An In-Depth Tutorial on Constitutive Equations for Elastic Anisotropic Materials

Home , Constitutive equation, Orthotropic material

NASA/TM–2011-217314

Michael P. Nemeth Langley Research Center, Hampton, Virginia

December 2011 NASA STI Program . . . in Profile

Since its founding, NASA has been dedicated to  CONFERENCE PUBLICATION. Collected the advancement of aeronautics and space science. papers from scientific and technical The NASA scientific and technical information (STI) conferences, symposia, seminars, or other program plays a key part in helping NASA maintain meetings sponsored or co-sponsored by NASA. this important role.  SPECIAL PUBLICATION. Scientific, The NASA STI program operates under the technical, or historical information from NASA auspices of the Agency Chief Information Officer. It programs, projects, and missions, often collects, organizes, provides for archiving, and concerned with subjects having substantial disseminates NASA’s STI. The NASA STI program public interest. provides access to the NASA Aeronautics and Space Database and its public interface, the NASA Technical  TECHNICAL TRANSLATION. English- Report Server, thus providing one of the largest language translations of foreign scientific and collections of aeronautical and space science STI in technical material pertinent to NASA’s mission. the world. Results are published in both non-NASA channels and by NASA in the NASA STI Report Specialized services also include creating custom Series, which includes the following report types: thesauri, building customized databases, and organizing and publishing research results.  TECHNICAL PUBLICATION. Reports of completed research or a major significant phase For more information about the NASA STI of research that present the results of NASA program, see the following: programs and include extensive data or theoretical analysis. Includes compilations of  Access the NASA STI program home page at significant scientific and technical data and http://www.sti.nasa.gov information deemed to be of continuing reference value. NASA counterpart of peer-  E-mail your question via the Internet to reviewed formal professional papers, but having [email protected] less stringent limitations on manuscript length and extent of graphic presentations.  Fax your question to the NASA STI Help Desk at 443-757-5803  TECHNICAL MEMORANDUM. Scientific and technical findings that are preliminary or of  Phone the NASA STI Help Desk at specialized interest, e.g., quick release reports, 443-757-5802 working papers, and bibliographies that contain minimal annotation. Does not contain extensive  Write to: analysis. NASA STI Help Desk NASA Center for AeroSpace Information  CONTRACTOR REPORT. Scientific and 7115 Standard Drive technical findings by NASA-sponsored Hanover, MD 21076-1320 contractors and grantees.

NASA/TM–2011-217314

An In-Depth Tutorial on Constitutive Equations for Elastic Anisotropic Materials

Michael P. Nemeth Langley Research Center, Hampton, Virginia

National Aeronautics and Space Administration

Langley Research Center Hampton, Virginia 23681-2199

December 2011

Available from:

NASA Center for AeroSpace Information 7115 Standard Drive Hanover, MD 21076-1320 443-757-5802

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

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

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

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 x3 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

MONOCLINIC MATERIALS...... 200 Monoclinic Materials - Reflective Symmetry About

the Plane x1 = 0 ...... 201 Monoclinic Materials - Reflective Symmetry About

the Plane x2 = 0 ...... 216 Monoclinic Materials - Reflective Symmetry About

the Plane x3 = 0 ...... 224

ORTHOTROPIC MATERIALS ...... 232 Orthotropic Materials - Reflective Symmetry About the

Planes x1 = 0 and x2 = 0 ...... 233 Orthotropic Materials - Reflective Symmetry About the

Planes x1 = 0, x2 = 0, and x3 = 0 ...... 242 Constitutive Equations...... 244 Specially Orthotropic Materials...... 246 Generally Orthotropic Materials ...... 247

TRIGONAL MATERIALS ...... 254 Trigonal Materials - Reflective Symmetry About Planes that

Contain the x3 axis...... 255 Trigonal Materials - Reflective Symmetry About Planes that

Contain the x1 axis...... 283 Trigonal Materials - Reflective Symmetry About Planes that

Contain the x2 axis...... 316 Summary of Trigonal Materials ...... 322

TETRAGONAL MATERIALS ...... 324 Tetragonal Materials - Reflective Symmetry Planes that

Contain the x3 axis...... 330 Tetragonal Materials - Reflective Symmetry Planes that Contain

the x1 axis ...... 335 Tetragonal Materials - Reflective Symmetry Planes that

Contain the x2 axis...... 338

Summary of Tetragonal Materials ...... 344

TRANSVERSELY ISOTROPIC MATERIALS ...... 347

Transversely Isotropic Materials - Isotropy Plane x3 = 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

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

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.

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.

PREFATORY COMMENTS

11 ! ! ! ! !

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 provide stand-alone sections that can be used independently for various reasons

" Third, to serve as a comprehensive reference document

12 ! ! ! !

DEDICATION

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

" To James 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

" To The Men and Women of the NACA - The premier examples of government researchers

BASIC CONCEPTS AND NOTATIONS

14 ! ! ! ! ! !

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 vary 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

15 ! ! ! ! ! !

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)

16 ! !

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 Normal stress Deformed shape of an anisotropic solid an isotropic solid

17 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)

Shearing stress Deformed shape of Deformed shape of an anisotropic solid an isotropic solid

18 ! !

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

Deformed shape of Deformed shape of an anisotropic solid an isotropic solid

" All unrestrained thermal expansion is volumetric and uniform within a homogeneous, isotropic solid; not so in a homogeneous, generally anisotropic solid

19 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

20 NOTATION FOR STRESSES AND STRAINS

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

relative to standard rectangular Cartesian coordinates x 1, x 2, x 3

σ 33 x3 σ 13 σ 23

σ 22 x 2 O σ 12

σ x1 11 Shearing stresses Normal stresses

" The normal strains ε 11 , ε 22 , and ε 33 correspond to the normal stresses

σ 11 , σ 22 , and σ 33 , respectively

" The shearing strains 2ε 12 = γ 12 , 2ε 13 = γ 13 , and 2ε 23 = γ 23 correspond to the

shearing stresses σ 12 , σ 13 , and σ 23 , respectively

21 INDICIAL NOTATION

" The rules of indicial notation associated with Cartesian tensors are used herein

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

1 " 1 For example, ε ij = E + ν σ ij − νδ ijσ kk

" Latin indices take on the values 1, 2, 3 , and repeated latin indices

imply summation over this set; e. g.; σ kk = σ kk = σ 11 + σ 22 + σ 33 kΣΣ= 1

" Indices that are not summed in an equation are called free indices and take on the complete set of possible values

" The symbol δjk is known as the Kronecker Delta Symbol and is equal to one when j = k and is equal to zero otherwise

22 CONSTITUTIVE EQUATIONS FOR ISOTROPIC MATERIALS

23 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

1 2 1 + ν = + 2ε 12 = γ 12 = σ 12 ε 11 E σ 11 − ν σ 22 σ 33 E

1 2 1 + ν = + 2ε 13 = γ 13 = σ 13 ε 22 E σ 22 − ν σ 11 σ 33 E

1 2 1 + ν = + 2ε 23 = γ 23 = σ 23 ε 33 E σ 33 − ν σ 11 σ 22 E

= 1 or in indicial notation εεij E 1 + νν σσij − ννδδijσσkk

24 HOOKE’S LAW - CONCLUDED HOMOGENEOUS, ISOTROPIC, LINEAR-ELASTIC SOLID

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

E E E σ 11 = 1 − ν ε + ν ε 22 + ε 33 σ 12 = γ 12 = ε 12 1 + ν 1 − 2ν 11 2 1 + ν 1 + ν

E E E σ 22 = 1 − ν ε + ν ε 11 + ε 33 σ 13 = γ 13 = ε 13 1 + ν 1 − 2ν 22 2 1 + ν 1 + ν

E E E σ 33 = 1 − ν ε + ν ε 11 + ε 22 σ 23 = γ 23 = ε 23 1 + ν 1 − 2ν 33 2 1 + ν 1 + ν

or in indicial notation

= E 1 2 + σ ij (1 + ν)(1 − 2ν) − ν ε ij νδ ijε kk

" 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

25 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 εij at a point of a solid, subjected to thermomechanical loading, consists of

σ T mechanical strain εij and strain caused by free thermal expansion εij

σ " The mechanical strain εij 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

σ T σ 1 T " ε ij = ε ij + ε ij = 1 + = T T where ε ij E ν σ ij − νδ ijσ kk , ε ij α δ ij − ref , T is

the temperature ﬁeld, and Tref is the temperature ﬁeld at which the body is deemed stress and strain free (or negligible)

26 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, T = T x 1, x 2, x 3 and T ref = T ref x 1, x 2, x 3

" 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 α T − Tref 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

27 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

1 2 1 + ν = + + T T 2ε 12 = γ 12 = σ 12 ε 11 E σ 11 − ν σ 22 σ 33 α − ref E

1 2 1 + ν = + + T T 2ε 13 = γ 13 = σ 13 ε 22 E σ 22 − ν σ 11 σ 33 α − ref E

2 1 + = 1 + + T T 2 = = ν ε 33 E σ 33 − ν σ 11 σ 22 α − ref ε 23 γ 23 E σ 23

or in indicial notation

= 1 εεij E 1 + νν σσij − ννδδijσσkk + ααδδij T − Tref

28 THE DUHAMEL-NEUMANN LAW - CONTINUED HOMOGENEOUS, ISOTROPIC, LINEAR-THERMOELASTIC SOLID

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

E Eα T − Tref σ 11 = 1 − ν ε 11 + ν ε 22 + ε 33 − 1 + ν 1 − 2ν 1 − 2ν

E Eα T − Tref σ 22 = 1 − ν ε 22 + ν ε 11 + ε 33 − 1 + ν 1 − 2ν 1 − 2ν

E Eα T − Tref σ 33 = 1 − ν ε 33 + ν ε 11 + ε 22 − 1 + ν 1 − 2ν 1 − 2ν

E E E E σ 12 = γ 12 = ε 12 σ 13 = γ 13 = ε 13 2 1 + ν 1 + ν 2 1 + ν 1 + ν

E E σ 23 = γ 23 = ε 23 or in indicial notation 2 1 + ν 1 + ν

= E 1 2 + Eα T − Tref σ ij 1 + ν 1 − 2ν − ν ε ij νδ ijε kk − δ ij 1 − 2ν

29 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’s ratio and α is the coefficient of linear thermal expansion

E " E and ν are related by G = , where G is called the 2 1 + ν shear modulus or the modulus of rigidity

30 GENERALIZED HOOKE’S LAW FOR HOMOGENEOUS, ANISOTROPIC, LINEAR-ELASTIC SOLIDS

31 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.,

S1111 S1122 S1133 S1123 S1113 S1112 S1132 S1131 S1121 ε11 σ11 S2211 S2222 S2233 S2223 S2213 S2212 S2232 S2231 S2221 ε22 σ22 S3311 S3322 S3333 S3323 S3313 S3312 S3332 S3331 S3321 ε33 σ33

ε23 S2311 S2322 S2333 S2323 S2313 S2312 S2332 S2331 S2321 σ23

ε13 = S1311 S1322 S1333 S1323 S1313 S1312 S1332 S1331 S1321 σ13

ε12 S1211 S1222 S1233 S1223 S1213 S1212 S1232 S1231 S1221 σ12 ε32 S3211 S3222 S3233 S3223 S3213 S3212 S3232 S3231 S3221 σ32 ε31 σ31 S3111 S3122 S3133 S3123 S3113 S3112 S3132 S3131 S3121 ε21 σ21 S2111 S2122 S2133 S2123 S2113 S2112 S2132 S2131 S2121

or in indicial notation by εεij = S ijklσσkl

-1 2 " Note that Sijkl have units of stress ; e.g., in /lb

32 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

4 " Without further simplication, there are 3 (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

33 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

C1111 C1122 C1133 C1123 C1113 C1112 C1132 C1131 C1121 σ 11 ε11 C2211 C2222 C2233 C2223 C2213 C2212 C2232 C2231 C2221 σ 22 ε22 C3311 C3322 C3333 C3323 C3313 C3312 C3332 C3331 C3321 σ 33 ε33

σ 23 C2311 C2322 C2333 C2323 C2313 C2312 C2332 C2331 C2321 ε23

σ 13 = C1311 C1322 C1333 C1323 C1313 C1312 C1332 C1331 C1321 ε13

σ 12 C1211 C1222 C1233 C1223 C1213 C1212 C1232 C1231 C1221 ε12 ε32 σ 32 C3211 C3222 C3233 C3223 C3213 C3212 C3232 C3231 C3221 σ 31 ε31 C3111 C3122 C3133 C3123 C3113 C3112 C3132 C3131 C3121 σ 21 ε21 C2111 C2122 C2133 C2123 C2113 C2112 C2132 C2131 C2121

or in indicial notation by σσij = C ijklεεkl

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

34 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 ε ij = S ijkl σ kl can be reduced by enforcing symmetry of the stress and strain tensors

" ε ij = S ijkl σ kl is the same as ε ji = S jikl σ kl

" ε ij = ε ji gives S ijkl σ kl = S jikl σ kl , which implies S ijkl = S jikl for a general stress state at a point in a body

" σ kl = σ lk can be used to show that S ijkl = S ijlk

" S ijkl = S jikl and S ijkl = S ijlk yield 36 independent compliance coefﬁcients

35 GENERAL FORM OF HOOKE’S LAW - CONTINUED

" The proof that Sijkl = Sijlk is given as follows

" The constitutive equation ε ij = S ijkl σ kl can be expressed as ε ij = S ijlk σ lk because l and k are summation indices and interchanging them doesn’t alter the content of the equation

" Equating εij = Sijklσ kl and εij = Sijlkσ lk gives S ijkl σ kl = S ijlk σ lk

" Next, enforcing σ kl = σ lk gives S ijkl σ kl = S ijlk σ kl , which implies

S ijkl = S ijlk

for a general stress state at a point in a body

36 GENERAL FORM OF HOOKE’S LAW - CONTINUED

" The number of independent coefficients in σ ij = C ijkl ε kl can also be reduced directly by enforcing symmetry of the stress and strain tensors

" σ ij = C ijkl ε kl is the same as σ ji = C jikl ε kl

" σ ij = σ ji gives C ijkl ε kl = C jikl ε kl , which implies C ijkl = C jikl for a general strain state at a point in a body

" ε kl = ε lk can be used to show that C ijkl = C ijlk

" C ijkl = C jikl and C ijkl = C ijlk yield 36 independent stiffness coefﬁcients

" Cauchy’s generalized form of Hooke’s Law ends up with 36 independent compliance or stiffness coefﬁcients

37 GENERAL FORM OF HOOKE’S LAW - CONTINUED

" The expanded form of Cauchy’s generalized Hooke’s law ε ij = S ijkl σ kl is obtained as follows

" First, expanding the last summation index gives

ε ij = S ijk 1σ k1 + S ijk 2σ k2 + S ijk 3σ k3

" Then, expanding the summation index k gives

εij = Sij11σ 11 + Sij21σ 21 + Sij31σ 31 + Sij12σ 12 + Sij22σ 22 + Sij32σ 32 +

Sij13σ 13 + Sij23σ 23 + Sij33σ 33

" Next, enforcing σ kl = σ lk yields

εij = Sij11σ 11 + Sij22σ 22 + Sij33σ 33 +

Sij23 + Sij32 σ 23 + Sij13 + Sij31 σ 13 + Sij12 + Sij21 σ 12

38 GENERAL FORM OF HOOKE’S LAW - CONTINUED

" Then, enforcing the conditions S ijkl = S ijlk give the result

ε ij = S ij11 σ 11 + S ij22 σ 22 + S ij33 σ 33 + 2S ij23 σ 23 + 2S ij13 σ 13 + 2S ij12 σ 12

" 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 ε ij = S ijklσ kl :

S1111 S1122 S1133 2S1123 2S1113 2S1112 ε11 σ 11 S S S 2S 2S 2S ε22 2211 2222 2233 2223 2213 2212 σ 22

ε S3311 S S3333 2S3323 2S3313 2S3312 σ 33 = 3322 33 2ε23 2S2311 2S2322 2S2333 4S2323 4S2313 4S2312 σ 23 2ε13 σ 13 2S1311 2S1322 2S1333 4S1323 4S1313 4S1312 2ε12 σ 12 2S1211 2S1222 2S1233 4S1223 4S1213 4S1212

39 GENERAL FORM OF HOOKE’S LAW - CONTINUED

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

σ ij = C ijkl ε kl is obtained as follows

" First, expanding the last summation index gives

σ ij = C ijk 1ε k1 + C ijk 2ε k2 + C ijk 3ε k3

" Then, expanding the summmation index k gives

σ ij = Cij11ε11 + Cij21ε21 + Cij31ε31 + Cij12ε12 + Cij22ε22 + Cij32ε32 +

Cij13ε13 + Cij23ε23 + Cij33ε33

" Next, enforcing ε kl = ε lk yields

σ ij = Cij11ε11 + Cij22ε22 + Cij33ε33 +

Cij23 + Cij32 ε23 + Cij13 + Cij31 ε13 + Cij12 + Cij21 ε12

40 GENERAL FORM OF HOOKE’S LAW - CONTINUED

" Then, enforcing the conditions C ijkl = C ijlk give the result

σ ij = C ij11 ε 11 + C ij22 ε 22 + C ij33 ε 33 + 2C ij23 ε 23 + 2C ij13 ε 13 + 2C ij12 ε 12

" 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 σ ij = C ijklε kl :

C1111 C1122 C1133 C1123 C1113 C1112 σ 11 ε11 C C C C C C σ 22 2211 2222 2233 2223 2213 2212 ε22

σ C3311 C C C C C ε 33 = 3322 3333 3323 3313 3312 33 2 σ 23 C2311 C2322 C2333 C2323 C2313 C2312 ε23 σ 13 2ε13 C1311 C1322 C1333 C1323 C1313 C1312 σ 12 2ε12 C1211 C1222 C1233 C1223 C1213 C1212

41 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)

42 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 density U 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 by external forces is converted into elastic-strain energy that can be recovered upon unloading, thus a loaded body has the potential to perform work

43 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 dU 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

44 STRAIN-ENERGY DENSITY

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

d U = σ 11 dε 11 + σ 22 dε 22 + σ 33 dε 33 + 2σ 23 dε 23 + 2σ 13 dε 13 + 2σ 12 dε 12

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

σ ij = σ ij ε 11 ,ε 22 ,ε 33 ,ε 23 ,ε 13 ,ε 12

" This expression is written compactly in indicial form as

d U = σ ij ε pq dε ij

" The strain-energy density U is obtained by integrating dU over the deformation associated with a loading process, that starts at a strain- free state and ends at a particular strain state; that is,

ε pq U = σ ij ε 11,ε 22,ε 33,ε 23,ε 13,ε 12 dε ij = U ε pq 0

45 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

U = σ ij ε pq dε ij

" 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

U = σ ij ε pq dε ij = 0

" For the condition that U = 0 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 = dU = 0

46 STRAIN-ENERGY DENSITY - CONTINUED ILLUSTRATION OF ELASTIC LOADING-UNLOADING PROCESSES

Two Independent Loading Systems

A O

Stress

A O B

Elastic loading-unloading cycle A

C C B

O Strain B C

47 STRAIN-ENERGY DENSITY - CONTINUED ILLUSTRATION OF ELASTIC LOADING-UNLOADING PROCESSES

One Loading System

Stress A O A

A O

Elastic loading-unloading cycle O Strain

48 STRAIN-ENERGY DENSITY - CONTINUED

" Because U = U ε pq , it follows mathematically that an exact ∂U differential has the property that dU = dε ij ∂ε ij

" A function with this property is described in mathematics as a potential function, thus U is sometimes referred to as the elastic potential

∂U ∂UU " Equating d U = σ ij ε pq dε ij with dU = dε ij gives = σ ij ∂ε ij ∂ε ij

" 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 Green- elastic material

49 STRAIN-ENERGY DENSITY - CONCLUDED

" The statement U = dU = 0 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 condition for a function U = U ε pq to be path independent

2 2 ∂ ∂ is for the following condition to be valid: UU = UU ∂ε ij ∂ε kl ∂ε kl ∂ε ij

" This condition arises from the connection of the path integral with Stokes’ integral theorem

50 PROOF THAT Cijkl = Cklij

2 2 ∂ U ∂ U ∂U ∂σ ij ∂σ kl " First, note that = and = σ ij give = ∂ε ij ∂ε kl ∂ε kl ∂ε ij ∂ε ij ∂ε kl ∂ε ij

∂σ ij ∂ ∂ε rs " Then, σ ij = C ijrs ε rs gives = C ijrs ε rs = C ijrs = C ijrs δ rkδ sl = C ijkl ∂ε kl ∂ε kl ∂ε kl

∂σ kl ∂ ∂ε pq " Also, σ kl = C klpq ε pq gives = C klpq ε pq = C klpq = C klpq δ piδ qj = C klij ∂ε ij ∂ε ij ∂ε ij

∂σ ij ∂σ kl " Thus, = yields C ijkl = C klij , which reduces the number of ∂ε kl ∂ε ij independent stiffness coefficients to 21

51 ILLUSTRATION OF THE PATH-INDEPENDENCE CONDITION

" The function U = U ε 11 ,ε 22 ,ε 33 ,ε 23 ,ε 13 ,ε 12 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, F x 1, x 2

∂F ∂F " The chain rule of differentiation gives dF = dx 1 + dx 2 ∂x 1 ∂x 2

" The vector form of dF is given by

∂F ∂F dF = i 1 + i 2 • dx 1 i 1 + dx 2 i 2 ≡ ∇F • dx ∂x 1 ∂x 2

52 ILLUSTRATION OF THE PATH-INDEPENDENCE CONDITION CONTINUED

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

dF = 0 becomes ∇F •dx = 0 P P

" Recall that Stokes’ Theorem is given by g • dx = n • ∇ × g dA ∂S S where

" g x 1, x 2 is an arbitrary vector ﬁeld with continuous ﬁrst derivatives

" n x 1, x 2 is the unit-magnitude normal-vector ﬁeld for any smooth surface S enclosed by the curve ∂S

53 ILLUSTRATION OF THE PATH-INDEPENDENCE CONDITION CONTINUED

" Applying Stokes’ theorem to ∇F •dx = 0 gives P

∇F • dx = n • ∇ × ∇F dA = 0 P S(P)

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,

n • ∇ × ∇FF = 0

54 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 ∇ × ∇F = 0

∂ ∂ ∂F ∂F " Expanding ∇ × ∇F = 0 gives i 1 + i 2 × i 1 + i 2 = 0 ∂x 1 ∂x 2 ∂x 1 ∂x 2

2 2 ∂ F ∂ F which simpliﬁes to i 1 × i 2 + i 2 × i 1 = 0 ∂x 1∂x 2 ∂x 2∂x 1

2 2 ∂ F ∂ F " Simplifying further gives − i 1 × i 2 = 0 which yields ∂x 1∂x 2 ∂x 2∂x 1

2 2 ∂ ∂ the condition F = F ∂x 1∂x 2 ∂x 2∂x 1

55 ILLUSTRATION OF THE PATH-INDEPENDENCE CONDITION CONTINUED

2 2 ∂ F ∂ F " The condition = is, in fact, a statement of path ∂x 1∂x 2 ∂x 2∂x 1 independence at the local level, which is illustrated in the following ﬁgure

F (x 1 , x 2 ) in the neighborhood F(x1,x2) of point P ∂F F + dx2 A ∂x2 ∂F F + dx1 ∂x1 D

B C x2 i 1 i 2

P: (x1,x2) x1

56 ILLUSTRATION OF THE PATH-INDEPENDENCE CONDITION CONCLUDED

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

2 ∂FF ∂ ∂FF ∂FF ∂FF ∂ FF FF + dx 1 + FF + dx 1 dx 2 = FF + dx 1 + dx 2 + dx 1dx 2 ∂x 1 ∂x 2 ∂x 1 ∂x 1 ∂x 2 ∂x 2∂x 1

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

2 ∂FF ∂ ∂FF ∂FF ∂FF ∂ FF FF + dx 2 + FF + dx 2 dx 1 = FF + dx 2 + dx 1 + dx 2dx 1 ∂x 2 ∂x 1 ∂x 2 ∂x 2 ∂x 1 ∂x 1∂x 2

" For path independence, it follows that these two expressions must be equal, hence 2 2 ∂ ∂ FF = FF ∂x 1∂x 2 ∂x 2∂x 1

57 COMPLEMENTARY STRAIN-ENERGY DENSITY

" The symmetry condition Sijkl = 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 dU = σ ijdε ij 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 by using the

product rule of differentiation to get d σ ijε ij = σ ijdε ij + ε ijdσ ij

" In the part σ ijdε ij , the strains are taken as the independent variables

" In the part ε ijdσ ij , the stresses are taken as the independent variables

58 COMPLEMENTARY STRAIN-ENERGY DENSITY CONTINUED

" Next, the expression d σ ijε ij = σ ijdε ij + ε ijdσ ij is integrated from the initial stress- and strain- free state to the current stress and strain state; i.e.,

ε pq ε pq σ pq d σ ijε ij = σ ij ε pq dε ij + ε ij σ pq dσ ij 0 0 0

ε pq " In the term d σ ijε ij , it is presumed that the stresses are known as 0 functions of the strains

σ pq " This term can also be expressed as d σ ijε ij , where it is 0 presumed that the strains are known as functions of the stresses

" Both terms yield σ ijε ij , the product of the current values of the stresses and strains

59 COMPLEMENTARY STRAIN-ENERGY DENSITY CONCLUDED

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

σ pq density function U gives σ ijε ij = U ε pq + ε ij σ pq dσ ij 0

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

σ pq is deﬁned as U * = ε ij σ pq dσ ij such that σ ijε ij = U ε pq + U * σ pq 0

" Note that d U * = ε ij σ pq dσ ij

" The form U * σ pq = σ ijε ij − U ε pq 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

60 ILLUSTRATION OF ENERGY DENSITY FUNCTIONALS ONE-DIMENSIONAL CASE

= Single-parameter σεσε UU εε + UU* σσ loading system

Stress Stress Stress σ = σ(ε) dσ σ σ σ

= + ε = ε(σ) dε ε ε ε Strain Strain Strain ε σ Area = σε Area = σ(ε) dε = U(ε) Area = ε(σ) dσ = U*(σ) 0 0

61 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

∂U * " Thus, U * = dU * = 0 and dU * = dσ ij ∂σ ij

∂U * ∂UU * " Equating d U * = ε ij σ pq dσ ij with dU * = dσ ij gives = ε ij ∂σ ij ∂σ ij

" The equations given above indicate that the strain-stress relations are derivable from a potential function when the deformation process is elastic (hyperelastic material)

62 PROOF THAT Sijkl = Sklij

" The necessary and sufﬁcient conditions for U * = U * σ pq to be path 2 * 2 * independent are for the conditions ∂ U = ∂ U to be valid ∂σ ij ∂σ kl ∂σ kl ∂σ ij

2 2 ∂ U * ∂ U * ∂U * ∂ε ij ∂ε kl " First note that = and = ε ij give = ∂σ ij ∂σ kl ∂σ kl ∂σ ij ∂σ ij ∂σ kl ∂σ ij

∂ε ij ∂ ∂σ rs " Then, ε ij = S ijrs σ rs gives = S ijrs σ rs = S ijrs = S ijrs δ rkδ sl = S ijkl ∂σ kl ∂σ kl ∂σ kl

∂ε kl ∂ ∂σ pq " And, ε kl = S klpq σ pq gives = S klpq σ pq = S klpq = S klpq δ piδ qj = S klij ∂σ ij ∂σ ij ∂σ ij

∂ε ij ∂ε kl " Thus, = yields S ijkl = S klij , which reduces the number of ∂σ kl ∂σ ij independent compliance coefﬁcients to 21

63 STANDARD FORMS FOR GENERALIZED HOOKE’S LAW

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

S1111 S1122 S1133 2S1123 2S1113 2S1112 ε11 σ 11 S S S 2S 2S 2S ε22 1122 2222 2233 2223 2213 2212 σ 22

ε S1133 S S 2S3323 2S3313 2S σ 33 = 2233 3333 3312 33 2ε23 2S1123 2S2223 2S3323 4S2323 4S2313 4S2312 σ 23 2 σ 13 ε13 2S1113 2S2213 2S3313 4S2313 4S1313 4S1312 σ 12 2ε12 2S1112 2S2212 2S3312 4S2312 4S1312 4S1212

C1111 C1122 C1133 C1123 C1113 C1112 σ 11 ε11 C C C C C C σ 22 1122 2222 2233 2223 2213 2212 ε22

σ C1133 C C C C C ε 33 = 2233 3333 3323 3313 3312 33 σ 23 C1123 C2223 C3323 C2323 C2313 C2312 2ε23 σ 13 2 C1113 C2213 C3313 C2313 C1313 C1312 ε13 σ 12 2ε12 C1112 C2212 C3312 C2312 C1312 C1212

64 CLAPEYRON’S FORMULA

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

d U = σ ij ε pq dε ij is combined with σ ij = C ijkl ε kl to get dU = C ijkl ε kldε ij

1 d C = 1 C d + C d " Now consider, 2 ijklε ijε kl 2 ijkl ε ijε kl ijklε ij ε kl

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

1 d C = 1 C d + C d = 1 C + C d 2 ijklε ijε kl 2 ijklε kl ε ij klijε kl ε ij 2 ijkl klij ε kl ε ij

" By using the path-independence condition C ijkl = C klij , it follows that

1 d C = C d d = 1 d C 2 ijkl ε ijε kl ijkl ε kl ε ij and that U 2 ijkl ε ijε kl

65 CLAPEYRON’S FORMULA - CONCLUDED

= 1 C + K " Integrating the last expression gives U 2 ijkl ε ijε kl where K is a constant of integration

" Noting that U = 0 when the strain ﬁeld is zero-valued gives K = 0

= 1 " Next, using σ ij = C ijkl ε kl gives the desired result, UU 2 σ ijε ij

" This expression for the strain-energy density of a homogeneous, linear- elastic, anisotropic solid is attributed to B. P. E. Clapeyron (1799-1864)

1 " A similar procedure can be followed to show that UU * = 2 σ ijε ij = UU for a homogeneous, linear-elastic, anisotropic solid

66 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 non- negative-valued function for all physically admissible elastic strain states

ε pq " Hence, U = σ ij(ε pq) dε ij 0 for a nonlinear-elastic material 0 ≥≥

= 1C 0 " For a linear-elastic material, U 2 ijklε ijε kl ≥≥ must hold, which places some thermodynamic restrictions on the stiffness coefﬁcients that must hold for reversible (elastic) loading-unloading processes

67 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

T ε11 C1111 C1122 C1133 C1123 C1113 C1112 ε11

ε22 C1122 C2222 C2233 C2223 C2213 C2212 ε22 ε C C C C C C ε = 1 33 1133 2233 3333 3323 3313 3312 33 UU 2 2ε23 C1123 C2223 C3323 C2323 C2313 C2312 2ε23

2ε13 C1113 C2213 C3313 C2313 C1313 C1312 2ε13

2ε12 C1112 C2212 C3312 C2312 C1312 C1212 2ε12

" 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

68 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

69 GENERALIZED DUHAMEL-NEUMANN LAW FOR HOMOGENEOUS, ANISOTROPIC, LINEAR-ELASTIC SOLIDS

70 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

71 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 linear- thermoelastic constitutive equations

72 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 εij at a point of a solid, subjected to thermomechanical loading, consists of stress-induced

σ T mechanical strain εij and strain caused by free thermal expansion εij

σ " The mechanical strain εij 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

σ T σ T " ε ij = ε ij + ε ij where ε ij = S ijrs σ rs , ε ij = α ij T − T ref , T is the temperature

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

73 THE GENERALIZED DUHAMEL-NEUMANN LAW CONTINUED

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

S1111 S1122 S1133 S1123 S1113 S1112 S1132 S1131 S1121 ε11 σ11 α11 S2211 S2222 S2233 S2223 S2213 S2212 S2232 S2231 S2221 ε22 σ22 α22 S3311 S3322 S3333 S3323 S3313 S3312 S3332 S3331 S3321 ε33 σ33 α33

ε23 S2311 S2322 S2333 S2323 S2313 S2312 S2332 S2331 S2321 σ23 α23

ε13 = S1311 S1322 S1333 S1323 S1313 S1312 S1332 S1331 S1321 σ13 + α13 T − Tref

ε12 S1211 S1222 S1233 S1223 S1213 S1212 S1232 S1231 S1221 σ12 α12 ε32 α32 S3211 S3222 S3233 S3223 S3213 S3212 S3232 S3231 S3221 σ32 ε31 σ31 α31 S3111 S3122 S3133 S3123 S3113 S3112 S3132 S3131 S3121 ε21 σ21 α21 S2111 S2122 S2133 S2123 S2113 S2112 S2132 S2131 S2121

and in indicial form by ε ij = S ijklσ kl + α ij T − Tref

" 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)

74 THE GENERALIZED DUHAMEL-NEUMANN LAW CONTINUED

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

C1111 C1122 C1133 C1123 C1113 C1112 C1132 C1131 C1121 σ 11 ε11 α11 C2211 C2222 C2233 C2223 C2213 C2212 C2232 C2231 C2221 σ 22 ε22 α22 C3311 C3322 C3333 C3323 C3313 C3312 C3332 C3331 C3321 σ 33 ε33 α33

σ 23 C2311 C2322 C2333 C2323 C2313 C2312 C2332 C2331 C2321 ε23 α23 σ 13 = C1311 C1322 C1333 C1323 C1313 C1312 C1332 C1331 C1321 ε13 − α13 T − Tref

σ 12 C1211 C1222 C1233 C1223 C1213 C1212 C1232 C1231 C1221 ε12 α12 ε32 α32 σ 32 C3211 C3222 C3233 C3223 C3213 C3212 C3232 C3231 C3221 σ 31 ε31 α31 C3111 C3122 C3133 C3123 C3113 C3112 C3132 C3131 C3121 σ 21 ε21 α21 C2111 C2122 C2133 C2123 C2113 C2112 C2132 C2131 C2121

σ and in indicial form by σ ij = C ijkl ε kl − α kl T − Tref = C ijklε kl

" 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

75 THE GENERALIZED DUHAMEL-NEUMANN LAW CONTINUED

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

C1111 C1122 C1133 C1123 C1113 C1112 C1132 C1131 C1121 β 11 σ 11 ε11 C2211 C2222 C2233 C2223 C2213 C2212 C2232 C2231 C2221 β 22 σ 22 ε22 C3311 C3322 C3333 C3323 C3313 C3312 C3332 C3331 C3321 β 33 σ 33 ε33

σ 23 C2311 C2322 C2333 C2323 C2313 C2312 C2332 C2331 C2321 ε23 β 23 σ 13 = C1311 C1322 C1333 C1323 C1313 C1312 C1332 C1331 C1321 ε13 + β 13 T − Tref

σ 12 C1211 C1222 C1233 C1223 C1213 C1212 C1232 C1231 C1221 ε12 β 12 ε32 σ 32 C3211 C3222 C3233 C3223 C3213 C3212 C3232 C3231 C3221 β 32 σ 31 ε31 C3111 C3122 C3133 C3123 C3113 C3112 C3132 C3131 C3121 β 31 σ 21 ε21 C2111 C2122 C2133 C2123 C2113 C2112 C2132 C2131 C2121 β 21

and in indicial form by σσ ij = C ijkl εεkl + ββ ij T − Tref

" βij are called the thermal moduli

76 THE GENERALIZED DUHAMEL-NEUMANN LAW CONTINUED

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

Hooke’s law when T = Tref , the following symmetry conditions must hold

Sijkl = Sjikl Sijkl = Sijlk Sijkl = Sklij

Cijkl = Cjikl Cijkl = Cijlk Cijkl = Cklij which indicates 21 independent compliance or stiffness coefﬁcients

" Symmetry of the strain tensor also yields α ij = α ji , and reduces the number of independent coefﬁcients of linear thermal expansion from 9 to 6

" Likewise, symmetry of the stress tensor also yields β ij = β ji , and reduces the number of independent thermal moduli from 9 to 6

77 THE GENERALIZED DUHAMEL-NEUMANN LAW CONCLUDED

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

S S S 2S 2S 2S ε11 1111 1122 1133 1123 1113 1112 σ 11 α11

ε22 S1122 S2222 S2233 2S2223 2S2213 2S2212 σ 22 α22

ε33 S1133 S2233 S3333 2S3323 2S3313 2S3312 σ 33 α33 = + T − Tref 2ε23 2S1123 2S2223 2S3323 4S2323 4S2313 4S2312 σ 23 2α23

2ε13 2S1113 2S2213 2S3313 4S2313 4S1313 4S1312 σ 13 2α13 σ 12 2ε12 2S1112 2S2212 2S3312 4S2312 4S1312 4S1212 2α12

C C C C C C ε11 β σ 11 1111 1122 1133 1123 1113 1112 11

σ 22 C1122 C2222 C2233 C2223 C2213 C2212 ε22 β 22

σ 33 C1133 C2233 C3333 C3323 C3313 C3312 ε33 β 33 = + T − Tref σ 23 C1123 C2223 C3323 C2323 C2313 C2312 2ε23 β 23

σ 13 C1113 C2213 C3313 C2313 C1313 C1312 2ε13 β 13 σ 12 C1112 C2212 C3312 C2312 C1312 C1212 2ε12 β 12

78 EQUATIONS FOR THE THERMAL MODULI

" The thermal moduli βij are related to the coefﬁcients of linear thermal

expansion by β ij = − C ijklα kl or by

β 11 C1111 C1122 C1133 C1123 C1113 C1112 α11

β 22 C1122 C2222 C2233 C2223 C2213 C2212 α22

β 33 C C C C C C α = − 1133 2233 3333 3323 3313 3312 33 β 23 C1123 C2223 C3323 C2323 C2313 C2312 2α23

β 13 C1113 C2213 C3313 C2313 C1313 C1312 2α13

β 12 C1112 C2212 C3312 C2312 C1312 C1212 2α12

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

-1 " Similarly, αij have units of temperature

79 STRAIN-ENERGY DENSITY FOR THERMAL LOADING

" The symmetry relations Cijkl = Cklij 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 density U 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

σ σ σ σ σ σ dU = σ 11dε11 + σ 22dε22 + σ 33dε33 + 2σ 23dε23 + 2σ 13dε13 + 2σ 12dε12

σ T where the stress-induced, mechanical strains are given by ε ij = ε ij − ε ij and the stresses depend on the mechanical strains; that is,

σ σ σ σ σ σ σ ij = σ ij ε 11, ε 22, ε 33, ε 23, ε 13, ε 12

80 STRAIN-ENERGY DENSITY FOR THERMAL LOADING CONTINUED

σ T " It is important to emphasize that the mechanical strains ε ij = ε ij − ε ij include strains generated by thermal stresses associated with a nonuniform temperature ﬁeld or spatial variations in the coefﬁcients of thermal expansion

" The expression

σ σ σ σ σ σ dU = σ 11dε11 + σ 22dε22 + σ 33dε33 + 2σ 23dε23 + 2σ 13dε13 + 2σ 12dε12

is written compactly in indicial form as

σ σ dUU = σ ij ε pq dε ij

81 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,

σ ε pq σ σ σ UU = σ ij ε pq dε ij = UU ε pq 0

" Because no mechanical work of the internal forces within a body is lost during a conservative, elastic, thermomechanical loading-

σ σ unloading process, it follows that U = σ ij ε pq dε ij = 0 , which implies 2 2 ∂UU σ ∂ UU ∂ UU dUU = σ dε ij and σ σ = σ σ ∂ε ij ∂ε ij ∂ε kl ∂ε kl ∂ε ij

82 STRAIN-ENERGY DENSITY FOR THERMAL LOADING CONCLUDED

σ σ ∂U σ ∂U σ " Equating d U = σ ij ε pq dε ij and dU = σ dε ij gives σ = σ ij ε pq ∂ε ij ∂ε ij

2 2 ∂σ ij ∂σ kl ∂ U ∂ U ∂U σ " Then, σ σ = σ σ and σ = σ ij ε pq give σ = σ ∂ε ij ∂ε kl ∂ε kl ∂ε ij ∂ε ij ∂ε kl ∂ε ij

σ σ ∂σ ij ∂ σ ∂ε rs " σ ij = C ijrs ε rs gives σ = σ C ijrs ε rs = C ijrs σ = C ijrs δ rkδ sl = C ijkl ∂ε kl ∂ε kl ∂ε kl

σ σ ∂σ kl ∂ σ ∂ε pq " σ ij = C ijpq ε pq gives σ = σ C klpq ε pq = C klpq σ = C klpq δ piδ qj = C klij ∂ε ij ∂ε ij ∂ε ij

∂σ ij ∂σ kl " Thus, σ = σ yields C ijkl = C klij ∂εkl ∂εij

83 COMPLEMENTARY STRAIN-ENERGY DENSITY FOR THERMAL LOADING

" The symmetry condition Sijkl = Sklij is obtained by examining the complementary strain-energy density function U*

" An expression for U* is obtained by ﬁrst requiring that a one-to-one relationship exists between the stresses and strains, and by expressing

σ σ σ ijε ij = σ ij ε ij + α ij T − Tref or σ ijε ij = σ ijε ij + σ ijα ij T − Tref

" Next, the product rule of differentiation is used to get σ σ d σ ijε ij = σ ijdε ij + ε ij dσ ij + d σ ijα ij T − Tref

σ " In the part σ ij dεij , the stress-induced, mechanical strains are taken as the independent variables

σ " In the part εij dσ ij , the stresses are taken as the independent variables

84 COMPLEMENTARY STRAIN-ENERGY DENSITY FOR THERMAL LOADING - CONTINUED

σ σ " Next, the expression d σ ijε ij = σ ijdε ij + ε ij dσ ij + d σ ijα ij T − Tref is integrated from the initial stress- and strain- free state to the current stress and strain state; i.e.,

σ ε pq ε pq σ pq ε pq σ σ σ d σ ijε ij = σ ij ε pq dε ij + ε ij σ pq dσ ij + d σ ijα ij T − Tref 0 0 0 0

εpq

" In the term d σ ijε ij it is presumed that the stresses are known as 0 functions of the strains

σ pq

" This term can also be expressed as d σ ijεij , where it is 0 presumed that the strains are known as functions of the stresses

" Both terms yield σ ijε ij , the current values of the stresses and strains

85 COMPLEMENTARY STRAIN-ENERGY DENSITY FOR THERMAL LOADING - CONCLUDED

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

σ pq σ σ density function U gives σ ijε ij = U ε pq + ε ij σ pq dσ ij + σ ijα ij T − Tref 0

" The complementary strain-energy density function U* is deﬁned as

σ pq σ σ U * = ε ij σ pq dσ ij + σ ijα ij T − Tref such that σ ijε ij = U ε pq + U * σ pq ,T 0

σ " Note that d U * = ε ij dσ ij + d σ ijα ij T − Tref

σ " Legendre’s transformation takes the form U * σ pq ,T = σ ijε ij − U ε pq

" 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 one- dimensional stress and strain state

86 ILLUSTRATION OF THERMOELASTIC ENERGY DENSITIES ONE-DIMENSIONAL CASE

σεσε = UU εεσ + UU* σσ,T Stress Stress Stress σ = σ ε σ σ σ σ d σ

T T = + α − ref α T − Tref σ σ dε σ ε = ε σ

εσ ε εσ εσ ε Total strain Total strain Total strain

ε σ σ Area = Area = ε σ d σ = ε σ σ σε σ ε U Area = ε σ dσ + σα T − Tref = U * σ,T 0 0

Loading process = mechanical loading followed by thermal loading

87 ILLUSTRATION OF THERMOELASTIC ENERGY DENSITIES ONE-DIMENSIONAL CASE - CONTINUED σεσε = UU εεσ + UU* σσ,T

Stress Stress Stress σ σ σ = σ ε σ ε = ε σ

σ σ d σ σ εσ = + dε σ

σ α T − Tref ε ε α T − Tref ε Total strain Mechanical strain Total strain

σ ε σ σ σ σ σ Area = σε Area = σ ε dε = U ε Area = ε σ dσ + σα T − Tref = U * σ,T 0 0

Loading process = thermal loading followed by mechanical loading

88 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 strain- energy density function would depend on the loading process

∂U * ∂U * " Thus, U * σ pq,T = dU * = 0 and dU * = dσ ij + dT ∂σ ij ∂T

σ σ " d U * = ε ij dσ ij + d σ ijα ij T − Tref gives d U * = ε ij + α ij T − Tref dσ ij + σ ijα ijdT

which reduces to dUU* = ε ijdσ ij + σ ijα ijdT

89 ILLUSTRATION OF THERMOELASTIC ENERGY DENSITIES ONE-DIMENSIONAL CASE - CONCLUDED

∂U * ∂U * " Equating dU * = dσ ij + dT with d U * = ε ijdσ ij + σ ijα ijdT gives ∂σ ij ∂T

∂UU* σ ∂UU* = ε ij = ε ij + α ij T − Tref and = σ ijα ij ∂σ ij ∂T

" Now, U * σ pq,T = dU * = 0 implies, and is implied by, the conditions

2 2 2 * 2 * ∂ * ∂ * ∂ U = ∂ U and U = U ∂σ ij ∂σ kl ∂σ kl ∂σ ij ∂σ ij ∂T ∂T ∂σ ij

2 2 ∂ U * ∂ U * ∂U * ∂ε ij ∂ε kl " = and = ε ij give = ∂σ ij ∂σ kl ∂σ kl ∂σ ij ∂σ ij ∂σ kl ∂σ ij

90 PROOF THAT Sijkl = Sklij FOR THERMOELASTIC SOLIDS

2 2 ∂U * σ ∂U * ∂ U * ∂ U * " First, = ε ij + α ij T − Tref and = σ ijα ij satisfy = ∂σ ij ∂T ∂σ ij ∂T ∂T ∂σ ij identically

" ε ij = S ijrs σ rs + α ij T − Tref gives

∂ε ij ∂ ∂σ rs = S ijrs σ rs + α ij T − Tref = S ijrs = S ijrs δ rkδ sl = S ijkl ∂σ kl ∂σ kl ∂σ kl

" ε ij = S ijpq σ pq + α ij T − Tref gives

∂ε kl ∂ ∂σ pq = S klpq σ pq + α ij T − Tref = S klpq = S klpq δ piδ qj = S klij ∂σ ij ∂σ ij ∂σ ij

∂ε ij ∂ε kl " Thus, = yields S ijkl = S klij ∂σ kl ∂σ ij

91 CLAPEYRON’S FORMULA FOR THERMOELASTIC SOLIDS

1 σ " Clapeyron’s formula UU = 2σijεij 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 σ ε ij = ε ij − α ij T − Tref to get

1 1 UU = 2 σσ ijεεij − 2 σσ ijαα ij T − Tref

92 STRAIN-ENERGY DENSITY EXPRESSIONS

= 1 σ " By deﬁnition, U 2σ ijεij for a homogeneous, anisotropic, linear- thermoelastic solid

1 σ = S " Using ε ij ijrs σ rs gives UU = 2 S ijrs σσ ijσσ rs

2 " = 1 1 + For isotropic materials, U 2E ν σ ijσ ij − ν σ kk

" In expanded form,

1 2 2 2 ν U = σ 11 + σ 22 + σ 33 − σ 11σ 22 + σ 22σ 33 + σ 11σ 33 + 2E E 2 2 2 1 + ν σ + σ + σ E 12 23 13

93 STRAIN-ENERGY DENSITY EXPRESSIONS CONCLUDED

= C σ = 1 C σ σ " Using σ ij ijkl ε kl gives U 2 ijkl ε ijε kl

σ " Further, using ε ij = ε ij − α ijΘ , Θ = T − Tref and C klij = C ijkl gives

= 1 C 2 + 2 U 2 ijkl ε ijε kl − ε ijα klΘ α ijα klΘ

" For isotropic materials,

2 = E ε ε + E ε ν ε − αΘαΘ + 3E α 2Θ U 2 1 + ν ij ij 2 1 − 2ν kk 1 + ν kk 2 1 − 2ν

2 = νE ε + ε + ε U 2 1 + ν 1 − 2ν 11 22 33 2 2 2 2 2 2 + E ε + ε + ε + 2 ε + 2 ε + 2 ε 2 1 + ν 11 22 33 12 23 13 2 3α E 2 − αE ε + ε + ε Θ + Θ 1 − 2ν 11 22 33 2 1 − 2ν

94 ABRIDGED NOTATION AND ELASTIC CONSTANTS

95 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

β 11 β 1 ε11 α11 σ 11 σ 1 ε1 α1 β 22 β 2 σ 22 σ 2 ε22 ε2 α22 α2 σ 33 σ 3 ε33 ε3 β 33 β 3 α33 α3

σ 23 →→ σ 4 2 →→ ε4 β 23 →→ β 4 2 →→ α4 ε23 α23 σ 13 σ 5 ε5 α5 2ε13 β 13 β 5 2α13 σ 12 σ 6 ε6 α6 2ε12 β 12 β 6 2α12

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

96 ABRIDGED NOTATION FOR CONSTITUTIVE EQUATIONS CONTINUED

" The components of the compliance and stiffness tensors are expressed as

S1111 S1122 S1133 2S1123 2S1113 2S1112 S11 S12 S13 S14 S15 S16

S2211 S2222 S2233 2S2223 2S2213 2S2212 S21 S22 S23 S24 S25 S26

S3311 S3322 S3333 2S3323 2S3313 2S3312 S31 S32 S33 S34 S35 S36 2S 2S 2S 4S 4S 4S →→ S S S S S S 2311 2322 2333 2323 2313 2312 41 42 43 44 45 46

2S1311 2S1322 2S1333 4S1323 4S1313 4S1312 S51 S52 S53 S54 S55 S56

2S1211 2S1222 2S1233 4S1223 4S1213 4S1212 S61 S62 S63 S64 S65 S66

C1111 C1122 C1133 C1123 C1113 C1112 C11 C12 C13 C14 C15 C16

C2211 C2222 C2233 C2223 C2213 C2212 C21 C22 C23 C24 C25 C26

C3311 C3322 C3333 C3323 C3313 C3312 C31 C32 C33 C34 C35 C36 C C C C C C →→ C C C C C C 2311 2322 2333 2323 2313 2312 41 42 43 44 45 46

C1311 C1322 C1333 C1323 C1313 C1312 C51 C52 C53 C54 C55 C56

C1211 C1222 C1233 C1223 C1213 C1212 C61 C62 C63 C64 C65 C66

" Materials that can be characterized by the matrices given above when they are symmetric are said to possess complete Voigt symmetry

97 ABRIDGED NOTATION FOR CONSTITUTIVE EQUATIONS CONTINUED

" The constitutive equations are often expressed in a nontensorial, indicial form given by

ε i = S ijσ j + α i T − Tref σ i = C ijε j + β i T − Tref

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

ε = S σ + α T − Tref

and

σ = C ε − α T − Tref

σ = C ε + β T − Tref

98 ABRIDGED NOTATION FOR CONSTITUTIVE EQUATIONS CONTINUED

S 11 S 12 S 13 S 14 S 15 S 16 ε1 σ 1 α1

ε2 S 12 S 22 S 23 S 24 S 25 S 26 σ 2 α2

ε3 S 13 S 23 S 33 S 34 S 35 S 36 σ 3 α3 = + T − Tref ε4 S 14 S 24 S 34 S 44 S 45 S 46 σ 4 α4 ε5 α5 S 15 S 25 S 35 S 45 S 55 S 56 σ 5 ε6 σ 6 α6 S 16 S 26 S 36 S 46 S 56 S 66

ε = S σ + α T − Tref

C 11 C 12 C 13 C 14 C 15 C 16 β 1 σ 1 ε1

σ 2 C 12 C 22 C 23 C 24 C 25 C 26 ε2 β 2

σ 3 C 13 C 23 C 33 C 34 C 35 C 36 ε3 β 3 = + T − Tref σ 4 C 14 C 24 C 34 C 44 C 45 C 46 ε4 β 4 ε5 σ 5 C 15 C 25 C 35 C 45 C 55 C 56 β 5 σ 6 ε6 C 16 C 26 C 36 C 46 C 56 C 66 β 6

