Structural Analysis of Polymeric Composite Materials 0-8247-4871-9

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Although great care has been taken to provide accurate and current information, neither the author(s) nor the publisher, nor anyone else associated with this publication, shall be liable for any loss, damage, or liability directly or indirectly caused or alleged to be caused by this book. The material contained herein is not intended to provide specific advice or recommendations for any specific situation. Trademark notice: Product or corporate names may be trademarks or registered trademarks and are used only for identification and explanation without intent to infringe. Library of Congress Cataloging-in-Publication Data A catalog record for this book is available from the Library of Congress. ISBN: 0-8247-4717-8 This book is printed on acid-free paper. Headquarters Marcel Dekker, Inc., 270 Madison Avenue, New York, NY 10016, U.S.A. tel: 212-696-9000; fax: 212-685-4540 Distribution and Customer Service Marcel Dekker, Inc., Cimarron Road, Monticello, New York 12701, U.S.A. tel: 800-228-1160; fax: 845-796-1772 Eastern Hemisphere Distribution Marcel Dekker AG, Hutgasse 4, Postfach 812, CH-4001 Basel, Switzerland tel: 41-61-260-6300; fax: 41-61-260-6333 World Wide Web http://www.dekker.com The publisher offers discounts on this book when ordered in bulk quantities. For more information, write to Special Sales/Professional Marketing at the headquarters address above. Copyright n 2004 by Marcel Dekker, Inc. All Rights Reserved. Neither this book nor any part may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopying, microfilming, and recording, or by any information storage and retrieval system, without permission in writing from the publisher. Current printing (last digit): 10 9 8 7 6 5 4 3 2 1 PRINTED IN THE UNITED STATES OF AMERICA

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To my wife, Lisa, my daughters, Stephanie and Jennifer, my son, Daniel, my parents, James and Rita, and my brothers and sisters, Mathew, Ann, David, and Claire.

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Preface

This textbook is intended for use in a senior-year undergraduate or first-year graduate course devoted to structural analyses of laminated polymer-matrix composite materials and structures. Discussion is framed almost entirely at the macromechanical (structural) level. Micromechanical issues and analyses are discussed briefly but are not covered in detail. This allows an expanded coverage of the structural response of composite beams and plates, as compared to other introductory texts on composite materials. The text contains ample material for a semester-based (15-week) course. I have used a similar manuscript for several years to support two sequential quarter-based (10week) courses, supplementing this material with one or two laboratory sessions. Since laboratory exercises depend heavily on the equipment and materials available to the instructor, these lab sessions are not described in this book. It is assumed that the reader has already completed a sophomore- or junior-level course devoted to the mechanics of isotropic solids. I have made every effort to extend the concepts of this earlier coursework in a natural and easily understandable way to the study of anisotropic composites. Chapter 1 begins with a broad overview of the various types of commercially available metal, ceramic, and polymer-based composite materials. This chapter also includes an overview of polymer and fibrous materials and the manufacturing processes used to produce polymer composites and v

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structures. Chapter 2 is devoted to a review of force, stress, and strain tensors and how these tensors may be transformed from one coordinate system to another. Although these topics are normally covered in a course on the mechanics of isotropic solids, it has been my experience that a review of these fundamental concepts is almost always required before they can be correctly applied to the study of anisotropic composites. Various material properties required to predict the structural performance of anisotropic composites are introduced in Chapter 3, and the threedimensional, anisotropic form of Hooke’s law is developed in Chapter 4. The three-dimensional form of Hooke’s law is then reduced to the plane stress (two-dimensional) form in Chapter 5 and applied to unidirectional laminated composites. Rudimentary elements of plate theory are developed in Chapter 6 and combined with Hooke’s law, resulting in the analysis methodology commonly known as classical lamination theory (CLT). Chapter 7 describes composite failure modes and mechanisms, including a qualitative description of composite fatigue behaviors and free-edge effects. The chapter also presents methods of combining macroscopic failure criteria with CLT to predict first-ply and last-ply failure loads. Chapter 8 is devoted to statically determinate and indeterminate composite beams. The chapter begins with the observation that CLT reduces to fundamental beam theory (as studied during an earlier course on the mechanics of isotropic solids) when isotropic properties and appropriate dimensions are assumed. CLT is then used to predict the effective axial and bending rigidities of composite beams with various cross sections (rectangular, I-, T-, hat-, and box-beams). Chapters 9 through 11 address composite plates. Discussion is limited to symmetric rectangular composite plates, since this topic is extensive and a complete discussion of nonrectangular and/or nonsymmetric composite plates or shells deserves a separate text in itself. The equations that govern the behavior of symmetric and rectangular composite plates are developed in Chapter 9. Chapter 10 presents several closed-form solutions for specially orthotropic composite plates. Chapter 11 presents approximate numerical solutions for generally orthotropic plates (e.g., quasi-isotropic plates); this includes solutions for the deflections of transversely loaded plates as well as mechanical and/or thermal buckling due to in-plane loads. Three appendixes that include material referenced in the main body of the book complete the text, with the second briefly describing experimental methods used to measure in-plane composite properties. While many of these topics are covered in other introductory composites textbooks, I have included here certain pedagogical features that have, in my experience, facilitated and enhanced an understanding of the concepts. For example, the crucially important effects of environment—in particular,

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the effects due to changes in temperature and/or moisture content—are integrated throughout Chapters 3 through 11, rather than being confined to later chapters, as is often the case in composite textbooks. With the exception of Chapters 1 and 9, each chapter includes numerical example problems that illustrate the concepts presented. A solutions manual for all homework problems posed in the text is available for educators using the text in their courses. I have also created a suite of computer programs that implement the analyses discussed. These executable programs may be downloaded free of charge from the following website: http : ==depts:washington:edu=amtas=computer:html These programs are meant to enhance the text and are referenced at appropriate points throughout the book. Of course, composite computer programs are now widely available, both commercially and otherwise, so the reader may opt to use resources other than those downloaded from the website provided. Preparation of this book has been a demanding and lengthy endeavor. I sincerely hope that it will be a worthy addition to the composites literature. Mark E. Tuttle

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Acknowledgments

I have been fortunate to have worked with talented and inspirational colleagues throughout my academic and professional career. I would especially like to acknowledge the lifelong influence of my two major academic advisors, Professor Emeritus Halbert F. Brinson of the University of Houston (formerly of Virginia Polytechnic Institute and State University) and Professor John B. Ligon of Michigan Technological University. There are many others whom I would like to mention by name, but space does not allow it. To my friends and colleagues at Battelle Columbus Laboratories, Michigan Tech, Virginia Tech, the University of Washington, NASA–Langley Research Center, the Boeing Company, and the Society for Experimental Mechanics: thank you. Particular thanks are extended to Mr. Rob Albers of the Boeing Company, who read and gave helpful critiques of initial versions of Chapters 9 through 11. I would also like to thank the staff at Marcel Dekker, Inc.—in particular Michael Deters, Production Editor, and John Corrigan, Acquisitions Editor—for their very professional and competent help throughout preparation of the final manuscript. Finally, to the many undergraduate and graduate students who have taken my composites courses or worked with me during the pursuit of their degrees and who consequently struggled through and ‘‘edited’’ manuscript versions of this textbook: thank you for your help and patience. Mark E. Tuttle ix

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Contents

Preface Acknowledgments

1. Introduction 1. 2. 3. 4. 5. 6.

Basic Definitions Polymeric Materials Fibrous Materials Commercially Available Forms Manufacturing Processes The Scope of This Book References

2. Review of Force, Stress, and Strain Tensors 1. 2. 3.

The Force Vector Transformation of a Force Vector Normal Forces, Shear Forces, and Free-Body Diagrams 4. Definition of Stress

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5. 6. 7. 8. 9. 10. 11. 12. 13.

The Stress Tensor Transformation of the Stress Tensor Principal Stresses Plane Stress Definition of Strain The Strain Tensor Transformation of the Strain Tensor Principal Strains Strains Within a Plane Perpendicular to a Principal Strain Direction 14. Relating Strains to Displacement Fields 15. Computer Programs 3DROTATE and 2DROTATE Homework Problems References 3. Material Properties 1. 2. 3.

Anisotropic vs. Isotropic Materials Material Properties That Relate Stress to Strain Material Properties Relating Temperature to Strain 4. Material Properties Relating Moisture Content to Strain 5. Material Properties Relating Stress (or Strain) to Failure 6. Predicting Elastic Composite Properties Based on Constituents: The Rule of Mixtures Homework Problems References 4. Elastic Response of Anisotropic Materials 1. 2.

Strains Induced by Stress: Anisotropic Materials Strains Induced by Stress: Orthotropic and Transversely Isotropic Materials 3. Strains Induced by a Change in Temperature or Moisture Content 4. Strains Induced by Combined Effects of Stress, Temperature, and Moisture Homework Problems References

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5. Unidirectional Composite Laminates Subject to Plane Stress 1. 2. 3. 4. 5. 6. 7.

Unidirectional Composites Referenced to the Principal Material Coordinate System Unidirectional Composites Referenced to an Arbitrary Coordinate System Calculating Transformed Properties Using Material Invariants Effective Elastic Properties of a Unidirectional Composite Laminate Failure of Unidirectional Composites Referenced to the Principal Material Coordinate System Failure of Unidirectional Composites Referenced to an Arbitrary Coordinate System Computer Programs UNIDIR and UNIFAIL Homework Problems References

6. Thermomechanical Behavior of Multiangle Composite Laminates 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11.

Definition of a ‘‘Thin Plate’’ and Allowable Plate Loadings Plate Deformations: The Kirchhoff Hypothesis Principal Curvatures Standard Methods of Describing Composite Laminates Calculating Ply Strains and Stresses Classical Lamination Theory (CLT) Simplifications Due to Stacking Sequence Summary of CLT Calculations Effective Properties of a Composite Laminate Transformation of the ABD Matrix Computer Program CLT Homework Problems References

7. Predicting Failure of a Multiangle Composite Laminate 1. 2. 3. 4.

Preliminary Discussion Free-Edge Stresses Predicting Laminate Failure Using CLT Laminate First-Ply Failure Envelopes

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5.

Computer Program LAMFAIL Homework Problems References

8. Composite Beams 1. 2. 3. 4. 5. 6. 7. 8. 9.

Preliminary Discussion Comparing Classical Lamination Theory to Isotropic Beam Theory Types of Composite Beams Considered Effective Axial Rigidity of Rectangular Composite Beams Effective Flexural Rigidities of Rectangular Composite Beams Effective Axial and Flexural Rigidities for Thin-Walled Composite Beams Statically Determinate and Indeterminate Axially Loaded Composite Beams Statically Determinate and Indeterminate Transversely Loaded Composite Beams Computer Program BEAM Homework Problems References

9. The Governing Equations of Thin-Plate Theory 1. 2.

Preliminary Discussion The Equations of Equilibrium for Symmetric Laminates 3. Boundary Conditions 4. Representing Arbitrary Transverse Loads as a Fourier Series References 10. Some Exact Solutions for Specially Orthotropic Laminates 1. 2. 3. 4.

Equations of Equilibrium for a Specially Orthotropic Laminate In-Plane Displacement Fields in Specially Orthotropic Laminates Specially Orthotropic Laminates Subject to Simple Supports of Type S1 Specially Orthotropic Laminates Subject to Simple Supports of Type S4

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5.

6.

7.

8.

9.

Specially Orthotropic Laminates With Two Simply Supported Edges of Type S1 and Two Edges of Type S2 The Navier Solution Applied to a Specially Orthotropic Laminate Subject to Simple Supports of Type S4 Buckling of Rectangular Specially Orthotropic Laminates Subject to Simple Supports of Type S4 Thermal Buckling of Rectangular Specially Orthotropic Laminates Subject to Simple Supports of Type S1 Computer Program SPORTHO References

11. Some Approximate Solutions for Symmetric Laminates 1. 2. 3. 4.

Preliminary Discussion In-Plane Displacement Fields Potential Energy in a Thin Composite Plate Symmetric Composite Laminates Subject to Simple Supports of Type S4 5. Buckling of Symmetric Composite Plates Subject to Simple Supports of Type S4 6. Computer Program SYMM References

Appendixes A. Finding the Cube-Root of a Complex Number B. Experimental Methods Used to Measure In-Plane Properties E11, E22, r12, and G12 C. Tables of Beam Deflections and Slopes

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1 Introduction

A broad overview of modern composite materials is provided in this chapter. The chapter begins with a definition of what is meant by the phrase ‘‘composite material.’’ Separate sections devoted to polymeric materials, fibrous materials, commercially available forms, and manufacturing techniques are included. The chapter concludes with a section indicating the scope of the remaining chapters. 1

BASIC DEFINITIONS

This textbook is devoted to a special class of structural materials often called ‘‘advanced’’ composites. Just what is a ‘‘composite material’’? A casual definition might be: ‘‘A composite material is one in which two (or more) materials are bonded together to form a third material.’’ Although not incorrect, upon further reflection, it becomes clear that this definition is far too broad because it implies that essentially all materials can be considered as ‘‘composites.’’ For example, the (nominal) composition of the 2024 aluminum alloy is 93.5% Al, 4.4% Cu, 0.6% Mn, and 1.5% Mg [1]. Hence, according to the broad definition stated above, this common aluminum alloy could be considered as a ‘‘composite’’ because it consists of four materials (aluminum, copper, manganese, and magnesium) bonded together at the atomic level to form the 2024 alloy. In a similar sense, virtually all metal alloys, polymers, and

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ceramics satisfy this broad definition of a ‘‘composite’’ because all of these materials contain more than one type of elemental atom. An important characteristic that is missing in the initial broad definition is a consideration of physical scale. Another definition of a ‘‘composite material,’’ which includes a reference to a physical scale appropriate for present purposes, is as follows: A composite material is a material system consisting of two (or more) materials, which are distinct at a physical scale greater than about 110 6 m (1 Am) and which are bonded together at the atomic and/or molecular levels. As a point of reference, the diameter of human hair ranges from about 30 to 60 Am. Objects of this size are easily seen with the aid of an optical microscope. Hence, when composite materials are viewed under an optical microscope, the distinct constituent materials (or distinct material phases) that form the composite are easily distinguished. Structural composites typically consist of a high-strength, high-stiffness reinforcing material, embedded within a relatively low-strength, low-stiffness matrix material. Ideally, the reinforcing and matrix materials interact to produce a composite whose properties are superior to either of the two constituent materials alone. Many naturally occurring materials can be viewed as composites. A good example is wood and laminated wood products. Wood is a natural composite, with a readily apparent grain structure. Wood exhibits a higher stiffness and strength parallel to the grains than transverse to the grains. In laminated wood products (which range from the large laminated beams used in a church cathedral to a common sheet of plywood), relatively thin layers of wood are adhesively bonded together. In plywood, the layers are arranged such that the grain direction varies from one layer to the next. Therefore, this laminated wood product has a high stiffness/strength in more than one direction. Although composites have been used in a variety of structural applications for centuries, modern (or ‘‘advanced’’) composites are a relatively recent development, having been in existence for about 60 years. Modern composites may be classified according to the size or shape of the reinforcing material used. Four common classifications of reinforcements are: 

Particulates, which are roughly spherical particles with diameters typically ranging from about 1 to 100 Am  Whiskers, with lengths less than about 10 mm  Short (or ‘‘chopped’’) fibers, with a length ranging from about 10 to 200 mm  Continuous fibers, whose length are, in effect, infinite.

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Whiskers, short fibers, and continuous fibers all have very small diameters relative to their length; the diameter of these products ranges from about 5 to 200 Am. Distinctly different types of composites can be produced using any of the above reinforcements. For example, three types of composites based on continuous fibers are shown in Fig. 1: unidirectional composites, woven composites, and braided composites. In a unidirectional composite, all fibers are aligned in the same direction and embedded within a matrix material. In contrast, woven composites are formed by first weaving continuous fibers into a fabric and then embedding the fabric in a matrix. Hence, a single layer of a woven composite contains fibers in two orthogonal directions. In contrast, a single layer in a braided composite typically contains two or three

Figure 1 Different types of composites based on continuous fibers.

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nonorthogonal fiber directions. Braided composites are then formed by embedding the fabric in a matrix (additional discussion of these types of composites is provided in Sec. 4). As implied in Fig. 1, composite products based on continuous fibers are usually produced in the form of thin layers. A single layer of these products is called a lamina or ply. The thickness of a single ply formed using unidirectional fibers ranges from about 0.12 to 0.20 mm, whereas the thickness of a single ply of a woven or braided fabric ranges from about 0.25 to 0.40 mm. Obviously, a single composite ply is quite thin. To produce a composite structure with significant thickness, many plies are stacked together to form a composite laminate. Conceptually any number of plies may be used in the laminate, but in practice, the number of plies usually ranges from about 10 plies to (in unusual cases) perhaps as many as 200 plies. The fiber represents the reinforcing material in these composites. Hence, the orientation of the fibers is, in general, varied from one ply to the next, so as to provide high stiffness and strength in more than one direction (as is the case in plywood). It is also possible to use unidirectional, woven, and/or braided plies within the same laminate. For example, it is common to use a woven or braided fabric as the two outermost facesheets of a laminate, and to use unidirectional plies at interior positions. In all composites, the reinforcement is embedded within a matrix material. The matrix may be polymeric, metallic, or ceramic. In fact, composite materials are often classified on the basis of the matrix material used, rather than the reinforcing material. That is, modern composites can be categorized into three main types: polymer–matrix, metal–matrix, or ceramic– matrix composite materials. Usually, the reinforcing material governs the stiffness and strength of a composite. On the other hand, the matrix material usually governs thermal stability. Polymeric–matrix composites are used in applications involving relatively modest temperatures (service temperatures of, say, 200jC or less). Metal–matrix composites are used at temperatures up to about 700jC, while ceramic–matrix composites are used at ultra-high temperatures (up to about 1200jC or greater). The matrix also defines several additional characteristics of the composite material system. Some additional roles of the matrix are: 

To provide the physical form of the composite To bind the fibers together  To protect the fibers from aggressive (chemical) environments or mechanical damage (e.g., due to abrasion, for example)  To transfer and redistribute stresses between fibers, between plies, and in areas of discontinuities in load or geometry. 

To summarize the preceding discussion, there are many types of composite materials, both natural and man-made. Composites can be clas-

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sified according to the physical form of the reinforcing material (particulate, whisker, short fiber, or continuous fiber reinforcement), by the type of matrix material used (metal, ceramic, or polymeric matrix), by the orientation of the reinforcement (unidirectional, woven, or braided), or by some combination thereof. The temperature the composite material/structure will experience in service often dictates the type of composite used in a given application. The primary focus of this textbook is the structural analysis of polymeric composite materials and structures. Metal–matrix and ceramic–matrix composites will not be further discussed, although many of the analysis methods developed herein may be applied to these types of composites as well. Because our focus is the structural analysis of polymeric composites, we will not be greatly considered with the behavior of the individual constituent materials. That is, we will not be greatly concerned with the behavior of an unreinforced polymer, nor with the behavior of an individual reinforcing fiber. Instead, we will be concerned with the behavior of the composite formed by combining these two constituents. Nevertheless, a structural engineer who wishes to use polymeric composites effectively in practice must understand at least the rudiments of polymer and fiber science, in much the same way as an engineer working with metal alloys must understand at least the rudiments of metallurgy. Toward that end, a brief introduction to polymeric and fibrous materials is provided in Secs. 2 and 3, respectively. At the minimum, the reader should become acquainted with the terminology used to describe polymeric and fibrous materials because such terms have naturally been carried over to the polymeric composites technical community.

2

POLYMERIC MATERIALS

A brief introduction to polymeric materials is provided in this section. This introduction is necessarily incomplete. The reader interested in a more detailed discussion is referred to the many available texts and/or web-based resources devoted to modern polymers (e.g., see Refs. 1–4). 2.1 Basic Concepts The term ‘‘polymer’’ comes from the Greek words poly (meaning ‘‘many’’) and mers (meaning ‘‘units’’). Quite literally, a polymer consists of ‘‘many units.’’ Polymer molecules are made up of thousands of repeating chemical units and have molecular weights ranging from about 103 to 107. As an illustrative example, consider the single chemical mer shown in Fig. 2. This mer is called ethylene (or ethene), and consists of two carbon atoms and four hydrogen atoms. The two lines between the carbon (C) atoms

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Figure 2 The monomer ethylene.

indicate a double covalent bond,* whereas the single line between the hydrogen (H) and carbon atoms represents a single covalent bond. The chemical composition of the ethylene mer is sometimes written as C2H4 or CH2jCH2. Under proper conditions, the double covalent bond between the two carbon atoms can be converted to a single covalent bond, which allows each of the two carbon atoms to form a new covalent bond with a suitable neighboring atom. A suitable neighboring atom would be a carbon atom in a neighboring ethylene mer, for example. If ‘‘n’’ ethylene mers join together in this way, the chemical composition of the resulting molecule can be represented C2nH4n, where n is any positive integer. Hence, a ‘‘chain’’ of ethylene mers joins together to form the well-known polymer, polyethylene (literally, ‘‘many ethylenes’’), as shown in Fig. 3. The process of causing a monomer to chemically react and form a long molecule in this fashion is called polymerization. The single ethylene unit is an example of a monomer (‘‘one mer’’). At room temperature, a bulk sample of the ethylene monomer is a low-viscosity fluid. If two ethylene monomers bond together, the resulting chemical entity has two repeating units and is called a dimer. Similarly, the chemical entity formed by three repeating units is called a trimer. The molecular weight of a dimer is twice that of the monomer, the molecular weight of a trimer is three times that of the monomer, etc. Prior to polymerization, most polymeric materials exist as relatively low-viscosity fluids known as oligomers (‘‘a few mers’’). The individual molecules within an oligomer possess a range of molecular weights, typically containing perhaps 2–20 mers. It should be clear from the above discussion that a specific molecular weight cannot be assigned to a polymer. Rather, the molecules within a bulk sample of a polymer are of differing lengths and hence exhibit a range in molecular weight. The average molecular weight of a bulk sample of a polymer is increased as the polymerization process is initiated and progresses. Another

* As fully described in an introductory chemistry text, a ‘‘covalent bond’’ is formed when two atoms share an electron pair, so as to fill an incompletely filled valence level.

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Figure 3 The polymer polyethylene.

measure of the ‘‘size’’ of the polymeric molecule is the degree of polymerization, defined as the ratio of the average molecular weight of the polymer molecule divided by the molecular weight of the repeating chemical unit within the molecular chain. The average molecular weight of a polymeric sample (or, equivalently, the degree of polymerization) depends on the conditions under which it was polymerized. Now, all physical properties exhibited by a polymer (e.g., strength, stiffness, density, thermal expansion coefficient, etc.) are dictated by the average molecular weight. Therefore, a fundamental point that must be appreciated by the structural engineer is that the properties exhibited by any polymer (or any polymeric composite) depend on the circumstances under which it was polymerized. As a general rule, the volume of a bulk sample of a monomer decreases during polymerization. That is, the bulk sample shrinks as the polymerization process proceeds. This may have serious ramifications if the polymer is to be used in structural applications. For example, if a fiber(s) is embedded within the sample during the polymerization process (as is the case for some fiberreinforced polymeric composite systems), then shrinkage of the matrix causes residual stresses to develop during polymerization. This effect contributes to so-called cure stresses, which are present in most polymeric composites. As will be seen in later chapters, cure stresses arise from two primary sources. The first is shrinkage of the matrix during polymerization, as just described. The second is stresses that arise due to temperature effects. In this case, the composite is polymerized at an elevated temperature (say, 175jC) and then cooled to room temperature (20jC). The thermal expansion coefficient of the matrix is typically much higher than the fibers, and so during cooldown, the matrix is placed in tension while the fiber is placed in compression. Cure stresses due to shrinkage during cure and/or temperature effects can be quite high relative to the strength of the polymer itself, and ultimately contribute toward failure of the composite.

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2.2 Addition vs. Condensation Polymers Although in the case of polyethylene the repeat unit is equivalent to the original ethylene monomer, this is not always the case. In fact, in many instances, the repeat unit is derived from two (or more) monomers. A typical example is nylon 6,6. A typical polymerization process for this polymer is shown schematically in Fig. 4. Two monomers are used to produce nylon 6,6: hexamethylene diamine (chemical composition: C6H16N2) and adipic acid (chemical composition: CO2H(CH2)4CO2H). Note that the repeat unit of nylon 6,6 (hexamethylene adipamide) is not equivalent to either of the two original monomers. A low-molecular-weight by-product (i.e., H2O) is produced during the polymerization of nylon 6,6. This is characteristic of condensation polymers. That is, if both a high-molecular-weight polymer as well as a low-molecularweight by-product are formed during the polymerization process, the polymer is classified as a condensation polymer. Conversely, addition polymers are those for which no by-product is formed, which implies that all atoms present in the original monomer(s) occur somewhere within the repeat unit. Generally speaking, condensation polymers shrink to a greater extent during the polymerization process than do addition polymers. Residual stresses caused by shrinkage during polymerization are often a concern in structural composites, and hence difficulties with residual stresses can often be minimized if an addition polymer is used in these applications.

Figure 4 The polymer nylon 6,6.

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2.3 Molecular Structure The molecular structure of a fully polymerized polymer can be roughly grouped into one of three major types: linear, branched, or crosslinked polymers. The three types of molecular structure are shown schematically in Fig. 5. Linear polymers can be visualized as beads on a string, where each bead represents a repeat unit. It should again be emphasized that the length of these ‘‘strings’’ is enormous; if a typical linear molecule were scaled up to be 10 mm in diameter, it would be roughly 4 km long. In a bulk sample, these long macromolecules become entangled and twisted together, much like a bowl of cooked spaghetti. Obviously, as the molecular weight (i.e., the length) of the polymer molecule is increased, the number of entanglements is increased. At the macroscopical scale, the stiffness exhibited by a bulk polymer is directly related to its molecular weight and number of entanglements. If all of the repeat units within a linear polymer are identical, the polymer is called a homopolymer. Polyethylene is a good example of a linear homopolymer. However, it is possible to produce linear polymers that consist of two separate and distinct repeat units. Such materials are called copolymers. In linear random copolymers, the two distinct repeat units appear randomly along the backbone of the molecule. In contrast, for linear block copolymers, the two distinct repeat units form long continuous segments within the polymer chain. A good example of a common copolymer is acrylonitrile–butadiene–styrene, commonly known as ‘‘ABS.’’ The second major type of polymeric molecular structure is the branched polymer (see Fig. 5). In branched polymers, relatively short side chains are bonded to the primary backbone of the macromolecule by means of a covalent bond. As before, the stiffness of a bulk sample of a branched polymer is related to the number of entanglements between molecules. Because the branches greatly increase the number of entanglements, the macroscopical stiffness of a branched polymer will, in general, be greater than the macroscopical stiffness of a linear polymer of similar molecular weight. In many branched polymers, the branches consist of the same chemical repeat unit as the backbone of the molecular chain. However, in some cases, the branch may have a distinctly different chemical repeat unit than the main backbone of the molecule. Such materials are called graft copolymers. Finally, the third major type of molecular structure is the crosslinked or network polymer (see Fig. 5). During polymerization of such polymers, a crosslink (i.e., a covalent bond) is formed between individual molecular chains. Hence, once polymerization is complete, a vast molecular network is formed. In the limit, a single ‘‘molecule’’ can no longer be identified. A bulk sample of a highly crosslinked polymer may be thought of as a single molecule.

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Figure 5 Types of polymer molecular structure.

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Linear and branched polymers can be visualized as a bowl of cooked spaghetti. One can imagine that a single spaghetti noodle could be extracted without damage from the bowl if the noodle were pulled slowly and carefully, allowing the noodle to ‘‘slide’’ past its neighbors. In much the same way, an individual molecule could also be extracted (at least conceptually) from a bulk sample of a linear or branched polymer. However, this is not the case for a fully polymerized crosslinked polymer. Because the ‘‘individual’’ molecular chains within a crosslinked polymer are themselves linked together by covalent bonds, the entire molecular network can be considered to be a single molecule. Although regions of the chain in a crosslinked polymer may slide past each other, eventually, relative motion between segments is limited by the crosslinks between segments. 2.4 Thermoplastic vs. Thermoset Polymers Suppose a bulk sample of a linear or branched polymer exists as a solid material at room temperature and is subsequently heated. Due to the increase in thermal energy, the average distance between individual molecular chains is increased as temperature is increased. This results in an increase in molecular mobility and a decrease in macroscopical stiffness. That is, as the molecules move apart, both the forces of attraction between individual molecules as well as the degree of entanglement decrease, resulting in a decrease in stiffness at the macroscopical level. Eventually, a temperature is reached at which the polymeric molecules can slide freely past each other and the polymer ‘‘melts.’’ Typically, melting does not occur at a single temperature, but rather over a temperature range of about 15–20jC. A polymer that can be melted (i.e., a linear or branched polymer) is called a thermoplastic polymer. The molecular structure of a thermoplastic polymer may be amorphous or semicrystalline. The molecular structure of an amorphous thermoplastic is completely random (i.e., the molecular chains are randomly oriented and entangled, with no discernible pattern). In contrast, in a semicrystalline thermoplastic, there exist regions of highly ordered molecular arrays. An idealized representation of a crystalline region is shown in Fig. 6. As indicated, in the crystalline region, the main backbone of the molecular chain undulates back and forth such that the thickness of the crystalline region is usually (about) 100 A˚. The crystalline region may extend over an area with a length dimension ranging from (about) 1000 to 10,000 A˚. Hence, the crystalline regions are typically platelike. The high degree of order within the crystalline array allows for close molecular spacing, and hence exceptionally high bonding forces between molecules in the crystalline region. At the macroscale, a semicrystalline thermoplastic typically has a higher strength, stiffness, and density than an otherwise comparable

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Figure 6 An idealized representation of a crystalline region in a thermoplastic polymer.

amorphous thermoplastic. No thermoplastic is completely crystalline, however. Instead, regions of crystallinity are surrounded by amorphous regions, as shown schematically in Fig. 7. Most semicrystalline thermoplastic are 10–50% amorphous (by volume). As just described, a thermoplastic can be melted. In contrast, a crosslinked polymer cannot be melted. If a crosslinked polymer is heated, it will exhibit a decrease in stiffness at the structural level because the average distance between individual segments of the molecular network is, in fact, increased as temperature is increased. However, the crosslinks do not allow indefinite relative motion between segments, and eventually limit molecular motion. Therefore, a crosslinked polymer will not melt. Of course, if the temperature is raised high enough, the covalent bonds that form both the backbone of the molecular chains as well as the crosslinks are broken, chemical degradation occurs, and the polymer is destroyed. A polymer that cannot be melted (i.e., a crosslinked polymer) is called a thermoset polymer. Three more or less distinct conditions are recognized during polymerization of a thermoset polymer. The original resin or oligomer is typically a low-viscosity, low-molecular-weight fluid, containing molecules with perhaps 2–10 repeat units. A thermoset resin is said to be A-staged when in this form. As the polymerization process is initiated (by the introduction of a catalyst, by an increase in temperature, by the application of pressure, or by some combination thereof), molecular weight and viscosity increase rapidly. If the polymerization process is then halted in some manner (say, by suddenly reducing the temperature), the polymerization process will stop (or be

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Figure 7 Overall molecular structure of a semicrystalline thermoplastic, showing crystalline and amorphous regions.

dramatically slowed) and the polymer will exist in an intermediate stage. The thermoset resin is said to be B-staged when in this form. If the polymerization process is allowed to resume (say, by reheating) and continue until the maximum possible molecular weight has been reached, the thermoset is said to be C-staged (i.e., the polymer is fully polymerized). Suppliers of composites based on thermoset polymers initially B-stage their product and sell it to their customers in this form. This requires that the B-staged composite be stored by the customer for months at low temperatures (typically at temperatures below about 15jC or 0jF). Refrigeration is required so that the thermosetting resin does not polymerize beyond the Bstage during storage. The polymerization process is reinitiated and completed (i.e., the composite is C-staged) during the final fabrication of a composite part, typically through the application of heat and pressure. Most commercially available thermoset composites are C-staged (or ‘‘cured’’) at a temperature of either 120jC or 175jC. In contrast, composites based on thermoplastic polymers do not require refrigeration. In this case, the matrix is fully polymerized when delivered to the customer, and may be stored for months or years without degradation. Heat and pressure are applied during the final fabrication of a thermoplastic composite part, but no chemical reaction occurs. That is, heat is applied

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simply to soften/melt the thermoplastic matrix, and pressure is applied to insure consolidation of the composite part. The temperature required to soften/melt thermoplastic composites is usually 350jC or higher. Note that the temperatures required to process thermoplastic composites are significantly higher than those required to cure thermoset composites. 2.5 The Glass Transition Temperature Stiffness and strength are physical properties of obvious importance to the structural engineer. These properties are temperature-dependent for all materials, but this is particularly true for polymers. The effect of temperature on the stiffness of a polymer is summarized in Fig. 8. Thermoset and thermoplastic polymers exhibit the same general behavior, except that at high temperatures, thermoplastics melt whereas thermosets do not. All polymers exhibit a decrease in stiffness near a characteristic temperature called the glass transition temperature, Tg. At temperatures well below the Tg, polymer stiffness decreases slowly with an increase in temperature. At temperatures well below the Tg, polymers are ‘‘glassy’’ and brittle. In contrast, at temperatures well above the Tg, all polymers are ‘‘rubbery’’ and ductile. Thus, the Tg denotes the transition between glassy and rubbery behaviors. This transition is associated with a sudden increase in mobility of segments within the molecular chain, and typically occurs over a range of 10– 15jC. At temperatures well below the Tg, the polymer molecules are closely packed together and tightly bonded, and cannot easily slide past each other.

Figure 8 Effects of temperature on polymer stiffness.

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Consequently, the polymer is ‘‘glassy’’ and exhibits high stiffness and strength but relatively low ductility. Conversely, at temperatures well above the Tg, the molecular spacing is increased such that the molecular chains (or segments of those chains) are mobile and can readily slide past each other. Consequently, the polymer is ‘‘rubbery’’ and exhibits relatively lower stiffness and strength but higher ductility. As implied in Fig. 8, for amorphous thermoplastics, the change in stiffness (and other physical properties) that occurs as the Tg is reached may be one to two orders of magnitude. This astonishing decrease in stiffness occurs over a temperature range of only a few degrees. A similar change in properties occurs for semicrystalline thermoplastics and crosslinked thermosets, although, in general, they are less pronounced. If temperature is raised high enough, then a thermoplastic polymer will melt. The temperature region at which melting occurs is denoted Tm in Fig. 8, although, as previously discussed, a thermoplastic does not exhibit a unique melting temperature but rather melts over a temperature range. Young’s modulus tends toward zero as melting occurs, as implied in Fig. 8. Thermoset polymers cannot be melted, although the polymer is destroyed if temperature is raised to an excessively high level. The Tg has been illustrated in Fig. 8 by demonstrating the change in stiffness as temperature is increased. However, many other characteristic physical properties (density, strength, thermal expansion coefficient, heat capacity, etc.) also change sharply at this transition. Hence, the Tg can be measured by monitoring any of these physical properties as a function of temperature. The Tg exhibited by a few common polymers is listed in Table 1.

Table 1

Approximate Glass Transition Temperatures for Some Common Polymers Typical glass transition temperature

Polymer

jC

jF

Silicone rubber Polybutadiene Polyisoprene Nylon 6,6 Polyvinyl chloride (PVC) Acrylonitrile–butadiene–styrene Polystyrene Polyester Epoxy Polyetheretherketone (PEEK) Polyetherimide

123 85 50 50 85 90 100 150 175 200 215

190 120 60 122 185 195 210 300 350 400 420

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General character at room temperature (22jC or 70jF) Rubbery Rubbery Rubbery Rigid Rigid Rigid Rigid Glassy Glassy Glassy Glassy

Note that knowledge of the Tg allows an immediate assessment of the general nature of the polymer at room temperature.

3

FIBROUS MATERIALS

Reinforcing fibers are the major strengthening element in all polymeric composites. A brief introduction to these materials is presented in this section. The reader interested in additional details is referred to Refs. 5 and 6. Common continuous fiber materials are:      

Glass Aramid Graphite or carbon Polyethylene Boron Silicon carbide.

In all cases, the fiber diameters are quite small, ranging from about 5 to 12 Am for glass, aramid, or graphite fibers; from about 25 to 40 Am for polyethylene fibers; and from about 100 to 200 Am for boron and silicon carbide fibers. Some of the terminologies used to describe fibers will be defined here. The terms fiber and filament are used interchangeably. An end (also called a strand) is a collection of a given number of fibers gathered together. If the fibers are twisted, the collection of fibers is called a yarn. The ends are themselves gathered together to form a tow (also called roving). The fibers are usually coated with a size (also called a finish). The size is applied for several reasons, such as: 

To bind the fibers in the strand To lubricate the fibers during fabrication  To serve as a coupling and wetting agent to insure a satisfactory adhesive bond between the fiber and matrix materials. 

There may be thousands of filaments in a single tow and, in fact, tow sizes are often described in terms of thousands of filaments per tow. For example, a ‘‘6k tow’’ implies that the tow consists of 6000 individual fibers. Properties of several types of glass fibers, organic fibers, carbon fibers, and silicon carbide fibers will be briefly described in the following subsections. It will be seen that: 

Young’s modulus (stiffness) ranges from about 70 GPa for glass fibers to 700 GPa (or higher) for carbon fibers.

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Comparable tensile strengths can be obtained using any of the fibers listed.  Specific gravity ranges from about 0.97 for polyethylene fiber to about 2.5 for glass fibers.  Elongation at failure ranges from about 0.3% for some carbon fibers to about 5% for glass fibers. The mechanism responsible for these high stiffnesses and strengths differs from one fiber type to another. For glass fibers, the process of drawing to very small diameters simply reduces the number and size of flaws in the material, thus increasing strength. For example, glass fibers with a diameter of about 1 mm (0.04 in.) will commonly have a strength of about 170 MPa (25 ksi); but if this same glass is drawn to a diameter of 10 Am (0.0004 in.), a strength of about 3450 MPa (500 ksi) will be achieved. For organic fibers, strengthening is accomplished by stretching the fiber and thereby aligning the polymer molecules. This produces fibers that are themselves anisotropic. For carbon or graphite fibers, strengthening is accomplished by aligning the basal planes of adjoining crystals, also producing an anisotropic fiber. 3.1 Glass Fibers Glass fibers are fabricated by melting glass marbles at a temperature of about 1260jC (2300jF) and drawing the melt through platinum bushings followed by a rapid cooldown and secondary drawing. A sizing is applied to the fibers, which are then combined into a strand and wound onto a spool. The two major types of glass fibers are called ‘‘E-glass’’ and ‘‘S-glass.’’ E-glass (alumino borosilicate) is so named (‘‘e’’lectrical glass) because of its high electrical resistivity. S-glass (magnesium aluminosilicate) is so named (‘‘s’’tructural glass) because of its high tensile strength. Glass fibers are usually isotropic. Some mechanical and physical properties typical of E-glass and S-glass fibers are listed in Table 2. 3.2 Aramid Fibers The aramid polymer fiber produced by DuPont Corp. and marketed under the trade name Kevlark is perhaps the most widely used organic fiber. This fiber is based on poly( p-phenylene terephthalamide), which is a member of a family of polymers called ‘‘aramids.’’ Aramid fibers are formed by a condensation/elongation process. The resulting fibers are highly anisotropic because strong covalent bonds are formed in the fiber direction, whereas weak hydrogen bonds are formed in the transverse direction. This chemical

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Table 2

Typical Properties of Glass Fibers

Property Specific gravity Young’s modulus, GPa (Msi) Tensile strength, MPa (ksi) Tensile elongation, % Coefficient of thermal expansion, Am/m/jC (Ain./in./jF)

E-glass

S-glass

2.60 72 (10.5) 3450 (500) 4.8 5.0 (2.8)

2.50 87 (12.6) 4310 (625) 5.0 5.6 (3.1)

bonding arrangement results in the anisotropic behavior exhibited at the macroscale. That is, aramid fibers have very high tensile strength, stiffness, and toughness in the axial direction of the fiber, but relatively low tensile strength and stiffness in the transverse direction. Kevlar is commercially available in the following grades: 

Kevlar 149: a high-performance, aerospace grade fiber with the highest modulus of all Kevlar fibers  Kevlar 49: a high-performance, aerospace grade fiber with the highest strength of all Kevlar fibers  Kevlar 129: a relatively inexpensive fiber with a lower strength and stiffness than Kevlar 149 and 49, but with a higher percent elongation  Kevlar 29: a relatively inexpensive fiber with a strength and stiffness lower than Kevlar 129, but a higher percent elongation. Nominal mechanical and physical properties of Kevlar fibers are listed in Table 3. Of particular interest is the negative coefficient of thermal

Table 3

Typical Properties of Kevlar Fibers (All Properties in Axial Direction of Fiber)

Property Specific gravity Young’s modulus, GPa (Msi) Tensile strength, MPa (ksi) Tensile elongation, % Coefficient of thermal expansion, Am/m/jC (Ain./in./jF)

Kevlar 149

Kevlar 49

Kevlar 129

Kevlar 29

1.44 186 (27)

1.44 124 (18)

1.44 96 (13.9)

1.44 68 (9.8)

3440 (500)

3700 (535)

3380 (490)

2930 (425)

2.5 2.0 ( 1.1)

2.8 2.0 ( 1.1)

3.3 2.0 ( 1.1)

3.6 2.0 ( 1.1)

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expansion in the fiber direction. This implies that an increase in temperature produces a decrease in the length of a Kevlar 49 fiber. This behavior produced unexpected behavior during cooldown from cure in early applications of the fiber. Further research indicates that Kevlar 49 has a high positive coefficient of thermal expansion in the transverse direction, further demonstrating the anisotropic nature of the fiber. 3.3 Graphite and Carbon Fibers The terms ‘‘graphite’’ and ‘‘carbon’’ are often used interchangeably within the composites community. The elemental carbon content of either type of fiber is above 90%, and the stiffest and strongest fibers have carbon contents approaching 100%. Some effort has been made to standardize these terms by defining graphite fibers as those that have: 

A carbon content above 95% Been heat-treated at temperatures in excess of 1700jC (3100jF)  Been stretched during heat treatment to produce a high degree of preferred crystalline orientation  A Young’s modulus on the order of 345 GPa (50 Msi). 

Fibers that do not satisfy all of the above conditions are called carbon fibers under this standard. However, as stated above, in practice, this definition is not widely followed, and the terms ‘‘graphite’’ and ‘‘carbon’’ are often used interchangeably. Both graphite and carbon fibers are produced by thermal decomposition of an organic (i.e., polymeric) fiber or ‘‘precursor’’ at high pressures and temperatures. The three most common precursors are:  

Polyacrylonitrile (‘‘PAN’’) Pitch (a by-product produced during the petroleum distillation process)  Rayon.

The PAN fiber is probably the most widely used. Details of the specific steps followed during fabrication of a fiber are proprietary and can only be described in a general manner. During the fabrication process, the precursor is first drawn into a thread and then oxidized at about 260jC (500jF) to form crosslinks and an extended carbon network. The precursor is then subjected to a carbonization treatment, during which noncarboneous atoms are driven off. This step typically involves temperatures of approximately 700jC (1290jF), and is usually conducted in an inert atmosphere. Finally, during the graphitization process, the fibers are subjected to a combination of high temperature and tensile elongation.

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The maximum temperature reached during this step determines, in large part, the strength and/or stiffness that will be achieved. The graphite crystal is highly anisotropic, with strong covalent bonds in the basal plane and weak Van der Waals (‘‘secondary’’) bonds perpendicular to the basal plane. High strengths and stiffnesses are attained by causing the basal planes to be aligned in the fiber direction. This can be accomplished by controlled stretching of the precursor during fabrication. Major developmental efforts have resulted in the ability to produce fibers with a wide range of stiffnesses and strengths. Fibers are sometimes classified in terms of the stiffness (i.e., elastic modulus). Mechanical and physical properties typical of low-modulus, intermediate-modulus, and ultrahigh-modulus fibers are listed in Table 4. 3.4 Polyethylene Fibers A high-strength, high-modulus polyethylene fiber called Spectrak was developed at Allied Signal Technologies during the 1980s. Spectra is based on ultra-high-molecular-weight polyethylene (UHMWPE). It has a specific gravity of 0.97, meaning that it is the only reinforcing fiber available that is lighter than water. Spectra is available in three classifications (Spectra 900, 1000, and 2000) and several grades are available in each class. Nominal properties are listed in Table 5. The high specific strength of the fiber makes it particularly attractive for tensile applications. The glass transition temperature of UHMWPE is in the range of 20jC to 0jC, and hence the fiber is in the rubbery state at room temperatures and exhibits time-dependent (viscoelastic) behavior. This feature imparts outstanding impact resistance and toughness, but may lead to undesirable creep effects under long-term sustained loading. The melting temperature of the fiber is 147jC (297jF),

Table 4

Typical Properties of Commercially Available Graphite Fibers

Property Specific gravity Young’s modulus, GPa (Msi) Tensile strength, MPa (ksi) Elongation, % Coefficient of thermal expansion, Am/m/jC (Ain./in./jF)

Low modulus

Intermediate modulus

Ultra-high modulus

1.8 230 (34)

1.9 370 (53)

2.2 900 (130)

3450 (500)

2480 (360)

3800 (550)

1.1 0.4 ( 0.2)

0.5 0.5 ( 0.3)

0.4 0.5 ( 0.3)

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Table 5

Nominal Properties of Spectra Fibers

Property Specific gravity Young’s modulus, GPa (Msi) Tensile strength, MPa (ksi) Elongation, % Coefficient of thermal expansion, Am/m/jC (Ain./in./jF)

Spectra 900

Spectra 1000

Spectra 2000

0.97 70 (10)

0.97 105 (15)

0.97 115 (17)

2600 (380)

3200 (465)

3400 (490)

3.8 >70 (>38)

3.0 >70 (>38)

3.0 >70 (>38)

and hence the use of polyethylene fibers is limited to relatively modest temperatures. The thermal expansion coefficients of Spectra fibers have apparently not been measured. The values listed in Table 5 are estimated based on the properties of bulk high-molecular-weight polyethylene. 3.5 Boron Fibers Boron fibers were one of the first high-performance fibers available for use in composites. They are fabricated by depositing boron on a heated core using the vapor deposition process. Both tungsten and carbon fiber cores have been used. Boron fiber diameters range from 0.1 to 0.2 mm (0.004–0.008 in), which is an order of magnitude larger than glass, aramid, or graphite fibers. Boron fibers have a Young’s modulus of about 410 GPa (60 Msi) and a tensile strength of about 3450 MPa (500 ksi). The combination of a large diameter and high stiffness greatly restricts the bend radius to which the fiber may be subjected. On the other hand, a large fiber diameter and high modulus of elasticity contribute to excellent compressive performance for boronreinforced composites. 3.6 Ceramic Fibers The single most outstanding feature offered by ceramic fibers is that they are resistant to extremely high temperatures while still maintaining competitive structural properties. An example is a silicon carbide fiber marketed under the trade name SCS-Ultrak and fabricated by Specialty Materials, Inc. (Lowell, MA), which can operate at temperatures up to 1200jC (2190jF). This fiber has a modulus of 415 GPa (60 Msi), a strength of 5865 MPa (850 ksi), and a specific gravity of 3.0. A second example is an aluminum– boron–silica fiber fabricated by the 3M Company and marketed under the trade name Nextelk. This fiber is capable of operating at temperatures up

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to 1650jC (3000jF). It also exhibits excellent properties, with a modulus of 193 GPa (28 Msi), a strength of 2000 MPa (300 ksi), and a specific gravity of 3.03. Because ceramic fibers are normally used at temperatures far in excess of the useable temperature range of polymers, they are rarely used in polymeric composites. Ceramic fibers will not be further discussed in this book.

4

COMMERCIALLY AVAILABLE FORMS

4.1 Discontinuous Fibers Virtually all of the continuous fibers described in Sec. 3 are also available in the form of discontinuous fibers. Discontinuous fibers are embedded within a matrix, and may be randomly oriented (in which case the composite is isotropic at the macroscale) or may be oriented to some extent (in which case the composite is anisotropic at the macroscale). Orientation of discontinuous fibers, if it occurs, is usually induced during the fabrication process used to create the composite material/structure; fiber alignment often mirrors the flow direction during injection molding, for example. Discontinuous fibers are roughly classified according to length, as follows: 

Milled fibers are produced by grinding the continuous fiber into very short lengths. For example, milled graphite fibers are available with lengths ranging from about 0.3 to 3 mm (0.0012–0.12 in.), and milled glass fibers are available with lengths ranging from about 0.4 to 6 mm (0.0016–0.24 in.).  Chopped fibers (or strands) have a longer length than milled fibers, and composites produced using chopped fibers usually have higher strengths and stiffnesses than those produced using milled fibers. Chopped graphite fibers are available with lengths ranging from about 3 to 50 mm (0.–2.0 in.), while chopped glass fibers are available with lengths ranging from about 6 to 50 mm (0.24–2.0 in.). In general, the mechanical properties of a composite produced using discontinuous fibers (say, the strength or stiffness) are not as good as those that can be obtained using continuous fibers. However, discontinuous fibers allow the use of relatively inexpensive, high-speed manufacturing processes such as injection molding or compression molding, and have therefore been widely used in applications in which extremely high strength or stiffness is not required. One of the most widely used composites systems based on the use of discontinuous fibers is known generically as ‘‘sheet molding compound’’ (SMC). In its most common form, SMC consists of chopped glass fibers

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embedded within a thermosetting polyester resin. However, other resin systems (e.g., vinyl esters or epoxies) as well as other fibers (e.g., chopped graphite or aramid fibers) are occasionally used in SMC material systems. 4.2 Roving Spools Most continuous fiber types are available in the form of spools of roving (i.e., roving wound onto a cylindrical tube and ultimately resembling a large spool of thread). As mentioned in Sec. 3, roving is also known as tow. The size of tow (or roving) is usually expressed in terms of the number of fibers contained in a single tow. For example, a specific glass fiber might be available in the form of 2k, 3k, 6k, or 12k tow. In this case, the product is available in tows containing from 2000 to 12,000 fibers. Fibers purchased in this form are usually ‘‘dry’’ and are subsequently combined with a polymer, metal, or ceramic matrix during a subsequent manufacturing operation such as filament winding or pultrusion. 4.3 Woven Fabrics Most types of high-performance continuous fibers can be woven to form a fabric. Weaving is accomplished using looms specially modified for use with high-performance fibers, which are stiffer than those customarily used in the textile industry. Woven fabrics are produced in various widths up to about 120 cm (48 in.), and are available in (essentially) infinite lengths. Two terms associated with woven fabrics are: 

The tow or yarn running along the length of the fabric is called the warp. The warp direction is parallel to the long axis of the woven fabric.  The tow or yarn running perpendicular to the warp is called the fill tow (also called the weft or the woof tow). The fill direction is perpendicular to the warp direction. Some common fabric weaves are shown schematically in Fig. 9. The plane weave (Fig. 9a) is the simplest fabric pattern and is most commonly used. The plane weave is produced by repetitively weaving a given warp tow over one fill tow and under the next. The point at which a tow passes over/ under another tow is called a crossover point. The plane weave pattern results in a very stable and firm fabric that exhibits minimum distortion (e.g., fiber slippage) during handling. A family of woven fabric patterns known as satin weaves provide better drape characteristics than the plane weave pattern. That is, a satin weave is more pliable and will more readily conform to complex curved surfaces than plane weaves. In the crowfoot satin weave (Fig. 9b), one warp tow is woven

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Figure 9 Some common woven fabrics used with high-performance fibers. (From Ref. 7.)

over three successive fill tows and then under one fill tow. In the five-harness satin weave, one warp tow passes over four fill tows and then under one fill tow. Similarly, in an eight-harness satin weave, one warp tow passes over seven fill tows and then under one fill tow. The stiffness and strength of woven fabrics are typically less than that achieved with unidirectional fibers. This decrease is due to fiber waviness.

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That is, in any woven fabric, the tow is required to pass over/under one (or more) neighboring tow(s) at each crossover point, resulting in a pre-existing fiber waviness. Upon application of a tensile load, the fibers within a ply tend to straighten, resulting in a lower stiffness than would be achieved if the ply contained straight unidirectional fibers. Further, due to the weave pattern, the fibers are not allowed to straighten fully and are subjected to bending stresses, resulting in fiber failures at lower tensile loads than would otherwise be achieved if the ply contained unidirectional fibers. This effect is most pronounced in the case of simple weaves because each tow passes over/under each neighboring tow. For simple weaves, the through-thickness distribution of tow in the warp and fill directions is identical. Consequently, the strength and stiffness of simple weaves are usually identical in the warp and fill directions. In contrast, for satin weaves, the through-thickness distribution of tow is inherently asymmetric. Referring to Fig. 9a, for example, for the fiveharness satin weave pattern, the warp tow is primarily within the ‘‘top’’ half of the fabric (as sketched), whereas the fill tow is primarily within the lower half. The asymmetrical through-thickness distribution of tow causes a coupling between in-plane loading and bending deflections. That is, if a uniform tensile load is applied to the midplane of a single layer of a satin weave fabric, the fabric will not only stretch but will also deflect out of plane (i.e., bend). Similarly, the crossover points are not symmetrically located with respect to either the warp or fill directions. This causes a coupling between in-plane loading and in-plane shear strain. That is, an in-plane shear strain is induced if a uniform tensile load is applied to a single layer of a satin weave fabric [7]. A woven fabric is, in essence, a 2D structure consisting of orthogonal warp and fill tows interlaced within a plane. Weaving or stitching several layers of a woven fabric together can produce a woven structure with a significant thickness. Structures produced in this fashion are called ‘‘3D weaves.’’ 4.4 Braided Fabrics Note from Sec. 4.3 that woven fabrics contain reinforcing tow in two orthogonal directions—the warp and fill directions. In contrast, braided fabrics typically contain tow oriented in two (or more) nonorthogonal directions. Three common braiding patterns are shown in Fig. 10. It is apparent from this figure that a braided fabric contains bias tow that intersects at a total included angle 2a. The angle a is called either the braid angle or the bias angle. While the braid angle can be varied over a wide range, there is always some minimum and maximum possible value that depends on the width of the tow and details of the braiding equipment used. Note that if a=45j,

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Figure 10 Some common braided fabrics used with high-performance fibers. (From Ref. 7.)

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then the bias tows are in fact orthogonal and the braided fabric shown in Fig. 10a and b is equivalent to a woven fabric. A braided fabric is described using the designation nn, where n is the number of tows between crossover points. A 11-bias and a 22-bias braided fabric is shown in Fig. 10a and b, respectively. A 11 triaxial braided fabric is shown in Fig. 10c. In this case, a third axial tow is present. Braided fabrics can be produced in tubular form or as a flat braided fabric. A concise description of the equipment used to produce braided fabrics, as well as a discussion of the maximum and minimum braid angles that can be achieved, is given in Ref. 8. It is also possible to braid 3D structures, in which tows are braided to form a (fibrous) structure that is subsequently infused with a matrix to form the composite. Applications include ‘‘I’’ or ‘‘T’’ cross sections, typically used with resin transfer molding (RTM) to produce composite stiffeners or beams. In contrast to structures produced using fabrics (which may be unidirectional, woven, or braided fabrics), in a 3D braided structure, there are no recognizable layers. 4.5 Preimpregnated Products or ‘‘Prepreg’’ As is obvious from the preceding discussion, at some point during the fabrication of a polymer composite, the reinforcing fiber must be embedded within the polymeric matrix. One approach is to combine the fiber and resin during the manufacturing operation in which the final form of the composite structure is defined. Three manufacturing processes in which this approach is taken are filament winding (briefly described in Sec. 5.3), pultrusion (Sec. 5.4), and resin transfer molding (Sec. 5.5). An alternative approach is to combine the fiber and matrix in an intermediate step, resulting in an intermediate product. In this case, either individual tow or a thin fabric of tow (which may be woven or braided) is embedded within a polymeric matrix and delivered to the user in this form. Because the fibers have already been embedded within a polymeric matrix when delivered, the fibers are said to have been ‘‘preimpregnated’’ with resin, and products delivered in this condition are commonly known as ‘‘prepreg.’’ The method used to impregnate a large number of unidirectional tows with resin is illustrated in Fig. 11 [9]. As indicated, tows delivered from a large number of roving spools are arranged in a relatively narrow band. The tows are passed through a resin bath and then wound onto a roll. An inert backing sheet (called a scrim cloth) is placed between layers on the roll to maintain a physical separation between layers and to aid during subsequent handling and processing. The tows/fibers are subjected to various surface pretreatments just prior to entering the resin bath. The pretreatments are proprietary but are intended to cause good wetting of the fiber by the resin,

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Figure 11 (a) Method used to impregnate unidirectional tows with resin. (b) A 3-in. wide roll of ‘‘prepreg tape.’’

which ultimately helps ensure good adhesion between the fibers and polymer matrix in the cured composite. Products produced in this fashion are commonly known as ‘‘prepreg tape’’ (Fig. 11b). Prepreg tape is available in width ranging from about 75 to 1220 mm (3–48 in.). Prepreg fabrics, produced using either woven or braided fabrics instead of unidirectional tows, are produced using similar techniques and are also available in widths ranging from about 75 to 1220 mm. A variety of fabrication methods have been developed based on the use of prepreg materials. A few such techniques will be described in Sec. 5. The first commercially successful prepreg materials were based on Bstaged epoxy resins. As discussed in Sec. 2, in the B-staged condition, a thermoset resin has been partially polymerized, resulting in relatively high viscosity, which aids in handling B-staged prepreg materials. However, prepreg must be kept at low temperatures until used, otherwise the resin continues to polymerize and slowly harden. This required that the prepreg be shipped to the user in a refrigerated condition (for small amounts, this is

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often accomplished using insulated shipping containers and dry ice). Furthermore, the user must keep the stock of prepreg on hand refrigerated until used. Typically, storage temperatures are required to be 15jC (0jF) or below. In practice, the prepreg material stock is removed from freezer, the amount of prepreg necessary is removed from the roll of stock, and the remaining stock is returned to the freezer. Hence, the cumulative ‘‘out-time’’ that a given roll of prepreg stock has experienced (i.e., total amount of time a roll has been out of a freezer) must be monitored and recorded. The need to store thermoset prepreg in a refrigerated condition and to maintain accurate records of cumulative out-time is a significant disadvantage because these factors add significantly to the final cost of the composite structure. More recently, prepreg materials based on thermoplastic resins have become commercially available. In this case, the polymeric matrix is a fully cured thermoplastic polymer, and hence the prepreg does not require refrigeration during shipping or storage, which is a distinct advantage. Heat and/or pressure is applied during the final fabrication of a composite based on prepreg materials. In the case of thermoset prepregs, heat and pressure serve to complete the polymerization of the polymeric resin (i.e., the composite is ‘‘C-staged’’). For thermoplastic prepreg, the objective is not necessarily to complete the polymerization but rather to melt the thermoplastic matrix so as to consolidate individual plies within the laminate.

5

MANUFACTURING PROCESSES

Fiber-reinforced composites may be produced using metallic, ceramic, or polymeric matrices. However, polymeric composites are the primary focus of this book, and so techniques used to fabricate metal or ceramic composite structures will not be discussed. Even with this limitation, a complete review of the many different manufacturing processes used to produce polymeric composite materials and structures is beyond the scope of this presentation. Instead, only the most common manufacturing techniques will be described. 5.1 Layup Many composites are produced using the tapes or fabrics discussed in Sec. 4.1. These may be unidirectional tape, woven fabrics, or braided fabrics. These products are all relatively thin. ‘‘Layup’’ simply refers to the process of stacking several layers together, much like a deck of cards. Stacking several layers together produces a laminate of significant thickness. The most direct method of producing a multi-ply composite laminate is to simply stack the desired number of layers of fabric by hand, referred to as ‘‘hand layup.’’ The layers may consist of either ‘‘dry’’ fabrics (i.e., fabrics that have not yet been

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impregnated with a resin) or prepreg materials. As will be discussed in later chapters, fiber angles are typically varied from one ply to the next, so as to insure adequate stiffness in more than one direction. Whereas hand layup is simple and straightforward, it is labor-intensive and therefore costly. It can also be very cumbersome if a large structure is being produced, such as a fuselage panel intended for use in a modern commercial aircraft. Therefore, various computer-controlled machines that automate the process of assembling the ply stack using prepreg materials have been developed. These include tow placement and tape placement machines (see Fig. 12). In either case, a roll (or rolls) of prepreg material is mounted on the head of a computer-controlled robot arm or gantry. The appropriate number of layers of prepreg is placed on a tool surface automatically and in the desired orientation. Although the capital costs of modern tow placement or tape placement machines may be very high, overall, this approach is often less costly than hand layup if production quantities are sufficiently high. In the case of dry fabrics (which are usually either woven or braided fabrics), the stack must be impregnated with a low-viscosity polymeric resin following assembly of the fiber stack. Conceptually, this may be accomplished by pouring liquidous resin over the dry fiber stack and using a squeegee or

Figure 12 A computer-controlled tape-laying machine, used to produce the composite skin used in the vertical stabilizer for a Boeing 777 aircraft. (Copyright n The Boeing Company.)

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similar device to assist the resin to wet the fibers within the stack. This technique is commonly used in the recreational boat-building industry, for example. However, as can be imagined, it is very difficult to insure uniform penetration of the resin and wetting of the fibers through the thickness of the stack, to insure that no air pockets remain trapped in the stack, and to avoid distortion of the fiber patterns while forcing resin into the fibrous assembly. There are also potential health issues associated with continually exposing workers to nonpolymerized resins. Hence, the technique of impregnating a dry fiber stack using handheld tools such as squeegees is rarely employed in industries requiring low variability in stiffness and strength and/ or high volumes, such as the aerospace or automotive industries, for example. Alternate methods of impregnating a dry fiber stack with resin have been developed, such as resin transfer molding (discussed in Sec. 5.5). These alternate techniques result in a composite with a much more uniform matrix volume fraction and almost no void content. A major advantage of using prepreg materials, of course, is that the fibers have been impregnated with resin a priori. Therefore, it is much easier to maintain the desired matrix volume fraction and to avoid entrapped airpockets. Further, prepreg material based on a B-staged thermoset are typically ‘‘tacky’’ (i.e., prepreg materials adhere to neighboring plies much like common masking tape), and hence once a given ply has been placed in the desired orientation, it is less likely to move or be distorted relative to neighboring plies than is the case with dry fabrics. 5.2 Autoclave Process Cycles Following layup (which may be accomplished using hand layup, automatic tow placement or tape placement machines, or other techniques), the individual plies must be consolidated to form a solid laminate. Usually, consolidation occurs through the application of pressure and heat. Although a simple hot press can be used for this purpose, applying pressure and heat using an autoclave produces highest-quality composites. An autoclave is simply a closed pressure vessel that can be used to apply a precisely controlled and simultaneous cycle of vacuum, pressure, and elevated temperature to the laminate during the consolidation process. Although many variations exist, a typical assembly used to consolidate a laminate using an autoclave is shown in Fig. 13. Some of the details of the assembly are as follows: 

The final shape of the composite is defined by a rigid tool. A simple flat tool is shown in Fig. 13, but in practice, the tool is rarely flat but instead mirrors the contour(s) desired in the final product. For example, the surface of a tool used to produce the skin of an air-

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Figure 13 Typical assembly used to consolidate a polymeric composite laminate using an autoclave (expanded edge view).

plane wing would possess the curvature(s) necessary to provide an aerodynamic surface. Various materials may be used to produce the tool, including steel alloys, aluminum alloys, ceramics, or composite materials.  The tool surface is coated with a release agent. Various liquid or wax release agents are available, which are either sprayed or wiped onto the surface. The purpose of the release agent is to prevent adherence between the tool and the polymeric matrix.  A peel ply is placed next to both upper and lower surfaces of the composite laminate. The release ply does not develop a strong bond to the composite, and hence can be easily removed following consolidation. The peel ply may be porous or nonporous. Porous peel plies allow resin to pass through the ply and be adsorbed by an adjacent bleeder/breather cloth (see below). Note that the surface texture of the consolidated laminate will be a mirror image of the peel ply used. For example, Teflon-coated porous glass fabrics are often used as peel plies, and these fabrics have a clothlike surface texture. Hence, a composite laminate consolidated with such fabrics will exhibit a clothlike surface texture as well.  One or more layers of a breather/bleeder cloth is placed adjacent to the porous peel ply. The bleeder cloth has the texture of a rather stiff cotton ball. Its purpose is to allow any gases released to be vented (hence the adjective breather), and also to adsorb any resin that passed through the porous peel ply (hence the adjective bleeder). The breather/bleeder is usually a glass, polyester, or jute cloth.

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An edge dam is placed around the periphery of the laminate. The edge dam is intended to maintain the position and resin content of the laminate edges.  A pressure plate (also called a caul plate) is placed over the breather/ bleeder cloth. The pressure plate insures a uniform distribution of pressures over the surface of the laminate.  The entire assembly is sealed within a vacuum film or bag. Often this is a relatively thick (say, 5 mm) layer of silicone rubber. Sealant tape is used to adhere the vacuum film to the tool surface, providing a pressure-tight seal around the periphery of the vacuum film.  The volume within the vacuum bag is evacuated by means of a vacuum port, which is often permanently attached to the silicone rubber vacuum film. The vacuum port often features a quick-disconnect fitting, which allows for easy connection to a vacuum pump or line. Following vacuum bagging of the laminate, the assembly is placed within an autoclave, the autoclave is sealed, and the thermomechanical process cycle that will consolidate the composite is initiated. A bagged composite laminate being loaded into an autoclave is shown in Fig. 14. The thermomechanical process cycle imposed using an autoclave varies from one composite prepreg system to the next, and also depends on part

Figure 14 A vacuum-bagged composite skin used in the tail-section of a Boeing 777 aircraft, about to be loaded into a large autoclave. (Copyright n The Boeing Company.)

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configuration (e.g., part thickness). Recall that if the prepreg is based on a Bstage thermoset resin, then the autoclave is used to complete the polymerization of the resin (i.e., the composite is C-staged). Alternatively, if the prepreg is based on a thermoplastic, the pressure and heat applied during the autoclave cycle soften the matrix and insure polymer flow across the ply interfaces. The laminate is then solidified upon cooling. A typical cure cycle, suitable for use with standard thermosetting resin systems such as epoxies, is as follows: 









Draw and hold a vacuum within the vacuum bag, resulting in a pressure of roughly 100 kPa (14.7 psi) applied to the laminate. The vacuum is typically maintained for about 30 min, and is intended to remove any entrapped air or volatiles, and to hold the laminate in place. While maintaining a vacuum, increase the temperature from room temperature to about 120jC (250jF), at a rate of about 2.8jC/min (5jF/min). Maintain this temperature for 30 min. During this 30-min dwell time, any remaining air or other volatiles are removed. After 30 min, increase autoclave pressure from atmospheric to about 585 kPa (85 psi), at a rate of 21 kPa/min (3 psi/min). Release vacuum when autoclave pressure reaches 138 kPa (20 psi). When an autoclave pressure of 585 kPa is reached, increase the temperature from 120jC to 175jC (350jF), at a rate of about 2.8jC/ min (5jF/min). Maintain temperature at 175jC for 2 hr. Polymerization of the thermosetting resin matrix is completed during this 2-hr dwell. Cool to room temperature at a rate of about 2.8jC/min (5jF/min), release autoclave pressure, and remove cured laminate from the autoclave.

Process cycles used with thermoplastic prepregs are similar, except that higher temperatures (500jC or higher) are involved. 5.3 Filament Winding Filament winding is an automated process in which tow is wound onto a mandrel at controlled position and orientation. A filament winder being used to produce a small rocket motor case is shown in Fig. 15 [10]. During operation, the mandrel rotates about its axis, and a fiber carriage simultaneously moves in a controlled manner along the length of the mandrel. The angle at which fibers are placed on the mandrel surface is a function of the mandrel diameter, rate of mandrel rotation, and translational speed of the fiber carriage. The mandrel can include domed heads to accommodate fiber

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Figure 15 A carbon-epoxy pressure vessel being produced using the filamentwinding process and unidirectional prepreg. The prepreg used in this case is based on large 50k tows of Zoltek Panexo 35 carbon fiber. (a) A band of prepreg is wound onto a mandrel, forming a Fhj fiber pattern. (b) Eventually the mandrel is completely enclosed by one or more Fhj plies. (c) One or more 90j (‘‘hoop’’) plies are often added to the cylindrical region to resist the high hoop stresses induced in cylindrical pressure vessels. (Photos provided courtesy of Entec Composite Machines Inc., Salt Lake City, UT.) R

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turnaround at the ends, or to wind the domes of a cylindrical pressure vessel (as shown in Fig. 15). If dry tows are used, then the tow must pass through a liquid resin bath before being wound onto the mandrel. In this case, the process is often referred to as ‘‘wet’’ winding. Often fiber tension provides sufficient compaction of the laminate, and so no additional external pressure is required. If a thermosetting resin that cures at room temperature is used, following completion of the winding operation, the structure is simply left in the winder until polymerization is complete. Of course, if prepreg tow is used, then the tow is already impregnated with a resin and is not passed through a resin bath. This process is called ‘‘dry’’ winding. In this case, heat and pressure are normally required to complete polymerization of the resin (in the case of a thermosetting polymer matrix), or to cause resin flow and consolidation (in the case of a thermoplastic polymer matrix). The appropriate heat and pressure are usually applied using an autoclave. Filament winding machines are available in highly automated, numerically controlled models (costing millions of dollars), but high-quality filament winding can also be accomplished for simple shapes and patterns on inexpensive gear/chain-driven machines similar to a lathe. For simple wound shapes with open ends (such as tubes), the mandrel is usually a simple solid cylinder whose surface has been coated with a release agent. In this case, the mandrel is forced out of the internal cavity after consolidation of the composite. Mandrel design and configuration become more complex when a shape with restricted openings at the ends is produced (such as the pressure tank shown in Fig. 15). In these cases, the mandrel must somehow be removed after the part is consolidated. Several different types of mandrel designs are used in these cases, including: 

Soluble mandrels, which are made from a material that can be dissolved in some fashion after the cure process is complete. In this approach, the mandrel is cast and machined to the desired shape, the composite part is filament wound over the mandrel, the part is cured, and the mandrel is then simply dissolved. The wall of the composite structure must obviously have at least one opening, such that the dissolved (and now liquidous) mandrel material can be drained from the internal cavity. Soluble mandrels can be made from metallic alloys with suitably low-melting temperatures, eutectic salts, sand with water-soluble binders, or various plasters.  Removable (or collapsible) mandrels, which resemble giant 3D puzzles. That is, the entire mandrel can be taken apart piece by piece. The composite structure being wound must have at least one wall opening,

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which allows the mandrel pieces to be removed from the internal cavity after cure. Obviously, the mandrel is designed such that no single piece is larger than the available opening(s).  ‘‘Inflatable’’ mandrels, which take on the desired shape when pressurized and then are simply deflated and removed after winding and consolidation.  Metal or polymer liners, which are actually a modification of the inflatable mandrel concept. Liners can be described as metal or polymer ‘‘balloons’’ and remain in the filament wound vessel after cure. The liner does not contribute significantly to the strength or stiffness of the structure. In fact, the wall thickness of the liner is often so small that an internal pressure must be applied to the liner during the winding process to avoid buckling of the liner wall. Metal liners are almost always used in composite pressure vessels, where allowable leakage rates are very low, or in filament wound chemical storage tanks, where corrosive liquids are stored. 5.4 Pultrusion Pultrusion is a fabrication process in which continuous tows or fabrics impregnated with resin are pulled through a forming die, as shown schematically in Fig. 16 [10]. If dry tow or fabric is used, then the tow/fabric must

Figure 16 Sketch of a typical pultruder.

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pass through a resin bath prior to entering the forming die. In this case, the process is called ‘‘wet’’ pultrusion. If prepreg material is used, then there is no need for a resin bath and the process is called ‘‘dry’’ pultrusion. The cross-sectional shape is defined by the die and is therefore constant along the length of the part. The principal attraction of pultrusion is that very high production rates are possible, as compared to other composite manufacturing techniques. 5.5 Resin Transfer Molding In the resin transfer molding process, a dry fiber preform is placed within a cavity formed between two rigid molds, as shown in Fig. 17. The dry preform may consist of a 3D braided structure, or may be made by stitching together multiple layers of 2D woven or braided fabrics. Liquidous resin is forced into the cavity under pressure via a port located in the upper or lower mold halves. Air originally within the internal cavity (or other gases that evolve during cure of the resin) is allowed to escape via one or more air vents. Alternatively, a vacuum pump may be used to evacuate the internal cavity, which also assists in drawing the resin into the cavity. When a vacuum is used, the process is called ‘‘vacuum-assisted resin transfer molding’’ (VARTM). Both the upper and lower molds must be sufficiently rigid so as to resist the internal pressures applied and to maintain the desired shape of the internal cavity. Usually, the closed molds are placed within a press, which provides a clamping pressure to assist in keeping the molds closed. 6

THE SCOPE OF THIS BOOK

A broad overview of modern composite materials has been provided in the preceding sections. It should be clear from this discussion that modern

Figure 17 Picture/sketch of the RTM process.

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polymeric composite material systems are a multidisciplinary subject, involving topics drawn from polymer chemistry, fiber science, surface chemistry and adhesion, materials testing, structural analysis, and manufacturing techniques, to name a few. It is simply not possible to cover all of these topics in any depth in a single book. Accordingly, the material presented in this book represents a small fraction of the scientific and technological developments that have ultimately led to the successful use of modern composite material systems. Specifically, the focus of this text is the structural analysis of laminated, continuous-fiber polymeric composite materials and structures. Having identified the structural analysis of laminated continuous-fiber polymeric composites as our focus, we must make still another decision: At what physical scale should we frame our analysis? The importance of physical scale has already been discussed in Sec. 1 in conjunction with the very definition of a ‘‘composite material.’’ Specifically, we have defined a composite as a material system consisting of two (or more) materials, which are distinct at a physical scale greater than about 1 Am and which are bonded together at the atomic and/or molecular levels. Fibers commonly used in polymeric composites possess diameters ranging from about 5 to 40 Am (Sec. 3). Therefore, we could perform a structural analysis at a physical scale comparable to the fiber diameter. Alternatively, laminated polymeric composites consist of well-defined layers (called ‘‘plies’’) of fibers embedded in a polymeric matrix. The thickness of these layers ranges from about 0.125 to 0.250 mm (Sec. 4). We could therefore elect to begin a structural analysis at a physical scale comparable to the thickness of a single ply. A distinction is drawn between structural analyses that begin at these two different physical scales. Analyses that are framed at a physical scale corresponding to the fiber diameter (or below) are classified as micromechanics analyses, whereas those framed at a physical scale corresponding to a single ply thickness (or above) are classified as macromechanics analyses. This distinction is comparable to the traditional distinction between metallurgy and continuum mechanics. That is, metallurgy typically involves the study of the crystalline nature of metals and metal alloys, and is therefore framed at a physical scale roughly corresponding to atomic dimensions. A metallurgist might attempt to predict Young’s modulus* of a given metal alloy, based on knowledge of the constituent atoms and crystalline structure present in the alloy, for example. In contrast, continuum mechanics is formulated at a much larger physical scale, such that the existence of indi-

* The definition of various material properties of interest to the structural engineer, such as Young’s modulus, will be reviewed and discussed in greater detail in Chap. 3.

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vidual atoms is not recognized. In continuum mechanics, a metal or metal alloy is said to be ‘‘homogeneous’’ even though it actually consists of several different atomic species. A structural engineer wishing to apply a solution based on continuum mechanics would simply measure Young’s modulus exhibited by the metal alloy of interest, rather than try to predict it based on knowledge of the atomic crystalline structure. In much the same way, composite micromechanics analyses are concerned with the predicting properties of composites based on the particular fiber and matrix materials involved, the spacing and orientation of the fibers, the adhesion (or lack thereof ) between fiber and matrix, etc. For example, suppose that a unidirectional graphite–epoxy composite is to be produced by combining graphite fibers with a known Young’s modulus (Ef) and an epoxy matrix with a known Young’s modulus (Em). An analysis framed at a physical scale corresponding to the fiber diameter (i.e., a micromechanics analysis) is required to predict the Young’s modulus that will be exhibited by the composite (Ec) formed using these two constituents. In contrast, composite macromechanics analyses are framed at a physical scale corresponding to the ply thickness (or above). The existence or properties of individual fibers or the matrix material are not recognized (in a mathematical sense) in a macromechanics analysis. Instead, the ply is treated as a homogenous layer whose properties is identical at all points, although they differ in different directions. Details of fiber or matrix type, fiber spacing, fiber orientation, etc., are represented in a macromechanics analysis only indirectly, via properties defined for the composite ply as a whole, rather than as properties of the individual constituents. Micromechanics-based structural analyses will not be discussed in any detail. A simple micromechanics model that may be used to predict ply stiffnesses based on knowledge of fiber and matrix properties, called the rule of mixtures, will be developed in Sec. 6 in Chap. 3. However, the material devoted to micromechanics in this text is abbreviated and does not do justice to the many advances made in this area. The lack of emphasis on micromechanical topics is not meant to imply that such analyses are unimportant. Quite the contrary, micromechanics analyses are crucial during development of new composite material systems because it is only through a detailed understanding of the behavior of composites at this physical scale that new and improved materials can be created. Micromechanics has been minimized herein simply due to space restrictions. The reader interested in learning more about micromechanics is referred to several excellent texts that cover this topic in greater detail, a few of which are Refs. 5, 7, 11, and 14. Finally, then, the scope of this book is macromechanics-based structural analysis of laminated, continuous-fiber polymeric composites.

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REFERENCES 1. Rodriguez, F. Principles of Polymer Systems; 3rd Ed.; Hemisphere Publ. Co.: New York, 1989; ISBN 0-89116-176-7. 2. Young, R.J.; Lovell, P.A. Introduction to Polymers; 2nd Ed.; Chapman and Hall Publ. Co.: New York, 1991; ISBN 0-89116-176-7. 3. Strong, A.B. Plastics: Materials and Processing; 2nd Ed.; Prentice-Hall: Upper Saddle River, NJ, 2000. ISBN 0-13-021626-7. 4. The Macrogalleria (http://www.psrc.usm.edu/macrog/index.htm), website maintained by the University of Southern Mississippi and devoted to polymeric materials. 5. Watt, W., Perov, B.V., Eds. Strong Fibres; Vol. 1. In Handbook of Composite Materials; Kelly, A., Rabotnov, Y.N., Series Eds.; Elsevier Sci. Publ.: New York, NY, 1985; ISBN 0-444-87505-0. 6. Donnet, J.-B.; Wang, T.K.; Peng, J.C.M.; Reboillat, M. Carbon Fibers; 3rd Ed.; Marcel Dekker, Inc.: New York, NY, 1998; ISBN 0-8247-0172-0. 7. Cox, B.; Flanagan, G. Handbook of Analytical Methods for Textile Composites. NASA Contractor Report 4750. NASA-Langley Res. Ctr., Hampton, VA, 1997. 8. Hasselbrack, S.A.; Pederson, C.L.; Seferis, J.C. Evaluation of carbon–fiber reinforced thermoplastic matrices in a flat braid process. Polym. Compos. 1992, 13 (1), 38–46. 9. Kalpakjian, S. Manufacturing Processes for Engineering Materials; 3rd Ed.; Addison-Wesley Longman, Inc.: Menlo Park CA, 1997; ISBN 0-201-82370-5. 10. Schwartz, M.M. Composite Materials Handbook; New York, NY, McGraw-Hill Book Co.: 1983; ISBN 0-07-055743-8. 11. Hyer, M.W. Stress Analysis of Fiber-Reinforced Composite Materials; New York, NY,McGraw-Hill Book Co.: 1998; ISBN 0-07-016700-1. 12. Herakovich, C.T. Mechanics of Fibrous Composites; John Wiley and Sons: New York, NY, 1998; ISBN 0-471-10636-4. 13. Jones, R.M. Mechanics of Composite Materials; McGraw-Hill Book Co.: New York, NY, 1975; ISBN 0-07-032790-4. 14. Hull, D. An Introduction to Composite Materials; Cambridge University Press: Cambridge, UK, 1981; ISBN 0-521-23991-5.

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2 Review of Force, Stress, and Strain Tensors

In this chapter, the fundamental definitions of force vectors, stress tensors, and strain tensors are reviewed. The chapter begins with a discussion of force vectors, since the concept of ‘‘force’’ is encountered in everyday life and is therefore very intuitive. Separate sections devoted to stress tensors and strain tensors are then presented. Certain parallels will be drawn between force vectors and stress/strain tensors. An important underlying principal is that a tensor cannot be described in a mathematical sense until a specific coordinate system is selected for use. Also, a tensor cannot be properly described using only a single component of the tensor, i.e., all components of a tensor must be known in order to describe the tensor.

1 THE FORCE VECTOR Forces can be grouped into two broad categories: surface forces and body forces. Surface forces are those that act over a surface (as the name implies) and result from direct physical contact between two bodies. In contrast, body forces are those that act at a distance and do not result from direct physical contact of one body with another. The force of gravity is the most common type of body force. In this text, we are primarily concerned with surface forces; the effects of body forces (such as the weight of a structure) will be ignored.

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A force is a 3-D vector. A force is defined by a magnitude and a line of action. In SI units, the magnitude of a force is expressed in Newtons, abbreviated N, whereas in English units, the magnitude of a force is expressed in pounds-force, abbreviated lbf. A force vector F acting at a point P and referenced to a right-handed x–y–z coordinate system is shown in Fig. 1. Components of F acting parallel to the x–y–z coordinate axes, Fx, Fyy, and Fz, respectively, are also shown in the figure. The algebraic sign of each force component is defined in accordance with the algebraically positive direction of the corresponding coordinate axis. All force components shown in Fig. 1 are algebraically positive since each component ‘‘points’’ in the corresponding positive coordinate direction. The reader is likely to have encountered several different ways of expressing force vectors in a mathematical sense. Three methods will be described here. The first is called vector notation and involves the use of unit vectors. Unit vectors parallel to the x-, y-, and z-coordinate axes are typically labeled ıˆ, jˆ, and kˆ, respectively, and by definition have a magnitude equal to unity. A force vector F is written in vector notation as follows: F ¼ Fx iˆ þ Fy jˆ þ Fz kˆ The magnitude of the force is given by: qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi jF j ¼ Fx2 þ Fy2 þ F 2z

ð1Þ ð2Þ

Figure 1 A force vector F acting at point P. Force components Fx, Fy, and Fz acting parallel to the x–y–z coordinate axes, respectively, are also shown.

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A second method of defining a force vector is through the use of indicial notation. In this case, a subscript is used to denote individual components of the vectorial quantity: F ¼ ðFx ; Fy ; Fz Þ The subscript denotes the coordinate direction of each force component. One of the advantages of indicial notation is that it allows a shorthand notation to be used, as follows: F ¼ Fi ;

where i ¼ x; y; or z

ð3Þ

Note that a range has been explicitly specified for the subscript i in Eq. (3). That is, it is explicitly stated that the subscript i may take on values of x, y, or z. Usually, however, the range of a subscript(s) is not stated explicitly but rather is implied. For example, Eq. (3) is normally written simply as: F ¼ Fi where it is understood that the subscript i takes on values of x, y, and z. The third approach is called matrix notation. In this case, individual components of the force vector are listed within braces in the form of a column array: 8 9 < Fx = ð4Þ F ¼ Fy : ; Fz Indicial notation is sometimes combined with matrix notation as follows: F ¼ fFi g

ð5Þ

2 TRANSFORMATION OF A FORCE VECTOR One of the most common requirements in the study of mechanics is the need to describe a vector in more than one coordinate system. For example, suppose all components of a force vector Fi are known in one coordinate system (say, the x–y–z coordinate system), and it is desired to express this force vector in a second coordinate system (say, the xV–yV–zV coordinate system). In order to describe the force vector in the new coordinate system, we must calculate the components of the force parallel to the x V-, y V-, and zV-axes—that is, we must calculate FxV, FyV, and FzV. The process of relating force components in one coordinate system to those in another coordinate system is called the transformation of the force vector. This terminology is perhaps unfortunate in the sense that the force vector itself is not ‘‘transformed,’’ but rather our de-

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scription of the force vector transforms as we change from one coordinate system to another. It can be shown (1,2) that the force components in the xV–yV–zV coordinate system ( FxV, FyV, and FzV) are related to the components in the x–y–z coordinate system ( Fx, Fy, and Fz) according to: FxV ¼ cxVx Fx þ cx Vy Fy þ cxVz Fz Fy V ¼ cy Vx Fx þ cy Vy Fy þ cy Vz Fz

ð6aÞ

FzV ¼ czVx Fx þ czVy Fy þ czVz Fz

The terms ci Vj that appear in Eq. (6a) are called direction cosines and equal the cosine of the angle between the axes of the new and original coordinate systems. An angle of rotation is defined from the original x–y–z coordinate system to the new xV–yV–zV coordinate system. The algebraic sign of the angle of rotation is defined in accordance with the right-hand rule. Equation (6a) can be succinctly written using the summation convention as follows: FiV ¼ ci Vj Fj

ð6bÞ

Alternatively, these three equations can be written using matrix notation as: 8 9 2 38 9 cxVx cx Vy cxVz < Fx = < FxV = ¼ 4 cy Vx cy Vy cy Vz 5 Fy ð6cÞ F : yV ; : ; czV x czVy czVz FzV Fz

Note that although values of individual force components vary as we change from one coordinate system to another, the magnitude of the force vector [given by Eq. (2)] does not. The magnitude is independent of the coordinate system used and is called an invariant of the force tensor. Direction cosines relate unit vectors in the ‘‘new’’ and ‘‘old’’ coordinate systems. For example, a unit vector directed along the xV-axis (i.e., unit vector ıˆV) is related to the unit vectors in the x–y–z coordinate system as follows: iVˆ ¼ cx Vx iˆ þ cxVy jˆ þ cx Vz kˆ

ð7Þ

ðcx Vx Þ2 þ ðcxVy Þ2 þ ðcxVz Þ2 ¼ 1

ð8Þ

Since iˆ V is a unit vector, then in accordance with Eq. (2):

To this point, we have referred to a force as a vector. A force vector can also be called a force tensor. The term ‘‘tensor’’ refers to any quantity that transforms in a physically meaningful way from one Cartesian coordinate system to another. The rank of a tensor equals the number of subscripts that must be used to describe the tensor. A force can be described using a single

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subscript, Fi, and therefore a force is said to be a tensor of rank one, or equivalently, a first-order tensor. Equations (6a)–(6c) is called the transformation law for a first-order tensor. It is likely that the reader is already familiar with two other tensors: the stress tensor, rij, and the strain tensor, eij. The stress and strain tensors will be reviewed later in this chapter, but at this point, it can be noted that two subscripts are used to describe stress and strain tensors. Hence, stress and strain tensors are said to be tensors of rank two, or equivalently, second-order tensors. Example Problem 1 Given. All components of a force vector F are known in a given x–y–z coordinate system. It is desired to express this force in a new xU–yU–zU coordinate system, where the xU–yU–zU is generated from the original x–y– z coordinate system by the following two rotations (see Fig. 2): 

A rotation of h about the original z-axis (which defines an intermediate xV–yV–zV coordinate system), followed by  A rotation of b about the xV-axis (which defines the final xU–yU–zU coordinate system). Problem. (a) Determine the direction cosines ciUj relating the original x–y–z coordinate system to the new xU–yU–zU coordinate system; (b) obtain a general expression for the force vector F in the xU–yU–zU coordinate system; and (c) calculate numerical values of the force vector F in the xU–yU–zU coordinate system if h=20j, b=60j, and Fx=1000 N, Fy=200 N, Fz=600 N. Solution Part (a). One way to determine the direction cosines ciUj is to rotate unit vectors. In this approach, unit vectors are first rotated from the original x–y–z coordinate system to the intermediate xV–yV–zV coordinate system, and then from the xV–yV–zV system to the final xU–yU–zU coordinate system. Define a unit vector I that is aligned with the x-axis: I u ð1Þiˆ That is, vector I is a vector for which Ix=1, Iy=0, and Iz=0. The vector I can be rotated to the intermediate xV–yV–zV coordinate system using Eqs. (6a)–(6c): IxV ¼ cxV x Ix þ cxV y Iy þ cxV z Iz

IyV ¼ cyV x Ix þ cyV y Iy þ cyV z Iz

IzV ¼ czV x Ix þ czV y Iy þ czV z Iz

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Figure 2 Generation of the xW–yW–zW coordinate system from the x–y–z coordinate system.

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The direction cosines associated with a transformation from the x–y–z coordinate system to the intermediate xV–yV–zV coordinate system can be determined by inspection [see Fig. 2(a)] and are given by: cxV x ¼ cosine ðangle between xV- and x-axesÞ ¼ cos h

cxV y ¼ cosine ðangle between xV- and y-axesÞ ¼ cosð90B hÞ ¼ sin h cxV z ¼ cosine ðangle between xV- and z-axesÞ ¼ cosð90B Þ ¼ 0 cyV x ¼ cosine ðangle between the yV- and x-axesÞ ¼ cosð90B þ hÞ ¼

sin h

cyV y ¼ cosine ðangle between the yV- and y -axesÞ ¼ cos h

cyV z ¼ cosine ðangle between the yV- and z-axesÞ ¼ cosð90B Þ ¼ 0 czV x ¼ cosine ðangle between the zV- and x-axesÞ ¼ cosð90B Þ ¼ 0 czV y ¼ cosine ðangle between the zV- and y-axesÞ ¼ cosð90B Þ ¼ 0

czV z ¼ cosine ðangle between the zV- and z-axesÞ ¼ cosð0B Þ ¼ 1 Using these direction cosines:

IxV ¼ cxV x Ix þ cxV y Iy þ cxV z Iz ¼ ðcos hÞð1Þ þ ðsin hÞð0Þ þ ð0Þð0Þ ¼ cos h IyV ¼ cyV x Ix þ cyV y Iy þ cyV z Iz ¼ ð sin hÞð1Þ þ ðcos hÞð0Þ þ ð0Þð0Þ ¼ sin h IzV ¼ czV x Ix þ czV y Iy þ czV z Iz ¼ ð0Þð1Þ þ ð0Þð0Þ þ ð1Þð0Þ ¼ 0

Therefore, in the xV–yV–zV coordinate system, the vector I is written as: I ¼ ðcos hÞiˆ V þ ð sin hÞjˆ V Now define two additional unit vectors, one aligned with the original y-axis (vector J ) and one aligned with the original z-axis (vector K ), i.e., let J=(1) jˆ and K=(1) kˆ. Transforming these vectors to the xV–yV–zV coordinate system, again using the direction cosines listed above, results in: J ¼ ðsin hÞiˆ V þ ðcos hÞjˆ V K ¼ ð1Þ kˆ V We now rotate vectors I , J, and K from the intermediate xV–yV–zV coordinate system to the final xU–yU–zU coordinate system. The direction cosines associated with a transformation from the xV–yV–zV coordinate system to the final xU–yU–zU coordinate system are easily determined by inspection [see Fig. 2(b)] and are given by: cx Ux V ¼ 1 cy Ux V ¼ 0 cz Ux V ¼ 0

cx Uy V ¼ 0 cy Uy V ¼ cos b cz Uy V ¼ sin b

cx Uz V ¼ 0 cy Uz V ¼ sin b cz Uz V ¼ cos b

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These direction cosines together with Eqs. (6a)–(6c) can be used to rotate the vector I from the intermediate xV–yV–zV coordinate system to the final xU–yU–zU coordinate system: Ix W ¼ cxW xV IxV þ cxW yV IyV þ cxW zV IzV ¼ ð1Þðcos hÞ þ ð0Þð sin hÞ þ ð0Þð0Þ IxW ¼ cos h

IyW ¼ cyWxV IxV þ cyWyV IyV þ cyWzV IzV ¼ð0Þðcos hÞ þ ðcos bÞð sin hÞ þ ðsin bÞð0Þ

Iy W ¼

cos b sin h

IzW ¼ czWxV IxV þ czWyV IyV þ czWzV IzV ¼ð0Þðcos hÞ þ ð sin bÞð sin hÞ þ ðcosbÞð0Þ IzW ¼ sin b sin h

Therefore, in the final xU–yU–zU coordinate system, the vector I is written as: I ¼ ðcos hÞiˆ W þ ð cos b sin hÞjˆ W þ ðsin b sin hÞkˆ W

ðaÞ

Recall that in the original x–y–z coordinate system, I is simply a unit vector aligned with the original x-axis: I u (1)ıˆ. Therefore result (a) defines the direction cosines associated with the angle between the original x-axis and the final xU-, yU-, and zU-axes. That is: cxWx ¼ cosine ðangle between xU- and x-axesÞ ¼ cos h cyWx ¼ cosine ðangle between yU- and x-axesÞ ¼ cos b sin h czWx ¼ cosine ðangle between zU- and x-axesÞ ¼ sin b sin h

A similar procedure is used to rotate the unit vectors J and K from the intermediate xV–yV–zV coordinate system to the final xU–yU–zU coordinate system. These rotations result in: J ¼ ðsin hÞiˆ W þ ðcos b cos hÞjˆ W þ ð sin b cos hÞkˆ W K ¼ ð0Þiˆ W þ ðsin bÞjˆ W þ ðcos bÞkˆ W

ðbÞ ðcÞ

Since vector J is a unit vector aligned with the original y-axis, J=(1)jˆ, result (b) defines the direction cosines associated with the angle between the original y-axis and the final xU-, yU-, and zU-axes: cxWy ¼ cosine ðangle between xU- and y-axesÞ ¼ sin h

cyWy ¼ cosine ðangle between yU- and y-axesÞ ¼ cos b cos h

czWx ¼ cosine ðangle between zU- and y-axesÞ ¼

sin b cos h

Finally, result (c) defines the direction cosines associated with the angle between the original z-axis and the final xU-, yU-, and zU-axes: cxWz ¼ cosineð angle between xU- and z-axesÞ ¼ 0

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cyWz ¼ cosine ðangle between yU-and z-axesÞ ¼ sin b

czWz ¼ cosine ðangle between zU-and z-axesÞ ¼ cos b

Assembling the preceding results, the set of direction cosines relating the original x–y–z coordinate system to the final xU–yU–zU coordinate system can be written as: 2 3 2 3 cxWx cxWy cxWz cos h sin h 0 4 cyWx cyWy cyWz 5 ¼ 4 cos b sin h cos b cos h sin b 5 czWx czWy czWz sin b sin h sin b cos h cos b Part (b). Since the direction cosines have been determined, transformation of force vector F can be accomplished using any version of Eqs. (6a)–(6c). For example, using matrix notation, Eq. (6c): 9 2 8 38 9 cxWx cxWy cxWz < Fx = < FxW = F W ¼ 4 cyWx cyWy cyWz 5 Fy : ; : y ; czWx czWy czWz Fz FzW 2 38 9 cos h sin h 0 < Fx = ¼ 4 cos b sin h cos b cos h sin b 5 Fy : ; sin b sin h sin b cos h cos b Fz 9 9 8 8 ðcos hÞFx þ ðsin hÞFy = < FxW = < FyW ¼ ð cos b sin hÞFx þ ðcos b cos hÞFy þ ðsin bÞFz ; ; : : ðsin b sin hÞFx þ ð sin b cos hÞFy þ ðcos bÞFz FzW

Part (c). Using the specified numerical values and the results of Part (b): 9 9 8 8 ðcos 20B Þð1000 NÞ þ ðsin 20B Þ200 N = < FxW = < FyW ¼ ð cos 60B sin 20B Þð1000 NÞ þ ðcos 60B cos 20B Þð200 NÞ þ ðsin 60B Þð600 NÞ ; ; : : FzW ðsin 60B sin 20B Þð1000 NÞ þ ð sin 60B cos 20B Þð200 NÞ þ ðcos 60B Þð600 NÞ 9 8 8 9 < FxW = < 1008 N = F W ¼ 442:6 N : y ; : ; FzW 433:4 N

Using vector notation, F can now be expressed in the two different coordinate systems as: F ¼ ð1000 NÞiˆ þ ð200 NÞjˆ þ ð600 NÞkˆ or equivalently F ¼ ð1008 NÞiˆ W þ ð442:6 NÞjˆ W þ ð433:4 NÞkˆ W

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where ıˆ, jˆ, kˆ and ıˆ U, jˆ U, kˆ U are unit vectors in the x–y–z and xU–yU–zUcoordinate systems, respectively. Force vector F drawn in the x–y–z and xU–yU– zUcoordinate systems is shown in Fig. 3(a) and (b), respectively. The two descriptions of F are entirely equivalent. A convenient way of (partially) verifying this equivalence is to calculate the magnitude of the original and transformed force vectors. Since the magnitude is an invariant, it is inde-

Figure 3 Force vector F drawn in the x–y–z and xW–yW–zW coordinate systems.

