation axis or the unit vector n =  .  3 3 3  In conclusion, we can say that instead of two intermediate sequential rotations, as mentioned above, we can rotate the original yi = ( y1 , y2 , y3 ) at an angle of 120°, using right-hand-rule convention, about an axis parallel  3 3 3 , , to the unit vector   (see Figure 12.2). Readers should note  3 3 3  that for calculating the rotation matrix using Equation 12.7, values of the  components of n should be those of the unit vector. However, to obtain the direction of the rotation axis we can multiply the components of the unit vector by a constant—for this example the multiplier is 3 . As shown in Figure 12.2, the result is a vector, or (1,1,1), which is a vector along the  same direction as the unit vector n . This is consistent with the property of the eigenvalues. We can examine these results by using Equation 12.6 to   3 3 3 , , calculate the single rotation matrix, using = q 120° and n =  ,  3 3 3 

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Cartesian Tensor Transformation—Rotations • 73

2 2  3   3  1 1 or R11 =  − = 0, similarly R= R= 0.  + 1 −    cos (120° ) = 22 33 3 3  3    3   2 2  3   3   3 Also, we get= R12  e123  = ° ) 1.  +  d   cos (120° ) +   sin (120 12 −  3   3   = 0  3   =1 

Similarly,

we

receive

R= R= 1, 23 31

and

the

remaining

elements

R= R= R= 0. These results are identical with the one obtained using 21 32 13 sequential rotations.

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CHAPTER

13

COORDINATE INDEPENDENT GOVERNING EQUATIONS Reliable mathematical models, also referred to as governing equations, are important tools for engineering analysis. A reliable and validated mathematical model of a physical phenomenon is a set of algebraic relations among quantities and their various derivatives, such as Newton’s 2nd law of motion, equilibrium equations for momentum flux, Fourier’s law of heat flux, Fick’s law of mass flux, Navier-Stokes equations for flow of fluids, Maxwell’s ­electromagnetic equations, etc. These governing equations, along with some fundamental principles (like the 2nd law of thermodynamics, conservations of energy, mass, electric charge, etc.) form the foundation of engineering science and its applications. The quantities related to any physical phenomenon can be represented by tensors of different ranks stated in the relevant governing equations, such as force and acceleration vectors and mass of a body as a scalar, in Newton’s 2nd law of motion; gradient of temperature in Fourier’s law; divergence of ­velocity vector in fluid flow, etc. Sometimes, for analysis and design purposes, we need to have the relevant governing equations written in coordinate systems other than the Cartesian system. For example, for analyzing mechanical stresses in the wall of a cylindrical pressure vessel we prefer to choose the cylindrical polar coordinate system that is a natural fit to the shape of the vessel. This requirement has encouraged scientists and engineers to define various coordinate systems suitable for solving the governing equations in practice. In other words, and in the context of tensor analysis, we would like to use the relations derived and discussed in the

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76 • Tensor Analysis for Engineers, 3E previous sections to write down the terms involved in well-known governing equations in engineering in a form that is general enough for application in an arbitrary but well-defined coordinate system. In this section, we derive some new relations mostly used in engineering, in addition to those discussed in the previous sections. We hope that this helps readers in writing down similar coordinate independent terms involved in equations of their choice for their applications.

13.1 THE ACCELERATION VECTOR—CONTRAVARIANT COMPONENTS Newton’s 2nd law is a mathematical model for the motion mechanics of physical objects—specifically, it is the balance of applied forces and the rate of change of momentum, or mass times  acceleration. Velocity is the time rate of change of displacement vector ds (see Section 3). Therefore, in  an    ds i arbitrary coordinate system x , we can write the velocity vector v as v = dt for the corresponding time increment of dt . Using Equation 3.6, we have     ds  dx i . Since ds is a vector, hence v is a vector and we can write it in = v = ei dt dt terms of its contravariant component vi , or



dx i   i = v v= ei ei 13.1 dt Equation 13.1 clearly gives the contravariant component of the velocity dx i  ­vector as vi = . Acceleration vector a is the time derivative of the ­velocity dt vector, or   dv  = a = a i ei 13.2 dt i   dv d ( v ei )  dvi i dei = ei +v Using Equation 13.1, we can write= . To relate dt dt dt dt d dx j ∂ the time derivative to the space gradient we use chain rule , = dt  dt ∂x j

hence

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vj

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Coordinate Independent Governing Equations • 77





d ∂ = v j j 13.3 dt ∂x    dv   j ∂vi  i  j ∂ei  Substituting into Equation 13.2, we get = . + v v a = ei  v j  j  dt  ∂x   ∂x   ∂ei   ∂vi   But j = Γ kij ek , using Equation 10.5. Therefore,= a ei v j j + vi v j Γ kij ek and ∂x ∂x in the last term we interchange dummy indices i ↔ k and use the symmetry property of the Christoffel symbol to obtain    ∂vi = a  v j j + vk v j Γ ijk  ei 13.4  ∂x 

Comparing Equations 13.2 and 13.4, we get the contravariant component of the acceleration vector as

 ∂vi  = a i v j  j + vkΓ ijk  13.5  ∂x  ∂vi + vkΓ ijk , after using Equation 10.6, to ∂x j receive the compact form of Equation 13.5, as In addition, we can write v,i j =

a i = v j v,i j 13.6



Equation 13.6 recovers the familiar expression for acceleration in Cartesian ∂vi coordinates (i.e., a i = v j j ). ∂x We can also write the covariant component of the acceleration vector using a metric tensor, or ak = gik a i (see Section 8) for an arbitrary coordinate system. Now, we can write the Newton’s 2nd law in a general coordinate independent form. We choose to use the contravariant component of the acceleration vector; hence, for compatibility we must use the contravariant component of   the total force vector, F = F i ei, as well. Therefore, we can write

TAE3E.CH13_1pp.indd 77

= F i mv j

∂vi j i + mvk v= Γ jk mv j v,i j 13.7 j ∂x

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78 • Tensor Analysis for Engineers, 3E For example, in a 3D coordinate system we get three equations of motions for each dimension, with summation on indices j and k only, as 1  1 j ∂v = + mvk v j Γ1jk F mv  j ∂ x  2  2 j ∂v + mvk v j Γ 2jk 13.8 F = mv j ∂x  3  3 j ∂v = + mvk v j Γ 3jk F mv  j ∂x 



Note that in a Cartesian system the last term in Equation 13.7, associated with the Christoffel symbol, is zero and we recover the familiar form of Newton’s equation. But when we write this equation in a curvilinear coordinate system, the term associated with the Christoffel symbol is not necessarily equal to zero. This extra term acts like an inertia force and affects the path of moving objects. For example, in a spherical coordinate system, like Earth’s geometry, ∂vi mv j j . The left-hand-side we can write Equation 13.7 as F i − mvk v j Γ ijk = ∂x ∂vi can be treated as a new total force, or F i = F i − mvk v j Γ ijk = mv j j . The ∂x k j i i term mv v Γ jk is the Coriolis force, Fc = Fci mvk v j Γ ijk 13.9



For example, from the point of view of an observer at the north pole of the Earth an object with an initial velocity along a meridian towards the equator will shift to the right, due to the inertia force resulting from the Coriolis force, while to an observer at the South Pole it will shift to the left. Or we can say that a force equivalent to the negative of the Coriolis force is acting on the object.

13.2 THE ACCELERATION VECTOR—PHYSICAL COMPONENTS Physical components of acceleration vector can be obtained using Equation 6.2, or a ( i ) = hi a i. Substituting for a i , Equation 13.5 yields ∂vi dvi a ( i ) = hi v j j + hi v j vk Γ ijk = hi + hi v j vk Γ ijk. In the latter expression, ∂ x dt  =

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dvi dt

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Coordinate Independent Governing Equations • 79

we write the velocity components in terms of their physical components d ( v ( i ) / hi ) v( j) v( k) i + hi Γ jk . But the first as well to receive a ( i )= hi dt h j hk term can be expanded to get hi

d ( v ( i ) / hi ) dv ( i ) d (1 / hi ) . = + hi v ( i ) dt dt dt

dv ( i ) = v ( i ) and use Equation 13.3 to write dt v ( i ) v ( j ) ∂ ( hi ) d (1 / hi ) d ( hi ) 1 = − v( i) = − . Now substituting into hi v ( i ) ∂x j dt hi dt hi h j

Now, we define

the relation for a ( i ), we obtain

v ( i ) v ( j ) ∂hi hi v ( j ) v ( k ) i + Γ jk 13.10 a ( i) = v ( i ) − hi h j ∂x j h j hk Equation 13.10 gives the physical components of the acceleration vector in curvilinear coordinate systems. Note that for Equation 13.10 summation applies only on indices j and k , but not i.

13.3 THE ACCELERATION VECTOR IN ORTHOGONAL SYSTEMS—PHYSICAL COMPONENTS A useful application of Equation 13.10 is for orthogonal coordinate systems. As ij g= 0 for mentioned previously, for an orthogonal system we should have g= ij i ≠ j and we can use relations given by Equations 10.23–10.25 for Christoffel symbols. Therefore, expanding Equation 13.10, for example for i = 1, we v (1 ) − receive a (1 ) =

v (1 ) v ( j ) ∂h1 h1 h j

∂x

j

+

h1 v ( j ) v ( k ) h j hk

v (1 ) − ­summations on j and k yields a (1 ) =

Γ1jk , which after writing the

v (1 )  v (1 ) ∂h1 v ( 2 ) ∂h1 v ( 3 ) ∂h1  + +   h1  h1 ∂x1 h2 ∂x2 h3 ∂x3    sum on j

    h1 v (1 )  v (1 ) 1 v ( 2 ) 1 v ( 3 ) 1  h1 v ( 2 )  v (1 ) 1 v ( 2 ) 1 v ( 3 ) 1  + Γ11 + Γ12 + Γ13  +  Γ 21 + Γ 22 + Γ 23  +  h1  h1 h h3 h2  h1 h2 h3 2       sum on k , j =1 sum on k , j = 2  

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80 • Tensor Analysis for Engineers, 3E

  h1 v ( 3 )  v (1 ) 1 v ( 2 ) 1 v ( 3 ) 1  But from Equations 10.23-10.25, we Γ 31 + Γ 32 + Γ 33  .  h3  h1 h h3 2    sum on k , j = 3

1 ∂h 1 ∂h1 1 1 ∂h1 1 h ∂h receive Γ111 = 11 , Γ112 = , Γ13 = , Γ 22 = Γ121 = Γ131 = − 2 2 12 , 2 3 h1 ∂x h1 ∂x h1 ∂x ( h1 ) ∂x h ∂h and Γ133 = Γ132 =. 0 Substituting for the corresponding − 3 2 13 and Γ123 = ∂ x h ( 1) v (1 ) v ( 2 ) ∂h1 v (1 ) + + Christoffel symbols and simplifying, we receive a (1 ) = h1 h2 ∂x2 v (1 ) v ( 3 ) ∂h1 v ( 2 ) v ( 2 ) ∂h2 v ( 3 ) v ( 3 ) ∂h3 . By similar operation we can − − h1 h3 ∂x3 h1 h2 ∂x1 h1 h3 ∂x1 calculate the second and third components, i.e., a ( 2 ) and a ( 3 ) . Equation 13.11 lists all the physical components of acceleration vector in a 3D orthogonal coordinate system, as

 v (1 ) v ( 2 ) ∂h1 v (1 ) v ( 3 ) ∂h1 v ( 2 ) v ( 2 ) ∂h2 v ( 3 ) v ( 3 ) ∂h3 + − − v (1 ) +  a (1 ) = 2 3 1 ∂ ∂ ∂ h h x h h x h h x h1 h3 ∂x1 1 2 1 3 1 2   v (1 ) v ( 2 ) ∂h2 v ( 2 ) v ( 3 ) ∂h2 v (1 ) v (1 ) ∂h1 v ( 3 ) v ( 3 ) ∂h3 13.11 + − − v ( 2 ) + a ( 2 ) = 1 3 2 2 ∂ ∂ ∂ ∂ h h x h h x h h x h h x 1 2 2 3 1 2 2 3   v (1 ) v ( 3 ) ∂h3 v ( 2 ) v ( 3 ) ∂h3 v (1 ) v (1 ) ∂h1 v ( 2 ) v ( 2 ) ∂h2 v ( 3 ) + + − − a ( 3 ) = h1 h3 ∂x1 h2 h3 ∂x2 h1 h3 ∂x3 h2 h3 ∂x3  Relations given by Equation 13.11 can be written in index form as, ­summation on index j only,

v ( i ) v ( j ) ∂hi v ( j ) v ( j ) ∂h j 13.12 − a ( i) = v ( i ) + ∂x i hi h j ∂x j hi h j In the next section, we use Equation 13.12 to calculate an acceleration vector’s physical components in cylindrical and spherical coordinate systems. For a Cartesian system, Equation 13.12 recovers the familiar relation (i.e., d ( v( i)) ). a ( i) = dt

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Coordinate Independent Governing Equations • 81

13.3.1 Example: Acceleration vector physical components in cylindrical and spherical coordinate systems From the results obtained in Examples 8.1 and 8.2, we have the scale factors for cylindrical and spherical coordinates as ( hr , hq , hz ) = (1, r ,1 )

( h , h , h ) = (1, r, r sin j ), respectively. Recall that the cylindrical coordinates are ( x , x , x ) ≡ ( r ,q , z ) and those of a spherical system are ( x , x , x ) ≡ ( r,j ,q ) . Now, using Equation 13.11 we can write the physiand

r

j

q

1

1

2

2

3

3

cal components of the acceleration vector for cylindrical coordinates as v(2) v(2) v (1 ) v ( 2 ) a= , a= , and a ( 3 ) = v ( 3 ) . In terms (1) v (1) − ( 2 ) v ( 2 ) + r r vq2 vv  v − vq + r q , and of coordinate notation, we can write it as a= , a= r r q r r az = v z . Note that all components are the physical components of the corresponding vectors. Therefore, the acceleration vector is written as v2   a =  v r − q r 

 vr vq   e ( r ) +  vq + r  

   e (q ) + v z e ( z ) 

 Note that e ( i ) are unit vectors in cylindrical coordinates. We  can also write E the acceleration vector in terms of Cartesian unit vectors i, by substitut        e ( r ) cosq E1 + sin q E2, e (q ) = − sin q E1 + cosq E2, and e ( z ) = E3 (see ing for= Example 8.1), or



v2  vv    a =  v r − q  cosq −  vq + r q r  r  

    sin q   E1  

  v2  vv     +  v r − q  sin q +  vq + r q  cosq   E2 + v z E3 r  r    



13.13

In terms of coordinates themselves, we have   a =   r − rq2 cosq − rq + 2 rq sin q  E1     2 +   r − rq sin q + rq + 2 rq cosq  E2 + zE3  

(

)

(

(

)

)

)

(

)

)

Similarly, for a spherical coordinate system, using Equation 13.11, we can write the physical components of the acceleration vector for spherical vr vj vq2 cos j v2 v2 coordinates as a (1 ) = ar = v r − j − q , a ( 2 ) = , and − aj = vj + r r sin j r r

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82 • Tensor Analysis for Engineers, 3E vr vq vq vj cos j . Note that all components are the physical + r r sin j components of the corresponding vectors. Therefore, the acceleration vector is a(3) = aq = vq +

vj2 + vq2   a =  v r − r 

 vr vj vq2 cos j  −  e ( r ) +  vj + r r sin j  

  vr vq vq vj cos j +  e (j ) +  vq + r r sin j  

  e (q ) 

 Note that e ( i ) are unit vectors in spherical coordinates. In terms of the coordinates themselves, we have    a = r − rj 2 − r sin 2 jq2  e ( r ) +  rj + 2 rj − r sin j cos jq2  e (j )    cos j  e (q ) . +  rqsin j + 2 rq sin j + 2 rjq

We can also write the acceleration vector in terms of Cartesian unit      ­vectors Ei, by substituting for e ( r ) = sin j cosq E1 + sin j sin q E2 + cos j E3,        − sin q E1 + cosq E2 e (j ) = cos j cosq E1 + cos j sin q E2 − sin j E3, and e (q ) = (see Example 8.1), or  vj2 v2  vr vj vq2 cos j    a =  v r − − q  sin j cosq +  vj + − r r  r r sin j  

   sin q  E1   2    13.14 vj v2  vr vj vq2 cos j   vr vq vq vj cos j   +  v r − − q  sin j sin q +  vj + − +  cos j sin q +  vq +  cosq  E2 r sin j  r r  r r sin j  r      + ( cos j − sin j ) E3   vr vq vq vj cos j +  cos j cosq −  vq + r r sin j  

Equations 13.13 and 13.14 are useful relations for calculating acceleration vectors in cylindrical and spherical coordinate systems, respectively. Similar calculations can be performed for arbitrary orthogonal curvilinear systems.

13.4 SUBSTANTIAL TIME DERIVATIVES OF TENSORS Another form of derivative of tensors appearing in governing equations is the substantial derivative—also referred to as the total or convective time derivative. For a quantity like A —a scalar, vector, or tensor—that varies with space and time, we would like to collect all its derivatives. Let’s assume that we have a body of fluid moving in a fixed coordinate system, for e­ xample, a Cartesian system. We consider two scenarios: a) we “ride” the fluid and

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Coordinate Independent Governing Equations • 83

move with it and register its variation with respect to time, and b) we stay at a fixed location in space and register the variation of the fluid passing by with respect to time. In case (a) the changes are due to any variation of the fluid w.r.t time, since we move with it and relative space change is absent/null. In other words, the change w.r.t space is implicitly included. But in case (b) we may have changes w.r.t both time and space for the fluid in motion.

FIGURE 13.1  Local coordinates attached to a moving fluid body in a fixed Cartesian coordinate system.

Let’s consider function A = A ( yi , t ) , where A is the tensor quantity, yi is the fixed Cartesian coordinate system, and t is time (see Figure 13.1). Here, we d yi assume that the ratio = u i applies, or the change of space coordinates of dt the fluid w.r.t time is the same as the local fluid velocity. Now the total time DA derivative is given by the following = limit, Dt



TAE3E.CH13_1pp.indd 83

 ∂A ∂A ∂yi  lim + i   , or d yi , d t → 0 ∂t ∂y ∂t  

DA ∂A ∂A = + u i i 13.15 Dt ∂t ∂y  An immediate extension is to write the total derivative for A = A ij ei e j , a i mixed tensor of the second rank, in an arbitrary system x , or  i  j ∂ A ij ei e j DA ∂ A j ei e k 13.16 = +u Dt ∂t ∂x k

(

)

(

)

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84 • Tensor Analysis for Engineers, 3E Using Equation 10.14, we can write the second term on the right-hand-side  ∂ A ij ei e j ∂A ij k k i  j = u A j, k ei e . The first term is just equal to as u , if the coor∂ xk ∂t dinates x i are also fixed. Note that if the x i system is moving, we should consider the changes of the basis vectors with reference to time as well (see Equation 13.16). Therefore, we get the components (mixed contravariant/ covariant) of the substantial time derivative of A, as

(

)

( )

i i  DA  ∂ A j + u k A ij,k 13.17 =  ∂t  Dt  j

Note that A= i j, k

∂A ij

i + Γ km A mj − Γ nkj Ani .

∂x A specific case is to let A be equal to velocity itself, which results in the total derivative of vi , as k

i





i  Du  ∂u = + u ju ,ij 13.18   ∂t  Dt 

In orthogonal coordinate systems, we can write the physical components the convective time derivative of a vector as   DA  ∂A ( i ) v ( j ) ∂A ( i ) u ( i ) A ( j ) ∂hi +   ( i) = + ∂t h j ∂x j hi h j ∂x j  Dt  13.19 u ( j ) A ( j ) ∂h j , no sum on i − ∂x i hi h j Using Equation 13.19 we can write the total time derivative of vectors like velocity or acceleration. The following example demonstrates these calculations. 13.4.1 Example: Substantial time derivation of acceleration vectors—physical components From previous examples, we have ( hr , hq , hz ) = (1, r ,1 ) for cylindrical coordinate systems and ( hr , hj , hq ) = (1, r , r sin j ) for spherical coordinate systems. We can write the physical components of the total derivative of acceleration  vector a = ( ar , aq , az ) , in cylindrical coordinates, by using Equation 13.19, as

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Coordinate Independent Governing Equations • 85



  Da  ∂ar a ∂ar u q ∂ar ∂a + + vz r − vq q  = + vr  r ∂q r ∂r ∂z  Dt  r ∂t  Da  ∂aq ∂aq u q ∂aq ∂a u a + + u z q + q r 13.20   = + vr r ∂q r ∂t ∂r ∂z  Dt q  Da  ∂az ∂az u q ∂az ∂a + + vz z   = + vr ∂r r ∂q ∂z  Dt  z ∂t Similarly, for spherical coordinates we receive the physical components of the total derivative of acceleration vector as



 Da  ∂ar aj u a ∂ar u j ∂ar ∂ar + + q − vj − vq q   = + vr ∂r r ∂j r sin j ∂q r r  Dt  r ∂t  Da  ∂aj ∂aj u j ∂aj ∂aj u j ar u q aq cos j u + vr + + q + − 13.21   = ∂t ∂r r ∂j r sin j ∂q r r sin j  Dt j  Da  ∂aq ∂a u j ∂aq u ∂aq u q ar u q aj cos j  + vr q + + q + +  = ∂t ∂r r ∂j r sin j ∂q r r sin j  Dt q Note that in these relations (i.e., Equations 13.20 and 13.21) all components are the physical components of the corresponding vectors.

13.5 CONSERVATION EQUATIONS—COORDINATE INDEPENDENT FORMS A fundamental equation for the transformation of a tensor quantity per  unit volume, Ψ in an arbitrary control volume with a velocity field V, can be obtained using the Gauss divergence theorem and the law of the principle of conservation [11]. An integral equation defining the time rate and divergence of convective transformation balanced with the rate of production/destruction Ω of the quantity leads to the differential relation given by Equation 13.22, as

∂Ψ + ( Ψ V i ), i =Ω 13.22 ∂t The second term can be written as ( Ψ V i ), i= V i Ψ , i + Ψ V i, i since covariant differentiation follows the differentiation product rule. Therefore, L.H.S

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86 • Tensor Analysis for Engineers, 3E

of Equation 13.22 can be written as

∂Ψ DΨ i + V i Ψ , i + Ψ V= + Ψ V i, i . In ,i t Dt ∂  =

vector form, this equation reads

DΨ Dt

  DΨ DΨ + Ψ∇ ⋅ V = + Ψ V i, i = Ω 13.23 Dt Dt



Equation 13.22 (or equally Equation 13.23) can be considered the equation of motion for the control volume or, in general, the medium that the quantity Ψ transports in it. The source term is associated with the quantity in question: for example, the mass, momentum, energy, electric charge, etc. Now we consider mass conservation, or let Ψ =r where r is mass per unit volume or density of the quantity in motion. After substitution in Dr + r V i, i = Ω. If mass source is zero, then we get Equation 13.23, we get Dt the continuity equation, as Dr + r V i, i = 0 13.24 Dt

Note that „„

Dr ∂r = + V i r , i . Some specific cases can be observed: Dt ∂t

r = r ( t ) only, i.e., we have an inhomogeneous quantity whose density changes with time. In this case the continuity equation reads as ∂r ∂r + V i ( r , i ) + r V i, i = 0 , or + r V i, i = 0.  ∂t ∂t =0

„„

r = r ( x i ) only, i.e., we have an inhomogeneous quantity whose density changes from location to location in the continuum but remains constant with time at any given location. In this case the continuity equation reads i as  ∂r  + V i r + r V i = 0 , or ( r V ) = 0 . ,i ,i  ∂t   =0

„„

r is a constant, i.e., we have a homogeneous quantity with constant density at all locations and times in the continuum. In this case the  Dr  i 0 , or V i, i = 0. continuity equation reads as   + r V ,i = Dt   =0

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Coordinate Independent Governing Equations • 87

 Now we consider momentum conservation, or let Ψ =r V , i.e., momentum per unit volume of the quantity in motion. In this case Ψ is a vector quantity. Substituting in Equation 13.22, the i th contravariant component reads

∂(rVi ) ∂t

(

+ r V iV j

)

, j

= Ω i 13.25

Equation 13.25 is Newton’s 2nd law written in the coordinate ­independent form for a medium, for example a fluid moving in the continuum. The L.H.S is the inertia force and the R.H.S is the applied force on the material, per unit volume. Considering the material as a fluid, then the applied force could be, in general, a combination of hydrostatic, viscous, gravitational, ­electromagnetic, etc. forces.

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CHAPTER

14

COLLECTION OF RELATIONS FOR SELECTED COORDINATE SYSTEMS In this section, we provide a list of some commonly used coordinate systems as well as relations for an arbitrary orthogonal curvilinear coordinate system. We categorize the content based on the coordinate systems. All coordinates are considered in a 3D Euclidean space. Cartesian coordinate systems are fixed references consisting of three flat planes. The curvilinear coordinates may include one or more curved coordinate surfaces.

14.1  CARTESIAN COORDINATE SYSTEM       We use ( e (1 ) , e ( 2 ) , e ( 3 ) ) = Ex , Ey , Ez , ( A (1 ) , A ( 2 ) , A ( 3 ) ) = Ax , Ay , Az symbols for physical components. Note that in a Cartesian system, all contravariant, covariant, and physical components of a tensor quantity are identical, hence all indices are shown as subscripts.

(

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90 • Tensor Analysis for Engineers, 3E

TABLE 14.1  Relations for tensors and their related derivatives in Cartesian coordinate systems.

Coordinates  Basis/Unit vectors, Ei

( x, y, z ) ≡ ( X, Y, Z )

  

  

Scale factors, hi

( E , E , E ) ≡ ( i , j, k ), orthogonal ( h , h , h ) = (1,1,1)

Metric tensors, gij

g= g= g= 1 , the rest are null xx yy zz

Jacobian

 =1

x

y

x

y

z

z

d𝕍 = dxdydz      2 2 2 Line element and magnitude dS = dxEx + dyEy + dzEz dS = ( dx ) + ( dy ) + ( dz )     A = Ax , Ay , Az = Ax Ex + Ay Ey + Az Ez Vector components Unit volume

(

)

Christoffel symbols

null   A ⋅= B Ax Bx + Ay By + Az Bz Dot-product (two vectors)     A ×= B Ay Bz − Az By Ex + ( Az Bx − Ax Bz ) Ey Cross-product (two vectors)  + Ax By − Ay Bx Ez

(

)

(

)

Gradient vector

  ∂  ∂  ∂ = ∇ Ex + Ey + Ez ∂x ∂y ∂z

Gradient of a scalar, Ψ

  ∂Ψ  ∂Ψ  ∂Ψ = ∇Ψ Ex + Ey + Ez ∂x ∂y ∂z

 Gradient of a vector, A

 ∂Ax   ∂x    ∂Ax ∇A =   ∂y  ∂A  x  ∂z

∂Ay ∂x ∂Ay ∂y ∂Ay ∂z

∂Az   ∂x  ∂Az   ∂y  ∂Az   ∂z 

 Curl of a vector, A

   ∂Az ∂Ay    ∂Ax ∂Az ∇= ×A  − −  Ex +  ∂z  ∂x  ∂z  ∂y

 Divergence of a vector, A

  ∂Ax ∂Ay ∂Az ∇ ⋅= A + + ∂x ∂y ∂z

   ∂Ay ∂Ax   −  Ez  Ey +  ∂y    ∂x

(continued)

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  ∂2 Ψ ∂2 Ψ ∂2 Ψ ∇ 2 Ψ = ∇ ⋅ ∇Ψ = + 2 + 2 ∂x2 ∂y ∂z  2 ∇ A=

(

Laplacian of a scalar, Ψ

)

2 2 2  ∂ 2 Ax ∂ 2 Ax ∂ 2 Ax    ∂ Ay ∂ Ay ∂ Ay   + + + + E   Ey  2 +  x  ∂x2 ∂y2 ∂z2  ∂y2 ∂z2   ∂x   ∂ 2 A ∂ 2 Az ∂ 2 Az   +  2z + + 2  Ez ∂y2 ∂z   ∂x

 Laplacian of a vector, A

 ∂4 ∂4 ∂4 ∇ 4 = 4 + 4 + 4 ∂y ∂z  ∂x

Biharmonic operator

  ∂4 ∂4 ∂4   + 2 2 2 + 2 2 + 2 2    ∂x ∂y ∂y ∂z ∂x ∂z 

14.2  CYLINDRICAL COORDINATE SYSTEMS        We use ( e (1 ) , e ( 2 ) , e ( 3 ) ) = ( e ( r= A (1 ) , A ( 2 ) , A ( 3 ) ) ) , e (q ) , e ( z ) ) , A (=  ( A ( r ) , A (q ) , A ( z) ) to designate the unit vectors e ( i ) and physical components of vector A. Most of the expressions are written in terms of physical components. For Christoffel symbols, see Equation 10.28. TABLE 14.2  Relations for quantities and their related derivatives in cylindrical coordinate systems.

Coordinates (orthogonal)

( x , x , x ) ≡ ( r,q , z ) , q is the azimuth angle 1

2

3

Coordinate functions

x = r cos q

y = r sin q

z=z

Coordinate surfaces

x 2 + y2 = r 2, cylinders

y / x = tan q, planes

z = constant, planes

 Basis vectors, ei, covariant

 cos q  − r sin q  0        ( er , eq , ez ) =   sin q  ,  r cosq  , 0     0   0  1        

 Basis vectors, e i, contravariant

 cos q  − sin q / r  0    r q  z ( e , e , e ) =   sin q  ,  cosq / r  , 0    1    0   0      

Scale factors, hi

( hr , hq , hz ) = (1, r,1) (continued)

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92 • Tensor Analysis for Engineers, 3E

Unit vectors, physical components

 cos q  − sin q  0        = e r , e q , e z ( ( ) ( ) ( ) )   sin q  ,  cosq  , 0     0   0  1        

Metric tensors, gij and g ij

g= g= ( g rr )= ( g zz )= 1, g= rr zz qq null-system is orthogonal

Jacobian

 =r

Unit volume

d𝕍 = rdrdq dz

Line element and magnitude

    dS =dre ( r ) + rdq e (q ) + dze ( z )

−1

−1

( g )= qq

−1

 dS =

( dr )2 + ( rdq )2 + ( dz )2

    A = A ( r ) e ( r ) + A (q ) e (q ) + A ( z ) e ( z )

Vector components Christoffel symbols, based on Unit vectors

r Γˆ qq = −1 , Γˆ qrq = Γˆ qq r =, 1 the rest are null

Christoffel symbols, based on covariant Basis vectors

r Γqq = − r , Γqrq = Γqq r = 1 / r , the rest are null

Dot-product (two vectors) Cross-product (two vectors)

r 2 , the rest are

  = A ⋅ B A ( r ) B ( r ) + A (q ) B (q ) + A ( z ) B ( z )    = A × B  A (q ) B ( z ) − A ( z ) B (q )  e ( r )  +  A ( z ) B ( r ) − A ( r ) B ( z )  e (q )  +  A ( r ) B (q ) − A (q ) B ( r )  e ( z )

Gradient vector

  ∂ 1 ∂  ∂ = ∇ er + eq + ez ∂r r ∂q ∂z

Gradient of a scalar, Ψ

  ∂Ψ 1  ∂Ψ  ∂Ψ = ∇Ψ er + eq + ez ∂r r ∂q ∂z

Gradient of a vector, A

∂A ( r )   ∂r     ∂A ( r ) A (q ) = ∇A  − r  r ∂q ∂A ( r )   ∂z 

∂A (q ) ∂r ∂A (q ) A ( r ) + r ∂q r ∂A (q ) ∂z

∂A ( z )   ∂r  ∂A ( z )  r ∂q  ∂A ( z )  ∂z 

(continued)

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Collection of Relations for Selected Coordinate Systems • 93

   ∂A ( z ) ∂A (q )    ∂A ( r ) ∂A ( z )   = ∇×A  − −  e (r) +   e (q ) r q z ∂ ∂ ∂r     ∂z  ∂ ( rA (q ) ) ∂A ( r )   + −  e ( z)  r ∂r r ∂q  

 Curl of a vector, A

Divergence  of a vector, A

  ∂ ( rA ( r ) ) ∂A (q ) ∂A ( z ) = ∇⋅A + + ∂z r ∂r r ∂q

Laplacian operator

  1 ∂  ∂  1 ∂2 ∂2 ∇2 = ∇ ⋅ ∇ = r + +   r ∂r  ∂r  r 2 ∂q 2 ∂z2

Laplacian of a scalar, Ψ

  1 ∂  ∂Ψ  1 ∂ 2 Ψ ∂ 2 Ψ ∇ 2 Ψ = ∇ ⋅ ∇Ψ = + r + r ∂r  ∂r  r 2 ∂q 2 ∂z2

Laplacian  of a vector, A

Biharmonic operator

( ) (

)

  2 2 ∂A (q ) A ( r )   ∇ 2 A =∇ − 2  e (r)  A(r) − 2 r ∂q r    2 ∂A ( r ) A (q )    + ∇ 2 A (q ) + 2 − 2  e (q ) + ∇ 2 A ( z )  e ( z ) r ∂ q r   1 ∂  ∂  1 ∂  ∂   1 ∂2 ∇4 = r  + r r ∂r  ∂r  r ∂r  ∂r    r 2 ∂q 2

 1 ∂2  ∂4 + 4  2 2   r ∂q  ∂z

14.3  SPHERICAL COORDINATE SYSTEMS       We use ( e (1 ) , e ( 2 ) , e ( 3 ) ) = ( er , ej , eq ), ( A (1 ) , A ( 2 ) , A ( 3 ) ) = ( A ( r ) , A (j ) , A (q ) )   to designate the unit vectors e ( i ) and physical components of vector A. Most of the expressions are written in terms of physical components. For Christoffel symbols, see Equation 10.28. TABLE 14.3  Relations for quantities and their related derivatives in spherical coordinate systems.