σ = C ε + β T − Tref

99 ABRIDGED NOTATION FOR CONSTITUTIVE EQUATIONS CONTINUED

" Often, the following mixed-abridged notation is used

S 11 S 12 S 13 S 14 S 15 S 16 ε11 σ 11 α 11 S 12 S 22 S 23 S 24 S 25 S 26 ε22 σ 22 α 22 ε S S S S S S σ α 33 = 13 23 33 34 35 36 33 + 33 T T σ − ref 2ε23 S 14 S 24 S 34 S 44 S 45 S 46 23 2α 23

2ε13 σ 13 2α 13 S 15 S 25 S 35 S 45 S 55 S 56 σ 12 2ε12 2α 12 S 16 S 26 S 36 S 46 S 56 S 66

C 11 C 12 C 13 C 14 C 15 C 16 β σ 11 ε11 11 C 12 C 22 C 23 C 24 C 25 C 26 σ 22 ε22 β 22

σ 33 C 13 C 23 C 33 C 34 C 35 C 36 ε33 β 33 = + T − Tref σ 23 C 14 C 24 C 34 C 44 C 45 C 46 2ε23 β 23

σ 13 2ε13 β 13 C 15 C 25 C 35 C 45 C 55 C 56 σ 12 2ε12 β 12 C 16 C 26 C 36 C 46 C 56 C 66

100 ABRIDGED NOTATION FOR CONSTITUTIVE EQUATIONS CONCLUDED

" The thermal moduli are given by

C11 C12 C13 C14 C15 C16 β 1 α 1 C C C C C C β 2 12 22 23 24 25 26 α 2

β 3 C C C C C C α = − 13 23 33 34 35 36 3 β 4 C14 C24 C34 C44 C45 C46 2α 4

β 5 2α 5 C15 C25 C35 C45 C55 C56 β 6 2α 6 C16 C26 C36 C46 C56 C66

C11 C12 C13 C14 C15 C16 β 11 α 11 C C C C C C β 22 12 22 23 24 25 26 α 22

β 33 C C C C C C α = − 13 23 33 34 35 36 33 β 23 C14 C24 C34 C44 C45 C46 2α 23

β 13 2α 13 C15 C25 C35 C45 C55 C56 β 12 2α 12 C16 C26 C36 C46 C56 C66

101 CLAPEYRON’S FORMULA IN ABRIDGED NOTATION

" Clapeyron’s formula for the strain-energy density of a linear- 1 σ thermoelastic solid was given previously by UU = 2σijεij or 1 1 = T T σ UU 2 σ ijε ij − 2 σ ijα ij − ref where ε ij = ε ij − α ij T − Tref

" A convenient matrix form of Clapeyron’s formula is obtained by using the following notation

T σ T σ σ σ σ σ σ 2 2 2 σσ = σ11 σ22 σ33 σ23 σ13 σ12 εε = ε11 ε22 ε33 ε23 ε13 ε12

T T εε = ε11 ε22 ε33 2ε23 2ε13 2ε12 αα = α11 α22 α33 2α23 2α13 2α12 where the superscript T denotes matrix or vector transposition

σ " First, by inspection, it follows that ε = ε − α Θ where

Θ = T − Tref

102 CLAPEYRON’S FORMULA IN ABRIDGED NOTATION CONTINUED

σ σ + σ + σ + 2 σ + 2 σ + 2 σ " Next, noting that σijεij = σ11ε11 σ22ε22 σ33ε33 σ23 ε23 σ13 ε13 σ12 ε12 it

σ T σ σ T follows that σ ijε ij = σ ε = ε σ in matrix notation

" Therefore,Clapeyron’s formula for the strain-energy density of a linear-thermoelastic solid is given in matrix from by

T T T = 1 σ 1 1 UU 2 σ ε or UU = 2 σσ εε − 2 σσ αα ΘΘ

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

Θ σ " First, deﬁne σ = C α Θ such that σ = C ε becomes σ = C ε − σ Θ

103 CLAPEYRON’S FORMULA IN ABRIDGED NOTATION CONTINUED

Θ − 1 Θ " Then, σ = C α Θ gives α Θ = C σ

σ − 1 " Next, ε = ε − α Θ becomes ε σ = ε − C σ Θ

T T T − 1 Θ " Also, σ ε σ = σ ε − σ C σ

Θ T T Θ T " Transposition of σ = C ε − σ gives σ = ε C − σ , where symmetry of [C] has been used

T T T − 1 Θ " Using the last expression with σ ε σ = σ ε − σ C σ gives

T T T − 1 T − 1 σ ε σ = σ ε − ε C C σ Θ + σ Θ C σ Θ which simpliﬁes

T T T − 1 to σ ε σ = σ − σ Θ ε + σ Θ C σ Θ

104 CLAPEYRON’S FORMULA IN ABRIDGED NOTATION CONCLUDED

Θ Θ T − 1 Θ T 2 " Next, using σ = C α Θ gives σ C σ = α C α Θ

T σ T Θ Θ T − 1 Θ " Using the last equation with σ ε = σ − σ ε + σ C σ

T T T 2 gives σ ε σ = σ − σ Θ ε + α C α Θ

T " = 1 σ Therefore, Clapeyron’s formula U 2 σ ε becomes

T T Θ T 1 + 1 2 UU = 2 σσ − σσ εε 2 αα C αα Θ

105 PHYSICAL MEANING OF THE ELASTIC CONSTANTS

" The shaded terms shown below correspond to independent interaction between pure extensional stresses and strains

C 11 C12 C13 C14 C15 C16 S 11 S12 S13 S14 S15 S16

S12 S22 S23 S24 S25 S26 C12 C22 C23 C24 C25 C26

S13 S23 S33 S34 S35 S36 C13 C23 C33 C34 C35 C36

S14 S24 S34 S44 S45 S46 C14 C24 C34 C44 C45 C46

S15 S25 S35 S45 S55 S56 C15 C25 C35 C45 C55 C56

S16 S26 S36 S46 S56 S66 C16 C26 C36 C46 C56 C66

" The shaded terms shown below correspond to independent interaction between pure shearing stresses and strains

C 11 C12 C13 C14 C15 C16 S 11 S12 S13 S14 S15 S16

S12 S22 S23 S24 S25 S26 C12 C22 C23 C24 C25 C26

S13 S23 S33 S34 S35 S36 C13 C23 C33 C34 C35 C36

S14 S24 S34 S44 S45 S46 C14 C24 C34 C44 C45 C46

S15 S25 S35 S45 S55 S56 C15 C25 C35 C45 C55 C56

S16 S26 S36 S46 S56 S66 C16 C26 C36 C46 C56 C66

106 PHYSICAL MEANING OF THE ELASTIC CONSTANTS CONTINUED

" The shaded terms shown below correspond to coupling of interactions between pure extensional stresses and strains

C 11 C12 C13 C14 C15 C16 S 11 S12 S13 S14 S15 S16

S12 S22 S23 S24 S25 S26 C12 C22 C23 C24 C25 C26

S13 S23 S33 S34 S35 S36 C13 C23 C33 C34 C35 C36

S14 S24 S34 S44 S45 S46 C14 C24 C34 C44 C45 C46

S15 S25 S35 S45 S55 S56 C15 C25 C35 C45 C55 C56

S16 S26 S36 S46 S56 S66 C16 C26 C36 C46 C56 C66

" The shaded terms shown below correspond to coupling of interactions between pure shearing stresses and strains

C 11 C12 C13 C14 C15 C16 S 11 S12 S13 S14 S15 S16

S12 S22 S23 S24 S25 S26 C12 C22 C23 C24 C25 C26

S13 S23 S33 S34 S35 S36 C13 C23 C33 C34 C35 C36

S14 S24 S34 S44 S45 S46 C14 C24 C34 C44 C45 C46

S15 S25 S35 S45 S55 S56 C15 C25 C35 C45 C55 C56

S16 S26 S36 S46 S56 S66 C16 C26 C36 C46 C56 C66

107 PHYSICAL MEANING OF THE ELASTIC CONSTANTS CONCLUDED

" The shaded terms shown below correspond to coupling or interactions between extensional and shearing behavior

C 11 C12 C13 C14 C15 C16 S 11 S12 S13 S14 S15 S16

S12 S22 S23 S24 S25 S26 C12 C22 C23 C24 C25 C26

S13 S23 S33 S34 S35 S36 C13 C23 C33 C34 C35 C36

S14 S24 S34 S44 S45 S46 C14 C24 C34 C44 C45 C46

S15 S25 S35 S45 S55 S56 C15 C25 C35 C45 C55 C56

S16 S26 S36 S46 S56 S66 C16 C26 C36 C46 C56 C66

108 TRANSFORMATION EQUATIONS

109 TRANSFORMATION OF [C] AND [S]

" Consider a general orthogonal transformation between the Cartesian

coordinates x1, x2, x3 and x1′, x2′, x3′ 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 Anisotropic is considered, coordinate body, B i 3 translations are excluded i 3′ i 2′ P " Although two right-handed i 2 i coordinate systems are 1 i 1′ shown in the ﬁgure, there is no such restriction on the following development of the transformation equations

110 TRANSFORMATION OF [C] AND [S] - CONTINUED

" The general relationship between the two orthonormal bases is given by the following matrix representations

i 1′ a 1′1 a 1′2 a 1′3 i 1

" i 2′ = a 2′1 a 2′2 a 2′3 i 2 or, in abridged form, i′ = a i

i 3′ a 3′1 a 3′2 a 3′3 i 3

i 1 a 1′1 a 2′1 a 3′1 i 1′ − 1 " i 2 = a 1′2 a 2′2 a 3′2 i 2′ or, in abridged form, i = a i′

i 3 a 1′3 a 2′3 a 3′3 i 3′

− 1 T " Examining these two matrices indicates that a = a

" Transformations of this type are known as orthogonal transformations and preserve the lengths of, and the angles between, vectors

a = a a = a " a p′q ≡ i p′ • i q = i q • i p′ = a qp′ and p′q qp′ ≠≠ pq′ q′p

111 TRANSFORMATION OF [C] AND [S] - CONTINUED

" The matrix equation i′ = a i is expressed in indicial notation by

i k′ = ak′qi q

− 1

" Likewise, i = a i′ is expressed in indicial notation by i p = ar′pi r′

" Important relationships between the direction cosines ak′q are obtained

by enforcing the two orthonormality conditions i k′ • i p′ = δkp and

i m • i n = δmn

" These conditions yield the relationships

a k′qa p′q = δδkp and a q′ka q′p = δδkp

" Each indicial equation possesses six independent relations

112 TRANSFORMATION OF [C] AND [S] - CONTINUED

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

a a = k p ak′qap′q = δkp k p q′k q′p δkp

2 2 2 2 2 2 1 1 a1′1 + a1′2 + a1′3 = 1 1 1 a1′1 + a2′1 + a3′1 = 1

2 1 a2′1a1′1 + a2′2a1′2 + a2′3a1′3 = 0 2 1 a1′2a1′1 + a2′2a2′1 + a3′2a3′1 = 0

3 1 a3′1a1′1 + a3′2a1′2 + a3′3a1′3 = 0 3 1 a1′3a1′1 + a2′3a2′1 + a3′3a3′1 = 0

2 2 2 2 2 2 2 2 a2′1 + a2′2 + a2′3 = 1 2 2 a1′2 + a2′2 + a3′2 = 1

3 2 a3′1a2′1 + a3′2a2′2 + a3′3a2′3 = 0 3 2 a1′3a1′2 + a2′3a2′2 + a3′3a3′2 = 0

2 2 2 2 2 2 3 3 a3′1 + a3′2 + a3′3 = 1 3 3 a1′3 + a2′3 + a3′3 = 1

113 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

σ′ = Tσ σ and ε′ = Tε ε where

ε11 σ11 σ1 ε1

σ22 σ2 ε22 ε2

σ33 σ3 ε ε σ = = ε = 33 = 3 σ23 σ4 2ε23 ε4 σ13 σ5 ε5 2ε13 σ12 σ6 ε6 2ε12

σ 1′1′ ε1′1′ σ 1′ ε1′ σ 2′2′ σ 2′ ε2′2′ ε2′ σ 3′3′ σ 3′ ε3 3 ε ′ ′ 3′ and σ′ = = σ ε′ = = σ 2′3′ 4′ 2ε2′3′ ε4′ σ 5′ ε5 σ 1′3′ 2ε1′3′ ′ σ 6′ ε6′ σ 1′2′ 2ε1′2′

114 TRANSFORMATION OF [C] AND [S] - CONTINUED

2 2 2 a1′1 a1′2 a1′3 2a1′2a1′3 2a1′1a1′3 2a1′1a1′2 2 2 2 a2′1 a2′2 a2′3 2a2′2a2′3 2a2′1a2′3 2a2′1a2′2 2 2 2 a3′1 a3′2 a3′3 2a3′2a3′3 2a3′1a3′3 2a3′1a3′2 Tσ = 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

2 2 2 a1′1 a1′2 a1′3 a1′2a1′3 a1′1a1′3 a1′1a1′2 2 2 2 a2′1 a2′2 a2′3 a2′2a2′3 a2′1a2′3 a2′1a2′2 2 2 2 a3′1 a3′2 a3′3 a3′2a3′3 a3′1a3′3 a3′1a3′2 Tε = 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

a j′p ≡ i j′ • i p

" Inspection of the matrices shown above indicates that when the off- diagonal terms vanish, which happens for certain transformations, the two matrices are identical

115 TRANSFORMATION OF [C] AND [S] - CONTINUED

x3′ x3

a j′p ≡≡ i j′ • i p

i x 3′ i 2′ 2′ i 3 x2 i P 2 x2 i 1 P i 1′

x1 x1′ Initial coordinate frame Initial and new coordinate frame

116 TRANSFORMATION OF [C] AND [S] - CONTINUED

" Recall that the thermoelastic constitutive equations are expressed in symbolic form by

ε = S σ + α Θ and σ = C ε + β Θ

where Θ = T − Tref

" In the x 1′, x 2′, x 3′ coordinate frame the thermoelastic constitutive equations are expressed in symbolic form by

ε ′ = S′ σ′ + α′ Θ and σ ′ = C′ ε′ + β′ Θ

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

− 1 − 1 σ = C ε + β Θ becomes Tσ σ′ = C Tε ε′ + β Θ

117 ! ! !

TRANSFORMATION OF [C] AND [S] - CONTINUED

" Premultiplying by Tσ gives

− 1 σ′ = Tσ C Tε ε′ + Tσ β Θ

" Comparing this equation with σ ′ = C′ ε′ + β′ Θ it follows that

− 1 C′′ = Tσ C Tε and ββ′′ = Tσ ββ

− 1 " Thus, C = Tσ C′′ Tε

118 ! ! ! !

TRANSFORMATION OF [C] AND [S] - CONTINUED

" Next, by using the matrix form of the stress and strain transformation equations, ε = S σ + α Θ becomes

− 1 − 1 Tε ε′ = S Tσ σ′ + α Θ

" Premultiplying by Tε gives

− 1 ε′ = Tε S Tσ σ′ + Tε α Θ

" Comparing this equation with ε ′ = S′ σ′ + α′ Θ it follows that

− 1 S′′ = Tε S Tσ and αα′′ = Tε αα

− 1 − 1 " Thus, S′ = Tε S Tσ gives S = Tε S′ Tσ

119 TRANSFORMATION OF [C] AND [S] CONCLUDED

" In summary:

− 1 − 1 S′′ = Tε S Tσ S = Tε S′′ Tσ

− 1 − 1 C′′ = Tσ C Tε C = Tσ C′′ Tε

αα′′ = Tε αα ββ′′ = Tσ ββ

120 TRANSFORMATIONS FOR DEXTRAL ROTATIONS

ABOUT THE x3 AXIS

x3 ,x3′

a i′j ≡≡ i i′ • i j

i 3 , i 3′ x2′

i 2′ θ3 x2 i 1 i 2

i 1′

θ3 Plane x3 = 0 x 1′ x1

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

121 TRANSFORMATIONS FOR DEXTRAL ROTATIONS

ABOUT THE x3 AXIS - CONTINUED

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

2 2 cos θ3 sin θ3 0 0 0 2sinθ3cosθ3 2 2 σ 1′1′ σ 11 sin θ3 cos θ3 0 0 0 − 2sinθ3cosθ3 σ 2′2′ σ 22 0 0 1 0 0 0 σ 3 3 σ 33 ′ ′ = σ 2 3 σ 23 ′ ′ 0 0 0 cosθ3 − sinθ3 0 σ 1′3′ σ 13 0 0 0 sinθ3 cosθ3 0 σ 1′2′ σ 12 2 2 − sinθ3cosθ3 sinθ3cosθ3 0 0 0 cos θ3 − sin θ3

and is expressed symbolically by

σσ′′ = Tσ θ 3 σσ

122 TRANSFORMATIONS FOR DEXTRAL ROTATIONS

ABOUT THE x3 AXIS - CONTINUED

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

2 2 cos θ3 sin θ3 0 0 0 − 2sinθ3cosθ3 2 2 σ 11 σ 1′1′ sin θ3 cos θ3 0 0 0 2sinθ3cosθ3 σ 22 σ 2′2′ 0 0 1 0 0 0 σ 33 σ 3 3 = ′ ′ σ 23 σ 2 3 0 0 0 cosθ3 sinθ3 0 ′ ′ σ 13 σ 1′3′ 0 0 0 − sinθ3 cosθ3 0 σ 12 σ 1′2′ 2 2 sinθ3cosθ3 − sinθ3cosθ3 0 0 0 cos θ3 − sin θ3

and is expressed symbolically by

− 1 σσ = Tσ θ 3 σσ′′

− 1 − where Tσ θ 3 = Tσ θ 3

123 ! !

TRANSFORMATIONS FOR DEXTRAL ROTATIONS

ABOUT THE x3 AXIS - CONTINUED

" By using the second-order, symmetric tensor transformation equations, the transformation law for the vector of engineering strains is given by

2 2 cos #3 sin #3 0 0 0 sin#3cos#3 !1"1" 2 2 !11 sin #3 cos #3 0 0 0 $ sin#3cos#3 !2"2" !22 0 0 1 0 0 0 !3 3 !33 " " = 2!2"3" 0 0 0 cos#3 $ sin#3 0 2!23

2!1"3" 2!13 0 0 0 sin#3 cos#3 0 2 2!1"2" 2 2!12 $ 2sin#3cos#3 2sin#3cos#3 0 0 0 cos #3 $ sin #3

which is expressed symbolically by

!!"" = T! # 3 !!

T $ 1 $ " Note that T! # 3 = T% # 3 = T% # 3

124 TRANSFORMATIONS FOR DEXTRAL ROTATIONS

ABOUT THE x3 AXIS - CONTINUED

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

2 2 cos θ3 sin θ3 0 0 0 − sinθ3cosθ3

ε11 2 2 ε1′1′ sin θ3 cos θ3 0 0 0 sinθ3cosθ3 ε22 ε2′2′ 0 0 1 0 0 0 ε33 ε3′3′ = 2ε23 0 0 0 cosθ3 sinθ3 0 2ε2′3′ 2 ε13 2ε1′3′ 0 0 0 − sinθ3 cosθ3 0 2ε12 2 2 2ε1′2′ 2sinθ3cosθ3 − 2sinθ3cosθ3 0 0 0 cos θ3 − sin θ3

and is expressed symbolically by

− 1 εε = Tε θ 3 εε′′

− 1 T − where Tε θ 3 = Tε θ 3 = Tσ θ 3

125 TRANSFORMATIONS FOR DEXTRAL ROTATIONS

ABOUT THE x3 AXIS - CONTINUED

" The general expression for the transformation of the stiffness coefﬁcients and thermal moduli are

− 1 C′ = Tσ C Tε and β′ = Tσ β

− 1 T " Noting that Tε = Tσ for a dextral rotation about the x3 axis gives

T C′ = Tσ C Tσ

" Similarly, the general expression for the inverse transformation of the − 1 stiffness coefﬁcients is C = Tσ C′ Tε

− 1 T " Noting that Tσ = Tε for a dextral rotation about the x3 axis gives

T C = Tε C′ Tε

126 TRANSFORMATIONS FOR DEXTRAL ROTATIONS

ABOUT THE x3 AXIS - CONTINUED

" The general expression for the transformation of the compliance coefﬁcients and thermal expansion coefﬁcients are

− 1 S′ = Tε S Tσ and α′ = Tε α

− 1 T " Noting that Tσ = Tε for a dextral rotation about the x3 axis gives

T S′ = Tε S Tε

" Similarly, the general expression for the inverse transformation of the − 1 compliance coefﬁcients is S = Tε S′ Tσ

− 1 T " Noting that Tε = Tσ for a dextral rotation about the x3 axis gives

T S = Tσ S′ Tσ

127

TRANSFORMATIONS FOR DEXTRAL ROTATIONS ABOUT

THE x3 AXIS - SUMMARY

T T S! = T" S T" S = T# S! T# T T C! = T# C T# C = T" C! T"

2 2 cos $3 sin $3 0 0 0 2sin$3cos$3 2 2 sin $3 cos $3 0 0 0 % 2sin$3cos$3 0 0 1 0 0 0 T = # 0 0 0 cos$3 % sin$3 0

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

2 2 cos $3 sin $3 0 0 0 sin$3cos$3 2 2 sin $3 cos $3 0 0 0 % sin$3cos$3 0 0 1 0 0 0 T = " 0 0 0 cos$3 % sin$3 0

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

128

TRANSFORMATIONS FOR DEXTRAL ROTATIONS ABOUT

THE x3 AXIS - SUMMARY

T !" = T# ! ! = T$ !"

T %" = T$ % % = T# %"

2 2 cos &3 sin &3 0 0 0 2sin&3cos&3 2 2 sin &3 cos &3 0 0 0 ' 2sin&3cos&3 T = 0 0 1 0 0 0 $ 0 0 0 cos&3 ' sin&3 0

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

2 2 cos &3 sin &3 0 0 0 sin&3cos&3 2 2 sin &3 cos &3 0 0 0 ' sin&3cos&3 0 0 1 0 0 0 T = # 0 0 0 cos&3 ' sin&3 0

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

129 TRANSFORMATIONS FOR DEXTRAL ROTATIONS

ABOUT THE x3 AXIS - CONTINUED

" Let m = cosθ3 and n = sinθ3 for convenience

T " Performing the calculations given by C′ = Tσ C Tσ yields

4 2 2 2 2 4 C 1′1′ = m C 11 + 2m n C 12 + 2C 66 + 4mn m C 16 + n C 26 + n C 22

2 2 2 2 4 4 C 1′2′ = m n C 11 + C 22 − 4C 66 − 2m n m − n C 16 − C 26 + m + n C 12

2 2 C 1′3′ = m C 13 + n C 23 + 2mn C 36

3 2 2 3 C 1′4′ = m C 14 + m n 2C 46 − C 15 − mn 2C 56 − C 24 − n C 25

3 2 2 3 C 1′5′ = m C 15 + m n 2C 56 + C 14 + mn 2C 46 + C 25 + n C 24

2 2 2 3 C1′6′ = m m − 3n C16 − m n C11 − C12 − 2C66 3 2 2 2 + mn C22 − C12 − 2C66 − n n − 3m C26

130 TRANSFORMATIONS FOR DEXTRAL ROTATIONS

ABOUT THE x3 AXIS - CONTINUED

4 2 2 2 2 4 C 2′2′ = m C 22 + 2m n C 12 + 2C 66 − 4m n m C 26 + n C 16 + n C 11

2 2 C 2′3′ = m C 23 + n C 13 − 2m nC 36

3 2 2 3 C 2′4′ = m C 24 − m n 2C 46 + C 25 + mn 2C 56 + C 14 − n C 15

3 2 2 3 C 2′5′ = m C 25 − m n 2C 56 − C 24 − mn 2C 46 − C 15 + n C 14

2 2 2 3 C2′6′ = m m − 3n C26 + m n C22 − C12 − 2C66 3 2 2 2 − mn C11 − C12 − 2C66 − n n − 3m C16

C 3′3′ = C 33

C 3′4′ = m C 34 − nC 35

C 3′5′ = m C 35 + nC 34

131 TRANSFORMATIONS FOR DEXTRAL ROTATIONS

ABOUT THE x3 AXIS - CONTINUED

2 2 C 3′6′ = m − n C 36 + mn C 23 − C 13

2 2 C 4′4′ = m C 44 + n C 55 − 2m nC 45

2 2 C 4′5′ = m − n C 45 + mn C 44 − C 55

3 2 2 3 C 4′6′ = m C 46 − m n C 56 + C 14 − C 24 − mn C 46 − C 15 + C 25 + n C 56

2 2 C 5′5′ = m C 55 + n C 44 + 2mn C 45

3 2 2 3 C 5′6′ = m C 56 + m n C 46 + C 25 − C 15 − mn C 56 + C 14 − C 24 − n C 46

2 2 2 2 2 2 2 C 6′6′ = m n C 11 + C 22 − 2C 12 − 2mn m − n C 16 − C 26 + m − n C 66

132 TRANSFORMATIONS FOR DEXTRAL ROTATIONS

ABOUT THE x3 AXIS - CONTINUED

" Again, let m = cosθ3 and n = sinθ3

T " Performing the calculations given by C = Tε C′ Tε yields

4 2 2 2 2 4 C 11 = m C 1′1′ + 2m n C 1′2′ + 2C 6′6′ − 4m n m C 1′6′ + n C 2′6′ + n C 2′2′

2 2 2 2 4 4 C 12 = m n C 1′1′ + C 2′2′ − 4C 6′6′ + 2mn m − n C 1′6′ − C 2′6′ + m + n C 1′2′

2 2 C 13 = m C 1′3′ + n C 2′3′ − 2m nC 3′6′

3 2 2 3 C 14 = m C 1′4′ − m n 2C 4′6′ − C 1′5′ − mn 2C 5′6′ − C 2′4′ + n C 2′5′

3 2 2 3 C 15 = m C 1′5′ − m n 2C 5′6′ + C 1′4′ + mn 2C 4′6′ + C 2′5′ − n C 2′4′

2 2 2 3 C16 = m m − 3n C1′6′ + m n C1′1′ − C1′2′ − 2C6′6′ 3 2 2 2 − mn C2′2′ − C1′2′ − 2C6′6′ − n n − 3m C2′6′

133 TRANSFORMATIONS FOR DEXTRAL ROTATIONS

ABOUT THE x3 AXIS - CONTINUED

4 2 2 2 2 4 C 22 = m C 2′2′ + 2m n C 1′2′ + 2C 6′6′ + 4mn m C 2′6′ + n C 1′6′ + n C 1′1′

2 2 C 23 = m C 2′3′ + n C 1′3′ + 2mn C 3′6′

3 2 2 3 C 24 = m C 2′4′ + m n 2C 4′6′ + C 2′5′ + mn 2C 5′6′ + C 1′4′ + n C 1′5′

3 2 2 3 C 25 = m C 2′5′ + m n 2C 5′6′ − C 2′4′ − mn 2C 4′6′ − C 1′5′ − n C 1′4′

2 2 2 3 C26 = m m − 3n C2′6′ − m n C2′2′ − C1′2′ − 2C6′6′ 3 2 2 2 + mn C1′1′ − C1′2′ − 2C6′6′ − n n − 3m C1′6′

C 33 = C 3′3′ C 34 = m C 3′4′ + nC 3′5′ C 35 = m C 3′5′ − nC 3′4′

2 2 C 36 = m − n C 3′6′ − mn C 2′3′ − C 1′3′

2 2 C 44 = m C 4′4′ + n C 5′5′ + 2mn C 4′5′

134 TRANSFORMATIONS FOR DEXTRAL ROTATIONS

ABOUT THE x3 AXIS - CONTINUED

2 2 C 45 = m − n C 4′5′ − mn C 4′4′ − C 5′5′

3 2 2 3 C 46 = m C 4′6′ + m n C 5′6′ + C 1′4′ − C 2′4′ − mn C 4′6′ − C 1′5′ + C 2′5′ − n C 5′6′

2 2 C 55 = m C 5′5′ + n C 4′4′ − 2m nC 4′5′

3 2 2 3 C 56 = m C 5′6′ − m n C 4′6′ + C 2′5′ − C 1′5′ − mn C 5′6′ + C 1′4′ − C 2′4′ + n C 4′6′

2 2 2 2 2 2 2 C 66 = m n C 1′1′ + C 2′2′ − 2C 1′2′ + 2mn m − n C 1′6′ − C 2′6′ + m − n C 6′6′

" Note that C ′ and C can be expressed as

C′ = Tσ θ 3 C Tε − θ 3 and C = Tσ − θ 3 C′ Tε θ 3

" 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

135 TRANSFORMATIONS FOR DEXTRAL ROTATIONS

ABOUT THE x3 AXIS - CONTINUED

" Let m = cosθ3 and n = sinθ3

T " Performing the calculations given by S′ = Tε S Tε yields

4 2 2 2 2 4 S 1′1′ = m S 11 + m n 2S 12 + S 66 + 2m n m S 16 + n S 26 + n S 22

2 2 2 2 4 4 S 1′2′ = m n S 11 + S 22 − S 66 − mn m − n S 16 − S 26 + m + n S 12

2 2 S 1′3′ = m S 13 + n S 23 + mnS 36

3 2 2 3 S 1′4′ = m S 14 + m n S 46 − S 15 − mn S 56 − S 24 − n S 25

3 2 2 3 S 1′5′ = m S 15 + m n S 56 + S 14 + mn S 46 + S 25 + n S 24

2 2 2 3 S1′6′ = m m − 3n S16 − m n 2S11 − 2S12 − S66 3 2 2 2 + mn 2S22 − 2S12 − S66 − n n − 3m S26

136 TRANSFORMATIONS FOR DEXTRAL ROTATIONS

ABOUT THE x3 AXIS - CONTINUED

4 2 2 2 2 4 S 2′2′ = m S 22 + m n 2S 12 + S 66 − 2m n m S 26 + n S 16 + n S 11

2 2 S 2′3′ = m S 23 + n S 13 − mn S 36

3 2 2 3 S 2′4′ = m S 24 − m n S 46 + S 25 + mn S 56 + S 14 − n S 15

3 2 2 3 S 2′5′ = m S 25 − m n S 56 − S 24 − mn S 46 − S 15 + n S 14

2 2 2 3 S2′6′ = m m − 3n S26 + m n 2S22 − 2S12 − S66 3 2 2 2 − mn 2S11 − 2S12 − S66 − n n − 3m S16

S 3′3′ = S 33

S 3′4′ = m S 34 − nS 35

S 3′5′ = m S 35 + nS 34

137 TRANSFORMATIONS FOR DEXTRAL ROTATIONS

ABOUT THE x3 AXIS - CONTINUED

2 2 S 3′6′ = m − n S 36 + 2mn S 23 − S 13

2 2 S 4′4′ = m S 44 + n S 55 − 2m nS 45

2 2 S 4′5′ = m − n S 45 + mn S 44 − S 55

3 2 2 3 S 4′6′ = m S 46 − m n S 56 + 2S 14 − 2S 24 − mn S 46 − 2S 15 + 2S 25 + n S 56

2 2 S 5′5′ = m S 55 + n S 44 + 2mn S 45

3 2 2 3 S 5′6′ = m S 56 + m n S 46 + 2S 25 − 2S 15 − mn S 56 + 2S 14 − 2S 24 − n S 46

2 2 2 2 2 2 2 S 6′6′ = 4m n S 11 + S 22 − 2S 12 − 4mn m − n S 16 − S 26 + m − n S 66

138 TRANSFORMATIONS FOR DEXTRAL ROTATIONS

ABOUT THE x3 AXIS - CONTINUED

" Let m = cosθ3 and n = sinθ3

T " Performing the calculations given by S = Tσ S′ Tσ yields

4 2 2 2 2 4 S 11 = m S 1′1′ + m n 2S 1′2′ + S 6′6′ − 2m n m S 1′6′ + n S 2′6′ + n S 2′2′

2 2 2 2 4 4 S 12 = m n S 1′1′ + S 2′2′ − S 6′6′ + m n m − n S 1′6′ − S 2′6′ + m + n S 1′2′

2 2 S 13 = m S 1′3′ + n S 2′3′ − mn S 3′6′

3 2 2 3 S 14 = m S 1′4′ − m n S 4′6′ − S 1′5′ − mn S 5′6′ − S 2′4′ + n S 2′5′

3 2 2 3 S 15 = m S 1′5′ − m n S 5′6′ + S 1′4′ + mn S 4′6′ + S 2′5′ − n S 2′4′

2 2 2 3 S16 = m m − 3n S1′6′ + m n 2S1′1′ − 2S1′2′ − S6′6′ 3 2 2 2 − mn 2S2′2′ − 2S1′2′ − S6′6′ − n n − 3m S2′6′

139 TRANSFORMATIONS FOR DEXTRAL ROTATIONS

ABOUT THE x3 AXIS - CONTINUED

4 2 2 2 2 4 S 22 = m S 2′2′ + m n 2S 1′2′ + S 6′6′ + 2mn m S 2′6′ + n S 1′6′ + n S 1′1′

2 2 S 23 = m S 2′3′ + n S 1′3′ + mnS 3′6′

3 2 2 3 S 24 = m S 2′4′ + m n S 4′6′ + S 2′5′ + mn S 5′6′ + S 1′4′ + n S 1′5′

3 2 2 3 S 25 = m S 2′5′ + m n S 5′6′ − S 2′4′ − mn S 4′6′ − S 1′5′ − n S 1′4′

2 2 2 3 S26 = m m − 3n S2′6′ − m n 2S2′2′ − 2S1′2′ − S6′6′ 3 2 2 2 + mn 2S1′1′ − 2S1′2′ − S6′6′ − n n − 3m S1′6′

S 33 = S 3′3′ S 34 = m S 3′4′ + nS 3′5′ S 35 = m S 3′5′ − nS 3′4′

2 2 S 36 = m − n S 3′6′ − 2mn S 2′3′ − S 1′3′

2 2 S 44 = m S 4′4′ + n S 5′5′ + 2mnS 4′5′

140 TRANSFORMATIONS FOR DEXTRAL ROTATIONS

ABOUT THE x3 AXIS - CONTINUED

2 2 S 45 = m − n S 4′5′ − mn S 4′4′ − S 5′5′

3 2 2 3 S 46 = m S 4′6′ + m n S 5′6′ + 2S 1′4′ − 2S 2′4′ − mn S 4′6′ − 2S 1′5′ + 2S 2′5′ − n S 5′6′

2 2 S 55 = m S 5′5′ + n S 4′4′ − 2m nS 4′5′

3 2 2 3 S 56 = m S 5′6′ − m n S 4′6′ + 2S 2′5′ − 2S 1′5′ − mn S 5′6′ + 2S 1′4′ − 2S 2′4′ + n S 4′6′

2 2 2 2 2 2 2 S 66 = 4m n S 1′1′ + S 2′2′ − 2S 1′2′ + 4mn m − n S 1′6′ − S 2′6′ + m − n S 6′6′

" Note that S ′ and S can be expressed as

S′ = Tε θ 3 S Tσ − θ 3 and S = Tε − θ 3 S′ Tσ θ 3

" 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

141 TRANSFORMATIONS FOR DEXTRAL ROTATIONS

ABOUT THE x3 AXIS - CONTINUED

" Let m = cosθ3 and n = sinθ3

" Performing the calculations given by α′ = Tε α yields

2 2 α 1′1′ = m α 11 + 2m nα 12 + n α 22

2 2 α 2′2′ = m α 22 − 2m nα 12 + n α 11

α 3′3′ = α 33

α 2′3′ = m α 23 − nα 13

α 1′3′ = m α 13 + nα 23

2 2 α 1′2′ = m − n α 12 + m n α 22 − α 11

142 ! !

TRANSFORMATIONS FOR DEXTRAL ROTATIONS

ABOUT THE x3 AXIS - CONTINUED

" Let m = cosθ3 and n = sinθ3

T " Performing the calculations given by α = Tσ α′ yields

= m 2 2m n + n 2 α 11 α 1′1′ − α 1′2′ α 2′2′ Note that T 2 2 T θ 3 = T − θ 3 α 22 = m α 2′2′ + 2m nα 1′2′ + n α 1′1′ σ ε and hence

α 33 = α 3′3′ α = Tε − θ3 α′

α 23 = m α 2′3′ + nα 1′3′ So, the expressions given here

for α ij can be obtained from the

αi′ j′ α 13 = m α 1 3 − nα 2 3 those previously given for ′ ′ ′ ′ by switching the primed and

2 2 unprimed indices and replacing α 12 = m − n α 1′2′ − mn α 2′2′ − α 1′1′ n with -n

143 TRANSFORMATIONS FOR DEXTRAL ROTATIONS

ABOUT THE x3 AXIS - CONTINUED

" Let m = cosθ3 and n = sinθ3

" Performing the calculations given by β′ = Tσ β yields

2 2 β 1′1′ = m β 11 + 2m nβ 12 + n β 22

2 2 β 2′2′ = m β 22 − 2m nβ 12 + n β 11

β 3′3′ = β 33

β 2′3′ = m β 23 − nβ 13

β 1′3′ = m β 13 + nβ 23

2 2 β 1′2′ = m − n β 12 + m n β 22 − β 11

144 TRANSFORMATIONS FOR DEXTRAL ROTATIONS

ABOUT THE x3 AXIS - CONCLUDED

" Let m = cosθ3 and n = sinθ3

T " Performing the calculations given by β = Tε β′ yields

= m 2 2m n + n 2 β 11 β 1′1′ − β 1′2′ β 2′2′ Note that T 2 2 T θ 3 = T − θ 3 β 22 = m β 2′2′ + 2m nβ 1′2′ + n β 1′1′ ε σ and hence

β 33 = β 3′3′ β = Tσ − θ3 β′

β 23 = m β 2′3′ + nβ 1′3′ So, the expressions given here

for β ij can be obtained from the

β i′ j′ β 13 = m β 1 3 − nβ 2 3 those previously given for ′ ′ ′ ′ by switching the primed and

2 2 unprimed indices and replacing β 12 = m − n β 1′2′ − mn β 2′2′ − β 1′1′ n with -n

145 TRANSFORMATIONS FOR DEXTRAL ROTATIONS

ABOUT THE x1 AXIS

x3 x3′

θ1 Plane x1 = 0

i 3 x2′

i 3′ i 2′ θ1 x2

i 2

i 1 , i 1′ a i′j ≡≡ i i′ • i j

x1 , x1′

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

146 TRANSFORMATIONS FOR DEXTRAL ROTATIONS

ABOUT THE x1 AXIS - CONTINUED

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

1 0 0 0 0 0

2 2 σ 1′1′ σ 11 0 cos θ1 sin θ1 2sinθ1cosθ1 0 0 σ 2′2′ 2 2 σ 22 0 sin θ1 cos θ1 − 2sinθ1cosθ1 0 0 σ 3′3′ σ 33 = 2 2 σ 2 3 σ 23 ′ ′ 0 − sinθ1cosθ1 sinθ1cosθ1 cos θ1 – sin θ1 0 0 σ 1′3′ σ 13 0 0 0 0 cosθ1 − sinθ1 σ 1′2′ σ 12

0 0 0 0 sinθ1 cosθ1

and is expressed symbolically by

σσ′′ = Tσ θ 1 σσ

147 TRANSFORMATIONS FOR DEXTRAL ROTATIONS

ABOUT THE x1 AXIS - CONTINUED

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

1 0 0 0 0 0

2 2 σ 11 σ 1′1′ 0 cos θ1 sin θ1 − 2sinθ1cosθ1 0 0 σ 22 2 2 σ 2′2′ 0 sin θ1 cos θ1 2sinθ1cosθ1 0 0 σ 33 σ 3′3′ = 2 2 σ 23 σ 2 3 0 sinθ1cosθ1 − sinθ1cosθ1 cos θ1 − sin θ1 0 0 ′ ′ σ 13 σ 1′3′ 0 0 0 0 cosθ1 sinθ1 σ 12 σ 1′2′

0 0 0 0 − sinθ1 cosθ1

and is expressed symbolically by

− 1 σσ = Tσ θ 1 σσ′′

− 1 − where Tσ θ 1 = Tσ θ 1

148 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

1 0 0 0 0 0 2 ε1′1′ 2 ε11 0 cos θ1 sin θ1 sinθ1cosθ1 0 0 ε2 2 ε22 ′ ′ 2 2 0 sin θ1 cos θ1 − sinθ1cosθ1 0 0 ε3′3′ ε33 = 2 2 2ε2′3′ 0 − 2sinθ1cosθ1 2sinθ1cosθ1 cos θ1 − sin θ1 0 0 2ε23

2ε1′3′ 2ε13 0 0 0 0 cosθ1 − sinθ1 2ε1′2′ 2ε12 0 0 0 0 sinθ1 cosθ1

which is expressed symbolically by

εε′′ = Tε θ 1 εε

T − 1 − " Note that Tε θ 1 = Tσ θ 1 = Tσ θ 1

149 TRANSFORMATIONS FOR DEXTRAL ROTATIONS

ABOUT THE x1 AXIS - CONTINUED

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

1 0 0 0 0 0

ε11 2 2 ε1′1′ 0 cos θ1 sin θ1 − sinθ1cosθ1 0 0 ε22 ε2′2′ 2 2 0 sin θ1 cos θ1 sinθ1cosθ1 0 0 ε33 ε3′3′ = 2 2 2ε23 0 2sinθ1cosθ1 − 2sinθ1cosθ1 cos θ1 − sin θ1 0 0 2ε2′3′ 2 ε13 2ε1′3′ 0 0 0 0 cosθ1 sinθ1 2ε12 2ε1′2′ 0 0 0 0 − sinθ1 cosθ1

and is expressed symbolically by

− 1 εε = Tε θ 1 εε′′

− 1 T − where Tε θ 1 = Tε θ 1 = Tσ θ 1

150 TRANSFORMATIONS FOR DEXTRAL ROTATIONS

ABOUT THE x1 AXIS - CONTINUED

" The general expression for the transformation of the stiffness coefﬁcients and thermal moduli are

− 1 C′ = Tσ C Tε and β′ = Tσ β

− 1 T " Noting that Tε = Tσ for a dextral rotation about the x1 axis gives

T C′ = Tσ C Tσ

" Similarly, the general expression for the inverse transformation of the − 1 stiffness coefﬁcients is C = Tσ C′ Tε

− 1 T " Noting that Tσ = Tε for a dextral rotation about the x1 axis gives

T C = Tε C′ Tε

151 TRANSFORMATIONS FOR DEXTRAL ROTATIONS

ABOUT THE x1 AXIS - CONTINUED

" The general expression for the transformation of the compliance coefﬁcients and thermal expansion coefﬁcients are

− 1 S′ = Tε S Tσ and α′ = Tε α

− 1 T " Noting that Tσ = Tε for a dextral rotation about the x1 axis gives

T S′ = Tε S Tε

" Similarly, the general expression for the inverse transformation of the − 1 compliance coefﬁcients is S = Tε S′ Tσ

− 1 T " Noting that Tε = Tσ for a dextral rotation about the x1 axis gives

T S = Tσ S′ Tσ

152

TRANSFORMATIONS FOR DEXTRAL ROTATIONS ABOUT

THE x1 AXIS - SUMMARY

T T S! = T" S T" S = T# S! T# T T C! = T# C T# C = T" C! T"

1 0 0 0 0 0

2 2 0 cos $1 sin $1 2sin$1cos$1 0 0 2 2 0 sin $1 cos $1 % 2sin$1cos$1 0 0 T = 2 2 # 0 % sin$1cos$1 sin$1cos$1 cos $1 % sin $1 0 0

0 0 0 0 cos$1 % sin$1

0 0 0 0 sin$1 cos$1

1 0 0 0 0 0

2 2 0 cos $1 sin $1 sin$1cos$1 0 0 2 2 0 sin $1 cos $1 % sin$1cos$1 0 0 T = 2 2 " 0 % 2sin$1cos$1 2sin$1cos$1 cos $1 % sin $1 0 0

0 0 0 0 cos$1 % sin$1

0 0 0 0 sin$1 cos$1

153

TRANSFORMATIONS FOR DEXTRAL ROTATIONS ABOUT

THE x1 AXIS - SUMMARY

T !" = T# ! ! = T$ !"

T %" = T$ % % = T# %"

1 0 0 0 0 0

2 2 0 cos &1 sin &1 2sin&1cos&1 0 0 2 2 0 sin &1 cos &1 ' 2sin&1cos&1 0 0 T = 2 2 $ 0 ' sin&1cos&1 sin&1cos&1 cos &1 ' sin &1 0 0

0 0 0 0 cos&1 ' sin&1

0 0 0 0 sin&1 cos&1

1 0 0 0 0 0

2 2 0 cos &1 sin &1 sin&1cos&1 0 0 2 2 0 sin &1 cos &1 ' sin&1cos&1 0 0 T = 2 2 # 0 ' 2sin&1cos&1 2sin&1cos&1 cos &1 ' sin &1 0 0

0 0 0 0 cos&1 ' sin&1

0 0 0 0 sin&1 cos&1

154 TRANSFORMATIONS FOR DEXTRAL ROTATIONS

ABOUT THE x1 AXIS - CONTINUED

" Let m = cosθ1 and n = sinθ1

T " Performing the calculations given by C′ = Tσ C Tσ yields

C 1′1′ = C 11

2 2 C 1′2′ = m C 12 + n C 13 + 2mn C 14

2 2 C 1′3′ = m C 13 + n C 12 − 2m nC 14

2 2 C 1′4′ = m − n C 14 + mn C 13 − C 12

C 1′5′ = m C 15 − nC 16

C 1′6′ = m C 16 + nC 15

155 TRANSFORMATIONS FOR DEXTRAL ROTATIONS

ABOUT THE x1 AXIS - CONTINUED

4 2 2 2 2 4 C 2′2′ = m C 22 + 2m n C 23 + 2C 44 + 4mn m C 24 + n C 34 + n C 33

2 2 2 2 4 4 C 2′3′ = m n C 22 + C 33 − 4C 44 − 2m n m − n C 24 − C 34 + m + n C 23

2 2 2 3 C2′4′ = m m − 3n C24 − m n C22 − C23 − 2C44 3 2 2 2 + mn C33 − C23 − 2C44 − n n − 3m C34

3 2 2 3 C 2′5′ = m C 25 + m n 2C 45 − C 26 − mn 2C 46 − C 35 − n C 36

3 2 2 3 C 2′6′ = m C 26 + m n 2C 46 + C 25 + mn 2C 45 + C 36 + n C 35

4 2 2 2 2 4 C 3′3′ = m C 33 + 2m n C 23 + 2C 44 − 4m n m C 34 + n C 24 + n C 22

2 2 2 3 C3′4′ = m m − 3n C34 + m n C33 − C23 − 2C44 3 2 2 2 − mn C22 − C23 − 2C44 − n n − 3m C24

156 TRANSFORMATIONS FOR DEXTRAL ROTATIONS

ABOUT THE x1 AXIS - CONTINUED

3 2 2 3 C 3′5′ = m C 35 − m n 2C 45 + C 36 + mn 2C 46 + C 25 − n C 26

3 2 2 3 C 3′6′ = m C 36 − m n 2C 46 − C 35 − mn 2C 45 − C 26 + n C 25

2 2 2 2 2 2 2 C 4′4′ = m n C 22 + C 33 − 2C 23 − 2mn m − n C 24 − C 34 + m − n C 44

3 2 2 3 C 4′5′ = m C 45 − m n C 46 + C 25 − C 35 − mn C 45 − C 26 + C 36 + n C 46

3 2 2 3 C 4′6′ = m C 46 + m n C 45 + C 36 − C 26 − mn C 46 + C 25 − C 35 − n C 45

2 2 C 5′5′ = m C 55 + n C 66 − 2m nC 56

2 2 C 5′6′ = m − n C 56 + mn C 55 − C 66

2 2 C 6′6′ = m C 66 + n C 55 + 2mn C 56

157 TRANSFORMATIONS FOR DEXTRAL ROTATIONS

ABOUT THE x1 AXIS - CONTINUED

" Let m = cosθ1 and n = sinθ1

T " Performing the calculations given by C = Tε C′ Tε yields

C 11 = C 1′1′

2 2 C 12 = m C 1′2′ + n C 1′3′ − 2m nC 1′4′

2 2 C 13 = m C 1′3′ + n C 1′2′ + 2mn C 1′4′

2 2 C 14 = m − n C 1′4′ − mn C 1′3′ − C 1′2′

C 15 = m C 1′5′ + nC 1′6′

C 16 = m C 1′6′ − nC 1′5′

4 2 2 2 2 4 C 22 = m C 2′2′ + 2m n C 2′3′ + 2C 4′4′ − 4m n m C 2′4′ + n C 3′4′ + n C 3′3′

158 TRANSFORMATIONS FOR DEXTRAL ROTATIONS

ABOUT THE x1 AXIS - CONTINUED

2 2 2 2 4 4 C 23 = m n C 2′2′ + C 3′3′ − 4C 4′4′ + 2mn m − n C 2′4′ − C 3′4′ + m + n C 2′3′

2 2 2 3 C24 = m m − 3n C2′4′ + m n C2′2′ − C2′3′ − 2C4′4′ 3 2 2 2 − mn C3′3′ − C2′3′ − 2C4′4′ − n n − 3m C3′4′

3 2 2 3 C 25 = m C 2′5′ − m n 2C 4′5′ − C 2′6′ − mn 2C 4′6′ − C 3′5′ + n C 3′6′

3 2 2 3 C 26 = m C 2′6′ − m n 2C 4′6′ + C 2′5′ + mn 2C 4′5′ + C 3′6′ − n C 3′5′

4 2 2 2 2 4 C 33 = m C 3′3′ + 2m n C 2′3′ + 2C 4′4′ + 4mn m C 3′4′ + n C 2′4′ + n C 2′2′

2 2 2 3 C34 = m m − 3n C3′4′ − m n C3′3′ − C2′3′ − 2C4′4′ 3 2 2 2 + mn C2′2′ − C2′3′ − 2C4′4′ − n n − 3m C2′4′

3 2 2 3 C 35 = m C 3′5′ + m n 2C 4′5′ + C 3′6′ + mn 2C 4′6′ + C 2′5′ + n C 2′6′

3 2 2 3 C 36 = m C 3′6′ + m n 2C 4′6′ − C 3′5′ − mn 2C 4′5′ − C 2′6′ − n C 2′5′

159 TRANSFORMATIONS FOR DEXTRAL ROTATIONS

ABOUT THE x1 AXIS - CONTINUED

2 2 2 2 2 2 2 C 44 = m n C 2′2′ + C 3′3′ − 2C 2′3′ + 2mn m − n C 2′4′ − C 3′4′ + m − n C 4′4′

3 2 2 3 C 45 = m C 4′5′ + m n C 4′6′ + C 2′5′ − C 3′5′ − mn C 4′5′ − C 2′6′ + C 3′6′ − n C 4′6′

3 2 2 3 C 46 = m C 4′6′ − m n C 4′5′ + C 3′6′ − C 2′6′ − mn C 4′6′ + C 2′5′ − C 3′5′ + n C 4′5′

2 2 2 2 C 55 = m C 5′5′ + n C 6′6′ + 2mn C 5′6′ C 56 = m − n C 5′6′ − mn C 5′5′ − C 6′6′

2 2 C 66 = m C 6′6′ + n C 5′5′ − 2m nC 5′6′

" Note that C ′ and C can be expressed as

C′ = Tσ θ 1 C Tε − θ 1 and C = Tσ − θ 1 C′ Tε θ 1

" 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

160 TRANSFORMATIONS FOR DEXTRAL ROTATIONS

ABOUT THE x1 AXIS - CONTINUED

" Let m = cosθ1 and n = sinθ1

T " Performing the calculations given by S′ = Tε S Tε yields

S 1′1′ = S 11

2 2 S 1′2′ = m S 12 + n S 13 + mnS 14

2 2 S 1′3′ = m S 13 + n S 12 − mn S 14

2 2 S 1′4′ = m − n S 14 + 2mn S 13 − S 12

S 1′5′ = m S 15 − nS 16

S 1′6′ = m S 16 + nS 15

4 2 2 2 2 4 S 2′2′ = m S 22 + m n 2S 23 + S 44 + 2m n m S 24 + n S 34 + n S 33

161 TRANSFORMATIONS FOR DEXTRAL ROTATIONS

ABOUT THE x1 AXIS - CONTINUED

2 2 2 2 4 4 S 2′3′ = m n S 22 + S 33 − S 44 − mn m − n S 24 − S 34 + m + n S 23

2 2 2 3 S2′4′ = m m − 3n S24 − m n 2S22 − 2S23 − S44 3 2 2 2 + mn 2S33 − 2S23 − S44 − n n − 3m S34

3 2 2 3 S 2′5′ = m S 25 + m n S 45 − S 26 − mn S 46 − S 35 − n S 36

3 2 2 3 S 2′6′ = m S 26 + m n S 46 + S 25 + mn S 45 + S 36 + n S 35

4 2 2 2 2 4 S 3′3′ = m S 33 + m n 2S 23 + S 44 − 2m n m S 34 + n S 24 + n S 22

2 2 2 3 S3′4′ = m m − 3n S34 + m n 2S33 − 2S23 − S44 3 2 2 2 − mn 2S22 − 2S23 − S44 − n n − 3m S24

3 2 2 3 S 3′5′ = m S 35 − m n S 45 + S 36 + mn S 46 + S 25 − n S 26

162 TRANSFORMATIONS FOR DEXTRAL ROTATIONS

ABOUT THE x1 AXIS - CONTINUED

3 2 2 3 S 3′6′ = m S 36 − m n S 46 − S 35 − mn S 45 − S 26 + n S 25

2 2 2 2 2 2 2 S 4′4′ = 4m n S 22 + S 33 − 2S 23 − 4mn m − n S 24 − S 34 + m − n S 44

3 2 2 3 S 4′5′ = m S 45 − m n S 46 + 2S 25 − 2S 35 − mn S 45 − 2S 26 + 2S 36 + n S 46

3 2 2 3 S 4′6′ = m S 46 + m n S 45 + 2S 36 − 2S 26 − mn S 46 + 2S 25 − 2S 35 − n S 45

2 2 S 5′5′ = m S 55 + n S 66 − 2m nS 56

2 2 S 5′6′ = m − n S 56 + mn S 55 − S 66

2 2 S 6′6′ = m S 66 + n S 55 + 2mn S 56

163 TRANSFORMATIONS FOR DEXTRAL ROTATIONS

ABOUT THE x1 AXIS - CONTINUED

" Let m = cosθ1 and n = sinθ1

T " Performing the calculations given by S = Tσ S′ Tσ yields

S 11 = S 1′1′

2 2 S 12 = m S 1′2′ + n S 1′3′ − mn S 1′4′

2 2 S 13 = m S 1′3′ + n S 1′2′ + mnS 1′4′

2 2 S 14 = m − n S 1′4′ − 2mn S 1′3′ − S 1′2′

S 15 = m S 1′5′ + nS 1′6′

S 16 = m S 1′6′ − nS 1′5′

4 2 2 2 2 4 S 22 = m S 2′2′ + m n 2S 2′3′ + S 4′4′ − 2m n m S 2′4′ + n S 3′4′ + n S 3′3′

164 TRANSFORMATIONS FOR DEXTRAL ROTATIONS

ABOUT THE x1 AXIS - CONTINUED

2 2 2 2 4 4 S 23 = m n S 2′2′ + S 3′3′ − S 4′4′ + m n m − n S 2′4′ − S 3′4′ + m + n S 2′3′

2 2 2 3 S24 = m m − 3n S2′4′ + m n 2S2′2′ − 2S2′3′ − S4′4′ 3 2 2 2 − mn 2S3′3′ − 2S2′3′ − S4′4′ − n n − 3m S3′4′

3 2 2 3 S 25 = m S 2′5′ − m n S 4′5′ − S 2′6′ − mn S 4′6′ − S 3′5′ + n S 3′6′

3 2 2 3 S 26 = m S 2′6′ − m n S 4′6′ + S 2′5′ + mn S 4′5′ + S 3′6′ − n S 3′5′

4 2 2 2 2 4 S 33 = m S 3′3′ + m n 2S 2′3′ + S 4′4′ + 2mn m S 3′4′ + n S 2′4′ + n S 2′2′

2 2 2 3 S34 = m m − 3n S3′4′ − m n 2S3′3′ − 2S2′3′ − S4′4′ 3 2 2 2 + mn 2S2′2′ − 2S2′3′ − S4′4′ − n n − 3m S2′4′

3 2 2 3 S 35 = m S 3′5′ + m n S 4′5′ + S 3′6′ + mn S 4′6′ + S 2′5′ + n S 2′6′

3 2 2 3 S 36 = m S 3′6′ + m n S 4′6′ − S 3′5′ − mn S 4′5′ − S 2′6′ − n S 2′5′

165 TRANSFORMATIONS FOR DEXTRAL ROTATIONS

ABOUT THE x1 AXIS - CONTINUED

2 2 2 2 2 2 2 S 44 = 4m n S 2′2′ + S 3′3′ − 2S 2′3′ + 4mn m − n S 2′4′ − S 3′4′ + m − n S 4′4′

3 2 2 3 S 45 = m S 4′5′ + m n S 4′6′ + 2S 2′5′ − 2S 3′5′ − mn S 4′5′ − 2S 2′6′ + 2S 3′6′ − n S 4′6′

3 2 2 3 S 46 = m S 4′6′ − m n S 4′5′ + 2S 3′6′ − 2S 2′6′ − mn S 4′6′ + 2S 2′5′ − 2S 3′5′ + n S 4′5′

2 2 2 2 S 55 = m S 5′5′ + n S 6′6′ + 2mnS 5′6′ S 56 = m − n S 5′6′ − mn S 5′5′ − S 6′6′

2 2 S 66 = m S 6′6′ + n S 5′5′ − 2m nS 5′6′

" Note that S ′ and S can be expressed as

S′ = Tε θ 1 S Tσ − θ 1 and S = Tε − θ 1 S′ Tσ θ 1

" 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

166 TRANSFORMATIONS FOR DEXTRAL ROTATIONS

ABOUT THE x1 AXIS - CONTINUED

" Let m = cosθ1 and n = sinθ1

" Performing the calculations given by α′ = Tε α yields

α 1′1′ = α 11

2 2 α 2′2′ = m α 22 + 2m nα 23 + n α 33

2 2 α 3′3′ = m α 33 − 2m nα 23 + n α 22

2 2 α 2′3′ = m − n α 23 + m n α 33 − α 22

α 1′3′ = m α 13 − nα 12

α 1′2′ = m α 12 + nα 13

167 ! !