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pendent of the coordinate system used to describe the force vector. Using Eq. (2), the magnitude of the force vector in the x–y–z coordinate system is: qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi jF j ¼ F x2 þ F y2 þ F z2 ¼ ð1000 NÞ2 þ ð200 NÞ2 þ ð600 NÞ2 ¼ 1183 N The magnitude of the force vector in the xU–yU–zU coordinate system is: ffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 2  jF j ¼ ðFxWÞ þ FyW þðFzWÞ ¼ ð1008 NÞ2 þ ð442:6 NÞ2 þ ð433:4 NÞ2 ¼ 1183N ðagreesÞ

3 NORMAL FORCES, SHEAR FORCES, AND FREE-BODY DIAGRAMS A force F acting at an angle to a planar surface is shown in Fig. 4. Since force is a vector, it can always be decomposed into two force components, a normal force component and a shear force component. The line-of-action of the normal force component is orthogonal to the surface, whereas the line-of-action of the shear force component is tangent to the surface. Internal forces induced within a solid body by externally applied forces can be investigated with the aid of free-body diagrams. A simple example is shown in Fig. 5, which shows a straight circular rod with constant diameter subjected to two external forces of equal magnitude (R) but opposite direction. The internal force (F I, say) induced at any cross section of the rod can be investigated by making an imaginary cut along the plane of interest. Suppose an imaginary cut is made along plane a-a, which is perpendicular to the axis

Figure 4 A force F acting at an angle to a planar surface.

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Figure 5 The use of free-body diagrams to determine internal forces acting on planes a-a and b-b: (a) free-body diagram based on plane a–a–a–a perpendicular to rod axis; (b) free-body diagram based on plane b–b–b–b, inclined at angle h to rod axis.

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of the rod. The resulting free-body diagram for the lower half of the rod is shown in Fig. 5(a), where an x–y–z coordinate system has been assigned such that the x-axis is parallel to the rod axis, as shown. On the basis of this freebody diagram, it is concluded that an internal force F I = (R)ıˆ+(0)jˆ+(0)kˆ is induced at cross-section a-a. That is, only a normal force of magnitude R is induced at cross-section a-a, which has been defined to be perpendicular to the axis of the rod. On the other hand, the imaginary cut need not be made perpendicular to the axis of the rod. Suppose the imaginary cut is made along plane b-b, which is oriented at an angle of h with respect to the axis of the rod. The resulting free-body diagram for the lower half of the rod is shown in Fig. 5(b). A new xV–yV–zV coordinate system has been assigned so that the xV-axis is perpendicular to plane b-b and the zV-axis is coincident with the z-axis—that is, the xV–yV–zV coordinate system is generated from the x–y–z coordinate system by a rotation of h about the original z-axis. The internal force FI can be expressed with respect to the xV–yV–zV coordinate system by transforming FI from the x–y–z coordinate system to the xV–yV–zV coordinate system. This coordinate transformation is a special case of the transformation considered in Example Problem 1. The direction cosines now become (with b=0j): cxV x ¼ cos h

cyV x ¼ sin h czV x ¼ 0

cxV y ¼ sin h

cyV y ¼ cos h czV y ¼ 0

c xV z ¼ 0 cy V z ¼ 0 czV z ¼ 1

Applying Eqs. (6a)–(6c), we have: 9 2 8 38 9 38 9 2 c xV x c xV y c xV z < F x = cos h sin h 0 < R = < FxV = F V ¼ 4 cyV x cyV y cyV z 5 Fy ¼ 4 sin h cos h 0 5 0 : ; : y ; : ; cz V x czV y czV z 0 0 1 Fz FzV 0 8 9 < ðcos hÞR = ¼ ð sin hÞR : ; 0

In the xV–yV–zV coordinate system, the internal force is F I = (R cos h)ıˆV (R sin h)jˆV+(0)kˆV. Hence, by defining a coordinate system which is inclined to the axis of the rod, we conclude that both a normal force (R cos h) and a shear force (R sin h) are induced in the rod. Although the preceding discussion may seem simplistic, it has been included in order to demonstrate the following: 

A specific coordinate system must be specified before a force vector can be defined in a mathematical sense. In general, the coordinate

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system is defined by the imaginary cut(s) used to form the free-body diagram.  All components of a force must be specified to fully define the force vector. Furthermore, the individual components of a force change as the vector is transformed from one coordinate system to another. These two observations are valid for all tensors, not just for force vectors. In particular, these observations hold in the case of stress and strain tensors, which will be reviewed in the following sections.

4 DEFINITION OF STRESS There are two fundamental types of stress: normal stress and shear stress. Both types of stress are defined as a force divided by the area over which it acts. A general 3-D solid body subjected to a system of external forces is shown in Fig. 6(a). It is assumed that the body is in static equilibrium, that is, it is assumed that the sum of all external forces is zero, SF i = 0. These external forces induce internal forces acting within the body. In general, the internal forces will vary in both magnitude and direction throughout the body. An illustration of the variation of internal forces along a line within an internal plane is shown in Fig. 6(b). A small area (DA) isolated from this plane is shown in Fig. 6(c). Area DA is assumed to be ‘‘infinitesimally small.’’ That is, the area DA is small enough such that the internal forces acting over DA can be assumed to be of constant magnitude and direction. Therefore, the internal forces acting over DA can be represented by a force vector which can be decomposed into a normal force, N, and a shear force, V, as shown in Fig. 6(c). Normal stress (usually denoted r) and shear stress (usually denoted s) are defined as the force per unit area acting perpendicular and tangent to the area DA, respectively. That is, N DA!0 DA

r u lim

V DA!0 DA

s u lim

ð9Þ

Note that by definition, the area DA shrinks to zero: DA!0. Stresses r and s are therefore said to exist ‘‘at a point.’’ Also, since internal forces generally vary from point-to-point (as shown in Fig. 6), stresses also vary from pointto-point. Stress has units of force per unit area. In SI units, stress is reported in terms of Pascals (abbreviated Pa), where 1 Pa=1 N/m2. In English units, stress is reported in terms of pounds-force per square inch (abbreviated psi),

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Figure 6 A solid 3-D body in equilibrium.

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that is, 1 psi=1 lbf/in2. Conversion factors between the two systems of measurement are 1 psi=6895 Pa, or equivalently, 1 Pa=0.145010 3 psi. Common abbreviations used throughout this text are as follows: 1  103 Pa ¼ 1 kilo-Pascals ¼ 1 kPa

1  103 psi ¼ 1 kilo-psi ¼ 1 ksi

1  106 Pa ¼ 1 Mega-Pascals ¼ 1 MPa 1  106 psi ¼ 1 mega-psi ¼ 1 Msi 1  109 Pa ¼ 1 Giga-Pascals ¼ 1 GPa

5 THE STRESS TENSOR A general 3-D solid body subjected to a system of external forces is shown in Fig. 7(a). It is assumed that the body is in static equilibrium and that body forces are negligible, that is, it is assumed that the sum of all external forces is zero, SF i = 0. A free-body diagram of an infinitesimally small cube removed from the body is shown in Fig. 7(b). The cube is referenced to an x– y–z coordinate system, and the cube edges are aligned with these axes. The lengths of the cube edges are denoted dx, dy, and dz. Although (in general) internal forces are induced over all six faces of the cube, for clarity, the forces acting on only three faces have been shown. The force acting over each cube face can be decomposed into a normal force component and two shear force components, as illustrated in Fig. 7(c). Although each component could be identified with a single subscript (since force is a first-order tensor), for convenience, two subscripts have been used. The first subscript identifies the face over which the force is distributed, while the second subscript identifies the direction in which the force is oriented. For example, Nxx refers to a normal force component which is distributed over the x-face and which ‘‘points’’ in the x-direction. Similarly, Vzy refers to a shear force distributed over the z-face which ‘‘points’’ in the y-direction. Three stress components can now be defined for each cube face, in accordance with Eq. (9). For example, for the three faces of the infinitesimal element shown in Fig. 7: Stresses acting on the +x-face:       Vxy Nxx Vxz rxx ¼ lim sxy ¼ lim sxz ¼ lim dy;dz!0 dydz dy;dz!0 dydz dy;dz!0 dydz Stresses acting on the y-face:       Nyy Vyx Vyz ryy ¼ lim syx ¼ lim syz ¼ lim dx;dz!0 dxdz dx;dz!0 dxdz dx;dz!0 dxdz

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Figure 7 Free-body diagrams used to define stress induced in a solid body.

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Stresses acting on the +z-face: rzz ¼ lim



dx;dy!0

Nzz dxdy



szx ¼ lim



dx;dy!0

Vzx dxdy



szy ¼ lim



dx;dy!0

Vzy dxdy



Since three force components (and therefore three stress components) exist on each of the six faces of the cube, it would initially appear that there are 18 independent force (stress) components. However, it is easily shown that for static equilibrium to be maintained (assuming body forces are negligible): 

Normal forces acting on opposite faces of the infinitesimal element must be of equal magnitude and opposite direction, and  Shear forces acting within a plane of the element must be orientated either ‘‘tip-to-tip’’ [e.g., forces Vxz and Vzx in Fig. 7(c)] or ‘‘tail-totail’’ (e.g., forces Vzy and Vyz) and be of equal magnitude. That is, jVxy=Vyxj, jVxz=Vzxj, and jVyz=Vzyj. These restrictions reduce the number of independent force (stress) components from 18 to 6, as follows:

Independent force components

Independent stress components

Nxx Nyy Nzz Vxy ð¼ Vyx Þ Vxz ð¼ Vzx Þ Vyz ð¼ Vzy Þ

rxx ryy rzz sxy ð¼ syx Þ sxz ð¼ szx Þ syz ð¼ szy Þ

An infinitesimal element showing all stress components is shown in Fig. 8. We must next define the algebraic sign convention we will use to describe individual stress components. We first associate an algebraic sign with each face of the infinitesimal element. A cube face is positive if the outward unit normal of the face (that is, the unit normal pointing away from the interior of the element) points in a positive coordinate direction; otherwise, the face is negative. For example, face (ABCD) in Fig. 8 is a positive face, while face (CDEF) is a negative face.

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Figure 8 An infinitesimal stress element (all stress components shown in a positive sense).

Having identified the positive and negative faces of the element, a stress component is positive if: 

The stress component acts on a positive face and points in a positive coordinate direction, or if  The stress component acts on a negative face and points in a negative coordinate direction otherwise, the stress component is negative. This convention can be used to confirm that all stress components shown in Fig. 8 are algebraically positive. For example, to determine the algebraic sign of the normal stress rxx which acts on face ABCD in Fig. 8, note that (a) face ABCD is positive and (b) the normal stress rxx which acts on this

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face points in the positive y-direction. Therefore, rxx is positive. As a second example, the shear stress syz which acts on cube face CDEF is positive because (a) face CDEF is negative and (b) syz points in the negative z-direction. The preceding discussion shows that the state of stress at a point is defined by six components of stress: three normal stress components and three shear stress components. The state of stress is written using matrix notation as follows: 3 2 3 2 rxx sxy sxz rxx sxy sxz 4 syx ryy syz 5 ¼ 4 sxy ryy syz 5 ð10Þ sxz syz rzz szx szy rzz To express the state of stress using indicial notation, we must first make the following change in notation: sxy ! rxy

sxz ! rxz syx ! ryx

syz ! ryz

szx ! rzx szy ! rzy With this change, becomes: 2 rxx sxy 4 syx ryy szx szy

the matrix on the left side of the equality sign in Eq. (10) 2 3 rxx sxz syz 5 ! 4 rxy rxz rzz

rxy ryy ryz

3 rxz ryz 5 rzz

which can be succinctly written using indicial notation as: rij ;

i; j ¼ x; y; or z

ð11Þ

In Sec. 1, it was noted that a force vector is a first-order tensor since only one subscript is required to describe a force tensor, Fi. From Eq. (11), it is clear that stress is a second-order tensor (or equivalently, a tensor of rank two) since two subscripts are required to describe a state of stress. Example Problem 2 Given. The stress element referenced to an x–y–z coordinate system and subject to the stress components shown in Fig. 9.

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Figure 9 Stress components acting on an infinitesimal element (all stresses in MPa).

Determine. Label all stress components, including algebraic sign. Solution. The magnitude and algebraic sign of each stress component are determined using the sign convention defined above. The procedure will be illustrated using the stress components acting on face CDEF. First, note that face CDEF is a negative face since an outward unit normal for this face points in the negative y-direction. The normal stress which acts on face CDEF has a magnitude of 50 MPa and points in the positive y-direction. Hence, this stress component is negative and is labeled ryy= 50 MPa. One of the shear stress components acting on face CDEF has a magnitude of 75 MPa and points in the positive x-direction. Hence, this stress component is also negative and is

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labeled syx= 75 MPa (or equivalently, sxy= 75 MPa). Finally, the second shear force component acting on face CDEF has a magnitude of 50 MPa and points in the positive z-direction. Hence, this component is labeled syz= 50 MPa (or equivalently, szy= 50 MPa). Following this process for all faces of the element, the state of stress represented by the element shown in Fig. 9 can be written as: 2

rxx 4 syx szx

sxy ryy szy

3 2 sxz 100 MPa syz 5 ¼ 4 75 MPa rzz 30 MPa

75 MPa 50 MPa 50 MPa

3 30 MPa 50 MPa 5 25 MPa

6 TRANSFORMATION OF THE STRESS TENSOR In Sec. 5, the stress tensor was defined using a free-body diagram of an infinitesimal element removed from a 3-D body in static equilibrium. This concept is again illustrated in Fig. 10(a), which shows the stress element referenced to an x–y–z coordinate system. Now, the infinitesimal element need not be removed in the orientation shown in Fig. 10(a). An infinitesimal element removed from precisely the same point within the body but at a different orientation is shown in Fig. 10(b). This stress element is referenced to a new xV–yV–zV coordinate system. The state of stress at the point of interest is dictated by the external loads applied to the body and is independent of the coordinate system used to describe it. Hence, the stress tensor referenced to the xV–yV–zVcoordinate system is equivalent to the stress tensor referenced to the x–y–z coordinate system, although the direction and magnitude of individual stress components will differ. The process of relating stress components in one coordinate system to those in another is called the transformation of the stress tensor. This terminology is perhaps unfortunate in the sense that the state of stress itself is not ‘‘transformed,’’ but rather our description of the state of stress transforms as we change from one coordinate system to another. It can be shown (1,2) that the stress components in the new xV–yV–zV coordinate system (riVjV) are related to the components in the original x–y–z coordinate system (rij) according to: ri V j V ¼ ci V k cj V l rkl

where i; j; k; l ¼ x; y; z

or equivalently (using matrix notation): ½ri V j V Š ¼ ½ci V j Š½rij Š½ci V j ŠT

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ð12aÞ

Figure 10 Infinitesimal elements removed from the same point from a 3-D solid but in two different orientations.

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where [ciVj]T is the matrix form: 2 rx V x V r x V y V 4 ryV xV ryV yV rzV xV rzV yV

transpose of the direction cosine array. Writing in full 3 2 32 c xV x c xV y c xV z rxV zV rxx ryV zV 5 ¼ 4 cyV x cyV y cyV z 54 ryx rz V z V czV x cz V y czV z rzx 2 3 c xV x c y V x c z V x  4 cxV y cyV y czV y 5 c xV z c y V z c z V z

rxy ryy rzy

3 rxz ryz 5 rzz ð12bÞ

As discussed in Sec. 2, the terms ciVj which appear in Eqs. (12a) and (12b) are direction cosines and equal the cosine of the angle between the axes of the x–y–z and xV–yV–zV coordinate systems. Recall that the algebraic sign of an angle of rotation is defined in accordance with the right-hand rule and that angles are defined from the x–y–z coordinate system to xV–yV–zV coordinate system. Equations (12a) and (12b) are called the transformation law for a second-order tensor. If an analysis is being performed with the aid of a digital computer, which nowadays is almost always the case, then matrix notation [Eq. (12b)] most likely will be used to transform a stress tensor from one coordinate system to another. Conversely, if a stress transformation is to be accomplished using hand calculations, then indicial notation [Eq. (12a)] may be the preferred choice. To apply Eq. (12a), desired values are first specified for subscripts iV and jV, and then the terms on the right side of the equality are summed over the entire range of the remaining two subscripts, k and l. For example, suppose we wish to write the relationship between rxVzV and the stress components in the x–y–z coordinate system in expanded form. We first specify that iV=xV and jV=zV, and Eq. (12a) becomes: rxV zV ¼ cxV k czV l rkl

where k; l ¼ x; y; z

We then sum all terms on the right side of the equality by cycling through the entire range of k and l. In expanded form, we have: rxV zV ¼ cxV x czV x rxx þ cxV x czV y rxy þ cxV x czV z rxz þ cxV y czV x ryx þ cxV y czV y ryy þ cxV y czV z ryz þ cxV z czV x rzx þ cxV z czV y rzy þ cxV z czV z rzz

ð13Þ

Equations (12a) and (12b) show that the value of any individual stress component riVjV varies as the stress tensor is transformed from one coordinate system to another. However, it can be shown (1,2) that there are features of the total stress tensor that do not vary when the tensor is transformed from one coordinate system to another. These features are called the stress in-

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variants. For a second-order tensor, three independent stress invariants exist and are defined as follows: First stress invariant ¼ H ¼ rii ð14aÞ  1 rii rjj rij rij Second stress invariant ¼ U ¼ ð14bÞ 2  1 Third stress invariant ¼ W ¼ rii rjj rkk 3rii rjk rjk þ 2rij rjk rki 6 ð14cÞ Alternatively, by expanding these equations over the range i, j, k=x, y, z and simplifying, the stress invariants can be written as: First stress invariant ¼ H ¼ rxx þ ryy þ rzz

Second stress invariant ¼ U ¼ rxx ryy þ rxx rzz þ ryy rzz   r2xy þ r2xz þ r2yz

Third stress invariant ¼ W ¼ rxx ryy rzz

rxx r2yz

ð15aÞ ð15bÞ

ryy r2xz

rzz r2xy þ 2rxy rxz ryz

ð15cÞ

The three stress invariants are conceptually similar to the magnitude of a force tensor. That is, the value of the three stress invariants is independent of the coordinate used to describe the stress tensor, just as the magnitude of a force vector is independent of the coordinate system used to describe the force. This invariance will be illustrated in the following example problem. Example Problem 3 Given. A state of stress referenced to an x–y–z coordinate is known to be: 2 3 2 3 rxx rxy rxz 50 10 15 4 ryx ryy ryz 5 ¼ 4 10 25 30 5ðksiÞ rzx rzy rzz 15 30 5 It is desired to express this state of stress in an xU–yU–zU coordinate system, generated by the following two sequential rotations: (i) Rotation of h=20j about the original z-axis (which defines an intermediate xV–yV–zV coordinate system), followed by (ii) Rotation of b=35j about the xV-axis (which defines the final xU–yU– zU coordinate system). Problem. (a) Rotate the stress tensor to the xU–yU–zU coordinate system and (b) calculate the first, second, and third invariants of the stress tensor using

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both elements of the stress tensor referenced to the x–y–z coordinate system, rij, and elements of the stress tensor referenced to the xU–yU–zU coordinate system, riUjU. Solution Part (a). General expressions for direction cosines relating the x–y–z and xU–yU–zU coordinate systems were determined as a part of Example Problem 1. The direction cosines were found to be: cxWx ¼ cos h cxWy ¼ sin h

cxWz ¼ 0 cyWx ¼ cos b sin h cyWy ¼ cos b cos h

cyWz ¼ sin b

czWx ¼ sin b sin h czWy ¼ sin b cos h

czWz ¼ cos b

Since in this problem h=20j and b=35j, the numerical values of the direction cosines are: cxWx ¼ cos ð20B Þ ¼ 0:9397

cxWy ¼ sin ð20B Þ ¼ 0:3420

cxWz ¼ 0 cyWx ¼

cos ð35B Þsin ð20B Þ ¼

0:2802

cyWy ¼ cos ð35 Þcos ð20 Þ ¼ 0:7698 B

B

cyWz ¼ sin ð35B Þ ¼ 0:5736

czWx ¼ sin ð35B Þsin ð20B Þ ¼ 0:1962 czWy ¼

sin ð35B Þcos ð20B Þ ¼

0:5390

czWz ¼ cos ð35 Þ ¼ 0:8192 B

Each component of the transformed stress tensor is now found through the application of either Eq. (12a) or Eq. (12b). For example, if indicial notation is used, stress component rxUzU can be found using Eq. (13): rxWzW ¼ cxWx czWx rxx þ cxWx czWy rxy þ cxWx czWz rxz þ cxWy czWx ryx þ cxWy czWy ryy þ cxWy czWz ryz þ cxWz czWx rzx þ cxWz czWy rzy þ cxWz czWz rzz

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rxW zW ¼ ð0:9397Þð0:1962Þð50 ksiÞ þ ð0:9397Þð 0:5390Þð 10 ksiÞ

þð0:9397Þð0:8192Þð15 ksiÞ þ ð0:3420Þð0:1962Þð 10 ksiÞ þð0:3420Þð 0:5390Þð25 ksiÞ þ ð0:3420Þð0:8192Þð30 ksiÞ þð0Þð0:1962Þð15ksiÞþð0Þð 0:5390Þð30ksiÞþð0Þð0:8192Þð 5 ksiÞ

rxWzW ¼ 28:95 ksi

Alternatively, if matrix notation is used, then Eq. (12b) becomes: 2 3 2 32 rxWxW rxWyW rxWzW 0:9397 0:3420 0 50 10 4ryWxW ryWyW ryWzW5 ¼4 0:2802 0:7698 0:573654 10 25 rzWxW rzWyW rzWzW 0:1962 0:5390 0:8192 15 30 2 3 0:9397 0:2802 0:1962 4 0:3420 0:7698 0:53905 0 0:5736 0:8192

3 15 30 5 5

Completing the matrix multiplication indicated yields the following: 2 3 2 3 rxWxW rxWyW rxWzW 40:65 1:113 28:95 4 ryWxW ryWyW ryWzW 5 ¼ 4 1:113 43:08 10:60 5 ðksiÞ rzWxW rzWyW rzWzW 28:95 10:60 13:72

Notice that the value of rxUzU determined through matrix multiplication is identical to that obtained using indicial notation, as previously described. The stress element is shown in the original and final coordinate systems in Fig. 11. Part (b). The first, second, and third stress invariants will now be calculated using components of both rij and riUjU. It is expected that identical values will be obtained since the stress invariants are independent of the coordinate system. First stress invariant: x–y–z coordinate system: H ¼ rii ¼ rxx þ ryy þ rzz H ¼ ð50 þ 25 H ¼ 70 ksi

5Þ ksi

xU–yU–zU coordinate system: H ¼ ri Wi W ¼ rxWxW þ ryWyW þ rzWzW H ¼ ð40:65 þ 43:08

13:72Þ

H ¼ 70 ksi As expected, the first stress invariant is independent of the coordinate system.

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Figure 11 Stress tensor of Example Problem 3 referenced to two different coordinates (magnitude of all stress components in ksi).

Second stress invariant: x–y–z coordinate system: U¼

1 rii rjj 2

 rij rij ¼ rxx ryy þ rxx rzz þ ryy rzz   r2xy þ r2xz þ r2yz

n U ¼ ð50Þð25Þ þ ð50Þð 5Þ þ ð25Þð 5Þ o þ ð30Þ2 Š ðksiÞ2

U¼

½ð 10Þ2 þ ð15Þ2

350ðksiÞ2

xU–yU–zU coordinate system: U¼

1 r W Wr W W 2 ii jj

 ri Wj Wri Wj W

U ¼ rxWxWryWyW þ rxWxWrzWzW þ ryWyWrzWzW



 r2xWyW þ r2xWzW þ r2yWzW

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n U ¼ ð40:65Þð43:08Þ þ ð40:65Þð 13:72Þ þ ð43:08Þð 13:72Þ o ½ð1:113Þ2 þ ð28:95Þ2 þ ð 10:60Þ2 Š ðksiÞ2

U¼

350ðksiÞ2

As expected, the second stress invariant is independent of the coordinate system. Third stress invariant: x–y–z coordinate system:  1 W ¼ rii rjj rkk 3rii rjk rjk þ 2rij rjk rki 6 W ¼ rxx ryy rzz rxx r2yz ryy r2xz rzz r2xy þ 2rxy rxz ryz W ¼ ½ð50Þð25Þð 5Þ

ð50Þð30Þ2

ð25Þð15Þ2

ð 5Þð 10Þ2

þ 2ð 10Þð15Þð30ފ ðksiÞ3 W¼

65375ðksiÞ3

xU–yU–zU coordinate system:  1 W ¼ ri Wi Wrj Wj WrkWkW 3ri Wi Wrj WkWrj WkW þ 2ri Wj Wrj WkWrkWi W 6 W ¼ rxWxWryWyWrzWzW rxWxWr2yWzW ryWyWr2xWzW rzWzWr2xWyW þ 2rxWyWrxWzWryWzW W ¼ ½ð40:65Þð43:08Þð 13:72Þ  ð28:95Þ2

ð40:65Þð 10:60Þ2

ð43:08Þ

ð 13:72Þð1:113Þ2 þ 2ð1:113Þð28:95Þ

 ð 10:60ފ ðksiÞ3 W¼

65375ðksiÞ3

As expected, the third stress invariant is independent of the coordinate system. 7 PRINCIPAL STRESSES The definition of a stress tensor was reviewed in Sec. 5, and transformation of a stress tensor from one coordinate system to another was discussed in Sec. 6. It can be shown (1,2) that it is always possible to rotate the stress tensor to a special coordinate system in which no shear stresses exist. This coordinate system is called the principal stress coordinate system, and the normal stresses

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that exist in this coordinate system are called the principal stresses. In most textbooks, principal stresses are denoted r1, r2, and r3. However, as will be discussed later (see Fig. 2 in Chap. 3) in this textbook, the labels ‘‘1,’’ ‘‘2,’’ and ‘‘3’’ will be used to label the axes in a special coordinate system called the principal material coordinate system. Therefore, in this text, the axes associated with the principal stress coordinate system will be labeled p1, p2, and p3 axes, and the principal stresses will be denoted rp1, rp2, and rp3. Knowledge of the principal stresses induced in an isotropic structure is of extreme importance, primarily because failure of isotropic materials (e.g., yielding and/or fracture) can often be directly related to the magnitude(s) of the principal stresses. This is not the case for composite structures, however. As will be seen in later chapters, failure of composite structures is not governed by principal stresses, and hence the topic of principal stresses is only of occasional importance to the composite engineer. Principal stresses may be related to stress components in an x–y–z coordinate system using the free-body diagram shown in Fig. 12. It is assumed that plane ABC is one of the three principal planes (i.e., n = 1, 2, or 3) and therefore no shear stress exists on this plane. The line-of-action of principal stress rpn defines one axis of the principal stress coordinate system. The direction cosines between this principal axis and the x-, y-, and z- axes are cpnx, cpny, and cpnz, respectively. The surface area of triangle ABC is denoted AABC. The normal force acting over triangle ABC therefore equals (rpnAABC). The components of the normal force acting in the x-, y-, and z-directions equal cpnxrpnAABC, cpnyrpnAABC, and cpnzrpnAABC, respectively. The area of the other triangular faces are given by: Area of triangle ABD=cpnxAABC. Area of triangle ACD=cpnyAABC. Area of triangle BCD=cpnzAABC. Summing forces in the x-direction and equating to zero, we obtain: cpn x rpn AABC

rxx cpn x AABC

sxy cpn y AABC

sxz cpn z AABC ¼ 0

which can be reduced and simplified to: ðrpn

rxx Þcpn x

sxy cpn y

sxz cpn z ¼ 0

ð16aÞ

Similarly, summing forces in the y- and z-directions results in: sxy cpn x þ ðrpn

sxz cpn x

ryy Þcpn y

syz cpn y þ ðrpn

syz cpn z ¼ 0

rzz Þcpn z ¼ 0

ð16bÞ ð16cÞ

Equations (16a)–(16c) represent three linear homogeneous equations which must be satisfied simultaneously. Since direction cosines cpnx, cpny, and cpnz must also satisfy Eq. (8), and therefore cannot all equal zero, the solu-

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Figure 12 Free-body diagram used to relate stress components in the x–y–z coordinate system to a principal stress.

tion can be obtained by requiring that the determinant of the coefficients of cpnx, cpny, and cpnz equal zero:    ðrpn rxx Þ sxy sxz    sxy ðrpn ryy Þ syz  ¼ 0   sxz syz ðrpn rzz Þ 

Equating the determinant to zero results in the following cubic equation: r3pn

Hr2pn þ Urpn

W¼0

ð17Þ

where H, U, and W are the first, second, and third stress invariants, respectively, and have been previously listed as Eqs. (14a)–(14c) and (15a)–(15c).

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The three roots of the cubic equation (that is, the three principal stresses) may be found by application of the standard approach (3), as follows. Define:  1 3U H2 3  1 3 b¼ 2H 9HU þ 27W 27 sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3 b b2 a3 þ A¼ þ 2 4 27 sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3 b b2 a3 þ B¼ 2 4 27

a¼

The three principal stresses [i.e., the three roots of Eq. (17)] are then given by: 8 H > > AþBþ > > 3 > > > < A þ B pffiffiffiffiffiffiffi A B  H þ 3 þ rp1 ; rp2 ; rp3 ¼ ð18Þ > 2 2 3 > >   > > A þ B pffiffiffiffiffiffiffi A B H > > : 3 þ 2 2 3

By convention, the principal stresses are numbered such that rp1 is the algebraically greatest principal stress, whereas rp3 is the algebraically least. That is, rp1 > rp2>rp3. Two comments regarding the practical application of either Eq. (17) or Eq. (18) are appropriate. First, many handheld calculators and computer software packages feature standard routines to find the roots of nth-order polynomials. Hence, the three roots of Eq. (17) (that is, the three principal stresses) may often be found most conveniently through the use of these standard calculator routines or software packages, rather than through application of Eq. (18). The second comment is that if the principal stresses are to be calculated through application of Eq. (18), then calculation of constants A and B generally involves finding the cube root of a complex number. The need to find the cube root of a complex number is encountered infrequently, and hence the reader may not be aware of how to make such a calculation. For convenience, a process that may be used to find the cube root of a complex number has been included in Appendix A. In any event, once the principal stresses are determined, the three sets of direction cosines (which define the principal coordinate directions) are found by substituting the three principal stresses given by Eq. (18) into Eqs. (16a)– (16c) in turn. Since only two of Eqs. (16a)–(16c) are independent, Eq. (8) is

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used as a third independent equation involving the three unknown constants cpnx, cpny, and cpnz. The process of finding principal stresses and direction cosines will be demonstrated in the following example problem. Example Problem 4 Given. A state of stress referenced to an x–y–z coordinate is known to be: 3 2 3 2 rxx rxy rxz 50 10 15 4 ryx ryy ryz 5 ¼ 4 10 25 30 5 ðksiÞ rzx rzy rzz 15 30 5

Problem. Find (a) the principal stresses and (b) the direction cosines that define the principal stress coordinate system. Solution. This is the same stress tensor considered in Example Problem 3. As a part of that problem, the first, second, and third stress invariants were found to be: H=70 ksi U= 350 (ksi)2 W= 65375 (ksi)3 Part (a). Determining the principal stresses. In accordance with Eq. (17), the three principal stresses are the roots of the following cubic equation: r3

70r2

350r þ 65375 ¼ 0

As discussed earlier, the three roots of this equation can often be found most conveniently using appropriate handheld calculators or software packages. In this example solution, the roots will be found through application of Eq. 18. Following this process, we have: i  1h 1 3U H2 ¼ 3ð 350Þ ð70Þ2 ¼ 1983 a¼ 3 3  1 3 2H 9HU þ 27W b¼ 27 i 1h 2ð70Þ3 9ð70Þð 350Þ þ 27ð 65375Þ ¼ 31801 ¼ 27 vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi u u 3 2 3 3 t 48134 b b a ð48134Þ2 ð 1983Þ3 A¼ þ þ þ ¼ þ 2 2 4 27 4 27 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi p ¼ 3 15900 þ ið6010Þ

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We find that constant A equals the cube root of the complex number: z¼

15900 þ ið6010Þ

Following the process described in Appendix A, the modulus and argument of this complex number are: pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi r ¼ a2 þ b2 ¼ ð 15900Þ2 þ ð6010Þ2 ¼ 16998   6010 1 u ¼ tan ¼ tan 1 ð 0:37799Þ ¼ 0:3614 radðorÞ2:780 rad 15900 Since in this case: a¼

15900 < 0

b ¼ 6010 > 0

it is clear that the argument u corresponds to an angle in the second quadrant of the complex plane (refer to Fig. A.1). Hence, we select u=2.780 rad. Applying Eq. A.3, we find:   hui h u io pffiffiffiffiffiffiffiffiffiffiffiffi lnðrÞ n A ¼ 3 a þ ib ¼ exp cos þ i sin 3 3 3   ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi p lnð16998Þ A ¼ 3 15900 þ ið6010Þ ¼ exp 3 



2:780 2:780  cos þ i sin 3 3 A ¼ 25:71f0:6005 þ ið0:7996Þg A ¼ 15:44 þ ið20:56Þ Following an identical procedure, constant B is found to be: B ¼ 15:44

ið20:56Þ

We now apply Eq. (18) to find: 8 8 54:21 ksi AþBþH > > > > 3 > > > >   > > > > < A þ B pffiffiffiffiffiffiffi A B < H þ 3 þ rp1 ; rp2 ; rp3 ¼ ¼ 27:72 ksi 2 2 3 > > > >   > > > > > > A þ B pffiffiffiffiffiffiffi A B H > > : : 3 þ 2 2 3 43:51 ksi Hence, rp1=54.21 ksi, rp2=43.51 ksi, and rp3= 27.72 ksi.

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Part (b). Determining the direction cosines. The first two of Eqs. (16a)– (16c) and Eq. (8) will be used to form three independent equations in three unknowns. We have: ðrpn

rxx Þcpn x

sxy cpn y

sxy cpn x þ ðrpn 2

2

ryy Þcpn y

sxz cpn z ¼ 0

2

syz cpn z ¼ 0

ðcpn x Þ þ ðcpn y Þ þ ðcpn z Þ ¼ 1 Direction cosines for rp1. The three independent equations become: ð54:21

50Þcp1 x þ 10cp1 y

10cp1 x þ ð54:21 2

2

25Þcp1 y 2

15cp1 z ¼ 0

30cp1 z ¼ 0

ðcp1 x Þ þ ðcp1 y Þ þ ðcp1 z Þ ¼ 1 Solving simultaneously, we obtain: cp1 x ¼

0:9726

cp1 y ¼ 0:1666

cp 1 z ¼

0:1620

Direction cosines for rp2. The three independent equations become: ð43:51

50Þcp2 x þ 10cp2 y

10cp2 x þ ð43:51 2

2

25Þcp2 y 2

15cp2 z ¼ 0

30cp2 z ¼ 0

ðcp2 x Þ þ ðcp2 y Þ þ ðcp2 z Þ ¼ 1 Solving simultaneously, we obtain: cp2 x ¼

0:05466

cp 2 y ¼

cp 2 z ¼

0:8416

0:5738

Direction cosines for rp3. The three independent equations become: ð 27:72

50Þcp3 x þ 10cp3 y

10cp3 x þ ð 27:72 2

2

25Þcp3 y 2

15cp3 z ¼ 0

30cp3 z ¼ 0

ðcp3 x Þ þ ðcp3 y Þ þ ðcp3 z Þ ¼ 1 Solving simultaneously, we obtain: cp3 x ¼

0:8276

cp3 y ¼ 0:2259

cp3 z ¼ 0:5138

8 PLANE STRESS A stress tensor is defined by six components of stress: three normal stress components and three shear stress components. Now, in practice, a state of stress often encountered is one in which all stress components in one coordinate direction are zero. For example, suppose rzz=sxz=syz=0, as

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shown in Fig. 13(a). Since the three remaining nonzero stress components (rxx, ryy, and sxy) all lie within the x–y plane, such a condition is called a state of plane stress. Plane stress conditions occur most often because of the geometry of the structure of interest. Specifically, the plane stress condition usually exists in thin, platelike structures. Examples include the web of an Ibeam, the body panel of an automobile, or the skin of an airplane fuselage.

Figure 13 Stress elements subjected to a state of plane stress.

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In these instances, the stresses induced normal to the plane of the structure are very small compared to those induced within the plane of the structure. Hence, the small out-of-plane stresses are usually ignored, and attention is focused on the relatively high stress components acting within the plane of the structure. Since laminated composites are often used in the form of thin plates or shells, the plane stress assumption is widely applicable in composite structures and will be used throughout most of the analyses discussed in this textbook. Since the out-of-plane stresses are negligibly small, for convenience, an infinitesimal stress element subjected to plane stress will usually be drawn as a square rather than a cube, as shown in Fig. 13(b). Results discussed in earlier sections for general 3-D state of stress will now be specialized for the plane stress condition. It will be assumed that the nonzero stresses lie in the x–y plane (i.e., rzz=sxz=syz=0). This allows the remaining components of stress to be written in the form of a column array rather than a 3  3 array: 8 9 2 3 r rxx sxy 0 > = < xx > 6 7 4 sxy ryy 0 5 ! ryy > > ; : 0 0 0 sxy

Note that when a plane stress state is described, stress appears to be a firstorder tensor since (apparently) only three components of stress (rxx, ryy, andsxy) need be specified in order to describe the state of stress. This is, of course, not the case. Stress is a second-order tensor in all instances, and six components of stress must always be specified in order to define a state of stress. When we invoke the plane stress assumption, we have simply assumed a priori that three stress components (rzz, sxz, and syz) are zero. Recall that either Eq. (12a) or Eq. (12b) governs the transformation of a stress tensor from one coordinate system to another. Equation (12b) is repeated here for convenience: 3 32 2 3 2 c xV x c xV y c xV z rxV xV rxV yV rxV zV rxx rxy rxz 6 7 6r V V r V V r V V 7 6c V cyV y cyV z 7 54 ryx ryy ryz 5 4 yx yz 5 ¼ 4 yx yy rzV xV rzV yV rzV zV czV x cz V y cz V z rzx rzy rzz 2 3 cxV x cyV x czV x 6c V c V c V 7 4 xy ð12bÞ yy z y 5ðrepeatedÞ cxV z cyV z czV z When transformation of a plane stress tensor is considered, it will be assumed that the xV–yV–zV coordinate system is generated from the x–y–z

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system by a rotation h about the z-axis. That is, the z- and zV-axes are coincident, as shown in Fig. 14. In this case, the direction cosines are: cxV x ¼ cos ðhÞ cxV y ¼ cos ð90B

hÞ ¼ sin ðhÞ

cxV z ¼ cos ð90 Þ ¼ 0 cyV x ¼ cos ð90B þ hÞ ¼ B

cyV y ¼ cos ðhÞ

sin ðhÞ

cyV z ¼ cos ð90B Þ ¼ 0

czV x ¼ cos ð90B Þ ¼ 0 czV y ¼ cos ð90B Þ ¼ 0

czV z ¼ cos ð0B Þ ¼ 1

If we now (a) substitute these direction cosines into Eq. (12b), (b) label the shear stresses using the symbol s rather r, and (c) note that rzz= sxz=syz=0 by assumption, then Eq. (12b) becomes: 2 3 2 3 32 rxV xV sxV yV sxV zV rxx sxy 0 cos h sin h 0 6s V V r V V s V V7 6 7 76 ryy 0 5 4 yx y z 5 ¼ 4 sin h cos h 0 54 syx yy szV xV szV yV rzV zV 0 0 1 0 0 0 3 2 cos h sin h 0 7 6  4 sin h cos h 0 5 0

0

1

Completing the matrix multiplication indicated results in:

2

rxV xV 4 s yV xV s z V xV 2

s x V yV r yV yV s z V yV

3 s xV z V s yV z V 5 r zV zV

cos2 hrxx þ sin2 hryy þ 2cosh sin hsxy

6 2 ¼6 4 cosh sinhrxx þ cosh sinhryy þ ðcos h

sin2 hÞsxy

cosh sinhrxx þ cosh sinhryy þ ðcos2 h sin2 hrxx þ cos2 hryy

0

sin2 hÞsxy

2cosh sinhsxy

0

0

As would be expected, the out-of-plane stresses are zero: rzVzV=sxVzV=s yVzV=0. The remaining stress components are: rxV xV ¼ cos 2 ðhÞrxx þ sin 2 ðhÞryy þ 2cos ðhÞsin ðhÞsxy

ryV yV ¼ sin 2 ðhÞrxx þ cos 2 ðhÞryy sx V y V ¼

2cos ðhÞsin ðhÞsxy

cos ðhÞsin ðhÞrxx þ cos ðhÞsin ðhÞryy þ ½cos 2 ðhÞ

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3 0 7 07 5

ð19Þ

sin 2 ðhފsxy

Figure 14 Transformation of a plane stress element from one coordinate system to another.

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Equation (19) can be written using matrix notation as: 9 2 8 > cos2 ðhÞ = < rx V x V > 6 ryV yV ¼ 4 sin2 ðhÞ > > ; : sxV yV cos ðhÞsin ðhÞ

sin2 ðhÞ cos2 ðhÞ

3 2cos ðhÞ sin ðhÞ 7 2 cosðhÞ sin ðhÞ 5

cos ðhÞsin ðhÞ cos2 ðhÞ

sin2 ðhÞ

9 8 > = < rxx >  ryy > > ; : sxy

ð20Þ

It should be kept in mind that these results are valid only for a state of plane stress. Transformation of a general (3-D) stress/tensor should be performed using Eq. (12a) or (12b). The 3  3 array that appears in Eq. (20) is called the transformation matrix and is abbreviated as [T]: 3 2 cos2 ðhÞ sin2 ðhÞ 2cosðhÞsin ðhÞ 7 6 ð21Þ ½TŠ ¼ 6 sin2 ðhÞ cos2 ðhÞ 2cosðhÞsin ðhÞ 7 5 4 cos ðhÞsin ðhÞ

cos ðhÞ sin ðhÞ

cos2 ðhÞ

sin2 ðhÞ

The stress invariants [given by Eqs. (14a)–(14c) or Eqs. (15a)–(15c)] are considerably simplified in the case of plane stress. Since by definition rzz=sxz=syz=0, the stress invariants become: First stress invariant ¼ H ¼ rxx þ ryy

Second stress invariant ¼ U ¼ rxx ryy

s2xy

Third stress invariant ¼ W ¼ 0

ð22Þ

The principal stresses equal the roots of the cubic equation previously listed as Eq. (17). In the case of plane stress, this cubic equation becomes (since W=0): r3

Hr2 þ Ur ¼ 0

ð23Þ

Obviously, one root of Eq. (23) is r=0. This root corresponds to rzz and for present purposes will be labeled rp3 although it may not be the algebraically least principal stress. Thus, in the case of plane stress, the z-axis is a principal stress direction and rzz=rp3=0 is one of the three principal stresses. Since the three principal stress directions are orthogonal, this implies that the remaining two principal stress directions must lie within the x–y plane.

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Removing the known root from Eq. (23), we have the following quadratic equation: r2

Hr þ U ¼ 0

ð24Þ

The two roots of this quadratic equation (that is, the two remaining principal stresses, rp1 and rp2) may be found by application of the standard approach (3) and are given by: pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi i 1h ð25Þ HF H2 4U rp1 ; rp2 ¼ 2 Substituting Eq. (22) into Eq. (25) and simplifying yields the following: rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi r rxx þ ryy ryy 2 2 xx rp1 ; rp2 ¼ F ð26Þ þsxy 2 2

The angle hp between the x-axis and either the p1 or p2 axis is given by:   2sxy 1 ð27Þ hp ¼ arctan 2 rxx ryy Example Problem 5 Given. The plane stress element shown in Fig. 15(a). Problem. (a) Rotate the stress element to a new coordinate system oriented 25j clockwise from the x-axis, and redraw the stress element with all stress components properly oriented; (b) determine the principal stresses and principal stress coordinate system, and redraw the stress element with the principal stress components properly oriented. Solution Part (a). The following components of stress are implied by the stress element shown (note that the shear stress is algebraically negative, in accordance with the sign convention discussed in Sec. 5): rxx ¼ 70 MPa ryy ¼ 15 MPa sxy ¼

50 MPa

The stress element is to be rotated clockwise. That is, the +xV-axis is rotated away from the +y-axis. Applying the right-hand rule, it is clear that this is a negative rotation: h¼

25B

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Figure 15 Plane stress elements associated with Example Problem 5.

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Equation (20) becomes: 8 9 2 > < rxV xV > = 6 ryV yV ¼ 6 4 > :s V V > ; xy

cos2 ð 25B Þ

sin2 ð 25B Þ

2cosð 25B Þsinð 25B Þ

3

8 9 < 70 = 7 sin2 ð 25B Þ cos2 ð 25B Þ 2cosð 25B Þsinð 25B Þ 7 5: 15 ; 50 2 B B B B 2 B B cosð 25 Þsinð 25 Þ cosð 25 Þsinð 25 Þ cos ð 25 Þ sin ð 25 Þ 9 8 9 38 8 9 2 70 > 0:8214 0:1786 0:7660 > 98:5 > > > > > > > = 6 < rxV xV > < = < = 7 7 ryV yV ¼ 6 0:1786 0:8214 0:7660 5 15 ¼ 13:5 MPa 4 > > > > > > : ; > > > : ; > : ; sxV yV 0:3830 0:3830 0:6428 50 11:1

The rotated stress element is shown in Fig. 15(b).

The principal stresses are found through application of Eq. (26): s ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi  70 þ 15 70 15 2 rp1 ; rp2 ¼ þð 50Þ2 ¼ 42:5 F 57:1 MPa F 2 2 rp1 ¼ 99:6MPa

Part (b).

rp2 ¼

14:6MPa

The orientation of the principal stress coordinate system is given by Eq. (27):   1 2ð 50Þ hp ¼ arctan ¼ 31B 2 70 15 The stress element is shown in the principal stress coordinate system in Fig. 15(c).

9 DEFINITION OF STRAIN All materials deform to some extent when subjected to external forces and/or environmental changes. In essence, the state of strain is a measure of the magnitude and orientation of the deformations induced by these effects. As in the case of stress, there are two types of strain: normal strain and shear strain. The two types of strain can be visualized using the strain element shown in Fig. 16. Imagine that a perfect square has been physically drawn on a surface of interest. Initially, angle BABC is exactly p2 radians (i.e., initially BABC=90j) and sides AB and BC are of exactly equal lengths. Now suppose that some mechanism(s) causes the surface to deform. The mechanism(s) which causes the surface to deform need not be defined at

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Figure 16 2-D element used to illustrate normal and shear strains (deformations shown are greatly exaggerated for clarity).

this point, but might be external loading (i.e., stresses), a change in temperature, and/or (in the case of polymeric-based materials such as composites) the adsorption or desorption of water molecules. In any event, since the surface is deformed, the initially square element drawn on the surface is deformed as well. As shown in Fig. 16, point A moves to point AV and point C moves to point CV. It is assumed that the element remains a parallelogram, i.e., it is assumed that sides AVB and CVB remain straight lines after deformation. This assumption is valid if the element is infinitesimally small. In the present context, ‘‘infinitesimally small’’ implies that lengths AB and CB are small enough such that the deformed element may be treated as a parallelogram. Normal strain exx is defined as the change in length of AB divided by the original length of AB: exx ¼

DAB AB

ð28Þ

The change in length AB is given by: DAB ¼ ðAVB

ABÞ

From the figure, it can be seen that the projection of length AVB in the xdirection, that is, length AUB, is given by: AWB ¼ AVBcosðBAVBAÞ

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ð29Þ

If we now assume that BAVBA is ‘‘small,’’ then we can invoke the small-angle approximation,* which states that if BAVBA is expressed in radians and is less than about 0.1745 radians (about 10j), then: sinðBAVBAÞcBAVBA

tanðBAVBAÞcBAVBA

cosðBAVBAÞc1

Based on the small-angle approximation, Eq. (29) implies that AUBc AVB, and therefore the change in length of AB is approximately given by: DABcðAUB

ABÞ ¼ AUA

Equation (28) can now be written as: exx ¼

AUA AB

ð30Þ

In an entirely analogous manner, normal strain eyy is defined as the change in length of CB divided by the original length of CB: eyy ¼

DCB CB

Based on the small-angle approximation, the change in length of CB is approximately given by: DCB ¼ ðC VB

CBÞcC UC

and therefore: eyy ¼

C UC CB

ð31Þ

As before, the approximation for change in length CB is valid if angle BCUBC is small. Recall that the original element shown in Fig. 16 was assumed to be perfectly square and, in particular, that angle BAVBC is exactly p2 radians (i.e., initially BAVBC=90j). Engineering shear strain is defined as the change in angle BABC, expressed in radians: cxy ¼ DðBABCÞ ¼ BAVBA þ BC VBC

ð32Þ

The subscripts associated with a shear strain [e.g., subscripts xy in Eq. (32)] indicate that the shear strain represents the change in angle defined by line segments originally aligned with the x- and y-axes.

* The reader is encouraged to personally verify the ‘‘small-angle approximation.’’ For example, use a calculator to demonstrate that an angle of 5j equals 0.08727 rad, and that sin(0.08727 rad)=0.08716, tan(0.08727 rad)=0.08749, and cos(0.08727 rad)=0.99619. Therefore, in this example, the small angle approximation results in a maximum error of less than 1%.

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As discussed in the following sections, it is very convenient to describe a state of strain as a second-order tensor. However, in order to do so, we must use a slightly different definition of shear strain. Specifically, tensoral shear strain is defined as: exy ¼

1 c 2 xy

ð33Þ

Since engineering shear strain has been defined as the total change in angle BAVBC, tensoral shear strain is simply half this change in angle. The use of tensoral shear strain is convenient because it greatly simplifies the transformation of a state of strain from one coordinate system to another. The use of engineering shear strain is far more common in practice, however. In this text, tensoral shear strain will be used when convenient during initial mathematical manipulations of the strain tensor, but all final results will be converted to relations involving engineering shear strain. Although strains are unitless quantities, normal strains are usually reported in units of (length/length), and shear strains are usually reported in units of radians. The values of a strain is independent of the system of units used, e.g., 1 (m/m)=1 (in/in). Common abbreviations used throughout this text are as follows: 1  10 1  10

6 6

meter=meter ¼ 1micrometer=meter ¼ 1 Am=m ¼ 1 Ain=in radians ¼ 1 microradians ¼ 1 Arad

We must next define the algebraic sign convention used to describe individual strain components. The sign convention for normal strains is very straightforward and intuitive: a positive (or ‘‘tensile’’) normal strain is associated with an increase in length, while a negative (or ‘‘compressive’’) normal strain is associated with a decrease in length. To define the algebraic sign of a shear strain, we first identify the algebraic sign of each face of the infinitesimal strain element (the algebraic sign of face was defined in Sec. 5). An algebraically positive shear strain corresponds to a decrease in the angle between two positive faces, or equivalently, to a decrease in the angle between two negative faces. The above sign conventions can be used to confirm that all strains shown in Fig. 16 are algebraically positive. Example Problem 6 Given. The following two sets of strain components: exx ¼ 1000 Am=m Set1 : eyy ¼ 500 Am=m cxy ¼ 1500 Arad

exx ¼ 1000 Am=m Set2 : eyy ¼ 500 Am=m cxy ¼ 1500 Arad

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Figure 17 Strain elements associated with Example Problem 6 (not to scale).

Determine. Prepare sketches (not to scale) of the deformed strain elements represented by the two sets of strain components. Solution. The required sketches are shown in Fig. 17. Note that the only difference between the two sets of strain components is that in Set 1, cxy is algebraically positive, whereas in Set 2, cxy is algebraically negative. 10 THE STRAIN TENSOR A general 3-D solid body is shown in Fig. 18(a). An infinitesimally small cube isolated from an interior region of the body is shown in Fig. 18(b). The cube is referenced to an x–y–z coordinate system, and the cube edges are aligned with these axes. Now assume that the body is subjected to some mechanism(s) which causes the body to deform. The mechanism(s) which causes this deformation need not be defined at this point, but might be external loading (i.e., stresses), a change in temperature, the adsorption or desorption of water molecules (in the case of polymeric-based materials such as composites), or any combination thereof. Since the entire body is deformed, the internal infinitesimal cube is deformed into a parallelepiped, as shown in Fig. 18(c). It can be shown (1,2) that the state of strain experienced by the cube can be represented as a

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Figure 18 Infinitesimal elements used to illustrate the strain tensor.

Copyright © 2004 by Marcel Dekker, Inc. All Rights Reserved.

symmetric second-order tensor, involving six components of strain: three normal strains (exx, eyy, ezz) and three tensoral shear strains (exy, exz, eyz). These six strain components are defined in the same manner as those discussed in the preceding section. Normal strains exx, eyy, and ezz represent the change in length in the x-, y-, and z-directions, respectively. Tensoral shear strains exy, exz, and eyz represent the change in angle between cube edges initially aligned with the (x-,y-), (x-,z-), and ( y-,z-) axes, respectively. Using matrix notation, the strain tensor may be written as: 3 2 3 2 exx exy exz exx exy exz 4 eyx eyy eyz 5 ¼ 4 exy eyy eyz 5 ð34Þ exz eyz ezz ezx ezy ezz

Alternatively, the strain tensor can be succinctly written using indicial notation as: eij ;

i; j ¼ x; y; or z

ð35Þ

Note that if engineering shear strain is used, then Eq. (34) becomes 3 2 3 2 exx ðcxy =2Þ ðcxz =2Þ exx exy exz 7 6 6 eyy ðcyz =2Þ 7 5 4 exy eyy eyz 5 ¼ 4 ðcxy =2Þ exz eyz ezz ezz ðcxx =2Þ ðcyz =2Þ

If engineering shear strain is used, the strain tensor cannot be written using indicial notation [as in Eq. (35)] due to the 1/2 factor that appears in all offdiagonal positions. In Sec. 1, it was noted that a force vector is a first-order tensor since only one subscript is required to describe a force tensor, Fi. The fact that strain is a second-order tensor is evident from Eq. (35) since two subscripts are necessary to describe a state of strain.

11 TRANSFORMATION OF THE STRAIN TENSOR Since both stress and strain are second-order tensors, the transformation of the strain tensor from one coordinate system to another is analogous to the transformation of the stress tensor, as discussed in Sec. 6. For example, it can be shown (1,2) that the strain components in the xV–yV–zV coordinate system (eiVjV) are related to the components in x–y–z coordinate system (eij) according to: ei V j V ¼ ci V k cj V l ekl

where k; l ¼ x; y; z

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ð36aÞ

Alternatively, using matrix notation, the strain tensor transforms according to: ½ei V j V Š ¼ ½ci V j Š½eij Š½ci V j ŠT which expands as follows: 2

e xV xV 6e V V 4 yx e z V xV

e xV y V ey V y V e zV y V

3 2 32 e xV z V c xV x c xV y c xV z exx 7 7 6 6 eyV zV 5 ¼ 4 cyV x cyV y cyV z 54 eyx cz V x czV y czV z e zV zV ezx 3 2 c xV x c y V x c z V x 6c V c V c V 7 4 xy yy zy5 c xV z c y V z c z V z

exy eyy ezy

3 exz 7 eyz 5 ezz ð36bÞ

The terms ci Vj which appear in Eqs. (36a) and (36b) are direction cosines and equal the cosine of the angle between the axes of the xV–yV–zV and x–y–z coordinate systems. As was the case for the stress tensor, there are certain features of the strain tensor that do not vary when the tensor is transformed from one coordinate system to another. These features are called the strain invariants. Three independent strain invariants exist and are defined as follows: First strain invariant ¼ He ¼ eii

ð37aÞ

 1 eii ejj eij eij ð37bÞ 2  1 Third strain invariant ¼ We ¼ eii ejj ekk 3eii ejk ejk þ 2eij ejk eki ð37cÞ 6 Second strain invariant ¼ Ue ¼

Alternatively, by expanding these equations over the range i, j, k=x, y, z and simplifying, the strain invariants can be written as: First strain invariant ¼ He ¼ exx þ eyy þ ezz

ð38aÞ

Second strain invariant ¼ Ue ¼ exx eyy þ exx ezz þ eyy ezz   e2xy þ e2xz þ e2yz

Third strain invariant ¼ We ¼ exx eyy ezz

exx e2yz

eyy e2xz

ezz e2xy þ 2exy exz eyz

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ð38bÞ

ð38cÞ

Example Problem 7 Given. A state of strain referenced to an x–y–z coordinate is known to be: 3 3 2 2 1000 Am=m 500 Arad 250 Arad exx exy exz 7 7 6 6 1500 Am=m 750 Arad 5 4 eyx eyy eyz 5 ¼ 4 500 Arad ezx ezy ezz 250 Arad 750 Arad 2000 Am=m

It is desired to express this state of strain in an xU–yU–zU coordinate system, generated by: (i) Rotation of h=20j about the original z-axis (which defines an intermediate xV–yV–zV coordinate system), followed by (ii) Rotation of b=35j about the xV-axis (which defines the final xU– yU–zU coordinate system). (This coordinate transformation has been previously considered in Example Problem 3 and is shown in Fig. 10.) Problem (a) Rotate the strain tensor to the xU–yU–zU coordinate system and (b) Calculate the first, second, and third invariants of the strain tensor using both elements of the strain tensor referenced to the x–y–z coordinate system, eij, and elements of the strain tensor referenced to the xU–yU–zU coordinate system, ei Uj U.

Solution Part (a). General expressions for direction cosines relating the x–y–z and xU–yU–zU coordinate systems were determined as a part of Example Problem 1. Further, numerical values for the particular rotation h=20j and b=35j were determined in Example Problem 3 and were found to be: cxWx ¼ cos ð20B Þ ¼ 0:9397 cxWy ¼ sin ð20B Þ ¼ 0:3420 cxWz ¼ 0 cyWx ¼

cos ð35B Þsin ð20B Þ ¼

0:2802

cyWy ¼ cos ð35B Þcos ð20B Þ ¼ 0:7698 cyWz ¼ sin ð35B Þ ¼ 0:5736 czWx ¼ sin ð35B Þsin ð20B Þ ¼ 0:1962

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czWy ¼

sin ð35B Þcos ð20B Þ ¼

0:5390

czWz ¼ cos ð35B Þ ¼ 0:8192 Each component of the transformed strain tensor can now be found through application of Eq. (36a) or Eq. (36b). For example, setting iV=xU, jV=xU, and expanding Eq. (36a), strain component exUxU is given by: ex Ux U ¼ cx Ux cx Ux exx þ cx Ux cx Uy exy þ cx Ux cx Uz exz þ cx Uy cx Ux eyx þ cx Uy cx Uy eyy þ cx Uy cx Uz eyz þ cx Uz cx Ux ezx þ cx Uz cx Uy ezy þ cx Uz cx Uz ezz ex Ux U ¼ ð0:9397Þð0:9397Þð1000Þ þ ð0:9397Þð0:3420Þð500Þ þð0:9397Þð0Þð250Þ þ ð0:3420Þð0:9397Þð500Þ þð0:3420Þð0:3420Þð1500Þ þ ð0:3420Þð0Þð750Þ þð0Þð0:9397Þð250Þ þ ð0Þð0:3420Þð750Þ þð0Þð0Þð2000Þ

ex Ux U ¼ 1380 lm=m

Alternatively, if matrix notation is used, then Eq. (36b) becomes: 2 3 2 3 0:9397 0:3420 0 exUxU exUyU exUzU 6 7 6 7 4 eyUxU eyUyU eyUzU 5 ¼ 4 0:2802 0:7698 0:5736 5 ezUxU ezUyU ezUzU 0:1962 0:5390 0:8192 3 2 1000 500 250 7 6  4 500 1500 750 5 250

2

0:9397

6  4 0:3420 0

750

2000

0:2802

0:1962

0:7698 0:5736

3

7 0:5390 5

0:8192

Completing the matrix multiplication indicated, there results: 3 3 2 2 1380 Am=m 727 Arad 91 Arad exUxU exUyU exUzU 7 7 6 6 1991 Am=m 625 Arad 5 4 eyUxU eyUyU eyUzU 5 ¼ 4 727 Arad ezUxU ezUyU ezUzU 91 Arad 625 Arad 1129 Am=m

Notice that the value of exUxU determined through matrix multiplication is identical to that obtained using indicial notation, as expected.

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Part (b). The first, second, and third strain invariants will now be calculated using components of both eij and eiUjU. It is expected that identical values will be obtained since the strain invariants are independent of the coordinate system. First strain invariant: x–y–z coordinate system: He ¼ eii ¼ exx þ eyy þ ezz He ¼ ð1000 þ 1500 þ 2000Þ Am=m He ¼ 4500Am=m ¼ :004500 m=m xU–yU–zU coordinate system: He ¼ eiUiU ¼ exUxU þ eyUyU þ ezUzU He ¼ ð1380 þ 1991 þ 1129Þ Am=m He ¼ 4500Am=m ¼ 0:004500 m=m As expected, the first strain invariant is independent of the coordinate system. Second strain invariant: x–y–z coordinate system:    1 eii ejj eij eij ¼ exx eyy þ exx ezz þ eyy ezz e2xy þ e2xz þ e2yz 2

Ue ¼ ð1000Þð1500Þ þ ð1000Þð2000Þ þ ð1500Þð2000Þ Ue ¼

½ð500Þ2 þ ð250Þ2 þ ð750Þ2 Šg ðAm=mÞ2

U ¼ 5:625  106 ðAm=mÞ2 ¼ 5:625  10 6 ðm=mÞ2 xU–yU–zU coordinate system: Ue ¼

1 eiUiU ejUjU 2

eiUjU eiUjU



Ue ¼ exUxU eyUyU þ exUxU ezUzU þ eyUyU ezUzU

  e2xUyU þ e2xUzU þ e2yUzU

Ue ¼ fð1380Þð1991Þ þ ð1380Þð1129Þ þ ð1991Þð1129Þ ½ð727Þ2 þ ð91Þ2 þ ð625Þ2 Šg ðAm=mÞ2

Ue ¼ 5:625  106 ðAm=mÞ2 ¼ 5:625  10

6

ðm=mÞ2

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As expected, the second strain invariant is independent of the coordinate system. Third strain invariant: x–y–z coordinate system:  1 eii ejj ekk 3eii ejk ejk þ 2eij ejk eki 6 We ¼ exx eyy ezz exx e2yz eyy e2xz ezz e2xy þ 2exy exz eyz We ¼

We ¼ ½ð1000Þð1500Þð2000Þ

ð1000Þð750Þ2

ð1500Þð250Þ2

ð2000Þ

 ð500Þ2 þ 2ð500Þð250Þð750ފ ðAm=mÞ3 W ¼ 2:031  109 ðAm=mÞ3 ¼ 2:031  10

9

ðm=mÞ3

xU–yU–zU coordinate system:  1 eiUiU ejUjU ekUkU 3eiUiU ejUkU ejUkU þ 2eiUjU ejUkU ekUiU 6 We ¼ exUxU eyUyU ezUzU exUxU e2yUzU eyUyU e2xUzU ezUzU e2xUyU þ 2exUyU exUzU eyUzU We ¼

We ¼ ½ð1380Þð1991Þð1129Þ

ð1380Þð625Þ2

ð1991Þð91Þ2

ð1129Þ

 ð727Þ2 þ 2ð727Þð91Þð625ފ ðAm=mÞ3 W ¼ 2:031  109 ðAm=mÞ3 ¼ 2:031  10

9

ðm=mÞ3

As expected, the third stress invariant is independent of the coordinate system. 12 PRINCIPAL STRAINS The definition of the strain tensor was reviewed in Sec. 10, and transformation of the strain tensor from one coordinate system to another was discussed in Sec. 11. It can be shown (1,2) that it is always possible to rotate the strain tensor to a special coordinate system in which no shear strains exist. This coordinate system is called the principal strain coordinate system, and the normal strains that exist in this coordinate system are called principal strains. In most texts, the principal strains are denoted e1, e2, and e3. However, as will be discussed later (see Fig. 2 in Chap. 3), in this text, the axis labels ‘‘1,’’ ‘‘2,’’ and ‘‘3’’ will be used to refer to the principal material coordinate system rather than the directions of principal strain. Therefore,

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in this text, the axes associated with the principal strain coordinate system will be labeled p1, p2, and p3 axes, and the principal strains will be denoted ep1, ep2, and ep3. Since both stress and strain are second-order tensors, the principal strains can be found using an approach analogous to that used to find principal stresses. Specifically, it can be shown (1,2) that the principal strains must satisfy the following three simultaneous equations: ðepn

exx Þcpn x

exy cpn x þ ðepn exz cpn x

exy cpn y eyy Þcpn y

eyz cpn y þ ðepn

exz cpn z ¼ 0

eyz cpn z ¼ 0

ezz Þcpn z ¼ 0

ð39aÞ

ð39bÞ ð39cÞ

Since direction cosines cpnx, cpny, and cpnz must also satisfy Eq. (8), and therefore cannot all equal zero, the solution can be obtained by requiring that the determinant of the coefficients of cpnx, cpny, and cpnz equal zero:   exy exz   ðepn exx Þ     exy ðepn eyy Þ eyz  ¼ 0      exz eyz ðepn ezz Þ Equating the determinant to zero results in the following cubic equation: e3pn

He e2pn þ Ue epn

We ¼ 0

ð40Þ

where He, Ue, and We are the first, second, and third strain invariants, respectively, and have been previously listed as Eqs. (37a)–(37c) and (38a)– (38c). The three roots of the cubic equation (that is, the three principal strains) may be found by application of the standard approach (3), as follows: Define:  1 3Ue H2e 3  1 3 2He 9He Ue þ 27We b¼ 27 sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3 b b2 a3 þ þ A¼ 2 4 27 sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3 b b2 a 3 B¼ þ 2 4 27

a¼

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The three principal strains [i.e., the three roots of Eq. (40)] are then given by:

ep1 ; ep2 ; ep3 ¼

He AþBþ 3   p ffiffiffiffiffiffi ffi AþB A B He þ 3 þ 3 2 2   p ffiffiffiffiffiffi ffi AþB A B He 3 þ 2 2 3

8 > > > > > > > < > > > > > > > :

ð41Þ

By convention, the principal strains are numbered such that ep1 is the algebraically greatest principal strain, whereas ep3 is the algebraically least. That is, ep1>ep2>ep3. It is appropriate to note that many handheld calculators and computer software packages now feature standard routines to find the roots of nthorder polynomials. Hence, the roots of Eq. (40) may often be found more conveniently using these standard routines, rather than Eq. (41). The calculation of constants A and B that appear in Eq. (41) often involves finding the cube root of a complex number. The need to find the cube root of a complex number is encountered infrequently, and hence the reader may not be aware of how to make such a calculation. For convenience, a process that may be used to find the cube root of a complex number has been included in Appendix A. Once the principal strains are determined, the three sets of direction cosines (which define the principal coordinate directions) are found by substituting the three principal strains given by Eq. (41) into Eqs. (39a)– (39c) in turn. Since only two of Eqs. (39a)–(39c) are independent, Eq. (8) is used as a third independent equation involving the three unknown constants, cpnx, cpny, and cpnz. The process of finding principal strains and direction cosines will be demonstrated in the following example problem. Example Problem 8 Given. A state of strain referenced to an x–y–z coordinate is known to be: 2

3

2

1000 Am=m

exx

exy

exz

6 4 eyx ezx

eyy

7 6 eyz 5 ¼ 4 500 Arad ezz 250 Arad

ezy

500 Arad 1500 Am=m 750 Arad

250 Arad

3

7 750 Arad 5 2000 Am=m

Problem. Find (a) the principal strains and (b) the direction cosines that define the principal strain coordinate system.

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Solution. This is the same strain tensor considered in Example Problem 7. As a part of that problem, the first, second, and third strain invariants were found to be: He ¼ 0:004500 m=m

U ¼ 5:625  10 6 ðm=mÞ2

W ¼ 2:031  10 9 ðm=mÞ3 Part (a). Determining the principal strains. In accordance with Eq. (40), the three principal strains are the roots of the following cubic equation: e3pn

ð0:004500Þe2pn þ ð5:625  10 6 Þepn

ð2:031  10 9 Þ ¼ 0

Following the standard procedure for finding the roots of a cubic equation, we have: i  1h 1 3Ue H2e ¼ 3ð5:625  10 6 Þ ð0:004500Þ2 a¼ 3 3 1:125  10 6  1 3 2He 9He Ue þ 27We b¼ 27 1h 2ð0:004500Þ3 9ð0:004500Þð5:625  10 6 Þ b¼ 27 þ27ð2:031  10 9 ފ ¼ 3:435  10 10 sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3 b b2 a3 þ þ A¼ 4 27 2 vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi u u 3 10 t 3:435  10 ð 3:435  10 10 Þ2 ð 1:125  10 6 Þ3 þ þ ¼ 2 4 27 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3 A ¼ 1:718  10 10 þ ið1:524  10 10 Þ ¼

A ¼ 594:5  10 6 þ ið146:7  10 6 Þ sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3 b b2 a 3 B¼ þ 2 4 27 vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi u u 3 3:435  10 10 t ð 3:435  10 10 Þ2 ð 1:125  10 6 Þ3 ¼ þ 2 4 27

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B¼

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3 1:718  10 10 ið1:524  10 10 Þ

B ¼ 594:5  10

ep1 ; ep2 ; ep3 ¼

6

ið146:7  10 6 Þ

8 H 2689Am=m > > AþBþ > > > >  3  > < A þ B pffiffiffiffiffiffiffi A B H þ 3 þ ¼ 651Am=m 2 2 3 > >   > > > A þ B pffiffiffiffiffiffiffi A B H > > 3 þ : 2 2 3 1160Am=m

8 > > > > > > > < > > > > > > > :

Hence, ep1=2689 Am/m, ep2=1160 Am/m, and ep3=651 Am/m. Part (b).

Determining the direction cosines.

Equations (8), (39a), and (39b) will be used to form three independent equations in three unknowns. We have: ðepn

exx Þcpn x

exy cpn x þ ðepn

exy cpn y eyy Þcpn y

exz cpn z ¼ 0 eyz cpn z ¼ 0

ðcpn x Þ2 þ ðcpn y Þ2 þ ðcpn z Þ2 ¼ 1 Direction cosines for ep1. The three independent equations become: ð2689

1000Þcp1 x

500cp1 x þ ð2689

500cp1 y

250cp1 z ¼ 0 750cp1 z ¼ 0

1500Þcp1 y

ðcp1 x Þ2 þ ðcp1 y Þ2 þ ðcp1 z Þ2 ¼ 1 Solving simultaneously, we obtain: cp1 x ¼ 0:2872

cp1 y ¼ 0:5945

cp1 z ¼ 0:7511

Direction cosines for ep2. The three independent equations become: ð1160

1000Þcp2 x

500cp2 x þ ð1160

500cp2 y 1500Þcp2 y

250cp2 z ¼ 0 750cp2 z ¼ 0

ðcp2 x Þ2 þ ðcp2 y Þ2 þ ðcp2 z Þ2 ¼ 1 Solving simultaneously, we obtain: cp2 x ¼ 0:5960

cp2 y ¼ 0:5035 cp2 z ¼

0:6256

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Direction cosines for ep3. The three independent equations become: ð651

1000Þcp3 x

500cp3 x þ ð651

500cp3 y

250cp3 z ¼ 0

1500Þcp3 y

750cp3 z ¼ 0

ðcp3 x Þ2 þ ðcp3 y Þ2 þ ðcp3 z Þ2 ¼ 1 Solving simultaneously, we obtain: cp3 x ¼

0:7481

cp3 y ¼ 0:6286

cp3 z ¼

0:2128

13 STRAINS WITHIN A PLANE PERPENDICULAR TO A PRINCIPAL STRAIN DIRECTION It has been seen that a strain tensor is defined by six components of strain: three normal strain components and three shear strain components. Now, in practice, there are circumstances in which it is known a priori that both shear strain components in one direction are zero: exz=eyz=0, say (or equivalently, cxz=cyz=0). This implies that the z-axis is a principal strain axis. In these instances, we are primarily interested in the strains induced within the x–y plane, exz, eyy, and exy. Two different circumstances are encountered in which it is known a priori that the z-axis is a principal strain axis. In the first case, all three out-of-plane strain components in the zdirection are known a priori to equal zero. That is, it is known a priori that ezz=exz=eyz=0. Not only is the z-axis a principal strain axis in this case, but, in addition, the principal strain equals zero: ezz=ep3=0. Since the three remaining nonzero strain components (exx, eyy, and exy) all lie within the x–y plane, it is natural to call this condition a state of plane strain. Plane strain conditions occur most often because of the geometry of the structure of interest. Specifically, the plane strain condition usually exists in internal regions of very long (or very thick) structures. Examples include solid shafts or long dams. In these instances, the strains induced along the long axis of the structure are often negligibly small compared to those induced within the transverse plane of the structure. The second case in which the out-of-plane z-axis may be a principal axis is when a structure is subjected to a state of plane stress. As has been discussed in Sec. 8, the state of plane stress occurs most often in thin, platelike structures. In this case, the z-axis is a principal strain axis, and ezz is again one of the principal strains. However, in this second case, the out-of-plane normal strain does not, in general, equal zero: ezz p 0. It is emphasized that a state of plane stress usually, but not always, causes a state of strain in which the z-axis is a principal strain axis. This point

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will be further discussed in Chap. 4. It will be seen there that it is possible for a material to exhibit a coupling between in-plane stresses and out-of-plane shear strains. That is, in some cases, stresses acting within the x–y plane (rxx, ryy, and/or sxy) can cause out-of-plane shear strains (exz and/or eyz). In these instances, the out-of-plane z-axis is not a principal strain axis, although the out-of-plane stresses all equal zero. In any event, for present purposes, assume that it is known a priori that the out-of-plane z-axis is a principal strain axis, and we are primarily interested in the strains induced within the x–y plane, exx, eyy, and exy. We will write these strains in the form of a column array, rather than a 33 array: 9 9 8 8 2 3 2 3 e exx exy 0 exx ðcxy =2Þ 0 > = > < exx > = < xx > 6 7 6 7 eyy eyy 0 5 ! eyy ¼ 4 exy eyy 0 5 ¼ 4 ðcxy =2Þ > > > ; > : ; : exy 0 0 ezz 0 0 ezz cxy =2 Note that ezz does not appear in the column array. This is not of concern in the case of plane strain since, in this case, ezz=0. However, in the case of plane stress, it is important to remember that (in general) ezz p 0. Although in the following chapters we will be primarily interested in strains induced within the x–y plane, the reader is advised to remember that an out-of-plane strain ezz is also induced by a state of plane stress. The transformation of a general 3-D strain tensor has already been discussed in Sec. 11. The relations presented there will now be simplified for the case of transformation of strains within a plane. Recall that either Eq. (36a) or Eq. (36b) governs the transformation of a strain tensor from one coordinate system to another. Equation (36b) is repeated here for convenience: 3 32 3 2 2 e xV xV e xV y V e xV z V c xV x c xV y c xV z exx exy exz 7 6 6e V V e V V e V V 7 6c V cyV y cyV z 7 54 eyx eyy eyz 5 4 yx yz 5 ¼ 4 yx yy cz V x czV y czV z e z V xV e z V y V e z V z V ezx ezy ezz 3 2 c xV x c y V x c z V x 7 6 ðrepeatedÞð36bÞ  4 cxV y cyV y czV y 5 c xV z c y V z c z V z

Assuming that the xV–yV–zV coordinate system is generated from the x–y–z system by a rotation h about the z-axis, the direction cosines are: cxV x ¼ cos ðhÞ cxV y ¼ cos ð90B

hÞ ¼ sin ðhÞ

cxV z ¼ cos ð90 Þ ¼ 0 B

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cyV x ¼ cos ð90B þ hÞ ¼ cyV y ¼ cos ðhÞ

sin ðhÞ

cyV z ¼ cos ð90B Þ ¼ 0

czV x ¼ cos ð90B Þ ¼ 0 czV y ¼ cos ð90B Þ ¼ 0

czV z ¼ cos ð0B Þ ¼ 1

Substituting these direction cosines into Eq. (36b) and noting that by assumption, exz=eyz=0, we have: 32 2 3 2 3 cos h sin h 0 exx exy 0 exV xV exV yV exV zV 76 6e V V e V V e V V 7 6 7 eyy 0 5 4 yx y z 5 ¼ 4 sin h cos h 0 54 eyx yy ezV xV ezV yV ezV zV 0 0 ezz 0 0 1 2

cos h

6  4 sin h 0

sin h

cos h 0

0

3

7 05 1

Completing the matrix multiplication indicated results in: 2

3

e xV xV

e x V yV

exV zV

6 6 e yV xV 4 e z V xV

e yV yV

7 e yV z V 7 5 e zV zV

e z V yV

cos2 hexx þ sin2 heyy þ 2 cos h sin hexy ¼4 cos h sin hexx þ cos h sin heyy þ ðcos2 h sin2 hÞexy 0 2

3 cos h sin hexx þ cos h sin heyy þ ðcos2 h sin2 hÞexy 0 2 sin hexx þ cos2 heyy 2 cos h sin hexy 05 0 ezz

As would be expected, exVzV=eyVzV=0. The remaining strain components are: exV xV ¼ cos2 ðhÞexx þ sin2 ðhÞeyy þ 2 cosðhÞsinðhÞexy

eyV yV ¼ sin2 ðhÞexx þ cos2 ðhÞeyy exV yV ¼

2 cosðhÞsinðhÞexy 2

2

ð42Þ

cosðhÞsinðhÞexx þ cosðhÞsinðhÞeyy þ ½cos ðhÞ þ sin ðhފexy

ezV zV ¼ ezz

Tensoral shear strains were used in Eqs. (36a) and (36b) for mathematical convenience; that is, tensoral shear strains have been used so that rotation of the strain tensor could be accomplished using the normal transformation law for a second-order tensor. Since engineering shear strains are far more commonly used in practice, we will now convert our final results, Eq. (42), to ones which involve engineering shear strain (cxy). Recall from Sec. 9 that

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exy ¼ 12 cxy . Hence, to convert Eq. (42), simply replace exy with 12 cxy everywhere, resulting in: exV xV ¼ cos2 ðhÞexx þ sin2 ðhÞeyy þ cosðhÞsinðhÞcxy

eyV yV ¼ sin2 ðhÞexx þ cos2 ðhÞeyy cosðhÞsinðhÞcxy ð43Þ c xV y V c xy ¼ cosðhÞsinðhÞexx þ cosðhÞsinðhÞeyy þ ½cos2 ðhÞ þ sin2 ðhފ 2 2 ezV zV ¼ ezz Equation (43) relates the components of strain in two different coordinate systems within a single plane and will be used extensively throughout the remainder of this text. The first three of Eq. (43) can be written using matrix notation as: 8 9 3 e xV xV > 2 > > > cos2 ðhÞ sin2 ðhÞ 2 cosðhÞsinðhÞ < = ey V y V 7 6 ¼4 sin2 ðhÞ cos2 ðhÞ 2 cosðhÞsinðhÞ 5 > > c V V > xy ; > : cosðhÞsinðhÞ cosðhÞ sinðhÞ cos2 ðhÞ sin2 ðhÞ 2 9 8 exx > > > > = < eyy ð44Þ  > > > ; : cxy > 2 Compare Eq. (44) with Eq. (20). In particular, note that the transformation matrix, [T], which was previously encountered during the discussion of plane stress in Sec. 8, also appears in Eq. (44). The strain invariants [given by Eqs. (37a)–(37c) or Eqs. (38a)–(38c)] are considerably simplified when the out-of-plane z-axis is a principal axis. Since c by definition exz ¼ c2xz ¼ eyz ¼ 2yz ¼ 0, the strain invariants become: First strain invariant ¼ He ¼ exx þ eyy þ ezz Second strain invariant ¼ Ue ¼ exx eyy þ exx ezz þ eyy ezz Third strain invariant ¼ We ¼ exx eyy ezz

ezz

c4xy

c2xy 4

ð45Þ

4 The principal strains equal the roots of the cubic equation previously listed as Eq. (40). Substituting Eq. (45) into Eq. (40) reduces to: e3pn

ðexx þ eyy þ ezz Þe2pn þ ðexx eyy þ exx ezz þ eyy ezz ðexx eyy ezz

ezz

c2xy Þ¼0 4

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c2xy 4

Þepn ð46Þ

One root of Eq. (23) is epn=ezz. For present purposes, this root will be labeled ep3 although it may not be the algebraically least principal strain. In the case of plane strain, ep3=ezz=0. Removing the known root from Eq. (46), we have the following quadratic equation: e2pn

c2xy

Þ¼0 4 The two roots of this quadratic equation (that is, the two remaining principal strains, ep1 and ep2) may be found by application of the standard approach (3) and are given by: ffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi e 2  c 2 exx þ eyy e xx yy xy F ep1 ; ep2 ¼ ð47Þ þ 2 2 2 ðexx þ eyy Þepn þ ðexx eyy

The angle hpe between the x-axis and either the p1 or p2 axis is given by:   cxy 1 hpe ¼ arctan ð48Þ exx eyy 2 Example Problem 9 Given. A state of plane strain is known to consist of: exx ¼ 500 Am=m eyy ¼ 1000 Am=m cxy ¼

2500 Arad

Problem. (a) Prepare a rough sketch (not to scale) of the deformed strain element in the x–y coordinate system; (b) determine the strain components which correspond to an xV–yV coordinate system, oriented 25j CCW from the x–y coordinate system, and prepare a rough sketch (not to scale) of the deformed strain element in the xV–yV coordinate system; and (c) determine the principal strain components that exist within the x–y plane, and prepare a rough sketch (not to scale) of the deformed strain element in the principal strain coordinate system. Solution Part (a). A sketch showing the deformed strain element (not to scale) in the x–y coordinate system is shown in Fig. 19(a). Note that: The length of the element side parallel to the x-axis has increased (corresponding to the tensile strain exx=500 Am/m). The length of the element side parallel to the y-axis has decreased (corresponding to the compressive strain eyy= 1000 Am/m).

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Figure 19 Strain elements associated with Example Problem 9 (all deformations shown greatly exaggerated for clarity).

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The angle defined by x–y axes has increased (corresponding to the negative shear strain cxy= 2500 Arad). Part (b). Since the xV-axis is oriented 25j CCW from the x-axis, in accordance with the right-hand rule, the angle of rotation is positive, i.e., h=+25j. Substituting this angle and the given strain components in Eq. (44): 9 2 8 3 exV xV > > cos2 ð25B Þ sin2 ð25B Þ 2 cosð25B Þsinð25B Þ > >

c V V> > ; : xy> 2 B B B B 2 B B cosð25 Þsinð25 Þ cosð25 Þsinð25 Þ cos ð25 Þ sin ð25 Þ 2 8 9 500 > > > > > > = < 1000  > > > 2500 > > > ; : 2 Completing the matrix multiplication indicated results in: 9 8 9 8 exV xV > > 725 Am=m > > > > > > = = < < ey V y V ¼ 225 Am=m > > > > > > ; : ; > : c xV y V > 1378 Arad 2

A sketch showing the deformed strain element (not to scale) in the xV–yV coordinate system is shown in Fig. 19(b). Note that: The length of the element side parallel to the xV-axis has decreased (corresponding to the compressive strain exVxV= 725 Am/m). The length of the element side parallel to the yV-axis has increased (corresponding to the tensile strain eyVyV=225 Am/m). The angle defined by the xV–yV axes has increased (corresponding to the negative shear strain cxVyV=2756 Arad). The principal strains are found through application of Eq. (47): sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi    ffi 500 1000 500 þ 1000 2 2500 2 F þ ep1 ; ep2 ¼ 2 2 2

Part (c).

ep1 ¼ 1208 Am=m

ep2 ¼

1708 Am=m

The orientation of the principal strain coordinate system is given by Eq. (48):   1 2500 ¼ 29:5B hpe ¼ arctan 2 500 þ 1000

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A sketch showing the deformed strain element (not to scale) in the principal strain coordinate system is shown in Fig. 19(c). Note that: The length of the element side parallel to the p1-axis has increased (corresponding to the tensile principal strain ep1=1208 Am/m). The length of the element side parallel to the p2-axis has decreased (corresponding to the compressive principal strain ep2= 1708 Am/m). The angle defined by the principal strain axes has remained precisely p/2 radians (i.e., 90j) since in the principal strain coordinate system, the shear strain is zero.

14 RELATING STRAINS TO DISPLACEMENT FIELDS Most analyses considered in this text begin with the consideration of the displacement fields induced in the structure of interest. That is, mathematical expressions that describe the displacements induced at all points within a structure by external loading and/or environmental changes will be assumed or otherwise specified. Strains induced in the structure will then be inferred from these displacement fields. In the most general case, three displacement fields are involved. Specifically, these are the displacements in the x-, y-, and z-directions, typically denoted as the u-, v-, and w-displacement fields, respectively. In general, all three displacement fields are functions of x, y, and z: Displacements in the x-direction: u=u(x,y,z). Displacements in the y-direction: v=v(x,y,z). Displacements in the z-direction: w=w(x,y,z). However, if the out-of-plane z-axis is a principal strain axis, then u and v are (at most) functions of x and y only, while w is (at most) a function of z only. In this case: Displacements in the x-direction: u=u(x,y). Displacements in the y-direction: v=v(x,y). Displacements in the z-direction: w=w(z). A detailed derivation of the relationship between displacements and strains is beyond the scope of this review, and the interested reader is referred to Frederick and Chang (1) or Fung (2) for details. It can be shown that the relationship between displacement fields and the strain tensor depends upon the magnitude of derivatives of displacement fields (also called displacement gradients). If displacement gradients are arbitrarily large, then

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the associated level of strain is said to be finite, and each component of the strain tensor is related nonlinearly to displacement gradients as follows: "      # Bu 1 Bu 2 Bv 2 Bw 2 þ exx ¼ þ þ Bx 2 Bx Bx Bx "      # Bv 1 Bu 2 Bv 2 Bw 2 þ eyy ¼ þ þ By 2 By By By "      # Bw 1 Bu 2 Bv 2 Bw 2 þ ezz ¼ þ þ Bz 2 Bz Bz Bz          Bu Bv Bu Bu Bv Bv Bw Bw þ þ cxy ¼ þ þ By Bx Bx By Bx By Bx By          Bu Bw Bu Bu Bv Bv Bw Bw þ þ cxz ¼ þ þ Bz Bx Bx Bz Bx Bz Bx Bz          Bw Bv Bu Bu Bv Bv Bw Bw þ þ cyz ¼ þ þ By Bz By Bz By Bz By Bz The expressions listed above define what is known as Green’s strain tensor (also known as the Lagrangian strain tensor). In most cases encountered in practice, however, displacement gradients are very small, and consequently the products of displacement gradients are negligibly small and can be discarded. For example, it can usually be assumed that:  2  2  2    Bu Bv Bw Bu Bu c0; etc: c0 c0 c0 Bx Bx Bx Bx By When displacement gradients are very small, the level of strain is said to be infinitesimal, and each component of the strain tensor is linearly related to displacement gradients as follows: Bu Bx Bv eyy ¼ By Bw ezz ¼ Bz Bv Bu þ cxy ¼ Bx By exx ¼

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ð49aÞ ð49bÞ ð49cÞ ð49dÞ

Bw Bu þ Bx Bz Bw Bv þ cyz ¼ By Bz

ð49eÞ

cxz ¼

ð49f Þ

For most analyses considered in this text, we will assume that strains are infinitesimal and are related to displacement fields in accordance with Eqs. (49a)–(49f). The one exception occurs in Chap. 11, where it will be necessary to include nonlinear terms in the strain–displacement relationships. As stated above, most analyses begin with the consideration of the displacement fields induced in a structure of interest. Strain fields implied by these displacements are then calculated in accordance with Eqs. (49a)–(49f). This process insures that strain fields are consistent with displacements. Consider the opposite approach. Specifically, suppose that mathematical expressions for strain fields are assumed, perhaps on the basis of engineering judgment. In this case, it is possible that the assumed strain fields correspond to physically unrealistic displacement fields. For example, displacement fields inferred from assumed strain fields may imply that the solid body has voids and/or overlapping regions, a physically unrealistic circumstance. A system of six equations known as the compatibility conditions can be developed that guarantee that assumed expressions for the six components of strain do, in fact, correspond to physically reasonable displacement fields u(x,y,z), v(x,y,z), and w(x,y,z). To develop the compatibility conditions, differentiate Eq. (49d) twice, once with respect to x and once with respect to y. We obtain: B2 cxy B3 u B3 v ¼ þ 2 2 Bx By BxBy BxBy From Eqs. (49a) and (49b), it is easily seen that: B2 exx B3 u ¼ BxBy2 By2

B2 eyy B3 v ¼ Bx2 By Bx2

Combining these results, we see that expressions for the strain components exx, eyy, and cxy correspond to physically reasonable displacement fields (i.e., ‘‘are compatible’’) only if they satisfy: B2 cxy B2 exx B2 eyy ¼ þ BxBy By2 Bx2

ð50aÞ

Equation (50a) is the first compatibility condition. Following a similar procedure using Eqs. (49e) and (49f), we obtain: B2 cyz B2 eyy B2 ezz þ ¼ Bz2 By2 ByBz

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ð50bÞ

B2 cxz B2 exx B2 ezz ¼ þ BxBz Bz2 Bx2

ð50cÞ

These are the second and third compatibility conditions. Next, the following expressions are obtained using Eqs. (49a), (49d), (49e), and (49f), respectively: B3 u B2 exx ¼ BxByBz ByBz B2 cxy B3 u ¼ BxByBz BxBz

B3 v Bx2 Bz

B3 u B2 cxz ¼ BxByBz BxBy

B3 w Bx2 By

B2 cyz B3 w B3 v þ ¼ Bx2 By Bx2 Bz Bx2 Combining these four expressions, we find that assumed expressions for strain components exx, cxy, cxz, and cyz, are compatible if:   B2 exx B Bcxy Bcxz cyz 2 ¼ þ ð50dÞ By ByBz Bx Bz Bx

This is the fourth compatibility condition. The final two compatibility conditions are developed using a similar process and are given by:   B2 eyy B Bcyz Bcxz cxy ¼ þ ð50eÞ 2 BxBz By Bx By Bz   B2 ezz B Bcyz Bcxz cxy ¼ þ ð50f Þ 2 BxBy Bz Bx By Bz

15 COMPUTER PROGRAMS 3DROTATE AND 2DROTATE A review of the force, stress, and strain tensors has been presented in this chapter. These concepts will be applied routinely throughout the remainder of this text, as we develop a macromechanics-based analysis of structural composite materials and structures. It will be seen that the transformation of stress and strain tensors is of particular importance. Indeed, nearly all analyses of composite materials and structures presented herein require multiple transformations of stress and strain tensors from one coordinate system to another. Two computer programs, 3DROTATE and 2DROTATE, that can be used to perform transformations of force, stress, or strain tensors have been developed to accompany this text. These programs can also be downloaded

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at no cost from the following website: http://debts.washington.edu/amtas/ computer.html. Program 3DROTATE performs the calculations necessary to transform a force, stress, or strain tensor from the x–y–z coordinate system to the xU–yU–zU coordinate system, where the xU–yU–zU coordinate system is generated from the x–y–z coordinate system by (up to) three successive rotations. Derivation of the direction cosines that relate these two coordinate systems is left as a student exercise, but are listed in Homework Problem 2. Program 3DROTATE also calculates the angles between the x–y–z and xU–yU–zU coordinate axes, invariants of the force, stress, or strain tensors, and principal stresses and strains. All of the numerical results discussed in Example Problems 1, 3, and 7 can be obtained through the use of program 3DROTATE. The second program, 2DROTATE, can be used to rotate stresses within a plane (as discussed in Sec. 8) and/or strains within a plane (as discussed in Sec. 13). For the most part, thin platelike composite structures will be considered in this textbook. Therefore, it can usually be assumed that the direction normal to the surface of the composite is a direction of principal stress or strain. Hence, most of the stress or strain transformations considered in this text involve rotations within a plane. Most of the numerical results discussed in Example Problems 5 and 9 can be obtained through the use of program 2DROTATE.

HOMEWORK PROBLEMS In the following problems, the phrase ‘‘solve by hand’’ means that numerical solutions should be obtained using a pencil, paper, and nonprogrammable calculator. Solutions obtained by hand will then be compared to numerical results returned by appropriate computer programs. This process will insure understanding of the mathematical processes involved. 1. Solve part (c) of Example Problem 1 by hand based on the rotation angles listed below. In each case, calculate the magnitude of the transformed force vector. Confirm your calculations using program 3DROTATE. (a) h=60j (b) h=60j (c) h= 60j (d) h= 60j (e) h= 45j

b= 45j. b=45j. b= 45j. b=45j. b=60j.

