December 2011 

NASA/TM–2011-217314 

An In-Depth Tutorial on Constitutive Equations 

for Elastic Anisotropic Materials 

Michael P. Nemeth 

Langley Research Center, Hampton, Virginia 
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December 2011 

NASA/TM–2011-217314 

An In-Depth Tutorial on Constitutive Equations 

for Elastic Anisotropic Materials 

Michael P. Nemeth 

Langley Research Center, Hampton, Virginia 
Available from: 

NASA Center for AeroSpace Information 

7115 Standard Drive 

Hanover, MD 21076-1320 

443-757-5802 
 1 

CONTENTS

SUMMARY . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .9

PREFATORY COMMENTS. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .11

Motivation and Approach. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .12

Dedication . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .13

BASIC CONCEPTS AND NOTATIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .14

Basic Concepts. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .15

Basic Notions of Deformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .17

Notation for Stresses and Strains. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .21

Indicial Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .22

CONSTITUTIVE EQUATIONS FOR ISOTROPIC MATERIALS. . . . . . . . . . . .23

Hooke’s Law for a Homogeneous, Isotropic, Linear-Elastic Solid . . . .24
 2 

The Duhamel-Neumann Law for a Homogeneous, Isotropic, 

Linear-Thermoelastic Solid . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .26

GENERALIZED HOOKE’S LAW FOR HOMOGENEOUS, ANISOTROPIC, 

LINEAR-ELASTIC SOLIDS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .31

General Form of Hooke’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .32

Reduction to 21 Independent Constants . . . . . . . . . . . . . . . . . . . . . . . . .43

Strain-Energy Density . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .45

Proof that Cijkl = Cklij . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

51

Illustration of the Path-Independence Condition. . . . . . . . . . . . . . . . . . .52

Complementary Strain-Energy Density . . . . . . . . . . . . . . . . . . . . . . . . . .58

Illustration of Energy Density Functionals. . . . . . . . . . . . . . . . . . . . . . . .61

Proof that Sijkl = Sklij . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

63

Standard Forms for Generalized Hooke’s Law . . . . . . . . . . . . . . . . . . . .64

Clapeyron’s Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .65
 3 

Positive-Definiteness of the Strain-Energy Density Function . . . . . . . .67

GENERALIZED DUHAMEL-NEUMANN LAW FOR HOMOGENEOUS, 

ANISOTROPIC, LINEAR-ELASTIC SOLIDS. . . . . . . . . . . . . . . . . . . . . . . . . .70

The Generalized Duhamel-Neumann Law . . . . . . . . . . . . . . . . . . . . . . . .71

Equations for the Thermal Moduli. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .79

Strain-Energy Density For Thermal Loading . . . . . . . . . . . . . . . . . . . . . .80

Complementary Strain-Energy Density for Thermal Loading . . . . . . . .84

Illustration of Thermoelastic Energy Densities . . . . . . . . . . . . . . . . . . . .87

Proof that Sijkl = Sklij for Thermoelastic Solids . . . . . . . . . . . . . . . . . . . . .

91

Clapeyron’s Formula for Thermoelastic Solids. . . . . . . . . . . . . . . . . . . .92

Strain-Energy Density Expressions . . . . . . . . . . . . . . . . . . . . . . . . . . . . .93

ABRIDGED NOTATION AND ELASTIC CONSTANTS. . . . . . . . . . . . . . . . . .95

Abridged Notation for Constitutive Equations . . . . . . . . . . . . . . . . . . . .96
 4 

Clapeyron’s Formula In Abridged Notation . . . . . . . . . . . . . . . . . . . . . .102

Physical Meaning of the Elastic Constants . . . . . . . . . . . . . . . . . . . . . .106

TRANSFORMATION EQUATIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .109

Transformation of [C] and [S] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .110

Transformations for Dextral Rotations about the x

3 Axis . . . . . . . . . .121

Transformations for Dextral Rotations about the x1 Axis . . . . . . . . . .146

MATERIAL SYMMETRIES. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .177

Material Symmetries. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .178

Mathematical Characterization of Symmetry. . . . . . . . . . . . . . . . . . . . .180

Some Types of Symmetry in Two Dimensions . . . . . . . . . . . . . . . . . . .186

Some Types of Symmetry in Three Dimensions . . . . . . . . . . . . . . . . . .187

Criteria for Material Symmetry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .192

Classes of Material Symmetry. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .196
 5 

MONOCLINIC MATERIALS. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .200

Monoclinic Materials - Reflective Symmetry About

the Plane x1 = 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .201

Monoclinic Materials - Reflective Symmetry About 

the Plane x

2 = 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .216

Monoclinic Materials - Reflective Symmetry About 

the Plane x

3 = 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .224

ORTHOTROPIC MATERIALS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .232

Orthotropic Materials - Reflective Symmetry About the 

Planes x1 = 0 and x

2 = 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .233

Orthotropic Materials - Reflective Symmetry About the 

Planes x1 = 0, x

2 = 0, and x3 = 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .242

Constitutive Equations. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .244

Specially Orthotropic Materials. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .246

Generally Orthotropic Materials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .247
 6 

TRIGONAL MATERIALS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .254

Trigonal Materials - Reflective Symmetry About Planes that 

Contain the x

3 axis. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .255

Trigonal Materials - Reflective Symmetry About Planes that 

Contain the x1 axis. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .283

Trigonal Materials - Reflective Symmetry About Planes that 

Contain the x

2 axis. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .316

Summary of Trigonal Materials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .322

TETRAGONAL MATERIALS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .324

Tetragonal Materials - Reflective Symmetry Planes that 

Contain the x

3 axis. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .330

Tetragonal Materials - Reflective Symmetry Planes that Contain 

the x1 axis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .335

Tetragonal Materials - Reflective Symmetry Planes that 

Contain the x

2 axis. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .338
 7 

Summary of Tetragonal Materials. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .344

TRANSVERSELY ISOTROPIC MATERIALS . . . . . . . . . . . . . . . . . . . . . . . .347

Transversely Isotropic Materials - Isotropy Plane x

3 = 0 . . . . . . . . . . .348

Transversely Isotropic Materials - Isotropy Plane x1 = 0 . . . . . . . . . . .359

CUBIC MATERIALS. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .370

COMPLETELY ISOTROPIC MATERIALS. . . . . . . . . . . . . . . . . . . . . . . . . . .373

CLASSES OF MATERIAL SYMMETRY - SUMMARY OF 

INDEPENDENT MATERIAL CONSTANTS . . . . . . . . . . . . . . . . . . . . . . . . . .387

ENGINEERING CONSTANTS FOR ELASTIC MATERIALS. . . . . . . . . . . . .388

Constitutive Equations in Terms of Engineering Constants . . . . . . . .389

Engineering Constants of a Specially Orthotropic Material. . . . . . . . .421

Engineering Constants of a Transversely Isotropic Material. . . . . . . .425

Engineering Constants of a Generally Orthotropic Material . . . . . . . .431
 8 

REDUCED CONSTITUTIVE EQUATIONS. . . . . . . . . . . . . . . . . . . . . . . . . . .444

Constitutive Equations for Plane Stress . . . . . . . . . . . . . . . . . . . . . . . .445

Stress and Strain Transformation Equations for Plane Stress . . . . . .474

Transformed Constitutive Equations for Plane Stress . . . . . . . . . . . . .481

Constitutive Equations for Generalized Plane Stress. . . . . . . . . . . . . .497

Constitutive Equations for Inplane Deformations of Thin Plates . . . .509

Constitutive Equations for Plane Strain. . . . . . . . . . . . . . . . . . . . . . . . .533

Stress and Strain Transformation Equations for Plane Strain. . . . . . .557

Transformed Constitutive Equations for Plane Stress . . . . . . . . . . . . .563

LINES AND CURVES OF MATERIAL SYMMETRY . . . . . . . . . . . . . . . . . . .572

BIBLIOGRAPHY. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .577
 9 

SUMMARY

An in-depth tutorial on the thermoelastic constitutive equations for elastic,

anisotropic materials is presented. First, basic concepts are introduced that 

are used to characterize materials, and then notions about how anisotropic 

material deform are presented. Next, a common notation used to describe 

stresses and strains is given,followed by the rules of indicial notation used 

herein. Based on this notation, Hooke’s law and the Duhamel-Neuman law 

for isotropic materials are presented and discussed.

After discussing isotropic materials, the most general form of Hooke’s law 

for elastic anisotropic materials is presented and symmetry requirements 

that are based on symmetry of the stress and strain tensors are given.

Additional symmetry requirements are then identified based on the 

reversible nature of the strain energy and complimentary strain energy 

densities of elastic materials. A similar presentation is then given for the 

generalized Duhamel-Neuman law for elastic, anisotropic materials that 

includes thermal effects. Next, a common abridged notation for the 

constitutive equations is introduced and physical meanings of the elastic 

constants are discussed.
 10 

SUMMARY - CONCLUDED

As a prelude to establishing various material symmetries, the 

transformation equations for stress and strains are presented, the most 

general form of the transformation equations for the constitutive matrices 

are presented.Then, specialized transformation equations are presented for 

dextral rotations about the coordinate axes. Next, the concepts of material 

symmetry are introduced, the mathematical process used to describe 

symmetries is discussed, and examples are given. After describing the 

mathematics of symmetry, the criteria for the existence of material 

symmetries are presented and the classes of material symmetries are given.

Then, the invariance conditions and simplifications to the constitutive 

equations are presented for monoclinic, orthotropic, trigonal, tetragonal,

transversely isotropic, and completely isotropic materials.

After establishing a broad range of material symmetries, the engineering 

constants of fully anisotropic, elastic materials are derived from first 

principles and then specialized to several cases of practical importance.

Lastly, reduced constitutive equations are derived for states of plane stress,

generalized plane stress, plane strain and generalized plane strain.

Transformation equations are also derived for these special cases.
 11 

PREFATORY

COMMENTS
 12 

MOTIVATION AND APPROACH

" Knowledge of anisotropic materials has become prominent in the last 

few decades because of the applications of advanced, lightweight fiber-

reinforced composite materials to aircraft and spacecraft

" The material presented herein is redundant in several sections, by

design

" First, to reinforce concepts and enhance learning

" Second, to pr

ovide stand-alone sections that can be used 

independently for various reasons

" Third, to serve as a comprehensive reference document
 13 

DEDICATION

"

To Manuel Stein - Wisdom, knowledge, humility, and kindness 

incarnate

"

ToJames H. Starnes, Jr. - The embodiment of scientific thirst and 

curiosity, leadership, and professional excellence

"

To Harold G. Bush - The epitome of common sense and sound 

engineering practice, and a constant source of advice and 

entertainment

"

ToThe Men and Women of the NACA - The premier examples of 

government researchers
 14 

BASIC CONCEPTS

AND NOTATIONS
 15 

BASIC CONCEPTS

" The macroscopic physical, or material, properties of a body are 

specified by constitutive equations

" For 

example, a relationship between stress, strain, and temperature 

is commonly specified for solid materials

" The material properties of a solid, regardless of its shape, are 

generally 

functions of the coordinates of the material particles

" Solids for which the material properties v

ary pointwise are 

described as inhomogeneous (e.g., a bi-metallic strip)

" For homogeneous solids, the material properties are the same

for 

every particle of the solid 

" The material properties of a homogeneous solid are described 

mathematically as invariant with respect to coordinate-frame 

translations 
 16 

BASIC CONCEPTS - CONCLUDED

" A body is described as isotropic at a point if its properties at that point 

are independent of direction

" A body that is not isotropic is described, in the most general case,

as anisotropic

" A body that is isotropic at a given point is described mathematically as 

invariant with respect to coordinate-frame rotations (for that point)

" A body is described as homogeneous and isotropic if its properties are 

independent of direction, and identical, at every point of the body

" Distortion is defined as deformation that consists of a change in 

shape without a change in volume (pure shearing deformation) 

" Dilatation is defined as deformation that consists of a change in 

volume without a change in shape (pure expansion-contraction-type 

deformation) 
 17 

BASIC NOTIONS OF DEFORMATION

" Pure normal stresses acting within a homogeneous, isotropic solid

produce only volumetric, extensional (dilatational) deformations 

" The angle between 

every pair of intersecting material line elements,

that lie in the planes that are perpendicular to the normal stresses 

in the solid, remains unchanged during deformation (no shearing)

Deformed shape of

an isotropic solid Normal stress Deformed shape of an anisotropic solid
 18 

BASIC NOTIONS OF DEFORMATION - CONTINUED

" Pure shearing stresses acting within a homogeneous, isotropic solid

produce only distortional, shearing (deviatoric) deformations 

" The angle between every pair of intersecting material line elements 

that lie in the planes of the shearing stresses in the solid change 

during deformation, but the length of the line elements does not 

change (no dilatation)

Deformed shape of

an anisotropic solid

Deformed shape of

an isotropic solid

Shearing stress
 19 

BASIC NOTIONS OF DEFORMATION - CONTINUED

" Pure shearing stresses acting within a homogeneous, isotropic solid

produce only distortional, shearing (deviatoric) deformations that are 

only in the plane of the shearing stresses 

" All unrestrained thermal expansion is 

volumetric and uniform within 

a homogeneous, isotropic solid; not so in a homogeneous, generally 

anisotropic solid

Deformed shape of

an isotropic solid

Deformed shape of

an anisotropic solid
 20 

BASIC NOTIONS OF DEFORMATION - CONCLUDED

" Strains that are caused by unconstrained thermal expansions, that 

do not produce stresses, are defined as free thermal strains

" A solid is described as ideally elastic (usually just called elastic) 

when it recovers to its initial, stress- and strain-free configuration upon 

removal of the applied loads or temperature field

" For this case, there exists a one-to-one (unique) mathematical 

relationship between the stresses and strains that act within the 

loaded solid
 21 

NOTATION FOR STRESSES AND STRAINS

" In the development that follows, stresses and strains are defined 

relative to standard rectangular Cartesian coordinates

" The normal strains correspond to the normal stresses

, respectively

" The shearing strains correspond to the 

shearing stresses , respectively

x1, x2, x3

Shearing stresses

σ11

σ22

σ33

Normal stresses

σ12

σ13σ23

x1

x2

x3

O

ε11, ε22, andε33

σ11, σ22, andσ33

2ε12=γ12, 2ε13=γ13, and2ε23=γ23

σ12, σ13, andσ23
 22 

INDICIAL NOTATION

" The rules of indicial notation associated with Cartesian tensors are 

used herein

" In particular, all indices appear as subscripts, unless noted otherwise

" For example,

" Latin indices take on the values , and repeated latin indices 

imply summation over this set; e. g.;

" Indices that are not summed in an equation are called free indices and 

take on the complete set of possible values

" The symbol is known as the Kronecker Delta Symbol and is 

equal to one when j = k and is equal to zero otherwise 

εij=

1

E

1

+νσij−νδijσkk

1, 2, 3

σkk

=

Σ

σkk k = 1

3

=

σ11

+

σ22

+

σ33

δjk
 23 

CONSTITUTIVE EQUATIONS

FOR ISOTROPIC MATERIALS
 24 

HOOKE’S LAW

HOMOGENEOUS, ISOTROPIC, LINEAR-ELASTIC SOLID

" In the 17th century, Robert Hooke began developing a constitutive 

law for elastic, isotropic solids

" The concept of elastic deformation was introduced by Hooke in 

1676

" Hooke’s work led to the following equations that are in use today

 or in indicial notation 

ε11=

1

E

σ11−νσ22+σ33

2ε12=γ12=

21+ν

E

σ12

ε22=

1

E

σ22−νσ11+σ33

2ε13=γ13=

21+ν

E

σ13

ε33=

1

E

σ33−νσ11+σ22

2ε23=γ23=

21+ν

E

σ23

εij

=

1

E

1

+νσij−νδijσkk
 25 

HOOKE’S LAW - CONCLUDED

HOMOGENEOUS, ISOTROPIC, LINEAR-ELASTIC SOLID

" The inverted form of Hooke’s law is given by

or in indicial notation

" For these equations, it is important to remember that the strains are 

caused by the externally applied loads and displacements

" Strains of this type are called (stress-induced) mechanical 

strains and are the result of the internal stresses

σ11=

E

1+ν

1−2ν

1−νε11+νε22+ε33σ12=

E

21+ν

γ12=

E

1+ν

ε12

σ22=

E

1+ν

1−2ν

1−νε22+νε11+ε33σ13=

E

21+ν

γ13=

E

1+ν

ε13

σ33=

E

1+ν

1−2ν

1−νε33+νε11+ε22σ23=

E

21+ν

γ23=

E

1+ν

ε23

σij=

E

(1+ν)(1−2ν) 1

−2νεij+νδijεkk
 26 

THE DUHAMEL-NEUMANN LAW

HOMOGENEOUS, ISOTROPIC, LINEAR-THERMOELASTIC SOLID

" Hooke’s law was extended by J. M. C. Duhamel (circa 1838) and 

F. E. Neumann (circa 1888) to include the first-order effects of thermal 

loading

" This law is based, in part, on the premise that the total strain at a 

point of a solid, subjected to thermomechanical loading, consists of 

mechanical strain and strain caused by free thermal expansion

" The mechanical strain is the stress-induced strain caused by the 

externally applied loads and displacements, and the stress-induced

strain caused by nonuniformity in the temperature field or in the 

thermal expansion properties of the material 

" where , , T is 

the temperature field, and Tref is the temperature field at which the 

body is deemed stress and strain free (or negligible)

εij

εij

σ

εij

T

εij

σ

εij =εij

σ

+εij

T

εij

σ

=

1

E

1+νσij −νδijσkk

εij

T

=αδij T−Tref
 27 

THE DUHAMEL-NEUMANN LAW - CONTINUED

HOMOGENEOUS, ISOTROPIC, LINEAR-THERMOELASTIC SOLID

" The temperature fields T and Tref are, in general, functions of position 

within the body; that is, and 

" T is, in general, also time dependent and Tref is typically uniform,

with a value equal to a nominal ambient temperature

"Thermal stresses are caused by two effects:

" The spatial nonuniformity in the field and

" Geometric restraints that prevent stress-free thermal expansion 

" When a solid is subjected to a nonuniform temperature field or its 

thermal expansion properties vary, there arises a mismatch in the 

thermal expansion of neighboring material particles

" Internal, "thermal stresses" develop to maintain continuity of the 

material body, which induces mechanical strains 

T=T

x1, x2, x3 Tref

=Tref

x1, x2, x3

αT−Tref
 28 

THE DUHAMEL-NEUMANN LAW - CONTINUED

HOMOGENEOUS, ISOTROPIC, LINEAR-THERMOELASTIC SOLID

" The work of Hooke, Duhamel, and Neumann led to the following 

thermoelastic constitutive equations that are used today

or in indicial notation

ε11=

1

E

σ11−νσ22+σ33+αT−Tref

2ε12=γ12=

21+ν

E

σ12

ε22=

1

E

σ22−νσ11+σ33+αT−Tref

2ε13=γ13=

21+ν

E

σ13

ε33=

1

E

σ33−νσ11+σ22+αT−Tref

2ε23=γ23=

21+ν

E

σ23

εij

=

1

E

1+νσij−νδijσkk+αδijT−Tref
 29 

THE DUHAMEL-NEUMANN LAW - CONTINUED

HOMOGENEOUS, ISOTROPIC, LINEAR-THERMOELASTIC SOLID

" The inverted form of the Duhamel-Neumann law is given by

 or in indicial notation

σ11=

E

1+ν

1−2ν

1−νε11+νε22+ε33−

EαT−Tref

1−2ν

σ22=

E

1+ν

1−2ν

1−νε22+νε11+ε33−

EαT−Tref

1−2ν

σ33=

E

1+ν

1−2ν

1−νε33+νε11+ε22−

EαT−Tref

1−2ν

σ12=

E

21+ν

γ12=

E

1+ν

ε12σ13=

E

21+ν

γ13=

E

1+ν

ε13

σ23=

E

21+ν

γ23=

E

1+ν

ε23

σij=

E

1+ν

1−2ν

1−2νεij+νδijεkk−δij

EαT−Tref

1−2ν
 30 

THE DUHAMEL-NEUMANN LAW - CONCLUDED

HOMOGENEOUS, ISOTROPIC, LINEAR-THERMOELASTIC SOLID

" The constitutive equations show that an isotropic material is 

characterized fully by two independent elastic constants E and 

ν, and 

by one thermal expansion coefficient

α

" E is the modulus of elasticity, which is also called Young’s 

modulus and the elastic modulus 

"ν is Poisson’sratio and α is the coefficient of linear thermal 

expansion

" E and ν are related by , where G is called the 

shear modulus or the modulus of rigidity

G=

E

2

1+ν
 31 

GENERALIZED HOOKE’S LAW

FOR

HOMOGENEOUS,

ANISOTROPIC,

LINEAR-ELASTIC SOLIDS
 32 

GENERAL FORM OF HOOKE’S LAW

" The generalization of Hooke’s law to anisotropic materials is attributed 

to Cauchy (in 1829) and postulates that every component of the stress 

tensor is coupled linearly with every component of the strain tensor;

i.e.,

or in indicial notation by

" Note that Sijkl have units of stress-1; e.g., in

2/lb 

ε11

ε22

ε33

ε23

ε13

ε12

ε32

ε31

ε21

=

S1111

S1122S1133S1123S1113S1112S1132S1131S1121

S2211S2222S2233S2223S2213S2212S2232S2231S2221

S3311S3322S3333S3323S3313S3312S3332S3331S3321

S2311S2322S2333S2323S2313S2312S2332S2331S2321

S1311S1322S1333S1323S1313S1312S1332S1331S1321

S1211S1222S1233S1223S1213S1212S1232S1231S1221

S3211S3222S3233S3223S3213S3212S3232S3231S3221

S3111S3122S3133S3123S3113S3112S3132S3131S3121

S2111S2122S2133S2123S2113S2112S2132S2131S2121

σ11

σ22

σ33

σ23

σ13

σ12

σ32

σ31

σ21

εij

=

Sijklσkl
 33 

GENERAL FORM OF HOOKE’S LAW - CONTINUED

"Sijkl are called the components of the (4th-order) compliance tensor 

and are often called compliances or compliance coefficients

" Without further simplication, there are 3

4 (or 81) independent 

compliance coefficients that must be determined from experiments,

to fully characterize a given homogeneous material

" The previous equation indicates that each normal-stress component 

produces shearing strains in all three coordinate planes, in addition to 

three extensional strains

" Similarly, each shearing-stress component produces extensional 

strains along all three coordinate directions and shearing strains in the 

two planes perpendicular to the plane of the shearing stress, in addition 

to a shearing strain in the plane of the shearing stress
 34 

GENERAL FORM OF HOOKE’S LAW - CONTINUED

" Thus, dilatational deformation (expansion-contraction) and 

distortional deformation (shearing) are fully coupled in an 

anisotropic material, unlike common isotropic materials

" The inverted form of generalized Hooke’s law is given by

or in indicial notation by

" Note that Cijkl have units of stress; e.g., lb/in

2

σ11

σ22

σ33

σ23

σ13

σ12

σ32

σ31

σ21

=

C1111

C1122C1133C1123C1113C1112C1132C1131C1121

C2211C2222C2233C2223C2213C2212C2232C2231C2221

C3311C3322C3333C3323C3313C3312C3332C3331C3321

C2311C2322C2333C2323C2313C2312C2332C2331C2321

C1311C1322C1333C1323C1313C1312C1332C1331C1321

C1211C1222C1233C1223C1213C1212C1232C1231C1221

C3211C3222C3233C3223C3213C3212C3232C3231C3221

C3111C3122C3133C3123C3113C3112C3132C3131C3121

C2111C2122C2133C2123C2113C2112C2132C2131C2121

ε11

ε22

ε33

ε23

ε13

ε12

ε32

ε31

ε21

σij

=

Cijklεkl
 35 

GENERAL FORM OF HOOKE’S LAW - CONTINUED

"Cijkl are called the components of the (4th-order) elasticity or stiffness 

tensor and are often called stiffness coefficients

" Note that Sijkl and Cijkl are constants for a homogeneous material 

" The number of independent coefficients in can be reduced 

by enforcing symmetry of the stress and strain tensor

s

" is the same as 

" gives , which implies for a general 

stress state at a point in a body

" can be used to show that 

" and yield 36 independent compliance coefficients

εij =Sijklσkl

εij =Sijklσklεji =Sjiklσkl

εij =εji Sijklσkl=SjiklσklSijkl=Sjikl

σkl=σlkSijkl=Sijlk

Sijkl=SjiklSijkl=Sijlk
 36 

GENERAL FORM OF HOOKE’S LAW - CONTINUED

" The proof that Sijkl = Sijlk is given as follows 

" The constitutive equation can be expressed as 

because l and k are summation indices and inter

changing them 

doesn’t alter the content of the equation

" Equating and gives 

" Next, enforcing gives , which implies 

 for a 

general stress state at a point in a body

εij =Sijklσklεij =Sijlkσlk

εij =Sijklσklεij =Sijlkσlk

Sijklσkl=Sijlkσlk

σkl=σlkSijklσkl=Sijlkσkl

Sijkl

=

Sijlk
 37 

GENERAL FORM OF HOOKE’S LAW - CONTINUED

" The number of independent coefficients in can also be 

reduced directly by enforcing symmetry of the stress and strain tensor

s

" is the same as 

" gives , which implies for a general 

strain state at a point in a bod

y

" can be used to show that 

" and yield 36 independent stiffness coefficients

" Cauchy’s generalized form of Hooke’s Law ends up with 36

independent compliance or stiffness coefficients

σij =Cijklεkl

σij =Cijklεklσji =Cjiklεkl

σij =σji Cijklεkl=CjiklεklCijkl=Cjikl

εkl=εlkCijkl=Cijlk

Cijkl=CjiklCijkl=Cijlk
 38 

GENERAL FORM OF HOOKE’S LAW - CONTINUED

" The expanded form of Cauchy’s generalized Hooke’s law is 

obtained as follows

" First, expanding the last summation index gives

" Then, expanding the summation index k gives

" Next, enforcing yields

εij =Sijklσkl

εij =Sijk1σk1 +Sijk2σk2 +Sijk3σk3

εij =Sij11σ11+Sij21σ21+Sij31σ31+Sij12σ12+Sij22σ22+Sij32σ32+

Sij13

σ13+Sij23σ23+Sij33σ33

σkl=σlk

εij =Sij11σ11+Sij22σ22+Sij33σ33+

Sij23+Sij32σ23+Sij13+Sij31σ13+Sij12+Sij21σ12
 39 

GENERAL FORM OF HOOKE’S LAW - CONTINUED

" Then, enforcing the conditions give the result

" Applying this equation for all, independent values of the free indices i 

and j results in the following matrix representation of Cauchy’s

generalized Hooke’s law :

Sijkl=Sijlk

εij =Sij11σ11+Sij22σ22+Sij33σ33+2Sij23σ23+2Sij13σ13+2Sij12σ12

εij=Sijklσkl

ε11

ε22

ε33

2ε23

2ε13

2ε12

=

S1111

S1122S11332S11232S11132S1112

S2211S2222S22332S22232S22132S2212

S3311S3322S33332S33232S33132S3312

2S23112S23222S23334S23234S23134S2312

2S13112S13222S13334S13234S13134S1312

2S12112S12222S12334S12234S12134S1212

σ11

σ22

σ33

σ23

σ13

σ12
 40 

GENERAL FORM OF HOOKE’S LAW - CONTINUED

" Similarly, the expanded form of Cauchy’s generalized Hooke’s law 

 is obtained as follows

" First, expanding the last summation index gives

" Then, expanding the summmation index k gives

" Next, enforcing yields

σij =Cijklεkl

σij =Cijk1εk1 +Cijk2εk2 +Cijk3εk3

σij =Cij11ε11+Cij21ε21+Cij31ε31+Cij12ε12+Cij22ε22+Cij32ε32+

Cij13

ε13+Cij23ε23+Cij33ε33

εkl=εlk

σij =Cij11ε11+Cij22ε22+Cij33ε33+

Cij23+Cij32ε23+Cij13+Cij31ε13+Cij12+Cij21ε12
 41 

GENERAL FORM OF HOOKE’S LAW - CONTINUED

" Then, enforcing the conditions give the result

" Applying this equation for all, independent values of the free indices i 

and j results in the following matrix representation of Cauchy’s

generalized Hooke’s law :

Cijkl=Cijlk

σij =Cij11ε11+Cij22ε22+Cij33ε33+2Cij23ε23+2Cij13ε13+2Cij12ε12

σij=Cijklεkl

σ11

σ22

σ33

σ23

σ13

σ12

=

C1111

C1122C1133C1123C1113C1112

C2211C2222C2233C2223C2213C2212

C3311C3322C3333C3323C3313C3312

C2311C2322C2333C2323C2313C2312

C1311C1322C1333C1323C1313C1312

C1211C1222C1233C1223C1213C1212

ε11

ε22

ε33

2ε23

2ε13

2ε12
 42 

GENERAL FORM OF HOOKE’S LAW - CONTINUED

" The compliance coefficients Sijkl and the stiffness coefficients Cijkl

are described as components of a 

fourth-order tensor (field) because 

each are the components of a linear transformation that relates 

components of the second-order stress tensor (field) to components of 

the second-order strain tensor (field) 
 43 

REDUCTION TO 21 INDEPENDENT CONSTANTS

" The number of independent elastic, compliance and stiffness 

coefficients is reduced from 36 to 21 by enforcing the 

thermodynamic properties of reversible, elastic deformations

" The key quantity to be examined is the strain-energy density of 

an elastic solid 

" The reduction to 21 is attributed to George Green (1793-1841)

" The strain-energy densityU of a generally elastic solid is defined as 

the work of the internal stresses, done through stress-induced

mechanical deformations, that is stored in a loaded body

" In an ideally elastic solid, experimental evidence indicates that all of 

the work done byexternal forces is converted into elastic-strain energy 

that can be recovered upon unloading, thus a loaded body has the 

potential to perform work
 44 

REDUCTION TO 21 INDEPENDENT CONSTANTS

CONCLUDED

" The existence of a strain-energy density function for linear- and 

nonlinear-elastic materials can be shown directly by using the first and 

second laws of thermodynamics

" The term "density" is used herein to indicate that the strain energy 

is defined per unit volume of material

" The expression for the strain-energy density function U is obtained by

determining the strain-energy-density increment d

U associated with 

an infinitesimal change in the deformation of a body

"dU can be obtained directly from the laws of thermodynamics or by

determining the work done by the internal forces of a body on a 

differential volume element of material
 45 

STRAIN-ENERGY DENSITY

" The strain-energy-density increment dU is given by

 where the stresses depend on the mechanical strains; that is,

" This expression is written compactly in indicial form as

" The strain-energy density U is obtained by integrating dU over the 

deformation associated with a loading process, that starts at a strainfree state and ends at a particular strain state; that is,

d

U

=σ11dε11+σ22dε22+σ33dε33+2σ23dε23+2σ13dε13+2σ12dε12

σij =σij ε11,ε22,ε33,ε23,ε13,ε12

d

U

=σij εpqdεij

U

=σij ε11,ε22,ε33,ε23,ε13,ε12dεij

0

εpq

=

U

εpq
 46 

STRAIN-ENERGY DENSITY - CONTINUED

" For an arbitrary process that involves loading followed by total 

unloading, the strain-energy density 

U is given by the circuit integral

" In addition, because strain-energy density is not lost during an arbitrary 

elastic loading-unloading process (conservation of energy - first law 

of thermodynamics), it follows that 

" For the condition that for an elastic loading-unloading process to 

be true, it requires that there must exist a strain-energy density function 

U for which dU is an exact differential; that is,

U

=σij εpqdεij

U

=σij εpqdεij =0

U

=0

U

=d

U

=0
 47 

STRAIN-ENERGY DENSITY - CONTINUED

ILLUSTRATION OF ELASTIC LOADING-UNLOADING PROCESSES

Strain

Stress

Elastic loading-unloading cycle

O

O

A

A

A

B

B

B

C

C

C

O

Two Independent Loading Systems
 48 

STRAIN-ENERGY DENSITY - CONTINUED

ILLUSTRATION OF ELASTIC LOADING-UNLOADING PROCESSES

Strain

Stress

Elastic loading-unloading cycle

O

O

A

A

A

O

One Loading System
 49 

STRAIN-ENERGY DENSITY - CONTINUED

" Because , it follows mathematically that an exact 

differential has the property that 

" A function with this property is described in mathematics as a 

potential function, thus U is sometimes referred to as the elastic 

potential

" Equating with gives 

" The last equation on the right indicates that the stress-strain 

relations are derivable from a potential function when the 

deformation process is elastic

" A material of this type is called a hyperelastic or a Greenelastic material

U

=

U

εpq

d

U

=

∂

U

∂εij

dεij

d

U

=σij εpqdεij d

U

=

∂

U

∂εij

dεij

∂

U

∂εij

=σij
 50 

STRAIN-ENERGY DENSITY - CONCLUDED

" The statement also indicates that an arbitrary elastic 

loading-unloading process is a path-independent process

" This result arises because the integral of an exact differential

depends only on the limits of integration (end points of the 

process), according to the fundamental theorem of calculus

" A necessary conditionfor a function to be path independent

is for the following condition to be valid:

" This condition arises from the connection of the path integral with 

Stokes’ integral theorem

U

=d

U

=0

U

=

U

εpq

∂

2

U

∂εij∂εkl

=

∂

2

U

∂εkl∂εij
 51 

PROOF THAT Cijkl = Cklij

" First, note that and give 

" Then, gives 

" Also, gives 

" Thus, yields , which reduces the number of 

independent stiffness coefficients to 21

∂

2

U

∂εij ∂εkl

=

∂

2

U

∂εkl∂εij

∂

U

∂εij

=σij

∂σij

∂εkl

=

∂σkl

∂εij

σij =Cijrsεrs

∂σij

∂εkl

=

∂

∂εkl

Cijrsεrs=Cijrs

∂εrs

∂εkl

=Cijrsδrkδsl=Cijkl

σkl=Cklpqεpq

∂σkl

∂εij

=

∂

∂εij

Cklpqεpq=Cklpq

∂εpq

∂εij

=Cklpqδpiδqj=Cklij

∂σij

∂εkl

=

∂σkl

∂εij

Cijkl=Cklij
 52 

ILLUSTRATION OF THE PATH-INDEPENDENCE CONDITION

" The function can be viewed as an ordinary,

simply connected, continuous, smooth function of six independent 

variables

"To enable visualization of the path-independence condition, consider 

the case of a similar function of two independent variables,

" The chain rule of differentiation gives 

" The vector form of is given by

U

=

U

ε11,ε22,ε33,ε23,ε13,ε12

F

x1, x2

d

F

=

∂

F

∂x1

dx1 +

∂

F

∂x2

dx2

d

F

d

F

= i 1

∂

F

∂x1

+ i 2

∂

F

∂x2

•dx1i 1 +dx2i 2 ≡∇

F

•dx
 53 

ILLUSTRATION OF THE PATH-INDEPENDENCE CONDITION

CONTINUED

" Let P denote a path traversed in a loading-unloading cycle, then

 becomes 

" Recall that Stokes’Theorem is given by

where

" is an arbitrary vector field with continuous first derivatives

" is the unit-magnitude normal-vector field for any smooth 

surface S enclosed by the curve 

d

F

P

=0∇

F

•dx

P

=0

g ∂S•dx=n•∇×g

S

dA

g

x1, x2

n

x1, x2

∂S
 54 

ILLUSTRATION OF THE PATH-INDEPENDENCE CONDITION

CONTINUED

" Applying Stokes’ theorem to gives

 where 

S(P) is the surface enclosed by the path P

" For a simply connected region, the necessary and sufficient conditions 

for the line integral to vanish are given by the requirement that the 

integrand in the double integral vanish; that is,

∇

F

•dx

P

=0

∇

F

P

•dx=n•∇×∇

F

S(P)

dA=0

n•∇×∇

F

=0
 55 

ILLUSTRATION OF THE PATH-INDEPENDENCE CONDITION

CONTINUED

" Because the unit-magnitude normal-vector field for an arbitrary 

smooth surface 

S(P) is generally nonzero, the necessary and 

sufficient conditions for the line integral to vanish become 

" Expanding gives 

which simplifies to 

" Simplifying further gives which yields 

the condition 

∇×∇

F

=0

∇×∇

F

=0 i 1

∂

∂x1

+ i 2

∂

∂x2

× i 1

∂

F

∂x1

+ i 2

∂

F

∂x2

=0

∂

2

F

∂x1∂x2

i 1 × i 2 +

∂

2

F

∂x2∂x1

i 2 × i 1 =0

∂

2

F

∂x1∂x2

−

∂

2

F

∂x2∂x1

i 1 × i 2 =0

∂

2

F

∂x1∂x2

=

∂

2

F

∂x2∂x1
 56 

ILLUSTRATION OF THE PATH-INDEPENDENCE CONDITION

CONTINUED

" The condition is, in fact, a statement of path 

independence at the local level, which is illustrated in the following 

figure

∂

2

F

∂x1∂x2

=

∂

2

F

∂x2∂x1

P: (x1,x

2

)

i1 i2

x1

x2

F(x1,x2)

F(x1,x2 in the )

neighborhood

of point P 

F

+

∂

F

∂x2

dx2

F

+

∂

F

∂x1

dx1

A

B

C

D
 57 

ILLUSTRATION OF THE PATH-INDEPENDENCE CONDITION

CONCLUDED

" By following path ABC, the value of F at point C is given by

" By following path ADC, the value of F at point C is given by

" For path independence, it follows that these two expressions must be 

equal, hence

F

+

∂

F

∂x1

dx1 +

∂

∂x2

F

+

∂

F

∂x1

dx1 dx2 =

F

+

∂

F

∂x1

dx1 +

∂

F

∂x2

dx2 +

∂

2

F

∂x2∂x1

dx1dx2

F

+

∂

F

∂x2

dx2 +

∂

∂x1

F

+

∂

F

∂x2

dx2 dx1 =

F

+

∂

F

∂x2

dx2 +

∂

F

∂x1

dx1 +

∂

2

F

∂x1∂x2

dx2dx1

∂

2

F

∂x1∂x

2

=

∂

2

F

∂x

2

∂x1
 58 

COMPLEMENTARY STRAIN-ENERGY DENSITY

" The symmetry conditionSijkl = Sklij is obtained by examining the 

complementary strain-energy density functional 

U* 

" The strain-energy density functional U was obtained by expressing 

the stresses in terms of the strains and integrating from the 

initial stress- and strain- free state to the current strain state

" An expression for U* is obtained by first requiring that a one-to-one 

relationship exists between the stresses and strains, and b

y using the 

product rule of differentiation to get 

" In the part , the strains are taken as the independent variables

" In the part , the stresses are taken as the independent 

variables

d

U

=σijdεij

dσijεij =σijdεij +εijdσij

σijdεij

εijdσij
 59 

COMPLEMENTARY STRAIN-ENERGY DENSITY

CONTINUED

" Next, the expression is integrated from the 

initial stress- and strain- free state to the current stress and strain state;

i.e.,

" In the term , it is presumed that the stresses are known as 

functions of the strains

" This term can also be expressed as , where it is 

presumed that the strains are known as functions of the stresses

" Both terms yield , the product of the current values of the 

stresses and strains

dσijεij =σijdεij +εijdσij

dσijεij

0

εpq

=σij εpqdεij

0

εpq

+εij σpqdσij

0

σpq

dσijεij

0

εpq

dσijεij

0

σpq

σijεij
 60 

COMPLEMENTARY STRAIN-ENERGY DENSITY

CONCLUDED

" Using the previous expression and the definition of the strain-energy 

density function 

U gives 

" The complementary (or conjugate) strain-energy density functionU* 

is defined as such that 

" Note that 

" The form is known as the Legendre 

transformation

" The complementary or conjugate relationship of the strain-energy 

density function and the complementary strain-energy density function 

are illustrated on the next chart for a one-dimensional case

σijεij =

U

εpq+εij σpqdσij

0

σpq

U*

=εij σpqdσij

0

σpq

σijεij =

U

εpq+

U*

σpq

d

U*

=εij σpqdσij

U*

σpq=σijεij −

U

εpq
 61 

ILLUSTRATION OF ENERGY DENSITY FUNCTIONALS

ONE-DIMENSIONAL CASE

ε

σ

Area= σε

ε

σ

ε

σ

Strain Strain Strain

Stress Stress Stress

=

+

Area=σ(ε)dε 0

ε

=

U

(ε) Area=ε(σ)dσ 0

σ

=

U*(

σ)

dε

σ=σ(ε)

dσ

ε=ε(σ)

Single-parameter

loading system σε

=

U

ε+

U*

σ
 62 

ILLUSTRATION OF ENERGY DENSITY FUNCTIONALS

ONE-DIMENSIONAL CASE - CONCLUDED

" The previous figure indicates that because strain-energy density is not 

lost in an arbitrary elastic loading process, neither is the 

complementary strain-energy density 

" Thus, the complementary strain-energy density function is also 

path independent and conserved in an elastic loading-unloading 

process

" Thus, and 

" Equating with gives 

" The equations given above indicate that the strain-stress relations 

are derivable from a potential function when the deformation 

process is elastic (hyperelastic material) 

U*

=d

U*

=0

d

U*

=

∂

U*

∂σij

dσij

d

U*

=εij σpqdσij d

U*

=

∂

U*

∂σij

dσij

∂

U*

∂σij

=εij
 63 

PROOF THAT Sijkl = Sklij

" The necessary and sufficient conditions for to be path 

independent are for the conditions to be valid

" First note that and give 

" Then, gives 

" And, gives 

" Thus, yields , which reduces the number of 

independent compliance coefficients to 21

U*

=

U*

σpq

∂

2

U*

∂σij ∂

σkl

=

∂

2

U*

∂σkl

∂σij

∂

2

U*

∂σij ∂

σkl

=

∂

2

U*

∂σkl

∂σij

∂

U*

∂σij

=εij

∂εij

∂σkl

=

∂εkl

∂σij

εij =Sijrsσrs

∂εij

∂σkl

=

∂

∂σkl

Sijrsσrs=Sijrs

∂σrs

∂σkl

=Sijrsδrkδsl=Sijkl

εkl=Sklpqσpq

∂εkl

∂σij

=

∂

∂σij

Sklpqσpq=Sklpq

∂σpq

∂σij

=Sklpqδpiδqj=Sklij

∂εij

∂σkl

=

∂εkl

∂σij

Sijkl=Sklij
 64 

STANDARD FORMS FOR GENERALIZED HOOKE’S LAW

" The standard forms of the generalized Hooke’s law are now given by

ε11

ε22

ε33

2ε23

2ε13

2ε12

=

S1111

S1122S11332S11232S11132S1112

S1122S2222S22332S22232S22132S2212

S1133S2233S33332S33232S33132S3312

2S11232S22232S33234S23234S23134S2312

2S11132S22132S33134S23134S13134S1312

2S11122S22122S33124S23124S13124S1212

σ11

σ22

σ33

σ23

σ13

σ12

σ11

σ22

σ33

σ23

σ13

σ12

=

C1111

C1122C1133C1123C1113C1112

C1122C2222C2233C2223C2213C2212

C1133C2233C3333C3323C3313C3312

C1123C2223C3323C2323C2313C2312

C1113C2213C3313C2313C1313C1312

C1112C2212C3312C2312C1312C1212

ε11

ε22

ε33

2ε23

2ε13

2ε12
 65 

CLAPEYRON’S FORMULA

" For a linear-elastic solid, strain-energy-density increment 

 is combined with to get 

" Now consider,

" Because all indices are summation indices, this expression can be 

expressed as 

" By using the path-independence condition , it follows 

that 

 and that 

d

U

=σij εpqdεij σij =

Cijklεkl

d

U

=Cijklεkldεij

1

2

dCijklεijεkl=

1

2

Cijkldεijεkl+Cijklεijdεkl

1

2

dCijklεijεkl=

1

2

Cijklεkldεij +Cklijεkldεij =

1

2

Cijkl+Cklijεkldεij

Cijkl=Cklij

1

2

dCijklεijεkl=Cijklεkldεij d

U

=

1

2

dCijklεijεkl
 66 

CLAPEYRON’S FORMULA - CONCLUDED

" Integrating the last expression gives where K is a 

constant of integration

" Noting that when the strain field is zero-valued gives 

" Next, using gives the desired result,

" This expression for the strain-energy density of a homogeneous, linear-

elastic, anisotropic solid is attributed to B. P. E. Clapeyron (1799-1864)

" A similar procedure can be followed to show that 

for a homogeneous, linear-elastic, anisotropic solid

U

=

1

2

Cijklεijεkl+K

U

=0

K=0

σij =Cijklεkl

U

=

1

2

σijεij

U*

=

1

2

σijεij=

U
 67 

POSITIVE-DEFINITENESS OF THE STRAIN-ENERGY 

DENSITY FUNCTION

" The strain-energy density of a solid in its stress- and strain-free 

state is defined to be zero-valued

" As a solid deforms under load, it stores strain energy and develops 

the potential to perform work upon removal of the loads

" Thus, it follows that the strain-energy density function is a nonnegative-valued function for all physically admissible elastic strain 

states

" Hence, for a nonlinear-elastic material

" For a linear-elastic material,must hold, which places 

some thermodynamic restrictions on the stiffness coefficients that 

must hold for reversible (elastic) loading-unloading processes

U

=σij(εpq)dεij

0

εpq

≥

0

U

=

1

2

Cijklεijεkl

≥

0
 68 

POSITIVE-DEFINITENESS OF THE STRAIN-ENERGY 

DENSITY FUNCTION - CONTINUED

" The strain-energy density of a linear-elastic material can be 

expressed in matrix form by

"Positive-definiteness of the strain-energy density is satisfied by

positive-definiteness of the matrix containing the stiffness coefficients

" Enforcing positive-definiteness defines relationships that the 

stiffness coefficients must obey; e.g., all the diagonal elements of 

the matrix must be positive-valued

U

=

1

2

ε11

ε22

ε33

2ε23

2ε13

2ε12

T

C1111C1122C1133C1123C1113C1112

C1122C2222C2233C2223C2213C2212

C1133C2233C3333C3323C3313C3312

C1123C2223C3323C2323C2313C2312

C1113C2213C3313C2313C1313C1312

C1112C2212C3312C2312C1312C1212

ε11

ε22

ε33

2ε23

2ε13

2ε12
 69 

POSITIVE-DEFINITENESS OF THE STRAIN-ENERGY 

DENSITY FUNCTION - CONCLUDED

"Positive-definiteness of the strain-energy density is used in the linear 

theory of elasticity to establish:

" Uniqueness of solutions 

" The theorem of minimum potential energy

" The theorem of minimum complementary energy

" Some aspects of St. Venant’s principle 
 70 

GENERALIZED

DUHAMEL-NEUMANN LAW

FOR

HOMOGENEOUS,

ANISOTROPIC,

LINEAR-ELASTIC SOLIDS
 71 

THE GENERALIZED DUHAMEL-NEUMANN LAW

" In general, when an elastic solid is subjected to heating or cooling, the 

equations of elasticity are coupled with the equations of 

thermodynamics and heat transfer

" When the heat generated by deformations is negligible, the 

equations uncouple and the temperature field can be solved for 

independently of the structural deformations

" The temperature field becomes a known quantity (loading) in the 

solution of the linear thermoelasticity equations 

" In general, when an elastic solid is subjected to heating or cooling, the 

stress-strain relations depend on the temperature of the body

" The extent of the temperature dependence depends on the extent of 

the heating or cooling
 72 

THE GENERALIZED DUHAMEL-NEUMANN LAW

CONTINUED

"To obtain a simple working theory that is linear and that includes 

thermal effects, a constitutive law was developed with the following 

attributes:

"Thermal expansion effects are included

"Variations in the elastic constants and coefficients of thermal 

expansion with temperature are neglected

" Inertial effects associated with heating rates are neglected

" A relatively simple extension of Hooke’s law that predicted 

accurately experimentally observed phenomenon was the desired 

result

" The resulting equations are typically referred to as the linearthermoelastic constitutive equations
 73 

THE GENERALIZED DUHAMEL-NEUMANN LAW

CONTINUED

" The generalized Hooke’s law was extended by J. M. C. Duhamel

(1797- 1872) and F. E. Neumann (1798-1895) to include the first-order,

linear effects of thermal loading 

" This law states, in part, that the total strain at a point of a solid,

subjected to thermomechanical loading, consists of stress-induced

mechanical strain and strain caused by free thermal expansion

" The mechanical strain is the strain caused by the externally 

applied loads and displacements, and the strain caused by

nonuniformity in the temperature field or in the thermal 

expansion properties of the material, or both 

" where , , T is the temperature 

field, and Tref is the temperature field at which the body is stress 

and strain free

εij

εij

σ

εij

T

εij

σ

εij =εij

σ

+εij

T

εij

σ

=Sijrsσrs

εij

T

=αij T−Tref
 74 

THE GENERALIZED DUHAMEL-NEUMANN LAW

CONTINUED

" The general form of the Duhamel-Neumann law is given in expanded 

form by

 and in indicial form by

"Sijkl are the components of the (4th-order) compliance tensor, at T = Tref,

that appear in the generalized Hooke’s law and 

αij are the coefficients 

of linear thermal expansion (with units of temperature-1

)

ε11

ε22

ε33

ε23

ε13

ε12

ε32

ε31

ε21

=

S1111

S1122S1133S1123S1113S1112S1132S1131S1121

S2211S2222S2233S2223S2213S2212S2232S2231S2221

S3311S3322S3333S3323S3313S3312S3332S3331S3321

S2311S2322S2333S2323S2313S2312S2332S2331S2321

S1311S1322S1333S1323S1313S1312S1332S1331S1321

S1211S1222S1233S1223S1213S1212S1232S1231S1221

S3211S3222S3233S3223S3213S3212S3232S3231S3221

S3111S3122S3133S3123S3113S3112S3132S3131S3121

S2111S2122S2133S2123S2113S2112S2132S2131S2121

σ11

σ22

σ33

σ23

σ13

σ12

σ32

σ31

σ21

+

α11

α22

α33

α23

α13

α12

α32

α31

α21

T−Tref

εij=Sijklσkl+αijT−Tref
 75 

THE GENERALIZED DUHAMEL-NEUMANN LAW

CONTINUED

" The inverted form of the Duhamel-Neumann law is given in expanded 

form by

 and in indicial form by

"Cijkl are the components of the (4th-order) stiffness tensor, at T = Tref,

that appear in the generalized Hooke’s law and the column vector on 

right-hand side of the matrix equation contains the mechanical strains 

σ11

σ22

σ33

σ23

σ13

σ12

σ32

σ31

σ21

=

C1111

C1122C1133C1123C1113C1112C1132C1131C1121

C2211C2222C2233C2223C2213C2212C2232C2231C2221

C3311C3322C3333C3323C3313C3312C3332C3331C3321

C2311C2322C2333C2323C2313C2312C2332C2331C2321

C1311C1322C1333C1323C1313C1312C1332C1331C1321

C1211C1222C1233C1223C1213C1212C1232C1231C1221

C3211C3222C3233C3223C3213C3212C3232C3231C3221

C3111C3122C3133C3123C3113C3112C3132C3131C3121

C2111C2122C2133C2123C2113C2112C2132C2131C2121

ε11

ε22

ε33

ε23

ε13

ε12

ε32

ε31

ε21

−

α11

α22

α33

α23

α13

α12

α32

α31

α21

T−Tref

σij=Cijklεkl−αklT−Tref

=

Cijklεkl

σ
 76 

THE GENERALIZED DUHAMEL-NEUMANN LAW

CONTINUED

" The inverted form of the Duhamel-Neumann law is also expressed often 

in matrix form by

 and in indicial form by

"βij are called the thermal moduli

σ11

σ22

σ33

σ23

σ13

σ12

σ32

σ31

σ21

=

C1111

C1122C1133C1123C1113C1112C1132C1131C1121

C2211C2222C2233C2223C2213C2212C2232C2231C2221

C3311C3322C3333C3323C3313C3312C3332C3331C3321

C2311C2322C2333C2323C2313C2312C2332C2331C2321

C1311C1322C1333C1323C1313C1312C1332C1331C1321

C1211C1222C1233C1223C1213C1212C1232C1231C1221

C3211C3222C3233C3223C3213C3212C3232C3231C3221

C3111C3122C3133C3123C3113C3112C3132C3131C3121

C2111C2122C2133C2123C2113C2112C2132C2131C2121

ε11

ε22

ε33

ε23

ε13

ε12

ε32

ε31

ε21

+

β 11

β 22

β 33

β 23

β 13

β 12

β 32

β 31

β 21

T−Tref

σij=Cijklεkl+βij

T−Tref
 77 

THE GENERALIZED DUHAMEL-NEUMANN LAW

CONTINUED

" By noting that the Duhamel-Neumann law becomes the generalized 

Hooke’s law when , the following symmetry conditions must 

hold

 which indicates 21 independent compliance or stiffness coefficients

" Symmetry of the strain tensor also yields , and reduces the 

number of independent coefficients of linear thermal expansion from 

9 to 6

" Likewise, symmetry of the stress tensor also yields , and 

reduces the number of independent thermal moduli from 9 to 

6

T=Tref

Sijkl=SjiklSijkl=SijlkSijkl=Sklij

Cijkl=CjiklCijkl=CijlkCijkl=Cklij

αij =αji

βij =βji
 78 

THE GENERALIZED DUHAMEL-NEUMANN LAW

CONCLUDED

" The expanded forms of the Duhamel-Neumann law are now given by

ε11

ε22

ε33

2ε23

2ε13

2ε12

=

S1111

S1122S11332S11232S11132S1112

S1122S2222S22332S22232S22132S2212

S1133S2233S33332S33232S33132S3312

2S11232S22232S33234S23234S23134S2312

2S11132S22132S33134S23134S13134S1312

2S11122S22122S33124S23124S13124S1212

σ11

σ22

σ33

σ23

σ13

σ12

+

α11

α22

α33

2α23

2α13

2α12

T−Tref

σ11

σ22

σ33

σ23

σ13

σ12

=

C1111

C1122C1133C1123C1113C1112

C1122C2222C2233C2223C2213C2212

C1133C2233C3333C3323C3313C3312

C1123C2223C3323C2323C2313C2312

C1113C2213C3313C2313C1313C1312

C1112C2212C3312C2312C1312C1212

ε11

ε22

ε33

2ε23

2ε13

2ε12

+

β 11

β 22

β 33

β 23

β 13

β 12

T−Tref
 79 

EQUATIONS FOR THE THERMAL MODULI

" The thermal moduliβij are related to the coefficients of linear thermal 

expansion by or by

" Note that βij have units of stress/ temperature; e.g., lb/in

2

-

o

F

" Similarly, αij have units of temperature-1

βij=−Cijklαkl

β 11

β 22

β 33

β 23

β 13

β 12

=−

C1111

C1122C1133C1123C1113C1112

C1122C2222C2233C2223C2213C2212

C1133C2233C3333C3323C3313C3312

C1123C2223C3323C2323C2313C2312

C1113C2213C3313C2313C1313C1312

C1112C2212C3312C2312C1312C1212

α11

α22

α33

2α23

2α13

2α12
 80 

STRAIN-ENERGY DENSITY FOR THERMAL LOADING

" The symmetry relations for a thermoelastic solid can also be 

obtained from first principles by enforcing path independence of the 

strain-energy density function 

U

" The strain-energy densityU of a generally thermoelastic solid is 

defined as the work of the internal stresses done through 

mechanical deformations

" The strain-energy-density increment dU is given for this case by

 where the stress-induced, mechanical strains are given by

 and the stresses depend on the mechanical strains; that is,

Cijkl=Cklij

d

U

=σ11dε11

σ

+σ22dε22

σ

+σ33dε33

σ

+2σ23dε23

σ

+2σ13dε13

σ

+2σ12dε12

σ

εij

σ

=εij −εij

T

σij=σijε11

σ

, ε22

σ

, ε33

σ

, ε23

σ

, ε13

σ

, ε12

σ
 81 

STRAIN-ENERGY DENSITY FOR THERMAL LOADING

CONTINUED

" It is important to emphasize that the mechanical strains

include strains generated by thermal stresses associated with a 

nonuniform temperature field or spatial variations in the coefficients of 

thermal expansion

" The expression

 is written compactly in indicial form as

εij

σ

=εij −εij

T

d

U

=σ11dε11

σ

+σ22dε22

σ

+σ33dε33

σ

+2σ23dε23

σ

+2σ13dε13

σ

+2σ12dε12

σ

d

U

=σij εpq

σ

dεij

σ
 82 

STRAIN-ENERGY DENSITY FOR THERMAL LOADING

CONTINUED

" The strain-energy density U is obtained by integrating dU over the 

deformation associated with a thermomechanical loading process

that starts at a stress- and strain-free state and ends at a particular 

stress and strain state; that is,

" Because no mechanical work of the internal forces within a body is 

lost during a conservative, elastic, thermomechanical loadingunloading process, it follows that , which 

implies 

 and 

U

=σijεpq

σ

dεij

σ

0

εpq

σ

=

U

εpq

σ

U

=σij εpq

σ

dεij

σ

=0

d

U

=

∂

U

∂εij

σ

dεij

σ

∂

2

U

∂εij

σ

∂εkl

σ

=

∂

2

U

∂εkl

σ

∂εij

σ
 83 

STRAIN-ENERGY DENSITY FOR THERMAL LOADING

CONCLUDED

" Equating and gives 

" Then, and give 

" gives 

" gives 

" Thus, yields 

d

U

=σij εpq

σ

dεij

σ

d

U

=

∂

U

∂εij

σ

dεij

σ

∂

U

∂εij

σ

=σij εpq

σ

∂

2

U

∂εij

σ

∂εkl

σ

=

∂

2

U

∂εkl

σ

∂εij

σ

∂

U

∂εij

σ

=σij εpq

σ

∂σij

∂εkl

σ

=

∂σkl

∂εij

σ

σij =Cijrsεrs

σ

∂σij

∂εkl

σ

=

∂

∂εkl

σ

Cijrsεrs

σ

=Cijrs

∂εrs

σ

∂εkl

σ

=Cijrsδrkδsl=Cijkl

σij =Cijpqεpq

σ

∂σkl

∂εij

σ

=

∂

∂εij

σ

Cklpqεpq

σ

=Cklpq

∂εpq

σ

∂εij

σ

=Cklpqδpiδqj=Cklij

∂σij

∂εkl

σ

=

∂σkl

∂εij

σ

Cijkl=Cklij
 84 

COMPLEMENTARY STRAIN-ENERGY DENSITY FOR 

THERMAL LOADING

" The symmetry condition is obtained by examining the 

complementary strain-energy density functionU* 

" An expression for U* is obtained by first requiring that a one-to-one 

relationship

exists between the stresses and strains, and byexpressing

 or 

" Next, the product rule of differentiation is used to get 

" In the part , the stress-induced, mechanical strains are taken 

as the independent variables

" In the part , the stresses are taken as the independent 

variables

Sijkl=Sklij

σijεij =σij εij

σ

+αij T−Trefσijεij =σijεij

σ

+σijαij T−Tref

dσijεij =σijdεij

σ

+εij

σ

dσij +dσijαij T−Tref

σijdεij

σ

εij

σ

dσij
 85 

COMPLEMENTARY STRAIN-ENERGY DENSITY FOR 

THERMAL LOADING - CONTINUED

" Next, the expression is 

integrated from the initial stress- and strain- free state to the current 

stress and strain state; i.e.,

" In the term it is presumed that the stresses are known as 

functions of the strains

" This term can also be expressed as , where it is 

presumed that the strains are known as functions of the stresses

" Both terms yield , the current values of the stresses and 

strains

dσijεij =σijdεij

σ

+εij

σ

dσij +dσijαij T−Tref

dσijεij

0

εpq

=σij εpq

σ

dεij

σ

0

εpq

σ

+εij

σ

σpqdσij

0

σpq

+dσijαij T−Tref

0

εpq

dσijεij

0

εpq

dσijεij

0

σpq

σijεij
 86 

COMPLEMENTARY STRAIN-ENERGY DENSITY FOR 

THERMAL LOADING - CONCLUDED

" Using the previous expression and the definition of the strain-energy 

density function

U gives 

" The complementary strain-energy density functionU* is defined as 

 such that 

" Note that 

" Legendre’s transformation takes the form 

" The complementary relationship of the strain-energy density function 

and the complementary strain-energy density function for the 

thermoelastic case are illustrated on the next two charts for a onedimensional stress and strain state 