TRANSFORMATIONS FOR DEXTRAL ROTATIONS

ABOUT THE x1 AXIS - CONTINUED

" Let m = cosθ1 and n = sinθ1

T " Performing the calculations given by α = Tσ α′ yields

= α 11 α 1′1′ Note that T 2 2 − α 22 = m α 2′2′ − 2m nα 2′3′ + n α 3′3′ Tσ θ 1 = Tε θ 1 and hence 2 2 α 33 = m α 3′3′ + 2m nα 2′3′ + n α 2′2′ α = Tε − θ1 α′ 2 2 α 23 = m − n α 2′3′ − mn α 3′3′ − α 2′2′ So, the expressions given here

for α ij can be obtained from the

those previously given for α i′j ′ α 13 = m α 1 3 + nα 1 2 ′ ′ ′ ′ by switching the primed and unprimed indices and replacing α 12 = m α 1′2′ − nα 1′3′ n with -n

168 TRANSFORMATIONS FOR DEXTRAL ROTATIONS

ABOUT THE x1 AXIS - CONTINUED

" Let m = cosθ1 and n = sinθ1

" Performing the calculations given by β′ = Tσ β yields

β 1′1′ = β 11

2 2 β 2′2′ = m β 22 + 2m nβ 23 + n β 33

2 2 β 3′3′ = m β 33 − 2m nβ 23 + n β 22

2 2 β 2′3′ = m − n β 23 + m n β 33 − β 22

β 1′3′ = m β 13 − nβ 12

β 1′2′ = m β 12 + nβ 13

169 TRANSFORMATIONS FOR DEXTRAL ROTATIONS

ABOUT THE x1 AXIS - CONTINUED

" Let m = cosθ1 and n = sinθ1

T " Performing the calculations given by β = Tε β′ yields

= β 11 β 1′1′ Note that T 2 2 θ 1 − θ 1 β 22 = m β 2′2′ − 2m nβ 2′3′ + n β 3′3′ Tε = Tσ and hence 2 2 β 33 = m β 3′3′ + 2m nβ 2′3′ + n β 2′2′ β = Tσ − θ1 β′ 2 2 β 23 = m − n β 2′3′ − mn β 3′3′ − β 2′2′ So, the expressions given here

for β ij can be obtained from the

those previously given for β i ′ j ′ β 13 = m β 1 3 + nβ 1 2 ′ ′ ′ ′ by switching the primed and unprimed indices and replacing β 12 = m β 1′2′ − nβ 1′3′ n with -n

170 TRANSFORMATIONS FOR DEXTRAL ROTATIONS

ABOUT THE x1 AXIS - CONTINUED

" The algebra involved in computing T T C′ = Tσ C Tσ S′ = Tε S Tε α′ = Tε α T T T C = Tε C′ Tε S = Tσ S′ Tσ α = Tσ α′ T β′ = Tσ β β = Tε β′ is deﬁnitely tedious when done by hand

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

dextral rotations about the x3 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 available

171 TRANSFORMATIONS FOR DEXTRAL ROTATIONS

ABOUT THE x1 AXIS - CONTINUED

" Consider the case in which the transformation expressions are known

for dextral rotations about the x3 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 x3 axis into congruence with the following

ﬁgure shown for dextral rotations about the x1 axis

172 TRANSFORMATIONS FOR DEXTRAL ROTATIONS

ABOUT THE x1 AXIS - CONTINUED

x Plane x1 = 0 x 3 x3 ,x3′ 3′

θ Plane x3 = 0 1

i 3 x2′ i 3 i 3′ x2′ , i 3′ i 2′ i 2′ θ3 θ1 x x2 2 i 1 i 2 i 2 i 1′ i i 1 , 1′ θ3 x1 x x1′ 1 , x1′

173 TRANSFORMATIONS FOR DEXTRAL ROTATIONS

ABOUT THE x1 AXIS - CONTINUED

" Inspection of the ﬁgures indicates the following transformation of the indices: 1 → 2, 2 → 3, and 3 → 1

" 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 tensor notation 11 22 33 23, 32 31, 13 12, 21

matrix notation 1 2 3 4 5 6

" Using this information along with 1 → 2, 2 → 3, and 3 → 1 gives the relations: 4 → 5, 5 → 6, and 6 → 4

174 TRANSFORMATIONS FOR DEXTRAL ROTATIONS

ABOUT THE x1 AXIS - CONTINUED

" Likewise, the transformation of index pairs that appear in the abridged notation are given by

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

4 2 2 2 2 4 " Consider C1′1′ = m C11 + 2m n C12 + 2C66 + 4mn m C16 + n C26 + n C22 ,

where m = cosθ3 and n = sinθ3

" The transformation of indices gives m = cosθ1 , n = sinθ1 , and

4 2 2 2 2 4 C2′2′ = m C22 + 2m n C23 + 2C44 + 4mn m C24 + n C34 + n C33

175 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 x3 axis yields exactly the same expressions

given herein previously for dextral rotations about the x1 axis

176 MATERIAL SYMMETRIES

177 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

178 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

179 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

Five-fold Original symmetry position

A A A

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

180 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

360 deg " = 5 and the shape is said to possess five-fold symmetry

72 deg " A situation of particular interest herein A is the case when a geometric object A

possesses only two-fold symmetry

B " For this object, the line B-B is described B B B as a line of reflective symmetry (or mirror symmetry)

" For each of the geometric objects, A A undergoing the given rotations, a transformation occurs in which the objects appears unchanged; it is said to remain invariant under the transformation

181 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 y-axis " For the ﬁgure, a generic point P of the region R has the coordinates (x,y), with A respect to the coordinate frame shown Coordinates P (x,y) of point P " The coordinate domain is given by 1 1 1 1 B B x-axis − 2 w ≤ x ≤ 2 w and − 2 h ≤ y ≤ 2 h , where w and h are the width and height of the rectangle, respectively Region R

" For other geometries, curvilinear A coordinates may be more suitable

182 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 y-axis coordinates (x,y), for which Coordinates -axis 1 1 1 1 y (x,y) of point P − 2 w ≤ x ≤ 2 w and − 2 h ≤ y ≤ 2 h A -axis " This domain also describes the P x same rectangle with respect to B φ a rotated coordinate frame, as x-axis shown in the ﬁgure B Region R " For values of the angle φ equal to 180 and 360 deg, the rectangle is A brought into coincidence with the initial conﬁguration

183 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 x = x x,y and y = y x,y 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

184 MATHEMATICAL CHARACTERIZATION OF SYMMETRY CONCLUDED

2 2 " Consider the function F x, y = x + y and the transformation of coordinates given by x = − x and y = − y

" Applying the transformation of coordinates gives

2 2 F x, y → − x + − y = x 2 + y 2 = F (x , y ) = F − x, − y

2 2 2 2 " The structural rules given by F x, y = x + y and F (x , y ) = x + y are identical; thus, x = − x and y = − y deﬁne a symmetry transformation for the function

" The more common, and more succinct, way of describing the symmetry is given by writing F x, y = F − x, − y

185 SOME TYPES OF SYMMETRY IN TWO DIMENSIONS GRAPHS OF FUNCTIONS

f(x) f(x) f(x) = ax3 f(x) = ax2 + b

x x

Line of reflective symmetry Line of reflective antisymmetry f(-x) = f(x) f(-x) = -f(x)

186 SOME TYPES OF SYMMETRY IN THREE DIMENSIONS PLANE OF REFLECTIVE SYMMETRY

Tapered beam

Symmetry plane

187 SOME TYPES OF SYMMETRY IN THREE DIMENSIONS PLANE OF REFLECTIVE SYMMETRY F − x, y = F x, y

Contour plot of F(x, y) y-axis

F − x, y F x, y

x-axis

y F x, y = y cos πx sin π 1 x 1 0 y 2 2 2 − ≤ ≤ ≤ ≤

188 SOME TYPES OF SYMMETRY IN THREE DIMENSIONS PLANE OF REFLECTIVE ANTISYMMETRY F − x, y = − F x, y

y-axis Contour plot of F(x, y)

F − x, y F x, y

x-axis

y F x, y = y sin x sin π π 2 − 1 ≤ x ≤ 1 0 ≤ y ≤ 2

189 SOME TYPES OF SYMMETRY IN THREE DIMENSIONS CENTRAL POINT OF INVERSION SYMMETRY (POLAR SYMMETRY) F − x, − y = F x, y

y-axis Contour plot of F(x, y) F x, y

x-axis

F − x, − y 9y F x, y = sin x sin y π − 10 π

− 1 ≤ x ≤ 1 0 ≤ y ≤ 1

190 SOME TYPES OF SYMMETRY IN THREE DIMENSIONS CENTRAL POINT OF INVERSION ANTISYMMETRY F − x, − y = − F x, y

y-axis Contour plot of F(x, y) F x, y

x-axis

F − x, − y 9y F x, y = sin x sin 2 y π − 10 π

− 1 ≤ x ≤ 1 0 ≤ y ≤ 1

191 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 x 1, x 2, x 3 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 x2- axis

coordinates x 1, x 2, x 3

P " The stresses, strains, stiffness x1- axis and compliance matrices, thermal moduli, and thermal-expansion Body, coefﬁcients for this coordinate B x - axis system are σij, εij, [C], [S], βij, and 3

αij, respectively

192 CRITERIA FOR MATERIAL SYMMETRY - CONTINUED

" Recall that the abridged forms of the thermoelastic constitutive

equations for the material in the x 1, x 2, x 3 coordinates are given by

σ = C ε + β T − Tref or ε = S σ + α T − Tref

" Now, consider a general orthogonal transformation between the

rectangular Cartesian coordinates x 1, x 2, x 3 and x 1′′, x 2′′, x 3′′ , that deﬁne a generic point P of the material body

" There is no need to place the restriction that x 1′′, x 2′′, x 3′′ 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 σ i′j′, εi′j′, [C′], [S′], β i′j′, and αi′j′ , respectively

193 CRITERIA FOR MATERIAL SYMMETRY - CONTINUED

" The abridged forms of the thermoelastic constitutive equations for the

material in the x 1′′, x 2′′, x 3′′ coordinate system are given by

σ′ = C′ ε′ + β′ T − Tref or ε′ = S′ σ′ + α′ T − Tref

" Moreover, it was shown previously that, for the given arbitrary transformation (rotation) of coordinates,

− 1 C′ = Tσ C Tε β′ = Tσ β

− 1 S′ = Tε S Tσ α′ = Tε α

" 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

194 CRITERIA FOR MATERIAL SYMMETRY - CONCLUDED

" Now, for xk = xk x1′, x2′, x3′ 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 C′ must be invariant under the transformation given

− 1 − 1 by C′ = Tσ C Tε ; that is, C = Tσ C Tε must hold

− 1 " Similarly, S = Tε S Tσ , α = Tε α , and β = Tσ β must hold

" Collectively, these invariance conditions are the criteria for a state of material symmetry to exist, and are sufﬁcient conditions because

xk = xk x1′, x2′, x3′ is presumed to be a symmetry transformation

195 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)

196 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

197 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

198 CLASSES OF MATERIAL SYMMETRY PICTORIAL REPRESENTATIONS

60 deg

Orthotropic Triclinic Monoclinic Trigonal

45 deg

45 deg Tetragonal Transversely Cubic Isotropic isotropic

" The blue lines represent the edge of a symmetry plane

199 MONOCLINIC MATERIALS

200 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

a i i " The coordinate transformation for i′j ≡ i′ • j x x this symmetry is shown in the 3 , 3′

ﬁgure and is given by x 1′′ = − x 1 , Plane x1 = 0

x1′ x 2′′ = x 2 , and x 3′′ = x 3

i i " The corresponding matrix of 3 , 3′ i 1′ direction cosines is given by

i 1 1 0 0 i i x2 x2 a 1′1 a 1′2 a 1′3 − 2 , 2′ , ′

a 2′1 a 2′2 a 2′3 = 0 1 0

a 3′1 a 3′2 a 3′3 0 0 1

x1 Plane x3 = 0

201 MONOCLINIC MATERIALS

REFLECTIVE SYMMETRY ABOUT THE PLANE x1 = 0 (CONTINUED)

" For this special case, the general transformation matrix

2 2 2 a 1′1 a 1′2 a 1′3 2a 1′2a 1′3 2a 1′1a 1′3 2a 1′1a 1′2 2 2 2 a 2′1 a 2′2 a 2′3 2a 2′2a 2′3 2a 2′1a 2′3 2a 2′1a 2′2 2 2 2 a 3′1 a 3′2 a 3′3 2a 3′2a 3′3 2a 3′1a 3′3 2a 3′1a 3′2 Tσ = reduces to a 2′1a 3′1 a 2′2a 3′2 a 2′3a 3′3 a 2′2a 3′3 + a 2′3a 3′2 a 2′1a 3′3 + a 2′3a 3′1 a 2′1a 3′2 + a 2′2a 3′1

a 1′1a 3′1 a 1′2a 3′2 a 1′3a 3′3 a 1′2a 3′3 + a 1′3a 3′2 a 1′1a 3′3 + a 1′3a 3′1 a 1′1a 3′2 + a 1′2a 3′1

a 1′1a 2′1 a 1′2a 2′2 a 1′3a 2′3 a 1′2a 2′3 + a 1′3a 2′2 a 1′1a 2′3 + a 1′3a 2′1 a 1′1a 2′2 + a 1′2a 2′1

1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 T = the diagonal matrix σ 0 0 0 1 0 0 0 0 0 0 − 1 0 0 0 0 0 0 − 1

202 MONOCLINIC MATERIALS

REFLECTIVE SYMMETRY ABOUT THE PLANE x1 = 0 (CONTINUED)

" Moreover, because Tσ is a diagonal matrix, it follows that Tε = Tσ

" It is worthwhile to mention that Tσ 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 x 1, x 2, x 3 coordinates

σ 33 x3 σ 13 σ 23

σ 22 x2

σ 11 σ 12

x1 Normal stresses Shearing stresses

203 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 x 1′′, x 2′′, x 3′′ coordinates

σ3′3′ σ1′3′ σ2′3′

x3′

x1 ′ σ2′2′ σ x 1′2′ σ1′1′ 2′ Normal stresses Shearing stresses

" Direct comparison of the stresses yields the relationships

= = = σσ1′1′ σσ11 σσ2′2′ σσ22 σσ3′3′ σσ33

= = = σσ2′3′ σσ23 σ 1′3′ − σ 13 σ 1′2′ − σ 12

204 MONOCLINIC MATERIALS

REFLECTIVE SYMMETRY ABOUT THE PLANE x1 = 0 (CONTINUED)

" Expressing the relationships in matrix form gives 1 0 0 0 0 0

σ 1 1 σ 11 ′ ′ 0 1 0 0 0 0 σ 2′2′ σ 22 0 0 1 0 0 0 σ 3 3 σ 33 ′ ′ = σ 2′3′ 0 0 0 1 0 0 σ 23

σ 1 3 σ 13 ′ ′ 0 0 0 0 − 1 0 σ 1′2′ σ 12 0 0 0 0 0 − 1

1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 " = T T = Thus, σ′ σ σ gives σ 0 0 0 1 0 0 directly 0 0 0 0 − 1 0 0 0 0 0 0 − 1

205 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 C ′ must be

− 1 invariant under the transformation given by C′ = Tσ C Tε

− 1 − 1 " That is, C′ = Tσ C Tε must become C = Tσ C Tε

" A more convenient form of this invariance condition is obtained for this particular transformation as follows

" First, postmultiplying the last expression by Tε gives C Tε = Tσ C as the (sufﬁcient) condition for symmetry

" Next, noting that Tε = Tσ for this particular symmetry transformation,

it follows that C Tε = Tσ C becomes C Tσ = Tσ C

206 MONOCLINIC MATERIALS

REFLECTIVE SYMMETRY ABOUT THE PLANE x1 = 0 (CONTINUED)

T " Also, because Tσ = Tσ for this particular symmetry transformation T T T T and C = C , it follows that C Tσ = Tσ C = Tσ C

" Thus, the sufﬁcient condition for symmetry, C Tσ = Tσ C , becomes T

C Tσ = C Tσ ; that is, C Tσ must be a symmetric matrix

" Computing C Tσ gives

C11 C12 C13 C14 − C15 − C16 C11 C12 C13 C14 C15 C16 1 0 0 0 0 0

C12 C22 C23 C24 C25 C26 0 1 0 0 0 0 C12 C22 C23 C24 − C25 − C26

C13 C23 C33 C34 C35 C36 0 0 1 0 0 0 C13 C23 C33 C34 − C35 − C36 C Tσ = 0 0 0 1 0 0 = C14 C24 C34 C44 C45 C46 C14 C24 C34 C44 − C45 − C46 0 0 0 0 − 1 0 C15 C25 C35 C45 C55 C56 C15 C25 C35 C45 − C − C 0 0 0 0 0 − 1 55 56 C16 C26 C36 C46 C56 C66 C16 C26 C36 C46 − C56 − C66

207 MONOCLINIC MATERIALS

REFLECTIVE SYMMETRY ABOUT THE PLANE x1 = 0 (CONTINUED)

T " Enforcing C Tσ = C Tσ yields the following requirements on the stiffnesses in order to exhibit a state of elastic reﬂective symmetry

about the plane x1 = 0:

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

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

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

C11 C12 C13 C14 0 0

C12 C22 C23 C24 0 0 C C C C 0 0 13 23 33 34 which has 13 independent stiffnesses C14 C24 C34 C44 0 0

0 0 0 0 C55 C56

0 0 0 0 C56 C66

208 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 S ′ must be invariant

− 1 under the transformation given by S′ = Tε S Tσ

− 1 − 1 " That is, S′ = Tε S Tσ must become S = Tε S Tσ

" Postmultiplying the last expressing by Tσ gives S Tσ = Tε S as the (sufﬁcient) condition for symmetry

" Next, noting that Tε = Tσ for this particular transformation, it follows

that S Tσ = Tε S becomes S Tσ = Tσ S

209 MONOCLINIC MATERIALS

REFLECTIVE SYMMETRY ABOUT THE PLANE x1 = 0 (CONTINUED)

T " Like before, because Tσ = Tσ for this particular transformation and T T T T S = S , it follows that S Tσ = Tσ S = Tσ S

" Thus, the sufﬁcient condition for symmetry, S Tσ = Tσ S , becomes T

S Tσ = S Tσ ; that is, S Tσ must be a symmetric matrix

" Like for the stiffness matrix, computing S Tσ and enforcing T S Tσ = S Tσ yields the following requirements on the compliances in order to exhibit a state of elastic reﬂective symmetry

about the plane x1 = 0:

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

210 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

S 11 S 12 S 13 S 14 0 0

S 12 S 22 S 23 S 24 0 0 S S S S 0 0 13 23 33 34 which has 13 independent compliances S 14 S 24 S 34 S 44 0 0

0 0 0 0 S 55 S 56

0 0 0 0 S 56 S 66

" Comparison of the compliance and stiffness matrices for this case indicates that they have the same form

211 MONOCLINIC MATERIALS

REFLECTIVE SYMMETRY ABOUT THE PLANE x1 = 0 (CONTINUED)

" The requirements on the coefﬁcients of thermal expansion and the

thermal moduli for 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 α ′ must be invariant under the

transformation given by α′ = Tε α

" That is, α′ = Tε α must become α = Tε α

" Similarly, β ′ must be invariant under the transformation given by

β′ = Tσ β ; that is, β′ = Tσ β must become β = Tσ β

212 MONOCLINIC MATERIALS

REFLECTIVE SYMMETRY ABOUT THE PLANE x1 = 0 (CONTINUED)

" Computing α = Tε α gives

1 0 0 0 0 0 α11 α11 α 11 0 1 0 0 0 0 α 11 α22 α22 α 22 α 22 α33 0 0 1 0 0 0 α33 = which reduces to α 33 α 33 2α23 0 0 0 1 0 0 2α23 = 2α 23 2α 2 2 23 α13 0 0 0 0 − 1 0 α13 2 α 13 − 2α 13 2α12 2α12 0 0 0 0 0 − 1 2α 12 − 2α 12

" Thus, enforcing α = Tε α requires α12 = α13 = 0 in order for a state

of reﬂective symmetry about the plane x1 = 0 to exist

" Similarly, enforcing β = Tσ β requires β 12 = β 13 = 0 in order for a

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

213 MONOCLINIC MATERIALS

REFLECTIVE SYMMETRY ABOUT THE PLANE x1 = 0 (CONTINUED)

" Applying all the simpliﬁcations, the linear thermoelastic constitutive equations become

S 11 S 12 S 13 S 14 0 0 ε11 σ 11 α 11 S 12 S 22 S 23 S 24 0 0 ε22 σ 22 α 22 ε S S S S 0 0 σ α 33 = 13 23 33 34 33 + 33 T T σ − ref 2ε23 S 14 S 24 S 34 S 44 0 0 23 2α 23

2ε13 σ 13 0 0 0 0 0 S 55 S 56 σ 12 2ε12 0 0 0 0 0 S 56 S 66

C 11 C 12 C 13 C 14 0 0 β σ 11 ε11 11 C 12 C 22 C 23 C 24 0 0 σ 22 ε22 β 22

σ 33 C 13 C 23 C 33 C 34 0 0 ε33 β 33 = + T − Tref σ 23 C 14 C 24 C 34 C 44 0 0 2ε23 β 23

σ 13 2ε13 0 0 0 0 0 C 55 C 56 σ 12 2ε12 0 0 0 0 0 C 56 C 66

214 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

β 11 C 11 C 12 C 13 C 14 α 11

β C C C C α 22 22 = − 12 22 23 24 β 33 C 13 C 23 C 33 C 34 α 33

β 23 2α 23 C 14 C 24 C 34 C 44

" 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 x2 = 0

and x3 = 0

215 MONOCLINIC MATERIALS

REFLECTIVE SYMMETRY ABOUT THE PLANE x2 = 0

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

about the plane x2 = 0

" The coordinate transformation for x3 ,x3′ this symmetry is shown in the a i′j ≡ i i′ • i j

ﬁgure and is given by x 1′′ = x 1 ,

x 2′ = − x 2 , and x 3′′ = x 3

i 3 ,i 3′ " The corresponding matrix of

x i 2 x direction cosines is given by 2′ ′ 2

i 2 1 0 0 a 1′1 a 1′2 a 1′3 i 1 , i 1′ a 2′1 a 2′2 a 2′3 = 0 − 1 0 a a a 3′1 3′2 3′3 0 0 1 Plane x2 = 0

x1 , x1′ Plane x3 = 0

216 MONOCLINIC MATERIALS

REFLECTIVE SYMMETRY ABOUT THE PLANE x2 = 0 (CONTINUED)

" For this special case, the general transformation matrix

2 2 2 a1′1 a1′2 a1′3 2a1′2a1′3 2a1′1a1′3 2a1′1a1′2 2 2 2 a2′1 a2′2 a2′3 2a2′2a2′3 2a2′1a2′3 2a2′1a2′2 2 2 2 a3′1 a3′2 a3′3 2a3′2a3′3 2a3′1a3′3 2a3′1a3′2 Tσ = reduces to 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

1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 the diagonal matrix T = σ 0 0 0 − 1 0 0 0 0 0 0 1 0 0 0 0 0 0 − 1

217 MONOCLINIC MATERIALS

REFLECTIVE SYMMETRY ABOUT THE PLANE x2 = 0 (CONTINUED)

" Also, because Tσ is a diagonal matrix, it follows that Tε = Tσ

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

symmetry about the plane x2 = 0 to exist are found from

T T C Tσ = C Tσ S Tσ = S Tσ α = Tε α β = Tσ β

" Computing C Tσ gives

C11 C12 C13 − C14 C15 − C16 C11 C12 C13 C14 C15 C16 1 0 0 0 0 0

C12 C22 C23 C24 C25 C26 0 1 0 0 0 0 C12 C22 C23 − C24 C25 − C26

C C C C C C 0 0 1 0 0 0 C13 C23 C33 − C C35 − C C T = 13 23 33 34 35 36 = 34 36 σ 0 0 0 − 1 0 0 C14 C24 C34 C44 C45 C46 C14 C24 C34 − C44 C45 − C46 0 0 0 0 1 0 C15 C25 C35 C45 C55 C56 C15 C25 C35 − C C55 − C 0 0 0 0 0 − 1 45 56 C16 C26 C36 C46 C56 C66 C16 C26 C36 − C46 C56 − C66

218 MONOCLINIC MATERIALS

REFLECTIVE SYMMETRY ABOUT THE PLANE x2 = 0 (CONTINUED)

T " Enforcing C Tσ = C Tσ yields the following requirements on the stiffnesses:

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

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

elastic reﬂective symmetry about the plane x2 = 0 has the form

C11 C12 C13 0 C15 0

C12 C22 C23 0 C25 0 C C C 0 C 0 13 23 33 35 which also has 13 independent stiffnesses 0 0 0 C44 0 C46

C15 C25 C35 0 C55 0

0 0 0 C46 0 C66

219 MONOCLINIC MATERIALS

REFLECTIVE SYMMETRY ABOUT THE PLANE x2 = 0 (CONTINUED)

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

elastic reﬂective symmetry about the plane x2 = 0 has the form

S11 S12 S13 0 S15 0

S12 S22 S23 0 S25 0 S S S 0 S 0 13 23 33 35 which also has 13 independent compliances 0 0 0 S44 0 S46

S15 S25 S35 0 S55 0

0 0 0 S46 0 S66

220 MONOCLINIC MATERIALS

REFLECTIVE SYMMETRY ABOUT THE PLANE x2 = 0 (CONTINUED)

" Next, computing α = Tε α gives

1 0 0 0 0 0 α11 α11 α11 α11 α 0 1 0 0 0 0 α 22 22 α α22 0 0 1 0 0 0 22 α33 α33 α = which reduces to α33 33 2α23 0 0 0 − 1 0 0 2α23 = 2α23 − 2α23 2α13 0 0 0 0 1 0 2α13 2 α13 2α13 2α 2α 12 0 0 0 0 0 − 1 12 2α12 − 2α12

" Thus, enforcing α = Tε α requires α12 = α23 = 0 in order for a state

of elastic reﬂective symmetry about the plane x2 = 0 to exist

" Similarly, enforcing β = Tσ β requires β 12 = β 23 = 0

221 MONOCLINIC MATERIALS

REFLECTIVE SYMMETRY ABOUT THE PLANE x2 = 0 (CONTINUED)

" Applying all the simpliﬁcations, the linear thermoelastic constitutive equations become

S 11 S 12 S 13 0 S 15 0 α 11 ε11 σ 11 S 12 S 22 S 23 0 S 25 0 ε22 σ 22 α 22

ε33 S 13 S 23 S 33 0 S 35 0 σ 33 α 33 = + T − Tref 2ε23 0 0 0 S 44 0 S 46 σ 23 0 2 σ 13 2α ε13 S 15 S 25 S 35 0 S 55 0 13 σ 12 2ε12 0 0 0 0 S 46 0 S 66

C 11 C 12 C 13 0 C 15 0 β 11 σ 11 ε11 C 12 C 22 C 23 0 C 25 0 σ 22 ε22 β 22

σ 33 C 13 C 23 C 33 0 C 35 0 ε33 β 33 = + T − Tref σ 23 0 0 0 C 44 0 C 46 2ε23 0 σ 13 2 β C 15 C 25 C 35 0 C 55 0 ε13 13 σ 12 2ε12 0 0 0 0 C 46 0 C 66

222 MONOCLINIC MATERIALS

REFLECTIVE SYMMETRY ABOUT THE PLANE x2 = 0 (CONCLUDED)

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

β 11 C 11 C 12 C 13 C 15 α 11

β C C C C α 22 22 = − 12 22 23 25 β 33 C 13 C 23 C 33 C 35 α 33

β 13 2α 13 C 15 C 25 C 35 C 55

" The constitutive equations show that the normal stresses, or a temperature change, produce shearing deformations only in the plane

x2 = 0

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

and x3 = 0

223 MONOCLINIC MATERIALS

REFLECTIVE SYMMETRY ABOUT THE PLANE x3 = 0

" Now, consider the case in which the

material exhibits symmetry about the x3 Plane x2 = 0 plane x3 = 0 a i′j ≡ i i′ • i j " The coordinate transformation for this symmetry is shown in the ﬁgure and is i given by x1′ = x1 , x2′ = x2 , and x 3′ = − x 3 3

x2 ,x2′

" The corresponding matrix of i 1 , i 1′ i i direction cosines is given by 2 , 2′

i 3′ 1 0 0 a1′1 a1′2 a1′3 Plane x3 = 0 a2′1 a2′2 a2′3 = 0 1 0

a3′1 a3′2 a3′3 0 0 1 x − x1 , x1′ 3′

224 MONOCLINIC MATERIALS

REFLECTIVE SYMMETRY ABOUT THE PLANE x3 = 0 (CONTINUED)

" For this special case, the general transformation matrix

2 2 2 a1′1 a1′2 a1′3 2a1′2a1′3 2a1′1a1′3 2a1′1a1′2 2 2 2 a2′1 a2′2 a2′3 2a2′2a2′3 2a2′1a2′3 2a2′1a2′2 2 2 2 a3′1 a3′2 a3′3 2a3′2a3′3 2a3′1a3′3 2a3′1a3′2 Tσ = reduces to 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

1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 the diagonal matrix T = σ 0 0 0 − 1 0 0 0 0 0 0 − 1 0 0 0 0 0 0 1

225 MONOCLINIC MATERIALS

REFLECTIVE SYMMETRY ABOUT THE PLANE x3 = 0 (CONTINUED)

" Also, because Tσ is a diagonal matrix, it follows that Tε = Tσ

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

symmetry about the plane x3 = 0 to exist are found from

T T C Tσ = C Tσ S Tσ = S Tσ α = Tε α β = Tσ β

" Computing C Tσ gives

C11 C12 C13 − C14 − C15 C16 C11 C12 C13 C14 C15 C16 1 0 0 0 0 0

C12 C22 C23 C24 C25 C26 0 1 0 0 0 0 C12 C22 C23 − C24 − C25 C26

C C C C C C 0 0 1 0 0 0 C13 C23 C33 − C − C C36 C T = 13 23 33 34 35 36 = 34 35 σ 0 0 0 − 1 0 0 C14 C24 C34 C44 C45 C46 C14 C24 C34 − C44 − C45 C46 0 0 0 0 − 1 0 C15 C25 C35 C45 C55 C56 C15 C25 C35 − C − C C56 0 0 0 0 0 1 45 55 C16 C26 C36 C46 C56 C66 C16 C26 C36 − C46 − C56 C66

226 MONOCLINIC MATERIALS

REFLECTIVE SYMMETRY ABOUT THE PLANE x3 = 0 (CONTINUED)

T " Enforcing C Tσ = C Tσ yields the following requirements on the stiffnesses:

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

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

elastic reﬂective symmetry about the plane x3 = 0 has the form

C11 C12 C13 0 0 C16

C12 C22 C23 0 0 C26 C C C 0 0 C 13 23 33 36 which also has 13 independent stiffnesses 0 0 0 C44 C45 0

0 0 0 C45 C55 0

C16 C26 C36 0 0 C66

227 MONOCLINIC MATERIALS

REFLECTIVE SYMMETRY ABOUT THE PLANE x3 = 0 (CONTINUED)

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

elastic reﬂective symmetry about the plane x3 = 0 has the form

S11 S12 S13 0 0 S16

S12 S22 S23 0 0 S26 S S S 0 0 S 13 23 33 36 which also has 13 independent compliances 0 0 0 S44 S45 0

0 0 0 S45 S55 0

S16 S26 S36 0 0 S66

228 MONOCLINIC MATERIALS

REFLECTIVE SYMMETRY ABOUT THE PLANE x3 = 0 (CONTINUED)

" Next, computing α = Tε α gives

1 0 0 0 0 0 α11 α11 α11 α11 α 0 1 0 0 0 0 α 22 22 α α22 0 0 1 0 0 0 22 α33 α33 α = which reduces to α33 33 2α23 0 0 0 − 1 0 0 2α23 = 2α23 − 2α23 2α13 0 0 0 0 1 0 2α13 − 2α13 − 2α 2α 2α 13 12 0 0 0 0 0 1 12 2α12 2α12

" Thus, enforcing α = Tε α requires α13 = α23 = 0 in order for a state

of elastic reﬂective symmetry about the plane x3 = 0 to exist

" Similarly, enforcing β = Tσ β requires β 13 = β 23 = 0

229 MONOCLINIC MATERIALS

REFLECTIVE SYMMETRY ABOUT THE PLANE x3 = 0 (CONTINUED)

" Applying all the simpliﬁcations, the linear thermoelastic constitutive equations become

S 11 S 12 S 13 0 0 S 16 α 11 ε11 σ 11 S 12 S 22 S 23 0 0 S 26 ε22 σ 22 α 22

ε33 S 13 S 23 S 33 0 0 S 36 σ 33 α 33 = + T − Tref 2ε23 0 0 0 S 44 S 45 0 σ 23 0 2 σ 13 0 ε13 0 0 0 S 45 S 55 0 σ 12 2ε12 2α 12 S 16 S 26 S 36 0 0 S 66

C 11 C 12 C 13 0 0 C 16 β 11 σ 11 ε11 C 12 C 22 C 23 0 0 C 26 σ 22 ε22 β 22

σ 33 C 13 C 23 C 33 0 0 C 36 ε33 β 33 = + T − Tref σ 23 0 0 0 C 44 C 45 0 2ε23 0 σ 13 2 0 0 0 0 C 45 C 55 0 ε13 σ 12 2ε12 β 12 C 16 C 26 C 36 0 0 C 66

230 MONOCLINIC MATERIALS

REFLECTIVE SYMMETRY ABOUT THE PLANE x3 = 0 (CONCLUDED)

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

β 11 C 11 C 12 C 13 C 16 α 11

β C C C C α 22 22 = − 12 22 23 26 β 33 C 13 C 23 C 33 C 36 α 33

β 12 2α 12 C 16 C 26 C 36 C 66

" The constitutive equations show that the normal stresses, or a temperature change, produce shearing deformations only in the plane

x3 = 0

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

and x2 = 0

231 ORTHOTROPIC MATERIALS

232 ORTHOTROPIC MATERIALS

REFLECTIVE SYMMETRY ABOUT THE PLANES x1 = 0 AND x2 = 0

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

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

plane x2 = 0, which is perpendicular to the plane x1 = 0

" For this monoclinic material, it was shown previously that the material properties are given by

C 11 C 12 C 13 C 14 0 0 β σ 11 ε11 11 C 12 C 22 C 23 C 24 0 0 σ 22 ε22 β 22

σ 33 C 13 C 23 C 33 C 34 0 0 ε33 β 33 = + T − Tref σ 23 C 14 C 24 C 34 C 44 0 0 2ε23 β 23

σ 13 2ε13 0 0 0 0 0 C 55 C 56 σ 12 2ε12 0 0 0 0 0 C 56 C 66

and

233 ! !

ORTHOTROPIC MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT THE PLANES x1 = 0 AND x2 = 0

S 11 S 12 S 13 S 14 0 0 ε11 σ 11 α 11 S 12 S 22 S 23 S 24 0 0 ε22 σ 22 α 22 ε S S S S 0 0 σ α 33 = 13 23 33 34 33 + 33 T T σ − ref 2ε23 S 14 S 24 S 34 S 44 0 0 23 2α 23

2ε13 σ 13 0 0 0 0 0 S 55 S 56 σ 12 2ε12 0 0 0 0 0 S 56 S 66

" To determine the effects of the second symmetry plane, x2 = 0, the

coordinate transformation given by x1′ = x1 , x 2′ = − x 2 , and x3′ = x3 is applied to the material properties of the monoclinic material

" This process is the same as applying the transformation for

symmetry about the plane x2 = 0 in succession to applying the

transformation for symmetry about the plane x1 = 0 to the initial anisotropic-material properties

234 ORTHOTROPIC MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT THE PLANES x1 = 0 AND x2 = 0

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

x2 = 0, is obtained directly from the results given previously for a

monoclinic material that exhibits symmetry about the plane x2 = 0

" That is, the second coordinate x x a i • i 3 , 3′ transformation was given by i′j ≡ i′ j

x1′ = x1 , x 2′ = − x 2 , and x3′ = x3 Plane x2 = 0

" And, the corresponding matrix i i of direction cosines were Plane x3 = 0 3 , 3′

shown to be given by i 2′ x2 x2′ 1 0 0 a1′1 a1′2 a1′3 i 2 a a a = 0 1 0 2′1 2′2 2′3 − i 1 , i 1′

a3′1 a3′2 a3′3 0 0 1

x1 , x1′

235 ORTHOTROPIC MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT THE PLANES x1 = 0 AND x2 = 0

1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 " For this case, T = T = ε σ 0 0 0 − 1 0 0 0 0 0 0 1 0 0 0 0 0 0 − 1

" It was also shown that, for symmetry about the plane x2 = 0,

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

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

α12 = 0 , α23 = 0 , β 12 = 0 , and β 23 = 0

236 ORTHOTROPIC MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT THE PLANES x1 = 0 AND x2 = 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

C 11 C 12 C 13 0 0 0 β 11 σ 11 ε11 C 12 C 22 C 23 0 0 0 σ 22 ε22 β 22

σ 33 C 13 C 23 C 33 0 0 0 ε33 β 33 = + T − Tref and σ 23 0 0 0 C 44 0 0 2ε23 0 σ 13 2 0 0 0 0 0 C 55 0 ε13 σ 12 2ε12 0 0 0 0 0 0 C 66

237 ORTHOTROPIC MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT THE PLANES x1 = 0 AND x2 = 0

S 11 S 12 S 13 0 0 0 ε11 σ 11 α 11 S 12 S 22 S 23 0 0 0 ε22 σ 22 α 22

ε33 S 13 S 23 S 33 0 0 0 σ 33 α 33 = + T − Tref 2ε23 0 0 0 S 44 0 0 σ 23 0 2 σ 13 0 ε13 0 0 0 0 S 55 0 σ 12 0 2ε12 0 0 0 0 0 S 66

" It is worth pointing out at this x x point in the development, that the 3 , 3′

ai′j i i′ • i j single coordinate transformation ≡ x1′

given by x 1′ = − x 1 , x 2′ = − x 2 , and i x = x does not produce the 1′ 3′ 3 i 3 ,i 3′ x same result as the two successive 2′ x2

symmetry transformations i 2′ i 2

i 1

238 ORTHOTROPIC MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT THE PLANES x1 = 0 AND x2 = 0

" In particular, the corresponding matrix of direction cosines is given by 1 0 0 a1′1 a1′2 a1′3 −

a2′1 a2′2 a2′3 = 0 − 1 0

a3′1 a3′2 a3′3 0 0 1

" The corresponding stress and strain transformation matrices are

1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 T = T = σ ε 0 0 0 − 1 0 0 0 0 0 0 − 1 0 0 0 0 0 0 1

239 ORTHOTROPIC MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT THE PLANES x1 = 0 AND x2 = 0

− 1 " Next, the condition C = Tσ C Tε for a state of symmetry to exist at a point P of the body is applied

" Using that Tε = Tσ for this particular transformation, it was

− 1 shown that the condition C = Tσ C Tε simpliﬁes to the

condition that C Tσ must be a symmetric matrix

" Computing C Tσ gives

C11 C12 C13 − C14 − C15 C16 C11 C12 C13 C14 C15 C16 1 0 0 0 0 0

C12 C22 C23 C24 C25 C26 0 1 0 0 0 0 C12 C22 C23 − C24 − C25 C26

240 ORTHOTROPIC MATERIALS - CONCLUDED

REFLECTIVE SYMMETRY ABOUT THE PLANES x1 = 0 AND x2 = 0

T " Enforcing C Tσ = C Tσ yields the following requirements on the stiffnesses:

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

" Inspection of the conditions and comparison with the previous result for the two successive transformations shows that the results are different

241 ! ! ! !