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(f ) h= 45j (g) h=45j (h) h=45j

b= 60j. b=60j. b= 60j.

2. Consider an xVVV–yVVV–zVVV coordinate system, which is generated from an x–y–z coordinate system by the following three rotations: 

A rotation of about the original z-axis, which defines an intermediate xV–yV–zV coordinate system [see Fig. 2(a)], followed by  A rotation of b about the xV-axis, which defines an intermediate xVV– yVV–zVV coordinate system [see Fig. 2(b)], followed by  A rotation of w about the yVV-axis, which defines the final xVVV–yVVV–zVVV coordinate system. Show that the xVVV–yVVV–zVVV and x–y–z coordinate systems are related by the following direction cosines: 2

3 cxVVV x cxVVV y cxVVV z 4 cyVVV x cyVVV y cyVVV z 5 czVVV x czVVV y czVVV z 2 cos w cos h sin w sin b sin h cos w sinh þ sin w sin b cos h ¼4 cos b sin h cos b cos h sin w cos h þ cos w sin b sin h sin w sin h cos w sin b cos h

3 sin w cos b sin b 5 cos w cos b

3. The force vector discussed in Example Problem 1 is given by: F ¼ 1000iˆ þ 200jˆ þ 600kˆ Using Eq. (6c), express F in a new coordinate system defined by three successive rotations, as listed below, using the direction cosines listed in Problem 2. In each case, compare the magnitude of the transformed force vector to the magnitudes calculated in Example Problem 1. Solve these problems by hand and then confirm your calculations using program 3DROTATE. (a) h=60j (b) h=60j (c) h=60j (d) h=60j (e) h= 60j

b= 45j b= 45j b=45j b=45j b= 45j

w=25j. w= 25j. w= 25j. w=25j. w=25j.

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4.

Solve Example Problem 3 by hand using the following rotation angles: (a) h=20j (b) h= 20j (c) h= 20j

b= 35j. b=35j. b= 35j.

Confirm your calculations using program 3DROTATE. 5. Use Eq. (12a) to obtain an expression (in expanded form) for the following stress component [in each case, the expanded expression will be similar to Eq. (13)]: (a) rxVxV. (b) rxVyV. (c) ryVyV. (d) ryVzV. (e) rzVzV. 6. Use program 3DROTATE to determine the stress invariants for the stress tensor listed below, and compare to those determined in Example Problem 3. (Note: this stress tensor is similar to the one considered in Example Problem 3 except that the algebraic sign of all three normal stresses has been reversed.): 2

rxx 4 ryx rzx

rxy ryy rzy

3 2 rxz 50 ryz 5 ¼ 4 10 rzz 15

10 25 30

3 15 30 5 ðksiÞ 5

7. Use program 3DROTATE to determine the stress invariants for the stress tensor listed below, and compare to those determined in Example Problem 3. (Note: this stress tensor is similar to the one considered in Example Problem 3 except that the algebraic sign of all three shear stresses has been reversed.): 2

rxx 4 ryx rzx

rxy ryy rzy

3 2 rxz 50 ryz 5 ¼ 4 10 rzz 15

10 25 30

3 15 30 5 ðksiÞ 5

8. Use program 3DROTATE to determine the stress invariants for the stress tensor listed below, and compare to those determined in Example

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Problem 3. (Note: this stress tensor is similar to the one considered in Example Problem 3 except that the algebraic sign of all stress components has been reversed.): 2

rxx 4 ryx rzx

rxy ryy rzy

3 2 rxz 50 ryz 5 ¼ 4 10 rzz 15

10 25 30

3 15 30 5 ðksiÞ 5

9. Use program 3DROTATE to determine the strain invariants for the strain tensor listed below, and compare to those determined in Example Problem 7. (Note: this strain tensor is similar to the one considered in Example Problem 7 except that the algebraic sign of all shear strain components has been reversed.): 2

exx 6 4 eyx ezx 10.

3 2 1000 Am=m exz 7 6 eyz 5 ¼ 4 500 Arad ezz 250 Arad

500 Arad

250 Arad

1500 Am=m 750 Arad

3

7 750 Arad 5

2000 Am=m

Use program 3DROTATE to determine the strain invariants for the strain tensor listed below, and compare to those determined in Example Problem 7. (Note: this strain tensor is similar to the one considered in Example Problem 7 except that the algebraic sign of all normal strain components has been reversed.): 2

exx 6 4 eyx ezx 11.

exy eyy ezy

exy eyy ezy

3 2 1000 Am=m exz 7 6 eyz 5 ¼ 4 500 Arad ezz 250 Arad

500 Arad

250 Arad

1500 Am=m

750 Arad

750 Arad

2000 Am=m

3 7 5

Use program 3DROTATE to determine the strain invariants for the strain tensor listed below, and compare to those determined in Example Problem 7. (Note: this strain tensor is similar to the one considered in Example Problem 7 except that the algebraic sign of all strain components has been reversed.): 2

exx 6 4 eyx ezx

exy eyy ezy

3 2 1000 Am=m exz 7 6 eyz 5 ¼ 4 500 Arad ezz 250 Arad

500 Arad 1500 Am=m 750 Arad

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250 Arad

3

7 750 Arad 5

2000 Am=m

REFERENCES 1. Frederick, D.; Chang, T.S. Continuum Mechanics; Scientific Publishers, Inc: Cambridge, MA, 1972. 2. Fung, Y.C. A First Course In Continuum Mechanics; Prentice-Hall, Inc.: Englewood Cliffs, NJ, 1969. 3. Consult any handbook of mathematical functions and tables, for example, CRC Basic Mathematical Tables; Shelby, S.M. The Chemical Company: Cleveland, OH, 1970.

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3 Material Properties

In this chapter, various material properties required to predict the performance of composite structures are introduced. The chapter begins with a general discussion of isotropic vs. anisotropic material behaviors. It will be pointed out that most composites can be classified as either orthotropic or transversely isotropic materials. Sections devoted to those material properties of primary interest to structural engineers then follow. Specifically, separate sections are presented, which describe material properties that allow an engineer to: 

Relate stress to strain Relate temperature to strain  Relate moisture content to strain  Relate stress (or strain) to failure. 

In each section, material properties will first be defined for anisotropic materials. These general definitions will then be applied to the case of composites (i.e., they will be specialized for the case of orthotropic or transversely isotropic materials). 1 ANISOTROPIC VS. ISOTROPIC MATERIALS The phrase material property refers to a measurable constant that is characteristic of a particular material and can be used to relate two disparate quantities 117

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of interest. Material properties that describe the ability of a material to conduct electricity, to transmit (or reflect) visible light, to transfer heat, or to support mechanical loading, to name but a few, have been defined. Material properties of interest herein are those used by engineers during the design of composite structures. Two specific examples are Young’s modulus, E and Poisson’s ratio v. These two familiar material properties, which will be reviewed and further discussed in Sec. 2, are used to relate the stress and strain tensors. The adjectives ‘‘anisotropic’’ and ‘‘isotropic’’ indicate whether a material exhibits a single value for a given material property. More specifically, if the properties of a material are independent of direction within the material, then the material is said to be isotropic. Conversely, if the material properties vary with direction within the material, then the material is said to be anisotropic. To clarify this statement, suppose that three test specimens are machined from a large block at three different orientations, as shown in Fig. 1. The geometry of the three specimens is assumed to be identical, so that the only difference between specimens is the original orientation of each specimen within the ‘‘parent’’ block. Now suppose that the axial stiffness (i.e., Young’s modulus E ) is measured for each specimen. Young’s modulus measured using specimen 1 will be denoted Exx (i.e., subscripts are used to indicate the original orientation of specimen 1 within the parent block). Similarly, Young’s modulus measured using specimens 2 and 3 will be denoted Eyy and Ezz, respectively. If the parent block consists of an isotropic material, then Young’s modulus measured for each specimen will be identical (to within engineering accuracies)—for isotropic materials: Exx = Eyy = Ezz. In this case, an identical value of Young’s modulus is measured in the x-, y-, and z-directions, and is independent of direction within the material. In contrast, if the parent block is an anisotropic material, a different Young’s modulus will, in general, be measured for each specimen—for anisotropic materials: Exx p Eyy p Ezz. In this case, the value of Young’s modulus depends on the direction within the material the modulus is measured. For anisotropic materials, a similar dependence on direction can occur for any material property of interest (Poisson’s ratio, thermal expansion coefficients, ultimate strengths, etc). It is the microstructural features of a material that determine whether it exhibits isotropic or anisotropic behavior. Consequently, to classify a given material as isotropic or anisotropic, one must first define the physical scale of interest. For example, it is well known that metals and metal alloys are made up of individual grains, and that the atoms that exist within these grains are

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Figure 1 Illustration of method used to determine whether a material is isotropic or anisotropic.

arranged in well-defined crystalline arrays. The most common crystalline arrays are the body-centered cubic (BCC), the face-centered cubic (FCC), or the hexagonal close-packed (HCP) structure [1]. Due to the highly ordered and symmetrical atomic structures that exist within these arrays, an individual grain exhibits different properties in different directions, and hence is anisotropic. That is to say, if material properties are defined at a physical scale on

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the order of a grain diameter or smaller, then all metals or metal alloys must be defined as anisotropic materials. It does not necessarily follow, however, that a metal or metal alloy will exhibit anisotropic behavior at the structural level. This is because individual grains are typically very small, and the orientation of the atomic crystalline arrays usually varies randomly from one grain to the next. As a typical case, it is not uncommon for a steel alloy to exhibit an average grain diameter of 0.044 mm (0.0017 in.), or roughly 8200 grains/mm3 (134  106 grains/in.3) [1]. If the grains are randomly oriented, which is a common case, then at the structural level (say, at a physical scale >1 mm), the steel alloy will exhibit isotropic properties even though the constituent grains are anisotropic. Conversely, if a significant percentage of grains is caused to be oriented by some mechanism (such as cold rolling, for example), then the same steel alloy will be anisotropic at the structural level. Polymeric composites are anisotropic at the structural level, and the microstructural features that lead to this anisotropy are immediately apparent. Specifically, it is the uniform and symmetrical orientation of the reinforcing fibers within a ply that leads to anisotropic behavior. As a simple example, suppose two specimens are machined from a thin unidirectional composite plate consisting of high-strength fibers embedded within a relatively flexible polymeric matrix, as shown in Fig. 2a and b. Note that the coordinate system used to describe the plate has been labeled the 1–2–3 axes. In this case, the 1-axis is defined to be parallel to the fibers, the 2-axis is defined to lie within the plane of the plate and is perpendicular to the fibers, and the 3axis is defined to be normal to the plane of the plate. Note that fibers are arranged symmetrically about the 1–3 and 2–3 planes. The 1–2–3 coordinate system will henceforth be referred to as the principal material coordinate system. Referring to Fig. 2a, specimen 1 is machined such that the fibers are aligned with the long axis of the specimen, whereas in specimen 2, the fibers are perpendicular to the axis of the specimen. Obviously, Young’s modulus measured for these two specimens will be quite different. Specifically, the modulus measured for specimen 1 will approach that of the fibers, whereas the modulus measured for specimen 2 will approach that of the polymeric matrix. Therefore, E11> >E22, and Young’s modulus varies with direction within the material, satisfying the definition of an anisotropic material. The principal material coordinate system is not always aligned with the fiber direction, as shown in Fig. 2c and d. In this case, the thin composite plate is formed using a braided fabric. As discussed in Sec. 4 of Chap. 1, braided fabrics contain fibers oriented in two (or more) nonorthogonal directions. The three principal material coordinate axes lie within planes that are symmetrical with respect to the fiber array. As before, the 1- and 2-axes lie within the plane of the plate, and the 3-axis is defined normal to the plane of the plate.

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Figure 2 Illustration of the principal material coordinate system for thin composite laminates.

One of the most unusual features of anisotropic materials is that they can exhibit coupling between normal stresses and shear strains, as well as coupling between shear stress and normal strains. A physical explanation of how this coupling occurs in the case of a unidirectional composite is presented in Fig. 3. A specimen in which the unidirectional fibers are oriented at an angle of 45j with respect to the x-axis is shown, and a small square element from the gage region is isolated. Because the element is initially square and, in this example, the fibers are defined to be at an angle of 45j, fibers are parallel to diagonal AC of the element. In contrast, fibers are perpendicular to diagonal BD. This implies that the element is stiffer along diagonal AC than along diagonal BD.

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Figure 3 A 45j off-axis composite specimen used to explain the origin of coupling effects.

Now assume that a tensile stress is applied, causing the square element (as well as the specimen as a whole) to deform. Because the stiffness is higher along diagonal AC than along diagonal BD, the length of diagonal AC is increased to a lesser extent than that of diagonal BD. Hence, the initially square element deforms into a parallelogram, as shown in the figure. Note that: 

The length of the square element is increased in the x-direction (corresponding to a tensile strain, exx).  The length of the square element is decreased in the y-direction (corresponding to a compressive strain eyy, and associated with the Poisson effect).  BDAB is no longer p/2 rad, which indicates that a shear strain cxy has been induced.

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Hence, in this example, the normal stress rxx has induced two normal strains (exx and eyy) as well as a shear strain cxy. Hence, coupling exists between rxx and cxy, as stipulated. The couplings between normal stresses and shear strains (as well as couplings between shear stresses and normal strains) will be explained in a formal mathematical sense in Chap. 4. However, two important conclusions can be drawn from the physical explanation shown in Fig. 3. First, note that the specimen shown is subjected to normal stress rxx only. In particular, note that no shear stress exists in the x–y coordinate system (sxy = 0). Consequently, the in-plane principal stresses applied to the specimen are rp1 = rxx and rp2 = ryy = 0, and the principal stress coordinate system is defined by the x–y coordinate system. However, a shear strain does exist in the x–y coordinate system (cxy p 0). Consequently, the x–y coordinate system is not the principal strain coordinate system. We conclude, therefore, that the principal stress coordinate system is not aligned with the principal strain coordinate system. This is generally true for all anisotropic materials and is in direct contrast to the behavior of isotropic materials because for isotropic materials, the principal stress and principal strain coordinate systems are always coincident. Secondly, note that the physical argument used above to explain the origin of the coupling effect hinges on the fact that the fiber direction differs from the direction of the applied stress rxx. Specifically, the fibers are oriented 45j away from the direction of the applied stress rxx. If the fibers were aligned with either the x- or y-axis, then a coupling between rxx and cxy would not occur. We conclude that the unusual coupling effects exhibited by composites only occur if stress and strain are referenced to a nonprincipal material coordinate system. Anisotropic materials are classified according to the number of planes of symmetry defined by the microstructure. The principal material coordinate axes lie within the planes of symmetry. For example, in the case of unidirectional composites, three planes of symmetry can be defined: the 1–2 plane, the 1–3 plane, and the 2–3 plane. Composites fall within one of two classifications of anisotropic behavior. Specifically, composites are either orthotropic materials or transversely isotropic materials (the distinction between orthotropic and transversely isotropic materials will be further discussed in Sec. 2). During the composite structural analyses discussed in this text, the composite will be called ‘‘anisotropic’’ if the coordinate system of reference is a nonprincipal material coordinate system. Use of the term ‘‘anisotropic’’ will therefore signal the possibility of couplings between normal stresses and shear strains, and couplings between shear stresses and normal strains. If, instead, a structural analysis is referenced to the principal material coordinate system, the composite will be called either orthotropic or transversely isotropic.

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Although many kinds of material properties may be defined, in this text we are only interested in those properties commonly used by structural engineers. The properties needed to perform a structural analysis of composite structures will be defined in the following sections. In each section, a general definition of the material property will be given, suitable for use with anisotropic materials. That is, the properties of a composite material when referenced to a nonprincipal material coordinate system will be discussed first. These general definitions will then be specialized to the principal material coordinate system (i.e., they will be specialized for the case of orthotropic or transversely isotropic composites). Typical values of the properties discussed in this chapter measured at room temperatures are listed for glass/epoxy, Kevlar/epoxy, and graphite/ epoxy in Table 3. These properties do not represent the properties of any specific commercial composite material system, but rather should be viewed as typical values. Due to ongoing research and development activities within the industry, the properties of composites are improved more or less continuously. Therefore, the properties listed in Table 3 may not reflect those of currently available materials. The properties that appear in the table will be used in example and homework problems throughout the remainder of this text. 2 MATERIAL PROPERTIES THAT RELATE STRESS TO STRAIN Both stress and strain are second-order tensors, as discussed in Chap. 2. The material properties used to relate the stress and strain tensors are inferred from experimental measurements. Conceptually, two different experimental approaches may be taken. In the first approach, the material of interest is subjected to a well-defined stress tensor, and components of the resulting strain tensor are measured. In the second approach, the material of interest is subjected to a well-defined strain tensor, and the components of the resulting stress tensor are measured. From an experimental standpoint, it is far easier to impose a well-defined stress tensor than a well-defined strain tensor, and hence the first approach is almost always used in practice. Recall that there are two fundamental types of stress components: normal stress and shear stress. As a consequence, two fundamental types of tests are used to relate stress to strain—specifically, a test that involves the application of a known normal stress component and a test involving application of a known shear stress component. In either case, a stress tensor is imposed in which five of the six stress components equal zero, and the resulting six components of the strain tensor are measured. Tests that involve application of a known normal stress component are called uniaxial tests. In a typical case, a single normal stress component is

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applied to a test specimen (say, rxx), while insuring that the remaining five stress components are zero (ryy = rzz = sxy = sxz = syz = 0). The six components of strain caused by rxx are measured, which allows the calculation of various material properties relating a normal stress component to the strain tensor. In contrast, tests that involve application of a known shear stress component are called pure shear tests. In a typical case, a single shear stress component is applied to a test specimen (say, sxy), while insuring that the remaining five stress components are zero (rxx = ryy = rzz = sxz = syz = 0). The six components of strain caused by sxy are measured, which allows the calculation of various material properties relating a shear stress component to the strain tensor. A detailed description of experimental methods used to impose a specified state of stress is beyond the scope of the present discussion. It should be mentioned, however, that some of the stress states discussed below are very difficult to achieve in practice. For example, because composites are usually produced in the form of thin platelike structures, it is very difficult to impose well-defined stresses acting normal to the plane of the composite, or to measure the strain components induced normal to the plane of the composite by a given state of stress. For present purposes, these very real practical difficulties will be ignored. It will simply be assumed that the stress tensors discussed have been induced in the test specimen, and that methods to measure the resulting strain components involved are available. A number of international and industrial organizations publish annual test standards that describe available experimental arrangements in detail. Some of the best known standards are those published by the American Society of Testing and Materials (ASTM). ASTM tests standards that describe techniques to measure composite material properties relevant to the present discussion are listed in Table 1. There are also many composite test methods used routinely in industrial, governmental, or university composite laboratories that have not as yet been standardized by organizations such as the ASTM. New test methods are being developed continuously, and the reader should be alert for new methods and test standards as they become available. 2.1 Uniaxial Tests Referring to Fig. 1, suppose a uniaxial test is conducted using specimen 1 (i.e., material properties are measured in the x-direction). As the test proceeds, stress rxx is increased from zero to some maximal level, and the components of strain induced as a result of this stress are measured. An idealized plot of strain data collected during a uniaxial test of an anisotropic material is shown in Fig. 4, where it is has been assumed that the magnitude of stress is relatively

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Table 1

ASTM Test Standards for Determining Elastic Moduli of Polymeric Composites and Related Standards Designation D3039 D5450 D695 D3410

D5467

D5449

D3518

D5379 D4255 D5448

Related standards D5687 D638 D882 D4018 D2343

Title Standard Test Method for Tensile Properties of Polymer Matrix Composite Materials Standard Test Method for Transverse Tensile Properties of Hoop Wound Polymer Matrix Composite Cylinders Standard Test Method for Compressive Properties of Rigid Plastics Standard Test Method for Compressive Properties of Polymer Matrix Composite Materials with Unsupported Gage Section by Shear Loading Standard Test Method for Compressive Properties of Unidirectional Polymer Matrix Composites Using a Sandwich Beam Standard Test Method for Transverse Compressive Properties of Hoop Wound Polymer Matrix Composite Cylinders Standard Practice for In-Plane Shear Response of Polymer Matrix Composite Materials by Tensile Test of a +45j Laminate Standard Test Method for Shear Properties of Composite Materials by the V-Notched Beam Method Standard Guide for Testing In-Plane Shear Properties of Composite Laminates Standard Test Method for In-Plane Shear Properties of Hoop Wound Polymer Matrix Composite Cylinders

Standard Guide for Preparation of Flat Composite Panels with Processing Guidelines for Specimen Preparation Standard Test Method for Tensile Properties of Plastics Standard Test Method for Tensile Properties of Thin Plastic Sheeting Standard Test Methods for Properties of Continuous Filament Carbon and Graphite Fiber Tows Standard Test Method for Tensile Properties of Glass Fiber Strands, Yarns, and Rovings Used in Reinforced Plastics

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Figure 4 Idealized plot of the six strain components caused by the application a uniaxial stress rxx.

low, such that a linear relationship exists between the stress component rxx and the resulting strains. Strains induced at high nonlinear stress levels, including failure stresses, will be considered in Sec. 5. Note that for an anisotropic material, stress rxx will induce all six components of strain: exx, eyy, ezz, cxy, cxz, and cyz. This is not the case for isotropic materials; a uniaxial stress rxx applied to an isotropic material will not induce any shear strains (cxy = cxz = cyz = 0); furthermore, the transverse normal strains will be identical (eyy = ezz). Hence, for anisotropic material, there is an unusual coupling between normal stress and shear strain, which would not be expected based on previous experience with isotropic materials. As would be expected, as the magnitude of rxx is increased, the magnitude of all resulting strain components is also increased. Because stress rxx causes six distinct components of strain for an anisotropic material, six material properties must be defined in order to relate rxx to the resulting strains. Let us first consider material properties relating normal stress rxx to normal strains exx, eyy, and ezz. The relationship between rxx and normal strain exx is characterized by Young’s modulus Exx (also called the ‘‘modulus of elasticity’’): rxx Exx u e xx

ð1Þ

Young’s modulus is simply the slope of the rxx vs. exx curve shown in Fig. 4. In words, Young’s modulus is defined as ‘‘the normal stress rxx divided

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by the resulting normal strain exx, with all other stress components equal to zero.’’ Subscripts xx have been used to indicate the direction in which Young’s modulus has been measured. Because we have restricted our attention to the linear region of the stress–strain curve, Eq. (1) is only valid at relatively low, linear stress levels. The relationship between the two transverse strains (eyy and ezz) and exx is defined by Poisson’s ratio: eyy vxy u e xx

ezz vxz u e xx

ð2Þ

In words, Poisson’s ratio vxy (or vxz) is defined as ‘‘the negative of the transverse normal strain eyy (or ezz) divided by the axial normal strain exx, both of which are induced by stress rxx, with all other stresses equal to zero.’’ As before, subscripts have been used to indicate the uniaxial stress condition under which Poisson’s ratio is measured. The first subscript indicates the direction of stress, and the second subscript indicates the direction of transverse strain. For example, in the case of vxy, the first subscript x indicates that a uniaxial stress rxx has been applied, and the second subscript y indicates that transverse normal strain eyy has been used to calculate Poisson’s ratio. Combining Eqs. (1) and (2), a relationship between rxx and transverse strains eyy and ezz is obtained: eyy ¼

vxy rxx Exx

ezz ¼

vxz rxx Exx

ð3Þ

Now consider material properties relating normal strain exx to shear strains cxy, cxz, and cyz. Material properties relating normal strains to shear strains were discussed by Lekhnitski [2] and are called ‘‘coefficients of mutual influence of the second kind.’’ In this text, they will be denoted using the symbol g, and are defined as follows: cxy gxx;xy u e xx

c gxx;xz u e xz

xx

cyz gxx;yz u e

xx

ð4Þ

In words, the coefficient of mutual influence of the second kind gxx,xy (or gxx,xz, or gxx,yz) is defined as ‘‘the shear strain cxy (or cxz, or cyz) divided by the normal strain exx, both of which are induced by normal stress rxx, when all other stresses equal zero.’’ Subscripts have once again been used to indicate the stress condition under which the coefficient of mutual influence of the second kind is measured. The first set of subscripts indicates the direction of stress, and the second set of subscripts indicates the shear strain used to calculate the coefficient. For example, in the case of gxx,xy, the first two subscripts xx

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indicate that a normal stress rxx has been applied, and the second two subscripts xy indicate that cxy has been used to calculate the coefficient. Combining Eqs. (1) and (4), a relationship between rxx and shear strain cxy, cxz, or cyz is obtained: gxx;xy cxy ¼ e rxx xx

gxx;xz rxx cxz ¼ e xx

gxx;yz cyz ¼ e rxx xx

ð5Þ

Equations (1)–(5) define six properties measured in the x-direction, using specimen 1. Referring again to Fig. 1, analogous results are obtained when properties are measured in the y- and z-directions, using specimens 2 and 3: Properties Measured Using Specimen 2 (syy Applied) ryy 1 ðorÞ eyy ¼ ryy Eyy u e yy Eyy vyx exx ðorÞ exx ¼ vyx u e ryy yy Eyy vyz ezz ðorÞ ezz ¼ ryy vyz u e yy Eyy cxy gyy;xy ðorÞ cxy ¼ e ryy gyy;xy u e yy yy gyy;xz cxz gyy;xz u e ðorÞ cxz ¼ e ryy yy yy cyz gyy;yz gyy;yz u e ðorÞ cyz ¼ e ryy yy yy Properties Measured Using Specimen 3 (szz Applied): rzz 1 ðorÞ ezz ¼ rzz Ezz u e zz Ezz exx vzx vzx u e ðorÞ exx ¼ rzz zz Ezz eyy vzy ðorÞ eyy ¼ rzz vzy u e zz Ezz cxy gzz;xy gzz;xy u e ðorÞ cxy ¼ e rzz zz zz gzz;xz cxz gzz;xz u e ðorÞ cxz ¼ e rzz zz zz cyz gzz;yz ðorÞ cyz ¼ e rzz gzz;yz u e zz zz

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ð6Þ

ð7Þ

2.2 Pure Shear Tests If a pure shear stress (say, sxy) is applied to an anisotropic material, six components of strain will be induced. An idealized plot of strain data collected during a pure shear test of an anisotropic material is shown schematically in Fig. 5, where it is assumed that magnitude of shear stress is relatively low such that a linear relationship exists between the stress component sxy and the resulting strains. Strains induced at high nonlinear stress levels, including failure stresses, will be considered in Sec. 5. Once again, the stress–strain response of an anisotropic material differs markedly from that of an isotropic material. Specifically, for an anisotropic material, stress sxy will induce all six components of strain: exx, eyy, ezz, cxy, cxz, and cyz. If an isotropic material is subjected to a pure shear stress sxy, only one strain component is induced (cxy); all other strain components are zero (exx = eyy=ezz = cxz = cyz = 0). Hence, for anisotropic material, there is an unusual coupling between shear stress and normal strain, as well as an unusual coupling between shear stress in one plane (say, the x–y plane) and out-ofplane shear strains (cxz and cyz). Neither of these coupling effects occurs in isotropic materials. As would be expected, as the magnitude of sxy is increased during the test, the magnitude of the resulting strains is also increased. Because stress sxy causes six distinct components of strain, six material properties must be defined in order to relate sxy to the resulting strains.

Figure 5 Idealized plot of the six strain components caused by the application a pure shear stress sxy.

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Let us first consider material properties relating shear stress sxy to shear strains cxy,cxz, and cyz. The relationship between sxy and shear strain cxy is characterized by the shear modulus Gxy: sxy Gxy u c ð8Þ xy In words, the shear modulus is defined as ‘‘the shear stress sxy divided by the resulting shear strain cxy, with all other stress components equal to zero.’’ Because we have restricted our attention to linear stress levels, Eq. (8) is only valid at relatively low, linear shear stress levels. The relationship between transverse strains (cxz, cyz) and cxy is characterized by Chentsov coefficients, which will be denoted using the symbol l in this text: cyz c lxy;yz u c ð9Þ lxy;xz u cxz xy xy In words, the Chentsov coefficient lxy,xz (or lxy,yz) is defined as ‘‘the shear strain cxz (or cyz) divided by the shear strain cxy, both of which are induced by shear stress sxy, with all other stresses equal to zero.’’ The first set of subscripts indicates the stress component, and the second set of subscripts indicates the out-of-plane shear strain used to calculate the Chentsov coefficient. For example, in the case of lxy,xz the subscripts xy indicate that a pure shear stress sxy has been applied, and the second two subscripts xz indicate that cxz has been used to calculate the coefficient. A comparison between Eqs. (2) and (9) reveals that Chentsov coefficients are directly analogous to Poisson’s ratio. Poisson’s ratio is defined as a ratio of normal strains caused by a normal stress, whereas Chentsov coefficients are defined as a ratio of shear strains caused by a shear stress. Combining Eqs. (8) and (9), a relationship between sxy and shear strain cxz or cyz is obtained: lxy;xz lxy;yz cxz ¼ sxy cyz ¼ sxy ð10Þ Gxy Gxy Finally, consider material properties relating shear stress sxy to normal strains exx, eyy, and ezz. Material properties relating shear stress to normal strains were discussed by Lekhnitski [2] and are called ‘‘coefficients of mutual influence of the first kind.’’ In this text, they will be denoted using the symbol g, and are defined as follows: exx gxy;xx u c xy

eyy gxy;yy u c xy

ezz gxy;zz u c xy

ð11Þ

In words, the coefficient of mutual influence of the first kind gxy,xx (or gxy,yy, or gxy,zz) is defined as ‘‘the normal strain exx (or eyy, or ezz) divided by

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the shear strain cxy, both of which are induced by shear stress sxy, when all other stresses equal zero.’’ The first set of subscripts indicates the stress component applied, and the second set indicates the normal strain used to calculate the coefficient. For example, in the case of gxy,xx, the first subscripts xy indicate that shear stress sxy has been applied, and the second set of subscripts xx indicates that exx has been used to calculate the coefficient. Combining Eqs. (8) and (11), a relationship between sxy and normal strain exx, eyy, or ezz is obtained: gxy;xx gxy;yy gxy;zz sxy eyy ¼ sxy ezz ¼ sxy ð12Þ exx ¼ Gxy Gxy Gxy Equations (8)–(12) define six properties measured when a pure shear stress sxy is applied. Analogous material properties are defined during tests in which pure shear sxz or syz is applied. Properties Measured Using Pure Shear t xz sxz Gxz u c xz cxy lxz;xy u c xz cyz lxz;yz u c xz exx gxz;xx u c xz eyy gxz;yy u c xz ezz gxz;zz u c xz

1 sxz Gxz lxz;xy cxy ¼ sxz Gxz lxz;yz sxz cyz ¼ Gxz gxz;xx exx ¼ sxz Gxz gxz;yy eyy ¼ sxz Gxz gxz;zz ezz ¼ sxz Gxz

ðorÞ cxz ¼ ðorÞ ðorÞ ðorÞ ðorÞ ðorÞ

ð13Þ

Properties Measured Using Pure Shear t yz syz Gyz u c yz cxy lyz;xy u c yz cxz lyz;xz u c yz exx gyz;xx u c yz eyy gyz;yy u c yz ezz gyz;zz u c yz

ðorÞ ðorÞ ðorÞ ðorÞ ðorÞ ðorÞ

1 syz Gyz lyz;xy syz cxy ¼ Gyz lyz;xz cxz ¼ syz Gyz gyz;xx exx ¼ syz Gyz gyz;yy syz eyy ¼ Gyz gyz;zz syz ezz ¼ Gyz cyz ¼

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ð14Þ

2.3 Specialization to Orthotropic and Transversely Isotropic Composites As previously shown in Fig. 2a and b, for unidirectional composites, the principal material coordinate system is defined by the fiber direction. That is, the 1-axis is defined parallel to the fiber direction, the 2-axis is perpendicular to the fibers and lies within the plane of the composite, and the 3-axis is perpendicular to the fibers and lies out of plane. In other cases, such as the braided composite shown in Fig. 2c and d, the 1–2–3 principal material coordinate system is not aligned with the fiber direction, but is instead defined by planes of symmetry associated with the fiber architecture involved. For all composite fabrics based on continuous fibers and typically encountered in practice (i.e., unidirectional, woven, or braided fabrics), the principal material coordinate system is readily identified. We will now consider those properties that are measured when the composite is referenced to the principal material coordinate system. It will be seen later that properties of an anisotropic composite (i.e., a composite referenced to a nonprincipal material coordinate system) can always be related to those measured relative to the 1–2–3 coordinate system. To simplify our discussion, we will assume that the composite under consideration is a unidirectional composite, and hence that the 1–2–3 axes are parallel and perpendicular to the fibers. A stress element representing a unidirectional composite subjected to uniaxial tensile stress r11 is shown in Fig. 6. The deformed shape of the element is also shown. Note that: 

The element has increased in length in the l-direction, corresponding to a tensile strain e11.  The element has decreased in width in the 2- and 3-directions, corresponding to compressive strains e22 and e33, respectively.  The deformed element is a rectangular parallelepiped. That is, due to the symmetrical distribution of fibers with respect to the 1-, 2-, and 3coordinate axes, in the deformed condition, all angles remained p/2 rad (90j). Hence, all shear strains equal zero (c12=c13=c23=0). Applying Eqs. (1), (2), and (4), we have: r11 E11 u e 11

ð15aÞ

e22 v12 u e 11

ð15bÞ

e33 v13 u e 11

ð15cÞ

g11;12 ¼ g11;13 ¼ g11;23 ¼ 0

ð15dÞ

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Figure 6 Deformations induced in a unidirectional composite by uniaxial stress r11 (deformations are shown greatly exaggerated for clarity).

Because no shear strains are induced by r11, the coefficients of mutual influence of the second kind all equal zero. This is only true when the composite is referenced to the principal material coordinate system. That is, uniaxial stress acting in a nonprincipal coordinate system will cause a shear strain, as previously shown in Fig. 3, for example. Therefore, the coefficients of mutual influence of the second kind do not equal zero for anisotropic composites (i.e., if the composite is referenced to a nonprincipal material coordinate system). Methods of calculating composite material properties in nonprincipal coordinate systems will be presented in Chap. 4. Similarly, material properties measured when stress r22 is applied are: r22 E22 u e ð16aÞ 22 e11 v21 u e ð16bÞ 22 v23 u g22;12

e33 e22 ¼ g22;13 ¼ g22;23 ¼ 0

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ð16cÞ ð16dÞ

Once again, due to the symmetrical distribution of fibers, stress r22 does not induce any shear strains, so the coefficients of mutual influence of the second kind all equal zero. As previously mentioned, due to the thin platelike nature of composites, it is difficult in practice to apply a well-defined out-of-plane uniaxial stress r33, or to measure the resulting normal strain induced in the out-of-plane direction e33. Assuming that these practical difficulties are overcome, the material properties measured when stress r33 is applied are: r33 E33 u e ð17aÞ 33 e11 v31 u e ð17bÞ 33 e22 v32 u e ð17cÞ 33 ð17dÞ g33;12 ¼ g33;13 ¼ g33;23 ¼ 0 For unidirectional composites, both the 2- and 3-axes are defined to be perpendicular to the fibers, and hence properties measured in the 2- and 3directions are typically similar in magnitude. In fact, if the distribution of fibers in the 2- and 3-directions is identical at the microlevel, then properties measured in these directions will be equal: E22 = E33, v12 = v13, v21 = v31, and v23 = v32. If this occurs, then the composite is classified as a transversely isotropic material. In contrast, if the distribution of fibers differs in the 2- and 3-directions, or if the composite under consideration is a woven or braided composite, then properties measured in the 2- and 3-directions will not be identical and the composite is classified as an orthotropic material. Optical micrographs showing the fiber distribution in the 2–3 plane for a unidirectional graphite–polyimide laminate are shown in Fig. 7. Fig. 7a was taken at a magnification of 150, and shows the fiber distribution in four adjacent plies. The fiber angles are (from left to right) 0j, 45j, 90j, and 45j. Fig. 7b was obtained for the same laminate but at higher magnification (300), and shows fiber angles (from left to right) of 0j, 45j, and 90j. As indicated, for this laminate, thin resin-rich zone exists between plies. The thickness of the resin-rich zone varies from one laminate to the next, depending on the material system, stacking sequence, and processing conditions used to produce the laminate. If the resin-rich zone is very thin (say, less than about 1/10 the ply thickness) and if the fibers are uniformly distributed within the interior of each ply, the composite will respond as a transversely isotropic material. If these conditions do not exist (if the thickness of the resin-rich zone is an appreciable fraction of the ply thickness, or if the distribution of fibers in the 2- and 3-directions differs substantially), then E33 will differ from E22, and the composite will respond as an orthotropic material.

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Figure 7 Optical micrographs of fibers within several plies of a [0/45/90/ 45]2s graphite–polyimide (IM7/K3B) composite laminate. Note the resin-rich zone between plies.

Let us now consider properties measured through the application of a pure shear stress in the principal material coordinate system. A stress element representing a unidirectional composite subjected to a pure shear stress s12 is shown in Fig. 8. The deformed shape of the element is also shown. Note that: 

The angle originally defined by the 1–2 axes has decreased, corresponding to a positive shear strain c12.

Figure 8 Deformations induced in a unidirectional composite by pure shear stress s12.

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Due to the symmetrical distribution of fibers with respect to the 1-, 2-, and 3-coordinate axes, in the deformed condition, all remaining angles remained p/2 rad (90j). Hence, the remaining two shear strains equal zero (c13 = c23 = 0).  The length, width, and thickness of the element have not changed; hence, all normal strain are zero (e11 = e22 = e33 = 0). Applying Eqs. (8)–(10), we have: G12 u

s12 c12

ð18aÞ

l12;13 ¼ l12;23 ¼ 0

ð18bÞ

g12;11 ¼ g12;22 ¼ g12;33 ¼ 0

ð18cÞ

Because only c12 is induced by s12, the Chentsov coefficients as well as the coefficients of mutual influence of the first kind are all equal to zero. This is only true when the composite is referenced to the principal material coordinate system. That is, a shear stress acting in a nonprincipal material coordinate system will, in general, cause both normal strains and shear strains. Therefore, neither the Chentsov coefficients nor the coefficients of mutual influence of the first kind equal zero if the composite is referenced to a nonprincipal material coordinate system. Methods of calculating composite material properties in nonprincipal coordinate systems will be presented in Chap. 4. Once again, due to the thin platelike nature of composites, in practice, it is difficult to apply well-defined out-of-plane shear stress s13 or s23, or to measure the resulting shear strains induced in the out-of-plane direction c13 or c23. Assuming that these practical difficulties were overcome, the material properties measured when stress s13 is applied are: s13 c13

ð19aÞ

l13;12 ¼ l13;23 ¼ 0

ð19bÞ

g13;11 ¼ g13;22 ¼ g13;33 ¼ 0

ð19cÞ

G13 u

If the fibers are not uniformly distributed within the 2–3 plane, or if the composite is based on woven or braided fabrics, then the composite will behave as an orthotropic material and G12 p G13. If the composite is based on a unidirectional fabric and fibers are uniformly distributed, then the composite is transversely isotropic and G12 = G13.

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Following an analogous process, material properties measured when s23 is applied are: s23 G23 u ð20aÞ c23 l23;12 ¼ l23;23 ¼ 0

ð20bÞ

g23;11 ¼ g23;22 ¼ g23;33 ¼ 0

ð20cÞ

A total of 12 material properties have been defined above for orthotropic or transversely isotropic composites: three Young’s moduli (E11, E22, and E33), six Poisson’s ratios (v12, v13, v21, v23, v31, and v32), and three shear moduli ( G12, G13, and G23). However, it will be seen later that for orthotropic composites, only nine of these 12 properties are independent, and for transversely isotropic composites, only five of the 12 properties are independent. Therefore, only nine material properties must be measured to fully characterize the elastic response of orthotropic composites; for transversely isotropic composites, only five material properties must be measured. The number of material properties required in most practical engineering applications of composite is reduced further still. For reasons that will be explained later, it is usually appropriate to assume that a composite structure is subjected to a state of plane stress. Ultimately, this means that we only require material properties in one plane. Hence, whereas an orthotropic composite possesses nine distinct elastic material properties (and a transversely isotropic composite possesses five), in practice, only four of these properties are ordinarily required: E11, E22, v12, and G12. Most of the ASTM test standards listed in Table 1 describe techniques used to measure these properties. Also, a brief summary of common experimental methods used to measure in-plane properties is provided in Appendix B. Typical values for several composite material systems are listed in Table 3. As a final comment, an often overlooked fact is that the elastic properties of composites usually differ in tension and compression (in fact, this is true for many materials, not just for composites). For example, for polymeric composites, it is not uncommon for E22 measured in tension to differ by 10– 15% from that measured in compression. Materials that exhibit this behavior are called ‘‘bimodulus materials.’’ Although it is possible to account for these differences during a structural analysis (e.g., see Ref. 3), the bimodulus phenomenon is a significant complication and will not be accounted for herein. Throughout this text, it will be assumed that in-plane elastic properties E11, E22, v12, and G12 are identical in tension and compression. The reader should be aware that these differences usually exist, however. If in practice the measured response of a composite structure differs from the predicted

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behavior, the discrepancy may well be due to differences in elastic properties in tension vs. compression. 3 MATERIAL PROPERTIES RELATING TEMPERATURE TO STRAIN If an unconstrained anisotropic composite is subjected to a uniform change T T in temperature DT, six components of strain will be induced: exx , eTyy , ezz , T T T cxy , cxz , and cyz. The superscript ‘‘T’’ has been used to indicate that these strains are caused solely by a change in temperature. Note that three of these strains are shear strains; for anisotropic materials, a change in temperature will, in general, cause shear strains to develop. Strains induced solely by a change in temperature are referred to as ‘‘free thermal strains’’ or simply ‘‘thermal strains.’’ Properties that relate strains to temperature change are called coefficients of thermal expansion (CTEs). As previously discussed, it is the microstructural features of a material that determine whether it exhibits isotropic or anisotropic behavior. The contention that a change in temperature will induce shear strains may seem unusual (because isotropic materials do not exhibit such behavior), but can be easily explained in the case of unidirectional composites. An initially square unidirectional composite is shown in Fig. 9, where it has been assumed that the fibers are oriented at an angle of 45j with respect to the x-axis. Because the composite is initially square and, in this example, the fibers are defined to be at an angle of 45j, fibers are parallel to diagonal AC and are perpendicular to diagonal BD. Now, the coefficient of thermal expansion exhibited by highperformance fibers is typically very low (or even slightly negative), whereas for

Figure 9 Deformations caused in a 45j unidirectional composite by a uniform change in temperature DT (deformations are shown greatly exaggerated for clarity).

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most polymers, it is relatively high. For example, the coefficient of thermal expansion of graphite fibers is about 1 Am/m jC, whereas for epoxies, it is on the order of 30 Am/m jC. Therefore, assuming that the composite shown in Fig. 9 consists of a graphite/epoxy system, then an increase in temperature will cause a slight decrease in the length of diagonal AC, but will cause a relatively large increase in the length of diagonal BD. Hence, the initially square composite deforms into a parallelogram, as shown in the figure. The fact that angle BDAB has increased reveals that a shear strain cxy (in this case, a negative shear strain cxy) has been induced by the change in temperature DT. Hence, there is a coupling between the change in temperature and shear strains, as stipulated. Note that this physical explanation of the coupling between a uniform change in temperature and shear strain indicates that this coupling only occurs if strain is referenced to a nonprincipal material coordinate system. An idealized plot of the six strain components induced in an anisotropic composite by a change in temperature is shown in Fig. 10. As would be

Figure 10 Idealized plot of the six strain components caused by a change in temperature DT.

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expected, as DT is increased, the magnitude of all strain components also increases. For modest changes in temperature (say, DTr22 and r22 >r22 ).

6 PREDICTING ELASTIC COMPOSITE PROPERTIES BASED ON CONSTITUENTS: THE RULE OF MIXTURES Various material properties exhibited by composites at the structural level have been described in preceding sections. These properties are usually measured for a composite material of interest, using one or more of the ASTM (or equivalent) test standards listed in Tables 1 and 2. However, in practice, the need to predict composite material properties exhibited at the structural level also arises. That is, in practice, there is a need to predict composite properties at the structural level based on properties of the individual constituent materials (i.e., the fiber and matrix). As a typical example, suppose a new high-performance graphite fiber has recently been developed, and properties of the fiber itself have been measured. Naturally, the structural engineer is interested in determining whether this new fiber will lead to improvements in composite material properties at the structural level. The potential improvement in properties can, of course, be evaluated directly, by embedding the new fiber in a polymeric matrix of interest and by measuring the properties exhibited by the new composite material system. However, creating and testing the new material system in this fashion is timeconsuming and expensive. A need to estimate the properties that will be provided by the new fiber exists, so as to justify the time and money that will

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be invested during the development of the composite material system based on the new graphite fiber. As described in Sec. 6 of Chap. 1, an analysis performed at a physical scale corresponding to the fiber diameter is classified as a micromechanics analysis. In the present instance, we wish to use a micromechanics-based analysis to predict composite properties at the structural level. A simple micromechanics model that can be used to make this prediction is called the rule of mixtures and is developed as follows. Consider the representative composite element shown in Fig. 14. As indicated, the element consists of unidirectional fibers embedded within a polymeric matrix. The principal material coordinate system, labeled the 1– 2–3 coordinate system, is defined by the fiber direction. It is assumed that the fibers are evenly spaced, and that the matrix is perfectly bonded to the fiber. If a force F11 is applied to the element, as shown in Fig. 14a, the length of the element is increased by an amount DL and the width of the element is decreased by an amount DW. Force F11 is related to the average stress imposed in the 1-direction by F11=r11A, where A is the cross-sectional area of the element. Furthermore, the sum of forces present in the matrix and fibers must equal the total applied force, which implies: r11 A ¼ rf Af þ rm Am

ð27Þ

where Af is the total cross-sectional area of the fibers presented within the element and Am is the cross-sectional area of the matrix. The strain in the 1direction is associated with the change in length (DL), and is identical in fiber and matrix because the fiber and matrix are assumed to be perfectly bonded. That is: DL ef ¼ em ¼ e11 ¼ L Stresses are assumed to be related to strains according to: r11 ¼ e11 E11

rf ¼ ef Ef ¼ e11 Ef

rm ¼ em Em ¼ e11 Em

ð28aÞ

ð28bÞ ð28cÞ

The expressions for stresses rf and rm are only approximate. In reality, a triaxial state of stress is induced rather than a uniaxial stress state, as implied by Eqs. (28b) and (28c), due to the mismatch in fiber and matrix properties as well as the presence of adjacent fibers. Properly accounting for this (and other) complicating factors requires a rigorous analysis that is beyond the scope of the brief introduction presented here. Therefore, we will assume that

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Figure 14 Representative composite element used to derive rule-of-mixtures equations. (a) Composite element deformed by a load F11, acting parallel to the fiber direction. (b) Composite element deformed by a load F22, acting perpendicular to the fiber direction. (c) Composite element deformed by shear load F12.

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fiber and matrix stresses are given by Eqs. (28b) and (28c) despite their shortcomings. Substituting these expressions into Eq. (27) and rearranging, we find: E11 ¼ Ef

Af Am þ Em A A

This expression allows us to predict E11 based on properties of the constituents (Ef and Em) and the area fractions of fiber and matrix (Af/A) and (Am/A). If no voids are present, then: A ¼ Af þ Am Usually, rule-of-mixtures expressions are written in terms of volume fractions rather than area fractions. Volume fractions are given by: Vf ¼

Af A

Vm ¼

Am A Af ¼ ¼ ð1 A A

Vf Þ

where Vf is the volume fraction of fibers and Vm = (1 Vf) is the volume fraction of matrix material. Consequently, the predicted value of E11 based on the rule of mixtures approach is given by: E11 ¼ Em þ Vf ðEf

Em Þ

ð29Þ

Polymeric composites used in practice are typically produced with a fiber volume fraction Vf of about 0.65, although it can be lower (say, Vf = 0.30), depending on application and the manufacturing process used to consolidate the composite. Equation (29) shows that if Ef  Em (which is usually the case), then to a first approximation, E11cVfEf. The value of E11 is dictated primarily by the fiber modulus Ef and fiber volume fraction Vf. E11 is therefore called a fiber-dominated property of the composite. Now consider the Poisson effect exhibited by the composite element shown in Fig. 14a. As per our normal definition, the average Poisson ratio is defined as the negative of the transverse normal strain (e22) divided by axial normal strain (e11), both of which are caused by r11: v12 ¼

e22 e11

The transverse normal strain associated with the change in width of the entire element (DW) is given by: e22 ¼

DW W

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The change in width can also be written as the sum of the change in width of the fibers present in the element DWf and the change in width of the matrix present DWm. These are approximated as follows: DWf ¼

WVf vf ef ¼

DWm ¼

WVf vf e11

WVm vm em ¼

WVm vm e11

where vf and vm are Poisson ratios of the fiber and matrix, respectively. Hence, the transverse strain is given by: e22 ¼

DW ¼ W

½Vf vf þ ðVm Þvm Še11

Applying the definition of Poisson’s ratio for the composite as a whole, we have: v12 ¼

e22 e11

¼ V f vf þ V m vm

Noting as before that Vm=(1 Vf), the predicted value for Poisson’s ratio v12 based on the rule of mixtures becomes: v12 ¼ vm

Vf ðvm

vf Þ

ð30Þ

Measurement of Poisson’s ratio of the matrix material vm is a straightforward matter. However, measuring Poisson’s ratio of the fiber vf is more difficult due to the small fiber diameters involved. Experimentally measured values of vf are often unavailable, even for fibers widely used in practice. The data that are available imply that both vm and vf are algebraically positive, and also that vm>vf. Hence, Eq. (30) implies that the composite Poisson ratio v12 varies linearly with fiber volume fraction Vf, and that Poisson’s ratio of the composite is less than that of the matrix (v120. Assuming an identical fiber distribution in the 1–2 and 1–3 planes, then an identical analysis can be conducted to predict Poisson’s ratio v13, which will result in an identical expression: v13=v12. Next, consider prediction of the transverse modulus E22 based on the rule-of-mixtures approach. A composite element subjected to a force applied perpendicular to the fibers, force F22, is shown in Fig. 14b. This force is related to the average stress imposed in the 2-direction by F22 = r22A. We assume that an identical and uniform stress r22 is induced in both the fiber and matrix. Once again, this assumption is approximate at best; in reality, a triaxial state

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of stress is induced in both the fiber and matrix. Based on this assumption, the strains induced in the fiber and matrix perpendicular to the 1-axis are: r22 Ef r22 em ¼ Em

ef ¼

The transverse length represented by the fibers present in the element equals VfW, whereas the transverse length represented by the matrix equals VmW. Hence, the change in width W caused by the application of r22 is: DW ¼ ðVf WÞ

r22 r22 þ ðVm WÞ Ef Em

The average transverse strain caused by r22 is:

DW Vf Vm ¼ r22 e22 ¼ þ Ef Em W Young’s modulus E22 as predicted by the rule of mixtures therefore becomes: E22 ¼

r22 1 Ef Em ¼ ¼ Vf Vm e22 E V m f þ Ef Vm Ef þ Em

As before, if no voids are present, then Vm=(1 Vf), and we obtain: E22 ¼

Ef

Ef Em Vf ðEf Em Þ

ð31Þ

For most polymeric composite material systems, Ef> >Em. Nevertheless, Eq. (31) shows that E22 is dictated primarily by Em, and is only modestly affected by the fiber modulus Ef. Indeed, even in the limit (i.e., as Ef !l), the predicted value of E22 is only increased to:   Em E22 Ef !l ¼ 1 Vf

Because Vf is usually about 0.65, this result shows that E22 is still less than three times the matrix modulus Em, even if the composite is produced using a fiber whose stiffness is infinitely high (Ef !l). E22 is therefore called a matrix-dominated property of the composite. Assuming an identical fiber distribution in the 1–2 and 1–3 planes, then an identical analysis can be conducted to predict Young’s modulus in the 3-direction E33, resulting in an identical expression. Hence, E33=E22.

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As before, E33 is dictated primarily by Em and is a matrix-dominated property. Now consider the shear modulus G12. An element subjected to a pure shear force F12 is shown in Fig. 14c. This force is related to the average shear stress according to F12=s12A. In a rule-of-mixture analysis, it is assumed that an identical shear stress is induced in both the fibers and matrix regions. This assumption is approximate at best. Nevertheless, on the basis of this assumption, the shear strains induced in fiber and matrix are given by: s12 cf ¼ Gf s12 cm ¼ Gm where Gf and Gm are the shear moduli of the fiber and matrix, respectively. The total shear strain is given by:     s12 s12 c12 ¼ Vf cf þ Vm cm ¼ Vf þ Vm Gf Gm The shear modulus predicted by the rule of mixtures is then: G12 ¼

s12 1 Gf Gm ¼ ¼ V V c12 Gm Vf þ Gf Vm m f Gf þ Gm

Assuming no voids are present, then Vm=(1 Vf), and we obtain: G12 ¼

Gf

Gf Gm Vf ðGf Vm Þ

ð32Þ

Comparing Eq. (32) with Eq. (31), it is seen that the shear modulus is related to fiber and matrix properties in a manner similar to E22 and E33. The value of G12 is dictated primarily by the shear modulus of the matrix Gm, and is considered a matrix-dominated property. Assuming an identical fiber distribution in the 1–2 and 1–3 planes, then G13=G12. To summarize, the analysis presented above allows prediction of elastic moduli E11, v12=v13, E22=E33, and G12=G13, based on knowledge of the fiber modulus Ef, matrix modulus Em, and fiber volume fraction Vf. Although not presented here, a rule-of-mixture approach can also be used to predict thermal expansion coefficients a1 and a2, or moisture expansion coefficients b1 and b2 (e.g., see Chap. 3 of Ref. 11). The analysis presented above is only one of several micromechanicsbased models that have been proposed. The rule of mixtures is certainly the simplest approach, but unfortunately is often the least accurate. In general,

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E11 and v12=v13 are reasonably well predicted by Eqs. (29) and (30), respectively. However, matrix-dominated properties E22=E33 and G12=G13 are generally underpredicted by Eqs. (31) and (32). The accuracy of these predictions is not high because many important factors have not been accounted for. A partial listing of factors not accounted for include:       

The more or less random distribution and spacing of fibers present in a real composite The triaxial state of stress induced in both matrix and fiber due to the mismatch in fiber/matrix properties. Differences in fiber distribution in the 1–2 and 1–3 planes The adhesion (or lack thereof ) between fiber and matrix Variations in fiber cross-sections from one fiber to the next The presence of voids or other defects The anisotropic nature of many high-performance fibers (e.g., Young’s modulus parallel and transverse to the long axis of the fiber usually differs).

A rigorous closed-form analytical solution that accounts for all of these factors (as well as others) is probably impossible to obtain. Consequently, most advanced micromechanics analyses are performed numerically using finite element methods. Because the primary objective of this book is to investigate composite materials at the structural (i.e., macroscopic) level, only the simple rule of mixtures is presented herein. The reader interested in learning more about micromechanics analyses is referred to the many excellent texts that discuss this topic in greater detail, a few of which are listed here as Refs. 6, 9–13. HOMEWORK PROBLEMS 1. An orthotropic material is known to have the following elastic properties: Exx ¼ 100 GPa

Eyy ¼ 200 GPa

Ezz ¼ 75 GPa

vxy ¼ 0:20

vxz ¼

vyz ¼ 0:60

Gxy ¼ 60 GPa

Gxz ¼ 75 GPa

Gyz ¼ 50 GPa

gxx;xy ¼

gxx;xz ¼ 0:25

gxx;yz ¼ 0:30

gyy;xy ¼ 0:60

gyy;xz ¼ 0:75

gyy;yz ¼ 0:20

gzz;xy ¼

0:20

gzz;xz ¼

0:05

gzz;yz ¼

lxy;xz ¼

0:10

lxy;yz ¼

0:05

gxz;yz ¼ 0:10

0:30

0:25

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0:15

(a) What strains are induced if a uniaxial tensile stress rxx=300 MPa is applied? (b) What strains are induced if a uniaxial tensile stress ryy=300 MPa is applied? (c) What strains are induced if a uniaxial tensile stress rzz=300 MPa is applied? (d) What strains are induced if a pure shear stress sxy=100 MPa is applied? (e) What strains are induced if a pure shear stress sxz=100 MPa is applied? (f ) What strains are induced if a pure shear stress syz=100 MPa is applied? 2. An orthotropic material is known to have the following elastic properties: E11 ¼ 100 GPa

E22 ¼ 200 GPa

E33 ¼ 75 GPa

v12 ¼ 0:20

v13 ¼

v23 ¼ 0:60

G12 ¼ 60 GPa

G13 ¼ 75 GPa

0:25

G23 ¼ 50 GPa

(a) What strains are induced if a uniaxial tensile stress r11=300 MPa is applied? (b) What strains are induced if a uniaxial tensile stress r22=300 MPa is applied? (c) What strains are induced if a uniaxial tensile stress r33=300 MPa is applied? (d) What strains are induced if a pure shear stress s12=100 MPa is applied? (e) What strains are induced if a pure shear stress s13=100 MPa is applied? (f ) What strains are induced if a pure shear stress s23=100 MPa is applied? 3. A tensile specimen is machined from an anisotropic material. The specimen is referenced to an x–y–z coordinate system, as shown in Fig. 15a. The cross-section of the specimen is initially a ‘‘perfect’’ 55 mm square. In addition, ‘‘perfect’’ 55 mm squares are drawn on the x–y and x–z surfaces of the specimen, as shown. A uniaxial tensile stress rxx=700 MPa is then applied, causing the specimen to deform as shown in Fig. 15b–d. Determine the values of Exx, vxy, vxz, gxx,xy, gxx,xz, and gxx,yz that correspond to these deformations.

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Figure 15 Tensile specimen described in Problem 3 (deformations shown greatly exaggerated for clarity). (a) Tensile specimen machined from an anisotropic material. (b) Change in cross-section. (c) Change in dimensions on x–y face. (d) Change in dimensions on x–z face.

Table 4

The [0]8 Specimen (Width=1.251 in.; Thickness=0.048 in.)

Load (1 bf) 0 260 630 1220 1910 2600 4100

Axial strain (Ain./in.)

Trans strain (Ain./in.)

0 192 454 860 1335 1807 2784

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0 61 146 279 433 587 930

Table 5 The [90]16 Specimen (Width= 1.254 in.; Thickness=0.090 in.) Load (1 bf) 0 64 102 172 275 385

Axial strain (Ain./in.) 0 300 539 923 1489 2072

4. Load vs. strain data collected during two different composite tensile tests are shown in Tables 4 and 5. Use linear regression to determine the following properties for this composite material: (a) Determine E11 and v12 using the data collected using the [0]8 specimen (Table 4). (b) Determine E22 using the data collected using the [90]16 specimen (Table 5). (c) Determine the value of v21 for this composite material system.

Figure 16 Tensile specimen described in Problem 5.

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Figure 17 Deformed square described in Problem 6.

5. A thin tensile specimen is machined from a material with unknown properties. A ‘‘perfect’’ square with dimensions 5  5 mm is drawn on one surface of the specimen, as shown in Fig. 16. A tensile stress of 500 MPa is then applied, causing the square to deform. Determine Exx, vxy, and gxx,xy for this material. 6. A ‘‘perfect’’ square with dimensions 11 mm is drawn on the surface of a plate. The temperature of the plate is then uniformly increased by 300jC, causing the square to deform as shown in Fig. 17. Determine the corresponding strains exx, eyy, and exy, and coefficients of thermal expansion axx, ayy, and axy.

REFERENCES 1. Dieter, G.E. Mechanical Metallurgy; New York, NY: McGraw-Hill Inc., 1986; ISBN0-07-016893-8. 2. Lekhnitski, S.G. Theory of Elasticity of an Anisotropic Body; Holden-Day: San Francisco, 1963. 3. Bert, C.W.; Reddy, J.N.; Reddy, V.S.; Chao, W.C. Analysis of thick rectangular plates laminated of bimodulus composite materials. AIAA J. 1981, 19 (10), 1342– 1349. 4. Measurement of Thermal Expansion Coefficient, M-M Tech Note 513; Measurement Group, Inc.: Raleigh, NC, USA (available at the Measurement Group website at: http://www.measurementsgroup.com/guide/indexes/ tn_index.htm). 5. Standard Test Method for Moisture Absorption Properties and Equilibrium Conditioning of Polymer Matrix Composite Materials, Test Standard 5229;

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6. 7. 8. 9.

American Society for Testing and Materials: West Conshohocken, PA, USA (may also be accessed via the ASTM website at: http://www.astm.org/). Hyer, M.W. Stress Analysis of Fiber-Reinforced Composite Materials; McGrawHill Book Co.: New York, NY; 1998; ISBN 0-07-016700-1. Herakovich, C.T. Mechanics of Fibrous Composites; John Wiley and Sons: New York, NY, 1998; ISBN 0-471-10636-4. Hull, D. An Introduction to Composite Materials; Cambridge University Press: Cambridge, Great Britain, 1981; ISBN 0-521-23991-5. Gibson, R.F. Principles of Composite Material Mechanics; McGraw-Hill Inc.: New York, NY, 1994; ISBN0-07-023451-5.

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4 Elastic Response of Anisotropic Materials

In this chapter, we will consider the strains induced in anisotropic materials when subjected to arbitrary combinations of stress, uniform changes in temperature, and/or uniform changes in moisture content. The chapter begins with a consideration of the strains induced by stress under constant environmental conditions. A ‘‘generalized’’ form of Hooke’s law, which relates strain to stress for any anisotropic material, will be developed. Next, Hooke’s law will be specialized for two particular types of anisotropy. First, for orthotropic materials, and then for transversely isotropic materials. Attention will then be focussed on strains caused by uniform changes in temperature or moisture content. As before, relationships for anisotropic materials will be developed first, and will then be specialized to the case of orthotropic and transversely isotropic materials. Finally, the strains induced by the combined effects of stress, temperature, and moisture will be discussed. 1 STRAINS INDUCED BY STRESS: ANISOTROPIC MATERIALS A review of the stress and strain tensors has been provided in Chap. 2. Stress is a symmetrical second-order tensor. In tensoral notation, stress is written as 167

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rij, where subscripts i and j take on values of x, y, and z. Alternatively, the stress tensor can be written using matrix notation as: 2 3 rxx rxy rxz rij ¼ 4 ryx ryy ryz 5 ð1Þ rzx rzy rzz

Because the stress tensor is symmetrical (i.e., ryx=rxy, rzx=rxz, and rzy = ryz), only six independent stress components appear in Eq. (1). Similarly, strain is a symmetrical second-order tensor eij and can be written as: 3 2 2 3 ðexx Þ ðcxy =2Þ ðcxz =2Þ exx exy exz 7 6 6 ðeyy Þ ðcyz =2Þ 7 ð2Þ eij ¼ 4 eyx eyy eyz 5 ¼ 4 ðcyx =2Þ 5 =2Þ ðc =2Þ ðe Þ ðc ezx ezy ezz zz zx zy Only six independent strain components appear in Eq. (2). Also, the tensoral shear strain components equal one-half the more commonly used engineering shear strain components, that is, exy ¼ eyx ¼

cxy cyx cyz czy c c ¼ ; exz ¼ ezx ¼ xz ¼ zx ; eyz ¼ ezy ¼ ¼ 2 2 2 2 2 2

ð3Þ

For any elastic solid, the strain and stress tensors are related as follows (assuming temperature and moisture content remain constant): eij ¼ Sijkl rkl

ð4Þ

All subscripts that appear in Eq. (4) take on values of x, y, and z. Equation (4) is called generalized Hooke’s law, and is valid for any elastic solid under constant environmental conditions. It is seen that the strain and stress tensors are related via the fourth-order compliance tensor Sijkl. Because strains are unitless quantities, from Eq. (4), it is seen that the units of Sijkl are 1/ (stress) (i.e., either 1/Pa or 1/psi). Because the compliance tensor is described using four subscripts and because each subscript may take on three distinct value (e.g., x, y, or z), it would initially appear that 34=81 independent terms appear within the compliance tensor. However, due to symmetry of both the strain and stress tensors, it will be shown below that the compliance tensor consists of (at most) 36 material constants. It will be very convenient to express Eq. (4) using matrix notation. However, because Sijkl is a fourth-order tensor (and hence can be viewed as having ‘‘four dimensions’’), we cannot expand Sijkl as a two-dimensional

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matrix directly. To expand Eq. (4), we must first define the components of stress and strain using contracted notation, as follows: exx ! e1 eyy ! e2 ezz ! e3 cyz ¼ czy ! e4 cxz ¼ czx ! e5 cxy ¼ cyx ! e6

rxx ! r1 ryy ! r2 rzz ! r3 ryz ¼ rzy ! r4 rxz ¼ rzx ! r5 rxy ¼ ryx ! r6

ð5Þ

Notice that the symmetry of the strain and stress tensors (cyz=cxz, etc.) is embedded within the very definition of contracted notation. Also note that the shear strain components (e4, e5, and e6) represent engineering shear strains, rather than tensoral shear strains. Based on this change in notation, we can now write Eq. (4) as: ej ¼ Sij rj

where i; j ¼ 1 6

ð6Þ

In contracted notation, the strain and stress tensors are expressed with a single subscript (i.e., ei and rj), and hence in Eq. (6), they appear to be firstorder tensors. This is, of course, not the case. Both strain and stress are second-order tensors. We are able to write them as using contracted notation only because they are both symmetrical tensors. Similarly, contracted notation allows us to refer to individual components of the fourth-order compliance tensor expressed using only two subscripts. We will henceforth refer to Sij as the compliance matrix, and the use of contracted notation will be implied. Expanding Eq. (6), we have: 8 9 2 e1 > S11 > > > > > 6 > > > S21 e2 > > > > 6 > > = 6

6 S31 3 ¼6 6S > > e > 6 41 > 4> > > 6 > > > 4 S51 > e5 > > > > ; : > S61 e6

S12 S22 S32 S42 S52 S62

S13 S23 S33 S43 S53 S63

S14 S24 S34 S44 S54 S64

S15 S25 S35 S45 S55 S65

38 9 S16 > r1 > > > > > >r > S26 7 2> > > 7> > > > 7> < 7 r3 = S36 7 S46 7 > > r4 > 7> > > > 7> > > r5 > S56 5> > > > ; : > S66 r6

ð7Þ

In contracted notation, the compliance matrix has six rows and six columns, so it is now clear that it consists of 36 independent material constants (at most), as previously stated. Furthermore, through a consideration of strain energy, it can be shown [1] that the compliance matrix must

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itself be symmetrical. That is, all terms in symmetrical off-diagonal positions must be equal: S21 ¼ S12

S31 ¼ S13

S41 ¼ S14

S51 ¼ S15

S61 ¼ S16

S32 ¼ S23

S42 ¼ S24

S52 ¼ S25

S62 ¼ S26

S43 ¼ S34

S53 ¼ S35

S63 ¼ S36

S54 ¼ S45

S64 ¼ S46

ð8Þ S65 ¼ S56

Hence, although the compliance matrix for an anisotropic composite consists of 36 material constants, only 21 of these constants are independent. Substituting the original strain and stress terms (defined in Eq. (5)) into Eq. (7), we have: 9 9 2 8 38 exx > S11 S12 S13 S14 S15 S16 > rxx > > > > > > > > 6 > > 7> > > > > > > ryy > S S S S S S e 21 22 23 24 25 26 yy > > > > 7 6 > > > > > > > > 7 6 < < e = 6S S32 S33 S34 S35 S36 7 rzz = 31 zz 7 6 ð9Þ ¼6 7 r > cyz > > yz > > > > 6 S41 S42 S43 S44 S45 S46 7> > > > > > > 6 > 7> > > c > > > > > 4 S51 S52 S53 S54 S55 S56 5> > > > > rxz > > > xz > ; ; : : cxy S61 S62 S63 S64 S65 S66 rxy

The individual constants that appear in the compliance matrix can be easily related to the material properties defined in Chap. 3 by invoking the principal of superposition. That is, because we have restricted our attention to linear elastic behavior, an individual component of strain caused by several stress components acting simultaneously can be obtained by adding the strain caused by each stress component acting independently. For example, the strains exx caused by each stress component independently are given by: 1 rxx rxx causes ðfrom Eq: 3:1Þ exx ¼ Exx myx ryy causes ðfrom Eq: 3:6Þ exx ¼ ryy Eyy mzx rzz causes ðfrom Eq: 3:7Þ exx ¼ rzz Ezz gyz;xx syz syz causes ðfrom Eq: 3:14Þ exx ¼ Gyz gxz;xx sxz causes ðfrom Eq: 3:13Þ exx ¼ sxz Gxz gxy;xx sxy causes ðfrom Eq: 3:12Þ exx ¼ sxy Gxy

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To determine the strain exx induced if all stress components act simultaneously, simply add up the contribution to exx caused by each stress individually, to obtain: exx ¼

       gyz;xx myx 1 mzx rzz þ rxx þ syz ryy þ Ezz Exx Gyz Eyy     gxy;xx gxz;xx sxy þ sxz þ Gxy Gxz



ð10aÞ

Using an identical procedure, the remaining five strain components caused by an arbitrary combination of stresses are: eyy

       gxz;yy mxy mzy 1 ¼ ryy þ rxx þ rzz þ syz Eyy Exx Ezz Gyz     gxz;yy gxy;yy þ sxz þ sxy Gxz Gxy 

       gyz;zz myz mxz 1 ezz ¼ rxx þ rzz þ ryy þ syz Exx Ezz Eyy Gyz     gxy;zz gxz;zz sxy sxz þ þ Gxy Gxz 

cxy

ð10cÞ

       gxx;yz gyy;yz gzz;yz 1 syz rxx þ ryy þ rzz þ Gyz Exx Eyy Ezz     lxz;yz lxy;yz þ sxz þ sxy Gxz Gxy

ð10dÞ

       gyy;xz lyz;xz gxx;xz gzz;xz ¼ ryy þ syz rxx þ rzz þ Eyy Gyz Exx Ezz     lxy;xz 1 þ sxz þ sxy Gxz Gxy

ð10eÞ

       gxx;xy gyy;xy gzz;xy lyz;xy ¼ rxx þ ryy þ rzz þ syz Exx Eyy Ezz Gyz     lxz;xy 1 sxy sxz þ þ Gxy Gxz

ð10f Þ

cyz ¼

cxz

ð10bÞ







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Equation (10a)–(10f) can be assembled in matrix form:    38         9 2  8 9 gyz;xx gxy;xx gxz;xx myx 1 mzx > > > > > > > r e xx > > > 7> > 6 Exx > xx > > > G G E G E > > > > yz xy yy xz zz 6 > > > >           7 > > > > 7 6 > > > > > > > > g g g m m 1 7 6 xy zy yz;yy xz;yy xy;yy > > > > > > > > e r 7 6 > > > yy > yy > > > > 7 6 > > > > E E E G G G yy xx zz yz xz xy > > > > 7 6 > > > >             > > > > 7 > > 6 > > g g g m m 1 > > > > yz;zz xy;zz yz xz xz;zz > > > > 7 6 > > > > e r zz zz < < = 6 E = 7 6  xx   Eyy   Ezz   Gyz   Gxz   Gxy  7 pc ¼ 6 7 gyy;yz gzz;yz lxz;yz lxy;yz 7> > > 6 gxx;yz > 1 > > cyz > > > 6 > 7> syz > > > > > > > > > 7 6 > > > > > > > 6  Exx   Eyy   Ezz   Gyz   Gxz   Gxy  7> > > > > > > > > 7 6 > > > > g l l g g 1 > > > yy;xz yz;xz xy;xz > xx;xz zz;xz 7 6 > > > > c s > > > > 7 6 xz xz > > > 6 Exx > > > > > G E G G E 7 yy yz xy xz zz > > > > > > > > 7 6             > > > > > > > > 5 4 g g g l l 1 > > > > xx;xy yy;xy zz;xy yz;xy xz;xy > > : cxy > ; : sxy > ; Gxy Exx Eyy Ezz Gyz Gxz ð11Þ

By comparing Eqs. (9) and (11), it can be seen that the individual components of the compliance matrix are directly related to the material properties measured during uniaxial tests or pure shear tests: S11 ¼

1 Exx

S22 ¼

1 Eyy

S33 ¼

S44 ¼

1 Gyx

S55 ¼

1 Gxz

S66 ¼

mxy Exx myz S32 ¼ S23 ¼ Eyy gyy;yz S42 ¼ S24 ¼ Eyy gxx;xz S51 ¼ S15 ¼ Exx gzz;xz S53 ¼ S35 ¼ Ezz gxx;xy S61 ¼ S16 ¼ Exx gzz;xy S63 ¼ S36 ¼ Ezz lxz;xy S65 ¼ S56 ¼ Gxz S21 ¼ S12 ¼

¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼

myx Eyy mzy Ezz gyz;yy Gyz gxz;xx Gxz gxz;zz Gxz gxy;xx Gxy gxy;zz Gxy lxy;xz Gxy

mxz Exx gxx;yz S41 ¼ S14 ¼ Exx gzz;yz S43 ¼ S34 ¼ Ezz gyy;xz S52 ¼ S25 ¼ Eyy lyz;xz S54 ¼ S45 ¼ Eyz gyy;xy S62 ¼ S26 ¼ Eyy lyz;xy S64 ¼ S46 ¼ Eyz S31 ¼ S13 ¼

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¼ ¼ ¼ ¼ ¼ ¼ ¼

1 Ezz

1 Gxy mzx Ezz gyz;xx Gyz gyz;zz Gyz gxz;yy Gxz lxz;yz Gxz gxy;yy Gxy lxy;yz Gxy

ð12Þ

Because the compliance matrix must be symmetric, Eq. (12) shows that many of the properties of anisotropic materials are related through the following inverse relationships: myx mxz ¼ Exx Eyy myz mzy mxz mzx ¼ ¼ Exx Ezz Eyy Ezz gxx;yz gyz;xx gyy;yz gyz;yy ¼ ¼ Exx Gyz Eyy Gyz gxx;xz gxz;xx gyy;yz gyz;yy ¼ ¼ Exx Gxz Eyy Gyz gxx;xy gxy;xx gyy;xy gxy;yy ¼ ¼ Exx Gxy Eyy Gxy lxz;xy lxy;xz ¼ Gxz Gxy

gzz;yz gyz;zz ¼ Ezz Gyz gzz;xz gxz;zz ¼ Ezz Gxz gzz;xy gxy;zz ¼ Ezz Gxy

lyz;xz lxz;yz ¼ Gyz Gxz lyz;xy lxy;yz ¼ Gyz Gxy

ð13Þ

The inverse relationships are very significant from an experimental point of view because they dramatically reduce the number of tests that must be performed in order to determine the value of the many terms that appear within the compliance matrix of an anisotropic composite. Specifically, if the compliance matrix were not symmetrical, and hence if the inverse relationships did not exist, then 36 tests would be required to measure all components of the compliance matrix. The fact that the compliance matrix must be symmetrical reduces the number of tests required to 21. Of course, this is still a large number of tests. Fortunately, because the principal material coordinate system of composites is readily apparent, the elastic properties of composites are usually measured relative to the principal material coordinate system rather than an arbitrary (nonprincipal) coordinate system. As discussed in Sec. 2, this further reduces the number of tests required. The 21 terms within the compliance matrix of an anisotropic composite can then be calculated based on properties measured relative to the principal material coordinate system. Thus far, we have discussed Hooke’s law in the form of ‘‘strain–stress’’ relationships. That is, given values of the components of stress, we can calculate the resulting strains using Eq. (9) or Eq. (11), for example. In practice, we are often interested in the opposite problem. That is, a common circumstance in practice is that the components of strain induced in a structure have been measured, and we wish to calculate the stresses that caused these strains. In this case, we need a ‘‘stress–strain’’ form of Hooke’s law. The stress–strain form of Hooke’s law can be obtained by simply

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inverting previous results. For example, inverting Hooke’s law given by Eq. (6), we have: ri ¼ Cij ej

where i; j ¼ 1 6

ð14Þ

Cij is called the ‘‘stiffness matrix.’’* The stiffness matrix is the mathematical inverse of the compliance matrix Cij=Sij 1. In expanded form, Eq. (14) is written as: 8 9 2 38 9 C11 C12 C13 C14 C15 C16 > e1 > r1 > > > > > > > > 7> > >r > > 6C > > > > C C C C C e2 > 21 22 23 24 25 26 > > > > 7 6 2 > > > > > > > > 7 6 =
r4 > 4> > > > 6 C41 C42 C43 C44 C45 C46 7> > > > > > 6 > 7> > > > > > > 4 C51 C52 C53 C54 C55 C56 5> > r > > > > e5 > > > 5> > : ; : ; r6 e6 C61 C62 C63 C64 C65 C66 The stiffness matrix is symmetrical (C21=C12, C31=C13, etc.). The units of each stiffness term are the same as stress (either Pa or psi). 2 STRAINS INDUCED BY STRESS: ORTHOTROPIC AND TRANSVERSELY ISOTROPIC MATERIALS As discussed in Sec. 2 of Chap. 3, many of the unusual couplings between stress and strain exhibited by composites referenced to an arbitrary coordinate system do not occur if the stress and strain tensors are referenced to the principal material coordinate system. In this text, a material referenced to an arbitrary (nonprincipal) coordinate system is called ‘‘anisotropic’’ whereas if the same material is referenced to the principal material coordinate system, it is called either an ‘‘orthotropic’’ or ‘‘transversely isotropic’’ material. All of the following coupling terms are zero for orthotropic or transversely isotropic materials: 

Coefficients of mutual influence of the second kind: g11;12 ¼ g11;13 ¼ g11;23 ¼ g22;12 ¼ g22;13 ¼ g22;23 ¼ g33;12 ¼ g33;13 ¼ g33;23 ¼ 0

* The variable names assigned to the compliance and stiffness matrices in this chapter have evolved over many years and are widely used within the structural mechanics community. The reader should note that, unfortunately, the symbol ‘‘S’’ is customarily used to refer to the compliance matrix, whereas the symbol ‘‘C’’ is customarily used to refer to the stiffness matrix.

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Coefficients of mutual influence of the first kind: g12;11 ¼ g12;22 ¼ g12;23 ¼ g13;11 ¼ g13;22 ¼ g13;33 ¼ g23;11 ¼ g23;22 ¼ g23;33 ¼ 0



Chentsov coefficients: l12;13 ¼ l12;23 ¼ l13;12 ¼ l13;23 ¼ l23;12 ¼ l23;13 ¼ 0

Because these coupling terms do not exist, Hooke’s law for orthotropic or transversely isotropic materials is simplified considerably relative to that of an anisotropic material. For an orthotropic material, Hooke’s law becomes (compare with Eq. (11)): 9 8 > > > e11 > > > > > > > > > > > > > > > > > > > > > > > > e22 > > > > > > > > > > > > > > > > > > > > > > > e 33 > > =
> > 6 > > > 6 > c23 > 0 0 > > 6 > > > > 6 > > > > 6 > > > > > 6 > > > 6 > > > > 0 0 c > > 13 > 6 > 6 > > > > 6 > > > > > > > 6 > > > > 4 > > 0 0 ; : c12 > 1 E11





m31 E33





 m32 E33   1 E33 

0

0

0

0

0

0

0

0

0

0

0

0

1 G23

0

0

0

0

 

1 G13



0

0 

1 G12

9 38 > > > r11 > > > > 7> > > > 7> > > > > 7> > > > 7> > > > 7> > > r22 > 7> > > 7> > > > 7> > > > > 7> > > > > 7> > > > 7> r 33 > > = 7< 7 7 > 7> > > 7> > > s23 > > 7> > > > > 7> > > > 7> > > 7> > > > 7> > > > s13 > 7> > > 7> > > > 7> > > > > 7 > > > 5> > > > > > ; : s12 >

ð16Þ

Alternatively, Eq. (16) may be written as: S11

S12

S13

0

0

6 6 S12 6 6 6S 6 13 ¼6 6 > > > 6 0 c23 > > > > > 6 > > > > 6 > > 6 0 > c13 > > > > > 4 > > > > ; : c12 0

S22

S23

0

0

S23

S33

0

0

0

0

S44

0

0

0

0

S55

0

0

0

0

9 8 e11 > > > > > > > > > > > > e > > 22 > > > > > > > > > = < e33 >

2

0

38 9 r11 > > > > > 7> > > > > > > 0 7 r > > 22 7> > > > > 7> > > > > < 0 7 7 r33 = 7 7 > > s23 > 0 7> > > > 7> > > > 7> > > > s13 > 0 7 > > > 5> > > > > ; : s12 S66

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ð17Þ

The fact that the compliance matrix must be symmetrical (S21=S12, etc.) has been included in Eq. (17). Each compliance term in Eq. (17) is related to the more familiar engineering properties as follows: 1 E11 1 ¼ G23

1 E22 1 ¼ G13

1 E33 1 ¼ G12

S11 ¼

S22 ¼

S33 ¼

S44

S55

S66

m12 m21 ¼ E11 E22 m13 m31 ¼ ¼ E11 E33 m23 m32 ¼ ¼ E22 E33

ð18Þ

S21 ¼ S12 ¼ S31 ¼ S13 S32 ¼ S23

It can be seen that only nine independent material constants exist for an orthotropic material. The set of nine independent constants can be viewed as: ðS11 ; S22 ; S33 ; S44 ; S55 ; S66 ; S12 ; S13 ; and S23 Þ or, equivalently, as: ðE11 ; E22 ; E33 ; m12 ; m13 ; m23 ; G12 ; G13 ; and G23 Þ Equation (17) is the strain–stress form of Hooke’s law suitable for use with orthotropic materials. To obtain a stress–strain relationship, Eq. (17) is inverted, resulting in: 9 2 8 9 38 C11 C12 C13 0 0 0 > e11 > r11 > > > > > > > > > > > > e22 > 6 C12 C22 C23 0 0 0 7 > > > r22 > > > > > 7> 6 > > > > =

6 0 s23 > c23 > 0 0 C44 0 0 7> > > > > > > > > 7 6 > 4 >s > > > > c > 0 0 0 0 C55 0 5> > > > 13 > > > > : ; : 13 > ; s12 c 0 0 0 0 0 C66 12

Individual components within the stiffness matrix for an orthotropic material are related to the compliance terms as follows: S22 S33 S223 S S11 S33 S213 ¼ S 1 ¼ S44

C11 ¼

C12 ¼

C22

C23 ¼

C44

C55 ¼

S13 S23

S12 S33 S

S12 S13

S11 S23 S

1 S55

C13 ¼

S12 S33

S13 S22 S

S11 S22 S212 S 1 ¼ S66

C33 ¼ C66

ð20Þ

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where: S ¼ S11 S22 S33

S11 S223

S22 S213

S33 S212 þ 2S12 S13 S23

Alternatively, the stiffness terms may be calculated using the elastic properties described in Sec. 2 of Chap. 3: m223 E33 ÞE211 X ðm12 m23 þ m13 ÞE11 E22 E33 ¼ X ðm23 E11 þ m12 m13 E22 ÞE22 E33 ¼ X

C11 ¼ C13 C23

ðE22

C44 ¼ G23

C55 ¼ G13

ðm12 E22 þ m13 m23 E33 ÞE11 E22 X 2 ðE11 m13 E33 ÞE222 ¼ X ð21Þ ðE11 m212 E22 ÞE22 E33 ¼ X

C12 ¼ C22 C33

C66 ¼ G12

where: X ¼ E11 E22

m212 E222

m213 E22 E33

m223 E11 E33

2m12 m13 m23 E22 E33

Hooke’s law for transversely isotropic materials is simplified furthermore because in this case, E22=E33, m12=m13, m21=m31, m23=m32, and G12=G13. Also, it can also be easily shown that for transversely isotropic E22 . Hence, for transversely isotropic composites, Eq. composites, G23 ¼ 2ð1þm 23 Þ (16) reduces to:      38 9 1 m21 m21 > > > > 0 0 0 r > > 11 7> 6 E11 > E22 E22 > > 6 7 > >      > > 6 7 > > > m21 1 m32 7> 6 > > > 0 0 0 7> r 6 > 22 > > > 7> 6 E11 > E22 E22 > > 7> 6 >      > > > > 7 6 m m 1 > > 12 23 > 7 6 > r33 > 0 0 0 7> = < 6 E E22 E22 11 7 6   ¼6 7 > > 6 > 7> 2ð1 þ m23 Þ > > > > > s23 > > c23 > 6 0 0 0 0 0 7 > > > > > > > 7> G22 > > 6 > > > > > > 7 6 > > > >   > > > > 7> 6 > > > 1 > > 6 > > 7 > > > s c 13 > 0 7> 0 0 0 0 > > 13 > 6 > > > > > > > > G12 7 6 > > > > > > > > 7 6   > > > > > > > > 5 4 > > > >c > 1 > > ; : s12 > : 12 ; 0 0 0 0 0 G12 9 8 > > > > e > > 11 > > > > > > > > > > > > > > > > e > 22 > > > > > > > > > > > > > > > > > > = < e33 >

2 

which may also be written as: 9 2 8 S11 S12 S12 e11 > > > > > > > 6 > e22 > > > 6 S12 S22 S23 > > = 6S

6 12 S23 S22 33 ¼6 > 6 0 c23 > 0 0 > > > > > 6 > > > 4 c > > 0 0 0 > ; : 13 > c12 0 0 0

0 0 0 2ðS22 0 0

S23 Þ

0 0 0 0 S66 0

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9 38 0 > r11 > > > > > > r22 > 0 7 > > > 7> > > < 7 0 7 r33 = 7 s23 > 0 7> > > 7> > >s > > > > 0 5> 13 > > ; : s 12 S66

ð22Þ

ð23Þ

It can be seen that only five independent material constants exist for a transversely isotropic material. The set of five independent constants can be viewed as: ðS11 ; S22 ; S66 ; S12 ; and S23 Þ or, equivalently, as: ðE11 ; E22 ; m12 ; m23 ; and G12 Þ Equation (23) is the strain–stress form of Hooke’s law suitable for use with transversely isotropic composites. To obtain a stress–strain relationship, Eq. (23) is inverted, resulting in: 8 9 2 9 38 C11 C12 C12 0 0 0 > e11 > r11 > > > > > > > > > 7> 6C > > > > > > > > C C 0 0 0 e 7 > > 6 > r 12 22 23 22 > 22 > > > > > > > > 7 6 > > > 6 > < < = = 7 C C 0 0 0 C e r33 12 23 22 7 33 6 ¼6 7 > > c23 > 6 0 s23 > 0 7 0 0 2ðC22 C23 Þ 0 > > > > > > > 7> 6 > > > > > > > > 7 6 > > > > s c > > > > 0 0 0 0 C 0 5 4 13 66 13 > > > > > > > > ; ; : : s12 c12 0 0 0 0 0 C66

ð24Þ

Individual components within the stiffness matrix for a transversely isotropic composite are related to the compliance terms, as follows: S22 þ S33 X 2 S S11 S23 ¼ 12 XðS22 S23 Þ

C11 ¼ C23

S12 X 1 ¼ S66

C12 ¼ C66

C22 ¼

S11 S22 S212 XðS22 S23 Þ

ð25Þ

where: V ¼ S11 ðS22 þ S23 Þ

2S212

Alternatively, the stiffness terms may be calculated using the elastic properties described in Sec. 2 of Chap. 3: E211 ð1 m23 Þ m12 E11 E22 E22 ðE11 m212 E22 Þ C22 ¼ C12 ¼ X X Xð1 þ m23 Þ 2 E22 ðm23 E11 þ m12 E22 Þ C22 C23 E22 ¼ C66 ¼ C12 C44 ¼ ¼ 2 2ð1 þ m23 Þ Xð1 þ m23 Þ

C11 ¼ C23

ð26Þ

where: X ¼ E11 ð1

m23 Þ

2m212 E22

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Example Problem 1 The properties of a composite material are known to be: E11 ¼ 170 GPa m12 ¼ 0:30 G12 ¼ 13 GPa

E22 ¼ 10 GPa m13 ¼ 0:35 G13 ¼ 10 GPa

E33 ¼ 8 GPa m23 ¼ 0:40 G23 ¼ 8 GPa

Note that nine distinct material properties have been specified, indicating that this composite material is orthotropic. Determine the strains caused by the following state of stress: 9 8 9 8 r11 > > 350 MPa > > > > > > > > > > > > > > > > 35 MPa > r22 > > > > > > > > > > > > = < r = < 15 MPa > 33 ¼ > > s23 > > > 30 MPa > > > > > > > > > > > > > > > > > > > 10 MPa s 13 > > > > > > > > ; ; : : 25 MPa s12

Solution. Because the composite is orthotropic, strains are calculated using Eq. (17), where each term within the compliance matrix is calculated using Eq. (18): S11 ¼

1 1 5:88 ¼ ¼ E11 170 GPa 1012 Pa

S22 ¼

1 1 125:0 ¼ 12 ¼ E33 8 GPa 10 Pa 1 1 100 ¼ ¼ ¼ 12 G13 10 GPa 10 Pa

1 1 125:0 ¼ 12 ¼ G23 8 GPa 10 Pa 1 1 76:9 ¼ ¼ ¼ 12 G12 13 GPa 10 Pa

S33 ¼

S44 ¼

S55

S66

m12 0:030 1:76 ¼ ¼ 170 GPa 1012 Pa E11 m23 0:40 40:0 ¼ 12 ¼ ¼ 10 GPa 10 Pa E22

S21 ¼ S12 ¼ S32 ¼ S23

1 1 100:0 ¼ ¼ E22 10 GPa 1012 GPa

S31 ¼ S13 ¼

m13 0:35 2:06 ¼ ¼ 170 GPa 1012 Pa E11

In this case, Eq. (17) becomes: 9 2 8 5:88 e11 > > > > > > > > 6 1:76 e > > 22 > > = 6 < 6 2:06 e33 ¼6 6 > > 6 0 > c23 > > > > > c > 4 0 > > ; : 13 > 0 c12

1:76 100:0 40:0 0 0 0

2:06 40:0 125:0 0 0 0

0 0 0 0 0 0 125:0 0 0 100:0 0 0

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3 0 0 7 7  1 0 7 7 12 0 7 7 10 Pa 5 0 76:9

8 8 9 350 > > > > > > > > > 35 > > > > > > > > > < = < 15 6  ð10 PaÞ ¼ 30 > > > > > > > > > > > > 10 > > > > > > : ; : 25

9 1966 Am=m > > > 2284 Am=m > > > = 246 Am=m 3750 Arad > > > 1000 Arad > > > ; 1923 Arad

Example Problem 2 The properties of a composite material are known to be: E11 ¼ 25 Msi m12 ¼ m13 ¼ 0:30 G12 ¼ G13 ¼ 2:0 Msi

E22 ¼ E33 ¼ 1:5 Msi m23 ¼ 0:40

Note that only five distinct material properties have been specified, indicating that this composite material is transversely isotropic. Determine the strains caused by the following state of stress: 9 8 9 8 50 ksi > r11 > > > > > > > > > > > > > r22 > > > > 5 ksi > > > > > = = > < < 2 ksi r33 ¼ > > > > > 4 ksi > > s23 > > > > > > > s13 > > 1:5 ksi > > > > > > > : : ; > ; 3:5 ksi s12

Solution. Because the composite is transversely isotropic, strains are calculated using Eq. (22). Individual terms within the compliance matrix are: S11 ¼

1 1 40:0 ¼ ¼ E11 25 Msi 109 psi

S44 ¼

1 2ð1 þ m23 Þ 2ð1 þ 0:40Þ 1866 ¼ 9 ¼ ¼ G23 E22 1:5 Msi 10 psi

S55 ¼ S66 ¼

1 1 667 ¼ ¼ E22 1:5 Msi 109 psi

1 1 500 ¼ ¼ G12 2:0 Msi 109 psi

S12 ¼ S21 ¼ S13 ¼ S31 ¼ S23 ¼ S32 ¼

S22 ¼ S33 ¼

m12 0:30 12:0 ¼ 9 ¼ 25 Msi 10 psi E11

m23 0:40 266 ¼ ¼ 1:5 Msi 109 psi E22

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Equation (17) becomes: 9 2 8 3 40:0 12:0 12:0 0 0 0 e11 > > > > > 6 > > 12:0 667 266 0 0 0 7 e22 > > > > > 7  = 6 < 7 6 12:0 1 266 667 0 0 0 e33 7 6 ¼6 9 0 0 800 0 0 7 c23 > > > > 7 10 psi 6 0 > > > > 5 4 0 0 0 0 500 0 c > > > : 13 > ; 0 0 0 0 0 500 c12 8 8 9 9 50 > 1966 Ain:=in: > > > > > > > > > > > > > 5 > 2203 Ain:=in: > > > > > > > > > < = < = 2 596 Ain:=in: 3  ð10 psiÞ ¼ 4 > 3200 Arad > > > > > > > > > > > > > > 1:5 750 Arad > > > > > > > > > ; : : ; 3:5 1750 Arad

Example Problem 3

An orthotropic composite is subjected to a state of stress that causes the following state of strain: 9 9 8 8 1500 Am=m > e11 > > > > > > > > > > > > > 2000 Am=m > e22 > > > > > > > > > = = < < 1000 Am=m e33 ¼ c23 > > > 2500 Arad > > > > > > > 500 Arad > > > > >c > > > > > 13 > > > > ; ; : : 2000 Arad c12

Determine the stresses that caused these strains (use material properties listed in Example Problem 1). Solution. Because the composite is orthotropic, stresses are calculated using Eq. (19). The stiffness matrix can be obtained by: (a) inverting the compliance matrix determined as a part of Example Problem 1, (b) through the use of Eq. (20), or (c) through the use of Eq. (21). All three methods are entirely equivalent, and which procedure is selected for use is simply a matter of convenience. Equation (21) will be used in this example: X ¼ E11 E22

m212 E222

X ¼ fð170Þð10Þ

m213 E22 E33

ð0:30Þ2 ð10Þ2

m223 E11 E33 ð0:35Þ2 ð10Þð8Þ

2ð0:30Þð0:35Þð0:40Þð10Þð8ÞgðGPaÞ2

X ¼ 1457ðGPaÞ2

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2m12 m13 m23 E22 E33 ð0:40Þ2 ð170Þð8Þ

ð0:40Þ2 ð8 GPaÞgð170 GPaÞ2

m223 E33 ÞE211 fð10 GPaÞ ¼ X ¼ 172:98 GPa

C11 ¼

C12 ¼ ¼

ðE22

1457ðGPaÞ2

ðm12 E22 þ m13 m23 E33 ÞE11 E22 X fð0:30Þð10 GPaÞ þ ð0:35Þð0:40Þð8 GPaÞgð170 GPaÞð10 GPaÞ 1457ðGPaÞ2

¼ 4:808 GPa C13 ¼ ¼

ðm12 m23 þ m13 ÞE11 E22 E33 X fð0:30Þð0:40Þ þ ð0:35Þð170 GPaÞgð10 GPaÞð8 GPaÞ 1457ðGPaÞ2

¼ 4:387 GPa ðE11 þ m213 E33 ÞE222 fð170 GPaÞ ¼ X ¼ 11:602 GPa

C22 ¼

C23 ¼ ¼

ð0:35Þ2 ð8 GPaÞgð10 GPaÞ

1457ðGPaÞ2

ðm23 E11 þ m12 m13 E22 ÞE22 E33 X fð0:40Þð170 GPaÞ þ ð0:35Þð0:35Þð10 GPaÞgð10 GPaÞð8 GPaÞ 1457ðGPaÞ2

¼ 3:792 GPa C33 ¼ ¼

m212 E22 ÞE22 E33 X fð170 GPaÞ ð0:30Þ2 ð10 GPaÞgð10 GPaÞð8 GPaÞ ðE11

C44 ¼ G23 ¼ 8 GPa

1457ðGPaÞ2

C55 ¼ G13 ¼ 10 GPa

¼ 9:286 GPa

C66 ¼ G12 ¼ 13 GPa

Applying Eq. (19), the stresses are: 9 2 8 172:98 r11 > > > > > > > > 6 4:808 r > > 22 > > = 6 < 6 4:387 r33 ¼6 6 > > 6 0 > s23 > > > > > 4 0 > > > s13 > ; : 0 s12

4:808 11:602 3:792 0 0 0

4:387 3:792 9:286 0 0 0

0 0 0 8:0 0 0

0 0 0 0 10:0 0

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3 0 0 7 7 0 7 7ðGPaÞ 0 7 7 0 5 13:0

8 9 8 1500 Am=m > > > > > > > > > > > > 2000 Am=m > > > > > < = > < 1000 Am=m  ¼ 2500 Arad > > > > > > > > > > > > 500 Arad > > > > > ; > : : 2000 Arad

Example Problem 4

9 245:5 MPa > > > 19:78 MPa > > > = 10:29 MPa 20:0 MPa > > > 5:00 MPa > > > ; 26:0 MPa

A transversely isotropic composite is subjected to a state of stress that causes the following state of strain: 8 9 8 9 e11 > 1250 Ain:=in: > > > > > > > > > > > > > e22 > > > > 1000 Ain:=in: > > > > > < = > < = e33 500 Ain:=in: ¼ c23 > 2500 Arad > > > > > > > > > > > > > c13 > 1000 Arad > > > > > > > > > : ; : ; c12 2000 Arad

Determine the stresses that caused these strains (use material properties listed in Example Problem 2) Solution. Because the composite is transversely isotropic, stresses are calculated using Eq. (24). The stiffness matrix can be obtained by: (a) inverting the compliance matrix determined as a part of Example Problem 2, (b) through the use of Eq. (25), or (c) through the use of Eq. (26). All three methods are entirely equivalent, and which procedure is selected for use is simply a matter of convenience. Equation (26) will be used in this example: X ¼ E11 ð1

m23 Þ

X ¼ ð25 MsiÞð1 C11 ¼

2m212 E22 0:40Þ

2ð0:30Þ2 ð1:5 MsiÞ ¼ 14:73 Msi

E211 ð1 m23 Þ ð25 MsiÞ2 ð1 0:40Þ ¼ ¼ 25:46 Msi X 14:73 Msi

m12 E11 E22 ð0:30Þð25 MsiÞð1:5 MsiÞ ¼ ¼ 0:7637 Msi X 14:73 Msi

 2 E22 ðE11 m212 E22 Þ ð1:5 MsiÞ ð25 MsiÞ ð0:30Þ ð1:5 MsiÞ ¼ ¼ Xð1 þ m23 Þ 14:73 Msið1 þ 0:40Þ ¼ 1:809 Msi

C12 ¼ C22

C23 ¼

E22 ðm23 E11 þ m212 E22 Þ Xð1 þ m23 Þ

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¼

 ð1:5 MsiÞ ð0:40Þð25 MsiÞ þ ð0:30Þ2 ð1:5 MsiÞ ¼ 0:7372 Msi 14:73Msið1 þ 0:40Þ

C44 ¼

E22 ð1:5 MsiÞ ¼ ¼ 0:5357 Msi 2ð1 þ m23 Þ 2ð1 þ 0:40Þ

C66 ¼ G12 ¼ 2:0 Msi Applying Eq. (24), the stresses are: 8 9 2 r11 > 25:46 0:7637 0:7637 0 0 > > > > > > > 6 0:7637 1:809 0:07372 r 0 0 > > 22 > > < = 6 6 0:7637 0:7372 r33 1:809 0 0 ¼6 6 0 0 0:5357 0 > > s23 > > 6 0 > > > > 4 0 s 0 0 0 2:0 > > 13 > > ; : 0 0 0 0 0 s12 9 8 9 8 32:97 ksi > 1250 Ain:=in: > > > > > > > > > > > > > > 3:13 ksi > 1000 Ain:=in: > > > > > > > > = = < < 2:60 ksi 500 Ain:=in: ¼  2500 Arad > > > > > 1:34 ksi > > > > > > > > > 2:00 ksi > > 1000 Arad > > > > > > > > ; ; : : 4:00 ksi 2000 Arad

3 0 0 7 7 0 7 7ðMsiÞ 0 7 7 0 5 2:0

3 STRAINS INDUCED BY A CHANGE IN TEMPERATURE OR MOISTURE CONTENT Material properties relating strains to a uniform change in temperature and a uniform change in moisture content were defined in Secs. 3 and 4 of Chap. 3, respectively. For anisotropic materials, strains caused by a change in temperature are given by Eq. (3.22), repeated here for convenience: eTxx 6 eT 4 yx eTzx

eTxy eTyy eTzy

eM xx

cM xy

cM xz

bxx

bxy

bxz

6 cM 4 yx

eM yy

6 7 cM yz 5 ¼ ðDMÞ4 bxy

byy

byz 7 5

2

3 2 axx eTxz 6a eTyz 7 5 ¼ DT4 yx eTzz azx

axy ayy azy

3 axz ayz 7 5

ðrepeatedÞð3:22Þ

azz

Similarly, strains caused by a change in moisture content are given by Eq. (3.25), repeated here for convenience: 2

cM zx

cM zy

eM zz

3

2

bxz

byz

3

bzz

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ðrepeatedÞð3:25Þ

As before, the strain tensors must be symmetric. This allows the use of contracted notation, and hence Eqs. (3.22) and (3.25) can be written in the form of column arrays: 8 T 9 8 9 8 M9 8 9 axx > bxx > exx > exx > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > byy > eTyy > > > > > eM > > > > ayy > yy > > > > > > > > > > > > > > > > > > > > < eT > =

= < eM > =

= zz zz zz zz ¼ DT ðandÞ ¼ DT ð27Þ T M > > > > > cyx > > > ayz > > > cyx > > > byz > > > > > > > > > > > > > > > > > > > > > > > > cTxz > > > axz > > > cM > > bxz > > > > > > > > > xz > > > > > > > > > > > > > > : cT > ; : ; : cM > ; :b > ; axy xy xy xy In the case of an orthotropic material a12=a13=a23=b12=b13= b23=0, and Eq. (27) becomes: 8 9 8 M9 8 T 9 9 8 > a11 > e11 > e11 > b11 > > > > > > > > > > > > > > > > > > > T > > > > > > > > > > M> > > > > > > > > a b e e > > > > > > > > 22 22 22 22 > > > > > > > > > > > > > > > > > > > > > > > < eM =

> > > > 0 > cM cT23 > 0 > > > > > > > > 23 > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > >0 > > > > > > cT13 > > cM 0 > > > > > > > > 13 > > > > > > > > > > : M; : : T ; ; ; : > c12 c12 0 0

In addition to these simplifications, for a transversely isotropic material with symmetry in the 2–3 plane, a33=a22 and b33=b22. Hence, for a transversely isotropic material, Eq. (28) becomes: 8 9 8 T 9 a11 > > e11 > > > > > > > > > > > > > > eT > > > > > > a > > > > 22 22 > > > > > > > > > > > < eT = =

22 33 ¼ DT > > cT23 > 0 > > > > > > > > > > > > > > > > > > > > T > > > c13 > >0 > > > > > > > > : T ; > ; : > c12 0

ðandÞ

9 8 8 M9 b11 > e11 > > > > > > > > > > > > > > eM > > > > > b > > > > 22 22 > > > > > > > > > > > < eM = =

22 33 ¼ DT > > > 0 > cM > > > 23 > > > > > > > > > > > > > > > > M > > > 0 > > c13 > > > > > > > > : M; ; : 0 c12

ð29Þ

4 STRAINS INDUCED BY COMBINED EFFECTS OF STRESS, TEMPERATURE, AND MOISTURE The strains induced by stress under constant environmental conditions for anisotropic materials were discussed in Sec. 1, and a similar discussion for the case of orthotropic and transversely isotropic materials was presented in

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Sec. 2. Strains induced by a uniform change in temperature or moisture content in the absence of stress was discussed for all three material classifications in Sec. 3. We will now consider the strains induced if all of these mechanisms occur simultaneously. That is, we wish to consider the strains induced by the combined effects of stress, a uniform change in temperature, and a uniform change in moisture content. We will call this the total strain. Rigorously speaking, the total strain tensor (eij) is a nonlinear coupled function of these three mechanisms: eij ¼ fðr; T; MÞ The function f( ) is a nonlinear function of stress, temperature, and moisture content, even though we have limited our attention to linear relationships between strain and these three mechanisms. That is, we have defined: 

Young’s modulus as the slope of linear region of the stress–strain curve (Sec. 2 of Chap. 3)  The coefficient of thermal expansion (CTE) as the slope of the linear region of the strain–DT curve (Sec. 3 of Chap. 3)  The coefficient of moisture expansion (CME) as the slope of the linear region of the strain–DT curve (Sec. 4 of Chap. 3). Despite these assumptions of linearity, the strain response may still be a coupled function of stress, temperature, and moisture because a change in one variable may cause a change in the other two. For example, for all polymerbased materials, an increase in temperature will ordinarily cause a decrease in Young’s modulus. Similarly, an increase in moisture content often causes an increase in CTEs and a decrease in Young’s modulus. These coupling effects are ignored throughout this text. It is assumed that the strain response is an uncoupled function of stress, temperature, and moisture. For example, we assume that Young’s modulus is measured under some standard environmental condition (say, at room temperature and 0% moisture content), and that subsequent changes in temperature or moisture content are relatively modest such that Young’s modulus may be assumed to remain constant. Based on these assumptions, the total strain tensor induced in a structure is simply the sum of the strains induced by each of mechanism acting independently: eij ¼ erij þ eTij þ eM ij

ð30Þ

The superscripts j, T, and M used in Eq. (30) indicate that the individual components of strain are caused by the application of stress, by a uniform

Copyright © 2004 by Marcel Dekker, Inc. All Rights Reserved.

change in temperature, and by a uniform change in moisture content, respectively. Based on this assumption, for anisotropic materials, the total strain is obtained by superimposing Eqs. (9) and (27): 8 9 exx > > > > > > eyy > > > > >

=

2

S11 6 S21 6 6 S31 zz ¼6 6 S41 c > > yz > > 6 > > > > 4 S51 c > > xz > > ; : cxy S61

S12 S13 S14 S15 S22 S23 S24 S25 S32 S33 S34 S35 S42 S43 S44 S45 S52 S53 S54 S55 S62 S63 S64 S65 8 9 8 9 bxx > axx > > > > > > > > > > > ayy > > > byy > > > > > > > > = =

< azz zz þDT þ DM byz > ayz > > > > > > > > > > > > > > > b a > > > xz xz > > > > > : : ; ; bxy axy

9 38 rxx > S16 > > > > > ryy > > S26 7 > > > 7> < = r S36 7 zz 7 syz > S46 7 > > 7> > > > s S56 5> > xz > > > ; : sxy S66 ð31Þ

Equation (31) allows the prediction of the strains induced by the simultaneous effects of stress and uniform changes in temperature and/or moisture content. In practice, the inverse problem is often encountered. That is, a common circumstance is that the strains, the change in temperature, and the change in moisture content have been measured, and we wish to calculate stresses. This can be accomplished by inverting Eq. (31) according to the laws of matrix algebra, resulting in: 8 9 2 3 rxx > C11 C12 C13 C14 C15 C16 > > > >r > > 6 > C21 C22 C23 C24 C25 C26 7 > yy > > > 7 < = 6 6 rzz C31 C32 C33 C34 C35 C36 7 7 ¼6 6 C41 C42 C43 C44 C45 C46 7 syz > > > > 7 6 > > > > 4 C51 C52 C53 C54 C55 C56 5 > > sxz > > : ; sxy C61 C62 C63 C64 C65 C66 9 8 exx DTaxx DMbxx > > > > > > > > > eyy DTayy DMbyy > > > > > > > > > < ezz DTazz DMbzz =  ð32Þ > > > cyz DTayz DMbyz > > > > > > > cxz DTaxz DMbxz > > > > > > > > :c DTa DMb ; xy

xy

xy

where the stiffness matrix Cij=Sij 1, as discussed in Sec. 1.