σijεij =

U

εpq

σ

+εij

σ

σpqdσij

0

σpq

+σijαij T−Tref

U*

=εij

σ

σpqdσij

0

σpq

+σijαij T−Tref

σijεij =

U

εpq

σ

+

U*

σpq,T

d

U*

=εij

σ

dσij +dσijαij T−Tref

U*

σpq,T=σijεij −

U

εpq

σ
 87 

ILLUSTRATION OF THERMOELASTIC ENERGY DENSITIES

ONE-DIMENSIONAL CASE

σ

Area= σε

Total strain

Stress

=

+

dσ

ε

σ

ε

Total strain

ε

σ

dε

σ

Stress

Total strain

ε

σ

ε

Stress

σσ

Loading process = mechanical loading followed by thermal loading

σε=

U

ε

σ

+

U*

σ,T

Area=ε

σ

σdσ 0

σ

Area=σ

ε

+ σα

T−Tref

=

U*

σ,T σ

dε

σ

0

εσ

=

U

ε

σ

α

T−Tref

α

T−Tref

ε

σ

=

ε

σ

σ

σ=σ

ε

σ
 88 

ILLUSTRATION OF THERMOELASTIC ENERGY DENSITIES

ONE-DIMENSIONAL CASE - CONTINUED

σ

Area= σε

Total strain

Stress

=

+

dσ

ε

Mechanical strain

dε

σ

Stress

Total strain

ε

Stress

σ

σ

Loading process = thermal loading followed by mechanical loading

ε

σ

ε

σ

σε=

U

ε

σ

+

U*

σ,T

α

T−Tref

ε

σ

=

ε

σ

σ=σ

ε

σ σ

α

T−Tref

Area=σ

ε

σ

dε

σ

0

εσ

=

U

ε

σ Area

=ε

σ

σdσ 0

σ

+ σα

T−Tref

=

U*

σ,T
 89 

ILLUSTRATION OF THERMOELASTIC ENERGY DENSITIES

ONE-DIMENSIONAL CASE - CONTINUED

" The previous figures illustrate that path independence of the elastic 

loading-unloading process implies path independence of the 

complementary strain-energy density function

" If the material were inelastic, the quantity of complementary strainenergy density function would depend on the loading process

" Thus, and 

" gives 

 which reduces to 

U*

σpq,T=d

U*

=0

d

U*

=

∂

U*

∂σij

dσij +

∂

U*

∂T dT

d

U*

=εij

σ

dσij +dσijαij T−Trefd

U*

=εij

σ

+αij T−Trefdσij +σijαijdT

d

U*

=εijdσij+σijαijdT
 90 

ILLUSTRATION OF THERMOELASTIC ENERGY DENSITIES

ONE-DIMENSIONAL CASE - CONCLUDED

" Equating with gives 

 and 

" Now, implies, and is implied by, the conditions

 and 

" and give 

d

U*

=

∂

U*

∂σij

dσij +

∂

U*

∂T dT

d

U*

=εijdσij +σijαijdT

∂

U*

∂σij

=εij=εij

σ

+αijT−Tref

∂

U*

∂T

=σijαij

U*

σpq,T=d

U*

=0

∂

2

U*

∂σij ∂

σkl

=

∂

2

U*

∂σkl

∂σij

∂

2

U*

∂σij ∂

T

=

∂

2

U*

∂T∂σij

∂

2

U*

∂σij ∂

σkl

=

∂

2

U*

∂σkl

∂σij

∂

U*

∂σij

=εij

∂εij

∂σkl

=

∂εkl

∂σij
 91 

PROOF THAT Sijkl = Sklij FOR THERMOELASTIC SOLIDS

" First, and satisfy 

identically

" gives 

" gives 

" Thus, yields 

∂

U*

∂σij

=εij

σ

+αij T−Tref

∂

U*

∂T

=σijαij

∂

2

U*

∂σij ∂

T

=

∂

2

U*

∂T∂σij

εij =Sijrsσrs+αij T−Tref

∂εij

∂σkl

=

∂

∂σkl

Sijrsσrs+αij T−Tref=Sijrs

∂σrs

∂σkl

=Sijrsδrkδsl=Sijkl

εij =Sijpqσpq+αij T−Tref

∂εkl

∂σij

=

∂

∂σij

Sklpqσpq+αij T−Tref=Sklpq

∂σpq

∂σij

=Sklpqδpiδqj=Sklij

∂εij

∂σkl

=

∂εkl

∂σij

Sijkl=Sklij
 92 

CLAPEYRON’S FORMULA FOR THERMOELASTIC SOLIDS

" Clapeyron’s formula remains the same because the 

strain-energy density is based on stress-induced, mechanical work

" Clapeyron’s formula is expressed in terms of the total strains by using 

 to get 

U

=

1

2

σijεij

σ

εij

σ

=εij −αij T−Tref

U

=

1

2

σijεij−

1

2

σijαijT−Tref
 93 

STRAIN-ENERGY DENSITY EXPRESSIONS

" By definition, for a homogeneous, anisotropic, linear-

thermoelastic solid

" Using gives 

" For isotropic materials,

" In expanded form,

U

=

1

2

σijεij

σ

εij

σ

=Sijrsσrs

U

=

1

2

Sijrsσijσrs

U

=

1

2E

1+νσijσij −νσkk

2

U

=

1

2E

σ11

2

+σ22

2

+σ33

2

−

ν

E

σ11σ22+σ22σ33+σ11σ33+

1+ν

E

σ12

2

+σ23

2

+σ13

2
 94 

STRAIN-ENERGY DENSITY EXPRESSIONS

CONCLUDED

" Using gives 

" Further, using , and gives 

" For isotropic materials,

or

σij =Cijklεkl

σ

U

=

1

2

Cijklεij

σ

εkl

σ

εij

σ

=εij −αijΘΘ=T−TrefCklij=Cijkl

U

=

1

2

Cijklεijεkl−2εijαklΘ+αijαklΘ2

U

=

E

21

+ν

εijεij +

E

21

−

2

ν

εkk

ν

1

+ν

εkk− αΘ+

3E

21

−

2

ν

α2Θ2

U

=

νE

21

+ν

1

−

2

ν

ε11+ε22+ε33

2

+

E

21

+ν

ε11

2

+ε22

2

+ε33

2

+2ε12

2

+2ε23

2

+2ε13

2

−

α

E

1

−

2

ν

ε11+ε22+ε33Θ+

3

α2

E

21

−

2

ν

Θ2
 95 

ABRIDGED NOTATION

AND

ELASTIC CONSTANTS
 96 

ABRIDGED NOTATION FOR CONSTITUTIVE EQUATIONS

" The abridged notation presented subsequently is attributed to 

Woldemar Voigt (1850-1919), and was developed for expressing the 

constitutive equations in the simpler, more intuitive matrix notation

" The components of the stress, strain, thermal expansion, and thermal 

moduli tensors are written as column vectors

" The order of the elements is obtained from cyclic permutations

of the numbers 1, 2, and 3

" The term "tensor," as it is used today, was introduced byVoigt in 1899 

σ11

σ22

σ33

σ23

σ13

σ12

→

σ1

σ2

σ3

σ4

σ5

σ6

ε11

ε22

ε33

2ε23

2ε13

2ε12

→

ε1

ε2

ε3

ε4

ε5

ε6

β 11

β 22

β 33

β 23

β 13

β 12

→

β 1

β 2

β 3

β 4

β 5

β 6

α11

α22

α33

2α23

2α13

2α12

→

α1

α2

α3

α4

α5

α6
 97 

ABRIDGED NOTATION FOR CONSTITUTIVE EQUATIONS

CONTINUED

" The components of the compliance and stiffness tensors are 

expressed as

" Materials that can be characterized by the matrices given above when 

they are symmetric are said to possess complete Voigt symmetry

S1111S1122S11332S11232S11132S1112

S2211S2222S22332S22232S22132S2212

S3311S3322S33332S33232S33132S3312

2S23112S23222S23334S23234S23134S2312

2S13112S13222S13334S13234S13134S1312

2S12112S12222S12334S12234S12134S1212

→

S11S12S13S14S15S16

S21S22S23S24S25S26

S31S32S33S34S35S36

S41S42S43S44S45S46

S51S52S53S54S55S56

S61S62S63S64S65S66

C1111C1122C1133C1123C1113C1112

C2211C2222C2233C2223C2213C2212

C3311C3322C3333C3323C3313C3312

C2311C2322C2333C2323C2313C2312

C1311C1322C1333C1323C1313C1312

C1211C1222C1233C1223C1213C1212

→

C11C12C13C14C15C16

C21C22C23C24C25C26

C31C32C33C34C35C36

C41C42C43C44C45C46

C51C52C53C54C55C56

C61C62C63C64C65C66
 98 

ABRIDGED NOTATION FOR CONSTITUTIVE EQUATIONS

CONTINUED

" The constitutive equations are often expressed in a nontensorial,

indicial form given by

" Similarly, the constitutive equations are often expressed in matrix form

given by

and

or

εi =Sijσj +αi T

−Tref

σi =Cijεj +βi T

−Tref

ε=Sσ+α

T−Tref

σ=Cε−α

T−Tref

σ=Cε+β

T−Tref
 99 

ABRIDGED NOTATION FOR CONSTITUTIVE EQUATIONS

CONTINUED

ε1

ε2

ε3

ε4

ε5

ε6

=

S11

S12S13S14S15S16

S12S22S23S24S25S26

S13S23S33S34S35S36

S14S24S34S44S45S46

S15S25S35S45S55S56

S16S26S36S46S56S66

σ1

σ2

σ3

σ4

σ5

σ6

+

α1

α2

α3

α4

α5

α6

T−Tref

ε=Sσ+α

T−Tref

σ1

σ2

σ3

σ4

σ5

σ6

=

C11

C12C13C14C15C16

C12C22C23C24C25C26

C13C23C33C34C35C36

C14C24C34C44C45C46

C15C25C35C45C55C56

C16C26C36C46C56C66

ε1

ε2

ε3

ε4

ε5

ε6

+

β 1

β 2

β 3

β 4

β 5

β 6

T−Tref

σ=Cε+β

T−Tref
 100 

ABRIDGED NOTATION FOR CONSTITUTIVE EQUATIONS

CONTINUED

" Often, the following mixed-abridged notation is used

ε11

ε22

ε33

2ε23

2ε13

2ε12

=

S11

S12S13S14S15S16

S12S22S23S24S25S26

S13S23S33S34S35S36

S14S24S34S44S45S46

S15S25S35S45S55S56

S16S26S36S46S56S66

σ11

σ22

σ33

σ23

σ13

σ12

+

α11

α22

α33

2α23

2α13

2α12

T−Tref

σ11

σ22

σ33

σ23

σ13

σ12

=

C11

C12C13C14C15C16

C12C22C23C24C25C26

C13C23C33C34C35C36

C14C24C34C44C45C46

C15C25C35C45C55C56

C16C26C36C46C56C66

ε11

ε22

ε33

2ε23

2ε13

2ε12

+

β 11

β 22

β 33

β 23

β 13

β 12

T−Tref
 101 

ABRIDGED NOTATION FOR CONSTITUTIVE EQUATIONS

CONCLUDED

" The thermal moduli are given by

or

β 1

β 2

β 3

β 4

β 5

β 6

=−

C11C12C13C14C15C16

C12C22C23C24C25C26

C13C23C33C34C35C36

C14C24C34C44C45C46

C15C25C35C45C55C56

C16C26C36C46C56C66

α1

α2

α3

2

α4

2

α5

2

α6

β 11

β 22

β 33

β 23

β 13

β 12

=−

C11C12C13C14C15C16

C12C22C23C24C25C26

C13C23C33C34C35C36

C14C24C34C44C45C46

C15C25C35C45C55C56

C16C26C36C46C56C66

α11

α22

α33

2

α23

2

α13

2

α12
 102 

CLAPEYRON’S FORMULA IN ABRIDGED NOTATION

" Clapeyron’s formula for the strain-energy density of a linearthermoelastic solid was given previously b

y or 

 where 

" A convenient matrix form of Clapeyron’s formula is obtained by

using the following notation

 where the superscript T denotes matrix or vector transposition

" First, by inspection, it follows that where

U

=

1

2

σijεij

σ

U

=

1

2

σijεij−

1

2

σijαijT−Tref

εij

σ

=εij −αij T−Tref

σ

T

=

σ11σ22σ33σ23σ13σ12

ε

σ

T

=

ε11

σ

ε22

σ

ε33

σ

2ε23

σ

2ε13

σ

2ε12

σ

ε

T

=

ε11ε22ε332ε232ε132ε12

α

T

=

α11α22α332α232α132α12

ε

σ

=

ε

−

α

Θ

Θ=T−Tref
 103 

CLAPEYRON’S FORMULA IN ABRIDGED NOTATION

CONTINUED

" Next, noting that it 

follows that in matrix notation

" Therefore,Clapeyron’s formulafor the strain-energy density of a 

linear-thermoelastic solid is given in matrix from by

 or 

" An alternate form of Clapeyron’s formula is obtained as follows

" First, define such that becomes 

σijεij

σ

=σ11

ε11

σ

+

σ22

ε22

σ

+

σ33

ε33

σ

+

σ23

2ε23

σ

+

σ13

2ε13

σ

+

σ12

2ε12

σ

σijεij

σ

=σ

T

ε

σ

=ε

σ T

σ

U

=

1

2

σ

T

ε

σ

U

=

1

2

σ

T

ε−

1

2

σ

T

αΘ

σ

Θ

=Cα

Θ

σ=Cε

σ

σ=Cε−σ

Θ
 104 

CLAPEYRON’S FORMULA IN ABRIDGED NOTATION

CONTINUED

" Then, gives 

" Next, becomes 

" Also,

"Transposition of gives ,

where symmetry of [C] has been used

" Using the last expression with gives

 which simplifies

 to 

σ

Θ

=Cα

Θ

α

Θ

=C − 1σ

Θ

ε

σ

=

ε

−

α

Θ

ε

σ

=

ε

−

C − 1σ

Θ

σ

T

ε

σ

=σ

T

ε

−

σ

T

C − 1σ

Θ

σ=Cε−σ

Θ

σ

T

=ε

T

C−σ

Θ

T

σ

T

ε

σ

=σ

T

ε

−

σ

T

C − 1σ

Θ

σ

T

ε

σ

=σ

T

ε

−

ε

T

CC − 1σ

Θ

+

σ

Θ

T

C − 1σ

Θ

σ

T

ε

σ

=σ

T

−σ

Θ

ε

+

σ

Θ

T

C − 1σ

Θ
 105 

CLAPEYRON’S FORMULA IN ABRIDGED NOTATION

CONCLUDED

" Next, using gives 

" Using the last equation with 

gives 

" Therefore, Clapeyron’s formula becomes

σ

Θ

=Cα

Θ

σ

Θ

T

C − 1σ

Θ

=α

T

Cα

Θ2

σ

T

ε

σ

=σ

T

−σ

Θ

ε

+

σ

Θ

T

C − 1σ

Θ

σ

T

ε

σ

=σ

T

−σ

Θ

ε

+

α

T

Cα

Θ2

U

=

1

2

σ

T

ε

σ

U

=

1

2

σ

T

−σ

Θ

T

ε

+

1

2

α

T

Cα

Θ

2
 106 

PHYSICAL MEANING OF THE ELASTIC CONSTANTS

" The shaded terms shown below correspond to independent interaction 

between pure extensional stresses and strains

" The shaded terms shown below correspond to independent interaction 

between pure shearing stresses and strains

S11S12S13S14S15S16

S12S22S23S24S25S26

S13S23S33S34S35S36

S14S24S34S44S45S46

S15S25S35S45S55S56

S16S26S36S46S56S66

C11C12C13C14C15C16

C12C22C23C24C25C26

C13C23C33C34C35C36

C14C24C34C44C45C46

C15C25C35C45C55C56

C16C26C36C46C56C66

S11S12S13S14S15S16

S12S22S23S24S25S26

S13S23S33S34S35S36

S14S24S34S44S45S46

S15S25S35S45S55S56

S16S26S36S46S56S66

C11C12C13C14C15C16

C12C22C23C24C25C26

C13C23C33C34C35C36

C14C24C34C44C45C46

C15C25C35C45C55C56

C16C26C36C46C56C66
 107 

PHYSICAL MEANING OF THE ELASTIC CONSTANTS

CONTINUED

" The shaded terms shown below correspond to coupling of interactions 

between pure extensional stresses and strains

" The shaded terms shown below correspond to coupling of interactions 

between pure shearing stresses and strains

S11S12S13S14S15S16

S12S22S23S24S25S26

S13S23S33S34S35S36

S14S24S34S44S45S46

S15S25S35S45S55S56

S16S26S36S46S56S66

C11C12C13C14C15C16

C12C22C23C24C25C26

C13C23C33C34C35C36

C14C24C34C44C45C46

C15C25C35C45C55C56

C16C26C36C46C56C66

S11S12S13S14S15S16

S12S22S23S24S25S26

S13S23S33S34S35S36

S14S24S34S44S45S46

S15S25S35S45S55S56

S16S26S36S46S56S66

S11S12S13S14S15S16

S12S22S23S24S25S26

S13S23S33S34S35S36

S14S24S34S44S45S46

S15S25S35S45S55S56

S16S26S36S46S56S66

C11C12C13C14C15C16

C12C22C23C24C25C26

C13C23C33C34C35C36

C14C24C34C44C45C46

C15C25C35C45C55C56

C16C26C36C46C56C66
 108 

PHYSICAL MEANING OF THE ELASTIC CONSTANTS

CONCLUDED

" The shaded terms shown below correspond to coupling or interactions 

between extensional and shearing behavior

S11S12S13S14S15S16

S12S22S23S24S25S26

S13S23S33S34S35S36

S14S24S34S44S45S46

S15S25S35S45S55S56

S16S26S36S46S56S66

C11C12C13C14C15C16

C12C22C23C24C25C26

C13C23C33C34C35C36

C14C24C34C44C45C46

C15C25C35C45C55C56

C16C26C36C46C56C66
 109 

TRANSFORMATION

EQUATIONS
 110 

TRANSFORMATION OF [C] AND [S]

" Consider a general orthogonal transformation between the Cartesian 

coordinates and at a fixed point P of a body

" The orthonormal bases for the two coordinate systems, for an 

arbitrary point P of a body, are indicated on the figure below

" Because only a fixed point P

is considered, coordinate 

translations are excluded

" Although two right-handed

coordinate systems are 

shown in the figure, there is 

no such restriction on the 

following development of the 

transformation equations

x1, x2, x3 x1′, x2′, x3′

Anisotropic

body, B

P

i 1

i 2

i 3

i 1′

i 2′

i 3′
 111 

TRANSFORMATION OF [C] AND [S] - CONTINUED

" The general relationship between the two orthonormal bases is given 

by the following matrix representations

" or, in abridged form,

" or, in abridged form,

" Examining these two matrices indicates that 

"Transformations of this type are known as orthogonal 

transformations and preserve the lengths of, and the angles 

between, vectors

" and 

i 1′

i 2′

i 3′

=

a1′1 a1′2 a1′3

a2′1 a2′2 a2′3

a3′1 a3′2 a3′3

i 1

i 2

i 3

i′ =a i

i 1

i 2

i 3

=

a1′1 a2′1 a3′1

a1′2 a2′2 a3′2

a1′3 a2′3 a3′3

i 1′

i 2′

i 3′

i =a − 1i′

a

− 1

=a

T

a

p′q

≡ i p′ • i q

= i q• i p′ =aqp′ a

p′q

=aqp′ ≠apq′ =a

q′p
 112 

TRANSFORMATION OF [C] AND [S] - CONTINUED

" The matrix equation is expressed in indicial notation by

" Likewise, is expressed in indicial notation by

" Important relationships between the direction cosines are obtained 

by enforcing the two orthonormality conditions and 

" These conditions yield the relationships

 and 

" Each indicial equation possesses six independent relations 

i′ =a i

i k′ =ak′qi q

i =a

− 1

i′ i p

=ar′pir′

ak′q

i k′ • i p′ =δkp

i m• i n =δmn

a

k′q

a

p′q

=δkpa

q′k

a

q′p

=δkp
 113 

TRANSFORMATION OF [C] AND [S] - CONTINUED

" The total of twelve independent relations are given in tabular form 

below:

a1′1

2

+a1′2

2

+a1′3

2

=1

a2′1a1′1 +a2′2a1′2 +a2′3a1′3 =0

a3′1a1′1 +a3′2a1′2 +a3′3a1′3 =0

a2′1

2

+a2′2

2

+a2′3

2

=1

a3′1a2′1 +a3′2a2′2 +a3′3a2′3 =0

a3′1

2

+a3′2

2

+a3′3

2

=1

k p

ak′qa

p′q

=δkp

1 1

2 

1

3 

1

2 

2

3 2

3 

3

k p

1 1

2 

1

3 

1

2 

2

3 2

3 

3

a

q′kaq′p

=δkp

a1′1

2

+a2′1

2

+a3′1

2

=1

a1′2a1′1 +a2′2a2′1 +a3′2a3′1 =0

a1′3a1′1 +a2′3a2′1 +a3′3a3′1 =0

a1′2

2

+a2′2

2

+a3′2

2

=1

a1′3a1′2 +a2′3a2′2 +a3′3a3′2 =0

a1′3

2

+a2′3

2

+a3′3

2

=1
 114 

TRANSFORMATION OF [C] AND [S] - CONTINUED

" By using the abridged notation, matrix forms of the stress and strain 

transformation equations can be obtained that are given by

 and 

 where

 and

σ′ =T

σ

σε′ =T

ε

ε

σ=

σ11

σ22

σ33

σ23

σ13

σ12

=

σ1

σ2

σ3

σ4

σ5

σ6

ε=

ε11

ε22

ε33

2ε23

2ε13

2ε12

=

ε1

ε2

ε3

ε4

ε5

ε6

σ′ =

σ1′1′

σ2′2′

σ3′3′

σ2′3′

σ1′3′

σ1′2′

=

σ1′

σ2′

σ3′

σ4′

σ5′

σ6′

ε′ =

ε1′1′

ε2′2′

ε3′3′

2ε2′3′

2ε1′3′

2ε1′2′

=

ε1′

ε2′

ε3′

ε4′

ε5′

ε6′
 115 

TRANSFORMATION OF [C] AND [S] - CONTINUED

" Inspection of the matrices shown above indicates that when the offdiagonal terms vanish, which happens for certain transformations,

the two matrices are identical

T

σ

=

a1′1

2

a1′2

2

a1′3

2

2a1′2a1′3 2a1′1a1′3 2a1′1a1′2

a2′1

2

a2′2

2

a2′3

2

2a2′2a2′3 2a2′1a2′3 2a2′1a2′2

a3′1

2

a3′2

2

a3′3

2

2a3′2a3′3 2a3′1a3′3 2a3′1a3′2

a2′1a3′1 a2′2a3′2 a2′3a3′3 a2′2a3′3 +a2′3a3′2 a2′1a3′3 +a2′3a3′1 a2′1a3′2 +a2′2a3′1

a1′1a3′1 a1′2a3′2 a1′3a3′3 a1′2a3′3 +a1′3a3′2 a1′1a3′3 +a1′3a3′1 a1′1a3′2 +a1′2a3′1

a1′1a2′1 a1′2a2′2 a1′3a2′3 a1′2a2′3 +a1′3a2′2 a1′1a2′3 +a1′3a2′1 a1′1a2′2 +a1′2a2′1

T

ε

=

a1′1

2

a1′2

2

a1′3

2

a1′2a1′3 a1′1a1′3 a1′1a1′2

a2′1

2

a2′2

2

a2′3

2

a2′2a2′3 a2′1a2′3 a2′1a2′2

a3′1

2

a3′2

2

a3′3

2

a3′2a3′3 a3′1a3′3 a3′1a3′2

2a2′1a3′1 2a2′2a3′2 2a2′3a3′3 a2′2a3′3 +a2′3a3′2 a2′1a3′3 +a2′3a3′1 a2′1a3′2 +a2′2a3′1

2a1′1a3′1 2a1′2a3′2 2a1′3a3′3 a1′2a3′3 +a1′3a3′2 a1′1a3′3 +a1′3a3′1 a1′1a3′2 +a1′2a3′1

2a1′1a2′1 2a1′2a2′2 2a1′3a2′3 a1′2a2′3 +a1′3a2′2 a1′1a2′3 +a1′3a2′1 a1′1a2′2 +a1′2a2′1

aj′p

≡ ij′ • i p
 116 

TRANSFORMATION OF [C] AND [S] - CONTINUED

i 1

i 2

i 3

x1

x2

x3

x1

x2

x3

x1′

x2′

x3′

i 1′

i 2′

i 3′

Initial coordinate frame Initial and new 

coordinate frame

P

P

aj′p

≡ ij′ • i p
 117 

TRANSFORMATION OF [C] AND [S] - CONTINUED

" Recall that the thermoelastic constitutive equations are 

expressed in symbolic form by

 and 

 where 

" In the coordinate frame the thermoelastic constitutive 

equations are expressed in symbolic form by

 and 

" By using the matrix form of the stress and strain transformation 

equations,

 becomes 

ε=Sσ+αΘσ=Cε+βΘ

Θ=

T−Tref

x1′, x2′, x3′

ε′ =S′ σ′ +α′ Θσ′ =C′ ε′ +β′ Θ

σ=Cε+βΘ

T

σ

− 1

σ′ =CT

ε

− 1

ε′ +βΘ
 118 

TRANSFORMATION OF [C] AND [S] - CONTINUED

" Premultiplying b

y gives

" Comparing this equation with it follows 

that

 and 

" Thus,

T

σ

σ′ =T

σ

CT

ε

− 1

ε′ +T

σ

βΘ

σ′ =C′ ε′ +β′ Θ

C′ =T

σ

CT

ε

− 1

β′ =T

σ

β

C=T

σ

− 1

C′ T

ε
 119 

TRANSFORMATION OF [C] AND [S] - CONTINUED

" Next, b

y using the matrix form of the stress and strain transformation 

equations, becomes

" Premultiplying b

y gives

" Comparing this equation with it follows 

that

 and 

" Thus, gives 

ε=Sσ+αΘ

T

ε

− 1

ε′ =ST

σ

− 1

σ′ +αΘ

T

ε

ε′ =T

ε

ST

σ

− 1

σ′ +T

ε

αΘ

ε′ =S′ σ′ +α′ Θ

S′ =T

ε

ST

σ

− 1

α′ =T

ε

α

S′ =T

ε

ST

σ

− 1

S=T

ε

− 1

S′ T

σ
 120 

TRANSFORMATION OF [C] AND [S]

CONCLUDED

" In summary:

S′ =T

ε

ST

σ

− 1

S=T

ε

− 1

S′ T

σ

C′ =T

σ

CT

ε

− 1

C=T

σ

− 1

C′ T

ε

α′ =T

ε

αβ′ =T

σ

β
 121 

TRANSFORMATIONS FOR DEXTRAL ROTATIONS 

ABOUT THE x

3 AXIS

" The term "dextral" refers to a right-handed rotation

i 1 i 2

i 3

x1

x2

x3

x1′

x2′

x3′

i 1′

i 2′

i 3′

,

,

θ3

θ3

ai′j ≡ ii′ • ij

Plane x

3 = 0
 122 

TRANSFORMATIONS FOR DEXTRAL ROTATIONS 

ABOUT THE x

3 AXIS - CONTINUED

" The matrix form of the stress-tensor transformation law is given by

 and is expressed symbolically by

σ1′1′

σ2′2′

σ3′3′

σ2′3′

σ1′3′

σ1′2′

=

cos2

θ3 sin2θ3 0

002sin

θ3cosθ3

sin2θ3 cos2θ3 0

00−2sin

θ3cosθ3

001000

000 cos

θ3 − sinθ3 0

000 sin

θ3 cosθ3 0

− sin

θ3cosθ3 sinθ3cosθ3 0

00 cos2

θ3 − sin2θ3

σ11

σ22

σ33

σ23

σ13

σ12

σ′ =T

σ

θ

3

σ
 123 

TRANSFORMATIONS FOR DEXTRAL ROTATIONS 

ABOUT THE x

3 AXIS - CONTINUED

" Similarly, the matrix form of the inverse transformation law is given by

 and is expressed symbolically by

 where 

σ11

σ22

σ33

σ23

σ13

σ12

=

cos2

θ3 sin2θ3 0

00

−

2sin

θ3cosθ3

sin2θ3 cos2θ3 0

002sin

θ3cosθ3

001000

000 cos

θ3 sinθ3 0

000

− sinθ3 cosθ3 0

sinθ3cosθ3 − sinθ3cosθ3 0

00 cos2

θ3 − sin2θ3

σ1′1′

σ2′2′

σ3′3′

σ2′3′

σ1′3′

σ1′2′

σ=T

σ

θ

3

− 1

σ′

T

σ

θ

3

− 1

=T

σ

−

θ

3
 124 

TRANSFORMATIONS FOR DEXTRAL ROTATIONS 

ABOUT THE x

3 AXIS - CONTINUED

" By using the second-order, symmetric tensor transformation equations,

the transformation law for the vector of engineering strains is given by

 which is expressed symbolically by

" Note that 

!1"1"

!2"2"

!3"3"

2!2"3"

2!1"3"

2!1"2"

=

cos2#3 sin2#3 0

00 sin

#3cos#3

sin2#3 cos2#3 0

00$ sin

#3cos#3

001000

000 cos

#3 $ sin#3 0

000 sin

#3 cos#3 0

$

2sin#3cos#3 2sin#3cos#3 0

00 cos2

#3 $ sin2#3

!11

!22

!33

2!23

2!13

2!12

!" =T

!

#

3

!

T

!

#

3

T

=T

%

#

3

$ 1

=T

%

$

#

3
 125 

TRANSFORMATIONS FOR DEXTRAL ROTATIONS 

ABOUT THE x

3 AXIS - CONTINUED

" Similarly, the matrix form of the inverse transformation law is given by

 and is expressed symbolically by

 where 

ε11

ε22

ε33

2ε23

2ε13

2ε12

=

cos2θ3 sin2θ3 0

00

− sinθ3cosθ3

sin2θ3 cos2θ3 0

00 sin

θ3cosθ3

001000

000 cos

θ3 sinθ3 0

000

− sinθ3 cosθ3 0

2sinθ3cosθ3 −2sinθ3cosθ3 0

00 cos2

θ3 − sin2θ3

ε1′1′

ε2′2′

ε3′3′

2ε2′3′

2ε1′3′

2ε1′2′

ε=T

ε

θ

3

− 1

ε′

T

ε

θ

3

− 1

=T

ε

−

θ

3

=T

σ

θ

3

T
 126 

TRANSFORMATIONS FOR DEXTRAL ROTATIONS 

ABOUT THE x

3 AXIS - CONTINUED

" The general expression for the transformation of the stiffness 

coefficients and thermal moduli are

 and 

" Noting that for a dextral rotation about the x

3 axis gives

" Similarly, the general expression for the inverse transformation of the 

stiffness coefficients is 

" Noting that for a dextral rotation about the x

3 axis gives 

C′ =T

σ

CT

ε

− 1

β′ =T

σ

β

T

ε

− 1

=

T

σ

T

C′ =T

σ

CT

σ

T

C=T

σ

− 1

C′ T

ε

T

σ

− 1

=

T

ε

T

C=T

ε

T

C′ T

ε
 127 

TRANSFORMATIONS FOR DEXTRAL ROTATIONS 

ABOUT THE x

3 AXIS - CONTINUED

" The general expression for the transformation of the compliance 

coefficients and thermal expansion coefficients are

 and 

" Noting that for a dextral rotation about the x

3 axis gives

" Similarly, the general expression for the inverse transformation of the 

compliance coefficients is 

" Noting that for a dextral rotation about the x

3 axis gives 

S′ =T

ε

ST

σ

− 1

α′ =T

ε

α

T

σ

− 1

=

T

ε

T

S′ =T

ε

ST

ε

T

S=T

ε

− 1

S′ T

σ

T

ε

− 1

=

T

σ

T

S=T

σ

T

S′ T

σ
 128 

TRANSFORMATIONS FOR DEXTRAL ROTATIONS ABOUT 

THE x

3 AXIS - SUMMAR

Y

S! =T

"

ST

"

T

S=T

#

T

S! T

#

C! =T

#

CT

#

T

C=T

"

T

C! T

"

T

#

=

cos2$3 sin2$3 0

00 2sin

$3cos$3

sin2$3 cos2$3 0

00% 2sin

$3cos$3

00 1 000

000 cos

$3 % sin$3 0

000 sin

$3 cos$3 0

% sin

$3cos$3 sin$3cos$3 0

00 cos2

$3 % sin2$3

T

"

=

cos2$3 sin2$3 0

00 sin

$3cos$3

sin2$3 cos2$3 0

00% sin

$3cos$3

00 1 000

000 cos

$3 % sin$3 0

000 sin

$3 cos$3 0

% 2sin

$3cos$3 2sin$3cos$3 0

00 cos2

$3 % sin2$3
 129 

TRANSFORMATIONS FOR DEXTRAL ROTATIONS ABOUT 

THE x

3 AXIS - SUMMAR

Y

!" =T

#

!

!=T

$

T

!"

%" =T

$

%

%=T

#

T

%"

T

$

=

cos2&3 sin2&3 0

00 2sin

&3cos&3

sin2&3 cos2&3 0

00' 2sin

&3cos&3

00 1 000

000 cos

&3 ' sin&3 0

000 sin

&3 cos&3 0

' sin

&3cos&3 sin&3cos&3 0

00 cos2

&3 ' sin2&3

T

#

=

cos2&3 sin2&3 0

00 sin

&3cos&3

sin2&3 cos2&3 0

00' sin

&3cos&3

00 1 000

000 cos

&3 ' sin&3 0

000 sin

&3 cos&3 0

' 2sin

&3cos&3 2sin&3cos&3 0

00 cos2

&3 ' sin2&3
 130 

TRANSFORMATIONS FOR DEXTRAL ROTATIONS 

ABOUT THE x

3 AXIS - CONTINUED

" Let and for convenience 

"Performing the calculations given by yields

m= cosθ3 n= sinθ3

C′ =T

σ

CT

σ

T

C1′1′ =m4C11+2m2n2C12+2C66+4mnm2C16+n2C26+n4C22

C1′2′ =m2n2C11+C22−4C66−2mnm2−n2C16−C26+m4+n4C12

C1′3′ =m2C13+n2C23+2mnC36

C1′4′ =m3C14+m2n2C46−C15−mn22C56−C24−n3C25

C1′5′ =m3C15+m2n2C56+C14+mn22C46+C25+n3C24

C1′6′ =m2m2−3n2C16−m3nC11−C12−2C66

+mn3C22−C12−2C66−n2n2−3m2C26
 131 

TRANSFORMATIONS FOR DEXTRAL ROTATIONS 

ABOUT THE x

3 AXIS - CONTINUED

C2′2′ =m4C22+2m2n2C12+2C66−4mnm2C26+n2C16+n4C11

C2′3′ =m2C23+n2C13−2mnC36

C2′4′ =m3C24−m2n2C46+C25+mn22C56+C14−n3C15

C2′5′ =m3C25−m2n2C56−C24−mn22C46−C15+n3C14

C2′6′ =m2m2−3n2C26+m3nC22−C12−2C66

−mn3C11−C12−2C66−n2n2−3m2C16

C3′3′ =C33

C3′4′ =mC34−nC35

C3′5′ =mC35+nC34
 132 

TRANSFORMATIONS FOR DEXTRAL ROTATIONS 

ABOUT THE x

3 AXIS - CONTINUED

C3′6′ =m2−n2C36+mnC23−C13

C4′4′ =m2C44+n2C55−2mnC45

C4′5′ =m2−n2C45+mnC44−C55

C4′6′ =m3C46−m2nC56+C14−C24−mn2C46−C15+C25+n3C56

C5′5′ =m2C55+n2C44+2mnC45

C5′6′ =m3C56+m2nC46+C25−C15−mn2C56+C14−C24−n3C46

C6′6′ =m2n2C11+C22−2C12−2mnm2−n2C16−C26+m2−n2 2C66
 133 

TRANSFORMATIONS FOR DEXTRAL ROTATIONS 

ABOUT THE x

3 AXIS - CONTINUED

" Again, let and 

"Performing the calculations given by yields

m= cosθ3 n= sinθ3

C=T

ε

T

C′ T

ε

C11=m4C1′1′ +2m2n2C1′2′ +2C6′6′ −4mnm2C1′6′ +n2C2′6′ +n4C2′2′

C12=m2n2C1′1′ +C2′2′ −4C6′6′ +2mnm2−n2C1′6′ −C2′6′ +m4+n4C1′2′

C13=m2C1′3′ +n2C2′3′ −2mnC3′6′

C14=m3C1′4′ −m2n2C4′6′ −C1′5′ −mn22C5′6′ −C2′4′ +n3C2′5′

C15=m3C1′5′ −m2n2C5′6′ +C1′4′ +mn22C4′6′ +C2′5′ −n3C2′4′

C16=m2m2−3n2C1′6′ +m3nC1′1′ −C1′2′ −2C6′6′

−mn3C2′2′ −C1′2′ −2C6′6′ −n2n2−3m2C2′6′
 134 

TRANSFORMATIONS FOR DEXTRAL ROTATIONS 

ABOUT THE x

3 AXIS - CONTINUED

C22=m4C2′2′ +2m2n2C1′2′ +2C6′6′ +4mnm2C2′6′ +n2C1′6′ +n4C1′1′

C23=m2C2′3′ +n2C1′3′ +2mnC3′6′

C24=m3C2′4′ +m2n2C4′6′ +C2′5′ +mn22C5′6′ +C1′4′ +n3C1′5′

C25=m3C2′5′ +m2n2C5′6′ −C2′4′ −mn22C4′6′ −C1′5′ −n3C1′4′

C26=m2m2−3n2C2′6′ −m3nC2′2′ −C1′2′ −2C6′6′

+mn3C1′1′ −C1′2′ −2C6′6′ −n2n2−3m2C1′6′

C33=C3′3′ C34=mC3′4′ +nC3′5′ C35=mC3′5′ −nC3′4′

C36=m2−n2C3′6′ − mnC2′3′ −C1′3′

C44=m2C4′4′ +n2C5′5′ +2mnC4′5′
 135 

TRANSFORMATIONS FOR DEXTRAL ROTATIONS 

ABOUT THE x

3 AXIS - CONTINUED

" Note that and can be expressed as 

 and 

" Thus, one set of transformed stiffness expressions can be 

obtained from the other by simply interchanging the primed and 

unprimed indices and replacing n with -n

C45=m2−n2C4′5′ − mnC4′4′ −C5′5′

C46=m3C4′6′ +m2nC5′6′ +C1′4′ −C2′4′ −mn2C4′6′ −C1′5′ +C2′5′ −n3C5′6′

C55=m2C5′5′ +n2C4′4′ −2mnC4′5′

C56=m3C5′6′ −m2nC4′6′ +C2′5′ −C1′5′ −mn2C5′6′ +C1′4′ −C2′4′ +n3C4′6′

C66=m2n2C1′1′ +C2′2′ −2C1′2′ +2mnm2−n2C1′6′ −C2′6′ +m2−n2 2C6′6′

C′ C

C′ =T

σ

θ3 C

Tε −

θ3 C

=T

σ

−

θ3 C′ Tε θ3
 136 

TRANSFORMATIONS FOR DEXTRAL ROTATIONS 

ABOUT THE x

3 AXIS - CONTINUED

" Let and 

"Performing the calculations given by yields

m= cosθ3 n= sinθ3

S′ =T

ε

ST

ε

T

S1′1′ =m4S11+m2n22S12+S66+2mnm2S16+n2S26+n4S22

S1′2′ =m2n2S11+S22−S66− mnm2−n2S16−S26+m4+n4S12

S1′3′ =m2S13+n2S23+mnS36

S1′4′ =m3S14+m2nS46−S15−mn2S56−S24−n3S25

S1′5′ =m3S15+m2nS56+S14+mn2S46+S25+n3S24

S1′6′ =m2m2−3n2S16−m3n2S11−2S12−S66

+mn32S22−2S12−S66−n2n2−3m2S26
 137 

TRANSFORMATIONS FOR DEXTRAL ROTATIONS 

ABOUT THE x

3 AXIS - CONTINUED

S2′2′ =m4S22+m2n22S12+S66−2mnm2S26+n2S16+n4S11

S2′3′ =m2S23+n2S13− mnS36

S2′4′ =m3S24−m2nS46+S25+mn2S56+S14−n3S15

S2′5′ =m3S25−m2nS56−S24−mn2S46−S15+n3S14

S2′6′ =m2m2−3n2S26+m3n2S22−2S12−S66

−mn32S11−2S12−S66−n2n2−3m2S16

S3′3′ =S33

S3′4′ =mS34−nS35

S3′5′ =mS35+nS34
 138 

TRANSFORMATIONS FOR DEXTRAL ROTATIONS 

ABOUT THE x

3 AXIS - CONTINUED

S3′6′ =m2−n2S36+2mnS23−S13

S4′4′ =m2S44+n2S55−2mnS45

S4′5′ =m2−n2S45+mnS44−S55

S4′6′ =m3S46−m2nS56+2S14−2S24−mn2S46−2S15+2S25+n3S56

S5′5′ =m2S55+n2S44+2mnS45

S5′6′ =m3S56+m2nS46+2S25−2S15−mn2S56+2S14−2S24−n3S46

S6′6′ =4m2n2S11+S22−2S12−4mnm2−n2S16−S26+m2−n2 2S66
 139 

TRANSFORMATIONS FOR DEXTRAL ROTATIONS 

ABOUT THE x

3 AXIS - CONTINUED

" Let and 

"Performing the calculations given by yields

m= cosθ3 n= sinθ3

S=T

σ

T

S′ T

σ

S11=m4S1′1′ +m2n22S1′2′ +S6′6′ −2mnm2S1′6′ +n2S2′6′ +n4S2′2′

S12=m2n2S1′1′ +S2′2′ −S6′6′ +mnm2−n2S1′6′ −S2′6′ +m4+n4S1′2′

S13=m2S1′3′ +n2S2′3′ − mnS3′6′

S14=m3S1′4′ −m2nS4′6′ −S1′5′ −mn2S5′6′ −S2′4′ +n3S2′5′

S15=m3S1′5′ −m2nS5′6′ +S1′4′ +mn2S4′6′ +S2′5′ −n3S2′4′

S16=m2m2−3n2S1′6′ +m3n2S1′1′ −2S1′2′ −S6′6′

−mn32S2′2′ −2S1′2′ −S6′6′ −n2n2−3m2S2′6′
 140 

TRANSFORMATIONS FOR DEXTRAL ROTATIONS 

ABOUT THE x

3 AXIS - CONTINUED

S22=m4S2′2′ +m2n22S1′2′ +S6′6′ +2mnm2S2′6′ +n2S1′6′ +n4S1′1′

S23=m2S2′3′ +n2S1′3′ +mnS3′6′

S24=m3S2′4′ +m2nS4′6′ +S2′5′ +mn2S5′6′ +S1′4′ +n3S1′5′

S25=m3S2′5′ +m2nS5′6′ −S2′4′ −mn2S4′6′ −S1′5′ −n3S1′4′

S26=m2m2−3n2S2′6′ −m3n2S2′2′ −2S1′2′ −S6′6′

+mn32S1′1′ −2S1′2′ −S6′6′ −n2n2−3m2S1′6′

S33=S3′3′ S34=mS3′4′ +nS3′5′ S35=mS3′5′ −nS3′4′

S36=m2−n2S3′6′ −2mnS2′3′ −S1′3′

S44=m2S4′4′ +n2S5′5′ +2mnS4′5′
 141 

TRANSFORMATIONS FOR DEXTRAL ROTATIONS 

ABOUT THE x

3 AXIS - CONTINUED

" Note that and can be expressed as 

 and 

" Thus, one set of transformed compliance expressions can be 

obtained from the other by simply interchanging the primed and 

unprimed indices and replacing n with -n

S45=m2−n2S4′5′ − mnS4′4′ −S5′5′

S46=m3S4′6′ +m2nS5′6′ +2S1′4′ −2S2′4′ −mn2S4′6′ −2S1′5′ +2S2′5′ −n3S5′6′

S55=m2S5′5′ +n2S4′4′ −2mnS4′5′

S56=m3S5′6′ −m2nS4′6′ +2S2′5′ −2S1′5′ −mn2S5′6′ +2S1′4′ −2S2′4′ +n3S4′6′

S66=4m2n2S1′1′ +S2′2′ −2S1′2′ +4mnm2−n2S1′6′ −S2′6′ +m2−n2 2S6′6′

S′ S

S′ =Tε θ3 ST

σ

−

θ3 S

=Tε −

θ3 S′ T

σ

θ3
 142 

TRANSFORMATIONS FOR DEXTRAL ROTATIONS 

ABOUT THE x

3 AXIS - CONTINUED

" Let and 

"Performing the calculations given by yields

m= cosθ3 n= sinθ3

α′ =T

ε

α

α1′1′ =m2α11+2mnα12+n2α22

α2′2′ =m2α22−2mnα12+n2α11

α3′3′ =α33

α2′3′ =mα23−nα13

α1′3′ =mα13+nα23

α1′2′ =m2−n2α12+mnα22−α11
 143 

TRANSFORMATIONS FOR DEXTRAL ROTATIONS 

ABOUT THE x

3 AXIS - CONTINUED

" Let and 

"

Performing the calculations given by yields

m= cosθ3 n= sinθ3

α=T

σ

T

α′

Note that

So, the expressions given here

for can be obtained from the

those previously given for 

by switching the primed and

unprimed indices and replacing

n with -n

and hence

αi′j′

αij

T

σ

θ3

T

=T

ε

−

θ3

α=T

ε

−

θ3 α′

α11=m2α1′1′ −2mnα1′2′ +n2α2′2′

α22=m2α2′2′ +2mnα1′2′ +n2α1′1′

α33=α3′3′

α23=mα2′3′ +nα1′3′

α13=mα1′3′ −nα2′3′

α12=m2−n2α1′2′ − mnα2′2′ −α1′1′
 144 

TRANSFORMATIONS FOR DEXTRAL ROTATIONS 

ABOUT THE x

3 AXIS - CONTINUED

" Let and 

"Performing the calculations given by yields

m= cosθ3 n= sinθ3

β′ =T

σ

β

β 1′1′ =m2β 11+2mnβ 12+n2β 22

β 2′2′ =m2β 22−2mnβ 12+n2β 11

β 3′3′ =β 33

β 2′3′ =mβ 23−nβ 13

β 1′3′ =mβ 13+nβ 23

β 1′2′ =m2−n2β 12+mnβ 22−β 11
 145 

TRANSFORMATIONS FOR DEXTRAL ROTATIONS 

ABOUT THE x

3 AXIS - CONCLUDED

" Let and 

"Performing the calculations given by yields

m= cosθ3 n= sinθ3

β=T

ε

T

β′

Note that

So, the expressions given here

for can be obtained fr βij om the

those previously given for βi′j′

by switching the primed and

unprimed indices and replacing

n with -n

and hence

T

ε

θ3

T

=T

σ

−

θ3

β=T

σ

−

θ3 β′

β 11=m2β 1′1′ −2mnβ 1′2′ +n2β 2′2′

β 22=m2β 2′2′ +2mnβ 1′2′ +n2β 1′1′

β 33=β 3′3′

β 23=mβ 2′3′ +nβ 1′3′

β 13=mβ 1′3′ −nβ 2′3′

β 12=m2−n2β 1′2′ − mnβ 2′2′ −β 1′1′
 146 

TRANSFORMATIONS FOR DEXTRAL ROTATIONS 

ABOUT THE x1 AXIS

" The term "dextral" refers to a right-handed rotation 

i 1

i 2

i 3

x1

x2

x3

x1′

x2′

x3′

i 1′

i 2′

i 3′

,

,

θ1

θ1

ai′j ≡ ii′ • ij

Plane x1 = 0
 147 

TRANSFORMATIONS FOR DEXTRAL ROTATIONS 

ABOUT THE x1 AXIS - CONTINUED

" The matrix form of the stress-tensor transformation law is given by

 and is expressed symbolically by

σ1′1′

σ2′2′

σ3′3′

σ2′3′

σ1′3′

σ1′2′

=

100000

0 cos2

θ1 sin2θ1 2sinθ1cosθ1 0

0

0 sin2

θ1 cos2θ1 −

2sin

θ1cosθ1 0

0

0

− sinθ1cosθ1 sinθ1cosθ1 cos2θ1 – sin2θ1 0

0

0000 cos

θ1 − sinθ1

0000 sin

θ1 cosθ1

σ11

σ22

σ33

σ23

σ13

σ12

σ′ =T

σ

θ1 σ
 148 

TRANSFORMATIONS FOR DEXTRAL ROTATIONS 

ABOUT THE x1 AXIS - CONTINUED

" Similarly, the matrix form of the inverse transformation law is given by

 and is expressed symbolically by

 where 

σ11

σ22

σ33

σ23

σ13

σ12

=

100000

0 cos2

θ1 sin2θ1 −

2sin

θ1cosθ1 0

0

0 sin2

θ1 cos2θ1 2sinθ1cosθ1 0

0

0 sin

θ1cosθ1 − sinθ1cosθ1 cos2θ1 − sin2θ1 0

0

0000 cos

θ1 sinθ1

0000

− sinθ1 cosθ1

σ1′1′

σ2′2′

σ3′3′

σ2′3′

σ1′3′

σ1′2′

σ

=

T

σ

θ1

− 1

σ′

T

σ

θ1

− 1

=

T

σ

−

θ1
 149 

TRANSFORMATIONS FOR DEXTRAL ROTATIONS 

ABOUT THE x1 AXIS - CONTINUED

" By using the second-order, symmetric tensor transformation equations,

the transformation law for the vector of engineering strains is given by

 which is expressed symbolically by

" Note that 

ε1′1′

ε2′2′

ε3′3′

2ε2′3′

2ε1′3′

2ε1′2′

=

100000

0 cos2

θ1 sin2θ1 sinθ1cosθ1 0

0

0 sin2

θ1 cos2θ1 − sinθ1cosθ1 0

0

0

−2sinθ1cosθ1 2sinθ1cosθ1 cos2θ1 − sin2θ1 0

0

0000 cos

θ1 − sinθ1

0000 sin

θ1 cosθ1

ε11

ε22

ε33

2ε23

2ε13

2ε12

ε′ =T

ε

θ1 ε

T

ε

θ1

T

=

T

σ

θ1

− 1

=

T

σ

−

θ1
 150 

TRANSFORMATIONS FOR DEXTRAL ROTATIONS 

ABOUT THE x1 AXIS - CONTINUED

" Similarly, the matrix form of the inverse transformation law is given by

 and is expressed symbolically by

 where 

ε11

ε22

ε33

2ε23

2ε13

2ε12

=

100000

0 cos2

θ1 sin2θ1 − sinθ1cosθ1 0

0

0 sin2

θ1 cos2θ1 sinθ1cosθ1 0

0

0

2sinθ1cosθ1 −2sinθ1cosθ1 cos2θ1 − sin2θ1 0

0

0000 cos

θ1 sinθ1

0000

− sinθ1 cosθ1

ε1′1′

ε2′2′

ε3′3′

2ε2′3′

2ε1′3′

2ε1′2′

ε

=

T

ε

θ1

− 1

ε′

T

ε

θ1

− 1

=

T

ε

−

θ1 =

T

σ

θ1

T
 151 

TRANSFORMATIONS FOR DEXTRAL ROTATIONS 

ABOUT THE x1 AXIS - CONTINUED

" The general expression for the transformation of the stiffness 

coefficients and thermal moduli are

 and 

" Noting that for a dextral rotation about the x1 axis gives

" Similarly, the general expression for the inverse transformation of the 

stiffness coefficients is 

" Noting that for a dextral rotation about the x1 axis gives 

C′ =T

σ

CT

ε

− 1

β′ =T

σ

β

T

ε

− 1

=

T

σ

T

C′ =T

σ

CT

σ

T

C=T

σ

− 1

C′ T

ε

T

σ

− 1

=

T

ε

T

C=T

ε

T

C′ T

ε
 152 

TRANSFORMATIONS FOR DEXTRAL ROTATIONS 

ABOUT THE x1 AXIS - CONTINUED

" The general expression for the transformation of the compliance 

coefficients and thermal expansion coefficients are

 and 

" Noting that for a dextral rotation about the x1 axis gives

" Similarly, the general expression for the inverse transformation of the 

compliance coefficients is 

" Noting that for a dextral rotation about the x1 axis gives 

S′ =T

ε

ST

σ

− 1

α′ =T

ε

α

T

σ

− 1

=

T

ε

T

S′ =T

ε

ST

ε

T

S=T

ε

− 1

S′ T

σ

T

ε

− 1

=

T

σ

T

S=T

σ

T

S′ T

σ
 153 

TRANSFORMATIONS FOR DEXTRAL ROTATIONS ABOUT 

THE x1 AXIS - SUMMARY

S! =T

"

ST

"

T

S=T

#

T

S! T

#

C! =T

#

CT

#

T

C=T

"

T

C! T

"

T

#

=

1 00000

0 cos2

$1 sin2$1 2sin$1cos$1 0

0

0 sin2

$1 cos2$1 % 2sin$1cos$1 0

0

0

% sin$1cos$1 sin$1cos$1 cos2$1 % sin2$1 0

0

0000 cos

$1 % sin$1

0000 sin

$1 cos$1

T

"

=

1 00000

0 cos2

$1 sin2$1 sin$1cos$1 0

0

0 sin2

$1 cos2$1 % sin$1cos$1 0

0

0

% 2sin$1cos$1 2sin$1cos$1 cos2$1 % sin2$1 0

0

0000 cos

$1 % sin$1

0000 sin

$1 cos$1
 154 

TRANSFORMATIONS FOR DEXTRAL ROTATIONS ABOUT 

THE x1 AXIS - SUMMARY

!" =T

#

!