Coordinates (orthogonal) Coordinate functions Coordinate surfaces

( r,j ,q ), q azimuth and j polar angle x = r sin j cos q x2 + y2 + z2 = r 2, spheres

y = r sin j sin q

(x

z = r cos j

+ y2 ) / z 2 = tan 2 j, y / x = tan q, planes cones 2

(continued)

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94 • Tensor Analysis for Engineers, 3E

 Basis vectors, ei , covariant

 sin j cos q   r cos j cos q  − r sin j sin q      = e , e , e ( r j q )   sin j sin q  ,  r cos j sin q  ,  r sin j cosq      cos j   − r sin j   0      

 Basis vectors, e i , contravariant

 sin j cos q  cos j cos q  1   r  j q ( e , e , e ) =   sin j sin q  , 1r  cos j sin q  , r sin j   cos j   − sin j     

Scale factors, hi

( h , h , h ) = (1, r, r sin j )

Unit vectors, physical components

 sin j cos q  cos j cos q  − sin q        = e r , e j , e q ( ) ( ) ( ) ( )   sin j sin q  ,  cos j sin q  ,  cosq     cos j   − sin j   0        

r

= grr

Metric tensors, gij ij and g

j

q

( g= ) rr −1

1,= gjj

Jacobian

 = r 2 sin j

Unit volume

d𝕍 = r 2 sin j drdj dq    = dS dre ( r ) + rdj e (j )  + r sin j dq e (q )

Vector components

r Γjj = −r

Christoffel symbols, based on Unit vectors

r Γjj = −1

Cross-product (two vectors)

jj

−1

r 2= , gqq

 dS =

(g = ) ( r sin j ) , the rest qq

−1

2

( dr )2 + ( rdj )2 + ( r sin j dq )2

    A = A ( r ) e ( r ) + A (j ) e (j ) + A (q ) e (q )

Christoffel symbols, based on covariant Basis vectors

Dot-product (two vectors)

(g = )

are null-system is orthogonal

Line element and magnitude

− sin q      cos q    0   

Γjrj = Γqrq

− r sin 2 j

j Γqq = − sin j cos j

r Γqq = − sin j

j Γqq = − cos j

Γjrj = 1,

r Γqq =

=1/r

sin j Γqrq =

Γqjq = cot j Γqjq = cos j

  = A ⋅ B A ( r ) B ( r ) + A (j ) B (j ) + A (q ) B (q )

   = A × B  A (j ) B (q ) − A (q ) B (j )  e ( r )  +  A (q ) B ( r ) − A ( r ) B (q )  e (j )  +  A ( r ) B (j ) − A (j ) B ( r )  e (q )

(continued)

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Collection of Relations for Selected Coordinate Systems • 95

Gradient vector

  ∂  1 ∂ = ∇   er +   ∂r   r ∂j

  1 ∂   ej +   eq ∂ r sin j q  

 ∂Ψ   1 ∂Ψ    1 ∂Ψ   er +  = ∇Ψ  ej +   eq ∂r  r ∂j   r sin j ∂q 

Gradient of a scalar, Ψ

  ∇A =  ∂A ( r )  ∂r   ∂A ( r ) A (j ) −  r ∂j r   1 ∂A ( r ) A (q ) −  r  r sin j ∂q

Gradient of a vector, A

  = ∇×A Curl

∂A (j ) ∂r ∂A (j ) A ( r ) + r ∂j r 1 ∂A (j ) A (q ) − cot j r sin j ∂q r

∂A (q ) ∂r ∂A (q )

     r ∂j   1 ∂A (q ) A ( r ) A (j ) + + cot j  r sin j ∂q r r 

 ∂ ( A (q ) sin j ) ∂A (j )   −   e (r)  ∂j ∂q   1  1 ∂A ( r ) ∂ ( rA(q )   +  −  e (j ) r  sin j ∂q ∂r  1 r sin j

1  ∂ ( rA (j ) ) ∂A ( r )   +  −  e (q ) r  ∂r ∂j 

  1 ∂ ( r 2 A ( r ) ) 1 ∂ ( A (j ) sin j ) 1 ∂A (q ) Divergence  of a= ∇⋅A + + 2 vector, A ∂r ∂j r r sin j r sin j ∂q

Divergence of a rank tensor, A

 ∂ ( r 2 A ( rr ) )  1 ∂ ( A ( rj ) sin j ) A (jj ) + A (qq ) 1 ∂A ( rq )    e (r) = ∇⋅ A  + − + 2  ∂ ∂ j j j ∂q  sin sin r r r r r r   ∂ ( r 2 A ( rj ) )  1 ∂ ( A (jj ) sin j ) A ( rj ) 1 ∂A (jq ) A (qq ) 2nd cot j  e (j ) + + + + − 2   r r r sin r r r sin r j j j q ∂ ∂ ∂   2  ∂ ( r A ( rq ) ) ∂A (jq )  1 ∂A (qq ) A ( rq ) 2 A (jq ) cot j  e (q ) + + + + +   r 2 ∂r r ∂j r sin j ∂q r r  

Laplacian operator

  1 ∂  2 ∂  1 1 ∂  ∂  ∂2 ∇2 = ∇ ⋅ ∇ = 2 r + 2  sin j + 2 2 r ∂r  ∂r  r sin j ∂j  ∂j  r sin j ∂q 2

Laplacian of a scalar, Ψ

  1 ∂  2 ∂Ψ  1 ∂  ∂Ψ  ∇ 2 Ψ = ∇ ⋅ ∇Ψ = 2 r +  sin j  r ∂r  ∂r  r 2 sin j ∂j  ∂j  1 ∂2 Ψ + 2 2 r sin j ∂q 2

( ) (

)

(continued)

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Laplacian  of a vector, A

  2Ar  2   sin  A    2 A      2 A   2 A  r    2  2  er 2 r r sin r sin         A   2   A  r   2 cos  A       2 A    2 2   2  2 2  e   r  r r sin    sin    A   2 A  r  2 coss    A        2 A    2 2  2  2 2  e   r sin    r sin  r sin   

Biharmonic operator, ∇ 4

1 ∂  2 ∂  1 ∂  2 ∂   ∇4 = 2 r   r r ∂r  ∂r  r 2 ∂r  ∂r    ∂  ∂  1 ∂  ∂   1 + 2 sin j  2  sin j   r sin j ∂j  ∂j  r sin j ∂j  ∂j    1 1 ∂2  ∂2  + 2 2   r sin j ∂q 2  r 2 sin 2 j ∂q 2 

14.4  PARABOLIC COORDINATE SYSTEMS       We use ( e (1 ) , e ( 2 ) , e ( 3 ) ) = ( e (x ) , e (h ) , e (q ) ), ( A (1 ) , A ( 2 ) , A ( 3 ) ) = ( A (x ) , A (h ) , A (q ) ) to designate the unit vectors e ( i ) and physical components of vector A. Most of the expressions are written in terms of physical components. For Christoffel symbols, see Example 15.4. TABLE 14.4  Relations for quantities and their related derivatives in parabolic coordinate systems.

Coordinates (orthogonal) Coordinate functions

(x ,h ,q ) x = xh cos q

x 2 + y2 =

Coordinate surfaces

 Basis vectors, eicovariant

y = xh sin q

= z

(x

2

−h2)/2

x 2 + y2 =

−2x 2 ( z − x 2 / 2 ) ,

−2h 2 ( z − h 2 / 2 ) ,

paraboloids

paraboloids

y / x = tan q , planes

 h cos q  x cos q  −xh sin q      ( ex , eh , eq ) =  h sin q  , x sin q  ,  xh cosq      x   −h   0      

(continued)

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 Basis vectors, e i, contravariant

 h cos q   1   x h q ( e , e , e ) =   x 2 + h 2 h sin q  , x 2 +1 h 2  x     

Scale factors, hi

(h , h ,h ) = (x x

h

q

2

+ h 2 , x 2 + h 2 , xh

x cos q    1 x sin q  ,  −h  xh  

− sin q       cos q     0   

)

( e (x ) , e (h ) , e (q ) ) = Unit vectors, physical components

 1   2 2  x +h 

Metric tensors, gij and g ij

gxx = x 2 h 2 , gh h = x 2 h 2 , gqq = (xh ) , ( gxx ) =+ ( gh h ) =+ ( gqq ) = the rest are null

 Basis vectors, e i , contravariant

 h cos q  x cos q   1  1   x h q  ( e , e , e ) =   x 2 + h 2 h sin q  , x 2 + h 2 x sin q  , xh1  x   −h       

Jacobian

h cos q  1    h sin q  , 2 2   x +h  x 

x cos q  − sin q       x sin q  ,  cos q        −h   0  

−1

−1

−1

2

− sin q       cos q     0   

=  xh (x 2 + h 2 )

Unit volume

= d𝕍 xh (x 2 + h 2 ) dx dh dq  dS =

Line element and magnitude

   dS = x 2 + h 2  dx e (x ) + dh e (h )   + xh dq e (q )

Vector components

    A = A (x ) e (x ) + A (h ) e (h ) + A (q ) e (q )

(x

2

+ h 2 ) ( dx ) + (x 2 + h 2 ) ( dh ) + (xh dq ) 2

2

2

Γqxq = Γqqx Γxxx = −Γhxh

Γxxh = Γhxx

=1/x

x Christoffel symbols, q Γqq = Γhqq = Γhq q = Γqh = Γhxh = Γhh x = Γhh h based on covariant 2 2 2 2 2 2 −xh / (x + h ) −h x / (x + h ) = 1 / h Basis vectors = x / (x 2 += h2) h / (x 2 + h 2 ) q Γhq q = Γqh

= 1 /h

Dot-product (two vectors)

  = A ⋅ B A (x ) B (x ) + A (h ) B (h ) + A (q ) B (q )

(continued)

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98 • Tensor Analysis for Engineers, 3E

Cross-product (two vectors)

   = A × B  A (h ) B (q ) − A (q ) B (h )  e (x )  +  A (q ) B (x ) − A (x ) B (q )  e (h )  +  A (x ) B (h ) − A (h ) B (x )  e (q )

  ∂   ∂   1 ∂  1 1  ex +   eh +  ∇  Gradient= vector  ej 2 2 2 2      xh ∂j   x + h ∂r   x + h ∂h    1 ∂Ψ    1 ∂Ψ    1 ∂Ψ   Gradient = of a  ex +  e + e ∇Ψ  2 2 2 2     h  xh ∂j  j scalar, Ψ  x + h ∂r   x + h ∂h    ∇2 = ∇ ⋅ ∇ =

( )

Laplacian operator

∂  ∂  ∂ 1 1 x + 2 2 2 x (x + h ) ∂x  ∂x  h (x + h ) ∂h 2

+

1

(xh )2

  ∇ 2 Ψ = ∇ ⋅ ∇Ψ = Laplacian of a scalar, Ψ

Laplacian  of a vector, A

(

)

 ∂  h   ∂h 

∂2 ∂q 2

∂  ∂Ψ  ∂  ∂Ψ  1 1 x + h  2 2 2 x (x + h ) ∂x  ∂x  h (x + h ) ∂h  ∂h  2

+

1

(xh )2

∂2 Ψ ∂q 2

 ∇2 A = , use Equation 11.21.

For this coordinate system, we didn’t list all relations, since they are lengthy. But readers can use the relations given in the next section, as they are applicable to all orthogonal coordinate systems.

14.5 ORTHOGONAL CURVILINEAR COORDINATE SYSTEMS We use ( x1 , x2 , x3 ) ≡ ( x1 , x2 , x3 ) symbols for presentation. Note that all components listed in Table 14.5 are given as physical components. The functional relations for coordinate variables vs. those of Cartesian ones are required for calculating the values of quantities listed in Table 14.5.

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Collection of Relations for Selected Coordinate Systems • 99

TABLE 14.5  Relations for quantities and their related derivatives in curvilinear orthogonal coordinate systems.

( x1 , x2 , x3 ) ≡ ( x1 , x2 , x3 )

Coordinates (orthogonal)

 Basis vectors ei , covariant

( e1 , e2 , e3 )

Scale factors, hi

( h1 , h2 , h3 )

 Unit vectors e ( i ) , physical components

( e (1) , e ( 2 ) , e ( 3 ) ) = ( e1 / h1 , e2 / h2 , e3 / h3 )

  

−1 Metric tensors, gij = g11 (= g11 ) ij and g rest are null

, g22 ( h1 )= 2

 Basis vectors e i contravariant

( e , e , e ) = ( e g

Jacobian

 = h1 h2 h3

Unit volume

d𝕍 = dx1 dx2 dx3

1

2

3

1

11

g ) g ) (h = ) , g (= ( h ) , the (=

Christoffel symbols Dot-product (two vectors) Cross-product (two vectors)

2

33

3

 dS =

( h1 dx1 )

2

+ ( h2 dx2 ) + ( h3 dx3 ) 2

2

    A = A (1 ) e (1 ) + A ( 2 ) e ( 2 ) + A ( 3 ) e ( 3 )

(use Equations 10.20-10.25)   = A ⋅ B A (1 ) B (1 ) + A ( 2 ) B ( 2 ) + A ( 3 ) B ( 3 )    = A × B  A ( 2 ) B ( 3 ) − A ( 3 ) B ( 2 )  e (1 )  +  A ( 3 ) B (1 ) − A (1 ) B ( 3 )  e ( 2 )  +  A (1 ) B ( 2 ) − A ( 2 ) B (1 )  e ( 3 )

 ∇ Gradient vector= Gradient of a scalar, Ψ

33 −1

2

2

  , e2 g 22 , e3 g 33 )

   dS h1 dx1 e (1 ) + h2 dx2 e ( 2 ) Line element and=  magnitude + h3 dx3 e ( 3 ) Vector components

22 −1

 = ∇Ψ

1 ∂  1 ∂  1 ∂  e (1 ) + e (2) + e (3) h1 ∂x1 h2 ∂x2 h3 ∂x3 1 ∂Ψ  1 ∂Ψ  1 ∂Ψ  e (1 ) + e (2) + e (3) h1 ∂x1 h2 ∂x2 h3 ∂x3

(continued)

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100 • Tensor Analysis for Engineers, 3E

Gradient of a vector, A

       ∂A i k i  j ∇ A =∇ j A e j = ∇ j ( A i ei ) e j =  j + A Γ kj  ei e , see Example 15.13.  ∂x 

(

  = ∇×A  Curl of a vector, A

)

1 h2 h3 + +

 ∂ ( h3 A ( 3 ) ) ∂ ( h2 A ( 2 ) )   −   e (1 ) ∂x2 ∂x3  

1  ∂ ( h1 A (1 ) ) ∂ ( h3 A ( 3 ) )   −   e (2) ∂x3 ∂x1 h1 h3   1  ∂ ( h2 A ( 2 ) ) ∂ ( h1 A (1 ) )   −   e (3) ∂x1 ∂x2 h1 h2  

  1  ∂ ( h2 h3 A (1 ) ) ∂ ( h1 h3 A ( 2 ) ) ∂ ( h1 h2 A ( 3 ) )  Divergence  of a = ∇ ⋅A + +   vector, A h1 h2 h3  ∂x1 ∂x2 ∂x3 

Divergence of a 2nd rank tensor, A

 1  ∂ ( h2 h3 A (11 ) ) ∂ ( h1 h3 A ( 21 ) ) ∂ ( h1 h2 A ( 31 ) )   + +       ∂x1 ∂x2 ∂x3  h1 h2 h3   e (1 ) ∇ ⋅ A =  A A A A 12 31 22 33 ( ) ( ) ( ) ( ) h ∂ h h h ∂ ∂ ∂ 3 1 1 2 +  + − −  h1 h2 ∂x2  h1 h3 ∂x3 h1 h2 ∂x1 h1 h3 ∂x1  1  ∂ ( h2 h3 A (12 ) ) ∂ ( h1 h3 A ( 22 ) ) ∂ ( h1 h2 A ( 32 ) )   + +      ∂x1 ∂x2 ∂x3  h1 h2 h3   e (2) +  A A A A 23 12 33 11 ( ) ( ) ( ) ( ) h ∂ h ∂ h ∂ h ∂ 3 2 2 1 +  − − +  h2 h3 ∂x3  h1 h2 ∂x1 h2 h3 ∂x2 h1 h2 ∂x2  1  ∂ ( h2 h3 A (13 ) ) ∂ ( h1 h3 A ( 23 ) ) ∂ ( h1 h2 A ( 33 ) )   + +      ∂x1 ∂x2 ∂x3  h1 h2 h3   e (3) +  31 23 11 22 A A A A ( ) ( ) ( ) ( ) h h ∂ ∂ h h ∂ ∂ 3 3 1 2 +  + − −  h1 h3 ∂x1  h2 h3 ∂x2 h3 h1 ∂x3 h2 h3 ∂x3

  ∇ 2 Ψ = ∇ ⋅ ∇Ψ Laplacian of a scalar, Ψ =

(

)

1  ∂  h2 h3 ∂Ψ  ∂  h1 h3 ∂Ψ  ∂  h1 h2 ∂Ψ       +  + h1 h2 h3  ∂x1  h1 ∂x1  ∂x2  h2 ∂x2  ∂x3  h3 ∂x3  

(continued)

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Collection of Relations for Selected Coordinate Systems • 101

 ∇2 A =

  ∂  h2  ∂ ( h1 A1 ) ∂ ( h3 A3 )      −       ∂x1       1 ∂   1  ∂x3  h1 h3  ∂x3 ∇⋅A +     e1 h2 h3  ∂  h  ∂ ( h2 A2 ) ∂ ( h1 A1 )      h1 ∂x1 3 −    − ∂x  h h  ∂x ∂x2     2 1 2 1        ∂  h3  ∂ ( h2 A2 ) ∂ ( h1 A1 )     −        ∂x2       1 ∂   1  ∂x1  h1 h2  ∂x1 + ∇ ⋅A +    e2 h1 h3  ∂  h  ∂ ( h3 A3 ) ∂ ( h2 A2 )      h2 ∂x2 1 −     − ∂x3      ∂x3  h2 h3  ∂x2     ∂  h1  ∂ ( h3 A3 ) ∂ ( h2 A2 )      −       ∂x3       1 ∂   1  ∂x2  h2 h3  ∂x2 + ∇ ⋅A +    e3 h2 h1  ∂  h  ∂ ( h1 A1 ) ∂ ( h3 A3 )      h3 ∂x3 2 −     − ∂x1      ∂x1  h1 h3  ∂x3  

(

Laplacian  of a vector, A

)

(

)

(

)

    1 ∂ ∂ 1 ∂ ∂      ij g kl l  g ∇4 = j   k  i   ∂x  ∂x   ∂x  ∂x     ∇2   Biharmonic operator, ∇ 4

 ∂ h h ∂  1  ∂  h2 h3 ∂    2 3         ∂x1  h1 ∂x1  h1 h2 h3 ∂x1  h1 ∂x1       ∂  h1 h3 ∂  1 ∂  h1 h3 ∂     1  =   +      h1 h2 h3  ∂x2  h2 ∂x2  h1 h2 h3 ∂x2  h2 ∂x2       ∂  h1 h2 ∂      + ∂  h1 h2 ∂  1         ∂x3  h3 ∂x3  h1 h2 h3 ∂x3  h3 ∂x3    

Modern engineering software tools provide facilities to include curvilinear coordinate systems for computer modeling. For example, COMSOL Multiphysics® has a curvilinear coordinate interface for defining object ­orientation in an arbitrary system [12].

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CHAPTER

15

RIGID BODY ROTATION: EULER ANGLES, QUATERNIONS, AND ROTATION MATRIX In Chapter 12, we discussed Cartesian tensors and rotation of coordinate system using Rodrigues’ formula (see Equation 12.6). For practical applications, mainly rigid body rotation, in this chapter we would like to expand on this topic in relation to Euler angles and quaternions methods. A rigid body transformation can be composed of two parts: a) translation and b) rotation. The translation is usually referred to a vector connecting the origin of the coordinate system to the centre of mass. Therefore, we can always refer the rotation to the instantaneous center of mass and focus on the rotation for calculation and analysis of rigid body motion. In general, there are at least eight methods to represent rotation [13]. Among these three inter-related methods commonly used for calculating rigid body orientation after it goes through possible rotations in a 3D space. These are: 1) rotation matrix, 2) single axis-angle, and 3) quaternions. The first two methods are discussed in Chapter 12 and expanded in this chapter with the inclusion of Euler angles method. The third method is more robust and with less limitations compared to the other methods in terms of practical applications and is new to this second edition. When dealing with rotation of a rigid body, several confusions may arise in terms of definition of rotation matrix and rotation variables. To avoid these ambiguities, identifying two definitions is critical. First, the assumed

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104 • Tensor Analysis for Engineers, 3E transformation should be clear by the way the rotation matrix (see Equation 12.7) is meant to be, i.e., it is meant to relate the coordinates obtained after rotation to the original ones or vice versa. Second, how the rotation itself is performed, i.e., is it that the coordinates are rotated and quantities, like a vector, is kept fixed and we seek to find its new coordinates with reference to the rotated ones or is it that the quantities, like a vector, are rotated and the coordinates are kept fixed and we would like to calculate the coordinates of the rotated vector with reference to the fix-kept coordinates. In this book, we resolve these confusing issues by following the convention described in Chapter 12, i.e., rotation matrix relates the known/given coordinates prior to the rotation to those resulted from the rotation through the rotation matrix ( y′i = Rij y j ). Therefore, we assign a passive convention, see next section. Assuming this definition then the second definition- so-called active and passive rotations [14]- is discussed in next section. Readers should note that some authors/references may have different definition assumed (sometimes implicitly) for the rotation matrix (e.g., y j = R ji y′i , or using the transpose of the rotation matrix considered in this book). Also, readers should note that an important fundamental property of all methods mentioned above is that the rotation procedure reduces to the rotation of one vector about another vector. This can be shown by finding the equivalent single axis and angle for any combinations of rotations [14], see Chapter 12. Examples of practical applications of the methods mentioned above are: rigid body rotation and orientation, computer graphics and animation, aircraft maneuver control, robot arm control, signal processing, quantum mechanics, special and general relativity-among others, [15], [16].

15.1  ACTIVE AND PASSIVE ROTATIONS Following Section 12.1, let’s assume, without losing generality, a 2D coor-

dinate system ( y1 , y2 ) ≡ ( x, y ) being transformed to ( y1′ , y2′ ) ≡ ( x′, y′ ) through  rotation about the axis y3 ≡ z with unit vector n ≡ ni = ( 0,0,1) by an angle q , according to R.H.R. Therefore, using the rotation matrix Rij (see  x′   cosq sin q   x  Equation 12.7) we have   =     and z′ = z. Readers should  y′  − sin q cosq   y

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Rigid Body Rotation: Euler Angles, Quaternions, and Rotation Matrix • 105

note that the rotation matrix given here relates the rotated coordinates to the existing/current ones. Now if we have a vector V = Vx , Vy given in x − y system and making angle j with positive x axis then its components in x′ − y′  V   cosq sin q   Vx  system, in terms of Vx and Vy , are given as  x′  =     . This Vy′   − sin q cosq  Vy 

(

)

type of transformation for which the vector is kept fixed and the coordinate system rotates is called passive rotation. We can also perform the opposite operation by rotating the vector and keep the original coordinate system  fixed. Therefore, we keep the system x − y fixed and rotate the vector V by an angle equal to −q about the same rotation axis (i.e., ni = ( 0,0,1 ) ) to get  the vector V′. This type of transformation for which the vector is rotated, and the coordinate system is kept fixed is called active rotation. See Figure   Vx′  15.1. The components of the rotated vector are V ′ =  ′  and are equivalent Vy  V  to  x′  in their magnitudes. In other words, passive rotation through an Vy′  angle q is equivalent to corresponding active rotation through an angle equal to −q . For 3D rotations, we can expand the relationship between passive and active rotations by using the properties of rotation matrix, see Section 12.1, or   Rij (= n,q ) R ji ( n, −q )



15.1

Equation 15.1 states that the rotation matrix for a given angle and axis of rotation is equal to its transpose when the rotation angle is in opposite direction but about the same rotation axis. As mentioned in this book, we consider Rij, as given by Equation 12.7, for passive rotations and use its transpose (or inverse) for active rotations. 15.1.1  Example: Active and Passive Rotations-Numerical Example  Vector = V = Vx , Vy ( 3,4 ) is given in Cartesian coordinate system ( x, y ), as

(

)

shown in Figure 15.1. Show that passive and active rotations provide equivalent answers. The absolute value of angle of rotation is q = 30° and rotation is about positive z-axis.

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106 • Tensor Analysis for Engineers, 3E Solution:

  The magnitude of the vector V is V = 3 2 + 4 2 = 5 and it makes an angle 4 = j tan −1 =   53.13° with the x-axis. For passive rotation, we keep the vec3  tor V fixed and rotate the coordinate system ( x, y ) by 30° in the counter clockwise direction to get the resulted coordinate system ( x′, y′ ) . Therefore, the  cos30 sin 30   3 /2 rotation matrix is  =  − sin 30 cos30   −1/2

1/2  . The components of the 3 /2 

  3 /2 1/2  3  4.598  V vector V in the ( x′, y′ ) system are  x′  = =    . V 4 1.964 ′ y 3 /2        passive  −1/2 For active rotation, we keep the coordinate system ( x, y ) fixed and rotate the vector by 30° in the clockwise direction (or −30° ) to receive the vec ° tor V ′ = Vx′, Vy′ that makes an angle j=′ 53.13° − 30= 23.13° with

(

)

= = the x-axis. Therefore, the components are Vx′ 5cos ( 23.13 ) 4.598 V′ 4.598  = = and Vy′ 5sin ( 23.13 ) 1.964 , or  x  =   . Note that Vy′  1.964  active

the rotation matrix follows the relation given in Equation 15.1, or T  cos30 sin 30  cos30 − sin 30  ° 30 Rij ( 30° ) =  = R − = )  sin 30 cos30  . ji (   − sin 30 cos30    y

y

V

y

V x V







Passive

 x

Active

x

FIGURE 15.1  Passive (left) and active (right) transformations for a 2D coordinate system.

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Rigid Body Rotation: Euler Angles, Quaternions, and Rotation Matrix • 107

15.2  EULER ANGLES We expand the discussion presented in the previous section by considering 3D rotation of a rigid body, for example an airplane. With reference to Chapter 12, we consider the following. Let’s assume the airplane maneuvers in the sky, we can then relate its orientation, focusing on its rotation only, at any given time to its original orientation in a step-wise manner in two ways; 1) by relating its successive rotations to a fixed global Cartesian coordinate system or 2) by relating its orientations to its instantaneous body-attached coordinate system resulted from successive rotations. The latter is so-called Intrinsic, and the former is Extrinsic method. In this book, we use the intrinsic approach, [17], [18]. The extrinsic approach is more common for computer graphics applications. The concept of describing orientation of a rigid body by a limited number of successive rotations goes back to Euler. His theorem, the Euler’s rotation theorem [18], [19], [14] from a practical application point of view, states that any orientation of a rigid body can be calculated through a single rotation about a specified axis, with reference to its original orientation. This relates to what we showed and used, single axis-angle method, in Chapter 12, see Example 12.2.1. Further, Euler showed that the final orientation of a rigid body can be obtained through at most three successive rotations, so called Euler angles, [18], [20], [21]. The intrinsic method, as mentioned above, is mostly used for successive rotations. There exist several definitions for the body-attached coordinate system in terms of axis orientations. Here we use the convention mostly used in engineering by defining positive x-axis (roll) direction towards the front/cockpit, y-axis (pitch) towards the right wing, and z-axis (yaw) downward towards the ground along with positive rotations about each axis. See Figure 15.2, for which R.H.R. applies. Two immediate successive rotations around the same axis are not counted as two separate/independent rotations since we can equally have a single rotation equal to the algebraic sum of the two and hence the three-axis rotation reduces to two-axis rotation. Therefore, in 3D space, having ( x, y, z) coordinates, we can have 12 ( = 3 × 2 × 2) combinations of Euler angles as shown in Table 15.1.

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108 • Tensor Analysis for Engineers, 3E

FIGURE 15.2  Body attached coordinates and corresponding Euler angles. TABLE 15.1  Possible combination of Euler angles for rigid body rotations.

Roll group

Pitch group

Yaw group

xyz

xzy

yxz

yzx

zxy

zyx

(roll-pitchyaw)

(roll-yawpitch)

(pitch-rollyaw)

(pitch-yawroll)

(yaw-rollpitch)

(yaw-pitchroll)

xyx

xzx (roll-yawroll)

yxy

yzy

(pitch-rollpitch)

(pitch-yawpitch)

zxz (yaw-rollyaw)

(yaw-pitchyaw)

(roll-pitchroll)

zyz

Few points should be noted about the order of executing Euler angles. As previously explained the rotation matrix related to each successive rotation should be pre-multiplied by the previous ones. Considering the order of rotations as the reference. This is a direct consequence of our definition for rotation matrix that transforms previous axes to the current rotated ones. Conversely, if one defines the transpose of the rotation matrix given by Equation 12.7 (i.e., transforming current/rotated coordinates to the original ones) then the sequence of rotations through Euler angles and their multiplications of their corresponding rotation matrices to generate the final rotation matrix should be performed in the same order. For example, according to our definition, see Equation 12.7, considering the combination yxz the order of rotation

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Rigid Body Rotation: Euler Angles, Quaternions, and Rotation Matrix • 109

is pitch first, followed by roll and having yaw as the last rotation, or (pitch roll-yaw). With this definition, yxz should be read in backward order, yxz to get the correct final rotation matrix resulted from multiplication of rotation matrices combination, see Section 12.2. Therefore, the related rotation T

T

T

matrices multiplication order is  Rq 3   Rq 2   Rq1  =  Rq1   Rq 2   Rq 3  , where q 1 is the first angle of rotation (for example pitch), q 2 second (for example roll), and q 3 third (for example yaw). The resulted rotation matrix reads

[ R ] =  Rq

3

  Rq 2   Rq1  

15.2

In this book, we use the convention of index-order from 1 to 3 for successive rotations, i.e., first rotation expressed as q 1 and so on. Once the rotation matrix [ R ] is given, it is possible to calculate the corresponding Euler angles which are 12 sets, or mutations of 3 Euler angles corresponding to the rotation matrix, see Table 15.1. As discussed in Chapter 12, we can use rotation matrix [ R ] to calculate the single axis-angle rotation equivalent to three successive rotations by Euler angles. The single angle is given by Equation 12.8, repeated here for convenience, or Rii − 1 R11 + R22 + R33 1 = − . Recall that Rii is the trace of matrix [ R ]. cosq = 2 2 2 The single axis of rotation is the eigenvalue of [ R ], refer to Example 12.2.1. Another method, and may be a quicker one, for calculating the components  of a unit vector along the rotation axis, represented by the unit vector n ≡ nk, can be obtained by manipulating the diagonal elements of [ R ]. For exam2 ple, considering element R (1,1 ) = cosq + ( n1 ) (1 − cosq ), see Equation 12.7. R −1 R −1 2 After substituting for cosq we receive R (1,1 ) = ii + ( n1 )  1 − ii . 2 2   Factoring out n1 gives, after some manipulation, n1 =

2 R (1,1 ) − Rii + 1 . 3 − Rii

 Therefore, for component k of the vector n = ( n1 , n2 , n3 ) , parallel to the rotation axis, we receive

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110 • Tensor Analysis for Engineers, 3E



nk =

2 R ( k, k ) − Rii + 1 = , k 1,2,3  3 − Rii

15.3

As an example, using data given in Example 12.2.1, we receive the trace of the rotation matrix, Rii = 0 and R ( k, k ) = 0. Using Equation 15.3, we receive 1  nk = 1/ 3 , or n = ± (1,1,1) . This is identical to the results obtained in 3 Example 12.2.1. 15.2.1  Example: Euler Angles-Numerical Example An airplane undergoes a zyz (yaw-pitch-yaw) maneuver. The first yaw angle is q 1 = 30, pitch angle q 2 = 45, and the final yaw angle is q 3 = 15. Calculate its final orientation in the sky with reference to the earth fixed coordinates and the equivalent single axis-angle rotation. Solution: We calculate the rotation matrices, using Equation 12.6 and multiply them using Equation 15.2. First yaw rotation: the rotation is about z-axis with ni = ( 0,0,1 ) and q 1 = 30  3  1 0  cos30 sin 30 0   2 2     which gives  Rq1  = − = sin 30 cos30 0  . 1 3   − 0   0 0 1   2 2   0 1   0 Pitch rotation: this rotation is about current y-axis (resulted from the previous rotation) with ni = ( 0,1,0 ) and q 2 = 45 which gives  cos 45 0 − sin 45       Rq 2  = 0 1 0 =     sin 45 0 cos 45    

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2 2 0 2 2

0 − 1 0

2  2  0 .  2  2 

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Rigid Body Rotation: Euler Angles, Quaternions, and Rotation Matrix • 111

Last yaw rotation: this rotation is about current z-axis (resulted from the immediate previous rotation) with ni = ( 0,0,1 ) and q 3 = 15 which gives  3 +1  3 −1 0   cos15 sin15 0  2 2 2 2      Rq 3  = sin15 cos15 0 − =  . 3 −1 3 +1   − 0    0 0 1  2 2  2 2   0 0 1  For final rotation matrix, we first calculate  6 2  2   1 2  3 4 0  4 0 −   2 2 2 2     1 3  Rq 2   Rq1  =  0 1 0  1 = − 3 2 0  2   − 2 2  2  2 0   6 2 0 1    2 2   0 4  4 to pre-multiply  Rq 3  to this result, or  3 +1   2 2 [ R ] = 3 − 1 −  2 2  0



TAE3E.CH15_2pp.indd 111

 6  3 −1 0  4 2 2  1  3 +1 − 0  2 2 2  6 0 1    4

 3+ 2 + 3 − 6  8   −3 − 2 + 3 − 6 = 8   6  4 

 2 2  3 0 −   2  2  2  Rq 2   Rq1  =  0 1 0  1   − 2 2 2   2 − 0   2  2  2   0  0 . The final operation is  2   2 

2 4 3 2 2 4

−

1 2 3 2 0

2  2   0   2   2 

1+3 2 + 3 − 6 8 1+3 2 − 3 + 6 8 2 4

1+ 3   4  −1 + 3   4  2   2 

−

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 0  = 0  1 

112 • Tensor Analysis for Engineers, 3E



 0.4621 0.5657 −0.6830    ≅  −0.6415 0.7450 0.1830  .  0.6124 0.3536 0.7071 

The corresponding single rotation angle is, using Equation 12.8,  2 1 = −  = 62.8°. The elements of the unit vector along the single rotaq cos−1  2 4  2 × 0.4621 − 1.9142 + 1 tion axis are obtained using 15.3, or n= = 9.6 × 10 −2, 1 3 − 1.9142 2 × 0.7071 − 1.9142 + 1 +1 2 × 0.745 − 1.9142= = 0.6786 0.7282, and n3 = 3 − 1.9142 3 − 1.9142 0.6786. Answers are rounded. In other words, we can obtain the final orientation of the airplane by rotating it through an angle of 62.8° about an axis in the direction of nk = ( 0.096, 0.7282, 0.6786 ) with reference to the original coordinate system ( x, y, z ). The successive rotations are shown in Figure 15.3.

= n2 2 × 0.7071 − 1.9142 + 1 3 − 1.9142

FIGURE 15.3  Successive rotation of an airplane, Euler angels left-to-right, yaw=30, pitch=45, yaw=15 degrees.