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Following an analogous procedure, the strains induced in an orthotropic material by the combined effects of stress, a uniform change in temperature, and/or a uniform change in moisture content can be found by superimposing Eqs. (17) and (28): 8 9 2 e11 > S11 > > > > > > > 6 S12 e > > 22 > > < = 6 6 S13 e33 ¼6 6 > > c23 > > 6 0 > > > > 4 0 > > c13 > > : ; c12 0

9 38 r11 > S13 0 0 0 > > > > > r22 > > S23 0 0 0 7 > > > 7> < = 7 S33 0 0 0 7 r33 0 S44 0 0 7 s > > 7> > 23 > > 0 0 S55 0 5> s > > > > : 13 > ; 0 0 0 S66 s12 8 8 9 9 a11 > b11 > > > > > > > > > > > > > a22 > b22 > > > > > > > > > < < = = a33 b33 þDT þ DM 0 > 0 > > > > > > > > > > > > > > 0 0 > > > > > > > > > : : ; ; 0 0 S12 S22 S23 0 0 0

Inverting Eq. (33), we obtain: 8 9 2 r11 > 0 0 C11 C12 C13 > > > > 6 > > > > > r C C 0 0 C 22 > 12 22 23 > > > 6 >

= 6 6 C13 C23 C33 0 0 33 ¼6 6 > > s23 > 6 0 0 0 C44 0 > > > > 6 > > > > > 4 0 0 0 0 C s 55 > 13 > > > ; : 0 0 0 0 0 s12 8 9 e11 DTa11 DMb11 > > > > > > > > > e22 DTa22 DMb22 > > > > > = < e DTa DMb 33 33 33  > > c23 > > > > > > > > > > c 13 > > ; : c12

ð33Þ

3 0 0 7 7 7 0 7 7 0 7 7 7 0 5 C66

ð34Þ

As discussed in preceding chapters, an implicit assumption in Eqs. (33) and (34) is that the strain tensor, stress tensor, and material properties are all referenced to the principal material coordinate system of the orthotropic material (i.e., the 1–2–3 coordinate system). If an orthotropic is referenced to a nonprincipal material coordinate system, then the relation between strain, stress, temperature, and moisture content is given by Eq. (31) or Eq. (32).

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Finally, the strains induced in a transversely isotropic material by the combined effects of stress, a uniform change in temperature, and/or a uniform change in moisture content can be found by superimposing Eqs. (23) and (29): 9 2 8 9 38 S11 S12 S12 0 0 0 > r11 > e11 > > > > > > > > > 7> 6S > > > > > > > > S S 0 0 0 e 7 > > 6 > r 12 22 23 22 > 22 > > > > > 6 > > > 7> > > > < e = 6 S12 S23 S22 = < 7 0 0 0 r 33 33 7 6 ¼6 7 > > s23 > 6 0 c23 > 0 7 0 0 2ðS22 S23 Þ 0 > > > > > > > 7> 6 > > > > > > > > 7 6 > > > > s c > > > > 0 0 0 0 S 0 5 4 13 66 13 > > > > > > > > ; ; : : s12 c12 0 0 0 0 0 S66 9 8 9 8 a11 > b11 > > > > > > > > > > > > > b > a22 > > > > > > > > = = < 22 > < b22 a22 þDT þ DM 0 > 0 > > > > > > > > > > > > > > 0 > 0 > > > > > > > > ; ; : : 0 0

Inverting Eq. (35), we have: 8 9 2 r11 > 0 C11 C12 C13 > > > > > 6 > > C r C C 0 > > 22 12 22 23 > > < = 6 6 C12 C23 C22 r33 0 ¼6 6 0 s 0 0 ðC C23 Þ=2 > > 23 22 > > 6 > > > > 4 0 s 0 0 0 > > 13 > > : ; s12 0 0 0 0 9 8 e11 DTa11 DMb11 > > > > > > > e22 DTa22 DMb22 > > > > > = < e33 DTa22 DMb22  c23 > > > > > > > > c13 > > > > : ; c12

ð35Þ

0 0 0 0 C66 0

3 0 0 7 7 0 7 7 0 7 7 0 5 C66

ð36Þ

Once again, Eqs. (35) and (36) are valid only if referenced to the principal material coordinate system of the transversely isotropic material (i.e., the 1–2–3 coordinate system). If a transversely isotropic material is referenced to a nonprincipal material coordinate system, then the relation between strain, stress, temperature, and moisture content is given by Eq. (31) or Eq. (32).

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HOMEWORK PROBLEMS An anisotropic material with the following properties is considered in problems 1–4: Exx ¼ 100 GPa

Eyy ¼ 200 GPa

Ezz ¼ 75 GPa

vxy ¼ 0:20

vxz ¼

vyz ¼ 0:60

Gxy ¼ 60 GPa

Gxz ¼ 75 GPa

Gyz ¼ 50 GPa

gxx;xy ¼

gxx;xz ¼ 0:25

gxx;yz ¼ 0:30

gyy;xy ¼ 0:60

gyy;xz ¼ 0:75

gyy;yz ¼ 0:20

gzz;xy ¼

0:20

gzz;xz ¼

0:05

gzz;yz ¼

lxy;xz ¼

0:10

lxy;yz ¼

0:05

gxz;yz ¼ 0:10

0:30

axx ¼ 5 Am=m axy ¼

jC

5 Arad=jC

0:25

ayy ¼ 10 Am=m

jC

axz ¼ 15 Arad=jC

azz ¼ 20 Am=m

jC

ayz ¼ 25 Arad=jC

bxx ¼ 300 Am=m%M byy ¼ 60 Am=m%M bxy ¼ 150 Arad=%M

0:15

bzz ¼ 1200 Am=m%M

bxz ¼ 1000 Arad=%M byz ¼ 350 Arad=%M

1. Calculate the compliance matrix, Sij. 2. Calculate the stiffness matrix, Cij=Sij 1. (Perform this calculation using a suitable software package such as Maple, Matlab, Mathematica, etc.) 3. Consider the following stress tensor: 2

rxx rij ¼ 4 ryx rzx

rxy ryy rzy

3 2 rxz 75 ryz 5 ¼ 4 10 rzz 25

10 90 30

3 25 30 5 ðMPaÞ 25

(a) Calculate the strains induced by this stress tensor, assuming no change in temperature or moisture content (i.e., assume DT=DM=0). (b) Calculate the strains induced by this stress tensor and a temperature increase of 100jC (assume DM=0). (c) Calculate the strains induced by this stress tensor and a 2% increase in moisture content (assume DT=0). (d) Calculate the strains induced by this stress tensor, a temperature increase of 100jC, and a 2% increase in moisture content.

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4. Consider the following strains: exx ¼ 1500 Am=m cxy ¼ 750 Arad

2000 Am=m 500 Arad

eyy ¼ cxz ¼

ezz ¼ 1750 Am=m cyz ¼ 850 Arad

(a)

Calculate the stress tensor that caused these strains, assuming no change in temperature or moisture content (i.e., assuming DT=DM=0). (b) Calculate the stress tensor that caused these strains, if these strains were caused by the simultaneous effects of stress and a temperature decrease of 100jC (assume DM=0). (c) Calculate the stress tensor that caused these strains, if these strains were caused by the simultaneous effects of stress, a temperature decrease of 100jC, and a 2% increase in moisture content.

An orthotropic material with the following properties is considered in problems 5–8: E11 ¼ 100 GPa m12 ¼ 0:20 G12 ¼ 60 GPa a11 ¼ 1 Am=mjC b11 ¼ 100 Am=m%M

E22 ¼ 200 GPa m13 ¼ 0:25 G13 ¼ 75 GPa a22 ¼ 25 Am=mjC b22 ¼ 600 Am=m%M

E33 ¼ 75 GPa m23 ¼ 0:60 G23 ¼ 50 GPa a33 ¼ 15 Am=mjC b33 ¼ 1000 Am=m%M

5. Calculate the compliance matrix, Sij. 6. Calculate the stiffness matrix, Cij. 7. Consider the following stress tensor: r11

r12

6 rij ¼ 4 r21

r22

2

r31

r32

r13

3

2

75 7 r23 5 ¼ 4 10 25 r33

10 90 30

3 25 30 5 ðMPaÞ 25

(a)

Calculate the strains induced by this stress tensor, assuming no change in temperature or moisture content (i.e., assume DT= DM=0). (b) Calculate the strains induced by this stress tensor and a temperature increase of 100jC (assume DM=0). (c) Calculate the strains induced by this stress tensor and a 2% increase in moisture content (assume DT=0). (d) Calculate the strains induced by this stress tensor, a temperature increase of 100jC, and a 2% increase in moisture content.

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8. Consider the following strains: e11 ¼ 2000 Am=m c12 ¼ 750 Arad

e22 ¼ 3000 Am=m c13 ¼ 1000 Arad

e33 ¼ 1500 Am=m c23 ¼ 1250 Arad

(a)

Calculate the stress tensor that caused these strains, assuming no change in temperature or moisture content (i.e., assuming DT=DM=0). (b) Calculate the stress tensor that caused these strains, if these strains were caused by the simultaneous effects of stress and a temperature decrease of 100jC (assume DM=0). (c) Calculate the stress tensor that caused these strains, if these strains were caused by the simultaneous effects of stress, a temperature decrease of 100jC, and a 2% increase in moisture content.

REFERENCE 1. Jones, R.M. Mechanics of Composite Materials; Hemisphere Publ. Co.: New York, NY, 1975; ISBN 0-89116-490-1.

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5 Unidirectional Composite Laminates Subject to Plane Stress

This chapter is devoted to the elastic behavior and failure response of unidirectional composite laminates. In the present context, the term ‘‘unidirectional’’ is meant to imply that although the laminates considered may contain many plies, the principal material coordinate system in all plies is oriented in the same direction. It is also assumed that the laminate is thin and platelike, such that a state of plane stress exists. Four primary topics are addressed in this chapter. First, in Sec. 1, the elastic response of a unidirectional composite referenced to the principal material coordinate system and subject to a plane stress state will be described. This will lead to a so-called ‘‘reduced’’ form of Hooke’s law. Second, in Secs. 2 and 3, the elastic response of a unidirectional composite referenced to an arbitrary (nonprincipal) coordinate system is discussed, which will lead to the definition of the ‘‘transformed, reduced’’ form of Hooke’s law. The ‘‘effective’’ elastic properties of a unidirectional composite laminate are then discussed in Sec. 4. We then turn our attention to the prediction of composite failure. In Sec. 5, several macromechanics-based failure theories are defined and used to predict the failure of a unidirectional laminate referenced to the principal material coordinate system. These theories are then used to predict the failure of a unidirectional composite laminate referenced to an arbitrary (nonprincipal) coordinate system in Sec. 6. 193

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1 UNIDIRECTIONAL COMPOSITES REFERENCED TO THE PRINCIPAL MATERIAL COORDINATE SYSTEM The strains induced in an orthotropic material subjected to a general 3-D stress tensor, a uniform change in temperature, and/or a uniform change in moisture content were described in Chap. 4. The strain response is summarized by Eq. (33) of Chap. 4, repeated here for convenience: 9 9 2 9 8 8 8 9 38 b11 > a11 > 0 0 > S11 S12 S13 0 r11 > e11 > > > > > > > > > > > > > > > > > 7> > > 6 > > > >b > >a > > > > > > > > > > S S 0 0 0 S r > 7 > > > > > 6 > > e 22 12 22 23 22 22 22 > > > > > > > > > > > > > > > 7> 6 > > > > > > > >

> > > 6 0 c23 > s23 > 0 0 S44 0 0 7 0 > 0 > > > > > > > > > > > > > > > > > 7 6 > > > > 6 > > > > > > > > > > > > 7 > > > > > > > c s13 > 0 0 0 S55 0 5> > > > > 0 > 4 0 0 > 13 > > > > > > > > > > > > > > > > > : ; : : ; ; : ; c12 s 0 0 0 0 0 S66 0 0 12

Three mechanisms that contribute to the total strains appear in the above equation: strains caused by stress, strains caused by a uniform change in temperature (DT ), and strains caused by a uniform change in moisture content (DM ). Equation (33) of Chap. 4 is valid for any orthotropic material, as long as the strain tensor, stress tensor, and material properties are all referenced to the principal 1–2–3 coordinate system. Let us now consider the strains induced in an orthotropic material by a state of plane stress. Assuming that r33=s23=s13=0, Eq. (33) of Chap. 4 becomes: 8 8 9 9 8 9 9 2 38 b11 > a11 > S11 S12 S13 0 0 0 > r11 > e11 > > > > > > > > > > > > > > > > > > > 6 > 7> > > > > > > > > > > > > > > > b a S S S 0 0 0 r 7 > 6 > > > > > > > e 22 22 22 12 22 23 22 > > > > > > > > > > > > > > > > 7 6 > > > > > > > > = < 6 0 > 0 > > > > > > 0 7 0 0 S44 0 > > > > > > 6 > > > c23 > > 0 > > 0 > > 7> > > > > > > > > > > > > > > > > 7 6 > > > 0 0 0 S55 0 5> > > 4 0 > 0 > > > 0 > > c13 > > 0 > > > > > > > > > > > > > > > > > > ; ; : : : ; : ; c12 s12 0 0 0 0 0 S66 0 0 ð1Þ

Note that Eq. (1) shows that in the case of plane stress, the out-of-plane shear strains are always equal to zero (c23=c13=0). It is customary to write the expressions for the remaining four strain components as follows: 9 9 2 9 9 8 8 8 38 b S11 S12 0 > > > > = = = = < a11 > < e11 > < 11 > < r11 > 6 7 ð2aÞ e22 ¼ 4 S12 S22 0 5 r22 þ DT a22 þ DM b22 > > > > > > > > ; ; ; ; : : : : c12 s12 0 0 S66 0 0

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and e33 ¼ S13 r11 þ S23 r22 þ DTa33 þ DMb33

ð2bÞ

Equations (2a) and (2b) are called the reduced forms of Hooke’s law for an orthotropic composite. They are only valid for a state of plane stress and are called ‘‘reduced’’ laws because we have reduced the allowable stress tensor from three dimensions to two dimensions. The 33 array in Eq. (2a) is called the reduced compliance matrix. Note that despite the reduction from three to two dimensions, we have retained the subscripts used in the original compliance matrix. For example, the element that appears in the (3,3) position of the reduced compliance matrix is labeled S66. The definition of each compliance term is not altered by the reduction from three to two dimensions, and each term is still related to the more familiar engineering constants (E11, E22, v12, etc.) in accordance with Eq. (18) of Chap. 4. Inverting Eq. (2a), we obtain: 9 9 2 8 38 Q11 Q12 0 > > = = < e11 DTa11 DMb11 > < r11 > 6 0 7 r22 ¼ 4 Q12 Q22 ð3Þ 5 e22 DTa22 DMb22 > > > > ; ; : : s12 c12 0 0 Q66

The 33 array that appears in Eq. (3) is called the reduced stiffness matrix and equals the inverse of the reduced compliance matrix: 2

Q11 6Q 4 12 0

Q12 Q22 0

3 2 S11 0 7 6 0 5u4 S12 Q66 0

3

S12

0

S22

0 7 5

0

1

S66

ð4Þ

Note that we are now using a different symbol to denote stiffness. That is, the original 3-D stiffness matrix was denoted Cij (as in Eq. (19) of Chap. 4, for example), whereas the reduced stiffness matrix is denoted Qij. This change in notation is required because individual members of the reduced stiffness matrix are not equal to the corresponding members in the original stiffness matrix. That is, Q11 p C11, Q12 p C12, Q22 p C22, and Q66 p C66. Relations between Qij and Cij can be derived as follows. From Eq. (19) of Chap. 4, it can be seen that for an orthotropic material subjected to an arbitrary state of stress: r11 ¼ C11 e11 þ C12 e22 þ C13 e33

ð5aÞ

r33 ¼ C13 e11 þ C23 e22 þ C33 e33

ð5bÞ

and

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However, the reduced stiffness matrix relates stress to strain under conditions of plane stress by definition. Setting r33=0 and solving Eq. (5b) for e33, we have: e33 ¼

ðC13 e11 þ C23 e22 Þ C33

ð6Þ

The out-of-plane strain e33 must be related to in-plane strains e11 and e22 in accordance with Eq. (6), otherwise, a state of plane stress does not exist in the composite. Substituting Eq. (6) into Eq. (5a) and simplifying, we have:



C11 C33 C213 C12 C33 C13 C23 ð7Þ r11 ¼ e22 e11 þ C33 C33 On the other hand, from Eq. (3), r11 is given by (with DT=DM=0): r11 ¼ Q11 e11 þ Q12 e22

ð8Þ

Comparing Eqs. (7) and (8), it is immediately apparent that: C11 C33 C213 C33 C12 C33 C13 C23 ¼ C33

Q11 ¼

ð9aÞ

Q12

ð9bÞ

Using a similar procedure, it can be shown that: C22 C33 C223 C33 ¼ C66

Q22 ¼

ð9cÞ

Q66

ð9dÞ

In essence, the definition of elements of the Qij matrix differs from those of the Cij matrix because the Qij matrix is defined for plane stress conditions only, whereas Cij can be used for any stress state. Elements of the reduced stiffness matrix may be related to the elements of the compliance matrix by either substituting Eq. (20) of Chap. 4 in Eqs. (9a)–(9d), or by simply performing the matrix inversion indicated in Eq. (4). In either case, it will be found that: Q11 ¼ Q22 ¼

S22 S11 S22 S212 S11 S11 S22

S212

Q12 ¼ Q21 ¼ Q66 ¼

S12 S11 S22 S212

1 S66

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ð10Þ

Alternatively, the elements of the reduced stiffness matrix are related to more familiar engineering constants as follows: E211 m212 E22

Q11 ¼

E11

Q22 ¼

E11 E22 E11 m212 E22

Q12 ¼ Q21 ¼

m12 E11 E22 E11 m212 E22

Q66 ¼ G12

ð11Þ

Equations (1)–(11) were developed assuming that the composite is an orthotropic material. Now consider the response of a transversely isotropic composite subjected to a state of plane stress. As before, we assume that r33=s23=s13=0. From Eq. (34) of Chap. 4, it is seen that the out-of-plane shear strains equal zero (c23=c13=0), and the remaining four strains can be written as: 9 2 9 9 9 8 8 8 38 S11 S12 0 > > > > = = = = < b11 > < a11 > < e11 > < r11 > 6 7 e22 ¼ 4 S12 S22 0 5 r22 þ DT a22 þ DM b22 > > > > > > > > ; ; ; ; : : : : 0 0 S66 s12 c12 0 0

ð12aÞ

and e33 ¼ S12 r11 þ S23 r22 þ DTa22 þ DMb22

ð12bÞ

Comparing Eq. (12a) with Eq. (2a), it is seen that the relationship between in-plane strains (e11, e22, and c12) and in-plane stress components (r11, r22, and s12) is identical for orthotropic and transversely isotropic materials. In fact, identical results are obtained for the out-of-plane normal strain as well since for a transversely isotropic material, S13=S12, a33=a22, and b33=b22, and therefore Eq. (2b) is equivalent to Eq. (12b). Consequently, Eq. (3) can also be applied to a transversely isotropic material. Equations (9a)–(9d) are also still applicable except that for a transversely isotropic material (with symmetry in the 2–3 plane), C33=C22 and C13=C12. Inverting Eq. (12a), we obtain: 9 2 9 8 38 Q11 Q12 0 > > < r11 > = = < e11 DTa11 DMb11 > 6 0 7 r22 ¼ 4 Q12 Q22 5 e22 DTa22 DMb22 > > > > ; ; : : s12 c12 0 0 Q66 Since this result is identical to Eq. (3), it is seen once again that the relationship between in-plane strains (e11, e22, and c12) and in-plane stress components (r11, r22, and s12) is identical for orthotropic and transversely isotropic materials.

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Example Problem 1 Determine the strains induced in a unidirectional graphite–epoxy composite subjected to the in-plane stresses shown in Fig. 1. Assume that DT=DM=0 and use material properties listed in Table 3 of Chap. 3. Solution. The magnitude of each stress component is indicated in Fig. 1. Based on the sign conventions reviewed in Sec. 5 of Chap. 2, the algebraic sign of each stress component is: r11 ¼ 200 MPa

r22 ¼

30 MPa

s12 ¼

50 MPa

Strains can be calculated using either Eq. (2a) or Eq. (12a). Since DT= DM=0, we have: 9 2 9 8 38 S11 S12 0 > > = = < e11 > < r11 > 6 7 e22 ¼ 4 S12 S22 0 5 r22 > > > > ; ; : : 0 0 S66 s12 c12

Each term within the reduced compliance matrix is calculated using Eq. (18) of Chap. 4: 3 2 1 m12 0 9 9 6 E 8 78 E11 11 r11 > > 7> = < = 6 < e11 > 7 6 m 1 12 7 r22 e22 ¼ 6 0 7 6 > > > E22 ; : ; 6 E11 : 7> s c12 5 4 12 1 0 0 G12

Figure 1 Unidirectional composite subjected to in-plane stresses (magnitudes of in-plane stresses shown).

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From Table 3 of Chap. 3: E11 ¼ 170 GPa m12 ¼ 0:30

E22 ¼ 10 GPa G12 ¼ 13 GPa

Using these values: 9 2 8 1:76  10 12 5:88  10 12 > < e11 > = 6 e22 ¼ 4 1:76  10 12 100  10 12 > > ; : c12 0 0 9 8 6 > = < 200  10 >  30  106 ðPaÞ > > ; : 50  106

0 0 76:9  10

12

3 7 5



1 Pa



Completing the matrix multiplication indicated, we obtain: 9 9 8 8 > = < 1230 Am=m > = > < e11 > e22 ¼ 3350 Am=m > > > ; : ; > : c12 3850 Arad Example Problem 2

Determine the strain induced in a unidirectional graphite–epoxy composite subjected to (a) The in-plane stresses shown in Fig. 1. (b) A decrease in temperature DT= 155jC. (c) An increase in moisture content DM=0.5%. Use material properties listed in Table 3 of Chap. 3. Solution. This problem involves three different mechanisms that contribute to the total strain induced in the laminate: the applied stresses, the temperature change, and the change in moisture content. The total strains induced by all three mechanisms can be calculated using either Eq. (2a) or Eq. (12a): 9 9 9 2 9 8 8 8 38 S11 S12 0 > > > > = = < r11 > = < e11 > = < b11 > < a11 > 7 6 e22 ¼ 4 S12 S22 0 5 r22 þ DT a22 þ DM b22 > > > > > > > > ; ; : ; ; : : : 0 0 S66 s12 c12 0 0

Numerical values for the stresses and reduced compliance matrix are given in Example 1, and the linear thermal and moisture expansion coefficients for graphite–epoxy are (from Table 3 of Chap. 3): a11 ¼

0:9 Am=m=o C

a22 ¼ 27 Am=m=o C

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b11 ¼ 150 Am=m=%M

b22 ¼ 4800 Am=m=%M

Hence, Eq. 2(a) or Eq. (12a) becomes: 9 2 8 3 5:88  10 12 1:76  10 12 0 > = < e11 > 7 6 e22 ¼ 4 1:76  10 12 100  10 12 0 5 > > ; : 12 c12 0 0 76:9  10 8 8 9 9 6 6 > > < 0:9  10 > < 200  10 > = =  þ ð0:5Þ 30  106 þ ð 155Þ 27  10 6 > > > > : : ; 6; 0 50  10 9 8 6 > = < 150  10 >  4800  10 6 > > ; : 0

Completing the matrix multiplication indicated, we obtain: 9 9 8 9 8 9 8 8 > = < 75 Am=m > = > < 140 Am=m > = > < 1230 Am=m > = > < e11 > e22 ¼ 3350 Am=m þ 4185 Am=m þ 2400 Am=m > > > > > : ; ; > : ; > : ; > : 0 0 3850 Arad c12 8 9 > < 1445 Am=m > = ¼ 5135 Am=m > > : ; 3850 Arad

An implicit assumption in this problem is that the composite is free to expand or contract, as dictated by changes in temperature and/or moisture content. Consequently, neither DT nor DM affects the state of stress, but rather affects only the state of strain. Conversely, if the composite is not free to expand or contract, then a change in temperature and/or moisture content does contribute to the state of stress, as illustrated in Example Problem 3. Example Problem 3 A thin, unidirectional graphite–epoxy composite laminate is firmly mounted within an infinitely rigid square frame, as shown in Fig. 2. The coefficients of thermal and moisture expansion of the rigid frame equal zero. The composite is initially stress-free. Subsequently, however, the composite/frame assembly is subjected to a decrease in temperature DT= 155jC and an increase in moisture content DM=0.5%. Determine the stresses induced in the composite by this change in temperature and moisture content. Use the material properties listed in Table 3 of Chap. 3 and ignore the possibility that the thin composite will buckle if compressive stresses occur.

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Figure 2 Unidirectional composite laminate mounted within an infinitely rigid frame.

Solution. According to the problem statement, the frame is ‘‘infinitely rigid’’ and is made of a material whose thermal and moisture expansion coefficients equal zero. Consequently, the frame will retain its original shape, regardless of the temperature change, moisture change, or stresses imposed on the frame by the composite laminate. Furthermore, the composite is ‘‘firmly mounted’’ within the frame. Together, these stipulations imply that the composite does not change shape, although changes in temperature and moisture content have occurred. Consequently, the total strains experienced by the composite equal zero: 9 8 9 8 > =

= > < e11 > e22 ¼ 0 > > : > ; ; > : 0 c12

The stresses induced can be calculated using Eq. (3): 8 9 2 Q11 > = < r11 > 6Q r22 ¼ 4 12 > > ; : s12 0

Q12 Q22 0

38 0 > < e11 7 0 5 e22 > : Q66

DTa11 DTa22 c12

9 DMb11 > =

DMb22

> ;

The terms within the reduced stiffness matrix are calculated in accordance with Eq. (11): Q11 ¼

E11

E211 ð170 GPaÞ2 ¼ 170:9 GPa ¼ m212 E22 170 GPa ð0:30Þ2 ð10 GPaÞ

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Q12 ¼ Q21 ¼

m12 E11 E22 ð0:30Þð170 GPaÞð10 GPaÞ ¼ 2 E11 m12 E22 170 GPa ð0:30Þ2 ð10 GPaÞ

¼ 3:016 GPa Q22 ¼

E11 E22 ð170 GPaÞð10 GPaÞ ¼ ¼ 10:05 GPa E11 m212 E22 170 GPa ð0:30Þ2 ð10 GPaÞ

Q66 ¼ G12 ¼ 13 GPa

Hence, in this case, Eq. (3) becomes: 8 9 2 3 0 170:9  109 3:016  109 > < r11 > = 6 7 r22 ¼ 4 3:016  109 10:05  109 0 5 > > : ; s12 0 0 13:0  109    9 8 6 ð0:5Þ 150  10 6 = < 0 ð 155Þ 0:9  10    0 ð 155Þ 27  10 6 ð0:5Þ 4800  10 6 ; : 0 8 9 8 9 > < r11 > = > < 31:36 MPa > =  r22 ¼ 17:29 MPa > > > : ; > : ; 0 s12

The reader may initially view this example to be somewhat unrealistic. After all, the frame has been assumed to be ‘‘infinitely rigid.’’ In reality, an infinitely rigid material (i.e., one for which E!l) does not exist. Furthermore, it is assumed that the thermal expansion coefficient for the frame is zero, which is also not valid for most real materials (the assumption that the moisture expansion coefficient is zero is true for metals and metal alloys). Despite these unrealistic assumptions, this example problem illustrates a common occurrence in composite laminates. Specifically, most modern composites are cured at an elevated temperature (for example, many graphite–epoxy systems are cured at 175jCc350jF). The composite can normally be considered to be stress- and strain-free at the cure temperature. After the cure is complete, the composite is typically cooled to room temperatures, say 20jCc70jF, which corresponds to a temperature decrease of DT= 155jCc 280jF. Also, the moisture content immediately after the cure can usually be assumed to equal 0%, but will slowly increase following exposure to normal humidity levels over subsequent days or weeks. Although adsorption of moisture rarely causes a significant gain in weight (most composites adsorb a maximum of 3–4% moisture by weight, even if totally immersed in water), even this slight gain in moisture content

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can nevertheless cause significant strains to develop. If a DT and/or DM occur, and if the unidirectional composite is not free to expand or contract as dictated by these changes, then thermal- and/or moisture-induced stresses will develop. 2 UNIDIRECTIONAL COMPOSITES REFERENCED TO AN ARBITRARY COORDINATE SYSTEM A unidirectional composite referenced to two different coordinate systems is shown in Fig. 3. In Fig. 3(a), the composite is referenced to the principal material coordinate system (i.e., the 1–2 coordinate system), and in this case, either Eqs. (2a) and (2b) or Eq. (3) may be used to relate strains and stresses within the composite. In Fig. 3(b), however, the composite is referenced to an arbitrary (nonprincipal) x–y coordinate system. This is often called an ‘‘off-axis’’ specimen since the specimen is referenced to an arbitrary x–y coordinate system rather than the principal 1–2 coordinate system. Suppose we wish to relate strains and stresses referenced to the x–y coordinate system. For example, suppose we know the stresses rxx, ryy, and sxy, as well as the material properties referenced to the principal material coordinate system (E11, E22, m12, etc.), and wish to calculate strains exx, eyy, and cxy. In this case, neither Eqs. (2a) and (2b) nor Eq. (3) can be used directly since they require that the stresses and strains be referenced to the 1–2 coordinate system. We can perform this calculation using a three-step process. Specifically, we can: (a) Transform the known stresses from the x–y coordinate system to the 1–2 coordinate system (using Eq. 20 of Chap. 2, for example), which will give us the stress components r11, r22, and s12 that correspond to the known values of rxx, ryy, and sxy. (b) Apply Eq. (2a) to obtain in-plane strains e11, e22, and c12. (c) Transform the calculated strains (e11, e22, and c12) from the 1–2 coordinate system back to the x–y coordinate system, finally obtaining the desired strains exx, eyy, and cxy. This three-step process is a rigorously valid procedure. However, it will later be seen that during the analysis of a multi-angle composite laminate, the process of transforming strains/stresses from an arbitrary x–y coordinate system to the 1–2 coordinate system (and vice versa) must be performed for each ply in the laminate. Since this transformation is encountered so frequently, it becomes cumbersome to apply the three-step process for every ply. Instead, it is very convenient to simply develop a form of reduced Hooke’s law suitable for use in an arbitrary x–y coordinate system. In effect, we will

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Figure 3 A unidirectional composite referenced to two different coordinate systems.

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transform Hooke’s law from the 1–2 coordinate system to an arbitrary x–y coordinate system. To simplify our discussion, assume for the moment that no change in environment occurs, i.e., assume that DT=DM=0. In this case, Eq. (2a) becomes: 9 2 9 8 38 S11 S12 0 > > = = < e11 > < r11 > 7 6 e22 ¼ 4 S12 S22 0 5 r22 ð13Þ > > > > ; ; : : s12 0 0 S66 c12 Equation (13) can be rewritten as: 9 2 38 2 e 1 0 0 > S11 S12 < 11 > = 7 6 6 e 22 ¼ 4 S12 S22 40 1 05 > > : ; 0 0 2 0 0 c12 =2

9 38 > = < r11 > 7 0 5 r22 > > ; : s12 S66 0

ð14Þ

In Eq. (14), we have employed the so-called ‘‘Reuter matrix’’ (named after the person who suggested this approach [1]) to, in effect, divide the shear strain by a factor of (1/2) within the strain array. Let: 3 3 2 2 S11 S12 0 1 0 0 7 7 6 6 ðandÞ ½SŠ ¼ 4 S12 S22 0 5 ½RŠ ¼ 4 0 1 0 5 0 0 2 0 0 S66 so that Eq. (14) can be written in the following abbreviated form: 9 9 8 8 e > > = = < r11 > < 11 > e22 ½RŠ ¼ ½SŠ r22 > > > > ; ; : : s12 c12 =2

The transformation of strains within a plane from an x–y coordinate system to another xV–yV coordinate system was discussed in Sec. 13 of Chap. 2. Adopting Eq. (44) of Chap. 2 for our use here (i.e., using axes labels 1 and 2, rather than xV and yV, respectively), we have: 9 2 8 3 e > cos2 ðhÞ sin2 ðhÞ 2cosðhÞsinðhÞ = < 11 > 6 7 e22 ¼4 sin2 ðhÞ cos2 ðhÞ 2cosðhÞsinðhÞ 5 > > ; : cosðhÞsinðhÞ cosðhÞsinðhÞ cos2 ðhÞ sin2 ðhÞ c12 =2 9 9 8 8 > > = = < exx > < exx > eyy eyy ¼ ½T Š  > > ; ; : c =2 > : c =2 > xy xy

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Similarly, the transformation of stress within a plane was discussed in Sec. 8 of Chap. 2, and Eq. (20) of Chap. 2 can be adopted as follows: 9 2 8 3 cos2 ðhÞ sin2 ðhÞ 2cosðhÞsinðhÞ > = < r11 > 7 6 r22 ¼ 4 sin2 ðhÞ cos2 ðhÞ 2cosðhÞsinðhÞ 5 > > ; : s12 cosðhÞsinðhÞ cosðhÞsinðhÞ cos2 ðhÞ sin2 ðhÞ 8 8 9 9 > > = < rxx > = < rxx >  ryy ¼ ½T Š ryy > > > > : : ; ; sxy sxy

As pointed out in Chap. 2, the identical transformation matrix, [T ], is used to relate strains and stresses in the two coordinate systems: 2 3 cos2 ðhÞ sin2 ðhÞ 2cosðhÞsinðhÞ 6 7 ½T Š ¼ 4 sin2 ðhÞ cos2 ðhÞ 2cosðhÞsinðhÞ 5 cosðhÞsinðhÞ cosðhÞsinðhÞ cos2 ðhÞ

sin2 ðhÞ

Inserting these transformation relationships into Eq. (14), we have: 8 8 9 9 e > > < rxx > < xx > = = eyy ½R Š½T Š ¼ ¼ ½S Š½T Š ryy > > > > : ; : ; sxy cxy =2

ð15Þ

To simplify Eq. (15), first multiply both sides of Eq. (15) by the inverse of the Reuter matrix, [R] 1, and then by the inverse of the transformation matrix, [T ] 1: 9 8 9 8 e > > = = < xx > < rxx > eyy ¼ ½T Š 1 ½R Š 1 ½S Š½T Š ryy ð16Þ > > > > ; ; : : sxy cxy =2

where:

2

1

0

6 ½R Š 1 ¼ 4 0

1

0

and

2

6 ½T Š 1 ¼ 4

0

0

3

7 0 5 1=2

cos2 ðhÞ sin2 ðhÞ

cosðhÞsinðhÞ

sin2 ðhÞ

cos2 ðhÞ

cosðhÞsinðhÞ

2cosðhÞsinðhÞ

3

7 2cosðhÞsinðhÞ 5

cos2 ðhÞ

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sin2 ðhÞ

We next extract the factor of ( 1⁄2 ) from the shear strain within the strain array using the [R] 1 matrix: 9 9 8 8 > > = = < rxx > < exx > ð17Þ ½R Š 1 eyy ¼ ½T Š 1 ½R Š 1 ½S Š½T Š ryy > > > > ; : ; : cxy sxy

Multiplying both sides of Eq. (17) by the [R] matrix, we arrive at our final result: 9 9 8 8 > > = = < rxx > < exx > eyy ¼ ½RŠ½T Š 1 ½RŠ 1 ½SŠ½T Š ryy ð18Þ > > > > ; ; : : cxy sxy

Equation (18) represents an ‘‘off-axis’’ version of Hooke’s Law. That is, it relates the strains induced in the arbitrary x–y coordinate system (exx, eyy, and cxy) to the stresses in the same x–y coordinate system (rxx, ryy, and sxy) via material properties referenced to the principal 1–2 coordinate system (represented by the [S] matrix) and the fiber angle h (represented by the transformation matrix [T ]). Completing the matrix algebra indicated, we obtain: 9 9 8 8 > = =  > < rxx > < exx > eyy ¼ S ryy ð19Þ > > > > ; ; : : cxy sxy where:   S ¼ ½RŠ½T Š 1 ½RŠ 1 ½SŠ½T Š

Equation (19) is known as transformed reduced Hooke’s law. It is called ‘‘transformed’’ because Eq. (19) has been transformed from the 1–2 coordinate system to the x–y coordinate system, and ‘‘reduced’’ because we have reduced the allowable state of stress from three dimensions to two dimensions [i.e., Eq. (19) is only valid for a plane stress state]. Matrix [S ] is called the transformed reduced compliance matrix.* In expanded form, Eq. (19) is written as: 9 9 2 8 38 S11 S12 S16 > > = = < rxx > < exx > 6 7 eyy ¼ 4 S21 S22 S26 5 ryy ð20Þ > > > > ; ; : : sxy cxy S61 S62 S66 * In common parlance, the [ S] matrix is often called the ‘‘S-bar’’ matrix.

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where: S11 ¼ S11 cos4 h þ ð2S12 þ S66 Þcos2 h sin2 h þ S22 sin4 h   S12 ¼ S21 ¼ S12 cos4 h þ sin4 h þ ðS11 þ S22 S66 Þcos2 h sin2 h

S16 ¼ S61 ¼ ð2S11

2S12

S26 ¼ S62 ¼ ð2S11

2S12

S66 Þcos3 h sin h

ð2S22

S66 Þcos h sin3 h

ð2S22

2S12

S22 ¼ S11 sin4 h þ ð2S12 þ S66 Þcos2 h sin2 h þ S22 cos4 h S66 ¼ 2ð2S11 þ 2S22

4S12

S66 Þ cos h sin3 h

2S12 S66 Þcos3 h sin h   S66 Þ cos2 h sin2 h þ S66 cos4 h þ sin4 h

ð21Þ

Three important observations should be made regarding the transformed reduced compliance matrix. First, we have retained the original subscripts used in our earlier discussion of 3-D states of stress and strain. For example, the term that appears in the (3,3) position of the transformed reduced compliance matrix is labeled S66 . Secondly, the [S ] matrix is symmetric. Therefore, S 21 ¼ S 12 , S 61 ¼ S 16 , and S 62 ¼ S 26 , as indicated in Eq. (21). Third, the [S] matrix is fully populated. That is, (in general) none of the terms within the [S] matrix equal zero, in contrast to the [S] matrix where four off-diagonal terms equal zero [see Eq. (2a)]. This simply reveals the anisotropic nature of unidirectional composites. Following the convention adopted earlier, a unidirectional composite laminate referenced to the principal 1–2 coordinate system is referred to as an orthotropic (or transversely isotropic) material, whereas if the same material is referenced to an arbitrary (nonprincipal) x–y coordinate system, it is called an anisotropic material. Since neither S 16 nor S 26 is equal to zero for an anisotropic composite, a coupling exists between shear stress and normal strains. That is, a shear stress sxy will cause normal strains exx and eyy to occur, as indicated by Eq. (20). This coupling does not occur for orthotropic or transversely isotropic composites (or for that matter, for isotropic materials). Let us now include thermal and moisture strains in the transformed, reduced form of Hooke’s law. In the 1–2 coordinate system, in-plane thermal strains are given by: 8 T 9 8 9 e11 > a11 > > > > > > > = = < < ð22Þ eT22 ¼ DT a22 > > > > > > > ; : T > ; : 0 c12 As has been previously discussed, in the 1–2 coordinate system, a uniform change in temperature does not produce a shear strain, i.e., a12=0. Now, thermally induced strains can be transformed from one coordinate

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system to another in exactly the same way as mechanically induced strains are transformed. That is, we can relate thermal strains in the 1–2 coordinate system to the x–y coordinate system using the transformation matrix: 8 9 2 3 2 2 eT11 > > cos ð h Þ sin ð h Þ 2cos ð h Þsin ð h Þ > > = 6 < 7 ¼6 eT22 sin2 ðhÞ cos2 ðhÞ 2cosðhÞsinðhÞ 7 5 4 > > > ; : T > cosðhÞsinðhÞ cosðhÞsinðhÞ cos2 ðhÞ sin2 ðhÞ c12 =2 8 T 9 e > > > = < xx > eTyy  > > > : T > ; cxy =2 Inverting this expression, we have: 8 T 9 8 9 exx > > > eT > > > > = = < < 11 > eTyy ¼ ½T Š 1 eT22 > > > > > > ; ; : T > : > cxy =2 0 2 cos2 ðhÞ 6 2 ¼6 4 sin ðhÞ cosðhÞsinðhÞ 8 T 9 e > > > = < 11 >  eT22 > > > : T > ; c12 =2

sin2 ðhÞ

cos2 ðhÞ cosðhÞsinðhÞ

2cosðhÞsinðhÞ

3

7 2cosðhÞsinðhÞ 7 5 cos2 ðhÞ sin2 ðhÞ ð23Þ

Substituting Eq. (22) in this result, completing the matrix multiplication indicated, and simplifying the resulting expressions, we obtain: 9 8 9 8 axx > > > eT > > > > = = < xx > < eTyy ¼ DT ayy ð24Þ > > > > > > > : cT > : ; ; axy xy

where:

axx ¼ a11 cos2 ðhÞ þ a22 sin2 ðhÞ

ayy ¼ a11 sin2 ðhÞ þ a22 cos2 ðhÞ

axy ¼ 2cosðhÞsinðhÞða11

a22 Þ

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ð25Þ

In Sec. 6, we will define the properties of a unidirectional composite laminate when referenced to an arbitrary x–y coordinate system. As further discussed there, Eq. (25) defines the effective coefficients of thermal expansion for an anisotropic composite laminate. Note that for anisotropic composites, a coupling exists between a uniform change in temperature and shear strain. That is, a change in temperature causes shear strain cTxy as well as normal T strains exx and eTyy. In the 1–2 coordinate system, the in-plane strains caused by a uniform change in moisture content are given by: 9 8 8 M9 b > > = = < 11 > < e11 > ð26Þ ¼ DM b eM 22 22 > > > ; ; : : M> 0 c12 As was the case for thermal strains, we wish to express strains induced by DM in an arbitrary x–y coordinate system. Using the identical procedure as before, moisture strains in the x–y coordinate system are given by: 9 8 M9 8 b e > > = = < xx > < xx > b eM ¼ DM ð27Þ yy yy > > > ; ; : M> : bxy cxy

where:

bxx ¼ b11 cos2 ðhÞ þ b22 sin2 ðhÞ byy ¼ b11 sin2 ðhÞ þ b22 cos2 ðhÞ

bxy ¼ 2cosðhÞsinðhÞðb11

b22 Þ

ð28Þ

In Sec. 6, we will define the effective properties of a unidirectional composite laminate when referenced to an arbitrary x–y coordinate system. As further discussed there, Eq. (28) defines the effective coefficients of moisture expansion for an anisotropic composite laminate. We can now calculate the strains induced by the combined effects of stress, a uniform change in temperature, and a uniform change in moisture content. Specifically, adding Eqs. (20), (24), and (27) together, we obtain: 9 2 8 exx > > S11 > > = < 6 eyy ¼ 4 S12 > > > > ; : S16 cxy

S12 S22 S26

9 9 8 8 38 r 9 bxx > axx > > > xx > S16 > > > > > > > = = = < < < 7 ð29Þ S26 5 ryy þ DT ayy þ DM byy > > > > > > > > > > > > ; ; ; : : : S66 sxy axy bxy

Equation (29) allows the calculation of the in-plane strains induced by any combination of in-plane stresses, a uniform change in temperature, and a

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uniform change in moisture content. If instead we have measured the total strains induced by a known DT, DM, and unknown in-plane stresses, we can calculate the stresses that caused these strains by inverting Eq. (29) to obtain: 9 2 8 38 exx DTaxx DMb 9 rxx > xx > > > Q Q Q > > > > 11 12 16 = = < < 7 e 6 DTa DMb ryy ¼ 4 Q12 Q22 Q26 5 ð30Þ yy yy yy > > > > > > > > ; ; : : Q16 Q26 Q66 sxy cxy DTaxy DMbxy where:

Q11 ¼ Q11 cos4 h þ 2ðQ12 þ 2Q66 Þcos2 h sin2 h þ Q22 sin4 h   Q12 ¼ Q21 ¼ Q12 cos4 h þ sin4 h þ ðQ11 þ Q22 4Q66 Þcos2 h sin2 h Q16 ¼ Q61 ¼ ðQ11 ðQ22

Q12

Q12

2Q66 Þcos3 h sin h

2Q66 Þ cos h sin3 h

Q22 ¼ Q11 sin4 h þ 2ðQ12 þ 2Q66 Þcos2 h sin2 h þ Q22 cos4 h Q26 ¼ Q62 ¼ ðQ11 ðQ22

Q12

Q66 ¼ ðQ11 þ Q22

Q12

2Q66 Þcos h sin3 h

2Q66 Þcos3 h sin h 2Q12

  2Q66 Þcos2 h sin2 h þ Q66 cos4 h þ sin4 h

ð31Þ

The [ Q] matrix is called the transformed, reduced stiffness matrix.* This name again reminds us that we have reduced our analysis to the 2-D plane stress case, and that we have transformed Hooke’s law from the 1–2 coordinate system to an arbitrary x–y coordinate system. Note that the [ Q] matrix is (in general) fully populated. This reflects the anisotropic nature of unidirectional composites when referenced to a nonprincipal material coordinate system. Also, the [ Q] matrix is symmetric, so that Q21 ¼ Q12 , Q61 ¼ Q16 , and Q62 ¼ Q26 , as indicated in Eq. (31). The reader should note that the functional form of the equations that define the elements of the transformed reduced stiffness matrix, i.e., Eq. (31), is not identical to the functional form of the equations defining the elements of the transformed reduced compliance matrix, Eq. (21). That is, Eq. (31) cannot be transformed into Eq. (21) by a simple substitution of S11 for Q11, S12 for Q12, S22 for Q22, etc. This difference in functional form is due to the fact that we have defined both the stiffness and compliance matrices in terms of

* In common parlance, the [ Q] matrix is often called the ‘‘Q-bar’’ matrix.

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engineering shear strains, rather than tensoral shear strains. The use of Eqs. (29) and (30) to solve simple problems involving off-axis composite laminates is demonstrated in Example Problems 4 to 6. Example Problem 4 Determine the strains induced in the off-axis graphite–epoxy composite subjected to the in-plane stresses shown in Fig. 4. Assume that DT=DM=0 and use the material properties listed in Table 3 of Chap. 3. Solution. This problem is analogous to Example Problem 1 except that we are now considering the behavior of an off-axis composite. The magnitude of each stress component is indicated in Fig. 4. Based on the sign conventions reviewed in Sec. 5 of Chap. 2, the algebraic sign of each stress component is: rxx ¼ 200 MPa

ryy ¼

30 MPa

sxy ¼

50 MPa

Fiber angles are measured from the +x-axis to the +1-axis (or equivalently, from the +y-axis to the +2-axis). In accordance with the right-hand rule, the fiber angle in Fig. 4 is algebraically positive: h=+30j. Strains are calculated using Eq. (29). Since DT=DM=0, we have: 9 2 8 38 r 9 exx > > > xx > S S S 11 12 16 > > > > = < = < 6 7 eyy ¼ 4 S12 S22 S26 5 ryy > > > > > > > > ; ; : : S16 S26 S66 cxy sxy

Figure 4 A 30j off-axis composite subjected to in-plane stresses (magnitudes of in-plane stresses shown).

Copyright © 2004 by Marcel Dekker, Inc. All Rights Reserved.

Recall that the elements of the reduced compliance matrix, the [S] matrix, were calculated as a part of Example Problem 1. Each term within the transformed reduced compliance matrix, the [S ] matrix, can therefore be calculated via a straightforward application of Eq. (21). The calculation of S 11, for example, proceeds as follows: S11 ¼ S11 cos4 h þ ð2S12 þ S66 Þcos2 h sin2 h þ S22 sin4 h   S11 ¼ 5:88  10 12 cos4 ð30B Þ

   þ 2 1:76  10 12 þ 76:9  10 12 cos2 ð30B Þsin2 ð30B Þ   þ 100:0  10 12 sin4 ð30B Þ   S11 ¼ 23:32  10 12 Pa 1

Calculating the remaining elements of the S matrix in similar fashion and applying Eq. (29), we find: 9 2 8 e 23:32  10 12 4:327  10 12 > < xx > = 6 eyy ¼ 4 4:327  10 12 70:38  10 12 > ; :c > xy 33:72  10 12 47:79  10 12 9 8 200  106 >  > = < 1  30  106 ðPaÞ > Pa > ; : 50  106

33:72  10

47:79  10

101:3  10

12 12 12

3 7 5

Completing the matrix multiplication indicated, we obtain: 8 9 8 9 > < exx > = > < 6220 Am=m > = eyy ¼ 1144 Am=m > > > : ; > : ; cxy 10375 Arad

Example Problem 5

Determine the strains induced in the off-axis graphite–epoxy composite subjected to (a) the in-plane stresses shown in Fig. 4, (b) a decrease in temperature DT= 155jC, and (c) an increase in moisture content DM=0.5%. Use the material properties listed in Table 3 of Chap. 3. Solution. This problem involves three different mechanisms that contribute to the total strain induced in the laminate: the applied stresses, the temper-

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ature change, and the change in moisture content. The total strains induced by all three mechanisms can be calculated using Eq. (29): 9 2 8 S11 exx > > > > = 6 < eyy ¼ 6 4 S12 > > > > ; : cxy S16

S12 S22 S26

9 9 9 8 8 38 bxx > S16 > axx > rxx > > > > > > > > > = = = < < 7< 7 b yy S26 5 ryy þ DT ayy þ DM > > > > > > > > > > > > ; ; ; : : : s a b xy xy S66 xy

Numerical values for the stresses and transformed reduced compliance matrix are given in Example 4. The linear thermal and moisture expansion coefficients for graphite–epoxy, referenced to the x–y coordinate system, are calculated using Eqs. (25) and (28), respectively. The thermal expansion coefficients are:   axx ¼ a11 cos2 ðhÞ þ a22 sin2 ðhÞ ¼ 0:9  10 6 cos2 ð30jÞ   þ 27  10 6 sin2 ð30jÞ axx ¼ 6:1 Am=m=o C   ayy ¼ a11 sin2 ðhÞ þ a22 cos2 ðhÞ ¼ 0:9  10 6 sin2 ð30jÞ   þ 27  10 6 cos2 ð30jÞ ayy ¼ 20:0 Am=m=o C axy ¼ 2 cosðhÞsinðhÞða11 a22 Þ ¼ 2 cosð30jÞsinð30jÞ    0:9  10 6 27  10 6 axy ¼ 24:2 Arad=o C The moisture expansion coefficients are:   bxx ¼ b11 cos2 ðhÞ þ b22 sin2 ðhÞ ¼ 150  10 6 cos2 ð30jÞ   þ 4800  10 6 sin2 ð30jÞ bxx ¼ 1313 Am=m=%M

  byy ¼ b11 sin2 ðhÞ þ b22 cos2 ðhÞ ¼ 150  10 6 sin2 ð30jÞ   þ 4800  10 6 cos2 ð30jÞ byy ¼ 3638 Am=m=%M

bxy ¼ 2 cosðhÞsinðhÞðb11 bxy ¼

4027 Arad=%M

b22 Þ ¼ 2 cosð30jÞsinð30jÞ    150  10 6 4800  10 6

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Hence, Eq. (29) becomes: 9 2 8 23:32  10 12 > = < exx > 6 eyy ¼ 4 4:327  10 12 > > ; : cxy 33:72  10 12 

8 >

:

> :

4:327  10

70:38  10

12 12

33:72  10

47:79  10

12 12

3 7 5

47:79  10 12 101:3  10 12 9 8 9 6:1  10 6 > 200  106 > > = = < 6 30  106 þ ð 155Þ 20:0  10 þ ð0:5Þ > > ; ; : 24:2  10 6 > 6 50  10 9 1313  10 6 > = 3638  10 6 > ; 4027  10 6

Completing the matrix multiplication indicated, we obtain: 9 9 8 9 8 9 8 8 exx > > 657 Am=m > > 946 Am=m > > 6220 Am=m > > > > > > > = < < = > = > < < = > eyy ¼ þ 3100 Am=m þ 1819 Am=m 1144 Am=m > > > > > > > > > > > > ; : : ; > ; > : ; > :c > 2014 Arad 3751 Arad 10375 Arad xy 8 9 5931 Am=m > > > > < = ¼ 137 Am=m > > > > : ; 8638 Arad

An implicit assumption in this problem is that the composite is free to expand or contract, as dictated by changes in temperature and/or moisture content. Consequently, neither DT nor DM affects the state of stress, but rather affects only the state of strain. Conversely, if the composite is not free to expand or contract, then a change in temperature and/or moisture content does contribute to the state of stress, as illustrated in Example Problem 6. Example Problem 6 A thin off-axis graphite–epoxy composite laminate is firmly mounted within an infinitely rigid square frame, as shown in Fig. 5. The coefficients of thermal and moisture expansion of the rigid frame equal zero. The composite is initially stress-free. Subsequently, however, the composite/frame assembly is subjected to a decrease in temperature DT= 155jC and an increase in moisture content DM=0.5%. Determine the stresses induced in the composite by this change in temperature and moisture content. Ignore the possibility that the thin composite will buckle if compressive stresses occur.

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Figure 5 Off-axis composite laminate mounted within an infinitely rigid frame.

Solution. As discussed in Example Problem 3, the situation described in this problem is very often encountered in composite materials, despite somewhat unrealistic assumptions regarding an ‘‘infinitely rigid’’ frame with zero thermal expansion coefficients. Since the frame is ‘‘infinitely rigid’’ and is made of a material whose thermal and moisture expansion coefficients equal zero, the frame will retain its original shape. Since the off-axis composite is ‘‘firmly mounted’’ within the frame, the composite does not change shape, although changes in temperature and moisture content have occurred. Consequently, the total strains experienced by the composite equal zero: 9 8 9 8 0 > = = > < > < exx > eyy ¼ 0 > ; ; > : > :c > xy 0

The stresses induced can be calculated using Eq. (30): 9 2 8 Q11 > = < rxx > ryy ¼ 6 4 Q12 > > ; : sxy Q16

Q12 Q22 Q26

38 e > xx Q16 < 7 Q26 5 eyy > : Q66 cxy

DTaxx DTayy DTaxy

9 DMbxx > = DMbyy > ; DMbxy

Recall that the elements of the reduced stiffness matrix, the [ Q] matrix, were calculated as a part of Example Problem 3. Each term within the transformed

Copyright © 2004 by Marcel Dekker, Inc. All Rights Reserved.

reduced stiffness matrix, the [ Q ] matrix, can therefore be calculated via a straightforward application of Eq. (31). Calculation of Q11 , for example, proceeds as follows: Q11 ¼ Q11 cos4 h þ 2ðQ12 þ 2Q66 Þ cos2 h sin2 h þ Q22 sin4 h   Q11 ¼ 170:9  109 cos4 ð30jÞ

  þ 2 3:016  109 þ 2 13:0  109 cos2 ð30jÞsin2 ð30jÞ   þ 10:05  109 sin4 ð30jÞ

Q11 ¼ 107:6  109 Pa

Calculating the remaining elements of the [ Q] matrix in similar fashion, and using the thermal and moisture expansion coefficients referenced to the x–y coordinate system (calculated as a part of Example Problem 5), we find: 9 2 8 3 r 107:6  109 26:1  109 48:1  109 > = < xx > 7 6 ryy ¼ 4 26:1  109 27:2  109 21:5  109 5 > > ; : sxy 48:1  109 21:5  109 36:0  109     9 8 0 ð 155Þ 6:1  10 6 ð0:5Þ 1313  10 6 > > <     = ð0:5Þ 3638  10 6  0 ð 155Þ 20:0  10 6 >    > ; : ð0:5Þ 4027  10 6 0 ð 155Þ 24:2  10 6

8 9 8 9 r 19:0 MPa > > < xx > = > < = ryy ¼ 5:04 MPa > > > : ; > : ; 21:1 MPa sxy

3 CALCULATING TRANSFORMED PROPERTIES USING MATERIAL INVARIANTS The stresses and strains in a unidirectional composite referenced to an arbitrary x–y coordinate system may be related using either Eq. (29) or Eq. (30). Equation (29) involves the use of the transformed reduced compliance matrix, [S ], and individual elements within the [S ] matrix are calculated in accordance with Eq. (21). Alternatively, Eq. (30) involves the use of the transformed reduced stiffness matrix, [ Q], and individual elements of the [ Q] matrix are calculated in accordance with Eq. (31).

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Now, both Eqs. (21) and (31) involve trigonometric functions raised to a power (i.e., sin4 h, cos4 h, cos h sin3 h, etc.). These equations can be simplified somewhat through the use of the following trigonometric identities: 1 sin4 h ¼ ð3 4 cos 2h þ cos 4hÞ 8 1 cos4 h ¼ ð3 þ 4 cos 2h þ cos 4hÞ 8 1 cos h sin3 h ¼ ð2 sin 2h sin 4hÞ 8 1 cos2 h sin2 h ¼ ð1 cos 4hÞ 8 1 cos3 h sin h ¼ ð2 sin 2h þ sin 4hÞ 8

ð32Þ

For example, substituting these identities into Eq. (21) and simplifying result in: S11 ¼ US1 þ US2 cos 2h þ US3 cos4 h S12 ¼ S21 ¼ US4

US3 cos 4h

S16 ¼ S61 ¼ US2 sin 2h þ 2US3 sin 4h S22 ¼ US1

US2 cos 2h þ US3 cos 4h

S26 ¼ S62 ¼ US2 sin 2h S66 ¼ US5

ð33Þ

2US3 sin 4h

4US3 cos 4h

The terms UiS which appear in Eq. (33) are called compliance invariants and are defined as follows: US1 ¼ US2 ¼ US3 ¼ US4 ¼ US5 ¼

1 ð3S11 þ 3S22 þ 2S12 þ S66 Þ 8 1 ðS11 S22 Þ 2 1 ðS11 þ S22 2S12 S66 Þ 8 1 ðS11 þ S22 þ 6S12 S66 Þ 8 1 ðS11 þ S22 2S12 þ S66 Þ 2

ð34Þ

The superscript S is used to indicate that these quantities are calculated using members of the reduced compliance matrix [S]. They are called compliance

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invariants because they define the elements of the compliance matrix that are independent of coordinate system. In this sense, compliance invariants are analogous to stress and strain invariants, which were discussed in Secs. 6 and 11 of Chap. 2, respectively. In a similar manner, substituting the trigonometric identities listed as Eq. (32) into Eq. (31) and simplifying result in: Q Q Q11 ¼ UQ 1 þ U2 cos 2h þ U3 cos 4h

Q12 ¼ Q21 ¼ UQ UQ 4 3 cos 4h 1 Q16 ¼ Q61 ¼ UQ sin 2h þ UQ 3 sin 4h 2 2 Q Q22 ¼ UQ UQ 1 2 cos 2h þ U3 cos 4h 1 sin 2h UQ Q26 ¼ Q62 ¼ UQ 3 sin 4h 2 2 Q66 ¼ UQ UQ 5 3 cos 4h

ð35Þ

where the stiffness invariants, UiQ, are defined as: UQ 1 ¼ UQ 2 ¼ UQ 3 ¼ UQ 4 ¼ UQ 5 ¼

1 ð3Q11 þ 3Q22 þ 2Q12 þ 4Q66 Þ 8 1 ðQ11 Q22 Þ 2 1 ðQ11 þ Q22 2Q12 4Q66 Þ 8 1 ðQ11 þ Q22 þ 6Q12 4Q66 Þ 8 1 ðQ11 þ Q22 2Q12 þ Q66 Þ 8

ð36Þ

The superscript Q is used to indicate that these quantities are calculated using members of the reduced stiffness matrix [ Q]. The reader should note that the functional forms of the stiffness invariants defined in Eq. (36) are not identical to those of the compliance invariants defined in Eq. (34). The difference in functional form can be traced to the use of engineering shear strain rather than tensoral shear strain. A comparison of Eqs. (21) and (33) reveals that the use of compliance invariants does indeed simplify the calculation of elements of the [S] matrix, although, mathematically, the two equations are entirely equivalent. Similarly, a comparison of Eqs. (31) and (35) shows that the use of the stiffness invariants simplifies the calculation of the terms within the [Q] matrix. At this point, this simplification may seem to be a trivial matter since, in practice,

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elements of the [S ] and [Q] matrices are calculated with the aid of a digital computer and do not require hand calculation. However, in Chap. 6, it will be shown that in special circumstances, the use of stiffness invariants leads to a convenient method of transforming the stiffness of multi-angle composite laminates from one coordinate system to another. Hence, in these special cases, the use of the invariant formulation is advantageous. Further discussion of the invariant approach will be deferred to Chap. 6. Example Problem 7 Problem. Use the material invariants [i.e., Eqs. (33) and (34)] to calculate the transformed reduced compliance matrix for a 30j graphite–epoxy laminate. Use the material properties listed in Table 3 of Chap. 3. Solution. From Example Problem 1, the reduced compliance matrix for this material system is: 2 3 2 3   S11 S12 0 1:76  10 12 0 5:88  10 12 1 4 S12 S22 0 5 ¼ 4 1:76  10 12 100:0  10 12 5 0 Pa 0 0 S66 0 0 76:9  10 12 The compliance invariants may be calculated using these values and Eq. (34):  1 1h  US1 ¼ ð3S11 þ 3S22 þ 2S12 þ S66 Þ ¼ 3 5:88  10 12 8 8 i     12 þ3 100:0  10 þ 2 1:76  10 12 þ 76:9  10 12 ¼ 48:9  10 12 1 1 US2 ¼ ðS11 S22 Þ ¼ 5:88  10 2 2 ¼

47:1  10

12

100:0  10

12



1h 5:88  10 12 þ 100:0  10 8 i 76:9  10 12 ¼ 4:06  10 12 1 1h US4 ¼ ðS11 þ S22 þ 6S12 S66 Þ ¼ 5:88  10 12 þ 100:0  10 8 8 i   76:9  10 12 ¼ 2:296  10 12 þ6 1:76  10 12 1 1h US5 ¼ ðS11 þ S22 2S12 þ S66 Þ ¼ 5:88  10 12 þ 100:0  10 2 2 i   2 1:76  10 12 þ 76:9  10 12 ¼ 93:17  10 12 US3 ¼

1 ðS11 þ S22 2S12 8   2 1:76  10 12

12

S66 Þ ¼

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12

12

12

Using the first equation of Eq. (33) and setting h=30j:   S11 ¼ US1 þ US2 cos 2h þ US3 cos 4h ¼ 48:9  10 12  þ 47:1  10

¼ 23:3  10

12

12

Pa

  cosð60jÞ þ 4:06  10

12

1



cosð120jÞ

Similarly, using the second equation of Eq. (33) and setting h=30j: S12 ¼ S21 ¼ US4

US3 cos 4h

 ¼ 2:296  10 ¼ 4:33  10

12

12



Pa

1



4:06  10

12



cosð120jÞ

The additional terms within the [S] matrix are calculated using the rest of Eq. (33). A summary of our results is: 3 2 23:3  10 12 4:33  10 12 33:7  10 12   7 1 6 12 ½S Š ¼ 6 70:4  10 12 47:8  10 12 7 5 Pa 4 4:33  10 33:7  10 12 47:8  10 12 101:3  10 12 Note that the [S] matrix is identical to that calculated in Example Problem 4.

4 EFFECTIVE ELASTIC PROPERTIES OF A UNIDIRECTIONAL COMPOSITE LAMINATE The definitions of common engineering material properties were reviewed in Chap. 3. In this section, these concepts will be used to define the ‘‘effective’’ properties of a unidirectional composite laminates referenced to an arbitrary x–y coordinate system. 4.1 Effective Properties Relating Stress to Strain Let us first consider the elastic properties measured during uniaxial tests. Consider the unidirectional composite laminate subjected to a uniaxial stress rxx, as shown in Fig. 6. The strains induced in this laminate can be determined using Eq. (29). Assuming that DT=DM=0 (and hence that thermal and moisture strains are zero), and also noting that by definition, ryy=sxy=0, Eq. (29) becomes for this case: 8 9 2 e > S11 = < xx > 6 eyy ¼ 4 S12 > ; : > S16 cxy

S12 S22 S26

38 9 >rxx> S16 < = 7 0 S26 5 > ; : > S66 0

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Figure 6 Unidirectional composite laminate subjected to uniaxial stress rxx.

In-plane strains caused by uniaxial stress rxx are therefore given by: exx ¼ S11 rxx

ð37aÞ

eyy ¼ S12 rxx

ð37bÞ

cxy ¼ S16 rxx ;

ð37cÞ

In Sec. 2 of Chap. 3, Young’s modulus was defined as ‘‘the normal stress rxx divided by the resulting normal strain exx, with all other stress components equal zero.’’ Applying this definition to the unidirectional laminate shown in Fig. 6, Young’s modulus in the x-direction is given by: rxx rxx 1 Exx ¼ ¼ ð38aÞ ¼ exx S11 rxx S11 Inserting the relation for Q11 listed in Eq. (22), we have: Exx ¼

S11

cos4

1 h þ ð2S12 þ S66 Þ cos2 h sin2 h þ S22 sin4 h

ð38bÞ

Since each of the compliance terms (S11, S12, etc.) can also be related to the more familiar engineering constants using Eq. (18) of Chap. 4, Young’s modulus can also be written as: 1 Exx ¼ ð38cÞ   4 cos ðhÞ 1 2m12 sin4 h 2 2 þ cos h sin h þ E11 G12 E22 E11

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In Sec. 2 of Chap. 3, Poisson’s ratio vxy was defined as ‘‘the negative of the transverse normal strain eyy divided by the axial normal strain exx, both of which are induced by stress rxx, with all other stresses equal zero.’’ Poisson’s ratio for the unidirectional laminate shown in Fig. 6 is therefore given by: vxy ¼

eyy S12 ¼ exx S11

ð39aÞ

Using Eq. (22), this can be written as: 

vxy ¼

  S12 cos4 h þ sin4 h þ ðS11 þ S22

S66 Þ cos2 h sin2 h

S11 cos4 h þ ð2S12 þ S66 Þ cos2 h sin2 h þ S22 sin4 h



ð39bÞ

or equivalently, using Eq. (18) of Chap. 4:

vxy

   v12  4 1 1 1 4 þ cos h þ sin h cos2 h sin2 h E11 E22 G12 E11 ¼   cos4 ðhÞ 1 2v12 sin4 ðhÞ þ cos 2 ðhÞ sin2 ðhÞ þ E11 G12 E22 E11

ð39cÞ

In Sec. 2 of Chap. 3, the coefficient of mutual influence of the second kind gxx,xy was defined as ‘‘the shear strain cxy divided by the normal strain exx, both of which are induced by normal stress rxx, when all other stresses equal zero.’’ For a unidirectional composite laminate, gxx,xy is therefore given by: gxx;xy ¼

cxy S16 ¼ exx S11

ð40aÞ

which may be written as: gxx;xy ¼

ð2S11

2S12 S66 Þ cos3 h sin h ð2S22 2S12 S66 Þ cos h sin3 h S11 cos4 h þ ð2S12 þ S66 Þ cos2 h sin 2 h þ S22 sin4 h

or equivalently:

gxx;xy ¼



ð40bÞ

   2 2v12 1 2 2v12 1 þ þ cos3 ðhÞ sin ðhÞ cos ðhÞ sin2 ðhÞ E11 E11 E11 G12 E22 G12   cos4 ðhÞ 1 2v12 sin4 ðhÞ cos2 ðhÞ sin2 ðhÞ þ þ E11 G12 E22 E11

ð40cÞ

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An identical procedure can be employed to define the properties measured during a uniaxial test in which only ryy is applied. In this case, Eq. (29) becomes: 9 9 2 8 38 0 > S11 S12 S16 > > = = < exx > < 7 eyy ¼ 6 4 S12 S22 S26 5 ryy > > > > ; : ; : cxy S16 S26 S66 0

In-plane strains are: exx ¼ S12 ryy

ð41aÞ

eyy ¼ S22 ryy

ð41bÞ

cxy ¼ S26 ryy

ð41cÞ

These strains can be used to define the Young’s modulus Eyy, Poisson’s ratio vyx, and coefficient of mutual influence of the second kind gyy,xy: 1 S22 S12 vyx ¼ S22 S26 gyy;xy ¼ S22 Eyy ¼

ð42aÞ ð42bÞ ð42cÞ

If desired, these relations can be expanded in terms of compliances referenced to the 1–2 coordinate system using Eq. (22) or written in terms of measured engineering properties using Eq. (18) of Chap. 4. Next, consider the effective material properties measured during a pure shear test. A composite laminate subjected to pure shear stress sxy is shown in Fig. 7. Assuming that DT=DM=0, Eq. (29) becomes: 9 2 9 8 38 S11 S12 S16 > > = = < exx > < 0 > 7 eyy ¼ 6 4 S12 S22 S26 5 0 > > > > ; ; : : cxy sxy S16 S26 S66

Hence, the strains caused by a pure shear stress are given by: exx ¼ S16 sxy eyy ¼ S26 sxy

cxy ¼ S66 sxy

ð43aÞ

ð43bÞ ð43cÞ

In Sec. 2 of Chap. 3, the shear modulus was defined as ‘‘the shear stress sxy divided by the resulting shear strain cxy, with all other stress components

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Figure 7 Unidirectional composite subjected to pure shear stress sxy.

equal zero.’’ Applying this definition to the laminate shown in Fig. 7, the shear modulus referenced to the x–y coordinate axes is given by: Gxy ¼

sxy 1 ¼ cxy S66

ð44Þ

As before, this expression can be expanded in terms of compliances referenced to the 1–2 coordinate system using Eq. (21) or written in terms of measured engineering properties using Eq. (18) of Chap. 4. The coefficient of mutual influence of the first kind gxy,xx (or gxy,yy) was defined as ‘‘the normal strain exx (or eyy) divided by the shear strain cxy, both of which are induced by shear stress sxy, when all other stresses equal zero.’’ For a unidirectional composite laminate, the coefficient of mutual influence of the first kind gxy,xx is therefore given by: gxy;xx ¼

exx S16 ¼ cxy S66

ð45aÞ

while gxy,yy is given by: gxy;yy ¼

eyy S26 ¼ cxy S66

ð45bÞ

Chentsov coefficients were defined in Sec. 2 of Chap. 3 as ‘‘the shear strain cxz (or cyz) divided by the shear strain cxy, both of which are induced by

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shear stress sxy, with all other stresses equal zero.’’ For a thin composite laminate, the principal material coordinate system lies within the plane of the laminate, and hence there is no coupling between a shear stress acting within the x–y plane (sxy) and out-of-plane shear strains (cxz or cyz). Consequently, Chentsov coefficients are always equal to zero for thin composite laminates. 4.2 Effective Properties Relate Temperature or Moisture Content to Strain As discussed in Sec. 3 of Chap. 3, the linear coefficients of thermal expansion are measured by determining the strains induced by a uniform change in temperature and forming the following ratios: axx ¼

eTxx DT

ayy ¼

eTyy

axy ¼

DT

cTxy

ð46Þ

DT

The superscript T is included as a reminder that the strains involved are those caused by a change in temperature only. The strains induced in a unidirectional laminate subjected to a change in temperature can be determined using Eq. (29). Assuming that rxx=ryy=sxy=DM=0, Eq. (29) becomes for this case: 8 9 8 9 > > < exx > = < axx > = eyy ¼ DT ayy ð47Þ > > > :c > ; : ; a xy xy Hence, the thermal expansion coefficients for a unidirectional laminate are given by Eq. (25), repeated here for convenience: axx ¼ a11 cos2 ðhÞ þ a22 sin2 ðhÞ ayy ¼ a11 sin2 ðhÞ þ a22 cos2 ðhÞ axy ¼ 2 cosðhÞ sinðhÞða11

a22 Þ

Similarly, the linear coefficient of moisture expansion is measured by determining the strains induced by a uniform change in moisture content and forming the following ratios: bxx ¼

eM xx DM

byy ¼

eM yy DM

bxy ¼

cM xy DM

ð48Þ

The superscript M is included as a reminder that the strains involved are those caused by a change in moisture only. The strains induced in a unidirectional laminate subjected to a change in moisture content can be determined using

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Eq. (29). Assuming that rxx=ryy=sxy=DT=0, Eq. (29) becomes for this case: 9 9 8 8 b e > > = = < xx > < xx > eyy ¼ DM byy ð49Þ > > > > ; ; : : cxy bxy Hence, the moisture expansion coefficients for a unidirectional laminate are given by Eq. (28), repeated here for convenience: bxx ¼ b11 cos2 ðhÞ þ b22 sin2 ðhÞ byy ¼ b11 sin2 ðhÞ þ b22 cos2 ðhÞ bxy ¼ 2 cosðhÞ sinðhÞðb11

b22 Þ

Example Problem 8 Plot the effective properties listed below for a unidirectional hj graphite– epoxy laminate, for all fiber angles ranging 0j V h V 90j: (a) Effective Young’s moduli, Exx and Eyy. (b) Effective Poisson’s ratio, vxy and vyx. (c) Effective shear modulus Gxy. (d) Coefficients of mutual influence of the first kind, gxy,xx and gxy,yy. (e) Coefficients of mutual influence of the second kind, gxx,xy and gyy,xy. (f) Coefficients of thermal expansion, axx, ayy, and axy. (g) Coefficients of moisture expansion, bxx, byy, and bxy. Use the material properties listed in Table 3 of Chap. 3. Solution. Plots of the effective elastic properties for unidirectional hj graphite–epoxy laminates are presented in Figs. 8–14. 5 FAILURE OF UNIDIRECTIONAL COMPOSITES REFERENCED TO THE PRINCIPAL MATERIAL COORDINATE SYSTEM Fundamental material strengths for a unidirectional composite were discussed in Sec. 5 of Chap. 3. Recall that material strengths are measured under simple states of stress, usually either a uniaxial stress state or a pure shear stress state. Also, high-performance fibers are often very brittle, whereas modern polymeric matrices are fairly ductile. Consequently, most fiberreinforced polymeric composites exhibit brittle behavior in the 1-direction, qualitatively similar to Fig. 12(a) of Chap. 3, but relatively ductile behavior in the 2- and 3-directions, qualitatively similar to Fig. 12(b) and c of Chap. 3. A brief summary of experimental methods used to measure properties in the 1–2 plane is provided in Appendix B.

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Figure 8 A plot of the effective Young’s moduli Exx and Eyy for unidirectional graphite–epoxy laminates and fiber angles ranging over 0j V h V 90j.

Figure 9 A plot of the effective Poisson ratios vxy and vyx for unidirectional graphite–epoxy laminates and fiber angles ranging over 0j V h V 90j.

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Figure 10 A plot of the effective shear modulus Gxy for unidirectional graphite–epoxy laminates and fiber angles ranging over 0j V h V 90j.

Figure 11 A plot of the effective coefficients of mutual influence of the first kind gxy,xx and gxy,yy for unidirectional graphite–epoxy laminates and fiber angles ranging over 0j V h V 90j.

Copyright © 2004 by Marcel Dekker, Inc. All Rights Reserved.

Figure 12 A plot of the effective coefficients of mutual influence of the second kind gxx,xy and gyy,xy for unidirectional graphite–epoxy laminates and fiber angles ranging over 0j V h V 90j.

Figure 13 A plot of the effective coefficients of thermal expansion axx, ayy, and axy for unidirectional graphite–epoxy laminates and fiber angles ranging over 0j V h V 90j.

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Figure 14 A plot of the effective coefficients of moisture expansion bxx, byy, and bxy for unidirectional graphite–epoxy laminates and fiber angles ranging over 0j V h V 90j.

In this section, we will discuss failure criteria that are commonly used to predict failure of composites under general 3-D or 2-D states of stress. It will be assumed that the composite is brittle in the 1-direction, but ductile in the 2- and 3-directions. That is, in the fiber direction, ‘‘failure’’ is assumed to involve fracture, whereas transverse to the fibers, ‘‘failure’’ is assumed to involved yielding, defined on the basis of a % strain offset. Both orthotropic and transversely isotropic composites will be considered. For the orthotropic case, failure predictions will be based on the combinations of the following fundamental material strengths: 



fT fC , r11 . Fracture stress in the 1-direction: r11 yT yC yT yC y y Yield stress in the 2- and 3-directions: r22 , r22 , r33 , r33 , s12, s13, y s23.

If the composite is transversely isotropic, then the number of independent material strengths involved is reduced, since in this case: yT ryT 22 ¼ r33

yC ryC 22 ¼ r33

y sy12 ¼ s13

Recall that all of these strengths may vary with temperature.

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The need for ‘‘failure criteria’’ in engineering analysis and design is often misunderstood. In essence, the objective of any failure criterion is to account for potential coupling effects of individual stress components on the yielding and/or fracture phenomenon. This statement applies to both anisotropic and isotropic materials. To explain what is meant by the phrase ‘‘coupling effects between individual stress components,’’ consider the two different tests of unidirectional composite shown in Fig. 15. A composite subjected to uniaxial stress r11 is shown in Fig. 15(a). This is, of course, the very state of stress used fT (as during the measurement of the fundamental material strength r11

Figure 15 An illustration of what is meant by ‘‘coupling effects’’ of stress on the failure phenomenon.

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discussed in Sec. 5 of Chap. 3). In this case then, we do not need to invoke any failure criterion to predict when failure occurs: failure occurs when r11 is fT increased to r11 , by definition. However, consider a more general state of stress, such as the state of plane stress shown in Fig. 15(b). In this test, two additional components of stress (r22 and s12) are applied prior to the application of r11. It is assumed that the combination of r22 and s12 does not cause failure prior to the application of r11. While maintaining r22 and s12 at constant values, stress r11 is increased until failure occurs. It is for this more general state of stress that a failure criterion is required. That is, does the application r22 and/or s12 change the value of r11 at which failure occurs? ‘‘Coupling effects’’ refer to the fact that the application of r22 and/or s12 often does alter the value of r11 necessary to cause failure. If the test represented by fT Fig. 8(b) is conducted and r11p r11 at failure, then a coupling effect has occurred, i.e., the presence of r22 and s12 has changed the value of r11 necessary to cause failure. Conversely, if the test depicted in Fig. 15(b) is performed and fT r11=r11 at failure, then no coupling has occurred. Experimental measurements have shown that the coupling phenomenon is much more significant in some materials than it is in others. This is unfortunate because it implies that it is not possible to develop a ‘‘universal’’ failure criterion that can be applied to all materials. Furthermore, there is no way of predicting a priori whether coupling effects are pronounced for a given material or not. For metals, the general trend is that coupling effects are less pronounced in brittle materials (such as cast irons) than in ductile materials (such as aluminum alloys). The question as to whether this general trend holds in the case of composites is complicated by the fact that composites are (usually) brittle in the fiber direction but ductile transverse to the fiber. It is generally accepted that coupling effects do exist in composites, but at the present state of the art, one is well advised to perform experimental measurements to evaluate the level of coupling for each composite material system of interest. Many failure criteria applicable to composites have been proposed in the literature. Three of the most common will be described in the following subsections: the maximum stress failure criterion, the Tsai–Hill failure criterion, and the Tsai–Wu failure criterion. As will be seen, the maximum stress criterion does not account for coupling effects, whereas potential coupling effects are accounted for in the Tsai–Hill and Tsai–Wu criteria. 5.1 The Maximum Stress Failure Criterion According to this criterion, a given state of stress will not cause a unidirectional composite to fail if all of the following nine inequalities are satisfied: fT 1*rfC 11 < r11 < r11

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ð50Þ

ðandÞ

1*ryC 22

< r22 < ryT 22

ðandÞ 1*ryC 33

< r33 < ryT 33

ðandÞ

js12 j < sy12 ðandÞ

js13 j < sy13 ðandÞ

js23 j < sy23 According to the maximum stress failure criterion, failure (i.e., fracture in the fiber direction or yield-like behavior transverse to the fiber) is predicted strictly on the basis of individual stress components. Thus, failure is assumed to be independent of any coupling effects between individual stress components. In the case of plane stress (r33=s13=s23=0), the maximum stress failure criterion reduces to the following five inequalities: fT 1*rfC 11 < r11 < r11

ðandÞ yT 1*ryC 22 < r22 < r22

ð51Þ

ðandÞ

js12 j < sy12 The maximum stress failure criterion is most commonly applied in the form of Eq. (51) since in most cases, an individual composite ply can be assumed to be in a state of plane stress. 5.2 The Tsai–Hill Failure Criterion The von Mises yield criterion is widely used to predict yielding of isotropic metals and metal alloys.* In 1950, Hill [2] proposed a modified version of the

* The von Mises yield criterion is also mathematically equivalent to the ‘‘octahedral shear stress’’ and ‘‘distortional energy’’ yield criteria.