ORTHOTROPIC MATERIALS

REFLECTIVE SYMMETRY ABOUT THE PLANES x1 = 0, x2 = 0, AND x3 = 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 x3 = 0, in addition to symmetry

about the perpendicular planes x1 = 0 and x2 = 0

" The effect of this third symmetry transformation is obtained directly by applying the conditions obtained for a monoclinic that exhibits

symmetry about the plane x3 = 0 to the constitutive equations that were obtained previously for the two successive symmetry transformations

" The conditions for symmetry about the plane x3 = 0 are C14 = 0 ,

C15 = 0 , C24 = 0 , C25 = 0 , C34 = 0 , C35 = 0 , C46 = 0 , C56 = 0 , S14 = 0 ,

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

α23 = 0 , β 13 = 0 , and β 23 = 0

242 ORTHOTROPIC MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT THE PLANES x1 = 0, x2 = 0, AND x3 = 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

243 CONSTITUTIVE EQUATIONS

" The constitutive equations for a linear, thermoelastic, orthotropic material are given by

C 11 C 12 C 13 0 0 0 β 11 σ 11 ε11 C 12 C 22 C 23 0 0 0 σ 22 ε22 β 22

σ 33 C 13 C 23 C 33 0 0 0 ε33 β 33 = + T − Tref and σ 23 0 0 0 C 44 0 0 2ε23 0 σ 13 2 0 0 0 0 0 C 55 0 ε13 σ 12 2ε12 0 0 0 0 0 0 C 66

S 11 S 12 S 13 0 0 0 ε11 σ 11 α 11 S 12 S 22 S 23 0 0 0 ε22 σ 22 α 22

ε33 S 13 S 23 S 33 0 0 0 σ 33 α 33 = + T − Tref 2ε23 0 0 0 S 44 0 0 σ 23 0 2 σ 13 0 ε13 0 0 0 0 S 55 0 σ 12 0 2ε12 0 0 0 0 0 S 66

244 CONSTITUTIVE EQUATIONS - CONCLUDED

" The thermal moduli are given in terms of the coefﬁcients of thermal expansion by C C C β 11 11 12 13 α 11

β 22 = − C 12 C 22 C 23 α 22 α 33 β 33 C 13 C 23 C 33

" The constitutive equations show that extension and shearing are totally

uncoupled in the planes x1 = 0, x2 = 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

245 SPECIALLY ORTHOTROPIC MATERIALS

" Any material that is fully characterized by the following constitutive equations is deﬁned as a specially orthotropic material

C 11 C 12 C 13 0 0 0 β 11 σ 11 ε11 C 12 C 22 C 23 0 0 0 σ 22 ε22 β 22

σ 33 C 13 C 23 C 33 0 0 0 ε33 β 33 = + T − Tref and σ 23 0 0 0 C 44 0 0 2ε23 0 σ 13 2 0 0 0 0 0 C 55 0 ε13 σ 12 2ε12 0 0 0 0 0 0 C 66

S 11 S 12 S 13 0 0 0 ε11 σ 11 α 11 S 12 S 22 S 23 0 0 0 ε22 σ 22 α 22

ε33 S 13 S 23 S 33 0 0 0 σ 33 α 33 = + T − Tref 2ε23 0 0 0 S 44 0 0 σ 23 0 2 σ 13 0 ε13 0 0 0 0 S 55 0 σ 12 0 2ε12 0 0 0 0 0 S 66

" The corresponding coordinate system used to describe this material is deﬁned as the principal material coordinate system

246 GENERALLY ORTHOTROPIC MATERIALS

" For an arbitrary coordinate transformation, from a principal x 1,x 2,x 3

coordinate frame to a x 1′,x 2′,x 3′ coordinate frame, the transformation

matrices T ε and T σ are fully populated

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

solid are transformed from the x 1,x 2,x 3 coordinate frame to the

x 1′,x 2′,x 3′ 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

247 GENERALLY ORTHOTROPIC MATERIALS CONTINUED

" For a dextral rotation about the x3 - axis, with m = cosθ3 and

n = sinθ3 , the transformed elastic constants are given by

4 2 2 4 C 1′1′ = m C 11 + 2m n C 12 + 2C 66 + n C 22

2 2 4 4 C 1′2′ = m n C 11 + C 22 − 4C 66 + m + n C 12

2 2 C 1′3′ = m C 13 + n C 23 C 1′4′ = 0 C 1′5′ = 0

2 2 2 2 C 1′6′ = mn m − n C 12 + 2C 66 + mn n C 22 − m C 11

4 2 2 4 C 2′2′ = m C 22 + 2m n C 12 + 2C 66 + n C 11

2 2 C 2′3′ = m C 23 + n C 13 C 2′4′ = 0 C 2′5′ = 0

2 2 2 2 C 2′6′ = mn n − m C 12 + 2C 66 + mn m C 22 − n C 11

248 GENERALLY ORTHOTROPIC MATERIALS CONTINUED

C 3′3′ = C 33 C 3′4′ = 0 C 3′5′ = 0 C 3′6′ = mn C 23 − C 13

2 2 C 4′4′ = m C 44 + n C 55 C 4′5′ = mn C 44 − C 55 C 4′6′ = 0 C 5′6′ = 0

2 2 2 2 2 2 2 C 5′5′ = m C 55 + n C 44 C 6′6′ = m n C 11 + C 22 − 2C 12 + m − n C 66

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

C 1′1′ C1′2′ C1′3′ 0 0 C1′6′

C1′2′ C2′2′ C2′3′ 0 0 C2′6′

C1′3′ C2′3′ C3′3′ 0 0 C3′6′

0 0 0 C4′4′ C4′5′ 0

0 0 0 C4′5′ C5′5′ 0

C1′6′ C2′6′ C3′6′ 0 0 C6′6′

" For the x1′,x2′,x3′ coordinate frame, the specially orthotropic material appears to have the properties of a monoclinic material

249 GENERALLY ORTHOTROPIC MATERIALS CONTINUED

" Similarly,

4 2 2 4 S 1′1′ = m S 11 + m n 2S 12 + S 66 + n S 22

2 2 4 4 S 1′2′ = m n S 11 + S 22 − S 66 + m + n S 12

2 2 S 1′3′ = m S 13 + n S 23 S 1′4′ = 0 S 1′5′ = 0

2 2 2 2 S 1′6′ = mn m − n 2S 12 + S 66 + 2mn n S 22 − m S 11

4 2 2 4 2 2 S 2′2′ = m S 22 + m n 2S 12 + S 66 + n S 11 S 2′3′ = m S 23 + n S 13

S 2′4′ = 0 S 2′5′ = 0 S 3′3′ = S 33 S 3′4′ = 0 S 3′5′ = 0

2 2 2 2 S 2′6′ = mn n − m 2S 12 + S 66 + 2mn m S 22 − n S 11

S 3′6′ = 2mn S 23 − S 13

250 GENERALLY ORTHOTROPIC MATERIALS CONTINUED

2 2 S 4′4′ = m S 44 + n S 55 S 4′5′ = mn S 44 − S 55 S 4′6′ = 0

2 2 S 5′5′ = m S 55 + n S 44 S 5′6′ = 0

2 2 2 2 2 S 6′6′ = 4m n S 11 + S 22 − 2S 12 + m − n S 66

" The population of the matrix of transformed elastic compliances is given by

S 1′1′ S1′2′ S1′3′ 0 0 S1′6′

S1′2′ S2′2′ S2′3′ 0 0 S2′6′

S1′3′ S2′3′ S3′3′ 0 0 S3′6′

0 0 0 S4′4′ S4′5′ 0

0 0 0 S4′5′ S5′5′ 0

S1′6′ S2′6′ S3′6′ 0 0 S6′6′

" For the x1′,x2′,x3′ coordinate frame, the specially orthotropic material, again, appears to have the properties of a monoclinic material

251 GENERALLY ORTHOTROPIC MATERIALS CONTINUED

" For a dextral rotation about the x3 - axis, the transformed thermal- expansion coefﬁcients are given by

2 2 2 2 α1′1′ = m α11 + n α22 α2′2′ = m α22 + n α11 α3′3′ = α33

α2′3′ = 0 α1′3′ = 0 α 1′2′ = mn α 22 − α 11

" Similarly, the transformed thermal moduli are given by

2 2 2 2 β 1′1′ = m β 11 + n β 22 β 2′2′ = m β 22 + n β 11 β 3′3′ = β 33

β 2′3′ = 0 β 1′3′ = 0 β 1′2′ = mn β 22 − β 11

252 GENERALLY ORTHOTROPIC MATERIALS CONCLUDED

" Thus, for a dextral rotation about the x3 - axis, the transformed constitutive equations for a specially orthotropic material are given by

S 1′1′ S 1′2′ S 1′3′ 0 0 S 1′6′ ε 1′1′ σ 1′1′ α 1′1′ S S S 0 0 S ε 2′2′ 1′2′ 2′2′ 2′3′ 2′6′ σ 2′2′ α 2′2′

ε 3′3′ S 1′3′ S 2′3′ S 3′3′ 0 0 S 3′6′ σ 3′3′ α 3′3′ = + T – Tref 2ε 2′3′ 0 0 0 S 4′4′ S 4′5′ 0 σ 2′3′ 0

2ε 1 3 0 ′ ′ 0 0 0 S 4′5′ S 5′5′ 0 σ 1′3′ 2ε 1′2′ σ 1′2′ 2α 1′2′ S 1′6′ S 2′6′ S 3′6′ 0 0 S 6′6′

C 1′1′ C 1′2′ C 1′3′ 0 0 C 1′6′ σ 1′1′ ε 1′1′ β 1′1′ C C C 0 0 C σ 2′2′ 1′2′ 2′2′ 2′3′ 2′6′ ε 2′2′ β 2′2′

σ 3′3′ C 1′3′ C 2′3′ C 3′3′ 0 0 C 3′6′ ε 3′3′ β 3′3′ = + T – Tref σ 2′3′ 0 0 0 C 4′4′ C 4′5′ 0 2ε 2′3′ 0

2ε 1 3 0 σ 1′3′ 0 0 0 C 4′5′ C 5′5′ 0 ′ ′ σ 1′2′ 2ε 1′2′ β 1′2′ C 1′6′ C 2′6′ C 3′6′ 0 0 C 6′6′

253 TRIGONAL MATERIALS

254 TRIGONAL MATERIALS

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x3 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 a i′j ≡ i i′ • i j

x3 ,x3′ elastic symmetry whose normal n

lies in the plane x3 = 0 and makes an x2′

angle θ3 with the x1 axis, as shown in i 3 , i 3′

i 2′ the ﬁgure θ3

i 1 i 2 x2 " The angle θ3 is deﬁned to be in the π π π range − < θ 3 ≤ , because θ = − i 1′, n 2 2 3 2 θ 3 x π 1′ and θ 3 = deﬁne the same plane 2 x1

" In addition, let the x 1′, x 2′, x 3′ be the Plane of elastic symmetry coordinates used to deﬁne the material symmetry

255 TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x3 AXIS

" Speciﬁcally, for a plane of elastic symmetry given by x 1′ = 0 , the symmetry transformation is shown in the ﬁgure and is given by

x 1′′ = − x 1′ , x 2′′ = x 2′ , and x 3′′ = x 3′

" The corresponding matrix of x3′ ,x3′′ a i′j ≡ i i′ • i j direction cosines is given by

a1 1 a1 2 a1 3 − 1 0 0 ′′ ′ ′′ ′ ′′ ′ x1′′

a2′′1′ a2′′2′ a2′′3′ = 0 1 0

i i 3 a3′′1′ a3′′2′ a3′′3′ 0 0 1 3′ , ′′

i 1′′

" For the x 1′, x 2′, x 3′ coordinate x x i 2′ ,i 2′′ 2′, 2′′ system, the general constitutive i 1′ Plane x = 0 equations are expressed as 1′ follows

x1′

256 TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x3 AXIS

ε S 1′1′ S 1′2′ S 1′3′ S 1′4′ S 1′5′ S 1′6′ α 1′1′ σ 1′1′ 1′1′ S 1 2 S 2 2 S 2 3 S 2 4 S 2 5 S 2 6 ε2′2′ ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ σ 2′2′ α2′2′ ε3′3′ S 1′3′ S 2′3′ S 3′3′ S 3′4′ S 3′5′ S 3′6′ σ 3′3′ α3′3′ = + T − Tref 2 σ 2 3 2 ε2′3′ S 1′4′ S 2′4′ S 3′4′ S 4′4′ S 4′5′ S 4′6′ ′ ′ α2′3′ 2ε1 3 σ 1′3′ 2α1 3 ′ ′ S 1′5′ S 2′5′ S 3′5′ S 4′5′ S 5′5′ S 5′6′ ′ ′ σ 1′2′ 2ε1′2′ 2α1′2′ S 1′6′ S 2′6′ S 3′6′ S 4′6′ S 5′6′ S 6′6′

C 1′1′ C 1′2′ C 1′3′ C 1′4′ C 1′5′ C 1′6′ ε β σ 1′1′ 1′1′ 1′1′ C 1 2 C 2 2 C 2 3 C 2 4 C 2 5 C 2 6 σ 2′2′ ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ ε2′2′ β 2′2′ σ 3′3′ C 1′3′ C 2′3′ C 3′3′ C 3′4′ C 3′5′ C 3′6′ ε3′3′ β 3′3′ = + T − Tref or by σ 2 3 2 2 ′ ′ C 1′4′ C 2′4′ C 3′4′ C 4′4′ C 4′5′ C 4′6′ ε2′3′ β 2′3′ σ 1′3′ 2ε1 3 2β 1 3 C 1′5′ C 2′5′ C 3′5′ C 4′5′ C 5′5′ C 5′6′ ′ ′ ′ ′ σ 1′2′ 2ε1′2′ 2β 1′2′ C 1′6′ C 2′6′ C 3′6′ C 4′6′ C 5′6′ C 6′6′

ε ′ = S′ σ′ + α′ Θ and σ ′ = C′ ε′ + β′ Θ ,

where Θ = T − Tref

257 TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x3 AXIS

" For the x1′′, x2′′, x3′′ coordinate system, the constitutive equations are identical in form and are obtained by replacing the index pair i′j ′ with i′′ j ′′

" In the abridged notation,

εε ′′′′ = S′′′′ σσ′′′′ + αα′′′′ ΘΘ and

σσ′′′′ = C′′′′ εε′′′′ + ββ′′′′ ΘΘ , where Θ = T − Tref

" For the transformation of coordinates deﬁned by the symmetry

transformation (a reﬂection about the plane x1′ = 0 ),

r 1′ r 1′ σσ′′′′ = Tσ σσ′′ and εε′′′′ = Tε εε′′ where

258 !

TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x3 AXIS

1 0 0 0 0 0 0 1 0 0 0 0

r 1′ r 1′ 0 0 1 0 0 0 T = T = σ ε 0 0 0 1 0 0 0 0 0 0 − 1 0 0 0 0 0 0 − 1

" Likewise,

−1 −1 r 1′ r 1′ r 1′ r 1′ S′′ = Tε S′ Tσ S′ = Tε S′′ Tσ

−1 −1 r 1′ r 1′ r 1′ r 1′ C′′ = Tσ C′ Tε C′ = Tσ C′′ Tε

r 1′ r 1′ α′′′′ = Tε α′ β′′′′ = Tσ β′

259 ! ! !

TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x3 AXIS

" The conditions for invariance under the symmetry transformation are given by

− 1 − 1 r 1′ r 1′ r 1′ r 1′ S′ = Tε S′ Tσ C′ = Tσ C′ Tε

r 1′ r 1′ α′ = Tε α′ β′ = Tσ β′

" 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

x1 = 0

" Direct comparison reveals that the material is monoclinic with

respect to the plane x1′ = 0

260 TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x3 AXIS

" That is, direct comparison yields

S 1 1 S 1 2 S 1 3 S 1 4 0 0 ε ′ ′ ′ ′ ′ ′ ′ ′ α 1′1′ 1′1′ σ 1′1′ S 1 2 S 2 2 S 2 3 S 2 4 0 0 ε2′2′ ′ ′ ′ ′ ′ ′ ′ ′ σ 2′2′ α 2′2′ ε3′3′ S 1′3′ S 2′3′ S 3′3′ S 3′4′ 0 0 σ 3′3′ α 3′3′ = + T − Tref and 2 σ 2 3 ε2′3′ S 1′4′ S 2′4′ S 3′4′ S 4′4′ 0 0 ′ ′ 2α 2′3′ σ 2ε1 3 1′3′ 0 ′ ′ 0 0 0 0 S 5′5′ S 5′6′ σ 1′2′ 2ε1′2′ 0 0 0 0 0 S 5′6′ S 6′6′

C 1 1 C 1 2 C 1 3 C 1 4 0 0 ′ ′ ′ ′ ′ ′ ′ ′ ε β 1 1 σ 1′1′ 1′1′ ′ ′ C 1 2 C 2 2 C 2 3 C 2 4 0 0 σ 2′2′ ′ ′ ′ ′ ′ ′ ′ ′ ε2′2′ β 2′2′ σ 3′3′ C 1′3′ C 2′3′ C 3′3′ C 3′4′ 0 0 ε3′3′ β 3′3′ = + T − Tref σ 2 3 2 ′ ′ C 1′4′ C 2′4′ C 3′4′ C 4′4′ 0 0 ε2′3′ β 2′3′ σ 1′3′ 2ε1 3 0 0 0 0 0 C 5′5′ C 5′6′ ′ ′ σ 1′2′ 2ε1′2′ 0 0 0 0 0 C 5′6′ C 6′6′

261 TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x3 AXIS

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

thermal moduli, expressed in x1′, x2′, x3′ coordinates, are referred back

to the original x1, x2, x3 coordinates

" The transformation corresponds to a i′j ≡ i i′ • i j

x3 x3′ the dextral rotation about the x3 axis, , shown in the ﬁgure, and is given by

x2′ x 1′ = x 1cos θ 3 + x 2sin θ 3 , i 3 , i 3′

i 2′ θ3 x 2′ = − x 1sin θ 3 + x 2cos θ 3 , and x 3′ = x 3 ,

π π i 1 i x with − 2 < θ 3 ≤ 2 2 2

i 1′, n " The corresponding matrix of θ 3 x direction cosines is given by 1′

cos sin 0 x1 a1′1 a1′2 a1′3 θ 3 θ 3 Plane of elastic a2′1 a2′2 a2′3 = − sinθ 3 cosθ 3 0 symmetry a3′1 a3′2 a3′3 0 0 1

262 TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x3 AXIS

" The corresponding stress and strain transformation matrices were shown previously to be given by

2 2 cos θ3 sin θ3 0 0 0 2sinθ3cosθ3 2 2 sin θ3 cos θ3 0 0 0 − 2sinθ3cosθ3

0 0 1 0 0 0 = Tσ θ3 and 0 0 0 cosθ3 − sinθ3 0

0 0 0 sinθ3 cosθ3 0 2 2 − sinθ3cosθ3 sinθ3cosθ3 0 0 0 cos θ3 − sin θ3

2 2 cos θ3 sin θ3 0 0 0 sinθ3cosθ3

2 2 sin θ3 cos θ3 0 0 0 − sinθ3cosθ3

0 0 1 0 0 0 = Tε θ3 0 0 0 cosθ3 − sinθ3 0

0 0 0 sinθ3 cosθ3 0

2 2 − 2sinθ3cosθ3 2sinθ3cosθ3 0 0 0 cos θ3 − sin θ3

263 TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x3 AXIS

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

− 1 − 1 C′ = Tσ C Tε S′ = Tε S Tσ

β′ = Tσ β α′ = Tε α

" These laws transform the previously obtained invariance conditions on C ′ , S ′ , α ′ , and β ′ into invariance conditions on C , S , α , and β

" Speciﬁc expressions for these transformation laws (dextral rotation

about the x3 axis) were given previously for a fully anisotropic material (triclinic)

" Note that the matrices C and S are fully populated

" Also, the vectors α and β are fully populated

264 TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x3 AXIS

" Because of the invariance conditions, the matrices C ′ and S ′ have

C 1′1′ C 1′2′ C 1′3′ C 1′4′ 0 0 S 1′1′ S 1′2′ S 1′3′ S 1′4′ 0 0 C 1′2′ C 2′2′ C 2′3′ C 2′4′ 0 0 S 1′2′ S 2′2′ S 2′3′ S 2′4′ 0 0 C C C C 0 0 S S S S 0 0 the form 1′3′ 2′3′ 3′3′ 3′4′ and 1′3′ 2′3′ 3′3′ 3′4′ C 1′4′ C 2′4′ C 3′4′ C 4′4′ 0 0 S 1′4′ S 2′4′ S 3′4′ S 4′4′ 0 0 0 0 0 0 C 5′5′ C 5′6′ 0 0 0 0 S 5′5′ S 5′6′ 0 0 0 0 C 5′6′ C 6′6′ 0 0 0 0 S 5′6′ S 6′6′

α 1′1′ β 1′1′

α 2′2′ β 2′2′

α 3 3 β 3 3 " Also, the vectors α ′ and β ′ have the form ′ ′ and ′ ′ 2α 2′3′ β 2′3′ 0 0 0 0

" The corresponding forms of C , S , α , and β are obtained from

the transformation laws once a value for the angle θ3 is speciﬁed

265 TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x3 AXIS

" For example, specifying θ3 = 0 yields

C1 1 C1 2 C1 3 C1 4 0 0 ′ ′ ′ ′ ′ ′ ′ ′ C11 C12 C13 C14 C15 C16 β 1′1′ β 11

C1′2′ C2′2′ C2′3′ C2′4′ 0 0 C12 C22 C23 C24 C25 C26 β 2′2′ β 22

C1 3 C2 3 C3 3 C3 4 0 0 C C C C C C β 3 3 β 33 ′ ′ ′ ′ ′ ′ ′ ′ = 13 23 33 34 35 36 ′ ′ = C1′4′ C2′4′ C3′4′ C4′4′ 0 0 C14 C24 C34 C44 C45 C46 β 2′3′ β 23

β 13 0 0 0 0 C5′5′ C5′6′ C15 C25 C35 C45 C55 C56 0 0 β 12 0 0 0 0 C5′6′ C6′6′ C16 C26 C36 C46 C56 C66

S 1 1 S 1 2 S 1 3 S 1 4 0 0 S S S S S S ′ ′ ′ ′ ′ ′ ′ ′ 11 12 13 14 15 16 α1′1′ α11 S S S S 0 0 1′2′ 2′2′ 2′3′ 2′4′ S12 S22 S23 S24 S25 S26 α2′2′ α22

S 1 3 S 2 3 S 3 3 S 3 4 0 0 S13 S23 S33 S34 S35 S36 α3 3 α33 ′ ′ ′ ′ ′ ′ ′ ′ = ′ ′ = S 1′4′ S 2′4′ S 3′4′ S 4′4′ 0 0 S14 S24 S34 S44 S45 S46 2α2′3′ 2α23

2α13 0 0 0 0 S 5′5′ S 5′6′ S15 S25 S35 S45 S55 S56 0 0 2α12 0 0 0 0 S 5′6′ S 6′6′ S16 S26 S36 S46 S56 S66

" Enforcing the invariance conditions on C ′ gives

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

266 TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x3 AXIS

" Likewise, enforcing the invariance conditions on S ′ gives

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

" Enforcing the invariance conditions on α ′ and β ′ gives

α12 = α13 = 0 and β 12 = β 13 = 0

" These conditions are identical to the conditions previously obtained for

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

" = π Similarly, specifying θ3 2 and enforcing the invariance conditions on C ′ , S ′ , α ′ , and β ′ yields the conditions previously obtained for

a material that is monoclinic with respect to the plane x2 = 0

267 TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x3 AXIS

" π < π Now consider an arbitrary value for the angle θ3 such that − 2 θ 3 ≤ 2

" The previous example for θ3 = 0 shows that the invariance conditions on C , S , α , and β were obtained by the terms of C ′ , S ′ , α ′ , and β ′ that were zero valued

" Using the transformation equations for a dextral rotation about the x3 axis that were given previously for a (triclinic) fully anisotropic, elastic material gives the following results for the invariance conditions on C ′

3 2 2 3 C 1′5′ = 0: m C 15 + m n 2C 56 + C 14 + mn 2C 46 + C 25 + n C 24 = 0

2 2 2 3 C1′6′ = 0: m m − 3n C16 − m n C11 − C12 − 2C66 3 2 2 2 + mn C22 − C12 − 2C66 − n n − 3m C26 = 0

268 TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x3 AXIS

3 2 2 3 C 2′5′ = 0: m C 25 − m n 2C 56 − C 24 − mn 2C 46 − C 15 + n C 14 = 0

2 2 2 3 C2′6′ = 0: m m − 3n C26 + m n C22 − C12 − 2C66 3 2 2 2 − mn C11 − C12 − 2C66 − n n − 3m C16 = 0

C 3′5′ = 0: m C 35 + nC 34 = 0

2 2 C 3′6′ = 0: m − n C 36 + mn C 23 − C 13 = 0

2 2 C 4′5′ = 0: m − n C 45 + mn C 44 − C 55 = 0

3 2 2 3 C 4′6′ = 0: m C 46 − m n C 56 + C 14 − C 24 − mn C 46 − C 15 + C 25 + n C 56 = 0

with m = cosθ3 and n = sinθ3

" These conditions give 8 equations and 20 unknowns

269 TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x3 AXIS

" The invariance conditions on S ′ are given by

3 2 2 3 S 1′5′ = 0: m S 15 + m n S 56 + S 14 + mn S 46 + S 25 + n S 24 = 0

2 2 2 3 S1′6′ = 0: m m − 3n S16 − m n 2S11 − 2S12 − S66 3 2 2 2 + mn 2S22 − 2S12 − S66 − n n − 3m S26 = 0

3 2 2 3 S 2′5′ = 0: m S 25 − m n S 56 − S 24 − mn S 46 − S 15 + n S 14 = 0

2 2 2 3 S2′6′ = 0: m m − 3n S26 + m n 2S22 − 2S12 − S66 3 2 2 2 − mn 2S11 − 2S12 − S66 − n n − 3m S16 = 0

2 2 S 3′5′ = 0: m S 35 + nS 34 = 0 S 3′6′ = 0: m − n S 36 + 2mn S 23 − S 13 = 0

2 2 S 4′5′ = 0: m − n S 45 + mn S 44 − S 55 = 0

3 2 2 3 S 4′6′ = 0: m S 46 − m n S 56 + 2S 14 − 2S 24 − mn S 46 − 2S 15 + 2S 25 + n S 56 = 0

270 TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x3 AXIS

" The invariance conditions on α ′ and β ′ yield

2 2 α 1′3′ = 0: m α 13 + nα 23 = 0 α 1′2′ = 0: m − n α 12 + mn α 22 − α 11 = 0

2 2 β 1′3′ = 0: m β 13 + nβ 23 = 0 β 1′2′ = 0: m − n β 12 + mn β 22 − β 11 = 0

271 TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x3 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 x3 axis, it is convenient to select the ﬁrst plane to be given by θ3 = 0

" Thus, the material is monoclinic with respect to the plane x1 = 0 and, as a result, the following conditions hold

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

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

α12 = 0 α13 = 0 β 12 = 0 β 13 = 0

" These relations, and the fact that n = sinθ3 ≠ 0 for nonzero values π π of − 2 < θ 3 ≤ 2 , are used to simplify the previously given invariance conditions into the following three uncoupled groups

272 TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x3 AXIS

" Group 1 2 2 C 1′5′ = 0: m 2C 56 + C 14 + n C 24 = 0 m = cosθ3 n = sinθ3

2 2 C 2′5′ = 0: − m 2C 56 − C 24 + n C 14 = 0

2 2 C 4′6′ = 0: n C 56 − m C 56 + C 14 − C 24 = 0

" Group 2

C 3′5′ = 0: C 34 = 0

C 3′6′ = 0: m C 23 − C 13 = 0

C 4′5′ = 0: m C 44 − C 55 = 0

" Group 3

2 3 C 1′6′ = 0: mn C 22 − C 12 − 2C 66 − m C 11 − C 12 − 2C 66 = 0

3 2 C 2′6′ = 0: m C 22 − C 12 − 2C 66 − mn C 11 − C 12 − 2C 66 = 0

273 TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x3 AXIS

" The ﬁrst group can be written as 2 C 1′5′ = 0: − C 24 = m 2C 56 + C 14 − C 24 m = cosθ3 n = sinθ3

2 C 2′5′ = 0: C 14 = m 2C 56 + C 14 − C 24

2 C 4′6′ = 0: C 56 = m 2C 56 + C 14 − C 24

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

that the left-hand sides are equal; that is, C 56 = − C 24 = C 14 = Γ

2 " Each equation can be expressed as Γ = 4Γcos θ 3 , which is satisﬁed = 0 cos 2 = 1 by Γ and by θ3 4 π π " Γ = 0 implies C 56 = − C 24 = C 14 = 0 and − 2 < θ 3 ≤ 2

2 1 " cos = = π C = C C = C θ3 4 implies θ 3 ± 3 , 56 14 , and 24 − 14

274 TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x3 AXIS

" The third group of equations is simpliﬁed by ﬁrst adding the two equations together and then by subtracting the two equations

" Adding gives C 1′6′ + C 2′6′ = 0: C 22 − C 11 cos θ 3 = 0 , π π which is satisﬁed for all − 2 < θ 3 ≤ 2 if C22 = C11 , or for all π C22 ≠ C11 if θ3 = 2

" Subtracting gives

C 1′6′ − C 2′6′ = 0: 2C 12 + 4C 66 − C 11 − C 22 cos θ 3cos 2θ 3 = 0 ,

π π which is satisﬁed for all − 2 < θ 3 ≤ 2 if 2C 12 + 4C 66 − C 11 − C 22 = 0 , 2C + 4C C C 0 π π or for all 12 66 − 11 − 22 ≠ if θ3 = ± 4 or 2

275 TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x3 AXIS

" Group 4 2 2 S 1′5′ = 0: m S 56 + S 14 + n S 24 = 0 m = cosθ3 n = sinθ3

2 2 S 2′5′ = 0: − m S 56 − S 24 + n S 14 = 0

2 2 S 4′6′ = 0: − m S 56 + 2S 14 − 2S 24 + n S 56 = 0

" Group 5

S 3′5′ = 0: S 34 = 0

S 3′6′ = 0: m S 23 − S 13 = 0

S 4′5′ = 0: m S 44 − S 55 = 0

" Group 6

2 3 S 1′6′ = 0: mn 2S 22 − 2S 12 − S 66 − m 2S 11 − 2S 12 − S 66 = 0

3 2 S 2′6′ = 0: m 2S 22 − 2S 12 − S 66 − mn 2S 11 − 2S 12 − S 66 = 0

276 TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x3 AXIS

" Like for the stiffnesses, the fourth group can also be written as 2 S 1′5′ = 0: − S 24 = m S 56 + S 14 − S 24 m = cosθ3 n = sinθ3

2 S 2′5′ = 0: S 14 = m S 56 + S 14 − S 24

S = 0: 1 S = m 2 S + S S 4′6′ 2 56 56 14 − 24

" Because the right-hand side of the equations are identical, it follows 1 S = S = S = that the left-hand sides are equal; that is, 2 56 − 24 14 Δ

2 " Each equation can be expressed as Δ = 4Δcos θ3 , which is satisﬁed = 0 cos 2 = 1 by Δ and by θ3 4 π π " Δ = 0 implies S56 = S24 = S14 = 0 and − 2 < θ 3 ≤ 2

2 1 π " cos = = S = 2S S = S θ3 4 implies θ3 ± 3 , 56 14 , and 24 − 14

277 TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x3 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 S 1′6′ + S 2′6′ = 0: S 22 − S 11 cos θ 3 = 0 , π π which is satisﬁed for all − 2 < θ 3 ≤ 2 if S22 = S11 , or for all π S22 ≠ S11 if θ3 = 2

" Subtracting gives

S 1′6′ − S 2′6′ = 0: S 11 + S 22 − 2S 12 − S 66 cos θ 3cos 2θ 3 = 0 ,

π π which is satisﬁed for all − 2 < θ 3 ≤ 2 if 2S 12 + S 66 − S 11 − S 22 = 0 , 2S + S S S 0 π π or for all 12 66 − 11 − 22 ≠ if θ3 = ± 4 or 2

278 TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x3 AXIS

" The symmetry properties for trigonal materials arise from the cos 2 = 1 solution for the invariance conditions that are given by θ3 4

" For this solution, the planes of Symmetry x elastic symmetry are all parallel to plane for 2 θ = 0 the x3 axis and are given by 3 π θ3 = + 3 π θ3 = 0 and ± 3 Symmetry plane for

π 1 θ3 = + " = π m = cos π = ≠ 0 3 For θ3 ± 3 , ± 3 2 x1 π 3 = 0 n = sin = ≠ 0 θ3 and ± 3 ± 2

" Symmetry The stiffness equations in group 2 plane for π θ 3 = − 3 yield the invariance conditions π θ 3 = − 3 C = 0 , C = C , and C = C 34 23 13 55 44 Plan view of symmetry planes

279 TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x3 AXIS

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

conditions C56 = C14 and C 24 = − C 14

" Furthermore, the stiffness equations in group 3 yield the invariance C = C C = 1 C C conditions 22 11 and 66 2 11 − 12

" The compliance equations in group 5 yield the invariance conditions

S34 = 0 , S23 = S13 , and S55 = S44

" The compliance equations in group 4 yield the invariance conditions

S56 = 2S14 and S 24 = − S 14

" The compliance equations in group 6 yield the invariance conditions

S22 = S11 and S 66 = 2 S 11 − S 12

280 TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x3 AXIS

" The invariance conditions α1′3′ = 0: mα13 + nα23 = 0 and 2 2 α 1′2′ = 0: m − n α 12 + mn α 22 − α 11 = 0 yield

α23 = α13 = α12 = 0 and α22 = α11

" The invariance conditions β 1′3′ = 0: mβ 13 + nβ 23 = 0 and 2 2 β 1′2′ = 0: m − n β 12 + mn β 22 − β 11 = 0 yield

β 23 = β 13 = β 12 = 0 and β 22 = β 11

" Together, the invariance conditions yield the following constitutive equations for a trigonal material

281 TRIGONAL MATERIALS - CONCLUDED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x3 AXIS

C C C C 0 0 11 12 13 14 β 11 σ 11 ε11 C12 C C13 − C 0 0 σ 22 11 14 ε22 β 11 C C C 0 0 0 σ 33 13 13 33 ε33 β 33 = + T − Tref σ 23 C14 − C14 0 C44 0 0 2ε23 0 σ 13 0 0 0 0 C44 C14 2ε13 0 σ 12 C11 − C12 2ε12 0 0 0 0 0 C 14 2

S 11 S 12 S 13 S 14 0 0 ε11 σ 11 α11 S S S S 0 0 ε22 12 11 13 − 14 σ 22 α11

ε S 13 S 13 S 33 0 0 0 σ α 33 = 33 + 33 T T 0 − ref 2ε23 S 14 − S 14 0 S 44 0 0 σ 23 σ 13 0 2ε13 0 0 0 0 S 44 2S 14 2 σ 12 0 ε12 0 0 0 0 2S 14 2 S 11 − S 12

" Therefore, a trigonal material has six independent elastic constants and two independent thermal-expansion or thermal-compliance parameters

282 TRIGONAL MATERIALS

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x1 AXIS

a i • i " In particular, consider a plane of x1 ,x1′ i′j ≡ i′ j elastic symmetry whose normal n

lies in the plane x1 = 0 and makes an x3′ angle θ with the x axis, as shown in i 1 2 i 1 , 1′ i 3′ the ﬁgure θ1

i 2 i 3 x3 " The angle θ1 is deﬁned to be in the π π π range − < θ 1 ≤ , because θ 1 = − i 2′, n 2 2 2 θ1 π and θ1 = deﬁne the same plane 2 x 2′ x2

" In addition, let the x1 , x2 , x3 be the Plane of elastic ′ ′ ′ symmetry coordinates used to deﬁne the material symmetry

283 TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x1 AXIS

" Speciﬁcally, for a plane of elastic symmetry given by x2′ = 0 , the symmetry transformation is shown in the ﬁgure and is given by

x 1′′ = x 1′ , x 2′′ = − x 2′ , and x 3′′ = x 3′

" The corresponding matrix of x1′,x1′′ a i′j ≡ i i′ • i j direction cosines is given by

1 0 0 a1′′1′ a1′′2′ a1′′3′ x2′′

a2′′1′ a2′′2′ a2′′3′ = 0 − 1 0

i 1 i 1 a3′′1′ a3′′2′ a3′′3′ 0 0 1 ′ , ′′

i 2′′

" x 1 , x 2 , x 3 For the ′ ′ ′ coordinate x3 ,x3 i 3′ ,i 3′′ ′ ′′ system, the constitutive i 2′ equations are expressed as follows

x2′

284 TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x1 AXIS

ε ′ = S′ σ′ + α′ Θ and σ ′ = C′ ε′ + β′ Θ ,

where Θ = T − Tref

285 TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x1 AXIS

" For the x1′′, x2′′, x3′′ coordinate system, the constitutive equations are identical in form and are obtained by replacing the index pair i′j ′ with i′′ j ′′

" In the abridged notation,

εε ′′′′ = S′′′′ σσ′′′′ + αα′′′′ ΘΘ and

σσ′′′′ = C′′′′ εε′′′′ + ββ′′′′ ΘΘ , where Θ = T − Tref

" For the transformation of coordinates deﬁned by the symmetry

transformation (a reﬂection about the plane x2′ = 0 ),

r 2′ r 2′ σσ′′′′ = Tσ σσ′′ and εε′′′′ = Tε εε′′ where

286 TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x1 AXIS

1 0 0 0 0 0 0 1 0 0 0 0

r 2′ r 2′ 0 0 1 0 0 0 T = T = σ ε 0 0 0 − 1 0 0 0 0 0 0 1 0 0 0 0 0 0 − 1

" Likewise,

− 1 − 1 r 2′ r 2′ r 2′ r 2′ S′′ = Tε S′ Tσ S′ = Tε S′′ Tσ

− 1 − 1 r 2′ r 2′ r 2′ r 2′ C′′ = Tσ C′ Tε C′ = Tσ C′′ Tε

r 2′ r 2′ αα′′′′ = Tε αα′′ ββ′′′′ = Tσ ββ′′

287 TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x1 AXIS

" The conditions for invariance under the symmetry transformation are given by

− 1 − 1 r 2′ r 2′ r 2′ r 2′ S′ = Tε S′ Tσ C′ = Tσ C′ Tε

r 2′ r 2′ αα′′ = Tε αα′′ ββ′′ = Tσ ββ′′

x2 = 0

" Direct comparison reveals that the material is monoclinic with

respect to the plane x2′ = 0

288 TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x1 AXIS

" That is, direct comparison yields

S 1′1′ S 1′2′ S 1′3′ 0 S 1′5′ 0 ε α 1′1′ 1′1′ σ 1′1′ S 1′2′ S 2′2′ S 2′3′ 0 S 2′5′ 0 ε2′2′ σ 2′2′ α 2′2′

ε3′3′ S 1′3′ S 2′3′ S 3′3′ 0 S 3′5′ 0 σ 3′3′ α 3′3′ = + T − Tref and 2ε2′3′ 0 0 0 S 4′4′ 0 S 4′6′ σ 2′3′ 0 σ 2ε1 3 1′3′ 2α ′ ′ S 1′5′ S 2′5′ S 3′5′ 0 S 5′5′ 0 1′3′ σ 1′2′ 2ε1′2′ 0 0 0 0 S 4′6′ 0 S 6′6′

C 1′1′ C 1′2′ C 1′3′ 0 C 1′5′ 0 ε1 1 β 1′1′ σ 1′1′ ′ ′ C 1′2′ C 2′2′ C 2′3′ 0 C 2′5′ 0 σ 2′2′ ε2′2′ β 2′2′

σ 3′3′ C 1′3′ C 2′3′ C 3′3′ 0 C 3′5′ 0 ε3′3′ β 3′3′ = + T − Tref σ 2′3′ 0 0 0 C 4′4′ 0 C 4′6′ 2ε2′3′ 0 σ 1′3′ 2ε β C 1′5′ C 2′5′ C 3′5′ 0 C 5′5′ 0 1′3′ 1′3′ σ 1′2′ 2ε1′2′ 0 0 0 0 C 4′6′ 0 C 6′6′

289 TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x1 AXIS

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

thermal moduli, expressed in x1′, x2′, x3′ coordinates, are expressed in

the original x1, x2, x3 coordinates

a i i " The transformation corresponds to x1 ,x1′ i′j ≡ i′ • j

the dextral rotation about the x1 axis, shown in the ﬁgure, and is given by x 3′ x = x x = x cos + x sin 1′ 1 , 2′ 2 θ1 3 θ1 , and i i 1 , 1′ i 3′ x 3′ = − x 2sin θ 1 + x 3cos θ 1 , with θ1 π π − 2 < θ 1 ≤ 2 i 2 i 3 x3

" The corresponding matrix of i 2′, n

direction cosines is given by θ1

x2′ 1 0 0 a1′1 a1′2 a1′3 x2 Plane of elastic a2′1 a2′2 a2′3 = 0 cosθ 1 sinθ 1 symmetry a3′1 a3′2 a3′3 0 − sinθ 1 cosθ 1

290 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

1 0 0 0 0 0

2 2 0 cos θ1 sin θ1 2sinθ1cosθ1 0 0 2 2 0 sin θ1 cos θ1 − 2sinθ1cosθ1 0 0 = Tσ θ1 2 2 and 0 − sinθ1cosθ1 sinθ1cosθ1 cos θ1 − sin θ1 0 0

0 0 0 0 cosθ1 − sinθ1

0 0 0 0 sinθ1 cosθ1

1 0 0 0 0 0

2 2 0 cos θ1 sin θ1 sinθ1cosθ1 0 0 2 2 0 sin θ1 cos θ1 − sinθ1cosθ1 0 0 = Tε θ1 2 2 0 − 2sinθ1cosθ1 2sinθ1cosθ1 cos θ1 − sin θ1 0 0

0 0 0 0 cosθ1 − sinθ1

0 0 0 0 sinθ1 cosθ1

291 TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x1 AXIS

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

− 1 − 1 C′ = Tσ C Tε S′ = Tε S Tσ

β′ = Tσ β α′ = Tε α

" These laws transform the invariance conditions on C ′ , S ′ , α ′ , and β ′ into invariance conditions on C , S , α , and β

" 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 C and S , and the vectors α and β , are fully populated

292 TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x1 AXIS

" Because of the invariance conditions, the matrices C ′ and S ′ have

C 1′1′ C 1′2′ C 1′3′ 0 C 1′5′ 0 S 1′1′ S 1′2′ S 1′3′ 0 S 1′5′ 0 C 1′2′ C 2′2′ C 2′3′ 0 C 2′5′ 0 S 1′2′ S 2′2′ S 2′3′ 0 S 2′5′ 0 C C C 0 C 0 S S S 0 S 0 the form 1′3′ 2′3′ 3′3′ 3′5′ and 1′3′ 2′3′ 3′3′ 3′5′ 0 0 0 C 4′4′ 0 C 4′6′ 0 0 0 S 4′4′ 0 S 4′6′ C 1′5′ C 2′5′ C 3′5′ 0 C 5′5′ 0 S 1′5′ S 2′5′ S 3′5′ 0 S 5′5′ 0 0 0 0 C 4′6′ 0 C 6′6′ 0 0 0 S 4′6′ 0 S 6′6′

α 1′1′ β 1′1′

α 2′2′ β 2′2′

α 3 3 β 3 3 " Also, the vectors α ′ and β ′ have the form ′ ′ and ′ ′ 0 0

2α 1′3′ β 1′3′ 0 0

" The corresponding forms of C , S , α , and β are obtained from

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

293 TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x1 AXIS

" For example, specifying θ1 = 0 yields

C1 1 C1 2 C1 3 0 C1 5 0 ′ ′ ′ ′ ′ ′ ′ ′ C11 C12 C13 C14 C15 C16 β 1′1′ β 11

C1′2′ C2′2′ C2′3′ 0 C2′5′ 0 C12 C22 C23 C24 C25 C26 β 2′2′ β 22

C1 3 C2 3 C3 3 0 C3 5 0 C C C C C C β 3 3 β 33 ′ ′ ′ ′ ′ ′ ′ ′ = 13 23 33 34 35 36 ′ ′ = 0 0 0 C4′4′ 0 C4′6′ C14 C24 C34 C44 C45 C46 0 β 23

β 13 C1′5′ C2′5′ C3′5′ 0 C5′5′ 0 C15 C25 C35 C45 C55 C56 β 1′3′ β 12 0 0 0 C4′6′ 0 C6′6′ C16 C26 C36 C46 C56 C66 0

S 1 1 S 1 2 S 1 3 0 S 1 5 0 S S S S S S ′ ′ ′ ′ ′ ′ ′ ′ 11 12 13 14 15 16 α1′1′ α11 S S S 0 S 0 1′2′ 2′2′ 2′3′ 2′5′ S12 S22 S23 S24 S25 S26 α2′2′ α22

S 1 3 S 2 3 S 3 3 0 S 3 5 0 S13 S23 S33 S34 S35 S36 α3 3 α33 ′ ′ ′ ′ ′ ′ ′ ′ = ′ ′ = 0 0 0 S 4′4′ 0 S 4′6′ S14 S24 S34 S44 S45 S46 0 2α23

2α13 S 1′5′ S 2′5′ S 3′5′ 0 S 5′5′ 0 S15 S25 S35 S45 S55 S56 2α1′3′ 0 2α12 0 0 0 S 4′6′ 0 S 6′6′ S16 S26 S36 S46 S56 S66

" Enforcing the invariance conditions on C ′ gives

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

294 TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x1 AXIS

" Enforcing the invariance conditions on S ′ gives

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

" Enforcing the invariance conditions on α ′ and β ′ gives

α12 = α23 = 0 and β 12 = β 23 = 0

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

material that is monoclinic with respect to the plane x2 = 0

" = π Similarly, specifying θ1 2 and enforcing the invariance conditions on C ′ , S ′ , α ′ , and β ′ yields the conditions previously obtained for

a material that is monoclinic with respect to the plane x3 = 0

295 TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x1 AXIS

" π < π Now consider an arbitrary value for the angle θ1 such that − 2 θ 1 ≤ 2

" The previous example for θ1 = 0 showed that the invariance conditions on C , S , α , and β were obtained by the terms of C ′ , S ′ , α ′ , and β ′ 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 C ′

2 2 C 1′4′ = 0: m − n C 14 + mn C 13 − C 12 = 0

C 1′6′ = 0: m C 16 + nC 15 = 0

296 TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x1 AXIS

2 2 2 3 C2′4′ = 0: m m − 3n C24 − m n C22 − C23 − 2C44 3 2 2 2 + mn C33 − C23 − 2C44 − n n − 3m C34 = 0

3 2 2 3 C 2′6′ = 0: m C 26 + m n 2C 46 + C 25 + mn 2C 45 + C 36 + n C 35 = 0

2 2 2 3 C3′4′ = 0: m m − 3n C34 + m n C33 − C23 − 2C44 3 2 2 2 − mn C22 − C23 − 2C44 − n n − 3m C24 = 0

3 2 2 3 C 3′6′ = 0: m C 36 − m n 2C 46 − C 35 − mn 2C 45 − C 26 + n C 25 = 0

3 2 2 3 C 4′5′ = 0: m C 45 − m n C 46 + C 25 − C 35 − mn C 45 − C 26 + C 36 + n C 46 = 0

2 2 C 5′6′ = 0: m − n C 56 + mn C 55 − C 66 = 0

with m = cosθ1 and n = sinθ1

" These conditions give 8 equations and 20 unknowns

297 TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x1 AXIS

" The invariance conditions on S ′ are given by

2 2 S 1′4′ = 0: m − n S 14 + 2mn S 13 − S 12 = 0 S 1′6′ = 0: m S 16 + nS 15 = 0

2 2 2 3 S2′4′ = 0: m m − 3n S24 − m n 2S22 − 2S23 − S44 3 2 2 2 + mn 2S33 − 2S23 − S44 − n n − 3m S34 = 0

3 2 2 3 S 2′6′ = 0: m S 26 + m n S 46 + S 25 + mn S 45 + S 36 + n S 35 = 0

2 2 2 3 S3′4′ = 0: m m − 3n S34 + m n 2S33 − 2S23 − S44 3 2 2 2 − mn 2S22 − 2S23 − S44 − n n − 3m S24 = 0

3 2 2 3 S 3′6′ = 0: m S 36 − m n S 46 − S 35 − mn S 45 − S 26 + n S 25 = 0

3 2 2 3 S 4′5′ = 0: m S 45 − m n S 46 + 2S 25 − 2S 35 − mn S 45 − 2S 26 + 2S 36 + n S 46 = 0

2 2 S 5′6′ = 0: m − n S 56 + mn S 55 − S 66 = 0

298 TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x1 AXIS

" The invariance conditions on α ′ and β ′ yield

2 2 α 1′2′ = 0: m α 12 + nα 13 = 0 α 2′3′ = 0: m − n α 23 + mn α 33 − α 22 = 0

2 2 β 1′2′ = 0: m β 12 + nβ 13 = 0 β 2′3′ = 0: m − n β 23 + mn β 33 − β 22 = 0

299 TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x1 AXIS

the x1 axis, it is convenient to select the ﬁrst plane to be given by θ1 = 0

" Thus, the material is monoclinic with respect to the plane x2 = 0 and, as a result, the following conditions hold

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

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

α12 = 0 α23 = 0 β 12 = 0 β 23 = 0

" These relations, and the fact that n = sinθ1 ≠ 0 for nonzero values π π of − 2 < θ 1 ≤ 2 , are used to simplify the previously given invariance conditions into the following three uncoupled groups

300 TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x1 AXIS

" Group 1 2 2 C 2′6′ = 0: m 2C 46 + C 25 + n C 35 = 0 m = cosθ1 n = sinθ1

2 2 C 3′6′ = 0: − m 2C 46 − C 35 + n C 25 = 0

2 2 C 4′5′ = 0: − m C 46 + C 25 − C 35 + n C 46 = 0

" Group 2

C 1′4′ = 0: m C 13 − C 12 = 0

C 1′6′ = 0: C 15 = 0

C 5′6′ = 0: m C 55 − C 66 = 0

" Group 3

3 2 C 2′4′ = 0: − m C 22 − C 23 − 2C 44 + mn C 33 − C 23 − 2C 44 = 0

3 2 C 3′4′ = 0: m C 33 − C 23 − 2C 44 − mn C 22 − C 23 − 2C 44 = 0

301 TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x1 AXIS

" The ﬁrst group can be written as 2 C 2′6′ = 0: − C 35 = m 2C 46 + C 25 − C 35 m = cosθ1 n = sinθ1

2 C 3′6′ = 0: C 25 = m 2C 46 + C 25 − C 35

2 C 4′5′ = 0: C 46 = m 2C 46 + C 25 − C 35

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

that the left-hand sides are equal; that is, C 46 = − C 35 = C 25 = Γ

2 " Each equation can be expressed as Γ = 4Γcos θ1 , which is satisﬁed = 0 cos 2 = 1 by Γ and by θ1 4 π π " Γ = 0 implies C 46 = − C 35 = C 25 = 0 and − 2 < θ 1 ≤ 2

2 1 π " cos = = C = C C = C θ1 4 implies θ1 ± 3 , 46 25 , and 35 − 25

302 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 C 2′4′ + C 3′4′ = 0: C 33 − C 22 cos θ 1 = 0 , π π which is satisﬁed for all − 2 < θ 1 ≤ 2 if C33 = C22 , or for all π C33 ≠ C22 if θ1 = 2

" Subtracting gives

C 2′4′ − C 3′4′ = 0: 2C 23 + 4C 44 − C 22 − C 33 cos θ 1cos 2θ 1 = 0 ,

π π which is satisﬁed for all − 2 < θ 1 ≤ 2 if 2C 23 + 4C 44 − C 22 − C 33 = 0 , 2C + 4C C C 0 π π or for all 23 44 − 22 − 33 ≠ if θ1 = ± 4 or 2

303 TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x1 AXIS

" Group 4 2 2 S 2′6′ = 0: m S 46 + S 25 + n S 35 = 0 m = cosθ1 n = sinθ1

2 2 S 3′6′ = 0: − m S 46 − S 35 + n S 25 = 0

2 2 S 4′5′ = 0: − m S 46 + 2S 25 − 2S 35 + n S 46 = 0

" Group 5

S 1′4′ = 0: m S 13 − S 12 = 0

S 1′6′ = 0: S 15 = 0

S 5′6′ = 0: m S 55 − S 66 = 0

" Group 6

3 2 S 2′4′ = 0: − m 2S 22 − 2S 23 − S 44 + mn 2S 33 − 2S 23 − S 44 = 0

3 2 S 3′4′ = 0: m 2S 33 − 2S 23 − S 44 − mn 2S 22 − 2S 23 − S 44 = 0

304 TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x1 AXIS

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

2 S 2′6′ = 0: − S 35 = m S 46 + S 25 − S 35 m = cosθ1 n = sinθ1

2 S 3′6′ = 0: S 25 = m S 46 + S 25 − S 35

S = 0: 1 S = m 2 S + S S 4′5′ 2 46 46 25 − 35

" Because the right-hand side of the equations are identical, it follows 1 S = S = S = that the left-hand sides are equal; that is, 2 46 − 35 25 Δ

2 " Each equation can be expressed as Δ = 4Δcos θ3 , which is satisﬁed = 0 cos 2 = 1 by Δ and by θ1 4 π π " Δ = 0 implies S46 = S35 = S25 = 0 and − 2 < θ 1 ≤ 2

2 1 π " cos = = S = 2S S = S θ1 4 implies θ1 ± 3 , 46 25 , and 35 − 25

305 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 S 2′4′ + S 3′4′ = 0: S 33 − S 22 cos θ 1 = 0 , π π which is satisﬁed for all − 2 < θ 1 ≤ 2 if S33 = S22 , or for all π S33 ≠ S22 if θ1 = 2

" Subtracting gives

S 2′4′ − S 3′4′ = 0: 2S 23 + S 44 − S 22 − S 33 cos θ 1cos 2θ 1 = 0 ,

π π which is satisﬁed for all − 2 < θ 1 ≤ 2 if 2S 23 + S 44 − S 22 − S 33 = 0 , 2S + S S S 0 π π or for all 23 44 − 22 − 33 ≠ if θ1 = ± 4 or 2

306 TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x1 AXIS

" The symmetry properties for trigonal materials arise from the cos 2 = 1 solution for the invariance conditions that is given by θ1 4

" For this solution, the planes of Symmetry x elastic symmetry are all parallel to plane for 3 the x axis and are given by θ1 = 0 1 π θ1 = + 3 π θ1 = 0 and ± 3 Symmetry plane for

π θ1 = + π π 1 3 " = m = cos = ≠ 0 x2 For θ1 ± 3 , ± 3 2 π 3 = 0 n = sin = ≠ 0 θ1 and ± 3 ± 2

" The stiffness equations in group 2 Symmetry plane for π θ 1 = − yield the invariance conditions π 3 θ 1 = − 3 C = 0 , C = C , and C = C 15 13 12 66 55 Plan view of symmetry planes

307 TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x1 AXIS

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

conditions C46 = C25 and C 35 = − C 25

" Furthermore, the stiffness equations in group 3 yield the invariance C = C C = 1 C C conditions 33 22 and 44 2 22 − 23

" The compliance equations in group 5 yield the invariance conditions

S15 = 0 , S13 = S12 , and S66 = S55

" The compliance equations in group 4 yield the invariance conditions

S46 = 2S25 and S 35 = − S 25

" The compliance equations in group 6 yield the invariance conditions

S33 = S22 and S 44 = 2 S 22 − S 23

308 TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x1 AXIS

" The invariance conditions α 1′2′ = 0: m α 12 + nα 13 = 0 and

2 2 α 2′3′ = 0: m − n α 23 + mn α 33 − α 22 = 0 yield

α 23 = α 13 = α 12 = 0 and α 33 = α 22

" The invariance conditions β 1′2′ = 0: m β 12 + nβ 13 = 0 and

2 2 β 2′3′ = 0: m − n β 23 + mn β 33 − β 22 = 0 yield

β 23 = β 13 = β 12 = 0 and β 33 = β 22

" Together, the invariance conditions yield the following constitutive equations for a trigonal material

309 TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x1 AXIS

C11 C12 C12 0 0 0 β 11 σ 11 ε11 C12 C22 C23 0 C25 0 σ 22 ε22 β 22 C12 C23 C22 0 − C25 0 σ 33 ε33 β 22 = + T − Tref C22 − C23 σ 23 0 0 0 0 C 2ε23 0 2 25 σ 13 2ε13 0 σ 12 0 C25 − C25 0 C55 0 2ε12 0 0 0 0 C25 0 C55

S 11 S 12 S 12 0 0 0 ε11 σ 11 α11 S 12 S 22 S 23 0 S 25 0 ε22 σ 22 α22

ε33 S 12 S 23 S 22 0 − S 25 0 σ 33 α22 = + T − Tref 2ε23 0 0 0 2 S 22 − S 23 0 2S 25 σ 23 0 0 2ε13 σ 13 0 S 25 − S 25 0 S 55 0 σ 12 0 2ε12 0 0 0 2S 25 0 S 55

" Again, the trigonal material has six independent elastic constants and two independent thermal-expansion or thermal-compliance parameters