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Rigid Body Rotation: Euler Angles, Quaternions, and Rotation Matrix • 113

It seems useful to refer the readers to some animation tools for Euler angles. The following Apps, available through the following links, could be used for demonstrating the Euler angles. “Controlling Airplane Flight” [22], “Euler Angles for Space Shuttle” [23], and “Euler Angles” [24].

15.3  CATEGORIZING EULER ANGLES There are several ways possible to categorize Euler angles. With reference to Table 15.1, we observe that there are two sets of Euler angles combination. One set is those six combinations that contain repeated rotations (1st and 3rd) about similar (but not the same) axes ( xyx, xzx, yxy, yzy, zxz, zyz), the socalled proper Euler angles and another set that has different axis of rotations for all three rotations (xyz, xzy, yxz, yzx, zxy, zyx), the so-called Tait-Bryan angles. The latter is commonly used in engineering, specifically in aerospace discipline [18], [19]. It is useful to collect all Euler angles types and their corresponding rotation matrices, as shown in Table 15.2 and Table 15.3. The corresponding single-axis rotation matrix of each case can be calculated as well, like those in Section 15.2.1.

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yaw

roll

pitch

pitch

yaw

yaw

yzy

zxz

zyz

yaw

roll

roll

xzx

pitch

pitch

roll

xyx

q2

yxy

q1

yaw

yaw

pitch

pitch

roll

roll

q3

0 0  1  0 cos sin ( ) ( )   0 − sin ( ) cos ( ) 

Rotation matrixstep

Proper Euler Angles

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Successive rotation-angles

Roll: ni = (1,0,0 )

unit vectorrotation axis cos ( ) 0 − sin ( )    1 0   0  sin ( ) 0 cos ( ) 

=  xyx

=  xzx

=  yxy

=  yzy

=  zxz

=  zyz

0 0   cos (q 2 ) sin (q 2 ) 0  1 0 0  1    0 cos q sin q − sin q cos q 0 0 cos q sin ( ) ( ) ( ) (q1 )  ( ) ( ) 3 3  2 2 1   0 − sin (q 3 ) cos (q 3 )   0 0 1  0 − sin (q 1 ) cos (q 1 )    0 0   cos (q 1 ) 0 − sin (q 1 )  cos (q 3 ) 0 − sin (q 3 )  1    0 1 0 0 cos q sin 1 0 ( ) (q 2 )   0 2     sin (q 3 ) 0 cos (q 3 )  0 − sin (q 2 ) cos (q 2 )   sin (q 1 ) 0 cos (q 1 )      cos (q 3 ) 0 − sin (q 3 )   cos (q 2 ) sin (q 2 ) 0  cos (q 1 ) 0 − sin (q 1 )      1 0 1 0  0   − sin (q 2 ) cos (q 2 ) 0   0    0 0 1   sin (q 1 ) 0 cos (q 1 )   sin (q 3 ) 0 cos (q 3 )   0 0   cos (q 1 ) sin (q 1 ) 0   cos (q 3 ) sin (q 3 ) 0  1   − sin q cos q 0 0 cos q sin ( ) (q 2 )   − sin (q1 ) cos (q1 ) 0  ( ) ( ) 3 3 2      0 0 1  0 − sin (q 2 ) cos (q 2 )   0 0 1    cos (q 3 ) sin (q 3 ) 0  cos (q 2 ) 0 − sin (q 2 )   cos (q 1 ) sin (q 1 ) 0      1 0  − sin (q 3 ) cos (q 3 ) 0   0   − sin (q 1 ) cos (q 1 ) 0        0 0 1   sin (q 2 ) 0 cos (q 2 )   0 0 1 

Rotation matrix

 cos ( ) sin ( ) 0     − sin ( ) cos ( ) 0   0 0 1 

Yaw: ni = ( 0,0,1 )

0 0  cos (q 2 ) 0 − sin (q 2 )  1 0 0  1    0 cos q sin q 0 1 0 0 cos q sin ( ) (q1 )  ( ) ( ) 3 3  1   0 − sin (q 3 ) cos (q 3 )   sin (q 2 ) 0 cos (q 2 )  0 − sin (q 1 ) cos (q 1 )     

Matrices multiplication for successive rotations

Pitch: ni = ( 0,1,0 )

TABLE 15.2  Proper Euler angles and rotation matrices.

114 • Tensor Analysis for Engineers, 3E

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Rigid Body Rotation: Euler Angles, Quaternions, and Rotation Matrix • 115

where,

 xyx

cos (q 2 ) sin (q 1 ) sin (q 2 ) −cos (q 1 ) sin (q 2 )     cos (q 3 ) cos (q 1 ) cos (q 3 ) sin (q 1 )  sin (q 2 ) sin (q 3 )  − sin (q 3 ) cos (q 2 ) sin (q 1 ) + sin (q 3 ) cos (q 2 ) cos (q 1 )  =   − sin (q 3 ) cos (q 1 ) cos (q 3 ) cos (q 2 ) cos (q 1 )  cos (q 3 ) sin (q 2 )  − cos (q 3 ) cos (q 2 ) sin (q 1 ) − sin (q 3 ) sin (q 1 )  

Conversely; for a given single-axis rotation matrix the corresponding  1 − R 2 (1,1 )   R (1,2 )  −1 −1  , and Euler angles are: q1 = tan   , q 2 = tan  R 1,1 R 1,3 − ( ) ( )        R 2,1 ( ) q 3 = tan −1    R ( 3,1 ) 

 xzx

cos (q 2 ) cos (q 1 ) sin (q 2 ) sin (q 1 ) sin (q 2 )     − sin (q 1 ) sin (q 3 ) cos (q 1 ) sin (q 3 )  − sin (q 2 ) cos (q 3 )  = + cos (q 3 ) cos (q 2 ) cos (q 1 ) + sin (q 1 ) cos (q 2 ) cos (q 3 )    − sin (q 1 ) cos (q 3 ) cos (q 1 ) cos (q 3 )  sin (q 2 ) sin (q 3 )    − − cos q cos q sin q sin q cos q sin q ( ) ( ) ( ) ( ) ( ) ( ) 1 2 3 3 2 1  

Conversely; for a given single-axis rotation matrix the corresponding  1 − R 2 (1,1 )   R (1,3 )  −1   , and Euler angles are: q 1 = tan −1  , q tan =  2  R (1,1 )   R (1,2 )   R ( 3,1 )  q 3 = tan −1    − R ( 2,1 ) 

 yxy

cos (q 3 ) cos (q 1 ) sin (q 2 ) sin (q 3 ) − cos (q 3 ) sin (q 1 )     − cos (q 1 ) cos (q 2 ) sin (q 3 )   − sin (q 3 ) cos (q 2 ) sin (q 1 )  = sin (q 1 ) sin (q 2 ) cos (q 2 ) cos (q 1 ) sin (q 2 )   sin (q 3 ) cos (q 1 ) − sin (q 2 ) cos (q 3 ) cos (q 3 ) cos (q 2 ) cos (q 1 )    + cos (q ) cos (q ) sin (q ) − sin (q 3 ) sin (q 1 )  3 2 1 

Conversely; for a given single-axis rotation matrix the corresponding  1 − R 2 ( 2,2 )   R ( 2,1 )  −1   , and q tan = Euler angles are: q 1 = tan −1  ,  2  R ( 2,2 )   R ( 2,3 )   R (1,2 )  q 3 = tan −1    − R ( 3,2 ) 

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116 • Tensor Analysis for Engineers, 3E

 yzy

− sin (q 3 ) sin (q 1 ) sin (q 2 ) cos (q 3 ) − sin (q 3 ) cos (q 1 )     − cos (q 3 ) cos (q 2 ) sin (q 1 )   + cos (q 3 ) cos (q 2 ) cos (q 1 )  = − cos (q 1 ) sin (q 2 ) cos (q 2 ) sin (q 1 ) sin (q 2 )   sin (q 1 ) cos (q 3 ) sin (q 2 ) sin (q 3 ) − sin (q 3 ) cos (q 2 ) sin (q 1 )    + cos (q ) cos (q ) sin (q )  + cos (q 3 ) cos (q 1 ) 1 2 3  

Conversely; for a given single-axis rotation matrix the corresponding  1 − R 2 ( 2,2 )   R ( 2,3 )  −1 −1  , and Euler angles are: q 1 = tan   , q 2 = tan   R ( 2,2 )   − R ( 2,1 )   R ( 3,2 )  q 3 = tan −1    R (1,2 ) 

 zxz

cos (q 3 ) cos (q 1 ) cos (q 3 ) sin (q 1 ) sin (q 3 ) sin (q 2 )      − sin (q 3 ) cos (q 2 ) sin (q 1 ) + sin (q 3 ) cos (q 2 ) cos (q 1 )  = − sin (q 3 ) cos (q 1 ) cos (q 3 ) cos (q 2 ) cos (q 1 ) cos (q 3 ) sin (q 2 )    − sin (q 3 ) sin (q 1 )  − sin (q 1 ) cos (q 2 ) cos (q 3 )    sin (q 1 ) sin (q 2 ) − cos (q 1 ) sin (q 2 ) cos (q 2 )  

Conversely; for a given single-axis rotation matrix the corresponding  1 − R 2 ( 3,3 )   R ( 3,1 )  −1   , and Euler angles are: q 1 = tan −1  , tan q =  2  R ( 3,3 )   − R ( 3,2 )   R (1,3 )  q 3 = tan −1    R ( 2,3 ) 

 zyz

− sin (q 3 ) sin (q 1 ) sin (q 3 ) cos (q 1 ) − cos (q 3 ) sin (q 2 )      + cos (q 3 ) cos (q 2 ) cos (q 1 ) + sin (q 1 ) cos (q 2 ) cos (q 3 )  = − sin (q 1 ) cos (q 3 ) − sin (q 3 ) cos (q 2 ) sin (q 1 ) sin (q 3 ) sin (q 2 )    + cos (q 3 ) cos (q 1 )   − sin (q 3 ) cos (q 2 ) cos (q 1 )   cos (q 1 ) sin (q 2 ) sin (q 1 ) sin (q 2 ) cos (q 2 )  

Conversely; for a given single-axis rotation matrix the corresponding  1 − R 2 ( 3,3 )   R ( 3,2 )  −1   , and Euler angles are: q 1 = tan −1  , q tan =  2  R ( 3,3 )   R ( 3,1 )   R ( 2,3 )  q 3 = tan −1    − R (1,3 ) 

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zyx

zxy

yzx

yxz

xzy

xyz

yaw

yaw

pitch

pitch

roll

roll

q1

pitch

roll

yaw

roll

yaw

pitch

q2

roll

pitch

roll

yaw

pitch

yaw

q3

=  xyz

=  xzy

=  yxz

=  yzx

=  zxy

=  zyx

0 0  cos (q 3 ) 0 − sin (q 3 )   cos (q 2 ) sin (q 2 ) 0  1    0 1 0 − sin q cos q 0 0 cos q sin ( ) ( ) ( ) (q1 )  2 2 1     sin (q 3 ) 0 cos (q 3 )   0 0 1  0 − sin (q 1 ) cos (q 1 )    0 0  cos (q 1 ) 0 − sin (q 1 )   cos (q 3 ) sin (q 3 ) 0  1    sin q cos q 0 0 cos q sin 1 0 − ( ) (q 2 )   0 ( ) ( ) 3 3 2     0 0 1  0 − sin (q 2 ) cos (q 2 )   sin (q 1 ) 0 cos (q 1 )   0 0   cos (q 2 ) sin (q 2 ) 0  cos (q 1 ) 0 − sin (q 1 )  1   0 cos q sin 1 0 ( 3) (q 3 )   − sin (q 2 ) cos (q 2 ) 0   0   0 − sin (q 3 ) cos (q 3 )   0 0 1   sin (q 1 ) 0 cos (q 1 )    0 0   cos (q 1 ) sin (q 1 ) 0  cos (q 3 ) 0 − sin (q 3 )  1     1 0  0  0 cos (q 2 ) sin (q 2 )   − sin (q 1 ) cos (q 1 ) 0   sin (q 3 ) 0 cos (q 3 )  0 − sin (q 2 ) cos (q 2 )   0 0 1     0 0  cos (q 2 ) 0 − sin (q 2 )   cos (q 1 ) sin (q 1 ) 0  1    0 cos q sin 1 0 ( 3) (q 3 )   0    − sin (q 1 ) cos (q 1 ) 0  0 − sin (q 3 ) cos (q 3 )   sin (q 2 ) 0 cos (q 2 )   0 0 1    

Rotation matrix

 cos ( ) sin ( ) 0     − sin ( ) cos ( ) 0   0 0 1 

Yaw: ni = ( 0,0,1 )

0 0   cos (q 3 ) sin ( 3 ) 0  cos (q 2 ) 0 − sin (q 2 )  1    − sin q cos q 0 0 1 0 0 cos q sin ( ) (q1 )  ( ) ( ) 3 3 1     0 0 1   sin (q 2 ) 0 cos (q 2 )  0 − sin (q 1 ) cos (q 1 )  

Matrices multiplication for successive rotations

cos ( ) 0 − sin ( )    1 0   0  sin ( ) 0 cos ( ) 

0 0  1  0 cos sin ( ) ( )   0 − sin ( ) cos ( ) 

Rotation matrixstep

Successive rotation-angles

Pitch: ni = ( 0,1,0 )

Roll: ni = (1,0,0 )

Tait-Bryan Angles

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unit vectorrotation axis

TABLE 15.3  Tait-Bryan angles and rotation matrices.

Rigid Body Rotation: Euler Angles, Quaternions, and Rotation Matrix • 117

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118 • Tensor Analysis for Engineers, 3E where,

 xyz

cos (q 1 ) sin (q 3 ) sin (q 3 ) sin (q 1 )  cos (q 2 ) cos (q 3 )    + cos (q 3 ) sin (q 2 ) sin (q 1 ) − cos (q 3 ) sin (q 2 ) cos (q 1 )    =  − cos (q 2 ) sin (q 3 ) − sin (q 3 ) sin (q 2 ) sin (q 1 ) cos (q 3 ) sin (q 1 )   + cos (q 3 ) cos (q 1 ) + sin (q 3 ) sin (q 2 ) cos (q 1 )     sin (q 2 ) − cos (q 2 ) sin (q 1 ) cos (q 2 ) cos (q 1 )  

Conversely; for a given single-axis rotation matrix the correspond R ( 3,1 )   − R ( 3,2 )  −1 , ing Euler angles are: q 1 = tan −1   , q 2 = tan   1 − R 2 ( 3,1 )   R ( 3,3 )   − R ( 2,1 )  q 3 = tan −1    R (1,1 ) 

 xzy

sin (q 3 ) sin (q 1 ) − sin (q 3 ) cos (q 1 ) cos (q 3 ) cos (q 2 )    + cos (q 3 ) sin (q 2 ) cos (q 1 ) + cos (q 3 ) sin (q 2 ) sin (q 1 )    =  − sin (q 2 ) cos (q 1 ) cos (q 2 ) sin (q 1 ) cos (q 2 )   − cos (q 3 ) sin (q 1 ) sin (q 3 ) sin (q 2 ) sin (q 1 )   sin (q 3 ) cos (q 2 )   + cos (q 1 ) sin (q 2 ) sin (q 3 ) + cos (q 3 ) cos (q 1 )  

Conversely; for a given single-axis rotation matrix the correspond − R ( 2,1 )   R ( 2,3 )  −1  , q = tan ing Euler angles are: q 1 = tan −1  ,  2  1 − R 2 ( 2,1 )   R ( 2,2 )   R ( 3,1 )  q 3 = tan −1    R (1,1 ) 

 yxz

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cos (q 3 ) cos (q 1 ) cos (q 2 ) sin (q 3 ) − cos (q 3 ) sin (q 1 )     + sin (q 3 ) sin (q 2 ) cos (q 1 )   + sin (q 3 ) sin (q 2 ) sin (q 1 )  = − sin (q 3 ) cos (q 1 ) cos (q 2 ) cos (q 3 ) sin (q 3 ) sin (q 1 )   + cos (q 3 ) sin (q 2 ) cos (q 1 )   + cos (q 3 ) sin (q 2 ) sin (q 1 )   sin (q 1 ) cos (q 2 ) cos (q 1 ) cos (q 2 ) − sin (q 2 )  

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Rigid Body Rotation: Euler Angles, Quaternions, and Rotation Matrix • 119

Conversely; for a given single-axis rotation matrix the corresponding  − R ( 3,2 )   R ( 3,1 )  −1  , and Euler angles are: q 1 = tan −1   , q 2 = tan   1 − R 2 ( 3,2 )   R ( 3,3 )   R (1,2 )  q 3 = tan −1    R ( 2,2 ) 

 yzx

cos (q 1 ) cos (q 2 ) sin (q 2 ) − sin (q 1 ) cos (q 2 )     sin (q 3 ) sin (q 1 ) cos (q 3 ) cos (q 2 ) sin (q 3 ) cos (q 1 )   =  − cos (q 3 ) cos (q 2 ) cos (q 1 ) + cos (q 3 ) sin (q 2 ) sin (q 1 )    cos (q 3 ) sin (q 1 ) − sin (q 3 ) cos (q 2 ) − sin (q 3 ) sin (q 2 ) sin (q 1 )     + cos (q 3 ) cos (q 1 )  + cos (q 1 ) sin (q 2 ) sin (q 3 ) 

Conversely; for a given single-axis rotation matrix the corresponding  R (1,2 )   − R (1,3 )  −1  , and Euler angles are: q 1 = tan −1   , q 2 = tan   1 − R 2 (1,2 )   R (1,1 )   − R ( 3,2 )  q 3 = tan −1    R ( 2,2 ) 

 zxy

cos (q 1 ) cos (q 3 ) cos (q 3 ) sin (q 1 ) − cos (q 2 ) sin (q 3 )      − sin (q 1 ) sin (q 2 ) sin (q 3 ) + sin (q 3 ) sin (q 2 ) cos (q 1 )   = − sin (q 1 ) cos (q 2 ) cos (q 1 ) cos (q 2 ) sin (q 2 )   sin (q 3 ) cos (q 1 ) − cos (q 3 ) sin (q 2 ) cos (q 1 ) cos (q 2 ) cos (q 3 )    + sin (q ) sin (q ) cos (q )  + sin (q 1 ) sin (q 3 ) 1 2 3  

Conversely; for a given single-axis rotation matrix the corresponding  R ( 2,3 )   − R ( 2,1 )  −1  , and Euler angles are: q 1 = tan −1   , q 2 = tan   1 − R 2 ( 2,3 )   R ( 2,2 )   −R (1,3 )  q 3 = tan −1    R ( 3,3 ) 

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120 • Tensor Analysis for Engineers, 3E

 zyx

cos (q 1 ) cos (q 2 ) sin (q 1 ) cos (q 2 ) − sin (q 2 )     − cos (q 3 ) sin (q 1 ) cos (q 3 ) cos (q 1 ) cos (q 2 ) sin (q 3 )    =  + cos (q 1 ) sin (q 2 ) sin (q 3 ) + sin (q 3 ) sin (q 2 ) sin (q 1 )   sin (q 3 ) sin (q 1 ) − sin (q 3 ) cos (q 1 ) cos (q 2 ) cos (q 3 )    + cos (q ) sin (q ) cos (q ) + sin (q ) sin (q ) cos (q )  3 2 1 1 2 3  

Conversely; for a given single-axis rotation matrix the corresponding  − R (1,3 )   − R (1,2 )  −1  , and Euler angles are: q 1 = tan −1   , q 2 = tan   1 − R 2 (1,3 )   R (1,1 )   R ( 2,3 )  q 3 = tan −1    R ( 3,3 )  15.3.1  Example: Euler Angles- Yaw-Pitch-Yaw, Numerical example Repeat the example given in Section 15.2.1 using relation  zyz, given in Table 15.2. Solution: using Table 15.2, we have q 1 = 30 for first yaw rotation, q 2 = 45 for pitch rotation, and q 1 = 15 for third yaw rotation. Substituting in relation for  zyz gives

 zyz

− sin (15 ) sin ( 30 ) sin (15 ) cos ( 30 ) − cos (15 ) sin ( 45 )      + cos (15 ) cos ( 45 ) cos ( 30 ) + sin ( 30 ) cos ( 45 ) cos (15 )  = − sin ( 30 ) cos (15 ) − sin (15 ) cos ( 45 ) sin ( 30 ) sin (15 ) sin ( 45 )    + cos (15 ) cos ( 30 )  − sin (15 ) cos ( 30 ) cos ( 45 )    cos 30 sin 45 sin 30 sin 45 cos 45 ( ) ( ) ( ) ( ) ( )    0.4621 0.5657 −0.6830      −0.6415 0.7450 0.1830  . These results are identical with those  0.6124 0.3536 0.7071  from the example given in Section 15.2.1 Therefore,  x′     y′ =    z′ 

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 0.4621 0.5657 −0.6830   x      , where ( x′ , y′, z′) are coordinates  −0.6415 0.7450 0.1830   y    0.6124 0.3536 0.7071   z 

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Rigid Body Rotation: Euler Angles, Quaternions, and Rotation Matrix • 121

of a fixed given point in space with initial coordinates ( x, y, z). In other words, the initially body-attached axes ( x, y, z) orient themselves to the ( x′ , y′, z′) after the airplane goes through the given successive yaw-pitch-yaw rotations.

15.4  GIMBAL LOCK-EULER ANGLES LIMITATION For a wide range of applications, the Euler angles method provide us a useful tool for calculating a rigid body orientation. However, for certain values of rotation angles the process breaks down and ran into a mathematical ”singularity” situation. This is usually referred to gimbal lock. This happens when the rigid body loses one of its three rotational degrees of freedom and rotation of two axes coincide when the rotation about the third remaining p axis is some integer multiple of . To see the mathematical explanation 2 and implication of gimbal lock, we use one of the cases from Table 15.3, for example roll-pitch-yaw with final rotation matrix  xyz, repeated here for convenience.

 xyz

cos (q 1 ) sin (q 3 ) sin (q 3 ) sin (q 1 )  cos (q 2 ) cos (q 3 )    + cos (q 3 ) sin (q 2 ) sin (q 1 ) − cos (q 3 ) sin (q 2 ) cos (q 1 )    =  − cos (q 2 ) sin (q 3 ) − sin (q 3 ) sin (q 2 ) sin (q 1 ) cos (q 3 ) sin (q 1 )   + cos (q 3 ) cos (q 1 ) + sin (q 3 ) sin (q 2 ) cos (q 1 )     − cos (q 2 ) sin (q 1 ) sin (q 2 ) cos (q 2 ) cos (q 1 )  

Let the pitch angle be q 2 = metric simplifications,  xyz

p . Therefore, we receive, after some trigono2

0 sin (q 1 + q 3 ) −cos (q 1 + q 3 )    =0 cos (q 1 + q 3 ) sin (q 1 + q 3 )  . Writing the transformed coordi1  0 0

 x′  0 sin ( b ) −cos ( b )   x       nates, we receive  y′ = 0 cos ( b ) sin ( b )   y . Where b= q 1 + q 3.     z  0 0  z′  1

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122 • Tensor Analysis for Engineers, 3E Writing the system of equations = gives x′ y sin ( b ) − zcos= ( b ), y′ ycos ( b ) + z sin ( b ) = y′ ycos ( b ) + z sin ( b ), and z′ = x . Therefore, the rotated z-axis and x-axis are coincided, or so-called locked. Obviously, the roll angle q 1 and yaw angle q 3 can take on several different values but their changes are indistinguishable since angle b changes value if either of q 1 or q 3 changes. When pitch angle is equal p p to (or in general, some odd integer multiple of ) then the rigid body 2 2 degrees of freedom is reduced from three to two, since z′ = x and yaw and roll rotations provide the same result in terms of the rigid body orientation. Similarly, other rotation matrices as listed in Table 15.2 and Table 15.3 can be used to show their corresponding gimbal lock orientation, [25]. It seems useful to follow the successive rotations step-by-step as well, for helping with understanding the gimbal lock. The first rotation is roll by 0 0   x  x1  1      angle q 1 which results in orientation  y1  = 0 cos ( q 1 ) sin (q 1 )   y.      z1  0 − sin (q 1 ) cos (q 1 )   z  Obviously, these leaves x-axis intact, x1 = x and rotates y − z plane about x-axis. The second rotation is pitch by angle q 2 which results in orientation  x2  cos (q 2 ) 0 − sin (q 2 )   x1        . Obviously, these leaves y -axis intact, 1 0 1  y2  =  0   y1   z   sin (q ) 0 cos (q )   z  2 2  2   1 p y2 = y1 and rotates x1 − z1 plane about y1-axis. Now for the value of q 2 = 2  x2  0 0 −1  x1       we receive  y2  = 0 1 0   y1 , or x2 = − z1, y2 = y1, and z= x= x. 2 1  z  1 0 0   z   2   1 The latter relation shows that original x-axis is coinciding with z2 axis, or so-called gimbal-locked. The last rotation is yaw by angle q 3 which results  x3   cos (q 3 ) sin ( q 3 ) 0   x2      in orientation  y3  =  − sin (q 3 ) cos (q 3 ) 0   y2  . Obviously, these leaves   0 0 1   z2   z3   z= x= x and rotates x2 − y2 plane about z2-axis. z3-axis intact, z= 3 2 1

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Rigid Body Rotation: Euler Angles, Quaternions, and Rotation Matrix • 123

Performing the matrix multiplication using all three matrices, produces 0 0   x  x′   x3   cos (q 3 ) sin ( q 3 ) 0  0 0 −1 1          ′  y  =  y3  =  − sin (q 3 ) cos (q 3 ) 0  0 1 0  0 cos ( q 1 ) sin (q 1 )   y ,  z′   z   0 0 1  1 0 0  0 − sin (q 1 ) cos (q 1 )   z     3   x′  0 sin (q 1 + q 3 ) −cos (q 1 + q 3 )   x       or  y′ =0 cos (q 1 + q 3 ) sin (q 1 + q 3 )   y, or again showing the loss of  z′  1   z  0 0    one degrees of freedom out of three as z′ = x , or gimbal lock. To recover the lost degree of freedom due to gimbal lock situation we can perform manual maneuver, if/when possible to rotate the rigid body as smoothly as possible out of the gimbal lock position. This approach is not practical, and the control system should be programmed such that close to values of rotation angle that creates gimbal lock a warning sent out and users become aware of it. As shown in the animation video, cited above/below, we can use other more robust methods for removing gimbal lock. This method makes use of quaternions. We discuss the mathematics of quaternions and their applications in rigid body rotation and matrix transformation, in the following sections. Readers interested in discussion about gimbal lock can consult with references cited [14], [26], [18]. Readers may also find it helpful to consult with the following lecture series, specifically those ones related to rigid body kinematics [27]. Also using Wolfram Alpha, readers can visualize a given quaternion, along with some related computations. An example for animation of a gimbal lock position, which could be helpful to visualize this phenomenon, can be found at references [28] and [29].

15.5 QUATERNIONS-APPLICATIONS FOR RIGID BODY ROTATION A quaternion is defined as a 4D complex number. Mathematically, it is represented with a combination of a scalar and a vector-like complex number. Its development started when William R. Hamilton was visioning and working on extending the complex numbers to more than 2D space, a real and

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124 • Tensor Analysis for Engineers, 3E a complex numbers dimension. He first struggled adding one dimension to the complex number (i.e., defining it in 3D space) and ran into difficulties for performing basic algebraic operations, like multiplications, etc. For example, the product of two imaginary numbers ij is not defined in 3D space when having one real and two imaginary dimensions. Then he realized that adding an additional dimension to the complex numbers and defining it in 3D space would solve the problem, hence Quaternions are defined as complex numbers in 4D space, [30]. This newly discovered mathematical quantity, the quaternions, found to be very useful in several fields of science/ physics and engineering such as kinematics of rigid body orientation, robotics, aerospace, computer graphics/animation and gaming software, relativity (General and Special), quantum mechanics. In this section, we discuss their definition, algebraic operations, and their applications for rigid body rotation [18], [21]. 15.5.1 Definition-Quaternion A quaternion can be defined as the combination of a scalar and a vector, as  q = ( q0 , q ) . The vector-like part is in principle defined as a 3D complex number when using their multiplication relation defined as given by Equation 15.4. It is well-known that in 1843 Hamilton, excited with his spark discovery, engraved this expression as a message on Broome Bridge in Dublin, Ireland [31], [32].

i2 = j2 = k2 = ijk = −1 

15.4

Therefore, we can write a quaternion q as

q =q0 + iq1 + jq2 + kq3 ≡ ( q0 , q1 , q2 , q3 ) 

15.5

where, q0 is a real number and remaining terms are complex numbers. For example, 2 − i + 3 j + 2 k. Using Equation 15.5, we can write complex num= 0,1,0,0 ) , j ( 0,0,1,0 = bers i, j, k as quaternions, or i (= ) , and k ( 0,0,0,1). Note that the similarity between ( i, j, k), here defined as complex numbers      with those of unit vectors ( e1 , e2 , e3 ) ≡ i, j, k that justifies the vector-like representation of a quaternion and hence to some extent simplifies the mathematical expressions resulted from algebraic manipulation of quaternions. The relations for the product of these complex quantities follows the cross-product of unit vectors, except for those of identical unit vectors

(

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)

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Rigid Body Rotation: Euler Angles, Quaternions, and Rotation Matrix • 125

    1, etc.). Please note i2 = ii is an 0 and i ⋅ i = (e.g., i2 = −1, whereas i × i = expression representing arithmetic multiplication of a complex number i by itself, hence manipulation of quaternions is relatively easier than those of vectors.

All possible combinations of complex numbers products are listed in Table 15.4, with clockwise multiplication considered to be positive. From these relations we can conclude that these relations are not commutative, for − ij, etc. example, ij ≠ ji = TABLE 15.4  Hamilton relations, quaternions multiplication rule.

k

i

i = ( 0,1,0,0 )

j = ( 0,0,1,0 )

k = ( 0,0,0,1 )

i

i2 = −1

ij = k

ik = − j

j

ji = − k

j2 = −1

jk = i

k

ki = j

kj = − i

k2 = −1

+ j

In terms of geometrical visualization, it is difficult to vision a 4D space since we are accustomed to visioning 3D space intuitively. It may help to imagine the surface of a hypersphere as representation of a surface defined in 4D space, like a 2D surface defined in a 3D space [21]. Readers may find it useful to watch the video clip cited in reference [33], demonstrating some animations of the geometry of a quaternion in terms of hyperspheres. 15.5.1.1  Example: Derivation of Hamilton Quaternion Relations Using Equation 15.4, derive the identities given in Table 15.4. Recall that quaternions are not commutative. Solution: We start with ijk = −1 and multiply both sides by i to receive  i2 jk = − i . −1 Hence, jk = i. Similarly, kj = − i. Now multiply both sides by j to receive ji jk = − j. Hence, − i j2 k = − j or   ik = − j. Similarly, ki = j. − ij −1

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126 • Tensor Analysis for Engineers, 3E Now multiply both sides by k to receive ki k = k or  jk = − k. Hence, i kj  − ik − ij  k2 = k. Hence, ij = k. Similarly, ji = − k.

− jk

−1

15.5.2  Quaternions-Basic Algebraic Operations and Properties We consider two quaternions, q =q0 + iq1 + jq2 + kq3 and

p = p0 + ip1 + jp2 + kp3

p = p0 + ip1 + jp2 + kp3 . The summation/subtraction operation on these quantities q ± p= ( q0 ± p0 ) + i(q1 ± p1 ) + j(q2 ± p2 ) + k ( q3 ± p3 is straight forward and produces p= ( q0 ± p0 ) + i(q1 ± p1 ) + j(q2 ± p2 ) + k ( q3 ± p3 ) . Obviously, for more than two quaternions it can be extended to include as many as involved in the operation. However, the multiplication operation is not as straight forward. As a matter of fact, it was this operation that held Hamilton back for a while before he realized that a 4D space is required to properly define quaternions. The multiplication operation is as follow: qp = ( q0 + iq1 + jq2 + kq3 )( p0 + ip1 + jp2 + kp3 ) . Performing the term-by-term multiplications and using the relations given in Table 15.4, after some rearranging, we receive qp = q0 p0 − ( q1 p1 + q2 p2 + q3 p3 ) + q0 ( ip1 + jp2 + kp3 ) + p0 ( iq1 + jq2 + kq3 )



+ i ( q2 p3 − q3 p2 ) + j ( q3 p1 − q1 p3 ) + k ( q1 p2 − q2 p1 ) 15.6

This is a long expression and some simplifications is justified. We can make  use of properties of the vector-like part of the quaternion, q = iq1 + jq2 + kq3  and p = ip1 + jp2 + kp3, and their dot and cross products when the complex   numbers i, j, k are interpreted as unit vectors. Or q ⋅ p= q1 p1 + q2 p2 + q3 p3   × p i ( q2 p3 − q3 p2 ) + j ( q3 p1 − q1 p3 ) + k ( q1 p2 − q2 p1 ) . Substituting for and q= these expressions back into multiplication result (Equation 15.6), we receive       qp = ( q0 p0 − q ⋅ p ) + ( q0 p + p0 q + q × p ) 15.7       scalar

vector − complex number

    Equation 15.7 clearly demonstrates that qp ≠ pq, since q × p ≠ p × q.

The author’s caution is to note the dual interpretations of complex numbers i, j, k as unit vectors when the complex part of a quaternion   is treated as   2 0 for vec1 or i × i = a vector (for example i = −1 but we treat as i ⋅ i = tor operation). Readers should note that simply the results of the product operations are the same, not necessarily the term-by-term operation, hence Equation 15.7 provides us with the same answer as that of Equation 15.6.