Copyright © 2004 by Marcel Dekker, Inc. All Rights Reserved.

von Mises criterion for use with orthotropic metals. Subsequently, Tsai [3] applied this method to predict failure of unidirectional polymeric composites, and the resulting theory is now known within the polymeric composites community as either the ‘‘Tsai–Hill’’ failure criterion or as the ‘‘quadratic’’ failure criterion. For general 3-D states of stress, the Tsai–Hill criterion predicts that failure of an orthotropic composite will not occur if the following inequality is satisfied: ðr11 Þ2 ðr22 Þ2 ðr33 Þ2 ðs23 Þ2 ðs13 Þ2 ðs12 Þ2 þ þ þ  fT 2  2  2  y 2 þ  y 2 þ  y 2 r11 s23 s13 s12 ryT ryT 22 33 3 2 1 6 1 r11 r22 4  2 þ  2 fT r11 ryT 22 2

6 1 r11 r33 4  2 rfT 11

1 7  2 5 ryT 33

ð52Þ

3

1 1 7  2 þ  2 5 ryT ryT 22 33 3

2

1 1 7 6 1 r22 r33 4  2 þ  2 þ  2 5 < 1 fT yT yT r11 r22 r33 In the case of plane stress conditions (r33=s13=s23=0), the Tsai–Hill criterion reduces to: 3 2 ðr11 Þ2 ðr22 Þ2 ðs12 Þ2  fT 2 þ  2 þ  y 2 r11 s12 ryT 22

1 6 1 r11 r22 4  2 þ  2 fT yT r11 r22

1 7  2 5 < 1 yT r33

ð53Þ

It is interesting to note that according to the Tsai–Hill failure criterion, failure of orthotropic composites is sensitive to the out-of-plane strength term yT ), even for plane stress conditions. (r33 yT yT =r22 ), then Eq. If the composite is transversely isotropic (that is, if r33 (53) reduces to: ðr11 Þ2 ðr22 Þ2 ðs12 Þ2  fT 2 þ  2 þ  y 2 r11 s12 ryT 22

r11 r22  fT 2 < 1 r11

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ð54Þ

A potential advantage of the Tsai–Hill failure criterion is that coupling effects between individual stress components are accounted for, in contrast with the maximum stress failure criterion. On the other hand, most composites exhibit significantly different failure strengths in tension and compression (as indicated in Table 3 of Chap. 3), and so a shortcoming of the Tsai–Hill failure criterion is that it does not directly account for these differences. That is, an implicit assumption of the Tsai–Hill criterion (as well as the original von Mises criterion) is that failure strengths in tension and yT fT compression have equal magnitudes. Hence, only tensile strengths r11 , r22 , yT and r33 appear in Eqs. (53) and (54). Differences in tensile and compressive strengths can be accounted for ‘‘artificially’’ in the Tsai–Hill criterion by using the appropriate compressive strength if a stress component involved is compressive. Suppose, for example, that a failure prediction is required for a transversely isotropic composite subjected to three stress components r11, r22, and s12, and also that r11 is tensile but r22 is compressive. In such a case, the differences in tensile/compressive strengths can be accounted for by using fT the tensile strength in the 1-direction, r11 , but the compressive strength in yC the 2-direction, r22 . While the Tsai–Hill criterion can be modified in this way to account for differences in tensile and compressive strengths, it would be ideal if a failure criterion was available that accounts for both coupling effects and differences in tensile and compressive strengths ‘‘automatically.’’ One such criterion is the Tsai–Wu criterion, described in the next paragraph.

5.3 The Tsai–Wu Failure Criterion Tsai and Wu [4] developed their criterion by postulating that the strength of a unidirectional composite can be treated mathematically as a tensoral quantity, in much the same way as stress or strain tensors. For general 3-D states of stress, the Tsai–Wu criterion predicts that failure will not occur if the following inequality is satisfied: X1 r11 þ X2 r22 þ X3 r33 þ X11 r211 þ X22 r222 þ X33 r233 þ X44 s223 þX55 s213 þ X66 s212 þ 2X12 r11 r22 þ 2X13 r11 r33 þ 2X23 r22 r33 < 1

ð55Þ

Most of the constants that appear in this inequality (i.e., X1, X2, X3, X11, etc.) fT , can be determined based on fundamental strength measurements (i.e., r11 yT yC fC r11 , r22 , r22 , etc.). First, consider a uniaxial strength measurement in which only stress r11 is applied (that is, a test in which r11 p 0, r22 =r33 =s23=s13=s12=0). Stress r11 is increased monotonically from zero until

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fT failure occurs. If r11 is tensile, then at the moment of failure, r11=r11 , and the Tsai–Wu criterion reduces to:

 fT 2 X1 rfT 11 þ X11 r11 ¼ 1

fC Conversely, if r11 is compressive, then at the moment of failure, r11= r11 fC (where the measured compressive strength, r11 , is treated as an algebraically positive number), and the Tsai–Wu criterion reduces to:

X1 rfC 11 þ X11



2 rfC 11 ¼ 1

Solving for X1 and X11, we find: X1 ¼

1 rfT 11

1 rfC 11

X11 ¼

1 fC rfT 11 r11

ð56Þ

Similarly, if failure is measured during two uniaxial stress tests in which only r22 is applied, we find: X2 ¼

1

1

ryT 22

ryC 22

X22 ¼

1

ð57Þ

yC ryT 22 r22

Using measurements obtained during two tests in which only r33 is applied: X3 ¼

1

1

ryT 33

ryC 33

X33 ¼

1

ð58Þ

yC ryT 33 r33

Three additional constants are determined using measured shear strengths: X44 ¼



1 sy23

2

X55 ¼



1 sy13

2

X66 ¼



1 sy12

2

ð59Þ

Only three coefficients remain to be determined, X12, X13, and X23. Several methods of determining these coefficients have been suggested, but thus far, no one technique has gained widespread acceptance. Two methods that have been proposed will be discussed here. Conceptually, the most straightforward approach is through the use of additional biaxial testing. For example, X12 can be determined by conducting a biaxial test in which r11=r22=r and r33=s23=s13=s12=0. The magnitude of biaxial stresses (r) is increased until failure occurs (i.e., increased until either fracture or yielding occurs). For simplicity, let us assume that failure occurs due to yielding and denote the onset of yielding using the superscript y. At the moment of failure then, the stresses applied are r11=r22=ry and

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r33=s23=s13=s12=0. Substituting these values into the Tsai–Wu criterion and solving for X12 results in: i 1 h ð60Þ 1 ðX1 þ X2 Þry ðX11 þ X22 Þðry Þ2 X12 ¼ 2 2 ð ry Þ

At least conceptually, X13 and X23 can also be determined in a similar manner. Two additional biaxial tests to failure would be required, where in one test, r11=r33=r, and in the second test, r22=r33=r. These data would then allow the calculation of X13 and X23, respectively. In practice, however, these tests would be very difficult to perform. Since composites are usually quite thin, it is especially difficult to apply well-defined out-of-plane stress components (i.e., r33, s13, or s23). Hence, in most instances, determining X13 or X23 in this manner is impractical. A second approach is to assume that X12, X13, and X23 can be calculated as follows: 1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 X12 ¼ X11 X22 ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 fC yT yC 2 rfT 11 r11 r22 r22 1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 X13 ¼ X11 X33 ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð61Þ 2 fC yT yC 2 rfT 11 r11 r33 r33 1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 X23 ¼ X22 X33 ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 yT yC yT yC 2 r22 r22 r33 r33

The basis of this approach is that if Eq. (61) is enforced and isotropic yT yT yC yC fT fC strengths are assumed (i.e., if r11 =r11 =r22 =r22 =r33 =r33 =ry, and sy12 p ffiffi ffi ¼ sy13 ¼ sy23 ¼ ry = 3), then the Tsai–Wu criterion reduces to the original von Mises criterion for isotropic materials. This approach holds some intellectual appeal since it ‘‘makes sense’’ that a failure criterion proposed for use with an orthotropic material should reduce to a well-known isotropic yield criterion if isotropic strengths are assumed. It is also a convenient assumption since X12, X13, and X23 are now calculated using fundamental strength data and hence the need to perform any additional testing is avoided. However, there is little data available to assess the validity of these assumptions and so the accuracy of failure predictions obtained using this approach is unknown. As discussed earlier, in most practical applications, composites are subjected to a state of plane stress within the 1–2 plane. In this case, the Tsai–Wu criterion reduces to: X1 r11 þ X2 r22 þ X11 r211 þ X22 r222 þ X66 s212 þ 2X12 r11 r22 < 1

ð62Þ

Hence, in the plane stress case, six constants are involved, five of which can yT fT fC be calculated using readily available strength data (r11 , r11 , r22 , etc.). Only one problematic coefficient remains, X12. This term can be determined using

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an off-axis specimen (which is, in effect, a biaxial test). For example, suppose a uniaxial stress rxx is applied to a unidirectional composite specimen in which the fibers are oriented at h=45j with respect to the direction of loading. Under these conditions, the stresses in the 1–2 coordinate system are easily calculated: 9 38 8 9 2 > cos2 ð45jÞ sin2 ð45jÞ 2cosð45jÞsinð45jÞ = < rxx > < r11 = 7 6 r22 ¼ 4 sin2 ð45jÞ cos2 ð45jÞ 2cosð45jÞsinð45jÞ 5 0 > > : ; ; : s12 0 cosð45jÞsinð45jÞ cosð45jÞsinð45jÞ cos2 ð45jÞ sin2 ð45jÞ or

9 8 9 8 8 9 rxx cos2 ð45jÞ r11 > > > > > rxx =2 > > > > > = > < < = > < = ¼ r22 ¼ rxx =2 rxx sin2 ð45jÞ > > > > > > > > > > : ; > ; > ; : : s12 rxx =2 rxx cosð45jÞsinð45jÞ

The strength of the 45j off-axis specimen is measured by increasing stress rxx until failure occurs. Let us assume that failure occurs due to yielding y . At failure, the and denote the stress level at which failure occurs as rxx=rxx y ply stresses are r11=r22= s12=rxx/2. Substituting these stresses into Eq. (62) and solving for X12, we find:

 1  X12 ¼ 4 ryxx 2ðX1 þ X2 Þ þ ryxx ðX11 þ X22 þ X66 Þ ð63Þ y 2 2ðrxx Þ

While this example has been based on a 45j off-axis specimen, a similar approach can be used with any hj off-axis specimen. From an analytical standpoint, the Tsai–Wu failure criterion is an improvement over the other two failure criteria considered. First, unlike the maximum stress failure criterion, the coupling effects between individual stress components are accounted for in the Tsai–Wu criterion. Second, unlike the Tsai–Hill criterion, differences in tensile and compressive strengths are automatically and naturally accounted for via the X1, X11, X2, X22, X3, and X33 terms.

6 FAILURE OF UNIDIRECTIONAL COMPOSITES REFERENCED TO AN ARBITRARY COORDINATE SYSTEM In this section, the three failure criteria introduced in Sec. 5 will be used to predict failure of unidirectional composites subjected to a state of plane stress, where stress components rxx, ryy, and sxy are referenced to an arbitrary x–y coordinate system. There are, of course, an infinite number of different combinations of rxx, ryy, and sxy that (collectively) define a state of

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plane stress. For illustrative purposes, two simple stress states will be considered: first, a state of uniaxial stress (i.e., rxx p 0, ryy=rxy=0), and second, a state of pure shear (sxy p 0, rxx=ryy=0). Numerical results for a unidirectional graphite–epoxy composite will be used to facilitate these comparisons. The following failure strengths are taken from Table 3 of Chap. 3 and are typical for graphite–epoxy at room temperature: rfT 11 ¼ 1500 MPa

ryT 22 ¼ 50 MPa

rfC 11 ¼ 1200 MPa

ryC 22 ¼ 100 MPa

sy12 ¼ 75 MPa

6.1 Uniaxial Stress An off-axis composite ply subjected to a uniaxial stress rxx has been previously shown in Fig. 6. The stresses induced in the 1–2 coordinate system by stress rxx can be determined using Eq. (20) of Chap. 2: 9 2 8 9 38 r11 > rxx > cos2 ðhÞ sin2 ðhÞ 2 cosðhÞsinðhÞ > > > > > > = 6 < < = 7 2 2 7 0 r22 ¼ 6 sin ð h Þ cos ð h Þ 2 cos ð h Þsin ð h Þ 5> 4 > > > > > > > ; : : ; 2 2 0 s12 cosðhÞsinðhÞ cosðhÞsinðhÞ cos ðhÞ sin ðhÞ or equivalently:

r11 ¼ rxx cos2 ðhÞ

r22 ¼ rxx sin2 ðhÞ

s12 ¼

rxx cosðhÞsinðhÞ

ð64Þ

6.1.1 Maximum Stress Criterion Substituting Eq. (64) into Eq. (50), we obtain: rfC rfT 11 11 < rxx < 2 cos ðhÞ cos2 ðhÞ ryC 22 2

sin ðhÞ

ðandÞ

< rxx
> > > = = = < < < eyy ¼ eoyy þ z jyy > > > > > > > > > > ; ; ; > : : co > : cxy j xy xy

ð12Þ

Equation (12) is the primary result we require for present purposes from classical thin-plate theory. It allows us to calculate the infinitesimal in-plane strains (exx,eyy,cxy) induced at any position z through the thickness of the plate, based on the midplane strains (eoxx ; eoyy ; coxy ) and midplane curvatures (jxx,jyy,jxy). Note that this result is based strictly on the Kirchhoff hypothesis. We have made no assumptions regarding the mechanism(s) that caused the flat plate to deform. Hence, Eq. (12) is valid if the plate is deformed by a change in temperature, a change in moisture content, externally applied mechanical loads, or any combination thereof. Also, we have made no assumptions regarding material properties. Equation (12) is therefore valid for isotropic, transversely isotropic, orthotropic, or anisotropic thin plates. Example Problem 1 A thin plate with a thickness of 1 mm is subjected to mechanical loads, a change in temperature, and a change in moisture content. Strain gages are used to measure the surface strains induced in the plate. They are found to be: at z ¼

t=2 ¼

0:5 mm : exx ¼ 250 Am=m;

at z ¼ þt=2 ¼ þ0:5 mm : exx ¼

eyy

250 Am=m; eyy

1500 Am=m; cxy ¼ 1000 Arad 1100 Am=m; cxy ¼ 800 Arad

What midplane strains and curvatures are induced in the plate?

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Solution. To solve this problem, we simply apply Eq. (12) to both surfaces of the plate. For example, using the measured strains for exx, we have: at z ¼

t=2 ¼

0:0005 m :

at z ¼ þt=2 ¼ þ0:0005 m :

exx ¼ 250 Am=m ¼ eoxx exx ¼

250 Am=m ¼

eoxx

ð0:0005Þjxx þ ð0:0005Þjxx

Solving simultaneously, we find: eoxx ¼ 0 Am=m;

jxx ¼

0:50 rad=m

Using a similar approach utilizing the measured values for eyy and cxy, we find: eoyy ¼ coxy

1300 Am=m;

¼ 900 Arad;

jyy ¼ 0:40 rad=m jxy ¼

0:20 rad=m

3 PRINCIPAL CURVATURES In Sec. 2, we invoked the Kirchhoff hypothesis, according to which it is assumed that a straight line that is initially perpendicular to the midplane of the plate remains straight and perpendicular to the midplane after deformation. The Kirchhoff hypothesis has ultimately allowed us to calculate the inplane strains referenced to the x–y coordinate system (exx, eyy, and cxy) induced at any position z through the thickness of a thin plate, using either Eq. (11) or Eq. (12). These equations are valid for any combination of midplane strains (eoxx, eoyy, and coxy) and midplane curvatures (jxx, jyy, and jxy). In this section, we will consider a special case. Specifically, we will consider a state of deformation in which the midplane strains are zero: eoxx ¼ eoyy ¼ coxy ¼ 0. In this special case, Eq. (12) becomes: 8 9 8 9 > > = = < exx > < jxx > eyy ¼ z jyy ð13Þ > > > > ; ; : : cxy jxy

This state of deformation is known as pure bending. When a thin plate is in a state of pure bending, all midplane strains are zero (eoxx ¼ eoyy ¼ coxy ¼ 0) and the midplane of the plate is called the neutral surface. Equation (13) gives the in-plane strains referenced to the x–y coordinate system. Referring to Fig. 6, suppose we wish to express these strains relative to a new x V–y Vcoordinate system, obtained by rotating through angle a about the z-axis. As noted in Sec. 2, the Kirchhoff hypothesis implies that cxz=cyz=0. Therefore, the z-axis is a principal strain axis. Consequently, we can rotate inplane strains from the x–y coordinate system to the new x V–y V coordinate

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Figure 6 In-plane coordinate system xV–y V, obtained by rotating through angle a about the z-axis.

system using Eq. (44) of Chap. 2 (developed in Sec. 13 of Chap. 2), repeated here for convenience: 9 2 8 ex Vx V > > cos2 ðaÞ > > > > = 6 < ey Vy V 6 ¼6 sin2 ðaÞ > > 4 > > c x Vy V > > ; : cosðaÞ sinðaÞ 2

sin2 ðaÞ cos2 ðaÞ cosðaÞ sinðaÞ

9 38 exx > > > > > = < 7> 7 eyy 2cosðaÞ sinðaÞ 7 5> > > c > > ; : xy > 2 2 cos ðaÞ sin ðaÞ 2 2cosðaÞ sinðaÞ

ð2:44Þ

Substituting Eq. (13) into Eq. (44) of Chap. 2, we can write: 8 8 9 9 < ex Vx V = < jx Vx V = ey Vy V ¼ z jy Vy V :c : ; ; jx Vy V x Vy V

ð14Þ

where:

8 9 2 jx Vx V > > cos2 ðaÞ > > = 6 < jy Vy V ¼6 sin2 ðaÞ > jx Vy V > > 4 > ; : cosðaÞ sinðaÞ 2

9 38 jxx > > > > < 7 jyy = cos2 ðaÞ 2cosðaÞ sinðaÞ 7 5> j > > ; : xy > 2 2 cosðaÞ sinðaÞ cos ðaÞ sin ðaÞ 2 sin2 ðaÞ

2cosðaÞ sinðaÞ

ð15Þ

Note that midplane curvatures in the x V–yV coordinate system are related to curvatures in the x–y coordinate system by means of the familiar transformation matrix [T]. This reveals that midplane curvatures can be treated as a second-order tensor, and can be transformed from one coordinate system to

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another in exactly the same way as the strain tensor (or stress tensor) is transformed. The in-plane principal strains and the orientation of the principal strain coordinate system can also be determined using Eqs. (47) and (48) of Chap. 2, respectively, repeated here for convenience: s ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi  exx þ eyy exx þ eyy 2  cxy 2 ð2:47Þ F þ ep 1 ; ep 2 ¼ 2 2 2   cxy 1 hpe ¼ arctan ð2:48Þ 2 exx eyy Substituting Eq. (13) into Eq. (47) of Chap. 2, we find that for pure bending, the in-plane principal strains are given by: 2 3 sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi  2   ffi 2 ðj þ j Þ ðj þ j Þ j xx yy xx yy xy 5 F ð16Þ þ e p 1 ; e p 2 ¼ z4 2 2 2

Substituting Eq. (13) into Eq. (48) of Chap. 2, we find that the orientation of the principal strain coordinate system is given by:   jxy 1 hpe ¼ arctan ð17Þ 2 jxx jyy

Noting that jxx, jyy, and jxy are midplane values, Eq. (16) shows that principal strains are linear functions of z. In contrast, Eq. (17) shows that, for the case of pure bending, the orientation of the principal strain coordinate system is constant and does not vary with through-thickness position, even though the principal strains do vary with z. A simplified expression for the principal strains is obtained by writing Eq. (16) as: ep1 ¼ jp1 ep2 ¼ jp2

where jp1 and jp2 are called principal curvatures and are given by: 2 3 sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi  2   ffi 2 ðj þ j Þ ðj þ j Þ j xx yy xx yy xy 5 F jp1 ; jp2 ¼ z4 þ 2 2 2

ð18Þ

ð19Þ

For the case of pure bending, the principal curvatures occur in the same coordinate system as the principal strains. Hence, Eq. (17) gives the orientation of the coordinate system in which the principal curvatures exist. Because shear strain is zero in the principal strains coordinate system, jp1p2=0 as well. A physical interpretation of the preceding results can be obtained through sketches of deformed strain elements parallel to the x–y plane, as

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was done in Sec. 2.13 (in particular, refer to Sample Problem 2.9). A thin plate reference to an x–y–z coordinate system is shown in Fig. 7a. A rectangular 3D element cut out of this plate by two pairs of planes parallel to the x–z and y–z planes is also shown. The dimensions of the element in the x-direction and ydirection are dx and dy, respectively, whereas the height of the element equals the plate thickness t. Assuming this plate is subjected to pure bending, then the strains induced at any position z, relative to the x–y coordinate system, can be calculated using Eq. (13). Consider as representative examples three 2D strain elements parallel to the x–y plane and located at positions defined by: 

z=t/2 (element a–b–c–d, shown in Fig. 7b) z=0 (element e–f–g–h, shown in Fig. 7c)  z=+t/2 (element i–j–k–l, shown in Fig. 7d). 

In each case, we imagine a 2D strain element whose sides are parallel to the x-axis and y-axis prior to deformation. As the plate is deformed, the length of the element sides increases or decreases, in accordance with the algebraic sign of strains exx and eyy, and the angle between adjacent faces of

Figure 7 Illustration of strains induced at the three through-thickness positions z = t/2, 0, and +t/2 by pure bending (deformations shown greatly exaggerated for clarity).

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the element changes from p/2 rad (i.e., 90j), in accordance with the algebraic sign of cxy. First, consider strain element i–j–k–l, located at z=+t/2 (Fig. 7d). For a state of pure bending, the strains induced at this through thickness position are given by Eq. (13): exx jz ¼ t=2 ¼ tjxx =2 eyy jz ¼ t=2 ¼ tjyy =2 cxy jz ¼ t=2 ¼ tjxy =2

Assume for illustrative purposes that all curvatures are positive (jxx,jyy, jxy > 0). This implies that all strains induced at z=+t/2 are algebraically positive. A sketch of a deformed element that corresponds to these assumptions is shown (not to scale) in Fig. 7d. In the deformed condition, the lengths of the element sides have increased because exx and eyy are positive, and angle j–i–l has decreased because cxy is positive. Now consider strain element e–f–g–h, located at the midplane of the plate z=0. Because we have assumed a state of pure bending, the strains at the midplane are zero, and consequently element e–f–g–h is not deformed, as shown in Fig. 7c. Finally, consider strain element a–b–c–d, located at z= t/2 (Fig. 7b). Using Eq. (13), the strains induced at this position are: exx jz ¼ eyy jz ¼ cxy jz ¼

¼ tjxx =2 ¼ tjyy =2 t=2 ¼ tjxy =2 t=2 t=2

Because we have already assumed that all midplane curvatures are positive, these results show that all strains induced at z= t/2 are algebraically negative. A sketch of the deformed element that corresponds to this condition is shown (not to scale) in Fig. 7b. Note that in this case, the lengths of the element sides have decreased because exx and eyy are negative, and angle b–a–d has increased because cxy is negative. The deformed 2D strains elements shown in Fig. 7b–d are assembled to create a sketch of the entire 3D element in Fig. 8. Note that, in accordance with the Kirchhoff hypothesis, the four line segments that define the vertical edges of the element (line segments a–e–i, b–f–j, c–g–k, and d–h–l) remain straight lines after deformation. However, the transverse planes are no longer plane after deformation. For example, plane b–j–h–c has been twisted during deformation of the plate. Inspection of Figs. 7b–d and 8 reveals that transverse planes do not remain plane after deformation due to curvature jxy. That is, if jxy p 0, shear strain cxy varies with through-thickness position z, in accordance with Eq. (13). It is this through-thickness variation in cxy that leads to twisting of the transverse planes. For this reason, jxy is known as the twist curvature.

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Figure 8 A 3D strain element assembled from the 2D deformed elements shown in Fig. 7b–d (deformations shown greatly exaggerated for clarity).

We will now repeat this process for a rectangular 3D element referenced to the principal strain coordinate system, as shown in Fig. 9a. Once again, we assume that the plate is subjected to pure bending. The principal strains induced at any position z can therefore be calculated using Eq. (18). We consider three 2D strain elements located at through-thickness positions z= t/2, 0, and +t/2. A 2D sketch of the deformed strain elements at these three positions is shown in Fig. 9b–d. Because the element is aligned with the principal strain coordinate system, no shear strain is induced in any element (i.e., all corner angles equal p/2 rad before and after deformation). Assuming, for illustrative purposes, that both principal strains are positive (jp1,jp2>0), the principal strains induced at z=+t/2 are tensile (Fig. 9d), whereas the principal strains induced at z= t/2 are compressive (Fig. 9b). No deformations occur at z=0 because we have assumed pure bending and the midplane is therefore the neutral surface. The deformed 2D strains elements shown in Fig. 9b–d are assembled to create a sketch of the deformed 3D element in Fig. 10. As before, the four line segments that define the vertical edges of the element (line segments a–e–i, b–f–j, c–g–k, and d–h–l) remain straight lines after deformation. However, in contrast to Fig. 8, the planes in which these line segment lie remain plane after deformation. Twisting of these transverse planes does not occur. When referenced to the coordinate system in which the principal curvatures exist, the transverse planes of the strain element simply rotate about the neutral surface. A summary of the results presented in this section is as follows. We have found that curvatures can be treated as second-order tensors, and can be rotated from one coordinate system to another using the same process as that

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Figure 9 Illustration of principal strains induced at the three through-thickness positions z= t/2, 0, and +t/2 by pure bending; compare with Fig. 7 (deformations shown greatly exaggerated for clarity).

Figure 10 A 3D strain element assembled from 2D deformed elements referenced to the principal strain coordinate system; compare with Fig. 8 (deformations shown greatly exaggerated for clarity).

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used to transform the strain or stress tensors. For a thin plate governed by the Kirchhoff hypothesis, midplane curvatures transform according to Eq. (15). In general, three midplane curvatures are induced in a thin plate: jxx, jyy, and jxy. Curvature jxy is called the twist curvature because it represents a twisting of a plane transverse to the midplane of the plate. The principal curvatures jp1 and jp1 are the maximum and minimum curvatures, respectively, induced at a given point in a plate. Equation (17) gives the orientation of the coordinate system in which the principal curvatures exists, and no twisting occurs in this coordinate system (the twist curvature equals zero in the principal coordinate system). For a thin plate in pure bending, the orientation of the principal strain coordinate system is constant through the thickness of the plate, and the principal curvatures are induced in this coordinate system. The reader should note that the results in this section are valid for the special case of pure bending. Some of the results presented above are not valid for the case of general nonuniform plate bending. For example, if the midplane is not the neutral surface (i.e., if eoxx,eoyy,coxy p 0), then it can be shown that the orientation of the principal strain coordinate system is not constant but rather varies as a function of z. However, even in this more general case, midplane curvatures transform according to Eq. (15), and principal curvatures are given by Eq. (19). A more detailed discussion of principal strains and curvatures under general conditions will not be presented because these topics are not of immediate interest. The results presented in this section for pure bending will be applied in Chap. 8, where the topic of composite beams is considered. 4 STANDARD METHODS OF DESCRIBING COMPOSITE LAMINATES A magnified edge view of a thin composite laminate that contains n plies is shown in Fig. 11. The figure is similar to the edge view of a thin plate shown in Fig. 4, except that now the ply interface positions are shown. The thickness of ply k will be denoted tk. The origin of the x–z axes lies at the geometrical midsurface of the laminate, and so the outer surfaces of the laminate exist at z= t/2 and z=+t/2, where t equals total thickness of the laminate. Total thickness of the laminate equals the sum of all ply thicknesses: t ¼ Snk ¼ 1 tk . Note that a ply interface does not necessarily exist at the midplane of the laminate, as indicated in Fig. 11. We will require a method of specifying the coordinate position of each ply interface with respect to the laminate midplane. By convention, we will denote the coordinate position of the outermost laminate surface in the negative z-direction as position z0 (i.e., z0 u t/2). Note that z0 is always an algebraically negative number. The coordinate position of the interface

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Figure 11 An edge view of an n-ply laminate showing ply interface positions.

between plies 1 and 2 is denoted z1, and z1=z0+t1. Similarly, the coordinate position of the interface between plies 2 and 3 is denoted z2, and z2=z1+t2, etc. For an n-ply laminate, the outermost surface of the laminate in the positive z-direction is will be labeled zn; obviously, zn=+t/2. Note that in all cases, zn is an algebraically positive number. Also note that the total thickness of the laminate equals (zn z0), and the thickness of an individual ply k is tk=(zk zk 1). For example, the thickness of ply 2 is t2=(z2 z1). We also need a method of consistently describing the stacking sequence of a composite laminate. That is, we need to develop a method of indicating the orientation of the principal material coordinate system of each ply with respect to the x-axis, and the order in which they appear. As discussed in previous chapters, a ply may contain unidirectional fibers, or may consist of a woven or braided fabric. In these latter two cases, there are two or more fiber directions present within each ply, although the orientation of the principal material coordinate system is always evident due to the symmetrical pattern of the fiber architecture. For simplicity in the following discussion, it will be assumed that all plies are composed of unidirectional fibers. In this case, the angle between the principal material coordinate system and the x-axis is equivalent to the angle between the fibers and the x-axis. Hence, in the discussion to follow, we will simply refer to the ‘‘fiber angle’’ in each ply. It should be understood that this angle actually refers to the orientation of the principal material coordinate system. This terminology is adopted simply

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because the phrase ‘‘fiber angle’’ is more concise than the phrase ‘‘principal material coordinate system angle.’’ To describe the stacking sequence of a laminate, we list fiber angles within square brackets ‘‘[’’ and ‘‘]’’. The fiber angle (in degrees) of ply 1 is listed first, followed by the fiber angle of ply 2, ply 3, etc. Each fiber angle will be separated by a slash ‘‘/’’. For example, a four-ply laminate consisting of plies with fiber angles of 0j, 45j, 20j, and 90j is shown in Fig. 12a. This laminate is denoted [0/45/ 20/90]T. The subscript ‘‘T’’ has been used to indicate that the ‘‘total’’ laminate has been described (i.e., a fiber angle is listed for all plies within the laminate within the square brackets). In practice, it is common to encounter laminates with 10, 20, 30, or (in unusual cases) even hundreds of plies. In such cases, it becomes very tedious to list all fiber angles within the laminate. Fortunately, for many reasons (some of which will be described later in this chapter), composite laminates are usually designed with some systematic pattern of fiber angles, which allows us to abbreviate the listing of ply fiber angles that appear within the laminate. It is easiest to introduce these abbreviations with a series of examples. Consider the eightply laminate shown in Fig. 12b. In this case, the fiber angles are (starting from ply 1) 0j, +45j, 45j, 90j, 90j, 45j, +45j, and 0j. This is an example of a symmetrical laminate because the fiber angles are symmetrical about the laminate midplane. This laminate is denoted [0/F45/90]s. The subscript ‘‘s’’ indicates that the four fiber angles listed appear symmetrically about the midplane, and hence a total of eight plies exist within the laminate, even though only four angles are listed. A nine-ply laminate containing fiber angles 0j, 30j, 60j, 10j, 45j, 10j, 60j, 30j, and 0j is shown in Fig. 12c. This laminate is symmetrical about the geometrical midplane, but because an odd number of plies is present, the midplane passes through the center of the 45j ply (ply 5). This laminate is denoted [0/30/ 60/10/45]s. That is, a bar is used to indicate that the midplane passes through the 45j ply, and hence ‘‘4 1/2’’ plies exist symmetrically about the midplane of this laminate. A 10-ply laminate containing fiber angles 20j, 30j, 30j, 20j, 0j, 0j, 20j, 30j, 30j, and 20j is shown in Fig. 12d. This laminate is symmetrical about the midplane, but also contains a symmetrical pattern within both halves of the laminate. In this case, the laminate is denoted [(20/ 30)s/0]s. The subscript ‘‘s’’ appears twice: first to indicate that fiber angles 20j and 30j appear symmetrically within one-half of the laminate, and the second to indicate that the entire laminate is symmetrical about the midplane. A final example is the 10-ply laminate shown in Fig. 12e. In this case, the fiber angles are 20j, 30j, 20j, 30j, 0j, 0j, 30j, 20j, 30j, and 20j. This laminate is denoted [(20/ 30)2/0]s, where the subscript ‘‘2’’ indicates that the fiber pattern listed within the parentheses occurs twice. Note that this laminate is similar but not identical to that shown in Fig. 12d.

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Figure 12 Edge view of several composite laminates illustrating stacking sequences: (a) [0/45/ 20/90]T, (b) [0/F45/90]s, (c) [0/30/ 60/10/45]s, (d) [(20/ 30)s/0]s, (e) [(20/ 30)2/0]s.

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Sample Problem 2 A ½(0/F30/90)2/45/20Šs laminate is fabricated using a graphite–epoxy material system. Each ply has a thickness of 0.125 mm. Determine the number of plies in the laminate, the total laminate thickness, and the z-coordinate of each ply interface. Solution. An ordered listing of all fiber angles that appear in the laminate is as follows: ½0j; 30j; 30j; 90j; 0j; 30j; 30j; 90j; 45j; 20j; 45j; 90j; 30j; 30j; 0j; 90j; 30j; 30j; 0Š z ply 1

z ply 10 ðmidplaneÞ

z ply 19

The laminate contains a total of 19 plies, the fiber angles appear symmetrically about the midplane of the laminate, and the midplane passes through the center of the 20j ply. Because all plies are made of the same composite material system, they all have the same thickness. The total laminate thickness is therefore t = 19 (0.125 mm)=2.375 mm. Ply interface positions are: z0= t/2= 1.1875 mm z3=z2+t3= 0.8125 mm z6=z5+t6= 0.4375 mm z9=z8+t9= 0.0625 mm z12=z11+t12=0.3125 mm z15=z14+t15=0.6875 mm z18=z17+t18=1.0625 mm

z1=z0+t1= 1.0625 mm z4=z3+t4= 0.6875 mm z7=z6+t7= 0.3125 mm z10=z9+t10=0.0625 mm z13=z12+t13=0.4375 mm z16=z15+t16=0.8125 mm z19=z18+t19=1.1875 mm

z2=z1+t2= 0.9375 mm z5=z4+t5= 0.5625 mm z8=z7+t8= 0.1875 mm z11=z10+t11=0.1875 mm z14=z13+t14=0.5625 mm z17=z16+t17=0.9375 mm

Note that the total laminate thickness equals the difference between z19 and z0, as expected: t=z19 z0=1.1875 mm ( 1.1875 mm)=2.375 mm.

5 CALCULATING PLY STRAINS AND STRESSES The theory developed to this point allows calculation of the elastic strains and stresses present at any through-thickness position within a multiangle composite laminate subjected to known midplane strains and curvatures. A summary of how strains and stresses are calculated is as follows. Laminate Description:

A composite laminate is described by specifying:



The laminate stacking sequence (i.e., the number of plies within a laminate and the fiber angles of each ply)  The material properties and thickness of each ply.

Copyright © 2004 by Marcel Dekker, Inc. All Rights Reserved.

Note that the plies are not necessarily all of the same material type. For example, some plies within a laminate may be of graphite–epoxy whereas others may be glass–epoxy. Once the stacking sequence and thickness of each ply have been specified, the total laminate thickness and interface positions throughout the laminate may be determined, as previously illustrated in Fig. 11. Also, the transformed reduced stiffness matrix ½ Q Š can be calculated for each ply in accordance with Eq. (31) of Chap. 5. Ply Strains: Strains are calculated using the laminate strains and curvatures (eoxx ; eoyy ; coxy) and (jxx,jyy,jxy), respectively, in accordance with the Kirchhoff hypothesis (Eq. (12)). For example, the strains induced in a distance zk from the laminate midplane are: 8 9 8 o 9 8 9  e > > > < exx > = < xx > < jxx > = = eyy  ¼ eoyy þ zk jyy > > > > > : ; : o > : ; ; cxy z¼z cxy j xy k

Note that these strains are referenced to the x–y coordinate system. If desired, these strains can be rotated from the x–y coordinate system to the ‘‘local’’ 1–2 coordinate system for each ply (defined by the ply fiber angle) using Eq. (44) of Chap. 2: 9 8 9 8 < exx = < e11 = eyy e22 ¼ ½TŠk ; : ; : cxy =2 z¼zk c12 =2 z ¼ zk 9 38 2 e sin2 ðhÞ 2cosðhÞ sinðhÞ cos2 ðhÞ > = < xx > 7 6 eyy ¼4 sin2 ðhÞ cos2 ðhÞ 2cosðhÞ sinðhÞ 5 > > ; : cosðhÞ sinðhÞ cosðhÞ sinðhÞ cos2 ðhÞ sin2 ðhÞ k cxy =2 z ¼ zk

j

j

j

Ply Stresses: Once ply strains are determined, ply stresses are calculated using Hooke’s law, as discussed in Sec. 5.2. For example, the stresses induced at a distance zk from the laminate midplane are calculated using Eq. (30) of Chap. 5: 9 8 9 8 2 3   exx DTaxx DMbxx > rxx > > > Q11 Q12 Q16    > > > > = < = <   7 6   7 6 eyy DTayy DMbyy  ¼ 4 Q12 Q22 Q26 5 ryy  > > > >  > > > > ; : ; : sxy z¼z Q16 Q26 Q66 z¼z cxy DTaxy DMbxy z ¼ z k

k

k

Note that the material properties used in this calculation (specifically, ½ Q Š, aij, and bij) are properties of the ply that exists at position zk, and in particular are

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functions of the ply fiber angle h. Because fiber angle generally varies from one ply to the next, these material properties also vary from one ply to the next. If desired, stresses can be rotated from the x–y coordinate system to the ‘‘local’’ 1–2 coordinate system for each ply using Eq. (20) of Chap. 2: 8 9 8 9 < r11 = < rxx = r ¼ ½TŠz¼zk ryy : 22 ; : ; sxy z¼zk s12 z¼zk

j

j

A numerical example that illustrates these calculations is presented in the following Sample Problem. Sample Problem 3

Assume that the panel considered in Sample Problem 1 is actually an eight-ply [0/30/90/ 30]s graphite–epoxy laminate. Assume that the laminate was initially flat and stress-free (i.e., ignore possible preexisting stresses/strains due to temperature and/or moisture changes). Determine the strains and stresses induced at each ply interface. Use material properties listed in Table 3 of Chap. 3, and assume that the thickness of each ply is 0.125 mm. From Sample Problem 1, the midplane strains and curvatures are:

Solution.

eoxx ¼ 0 Am=m eoyy coxy

¼

1300 Am=m

¼ 900 Arad

jxx ¼

0:50 rad=m

jxy ¼

0:20 rad=m

jyy ¼ 0:40 rad=m

To determine ply interface positions, first note that the total laminate thickness is: t ¼ ð8 pliesÞð0:125 mmÞ ¼ 1:0 mm ¼ 0:001 m

A total of nine ply interface positions must be determined because there are eight plies in the laminate. Following the numbering scheme discussed in Sec. 4 and referring to Fig. 11, ply interface positions are: z0= t/2= (0.001 m)= 0.000500 m z1=z0+t1= 0.000500 m+0.000125 m= 0.000375 m z2=z1+t2= 0.000375 m+0.000125 m= 0.000250 m z3=z2+t3= 0.000250 m+0.000125 m= 0.000125 m z4=z3+t4= .000125 m+0.000125 m=0.000000 m z5=z4+t5=0.000000 m+0.000125 m=0.000125 m z6=z5+t6=0.000125 m+0.000125 m=0.000250 m z7=z6+t7=0.000250 m+0.000125 m=0.000375 m z8=z7+t7=0.000375 m+0.000125 m=0.000500 m

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Strain Calculations. Strains are calculated using Eq. (12), and can be determined at any through-thickness position. Usually, strains of greatest interest are those induced at the ply interface locations. For example, strains present at the outer surface of ply 1 (i.e., strains present at zo= 0.000500 m) are: 9 9 8 8o 9 8 9 8 8 9 0 e e 0:50 rad=m > > > j > > > > > = < = < xx > = > < xx> = < xx> = < eyy 1300  10 6 m=m þ ð 0:000500 mÞ 0:40 rad=m ¼ eoyy þ z0 jyy ¼ > > > > > > ; : ; > ; > ; : o > : > : > > ; : cxy cxy z¼z0 jxy 0:20 rad=m 900  10 6 m=m

j

9 8 9 8 > < exx = = < 250 Am=m > eyy ¼ 1500 Am=m ; : > > ; : cxy z¼z 1000 Arad 0

j

Similarly, strains present at the interface between plies 1 and 2 (i.e., strains present at z1= 0.000375 m) are: 8 o 9 8 9 8 8 9 9 8 9 e > e 0:50 rad=m > 0 > j > > > > > < xx > < < xx > < xx > = > = = = < = eyy ¼ eoyy þ z1 jyy ¼ 1300  10 6 m=m þ ð 0:000375 mÞ 0:40 rad=m > > > > > > : : > : > > > ; > ; ; : > ; cxy z¼z1 : coxy ; jxy 900  10 6 m=m 0:20 rad=m

j

9 8 9 8 > < exx = = < 188 Am=m > eyy ¼ 1450 Am=m ; : > > cxy z¼z1 : 975 Arad ;

j

Strains present at all remaining interfaces are calculated in exactly the same fashion. Strains calculated at all ply interfaces are summarized in Table 1 and are plotted in Fig. 13. Note that all three strain components (exx, eyy, and cxy) are predicted to be linearly distributed through the plate thickness. This linear distribution is a direct consequence of the Kirchhoff hypothesis, which is a good approximation as long as the plate is ‘‘thin.’’ In fact, identical strain distributions would be predicted for any thin plate subjected to the midplane strains and curvatures specified in Sample Problem 1. For example, we would predict the identical strains if an aluminum plate were under consideration rather than a laminated composite plate. The strains listed in Table 1 and plotted in Fig. 13 are referenced to the global x–y coordinate system. As will be seen, knowledge of ply strains referenced to the local 1–2 coordinate system (defined by the fiber angle within each ply) is often required. Transformation of the strain tensor from one

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Table 1 Ply Interface Strains in a [0/ 30/90/30]s Graphite-Epoxy Laminate Subjected to the Midplane Strains and Curvatures Discussed in Sample Problem 1 z-coordinate (mm)

exx (Am/m)

0.500 0.375 0.250 0.125 0.0 0.125 0.250 0.375 0.500

eyy (Am/m)

250 188 125 62 0 62 125 188 250

1500 145 1400 1350 1300 1250 1200 1150 1100

cxy (Arad) 1000 975 950 925 900 875 850 825 800

Strains are referenced to the x–y coordinate system.

coordinate system to another was reviewed in Chap. 2 and, in particular, strains can be rotated from the x–y coordinate system to the 1–2 coordinate system using Eq. (44) of Chap. 2. In practice, strains are usually calculated at both the ‘‘top’’ and ‘‘bottom’’ interface for each ply. Example calculations for plies 1 and 2 are listed below: Ply 1. Because h1=0j, the x–y and 1–2 coordinate systems are coincident, and therefore the description of the strain tensor is identical in both coordinate systems. This can be confirmed through application of Eq. (44) of Chap. 2: Top interface: 38 9 9 ply 1 2 8 cos2 h1 sin2 h1 2 cos h1 sin h1 > exx > ply 1 > = < = < e11 > 7 6 2 2 7 e e22 yy ¼6 sin h cos h 2 cos h sin h 1 1 1 15 4 > > > > ; ; : : 2 c12 =2 z ¼ z0 cxy =2 z ¼ z0 2 cos h1 sin h1 cos h1 sin h1 cos h1 sin h1

j

9 ply 1 8 2 1 > = < e11 > 6 e22 ¼ 40 > > ; : c12 =2 z ¼ z0 0

j j

0 1 0

9 38 250 Am=m > ply1 0 > = < 7 1500 Am=m 05 > > ; : 1 ð1000 AradÞ=2 z ¼ z0

9 ply 1 9 ply 1 8 8 > > = = < e11 > < 250 Am=m > e22 ¼ 1500 Am=m > > > > ; ; : : c12 =2 z ¼ z0 1000 Arad z ¼ z0

j

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j

j

Figure 13 Through-thickness strain plots dictated by the midplane strains and curvatures discussed in Sample Problem 1. Strains are referenced to the x–y coordinate system.

Copyright © 2004 by Marcel Dekker, Inc. All Rights Reserved.

Bottom interface: 9 ply 1 2 8 cos2 h1 > = < e11 > 6 e22 ¼6 sin2 h1 4 > > ; : c12 =2 z ¼ z1 cos h sin h

sin2 h1

38 9 ply 1 exx > > = < 7 eyy cos2 h1 2cos h1 sin h1 7 5> > ; : cxy =2 z ¼ z1 2 2 cos h1 sin h1 1 1 cos h1 sin h1 9 9 ply 1 8 2 38 188 Am=m > ply 1 1 0 0 > > = < = < e11 > 6 7 e22 1450 Am=m ¼ 40 1 05 > > > > ; : ; : c12 =2 z ¼ z1 0 0 1 ð975 AradÞ=2 z ¼ z1 9 ply 1 9 ply 1 8 8 > > = = < e11 > < 188 Am=m > e22 ¼ 1450 Am=m > > > > ; : ; : c12 z ¼ z1 975 Arad z ¼ z1

j

j

j

2cos h1 sin h1

j

j

j

Ply 2. In this case, h2=30j and consequently the description of strain in the x–y and 1–2 coordinate systems differs substantially. Applying Eq. (44) of Chap. 2, we have: Top interface: 38 9 9 ply 2 2 8 cos2 h2 sin2 h2 2 cos h2 sin h2 > exx > ply 2 e > > 11 = < = < 7 6 eyy e22 ¼6 sin2 h2 cos2 h2 2 cos h2 sin h2 7 5 4 > > > > ; : ; : c12 =2 z ¼ z1 cxy =2 z ¼ z1 cos h sin h cos h sin h cos2 h sin2 h

j

2

2

2

2

2

j

2

3 8 9 ply 2 2 cos2 ð30jÞ sin2 ð30jÞ 2 cosð30jÞ sinð30jÞ > < e11 > = 6 7 e22 ¼6 cos2 ð30jÞ 2 cosð30jÞ sinð30jÞ 7 sin2 ð30jÞ 4 5 > > : ; 2 c12 =2 z ¼ z1 2 cosð30jÞ sinð30jÞ cosð30jÞ sinð30jÞ cos ð30jÞ sin ð30jÞ 9 8 188 Am=m > ply 2 > = <  1450 Am=m > > ; : ð975 AradÞ=2 z ¼ z1 9 9 ply 2 2 8 38 188 Am=m > ply 2 0:750 0:250 0:866 > > < = = < e11 > 7 6 e22 1450 Am=m ¼ 4 0:250 0:750 0:866 5 > > > > : ; ; : c12 =2 z ¼ z1 0:433 0:433 0:500 ð975 AradÞ=2 z ¼ z1

j

j

j

9 ply 2 9 ply 2 8 8 > > = < e11 > < 200 Am=m > = ¼ e22 1463 Am=m > > > > ; : ; : c12 z ¼ z1 931 Arad z ¼ z1

j

j

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j

Bottom interface: 9 ply 2 2 8 cos2 h2 e > > 11 = < 6 e22 ¼6 sin2 h2 4 > > ; : c12 =2 z ¼ z2 cos h sin h

j

2

sin2 h2 cos2 h2 2

cos h2 sin h2

38 9 exx > ply 2 > < 7 e = yy 2 cos h2 sin h2 7 5> ; :c =2> xy z ¼ z2 cos2 h2 sin2 h2 2 cos h2 sin h2

j

3 8 9ply 2 2  sin2 ð30jÞ 2 cosð30jÞ sinð30jÞ cos2 ð30jÞ > < e11 > = 7 6  e22 ¼6  sin2 ð30jÞ cos2 ð30jÞ 2 cosð30jÞ sinð30jÞ 7 5 4  > > : ; 2 c12 =2 z ¼ z2 2 cosð30jÞ sinð30jÞ cosð30jÞ sinð30jÞ cos ð30jÞ sin ð30jÞ 9 8 125 Am=m > ply 2 > = <  1400 Am=m > > ; : ð950 AradÞ=2 z ¼ z2 9 9 ply 2 2 8 38 125 Am=m > ply 2 0:750 0:250 0:866 > > = < = < e11 > 7 6 e22 1400 Am=m ¼ 4 0:250 0:750 0:866 5 > > > > ; : ; : c12 =2 z ¼ z2 0:433 0:433 0:500 ð950 AradÞ=2 z ¼ z2 9 ply 2 8 8 9 ply 2 > > = < e11 > < 155 Am=m > = e22 ¼ 1430 Am=m > > > > ; : : ; c12 z ¼ z2 846 Arad z ¼ z2

j

j

j

j

j

Ply strains referenced to the local 1–2 coordinate systems at all interface locations are summarized in Table 2 and plotted in Fig. 14. Comparing Figs. 12 and 13, it is apparent that the through-thickness strain distributions no longer appear linear or continuous when referenced to the 1–2 coordinate system. This is of course illusionary, in the sense that strains appear to be discontinuous only because the coordinate system used to describe the throughthickness strain is varied from one ply to the next.

Stress Calculations. Because strains are now known at all ply interface positions, we can calculate stresses at these locations using Eq. (30) of Chap. 5, with DT=DM=0. During these calculations, we will require the transformed reduced stiffness matrix for each ply. Using graphite–epoxy material properties from Table 2 of Chap. 3 and Eqs. (11) and (31) of Chap. 5, we find: For 0j plies: 2 3 170:9  109 3:016  109 0 6 7   9 9 7ðPaÞ Q 0j plies ¼ 6 0 4 3:016  10 10:05  10 5 9 0 0 13:00  10

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Table 2 Ply Interface Strains in a [0/-30/90/30]s Graphite-Epoxy Laminate Subjected to the Midplane Strains and Curvatures Discussed in Sample Problem 1 Ply number

z-coordinate (mm)

Ply 1

0.500 0.375 0.375 0.250 0.250 0.125 0.125 0.000 0.000 0.125 0.125 0.250 0.250 0.375 0.375 0.500

Ply 2 Ply 3 Ply 4 Ply 5 Ply 6 Ply 7 Ply 8

e11 (Am/m)

e22 (Am/m)

250 188 200 155 1400 1350 691 715 715 738 1250 1200 26 71 188 250

c12 (Arad)

1500 1450 1463 1430 125 63 596 585 585 574 62 125 1299 1267 1150 1100

1000 975 931 846 950 925 1686 1576 1576 1466 875 850 506 421 825 800

Strains are referenced to the 1–2 coordinate system local to individual plies.

For 30j plies: 2

  Q 30j

plies

For

plies

26:06  109

48:3  109

21:52  109

6 9 ¼6 4 26:06  10

For 90j plies: 2

  Q 90j

107:6  109

10:05  109

6 9 ¼6 4 3:016  10 0

30j plies: 2 107:6  109 6   9 Q 30j plies ¼ 6 4 26:06  10 48:3  109

27:22  109

3:016  109

170:9  109 0

26:06  109

27:22  109

21:52  109

48:3  109

3

7 21:52  109 7 5ðPaÞ 36:05  109 0 0 13:00  109

3

7 7ðPaÞ 5

48:3  109

3

7 21:52  109 7 5ðPaÞ

36:05  109

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Figure 14 Through-thickness strain plots dictated by the midplane strains and curvatures discussed in Sample Problem 1. Strains are referenced to the 1–2 coordinate system.

Stresses present at the outer surface of ply 1 (i.e., strains present at z0= 0.000500 m) can now be calculated: 8 9 8 9 ply 1 2 3 Q11 Q12 Q16 ply 1 > > < exx > < rxx > = = 6 7 eyy ¼ 4 Q12 Q22 Q26 5 ryy > > > :c > : ; ; sxy z ¼ z0 xy Q Q Q z ¼ z0 z ¼ z0

j

16

8 9 ply 1 2 170:9  109 > =

6 9 ¼6 ryy 4 3:016  10 > ; : > sxy z ¼ z0 0

j

26

66

j

3:016  109

10:05  109 0

9 9 ply 1 8 8 > > = = < 38:2 MPa > < rxx > ¼ 14:3 MPa ryy > > > > ; ; : : sxy z ¼ z0 13 MPa

j

j

38 9 > 250 Am=m > = 7< 7 0 5> 1500 Am=m > ; : 1000 Arad 13:00  109 0

To calculate stresses at the interface between plies 1 and 2 (i.e., at z1 = 0.000375 m), we must specify whether we are interested in the stresses within ply 1 or ply 2. That is, according to our idealized model, a ply interface is treated as a plane of discontinuity in material properties. Ply 1 ‘‘ends’’ at z = z1( ), whereas ply 2 ‘‘begins’’ at z=z1(+). Hence, the stresses within ply 1 at z = z1( ) are: 3 9 9 ply 1 2 Q 8 Q12 Q16 ply 1 8 exx > 11 r > > > xx = = < < 7 6 Q12 Q22 Q26 7 e ryy ¼6 yy 5 4 > > > > ; ; : : Q16 Q26 Q66 sxy z ¼ z1 cxy z ¼ z1

j

8 9 ply 1 2 170:9  109 > =

6 9 ryy ¼6 4 3:016  10 > > : ; sxy z ¼ z1 0

j

j

z ¼ z1

3:016  109

10:05  109 0

9 9 ply 1 8 8 > > = = < rxx > < 27:8 MPa > ryy ¼ 14:0 MPa > > > > ; : ; : sxy z ¼ z1 12:7 MPa

j

j

38 9 188 Am=m > > < = 7 7 0 5> 1450 Am=m > ; : 975 Arad 13:00  109 0

The stresses within ply 2 (a 30j ply) at z=z1(+) are: 3 9 ply 2 2 9 8 8 Q11 Q12 Q16 ply 2 > exx > > = = < rxx > < 7 6 7 e ¼6 Q Q Q ryy yy 22 26 5 4 12 > > > > ; ; : : sxy z ¼ z1 Q16 Q26 Q66 z ¼ z1 cxy z ¼ z1

j

j

j

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8 9 ply > =

ryy > ; : sxy >

j

2

z ¼ z1

2

107:6  109

6 9 ¼6 4 26:06  10 48:3  109

26:06  109

27:22  109

21:52  109

9 9 ply 2 8 8 > > < rxx > = < 29:5 MPa > = ryy ¼ 13:6 MPa > > > > : ; : ; sxy z ¼ z1 13:0 MPa

j

9 38 188 Am=m > > > > = < 7 1450 Am=m 21:52  109 7 5> > > 975 Arad > ; 9 : 36:05  10 48:3  109

Stresses are calculated at all remaining ply interfaces in exactly the same fashion. Ply interface stresses are summarized in Table 3 and are plotted in Fig. 15. Obviously, stresses are not linearly distributed through the thickness of the laminate, even when referenced to the global x–y coordinate system. In general, all stress components exhibit a sudden discontinuous change at all ply interface positions. The abrupt in stresses at ply interfaces is due to the  change  Q matrix from one ply to the next. In turn, the discontinuous change in the   discontinuous change in Q occurs because the fiber angle (in general) changes from one ply to the next. Indeed, in this example problem, the same fiber angle occurs in only two adjacent plies (namely, plies 4 and 5, both of which have a fiber angle of 30j), and inspection of Fig. 15 shows that the Table 3

Ply Interface Stresses in a [0/30/90/30]s Graphite-Epoxy Laminate Subjected to the Midplane Strains and Curvatures Discussed in Sample Problem 1 Ply number Ply 1 Ply 2 Ply 3 Ply 4 Ply 5 Ply 6 Ply 7 Ply 8

z-coordinate (mm) 0.500 0.375 0.375 0.250 0.250 0.125 0.125 0.000 0.000 0.125 0.125 0.250 0.250 0.375 0.375 0.500

rxx (MPa) 38.2 27.8 29.3 22.7 2.97 3.44 73.0 77.2 77.2 81.4 4.40 4.90 3.82 10.4 35.5 46.0

ryy (MPa) 14.3 14.0 13.6 14.4 239. 231 55.0 54.7 54.7 54.5 214 205 17.6 18.4 12.1 11.8

sxy (MPa) 13.0 12.7 13.0 10.1 12.4 12.0 59.4 60.4 60.4 61.4 11.4 11.0 1.20 4.03 10.7 10.4

Stresses are referenced to the x–y coordinate system.

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Q (MPa) 23.9 13.8 15.7 8.3 242 234 128 132 132 136 218 210 21.4 28.8 47.6 57.8

A (MPa)2 715 550 567 429 556 651 487 575 575 666 812 884 65.8 175 315 435

Figure 15 Through-thickness stress plots predicted for a [0/30/90/ 30]s graphite-epoxy laminate subjected to the midplane strains and curvatures discussed in Sample Problem 1. Stresses are referenced to the x–y coordinate system.

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interface between plies 4 and 5 is the only interface for which the stresses do not change abruptly. It has been mentioned that the linear strain distributions shown in Fig. 13 would be the same for any thin plate, regardless of the material the plate is made of. The same statement cannot be made for stress distributions. In general, through-thickness stress distributions for isotropic plates (e.g., an isotropic aluminum plate) are linear and continuous, unless high nonlinear stresses occur, in which case the stress distribution may not be linear but will nevertheless be continuous. In contrast, the stress distributions in laminated composite plates are usually discontinuous. The only conditions under which a linear and continuous stress distribution is encountered is when:  (a) the laminate is subjected to elastic stress/strain levels, and (b) when the Q matrix does not vary from one ply to the next (i.e., for unidirectional laminates in which the fiber angle does not vary from one ply to the next). Knowledge of ply stresses referenced to the local 1–2 coordinate system (defined by the fiber angle within each ply) is often required. Transformation of the stress tensor from one coordinate system to another was reviewed in Chap. 2 and, in particular, stresses can be rotated from the x–y coordinate system to the 1–2 coordinate system using Eq. (20) of Chap. 2. Typically, stresses are calculated at both the ‘‘top’’ and ‘‘bottom’’ interfaces for all plies. For example, rotation of the ply stresses that exist within ply 2 at the interface between plies 1 and 2 (i.e., at z=z1= 0.375 mm) proceeds as follows: 38 8 9ply 2 2 9 cos2 h2 sin2 h2 2 cos h2 sin h2 < rxx =ply 2

> = = < rxx > < exx > 6 7 eyy ¼ 4 Q21 Q22 Q26 5 ð20Þ ryy > > > > ; ; : : cxy k sxy k Q Q Q 61

62

66

k

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The subscript k in Eq. (20) indicates that the stresses, transformed reduced stiffness matrix, and strains are all for ply number k, where 1 V k V n and n = number of plies in the laminate. From Eq. (20), the stress rxx induced in ply k is:

 ðrxx Þk ¼ Q11 exx þ Q12 eyy þ Q16 cxy k

Substituting this relationship into Eq. (1a), we have: Z t=2

 Q11 exx þ Q12 eyy þ Q16 cxy k dz Nxx ¼

ð21Þ

t=2

The strains induced in ply k can be related to the midplane strains and curvatures via the Kirchhoff hypothesis, in accordance with Eq. (11) or Eq. (12). Substituting Eqs. (11) and (12) into Eq. (21), we obtain: Z t=2 n o Nxx ¼ Q11 eoxx þ Q12 eoyy þ Q16 coxy þ zQ11 jxx þ zQ12 jyy þ zQ16 jxy dz t=2

ð22Þ

We cannot integrate Eq. (22) directly because the integrand is a discontinuous function of z. That is, the transformed reduced stiffness terms Q11 , Q12 , and Q16 are all directly related to the ply material properties and fiber angle h (see Eq. (31) of Chap. 5). Because the ply material and/or fiber angle may change from one ply to the next, the transformed reduced stiffness terms also change, and hence are discontinuous functions of z. Note, however, that the midplane strains and curvatures are not functions of z, but instead are constants for a given laminate. Hence, they may be brought out from under the integral sign. Equation (15) can therefore be broken into six individual integrals: Z t=2 Z t=2



 o o Q11 k dz þ eyy Q12 k dz Nxx ¼ exx t=2

t=2

Z

t=2

Z þ jyy

t=2

þ coxy

t=2

t=2

Q16



k

Z

dz þ jxx

t=2

t=2

Q11



k

dz

ð23Þ



 z Q12 k dz þ jxy z Q16 k dz

Because the transformed stiffness terms are constant over each ply thickness, each of the six integrals in Eq. (23) can be evaluated in a ‘‘piecewise’’ fashion:  Z Z Z   z1   z2   z3 o Nxx ¼ exx Q11 1 dz þ Q11 2 dz þ Q11 3 dz þ : : : 

þ Q11



Z

n 1

z0 zn 1 zn

2



dz þ Q11

Z  n

z1 zn

z2

dz zn

1



Copyright © 2004 by Marcel Dekker, Inc. All Rights Reserved.

þeoyy 





þ Q12 þcoxy





þ Q16 þjxx 

þjyy

Z







Z



2

zn



zn

2

zn

Z

n 1



Q12





Z 1

Q16



Z

n 1

1

zn

2

1

zn zn

2

1

zn

2

z1

z3

dz

2

z1

zn

Z 



dz

n

zn

1

z2



zdz þ Q16

z1

zn 1 Z  z2 2

z1

zn zn

Z  n

2

1

z2



z1

zdz 1





zdz þ Q16

zn zn

z2

Z   zdz þ Q12 3 zdz

Z 

z2



zdz

n

zdz þ Q16



3

Z   zdz þ Q11 3

zn

Z 

 zdz þ Q12



Z   dz þ Q16 3

zn

Z   zdz þ Q12 n

z0

z3



dz þ Q12

zn 1 Z  z2

z1

1

Z 

n

zdz þ Q11

z0

z2

Z   zdz þ Q11 2

z1

Z  zn



dz þ Q16

z0

1

Z 

 dz þ Q16

z1

1

Z 



dz þ Q12

z0

Z  zn

Z   dz þ Q12 2

z1

1

1

Q11



z0

Z 

n 1



þ Q16

zn

Q16

  þ Q12 n þjxy

1

1

n 1



þ Q11

Q12

z1

Z 



Z  3

dz þ : : :

dz þ : : :

z3 z2

z3 z2

z3 z2

zdz þ : : :

zdz þ : : :

zdz þ : : : ð24Þ

Although Eq. (24) may appear daunting at first, closer inspection reveals that evaluation of Eq. (24) is actually a simple matter. All integrals that appear in Eq. (24) are of one of the following two forms, both of which are easily evaluated: Z zk dz ¼ ðzk zk 1 Þ zk

1

or

Z

zk zk

1

zdz ¼

1 2 ðz 2 k

z2k 1 Þ

Hence, evaluating all integrals that appear in Eq. (24), we obtain: n      Nxx ¼ eoxx Q11 1 ½z1 z0 Š þ Q11 2 ½z2 z1 Š þ Q11 3 ½z3 z2 Š þ : : : o   þ Q11 n ½zn zn 1 Š n      þeoyy Q12 1 ½z1 z0 Š þ Q12 2 ½z2 z1 Š þ Q12 3 ½z3 z2 Š þ : : :

Copyright © 2004 by Marcel Dekker, Inc. All Rights Reserved.

o   þ Q12 n ½zn zn 1 Š n    þcoxy Q16 1 ½z1 z0 Š þ Q16 2 ½z2 o   þ Q16 n ½zn zn 1 Š

  z1 Š þ Q16 3 ½z3

n       1 þ jxx Q11 1 z21 z20 þ Q11 2 z22 2 o    : þ : : þ Q11 n z2n z2n 1 n       1 þ jyy Q12 1 z21 z20 þ Q12 2 z22 2 o    : þ : : þ Q12 n z2n z2n 1 n       1 þ jxy Q16 1 z21 z20 þ Q16 2 z22 2 o    þ . . . þ Q16 n z2n z2n 1

z2 Š þ : : :

    z21 þ Q11 3 z23

z22

    z21 þ Q12 3 z23

z22



    z21 þ Q16 3 z23

z22





ð25Þ

Equation (25) can be simplified substantially by defining the following terms: n      A11 ¼ Q11 1 ½z1 z0 Š þ Q11 2 ½z2 z1 Š þ Q11 3 ½z3 z2 Š þ : : : o   þ Q11 n ½zn zn 1 Š n      A12 ¼ Q12 1 ½z1 z0 Š þ Q12 2 ½z2 z1 Š þ Q12 3 ½z3 z2 Š þ : : : o   þ Q12 n ½zn zn 1 Š n      A16 ¼ Q16 1 ½z1 z0 Š þ Q16 2 ½z2 z1 Š þ Q16 3 ½z3 z2 Š þ : : : o   þ Q16 n ½zn zn 1 Š       1 n Q11 1 z21 z20 þ Q11 2 z22 2 o    þ Q11 n z2n z2n 1       1 n Q12 1 z21 z20 þ Q12 2 z22 ¼ 2 o    þ Q12 n z2n z2n 1       1 n Q16 1 z21 z20 þ Q16 2 z22 ¼ 2 o    þ Q16 n z2n z2n 1

B11 ¼

    z21 þ Q11 3 z23

 z22 þ : : :

B12

    z21 þ Q12 3 z23

 z22 þ : : :

    z21 þ Q16 3 z23

 z22 þ : : :

B16

Copyright © 2004 by Marcel Dekker, Inc. All Rights Reserved.

With these definitions, Eq. (25), becomes: Nxx ¼ A11 eoxx þ A12 eoyy þ A16 coxy þ B11 jxx þ B12 jyy þ B16 jxy

ð26aÞ

Following an entirely analogous procedure for stress resultants Nyy and Nxy, it can be shown that: Nyy ¼ A21 eoxx þ A22 eoyy þ A26 coxy þ B21 jxx þ B22 jyy þ B26 jxy

ð26bÞ

þ B61 jxx þ B62 jyy þ B66 jxy

ð26cÞ

Nxy ¼

A61 eoxx

þ

A62 eoyy

þ

A66 coxy

where: Aij ¼ Bij ¼

n X  Qij k ðzk

zk 1 Þ

k¼1

n   2 1X Q z 2 k¼1 ij k k

z2k

1

ð27aÞ 

ð27bÞ

and i, j = 1, 2, or 6. Because subscripts i and j may take on one of three values, both Aij and Bij can be written as 3  3 matrices. Also, recall that the transformed reduced stiffness matrix is symmetrical (see Eq. (31) of Chap. 5). Hence, both Aij and Bij are also symmetrical: 3 2 3 2 A11 A12 A16 A11 A12 A16 7 6 7 6 Aij ¼ 4 A21 A22 A26 5 ¼ 4 A12 A22 A26 5 A61

2

A62

A66

3

2

A16

A26

A66

3

B11

B12

B16

B11

B12

B16

6 Bij ¼ 4 B21

B22

7 6 B26 5 ¼ 4 B12

B22

7 B26 5

B61

B62

B66

B16

B26

B66

Equation (26a) (26b) (26c) can be written in matrix form as follows: 8 o 9 e > > > > > xx > o > 9 2 8 > > 3> e > > yy > > N A A A B B B > > > xx 11 12 16 11 12 16 = < < o > = 7 cxy 6 Nyy ¼ 4 A12 A22 A26 B12 B22 B26 5 > > > ; > jxx > : > > > > Nxy A16 A26 A66 B16 B26 B66 > > > > > j yy > > > > : ; jxy

ð28Þ

To summarize our results to this point, Eq. (28) relates the stress resultants applied to a composite laminate to the resulting midplane strains and curvatures via the Aij and Bij matrices. The values of each term within the Aij and Bij matrices depend on the material properties and fiber angle of each ply (i.e., they depend on terms within the Qij matrix) as well as the stacking

Copyright © 2004 by Marcel Dekker, Inc. All Rights Reserved.

sequence (i.e., the distance zk of each ply from the laminate midplane), in accordance with Eq. (27a) (27b). In practice then, if the midplane strains and curvatures induced in a laminate under constant environmental conditions are measured, then the stress resultants that caused these strains and curvatures can be calculated using Eq. (28). This entire process must now be repeated for the moment resultants. The stresses rxx induced in a thin composite laminate are related to moment resultant Mxx in accordance with Eq. (2a), repeated here for convenience: Z t=2 rxx zdz ðrepeatedÞ ð2aÞ Mxx ¼ t=2

Substituting the expression for rxx from Eq. (20), we obtain: Z t=2

 Q11 exx þ Q12 eyy þ Q16 cxy k zdz Mxx ¼ t=2

ð29Þ

Each strain that appears in Eq. (29) can be related to the midplane strains and curvatures via the Kirchhoff hypothesis. Hence, substituting either Eq. (11) or Eq. (12), we have: Z t=2 n o Mxx ¼ zQ11 eoxx þ zQ12 eoyy þ zQ16 coxy þ z2 Q11 jxx þ z2 Q12 jyy þ z2 Q16 jxy dz t=2

ð30Þ

Equation (30) is similar to Eq. (22). Once again, this integral cannot be evaluated directly because the integrand is a discontinuous function of z. However, (a) noting that the midplane strains and curvatures are not functions of z and can be brought outside the integral sign, and then (b) evaluating the integral in a ‘‘piecewise’’ fashion through the thickness of the laminate, we obtain:  Z Z Z   z1   z2   z3 o Mxx ¼ exx Q11 1 zdz þ Q11 2 zdz þ Q11 3 zdz þ : : : : 

þ Q11 þeoyy 





þ Q12 þcoxy



z0 zn

n 1

Q12





Z

Z

zn

zn

2

Q16

zn

Z  1



zdz þ Q11

z1

1

n 1



2

Z  1

z1 zn

1

z0



zdz þ Q12

z1 z0

zdz

n

zdz þ Q12 

z2

Z 

zn 1 Z z2  2

Z 

 zdz þ Q16

z1

Z 

z3

Z   zdz þ Q16 3

z3

zdz

zn 1 Z  z2 2

z1



zdz þ Q12

zn

n





Copyright © 2004 by Marcel Dekker, Inc. All Rights Reserved.

3

z2

z2

zdz þ : : : :

zdz þ : : : :

  þ Q16 n þjxx







Z 1

zn zn





Z 1

Q16

  þ Q16 n

Z 1

1

2

1

zn

2

1

zn

zn zn



zdz þ Q16 z1

Z 

Z   Q12 1

  þ Q12 n þjxy

1

Q11

  þ Q11 n þjyy

Z

z0

1

2

zn

1

zn

2



z dz þ Q12 z0

zn

1

z2

Z  2

Z   z dz þ Q11 3

z1

2

z dz

n

zn 1 Z  z2

Z  n

2

z1

zn

2

1

z2 z1

zn zn



Z   z2 dz þ Q12 3 z dz

zn

Z   z2 dz þ Q16 2

Z   z2 dz þ Q16 n



2

zn

Z 

 z2 dz þ Q12

z1

1



z dz þ Q11

z0

Z 



zdz

n

z dz þ Q11

2

z1

2

2

zn

Z 



Z   z2 dz þ Q16 3 z2 dz

1

z3 z2

z3 z2

z3 z2

z2 dz þ : : : :

z2 dz þ : : : :

z2 dz þ : : : :



ð31Þ

The piecewise integrals that appear in Eq. (31) are of one of the following two forms, both of which are easily evaluated: Z zk  1 zdz ¼ z2k z2k 1 2 zk 1

or

Z

zk zk

1

z2 dz ¼

1 3 z 3 k

z3k

1



Hence, evaluating all integrals, we obtain: Mxx ¼

      1 o n exx Q11 1 z21 z20 þ Q11 2 z22 2 o    2 z z2 þ: : : : þ Q 11 n

n

n 1

    1 z20 þ Q12 2 z22 þ eoyy 2 o    : þ : : : þ Q12 n z2n z2n 1       1 n þ coxy Q16 1 z21 z20 þ Q16 2 z22 2 o    : þ : : : þ Q16 n z2n z2n 1 n

  Q12 1 z21

    z21 þ Q11 3 z23

z22



    z21 þ Q12 3 z23

z22

    z21 þ Q16 3 z23

z22

Copyright © 2004 by Marcel Dekker, Inc. All Rights Reserved.

 

n   1 þ jxx Q11 1 z31 3    3 z þ: : : : þ Q 11 n

    z30 þ Q11 2 z32 o z3n 1

n

  1 þ jyy Q12 1 z31 3    3 : þ :::þ Q z n

12 n

    z30 þ Q12 2 z32 o z3n 1

n

n   1 þ jxy Q16 1 z31 3    þ : : : : þ Q16 n z3n

    z30 þ Q16 2 z32 o z3n 1

    z31 þ Q11 3 z33

z32

    z31 þ Q12 3 z33

z32



    z31 þ Q16 3 z33

z32





ð32Þ

The first three quantities on the right-hand side of the equality sign involve the previously defined terms B11, B12, and B16. We now define three new terms, associated with the last three quantities:            1 n D11 ¼ Q11 1 z31 z30 þ Q11 2 z32 z31 þ Q11 3 z33 z32 þ : : : : 3    o þ Q11 n z3n z3n 1            1 n D12 ¼ Q12 1 z31 z30 þ Q12 2 z32 z31 þ Q12 3 z33 z32 þ : : : : 3 o    þ Q12 n z3n z3n 1            1 n D16 ¼ Q16 1 z31 z30 þ Q16 2 z32 z31 þ Q16 3 z33 z32 þ : : : : 3    o þ Q16 n z3n z3n 1 Hence, Eq. (32) can be written in the following simplified form:

Mxx ¼ B11 eoxx þ B12 eoyy þ B16 coxy þ D11 jxx þ D12 jyy þ D16 jxy

ð33aÞ

Following an entirely equivalent procedure for Myy and Mxy, it can be shown that: Myy ¼ B21 eoxx þ B12 eoyy þ B26 coxy þ D21 jxx þ D22 jyy þ D26 jxy

ð33bÞ

Mxy ¼ B61 eoxx þ B62 eoyy þ B66 coxy þ D61 jxx þ D62 jyy þ D66 jxy

ð33cÞ

The Bij terms that appear in Eqs. (33a) (33b) (33c) have been previously encountered and are given by Eq. (27b). The new terms Dij are given by: Dij ¼

n   1X Qij k z3k 3 k¼1

z3k

1



Copyright © 2004 by Marcel Dekker, Inc. All Rights Reserved.

ð34Þ

The Dij terms can be written as a symmetrical 3  3 matrix: 3 2 3 2 D11 D12 D16 D11 D12 D16 7 6 7 6 Dij ¼ 4 D21 D22 D26 5 ¼ 4 D12 D22 D26 5 D61

D62

D66

D16

D26

D66

Equations (33) can be written in matrix form as follows: 8 o 9 e > > > > > > xx o > 9 2 8 > > 3> e > > yy > > B11 B12 B16 D11 D12 D16 > > = = < o > < Mxx > 7 cxy 6 Myy ¼ 4 B12 B22 B26 D12 D22 D26 5 > > > > ; > : > > jxx > > Mxy B16 B26 B66 D16 D26 D66 > > > > jyy > > > > > ; : jxy

ð35Þ

Equation (35) relates the moment resultants applied to a composite laminate to the resulting midplane strains and curvatures via the Bij and Dij matrices. The value of each term within the Bij and Dij matrices depends on the material properties and fiber angle of each ply (i.e., they depend on terms within the Qij matrix) as well as the stacking sequence (i.e., the distance zk of each ply from the laminate midplane), in accordance with Eqs. (27b) and (34). In practice then, if the midplane strains and curvatures induced in a laminate under constant environmental conditions are measured, then the moment resultants that caused these strains and curvatures can be calculated using Eq. (35). It is customary to combine Eqs. (28) and (35) and express them together in matrix form: 8 9 2 38 o 9 Nxx > A11 A12 A16 B11 B12 B16 > exx > > > > > > > > > > 7 6 > > > > > > > > o N A A A B B B 7 > 6 > > yy > 12 22 26 12 22 26 > eyy > > > > > > 6 > 7> > > > < Nxy = 6 A < = 7 o A A B B B 26 66 16 26 66 7 6 16 cxy ¼6 ð36Þ 7 > 6 B11 B12 B16 D11 D12 D16 7> Mxx > > > > > jxx > > > > > 7 6 > > > > > > 7> > > jyy > > 6 > > > > Myy > > > 4 B12 B22 B26 D12 D22 D26 5> > > > > > : : ; ; jxy Mxy B16 B26 B66 D16 D26 D66 Equation (36) will sometimes be written in abbreviated form as: 

 o N A B e ¼ j M B D

The 6  6 array that appears in Eq. (36) is called the ‘‘ABD matrix.’’ Because each of the individual matrices that make up the total ABD matrix is in itself symmetrical (e.g., A12=A21, B12=B21, D12=D21, etc.), the entire ABD matrix is also symmetrical.

Copyright © 2004 by Marcel Dekker, Inc. All Rights Reserved.

It should be noted that the above results are applicable to thin laminates fabricated using any combination of ply materials. Because the Aij, Bij, and Dij matrices are each calculated based on a summation over all plies, and individual ply properties (represented by the Qij matrix) are embedded within these summations, both ply material type and fiber angle can vary from one ply to the next. Hence, the ABD matrix for any thin plate can be calculated using (Eq. (27a), (27b), and (34). For example, the ABD matrix for ‘‘hybrid’’ laminates (i.e., laminates fabricated using two different prepreg material systems) are calculated using (Eq. (27a), (27b), and (34). The Aij matrix relates in-plane stress resultants to in-plane midplane strains. For this reason, the Aij terms are called extensional stiffnesses. Similarly, the Dij matrix relates moment resultants to midplane curvatures, and elements within the Dij matrix are therefore called bending stiffness. The Bij matrix relates in-plane stress resultants to midplane curvatures, and also relates moment resultants to the in-plane midplane strains. The Bij terms are called coupling stiffnesses. For an isotropic plate, the coupling stiffnesses are always zero. The stress and moment resultants can be thought of as ‘‘stress-like’’ quantities because they are directly related to the stresses through the thickness of the laminate via Eqs. (1) and (2). On the other hand, the midplane strains and curvatures are ‘‘strain-like’’ quantities because they can be used to calculate the strains at any position through the thickness of the laminate via Eqs. (11) and (12). Hence, Eq. (36) relates ‘‘stress-like’’ quantities to ‘‘strainlike’’ quantities, and in this sense can be thought of as ‘‘Hooke’s law’’ for a composite laminate. Equation (36) is in convenient form if we measure midplane strains and curvatures and wish to calculate the stress and moment resultants that caused these strains and curvatures. Suppose, instead, that the stress and moment resultants are known and we wish to calculate the midplane strain and curvatures that will be caused by these known loads. In this case, we must invert Eq. (36) to obtain a relationship of the form:  o

 A B 1 N e ð37Þ ¼ j M B D In this text, the inverse of the ABD matrix will be called the abd matrix: " #

a b A B 1 ¼ B D b d Methods of inverting the [ABD] matrix analytically are discussed in several composite texts, including Refs. 1, 2, and 3. However, in practice, the ABD matrix is most often inverted numerically with the aid of a digital

Copyright © 2004 by Marcel Dekker, Inc. All Rights Reserved.

computer because many commercial software packages (e.g., MATLAB, Maple, Mathematica, etc.) that can invert a 66 matrix routinely are available nowadays. Written out in full, Eq. (37) is: 8 o 9 2 9 38 a11 a12 a16 b11 b12 b16 > Nxx > exx > > > > > > > > 6 > 7> > > > > > > > o > a a a b b b N 7 6 > > > > 12 22 26 21 22 26 yy e > > > yy > > > > > 7 6 > > > < o = 6 a16 a26 a66 b61 b62 b66 7< Nxy > = 7 6 cxy ¼6 ð38Þ 7 > 6 b11 b21 b61 d11 d12 d16 7> Mxx > > > > jxx > > > > > 7> 6 > > > > > > 6 7> M > > > 4 b12 b22 b62 d12 d22 d26 5> > > jyy > > yy > > > > > > > > > ; ; : : jxy M b16 b26 b66 d16 d26 d66 xy

The reader should carefully inspect the subscripts used in Eq. (38). Note that the [abd] matrix is symmetrical. Furthermore, the individual 3  3 matrices that appear in the upper left-hand quadrant and lower right-hand quadrant of the [abd] matrix, aij and dij, respectively, are also symmetrical. However, the 3  3 matrix that appears in the upper right-hand quadrant is not symmetrical (b12 p b21, b16 p b61, and b26 p b62). Also, the 3  3 matrix in the lower lefthand quadrant is the transpose of the 3  3 matrix that appears in the upper right-hand quadrant. Example Problem 4 Determine the [ABD] and [abd] matrices for a [30/0/90]T graphite-epoxy laminate. Use material properties listed for graphite-epoxy in Table 3 of Chap. 3, and assume that each ply has a thickness of 0.125 mm. Solution. A side view of the laminate is shown in Fig. 17. The total laminate thickness t = 3 (0.125 mm) = 0.375 mm. Because all three plies are of the same material, the thickness of each ply is identical: t1 = t2 = t3 = 0.125 mm.

Figure 17 Side view of the [30/0/90]T laminate considered in Sample Problem 4.

Copyright © 2004 by Marcel Dekker, Inc. All Rights Reserved.

Note that because an odd number of plies are used, the origin of the x–y–z coordinate system exists at the midplane of ply 2. The ply interface coordinates can be calculated as: z0 ¼

t=2 ¼

ð0:375 mmÞ=2 ¼

0:1875 mm ¼

0:0001875 m

z1 ¼ z0 þ t 1 ¼

0:1875 mm þ 0:125 mm ¼

0:0625 mm ¼

z2 ¼ z1 þ t 2 ¼

0:0625 mm þ 0:125 mm ¼ 0:0625 mm ¼ 0:0000625 m

0:0000625 m

z3 ¼ z2 þ t3 ¼ 0:0625 mm þ 0:125 mm ¼ 0:1875 mm ¼ 0:0001875 m We will also require the transformed reduced stiffness matrix for each ply. Elements of the [Q]k matrices are calculated using Eq. (31) of Chap. 5* and are equal to: For ply 1 (the 30j ply):



Q



30j ply

2

Q11

6 6 ¼ 6 Q12 4 Q16

2

Q12 Q22 Q26

107:6  109

6 9 ¼6 4 26:06  10 48:13  109

Q16

3

7 7 Q26 7 5 Q66

26:06  109

27:22  109

21:52  109

48:13  109

3

7 21:52  109 7 5ðPaÞ

36:05  109

For ply 2 (the 0j ply):



Q



0B ply

2

Q11

Q12

Q16

3

7 6 7 6 ¼ 6 Q12 Q22 Q26 7 5 4 Q16 Q26 Q66 2 107:9  109 3:016  109 6 9 9 ¼6 4 3:016  10 10:05  10 0 0

0 0 13:00  109

3

7 7ðPaÞ 5

  * The Q matrix for a 30j graphite-epoxy ply was calculated as a part of Example Problem 5.6 of Chap. 5.

Copyright © 2004 by Marcel Dekker, Inc. All Rights Reserved.

For ply 3 (the 90j ply): 3 2 Q11 Q12 Q16 7 6   7 6 Q 90B ply ¼ 6 Q12 Q22 Q26 7 5 4 Q16 Q26 Q66 2 10:05  109 3:016  109 6 9 9 ¼6 4 3:016  10 170:9  10 0 0

0 0 13:00  109

3

7 7ðPaÞ 5

We can now calculate each member of the Aij, Bij, and Dij matrices, in accordance with (Eq. (27a), (27b), and (34), respectively. .

Using Eq. (27a), element A11 is calculated as follows:

A11 ¼

3 X

k¼1

Q11



k

ð zk

zk 1 Þ





 A11 ¼ Q11 1 ðz1 z0 Þ þ Q11 2 ðz2 z1 Þ þ Q11 3 ðz3 z2 Þ



 A11 ¼ 107:6  109 ð :0000625 þ 0:0001875Þ þ 170:9  109

  ð0:0000625 þ 0:0000625Þ þ 10:05  109  ð0:0001875 6

A11 ¼ 36:07  10 Pa

0:0000625Þ m

The remaining elements of the Aij matrix are found in similar fashion: 2 3 36:07 4:012 6:016 6 7 Aij ¼ 4 4:012 26:02 2:690 5  106 ðPa mÞ 6:016 2:690

.

7:756

Using Eq. (27b), element B11 is calculated as follows:

3    2 1X z Q z2k 1 2 k ¼ 1 11 k k         1 h Q11 1 z21 z20 þ Q11 2 z22 z21 þ Q11 3 z23 ¼ 2 o n 1 h ¼ 107:6  109 ð :0000625Þ2 þ ð 0:0001875Þ2 2 o

n þ 170:9  109 ð0:0000625Þ2 ð 0:0000625Þ2 oi

n þ 10:05  109 ð0:0001875Þ2 ð0:0000625Þ2

B11 ¼ B11 B11

Copyright © 2004 by Marcel Dekker, Inc. All Rights Reserved.

z22

i

1:524  103 Pa

B11 ¼

m2

The remaining elements of the Bij matrix are found in similar fashion: 2

1:524

6 Bij ¼ 4 0:3601 0:7521

3

0:3601

0:7521

2:245

 7 0:3362 5  103 Pa

0:3362

m2

0:3601



In passing, in this example, it appears that B12 is numerically equal to B66. This is not true, in general. In this problem, the apparent numerical equivalence is due to the fact that only four significant digits have been used. Nevertheless, for laminates produced using a single material system, it is often (but not always) the case that B12 c B66. This common occurrence can be traced to the fact the functional form and magnitude of Q 12 and Q 66 are similar (see Eq. (31) of Chap. 5). Because B12 and B66 are directly related to Q12 and Q66, respectively, their values are often nearly identical. Also, in Sec. 6.2, it will be seen that all elements within the Bij matrix are zero for symmetrical laminates. Hence, for symmetrical laminates, these two terms are, in fact, numerically equal, that is, B12=B66=0 for symmetrical laminates. .

Using Eq. (34), element D11 is calculated as follows:

D11 ¼

3   1X Q11 k z3k 3 k¼1

z3k

1



  i       1 h Q11 1 z31 z30 þ Q11 2 z32 z31 þ Q11 3 z33 z32 3 o n 1 h 107:6  109 ð :0000625Þ3 ð 0:0001875Þ3 ¼ 3 o

 n þ 170:9  109 þ ð0:0000625Þ3 ð 0:0000625Þ3

D11 ¼ D11

n þ 10:05  109 ð0:0001875Þ3

D11 ¼ 0:2767 Pa

ð0:0000625Þ3

m3

oi

The remaining elements of the Dij matrix are found in similar fashion: 2

0:2767 0:0620

6 Dij ¼ 6 4 0:0620 2:513

0:1018 0:0455

0:1018

3

7 0:0455 7 5 Pa

0:1059

m3



Copyright © 2004 by Marcel Dekker, Inc. All Rights Reserved.

The [ABD] matrix can now be assembled: 2 36:07  106 4:012  106 6:016  106 6 6 4:012  106 26:02  106 2:690  106 6 6 6 6:016  106 2:690  106 7:756  106 ½ABDŠ ¼ 6 6 6 1524 360:1 752:1 6 6 360:1 2245 336:2 4 752:1 336:2 360:1

1524

360:1

360:1

2245

752:1

336:2

0:2767

0:0620

0:0620

2:513

0:1018

0:0455

752:1

3

7 336:2 7 7 7 360:1 7 7 7 0:1018 7 7 0:0455 7 5 0:1059

The [abd] matrix is obtained by inverting the [ABD] matrix, and is found to be: 2

6 6 6 6 6 6 6 ½abdŠ ¼ 6 6 6 6 6 6 4

3:757  10

8

1:964  10

9

1:038  10

8

1:440  10

4

3:905  10

6

1:964  10

9

1:037  10

7

4:234  10

8

1:866  10

5

6:361  10

4

1:038  10

8

4:234  10

8

2:004  10

7

3:661  10

4

3:251  10

4

1:440  10

4

1:866  10

5

3:661  10

4

7:064

3:122  10

2

3:905  10

6

6:361  10

4

3:251  10

4

3:122  10

8:513  10

5

4:268  104

1:851  10

5

4:572

2

6:429 3:620

8:513  10

A [30/0/90]T graphite-epoxy laminate is subjected to the following stress and moment resultants: Mxx ¼ 1 N

Nyy ¼

m=m Myy ¼

10 kN=m 1N

m=m

Nxy ¼ 0 N=m

Mxy ¼ 0 N

m=m

Determine the following quantities caused by these stress and moment resultants: (a) Midplane strains and curvatures (b) Ply strains relative to the x–y coordinate system (c) Ply stresses relative to the x–y coordinate system. Use material properties listed for graphite-epoxy in Table 3 of Chap. 3 and assume that each ply has a thickness of 0.125 mm. Solution. Note that this is the same laminate considered in Example Problem 4. A side view of the laminate appears in Fig. 17. (a) Midplane strains and curvatures. The [abd] matrix for this laminate was calculated as a part of Example Problem 4. Hence, midplane strains

Copyright © 2004 by Marcel Dekker, Inc. All Rights Reserved.

3

7 4:268  104 7 7 7 57 1:851  10 7 7 7 7 4:572 7 7 7 3:620 7 5 17:41

Example Problem 5

Nxx ¼ 50 kN=m

5

and curvature may be obtained through application of Eq. (38), which becomes: 8 o 9 2 exx > > > > > > > > > 6 > > o > > 6 > 6 > e > > yy > > > 6 > > > 6 > > = 6 < co > 6 xy ¼6 > 6 > > jxx > > > 6 > > > > 6 > > > 6 > > > > > 6 j yy > > > > > 4 > > > ; : jxy

3:757  10

8

1:964  10

9

1:038  10

8

1:440  10

4

3:905  10

6

1:964  10

9

1:037  10

7

4:234  10

8

1:866  10

5

6:361  10

4

1:038  10

8

4:234  10

8

2:004  10

7

3:661  10

4

3:251  10

4

1:440  10

4

1:866  10

5

3:661  10

4

7:064

3:122  10

2

3:905  10

6

6:361  10

4

3:251  10

4

3:122  10

8:513  10

5

1:851  10

5

4:572

8 50  103 > > > > > > > 10  10 > > > > > < 0  > > 1 > > > > > > 1 > > > > : 0

4:628  10

4

2

6:429 3:620

8:513  10

9 > > > > > 3> > > > > > > = > > > > > > > > > > > > ;

(b) Ply strains relative to the x–y coordinate system. Ply strains may now be calculated using Eq. (12). For example, strains present at the outer surface of ply 1 (i.e., strains present at z0 = 0.0001875 m) are:

j

z ¼ z0

8 8 o 9 8 9 > 2038  10 e > = < < jxx = > < xx > ¼ eoyy þ z0 jyy ¼ 518  10 > > > : ; > : o ; : jxy cxy 55  10

9 8 < exx = e : yy ; cxy

j

z ¼ z0

8 9 > < 677 Am=m > = ¼ 536 Am=m > > : ; 194 Arad

9 9 8 m=m > 14:48 rad=m > > > = = < 6 þ ð 0:0001875mÞ 0:096 rad=m m=m > > > ; : > ; 6 1:328 rad=m m=m 6

Strains calculated at the remaining ply interface positions are summarized in Table 5.   (c) Ply stresses relative to the x–y coordinate system. The Q matrix for all plies was calculated as a part of Example Problem 4. Ply stresses may now be

Copyright © 2004 by Marcel Dekker, Inc. All Rights Reserved.

3

7 7 4:628  104 7 7 7 1:851  10 5 7 7 7 7 4:572 7 7 7 7 3:620 5 17:41

Completing this matrix multiplication, we obtain: 9 8 o 9 8 2039 Am=m > exx > > > > > > > > > > > > > o > > > > > 518 Am=m > > > > > e > > > > yy > > > > = < < 55 Arad = o cxy ¼ > > 14:48 m 1 > > > > > jxx > > > > > > > > > > > > 1 > > > > > 0:096 m j > > > > yy > > > ; ; > : : jxy 1:323 m 1

9 8 > = < exx > eyy > :c > ; xy

5

Table 5

Ply Interface Strains in a [30/0/90] Graphite-Epoxy Laminate Caused by the Stress and Moment Resultants Specified in Example Problem 5 z-coordinate (mm)

exx (Am/m)

0.1875 0.0625 0.0625 0.1875

eyy (Am/m)

677 1133 2943 4753

536 524 512 500

cxy (Arad) 194 28 137 303

Strains are referenced to the x–y coordinate system.

calculated using Eq. (30) of Chap. 5, with DT = DM = 0. The stresses present at the outer surface of ply 1 (i.e., at z=z0) are: 8 9 ply 1 2 3 9 Q11 Q12 Q16 ply 1 8 > < exx = < rxx > = 6 7 eyy ryy ¼ 4 Q12 Q22 Q26 5 :c ; > > : ; xy sxy z ¼ z0 Q16 Q26 Q66 z ¼ z0 z ¼ z0 9 ply 1 9 8 38 2 677  10 6 > 107:6  109 26:06  109 48:13  109 > > = = < rxx > < 7 6 ryy 536  10 6 ¼ 4 26:06  109 27:22  109 21:52  109 5 > > > > ; ; : : sxy z ¼ z0 48:13  109 21:52  109 36:05  109 194  10 6 9 9 ply 1 8 8 > > = = < 77:5 MPa > < rxx > 28:1 MPa ryy ¼ > > > > ; ; : : sxy z ¼ z0 37:1 MPa

j

j

j

j

j

Stresses calculated at remaining ply interface positions are summarized in Table 6.

Table 6 Ply Interface Stresses in a [30/0/90]T Graphite-Epoxy Laminate Caused by the Stress and Moment Resultants Specified in Example Problem 5 Ply number Ply 1 Ply 2 Ply 3

z-coordinate (mm) 0.1875 0.0625 0.0625 0.0625 0.0625 0.1875

rxx (MPa)

ryy (MPa)

77.5 109.7 192.1 501.5 28.0 46.3

Stresses are referenced to the x–y coordinate system.

Copyright © 2004 by Marcel Dekker, Inc. All Rights Reserved.

28.1 15.9 1.85 3.73 78.6 71.1

sxy (MPa) 37.1 44.3 0.366 1.78 1.78 3.93

6.2 Including Changes in Environmental Conditions Recall that we simplified the analysis leading up to Eq. (36) by assuming that DT=DM=0. We will now consider how to predict the behavior of a laminate subjected to a change in temperature and/or moisture content as well as external mechanical loads. To begin, the stresses in any ply (say, in ply number k) are related to ply strains in accordance with Eq. (30) of Chap. 5: 9 2 9 8 3 8 Q11 Q12 Q16 > exx DTaxx DMbxx > rxx > > > > > > = 6 = < < 7 7 e DTa DMb ryy ¼6 ðrepeatedÞ ð5:30Þ Q Q Q yy yy yy 22 26 5 > > 4 21 > > > > > > ; ; : : cxy DTaxy DMbxy k sxy k Q Q Q 61

62

66

k

Stress rxx in ply k is given by:

rxx ¼ Q11 fexx DTaxx DMbxx g þ Q12 eyy

 þ Q16 cxy DTaxy DMbxy

DTayy

DMbxx



ð39Þ

Stress resultant Nxx is related to rxx via Eq. (1a). Substituting Eq. (39) into Eq. (1a), we have: Z t=2

 Nxx ¼ Q11 exx þ Q12 eyy þ Q16 cxy k dz t=2

Z DT

Z DM

t=2

t=2 t=2

t=2

Q11 axx þ Q12 ayy þ Q16 cxy



Q11 bxx þ Q12 byy þ Q16 bxy

k

ð40Þ

dz



k

dz

The first integral on the right-hand side of the equality sign is identical to Eq. (21), and after evaluation (using the same techniques as previously described) will result in Eq. (26a). The second and third integrals were not previously encountered because they involve DT and DM, which were previously assumed to equal zero. Using methods similar to those used previously, it can be shown that the second integral may be written as: Z t=2

 DT Q11 axx þ Q12 ayy þ Q16 cxy k dz t=2

¼ DT

n  X

k¼1

 Q11 axx þ Q12 ayy þ Q16 axy k ½zk

zk 1 Š



T . This quantity is called a thermal stress resultant, and will be denoted Nxx That is, n  X   Q11 axx þ Q12 ayy þ Q16 axy k ½zk zk 1 Š ð41aÞ NTxx ¼ DT

k¼1

Copyright © 2004 by Marcel Dekker, Inc. All Rights Reserved.