!=T

$

T

!"

%" =T

$

%

%=T

#

T

%"

T

$

=

1 00000

0 cos2

&1 sin2&1 2sin&1cos&1 0

0

0 sin2

&1 cos2&1 ' 2sin&1cos&1 0

0

0

' sin&1cos&1 sin&1cos&1 cos2&1 ' sin2&1 0

0

0000 cos

&1 ' sin&1

0000 sin

&1 cos&1

T

#

=

1 00000

0 cos2

&1 sin2&1 sin&1cos&1 0

0

0 sin2

&1 cos2&1 ' sin&1cos&1 0

0

0

' 2sin&1cos&1 2sin&1cos&1 cos2&1 ' sin2&1 0

0

0000 cos

&1 ' sin&1

0000 sin

&1 cos&1
 155 

TRANSFORMATIONS FOR DEXTRAL ROTATIONS 

ABOUT THE x1 AXIS - CONTINUED

" Let and 

"Performing the calculations given by yields

m= cosθ1 n= sinθ1

C′ =T

σ

CT

σ

T

C1′1′ =C11

C1′2′ =m2C12+n2C13+2mnC14

C1′3′ =m2C13+n2C12−2mnC14

C1′4′ =m2−n2C14+mnC13−C12

C1′5′ =mC15−nC16

C1′6′ =mC16+nC15
 156 

TRANSFORMATIONS FOR DEXTRAL ROTATIONS 

ABOUT THE x1 AXIS - CONTINUED

C2′2′ =m4C22+2m2n2C23+2C44+4mnm2C24+n2C34+n4C33

C2′3′ =m2n2C22+C33−4C44−2mnm2−n2C24−C34+m4+n4C23

C2′4′ =m2m2−3n2C24−m3nC22−C23−2C44

+mn3C33−C23−2C44−n2n2−3m2C34

C2′5′ =m3C25+m2n2C45−C26−mn22C46−C35−n3C36

C2′6′ =m3C26+m2n2C46+C25+mn22C45+C36+n3C35

C3′3′ =m4C33+2m2n2C23+2C44−4mnm2C34+n2C24+n4C22

C3′4′ =m2m2−3n2C34+m3nC33−C23−2C44

−mn3C22−C23−2C44−n2n2−3m2C24
 157 

TRANSFORMATIONS FOR DEXTRAL ROTATIONS 

ABOUT THE x1 AXIS - CONTINUED

C3′5′ =m3C35−m2n2C45+C36+mn22C46+C25−n3C26

C3′6′ =m3C36−m2n2C46−C35−mn22C45−C26+n3C25

C4′4′ =m2n2C22+C33−2C23−2mnm2−n2C24−C34+m2−n2 2C44

C4′5′ =m3C45−m2nC46+C25−C35−mn2C45−C26+C36+n3C46

C4′6′ =m3C46+m2nC45+C36−C26−mn2C46+C25−C35−n3C45

C5′5′ =m2C55+n2C66−2mnC56

C5′6′ =m2−n2C56+mnC55−C66

C6′6′ =m2C66+n2C55+2mnC56
 158 

TRANSFORMATIONS FOR DEXTRAL ROTATIONS 

ABOUT THE x1 AXIS - CONTINUED

" Let and 

"Performing the calculations given by yields

m= cosθ1 n= sinθ1

C=T

ε

T

C′ T

ε

C11=C1′1′

C12=m2C1′2′ +n2C1′3′ −2mnC1′4′

C13=m2C1′3′ +n2C1′2′ +2mnC1′4′

C14=m2−n2C1′4′ − mnC1′3′ −C1′2′

C15=mC1′5′ +nC1′6′

C16=mC1′6′ −nC1′5′

C22=m4C2′2′ +2m2n2C2′3′ +2C4′4′ −4mnm2C2′4′ +n2C3′4′ +n4C3′3′
 159 

TRANSFORMATIONS FOR DEXTRAL ROTATIONS 

ABOUT THE x1 AXIS - CONTINUED

C23=m2n2C2′2′ +C3′3′ −4C4′4′ +2mnm2−n2C2′4′ −C3′4′ +m4+n4C2′3′

C24=m2m2−3n2C2′4′ +m3nC2′2′ −C2′3′ −2C4′4′

−mn3C3′3′ −C2′3′ −2C4′4′ −n2n2−3m2C3′4′

C25=m3C2′5′ −m2n2C4′5′ −C2′6′ −mn22C4′6′ −C3′5′ +n3C3′6′

C26=m3C2′6′ −m2n2C4′6′ +C2′5′ +mn22C4′5′ +C3′6′ −n3C3′5′

C33=m4C3′3′ +2m2n2C2′3′ +2C4′4′ +4mnm2C3′4′ +n2C2′4′ +n4C2′2′

C34=m2m2−3n2C3′4′ −m3nC3′3′ −C2′3′ −2C4′4′

+mn3C2′2′ −C2′3′ −2C4′4′ −n2n2−3m2C2′4′

C35=m3C3′5′ +m2n2C4′5′ +C3′6′ +mn22C4′6′ +C2′5′ +n3C2′6′

C36=m3C3′6′ +m2n2C4′6′ −C3′5′ −mn22C4′5′ −C2′6′ −n3C2′5′
 160 

TRANSFORMATIONS FOR DEXTRAL ROTATIONS 

ABOUT THE x1 AXIS - CONTINUED

" Note that and can be expressed as 

 and 

" Thus, one set of transformed stiffness expressions can be 

obtained from the other by simply interchanging the primed and 

unprimed indices and replacing n with -n

C44=m2n2C2′2′ +C3′3′ −2C2′3′ +2mnm2−n2C2′4′ −C3′4′ +m2−n2 2C4′4′

C45=m3C4′5′ +m2nC4′6′ +C2′5′ −C3′5′ −mn2C4′5′ −C2′6′ +C3′6′ −n3C4′6′

C46=m3C4′6′ −m2nC4′5′ +C3′6′ −C2′6′ −mn2C4′6′ +C2′5′ −C3′5′ +n3C4′5′

C55=m2C5′5′ +n2C6′6′ +2mnC5′6′ C56

=m2−n2C5′6′ − mnC5′5′ −C6′6′

C66=m2C6′6′ +n2C5′5′ −2mnC5′6′

C′ C

C′ =T

σ

θ1 C

Tε −

θ1 C

=T

σ

−

θ1 C′ Tε θ1
 161 

TRANSFORMATIONS FOR DEXTRAL ROTATIONS 

ABOUT THE x1 AXIS - CONTINUED

" Let and 

"Performing the calculations given by yields

m= cosθ1 n= sinθ1

S′ =T

ε

ST

ε

T

S1′1′ =S11

S1′2′ =m2S12+n2S13+mnS14

S1′3′ =m2S13+n2S12− mnS14

S1′4′ =m2−n2S14+2mnS13−S12

S1′5′ =mS15−nS16

S1′6′ =mS16+nS15

S2′2′ =m4S22+m2n22S23+S44+2mnm2S24+n2S34+n4S33
 162 

TRANSFORMATIONS FOR DEXTRAL ROTATIONS 

ABOUT THE x1 AXIS - CONTINUED

S2′3′ =m2n2S22+S33−S44− mnm2−n2S24−S34+m4+n4S23

S2′4′ =m2m2−3n2S24−m3n2S22−2S23−S44

+mn32S33−2S23−S44−n2n2−3m2S34

S2′5′ =m3S25+m2nS45−S26−mn2S46−S35−n3S36

S2′6′ =m3S26+m2nS46+S25+mn2S45+S36+n3S35

S3′3′ =m4S33+m2n22S23+S44−2mnm2S34+n2S24+n4S22

S3′4′ =m2m2−3n2S34+m3n2S33−2S23−S44

−mn32S22−2S23−S44−n2n2−3m2S24

S3′5′ =m3S35−m2nS45+S36+mn2S46+S25−n3S26
 163 

TRANSFORMATIONS FOR DEXTRAL ROTATIONS 

ABOUT THE x1 AXIS - CONTINUED

S3′6′ =m3S36−m2nS46−S35−mn2S45−S26+n3S25

S4′4′ =4m2n2S22+S33−2S23−4mnm2−n2S24−S34+m2−n2 2S44

S4′5′ =m3S45−m2nS46+2S25−2S35−mn2S45−2S26+2S36+n3S46

S4′6′ =m3S46+m2nS45+2S36−2S26−mn2S46+2S25−2S35−n3S45

S5′5′ =m2S55+n2S66−2mnS56

S5′6′ =m2−n2S56+mnS55−S66

S6′6′ =m2S66+n2S55+2mnS56
 164 

TRANSFORMATIONS FOR DEXTRAL ROTATIONS 

ABOUT THE x1 AXIS - CONTINUED

" Let and 

"Performing the calculations given by yields

m= cosθ1 n= sinθ1

S=T

σ

T

S′ T

σ

S11=S1′1′

S12=m2S1′2′ +n2S1′3′ − mnS1′4′

S13=m2S1′3′ +n2S1′2′ +mnS1′4′

S14=m2−n2S1′4′ −2mnS1′3′ −S1′2′

S15=mS1′5′ +nS1′6′

S16=mS1′6′ −nS1′5′

S22=m4S2′2′ +m2n22S2′3′ +S4′4′ −2mnm2S2′4′ +n2S3′4′ +n4S3′3′
 165 

TRANSFORMATIONS FOR DEXTRAL ROTATIONS 

ABOUT THE x1 AXIS - CONTINUED

S23=m2n2S2′2′ +S3′3′ −S4′4′ +mnm2−n2S2′4′ −S3′4′ +m4+n4S2′3′

S24=m2m2−3n2S2′4′ +m3n2S2′2′ −2S2′3′ −S4′4′

−mn32S3′3′ −2S2′3′ −S4′4′ −n2n2−3m2S3′4′

S25=m3S2′5′ −m2nS4′5′ −S2′6′ −mn2S4′6′ −S3′5′ +n3S3′6′

S26=m3S2′6′ −m2nS4′6′ +S2′5′ +mn2S4′5′ +S3′6′ −n3S3′5′

S33=m4S3′3′ +m2n22S2′3′ +S4′4′ +2mnm2S3′4′ +n2S2′4′ +n4S2′2′

S34=m2m2−3n2S3′4′ −m3n2S3′3′ −2S2′3′ −S4′4′

+mn32S2′2′ −2S2′3′ −S4′4′ −n2n2−3m2S2′4′

S35=m3S3′5′ +m2nS4′5′ +S3′6′ +mn2S4′6′ +S2′5′ +n3S2′6′

S36=m3S3′6′ +m2nS4′6′ −S3′5′ −mn2S4′5′ −S2′6′ −n3S2′5′
 166 

TRANSFORMATIONS FOR DEXTRAL ROTATIONS 

ABOUT THE x1 AXIS - CONTINUED

" Note that and can be expressed as 

 and 

" Thus, one set of transformed compliance expressions can be 

obtained from the other by simply interchanging the primed and 

unprimed indices and replacing n with -n

S44=4m2n2S2′2′ +S3′3′ −2S2′3′ +4mnm2−n2S2′4′ −S3′4′ +m2−n2 2S4′4′

S45=m3S4′5′ +m2nS4′6′ +2S2′5′ −2S3′5′ −mn2S4′5′ −2S2′6′ +2S3′6′ −n3S4′6′

S46=m3S4′6′ −m2nS4′5′ +2S3′6′ −2S2′6′ −mn2S4′6′ +2S2′5′ −2S3′5′ +n3S4′5′

S55=m2S5′5′ +n2S6′6′ +2mnS5′6′ S56

=m2−n2S5′6′ − mnS5′5′ −S6′6′

S66=m2S6′6′ +n2S5′5′ −2mnS5′6′

S′ S

S′ =Tε θ1 ST

σ

−

θ1 S

=Tε −

θ1 S′ T

σ

θ1
 167 

TRANSFORMATIONS FOR DEXTRAL ROTATIONS 

ABOUT THE x1 AXIS - CONTINUED

" Let and 

"Performing the calculations given by yields

m= cosθ1 n= sinθ1

α′ =T

ε

α

α1′1′ =α11

α2′2′ =m2α22+2mnα23+n2α33

α3′3′ =m2α33−2mnα23+n2α22

α2′3′ =m2−n2α23+mnα33−α22

α1′3′ =mα13−nα12

α1′2′ =mα12+nα13
 168 

TRANSFORMATIONS FOR DEXTRAL ROTATIONS 

ABOUT THE x1 AXIS - CONTINUED

" Let and 

"Performing the calculations given by yields

m= cosθ1 n= sinθ1

α=T

σ

T

α′

Note that

So, the expressions given here

for can be obtained from the

those previously given for 

by switching the primed and

unprimed indices and replacing

n with -n

and hence

αi′j′

αij

T

σ

θ1

T

=T

ε

−

θ1

α=T

ε

−

θ1 α′

α11=α1′1′

α22=m2α2′2′ −2mnα2′3′ +n2α3′3′

α33=m2α3′3′ +2mnα2′3′ +n2α2′2′

α23=m2−n2α2′3′ − mnα3′3′ −α2′2′

α13=mα1′3′ +nα1′2′

α12=mα1′2′ −nα1′3′
 169 

TRANSFORMATIONS FOR DEXTRAL ROTATIONS 

ABOUT THE x1 AXIS - CONTINUED

" Let and 

"Performing the calculations given by yields

m= cosθ1 n= sinθ1

β′ =T

σ

β

β 1′1′ =β 11

β 2′2′ =m2β 22+2mnβ 23+n2β 33

β 3′3′ =m2β 33−2mnβ 23+n2β 22

β 2′3′ =m2−n2β 23+mnβ 33−β 22

β 1′3′ =mβ 13−nβ 12

β 1′2′ =mβ 12+nβ 13
 170 

TRANSFORMATIONS FOR DEXTRAL ROTATIONS 

ABOUT THE x1 AXIS - CONTINUED

" Let and 

"Performing the calculations given by yields

m= cosθ1 n= sinθ1

β=T

ε

T

β′

Note that

So, the expressions given here

for can be obtained fr βij om the

those previously given for βi′j′

by switching the primed and

unprimed indices and replacing

n with -n

and hence

T

ε

θ1

T

=T

σ

−

θ1

β=T

σ

−

θ1 β′

β 11=β 1′1′

β 22=m2β 2′2′ −2mnβ 2′3′ +n2β 3′3′

β 33=m2β 3′3′ +2mnβ 2′3′ +n2β 2′2′

β 23=m2−n2β 2′3′ − mnβ 3′3′ −β 2′2′

β 13=mβ 1′3′ +nβ 1′2′

β 12=mβ 1′2′ −nβ 1′3′
 171 

TRANSFORMATIONS FOR DEXTRAL ROTATIONS 

ABOUT THE x1 AXIS - CONTINUED

" The algebra involved in computing

 is definitely tedious when done by hand

" When the expressions for these matrix operations are known for either 

dextral rotations about the x

3 axis or about the x2 axis, a simpler and 

much less tedious method for obtaining the transformed stiffnesses,

compliances, thermal moduli, and thermal-expansion coefficients for 

dextral rotations about the x1 axis is a

vailable

C′ =T

σ

CT

σ

T

S′ =T

ε

ST

ε

T

α′ =T

ε

α

C=T

ε

T

C′ T

ε

S=T

σ

T

S′ T

σ

α=T

σ

T

α′

β′ =T

σ

β

β=T

ε

T

β′
 172 

TRANSFORMATIONS FOR DEXTRAL ROTATIONS 

ABOUT THE x1 AXIS - CONTINUED

" Consider the case in which the transformation expressions are known 

for dextral rotations about the x

3 axis and one wishes to find similar 

expressions for dextral rotations about the x1 axis

" The desired tranformation equations are found by simply determining 

the renumbering of the indices that brings the following figure shown 

for dextral rotations about the x

3 axis into congruence with the 

following 

figure shown for dextral rotations about the x1 axis
 173 

TRANSFORMATIONS FOR DEXTRAL ROTATIONS 

ABOUT THE x1 AXIS - CONTINUED

i 1

i 2

i 3

x1

x2

x3

x1′

x2′

x3′

i 1′

i 2′

i 3′

,

,

θ1

θ1

i 1 i 2

i 3

x1

x2

x3

x1′

x2′

x3′

i 1′

i 2′

i 3′

,

,

θ3

θ3

Plane x1 = 0

Plane x

3 = 0
 174 

TRANSFORMATIONS FOR DEXTRAL ROTATIONS 

ABOUT THE x1 AXIS - CONTINUED

" Inspection of the figures indicates the following transformation of the 

indices:

" Next, it must be realized that the exchanging of indices must be used 

with the indices of tensors to determine the indices used with the 

abridged notation (matrix)

" The following index pairs relate the tensor indices to the matrix 

indices

" Using this information along with gives the 

relations:

1→2, 2→3, and3→1

tensor notation 11 22 33 23, 3231, 13 12, 21

matrix notation123456

1→2, 2→3, and3→1

4→5, 5→6, and6→4
 175 

TRANSFORMATIONS FOR DEXTRAL ROTATIONS 

ABOUT THE x1 AXIS - CONTINUED

" Likewise, the transformation of index pairs that appear in the abridged 

notation are given by

" Consider ,

where and 

" The transformation of indices gives , , and 

11→ 22

12→ 23 22→ 33

13→ 12 23→ 13 33→ 11

14→ 25 24→ 35 34→ 15 44→ 55

15→ 26 25→ 36 35→ 16 45→ 56 55→ 66

16→ 24 26→ 34 36→ 14 46→ 45 56→ 46 66→ 44

C1′1′ =m

4

C11+2m2n2 C12+2C66+4mnm2C16+n2C26+n

4

C22

m= cosθ3 n= sinθ3

m= cosθ1 n= sinθ1

C2′2′ =m

4

C22+2m2n2 C23+2C44+4mnm2C24+n2C34+n

4

C33
 176 

TRANSFORMATIONS FOR DEXTRAL ROTATIONS 

ABOUT THE x1 AXIS - CONCLUDED

" Applying the index transformation to the transformed stiffnesses,

compliances, thermal moduli, and thermal-expansion coefficients for 

dextral rotations about the x

3 axis yields exactly the same expressions 

given herein previously for dextral rotations about the x1 axis
 177 

MATERIAL SYMMETRIES
 178 

MATERIAL SYMMETRIES

" The next logical step in the development of linear thermoelastic 

constitutive equations is the search for analytical conditions for which 

dilatation and distortion uncouple

" For example, experience with common metals indicates that there 

are classes of materials for which dilatation and distortion 

uncouple

" Also, from a practical viewpoint, there is a need to find ways to 

minimize the number of laboratory experiments needed to fully 

characterize a given material

"Together, these considerations suggest a need for a systematic way to 

reduce the number of independent elastic constants and the number of 

independent thermal-expansion coefficients 

" Previously, it was shown herein that there exists 21 independent elastic 

constants for an elastic anisotropic material - a finding that is 

substantiated by experimental evidence
 179 

MATERIAL SYMMETRIES - CONCLUDED

" However, the number of independent constants needed to fully 

characterize an anisotropic material was the subject of a lengthy 

controversy

" In the early to mid 19th century, A. L. Cauchy (1789-1857) and 

S. D. Poisson (1781-1840) formulated specialized mathematical 

models of the molecular interaction in solids, and argued that the 

number of independent constants could not exceed 15 

" Investigations have indicated that when a solid exhibits a geometry 

symmetry, the elastic properties are identical in certain directions

" However, experience has shown that geometric symmetry is not 

equivalent to elastic symmetry; that is, it is possible to have elastic 

symmetry in directions that do not exhibit geometric symmetry

"Just as concepts of symmetry are used to reduce complexity of 

geometric objects, they are also used to reduce the complexity of 

material properties
 180 

MATHEMATICAL CHARACTERIZATION OF SYMMETRY

" The commonplace notion of symmetry is usually concerned with 

geometric objects

" For example, a two-dimensional geometric object may possess a shape 

that can be rotated about a central point by a finite angle with no 

appearent change in shape

A

A

A

A

A

A

A

A

A

A

A

A

Original 

position

72 deg. 144 deg. 216 deg. 288 deg. 360 deg.

Five-fold

symmetry
 181 

MATHEMATICAL CHARACTERIZATION OF SYMMETRY

CONTINUED

" For the object shown in the previous figure, it can be rotated 

incrementally by 72 degrees into the identical shape

" and the shape is said to possess five-fold symmetry

" A situation of particular interest herein 

is the case when a geometric object 

possesses only two-fold symmetry

" For this object, the line B-B is described 

as a line of reflective symmetry

(or mirror symmetry)

" For each of the geometric objects,

undergoing the given rotations, a 

transformation occurs in which the objects appears unchanged; it is 

said to remain invariant under the transformation

360 deg

72 deg

=5

A

A

BB

A

A

BB
 182 

MATHEMATICAL CHARACTERIZATION OF SYMMETRY

CONTINUED

" For any geometric object in three-dimensional Euclidean space, the 

object can be represented by a set of points, whose position in space 

can be determined by a coordinate frame and a coordinate domain

for the set of points

" For the figure, a generic point P of the 

region R has the coordinates (x,y), with 

respect to the coordinate frame shown

" The coordinate domain is given by

 and ,

where w and h are the width and height 

of the rectangle, respectively

" For other geometries, curvilinear 

coordinates may be more suitable

A

x-axis

y-axis

P

Coordinates

(x,y) of point P

Region R

BB

A

−

1

2

w≤x≤

1

2

w−

1

2

h≤y≤

1

2

h
 183 

MATHEMATICAL CHARACTERIZATION OF SYMMETRY

CONTINUED

" For the purpose of investigating symmetry, it is convenient to place the 

origin at the central point of the rectangle

" Now consider a second set of 

coordinates (

x,y), for which 

 and 

" This domain also describes the 

same rectangle with respect to 

a rotated coordinate frame, as 

shown in the figure

" For values of the angle φ equal to 

180 and 360 deg, the rectangle is 

brought into coincidence with the 

initial configuration

x-axis

y-axis

P

Coordinates

(x,y) of point P

Region R

x-axis

y-axis

B

φ

B

A

A

−

1

2

w≤

x

≤

1

2

w−

1

2

h≤

y

≤

1

2

h
 184 

MATHEMATICAL CHARACTERIZATION OF SYMMETRY

CONTINUED

" For φ = 180 deg and φ = 360 deg, the geometric shape is invariant, and 

the transformation of coordinates given symbolically by and 

 is called a symmetry transformation for the rectangle

" Obviously, this process of characterizing symmetry is easily extended 

to three dimensions

" Moreover, the functional characterization of symmetry in geometric 

shapes can be extended intuitively to symmetry in functions

" A transformation of coordinates that leaves the structural form of 

the rule that defines a given function unchanged (invariant) is 

defined as a symmetry transformation for that function

" The use of algebraic structure and sets of transformations for 

quantifying symmetry in (real and abstract) objects is part of a branch 

of mathematics known as group theory

x

=

x

x,y

y

=

y

x,y
 185 

MATHEMATICAL CHARACTERIZATION OF SYMMETRY

CONCLUDED

" Consider the function and the transformation of 

coordinates given by and 

" Applying the transformation of coordinates gives

" The structural rules given by and 

are identical; thus, and define a symmetry 

transformation for the function

" The more common, and more succinct, way of describing the 

symmetry is given by writing 

Fx, y=x2+y2

x=−

x

y=−

y

Fx, y→−

x 2

+−

y 2

=

x

2

+

y

2

=

F(x, y

)=F−x, −y

Fx, y=x2+y2

F(x, y

)=

x

2

+

y

2

x=−

x

y=−

y

Fx, y=F−x, −y
 186 

SOME TYPES OF SYMMETRY IN TWO DIMENSIONS

GRAPHS OF FUNCTIONS

f(-x) = f(x)

f(x) = ax

2 + b

f(x)

x

Line of reflective symmetry Line of reflective antisymmetry

f(-x) = -f(x)

f(x) = ax

3 f(x)

x
 187 

SOME TYPES OF SYMMETRY IN THREE DIMENSIONS

PLANE OF REFLECTIVE SYMMETRY

Symmetry plane

Tapered beam
 188 

SOME TYPES OF SYMMETRY IN THREE DIMENSIONS

PLANE OF REFLECTIVE SYMMETRY

F−x, y=Fx, y

y-axis

x-axis

Fx, y

−1≤x≤1

F−x, y

Fx, y=y cos

0≤y≤2

πx

2 sin

πy

2

Contour plot of F(x, y)
 189 

SOME TYPES OF SYMMETRY IN THREE DIMENSIONS

PLANE OF REFLECTIVE ANTISYMMETRY

F−x, y=−Fx, y

y-axis

x-axis

Fx, y

−1≤x≤1

F−x, y

Fx, y=y sinπx sin

0≤y≤2

πy

2

Contour plot of F(x, y)
 190 

SOME TYPES OF SYMMETRY IN THREE DIMENSIONS

CENTRAL POINT OF INVERSION SYMMETRY (POLAR SYMMETRY)

F−x, −y=Fx, y

y-axis

x-axis

Fx, y

F−x, −y

Fx, y= sinπx−

9y

10 sin

πy

−1≤x≤1

0≤y≤1

Contour plot of F(x, y)
 191 

SOME TYPES OF SYMMETRY IN THREE DIMENSIONS

CENTRAL POINT OF INVERSION ANTISYMMETRY

F−x, −y=−Fx, y

y-axis

x-axis

Fx, y

F−x, −y

Fx, y= sinπx−

9y

10 sin

2πy

−1≤x≤10≤y≤1

Contour plot of F(x, y)
 192 

CRITERIA FOR MATERIAL SYMMETRY

"To define the conditions on the stiffness or compliance coefficients for 

a given type of symmetry to exist, one must first realize that the 

stresses, strains, and stiffness or compliance coefficients are functions 

of position within a given material body

" Let the coordinates and the corresponding coordinate 

frame be a coordinate system for a material body and its properties

" The point P of the material body 

shown in the figure has 

coordinates 

" The stresses, strains, stiffness 

and compliance matrices, thermal 

moduli, and thermal-expansion 

coefficients for this coordinate 

system are σij, εij, [C], [S],

βij, and 

αij, respectivel

y

x1, x2, x3

Body, B

P

x1- axis

x

2- axis

x

3- axis

x1, x2, x3
 193 

CRITERIA FOR MATERIAL SYMMETRY - CONTINUED

" Recall that the abridged forms of the thermoelastic constitutive 

equations for the material in the coordinates are given by

 or 

" Now, consider a general orthogonal transformation between the 

rectangular Cartesian coordinates and , that 

define a generic point P of the material body

" There is no need to place the restriction that be the 

coordinates of a right-handed (dextral) coordinate system

" The stresses, strains, stiffness and compliance matrices, thermal 

moduli, and thermal-expansion coefficients for this coordinate 

system are , respectively

x1, x2, x3

σ=Cε+β

T−Tref

ε=Sσ+α

T−Tref

x1, x2, x3 x1′, x2′, x3′

x1′, x2′, x3′

σi′j′, εi′j′, [C′], [S′],βi′j′, andαi′j′
 194 

CRITERIA FOR MATERIAL SYMMETRY - CONTINUED

" The abridged forms of the thermoelastic constitutive equations for the 

material in the coordinate system are given by

 or 

" Moreover, it was shown previously that, for the given arbitrary 

transformation (rotation) of coordinates,

" When the mathematical description of the material properties are 

identical for two different coordinate systems (reference frames), a 

certain type of symmetry exists, whose character depends on the type 

of transformation between the two coordinates systems

x1′, x2′, x3′

σ′ =C′ ε′ +β′ T

−Tref

ε′ =S′ σ′ +α′ T

−Tref

C′ =T

σ

CT

ε

− 1

β′ =T

σ

β

S′ =T

ε

ST

σ

− 1

α′ =T

ε

α
 195 

CRITERIA FOR MATERIAL SYMMETRY - CONCLUDED

" Now, for to define a symmetry transformation, such 

that a predetermined state of symmetry exist at a point 

P of the body,

the structural form (rule) of the constitutive equations must remain 

invariant; specifically:

" The matrix must be invariant under the transformation given 

by ; that is, must hold

" Similarly, , , and must 

hold

" Collectively, these invariance conditions are the criteria for a state of 

material symmetry to exist, and are sufficient conditions because 

 is presumed to be a symmetry transformation

xk =xk x1′, x2′, x3′

C′

C′ =T

σ

CTε

− 1

C=T

σ

CTε

− 1

S=Tε ST

σ

− 1

α=Tε αβ=T

σ

β

xk =xk x1′, x2′, x3′
 196 

CLASSES OF MATERIAL SYMMETRY

" Presently, there exists eight distinct classes of elastic-material 

symmetry

" Many of these classes were discovered while studying the 

composition of various crystals

" The classes are distinguished by the number of, and orientation of,

planes of elastic symmetry 

" A plane of elastic symmetry, at a point of an elastic material body, is 

defined as a plane for which the material exhibits reflective symmetry

" A plane of isotropy, at a point of an elastic material body, is defined 

as a plane for which there exists an infinite number of perpendicular 

planes of elastic symmetry (also called axisymmetry)
 197 

CLASSES OF MATERIAL SYMMETRY

CONTINUED

" The eight distinct classes of elastic-material symmetry are given by:

"Triclinic materials - no inherent symmetry (fully anisotropic)

" Monoclinic materials - one plane of symmetry

" Orthotropic materials - three perpendicular planes of symmetry 

"Trigonal materials - three aligned planes of symmetry that are 

spaced 60 degrees apart

"Tetragonal materials - four aligned planes of symmetry that are 

spaced 45 degrees apart and that are all perpendicular to one 

additional symmetry plane

"Transversely isotropic materials - one plane of isotropy that is 

perpendicular to two other mutually perpendicular symmetry 

planes
 198 

CLASSES OF MATERIAL SYMMETRY

CONTINUED

" Cubic materials - three mutually perpendicular planes of symmetry 

and six additional symmetry planes, in which two of the six are 

aligned with one of the perpendicular planes and intersect it at 45 

degrees

" Completely isotropic materials - an infinite number of planes of 

isotropy exist

" There are four classes of elastic materials that are of great practical 

importance in engineering

" These classes of materials are monoclinic, orthotropic,

transversely isotropic, and isotropic materials
 199 

CLASSES OF MATERIAL SYMMETRY

PICTORIAL REPRESENTATIONS

" The blue lines represent the edge of a symmetry plane

Triclinic MonoclinicTrigonal

60 deg

Orthotropic

TetragonalTransversely Cubic

isotropic

Isotropic

45 deg

45 deg
 200 

MONOCLINIC

MATERIALS
 201 

MONOCLINIC MATERIALS

REFLECTIVE SYMMETRY ABOUT THE PLANE x1 = 0

" First, consider the case in which the material exhibits elastic symmetry

about the plane x1 = 0

" The coordinate transformation for 

this symmetry is shown in the 

figure and is given by ,

, and 

" The corresponding matrix of 

direction cosines is given by

i 1 i 2

i 3

x1

x2

x3

x1′

x2′

x3′

i 1′

i 2′

i 3′

,

,

, ,

ai′j ≡ ii′ • ij

Plane x

3 = 0

Plane x1 x1′ =

−x1 = 0

x2′ =x2 x3′ =x3

a1′1 a1′2 a1′3

a2′1 a2′2 a2′3

a3′1 a3′2 a3′3

=

−100

010

001
 202 

MONOCLINIC MATERIALS

REFLECTIVE SYMMETRY ABOUT THE PLANE x1 = 0 (CONTINUED)

" For this special case, the general transformation matrix

reduces to

 the diagonal matrix 

T

σ

=

a1′1

2

a1′2

2

a1′3

2

2a1′2a1′3 2a1′1a1′3 2a1′1a1′2

a2′1

2

a2′2

2

a2′3

2

2a2′2a2′3 2a2′1a2′3 2a2′1a2′2

a3′1

2

a3′2

2

a3′3

2

2a3′2a3′3 2a3′1a3′3 2a3′1a3′2

a2′1a3′1 a2′2a3′2 a2′3a3′3 a2′2a3′3 +a2′3a3′2 a2′1a3′3 +a2′3a3′1 a2′1a3′2 +a2′2a3′1

a1′1a3′1 a1′2a3′2 a1′3a3′3 a1′2a3′3 +a1′3a3′2 a1′1a3′3 +a1′3a3′1 a1′1a3′2 +a1′2a3′1

a1′1a2′1 a1′2a2′2 a1′3a2′3 a1′2a2′3 +a1′3a2′2 a1′1a2′3 +a1′3a2′1 a1′1a2′2 +a1′2a2′1

T

σ

=

100000

010000

001000

000100

0000−10

00000−1
 203 

MONOCLINIC MATERIALS

REFLECTIVE SYMMETRY ABOUT THE PLANE x1 = 0 (CONTINUED)

" Moreover, because is a diagonal matrix, it follows that 

" It is worthwhile to mention that can be deduced directly, and 

quickly, by a direction comparison of the positive-valued stresses that 

act on a differential volume element

" First, sketch the positive-valued stresses that act on a differential 

volume element when described by the coordinates

T

σ

T

ε

=T

σ

T

σ

x1, x2, x3

Shearing stresses

σ11

σ22

σ33

Normal stresses

σ12

σ13σ23

x1

x2

x3
 204 

MONOCLINIC MATERIALS

REFLECTIVE SYMMETRY ABOUT THE PLANE x1 = 0 (CONTINUED)

" Then, sketch the positive-valued stresses that act on a differential 

volume element when described by the coordinates

" Direct comparison of the stresses yields the relationships

x1′, x2′, x3′

Normal stresses Shearing stresses

x1′

x2′

x3′

σ1′1′

σ2′2′

σ3′3′

σ1′2′

σ1′3′ σ2′3′

σ1′1′ =σ11σ2′2′ =σ22σ3′3′ =σ33

σ2′3′ =σ23

σ1′3′ =

−

σ13σ1′2′ =

−

σ12
 205 

MONOCLINIC MATERIALS

REFLECTIVE SYMMETRY ABOUT THE PLANE x1 = 0 (CONTINUED)

" Expressing the relationships in matrix form gives

" Thus, gives directly

σ1′1′

σ2′2′

σ3′3′

σ2′3′

σ1′3′

σ1′2′

=

100000

010000

001000

000100

0000−10

00000−1

σ11

σ22

σ33

σ23

σ13

σ12

σ′ =T

σ

σ

T

σ

=

100000

010000

001000

000100

0000−10

00000−1
 206 

MONOCLINIC MATERIALS

REFLECTIVE SYMMETRY ABOUT THE PLANE x1 = 0 (CONTINUED)

" Now, for a state of reflective symmetry about the plane x1 = 0 to exist at 

a point P of the body, it was shown herein that the matrix must be 

invariant under the transformation given by

" That is, must become 

" A more convenient form of this invariance condition is obtained for this 

particular transformation as follows

" First, postmultiplying the last expression by gives 

as the (sufficient) condition for symmetry

" Next, noting that for this particular symmetry transformation,

it follows that becomes 

C′

C′ =T

σ

CTε

− 1

C′ =T

σ

CTε

− 1

C=T

σ

CTε

− 1

Tε CTε =T

σ

C

Tε =T

σ

CTε =T

σ

CCT

σ

=T

σ

C
 207 

MONOCLINIC MATERIALS

REFLECTIVE SYMMETRY ABOUT THE PLANE x1 = 0 (CONTINUED)

" Also, because for this particular symmetry transformation 

and , it follows that 

" Thus, the sufficient condition for symmetry, , becomes 

; that is,must be a symmetric matrix

" Computing gives

T

σ

=T

σ

T

C=C

T

CT

σ

T

=T

σ

T

C

T

=T

σ

C

CT

σ

=T

σ

C

CT

σ

=CT

σ

T

CT

σ

CT

σ

CT

σ

=

C11C12C13C14C15C16

C12C22C23C24C25C26

C13C23C33C34C35C36

C14C24C34C44C45C46

C15C25C35C45C55C56

C16C26C36C46C56C66

1 00000

0 1 0000

00 1 000

000 1 00

0000− 1 0

00000− 1

=

C11C12C13C14−C15−C16

C12C22C23C24−C25−C26

C13C23C33C34−C35−C36

C14C24C34C44−C45−C46

C15C25C35C45−C55−C56

C16C26C36C46−C56−C66
 208 

MONOCLINIC MATERIALS

REFLECTIVE SYMMETRY ABOUT THE PLANE x1 = 0 (CONTINUED)

" Enforcing yields the following requirements on the 

stiffnesses in order to exhibit a state of elastic reflective symmetry

about the plane x1 = 0:

, , , , , , , and 

" Thus, the stiffness matrix for a monoclinic material, which exhibits 

elastic reflective symmetry about the plane x1 = 0, has the form

 which has 13 independent stiffnesses

CT

σ

=CT

σ

T

C15=0C16=0C25=0C26=0C35=0C36=0C45=0C46=0

C11C12C13C1400

C12

C22C23C2400

C13

C23C33C3400

C14

C24C34C4400

0000C55

C56

0000C56C66
 209 

MONOCLINIC MATERIALS

REFLECTIVE SYMMETRY ABOUT THE PLANE x1 = 0 (CONTINUED)

" Similarly, for a state of elastic reflective symmetry about the plane x1 = 0 

to exist at a point P of the body, the matrix must be invariant

under the transformation given by

" That is, must become 

"Postmultiplying the last expressing by gives as the 

(sufficient) condition for symmetry

" Next, noting that for this particular transformation, it follows 

that becomes 

S′

S′ =Tε ST

σ

− 1

S′ =Tε ST

σ

− 1

S=Tε ST

σ

− 1

T

σ

ST

σ

=Tε S

Tε =T

σ

ST

σ

=Tε SST

σ

=T

σ

S
 210 

MONOCLINIC MATERIALS

REFLECTIVE SYMMETRY ABOUT THE PLANE x1 = 0 (CONTINUED)

" Like before, because for this particular transformation and 

, it follows that 

" Thus, the sufficient condition for symmetry, , becomes 

; that is,must be a symmetric matrix

" Like for the stiffness matrix, computing and enforcing 

 yields the following requirements on the 

compliances in order to exhibit a state of elastic reflective symmetry

about the plane x1 = 0:

, , , , , , , and 

T

σ

=T

σ

T

S=S

T

ST

σ

T

=T

σ

T

S

T

=T

σ

S

ST

σ

=T

σ

S

ST

σ

=ST

σ

T

ST

σ

ST

σ

ST

σ

=ST

σ

T

S15=0S16=0S25=0S26=0S35=0S36=0S45=0S46=0
 211 

MONOCLINIC MATERIALS

REFLECTIVE SYMMETRY ABOUT THE PLANE x1 = 0 (CONTINUED)

" Thus, the compliance matrix for a monoclinic material that a state of 

elastic reflective symmetry about the plane x1 = 0 has the form

 which has 13 independent compliances

" Comparison of the compliance and stiffness matrices for this case 

indicates that they have the same form

S11S12S13S1400

S12S22S23S2400

S13

S23S33S3400

S14

S24S34S4400

0000S55

S56

0000S56S66
 212 

MONOCLINIC MATERIALS

REFLECTIVE SYMMETRY ABOUT THE PLANE x1 = 0 (CONTINUED)

" The requirements on the coefficients of thermal expansion and the 

thermal modulifor a state of reflective symmetry about the plane 

x1 = 0 

to exist are simpler than those for the stiffnesses and compliances

" The requirements on the coefficients of thermal expansion are given by

the requirement that the vector must be invariant under the 

transformation given by

" That is, must become 

" Similarly, must be invariant under the transformation given by

; that is, must become 

α′

α′ =Tε α

α′ =Tε αα=Tε α

β′

β′ =T

σ

ββ′ =T

σ

ββ=T

σ

β
 213 

MONOCLINIC MATERIALS

REFLECTIVE SYMMETRY ABOUT THE PLANE x1 = 0 (CONTINUED)

" Computing gives 

 which reduces to 

" Thus, enforcing requires in order for a state 

of reflective symmetry about the plane x1 = 0 to exist

" Similarly, enforcing requires in order for a 

state of reflective symmetry about the plane x1 = 0 to exist

α=Tε α

α11

α22

α33

2α23

2α13

2α12

=

100000

010000

001000

000100

0000−10

00000−1

α11

α22

α33

2α23

2α13

2α12

α11

α22

α33

2α23

2α13

2α12

=

α11

α22

α33

2α23

−2α13

−2α12

α=Tε α

α12=α13=0

β=T

σ

β

β 12=β 13=0
 214 

MONOCLINIC MATERIALS

REFLECTIVE SYMMETRY ABOUT THE PLANE x1 = 0 (CONTINUED)

" Applying all the simplifications, the linear thermoelastic 

constitutive equations become

ε11

ε22

ε33

2ε23

2ε13

2ε12

=

S11

S12S13S1400

S12

S22S23S2400

S13

S23S33S3400

S14S24S34S4400

0000S55

S56

0000S56S66

σ11

σ22

σ33

σ23

σ13

σ12

+

α11

α22

α33

2α23

0

0

T−Tref

σ11

σ22

σ33

σ23

σ13

σ12

=

C11

C12C13C1400

C12C22C23C2400

C13

C23C33C3400

C14C24C34C4400

0000C55

C56

0000C56C66

ε11

ε22

ε33

2ε23

2ε13

2ε12

+

β 11

β 22

β 33

β 23

0

0

T−Tref
 215 

MONOCLINIC MATERIALS

REFLECTIVE SYMMETRY ABOUT THE PLANE x1 = 0 (CONCLUDED)

" Finally, the nonzero thermal moduli are given in terms of the 

coefficients of thermal expansion by

" The constitutive equations show that the normal stresses, or a 

temperature change, produce shearing deformations only in the plane 

x1 = 0 

" Extension and shearing are totally uncoupled in the planes x

2 = 0 

and x

3 = 0

β 11

β 22

β 33

β 23

=−

C11C12C13C14

C12C22C23C24

C13C23C33C34

C14C24C34C44

α11

α22

α33

2

α23
 216 

MONOCLINIC MATERIALS

REFLECTIVE SYMMETRY ABOUT THE PLANE x

2 = 0

" Next, consider the case in which the material exhibits elastic symmetry 

about the plane x

2 = 0

" The coordinate transformation for 

this symmetry is shown in the 

figure and is given by ,

, and 

" The corresponding matrix of 

direction cosines is given by

i 2

i 3

x1

x2

x3

x1′

x2′

x3′

i 2′

i 3′

,

,

,

i 1 i , 1′

ai′j ≡ ii′ • ij

Plane x

3 = 0

Plane x

2 = 0

x1′ =x1

x2′ =−x2 x3′ =

x3

a1′1 a1′2 a1′3

a2′1 a2′2 a2′3

a3′1 a3′2 a3′3

=

100

0−10

001
 217 

MONOCLINIC MATERIALS

REFLECTIVE SYMMETRY ABOUT THE PLANE x

2 = 0 (CONTINUED)

" For this special case, the general transformation matrix

reduces to

 the diagonal matrix 

T

σ

=

a1′1

2

a1′2

2

a1′3

2

2a1′2a1′3 2a1′1a1′3 2a1′1a1′2

a2′1

2

a2′2

2

a2′3

2

2a2′2a2′3 2a2′1a2′3 2a2′1a2′2

a3′1

2

a3′2

2

a3′3

2

2a3′2a3′3 2a3′1a3′3 2a3′1a3′2

a2′1a3′1 a2′2a3′2 a2′3a3′3 a2′2a3′3 +a2′3a3′2 a2′1a3′3 +a2′3a3′1 a2′1a3′2 +a2′2a3′1

a1′1a3′1 a1′2a3′2 a1′3a3′3 a1′2a3′3 +a1′3a3′2 a1′1a3′3 +a1′3a3′1 a1′1a3′2 +a1′2a3′1

a1′1a2′1 a1′2a2′2 a1′3a2′3 a1′2a2′3 +a1′3a2′2 a1′1a2′3 +a1′3a2′1 a1′1a2′2 +a1′2a2′1

T

σ

=

100000

010000

001000

000−100

000010

00000−1
 218 

MONOCLINIC MATERIALS

REFLECTIVE SYMMETRY ABOUT THE PLANE x

2 = 0 (CONTINUED)

" Also, because is a diagonal matrix, it follows that 

" Like before, the sufficient conditions for a state of elastic reflective 

symmetry about the plane x

2 = 0 to exist are found from 

" Computing gives

T

σ

T

ε

=T

σ

CT

σ

=CT

σ

T

ST

σ

=ST

σ

T

α=Tε αβ=T

σ

β

CT

σ

CT

σ

=

C11C12C13C14C15C16

C12C22C23C24C25C26

C13C23C33C34C35C36

C14C24C34C44C45C46

C15C25C35C45C55C56

C16C26C36C46C56C66

1 00000

0 1 0000

00 1 000

000− 1 00

0000 1 0

00000− 1

=

C11C12C13−C14C15−C16

C12C22C23−C24C25−C26

C13C23C33−C34C35−C36

C14C24C34−C44C45−C46

C15C25C35−C45C55−C56

C16C26C36−C46C56−C66
 219 

MONOCLINIC MATERIALS

REFLECTIVE SYMMETRY ABOUT THE PLANE x

2 = 0 (CONTINUED)

" Enforcing yields the following requirements on the 

stiffnesses:

, , , , , , , and 

" Thus, the stiffness matrix for a monoclinic material that a state of 

elastic reflective symmetry about the plane x

2 = 0 has the form

 which also has 13 independent stiffnesses

CT

σ

=CT

σ

T

C14=0C16=0C24=0C26=0C34=0C36=0C45=0C56=0

C11C12C130C150

C12

C22C230C250

C13

C23C330C350

000C44

0C46

C15C25C350C550

000C46

0C66
 220 

MONOCLINIC MATERIALS

REFLECTIVE SYMMETRY ABOUT THE PLANE x

2 = 0 (CONTINUED)

" Likewise, the compliance matrix for a monoclinic material that a state of 

elastic reflective symmetry about the plane x

2 = 0 has the form

which also has 13 independent compliances

S11S12S130S150

S12

S22S230S250

S13

S23S330S350

000S44

0S46

S15S25S350S550

000S46

0S66
 221 

MONOCLINIC MATERIALS

REFLECTIVE SYMMETRY ABOUT THE PLANE x

2 = 0 (CONTINUED)

" Next, computing gives 

 which reduces to 

" Thus, enforcing requires in order for a state 

of elastic reflective symmetry about the plane x

2 = 0 to exist

" Similarly, enforcing requires 

α=Tε α

α11

α22

α33

2α23

2α13

2α12

=

100000

010000

001000

000−100

000010

00000−1

α11

α22

α33

2α23

2α13

2α12

α11

α22

α33

2α23

2α13

2α12

=

α11

α22

α33

−2α23

2α13

−2α12

α=Tε α

α12=α23=0

β=T

σ

β

β 12=β 23=0
 222 

MONOCLINIC MATERIALS

REFLECTIVE SYMMETRY ABOUT THE PLANE x

2 = 0 (CONTINUED)

" Applying all the simplifications, the linear thermoelastic 

constitutive equations become

ε11

ε22

ε33

2ε23

2ε13

2ε12

=

S11

S12S130S150

S12

S22S230S250

S13

S23S330S350

000S440S46

S15S25S350S550

000S460S66

σ11

σ22

σ33

σ23

σ13

σ12

+

α11

α22

α33

0

2α13

0

T−Tref

σ11

σ22

σ33

σ23

σ13

σ12

=

C11

C12C130C150

C12

C22C230C250

C13C23C330C350

000C44

0C46

C15C25C350C550

000C46

0C66

ε11

ε22

ε33

2ε23

2ε13

2ε12

+

β 11

β 22

β 33

0

β 13

0

T−Tref
 223 

MONOCLINIC MATERIALS

REFLECTIVE SYMMETRY ABOUT THE PLANE x

2 = 0 (CONCLUDED)

" The nonzero thermal moduli are given in terms of the coefficients of 

thermal expansion by

" The constitutive equations show that the normal stresses, or a 

temperature change, produce shearing deformations only in the plane 

x

2 = 0 

" Extension and shearing are totally uncoupled in the planes x1 = 0 

and x

3 = 0

β 11

β 22

β 33

β 13

=−

C11

C12C13C15

C12C22C23C25

C13C23C33C35

C15C25C35C55

α11

α22

α33

2α13
 224 

MONOCLINIC MATERIALS

REFLECTIVE SYMMETRY ABOUT THE PLANE x

3 = 0

" Now, consider the case in which the 

material exhibits symmetry about the 

plane x

3 = 0

" The coordinate transformation for this 

symmetry is shown in the figure and is 

given by , , and 

" The corresponding matrix of 

direction cosines is given by i 2

i 3

x2

x3

x2′

x3′

i 2′

i 3′

,

,

x1 ,x1′

i 1 i , 1′

ai′j ≡ ii′ • ij

Plane x

2 = 0

Plane x

3 = 0

x1′ =x1 x2′ =x2 x3′ =

−x3

a1′1 a1′2 a1′3

a2′1 a2′2 a2′3

a3′1 a3′2 a3′3

=

100

010

00−1
 225 

MONOCLINIC MATERIALS

REFLECTIVE SYMMETRY ABOUT THE PLANE x

3 = 0 (CONTINUED)

" For this special case, the general transformation matrix

reduces to

 the diagonal matrix 

T

σ

=

a1′1

2

a1′2

2

a1′3

2

2a1′2a1′3 2a1′1a1′3 2a1′1a1′2

a2′1

2

a2′2

2

a2′3

2

2a2′2a2′3 2a2′1a2′3 2a2′1a2′2

a3′1

2

a3′2

2

a3′3

2

2a3′2a3′3 2a3′1a3′3 2a3′1a3′2

a2′1a3′1 a2′2a3′2 a2′3a3′3 a2′2a3′3 +a2′3a3′2 a2′1a3′3 +a2′3a3′1 a2′1a3′2 +a2′2a3′1

a1′1a3′1 a1′2a3′2 a1′3a3′3 a1′2a3′3 +a1′3a3′2 a1′1a3′3 +a1′3a3′1 a1′1a3′2 +a1′2a3′1

a1′1a2′1 a1′2a2′2 a1′3a2′3 a1′2a2′3 +a1′3a2′2 a1′1a2′3 +a1′3a2′1 a1′1a2′2 +a1′2a2′1

T

σ

=

100000

010000

001000

000−100

0000−10

000001
 226 

MONOCLINIC MATERIALS

REFLECTIVE SYMMETRY ABOUT THE PLANE x

3 = 0 (CONTINUED)

" Also, because is a diagonal matrix, it follows that 

" Like before, the sufficient conditions for a state of elastic reflective 

symmetry about the plane x

3 = 0 to exist are found from 

" Computing gives

T

σ

T

ε

=T

σ

CT

σ

=CT

σ

T

ST

σ

=ST

σ

T

α=Tε αβ=T

σ

β

CT

σ

CT

σ

=

C11C12C13C14C15C16

C12C22C23C24C25C26

C13C23C33C34C35C36

C14C24C34C44C45C46

C15C25C35C45C55C56

C16C26C36C46C56C66

1 00000

0 1 0000

00 1 000

000− 1 00

0000− 1 0

00000 1

=

C11C12C13−C14−C15C16

C12C22C23−C24−C25C26

C13C23C33−C34−C35C36

C14C24C34−C44−C45C46

C15C25C35−C45−C55C56

C16C26C36−C46−C56C66
 227 

MONOCLINIC MATERIALS

REFLECTIVE SYMMETRY ABOUT THE PLANE x

3 = 0 (CONTINUED)

" Enforcing yields the following requirements on the 

stiffnesses:

, , , , , , , and 

" Thus, the stiffness matrix for a monoclinic material that a state of 

elastic reflective symmetry about the plane x

3 = 0 has the form

 which also has 13 independent stiffnesses

CT

σ

=CT

σ

T

C14=0C15=0C24=0C25=0C34=0C35=0C46=0C56=0

C11C12C1300C16

C12C22C2300C26

C13C23C3300C36

000C44C450

000C45

C550

C16

C26C3600C66
 228 

MONOCLINIC MATERIALS

REFLECTIVE SYMMETRY ABOUT THE PLANE x

3 = 0 (CONTINUED)

" Likewise, the compliance matrix for a monoclinic material that a state of 

elastic reflective symmetry about the plane x

3 = 0 has the form

which also has 13 independent compliances

S11S12S1300S16

S12S22S2300S26

S13S23S3300S36

000S44S450

000S45

S550

S16

S26S3600S66
 229 

MONOCLINIC MATERIALS

REFLECTIVE SYMMETRY ABOUT THE PLANE x

3 = 0 (CONTINUED)

" Next, computing gives 

 which reduces to 

" Thus, enforcing requires in order for a state 

of elastic reflective symmetry about the plane x

3 = 0 to exist

" Similarly, enforcing requires 

α=Tε α

α11

α22

α33

2α23

2α13

2α12

=

100000

010000

001000

000−100

0000−10

000001

α11

α22

α33

2α23

2α13

2α12

α11

α22

α33

2α23

2α13

2α12

=

α11

α22

α33

−2α23

−2α13

2α12

α=Tε α

α13=α23=0

β=T

σ

β

β 13=β 23=0
 230 

MONOCLINIC MATERIALS

REFLECTIVE SYMMETRY ABOUT THE PLANE x

3 = 0 (CONTINUED)

" Applying all the simplifications, the linear thermoelastic 

constitutive equations become

ε11

ε22

ε33

2ε23

2ε13

2ε12

=

S11

S12S1300S16

S12S22S2300S26

S13S23S3300S36

000S44S450

000S45

S550

S16

S26S3600S66

σ11

σ22

σ33

σ23

σ13

σ12

+

α11

α22

α33

0

0

2α12

T−Tref

σ11

σ22

σ33

σ23

σ13

σ12

=

C11

C12C1300C16

C12C22C2300C26

C13C23C3300C36

000C44C450

000C45

C550

C16

C26C3600C66

ε11

ε22

ε33

2ε23

2ε13

2ε12

+

β 11

β 22

β 33

0

0

β 12

T−Tref
 231 

MONOCLINIC MATERIALS

REFLECTIVE SYMMETRY ABOUT THE PLANE x

3 = 0 (CONCLUDED)