310 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 x3 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 x3 axis into congruence with the adjacent ﬁgure shown below for symmetry planes

that contain the x1 axis

311 TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x1 AXIS

x Plane 3 Plane Plane x1 Plane θ3 = +π/3 θ1 = +π/3 θ3 = -π/3 θ1 = -π/3

x2 x3

Plane Plane

θ3 = 0 θ1 = 0 x 1 x2

Symmetry planes that Symmetry planes that

contain the x3 axis contain the x1 axis

312 TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x1 AXIS

" Inspection of the ﬁgures indicates the following transformation of the indices: 1 → 2, 2 → 3, and 3 → 1

" 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 tensor notation 11 22 33 23, 32 31, 13 12, 21

matrix notation 1 2 3 4 5 6

" Using this information along with 1 → 2, 2 → 3, and 3 → 1 gives the relations: 4 → 5, 5 → 6, and 6 → 4

313 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 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

" Consider the following constitutive equations for a trigonal material that

has reﬂective symmetry planes that contain the x3 axis

C11 C12 C13 C14 0 0 ε11 β 11 σ11 C12 C11 C13 − C14 0 0 σ22 ε22 β 11

σ33 C13 C13 C33 0 0 0 ε33 β 33 = + T − Tref σ23 C14 − C14 0 C44 0 0 2ε23 0 σ13 2 0 0 0 0 0 C44 C14 ε13 σ12 1 2ε12 0 0 0 0 0 C14 2 C11 − C12

314 TRIGONAL MATERIALS - CONCLUDED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x1 AXIS

" Applying the index transformation to these constitutive equations gives C C C C 0 0 22 23 12 25 β 22 σ 22 ε22 C 23 C C 12 − C 0 0 σ 33 22 25 ε33 β 22

σ 11 C 12 C 12 C 11 0 0 0 ε11 β 11 = + T − Tref σ 13 C 25 − C 25 0 C 55 0 0 2ε13 0 σ 12 2 0 0 0 0 0 C 55 C 25 ε12 σ 23 2ε 0 0 0 0 0 C 1 C − C 23 25 2 22 23

" Reordering these equations into the standard form yields

C11 C12 C12 0 0 0 β 11 σ 11 ε11 C12 C22 C23 0 C25 0 σ 22 ε22 β 22 C C C 0 C 0 σ 12 23 22 − 25 ε β 22 33 33 = 1 + T − Tref σ 23 0 0 0 C − C 0 C 2ε23 0 2 22 23 25 σ 13 2ε13 0 0 C25 − C25 0 C55 0 σ 12 2ε12 0 0 0 0 C25 0 C55

which are identical to the corresponding equations previously given

315 TRIGONAL MATERIALS

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x2 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 x3 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 x3 axis into congruence with the adjacent ﬁgure shown below for symmetry planes

that contain the x2 axis

316 TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x2 AXIS

x Plane 3 Plane Plane x2 Plane θ3 = +π/3 θ2 = +π/3 θ3 = -π/3 θ2 = -π/3

x2 x1

Plane Plane

θ3 = 0 θ2 = 0

x1 x3

Symmetry planes that Symmetry planes that

contain the x3 axis contain the x2 axis

317 TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x2 AXIS

" Inspection of the ﬁgures indicates the following transformation of the indices: 1 → 3, 2 → 1, and 3 → 2

" 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 tensor notation 11 22 33 23, 32 31, 13 12, 21

matrix notation 1 2 3 4 5 6

" Using this information along with 1 → 3, 2 → 1, and 3 → 2 gives the relations: 4 → 6, 5 → 4, and 6 → 5

318 TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x2 AXIS

" Likewise, the transformation of index pairs that appear in the abridged notation are given by 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

" Consider the following constitutive equations for a trigonal material that

has reﬂective symmetry planes that contain the x3 axis

C11 C12 C13 C14 0 0 ε11 β 11 σ11 C12 C11 C13 − C14 0 0 σ22 ε22 β 11

σ33 C13 C13 C33 0 0 0 ε33 β 33 = + T − Tref σ23 C14 − C14 0 C44 0 0 2ε23 0 σ13 2 0 0 0 0 0 C44 C14 ε13 σ12 1 2ε12 0 0 0 0 0 C14 2 C11 − C12

319 TRIGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x2 AXIS

" Applying the index transformation to these constitutive equations gives

C C C C 0 0 33 13 23 36 β 33 σ 33 ε33 C13 C C23 − C 0 0 σ 11 33 36 ε11 β 33

σ 22 C23 C23 C22 0 0 0 ε22 β 22 = + T − Tref σ 12 C36 − C36 0 C66 0 0 2ε12 0 σ 23 2 0 0 0 0 0 C66 C36 ε23 σ 13 2ε 0 0 0 0 0 C 1 C − C 13 36 2 33 13

" Reordering these equations into the standard form yields

C33 C23 C13 0 0 − C 36 β 33 σ 11 ε11 C23 C22 C23 0 0 0 σ 22 ε22 β 22

σ 33 C13 C23 C33 0 0 C36 ε33 β 33 = + T − Tref σ 23 0 0 0 C66 C36 0 2ε23 0 σ 13 1 2ε13 0 0 0 0 C36 C33 − C13 0 σ 12 2 2ε12 0

− C36 0 C36 0 0 C66

320 TRIGONAL MATERIALS - CONCLUDED

REFLECTIVE SYMMETRY ABOUT PLANES THAT CONTAIN THE x2 AXIS

" Applying the same procedure to the inverse equations yields

S33 S23 S13 0 0 − S36 ε11 σ11 α33 S23 S22 S23 0 0 0 ε22 σ22 α33

ε33 S13 S23 S33 0 0 S36 σ33 α22 = + T − Tref 2ε23 0 0 0 S66 2S36 0 σ23 0 0 2ε13 σ13 0 0 0 2S36 2 S33 − S13 0 σ12 0 2ε12 − S36 0 S36 0 0 S66

321 SUMMARY OF TRIGONAL MATERIALS

C11 C12 C12 0 0 0 ε 11 β 11 σ11 C12 C22 C23 0 C25 0 σ22 ε 22 β 22

C12 C23 C22 0 − C 0 σ33 25 ε 33 β 22 x axis = 1 + T − Tref 1 σ23 0 0 0 0 2 C22 − C23 0 C25 2ε 23 σ13 2ε 13 0 0 C25 − C25 0 C55 0 σ12 2ε 12 0 0 0 0 C25 0 C55

C33 C23 C13 0 0 − C36 ε11 β 33 σ11 C23 C22 C23 0 0 0 σ22 ε22 β 22

C13 C23 C33 0 0 C36 β 33 σ33 = ε33 + T T − ref x2 axis σ23 0 0 0 C66 C36 0 2ε23 0 σ13 1 2ε13 0 0 0 0 C36 C33 − C13 0 σ12 2 2ε12 0 − C36 0 C36 0 0 C66

C11 C12 C13 C14 0 0 ε11 β 11 σ11 C12 C11 C13 − C14 0 0 σ22 ε22 β 11

σ33 C13 C13 C33 0 0 0 ε33 β 33 = + T − Tref x3 axis σ23 C14 − C14 0 C44 0 0 2ε23 0 σ13 2 0 0 0 0 0 C44 C14 ε13 σ12 1 2ε12 0 0 0 0 0 C14 2 C11 − C12

322 SUMMARY OF TRIGONAL MATERIALS - CONCLUDED

S11 S12 S12 0 0 0 ε11 σ11 α11 S12 S22 S23 0 S25 0 ε22 σ22 α22

ε33 S12 S23 S22 0 − S25 0 σ33 α22 = + T − Tref x1 axis 2ε23 0 0 0 2 S22 − S23 0 2S25 σ23 0 σ13 2ε13 0 0 S25 − S25 0 S55 0 σ12 0 2ε12 0 0 0 2S25 0 S55

S33 S23 S13 0 0 − S36 ε11 σ11 α33 S23 S22 S23 0 0 0 ε22 σ22 α33 S S S 0 0 S ε33 = 13 23 33 36 σ33 + α22 T T − ref x2 axis 2ε23 0 0 0 S66 2S36 0 σ23 0 σ13 2ε13 0 0 0 0 2S36 2 S33 − S13 0 σ12 0 2ε12 − S36 0 S36 0 0 S66

S 11 S 12 S 13 S 14 0 0 ε 11 σ 11 α 11 S 12 S 11 S 13 − S 14 0 0 ε 22 σ 22 α 11 S S S 0 0 0 ε 33 13 13 33 σ 33 α 33 x axis = + T − Tref 3 2ε 23 S 14 − S 14 0 S 44 0 0 σ 23 0 σ 13 0 2ε 13 0 0 0 0 S 44 2S 14 σ 12 0 2ε 12 0 0 0 0 2S 14 2 S 11 − S 12

323 TETRAGONAL MATERIALS

324 TETRAGONAL MATERIALS

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

containing the x3 axis, special invariance conditions arise for values of θ = ± π 3 4 , in addition to θ3 = 0

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

containing the x1 axis, special invariance conditions arise for values of θ = ± π 1 4 , in addition to θ1 = 0

" Furthermore, for a single plane of elastic symmetry containing the x2 θ = ± π axis, special invariance conditions also arise for values of 2 4 , in

addition to θ2 = 0

" 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

325 TETRAGONAL MATERIALS - CONTINUED

" Speciﬁcally, consider a material with three planes of elastic symmetry θ = 0 and ± π given by 3 4

Plane x3 Plane π θ = -π/4 θ3 = +π/4 " θ = + θ3 π By taking 1 2 as an additional plane of elastic symmetry, the fact that Plane π θ = + θ = +π/2 the planes θ3 = 0 and 1 2 are 1

perpendicular implies the existence of x2

a ﬁfth symmetry plane given by θ1 = 0

" This fact was shown previously for Plane θ = 0 specially orthotropic materials; θ3 Plane θ1 = 0 that is, two perpendicular planes x1 of elastic symmetry imply a third perpendicular plane of elastic symmetry

" Thus, there exists five planes of elastic symmetry, four of which θ = + π are perpendicular to plane 1 2

326 TETRAGONAL MATERIALS - CONTINUED

" Next, consider a material with three planes of elastic symmetry given by θ = 0 and ± π 1 4

Plane θ1 = 0

" By taking θ3 = 0 as an additional plane of elastic symmetry, the fact Plane that the planes θ1 = 0 and θ3 = 0 are θ1 = +π/2 perpendicular implies the existence x2 of a ﬁfth symmetry plane given by Plane π θ = -π/4 θ = + θ1 π 1 2 Plane

θ3 = 0 Plane x1 θ = +π/4 " Thus, there exists five planes of 1 elastic symmetry, four of which

are perpendicular to plane θ3 = 0

327 TETRAGONAL MATERIALS - CONTINUED

" Finally, consider a material with three planes of elastic symmetry given θ = 0 and ± π by 2 4

x3 Plane θ = 0 θ3 Plane θ3 = +π/2 π " θ = + By taking 3 2 as an additional plane of elastic symmetry, the fact θ = + π Plane that the planes 3 2 and θ2 = 0 θ2 = 0 x2 are perpendicular implies the existence of a ﬁfth symmetry plane Plane given by θ3 = 0 θ2 = -π/4 Plane x1 θ2 = +π/4 " Thus, there exists five planes of elastic symmetry θ = + π , four of which are perpendicular to plane 3 2

328 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:

π π " θ = 0 and ± θ = + Planes of elastic symmetry given by 3 4 and 1 2

π " θ = 0 and ± θ = 0 Planes of elastic symmetry given by 1 4 and 3

π π " θ = 0 and ± θ = + Planes of elastic symmetry given by 2 4 and 3 2

" 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 θ = ± π θ = ± π θ = ± π invariance conditions for 1 4 , 2 4 , or 3 4

329 TETRAGONAL MATERIALS

REFLECTIVE SYMMETRY PLANES THAT CONTAIN THE x3 AXIS

" Consider a tetragonal material with planes of reﬂective symmetry θ = 0 and ± π θ = + π deﬁned by 3 4 and by 1 2

" For a specially orthotropic material, it was shown previously that the constitutive equations are given by

C11 C12 C13 0 0 0 β 11 σ 11 ε11

σ 22 C12 C22 C23 0 0 0 ε22 β 22

σ 33 C13 C23 C33 0 0 0 ε33 β 33 = + T − Tref and σ 23 0 0 0 C44 0 0 2ε23 0

σ 13 0 0 0 0 C55 0 2ε13 0 σ 12 0 0 0 0 0 C66 2ε12 0

S 11 S 12 S 13 0 0 0 ε11 σ 11 α11

ε22 S 12 S 22 S 23 0 0 0 σ 22 α22

ε33 S 13 S 23 S 33 0 0 0 σ 33 α33 = + T − Tref 2ε23 0 0 0 S 44 0 0 σ 23 0 0 2ε13 0 0 0 0 S 55 0 σ 13 σ 12 0 2ε12 0 0 0 0 0 S 66

330 TETRAGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY PLANES THAT CONTAIN THE x3 AXIS

" For this case, the additional conditions on the constitutive equations θ = ± π are obtained by enforcing symmetry planes given by 3 4

π 2 2 " θ = ± m = cos π = ≠ 0 n = sin π = ≠ 0 For 3 4 , ± 4 2 and ± 4 ± 2

π " θ = ± With 3 4 , the ﬁrst group of stiffness equations previously given herein for an arbitrary plane of symmetry, deﬁned by constant values of

θ3, requires Γ = 0

" Recall that Γ = 0 implies C 56 = − C 24 = C 14 = 0 for arbitrary values of π π the angle − 2 < θ 3 ≤ 2

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

of symmetry, deﬁned by constant values of θ3, requires C34 = 0,

C23 = C13, and C55 = C44

331 TETRAGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY PLANES THAT CONTAIN THE x3 AXIS

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

condition C22 = C11

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

conditions S34 = 0 , S23 = S13 , and S55 = S44

" The compliance equations in group 4 yield the invariance conditions

S56 = S24 = S14 = 0

" The compliance equations in group 6 yield the invariance condition

S22 = S11

332 TETRAGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY PLANES THAT CONTAIN THE x3 AXIS

" The invariance conditions α1′3′ = 0: mα13 + nα23 = 0 and 2 2 α 1′2′ = 0: m − n α 12 + mn α 22 − α 11 = 0 yield

α23 = α13 = α12 = 0 and α22 = α11

" The invariance conditions β 1′3′ = 0: mβ 13 + nβ 23 = 0 and 2 2 β 1′2′ = 0: m − n β 12 + mn β 22 − β 11 = 0 yield

β 23 = β 13 = β 12 = 0 and β 22 = β 11

" Together, the invariance conditions yield the following constitutive equations for a tetragonal material

333 TETRAGONAL MATERIALS - CONCLUDED

REFLECTIVE SYMMETRY PLANES THAT CONTAIN THE x3 AXIS

C11 C12 C13 0 0 0 β 11 σ 11 ε11 C C C 0 0 0 σ 22 12 11 13 ε22 β 11 C C C 0 0 0 σ 33 13 13 33 ε33 β 33 = + T − Tref σ 23 0 0 0 C44 0 0 2ε23 0

σ 13 0 0 0 0 C44 0 2ε13 0 σ 12 0 0 0 0 0 C66 2ε12 0

S11 S12 S13 0 0 0 ε11 σ 11 α11 S S S 0 0 0 ε22 12 11 13 σ 22 α11

ε33 S13 S13 S33 0 0 0 σ 33 α33 = + T − Tref 2ε23 0 0 0 S44 0 0 σ 23 0 0 2ε13 0 0 0 0 S44 0 σ 13 σ 12 0 2ε12 0 0 0 0 0 S66

" Therefore, a tetragonal material has six independent elastic constants and two independent thermal-expansion or thermal- compliance parameters

334 TETRAGONAL MATERIALS

REFLECTIVE SYMMETRY PLANES THAT CONTAIN THE x1 AXIS

" Next, consider a tetragonal material with planes of reﬂective θ = 0 and ± π symmetry deﬁned by 1 4 and by θ3 = 0

π 2 2 " θ = ± m = cos π = ≠ 0 n = sin π = ≠ 0 For 1 4 , ± 4 2 and ± 4 ± 2

" 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 C46 = C35 = C25 = 0

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

symmetry, deﬁned by constant values of θ1, requires C15 = 0, C13 = C12,

and C66 = C55

" The stiffness equations in group 3 yield the invariance condition

C33 = C22

335 TETRAGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY PLANES THAT CONTAIN THE x1 AXIS

" The invariance conditions β 1′2′ = 0: mβ 12 + nβ 13 = 0 and 2 2 β 2′3′ = 0: m − n β 23 + m n β 33 − β 22 = 0 yield

β 23 = β 13 = β 12 = 0 and β 33 = β 22

" The resulting constitutive equation, obtained by simplifying the constitutive equation for a specialy orthotropic material, is given by

C11 C12 C12 0 0 0 β 11 σ 11 ε11 C C C 0 0 0 σ 22 12 22 23 ε22 β 22 C C C 0 0 0 σ 33 12 23 22 ε33 β 22 = + T − Tref σ 23 0 0 0 C44 0 0 2ε23 0

σ 13 0 0 0 0 C55 0 2ε13 0 σ 12 0 0 0 0 0 C55 2ε12 0

336 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

S11 S12 S12 0 0 0 ε11 σ 11 α11 S S S 0 0 0 ε22 12 22 23 σ 22 α22

ε33 S12 S23 S22 0 0 0 σ 33 α22 = + T − Tref 2ε23 0 0 0 S44 0 0 σ 23 0 0 2ε13 0 0 0 0 S55 0 σ 13 σ 12 0 2ε12 0 0 0 0 0 S55

337 TETRAGONAL MATERIALS

REFLECTIVE SYMMETRY PLANES THAT CONTAIN THE x2 AXIS

" Finally, consider a tetragonal material with planes of reﬂective θ = 0 and ± π θ = + π symmetry deﬁned by 2 4 and by 3 2

" 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 x3 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 x3 axis into congruence with the adjacent ﬁgure shown below with four symmetry

planes that contain the x2 axis

338 TETRAGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY PLANES THAT CONTAIN THE x2 AXIS

x Plane 3 Plane Plane x2 Plane θ3 = +π/3 θ2 = +π/3 θ3 = -π/3 θ2 = -π/3

Plane Plane

θ1 = +π/2 θ3 = +π/2

x2 x1

Plane Plane

θ3 = 0 θ2 = 0 Plane Plane x1 θ = +π/2 θ3 = +π/2 x3 2

Four symmetry planes Four symmetry planes

that contain the x3 axis that contain the x2 axis

339 TETRAGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY PLANES THAT CONTAIN THE x2 AXIS

" Inspection of the ﬁgures indicates the following transformation of the indices: 1 → 3, 2 → 1, and 3 → 2

" 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 tensor notation 11 22 33 23, 32 31, 13 12, 21

matrix notation 1 2 3 4 5 6

" Using this information along with 1 → 3, 2 → 1, and 3 → 2 gives the relations: 4 → 6, 5 → 4, and 6 → 5

340 TETRAGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY PLANES THAT CONTAIN THE x2 AXIS

" Likewise, the transformation of index pairs that appear in the abridged notation are given by 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

" Consider the following constitutive equations for a tetragonal material

that has reﬂective symmetry planes that contain the x3 axis

C11 C12 C13 0 0 0 β 11 σ 11 ε11

σ 22 C12 C11 C13 0 0 0 ε22 β 11

σ 33 C13 C13 C33 0 0 0 ε33 β 33 = + T − Tref σ 23 0 0 0 C44 0 0 2ε23 0

σ 13 0 0 0 0 C44 0 2ε13 0 σ 12 0 0 0 0 0 C66 2ε12 0

341 TETRAGONAL MATERIALS - CONTINUED

REFLECTIVE SYMMETRY PLANES THAT CONTAIN THE x2 AXIS

" Applying the index transformation to these constitutive equations gives

C33 C13 C23 0 0 0 β 33 σ 33 ε33 C C C 0 0 0 σ 11 13 33 23 ε11 β 33 C C C 0 0 0 σ 22 23 23 22 ε22 β 22 = + T − Tref σ 12 0 0 0 C66 0 0 2ε12 0

σ 23 0 0 0 0 C66 0 2ε23 0 σ 13 0 0 0 0 0 C55 2ε13 0

" Reordering these equations into the standard form yields

C33 C23 C13 0 0 0 β 33 σ 11 ε11 C C C 0 0 0 σ 22 23 22 23 ε22 β 22 C C C 0 0 0 σ 33 13 23 33 ε33 β 33 = + T − Tref σ 23 0 0 0 C66 0 0 2ε23 0

σ 13 0 0 0 0 C55 0 2ε13 0 σ 12 0 0 0 0 0 C66 2ε12 0

342 TETRAGONAL MATERIALS - CONCLUDED

REFLECTIVE SYMMETRY PLANES THAT CONTAIN THE x2 AXIS

" Applying the same procedure to the inverse equations yields

ε11 S33 S23 S13 0 0 0 σ 11 α33 S S S 0 0 0 ε22 23 22 23 σ 22 α22

ε33 S13 S23 S33 0 0 0 σ 33 α33 = + T − Tref 2ε23 0 0 0 S66 0 0 σ 23 0 0 2ε13 0 0 0 0 S55 0 σ 13 σ 12 0 2ε12 0 0 0 0 0 S66

343 SUMMARY OF TETRAGONAL MATERIALS

C11 C12 C12 0 0 0 β 11 σ 11 ε11

σ 22 C12 C22 C23 0 0 0 ε22 β 22

σ 33 C12 C23 C22 0 0 0 ε33 β 22 = + T − Tref x1 axis σ 23 0 0 0 C44 0 0 2ε23 0

σ 13 0 0 0 0 C55 0 2ε13 0 σ 12 0 0 0 0 0 C55 2ε12 0

C33 C23 C13 0 0 0 β 33 σ 11 ε11

σ 22 C23 C22 C23 0 0 0 ε22 β 22

σ 33 C13 C23 C33 0 0 0 ε33 β 33 = + T − Tref x2 axis σ 23 0 0 0 C66 0 0 2ε23 0

σ 13 0 0 0 0 C55 0 2ε13 0 σ 12 0 0 0 0 0 C66 2ε12 0

C11 C12 C13 0 0 0 β 11 σ 11 ε11

σ 22 C12 C11 C13 0 0 0 ε22 β 11

σ 33 C13 C13 C33 0 0 0 ε33 β 33 = + T − Tref x3 axis σ 23 0 0 0 C44 0 0 2ε23 0

σ 13 0 0 0 0 C44 0 2ε13 0 σ 12 0 0 0 0 0 C66 2ε12 0

344 SUMMARY OF TETRAGONAL MATERIALS - CONTINUED

S 11 S 12 S 12 0 0 0 ε11 σ 11 α11

ε22 S 12 S 22 S 23 0 0 0 σ 22 α22

ε33 S 12 S 23 S 22 0 0 0 σ 33 α22 = + T − Tref x1 axis 2ε23 0 0 0 S 44 0 0 σ 23 0 0 2ε13 0 0 0 0 S 55 0 σ 13 σ 12 0 2ε12 0 0 0 0 0 S 55

S 33 S 23 S 13 0 0 0 ε11 σ 11 α33

ε22 S 23 S 22 S 23 0 0 0 σ 22 α22

ε33 S 13 S 23 S 33 0 0 0 σ 33 α33 = + T − Tref x2 axis 2ε23 0 0 0 S 66 0 0 σ 23 0 0 2ε13 0 0 0 0 S 55 0 σ 13 σ 12 0 2ε12 0 0 0 0 0 S 66

S 11 S 12 S 13 0 0 0 ε11 σ 11 α11

ε22 S 12 S 11 S 13 0 0 0 σ 22 α11

ε33 S 13 S 13 S 33 0 0 0 σ 33 α33 = + T − Tref x3 axis 2ε23 0 0 0 S 44 0 0 σ 23 0 0 2ε13 0 0 0 0 S 44 0 σ 13 σ 12 0 2ε12 0 0 0 0 0 S 66

345 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

346 TRANSVERSELY ISOTROPIC MATERIALS

347 TRANSVERSELY ISOTROPIC MATERIALS

ISOTROPY PLANE x3 = 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 x3 = 0

" For the specially orthotropic material, it was shown previously that the material properties are given by

C11 C12 C13 0 0 0 ε 11 β 11 σ 11 C C C 0 0 0 σ 22 12 22 23 ε 22 β 22

σ 33 C13 C23 C33 0 0 0 ε 33 β 33 = + T − Tref and σ 23 0 0 0 C44 0 0 2ε 23 0 σ 13 0 0 0 0 C55 0 2ε 13 0 σ 12 0 0 0 0 0 C66 2ε 12 0

ε 11 S11 S12 S13 0 0 0 σ11 α 11 S S S 0 0 0 ε 22 12 22 23 σ22 α 22

ε 33 S13 S23 S33 0 0 0 σ33 α 33 = + T − Tref 2ε 23 0 0 0 S 44 0 0 σ23 0 σ13 2ε 13 0 0 0 0 S 55 0 0 σ12 0 2ε 12 0 0 0 0 0 S 66

348 TRANSVERSELY ISOTROPIC MATERIALS

ISOTROPY PLANE x3 = 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 x3 = 0

" Equivalently, the constitutive matrices and vectors must invariant

with respect to dextral rotations about the x3 axis

x x " The coordinate transformation for this 3 , 3′ a i′j ≡ i i′ • i j symmetry is shown in the ﬁgure and x 2′ is given by x1′ = x1cosθ3 + x2sinθ3 , i 3 , i 3′

i 2′ x 2′ = − x 1sin θ 3 + x 2cos θ 3 , and x3′ = x3 , θ3

i 1 with 0 ≤ θ3 < 2π i 2 x2

i 1′ " The corresponding matrix of θ3 direction cosines is given by x 1′ Plane of isotropy cos sin 0 x1 a1′1 a1′2 a1′3 θ 3 θ 3

a2′1 a2′2 a2′3 = − sinθ 3 cosθ 3 0

a3′1 a3′2 a3′3 0 0 1

349 TRANSVERSELY ISOTROPIC MATERIALS

ISOTROPY PLANE x3 = 0 (CONTINUED)

" The corresponding stress and strain transformation matrices were shown previously to be given by

2 2 cos θ3 sin θ3 0 0 0 2sinθ3cosθ3 2 2 sin θ3 cos θ3 0 0 0 − 2sinθ3cosθ3

0 0 1 0 0 0 = Tσ θ3 and 0 0 0 cosθ3 − sinθ3 0

0 0 0 sinθ3 cosθ3 0 2 2 − sinθ3cosθ3 sinθ3cosθ3 0 0 0 cos θ3 − sin θ3

2 2 cos θ3 sin θ3 0 0 0 sinθ3cosθ3 2 2 sin θ3 cos θ3 0 0 0 − sinθ3cosθ3

0 0 1 0 0 0 = Tε θ3 0 0 0 cosθ3 − sinθ3 0

0 0 0 sinθ3 cosθ3 0 2 2 − 2sinθ3cosθ3 2sinθ3cosθ3 0 0 0 cos θ3 − sin θ3

350 TRANSVERSELY ISOTROPIC MATERIALS

ISOTROPY PLANE x3 = 0 (CONTINUED)

" Speciﬁc expressions that result from the transformation − 1 C′ = Tσ C Tε were given previously for an anisotropic material

− 1 " For a specially orthotropic material, C′ = Tσ C Tε yields the expressions given previously for a generally orthotropic material; that is,

4 2 2 4 C 1′1′ = m C 11 + 2m n C 12 + 2C 66 + n C 22

2 2 4 4 C 1′2′ = m n C 11 + C 22 − 4C 66 + m + n C 12

2 2 C 1′3′ = m C 13 + n C 23 C 1′4′ = 0 C 1′5′ = 0

2 2 2 2 C 1′6′ = mn m − n C 12 + 2C 66 + mn n C 22 − m C 11

4 2 2 4 C 2′2′ = m C 22 + 2m n C 12 + 2C 66 + n C 11

351 TRANSVERSELY ISOTROPIC MATERIALS

ISOTROPY PLANE x3 = 0 (CONTINUED)

2 2 C 2′3′ = m C 23 + n C 13 C 2′4′ = 0 C 2′5′ = 0

2 2 2 2 C 2′6′ = mn n − m C 12 + 2C 66 + mn m C 22 − n C 11

C 3′3′ = C 33 C 3′4′ = 0 C 3′5′ = 0 C 3′6′ = mn C 23 − C 13

2 2 C 4′4′ = m C 44 + n C 55 C 4′5′ = mn C 44 − C 55 C 4′6′ = 0 C 5′6′ = 0

2 2 2 2 2 2 2 C 5′5′ = m C 55 + n C 44 C 6′6′ = m n C 11 + C 22 − 2C 12 + m − n C 66

with m = cosθ3 and n = sinθ3

− 1 " Next, the invariance condition C = Tσ C Tε is enforced, which

implies the conditions Cr′s′ = Crs , where r, s ∈ 1,2, 3, 4, 5, 6

352 TRANSVERSELY ISOTROPIC MATERIALS

ISOTROPY PLANE x3 = 0 (CONTINUED)

4 2 2 4 " Enforcing C1′1′ = C11 gives C11 = m C11 + 2m n C12 + 2C66 + n C22 , which

can only be satisﬁed in a general sense (arbitrary values for θ3) if

C 11 = C 22 and C 12 + 2C 66 = C 11

2 2 " Similarly, enforcing C1′3′ = C13 gives C13 = m C13 + n C23 , which yields

C 23 = C 13

2 2 " Likewise, enforcing C4′4′ = C44 gives C44 = m C44 + n C55 , which yields

C 55 = C 44

" Finally, by substituting C 11 = C 22 , C 12 + 2C 66 = C 11 , C 23 = C 13 , and

C 55 = C 44 into the remaining expressions for Cr′s′ and enforcing

Cr′s′ = Crs , it is found the remaining expressions are identically satisﬁed

353 TRANSVERSELY ISOTROPIC MATERIALS

ISOTROPY PLANE x3 = 0 (CONTINUED)

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

isotropic in the plane x3 = 0 has the form

C11 C12 C13 0 0 0

C12 C11 C13 0 0 0 C C C 0 0 0 13 13 33 which has 5 independent stiffnesses

0 0 0 C44 0 0

0 0 0 0 C44 0 0 0 0 0 0 1 C C 2 11 − 12

" 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 − 1

condition S = Tε S Tσ or Sr′s′ = Srs yields similar results

354 TRANSVERSELY ISOTROPIC MATERIALS

ISOTROPY PLANE x3 = 0 (CONTINUED)

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

isotropic in the plane x3 = 0 has the form

S11 S12 S13 0 0 0

S12 S11 S13 0 0 0 S S S 0 0 0 13 13 33 which has 5 independent compliances

0 0 0 S44 0 0

0 0 0 0 S44 0 0 0 0 0 0 1 S S 2 11 − 12

355 TRANSVERSELY ISOTROPIC MATERIALS

ISOTROPY PLANE x3 = 0 (CONTINUED)

" For a specially orthotropic material, α′ = Tε α yields the expressions for the transformed thermal-expansion coefﬁcients, given previously for a generally orthotropic material; that is,

2 2 2 2 α1′1′ = m α11 + n α22 α2′2′ = m α22 + n α11 α3′3′ = α33

α 2′3′ = 0 α 1′3′ = 0 α 1′2′ = mn α 22 − α 11

" Enforcing the invariance condition α = Tε α or αr′s′ = αrs yields

the requirements that α 22 = α 11 and α 12 = 0

" 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 β = Tσ β or β r′s′ = β rs yields similar results; that is,

β 22 = β 11 and β 12 = 0

356 TRANSVERSELY ISOTROPIC MATERIALS

ISOTROPY PLANE x3 = 0 (CONTINUED)

" Applying all these simpliﬁcations, the linear thermoelastic constitutive equations become

C11 C12 C13 0 0 0 β 11 σ 11 ε11 C12 C11 C13 0 0 0 σ 22 ε22 β 11 C C C 0 0 0 σ 33 13 13 33 ε33 β 33 and = + T − Tref σ 23 0 0 0 C44 0 0 2ε23 0 σ 13 0 0 0 0 0 C44 0 2ε13 σ 12 1 2ε12 0 0 0 0 0 0 2 C11 − C12

S11 S12 S13 0 0 0 ε11 σ 11 α11 S12 S11 S13 0 0 0 ε22 σ 22 α11 S S S 0 0 0 ε33 13 13 33 σ 33 α33 = + T − Tref 2ε23 0 0 0 S44 0 0 σ 23 0

σ 13 0 2ε13 0 0 0 0 S44 0 σ 12 0 2ε12 1 0 0 0 0 0 2 S11 − S12

357 TRANSVERSELY ISOTROPIC MATERIALS

ISOTROPY PLANE x3 = 0 (CONCLUDED)

β 11 C 11 + C 12 C 13 α = − 11 " Likewise α β 33 2C 13 C 33 33

" 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 = 0, π planes of reﬂective symmetry given by θ3 ± 6 , have the same number of independent elastic constants and thermal parameters as the transversely isotropic material

358 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

C11 C12 C13 0 0 0 ε 11 β 11 σ 11 C C C 0 0 0 σ 22 12 22 23 ε 22 β 22

σ 33 C13 C23 C33 0 0 0 ε 33 β 33 = + T − Tref and σ 23 0 0 0 C44 0 0 2ε 23 0 σ 13 0 0 0 0 C55 0 2ε 13 0 σ 12 0 0 0 0 0 C66 2ε 12 0

ε 11 S11 S12 S13 0 0 0 σ11 α 11 S S S 0 0 0 ε 22 12 22 23 σ22 α 22

ε 33 S13 S23 S33 0 0 0 σ33 α 33 = + T − Tref 2ε 23 0 0 0 S 44 0 0 σ23 0 σ13 2ε 13 0 0 0 0 S 55 0 0 σ12 0 2ε 12 0 0 0 0 0 S 66

359 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 Plane of x3 " The coordinate transformation for this x3′ isotropy symmetry is shown in the ﬁgure and is x = x x = x cos + x sin given by 1′ 1 , 2′ 2 θ1 3 θ1 , θ1

and x 3′ = − x 2sin θ 1 + x 3cos θ 1 , with

0 ≤ θ1 < 2π i 3 x2′

i 3′ i 2′ θ1 " The corresponding matrix of direction x2 cosines is given by i 2 1 0 0 a1′1 a1′2 a1′3 i 1 , i 1′

a2′1 a2′2 a2′3 = 0 cosθ 1 sinθ 1 a i′j ≡ i i′ • i j a a a 3′1 3′2 3′3 0 − sinθ 1 cosθ 1 x1 , x1′

360 TRANSVERSELY ISOTROPIC MATERIALS

ISOTROPY PLANE x1 = 0 (CONTINUED)

" The corresponding stress and strain transformation matrices were shown previously to be given by

1 0 0 0 0 0

2 2 0 cos θ1 sin θ1 2sinθ1cosθ1 0 0 2 2 0 sin θ1 cos θ1 − 2sinθ1cosθ1 0 0 and Tσ = 2 2 0 − sinθ1cosθ1 sinθ1cosθ1 cos θ1 − sin θ1 0 0

0 0 0 0 cosθ1 − sinθ1

0 0 0 0 sinθ1 cosθ1

1 0 0 0 0 0

2 2 0 cos θ1 sin θ1 sinθ1cosθ1 0 0 2 2 0 sin θ1 cos θ1 − sinθ1cosθ1 0 0

Tε = 2 2 0 − 2sinθ1cosθ1 2sinθ1cosθ1 cos θ1 − sin θ1 0 0

0 0 0 0 cosθ1 − sinθ1

0 0 0 0 sinθ1 cosθ1

361 TRANSVERSELY ISOTROPIC MATERIALS

ISOTROPY PLANE x1 = 0 (CONTINUED)

" Speciﬁc expressions that result from the transformation − 1 C′ = Tσ C Tε were given previously for an anisotropic material

− 1 " For a specially orthotropic material, C′ = Tσ C Tε yields the expressions for a generally orthotropic material; that is,

2 2 2 2 C1′1′ = C11 C1′2′ = m C12 + n C13 C1′3′ = m C13 + n C12

C 1′4′ = mn C 13 − C 12 C1′5′ = 0 C1′6′ = 0

4 2 2 4 C2′2′ = m C22 + 2m n C23 + 2C44 + n C33

2 2 4 4 C 2′3′ = m n C 22 + C 33 − 4C 44 + m + n C 23

3 3 C 2′4′ = mn C 33 − C 23 − 2C 44 − m n C 22 − C 23 − 2C 44

362 TRANSVERSELY ISOTROPIC MATERIALS

ISOTROPY PLANE x1 = 0 (CONTINUED)

4 2 2 4 C2′5′ = 0 C2′6′ = 0 C3′3′ = m C33 + 2m n C23 + 2C44 + n C22

3 3 C 3′4′ = m n C 33 − C 23 − 2C 44 − mn C 22 − C 23 − 2C 44

2 2 2 2 2

C3′5′ = 0 C3′6′ = 0 C 4′4′ = m n C 22 + C 33 − 2C 23 + m − n C 44

2 2 C4′5′ = 0 C4′6′ = 0 C5′5′ = m C55 + n C66

2 2 C 5′6′ = mn C 55 − C 66 C6′6′ = m C66 + n C55

with m = cosθ1 and n = sinθ1

− 1 " Next, the invariance condition C = Tσ C Tε is enforced, which

implies the conditions Cr′s′ = Crs , where r, s ∈ 1,2, 3, 4, 5, 6

363 TRANSVERSELY ISOTROPIC MATERIALS

ISOTROPY PLANE x1 = 0 (CONTINUED)

4 2 2 4 " Enforcing C2′2′ = C22 gives C22 = m C22 + 2m n C23 + 2C44 + n C33 , which

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

C 33 = C 22 and C 23 + 2C 44 = C 22

2 2 " Similarly, enforcing C1′3′ = C13 gives C13 = m C13 + n C12 , which yields

C 12 = C 13

2 2 " Likewise, enforcing C5′5′ = C55 gives C55 = m C55 + n C66 , which yields

C 66 = C 55

" Finally, by substituting C 33 = C 22 , C 23 + 2C 44 = C 22 , C 12 = C 13 , and

C 66 = C 55 into the remaining expressions for Cr′s′ and enforcing

Cr′s′ = Crs , it is found the remaining expressions are identically satisﬁed

364 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

C11 C12 C12 0 0 0

C12 C22 C23 0 0 0 C C C 0 0 0 12 23 22 which has 5 independent stiffnesses 0 0 0 1 C C 0 0 2 22 − 23

0 0 0 0 C55 0

0 0 0 0 0 C55

condition S = Tε S Tσ or Sr′s′ = Srs yields similar results

365 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

S11 S12 S12 0 0 0

S12 S22 S23 0 0 0 S S S 0 0 0 12 23 22 which has 5 independent compliances 0 0 0 1 S S 0 0 2 22 − 23

0 0 0 0 S55 0

0 0 0 0 0 S55

366 TRANSVERSELY ISOTROPIC MATERIALS

ISOTROPY PLANE x1 = 0 (CONTINUED)

" For a specially orthotropic material, α′ = Tε α yields the expressions for the transformed thermal-expansion coefﬁcients for a generally orthotropic material; that is,

2 2 2 2 α1′1′ = α11 α2′2′ = m α22 + n α33 α3′3′ = m α33 + n α22

α 2′3′ = mn α 33 − α 22 α1′3′ = 0 α1′2′ = 0

" Enforcing the invariance condition α = Tε α or αr′s′ = αrs yields

the requirements that α 33 = α 22 and α 23 = 0

" 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 β = Tσ β or β r′s′ = β rs yields similar results; that is,

β 33 = β 22 and β 23 = 0

367 TRANSVERSELY ISOTROPIC MATERIALS

ISOTROPY PLANE x1 = 0 (CONTINUED)

" Applying all these simpliﬁcations, the linear thermoelastic constitutive equations become

C11 C12 C12 0 0 0 β 11 σ 11 ε11 C12 C22 C23 0 0 0 σ 22 ε22 β 22

σ C12 C23 C22 0 0 0 ε β 22 33 = 33 + T − T and σ 1 0 ref 23 0 0 0 2 C22 − C23 0 0 2ε23 σ 13 0 0 0 0 0 C55 0 2ε13 σ 12 2ε12 0 0 0 0 0 0 C55

S11 S12 S12 0 0 0 ε11 σ 11 α11 S12 S22 S23 0 0 0 ε22 σ 22 α22 ε S12 S23 S22 0 0 0 σ α 33 = 33 + 22 T − T 1 σ 0 ref 2ε23 0 0 0 2 S22 − S23 0 0 23 σ 13 0 2ε13 0 0 0 0 S55 0 σ 12 0 2ε12 0 0 0 0 0 S55

368 TRANSVERSELY ISOTROPIC MATERIALS

ISOTROPY PLANE x1 = 0 (CONCLUDED)

β 11 C 11 2C 12 α " Likewise = − 11 α 22 β 22 C 12 C 22 + C 23

" It is interesting to point out that hexagonal materials, deﬁned by = 0, π planes of reﬂective symmetry given by θ1 ± 6 , have the same number of independent elastic constants and thermal parameters as the transversely isotropic material

369 CUBIC MATERIALS

370 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

x3 " 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 x2 = 0, π symmetry that are given by θ1 ± 4 , = 0, π = 0, π θ2 ± 4 , and θ3 ± 4

" Expressions for the constitutive x equations are obtained directly by 1 enforcing the three sets of invariance conditions for the previously given cases for tetragonal materials sequentially

371 CUBIC MATERIALS - CONCLUDED

The resulting constitutive equations are given by

C11 C12 C12 0 0 0 β 11 σ 11 ε11 C C C 0 0 0 σ 22 12 11 12 ε22 β 11 C C C 0 0 0 σ 33 12 12 11 ε33 β 11 = + T − Tref σ 23 0 0 0 C44 0 0 2ε23 0

σ 13 0 0 0 0 C44 0 2ε13 0 σ 12 0 0 0 0 0 C44 2ε12 0

S11 S12 S12 0 0 0 ε11 σ 11 α11 S S S 0 0 0 ε22 12 11 12 σ 22 α11

ε33 S12 S12 S11 0 0 0 σ 33 α11 = + T − Tref 2ε23 0 0 0 S44 0 0 σ 23 0 0 2ε13 0 0 0 0 S44 0 σ 13 σ 12 0 2ε12 0 0 0 0 0 S44

" Therefore, a cubic material has three independent elastic constants and one independent thermal-expansion or thermal-compliance parameter

372 COMPLETELY ISOTROPIC MATERIALS

373 COMPLETELY ISOTROPIC MATERIALS

" First, consider the case in which both the planes x1 = 0 and x2 = 0 are planes of isotropy

" Applying the results of both of the corresponding symmetry transformations successively yields the following constitutive equations

C11 C12 C12 0 0 0 β 11 σ 11 ε11 C12 C11 C12 0 0 0 σ 22 ε22 β 11

σ C12 C12 C11 0 0 0 ε β 11 33 = 33 + T − T σ 1 0 ref 23 0 0 0 2 C11 − C12 0 0 2ε23 σ 13 1 2ε 0 0 0 0 0 2 C11 − C12 0 13 σ 12 1 2ε12 0 0 0 0 0 0 2 C11 − C12

" These equations possess two independent elastic stiffnesses and one independent thermal moduli; the same as a completely isotropic material

374 COMPLETELY ISOTROPIC MATERIALS (CONTINUED)

" Similarly,

S11 S12 S12 0 0 0 ε11 σ 11 α11 S12 S11 S12 0 0 0 ε22 σ 22 α11 ε S12 S12 S11 0 0 0 σ α 33 = 33 + 11 T T 1 σ 0 − ref 2ε23 0 0 0 2 S11 − S12 0 0 23 2ε 1 σ 13 0 13 0 0 0 0 2 S11 − S12 0 σ 12 0 2ε12 1 0 0 0 0 0 2 S11 − S12

β 11 = − C 11 + 2C 12 α 11

" Now consider a general transformation of the stiffness matrix previously obtained for a material with two perpendicular planes of isotropy

375 COMPLETELY ISOTROPIC MATERIALS (CONTINUED)

− 1 " The general transformation law is given by C′ = Tσ C Tε , where

2 2 2 a1′1 a1′2 a1′3 2a1′2a1′3 2a1′1a1′3 2a1′1a1′2 2 2 2 a2′1 a2′2 a2′3 2a2′2a2′3 2a2′1a2′3 2a2′1a2′2 2 2 2 a3′1 a3′2 a3′3 2a3′2a3′3 2a3′1a3′3 2a3′1a3′2 Tσ = and 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

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

376 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:

a a = k p ak′qap′q = δkp k p q′k q′p δkp

2 2 2 2 2 2 1 1 a1′1 + a1′2 + a1′3 = 1 1 1 a1′1 + a2′1 + a3′1 = 1

2 1 a2′1a1′1 + a2′2a1′2 + a2′3a1′3 = 0 2 1 a1′2a1′1 + a2′2a2′1 + a3′2a3′1 = 0

3 1 a3′1a1′1 + a3′2a1′2 + a3′3a1′3 = 0 3 1 a1′3a1′1 + a2′3a2′1 + a3′3a3′1 = 0

2 2 2 2 2 2 2 2 a2′1 + a2′2 + a2′3 = 1 2 2 a1′2 + a2′2 + a3′2 = 1

3 2 a3′1a2′1 + a3′2a2′2 + a3′3a2′3 = 0 3 2 a1′3a1′2 + a2′3a2′2 + a3′3a3′2 = 0

2 2 2 2 2 2 3 3 a3′1 + a3′2 + a3′3 = 1 3 3 a1′3 + a2′3 + a3′3 = 1

377 COMPLETELY ISOTROPIC MATERIALS (CONTINUED)

" For the general transformation to be a symmetry transformation, − 1 − 1 C′ = Tσ C Tε becomes C = Tσ C Tε

" It is convenient to rewrite this expression as C Tε − Tσ C = 0 , C = 1 C C where 44 2 11 − 12 ,

C C C 0 0 0 2 2 2 11 12 12 a1′1 a1′2 a1′3 a1′2a1′3 a1′1a1′3 a1′1a1′2 2 2 2 C 12 C 11 C 12 0 0 0 a2′1 a 2′2 a2′3 a2′2a2′3 a2′1a2′3 a2′1a2′2 2 2 2 C 12 C 12 C 11 0 0 0 a3′1 a3′2 a3′3 a3′2a3′3 a3′1a3′3 a3′1a3′2 C Tε = 2a a 2a a 2a a a a + a a a a + a a a a + a a and 0 0 0 C 44 0 0 2′1 3′1 2′2 3′2 2′3 3′3 2′2 3′3 2′3 3′2 2′1 3′3 2′3 3′1 2′1 3′2 2′2 3′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 0 0 0 0 C 44 0 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 0 0 0 0 0 C 44

2 2 2 a1′1 a1′2 a1′3 2a1′2a1′3 2a1′1a1′3 2a1′1a1′2 2 2 2 C 11 C 12 C 12 0 0 0 a2′1 a2′2 a2′3 2a2′2a2′3 2a2′1a2′3 2a2′1a2′2 C 12 C 11 C 12 0 0 0 a 2 a 2 a 2 2a a 2a a 2a a 3′1 3′2 3′3 3′2 3′3 3′1 3′3 3′1 3′2 C 12 C 12 C 11 0 0 0 Tσ C = 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 0 0 0 C 44 0 0

0 0 0 0 C 44 0 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

0 0 0 0 0 C 44 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

378 COMPLETELY ISOTROPIC MATERIALS (CONTINUED)

" Performing the calculations for each element of the matrix equation

C Tε − Tσ C = 0 gives

2 2 2 2 11 : C 12 a 2′1 + a 3′1 − a 1′2 − a 1′3 = 0

2 2 2 2 12 : C 12 a 2′2 + a 3′2 − a 1′2 − a 1′3 = 0

2 2 2 2 13 : C 12 a 2′3 + a 3′3 − a 1′1 − a 1′2 = 0

14 : C 12 a 1′3a 1′2 + a 2′3a 2′2 + a 3′3a 3′2 = 0

15 : C 12 a 1′3a 1′1 + a 2′3a 2′1 + a 3′3a 3′1 = 0

16 : C 12 a 1′2a 1′1 + a 2′2a 2′1 + a 3′2a 3′1 = 0

2 2 2 2 21 : C 12 a 1′1 + a 3′1 − a 2′2 − a 2′3 = 0

379 COMPLETELY ISOTROPIC MATERIALS (CONTINUED)

2 2 2 2 22 : C 12 a 1′2 + a 3′2 − a 2′1 − a 2′3 = 0

2 2 2 2 23 : C 12 a 1′3 + a 3′3 − a 2′1 − a 2′2 = 0

24 : C 12 a 1′3a 1′2 + a 2′3a 2′2 + a 3′3a 3′2 = 0

25 : C 12 a 1′3a 1′1 + a 2′3a 2′1 + a 3′3a 3′1 = 0

26 : C 12 a 1′2a 1′1 + a 2′2a 2′1 + a 3′2a 3′1 = 0

2 2 2 2 31 : C 12 a 1′1 + a 2′1 − a 3′2 − a 3′3 = 0

2 2 2 2 32 : C 12 a 1′2 + a 2′2 − a 3′1 − a 3′3 = 0

2 2 2 2 33 : C 12 a 1′3 + a 2′3 − a 3′1 − a 3′2 = 0

380 COMPLETELY ISOTROPIC MATERIALS (CONTINUED)

34: C12 a1′3a1′2 + a2′3a2′2 + a3′3a3′2 = 0

35: C12 a1′3a1′1 + a2′3a2′1 + a3′3a3′1 = 0

36: C12 a1′2a1′1 + a2′2a2′1 + a3′2a3′1 = 0

41: C12 a3′1a2′1 + a3′2a2′2 + a3′3a2′3 = 0

42: C12 a3′1a2′1 + a3′2a2′2 + a3′3a2′3 = 0

43: C12 a3′1a2′1 + a3′2a2′2 + a3′3a2′3 = 0

51: C12 a3′1a1′1 + a3′2a1′2 + a3′3a1′3 = 0

52: C12 a3′1a1′1 + a3′2a1′2 + a3′3a1′3 = 0

53: C12 a3′1a1′1 + a3′2a1′2 + a3′3a1′3 = 0

381 COMPLETELY ISOTROPIC MATERIALS (CONTINUED)

61 : C 12 a 2′1a 1′1 + a 2′2a 1′2 + a 2′3a 1′3 = 0

62 : C 12 a 2′1a 1′1 + a 2′2a 1′2 + a 2′3a 1′3 = 0

63 : C 12 a 2′1a 1′1 + a 2′2a 1′2 + a 2′3a 1′3 = 0

" The elements of the matrix equation C Tε − Tσ C = 0 that are not listed above are satisﬁed identically

2 2 2 2 " Now, consider 11 : C 12 a 2′1 + a 3′1 − a 1′2 − a 1′3 = 0

2 2 2 " The condition a 1′1 + a 1′2 + a 1′3 = 1 given in the previous table

2 2 2 yields a 1′2 + a 1′3 = 1 − a 1′1 ; hence,

2 2 2 11 : C 12 a 1′1 + a 2′1 + a 3′1 − 1 = 0

382 COMPLETELY ISOTROPIC MATERIALS (CONTINUED)

2 2 2 " Next, using the condition a1′1 + a2′1 + a3′1 = 1 , given in the 2 2 2 previous table, shows that 11 : C 12 a 1′1 + a 2′1 + a 3′1 − 1 = 0 is identically satisﬁed

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

conditions ak′qap′q = δkp and aq′kaq′p = δkp given in the previous table, it can

be shown that the invariance condition C Tε − Tσ C = 0 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