Similarly, the third integral in Eq. (40) can be evaluated to give the moisture stress resultant, denoted NM xx: n n o X   NM ½ z z Š ð42aÞ Q b þ Q b þ Q b ¼ DM k k 1 11 xx 12 yy 16 xy k xx k¼1

Hence, after evaluating all integrals, Eq. (40) may be written as:

Nxx ¼ A11 eoxx þ A12 eoyy þ A16 coxy þ B11 jxx þ B12 jyy þ B16 jxy NTxx

NM xx

ð43aÞ

This result should be compared to Eq. (26a). It will be seen that the inclusion of temperature and/or moisture changes has resulted in the addition of two new terms (NTxx and NM xx); otherwise, our earlier results remain unchanged. If an analogous procedure is now followed for the remaining stress and moment resultants, using Eqs. (1b), (1c), (2a), (2b), and (2c), five additional thermal stress/moment resultants and five additional moisture stress/moment resultants will be identified, as follows: n  X   NTyy u DT ð41bÞ Q12 axx þ Q22 ayy þ Q26 axy k ½zk zk 1 Š NTxy u DT

k¼1 n  X

 Q16 axx þ Q26 ayy þ Q66 axy k ½zk

k¼1 n  X

  DT Q11 axx þ Q12 ayy þ Q16 axy k z2k 2 k¼1 n    DT X Q12 axx þ Q22 ayy þ Q26 axy k z2k u 2 k¼1

zk 1 Š



z2k

1

z2k

1

MTxy u

z2k

1

NM yy

zk 1 Š

MTxx u MTyy

n    DT X Q16 axx þ Q26 ayy þ Q66 axy k z2k 2 k¼1 n n X  Q12 bxx þ Q22 byy þ Q26 bxy k ½zk u DM

NM xy u DM

MM xy

 Q16 bxx þ Q26 byy þ Q66 bxy k ½zk

k¼1 n n X

  DM Q11 bxx þ Q12 byy þ Q16 bxy k z2k 2 k¼1 n n   DM X u Q12 bxx þ Q22 byy þ Q26 bxy k z2k 2 k¼1 n n   DM X Q16 bxx þ Q26 byy þ Q66 bxy k z2k u 2 k¼1

MM xx u MM yy

k¼1 n n X

Copyright © 2004 by Marcel Dekker, Inc. All Rights Reserved.

ð41cÞ



ð41dÞ



ð41eÞ



ð41f Þ

o

ð42bÞ

o zk 1 Š

ð42cÞ

z2k z2k z2k

1

1

1

o

o

o

ð42dÞ ð42eÞ ð42f Þ

In each case, the corresponding thermal and moisture resultants will be subtracted from the right-hand side of Equations (26a) (26b) (26c) and (33a) (33b) (33c)) and (26a) (26b) (26c) and (33a) (33b) (33c). Finally, the response of a composite laminate subjected to mechanical loads, a change in temperature, and a change in moisture content can be written in a form similar to Eq. (36): 8 9 2 9 8 T 9 8 M 9 38 Nxx > Nxx > > Nxx > A11 A12 A16 B11 B12 B16 > eoxx > > > > > > > > > > > > > > > > > > > > > > 7> 6 > > > > > > > T M > > > > > > > > > o N N 7 6 > > > > > > > N A A A B B B yy > 12 22 26 12 22 26 > eyy > yy > yy > > > > > > > > > > > > > 6 > > 7> > > > > > > > > > 7> T > M > = = < < < Nxy > = 6 < = o N N 6A16 A26 A66 B16 B26 B66 7 cxy xy xy 7 ¼6 7 6 > > > > MTxx > MM Mxx > > > > > > > jxx > > 6B11 B12 B16 D11 D12 D16 7> xx > > > > > > > > > > > > > > > > 7> 6 > > > > > > > > > > > > 7> T > M> > > > > >Myy > > 6 > > j M M yy B B B D D D > > > > > > > 5 4 12 22 26 12 22 26 > yy > yy > > > > > > > > > > > > > > > T > ; > M> ; > : : ; ; : : j M M xy Mxy B16 B26 B66 D16 D26 D66 xy xy ð44Þ

Equation (44) will sometimes be abbreviated as: 

 o ( T ) ( M ) N N N A B e ¼ j M B D MM MT Equation (44) is comparable to Eq. (36), except we have now included the effects due to a change in temperature and/or moisture content. Equation (44) can be viewed as ‘‘Hooke’s law’’ for a composite laminate, in the sense that it may be used to relate stresslike quantities (i.e., stress and moment resultants) to strainlike quantities (i.e., midplane strains and curvatures). Inverting Eq. (44), we obtain: 8 o 9 2 a11 exx > > > > 6 > > > > 6 a12 > > > eoyy > 6 > > > > > > 6 > > < co = 6 a16 6 xy ¼6 >j > 6 b11 > > xx > > 6 > > > > 6 > > j > > 6 b12 yy > > > > 4 > > : ; jxy b16

a12

a16

b11

b12

a22

a26

b21

b22

a26

a66

b61

b62

b21

b61

d11

d12

b22

b62

d12

d22

b26

b66

d16

d26

where, as before:

a

b

b

d



¼



A B

B D



38 9 b16 > Nxx þ NTxx þ NM xx > > 7> > > > > N þ NT þ NM > > b26 7 > 7> yy > yy yy > > > > 7> > < N þ NT þ NM > = b66 7 xy 7 xy xy 7 > M þ MT þ MM > d16 7 > > 7> xx xx > > xx > > 7> > T M > > > d26 7 > M þ M þ M yy > > 5> yy yy > > : T M ; M þ M þ M xy d66 xy xy

1

Copyright © 2004 by Marcel Dekker, Inc. All Rights Reserved.

ð45Þ

A subtlety embedded fact within the preceding discussion is that most composites are subjected to a significant state of stress prior to the application of any external mechanical loading. That is, most modern composite material systems are cured at an elevated temperature (common cure temperatures are either 120jC or 175jC), and are nominally stress-free at the cure temperature. Once the polymerization process is complete, the composite is cooled to room temperatures (say, 20jC) and, consequently, the composite experiences a uniform change in temperature of DT = 100jC or 155jC during cooldown. In general, this change in temperature results in thermal stress and/or moment resultants to develop, causing thermal stresses within all plies of the laminate. These thermal stresses can be quite high, and contribute toward failure of the laminate.* A further complicating factor is related to measurement of strains. In most practical situations, strain measurement devices (e.g., resistance foil strain gages) are bonded to a composite material or structure after cooldown to room temperature. Hence, in practice, the reference state of a strain measurement device mounted on a laminate at room temperature does not necessarily correspond to the stress-free (or strain-free) state of the composite. This complication will be further explored in Chap. 7. At this point, it will simply be noted that a significant difficulty arises when prediction of nonlinear behavior (or more generally, the prediction of composite failure) is required based on measured laminate strains. Example Problem 6 A [30/0/90]T graphite-epoxy laminate is cured at 175jC and then cooled to room temperature (20 jC). Determine: (a) Midplane strains and curvatures (b) Ply strains relative to the x–y coordinate system (c) Ply stresses relative to the x–y coordinate system which are induced during cooldown. Use material properties listed for graphite-epoxy in Table 3 of Chap. 3, assume that each ply has a thickness of 0.125 mm, and assume no change in moisture content (i.e., assume DM=0). * Determination of the ‘‘stress-free temperature’’ is actually more complex than is implied here. It is true that thermal stresses begin to develop as cooldown begins, but because polymeric materials exhibit viscoelastic characteristics at these elevated temperatures, the matrix will creep, initially relieving thermal stresses somewhat. As temperature is decreased further, the viscoelastic nature of the matrix is rapidly decreased, and thermal stresses develop as described. A second factor is that all polymers exhibit some shrinkage during the polymerization process (see Sec. 1.2), and this shrinkage results in additional stresses similar to thermal stresses. As a rule of thumb, the stress-free temperature is often estimated to be 20–50jC below the final cure temperature. Nevertheless, this complication will be ignored in this text; it will be assumed that the final cure temperature defines the stress-free temperature.

Copyright © 2004 by Marcel Dekker, Inc. All Rights Reserved.

Solution. Note that this is the same laminate considered in Sample Problem 4. A side view of the laminate appears in Fig. 17. (a) Midplane strains and curvatures. The laminate has experienced a change in temperature DT = (20–175) = 155jC and, consequently, is subjected thermal stress and moment resultants. However, no external loads are applied and there has been no change in moisture content; therefore, the stress and moment resultants and the moisture stress and moment results are zero: 9 8 M 9 8 9 8 0> Nxx > Nxx > > > > > > > > > > > > > > > > > > > > > > > > > > > > M > > > > > > N 0 N yy > > > > > > yy > > > > > > > > > > > > > > > > > =

xy ¼ ¼ > > > > 0> MM Mxx > > > > > > xx > > > > > > > > > > > > > > > > > > > M > > > > > > > > > > > 0 M M yy > > > > yy > > > > > > > > > > > > > > ; :M ; : MM ; : 0 > xy xy The effective thermal expansion coefficients for each ply are calculated using Eq. (25) of Chap. 5, repeated here for convenience: axx ¼ a11 cos2 ðhÞ þ a22 sin2 ðhÞ ayy ¼ a11 sin2 ðhÞ þ a22 cos2 ðhÞ axy ¼ 2cosðhÞsinðhÞða11

ð5:25Þ

a22 Þ

From Table 3 of Chap. 3, the thermal expansion coefficients for graphiteepoxy (relative to the 1–2 coordinate system) are a11 = 0.9 Am/m jC and a22 = 27Am/m jC. Therefore: For ply 1 (the 30j ply): 1Þ aðxx ¼ ð 0:9 Am=m

aðyy1Þ aðxy1Þ

B

¼ 6:08 Am=m

B

¼ 20:0 Am=m

B

¼ ð 0:9 Am=m

CÞ cos2 ð30B Þ þ ð27 Am=m

B

CÞ sin2 ð30B Þ

CÞ sin2 ð30B Þ þ ð27 Am=m

B

CÞ cos2 ð30B Þ

C B

C

¼ 2cosð30Þ sinð30Þ½ð 0:9

27ÞAm=m

B

CŠ ¼

24:2 Arad=B C

For ply 2 (the 0j ply): 2Þ ¼ ð 0:9 Am=m aðxx

aðyy2Þ aðxy2Þ

¼

0:9 Am=m

¼ ð 0:9 Am=m

¼ 27:0 Am=m

B B

B

CÞ sin2 ð0B Þ

CÞ sin2 ð0B Þ þ ð27 Am=m

B

CÞ cos2 ð0B Þ

C B

B

CÞ cos2 ð0B Þ þ ð27 Am=m

C

¼ 2cosð0 Þ sinð0 Þ½ð 0:9 B

B

27ÞAm=m

B

CŠ ¼ 0 Arad=B C

Copyright © 2004 by Marcel Dekker, Inc. All Rights Reserved.

For ply 3 (the 90j ply): 3Þ aðxx ¼ ð 0:9 Am=m

¼ 27:0 Am=m

aðyy3Þ

¼ ð 0:9 Am=m

¼ aðxy3Þ

0:9 Am=m

B B

CÞ cos2 ð90B Þ þ ð27 Am=m

B

CÞ sin2 ð90B Þ

CÞ sin2 ð90B Þ þ ð27 Am=m

B

CÞ cos2 ð90B Þ

C B

B

C

¼ 2cosð90 Þ sinð90B Þ½ð 0:9 B

27ÞAm=m

B

CŠ ¼ 0 Arad=B C

Both the ply interface positions as well as the Qij matrices for each ply were calculated as a part of Example Problem 4. Hence, we now have all the information needed to calculate the thermal stress and moment resultants, using Eqs. (41a)–(41f). For example, Eq. (41a) is evaluated as follows: n  X   NTxx u DT Q11 axx þ Q12 ayy þ Q16 axy k ½zk zk 1 Š k¼1

  Q11 axx þ Q12 ayy þ Q16 axy 1 ½z1 z0 Š NTxx ¼ DT    þ Q11 axx þ Q12 ayy þ Q16 axy 2 ½z2 z1 Š  o  þ Q11 axx þ Q12 ayy þ Q16 axy 3 ½z3 z2 Š n

n      NTxx ¼ ð 155Þ 107:6  109 6:08  10 6 þ 26:06  109 20:0  10 6     þ 48:13  109 24:2  10 6 ð 0:0625 þ 0:1875Þ  10 3        0:9  10 6 þ 3:016  109 27:0  10 6 þ 170:9  109      þð0Þð0Þ ð0:0625 þ 0:0625Þ  10 3 þ 10:05  109 27  10 6     o þ 3:016  109 0:9  10 6 þ ð0Þð0Þ ð0:1875 0:0625Þ  10 3 NTxx ¼ 4060 N=m

The remaining thermal stress and moment resultants are calculated in similar fashion, eventually resulting in: 8 T 9 8 9 Nxx > 4060 N=m > > > > > > > > > > > > > > T > > > > > > > N 7360 N=m > > > > yy > > > > > > > > > > > = < NT = < 2860 N=m > xy ¼ > > MTxx > 0:62 N m=m > > > > > > > > > > > > > > > > > T > > > > > > > 0:62 N m=m > > Myy > > > > > > > ; > ; : : MT > 0:36 N m=m xy

Copyright © 2004 by Marcel Dekker, Inc. All Rights Reserved.

We can now calculate midplane strains and curvature using Eq. (45), which becomes:* 8 o 9 2 e > > > > > xx > 6 > > > > > 6 > > 6 > eoyy > > > > > > 6 > > > > 6 > = 6 < co > 6 xy ¼6 > > 6 > > j > > 6 xx > > > > 6 > > > > 6 > > > > 6 jyy > > > > 4 > > > > : ; jxy



3:757  10

8

1:964  10

9

1:038  10

8

1:440  10

4

3:905  10

6

1:964  10

9

1:037  10

7

4:234  10

8

1:866  10

5

6:361  10

4

1:038  10

8

4:234  10

8

2:004  10

7

3:661  10

4

3:251  10

4

1:440  10

4

1:866  10

5

3:661  10

4

7:064

3:122  10

2

3:905  10

6

6:361  10

4

3:251  10

4

3:122  10

8:513  10 9 8 4060 > > > > > > > > > > > 7360 > > > > > > > > > > = < 2860 >

5

4:628  104

1:851  10

5

4:572

> > > > > > > > > > > :

2

6:429 3:620

8:513  10

9 8 o 9 8 285 Am=m > exx > > > > > > > > > o > > > > > > > > > 1424 Am=m > e > > > > yy > > > > > > > < o = < 908 Arad > = cxy ¼ > > 2:16 m 1 > jxx > > > > > > > > > > > > > > > > > 1 > > > > 10:9 m j > > > > yy > : > > > ; : ; 1 jxy 9:4 m

(b) Ply strains relative to the x–y coordinate system. Ply strains may now be calculated using Eq. (12). For example, strains present at the outer surface of ply 1 (i.e., strains present at zo= 0.0001875 m) are: 9 8 8 o 9 8 e j 285  10 > > = = > < xx > < xx > < o ¼ eyy þ z0 jyy ¼ 1424  10 > > > ; ; > : o > : : cxy jxy 908  10 z ¼ z0

xy

9 8 > = < 2:16 rad=m > þ ð 0:0001875 mÞ 10:9 rad=m > > ; : 9:4 rad=m

6

9 m=m > =

6

m=m

6

m=m

> ;

* The [abd] matrix for a [30/0/90]T graphite-epoxy laminate was calculated in Sample Problem 3.

Copyright © 2004 by Marcel Dekker, Inc. All Rights Reserved.

3

7 4:628  104 7 7 7 7 1:851  10 5 7 7 7 7 4:572 7 7 7 7 3:620 5 17:41

> 0:62 > > > > > > 0:62 > > > > ; 0:36

9 8  e > = < xx > eyy  > > ; : c

5

8 9 < exx = eyy  : c ; xy

8 9 < 120 Am=m = ¼ 3468 Am=m : ; 2672 Am=m z ¼ z0

Strains calculated at the remaining ply interface positions are summarized in Table 7. (c) Ply stresses relative to the x–y coordinate system. Ply stresses may now be calculated using Eq. (30) of Chap. 5, with DM=0. The stresses present at the outer surface of ply 1 are (i.e., at z=z0): 8 9 9ply 1 2 3ply 1 8   exx DTaxx > Q11 Q12 Q16  rxx > > >  > > > > < < = = 7 6 Q Q22 Q26 7 eyy DTayy  ryy  ¼6 5 4 12 > > > >  > > > > ; ; : : Q16 Q26 Q66  cxy DTaxy  sxy  z ¼ z0

z ¼ z0

z ¼ z0

9ply 1 2 8 3  107:6  109 26:06  109 48:13  109 > = < rxx >  7 6 ryy  ¼ 4 26:06  109 27:22  109 21:52  109 5  > > ; : sxy z¼z0 48:13  109 21:52  109 36:05  109 9 8 > ½ð120Þ ð 155Þð6:08ފ  10 6 > > > = < 6  ½ð 3468Þ ð 155Þð20:0ފ  10 > > > > ; : ½ð2672Þ ð 155Þð 24:2ފ  10 6

8 9ply 1 8 53 MPa 9 > > < rxx = < =  ryy  ¼ 5:2 MPa : ; > > sxy z ¼ z0 : 4:8 MPa ;

Stresses calculated at the remaining plies and ply interface positions are summarized in Table 8.

Table 7 Ply Interface Strains in a [30/0/90]T Graphite-Epoxy Laminate Caused by a Cooldown from 175jC to 20jC z-coordinate (mm) 0.1875 0.0625 0.0625 0.1875

exx (Am/m) 120 150 420 690

eyy (Am/m) 3468 2100 750 620

Strains are referenced to the x–y coordinate system.

Copyright © 2004 by Marcel Dekker, Inc. All Rights Reserved.

cxy (Arad) 2672 1500 320 860

Table 8

Ply Interface Stresses in a [30/0/90]T Graphite-Epoxy Laminate Caused by a Cooldown from 175jC to 20jC

Ply number

z-coordinate (mm)

Ply 1

0.1875 0.0625 0.0625 0.0625 0.0625 0.1875

Ply 2 Ply 3

rxx (MPa)

ryy (MPa)

53 3.1 43 85 35 37

5.2 0.53 20 33 140 92

sxy (MPa) 4.8 21 19 4.1 4.1 11

Stresses are referenced to the x–y coordinate system.

Example Problem 7 A [30/0/90]T graphite-epoxy laminate is cured at 175 jC and cooled to room temperature (20 jC). Initially, the moisture content of the laminate is zero. However, the laminate is subjected to a humid environment for several weeks, resulting in an increase of moisture content of 0.5% (by weight). Determine: (a) Midplane strains and curvatures (b) Ply strains relative to the x–y coordinate system (c) Ply stresses relative to the x–y coordinate system which are present following the increase in moisture content. Use material properties listed for graphite-epoxy in Table 3 of Chap. 3, and assume that each ply has a thickness of 0.125 mm. Solution. Note that this is the same laminate considered in Sample Problem 6, and the midplane strains and curvatures, ply strains, and ply stresses that will be induced immediately upon cooldown by the change in temperature have already been calculated. These quantities will all be modified due to the slow diffusion of water molecules into the epoxy matrix. (a) Midplane strains and curvatures. The laminate has experienced a change in moisture content DM=+0.5% and, consequently, is subjected moisture stress and moment resultants. The effective moisture expansion coefficients for each ply are calculated using Eq. (28) of Chap. 5, repeated here for convenience: bxx ¼ b11 cos2 ðhÞ þ b22 sin2 ðhÞ byy ¼ b11 sin2 ðhÞ þ b22 cos2 ðhÞ bxy ¼ 2cosðhÞsinðhÞðb11

b22 Þ

Copyright © 2004 by Marcel Dekker, Inc. All Rights Reserved.

ð5:28Þ

From Table 3 of Chap. 3, the moisture expansion coefficients for graphiteepoxy (relative to the 1–2 coordinate system) are b11=150 Am/m %M and b22=4800 Am/m %M. Therefore: For ply 1 (the 30j ply): ð1Þ bxx ¼ ð150 Am=m

%MÞ cos2 ð30B Þ þ ð4800 Am=m

ð1Þ byy

%MÞsin2 ð30B Þ þ ð4800 Am=m

ð1Þ bxy

¼ 1310 Am=m ¼ ð150 Am=m

¼ 3640 Am=m

%M

%M

¼ 2cosð30Þsinð30Þ½ð150

4800ÞAm=m

%MÞ sin2 ð30B Þ %MÞcos2 ð30B Þ

%MŠ ¼ 4030 Arad=%M

For ply 2 (the 0j ply): ð2Þ bxx ¼ ð150 Am=m ð2Þ byy

ð2Þ bxy

¼ 150 Am=m

¼ ð150 Am=m

¼ 4800 Am=m

%MÞ cos2 ð0B Þ þ ð4800 Am=m

%MÞ sin2 ð0B Þ

%MÞ sin2 ð0B Þ þ ð4800 Am=m

%MÞ cos2 ð0B Þ

%M

%M

¼ 2cosð0 Þ sinð0 Þ½ð150 B

B

4800ÞAm=m

%MŠ ¼ 0 Arad=%M

For ply 3 (the 90j ply): ð3Þ ¼ ð150 Am=m bxx

%MÞ cos2 ð90B Þ þ ð4800 Am=m

%MÞ sin2 ð90B Þ

ð3Þ byy

%MÞ sin2 ð90B Þ þ ð4800 Am=m

%MÞ cos2 ð90B Þ

ð3Þ bxy

¼ 4800 Am=m

¼ ð150 Am=m ¼ 150 Am=m

%M

%M

¼ 2cosð90 Þsinð90B Þ½ð150 B

4800Þ Am=m

%MŠ ¼ 0 Arad=%M

Both the ply interface positions as well at the [Q] matrices for each ply were calculated as a part of Example Problem 4. Hence, we now have all the information needed to calculate the moisture stress and moment resultants, using Eqs. (42a)–(42f). For example, Eq. (42a) is evaluated as follows: 3 n o X  NM Q11 bxx þ Q12 byy þ Q16 bxy k ½zk zk 1 Š xx ¼ DM k¼1

n

  NM Q11 bxx þ Q12 byy þ Q16 bxy 1 ½z1 z0 Š xx ¼ DM    þ Q11 bxx þ Q12 byy þ Q16 bxy 2 ½z2 z1 Š  o  þ Q11 bxx þ Q12 byy þ Q16 bxy 3 ½z3 z2 Š

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9 6 NM xx ¼ ðþ0:5Þ ð½ð107:6  10 Þð1312  10 Þ

þð26:06  109 Þð3638  10 6 Þ þ ð48:13  109 Þð 4027  10 6 Þ

½ð 0:0625 þ 0:1875Þ  10 3 ŠÞ þ ð½ð170:9  109 Þð150  10 6 ފ

þð3:016  109 Þð4800  10 6 Þ þ ð0Þð0ފ½ð0:0625 þ 0:0625Þ  10 3 ŠÞ þ ð½ð10:05  109 Þð4800  10 6 Þ þ ð3:016  109 Þ   ð150  10 6 Þ þ ð0Þð0ފ½ð0:1875 0:0625Þ  10 3 ŠÞ

NTxx ¼ 8190 N=M

The remaining thermal stress and moment resultants are calculated in similar fashion, eventually resulting in: 8 9 9 8 NTxx > > 8190 N=m > > > > > > > > > > > T > > > 8460 N=m > > > > > > > > > > > Nyy > > > > > > > > > > > > > T =

> 0:05 N m=m > > > > > > > > > Mxx > > > > > > > > > > > > T > > > > > 0:05 N m=m M > > > > yy > > > > ; > : > > > : MT ; 0:03 N m=m xy

We can now calculate midplane strains and curvatures using Eq. (45), which becomes:*

8 o 9 2 exx > > > > > > > 6 > > 6 > > eoyy > > > > > > 6 > > = 6 < o > 6 cxy ¼6 6 > > > jxx > 6 > > > > 6 > > > > 6 > > j > > 4 yy > > > > ; : jxy



3:757  10 1:964  10 1:038  10 1:440  10 3:905  10

8 9 8 4 6

1:964  10 1:037  10 4:234  10 1:866  10 6:361  10

8:513  10 5 4:628  104 9 8 4060 þ 8190 > > > > > > > 7360 þ 8460 > > > > > > > > > > = < 2860 233 > > > > > > > > > > :

9 7 8 5 4

1:038  10 4:234  10 2:004  10 3:661  10 3:251  10 1:851  10

8 8 7 4 4 5

1:440  10 1:866  10 3:661  10

4 5 4

7:064 3:122  10

2

4:572

3:905  10 6:361  10 3:251  10 3:122  10 6:429 3:620

6 4 4 2

8:153  10

3

7 4:628  104 7 7 7 1:851  10 5 7 7 7 4:572 7 7 7 3:620 5

17:41

0:62 þ 0:05 > > > > > > 0:62 0:05 > > > ; 0:36 þ 0:03

* The [abd] matrix for a [30/0/90]T graphite-epoxy laminate was calculated in Sample Problem 3, and the thermal stress and moment resultants were calculated in Sample Problem 5.

Copyright © 2004 by Marcel Dekker, Inc. All Rights Reserved.

5

9 8 o 9 8 18 Am=m > exx > > > > > > > > > > > > o > > > > > > 509 Am=m > e > > > > yy > > > > > > > > = = < < 420 Arad o cxy ¼ > > 1:0 m 1 > jxx > > > > > > > > > > > > > > > > > 1 > > > > 5:0 m j yy > > > > > > > > ; : ; : jxy 4:4 m 1

(b) Ply strains relative to the x–y coordinate system. Ply strains may now be calculated using Eq. (12). For example, strains present at the outer surface of ply 1 (i.e., strains present at z0= 0.0001875 m) are: 9 8 e  > = < xx > eyy  > > ; : c xy

9 8 8 o 9 9 8 exx > jxx > > > 18  10 6 m=m > > > > > > > < = < < = > = ¼ eoyy þ z0 jyy ¼ 509  10 6 m=m þ ð 0:0001875 mÞ > > > > > > > > > > ; > ; ; : : : o > jxy cxy 420  10 6 m=m z¼z0 8 9 1:0 rad=m > > > > < =  5:0 rad=m > > > > : ; 4:4 rad=m

8 9  > < exx > = eyy  > : cxy > ;

8 9 206 Am=m > > = < ¼ 1450 Am=m > > ; : 1240 Am=m z¼z0

Strains calculated at the remaining ply interface positions are summarized in Table 9.

Table 9 Ply Interface Strains in a [30/0/90]T Graphite-Epoxy Laminate Caused by the Combined Effects of Cooldown from 175jC to 20jC and an Increase in Moisture Content of +0.5% z-coordinate (mm) 0.1875 0.0625 0.0625 0.1875

exx (Am/m) 206 80 44 170

eyy (Am/m) 1450 820 190 440

Strains are referenced to the x–y coordinate system.

Copyright © 2004 by Marcel Dekker, Inc. All Rights Reserved.

cxy (Arad) 1240 690 150 400

(c) Ply stresses relative to the x–y coordinate system. Ply stresses may now be calculated using Eq. (30) of Chap. 5. The stresses present at the outer surface of ply 1 are (i.e., at z=z0): 9ply 1 2 8 3ply 1 8 9  rxx > > Q11 Q12 Q16  exx DTaxx DMbxx > > > > = = < < 7 6  7 e DTa DMb ryy  ¼6  Q Q Q yy yy yy 22 26 5 4 12 > > > >  > ; : > ; : sxy Q16 Q26 Q66 z ¼ z0 cxy DTaxy DMbxy z ¼ z0 z ¼ z0 2 9ply 1 8 107:6  109  r < xx = 6 9 ryy  ¼6 4 26:06  10 ; : sxy z ¼ z0 48:13  109 

8 > >
> =

½ð 1450Þ ð 155Þð20:0Þ ð0:5Þð3638ފ  10 6 > > > ; : ½ð1240Þ ð 155Þð 24:2Þ ð0:5Þð 4027ފ  10 6 >

9ply 1 8 25 MPa 9 8  > > > > > = = < < rxx >  ryy  2:4 MPa ¼ > > > ; > : sxy > ; : 2:2 MPa > z ¼ z0

Stresses calculated at the remaining plies and ply interface positions are summarized in Table 10. A comparison of the results obtained in Example Problems 6 and 7 leads to the following observation: The initial ply stresses and strains caused by cooldown from cure temperatures to room temperatures are partially relieved Table 10 Ply Interface Stresses in a [30/0/90]T Graphite-Epoxy Laminate Caused by the Combined Effects of Cooldown from 175jC to 20jC and an Increase in Moisture Content of +0.5% Ply number Ply 1 Ply 2 Ply 3

z-coordinate (mm) 0.1875 0.0625 0.0625 0.0625 0.0625 0.1875

rxx (MPa)

ryy (MPa)

25 1.4 20 40 16 17

Stressed are referenced to the x–y coordinate system.

Copyright © 2004 by Marcel Dekker, Inc. All Rights Reserved.

2.2 0.24 9.3 15 65 43

sxy (MPa) 2.2 9.9 9.0 1.9 1.9 5.2

by the subsequent adsorption of moisture. Although the interaction between temperature and moisture effects obviously depends on the details of the situation (material properties involved, stacking sequence, magnitudes of DT and DM, etc.), this observation is often true. That is, the thermal stresses predicted to develop in a multiangle laminate during cooldown are usually predicted to be relieved somewhat by subsequent adsorption of moisture. 7 SIMPLIFICATIONS DUE TO STACKING SEQUENCE Eqs. (44) and (45) summarize the response of a multiangle composite laminate due to the combined effects of uniform mechanical loading, uniform changes in temperature, and/or uniform changes in moisture content. The primary objective of this section is to show that these equations may be substantially simplified through proper selection of the laminate stacking sequence. Before these simplifications are discussed, however, it is illustrative to consider the simplest case of all-specifically, let us consider Eqs. (44) and (45) when applied to a plate of total thickness t made from an isotropic material. Recall that for isotropic materials, all properties are independent of direction. For present purposes, let: E11 ¼ E22 ¼ E

m12 ¼ m21 ¼ m G12 ¼ G

a11 ¼ a22 ¼ a b11 ¼ b22 ¼ b

Also recall that for isotropic materials, only two of the elastic moduli are independent. That is: G¼

E 2ð1 þ mÞ

If these interrelations between material properties are enforced, then Eq. (44) reduces to: A11 6 6 A12 6 6 6 6 0 6 6 ¼6 > > 6 0 M > > xx > > 6 > > > > 6 > > > > 0 > 6 > Myy > > 6 > > > > ; 6 : 4 Mxy 0 8 Nxx > > > > > > Nyy > > > > > < Nxy

9 > > > > > > > > > > > =

2

A12

0

0

0

0

A11

0

0

0

0

0

0

D11

D12

0 0



A11

A12 2 0



0

0

D12

D11

0

0

0

0



D11

3

8 o 9 7> > > exx > > 7> > > > 7> > > eo > > 7> > yy > 7> > > > > > > 7 > 0 = 7 < co > 7 xy 7 0 > 7> > > > jxx > 7> > > 7> > > 7> 0 > >j > 7> yy > > > >  7> > > > D12 5> ; : jxy 2

Copyright © 2004 by Marcel Dekker, Inc. All Rights Reserved.

8 T9 N > > > > > > > > > > > T> > > > > N > > > > > > > > > = < 0 > > > > > > 0 > > > > > > > > > > > > > > 0 > > > > > > > ; : 0

8 M9 N > > > > > > > > > > > M> > > > > N > > > > > > > > > = < 0 > > > > 0 > > > > > > > > > > > > > > > 0 > > > > > > > > ; : 0

(46)

where: A11 ¼

Et 1 m2

A12 ¼ mA11 ¼ Et3 12ð1 m2 Þ

D11 ¼ D22 ¼ D66 ¼

ðD11

2

D12 Þ

¼

mEt 1 m2

D12 ¼ mD11 ¼

mEt3 12ð1 m2 Þ

Et3 24ð1 þ mÞ



Eta NT ¼ DT ð1 mÞ

NM ¼ DM



Etb ð1 mÞ



The constant D11 is often called the flexural rigidity of an isotropic plate. Taking the inverse of Eq. (46), we find: 8 o 9 2a 11 exx > > > > 6 > > > > > a12 o > > 6 > > > 6 > eyy > > = 6 < o > 0 cxy ¼ 6 6 6 0 > > > > 6 > jxx > > > 6 > > > > 6 0 j > > yy > > 4 > > ; : jxy 0

a12

0

0

0

0

a11

0

0

0

0

0

0

0

d11

d12

0 0

0 0

2ða11

0

a12 Þ

0

0

d12

d11

0

0

0

0

2ðd11

where:

a11 ¼

1 1 ¼ A11 ð1 m2 Þ Et

d11 ¼ d22 ¼ d66 ¼ 2ðd11

a12 ¼

1 12 ¼ D11 ð1 m2 Þ Et3 d12 Þ ¼

ma11

d12 ¼

9 38 Nxx þ NT þ NM > > > > > > 7> > T M> 7> > > N þ N þ N > > yy > 7> > > > 7> = < 7 N xy 7 7> > Mxx > 7> > > > 7> > > > > 7> > M > yy 5> > > > > ; : Mxy d12 Þ

m A11 ð1 md11 ¼

m2 Þ

¼

m Et

m D11 ð1

m2 Þ

¼

12m Et3

24ð1 þ mÞ Et3

Comparing Eqs. (44) and (45) with Eqs. (46) and (47), it is apparent that multiangle composite laminates may exhibit unusual coupling effects, as compared to the more familiar behavior of isotropic plates. For example, referring to Eq. (45), it can be seen that application of a normal stress resultant Nxx will (in general) induce a midplane shear strain cjxy and curvatures jxx,

Copyright © 2004 by Marcel Dekker, Inc. All Rights Reserved.

ð47Þ

jyy, and jxy, due to the presence of the a16, b11, b12, and b16 terms, respectively. Physically, these means that a uniform in-plane uniaxial loading will cause inplane shear strains as well as out-of-plane curvatures in a composite plate (i.e., the plate will bend). These couplings do not exist for isotropic panels, as indicated by Eq. (47). Unusual couplings between thermal resultants and laminate strains and curvatures also exist, and are immediately apparent in practice. As previously discussed, most modern composite material systems are cured at elevated temperatures and are subsequently cooled to room temperatures. Therefore, thermal stress and moment resultants (NijT and MijT, respectively) develop during cooldown. Equation (45) shows that the thermal stress resultants NijT and MijT will cause curvatures jxx, jyy, and jxy to develop upon cooldown. Physically, this means that (in general) a composite laminate that is flat at the elevated cure temperature will bend and warp as it is cooled to room temperature. Coupling effects due to moisture stress and moment resultants, NijT and MijT, are analogous to those associated with thermal stress and moment resultants. Thus, even if a composite laminate is cured and used at the same temperature (so that NijT=MijT=0), the laminate may still bend or warp if the surrounding humidity causes the moisture content of the laminate to change with time. Because coupling effects greatly complicate the design of composite structures, it is of interest to determine whether these coupling effects can be reduced or eliminated. It will be seen that it is indeed possible to reduce or eliminate many of these coupling effects through proper selection of the laminate stacking sequence. Common stacking sequences used to eliminate coupling effects are described in separate sections below. 7.1 Symmetrical Laminates A symmetrical laminate is one that possesses both geometrical and material symmetry about the midplane. In a symmetrical laminate, plies located symmetrically about the laminate midplane are of the same material, have the same thickness, and have the same fiber angle. Several examples of symmetrical stacking sequences have been previously shown in Fig. 12. For a symmetrical n-ply laminate, the material and fiber angle used in ply 1 is identical to that used in ply n, the material and fiber angle used in ply 2 is identical to that used ply n-1, etc. Use of a symmetrical stacking sequence results in three major simplifications to Eqs. (44) and (45). Specifically, for a symmetrical laminate: All coupling stiffnesses equal zero (Bij=0). All thermal moment resultants equal zero (MijT=0). All moisture moment resultants equal zero (MijM=0).

Copyright © 2004 by Marcel Dekker, Inc. All Rights Reserved.

To demonstrate that coupling stiffnesses are zero for a symmetrical laminate, consider the coupling stiffness B11. From Eq. (27b), B11 is given by: B11 ¼

n   1X Q11 k z2k 2 k¼1

z2k

1



In expanded form, B11 is given by: B11 ¼

           1 n Q11 1 z21 z20 þ Q11 2 z22 z21 þ Q11 3 z23 z22 2         þ : : : þ Q11 n 2 z2n 2 z2n 3 þ Q11 n 1 z2n 1 z2n 2     þ Q11 n z2n z2n 1 g

Because the laminate is assumed to be symmetrical, it must be that:     Q11 1 ¼ Q11 n     Q11 2 ¼ Q11 n 1     Q11 3 ¼ Q11 n 2


etc:

ð48Þ

ð49aÞ

Also, due to symmetry, the ply interface positions are located symmetrically about the midplane, and hence (recalling that z00): z0 ¼ zn z1 ¼ zn

z2 ¼ zn


etc:

1 2

ð49bÞ

Together, the relations listed as Eqs. (49a) and (49b) imply that for any symmetrical laminate:         Q11 1 z21 z20 ¼ Q11 n z2n z2n 1         ð50Þ Q11 2 z22 z21 ¼ Q11 n 1 z2n 1 z2n 2       2   Q11 3 z3 z22 ¼ Q11 n 2 z2n 2 z2n 3


etc:

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Substituting Eq. (50) into Eq. (48), it is seen that B11 = 0. Similar results may be demonstrated for all other coupling stiffnesses, and hence Bij = 0 for any symmetrical laminate, as stated. To demonstrate that thermal moment resultants are zero for a symmetT . From Eq. (42d), rical laminate, consider the thermal moment resultant Mxx T Mxx is given by: MTxx ¼

n    DT X Q11 axx þ Q12 ayy þ Q16 axy k z2k 2 k¼1

T In expanded form, Mxx is given by: (   DT  Q11 axx þ Q12 ayy þ Q16 axy 1 z21 MTxx ¼ 2

z2k

z20

   þ Q11 axx þ Q12 ayy þ Q16 axy 2 z22    þ Q11 axx þ Q12 ayy þ Q16 axy 3 z23

þ......    þ Q11 axx þ Q12 ayy þ Q16 axy n 2 z2n    þ Q11 axx þ Q12 ayy þ Q16 axy n 1 z2n    þ Q11 axx þ Q12 ayy þ Q16 axy n z2n

1





z21 z22

  z2n

2 1

z2n

1

ð51Þ 3

z2n 2 ) 

 

Because the laminate is assumed to be symmetrical, it must be that:   2 z 1 1    2 Q11 axx þ Q12 ayy þ Q16 axy 2 z2    Q11 axx þ Q12 ayy þ Q16 axy 3 z23 

Q11 axx þ Q12 ayy þ Q16 axy

 z20 ¼  z21 ¼  z22 ¼

  2  z z2n 1 n n     Q11 axx þ Q12 ayy þ Q16 axy n 1 z2n 1 z2n 2     Q11 axx þ Q12 ayy þ Q16 axy n 2 z2n 2 z2n 3 

Q11 axx þ Q12 ayy þ Q16 axy

(52)

T =0 for a symmetrical Substituting Eq. (52) into Eq. (51), it is seen that Mxx T T laminate. Similar results may be demonstrated for Myy and Mxy , and hence all thermal moment resultants equal zero for any symmetrical laminate, as stated. Therefore, a symmetrical laminate will not bend or warp when subjected to a uniform change in temperature. In particular, a symmetrical laminate will not warp or bend during cooldown from the cure temperature.

Copyright © 2004 by Marcel Dekker, Inc. All Rights Reserved.

An identical procedure may be used to prove that all moisture moment resultants are zero for a symmetrical laminate. In summary then, for symmetrical laminates, Eqs. (44) and (45) reduce to: 9 2 8 38 eo 9 8 T 9 8 M 9 Nxx > Nxx > > > A11 A12 A16 0 0 0 > Nxx > xx > > > > > > > > > > > > > > > > > > > > > > > 7 6 > > > > > > > > o T M> > > > > > > > A12 A22 A26 0 0 0 7> N e N N yy > > > > > > > 6 yy > yy > yy > > > > > > > > > > > > > > > 7 6 = = < < N = 6A < < = T M o 7 A26 A66 0 0 0 xy 16 N N c xy xy 7 xy ¼6 7 6 > > > Mxx > 0 0 D11 D12 D16 7> > > > > jxx > > > > 0 > > 0 > 6 0 > > > > > > > > > > > > > > > 7> 6 > > > > > > > > > > > > > > 5> 4 0 Myy > 0 0 D D D 12 22 26 j > > > > > > > > 0 0 yy > > > > > > > > > > > > ; : > > : ; : ; ; : Mxy 0 0 0 D16 D26 D66 jxy 0 0 (53a)

8 o 9 2 a11 exx > > > 6 > > > > eo > > > 6 a12 > > > > 6 = < yy a coxy ¼ 6 6 16 6 0 > > > > 6 > jxx > > > 6 > > 4 0 > jyy > > > ; : jxy 0

a12 a22 a26

a16 a26 a66

0 0 0

0 0 0

0

0

d11

d12

0

0

d12

d22

0

0

d16

d26

9 38 Nxx þ NTxx þ NM > 0 > xx > > > > > > 7 > > T M > 0 7> N þ N þ N yy > yy yy > > > > 7> < = M T 7 0 N þ N þ N xy xy xy 7 7 > d16 7> Mxx > > > > > 7> > > > d26 5> > M > yy > > > > ; : d66 Mxy

ð53bÞ

Due to these dramatical simplifications, symmetrical laminates are almost always used in practice. In those rare circumstances in which the use of a nonsymmetrical laminate is required for some reason, it is best to place the nonsymmetrical ply (or plies) at or near the laminate midplane, which will minimize the coupling stiffnesses and thermal and moisture moment resultants. It is noted in passing that unidirectional composite laminates [h]n are symmetrical and hence Bij=MijT=MijM=0 for unidirectional laminates. In T addition, for the case of [0j]n or [90j]n laminates, A16=A26=D16=D26=Nxy M =Nxy ¼ 0. 7.2 Cross-Ply Laminates Composite laminates that contain plies with fiber angles of 0j or 90j only are called ‘‘cross-ply’’ laminates. Inspection of Eq. (31) of Chap. 5 reveals that for any 0j or 90j ply, Q16=Q26 ¼ 0. From (Eq. (27a) (27b) and (34) it is seen that A16, A26, B16, B26, D16, and D26 all involve a summation of terms involving Q16 and Q26 , and hence all of these terms also equal zero for cross-ply laminates. Furthermore, from Eq. (25) of Chap. 5, it is seen that axy=0 for any 0j or 90j ply, and from Eq. (38) of Chap. 5 that bxy=0 for any 0j or 90j ply. From Eqs. T T M M (42c), (42d), (42e), and (42f), it is seen that Nxy ¼ Mxy ¼ Nxy ¼ Mxy ¼ 0 for

Copyright © 2004 by Marcel Dekker, Inc. All Rights Reserved.

cross-ply laminates (because Q16 ¼ Q26 ¼ axy ¼ bxy ¼ 0Þ . For cross-ply laminates then, Eqs. (44) and (45) reduce to: 9 2 8 38 eo 9 8 T 9 8 M 9 > Nxx > Nxx > A11 A12 0 B11 B12 0 > Nxx > xx > > > > > > > > > > > > > > > > > 6 > > > o > > > > > 7 > > > > > > > T > M > > > > > > > > A A 0 B B 0 N e N N 22 12 22 yy > > > > 7> 6 12 yy > yy > yy > > > > > > > > > > > > > > > > > 7 6

> Mxx > 0 D11 D12 0 7 > > > jxx > > > Mxx > > Mxx > > > > 7> 6 B11 B12 > > > > > 6 > > > > T > > M> > > > 7> > > > > > > > > > > > > > > > 5 4 M B 0 D D 0 B yy > Myy > Myy > 12 22 12 22 > > jyy > > > > > > > > > > > > ; : > > > > > > Mxy 0 0 B66 0 0 D66 : jxy ; : 0 ; : 0 ; and 8 o 9 2 exx > a11 > > > > > 6 > o > > > > > a12 e > > yy > > 6 > < o > = 6 6 0 cxy ¼6 6b > > > jxx > 6 11 > > > > 6 > > > > 4 b12 > > jyy > > > > : ; 0 jxy

ð54aÞ

a12

0

b11

b12

a22

0

b21

b22

0

a66

0

0

b21

0

d11

d12

b22

0

d12

d22

0

b66

0

0

38 N þ NT þ NM 9 > xx xx xx > > > > > > T M > > > > > N þ N þ N 0 7 yy 7> > yy yy > > > 7> < = Nxy b66 7 7 > 0 7 > Mxx þ MTxx þ MM 7> xx > > > > 7> > > > > T M 5 0 > > M þ M þ M yy > yy yy > > > : ; d66 Mxy 0

ð54bÞ

If a cross-ply laminate is also symmetrical, then all remaining coupling stiffnesses equal zero, as well as all remaining thermal and moisture moment resultants. Hence, for symmetrical cross-ply laminates, Eqs. (54a) and (54b) are further simplified to: 8 9 2 38 eo 9 8 T 9 8 M 9 > > Nxx > Nxx > 0 0 0 0 > A11 A12 Nxx > xx > > > > > > > > > > > > > > > > > > > > > > > > 7 6 > > > > > > > o T M> > > > > > > > A12 A22 0 0 0 0 7> N e N N yy > > > > > > > 6 yy > yy > yy > > > > > > > > > > 6 > > > > 7> < Nxy = 6 0 0 A66 0 0 0 7< co = < 0 = < 0 = 7 xy ¼6 6 0 > > > Mxx > 0 7 0 0 D11 D12 0 > 0 > > > > > > > > > jxx > 7> 6 > > > > > > > > > > > > > > > 7> 6 > > > > > > > > > > > > > > 5> 4 0 Myy > D 0 0 0 D 0 0 12 22 j > > > > > > > > yy > > > > > > > > > : > : > > ; > > : ; ; ; : Mxy 0 0 0 0 0 D66 0 0 jxy ð55aÞ

(and) 8 o 9 2 a11 e > > > > > > xx > > 6 > eo > > > 6 a12 > > = 6 0 < yy o cxy ¼ 6 6 > 6 0 > j > xx > > > > 6 > > 4 0 jyy > > > > > ; : jxy 0

a12 a22 0 0 0 0

0 0 a66 0 0 0

0 0 0 d11 d12 0

0 0 0 d12 d22 0

9 38 0 > Nxx þ NTxx þ NM > xx > > > > > > > 7 > T M 0 7> > N þ N þ N yy > yy yy > > > = < 7 0 7 Nxy 7 > 0 7> Mxx > > > 7> > > > 0 5> > > M yy > > > > ; : d66 Mxy

Copyright © 2004 by Marcel Dekker, Inc. All Rights Reserved.

ð55bÞ

Note from Eqs. (55a) and (55b) that symmetrical cross-ply laminates do not exhibit any coupling stiffnesses. That is: A16 ¼ A26 ¼ D16 ¼ D26 ¼ Bij ¼ 0 or, equivalently, a16 ¼ a26 ¼ d16 ¼ d26 ¼ bij ¼ 0 Laminates that do not possess these coupling stiffnesses are called specially orthotropic laminates. The fact that symmetrical cross-ply laminates are specially orthotropic will become an important factor in Chaps. 9 and 10, where the mechanical response of such laminates subjected to varying loads will be considered. 7.3 Balanced Laminates A laminate is ‘‘balanced’’ if, for every ply with fiber angle h, there exists a second ply whose fiber angle is h. The two plies must be otherwise identical, (i.e. they must be composed of the same material and have the same thickness). Inspection of Eq. (31) of Chap. 5 reveals that the Q16 and Q26 terms for these balanced plies will always be of equal magnitude but opposite algebraic sign (i.e., Q16 jh ¼ Q16 j h and Q26 jh ¼ Q26 j h Þ. Consequently, from Eq. (27a), A16=A26=0 for a balanced laminate. Further, from Eqs. (25) and (28) of Chap. 5, it is seen that axy and bxy for the two balanced plies will always be of equal magnitude but opposite algebraic sign (i.e., axy jh ¼ axy j h and bxy jh T ¼ bxy j h Þ . Consequently, from Eqs. (42c) and (43), Nxy =NM xy=0 for a balanced laminate. Equations (44) and (45) are therefore simplified to: 8 9 8 9 2 3 eoxx > 8 NT 9 8 NM 9 > > > xx > Nxx > xx > 0 B11 B12 B16 > A11 A12 > > > > > > > > > > > > > > > > > > > > o > T > M > > Nyy > > 6 A12 A22 7 > > > > > > e N N 0 B B B > > > > > > > yy > yy > 12 22 26 7> yy > > > > > > 6 > > > > > > > = 6 0

= < < = < 7 o 0 A66 B16 B26 B66 7 cxy 0 0 = xy 6 ¼6 7 MTxx > MM > > > 6 B11 B12 B16 D11 D12 D16 7> Mxx > xx > > > > > > > > jxx > > > > > > > > > 7> 6 > > > > > > > T M> > > > > > > > > 5 4 M B12 B22 B26 D12 D22 D26 > M M > > > > > > yy > yy yy > > > > j > > > > yy > : ; > > > > > ; : > > : T M ; Mxy B16 B26 B66 D16 D26 D66 : j ; M M xy xy xy ð56aÞ

and

8 o 9 2 a11 exx > > > 6 > > > > eo > a12 > > > 6 > yy > = 6 < o > 6 cxy ¼ 6 0 6b > > > jxx > 6 11 > > > > 6 > jyy > > > 4 b12 > > ; : jxy b16

0

b11

b12

a22

0

b21

b22

0

a66

b61

b62

b12

b61

d11

d12

b22

b62

d12

d22

b26

b66

d16

d26

a12

9 8 3 Nxx þ NT þ NM > > xx xx b16 > > > > > > > T M > 7 > > N þ N þ N > yy b26 7> yy yy > > > > > 7> = < Nxy b66 7 7 T M d16 7 > > Mxx þ Mxx þ Mxx > 7> > > > 7> > > Myy þ MT þ MM > d26 5> > yy yy > > > > > > > : T M d66 M þM þM ; xy

Copyright © 2004 by Marcel Dekker, Inc. All Rights Reserved.

xy

xy

ð56bÞ

If a balanced laminate is also symmetrical, then Eqs. (56a) and (56b) are simplified to: 9 2 8 A11 Nxx > > > > > > 6A > > > > N 6 12 > yy > > > > 6 > > = 6 0 < Nxy > 6 ¼6 M > > 6 0 xx > > > > 6 > > > > M > 6 > yy > > 4 0 > > > > : Mxy ; 0

A12 A22 0 0 0 0

0 0 A66 0 0 0

0 0 0 D11 D12 D16

0 0 0 D12 D22 D26

3 0 8 eo 9 > > > xx > 0 7 7> > > eo > > yy > > 7> > < o > = 0 7 7 cxy 7 D16 7> jxx > > > > 7> > > > jyy > D26 7 > > > > 5: ; j xy D 66

8 T 9 Nxx > > > > T > > > > > Nyy > > > > > > > > < 0 > = 0 > > > > > > > > > > > 0 > > > > > > : 0 > ;

8 M9 Nxx > > > > M> > > > > Nyy > > > > > > > > < 0 > = 0 > > > > > > > > > > > 0 > > > > > > : 0 > ; ð57aÞ

and 8 o 9 2 a11 exx > > > > > 6 a12 > > o > > 6 > e > yy > > < o > = 6 0 cxy ¼ 6 6 6 0 > > j > > xx > 6 > > > > > 6 > > jyy > > 4 0 : ; jxy 0

a12 a22 0

0 0 a66

0 0 0

0 0 0

0

0

d11

d12

0

0

d12

d22

0

0

d16

d26

9 38 Nxx þ NTxx þ NM > xx > > > > > 7> > > Nyy þ NTyy þ NM > 7> yy > > > > 7> = 7< N xy 7 7 > d16 7> Mxx > > > > > 7> > > > d26 5> M > > yy > > ; : d66 Mxy 0 0 0

ð57bÞ

7.4 Balanced Angle-Ply Laminates The definition of a ‘‘balanced’’ laminate was given in Sec. 7.4. A balanced angle-ply laminate is really just a special class of balanced laminates; it is discussed in a separate subsection because of an additional simplification that occurs. All plies in an angle-ply laminate have a fiber angle of the same magnitude. That is, all plies within an angle-ply laminate have a fiber angle of either+h or h, where the value h is the same for all plies. In general, for an angle-ply laminate, the number of plies with fiber angle+h may differ from the number of plies with fiber angle h. However, a balanced angle-ply laminate must have an equal number of plies with fiber angle+h and h, so as to satisfy the preceding definition of a balanced laminate. Carefully note the distinction between a balanced laminate and balanced angle-ply laminate. A balanced laminate may involve more than one ‘‘distinct’’ fiber angle. For example, a [35/65/-35/-65]T laminate is balanced (although not symmetrical), and A16=A26=0 for this laminate. In contrast, a balanced angle-ply laminate may involve only one ‘‘distinct’’ angle. A stacking sequence of either [35/-35/

Copyright © 2004 by Marcel Dekker, Inc. All Rights Reserved.

35/-35]T or [65/65/-65/-65]T result in balanced angle-ply laminates, for example. The following simplifications occur for a balanced angle-ply laminate: B11 ¼ B22 ¼ B66 ¼ MTxx ¼ MTyy ¼ MTxx ¼ MTyy ¼ 0 In addition, the simplifications that exist for any balanced laminate T M (A16=A26=Nxy =Nxy =0) also occur for a balanced angle-ply laminate. Equations (44) and (45) are therefore simplified to: 9 2 8 A11 Nxx > > > > > > 6 > > > A12 Nyy > > > > 6 > > = 6

6 0 xy ¼6 6 0 > > M > 6 > xx > > > 6 > > > 4 0 > Myy > > > > > ; : Mxy B16

A12 A22 0 0 0 B26

0 0 A66 B16 B26 0

0 0 B16 D11 D12 D16

0 0 B26 D12 D22 D26

3 B16 8 eo 9 > > > > > xx > B26 7 o > > > 7> e > yy > > 7> = < 7 0 7 coxy 7 D16 7> > > jxx > > > > 7> > > jyy > D26 5> > > ; : j xy D 66

8 T 9 Nxx > > > > > > > > > > > NTyy > > > > > > < 0 > = > 0 > > > > > > > > > 0 > > > > > > > : T > ; Mxy

8 M9 Nxx > > > > > > > > > > > > NM > > yy > > > < 0 > = > 0 > > > > > > > > > 0 > > > > > > > : M> ; Mxy ð58aÞ

and 2 8 o 9 a11 e xx > > 6 > > > > > eo > 6a > > 6 12 > yy > > = 6 0 < o > cxy ¼ 6 6 > 6 0 > j > > 6 > xx > > > > > 6 > jyy > > > ; 4 0 : jxy b

16

a12

0

0

0

a22

0

0

0

0

a66

b16

b62

0

b61

d11

d12

0

b62

d12

d22

b26

0

d16

d26

9 38 Nxx þ NTxx þ NM b16 > xx > > > > 7> > > > b26 7> > Nyy þ NTyy þ NM > yy > > > 7> > > < = 7 Nxy b66 7 7 > > Mxx d16 7 > > > 7> > > > 7> > M d26 5> yy > > > > > > : ; T M d66 Mxy þ Mxy þ Mxy

ð58bÞ

If a balanced angle-ply laminate is also symmetrical, then Eqs. (58a) and (58b) are simplified still further to: 8 9 2 A11 Nxx > > > > 6 > > > > > 6 A12 Nyy > > > > > > > 6 < Nxy = 6 6 0 ¼6 > 6 0 Mxx > > > > > 6 > > > > 6 > > M > yy > > 4 0 > : ; Mxy 0

0

0

A22

0

0

0

0

A66

0

0

A12

0

0

0

D11

D12

0

0

D12

D22

0

0

D16

D26

3 8 o 9 > exx > 7> > 0 7> > eo > > > > > 7> yy > < o > = 0 7 7 cxy 7 > jxx > D16 7 > > > 7> > > > > > > j D26 7 yy > > 5: ; j xy D 0

66

Copyright © 2004 by Marcel Dekker, Inc. All Rights Reserved.

8 T 9 N > > > > xx > > > > > > > NTyy > > > > > > > < 0 = > 0 > > > > > > > > > > > 0 > > > > > > : ; 0

8 M9 N > > > > xx > > > > > > > > NM > yy > > > > > < 0 = > 0 > > > > > > > > > > > 0 > > > > > > : ; 0 ð59aÞ

and 8 o 9 2a 11 e > > > 6 > xx > > > > > 6 a12 > > eoyy > > > > > 6 > < o > = 6 6 0 cxy ¼6 6 0 > > > 6 jxx > > > > > 6 > > > > > 6 > jyy > > 4 0 > > ; : jxy 0

a12

0

0

0

a22

0

0

0

0

a66

0

0

0

0

d11

d12

0

0

d12

d22

0

0

d16

d26

9 38 Nxx þ NTxx þ NM > xx > > > > > 7> > T M > > > 0 7> N þ N þ N yy yy yy > 7> > > > > > 7> < = 0 7 Nxy 7 > > d16 7 > Mxx 7> > > > 7> > > > > 7 > > d26 5> M > yy > > > > ; : d66 Mxy 0

ð59bÞ

7.5 Quasi-Isotropic Laminates A quasi-isotropic laminate is one that satisfies the following conditions: 

Three or more distinct fiber angles must be present within a laminate. The number of distinct fiber angles will be denoted m, and hence if a laminate is quasi-isotropic, then m z 3.  The m distinct fiber angles must appear at equal increments of (180/ m) degrees.  An equal number of plies must be present at each of the m distinct fiber angles. It can be shown (see Ref. 2) that if the three conditions are met, then members of the Aij matrix for the laminate are related as follows: A11 ¼ A22

1 ðA11 2

A66 ¼

A12 Þ

A16 ¼ A26 ¼ 0

Now, these same relations between extensional stiffness also hold for an isotropic plate (see Eq. (46)). Hence, laminates that satisfy the above conditions are called quasi-isotropic laminates. Also, for a quasi-isotropic T T M M laminate, Nxy ¼ Mxy ¼ Nxy ¼ Mxy ¼ 0 . In this case, Eqs. (44) and (45) reduce to: 8 N > > > xx > > > Nyy > > > > > > < Nxy

9 > > > > > > > > > > > > =

A11

A12

6 6 A12 6 6 6 6 ¼6 0 > > 6 M > > xx > > 6 B11 > > > > 6 > > > > > > Myy > 6 > 4 B12 > > > > > > : ; B16 Mxy

A22

2

0 B12

0 0 

A11

A12

2 B16



B11

B12

B12

B22

B16

B26

D11

D12

B22

B26

D12

D22

B26

B66

D16

D26

38 o 9 e > B16 > > > xx > > > 7> > > eo > > B26 7> > > > yy 7> > > > 7> > o > > = 7< B66 7 cxy 7 7> j > > xx > > D16 7> > > > 7> > > > > 7 > > j D26 5> > yy > > > > > > : ; D66 jxy

Copyright © 2004 by Marcel Dekker, Inc. All Rights Reserved.

8 T Nxx > > > > > T > > > Nyy > > > < 0

9 > > > > > > > > > > > =

> MTxx > > > > > > > > > > > MT > > > > > > yy > > > > : ; 0

8 M Nxx > > > > > > M > > Nyy > > > < 0

9 > > > > > > > > > > > =

> > MM > xx > > > > > > > > > MM > > > > > > yy > > > > ; : 0

ð60aÞ

2 8 o 9 a11 e xx > > 6a > > > > > 6 12 eoyy > > > > 6 > > = 6 0 < o > cxy 6 ¼6 > 6 b11 jxx > > > > > 6 > > > > 6b j yy > > > > : j ; 4 12 xy b16

0

a12

0

a22 0 b21

b11

2ða11

b61

b12

b21

b22

a12 Þ b61

b62

d11

d12

b22

b62

d12

d22

b26

b66

d16

d26

9 38 > Nxx þ NTxx þ NM > b16 > xx > > > > > 7 > > T M > > b26 7> N þ N þ N yy > yy yy > > > 7> > > = < N b66 7 xy 7 7 d16 7> Mxx þ MTxx þ MM xx > > > > 7> > > > 7 T M > d26 5> > M þ M þ M > > yy yy yy > > > > > > ; d66 : Mxy ð60bÞ

The simplest possible quasi-isotropic laminate contains three plies, oriented at equal increments of (180j/3)=60j. For example, [0/60/ 60]T or [60/0/ 60]T laminates are quasi-isotropic. Probably the most common quasiisotropic laminate involves four distinct fiber angles (m=4). These laminates must have ply angles oriented at increments of (180j/4)=45j. Typical stacking sequences in this case are [0/45/90/ 45]T or [45/0/ 45/90]T. Although the extensional stiffnesses for these laminates are ‘‘quasi-isotropic’’, they still exhibit coupling stiffnesses (i.e., Bij p 0); furthermore, the bending stiffnesses are not isotropic (e.g., D11 p D22). If a quasi-isotropic laminate is also symmetrical (e.g., [0/45/90/ 45]s), then the coupling stiffnesses Bij=0, and all remaining thermal moment resultants and moisture moment resultants equal zero. Hence, Eqs. (60a) and (60b) are simplified still further: 8 Nxx > > > > > > Nyy > > > > < Nxy

9 > > > > > > > > > > =

A11

A12

6A 6 12 6 6 6 0 ¼6 6 > Mxx > > > 6 0 > > > > 6 > > > > 6 > Myy > > > 4 0 > > > > ; : Mxy 0

A11

8 o 9 2 exx > > a11 > > > > > 6a > > o > > > e > 6 12 > > yy > > 6 > > = 6 0 < co > 6 xy ¼6 > > j 6 0 xx > > > > 6 > > > > 6 > > 4 0 > jyy > > > > > > 0 ; : jxy >

a12 a11

2

0 0

0

0

0

0

0

0

0

D11

D12

0 

A11

A12 2 0



0

0

D12

D22

0

0

D16

D26

0 0

0 0 2ða11

0

a12 Þ

0 0

0 0

0

0

d11

d12

0

0

d12

d22

0

0

d16

d26

38 o 9 exx > > > > > > > > > o > 0 7 7> > e yy > > > 7> > > > 7> < o = 7 0 7 cxy 7> > > jxx > > D16 7 > > 7> > > > 7> > > j > 5 D26 > yy > > > > ; : jxy D66 0

8 T 9 N > > > > > xx > > > > > > NTyy > > > > > > > > = < 0 > > > 0 > > > > > > > > > > > 0 > > > > > > > ; : 0

8 M9 N > > > > > xx > > > > > > > NM > > yy > > > > > = < 0 > > > 0 > > > > > > > > > > > 0 > > > > > > > ; : 0

ð61aÞ

9 38 0 > Nxx þ NTxx þ NM xx > > > > > > T M > > 0 7 7> > N þ N þ N yy > yy yy > > > 7> > > = < 0 7 Nxy 7 ð61bÞ 7 d16 7> Mxx > > > > 7> > > > > d26 7 > Myy > > 5> > > > > ; d66 : Mxy

Copyright © 2004 by Marcel Dekker, Inc. All Rights Reserved.

Again, note that only the in-plane extensional stiffnesses are ‘‘isotropic’’; the bending stiffnesses for a symmetrical quasi-isotropic laminate are not related in the same manner as in an isotropic plate. It is noted in passing that symmetrical quasi-isotropic laminates are not specially orthotropic, in contrast with symmetrical cross-ply laminates. That is, for symmetrical quasi-isotropic laminates, D16,D26,d16,d26 p 0. 8 SUMMARY OF CLT CALCULATIONS At this point, it is helpful to summarize the calculations steps involved in a composites analysis based on classical lamination theory. Two different analysis requirements are commonly encountered. In the first case, the loads applied to the composite laminate are known, and the objective of the analysis is to determine the ply stresses and strains that will be induced by these loads. This situation is usually encountered during the design process. For example, suppose the wing of a new airplane is being designed, and the skin is to be fabricated using a composite laminate. Typically, the loads that must be supported by the skin, temperatures extremes that will be encountered in service, and the moisture content that may develop over long times are known (having been calculated on the basis of separate analyses), and are supplied as input data to the design engineer. The objective is to design the laminate such that the ply stresses and strains induced by the given mechanical and environmental loads will remain within safe limits. A second case is when the strains induced in a laminate are known, and the objective of the analysis is to determine the loads that caused these strains. This situation is commonly encountered during stress analysis of an existing structure or prototype. For example, suppose a prototype of the wing described in the preceding paragraph has been built, and a stress analyst is assigned to evaluate its performance. During the evaluation process, the entire wing structure is subjected to various aerodynamic maneuvers and/or other loading conditions expected to be encountered during the service life of the aircraft. The strains, temperatures, and changes in moisture content are all measured. The objective is now to deduce the loads that caused these measured strains, as well as to determine the strains and stresses induced within individual plies. A slightly different calculation path is used in these two different scenarios. The calculation procedures will be summarized in Secs. 8.1 and 8.2. 8.1 A CLT Analysis When Loads Are Known The calculation steps followed when the applied loads are known are summarized below:

Copyright © 2004 by Marcel Dekker, Inc. All Rights Reserved.

1. Define the problem: (a) Specify the number of different ply materials to be used in the laminate. (b) Specify the elastic properties for each ply material used (i.e., specify E11, E22, v12, G12, a11, a22, b11, and b22 for each ply material). (c) Specify the laminate description (i.e., specify the number of plies n and the ply material and fiber angle for each ply). (d) Specify the mechanical and environmental loads applied to the laminate (specify Nxx, Nyy, Nxy, Mxx, Myy, Mxy, DT, and DM). 2. Calculate the ABD matrix: (a) Calculate the reduced stiffness matrix Qij for each material used in the laminate (using Eq. (11) of Chap. 5). (b) Calculate the transformed reduced stiffness matrix Qij for each ply based on the appropriate reduced stiffness matrix and ply fiber angle (using Eq. (31) of Chap. 5). (c) Calculate the Aij, Bij, and Dij matrices, using (Eq. (27a), (27b), and (34), respectively. (d) Assemble the ABD matrix. 3. Calculate the inverse of the ABD matrix: abd=ABD 1. 4. Calculate the thermal and moisture stress and moment resultants: (a) Calculate the effective thermal and moisture expansion coefficients for each ply, using Eqs. (25) and (28) of Chap. 5, respectively. (b) Calculate the thermal and moisture stress and moment resultants using Eqs. (42a) (42b) (42c) (42d) (42e) (42f) (43), respectively. 5. Calculate the midplane strains and curvatures induced in the laminate, using Eq. (45). 6. For each ply: (a) Calculate the ply strains in the x–y coordinate system, using Eq. (12). Strains may be calculated at any desired position z, but most often they are calculated at the ply interface positions (i.e., at positions z0,z1,z2, z3,. . ., zn). Strains are also often transformed to the local 1–2 coordinate system, defined by the fiber angle within each ply. (b) Calculate the ply stresses in the x–y coordinate system, using Eq. (30) of Chap. 5. Once again, whereas stresses may be calculated at any desired position z, most often they are calculated at the ply interface positions (i.e., at positions z0, z1,z2,z3, . . ., zn). Stresses

Copyright © 2004 by Marcel Dekker, Inc. All Rights Reserved.

are also often transformed to the local 1–2 coordinate system, defined by the fiber angle within each ply. 8.2 A CLT Analysis When Midplane Strains and Curvatures are Known The calculation steps followed when midplane strains and curvatures are known are summarized below: 1. Define the problem: (a) Specify the number of different ply materials to be used in the laminate. (b) Specify the elastic properties for each ply material used (i.e., specify E11, E22, v12, G12, a11, a22, b11, and b22 for each ply material). (c) Specify the laminate description (i.e., specify the number of plies n and the ply material and fiber angle for each ply). (d) Specify the midplane strains and curvatures and environmental loads experienced by the laminate (specify ejxx, ejyy, cjxy, jxx, jyy, jxy, DT, and DM). 2. Calculate the ABD matrix: (a) Calculate the reduced stiffness matrix Qij for each material used in the laminate (using Eq. (11) of Chap. 5). (b) Calculate the transformed reduced stiffness matrix Q ij for each ply based on the appropriate reduced stiffness matrix and ply fiber angle (using Eq. (31) of Chap. 5). (c) Calculate the Aij, Bij, and Dij matrices, using (Eq. (27a), (27b), and (34), respectively. (d) Assemble the ABD matrix. 3. Calculate the thermal and moisture stress and moment resultants: (a) Calculate the effective thermal and moisture expansion coefficients for each ply, using Eqs. (25) and (28) of Chap. 5, respectively. (b) Calculate the thermal and moisture stress and moment resultants using Eqs. (42a–42f) (43), respectively. 4. Calculate the applied stress and moment resultants, using Eq. (44). 5. For each ply: (a) Calculate the ply strains in the x–y coordinate system, using Eq. (12). Strains may be calculated at any desired position z, but most often they are calculated at the ply interface positions (i.e., at

Copyright © 2004 by Marcel Dekker, Inc. All Rights Reserved.

positions z0, z1,z2,z3, . . . ,zn). Strains are also often transformed to the local 1–2 coordinate system, defined by the fiber angle within each ply. (b) Calculate the ply stresses in the x–y coordinate system, using Eq. (30) of Chap. 5. Once again, whereas stresses may be calculated at any desired position z, most often they are calculated at the ply interface positions (i.e., at positions z0,z1,z2,z3, . . . ,zn). Stresses are also often transformed to the local 1–2 coordinate system, defined by the fiber angle within each ply.

9 EFFECTIVE PROPERTIES OF A COMPOSITE LAMINATE The definitions of common engineering material properties were reviewed in Chap. 3. In this section, these concepts will be used to define the ‘‘effective’’ properties of multiangle composite laminates. Before we begin this discussion, it is pertinent to note that most of the properties defined in Chap. 3 do not take into account the peculiar coupling effects exhibited by general composite laminates. As a typical example, consider the definition of the thermal expansion coefficient. As discussed in Sec. 3.3, the thermal expansion coefficient is defined as the ratio of a thermal strain divided by the temperature change that caused that strain. Six different thermal expansion coefficients may be defined for an anisotropic material (see Eq. (21) of Chap. 3). For example, if a uniform temperature change (DT ) causes a normal strain in the T x-direction (exx ), then the thermal expansion coefficient associated with the xdirection is: axx ¼

eTxx DT

However, this definition does not anticipate the out-of-plane coupling effects exhibited by general composite laminates. That is, for a general composite laminate, a temperature change DT causes both midplane strain ejxx and midplane curvature jxx, and consequently eTxx varies through the laminate thickness. In this case then, what strain should be used to calculate axx? A reasonable approach is to define axx based on the thermal strain induced at the laminate midplane, axx=ejxx /DT, regardless of whether or not a curvature is induced as well. A second thermal property can then be defined as the ratio of midplane curvature to temperature change (jxx/DT ). Although several new properties representing unusual coupling effects can be defined in this manner, this topic will not be pursued here. Recall that coupling effects occur in composite laminates with arbitrary stacking sequences because Bij p 0. Symmetrical laminates are almost always used in practice

Copyright © 2004 by Marcel Dekker, Inc. All Rights Reserved.

so as to insure that Bij = 0. Consequently, these couplings are rarely encountered in practice. The complications due to out-of-plane coupling effects present in general laminates will be ignored in this section. The effective properties for symmetrical composite laminates are discussed in the following subsections. 9.1 Effective Properties Relating Stress to Strain The standard definitions of material properties used to relate stress to strain were discussed in Sec. 3.2. Three types of properties are measured during uniaxial tests: Young’s modulus, Poisson’s ratios, and coefficients of mutual influence of the second kind. Although these properties are usually measured using uniaxial tests, they can also be measured based on strains induced by pure bending. Properties measured for isotropic materials during uniaxial tests are identical to those measured during pure bending tests. For example, for an isotropic material, the value of Young’s modulus measured during a uniaxial test is identical to that measured in pure bending. However, for composite materials, the properties measured during uniaxial tests differ substantially from those measured during pure bending tests. For example, Young’s modulus of a composite as measured during a uniaxial test is substantially different than that measured in pure bending. Therefore, we must distinguish between the two. In the following discussion, those properties measured through application of in-plane loads are called ‘‘extensional’’ properties. In contrast, properties during a pure bending test are called ‘‘flexural’’ properties. Extensional Properties Consider a symmetrical composite laminate subjected to uniaxial loading Nxx, as shown in Fig. 18. The in-plane strains induced in this laminate can be determined using classical lamination theory, in accordance with Eq. (45). Assuming DT=DM=0 (and hence that thermal and moisture stress and moment resultants are zero), and also noting that by definition Nyy=Nxy= Myy=Myy=Mxy=0, Eq. (45) becomes (for symmetrical laminates): 8 o 9 2 9 38 exx > 0 0 > Nxx > a11 a12 a12 0 > > > > > > > > > > 6 a12 a22 a26 0 0 0 7> eoyy > 0 > > > > > > > > > 7 6 < < o = 6a = 0 0 7 0 cxy ¼ 6 16 a26 a66 0 7 6 0 0 0 d11 d12 d16 7 0 > > jxx > > > > > 7> 6 > > > > > > > > 5 4 > > > 0 0 0 d d d j 12 22 26 > 0 > > > > yy ; : : ; 0 0 0 0 d16 d26 d66 jxy The midplane strains caused by uniaxial loading are therefore given by: eoxx ¼ a11 Nxx

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ð62aÞ

Figure 18 A symmetrical composite laminate subjected to uniaxial load Nxx.

eoyy ¼ a12 Nxx

ð62bÞ

coxy ¼ a16 Nxx

ð62cÞ

Recalling that Nxx is defined as a constant load/unit plate length (with units of N/m or lbf/in.), the effective (or nominal) normal stress rxx applied to the laminate is given by rxx =Nxx/t, where t is the total laminate thickness. In Sec. 3.2, Young’s modulus was defined as ‘‘the normal stress rxx divided by the resulting normal strain exx, with all other stress components equal to zero.’’ Applying the standard definition to the laminate shown in Fig. 18, the effective extensional Young’s modulus in the x-direction is given by: ex

Exx ¼

rxx ðNxx =tÞ 1 ¼ ¼ o exx ða11 Nxx Þ ta11

ð63Þ

The superscript ‘‘ex’’ is used to denote that this property is Young’s modulus measured in extension. In Sec. 3.2, Poisson’s ratio vxy was defined as ‘‘the negative of the transverse normal strain eyy divided by the axial normal strain exx, both of which are induced by stress rxx, with all other stresses equal to zero.’’ The effective Poisson’s ratio in extension for the laminate shown in Fig. 18 is given by: m ex xy ¼

eoyy eoxx

¼

a12 Nxx a12 ¼ a11 Nxx a11

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ð64Þ

Once again, a superscript ‘‘ex’’ has been used to indicate that this Poisson’s ratio is measured in extension. The coefficient of mutual influence of the second kind gxx,xy was defined as ‘‘the shear strain cxy divided by the axial normal strain exx, both of which are induced by normal stress rxx, when all other stresses equal zero.’’ For a composite laminate, the effective coefficient of mutual influence ex of the second kind gxx;xy is therefore given by: ex gxx;xy ¼

coxy a16 Nxx a16 ¼ ¼ o exx a11 Nxx a11

ð65Þ

An identical procedure can be employed to define properties measured during a uniaxial test in which only ryy =Nyy/t is applied. In this case, Eq. (45) becomes: 8 o 9 2 a11 exx > > > > > 6 a12 > > o > > > eyy > > > 6 > > = 6a < o > cxy ¼ 6 6 16 6 0 > jxx > > > 6 > > > > 6 > > > > 4 0 > > > jyy > ; : jxy 0

a12 a22 a26

a12 a26 a66

0 0 0

0 0 0

0

0

d11

d12

0

0

d12

d22

0

0

d16

d26

Midplane strains induced are therefore:

0 0 0

9 38 0 > > > > 7> > >N > yy > > > 7> > > > 7> < 7 0 = 7 d16 7 > > 0 > 7> > > > 7> > > > > 5 0 d26 > > > > ; : 0 d66

eoxx ¼ a12 Nyy

ð66aÞ

eoyy ¼ a22 Nyy

ð66bÞ

coxy ¼ a26 Nyy

ð66cÞ

These strains can be used to define the effective extensional Young’s modex ex , and coefficient of mutual influence of the seculus, Eyy , Poisson’s ratio, myx ex ond kind, g yy,xy : ex

Eyy ¼ ex myx ¼ ex gyy;xy

1 ta22 a12 a22 a26 ¼ a22

ð67aÞ ð67bÞ ð67cÞ

Next, consider the effective material properties measured during a pure shear test. A symmetrical composite laminate subjected to pure shear loading

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Nxy is shown in Fig. 19. Because the load Nxy is applied within the plane of the laminate, properties measured during a pure shear test are included in this discussion of ‘‘extensional’’ properties, even though there is no common counterpart measured in a test involving out-of-plane loading. Assuming DT=DM=0, Eq. (45) becomes: 8 o 9 2 a11 e > > > > > > xx 6 > > > a12 eoyy > > > > 6 > > = 6 < o > a cxy ¼ 6 6 16 6 0 > > jxx > > 6 > > > > 6 > > > > 4 0 j > > > yy > ; : jxy 0

a12 a22 a26

a12 a26 a66

0 0 0

0 0 0

0

0

d11

d12

0

0

d12

d22

0

0

d16

d26

0 0 0

9 38 0 > > > > > > 7> > > 0 7> > > > > > 7> < 7 Nxy = 7 d16 7 > 0 > > 7> > > > 7> > > > 0 d26 5> > > > > ; : 0 d66

Hence, the midplane strains caused by pure shear loading are given by: eoxx ¼ a16 Nxy

ð68aÞ

eoyy ¼ a26 Nxy

ð68bÞ

coxy ¼ a66 Nxy

ð68cÞ

The effective (or nominal) shear stress sxy applied to the laminate is given by sxy=Nxy/t, where t is the total laminate thickness. In Sec. 3.2, the shear modulus was defined as ‘‘the shear stress sxy divided by the resulting shear strain cxy, with all other stress components

Figure 19 A symmetrical composite laminate subjected to shear load Nxy.

Copyright © 2004 by Marcel Dekker, Inc. All Rights Reserved.

equal to zero.’’ Applying this definition to the laminate shown in Fig. 19, the effective shear modulus referenced to the x–y coordinate axes is given by:   Nxy =t sxy 1 ex ¼ ð69Þ Gxy ¼ o ¼  ta66 cxy a66 Nxy The coefficient of mutual influence of the first kind gxy,xx (or gxy,yy) was defined as ‘‘the normal strain exx (or eyy) divided by the shear strain cxy, both of which are induced by shear stress sxy, when all other stresses equal zero.’’ For a composite laminate, the effective coefficient of mutual influence of the ex is therefore given by: first kind gxy;xx ex gxy;xx ¼

eoxx a16 Nxx a16 ¼ ¼ coxy a66 Nxy a66

ð70aÞ

ex is given by: whereas gxy;yy ex gxy;yy ¼

eoyy coxy

¼

a26 a66

ð70bÞ

Recall that during the derivation of classical lamination theory, we assumed that a state of plane stress exists within a thin composite laminate. This assumption implies that out-of-plane shear strains (cjxz and cjyz ) are equal to zero. Consequently, Chentsov coefficients, which were defined in Sec. 3.2, are always equal to zero for thin composite laminates. The effective properties of a laminate obey the inverse relations followed by any anisotropic plate, which are defined in Eq. (13) of Chap. 4. In particular: ex mxy ex Exx

¼

ex myx

ex gxx;xy

ex Eyy

ex Exx

¼

ex gxy;xx

ex gyy;xy

ex Gxy

ex Eyy

¼

ex gxy;yy ex

Gxy

Cross-ply, balanced, and balanced angle-ply laminates were discussed in Secs. 7.2, 7.3, and 7.4, respectively. Recall that a16=a26=0 for these types of stacking sequences. Therefore, coefficients of mutual influence of the first and second kind always equal zero for cross-ply, balanced, and balanced angle-ply laminates. Finally, in the case of a quasi-isotropic laminate (disex cussed in Sec. 7.4), it will be found that the effective in-plane moduli Exx = ex ex ex ex Eyy , gxy =gyx , and Gxy are related in the same manner as for an isotropic plate: ex

ex Gxy

E ¼  xx  ex 2 1 þ mxy

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Flexural Properties All of the effective properties described in Sec. 9.1.1 are determined by subjecting the composite laminate to a load whose line of action lies within the plane of the plate, and by measuring the resulting strains. Properties measured in this way are called effective extensional properties. In this section, we consider properties measured by applying a pure bending moment and measuring the resulting strains. Properties measured under a state of pure bending are called effective flexural properties. Effective flexural properties do not follow from the fundamental definitions of material properties reviewed in Sec. 3.2. Rather, effective flexural properties represent the response of a structure subjected to bending, and are defined in analogy to those exhibited by an isotropic plate. As discussed in Sec. 7, the bending stiffnesses (i.e., the Dij matrix) for an isotropic plate are given by: Et3 12ð1 m2 Þ mEt3 ¼ 12ð1 m2 Þ D12 Þ Et3 ¼ 2 24ð1 þ mÞ

D11 ¼ D22 ¼ D12 ¼ mD11 D66 ¼

ðD11

Also, for an isotropic plate, the inverse of the Dij matrix (i.e., the dij matrix) is given by: 1 12 ¼ d11 ¼ d22 ¼ D11 ð1 m2 Þ Et3 m 12m ¼ d12 ¼ md11 ¼ D11 ð1 m2 Þ Et3 24ð1 þ mÞ d66 ¼ 2ðd11 d12 Þ ¼ Et3 Thus, for isotropic plates, both the Dij and dij matrices can be calculated directly from material properties E and v and the plate thickness t. Given these results, one might anticipate that the Dij and dij matrices for a composite laminate could be calculated in a similar manner, using the total laminate ex ex ex ex , and gyx . This is thickness t and effective extensional properties Exx , Eyy , gxy not the case. The bending stiffnesses of a composite laminate (i.e., the Dij matrix) cannot, in general, be calculated on the basis of effective extensional properties. For example, in general: ex

D11 p

12 1

Exx t3  2

mex xy

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for a composite laminate. In essence, this cannot be done because the flexural properties of a laminate are dictated, in part, by the laminate stacking sequence. In contrast, the effective extensional properties are independent of the stacking sequence. Hence, effective flexural properties cannot be directly related to effective extensional properties. As stated above, effective flexural properties are defined for a state of pure bending. Note from the above discussion that for an isotropic plate: d11 ¼

12 Et3

d12 ¼

md11

The effective Young’s moduli and Poisson ratios exhibited by a composite fl fl fl fl laminate in flexure, denoted Exx , E yy , gxy , and gyx , can therefore be defined as follows: fl

12 d11 t3 d12 ¼ d11

fl

12 d22 t3 d12 ¼ d22

Exx ¼

Eyy ¼

fl mxy

fl myx

Although the effective flexural Young’s moduli and Poisson ratios differ from the corresponding effective extensional properties, direct substitution will shown that the flexural properties obey the inverse relations: fl mxy fl

Exx

¼

fl myx fl

Eyy

It is also interesting to note that d11 p d22 for a quasi-isotropic laminate. fl fl fl fl p myx ; even for a quasi-isotropic laminate. Consequently, Exx p Eyy ; mxy The effective flexural coefficient of mutual influence of the second kind fl can be defined as ‘‘the midplane shear strain cjxx divided by the midplane gxx;xy normal strain ejxx, both of which are induced by moment resultant Mxx, when all other stress resultants equal zero.’’ For a composite laminate, the effective fl is therefore flexural coefficient of mutual influence of the second kind gxx,xy given by: gflxx;xy ¼

coxy eoxx

¼

d16 Mxx d16 ¼ d11 Mxx d11

An identical procedure can be employed to define the effective flexural fl , a property measured coefficient of mutual influence of the second kind gyy,xy during a test in which only Myy is applied: gflyy;xy ¼

coxy d26 Myy d26 ¼ ¼ d22 Myy d22 eoyy

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It is mentioned in passing that flexural coefficients of mutual influence of the second kind are not encountered during study of isotropic plates because d16=d26=0 for isotropic plates. The effective flexural properties will be very useful during the study of composite beams, considered in Chap. 8. 9.2 Effective Properties Relating Temperature or Moisture Content to Strain As discussed in Sec. 3.3, linear coefficients of thermal expansion are measured by determining the strains induced by a uniform change in temperature, and forming the following ratios: axx ¼

eTxx DT

ayy ¼

eTyy DT

axy ¼

cTxy DT

ð71Þ

The superscript ‘‘T’’ is included as a reminder that the strains involved are those caused by a change in temperature only. The midplane strains and curvatures induced by a uniform temperature change may be calculated according to Eq. (45), which becomes (for a symmetrical laminate and for DM ¼ Nxx ¼ Nyy ¼ Nxy ¼ Mxx ¼ Myy ¼ Mxy ¼ 0): 8 o 9 2 38 T 9 N > > a a a 0 0 0 e 11 12 12 > > > xx > > > xx > > > 7> > 6 a12 a22 a26 0 > NT > > o > > > > > > 0 0 e > > > > 7 6 yy yy > > > > > > 7> = 6 < o > < T = 7 6 a a a 0 0 0 16 26 66 cxy ¼ 6 7 Nxy ð72Þ 6 0 > 0 0 d11 d12 d16 7 jxx > > > > > 0 > 7> 6 > > > > > > > 7> 6 > > > > > 4 0 > > 0 0 d12 d22 d26 5> > > > > jyy > > 0 > > ; : > > : ; jxy 0 0 0 d16 d26 d66 0 Substituting the midplane strains indicated by Eq. (72) into Eq. (71), the effective linear thermal expansion coefficients for a general laminate are: i 1 h a11 NTxx þ a12 NTyy þ a16 NTxy axx ¼ DT i 1 h a12 NTxx þ a22 NTyy þ a26 NTxy ð73aÞ ayy ¼ DT h i 1 a16 NTxx þ a26 NTyy þ a66 NTxy axy ¼ DT It is noted in passing that these results are for a symmetrical laminate and can therefore be inverted to give:   NTxx ¼ DT A11 axx þ A12 ayy þ A16 axy   NTyy ¼ DT A12 axx þ A22 ayy þ A26 axy

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ð73bÞ

  NTxy ¼ DT A16 axx þ A26 ayy þ A66 axy

The effective linear coefficient of moisture expansion is measured by determining the strains induced by a uniform change in moisture content, and forming the following ratios: bxx ¼

eM xx DM

byy ¼

eM yy DM

bxy ¼

cM xy DM

ð74Þ

The superscript ‘‘M’’ in the above equations is included as a reminder that the strains involved are those caused by a change in moisture only. The strains induced by a change in moisture content (only) may be calculated according to Eq. (45), which becomes (for a symmetrical laminate and for DT=Nxx=Nyy =Nxy=Mxx=Myy=Mxy=0): 8 o 9 2 a11 exx > > > > > > 6 > > o > a12 eyy > > > > > 6 > < o > = 6 6 a cxy ¼ 6 16 6 0 > > jxx > > 6 > > > > 6 > > 4 0 > jyy > > > > > ; : jxy 0

a12 a22 a26

a12 a26 a66

0 0 0

0 0 0

0

0

d11

d12

0

0

d12

d22

0

0

d16

d26

38 M 9 > > > Nxx > 7> > > M> > N > 7> yy > > > > 7> 7< NM = 7 xy d16 7 > > 0 > 7> > > > 7> > > > d26 5> 0 > > > > ; : d66 0 0 0 0

ð75Þ

Substituting the midplane strains indicated by Eq. (75) into Eq. (74), the effective linear moisture expansion coefficients for a general laminate are: i 1 h M M a11 NM þ a N þ a N 12 16 xx yy xy DM i 1 h M M a12 NM byy ¼ xx þ a22 Nyy þ a26 Nxy DM i 1 h M M a16 NM bxy ¼ þ a N þ a N 26 yy 66 xy xx DM bxx ¼

ð76aÞ

Because these results are for a symmetrical laminate, they can be inverted to give:   NM xx ¼ DM A11 bxx þ A12 byy þ A16 bxy   NM yy ¼ DM A12 bxx þ A22 byy þ A26 bxy   NM xy ¼ DM A16 bxx þ A26 byy þ A66 bxy

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ð76bÞ

Sample Problem 8 Determine the effective extensional and flexural moduli, thermal expansion coefficients, and moisture expansion coefficients for a [30/0/90]s graphiteepoxy laminate. Use material properties listed for graphite-epoxy in Table 3 of Chap. 3, and assume that each ply has a thickness of 0.125 mm. Solution. As described in this section, effective moduli are calculated using various elements of the [abd] matrix. A six-ply symmetrical laminate is considered in this problem. The total laminate thickness is t=6(0.000125 m)=0.000750 m. Using methods discussed in Sec. 6, the [ABD] matrix is determined to be: 3 2 72:2  106 8:02  106 12:0  106 0 0 0 6 8:02  106 52:0  106 5:38  106 0 0 0 7 7 6 7 6 6 12:0  106 5:38  106 15:5  106 0 0 0 7 7 6 ½ABDŠ ¼ 6 7 6 0 0 0 4:23 0:676 1:19 7 7 6 6 0 0 0 0:676 0:988 0:532 7 5 4 0 0 0 1:19 0:532 1:03 Because the laminate is symmetrical, all elements of the Bij matrix are zero, as expected. We obtain the [abd] by inverting the [ABD] numerically: 16:0  10 9 6 6 1:23  10 9 6 6 6 6 12:0  10 9 ½abdŠ ¼ 6 6 0 6 6 6 6 0 4 2

0

9

1:23  10

9

12:0  10

0

0 0

20:0  10

9

6:0  10

9

0

6:0  10

9

75:8  10

9

0

0

0

0

0

0

0

3:51  10

0

2:92  10

2:92  10

2

1:408

3:92  10

1

6:96  10

7 7 7 7 7 0 7 7 17 3:92  10 7 7 7 6:96  10 1 7 5 0

0 1

2

1

1:79

The effective extensional moduli of the laminate can now be calculated using Eqs. (63)–(70b): ex

1 1 ¼ 83:3 GPa ¼ ta11 ð0:000750Þð16:0  10 9 Þ

ex

1 1 ¼ 66:7 GPa ¼ ta22 ð0:000750Þð20  10 9 Þ

ex

1 1 ¼ 17:6 GPa ¼ ta66 ð0:000750Þð75:8  10 9 Þ

Exx ¼ Eyy ¼ Gxy ¼

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3

mex xy

a12 ¼ ¼ a11

mex yx ¼

a12 ¼ a22

 1:23  10 9 ¼ 0:077 16:0  10 9   1:23  10 9 ¼ 0:061 20  10 9 

gex xx;xy ¼

a16 12  10 9 ¼ ¼ a11 16  10 9

0:75

gex yy;xy ¼

a26 6:0  10 9 ¼ ¼ a22 20  10 9

0:30

gex xy;xx ¼

a16 12:0  10 9 ¼ ¼ a66 75:8  10 9

gex xy;yy ¼

a26 6:0  10 ¼ a66 75:8  10

0:16

9 9

¼

0:079

Effective flexural properties are found to be: E flxx ¼

12 12 ¼ 81:0 GPa ¼ 3 d11 t ð0:351Þð0:000750Þ3

E flyy ¼

12 12 ¼ 20:2 GPa ¼ d22 t3 ð1:408Þð0:000750Þ3

mflxy ¼

d12 2:92  10 ¼ d11 0:351

2

mflyx ¼

d12 2:92  10 ¼ d22 1:408

2

¼ 0:083 ¼ 0:021

gflxx;xy ¼

d16 0:392 ¼ ¼ d11 0:351

1:12

gflyy;xy ¼

d26 0:696 ¼ ¼ 1:408 d22

0:49

Note that the values of extensional properties are quite different from analogous flexural properties. The thermal stress resultants associated with a given change in temperature must be determined in order to calculate the effective thermal expansion coefficients. Numerically speaking, any change in temperature can be used, but for present purposes, a unit change in temperature will be assumed (i.e.,

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DT=1). Using (42a–(42f), the thermal stress resultants associated with DT=1 are: NTxx jDT¼1 ¼ 52:3 N=m

NTyy jDT¼1 ¼ 94:9 N=m

NTxy jDT¼1 ¼

36:9 N=m

The effective thermal expansion coefficient axx can now be calculated in accordance with Eq. (73a): axx ¼ axx ¼

i 1 h a11 NTxx þ a12 NTyy þ a16 NTxy DT 1  16:0  10 9 ð52:3Þ ð 1Þ

1:23  10 9 ð94:9Þ

12:0  10 9 ð 36:9ފ

axx ¼ 1:16 Am=m

B

C

Using an equivalent procedure: ayy ¼ 2:06 Am=m

B

C

axy ¼

4:00 Arad=o C

Finally, moisture stress resultants associated with a given change in moisture content must be determined in order to calculate the effective moisture expansion coefficients. Numerically speaking, any change moisture content can be used, but for present purposes, a unit change in content will be assumed (i.e., DM=1). Using Eq. (43), the moisture stress resultants associated with DM=1% are: M NM xx jDM¼1 ¼ 32; 800 N=m Nyy jDM¼1 ¼ 33; 800 N=m

NM xy jDM¼1 ¼

930 N=m

Applying Eqs. (76a) and (76b), we find: bxx ¼ 494 Am=m %M

byy ¼ 643 Am=m %M

bxy ¼

667 Arad %M

10 TRANSFORMATION OF THE ABD MATRIX The Aij, Bij, and Dij matrices all involve a summation over the thickness of the laminate in accordance with (Eq. (27a) (27b) and (34), repeated here for convenience: Aij ¼

n X k¼1

Qij



k

ð zk

zk 1 Þ

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ðrepeatedÞ ð27aÞ

Bij ¼

n   1X Qij k z2k 2 k¼1

Dij ¼

n   1X Qij k z3k 3 k¼1

z2k

1

z3k

1



ðrepeatedÞ ð27bÞ



ðrepeatedÞ ð34Þ

As discussed in Sec. 5.2, Qij may be calculated using Eq. (31) of Chap. 5, which involves use of the Qij matrix and various trigonometrical functions raised to a power (e.g., cos4 h, cos2 h sin2h, sin4 h, etc.). Alternatively, as discussed in Sec. 5.3, Qij may be calculated using Eq. (35) of Chap. 5, which involves the use of material invariants UiQ and trigonometrical functions, whose arguments involve fiber angles multiplied by a constant (i.e., cos 2h, cos 4h, sin 2h, or sin 4h). Although either approach is mathematically equivalent, in some circumstances, the use of material invariants (Eq. (35) of Chap. 5) can be advantageous. Specifically, if all plies within the laminate are of the material type, then use of Eq. (35) of Chap. 5 leads to the ability to easily transform the ABD matrix from one coordinate system to another. To aid in our development, define the following ‘‘geometry factors,’’ which are related to the fiber angles and ply interface positions: n n X X A cos 2hk ðzk zk 1 Þ ð z z Þ ¼ t V ¼ ¼ VA k k 1 1 0 k¼1

k¼1

VA 2 ¼ VA 4 ¼ VB0 ¼ VB2 ¼ VB4 ¼ VD 0 ¼ VD 2 ¼

n X

k¼1 n X

VA 3 ¼

sin 2hk ðzk

zk 1 Þ

sin 4hk ðzk

zk 1 Þ ¼ t

k¼1 n  X

1 z2 2 k¼1 k

z2k

1



n  1X sin 2hk z2k 2 k¼1 n  1X sin 4hk z2k 2 k¼1 n  1X z3 3 k¼1 k

z3k

1

n  1X sin 2hk z3k 3 k¼1



VB1 ¼

¼0 z2k

1

z2k

1

¼

t3 12

z3k

1



VB3 ¼

 VD 1 ¼



VD 3 ¼

n X

k¼1

cos 4hk ðzk

n  1X cos 2hk z2k 2 k¼1 n  1X cos 4hk z2k 2 k¼1

n  1X cos 2hk z3k 3 k¼1

n  1X cos 4hk z3k 3 k¼1

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zk 1 Þ

z2k

1

z2k

1

z3k

1

z3k

1

ð77Þ

 

ð78Þ

 

ð79Þ

VD 4 ¼

n  1X sin 4hk z3k 3 k¼1

z3k

1



Next, consider the steps necessary to calculate the A11 term. In this case, Eq. (27a) becomes: A11 ¼

n X  Qij k ðzk

zk 1 Þ

k¼1

ð80Þ

Using the invariant formulation the expression for Q 11 listed in Eq. (35) of Chap. 5 may be substituted, resulting in: A11 ¼

n n o X Q Q UQ ð zk 1 þ U2 cos 2h þ U3 cos 4h k

k¼1

zk 1 Þ

ð81Þ

where U1Q, U2Q, and U3Q are the stiffness invariants defined in Eq. (36) of Chap. 5. If all plies within the laminate are composed of the same material, then stiffness invariants are constant for the laminate and only fiber angle h varies from one ply to the next. Therefore, Eq. (81) can be rewritten as: A11 ¼ UQ 1

n X

ð zk

k¼1

n X Q

þ U3

k¼1

zk 1 Þ þ U Q 2

cos 4hk ðzk

n X

k¼1

cos 2hk ðzk

zk 1 Þ ð82Þ

zk 1 Þ

A A The geometry factors VA 0 , V1 , and V3 appear in Eq. (82), and hence the equation may be written as:

Q A Q A A A11 ¼ UQ 1 V0 þ U2 V1 þ U3 V3

ð83aÞ

Following an identical procedure, the remaining elements of the Aij matrix are given by: A A22 ¼ UQ 1 V0

Q A A UQ 2 V1 þ U3 V3

A A12 ¼ A21 ¼ UQ 4 V0

A66 ¼

A UQ 5 V0

A UQ 3 V3

A UQ 3 V3

1 Q A A U V þ UQ 3 V4 2 2 2 1 A ¼ UQ VA UQ 3 V4 2 2 2

ð83bÞ ð83cÞ ð83dÞ

A16 ¼ A61 ¼

ð83eÞ

A26 ¼ A62

ð83f Þ

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Analogous procedures can be used to calculate members of the Bij and Dij matrices. It can be shown that elements of the Bij matrix are given by: Q B Q B B B11 ¼ UQ 1 V0 þ U2 V1 þ U3 V3

B22 ¼ B12 ¼

B66 ¼

ð84aÞ

B UQ 1 V0

Q B B UQ 2 V1 þ U3 V3 B B UQ B21 ¼ UQ 3 V3 4 V0 B B UQ UQ 5 V0 3 V3

ð84bÞ ð84cÞ

ð84dÞ

1 B VB þ UQ B16 ¼ B61 ¼ UQ 3 V4 2 2 2 1 B VB UQ B26 ¼ B62 ¼ UQ 3 V4 2 2 2 Members of the Dij matrix are given by:

ð84eÞ ð84f Þ

Q D Q D D D11 ¼ UQ 1 V0 þ U2 V1 þ U3 V3

D22 ¼

D12 ¼

D66 ¼

ð85aÞ

Q D D UQ 2 V1 þ U3 V3 D D UQ D21 ¼ UQ 3 V3 4 V0 D D UQ UQ 5 V0 3 V3 D UQ 1 V0

D16 ¼ D61 D26 ¼ D62

ð85bÞ ð85cÞ

ð85dÞ

1 VD þ UQ VD ¼ UQ 4 2 2 2 1 Q D D ¼ U2 V2 UQ 3 V4 2

ð85eÞ ð85f Þ

Let us now consider transformation of the ABD matrix from one coordinate system to another. A multiangle composite laminate referenced to an x–y coordinate system is shown in Fig. 20. It is assumed that the ABD matrix for this laminate has been calculated and is known based on fiber angles referenced to the x–y coordinate system. Now suppose that a different coordinate system is of interest, the x V–yV coordinate system, orientated h degrees counterclockwise from the original x–y coordinate system. The transformed stiffness matrices referenced to the x V–yV coordinate system will U U U be labeled Aij, Bij, and Dij. Inspection of Fig. 20 shows that a ply with fiber angle hk relative to the original x-axis will form an angle (hk /) relative to the xV-axis. Element A 11 may therefore be calculated using Eqs. (83a)–(83f) by substituting (hk /) for hk. Hence: n n X X A 11 ¼ U1Q ½cos 2ðhk /ފðzk zk 1 Þ ð86Þ ðzk zk 1 Þ þ UQ 2 k¼1 n X ½cos þUQ 3 k¼1

k¼1

4ð hk

/ފðzk

zk 1 Þ

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Figure 20 A general composite laminate, showing the ‘‘original’’ x–y coordinate system and the ‘‘new xV–yV coordinate system.