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Rigid Body Rotation: Euler Angles, Quaternions, and Rotation Matrix • 127

However, Equation 15.7 is presented in more compact memorable form of the quaternions’ product expression.  It is useful to define conjugate of the quaternion, q∗ = ( q0 − q ) = q0 − ( iq1 + jq2 +

 q∗ = ( q0 − q ) = q0 − ( iq1 + jq2 + kq3 ). We can use Equation 15.7, to show that 2 qq∗ = q∗ q = q02 + q = q02 + q12 + q22 + q32 . This quantity is equal to the square of the norm or magnitude of the quaternion q , or

q=

qq∗ =

q02 + q12 + q22 + q32 = q∗ 15.8

q =Q0 + iQ1 + jQ2 + kQ3 is defined as a quaternion q with magnitude/norm equal to unity, with having QQ∗ = 1. Therefore, for a given unit quaternion its inverse and conjugate are equal, Q −1 = Q∗ . Also, 2 since q∗ q = q , see Equation 15.8, we can define the inverse of the quaterq∗ nion as q−1 = 2 . Therefore, the division operation of two quaternions can q be performed. This, however, has two forms since the quaternions are noncommutative. For example, the ratio q (assuming p ≠ 0) can be written as p −1 −1 qp or p q. Users should define which quotient is desired for their related calculations. A unit quaternion, Q =

15.5.2.1  Quaternions Algebraic Operations-Numerical Example Given q = 3 + i − 2 j + k and p = 2 − i + 2 j + 3 k , compute q + p, qp, their corresponding conjugate, norm, inverse, and unit quaternion. Solution: For the prodq + p =( 3 + i − 2 j + k ) + ( 2 − i + 2 j + 3 k ) =5 + 4 k . uct we calculate the dot and cross products of the correspond  ing vectors, or and q ⋅ p =(1, −2,1 ) ⋅ ( −1,2,3 ) =−1 − 4 + 3 =−2   q × p =emnl qn pl =−8 i − 4 j. Therefore, using Equation 15.7, we have

qp =[ 6 + 2 ] + −3 i + 6 j + 9 k + 2 i − 4 j + 2 k − 8 i − 4 j = 8 − 9 i − 2 j + 11k . The conjugates are q∗ = 3 − ( i − 2 j + k ) and p∗ = 2 − ( − i + 2 j + 3 k ) . The p = 4 + 1 + 4 + 9= 3 2. norm and q = 9 + 1 + 4 + 1 = 15

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128 • Tensor Analysis for Engineers, 3E

The

inverse

q∗ 3 − i + 2 j − k 1 1 2 1 = − i+ q−1 = 2 = j− k 15 5 15 15 15 q

and

p∗ 2 + i − 2 j − 3 k 1 1 1 1 p−1 = 2 = = + i − j − k. The unit quaternions are, 18 9 18 9 6 p 3+ i−2 j+k 2 − i + 2 j + 3k 15 15 2 15 15 and 2 2 2 Q= i− j+ k = + = − P= i+ j+ 5 15 15 15 3 6 3 15 3 2 2 − i + 2 j + 3k 2 2 2 2 P= = − i+ j+ k. 3 6 3 2 3 2 Having the algebraic rules defined for quaternions, in the following sections we present their application for coordinates system and rigid body rotation. Since quaternions are basically an extension to the 2D complex numbers, first we discuss the relation between 2D complex numbers and rotation matrix followed by similar relations for 4D quaternions. 15.5.3  Complex Numbers and Rotation Matrix Complex numbers can be used to represent rotation of a rigid body when Euler formula is used. Euler formula relates an exponential complex number/variable to periodic sinusoidal functions [25], or

= e ia cos a + i sin a 15.9

where i= −1 . For example, in a 2D space, we can define a complex variable z= z ( x, y )= x + iy representing a vector drawn from coordinates’ ori y gin to the point z on the x − y plane and making an angle j = tan −1   ,  x phase/argument, with the positive x-axis and having the magnitude z = zz∗ = ( x + iy )( x − iy ) . Writing down the components, we get x = z cos j and y = z sin j . Or z = z ( cosj + i sinj ) = z e ij . Now, we operiq = z e ij z e i(j +q ) . Note that the magniate the unit vector e iq on z, to get z′ e= tude of the original vector is preserved, but its phase angle changes to (j + q ) . Expanding this expression, using Equation 15.9, we receive the components x cosq − y sin q + i ( x sin q + y cosq ) , of transformed vector z′ = ( x + iy) eiq = ′  x  cosq − sin q   x  as (omitting i)   =     . This is identical to active rotation  y′  sin q cosq   y operation discussed in Section 15.1. Similarly, operating e− iq on z gives the equivalent of passive rotation operation.

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Rigid Body Rotation: Euler Angles, Quaternions, and Rotation Matrix • 129

15.5.4  Quaternions and Rotation To extend the application of complex numbers to 3D rotations, we make use of the process/method mentioned in the previous section and include quaternions in the discussion. First, we note that an arbitrary vector,  ˆ ˆ ˆ V = iV1 + jV2 + kV3 = ( V1 , V2 , V3 ) can be considered as a quaternion whose real part is zero, i.e., pure quaternions. Similarly, a real quaternion is a quaternion whose vector part is null with non-zero real part. Now, we consider a unit 2  quaternion Q = Q0 + Q. Therefore, Q = Q02 + Q = Q02 + Q12 + Q22 + Q32 = 1 which implies that we can define an angle b such that

 Q0 = cos b 15.10    Q = sin b   Q  Q n = We also define a unit vector as = using the vector  Q Q12 + Q22 + Q32  part (i.e., vector Q ) of the unit quaternion Q. Therefore, we can write the  unit quaternion as = Q cos b + n sin b . Now, we can write as an extension to Euler formula, see Equation 15.9, the so-called De Moivre’s formula for quaternion Q, [26]



  nb = Q e= cos b + n sin b 15.11

 Note that n is a 3D imaginary number treated as a vector-like quantity calculated from the imaginary part of the unit quaternion. Using Equation 15.10,  Q12 + Q22 + Q32    and since n = ( n1 , n2 , n3 ) is defined we can write b = tan −1    Q0   2 1. as a unit vector we have n1 + n22 + n32 =

Recall that the rotation matrix/operation preserves the magnitude of the ­vector which goes under rotation. In other words, magnitude of a vector under rotation is invariant (see Sections 12.2 and 15.5.3). Therefore, for using quaternions in rotation operation we look for an operator which is a combination of relevant quaternion and maintains the magnitude of the vector on which it operates. The quaternion operation on a vector is done in a two-step process, using a unit quaternion and its inverse/conjugate to make sure that the result is a vector with preserved magnitude. For this purpose,  ∗  −1 we use the unit quaternion Q and form the combination QVQ = QVQ to   transform vector V to V ′ by rotation, or

(

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130 • Tensor Analysis for Engineers, 3E



  ∗    V ′ QVQ enb Ve− nb 15.12 = =

As mentioned, the reason that we use unit quaternion and its inverse/conjugate  is simply because the combination QV , as much as it may intuitively seem the correct combination, it doesn’t result in a new vector and nor maintaining the  magnitude of V , under rotation operation. This can be seen using Equation 15.7   Q0 , Q1 , Q2 , Q3 ) ( 0, V1 , V2 , V3 ) and writing V as= a pure quaternion. Or QV (=      −Q ⋅ V + Q0 V + Q × V , which is clearly a quaternion quantity and not a vector   due to having a scalar part (i.e. Q ⋅ V ).    Now, after expanding Equation 15.12, using Equation 15.11, we have QVQ∗ =( cos b + n sin b ) V ( co        = V cos b + nV sin b ( cos b − n sin b ) = QVQ∗ =( cos b + n sin b ) V ( cos b − n sin b ) =       V cos2 b − nVn sin 2 b + nV − Vn sin b cos b . But using vector identities, we            can write nVn = V − 2 n ⋅ V n and nV − Vn =2 n ×V . After substituting we

( ) ( )

(

)

receive, in terms of double angle 2b ,  ∗       QVQ = V cos2 b + (1 − cos2 b ) n ⋅ V n + n × V sin 2 b

(

) (

)

Therefore, we can write Equation 15.12 in index notation as (note that in Equation 15.13 subscripts i, j, k are simply indices, ranging from 1 to 3 and should not be confused with the imaginary numbers)

= Vi′ Vi cos2 b + (1 − cos2 b ) ni n j V j + eijk n j Vk sin 2 b 15.13 Writing Equation 15.13 in matrix form, let q = 2 b , produces

V1′   cos q + n1 n1 (1 − cos q )    V ′  = n n (1 − cos q ) + n3 sin q  2   1 2 V3′   n1 n3 (1 − cos q ) − n2 sin q

n1 n2 (1 − cos q ) − n3 sin q cos q + n2 n2 (1 − cos q ) n2 n3 (1 − cos q ) + n1 sin q

n1 n3 (1 − cos q ) + n2 sin q  V1    n2 n3 (1 − cos q ) − n1 sin q  V2  cos q + n3 n3 (1 − cos q )  V3 

15.14 This is identical to using Rodrigues equation when rotation angle is measured in the opposite direction, or the transpose of rotation matrix given in Equation 12.7. In other words, the operation performed  according to Equations 15.12 or 15.14 gives active rotation of vector V through angle  q = 2 b about unit vector n. Therefore, the quaternions should be written as half-angle of the rotation angle (i.e., b = q /2), Or

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Rigid Body Rotation: Euler Angles, Quaternions, and Rotation Matrix • 131



  Q q rotation (2= = b )quaternion 2 tan −1    Q0     Q n = 2 Q1 + Q22 + Q32 

    15.15

Readers may recall that for active rotation we keep the coordinate system fixed and rotate the vector, see Section 15.1. From the above-mentioned derivation, we can conclude the following theorem (note that Q −1 = Q∗):   = Q0 + Q, vector For any given vector V and unit quaternion Q   QVQ −1 is equivalent to the active rotation of the vector V through  Q    about an axis parallel to Q  or equivaan angle q = 2 tan −1   Q0        Q lently the unit vector n =   Q 



Similarly, it can be shown that the combination     Q∗ VQ = e− nb Venb 15.16  is equivalent to a passive rotation of vector V . Two points worth mentioning, about quaternions and rotations. 1) It takes two quaternions, the unit quaternion itself and its inverse/conjugate, to perform a rotation for a rigid body. Each quaternion carries half of the total q angle of rotation, i.e., b = . 2) Further transformation follows the same 2  rule. For example, the vector Q∗ VQ can be transformed again using unit quaternion P representing a second rotation, or



  ∗  P ∗ Q∗ VQ P = ( QP ) V ( QP ) = W ∗ VW 15.17       nd

(

)

st

1 rotation

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132 • Tensor Analysis for Engineers, 3E As seen from Equation 15.17, we can combine the related two quaternions as their product W = QP and perform the operation on the desired vector as a single rotation. The first rotation corresponds to Q and the second one to P. Similarly, for an active rotation we can write the following relation for two successive rotations,    ∗ P QVQ∗ P ∗ = ( PQ ) V ( PQ ) = V  ∗      15.18 nd

(



)

st

1 rotation

2

rotation

Example: Show that following theorem is correct.   = Q0 + Q , comFor any given vector V and unit quaternion Q   ponents of vector Q −1 VQ are equivalent to those of vector V in the rotated coordinates (i.e., passive rotation) by an angle  Q    about an axis parallel to the vector Q  or equivq = 2 tan−1   Q0       Q  alently unit vector n =  Q  Solution:

Note that quaternions should be written as half of the total rotation angle,    q     or b = , Q−1 VQ =( cos b − n sin b ) V ( cos b + n sin b ) == ( V cos b − nV sin b ) ( cos b + n sin b ) = 2          = V cos2 b − nVn sin 2 b − nV − Vn sin b cos b . However, sin b ) ( cos b + n sin b ) = ( V cos b − nV            using vector identities, we can write nVn = V − 2 n ⋅ V n and nV − Vn = 2 n × V        ∗ nV − Vn = 2 n × V . Substituting, we obtain, in terms of double angle 2b , Q= VQ V cos2 b + (1 − cos2 b        ∗ Q= VQ V cos2 b + (1 − cos2 b ) n ⋅ V n − n × V sin 2 b . Or in index ­notation form

(

(

) (

(

)

)

)

we have (note that in Equations 15.13 and 15.19 subscripts i, j, k are simply indices range from 1 to 3 and should not be confused with the ­imaginary  numbers)

= Vx′i Vi cos2 b + (1 − cos2 b ) ni n j V j − eijk n j Vk sin 2 b 15.19 Writing Equation 15.19 in matrix form, let q = 2 b , produces

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Rigid Body Rotation: Euler Angles, Quaternions, and Rotation Matrix • 133

Vx′   cos q + n1 n1 (1 − cos q )    Vy′  =  n1 n2 (1 − cos q ) − n3 sin q    Vz′   n1 n3 (1 − cos q ) + n2 sin q

n1 n2 (1 − cos q ) + n3 sin q cos q + n2 n2 (1 − cos q ) n2 n3 (1 − cos q ) − n1 sin q

n1 n3 (1 − cos q ) − n2 sin q  V1    n2 n3 (1 − cos q ) + n1 sin q  V2  cos q + n3 n3 (1 − cos q )  V3 

15.20 This is identical to the rotation matrix given in Equation 12.7. In other   words, the operation given by Q −1 VQ gives the components of vector V  when coordinates are rotated through angle q = 2 b about unit vector n , see Equation 15.15. Readers may recall that for passive rotation we keep the vector fixed and rotate the coordinate system, see Section 15.1. Example: Rotation using quaternions-numerical example   2p Consider the rotation of vector V = i= (1,0,0 ) by an angle q = about an 3  axis in the direction r = (1,1,1 ). Find the transformed vector assuming active and passive rotations using following methods: a. Rodrigues’ formula b. quaternions Solution:

 r  1 1 1  , , =  . r  3 3 3 The transformation matrix is, using Equations 15.13 or 12.7, for active

 a. The unit vector related to the rotation axis is = n

transformation (vector rotates and coordinates are fixed) gives (note that 2p

when 12.7) q2p= − 1  for active 2p  rotation, 1 2p plugged 1 2in p Equation 1 2p  1 2p   sin sin   1 − cos −  1 − cos +  cos 3 + 33 1 − cos 3  3 3 3 3 3 3 3 3        1  V1′   2p  1 2p 2p 1  2p  1 2p  1 2p      1  sin cos sin +  1 − cos 0  V2′  =   1 − cos +   1 − cos − 3  3 3 3 3  3 3  3    3 3 V ′   3  0  3  1 2p  1 2p 1  2p  1 2p 2p 1  2p      +  1 − cos sin cos  1 − cos +     1 − cos 3 2p−  sin 2p  3 2p  − 13 sin 3 1  13 − cos  2p3 1  13 − cos 2p 3+ 13 sin 2p  3     3+ 1  1 − cos cos       3 3 3  3 3  3 3 3  3  3 3  1  V1′   2p  1 2p 2p 1  2p  1 2p  1 2p      1  sin cos +  1 − cos sin V2′  =   1 − cos 0  +   1 − cos − 3  3 3 3 3  3 3  3    3 3 V ′   3  0  3  1 2p  1 2p 1  2p  1 2p 2p 1  2p      1 cos sin 1 cos sin cos 1 cos − − − + + −        3  3 3 3  3 3 3 3   3 3 3 

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134 • Tensor Analysis for Engineers, 3E Or ìV1¢ ü é0 0 1 ù ì1 ü ì0 ü  ï ï ê úï ï ï ï íV2¢ ý = ê1 0 0 ú í0 ý = í1 ý . Hence active rotation of vector i gives vector ï V ¢ ï ê 0 1 0 ú ï0 ï ï0 ï î 3þ ë ûî þ î þ  j. Similarly, the passive rotation (coordinates rotate and vector is fixed) gives 2p (note 2that for passive rotation, when plugged in Equation 12.7) 2p ù p 1qæ= 3 2p ö 1æ 2p ö 1 2p 1 æ 2p ö 1 é cos 1 cos sin 1 cos sin + ç 1 - cos + ÷ ç ÷ ç ÷ .

ê 3 3è 3 ø 3è 3 ø 3 3è 3 ø 3 ú 3 3 ú ìVx¢ ü ê 2p ö 1 2p 2p 1 æ 2p ö 1æ 2p ö 1 2p ú ï ï ê1 æ = sin cos sin + ç 1 - cos íVy¢ ý = ê ç 1 - cos ÷÷ ç 1 - cos ÷+ 3 ø 3 3 3è 3 ø 3è 3 ø 3 úú 3 3 ï ï ê3 è îVz¢ þ ê 1 æ 2p ö 1 2p 1 æ 2p ö 1 2p 2p 1 æ 2p ö ú 1 - cos sin 1 - cos + 1 - cos sin cos + 2pëê 3 çè1 æ 2÷øp ö 3 1 æ 3 3 2çè p ö 1 3 ÷ø 2p 3 1 æ 3 2p ö 3 1 3 çè 2p ù 3 ø÷ ûú é 3 sin sin ç 1 - cos ÷+ ç 1 - cos ÷ê cos 3 + 3 ç 1 - cos 3 ÷ 3è 3 ø 3 3è 3 ø 3 ú 3 3 è ø ú ìVx¢ ü ê 2p ö 1 2p 2p 1 æ 2p ö 1æ 2p ö 1 2p ú ï ï ê1 æ sin cos sin = + ç 1 - cos íVy¢ ý = ê ç 1 - cos ÷÷ ç 1 - cos ÷+ 3 ø 3 3 3è 3 ø 3è 3 ø 3 úú 3 3 ï ï ê3 è îVz¢ þ ê 1 æ 2p ö 1 2p 1 æ 2p ö 1 2p 2p 1 æ 2p ö ú sin sin cos + ç 1 - cos ç 1 - cos ÷÷ ê 3 ç 1 - cos 3 ÷ + 3 3è 3 ø 3 3 3è 3 ø úû 3 3 ø ë è

é 0 1 0 ù ì1 ü ì 0 ü   ê ú ï ï ï ï . Hence passive rotation of vector i gives vector k . 0 0 1 0 = 0 ê úí ý í ý êë1 0 0 úû îï0 þï îï1 þï

b. When quaternions are used for rotation note that half-angle pö æ should be used ç b = q / 2 = ÷. The unit quaternion reads, 3ø è p  p 1 æ i j k ö 3 1 i j k using Equation 15.11, Q = cos + n sin = + ç + + = + + + ÷ 3 3 2 è 3 2 2 2 2 3 3ø 2  i j k j k ö 3 1 i j k . Hence, Q = 1 and Q æ i = + + . For active rota0 + + = + + + ç ÷ 2 2 2 2 2 2 2 2 3 3ø 2 è 3  * æ1 i j kö æ1 i j kö tion, we have QVQ = ç + + + ÷ ( 0,1,0,0 ) ç - - - ÷ = è2 2 2 2ø è2 2 2 2ø æ 1 i k j öæ 1 i j k ö . Similarly, for passive rotation ç - 2 + 2 - 2 + 2 ÷ç 2 - 2 - 2 - 2 ÷ = j è øè ø  j i k 1 æ ö ˆ æ 1 i j k ö we have Q* VQ = ç - - - ÷ i ç + + + ÷ = k . Readers can è2 2 2 2ø è2 2 2 2ø instead use Equations 15.14 and 15.20, respectively.

()

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2

Rigid Body Rotation: Euler Angles, Quaternions, and Rotation Matrix • 135

15.6  FROM A GIVEN QUATERNION TO ROTATION MATRIX In this section, we would like to derive the rotation matrix from a given quaternion related to a rigid body rotation. For a given unit quaternion, we can use Equations 15.14 in conjunction with Equation 15.15 for active rotation and Equation 15.20 for passive rotation for calculating the equivalent rotation matrices. However, equivalently we can directly calculate the rotation matrix by implementing quaternions algebraic product rules on QVQ∗, for  example, for a given vector V = iV1 + jV2 + kV3.  For a given unit quaternion, we have Q = Q0 + Q = Q0 + iQ1 + jQ2 + kQ3  along with Q = Q02 + Q12 + Q22 + Q32 = 1. Therefore, the combination QVQ∗ can be written as follow.  QVQ∗ = ( Q0 + iQ1 + jQ2 + kQ3 ) ( iV1 + jV2 + kV3 )( Q0 − iQ1 − jQ2 − kQ3 ) .  Performing the multiplication on the first two brackets, produces QVQ∗ = −  ( Q1 V1 + Q2 V2 + Q3 V3 ) + i ( Q0 V1 + Q2 V3 − Q3 V2 ) + j ( Q0 V2 − Q1 V3 + Q3 V1 ) + k ( Q0 V3 + Q1 V2 −

) + j ( Q0 V2 − Q1 V3 + Q3 V1 ) + k ( Q0 V3 + Q1 V2 − Q2 V1 ) ( Q0 − iQ1 −

jQ2 − kQ3 ) . Performing the multiplica-

tion on the last bracket produces  ∗ QVQ = i  V1 ( Q02 + Q12 − Q22 − Q32 ) + 2 V2 ( Q1Q2 − Q0 Q3 ) + 2 V3 ( Q0 Q2 + Q1Q3 )  + j 2 V1 ( Q0 Q3 + Q1Q2 ) + V2 ( Q02 − Q12 + Q22 − Q32 ) + 2 V3 ( Q2 Q3 − Q0 Q1 )  +

k 2 V1 ( Q1Q3 − Q0 Q2 ) + 2 V2 ( Q0 Q1 + Q2 Q3 ) + V3 ( Q02 − Q12 − Q22 + Q32 ) . Now

1 we can substitute for −Q22 − Q32 = Q02 + Q12 − 1, using Q02 + Q12 + Q22 + Q32 =

−Q12 − Q32 = Q02 + Q22 − 1, and −Q12 − Q22 = Q02 + Q32 − 1, to obtain the rotation matrix as: (1st row is the coefficients of i, 2nd row those of j, and 3rd row those of k) Q02 + Q12 − 0.5 Q1Q2 − Q0 Q3 Q0 Q2 + Q1Q3  = [ Ractive ] 2  Q0Q3 + Q1Q2 Q02 + Q22 − 0.5 Q2Q3 − Q0Q1  15.21  Q1Q3 − Q0 Q2 Q0 Q1 + Q2 Q3 Q02 + Q32 − 0.5   



Or  V1′   V1      V2′  = [ Ractive ] V2  V ′  V   3  3

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136 • Tensor Analysis for Engineers, 3E  Similarly, we can calculate the passive rotation matrix using Q∗ VQ, to receive

Q02 + Q12 − 0.5 Q0 Q3 + Q1Q2 Q1Q3 − Q0 Q2   Rpassive  2  Q1Q2 − Q0 Q3 Q02 + Q22 − 0.5 Q0 Q1 + Q2 Q3  15.22 =    Q0 Q2 + Q1Q3 Q2 Q3 − Q0 Q1 Q02 + Q32 − 0.5   



Or  Vx′   V1       Vy′  =  Rpassive  V2  V  V   z′   3 Reader may note that  Rpassive  = [ Ractive ] , as expected. Readers should also note that elements of the above-mentioned matrices are calculated using related unit quaternions. T

Example: Calculating rotation matrix from a given quaternion-numerical example The quaternion q = 3 + i − 2 j + k is given. Find the equivalent rotation matrix for both active and passive rotation scenarios. For each scenario calculate the equivalent single axis and angle of rotation and compare the results. Solution: We first check if the given quaternion is a unit quaternion by calculating its norm/magnitude. Hence for q = 3 + i − 2 j + k we have q = 9 + 1 + 4 + 1 = 15 ≠ 1. Therefore, we need to calculate the corre1 1 −2 1   3 sponding unit quaternion = Q + k)  , , , ( 3 + i − 2 j= . 15  15 15 15 15  3 Now using Equation 15.21 for active rotation and plugging in for Q0 = , 15 −2 1 1 , Q2 = , and Q3 = we receive Q1 = 15 15 15

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Rigid Body Rotation: Euler Angles, Quaternions, and Rotation Matrix • 137

2 2 1 1 2 3 6 1  1 9 − −  − +   15 + 15 − 2 − 15 − 15  3 3 3 15 15     2 11 2 3 2 9 4 1 2 3    − . = − + − − −= [ Ractive ] 2  15 15 3 15 15 15 15 2 15 15    1  14 2 1  6 3 2 9 1 1    + − + −  3  15 15 15 15 2   15 15  15 15 The single axis of rotation is the vector corresponding to the vector part of the  Q   1 −2 1  −1   unit quaternion, or Q =  , ,  and the angle is 2 tan  Q  =  15 15 15   0  6 ° 2 tan −1   = 78.46 , see Equation 15.15. Note that the corresponding  3  unit vector is   1 , −2 , 1  Q  15 15 15   1 −2 1  .  , , = = =  n active   1 4 1 Q  6 6 6 + + 15 15 15 Similarly, for passive rotation, using Equation 15.22, we find 2 14   1  3 15 15    2 11 2   Rpassive  =  −    3 15 15  . We use the resulted rotation matrix to calcu 2 2 1 −  − 3 3   3  21  − 1 ° late the single angle of rotation, or cos  15  = 78.46 , see Equation  2    12.8. The axis of rotation is the vector part of the conjugate unit −1

quaternion

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= Q∗

1 −1 2 −1   3 − k)  , , , ( 3 − i + 2 j= , 15  15 15 15 15 

or

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e

138 • Tensor Analysis for Engineers, 3E

2 −1   −1 , ,  ∗  −1 2 −1     15 15   −1 2 −1   Q = , ,  15=  with the unit = vector n , , =  n act passive  15 15 15  1 4 1  6 6 6 + + 2 −1   −1 15 15 15 , ,   15 15   −1 2 −1    15= , , =  n active . 1 4 1  6 6 6 + + 15 15 15

15.7  FROM A GIVEN ROTATION MATRIX TO QUATERNION In this section, we would like to derive the equivalent quaternion from a given rotation matrix related to a rigid body rotation. Having the rotation matrix, we can use the relations obtained in the previous section to obtain the corresponding quaternion. Let us consider a passive rotation case, given by Equation 15.22. Note that the trace for both active and passive rotation matrices are equal in their values. The trace of the rotation matrix is Rtrace = 2 ( 3Q02 + Q12 + Q22 + Q32 − 1.5= )   2  2Q02 + Q02 + Q12 + Q22 + Q32 − 1.5  = 4Q02 − 1. Therefore, we have the real     1   part of the quaternion as Q0 = ±



1 + Rtrace 2

15.23

Knowing the real part, we can calculate the elements of complex  part Q using the diagonal elements of rotation matrix. Therefore, 2 R (1,1 ) − Rtrace + 1 R (1,1 ) − 2Q02 + 1 R (1,1 ) = 2Q02 + 2Q12 − 1 or Q1 = ± = . 2 2 Similarly, q2 = ±

2 R ( 2,2 ) − Rtrace + 1

, and q3 = ±

2 R ( 3,3 ) − Rtrace + 1

2 2 Therefore, component m of the complex part of the quaternion reads

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.

2 R ( m, m ) − Rtrace + 1 , m= 1,2,3 15.24 Qm = ± 2

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Rigid Body Rotation: Euler Angles, Quaternions, and Rotation Matrix • 139

From Equations 15.23 and 15.24, it is shown that for a given rotation matrix we get two unit quaternions that describes the rotation, or Q = Q0 + Q1 + Q2 + Q3 −Q = −Q0 − Q1 − Q2 − Q3 . Here, Q represents the unit quaternion and P = for a positive (R.H.R.) rotation about the axis of rotation and P represents the rotation about the same axis but in opposite direction, [27]. Example: Calculating quaternion from a given rotation matrix-numerical example 0 0 1    The rotation matrix R = 1 0 0  is given. Find the corresponding quaternions. 0 1 0  Solution: Using Equation 15.23, we receive the real part of the corresponding unit 1+0 1 ± = ± . The vector part can be calculated using quaternion as Q0 = 2 2 0 −0 +1 1 Q2 = Q3 = ± = ± . Therefore the quaterEquation 15.24, or Q1 = 2 2 1 1 1 1  1 1 1 1 − qˆ = nions are Q =  , , ,  and Q =  − ,− ,− ,−  . 2 2 2 2  2 2 2 2

15.8  FROM EULER ANGLES TO A QUATERNION In this section, we would like to derive the quaternion from given Euler angles related to a rigid body rotation. Using the relations derived in the previous section, we can find corresponding quaternions for Euler angles. For example, for 0 0 1   a roll rotation by angle q roll about x-axis, we have 0 cos ( q roll ) sin (q roll )  , 0 − sin (q roll ) cos ( q roll )  see Table 15.2, as the rotation matrix. Therefore, using Equation 15.23, 1 + 1 + 2 cos (q roll ) 2 − 2 cos (q roll ) − 1 + 1 q q cos roll , Q1 = Q0 = = = sin roll , 2 2 2 2 2 cosq roll − 2 cos (q roll ) − 1 + 1 = Q = = 0. This offers the correand Q 2 3 2 sponding unit quaternion equivalent to the roll rotation as

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140 • Tensor Analysis for Engineers, 3E

= Qq roll cos

q roll q 15.25 + i sin roll 2 2

Similarly, the unit quaternions for pitch and yaw rotations are = Qq pitch cos



q pitch 2

+ j sin

q pitch 2

15.26

and

= Qq yaw cos

q yaw 2

+ k sin

q yaw 2

15.27

Similarly, we can obtain 12 quaternions related to 12 Euler angles combinations. This can be achieved by equating the elements of each rotation matrix given in Table 15.2 and Table 15.3 with Equation 15.22. The result would be nine equations, which can be solved for corresponding quaternion real and imaginary parts [27]. However, in practice it is more efficient to calculate the final rotation matrix and extract the corresponding quaternions, as described in Section 15.7. Another method of extracting the quaternion related to a given set of Euler angles (without calculating the single-axis rotation matrix) is to use Equation 15.17 in conjunction with Equations 15.25 through 15.27.

15.9  PUTTING IT ALL TOGETHER It seems useful to collect all related methods and their interrelations, see Table 15.5. In this table, we list the related equations for calculating parameters of a desired method from those of a given method, directly. For example, S→E indicates the direct relations for obtaining Euler angles from a given rotation matrix. Readers should note that it is always possible to calculate the rotation matrix using one of the three methods (i.e., Euler’s angle, quaternions, single axis-angle) and then use it for calculating the parameters related to another method.

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TAE3E.CH15_2pp.indd 141

E→R

Eq. 12.7 S→R→Q

S→R

S→Q

E→Q

E→S

S→R→E

S→E

E-Euler angles

procedure

S-single axis-angle

E→R→Q

Table 15- & Table 15-3.

E→R→S

procedure

TABLE 15.5  Rigid body rotation direct relations and methods.

Q→E

Q→R

Q→S

Q→R→E

Eq. 15.21 &15.22

Eq. 15.15

Q-quaternions procedure

R→Q

R→E

R→S

R-rotation matrix

Eq. 15.23, 15.24 & 15.24

Table 15- & Table 15-3.

Eq. 12.8, 15.3

procedure

Rigid Body Rotation: Euler Angles, Quaternions, and Rotation Matrix • 141

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CHAPTER

16

MECHANICAL STRESS TRANSFORMATION: ANALYTICAL AND MOHR’S CIRCLE METHODS In Chapters 12 and 15, we discussed the transformation of Cartesian tensors and their applications for rigid body rotations. In this section, we would like to expand on this topic in relation to transformation of mechanical-stress type tensors and an equivalent graphical method called Mohr’s circle. In engineering, there are quantities such as mechanical stress, strain, and moment of inertia that sometimes need to be transformed for material failure analysis and design purposes. The transformations follow the Cartesian tensors rule discussed in Chapters 12 and 7.  xx  xy  xz    We define a 3D mechanical stress tensor, in matrix form, as  yx  yy  yz   zx  zy  zz    with the first index of each element of the matrix being the direction of the plane outward normal and the second index the direction of the force acting on that plane. The positive stress is hence defined considering the outward normal and the coordinates selected (see Figs. 16.1 and 16.2). To simplify the notations, the stresses in the Normal direction are given with only one index (i.e.,  xx   x ,  yy   y , and  zz   z ) and the shear stresses are assumed to be symmetric (i.e.,  xy   yx   xy ,  xz   zx   xz , and  yz   zy   yz ). Therefore,

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144 • Tensor Analysis for Engineers, 3E

 x  xy  xz    we have the symmetric stress matrix/tensor as    y  yz  , with six independent elements.     z  The coordinate system is arbitrary, and we define them to perform calculations in a consistent manner. We define the Cartesian coordinates at the origin in any desired direction, at a point of interest in the material geometrical domain. In the mechanics of solids, certain orientations of the coordinates system are preferred, since some materials show failure due the maximum normal or maximum shear stresses. Obviously, under any transformation, the magnitude of the stress tensor quantity does not change, only its components change.

16.1 PLANE STRESS CONDITION In practice, it is sometimes justifiable to make assumptions for reducing 3D mechanical stress to a 2D one. These assumptions are as follows: „

„

„

„

Stress elements are applied in a plane or are planar (i.e., no, or negligible out-of-plane stress). The Variation of the stresses are uniform across the thickness of the plane/plate. The Principle of complementary shear stress applies, i.e., shear acting on perpendicular surfaces, but in the plane, these are identical in magnitude, because of equilibrium. The stress matrix/tensor is symmetric.

For example, we consider the x–y plane (see Figure 16.2) as the plane on which the stresses are acting. Therefore, we have all stress components equal to zero in the z-direction. Thus, the symmetric stress tensor is  x  xy 0     x  xy    0 xy y     xy  y  , with three independent elements for the plane   0 0 0   mechanical stress tensor, i.e.,  x , y , xy . 16.1.1 Plane Stress Transformation In general, we can transform the Cartesian coordinates from their original position to any desired position by a series of translations and rotations. For

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Mechanical Stress Transformation: Analytical and Mohr’s Circle Methods • 145

FIGURE 16.1  Plane stress sketch acting on the x-y plane with positive stress components.

a stress state given at a point, we are interested in rotation of axes about that point, see Figure 16.2, which also shows positive sign convention for stress components acting on the stress element faces. Therefore, a stress component is considered to be positive when acting on a plane with positive (negative) outward normal and in the positive (negative) direction relative to the related axis of assigned coordinate system.

FIGURE 16.2  Positive sign convention for stress components; rotation is in the counterclockwise direction or normal to the plane of the acting stress components.

The stress components (or vectors acting on the stress element surfaces) can be transformed using Equation 16.1 (repeated here for convivence, see Equation 12.6), where Rij is the components of transformation matrix  (i, j = 1, 2), ni is the components of the normal to the plane n, where stresses are acting, δ ij is the Dirac-delta function, and eijk is the permutation symbol.  Rotation direction θ is in the normal vector n direction.