383 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 β′ = Tσ β

" For the general transformation to be a symmetry transformation,

β′ = Tσ β becomes β − Tσ β = 0

" The expanded form is given by

2 2 2 β 11 a 1′1 a 1′2 a 1′3 2a 1′2a 1′3 2a 1′1a 1′3 2a 1′1a 1′2 β 11 2 2 2 β 11 a 2′1 a 2′2 a 2′3 2a 2′2a 2′3 2a 2′1a 2′3 2a 2′1a 2′2 β 11 2 2 2 β a 3′1 a 3′2 a 3′3 2a 3′2a 3′3 2a 3′1a 3′3 2a 3′1a 3′2 β 11 − 11 = 0 0 a 2′1a 3′1 a 2′2a 3′2 a 2′3a 3′3 a 2′2a 3′3 + a 2′3a 3′2 a 2′1a 3′3 + a 2′3a 3′1 a 2′1a 3′2 + a 2′2a 3′1 0

0 a 1′1a 3′1 a 1′2a 3′2 a 1′3a 3′3 a 1′2a 3′3 + a 1′3a 3′2 a 1′1a 3′3 + a 1′3a 3′1 a 1′1a 3′2 + a 1′2a 3′1 0

0 a 1′1a 2′1 a 1′2a 2′2 a 1′3a 2′3 a 1′2a 2′3 + a 1′3a 2′2 a 1′1a 2′3 + a 1′3a 2′1 a 1′1a 2′2 + a 1′2a 2′1 0

384 COMPLETELY ISOTROPIC MATERIALS (CONTINUED)

" By using direct substitution of the conditions ak′qap′q = δkp and

aq′kaq′p = δkp given in the previous table, it can be shown that the

invariance condition β − Tσ β = 0 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 − 1 given by S′ = Tε S Tσ and α′ = Tε α , respectively

" For the general transformations to be symmetry transformations, − 1 S′ = Tε S Tσ becomes Tε S − S Tσ = 0 and

α′ = Tε α becomes α − Tε α = 0

385 COMPLETELY ISOTROPIC MATERIALS (CONCLUDED)

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

Tε S − S Tσ = 0 , and using the conditions ak′qap′q = δkp and

aq′kaq′p = δkp given in the previous table, it can be shown that the invariance condition is satisﬁed identically

" Similarly, the invariance condition α − Tε α = 0 is also satisﬁed identically

" Therefore, an isotropic material has two independent compliances and one coefﬁcient of thermal expansion

386 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

387 ENGINEERING CONSTANTS FOR ELASTIC MATERIALS

388 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

S 11 S 12 S 13 S 14 S 15 S 16 ε11 σ 11 α 11 S 21 S 22 S 23 S 24 S 25 S 26 ε22 σ 22 α 22 ε S S S S S S σ α 33 = 31 32 33 34 35 36 33 + 33 T T σ − ref 2ε23 S 41 S 42 S 43 S 44 S 45 S 46 23 2α 23

2ε13 σ 13 2α 13 S 51 S 52 S 53 S 54 S 55 S 56 σ 12 2ε12 2α 12 S 61 S 62 S 63 S 64 S 65 S 66

389 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

normal stress σ jj " In general, E j ≡ normal strain ε jj caused by σ jj

390 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

shearing stress σ ij " In general, G ij ≡ shearing strain 2ε ij caused by σ ij

391 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 by

a normal stress applied in the xi - coordinate direction

normal strain ε jj caused by normal stress σ ii " In general, ν ij ≡ − normal strain ε ii caused by σ ii

392 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 xk - coordinate direction induced by

a shearing stress applied in the xi - xj coordinate plane

normal strain ε kk caused by shearing stress σ ij " That is, η k, ij ≡ shearing strain 2ε ij caused by shearing stress σ ij

393 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. Rabinovich (circa 1946)

" In particular, the symbol ηij,k is used herein to relate the shearing

strain in the xi - xj coordinate plane induced by the action of a

normal stress applied in the xk - coordinate direction

shearing strain 2ε ij caused by normal stress σ kk " That is, η ij, k = normal strain ε kk caused by σ kk

394 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 by a shearing stress

applied in the xk - xl coordinate plane

shearing 2ε ij caused by shearing stress σ kl " That is, µ ij, kl ≡ shearing strain 2ε kl caused by shearing stress σ kl

" Note that µij, kl = µji, kl = µji, lk = µij, lk because of symmetry of 2εij and σkl

395 CONSTITUTIVE EQUATIONS IN TERMS OF ENGINEERING CONSTANTS - CONTINUED

" Now, consider a parallelopiped of homogeneous material that is subjected to only a constant value of σ 11 σ11 and no thermal loading, like a stress state that might be exist in an experiment like a tensile test

" For this case, σ 11

S11 S12 S13 S14 S15 S16 ε11 σ 11

ε22 S21 S22 S23 S24 S25 S26 0

ε33 S31 S32 S33 S34 S35 S36 0 = 2ε23 S41 S42 S43 S44 S45 S46 0 2 ε13 S51 S52 S53 S54 S55 S56 0 2ε12 0 S61 S62 S63 S64 S65 S66

" This equation indicates that, in general, the parallelopiped will

extend in the x1 - coordinate direction, expand or contract in the x2 -

and x3 - coordinate directions, and shear in each face for this very simple state of uniaxial stress

396 CONSTITUTIVE EQUATIONS IN TERMS OF ENGINEERING CONSTANTS - CONTINUED

σ 11 " From the deﬁnition of the elastic moduli, it follows that E1 = ε11

" Noting that the previous matrix equation gives ε11 = S11σ 11 for this 1 simple state of stress, it also follows that S 11 = E 1

" Next, from the deﬁnition for the Poisson’s ratios, it follows that

ε 22 ε 33 ν 12 = − and ν 13 = − for this simple state of stress ε 11 ε 11

" Noting that the previous matrix equation also gives ε22 = S21σ 11 and

ε33 = S31σ 11 for this simple state of stress, and using ε11 = S11σ 11 , it

S 21 S 31 also follows that ν 12 = − and ν 13 = − S 11 S 11

397 CONSTITUTIVE EQUATIONS IN TERMS OF ENGINEERING CONSTANTS - CONTINUED

1 ν 12 ν 13 " Using S 11 = gives the results S 21 = − and S 31 = − E 1 E 1 E 1

" Next, from the deﬁnition for the coefﬁcients of interaction of the second

2ε23 2ε13 2ε12 kind, it follows that η 23, 1 = , η 13, 1 = , and η 12, 1 = for this ε11 ε11 ε11 simple state of stress

" Noting that the previous matrix equation also gives 2ε 23 = S41σ 11 ,

2ε 13 = S51σ 11 , and 2ε 12 = S61σ 11 for this simple state of stress, and

using ε11 = S11σ 11 , it also follows that

S41 S51 S61 η 23, 1 = , η 13, 1 = , and η 12, 1 = S11 S11 S11

398 CONSTITUTIVE EQUATIONS IN TERMS OF ENGINEERING CONSTANTS - CONTINUED

1 " Using S 11 = gives the results E 1

η 23, 1 η 13, 1 η 12, 1 S 41 = , S 51 = , and S 61 = E 1 E 1 E 1

" Now, consider a parallelopiped of homogeneous material that is subjected to only a constant value of

σ22 and no thermal loading σ 22 σ 22

" For this case,

S S S S S S ε11 11 12 13 14 15 16 0

ε22 S21 S22 S23 S24 S25 S26 σ 22

ε33 S31 S32 S33 S34 S35 S36 0 = 2ε23 S41 S42 S43 S44 S45 S46 0 2 ε13 S51 S52 S53 S54 S55 S56 0 2ε12 S61 S62 S63 S64 S65 S66 0

399 CONSTITUTIVE EQUATIONS IN TERMS OF ENGINEERING CONSTANTS - CONTINUED

" This equation indicates that, in general, the parallelopiped will

extend in the x2 - coordinate direction, expand or contract in the x1 -

and x3 - coordinate directions, and shear in each face for this very simple state of uniaxial stress

σ 22 " From the deﬁnition of the elastic moduli, it follows that E2 = ε22

" Noting that the previous matrix equation gives ε22 = S22σ 22 for this 1 simple state of stress, it also follows that S 22 = E 2

" Next, from the deﬁnition for the Poisson’s ratios, it follows that

ε 11 ε 33 ν 21 = − and ν 23 = − for this simple state of stress ε 22 ε 22

400 CONSTITUTIVE EQUATIONS IN TERMS OF ENGINEERING CONSTANTS - CONTINUED

" Noting that the previous matrix equation also gives ε11 = S12σ 22 and

ε33 = S32σ 22 for this simple state of stress, and using ε22 = S22σ 22 , it

S 12 S 32 also follows that ν 21 = − and ν 23 = − S 22 S 22

1 ν 21 ν 23 " Using S 22 = gives the results S 12 = − and S 32 = − E 2 E 2 E 2

" Next, from the deﬁnition for the coefﬁcients of interaction of the second

2ε23 2ε13 2ε12 kind, it follows that η 23, 2 = , η 13, 2 = , and η 12, 2 = for this ε22 ε22 ε22 simple state of stress

401 CONSTITUTIVE EQUATIONS IN TERMS OF ENGINEERING CONSTANTS - CONTINUED

" Noting that the previous matrix equation also gives 2ε 23 = S42σ 22 ,

2ε 13 = S52σ 22 , and 2ε 12 = S62σ 22 for this simple state of stress, and

using ε22 = S22σ 22 , it also follows that

S42 S52 S62 η 23, 2 = , η 13, 2 = , and η 12, 2 = S22 S22 S22

1 " Using S 22 = gives the results E 2

η 23, 2 η 13, 2 η 12, 2 S 42 = , S 52 = , and S 62 = E 2 E 2 E 2

402 CONSTITUTIVE EQUATIONS IN TERMS OF ENGINEERING CONSTANTS - CONTINUED

" Now, consider a parallelopiped of homogeneous material σ 33 that is subjected to only a constant value of σ33 and no thermal loading

" For this case, σ 33 S S S S S S ε11 11 12 13 14 15 16 0

ε22 S21 S22 S23 S24 S25 S26 0

ε33 S31 S32 S33 S34 S35 S36 σ 33 = 2ε23 S41 S42 S43 S44 S45 S46 0 2 ε13 S51 S52 S53 S54 S55 S56 0 2ε12 S61 S62 S63 S64 S65 S66 0

" This equation indicates that, in general, the parallelopiped will

extend in the x3 - coordinate direction, expand or contract in the x1 -

and x2 - coordinate directions, and shear in each face for this very simple state of uniaxial stress

403 CONSTITUTIVE EQUATIONS IN TERMS OF ENGINEERING CONSTANTS - CONTINUED

σ 33 " E3 = From the deﬁnition of the elastic moduli, it follows that ε33

" Noting that the previous matrix equation gives ε33 = S33σ 33 for this 1 simple state of stress, it also follows that S 33 = E 3

" Next, from the deﬁnition for the Poisson’s ratios, it follows that

ε 11 ε 22 ν 31 = − and ν 32 = − for this simple state of stress ε 33 ε 33

" Noting that the previous matrix equation also gives ε11 = S13σ 33

and ε22 = S23σ 33 for this simple state of stress, and using

S 13 S 23

ε33 = S33σ 33 , it also follows that ν 31 = − and ν 32 = − S 33 S 33

404 CONSTITUTIVE EQUATIONS IN TERMS OF ENGINEERING CONSTANTS - CONTINUED

1 ν 31 ν 32 " Using S 33 = gives the results S 13 = − and S 23 = − E 3 E 3 E 3

" Next, from the deﬁnition for the coefﬁcients of interaction of the second

2ε23 2ε13 2ε12 η 23, 3 = η 13, 3 = η 12, 3 = kind, it follows that ε33 , ε33 , and ε33 for this simple state of stress

" Noting that the previous matrix equation also gives 2ε 23 = S43σ 33 ,

2ε 13 = S53σ 33 , and 2ε 12 = S63σ 33 for this simple state of stress, and

using ε33 = S33σ 33 , it also follows that

S43 S53 S63 η 23, 3 = , η 13, 3 = , and η 12, 3 = S33 S33 S33

405 CONSTITUTIVE EQUATIONS IN TERMS OF ENGINEERING CONSTANTS - CONTINUED

1 " Using S 33 = gives the results E 3

η 23, 3 η 13, 3 η 12, 3 S 43 = , S 53 = , and S 63 = E 3 E 3 E 3

" Now, consider a parallelopiped of homogeneous

material that is subjected to only a constant value of σ 23 σ23 and no thermal loading

" For this case,

S S S S S S 0 ε11 11 12 13 14 15 16

ε22 S21 S22 S23 S24 S25 S26 0

ε33 S31 S32 S33 S34 S35 S36 0 = 2ε23 S41 S42 S43 S44 S45 S46 σ 23 2 ε13 S51 S52 S53 S54 S55 S56 0 2ε12 S61 S62 S63 S64 S65 S66 0

406 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

σ 23 " From the deﬁnition of the shear moduli, it follows that G23 = 2ε23

" Noting that the previous matrix equation gives 2ε 23 = S44σ 23 for this 1 simple state of stress, it also follows that S 44 = G 23

" Next, from the definition for the coefficients of interaction of the first

ε11 ε22 ε33 kind, it follows that η 1, 23 = , η 2, 23 = , and η 3, 23 = for this 2ε23 2ε23 2ε23 simple state of stress

407 CONSTITUTIVE EQUATIONS IN TERMS OF ENGINEERING CONSTANTS - CONTINUED

" Noting that the previous matrix equation also gives ε11 = S14σ 23 ,

ε22 = S24σ 23 , and ε33 = S34σ 23 f or this simple state of stress, and

using 2ε 23 = S44σ 23 , it also follows that

S14 S24 S34 η 1, 23 = , η 2, 23 = , and η 3, 23 = S44 S44 S44

1 " Using S 44 = gives the results G 23

η 1, 23 η 2, 23 η 3, 23 S 14 = , S 24 = , and S 34 = G 23 G 23 G 23

408 CONSTITUTIVE EQUATIONS IN TERMS OF ENGINEERING CONSTANTS - CONTINUED

" Next, from the deﬁnition for the Chentsov’s coefﬁcients, it follows that

2ε13 2ε12 µ13, 23 = and µ12, 23 = for this simple state of stress 2ε23 2ε23

" Noting that the previous matrix equation also gives 2ε 13 = S54σ 23

and 2ε 12 = S64σ 23 for this simple state of stress, and using

S54 S64 2ε 23 = S44σ 23 , it also follows that µ13, 23 = and µ12, 23 = S44 S44

1 " Using S 44 = gives the results G 23

µ 13, 23 µ 12, 23 S 54 = and S 64 = G 23 G 23

409 CONSTITUTIVE EQUATIONS IN TERMS OF ENGINEERING CONSTANTS - CONTINUED

" Now, consider a parallelopiped of homogeneous σ 13 material that is subjected to only a constant value of

σ13 and no thermal loading

" For this case,

S S S S S S ε11 11 12 13 14 15 16 0

ε22 S21 S22 S23 S24 S25 S26 0

ε33 S31 S32 S33 S34 S35 S36 0 = 2ε23 S41 S42 S43 S44 S45 S46 0 2 ε13 S51 S52 S53 S54 S55 S56 σ 13 2ε12 S61 S62 S63 S64 S65 S66 0

410 CONSTITUTIVE EQUATIONS IN TERMS OF ENGINEERING CONSTANTS - CONTINUED

σ 13 " From the deﬁnition of the shear moduli, it follows that G13 = 2ε13

" Noting that the previous matrix equation gives 2ε 13 = S55σ 13 for this 1 simple state of stress, it also follows that S 55 = G 13

" Next, from the definition for the coefficients of interaction of the first

ε11 ε22 ε33 kind, it follows that η 1, 13 = , η 2, 13 = , and η 3, 13 = for this 2ε13 2ε13 2ε13 simple state of stress

411 CONSTITUTIVE EQUATIONS IN TERMS OF ENGINEERING CONSTANTS - CONTINUED

" Noting that the previous matrix equation also gives ε11 = S15σ 13 ,

ε22 = S25σ 13 , and ε33 = S35σ 13 for this simple state of stress, and

using 2ε 13 = S 55σ 13 , it also follows that

S15 S25 S35 η 1, 13 = , η 2, 13 = , and η 3, 13 = S55 S55 S55

1 " Using S 55 = gives the results G 13

η 1, 13 η 2, 13 η 3, 13 S 15 = , S 25 = , and S 35 = G 13 G 13 G 13

412 CONSTITUTIVE EQUATIONS IN TERMS OF ENGINEERING CONSTANTS - CONTINUED

" Next, from the deﬁnition for the Chentsov’s coefﬁcients, it follows that

2ε23 2ε12 µ23, 13 = and µ12, 13 = for this simple state of stress 2ε13 2ε13

" Noting that the previous matrix equation also gives 2ε 23 = S45σ 13

and 2ε 12 = S65σ 13 for this simple state of stress, and using

S45 S65 2ε 13 = S 55σ 13 , it also follows that µ23, 13 = and µ12, 13 = S55 S55

1 " Using S 55 = gives the results G 13

µ 23, 13 µ 12, 13 S 45 = and S 65 = G 13 G 13

413 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,

σ 12 S S S S S S ε11 11 12 13 14 15 16 0

ε22 S21 S22 S23 S24 S25 S26 0

ε33 S31 S32 S33 S34 S35 S36 0 = 2ε23 S41 S42 S43 S44 S45 S46 0 2 ε13 S51 S52 S53 S54 S55 S56 0 2ε12 S61 S62 S63 S64 S65 S66 σ 12

414 CONSTITUTIVE EQUATIONS IN TERMS OF ENGINEERING CONSTANTS - CONTINUED

σ 12 " From the deﬁnition of the shear moduli, it follows that G12 = 2ε12

" Noting that the previous matrix equation gives 2ε 12 = S66σ 12 for this 1 simple state of stress, it also follows that S 66 = G 12

" Next, from the definition for the coefficients of interaction of the first

ε11 ε22 ε33 kind, it follows that η 1, 12 = , η 2, 12 = , and η 3, 12 = for this 2ε12 2ε12 2ε12 simple state of stress

415 CONSTITUTIVE EQUATIONS IN TERMS OF ENGINEERING CONSTANTS - CONTINUED

" Noting that the previous matrix equation also gives ε11 = S16σ 12 ,

ε22 = S26σ 12 , and ε33 = S36σ 12 for this simple state of stress, and

using 2ε 12 = S 66σ 12 , it also follows that

S16 S26 S36 η 1, 12 = , η 2, 12 = , and η 3, 12 = S66 S66 S66

1 " Using S 66 = gives the results G 12

η 1, 12 η 2, 12 η 3, 12 S 16 = , S 26 = , and S 36 = G 12 G 12 G 12

416 CONSTITUTIVE EQUATIONS IN TERMS OF ENGINEERING CONSTANTS - CONTINUED

" Next, from the deﬁnition for the Chentsov’s coefﬁcients, it follows that

2ε23 2ε13 µ23, 12 = and µ13, 12 = for this simple state of stress 2ε12 2ε12

" Noting that the previous matrix equation also gives 2ε 23 = S46σ 12

and 2ε 13 = S56σ 12 for this simple state of stress, and using

S46 S56 2ε 12 = S 66σ 12 , it also follows that µ23, 12 = and µ13, 12 = S66 S66

1 " Using S 66 = gives the results G 12

µ 23, 12 µ 13, 12 S 46 = and S 56 = G 12 G 12

417 CONSTITUTIVE EQUATIONS IN TERMS OF ENGINEERING CONSTANTS - CONTINUED

" Using all of the derived expressions for Sij, the constitutive equation

ε 11 S 11 S 12 S 13 S 14 S 15 S 16 α 11 σ 11

ε 22 S 21 S 22 S 23 S 24 S 25 S 26 σ 22 α 22

ε 33 S 31 S 32 S 33 S 34 S 35 S 36 σ 33 α 33 = + T − Tref becomes 2ε 23 S 41 S 42 S 43 S 44 S 45 S 46 σ 23 2α 23 σ 13 2ε 13 S 51 S 52 S 53 S 54 S 55 S 56 2α 13 σ 12 2ε 12 S 61 S 62 S 63 S 64 S 65 S 66 2α 12

ν21 ν31 η 1, 23 η 1, 13 η 1,12 1 − − E1 E2 E3 G23 G13 G12

ν12 ν32 η 2, 23 η 2, 13 η 2,12 − 1 − ε11 E1 E2 E3 G23 G13 G12 σ 11 α11

ε22 ν13 ν23 η 3, 23 η 3, 13 η 3,12 α22 − − 1 σ 22 ε33 E1 E2 E3 G23 G13 G12 σ 33 α33 = + T − Tref 2ε23 η 23, 1 η 23, 2 η 23, 3 1 µ 23, 13 µ 23, 12 σ 23 2α23

2ε13 E1 E2 E3 G23 G13 G12 σ 13 2α13

η 13, 1 η 13, 2 η 13, 3 µ 13, 23 µ 13, 12 σ 12 2ε12 1 2α12

E1 E2 E3 G23 G13 G12

η 12, 1 η 12, 2 η 12, 3 µ 12, 23 µ 12, 13 1

E1 E2 E3 G23 G13 G12

418 CONSTITUTIVE EQUATIONS IN TERMS OF ENGINEERING CONSTANTS - CONTINUED

" Relationships between the engineering constants are obtained by enforcing symmetry of the matrix [S]; that is:

ν21 ν12 ν31 ν13 η 1, 23 η 23, 1 S12 = S21 →→ = S13 = S31 →→ = S14 = S41 →→ = E2 E1 E3 E1 G23 E1

η 1, 13 η 13, 1 η 1, 12 η 12, 1 ν32 ν23 S15 = S51 →→ = S16 = S61 →→ = S23 = S32 →→ = G13 E1 G12 E1 E3 E2

η 2, 23 η 23, 2 η 2, 13 η 13, 2 η 2, 12 η 12, 2 S24 = S42 →→ = S25 = S52 →→ = S26 = S62 →→ = G23 E2 G13 E2 G12 E2

η 3, 23 η 23, 3 η 3, 13 η 13, 3 η 3, 12 η 12, 3 S34 = S43 →→ = S35 = S53 →→ = S36 = S63 →→ = G23 E3 G13 E3 G12 E3

µ23, 13 µ13, 23 µ23, 12 µ12, 23 µ13, 12 µ12, 13 S45 = S54 →→ = S46 = S64 →→ = S56 = S65 →→ = G13 G23 G12 G23 G12 G13

419 CONSTITUTIVE EQUATIONS IN TERMS OF ENGINEERING CONSTANTS - CONCLUDED

" Using the previous symmetry conditions, the constitutive equations are expressed as

1 ε11 = σ 11 − ν12σ 22 − ν13σ 33 + η 23, 1σ 23 + η 13, 1σ 13 + η 12, 1σ 12 + α11 T − Tref E1

1 ε22 = − ν21σ 11 + σ 22 − ν23σ 33 + η 23, 2σ 23 + η 13, 2σ 13 + η 12, 2σ 12 + α22 T − Tref E2

1 ε33 = − ν31σ 11 − ν32σ 22 + σ 33 + η 23, 3σ 23 + η 13, 3σ 13 + η 12, 3σ 12 + α33 T − T E ref 3

2 1 ε23 = η 1, 23σ 11 + η 2, 23σ 22 + η 3, 23σ 33 + σ 23 + µ13, 23σ 13 + µ12, 23σ 12 + 2α23 T − Tref G23

2 1 ε13 = η 1, 13σ 11 + η 2, 13σ 22 + η 3, 13σ 33 + µ23, 13σ 23 + σ 13 + µ12, 13σ 12 + 2α13 T − Tref G13

2 1 ε12 = η 1, 12σ 11 + η 2, 12σ 22 + η 3, 12σ 33 + µ23, 12σ 23 + µ13, 12σ 13 + σ 12 + 2α12 T − Tref G12

420 ENGINEERING CONSTANTS OF A SPECIALLY ORTHOTROPIC MATERIAL

" The engineering constants for an anisotropic material have been

presented previously herein, for an x 1,x2,x3 coordinate frame

" The subset of engineering constants for a specially orthotropic material are given by

1 S 12 1 E1 = ν 12 = − G23 = S11 S 11 S44

1 S 13 1 E2 = ν 13 = − G13 = S22 S 11 S55

1 S 23 1 E3 = = G12 = S ν 23 − S 33 S 22 66

421 ENGINEERING CONSTANTS OF A SPECIALLY ORTHOTROPIC MATERIAL - CONTINUED

" Substituting these expressions into

S11 S12 S13 0 0 0

ε11 σ 11 α11 S12 S22 S23 0 0 0 ε22 σ 22 α22

ε33 S13 S23 S33 0 0 0 σ 33 α33 = + T − Tref gives σ 23 0 2ε23 0 0 0 S 0 0 44 0 2ε13 σ 13 0 0 0 0 S55 0 σ 12 0 2ε12

0 0 0 0 0 S66

ν21 ν31 1 − − 0 0 0 E1 E2 E3

ν12 ν32 − 1 − 0 0 0 E1 E2 E3 ε11 σ 11 α11 ν13 ν23 1 ε22 − − 0 0 0 σ 22 α22 ε E1 E2 E3 σ α 33 = 33 33 1 + T − Tref 2ε23 0 0 0 0 0 σ 23 0 G23 0 2ε13 σ 13 1 σ 12 0 2ε12 0 0 0 0 0 G13 0 0 0 0 0 1 G12

422 ENGINEERING CONSTANTS OF A SPECIALLY ORTHOTROPIC MATERIAL - CONTINUED

" Inverting these matrix equations yields

C 11 C 12 C 13 0 0 0 β 11 σ 11 ε11 C 12 C 22 C 23 0 0 0 σ 22 ε22 β 22

σ 33 C 13 C 23 C 33 0 0 0 ε33 β 33 = + T − Tref where σ 23 0 0 0 C 44 0 0 2ε23 0 σ 13 2 0 0 0 0 0 C 55 0 ε13 σ 12 2ε12 0 0 0 0 0 0 C 66

E 1 E 1 E 2 C 11 = 1 − ν 23 ν 32 C 12 = ν 21 + ν 23 ν 31 = ν 12 + ν 13 ν 32 Δ Δ Δ

E 1 E 3 E 2 C 13 = ν 31 + ν 21 ν 32 = ν 13 + ν 12 ν 23 C = 1 − ν ν Δ Δ 22 Δ 13 31 E E E C = 2 ν + ν ν = 3 ν + ν ν C = 3 1 − ν ν 23 Δ 32 12 31 Δ 23 13 21 33 Δ 12 21

C 44 = G 23 C 55 = G 13 C 66 = G 12

Δ = 1 − ν 12 ν 21 − ν 23 ν 32 − ν 13 ν 31 − 2ν 21 ν 32 ν 13

423 ENGINEERING CONSTANTS OF A SPECIALLY ORTHOTROPIC MATERIAL - CONCLUDED

" And, where the thermal moduli are given by

α 11 1 − ν 23 ν 32 + α 22 ν 21 + ν 23 ν 31 + α 33 ν 31 + ν 21 ν 32 β 11 = − E 1 1 − ν 12 ν 21 − ν 23 ν 32 − ν 13 ν 31 − 2ν 21 ν 32 ν 13

α 11 ν 12 + ν 13 ν 32 + α 22 1 – ν 13 ν 31 + α 33 ν 32 + ν 12 ν 31 β 22 = − E 2 1 − ν 12 ν 21 − ν 23 ν 32 − ν 13 ν 31 − 2ν 21 ν 32 ν 13

α 11 ν 13 + ν 12 ν 23 + α 22 ν 23 + ν 13 ν 21 + α 33 1 − ν 12 ν 21 β 33 = − E 3 1 − ν 12 ν 21 − ν 23 ν 32 − ν 13 ν 31 − 2ν 21 ν 32 ν 13

424 ENGINEERING CONSTANTS OF A TRANSVERSELY ISOTROPIC MATERIAL

" Consider the constitutive equations for a specially orthotropic material

S 11 S 12 S 13 0 0 0 ε 11 σ 11 α 11 S 12 S 22 S 23 0 0 0 ε 22 σ 22 α 22

ε 33 S 13 S 23 S 33 0 0 0 σ 33 α 33 = + T − Tref and 2ε 23 0 0 0 S 44 0 0 σ 23 0 σ 13 0 2ε 13 0 0 0 0 S 55 0 σ 12 0 2ε 12 0 0 0 0 0 S 66

ν21 ν31 1 − − 0 0 0 E1 E2 E3

425 ENGINEERING CONSTANTS OF A TRANSVERSELY ISOTROPIC MATERIAL - CONTINUED

" For a transversely isotropic material with the plane of isotropy given by

x3 = 0, the constitutitve equations have the following forms

S11 S12 S13 0 0 0 ε11 σ 11 α 11 S12 S11 S13 0 0 0 ε22 σ 22 α 11

ε33 S13 S13 S33 0 0 0 σ 33 α 33 = + T − Tref 2ε23 0 0 0 S44 0 0 σ 23 0 σ 13 0 2ε13 0 0 0 0 S44 0 σ 12 0 2ε12 0 0 0 0 0 2 S 11 − S12

426 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

ν31 1 − ν − 0 0 0 E E E′ ν 1 ν31 ε − − 0 0 0 11 E E E′ σ 11 α ε22 ν13 ν13 σ − − 1 0 0 0 22 α ε E E E′ σ 33 = 33 α′ 1 + T − Tref 2ε23 0 0 0 0 0 σ 23 0 G′ 0 2ε σ 13 13 1 0 0 0 0 0 0 σ 12 2ε12 G′ 0 0 0 0 0 1 G

E where E = E ≡ E , ν = ν ≡ ν , G ≡ G = , α = α ≡ α , 1 2 12 21 12 2 1 + ν 11 22

E 3 ≡ E′ , G 13 = G23 ≡ G′ , and α33 ≡ α′

427 ENGINEERING CONSTANTS OF A TRANSVERSELY ISOTROPIC MATERIAL - CONTINUED

" In this matrix equation; E, ν, G, and α 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 ν12 = ν21 ≡ ν characterizes contractions in the plane of isotropy that result from only tensile stresses applied in that plane

" E′ , G′, and α′ 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 σ 33 applied perpendicular to the plane of

ε 11 ε 22 isotropy, the contractions are characterized by = = − ν 31 ≡ − ν′ ε 33 ε 33

" = = E Symmetry of the compliance matrix yields ν 13 ν 23 ν′ E′

428 ENGINEERING CONSTANTS OF A TRANSVERSELY ISOTROPIC MATERIAL - CONTINUED

" Finally, the constitutive equations for a transversely isotropic material,

with the plane of isotropy given by x3 = 0, are expressed by

1 − ν − ν′ 0 0 0 E E E′ ε − ν 1 − ν′ 0 0 0 11 E E E′ σ 11 α ε22 σ − ν′ − ν′ 1 0 0 0 22 α ε E′ E′ E′ σ 33 = 33 α′ 1 + T − Tref 2ε23 0 0 0 0 0 σ 23 0 G′ 0 2ε σ 13 13 1 0 0 0 0 0 0 σ 12 2ε12 G′ 0 0 0 0 0 1 G

" The ﬁve independent elastic constants are E, ν, E′, ν′, and G′

" The two independent thermal parameters are α and α′

429 ENGINEERING CONSTANTS OF A TRANSVERSELY ISOTROPIC MATERIAL - CONCLUDED

" The inverted form of the previous matrix constitutive equation is given by

C11 C12 C13 0 0 0 β 11 σ11 ε11 C12 C11 C13 0 0 0 σ22 ε22 β 11

σ33 C13 C13 C33 0 0 0 ε33 β 33 = + T − Tref where σ23 0 0 0 G′ 0 0 2ε23 0 σ 13 0 0 0 0 G′ 0 2ε13 0 σ12 0 0 0 0 0 G 2ε12 0

E 2 E 2 1 − ν′ ν + ν′ ν′E C = E E′ C = E E′ C = 11 2 12 2 13 E 2 1 + ν 1 − ν − 2 E ν′ 1 + ν 1 − ν − 2 E ν′ 1 − ν − 2 ν′ E′ E′ E′

E 1 α′ Eα 2ν′ + ′α′ 1 − ν E′ − ν Eα 1 + α ν′ Eα C 33 = 2 β 11 = − 2 β 33 = − 2 1 − ν − 2 E ν′ 1 − ν − 2 E ν′ 1 − ν − 2 E ν′ E′ E′ E′

430 ENGINEERING CONSTANTS OF A GENERALLY ORTHOTROPIC MATERIAL

" Results presented herein indicate that, for an x 1,x2,x3 coordinate frame, the engineering constants of an anisotropic material are given by

S S S S E = 1 12 G = 1 14 15 16 1 ν 12 = − 23 η 23, 1 = η 13, 1 = η 12, 1 = S11 S 11 S44 S11 S11 S11

1 S 13 1 S24 S25 S26 E2 = ν 13 = − G13 = η 23, 2 = η 13, 2 = η 12, 2 = S22 S 11 S55 S22 S22 S22

1 S 23 1 S34 S35 S36 E3 = = G12 = η 23, 3 = η 13, 3 = η 12, 3 = S ν 23 − S 33 S 22 66 S33 S33 S33

431 ENGINEERING CONSTANTS OF A GENERALLY ORTHOTROPIC MATERIAL - CONTINUED and

S14 S24 S34 S45 S46 η 1, 23 = η 2, 23 = η 3, 23 = µ13, 23 = µ12, 23 = S44 S44 S44 S44 S44

S15 S25 S35 S45 S56 η 1, 13 = η 2, 13 = η 3, 13 = µ23, 13 = µ12, 13 = S55 S55 S55 S55 S55

S16 S26 S36 S46 S56 η 1, 12 = η 2, 12 = η 3, 12 = µ23, 12 = µ13, 12 = S66 S66 S66 S66 S66

432 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 x3 - axis m = cos θ 3 and n = sin θ 3 become

1 4 2 2 E 2 E 1 4 E 1 ν 13 2 2 ν 23 E 1 S 1′1′ = m + m n − 2ν 12 + n S 1′3′ = − m + n E 1 G 12 E 2 E 2 E 1 ν 13 E 2

1 4 2 2 E 2 E 2 4 E 2 ν 23 2 2 ν 13 E 2 S 2′2′ = m + m n − 2ν 12 + n S 2′3′ = − m + n E 2 G 12 E 1 E 1 E 2 ν 23 E 1

ν 12 4 4 2 2 1 E 1 E 2 E 1 1 S 1′2′ = − m + n − m n 1 + − S 3′3′ = E 1 ν 12 E 2 G 12 E 2 E 3

mn 2 2 E 2 E 1 2 E 1 2 S 1′6′ = m − n − 2ν 12 + 2 n − m E 1 G 12 E 2 E 2

433 ENGINEERING CONSTANTS OF A GENERALLY ORTHOTROPIC MATERIAL - CONTINUED

and

mn 2 2 E 2 E 2 2 2 E 2 S 2′6′ = n − m − 2ν 12 + 2 m − n E 2 G 12 E 1 E 1

2mn E 3 E 1 1 2 2 G 23 S 3′6′ = ν 13 − ν 23 S 4′4′ = m + n E 3 E 1 E 2 G 23 G 13

mn G 23 1 2 2 G 13 S 4′5′ = 1 − S 5′5′ = m + n G 23 G 13 G 13 G 23

2 1 2 2 G 12 E 2 E 1 2 2 S 6′6′ = 4m n 1 + 2ν 12 + + m − n G 12 E 2 E 1 E 2

434 ENGINEERING CONSTANTS OF A GENERALLY ORTHOTROPIC MATERIAL - CONTINUED

" The matrix form of the constitutive equations is given by

S 1′1′ S 1′2′ S 1′3′ 0 0 S 1′6′ ε 1′1′ σ 1′1′ α 1′1′ S S S 0 0 S ε 2′2′ 1′2′ 2′2′ 2′3′ 2′6′ σ 2′2′ α 2′2′

ε 3′3′ S 1′3′ S 2′3′ S 3′3′ 0 0 S 3′6′ σ 3′3′ α 3′3′ = + T – Tref 2ε 2′3′ 0 0 0 S 4′4′ S 4′5′ 0 σ 2′3′ 0

2ε 1 3 0 ′ ′ 0 0 0 S 4′5′ S 5′5′ 0 σ 1′3′ 2ε 1′2′ σ 1′2′ 2α 1′2′ S 1′6′ S 2′6′ S 3′6′ 0 0 S 6′6′

435 !

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 x1′,x2′,x3′ coordinate frame, the corresponding engineering constants are given by

S 1 1′2′ 1 S1′4′ S1′5′ S1′6′ E1′ = ν1′2′ = − G2′3′ = η 2′3′, 1′ = η 1′3′, 1′ = η 1′2′, 1′ = S1′1′ S1′1′ S4′4′ S1′1′ S1′1′ S1′1′

S 1 1′3′ 1 S2′4′ S2′5′ S2′6′ E2′ = ν1′3′ = − G1′3′ = η 2′3′, 2′ = η 1′3′, 2′ = η 1′2′, 2′ = S2′2′ S1′1′ S5′5′ S2′2′ S2′2′ S2′2′

S 1 2′3′ 1 S3′4′ S3′5′ S3′6′ E3′ = ν2′3′ = − G1′2′ = η 2′3′, 3′ = η 1′3′, 3′ = η 1′2′, 3′ = S3′3′ S2′2′ S6′6′ S3′3′ S3′3′ S3′3′

436 ENGINEERING CONSTANTS OF A GENERALLY ORTHOTROPIC MATERIAL - CONTINUED

and

S S S S 1′4′ S2′4′ 3′4′ 4′5′ 4′6′ η 1′, 2′3′ = η = η 3′, 2′3′ = µ1′3′, 2′3′ = µ1′2′, 2′3′ = S 2′, 2′3′ S S S 4′4′ S4′4′ 4′4′ 4′4′ 4′4′

S1′5′ S2′5′ S3′5′ S4′5′ S5′6′ η 1′, 1′3′ = η 2′, 1′3′ = η 3′, 1′3′ = µ2′3′, 1′3′ = µ1′2′, 1′3′ = S5′5′ S5′5′ S5′5′ S5′5′ S5′5′

S1′6′ S2′6′ S3′6′ S4′6′ S5′6′ η 1′, 1′2′ = η 2′, 1′2′ = η 3′, 1′2′ = µ2′3′, 1′2′ = µ1′3′, 1′2′ = S6′6′ S6′6′ S6′6′ S6′6′ S6′6′

" It is important to note that because the transformed compliance matrix has the same structure as a monoclinic material, the conditions

S1 ′4′ = S1′5′ = S2′4′ = S2′5′ = S3′4′ = S3′5′ = S4′6′ = S5′6′ = 0 are valid

437 ENGINEERING CONSTANTS OF A GENERALLY ORTHOTROPIC MATERIAL- CONTINUED

ε 1′1′ S 1′1′ S 1′2′ S 1′3′ 0 0 S 1′6′ σ 1′1′ α 1′1′

ε 2′2′ S 1′2′ S 2′2′ S 2′3′ 0 0 S 2′6′ σ 2′2′ α 2′2′

ε 3′3′ S 1′3′ S 2′3′ S 3′3′ 0 0 S 3′6′ σ 3′3′ α 3′3′ " Thus = + T − Tref becomes 2ε 2′3′ 0 0 0 S 4′4′ S 4′5′ 0 σ 2′3′ 0

2ε 1′3′ 0 0 0 S 4′5′ S 5′5′ 0 σ 1′3′ 0

2ε 1′2′ S 1′6′ S 2′6′ S 3′6′ 0 0 S 6′6′ σ 1′2′ 2α 1′2′

ν 2 1 ν 3 1 η 1 ,1 2 1 − ′ ′ − ′ ′ 0 0 ′ ′ ′ E 1′ E 2′ E 3′ G 1′2′

ν 1 2 ν 3 2 η 2 ,1 2 − ′ ′ 1 − ′ ′ 0 0 ′ ′ ′ E E E G ε 1′1′ 1′ 2′ 3′ 1′2′ σ 1′1′ α 1′1′ ν 1′3′ ν 2′3′ 1 η 3′,1′2′ ε 2′2′ − − 0 0 σ 2′2′ α 2′2′ E 1′ E 2′ E 3′ G 1′2′ ε 3′3′ σ 3′3′ α 3′3′ = + T − Tref 1 µ 2′3′, 1′3′ 2ε 2′3′ 0 0 0 0 σ 2′3′ 0 G 2′3′ G 1′3′ 2ε 1′3′ σ 1′3′ 0 µ 1′3′, 2′3′ 1 2ε 1′2′ 0 0 0 0 σ 1′2′ 2α 1′2′ G 2′3′ G 1′3′

η 1 2 , 1 η 1 2 , 2 η 1 2 , 3 ′ ′ ′ ′ ′ ′ ′ ′ ′ 0 0 1 E 1′ E 2′ E 3′ G 1′2′

438 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

− 1 E 1′ 4 2 2 E 2 E 1 4 E 1 = m + m n − 2ν 12 + n E 1 G 12 E 2 E 2

− 1 E 2′ 4 2 2 E 2 E 2 4 E 2 E 3′ = m + m n − 2ν 12 + n = 1 E 2 G 12 E 1 E 1 E 3

− 1 − 1 G 2 2 G 23 G 2 2 G 13 2′3′ = m + n 1′3′ = m + n G 23 G 13 G 13 G 23

− 1 2 2 G 1′2′ 2 G 12 E 2 E 1 2 2 = 4m n 1 + 2ν 12 + + m − n G 12 E 2 E 1 E 2

439 ENGINEERING CONSTANTS OF A GENERALLY ORTHOTROPIC MATERIAL - CONTINUED

" Similarly, the effective Poisson’s ratios of a generally orthotropic material are given by

4 4 2 2 1 E E E m + n − m n 1 + 1 − 2 1 ν ν 12 E 2 G 12 E 2 1′2′ = ν 12 4 2 2 E 2 E 1 4 E 1 m + m n − 2ν 12 + n G 12 E 2 E 2

2 2 ν E m + n 23 1 ν ν 13 E 2 1′3′ = ν 13 4 2 2 E 2 E 1 4 E 1 m + m n − 2ν 12 + n G 12 E 2 E 2

2 2 ν E m + n 13 2 ν ν 23 E 1 2′3′ = ν 23 4 2 2 E 2 E 2 4 E 2 m + m n − 2ν 12 + n G 12 E 1 E 1

440 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

2 2 E 2 E 1 2 E 1 2 m − n − 2ν 12 + 2 n − m G 12 E 2 G 12 E 2 E 2 η 1′, 1′2′ = mn 2 E 2 E 1 2 2 G 12 E 2 E 1 2 2 4m n 1 + 2ν 12 + + m − n E 2 E 1 E 2

2 2 E 2 E 2 2 2 E 2 n − m − 2ν 12 + 2 m − n G 12 G 12 E 1 E 1 η 2′, 1′2′ = mn 2 E 2 2 2 G 12 E 2 E 1 2 2 4m n 1 + 2ν 12 + + m − n E 2 E 1 E 2

E 1 2 ν 13 − ν 23 G 12 E 2 E 3 E 2 η 3′, 1′2′ = 2mn 2 E 2 E 1 E 1 2 2 G 12 E 2 E 1 2 2 4m n 1 + 2ν 12 + + m − n E 2 E 1 E 2

441 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

2 2 E 2 E 1 2 E 1 2 m − n − 2ν 12 + 2 n − m G 12 E 2 E 2 η 1′2′, 1′ = mn 4 2 2 E 2 E 1 4 E 1 m + m n − 2ν 12 + n G 12 E 2 E 2

2 2 E 2 E 2 2 2 E 2 n − m − 2ν 12 + 2 m − n G 12 E 1 E 1 η 1′2′, 2′ = mn 4 2 2 E 2 E 2 4 E 2 m + m n − 2ν 12 + n G 12 E 1 E 1

E 3 E 1 η 1′2′, 3′ = 2mn ν 13 − ν 23 E 1 E 2

442 ENGINEERING CONSTANTS OF A GENERALLY ORTHOTROPIC MATERIAL - CONCLUDED

" The nonzero effective Chentsov coefficients for a generally orthotropic material are given by

G G 1 − 23 1 − 23 G 13 G 13 G 13 µ 1′3′, 2′3′ = mn µ 2′3′, 1′3′ = mn 2 2 G G 2 2 G m + n 23 23 m + n 13 G 13 G 23

443 REDUCED CONSTITUTIVE EQUATIONS

444 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 x the stresses in a thin, ﬂat body, 3 that are normal to the plane of ﬂatness, are negligible compared to the other x stresses 2

" This simpliﬁcation is x commonly referred to as the 1 plane-stress assumption

" For a state of plane stress in a homogeneous, anisotropic solid, with

respect to the x1 - x2 plane, the stress ﬁeld is approximated such that

σ 33 = σ 23 = σ 13 = 0

445 CONSTITUTIVE EQUATIONS FOR PLANE STRESS CONTINUED

" For this special case, the general matrix constitutive equation

2ε13 σ 13 2α 13 S 15 S 25 S 35 S 45 S 55 S 56 σ 12 2ε12 2α 12 S 16 S 26 S 36 S 46 S 56 S 66 uncouples directly into

S S S ε 11 11 12 16 σ 11 α 11

ε 22 = S 12 S 22 S 26 σ 22 + α 22 T − Tref and

2 σ 12 2 ε 12 S 16 S 26 S 66 α 12

S S S ε 33 13 23 36 σ 11 α 33

2ε 23 = S 14 S 24 S 46 σ 22 + 2α 23 T − Tref

2 σ 12 2 ε 13 S 15 S 25 S 56 α 13

446 CONSTITUTIVE EQUATIONS FOR PLANE STRESS CONTINUED

" In terms of the engineering constants, the plane-stress constitutive equations for a homogeneous, anisotropic material are given by

ν 21 η 1,12 1 − E 1 E 2 G 12 ε 11 σ 11 α 11 ν 12 1 η 2,12 ε 22 = − σ 22 + α 22 T − Tref E 1 E 2 G 12 2ε 12 σ 12 2α 12 η 12,1 η 12,2 1

E 1 E 2 G 12

ν 13 ν 23 η 3,12 − − E 1 E 2 G 12 ε 33 σ 11 α 33 η 23, 1 η 23, 2 µ 23, 12 2ε 23 = σ 22 + 2α 23 T − Tref E 1 E 2 G 12 2ε 13 σ 12 2α 13 η 13, 1 η 13, 2 µ 13, 12

E 1 E 2 G 12

447 CONSTITUTIVE EQUATIONS FOR PLANE STRESS CONTINUED

" However, simpliﬁcation of the following general constitutive equation is not as easy

C 11 C 12 C 13 C 14 C 15 C 16 β σ 11 ε11 11 C 12 C 22 C 23 C 24 C 25 C 26 σ 22 ε22 β 22

σ C C C C C C ε β 33 33 = 13 23 33 34 35 36 33 + T T σ − ref 23 C 14 C 24 C 34 C 44 C 45 C 46 2ε23 β 23

σ 13 2ε13 β 13 C 15 C 25 C 35 C 45 C 55 C 56 σ 12 2ε12 β 12 C 16 C 26 C 36 C 46 C 56 C 66

" First, the equation given above is expressed as

C11 C12 C13 C14 C15 C16 σ 11 ε11 α 11 C12 C22 C23 C24 C25 C26 σ 22 ε22 α 22 σ C C C C C C ε α 33 = 13 23 33 34 35 36 33 33 T T σ − − ref 23 C14 C24 C34 C44 C45 C46 2ε23 2α 23

σ 13 2ε13 2α 13 C15 C25 C35 C45 C55 C56 σ 12 2ε12 2α 12 C16 C26 C36 C46 C56 C66

448 CONSTITUTIVE EQUATIONS FOR PLANE STRESS CONTINUED

" Then, using σ 33 = σ 23 = σ 13 = 0 gives

C11 C12 C13 C14 C15 C16 σ 11 ε11 α 11 C12 C22 C23 C24 C25 C26 σ 22 ε22 α 22

0 C13 C23 C33 C34 C35 C36 ε33 α 33 = − T − Tref 0 C14 C24 C34 C44 C45 C46 2ε23 2α 23

0 2ε13 2α 13 C15 C25 C35 C45 C55 C56 σ 12 2ε12 2α 12 C16 C26 C36 C46 C56 C66

" Rearranging the rows and columns into a convenient form gives

C 11 C 12 C 16 C 13 C 14 C 15 σ ε11 α 11 11 C C C C C C σ 22 12 22 26 23 24 25 ε22 α 22

σ C C C C C C 2ε12 2α 12 12 = 16 26 66 36 46 56 T T 0 − − ref C 13 C 23 C 36 C 33 C 34 C 35 ε33 α 33 0 2ε23 2α 23 C 14 C 24 C 46 C 34 C 44 C 45 0 2ε13 2α 13 C 15 C 25 C 56 C 35 C 45 C 55

449 CONSTITUTIVE EQUATIONS FOR PLANE STRESS CONTINUED

" For convenience, let the mechanical strains be denoted by

σ ε11 ε11 α11 σ ε22 ε22 α22 σ 2 2 2 ε12 ε12 α12 σ = − T − Tref ε33 ε33 α33 σ 2 2 2ε23 ε23 α23 σ 2ε13 2α13 2ε13

" The previous matrix constitutive equation becomes

σ C 11 C 12 C 16 C 13 C 14 C 15 ε 11 σ 11 C C C C C C εσ σ 22 12 22 26 23 24 25 22 σ C C C C C C 2ε σ 12 = 16 26 66 36 46 56 12 0 σ C 13 C 23 C 36 C 33 C 34 C 35 ε 33 0 σ C 14 C 24 C 46 C 34 C 44 C 45 2ε 23 0 σ 2 C 15 C 25 C 56 C 35 C 45 C 55 ε 13

450 CONSTITUTIVE EQUATIONS FOR PLANE STRESS CONTINUED

" Next, the matrix constitutive equation

σ C 11 C 12 C 16 C 13 C 14 C 15 ε 11 σ 11 C C C C C C εσ σ 22 12 22 26 23 24 25 22 σ C C C C C C 2ε σ 12 = 16 26 66 36 46 56 12 0 σ is separated to get C 13 C 23 C 36 C 33 C 34 C 35 ε 33 0 σ C 14 C 24 C 46 C 34 C 44 C 45 2ε 23 0 σ 2 C 15 C 25 C 56 C 35 C 45 C 55 ε 13

C C C σ C C C σ σ 11 11 12 16 ε 11 13 14 15 ε 33 σ σ σ 22 = C 12 C 22 C 26 ε 22 + C 23 C 24 C 25 2ε 23 and σ σ σ 12 C 16 C 26 C 66 2ε 12 C 36 C 46 C 56 2ε 13

σ σ 0 C 13 C 23 C 36 ε 11 C 33 C 34 C 35 ε 33 σ σ 0 = C 14 C 24 C 46 ε 22 + C 34 C 44 C 45 2ε 23 0 σ σ C 15 C 25 C 56 2ε 12 C 35 C 45 C 55 2ε 13

451 CONSTITUTIVE EQUATIONS FOR PLANE STRESS CONTINUED

σ σ σ " Solving the previous homogeneous equation for ε 33 , 2ε 23 , and 2ε 13 gives

− 1 σ σ ε 33 C 33 C 34 C 35 C 13 C 23 C 36 ε 11 σ σ 2ε 23 = − C 34 C 44 C 45 C 14 C 24 C 46 ε 22 σ σ 2ε 13 C 35 C 45 C 55 C 15 C 25 C 56 2ε 12

σ σ σ " Back substitution of the column vector containing ε 33 , 2ε 23 , and 2ε 13 into

C C C σ C C C σ σ 11 11 12 16 ε 11 13 14 15 ε 33 σ σ σ 22 = C 12 C 22 C 26 ε 22 + C 23 C 24 C 25 2ε 23 σ σ σ 12 C 16 C 26 C 66 2ε 12 C 36 C 46 C 56 2ε 13

yields the result

Q Q Q σ σ 11 11 12 16 ε 11 σ σ 22 = Q 12 Q 22 Q 26 ε 22 σ σ 12 Q 16 Q 26 Q 66 2ε 12