Recalling the general trigonometrical identity: cosða

bÞ ¼ cos a cos b þ sin a sin b

Equation (86) can be written as:  Q A A A A11 ¼ UQ 1 V0 þ U2 V1 cosð2/Þ þ V2 sinð2/Þ

 A A þUQ 3 V3 cosð4/Þ þ V4 sinð4/Þ

ð87aÞ

The geometry factors ViA have been defined in Eq. (77). A similar procedure can be applied to all remaining elements of the Aij matrix as well as the Bij and Dij matrices:

A  A A A22 ¼ UQ UQ 1 V0 2 V1 cosð2/Þ þ V2 sinð2/Þ ð87bÞ

A  A þUQ 3 V3 cosð4/Þ þ V4 sinð4/Þ

A  A A A12 ¼ A21 ¼ UQ UQ ð87cÞ 4 V0 3 V3 cosð4/Þ þ V sinð4/Þ

 A A A A66 ¼ UQ UQ ð87dÞ 5 V0 3 V3 cosð4/Þ þ V4 sinð4/Þ  1 Q A U2 V2 cosð2/Þ VA 1 sinð2/Þ 2  A VA þUQ 3 sinð4/Þ 3 V4 cosð4/Þ

A16 ¼ A61 ¼

A26 ¼ A62 ¼

 1 Q A U2 V2 cosð2/Þ VA 1 sinð2/Þ 2  A UQ VA 3 sinð4/Þ 3 V4 cosð4/Þ

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ð87eÞ

ð87f Þ

 B11 ¼ U1Q VB0 þ U2Q V1B cosð2/Þ þ VB2 sinð2/Þ

 þ U3Q V3B cosð4/Þ þ V4B sinð4/Þ

 B22 ¼ U1Q V0B U2Q V1B cosð2/Þ þ V2B sinð2/Þ

 þU3Q V3B cosð4/Þ þ V4B sinð4/Þ

 B12 ¼ B21 ¼ U Q V0B U3Q V3B cosð4/Þ þ V4B sinð4/Þ

 B66 ¼ U5Q V B U3Q V3B cosð4/Þ þ V4B sinð4/Þ B16 ¼ B61 ¼

D22

D12 D66

 V1B sinð2/Þ  V3B sinð4/Þ

 1 Q B U2 V2 cosð2/Þ V1B sinð2/Þ 2

 U3Q V4B cosð4/Þ V3B sinð4/Þ

 ¼ U1Q V0D þ U2Q V1D cosð2/Þ þ V2D sinð2/Þ

 þU3Q V3D cosð4/Þ þ V4D sinð4/Þ

 ¼ U Q V0D U2Q V1D cosð2/Þ þ V2D sinð2/Þ

 þU3Q V3D cosð4/Þ þ V4D sinð4/Þ

 ¼ D21 ¼ U4Q V D U3Q V3D cosð4/Þ þ V4D sinð4/Þ

 ¼ U5Q V0D U3Q V3D cosð4/Þ þ V4D sinð4/Þ

B26 ¼ B62 ¼ D11

1 Q B U V2 cosð2/Þ 2 2

þU3Q V4B cosð4/Þ

D16 ¼ D61 ¼

D26 ¼ D62 ¼

1 Q D U V2 cosð2/Þ 2 2

þU3Q V4D cosð4/Þ

V1D sinð2/Þ

V3D

sinð4/Þ





 1 Q D U V2 cosð2/Þ V1D sinð2/Þ 2 2

 U3Q V4D cosð4/Þ V3D sinð4/Þ

ð88aÞ

ð88bÞ ð88cÞ ð88dÞ ð88eÞ

ð88f Þ ð89aÞ ð89bÞ ð89cÞ ð89dÞ ð89eÞ

ð89f Þ

It is again emphasized that all new results presented in this section (in particular, Eqs. (82a)–(89f ) are valid only if all plies within the laminate are of the same material type. In many cases, this is a severe restriction. These equations cannot be used to calculate the ABD matrix for a hybrid composite

Copyright © 2004 by Marcel Dekker, Inc. All Rights Reserved.

laminate, for example. On the other hand, an advantage of the invariant approach is the ability to easily rotate the ABD matrix from one coordinate system to another using Eqs. (88a)–(89f), so in the proper circumstances, the invariant approach is convenient. 11 COMPUTER PROGRAM CLT The computer program CLT has been developed to supplement the material presented in this chapter. This program can be downloaded at no cost from the following website: http://depts.washington.edu/amtas/computer.html. Program CLT can be used to recreate all numerical results discussed in the Example Problems presented in this chapter. In essence, the calculation steps described in Sec. 8 are implemented in this program. The user may select two different analysis paths. One analysis path corresponds to the case in which the loads applied to the laminate are specified, whereas the second corresponds to the case in which midplane strains and curvatures applied to the laminate are specified. In either case, the user must provide various numerical values required during the calculations performed. The user must define these values using a consistent set of units. For example, the user must input elastic moduli, thermal expansion coefficients, and moisture expansion coefficients for the composite material system(s) of interest. Using the properties listed in Table 3 of Chap. 3 and based on the SI system of units, the following numerical values would be inputted for graphite-epoxy: E11 ¼ 170  109 Pa a11 ¼

0:9  10

b11 ¼ 150:0  10

E22 ¼ 10  109 Pa 6

m=m

6

m=m

B

C

G12 ¼ 13  109 Pa

m12 ¼ 0:30

a11 ¼ 27:0  10

%M b22 ¼ 4800  10

6

m=m 6

m=m

B

C %M

If the analysis requires the user to input numerical values for stress and moment resultants, then stress resultants must be inputted in Newtons per meter, and moment resultants must be inputted in Newton meters per meter. Typical value would be Nxx=150103 N/m and Mxx=5 N m/m. If, instead, the analysis requires the user to input numerical values for midplane strains and curvatures, then strains must be inputted in meters per meter (not in Am/ m) and curvatures must be inputted in per meter. Typical value would be ejxx=200010 6 m/m=0.002000 m/m and jxx=0.5 m 1. All temperatures would be inputted in degrees Celsius. Ply thicknesses must be inputted in meters (not millimeters). A typical value would be t k =0.000125 m (corresponding to a ply thickness of 0.125 mm).

Copyright © 2004 by Marcel Dekker, Inc. All Rights Reserved.

In contrast, if the English system of units were used, then the following numerical values would be inputted for the same graphite-epoxy material system: E11 ¼ 25:0  106 psi a11 ¼

0:5  10

b11 ¼ 150:0  10

E22 ¼ 1:5  106 psi 6

in=in

6

in=in

B

F %M

m12 ¼ 0:30

a11 ¼ 15  10

6

G12 ¼ 1:9  106 psi in=in

b22 ¼ 4800  10

6

F

B

in=in

%M

Stress resultants must be inputted in pound-forces per inch, and moment resultants must be inputted in pound-force inches per inch. Typical value would be Nxx=1000 lbf/in. and Mxx=1 lbf in./in. If, instead, the analysis requires the user to input numerical values for midplane strains and curvatures, then strains must be inputted in inches per inch and curvatures must be inputted in per inch. Typical value would be ejxx=2000  10 6 in./ in. = 0.002000 in./in. and jxx = 0.01 in 1. All temperatures would be inputted in degrees Fahrenheit. Ply thicknesses must be inputted in inches. A typical value would be tk = 0.005 in. HOMEWORK PROBLEMS Notes: (a) In the following problems, the phrase ‘‘by hand calculation’’ means that solutions are to be obtained using a calculator, pencil, and paper. (b) The computer programs UNIDIR and CLT are referenced in some of the following problems. As described in Sec. 11, these programs can be downloaded from the following website: http://depts.washington.edu/amtas/ computer.html. 1. Three-element strain gage rosettes are mounted on opposite sides of a [0/ F30]s graphite-epoxy laminate, as shown in Fig. 21. An individual ply has a thickness of 0.005 in. The laminate is then subjected to an unknown system of forces. The strains measured by strain gage rosette 1 are: exx ¼ 2000 Ain:=in:

eyy ¼ 500 Ain:=in:

cxy ¼

1000 Arad

Similarly, the strains measured by strain gage rosette 2 are: exx ¼ 3000 Ain:=in: eyy ¼

2000 Ain:=in:

cxy ¼

1000 Arad

Determine by hand calculation: (a) Midplane strains and curvatures induced in the laminate (b) Strains (exx,eyy,cxy) induced at the interface between plies 1 and 2.

Copyright © 2004 by Marcel Dekker, Inc. All Rights Reserved.

Figure 21 Edge view of the composite laminate described in Problem 1.

2. Imagine that a new room temperature cure graphite-epoxy prepreg system has been developed. A [0/F30]s laminate is produced at room temperature using this material, and is therefore initially stress-free and strain-free. Strain gage rosettes are mounted on opposite sides of the laminate, as shown in Fig. 21. The laminate is then subjected to an unknown system of forces, whereas temperature and moisture content remain constant. The strains measured by rosette 1 are: exx ¼ 500 Ain:=in:

eyy ¼ 1000 Ain:=in:

cxy ¼ 750 Arad

Similarly, the strains measured by rosette 2 are: 2000 Ain:=in: cxy ¼ 750 Arad   Obtain numerical values for all elements of the Q matrices for plies 1 and 2 using program UNIDIR and properties listed in Table 3 of Chap. 3. Assume that an individual ply has a thickness of 0.005 in. Then determine the following by hand calculation: exx ¼

1000 in:=in:

eyy ¼

(a) Midplane strains and curvatures induced in the laminate (b) Strains (exx,eyy,cxy) induced at the interface between plies 1 and 2 (c) Stresses (rxx,ryy,sxy) induced in ply 1, at the interface between plies 1 and 2 (d) Stresses (rxx,ryy,sxy) induced in ply 2, at the interface between plies 1 and 2.

Copyright © 2004 by Marcel Dekker, Inc. All Rights Reserved.

3. Repeat Problem 2 for a glass-epoxy system. 4. Repeat Problem 2 for a Kevlar-epoxy system. 5. Imagine that a new room temperature cure graphite-epoxy prepreg system has been developed. A [0/F30]s laminate is produced at room temperature using this material, and is therefore initially stress-free and strain-free. Strain gage rosettes are mounted on opposite sides of the laminate, as shown in Fig. 21. The laminate is then subjected to an unknown system of forces and a temperature increase of 250jF (moisture content remain constant). The strains measured by rosette 1 are: exx ¼ 500 Ain:=in:

eyy ¼ 1000 Ain:=in:

Similarly, the strains measured by rosette 2 are: exx ¼

1000 Ain:=in:

eyy ¼

cxy ¼ 750 Arad

2000 Ain:=in:

cxy ¼ 750 Arad

Obtain numerical values for all elements of the [Q] matrices for plies 1 and 2 using program UNIDIR and properties listed in Table 3 of Chap. 3. Assume that an individual ply has a thickness of 0.005 in. Then determine the following by hand calculation: (a) Midplane strains and curvatures induced in the laminate (b) Strains (exx,eyy,cxy) induced at the interface between plies 1 and 2 (c) Stresses (rxx,ryy,sxy) induced in ply 1, at the interface between plies 1 and 2 (d) Stresses (rxx,ryy,sxy) induced in ply 2, at the interface between plies 1 and 2. 6. Repeat Problem 5 for a glass-epoxy system. 7. Repeat Problem 5 for a Kevlar-epoxy system. 8. An engineer is designing a structure that involves a [0/10/90]s graphiteepoxy composite laminate that will be cured at 350jF. During service, the structure must support a load of 1000 lbf, and will experience a change of temperature of 150jF in a dry environment. Based on the cure temperature and expected service conditions, the engineer predicts that the laminate will experience the following midplane strains and curvatures: eoxx ¼ 1500 Ain:=in: eoyy ¼ 2200 Ain:=in: coxy ¼

1000 Arad

Copyright © 2004 by Marcel Dekker, Inc. All Rights Reserved.

jxx ¼ jyy ¼ jxy ¼ 0

  Obtain numerical values for all elements of the Q matrix for ply 3 using program UNIDIR and properties listed in Table 3 of Chap. 3. Assume that an individual ply has a thickness of 0.005 in. Then determine the stresses induced in ply 3 relative to the 1-2 coordinate system by hand calculation. 9. Repeat Problem 8 for a glass-epoxy structure. 10. Repeat Problem 8 for a Kevlar-epoxy structure. 11. A [0/F30/90]s graphite-epoxy laminate is cured at 350jF and then cooled to room temperature (70jF). Use hand calculation to determine the following thermal resultants induced during cooling (use program UNIDIR and properties listed in Table 3 of Chap. 3 to determine   elements of the Q matrix as necessary): T (a) Nxx T (b) Nyy T (c) Nxy T (d) Mxx (e) MTyy T: (f) Mxy

12. Repeat Problem 11 for a glass-epoxy structure. 13. Repeat Problem 11 for a Kevlar-epoxy structure. Note the following: A [0/F30/90]s graphite-epoxy laminate will be considered in Problems 14–18. Assuming a ply thickness of 0.125 mm and using properties listed in Table 3 of Chap. 3 (SI units), the [Q] matrix for ply 2 is: 2 3 10:76E10 2:606E10 4:813E10 6 7 4 2:606E10 2:722E10 2:152E10 5 4:813E10 2:152E10 3:605E10

The [ABD] matrix for this laminate is: 2 9:906E7 1:454E7 0 0 6 1:454E7 5:885E7 0 0 6 6 6 0 0 2:452E7 0 6 6 0 0 0 11:89 6 6 4 0 0 0 1:032 0 0 0 0:7521

0 0 0 1:032 1:628 0:3362

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3 0 0 7 7 7 0 7 7 0:7521 7 7 7 0:3362 5 1:864

The [abd] matrix for this laminate is: 2

1:048E

6 0:2558E 6 6 6 0 6 6 0 6 6 6 0 4 0

8

0:2558E 8

1:763E

8 8

0

0

0

0

0

0

0

0

0

4:078E

8

0 9:029E

0

0

0

2

0

0

5:160E

2

0

0

2:713E

2

5:160E

0 2

0:6674 9958E

2:713E 9:958E

2

0:5655

If the laminate were cured at 175jC and then cooled to room temperatures (20jC), the thermal resultants induced during cooldown would be: NTxx ¼

NTyy

¼

8607 N=m 21; 825 N=m

NTxy ¼ MTxx ¼ MTyy ¼ MTxy ¼ 0 If the moisture content of the laminate immediately after cure is 0%, but over the course of several months is increased to 0.5%, then the moisture resultants induced by this slow moisture adsorption are: NM xx ¼ 6092 N=m

NM yy ¼ 6098 N=m

M M M NM xy ¼ Mxx ¼ Myy ¼ Mxy ¼ 0

14. A [0/F30/90]s graphite-epoxy laminate is cured at 175jC and then cooled to room temperatures (20jC). Use hand calculation to determine the following: (a) Midplane strains and curvatures induced during cooldown (b) Strains (exx,eyy,cxy) induced in ply 2 during cooldown (c) Stresses (rxx,ryy,sxy) induced in ply 2 during cooldown (d) Stresses (r11,r22,s12) induced in ply 2 during cooldown 15. A [0/F30/90]s graphite-epoxy laminate is cured at 175jC and then cooled to room temperatures (20jC). Although the moisture content immediately after cure was 0%, the laminate is stored in a humid environment and, over the course of several months, the moisture content is increased to 0.5%. Use hand calculation to determine the following: (a) Midplane strains and curvatures induced after moisture content increased to 0.5%

Copyright © 2004 by Marcel Dekker, Inc. All Rights Reserved.

3

7 7 7 7 7 27 7 7 27 5

(b) Strains (exx,eyy,cxy) induced in ply 2 after moisture content increased to 0.5% (c) Stresses (rxx,ryy,sxy) induced in ply 2 after moisture content increased to 0.5% (d) Stresses (r11,r22,s12) induced in ply 2 after moisture content increased to 0.5% 16. A [0/F30/90]s graphite-epoxy laminate is subjected to the following loads: Nxx ¼ Nyy ¼ Nxy ¼ 10; 000 N=m

Mxx ¼ Myy ¼ Mxy ¼ 0

Ignoring thermal and moisture effects, use hand calculation to determine the following: (a) Midplane strains and curvatures (b) Strains (exx, eyy, cxy) induced in ply 2 (c) Stresses (rxx, ryy,sxy) induced in ply 2 (d) Stresses (r11,r22, s12) induced in ply 2. 17. A [0/F30/90]s graphite-epoxy laminate is cured at 175jC and then cooled to room temperatures (20jC). The moisture content immediately after cure was 0%. However, the laminate is stored in a humid environment and, over the course of several months, the moisture content is increased to 0.5%. The laminate is then subjected to the following loads: Nxx ¼ Nyy ¼ Nxy ¼ 10; 000 N=m

Mxx ¼ Myy ¼ Mxy ¼ 0

Use hand calculation to determine the following: (a) Midplane strains and curvatures (b) Strains (exx,eyy,cxy) induced in ply 2 (c) Stresses (rxx,ryy,sxy) induced in ply 2 (d) Stresses (r11,r22,s12) induced in ply 2. 18. A 10  10 in.2 [0/F30/90]s graphite-epoxy laminate is supported between three infinitely rigid walls and rollers, as shown in Fig. 22. A load Nxx= 7500 N/m is applied to the plate. Ignoring thermal effects, moisture effects, and the possibility of buckling, use hand calculation to determine Nxx, Nyy, Nxy, ejxx, ejyy, and cjxx. 19. A [20/65/ 25]s graphite-epoxy laminate is cured at 175jC and then cooled to room temperatures (20jC). Moisture content remains at 0%. The following loads are then applied: Nxx ¼ 30 kN=m

Nxx ¼ 10 N m=m

Nyy ¼

7 kN=m

Myy ¼ Mxy ¼ 0

Nxy ¼ 0

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Figure 22 A [0/F30/90]s graphite-epoxy laminate described in Problem 18.

Using properties listed in Table 3 of Chap. 3 and assuming ply thicknesses of 0.125 mm: (a) Determine all ply strains and stresses using program CLT. (b) Prepare plots similar to Fig. 13, showing the through-thickness variation of strains exx, eyy, and cxy. (c) Prepare plots similar to Fig. 14, showing the through-thickness variation of strains e11, e22, and c12. (d) Prepare plots similar to Fig. 15, showing the through-thickness variation of stresses rxx, ryy, and sxy. (e) Prepare plots similar to Fig. 16, showing the through-thickness variation of stresses r11, r22, and s12. 20. Repeat Problem 19 for a glass-epoxy laminate. 21. Repeat Problem 19 for a Kevlar-epoxy laminate. 22. A [20/65/-25]s ‘‘hybrid’’ laminate is cured at 175jC and then cooled to room temperatures (20jC). Plies 1 and 6 are graphite-epoxy, plies 2 and 5 are glass-epoxy, and plies 3 and 4 are Kevlar-epoxy. The following loads are then applied: Nxx ¼ 30 kN=m

Nxx ¼ 10 N m=m

Nyy ¼

7 kN=m

Myy ¼ Mxy ¼ 0

Nxy ¼ 0

Moisture content remains constant at 0%. Using properties listed in Table 3 of Chap. 3 and assuming that the graphite-epoxy, glass-epoxy,

Copyright © 2004 by Marcel Dekker, Inc. All Rights Reserved.

and Kevlar-epoxy plies have thicknesses of 0.125, 0.200, and 0.15 mm, respectively: (a) Determine all ply strains and stresses using program CLT. (b) Prepare plots similar to Fig. 13, showing the through-thickness variation of strains exx, eyy, and cxy. (c) Prepare plots similar to Fig. 14, showing the through-thickness variation of strains e11, e22, and c12. (d) Prepare plots similar to Fig. 15, showing the through-thickness variation of stresses rxx, ryy, and sxy. (e) Prepare plots similar to Fig. 16, showing the through-thickness variation of stresses r11, r22, and s12. REFERENCES 1. Jones, R.M. Mechanics of Composite Materials; Hemisphere Publishing Corporation: New York, NY. ISBN 0-89116-490-1, 1975. 2. Tsai, S.W.; Hanh, H.T. Introduction to Composite Material; Technomic Publishing Co. ISBN 0-87762-288-4, 1980. 3. Halpin, J.C. Primer on Composite Materials: Analysis; Technomic Publishing Co. ISBN 87762-349-X, 1984.

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7 Predicting Failure of a Multiangle Composite Laminate

Ideally, the objective of this chapter would be to describe the analytical tools and/or methodologies available to accurately predict the yielding and fracture of multiangle composite laminates under general thermomechanical loading conditions. Unfortunately, we will not be able to reach this objective. As will be seen, predicting the fracture of multiangle composite laminates under general loading conditions and with a high degree of accuracy is still beyond the state of the art, despite extensive research efforts undertaken over the past several decades. Of course, fracture predictions for the ‘‘simple’’ case of an isotropic metallic structure under general loading conditions are not completely reliable either, despite more than a century of effort. In any case, predicting the fracture of polymeric composite structure has proven to be a formidable challenge, and methods of predicting this phenomenon remain an active area of research. The difficulties encountered are many and varied. Some of the most common factors involved are summarized in the following introductory section, and a more detailed discussion of several of these factors will be presented in following sections. However, the reader should be aware from the outset that composite fracture predictions are currently a blend of engineering art and science. The reader is advised to keep abreast of new literature in this area. 375

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1 PRELIMINARY DISCUSSION 1.1 Yielding and Fracture of Isotropic Metals vs. Unidirectional Polymeric Composites Let us begin our discussion of the difficulties posed by composites by contrasting the initial nonlinear behavior of isotropic metals and metal alloys to that of a unidirectional composite laminate. As is well known, the initial nonlinear deformation (i.e., yielding) of a metal or metal alloy can be explained based on the formation and subsequent motion and coalescence of various types of imperfections within the crystalline atomic structure of the metal/metal alloy (e.g., dislocations). Nonlinear behavior of polymeric composites (here loosely termed ‘‘yielding’’) can also be explained on the basis of atomic and/or molecular mechanisms. However, a composite consists of at least two constituents (the reinforcing fiber and the polymeric matrix), and the atomic and/or molecular structure of these two constituents differs substantially. Furthermore, the molecular structure of the polymeric matrix in the immediate vicinity of the fiber/matrix interface usually differs from the molecular structure at positions removed from the fiber. That is, the molecular structure developed during polymerization of a polymer in bulk differs from that developed in the region immediately adjacent to the surface of the fiber. The region near the fiber is often referred to as the interphase, rather than the ‘‘interface,’’ and the mechanical properties exhibited by the polymer in the interphase differ from bulk properties. Hence, from a continuummechanics point of view, three more or less distinct ‘‘materials’’ can be defined: the fiber, the matrix, and the fiber–matrix interphase. Consider the case of graphite–epoxy. As discussed in Sec. 3 of Chap. 1, graphite is a highly ordered crystalline structure, and the high strength and the high stiffness exhibited by a graphite fiber at the macroscale are achieved by aligning the basal planes of the graphite crystal with the axis of the fiber. The macroscopical stress–strain response of an individual graphite fiber is almost perfectly linear up to final fracture. In contrast, epoxy is a cross-linked thermoset and generally does not possess a crystalline structure at the atomic level (see Sec. 2 of Chap. 1). The stress–strain response of epoxy becomes nonlinear when stresses are high enough to cause segments of the overall molecular structure to slide relative to one another. At the macroscopical level, epoxy exhibits a ductile stress–strain response if cross-link density is low (i.e., relatively low stress levels can cause segments of the molecular structure to slide past one another), but becomes increasingly brittle as cross-link density is increased. The macroscopical stress–strain response of the interphase is difficult to measure (or even define), but is likely somewhere between that of the fiber and matrix. The stress necessary to cause dislocations or other imperfections to develop or move within the crystalline lattice of the graphite

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fiber are far higher than those necessary to cause molecular segments to slide in the epoxy matrix. The initial nonlinear deformations exhibited by a graphite–epoxy composite are therefore almost entirely initiated within the polymeric matrix. As a metal or metal alloy is loaded beyond the yield point, the crystalline imperfections coalesce to form microcracks, often located (at least initially) at grain boundaries. As loading is further increased (and/or if loading fluctuates with time, as in fatigue loading), these microcracks grow in size until they may be observed with a low-power microscope or with the naked eye. The principal stresses present in the isotropic structure ultimately govern the rate at which cracks form and grow. Furthermore, the orientation of the principal stress coordinate system governs the direction of crack propagation. The final fracture of metals and metal alloys occurs when the crack reaches a critical length, at which point the crack generally propagates at a high rate of speed and the metal fractures. In a composite, the fracture process is also initiated when one or more microcracks are formed. Although a crack may form in either the graphite fiber or the epoxy matrix, the ‘‘imperfections’’ that lead to the formation of a crack in these two mediums are wholly different. In addition, the stress levels necessary to cause crack growth in a graphite fiber are at least an order of magnitude higher than that necessary to cause crack growth in epoxy. Consequently, cracks are far more likely to form in the polymeric matrix than in the fiber. The orientation of matrix crack(s) that forms in a unidirectional laminate is invariably related to the fiber direction (e.g., cracks form either parallel or perpendicular to the fibers) and is independent of the principal stress and principal stress coordinate system when defined at the macroscale. The following types of cracks are observed in unidirectional composite laminates: 

Matrix cracks. These are cracks that occur in the polymeric matrix, at some distance from the fiber–matrix interface. Matrix cracks generally occur in planes either parallel or perpendicular to the fiber direction.  Fiber–matrix debonding. In this case, the crack has formed in the interphase region, and a (nonplanar) crack extends around the periphery of the fiber.  Fiber cracks. These are cracks that occur in the fiber itself. Fiber cracks almost always occur in a plane perpendicular to the axis of the fiber, and extend across the entire width of the fiber. Another complication is time dependency. Under most conditions, yielding and crack growth in metals and metal alloys, at least those used in load-bearing structural applications, can be considered to be independent of

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time at room temperatures.* For example, if a tensile stress is applied to a metal structure and initially causes yielding, no further yielding occurs if the tensile stress is held constant for long times. Similarly, if a tensile stress that causes some (noncatastrophic) amount of crack growth is applied, no further crack growth occurs if the tensile stress is held constant for long times.y Graphite fibers are also time-independent. In contrast, yielding and crack growth in polymers are time-dependent phenomena, even at room temperatures. An increase in temperature and/or an increase in moisture content further accentuates the time dependency. Hence, polymer–matrix composites are time-dependent materials. If a tensile load is applied to a unidirectional [90j]n composite and held constant for a long period of time, the resulting strain will slowly increase (i.e., the composite exhibits a creep response). Similarly, if a tensile stress is applied and held constant, the composite may eventually fail due to slow crack growth (often called a ‘‘creep-to-rupture’’ failure). Although the time-dependent behavior exhibited by modern polymeric composites for short times at room temperature is usually minimal, these effects must be considered if a composite structure is intended for years or tens of years of service, or if the structure will be exposed to elevated temperatures. 1.2 Failure of Multiangle Composite Laminates The preceding discussion concerned the failure of unidirectional composites. Recall that several macromechanics-based failure criteria were discussed and applied to unidirectional composites in Secs. 5 and 6 of Chap. 5. Predictions obtained using these criteria are based on fracture/yield strengths measured during simple uniaxial and pure shear tests. Although these macromechanical criteria cannot be used to capture the details of crack coalescence and growth, they may nevertheless be used to predict the failure of a unidirectional composite laminate subjected to an arbitrary state of plane stress (assuming timedependent effects are not pronounced) with an acceptable degree of accuracy. Predicting the failure of multiangle composite laminates is far more difficult than for unidirectional composites, however. One difficulty is that a single macroscopical crack or flaw has no measurable impact on the overall mechanical response of a multiangle composite laminate. This is in direct

* Roughly, yielding and crack growth in metals/metal alloys become time-dependent if the temperature exceeds one-half the melting temperature, measured on an absolute scale (Kelvin or Rankine). y The significance and importance of crack growth, as related to yielding and fracture of metallic structures, began to be recognized in the 1930s and 1940s, and have ultimately led to the development of the branch of engineering known as fracture mechanics [1].

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contrast to the case of a unidirectional composite (e.g., a [90]n laminate), in which the final fracture is ultimately governed by a single (or, at most, a few) crack or flaw. Hence, in the case of a multiangle laminate, the successful prediction of a single-ply fracture does not necessarily lead to a successful prediction of overall laminate fracture. As will be seen, the final fracture of a multiangle laminate does not occur until hundreds or even thousands of cracks and other flaws have formed. A second factor is that multiangle laminates are subject to failure modes that do not exist in unidirectional laminates. For example, multiangle laminates are subject to delamination failures. A delamination occurs when the bond between adjacent plies fails, such that a crack forms in a plane parallel to the plies. The initiation of delamination failures is often attributed to freeedge stresses, discussed in Sec. 2. As will be seen, free-edge stresses occur whenever adjacent plies possess differing Poisson ratios or coefficients of mutual influence. Thirdly, pre-existing thermal and/or moisture stresses occur in multiangle laminates due to a mismatch in effective thermal expansion and moisture expansion coefficients from one ply to the next. As discussed in previous chapters, thermal and moisture stresses are substantial and contribute toward the failure of individual plies and the final fracture of the laminate as a whole. The many difficulties encountered when attempting to predict the failure of a multiangle laminate can be summarized by considering the damage induced within an eight-ply symmetrical quasi-isotropic [0/45/90/ 45]s laminate subjected to tensile loading. A highly idealized (but more or less representative) response is shown in Fig. 1. It is assumed that the laminate is defect-free prior to loading, which implies that any pre-existing thermal and/or moisture stresses are not high enough to have caused any ply failures or other defects. It is also assumed that the laminate is tested under conditions in which time-dependent factors are not an issue. A uniaxial tensile load Nxx (or, equivalently, a tensile effective stress r¯ xx ¼ Nxx =t) is applied and steadily increased until the final laminate fracture occurs. An axial strain exx is, of course, induced as r¯ xx is increased from zero. The initial slope of the r¯ xx vs: exx represents the effective Young’s modulus of the laminate, E¯ xx, and can be predicted using classical lamination theory (CLT), as discussed earlier. As r¯ xx is increased, individual ply stresses are increased as well. Eventually, ply stresses are increased to the point that ply stresses are no longer linearly related to ply strains (i.e., the ply ‘‘yields’’) (see Sec. 5 of Chap. 3). At somewhat higher load levels, cracks begin to form in one (or more) ply. For a [0/45/90/ 45]s laminate subjected to uniaxial tensile loading, the first plies to yield and then crack are almost always the 90j plies; that is, the 90j plies yield when r¯ xx ¼ Nxx =t has been increased to a critical

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Figure 1 An idealized stress–strain plot for a [0/45/90/ 45]s laminate, showing the evolution of internal damage.

level. The difference between the load level at which yielding is initiated and the level at which cracks begin to form within the 90j plies depends on the level of ductility exhibited by the particular composite material system. In this text, ply yielding is considered to be a form of composite failure, and the effective stress level necessary to cause yielding of the 90j plies is called the first-ply failure stress. Hence, cracks begin to form in the 90j plies at load levels above the first-ply failure stress. Note that if the laminate were a unidirectional [90]n laminate, then final fracture would occur as soon as a single crack formed within the 90j plies. Because the laminate is instead a [0/ 45/90/ 45]s laminate and because no fractures have yet occurred within the F45j or 0j plies, the laminate as a whole can support higher stress/strain levels. Further, experimentally, it has been shown that many cracks form within the 90j plies as the load is increased, and the distance between cracks is (approximately) constant. The characteristic spacing between cracks within

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the 90j plies can be predicted on the basis of so-called shear lag models. A number of researchers have developed shear lag models. A detailed discussion of such models will not be presented here, but the interested reader should refer to Refs. 2–5 for representative example analyses. Once the 90j plies have cracked/fractured, they can no longer contribute fully to the effective stiffness of the laminate. Hence, it would be expected that the slope of the r¯ xx vs: exx curve (i.e., the effective Young’s modulus Exx) would decrease as cracks form in the 90j plies. Although such a decrease has been observed in practice, it is often barely discernible (a pronounced decrease in slope has been shown in Fig. 1 for illustrative purposes; in practice, the change in slope is far less than that implied in the figure).* The failure of the 90j plies has little impact on the effective Young’s modulus because the stiffness of these plies in the x-direction, relative to the stiffness of the F45j and 0j plies, is very low even before failure occurs. As the effective stress level is further increased, cracks eventually begin to form within the F45j plies. Once again, many cracks form within the F45j plies, and these cracks tend toward a characteristic spacing that may be predicted based on a shear lag model. At this elevated stress level, extensive matrix cracking has been induced within the 90j and F45j plies, and yet no significant damage has yet occurred within the 0j plies. Hence, the laminate as a whole can support still higher effective stress levels. An additional decrease in slope (i.e., a decrease in apparent Young’s modulus) occurs as cracks develop in the F45j plies. Thus far, the cracks that have formed within the 90j and F45j plies all lie within planes perpendicular to the x–y plane. As the effective stress is increased further still, delamination failures begin to develop. That is, matrix cracks begin to form between plies, and these new matrix cracks lie within planes that are parallel to the x–y plane. The initiation of delamination is often attributed to free-edge stresses, which will be further discussed in Sec. 2. The delaminated regions grow in size as the stress is increased and eventually coalesce, such that a delaminated region may extend across the entire width of the specimen. At still higher effective stress levels, matrix cracks begin to form within the 0j plies (often referred to as ‘‘splitting’’). These cracks lie within a plane perpendicular to the x–y plane. The final laminate fracture is precipitated by fiber failures within the 0j plies. The effective stress level at which final fracture occurs is often called the last-ply failure stress. At the final fracture, the laminate usually fractures into many fragments due to the extensive and pre-existing matrix cracks and

* It is interesting to note, however, that a pronounced decrease in the effective Poisson ratio vxy occurs due to transverse cracking of the 90j plies [4].

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delamination that occurred at lower stress levels, as well as the large amount of energy release associated with fiber failure. Two other forms of damage encountered in composites are fiber–matrix debonding and fiber microbuckling. In the case of fiber–matrix debonding, a crack forms around the periphery of a fiber, so that the load can no longer be transferred from the matrix to the fiber. In the case of fiber microbuckling, the fibers within a ply that experiences compressive stresses in the fiber direction buckle. This reduces the compressive stiffness exhibited by the ply and ultimately leads to failure of the fibers due to the bending stresses induced. Whereas the sequence of damage events depicted in Fig. 1 is for the particular case of a [0/45/90/ 45]s laminate subjected to uniaxial tensile loading, this sequence of events is more or less representative of all multiangle laminates. That is, multiangle laminates subjected to a monotonically increasing but otherwise arbitrary loading condition typically experience yielding and significant internal damage at load levels far below that required to cause final fracture. It should now be clear that ‘‘damage’’ refers to many failure events including matrix cracks, delaminations, fiber–matrix debonding, fiber microbuckling, fiber cracks, etc. Hence, a fundamental question arises during the design of composite structures: What is ‘‘failure’’ and what is a ‘‘safe’’ effective stress level? There is no single answer to this question, and the definition of ‘‘failure’’ as well as selection of a design failure stress often depends on details of the structural application. A partial list of the factors involved includes: 

The intended service life of the structure. A structure intended for years of service (such as a highway bridge or an airplane fuselage) will generally be designed to a much more conservative failure stress level compared to a structure intended for a single use (such as a rocket motor case).  The cyclical nature of the applied loading. A structure subjected to cyclical fatigue loading will generally be designed to a more conservative failure stress level compared to a structure that will experience static or slowly varying loads during service.  The consequences of structural failure. A structure whose failure will result in the loss of life and/or extensive property damage will generally be designed to a more conservative failure stress level compared to a structure whose failure is of lesser consequence. In practice, the maximum allowed effective stress level used during the design process is also related to our ability to accurately predict the loads induced by all possible service conditions. This factor is governed by the complexity of the structure being designed and/or the service environment. For example, compare the design of a simple pressure vessel intended to store

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a pressurized gas (the body of a fire extinguisher, perhaps) to that of a pressurized airplane fuselage. In the case of a fire extinguisher, the primary service loading involved is due to the internal gas pressure, and over most of the service life of this structure the internal pressure varies slowly, if at all. Hence, in this case, it is possible to predict the loads/stresses experienced by the structure during service quite accurately. On the other hand, although a pressurized composite airplane fuselage may be (roughly) thought of as a ‘‘pressure vessel,’’ the service loading conditions the structure must withstand are very complex and difficult to quantify precisely. In this case, the ‘‘pressure vessel’’ must accommodate not only internal pressures but also direct aerodynamic loading and loads transferred from the wings and landing gear to the fuselage. Furthermore, the fuselage will experience distinctly different loading conditions during takeoff, climb, cruising, descent, landing, and other aircraft maneuvers. The design of an aircraft fuselage must also accommodate the need for passenger/cargo doors, windows, wing attachments, etc. Taken together, these many factors imply that it is very difficult to precisely predict the loads/stresses induced in a fuselage under all conditions that will be encountered in service and, consequently, the design of a fuselage is based on relatively conservative design philosophies. Again, what is a ‘‘failure’’? Referring to Fig. 1, at least two limiting design philosophies may be defined. First, the laminate failure stress may be defined as the effective stress level required to initiate a ply failure of any kind. ‘‘Ply failure’’ is meant to imply either yielding or fiber fractures in one or more plies. The laminate failure stress or load defined in this manner is called the first-ply failure stress or first ply failure load. This is a conservative design approach (in some cases overly conservative) because for most multiangle laminates, it will be found that first-ply failure is initiated due to yielding of one or more plies rather than fiber fracture. Most new-generation polymeric composite material systems are relatively ductile, and ply yielding occurs at stress levels well below that necessary to cause any matrix cracks or fiber fractures. Hence, the ultimate load-carrying capacity of most multiangle laminates is well above the first-ply failure load. Still, a conservative design philosophy is appropriate when failure of the composite structure will result in the loss of life and/or extensive property damage; thus, in many instances, the use of the first-ply failure approach may be appropriate. The second limiting philosophy is to define the laminate failure stress as the effective stress level that causes final catastrophic fracture; in this case, the laminate failure stress is defined as the last-ply failure stress. Obviously, this latter definition is far less conservative and should be employed with caution and only under special circumstances. The design of any composite structure based on last-ply design philosophies should incorporate a generous factor of safety.

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2 FREE-EDGE STRESSES 2.1 The Origins of Free-Edge Stresses A thin multiangle (and/or multimaterial) composite laminate subjected to a uniaxial load Nxx is shown in Fig. 2. The thickness and width of the laminate are denoted t and 2b, respectively. The unloaded edges of the laminate, defined by y = Fb, are called ‘‘free edges’’ because no external forces are applied to these edges. The phrase ‘‘free-edge stresses’’ (also called ‘‘interlaminar stresses’’) refers to the stresses induced at and near a free edge. As will be seen, a highly three-dimensional state of stress is, in general, induced near a free edge. Furthermore, these free-edge stresses are developed because of a mismatch in the properties of adjacent plies within a laminate. Before discussing free-edge stresses directly, consider the ply stresses that would be predicted on the basis of classical lamination theory. The material and/or fiber angle is assumed to vary from one ply to the next for the laminate shown in Fig. 2. In other words, the effective material properties of each ply (in particular, the effective Poisson’s ratio vxy and the effective coefficient of mutual influence of the second kind gxx,xy) varies from one ply to the next. Due to these variations, the state of stress predicted for any ply based on a CLT analysis will, in general, include all three in-plane stress components. That is, the state of stress that is predicted for ply k on the basis of CLT generally includes stress components (rxx, ryy, and sxy)k. Hence, even for a

Figure 2 A thin multiangle (and/or multimaterial) composite laminate subjected to a uniaxial load Nxx.

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simple tensile specimen, a uniaxial load in the x-direction (Nxx) is predicted to cause stresses ryy and sxy, as well as stress rxx. Referring again to Fig. 2, note that although it may be possible for ryy and sxy to exist in the central regions of the laminate, neither ryy nor sxy can possibly exist at the free edge because neither Nyy nor Nxy is applied to the laminate. Therefore, even if ryy and sxy are induced in the central regions of the laminate, as y ! Fb, it must be that ryy ! 0 and sxy ! 0. Hence, CLT predictions must be incorrect, at least for regions near the free edge. To further investigate these observations, we will consider CLT predictions for a very simple laminate, where it is easily seen that the stresses predicted by CLT are reasonable for interior regions of the laminate, but are invalid near a free edge. This example will also give some physical insights into why a three-dimensional state of stress exists near a free edge. Consider a hybrid three-ply [0j]3 laminate subjected to uniaxial tensile load Nxx (only), as shown in Fig. 3a. The laminate is assumed to consist of three 0j plies, but the material used in plies 1 and 3 differs from that used ð1Þ ð2Þ ð1Þ ð2Þ ð1Þ ð2Þ in ply 2. Furthermore, assume that E11 ¼ E11 ; E22 ¼ E22 ; G12 ¼ G12 , but ð1Þ ð2Þ that v12 > v12 . In this hypothetical case then, a mismatch in Poisson’s ratio exists between plies, but all other material properties are exactly identical. In addition, the coefficient of mutual influence of the second kind equals zero for all three laminates because the 1–2 axes are coincident with the x–y axis for all three plies. We assume that the thickness of each ply is identical (say, t1 = t2 = t3 = tp), so that the total laminate thickness is t = 3tp. Based strictly on physical reasoning, it is clear that all plies will experience a tensile axial stress and strain (rxx and  xx) because the overall loading applied to the laminate (Nxx) has been assumed to be tensile. If the three plies are firmly bonded together, it is reasonable to assume that the tensile loading will cause a uniform contraction in the transverse y-direction due to the Poisson effect. On the other hand, if the three plies were not bonded together, plies 1 and 3 would contract to a greater extent than ply 2 because it has been assumed that Poisson’s ratio of the material used in these plies is ð1Þ ð2Þ greater than of ply 2: v12 > v12 . Thus, if the plies were not bonded together, then the magnitude of the compressive transverse strains induced in plies 1 and 3 would be greater than that in ply 2: |eyy|plies 1 and 3>|eyy|ply 2. The transverse contractions that would occur if the three plies were not bonded together are shown schematically (and highly exaggerated) in Fig. 3b. Of course, in reality, the three plies are bonded together, and hence we assume that all three plies contract by an equal amount: jeyy jplies 1 and 3 ¼ jeyy jply 2 : Because ply 2 is forced to contract to a greater extent than it would otherwise, in the bonded case, it is expected that a transverse compressive stress will be induced in ply 2. On the other hand, plies 1 and 3 contract to a lesser extent

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Figure 3 A hybrid three-ply 0j laminate subjected to a uniaxial tensile loading. It is assumed that plies 1 and 3 have a Poisson ratio that is numerically greater than ply 2, but all other properties are identical.

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that they would otherwise, and hence it is expected that a transverse tensile stress will be induced in plies 1 and 3. CLT calculations confirm these expectations. Using the procedures described in Chap. 6 (and assuming that DT = DM = 0), it can be shown* that the stresses induced in each ply under these conditions are: n h io ð2Þ ð2Þ ð1Þ 3E11 E22 v12 2v12 þ v12 Nxx rxx jplies 1 and 3 ¼  ð1aÞ hn o n oi2 ð1Þ ð2Þ tp 9E11 E22 v12 þ 2 v12 ryy jplies

1 and 3

¼

sxy jplies

1 and 3

¼0

rxx jply

ryy jply

2

2

¼

¼

 tp 9E11

n

3E11  tp 9E11 

tp 9E11

¼

h

i ð2Þ v12 E22 Nxx hn o n oi2 ð1Þ ð2Þ E22 v12 þ 2 v12

ð1Þ

v12

h io ð1Þ ð2Þ ð1Þ E22 v12 2v12 þ v12 Nxx hn o n oi2 ð1Þ ð2Þ E22 v12 þ 2 v12

h i ð1Þ ð2Þ 2 v12 v12 E22 Nxx hn o n oi2 ð1Þ ð2Þ E22 v12 þ 2 v12

2ryy jplies

1 and 3

sxy jply 2 ¼ 0

ð1bÞ ð1cÞ ð2aÞ

ð2bÞ ð2cÞ

It is interesting to note that Eqs. (1a) and (2a) show that the axial stress rxx induced in plies 1 and 3 is not precisely equal to that induced in ply 2 ð1Þ ð2Þ because v12 p v12 . Of greater immediate interest, however, is that Eqs. (1b) and (2b) predict that a transverse stress ryy is induced in all three plies, again ð1Þ ð2Þ because v12 p v12 . Hence, a load in the x-direction (Nxx) is predicted to cause a normal stress in the y-direction. The transverse stress induced in ply 2 is predicted to be compressive (as expected) and with a magnitude twice as high as the tensile transverse stress induced in plies 1 and 3. These transverse stresses occur solely because of the mismatch in Poisson’s ratio. That is, if ð1Þ ð2Þ v12 ¼ v12 then from Eqs. (1b) and (2b), the transverse stress in all three plies becomes ryy = 0.

* Because this is a relatively simple laminate, Eqs. (1) and (2) can be confirmed by hand calculations.

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The physical reasoning described above is simple and compelling; furthermore, the expected algebraic sign of the transverse stresses expected in each ply is confirmed by CLT calculations, lending credence to the CLT approach. Nevertheless, as pointed out earlier, a transverse stress ryy cannot possibly exist at the free edge of the laminate because no transverse load Nyy is applied to the laminate. A free-body diagram of an individual ply can now be used to understand why free-edge stresses develop and are three-dimensional. A cross-section within the y–z plane of the three-ply laminate and a free-body diagram of a section removed from ply 3 is shown in Fig. 4. A force Fyy (associated with stress ryy) acts on the left side of the free-body diagram. However, an

Figure 4 A free-body diagram of a section removed from ply 3 of the hybrid three-ply laminate shown in Fig. 3.

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equilibrating force Fyy cannot be present on the right side of the free-body diagram because no load is applied to the free edge of the laminate. Therefore, a shear force Vzy must develop on the interior surface of ply 3 (i.e., the ‘‘lower’’ surface of ply 3, as shown in Fig. 4), so as to maintain force equilibrium in the y-direction (SFy = 0). The fact that a shear force Vzy is present implies, of course, that a shear stress syz = szy is present near the free edge. Note, however, that it must be that Vzy ! 0 as y !Fb because no shear force can exist at the free edge; Vzy can only be nonzero at interior regions of ply 3. Also, the normal force acting on the left side Fyy is acting through the centroid of the free-body diagram, whereas the shear force on the right side Vzy is present on the lower surface. These two forces therefore induce a bending moment about the x-axis because they are not colinear. The only way that this bending moment can be reacted so as to satisfy moment equilibrium (i.e., so as to maintain SMx = 0) is if a force acting normal to the x–y plane ( Fzz) also develops. The fact that a normal force Fzz is present implies, of course, that a normal stress rzz is present. Numerical and analytical solutions (discussed below) indicate that Fzz (and hence rzz) reaches a maximum value at the free edge ( y =Fb), as shown schematically in Fig. 4. Because the only external force applied to the laminate as a whole is Nxx, internal force Fzz (or, equivalently, stress rzz) must be self-equilibrating: mby ¼ 0 rzz dxdy ¼ 0. This implies that rzz must undergo a change in algebraic sign at regions near the free edge. If, for example, rzz is tensile at y = b, it must become compressive at some distance from the free edge so as to maintain static equilibrium. This simple example illustrates that free-edge stresses rzz and szy are caused by a uniaxial tensile loading Nxx. The magnitude and distribution of rzz and szy cannot be determined solely on the basis of the equations of equilibrium. Nevertheless, the fact that they must exist can be appreciated through consideration of the free-body diagram shown in Fig. 4. For the simple hybrid [0j]3 laminate considered, the coefficient of mutual influence of the second kind gxx,xy=0, and therefore according to CLT, a uniaxial load Nxx does not cause an in-plane shear stress (sxy) to be induced at interior regions in any ply. However, for more general laminates, gxx,xy p 0, and a uniaxial loading Nxx will cause both ryy and sxy at interior regions of the laminate. In these cases, the state of stress near a free edge involves all six components of stress (rxx, ryy, rzz, syz, sxz, and sxy). Once again, it can be said that free-edge stresses develop because of a mismatch in the properties of adjacent plies within a laminate. 2.2 Numerical and Analytical Studies of Free-Edge Stresses Free-edge stresses have a profound influence on the delamination failure of multiangle composite laminates, and have therefore been the topic of a great

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many research papers. The existence of free-edge strains in multiangle composite laminates has been well documented experimentally [5–8]. Methods used to predict free-edge stresses may be roughly grouped into numerical solutions [8–14] and approximate analytical solutions [15–21]. With the exception of Ref. 9 (summarized below), all of the numerical solutions referenced here were based on the finite-element method. A detailed discussion of the numerical methods and/or approximate analytical tools that have been used to study free-edge stresses is beyond the scope of this book. The results of these analyses will simply be summarized, and the interested reader is referred to the original references if additional details are desired. The studies that have led to our current understanding of free-edge stresses will be discussed more or less in chronological order. Pipes and Pagano [10] presented the first numerical study of free-edge stresses in 1970. Their analysis was based on the method of finite differences. They investigated free-edge stresses induced in four-ply symmetrical angle-ply laminates having the general stacking sequence [h/ h]s and subjected to a remote uniform axial extension, which implies that, at remote locations, the laminate is subjected to a uniform axial tensile strain exx. The following properties, which were typical of transversely isotropic graphite–epoxy material systems at the time, were assumed in their study: E11 ¼ 138 GPa ð20 MsiÞ E22 ¼ E33 ¼ 14:5 GPa ð2:1 MsiÞ

G12 ¼ G23 ¼ G13 ¼ 5:86 GPa ð0:85 MsiÞ v12 ¼ v23 ¼ v31 ¼ 0:21

The thickness of each ply was denoted ho. In this text, the total laminate thickness has been denoted t; thus, in the Pipes and Pagano study, the total laminate thickness was t = 4ho. The width of the laminate was denoted 2b. Laminates with the three different width-to-laminate thickness ratios of 2, 4, and 6 were studied (i.e., laminates were studied with width-to-ply thickness ratios of b/h = 4, 8, and 12, respectively). Free-edge stresses were found to be independent of width-to-laminate thickness ratio over the range considered, and most of the results presented in Ref. 9 are for laminates with b/ho = 8 (i.e., for laminates whose width is four times as great as their thickness). A summary of ply stresses calculated by Pipes and Pagano for a [45/ 45]s laminate, at z = hþ o (i.e., just within the +45j ply), is presented in Fig. 5. In this figure, the ply stresses are plotted vs. normalized position y/b within the laminate, where the position y/b = 0 corresponds to the centerline of the laminate and y/b = 1.0 corresponds to the free edge (see Fig. 2). Using the material properties listed above, stresses predicted by a CLT analysis for

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Figure 5 Ply stresses calculated for a graphite–epoxy [45/ 45]s laminate, at the through-thickness position z = h+ 0 . (From Ref. 9, with permission from Sage Publications Ltd.)

the 45j plies in a [45/ 45]s laminate subjected to a uniform axial strain exx are as follows: rxx ¼ 20:4 GPa ð2:96 MsiÞ exx ryy ¼ 0 sxy ¼ 7:93 GPa ð1:15 MsiÞ exx

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Because CLT is based on the assumption of plane stress, for a CLT calculation, the remaining stress components are zero by assumption: rzz=sxz=syz=0. Referring to Fig. 5, at interior regions of the laminate (say, for y/b0.5), the state of stress becomes three-dimensional and finite-difference results do not agree with CLT calculations. The distribution of each of the six possible components of stress as predicted by the Pipes and Pagano analysis may be summarized as follows: 











As the free edge is approached, the normalized axial stress rxx/exx is predicted to increase very slightly from the CLT value, reaching a maximum value of about 21.0 GPa (3.05 Msi) at y/bc0.7 It then decreases to a finite value of about 15.9 GPa (2.3 Msi) at the free edge. A normalized transverse stress ryy/exx is predicted to develop at y/ bc0.5, reaching a maximum value of about 1.3 GPa (0.18 Msi) at y/ bc0.85. It then decreases back to zero at the free edge as it must to satisfy equilibrium at that point. A normalized out-of-plane stress rzz/exx is predicted to occur in regions very near the free edge; rzz/exx is essentially zero for y/ b xx > > > > > > > > > > 6 > 7> > > > T M > o > > > > > N þ N þ N e a a a b b b yy 7 > > > > 6 12 22 26 21 22 26 yy yy > > > > yy > > > 6 7> > > > T M > < < co = 6 a = 7 N þ N þ N a a b b b xy 16 26 66 61 62 66 xy xy 7 6 xy ¼6 7 > > 6 b11 b21 b61 d11 d12 d16 7> Mxx þ MTxx þ MM jxx > > > xx > > > > > 7> 6 > > > > > > > > 7> 6 T M > > > > > > M þ M þ M 5 4 j b b b d d d yy 12 22 62 12 22 26 > > > > > yy > yy yy > > > > > > ; ; : M þ MT þ MM > : jxy b16 b26 b66 d16 d26 d66 xy xy xy

Since our current objective is to compare CLT to isotropic beam theory, we will simplify our discussion by ignoring the effects of changes in temperature and moisture content (i.e., let NijT=MijT=NijM=MijM=0). For an isotropic beam or plate, substantial simplifications occur in the [abd] matrix, as was already discussed in Sec. 7 of Chap. 6. Noting that we have already specified that Nyy=Nxy=Myy=Mxy=0, for an isotropic beam, Eq. (45) of Chap. 6 reduces to: 8 o 9 2 a11 e > > > > > 6 > xx > o > > > a 12 eyy > > > 6 > > = 6 < o > 6 0 cxy ¼ 6 > 6 > j > 6 0 xx > > > > > 6 > > > > 4 0 > > > jyy > ; : jxy 0

a12 a11 0 0 0 0

0 0 2ða11 0 0

a12 Þ

0

0 0

0 0

0 d11 d12

0 d12 d11

0

0

38 b 9 Nxx =b > > > > > > 7> > > 0 > 7> > > > > 7> < 7 0 = 0 7 7> b > 0 M =b > 7> > > xx > 7> > > > > 5 0 > > > 0 > : ; 2ðd11 d12 Þ 0 0 0

where for an isotropic material, the elements of the [abd] matrix reduce to (see Sec. 7 of Chap. 6): 1 Et 12 ¼ Et3

a11 ¼ d11

d66 ¼ 2ðd11

v Et 12v d12 ¼ Et3 24ð1 þ vÞ d12 Þ ¼ Et3 a12 ¼

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b Let us first consider the midplane strains induced by Nxx alone (that is, let b b Mxx = 0). From the above result, the axial midplane strain induced by Nxx (only) is given by:   b Nbxx 1 Nxx o exx ¼ ða11 Þ ¼ Et b b

Since the cross-sectional area of the rectangular beam is A = (t)(b), we can write this result as:

Nbxx rxx ¼ ð1aÞ AE E b The quantity rxx = Nxx /A is simply the uniaxial stress induced in a prismatic isotropic beam subjected to an axial load. Hence, CLT reduces to the uniaxial form of Hooke’s Law for an isotropic material. The product AE is known as the axial rigidity of an isotropic beam. Similarly, the transverse midplane strain is given by:  v  Nb Nb vNbxx vrxx xx ¼ ¼ ¼ veoxx ð1bÞ eoyy ¼ ða12 Þ xx ¼ b b AE E Et Hence, for uniform axial loading, the transverse strain eoyy is related to the o via Poisson’s ratio, as expected for an isotropic beam subaxial strain exx jected to a state of uniaxial stress. Finally, no midplane shear strain is predicted (also as expected for an isotropic beam): eoxx ¼

coxy ¼ 0

ð1cÞ

It is mentioned in passing that the stress tensor induced in the beam is uniaxial (i.e., rxx p 0, ryy = rzz = sxy = sxz = syz = 0). That is, the principal stresses are rp1 = rxx, rp2 = rp3 = 0, and the principal stress coordinate system is coincident with the x–y–z coordinate system. Similarly, since shear strain is zero, the principal strain coordinate system is coincident with the o x–y–z coordinate system and ep1 = exx , ep2 = eoyy . The principal stress and principal strain coordinate systems are therefore coincident, which is always the case for isotropic materials. b Now consider the midplane strains and curvatures induced by Mxx b alone (that is, let Nxx = 0). In this case, CLT predicts that all midplane strains o o o = eyy = cxy = 0). Therefore, the midplane represents the neutral are zero (exx surface, as expected for an isotropic beam subjected to pure bending. The midplane curvatures are:  b   b  Mxx Mxx jxx ¼ d11 jyy ¼ d12 jxy ¼ 0 b b

The fact that the twist curvature jxy is zero implies that the x–z and y–z planes are the principal planes of curvature, as expected for a prismatic isotropic

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beam with symmetric rectangular cross section. Substituting the reduced forms for d11 and d12 for an isotropic material, we find:     12 Mbxx 12v Mbxx jxx ¼ 3 jyy ¼ Et Et3 b b The moment of inertia for a rectangular cross section of width b and height t is I = bt3/12. Therefore the above expressions can be written as: jxx ¼ jyy ¼

Mbxx EI vMbxx EI

ð2aÞ ð2bÞ

Equations (2a) and (2b) are the well-known moment-curvature equations for isotropic beams. The product EI is known as the flexural rigidity of an isotropic beam. Together, Eqs. (2a) and (2b) imply that jyy = –vjxx, which shows that anticlastic bending of an isotropic beam with symmetric rectangular cross section is predicted by CLT and, further, that jyy is related to jxx via Poisson’s ratio, as expected. The through-thickness variation in normal strains exx and eyy is given by:  b  Mxx o exx ¼ exx þ zjxx ¼ z EI  b  vMxx eyy ¼ eoyy þ zjyy ¼ z EI Through-thickness variation in stresses can be calculated using Eq. (30) in Chap. 5, which becomes (assuming isotropic properties and also that DT = DM = 0): 3 2 E vE 9 9 8 8 0 exx > rxx > 7> > 6 ð1 v2 Þ ð1 v2 Þ > > > > > > > > 7 6 > > > = 6 vE = < < 7> E 7 6 e 0 ryy ¼ 6 7 yy 2 2 > > > 7> 6 ð1 v Þ ð1 v Þ > > > > > > > 7> 6 > > > ; 4 ; : : 5> E c sxy xy 0 0 2ð1 þ vÞ 38 2 9 E vE zMbxx > > 0 > > 7> 6 ð1 v2 Þ ð1 v2 Þ > EI > > > 7> 6 > > = 7< 6 vE E 7 6 b zvM 0 ¼6 7 xx > 7> 6 ð1 v2 Þ ð1 v2 Þ > > 7> 6 EI > > > > > 5 4 E > > : ; 0 0 0 2ð1 þ vÞ

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Upon completing the matrix multiplication indicated above, we find: 9 8 b 9 8 rxx > > > Mxx z > > > > < = = > < I ryy ¼ > > > 0 > > > > : ; ; > : sxy 0

Hence, the CLT analysis predicts that ryy and sxy are both zero, and that the b (only) is given by the axial stress rxx induced in an isotropic beam by Mxx familiar flexure formula: rxx ¼

Mbxx z I

ð3Þ

As before, since only an axial stress is induced, the stress tensor is everywhere uniaxial and principal stresses are rp1 = rxx, rp2 = rp3 = 0. The principal stress coordinate system is coincident with the x–y–z coordinate system. Similarly, since shear strain is zero, the principal strain coordinate o system is coincident with the x–y–z coordinate system and ep1 = exx , ep2 = eoyy . We again conclude that the principal stress and principal strain coordinate systems are coincident, which is always the case for isotropic materials. Based on the above, we conclude that the analysis represented by CLT can be used to represent an isotropic beam with rectangular cross section b b subjected to an axial load Nxx and bending moment Mxx , both of which are b constant along the length of the beam. If Nxx = 0, then the loading condition represented by CLT corresponds to a state of pure bending. Now, there is an additional fundamental result from traditional beam theory that cannot be recovered by applying CLT to an isotropic beam. Specifically, traditional beam theory allows one to calculate the shear stresses induced in a beam. Recall that shear stresses sxz are given by the so-called shear formula (1–5): sxz ¼

VQ Ib

where V is the shear force present at a specified beam cross section, Q is the first moment of an area about the neutral axis (for a rectangular beam Q = b[(t2/4)–z2]/2), I is the area moment of inertia of the entire cross section about the neutral axis, and b is the width of the beam. The reason that CLT cannot be used to predict shear stress sxz is due to the fundamental b is constant along the length of assumption that the bending moment Mxx the beam. That is, recall from earlier studies [see, for example, Refs. (1–5)] that b the shear force V is related to the bending moment Mxx according to: V¼

dMbxx dx

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b b Since Mxx is assumed constant in a CLT analysis, V = dMxx /dx = 0. Therefore a CLT analysis implies that shear stress sxz = 0 due to the assumed loading conditions. The discussion presented in this section has been intended to show that CLT is entirely consistent with traditional isotropic beam theory. The reader should carefully note, however, that in this discussion, CLT has been specialized to the case of isotropic beams. The behavior of composite beams will be discussed in following sections. It will be seen that in some ways, the behavior of composite and isotropic beams is similar, but in others, they are quite different. In particular, for composite beams, the flexure formula [Eq. (3)] is only valid if the beam is oriented in a specific way with respect to the applied loads. Also, for composite beams, the principal stress and principal strain coordinate systems are not (in general) coincident.

3 TYPES OF COMPOSITE BEAMS CONSIDERED The types of composite beams considered in this chapter are summarized in b b Fig. 1. An externally applied axial load Nxx , bending moment Mxx , and transverse distributed load q(x) are also shown. Both the beam cross section and applied loads are referenced to an x–y–z coordinate system, where the

Figure 1 Composite beams with various cross sections. (a) Rectangular beam, plies orthogonal to plane of loading. (b) Rectangular beam, plies parallel to plane of loading. (c) I-beam. (d) T-beam. (e) Hat-beam. (f) Box-beam.

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y- and z-axes are orthogonal to the long axis of the beam. Bending moment Mxxb and transverse load q(x) act within the x–z plane, and hence the x–z plane is called the plane of loading. The type of ‘‘composite beam’’ implied during earlier sections is shown in Fig. 1(a). In this case, the beam cross section is rectangular, and all ply interfaces are orthogonal to the plane of loading. In contrast, a composite beam with rectangular cross section but in a decidedly different orientation with respect to the plane of loading is shown in Fig. 1(b). In this case, the ply interfaces are parallel to the plane of loading. The effective axial rigidity exhibited by a composite beam is identical in either orientation. As will be seen, however, the effective flexural rigidity of a rectangular composite beam will differ substantially depending on whether the beam is orientated as shown in Fig. 1(a) or (b). It is of course possible to manufacture and use composite beams with rectangular cross sections. However, in practice, it is far more common to use thin-walled composite beams, such as those shown in Fig. 1(c)–(f ). Beams with thin-walled cross sections are more commonly used because they provide far higher flexural rigidities per unit weight than a solid rectangular beam. The method used to study thin-walled composite beams herein is similar to that described by Swanson (8). The general approach is to approximate the cross sections shown in Fig. 1(c)–(f ) as an assembly of flat rectangular laminates. The stresses and strains induced in each region of the cross section will then be determined based on CLT. In all cases, the plane of loading is defined as the x–z plane. Those regions of a beam cross section that are parallel to the plane of loading will be called the ‘‘web laminate’’ (or ‘‘web laminates’’), whereas those regions that are orthogonal to the plane of loading will be called the ‘‘flange laminate’’ (or ‘‘flange laminates’’). Thus, for example, a composite I-beam has one web laminate and two flange laminates, whereas a composite box-beam has two flange laminates and two web laminates. Several restrictions are placed on the types of composite beams considered. First and foremost, only beams in which all flange and web laminates are produced using symmetric stacking sequences will be considered. This restriction is imposed to avoid complications due to thermal and moisture effects. That is, if a beam is produced using nonsymmetric web or flange laminates, then substantial thermal and/or moisture moment resultants (MijT and MijM) are induced, which may lead to substantial warping of the beam, even before application of any external load. For this reason, composite beams used in practice are almost always produced using laminates based on a symmetric stacking sequence. As discussed in Sec. 7.1 of Chap. 6, for a symmetric stacking sequence, MijT = MijM = 0, and hence this complication is eliminated. Secondly, we require that all beam cross sections possess both geometric and material symmetry about the plane of loading (the x–z plane). Having

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already stipulated that all laminates are symmetric, then the requirement of geometric and material symmetry about the x–z plane is achieved automatically for solid rectangular composite beams and in these cases requires no further discussion. However, symmetry about the x–z plane implies additional restrictions for thin-walled cross sections involving multiple web laminates. For cross sections with a single web laminate (an I-beam or T-beam), symmetry about the x–z plane is automatically satisfied if the web laminate is symmetric. However, for cross sections with two web laminates (a hat- or box-beam), the two symmetric web laminate must represent a set of fiber angles that are symmetric about the x–z plane, as well as being symmetric about the local web laminate midplanes. Note that symmetry about the x–z plane places no restrictions on the flange laminates. Thus, the top and bottom flanges in an I-, hat-, or box-beam are allowed to have a different stacking sequence and thickness, although they must both be symmetric about their respective midplanes (the stacking sequences and widths of the two bottom flanges in a hat-beam must be identical, however). The beam cross section may therefore be nonsymmetrical about the x–y plane. In practice, it can be problematic to produce thin-walled composite beams that feature the symmetries just described. Difficulties may arise due to the manufacturing process used to produce the beam. Methods of producing a thin-walled composite beam using unidirectional pre-preg tapes will be discussed to illustrate the practical difficulties encountered. First, consider the processes that could be used to produce a composite T-beam. Two possibilities are shown in Fig. 2. In Fig. 2(a), the cross section is formed by first curing the flange and web laminates separately. The two laminates are then bonded together in a second operation to form the desired T-cross section, as indicated. An advantage to this approach is that the stacking sequences used in the web and flange laminates are completely independent. Since any stacking sequence can be used in either laminate, it is easy to produce a T-beam with the required symmetric stacking sequence in both web and flange laminates in this manner and to insure symmetry about the x–z plane. However, a distinct disadvantage is that the web and flange laminates are joined solely by the adhesive bond—no continuous fibers cross the junction between web and flange. Hence, this approach is likely to result in a relatively low-strength T-beam. A more common method of producing a composite T-beam is illustrated in Fig. 2(b). In this case, the plies that exist within the web laminate are extended into (and become part of ) the flange laminate. Additional plies, which span the width of the flange, are added to complete the flange laminate. During layup, an internal ‘‘v-shaped’’ cavity is formed near the web–flange junction. This cavity is filled with some filler material (often pre-impregnated unidirectional tow) prior to curing the

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Figure 2 Methods to produce a composite T-beam using unidirectional pre-preg tape. (a) Forming a T-cross section by bonding together web and flange laminates. (b) Forming a T-cross section by extending web plies into the flange.

T-beam. This second approach of producing a T-beam has the distinct advantage of providing continuous fibers across the junction between the web and flange. However, the stacking sequences used in the web and flange laminates are no longer independent. Since the flange laminate is required to be symmetric (at least for the analysis presented in this section), the fiber angles used in the web and inner portions of the flange laminates must be repeated on the outer surface of the flange. Furthermore, if the web is produced using unidirectional pre-preg tape, then it is impossible to produce a T-beam that has both a symmetric web and a symmetric flange laminate,

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unless only 0j or 90j plies are used in the web. This difficulty may not be immediately obvious, but is illustrated in Fig. 3. The figure shows the formation of a symmetric 2-ply web laminate, that is, a [hj]s web laminate (where h p 0j or 90j). As indicated, extending the web plies into the flange results in a single flange ply in which the fiber angle is +hj on one side of the flange–web junction, but –hj on the other side. Hence, it is not possible to add additional plies across the entire width of the flange to produce both a symmetric laminate, unless web fiber angles are restricted to either h = 0j or 90j. In some instances, it may not be acceptable to restrict web angles to 0j or 90j since this restriction results in a web laminate with relatively low shear stiffness. A similar difficulty is encountered when producing a beam with two web laminates such as a hat- or box-beam. A single ply that becomes part of both web laminates and both flange laminates in a box-beam is shown in Fig. 4. In this case, it is not possible to produce web laminates that are symmetric about the x–z plane since a web ply with a fiber angle of +hj on one side of the x–z plane becomes a ply with a fiber angle of –hj on the other side of the x–z plane.

Figure 3 Illustration of why unidirectional fabrics or pre-pregs cannot be used to produce web and flange laminates that are both symmetric for T- or I-beams.

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Figure 4 Illustration of why unidirectional fabrics or pre-pregs cannot be used to produce web laminates that are symmetric about the x–z plane.

As before, the only way this can be avoided (if the beam is produced using unidirectional pre-preg tape) is to use fiber angles of h = 0j or 90j in the web, which may lead to web laminates with unacceptably low shear stiffness. These difficulties in achieving symmetry are avoided if the web is produced using a woven or braided pre-preg fabrics rather than unidirectional pre-preg, as shown in Fig. 5. Recall from Sec. 4 of Chap. 1 that a single ply of a woven or braided fabric features two or three fiber directions, oriented symmetrically about the warp direction. Since each individual woven or braided ply features a symmetric Fhj fiber pattern, plies extending from the web into the flange (as in a T-beam) or completely around the circumference of the cross section (as in a box-beam) retain the identical F hj fiber pattern in both web and flange laminates. Note that the use of a woven or braided ply to form the web laminate(s) does not preclude the use of additional unidirectional plies in the flange laminate(s). Finally, objectionable levels of thermal warping may still occur for some thin-walled composite beams, even though they satisfy all of the symmetry requirements described above. Thermal warping may occur because we have allowed the beam cross section to be nonsymmetric about the x–y plane. Suppose, for example, that an I-beam is produced in which the (symmetric) top flange is produced using a very different stacking sequence than the (symmetric) bottom flange. This implies that the effective thermal expansion coefficient axx of the top flange may be very different than that of the bottom flange. Consequently, if the beam experiences a uniform change in temper-

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Figure 5 The use of woven or braided fabrics or pre-preg avoids the difficulties shown in Figs. 3 and 4.

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ature, then the top flange will tend to expand or contract at a different rate than the bottom flange, causing the beam to warp. The symmetric web laminate will tend to restrict such warping, but nevertheless, this effect may be significant depending on details of a given beam design. This possibility will not be addressed in this text, although the theory presented can be easily modified to account for this effect. 4 EFFECTIVE AXIAL RIGIDITY OF RECTANGULAR COMPOSITE BEAMS b A rectangular composite beam subjected to axial load Nxx is shown in Fig. 6. The width and thickness of the beam cross section are labeled b and t, respectively. We will now use CLT to predict the mechanical response of this beam to pure axial loading. As mentioned earlier, significant free-edge stresses may be present (depending on the stacking sequence and material system used to produce the beam) but will be neglected in the following discussion. Since the beam is assumed to be symmetric and the only applied load b is Nxx , the midplane strains and curvatures induced in a symmetric composite beam can be predicted using Eq. (45) of Chap. 6, which becomes (for Nyy=Nxy=Mxx=Myy=Mxy=bij=MijT=MijM=0):

8 o 9 2 a11 exx > > > > > > 6 > > > a12 eoyy > > > > 6 > > = 6 < o > a cxy ¼ 6 6 16 6 0 > > > > 6 > jxx > > > > 6 > > j > 4 0 > yy > > > ; : jxy 0

a12 a22 a26

a16 a26 a66

0 0 0

0 0 0

0

0

d11

d12

0

0

d12

d22

0

0

d16

d26

9 38 b ðNxx =bÞ þ NTxx þ NM > xx > > > > > > 7> > > T M > > N þ N > 7> yy yy > > > 7> = < T M 7 N þ N xy xy 7 > d16 7 > > 0 7> > > > 7> > > > d26 5> > > 0 > > > > ; : d66 0 0 0 0

ð4Þ

Note that thermal and moisture stress resultants are (in general) present in the symmetric composite beam, even though the only externally applied b load is Nxx . As discussed in earlier chapters, thermal and moisture resultants contribute substantially to the strains and stresses induced in each ply and may lead to premature failure of a composite beam if not properly accounted for. The preexisting ply stresses and strains are of course changed upon b . From Eq. (4), it is easy to see application of an external axial load Nxx b (only) is that the incremental change in midplane strains caused by Nxx given by: eoxx ¼

a11 b N b xx

eoyy ¼

a12 b N b xx

coxy ¼

a16 b N b xx

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ð5Þ

Figure 6 A composite beam with cross-section bt, subjected to axial load b Nxx .

The effective extensional Young’s modulus, Poisson’s ratio, and coefficient of mutual influence of the second kind for a composite laminate were all defined as follows in Sec. 9.1.1 of Chap. 6: ex

E xx ¼

1 ta11

ex vxy ¼

a12 a11

ex gxx;xy ¼

a16 a11

Substituting these effective properties into Eq. (5), we find: Nbxx Nbxx rxx ex ¼ ex ¼ ex E xx ðtbÞE xx AE xx ! ! Nbxx ex ex rxx ¼ vxy ¼ vxy ¼ ex ex AE xx E xx

eoxx ¼ eoyy

coxy

¼

ex gxx;xy

Nbxx ex AE xx

!

¼

ex gxx;xy

rxx ex E xx

!

ð6aÞ ex o vxy exx

ex eoxx ¼ gxx;xy

ð6bÞ ð6cÞ

From Eq. 6(a), we see that the effective axial rigidity of a rectangular composite beam can be written as: 

ex  ex AE xx ¼ ðtbÞE xx

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ð7aÞ

where A is the cross-sectional area of the beam and E ex xx is the effective extensional Young’s modulus of the laminate (defined in Sec. 9.1.1 of Chap. 6). Alternatively, the effective axial rigidity may be written as:    b ex  AE xx ¼ ð7bÞ a11

Although the beam shown in Fig. 6 is oriented such that ply interfaces are orthogonal to the x–z plane, the analysis is also applicable to a beam oriented such that ply interfaces are parallel to the x–z plane, as shown in Fig. 1(b). Hence, Eqs. (7a) and (7b) give the effective axial rigidity of a symmetric rectangular composite beam, regardless of ply orientation. Note that Eqs. (6a)–(6c) are directly analogous to similar relations for an isotropic beam, as given by Eqs. (1a) and (1b). Since only an axial load is applied, the effective stress tensor is uniaxial (i.e, rxx p 0; ryy ¼ rzz ¼ sxy ¼ sxz ¼ syz ¼ 0Þ. That is to say, the effective principal stresses are rp1 = rxx ; rp2 = rp3 ¼ 0, and the effective principal stress coordinate system is coincident with the x–y–z coordinate system. However, for a composite beam, an axial stress jxx will (in general) cause a shear strain coxy due to the presence of ex ). Therefore, the principal a16 (or equivalently, due to the presence of Dxx,xy strain coordinate system is not, in general, coincident with the effective principal stress coordinate system. Recall from Sec. 7 of Chap. 6 that a16=0 for the following commonly used stacking sequence: 

Cross-ply. Balanced.  Balanced angle-ply.  Quasi-isotropic. 

If an axially loaded symmetric composite beam is produced using any of these stacking sequences, then shear strain coxy is zero and the effective principal stress and principal strain coordinate systems are coincident. 5 EFFECTIVE FLEXURAL RIGIDITIES OF RECTANGULAR COMPOSITE BEAMS An initially straight beam subjected to an external bending moment of equal magnitude at either end is said to be in state of pure bending. The flexural rigidity of a beam is defined in pure bending. In this section, the effective flexural rigidities of composite beams with rectangular cross sections will be determined. In order to do so, we must consider two different cases. In the first case, we consider a beam oriented such that ply interfaces are orthogonal to

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the plane of loading, as previously shown in Fig. 1(a). In the second case, we consider a beam for which ply interfaces are parallel to the plane of loading, as was shown in Fig. 1(b). 5.1 Effective Flexural Rigidity of Rectangular Composite Beams with Ply Interfaces Orthogonal to the Plane of Loading A composite beam oriented such that ply interfaces are orthogonal to the plane of loading is shown in Fig. 7. The height and width of the beam cross section are denoted t and b, respectively. We will now use CLT to predict the mechanical response of this beam in pure bending. As mentioned earlier, significant free-edge stresses may be present (depending on the stacking

Figure 7 A composite beam with ply interfaces orthogonal to the x–z plane, subjected to pure bending. (a) A composite beam subjected to pure bending. (b) Cutaway view of composite beam at cross-section A–A.

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sequence and material system used to produce the beam) but will be neglected in the following discussion. Since we have assumed that the beam is symmetric and subjected to b only, the midplane strains and curvatures can be predicted using Eq. Mxx (45) of Chap. 6, which becomes (for bij ¼ Nxx ¼ Nyy ¼ Nxy ¼ Myy ¼ Mxy ¼ b MTij ¼ MM ij ¼ 0 and Mxx=Mxx /b): 8 o 9 2 a11 exx > > > > > 6 a12 > o > > > > e > 6 > > > yy > > < = 6a coxy ¼ 6 6 16 6 0 > > j > xx > 6 > > > > 6 > > > > 4 0 j yy > > > > : ; jxy 0

a12 a22 a26

a16 a26 a66

0 0 0

0 0 0

0

0

d11

d12

0

0

d12

d22

0

0

d16

d26

9 38 T Nxx þ NM > xx > > > 7> > NT þ NM > > > > 7> yy > > > > Tyy 7> < 7 N þ NM = xy xy 7 d16 7 > > Mbxx =b > 7> > > > 7> > > > d26 5> 0 > > > > ; : 0 d66 0 0 0

ð8Þ

Thermal and moisture stress resultants NijT and NijM are (in general) present in the symmetric composite beam. As discussed in earlier chapters, thermal and moisture stress resultants contribute substantially to the strains and stresses induced in each ply and may lead to premature failure of a composite beam if not properly accounted for. The thermal and moisture stress resultants do not influence the initial elastic response of the beam, however. From Eq. (8), it is easy to see that midplane strains are not changed b upon the application of Mxx since bij=0. Hence, the midplane of the beam cross section represents the neutral surface in the sense that application of b does not contribute to (or alter) preexisting midplane strains. As noted in Mxx Sec. 2, the neutral surface for an isotropic beam with rectangular cross section is also coincident with the midplane. From Eq. (8), the midplane curvatures are given by:  b   b   b  Mxx Mxx Mxx jxx ¼ d11 jyy ¼ d12 jxy ¼ d16 ð9Þ b b b The fact that the twist curvature jxy p 0 (in general) shows that the x–z and y–z planes are not the principal planes of curvature. This is in direct contrast to an isotropic beam since for a prismatic isotropic beam with symmetric rectangular cross section, the x–z and y–z planes represent the principal planes of curvature. The effective flexural Young’s modulus, Poisson’s ratio, and coefficient of mutual influence of the second kind for a composite laminate were all defined as follows in Sec. 9.1 of Chap. 6: fl

E xx ¼

12 d11 t3

fl vxy ¼

d12 d11

fl gxx;xy ¼

d16 d11

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Substituting these effective properties into Eq. (9), we find: jxx

12 ¼ fl 3 E xx t



b Mxx b



jyy ¼

fl 12vxy fl 3 E xx t



b Mxx b



jxy ¼

fl 12gxx;xy fl 3 E xx t



Mbxx b



The moment of inertia for a rectangular cross section of width b and height t is I = bt3/12. Therefore the above expressions can be written as: jxx ¼ jyy ¼ jxy ¼

b Mxx fl I E xx

fl vxy Mbxx fl I E xx b fl gxx;xy Mxx

fl I E xx

ð10aÞ ð10bÞ ð10cÞ

Equations (10a)–(10c) are the moment-curvature equations for a composite beam with ply orientation orthogonal to the plane of loading and should be compared with analogous results for isotropic beams [Eqs. (2a) and (2b)]. The product E flxx I represents the effective flexural rigidity for a composite beam in this orientation. That is, the effective flexural rigidity of a rectangular composite beam with ply orientation orthogonal to the plane of loading can be written as:   fl  fl  bt3 E xx b ¼ IE xx ¼ ð11Þ d11 12

where E flxx is the effective flexural Young’s modulus of the laminate (defined in Sec. 9.1.2 of Chap. 6). Equations (10a) and (10b) also imply jyy = –vxyfl jxx, which shows that a composite beam with rectangular cross section will exhibit anticlastic bending in the same manner as an isotropic beam. Equations (10a) and fl (10c) imply jxy = gxx,xy jxx, which shows that a twist curvature is induced in a fl = 0. A midplane twist curvature jxy does composite laminate, unless gxx,xy fl not occur for isotropic beams subjected to pure bending. Recalling that gxx,xy fl = d16/d11, it is seen that g xx,xy is zero only for specially orthotropic laminates (i.e., if d16=0). This rarely occurs for composite beams used in practice. The only common stacking sequences that lead to a specially orthotropic beam are unidirectional [0]n or [90]n stacks or symmetric cross-ply [0/90]ns stacks. The first two stacking sequences are essentially never used in practice because they possess significant strength and stiffness in one direction only. Cross-ply beams may be used on occasion but suffer from very low shear stiffness. Beams produced using any other stacking sequence, for example, a symmetric

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quasi-isotropic stacking sequence, will exhibit a twist curvature when subjected to a pure bending. Based on the above definitions, we see that through-thickness variation in strains exx, eyy, and cxy induced by pure bending is given by:  b  Mxx exx ðzÞ ¼ zjxx ¼ z ð12aÞ E flxx I ! fl Mb vxy xx ð12bÞ eyy ðzÞ ¼ zjyy ¼ z fl E xx I ! fl gxx;xy Mbxx cxy ðzÞ ¼ zjxy ¼ z ð12cÞ fl E xx I Strains are predicted to vary as linear functions of z, as expected from results in earlier chapters. Note, however, that ply stresses rxx, ryy, and sxy are not linear functions of z. It is not possible to develop a ‘‘flexure formula’’ for use with composite beams with ply orientation orthogonal to the plane of loading. Rather, stresses must be calculated using Hooke’s Law for an anisotropic material and based on the ply strain distributions given by Eq. (11), as discussed in earlier chapters. 5.2 Effective Flexural Rigidity of Rectangular Composite Beams with Ply Interfaces Parallel to the Plane of Loading We now consider a composite beam that is oriented such that the ply interfaces are parallel to the plane of loading. Such a beam in pure bending is shown in Fig. 8. We will continue to label the beam cross section using the same symbols as in earlier sections. Hence, the beam depth is denoted b, whereas the beam width (which equals the thickness of the composite laminate) is denoted t (compare Figs. 7 and 8). As drawn in Fig. 8, plies are numbered from left to right. That is, the outermost surface of ply 1 exists at zV=–t/2, whereas the outermost surface of ply n exists at zV=+t/2. Ply fiber angles are referenced to the +x V-axes and are measured positive from the +x V-axis towards the +yV-axis, in accordance with the right-hand rule. Note that the bending moment is referenced to the x–y–z coordinate system, whereas the ply stacking sequence is referenced to a ‘‘new’’ x V–y V–z V coordinate system. As is apparent from Fig. 8, the +z V- and +yV-axes are coincident, as are the +x- and +x V-axes. The +y- and +zV-axes are parallel, but the +zV-direction is opposite to the +y-direction. The reader should carefully consider the two coordinate systems shown since they represent a change in nomenclature from our earlier discussion. This is a subtle but b potentially confusing change. For example, the bending moment Mxx shown

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Figure 8 A composite beam with ply interfaces parallel to the x–z plane, subjected to pure bending. (a) A composite beam subjected to pure bending. (b) Cutaway view of composite beam at cross-section A-A.

in Fig. 8 causes a very different state of stress in the composite beam than does b the bending moment shown in Fig. 7 (which is also denoted Mxx ). This difference is of course due to the change in orientation of the composite beam. Throughout this chapter, we assume that the stacking sequence is symmetric. Hence, both the beam cross section as well as beam material b properties are symmetric about the x–z plane. Since the moment Mxx is constant along the length of the beam, all beam cross sections must deform in an identical manner. Under these conditions, the long axis of the beam must deform into a circular arc with a radius of curvature rxx, as shown in Fig. 9. Hence, beam cross sections that are initially plane and perpendicular to the axis of the beam must remain plane and perpendicular to the deformed axis of the beam following loading. These are precisely the same conditions encoun-

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Figure 9 Deformations induced in a composite beam with ply interfaces parallel to the plane of loading, subjected to pure bending (deformations shown greatly exaggerated). (a) A composite beam subjected to pure bending. (b) Deformed beam.