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146 • Tensor Analysis for Engineers, 3E







Rij  ni n j   ij  ni n j cos  eijk nk sin  16.1 Therefore, for a rotation of the x-y coordinates about the positive z-axis  direction, we have n   0, 0, 1  and R11  cos , R12   R21  sin  , and R22  cos . We used the properties of δ ij (equal to 1 for identical indices, otherwise it is zero, i.e.,  11   22  1 and  12   21  0 ) and eijk (1 for even permutations, -1 for odd permutations of the indices, and zero if any of the indices is repeated, i.e., e123  e231  e312   e132   e321   e213  1 ). We write the Rij in matrix form, as  cos Rij     sin 

sin   cos 

The components of the transformed stress can be written as (see Chapter 7)  cos sin    x  xy   cos sin    ij  Rik R jl kl   i, j, k  1, 2      sin  cos   xy  y    sin  cos  16.2 T

  x  xy  where  ij    , using the notations from Figure 16.2. Expanding  xy  y  Equation 16.2 using the Einstein summation convention for the indices (or performing the successive matrix multiplications), gives



 x   x cos2    y sin 2   2 xy sin  cos  2 2  xy   x sin  cos   y sin  cos   xy  cos   sin     2 2  y   x sin    y cos   2 xy sin  cos

16.3

Writing Equation 16.3 in terms of the trigonometric functions of 2θ , gives



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x y x y     x   cos 2   xy sin 2 2  2    x y   n 2   xy cos 2  xy     sin  2     y x y   y  x   cos 2   xy sin 2  2   2 

16.4

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Mechanical Stress Transformation: Analytical and Mohr’s Circle Methods • 147

 x ,  y , and  xy are the transformed stress components in the x  y coordinates, resulting from a counterclockwise (according to R.H.R.) rotation of the x − y coordinates by angle θ . Note that for   0, transformed stress components are identical to of the original stresses and for    / 2, only shear stress component changes sign. A spreadsheet can be built to graph these equations for a range of angles, for example   0   . The data for stress at a point, or initial state of stress, is required for the transformation calculation. For example, for  x  7 Mpa ,  y  2 Mpa , and  xy  5 Mpa , the transformed stress components are shown in Figure 16.3. 15

Stress

10

5

0 0

10

20

30

40

50

60

70

80

90

100 110 120 130 140 150 160 170 180

-5

-10

, rotation of axes in degrees Normal stress - Sx'

Shear stress- Txy'

Sy'

FIGURE 16.3  Variations of stresses at a point vs coordinates’ rotation angle θ , Sx   x and Sy   y .

Observations from Figure 16.3: „

„

„

„

„

TAE3E.CH16_2pp.indd 147

The variations are repeated after    . Hence, we don’t need to graph the variations out of this range, i.e.,  0,   .  y is out of phase with respect to  x by π / 2 . Therefore, we focus on the variation of  x .  x and  xy have maximum/minimum values. For this example, we have  1  10.09 MPa and  2  1.09 MPa . The maximum value of  x occurs at a specific angle θ , for which  xy vanishes. For this example, we have   31.72 . The angle θ for which the value of  x is a maximum and that of the absolute value of  xy (i.e.,   76.72 ) are out of phase by π / 4 .

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148 • Tensor Analysis for Engineers, 3E 16.1.2 Analysis of Transformed Stresses: Analytical Method Definitions: „

„

„

„

The maximum and minimum values of normal stress components (i.e.,  x and  y ) are called principal stresses. The orientations or the values of θ for which we get the principal stresses are called principal directions. The orientations or the values of θ for which we get the absolute maximum shear stresses are called maximum shear stress directions. The absolute maximum shear stress for cases when both principal stresses have the same sign requires the inclusion of the zero-value stress acting out of plane, i.e.,  3  0 . Or  max  max(( ABS( 1   2 ) / 2, ABS( 2  0) / 2, ABS( 3  0) / 2)) .

16.1.3 Principal Stresses and Direction d x For the max/min of  x we should have  0; after using Equation 16.4, d  we have





x



  y sin 2  2 xy cos 2  0

From Equation 16.5, we get tan 2  relation, and similarly tan   2  

2 xy

x y 2 xy

x y

16.5

. The angle that satisfies this

, represents the principal direc-

tions θ1 and θ 2 at which the normal stress  x is a maximum or minimum, respectively. Therefore, we have

 2 xy    2 xy  1 1 1  tan 1   and  2  tan 1    16.6 x y  x y  2 2 2    

We refer to θ1 as the principal direction, θ P . For calculating the values of the principal stresses, we substitute Equation 16.6 into the relation for  x from Equation 16.4. Equivalently, and for simplifying the algebra, we can eliminate θ between Equation 16.5 and the relation for  x . This can be done by adding the squares of these two relations, or Equation 16.4 can be manipulated by eliminating θ between the relations for  x and  xy . The result, after some

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Mechanical Stress Transformation: Analytical and Mohr’s Circle Methods • 149

2

x y   x y  2 manipulations, is   x       xy . Therefore, the prin2 2     cipal stresses are σ 1 and σ 2 (usually, the symbol σ 1 is used for the largest algebraic value between the principal stresses) are

1



at  P 1

and  2



at  2





x y 2

x y 2

2

x y  2      xy   2 

16.7

2

x y  2      xy  2  

16.8

Note that the first term in Equation 16.7 and Equation 16.8 is the averaged normal stresses. The second term in these equations is equal to the radius of Mohr’s circle (see further sections for more explanation about Mohr’s circle). It is evident that σ 1 corresponds to the orientation given by θ1 and σ 2 corresponds to the orientation given by θ 2 . Note that  1

2

at  2

  x

min

  y

max

at 1

  x

max

  y

, as shown in Figure 16.3 and Figure 16.4.

min

and

FIGURE 16.4  Principal stresses σ 1 and σ 2 with orientation angle θ p after transformation.

16.1.4 Maximum Shear Stress and Direction d xy  0 , and For calculating the max/min of  xy , we use the requirement d  after using Equation 16.4, we have

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150 • Tensor Analysis for Engineers, 3E





 x   y cos 2  2 xy sin 2  0 16.9 From Equation 16.9, we get tan 2  



this relation, and similarly tan   2   

x

y

2 xy



x

 . The angle that satisfies

y

2 xy

 , represents the direc-

tions 1 and  2 at which the shear stress  xy is the maximum or minimum, respectively. By definition, we refer to τ max as the absolute value of the max/ min shear stress. Therefore, we have  x y   x y   1 1 1  tan 1    and  2  tan 1      2 2 xy  2 2 2 xy    16.10 We refer to 1 as the maximum shear stress direction, θS . For calculating the values of the maximum shear, τ max we substitute Equation 10 into the relation for  xy from Equation 16.4. Equivalently, and for simplifying the algebra, we can eliminate θ between Equation 9 and  xy . This can be done by subtracting the squares of these two relations. The result, after some manipulations, is 2

x y  2  max   xy      xy  2  16.11 1 Note that  max  ( 1   2 ), see Equation 16.7 and Equation 16.8, and is 2 equal to the radius of the Mohr’s circle. At the maximum shear stress direction/plane, the normal stresses are not zero in general. These normal stresses can be found by substituting 1 into  x (see Equation 16.10 and Equation 16.4) or equivalently, eliminate θ between  2 Equation 16.9 and Equation 16.4, to get     avg  1 (see Figure 16.5). 2

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Mechanical Stress Transformation: Analytical and Mohr’s Circle Methods • 151

FIGURE 16.5  Maximum shear stress and direction after transformation.

16.1.5 Relation between Principal and Maximum Shear Directions A very important relation between the principal and maximum shear directions exists and can be obtained by using Equation 16.6 and Equation 16.10,  2 xy  or forming the multiplication relation, tan 21  tan 21’    x y     x y        1. Therefore, tan  21    cot  21   tan    21  , or    2  2 xy     

(using 1   P and 1  S );

  P  S  4 16.12 Conclusions: „ „

„ „ „

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At the principal plane/direction, the shear stress is zero. At the maximum shear stress direction/plane, the average of the normal stresses acts. The planes of the max. and min. normal stresses are perpendicular. The Planes of the max. and min. shear stresses are perpendicular. The principal and maximum shear stress planes make a 45° angle with each other.

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152 • Tensor Analysis for Engineers, 3E

16.2 PRINCIPAL STRESSES AND DIRECTIONS: EIGENVALUES AND EIGENVECTORS As mentioned in Chapter 12, and instead of the analytical method using Rij and derivations described in the previous sections, we can find the principal stresses by calculating the eigenvalues of the stress tensor and the principal directions by calculating the corresponding eigenvectors. Let’s consider a  x  xy  2D-stress state as   and modify it by subtracting a multiple of the  xy  y 

 xy   x  xy  1 0   x    unity matrix, or  , where λ is an     y    0 1    xy  xy  y  arbitrary constant. Now we calculate the determinant of the resulted matrix, x   xy which yields   x     y     xy2 . A critical question  xy y  





could be asked (since λ is arbitrary): at what values of λ do the calculated determinants vanish? The answer can be found by solving the quadratic equation  2   x   y    x y   xy2  0 . The two solutions are





2 1  1,2   x   y   x   y  4  xy2   x y  . After expanding the expres2  sion that exists under the radical and some algebraic manipulations, we get





x y





2

x y  2 1,2       xy 2 2   16.13 Equation 16.13 covers Equation 16.7 and Equation 16.8, or the principal stresses. In other words, 1   1 and 2   2 . From mathematics, we have λ1 and λ2 defined as the eigenvalues of the 2D stress matrix considered (or any 2D symmetric tensor). Therefore, by calculating the principal stresses, we can calculate the eigenvalues of the stress matrix.  We can also determine the principal direction, n = (nx , ny ), where nx and ny are components of the normal unit vector to the principal plane (see  Figure 16.6). Therefore, we have n2x  ny2  1 . To find the vector n, we sub nx   x  xy   nx   1   . The stitute λ1 into the system of equations      xy  y   ny   ny 

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Mechanical Stress Transformation: Analytical and Mohr’s Circle Methods • 153

 nx  non-­trivial solution for the system of equations gives   . A similar opera ny 1  nx  tion using λ2 gives   .  ny  2

These are the eigenvalues of the stress matrix and are equal to the cosθ1 and cosθ 2 (see Equation 16.6), after normalization. For related detailed calculations, readers can see Section 12.2.1.

FIGURE 16.6  A plane with normal unit vector and its components.

16.3 ANALYSIS OF TRANSFORMED STRESSES: MOHR’S CIRCLE GRAPHICAL METHOD Mohr’s circle method is a graphical approach based on the geometrical representation of transformed stresses. By eliminating θ between the relations for  x and  xy from Equation 16.4, we get 2

2

x y   x y  2 2   x     xy      xy 16.14 2    2 



Let’s define the following quantities: 2   y  2  R   x    xy  2   16.15   x y  avg  2 



then Equation 16.14 gives the algebraic equation of a circle with radius R with its center at (σ avg , 0) in a    coordinate system (i.e., normal stresses on the abscissa and shear stresses on the ordinate). This circle is called Mohr’s circle.1 Therefore, stress components lie on Mohr’s circle for arbitrary rotation of initial coordinate system for any given stress element.   Christian Otto Mohr (1835-1918).

1

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154 • Tensor Analysis for Engineers, 3E Mohr suggested a graphical method for analyzing stress tensor transformation specifically used for plane stress before the computer power and resources became available for making graphs of the stress variations (see Figure 16.3). However, this method (and its versions) is still in use, mainly for its simplicity, merits for on visualizing stress transformation, principal stresses, and helping with understanding the core concept. 16.3.1 Mohr’s Circle Construction and Application Application of Mohr’s graphical method using the circle given by Equation 16.14 consists of two parts: 1. The construction of Mohr’s circle for a given state of stress (i.e., stress element) 2. Extracting information from the constructed Mohr’s circle, e.g., principal stresses and directions, and maximum shear stresses and directions In general, there are two approaches for constructing Mohr’s circle for a given stress element and extracting information from it. We refer to these methods as the 2θ -approach (when rotation is measured with reference to the center of the Mohr’s circle) and the Pole-point approach [37] (when rotation is measured with reference to a specific point on the circumference of the center of the Mohr’s circle). The former is the common approach, with the consideration of a 2θ rotation in Mohr-space corresponding to a θ rotation in the physical-space, or the rotation of coordinate system x – y. This can be easily verified by Equation 16.4. This mathematical requirement (i.e., 2θ correspondence) is sometimes confusing and could cause computational and interpretational errors in practice. The Pole-point approach is more straightforward since it focuses on the circumference of the Mohr’s circle in terms of the rotation angle. Therefore, a rotation by angle θ is equivalent in physical and Mohr’s circle spaces. In further sections, we will demonstrate the latter using several examples. 16.3.2 Sign Convention for Shear Stress in Mohr’s Circle For constructing a Mohr’s circle, attention should be paid to the sign of the shear stresses. The sign conventions given in Figure 16.2 for normal stresses in physical space are treated in the same way in a Mohr’s circle. However, the shear stresses are treated differently compared to the physical space (see Table 16.1). In the Mohr-circle space, we consider the shear stresses as positive when they result in a clockwise rotation (i.e., the

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resultant moment out of the shear forces couple acting on the opposite sides of the stress element is in the clockwise direction) and negative when they are in counterclockwise direction (i.e., the resultant moment out of the shear forces couple acting on the opposite sides of the stress element is in the counterclockwise direction). This sign convention, along with the positive τ -axis being upward, gives the same rotation directions for both the Mohr’s circle and stress element defined in the physical space. Note that in the physical space, the positive sign convention for the shear stress remains, as discussed previously, i.e., according to outward normal to the plane and the selected coordinate system. This sign convention is merely for the construction of a Mohr’s circle and should not be confused with the positive sign conventions given in Figure 16.2 for shear stresses. In other words, the sign convention in the physical space is different from that of in the Mohr-circle space, if shear stresses are involved. Further numerical examples will clarify this point. TABLE 16.1  Sign convention used for plane shear stress τ xy in physical vs Mohr’s circle space.

Stress state/shear

Physical space, rotation ( x, y)

Mohr’s circle space, rotation

Coordinates and stress element

( , )

+

−

−

+

+

+

−

−

16.3.3 Why are there Different Sign Conventions for the Shear Stress in a Mohr’s Circle? There are four possible options for defining a sign convention for shear stresses when using the Mohr’s circle method. When considering a stress element

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156 • Tensor Analysis for Engineers, 3E and coordinate system x – y, see Figure 16.2 , the shear stress components acting on the vertical faces (i.e., with its outward normal parallel to the x-axis) create a counterclockwise moment and similarly those acting on the horizontal faces (i.e., with its normal parallel to the y-axis) create a clockwise moment. Obviously, as mentioned, these two moments should be equal in magnitude to satisfy the equilibrium requirement related to the stress element. Which direction (i.e., clockwise or counterclockwise) should be considered as being positive in Mohr’s circle space? The short answer is “either” can be used for defining a sign convention. However, considering the coordinate system σ -τ , the positive direction of τ -axis could be considered upward or downward, as well. Therefore, we end up with four possible options for defining a sign convention for the shear stress in the Mohr’s circle; two options resulted from the directions of rotation and two from the τ -axis orientation. The criterion for choosing from these four possible options comes from the preferred choice that we would like to have rotations implemented to the stress element coordinates (i.e., x-y) and equivalently to its representative Mohr’s circle being in the same direction. For example, if we rotate the x-y coordinates of the stress element in the counterclockwise direction, then we prefer to have the corresponding representative points move on the Mohr’s circle perimeter in the counterclockwise direction, as well. Therefore, only two options (i.e., 1 and 4) out of four give the desired outcome, as shown in Table 16.2 and Table 16.3. TABLE 16.2  Possible options for shear stress sign convention in Mohr’s circle.

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Option

Shear stress Coordinates, sign, if selected if selected

Mohr’s circle

Implemented rotation (physical vs. Mohr’s)

1

CW (+)

C1

compatible

2

CCW (+)

C2

incompatible

3

CW (+)

C3

Incompatible

4

CCW (+)

C4

compatible

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TABLE 16.3  Corresponding Mohr’s circles for the options listed in Table 2.

C1

C2

C3

C4

As mentioned in the previous section, among the two preferred options (i.e., 1 and 4), we select option 1 for defining the sign convention for the shear stresses in the Mohr’s circle. Some authors [38] consider option 4, instead. For all cases we considered here, the stress element is assumed to be the one shown in Figure 16.2, with positive stress elements  x , y , and  xy defined in the physical space. To help with understanding the shear stress sign convention and its preferred selection, we consider option 2 (for example, see Table 16.2). For this option, we have selected the CCW direction and upward τ -axis as being positive. Therefore, side A (i.e., the vertical side of the stress element) is represented with coordinates A( x , xy ) on the Mohr’s circle. Similarly, side B (i.e., the horizontal side of the stress element) is represented with coordinates B( y ,  xy ) on the Mohr’s circle. The resulting Mohr’s circle is shown as designated with C2 in Table 16.3. Now, for orienting the line segment A − B in the principal direction (i.e., null shear stress), it should rotate CW, as shown with the curved arrow. This rotation is in the opposite direction of the required equivalent rotation for the stress element in the physical space (i.e., CCW according to R.H.R.) in order to end up with the principal direction. The inconsistency of these rotations in the physical space and corresponding Mohr’s circle is not preferred. Following a similar analysis, option 3 comes out to be inconsistent, but options 1 and 4 are consistent (see Table 16.3). 16.3.4 Pole-Point Method: Numerical Example For this example, we use the same data given for the graph as sketched in Figure 16.3 and repeated here for the stress element shown in Figure 16.7. The Pole-point method is described in a step-by-step manner for constructing the corresponding Mohr circle and extracting information from it. We make use of GeoGebra2 (a free software tool, but not required) for ­having

 https://www.geogebra.org/.

2

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158 • Tensor Analysis for Engineers, 3E our sketches drawn to scale and organized, as well as using its algebraic facilities/tools. Please refer to Figure 16.7 and Figure 16.8 while following the steps given below:

FIGURE 16.7  Stress element data given in physical space.

1. Calculate the data for the state of stress at the desired point in the domain of the problem. For example, we assume that in this data set,  x  7 Mpa ,  y  2 Mpa , and  xy  5 Mpa are given and defined in the physical space of a material domain. 2. Draw the stress element and identify the stress components on its corresponding faces using the sign conventions given in Table 16.1. Identify a vertical face by label A and a horizontal face by label B, as shown in Figure 16.7. 3. Launch GeoGebra, either cloud based or an installed version. Consider the horizontal axis representing the normal stresses, σ , and vertical axis shear stresses, τ (see Figure 16.8). Note that the positive shear stress axis for Mohr’s circle is considered to be upward. 4. Create two points: A with the coordinates A  7,5  and B  2, 5  . Notice the negative sign for the shear stress acting on face A, since it is in the counterclockwise direction, and positive for shear stress acting on face B, since it is in the clockwise direction. 5. Connect points A and B, using a line segment, AB. 6. Find the point intersecting the line AB with the σ -axis (i.e., the horizontal axis). This is the center point of the corresponding Mohr’s circle, or C (4.5,0), with  avg  4.5 MPa . 7. Build Mohr’s circle using points C and A (or B). The equation of this 2 circle is   4.5    2  31.25 (i.e., 31.25 = 5.59 the radius of the Mohr’s circle, rounded with two-decimal point accuracy). Each point

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on the Mohr’s circle represents a state of stress with the corresponding orientation. For example, point B represents the original stresses (2, 5) acting on face B. 8. To find the principal stresses, find the intersections of the circle with σ -axis, i.e., two points with zero shear stress values. We get S1  10.09, 0  and S2   1.09, 0  , hence, the principal stresses (or maximum and minimum normal stresses) are  1  10.09 MPa and  2  1.09 MPa , respectively. 9. To find the maximum/minimum shear stresses, draw a line parallel to the τ -axis from the center point C and find its intersections with the circle, or points Tmax   4.5, 5.59  and Tmin   4.5, 5.59  . Hence, maximum shear stress is  max  5.59 MPa .

To find the corresponding orientations for the principal and max shear stresses, we use the Pole-point approach. Every Mohr circle has only one pole3, with reference to the planes on which stresses are acting. To find the pole for our Mohr circle, we can select any state point, for example face A, and draw a line from point A on the Mohr circle parallel to the Normal to face A (i.e., a horizontal line). The intersection point of this line with Mohr circle is the Pole of the Mohr’s circle.

10. We select face A and from the corresponding point A on the circle, draw a line parallel to its outward Normal. This line intersects the circle at a specific/unique point or, Pole  2,5  . This point is the Pole of Mohr’s circle. Note that drawing a line from point B parallel to outward Normal to face B (i.e., a vertical line) would intersect the circle at the same pole point, since, pole point is unique for a given Mohr’s circle.

A line connecting the Pole point and any other selected point on the Mohr’s circle is parallel to the outward Normal of the face that corresponds to the selected point. The coordinates of the selected point are the normal and shear stresses acting on the corresponding face.

11. Connect the Pole point to S1 and S2. These lines are parallel to the principal directions, respectively. Therefore, original stress element should be rotated by an angle 31.72o, counterclockwise to be oriented in the principal direction with principal stress S1 = 10.09 acting along  Here, we consider the Pole with reference and parallel to the outward Normal to the plane instead of the Pole point parallel to the plane itself. (http://www.mech.utah. edu/~brannon/public/Mohrs_Circle.pdf).

3

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160 • Tensor Analysis for Engineers, 3E the ­normal direction or the line segment Pole-S1 and S2  1.09 acting along the normal direction or the line segment Pole-S2. 12. Similarly, for maximum shear stress direction, connect the Pole point to points representing Tmax or Tmin. These lines are parallel to the maximum shear stress directions, respectively. Therefore, the original stress element should be rotated by an angle 45  31.72  76.72 deg., counterclockwise. Note that Tmax = 5.59 acts on a face with its outward normal parallel to the line segment Pole-Tmax and Tmin  5.59 acts on a face with its outward normal parallel to the line segment Pole-Tmin. As mentioned, this approach produces accurate and fast results for constructing a Mohr’s circle and extracting principal stresses and directions from it. We believe that readers, when they become familiar with this approach, will find it more useful and effective than the 2θ approach.

FIGURE 16.8  Mohr’s circle construction and information extraction, using Pole-point approach.

16.3.5 2θ-Method: Numerical Example The Mohr’s circle center and radius are given by Equation 16.15. Therefore, we can draw the Mohr’s circle using these data values. It is also possible to draw the circle using data on face A (or B) and the center point. The circle would

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be the same as the obtained previously (see Figure 16.8). Then the principal stresses are

 1   avg  R and

 2   avg  R 

16.16

With reference to Figure 8 and using Equation 6, the principal orientation θ p  2 xy  1 is  p  tan 1  . x y  2   It is also possible to use the trigonometric relations from the Mohr’s circle. The results for the principal, maximum shear, and their corresponding directions are the same as those found in the previous section, using the Polepoint method. Solution: We assume the same data set;  x  7 Mpa ,  y  2 Mpa , and  xy  5 Mpa , defined in physical space (see Figure 7). 1. Calculate the circle’s x y 7  2  avg    4.5 . 2 2

center

point

σ -coordinate,

or

2. Identify the center as point C   4.5, 0  on the σ -axis. 3. Identify point A   7, 5  . Similarly, point B can be used instead. 4. Draw a Mohr’s circle with its center at point C and its radius equal to CA. 5. Calculate, or measure, the radius of the Mohr’s circle as 2

2 x y  72 2 2 R         5  5.59 . xy 2 2    

6. Calculate, or measure, the principal stresses as  1   avg  R  4.5  5.59  10.09 and  2   avg  R  4.5  5.59  1.09 . 7. The maximum shear, τ max is equal to R, or  max  5.59 . 8. Calculate the principal and the maximum shear orientations either using the formula or trigonometric relations using the constructed Mohr’s  2 xy  1  25  circle. Or,  p  tan 1    0.5(tan 1    31.72 , face A   2 72 x y  should be rotated in the counterclockwise direction to get the principal

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162 • Tensor Analysis for Engineers, 3E stress state. For the maximum shear stress orientation, face A should be oriented in 31.72  45  76.72 in a counterclockwise direction. All orientations are with reference to the original stress element’s face A and its outward Normal.

Note that if the orientation angles are measured or calculated using the trigonometric relation of Mohr’s circle, attention should be paid to the fact that the angles related to the Mohr’s circle are double the actual amounts, i.e., 2θ -angle rotation for a physical θ -angle rotation.

16.3.6 Eigenvalue and Eigenvector Method: Numerical Example In this example, we use the analytical method of the eigenvalue/eigenvector to calculate the principal stresses and directions for the stress tensor, 7 5   5 2  MPa. The results can be compared with those obtained in previous   sections (for example, with those given in Figure 16.8). Solution: The eigenvalues are, using Equation 16.13, 1  10.09 MPa and 2  1.09 MPa. These are identical to the principal stresses obtained (see Figure 16.8). 13.09   nx   The eigenvectors are    . We normalize this vector by  ny  10.09  8.09  1

dividing it with 8.09, then

1 1.618    is the unit vector representing 3.618  1 

the principal plane that λ1 is acting on (see Figure 16.6). Now we calculate  1.618  0 the angle, or cos1    31.72 . This is the angle that the Normal to  3.618  the principal plane (corresponding to 10.09 MPa) makes with the horizontal 1  1  0 x-axis. Note that sin 1  is the y-component   31.72 , since 3.618  3.618  of the unit vector. Similarly, the eigenvector corresponding to λ2 can be cal nx   1.91  1  1   culated as   , which gives   . This is the  3.618 1.618   ny  1.09 3.09  2

unit vector representing the principal plane that λ2 is acting on. Now we

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Mechanical Stress Transformation: Analytical and Mohr’s Circle Methods • 163

 1.618  0 calculate the angle, or sin 1    58.28 (i.e., complementary angle  3.618  to 31.72° , or 31.72  58.28 0  90). This is the angle that the Normal to the principal plane (corresponding to –1.09 MPa) makes with the horizontal negative direction of x-axis. The negative sign is inconsequential, since we can multiply the eigenvector by -1, without losing generality. The eigenvectors or principal directions are identical to those given in Figure 16.8. 16.3.7 Mohr’s Circle: 3D and Special Cases Special cases arise in 2D plane stress states when the principal stresses have the same sign (either positive or negative, i.e., sign  1   2   0 ). In these cases, the value of the maximum shear stress is affected, and attention should be given to calculating the absolute maximum shear stress and for the proper and safe design when using the Mohr’s circle. This topic is usually referred to as the “3D Mohr’s circle or Mohr’s 3-circle”. Considering the 2D plane stress state defined in x–y, as a subset of 3D state with the stresses acting on the plane Normal to z-direction to be null, we  x  xy 0   x  xy    have, as mentioned previously,     xy  y 0  . Calculating all   y  xy  0 0 0  three principal stresses gives σ 1 ,σ 2 , and 0 . Therefore, we have three principal stresses with the inclusion of  3  0 . We discuss these special cases in the following sections. It is customary to designate the normal stresses by their algebraic values from largest to smallest, or  1   2   3 . Case 1:  1   2   3  0 In this case, both non-zero principal stresses σ 1 and σ 2 have positive signs. The Mohr’s circle for this case, for example, is sketched in Figure 16.9, in the order of the Mohr’s circle in the x–y plane, y–z plane, and x–z plane. As shown, the maximum shear stress should be calculated from the Mohr’s circle constructed with  1  0 and  3  0 as the principal stresses, i.e., the outer circle. Therefore, the absolute maximum shear stress corresponds to

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164 • Tensor Analysis for Engineers, 3E

1 , comparable to the other relative ­maximum 2 shear stress values (corresponding to the x–y and y–z planes). This shear stress should be used for design and failure analysis, and usually is referred to as the absolute maximum stress. the x-z plane, defined as  max 

FIGURE 16.9  3D Mohr’s circle for Case 1 with all non-zero principal stresses having positive signs.

Case 2:  1  0   2   3 In this case, both non-zero principal stresses, σ 2 and σ 3 , have negative signs. The Mohr’s circle for this case, for example, is sketched in Figure 16.10, in the order of the Mohr’s circle in the y–z plane, x–y plane, and x–z plane. As shown, the maximum shear stress should be calculated from the Mohr’s circle constructed with  1  0 and  3  0 as the principal stresses, i.e., the outer circle. Therefore, the absolute maximum shear stress corresponds to  3 the x–z plane, defined as  max  , comparable to the other relative maxi2 mum shear stress values (corresponding to the y–z and x–y planes). This shear stress should be used for design and failure analysis, and usually is referred to as the absolute maximum stress.

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FIGURE 16.10  3D Mohr’s circle for Case 2 with all non-zero principal stresses having negative signs.

16.3.8 3D Mohr’s Circle: Numerical Example 1 For a state of stress given as  x  80 Mpa ,  y  40 Mpa , and  xy  30 Mpa , defined in the physical space, calculate the principal stresses, maximum shear, and their corresponding directions. Use the Pole-point approach and draw the corresponding stress element for the resulting principal stress state. Solution: Considering sides A (with its normal in the x-axis direction) and B (with its normal in the y-axis direction) of a stress element, we have A(80,–30) and B(40,30) as coordinates of the corresponding points on the Mohr’s circle. Now, we draw the Mohr’s circle with the diameter AB and find the intersections of the circle with the σ -axis, i.e., points S1  96.06, 0  and S2  23.94, 0  , representing the principal stresses. Therefore, the principal stresses are  1  96.06 MPa and  2  23.94 MPa. Note that the corresponding maximum shear stress equal to the radius of this circle is given by point K, or 36.06 MPa. Since both principal stresses are positive (i.e., Case 1, as mentioned previously), we should draw the outer Mohr’s circle (passing through points S1 and the origin) to the get the absolute maximum shear stress as

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166 • Tensor Analysis for Engineers, 3E

 1 96.06   48.03 (see Figure 11). The outer Mohr’s circle corre2 2 sponds to the  1  96.06 MPa and  3  0 .

 max 

After finding the pole at (40,–30) by drawing a horizontal line from point A and finding its intersection with the Mohr’s circle, the principal directions can be obtained by rotating the original stress element by an angle equal to 28.15° in the counterclockwise direction with respect to the x-axis, as shown in Figure 16.11. The direction of the maximum shear is at 45° with respect to coordinates 1-2-3, defined in the principal directions, and rotated counterclockwise about the 2-axis (i.e., in the 1-3 plane).

FIGURE 16.11  3D Mohr’s circle for the numerical example for Case 1 with all non-zero principal stresses having positive signs.

16.3.9 3D Mohr’s Circle: Numerical Example 2 For a state of stress given as  x  50 Mpa ,  y  180 Mpa , and  xy  30 Mpa , defined in the physical space, calculate the principal stresses, maximum shear, and their corresponding directions. Use the Pole-point approach and draw the corresponding stress element for the resulting principal stress state.

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Solution: Considering sides A (with its Normal in the x-axis direction) and B (with its Normal in the y-axis direction) of a stress element we have A(–50,–30) and B(–180,30) as coordinates of the corresponding points on the Mohr’s circle. Now, we draw the Mohr’s circle with the diameter AB and find the intersections of the circle with the σ -axis, i.e., points S2  43.411, 0  and S3  186.589, 0  , representing the principal stresses. Therefore, the principal stresses are  2  43.411 MPa and  3  186.589 MPa . Note that the corresponding maximum shear stress equal to the radius of this circle is given by point G, or 71.589 MPa. Since both principal stresses are negative (i.e., Case 2, as mentioned previously), we should draw the outer Mohr’s circle (passing through points S3 and the origin) to the get the absolute maximum  3 186.589 shear stress as  max    93.295 (see Figure 16.12). The outer 2 2 Mohr’s circle corresponds to the  1  0 MPa and  3  186.589 MPa . After finding the pole at (–180,–30) by drawing a horizontal line from point A and finding its intersection with the Mohr’s circle, the principal directions can be obtained by rotating the original stress element by an angle equal to 12.39° in the counterclockwise direction with respect to the x-axis, as shown in Figure 16.12. The direction of maximum shear is at 45° with respect to coordinates 1-2-3, defined in the principal directions, and rotated counterclockwise at about the 2-axis (i.e., in the 1-3 plane).

FIGURE 16.12  3D Mohr’s circle for the numerical example for Case 2 with all non-zero principal stresses having negative signs.

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168 • Tensor Analysis for Engineers, 3E 16.3.10  Stress State on Selected Plane: Numerical Example In practice, it is sometimes desirable to calculate the stresses on a plane with a specified normal direction, such as stresses in the direction of the wood grain/fiber and along a weld (see Figure 16.13). A Mohr’s circle can be used to visualize and calculate the acting stresses for a desirable direction with respect to the reference’s original direction of the stress element.

FIGURE 16.13  Sketch for stress element on a specified plane.

For a state of stress given as  x  400 Mpa ,  y  300 Mpa , and  xy  150 Mpa , defined in the physical space, find the state of stress along a welded seam at an angle of 25° CCW from x-axis. The reference stress element is aligned with the x–y coordinates. Solution: We use the pole-point method to construct the Mohr’s circle using A(400, –150) and B(–300, 150), representing the reference stress element. By drawing a horizontal line passing through A, we get its intersection with the Mohr’s circle as the Pole(–300,–150). Now we draw a line from the Pole, making an angle equal to 25°, the desired angle, with the line A-Pole. This line intersects with the circle at point F(398.88, 171.7), representing the normal and shear stresses acting on the stress element surface perpendicular to it. Connecting point F to the center C and extending it to intersect with the circle, we find point H(–298.88,–171.7), representing the normal and shear stresses acting on the stress element surface perpendicular to the seam line. The reference and desired stress elements are shown in Figure 16.14.

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FIGURE 16.14  Mohr’s circle for the numerical example calculating stresses on a specified plane.

16.4 3D STRESS TRANSFORMATION AND ANALYSIS Mohr’s graphical method could be extended to 3D and non-symmetric stress tensors. However, the analytical method, along with computer aided computations, is more efficient for practical applications when solving 3D problems. As mentioned, a Mohr’s circle has merits for 2D-stress transformation with physical insights and educational purposes. For example, once we have the principal stresses, we can construct the corresponding Mohr’s 3-circle diagram. In this section, we discuss 3D stress transformation for mechanical stress-type quantities. General formulation In previous sections, we covered graphical and analytical methods for 2D stress transformation and analysis. These methods could be applied to any symmetric square tensor, like strain and moment of inertia. In this section, we extend the analysis and derive more general relations with an emphasis

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170 • Tensor Analysis for Engineers, 3E on 3D tensors which are used to determine the strength of materials and applied elasticity. For this treatment, we use the index notation along with the Einstein summation convention, unless explicitly specified, for example,  xx   11 ,  yy   22 ,  xz   13 ,  ii   11   22   33 , and so forth. 16.4.1 Traction Force and Its Normal Extremum-Principal Stresses A plane can be represented by its unit outward normal vector   n  ni   n1 , n2 , n3  . Therefore, the traction force T  Ti   T1 , T2 , T3  acting on this plane (not necessarily normal/perpendicular to it) as a result of stress tensor σ ij is Ti   ij n j (Cauchy’s formula). To find the magnitude of the normal component of traction N acting on the plane, we find the inner product of the traction and the plane’s outward normal vector, as shown in Figure 16.15, or  N  N  Ti ni   ij ni n j 16.17

FIGURE 16.15  Definition of traction force and normal component.