452 CONSTITUTIVE EQUATIONS FOR PLANE STRESS CONTINUED

" The matrix with the subscripted Q terms is given by

− 1 Q 11 Q 12 Q 16 C 11 C 12 C 16 C 13 C 14 C 15 C 33 C 34 C 35 C 13 C 23 C 36

Q 12 Q 22 Q 26 = C 12 C 22 C 26 − C 23 C 24 C 25 C 34 C 44 C 45 C 14 C 24 C 46

Q 16 Q 26 Q 66 C 16 C 26 C 66 C 36 C 46 C 56 C 35 C 45 C 55 C 15 C 25 C 56

" Next, expressing the mechanical strains in terms of the total strains and the strains caused by free thermal expansion results in

Q Q Q σ 11 11 12 16 ε 11 α 11

σ 22 = Q 12 Q 22 Q 26 ε 22 − α 22 T − Tref

σ 12 2 2 Q 16 Q 26 Q 66 ε 12 α 12

" This equation is manipulated further by deﬁning

β 11 Q 11 Q 12 Q 16 α 11

β 22 = − Q 12 Q 22 Q 26 α 22 2 β 12 Q 16 Q 26 Q 66 α 12

453 CONSTITUTIVE EQUATIONS FOR PLANE STRESS CONTINUED

" Thus, the constitutive equations for plane stress, in terms of stiffness coefﬁcients and thermal moduli, become

Q Q Q β 11 σ 11 11 12 16 ε 11 σ 22 = Q 12 Q 22 Q 26 ε 22 + β 22 T − Tref where σ 12 2 Q 16 Q 26 Q 66 ε 12 β 12

− 1 Q 11 Q 12 Q 16 C 11 C 12 C 16 C 13 C 14 C 15 C 33 C 34 C 35 C 13 C 23 C 36

Q 12 Q 22 Q 26 = C 12 C 22 C 26 − C 23 C 24 C 25 C 34 C 44 C 45 C 14 C 24 C 46

Q 16 Q 26 Q 66 C 16 C 26 C 66 C 36 C 46 C 56 C 35 C 45 C 55 C 15 C 25 C 56

β 11 Q 11 Q 12 Q 16 α 11

and β 22 = − Q 12 Q 22 Q 26 α 22 2 β 12 Q 16 Q 26 Q 66 α 12

" The Q ij and β ij are called the reduced stiffness coefficients and reduced thermal moduli, respectively

454 CONSTITUTIVE EQUATIONS FOR PLANE STRESS CONTINUED

" The relationship between the reduced stiffnesses and the compliances is obtained by ﬁrst considering the previously derived equation

S S S ε 11 11 12 16 σ 11 α 11

ε 22 = S 12 S 22 S 26 σ 22 + α 22 T − Tref

2 σ 12 2 ε 12 S 16 S 26 S 66 α 12

" Inverting gives

− 1 S S S σ 11 11 12 16 ε 11 α 11

σ 22 = S 12 S 22 S 26 ε 22 − α 22 T − Tref σ 12 2 2 S 16 S 26 S 66 ε 12 α 12

455 CONSTITUTIVE EQUATIONS FOR PLANE STRESS CONTINUED

" Comparing

− 1 S S S σ 11 11 12 16 ε 11 α 11

σ 22 = S 12 S 22 S 26 ε 22 − α 22 T − Tref with σ 12 2 2 S 16 S 26 S 66 ε 12 α 12

Q Q Q β 11 σ 11 11 12 16 ε 11

σ 22 = Q 12 Q 22 Q 26 ε 22 + β 22 T − Tref indicates that

σ 12 2 Q 16 Q 26 Q 66 ε 12 β 12

− 1 Q 11 Q 12 Q 16 S 11 S 12 S 16

Q 12 Q 22 Q 26 = S 12 S 22 S 26

Q 16 Q 26 Q 66 S 16 S 26 S 66

456 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

C11 C12 C13 C14 C15 C16 β 11 α 11 C C C C C C β 22 12 22 23 24 25 26 α 22

β 33 C C C C C C α = − 13 23 33 34 35 36 33 β 23 C14 C24 C34 C44 C45 C46 2α 23

β 13 2α 13 C15 C25 C35 C45 C55 C56 β 12 2α 12 C16 C26 C36 C46 C56 C66

" This matrix equation can be separated into

β 11 C 11 C 12 C 16 α 11 C 13 C 14 C 15 α 33

β 22 = − C 12 C 22 C 26 α 22 − C 23 C 24 C 25 2α 23 and 2 2 β 12 C 16 C 26 C 66 α 12 C 36 C 46 C 56 α 13

β 33 C 13 C 23 C 36 α 11 C 33 C 34 C 35 α 33

β 23 = − C 14 C 24 C 46 α 22 − C 34 C 44 C 45 2α 23 2 2 β 13 C 15 C 25 C 56 α 12 C 35 C 45 C 55 α 13

457 CONSTITUTIVE EQUATIONS FOR PLANE STRESS CONTINUED

" Solving the matrix equation

β 33 C 13 C 23 C 36 α 11 C 33 C 34 C 35 α 33

β 23 = − C 14 C 24 C 46 α 22 − C 34 C 44 C 45 2α 23 2 2 β 13 C 15 C 25 C 56 α 12 C 35 C 45 C 55 α 13

for the column vector containing 2α 13, 2α 23, and α 33 gives

− 1 − 1

α 33 C 33 C 34 C 35 β 33 C 33 C 34 C 35 C 13 C 23 C 36 α 11

2α 23 = − C 34 C 44 C 45 β 23 − C 34 C 44 C 45 C 14 C 24 C 46 α 22 2 2 α 13 C 35 C 45 C 55 β 13 C 35 C 45 C 55 C 15 C 25 C 56 α 12

458 CONSTITUTIVE EQUATIONS FOR PLANE STRESS CONTINUED

" Substituting

− 1 − 1

α 33 C 33 C 34 C 35 β 33 C 33 C 34 C 35 C 13 C 23 C 36 α 11

2α 23 = − C 34 C 44 C 45 β 23 − C 34 C 44 C 45 C 14 C 24 C 46 α 22 2 2 α 13 C 35 C 45 C 55 β 13 C 35 C 45 C 55 C 15 C 25 C 56 α 12

into

β 11 C 11 C 12 C 16 α 11 C 13 C 14 C 15 α 33

β 22 = − C 12 C 22 C 26 α 22 − C 23 C 24 C 25 2α 23 gives 2 2 β 12 C 16 C 26 C 66 α 12 C 36 C 46 C 56 α 13

− 1

β 11 C 13 C 14 C 15 C 33 C 34 C 35 β 33

β 22 − C 23 C 24 C 25 C 34 C 44 C 45 β 23 =

β 12 C 36 C 46 C 56 C 35 C 45 C 55 β 13 − 1

C 11 C 12 C 16 C 13 C 14 C 15 C 33 C 34 C 35 C 13 C 23 C 36 α 11

− C 12 C 22 C 26 − C 23 C 24 C 25 C 34 C 44 C 45 C 14 C 24 C 46 α 22 2 C 16 C 26 C 66 C 36 C 46 C 56 C 35 C 45 C 55 C 15 C 25 C 56 α 12

459 CONSTITUTIVE EQUATIONS FOR PLANE STRESS CONTINUED

" Noting that

− 1 Q 11 Q 12 Q 16 C 11 C 12 C 16 C 13 C 14 C 15 C 33 C 34 C 35 C 13 C 23 C 36

Q 12 Q 22 Q 26 = C 12 C 22 C 26 − C 23 C 24 C 25 C 34 C 44 C 45 C 14 C 24 C 46 ,

Q 16 Q 26 Q 66 C 16 C 26 C 66 C 36 C 46 C 56 C 35 C 45 C 55 C 15 C 25 C 56

− 1

β 11 C 13 C 14 C 15 C 33 C 34 C 35 β 33

β 22 − C 23 C 24 C 25 C 34 C 44 C 45 β 23 =

β 12 C 36 C 46 C 56 C 35 C 45 C 55 β 13 − 1

C 11 C 12 C 16 C 13 C 14 C 15 C 33 C 34 C 35 C 13 C 23 C 36 α 11

− C 12 C 22 C 26 − C 23 C 24 C 25 C 34 C 44 C 45 C 14 C 24 C 46 α 22 2 C 16 C 26 C 66 C 36 C 46 C 56 C 35 C 45 C 55 C 15 C 25 C 56 α 12

gives

− 1

β 11 C 13 C 14 C 15 C 33 C 34 C 35 β 33 Q 11 Q 12 Q 16 α 11

β 22 − C 23 C 24 C 25 C 34 C 44 C 45 β 23 = − Q 12 Q 22 Q 26 α 22 2 β 12 C 36 C 46 C 56 C 35 C 45 C 55 β 13 Q 16 Q 26 Q 66 α 12

460 CONSTITUTIVE EQUATIONS FOR PLANE STRESS CONTINUED

" Finally, noting that

β 11 Q 11 Q 12 Q 16 α 11

β 22 = − Q 12 Q 22 Q 26 α 22 , the expression 2 β 12 Q 16 Q 26 Q 66 α 12

− 1

β 11 C 13 C 14 C 15 C 33 C 34 C 35 β 33 Q 11 Q 12 Q 16 α 11

β 22 − C 23 C 24 C 25 C 34 C 44 C 45 β 23 = − Q 12 Q 22 Q 26 α 22 gives 2 β 12 C 36 C 46 C 56 C 35 C 45 C 55 β 13 Q 16 Q 26 Q 66 α 12

− 1 C C C C C C β 11 β 11 13 14 15 33 34 35 β 33

β 22 = β 22 − C 23 C 24 C 25 C 34 C 44 C 45 β 23 β β 12 β 13 12 C 36 C 46 C 56 C 35 C 45 C 55

461 CONSTITUTIVE EQUATIONS FOR PLANE STRESS CONTINUED

" For a material that is monoclinic, with x3 = 0 being a plane of reﬂective symmetry, the plane-stress constitutive equations become

S S S ε 11 11 12 16 σ 11 α 11

ε 22 = S 12 S 22 S 26 σ 22 + α 22 T − Tref and

2 σ 12 2 ε 12 S 16 S 26 S 66 α 12

S S S ε 33 13 23 36 σ 11 α 33

2ε 23 = 0 0 0 σ 22 + 0 T − Tref 2ε 13 0 0 0 σ 12 0

" The last matrix equation reduces to

ε 33 = S 13 σ 11 + S 23 σ 22 + S 36 σ 12 + α 33 T − Tref

462 CONSTITUTIVE EQUATIONS FOR PLANE STRESS CONTINUED

" In terms of engineering constants,

ν 21 η 1,12 1 − E 1 E 2 G 12 ε 11 σ 11 α 11 ν 12 1 η 2,12 ε 22 = − σ 22 + α 22 T − Tref E 1 E 2 G 12 2ε 12 σ 12 2α 12 η 12,1 η 12,2 1

E 1 E 2 G 12

ν 13 ν 23 η 12, 3 ε 33 = − σ 11 − σ 22 + σ 33 + α 33 T − Tref E 1 E 2 E 3

463 CONSTITUTIVE EQUATIONS FOR PLANE STRESS CONTINUED

" Similarly, for a material that is monoclinic, with x3 = 0 being a plane of reﬂective symmetry, the plane-stress constitutive matrix becomes

− 1 Q 11 Q 12 Q 16 C 11 C 12 C 16 C 13 0 0 C 33 0 0 C 13 C 23 C 36

Q 12 Q 22 Q 26 = C 12 C 22 C 26 − C 23 0 0 0 C 44 C 45 0 0 0

Q 16 Q 26 Q 66 C 16 C 26 C 66 C 36 0 0 0 C 45 C 55 0 0 0

which simpliﬁes to

C13C13 C13C23 C13C36 C11 − C12 − C16 − C33 C33 C33 Q11 Q12 Q16 C13C23 C23C23 C23C36 Q12 Q22 Q26 = C12 − C22 − C26 − C33 C33 C33 Q16 Q26 Q66 C13C36 C23C36 C36C36 C16 − C26 − C66 − C33 C33 C33

464 CONSTITUTIVE EQUATIONS FOR PLANE STRESS CONTINUED

" The remaining matrix equation for the strains, given by

− 1 σ σ ε 33 C 33 C 34 C 35 C 13 C 23 C 36 ε 11 σ σ 2ε 23 = − C 34 C 44 C 45 C 14 C 24 C 46 ε 22 becomes σ σ 2ε 13 C 35 C 45 C 55 C 15 C 25 C 56 2ε 12

− 1 σ σ ε 33 C 33 0 0 C 13 C 23 C 36 ε 11 σ σ 2ε 23 = − 0 C 44 C 45 0 0 0 ε 22 which reduces to σ σ 2ε 13 0 C 45 C 55 0 0 0 2ε 12

σ 1 σ σ σ σ σ ε 33 = − C 13ε 11 + C 23ε 22 + 2C 36ε 12 and 2ε 23 = 2ε 13 = 0 C 33

465 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

1 ν 21 η 1,12 − 1 − Q 11 Q 12 Q 16 S 11 S 12 S 16 S S S E 1 E 2 G 12 11 12 16 Q Q Q = S S S ν 12 1 η 2,12 and 12 22 26 12 22 26 S 12 S 22 S 26 = − E 1 E 2 G 12 Q 16 Q 26 Q 66 S 16 S 26 S 66 S 16 S 26 S 66 η 12,1 η 12,2 1

E 1 E 2 G 12

" Inverting the matrix of compliances yields

E 1 E 1 E 2 Q 11 = 1 − η 12,2η 2,12 Q 12 = ν 21 + η 12,2η 1,12 = ν 12 + η 12,1η 2,12 Δ Δ Δ

E 1 G 12 E 2 Q 16 = − η 1,12 + ν 21η 2,12 = − η 12,1 + ν 12η 12,2 Q = 1 − η η Δ Δ 22 Δ 12,1 1,12 E G G Q = − 2 η + ν η = − 12 η + ν η Q = 12 1 − ν ν 26 Δ 2,12 12 1,12 Δ 12,2 21 12,1 66 Δ 12 21

Δ = 1 − ν 12 ν 21 − η 12 ,1 η 1,12 + ν 21 η 2,12 − η 12 ,2 η 2,12 + ν 12 η 1,12

466 CONSTITUTIVE EQUATIONS FOR PLANE STRESS CONTINUED

β 11 Q 11 Q 12 Q 16 α 11

" Similarly, the expression β 22 = − Q 12 Q 22 Q 26 α 22 gives 2 β 12 Q 16 Q 26 Q 66 α 12

1 β 11 = − E 1α 11 1 − η 12,2η 2,12 + E 2α 22 ν 12 + η 12,1η 2,12 − 2G 12α 12 η 12,1 + ν 12η 12,2 Δ 1 β 22 = − E 1α 11 ν 21 + η 12,2η 1,12 + E 2α 22 1 − η 12,1η 1,12 − 2G 12α 12 η 12,2 + ν 21η 12,1 Δ 1 β 12 = E 1α 11 η 1,12 + ν 21η 2,12 + E 2α 22 η 2,12 + ν 12η 1,12 − 2G 12α 12 1 − ν 12ν 21 Δ

467 CONSTITUTIVE EQUATIONS FOR PLANE STRESS CONTINUED

" For a specially orthotropic material, the plane-stress constitutive equations become

S S 0 ε 11 11 12 σ 11 α 11

ε 22 = S 12 S 22 0 σ 22 + α 22 T − Tref and

2 σ 12 ε 12 0 0 S 66 0

ε 33 = S 13 σ 11 + S 23 σ 22 + α 33 T − Tref and 2ε 23 = 2ε 13 = 0

" In terms of engineering constants,

ν21 1 − 0 E1 E2 ε11 σ 11 α11 ν12 1 ε22 = − 0 σ 22 + α22 T − Tref E1 E2 2ε12 σ 12 0 0 0 1 G12

ν 13 ν 23 2ε 23 = 2ε 13 = 0 ε 33 = − σ 11 − σ 22 + α 33 T − Tref and E 1 E 2

468 CONSTITUTIVE EQUATIONS FOR PLANE STRESS CONTINUED

" Similarly, for a specially orthotropic material, the plane-stress constitutive equations become

C C 0 σ C 0 0 σ σ 11 11 12 ε 11 13 ε 33 σ σ 2 σ 22 = C 12 C 22 0 ε 22 + C 23 0 0 ε 23 and σ σ σ 12 2 2 0 0 C 66 ε 12 0 0 0 ε 13

C C 0 σ C 0 0 σ 0 13 23 ε 11 33 ε 33 σ σ 2 0 = 0 0 0 ε 22 + 0 C 44 0 ε 23 σ σ 0 2 2 0 0 0 ε 12 0 0 C 55 ε 13

" The last matrix equation gives

σ 1 σ σ ε 33 = − C 13ε 11 + C 23ε 22 and 2ε 23 = 2ε 13 = 0 C 33

469 CONSTITUTIVE EQUATIONS FOR PLANE STRESS CONTINUED

σ σ σ " Using the previous expressions for ε 33 , 2ε 23 , and 2ε 13 , the constitutive equations for plane stress, in terms of stiffness coefﬁcients, become

Q Q 0 σ Q Q 0 σ 11 11 12 ε 11 11 12 ε 11 α 11 σ σ 22 = Q 12 Q 22 0 ε 22 = Q 12 Q 22 0 ε 22 − α 22 T − Tref σ σ 12 2 2ε 12 0 0 Q 66 ε 12 0 0 Q 66 0

where

C13C13 C13C23 C11 − C12 − 0 C33 C33 Q11 Q12 0 C13C23 C23C23 Q12 Q22 0 = C12 − C22 − 0 C33 C33 0 0 Q66

0 0 C66

470 CONSTITUTIVE EQUATIONS FOR PLANE STRESS CONTINUED

" Next, noting that

β 11 Q 11 Q 12 0 α 11

β 22 = − Q 12 Q 22 0 α 22 for a specially orthotropic solid

β 12 0 0 Q 66 0

Q Q 0 β 11 σ 11 11 12 ε 11

gives σ 22 = Q 12 Q 22 0 ε 22 + β 22 T − Tref

σ 12 2ε 12 0 0 Q 66 0

" Now, note that the general expression for the regular thermal moduli simpliﬁes to

C11 C12 C13 0 0 0 β 11 α 11 C C C 0 0 0 β 22 12 22 23 α 22

β C13 C23 C33 0 0 0 α 33 = − 33 β 23 0 0 0 C 44 0 0 0 β 0 13 0 0 0 0 C 55 0 β 12 0 0 0 0 0 0 C 66

471 CONSTITUTIVE EQUATIONS FOR PLANE STRESS CONTINUED

" In addition,

− 1

β 11 β 11 C 13 C 14 C 15 C 33 C 34 C 35 β 33

β 22 = β 22 − C 23 C 24 C 25 C 34 C 44 C 45 β 23 simpliﬁes to

β 12 β 12 C 36 C 46 C 56 C 35 C 45 C 55 β 13

− 1

β 11 β 11 C 13 0 0 C 33 0 0 β 33

β 22 = β 22 − C 23 0 0 0 C 44 0 0 which yields 0 0 β 12 0 0 0 0 0 C 55

C 13 C 23 β 11 = β 11 − β 33 β 22 = β 22 − β 33 β 12 = 0 C 33 C 33

472 CONSTITUTIVE EQUATIONS FOR PLANE STRESS CONCLUDED

" In terms of the engineering constants, the reduced stiffness coefﬁcients and are given by

E 1 ν 12E 2 ν 21E 1 Q 11 = Q 12 = = 1 − ν 12ν 21 1 − ν 12ν 21 1 − ν 12ν 21

E 2 Q 22 = Q 66 = G 12 1 − ν 12ν 21

" Similarly, the nonzero reduced thermal moduli are given by

α 11 + α 22 ν 21 α 22 + α 11 ν 12 β 11 = − E 1 β 22 = − E 2 1 − ν 12ν 21 1 − ν 12ν 21

473 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

x3 ,x3′

i 3 ,i 3′ x2′

i 2′ θ3 x2 i 1 i 2

i 1′

θ3

x 1′ x1

474 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

2 2 cos θ3 sin θ3 0 0 0 2sinθ3cosθ3 2 2 σ1′1′ sin θ3 cos θ3 0 0 0 − 2sinθ3cosθ3 σ11 σ2′2′ σ22 0 0 1 0 0 0 σ3′3′ = σ33 σ2′3′ σ23 0 0 0 cosθ3 − sinθ3 0 σ1′3′ σ13 0 0 0 sinθ3 cosθ3 0 σ1′2′ σ12 2 2 − sinθ3cosθ3 sinθ3cosθ3 0 0 0 cos θ3 − sin θ3

" Substituting σ 33 = σ 13 = σ 23 = 0 into this equation yields

2 2 cos θ3 sin θ3 2sinθ3cosθ3

σ 1′1′ σ 11 0 2 2 σ 3′3′ σ 2′2′ = sin θ3 cos θ3 − 2sinθ3cosθ3 σ 22 and σ 2′3′ = 0

2 2 1 3 σ 1′2′ σ 12 σ ′ ′ 0 − sinθ3cosθ3 sinθ3cosθ3 cos θ3 − sin θ3

475 STRESS AND STRAIN TRANSFORMATION EQUATIONS FOR PLANE STRESS - CONTINUED

" Thus, the stress-transformation law for a state of plane stress and a

dextral rotation about the x3 axis is given by

2 2 cos θ3 sin θ3 2sinθ3cosθ3 σ 1′1′ σ 11 2 2 σ 2′2′ = sin θ3 cos θ3 − 2sinθ3cosθ3 σ 22 2 2 σ 1′2′ σ 12 − sinθ3cosθ3 sinθ3cosθ3 cos θ3 − sin θ3

" This law is expressed symbolically by Σ′′ = Tσ θ 3 Σ

σ 1′1′ σ 11 where Σ′′ ≡≡ σ 2′2′ , Σ ≡≡ σ 22 , and σ 1′2′ σ 12

2 2 cos θ3 sin θ3 2sinθ3cosθ3

2 2 sin θ cos θ − 2sinθ cosθ Tσ θ3 ≡≡ 3 3 3 3 2 2 − sinθ3cosθ3 sinθ3cosθ3 cos θ3 − sin θ3

476 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

2 2 cos θ3 sin θ3 0 0 0 − 2sinθ3cosθ3 2 2 σ11 sin θ3 cos θ3 0 0 0 2sinθ3cosθ3 σ1′1′ σ22 σ2′2′ 0 0 1 0 0 0 σ33 = σ3′3′ σ23 σ2′3′ 0 0 0 cosθ3 sinθ3 0 σ13 σ1′3′ 0 0 0 − sinθ3 cosθ3 0 σ12 σ1′2′ 2 2 sinθ3cosθ3 − sinθ3cosθ3 0 0 0 cos θ3 − sin θ3

− 1 " Following a similar process gives Σ = Tσ θ 3 Σ′′ where

2 2 cos θ sin θ − 2sinθ cosθ − 1 3 3 3 3 2 2 – sin θ cos θ 2sinθ cosθ Tσ θ3 = Tσ θ3 = 3 3 3 3 2 2 sinθ 3cosθ 3 − sinθ 3cosθ 3 cos θ 3 − sin θ 3

477 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

2 2 cos θ3 sin θ3 0 0 0 sinθ3cosθ3

2 2 ε1′1′ ε11 sin θ3 cos θ3 0 0 0 − sinθ3cosθ3 ε2′2′ ε22 0 0 1 0 0 0 ε33 ε3′3′ = 2ε2′3′ 0 0 0 cosθ3 − sinθ3 0 2ε23

2ε1′3′ 2ε13 0 0 0 sinθ3 cosθ3 0 2 12 2ε1′2′ 2 2 ε − 2sinθ3cosθ3 2sinθ3cosθ3 0 0 0 cos θ3 − sin θ3

" This matrix equation can be partitioned into the following parts

2 2 cos θ 3 sin θ 3 sinθ 3cosθ 3 ε1′1′ ε11 2 2 ε2′2′ = sin θ 3 cos θ 3 − sinθ 3cosθ 3 ε22 , 2 2 2ε1′2′ 2ε12 − 2sinθ 3cosθ 3 2sinθ 3cosθ 3 cos θ 3 − sin θ 3

cos − sin 2ε2′3′ θ 3 θ 3 2 = ε23 = 2 and ε 3′3′ ε 33 2ε1′3′ sinθ 3 cosθ 3 ε13

478 STRESS AND STRAIN TRANSFORMATION EQUATIONS FOR PLANE STRESS - CONTINUED

" The ﬁrst law is expressed symbolically by E′′ = Tε θ 3 E

ε 1′1′ ε 11 where E′′ ≡≡ ε 2′2′ , E ≡≡ ε 22 , and 2ε 1′2′ 2ε 12

2 2 cos θ3 sin θ3 sinθ3cosθ3

2 2 sin θ cos θ − sinθ cosθ Tε θ3 ≡≡ 3 3 3 3 2 2 − 2sinθ3cosθ3 2sinθ3cosθ3 cos θ3 − sin θ3

" A useful relationship that is easily veriﬁed is given by

T − 1 − Tε θ 3 = Tσ θ 3 = Tσ θ 3

479 STRESS AND STRAIN TRANSFORMATION EQUATIONS FOR PLANE STRESS - CONCLUDED

" Similarly, the matrix form of the inverse strain-transformation law was also given previously as

2 2 cos θ3 sin θ3 0 0 0 − sinθ3cosθ3

2 2 1 1 ε11 ε ′ ′ sin θ3 cos θ3 0 0 0 sinθ3cosθ3 ε22 ε2′2′ 0 0 1 0 0 0 ε33 ε3′3′ = 2 ε23 0 0 0 cosθ3 sinθ3 0 2ε2′3′ 2 ε13 2ε1′3′ 0 0 0 − sinθ3 cosθ3 0 2ε12 2 2 2ε1′2′ 2sinθ3cosθ3 − 2sinθ3cosθ3 0 0 0 cos θ3 − sin θ3

− 1 " From this matrix equation, it follows that E = Tε θ 3 E′′

− 1 T − where Tε θ 3 = Tε θ 3 = Tσ θ 3

480 TRANSFORMED CONSTITUTIVE EQUATIONS FOR PLANE STRESS

" The two matrix equations that resulted from the reduction for a state of plane stress are

S S S ε 11 11 12 16 σ 11 α 11

ε 22 = S 12 S 22 S 26 σ 22 + α 22 T − Tref and

2 σ 12 2 ε 12 S 16 S 26 S 66 α 12

Q Q Q β 11 σ 11 11 12 16 ε 11

σ 22 = Q 12 Q 22 Q 26 ε 22 + β 22 T − Tref , where

σ 12 2 Q 16 Q 26 Q 66 ε 12 β 12

− 1

β 11 Q 11 Q 12 Q 16 α 11 Q 11 Q 12 Q 16 S 11 S 12 S 16

β 22 = − Q 12 Q 22 Q 26 α 22 and Q 12 Q 22 Q 26 = S 12 S 22 S 26 2 β 12 Q 16 Q 26 Q 66 α 12 Q 16 Q 26 Q 66 S 16 S 26 S 66

" Also, σ 33 = σ 13 = σ 23 = 0 and generally ε33 ≠ ε13 ≠ ε23 ≠ 0

481 TRANSFORMED CONSTITUTIVE EQUATIONS FOR PLANE STRESS - CONTINUED

" In terms of another set of coordinates x1′, x2′, x3′ that correspond to a

dextral rotation about the x3 axis, the constitutive equations must also have the forms given as

ε 1′1′ S 1′1′ S 1′2′ S 1′6′ σ 1′1′ α 1′1′

ε 2′2′ = S 1′2′ S 2′2′ S 2′6′ σ 2′2′ + α 2′2′ T − Tref and

2ε 1′2′ S 1′6′ S 2′6′ S 6′6′ σ 1′2′ 2α 1′2′

β 1′1′ σ 1′1′ Q 1′1′ Q 1′2′ Q 1′6′ ε 1′1′

σ 2′2′ = Q 1′2′ Q 2′2′ Q 2′6′ ε 2′2′ + β 2′2′ T − Tref , where

σ 1′2′ Q Q Q 2 1′6′ 2′6′ 6′6′ ε 1′2′ β 1′2′

− 1 β 1′1′ Q 1′1′ Q 1′2′ Q 1′6′ α 1′1′ Q 1′1′ Q 1′2′ Q 1′6′ S 1′1′ S 1′2′ S 1′6′

β 2′2′ = − Q 1′2′ Q 2′2′ Q 2′6′ α 2′2′ and Q 1′2′ Q 2′2′ Q 2′6′ = S 1′2′ S 2′2′ S 2′6′ Q Q Q 2 Q Q Q S S S β 1′2′ 1′6′ 2′6′ 6′6′ α 1′2′ 1′6′ 2′6′ 6′6′ 1′6′ 2′6′ 6′6′

482 TRANSFORMED CONSTITUTIVE EQUATIONS FOR PLANE STRESS - CONTINUED

" For convenience, let

S 11 S 12 S 16 Q 11 Q 12 Q 16 S = S 12 S 22 S 26 Q = Q 12 Q 22 Q 26 S 16 S 26 S 66 Q 16 Q 26 Q 66

β 11 α 11 β αα = α 22 β = 22 2α 12 β 12

S 1′1′ S 1′2′ S 1′6′ Q 1′1′ Q 1′2′ Q 1′6′ S S S Q Q Q S′′ = 1′2′ 2′2′ 2′6′ Q′′ = 1′2′ 2′2′ 2′6′ S 1′6′ S 2′6′ S 6′6′ Q 1′6′ Q 2′6′ Q 6′6′

β 1′1′ α 1′1′ β αα′′ = α 2′2′ β′′ = 2′2′ 2 α 1′2′ β 1′2′

483 TRANSFORMED CONSTITUTIVE EQUATIONS FOR PLANE STRESS - CONTINUED

" With this notation, the two sets of thermoelastic constitutive equations are expressed in symbolic form by

E = S Σ + α Θ and Σ = Q E + β Θ

E′ = S′ Σ′ + α′ Θ and Σ′ = Q′ E′ + β′ Θ

where Θ ≡ T − Tref

" By using the matrix form of the stress and strain transformation equations for plane stress, Σ = Q E + β Θ becomes

–1 –1 Tσ Σ′ = Q Tε E′ + β Θ

484 TRANSFORMED CONSTITUTIVE EQUATIONS FOR PLANE STRESS - CONTINUED

" Premultiplying by Tσ gives

–1 Σ′ = Tσ Q Tε E′ + Tσ β Θ

" Comparing this equation with Σ′ = Q′ E′ + β′ Θ it follows that

–1 = Q′ = Tσ Q Tε and β′ Tσ β

–1 " Rearranging Q′ = Tσ Q Tε gives

–1 Q = Tσ Q′ Tε

485 TRANSFORMED CONSTITUTIVE EQUATIONS FOR PLANE STRESS - CONTINUED

" Next, by using the matrix form of the stress and strain transformation equations for plane stress, E = S Σ + α Θ becomes

–1 –1 Tε E′ = S Tσ Σ′ + α Θ

" Premultiplying by Tε gives

–1 E′ = Tε S Tσ Σ′ + Tε α Θ

" By comparing this equation with E′ = S′ σ′ + α′ Θ it follows that

–1 = S′ = Tε S Tσ and α′ Tε α

486 TRANSFORMED CONSTITUTIVE EQUATIONS FOR PLANE STRESS - CONTINUED

–1 –1 = " Rearranging S′ Tε S Tσ gives S = Tε S′ Tσ

" Noting that for a dextral rotation about the x3 axis,

–1 T –1 T Tε = Tσ and Tσ = Tε it follows that

T T

S′ = Tε S Tε S = Tσ S′ Tσ

T T Q′ = Tσ Q Tσ Q = Tε Q′ Tε

α′ = Tε α α = Tσ α′

β′ = Tσ β β = Tε β′

487 TRANSFORMED CONSTITUTIVE EQUATIONS FOR PLANE STRESS - CONTINUED with

2 2 cos θ3 sin θ3 2sinθ3cosθ3

2 2 sin θ cos θ − 2sinθ cosθ Tσ θ3 ≡≡ 3 3 3 3 and 2 2 − sinθ3cosθ3 sinθ3cosθ3 cos θ3 − sin θ3

2 2 cos θ3 sin θ3 sinθ3cosθ3

2 2 sin θ cos θ − sinθ cosθ Tε θ3 ≡≡ 3 3 3 3 2 2 − 2sinθ3cosθ3 2sinθ3cosθ3 cos θ3 − sin θ3

488 TRANSFORMED CONSTITUTIVE EQUATIONS FOR PLANE STRESS - CONTINUED

T = " Q ′ Tσ Q Tσ , with m = cosθ3 and n = sinθ3 , yields

4 2 2 2 2 4 Q 1′1′ = m Q 11 + 2m n Q 12 + 2Q 66 + 4mn m Q 16 + n Q 26 + n Q 22

2 2 2 2 4 4 Q 1′2′ = m n Q 11 + Q 22 − 4Q 66 − 2m n m − n Q 16 − Q 26 + m + n Q 12

2 2 2 3 Q1′6′ = m m − 3n Q16 − m n Q11 − Q12 − 2Q66 3 2 2 2 + mn Q22 − Q12 − 2Q66 − n n − 3m Q26

4 2 2 2 2 4 Q 2′2′ = m Q 22 + 2m n Q 12 + 2Q 66 − 4m n m Q 26 + n Q 16 + n Q 11

2 2 2 3 Q2′6′ = m m − 3n Q26 + m n Q22 − Q12 − 2Q66 3 2 2 2 − mn Q11 − Q12 − 2Q66 − n n − 3m Q16

2 2 2 2 2 2 2 Q 6′6′ = m n Q 11 + Q 22 − 2Q 12 − 2mn m − n Q 16 − Q 26 + m − n Q 66

489 TRANSFORMED CONSTITUTIVE EQUATIONS FOR PLANE STRESS - CONTINUED

T = " Q Tε Q ′ Tε , with m = cosθ3 and n = sinθ3 , yields

4 2 2 2 2 4 Q 11 = m Q 1′1′ + 2m n Q 1′2′ + 2Q 6′6′ − 4m n m Q 1′6′ + n Q 2′6′ + n Q 2′2′

2 2 2 2 4 4 Q 12 = m n Q 1′1′ + Q 2′2′ − 4Q 6′6′ + 2m n m − n Q 1′6′ − Q 2′6′ + m + n Q 1′2′

2 2 2 3 Q16 = m m − 3n Q1′6′ + m n Q1′1′ − Q1′2′ − 2Q6′6′ 3 2 2 2 − mn Q2′2′ − Q1′2′ − 2Q6′6′ − n n − 3m Q2′6′

4 2 2 2 2 4 Q 22 = m Q 2′2′ + 2m n Q 1′2′ + 2Q 6′6′ + 4m n m Q 2′6′ + n Q 1′6′ + n Q 1′1′

2 2 2 3 Q26 = m m − 3n Q2′6′ − m n Q2′2′ − Q1′2′ − 2Q6′6′ 3 2 2 2 + mn Q1′1′ − Q1′2′ − 2Q6′6′ − n n − 3m Q1′6′

2 2 2 2 2 2 2 Q 66 = m n Q 1′1′ + Q 2′2′ − 2Q 1′2′ + 2mn m − n Q 1′6′ − Q 2′6′ + m − n Q 6′6′

490 TRANSFORMED CONSTITUTIVE EQUATIONS FOR PLANE STRESS - CONTINUED

T = " S′ Tε S Tε , with m = cosθ3 and n = sinθ3 , yields

4 2 2 2 2 4 S 1′1′ = m S 11 + m n 2S 12 + S 66 + 2m n m S 16 + n S 26 + n S 22

2 2 2 2 4 4 S 1′2′ = m n S 11 + S 22 − S 66 − mn m − n S 16 − S 26 + m + n S 12

2 2 2 3 S1′6′ = m m − 3n S16 − m n 2S11 − 2S12 − S66 3 2 2 2 + mn 2S22 − 2S12 − S66 − n n − 3m S26

4 2 2 2 2 4 S 2′2′ = m S 22 + m n 2S 12 + S 66 − 2m n m S 26 + n S 16 + n S 11

2 2 2 3 S2′6′ = m m − 3n S26 + m n 2S22 − 2S12 − S66 3 2 2 2 − mn 2S11 − 2S12 − S66 − n n − 3m S16

2 2 2 2 2 2 2 S 6′6′ = 4m n S 11 + S 22 − 2S 12 − 4mn m − n S 16 − S 26 + m − n S 66

491 TRANSFORMED CONSTITUTIVE EQUATIONS FOR PLANE STRESS - CONTINUED

T = " S Tσ S′ Tσ , with m = cosθ3 and n = sinθ3 , yields

4 2 2 2 2 4 S 11 = m S 1′1′ + m n 2S 1′2′ + S 6′6′ − 2m n m S 1′6′ + n S 2′6′ + n S 2′2′

2 2 2 2 4 4 S 12 = m n S 1′1′ + S 2′2′ − S 6′6′ + m n m − n S 1′6′ − S 2′6′ + m + n S 1′2′

2 2 2 3 S16 = m m − 3n S1′6′ + m n 2S1′1′ − 2S1′2′ − S6′6′ 3 2 2 2 − mn 2S2′2′ − 2S1′2′ − S6′6′ − n n − 3m S2′6′

4 2 2 2 2 4 S 22 = m S 2′2′ + m n 2S 1′2′ + S 6′6′ + 2mn m S 2′6′ + n S 1′6′ + n S 1′1′

2 2 2 3 S26 = m m − 3n S2′6′ − m n 2S2′2′ − 2S1′2′ − S6′6′ 3 2 2 2 + mn 2S1′1′ − 2S1′2′ − S6′6′ − n n − 3m S1′6′

2 2 2 2 2 2 2 S 66 = 4m n S 1′1′ + S 2′2′ − 2S 1′2′ + 4mn m − n S 1′6′ − S 2′6′ + m − n S 6′6′

492 TRANSFORMED CONSTITUTIVE EQUATIONS FOR PLANE STRESS - CONTINUED

" Note that Q′ and Q can be expressed as

Q′ = Tσ θ 3 Q Tε − θ 3 and Q = Tσ − θ 3 Q′ Tε θ 3

" 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, S′ and S can be expressed as

S′ = T ε θ 3 S T σ − θ 3 and S = T ε − θ 3 S′ T σ θ 3

" 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

493 TRANSFORMED CONSTITUTIVE EQUATIONS FOR PLANE STRESS - CONTINUED

= " α′ Tε α , with m = cosθ3 and n = sinθ3 , yields

2 2 2 2 α 1′1′ = m α 11 + 2m nα 12 + n α 22 α 2′2′ = m α 22 − 2m nα 12 + n α 11

2 2 α 1′2′ = m − n α 12 + m n α 22 − α 11

T " Similarly, α = Tσ α′ gives

2 2 2 2 α 11 = m α 1′1′ + 2m nα 1′2′ + n α 2′2′ α 22 = m α 2′2′ − 2m nα 1′2′ + n α 1′1′

2 2 α 12 = m − n α 1′2′ + m n α 2′2′ − α 1′1′

494 TRANSFORMED CONSTITUTIVE EQUATIONS FOR PLANE STRESS - CONTINUED

= " β′ Tσ β , with m = cosθ3 and n = sinθ3 , yields

2 2 2 2 β 1′1′ = m β 11 + 2m nβ 12 + n β 22 β 2′2′ = m β 22 − 2m nβ 12 + n β 11

2 2 β 1′2′ = m − n β 12 + mn β 22 − β 11

T " Similarly, β = Tε β′ gives

2 2 2 2 β 11 = m β 1′1′ − 2m nβ 1′2′ + n β 2′2′ β 22 = m β 2′2′ + 2m nβ 1′2′ + n β 1′1′

2 2 β 12 = m − n β 1′2′ − mn β 2′2′ − β 1′1′

495 TRANSFORMED CONSTITUTIVE EQUATIONS FOR PLANE STRESS - CONCLUDED

" Note that α′ and α can be expressed as

α ′ = T ε θ 3 α and α = T ε − θ 3 α ′

" 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, β′ and β can be expressed as

β′ = T σ θ 3 β and β = T σ − θ 3 β′

" 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

496 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 x3 = 0 shown in the figure, significant simplifications can be made to the equations governing the elastic response x " Moreover, these dominant response 3 quantities are presumed to vary symmetrically through the plate h h thickness, given by − ≤ x ≤ + , x2 2 3 2 h such that no significant bending deformations are exhibited by the x plate 1

" In contrast to the plane stress approximations previously presented herein, when these conditions exist, with respect to the through-the- thickness variations, the plate response is described as a state of generalized plane stress

497 CONSTITUTIVE EQUATIONS FOR GENERALIZED PLANE STRESS - CONTINUED

" For a state of generalized plane stress in an anisotropic solid, with

respect to the plane x3 = 0, the displacement ﬁelds in the x1-, x2-, and x3- coordinate directions are approximated by averaging the through-the- thickness variations as follows

+ h + h 1 2 1 2 U 1 x 1, x 2 = u 1 x 1, x 2, x 3 dx 3 U 2 x 1, x 2 = u 2 x 1, x 2, x 3 dx 3 h – h h – h 2 2

+ h 1 2 and U 3 x 1, x 2 = u 3 x 1, x 2, x 3 dx 3 = 0 h – h 2

" To exclude bending deformations, U 3 x 1, x 2 = 0 is required for the out-of-plane displacement ﬁeld

" Note that allowing u 3 x 1, x 2, x 3 = x 3 ε , where ε is a constant

satisﬁes U 3 x 1, x 2 = 0

498 CONSTITUTIVE EQUATIONS FOR GENERALIZED PLANE STRESS - CONTINUED

" The strain-displacement relations of the linear theory of elasticity, given

1 ∂u i ∂u j by ε ij x 1, x 2, x 3 = + , are approximated as follows 2 ∂x j ∂x i

+ h 1 2 x 1, x 2 = x 1, x 2, x 3 dx " First, average strains are deﬁned by ε ij h ε ij 3 – h 2

1 ∂u i ∂u j " Substituting ε ij x 1, x 2, x 3 = + into this expression and using 2 ∂x j ∂x i the deﬁnitions for the average displacements yields

U ∂U j x , x 1 ∂ i ε ij 1 2 = + , 2ε 13 = 2ε 23 = 0 and ε 33 = ε 2 ∂x j ∂x i

" Thus, ε represent a uniform through-the-thickness normal strain

499 CONSTITUTIVE EQUATIONS FOR GENERALIZED PLANE STRESS - CONTINUED

" Next, the stress ﬁeld is approximated such that σ 33 = σ 23 = σ 13 = 0

" For this case, the stresses in a thin, ﬂat body, that are normal to the

plane x3 = 0, are presumed negligible compared to the other stresses

" In addition, average stresses are deﬁned by

+ h 1 2 σ ij x 1, x 2 = σ ij x 1, x 2, x 3 dx 3 h – h 2

" The conditions on the presumed stress ﬁeld are satisﬁed by the previously derived plane-stress constitutive equations given in the form

Q Q Q σ 11 11 12 16 ε 11 α 11

σ 22 = Q 12 Q 22 Q 26 ε 22 − α 22 T − Tref and

σ 12 2 2 Q 16 Q 26 Q 66 ε 12 α 12

500 CONSTITUTIVE EQUATIONS FOR GENERALIZED PLANE STRESS - CONTINUED

S S S ε 11 11 12 16 σ 11 α 11

ε 22 = S 12 S 22 S 26 σ 22 + α 22 T − Tref

2 σ 12 2 ε 12 S 16 S 26 S 66 α 12

S S S ε 33 13 23 36 σ 11 α 33

2ε 23 = S 14 S 24 S 46 σ 22 + 2α 23 T − Tref

2 σ 12 2 ε 13 S 15 S 25 S 56 α 13

" That is, substituting σ 33 = σ 23 = σ 13 = 0 into the general form of the constitutive equations and simplifying yields the same plane stress constitutive equation given above

501 CONSTITUTIVE EQUATIONS FOR GENERALIZED PLANE STRESS - CONTINUED

" Restricting the plate to homogeneous construction, integration of the constitutive equations through the plate thickness yields

11 h σ Q 11 Q 12 Q 16 ε Q 11 Q 12 Q 16 α 11 + 11 1 2 σ 22 = Q 12 Q 22 Q 26 ε 22 − Q 12 Q 22 Q 26 α 22 T − Tref dx 3 h – h σ 12 2α 2 Q 16 Q 26 Q 66 2 Q 16 Q 26 Q 66 12 ε 12

" 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

+ h 1 2 Θ x 1, x 2 = T − Tref dx 3 h – h 2 such that the constitutive equations become

502 CONSTITUTIVE EQUATIONS FOR GENERALIZED PLANE STRESS - CONTINUED

11 σ 11 Q 11 Q 12 Q 16 ε α 11

σ 22 = Q 12 Q 22 Q 26 ε 22 − α 22 Θ σ 12 2α Q 16 Q 26 Q 66 2 12 ε 12

11 ε S 11 S 12 S 16 σ 11 α 11

" Similarly, ε 22 = S 12 S 22 S 26 σ 22 + α 22 Θ and σ 12 2α 2 S 16 S 26 S 66 12 ε 12

33 ε S 13 S 23 S 36 σ 11 α 33 2 ε 23 = S 14 S 24 S 46 σ 22 + α 23 Θ σ 12 2α 2 S 15 S 25 S 56 13 ε 13

" Inspection of the last matrix equation reveals that the material must be

monoclinic, with the plane x3 = 0 being a plane of reﬂective symmetry, in

order to satisfy the kinematic hypothesis 2ε 13 = 2ε 23 = 0 and ε 33 = ε

503 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

33 ε S 13 S 23 S 36 σ 11 α 33 2 " Thus, ε 23 = S 14 S 24 S 46 σ 22 + α 23 Θ becomes σ 12 2α 2 S 15 S 25 S 56 13 ε 13

33 ε S 13 S 23 S 36 σ 11 α 33

ε 23 = 0 0 0 σ 22 + 0 Θ , which reduces to 2 0 0 0 σ 12 0 ε 13

= + + + εε S 13σ 11 S 23σ 22 S 36σ 12 α 33Θ

504 CONSTITUTIVE EQUATIONS FOR GENERALIZED PLANE STRESS - CONTINUED

ν 13 ν 23 η 12, 3 " In terms of engineering constants, ε = − σ 11 − σ 22 + σ 33 + α 33Θ and E 1 E 2 E 3

ν 21 η 1,12 1 − 11 ε E 1 E 2 G 12 σ 11 α 11 ν 12 η 2,12 = − 1 ε 22 σ 22 + α 22 Θ E 1 E 2 G 12

η 12,1 η 12,2 2 2 1 σ 12 α 12 ε 12 E 1 E 2 G 12

" Now consider the previously derived stress-transformation law for

a state of plane stress and a dextral rotation about the x3 axis is given by

2 2 cos θ3 sin θ3 2sinθ3cosθ3 σ 1′1′ σ 11 2 2 σ 2′2′ = sin θ3 cos θ3 − 2sinθ3cosθ3 σ 22 2 2 σ 1′2′ σ 12 − sinθ3cosθ3 sinθ3cosθ3 cos θ3 − sin θ3

" This law was expressed symbolically by Σ′′ = Tσ θ 3 Σ

505 CONSTITUTIVE EQUATIONS FOR GENERALIZED PLANE STRESS - CONTINUED

σ 1′1′ σ 11 where Σ′′ ≡≡ σ 2′2′ , Σ ≡≡ σ 22 , and σ 1′2′ σ 12

2 2 cos θ3 sin θ3 2sinθ3cosθ3

2 2 sin θ cos θ − 2sinθ cosθ Tσ θ3 ≡≡ 3 3 3 3 2 2 − sinθ3cosθ3 sinθ3cosθ3 cos θ3 − sin θ3

" Integrating the stress transformation law over the plate thickness yields

the trans formation law Σ′′ = Tσ θ 3 Σ , where

+ h + h 2 2 σ 11 σ 1 1 1 1 ′ ′ σ 22 dx3 σ 2 2 dx3 Σ ≡≡ h and Σ′′ ≡≡ h ′ ′ σ 12 σ 1′2′ – h – h 2 2

506 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 x3 axis is given by

E′′ = Tε θ 3 E

ε 1′1′ ε 11 where E′′ ≡≡ ε 2′2′ , E ≡≡ ε 22 , and 2ε 1′2′ 2ε 12

2 2 cos θ3 sin θ3 sinθ3cosθ3

2 2 sin θ cos θ − sinθ cosθ Tε θ3 ≡≡ 3 3 3 3 2 2 − 2sinθ3cosθ3 2sinθ3cosθ3 cos θ3 − sin θ3

507 CONSTITUTIVE EQUATIONS FOR GENERALIZED PLANE STRESS - CONCLUDED

" Integrating the strain transformation law over the plate thickness yields

the transformation law E′′ = Tε θ 3 E , where

+ h + h 2 2 ε 11 ε 1 1 1 1 ′ ′ ε 22 dx 3 ε 2 2 dx 3 E ≡≡ h and E′′ ≡≡ h ′ ′ 2ε 12 2ε 1′2′ – h – h 2 2

" 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

508 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, x3 flat body, that are normal to the x1 - x2 plane shown in the figure, are presumed negligible compared to the other stresses x2 h " Thus, the stress field is

σ 13 = 0 approximated such that , x1 = 0 = 0 h x + h σ 23 , and σ 33 − 2 ≤ 3 ≤ 2

x ,x ,x x ,x ,x x ,x ,x " But, σ 11 = σ 11 1 2 3 , σ 22 = σ 22 1 2 3 , and σ 12 = σ 12 1 2 3 are permitted because of through-the thickness nonhomogeneity that is presumed to exist

509 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

Q Q Q σ 11 11 12 16 ε 11 α 11

σ 22 = Q 12 Q 22 Q 26 ε 22 − α 22 T − Tref

σ 12 2 2 Q 16 Q 26 Q 66 ε 12 α 12

" In the present formulation, the total strains are presumed to be uniform through the plate thickness; that is,

x ,x x ,x x ,x ε11 = ε11 1 2 , ε22 = ε22 1 2 , and ε12 = ε12 1 2

" 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 x3 coordinate

510 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 x3 = 0 and then to require that the following integral be valid

+ h 2 σ 11 0

σ 22 x3 dx3 = 0

σ 12 0 – h 2

" This matrix integral equation then requires the following through-the- thickness symmetry conditions on the stress ﬁeld

σ 11 x 1, x 2, − x 3 = σ 11 x 1, x 2, x 3 , σ 22 x 1, x 2, − x 3 = σ 22 x 1, x 2, x 3 , and

σ 12 x 1, x 2, − x 3 = σ 12 x 1, x 2, x 3

511 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

h + h + 2 2 Q 11 Q 12 Q 16 ε11 Q 11 Q 12 Q 16 α 11

Q 12 Q 22 Q 26 x 3 dx 3 ε22 − Q 12 Q 22 Q 26 α 22 x 3 dx 3 T − Tref = 0 Q Q Q 2 Q Q Q 2 h 16 26 66 ε12 16 26 66 α 12 − − h 2 2

" Next, note that for arbitrary strain and temperature ﬁelds, the two integrals in the above equation must vanish independently; that is,