" The nonzero thermal moduli are given in terms of the coefficients of 

thermal expansion by

" The constitutive equations show that the normal stresses, or a 

temperature change, produce shearing deformations only in the plane 

x

3 = 0 

" Extension and shearing are totally uncoupled in the planes x1 = 0 

and x

2 = 0

β 11

β 22

β 33

β 12

=−

C11

C12C13C16

C12C22C23C26

C13C23C33C36

C16C26C36C66

α11

α22

α33

2α12
 232 

ORTHOTROPIC

MATERIALS
 233 

ORTHOTROPIC MATERIALS

REFLECTIVE SYMMETRY ABOUT THE PLANES x1 = 0 AND x

2 = 0

" Now, consider the case in which a monoclinic material, which exhibits 

symmetry about the plane x1 = 0, also exhibits symmetry about the 

plane x

2 = 0, which is perpendicular to the plane x1 = 0

" For this monoclinic material, it was shown previously that the material 

properties are given by

 and

σ11

σ22

σ33

σ23

σ13

σ12

=

C11

C12C13C1400

C12

C22C23C2400

C13C23C33C3400

C14

C24C34C4400

0000C55

C56

0000C56C66

ε11

ε22

ε33

2ε23

2ε13

2ε12

+

β 11

β 22

β 33

β 23

0

0

T−Tref
 234 

ORTHOTROPIC MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT THE PLANES x1 = 0 AND x

2 = 0

"

To determine the effects of the second symmetry plane, x

2 = 0, the 

coordinate transformation given by , , and is 

applied to the material properties of the monoclinic material

" This process is the same as applying the transformation for 

symmetry about the plane x

2 = 0 in succession to applying the 

transformation for symmetry about the plane x1 = 0 to the initial 

anisotropic-material properties

ε11

ε22

ε33

2ε23

2ε13

2ε12

=

S11

S12S13S1400

S12

S22S23S2400

S13

S23S33S3400

S14

S24S34S4400

0000S55

S56

0000S56S66

σ11

σ22

σ33

σ23

σ13

σ12

+

α11

α22

α33

2α23

0

0

T−Tref

x1′ =x1 x2′ =

−x2 x3′ =

x3
 235 

ORTHOTROPIC MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT THE PLANES x1 = 0 AND x

2 = 0

" The effect of the second transformation, for symmetry about the plane 

x

2 = 0, is obtained directly from the results given previously for a 

monoclinic material that exhibits symmetry about the plane x

2 = 0

" That is, the second coordinate 

transformation was given by

, , and 

" And, the corresponding matrix 

of direction cosines were 

shown to be given by

i 2

i 3

x1

x2

x3

x1′

x2′

x3′

i 2′

i 3′

,

,

,

i 1 i , 1′

ai′j ≡ ii′ • ij

Plane x

3 = 0

Plane x

2 = 0

x1′ =x1 x2′ =

−x2 x3′ =

x3

a1′1 a1′2 a1′3

a2′1 a2′2 a2′3

a3′1 a3′2 a3′3

=

100

0−10

001
 236 

ORTHOTROPIC MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT THE PLANES x1 = 0 AND x

2 = 0

" For this case,

" It was also shown that, for symmetry about the plane x

2 = 0,

, , , , , , , ,

, , , , , , , ,

, , , and 

T

ε

=

T

σ

=

100000

010000

001000

000−100

000010

00000−1

C14=0C16=0C24=0C26=0C34=0C36=0C45=0C56=0

S14=0S16=0S24=0S26=0S34=0S36=0S45=0S56=0

α12=0α23=0β 12=0β 23=0
 237 

ORTHOTROPIC MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT THE PLANES x1 = 0 AND x

2 = 0

" The net effects of applying the two symmetry transformations 

successively is obtained by applying the conditions given on the 

previous page to the constitutive equations for the monoclinic material

that exhibits symmetry about the plane x1 = 0

" This process yields 

 and

σ11

σ22

σ33

σ23

σ13

σ12

=

C11

C12C13000

C12

C22C23000

C13C23C33000

000C44

00

0000C55

0

00000C66

ε11

ε22

ε33

2ε23

2ε13

2ε12

+

β 11

β 22

β 33

0

0

0

T−Tref
 238 

ORTHOTROPIC MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT THE PLANES x1 = 0 AND x

2 = 0

" It is worth pointing out at this 

point in the development, that the 

single coordinate transformation 

given by , , and 

 does not produce the 

same result as the two successive 

symmetry transformations

ε11

ε22

ε33

2ε23

2ε13

2ε12

=

S11

S12S13000

S12

S22S23000

S13

S23S33000

000S4400

0000S55

0

00000S66

σ11

σ22

σ33

σ23

σ13

σ12

+

α11

α22

α33

0

0

0

T−Tref

i 2

i 3

x1

x2

x3

x2′

x3′

i 2′

i 3′

,

,

ai′j ≡ ii′ • ij

i 1

i 1′

x1′

x1′ =−x1 x2′ =−x2

x3′ =x3
 239 

ORTHOTROPIC MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT THE PLANES x1 = 0 AND x

2 = 0

" In particular, the corresponding matrix of direction cosines is given 

by

" The corresponding stress and strain transformation matrices are

a1′1 a1′2 a1′3

a2′1 a2′2 a2′3

a3′1 a3′2 a3′3

=

−100

0−10

001

T

σ

=

T

ε

=

100000

010000

001000

000−100

0000−10

000001
 240 

ORTHOTROPIC MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT THE PLANES x1 = 0 AND x

2 = 0

" Next, the condition for a state of symmetry to exist at 

a point P of the body is applied

" Using that for this particular transformation, it was 

shown that the condition simplifies to the 

condition that must be a symmetric matrix

" Computing gives

C=T

σ

CTε

− 1

Tε =T

σ

C=T

σ

CTε

− 1

CT

σ

CT

σ

CT

σ

=

C11C12C13C14C15C16

C12C22C23C24C25C26

C13C23C33C34C35C36

C14C24C34C44C45C46

C15C25C35C45C55C56

C16C26C36C46C56C66

1 00000

0 1 0000

00 1 000

000− 1 00

0000− 1 0

00000 1

=

C11C12C13−C14−C15C16

C12C22C23−C24−C25C26

C13C23C33−C34−C35C36

C14C24C34−C44−C45C46

C15C25C35−C45−C55C56

C16C26C36−C46−C56C66
 241 

ORTHOTROPIC MATERIALS - CONCLUDED

REFLECTIVE SYMMETRY ABOUT THE PLANES x1 = 0 AND x

2 = 0

" Enforcing yields the following requirements on the 

stiffnesses:

, , , , , , , and 

" Inspection of the conditions and comparison with the previous result 

for the two successive transformations shows that the results are 

different

CT

σ

=CT

σ

T

C14=0

C15=0

C24=0

C25=0

C34=0

C35=0C46=0

C56=0
 242 

ORTHOTROPIC MATERIALS

REFLECTIVE SYMMETRY ABOUT THE PLANES x1 =0, x

2

=0, ANDx

3

=0

" The next step in the analysis of an orthotropic material is to consider 

the effects of yet a third, successive transformation

" That is, a transformation for which the material exhibits symmetry 

about the perpendicular plane x

3 = 0, in addition to symmetry 

about the perpendicular planes x1 = 0 and x

2 = 0 

" The effect of this third symmetry transformation is obtained directly b

y

applying the conditions obtained for a monoclinic that exhibits 

symmetry about the plane x

3 = 0 to the constitutive equations that were 

obtained previously for the two successive symmetry transformations

" The conditions for symmetry about the plane x

3 = 0 are ,

, , , , , , , ,

, , , , , , , ,

, , and 

C14=0

C15=0C24=0C25=0C34=0C35=0C46=0C56=0

S14=0

S15=0S24=0S25=0S34=0S35=0S46=0S56=0α13=0

α23=0β 13=0β 23=0
 243 

ORTHOTROPIC MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT THE PLANES x1 =

0, x

2

=0, ANDx

3

=0

" Examination of these conditions indicates that the third successive 

transformation yields no new conditions on the constitutive equations 

that are not obtained from the first two successive transformations

" Therefore, two perpendiular planes of material symmetry imply the 

existence of a third mutually perpendicular plane

" An orthotropic material (that is, an orthogonally anisotropic

material) is defined as a material that has three mutually perpendicular 

planes of elastic symmetry 

" An orthotropic material has 9 independent stiffnesses, 9 

independent compliances, 3 independent coefficients of

thermal expansion, and 3 independent thermal moduli
 244 

CONSTITUTIVE EQUATIONS

" The constitutive equations for a linear, thermoelastic, orthotropic 

material are given by

 and

σ11

σ22

σ33

σ23

σ13

σ12

=

C11

C12C13000

C12

C22C23000

C13

C23C33000

000C44

00

0000C550

00000C66

ε11

ε22

ε33

2ε23

2ε13

2ε12

+

β 11

β 22

β 33

0

0

0

T−Tref

ε11

ε22

ε33

2ε23

2ε13

2ε12

=

S11

S12S13000

S12

S22S23000

S13S23S33000

000S44

00

0000S55

0

00000S66

σ11

σ22

σ33

σ23

σ13

σ12

+

α11

α22

α33

0

0

0

T−Tref
 245 

CONSTITUTIVE EQUATIONS - CONCLUDED

" The thermal moduli are given in terms of the coefficients of thermal 

expansion by

" The constitutive equations show that extension and shearing are totally 

uncoupled in the planes x1 = 0, x

2 = 0, and x3 = 0

" When using orthotropic materials with various directional orientations 

in a structure, the structural coordinate system must be distinguished 

from the coordinate systems of the orthotropic materials

" The structural coordinate system is typically picked to facilitate the 

geometric representation of the structure

" For convenience, the coordinate system of an orthotropic material with 

the previously derived constitutive equations is defined as the 

principal material coordinate system and the material is referred 

to as a specially orthotropic material

β 11

β 22

β 33

=−

C11C12C13

C12C22C23

C13C23C33

α11

α22

α33
 246 

SPECIALLY ORTHOTROPIC MATERIALS

" Any material that is fully characterized by the following constitutive 

equations is defined as a specially orthotropic material

 and

" The corresponding coordinate system used to describe this material is 

defined as the principal material coordinate system

σ11

σ22

σ33

σ23

σ13

σ12

=

C11

C12C13000

C12

C22C23000

C13

C23C33000

000C4400

0000C55

0

00000C66

ε11

ε22

ε33

2ε23

2ε13

2ε12

+

β 11

β 22

β 33

0

0

0

T−Tref

ε11

ε22

ε33

2ε23

2ε13

2ε12

=

S11

S12S13000

S12

S22S23000

S13S23S33000

000S44

00

0000S55

0

00000S66

σ11

σ22

σ33

σ23

σ13

σ12

+

α11

α22

α33

0

0

0

T−Tref
 247 

GENERALLY ORTHOTROPIC MATERIALS

" For an arbitrary coordinate transformation, from a principal 

coordinate frame to a coordinate frame, the transformation 

matrices and are fully populated

" Thus, when the elastic stiffness coefficients of a specially orthotropic 

solid are transformed from the coordinate frame to the 

 coordinate frame, the matrices of transformed elastic 

constants are also fully populated

"To an observer, without prior knowledge of the material, the solid 

appears to be anisotropic

" When a coordinate frame exists for a solid in which it is specially 

orthotropic, the material is referred to generally orthotropic, to 

distinguish it from an anisotropic material

x1,x2,x3

x1′,x2′,x3′

Tε T

σ

x1,x2,x3

x1′,x2′,x3′
 248 

GENERALLY ORTHOTROPIC MATERIALS

CONTINUED

" For a dextral rotation about the x

3 - axis, with and 

, the transformed elastic constants are given by

m= cosθ3

n= sinθ3

C1′1′ =m4C11+2m2n2C12+2C66+n4C22

C1′2′ =m2n2C11+C22−4C66+m4+n4C12

C1′3′ =m2C13+n2C23

C1′4′ =0C1′5′ =0

C1′6′ =mnm2−n2C12+2C66+mnn2C22−m2C11

C2′2′ =m4C22+2m2n2C12+2C66+n4C11

C2′3′ =m2C23+n2C13

C2′4′ =0C2′5′ =0

C2′6′ =mnn2−m2C12+2C66+mnm2C22−n2C11
 249 

GENERALLY ORTHOTROPIC MATERIALS

CONTINUED

" The population of the matrix of transformed elastic stiffnesses is given 

by

" For the coordinate frame, the specially orthotropic material 

appears to have the properties of a monoclinic material

C3′3′ =C33C3′4′ =0C3′5′ =0C3′6′ =mnC23−C13

C4′4′ =m2C44+n2C55

C4′5′ =mnC44−C55C4′6′ =0C5′6′ =0

C5′5′ =m2C55+n2C44

C6′6′ =m2n2C11+C22−2C12+m2−n2 2C66

C1′1′ C1′2′ C1′3′ 00C1′6′

C1′2′ C2′2′ C2′3′ 00C2′6′

C1′3′ C2′3′ C3′3′ 00C3′6′

000C

4′4′ C4′5′ 0

000C

4′5′ C5′5′ 0

C1′6′ C2′6′ C3′6′ 0

0C6′6′

x1′,x2′,x3′
 250 

GENERALLY ORTHOTROPIC MATERIALS

CONTINUED

" Similarly,

S1′1′ =m4S11+m2n22S12+S66+n4S22

S1′2′ =m2n2S11+S22−S66+m4+n4S12

S1′3′ =m2S13+n2S23

S1′4′ =0S1′5′ =0

S1′6′ =mnm2−n22S12+S66+2mnn2S22−m2S11

S2′2′ =m4S22+m2n22S12+S66+n4S11

S2′3′ =m2S23+n2S13

S2′4′ =0S2′5′ =0S3′3′ =S33S3′4′ =0S3′5′ =0

S2′6′ =mnn2−m22S12+S66+2mnm2S22−n2S11

S3′6′ =2mnS23−S13
 251 

GENERALLY ORTHOTROPIC MATERIALS

CONTINUED

" The population of the matrix of transformed elastic compliances is 

given by

" For the coordinate frame, the specially orthotropic material,

again, appears to have the properties of a monoclinic material

S4′4′ =m2S44+n2S55

S4′5′ =mnS44−S55S4′6′ =0

S5′5′ =m2S55+n2S44

S5′6′ =0

S6′6′ =4m2n2S11+S22−2S12+m2−n2 2S66

S1′1′ S1′2′ S1′3′ 00S1′6′

S1′2′ S2′2′ S2′3′ 00S2′6′

S1′3′ S2′3′ S3′3′ 00S3′6′

000S

4′4′ S4′5′ 0

000S

4′5′ S5′5′ 0

S1′6′ S2′6′ S3′6′ 0

0S6′6′

x1′,x2′,x3′
 252 

GENERALLY ORTHOTROPIC MATERIALS

CONTINUED

" For a dextral rotation about the x

3 - axis, the transformed thermalexpansion coefficients are given b

y

" Similarly, the transformed thermal moduli are given by

α1′1′ =m2α11+n2α22α2′2′ =m2α22+n2α11

α3′3′ =α33

α2′3′ =0α1′3′ =0

α1′2′ =mnα22−α11

β 1′1′ =m2β 11+n2β 22β 2′2′ =m2β 22+n2β 11

β 3′3′ =β 33

β 2′3′ =0β 1′3′ =0

β 1′2′ =mnβ 22−β 11
 253 

GENERALLY ORTHOTROPIC MATERIALS

CONCLUDED

" Thus, for a dextral rotation about the x

3 - axis, the transformed 

constitutive equations for a specially orthotropic material are given by

ε1′1′

ε2′2′

ε3′3′

2ε2′3′

2ε1′3′

2ε1′2′

=

S1′1′ S1′2′ S1′3′ 0

0S1′6′

S1′2′ S2′2′ S2′3′ 00S2′6′

S1′3′ S2′3′ S3′3′ 00S3′6′

000S4′4′ S4′5′ 0

000S4′5′ S5′5′ 0

S1′6′ S2′6′ S3′6′ 00S6′6′

σ1′1′

σ2′2′

σ3′3′

σ2′3′

σ1′3′

σ1′2′

+

α1′1′

α2′2′

α3′3′

0

0

2α1′2′

T–Tref

σ1′1′

σ2′2′

σ3′3′

σ2′3′

σ1′3′

σ1′2′

=

C1′1′ C1′2′ C1′3′ 0

0C1′6′

C1′2′ C2′2′ C2′3′ 00C2′6′

C1′3′ C2′3′ C3′3′ 00C3′6′

000C4′4′ C4′5′ 0

000C4′5′ C5′5′ 0

C1′6′ C2′6′ C3′6′ 00C6′6′

ε1′1′

ε2′2′

ε3′3′

2ε2′3′

2ε1′3′

2ε1′2′

+

β 1′1′

β 2′2′

β 3′3′

0

0

β 1′2′

T–Tref
 254 

TRIGONAL

MATERIALS
 255 

TRIGONAL MATERIALS

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x

3 AXIS

"To determine the conditions on the constitutive equations for trigonal 

materials, it is necessary to consider a plane of elastic symmetry that 

is oriented arbitrarily, with respect to two of the coordinate axes

" In particular, consider a plane of

elastic symmetry whose normal 

lies in the plane x

3 = 0 and makes an 

angle θ

3 with the x1 axis, as shown in 

the figure

" The angle θ

3 is defined to be in the 

range , because 

and define the same plane

" In addition, let the be the 

coordinates used to define the 

material symmetry

i 1 i 2

i 3

x1

x2

x3

x1′

x2′

x3′

i 3′

,

,

θ3

θ3

i 1′, n

i 2′

Plane of elastic

symmetry

ai′j ≡ ii′ • ij

n

−

π

2

<θ3 ≤

π

2

θ3 =−

π

2

θ3 =

π

2

x1′, x2′, x3′
 256 

TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x

3 AXIS

" Specifically, for a plane of elastic symmetry given by , the 

symmetry transformation is shown in the figure and is given by

, , and 

" The corresponding matrix of 

direction cosines is given by

" For the coordinate 

system, the general constitutive 

equations are expressed as 

follows

x1′ =0

x1′′ =−x1′ x2′′ =x2′ x3′′ =x3′

x1′

x2′

i 1′

i 2′

i 3′

,

,

,i 2′′

i 3′′

i 1′′

x3′ x3′′

,x2′′

x1′′

ai′j ≡ ii′ • ij

Planex1′ =0

a1′′1′ a1′′2′ a1′′3′

a2′′1′ a2′′2′ a2′′3′

a3′′1′ a3′′2′ a3′′3′

=

−100

010

001

x1′, x2′, x3′
 257 

TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x

3 AXIS

orby

 and ,

 where 

ε1′1′

ε2′2′

ε3′3′

2ε2′3′

2ε1′3′

2ε1′2′

=

S1′1′ S1′2′ S1′3′ S1′4′ S1′5′ S1′6′

S1′2′ S2′2′ S2′3′ S2′4′ S2′5′ S2′6′

S1′3′ S2′3′ S3′3′ S3′4′ S3′5′ S3′6′

S1′4′ S2′4′ S3′4′ S4′4′ S4′5′ S4′6′

S1′5′ S2′5′ S3′5′ S4′5′ S5′5′ S5′6′

S1′6′ S2′6′ S3′6′ S4′6′ S5′6′ S6′6′

σ1′1′

σ2′2′

σ3′3′

σ2′3′

σ1′3′

σ1′2′

+

α1′1′

α2′2′

α3′3′

2α2′3′

2α1′3′

2α1′2′

T−Tref

σ1′1′

σ2′2′

σ3′3′

σ2′3′

σ1′3′

σ1′2′

=

C1′1′ C1′2′ C1′3′ C1′4′ C1′5′ C1′6′

C1′2′ C2′2′ C2′3′ C2′4′ C2′5′ C2′6′

C1′3′ C2′3′ C3′3′ C3′4′ C3′5′ C3′6′

C1′4′ C2′4′ C3′4′ C4′4′ C4′5′ C4′6′

C1′5′ C2′5′ C3′5′ C4′5′ C5′5′ C5′6′

C1′6′ C2′6′ C3′6′ C4′6′ C5′6′ C6′6′

ε1′1′

ε2′2′

ε3′3′

2ε2′3′

2ε1′3′

2ε1′2′

+

β 1′1′

β 2′2′

β 3′3′

2β2′3′

2β1′3′

2β1′2′

T−Tref

ε′ =S′ σ′ +α′ Θσ′ =C′ ε′ +β′ Θ

Θ=

T−Tref
 258 

TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x

3 AXIS

" For the coordinate system, the constitutive equations are 

identical in form and are obtained by replacing the index pair with 

" In the abridged notation,

 and

, where 

" For the transformation of coordinates defined by the symmetry 

transformation (a reflection about the plane ),

 and where

x1′′, x2′′, x3′′

i′j′

i′′j′′

ε′′=S′′σ′′+α′′Θ

σ′′=C′′ε′′+β′′Θ

Θ=

T−Tref

x1′ =0

σ′′=T

σ

r 1′ σ′ ε′′

=T

ε

r 1′ ε′
 259 

TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x

3 AXIS

" Likewise,

T

σ

r 1′

=

T

ε

r 1′

=

100000

010000

001000

000100

0000−10

00000−1

S′′=T

ε

r 1′

S′ T

σ

r 1′

−1

S′ =T

ε

r 1′

−1

S′′T

σ

r 1′

C′′=T

σ

r 1′

C′ T

ε

r 1′

−1

C′ =T

σ

r 1′

−1

C′′T

ε

r 1′

α′′=T

ε

r 1′

α′ β′′=T

σ

r 1′

β′
 260 

TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x

3 AXIS

" The conditions for invariance under the symmetry transformation 

are given by

" Rather than calculating the outcome of the in

variance conditions, the 

outcome can be found by direct comparison with the results given 

previously for a material that is monoclinic with respect to the plane 

x1 = 0 

" Direct comparison reveals that the material is monoclinic with 

respect to the plane 

S′ =T

ε

r 1′

S′ T

σ

r 1′

− 1

C′ =T

σ

r 1′

C′ T

ε

r 1′

− 1

α′ =T

ε

r 1′

α′ β′ =T

σ

r 1′

β′

x1′ =0
 261 

TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x

3 AXIS

" That is, direct comparison yields

 and

ε1′1′

ε2′2′

ε3′3′

2ε2′3′

2ε1′3′

2ε1′2′

=

S1′1′ S1′2′ S1′3′ S1′4′ 0

0

S1′2′ S2′2′ S2′3′ S2′4′ 0

0

S1′3′ S2′3′ S3′3′ S3′4′ 0

0

S1′4′ S2′4′ S3′4′ S4′4′ 0

0

0000S5′5′ S5′6′

0000S5′6′ S6′6′

σ1′1′

σ2′2′

σ3′3′

σ2′3′

σ1′3′

σ1′2′

+

α1′1′

α2′2′

α3′3′

2α2′3′

0

0

T−Tref

σ1′1′

σ2′2′

σ3′3′

σ2′3′

σ1′3′

σ1′2′

=

C1′1′ C1′2′ C1′3′ C1′4′ 0

0

C1′2′ C2′2′ C2′3′ C2′4′ 0

0

C1′3′ C2′3′ C3′3′ C3′4′ 0

0

C1′4′ C2′4′ C3′4′ C4′4′ 0

0

0000C5′5′ C5′6′

0000C5′6′ C6′6′

ε1′1′

ε2′2′

ε3′3′

2ε2′3′

2ε1′3′

2ε1′2′

+

β 1′1′

β 2′2′

β 3′3′

β 2′3′

0

0

T−Tref
 262 

TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x

3 AXIS

" Next, the compliances, stiffnesses, thermal-expansion coefficients, and 

thermal moduli,expressed in coordinates, are referred back 

to the original coordinates

" The transformation corresponds to 

the dextral rotation about the x

3 axis,

shown in the figure, and is given by

,

, and ,

with 

" The corresponding matrix of 

direction cosines is given by

x1′, x2′, x3′

x1, x2, x3

i 1 i 2

i 3

x1

x2

x3

x1′

x2′

x3′

i 3′

,

,

θ3

θ3

i 1′, n

i 2′

Plane of elastic

symmetry

ai′j ≡ ii′ • ij

x1′ =x1cosθ3 +x2sinθ3

x2′ =−x1sinθ3 +x2cosθ3 x3′ =x3

−

π

2

<θ3 ≤

π

2

a1′1 a1′2 a1′3

a2′1 a2′2 a2′3

a3′1 a3′2 a3′3

=

cosθ3 sin

θ3 0

− sinθ3 cos

θ3 0

001
 263 

TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x

3 AXIS

" The corresponding stress and strain transformation matrices were 

shown previously to be given by

T

σ

θ3 = and

cos2θ3 sin2θ3 0

00 2sin

θ3cosθ3

sin2θ3 cos2θ3 0

00− 2sin

θ3cosθ3

00 1 000

000 cos

θ3 − sinθ3 0

000 sin

θ3 cosθ3 0

− sin

θ3cosθ3 sinθ3cosθ3 0

00 cos2

θ3 − sin2θ3

T

ε

θ3 =

cos2θ3 sin2θ3 0

00 sin

θ3cosθ3

sin2θ3 cos2θ3 0

00− sin

θ3cosθ3

00 1 000

000 cos

θ3 − sinθ3 0

000 sin

θ3 cosθ3 0

− 2sin

θ3cosθ3 2sinθ3cosθ3 0

00 cos2

θ3 − sin2θ3
 264 

TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x

3 AXIS

" The transformation laws for the compliances, stiffnesses, thermalexpansion coefficients, and thermal moduli have been given as 

" These laws transform the previously obtained invariance conditions on 

, , , and into invariance conditions on , , ,

and 

" Specific expressions for these transformation laws (dextral rotation 

about the x

3 axis) were given previously for a fully anisotropic material 

(triclinic) 

" Note that the matrices and are fully populated

" Also, the vectors and are fully populated

C′ =T

σ

CTε

− 1

S′ =Tε ST

σ

− 1

β′ =T

σ

βα′ =Tε α

C′ S′ α′ β′ CSα

β

CS

αβ
 265 

TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x

3 AXIS

" Because of the invariance conditions, the matrices and have 

the form and 

" Also, the vectors and have the form and 

" The corresponding forms of , , , and are obtained from 

the transformation laws once a value for the angle θ

3 is specified 

C′ S′

C1′1′ C1′2′ C1′3′ C1′4′ 00

C1′2′ C2′2′ C2′3′ C2′4′ 0

0

C1′3′ C2′3′ C3′3′ C3′4′ 0

0

C1′4′ C2′4′ C3′4′ C

4′4′ 0

0

0000C5′5′ C5′6′

0000C5′6′ C6′6′

S1′1′ S1′2′ S1′3′ S1′4′ 00

S1′2′ S2′2′ S2′3′ S2′4′ 0

0

S1′3′ S2′3′ S3′3′ S3′4′ 0

0

S1′4′ S2′4′ S3′4′ S

4′4′ 0

0

0000S5′5′ S5′6′

0000S5′6′ S6′6′

α′ β′

α1′1′

α2′2′

α3′3′

2α2′3′

0

0

β 1′1′

β 2′2′

β 3′3′

β 2′3′

0

0

CSαβ
 266 

TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x

3 AXIS

" For example, specifying yields 

" Enforcing the invariance conditions on gives 

, , , , , , , and 

θ3 =0

C1′1′ C1′2′ C1′3′ C1′4′ 00

C1′2′ C2′2′ C2′3′ C2′4′ 0

0

C1′3′ C2′3′ C3′3′ C3′4′ 0

0

C1′4′ C2′4′ C3′4′ C4′4′ 0

0

0000C5′5′ C5′6′

0000C5′6′ C6′6′

=

C11

C12C13C14C15C16

C12C22C23C24C25C26

C13C23C33C34C35C36

C14C24C34C44C45C46

C15C25C35C45C55C56

C16C26C36C46C56C66

β 1′1′

β 2′2′

β 3′3′

β 2′3′

0

0

=

β 11

β 22

β 33

β 23

β 13

β 12

S1′1′ S1′2′ S1′3′ S1′4′ 00

S1′2′ S2′2′ S2′3′ S2′4′ 0

0

S1′3′ S2′3′ S3′3′ S3′4′ 0

0

S1′4′ S2′4′ S3′4′ S4′4′ 0

0

0000S5′5′ S5′6′

0000S5′6′ S6′6′

=

S11

S12S13S14S15S16

S12S22S23S24S25S26

S13S23S33S34S35S36

S14S24S34S44S45S46

S15S25S35S45S55S56

S16S26S36S46S56S66

α1′1′

α2′2′

α3′3′

2α2′3′

0

0

=

α11

α22

α33

2α23

2α13

2α12

C′

C15=0C16=0C25=0C26=0C35=0C36=0C45=0C46=0
 267 

TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x

3 AXIS

" Likewise, enforcing the invariance conditions on gives 

, , , , , , , and 

" Enforcing the invariance conditions on and gives

 and 

" These conditions are identical to the conditions previously obtained for 

a material that is monoclinic with respect to the plane x1 = 0

" Similarly, specifying and enforcing the invariance conditions on 

, , , and yields the conditions previously obtained for 

a material that is monoclinic with respect to the plane x

2 = 0

S′

S15=0S16=0S25=0S26=0S35=0S36=0S45=0S46=0

α′ β′

α12=α13=0β 12=β 13=0

θ3 =

π

2

C′ S′ α′ β′
 268 

TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x

3 AXIS

" Now consider an arbitrary value for the angle θ

3 such that 

" The previous example for shows that the invariance conditions 

on , , , and were obtained by the terms of , ,

, and that were zero valued

" Using the transformation equations for a dextral rotation about the x

3

axis that were given previously for a (triclinic) fully anisotropic, elastic 

material gives the following results for the invariance conditions on 

−

π

2

<θ3 ≤

π

2

θ3 =0

CSαβC′ S′

α′ β′

C′

C1′5′ =0: m3C15+m2n2C56+C14+mn22C46+C25+n3C24=0

C1′6′ =0: m2m2−3n2C16−m3nC11−C12−2C66

+mn3C22−C12−2C66−n2n2−3m2C26=0
 269 

TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x

3 AXIS

 with and 

" These conditions give 8 equations and 20 unknowns

C2′5′ =0: m3C25−m2n2C56−C24−mn22C46−C15+n3C14=0

C2′6′ =0: m2m2−3n2C26+m3nC22−C12−2C66

−mn3C11−C12−2C66−n2n2−3m2C16=0

C3′5′ =0: mC35+nC34=0

C3′6′ =0: m2−n2C36+mnC23−C13=0

C4′5′ =0: m2−n2C45+mnC44−C55=0

C4′6′ =0: m3C46−m2nC56+C14−C24−mn2C46−C15+C25+n3C56=0

m= cosθ3 n= sinθ3
 270 

TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x

3 AXIS

" The invariance conditions on 

S′ are given b

y

S1′5′ =0: m3S15+m2nS56+S14+mn2S46+S25+n3S24=0

S1′6′ =0: m2m2−3n2S16−m3n2S11−2S12−S66

+mn32S22−2S12−S66−n2n2−3m2S26=0

S2′5′ =0: m3S25−m2nS56−S24−mn2S46−S15+n3S14=0

S2′6′ =0: m2m2−3n2S26+m3n2S22−2S12−S66

−mn32S11−2S12−S66−n2n2−3m2S16=0

S3′5′ =0: mS35+nS34=0

S3′6′ =0: m2−n2S36+2mnS23−S13=0

S4′5′ =0: m2−n2S45+mnS44−S55=0

S4′6′ =0: m3S46−m2nS56+2S14−2S24−mn2S46−2S15+2S25+n3S56=0
 271 

TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x

3 AXIS

" The invariance conditions on 

α′ and β′ yield

α1′3′ =0: mα13+nα23=0

α1′2′ =0: m2−n2α12+mnα22−α11=0

β 1′3′ =0: mβ 13+nβ 23=0

β 1′2′ =0: m2−n2β 12+mnβ 22−β 11=0
 272 

TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x

3 AXIS

" In determining the restrictions on the compliances, stiffnesses,

thermal-expansion coefficients, and thermal moduli for materials that 

possess more than one plane of elastic symmetry, all of which contain 

the x

3 axis, it is convenient to select the first plane to be given b

y

" Thus, the material is monoclinic with respect to the plane x1 = 0

and, as a result, the following conditions hold

" These relations, and the fact that for nonzero values 

of , are used to simplify the previously given invariance 

conditions into the following three uncoupled groups

θ3 =0

C15=0C16=0C25=0C26=0C35=0C36=0C45=0C46=0

S15=0S16=0S25=0S26=0S35=0S36=0S45=0S46=0

α12=0α13=0β 12=0β 13=0

n= sinθ3 ≠0

−

π

2

<θ3 ≤

π

2
 273 

TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x

3 AXIS

" Group 1

" Group 2

" Group 3

C1′5′ =0: m22C56+C14+n2C24=0

m= cosθ3 n= sinθ3

C2′5′ =0: −m22C56−C24+n2C14=0

C4′6′ =0: n2C56−m2C56+C14−C24=0

C3′5′ =0: C34=0

C3′6′ =0: mC23−C13=0

C4′5′ =0: mC44−C55=0

C1′6′ =0: mn2C22−C12−2C66−m3C11−C12−2C66=0

C2′6′ =0: m3C22−C12−2C66−mn2C11−C12−2C66=0
 274 

TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x

3 AXIS

" The first group can be written as

" Because the right-hand side of the equations are identical, it follows 

that the left-hand sides are equal; that is,

" Each equation can be expressed as , which is satisfied 

by and by

" implies and 

" implies , , and 

C1′5′ =0: −C24=m22C56+C14−C24

m= cosθ3 n= sinθ3

C2′5′ =0: C14=m22C56+C14−C24

C4′6′ =0: C56=m22C56+C14−C24

C56=−C24=C14=Γ

Γ=4Γcos2θ3

Γ=0 cos2

θ3 =

1

4

Γ=0

C56=−C24=C14=0

−

π

2

<θ3 ≤

π

2

cos2θ3 =

1

4

θ3 =±

π

3

C56=C14

C24=−C14
 275 

TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x

3 AXIS

" The third group of equations is simplified by first adding the two 

equations together and then by subtracting the two equations

" Adding gives ,

 which is satisfied for all if , or for all 

 if 

" Subtracting gives

,

 which is satisfied for all if ,

 or for all if 

C1′6′ +C2′6′ =0: C22−C11 cosθ3 =0

−

π

2

<θ3 ≤

π

2

C22=C11

C22≠C11

θ3 =

π

2

C1′6′ −C2′6′ =0: 2C12+4C66−C11−C22 cosθ3cos2θ3 =0

−

π

2

<θ3 ≤

π

2

2C12+4C66−C11−C22=0

2C12+4C66−C11−C22≠0

θ3 =±

π

4 or

π

2
 276 

TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x

3 AXIS

" Group 4

" Group 5

" Group 6

S1′5′ =0: m2S56+S14+n2S24=0

m= cosθ3 n= sinθ3

S2′5′ =0: −m2S56−S24+n2S14=0

S4′6′ =0: −m2S56+2S14−2S24+n2S56=0

S3′5′ =0: S34=0

S3′6′ =0: mS23−S13=0

S4′5′ =0: mS44−S55=0

S1′6′ =0: mn22S22−2S12−S66−m32S11−2S12−S66=0

S2′6′ =0: m32S22−2S12−S66−mn22S11−2S12−S66=0
 277 

TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x

3 AXIS

" Like for the stiffnesses, the fourth group can also be written as

" Because the right-hand side of the equations are identical, it follows 

that the left-hand sides are equal; that is,

" Each equation can be expressed as , which is satisfied 

by and by

" implies and 

" implies , , and 

S1′5′ =0: −S24=m2S56+S14−S24

m= cosθ3 n= sinθ3

S2′5′ =0: S14=m2S56+S14−S24

S4′6′ =0: 1

2

S56=m2S56+S14−S24

1

2

S56=−S24=S14=Δ

Δ=4Δcos2θ3

Δ=0 cos2

θ3 =

1

4

Δ=0S56=S24=S14=0

−

π

2

<θ3 ≤

π

2

cos2θ3 =

1

4

θ3 =±

π

3

S56=2S14

S24=−S14
 278 

TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x

3 AXIS

" The sixth group of equations is also simplified by first adding the two 

equations together and then by subtracting the two equations

" Adding gives ,

 which is satisfied for all if , or for all 

 if 

" Subtracting gives

,

 which is satisfied for all if ,

 or for all if 

S1′6′ +S2′6′ =0: S22−S11 cosθ3 =0

−

π

2

<θ3 ≤

π

2

S22=S11

S22≠S11

θ3 =

π

2

S1′6′ −S2′6′ =0: S11+S22−2S12−S66 cosθ3cos2θ3 =0

−

π

2

<θ3 ≤

π

2

2S12+S66−S11−S22=0

2S12+S66−S11−S22≠0

θ3 =±

π

4 or

π

2
 279 

TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x

3 AXIS

" The symmetry properties for trigonal materials arise from the 

solution for the invariance conditions that are given by

" For this solution, the planes of 

elastic symmetry are all parallel to 

the x

3 axis and are given b

y

" For ,

and 

" The stiffness equations in group 2 

yield the invariance conditions 

, , and 

cos2θ3 =

1

4

x1

x2

θ3 = 0

Symmetry

plane for

θ3 =+

π

3

Symmetry

plane for

θ3 = 0

Symmetry

plane for

θ3 =+

π

3

Plan view of symmetry planes

θ3 =−

π

3

θ3 =−

π

3

θ3 =0 and±

π

3

θ3 =±

π

3

m= cos±

π

3

=

1

2

≠0

n= sin±

π

3

=±

3

2

≠0

C34=0C23=C13C55=C44
 280 

TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x

3 AXIS

" Likewise, the stiffness equations in group 1 yield the invariance 

conditions and 

" Furthermore, the stiffness equations in group 3 yield the invariance 

conditions and 

" The compliance equations in group 5 yield the invariance conditions 

, , and 

" The compliance equations in group 4 yield the invariance conditions 

 and 

" The compliance equations in group 6 yield the invariance conditions 

 and 

C56=C14

C24=−C14

C22=C11

C66=

1

2

C11−C12

S34=0S23=S13S55=S44

S56=2S14

S24=−S14

S22=S11

S66=2S11−S12
 281 

TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x

3 AXIS

" The invariance conditions and 

 yield

 and 

" The invariance conditions and 

 yield 

 and 

"Together, the invariance conditions yield the following constitutive 

equations for a trigonal material

α1′3′ =0: mα13+nα23=0

α1′2′ =0: m2−n2α12+mnα22−α11=0

α23=α13=α12=0α22=α11

β 1′3′ =0: mβ 13+nβ 23=0

β 1′2′ =0: m2−n2β 12+mnβ 22−β 11=0

β 23=β 13=β 12=0β 22=β 11
 282 

TRIGONAL MATERIALS - CONCLUDED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x

3 AXIS

" Therefore, a trigonal material has six independent elastic constants 

and two independent thermal-expansion or thermal-compliance 

parameters

σ11

σ22

σ33

σ23

σ13

σ12

=

C11C12C13C1400

C12

C11C13−C1400

C13

C13C33000

C14

−C140C4400

0000C44

C14

0000C14

C11−C12

2

ε11

ε22

ε33

2ε23

2ε13

2ε12

+

β 11

β 11

β 33

0

0

0

T−Tref

ε11

ε22

ε33

2ε23

2ε13

2ε12

=

S11S12S13S1400

S12

S11S13−S1400

S13

S13S33000

S14

−S140S4400

0000S44

2S14

00002S142S11−S12

σ11

σ22

σ33

σ23

σ13

σ12

+

α11

α11

α33

0

0

0

T−Tref
 283 

TRIGONAL MATERIALS

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x1 AXIS

"To determine the conditions on the constitutive equations for trigonal 

materials, it is necessary to consider a plane of elastic symmetry that 

is oriented arbitrarily, with respect to two of the coordinate axes

" In particular, consider a plane of

elastic symmetry whose normal 

lies in the plane x1 = 0 and makes an 

angle θ1 with the x

2 axis, as shown in 

the figure

" The angle θ1 is defined to be in the 

range , because 

and define the same plane

" In addition, let the be the 

coordinates used to define the 

material symmetry

i 1

i 2 i 3

x1

x2

x3

x1′

x2′

x3′

,

,

Plane of elastic

symmetry

i 3′

i 1′

i 2′, n

θ1

θ1

ai′j ≡ ii′ • ij

n

−

π

2

<θ1 ≤

π

2

θ1 =−

π

2

θ1 =

π

2

x1′, x2′, x3′
 284 

TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x1 AXIS

" Specifically, for a plane of elastic symmetry given by , the 

symmetry transformation is shown in the figure and is given by

, , and 

" The corresponding matrix of 

direction cosines is given by

" For the coordinate 

system, the constitutive 

equations are expressed as 

follows

x2′ =0

x1′′ =x1′ x2′′ =−x2′ x3′′ =x3′

x2′

i 1′

i 2′

i 3′

,

,

i 2′′

i 3′′

i 1′′

x3′′

x2′′

,

x1′ x1′′ ,

x3′

ai′j ≡ ii′ • ij

a1′′1′ a1′′2′ a1′′3′

a2′′1′ a2′′2′ a2′′3′

a3′′1′ a3′′2′ a3′′3′

=

100

0−10

001

x1′, x2′, x3′
 285 

TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x1 AXIS

orby

 and ,

 where 

ε1′1′

ε2′2′

ε3′3′

2ε2′3′

2ε1′3′

2ε1′2′

=

S1′1′ S1′2′ S1′3′ S1′4′ S1′5′ S1′6′

S1′2′ S2′2′ S2′3′ S2′4′ S2′5′ S2′6′

S1′3′ S2′3′ S3′3′ S3′4′ S3′5′ S3′6′

S1′4′ S2′4′ S3′4′ S4′4′ S4′5′ S4′6′

S1′5′ S2′5′ S3′5′ S4′5′ S5′5′ S5′6′

S1′6′ S2′6′ S3′6′ S4′6′ S5′6′ S6′6′

σ1′1′

σ2′2′

σ3′3′

σ2′3′

σ1′3′

σ1′2′

+

α1′1′

α2′2′

α3′3′

2α2′3′

2α1′3′

2α1′2′

T−Tref

σ1′1′

σ2′2′

σ3′3′

σ2′3′

σ1′3′

σ1′2′

=

C1′1′ C1′2′ C1′3′ C1′4′ C1′5′ C1′6′

C1′2′ C2′2′ C2′3′ C2′4′ C2′5′ C2′6′

C1′3′ C2′3′ C3′3′ C3′4′ C3′5′ C3′6′

C1′4′ C2′4′ C3′4′ C4′4′ C4′5′ C4′6′

C1′5′ C2′5′ C3′5′ C4′5′ C5′5′ C5′6′

C1′6′ C2′6′ C3′6′ C4′6′ C5′6′ C6′6′

ε1′1′

ε2′2′

ε3′3′

2ε2′3′

2ε1′3′

2ε1′2′

+

β 1′1′

β 2′2′

β 3′3′

2β2′3′

2β1′3′

2β1′2′

T−Tref

ε′ =S′ σ′ +α′ Θσ′ =C′ ε′ +β′ Θ

Θ=

T−Tref
 286 

TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x1 AXIS

" For the coordinate system, the constitutive equations are 

identical in form and are obtained by replacing the index pair with 

" In the abridged notation,

 and

, where 

" For the transformation of coordinates defined by the symmetry 

transformation (a reflection about the plane ),

 and where

x1′′, x2′′, x3′′

i′j′

i′′j′′

ε′′=S′′σ′′+α′′Θ

σ′′=C′′ε′′+β′′Θ

Θ=

T−Tref

x2′ =0

σ′′=T

σ

r 2′ σ′ ε′′

=T

ε

r 2′ ε′
 287 

TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x1 AXIS

" Likewise,

T

σ

r 2′

=

T

ε

r 2′

=

100000

010000

001000

000−100

000010

00000−1

S′′=T

ε

r 2′

S′ T

σ

r 2′

− 1

S′ =T

ε

r 2′

− 1

S′′T

σ

r 2′

C′′=T

σ

r 2′

C′ T

ε

r 2′

− 1

C′ =T

σ

r 2′

− 1

C′′T

ε

r 2′

α′′=T

ε

r 2′ α′ β′′

=T

σ

r 2′ β′
 288 

TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x1 AXIS

" The conditions for invariance under the symmetry transformation 

are given by

" Rather than calculating the outcome of the invariance conditions, the 

outcome can be found by direct comparison with the results given 

previously for a material that is monoclinic with respect to the plane 

x

2 = 0 

" Direct comparison reveals that the material is monoclinic with 

respect to the plane 

S′ =T

ε

r 2′

S′ T

σ

r 2′

− 1

C′ =T

σ

r 2′

C′ T

ε

r 2′

− 1

α′ =T

ε

r 2′ α′ β′ =

T

σ

r 2′ β′

x2′ =0
 289 

TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x1 AXIS

" That is, direct comparison yields

 and

ε1′1′

ε2′2′

ε3′3′

2ε2′3′

2ε1′3′

2ε1′2′

=

S1′1′ S1′2′ S1′3′ 0

S1′5′ 0

S1′2′ S2′2′ S2′3′ 0

S2′5′ 0

S1′3′ S2′3′ S3′3′ 0

S3′5′ 0

000S4′4′ 0

S4′6′

S1′5′ S2′5′ S3′5′ 0S5′5′ 0

000S4′6′ 0

S6′6′

σ1′1′

σ2′2′

σ3′3′

σ2′3′

σ1′3′

σ1′2′

+

α1′1′

α2′2′

α3′3′

0

2α1′3′

0

T−Tref

σ1′1′

σ2′2′

σ3′3′

σ2′3′

σ1′3′

σ1′2′

=

C1′1′ C1′2′ C1′3′ 0

C1′5′ 0

C1′2′ C2′2′ C2′3′ 0

C2′5′ 0

C1′3′ C2′3′ C3′3′ 0

C3′5′ 0

000C4′4′ 0

C4′6′

C1′5′ C2′5′ C3′5′ 0C5′5′ 0

000C4′6′ 0

C6′6′

ε1′1′

ε2′2′

ε3′3′

2ε2′3′

2ε1′3′

2ε1′2′

+

β 1′1′

β 2′2′

β 3′3′

0

β 1′3′

0

T−Tref
 290 

TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x1 AXIS

" Next, the compliances, stiffnesses, thermal-expansion coefficients, and 

thermal moduli, expressed in coordinates, are expressed in 

the original coordinates

" The transformation corresponds to 

the dextral rotation about the x1 axis,

shown in the figure, and is given by

, , and 

, with 

" The corresponding matrix of 

direction cosines is given by

x1′, x2′, x3′

x1, x2, x3

i 1

i 2 i 3

x1

x2

x3

x1′

x2′

x3′

,

,

Plane of elastic

symmetry

i 3′

i 1′

i 2′, n

θ1

θ1

ai′j ≡ ii′ • ij

x1′ =x1 x2′ =x2cosθ1 +x3sinθ1

x3′ =−x2sinθ1 +x3cosθ1

−

π

2

<θ1 ≤

π

2

a1′1 a1′2 a1′3

a2′1 a2′2 a2′3

a3′1 a3′2 a3′3

=

100

0 cosθ1 sinθ1

0− sinθ1 cosθ1
 291 

TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x1 AXIS

" The corresponding stress and strain transformation matrices were 

shown previously to be given by

T and

σ

θ1 =

100000

0 cos2

θ1 sin2θ1 2sinθ1cosθ1 0

0

0 sin2

θ1 cos2θ1 −

2sin

θ1cosθ1 0

0

0

− sinθ1cosθ1 sinθ1cosθ1 cos2θ1 − sin2θ1 0

0

0000 cos

θ1 − sinθ1

0000 sin

θ1 cosθ1

T

ε

θ1 =

100000

0 cos2

θ1 sin2θ1 sinθ1cosθ1 0

0

0 sin2

θ1 cos2θ1 − sinθ1cosθ1 0

0

0

−2sinθ1cosθ1 2sinθ1cosθ1 cos2θ1 − sin2θ1 0

0

0000 cos

θ1 − sinθ1

0000 sin

θ1 cosθ1
 292 

TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x1 AXIS

" The transformation laws for the compliances, stiffnesses, thermalexpansion coefficients, and thermal moduli have been given as 

" These laws transform the invariance conditions on , , , and 

 into invariance conditions on , , , and 

" Specific expressions for these transformation laws (dextral rotation 

about the x1 axis) were given previously for a fully anisotropic material 

(triclinic) 