Expanding Equation 16.17 gives the magnitude of normal force as N  T1 n1  T2 n2  T3 n3 , a scalar quantity. Therefore, the magnitude of shear stress component aligned with the surface of the plane is given as  2 2 2      Ti    Ti ni   T12  T22  T32   T1 n1  T2 n2  T3 n3  . Note that τ can be written in tensor notation, as

   ij ik n j nk   ij kl ni n j nk nl 16.18 Collecting these relations in Equation 16.19 and Equation 16.20 gives

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 N  T1 n1  T2 n2  T3 n3  2 2 2 2    T1  T2  T3   T1 n1  T2 n2  T3 n3 

16.19

where,  T1   11 n1   12 n2   13 n3  T2   21 n1   22 n2   23 n3 T   n   n   n 16.20 31 1 32 2 33 3  3 Now, we would like to find the extremum value of N (or the principal stresses). In other words, we seek a specific orientation of ni (or its corresponding plane) for which we get the max/min of N. This is done by using the method of the Lagrange multiplier. Therefore, we form a linear combination of N and the constraint, ni ni = 1 (or the condition that the Normal is aligned with the direction that gives the maximum N, hence its inner product with itself is unity). The linear combination reads    ij ni n j    ni ni  1  where λ (or – λ ) is a constant, the Lagrange multiplier. Note that with reference to principal orientation  = N , since the Normal to the principal plane is in the desired direction and hence the value in the bracket  vanishes. Alternatively, to find the extremum of  we let  0. For ni this, we rewrite  equivalently as,    ij ni n j   (ni n j ij  1) . Therefore,    ni   ij n j   n j ij  0 , or ni    n   ij

ij

j

0

Aij 16.21

equations for the unknown n j ≠ 0 . In  1 j   n1            0 . For a symmetric   jj     n j  tensor, off-diagonal symmetric elements of the matrix are the same,  ij   ji . Obviously, a null vector n j = 0 is a trivial solution, but it does not satisfy the Equation 16.21 gives a system of  11    matrix form, it is written as     j1 

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constraint ni ni = 1 nor does it have physical significance! We are looking for a non-zero normal vector. Therefore, we the determinant of the coefficient matrix must be zero, or det  ij   ij  detAij  Aij  0. The determinant4





can be calculated using the permutation symbol eijk, or = Aij e= 0 ijk Ai1 A j 2 Ak 3 16.22 For simplicity when writing the derivations (without losing generality), we consider a 3D stress tensor. Hence, assuming Aij as a 3 × 3 matrix, the expansion of the determinant, Equation 16.22, gives  11     22     33      21 32 13   31 12 23   11    32 23   31  22    13   21 12  33     0 . This is a cubic equation in terms of λ , that, after rearranging the terms, reads   ii jj   ij ji   3   ii  2       ij  0 2   16.23 Equation 16.23 has a special place in the tensor analysis subject area; hence it is given a special name, i.e., the characteristic equation of the stress tensor. In other words, each tensor has a unique characteristic equation. The solutions of this equation are the eigenvalues and corresponding ni are the eigenvectors of the stress tensor. Eigenvalues are the principal stresses (i.e., 1   1 , 1   2 , 3   3 ) and the eigenvectors are the unit normal of the principal planes, the principal directions. The scalar coefficients of the characteristic equation also have definite interpretations and are called the stress invariants Ii (i.e., they remain intact under transformation), or I   def ii  trace of  ij  1  2  3 1  ( ii jj   ij ji ) def  sum of diagonal minors of  ij =1 2 +2 3 +1 3  I2  2  I |  | def ij  determinant of  ij  1 2 3  3 16.24 For a given tensor, its invariants do not change under transformation, i.e., they are independent of rotation angle, θ . Therefore, for a given symmetric stress   For higher order tensors, see Section 9.1.

4

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Mechanical Stress Transformation: Analytical and Mohr’s Circle Methods • 173

 x  xy  xz    tensor    y  yz  , using Equation 16.24 after transformation to orient it     z   1 0 0    with the principal direction, we have    2 0  and the invariants are     3 





I1   x   y   z   1   2   3 , I2   x y   x z   y z   xy2   xz2   yz2  2  1 2   2 3   3 1 , and I3   x y z  2 xy xz yz   x yz2   y cxz   z xy2   1 2 3 . Here, the transformed stresses are given in terms of the principal stresses (i.e., the original stress matrix is given with reference to the principal directions 1–2–3 resulting from the rotated original x–y–z coordinates). For a symmetric stress tensor, the principal stresses are real numbers.





16.4.2 Traction Force and Its In-Plane Extremum - Principal Shear Stresses A similar approach to the calculation of the principal stresses can be employed for finding the extremum value of shear force’s magnitude, τ . Therefore, for a given stress tensor at a point, σ ij and assuming a unit normal  vector n s  nis   n1s , n2s , n3s  , we can write the expression for a scalar  as the linear combination of τ

and the required condition nis nis = 1, or

   ij ik n sj nks   ij kl nis n sj nks nls    nis nis  1  . To find the extremum value

of  , we can write

      0 . The result is a cubic equation, in n1s n2s n3s

terms of the Lagrange multiplier γ with its three solutions as the principal shear stresses. The corresponding principal planes can then be obtained for each value of the principal shear stress. To simplify the derivation, without losing generality, we consider the coordinate axes to be aligned with the principal directions ni   n1 , n2 , n3  , as calculated for the principal stresses. Note that on these coordinate planes the shear stress is zero. In terms of the principal stresses (i.e., σ 1 ,σ 2 ,σ 3 ), using Equation 16.19, we get

 2   12 n12   22 n22   32 n32   1 n12   2 n22   3 n32  16.25 2

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174 • Tensor Analysis for Engineers, 3E

The unit vector should satisfy n12  n22  n32  1 . We eliminate n3 between the last relation and Equation 16.25 to get

 2   12   32  n12   22   32  n22   32   1   3 )n12   2   3  n22   3  16.26 2

 2  2 Now, to find the extremum value of τ , we have   0 , which yield, n1 n2 after some manipulations

 

 

  1   3  n1  1   3   2  1   3  n12   2   3  n22   0     2 2  2   3  n2  2   3   2  1   3  n1   2   3  n2   0 16.27 In order to have the relations given by Equation 16.27 satisfied, the following cases are considered: a. n= n= 0 , then from n12  n22  n32  1 , we get n3 = 1 . Using these val1 2 ues, and after plugging them into Equation 16.25, we get   0. This result confirms that the shear stress acting on the principal plane with Normal (0,0,1) is null. b. n1  n2  0 , then from the first relation in Equation 16.27, we get 2 2  1   3  1  2 n12   0 . Therefore, n1   and from n12  n22  n32  1 , 2 2 we get n3   . Finally, the principal shear stress planes are given 2  2 2 with normal vectors ni    , 0,   . Using these values, and after 2   2 1 plugging them into Equation 16.25, we get     1   3  . Note that 2 π the unit vector makes an angle of with the coordinate axes 1-3. 4 c. n2  n1  0 . Then, from the second relation in Equation 16.27, we get 2 2  2   3  1  2 n22   0 . Therefore, n2   and from n12  n22  n32  1 , 2 2 we get n3   . Finally, the principal shear stress planes are given 2

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Mechanical Stress Transformation: Analytical and Mohr’s Circle Methods • 175

 2 2 with normal vectors ni   0,  ,  . Using these values, and after 2 2   1 plugging them into Equation 16.25, we get     2   3  . Note that 2 π the unit vector makes an angle of with the coordinate axes 2-3. 4

We eliminate n2 from Equation 16.25 using n12  n22  n32  1 to get

 2   12   22  n12   22   32  n32   22   1   2 )n12   3   2  n32   2  16.28 2



 2  2 Now, to find the extremum value of τ , we have   0 , which n1 n3 yields, after some manipulations,

 

 

  1   2  n1  1   2   2  1   2  n12   3   2  n32   0     2 2  2   3  n3  3   2   2  1   2  n1   3   2  n3   0 16.29

In order to have the relations given by Equation 16.27 satisfied, following cases are considered:

d. n= n= 0 , then from n12  n22  n32  1 , we get n2 = 1. Using these values, 1 3 and after plugging them into Equation 16.25, we get   0. This confirms that the shear stress acting on the principal plane with the Normal (0,1,0) is null. e. n1  n3  0 , then from the first relation in Equation 16.29, we get 2 2  1   2  1  2 n12   0 . Therefore, n1   and from n12  n22  n32  1 , 2 2 we get n2   . Finally, the principal shear stress planes are given 2  2 2  with normal vectors ni    , , 0  . Using these values, and after 2   2 1 plugging them into Equation 16.25, we get     1   2  . Note that 2 π the unit vector makes an angle of with the coordinate axes 1-2. 4

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f. n3  n1  0 , then from the second relation in Equation 16.29, we get 2 2  2   3  1  2 n32   0 . Therefore, n3   and from n12  n22  n32  1 , 2 2 we get n2   . Finally, the principal shear stress planes are given 2  2 2 with normal vectors ni   0,  ,  . Using these values, and after 2 2   1 plugging them into Equation 16.25, we get     2   3  . Note that 2 π the unit vector makes an angle of with the coordinate axes 2-3. 4

Collecting the results obtained for maximum shear stress, we have 1   12  2  1   2  1   23   2   3 2  1  16.30  31  2  3   1

16.4.3 Principal Stresses for a 3D Stress Tensor: Numerical Example 7 5 0    A stress tensor is given as  5 2 0  . Calculate the principal stresses using 0 0 3  the eigenvalues of the matrix given. Solution: We calculate the stress invariants, first: I1 = 7+2-3=6, I2=–6–21–11=–38, and I3=7(–6)–5(–15)=33. The characteristic equation is  3  6 2  38  33  0 . 95 5  10.09 , 2 95 5 2   2   1.09 , and 3   3  3 . These are the principal stresses 2 for the given stress state. This

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equation

has

three

real

roots;

1   1 

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Mechanical Stress Transformation: Analytical and Mohr’s Circle Methods • 177

16.5 PRINCIPAL DIRECTIONS: EIGENVECTORS For each eigenvalue, we can calculate its corresponding plane normal vector using Equation 16.21. The system of equations, along with the condition for unit normal, or nini = 1 are solvable for non-trivial solution. To elaborate the calculation steps, we write Equation 16.21 in matrix form for a 3 × 3 symmetric tensor, or

 12  13   n1i   11  i 2 2 2     22  i  23   n2i   0 and  n1i   n2i   n3i   1   12   13  23  33  i   n3i       16.31 Aij

where n1i is the x-component, n2i the y-component, and n3i the z-­component of unit vector representing the principal plane, where principal stress component λi acts. The superscript (for nij ) connects to the eigenvalue and the subscript connects to the vector component. Similar equations can be obtained for the remaining eigenvalues. It should be mentioned that one of the equations resulting from working out Equation 16.31 is redundant and hence only two of these equations are independent. Therefore, extra information, i.e.,  n1i    n2i    n3i   1 , is required to find a definite solution. 2

2

2

A procedure can be developed to make the eigenvector calculations [39], [40] more systematic and less tedious. This is based on the properties of the unit normal vector and application of Cramer’s rule for solving Equation 16.31. This procedure is as follows: 1. Form Matrix Aij, using Equation 16.31, for each value of λi . i.e., three systems of equations for three eigenvalues or principal stresses. 2. Calculate the cofactors of the determinant of Aij (or coefficient matrix from Equation 16.31) on the elements of the first row; designated by ai, bi, ci for i = 1,2,3. Note that no summation on indices is required and these quantities are merely scalars:

ai 

 22  i  23  12  23  12  22  i , bi   , ci   23  33  i  13  33  i  13  23

16.32

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178 • Tensor Analysis for Engineers, 3E

3. Calculate K i   ai2  bi2  ci2 

0.5

4. The components of eigenvector can be calculated, using the following relations; n1i  ai  K i , n2i  bi  K i , n3i  ci  K i 16.33 5. Repeat the calculation, i.e., steps 2-4, for remaining eigenvalues, λi ’ s. 16.5.1 Principal Directions for a 3D Stress Tensor: Numerical Example Calculate the eigenvectors for the stress tensor given in the previous example. Solution: We follow the above-mentioned steps. For 1  10.09 Mpa , we 5 0   3.09 8.09 0    105.8981 , 8.09 get Aij   5 0  . Then a1  0 13.09  0 13.09  0 5 0 b1    65.45 , c1 = 0. Then K1 = (105.89812 + 65.452)–0.5 0 13.09 = 0.008033. Therefore, n11 = nx = 105.8981 × 0.008033 = 0.8506, n21 = ny = 65.45 × 0.008033 = 0.52574, n31 = nz = 0. This gives θ1 = cos–1 0.8506 = 31.72°. Similarly, for 2  1.09 Mpa and 3  3 Mpa the corresponding eigenvectors can be calculated and are shown in Table 16.4. TABLE 16.4  Results for the eigenvalues and corresponding eigenvectors given in numerical examples.

Principal stress Principal direction

λ1

= 10.09 Mpa

λ2

= –1.09 Mpa –0.522

λ3

= –3 Mpa

nx

0.8506

ny

0.52574

0.8529

0 0

nz

0

0

1

b= c= K= 0, Note that for 3  3 Mpa , the process breaks down, since a= 3 3 3 3 but since the unit vector constraints gives nz = 1, we can calculate 3   33  3 .

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Mechanical Stress Transformation: Analytical and Mohr’s Circle Methods • 179

16.5.2 Eigenvalues and Eigenvectors of an Anti-Symmetric Stress Tensor: Numerical Example Calculate the principal stresses and direction for the anti-symmetric stress    3 2 1    1 4  tensor, 3  .  4 5 2   7 5  3  Answer: Use the procedure/steps to solve the problem, as in the previous example. Principal stress Principal direction

 1  0.521 Mpa

 2  1.727 Mpa

–0.04

 3  5.502 Mpa

nx

0.0833

0.441

ny

0.5309

0.4263

–0.106

nz

0.8428

0.904

0.891

16.5.3 Normal and Shear Stresses on a Specified Plane: Numerical Example 1  12 3 5    For a given 3D stress  3 7 10  MPa, find the principal stresses. Also  5 10 11 find the normal and shear stress components on a plane with the unit normal 1 1,1,1 with reference in the x–y–z coordinate system. 3 Solution: The stress invariants are I1  12  7  11  8 , I2 = (12 × 7 – 9) + (–12 × 11 – 25) + (–7 × 11 – 100) = –259, and I3  12

7 10 3 10 3 7 3 5 10 11 5 11 5 10

= – 2500. Therefore, the characteristic equation (see Equation 16.23) is  3  8 2  259  2500  0 . The three solutions for this equation are the principal stresses,  1  13.415, 2  11.21 , and  3  16.625 . Figure 16.16 shows the 3D Mohr’s diagram for this example.

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FIGURE 16.16  Mohr’s 3-circle diagram.

 Now, we calculate the components of the traction force, T on the given 1 plane with the unit normal  n1 , n2 , n3   1,1,1 . Using Equation 16.20, 3 1 14 1 14 these components are T1  12  3  5   , T2   3  7  10   , 3 3 3 3 1 4 and T3   5  10  11   . Hence, the magnitude of the traction force 3 3  408 is T  T12  T22  T32  . 3 The magnitude of the normal component of the traction force (see 14 1 14 1 4 1 32 Equation 16.20) is N  Ti ni        . 3 3 3 3 3 3 3 The magnitude of the shear stress on the given plane (see Equation 16.19) is 2



408  32      4.714 . 3  3 

Table 16.5 summarizes the calculations.

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TABLE 16.5  Summary of calculation results for Example 1.

Traction com- Plane’s normal, ponents, Ti components ni

Ti 2

Tini

normal force 10.67 N

14 3

1 3

65.33

4.67

shear force 4.714 N

14 3

1 3

65.33

4.67

 12 3 5     ij   3 7 10   5 10 11

4 3

1 3

5.33

1.33

Sum=136

Ti   ij n j

Sum=10.67

16.5.4 Normal and Shear Stresses on a Specified Plane: Numerical Example 2  19 4.7 6.45    For a given 3D stress  4.7 4.6 11.8  MPa, find the principal stresses.  6.45 11.8 8.3  Also find the normal and shear stress components on a plane with the unit 1 normal 1,1, 0  . 2 Solution: The stress invariants are I1  19  4.6  8.3  22.7 , I2 = (–19 × 4.6 – 4.72) + (19 × 8.3 – 6.452) + (–4.6 × 8.3 – 11.82) = – 170.8125, and I3  19

4.6 11.8 4.7 11.8 4.7 4.6  4.7  6.45  2647.5215 . Therefore, 11.8 8.3 6.45 8.3 6.45 11.8

the characteristic equation (see Equation 16.23) is λ 3 + 22.7 λ 2 – 170.8125 λ – 2647.5215 = 0. The three solutions for this equation are the principal stresses,  1  11.6178,  2  9.00151, and  3  25.3163 . Figure 16.17 shows the 3D Mohr’s diagram for this example.

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FIGURE 16.17  3D Mohr’s diagram.

 Now, we calculate the components of the traction force, T on the given 1 plane with the unit normal  n1 , n2 , n3   1,1, 0  . Using Equation 16.20, 2 1 23.7 1 these components are T1  , T2   19  4.7    4.7  4.6  2 2 2 1 18.25 0.1 , and T3  . Hence, the magnitude of the   6.45  11.8   2 2 2  traction force is T  T12  T22  T32  447.38125  21.1514 . The magnitude of the normal component of the traction force (see 23.7 0.1 Equation 16.20) is N  Ti ni    11.9 . 2 2 The magnitude of the shear stress on the given plane (see Equation 16.19) is

  613.9125  11.9   21.7325 . 2

Table 16.6 summarizes the calculations.

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TABLE 16.6  Summary of calculation results for Example 2.

Traction Plane’s norcompo- mal components Ti nents ni

Ti 2

−23.7 2

1 2

280.845

−0.1 2

1 2

0.005

18.25 2

0

Tini

Normal force –11.9 N

−23.7 2

shear force 21.7325 N

−

0.1 2

333.0625

0

Sum=613.9125

Sum=141.61

 19 4.7 6.45     ij   4.7 4.6 11.8   6.45 11.8 8.3  Ti   ij n j

16.6 OCTAHEDRAL STRESSES IN PRINCIPAL COORDINATE SYSTEM Considering a coordinate system aligned with the principal directions (i.e., x-axis is aligned with the Normal to the plane upon which σ 1 acting, and so forth), we define a plane with an equal inclination with the positive x,y, and z axes. This is the octahedral plane, since eight octahedral planes form an octahedron with reference to the principal axes/directions (see Figure 16.18).

FIGURE 16.18  Octahedral planes form a principal coordinate system.

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184 • Tensor Analysis for Engineers, 3E We consider, for example and for this analysis, the plane with the normal unit 1 vector 1,1,1 . We would like to calculate the normal and shear stress com3 ponents on this plane when the stress tensor is given, in terms of the principal  1 0 0    stresses, as  0  2 0  . Therefore, the traction force components acting on  0 0  3  1 1 1  1 , T2   3 . The  2 , and T3  3 3 3 normal component of the traction force, using Equation 16.19, is 1 N  N oct   1   2   3  and the shear component is τ = τoct = 3 the plane, using Equation 20, are T1 

1 1 2  1   22   32   19  1   2   3 2  3 3

 1   2 

2

  2   3    3   1  . 2

2

Rewriting the N and τ components in terms of the stress invariants I1   1   2   3 (i.e., the trace of the stress tensor) and I2   1 2   1 3   2 3 (i.e., the diagonal minors of the stress tensor) gives, after some manipulations, 1 1   N oct  3  1   2   3   3 I1    1    2     2     2  2 I 2  3 I 1 2 1 2 2 3 3 1  oct 3 3 16.34 Equation 16.34 clearly shows that octahedral stresses are invariants of the stress tensor, since I1 and I2 are also invariants. The octahedral shear stress τ oct is the maximum shear stress on the octahedral plane, but not the absolute maximum shear stress that can occur in the medium. Using Equation 16.30 and Equation 16.34, we can relate the octahedral shear stress with the maximum shear values, after substitutions, as 2 2  oct   12   232   312 3 16.35 Please recall that, in Equation 16.35, τ 12 is the absolute value of the maximum shear stress in 1–2 planes, and so forth.

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Noteworthy: Normal stresses of magnitude N oct act on all eight octahedral planes. Therefore, if they are acting by themselves alone, they tend to compress or enlarge the octahedron but not distort it. For this reason, they are also called hydrostatic. „ Shear stresses of magnitude τ oct act on all eight octahedral planes. Therefore, if they are acting by themselves alone, they tend to distort the octahedron without changing its volume. „ The octahedral shear stress is smaller than the maximum/principal shear stress. However, it is influenced by all three principal shear stresses (see Equation 16.35). „ The octahedral shear stress is considered a criterion for predicting the yielding of a stressed material, for example, von Mises yield criterion.5  x  xy  xz    Rewriting Equation 16.34 in terms of    y  yz  components, using     z  „

Equation 16.24, gives 1   N oct  3  x   y   z  2 2 2   1      y   z   z   x   6  xy2   x2z   yz2 oct x y  3 16.36







 







which gives the octahedral stresses in terms of the stress components given at a point in an arbitrary coordinate system. 16.6.1 Octahedral Stresses: Numerical Example  50 20 10    For the stress tensor given as  20 100 60  MPa , find the octahedral nor10 60 50  mal and shear stress components. Consider the octahedral plane with the 1 unit normal 1,1,1 , defined in the principal coordinate system. 3   Richard Edler von Mises (1883-1953).

5

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Solution: We can find the principal stresses and then calculate the octahedral stress components. However, we can use Equation 16.34 in conjunction with the stress matrix. Note that these two methods are equivalent, since the stress invariants are independent of the coordinates’ rotation. The trace of the 50 20 50 10  stress matrix is I1  50  100  50  200 , and I2  + 20 100 10 50 100 60 I1 200 1 = 8400 . Therefore, N oct= = = 66.667 and  oct  2 I12  6 I2 60 50 3 3 3 1 2 = 2  200   6  8400   57.349 . Readers may want to compare these 3 results against those obtained using the corresponding principal stresses  1  145.176 ,  2  45.0354 , and  3  9.7889 .

16.7 OCTAHEDRAL STRESSES AND DEVIATORIC STRESSES The octahedral stresses guide us to write the stress tensor in two parts: a hydrostatic part and a pure shear part. Following this guide, we subtract the N 0 0    normal stress N ij   0 N 0  , a diagonal matrix, from the symmetric  0 0 N   x  xy  xz    stress tensor  ij     y  yz  and write it down, in index notation, as     z   I Sij   ij  N ij . But N ij  1  ij  kk  ij (see Equation 16.36). Therefore, we 3 3 have

 xy  xz   x  N 1  Sij   ij   ij kk    y  N  yz  3     z  N  16.37

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Mechanical Stress Transformation: Analytical and Mohr’s Circle Methods • 187

where Sij is called the deviatoric stress, a pure shear stress by definition. Note that the trace (or its first invariant J1) of the deviatoric stress is null, or 1 Sii   ii   ii  kk   ii   kk  0 . The second invariant J2 and third invariant 3 3 J3 can be found [41] in a similar way described previously using the Lagrange multiplier method. The results, in terms of the principal values S1, S2, and S3 are given as   J1  Sii  S1  S2  S3  0  1 1 2  2 2  J2  Sij S ji   S1  S2  S3  2 2  1   J3  3 Sij S jk Ski  S1S2 S3 16.38 It is useful to have the invariants of the deviatoric stress in terms of the invariants of the stress tensor itself, or   J1  0  1 2   J2   I1  3 I2  3  1  3  J3  27  2 I1  9 I1 I2  27 I3  16.39 Using the invariants J1, J2, and J3, we can write the characteristic equation for the deviatoric stress tensor in terms of its principal variable s, as s3  J2 s  J3  0 16.40 The solutions obtained by solving Equation 16.40 are the principal deviatoric stresses S1, S2, and S3.

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16.8 VON MISES YIELD CRITERION VS OCTAHEDRAL SHEAR STRESS In the engineering mechanics of materials, several failure theories [42] are used for defining the required criteria for materials’ yielding and failure. Usually, yielding, or yield strength, is defined and considered for ductile materials, and ultimate strength is for brittle materials. These theories, in general, are based on measuring the maximum state of stress developed in the physical domain of a material against unidirectional tensile data, as the defined criterion. For example, Tresca [43] defined the absolute maximum shear stress against that of shear stress when the same material yields under uniaxial tension. Similarly, von Mises [44] defined the maximum-distortionenergy theory. The theory developed by von Mises (and its related versions) is widely used and assumes that yielding will occur when the second invariant of the deviatoric stress tensor, J2 reaches a critical value. Later, Nadai [45] suggested that materials yield when the octahedral shear stress reaches a critical value. These developments suggest that having a relationship between octahedral shear stress, Equation 16.34, and the second deviatoric stress invariant, Equation 16.39, would be useful. Therefore, we can write 2 2 2 I1  3 I2  3 J2 . The scalar quantity 3 3 von Mises yield measuring stress, or

 oct 

3 J2 is defined as the

3  vm  3 J2   oct 2 16.41 When σ vm reaches the yield strength of the material at hand, σ y is measured in the uniaxial tension, and then the material fails/yields. Or

 vm  3 J2   y In order to have an explicit relation for σ vm , we use Equation 16.41 and Equation 16.36 to write the von Mises yield criterion in terms of the stress tensor components:

 vm 



x

y

   2

y

z



2

  z   x 

2



 3  xy2   xz2   yz2



2 16.42

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Mechanical Stress Transformation: Analytical and Mohr’s Circle Methods • 189

Similarly, in terms of the principal stresses using Equation 16.34, we can write this as

 vm 

 1   2 

2

  2   3    3   1  2

2

2

16.43 For plane stress cases (i.e.,  z   xz   yz  0 ), the von Mises criterion reduces to the relation shown by Equation 16.44:

 vm   12   22   1 2   x2   y2   x y  3 xy2   y

 plane stress 

16.44 For pure torsion (i.e., all stress components are null, except  xy  0), the von Mises criterion reduces to the relation shown by Equation 16.45:

 vm  3 xy   y  pure torsion  16.45 Therefore, pure torsion yielding occurs when the shear stress τ xy reaches 1 = 57.7% of σ y . 3

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CHAPTER

17

THE WORKED EXAMPLES In this section, we present several worked-out examples related to the topics covered in the previous sections.

17.1 EXAMPLE: EINSTEIN SUMMATION CONVENTIONS Write out the expanded expression in full detail for N = 3 dimensions for,   ∂x′ j   = A dx i  ei − i e′ j  . ∂x   Solution:

  ∂x′ j   Summation applies on both indices i and j , hence A = dx i  ei − i e′ j  = ∂x   j  1  2  3      ∂x′  ∂x′ ∂x′ ∂x′     dx i ei − dx i i e′ j = dx1 e1 + dx2 e2 + dx3 e3 − dx1  1 e1′ + 1 e2′ + 1 e3′  − ∂x ∂x ∂x  ∂x   ∂x′1  ∂x′2  ∂x′3    ∂x′1  ∂x′2  ∂x′3   dx2  2 e1′ + 2 e2′ + 2 e3′  − dx3  3 e1′ + 3 e2′ + 3 e3′  . ∂x ∂x ∂x ∂x  ∂x   ∂x  1  2   ∂x′  ∂x′  ∂x′3   Re-arranging the terms, we receive A dx1  e1 − 1 e1′ − 1 e2′ − 1 e3′  + = ∂x ∂x ∂x     ∂x′1  ∂x′2  ∂x′3     ∂x′1  ∂x′2  ∂x′3   dx2  e2 − 2 e1′ − 2 e2′ − 2 e3′  + dx3  e3 − 3 e1′ − 3 e2′ − 3 e3′  . ∂x ∂x ∂x ∂x ∂x ∂x    

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192 • Tensor Analysis for Engineers, 3E

17.2 EXAMPLE: CONVERSION FROM VECTOR TO INDEX NOTATIONS Write out the following expressions written in vector form in index notation and expand for 3D coordinates.   2 a. ∇ ⋅ A   b. A ⋅ ∇ 2 A     c. A ⋅ ∇ ∇ ⋅ A

(

)

(

)

( (

))

  d. ∇ ⋅ A ( ∇ 2 Φ )  e. ∇ 2 A ⋅ ( ∇Φ )   f. A ⋅ ∇ ( ∇ 2 Φ )

(

)

( (

) )

Solution:   2 2 k a. ∇ ⋅ A= A j j A = A 1 + A22 + A 33 ) ′ ′ k ( ′1 ′ ′   2 = A A k= Ak ′⋅ jj A1 A1′ ⋅ jj + A2 A2 ′ ⋅ jj + A3= A3 ′⋅ jj A1 A1 ′⋅ 11 + A1 ′⋅ 22 + A1′ ⋅33 b. A ⋅ ∇

(

)

(

)

(

)

(

(

+ A2 A1′ ⋅ 11 + A1′ ⋅22 + A1′ ⋅33 + A3 A1′ ⋅11 + A1′ 2⋅2 + A1′ ⋅33

(

)

    j k 3 k ⋅A A= A′kj A1 A′kk1 + A2 A′kk 2 + A= A′k 3 A1 A1′11 + A2′21 + A3′31 c. A ⋅ ∇ ∇=

( (

))

(

) (

+ A2 A1′11 + A2′21 + A3′31 + A3 A1′11 + A2′21 + A3′31

)

)

)

  k k k k 1 2 3 d. ∇ ⋅ A ( ∇ 2 Φ= ) A′j j ( ∇ kΦ )= ′ ⋅ A′1 ( ∇ k Φ ) ′ ⋅ + A′2 ( ∇ k Φ )′ ⋅ + A′3 ( ∇ k Φ )= ′⋅ 1 2 3 1 2 3 A ′11 ( ∇1 Φ )′ ⋅ + ( ∇ 2 Φ ) ′⋅ + ( ∇ 3 Φ )′ ⋅  + A ′22 ( ∇1 Φ ) ′⋅ + ( ∇ 2 Φ ) ′⋅ + ( ∇ 3 Φ ) ′⋅  +    

(

)

1 2 3 A ′33 ( ∇1 Φ ) ′⋅ + ( ∇ 2 Φ ) ′⋅ + ( ∇ 3 Φ ) ′⋅   

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The Worked Examples • 193

 e. ∇ 2 A ⋅ ( ∇Φ = )

(

) (A

= =

k

( ∇ k Φ ) )′ j ⋅ j

( A (∇ Φ ) + A (∇ Φ ) + A (∇ Φ ))′ ( A (∇ Φ ) + A (∇ Φ ) + A (∇ Φ ))′ + ( A (∇ Φ ) + A (∇ Φ ) + A (∇ Φ )) ′ + ( A (∇ Φ ) + A (∇ Φ ) + A (∇ Φ )) ′ 1

2

1

1

2

1

1 1

2

1

3

2

2

1

3

2

2 2

3 3

j

3

j⋅

3

1 1⋅

3

2 2⋅

3

3 3⋅

  k f. A ⋅ ∇ ( ∇ 2 Φ )= A j ( ∇ k Φ ) ⋅ j

(

)

′

= A1 ( ∇ k Φ )

k + A 3 ( ∇ k Φ ) ⋅3 ′⋅ 2 ′ 3 1  1 2  = A ( ∇1 Φ ) ⋅ 1 + ( ∇ 2 Φ ) ⋅ 1 + ( ∇ 3 Φ ) ⋅1  ′ ′ ′  1 2 3 2  + A ( ∇1 Φ ) ⋅2 + ( ∇ 2 Φ ) ⋅2 + ( ∇ 3 Φ ) ⋅2   ′ ′ ′  1 2 3 + A3 ( ∇1 Φ ) ⋅ 3 + ( ∇ 2 Φ ) ⋅ 3 + ( ∇ 3 Φ ) ⋅3   ′ ′ ′  k

′⋅ 1

+ A2 ( ∇ k Φ )

k

17.3 EXAMPLE: OBLIQUE RECTILINEAR COORDINATE SYSTEMS Consider a 2D oblique/slanted coordinate system, x i in which the coordinate axes are not at right angles. With reference to the Cartesian coordinate sys tem yi, as shown in Figure 17.1, for a given vector A find: a. Covariant and contravariant components; Ai and A i   b. Covariant and contravariant basis vectors; ei and e i with sketch on the coordinate system      i i c. Show that = A A= ei A= Ac1 E1 + Ac 2 E2 ; where E1 and E2 are ie ­Cartesian unit vectors   d. Verify that e i ⋅ e j = d ji and sketch all basis vectors e. Metric tensors     f. Unit vectors ei / ei , or scale factor, and e i / e i

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194 • Tensor Analysis for Engineers, 3E

FIGURE 17.1  Oblique coordinate system x i and Cartesian coordinate system y i.