+ h + h 2 2 Q 11 Q 12 Q 16 Q 11 Q 12 Q 16 α 11

Q 12 Q 22 Q 26 x3 dx3 = 0 and Q 12 Q 22 Q 26 α 22 x 3 dx 3 = 0

Q 16 Q 26 Q 66 Q 16 Q 26 Q 66 2α 12 − h − h 2 2

512 CONSTITUTIVE EQUATIONS FOR INPLANE DEFORMATIONS OF THIN PLATES - CONTINUED

+ h 2 Q 11 Q 12 Q 16

" The integral Q 12 Q 22 Q 26 x3 dx3 = 0 requires that the reduced

Q 16 Q 26 Q 66 − h 2

stiffness coefﬁcients be symmetric functions about the plane x3 = 0

" Thus, the full set of stiffness and compliance coefﬁcients, Cij and

Sij, must be symmetric about the plane x3 = 0; i. e., monoclinic

+ h 2 Q 11 Q 12 Q 16 α 11

" The integral Q 12 Q 22 Q 26 α 22 x 3 dx 3 = 0 and the symmetry

Q 16 Q 26 Q 66 2α 12 − h 2 requirement on the reduced stiffness coefﬁcients require that the coefﬁcients of thermal expansion be symmetric functions about the

plane x3 = 0

513 CONSTITUTIVE EQUATIONS FOR INPLANE DEFORMATIONS OF THIN PLATES - CONTINUED

" Recall, that for a material that is monoclinic with respect to the plane

x3 = 0, the general form of the constitutive equations is

S 11 S 12 S 13 0 0 S 16 α 11 ε11 σ 11 S 12 S 22 S 23 0 0 S 26 ε22 σ 22 α 22

ε33 S 13 S 23 S 33 0 0 S 36 σ 33 α 33 = + T − Tref 2ε23 0 0 0 S 44 S 45 0 σ 23 0 2 σ 13 0 ε13 0 0 0 S 45 S 55 0 σ 12 2ε12 2α 12 S 16 S 26 S 36 0 0 S 66

C 11 C 12 C 13 0 0 C 16 β 11 σ 11 ε11 C 12 C 22 C 23 0 0 C 26 σ 22 ε22 β 22

σ 33 C 13 C 23 C 33 0 0 C 36 ε33 β 33 = + T − Tref σ 23 0 0 0 C 44 C 45 0 2ε23 0 σ 13 2 0 0 0 0 C 45 C 55 0 ε13 σ 12 2ε12 β 12 C 16 C 26 C 36 0 0 C 66

514 CONSTITUTIVE EQUATIONS FOR INPLANE DEFORMATIONS OF THIN PLATES - CONTINUED

" In addition,

1 ν 21 η 1,12 − − 1 E 1 E 2 G 12 S 11 S 12 S 16 Q 11 Q 12 Q 16 S 11 S 12 S 16 ν 12 1 η 2,12 S 12 S 22 S 26 = − and Q 12 Q 22 Q 26 = S 12 S 22 S 26 E 1 E 2 G 12 S 16 S 26 S 66 Q 16 Q 26 Q 66 S 16 S 26 S 66 η 12,1 η 12,2 1

E 1 E 2 G 12

515 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

+ h 2 N 11 σ 11

N 22 ≡ σ 22 dx3

N 12 σ 12 − h 2

" Substituting

Q Q Q σ 11 11 12 16 ε 11 α 11

σ 22 = Q 12 Q 22 Q 26 ε 22 − α 22 T − Tref

σ 12 2 2 Q 16 Q 26 Q 66 ε 12 α 12

into the expressions for the stress resultants and performing the integration yields a two-dimensional constitutive equation

516 CONSTITUTIVE EQUATIONS FOR INPLANE DEFORMATIONS OF THIN PLATES - CONTINUED

" The two-dimensional constitutive equation is given by

β N 11 A 11 A 12 A 16 ε 11 11 N = 22 A 12 A 22 A 26 ε 22 + β 22 T − Tref N 12 A A A 2ε 12 16 26 66 β 12 where

+ h + h 2 β 2 A 11 A 12 A 16 Q 11 Q 12 Q 16 11 Q11 Q12 Q16 α11 Q Q Q dx and Q Q Q dx A 12 A 22 A 26 ≡ 12 22 26 3 β 22 ≡ − 12 22 26 α22 3 Q Q Q Q Q Q 2 A 16 A 26 A 66 16 26 66 β 16 26 66 α12 − h 12 − h 2 2

β " The Aij and ij are called the inplane, plate stiffness coefﬁcients and thermal moduli, respectively

517 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

− 1 − 1 β ε 11 A 11 A 12 A 16 N 11 A 11 A 12 A 16 11 N ε 22 = A 12 A 22 A 26 22 − A 12 A 22 A 26 β 22 T − Tref 2ε 12 A A A N 12 A A A 16 26 66 16 26 66 β 12

" Next, the inverted matrix equation is manipulated to look like the plane- stress constitutive equations

S S S ε 11 11 12 16 σ 11 α 11

ε 22 = S 12 S 22 S 26 σ 22 + α 22 T − Tref

2 σ 12 2 ε 12 S 16 S 26 S 66 α 12

" That is, a set of overall plate coefﬁcients of thermal expansion are deﬁned to make the construction of the equations parallel

518 CONSTITUTIVE EQUATIONS FOR INPLANE DEFORMATIONS OF THIN PLATES - CONTINUED

" The overall plate coefﬁcients of thermal expansion are deﬁned as

− 1 − 1 + h β 2 α 11 A 11 A 12 A 16 11 A 11 A 12 A 16 Q 11 Q 12 Q 16 α 11 = Q Q Q dx α 22 = A 12 A 22 A 26 β 22 A 12 A 22 A 26 12 22 26 α 22 3 2 Q Q Q 2 α 12 A 16 A 26 A 66 β A 16 A 26 A 66 16 26 66 α 12 12 − h 2

so that the two-dimensional constitutive equation is given by

N 11 A 11 A 12 A 16 ε 11 α 11

N 22 = A 12 A 22 A 26 ε 22 − α 22 T − Tref or N 2 2 12 A 16 A 26 A 66 ε 12 α 12

− 1

ε 11 A 11 A 12 A 16 N 11 α 11

ε 22 = A 12 A 22 A 26 N 22 + α 22 T − Tref 2 N 2 ε 12 A 16 A 26 A 66 12 α 12

519 CONSTITUTIVE EQUATIONS FOR INPLANE DEFORMATIONS OF THIN PLATES - CONTINUED

− 1 A 11 A 12 A 16 a 11 a 12 a 16

" For convenience, let A 12 A 22 A 26 = a 12 a 22 a 26 so that

A 16 A 26 A 66 a 16 a 26 a 66

ε 11 a 11 a 12 a 16 N 11 α 11

ε 22 = a 12 a 22 a 26 N 22 + α 22 T − Tref

2ε 12 a 16 a 26 a 66 N 12 2α 12

" The subscripted a terms are plate compliances that are given by

2 A 22A 66 − A 26 A 16A 26 − A 12A 66 A 12A 26 − A 16A 22 a 11 = a 12 = a 16 = A A A

2 2 A 11A 66 − A 16 A 12A 16 − A 11A 26 A 11A 22 − A 12 a 22 = a 26 = a 66 = A A A

2 2 2 where A = A 11 A 22 − A 12 A 66 − A 11 A 26 − A 22 A 16 + 2A 12 A 16 A 26

520 CONSTITUTIVE EQUATIONS FOR INPLANE DEFORMATIONS OF THIN PLATES - CONTINUED

" Once a given boundary-value problem is solved, the strain ﬁelds

x ,x x ,x 2 x ,x ε11 1 2 , ε22 1 2 , and ε 12 1 2 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

Q Q Q σ 11 11 12 16 ε 11 α 11

σ 22 = Q 12 Q 22 Q 26 ε 22 − α 22 T − Tref

σ 12 2 2 Q 16 Q 26 Q 66 ε 12 α 12

" The other strains are given by

ε 33 S S S 13 23 36 σ 11 α 33 2 ε 23 = 0 0 0 σ 22 + 0 T − Tref which reduce to 2 0 0 0 σ 12 0 ε 13

2 2 ε 23 = ε 13 = 0 and ε 33 = S 13 σ 11 + S 23 σ 22 + S 36 σ 12 + α 33 T − Tref

521 CONSTITUTIVE EQUATIONS FOR INPLANE DEFORMATIONS OF THIN PLATES - CONTINUED

" Now consider a transformation of coordinates that corresponds to a

dextral rotation about the x3 axis

" The stress-transformation law for a state of plane stress has been given by

σ 1′1′ σ 11

σ 2 2 σ 22 Σ′′ = Tσ θ 3 Σ where Σ′′ ≡≡ ′ ′ , Σ ≡≡ , and σ 1′2′ σ 12

2 2 cos θ3 sin θ3 2sinθ3cosθ3

2 2 sin θ cos θ − 2sinθ cosθ Tσ θ3 ≡≡ 3 3 3 3 2 2 − sinθ3cosθ3 sinθ3cosθ3 cos θ3 − sin θ3

522 CONSTITUTIVE EQUATIONS FOR INPLANE DEFORMATIONS OF THIN PLATES - CONTINUED

" Integrating the stress transformation law over the plate thickness yields

the transformation law NN′′ = Tσ θ 3 NN , where

+ h + h 2 2 N 11 σ 11 N1′1′ σ 1′1′

N dx N2′2′ σ 2′2′ dx N ≡ 22 ≡ σ 22 3 and N′′ ≡ ≡ 3 N 12 σ 12 N1′2′ σ 1′2′ – h – h 2 2

" Likewise, the previously derived strain-transformation law for a

state of plane stress and a dextral rotation about the x3 axis is given by

ε 1 1 ε 11 = ′ ′ E′′ Tε θ 3 E where E′′ ≡≡ ε 2′2′ , E ≡≡ ε 22 , 2ε 1′2′ 2ε 12

523 CONSTITUTIVE EQUATIONS FOR INPLANE DEFORMATIONS OF THIN PLATES - CONTINUED

2 2 cos θ3 sin θ3 sinθ3cosθ3

2 2 sin θ cos θ − sinθ cosθ and Tε θ3 ≡≡ 3 3 3 3 ; and 2 2 − 2sinθ3cosθ3 2sinθ3cosθ3 cos θ3 − sin θ3

remains unchanged for thin, nonhomogeneous plate

A 11 A 12 A 16 Q11 Q12 Q16

" For convenience, let A = A 12 A 22 A 26 and Q = Q12 Q22 Q26

A 16 A 26 A 66 Q16 Q26 Q66

+ h 2 such that A = Q dx 3 − h 2

524 CONSTITUTIVE EQUATIONS FOR INPLANE DEFORMATIONS OF THIN PLATES - CONTINUED

A 1′1′ A 1′2′ A 1′6′ Q 1′1′ Q 1′2′ Q 1′6′

" Similarly, let A′ ≡ A 1′2′ A 2′2′ A 2′6′ and Q′ ≡ Q 1′2′ Q 2′2′ Q 2′6′

A 1′6′ A 2′6′ A 6′6′ Q 1′6′ Q 2′6′ Q 6′6′

+ h 2 such that A′ = Q′ dx 3 − h 2

" The transformation laws that relate A and A′ are obtained by using the following plane-stress transformation equations

T T Q′ = Tσ Q Tσ Q = Tε Q′ Tε

525 CONSTITUTIVE EQUATIONS FOR INPLANE DEFORMATIONS OF THIN PLATES - CONTINUED

T T " Integation of Q′ = Tσ Q Tσ and Q = Tε Q′ Tε over the

T T plate thickness yields A′ = Tσ A Tσ and A = Tε A′ Tε

+ h + h 2 − 1 2 − 1 " From A = Q dx 3 , it follows that A = Q dx 3 h − − h 2 2

S 11 S 12 S 16 a 11 a 12 a 16 − 1 − 1 " Noting that a = a 12 a 22 a 26 ≡ A and S = S 12 S 22 S 26 ≡ Q ,

a 16 a 26 a 66 S 16 S 26 S 66

+ h 2 it follows that a = S dx 3 − h 2

526 CONSTITUTIVE EQUATIONS FOR INPLANE DEFORMATIONS OF THIN PLATES - CONTINUED

S 1′1′ S 1′2′ S 1′6′ a 1′1′ a 1′2′ a 1′6′ − 1 − 1 S S S " Similarly a′ = a 1′2′ a 2′2′ a 2′6′ ≡ A′ , S′ = 1′2′ 2′2′ 2′6′ ≡ Q′ ,

a 1′6′ a 2′6′ a 6′6′ S 1′6′ S 2′6′ S 6′6′

+ h 2 and a′ = S′ dx 3 − h 2

" The transformation laws that relate a and a′ are obtained by using the following plane-stress transformation equations

T T

S′ = Tε S Tε S = Tσ S′ Tσ

527 CONSTITUTIVE EQUATIONS FOR INPLANE DEFORMATIONS OF THIN PLATES - CONTINUED

T T " Integration of S′ = Tε S Tε and S = Tσ S′ Tσ over the

T T plate thickness yields a′ = Tε a Tε and a = Tσ a′ Tσ

" Now consider the thermal moduli of the plate given by

+ h β 2 11 Q 11 Q 12 Q 16 α 11

β ≡ − Q 12 Q 22 Q 26 α 22 dx3 ββ ≡ 22 β Q 16 Q 26 Q 66 2α 12 12 − h 2

β 11 Q11 Q12 Q16 α11

" Noting that β ≡ β 22 = − Q12 Q22 Q26 α22 for plane stress, it 2 β 12 Q16 Q26 Q66 α12

+ h + h 2 2 follows that β = β dx3 and that β′ = β′ dx3 − h − h 2 2

528 CONSTITUTIVE EQUATIONS FOR INPLANE DEFORMATIONS OF THIN PLATES - CONTINUED

" The transformation laws that relate β and β′ are obtained by using the following plane-stress transformation equations

β′ = Tσ β β = Tε β′

" Integration of these equations over the plate thickness yields T ′ ′ β = Tσ β and β = Tε β

" Next, recall that the overall plate coefﬁcients of thermal expansion have been given by

β α 11 a 11 a 12 a 16 11

α ≡ α 22 = a 12 a 22 a 26 β 22 = a β

2α 12 a 16 a 26 a 66 β 12

529 CONSTITUTIVE EQUATIONS FOR INPLANE DEFORMATIONS OF THIN PLATES - CONTINUED

β α 1′1′ a 1′1′ a 1′2′ a 1′6′ 1′1′ " It follows logically that α = a a a β = α′ = 2′2′ 1′2′ 2′2′ 2′6′ 2′2′ a′ β′ 2 α 1′2′ a 1′6′ a 2′6′ a 6′6′ β 1′2′

T " Substituting a′ = Tε a Tε and β′ = Tσ β into α′ = a′ β′

T gives α′ = Tε a Tε Tσ β

T − 1 " Next, using Tε = Tσ gives α′ = Tε a β

′ " Then using α = a β gives the result α = Tε α

T − 1 T = ′ " Inverting this result and using Tε = Tσ gives α Tσ α

530 CONSTITUTIVE EQUATIONS FOR INPLANE DEFORMATIONS OF THIN PLATES - CONTINUED

" In summary:

T T A′ = Tσ A Tσ A = Tε A′ Tε

T T

a′ = Tε a Tε a = Tσ a′ Tσ

T ′ ′ β = Tσ β β = Tε β

T ′ ′ α = Tε α α = Tσ α

" 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

531 CONSTITUTIVE EQUATIONS FOR INPLANE DEFORMATIONS OF THIN PLATES - CONCLUDED

T = " For plane stress, Q ′ Tσ Q Tσ , with m = cosθ3 and

n = sinθ3 , gave

4 2 2 2 2 4 Q 1′1′ = m Q 11 + 2m n Q 12 + 2Q 66 + 4mn m Q 16 + n Q 26 + n Q 22

T " Thus, by similarity, A′ = Tσ A Tσ gives

4 2 2 2 2 4 A 1′1′ = m A 11 + 2m n A 12 + 2A 66 + 4mn m A 16 + n A 26 + n A 22

" The other transformation equations are obtained in a similar manner

532 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

x3 " One such assumption is that the strains in a slender, prismatic body, that distort the cross- sectional planes, are negligible compared to the other strains

" This simpliﬁcation is commonly referred to as the generalized plane-strain assumption

x2 " For a state of generalized plane strain in a homogeneous, anisotropic solid, with respect to x1 the x1 - x2 plane, the strain ﬁeld is approximated

such that ε 33 − α 33 T − Tref = ε and 2ε 23 = 2ε 13 = 0 , where ε is a constant

533 CONSTITUTIVE EQUATIONS FOR PLANE STRAIN CONTINUED

" For this special case, the general constitutive equation

C11 C12 C13 C14 C15 C16 σ 11 ε11 α 11 C12 C22 C23 C24 C25 C26 σ 22 ε22 α 22 σ C C C C C C ε α 33 = 13 23 33 34 35 36 33 33 T T σ − − ref 23 C14 C24 C34 C44 C45 C46 2ε23 2α 23

σ 13 2ε13 2α 13 C15 C25 C35 C45 C55 C56 σ 12 2ε12 2α 12 C16 C26 C36 C46 C56 C66 uncouples directly into

C C C C σ 11 11 12 16 ε 11 α 11 13

σ 22 = C 12 C 22 C 26 ε 22 − α 22 T − Tref + C 23 ε

σ 12 2 2 C 16 C 26 C 66 ε 12 α 12 C 36 and C C C C σ 33 13 23 36 ε 11 α 11 33

σ 23 = C 14 C 24 C 46 ε 22 − α 22 T − Tref + C 34 ε

σ 13 2 2 C 15 C 25 C 56 ε 12 α 12 C 35

534 CONSTITUTIVE EQUATIONS FOR PLANE STRAIN CONTINUED

" This equation is manipulated further by deﬁning thermal moduli

β 11 C 11 C 12 C 16 α 11 β 33 C 13 C 23 C 36 α 11

β 22 = − C 12 C 22 C 26 α 22 and β 23 = − C 14 C 24 C 46 α 22 2 2 β 12 C 16 C 26 C 66 α 12 β 13 C 15 C 25 C 56 α 12

such that

C C C β 11 C σ 11 11 12 16 ε 11 13

σ 22 = C 12 C 22 C 26 ε 22 + β 22 T − Tref + C 23 ε and

σ 12 2 C 16 C 26 C 66 ε 12 β 12 C 36

C C C β 33 C σ 33 13 23 36 ε 11 33

σ 23 = C 14 C 24 C 46 ε 22 + β 23 T − Tref + C 34 ε

σ 13 2 C 15 C 25 C 56 ε 12 β 13 C 35

" Note that the stiffness coefﬁcients are obtained by inverting the fully populated compliance matrix - a nontrivial task

535 CONSTITUTIVE EQUATIONS FOR PLANE STRAIN CONTINUED

" Simpliﬁcation of the following general constitutive equation is not as easy

2ε13 σ 13 2α 13 S 15 S 25 S 35 S 45 S 55 S 56 σ 12 2ε12 2α 12 S 16 S 26 S 36 S 46 S 56 S 66

" First, the equation given above is expressed as

S 11 S 12 S 13 S 14 S 15 S 16 ε11 α 11 σ 11 S 12 S 22 S 23 S 24 S 25 S 26 ε22 α 22 σ 22 ε α S S S S S S σ 33 33 T T = 13 23 33 34 35 36 33 − − ref σ 2ε23 2α 23 S 14 S 24 S 34 S 44 S 45 S 46 23

2ε13 2α 13 σ 13 S 15 S 25 S 35 S 45 S 55 S 56 σ 12 2ε12 2α 12 S 16 S 26 S 36 S 46 S 56 S 66

536 CONSTITUTIVE EQUATIONS FOR PLANE STRAIN CONTINUED

" Then, elimination of the transverse-shearing strains gives

S 11 S 12 S 13 S 14 S 15 S 16 ε11 α 11 σ 11 S 12 S 22 S 23 S 24 S 25 S 26 ε22 α 22 σ 22

ε33 α 33 S S S S S S σ T T = 13 23 33 34 35 36 33 − − ref σ 0 0 S 14 S 24 S 34 S 44 S 45 S 46 23 0 0 σ 13 S 15 S 25 S 35 S 45 S 55 S 56 σ 12 2ε12 2α 12 S 16 S 26 S 36 S 46 S 56 S 66

" Rearranging the rows and columns into a convenient form gives

S11 S12 S16 S13 S14 S15 ε11 α11 σ 11 S12 S22 S26 S23 S24 S25 ε22 α22 σ 22

2ε12 2α12 S S S S S S σ T T = 16 26 66 36 46 56 12 − − ref σ ε33 α33 S13 S23 S36 S33 S34 S35 33 0 0 σ 23 S14 S24 S46 S34 S44 S45 0 0 σ 13 S15 S25 S56 S35 S45 S55

537 CONSTITUTIVE EQUATIONS FOR PLANE STRAIN CONTINUED

" For convenience, let the mechanical strains be denoted by εσ 11 ε11 α11 σ ε22 ε22 α22 σ 2ε12 2ε12 2α12 σ = − T − Tref ε33 ε33 α33 0 0 0 0 0 0

" Using ε 33 − α 33 T − Tref = ε , the previous matrix constitutive equation becomes

σ ε11 S11 S12 S16 S13 S14 S15 σ σ 11 ε22 S12 S22 S26 S23 S24 S25 σ σ 22 2ε12 S S S S S S σ = 16 26 66 36 46 56 12 σ ε S13 S23 S36 S33 S34 S35 33 0 S S S S S S σ 23 14 24 46 34 44 45 σ 0 13 S15 S25 S56 S35 S45 S55

538 CONSTITUTIVE EQUATIONS FOR PLANE STRAIN CONTINUED

" Next, the matrix constitutive equation

σ ε11 S11 S12 S16 S13 S14 S15 σ σ 11 ε22 S12 S22 S26 S23 S24 S25 σ σ 22 2ε12 S S S S S S σ = 16 26 66 36 46 56 12 is separated to get σ ε S13 S23 S36 S33 S34 S35 33 0 S S S S S S σ 23 14 24 46 34 44 45 σ 0 13 S15 S25 S56 S35 S45 S55

σ S S S S S S ε 11 11 12 16 σ 11 13 14 15 σ 33 σ ε 22 = S 12 S 22 S 26 σ 22 + S 23 S 24 S 25 σ 23 and σ σ 12 σ 13 2ε 12 S 16 S 26 S 66 S 36 S 46 S 56

S S S S S S ε 13 23 36 σ 11 33 34 35 σ 33 0 = S 14 S 24 S 46 σ 22 + S 34 S 44 S 45 σ 23 0 σ 12 σ 13 S 15 S 25 S 56 S 35 S 45 S 55

539 CONSTITUTIVE EQUATIONS FOR PLANE STRAIN CONTINUED

, " Solving the previous equation with ε for σ 33 σ 23, and σ 13 gives

− 1 − 1 S S S S S S S S S σ 33 33 34 35 ε 33 34 35 13 23 36 σ 11 σ 23 = S 34 S 44 S 45 0 − S 34 S 44 S 45 S 14 S 24 S 46 σ 22 σ 13 0 σ 12 S 35 S 45 S 55 S 35 S 45 S 55 S 15 S 25 S 56

" Back substitution of the vector containing σ 33 , σ 23, and σ 13 into

σ S S S S S S ε 11 11 12 16 σ 11 13 14 15 σ 33 σ ε 22 = S 12 S 22 S 26 σ 22 + S 23 S 24 S 25 σ 23 σ σ 12 σ 13 2ε 12 S 16 S 26 S 66 S 36 S 46 S 56

yields the result

σ ε 11 s 11 s 12 s 16 σ 11 S 11 σ + ε 22 = s 12 s 22 s 26 σ 22 S 22 ε σ σ 2ε 12 12 S 12 s 16 s 26 s 66

540 CONSTITUTIVE EQUATIONS FOR PLANE STRAIN CONTINUED

" The matrix with the subscripted s terms is given by − 1 s 11 s 12 s 16 S 11 S 12 S 16 S 13 S 14 S 15 S 33 S 34 S 35 S 13 S 23 S 36

s 12 s 22 s 26 = S 12 S 22 S 26 − S 23 S 24 S 25 S 34 S 44 S 45 S 14 S 24 S 46 S S S S S S S S S S S S s 16 s 26 s 66 16 26 66 36 46 56 35 45 55 15 25 56

" The vector with the subscripted S terms is given by

− 1 S 11 S 13 S 14 S 15 S 33 S 34 S 35 1 S 22 = S 23 S 24 S 25 S 34 S 44 S 45 0 0 S 12 S 36 S 46 S 56 S 35 S 45 S 55

" Next, expressing the mechanical strains in terms of the total strains and the strains caused by free thermal expansion results in

ε 11 s 11 s 12 s 16 σ 11 α 11 S 11 + ε 22 = s 12 s 22 s 26 σ 22 α 22 T − Tref + S 22 ε 2 σ 2 ε 12 12 α 12 S 12 s 16 s 26 s 66

541 ! ! !

CONSTITUTIVE EQUATIONS FOR PLANE STRAIN CONTINUED

" Thus, the constitutive equations for generalized plane strain, in terms of compliance coefﬁcients and thermal-expansion coefﬁcients, become

! 11 s 11 s 12 s 16 " 11 # 11 S 11 + ! 22 = s 12 s 22 s 26 " 22 # 22 T $ Tref + S 22 ! where 2 " 12 2 ! 12 # 12 S 12 s 16 s 26 s 66

$ 1 s 11 s 12 s 16 S 11 S 12 S 16 S 13 S 14 S 15 S 33 S 34 S 35 S 13 S 23 S 36

s 12 s 22 s 26 = S 12 S 22 S 26 $ S 23 S 24 S 25 S 34 S 44 S 45 S 14 S 24 S 46 S S S S S S S S S S S S s 16 s 26 s 66 16 26 66 36 46 56 35 45 55 15 25 56

$ 1 S 11 S 13 S 14 S 15 S 33 S 34 S 35 1 and S 22 = S 23 S 24 S 25 S 34 S 44 S 45 0 0 S 12 S 36 S 46 S 56 S 35 S 45 S 55

" The s ij are called the reduced compliance coefficients " When ! = 0 , the state of strain reduces to that known simply as plane strain

542 CONSTITUTIVE EQUATIONS FOR PLANE STRAIN CONTINUED

" Once a given generalized-plane-strain or regular-plane-strain boundary- value problem is solved by using the following equation, the stresses and strains in the following equation are known

ε 11 s 11 s 12 s 16 σ 11 α 11 S 11 + ε 22 = s 12 s 22 s 26 σ 22 α 22 T − Tref + S 22 ε 2 σ 2 ε 12 12 α 12 S 12 s 16 s 26 s 66

" The other stresses are then given by

− 1 − 1 S S S S S S S S S σ 33 33 34 35 ε 33 34 35 13 23 36 σ 11 σ 23 = S 34 S 44 S 45 0 − S 34 S 44 S 45 S 14 S 24 S 46 σ 22 σ 13 0 σ 12 S 35 S 45 S 55 S 35 S 45 S 55 S 15 S 25 S 56

543 CONSTITUTIVE EQUATIONS FOR PLANE STRAIN CONTINUED

" The relationship between the reduced compliances and the stiffnesses is obtained by ﬁrst considering the equation

C C C C σ 11 11 12 16 ε 11 α 11 13

σ 22 = C 12 C 22 C 26 ε 22 − α 22 T − Tref + C 23 ε

σ 12 2 2 C 16 C 26 C 66 ε 12 α 12 C 36

" Inverting gives

− 1 − 1 C C C C C C C ε 11 11 12 16 σ 11 α 11 11 12 16 13

ε 22 = C 12 C 22 C 26 σ 22 + α 22 T − Tref − C 12 C 22 C 26 C 23 ε 2 σ 12 2 ε 12 C 16 C 26 C 66 α 12 C 16 C 26 C 66 C 36

544 CONSTITUTIVE EQUATIONS FOR PLANE STRAIN CONTINUED

" Comparing

− 1 − 1 C C C C C C C ε 11 11 12 16 σ 11 α 11 11 12 16 13

ε 22 = C 12 C 22 C 26 σ 22 + α 22 T − Tref − C 12 C 22 C 26 C 23 ε 2 σ 12 2 ε 12 C 16 C 26 C 66 α 12 C 16 C 26 C 66 C 36

with

ε 11 s 11 s 12 s 16 σ 11 α 11 S 11 + ε 22 = s 12 s 22 s 26 σ 22 α 22 T − Tref + S 22 ε indicates that 2 σ 2 ε 12 12 α 12 S 12 s 16 s 26 s 66

− 1 − 1 s 11 s 12 s 16 C 11 C 12 C 16 S 11 C 11 C 12 C 16 C 13

s 12 s 22 s 26 = C 12 C 22 C 26 and S 22 = − C 12 C 22 C 26 C 23 C C C C C C C s 16 s 26 s 66 16 26 66 S 12 16 26 66 36

545 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

β 11 C 11 C 12 C 16 α 11 β 33 C 13 C 23 C 36 α 11

β 22 = − C 12 C 22 C 26 α 22 and β 23 = − 0 0 0 α 22 2 2 β 12 C 16 C 26 C 66 α 12 β 13 0 0 0 α 12

" The generalized-plane-strain constitutive equations become

C C C β 11 C σ 11 11 12 16 ε 11 13

σ 22 = C 12 C 22 C 26 ε 22 + β 22 T − Tref + C 23 ε and

σ 12 2 C 16 C 26 C 66 ε 12 β 12 C 36

C C C β 33 C σ 33 13 23 36 ε 11 33

σ 23 = 0 0 0 ε 22 + 0 T − Tref + 0 ε σ 13 0 0 0 2ε 12 0 0

546 CONSTITUTIVE EQUATIONS FOR PLANE STRAIN CONTINUED

" The last equation is reduced further to give σ 23 = σ 13 = 0 and

σ 33 = C 13 ε 11 − α 11 T − Tref + C 23 ε 22 − α 22 T − Tref + 2C 36 ε 12 − α 12 T − Tref + C 33 ε

" Likewise, for a material that is monoclinic in the plane of the cross- section of the body,

− 1 s 11 s 12 s 16 S 11 S 12 S 16 S 13 0 0 S 33 0 0 S 13 S 23 S 36

s 12 s 22 s 26 = S 12 S 22 S 26 − S 23 0 0 0 S 44 S 45 0 0 0 S S S S 0 0 0 S S 0 0 0 s 16 s 26 s 66 16 26 66 36 45 55 which simpliﬁes to

S 13S 13 S 13S 23 S 13S 36 S 11 − S 12 − S 16 − S 33 S 33 S 33 s 11 s 12 s 16 S S S S S S S 13 23 S 23 23 S 23 36 s 12 s 22 s 26 = 12 − 22 − 26 − S 33 S 33 S 33

s 16 s 26 s 66 S 13S 36 S 23S 36 S 36S 36 S 16 − S 26 − S 66 − S 33 S 33 S 33

547 CONSTITUTIVE EQUATIONS FOR PLANE STRAIN CONTINUED

" Similarly,

− 1 S 11 S 13 S 14 S 15 S 33 S 34 S 35 1 S 22 = S 23 S 24 S 25 S 34 S 44 S 45 0 simpliﬁes to 0 S 12 S 36 S 46 S 56 S 35 S 45 S 55

− 1 T S 11 S 13 0 0 S 33 0 0 1 T S13 S23 S36 S 22 = S 23 0 0 0 S 44 S 45 0 or S 11 S 22 S 12 = 0 S33 S33 S33 S 12 S 36 0 0 0 S 45 S 55

" The other stresses are then given by

− 1 − 1 S 0 0 S 0 0 S S S σ 33 33 ε 33 13 23 36 σ 11 σ 23 = 0 S 44 S 45 0 − 0 S 44 S 45 0 0 0 σ 22 σ 13 0 σ 12 0 S 45 S 55 0 S 45 S 55 0 0 0

1 = = 0 which reduce to σ 33 = ε − S 13σ 11 − S 23σ 22 − S 36σ 12 and σ 23 σ 13 S 33

548 CONSTITUTIVE EQUATIONS FOR PLANE STRAIN CONTINUED

" For a specially orthotropic material, the generalized-plane-strain thermal moduli reduce to

β 11 C 11 C 12 0 α 11 β 33 C 13 C 23 0 α 11

β 22 = − C 12 C 22 0 α 22 and β 23 = − 0 0 0 α 22

β 12 0 0 C 66 0 β 13 0 0 0 0

" The generalized-plane-strain constitutive equations become

C C 0 β 11 C σ 11 11 12 ε 11 13

σ 22 = C 12 C 22 0 ε 22 + β 22 T − Tref + C 23 ε and

σ 12 2 0 0 C 66 ε 12 0 0

C C 0 β 33 C σ 33 13 23 ε 11 33

σ 23 = 0 0 0 ε 22 + 0 T − Tref + 0 ε σ 13 0 0 0 2ε 12 0 0

549 CONSTITUTIVE EQUATIONS FOR PLANE STRAIN CONTINUED

" The last equation is reduced further to give σ 23 = σ 13 = 0 and

σ 33 = C 13 ε 11 − α 11 T − Tref + C 23 ε 22 − α 22 T − Tref + C 33 ε

" In terms of engineering constants,

E 1 E 2 E 3 C = 1 − ν ν C = 1 − ν ν C 33 = 1 − ν 12 ν 21 11 Δ 23 32 22 Δ 13 31 Δ

E 1 E 2 C 12 = ν 21 + ν 23 ν 31 = ν 12 + ν 13 ν 32 C 66 = G 12 Δ Δ E E C = 1 ν + ν ν = 3 ν + ν ν 13 Δ 31 21 32 Δ 13 12 23 E E C = 2 ν + ν ν = 3 ν + ν ν 23 Δ 32 12 31 Δ 23 13 21

where Δ = 1 − ν 12 ν 21 − ν 23 ν 32 − ν 13 ν 31 − 2ν 21 ν 32 ν 13

550 CONSTITUTIVE EQUATIONS FOR PLANE STRAIN CONTINUED

" The nonzero thermal moduli are expressed as

E 1 1 − ν 23 ν 32 α 11 + ν 21 + ν 23 ν 31 α 22 β 11 = − 1 − ν 12 ν 21 − ν 23 ν 32 − ν 13 ν 31 − 2ν 21 ν 32 ν 13

E 2 ν 12 + ν 13 ν 32 α 11 + 1 – ν 13 ν 31 α 22 β 22 = − 1 − ν 12 ν 21 − ν 23 ν 32 − ν 13 ν 31 − 2ν 21 ν 32 ν 13

E 3 ν 31 + ν 21 ν 32 α 11 + ν 23 + ν 13 ν 21 α 22 β 33 = − 1 − ν 12 ν 21 − ν 23 ν 32 − ν 13 ν 31 − 2ν 21 ν 32 ν 13

551 CONSTITUTIVE EQUATIONS FOR PLANE STRAIN CONTINUED

" Similarly, for a specially orthotropic material, the generalized-plane- strain constitutive matrix becomes

− 1 s 11 s 12 s 16 S 11 S 12 0 S 13 0 0 S 33 0 0 S 13 S 23 0

s 12 s 22 s 26 = S 12 S 22 0 − S 23 0 0 0 S 44 0 0 0 0 0 0 S 0 0 0 0 0 S 0 0 0 s 16 s 26 s 66 66 55

which simpliﬁes to

S13S13 S13S23 S11 − S12 − 0 S33 S33 s 11 s 12 0 S13S23 S23S23 s 12 s 22 0 = S12 − S22 − 0 S33 S33 0 0 s 66

0 0 S66

552 CONSTITUTIVE EQUATIONS FOR PLANE STRAIN CONTINUED

" Also − 1 s 11 s 12 0 C 11 C 12 0

s 12 s 22 0 = C 12 C 22 0 0 0 C 0 0 s 66 66

" Similarly,

− 1 S 11 S 13 S 14 S 15 S 33 S 34 S 35 1 S 22 = S 23 S 24 S 25 S 34 S 44 S 45 0 simpliﬁes to 0 S 12 S 36 S 46 S 56 S 35 S 45 S 55

− 1 T S 13 0 0 S 33 0 0 S 11 1 T S 13 S 23 S 22 = S 23 0 0 0 S 44 0 0 or S 11 S 22 S 12 = 0 0 S 33 S 33 S 12 0 0 0 0 0 S 55

553 CONSTITUTIVE EQUATIONS FOR PLANE STRAIN CONTINUED

" Finally, the generalized-plane-strain constitutive equations for a specially orthotropic material become

0 ε 11 s 11 s 12 σ 11 α 11 S 11 + ε 22 = s 12 s 22 0 σ 22 α 22 T − Tref + S 22 ε 2ε 12 σ 12 0 0 0 0 s 66

S13S13 S13S23 S11 − S12 − 0 S33 S33 s 11 s 12 0 S13S23 S23S23 s 12 s 22 0 = S12 − S22 − 0 S33 S33 0 0 s 66

0 0 S66

T T S 13 S 23 S 11 S 22 S 12 = 0 S 33 S 33

554 CONSTITUTIVE EQUATIONS FOR PLANE STRAIN CONTINUED

" The matrix equation for the other stresses becomes

− 1 − 1 S 0 0 S 0 0 S S 0 σ 33 33 ε 33 13 23 σ 11 σ 23 = 0 S 44 0 0 − 0 S 44 0 0 0 0 σ 22 σ 13 0 σ 12 0 0 S 55 0 0 S 55 0 0 0

1 which reduces to σ 33 = ε − S 13σ 11 − S 23σ 22 and σ 23 = σ 13 = 0 S 33

555 CONSTITUTIVE EQUATIONS FOR PLANE STRAIN CONCLUDED

" In terms of the engineering constants, the reduced compliance coefficients and are given by

1 − ν 13ν 31 ν 12 + ν 13ν 32 ν 21 + ν 23ν 31 s 11 = s 12 = − = − E 1 E 1 E 2

1 − ν 23ν 32 1 s 22 = s 66 = E 2 G 12

" Similarly,

S 11 = − ν 31 S 22 = − ν 32 S 12 = 0 and

σ 33 = E 3ε + ν 31 σ 11 + ν 32 σ 22

556 STRESS AND STRAIN TRANSFORMATION EQUATIONS FOR PLANE STRAIN

" The two primary constitutive equations of generalized plane strain are given by

C C C β 11 C σ 11 11 12 16 ε 11 13

σ 22 = C 12 C 22 C 26 ε 22 + β 22 T − Tref + C 23 ε and

σ 12 2 C 16 C 26 C 66 ε 12 β 12 C 36

ε 11 s 11 s 12 s 16 σ 11 α 11 S 11 + ε 22 = s 12 s 22 s 26 σ 22 α 22 T − Tref + S 22 ε 2 σ 12 2 ε 12 α 12 S 12 s 16 s 26 s 66

" To obtain transformation equations for the constitutive terms appearing in these equations, transformation equations that relate

σ 1′1′ σ 11 Σ′′ ≡≡ σ 2′2′ to Σ ≡≡ σ 22 are needed σ 1′2′ σ 12

557 STRESS AND STRAIN TRANSFORMATION EQUATIONS FOR PLANE STRAIN - CONTINUED

" Likewise transformation equations that relate

ε 1′1′ ε 11 E′′ ≡≡ ε 2′2′ to E ≡≡ ε 22 are needed 2ε 1′2′ 2ε 12

" Consider the dextral (right-handed) rotation of coordinate frames shown in the ﬁgure

x3 ,x3′

i 3 ,i 3′ x2′

i 2′ θ3 x2 i 1 i 2

i 1′

θ3 x1

x1′

558 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

2 2 cos θ3 sin θ3 2sinθ3cosθ3 σ 1′1′ σ 11 2 2 σ 2′2′ = sin θ3 cos θ3 − 2sinθ3cosθ3 σ 22 2 2 σ 1′2′ σ 12 − sinθ3cosθ3 sinθ3cosθ3 cos θ3 − sin θ3

559 STRESS AND STRAIN TRANSFORMATION EQUATIONS FOR PLANE STRAIN - CONTINUED

" Thus, the stress-transformation law for a state of generalized plane

strain and a dextral rotation about the x3 axis is identical to that for the corresponding plane-stress case and is given by

Σ′′ = Tσ θ 3 Σ where

2 2 cos θ3 sin θ3 2sinθ3cosθ3

2 2 sin θ cos θ − 2sinθ cosθ Tσ θ3 ≡≡ 3 3 3 3 2 2 − sinθ3cosθ3 sinθ3cosθ3 cos θ3 − sin θ3

− 1 " Likewise, the inverse is given by Σ = Tσ θ 3 Σ′′ where

2 2 cos θ sin θ − 2sinθ cosθ − 1 3 3 3 3 2 2 – sin θ cos θ 2sinθ cosθ Tσ θ3 = Tσ θ3 = 3 3 3 3 2 2 sinθ 3cosθ 3 − sinθ 3cosθ 3 cos θ 3 − sin θ 3

560 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

2 2 cos θ3 sin θ3 0 0 0 sinθ3cosθ3

2 2 ε1′1′ ε11 sin θ3 cos θ3 0 0 0 − sinθ3cosθ3 ε2′2′ ε22 0 0 1 0 0 0 ε33 ε3′3′ = 2ε2′3′ 0 0 0 cosθ3 − sinθ3 0 2ε23

2ε1′3′ 2ε13 0 0 0 sinθ3 cosθ3 0 2 12 2ε1′2′ 2 2 ε − 2sinθ3cosθ3 2sinθ3cosθ3 0 0 0 cos θ3 − sin θ3

" Inspection of this matrix equation reveals

2 2 cos θ 3 sin θ 3 sinθ 3cosθ 3 ε1′1′ ε11 2 2 ε2′2′ = sin θ 3 cos θ 3 − sinθ 3cosθ 3 ε22 2 2 2ε1′2′ 2ε12 − 2sinθ 3cosθ 3 2sinθ 3cosθ 3 cos θ 3 − sin θ 3

561 STRESS AND STRAIN TRANSFORMATION EQUATIONS FOR PLANE STRAIN - CONCLUDED

" The last matrix equation is expressed symbolically by

E′′ = Tε θ 3 E where

2 2 cos θ3 sin θ3 sinθ3cosθ3

2 2 sin θ cos θ − sinθ cosθ Tε θ3 ≡≡ 3 3 3 3 2 2 − 2sinθ3cosθ3 2sinθ3cosθ3 cos θ3 − sin θ3

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

− 1 E = Tε θ 3 E′′

562 TRANSFORMED CONSTITUTIVE EQUATIONS FOR PLANE STRAIN

" In terms of another set of coordinates x1′, x2′, x3′ that correspond to a

dextral rotation about the x3 axis, the constitutive equations must also have the forms given as

σ 1′1′ C 1′1′ C 1′2′ C 1′6′ ε 1′1′ β 1′1′ C 1′3′

σ 2′2′ = C 1′2′ C 2′2′ C 2′6′ ε 2′2′ + β 2′2′ T − Tref + C 2′3′ ε and

σ 1′2′ C 1′6′ C 2′6′ C 6′6′ 2ε 1′2′ β 1′2′ C 3′6′

ε 1′1′ s 1′1′ s 1′2′ s 1′6′ σ 1′1′ α 1′1′ S 1′1′ + ε 2′2′ = s 1′2′ s 2′2′ s 2′6′ σ 2′2′ α 2′2′ T − Tref + S 2′2′ ε 2ε σ 1′2′ 2α S 1 2 1′2′ s 1′6′ s 2′6′ s 6′6′ 1′2′ ′ ′

− 1 β 1′1′ C 1′1′ Q 1′2′ C 1′6′ α 1′1′ s 1′1′ s 1′2′ s 1′6′ C 1′1′ C 1′2′ C 1′6′

β 2′2′ = − C 1′2′ C 2′2′ C 2′6′ α 2′2′ s 1′2′ s 2′2′ s 2′6′ = C 1′2′ C 2′2′ C 2′6′ β C C C 2 C C C 1′2′ 1′6′ 2′6′ 6′6′ α 1′2′ s 1′6′ s 2′6′ s 6′6′ 1′6′ 2′6′ 6′6′

563 TRANSFORMED CONSTITUTIVE EQUATIONS FOR PLANE STRAIN - CONTINUED

− 1

S 1′1′ C 1′1′ C 1′2′ C 1′6′ C 1′3′

and S 2′2′ = − C 1′2′ C 2′2′ C 2′6′ C 2′3′

S 1′2′ C 1′6′ C 2′6′ C 6′6′ C 3′6′

" For convenience, let

s 11 s 12 s 16 C 11 C 12 C 16 C 13 C C C C ss = s 12 s 22 s 26 C = 12 22 26 C = 23 C C C C s 16 s 26 s 66 16 26 66 36

α 11 β 11 S 11

α 22 β 22 S 22 T T αα = β = SS = Θ ≡ − ref 2α 12 β 12 S 12 such that the corresponding constitutive equations are given by E = ss Σ + α Θ + S ε and Σ = C E + β Θ + C ε

564 TRANSFORMED CONSTITUTIVE EQUATIONS FOR PLANE STRAIN - CONTINUED

" Additionally, let

s 1′1′ s 1′2′ s 1′6′ C 1′1′ C 1′2′ C 1′6′ C1′3′ C C C C ss′′ = s 1′2′ s 2′2′ s 2′6′ C′′ = 1′2′ 2′2′ 2′6′ C′′ = 2′3′ C C C C s 1′6′ s 2′6′ s 6′6′ 1′6′ 2′6′ 6′6′ 3′6′

β 1′1′ α 1′1′ S 1′1′ β αα′′ = α 2′2′ β′′ = 2′2′ ′′ = S 2′2′ 2 SS α 1′2′ β 1′2′ S 1′2′

such that the corresponding constitutive equations are given by

E′ = ss′ Σ′ + α′ Θ + S′ ε and

Σ′ = C′ E′ + β ′ Θ + C′ ε

565 TRANSFORMED CONSTITUTIVE EQUATIONS FOR PLANE STRAIN - CONTINUED

− 1 − 1 " Substituting Σ = Tσ Σ′ and E = Tε E′ into Σ = C E + β Θ + C ε gives

− 1 − 1 Tσ Σ′ = C Tε E′ + β Θ + C ε

" Premultiplying by Tσ gives

− 1 Σ′ = Tσ C Tε E′ + Tσ β Θ + Tσ C ε

" Comparing this equation with Σ′ = C′ E′ + β ′ Θ + C′ ε it follows that − 1 C′ = Tσ C Tε , β ′ = Tσ β , and C′ = Tσ C

566 TRANSFORMED CONSTITUTIVE EQUATIONS FOR PLANE STRAIN - CONTINUED

− 1 − 1 " Rearranging C′ = Tσ C Tε gives C = Tσ C′ Tε

− 1 − 1 " Likewise, β = Tσ β ′ and C = Tσ C′

− 1 − 1 " Next, substituting Σ = Tσ Σ′ and E = Tε E′ into

E = ss Σ + α Θ + S ε gives

− 1 − 1 + ε Tε E′ = ss Tσ Σ′ + α Θ S

" Premultiplying by Tε gives

− 1 + ε E′ = Tε ss Tσ Σ′ + Tε α Θ Tε S

567 TRANSFORMED CONSTITUTIVE EQUATIONS FOR PLANE STRAIN - CONTINUED

" Comparing the last equation with E′ = ss′ Σ′ + α′ Θ + S′ ε reveals that

− 1 = ss′ Tε ss Tσ , α′ = Tε α , and S′ = Tε S

− 1 − 1 = " Rearranging ss′ = Tε ss Tσ gives ss Tε ss′ Tσ

− 1 − 1 " In addition, α = Tε α′ and S = Tε S′

568 TRANSFORMED CONSTITUTIVE EQUATIONS FOR PLANE STRAIN - CONTINUED

" Noting that for a dextral rotation about the x3 axis, − 1 T − 1 T Tε = Tσ and Tσ = Tε it follows that

T T = = ss′ Tε ss Tε ss Tσ ss′ Tσ

T T C′ = Tσ C Tσ C = Tε C′ Tε

α′ = Tε α α = Tσ α′

β ′ = Tσ β β = Tε β ′

C′ = Tσ C C = Tε C′

S′ = Tε S S = Tσ S′

569 TRANSFORMED CONSTITUTIVE EQUATIONS FOR PLANE STRAIN - CONTINUED

with

2 2 cos θ3 sin θ3 2sinθ3cosθ3

2 2 sin θ cos θ − 2sinθ cosθ Tσ θ3 ≡≡ 3 3 3 3 and 2 2 − sinθ3cosθ3 sinθ3cosθ3 cos θ3 − sin θ3

2 2 cos θ3 sin θ3 sinθ3cosθ3

2 2 sin θ cos θ − sinθ cosθ Tε θ3 ≡≡ 3 3 3 3 2 2 − 2sinθ3cosθ3 2sinθ3cosθ3 cos θ3 − sin θ3

" 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

570 TRANSFORMED CONSTITUTIVE EQUATIONS FOR PLANE STRAIN - CONCLUDED

T = " For plane stress, Q ′ Tσ Q Tσ , with m = cosθ3 and

n = sinθ3 , gave

4 2 2 2 2 4 Q 1′1′ = m Q 11 + 2m n Q 12 + 2Q 66 + 4mn m Q 16 + n Q 26 + n Q 22

T " Thus, by similarity, C′ = Tσ C Tσ gives

4 2 2 2 2 4 C 1′1′ = m C 11 + 2m n C 12 + 2C 66 + 4mn m C 16 + n C 26 + n C 22

" The other transformation equations are obtained in a similar manner

571 LINES AND CURVES OF MATERIAL SYMMETRY

572 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

573 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

574 ! ! ! ! !

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

575 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

Plane of material Curve of material symmetry at point P symmetry Tangent vector at point P P

Q Plane of material symmetry at point Q Coordinate curve C Tangent vector Surface of at point Q material points Coordinate curve

576 BIBLIOGRAPHY

577 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.: Advanced 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.

578 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, Part 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., Englewood Cliffs, New Jersey, 1965.

Graff, K. F.: Wave 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.

579 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 Engineers. CRC Press, Inc., 1992.

580 BIBLIOGRAPHY - CONTINUED

Noor, A. K.; Kim, Y. H., and Peters, J. M.: Strain Energy of Thermally- Stressed 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.

581 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.

582 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.

583 Form Approved REPORT DOCUMENTATION PAGE OMB No. 0704-0188 The public reporting burden for this collection of information is estimated to average 1 hour per response, including the time for reviewing instructions, searching existing data sources, gathering and maintaining the data needed, and completing and reviewing the collection of information. Send comments regarding this burden estimate or any other aspect of this collection of information, including suggestions for reducing this burden, to Department of Defense, Washington Headquarters Services, Directorate for Information Operations and Reports (0704-0188), 1215 Jefferson Davis Highway, Suite 1204, Arlington, VA 22202-4302. Respondents should be aware that notwithstanding any other provision of law, no person shall be subject to any penalty for failing to comply with a collection of information if it does not display a currently valid OMB control number. PLEASE DO NOT RETURN YOUR FORM TO THE ABOVE ADDRESS. 1. REPORT DATE (DD-MM-YYYY) 2. REPORT TYPE 3. DATES COVERED (From - To) 01- 12 - 2011 Technical Memorandum 4. TITLE AND SUBTITLE 5a. CONTRACT NUMBER

An In-depth Tutorial on Constitutive Equations for Elastic Anisotropic 5b. GRANT NUMBER Materials

5c. PROGRAM ELEMENT NUMBER

6. AUTHOR(S) 5d. PROJECT NUMBER

Nemeth, Michael P. 5e. TASK NUMBER

5f. WORK UNIT NUMBER 869021.04.07.01.13 7. PERFORMING ORGANIZATION NAME(S) AND ADDRESS(ES) 8. PERFORMING ORGANIZATION REPORT NUMBER NASA Langley Research Center Hampton, VA 23681-2199 L-20083

9. SPONSORING/MONITORING AGENCY NAME(S) AND ADDRESS(ES) 10. SPONSOR/MONITOR'S ACRONYM(S) National Aeronautics and Space Administration NASA Washington, DC 20546-0001 11. SPONSOR/MONITOR'S REPORT NUMBER(S) NASA/TM-2011-217314 12. DISTRIBUTION/AVAILABILITY STATEMENT Unclassified Unlimited Subject Category 39 Availability: NASA CASI (443) 757-5802 13. SUPPLEMENTARY NOTES

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

16. SECURITY CLASSIFICATION OF: 17. LIMITATION OF 18. NUMBER 19a. NAME OF RESPONSIBLE PERSON ABSTRACT OF a. REPORT b. ABSTRACT c. THIS PAGE PAGES STI Help Desk (email: [email protected]) 19b. TELEPHONE NUMBER (Include area code) U U U UU 588 (443) 757-5802 Standard Form 298 (Rev. 8-98) Prescribed by ANSI Std. Z39.18