" Note that the matrices and , and the vectors and ,

are fully populated

C′ =T

σ

CTε

− 1

S′ =Tε ST

σ

− 1

β′ =T

σ

βα′ =Tε α

C′ S′ α′

β′ CSαβ

CSαβ
 293 

TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x1 AXIS

" Because of the invariance conditions, the matrices and have 

the form and 

" Also, the vectors and have the form and 

" The corresponding forms of , , , and are obtained from 

the transformation laws once a value for the angle θ1 is specified 

C′ S′

C1′1′ C1′2′ C1′3′ 0C1′5′ 0

C1′2′ C2′2′ C2′3′ 0

C2′5′ 0

C1′3′ C2′3′ C3′3′ 0

C3′5′ 0

000C

4′4′ 0

C

4′6′

C1′5′ C2′5′ C3′5′ 0C5′5′ 0

000C

4′6′ 0

C6′6′

S1′1′ S1′2′ S1′3′ 0S1′5′ 0

S1′2′ S2′2′ S2′3′ 0

S2′5′ 0

S1′3′ S2′3′ S3′3′ 0

S3′5′ 0

000S

4′4′ 0

S

4′6′

S1′5′ S2′5′ S3′5′ 0S5′5′ 0

000S

4′6′ 0

S6′6′

α′ β′

α1′1′

α2′2′

α3′3′

0

2α1′3′

0

β 1′1′

β 2′2′

β 3′3′

0

β 1′3′

0

CSαβ
 294 

TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x1 AXIS

" For example, specifying yields 

" Enforcing the invariance conditions on gives 

, , , , , , , and 

θ1 =0

C1′1′ C1′2′ C1′3′ 0C1′5′ 0

C1′2′ C2′2′ C2′3′ 0

C2′5′ 0

C1′3′ C2′3′ C3′3′ 0

C3′5′ 0

000C4′4′ 0

C4′6′

C1′5′ C2′5′ C3′5′ 0C5′5′ 0

000C4′6′ 0

C6′6′

=

C11

C12C13C14C15C16

C12C22C23C24C25C26

C13C23C33C34C35C36

C14C24C34C44C45C46

C15C25C35C45C55C56

C16C26C36C46C56C66

β 1′1′

β 2′2′

β 3′3′

0

β 1′3′

0

=

β 11

β 22

β 33

β 23

β 13

β 12

S1′1′ S1′2′ S1′3′ 0S1′5′ 0

S1′2′ S2′2′ S2′3′ 0

S2′5′ 0

S1′3′ S2′3′ S3′3′ 0

S3′5′ 0

000S4′4′ 0

S4′6′

S1′5′ S2′5′ S3′5′ 0S5′5′ 0

000S4′6′ 0

S6′6′

=

S11

S12S13S14S15S16

S12S22S23S24S25S26

S13S23S33S34S35S36

S14S24S34S44S45S46

S15S25S35S45S55S56

S16S26S36S46S56S66

α1′1′

α2′2′

α3′3′

0

2α1′3′

0

=

α11

α22

α33

2α23

2α13

2α12

C′

C14=0C16=0C24=0C26=0C34=0C36=0C45=0C56=0
 295 

TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x1 AXIS

" Enforcing the invariance conditions on gives 

, , , , , , , and 

" Enforcing the invariance conditions on and gives

 and 

" The conditions are identical to the conditions previously obtained for a 

material that is monoclinic with respect to the plane x

2 = 0

" Similarly, specifying and enforcing the invariance conditions on 

, , , and yields the conditions previously obtained for 

a material that is monoclinic with respect to the plane x

3 = 0

S′

S14=0

S16=0

S24=0

S26=0

S34=0

S36=0S45=0

S56=0

α′ β′

α12=α23=0β 12=β 23=0

θ1 =

π

2

C′ S′ α′ β′
 296 

TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x1 AXIS

" Now consider an arbitrary value for the angle θ1 such that 

" The previous example for showed that the invariance conditions 

on , , , and were obtained by the terms of , ,

, and that were zero valued

" Using the transformation equations for a dextral rotation about the x1

axis that were given previously for a (triclinic) fully anisotropic, elastic 

material gives the following results for the invariance conditions on 

−

π

2

<θ1 ≤

π

2

θ1 =0

CSαβC′ S′

α′ β′

C′

C1′4′ =0: m2−n2C14+mnC13−C12=0

C1′6′ =0: mC16+nC15=0
 297 

TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x1 AXIS

 with and 

" These conditions give 8 equations and 20 unknowns

C2′4′ =0: m2m2−3n2C24−m3nC22−C23−2C44

+mn3C33−C23−2C44−n2n2−3m2C34=0

C2′6′ =0: m3C26+m2n2C46+C25+mn22C45+C36+n3C35=0

C3′4′ =0: m2m2−3n2C34+m3nC33−C23−2C44

−mn3C22−C23−2C44−n2n2−3m2C24=0

C3′6′ =0: m3C36−m2n2C46−C35−mn22C45−C26+n3C25=0

C4′5′ =0: m3C45−m2nC46+C25−C35−mn2C45−C26+C36+n3C46=0

C5′6′ =0: m2−n2C56+mnC55−C66=0

m= cosθ1 n= sinθ1
 298 

TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x1 AXIS

" The invariance conditions on 

S′ are given b

y

S1′4′ =0: m2−n2S14+2mnS13−S12=0

S1′6′ =0: mS16+nS15=0

S2′4′ =0: m2m2−3n2S24−m3n2S22−2S23−S44

+mn32S33−2S23−S44−n2n2−3m2S34=0

S2′6′ =0: m3S26+m2nS46+S25+mn2S45+S36+n3S35=0

S3′4′ =0: m2m2−3n2S34+m3n2S33−2S23−S44

−mn32S22−2S23−S44−n2n2−3m2S24=0

S3′6′ =0: m3S36−m2nS46−S35−mn2S45−S26+n3S25=0

S4′5′ =0: m3S45−m2nS46+2S25−2S35−mn2S45−2S26+2S36+n3S46=0

S5′6′ =0: m2−n2S56+mnS55−S66=0
 299 

TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x1 AXIS

" The invariance conditions on 

α′ and β′ yield

α1′2′ =0: mα12+nα13=0

α2′3′ =0: m2−n2α23+mnα33−α22=0

β 1′2′ =0: mβ 12+nβ 13=0

β 2′3′ =0: m2−n2β 23+mnβ 33−β 22=0
 300 

TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x1 AXIS

" In determining the restrictions on the compliances, stiffnesses,

thermal-expansion coefficients, and thermal moduli for materials that 

possess more than one plane of elastic symmetry, all of which contain 

the x1 axis, it is convenient to select the first plane to be given b

y

" Thus, the material is monoclinic with respect to the plane x

2 = 0

and, as a result, the following conditions hold

" These relations, and the fact that for nonzero values 

of , are used to simplify the previously given invariance 

conditions into the following three uncoupled groups

θ1 =0

C14=0C16=0C24=0C26=0C34=0C36=0C45=0C56=0

S14=0

S16=0

S24=0

S26=0

S34=0

S36=0S45=0

S56=0

α12=0α23=0β 12=0β 23=0

n= sinθ1 ≠0

−

π

2

<θ1 ≤

π

2
 301 

TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x1 AXIS

" Group 1

" Group 2

" Group 3

C2′6′ =0: m22C46+C25+n2C35=0

m= cosθ1 n= sinθ1

C3′6′ =0: −m22C46−C35+n2C25=0

C4′5′ =0: −m2C46+C25−C35+n2C46=0

C1′4′ =0: mC13−C12=0

C1′6′ =0: C15=0

C5′6′ =0: mC55−C66=0

C2′4′ =0: −m3C22−C23−2C44+mn2C33−C23−2C44=0

C3′4′ =0: m3C33−C23−2C44−mn2C22−C23−2C44=0
 302 

TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x1 AXIS

" The first group can be written as

" Because the right-hand side of the equations are identical, it follows 

that the left-hand sides are equal; that is,

" Each equation can be expressed as , which is satisfied 

by and by

" implies and 

" implies , , and 

C2′6′ =0: −C35=m22C46+C25−C35

m= cosθ1 n= sinθ1

C3′6′ =0: C25=m22C46+C25−C35

C4′5′ =0: C46=m22C46+C25−C35

C46=−C35=C25=Γ

Γ=4Γcos2θ1

Γ=0 cos2

θ1 =

1

4

Γ=0

C46=−C35=C25=0

−

π

2

<θ1 ≤

π

2

cos2θ1 =

1

4

θ1 =±

π

3

C46=C25

C35=−C25
 303 

TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x1 AXIS

" The third group of equations is simplified by first adding the two 

equations together and then by subtracting the two equations

" Adding gives ,

 which is satisfied for all if , or for all

 if 

" Subtracting gives 

,

 which is satisfied for all if ,

 or for all if 

C2′4′ +C3′4′ =0: C33−C22 cosθ1 =0

−

π

2

<θ1 ≤

π

2

C33=C22

C33≠C22

θ1 =

π

2

C2′4′ −C3′4′ =0: 2C23+4C44−C22−C33 cosθ1cos2θ1 =0

−

π

2

<θ1 ≤

π

2

2C23+4C44−C22−C33=0

2C23+4C44−C22−C33≠0

θ1 =±

π

4 or

π

2
 304 

TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x1 AXIS

" Group 4

" Group 5

" Group 6

S2′6′ =0: m2S46+S25+n2S35=0

m= cosθ1 n= sinθ1

S3′6′ =0: −m2S46−S35+n2S25=0

S4′5′ =0: −m2S46+2S25−2S35+n2S46=0

S1′4′ =0: mS13−S12=0

S1′6′ =0: S15=0

S5′6′ =0: mS55−S66=0

S2′4′ =0: −m32S22−2S23−S44+mn22S33−2S23−S44=0

S3′4′ =0: m32S33−2S23−S44−mn22S22−2S23−S44=0
 305 

TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x1 AXIS

" Like for the stiffnesses, the fourth group can also be written as

" Because the right-hand side of the equations are identical, it follows 

that the left-hand sides are equal; that is,

" Each equation can be expressed as , which is satisfied 

by and by

" implies and 

" implies , , and 

S2′6′ =0: −S35=m2S46+S25−S35

m= cosθ1 n= sinθ1

S3′6′ =0: S25=m2S46+S25−S35

S4′5′ =0: 1

2

S46=m2S46+S25−S35

1

2

S46=−S35=S25=Δ

Δ=4Δcos2θ3

Δ=0 cos2

θ1 =

1

4

Δ=0S46=S35=S25=0

−

π

2

<θ1 ≤

π

2

cos2θ1 =

1

4

θ1 =±

π

3

S46=2S25

S35=−S25
 306 

TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x1 AXIS

" The sixth group of equations is also simplified by first adding the two 

equations together and then by subtracting the two equations

" Adding gives ,

 which is satisfied for all if , or for all

 if 

" Subtracting gives 

,

 which is satisfied for all if ,

 or for all if 

S2′4′ +S3′4′ =0: S33−S22 cosθ1 =0

−

π

2

<θ1 ≤

π

2

S33=S22

S33≠S22

θ1 =

π

2

S2′4′ −S3′4′ =0: 2S23+S44−S22−S33 cosθ1cos2θ1 =0

−

π

2

<θ1 ≤

π

2

2S23+S44−S22−S33=0

2S23+S44−S22−S33≠0

θ1 =±

π

4 or

π

2
 307 

TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x1 AXIS

" The symmetry properties for trigonal materials arise from the 

solution for the invariance conditions that is given by

" For this solution, the planes of 

elastic symmetry are all parallel to 

the x1 axis and are given b

y

" For ,

and 

" The stiffness equations in group 2 

yield the invariance conditions 

, , and 

cos2θ1 =

1

4

θ1 = 0

Symmetry

plane for

Symmetry

plane for

θ1 = 0

Symmetry

plane for

Plan view of symmetry planes

x3

x2

θ1 =+

π

3

θ1 =+

π

3

θ1 =−

π

3

θ1 =−

π

3

θ1 =0 and±

π

3

θ1 =±

π

3

m= cos±

π

3

=

1

2

≠0

n= sin±

π

3

=±

3

2

≠0

C15=0C13=C12C66=C55
 308 

TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x1 AXIS

" Likewise, the stiffness equations in group 1 yield the invariance 

conditions and 

" Furthermore, the stiffness equations in group 3 yield the invariance 

conditions and 

" The compliance equations in group 5 yield the invariance conditions 

, , and 

" The compliance equations in group 4 yield the invariance conditions 

 and 

" The compliance equations in group 6 yield the invariance conditions 

 and 

C46=C25

C35=−C25

C33=C22

C44=

1

2

C22−C23

S15=0S13=S12S66=S55

S46=2S25

S35=−S25

S33=S22

S44=2S22−S23
 309 

TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x1 AXIS

" The invariance conditions and 

 yield 

 and 

" The invariance conditions and 

 yield 

 and 

"Together, the invariance conditions yield the following constitutive 

equations for a trigonal material

α1′2′ =0: mα12+nα13=0

α2′3′ =0: m2−n2α23+mnα33−α22=0

α23=α13=α12=0α33=α22

β 1′2′ =0: mβ 12+nβ 13=0

β 2′3′ =0: m2−n2β 23+mnβ 33−β 22=0

β 23=β 13=β 12=0β 33=β 22
 310 

TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x1 AXIS

" Again, the trigonal material has six independent elastic constants 

and two independent thermal-expansion or thermal-compliance 

parameters

σ11

σ22

σ33

σ23

σ13

σ12

=

C11C12C12000

C12

C22C230C250

C12

C23C220−C250

000

C22

−C23

2

0C25

0C25−C250C550

000C25

0C55

ε11

ε22

ε33

2ε23

2ε13

2ε12

+

β 11

β 22

β 22

0

0

0

T−Tref

ε11

ε22

ε33

2ε23

2ε13

2ε12

=

S11S12S12000

S12

S22S230S250

S12

S23S220−S250

0002S22

−S2302S25

0S25−S250S550

0002S25

0S55

σ11

σ22

σ33

σ23

σ13

σ12

+

α11

α22

α22

0

0

0

T−Tref
 311 

TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x1 AXIS

" The previous derivation of the constitutive equations for a trigonal 

material that has reflective symmetry planes that contain the x1 axis is 

quite tedious

" These constitutive equations can be derived in alternate manner by

using the corresponding equations given first for a trigonal material 

that has reflective symmetry planes that contain the x

3 axis, along with 

a juxtaposition of indices

" That is, the desired constitutive equations are found by simply 

determining the renumbering of the indices that brings the figure 

shown below for symmetry planes that contain the x

3 axis into 

congruence with the adjacent figure shown below for symmetry planes 

that contain the x1 axis
 312 

TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x1 AXIS

x1

x2

x3

Plane

θ3 = 0

Plane

θ3 = -

π/3

Plane

θ3 = +

π/3

x1

x2

x3

Plane

θ1 = 0

Plane

θ1 = -

π/3

Plane

θ1 = +

π/3

Symmetry planes that 

contain the x

3 axis

Symmetry planes that 

contain the x1 axis
 313 

TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x1 AXIS

" Inspection of the figures indicates the following transformation of the 

indices:

" Next, it must be realized that the exchanging of indices must be used 

with the indices of tensors to determine the indices used with the 

abridged notation (matrix)

" The following index pairs relate the tensor indices to the matrix 

indices

" Using this information along with gives the 

relations:

1→2, 2→3, and3→1

tensor notation 11 22 33 23, 3231, 13 12, 21

matrix notation123456

1→2, 2→3, and3→1

4→5, 5→6, and6→4
 314 

TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x1 AXIS

" Likewise, the transformation of index pairs that appear in the abridged 

notation are given by

" Consider the following constitutive equations for a trigonal material that 

has reflective symmetry planes that contain the x

3 axis

11→ 22

12→ 23 22→ 33

13→ 12 23→ 13 33→ 11

14→ 25 24→ 35 34→ 15 44→ 55

15→ 26 25→ 36 35→ 16 45→ 56 55→ 66

16→ 24 26→ 34 36→ 14 46→ 45 56→ 46 66→ 44

σ11

σ22

σ33

σ23

σ13

σ12

=

C11C12C13C1400

C12

C11C13−C1400

C13

C13C33000

C14

−C140C4400

0000C44

C14

0000C14

1

2

C11−C12

ε11

ε22

ε33

2ε23

2ε13

2ε12

+

β 11

β 11

β 33

0

0

0

T−Tref
 315 

TRIGONAL MATERIALS - CONCLUDED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x1 AXIS

" Applying the index transformation to these constitutive equations gives

" Reordering these equations into the standard form yields

 which are identical to the corresponding equations previously given

σ22

σ33

σ11

σ13

σ12

σ23

=

C22C23C12C2500

C23

C22C12−C2500

C12

C12C11000

C25

−C250C5500

0000C55

C25

0000C25

1

2

C22−C23

ε22

ε33

ε11

2ε13

2ε12

2ε23

+

β 22

β 22

β 11

0

0

0

T−Tref

σ11

σ22

σ33

σ23

σ13

σ12

=

C11C12C12000

C12

C22C230C250

C12

C23C220−C250

000

1

2

C22−C230C25

0C25−C250C550

000C25

0C55

ε11

ε22

ε33

2ε23

2ε13

2ε12

+

β 11

β 22

β 22

0

0

0

T−Tref
 316 

TRIGONAL MATERIALS

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x

2 AXIS

" These constitutive equations are derived by using the corresponding 

equations given first for a trigonal material that has reflective symmetry 

planes that contain the x

3 axis, along with a juxtaposition of indices

" That is, the desired constitutive equations are found by simply 

determining the renumbering of the indices that brings the figure 

shown below for symmetry planes that contain the x

3 axis into 

congruence with the adjacent figure shown below for symmetry planes 

that contain the x

2 axis
 317 

TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x

2 AXIS

x1

x2

x3

Plane

θ3 = 0

Plane

θ3 = -

π/3

Plane

θ3 = +

π/3

x1

x2

x3

Plane

θ2 = 0

Plane

θ2 = -

π/3

Plane

θ2 = +

π/3

Symmetry planes that 

contain the x

3 axis

Symmetry planes that 

contain the x

2 axis
 318 

TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x

2 AXIS

" Inspection of the figures indicates the following transformation of the 

indices:

" Next, it must be realized that the exchanging of indices must be used 

with the indices of tensors to determine the indices used with the 

abridged notation (matrix)

" The following index pairs relate the tensor indices to the matrix 

indices

" Using this information along with gives the 

relations:

1→3, 2→1, and3→2

tensor notation 11 22 33 23, 3231, 13 12, 21

matrix notation123456

1→3, 2→1, and3→2

4→6, 5→4, and6→5
 319 

TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x

2 AXIS

" Likewise, the transformation of index pairs that appear in the abridged 

notation are given by

" Consider the following constitutive equations for a trigonal material that 

has reflective symmetry planes that contain the x

3 axis

11→ 33

12→ 13 22→ 11

13→ 23 23→ 12 33→ 22

14→ 36 24→ 16 34→ 26 44→ 66

15→ 34 25→ 14 35→ 24 45→ 46 55→ 44

16→ 35 26→ 15 36→ 25 46→ 56 56→ 45 66→ 55

σ11

σ22

σ33

σ23

σ13

σ12

=

C11C12C13C1400

C12

C11C13−C1400

C13

C13C33000

C14

−C140C4400

0000C44

C14

0000C14

1

2

C11−C12

ε11

ε22

ε33

2ε23

2ε13

2ε12

+

β 11

β 11

β 33

0

0

0

T−Tref
 320 

TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x

2 AXIS

" Applying the index transformation to these constitutive equations gives

" Reordering these equations into the standard form yields

σ33

σ11

σ22

σ12

σ23

σ13

=

C33C13C23C3600

C13

C33C23−C3600

C23

C23C22000

C36

−C360C6600

0000C66

C36

0000C36

1

2

C33−C13

ε33

ε11

ε22

2ε12

2ε23

2ε13

+

β 33

β 33

β 22

0

0

0

T−Tref

σ11

σ22

σ33

σ23

σ13

σ12

=

C33C23C1300−C36

C23C22C23000

C13

C23C3300C36

000C66C360

000C36

1

2

C33−C130

−C36

0C3600C66

ε11

ε22

ε33

2ε23

2ε13

2ε12

+

β 33

β 22

β 33

0

0

0

T−Tref
 321 

TRIGONAL MATERIALS - CONCLUDED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x

2 AXIS

" Applying the same procedure to the inverse equations yields

ε11

ε22

ε33

2ε23

2ε13

2ε12

=

S33S23S1300−S36

S23S22S23000

S13

S23S3300S36

000S662S360

0002S36

2S33−S130

−S36

0S3600S66

σ11

σ22

σ33

σ23

σ13

σ12

+

α33

α33

α22

0

0

0

T−Tref
 322 

SUMMARY OF TRIGONAL MATERIALS

 x1 axis

 x

2 axis

 x

3 axis

σ11

σ22

σ33

σ23

σ13

σ12

=

C11C12C12000

C12

C22C230C250

C12

C23C220−C250

000 1

2

C22−C230C25

0C25−C250C550

000C25

0C55

ε11

ε22

ε33

2ε23

2ε13

2ε12

+

β 11

β 22

β 22

0

0

0

T−Tref

σ11

σ22

σ33

σ23

σ13

σ12

=

C33C23C1300−C36

C23C22C23000

C13

C23C3300C36

000C66C360

000C36

1

2

C33−C130

−C36

0C3600C66

ε11

ε22

ε33

2ε23

2ε13

2ε12

+

β 33

β 22

β 33

0

0

0

T−Tref

σ11

σ22

σ33

σ23

σ13

σ12

=

C11C12C13C1400

C12

C11C13−C1400

C13

C13C33000

C14

−C140C4400

0000C44

C14

0000C14

1

2

C11−C12

ε11

ε22

ε33

2ε23

2ε13

2ε12

+

β 11

β 11

β 33

0

0

0

T−Tref
 323 

SUMMARY OF TRIGONAL MATERIALS - CONCLUDED

 x1 axis

 x

2 axis

 x

3 axis

ε11

ε22

ε33

2ε23

2ε13

2ε12

=

S11S12S12000

S12

S22S230S250

S12

S23S220−S250

0002S22

−S2302S25

0S25−S250S550

0002S25

0S55

σ11

σ22

σ33

σ23

σ13

σ12

+

α11

α22

α22

0

0

0

T−Tref

ε11

ε22

ε33

2ε23

2ε13

2ε12

=

S33S23S1300−S36

S23S22S23000

S13

S23S3300S36

000S662S360

0002S36

2S33−S130

−S36

0S3600S66

σ11

σ22

σ33

σ23

σ13

σ12

+

α33

α33

α22

0

0

0

T−Tref

ε11

ε22

ε33

2ε23

2ε13

2ε12

=

S11S12S13S1400

S12

S11S13−S1400

S13

S13S33000

S14

−S140S4400

0000S44

2S14

00002S142S11−S12

σ11

σ22

σ33

σ23

σ13

σ12

+

α11

α11

α33

0

0

0

T−Tref
 324 

TETRAGONAL

MATERIALS
 325 

TETRAGONAL MATERIALS

" Previously, it was shown that for a single plane of elastic symmetry 

containing the x

3 axis, special in

variance conditions arise for values of 

, in addition to 

" Likewise, it was shown that for a single plane of elastic symmetry 

containing the x1 axis, special in

variance conditions arise for values of 

, in addition to 

" Furthermore, for a single plane of elastic symmetry containing the x

2

axis, special invariance conditions also arise for values of , in 

addition to 

" Three particular cases of interest arise from these invariance conditions 

in which there exists five planes of elastic symmetry, four of which 

are perpendicular to the fifth plane

θ

3 = ±

π

4

θ

3 = 0

θ1 = ±

π

4

θ1 = 0

θ

2 = ±

π

4

θ

2 = 0
 326 

TETRAGONAL MATERIALS - CONTINUED

" Specifically, consider a material with three planes of elastic symmetry 

given by

" By taking as an additional 

plane of elastic symmetry, the fact that 

the planes and are 

perpendicular implies the existence of 

a fifth symmetry plane given by

" This fact was shown previously for 

specially orthotropic materials;

that is, two perpendicular planes 

of elastic symmetry imply a third 

perpendicular plane of elastic symmetry

" Thus, there exists five planes of elastic symmetry, four of which 

are perpendicular to plane 

θ

3 = 0 and ±

π

4

x1

x2

x3

Plane

θ3 = 0

Plane

θ1 = +

π/2

Plane

θ3 = -

π/4

Plane

θ3 = +

π/4

Plane θ1 = 0

θ1 = +

π

2

θ

3 = 0

θ1 = +

π

2

θ1 = 0

θ1 = +

π

2
 327 

TETRAGONAL MATERIALS - CONTINUED

" Next, consider a material with three planes of elastic symmetry given by

" By taking as an additional 

plane of elastic symmetry, the fact 

that the planes and are 

perpendicular implies the existence 

of a fifth symmetry plane given by

" Thus, there exists five planes of

elastic symmetry, four of which 

are perpendicular to plane 

θ1 = 0 and ±

π

4

x1

x2

x3

Plane

θ3 = 0

Plane

θ1 = -

π/4

Plane

θ1 = +

π/4

Plane θ1 = 0

Plane

θ1 = +

π/2

θ

3 = 0

θ1 = 0θ

3 = 0

θ1 = +

π

2

θ

3 = 0
 328 

TETRAGONAL MATERIALS - CONTINUED

" Finally, consider a material with three planes of elastic symmetry given 

by

" By taking as an additional 

plane of elastic symmetry, the fact 

that the planes and 

are perpendicular implies the 

existence of a fifth symmetry plane 

given by

" Thus, there exists five planes of

elastic symmetry, four of which are perpendicular to plane 

θ

2 = 0 and ±

π

4

x1

x2

x3

Plane

θ2 = -

π/4

Plane

θ2 = +π/4

Plane θ3 Plane 

θ3 = 0 = +

π/2

Plane 

θ2 = 0

θ

3 = +

π

2

θ

3 = +

π

2

θ

2 = 0

θ

3 = 0

θ

3 = +

π

2
 329 

TETRAGONAL MATERIALS - CONCLUDED

" When there exists five planes of elastic symmetry, in which four 

planes are perpendicular to a fifth plane, the material is classified as a 

tetragonal material

" Three specific cases are defined as follows:

" Planes of elastic symmetry given by and 

" Planes of elastic symmetry given by and 

" Planes of elastic symmetry given by and 

" For each of these three arrangements of symmetry planes, the 

corresponding constitutive equations can be derived directly from 

those for a specially orthotropic material by enforcing the 

invariance conditionsfor , , or 

θ

3 = 0 and ±

π

4

θ1 = +

π

2

θ1 = 0 and ±

π

4

θ

3 = 0

θ

2 = 0 and ±

π

4

θ

3 = +

π

2

θ1 = ±

π

4

θ

2 = ±

π

4

θ

3 = ±

π

4
 330 

TETRAGONAL MATERIALS

REFLECTIVE SYMMETRY PLANES THAT CONTAIN THE x

3 AXIS

" Consider a tetragonal material with planes of reflective symmetry 

defined by and by

" For a specially orthotropic material, it was shown previously that the 

constitutive equations are given by

 and

θ

3 = 0 and ±

π

4

θ1 = +

π

2

σ11

σ22

σ33

σ23

σ13

σ12

=

C11C12C13000

C12

C22C23000

C13

C23C33000

000C44

00

0000C55

0

00000C66

ε11

ε22

ε33

2ε23

2ε13

2ε12

+

β 11

β 22

β 33

0

0

0

T−Tref

ε11

ε22

ε33

2ε23

2ε13

2ε12

=

S11S12S13000

S12

S22S23000

S13

S23S33000

000S44

00

0000S55

0

00000S66

σ11

σ22

σ33

σ23

σ13

σ12

+

α11

α22

α33

0

0

0

T−Tref
 331 

TETRAGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY PLANES THAT CONTAIN THE x

3 AXIS

" For this case, the additional conditions on the constitutive equations 

are obtained by enforcing symmetry planes given by

" For , and 

" With , the first group of stiffness equations previously given 

herein for an arbitrary plane of symmetry, defined by constant values of 

θ

3, requires 

" Recall that implies for arbitrary values of 

the angle 

" Similarly, the second group of stiffness equations for an arbitrary plane 

of symmetry, defined by constant values of θ

3, requires ,

, and 

θ

3 = ±

π

4

θ

3 = ±

π

4

m= cos±

π

4

=

2

2

≠0

n= sin±

π

4

=±

2

2

≠0

θ

3 = ±

π

4

Γ=0

Γ=0

C56=−C24=C14=0

−

π

2

<θ3 ≤

π

2

C34 = 0

C23 = C13C55 = C44
 332 

TETRAGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY PLANES THAT CONTAIN THE x

3 AXIS

" Furthermore, the stiffness equations in group 3 yield the invariance 

condition 

" Likewise, the compliance equations in group 5 yield the invariance 

conditions , , and 

" The compliance equations in group 4 yield the invariance conditions 

" The compliance equations in group 6 yield the invariance condition 

C22=C11

S34=0S23=S13S55=S44

S56 = S24 = S14 = 0

S22=S11
 333 

TETRAGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY PLANES THAT CONTAIN THE x

3 AXIS

" The invariance conditions and 

 yield

 and 

" The invariance conditions and 

 yield 

 and 

"Together, the invariance conditions yield the following constitutive 

equations for a tetragonal material

α1′3′ =0: mα13+nα23=0

α1′2′ =0: m2−n2α12+mnα22−α11=0

α23=α13=α12=0α22=α11

β 1′3′ =0: mβ 13+nβ 23=0

β 1′2′ =0: m2−n2β 12+mnβ 22−β 11=0

β 23=β 13=β 12=0β 22=β 11
 334 

TETRAGONAL MATERIALS - CONCLUDED

REFLECTIVE SYMMETRY PLANES THAT CONTAIN THE x

3 AXIS

" Therefore, a tetragonal material has six independent elastic 

constants and two independent thermal-expansion or thermalcompliance parameter

s

σ11

σ22

σ33

σ23

σ13

σ12

=

C11C12C13000

C12

C11C13000

C13C13C33000

000C44

00

0000C44

0

00000C66

ε11

ε22

ε33

2ε23

2ε13

2ε12

+

β 11

β 11

β 33

0

0

0

T−Tref

ε11

ε22

ε33

2ε23

2ε13

2ε12

=

S11S12S13000

S12S11S13000

S13

S13S33000

000S44

00

0000S44

0

00000S66

σ11

σ22

σ33

σ23

σ13

σ12

+

α11

α11

α33

0

0

0

T−Tref
 335 

TETRAGONAL MATERIALS

REFLECTIVE SYMMETRY PLANES THAT CONTAIN THE x1 AXIS

" Next, consider a tetragonal material with planes of reflective 

symmetry defined by and by

" For , and 

" The first group of stiffness equations previously given herein for an 

arbitrary plane of symmetry, defined by constant values of θ1, yields the 

invariance conditions 

" The second group of stiffness equations for an arbitrary plane of 

symmetry, defined by constant values of θ1, requires , ,

and 

" The stiffness equations in group 3 yield the invariance condition 

θ1 = 0 and ±

π

4

θ

3 = 0

θ1 = ±

π

4

m= cos±

π

4

=

2

2

≠0

n= sin±

π

4

=±

2

2

≠0

C46 = C35 = C25 = 0

C15 = 0C13 = C12

C66 = C55

C33 = C22
 336 

TETRAGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY PLANES THAT CONTAIN THE x1 AXIS

" The invariance conditions and 

 yield 

 and 

" The resulting constitutive equation, obtained by simplifying the 

constitutive equation for a specialy orthotropic material, is given by

β 1′2′ =0: mβ 12+nβ 13=0

β 2′3′ =0: m2−n2β 23+mnβ 33−β 22=0

β 23=β 13=β 12=0β 33=β 22

σ11

σ22

σ33

σ23

σ13

σ12

=

C11C12C12000

C12

C22C23000

C12C23C22000

000C44

00

0000C55

0

00000C55

ε11

ε22

ε33

2ε23

2ε13

2ε12

+

β 11

β 22

β 22

0

0

0

T−Tref
 337 

TETRAGONAL MATERIALS - CONCLUDED

REFLECTIVE SYMMETRY PLANES THAT CONTAIN THE x1 AXIS

" Applying the same process to the equations for the compliances and 

coefficients of thermal expansion yields

ε11

ε22

ε33

2ε23

2ε13

2ε12

=

S11S12S12000

S12S22S23000

S12S23S22000

000S44

00

0000S55

0

00000S55

σ11

σ22

σ33

σ23

σ13

σ12

+

α11

α22

α22

0

0

0

T−Tref
 338 

TETRAGONAL MATERIALS

REFLECTIVE SYMMETRY PLANES THAT CONTAIN THE x

2 AXIS

" Finally, consider a tetragonal material with planes of reflective 

symmetry defined by and by

" The constitutive equations are derived by using the corresponding 

equations given previously for a tetragonal material that has four 

reflective-symmetry planes that contain the x

3 axis, along with a 

juxtaposition of indices

" In particular, the desired constitutive equations are found by simply 

determining the renumbering of the indices that brings the figure 

shown below with four symmetry planes that contain the x

3 axis into 

congruence with the adjacent figure shown below with four symmetry 

planes that contain the x

2 axis

θ

2 = 0 and ±

π

4

θ

3 = +

π

2
 339 

TETRAGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY PLANES THAT CONTAIN THE x

2 AXIS

x1

x2

x3

Plane

θ3 = 0

Plane

θ3 = -

π/3

Plane

θ3 = +

π/3

x1

x2

x3

Plane

θ2 = 0

Plane

θ2 = -

π/3

Plane

θ2 = +

π/3

Four symmetry planes 

that contain the x

3 axis

Four symmetry planes

that contain the x

2 axis

Plane

θ1 = +

π/2

Plane

θ3 = +

π/2

Plane

θ3 = +

π/2

Plane

θ2 = +

π/2
 340 

TETRAGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY PLANES THAT CONTAIN THE x

2 AXIS

" Inspection of the figures indicates the following transformation of the 

indices:

" Next, it must be realized that the exchanging of indices must be used 

with the indices of tensors to determine the indices used with the 

abridged notation (matrix)

" The following index pairs relate the tensor indices to the matrix 

indices

" Using this information along with gives the 

relations:

1→3, 2→1, and3→2

tensor notation 11 22 33 23, 3231, 13 12, 21

matrix notation123456

1→3, 2→1, and3→2

4→6, 5→4, and6→5
 341 

TETRAGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY PLANES THAT CONTAIN THE x

2 AXIS

" Likewise, the transformation of index pairs that appear in the abridged 

notation are given by

" Consider the following constitutive equations for a tetragonal material 

that has reflective symmetry planes that contain the x

3 axis

11→ 33

12→ 13 22→ 11

13→ 23 23→ 12 33→ 22

14→ 36 24→ 16 34→ 26 44→ 66

15→ 34 25→ 14 35→ 24 45→ 46 55→ 44

16→ 35 26→ 15 36→ 25 46→ 56 56→ 45 66→ 55

σ11

σ22

σ33

σ23

σ13

σ12

=

C11C12C13000

C12

C11C13000

C13

C13C33000

000C44

00

0000C44

0

00000C66

ε11

ε22

ε33

2ε23

2ε13

2ε12

+

β 11

β 11

β 33

0

0

0

T−Tref
 342 

TETRAGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY PLANES THAT CONTAIN THE x

2 AXIS

" Applying the index transformation to these constitutive equations gives

" Reordering these equations into the standard form yields

σ33

σ11

σ22

σ12

σ23

σ13

=

C33C13C23000

C13

C33C23000

C23

C23C22000

000C66

00

0000C66

0

00000C55

ε33

ε11

ε22

2ε12

2ε23

2ε13

+

β 33

β 33

β 22

0

0

0

T−Tref

σ11

σ22

σ33

σ23

σ13

σ12

=

C33C23C13000

C23

C22C23000

C13C23C33000

000C66

00

0000C55

0

00000C66

ε11

ε22

ε33

2ε23

2ε13

2ε12

+

β 33

β 22

β 33

0

0

0

T−Tref
 343 

TETRAGONAL MATERIALS - CONCLUDED

REFLECTIVE SYMMETRY PLANES THAT CONTAIN THE x

2 AXIS

" Applying the same procedure to the inverse equations yields

ε11

ε22

ε33

2ε23

2ε13

2ε12

=

S33S23S13000

S23S22S23000

S13

S23S33000

000S6600

0000S55

0

00000S66

σ11

σ22

σ33

σ23

σ13

σ12

+

α33

α22

α33

0

0

0

T−Tref
 344 

SUMMARY OF TETRAGONAL MATERIALS

 x1 axis

 x

2 axis

 x

3 axis

σ11

σ22

σ33

σ23

σ13

σ12

=

C11C12C12000

C12

C22C23000

C12

C23C22000

000C44

00

0000C55

0

00000C55

ε11

ε22

ε33

2ε23

2ε13

2ε12

+

β 11

β 22

β 22

0

0

0

T−Tref

σ11

σ22

σ33

σ23

σ13

σ12

=

C33C23C13000

C23

C22C23000

C13

C23C33000

000C66

00

0000C55

0

00000C66

ε11

ε22

ε33

2ε23

2ε13

2ε12

+

β 33

β 22

β 33

0

0

0

T−Tref

σ11

σ22

σ33

σ23

σ13

σ12

=

C11C12C13000

C12

C11C13000

C13

C13C33000

000C44

00

0000C44

0

00000C66

ε11

ε22

ε33

2ε23

2ε13

2ε12

+

β 11

β 11

β 33

0

0

0

T−Tref
 345 

SUMMARY OF TETRAGONAL MATERIALS - CONTINUED

 x1 axis

 x

2 axis

 x

3 axis

ε11

ε22

ε33

2ε23

2ε13

2ε12

=

S11S12S12000

S12

S22S23000

S12

S23S22000

000S44

00

0000S55

0

00000S55

σ11

σ22

σ33

σ23

σ13

σ12

+

α11

α22

α22

0

0

0

T−Tref

ε11

ε22

ε33

2ε23

2ε13

2ε12

=

S33S23S13000

S23

S22S23000

S13

S23S33000

000S66

00

0000S55

0

00000S66

σ11

σ22

σ33

σ23

σ13

σ12

+

α33

α22

α33

0

0

0

T−Tref

ε11

ε22

ε33

2ε23

2ε13

2ε12

=

S11S12S13000

S12

S11S13000

S13

S13S33000

000S44

00

0000S44

0

00000S66

σ11

σ22

σ33

σ23

σ13

σ12

+

α11

α11

α33

0

0

0

T−Tref
 346 

SUMMARY OF TETRAGONAL MATERIALS

CONCLUDED

" Inspection of the equations shows that a tetragonal material is a 

specially orthotropic material in which the properties associated with 

two of the coordinate directions are identical
 347 

TRANSVERSELY ISOTROPIC

MATERIALS
 348 

TRANSVERSELY ISOTROPIC MATERIALS

ISOTROPY PLANE x

3 = 0

" Now, consider the case in which a specially orthotropic material, which 

exhibits symmetry about the perpendicular principal-coordinate planes,

also exhibits isotropy in the plane x

3 = 0

" For the specially orthotropic material, it was shown previously that the 

material properties are given by

 and

σ11

σ22

σ33

σ23

σ13

σ12

=

C11C12C13000

C12

C22C23000

C13

C23C33000

000C44

00

0000C55

0

00000C66

ε11

ε22

ε33

2ε23

2ε13

2ε12

+

β 11

β 22

β 33

0

0

0

T−Tref

ε11

ε22

ε33

2ε23

2ε13

2ε12

=

S11S12S13000

S12

S22S23000

S13

S23S33000

000S44

00

0000S55

0

00000S66

σ11

σ22

σ33

σ23

σ13

σ12

+

α11

α22

α33

0

0

0

T−Tref
 349 

TRANSVERSELY ISOTROPIC MATERIALS

ISOTROPY PLANE x

3 = 0 (CONTINUED)

" For this type of symmetry to exist, there must be an infinite number of 

elastic symmetry planes that are perpendicular to the plane x

3 = 0

" Equivalently, the constitutive matrices and vectors must invariant 

with respect to dextral rotations about the x

3 axis 

" The coordinate transformation for this 

symmetry is shown in the figure and 

is given by ,

, and ,

with 

" The corresponding matrix of 

direction cosines is given by

i 1 i 2

i 3

x1

x2

x3

x1′

x2′

x3′

i 1′

i 2′

i 3′

,

,

θ3

θ3

Plane of isotropy

ai′j ≡ ii′ • ij

x1′ =x1cosθ3 +x2sinθ3

x2′ =−x1sinθ3 +x2cosθ3 x3′ =

x3

0≤θ3 <2π

a1′1 a1′2 a1′3

a2′1 a2′2 a2′3

a3′1 a3′2 a3′3

=

cosθ3 sin

θ3 0

− sinθ3 cos

θ3 0

001
 350 

TRANSVERSELY ISOTROPIC MATERIALS

ISOTROPY PLANE x

3 = 0 (CONTINUED)

" The corresponding stress and strain transformation matrices were 

shown previously to be given by

T

σ

θ3 = and

cos2θ3 sin2θ3 0

00 2sin

θ3cosθ3

sin2θ3 cos2θ3 0

00− 2sin

θ3cosθ3

00 1 000

000 cos

θ3 − sinθ3 0

000 sin

θ3 cosθ3 0

− sin

θ3cosθ3 sinθ3cosθ3 0

00 cos2

θ3 − sin2θ3

T

ε

θ3 =

cos2θ3 sin2θ3 0

00 sin

θ3cosθ3

sin2θ3 cos2θ3 0

00− sin

θ3cosθ3

00 1 000

000 cos

θ3 − sinθ3 0

000 sin

θ3 cosθ3 0

− 2sin

θ3cosθ3 2sinθ3cosθ3 0

00 cos2

θ3 − sin2θ3
 351 

TRANSVERSELY ISOTROPIC MATERIALS

ISOTROPY PLANE x

3 = 0 (CONTINUED)

" Specific expressions that result from the transformation 

 were given previously for an anisotropic material 

" For a specially orthotropic material, yields the 

expressions given previously for a generally orthotropic material;

that is,

C′ =T

σ

CTε

− 1

C′ =T

σ

CTε

− 1

C1′1′ =m4C11+2m2n2C12+2C66+n4C22

C1′2′ =m2n2C11+C22−4C66+m4+n4C12

C1′3′ =m2C13+n2C23

C1′4′ =0C1′5′ =0

C1′6′ =mnm2−n2C12+2C66+mnn2C22−m2C11

C2′2′ =m4C22+2m2n2C12+2C66+n4C11
 352 

TRANSVERSELY ISOTROPIC MATERIALS

ISOTROPY PLANE x

3 = 0 (CONTINUED)

 with and 

" Next, the invariance condition is enforced, which 

implies the conditions , where 

C2′3′ =m2C23+n2C13

C2′4′ =0C2′5′ =0

C2′6′ =mnn2−m2C12+2C66+mnm2C22−n2C11

C3′3′ =C33C3′4′ =0C3′5′ =0C3′6′ =mnC23−C13

C4′4′ =m2C44+n2C55

C4′5′ =mnC44−C55C4′6′ =0C5′6′ =0

C5′5′ =m2C55+n2C44

C6′6′ =m2n2C11+C22−2C12+m2−n2 2C66

m= cosθ3 n= sinθ3

C=T

σ

CTε

− 1

Cr′s′ =Crs

r, s∈1,2, 3, 4, 5, 6
 353 

TRANSVERSELY ISOTROPIC MATERIALS

ISOTROPY PLANE x

3 = 0 (CONTINUED)

" Enforcing gives , which 

can only be satisfied in a general sense (arbitrary values for θ

3) if 

 and 

" Similarly, enforcing gives , which yields 

" Likewise, enforcing gives , which yields 

" Finally, by substituting , , , and 

 into the remaining expressions for and enforcing 

, it is found the remaining expressions are identically satisfied

C1′1′ =C11

C11=m

4

C11+2m2n2 C12+2C66+n

4

C22

C11=C22C12+2C66=C11

C1′3′ =C13C13=m2C13+n2C23

C23=C13

C

4′4′ =

C44C44=m2C44+n2C55

C55=C44

C11=C22C12+2C66=C11C23=C13

C55=C44

Cr′s′

Cr′s′ =Crs
 354 

TRANSVERSELY ISOTROPIC MATERIALS

ISOTROPY PLANE x

3 = 0 (CONTINUED)

" Thus, the stiffness matrix for a transversely isotropic material that is 

isotropic in the plane x

3 = 0 has the form

which has 5 independent stiffnesses

" Following the same procedure for the compliance coefficients by using 

the transformed-compliance expressions previously given for a 

generally orthotropic material and enforcing the invariance 

condition or yields similar results

C11C12C13000

C12

C11C13000

C13C13C33000

000C4400

0000C440

00000

1

2

C11−C12

S=Tε ST

σ

− 1

Sr′s′ =Srs
 355 

TRANSVERSELY ISOTROPIC MATERIALS

ISOTROPY PLANE x

3 = 0 (CONTINUED)

" Thus, the compliance matrix for a transversely isotropic material that is 

isotropic in the plane x

3 = 0 has the form

which has 5 independent compliances

S11S12S13000

S12

S11S13000

S13S13S33000

000S4400

0000S440

00000

1

2

S11−S12
 356 

TRANSVERSELY ISOTROPIC MATERIALS

ISOTROPY PLANE x

3 = 0 (CONTINUED)

" For a specially orthotropic material, yields the 

expressions for the transformed thermal-expansion coefficients, given 

previously for a generally orthotropic material; that is,

" Enforcing the invariance condition or yields 

the requirements that and 

" Following the same procedure for the thermal moduli by using the 

transformed-thermal-moduli expressions previously given for a 

generally orthotropic material and enforcing the invariance 

condition or yields similar results; that is,

 and 

α′ =Tε α

α1′1′ =m2α11+n2α22α2′2′ =m2α22+n2α11

α3′3′ =α33

α2′3′ =0α1′3′ =0α1′2′ =mnα22−α11

α=Tε α

αr′s′ =αrs

α22=α11α12=0

β=T

σ

β

βr′s′ =βrs

β 22=β 11β 12=0
 357 

TRANSVERSELY ISOTROPIC MATERIALS

ISOTROPY PLANE x

3 = 0 (CONTINUED)

" Applying all these simplifications, the linear thermoelastic constitutive 

equations become

and

σ11

σ22

σ33

σ23

σ13

σ12

=

C11C12C13000

C12

C11C13000

C13

C13C33000

000C44

00

0000C44

0

00000 1

2

C11−C12

ε11

ε22

ε33

2ε23

2ε13

2ε12

+

β 11

β 11

β 33

0

0

0

T−Tref

ε11

ε22

ε33

2ε23

2ε13

2ε12

=

S11S12S13000

S12

S11S13000

S13

S13S33000

000S44

00

0000S44

0

00000 1

2

S11−S12

σ11

σ22

σ33

σ23

σ13

σ12

+

α11

α11

α33

0

0

0

T−Tref
 358 

TRANSVERSELY ISOTROPIC MATERIALS

ISOTROPY PLANE x

3 = 0 (CONCLUDED)

" Likewise 

" Thus, the transversely isotropic material considered has 5 

independent stiffness coefficients, 5 independent compliance 

coefficients, 2 independent thermal-expansion coefficients, and 2 

independent thermal-moduli coefficients

" It is interesting to point out that hexagonal materials, defined by

planes of reflective symmetry given by , have the same 

number of independent elastic constants and thermal parameters as 

the transversely isotropic material

β 11

β 33

=−

C11+C12C13

2C13C33

α11

α33

θ3 =0, ±

π

6
 359 

TRANSVERSELY ISOTROPIC MATERIALS

ISOTROPY PLANE x1 = 0

" Now, consider the case in which a specially orthotropic material,

which exhibits symmetry about the perpendicular principal-coordinate 

planes, also exhibits isotropy in the plane x1 = 0

" For the specially orthotropic material, it was shown previously that the 

material properties are given by

 and

σ11

σ22

σ33

σ23

σ13

σ12

=

C11C12C13000

C12

C22C23000

C13

C23C33000

000C44

00

0000C55

0

00000C66

ε11

ε22

ε33

2ε23

2ε13

2ε12

+

β 11

β 22

β 33

0

0

0

T−Tref

ε11

ε22

ε33

2ε23

2ε13

2ε12

=

S11S12S13000

S12

S22S23000

S13

S23S33000

000S44

00

0000S55

0

00000S66

σ11

σ22

σ33

σ23

σ13

σ12

+

α11

α22

α33

0

0

0

T−Tref
 360 

TRANSVERSELY ISOTROPIC MATERIALS

ISOTROPY PLANE x1 = 0 (CONTINUED)

" For this type of symmetry to exist, there must be an infinite number of 

elastic symmetry planes that are perpendicular to the plane x1 = 0

" Equivalently, the constitutive matrices and vectors must invariant 

with respect to dextral rotations about the x1 axis 

" The coordinate transformation for this 

symmetry is shown in the figure and is 

given by , ,

and , with 

" The corresponding matrix of direction 

cosines is given by

i 1

i 2

i 3

x1

x2

x3

x1′

x2′

x3′

i 1′

i 2′

i 3′

,

,

θ1

θ1

Plane of 

isotropy

ai′j ≡ ii′ • ij

x1′ =x1 x2′ =x2cosθ1 +x3sinθ1

x3′ =−x2sinθ1 +x3cosθ1

0≤θ1 <2π

a1′1 a1′2 a1′3

a2′1 a2′2 a2′3

a3′1 a3′2 a3′3

=

100

0cosθ1 sinθ1

0− sinθ1 cosθ1
 361 

TRANSVERSELY ISOTROPIC MATERIALS

ISOTROPY PLANE x1 = 0 (CONTINUED)

" The corresponding stress and strain transformation matrices were 

shown previously to be given by

T and

σ

=

1 00000

0 cos2

θ1 sin2θ1 2sinθ1cosθ1 0

0

0 sin2

θ1 cos2θ1 −

2sin

θ1cosθ1 0

0

0

− sinθ1cosθ1 sinθ1cosθ1 cos2θ1 − sin2θ1 0

0

0000 cos

θ1 − sinθ1

0000 sin

θ1 cosθ1

T

ε

=

1 00000

0 cos2

θ1 sin2θ1 sinθ1cosθ1 0

0

0 sin2

θ1 cos2θ1 − sinθ1cosθ1 0

0

0

− 2sinθ1cosθ1 2sinθ1cosθ1 cos2θ1 − sin2θ1 0

0

0000 cos

θ1 − sinθ1

0000 sin

θ1 cosθ1
 362 

TRANSVERSELY ISOTROPIC MATERIALS

ISOTROPY PLANE x1 = 0 (CONTINUED)

" Specific expressions that result from the transformation 

 were given previously for an anisotropic material 

" For a specially orthotropic material, yields the 

expressions for a generally orthotropic material; that is,

C′ =T

σ

CTε

− 1

C′ =T

σ

CTε

− 1

C1′1′ =C11C1′2′ =m2C12+n2C13C1′3′ =m2C13+n2C12

C1′4′ =mnC13−C12

C1′5′ =0C1′6′ =0

C2′2′ =m

4

C22+2m2n2 C23+2C44+n

4

C33

C2′3′ =m2n2C22+C33−4C44+m4+n4C23

C2′4′ =mn3C33−C23−2C44−m3nC22−C23−2C44
 363 

TRANSVERSELY ISOTROPIC MATERIALS

ISOTROPY PLANE x1 = 0 (CONTINUED)

 with and 

" Next, the invariance condition is enforced, which 

implies the conditions , where 

C2′5′ =0C2′6′ =0

C3′3′ =m

4

C33+2m2n2 C23+2C44+n

4

C22

C3′4′ =m3nC33−C23−2C44−mn3C22−C23−2C44

C3′5′ =0C3′6′ =0

C4′4′ =m2n2C22+C33−2C23+m2−n2 2C44

C

4′5′ =

0C

4′6′ =

0C5′5′ =m2C55+n2C66

C5′6′ =mnC55−C66

C6′6′ =m2C66+n2C55

m= cosθ1 n= sinθ1

C=T

σ

CTε

− 1

Cr′s′ =Crs

r, s∈1,2, 3, 4, 5, 6
 364 

TRANSVERSELY ISOTROPIC MATERIALS

ISOTROPY PLANE x1 = 0 (CONTINUED)

" Enforcing gives , which 

can only be satisfied in a general sense (arbitrary values for θ1) if 

 and 

" Similarly, enforcing gives , which yields 

" Likewise, enforcing gives , which yields 

" Finally, by substituting , , , and 

 into the remaining expressions for and enforcing 

, it is found the remaining expressions are identically satisfied

C2′2′ =C22

C22=m

4

C22+2m2n2 C23+2C44+n

4

C33

C33=C22C23+2C44=C22

C1′3′ =C13C13=m2C13+n2C12

C12=C13

C5′5′ =C55C55=m2C55+n2C66

C66=C55

C33=C22C23+2C44=C22C12=C13

C66=C55

Cr′s′

Cr′s′ =Crs
 365 

TRANSVERSELY ISOTROPIC MATERIALS

ISOTROPY PLANE x1 = 0 (CONTINUED)

" Thus, the stiffness matrix for a transversely isotropic material that is 

isotropic in the plane x1 = 0 has the form

which has 5 independent stiffnesses

" Following the same procedure for the compliance coefficients by using 

the transformed-compliance expressions previously given for a 

generally orthotropic material and enforcing the invariance 

condition or yields similar results

C11C12C12000

C12

C22C23000

C12C23C22000

000

1

2

C22−C2300

0000C550

00000C55

S=Tε ST

σ

− 1

Sr′s′ =Srs
 366 

TRANSVERSELY ISOTROPIC MATERIALS

ISOTROPY PLANE x1 = 0 (CONTINUED)

" Thus, the compliance matrix for a transversely isotropic material that is 

isotropic in the plane x1 = 0 has the form

which has 5 independent compliances

S11S12S12000

S12

S22S23000

S12S23S22000

000

1

2

S22−S2300

0000S550

00000S55
 367 

TRANSVERSELY ISOTROPIC MATERIALS

ISOTROPY PLANE x1 = 0 (CONTINUED)

" For a specially orthotropic material, yields the 

expressions for the transformed thermal-expansion coefficients for a 

generally orthotropic material; that is,

" Enforcing the invariance condition or yields 

the requirements that and 

" Following the same procedure for the thermal moduli by using the 

transformed-thermal-moduli expressions previously given for a 

generally orthotropic material and enforcing the invariance 

condition or yields similar results; that is,

 and 

α′ =Tε α

α1′1′ =α11α2′2′ =m2α22+n2α33α3′3′ =m2α33+n2α22

α2′3′ =mnα33−α22

α1′3′ =0α1′2′ =0

α=Tε α

αr′s′ =αrs

α33=α22α23=0

β=T

σ

β

βr′s′ =βrs

β 33=β 22β 23=0
 368 

TRANSVERSELY ISOTROPIC MATERIALS

ISOTROPY PLANE x1 = 0 (CONTINUED)

" Applying all these simplifications, the linear thermoelastic constitutive 

equations become

 and

σ11

σ22

σ33

σ23

σ13

σ12

=

C11C12C12000

C12

C22C23000

C12

C23C22000

000 1

2

C22−C2300

0000C55

0

00000C55

ε11

ε22

ε33

2ε23

2ε13

2ε12

+

β 11

β 22

β 22

0

0

0

T−Tref

ε11

ε22

ε33

2ε23

2ε13

2ε12

=

S11S12S12000

S12

S22S23000

S12

S23S22000

000 1

2

S22−S2300

0000S55

0

00000S55

σ11

σ22

σ33

σ23

σ13

σ12

+

α11

α22

α22

0

0

0

T−Tref
 369 

TRANSVERSELY ISOTROPIC MATERIALS

ISOTROPY PLANE x1 = 0 (CONCLUDED)

" Likewise 

" Thus, the transversely isotropic material considered has 5 independent 

stiffness coefficients, 5 independent compliance coefficients, 2 

independent thermal-expansion coefficients, and 2 independent 

thermal-moduli coefficients

" It is interesting to point out that hexagonal materials, defined by

planes of reflective symmetry given by , have the same 

number of independent elastic constants and thermal parameters as 

the transversely isotropic material

β 11

β 22

=−

C112C12

C12C22+C23

α11

α22

θ1 =0, ±

π

6
 370 

CUBIC

MATERIALS
 371 

CUBIC MATERIALS

" A cubic material arises from the invariance conditions that are obtained 

for a single plane of elastic reflective symmetry that contains a 

coordinate axis

" Precisely, a cubic material is obtained by

enforcing the invariance conditions for a 

tetragonal material, for each of the three 

coordinate axes

" That is, there are nine planes of elastic 

symmetry that are given by ,

, and 

" Expressions for the constitutive 

equations are obtained directly by

enforcing the three sets of invariance conditions for the previously 

given cases for tetragonal materials sequentially

x1

x2

x3

θ1 =0, ±

π

4

θ2 =0, ±

π

4

θ3 =0, ±

π

4
 372 

CUBIC MATERIALS - CONCLUDED

The resulting constitutive equations are given by

" Therefore, a cubic material has three independent elastic constants 

and one independent thermal-expansion or thermal-compliance 

parameter

σ11

σ22

σ33

σ23

σ13

σ12

=

C11C12C12000

C12

C11C12000

C12C12C11000

000C44

00

0000C44

0

00000C44

ε11

ε22

ε33

2ε23

2ε13

2ε12

+

β 11

β 11

β 11

0

0

0

T−Tref

ε11

ε22

ε33

2ε23

2ε13

2ε12

=

S11S12S12000

S12S11S12000

S12S12S11000

000S4400

0000S44

0

00000S44

σ11

σ22

σ33

σ23

σ13

σ12

+

α11

α11

α11

0

0

0

T−Tref
 373 

COMPLETELY ISOTROPIC

MATERIALS
 374 

COMPLETELY ISOTROPIC MATERIALS

" First, consider the case in which both the planes x1 = 0 and x

2 = 0 are 

planes of isotropy

" Applying the results of both of the corresponding symmetry 

transformations successively yields the following constitutive 

equations

" These equations possess two independent elastic stiffnesses and 

one independent thermal moduli; the same as a completely 

isotropic material

σ11

σ22

σ33

σ23

σ13

σ12

=

C11C12C12000

C12

C11C12000

C12

C12C11000

000 1

2

C11−C1200

0000 1

2

C11−C120

00000 1

2

C11−C12

ε11

ε22

ε33

2ε23

2ε13

2ε12

+

β 11

β 11

β 11

0

0

0

T−Tref
 375 

COMPLETELY ISOTROPIC MATERIALS

(CONTINUED)

" Similarly,

" Now consider a general transformation of the stiffness matrix

previously obtained for a material with two perpendicular planes of 

isotropy

ε11

ε22

ε33

2ε23

2ε13

2ε12

=

S11S12S12000

S12

S11S12000

S12

S12S11000

000 1

2

S11−S1200

0000 1

2

S11−S120

00000 1

2

S11−S12

σ11

σ22

σ33

σ23

σ13

σ12

+

α11

α11

α11

0

0

0

T−Tref

β 11=−

C11+2C12

α11
 376 

COMPLETELY ISOTROPIC MATERIALS

(CONTINUED)

" The general transformation law is given by , where

 and

C′ =T

σ

CTε

− 1

T

σ

=

a1′1

2

a1′2

2

a1′3

2

2a1′2a1′3 2a1′1a1′3 2a1′1a1′2

a2′1

2

a2′2

2

a2′3

2

2a2′2a2′3 2a2′1a2′3 2a2′1a2′2

a3′1

2

a3′2

2

a3′3

2

2a3′2a3′3 2a3′1a3′3 2a3′1a3′2

a2′1a3′1 a2′2a3′2 a2′3a3′3 a2′2a3′3 +a2′3a3′2 a2′1a3′3 +a2′3a3′1 a2′1a3′2 +a2′2a3′1

a1′1a3′1 a1′2a3′2 a1′3a3′3 a1′2a3′3 +a1′3a3′2 a1′1a3′3 +a1′3a3′1 a1′1a3′2 +a1′2a3′1

a1′1a2′1 a1′2a2′2 a1′3a2′3 a1′2a2′3 +a1′3a2′2 a1′1a2′3 +a1′3a2′1 a1′1a2′2 +a1′2a2′1

T

ε

=

a1′1

2

a1′2

2

a1′3

2

a1′2a1′3 a1′1a1′3 a1′1a1′2

a2′1

2

a2′2

2

a2′3

2

a2′2a2′3 a2′1a2′3 a2′1a2′2

a3′1

2

a3′2

2

a3′3

2

a3′2a3′3 a3′1a3′3 a3′1a3′2

2a2′1a3′1 2a2′2a3′2 2a2′3a3′3 a2′2a3′3 +a2′3a3′2 a2′1a3′3 +a2′3a3′1 a2′1a3′2 +a2′2a3′1

2a1′1a3′1 2a1′2a3′2 2a1′3a3′3 a1′2a3′3 +a1′3a3′2 a1′1a3′3 +a1′3a3′1 a1′1a3′2 +a1′2a3′1

2a1′1a2′1 2a1′2a2′2 2a1′3a2′3 a1′2a2′3 +a1′3a2′2 a1′1a2′3 +a1′3a2′1 a1′1a2′2 +a1′2a2′1
 377 

COMPLETELY ISOTROPIC MATERIALS

(CONTINUED)

" In addition, the direction cosines that appear in the transformation 

matrices satisfy the following conditions, that are given in expanded,

tabular form:

a1′1

2

+a1′2

2

+a1′3

2

=1

a2′1a1′1 +a2′2a1′2 +a2′3a1′3 =0

a3′1a1′1 +a3′2a1′2 +a3′3a1′3 =0

a2′1

2

+a2′2

2

+a2′3

2

=1

a3′1a2′1 +a3′2a2′2 +a3′3a2′3 =0

a3′1

2

+a3′2

2

+a3′3

2

=1

k p

ak′qa

p′q

=δkp

1 1

2 1

3 

1

2 

2

3 2

3 

3

k p

1 1

2 1

3 

1

2 

2

3 2

3 

3

a

q′kaq′p

=δkp

a1′1

2

+a2′1

2

+a3′1

2

=1

a1′2a1′1 +a2′2a2′1 +a3′2a3′1 =0

a1′3a1′1 +a2′3a2′1 +a3′3a3′1 =0

a1′2

2

+a2′2

2

+a3′2

2

=1

a1′3a1′2 +a2′3a2′2 +a3′3a3′2 =0

a1′3

2

+a2′3

2

+a3′3

2

=1
 378 

COMPLETELY ISOTROPIC MATERIALS

(CONTINUED)