Solution: Assume axes x1 and x2 make angles a and b with Cartesian axis y1 , respectively. Therefore, we can write the following functional relations between the oblique and Cartesian systems:  y1 x1 cos a + x2 cos b = , or inversely  2 1 2 =  y x sin a + x sin b

1 − csc (a − b )  y1 sin b − y2 cos b   x =  2 csc (a − b )  y1 sin a − y2 cos a   x =

 a. Covariant components of A are obtained by drawing perpendicular

lines to the coordinate axes. We have Ai =

∂y j ∂y1 A or A1 Ac1 + cj, = ∂x i ∂x1

∂y2 ∂y1 ∂y2 = A cos a A + sin a A . Similarly, A A Ac 2 = cos b Ac1 + = + 2 1 2 c c c 2 c1 ∂x1 ∂x2 ∂x2  sin b Ac 2 . The contravariant components of A are obtained by d ­ rawing parallel lines to the coordinate axes. We have A i =

∂x i Acj , or A1 = j ∂y

∂x1 ∂x1 A Ac 2 = + − sin b csc (a − b ) Ac1 + cos b csc (a − b ) Ac 2. Similarly, c1 ∂y1 ∂y2

∂x2 ∂x2 A Ac 2 = sin a csc (a − b ) Ac1 − cos a csc (a − b ) Ac 2 . Note + c 1 ∂y1 ∂y2 that for the Cartesian system both covariant and contravariant components are identical. A2 =

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The Worked Examples • 195

 ∂y  b. The covariant basis vectors for oblique system are ei = i E j , ∂x 1 2 1     ∂ ∂ y y   ∂y  or e1 = 1 E1 + 1 E2 = cos a E1 + sin a E2. Similarly,= e2 E1 + ∂x ∂x ∂x2 j

  ∂y2   ∂x i  = E2 cos b E1 + sin b E2. The contravariant basis vectors are e i = j E j, 2 ∂y ∂x

  ∂x1  ∂x1   e1 = E E sin csc E cos csc E b a b b a b + = − − + − ( ) ( ) 1 2 1 2. ∂y1 ∂y2   ∂x2  ∂x2   Similarly, e 2 = E1 + 2 E2 = sin a csc (a − b ) E1 − cos a csc (a − b ) E2 . 1 ∂y ∂y Note that for the Cartesian system both covariant and contravariant basis vectors are identical and are unit vectors, i.e. Ei = 1 .     c. A i ei = A1 e1 + A2 e2 = [− sin b csc (a − b ) Ac1 + cos b csc (a − b ) Ac 2 ][cos a E1 +    sin a E2 ] + [sin a csc (a − b ) Ac1 − cos a csc (a − b ) Ac 2 ][cos b E1 + sin b E2 ].  After some manipulations and rearranging the terms, we find A i ei =   i 1 2 Ac1 E1 + Ac 2 E2 . Similarly, Ai e = A1 e + A2 e = [cos a Ac1 + sin a Ac 2 ]   [− sin b csc (a − b ) E1 + cos b csc (a − b ) E2 ] + [cos b Ac1 + sin b Ac 2 ][sin a   csc (a − b ) E1 − cos a csc (a − b ) E2 ]. After some manipulations and    ­rearranging the terms, we find= Ai e i Ac1 E1 + Ac 2 E2. or

d. Dot-product of covariant and contravariant basis vectors can cos a sin a sin a cos a     now be written as e1 ⋅ e 2 = e 2 ⋅ e1 = − = 0 and sin (a − b ) sin (a − b ) cos b sin b sin b cos b   1  e2 ⋅ e 1 = e ⋅ e2 = − + = 0 . These results consin (a − b ) sin (a − b )     firm that e1 ⊥ e 2 and e2 ⊥ e 1 , as shown in Figure 17.2. Also, we can cos b sin a cos a sin b sin a cos b     write e1 ⋅ e 1 = e2 ⋅ e 2 − − + = 1 and = sin (a − b ) sin (a − b ) sin (a − b ) sin b cos a   d ji . = 1. These results verify that e i ⋅ e j = sin (a − b )

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196 • Tensor Analysis for Engineers, 3E

FIGURE 17.2  Covariant and contravariant basis vectors in oblique coordinate systems.

e. The

doubly

covariant

     e1 ⋅ e1 e1 ⋅ e2  metric = tensor gij =       e2 ⋅ e1 e2 ⋅ e2 

1 cos (a − b )    . We can calculate doubly contravariant metric 1 cos (a − b )       e1 ⋅ e1 e1 ⋅ e 2  ij = g   2  1 = tensor by inverting the gij , or using csc 2 (a − b ) 2 2  ⋅ ⋅ e e e e   1 − cos (a − b )   i   . These results confirm that x is a non− cos a − b 1 ( )   orthogonal system, since off-diagonal elements of metric tensors are not equal to zero for all values of angles a and b with the condition that p (a − b ) ≠ ( 2 k + 1) . where, k is an integer. 2    e1 f. Unit vectors along the covariant basis vectors are; = cos a E1 + sin a E2  e1    e2 and = cos b E1 + sin b E2. The unit vectors along the contravariant basis  e2       e1 e1 e2 e2 = − sin b E + cos b E vectors are;  1 = and = =  1 2     e e2 e1 ⋅ e1 e2 ⋅ e2   sin a E1 − cos a E2 .

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The Worked Examples • 197

17.4 EXAMPLE: QUANTITIES RELATED TO PARABOLIC COORDINATE SYSTEM Consider a parabolic coordinate system ( x1 , x2 , x3 ) ≡ (x ,h ,q ) given in terms of Cartesian ( y1 , y2 , y3 ) ≡ ( X , Y, Z ), as shown in Figure 17.3. The functional relations are

FIGURE 17.3  Parabolic coordinate system.

 X = xh cosq     . Find the basis vectors e1, e2, e3 for the parabolic coordi Y = xh sin q  Z (x 2 − h 2 ) / 2 = nate system in terms of the Cartesian unit vectors. Also find the scale factors, unit vectors, metric tensors (covariant and contravariant), Jacobian, volume element, and Christoffel symbols of the 2nd kind. Solution:

j  ∂y     The covariant basis vector ( ex , eh , eq ) reads, using ek = k E j ∂x      ex = h cosq E1 + h sin q E2 + x E3       eh = x cosq E1 + x sin q E2 − h E3    −xh sin q E1 + xh cosq E2  eq =

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198 • Tensor Analysis for Engineers, 3E   e= x 2 + h 2 , h= e= The scale factors are h= x x h h   The unit vectors are, using e ( k ) = ek / hk ,

x 2 + h 2 , and hq = xh.

    e (x ) = h cosq E1 + h sin q E2 + x E3 / x 2 + h 2      2 2  e (h ) = x cosq E1 + x sin q E2 − h E3 / x + h    − sin q E1 + cosq E2  e (q ) = 

( (

Covariant

metric

) )

tensor

reads,

using

  gij= ei ⋅ e j ,

x 2 + h 2 0 0    2 2 = gij  0 x +h 0 . Therefore, the parabolic coordinate sys  0 (xh )2   0 tem is an orthogonal one, since metric tensor is a diagonal matrix with null off-diagonal elements. The contravariant metric tensor is the inverse (x 2 + h 2 )−1 0 0    −1 ij 2 2 −1  = g = gij 0 0 . of the covariant one, or (x + h )   −2   0 0 (xh )    g Jacobian is the square root of the determinant of ij , or the product of all scale

( )

factors,= 

= gij xh (x 2 + h 2 ) . The volume element is d   d d d 

g mk  ∂g jk ∂gik ∂gij  + j − k  ∂x ∂x  2  ∂x i g11  ∂g11  x = Γ111 = to calculate the Christoffel symbols. Therefore, ,   2 2  ∂x  x + h 2

  2   2  d d d . We use Equation 10.20,= or Γ mij

= Γ112

g11  ∂g11  h = ,   2 2  ∂h  x + h 2

2 = Γ12

g 22  ∂g22  x = ,   2 2  ∂x  x + h 2

TAE3E.CH17_1pp.indd 198

 ∂g11  h  − ∂h  =− x 2 + h 2 ,   33 g  ∂g33  1 3 2 3 3 Γ13 = , Γ13 = Γ11 = Γ12 =, 0 = 2  ∂x  x Γ113 = 0,

g 22 2 Γ11 = 2

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The Worked Examples • 199

 ∂g22  x − 2 ,  − ∂x  = x +h 2   g 22  ∂g22  h = Γ 222 = ,   2 2  ∂h  x + h 2 g11 Γ122 = 2

= Γ 323

Γ 223 = 0,

g 33  ∂g33  1 3 = , Γ 33 = 0. In matrix form, we have 2  ∂h  h

 Γxxx  Γx = Γhxx x  Γqx 

 Γhxx  Γh = Γhh x  Γhqx 

Γxxh Γhxh x Γqh

Γhxh Γhh h Γhqh

 x  2 x +h 2 x   Γxq   h Γhxq  =  2 2 x  x + h Γqq    0   h − 2 x +h 2 Γhxq   x   Γhh q  =  2 x +h 2 Γhqq    0  

 Γqxx  Γq = Γhq x  Γqqx 

TAE3E.CH17_1pp.indd 199

 ∂g33  xh 2 − = − ,  ∂x  x 2 +h 2   g 22  ∂g  hx2 Γ 233 =  − 33  = − 2 , x +h 2 2  ∂h  g11 Γ133 = 2

Γ123 = 0,

Γqxh Γhq h Γqqh

h x +h 2 x − 2 x +h 2 2

0 x x +h 2 h 2 x +h 2

 0 q Γxq     Γhq q  =  0 Γqqq   1 x 

2

0

0 0 1 h

    0   xh 2  − 2  x +h 2  0

    0   2 hx  − 2  x +h 2  0

1 x  1 h   0 

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200 • Tensor Analysis for Engineers, 3E

17.5 EXAMPLE: QUANTITIES RELATED TO BI-POLAR COORDINATE SYSTEMS Consider a bi-polar coordinate system ( x1 , x2 , x3 ) ≡ (x ,h , z ) given in terms

of Cartesian ( y1 , y2 , y3 ) ≡ ( X , Y, Z ) , as shown in Figure 17.4. The functional a sinhh   X = coshh − cos x  a sin x     relations are Y = . Find the basis vectors e1 , e2 , e3 for the coshh − cos x  Z = z   bi-polar coordinate system in terms of the Cartesian unit vectors. Also find the scale factors, unit vectors, metric tensors (covariant and contravariant), Jacobian, volume element, and Christoffel symbols of the 2nd kind.

FIGURE 17.4  Bipolar coordinates: x and h iso-surfaces (Foci are located at (−1, 0) and (1, 0)1.

 (https://commons.wikimedia.org/wiki/File:Bipolar_isosurfaces.png)

1

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The Worked Examples • 201

Solution:

j  ∂y     The covariant basis vector ( ex , eh , ez ) reads, using ek = k E j ∂x

 − a sinhh sin x  a ( coshh cos x − 1)  E1 + E2 =  ex 2 2 cosh h cos x cosh h cos x − − ( ) ( )    a (1 − coshh cos x )  a sinhh sin x  E1 − E2 =  eh 2 ( coshh − cos x )2 ( coshh − cos x )     ez = E3   e= The scale factors, after simplification, are h= x x a coshh − cos x

a , h= e= h coshh − cos x h   , and hz = 1. The unit vectors are, using e ( k ) = ek / hk,

− sinhh sin x  coshh cos x − 1   e x E1 + E2 = ( )  coshh − cos x coshh − cos x  1 − coshh cos x  sinhh sin x   = E1 − E2  e (h ) coshh − cos x coshh − cos x    e ( z ) = E 3   The

covariant

 a2  2  ( coshh − cos x )  0 gij =    0  

metric

tensor 0 a2

( coshh − cos x )2 0

reads,

using

  gij= ei ⋅ e j ,

 0   0  . Therefore, the parabolic   1 

coordinate system is an orthogonal one, since metric tensor is a diagonal

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202 • Tensor Analysis for Engineers, 3E matrix with null off-diagonal elements. The contravariant metric tensor is the inverse of the covariant one, or  ( coshh − cos x )2  0 0  2 a   2   −1 ( coshh − cos x )   . Jacobian is the g ij = gij = 0 0 a2    0 0 1     square root of the determinant of gij , or the product of all scale factors, a2 dx dh dq . The volume element= is d  = =  = gij ( coshh − cos x )2

( )

a2

( coshh − cos x )2

dx dh dz. Similarly, as given in Example 17.5, the Christoffel

symbols can be calculated as

 Γxxx  Γx = Γhxx  Γxzx 

 Γhxx  Γh = Γhh x  Γhzx 

Γxxh Γhxh Γxzh

Γhxh Γhh h Γhzh

sin x   cos x − coshh Γxx z   sinhh  Γhx z  =  cos x − coshh Γxzz   0  

sinhh cos x − coshh − sin x cos x − coshh 0

 0  0   0 

 − sinhh  cos x − coshh Γhx z   sin x  Γhh z  =  cos x − coshh Γhzz   0  

sin x cos x − coshh sinhh cos x − coshh 0

 0  0   0 

Γ z =[ 0 ]

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The Worked Examples • 203

17.6 EXAMPLE: APPLICATION OF CONTRAVARIANT METRIC TENSORS Using the property of contravariant metric tensors, show that the following relation holds, g ile ljk A j Bk = e ijk A j Bk. Solution: Knowing that metric tensors replace repeated indices with the free index while raising (contravariant metric tensor) or lowering (covariant metric tensor) them. We can write the expression in the L.H.S of the above relation as, g ile ljk A j Bk =  jki A j Bk. But we  can write A j = g jn An and Bk = g km Bm , substituting, gives g ile ljk A j Bk =  jn jn km i km i in km inm  jk g= An g Bm  jk g= g An Bm = AB  An Bm , after changing k g   n m  dummy indices, i.e., n → j and m → k, we get  inm An Bm =  ijk A j Bk. Therefore, the given relation holds.

17.7 EXAMPLE: DOT AND CROSS PRODUCTS IN CYLINDRICAL AND SPHERICAL COORDINATES   Given that A ( i ) and B ( i ) are the physical components of vectors A and B, respectively, write  down the expressions   for physical components of the cross-product, A × B, and dot-product A ⋅ B in terms of A ( i ) and B ( i ), for the following coordinate systems (assume a 3D space, N = 3 ):

a. Cylindrical polar ( x1 , x2 , x3 ) ≡ ( r ,q , z ) b. Spherical polar ( x1 , x2 , x3 ) ≡ ( r ,j ,q ) Solution:

  The dot-product and cross-product relations are A ⋅ B = A i Bi, C i =  ijk A j Bk (contravariant component). We use the relations between covariant/­ contravariant and physical components, or Ai = hi A ( i ) and A i = A ( i ) / hi .

a. For cylindrical coordinates, we have the scale factors as ( h1 , h2 , h3 ) = (1, r,1) and the Jacobian is  = r (see Section 11.6.1). We = B1 h= B (= 1 ) , B2 h= rB ( 2= can write B(3). ) , B3 h= 1 B (1 ) 2B(2) 3B(3)

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204 • Tensor Analysis for Engineers, 3E A(2) A(2) A (1 ) A(3) ,= A2 = = A (1 ) , = A3 = A ( 3 ) . h2 r h1 h3 Therefore, the dot-product reads   A⋅B = A i B=i A1 B1 + A2 B2 + A3 B=3 A (1 ) B (1 ) + A ( 2 ) B ( 2 ) + A ( 3 ) B ( 3 ) , a scalar quantity. The physical component of the cross-product i hi ijk A j Bk, no-summation on index i. Expanding is C= ( i ) h= iC e1 jk 1 the expression gives C= h1 1 jk A= A= (1) h= 1C j Bk j Bk  1 A ( 2 ) B ( 3 ) − A ( 3 ) B ( 2 ), and C ( 2 ) =  h2 A ( 2 ) h3 B ( 3 ) − h3 A ( 3 ) h2 B ( 2 )  = r e2 jk h2 C2 = r 2 jk A j Bk = r A j Bk =  h3 A ( 3 ) h1 B (1 ) − h1 A (1 ) h3 B ( 3 )  =  e3 jk 3 jk A ( 3 ) B (1 ) − A (1 ) B ( 3 ) , and C (= 3 ) = h3 C3 h= A j Bk A j Bk = 3  1 A (1 ) B ( 2 ) − A ( 2 ) B (1 ) .  h1 A (1 ) h2 B ( 2 ) − h2 A ( 2 ) h1 B (1 )  = r

Similarly, = A1

b. For spherical coordinates, we have the scale factors as ( h1 , h2 , h3 ) = (1, r, r sin j ) and the Jacobian is  = r 2 sin j , see section 11.6.1. We can write = B1 h= B (1= rB ( 2= r sin(ϕ)B(3). ) , B2 h= ) , B3 h= 1 B (1 ) 2B(2) 3B(3) A (1 ) A(2) A(2) A(3) A(3) = , A2 = , A3 = . = A (1 ) = h3 r sin(j ) h1 h2 r Therefore, the dot-product reads   A⋅B = A i B=i A1 B1 + A2 B2 + A3 B= A (1 ) B (1 ) + A ( 2 ) B ( 2 ) + A ( 3 ) B ( 3 ) , 3 a scalar quantity.

Similarly,= A1

i The physical component of the cross-product is C= hi ijk A j Bk , ( i ) h= iC 1 no-summation on index i. Expanding the expression gives C= (1) h= 1C e1 jk 1 h1 1 jk A j Bk = A j Bk =  h2 A ( 2 ) h3 B ( 3 ) − h3 A ( 3 ) h2 B ( 2 )  = 2  r sin j 

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The Worked Examples • 205

A (2) B(3) − A (3) B(2),

and

2 C= r 2 jk A= r ( 2 ) h= 2C j Bk

e2 jk 

A= j Bk

A ( 3 ) B (1 ) − A (1 ) B ( 3 ) , and C ( 3 ) =  h3 A ( 3 ) h1 B (1 ) − h1 A (1 ) h3 B ( 3 )  = e3 jk 1 h3 C3 = h3  3 jk A j Bk = r sin j A j Bk =  h1 A (1 ) h2 B ( 2 ) − h2 A ( 2 ) h1 B (1 )  = r  A (1 ) B ( 2 ) − A ( 2 ) B (1 ) .

17.8 EXAMPLE: RELATION BETWEEN JACOBIAN AND METRIC TENSOR DETERMINANTS Show that determinant g of metric tensor gij is equal to the square of Jacobian  , in a coordinate system, when measured against Cartesian system. Solution: Let’s consider two arbitrary system x i and x′i . We can write the metric ten∂x k ∂x l gkl , since it is a doubly covariant tensor. ∂x′i ∂x′ j ∂x k ∂x k ∂x l g Taking determinant of the both sides gives g′ij = . But kl ∂x′i ∂x′i ∂x′ j

sor transformation as g′ij =

∂x l is the Jacobian of transformation, hence g′ij =  2 gkl . Rewriting j ∂x′ g′ the expression and using gkl = g and g′ij = g′ , we get g′ =  2 g , or  = . g

or

Now if the original x i system is the Cartesian system, then gcartesian = 1 and we receive g′ =  2.

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17.9 EXAMPLE: DETERMINANT OF METRIC TENSORS USING DISPLACEMENT VECTORS For an arbitrary coordinate system x i  ( , ,  ) the magnitude of displace­ ment vector is given by ds2 = 5 dx 2 + 3 dh 2 + 7 dz 2 − 4 dx dh − 8 dx dz + 2 dh dz . Find g and g ij (i.e., the determinant of covariant metric tensor and the contravariant metric tensor). Solution: We have ds2 = gij dx i dx j, expanding for the given coordinate system and using gij = g ji, gives ds2 = g11 dx 2 + g22 dh 2 + g33 dz 2 + 2 g12 dx dh + 2 g13 dx dz + 2g23 dh dz . Therefore, after comparison we receive g11 = 5 , g22 = 3 , g33 = 7 , g12 = −2 , g13 = −4 , and g23 = 1 . The matrix form of covariant metric tensor  5 −2 −4  is gij =  −2 3 1  . Determinant of gij = g =5 ( 21 − 1 ) + 2 ( −14 + 4 ) −  −4 1 7  4 ( −2 + 12 ) =40. Note that the given system is not orthogonal, since off-­

diagonal elements of gij are not zero. For contravariant metric tensor, we find  5 −2 −4   5 −2 −4      ij the inverse of gij =  −2 3 1  , or g =  −2 3 1   −4 1 7   −4 1 7 

ij This is calculated using the formula g=

−1

1 1 1 2 4 4    1 19 3  . =  4 40 40   1 3 11     4 40 40 

Cij cofactor of gij , for = g determinante of gij

3 1 5 −2 − −4 1 C23 C11 1 7 21 − 1 −5 + 8 3 23 11 ­example; g= , = = = = = = 20, g= 40 40 40 40 40 40 40 etc.

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17.10 EXAMPLE: DETERMINANT OF A 4 × 4 MATRIX USING PERMUTATION SYMBOLS  5 8 15 21  4 10 13 11  , using Equation 9.13. Calculate determinant of matrix M =   3 6 9 12    2 4 7 1  Solution: We expand based on row, hence M = eijkl M1 i M2 j M3 k M4 l. Expanding we get M = e1234M11M22M33M44 + e1243M11M22M34M43 + e1342M11M23M34M42 + e1324M11M23M32M44 + e1423M11M24M32M43 + e1432M11M24M33M42 + e2134M12M21M33M44 + e2143M12M21M34M43 + e2341M12M23M34M41 + e2314M12M23M31M44 + e2413M12M24M31M43 + e2431M12M24M33M41 + e3124M13M21M32M44 + e3142M13M21M34M42 + e3241M13M22M34M41 + e3214M13M22M31M44 + e3412M13M24M31M42 + e3421M13M24M32M41 + e4123M14M21M32M43 + e4132M14M21M33M42 + e4231M14M22M33M41 + e4213M14M22M31M43   +  e4312M14M23M31M42 +   e4321M14M23M32M41. Substituting for the values of the matrix elements and the permutation ­symbols (i.e., +1 for even and −1 for odd number of interchanges for indices), we receive M = 18. Note that the number of terms out of expansion is equal to 4! = 4 × 3 × 2 = 24, or in general N ! for an N × N matrix.

17.11  EXAMPLE: TIME DERIVATIVES OF THE JACOBIAN  1 ∂   Show that the time derivative of the Jacobian reads = ∇ ⋅ V, where V  ∂t is the velocity vector. Solution: Considering the transformation from coordinate system x i to that of x′i, we can write the Jacobian as the determinant of the transformation, using ∂x1 ∂x2 ∂x3 . Taking the time derivative, we have ∂x′k ∂x′n ∂x′l  ∂  ∂x1  ∂x2 ∂x3 ∂x1 ∂  ∂x2  ∂x3 ∂x1 ∂x2 ∂  ∂x3   ∂ = e knl  k  + k  n  l+ k  . l n  n l  ∂t  ∂x′  ∂t  ∂x′ ∂x′ ∂x′ ∂x′  ∂t  ∂x′ ∂x′ ∂x′ ∂x′  ∂t  

Equation 4.4, or  = e knl

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208 • Tensor Analysis for Engineers, 3E

But we can write

∂x1 ∂x2 ∂x3 = V 1, = V 2, = V 3, i.e., contravariant components ∂t ∂t ∂t

of velocity vector. Hence, using chain-rule or e

knl

∂x i ∂x i ∂x j we can write ∂ = = j n n ∂x′ ∂x ∂x′ ∂t

   ∂V 1 ∂x i ∂x2 ∂x j ∂x3 ∂x p ∂x1 ∂x j ∂V 2 ∂x i ∂x3 ∂x p ∂x1 ∂x j ∂x2 ∂x i ∂V 3 ∂x p  . + j  i ′k j ′n p ′l + j ′k ′l  ∂x ∂x ∂x ∂x ∂x ∂x  ∂x ∂x ∂x i ∂x′n ∂x p ∂x′l  ∂x ∂x′k ∂x i ∂x′n ∂x p ∂x   ≠ 0; i = 1 & j = 2 & p= 3 ≠ 0; j = 1 & i = 2 & p= 3 ≠ 0; j = 1 & i = 2 & p = 3  

Therefore, after rearranging the terms, in each group, we receive ∂ ∂V 1  knl ∂x1 ∂x2 ∂x3  ∂V 2  knl ∂x1 ∂x2 ∂x3  ∂V 3  knl ∂x1 ∂x2 ∂x3  = 1 e + e + e . ∂t ∂x  ∂x′k ∂x′n ∂x′l  ∂x2  ∂x′k ∂x′n ∂x′l  ∂x3  ∂x′k ∂x′n ∂x′l 

But the terms in each bracket is the Jacobian, and hence we receive

   1  ∂V i    ∂V 1 ∂V 2 ∂V 3  ∂ =   1 + 2 + 3 =     i  =  ∇ ⋅ V ,   ∂x   ∂t ∂x ∂x   ∂x     

which

∇⋅V

1 ∂   yields = ∇ ⋅ V. For divergence of a vector, see Equation 11.12.  ∂t

17.12 EXAMPLE: COVARIANT DERIVATIVES OF A CONSTANT VECTOR  Show that for a constant vector C, the covariant derivative C ,i j is not ­necessarily zero. Find the expression for its derivative.

Solution:

 ∂    We have C = C i ei ; writing its covariant derivative, we receive C i ei ) e j . j ( ∂x i   ∂ ∂C    i ∂ The covariant component then reads as j ( C i ei ) = e =  j  ei + C j ( i) ∂x ∂ x ∂ x     ∂C i     ∂C i i k k i  Γ= e ei  j + C Γ kj  . Now, since C is a constant, the first  j  ei + C ijk  ∂x   ∂x  i↔ k   C′ ij

term in the bracket vanishes but the second term does not. Therefore, we i receive C′= C k Γ kji . j

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17.13 EXAMPLE: COVARIANT DERIVATIVES OF PHYSICAL COMPONENTS OF A VECTOR   i A A= ei A ( i ) e ( i ), write down the expression for A ( i ) j. For the vector= ′ Use the of the vector A in terms its physical components,  covariant derivative   or ∇ j A = ∇ j ( A ( i) e ( i)) = e ( i) A ( i) j . ′ Solution: The covariant derivative, in terms of contravariant component of the   ∂A i  k i  ∇ j ( A i ei ) = vector is ∇ j A =  j + A Γ kj  ei . Substituting for contravariant  ∂x     i components A = A ( i ) / hi and covariant basis vector ei = hi e ( i ) , we get ∇ j A =   A ( i)    ∂ h     i  + A ( k ) Γ i  h e (= i) kj  ∂x j  i hk    

 ∂ ( A ( i ) ) A ( i ) ∂ ( hi ) h i  − + A ( k ) Γ kji  e ( i ) .  j j  hi ∂x hk  ∂x 

∂ ( A ( i ) ) A ( i ) ∂ ( hi ) hi A ( i) − + A ( k ) Γ kji =∇ j A ( i ) − ∇ j hi + Therefore, A ( i ), j = j j ∂x hi ∂x hk hi

hi A ( k ) Γ kji . Note that the Christoffel symbol could be written in terms of hk  physical components/ Unit vectors, i.e., Γ kji could be written in terms of e ( i ) , designated by Γˆ i (see Example 10.4.1). kj

17.14 EXAMPLE: CONTINUITY EQUATIONS IN SEVERAL COORDINATE SYSTEMS The continuity for a fluid with density r and velocity V i is given as ∂r 1 ∂ + 0. Write this equation in tensor notation for N = 3, and (r V i ) = ∂t  ∂x i express the continuity equation in: a. Cartesian coordinates ( x, y, z) with velocity physical components given as  V = (Vx , Vy , Vz ) .

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210 • Tensor Analysis for Engineers, 3E b. Cylindrical coordinates (r,q , z) with velocity physical components given  as V = (Vr , Vq , Vz ). c. Spherical coordinates (r ,j ,q ) with velocity physical components given as  V = (Vr , Vj , Vq ) . Solution:

∂r 1  ∂ ∂ ∂  +  1 ( r V 1 ) + 2 ( r V 2 ) + 3 ( r V 3 )  = 0. ∂t   ∂x ∂x ∂x  a. Jacobian for a Cartesian system is unity and physical and contravariant components are identical, hence the continuity equation reads, Summation on i gives,

 ∂r  ∂ ∂ ∂ +  ( r Vx ) + r Vy + ( r Vz )  = 0. ∂t  ∂x ∂y ∂z 

(

)

b. Jacobian for cylindrical systems is  = r and V 1 = Vr , V 2 = Vq / r , V 3 = Vz . Hence, the continuity reads, after some manipulations, ∂r 1 ∂ 1 ∂ ∂ 0. + ( r r Vr ) + ( r Vq ) + ( r Vz ) = r ∂q ∂t r ∂r ∂z c. Jacobian for spherical systems is  = r 2 sin j and V 1 = Vr, V 2 = Vj / r, V 3 = Vq / ( r sin j ). Hence, the continuity reads, after some manipulations, ∂r 1 + 2 ∂t r sin j

∂ 2  ∂ ∂ 0.  ∂r ( r sin jr Vr ) + ∂j ( r sin jr Vj ) + ∂q ( r r Vq )  =  

17.15  EXAMPLE: 4D SPHERICAL COORDINATE SYSTEMS A 4D spherical system x i ≡ (r,y ,q ,j ) with two polar angles y and q range from 0 to p and azimuth angle j from 0 to 2p , is related to the 4D Cartesian coordinates yi ≡ ( x, y, z, w) as  x = r siny sin q cos j  y = r siny sin q sin j    z = r siny cosq  w = r cosy a. Find the Jacobian for the spherical system and volume of a 4D sphere in this coordinate system. b. Find the expression for the Laplacian of a scalar in the spherical coordinates and write out it in detail.

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Solution:

∂ ( x, y, z, w ) = 1. Jacobian reads  ∂= ( r,y ,q ,j )

siny sin q cos j siny sin q sin j siny cosq cosy

r cosy sin q cos j r cosy sin q sin j r cosy cosq − r siny

∂x ∂r ∂y ∂r ∂z ∂r ∂w ∂r

∂x ∂x ∂x ∂y ∂q ∂j ∂y ∂y ∂y ∂y ∂q ∂j = ∂z ∂z ∂z ∂y ∂q ∂j ∂w ∂w ∂w ∂y ∂q ∂j

r siny sin q cos j r siny cosq sin j − r siny sin q

− r siny sin q sin j r siny sin q cos j 0 0

0

=

− r sin y sin q . 3

The

2

differential

volume

is

d  = drdy dq dj,

with

0 ≤q &y ≤p

and 0 ≤ j ≤ 2p . Volume is then a quadruplet integral of d , or R

p

p

2p

p

1 − cos2j p 2 R 4 4 . = = d j p R ∫ ∫ 2 2 0 0 0 0 0 1 2. The expression for the Laplacian of scalar S reads ∇ 2 S = ∇ i g ij ∇ j S .  This coordinate system is orthogonal, since gij = 0 for i ≠ j. For example 3 2 = ∫ ( −r ) ∫ sin y dy ∫ sinq dq

(

)

  g12 = e1 ⋅ e2 = r siny cosy sin 2 q cos2 q − r siny cosy sin 2 q sin 2 j + r siny cosy

1 = hi2 . For calculating cos2 q − r siny cosy = 0 . Therefore, g= ii g ii   gii= ei ⋅ ei , need the covariant basis vectors, or      = e1 siny sin q cos j E1 + siny sin q sin j E2 + siny cosq E3 + cosy E4      e2 r cosy sin q cos j E1 + r cosy sin q sin j E2 + r cosy cosq E3 − r siny E4 =      e3 = r siny cosq cos j E1 + r siny cosq sin j E2 − r siny sin q E3    − r siny sin q sin j E1 + r siny sin q cos j E2  e4 =

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212 • Tensor Analysis for Engineers, 3E

which yields, g11 = 1, g22 = r 2 , g33 = r 2 sin 2 y , g44 = r 2 sin 2 y sin 2 q . Now, ∇2S =

1 1 1 1 ∇ i g ij ∇ j S = ∇ 2 S = ∇ i ( g ii∇ iS ) = ∇1 ( g11∇1S ) + ∇ 2 ( g 22 ∇ 2 S ) +    

(

)

1 1 ∇ 3 ( g 33 ∇ 3 S ) + ∇ 4 ( g 44 ∇ 4 S ) . Each term reads, after substitution   −1 ∂  3 2 ∂S  3 ∂S ∂ 2 S 1 ∇1 ( g11∇1S )= 3 2 +  − r sin y sin q =  r sin y sin q ∂r  ∂r  r ∂r ∂r 2 −1 ∂  −2 3 2 ∂S  1 = ∇ 2 ( g 22 ∇ 2 S )  − r r sin y sin q  3 2 ∂y   r sin y sin q ∂y  ∂S ∂ 2 S  1 = 2  coty +  ∂y ∂y 2  r  1 −1 ∂  − r 3 sin 2 y sin q ∂S  ∇ 3 ( g 33 ∇ 3 S ) =   r 3 sin 2 y sin q ∂q  r 2 sin 2 y  ∂q   1 ∂S ∂ 2 S  = 2 2  cot q +  ∂q ∂q 2  r sin y  1 −1 ∂  − r 3 sin 2 y sin q ∂S  ∇ 4 ( g 44 ∇ 4 S ) =   r 3 sin 2 y sin q ∂j  r 2 sin 2 y sin 2 q ∂j   ∂ 2S 1 = 2 2 r sin y sin 2 q ∂j 2 Hence, ∇ 2 S=

 3 ∂S ∂ 2 S 1  ∂S ∂ 2 S  1 ∂S ∂ 2 S  + 2 + 2  coty + + 2 2  cot q + + 2  r ∂r ∂r r  ∂y ∂y  r sin y  ∂q ∂q 2 

∂ 2S 1 . r 2 sin 2 y sin 2 q ∂j 2

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17.16 EXAMPLE: COMPLEX DOUBLE DOT-CROSS PRODUCT EXPRESSIONS × ×     The expression ∇ A ∇  A is given. The notations mean that the two gradient ×

×

vectors (marked by × ) are cross product to each other as well as the two  vectors A, then the resulted vectors of the cross-product operations are dotted. Write the expression in tensor notation and simplify the result. Solution: The two cross-product results are dotted; hence we require the contravariant component of the cross-product of the gradient’s and the covariant  component of the cross-product of the vectors A . Therefore, we can write × ×     ∇ e ijk e ilm A l A mk . Expanding the resulted expression yields, A ∇ A= ′ ′j × ×

(

)

(

)

e ijke ilm ( A l A mk ) j = d l jd mk ( A l A mk ) j − d mj d lk ( A l A mk ) j = A l A kk ) l − A k A kj j. ′ ′ ′ ′ ′ ′ ( ′ ′ ′ ′ × ×     Performing the covariant derivative operations gives ∇ A A ll A kk + A ∇ = ′ ′ × × A l A kkl − A kj A kj − A k A jkj = A l l A kk − A k j A j k . The first term of the result can ′ ′ ′  ′ ′ ′ ′ ′ k→ l, j→ k   2 be written in vector notation as A l l A kk = ∇ ⋅ A . Therefore, we finally have ′ ′ × ×       2 j k ∇ A ∇  A = ∇ ⋅ A − A ′ j A ′k . Using the method given in this example, apply × × it to the following expressions:

(

(

)

)

× ×      ijk l = A∇ e ilm A l Ak m⋅ d l jd mk A l Ak m⋅ j − d mj d lk A = Ak m⋅ a. ∇  A e = ′ ′ ′ ′ ′j j ′ × ×

(

( A A ′ )′ j

k

k ⋅

j

)

(

)

) ( A A ′ )′ −

(

j

k k ⋅

j

= A j j Ak ⋅k + A j Ak ⋅ kj − A k j Ak ⋅ j − A k Ak ⋅ jj . In vector notation, the ′ ′ ′ ′ ′ ′ × ×     

 



(

  

)

resulted expression can be written as ∇  A∇  A = (∇ ⋅ A ) + A ⋅ ∇ (∇ ⋅ A) − × ×   A k j Ak ⋅ j − A ⋅ ∇ 2 A . ′ ′

(

TAE3E.CH17_1pp.indd 213

)

2

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214 • Tensor Analysis for Engineers, 3E

    l l j b. ∇∇ A A e ijke ilm ( A m A= d l jd mk − d mj d lk ( A m A= ( Ak Ak )′⋅ j − k) ⋅j k) ⋅j  =  ′ ′ × ×       2 k A j Ak ⋅ j = ∇ 2 A2 − 2 A ⋅ ∇ ∇ ⋅ A − A j⋅k Ak j − ∇ ⋅ A . ′ ′ ′ ×

(

×

(

( (

)

)

))

(

)

Note that vector-type notation is not sufficient to express some of the terms in these expressions—another reason to use index/tensor notation for its generality.