" For the general transformation to be a symmetry transformation,

 becomes 

" It is convenient to rewrite this expression as ,

where ,

 and

C′ =T

σ

CTε

− 1

C=T

σ

CTε

− 1

CTε −T

σ

C=0

C44=

1

2

C11−C12

CTε =

C11C12C12000

C12

C11C12000

C12

C12C11000

000C44

00

0000C44

0

00000C44

a1′1

2 a1′22 a1′32 a1′2a1′3 a1′1a1′3 a1′1a1′2

a2′1

2 a2′2 2 a2′32 a2′2a2′3 a2′1a2′3 a2′1a2′2

a3′1

2 a3′22 a3′32 a3′2a3′3 a3′1a3′3 a3′1a3′2

2a2′1a3′1 2a2′2a3′2 2a2′3a3′3 a2′2a3′3 +a2′3a3′2 a2′1a3′3 +a2′3a3′1 a2′1a3′2 +a2′2a3′1

2a1′1a3′1 2a1′2a3′2 2a1′3a3′3 a1′2a3′3 +a1′3a3′2 a1′1a3′3 +a1′3a3′1 a1′1a3′2 +a1′2a3′1

2a1′1a2′1 2a1′2a2′2 2a1′3a2′3 a1′2a2′3 +a1′3a2′2 a1′1a2′3 +a1′3a2′1 a1′1a2′2 +a1′2a2′1

T

σ

C=

a1′1

2

a1′2

2

a1′3

2

2a1′2a1′3 2a1′1a1′3 2a1′1a1′2

a2′1

2

a2′2

2

a2′3

2

2a2′2a2′3 2a2′1a2′3 2a2′1a2′2

a3′1

2

a3′2

2

a3′3

2

2a3′2a3′3 2a3′1a3′3 2a3′1a3′2

a2′1a3′1 a2′2a3′2 a2′3a3′3 a2′2a3′3 +a2′3a3′2 a2′1a3′3 +a2′3a3′1 a2′1a3′2 +a2′2a3′1

a1′1a3′1 a1′2a3′2 a1′3a3′3 a1′2a3′3 +a1′3a3′2 a1′1a3′3 +a1′3a3′1 a1′1a3′2 +a1′2a3′1

a1′1a2′1 a1′2a2′2 a1′3a2′3 a1′2a2′3 +a1′3a2′2 a1′1a2′3 +a1′3a2′1 a1′1a2′2 +a1′2a2′1

C11C12C12000

C12

C11C12000

C12

C12C11000

000C44

00

0000C44

0

00000C44
 379 

COMPLETELY ISOTROPIC MATERIALS

(CONTINUED)

"Performing the calculations for each element of the matrix equation 

CTε −T

σ

C=0 gives

11: C12a2′1

2

+a3′1

2

−a1′2

2

−a1′3

2

=0

12: C12a2′2

2

+a3′2

2

−a1′2

2

−a1′3

2

=0

13: C12a2′3

2

+a3′3

2

−a1′1

2

−a1′2

2

=0

14: C12a1′3a1′2 +a2′3a2′2 +a3′3a3′2 =0

15: C12a1′3a1′1 +a2′3a2′1 +a3′3a3′1 =0

16: C12a1′2a1′1 +a2′2a2′1 +a3′2a3′1 =0

21: C12a1′1

2

+a3′1

2

−a2′2

2

−a2′3

2

=0
 380 

COMPLETELY ISOTROPIC MATERIALS

(CONTINUED)

22: C12a1′2

2

+a3′2

2

−a2′1

2

−a2′3

2

=0

23: C12a1′3

2

+a3′3

2

−a2′1

2

−a2′2

2

=0

24: C12a1′3a1′2 +a2′3a2′2 +a3′3a3′2 =0

25: C12a1′3a1′1 +a2′3a2′1 +a3′3a3′1 =0

26: C12a1′2a1′1 +a2′2a2′1 +a3′2a3′1 =0

31: C12a1′1

2

+a2′1

2

−a3′2

2

−a3′3

2

=0

32: C12a1′2

2

+a2′2

2

−a3′1

2

−a3′3

2

=0

33: C12a1′3

2

+a2′3

2

−a3′1

2

−a3′2

2

=0
 381 

COMPLETELY ISOTROPIC MATERIALS

(CONTINUED)

34: C12a1′3a1′2 +a2′3a2′2 +a3′3a3′2 =0

35: C12a1′3a1′1 +a2′3a2′1 +a3′3a3′1 =0

36: C12a1′2a1′1 +a2′2a2′1 +a3′2a3′1 =0

41: C12a3′1a2′1 +a3′2a2′2 +a3′3a2′3 =0

42: C12a3′1a2′1 +a3′2a2′2 +a3′3a2′3 =0

43: C12a3′1a2′1 +a3′2a2′2 +a3′3a2′3 =0

51: C12a3′1a1′1 +a3′2a1′2 +a3′3a1′3 =0

52: C12a3′1a1′1 +a3′2a1′2 +a3′3a1′3 =0

53: C12a3′1a1′1 +a3′2a1′2 +a3′3a1′3 =0
 382 

COMPLETELY ISOTROPIC MATERIALS

(CONTINUED)

" The elements of the matrix equation that are not 

listed above are satisfied identically

" Now, consider 

" The condition given in the previous table 

yields ; hence,

61: C12a2′1a1′1 +a2′2a1′2 +a2′3a1′3 =0

62: C12a2′1a1′1 +a2′2a1′2 +a2′3a1′3 =0

63: C12a2′1a1′1 +a2′2a1′2 +a2′3a1′3 =0

CTε −T

σ

C=0

11: C12a2′1

2

+a3′1

2

−a1′2

2

−a1′3

2

=0

a1′1

2

+a1′2

2

+a1′3

2

=1

a1′2

2

+a1′3

2

=1−a1′1

2

11: C12a1′1

2

+a2′1

2

+a3′1

2

−1=0
 383 

COMPLETELY ISOTROPIC MATERIALS

(CONTINUED)

" Next, using the condition , given in the 

previous table, shows that is 

identically satisfied

" By following a similar procedure or by using direct substitution of the 

conditions and given in the previous table, it can 

be shown that the invariance condition is 

identically satisfied

" Therefore, two orthogonal planes of isotropy imply that every plane is a 

plane of isotropy because a symmetry transformation 

for any plane can 

be obtained from the general transformation

" An isotropic material has two independent stiffnesses and is the 

simplest known material, with no dependence on direction

a1′1

2

+a2′1

2

+a3′1

2

=1

11: C12a1′1

2

+a2′1

2

+a3′1

2

−1=0

ak′qa

p′q

=δkpa

q′kaq′p

=δkp

CTε −T

σ

C=0
 384 

COMPLETELY ISOTROPIC MATERIALS

(CONTINUED)

" Now consider a general transformation of the thermal moduli

previously obtained for a material with two perpendicular plane of 

isotropy

" The transformation law is given by

" For the general transformation to be a symmetry transformation,

 becomes 

" The expanded form is given by

β′ =T

σ

β

β′ =T

σ

ββ−T

σ

β=0

β 11

β 11

β 11

0

0

0

−

a1′1

2

a1′2

2

a1′3

2

2a1′2a1′3 2a1′1a1′3 2a1′1a1′2

a2′1

2

a2′2

2

a2′3

2

2a2′2a2′3 2a2′1a2′3 2a2′1a2′2

a3′1

2

a3′2

2

a3′3

2

2a3′2a3′3 2a3′1a3′3 2a3′1a3′2

a2′1a3′1 a2′2a3′2 a2′3a3′3 a2′2a3′3 +a2′3a3′2 a2′1a3′3 +a2′3a3′1 a2′1a3′2 +a2′2a3′1

a1′1a3′1 a1′2a3′2 a1′3a3′3 a1′2a3′3 +a1′3a3′2 a1′1a3′3 +a1′3a3′1 a1′1a3′2 +a1′2a3′1

a1′1a2′1 a1′2a2′2 a1′3a2′3 a1′2a2′3 +a1′3a2′2 a1′1a2′3 +a1′3a2′1 a1′1a2′2 +a1′2a2′1

β 11

β 11

β 11

0

0

0

=0
 385 

COMPLETELY ISOTROPIC MATERIALS

(CONTINUED)

" By using direct substitution of the conditions and 

 given in the previous table, it can be shown that the 

invariance condition is identically satisfied

" Therefore, an isotropic material has one independent thermal 

moduli

" For general transformations of the compliance matrix and the 

thermal-expansion coefficents previously obtained for a material 

with two perpendicular plane of isotropy, the transformation laws are 

given by and , respectively

" For the general transformations to be symmetry transformations,

 becomes and 

 becomes 

ak′qa

p′q

=δkp

a

q′kaq′p

=δkp

β−T

σ

β=0

S′ =Tε ST

σ

− 1

α′ =Tε α

S′ =Tε ST

σ

− 1

Tε S−ST

σ

=

0

α′ =Tε αα−Tε α=0
 386 

COMPLETELY ISOTROPIC MATERIALS

(CONCLUDED)

" By performing the calculations for each element of the matrix equation 

, and using the conditions and 

 given in the previous table, it can be shown that the 

invariance condition is satisfied identically

" Similarly, the invariance condition is also satisfied 

identically

" Therefore, an isotropic material has two independent compliances 

and one coefficient of thermal expansion

Tε S−ST

σ

=

0

ak′qa

p′q

=δkp

a

q′kaq′p

=δkp

α−Tε α=0
 387 

CLASSES OF MATERIAL SYMMETRY

SUMMARY OF INDEPENDENT MATERIAL CONSTANTS

" The eight distinct classes of elastic-material symmetry are classified 

by the number of independent material constants as follows:

"Triclinic materials - 21 elastic, 6 thermal

" Monoclinic materials - 13 elastic, 4 thermal

" Orthotropic materials - 9 elastic, 3 thermal

"Trigonal materials - 6 elastic, 2 thermal

"Tetragonal materials - 6 elastic, 2 thermal

"Transversely isotropic materials - 5 elastic, 2 thermal

" Cubic materials - 3 elastic, 1 thermal

" Completely isotropic materials - 2 elastic, 1 thermal
 388 

ENGINEERING CONSTANTS

FOR ELASTIC MATERIALS
 389 

CONSTITUTIVE EQUATIONS IN TERMS OF 

ENGINEERING CONSTANTS

" The compliances of a homogeneous, elastic, anisotropic solid are 

usually expressed in terms of engineering constants when practical 

applications are under consideration

" These constants are determined from experiments

" 21 independent elastic constants imply 21 separate experiments

"To determine expressions for the compliances in terms of engineering 

constants, it is useful to examine the meaning of each term in the 

general, unsymmetric compliance matrix given below

ε11

ε22

ε33

2ε23

2ε13

2ε12

=

S11

S12S13S14S15S16

S21S22S23S24S25S26

S31S32S33S34S35S36

S41S42S43S44S45S46

S51S52S53S54S55S56

S61S62S63S64S65S66

σ11

σ22

σ33

σ23

σ13

σ12

+

α11

α22

α33

2α23

2α13

2α12

T−Tref
 390 

CONSTITUTIVE EQUATIONS IN TERMS OF 

ENGINEERING CONSTANTS - CONTINUED

" It is important to remember, that stresses cannot be measured in a 

laboratory experiment; only strains

" The terms S11, S22, and S33 relate the normal strain to the corresponding 

normal stress

" The engineering constants used to represent these relationships 

are called elastic moduli or moduli of elasticity 

" In particular, the symbol Ej is used herein to denote the elastic 

modulus in the xj - coordinate direction

" In general,

Ej ≡ normal stress

σjj

normal strain

εjj caused byσjj
 391 

CONSTITUTIVE EQUATIONS IN TERMS OF 

ENGINEERING CONSTANTS - CONTINUED

" The terms S44, S55, and S66 relate the shearing strain in each coordinate 

plane to the corresponding shearing stress

" The engineering constants used to represent these relationships 

are called shear moduli or moduli of rigidity 

" In particular, the symbol Gij is used herein to denote the shear 

modulus in the xi - xj coordinate plane

" In general,

Gij ≡ shearing

stress

σij

shearingstrain

2εij caused byσij
 392 

CONSTITUTIVE EQUATIONS IN TERMS OF 

ENGINEERING CONSTANTS - CONTINUED

" The terms S12, S21, S13, S31, S23, and S32 relate the lateral contraction or 

expansion to the expansion or contraction of the solid, in the direction 

of a given normal stress

" The engineering constants used to represent these relationships 

are called Poisson’s ratios

" In particular, the symbol νij is used herein to denote the lateral 

contraction or expansion in the xj - coordinate direction caused b

y

a normal stress applied in the xi - coordinate direction

" In general,

νij ≡−

normal strain

εjj caused by normal stressσii

normal strain

εii caused byσii
 393 

CONSTITUTIVE EQUATIONS IN TERMS OF 

ENGINEERING CONSTANTS - CONTINUED

" The terms S14, S15, S16, S24, S25, S26, S34, S35, and S36 relate normal strains 

to shearing stresses

" The engineering constants used to represent these relationships 

are generalizations of Poisson’s ratios and are called coefficients 

of interaction (or mutual influence) of the first kind, and are 

attributed to A. L. Rabinovich

" In particular, the symbol ηk,ij is used herein to relate the 

contraction or expansion in the x

k - coordinate direction induced 

by

a shearing stress applied in the xi - xj coordinate plane

" That is,

ηk, ij ≡

normal strain

εkk caused by shearing stressσij

shearing strain

2εij caused by shearing stressσij
 394 

CONSTITUTIVE EQUATIONS IN TERMS OF 

ENGINEERING CONSTANTS - CONTINUED

" The terms S41, S42, S43, S51, S52, S53, S61, S62, and S63 relate shearing 

strains to normal stresses

" The engineering constants used to represent these relationships 

are also generalizations of Poisson’s ratios and are called 

coefficients of interaction (or mutual influence) of the 

second kind, and are also attributed to A. L. Rabinovic

h (circa 

1946)

" In particular, the symbol ηij,k is used herein to relate the shearing 

strain in the xi - xj coordinate plane induced b

y the action of a 

normal stress applied in the x

k - coordinate direction

" That is,

ηij, k =

shearing strain

2εij caused by normal stressσkk

normal strain

εkk caused byσkk
 395 

CONSTITUTIVE EQUATIONS IN TERMS OF 

ENGINEERING CONSTANTS - CONTINUED

" The terms S45, S46, S54, S56, S64, and S65 relate shearing strains to 

noncorresponding shearing stresses

" The engineering constants used to represent these relationships 

are called Chentsov’s coefficients, and are attributed to N. G.

Chentsov

" In particular, the symbol μij,kl is used herein to relate the shearing 

strain in the xi - xj coordinate plane induced b

y a shearing stress 

applied in the x

k - xl coordinate plane

" That is,

" Note that because of symmetry of 2εij and σkl

μij, kl ≡

shearing

2εij caused by shearing

stress

σkl

shearingstrain

2εkl caused by shearing

stress

σkl

μij, kl =μji, kl =μji, lk =μij, lk
 396 

CONSTITUTIVE EQUATIONS IN TERMS OF 

ENGINEERING CONSTANTS - CONTINUED

" Now, consider a parallelopiped of homogeneous 

material that is subjected to only a constant value of 

σ11 and no thermal loading, like a stress state that 

might be exist in an experiment like a tensile test

" For this case,

" This equation indicates that, in general, the parallelopiped will 

extend in the x1 - coordinate direction,

expand or contract in the x

2 - 

and x

3 - coordinate directions, and shear in each face for this very 

simple state of uniaxial stress

σ11

σ11

ε11

ε22

ε33

2ε23

2ε13

2ε12

=

S11

S12S13S14S15S16

S21S22S23S24S25S26

S31S32S33S34S35S36

S41S42S43S44S45S46

S51S52S53S54S55S56

S61S62S63S64S65S66

σ11

0

0

0

0

0
 397 

CONSTITUTIVE EQUATIONS IN TERMS OF 

ENGINEERING CONSTANTS - CONTINUED

" From the definition of the elastic moduli, it follows that 

" Noting that the previous matrix equation gives for this 

simple state of stress, it also follows that 

" Next, from the definition for the Poisson’s ratios, it follows that 

 and for this simple state of stress

" Noting that the previous matrix equation also gives and 

 for this simple state of stress, and using , it 

also follows that and 

E1 =

σ11

ε11

ε11=

S11

σ11

S11=

1

E1

ν12=−

ε22

ε11

ν13=−

ε33

ε11

ε22=

S21

σ11

ε33=

S31

σ11ε11=

S11

σ11

ν12=−

S21

S11

ν13=−

S31

S11
 398 

CONSTITUTIVE EQUATIONS IN TERMS OF 

ENGINEERING CONSTANTS - CONTINUED

" Using gives the results and 

" Next, from the definition for the coefficients of interaction of the second 

kind, it follows that , , and for this 

simple state of stress

" Noting that the previous matrix equation also gives ,

, and for this simple state of stress, and 

using , it also follows that

, , and 

S11=

1

E1

S21=−

ν12

E1

S31=−

ν13

E1

η23, 1 =

2ε23

ε11

η13, 1 =

2ε13

ε11

η12, 1 =

2ε12

ε11

2ε23=

S41

σ11

2ε13=

S51

σ112ε12=

S61

σ11

ε11=

S11

σ11

η23, 1 =

S41

S11

η13, 1 =

S51

S11

η12, 1 =

S61

S11
 399 

CONSTITUTIVE EQUATIONS IN TERMS OF 

ENGINEERING CONSTANTS - CONTINUED

" Using gives the results

, , and 

" Now, consider a parallelopiped of homogeneous 

material that is subjected to only a constant value of 

σ22 and no thermal loading

" For this case,

S11=

1

E1

S41=

η23, 1

E1

S51=

η13, 1

E1

S61=

η12, 1

E1

σ22

σ22

ε11

ε22

ε33

2ε23

2ε13

2ε12

=

S11

S12S13S14S15S16

S21S22S23S24S25S26

S31S32S33S34S35S36

S41S42S43S44S45S46

S51S52S53S54S55S56

S61S62S63S64S65S66

0

σ22

0

0

0

0
 400 

CONSTITUTIVE EQUATIONS IN TERMS OF 

ENGINEERING CONSTANTS - CONTINUED

" This equation indicates that, in general, the parallelopiped will 

extend in the x

2 - coordinate direction,

expand or contract in the x1 - 

and x

3 - coordinate directions, and shear in each face for this very 

simple state of uniaxial stress

" From the definition of the elastic moduli, it follows that 

" Noting that the previous matrix equation gives for this 

simple state of stress, it also follows that 

" Next, from the definition for the Poisson’s ratios, it follows that 

 and for this simple state of stress

E2 =

σ22

ε22

ε22=

S22

σ22

S22=

1

E2

ν21=−

ε11

ε22

ν23=−

ε33

ε22
 401 

CONSTITUTIVE EQUATIONS IN TERMS OF 

ENGINEERING CONSTANTS - CONTINUED

" Noting that the previous matrix equation also gives and 

 for this simple state of stress, and using , it 

also follows that and 

" Using gives the results and 

" Next, from the definition for the coefficients of interaction of the second 

kind, it follows that , , and for this 

simple state of stress

ε11=

S12

σ22

ε33=

S32

σ22ε22=

S22

σ22

ν21=−

S12

S22

ν23=−

S32

S22

S22=

1

E2

S12=−

ν21

E2

S32=−

ν23

E2

η23, 2 =

2ε23

ε22

η13, 2 =

2ε13

ε22

η12, 2 =

2ε12

ε22
 402 

CONSTITUTIVE EQUATIONS IN TERMS OF 

ENGINEERING CONSTANTS - CONTINUED

" Noting that the previous matrix equation also gives ,

, and for this simple state of stress, and 

using , it also follows that

, , and 

" Using gives the results

, , and 

2ε23=

S42

σ22

2ε13=

S52

σ222ε12=

S62

σ22

ε22=

S22

σ22

η23, 2 =

S42

S22

η13, 2 =

S52

S22

η12, 2 =

S62

S22

S22=

1

E2

S42=

η23, 2

E2

S52=

η13, 2

E2

S62=

η12, 2

E2
 403 

CONSTITUTIVE EQUATIONS IN TERMS OF 

ENGINEERING CONSTANTS - CONTINUED

" Now, consider a parallelopiped of homogeneous material 

that is subjected to only a constant value of σ33 and no 

thermal loading

" For this case,

" This equation indicates that, in general, the parallelopiped will 

extend in the x

3 - coordinate direction,

expand or contract in the x1 - 

and x

2 - coordinate directions, and shear in each face for this very 

simple state of uniaxial stress

σ33

σ33

ε11

ε22

ε33

2ε23

2ε13

2ε12

=

S11

S12S13S14S15S16

S21S22S23S24S25S26

S31S32S33S34S35S36

S41S42S43S44S45S46

S51S52S53S54S55S56

S61S62S63S64S65S66

0

0

σ33

0

0

0
 404 

CONSTITUTIVE EQUATIONS IN TERMS OF 

ENGINEERING CONSTANTS - CONTINUED

" From the definition of the elastic moduli, it follows that 

" Noting that the previous matrix equation gives for this 

simple state of stress, it also follows that 

" Next, from the definition for the Poisson’s ratios, it follows that 

 and for this simple state of stress

" Noting that the previous matrix equation also gives 

and for this simple state of stress, and using 

, it also follows that and 

E3 =

σ33

ε33

ε33=

S33

σ33

S33=

1

E3

ν31=−

ε11

ε33

ν32=−

ε22

ε33

ε11=

S13

σ33

ε22=

S23

σ33

ε33=

S33

σ33

ν31=−

S13

S33

ν32=−

S23

S33
 405 

CONSTITUTIVE EQUATIONS IN TERMS OF 

ENGINEERING CONSTANTS - CONTINUED

" Using gives the results and 

" Next, from the definition for the coefficients of interaction of the second 

kind, it follows that , , and for this 

simple state of stress

" Noting that the previous matrix equation also gives ,

, and for this simple state of stress, and 

using , it also follows that

, , and 

S33=

1

E3

S13=−

ν31

E3

S23=−

ν32

E3

η23, 3 =

2ε23

ε33

η13, 3 =

2ε13

ε33

η12, 3 =

2ε12

ε33

2ε23=

S43

σ33

2ε13=

S53

σ332ε12=

S63

σ33

ε33=

S33

σ33

η23, 3 =

S43

S33

η13, 3 =

S53

S33

η12, 3 =

S63

S33
 406 

CONSTITUTIVE EQUATIONS IN TERMS OF 

ENGINEERING CONSTANTS - CONTINUED

" Using gives the results

, , and 

" Now, consider a parallelopiped of homogeneous 

material that is subjected to only a constant value of 

σ23 and no thermal loading

" For this case,

S33=

1

E3

S43=

η23, 3

E3

S53=

η13, 3

E3

S63=

η12, 3

E3

σ23

ε11

ε22

ε33

2ε23

2ε13

2ε12

=

S11

S12S13S14S15S16

S21S22S23S24S25S26

S31S32S33S34S35S36

S41S42S43S44S45S46

S51S52S53S54S55S56

S61S62S63S64S65S66

0

0

0

σ23

0

0
 407 

CONSTITUTIVE EQUATIONS IN TERMS OF 

ENGINEERING CONSTANTS - CONTINUED

" This equation also indicates that, in general, the parallelopiped will 

expand or contract along all three coordinate directions, and shear 

in each face for this very simple state of pure shearing stress

" From the definition of the shear moduli, it follows that 

" Noting that the previous matrix equation gives for this 

simple state of stress, it also follows that 

" Next, from the definition for the coefficients of interaction of the first 

kind, it follows that , , and for this 

simple state of stress

G23=

σ23

2ε23

2ε23=

S44

σ23

S44=

1

G23

η1, 23 =

ε11

2ε23

η2, 23 =

ε22

2ε23

η3, 23 =

ε33

2ε23
 408 

CONSTITUTIVE EQUATIONS IN TERMS OF 

ENGINEERING CONSTANTS - CONTINUED

" Noting that the previous matrix equation also gives ,

, and f or this simple state of stress, and 

using , it also follows that

, , and 

" Using gives the results

, , and 

ε11=

S14

σ23

ε22=

S24

σ23ε33=

S34

σ23

2ε23=

S44

σ23

η1, 23 =

S14

S44

η2, 23 =

S24

S44

η3, 23 =

S34

S44

S44=

1

G23

S14=

η1, 23

G23

S24=

η2, 23

G23

S34=

η3, 23

G23
 409 

CONSTITUTIVE EQUATIONS IN TERMS OF 

ENGINEERING CONSTANTS - CONTINUED

" Next, from the definition for the Chentsov’s coefficients, it follows that 

 and for this simple state of stress

" Noting that the previous matrix equation also gives 

and for this simple state of stress, and using 

, it also follows that and 

" Using gives the results

 and 

μ13, 23 =

2ε13

2ε23

μ12, 23 =

2ε12

2ε23

2ε13=

S54

σ23

2ε12=

S64

σ23

2ε23=

S44

σ23

μ13, 23 =

S54

S44

μ12, 23 =

S64

S44

S44=

1

G23

S54=

μ13, 23

G23

S64=

μ12, 23

G23
 410 

CONSTITUTIVE EQUATIONS IN TERMS OF 

ENGINEERING CONSTANTS - CONTINUED

" Now, consider a parallelopiped of homogeneous 

material that is subjected to only a constant value of 

σ13 and no thermal loading

" For this case,

" This equation also indicates that, in general, the parallelopiped will 

expand or contract along all three coordinate directions, and shear 

in each face for this very simple state of pure shearing stress

σ13

ε11

ε22

ε33

2ε23

2ε13

2ε12

=

S11

S12S13S14S15S16

S21S22S23S24S25S26

S31S32S33S34S35S36

S41S42S43S44S45S46

S51S52S53S54S55S56

S61S62S63S64S65S66

0

0

0

0

σ13

0
 411 

CONSTITUTIVE EQUATIONS IN TERMS OF 

ENGINEERING CONSTANTS - CONTINUED

" From the definition of the shear moduli, it follows that 

" Noting that the previous matrix equation gives for this 

simple state of stress, it also follows that 

" Next, from the definition for the coefficients of interaction of the first 

kind, it follows that , , and for this 

simple state of stress

G13=

σ13

2ε13

2ε13=

S55

σ13

S55=

1

G13

η1, 13 =

ε11

2ε13

η2, 13 =

ε22

2ε13

η3, 13 =

ε33

2ε13
 412 

CONSTITUTIVE EQUATIONS IN TERMS OF 

ENGINEERING CONSTANTS - CONTINUED

" Noting that the previous matrix equation also gives ,

, and for this simple state of stress, and 

using , it also follows that

, , and 

" Using gives the results

, , and 

ε11=

S15

σ13

ε22=

S25

σ13ε33=

S35

σ13

2ε13=

S55

σ13

η1, 13 =

S15

S55

η2, 13 =

S25

S55

η3, 13 =

S35

S55

S55=

1

G13

S15=

η1, 13

G13

S25=

η2, 13

G13

S35=

η3, 13

G13
 413 

CONSTITUTIVE EQUATIONS IN TERMS OF 

ENGINEERING CONSTANTS - CONTINUED

" Next, from the definition for the Chentsov’s coefficients, it follows that 

 and for this simple state of stress

" Noting that the previous matrix equation also gives 

and for this simple state of stress, and using 

, it also follows that and 

" Using gives the results

 and 

μ23, 13 =

2ε23

2ε13

μ12, 13 =

2ε12

2ε13

2ε23=

S45

σ13

2ε12=

S65

σ13

2ε13=

S55

σ13

μ23, 13 =

S45

S55

μ12, 13 =

S65

S55

S55=

1

G13

S45=

μ23, 13

G13

S65=

μ12, 13

G13
 414 

CONSTITUTIVE EQUATIONS IN TERMS OF 

ENGINEERING CONSTANTS - CONTINUED

" Finally, consider a parallelopiped of homogeneous 

material that is subjected to only a constant value of 

σ12 and no thermal loading

" For this case,

" This equation also indicates that, in general, the parallelopiped will 

expand or contract along all three coordinate directions, and shear 

in each face for this very simple state of pure shearing stress

σ12

ε11

ε22

ε33

2ε23

2ε13

2ε12

=

S11

S12S13S14S15S16

S21S22S23S24S25S26

S31S32S33S34S35S36

S41S42S43S44S45S46

S51S52S53S54S55S56

S61S62S63S64S65S66

0

0

0

0

0

σ12
 415 

CONSTITUTIVE EQUATIONS IN TERMS OF 

ENGINEERING CONSTANTS - CONTINUED

" From the definition of the shear moduli, it follows that 

" Noting that the previous matrix equation gives for this 

simple state of stress, it also follows that 

" Next, from the definition for the coefficients of interaction of the first 

kind, it follows that , , and for this 

simple state of stress

G12=

σ12

2ε12

2ε12=

S66

σ12

S66=

1

G12

η1, 12 =

ε11

2ε12

η2, 12 =

ε22

2ε12

η3, 12 =

ε33

2ε12
 416 

CONSTITUTIVE EQUATIONS IN TERMS OF 

ENGINEERING CONSTANTS - CONTINUED

" Noting that the previous matrix equation also gives ,

, and for this simple state of stress, and 

using , it also follows that

, , and 

" Using gives the results

, , and 

ε11=

S16

σ12

ε22=

S26

σ12

ε33=

S36

σ12

2ε12=

S66

σ12

η1, 12 =

S16

S66

η2, 12 =

S26

S66

η3, 12 =

S36

S66

S66=

1

G12

S16=

η1, 12

G12

S26=

η2, 12

G12

S36=

η3, 12

G12
 417 

CONSTITUTIVE EQUATIONS IN TERMS OF 

ENGINEERING CONSTANTS - CONTINUED

" Next, from the definition for the Chentsov’s coefficients, it follows that 

 and for this simple state of stress

" Noting that the previous matrix equation also gives 

and for this simple state of stress, and using 

, it also follows that and 

" Using gives the results

 and 

μ23, 12 =

2ε23

2ε12

μ13, 12 =

2ε13

2ε12

2ε23=

S46

σ12

2ε13=

S56

σ12

2ε12=

S66

σ12

μ23, 12 =

S46

S66

μ13, 12 =

S56

S66

S66=

1

G12

S46=

μ23, 12

G12

S56=

μ13, 12

G12
 418 

CONSTITUTIVE EQUATIONS IN TERMS OF 

ENGINEERING CONSTANTS - CONTINUED

" Using all of the derived expressions for Sij, the constitutive equation

 becomes

ε11

ε22

ε33

2ε23

2ε13

2ε12

=

S11S12S13S14S15S16

S21S22S23S24S25S26

S31S32S33S34S35S36

S41S42S43S44S45S46

S51S52S53S54S55S56

S61S62S63S64S65S66

σ11

σ22

σ33

σ23

σ13

σ12

+

α11

α22

α33

2α23

2α13

2α12

T−Tref

ε11

ε22

ε33

2ε23

2ε13

2ε12

=

1

E1

−

ν21

E2

−

ν31

E3

η1, 23

G23

η1, 13

G13

η1,12

G12

−

ν12

E1

1

E2

−

ν32

E3

η2, 23

G23

η2, 13

G13

η2,12

G12

−

ν13

E1

−

ν23

E2

1

E3

η3, 23

G23

η3, 13

G13

η3,12

G12

η23, 1

E1

η23, 2

E2

η23, 3

E3

1

G23

μ23, 13

G13

μ23, 12

G12

η13, 1

E1

η13, 2

E2

η13, 3

E3

μ13, 23

G23

1

G13

μ13, 12

G12

η12, 1

E1

η12, 2

E2

η12, 3

E3

μ12, 23

G23

μ12, 13

G13

1

G12

σ11

σ22

σ33

σ23

σ13

σ12

+

α11

α22

α33

2α23

2α13

2α12

T−Tref
 419 

CONSTITUTIVE EQUATIONS IN TERMS OF 

ENGINEERING CONSTANTS - CONTINUED

" Relationships between the engineering constants are obtained by

enforcing symmetry of the matrix [S]; that is:

S12=S21

→

ν21

E2

=

ν12

E1

S13=S31

→

ν31

E3

=

ν13

E1

S14=S41

→

η1, 23

G23

=

η23, 1

E1

S15=S51

→

η1, 13

G13

=

η13, 1

E1

S16=S61

→

η1, 12

G12

=

η12, 1

E1

S23=S32

→

ν32

E3

=

ν23

E2

S24=S42

→

η2, 23

G23

=

η23, 2

E2

S25=S52

→

η2, 13

G13

=

η13, 2

E2

S26=S62

→

η2, 12

G12

=

η12, 2

E2

S34=S43

→

η3, 23

G23

=

η23, 3

E3

S35=S53

→

η3, 13

G13

=

η13, 3

E3

S36=S63

→

η3, 12

G12

=

η12, 3

E3

S45=S54

→

μ23, 13

G13

=

μ13, 23

G23

S46=S64

→

μ23, 12

G12

=

μ12, 23

G23

S56=S65

→

μ13, 12

G12

=

μ12, 13

G13
 420 

CONSTITUTIVE EQUATIONS IN TERMS OF 

ENGINEERING CONSTANTS - CONCLUDED

" Using the previous symmetry conditions, the constitutive equations are 

expressed as

ε11=

1

E1

σ11−ν12σ22−ν13σ33+η23, 1σ23+η13, 1σ13+η12, 1σ12+α11

T−Tref

ε22=

1

E2

−ν21σ11+σ22−ν23σ33+η23, 2σ23+η13, 2σ13+η12, 2σ12+α22

T−Tref

ε33=

1

E3

−ν31σ11−ν32σ22+σ33+η23, 3σ23+η13, 3σ13+η12, 3σ12+α33

T−Tref

2

ε23=

1

G23

η1, 23σ11+η2, 23σ22+η3, 23σ33+σ23+μ13, 23σ13+μ12, 23σ12+

2

α23

T−Tref

2

ε13=

1

G13

η1, 13σ11+η2, 13σ22+η3, 13σ33+μ23, 13σ23+σ13+μ12, 13σ12+

2

α13

T−Tref

2

ε12=

1

G12

η1, 12σ11+η2, 12σ22+η3, 12σ33+μ23, 12σ23+μ13, 12σ13+σ12+

2

α12

T−Tref
 421 

ENGINEERING CONSTANTS OF A SPECIALLY 

ORTHOTROPIC MATERIAL

" The engineering constants for an anisotropic material have been 

presented previously herein, for an coordinate frame

" The subset of engineering constants for a specially orthotropic 

material are given by

x1,x2,x3

E1 =

1

S11

E2 =

1

S22

E3 =

1

S33

G23=

1

S44

G13=

1

S55

G12=

1

S66

ν12=−

S12

S11

ν13=−

S13

S11

ν23=−

S23

S22
 422 

ENGINEERING CONSTANTS OF A SPECIALLY 

ORTHOTROPIC MATERIAL - CONTINUED

" Substituting these expressions into 

 gives

ε11

ε22

ε33

2ε23

2ε13

2ε12

=

S11

S12S13000

S12

S22S23000

S13

S23S33000

000S44

00

0000S55

0

00000S66

σ11

σ22

σ33

σ23

σ13

σ12

+

α11

α22

α33

0

0

0

T−Tref

ε11

ε22

ε33

2ε23

2ε13

2ε12

=

1

E1

−

ν21

E2

−

ν31

E3

000

−

ν12

E1

1

E2

−

ν32

E3

000

−

ν13

E1

−

ν23

E2

1

E3

000

000

1

G23

00

0000

1

G13

0

00000

1

G12

σ11

σ22

σ33

σ23

σ13

σ12

+

α11

α22

α33

0

0

0

T−Tref
 423 

ENGINEERING CONSTANTS OF A SPECIALLY 

ORTHOTROPIC MATERIAL - CONTINUED

" Inverting these matrix equations yields

 where

σ11

σ22

σ33

σ23

σ13

σ12

=

C11

C12C13000

C12

C22C23000

C13

C23C33000

000C4400

0000C55

0

00000C66

ε11

ε22

ε33

2ε23

2ε13

2ε12

+

β 11

β 22

β 33

0

0

0

T−Tref

C11=

E1

Δ

1−ν23ν32C12=

E1

Δ

ν21+ν23ν31=

E2

Δ

ν12+ν13ν32

C13=

E1

Δ

ν31+ν21ν32=

E3

Δ

ν13+ν12ν23

C22=

E2

Δ

1−ν13ν31

C23=

E2

Δ

ν32+ν12ν31=

E3

Δ

ν23+ν13ν21C33=

E3

Δ

1−ν12ν21

C44=G23C55=G13C66=G12

Δ=1−ν12ν21−ν23ν32−ν13ν31−2ν21ν32ν13
 424 

ENGINEERING CONSTANTS OF A SPECIALLY 

ORTHOTROPIC MATERIAL - CONCLUDED

" And, where the thermal moduli are given by

β 11=−E1

α111−ν23ν32+α22ν21+ν23ν31+α33ν31+ν21ν32

1−ν12ν21−ν23ν32−ν13ν31−2ν21ν32ν13

β 22=−E2

α11ν12+ν13ν32+α221–ν13ν31+α33ν32+ν12ν31

1−ν12ν21−ν23ν32−ν13ν31−2ν21ν32ν13

β 33=−E3

α11ν13+ν12ν23+α22ν23+ν13ν21+α331−ν12ν21

1−ν12ν21−ν23ν32−ν13ν31−2ν21ν32ν13
 425 

ENGINEERING CONSTANTS OF A TRANSVERSELY 

ISOTROPIC MATERIAL

" Consider the constitutive equations for a specially orthotropic material

 and

ε11

ε22

ε33

2ε23

2ε13

2ε12

=

S11

S12S13000

S12

S22S23000

S13

S23S33000

000S44

00

0000S55

0

00000S66

σ11

σ22

σ33

σ23

σ13

σ12

+

α11

α22

α33

0

0

0

T−Tref

ε11

ε22

ε33

2ε23

2ε13

2ε12

=

1

E1

−

ν21

E2

−

ν31

E3

000

−

ν12

E1

1

E2

−

ν32

E3

000

−

ν13

E1

−

ν23

E2

1

E3

000

000

1

G23

00

0000

1

G13

0

00000

1

G12

σ11

σ22

σ33

σ23

σ13

σ12

+

α11

α22

α33

0

0

0

T−Tref
 426 

ENGINEERING CONSTANTS OF A TRANSVERSELY 

ISOTROPIC MATERIAL - CONTINUED

" For a transversely isotropic material with the plane of isotropy given by

x

3 = 0, the constitutitve equations have the following forms

and

σ11

σ22

σ33

σ23

σ13

σ12

=

C11C12C13000

C12

C11C13000

C13

C13C33000

000C44

00

0000C44

0

00000 1

2

C11−C12

ε11

ε22

ε33

2ε23

2ε13

2ε12

+

β 11

β 11

β 33

0

0

0

T−Tref

ε11

ε22

ε33

2ε23

2ε13

2ε12

=

S11S12S13000

S12

S11S13000

S13

S13S33000

000S44

00

0000S44

0

000002S11

−S12

σ11

σ22

σ33

σ23

σ13

σ12

+

α11

α11

α33

0

0

0

T−Tref
 427 

ENGINEERING CONSTANTS OF A TRANSVERSELY 

ISOTROPIC MATERIAL - CONTINUED

" Using the conditions on the compliance coefficients required for a 

transversely isotropic material, the matrix equation for specially 

orthotropic materials, given in terms of engineering constants, can be 

expressed as

 where , , , ,

, , and 

ε11

ε22

ε33

2ε23

2ε13

2ε12

=

1

E

−

ν

E

−

ν31

E′ 0

00

−

ν

E

1

E

−

ν31

E′ 0

00

−

ν13

E

−

ν13

E

1

E′ 0

00

000

1

G′ 0

0

0000

1

G′ 0

00000

1

G

σ11

σ22

σ33

σ23

σ13

σ12

+

α

α

α′

0

0

0

T−Tref

E1 =E2 ≡Eν12=ν21≡ν

G12≡G=

E

21+ν

α11=α22≡α

E3 ≡E′ G13=G23≡G′ α33≡α′
 428 

ENGINEERING CONSTANTS OF A TRANSVERSELY 

ISOTROPIC MATERIAL - CONTINUED

" In this matrix equation; are the Young’s modulus, the 

Poisson’s ratio, the shear modulus, and the coefficient of thermal 

expansion of the material in the plane of isotropy

" Note that characterizes contractions in the plane of 

isotropy that result from only tensile stresses applied in that plane

" are the Young’s modulus, the shear modulus, and the 

coefficient of thermal expansion of the material in the plane 

perpendicular to the plane of isotropy

" For only a tensile stress applied perpendicular to the plane of 

isotropy, the contractions are characterized by

" Symmetry of the compliance matrix yields 

E,ν, G, andα

ν12=ν21≡ν

E′, G′, andα′

σ33

ε11

ε33

=

ε22

ε33

=−ν31≡−ν′

ν13=ν23=ν′ E

E′
 429 

ENGINEERING CONSTANTS OF A TRANSVERSELY 

ISOTROPIC MATERIAL - CONTINUED

" Finally, the constitutive equations for a transversely isotropic material,

with the plane of isotropy given by x

3 = 0, are expressed b

y

" The five independent elastic constants are 

" The two independent thermal parameters are 

ε11

ε22

ε33

2ε23

2ε13

2ε12

=

1

E

−

ν

E

−

ν′

E′ 0

00

−

ν

E

1

E

−

ν′

E′ 0

00

−

ν′

E′ −

ν′

E′

1

E′ 0

00

000

1

G′ 0

0

0000

1

G′ 0

00000

1

G

σ11

σ22

σ33

σ23

σ13

σ12

+

α

α

α′

0

0

0

T−Tref

E,ν, E′, ν′, andG′

α andα′
 430 

ENGINEERING CONSTANTS OF A TRANSVERSELY 

ISOTROPIC MATERIAL - CONCLUDED

" The inverted form of the previous matrix constitutive equation is given 

by

 where

σ11

σ22

σ33

σ23

σ13

σ12

=

C11C12C13000

C12

C11C13000

C13

C13C33000

000G′ 00

0000G′ 0

00000G

ε11

ε22

ε33

2ε23

2ε13

2ε12

+

β 11

β 11

β 33

0

0

0

T−Tref

C11=

E

1+ν

1−

E

E′

ν′

2

1−

ν

−2

E

E′

ν′

2 C12

=

E

1+ν

ν

+

E

E′

ν′

2

1−

ν

−2

E

E′

ν′

2 C13

=

ν′E

1−

ν

−2

E

E′

ν′

2

C33=

E′ 1

−ν

1−

ν

−2

E

E′

ν′

2 β 11

=−

Eα

1+

α′

α

ν′

1−

ν

−2

E

E′

ν′

2 β 33

=−

Eα

2

ν′ +

E′α′ E

α

1−

ν

1−

ν

−2

E

E′

ν′

2
 431 

ENGINEERING CONSTANTS OF A GENERALLY 

ORTHOTROPIC MATERIAL

" Results presented herein indicate that, for an coordinate 

frame, the engineering constants of an anisotropic material are given by

x1,x2,x3

E1 =

1

S11

E2 =

1

S22

E3 =

1

S33

G23=

1

S44

G13=

1

S55

G12=

1

S66

η23, 1 =

S14

S11

η13, 1 =

S15

S11

η12, 1 =

S16

S11

η23, 2 =

S24

S22

η13, 2 =

S25

S22

η12, 2 =

S26

S22

η23, 3 =

S34

S33

η13, 3 =

S35

S33

η12, 3 =

S36

S33

ν12=−

S12

S11

ν13=−

S13

S11

ν23=−

S23

S22
 432 

ENGINEERING CONSTANTS OF A GENERALLY 

ORTHOTROPIC MATERIAL - CONTINUED

 and

η1, 23 =

S14

S44

η2, 23 =

S24

S44

η3, 23 =

S34

S44

η1, 13 =

S15

S55

η2, 13 =

S25

S55

η3, 13 =

S35

S55

η1, 12 =

S16

S66

η2, 12 =

S26

S66

η3, 12 =

S36

S66

μ23, 13 =

S45

S55

μ23, 12 =

S46

S66

μ13, 12 =

S56

S66

μ13, 23 =

S45

S44

μ12, 23 =

S46

S44

μ12, 13 =

S56

S55
 433 

ENGINEERING CONSTANTS OF A GENERALLY 

ORTHOTROPIC MATERIAL - CONTINUED

" Using these expressions, the nonzero transformed elastic compliances 

for a specially orthotropic material transformed by a dextral rotation

about the x

3 - axis 

m= cosθ3 andn= sinθ3 become

S1′1′ =

1

E1

m4+m2n2 E2

G12

E1

E2

−2ν12+n4E1

E2

S1′3′ =−

ν13

E1

m2+n2 ν23

ν13

E1

E2

S2′2′ =

1

E2

m4+m2n2 E2

G12

−2ν12

E2

E1

+n4E2

E1

S2′3′ =−

ν23

E2

m2+n2 ν13

ν23

E2

E1

S1′2′ =−

ν12

E1

m4+n4−m2n2 1

ν12

1+

E1

E2

−

E2

G12

E1

E2

S3′3′ =

1

E3

S1′6′ = mn

E1

m2−n2 E2

G12

E1

E2

−2ν12+2n2E1

E2

−m2
 434 

ENGINEERING CONSTANTS OF A GENERALLY 

ORTHOTROPIC MATERIAL - CONTINUED

 and

S2′6′ = mn

E2

n2−m2 E2

G12

−2ν12

E2

E1

+2m2−n2E2

E1

S3′6′ =

2mn

E3

E3

E1

ν13−ν23

E1

E2

S4′4′ =

1

G23

m2+n2G23

G13

S4′5′ = mn

G23

1−

G23

G13

S5′5′ =

1

G13

m2+n2G13

G23

S6′6′ =

1

G12

4m2n2G12

E2

E2

E1

1+2ν12+

E1

E2

+m2−n2 2
 435 

ENGINEERING CONSTANTS OF A GENERALLY 

ORTHOTROPIC MATERIAL - CONTINUED

" The matrix form of the constitutive equations is given by

ε1′1′

ε2′2′

ε3′3′

2ε2′3′

2ε1′3′

2ε1′2′

=

S1′1′ S1′2′ S1′3′ 0

0S1′6′

S1′2′ S2′2′ S2′3′ 00S2′6′

S1′3′ S2′3′ S3′3′ 00S3′6′

000S4′4′ S4′5′ 0

000S4′5′ S5′5′ 0

S1′6′ S2′6′ S3′6′ 00S6′6′

σ1′1′

σ2′2′

σ3′3′

σ2′3′

σ1′3′

σ1′2′

+

α1′1′

α2′2′

α3′3′

0

0

2α1′2′

T–Tref
 436 

ENGINEERING CONSTANTS OF A GENERALLY 

ORTHOTROPIC MATERIAL - CONTINUED

" Next,

by inspection of the previously given engineering constants for an 

anisotropic material, it follows that, for an coordinate frame,

the corresponding engineering constants are given by

x1′,x2′,x3′

E1′ =

1

S1′1′

ν1′2′ =−

S1′2′

S1′1′

G2′3′ =

1

S

4′4′

η2′3′, 1′ =

S1′4′

S1′1′

η1′3′, 1′ =

S1′5′

S1′1′

η1′2′, 1′ =

S1′6′

S1′1′

E2′ =

1

S2′2′

ν1′3′ =−

S1′3′

S1′1′

G1′3′ =

1

S5′5′

η2′3′, 2′ =

S2′4′

S2′2′

η1′3′, 2′ =

S2′5′

S2′2′

η1′2′, 2′ =

S2′6′

S2′2′

E3′ =

1

S3′3′

ν2′3′ =−

S2′3′

S2′2′

G1′2′ =

1

S6′6′

η2′3′, 3′ =

S3′4′

S3′3′

η1′3′, 3′ =

S3′5′

S3′3′

η1′2′, 3′ =

S3′6′

S3′3′
 437 

ENGINEERING CONSTANTS OF A GENERALLY 

ORTHOTROPIC MATERIAL - CONTINUED

 and

" It is important to note that because the transformed compliance matrix 

has the same structure as a monoclinic material, the conditions 

 are valid

η1′, 2′3′ =

S1′4′

S

4′4′

η2′, 2′3′ =

S2′4′

S

4′4′

η3′, 2′3′ =

S3′4′

S

4′4′

μ1′3′, 2′3′ =

S

4′5′

S

4′4′

μ1′2′, 2′3′ =

S

4′6′

S

4′4′

η1′, 1′3′ =

S1′5′

S5′5′

η2′, 1′3′ =

S2′5′

S5′5′

η3′, 1′3′ =

S3′5′

S5′5′

μ2′3′, 1′3′ =

S

4′5′

S5′5′

μ1′2′, 1′3′ =

S5′6′

S5′5′

η1′, 1′2′ =

S1′6′

S6′6′

η2′, 1′2′ =

S2′6′

S6′6′

η3′, 1′2′ =

S3′6′

S6′6′

μ2′3′, 1′2′ =

S

4′6′

S6′6′

μ1′3′, 1′2′ =

S5′6′

S6′6′

S1′4′ =S1′5′ =S2′4′ =S2′5′ =S3′4′ =S3′5′ =S

4′6′ =

S5′6′ =0
 438 

ENGINEERING CONSTANTS OF A GENERALLY 

ORTHOTROPIC MATERIAL- CONTINUED

" Thus becomes

ε1′1′

ε2′2′

ε3′3′

2ε2′3′

2ε1′3′

2ε1′2′

=

S1′1′ S1′2′ S1′3′ 00S1′6′

S1′2′ S2′2′ S2′3′ 00S2′6′

S1′3′ S2′3′ S3′3′ 00S3′6′

000S4′4′ S4′5′ 0

000S4′5′ S5′5′ 0

S1′6′ S2′6′ S3′6′ 0

0S6′6′

σ1′1′

σ2′2′

σ3′3′

σ2′3′

σ1′3′

σ1′2′

+

α1′1′

α2′2′

α3′3′

0

0

2α1′2′

T−Tref

ε1′1′

ε2′2′

ε3′3′

2ε2′3′

2ε1′3′

2ε1′2′

=

1

E1′

−

ν2′1′

E2′

−

ν3′1′

E3′

00

η1′,1′2′

G1′2′

−

ν1′2′

E1′

1

E2′

−

ν3′2′

E3′

00

η2′,1′2′

G1′2′

−

ν1′3′

E1′

−

ν2′3′

E2′

1

E3′

00

η3′,1′2′

G1′2′

000

1

G2′3′

μ2′3′, 1′3′

G1′3′

0

000

μ1′3′, 2′3′

G2′3′

1

G1′3′

0

η1′2′, 1′

E1′

η1′2′, 2′

E2′

η1′2′, 3′

E3′

00

1

G1′2′

σ1′1′

σ2′2′

σ3′3′

σ2′3′

σ1′3′

σ1′2′

+

α1′1′

α2′2′

α3′3′

0

0

2α1′2′

T−Tref
 439 

ENGINEERING CONSTANTS OF A GENERALLY 

ORTHOTROPIC MATERIAL - CONTINUED

" By using the expressions on the previous few pages, the effective 

elastic and shear moduli of a generally orthotropic material are 

given by

E1′

E1

=m4+m2n2 E2

G12

E1

E2

−2ν12+n4E1

E2

− 1

E2′

E2

=m4+m2n2 E2

G12

−2ν12

E2

E1

+n4E2

E1

− 1

E3′

E3

=1

G2′3′

G23

=m2+n2G23

G13

− 1

G1′3′

G13

=m2+n2G13

G23

− 1

G1′2′

G12

=4m2n2G12

E2

E2

E1

1+2ν12+

E1

E2

+m2−n2 2

− 1
 440 

ENGINEERING CONSTANTS OF A GENERALLY 

ORTHOTROPIC MATERIAL - CONTINUED

" Similarly, the effective Poisson’s ratios of a generally orthotropic 

material are given by

ν1′2′

ν12

=

m4

+n4−m2n2 1

ν12

1+

E1

E2

−

E2

G12

E1

E2

m4+m2n2 E2

G12

E1

E2

−2ν12+n4E1

E2

ν1′3′

ν13

=

m2

+n2 ν23

ν13

E1

E2

m4+m2n2 E2

G12

E1

E2

−2ν12+n4E1

E2

ν2′3′

ν23

=

m2

+n2 ν13

ν23

E2

E1

m4+m2n2 E2

G12

−2ν12

E2

E1

+n4E2

E1
 441 

ENGINEERING CONSTANTS OF A GENERALLY 

ORTHOTROPIC MATERIAL - CONTINUED

" Similarly, the nonzero effective coefficients of mutual influence 

of the first kind for a generally orthotropic material are given by

η1′, 1′2′ =mn

G12

E2

E2

E1

m2−n2 E2

G12

E1

E2

−2ν12+2n2E1

E2

−m2

4m2n2G12

E2

E2

E1

1+2ν12+

E1

E2

+m2−n2 2

η2′, 1′2′ =mn

G12

E2

n2−m2 E2

G12

−2ν12

E2

E1

+2m2−n2E2

E1

4m2n2G12

E2

E2

E1

1+2ν12+

E1

E2

+m2−n2 2

η3′, 1′2′ =2mn

G12

E2

E2

E1

E3

E1

2 ν13

−ν23

E1

E2

4m2n2G12

E2

E2

E1

1+2ν12+

E1

E2

+m2−n2 2
 442 

ENGINEERING CONSTANTS OF A GENERALLY 

ORTHOTROPIC MATERIAL - CONTINUED

" The nonzero effective coefficients of mutual influence of the 

second kind for a generally orthotropic material are given by

η1′2′, 1′ =mn

m2

−n2 E2

G12

E1

E2

−2ν12+2n2E1

E2

−m2

m4+m2n2 E2

G12

E1

E2

−2ν12+n4E1

E2

η1′2′, 2′ =mn

n2

−m2 E2

G12

−2ν12

E2

E1

+2m2−n2E2

E1

m4+m2n2 E2

G12

−2ν12

E2

E1

+n4E2

E1

η1′2′, 3′ =2mn

E3

E1

ν13−ν23

E1

E2
 443 

ENGINEERING CONSTANTS OF A GENERALLY 

ORTHOTROPIC MATERIAL - CONCLUDED

" The nonzero effective Chentsov coefficients for a generally 

orthotropic material are given by

μ1′3′, 2′3′ =mn

1−

G23

G13

m2+n2G23

G13

μ2′3′, 1′3′ =mn

G13

G23

1−

G23

G13

m2+n2G13

G23
 444 

REDUCED CONSTITUTIVE

EQUATIONS
 445 

CONSTITUTIVE EQUATIONS FOR PLANE STRESS

" When, analyzing solids that are relatively flat and thin, simplifying 

assumptions are made about the stress state to facilitate analytical 

solution of practical problems

" One such assumption is that 

the stresses in a thin, flat body,

that are normal to the plane of 

flatness, are negligible 

compared to the other 

stresses 

" This simplification is 

commonly referred to as the 

plane-stress assumption

" For a state of plane stress in a homogeneous, anisotropic solid, with 

respect to the x1 - x

2 plane, the stress field is appr

oximated such that 

x1

x2

x3

σ33=σ23=σ13=0
 446 

CONSTITUTIVE EQUATIONS FOR PLANE STRESS

CONTINUED

" For this special case, the general matrix constitutive equation

 uncouples directly into

 and

ε11

ε22

ε33

2

ε23

2

ε13

2

ε12

=

S11

S12S13S14S15S16

S12S22S23S24S25S26

S13S23S33S34S35S36

S14S24S34S44S45S46

S15S25S35S45S55S56

S16S26S36S46S56S66

σ11

σ22

σ33

σ23

σ13

σ12

+

α11

α22

α33

2

α23

2

α13

2

α12

T−Tref

ε11

ε22

2

ε12

=

S11

S12S16

S12S22S26

S16S26S66

σ11

σ22

σ12

+

α11

α22

2

α12

T−Tref

ε33

2

ε23

2

ε13

=

S13

S23S36

S14S24S46

S15S25S56

σ11

σ22

σ12

+

α33

2

α23

2

α13

T−Tref
 447 

CONSTITUTIVE EQUATIONS FOR PLANE STRESS

CONTINUED

" In terms of the engineering constants, the plane-stress constitutive 

equations for a homogeneous, anisotropic material are given b

y

ε11

ε22

2

ε12

=

1

E1

−

ν21

E2

η1,12

G12

−

ν12

E1

1

E2

η2,12

G12

η12,1

E1

η12,2

E2

1

G12

σ11

σ22

σ12

+

α11

α22

2

α12

T−Tref

ε33

2

ε23

2

ε13

=

−

ν13

E1

−

ν23

E2

η3,12

G12

η23, 1

E1

η23, 2

E2

μ23, 12

G12

η13, 1

E1

η13, 2

E2

μ13, 12

G12

σ11

σ22

σ12

+

α33

2

α23

2

α13

T−Tref
 448 

CONSTITUTIVE EQUATIONS FOR PLANE STRESS

CONTINUED

" However, simplification of the following general constitutive equation is 

not as easy

" First, the equation given above is expressed as

σ11

σ22

σ33

σ23

σ13

σ12

=

C11

C12C13C14C15C16

C12C22C23C24C25C26

C13C23C33C34C35C36

C14C24C34C44C45C46

C15C25C35C45C55C56

C16C26C36C46C56C66

ε11

ε22

ε33

2

ε23

2

ε13

2

ε12

+

β 11

β 22

β 33

β 23

β 13

β 12

T−Tref

σ11

σ22

σ33

σ23

σ13

σ12

=

C11

C12C13C14C15C16

C12C22C23C24C25C26

C13C23C33C34C35C36

C14C24C34C44C45C46

C15C25C35C45C55C56

C16C26C36C46C56C66

ε11

ε22

ε33

2

ε23

2

ε13

2

ε12

−

α11

α22

α33

2

α23

2

α13

2

α12

T−Tref
 449 

CONSTITUTIVE EQUATIONS FOR PLANE STRESS

CONTINUED

" Then, using gives

" Rearranging the rows and columns into a convenient form gives

σ33=σ23=σ13=0

σ11

σ22

0

0

0

σ12

=

C11

C12C13C14C15C16

C12C22C23C24C25C26

C13C23C33C34C35C36

C14C24C34C44C45C46

C15C25C35C45C55C56

C16C26C36C46C56C66

ε11

ε22

ε33

2

ε23

2

ε13

2

ε12

−

α11

α22

α33

2

α23

2

α13

2

α12

T−Tref

σ11

σ22

σ12

0

0

0

=

C11

C12C16C13C14C15

C12C22C26C23C24C25

C16C26C66C36C46C56

C13C23C36C33C34C35

C14C24C46C34C44C45

C15C25C56C35C45C55

ε11

ε22

2

ε12

ε33

2

ε23

2

ε13

−

α11

α22

2

α12

α33

2

α23

2

α13

T−Tref
 450 

CONSTITUTIVE EQUATIONS FOR PLANE STRESS

CONTINUED

" For convenience, let the mechanical strains be denoted by

" The previous matrix constitutive equation becomes

ε11

σ

ε22

σ

2

ε12

σ

ε33

σ

2

ε23

σ

2

ε13

σ

=

ε11

ε22

2

ε12

ε33

2

ε23

2

ε13

−

α11

α22

2

α12

α33

2

α23

2

α13

T−Tref

σ11

σ22

σ12

0

0

0

=

C11

C12C16C13C14C15

C12C22C26C23C24C25

C16C26C66C36C46C56

C13C23C36C33C34C35

C14C24C46C34C44C45

C15C25C56C35C45C55

ε11

σ

ε22

σ

2

ε12

σ

ε33

σ

2

ε23

σ

2

ε13

σ
 451 

CONSTITUTIVE EQUATIONS FOR PLANE STRESS

CONTINUED

" Next, the matrix constitutive equation

 is separated to get

 and

σ11

σ22

σ12

0

0

0

=

C11

C12C16C13C14C15

C12C22C26C23C24C25

C16C26C66C36C46C56

C13C23C36C33C34C35

C14C24C46C34C44C45

C15C25C56C35C45C55

ε11

σ

ε22

σ

2

ε12

σ

ε33

σ

2

ε23

σ

2

ε13

σ

σ11

σ22

σ12

=

C11

C12C16

C12C22C26

C16C26C66

ε11

σ

ε22

σ

2

ε12

σ

+

C13

C14C15

C23C24C25

C36C46C56

ε33

σ

2

ε23

σ

2

ε13

σ

0

0

0

=

C13

C23C36

C14C24C46

C15C25C56

ε11

σ

ε22

σ

2

ε12

σ

+

C33

C34C35

C34C44C45

C35C45C55

ε33

σ

2

ε23

σ

2

ε13

σ
 452 

CONSTITUTIVE EQUATIONS FOR PLANE STRESS

CONTINUED

" Solving the previous homogeneous equation for gives

" Back substitution of the column vector containing into 

 yields the result

ε33

σ , 2

ε23

σ

, and

2

ε13

σ

ε33

σ

2

ε23

σ

2

ε13

σ

=−

C33C34C35

C34C44C45

C35C45C55

− 1

C13C23C36

C14C24C46

C15C25C56

ε11

σ

ε22

σ

2

ε12

σ

ε33

σ , 2

ε23

σ

, and

2

ε13

σ

σ11

σ22

σ12

=

C11

C12C16

C12C22C26

C16C26C66

ε11

σ

ε22

σ

2

ε12

σ

+

C13

C14C15

C23C24C25

C36C46C56

ε33

σ

2

ε23

σ

2

ε13

σ

σ11

σ22

σ12

=

Q11

Q12Q16

Q12Q22Q26

Q16Q26Q66

ε11

σ

ε22

σ

2

ε12

σ
 453 

CONSTITUTIVE EQUATIONS FOR PLANE STRESS

CONTINUED

" The matrix with the subscripted Q terms is given by

" Next, expressing the mechanical strains in terms of the total strains 

and the strains caused by free thermal expansion results in

" This equation is manipulated further by defining

Q11Q12Q16

Q12Q22Q26

Q16Q26Q66

=

C11

C12C16

C12C22C26

C16C26C66

−

C13C14C15

C23C24C25

C36C46C56

C33C34C35

C34C44C45

C35C45C55

− 1

C13C23C36

C14C24C46

C15C25C56

σ11

σ22

σ12

=

Q11

Q12Q16

Q12Q22Q26

Q16Q26Q66

ε11

ε22

2

ε12

−

α11

α22

2

α12

T−Tref

β 11

β 22

β 12

=−

Q11

Q12Q16

Q12Q22Q26

Q16Q26Q66

α11

α22

2

α12
 454 

CONSTITUTIVE EQUATIONS FOR PLANE STRESS

CONTINUED

" Thus, the constitutive equations for plane stress, in terms of 

stiffness coefficients and thermal moduli, become

 where

 and 

" The and are called the reduced stiffness coefficients and 

reduced thermal moduli, respectively

σ11

σ22

σ12

=

Q11

Q12Q16

Q12Q22Q26

Q16Q26Q66

ε11

ε22

2

ε12

+

β 11

β 22

β 12

T−Tref

Q11Q12Q16

Q12Q22Q26

Q16Q26Q66

=

C11

C12C16

C12C22C26

C16C26C66

−

C13C14C15

C23C24C25

C36C46C56

C33C34C35

C34C44C45

C35C45C55

− 1

C13C23C36

C14C24C46

C15C25C56

β 11

β 22

β 12

=−

Q11

Q12Q16

Q12Q22Q26

Q16Q26Q66

α11

α22

2

α12

Qij βij
 455 

CONSTITUTIVE EQUATIONS FOR PLANE STRESS

CONTINUED

" The relationship between the reduced stiffnesses and the compliances

is obtained by first considering the previously derived equation

" Inverting gives

ε11

ε22

2

ε12

=

S11

S12S16

S12S22S26

S16S26S66

σ11

σ22

σ12

+

α11

α22

2

α12

T−Tref

σ11

σ22

σ12

=

S11S12S16

S12S22S26

S16S26S66

− 1

ε11

ε22

2

ε12

−

α11

α22

2

α12

T−Tref
 456 

CONSTITUTIVE EQUATIONS FOR PLANE STRESS

CONTINUED

" Comparing 

 with

 indicates that

σ11

σ22

σ12

=

S11S12S16

S12S22S26

S16S26S66

− 1

ε11

ε22

2

ε12

−

α11

α22

2

α12

T−Tref

σ11

σ22

σ12

=

Q11

Q12Q16

Q12Q22Q26

Q16Q26Q66

ε11

ε22

2

ε12

+

β 11

β 22

β 12

T−Tref

Q11Q12Q16

Q12Q22Q26

Q16Q26Q66

=

S11

S12S16

S12S22S26

S16S26S66

− 1
 457 

CONSTITUTIVE EQUATIONS FOR PLANE STRESS

CONTINUED

" The relationship between the reduced thermal moduli and the regular 

thermal moduli is obtained by first considering the equation

" This matrix equation can be separated into

 and 

β 11

β 22

β 33

β 23

β 13

β 12

=−

C11C12C13C14C15C16

C12C22C23C24C25C26

C13C23C33C34C35C36

C14C24C34C44C45C46

C15C25C35C45C55C56

C16C26C36C46C56C66

α11

α22

α33

2

α23

2

α13

2

α12

β 11

β 22

β 12

=−

C11

C12C16

C12C22C26

C16C26C66

α11

α22

2

α12

−

C13

C14C15

C23C24C25

C36C46C56

α33

2

α23

2

α13

β 33

β 23

β 13

=−

C13

C23C36

C14C24C46

C15C25C56

α11

α22

2

α12

−

C33

C34C35

C34C44C45

C35C45C55

α33

2

α23

2

α13
 458 

CONSTITUTIVE EQUATIONS FOR PLANE STRESS

CONTINUED

" Solving the matrix equation

 for the column vector containing gives

β 33

β 23

β 13

=−

C13

C23C36

C14C24C46

C15C25C56

α11

α22

2

α12

−

C33

C34C35

C34C44C45

C35C45C55

α33

2

α23

2

α13

2

α13, 2α23, andα33

α33

2

α23

2

α13

=−

C33C34C35

C34C44C45

C35C45C55

− 1

β 33

β 23

β 13

−

C33

C34C35

C34C44C45

C35C45C55

− 1

C13C23C36

C14C24C46

C15C25C56

α11

α22

2

α12
 459 

CONSTITUTIVE EQUATIONS FOR PLANE STRESS

CONTINUED

" Substituting

 into

 gives

α33

2

α23

2

α13

=−

C33C34C35

C34C44C45

C35C45C55

− 1

β 33

β 23

β 13

−

C33

C34C35

C34C44C45

C35C45C55

− 1

C13C23C36

C14C24C46

C15C25C56

α11

α22

2

α12

β 11

β 22

β 12

=−

C11

C12C16

C12C22C26

C16C26C66

α11

α22

2

α12

−

C13

C14C15

C23C24C25

C36C46C56

α33

2

α23

2

α13

β 11

β 22

β 12

−

C13C14C15

C23C24C25

C36C46C56

C33C34C35

C34C44C45

C35C45C55

− 1

β 33

β 23

β 13

=

−

C11

C12C16

C12C22C26

C16C26C66

−

C13C14C15

C23C24C25

C36C46C56

C33C34C35

C34C44C45

C35C45C55

− 1

C13C23C36

C14C24C46

C15C25C56

α11

α22

2

α12
 460 

CONSTITUTIVE EQUATIONS FOR PLANE STRESS

CONTINUED

" Noting that

,

 gives

Q11Q12Q16

Q12Q22Q26

Q16Q26Q66

=

C11

C12C16

C12C22C26

C16C26C66

−

C13C14C15

C23C24C25

C36C46C56

C33C34C35

C34C44C45

C35C45C55

− 1

C13C23C36

C14C24C46

C15C25C56

β 11

β 22

β 12

−

C13C14C15

C23C24C25

C36C46C56

C33C34C35

C34C44C45

C35C45C55

− 1

β 33

β 23

β 13

=

−

C11

C12C16

C12C22C26

C16C26C66

−

C13C14C15

C23C24C25

C36C46C56

C33C34C35

C34C44C45

C35C45C55

− 1

C13C23C36

C14C24C46

C15C25C56

α11

α22

2

α12

β 11

β 22

β 12

−

C13C14C15

C23C24C25

C36C46C56

C33C34C35

C34C44C45

C35C45C55

− 1

β 33

β 23

β 13

=−

Q11Q12Q16

Q12Q22Q26

Q16Q26Q66

α11

α22

2

α12
 461 

CONSTITUTIVE EQUATIONS FOR PLANE STRESS

CONTINUED

" Finally, noting that

, the expression

 gives

β 11

β 22

β 12

=−

Q11

Q12Q16

Q12Q22Q26

Q16Q26Q66

α11

α22

2

α12

β 11

β 22

β 12

−

C13C14C15

C23C24C25

C36C46C56

C33C34C35

C34C44C45

C35C45C55

− 1

β 33

β 23

β 13

=−

Q11Q12Q16

Q12Q22Q26

Q16Q26Q66

α11

α22

2

α12

β 11

β 22

β 12

=

β 11

β 22

β 12

−

C13C14C15

C23C24C25

C36C46C56

C33C34C35

C34C44C45

C35C45C55

−1

β 33

β 23

β 13
 462 

 CONSTITUTIVE EQUATIONS FOR PLANE STRESS

CONTINUED

" For a material that is monoclinic, with x

3 = 0 being a plane of reflective 

symmetry, the plane-stress constitutive equations become

 and

" The last matrix equation reduces to 

ε11

ε22

2

ε12

=

S11

S12S16

S12S22S26

S16S26S66

σ11

σ22

σ12

+

α11

α22

2

α12

T−Tref

ε33

2

ε23

2

ε13

=

S13

S23S36

000

000

σ11

σ22

σ12

+

α33

0

0

T−Tref

ε33=

S13

σ11+

S23

σ22+

S36

σ12+α33

T−Tref
 463 

 CONSTITUTIVE EQUATIONS FOR PLANE STRESS

CONTINUED

" In terms of engineering constants,

ε11

ε22

2

ε12

=

1

E1

−

ν21

E2

η1,12

G12

−

ν12

E1

1

E2

η2,12

G12

η12,1

E1

η12,2

E2

1

G12

σ11

σ22

σ12

+

α11

α22

2

α12

T−Tref

ε33=

−

ν13

E1

σ11

−

ν23

E2

σ22+

η12, 3

E3

σ33+α33

T−Tref
 464 

CONSTITUTIVE EQUATIONS FOR PLANE STRESS

CONTINUED

" Similarly, for a material that is monoclinic, with x

3 = 0 being a plane of 

reflective symmetry, the plane-stress constitutive matrix becomes

 which simplifies to

Q11Q12Q16

Q12Q22Q26

Q16Q26Q66

=

C11

C12C16

C12C22C26

C16C26C66

−

C1300

C2300

C3600

C3300

0C44C45

0C45C55

− 1

C13C23C36

000

000

Q11Q12Q16

Q12Q22Q26

Q16Q26Q66

=

C11

−

C13

C13

C33

C12−

C13

C23

C33

C16−

C13

C36

C33

C12−

C13

C23

C33

C22−

C23

C23

C33

C26−

C23

C36

C33

C16−

C13

C36

C33

C26−

C23

C36

C33

C66−

C36

C36

C33
 465 

CONSTITUTIVE EQUATIONS FOR PLANE STRESS

CONTINUED

" The remaining matrix equation for the strains, given by

 becomes

 which reduces to

 and 

ε33

σ

2

ε23

σ

2

ε13

σ

=−

C33C34C35

C34C44C45

C35C45C55

− 1

C13C23C36

C14C24C46

C15C25C56

ε11

σ

ε22

σ

2

ε12

σ

ε33

σ

2

ε23

σ

2

ε13

σ

=−

C3300

0C44C45

0C45C55

− 1

C13C23C36

000

000

ε11

σ

ε22

σ

2

ε12

σ

ε33

σ

=−

1

C33

C13

ε11

σ

+C23

ε22

σ

+2C36

ε12

σ

2

ε23

σ

=

2

ε13

σ

=0
 466 

CONSTITUTIVE EQUATIONS FOR PLANE STRESS

CONTINUED

" Expressions for the reduced stiffnesses in terms of engineering 

constants are obtained directly by using the previously obtained results

 and 

" Inverting the matrix of compliances yields

S11S12S16

S12S22S26

S16S26S66

=

1

E1

−

ν21

E2

η1,12

G12

−

ν12

E1

1

E2

η2,12

G12

η12,1

E1

η12,2

E2

1

G12

Q11Q12Q16

Q12Q22Q26

Q16Q26Q66

=

S11

S12S16

S12S22S26

S16S26S66

− 1

Q11=

E1

Δ

1−η12,2η2,12 Q12=

E1

Δ

ν21+η12,2η1,12 =

E2

Δ

ν12+η12,1η2,12

Q16=−

E1

Δ

η1,12 +ν21η2,12 =−

G12

Δ

η12,1 +ν12η12,2 Q22

=

E2

Δ

1−η12,1η1,12

Q26=−

E2

Δ

η2,12 +ν12η1,12 =−

G12

Δ

η12,2 +ν21η12,1 Q66=

G12

Δ

1−ν12ν21

Δ=1−ν12ν21−η12,1 η1,12 +ν21η2,12 −η12,2 η2,12 +ν12η1,12
 467 

CONSTITUTIVE EQUATIONS FOR PLANE STRESS

CONTINUED

" Similarly, the expression gives

β 11

β 22

β 12

=−

Q11

Q12Q16

Q12Q22Q26

Q16Q26Q66

α11

α22

2

α12

β 11=−

1

Δ

E1α111−η12,2η2,12 +E2α22ν12+η12,1η2,12 −2G12

α12

η12,1 +ν12η12,2

β 22=−

1

Δ

E1α11ν21+η12,2η1,12 +E2α221−η12,1η1,12 −2G12

α12

η12,2 +ν21η12,1

β 12=

1

Δ

E1α11η1,12 +ν21η2,12 +E2α22η2,12 +ν12η1,12 −2G12

α12

1−ν12ν21
 468 

 CONSTITUTIVE EQUATIONS FOR PLANE STRESS

CONTINUED

" For a specially orthotropic material, the plane-stress constitutive 

equations become

 and

 and 

" In terms of engineering constants,

 and 

ε11

ε22

2

ε12

=

S11

S120

S12S220

00S66

σ11

σ22

σ12

+

α11

α22

0

T−Tref

ε33=

S13

σ11+

S23

σ22+α33

T−Tref

2

ε23

=

2

ε13

=0

ε11

ε22

2

ε12

=

1

E1

−

ν21

E2

0

−

ν12

E1

1

E2

0

00

1

G12

σ11

σ22

σ12

+

α11

α22

0

T−Tref

ε33=

−

ν13

E1

σ11

−

ν23

E2

σ22+α33

T−Tref

2

ε23

=

2

ε13

=0
 469 

CONSTITUTIVE EQUATIONS FOR PLANE STRESS

CONTINUED

" Similarly, for a specially orthotropic material, the plane-stress 

constitutive equations become

 and

" The last matrix equation gives

 and 

σ11

σ22

σ12

=

C11

C120

C12

C220

00C66

ε11

σ

ε22

σ

2

ε12

σ

+

C13

00

C23

00

000

ε33

σ

2

ε23

σ

2

ε13

σ

0

0

0

=

C13

C230

000

000

ε11

σ

ε22

σ

2

ε12

σ

+

C33

00

0C44

0

00C55

ε33

σ

2

ε23

σ

2

ε13

σ

ε33

σ

=−

1

C33

C13

ε11

σ

+C23

ε22

σ

2

ε23

=

2

ε13

=0
 470 

CONSTITUTIVE EQUATIONS FOR PLANE STRESS

CONTINUED

" Using the previous expressions for , the constitutive 

equations for plane stress, in terms of stiffness coefficients, become

 where

ε33

σ , 2

ε23

σ

, and

2

ε13

σ

σ11

σ22

σ12

=

Q11

Q120

Q12

Q220

00Q66

ε11

σ

ε22

σ

2

ε12

σ

=

Q11

Q120

Q12

Q220

00Q66

ε11

ε22

2

ε12

−

α11

α22

0

T−Tref

Q11Q120

Q12Q220

00Q66

=

C11

−

C13

C13

C33

C12−

C13

C23

C33

0

C12

−

C13

C23

C33

C22−

C23

C23

C33

0

00C66
 471 

CONSTITUTIVE EQUATIONS FOR PLANE STRESS

CONTINUED

" Next, noting that 

 for a specially orthotropic solid

 gives 

" Now, note that the general expression for the regular thermal moduli 

simplifies to

β 11

β 22

β 12

=−

Q11

Q120

Q12

Q220

00Q66

α11

α22

0

σ11

σ22

σ12

=

Q11

Q120

Q12

Q220

00Q66

ε11

ε22

2

ε12

+

β 11

β 22

0

T−Tref

β 11

β 22

β 33

β 23

β 13

β 12

=−

C11C12C13000

C12C22C23000

C13C23C33000

000C44

00

0000C55

0

00000C66

α11

α22

α33

0

0

0
 472 

CONSTITUTIVE EQUATIONS FOR PLANE STRESS

CONTINUED

" In addition,

 simplifies to

 which yields

β 11

β 22

β 12

=

β 11

β 22

β 12

−

C13C14C15

C23C24C25

C36C46C56

C33C34C35

C34C44C45

C35C45C55

− 1

β 33

β 23

β 13

β 11

β 22

β 12

=

β 11

β 22

0

−

C1300

C2300

000

C3300

0C440

00C55

− 1

β 33

0

0

β 11=β 11

−

C13

C33

β 33β 22=β 22

−

C23

C33

β 33

β 12=

0
 473 

CONSTITUTIVE EQUATIONS FOR PLANE STRESS

CONCLUDED

" In terms of the engineering constants, the reduced stiffness 

coefficients and are given by

" Similarly, the nonzero reduced thermal moduli are given by

Q11=

E1

1−ν12ν21

Q12=

ν12

E2

1−ν12ν21

=

ν21

E1

1−ν12ν21

Q22=

E2

1−ν12ν21

Q66=G12

β 11=−E1

α11+α22ν21

1−ν12ν21

β 22=−E2

α22+α11ν12

1−ν12ν21
 474 

STRESS AND STRAIN TRANSFORMATION EQUATIONS 

FOR PLANE STRESS

" Often, the relationship between the planar stresses and strains that are 

defined relative to two different coordinate systems is needed

" Consider the dextral (right-handed) rotation of coordinate frames 

shown in the figure

i 1 i 2

i 3

x1

x2

x3

x1′

x2′

x3′

i 1′

i 2′

i 3′

,

,

θ3

θ3
 475 

STRESS AND STRAIN TRANSFORMATION EQUATIONS 

FOR PLANE STRESS - CONTINUED

" Previously, the matrix form of the stress-transformation law for this 

specific transformation was given as

" Substituting into this equation yields

and 

σ1′1′

σ2′2′

σ3′3′

σ2′3′

σ1′3′

σ1′2′

=

cos2

θ3 sin2θ3 0

002sin

θ3cosθ3

sin2θ3 cos2θ3 0

00−2sin

θ3cosθ3

001000

000 cos

θ3 − sinθ3 0

000 sin

θ3 cosθ3 0

− sin

θ3cosθ3 sinθ3cosθ3 0

00 cos2

θ3 − sin2θ3

σ11

σ22

σ33

σ23

σ13

σ12

σ33=σ13=σ23=0

σ1′1′

σ2′2′

σ1′2′

=

cos2

θ3 sin2θ3 2sinθ3cosθ3

sin2θ3 cos2θ3 −

2sin

θ3cosθ3

− sin

θ3cosθ3 sinθ3cosθ3 cos2θ3 − sin2θ3

σ11

σ22

σ12

σ3′3′

σ2′3′

σ1′3′

=

0

0

0
 476 

STRESS AND STRAIN TRANSFORMATION EQUATIONS 

FOR PLANE STRESS - CONTINUED

" Thus, the stress-transformation lawfor a state of plane stress and a 

dextral rotation about the x

3 axis is given b

y

" This law is expressed symbolically by

 where , , and 

σ1′1′

σ2′2′

σ1′2′

=

cos2

θ3 sin2θ3 2sinθ3cosθ3

sin2θ3 cos2θ3 −

2sin

θ3cosθ3

− sin

θ3cosθ3 sinθ3cosθ3 cos2θ3 − sin2θ3

σ11

σ22

σ12

Σ′ =T

σ

θ

3

Σ

Σ′ ≡

σ1′1′

σ2′2′

σ1′2′

Σ≡

σ11

σ22

σ12

T

σ

θ

3

≡

cos2θ3 sin2θ3 2sinθ3cosθ3

sin2θ3 cos2θ3 −

2sin

θ3cosθ3

− sin

θ3cosθ3 sinθ3cosθ3 cos2θ3 − sin2θ3
 477 

STRESS AND STRAIN TRANSFORMATION EQUATIONS 

FOR PLANE STRESS - CONTINUED

" Similarly, the matrix form of the inverse stress-tensor transformation 

law was also given previously as

" Following a similar process gives where

σ11

σ22

σ33

σ23

σ13

σ12

=

cos2

θ3 sin2θ3 0

00

−

2sin

θ3cosθ3

sin2θ3 cos2θ3 0

002sin

θ3cosθ3

001000

000 cos

θ3 sinθ3 0

000

− sinθ3 cosθ3 0

sinθ3cosθ3 − sinθ3cosθ3 0

00 cos2

θ3 − sin2θ3

σ1′1′

σ2′2′

σ3′3′

σ2′3′

σ1′3′

σ1′2′

Σ=T

σ

θ

3

− 1

Σ′

T

σ

θ

3

− 1

=

T

σ

–

θ

3

=

cos2θ3 sin2θ3 −

2sin

θ3cosθ3

sin2θ3 cos2θ3 2sinθ3cosθ3

sinθ3cosθ3 − sinθ3cosθ3 cos2θ3 − sin2θ3
 478 

STRESS AND STRAIN TRANSFORMATION EQUATIONS 

FOR PLANE STRESS - CONTINUED

" Previously, the matrix form of the strain-transformation law for this 

specific transformation was given as

" This matrix equation can be partitioned into the following parts

,

 and 

ε1′1′

ε2′2′

ε3′3′

2ε2′3′

2ε1′3′

2ε1′2′

=

cos2θ3 sin2θ3 0

00 sin

θ3cosθ3

sin2θ3 cos2θ3 0

00− sin

θ3cosθ3

00 1 000

000 cos

θ3 − sinθ3 0

000 sin

θ3 cosθ3 0

−2sin

θ3cosθ3 2sinθ3cosθ3 0

00 cos2

θ3 − sin2θ3

ε11

ε22

ε33

2ε23

2ε13

2ε12

ε1′1′

ε2′2′

2

ε1′2′

=

cos2θ3 sin2θ3 sinθ3cosθ3

sin2θ3 cos2θ3 − sinθ3cosθ3

−2sin

θ3cosθ3 2sinθ3cosθ3 cos2θ3 − sin2θ3

ε11

ε22

2

ε12

2

ε2′3′

2

ε1′3′

=

cosθ3 − sinθ3

sinθ3 cosθ3

2

ε23

2

ε13

ε3′3′ =ε33
 479 

STRESS AND STRAIN TRANSFORMATION EQUATIONS 

FOR PLANE STRESS - CONTINUED

" The first law is expressed symbolically by

 where , , and 

" A useful relationship that is easily verified is given by

E′ =T

ε

θ

3

E

E′ ≡

ε1′1′

ε2′2′

2

ε1′2′

E≡

ε11

ε22

2

ε12

T

ε

θ

3

≡

cos2θ3 sin2θ3 sinθ3cosθ3

sin2θ3 cos2θ3 − sinθ3cosθ3

−2sin

θ3cosθ3 2sinθ3cosθ3 cos2θ3 − sin2θ3

T

ε

θ

3

T

=

T

σ

θ

3

− 1

=

T

σ

−

θ

3
 480 

STRESS AND STRAIN TRANSFORMATION EQUATIONS 

FOR PLANE STRESS - CONCLUDED

" Similarly, the matrix form of the inverse strain-transformation law was 

also given previously as

" From this matrix equation, it follows that 

 where 

ε11

ε22

ε33

2ε23

2ε13

2ε12

=

cos2θ3 sin2θ3 0

00

− sinθ3cosθ3

sin2θ3 cos2θ3 0

00 sin

θ3cosθ3

001000

000 cos

θ3 sinθ3 0

000

− sinθ3 cosθ3 0

2sinθ3cosθ3 −2sinθ3cosθ3 0

00 cos2

θ3 − sin2θ3

ε1′1′

ε2′2′

ε3′3′

2ε2′3′

2ε1′3′

2ε1′2′

E

=

T

ε

θ

3

− 1

E′

T

ε

θ

3

− 1

=

T

ε

−

θ

3

=

T

σ

θ

3

T
 481 

TRANSFORMED CONSTITUTIVE EQUATIONS FOR 

PLANE STRESS

" The two matrix equations that resulted from the reduction for a state of 

plane stress are

 and

, where

 and 

" Also, and generally 

ε11

ε22

2

ε12

=

S11

S12S16

S12S22S26

S16S26S66

σ11

σ22

σ12

+

α11

α22

2

α12

T−Tref

σ11

σ22

σ12

=

Q11

Q12Q16

Q12Q22Q26

Q16Q26Q66

ε11

ε22

2

ε12

+

β 11

β 22

β 12

T−Tref

β 11

β 22

β 12

=−

Q11

Q12Q16

Q12Q22Q26

Q16Q26Q66

α11

α22

2

α12

Q11Q12Q16

Q12Q22Q26

Q16Q26Q66

=

S11

S12S16

S12S22S26

S16S26S66

− 1

σ33=σ13=σ23=0ε33≠ε13≠ε23≠0
 482 

TRANSFORMED CONSTITUTIVE EQUATIONS FOR 

PLANE STRESS - CONTINUED

" In terms of another set of coordinates that correspond to a 

dextral rotation about the x

3 axis, the constitutive equations must also 

have the forms given as

 and

, where

 and 

x1′, x2′, x3′

ε1′1′

ε2′2′

2

ε1′2′

=

S1′1′ S1′2′ S1′6′

S1′2′ S2′2′ S2′6′

S1′6′ S2′6′ S6′6′

σ1′1′

σ2′2′

σ1′2′

+

α1′1′

α2′2′

2

α1′2′

T−Tref

σ1′1′

σ2′2′

σ1′2′

=

Q1′1′ Q1′2′ Q1′6′

Q1′2′ Q2′2′ Q2′6′

Q1′6′ Q2′6′ Q6′6′

ε1′1′

ε2′2′

2

ε1′2′

+

β 1′1′

β 2′2′

β 1′2′

T−Tref

β 1′1′

β 2′2′

β 1′2′

=−

Q1′1′ Q1′2′ Q1′6′

Q1′2′ Q2′2′ Q2′6′

Q1′6′ Q2′6′ Q6′6′

α1′1′

α2′2′

2

α1′2′

Q1′1′ Q1′2′ Q1′6′

Q1′2′ Q2′2′ Q2′6′

Q1′6′ Q2′6′ Q6′6′

=

S1′1′ S1′2′ S1′6′

S1′2′ S2′2′ S2′6′

S1′6′ S2′6′ S6′6′

− 1
 483 

TRANSFORMED CONSTITUTIVE EQUATIONS FOR 

PLANE STRESS - CONTINUED

" For convenience, let

S=

S11S12S16

S12S22S26

S16S26S66

Q=

Q11Q12Q16

Q12Q22Q26

Q16Q26Q66

α=

α11

α22

2

α12

β=

β 11

β 22

β 12

S′ =

S1′1′ S1′2′ S1′6′

S1′2′ S2′2′ S2′6′

S1′6′ S2′6′ S6′6′

Q′ =

Q1′1′ Q1′2′ Q1′6′

Q1′2′ Q2′2′ Q2′6′

Q1′6′ Q2′6′ Q6′6′

α′ =

α1′1′

α2′2′

2

α1′2′

β′ =

β 1′1′

β 2′2′

β 1′2′
 484 

TRANSFORMED CONSTITUTIVE EQUATIONS FOR 

PLANE STRESS - CONTINUED

" With this notation, the two sets of thermoelastic constitutive equations 

are expressed in symbolic form by

 and 

 and 

 where 

" By using the matrix form of the stress and strain transformation 

equations for plane stress, becomes

E=SΣ+αΘ

Σ=QE+βΘ

E′ =S′ Σ′ +α′ Θ

Σ′ =Q′ E′ +β′ Θ

Θ

≡T−Tref

Σ=QE+βΘ

T

σ

–1

Σ′ =Q

T

ε

–1

E′ +βΘ
 485 

TRANSFORMED CONSTITUTIVE EQUATIONS FOR 

PLANE STRESS - CONTINUED

" Premultiplying by gives

" Comparing this equation with it follows 

that

 and 

" Rearranging gives 

T

σ

Σ′ =

T

σ

Q

T

ε

–1

E′ +

T

σ

βΘ

Σ′ =Q′ E′ +β′ Θ

Q′ =

T

σ

Q

T

ε

–1

β′ =

T

σ

β

Q′ =

T

σ

Q

T

ε

–1

Q=

T

σ

–1

Q′ T

ε
 486 

TRANSFORMED CONSTITUTIVE EQUATIONS FOR 

PLANE STRESS - CONTINUED

" Next, by using the matrix form of the stress and strain transformation 

equations for plane stress, becomes

" Premultiplying by gives

" By comparing this equation with it follows 

that

 and 

E=SΣ+αΘ

T

ε

–1

E′ =

ST

σ

–1

Σ′ +αΘ

T

ε

E′ =

T

ε

ST

σ

–1

Σ′ +

T

ε

αΘ

E′ =S′ σ′ +α′ Θ

S′ =T

ε

ST

σ

–1

α′ =

T

ε

α
 487 

TRANSFORMED CONSTITUTIVE EQUATIONS FOR 

PLANE STRESS - CONTINUED

" Rearranging gives 

" Noting that for a dextral rotation about the x

3 axis,

 and it follows that

S′ =Tε ST

σ

–1

S

=

T

ε

–1

S′ T

σ

T

ε

–1

=T

σ

T

T

σ

–1

=T

ε

T

S′ =T

ε

ST

ε

T

S

=

T

σ

T

S′ T

σ

Q′ =T

σ

QT

σ

T

Q

=

T

ε

T

Q′ T

ε

α′ =

T

ε

α

α=

T

σ

T

α′

β′ =

T

σ

β

β=

T

ε

T

β′
 488 

TRANSFORMED CONSTITUTIVE EQUATIONS FOR 

PLANE STRESS - CONTINUED

 with

T

σ

θ

3

≡ and

cos2θ3 sin2θ3 2sinθ3cosθ3

sin2θ3 cos2θ3 −

2sin

θ3cosθ3

− sin

θ3cosθ3 sinθ3cosθ3 cos2θ3 − sin2θ3

T

ε

θ

3

≡

cos2θ3 sin2θ3 sinθ3cosθ3

sin2θ3 cos2θ3 − sinθ3cosθ3

−2sin

θ3cosθ3 2sinθ3cosθ3 cos2θ3 − sin2θ3
 489 

TRANSFORMED CONSTITUTIVE EQUATIONS FOR 

PLANE STRESS - CONTINUED

"

Q′ =T

σ

QT

σ , with and , yields

T

m= cosθ3 n= sinθ3

Q1′1′ =m4Q11+2m2n2Q12+2Q66+4mnm2Q16+n2Q26+n4Q22

Q1′2′ =m2n2Q11+Q22−4Q66−2mnm2−n2Q16−Q26+m4+n4Q12

Q1′6′ =m2m2−3n2Q16−m3nQ11−Q12−2Q66

+mn3Q22−Q12−2Q66−n2n2−3m2Q26

Q2′2′ =m4Q22+2m2n2Q12+2Q66−4mnm2Q26+n2Q16+n4Q11

Q2′6′ =m2m2−3n2Q26+m3nQ22−Q12−2Q66

−mn3Q11−Q12−2Q66−n2n2−3m2Q16

Q6′6′ =m2n2Q11+Q22−2Q12−2mnm2−n2Q16−Q26+m2−n2 2Q66
 490 

TRANSFORMED CONSTITUTIVE EQUATIONS FOR 

PLANE STRESS - CONTINUED

"

Q

=

T

ε , with and , yields

T

Q′ T

ε

m= cosθ3 n= sinθ3

Q11=m4Q1′1′ +2m2n2Q1′2′ +2Q6′6′ −4mnm2Q1′6′ +n2Q2′6′ +n4Q2′2′

Q12=m2n2Q1′1′ +Q2′2′ −4Q6′6′ +2mnm2−n2Q1′6′ −Q2′6′ +m4+n4Q1′2′

Q16=m2m2−3n2Q1′6′ +m3nQ1′1′ −Q1′2′ −2Q6′6′

−mn3Q2′2′ −Q1′2′ −2Q6′6′ −n2n2−3m2Q2′6′

Q22=m4Q2′2′ +2m2n2Q1′2′ +2Q6′6′ +4mnm2Q2′6′ +n2Q1′6′ +n4Q1′1′

Q26=m2m2−3n2Q2′6′ −m3nQ2′2′ −Q1′2′ −2Q6′6′

+mn3Q1′1′ −Q1′2′ −2Q6′6′ −n2n2−3m2Q1′6′

Q66=m2n2Q1′1′ +Q2′2′ −2Q1′2′ +2mnm2−n2Q1′6′ −Q2′6′ +m2−n2 2Q6′6′
 491 

TRANSFORMED CONSTITUTIVE EQUATIONS FOR 

PLANE STRESS - CONTINUED

"

S′ =T

ε

ST

ε , with and , yields

T

m= cosθ3 n= sinθ3

S1′1′ =m4S11+m2n22S12+S66+2mnm2S16+n2S26+n4S22

S1′2′ =m2n2S11+S22−S66− mnm2−n2S16−S26+m4+n4S12

S1′6′ =m2m2−3n2S16−m3n2S11−2S12−S66

+mn32S22−2S12−S66−n2n2−3m2S26

S2′2′ =m4S22+m2n22S12+S66−2mnm2S26+n2S16+n4S11

S2′6′ =m2m2−3n2S26+m3n2S22−2S12−S66

−mn32S11−2S12−S66−n2n2−3m2S16

S6′6′ =4m2n2S11+S22−2S12−4mnm2−n2S16−S26+m2−n2 2S66
 492 

TRANSFORMED CONSTITUTIVE EQUATIONS FOR 

PLANE STRESS - CONTINUED

"

S

=

T

σ , with and , yields

T

S′ T

σ

m= cosθ3 n= sinθ3

S11=m4S1′1′ +m2n22S1′2′ +S6′6′ −2mnm2S1′6′ +n2S2′6′ +n4S2′2′

S12=m2n2S1′1′ +S2′2′ −S6′6′ +mnm2−n2S1′6′ −S2′6′ +m4+n4S1′2′

S16=m2m2−3n2S1′6′ +m3n2S1′1′ −2S1′2′ −S6′6′

−mn32S2′2′ −2S1′2′ −S6′6′ −n2n2−3m2S2′6′

S22=m4S2′2′ +m2n22S1′2′ +S6′6′ +2mnm2S2′6′ +n2S1′6′ +n4S1′1′

S26=m2m2−3n2S2′6′ −m3n2S2′2′ −2S1′2′ −S6′6′

+mn32S1′1′ −2S1′2′ −S6′6′ −n2n2−3m2S1′6′

S66=4m2n2S1′1′ +S2′2′ −2S1′2′ +4mnm2−n2S1′6′ −S2′6′ +m2−n2 2S6′6′
 493 

TRANSFORMED CONSTITUTIVE EQUATIONS FOR 

PLANE STRESS - CONTINUED

" Note that and can be expressed as 

 and 

" Thus, one set of transformed stiffness expressions can be 

obtained from the other by simply interchanging the primed and 

unprimed indices and replacing n with -n

" Likewise, and can be expressed as 

 and 

" Thus, one set of transformed compliance expressions can be 

obtained from the other by simply interchanging the primed and 

unprimed indices and replacing n with -n

Q′ Q

Q′ =T

σ

θ3 Q

Tε −

θ3 Q

=T

σ

−θ3 Q′ Tε θ3

S′ S

S′ =Tε θ3 ST

σ

−θ3 S

=

Tε −

θ3 S′ T

σ

θ3
 494 

TRANSFORMED CONSTITUTIVE EQUATIONS FOR 

PLANE STRESS - CONTINUED

" , with and , yields

" Similarly, gives

α′ =

T

ε

α

m= cosθ3 n= sinθ3

α1′1′ =m2α11+2mnα12+n2α22α2′2′ =m2α22−2mnα12+n2α11

α1′2′ =m2−n2α12+mnα22−α11

α=

T

σ

T

α′

α11=m2α1′1′ +2mnα1′2′ +n2α2′2′ α22=m2α2′2′ −2mnα1′2′ +n2α1′1′

α12=m2−n2α1′2′ +mnα2′2′ −α1′1′
 495 

TRANSFORMED CONSTITUTIVE EQUATIONS FOR 

PLANE STRESS - CONTINUED

" , with and , yields

" Similarly, gives

β′ =

T

σ

β

m= cosθ3 n= sinθ3

β 1′1′ =m2β 11+2mnβ 12+n2β 22β 2′2′ =m2β 22−2mnβ 12+n2β 11

β 1′2′ =m2−n2β 12+mnβ 22−β 11

β=

T

ε

T

β′

β 11=m2β 1′1′ −2mnβ 1′2′ +n2β 2′2′ β 22=m2β 2′2′ +2mnβ 1′2′ +n2β 1′1′

β 12=m2−n2β 1′2′ − mnβ 2′2′ −β 1′1′
 496 

TRANSFORMED CONSTITUTIVE EQUATIONS FOR 

PLANE STRESS - CONCLUDED

" Note that and can be expressed as 

 and 

" Thus, one set of transformed thermal-expansion expressions can 

be obtained from the other by simply interchanging the primed and 

unprimed indices and replacing n with -n

" Likewise, and can be expressed as 

 and 

" Thus, one set of transformed thermal-compliance expressions can 

be obtained from the other by simply interchanging the primed and 

unprimed indices and replacing n with -n

α′ α

α′ =

Tε θ3 α

α=

Tε −

θ3 α′

β′ β

β′ =

T

σ

θ3 β

β=

T

σ

−θ3 β′
 497 

CONSTITUTIVE EQUATIONS FOR GENERALIZED 

PLANE STRESS

" When relatively thin plates, with uniform thickness h, are supported 

and subjected to inplane loads such that the dominant stresses,

strains, and displacements act only in planes parallel to the plane x

3 = 0 

shown in the figure, significant simplifications can be made to the 

equations governing the elastic response

" Moreover, these dominant response 

quantities are presumed to vary

symmetrically through the plate 

thickness, given by ,

such that no significant bending 

deformations are exhibited by the 

plate

" In contrast to the plane stress approximations previously presented 

herein, when these conditions exist, with respect to the through-thethickness variations, the plate response is described as a state of 

generalized plane stress

x1

x2

x3

h

−

h

2

≤x3 ≤+

h

2
 498 

CONSTITUTIVE EQUATIONS FOR GENERALIZED 

PLANE STRESS - CONTINUED

" For a state of generalized plane stress in an anisotropic solid, with 

respect to the plane x

3 = 0, the displacement fields in the 

x1-,x

2-, and 

x

3

-

coordinate directions are approximated by averaging the through-thethickness variations as follows

 and 

"To exclude bending deformations, is required for the 

out-of-plane displacement field

" Note that allowing , where is a constant 

satisfies 

U1 x1, x2 =

1

h

u1 x1, x2, x3 d

x3

–

h

2

+

h

2

U2 x1, x2 =

1

h

u2 x1, x2, x3 d

x3

–

h

2

+

h

2

U3 x1, x2 =

1

h

u3 x1, x2, x3 d

x3

–

h

2

+

h

2

=0

U3 x1, x2 =0

u3 x1, x2, x3 =x3 ε

ε

U3 x1, x2 =0
 499 

CONSTITUTIVE EQUATIONS FOR GENERALIZED 

PLANE STRESS - CONTINUED

" The strain-displacement relations of the linear theory of elasticity, given 

by , are approximated as follows

" First, average strains are defined by

" Substituting into this expression and using 

the definitions for the average displacements yields

, and 

" Thus,ε represent a uniform through-the-thickness normal strain

εij

x1, x2, x3 =

1

2

∂ui

∂xj

+

∂uj

∂xi

εij

x1, x2 =

1

h

εij

x1, x2, x3 d

x3

–

h

2

+

h

2

εij

x1, x2, x3 =

1

2

∂ui

∂xj

+

∂uj

∂xi

εij

x1, x2 =

1

2

∂Ui

∂xj

+

∂Uj

∂xi

2

ε13

=2

ε23

=0

ε33

=

ε
 500 

CONSTITUTIVE EQUATIONS FOR GENERALIZED 

PLANE STRESS - CONTINUED

" Next, the stress field is approximated such that 

" For this case, the stresses in a thin, flat body, that are normal to the 

plane x

3 = 0, are presumed negligible compared to the other 

stresses 

" In addition, average stresses are defined by

" The conditions on the presumed stress field are satisfied by the 

previously derived plane-stress constitutive equations given in the form

 and

σ33=σ23=σ13=0

σij

x1, x2 =

1

h

σij

x1, x2, x3 d

x3

–

h

2

+

h

2

σ11

σ22

σ12

=

Q11Q12Q16

Q12Q22Q26

Q16Q26Q66

ε11

ε22

2

ε12

−

α11

α22

2

α12

T−Tref
 501 

CONSTITUTIVE EQUATIONS FOR GENERALIZED 

PLANE STRESS - CONTINUED

" That is, substituting into the general form of the 

constitutive equations and simplifying yields the same plane stress 

constitutive equation given above

ε11

ε22

2

ε12

=

S11

S12S16

S12S22S26

S16S26S66

σ11

σ22

σ12

+

α11

α22

2

α12

T−Tref

ε33

2

ε23

2

ε13

=

S13

S23S36

S14S24S46

S15S25S56

σ11

σ22

σ12

+

α33

2

α23

2

α13

T−Tref

σ33=σ23=σ13=0
 502 

CONSTITUTIVE EQUATIONS FOR GENERALIZED 

PLANE STRESS - CONTINUED

" Restricting the plate to homogeneous construction, integration of the 

constitutive equations through the plate thickness yields

" Next, the temperature change T - Tref is presumed to vary symmetrically 

through the plate thickness so as not to cause bending deformations

" The average temperature change is defined as

 such that the constitutive equations become

σ11

σ22

σ12

=

Q11Q12Q16

Q12Q22Q26

Q16Q26Q66

ε11

ε22

2

ε12

−

Q11

Q12Q16

Q12Q22Q26

Q16Q26Q66

α11

α22

2

α12

1

h

T−Tref

dx3

–

h

2

+

h

2

Θ

x1, x2 =

1

h

T−Tref

dx3

–

h

2

+

h

2
 503 

CONSTITUTIVE EQUATIONS FOR GENERALIZED 

PLANE STRESS - CONTINUED

" Similarly, and

" Inspection of the last matrix equation reveals that the material must be 

monoclinic, with the plane x

3 = 0 being a plane of reflective symmetr

y, in 

order to satisfy the kinematic hypothesis and 

σ11

σ22

σ12

=

Q11Q12Q16

Q12Q22Q26

Q16Q26Q66

ε11

ε22

2

ε12

−

α11

α22

2

α12

Θ

ε11

ε22

2

ε12

=

S11

S12S16

S12S22S26

S16S26S66

σ11

σ22

σ12

+

α11

α22

2

α12

Θ

ε33

ε23

2

ε13

=

S13

S23S36

S14S24S46

S15S25S56

σ11

σ22

σ12

+

α33

2

α23

2

α13

Θ

2

ε13

=2

ε23

=0

ε33

=

ε
 504 

CONSTITUTIVE EQUATIONS FOR GENERALIZED 

PLANE STRESS - CONTINUED

" Note that the symmetry requirement on the material properties is 

consistent with the symmetry requirements imposed up front on the 

displacements, stresses, and strains

" Thus, becomes

, whichreducesto

ε33

ε23

2

ε13

=

S13

S23S36

S14S24S46

S15S25S56

σ11

σ22

σ12

+

α33

2

α23

2

α13

Θ

ε33

ε23

2

ε13

=

S13

S23S36

000

000

σ11

σ22

σ12

+

α33

0

0

Θ

ε

=

S13

σ11+

S23

σ22+

S36

σ12+α33

Θ
 505 

CONSTITUTIVE EQUATIONS FOR GENERALIZED 

PLANE STRESS - CONTINUED

" In terms of engineering constants, and

" Now consider the previously derived stress-transformation law for 

a state of plane stress and a dextral rotation about the x

3 axis is given 

by

" This law was expressed symbolically by

ε

=

−

ν13

E1

σ11

−

ν23

E2

σ22+

η12, 3

E3

σ33+α33

Θ

ε11

ε22

2

ε12

=

1

E1

−

ν21

E2

η1,12

G12

−

ν12

E1

1

E2

η2,12

G12

η12,1

E1

η12,2

E2

1

G12

σ11

σ22

σ12

+

α11

α22

2

α12

Θ

σ1′1′

σ2′2′

σ1′2′

=

cos2

θ3 sin2θ3 2sinθ3cosθ3

sin2θ3 cos2θ3 −

2sin

θ3cosθ3

− sin

θ3cosθ3 sinθ3cosθ3 cos2θ3 − sin2θ3

σ11

σ22

σ12

Σ′ =T

σ

θ

3

Σ
 506 

CONSTITUTIVE EQUATIONS FOR GENERALIZED 

PLANE STRESS - CONTINUED

 where , , and 

" Integrating the stress transformation law over the plate thickness yields

 the trans formation law , where 

 and 

Σ′ ≡

σ1′1′

σ2′2′

σ1′2′

Σ≡

σ11

σ22

σ12

T

σ

θ

3

≡

cos2θ3 sin2θ3 2sinθ3cosθ3

sin2θ3 cos2θ3 −

2sin

θ3cosθ3

− sin

θ3cosθ3 sinθ3cosθ3 cos2θ3 − sin2θ3

Σ′ =T

σ

θ

3

Σ

Σ≡

1

h

σ11

σ22

σ12

dx3

–

h

2

+

h

2

Σ′ ≡

1

h

σ1′1′

σ2′2′

σ1′2′

dx3

–

h

2

+

h

2
 507 

CONSTITUTIVE EQUATIONS FOR GENERALIZED 

PLANE STRESS - CONTINUED

" Likewise, the previously derived strain-transformation law for a 

state of plane stress and a dextral rotation about the x

3 axis is given b

y

 where , , and 

E′ =T

ε

θ

3

E

E′ ≡

ε1′1′

ε2′2′

2

ε1′2′

E≡

ε11

ε22

2

ε12

T

ε

θ

3

≡

cos2θ3 sin2θ3 sinθ3cosθ3

sin2θ3 cos2θ3 − sinθ3cosθ3

−2sin

θ3cosθ3 2sinθ3cosθ3 cos2θ3 − sin2θ3
 508 

CONSTITUTIVE EQUATIONS FOR GENERALIZED 

PLANE STRESS - CONCLUDED

" Integrating the strain transformation law over the plate thickness yields

 the transformation law , where 

 and 

" Comparing the stress and strain transformation equations and the 

constitutive equations for generalized plane stress with the 

corresponding equations for plane stress, it is seen that they have 

identical structure

" As a result, the constitutive equations for generalized plane 

stress transform in exactly the same way as those for plane 

stress

E′ =T

ε

θ

3

E

E≡

1

h

ε11

ε22

2

ε12

dx3

–

h

2

+

h

2

E′ ≡

1

h

ε1′1′

ε2′2′

2

ε1′2′

dx3

–

h

2

+

h

2
 509 

CONSTITUTIVE EQUATIONS FOR INPLANE 

DEFORMATIONS OF THIN PLATES

" Another practical case of interest that is similar to generalized plane 

stress is the case of thin, nonhomogeneous plates that undergo inplane 

deformations without any bending deformations

" For this case, the stresses in a thin,

flat body, that are normal to the x1 - 

x

2

plane shown in the figure, are 

presumed negligible compared to the 

other stresses 

" Thus, the stress field is 

approximated such that ,

, and 

" But, , , and are permitted 

because of through-the thickness nonhomogeneity that is presumed to 

exist

x1

x2

x3

h

−

h

2

≤x3 ≤+

h

2

σ13=0

σ23=0σ33=0

σ11=σ11

x1,x2,x3 σ22

=σ22

x1,x2,x3 σ12

=σ12

x1,x2,x3
 510 

CONSTITUTIVE EQUATIONS FOR INPLANE 

DEFORMATIONS OF THIN PLATES - CONTINUED

" In addition, the plate-like body is required to have a uniform thickness

h and is not allowed to bend when subjected to inplane loads

" The conditions on the stress field are satisfied by the previously 

derived plane-stress constitutive equations given in the form

" In the present formulation, the total strains are presumed to be uniform 

through the plate thickness; that is,

, , and 

" However, the plate is allowed to be nonhomogeneous through the 

thickness such that the reduced stiffness coefficients and the 

coefficients of thermal expansion can vary with the x

3 coordinate

σ11

σ22

σ12

=

Q11Q12Q16

Q12Q22Q26

Q16Q26Q66

ε11

ε22

2

ε12

−

α11

α22

2

α12

T−Tref

ε11=ε11

x1,x2 ε22

=ε22

x1,x2 ε12

=ε12

x1,x2
 511 

CONSTITUTIVE EQUATIONS FOR INPLANE 

DEFORMATIONS OF THIN PLATES - CONTINUED

" Because of the plate’s thinness, the temperature change T - Tref is 

presumed to be uniform through the thickness

" The requirement that the plate not bend is fulfilled by picking the middle 

surface of the plate to correspond to x

3 = 0 and then to require that 

the following integral be valid

" This matrix integral equation then requires the following through-thethickness symmetry conditions on the stress field