17.17 EXAMPLE: COVARIANT DERIVATIVES OF METRIC TENSORS Show that the derivative of metric tensors (i.e., gij and g ij ) is equal to zero. Solution: Since metric tensor is a constant in the Cartesian coordinate system, hence its derivative is null and will be the same for any coordinate system. ∂gij To show this, we use Equation 10.14, or gij = − Γ nik gnj − Γ njk gni . Now k ′ ∂x k using Equation 10.19 we can expand the last two terms on the R.H.S, or ∂gij 1  ∂gij ∂gkj ∂g  1  ∂gij ∂g ∂gkj  gij = k −  k + i − kij  −  k + kij − i  . The terms on ′k ∂x ∂x ∂x  2  ∂x ∂x ∂x  2  ∂x

( )

( )

( )

the R.H.S cancel out and the expression is equal to zero, hence gij = 0 , ′k and similarly g ij = 0. ′k

( )

17.18 EXAMPLE: ACTIVE ROTATION USING SINGLE-AXIS AND QUATERNIONS METHODS  Consider the vector OA = (1, −1, 2 ) connecting the origin to point A in the x , y , z coordinate system. Now we rotate this vector by an angle of 60° ( )   1 3 about an axis in the direction of unit vector n =  0, ,  . Calculate the  2 2   resulted vector OA′ , using a) rotation matrix with Equation 12.7, and b)

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The Worked Examples • 215

quaternion method with Equation 15.12. Note that the rotation angle here is given for an active rotation (i.e., vector rotates but coordinate system is kept fix). Solution:

 a. Since the vector OA is rotated by 60 degrees, active rotation, then we use q = −60° in Equation 12.7. Therefore, for n1 = 0,  1   2  3 n2 = 1 / 2, and n3 = 3 / 2 we receive R =   4  3 −  4 The rotated vector component in the ( x, y, z ) is  1   2  3   4  3 −  4

3 4 5 8 3 8

−

3  4  1 3     −1= 8   2 7    8 

3 4 5 8 3 8

−

 1 3 3  + +   10 + 4 3   2 4 2    3 5 3  1  − + =    1+ 2 3 ,  4 8 4  8  14 − 3 3   3 3 7 − +  − 8 4  4

3  4  3 . 8  7   8 

given by

or

  10 + 4 3 1 + 2 3 14 − 3 3  OA′ = , , . 8 8 8   b. Using the active quaternion operator, see Equation 15.12, we need to have the corresponding quaternion based on half of the rotation angle, 1 3  3 1 3 60° + j+ k  sin 30 = k. Now, we = 30°, Q = cos30 +  j + 2 2  2 4 4 2     n

  ∗  3 1  3 1 3  3  + j+ k  ( i − j + 2k)  − j− k= calculate OA=′ QOAQ=   4  4   2 4  2 4

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216 • Tensor Analysis for Engineers, 3E

1 3 1 3 3  3 10 + 4 3 1+2 3 14 − 3 3 1   + + + 2k) j −  − 3  k  ( i − j= i+ j+ k  − i− 4 2 2 4 4 4 8 8 8         10 + 4 3 1 + 2 3 14 − 3 3  10 + 4 3 1+2 3 14 − 3 3   Or OA ′ = , , . k i j 2 k i j k . − = + + + )  ( 8 8 8 8 8 8     Alternatively, we can use Equation 15.14. Readers may want to use Wolfram Alpha (www.wolframalpha.com) to visualize the quaternion as a rotation operator. After accessing the wolfram Web site, enter the phrase “quaternions” in the space provided and use the format provided similar to those examples given on the page.

17.19 EXAMPLE: PASSIVE ROTATION USING SINGLE-AXIS AND QUATERNIONS METHODS Repeat Example 17.18 for a passive rotation of an angle 45˚. Having passive  rotation we keep vector OA fixed and rotate coordinates ( x, y, z ). Repeating  = the data here, for convenience, OA

  1 3 (1, −1, 2 ) , n =  0, ,  . 2 2 



Solution:

 a. Since the vector OA is rotated by 45 degrees, passive rotation, then we use q = 45° in Equation 12.7. Therefore, for n1 = 0, n2 = 1 / 2 , and  2   2  6 n3 = 3 / 2 we receive R =  −  4  2   4

6 4 2+3 2 8 2 3− 6 8

components in the rotated coordinates

TAE3E.CH17_1pp.indd 216

2   4  2 3− 6  . The vector 8  6+ 2   8  −

( x′, y′, z′)

is given by

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The Worked Examples • 217

 2   2  6 −  4  2   4

6 4 2+3 2 8 2 3− 6 8

 2    4  1   2 3 − 6     −1 = − 8  2   6 + 2      8   −

 2 6 2 − −  2 4 2  −2  6 2 + 3 2 2 3 − 6  1  − +  = −2 − 3 2 + 4 4 8 4  8  12 + 4 2 − 2 2 2 3 − 6 6+ 2  − +  4 8 4 

  2 6 2 − −   2 4 2   −2 6     −2 6 −2 − 3 2 + 4 3 − 4 6 12 + 4 2 − 2 3 +   6 2 + 3 2 2 3 − 6  1  , = − + −  = −2 − 3 2 + 4 3 − 4 6  , or OA′  8 , 8 8 4 8 4 8       2 2 3 − 6 6+ 2   12 + 4 2 − 2 3 + 6  − +   8 4  4  −2 6 −2 − 3 2 + 4 3 − 4 6 12 + 4 2 − 2 3 + 6  , ,   ≅ ( −0.61237, −1.13905,2.0803 ). 8 8  8 

∗

b. The quaternion related to this rotation, using half of the rotation angle, is Q = cos

p 1 3  p + j+ k  sin = 8 2 2  8    

2+ 2 2− 2 6−3 2 + j+ k. 2 4 4

n

Now,

we

calculate

 2+ 2  2− 2 6−3 2  − OAQ  j− k  ( i − j + 2k)   2 4 4  

 2+ 2   2− 2 6−3 2  ∗ = (OA )′ Q= − OAQ  j− k( i −   2 4 4    2+ 2 2− 2 6−3 2   + j+ k  ≅.   2 4 4  

( 0.92388 − 0.191342 j − 0.331414 k )( i − j + 2 k )( 0.92389 + 0.191342 j + 0.331414k ) = ( 0.47149 + 0.20979 i − 1.2553 j + 2.03912 k ) ( 0.92389 +  0.191342 j + 0.331414 k ) = −0.61237 i − 1.13905 j + 2.0803 k. Or (OA )′ =  (OA )′ =

TAE3E.CH17_1pp.indd 217

( −0.61237, −1.13

( −0.61237, −1.13907,2.08032 ) . Alternatively, we can use Equation 15.20.

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218 • Tensor Analysis for Engineers, 3E

17.20 EXAMPLE: SUCCESSIVE ROTATIONS USING QUATERNIONS METHOD Use data given in Example 17.18 and rotate the resulted vector, actively in that example through a successive rotation of an angle 45°. For this operation use the quaternion method. Solution:

From Example 17.18 we have the resulted vector   10 + 4 3 1 + 2 3 14 − 3 3    1 3 OA′ =  , ,  and the unit vector n =  0, ,  8 8 8  2 2    along the axis of rotation. The follow up rotation quaternion is p related to rotation angle of 45° = . Hence, using half-angle we have 4 p 1 3  p 2+ 2 2− 2 6−3 2 P = cos +  j + k  sin = + j+ k . Now, 8 2 2  8 2 4 4     n   ∗  2 + 2 2− 2 ′P  using Equation 15.18 we can write = = + OA″ POA  2 4    ∗  2 + 2 2− 2 6 − 3 2   10 + 4 3 1+2 3 14 − 3 3   2 + 2 2− 2 ′P  = = + OA″ POA j+ k i+ j+ k −    2 4 4 8 8 8 2 4     1+2 3 14 − 3 3   2 + 2 2− 2 6−3 2  j+ k − j− k  ≅ 1.5436 i + 1.8708 j + 0.3425 k.  8 8 2 4 4     We can also use the original vector OA with the dual-quaternion as the product of the two corresponding quaternions, see  ∗ Equation  15.18, or ( PQ ) OA ( PQ ) . Note that the Q belongs to the first rotation (60° rotation) and P to the second one (45° rotation). But

 2+ 2 2− 2 6−3 2  3 1 3  = + + j+ PQ  j+ k  k  ≅ 0.60876 + 0.39668 j + 0.68706 k   2 2 4 4 4 4    ∗ 0.60876 + 0.39668 j + 0.68706 k . Therefore, its conjugate = ( PQ ) 0.60876 − 0.39668 j   ∗ −0.68706k . Now, we can write OA″ =( PQ ) OA ( PQ ) =(0.60876 + 0.39668 j + 0.68706 k) ( i − j + 2 k )( 0.60876 − 0.39668 j − 0.68706 k ) =1.5436 i + 1.8708 j + 0.3425 k 1.5436 i + 1.8708 j + 0.3425 k. As seen the two methods provide us with ­identical results.

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CHAPTER

18

EXERCISES 1. Write out the following expression in full detail for N = 3 dimensions using Einstein summation convention:  a. A = A ij ei e j   A i B j gij b. A ⋅ B = 1 ∂  ∂y  c. ∇ 2y = i  g ij j   ∂x  ∂x 

d. C ( i ) = hi e. Akij, n 

eijk ∂ ( hk A ( k ) ) 

∂x j

Akij j i   nm Akjm   nm Akim   mnk Amij  x n

2. By rotating bipolar coordinates about X-axis (the axis where foci are located on) we obtain the bi-spherical coordinate system, ( x1 , x2 , x3 ) ≡ (x ,h ,j )  a sin x cos j  X = coshh − cos x  a sin x sin j  1 2 3 given in terms of Cartesian ( y , y , y ) ≡ ( X , Y, Z ), as  Y = . coshh − cos x   a sinhh Z = coshh − cos x     Find the basis vectors e1 , e2, e3 for the bi-spherical coordinate system in terms of the Cartesian unit vectors. Also find the scale factors, unit vectors, metric tensors (covariant and contravariant), Jacobian, volume element, and Christoffel symbols of the 1st kind.

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220 • Tensor Analysis for Engineers, 3E

FIGURE 18.1  Bi-spherical coordinate system.

3. By rotating bipolar coordinates about Y-axis (the axis perpendicular to the line connecting the foci) we obtain a toroidal coordinate system,

( x , x , x ) ≡ (x ,h ,j ) given in terms of Cartesian ( y , y , y ) ≡ ( X, Y, Z ) , 1

2

3

1

2

3

 a sinhh cos j  X = coshh − cos x  a sinhh sin j     as  Y = . Find the basis vectors e1, e2 , e3 for the toroidal − cosh h cos x   a sin x Z = coshh − cos x  coordinate system in terms of the Cartesian unit vectors. Also find the scale factors, unit vectors, metric tensors (covariant and contravariant), Jacobian, volume element, and Christoffel symbols of the 2nd kind.

FIGURE 18.2  Toroidal coordinate system.

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Exercises • 221

  4. Having displacement vector r = yi Ei defined in Cartesian coordinate  system yi, express r in terms of the contravariant and physical components in cylindrical, spherical, bi-polar, bi-spherical, and toroidal coordinate systems. Use the definitions of these systems as given in previous examples and exercises.

= gii 1= / g ii hi2 5. Show that for an orthogonal coordinate system the relation holds, i.e., the diagonal elements of covariant and contravariant metric tensors. Note, no sum on index i. 6. For an orthogonal coordinate system, show that the following relations holds for Christoffel symbols of the 2nd kind (all indices are set to be ­different): Γ kij = 0, Γ iij =−

hi ∂hi i 1 ∂hi i 1 ∂hi , Γ ij = , Γ ii = . hi ∂x j hi ∂x i h2j ∂x j

7. Find the relations for Christoffel symbols of the 1st kind for cylindrical and spherical coordinate systems. Use results from the example given in Section 10.4.1. 8. Show that covariant differentiation obeys the same rule as does ordinary

(

)

j differentiation, when operating on products, e.g., A i B = A′ik B j + A i B′ kj . ′k

9. Write down the expression for the covariant derivative of a tensor of rank ij five, Akm , using hints from Section 10.14. 10. Show that metric tensors behave like a constant under a covariant dif-

(

)

(

= A′ik gij and Ai g ij

)

= Ai′ k g ij .  11. Show that the recursive relation for the curl of a vector A reads   A ( n ) = −∇ 2 A ( n − 2 ), for n > 2. where n is the number of curl operations        × ∇ ×  × ∇ × A (note, A ( 0 ) = A ). and A ( n ) = ∇ 

ferentiation operation, i.e., A i gij

′k

′k

n times

12. Using the recursive relation for the curl of a vector given in Exercise 11, show that the recursive relation for the biharmonic operator reads   A ( n) = ∇ 4 A ( n − 4 ) , for n > 4 . 13. Using the results of Exercises 11 and 12, show that the recursive rela  tions for higher order operator ∇ 2 n reads A ( 2 n + 1 ) =− ( 1 ) n ∇ 2 n A (1 )   and A ( 2 n ) = −∇ 2 n− 2 A ( 2 ) .

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222 • Tensor Analysis for Engineers, 3E 14. For the 4D spherical coordinate system from Example 15.15, calculate the Christoffel symbols: Γ rrr   r Γ =    

Γ rry r Γyy

Γ rrq r Γyq

 

r Γqq

Γqrr   q Γ =    

Γqry q Γyy

Γqrq q Γyq

 

Γqqq





Γyrr Γ rrj   r  Γyj , Γy =  r   Γqj   r Γjj   

Γyry y Γyy

Γyrq y Γyq

 

Γyqq

Γqrj  Γjrr  q  Γyj  , j  Γ =  Γqqj    q Γjj   

Γjry j Γyy

Γjrq j Γyq

 

j Γqq





Γyrj  y  Γyj , Γyqj   Γyjj  Γjrj  j  Γyj . j  Γqj  Γjjj 

15. Show that the following expressions simplify as given (see Example 15.16): ×    ×       k j jk 2 A = ∇ A A A + A ⋅ ∇ A − A A − A ⋅ ∇ ∇⋅A a. ∇ ′ k j k j ′ ′ ⋅ ′ ⋅  × × ×  ×      j 2 k b. ∇  A ∇∇  Φ = ∇ ⋅ A ( ∇ Φ ) − A′ j ( ∇ k Φ )′ ⋅

(

×

(

×

( (

)

))

)

×   ×          2 jk 2 2 c. ∇∇  Φ =∇ A ⋅ ( ∇Φ ) − ∇ ∇ ⋅ A ⋅ ( ∇Φ ) − A′ ⋅ ( ∇ k Φ ) − ∇ ⋅ A ( ∇ Φ ) − A ⋅ ∇ ( ∇ Φ )  A∇ × ×      jk 2 ⋅ A ⋅ ( ∇Φ ) − A′ ⋅ ( ∇ k Φ ) − ∇ ⋅ A ( ∇ Φ ) − A ⋅ ∇ ( ∇ 2 Φ )

(

)

(

)

(

)

(

)

(

)

(

)

16. Write the below expressions in index notation:      a. V ∇ ⋅ A + ∇ ⋅ V A    b. A ⋅ B × C     c. A ⋅ ∇ ∇ ⋅ A

(

)

(

(

)

)

( (

))

17. Show that the following relations holds: d li

d mi

a. ε ijk ε lmn =d l j d mj d lk d mk

d ni ijk k ijk = d mj d nk − d njd mk ,  c. ε ε ijn =2d n d nj ,  b. ε ε imn d nk

d. ε ijk ε ijk =6

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)

Exercises • 223

18. Using Equation 9.13, write the determinant for a 5 × 5 matrix. 19. Use Equation 13.19 to find physical components of the convective time derivative of a velocity vector, and express the results for (3D) (hint: see Example 15.14): a. Cartesian coordinates b. Cylindrical coordinates c. Spherical coordinates 20. Show that the non-zero Christoffel symbols of the 1st kind for cylindrical and spherical coordinates read: a. Cylindrical coordinates: Γ rq ,q = r and Γqq ,r = −r . r2 − r sin 2 j , Γqq ,j = − sin 2j , − r , Γqq , r = b. Spherical coordinates: Γjj ,r = 2 r2 2 Γ rj ,j = r and Γ rq ,q = r sin j , Γjq ,q =sin 2j 2 21. Using cross-product expression for two vectors, show that ( is the ­Jacobian): a. ijk =

1 mnl  gmi gnj glk 

1 ijk mnl 2 b.  =   gmi gnj gkl 6 22. Calculate the single equivalent rotation angle and rotation axis for transforming Cartesian system ( y1 , y2 , y3 ) to ( y′1 , y′2 , y′3 ) such that y′1 coincides with y1, y′2 coincides with y3 , and y′3 coincides with y1 .

23. Find the rotation matrix for transforming ( y1 , y2 , y3 ) : a. Rotate about (0, 0, 1) for 45° b. Then, rotate about (1, 0, 0) for 60° c. Then, rotate about (0, 1, 0) for 75°

d. Find the final rotation matrix and equivalent rotation angle e. Find equivalent axis of rotation

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224 • Tensor Analysis for Engineers, 3E  ˆ Q + Q and 24. Show the following relations hold for unit quaternion Q = 0  arbitrary vector V .         ˆ ˆ ∗ = Q2 − Q 2 V + 2 Q ⋅ V Q + 2Q Q × V , active rotation a. QVQ 0 0  b. Qˆ ∗ VQˆ =

)

(

( 2Q

2 0

(

)

(

)

      − 1 ) V + 2 V ⋅ Q Q + 2Q0 V × Q , passive rotation

(

)

(

)

25. Derive The rotation matrix given by Equation 15.22, using unit quater nion Qˆ and arbitrary vector V . 26. Using Example 16.18 rotation matrix result, calculate the corresponding Euler angles. 27. Using Example 16.19 rotation matrix result, calculate the corresponding Euler angles. 28. Using the quaternions  method as shown in Example 16.20, calculate the  = (1, −1,2 ) after two successive rotations. The components of vector OA   1 3 first rotation is by an angle of 80° about the unit vector n =  0, ,   2 2  and the second rotation is by an angle of 50° about the unit vector

  3 1  n= , ,0  . Perform the calculations for both active and passive  2 2  rotation scenarios. 29. For Exercise 28, calculate the corresponding rotation matrix using the product of the quaternions involved. Also using the resulted rotation matrix calculate the equivalent axis and angles of single rotation. 30. A pilot maneuvers its airplane through a pitch-roll-yaw series or rotations. Using the relation for the corresponding rotation matrix  yxz analyze the gimbal lock orientation when roll angle is equal to 3p . 2 31. Show that quaternion operation for active rotation ( QVQ∗ ) preserves dot-products and triple products, hence it is a rotation. 32. Repeat the Exercise 16.20 considering the rotation as a passive one (i.e., coordinates are rotated).

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Exercises • 225

33. A 2D stress element in the x−y plane is given as σx=−820 MPa, σy=−330 MPa, and τxy=100 (CCW) MPa. The yield strength of the material is σyield=530 MPa, resulted from uniaxial tension measurement tests. Answer the following questions: a. Draw the stress element including the stress elements acting on its faces and write down the stress matrix. b. Draw the corresponding Mohr’s circle. Note that the shear stress data is given with reference to the positive plane with its outward normal in the direction of x-axis. c. Find principal stresses and the corresponding orientation. d. Find maximum shear stress and the corresponding orientation. e. Draw the stress elements for the cases (c) and (d). f. Using Tresca failure criterium, investigate if the material fails? (Hint: compare maximum shear stress against σyield/2). g. Using von Mises failure criterium, investigate if the material fails? (Hint: compare von Mises stress against σyield). 34. Using data from problem 33, build an Excel sheet to answer all questions listed. The sheet should be interactive to take in new data, draws the Mohr’s circle, and plots related to the stress’s transformation for a range of rotation, θ = 0, π. Note that rotation is about z-axis. (Hint: see ­Equation 16.4 and Figure 16.3). 35. A 3D stress element, é s x t xy 0 ù ê ú s ij = ê tyx s y 0 ú = êë 0 0 s z úû

in the é 20 ê ê -40 êë 0

x − y − z coordinate system, is given as -40 0 ù ú -40 0 ú MPa. The yield strength of the 0 100 úû material is σyield = 240 MPa, resulted from uniaxial tension measurement tests. Answer the following questions: a. What are the normal and shear stresses on a plane with outward unit normal ni = (0.9,0.173,0.41). b. Draw the Mohr’s circles.

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226 • Tensor Analysis for Engineers, 3E c. Find principal stresses using Mohr’s circle and by calculating the eigenvalues of the stress tensor. Compare the corresponding results. d. Find principal direction using Mohr’s circle and by calculating the eigenvectors of the stress tensor. Compare the corresponding results. e. Find maximum shear stress using Mohr’s circle and by calculating the eigenvalues of the stress tensor. Compare the corresponding results. [Hint: use Equation 16.30] f. Find octahedral normal and shear stresses. Compare these results against the maximum principal and shear stresses. g. Find invariants of the deviatoric stress tensor and compare the second invariant J2 with von Mises stress measuring stress. [Hint: use Equation 16.41]. h. Using Tresca failure criterium, investigate if the material fails? (Hint: compare maximum shear stress against σyield/2). i. Using von Mises failure criterium, investigate if the material fails? (Hint: compare von Mises stress against σyield).  σ x τ xy 0   50 20 10  36. Repeat problem 35 for σ ij =  τ yx σ y 0  =  20 100 60  MPa.      0 0 σ z  10 60 50  37. Repeat problem 33 with σx=170 MPa, σy=20 MPa, and τxy=100 MPa (CW). Also find stress components when the stress element is rotated 75° CCW, with respect to the x-axis.

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REFERENCES 1. G. Ricci and T. Levi-Civita. “Méthodes de calcul differéntiel absolu et leurs applications,” Math. Ann., vol. 54, 1901. 2. Y. I. Dimitrienko, Tensor Analysis and Nonlinear Tensor Functions, Springer, 2002. 3. D. Leigh, Nonlinear Continuum Mechanics: An Introduction to Continuum Physics and Mathematical Theory of the Nonlinear Mechanical Behavior of Materials, McGraw-Hill, 1968. 4. I. Sokolnikoff, Tensor Analysis: Theory and Applications, John Wiley & Sons, 1951. 5. P. Grinfeld, Introduction to Tensor Analysis and the Calculus of Moving Surfaces, Springer, 2010. 6. R. J. Kee, M. E. Coltrin, and P. Glarborg, Chemically Reactive Flow: Theory & Practice, John Wiley & Sons, 2033. 7. G. G. Bach, Tensor Analysis and Applications (lecture notes), Montreal, Canada: Mech. Eng., McGill University, 1979. 8. E. Christoffel, “Über die Transformation der homogenen Differentialausdrücke zweiten Grades,” Journal für die reine und angewandte Mathematik, vol. B, pp. 46–70, 1869. 9. O. Rodrigues, “Des lois géometriques qui regissent les déplacements d’ un systéme solide dans l’ espace, et de la variation des coordonnées provenant de ces déplacement considérées indépendant des causes qui peuvent les produire,” J. Math. Pures Appl., vol. 5, pp. 380–440, 1840. 10. E. Kreyszig, Advanced Engineering Mathematics, 10th ed., John Wiley & Sons, 2011. 11. W. Z.U.A., Fluid Dynamics: Theoretical and Computational Approaches, 3rd ed., CRC Press, 2005. 12. “Anisotropic Heat Transfer through Woven Carbon Fibers,” COMSOL, Application ID: 16709. 13. B. K. Horn, “Some Notes on Unit Quaternions and Rotation,” 2001. 14. D. Koks, Explorations in Mathematical Physics: The Concepts Behind an Elegant Language, Springer, 2006.

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228 • Tensor Analysis for Engineers, 3E

15. Erik B. Dam, M. Koch, and M. Lillholm, “Quaternion, Interpolation and Animation,” Department of Computer Science, University of Copenhagen, Denmark, 1998. 16. S. Oldenburger, “Applications of Quaternions. Written project of the course “Problem Solving Techniques in Applied Computer Science,” Department of Computer Science, Iowa State University, 2005. 17. H. Goldstein, C. Poole, and J. Safko, Classical Mechanics, 3rd Ed., Addison Wesley, 2000. 18. J. Kuipers, Quaternions and Rotation: A Primer with Applications to Orbits, Aerospace and Virtual Reality, Princeton University Press, 1999. 19. R. Brannon, Rotation, Reflection, and Frame Changes: Orthogonal tensors in computational engineering mechanics, IOP Publishing Ltd, 2018. 20. W. Heard, Rigid Body Mechanics: Mathematics, Physics and Applications, WILEYVCH Verlag GmbH & Co, 2005. 21. J. Vince, Quaternions for Computer Graphics, Hardcover, Springer, 2011. 22. E. Zeleny, “Controling Airplane Flight,” Wolfram Demonstration Project, 2011. 23. S. Blinder, “Euler Angles for Space Shuttle,” Wolfram Demonstrations Project, 2011. 24. F. W. Strauch, “Euler Angles,” Wolfram Demonstrations Project, 2011. 25. D. Hosier, “Avoiding Gimbal Lock in a Trajectory Simulation,” U.S. Army Armament Research Development and Engineering Center, New Hersey, 2018. 26. D. Hoag, “Apollo Guidance and Navigation: Consideration of Apollo IMU Gimbal Lock,” Apollo Lunar Surface Journal, Cambridge, MIT, 1963. 27. H. Schaub, “Spacecraft Dynamics & Control,” University of Colorado Boulder, 2019. 28. GuerrillaCG, “Euler (gimbal lock) Explained,” 2009. 29. “Choose Rotation Order and Gimbal Lock | Blender 2.8x – Rigging Tutorial,” 2019. 30. W. R. Hamilton, “On quaternions or a new system of imaginaries in algebra,” Philosophical Magazine, link to David R. Wilkins collection at Trinity College, Dublin, Dublin. 31. A. J. Hanson, “Quaternion Application,” School of Informatics and Computing , Indiana University, 2012. 32. L.C. Bruno and L.W. Baker, “Math and mathematicians : the history of math discoveries around the world,” Detroit, Mich. : U X L, 1999. 33. YouTube, “Visualizing 4D Geometry – A Journey Into the 4th Dimension [Part 2],” 2017. 34. E. W. Weisstein, “Euler Formula From MathWorld—A Wolfram Web Resource”. 35. E. Cho, “De Moivre’s Formula for Quaternions,” Appl. Math. Lett, vol. 11, no. 6, pp. 33–35, 1998. 36. D. Henderson, “Euler Angles, Quaternions, A140 Transformation Matrices for Space Shuttle Analysis,” NASA-McDonnell Douglas Technical Services, 1977. 37. Allison, I. (1984) The pole of the Mohr diagram. J. Struct. Geol. 6, 331-333. 38. Hibbeler, R., Mechanics of Materials, 10th edition, 2016, Pearson. 39. Messal, E.F., Finding true maximum shear stress, Mach. Des., 1978, pp. 166-169. 40. Ugural, A.C. and Fenster, S.K., Advanced Mechanics of Materials and Applied Elasticity, 5th edition. Prentice Hall. 2012.

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References • 229

41. Constitutive Equations for Engineering Materials Elasticity and Modeling, Wai-Fah Chen, Atef F. Saleeb, 1994, Elsevier. 42. Mechanics of Solids, A. K. Singh, 2007, PHI Learning Private Limited. 43. Henri Édouard Tresca (1814 – 1885). 44. von Mises, R. (1913) “Mechanik der Festen Korper im Plastisch Deformablen Zustand,” Nachr. Ges. Wiss. Gottingen, pp. 582. 45. Nadai, Arpad. Mechanics of Solids, A. K. Singh, 2007, PHI Learning Private Limited. 2nd ed. New York: McGraw Hill Book Co., 1950.

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Index A acceleration vector contravariant components of, 76–78 physical components in, 78–79 cylindrical coordinate systems, 81–82 orthogonal systems, 79–82 spherical coordinate systems, 81–82 substantial time derivation, 84–85 active rotations, 104–105 definition, 105 example, 105–107 arbitrary coordinate systems, 4, 31

B basis vectors, 5–7 biharmonic operations on tensors, 58–59

C Cartesian coordinate system, 3–4, 89–91 equivalent single rotation to sequential rotations of, 70–73 fluid in motion, 83 Cartesian tensors, 65–66 equivalent single rotation, 67–73 rotation matrix, 67 characteristic equation, 70 Christoffel symbols, 52, 58, 59, 77–80 of 2nd kind, 35–39 for cylindrical coordinates, 44–46 for 2D polar coordinate system, 38 relations and properties, 41–44 for spherical coordinates, 46–49 of 1st kind, relations and properties, 41–44

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complex numbers, 128 component notation, 2 connection coefficients, 37 conservation equations, coordinate independent forms, 85–87 contravariant components, 31, 33 of acceleration vector, 76–78 of curl, 52 and transformations, 9–12 contravariant derivatives of vectors, 39–40 contravariant metric tensor, 22 contravariant permutation tensors, 32 coordinate systems Cartesian, 89–91 cylindrical, 91–93 definition, 3–4 orthogonal curvilinear, 98–101 parabolic, 96–98 spherical, 93–96 covariant components, and transformations, 13–14 covariant derivatives of mixed tensor, 40 of vectors, Christoffel symbols of the 2nd kind, 35–39 covariant/contravariant basis vectors, 22 covariant permutation tensors, 32 cross-product, 29–33, 53 curl operations on tensors, 51–54 cylindrical coordinate systems, 3, 91–93 acceleration vector physical components in, 81–82 Christoffel symbols for, 44–46 example, 23–25

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232 • Index

metric tensor, 23–25 physical components of Laplacian of a vector in, 60–61

D determinant of a 3 × 3 matrix using permutation symbols, 34 divergence operation on tensors, 55–57 3D mechanical stress tensor, in matrix form, 143 3D Mohr’s circle, 163–164, 180 dot-product, 29–33, 57. See also cross-product

E eigenvalues and eigenvectors, 67–73, 152–153 Einstein summation convention, 2 equivalent single rotation, 67–73 Euler angles, 107–110 categorizing, 113–120 derive the quaternion from given, 139–140 example, 110–113, 120–121 for rigid body rotations, 108

F first principal scalar invariant, 70

G Gauss divergence theorem, 85 GeoGebra, 157 gimbal lock, 121 gimbal lock-Euler angles limitation, 121–123

H Hamilton quaternion relations, derivation of, 125–126

I index notation, 2 invariant, 6

J Jacobian and metric tensor determinants, relation between, 157 time derivatives of , 159–160

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L Lagrange multiplier method, 69 Laplacian of a vector, physical components of, 59–60 cylindrical systems, 60–61 spherical systems, 61–63 Laplacian operations on tensors, 57–58 Levi Civita symbol, 30

M maximum-distortion-energy theory, 188 maximum shear stress directions, 148 vs. principal stresses and direction, 151 mechanical stress transformation 3D stress transformation and analysis eigenvector calculations, 177–183 general formulation, 169–173 traction force and its in-plane extremum, 173–176 Mohr’s circle graphical method, 153–169 octahedral stresses and deviatoric stress, 186–187 in principal coordinate system, 183–186 vs. von Mises yield criterion, 188–189 plane stress condition assumptions, 144 maximum shear stress directions, 150–151 principal stresses and direction, 148–149 stress components transformation, 144–147 metric tensor, 18–22 cylindrical coordinate systems, 23–25 spherical coordinate systems, 25–27 mixed tensor, 18 covariant derivatives of, 40 Mohr’s 3 circle. See 3D Mohr’s circle Mohr’s circle graphical method, 143, 153 3D Mohr’s circle and special cases, 163–164 example, 165–167 stress state on selected plane, 168 2θ-method, 154, 160–162 Pole-point method, 154, 157–160 sign conventions, 154–157

N Newton’s 2nd law, 74, 76–78, 87

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Index • 233

O orthogonal Cartesian system, 3, 30 orthogonal coordinate systems, 33 acceleration vector physical components in, 79–82 curvilinear, 98–101

P parabolic coordinate systems, 96–98 passive rotations, 104–105, 105–107 definition, 105 example, 105–107 permutation symbols, determinant of a 3 × 3 matrix using, 34 physical components of acceleration vector in, 78–79 cylindrical coordinate systems, 81–82 orthogonal systems, 79–82 spherical coordinate systems, 81–82 substantial time derivation, 84–85 curl of tensors-3D orthogonal systems, 54–55 of Laplacian of a vector cylindrical systems, 60–61 spherical systems, 61–63 and transformations, 15–16 principal stresses and direction, 148–149 eigenvalues and eigenvectors, 152–153 example, 162–163 proper Euler angles, 113, 114 Pythagorean theorem, 6, 18, 19

Q quaternion algebraic operations and properties, 126–127 example, 127–128 and rotation, 129–132 definition, 123–125 derive the rotation matrix from a given, 135–138 quaternions multiplication rule, 125

R right-hand-rule convention, 67, 72 rigid body rotations Euler angles for, 108

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quaternions-applications for, 123–134 direct relations and methods, 141 Rodrigues’ rotation formula, 67 rotation matrix, 67, 114, 117, 128 calculating, 72 derive the equivalent quaternion from, 138–139 rotation, quaternion and, 129–132 example, 133–134

S scale factor, 7 second principal scalar invariant, 70 spherical coordinate systems, 3, 93–96 acceleration vector physical components in, 81–82 for Christoffel symbols, 46–49 example, 25–27 metric tensor, 25–27 physical components of Laplacian of a vector in, 61–63 substantial time derivatives of tensors, 82–85

T Tait-Bryan angles, 113, 117 third principal scalar invariant, 70 total/convective time derivative, 82 traction force and normal component, definition of, 170 transformation coefficient, 10, 11 transformations, 15–16 contravariant components and, 9–12 covariant components and, 13–14 physical components and, 15–16

U unit vector, 6, 7

W worked-out examples active rotation using single-axis and quaternions methods, 214–216 application of contravariant metric tensors, 203 complex double dot-cross product expressions, 213–214

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234 • Index

continuity equations in several coordinate systems, 209–210 conversion from vector to index notations, 192–193 covariant derivatives of a constant vector, 208 covariant derivatives of metric tensors, 214 covariant derivatives of physical components of a vector, 209 determinant of a 4 × 4 matrix using permutation symbols, 207 determinant of metric tensors using displacement vectors, 206 dot and cross products in cylindrical and spherical coordinates, 203–205 Einstein summation convention, 191

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4D spherical coordinate systems, 210–212 oblique rectilinear coordinate systems, 193–196 passive rotation using single-axis and quaternions methods, 216–217 quantities related to bi-polar coordinate systems, 200–202 quantities related to parabolic coordinate system, 197–199 relation between Jacobian and metric tensor determinants, 205 successive rotations using quaternions method, 218 time derivatives of the Jacobian, 207–208

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