, , and

σ11

σ22

σ12

x3 dx3

–

h

2

+

h

2

=

0

0

0

σ11

x1, x2, −x3 =

σ11

x1, x2, x3 σ22x1, x2, −x3 =

σ22

x1, x2, x3

σ12

x1, x2, −x3 =

σ12

x1, x2, x3
 512 

CONSTITUTIVE EQUATIONS FOR INPLANE 

DEFORMATIONS OF THIN PLATES - CONTINUED

" The requirement that the plate not bend also places requirements on 

material properties

" First, note that substituting the plane stress constitutive equation into 

the previously stated integral equation yields

" Next, note that for arbitrary strain and temperature fields, the two 

integrals in the above equation must vanish independently; that is,

 and 

Q11Q12Q16

Q12Q22Q26

Q16Q26Q66

x3 dx3

−

h

2

+

h

2

ε11

ε22

2

ε12

−

Q11Q12Q16

Q12Q22Q26

Q16Q26Q66

α11

α22

2

α12

x3 dx3 T

−Tref

−

h

2

+

h

2

=0

Q11Q12Q16

Q12Q22Q26

Q16Q26Q66

x3 dx3

−

h

2

+

h

2

=0

Q11Q12Q16

Q12Q22Q26

Q16Q26Q66

α11

α22

2

α12

x3 dx3

−

h

2

+

h

2

=0
 513 

CONSTITUTIVE EQUATIONS FOR INPLANE 

DEFORMATIONS OF THIN PLATES - CONTINUED

" The integral requires that the reduced 

stiffness coefficients be symmetric functions about the plane x

3 = 0 

" Thus, the full set of stiffness and compliance coefficients, Cij and 

Sij, must be symmetric about the plane x

3 = 0; i. e., monoclinic

" The integral and the symmetry 

requirement on the reduced stiffness coefficients require that the 

coefficients of thermal expansion be symmetric functions about the 

plane x

3 = 0 

Q11Q12Q16

Q12Q22Q26

Q16Q26Q66

x3 dx3

−

h

2

+

h

2

=0

Q11Q12Q16

Q12Q22Q26

Q16Q26Q66

α11

α22

2

α12

x3 dx3

−

h

2

+

h

2

=0
 514 

CONSTITUTIVE EQUATIONS FOR INPLANE 

DEFORMATIONS OF THIN PLATES - CONTINUED

" Recall, that for a material that is monoclinic with respect to the plane 

x

3 = 0, the general form of the constitutive equations is

ε11

ε22

ε33

2ε23

2ε13

2ε12

=

S11

S12S1300S16

S12S22S2300S26

S13S23S3300S36

000S44S450

000S45

S550

S16

S26S3600S66

σ11

σ22

σ33

σ23

σ13

σ12

+

α11

α22

α33

0

0

2α12

T−Tref

σ11

σ22

σ33

σ23

σ13

σ12

=

C11C12C1300C16

C12C22C2300C26

C13C23C3300C36

000C44C450

000C45

C550

C16C26C3600C66

ε11

ε22

ε33

2ε23

2ε13

2ε12

+

β 11

β 22

β 33

0

0

β 12

T−Tref
 515 

CONSTITUTIVE EQUATIONS FOR INPLANE 

DEFORMATIONS OF THIN PLATES - CONTINUED

" In addition,

 and 

Q11Q12Q16

Q12Q22Q26

Q16Q26Q66

=

C11

−

C13

C13

C33

C12−

C13

C23

C33

C16−

C13

C36

C33

C12−

C13

C23

C33

C22−

C23

C23

C33

C26−

C23

C36

C33

C16−

C13

C36

C33

C26−

C23

C36

C33

C66−

C36

C36

C33

S11S12S16

S12S22S26

S16S26S66

=

1

E1

−

ν21

E2

η1,12

G12

−

ν12

E1

1

E2

η2,12

G12

η12,1

E1

η12,2

E2

1

G12

Q11Q12Q16

Q12Q22Q26

Q16Q26Q66

=

S11

S12S16

S12S22S26

S16S26S66

− 1
 516 

CONSTITUTIVE EQUATIONS FOR INPLANE 

DEFORMATIONS OF THIN PLATES - CONTINUED

"To facilitate the formulation of a two-dimensional boundary-value 

problem, the through-the-thickness functional dependence of the stress 

field is eliminated by introducing the following stress resultants

" Substituting

 into the expressions for the stress resultants and performing the

 integration yields a two-dimensional constitutive equation

N11

N22

N12

≡

σ11

σ22

σ12

dx3

−

h

2

+

h

2

σ11

σ22

σ12

=

Q11Q12Q16

Q12Q22Q26

Q16Q26Q66

ε11

ε22

2

ε12

−

α11

α22

2

α12

T−Tref
 517 

CONSTITUTIVE EQUATIONS FOR INPLANE 

DEFORMATIONS OF THIN PLATES - CONTINUED

" The two-dimensional constitutive equation is given by

 where

 and 

" The Aij and are called the inplane, plate stiffness coefficients and 

thermal moduli, respectively

N11

N22

N12

=

A11A12A16

A12A22A26

A16A26A66

ε11

ε22

2

ε12

+

β 11

β 22

β 12

T−Tref

A11A12A16

A12A22A26

A16A26A66

≡

Q11Q12Q16

Q12Q22Q26

Q16Q26Q66

dx3

−

h

2

+

h

2

β 11

β 22

β 12

≡

−

Q11

Q12Q16

Q12Q22Q26

Q16Q26Q66

α11

α22

2

α12

dx3

−

h

2

+

h

2

βij
 518 

CONSTITUTIVE EQUATIONS FOR INPLANE 

DEFORMATIONS OF THIN PLATES - CONTINUED

" A more convenient form of the two-dimensional constitutive equation is 

obtained by first inverting it to obtain

" Next, the inverted matrix equation is manipulated to look like the planestress constitutive equations

" That is, a set of overall plate coefficients of thermal expansion are 

defined to make the construction of the equations parallel

ε11

ε22

2

ε12

=

A11

A12A16

A12A22A26

A16A26A66

− 1

N11

N22

N12

−

A11

A12A16

A12A22A26

A16A26A66

− 1

β 11

β 22

β 12

T−Tref

ε11

ε22

2

ε12

=

S11

S12S16

S12S22S26

S16S26S66

σ11

σ22

σ12

+

α11

α22

2

α12

T−Tref
 519 

CONSTITUTIVE EQUATIONS FOR INPLANE 

DEFORMATIONS OF THIN PLATES - CONTINUED

" The overall plate coefficients of thermal expansion are defined as

 so that the two-dimensional constitutive equation is given by

 or

α11

α22

2

α12

=

A11

A12A16

A12A22A26

A16A26A66

− 1

β 11

β 22

β 12

=

A11A12A16

A12A22A26

A16A26A66

− 1

Q11Q12Q16

Q12Q22Q26

Q16Q26Q66

α11

α22

2

α12

dx3

−

h

2

+

h

2

N11

N22

N12

=

A11A12A16

A12A22A26

A16A26A66

ε11

ε22

2

ε12

−

α11

α22

2

α12

T−Tref

ε11

ε22

2

ε12

=

A11

A12A16

A12A22A26

A16A26A66

− 1

N11

N22

N12

+

α11

α22

2

α12

T−Tref
 520 

CONSTITUTIVE EQUATIONS FOR INPLANE 

DEFORMATIONS OF THIN PLATES - CONTINUED

" For convenience, let so that

" The subscripted a terms are plate compliances that are given by

 where 

A11A12A16

A12A22A26

A16A26A66

− 1

=

a11

a12a16

a12a22a26

a16a26a66

ε11

ε22

2

ε12

=

a11

a12a16

a12a22a26

a16a26a66

N11

N22

N12

+

α11

α22

2

α12

T−Tref

a11=

A22A66−A26

2

A

a12=

A16

A26−A12A66

A

a16=

A12

A26−A16A22

A

a22=

A11A66−A16

2

A

a26=

A12

A16−A11A26

A

a66=

A11A22−A12

2

A

A=A11A22−A12

2

A66−A11A26

2

−A22A16

2

+2A12A16A26
 521 

CONSTITUTIVE EQUATIONS FOR INPLANE 

DEFORMATIONS OF THIN PLATES - CONTINUED

" Once a given boundary-value problem is solved, the strain fields 

, , and are known

" The stresses at any point of the body are found by substituting the 

strain fields and the coordinates of the given point into

" The other strains are given by

 which reduce to

 and 

ε11

x1,x2 ε22x1,x2 2

ε12

x1,x2

σ11

σ22

σ12

=

Q11Q12Q16

Q12Q22Q26

Q16Q26Q66

ε11

ε22

2

ε12

−

α11

α22

2

α12

T−Tref

ε33

2

ε23

2

ε13

=

S13

S23S36

000

000

σ11

σ22

σ12

+

α33

0

0

T−Tref

2

ε23

=

2

ε13

=0

ε33

=

S13

σ11+

S23

σ22+

S36

σ12+α33

T−Tref
 522 

CONSTITUTIVE EQUATIONS FOR INPLANE 

DEFORMATIONS OF THIN PLATES - CONTINUED

" Now consider a transformation of coordinates that corresponds to a 

dextral rotation about the x

3 axis

" The stress-transformation law for a state of plane stress has been 

given by

Σ′ =T

σ

θ

3

Σ where 

Σ′ ≡ , , and

σ1′1′

σ2′2′

σ1′2′

Σ≡

σ11

σ22

σ12

T

σ

θ

3

≡

cos2θ3 sin2θ3 2sinθ3cosθ3

sin2θ3 cos2θ3 −

2sin

θ3cosθ3

− sin

θ3cosθ3 sinθ3cosθ3 cos2θ3 − sin2θ3
 523 

CONSTITUTIVE EQUATIONS FOR INPLANE 

DEFORMATIONS OF THIN PLATES - CONTINUED

" Integrating the stress transformation law over the plate thickness yields

 the transformation law , where 

 and 

" Likewise, the previously derived strain-transformation law for a 

state of plane stress and a dextral rotation about the x

3 axis is given b

y

 where , ,

N′ =T

σ

θ

3

N

N

≡

N11

N22

N12

≡

σ11

σ22

σ12

dx3

–

h

2

+

h

2

N′ ≡

N1′1′

N2′2′

N1′2′

≡

σ1′1′

σ2′2′

σ1′2′

dx3

–

h

2

+

h

2

E′ =T

ε

θ

3

E

E′ ≡

ε1′1′

ε2′2′

2

ε1′2′

E≡

ε11

ε22

2

ε12
 524 

CONSTITUTIVE EQUATIONS FOR INPLANE 

DEFORMATIONS OF THIN PLATES - CONTINUED

 and ; and 

 remains unchanged for thin, nonhomogeneous plate

" For convenience, let and 

 such that 

T

ε

θ

3

≡

cos2θ3 sin2θ3 sinθ3cosθ3

sin2θ3 cos2θ3 − sinθ3cosθ3

−2sin

θ3cosθ3 2sinθ3cosθ3 cos2θ3 − sin2θ3

A=

A11A12A16

A12A22A26

A16A26A66

Q

=

Q11

Q12Q16

Q12Q22Q26

Q16Q26Q66

A=Q

dx3

−

h

2

+

h

2
 525 

CONSTITUTIVE EQUATIONS FOR INPLANE 

DEFORMATIONS OF THIN PLATES - CONTINUED

" Similarly, let and 

 such that 

" The transformation laws that relate and are obtained by using 

the following plane-stress transformation equations

A′ ≡

A1′1′ A1′2′ A1′6′

A1′2′ A2′2′ A2′6′

A1′6′ A2′6′ A6′6′

Q′ ≡

Q1′1′ Q1′2′ Q1′6′

Q1′2′ Q2′2′ Q2′6′

Q1′6′ Q2′6′ Q6′6′

A′ =Q′ d

x3

−

h

2

+

h

2

A

A′

Q′ =T

σ

QT

σ

T

Q

=

T

ε

T

Q′ T

ε
 526 

CONSTITUTIVE EQUATIONS FOR INPLANE 

DEFORMATIONS OF THIN PLATES - CONTINUED

" Integation of and over the 

plate thickness yields and 

" From , it follows that 

" Noting that and ,

 it follows that 

Q′ =T

σ

QT

σ

T

Q

=

Tε

T

Q′ Tε

A′ =T

σ

AT

σ

T

A

=

Tε

T

A′ Tε

A=Q

dx3

−

h

2

+

h

2

A

− 1

=Q

− 1

dx3

−

h

2

+

h

2

a=

a11a12a16

a12a22a26

a16a26a66

≡

A

− 1

S=

S11S12S16

S12S22S26

S16S26S66

≡

Q

− 1

a=S

dx3

−

h

2

+

h

2
 527 

CONSTITUTIVE EQUATIONS FOR INPLANE 

DEFORMATIONS OF THIN PLATES - CONTINUED

" Similarly , ,

 and 

" The transformation laws that relate and are obtained by using 

the following plane-stress transformation equations

a′ =

a1′1′ a1′2′ a1′6′

a1′2′ a2′2′ a2′6′

a1′6′ a2′6′ a6′6′

≡

A′

− 1

S′ =

S1′1′ S1′2′ S1′6′

S1′2′ S2′2′ S2′6′

S1′6′ S2′6′ S6′6′

≡

Q′

− 1

a′ =S′ d

x3

−

h

2

+

h

2

aa′

S′ =T

ε

ST

ε

T

S

=

T

σ

T

S′ T

σ
 528 

CONSTITUTIVE EQUATIONS FOR INPLANE 

DEFORMATIONS OF THIN PLATES - CONTINUED

" Integration of and over the 

plate thickness yields and 

" Now consider the thermal moduli of the plate given by

" Noting that for plane stress, it 

follows that and that 

S′ =Tε STε

T

S

=

T

σ

T

S′ T

σ

a′ =Tε aTε

T

a

=

T

σ

T

a′ T

σ

β

≡

β 11

β 22

β 12

≡

−

Q11

Q12Q16

Q12Q22Q26

Q16Q26Q66

α11

α22

2

α12

dx3

−

h

2

+

h

2

β≡

β 11

β 22

β 12

=−

Q11

Q12Q16

Q12Q22Q26

Q16Q26Q66

α11

α22

2

α12

β

=

β

dx3

−

h

2

+

h

2

β′ =β′ d

x3

−

h

2

+

h

2
 529 

CONSTITUTIVE EQUATIONS FOR INPLANE 

DEFORMATIONS OF THIN PLATES - CONTINUED

" The transformation laws that relate and are obtained by

using the following plane-stress transformation equations

" Integration of these equations over the plate thickness yields 

 and 

" Next, recall that the overall plate coefficients of thermal expansion

have been given by

ββ′

β′ =

T

σ

β

β=

T

ε

T

β′

β′ =

T

σ

β

β=

T

ε

T

β′

α≡

α11

α22

2

α12

=

a11

a12a16

a12a22a26

a16a26a66

β 11

β 22

β 12

=

aβ
 530 

CONSTITUTIVE EQUATIONS FOR INPLANE 

DEFORMATIONS OF THIN PLATES - CONTINUED

" It follows logically that 

" Substituting and into 

gives 

" Next, using gives 

" Then using gives the result 

" Inverting this result and using gives 

α′ =

α1′1′

α2′2′

2

α1′2′

=

a1′1′ a1′2′ a1′6′

a1′2′ a2′2′ a2′6′

a1′6′ a2′6′ a6′6′

β 1′1′

β 2′2′

β 1′2′

=

a′ β′

a′ =Tε aTε

T

β′ =

T

σ

βα′ =

a′ β′

α′ =

Tε aTε

T

T

σ

β

Tε

T

=

T

σ

− 1

α′ =

Tε a

β

α=

a

β

α′ =

T

ε

α

Tε

− 1

=

T

σ

T

α=

T

σ

T

α′
 531 

CONSTITUTIVE EQUATIONS FOR INPLANE 

DEFORMATIONS OF THIN PLATES - CONTINUED

" In summary:

" Comparing these equations with those of the plane stress case reveals 

that the specific transformation equations can be obtained from those 

given previously for plane stress as follows

A′ =T

σ

AT

σ

T

A

=

T

ε

T

A′ T

ε

a′ =T

ε

aT

ε

T

a

=

T

σ

T

a′ T

σ

β′ =

T

σ

β

β=

T

ε

T

β′

α′ =

T

ε

α

α=

T

σ

T

α′
 532 

CONSTITUTIVE EQUATIONS FOR INPLANE 

DEFORMATIONS OF THIN PLATES - CONCLUDED

" For plane stress, , with and 

, gave

" Thus, by similarity, gives 

" The other transformation equations are obtained in a similar manner

Q′ =T

σ

QT

σ

T

m= cosθ3

n= sinθ3

Q1′1′ =m4Q11+2m2n2Q12+2Q66+4mnm2Q16+n2Q26+n4Q22

A′ =T

σ

AT

σ

T

A1′1′ =m4A11+2m2n2A12+2A66+4mnm2A16+n2A26+n4A22
 533 

CONSTITUTIVE EQUATIONS FOR PLANE STRAIN

" When, analyzing solids that are relatively prismatic and slender,

simplifying assumptions are made about the strain state to facilitate 

analytical solution of practical problems

" One such assumption is that the strains in a 

slender, prismatic body, that distort the crosssectional planes, are negligible compared to 

the other strains 

" This simplification is commonly referred to as 

the generalized plane-strain assumption

" For a state of generalized plane strain in a 

homogeneous, anisotropic solid, with respect to 

the x1 - x

2 plane, the strain field is appr

oximated 

such that and ,

where ε is a constant

x1

x2

x3

ε33−α33

T−Tref

=ε

2

ε23=

2

ε13=0
 534 

CONSTITUTIVE EQUATIONS FOR PLANE STRAIN

CONTINUED

" For this special case, the general constitutive equation

 uncouples directly into

 and

σ11

σ22

σ33

σ23

σ13

σ12

=

C11

C12C13C14C15C16

C12C22C23C24C25C26

C13C23C33C34C35C36

C14C24C34C44C45C46

C15C25C35C45C55C56

C16C26C36C46C56C66

ε11

ε22

ε33

2

ε23

2

ε13

2

ε12

−

α11

α22

α33

2

α23

2

α13

2

α12

T−Tref

σ11

σ22

σ12

=

C11

C12C16

C12C22C26

C16C26C66

ε11

ε22

2

ε12

−

α11

α22

2

α12

T−Tref+

C13

C23

C36

ε

σ33

σ23

σ13

=

C13

C23C36

C14C24C46

C15C25C56

ε11

ε22

2

ε12

−

α11

α22

2

α12

T−Tref+

C33

C34

C35

ε
 535 

CONSTITUTIVE EQUATIONS FOR PLANE STRAIN

CONTINUED

" This equation is manipulated further by defining thermal moduli

 and 

 such that

 and

" Note that the stiffness coefficients are obtained by inverting the fully 

populated compliance matrix - a nontrivial task

β 11

β 22

β 12

=−

C11

C12C16

C12C22C26

C16C26C66

α11

α22

2

α12

β 33

β 23

β 13

=−

C13

C23C36

C14C24C46

C15C25C56

α11

α22

2

α12

σ11

σ22

σ12

=

C11

C12C16

C12C22C26

C16C26C66

ε11

ε22

2

ε12

+

β 11

β 22

β 12

T−Tref+

C13

C23

C36

ε

σ33

σ23

σ13

=

C13

C23C36

C14C24C46

C15C25C56

ε11

ε22

2

ε12

+

β 33

β 23

β 13

T−Tref+

C33

C34

C35

ε
 536 

CONSTITUTIVE EQUATIONS FOR PLANE STRAIN

CONTINUED

" Simplification of the following general constitutive equation is not as 

easy

" First, the equation given above is expressed as

ε11

ε22

ε33

2

ε23

2

ε13

2

ε12

=

S11

S12S13S14S15S16

S12S22S23S24S25S26

S13S23S33S34S35S36

S14S24S34S44S45S46

S15S25S35S45S55S56

S16S26S36S46S56S66

σ11

σ22

σ33

σ23

σ13

σ12

+

α11

α22

α33

2

α23

2

α13

2

α12

T−Tref

ε11

ε22

ε33

2

ε23

2

ε13

2

ε12

−

α11

α22

α33

2

α23

2

α13

2

α12

T−Tref=

S11

S12S13S14S15S16

S12S22S23S24S25S26

S13S23S33S34S35S36

S14S24S34S44S45S46

S15S25S35S45S55S56

S16S26S36S46S56S66

σ11

σ22

σ33

σ23

σ13

σ12
 537 

CONSTITUTIVE EQUATIONS FOR PLANE STRAIN

CONTINUED

" Then, elimination of the transverse-shearing strains gives

" Rearranging the rows and columns into a convenient form gives

ε11

ε22

ε33

0

0

2

ε12

−

α11

α22

α33

0

0

2

α12

T−Tref=

S11

S12S13S14S15S16

S12S22S23S24S25S26

S13S23S33S34S35S36

S14S24S34S44S45S46

S15S25S35S45S55S56

S16S26S36S46S56S66

σ11

σ22

σ33

σ23

σ13

σ12

ε11

ε22

2

ε12

ε33

0

0

−

α11

α22

2

α12

α33

0

0

T−Tref=

S11

S12S16S13S14S15

S12S22S26S23S24S25

S16S26S66S36S46S56

S13S23S36S33S34S35

S14S24S46S34S44S45

S15S25S56S35S45S55

σ11

σ22

σ12

σ33

σ23

σ13
 538 

CONSTITUTIVE EQUATIONS FOR PLANE STRAIN

CONTINUED

" For convenience, let the mechanical strains be denoted by

" Using , the previous matrix constitutive equation 

becomes

ε11

σ

ε22

σ

2

ε12

σ

ε33

σ

0

0

=

ε11

ε22

2

ε12

ε33

0

0

−

α11

α22

2

α12

α33

0

0

T−Tref

ε33−α33

T−Tref

=ε

ε11

σ

ε22

σ

2

ε12

σ

ε

0

0

=

S11

S12S16S13S14S15

S12S22S26S23S24S25

S16S26S66S36S46S56

S13S23S36S33S34S35

S14S24S46S34S44S45

S15S25S56S35S45S55

σ11

σ22

σ12

σ33

σ23

σ13
 539 

CONSTITUTIVE EQUATIONS FOR PLANE STRAIN

CONTINUED

" Next, the matrix constitutive equation

 is separated to get

 and

ε11

σ

ε22

σ

2

ε12

σ

ε

0

0

=

S11

S12S16S13S14S15

S12S22S26S23S24S25

S16S26S66S36S46S56

S13S23S36S33S34S35

S14S24S46S34S44S45

S15S25S56S35S45S55

σ11

σ22

σ12

σ33

σ23

σ13

ε11

σ

ε22

σ

2

ε12

σ

=

S11

S12S16

S12S22S26

S16S26S66

σ11

σ22

σ12

+

S13

S14S15

S23S24S25

S36S46S56

σ33

σ23

σ13

ε

0

0

=

S13

S23S36

S14S24S46

S15S25S56

σ11

σ22

σ12

+

S33

S34S35

S34S44S45

S35S45S55

σ33

σ23

σ13
 540 

CONSTITUTIVE EQUATIONS FOR PLANE STRAIN

CONTINUED

" Solving the previous equation with for gives

" Back substitution of the vector containing into 

 yields the result

ε

σ33, σ23, andσ13

σ33

σ23

σ13

=

S33S34S35

S34S44S45

S35S45S55

− 1

ε

0

0

−

S33S34S35

S34S44S45

S35S45S55

− 1

S13S23S36

S14S24S46

S15S25S56

σ11

σ22

σ12

σ33, σ23, andσ13

ε11

σ

ε22

σ

2

ε12

σ

=

S11

S12S16

S12S22S26

S16S26S66

σ11

σ22

σ12

+

S13

S14S15

S23S24S25

S36S46S56

σ33

σ23

σ13

ε11

σ

ε22

σ

2

ε12

σ

=

s 11s 12s 16

s 12s 22s 26

s 16s 26s 66

σ11

σ22

σ12

+

S11

S22

S12

ε
 541 

CONSTITUTIVE EQUATIONS FOR PLANE STRAIN

CONTINUED

" The matrix with the subscripted s terms is given by

" The vector with the subscripted S terms is given by

" Next,expressing the mechanical strains in terms of the total strains and 

the strains caused by free thermal expansion results in

s 11s 12s 16

s 12s 22s 26

s 16s 26s 66

=

S11

S12S16

S12S22S26

S16S26S66

−

S13S14S15

S23S24S25

S36S46S56

S33S34S35

S34S44S45

S35S45S55

− 1

S13S23S36

S14S24S46

S15S25S56

S11

S22

S12

=

S13

S14S15

S23S24S25

S36S46S56

S33S34S35

S34S44S45

S35S45S55

− 1

1

0

0

ε11

ε22

2

ε12

=

s 11s 12s 16

s 12s 22s 26

s 16s 26s 66

σ11

σ22

σ12

+

α11

α22

2

α12

T−Tref+

S11

S22

S12

ε
 542 

CONSTITUTIVE EQUATIONS FOR PLANE STRAIN

CONTINUED

" Thus, the constitutive equations for generalized plane strain, in terms 

of compliance coefficients and thermal-expansion coefficients, become

 where

 and 

" The 

sij are called the 

reduced compliance coefficients

" When , the state of strain reduces to that known simply as plane 

strain

!11

!22

2

!12

=

s 11s 12s 16

s 12s 22s 26

s 16s 26s 66

"11

"22

"12

+

#11

#22

2

#12

T$Tref+

S11

S22

S12

!

s 11s 12s 16

s 12s 22s 26

s 16s 26s 66

=

S11

S12S16

S12S22S26

S16S26S66

$

S13S14S15

S23S24S25

S36S46S56

S33S34S35

S34S44S45

S35S45S55

$ 1

S13S23S36

S14S24S46

S15S25S56

S11

S22

S12

=

S13

S14S15

S23S24S25

S36S46S56

S33S34S35

S34S44S45

S35S45S55

$ 1

1

0

0

!=0
 543 

CONSTITUTIVE EQUATIONS FOR PLANE STRAIN

CONTINUED

" Once a given generalized-plane-strain or regular-plane-strain boundaryvalue problem is solved b

y using the following equation, the stresses 

and strains in the following equation are known

" The other stresses are then given by

ε11

ε22

2

ε12

=

s 11s 12s 16

s 12s 22s 26

s 16s 26s 66

σ11

σ22

σ12

+

α11

α22

2

α12

T−Tref+

S11

S22

S12

ε

σ33

σ23

σ13

=

S33S34S35

S34S44S45

S35S45S55

− 1

ε

0

0

−

S33S34S35

S34S44S45

S35S45S55

− 1

S13S23S36

S14S24S46

S15S25S56

σ11

σ22

σ12
 544 

CONSTITUTIVE EQUATIONS FOR PLANE STRAIN

CONTINUED

" The relationship between the reduced compliances and the stiffnesses 

is obtained by first considering the equation

" Inverting gives

σ11

σ22

σ12

=

C11

C12C16

C12C22C26

C16C26C66

ε11

ε22

2

ε12

−

α11

α22

2

α12

T−Tref+

C13

C23

C36

ε

ε11

ε22

2

ε12

=

C11

C12C16

C12C22C26

C16C26C66

− 1

σ11

σ22

σ12

+

α11

α22

2

α12

T−Tref−

C11

C12C16

C12C22C26

C16C26C66

− 1

C13

C23

C36

ε
 545 

CONSTITUTIVE EQUATIONS FOR PLANE STRAIN

CONTINUED

" Comparing 

 with

 indicates that

 and 

ε11

ε22

2

ε12

=

C11

C12C16

C12C22C26

C16C26C66

− 1

σ11

σ22

σ12

+

α11

α22

2

α12

T−Tref−

C11

C12C16

C12C22C26

C16C26C66

− 1

C13

C23

C36

ε

ε11

ε22

2

ε12

=

s 11s 12s 16

s 12s 22s 26

s 16s 26s 66

σ11

σ22

σ12

+

α11

α22

2

α12

T−Tref+

S11

S22

S12

ε

s 11s 12s 16

s 12s 22s 26

s 16s 26s 66

=

C11

C12C16

C12C22C26

C16C26C66

− 1

S11

S22

S12

=−

C11

C12C16

C12C22C26

C16C26C66

− 1

C13

C23

C36
 546 

 CONSTITUTIVE EQUATIONS FOR PLANE STRAIN

CONTINUED

" For a material that is monoclinic in the plane of the cross-section 

x3 = 

0 of the body, the generalized-plane-strain thermal moduli reduce 

to

 and 

" The generalized-plane-strain constitutive equations become

 and

β 11

β 22

β 12

=−

C11

C12C16

C12C22C26

C16C26C66

α11

α22

2

α12

β 33

β 23

β 13

=−

C13

C23C36

000

000

α11

α22

2

α12

σ11

σ22

σ12

=

C11

C12C16

C12C22C26

C16C26C66

ε11

ε22

2

ε12

+

β 11

β 22

β 12

T−Tref+

C13

C23

C36

ε

σ33

σ23

σ13

=

C13

C23C36

000

000

ε11

ε22

2

ε12

+

β 33

0

0

T−Tref+

C33

0

0

ε
 547 

 CONSTITUTIVE EQUATIONS FOR PLANE STRAIN

CONTINUED

" The last equation is reduced further to give and

" Likewise, for a material that is monoclinic in the plane of the crosssection of the bod

y,

 which simplifies to

σ23=σ13=0

σ33=

C13

ε11−α11

T−Tref+C23

ε22−α22

T−Tref+2C36

ε12−α12

T−Tref+C33

ε

s 11s 12s 16

s 12s 22s 26

s 16s 26s 66

=

S11

S12S16

S12S22S26

S16S26S66

−

S1300

S23

00

S36

00

S3300

0S44

S45

0S45S55

− 1

S13S23S36

000

000

s 11s 12s 16

s 12s 22s 26

s 16s 26s 66

=

S11−

S13S13

S33

S12−

S13S23

S33

S16−

S13S36

S33

S12−

S13S23

S33

S22−

S23S23

S33

S26−

S23S36

S33

S16−

S13S36

S33

S26−

S23S36

S33

S66−

S36S36

S33
 548 

CONSTITUTIVE EQUATIONS FOR PLANE STRAIN

CONTINUED

" Similarly,

 simplifies to

 or 

" The other stresses are then given by

 which reduce to and 

S11

S22

S12

=

S13

S14S15

S23S24S25

S36S46S56

S33S34S35

S34S44S45

S35S45S55

− 1

1

0

0

S11

S22

S12

=

S13

00

S23

00

S36

00

S3300

0S44

S45

0S45S55

− 1

1

0

0

S11S22S12

T

=

S13

S33

S23

S33

S36

S33

T

σ33

σ23

σ13

=

S3300

0S44

S45

0S45S55

− 1

ε

0

0

−

S3300

0S44

S45

0S45S55

− 1

S13S23S36

000

000

σ11

σ22

σ12

σ33=

1

S33

ε−

S13

σ11

−S23

σ22

−S36

σ12

σ23=σ13=0
 549 

 CONSTITUTIVE EQUATIONS FOR PLANE STRAIN

CONTINUED

" For a specially orthotropic material, the generalized-plane-strain

thermal moduli reduce to

 and 

" The generalized-plane-strain constitutive equations become

 and

β 11

β 22

β 12

=−

C11

C120

C12

C220

00C66

α11

α22

0

β 33

β 23

β 13

=−

C13

C230

000

000

α11

α22

0

σ11

σ22

σ12

=

C11

C120

C12

C220

00C66

ε11

ε22

2

ε12

+

β 11

β 22

0

T−Tref+

C13

C23

0

ε

σ33

σ23

σ13

=

C13

C230

000

000

ε11

ε22

2

ε12

+

β 33

0

0

T−Tref+

C33

0

0

ε
 550 

 CONSTITUTIVE EQUATIONS FOR PLANE STRAIN

CONTINUED

" The last equation is reduced further to give and

" In terms of engineering constants,

 where 

σ23=σ13=0

σ33=

C13

ε11−α11

T−Tref+C23

ε22−α22

T−Tref+C33

ε

C11=

E1

Δ

1−ν23ν32C22=

E2

Δ

1−ν13ν31

C33=

E3

Δ

1−ν12ν21

C12=

E1

Δ

ν21+ν23ν31=

E2

Δ

ν12+ν13ν32

C66=G12

C13=

E1

Δ

ν31+ν21ν32=

E3

Δ

ν13+ν12ν23

C23=

E2

Δ

ν32+ν12ν31=

E3

Δ

ν23+ν13ν21

Δ=1−ν12ν21−ν23ν32−ν13ν31−2ν21ν32ν13
 551 

 CONSTITUTIVE EQUATIONS FOR PLANE STRAIN

CONTINUED

" The nonzero thermal moduli are expressed as

β 11=−

E1 1−ν23ν32

α11+

ν21+ν23ν31

α22

1−ν12ν21−ν23ν32−ν13ν31−2ν21ν32ν13

β 22=−

E2 ν12+ν13ν32

α11+

1–ν13ν31

α22

1−ν12ν21−ν23ν32−ν13ν31−2ν21ν32ν13

β 33=−

E3 ν31+ν21ν32

α11+

ν23+ν13ν21

α22

1−ν12ν21−ν23ν32−ν13ν31−2ν21ν32ν13
 552 

CONSTITUTIVE EQUATIONS FOR PLANE STRAIN

CONTINUED

" Similarly, for a specially orthotropic material, the generalized-planestrain constitutive matrix becomes

 which simplifies to

s 11s 12s 16

s 12s 22s 26

s 16s 26s 66

=

S11

S120

S12

S220

00S66

−

S1300

S23

00

000

S3300

0S44

0

00S55

− 1

S13S230

000

000

s 11s 12

0

s 12s 22

0

00

s 66

=

S11

−

S13

S13

S33

S12−

S13

S23

S33

0

S12

−

S13

S23

S33

S22−

S23

S23

S33

0

00S66
 553 

CONSTITUTIVE EQUATIONS FOR PLANE STRAIN

CONTINUED

" Also

" Similarly,

 simplifies to

 or 

s 11s 12

0

s 12s 22

0

00

s 66

=

C11C120

C12C220

00C66

− 1

S11

S22

S12

=

S13

S14S15

S23S24S25

S36S46S56

S33S34S35

S34S44S45

S35S45S55

− 1

1

0

0

S11

S22

S12

=

S13

00

S23

00

000

S3300

0S44

0

00S55

− 1

1

0

0

S11S22S12

T

=

S13

S33

S23

S33

0

T
 554 

CONSTITUTIVE EQUATIONS FOR PLANE STRAIN

CONTINUED

" Finally, the generalized-plane-strain constitutive equations for a 

specially orthotropic material become

ε11

ε22

2

ε12

=

s 11s 12

0

s 12s 22

0

00

s 66

σ11

σ22

σ12

+

α11

α22

0

T−Tref+

S11

S22

0

ε

s 11s 12

0

s 12s 22

0

00

s 66

=

S11

−

S13

S13

S33

S12−

S13

S23

S33

0

S12

−

S13

S23

S33

S22−

S23

S23

S33

0

00S66

S11S22S12

T

=

S13

S33

S23

S33

0

T
 555 

CONSTITUTIVE EQUATIONS FOR PLANE STRAIN

CONTINUED

" The matrix equation for the other stresses becomes

 which reduces to and 

σ33

σ23

σ13

=

S3300

0S44

0

00S55

− 1

ε

0

0

−

S3300

0S44

0

00S55

− 1

S13S230

000

000

σ11

σ22

σ12

σ33=

1

S33

ε−

S13

σ11

−S23

σ22

σ23=σ13=0
 556 

CONSTITUTIVE EQUATIONS FOR PLANE STRAIN

CONCLUDED

" In terms of the engineering constants, the reduced compliance 

coefficients and are given by

" Similarly,

 and

s 11

=

1−ν13

ν31

E1

s 12

=−

ν12

+ν13ν32

E1

=−

ν21

+ν23ν31

E2

s 22

=

1−ν23

ν32

E2

s 66

=

1

G12

S11=

−

ν31S22=

−

ν32S12=0

σ33=

E3ε+ν31

σ11

+ν32

σ22
 557 

STRESS AND STRAIN TRANSFORMATION EQUATIONS 

FOR PLANE STRAIN

" The two primary constitutive equations of generalized plane strain are 

given by

 and

"To obtain transformation equations for the constitutive terms appearing 

in these equations, transformation equations that relate

 to are needed

σ11

σ22

σ12

=

C11

C12C16

C12C22C26

C16C26C66

ε11

ε22

2

ε12

+

β 11

β 22

β 12

T−Tref+

C13

C23

C36

ε

ε11

ε22

2

ε12

=

s 11s 12s 16

s 12s 22s 26

s 16s 26s 66

σ11

σ22

σ12

+

α11

α22

2

α12

T−Tref+

S11

S22

S12

ε

Σ′ ≡

σ1′1′

σ2′2′

σ1′2′

Σ≡

σ11

σ22

σ12
 558 

STRESS AND STRAIN TRANSFORMATION EQUATIONS 

FOR PLANE STRAIN - CONTINUED

" Likewise transformation equations that relate

 to are needed

" Consider the dextral (right-handed) rotation of coordinate frames 

shown in the figure

E′ ≡

ε1′1′

ε2′2′

2

ε1′2′

E≡

ε11

ε22

2

ε12

i 1 i 2

i 3

x1

x2

x3

x1′

x2′

x3′

i 1′

i 2′

i 3′

,

,

θ3

θ3
 559 

STRESS AND STRAIN TRANSFORMATION EQUATIONS 

FOR PLANE STRAIN - CONTINUED

" Previously, the matrix form of the stress-transformation law for this 

specific transformation was given as

" By inspection, it follows that

σ1′1′

σ2′2′

σ3′3′

σ2′3′

σ1′3′

σ1′2′

=

cos2

θ3 sin2θ3 0

002sin

θ3cosθ3

sin2θ3 cos2θ3 0

00−2sin

θ3cosθ3

001000

000 cos

θ3 − sinθ3 0

000 sin

θ3 cosθ3 0

− sin

θ3cosθ3 sinθ3cosθ3 0

00 cos2

θ3 − sin2θ3

σ11

σ22

σ33

σ23

σ13

σ12

σ1′1′

σ2′2′

σ1′2′

=

cos2

θ3 sin2θ3 2sinθ3cosθ3

sin2θ3 cos2θ3 −

2sin

θ3cosθ3

− sin

θ3cosθ3 sinθ3cosθ3 cos2θ3 − sin2θ3

σ11

σ22

σ12
 560 

STRESS AND STRAIN TRANSFORMATION EQUATIONS 

FOR PLANE STRAIN - CONTINUED

" Thus, the stress-transformation lawfor a state of generalized plane 

strain and a dextral rotation about the x

3 axis is identical to that for the 

corresponding plane-stress case and is given by

 where

" Likewise, the inverse is given by where

Σ′ =T

σ

θ

3

Σ

T

σ

θ

3

≡

cos2θ3 sin2θ3 2sinθ3cosθ3

sin2θ3 cos2θ3 −

2sin

θ3cosθ3

− sin

θ3cosθ3 sinθ3cosθ3 cos2θ3 − sin2θ3

Σ=T

σ

θ

3

− 1

Σ′

T

σ

θ

3

− 1

=

T

σ

–

θ

3

=

cos2θ3 sin2θ3 −

2sin

θ3cosθ3

sin2θ3 cos2θ3 2sinθ3cosθ3

sinθ3cosθ3 − sinθ3cosθ3 cos2θ3 − sin2θ3
 561 

STRESS AND STRAIN TRANSFORMATION EQUATIONS 

FOR PLANE STRAIN - CONTINUED

" Previously, the matrix form of the strain-transformation law for this 

specific transformation was given as

" Inspection of this matrix equation reveals

ε1′1′

ε2′2′

ε3′3′

2ε2′3′

2ε1′3′

2ε1′2′

=

cos2θ3 sin2θ3 0

00 sin

θ3cosθ3

sin2θ3 cos2θ3 0

00− sin

θ3cosθ3

00 1 000

000 cos

θ3 − sinθ3 0

000 sin

θ3 cosθ3 0

−2sin

θ3cosθ3 2sinθ3cosθ3 0

00 cos2

θ3 − sin2θ3

ε11

ε22

ε33

2ε23

2ε13

2ε12

ε1′1′

ε2′2′

2

ε1′2′

=

cos2θ3 sin2θ3 sinθ3cosθ3

sin2θ3 cos2θ3 − sinθ3cosθ3

−2sin

θ3cosθ3 2sinθ3cosθ3 cos2θ3 − sin2θ3

ε11

ε22

2

ε12
 562 

STRESS AND STRAIN TRANSFORMATION EQUATIONS 

FOR PLANE STRAIN - CONCLUDED

" The last matrix equation is expressed symbolically by

 where

" Similarly, the matrix form of the inverse is given by

E′ =T

ε

θ

3

E

T

ε

θ

3

≡

cos2θ3 sin2θ3 sinθ3cosθ3

sin2θ3 cos2θ3 − sinθ3cosθ3

−2sin

θ3cosθ3 2sinθ3cosθ3 cos2θ3 − sin2θ3

E

=

T

ε

θ

3

− 1

E′
 563 

TRANSFORMED CONSTITUTIVE EQUATIONS FOR 

PLANE STRAIN

" In terms of another set of coordinates that correspond to a 

dextral rotation about the x

3 axis, the constitutive equations must also 

have the forms given as

 and

x1′, x2′, x3′

σ1′1′

σ2′2′

σ1′2′

=

C1′1′ C1′2′ C1′6′

C1′2′ C2′2′ C2′6′

C1′6′ C2′6′ C6′6′

ε1′1′

ε2′2′

2

ε1′2′

+

β 1′1′

β 2′2′

β 1′2′

T−Tref+

C1′3′

C2′3′

C3′6′

ε

ε1′1′

ε2′2′

2

ε1′2′

=

s 1′1′ s 1′2′ s 1′6′

s 1′2′ s 2′2′ s 2′6′

s 1′6′ s 2′6′ s 6′6′

σ1′1′

σ2′2′

σ1′2′

+

α1′1′

α2′2′

2

α1′2′

T−Tref+

S1′1′

S2′2′

S1′2′

ε

β 1′1′

β 2′2′

β 1′2′

=−

C1′1′ Q1′2′ C1′6′

C1′2′ C2′2′ C2′6′

C1′6′ C2′6′ C6′6′

α1′1′

α2′2′

2

α1′2′

s 1′1′ s 1′2′ s 1′6′

s 1′2′ s 2′2′ s 2′6′

s 1′6′ s 2′6′ s 6′6′

=

C1′1′ C1′2′ C1′6′

C1′2′ C2′2′ C2′6′

C1′6′ C2′6′ C6′6′

− 1
 564 

TRANSFORMED CONSTITUTIVE EQUATIONS FOR 

PLANE STRAIN - CONTINUED

 and 

" For convenience, let

 such that the corresponding constitutive equations are given by

 and 

S1′1′

S2′2′

S1′2′

=−

C1′1′ C1′2′ C1′6′

C1′2′ C2′2′ C2′6′

C1′6′ C2′6′ C6′6′

− 1

C1′3′

C2′3′

C3′6′

s

=

s 11s 12s 16

s 12s 22s 26

s 16s 26s 66

C=

C11C12C16

C12C22C26

C16C26C66

C=

C13

C23

C36

α=

α11

α22

2

α12

β=

β 11

β 22

β 12

S

=

S11

S22

S12

Θ

≡T−Tref

E=

s

Σ+αΘ

+

S

ε

Σ=CE+βΘ

+

C

ε
 565 

TRANSFORMED CONSTITUTIVE EQUATIONS FOR 

PLANE STRAIN - CONTINUED

" Additionally, let

 such that the corresponding constitutive equations are given by

 and

s′ =

s 1′1′ s 1′2′ s 1′6′

s 1′2′ s 2′2′ s 2′6′

s 1′6′ s 2′6′ s 6′6′

C′ =

C1′1′ C1′2′ C1′6′

C1′2′ C2′2′ C2′6′

C1′6′ C2′6′ C6′6′

C′ =

C1′3′

C2′3′

C3′6′

α′ =

α1′1′

α2′2′

2

α1′2′

β′ =

β 1′1′

β 2′2′

β 1′2′

S′ =

S1′1′

S2′2′

S1′2′

E′ =

s′ Σ′ +

α′ Θ

+

S′ ε

Σ′ =C′ E′ +β′ Θ

+

C′ ε
 566 

TRANSFORMED CONSTITUTIVE EQUATIONS FOR 

PLANE STRAIN - CONTINUED

" Substituting and into 

 gives

" Premultiplying by gives

" Comparing this equation with it 

follows that

, , and 

Σ=T

σ

− 1

Σ′ E

=

T

ε

− 1

E′

Σ=CE+βΘ

+

C

ε

T

σ

− 1

Σ′ =CT

ε

− 1

E′ +βΘ

+

C

ε

T

σ

Σ′ =T

σ

CT

ε

− 1

E′ +T

σ

βΘ

+

T

σ

C

ε

Σ′ =C′ E′ +β′ Θ

+

C′ ε

C′ =T

σ

CT

ε

− 1

β′ =T

σ

β

C′ =T

σ

C
 567 

TRANSFORMED CONSTITUTIVE EQUATIONS FOR 

PLANE STRAIN - CONTINUED

" Rearranging gives 

" Likewise, and 

" Next, substituting and into 

 gives

" Premultiplying by gives

C′ =T

σ

CT

ε

− 1

C=T

σ

− 1

C′ T

ε

β

=

T

σ

− 1

β′ C

=

T

σ

− 1

C′

Σ=T

σ

− 1

Σ′ E

=

T

ε

− 1

E′

E=

s

Σ+αΘ

+

S

ε

T

ε

− 1

E′ =

s

T

σ

− 1

Σ′ +αΘ

+

S

ε

Tε

E′ =T

ε

s

T

σ

− 1

Σ′ +T

ε

αΘ

+

T

ε

S

ε
 568 

TRANSFORMED CONSTITUTIVE EQUATIONS FOR 

PLANE STRAIN - CONTINUED

" Comparing the last equation with 

reveals that

, , and 

" Rearranging gives 

" In addition, and 

E′ =

s′ Σ′ +

α′ Θ

+

S′ ε

s′ =

T

ε

s

T

σ

− 1

α′ =T

ε

αS′ =T

ε

S

s′ =

T

ε

s

T

σ

− 1

s

=

Tε

− 1

s′ T

σ

α=Tε

− 1

α′ S=Tε

− 1

S′
 569 

TRANSFORMED CONSTITUTIVE EQUATIONS FOR 

PLANE STRAIN - CONTINUED

" Noting that for a dextral rotation about the x

3 axis,

Tε and it follows that − 1=T

σ

T

T

σ

− 1

=Tε

T

s′ =

T

ε

s

T

ε

T

s

=

T

σ

T

s′ T

σ

C′ =T

σ

CT

σ

T

C

=

T

ε

T

C′ T

ε

α′ =

T

ε

α

α=

T

σ

T

α′

β′ =

T

σ

β

β=

T

ε

T

β′

C′ =

T

σ

C

C=

T

ε

T

C′

S′ =

T

ε

S

S=

T

σ

T

S′
 570 

TRANSFORMED CONSTITUTIVE EQUATIONS FOR 

PLANE STRAIN - CONTINUED

 with

 and

" Comparing these equations with those of the plane stress case reveals 

that the specific transformation equations can be obtained from those 

given previously for plane stress as follows

T

σ

θ

3

≡

cos2θ3 sin2θ3 2sinθ3cosθ3

sin2θ3 cos2θ3 −

2sin

θ3cosθ3

− sin

θ3cosθ3 sinθ3cosθ3 cos2θ3 − sin2θ3

T

ε

θ

3

≡

cos2θ3 sin2θ3 sinθ3cosθ3

sin2θ3 cos2θ3 − sinθ3cosθ3

−2sin

θ3cosθ3 2sinθ3cosθ3 cos2θ3 − sin2θ3
 571 

TRANSFORMED CONSTITUTIVE EQUATIONS FOR 

PLANE STRAIN - CONCLUDED

" For plane stress, , with and 

, gave

" Thus, by similarity, gives 

" The other transformation equations are obtained in a similar manner

Q′ =T

σ

QT

σ

T

m= cosθ3

n= sinθ3

Q1′1′ =m4Q11+2m2n2Q12+2Q66+4mnm2Q16+n2Q26+n4Q22

C′ =T

σ

CT

σ

T

C1′1′ =m4C11+2m2n2C12+2C66+4mnm2C16+n2C26+n4C22
 572 

LINES AND CURVES

OF

MATERIAL SYMMETRY
 573 

LINES AND CURVES OF MATERIAL SYMMETRY

" Up to this point of the present study, a local view of material symmetry

at a specific point P of a material body has been examined

" The conditions for the existence of various types of material 

symmetries have been ascertained by using reflective-symmetry 

transformations based on rectangular Cartesian coordinate frames that 

are local to the point P

" For some homogeneous materials, the corresponding planes of 

material symmetry for every point of the material body are aligned such 

that it is possible to define at least one straight line whose tangent is 

perpendicular to each corresponding symmetry plane

" This line is called a principal material direction

" When more than one distinct line connects sets of contiguous material 

points with identical planes of reflective symmetry, more than one 

principal material direction exists
 574 

LINES AND CURVES OF MATERIAL SYMMETRY

CONTINUED

" For example, a homogeneous orthotropic material possesses three 

principal material directions that are mutually perpendicular

" One important point illustrated herein is that the constitutive equations 

become simpler when a material symmetry plane exists, and that the 

inherent simplicity becomes hidden when the constitutive equations 

are expressed in terms of another rectangular Cartesian coordinate 

frame whose axes are oriented differently

" For example, one coordinate frame exists for a generally 

orthotropic material in which the constitutive equations correspond 

to a specially orthotropic material

" In addition,by defining a global rectangular Cartesian coordinate frame 

with at least one axis parallel to a principal material direction, the 

simplicity of the constitutive equations can be exploited to simplify the 

corresponding boundary-value problem
 575 

LINES AND CURVES OF MATERIAL SYMMETRY

CONTINUED

" In a more general scenario, a contiguous set of material points may 

exist with a plane of elastic symmetry, at a given point, that is 

perpendicular to the tangent to a curve, at that point

" For this case, the principal material direction follows a smooth 

curve, referred to herein as a material symmetry curve

" For this case, a curvilinear coordinate system can be defined in which 

one coordinate curve coincides with a material symmetry curve

" For example, consider an orthotropic material in which the three 

perpendicular principal directions at a given point of the body 

coincide with radial, circumferential, and axial directions of a 

cylindrical coordinate system

" It is important to emphasize that the rectangular Cartesian coordinate 

frames used to define symmetry transformations are local frames 

associated with a material point and not a global coordinate frame
 576 

LINES AND CURVES OF MATERIAL SYMMETRY

CONCLUDED

" Thus, at a given point of a material symmetry curve, one should 

envision a local rectangular Cartesian coordinate frames upon which all 

symmetry conditions associated with that point are deduced

P

C

Q

Surface of

material

points

Plane of material

symmetry at point P

Plane of material

symmetry at point Q

Curve of material

symmetry

Tangent vector 

at point P

Tangent vector 

at point Q

Coordinate curve

Coordinate curve
 577 

BIBLIOGRAPHY
 578 

BIBLIOGRAPHY

Allen, D. H. and Haisler, W. E.: Introduction to Aerospace Structural 

Analysis. John Wiley & Sons, Inc., New York, 1985.

Berthelot, J.-M.: Composite Materials - Mechanical Behavior and Structural 

Analysis. Springer-Verlag New York, Inc., New York, 1999.

Bisplinghoff, R. L.; Mar, J. W.; and Pian,T. H. H.: Statics of Deformable 

Solids. Dover Publications, Inc., New York, 1990.

Boley, B. A. and Weiner, J. H.: Theory of Thermal Stresses. Dover 

Publications, Inc., New York, 1988.

Boresi, A. P.; Sidebottom, O. M.; Seely, F. B.; and Smith, J. O.: Ad

vanced 

Mechanics of Materials.Third Edition, John Wiley & Sons, Inc., New York,

1978.

Chadwick, P.;Vianello, M.; and Cowin, S. C.: A New Proof that the Number 

of Linear Elastic Symmetries is Eight. Journal of the Mechanics and 

Physics of Solids,Vol. 49, 2001, pp. 2471-2492.
 579 

BIBLIOGRAPHY - CONTINUED

Chandrasekharaiah, D. S. and Debnath, L.: Continuum Mechanics.

Academic Press, Inc, New York, 1994.

Cowin, S. C.: Properties of the Anisotropic Elasticity Tensor.The Quarterly 

Journal of Mechanics and Applied Mathematics,Vol. 42, P

art 2, May 1989,

pp. 249-266.

Donaldson, B. K.: Analysis of Aircraft Structures. McGraw-Hill, Inc., New 

York, 1993.

Fung,Y. C.: Foundations of Solid Mechanics. Prentice-Hall, Inc., Engle

wood 

Cliffs, New Jersey, 1965.

Graff, K. F.: W

ave Motion in Elastic Solids. Dover Publications, Inc., New 

York, 1991.

Green, A. E. and Zerna,W.: Theoretical Elasticity. Second Edition, Dover 

Publications, Inc., New York, 1992, pp. 148-159.
 580 

BIBLIOGRAPHY - CONTINUED

Hearmon, R. F. S.: An Introduction to Applied Anisotropic Elasticity. Oxford 

University Press, 1961.

Herakovich, C.T. and Aboudi, J.:Thermal Effects in Composites, Chapter 1,

Thermal Stresses I. Vol. 5, R. B. Hetnarski, Ed., Lastran Corporation,

Rochester, New York, 1999, pp. 1-142.

Hetnarski, R. B.: Basic Equations of the Theory of Thermal Stresses,

Chapter 1,Thermal Stresses I. Vol. 1, R. B. Hetnarski, Ed., North-Holland 

Publishing Co.,The Netherlands, 1986, pp. 1-21.

Khan, A. S. and Huang, S.: Continuum Theory of Plasticity. John Wiley & 

Sons, Inc., New York, 1995.

Malvern, L. E.: Introduction to the Mechanics of a Continuous Medium.

Prentice-Hall, Inc., Englewood Cliffs, New Jersey, 1969.

Mase, G. E. and G.T. Mase: Continuum Mechanics for Engineer

s. CRC 

Press, Inc., 1992.
 581 

BIBLIOGRAPHY - CONTINUED

Noor, A. K.; Kim,Y. H., and Peters, J. M.: Strain Energy of ThermallyStressed Multilayer Panels and its Sensitivity Coefficients, Non-Classical 

Problems of the Theory and Behavior of Structures Exposed to Complex 

Environmental Conditions. L. Librescu, Ed., AMD-Vol. 164, ASME, 1993, pp.

33-43.

Nowinski, J. L.: Theory of Thermoelasticity with Applications. Sijthoff and 

Noordhoff International Publishers, Alphen aan den Rijn,The Netherlands,

1978.

Padovan, J.: Anisotropic Thermal Stress Analysis, Chapter 3,Thermal 

Stresses I. Vol. 1, R. B. Hetnarski, Ed., North-Holland Publishing Co.,The 

Netherlands, 1986, pp. 143-262.

Pilkey, W. D. and Wunderlich,W.: Mechanics of Structures - 

Variational and 

Computational Methods. CRC Press, Inc., 1994.
 582 

BIBLIOGRAPHY - CONTINUED

Reddy, J. N.: Energy and Variational Methods in Applied Mechanics - With 

an Introduction to the Finite Element Method. John Wiley & Sons, Inc., New 

York, 1984.

Rosen,J.: Symmetry Discovered - Concepts and Applications in Nature and 

Science. Dover Publications, Inc., New York, 1998.

Saada, A. S.: Elasticity Theory and Applications. Second Edition, R. E.

Krieger Publishing Co., Malibar, Florida, 1993.

Shames, I. H.: Mechanics of Deformable Solids. R. E. Krieger Publishing 

Co., Malibar, Florida, 1964.

Shames, I. H. and Dym, C. L.: Energy and Finite Element Methods in 

Structural Mechanics. Hemisphere Publishing Corp., New York, 1985.

Sokolnikoff, I. S.: Mathematical Theory of Elasticity. Second Edition,

McGraw-Hill Book Company, Inc., New York, 1956.
 583 

BIBLIOGRAPHY - CONTINUED

Tauchert,T. R.: Energy Principles in Structural Mechanics. R. E. Krieger 

Publishing Co., Malibar, Florida, 1981.

Ting,T. C.T.: Anisotropic Elasticity - Theory and Applications. Oxford 

University Press, Inc., New York, 1996.

Washizu, K.:Variational Methods in Elasticity and Plasticity. Third Edition,

Pergamon Press, Ltd., Elmsford, NewYork, 1982.

Wempner, G.: Mechanics of Solids with Applications to Thin Bodies.

Sijthoff and Noordhoff International Publishers, Alphen aan den Rijn,The 

Netherlands, 1981.
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 4. TITLE AND SUBTITLE

An In-depth Tutorial on Constitutive Equations for Elastic Anisotropic 

Materials

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 6. AUTHOR(S)

Nemeth, Michael P.

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NASA Langley Research Center

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14. ABSTRACT

An in-depth tutorial on the constitutive equations for elastic, anisotropic materials is presented. Basic concepts are introduced that are used to characterize 

materials, and notions about how anisotropic material deform are presented. Hooke’s law and the Duhamel-Neuman law for isotropic materials are presented 

and discussed. Then, the most general form of Hooke’s law for elastic anisotropic materials is presented and symmetry requirements are given. A similar 

presentation is also given for the generalized Duhamel-Neuman law for elastic, anisotropic materials that includes thermal effects. Transformation equations 

for stress and strains are presented and the most general form of the transformation equations for the constitutive matrices are given. Then, specialized 

transformation equations are presented for dextral roatations about the coordinate axes. Next, concepts of material symmetry are introduced and criteria for 

material symmetries are presesented. Additionally, engineering constants of fully anisotropic, elastic materials are derived from first principles and the 

specialized to several cases of practical importance.

15. SUBJECT TERMS

Constitutive equations; anisotropic materials; engineering constants; plane strain; plane stress; strain; stress

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