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tered in pure bending of isotropic beams with symmetric rectangular cross sections. Hence, many conclusions based on isotropic beam theory are also applicable to the composite beam shown in Figs. 8 and 9. In particular, the neutral axis must pass through the centroid of the beam cross section, and axial normal strains are given by: exx ¼ zjxx

ð13Þ

eoxx ¼ eox Vx V ¼ zjxx

ð14Þ

where curvature jxx = 1/rxx. Hence, axial strain is a maximum at z = Fb/2 and is zero at z = 0, the neutral axis. A subtle point is that in this case, the curvature jxx represents bending in the x–z plane (the plane of loading) and is not comparable to the curvature jxx discussed previously. The curvature previously discussed occurs in the xV–z V plane and would now be labeled jx Vx V. Since we have assumed both a symmetric stacking sequence and symmetric rectangular cross section, the axial strain induced at any position z will be uniform across the width of the beam. Hence, the axial strain given by Eq. (13) represents the strain induced at the midplane of the laminate: We also note that for the loading condition shown in Fig. 8: Ny Vy V ¼ Nx Vy V ¼ Mx Vx V ¼ My Vy V ¼ Mx Vy V ¼ 0: Therefore, Eq. (45) of Chap. 6 becomes: 8 o 9 2 a11 a12 a16 0 ex Vx V > > > > > > o > > 6 a12 a22 a26 0 > > e > 6 > > > 6 > y Vy V > = < o a a26 a66 0 cx Vy V ¼ 6 6 16 6 > 0 0 d11 jx VxV > > > 6 0 > > > > 6 > > > > 4 0 0 0 d12 jy Vy V > > > > ; : jx Vy V 0 0 0 d16

0 0 0 d12 d22 d26

9 38 Nx Vx V þ NxTVx V þ NxMVx V > > > > > 7> > > M T > þ N N > > 7> y Vy V y Vy V > > > > 7< = 7 T M N þ N 7 x Vy V x Vy V > d16 7 0 > > 7> > > > 7> > > > > d26 5> 0 > > > : ; d66 0 0 0 0

Our current objective is to evaluate the effective flexural rigidity of the composite beam, so we ignore strains caused by thermal or moisture stress resultants. Hence, the midplane strain caused by application of stress resultant NxVxV=Nxx is: eox Vx V ¼ eoxx ¼ a11 Nx Vx V ¼ a11 Nxx Combining this result with Eq. (14), we find:   jxx Nxx ¼ z a11

ð15Þ

Now consider a free-body diagram showing the internal distribution of stress resultant Nxx at arbitrary cross-section A–A, as shown in Fig. 10. Since the

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Figure 10 Free-body diagram used to relate external bending moment Mob to internal stress resultant NxVxV.

beam is in static equilibrium, the externally applied moment Mob must be balanced by the distribution of internal stress resultant Nxx. Recalling that the units of Nxx are force-per-length (e.g., N/m), the incremental axial force acting over an infinitesimal strip of height dz, located at an arbitrary distance z from the neutral axis, is given by: dNbxx ¼ Nxx dz Since this force acts at distance ‘‘z’’ from the neutral axis, the incremental internal moment associated with this force is: dMbxx ¼ dNbxx z ¼ Nxx zdz The total moment is obtained by integrating over the height of the beam: Mbxx

¼

þb=2

m

b=2

Nxx zdz ¼

þb=2 

m

b=2

 jxx 2 z dz a11

Neither jxx nor a11 is a function of z, so they can be removed from the integral sign. Completing the integration indicated, we find: Mbxx ¼

jxx b3 12a11

The effective extensional Young’s modulus is given by Eq. (63) of Chap. 6, repeated here for convenience: ex

E xx ¼

1 ta11

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b Our expression for Mxx can therefore be written as:

ex

Mbxx

jxx E xx tb3 ¼ 12

Noting that the area moment of inertia about the y-axis for the rectangular cross section is I = tb3/12, we can rearrange this result to find: jxx ¼

Mbxx ex E xx I

ð16Þ

Equation (16) is a moment-curvature equation for a composite beam with ex ply orientation parallel to the plane of loading. The product E xx I represents the effective flexural rigidity of a composite beam in this orientation. That is, the effective flexural rigidity of a rectangular composite beam with ply orientation parallel to the plane of loading can be written as: 

ex tb3 E xx ex  ¼ IE xx ¼ 12



b3 12a11



ð17Þ

ex where E xx is the effective extensional modulus of the laminate (defined in Sec. 9.1.1 of Chap. 6). Equation (17) should be compared with our earlier expression for the effective flexural rigidity of a composite beam with plies orthogonal to the plane of loading Eq. (11). It is seen that the flexural rigidity of a composite beam with plies parallel to the plane of loading is dominated by ex extensional stiffnesses (i.e., E xx or a11). In contrast, the flexural rigidity of a beam with plies orthogonal to the plane of loading is dominated by flexural fl or d11). stiffnesses (i.e., E xx Anticlastic bending will occur for the composite beam shown in Fig. 10. Normal strains eyy=ez Vz V will vary linearly with distance from the neutral axis. Assuming a positive bending moment, then in regions below the neutral axis (i.e., for z = y V > 0), the transverse normal strains will be compressive, whereas in regions above the neutral axis (for z = y V0, then the bending moment Mob is algebraically negative and would act in the sense opposite to that shown in Fig. 22(b). The internal shear and bending moment induced at any cross section located at arbitrary position x can now be determined using the free-body diagram shown in Fig. 23. Summing forces in the z-direction, we find: qo 2 b ðL Vxz x2 Þ ðxÞ ¼ 2L

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Figure 23 Free-body diagram used to determine internal shear and bending b b moment, Vxz and Mxx , respectively, acting at an arbitrary cross section located at position x.

b A so-called shear force diagram is created by plotting Vxz (x), as shown in Fig. 24. Similarly, summing moments about an axis passing through the lefthand side of the free-body diagram shown in Fig. 23 and parallel to the y-axis, we find: qo b ð 2L3 þ 3L2 x x3 Þ Mxx ðxÞ ¼ 6L b A so-called bending moment diagram is created by plotting Mxx (x), as shown b in Fig. 24. Substituting the above expression for the bending moment Mxx (x) into Eq. (44), we find:

d2 w ¼ dx2

qo ð 2L3 þ 3L2 x 6LðIE xx Þ

Integrating once results in:  dw qo 3L2 x2 ¼ 2L3 x þ dx 2 6LðIE xx Þ

x3 Þ

x4 4



þ C1

where C1 is a constant of integration. Since the beam is clamped at the left end, the slope must equal zero there: dw/dx = 0 at x = 0. Enforcing this boundary condition, we find that the constant of integration must equal zero: C1 ¼ 0

Performing a second integration and simplifying, we obtain: w¼

qo x2 ð20L3 120LðIE xx Þ

10L2 x þ x3 Þ þ C2

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Figure 24 Shear and bending moment diagrams for the cantilevered beam considered in Example Problem 4.

where C2 is a second constant of integration. Since the beam is clamped at the left end, the deflection must also equal zero there: w = 0 at x = 0. Enforcing this second boundary condition, we find that the constant of integration must again equal zero: C2 = 0 Hence, we find that beam deflections are given by: w¼

q o x2 ð20L3 120LðIE xx Þ

10L2 x þ x3 Þ

ðbÞ

Note that this result is included in the list of solutions tabulated in Appendix C, Table C.1. Part (b). The effective flexural rigidity of the box-beam shown in Fig. 14(c) was determined as a part of Example Problem 1 and was found to be: ðIE xx Þ ¼ 8:43  103 N m2

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Substituting this and the other specified numerical values (L = 1 m, qo = 200 N/m) into expression (b) above, we find that the transverse deflections are predicted to be (in millimeters):   wðxÞ ¼ ð0:1977x2 Þ 20L3 10L2 x þ x3

A plot of these beam deflections is shown in Fig. 25. Note that both the deflection and slope are zero at the clamped end, as dictated by this boundary condition. A maximum deflection of 2.17 mm is predicted to occur at the free end. Example Problem 5 Suppose the cantilevered graphite–epoxy beam considered in Example Problem 4 is subjected to both a bending moment Mob applied at the free end and a linearly increasing distributed load qðxÞ ¼ ðqo =LÞx , as shown in Fig. 26. Perform the following: (a)

Use the method of superposition to obtain an analytical expression giving the predicted deflection w(z). (b) Plot numerical values of the predicted deflections, assuming the following beam length and loading: L¼1m

qo ¼ 200 N=m Mbo ¼ 40 N m

Figure 25 Deflections predicted for the graphite–epoxy composite box-beam considered in Example Problem 4.

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Figure 26 Cantilevered graphite–epoxy beam considered in Example Problem 5.

Solution Part (a). Deflections induced by a concentrated moment Mob and distributed load q(x) when acting separately are included in Table C.1. The deflections due to Mob acting alone are given by:   wðxÞ

Mbo

¼

Mbo x2 2ðIE xx Þ

The deflections due to the distributed load are given by:   wðxÞ

qðxÞ

¼

qo x 2 ð20L3 120LðIE xx Þ

10L2 x þ x3 Þ

Applying the principle of superposition, the beam deflections caused by both load components acting simultaneously are simply the sum of these two solutions:

    Mbo x2 q o x2   wðxÞ ¼ wðxÞ b þ wðxÞ ¼ ð20L3 10L2 x þ x3 Þ þ qðxÞ Mo 2ðIE xx Þ 120LðIE xx Þ Part (b). The distributed load q(x) will tend to deflect the beam downwards (i.e., in the positive z-direction), whereas Mob will tend to deflect the beam upwards (in the negative z-direction). Using the values specified for beam dimensions, material properties, and loads, deflections are given (in millimeters) by: wðxÞ ¼ ð0:1977x2 Þð20

10x þ x3 Þ

2:372x2

which can be rearranged as: wðxÞ ¼ 1:582x2

1:977x3 þ 0:1977x5

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A plot of predicted deflections is shown in Fig. 27. As before, both the deflection and slope are zero at the left end, as required by the clamped boundary condition. A locally maximum positive (downwards) deflection of 0.160 mm occurs at an axial position of x = 0.563 m. The globally maximum deflection occurs at the free end, where a negative (upward) deflection of 0.198 mm occurs. Example Problem 6 Consider a composite beam supported as previously shown in Fig. 21. (a) Obtain analytical expressions for the reaction forces. (b) Assume that the beam has the box cross section previously shown in Fig. 14(c) and length L=1 m. Obtain numerical values for reaction forces and plot transverse beam deflections if a = 3 L/4 = 0.75 m and P = 1000 N. Solution Part (a). The beam shown in Fig. 21 is statically indeterminate to the first degree since there are three unknown reaction forces. Therefore one of the unknown reaction forces must be selected to be a redundant force. Anyone of ðAÞ ðBÞ the reaction forces (Rz , Rz , or Mob) can be treated as the redundant force. ðBÞ For present purposes, Rz will be treated as the redundant. This problem can be solved based on the method of superposition, as summarized in Fig. 28. The first step is to determine the beam deflections that would occur if the redundant force was removed. Removal of redundant ðBÞ reaction force Rz implies that the roller support at the right end is removed, as shown in Fig. 28(b). The process of obtaining beam deflections when the redundant force is removed is often called the reduced problem. From Appen-

Figure 27 Deflections predicted for the beam considered in Example Problem 5.

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Figure 28 Summary of the method of superposition applied to the indeterminate beam considered in Example Problem 4.

dix C Table C.1 we find that beam deflections over the range a V x V L associated with the reduced problem are given by:   wðxÞ

reduced

¼

Pa2 ð3x 6ðIE xx Þ

aÞ

The next step is to determine beam deflections that would be caused if only the redundant force was applied, as shown in Fig. 28(c). From Table C.1, ðBÞ we find that the beam deflections associated with the redundant force Rz are given by:   wðxÞ

ðBÞ

Rz

¼

ðBÞ

Rz x2 ð3L 6ðIE xx Þ

xÞ

This expression is valid for any axial position 0 V x V L. We now superimpose the deflections associated with the reduced and redundant problems. In this case, we have, for a V x V L:   wðxÞ ¼ wðxÞ

reduced

  þ wðxÞ

ðBÞ

Rz

¼

h 1 Pa2 ð3x 6ðIE xx Þ

aÞ þ RzðBÞ x2 ð3L

i xÞ

ðaÞ

We require that the total beam deflection equals zero at the right end (at x=L) since the beam is supported by a roller at that point. Hence: wðx ¼ LÞ ¼

h 1 Pa2 ð3L 6ðIE xx Þ

2 aÞ þ RðBÞ z ðLÞ ð3L

i LÞ ¼ 0

which reduces to: 3 2 2RðBÞ z L ¼ Pa ða

3LÞ

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ðbÞ

Expression (b) represents an equation of compatibility. Only one such equation is necessary in this problem since the beam is indeterminate to the first degree. Next, we enforce the equations of equilibrium, SFz = 0 and SMy = 0, resulting in (respectively): ðBÞ RðAÞ z þ Rz ¼

b RðBÞ z L þ Mo ¼

P

ðcÞ

Pa

ðdÞ

where Eq. (d) was obtained by summing moments about the left end of the beam (at x = 0). Expressions (b), (c), and (d) represent three simultaneous equations in terms of the three unknown reaction forces. They can be rewritten in matrix form as: 9 38 ðAÞ 9 8 2 2 Pa ða 3LÞ > 0 2L3 0 > > Rz > > > > > > > > > > > > 7> 6 = 7< ðBÞ = < 6 7 61 ¼ 1 0 P R 7> z > > 6 > > > 5> 4 > > > > > > > ; > : b > ; : 0 L 1 Pa Mo Solving this system of equations results in: RðAÞ ¼ z RðBÞ z ¼ Mbo ¼

P ða3 þ 2L3 2L3 Pa2 ð3L aÞ 2L3 Pa 2 ða þ 2L2 2L2

3a2 LÞ

ðeÞ ðf Þ ðgÞ

3aLÞ

Part (b). Based on the specified numerical values, the reaction forces are: RðAÞ ¼ z RðBÞ z ¼ Mbo ¼ ¼

ð1000NÞ h ð0:75mÞ3 þ 2ð1mÞ3 2ð1mÞ3

ð1000NÞð0:75mÞ2 2ð1mÞ3

½3ð1mÞ

i 3ð0:75mÞ2 ð1mÞ ¼

ð0:75mފ ¼

ð1000NÞð0:75mÞ h ð0:75mÞ2 þ 2ð1mÞ2 2 2ð1mÞ 117Nm

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367N

633N

3ð0:75mÞð1mÞ

i

ðAÞ

ðBÞ

Note that the calculated values for Rz , Rz , and Mob are all algebraically negative, which indicates that these reaction forces are acting in a sense opposite to that shown in Fig. 21. To plot beam deflections, note that expression (a), developed in the first part of this example problem, gives the deflection over the range a V x V L. Hence, we need an additional expression for beam deflections that is valid over 0 V x V a. From Table C.1, we find that beam deflections over the range 0 V x V a associated with the reduced problem are given by:   wðxÞ

reduced

¼

Px2 ð3a 6ðIE xx Þ

xÞ

As before, from Table C.1, we find that the beam deflections associated with the redundant force R (B) z are given by:   wðxÞ

ðBÞ

Rz

¼

ðBÞ

Rz x2 ð3L 6ðIE xx Þ

xÞ

This result is valid for any axial position 0 V x V L. Superimposing these two results, we obtain an expression for the total beam deflection, valid for 0 V x V a:   wðxÞ ¼ wðxÞ

reduced

  þ wðxÞ

ðBÞ

Rz

¼

h 1 Px2 ð3a 6ðIE xx Þ

xÞ þ RzðBÞ x2 ð3L

xÞ

i

ðhÞ

Figure 29 Deflections predicted for the beam considered in Example Problem 6.

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Recall from Example Problem 1 that the effective flexural rigidity for this beam is ( IE xx) = 8.43103 N m2. Substituting this value (and the other numerical values specific to this problem) into expressions (a) and (h) and plotting over the appropriate ranges, results in Fig. 29. As dictated by the specified boundary conditions, the beam deflection is zero at either end of the beam, and the slope of the beam is zero at the left end. A maximum positive (downward) deflection of 0.94 mm is predicted to occur at the axial location x = 0.64 m.

9 COMPUTER PROGRAM BEAM The computer program BEAM has been developed to supplement the material presented in 4 Secs. 5 Secs. 6 of this chapter. This program can also be downloaded at no cost from the following website: http://depts.washington.edu/ amtas/computer.html. Program BEAM can be used to calculate the centroidal location, effective axial rigidity, and effective flexural rigidity of a composite beam with either a rectangular cross section or with any of the thin-walled cross sections shown in Tables 1–3. The program prompts the user to input all information necessary to perform these calculations. Properties of up to five different materials may be defined. The user must input various numerical values using a consistent set of units. For example, the user must input elastic moduli for the composite material system(s) of interest. Using the properties listed in Table 3 of Chap. 3 and based on the SI system of units, the following numerical values would be input for graphite–epoxy: E11 ¼ 170  109 Pa 9

G12 ¼ 13  10 Pa

E22 ¼ 10  109 Pa

v12 ¼ 0:30

Since 1 Pa = 1 N/m2, all lengths must be input in meters. For example, ply thicknesses must be input in meters (not millimeters). A typical value would be tk = 0.000125 m (corresponding to a ply thickness of 0.125 mm). Similarly, if an I-beam that involves a 50-mm-wide flange laminate was under consideration, then the width of the flange must be input as 0.050 m. If the English system of units was used, then the following numerical values would be input for the same graphite–epoxy material system: E11 ¼ 25:0  106 psi 6

G12 ¼ 1:9  10 psi

E22 ¼ 1:5  106 psi

v12 ¼ 0:30

All lengths would be input in inches. A typical ply thickness would be tk = 0.005 in., and for an I-beam, a typical flange width would be 2.0 in.

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Having determined the effective axial and flexural rigidities of a composite beam of interest, then beam deflections caused by axial loading or transverse loading can be calculated using the methods discussed in Secs. 7 and 8, respectively. HOMEWORK PROBLEMS Notes: Computer programs CLT and BEAM are used in the following problems. These programs can be downloaded from the following website: http://depts.washington.edu/amtas/computer.html. 1. A [(02/F30/F45/F60)3/902]s composite beam with rectangular cross section is shown in Fig. 30. Assume this laminate is produced using a graphite–epoxy pre-preg with a thickness of 0.125 mm and properties listed in Table 3 of Chap. 3: (a)

Use the program CLT to determine numerical values of the [abd] matrix for this stacking sequence. (b) Use hand calculations to determine: (i) the effective axial rigidity of the beam, (ii) the effective flexural rigidity of the beam when ply interfaces are orthogonal to the plane of loading [Fig. 30(a)], and (iii) the effective flexural rigidity of the beam when ply interfaces are parallel to the plane of loading [Fig. 30(b)]. (c) Use the program BEAM to determine the effective axial and flexural rigidities of the beam and compare these results with your hand calculations obtained in step (b). 2. Repeat Problem 1, except assume that the beam is produced using a glass/epoxy pre-preg with a thickness of 0.150 mm. 3. Repeat Problem 1, except assume that the beam is produced using a Kevlar/epoxy pre-preg with a thickness of 0.125 mm. 4. Repeat Problem 1, except assume that the 0j, +30j, and 30j plies are produced using a glass/epoxy pre-preg with a thickness of 0.150 mm, whereas the remaining plies are produced using a graphite–epoxy prepreg with a thickness of 0.125 mm. 5. A composite I-beam is shown in Fig. 31. Assume the beam is produced using a graphite–epoxy pre-preg with a thickness of 0.125 mm and properties listed in Table 3 of Chap. 3. Also, the stacking sequences of the top flange laminate, web laminate, and bottom flange laminate are [(02/902/02)2/452/0/–452]s, [02/902/02]2s, and [(02/902/02)2/452/0/–452]s, respectively (note that Fig. 31 implies that the web plies extend into the top and bottom flange laminates).

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Figure 30 Composite beam with rectangular cross section described in Problem 1 (not all ply interfaces shown).

(a) Determine the centroid location, effective axial rigidity, and effective flexural rigidity of the beam using hand calculation and appropriate expressions from Tables 1–3. Determine the effective elastic moduli involved using program CLT. (b) Use the program BEAM to calculate the centroid location, effective axial rigidity, and effective flexural rigidity of the beam, and compare these results with your hand calculations obtained in step (a).

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Figure 31 Composite I-beam described in Problem 5 (not all ply interfaces shown).

6. Repeat Problem 5, except assume that the beam is produced using a glass/epoxy pre-preg with a thickness of 0.150 mm. 7. Repeat Problem 5, except assume that the beam is produced using a Kevlar/epoxy pre-preg with a thickness of 0.125 mm. 8. Repeat Problem 5, except assume that the stacking sequences of the top and bottom flange laminates are [(02/902/02)2/452/0/–452]s and [02/902/ 02]2s, respectively. 9. Refer to Example Problem 4, part (b). Suppose the maximum deflection of the cantilevered beam must be reduced by a factor of 2. That is, the maximum deflection must be reduced to 1.085 mm (or less). Deflections will be reduced by adding 0j plies to the top and bottom flange laminates of the box-beam. The stacking sequence of the flange laminates will then be of the type [(0/90)2s/F45/0n]s. Determine the value of n necessary to reduce deflections to the desired level. 10. Refer to Example Problem 5, what maximum deflection occurs if a negative bending moment is applied, i.e., if Mob = 40 N m? REFERENCES 1. Gere, J.M.; Timoshenko, S.P. Mechanics of Materials, 4th Ed.; PWS Publishing Co.: Boston, MA, ISBN 0-534-93429-3, 1997. 2. Craig, R.R. Mechanics of Materials. John Wiley and Sons: New York, NY, ISBN 0-471-50284-7, 1996. 3. Hibbeler, R.C. Mechanics of Materials, 4th Ed.; Prentice Hall: Upper Saddle River, New Jersey, ISBN 0-13-016467-4, 2000.

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4. Bedford, A.; Liechti, K.M. Mechanics of Materials. Prentice Hall: Upper Saddle River, New Jersey, ISBN 0-201-89552-8, 2000. 5. Benham, P.P.; Crawford, R.J.; Armstrong, C.G. Mechanics of Engineering Materials, 2nd Ed.; Longman Group Limited: Essex, ISBN 0-582-25164-8, 1996. 6. Timoshenko, S.; Goodier, J.N. Theory of Elasticity, 3rd Ed.; McGraw-Hill Book Company: New York, NY, Section 124, ISBN 07-064720-8, 1970. 7. Roark, R.J.; Young, W.C. Formulas for Stress and Strain, 6th Ed.; McGraw-Hill Book Company: New York, NY, ISBN 0-07-072541-1, 1989. 8. Swanson, S.R. Introduction to Design and Analysis with Advanced Composite Materials. Prentice-Hall Inc.: Upper Saddle River, New Jersey, ISBN 0-02418554-X,1997.

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9 The Governing Equations of Thin-Plate Theory

In earlier chapters we required that stress and moment resultants applied to the edge of a thin plate were constant and uniform. In this and following chapters we will relax this requirement. That is, we consider problems in which loads vary along the edge(s) of a plate. To predict stress, strain, or deflections induced in a thin plate by varying loads we must first develop equations that govern the behavior of a thin composite plate. These are called the ‘‘governing equations,’’ and are derived in this chapter. It turns out that the governing equations for composite laminates with arbitrary stacking sequences become quite lengthy. Therefore we limit our discussion to symmetric composite laminates, which results in a substantial simplification of the governing equations. This limitation is reasonable, as most composites used in practice are symmetric. 1 PRELIMINARY DISCUSSION Thin-plate theory as applied to isotropic plates was developed throughout the 19th and 20th centuries and is now very well established. Many textbooks devoted to isotropic plates have been published; two typical examples are Timoshenko and Woinowsky-Krieger (1) and Ugural (2). Although thin plate theory is also applicable to anisotropic plates, relatively few texts devoted to 487

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this topic have appeared. Two of the best known references describing the application of thin plate theory to composite laminates (or other anisotropic plates) are those by Whitney (3) and Turvey and Marshall (4). A complete expose´ of thin-plate theory as applied to composites is beyond the scope of this textbook. Rather, the objective of this and following chapters is to introduce the fundamental equations that govern the behavior of thin composite plates and to present solutions to a few selected problems. It is hoped that this discussion will prepare the reader for more advanced studies in this area, including those presented in Whitney (3) and Turvey and Marshall (4). Thin plate problems may involve many different types of boundary conditions. In this text, discussion is limited to rectangular composite laminates with symmetric stacking sequences, subjected to simply supported boundary conditions along all four edges. The range of problems considered herein is therefore not as extensive as is presented elsewhere. The reader interested in a more detailed discussion of symmetric or nonsymmetric rectangular panels subject to alternate boundary conditions and/or a discussion of other panel shapes (e.g., elliptical composite panels) is referred to Whitney (3) and Turvey and Marshall (4). The rudiments of thin plate theory have already been applied in Chapter 6. Specifically, during the development of Classical Lamination Theory (CLT) we made several assumptions, two of which were: 

All plies within a thin laminate are subjected to a state of plane stress (rzz = sxz = syz = 0), and  The Kirchhoff hypothesis is valid: a straight line that is initially perpendicular to the midplane of a thin plate is assumed to remain straight and perpendicular to the midplane after deformation.

These assumptions are central to classical thin plate theory, and we will continue to make these two assumptions throughout the analyses presented in this and following chapters. However, in Chapter 6 we also assumed that the external loads applied to the laminate (i.e., stress and moment resultants Nxx, Nyy, Nxy, Mxx, Myy, Mxy) were constant and uniformly distributed along the edge of the plate. Because the stress and moment resultants applied to the edge of the laminate were assumed to be uniform and constant, the resultants induced at all interior regions of the laminate were also constant and identically equal to the edge loads. Therefore there was no need to distinguish between the resultants applied to the edge of a laminate and the resultants induced at any interior point. In reality, uniform edge loads rarely occur in practice. That is, the loads applied to a composite laminate in a real structure typically vary over the length and/or width of the laminate. In this chapter we will consider

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conditions in which nonuniform resultants are applied to the edge of the plate. Because the edge loads will now be allowed to vary, the stress and moment resultants induced at internal regions of the laminate will also vary and do not necessarily equal the stress and moment resultants applied along the edge of the laminate. We must therefore be careful to distinguish between the stress and moment resultants applied to the edge of the laminate and the resulting stress and moment resultants induced at internal regions of the laminate. We will also include an additional type of loading in our analysis. Specifically, we will include the possibility that a distributed load acts perpendicular to the surface of the laminate. This transverse load can be visualized as a transverse pressure, and will be denoted q(x,y). A formal mathematical definition of the stress and moment resultants applied to the edge of the laminate will be given in Sec. 3. However, we will introduce some of the nomenclature used to describe external edge loads here. Examples of externally applied loads that vary over the edge of a laminate are shown schematically in Figs. 1 and 2. As in earlier chapters, we assume the laminate is rectangular with plate edges parallel to the x- and y-axes. A sketch showing stress resultants Nxx acting on opposite edges of a thin laminate is presented in Fig. 1a. Because Nxx is a load applied to the two plate edges that are parallel to the y-axis, Nxx cannot vary as a function of x. Therefore along the plate edge the externally applied resultant Nxx is either a constant or, at most, a function of y only: Nxx ¼ Nxx ðyÞ: There is no reason to expect that an identical distribution of loading is present on opposite sides of laminate. Hence we must distinguish between the resultant Nxx( y) applied to the ‘‘negative’’ laminate edge (i.e., the edge whose outward normal ‘‘points’’ in the negative x-direction) and the resultant Nxx( y) applied to the ‘‘positive’’ laminate edge. Therefore the load applied ( x) ( y), whereas the load applied to to the negative x-edge will be labeled Nxx (+x) the positive x-edge will be labeled Nxx ( y). Although not shown in Fig. 1a, stress resultant Nxy and moment resultants Mxx and Mxy may also be applied to the two laminate edges parallel to the y-axis. These additional edge loads will be labeled: Resultants acting on the negative x-edge: ( x) ( y) Nxy ( x) ( y) Mxx ( x) ( y) Mxy

Resultants acting on the positive x-edge: (+x) Nxy ( y) (+x) Mxx ( y) (+x) Mxy ( y)

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Figure 1 Illustration of varying stress resultants acting on opposite edges of a (+) thin plate. (a) Variation of stress resultants Nxx (y). (b) Variation of stress (+) resultants Nyy (x).

It is emphasized that the resultants applied to the edges parallel to the y-axis are either constants or are functions of y only. This will become an important point in following discussions. In a similar manner, a stress resultant Nyy may be applied to the two plate edges parallel to the x-axis, as shown in Fig. 1b. In this case the plate edge is parallel to the x-axis, and hence the stress resultants applied along these edges are functions of x only. Stress and moment resultants Nyx, Myy,

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Figure 2 Illustration of varying transverse load q(x,y) acting over the surface of a thin plate and perpendicular to the x–y plane and the resulting out-of-plane (Fy) shear stress resultants V (Fx) xz ( y) and V yz (x).

and Myx also act on these plate edges. As before, we must distinguish between loads applied to the negative and positive y-edge:

Resultants acting on the negative y-edge: N (yyy)(x) N (yx y)(x) M (yy y)(x) ( y) M yx (x)

Resultants acting on the positive y-edge: N (+y) yy (x) N (+y) yx (x) M (+y) yy (x) (+y) M yx (x)

Because in our earlier analysis we assumed that stress resultants were constant and uniformly distributed along the edge of the plate, there was no need to distinguish between shear stress resultants acting on adjacent faces. That is, an assumption throughout Chapters 6 and 7 was that Nxy = Nyx. We can no longer make this assumption. For example, the variation of the shear stress resultant over the finite length of the positive x-face of the laminate, (+x) ( y), is independent of the variation of the shear stress resultant applied Nxy (+y) over the finite length of the positive y-face, Nyx (x). Similarly, the variation (+x) ( y), is now of the twisting moment applied to the positive x-face, Mxy (+y) (x). independent of the twisting moment applied to the positive y-face, Myx

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Exceptions to this statement occur at the four corners of the laminate. For example, at the corner (x = a, y = b): NðxyþxÞ ðbÞ ¼ NðyxþyÞ ðaÞ

and

MðxyþxÞ ðbÞ ¼ MðyxþyÞ ðaÞ

A distributed transverse force acting over the surface of the laminate, q(x,y), is shown in Fig. 2. The transverse force q(x,y) has units of force per area, such as N/m2 = Pascals or lbf/in2 = psi. Because q(x,y) is a distributed force acting in the z-direction, shear stress resultants Vxz( y) and/or Vyz(x) must be present along one or more edges of the laminate, so as to maintain static (Fx) (Fy) ( y) and Vyz (x) are given by: equilibrium (i.e., to maintain AFz = 0). Vxz Z t=2 Z t=2 FxÞ ¼ sðxzFxÞ dz V ðyzFyÞ ¼ sðyzFyÞ dz ð1Þ V ðxz t=2

t=2

(Fx) ( y), The reader should immediately object to the inclusion of q(x,y), Vxz ðFyÞ and V yz ðxÞ in our analysis. After all, they violate our assumption of plane stress. That is, the presence of q(x,y) implies that stress rzz will be induced in the plate, and furthermore Eq. (1) implies that shear stresses sxz and syz will also be induced. These objections are entirely valid. If a transverse load q(x,y) is applied then a 3-D state of stress is, in fact, induced in the plate. However, if the plate is thin, then the magnitudes of the out-of-plane stresses are far lower than the magnitudes of the in-plane stresses: rzz, sxz, syz > > > > > > > 6 a12 a22 0 > Nyy þ NT þ NM > > > eoyy > 0 0 0 7 > > > > yy yy > > > > 7 6 < o = 6 = < 7 0 0 a 0 0 0 66 cxy ¼ 6 Nxy 7 7 6 0 0 d11 d12 0 7> > > jxx > 0 > > > > 6 0 > > > > > > > 5> 4 0 0 0 d d 0 > > > > 12 22 j 0 > > > > yy ; : : ; 0 0 0 0 0 d66 jxy 0

As in earlier chapters, we assume infinitesimal strain levels. Therefore midplane strains are related to midplane displacements according to Eq. (10) of Chap. 6: eoxx ¼

@uo @x

eoyy ¼

@vo @y

coxy ¼

@uo @vo þ @y @x

Consequently, midplane displacement fields are given by: @uo T M ¼ a11 ðNxx þ NTxx þ NM xx Þ þ a12 ðNyy þ Nyy þ Nyy Þ @x

ð2aÞ

@vo T M ¼ a12 ðNxx þ NTxx þ NM xx Þ þ a22 ðNyy þ Nyy þ Nyy Þ @y

ð2bÞ

@uo @vo þ ¼ a66 ðNxy Þ @y @x

ð2cÞ

Integrating Eq. (2a) with respect to x, we find: h i T M uo ðx; yÞ ¼ a11 ðNxx þNTxx þNM Þþ a ðN þ N þ N Þ 12 yy xx yy yy x þ f1 ðyÞ þ k1

Where f1( y) is an unknown function of y (only) and k1 is an unknown constant of integration. Similarly, integrating Eq. (2b) with respect to y we find: h i T M vo ðx; yÞ ¼ a12 ðNxx þNTxx þNM Þ þ a ðN þ N þ N Þ 22 yy xx yy yy yþf2 ðxÞ þ k2

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where f2(x) is an unknown function of x (only) and k2 is a second unknown constant. Without a loss in generality, we assume that midplane displacements are zero at the origin (i.e., let uo = vo = 0 at x = y = 0), and consequently we conclude k1 = k2 = 0. Substituting the above expressions for uo(x,y) and vo(x,y) into Eq. (2c), we find: @uo @vo @f1 @f2 þ ¼ þ ¼ a66 Nxy @y @x @y @x Because the terms that appear on the right side of the equality (i.e., a66 and Nxy) are known constants, it follows that f1 and f2 must be at most linear functions of y and x, respectively: f1 ðyÞ ¼ k3 y

f2 ðxÞ ¼ k4 x Hence we can write: k3 þ k4 ¼ a66 Nxy Constants k3 and k4 can take on any value as long as they sum to the product (a66Nxy). To determine particular values convenient for our use, we now require that the infinitesimal rotation vector in the x–y plane, xxy (which represents rigid body motion of the plate), is zero. The infinitesimal rotation vector is given by [see Ref. 3]:   1 @uo @vo xxy ¼ @x 2 @y Requiring that xxy=0 leads to: k3 ¼ k4 ¼

1 a66 Nxy 2

Combining the preceding results, we conclude that in-plane midplane displacement fields induced in a symmetric specially orthotropic composite panel by the combination of uniform in-plane stress resultants, a uniform change in te mperature, and/or a uniform change in moisture contents are given by: h i 1 T M Þ þ a ðN þ N þ N Þ uo ðx; yÞ ¼ a11 ðNxx þ NTxx þ NM 12 yy xx yy yy x þ 2  ða66 Nxy Þy

vo ðx; yÞ ¼

 1 ða66 Nxy Þx þ a12 ðNxx þ NTxx þ NM xx Þ 2 þ a22 ðNyy þ NTyy þ NM yy ފy

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These expressions can be simplified through the use of the effective thermal expansion coefficients of the laminate, defined by Eq. (73a) of Chap. 6, and the effective moisture expansion coefficients, defined by Eq. (76a) of Chap. 6. For a specially orthotropic laminate, these become: i i 1 h 1 h a11 NTxx þ a12 NTyy a12 NTxx þ a22 NTyy axx ¼ ayy ¼ axy ¼ 0 DT DT i i 1 h 1 h M M a11 NM a12 NM bxx ¼ byy ¼ bxy ¼ 0 xx þ a12 Nyy xx þ a22 Nyy DM DM Hence we see that the midplane displacement fields can be written as:   1 uo ðx; yÞ ¼ a11 Nxx þ a12 Nyy þ DTaxx þ DMbxx x þ ða66 Nxy Þy 2   1 vo ðx; yÞ ¼ ða66 Nxy Þx þ a12 Nxx þ a22 Nyy þ DTayy þ DMbyy y 2

ð3aÞ ð3bÞ

Note that the displacement fields are independent of the transverse load q(x,y). As pointed out in Sec. 2 of Chap. 9, uo(x,y) and vo(x,y) are predicted to be independent of q(x,y) because we have assumed displacement gradients are small, such that gradients squared can be ignored [e.g., (@w/@x)2 c 0]. Analyses that account for large displacement gradients are not considered in this text. If we had included large gradients in our analysis, then expressions for uo(x,y) and vo(x,y) corresponding to Eqs. (3a) and (3b) would depend on transverse load q(x,y). Direct substitution of Eqs. (3a) and (3b) will reveal that these equations satisfy the equations of equilibrium, Eqs. (1a) and (1b). In the following sections, we will use these expressions to specify in-plane displacement fields associated with simple supports of type S1 through S4.

3 SPECIALLY ORTHOTROPIC LAMINATES SUBJECT TO SIMPLE SUPPORTS OF TYPE S1 In this section, we consider the specially orthotropic plate shown in Fig. 1. The plate is assumed to be rectangular with thickness t and in-plane dimensions a  b. All four edges of the plate are subject to simple supports of type S1, where it is assumed that the laminate was mounted in the structure that imposes type S1 supports following cooldown from the cure temperature to room temperature. After assembly, the plate is subjected to a uniform change in temperature and a transverse loading that varies over the x–y plane according to:  px   py  qðx; yÞ ¼ qo sin sin ð4Þ a b

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Figure 1 Thin rectangular plate of thickness t and in-plane dimensions a  b, subjected to a transverse load q(x,y) = qo {sin[(px)/a]}{sin[(py)/b]}. All four edges of the plate are subject to simple supports of type S1.

We will not consider a change in moisture content (i.e., let DM = 0), although from earlier discussion it should be clear that a change in moisture content can be accounted for (in a mathematical sense) using the same techniques used to model uniform changes in temperature. Let us define DT c as the change in temperature from cure to room temperature: DT c ¼ TRT

TC

where TRT = room temperature and TC = the cure temperature. For a symmetric specially orthotropic laminate, the in-plane displacements caused by DT c can be calculated using Eqs. (3a) and (3b) with Nxx = Nyy = Nxy = 0: uco ðx; yÞ ¼ ðaxx DT c Þx   vco ðx; yÞ ¼ ayy DT c y

ð5aÞ ð5bÞ

These displacements are induced before assembly of the simply supported plate. Note that these in-plane displacement fields satisfy the first two equations of equilibrium, Eqs. (1a) and (1b). A type S1 simple support will simply maintain these displacements during subsequent loading and/or

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temperature changes. Therefore type S1 boundary conditions for all four edges of the plate become: For x ¼ 0

For x ¼ a :

* ð0; yÞ ¼ D Mxx 11

@2w @2w þ D12 2 ¼ 0 2 @x @y ð xÞ uo ð0; yÞ ¼ uo ðyÞ ¼ 0   ð xÞ vo ð0; yÞ ¼ vo ðyÞ ¼ ayy DT c y

@2w @2w þ D12 2 ¼ 0 2 @x @y ðþxÞ uo ða; yÞ ¼ uo ðyÞ ¼ ðaxx DT c Þa   ðþxÞ vo ða; yÞ ¼ vo ðyÞ ¼ ayy DT c y

For y ¼ 0

For y ¼ b :

wð0; yÞ ¼ 0

wðx; 0Þ ¼ 0

wða; yÞ ¼ 0

2

wðx; bÞ ¼ 0

2 2 @ w @ w * ðx; bÞ ¼ D @ w þ D @ w ¼ 0 þ D ¼ 0 M 11 12 12 yy @x2 @y2 @x2 @y2 ðþxÞ ð yÞ c uo ðx; 0Þ ¼ uo ðxÞ ¼ ðaxx DT Þx uo ðx; bÞ ¼ uo ðxÞ ¼ ðaxx DT c Þx   ðþxÞ ð yÞ vo ðx; bÞ ¼ vo ðyÞ ¼ ayy DT c b vo ðx; 0Þ ¼ vo ðxÞ ¼ 0

* ðx; 0Þ ¼ D Myy 11

2

* ða; yÞ ¼ D Mxx 11

We wish to determine the out-of-plane displacement field w(x,y) that satisfies these boundary conditions as well as the equations of equilibrium, when the plate is subjected to a transverse loading q(x,y) and/or a further change in temperature. Guided by the functional form of the transverse pressure [i.e., Eq. (4)], we assume that the out-of-plane displacement field is given by: wðx; yÞ ¼ c sin

 px 

 py  sin a b

ð6Þ

where c is an unknown constant. Substituting this assumed form into the boundary conditions will reveal that they are identically satisfied for any value of c. Hence the value of constant c must be determined by enforcing the third equation of equilibrium, Eq. (1c). We perform the following operations: (a) substitute Eqs. (4), (5a), (5b), and (6) into the third equation of equilibrium, Eq. (1c); T T and Nyy in terms of effec(b) write the thermal stress resultants Nxx tive thermal expansion coefficients [using Eq. (73b) of Chap. 6]; and then (c) solve the resulting expression for constant c.

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Following this process we find: c¼

qo  

c  1   D 2 D ð DT DT Þ 1  11 22 A11 axx þ A12 ayy þ 2 A12 axx þ A22 ayy p4 4 þ 2 2 ðD12 þ 2D66 Þ þ 4 þ 2 2 a b p a b a b

ð7aÞ

Notice that the temperature change as defined in earlier chapters (DT ) appears in Eq. (7a). That is: DT ¼ ðcurrent temperatureÞ

ðcure temperatureÞ

Also note that if the temperature is not changed following assembly in the type S1 simple supports, then DT = DT c, and the effects of temperature cancel in Eq. (7a). Results from thin-plate theory are often expressed in terms of the socalled plate aspect ratio, R = a/b. Equation (7a) can be rewritten using the aspect ratio as follows: c¼

qo R4 b4

   ðDT DT Þa2  2 2 a A þ R A þ a A þ R A p4 D11 þ 2R2 ðD12 þ 2D66 Þ þ R4 D22 þ xx 11 12 yy 12 22 p2 c



ð7bÞ

The predicted out-of-plane deflection is obtained by combining either Eq. (7a) or (7b) with Eq. (6). Using Eq. (7b) for example, we have: wðx; yÞ ¼

qo R4 b4 sinðpx=aÞsinðpy=bÞ

   ðDT c DT Þa2  2 2 p4 D11 þ 2R2 ðD12 þ 2D66 Þ þ R4 D22 þ a A þ R A a A þ R A þ xx 11 12 yy 12 22 p2

ð8Þ

Equations (5a), (5b), and (8) give the predicted displacement field induced in the plate and represent the solution to this problem. To summarize, we have considered a symmetric specially orthotropic plate subjected to type S1 simple-supports. We have assumed that the plate is mounted within simple supports while at room temperature. The laminate has therefore likely experienced a change in temperature prior to assembly because modern composites are typically cured at an elevated temperature. The change in temperature associated with cooldown to room temperature is represented by DT c. After assembly, the plate is subjected to a sinusoidally varying transverse loading and/or the temperature is changed away from room temperature. The resulting displacement fields are given by Eqs. (5a), (5b), and (8). A typical application of this solution is discussed in Sample Problem 1.

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As a closing comment, it should be kept in mind that we have not yet considered the possibility of buckling. The solution presented here is not valid if the change in temperature is such that the resulting in-plane stress resultants are compressive and have a magnitude large enough to cause buckling. The phenomenon of buckling induced by a change in temperature is called ‘‘thermal buckling’’ and will be discussed in Sec. 8. Sample Problem 1 A [02/90)2]s graphite–epoxy laminate is cured at 175jC and then cooled to room temperature (20jC). After cooling, the flat laminate is trimmed to inplane dimensions of 300  150 mm and mounted in an assembly that provides type S1 simple supports along all four edges. The x axis is defined parallel to the 300 mm edge (i.e., a = 0.3 m; b = 0.15 m). The laminate is then subjected to a transverse pressure given by q(x,y) = 40 {sin[(px)/a]}{sin[(py)/b]} (kPa) and a uniform temperature change. No change in moisture content occurs (DM = 0). Plot the maximum out-of-plane displacement as a function of temperature, over the range 50jC < T < 20jC. Use the properties listed for graphite–epoxy in Table 3 of Chap. 3, and assume each ply has a thickness of 0.125 mm. Solution. The rectangular plate is a 12-ply laminate with total thickness t = 12(0.125 mm) = 1.5 mm and aspect ratio R = a/b = (0.3 m)/(0.15 m) = 2.0. Out-of-plane displacements are given by Eq. (8), and hence elements of the [ABD] matrix are required. Based on the properties listed in Table 3 of Chap. 3 for graphite–epoxy and the specified stacking sequence, the [ABD] matrix is: 2

1:76  106

6 6 4:52  106 6 6 0 ½ABDŠ ¼ 6 6 0 6 6 4 0 0

4:52  106

95:6  106 0 0 0 0

3

0

0

0

0

0

0

0

19:5  106 0 0 0

0 40:1 0:848 0

7 0 7 7 0 7 7 7 0 7 7 0 5

0 0:848 10:8 0 3:66

where the units of Aij are Pa m and the units of Dij are Pa m3. We also require the effective thermal expansion coefficients. Based on the properties listed in Table 3 of Chap. 3 for graphite–epoxy and the specified stacking sequence, these are: axx ¼ 0:29 lm=m

jC

ayy ¼ 2:44 lm=m

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jC

Because the laminate is specially orthotropic, the effective shear thermal expansion coefficient is zero (axy = 0). The cooldown from the cure temperature to room temperature is: DT c ¼ ð20jCÞ

ð175jCÞ ¼

155jC

Following assembly, the temperature ranges from room temperature to as low as 50jC. Therefore: 255jC < DT
a a : ; if ðm þ iÞ is odd p½m2 i2 Š 8 0; if ðn þ jÞ is even >     < npx jpx sin cos dy ¼ 2nb > b b : ; if ðn þ jÞ is odd p½n2 j2 Š 

On the basis of these identities, we find after integration and evaluation:

Z Z 1 b 2 0

a

4D16 0



B2 w Bx2



 B2 w dxdy BxBy

 M N X M X N X X 2p2 ¼ 2 D16 cmn cij a ½m2 m ¼ 1n ¼ 1i ¼ 1j ¼ 1

m2 nij i 2 Š½n2

where: ðMIÞ ¼ ½ð 1Þm ð 1Þi

1Š

ðNJ Þ ¼ ½ð 1Þ ð 1Þ

1Š

n

j

Notice that: ðMI Þ ¼

ðNJ Þ ¼

8 < 0; if ðm þ iÞ is even

2; if ðm þ iÞ is odd 8 < 0; if ðn þ jÞ is even :

:

2; if ðn þ jÞ is odd

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j 2Š



ðMIÞðNJ Þ

The final term that appears in Eq. (17) is evaluated in a similar manner and becomes:  2  2  Z Z 1 b a B w B w 4D26 dxdy 2 0 0 By2 BxBy  M N X M X N X X 2p2 mn3 ij ¼ 2 D26 cmn cij ðMIÞðNJ Þ ½m2 i 2 Š½n2 j 2 Š b m ¼ 1n ¼ 1i ¼ 1j ¼ 1 We have now integrated all terms that appear in Eq. (17). Combining these results and rearranging, the integrated form of Eq. (17) can be written:

M N  4 X X p 2 bm4 2m2 n2 an4 cmn D þ ðD þ 2D Þ þ D UII ¼ 11 12 66 22 8 a3 ab b3 m ¼ 1n ¼ 1  M X N  X ij 2 2p mncmn cij ð18Þ ð m 2 i 2 Þ ð n2 j 2 Þ i ¼ 1j ¼ 1 

m2 n2  2 D16 þ 2 D26 ðMIÞðNJÞ a b 

3.3 Evaluation of Strain Energy Component UIII Strain energy component UIII represents the change in strain energy associated with stress resultants Nij caused by bending. We assume that the stress resultants Nij that exist prior to bending remain constant; only in-plane strains are changed during bending. We must therefore evaluate the change in inplane strains caused by the development of out-of-plane displacements w(x,y). The change in midplane strain e0xx may be determined via Fig. 1. The figure shows an element of length dx, which represents an infinitesimal element of the midplane that has already been deformed by stress resultants Nij during step 1, as previously discussed. The length dx is further increased to length dxV if the plate is deflected out of plane. From Fig. 1, we see that: " #1=2 "  2  2 #1=2 Bw Bw 2 2 ¼ dx 1 þ dx dxV ¼ dx þ Bx Bx The quantity within the square bracket and raised to the 1/2 power can be expanded in terms of a binomial power series expansion. A general statement of this series expansion is given by (5):        1 1 1 2 1 1 3 3 ð1 þ nÞ1=2 ¼ 1 þ n n þ n þ ... 2 2 4 2 4 6

Copyright © 2004 by Marcel Dekker, Inc. All Rights Reserved.

Figure 1 Sketch used to determine change in in-plane strain caused by out-ofplane deflections.

We adopt this general formula for our use by retaining only the first two terms and letting n = (Bw/Bx)2. Hence: "  2 #1=2  2 Bw 1 Bw 1þ c1 þ Bx 2 Bx With this approximation, the new length of the element is given by: "   # 1 Bw 2 dx V ¼ dx 1 þ 2 Bx The change in-plane strain e0xx caused by out-of-plane displacement w(x,y) 0;b is labeled e0;b xx . Based on the above, exx is given by:  2 dx V dx 1 Bw 0;b ¼ exx ¼ ð19aÞ dx 2 Bx Using a similar approach, the change in in-plane strains e0yy and c0xy caused by out-of-plane displacement w(x,y) are given by: 0;b eyy ¼

 2 1 Bw 2 By

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ð19bÞ

0;b cxy

¼



Bw Bx



Bw By



ð19cÞ

We can now calculate the strain energy associated with the change in in-plane strains caused by bending. We assume in-plane stress resultants Nij remain constant as bending develops. Consider, for example, the force represented by stress resultant Nxx acting over an infinitesimal element with length and width dx and dy. The x-directed force is (Nxxdy), and the distance through 0;b which this force moves as bending develops equals exx dx. The incremental strain energy associated with this force and caused by bending is therefore 0;b dUIII=(Nxxdy)(exx dx). Analogous expressions hold for stress resultants Nyy and Nxy. Hence, strain energy UIII is given by: ZZ h  0;b 0;b UIII ¼ Nxx exx þ Nyy e0;b ð20Þ yy þ Nxy cxy dxdy Equation (20) represents a general expression for UIII. Recall that during application of the Ritz method, we are able to specify geometric boundary conditions directly, but are not able to specify static boundary conditions, at least directly. Therefore, in present form, Eq. (20) is inconvenient for use with the Ritz method. That is, we wish to express stress resultants Nxx, Nyy, and Nxy in terms of displacement fields, which will ultimately allow us to specify geometric boundary conditions that represent known values of Nxx, Nyy, and Nxy. From Eq. (44) of Chap. 6, we can write (for Bij = DM = 0) Nxx ¼ A11 e0xx þ A12 e0yy þ A16 c0xy NTxx       Buo Bvo Buo Bvo þ ¼ A11 þ A12 þ A16 Bx By By Bx

NTxx

Similarly, Nyy ¼ A12



Buo Bx



Nxy ¼ A16



Buo Bx



þ A22



Bvo By



þ A26



Bvo By



þ A26



Buo Bvo þ By Bx



NTyy

þ A66



Buo Bvo þ By Bx



NTxy

Substituting these expressions as well as Eqs. (19a,b,c) into Eq. (20), we obtain:      Z Z   1 Buo Bvo Buo Bvo UIII ¼ þ A11 þ A12 þ A16 NTxx 2 Bx By By Bx

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        Bw 2 Buo Bvo Buo Bvo T þ  þ A12 þ A22 þ A26 Nyy Bx Bx By By Bx       2   Bw Buo Bvo Buo Bvo þ  þ A26 þ A66 NTxy þ2 A16 By Bx By By Bx   

Bw Bw  dxdy ð21Þ Bx By 

We will now integrate Eq. (21) for the class of problems considered in this text. We assume the plate is simply supported and that out-of-plane displacements are given by Eq. (6). We also assume the laminate is symmetric and subjected to uniform stress resultants Nij and/or a temperature change DT. For these conditions, the in-plane displacement fields are given by Eqs. (7). Substituting Eqs. (6) and (7) into Eq. (21) [and utilizing the simplifying change in notation introduced as Eqs. (15)], we have: UIII

Z Z 1 b ¼ 2 0 

a 0

"

M X

A11 C1 þ A12 C2 þ 2A16 C3

N X

cmn

m ¼ 1n ¼ 1





m ¼ 1n ¼ 1

cmn





    !2 mp mpx npy cos sin a a b

n þ A12 C1 þ A22 C2 þ 2A26 C3 M N X X

NTxx

NTyy

o

    !2 np mpx npy sin cos b a b

n o þ 2 A16 C1 þ A26 C2 þ 2A66 C3 NTxy (     )  M N X X mp mpx npy cmn cos sin  a a b m ¼ 1n ¼ 1 

     !# jp ipx jpy dxdy sin cos cij b a b ¼ 1

M X N X

i ¼ 1j

Integration of this expression is simplified through the use of the trigonometric identities listed in preceding section. We obtain: UIII ¼

M N  2 X X p

m ¼ 1n ¼ 1

8

c2mn



m2 b  A11 C1 þ A12 C2 þ 2A16 C3 a

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NTxx





n2 a  T A12 C1 þ A22 C2 þ 2A26 C3 Nyy þ b h i þ 2m2 ncmn A16 C1 þ A26 C2 þ 2A66 C3 NTxy 

M X N  X j cij  ðMIÞðNJ Þ ðm2 i 2 Þðn2 j 2 Þ i ¼ 1j ¼ 1 This result can be further simplified by substituting Eq. (73b) of Chap. 6, which give the thermal stress resultants in terms of elements of the [A] matrix, and Eq. (5). We finally find:

M N  2 X X p 2 m2 b n2 a cmn Nxx þ Nyy þ 2m2 ncmn Nxy UIII ¼ a b 8 m ¼ 1n ¼ 1 ð22Þ 

M X N  X j cij  ðMI ÞðNJ Þ ðm2 i 2 Þðn2 j 2 Þ i ¼ 1j ¼ 1 As an aside, it is interesting to note that the change in in-plane strains given by Eqs. (19a,b,c) are similar to the nonlinear terms that appear in Green’s strain tensor, mentioned in Sec. 14 of Chap. 2. Recall that Green’s strain tensor represents a definition of ‘‘finite’’ strains, which must be accounted for during analyses involving large displacement gradients. For example, finite strain exx is defined as: "      # Bu 1 Bu 2 Bv 2 Bw 2 þ exx ¼ þ þ Bx 2 Bx Bx Bx 0;b Because we have incorporated the term exx = 1/2(Bw/Bx)2 during our calculation of UIII, it would be natural to conclude that our analysis is valid for finite strain levels. This conclusion would be incorrect. Our analysis is based on infinitesimal strains, despite the inclusion of Eq. (18) during calculation of UIII. To perform an analysis based on energy methods that accounts for finite strain levels, we would need to develop new expressions comparable to Eqs. (16), (18), and (20) (i.e., our current expressions for UI, UII, and UIII, respectively), based on the Green strain tensor. In turn, this would require a new derivation of results from Chap. 6; that is, an analysis based on finite strains would require a rederivation of CLT. Such an analysis is beyond the scope of this textbook and will not be discussed.

3.4 Evaluation of Work Done by Transverse Loads The work done by the transverse load q(x,y) is denoted W and is calculated in accordance with Eq. (8). Only simply supported plates are considered in

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this text, and w(x,y) is assumed to be given by Eq. (6) in all cases. Therefore, the work done by transverse loads is given by: (    ) ZZ n M N o X X mpx npx W¼ qðx; yÞ dxdy ð23Þ sin cmn sin a b m ¼ 1n ¼ 1 Integration of Eq. (23) depends on the functional form of transverse load q(x,y). Problems involving various types of transverse loads will be considered in the following sections. Equation (23) will be integrated as needed during the discussion to follow. 4 SYMMETRIC COMPOSITE LAMINATES SUBJECT TO SIMPLE SUPPORTS OF TYPE S4 In this section, the transverse deflections of simply supported symmetric composite panels will be predicted on the basis of a Ritz analysis. The general approach is to first obtain an expression for the total potential energy P of the plate, given by: P ¼ UI þ UII þ UIII

W

In general, P is a function of the elastic properties of the plate, plate dimensions, and midplane displacement fields uo(x,y), vo(x,y), and w(x,y). For the problems considered herein, in-plane displacement fields uo(x,y) and vo(x,y) are known, while the out-of-plane displacement field w(x,y) is unknown. The out-of-plane displacement field is assumed to be of the form: wðx; yÞ ¼

M N X X

m ¼ 1n

    mpx npy cmn sin sin a b ¼ 1

The magnitude of out-of-plane deflections are obtained by applying the principle of minimum potential energy, which requires:  BP m ¼ 1; 2; . . . ; M ¼0 n ¼ 1; 2; . . . ; N Bcmn This process leads to (M  N) equations that must be satisfied simultaneously. Hence, by solving these equations for constants cmn, the magnitude of out-of-plane displacements that corresponds to the state of minimum potential energy is determined and the problem is solved. Equations for strain energy components UI, UII, and UIII are identical for all problems considered herein and were developed in Sec. 3. The work done by the transverse load W depends on the nature of the applied load. Solutions for a few common transverse loads are presented in the following subsections.

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4.1 Deflections Due to a Uniform Transverse Load Consider a composite plate subjected to a constant and uniform transverse load, q(x,y)=qo. In this case, Eq. (23) becomes:    ) Z bZ a ( X M N X mpx npx cmn sin W ¼ qo sin dxdy a b 0 0 m ¼ 1n ¼ 1 During integration of this expression, we note the following: 8 >   Z a < 0; if m is even mpx sin dx ¼ 2a > a 0 : ; if m is odd mp 8 >   Z b < 0; if n is even npx sin dy ¼ 2b > b 0 : ; if n is odd np

Hence, the work done by a uniform transverse load can be written: M N  ih i X X abqo cmn h m n W¼ 1 ð 1Þ 1 ð 1Þ p2 mn m ¼ 1n ¼ 1

ð24Þ

The total potential energy can now be obtained by combining Eqs. (16), (18), (22), and (24): P¼



þ

þ

  C1 Nxx þ C2 Nyy þ 2C3 Nxy ðabÞ

f

f



N  4 M X X p 2 bm4 2m2 n2 an4 þ 2D Þ þ D þ D cmn ðD 12 66 11 22 8 a3 ab b3 m ¼ 1n ¼ 1 2p2 mncmn

N  M X X cij

i ¼ 1j ¼ 1

M N  2 X X p

m ¼ 1n

(

ij ðm 2

i 2 Þðn 2

m2 b n2 a Nxx þ Nyy 8 a b ¼ 1 M N X X  þ2m2 ncmn Nxy cij ðm2 i ¼ 1j ¼ 1 c2mn



  m2 n2 D16 þ 2 D26 ðMI ÞðNJ Þ 2 a b



N  M X X abqo cmn

m ¼ 1n ¼ 1

j 2Þ

p2 mn

m

ð 1Þ

1



n

ð 1Þ

j i 2 Þðn2 1

j 2Þ

g



ðMI ÞðNJ Þ

)

g

ð25Þ

The four individual energy components UI, UII, UIII and W are shown within the large braces in Eq. (25). This expression is unwieldy so a change in

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notation will be made to facilitate inspection of the mathematical structure of P. Toward that end, define the following constants: p4 b D11 8a3 p2 b Nxx F4 ¼ 8a 2p2 F7 ¼ 2 D26 b F1 ¼

2p4 ðD12 þ 2D66 Þ 8ab p2 a Nyy F5 ¼ 8b

p4 a D22 8b3 2p2 F6 ¼ 2 D16 a abqo F9 ¼ p2 F3 ¼

F2 ¼

F8 ¼ 2Nxy

Note that these terms are all known constants for a given laminate and loading condition. On the basis of these definitions, Eq. (25) can be rearranged as follows:  4 2 2 4 2 2 2 cmn F1 m þ F2 m n þ F3 n þ F4 m þ F5 n

P ¼ UI þ

M N X X

f

m ¼ 1n ¼ 1

þ cmn þ cmn



F9 ð 1Þm mn ( N M X X i ¼ 1j ¼ 1

1



ðm2

ð 1Þn

cij 2 i Þðn2

1



j 2Þ



) 3 3 2  F6 ðm nijÞ þ F7 ðmn ijÞ þ F8 ðm njÞ ðMIÞðNJÞ

g

ð26Þ

To further explore the Ritz method, we must now expand the expression for P, based on some specified values of M and N. In general, the accuracy of the Ritz approach is improved as M and N are increased. Although not required, it is usual practice to let M=N, which means that the number of terms with xand y- dependency in Eq. (6) is identical. Often, 100 terms or more (M=N=10, or more) are necessary to obtain a reasonable convergence of the Ritz solution. Writing down the expanded form of P based on values of M and N as high as 10 is obviously untenable. For purposes of illustration, we will expand P using M=N=2, which will allow us to explore the essential elements of the Ritz analysis. Hence, expanding our Eq. (26) based on M=N=2, we find: P ¼ UI þ c211 ðF1 þ F2 þ F3 þ F4 þ F5 Þ þ c212 ðF1 þ 4F2 þ 16F3 þ F4 þ 4F5 Þ þ c221 ð16F1 þ 4F2 þ F3 þ 4F4 þ F5 Þ þ 4c222 ð4F1 þ 4F2 þ 4F3 þ F4 þ F5 Þ

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ð27Þ

þ

40 c11 c22 ð2F6 þ 2F7 þ F8 Þ 9 40 c12 c21 ð2F6 þ 2F7 þ F8 Þ 9

4c11 F9

Two important features of Eq. (27) should be noted. First, the expression for P is a second-order polynomial in terms of the unknown coefficients cmn. Second, the term involving F9 is a linear function of cmn. Referring to the definitions of F1 through F9 listed above, it is seen that F9 is the only term related to the transverse load qo. Hence, that portion of the total expression for P which is related to the transverse loading is a linear function of the unknown coefficients cmn. These two characteristics always hold for P, regardless of the values of M and N or the nature of the transverse loading. That is, P is always a second-order polynomial in cmn, and terms involving the transverse load are always linear functions of cmn. Proceeding with the Ritz analysis, we now apply the principle of minimum potential energy. That is, we wish to identify the particular values of cmn dictated by: BP ¼0 Bcmn Because M=N=2 in this example, we must take four partial derivatives of P, each of which will represent an independent equation that is then equated to zero. For example, taking the derivative of P with respect to c11 and equating to zero, we have: BP 40 ¼ 2ðF1 þ F2 þ F3 þ F4 þ F5 Þc11 þ ð2F6 þ 2F7 þ F8 Þc22 Bc11 9

4F9 ¼ 0

Three additional equations are also formed (BP/Bc12=BP/Bc21=BP/ Bc22=0). The four equations can be represented using matrix notation as shown in Fig. 2(a). These equations may be easily solved for coefficients cmn by multiplying both sides of the equation by the inverse of the [4  4] array, as shown in Fig. 2(b). Once coefficients cmn have been determined, the out-ofplane deflections can be calculated using Eq. (6) and the problem is solved. Referring to the definitions of F1 through F9 listed above, it is seen that normal stress resultants Nxx and Nyy appear only in terms F4 and F5, respectively, while the shear stress resultant Nxy appears only in term F8. Inspection of the [4  4] array shown on the left side of the equality in Fig. 2(a) reveals that F4 and F5 appear only along the main diagonal of the matrix, whereas F8 appears only in off-diagonal positions. This pattern always occurs, regardless of the value of the value of M and N or the nature of the transverse loading—the normal stress resultants appear only along the main

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Copyright © 2004 by Marcel Dekker, Inc. All Rights Reserved. Figure 2 Summary of the solution obtained for a simply supported laminate subjected to a uniform transverse pressure using a Ritz analysis and M = N = 2. (a) The set of simultaneous equations obtained by enforcing BP/Bcmn= 0, for M = N = 2. (b) Solving the set of simultaneous equations shown in part (a).

diagonal, while the shear stress resultant appears only in off-diagonal positions. In summary, Fig. 2 represents the solution for a simply supported symmetric composite panel subjected to a uniform transverse pressure qo, a uniform in-plane stress resultants Nxx, Nyy, and Nxy, and a uniform change in temperature, DT, based on M = N = 2. Depending on details of a specific problem, the number of terms required to insure convergence may be substantially higher than only four terms; it is not uncommon to require 100 terms or more (M = N = 10 or more). Obviously, the solution process presented above is rarely (if ever) performed by hand calculation. Rather, a computerbased routine is typically used to expand Eq. (26) [or, equivalently, Eq. (25)] for specified values of M and N, to perform the required partial differentiation, to determine the inverse of the resulting [M  N] matrix, and to complete the final matrix multiplication that gives the coefficients cmn. Typical solutions obtained on the basis of the Ritz approach are illustrated in the following three sample problems. Sample Problem 1 A [(F45/0)2]s graphite-epoxy laminate is cured at 175jC and then cooled to room temperature (20jC). After cooling, the flat laminate is trimmed to inplane dimensions of 300150 mm and is mounted in an assembly that provides type S4 simple supports along all four edges. The x axis is defined parallel to the 300-mm edge (i.e., a = 0.3 m, b = 0.15 m). The laminate is then subjected to a uniform transverse load q(x,y) = 30 kPa. No in-plane loads are applied (i.e., Nxx = Nyy = Nxy = 0). Determine the maximum out-of-plane displacement based on a Ritz analysis and plot the out-of-plane displacement field. Use the properties listed for graphite-epoxy in Table 3 of Chap. 3 and assume each ply has a thickness of 0.125 mm. Solution. Based on the properties listed in Table 3 of Chap. 3 for graphiteepoxy, the [ABD] matrix for a [(F45/0)2]s laminate is: 3 2 0 0 0 0 145:2  106 35:3  106 6 35:3  106 64:8  106 0 0 0 0 7 7 6 6 6 0 0 50:2  10 0 0 0 7 7 6 ½ABDŠ ¼ 6 0 0 0 22:3 7:97 2:20 7 7 6 4 0 0 0 7:97 14:3 2:20 5 0 0 0 2:20 2:20 10:8

where the units of Aij are Pa m and the units of Dij are Pa m3. Notice that neither D16 nor D26 equals zero; hence, the laminate is generally orthotropic. The 12-ply laminate has a total thickness t = 1.5 mm and the aspect ratio R = a/b = 2.0. The computer program SYMM (described in Sec. 6) can be used to perform the required Ritz analysis. Several analyses were performed using

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increasing values of M (and N ) to evaluate whether the solution has converged to a reasonably constant value. Solutions were obtained using values of M (and N ) ranging from 1 through 10 (i.e., analyses were performed in which the number of terms used to describe the displacement field ranged from 1 through 100). Maximum predicted displacement is plotted as a function of M and N in Fig. 3. As indicated, the maximum displacement converges to a value of 8.03 mm when M=N=10. A contour plot of out-of-plane displacements predicted using M=N= 10 is shown in Fig. 4. As would be expected, the maximum displacement occurs at the center of the plate (i.e., at x = 150 mm, y = 75 mm). Careful examination of these contours will reveal that the contours are very slightly distorted. This distortion (which is barely discernible in Fig. 4) occurs because the plate is generally orthotropic. That is, for a [(F45/0)2]s laminate D16, D26 p 0. However, for this problem, the magnitudes of D16 and D26 (relative to D11 and D22) are very small. Specifically, for the laminate considered in this problem D16/D11 = D26D11 = 0.0986 and D16/D22 = D26 D22 = 0.153. Consequently, distortion of out-of-plane displacements is very slight. The out-of-plane displacement induced by a uniform transverse load applied to a laminate with relatively higher values of D16 and D26 is considered in Sample Problem 3. As will be seen, the distortion of displacement contours is much more pronounced in that case due to the relatively higher values of D16 and D26.

Figure 3 Convergence of predicted plate deflections based on a Ritz analysis as M and N are increased from 1 to 10 (i.e., as the number of terms is increased from 1 to 100).

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Figure 4 A contour plot of out-of-plane displacements for the [(F45/0)2]s laminate considered in Sample Problem 1.

Sample Problem 2 The [(F45/0)2]s graphite-epoxy laminate described in Sample Problem 1 is again subjected to type S4 simple supports along all four edges and a uniform transverse load q(x,y) = 30 kPa. However, the plate is now also subjected to uniform in-plane stress resultants Nxx = Nyy. Use a Ritz analysis to determine the maximum out-of-plane displacement for 0 < N xx = N yy< 100 kN/m. Solution. As was the case for Sample Problem 1, solutions for this problem can be obtained using program SYMM (described in Sec. 6). Multiple solutions were obtained using the specified range in Nxx and Nyy, and 100 terms were used in the displacement field in all cases. Results are summarized in Fig. 5. The maximum displacement occurs at the center of the plate (i.e., at x = 150 mm, y = 75 mm). As would be expected, the in-plane tensile stress resultants tend to reduce out-of-plane displacement. For Nxx = Nyy = 0, a maximum out-of-plane displacement of 8.03 mm is predicted. In contrast, if tensile stress resultants Nxx = Nyy = 100 kN/m are applied, the maximum out-of-plane displacement is reduced to 0.71 mm.

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Figure 5 Predicted maximum deflections of the plate considered in Sample Problem 2 (M=N=10 in all cases).

Sample Problem 3 A [25j]12 graphite-epoxy laminate is trimmed to in-plane dimensions of 300150 mm and is mounted in an assembly that provides type S4 simple supports along all four edges. The laminate is then subjected to a uniform transverse load q(x,y) = 30 kPa. No in-plane loads are applied (i.e., Nxx= Nyy = Nxy = 0). Determine the maximum out-of-plane displacement based on a Ritz analysis and plot the out-of-plane displacement field. Use the properties listed for graphite-epoxy in Table 3 of Chap. 3, and assume each ply has a thickness of 0.125 mm. Solution. Note that the plate has an aspect ratio R = 150/300 = 2.0, as was the case for the laminates considered in Sample Problems 1 and 2. A rather unusual fiber angle of 25j has been selected for consideration in this problem because it results in high relative values of D16 and D26, resulting in an interesting distortion of the predicted out-of-plane displacement field. Specifically, for this laminate: D16 =D11 ¼ 0:370

D16 =D22 ¼ 2:21

D26 =D11 ¼ 0:126

D26 =D22 ¼ 0:755

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These relative values of D16 and D26 are quite high, at least as compared to those exhibited by the [(F45/0)2]s laminate considered in Sample Problem 1. A solution for this problem was obtained using program SYMM, using M=N=10. A maximum displacement of 16.1 mm is predicted to occur at the center of the plate. A contour plot of out-of-plane displacements is shown in Fig. 6. Distortion of the displacement field due to the generally orthotropic nature of the [25j]12 panel is obvious, especially when compared to the very slightly distorted pattern for a [(F45/0)2]s laminate, previously shown in Fig. 4. 4.2 Deflections Due to a Sinusoidal Transverse Load Consider a simply supported symmetric composite plate subjected to a transverse load that varies sinusoidally over the surface of the plate:  px   py  qðx; yÞ ¼ qo sin sin a b

Figure 6 A contour plot of out-of-plane displacements for the [25]12 laminate considered in Sample Problem 3.

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In this case, the work done by the transverse load, calculated using Eq. (23), is given by: ) Z bZ a n    o(X M X N  mpx   npx  px py dxdy sin sin sin cmn sin W ¼ qo a b a b 0 0 m¼1 n¼1 During integration of this expression, we note the following: 8 Z a < 0; if m p 1  px   mpx  sin dx ¼ a sin : ; if m ¼ 1 a a 0 2 8 > Z b < 0; if n p 1  py   npx  sin sin dy ¼ b > b b 0 : ; if n ¼ 1 2 Hence, the work done by a sinusoidal transverse load is simply: qo abc11 4 The total potential energy is P=UI+UII+UIII+W. Our earlier expressions for UI, UII, and UIII are not altered by the change in transverse load and are given by Eqs. (16), (18), and (22), respectively. Hence, the total potential energy is: W¼

P¼



C1 Nxx þ C2 Nyy þ 2C3 Nxy ðabÞ

þ

þ

f

f



g



M N  4 X X p 2 bm4 2m2 n2 an4 cmn D þ ðD þ 2D Þ þ D 11 12 66 22 8 a3 ab b3 m ¼ 1n ¼ 1   M X N X ij 2p2 mncmn cij 2 2 Š½n 2 m ½ i j 2Š i ¼ 1j ¼ 1 





m2 n2 ðMI ÞðNJ Þ D þ D 16 26 a2 b2



M N  2 X X p 2 m2 b n2 a Nxx þ Nxx cmn a b 8 m ¼ 1n ¼ 1 M X N  X j cij þ 2m2 ncmn Nxy 2 2 Þðn2 ðm i i ¼ 1j ¼ 1

qo abc11 4



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j 2Þ



ðMI ÞðNJ Þ

ð28Þ

g



As before, the next step is to apply the principle of minimum potential energy:  BP m ¼ 1; 2; . . . ; M ¼0 n ¼ 1; 2; . . . ; N Bcmn

This process leads to (M  N) equations (similar in form to those shown in Fig. 2) that must be satisfied simultaneously. Hence, by solving these equations for constants cmn, the out-of-plane displacement caused by a transverse load that varies sinusoidally over the surface of the plate can be calculated using Eq. (6), and the problem is solved. 4.3 Deflections Due to a Transverse Load Distributed Over an Interior Region

A composite panel subjected to a transverse force P uniformly distributed over an internal rectangular region of dimensions ai  bi was previously shown in Fig. 10 of Chap. 9. The center of the interior region is located at x = n and y = g. As discussed in Sec. 4 of Chap. 9, this loading can be expressed in terms of a Fourier series expansion:       l l  X 16P X 1 mpn npg mpai sin qðx; yÞ ¼ 2 sin sin p ai bi m ¼ 1 n ¼ 1 mn a b 2a       npbi mpx npy sin sin  sin 2b a b Substituting the above into the expression representing the work done by transverse loads W [Eq. (23)] and integrating, it will be found: W¼

     M X N  4abP X cmn mpn npg   mpai  npbi sin sin sin sin 2a 2b p2 ai bi m¼1 n¼1 mn a b

The total potential energy is P=UI+UII+UIII+W. Our earlier expressions for UI, UII, and UIII are not altered by the change in transverse load and are given by Eqs. (16), (18), and (22), respectively. Hence, the total potential energy is: P¼



þ

  C1 Nxx þ C2 Nyy þ 2C3 Nxy ðabÞ

f

M X N  4 X p

bm4 2m2 n2 an4 D11 þ ðD12 þ 2D66 Þ þ 3 D22 3 8 a ab b m ¼ 1 n¼1  M X N  X ij cij 2p2 mncmn 2 2 Š½n2 ½ i j 2Š m i ¼ 1j ¼ 1  2 

m n2  2 D16 þ 2 D26 ðMI ÞðNJ Þ a b c2mn



Copyright © 2004 by Marcel Dekker, Inc. All Rights Reserved.

g



þ

f



M X N  2 X p 2 m2 b n2 a Nxx þ Nyy cmn a b 8 m ¼1n ¼1 þ2m2 ncmn Nxy

(

M X N  X cij

i ¼ 1j ¼ 1

ðm2

j i 2 Þðn2

j 2Þ



ðMI ÞðNJ Þ

g

ð29Þ

        ) M X N  4abP X cmn mpn npg mpai npbi sin sin sin sin 2a 2b p2 ai bi m ¼ 1 n ¼ 1 mn a b

As before, the next step is to apply the principle of minimum potential energy: BP ¼0 Bcmn



m ¼ 1; 2; . . . ; M n ¼ 1; 2; . . . ; N

This process leads to (M  N ) equations (similar in form to those shown in Fig. 2) that must be satisfied simultaneously. Hence, by solving these equations for constants cmn, the out-of-plane displacements caused by a transverse force P uniformly distributed over an interior rectangular region of dimensions ai  bi can be calculated using Eq. (6), and the problem is solved. 4.4 Deflections Due to a Transverse Point Load The work done by a concentrated point load P applied at x = n and y = g can be obtained by allowing ai ! 0, bi ! 0 in Eq. (29). In the limit, we obtain: W¼P

   

M X N X mpn npg cmn sin sin a b m ¼ 1n ¼ 1

The total potential energy is: P¼

   C1 Nxx þ C2 Nyy þ 2C3 Nxy ðabÞ

þ

f

N  4 M X X p

m ¼1n ¼ 1

8

c2mn



bm4 2m2 n2 an4 D þ þ 2D Þ þ D22 ðD 11 12 66 a3 ab b3

M X N  X cij

ij 2 2 Š½n 2 ½ m i i ¼1j ¼ 1  2 

m n2  2 D16 þ 2 D26 ðMI ÞðNJ Þ a b 2p2 mncmn

j 2Š

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g

þ

f

(



M X N  2 X p 2 m2 b n2 a Nxx þ Nyy cmn a b 8 m ¼1n ¼1

P

þ2m2 ncmn Nxy M X N X

M X N  X cij

i ¼1j ¼ 1

cmn sin

m ¼1n ¼1



ðm2

j i 2 Þðn2

j 2Þ

g



ðMI ÞðNJ Þ

   ) mpn npg sin a b

ð30Þ

We apply the principle of minimum potential energy:  BP m ¼ 1; 2; . . . ; M ¼0 n ¼ 1; 2; . . . ; N Bcmn This process leads to (M  N) equations (similar in form to those shown in Fig. 2) that must be satisfied simultaneously. Hence, by solving these equations for constants cmn, the out-of-plane displacements caused by a concentrated load P applied at x = n and y = g can be calculated using Eq. (6), and the problem is solved. 5 BUCKLING OF SYMMETRIC COMPOSITE PLATES SUBJECT TO SIMPLE SUPPORTS OF TYPE S4 In this section, buckling of type S4 simply supported symmetric composite panels will be considered. A brief explanation of what is meant by ‘‘buckling’’ was presented in Sec. 4 of Chap. 10. In essence, the term buckling refers to the fact that (under the proper circumstances) in-plane stress resultants Nxx, Nyy, and/or Nxy can cause out-of-plane displacements w(x,y). A coupling between in-plane loads and out-of-plane displacement is of course expected for nonsymmetric laminates because Bij p 0 in this case. We have limited discussion to symmetric laminates, however, so we are considered with a coupling between in-plane loads and out-of-plane displacement that is not predicted by the CLT analysis developed in Chap. 6. The buckling phenomenon can be explained on the basis of the principle of minimum potential energy as follows. Based on our earlier discussion, the total potential energy P of an initially flat plate subjected to inplane loading is given by: P ¼ UI þ UII þ UIII Note that the work term W no longer appears in our expression for P because we assume that no transverse loads are applied. General expressions for strain energy components UI, UII, UIII for a composite panel have been

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presented as Eqs. (14), (17), and (21), respectively. Inspection of these equations reveals that UI is a function of in-plane displacements uo(x,y) and vo(x,y), that UII is a function of out-of-plane displacements w(x,y), and that UIII is a function of uo, vo, and w. If an initially flat plate is subjected to relatively low levels of in-plane loads, then only in-plane displacements occur; that is, the plate remains flat because at low load levels, the flat configuration corresponds to the state of minimum potential energy. In this case, UI is the only nonzero strain energy component: UII = UIII = 0 because w(x,y) = 0. However, as the magnitude of in-plane loading is increased, then the configuration that corresponds to the state of minimum potential energy may no longer be flat but rather bent; the configuration that corresponds to the state of minimum potential energy involves out-of-plane displacement w(x,y) and the plate ‘‘buckles.’’ Once bucking occurs, then UI, UII, UIII p 0. In Secs. 3 and 4, we developed expressions for the potential energy of a simply supported symmetric composite panel subjected to both in-plane and transverse loads. These earlier results can be adopted for present purposes by simply discarding those terms involving the transverse load. Hence, from Eq. (26), we can immediately write: " N M X X

 c2mn F1 m4 þ F2 m2 n2 þ F3 n4 þ F4 m2 þ F5 n2 P ¼ UI þ m ¼ 1n ¼ 1

(

M X N X

cij þcmn 2 2 ðm i Þðn2 i ¼ 1j ¼ 1

)# þF8 ðm2 njÞ ðMI ÞðNJ Þ

j 2Þ



F6 ðm3 nijÞ þ F7 ðmn3 ijÞ ð31Þ

The constants F1 through F8 were defined in Sec. 4, in conjunction with Eq. (26). Note that one constant, namely, F9, no longer appears. This term has been dropped because it represents a transverse load that has now been assumed to be zero. We can predict the onset of buckling by applying the principle of minimum potential energy:  BP m ¼ 1; 2; . . . ; M ¼0 n ¼ 1; 2; . . . ; N Bcmn This process leads to (M  N) equations that must be satisfied simultaneously. For purposes of illustration, consider the set of equations that are obtained using M = N = 2. In this case, we obtain the same set of equations as those previously illustrated in Fig. 2(a), except that now F9 = 0. Referring to Fig. 2(a), we see that in the present case, all terms on the right side of the equality are zero.

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We must now specify the loading condition of interest. That is, we must specify the loading conditions under which buckling is to be predicted. There are many possibilities although we have limited discussion in this chapter to type S4 simple supports. For example, we may wish to determine the value of Nxx that will cause buckling, given some constant values for Nyy and Nxy. Alternatively, we may wish to determine the value of Nxy that will cause buckling, given some constant values for Nxx and Nyy. A third possibility is a loading situation in which all three resultants are increased proportionately (i.e., Nyy = k1Nxx and Nxy = k2Nxx, where k1 and k2 are known constants). In this case, we are interested in buckling caused by a simultaneous increase in Nxx, Nyy, and Nxy. For illustrative purposes, let us assume that we are interested in buckling caused by uniaxial loading, i.e., for the loading condition Nxx p 0, Nyy = Nxy = 0. In this case, we have: F5 ¼

p2 a Nyy ¼ 0 8b

F8 ¼ 2Nxy ¼ 0

Also, F9=0, because no transverse load is applied. For this situation, the set of equations shown in Fig. 2(a) reduce to: 3 40 0 0 ð2F6 þ 2F7 Þ 7 6 2ðF1 þ F2 þ F3 þ F4 Þ 9 6 7 7 6 40 6 7 ð2F þ 2F Þ 0 þ 4F þ 16F þ F Þ 0 2ðF 6 7 1 2 3 4 7 6 9 6 7 7 6 40 7 6 0 ð2F6 þ 2F7 Þ 2ð16F1 þ 4F2 þ F3 þ 4F4 Þ 0 7 6 9 7 6 5 4 40 ð2F6 þ 2F7 Þ 0 0 8ð4F1 þ 4F2 þ 4F3 þ F4 Þ 9 9 8 9 8 0> c11 > > > > > > > > > > > > > > > > > > > > > > =

= > < c12 > ¼  > > >0> > > > c > > > > > > > > 21 > > > > > > > > > ; > : ; : > c22 0 2

The unknown buckling load is contained in the term F4 = p2bNxx/8a. We can therefore rearrange the above expression by bringing all terms involving F4 to the right side of the equality: 2

6 2ðF1 þ F2 þ F3 Þ 6 6 6 0 6 6 6 6 0 6 6 4 40 ð2F6 þ 2F7 Þ 9

0 2ðF1 þ 4F2 þ 16F3 Þ 40 ð2F6 þ 2F7 Þ 9 0

0 40 ð2F6 þ 2F7 Þ 9

40 ð2F6 þ 2F7 Þ 9 0

2ð16F1 þ 4F2 þ F3 Þ

0

0

8ð4F1 þ 4F2 þ 4F3 Þ

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3 7 7 7 7 7 7 7 7 7 7 5

8 8 9 2 c11 > > > > > > > > > > > > > >

= < 0 12  ¼ F4 > > > > > > > > c21 > 0 > > > > > > : ; : c22 0 8 > p2 b > > 0 > > 4a > > < p2 b ¼ Nxx 0 > a > > > > > > : 0 0

0 2

0 0

9 98 0 >> c11 > > > > > > > > > > > > > = < 0 c = 12

>> > > > > > c21 > 0 8 0 > > > > > > > ; ;: c22 0 0 8 98 9 ð32Þ >> c11 > > > > 0 0> >> > > > > > >c > > => < 12 > = 0 0 >> c21 > > > > > > > > > > 2 > > > > > > p b > : ; ; 0 c22 a Equation (32) is in the form of a so-called generalized eigenvalue problem. In general, there are (M  N) eigenvalues that satisfy a generalized eigenvalue problem; for each eigenvalue, there exists a corresponding eigenvector. For this particular example, the eigenvalues corresponds to Nxx, and the eigenvector corresponds to the column matrix containing coefficients cij. Because we have assumed M = N = 2, there are four values of Nxx that will satisfy Eq. (32). The eigenvalue with lowest magnitude represents the critical c , and the coefficients cij that correspond to this eigenvalue buckling load Nxx N represent the critical buckling mode because wðx; yÞ ¼ SM m ¼ 1 Sn ¼ 1 cmn sin (mkx/a) sin (mky/b). As in all analyses based on the Ritz method, several predictions of buckling load should be obtained using increased values of M and N to insure convergence of the predicted buckling load and mode. Equation (32) represents the generalized eigenvalue problem for the case Nxx p 0, Nyy = Nxy = 0, and M = N = 2. As a second example, consider buckling caused by a pure shear load. That is, assume Nxy p 0, Nxx=Nyy=0. We now have F4 = F5 = F9 = 0. Following a process identical to that described above, we arrive at the following generalized eigenvalue problem: 2

0 6 2ðF1 þ F2 þ F3 Þ 6 6 6 0 2ðF1 þ 4F2 þ 16F3 Þ 6 6 6 40 6 ð2F6 þ 2F7 þ F8 Þ 0 6 9 6 4 40 ð2F6 þ 2F7 Þ 0 9 9 8 8 c11 > > > > > > > > 0 0 0 > > > > > > > > > > > > > > > 80 c 12 > > > < 0 = < 0 9 ¼ Nxy  80 > > > > > c21 > 0 0 > > > > > > > > > 9 > > > > > > 80 > > > : > > ; : 0 0 c22 9

0 40 ð2F6 þ 2F7 Þ 9

40 ð2F6 þ 2F7 Þ 9 0

2ð16F1 þ 4F2 þ F3 Þ

0

0

8ð4F1 þ 4F2 þ 4F3 Þ

98 c 9 80 >> > 11 > > > > >> > > > > 9 > > > > > > > > > > c 12 >< > > = 0 =

>> > >c > 21 > 0 > > > > > > > > > > > > > > > >> > > ; : 0 ;> c22

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3 7 7 7 7 7 7 7 7 7 7 5

ð33Þ

Once again, there will be four eigenvalue/eigenvector pairs that satisfy Eq. (33) because we have assumed M = N = 2. The eigenvalue with lowest c magnitude represents the critical buckling shear load Nxy and the corresponding eigenvector represents the critical buckling mode. 6 COMPUTER PROGRAM SYMM The solutions described in this chapter are implemented in a computer program called SYMM. This program can be downloaded at no cost from the following website: http://depts.washington.edu/amtas/computer.html. Program SYMM is based on the Ritz method and is applicable to any symmetric composite laminate. The program prompts the user to input all information necessary to perform these calculations. Properties of up to five different materials may be defined. The user must input various numerical values using a consistent set of units. For example, the user must input elastic moduli for the composite material system(s) of interest. Using the properties listed in Table 3 of Chap. 3 and based on the SI system of units, the following numerical values would be input for graphite-epoxy: E11 ¼ 170  109 Pa 9

E22 ¼ 10  109 Pa

v12 ¼ 0:30

G12 ¼ 13  10 Pa

Because 1 Pa=1 N/m2, all lengths must be input in meters. For example, ply thicknesses must be input in meters (not millimeters). A typical value would be tk=0.000125 m (corresponding to a ply thickness of 0.125 mm). Similarly, if an analysis of a plate with a length and width of 500300 cm were being performed, then the length and width of the plate must be input as 5.00 and 3.00 m, respectively. If the English system of units were used, then the following numerical values would be input for the same graphite-epoxy material system: E11 ¼ 25:0  106 psi

E22 ¼ 1:5  106 psi

v12 ¼ 0:30

G12 ¼ 1:9  106 psi All lengths would be input in inches. A typical ply thickness might be tk=0.005 in., and the length and width of a plate might be 36 and 20 in., for example. REFERENCES 1. Whitney, J.M. Structural Analysis of Laminated Anisotropic Plates. Technomic Pub Co.: Lancaster, PA, ISBN 87762-518-2.

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2. Turvey, G.J.; Marshall, I.H., Eds. Buckling and Postbuckling of Composite Plates. Chapman and Hall: New York, NY, 1995. 3. Fung, Y.C. A First Course in Continuum Mechanics. Prentice-Hall: Englewood Cliffs, NJ, 1969. 4. Timoshenko, S.; Woinowsky-Krieger. Theory of Plates and Shells. McGraw-Hill Book Co.: New York, NY, 1987. ISBN 0-07-0647798. 5. The binomial series expansion can be found in most reference books devoted to mathematical tables and formulas. See, for example, Korn, G.A.; Korn, T.M. Mathematical Handbook for Scientists and Engineers. 2nd ed. McGraw-Hill Book Co.: New York, NY, 1968. Table E-6.

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Appendix A Finding the Cube-Root of a Complex Number

The process of finding the principal stresses and principal strains is discussed in Section 7 and 12 of Sec. 2, respectively. The approach described in these sections ultimately leads to the requirement of calculating the cube root of a complex number. The nth-root of a complex number can be determined using ‘‘DeMoivre’s theorem,’’ which is explained in many advanced calculus books. A complete review of this topic is beyond the scope of the present discussion. The explanation given below is a specialization of DeMoivre’s theorem to find the cube root of a complex number. Any complex number z can be written as: z ¼ h þ iv where: iu

pffiffiffiffiffiffi i

h ¼ the ‘‘real part’’ of z v ¼ the ‘‘imaginary part’’ of z The modulus (r) and argument (u) of a complex number are defined as: pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi r ¼ h2 þ v 2 ðA:1Þ 617

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u ¼ tan

1

ðv=hÞ

ðA:2Þ

The relationship between the modulus, argument, and components h and v can be visualized by plotting the number z in the complex plane, as shown in Fig. 1. Based on the above definitions, DeMoivre’s theorem can be used to show that the cube root of a complex number z = h+iv is given by:   h u io ffiffiffiffiffiffiffiffiffiffiffiffi ffiffiffi p p lnðrÞ n h u i 3 3 z ¼ h þ iv ¼ exp cos þ isin ðA:3Þ 3 3 3 Two important cautions should be noted: 

The value of u used in Eq. (A.3) must be expressed in radians (not in degrees).  Referring to Fig. 1, note that the function tan 1(v/h) corresponds to two different arguments (that is, two different angles), which occur in opposite quadrants. For example, assume that both h and v are positive for some complex number z = h+iv (this is the situation illustrated in Fig. 1). Referring to Fig. 1, it is clear that in this case, argument u is an angle in the first quadrant of the complex plane: p 0 V u radians 2 In contrast, consider a different complex number z V, where z V= 1* z= h iv. The modulus r for complex numbers z and zV [calculated using Eq.

Figure 1 The number z plotted in the complex plane, showing the real part (h), imaginary part (v), the modulus (r), and the argument (u).

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(A.1)] is identical. However, the argument of z V, angle uV say, is an angle in the third quadrant of the complex plane: 3p radians p rad V uV V 2 Hence, when calculating the argument of a complex number using Eq. (A.2), one must insure that the angle returned by the tan 1( ) function corresponds to the proper quadrant, as determined by the algebraic signs of h and v.

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Appendix B Experimental Methods Used to Measure In-Plane Properties E11, E22, v12, and G12

An abbreviated discussion of tests used to measure tensile in-plane properties E11, E22, v12, and G12 is provided in this Appendix. The fact that this discussion is limited to tensile test methods is significant and should be noted by the reader. First, since E11, E22, and v12 often differ in tension and compression, in practice both tensile and compressive values should be measured. Specimen geometries used during compression testing generally differ from the tensile test specimen geometries described herein. Secondly, it is more-or-less implied that the specimens described herein are produced using pre-preg tape. As opposed to resin-transfer molding or filament winding, for example. Third, material properties can also be inferred from flexure tests, and these will not be discussed herein. The reader interested in a more detailed discussion of the experimental methods used to measure composite properties should consult Refs. (1–3) or the appropriate ASTM test standards (many of which are listed in Tables 1 and 2 of Chapter 3). The most widely used tensile test specimen geometry is based on ASTM Standard 3039 ‘‘Test Method for Tensile Properties of Polymer Matrix Composite Materials.’’ Typical specimens that conform to this standard are shown in Figs. 1–4. As indicated, tensile specimens are flat and straight-sided with rectangular cross section. Very often adhesively bonded end tabs are used. The [0]n specimen, the ‘‘off-axis’’ [h]n specimen, and the [90]n specimen 621

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Figure 1 A [0]n tensile specimen, showing typical dimensions and adhesively bonded end tabs.

shown in Figs. 1–3 are equipped with adhesively bonded end tabs. As the use of bonded end-tabs increases specimen preparation time and cost, they are only used when necessary. In general, end tabs are used when  

the fracture stress or strain is to be measured, or if it is found that the specimen fails within the grip region if bonded end tabs are not used

End tabs used with high-strength specimens (such as the [0]n or [h]n specimens shown in Figs. 1 and 2, respectively) must be beveled at a shallow angle to avoid specimen failure at the end of the tab. The shallow bevel provides for a smooth transfer of load from the grip region to the gage region of the

Copyright © 2004 by Marcel Dekker, Inc. All Rights Reserved.

Figure 2 A [h]n tensile specimen, showing typical dimensions and adhesively bonded end tabs.

specimen. The bevel angle is usually about 5–7j, although angles as high as 30j are used on occasion. Use of a shallow bevel angle becomes less important when testing low-strength specimens, and in fact end tabs with a bevel angle of 90j are often used when testing [90]n specimens (Figs. 3). As a general rule all tensile test specimens are long and narrow. That is, they have a high ‘‘aspect ratio.’’ Aspect ratio equals specimen length (where specimen length is defined as the tab-to-tab distance) to specimen width.

Copyright © 2004 by Marcel Dekker, Inc. All Rights Reserved.

Figure 3 A [90]n tensile specimen, showing typical dimensions and adhesively bonded end tabs.

However, high-strength composite specimens must have an exceptionally high aspect ratio. For example, the aspect ratios of the [0]n and [h]n specimens shown in Figs. 1 and 2 are 9.3 and 12.0, respectively. In contrast, the aspect ratio of the relatively low-strength [90]n specimen (Fig. 3) is 5.0. Bonded end tabs are often not necessary when testing symmetric and balanced laminates, such as the [F45]ns specimen shown in Fig. 4. In these cases so-called ‘‘friction’’ tabs may be used. A friction tab is held in place by the pressure of the grips. Occasionally, an abrasive paper (such as emery cloth) is placed between the tab and surface of the specimen to enhance friction. End tabs (either bonded or friction) are most often made using a cross-ply [0/90]ns E-glass/polymer composite, although end tabs made using steel or aluminum have also been successfully used.

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Figure 4 A [F45]ns tensile specimen, showing typical dimensions.

Strains induced during a tensile test are measured using either bonded resistance strain gages or extensometers. Properties E11 and v12 are measured using a [0]n specimen while E22 is measured using a [90]n specimen. It is possible to measure v21 as well, using a [90]n specimen. However, for most advanced unidirectional composites the numerical value of v21 (the so-called ‘‘minor’’ Poisson ratio) is very small and is at least an order of magnitude smaller than the ‘‘major’’ Poisson ratio v12. For example, for graphite/epoxy the major Poisson ratio v12c0.3 whereas the minor Poisson ratio v12c0.01. The small value of v21 can lead to a relatively high measurement error.

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Consequently, v21 is rarely measured in practice. If the value of v21 is required it is calculated using the inverse relation: v21 ¼ v12

E22 E11

Several techniques by which the shear modulus G12 can be measured using a tensile specimen are available. One approach is based on an off-axis [h]n specimen. Referring to Eq. (38c) in Chapter 5, the axial modulus Exx of such a specimen is given by: Exx ¼

4

cos ðhÞ þ E11



1 G12

1  2v12 sin4 ðhÞ cos2 ðhÞsin2 ðhÞ þ E22 E11

Solving this equation for G12, we obtain: G12 ¼

E11 E22

Exx



Exx E11 E22 cos2 ðhÞsin2 ðhÞ E22 cos4 ðhÞ þ 2v12 E22 cos2 ðhÞsin2 ðhÞ

E11 sin4 ðhÞ

 ðB:1Þ

Thus if Exx is measured for a [h]n specimen (where h is known), and assuming that E11, v12, and E22 have also been measured and are known, then the shear modulus G12 can be calculated using Eq. (B.1). For example, if h = 45j (the most common case) then Eq. (B.1) reduces to: G12 ¼

4E11 E22

Exx E11 E22 Exx ½E22 þ 2v12 E22

E11 Š

ðB:2Þ

An alternate method is based on the use of a [F45]ns specimen. In this case the axial and transverse strains (exx and eyy) induced by a uniaxial tensile stress (rxx) are measured, usually using biaxial strains gages. For this stacking sequence it can be shown (4) that the shear stress (s12) and shear strain (c12) induced in the +45j plies are given by: s12 Aþ45j ¼

c12 Aþ45j ¼

rxx =2 ðexx

eyy Þ

In contrast, the shear stress and strain induced in the by: s12 A

45j

c12 A

45j

¼ rxx =2 ¼ ðexx

eyy Þ

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45j plies are given

Hence for the [F45]ns stacking sequence the shear modulus is given by: G12 ¼

s12 rxx ¼ c12 2ðexx eyy Þ

REFERENCES 1. Whitney, J.M.; Daniel, I.M.; Pipes, R.B. Experimental Mechanics of Fiber Reinforced Composite Materials, 2nd Ed; Bethel, CT: SEM Monograph 4 Society for Experimental Mechanics (ISBN 0-912053-01-1). 2. Carlsson, L.A.; Pipes, R.B. Experimental Characterization of Advanced Composite Materials, 2nd Ed; Lancaster, PA: Technomic Pub Co. (ISBN 1-56676433-5), 1997. 3. Manual on Experimental Methods of Mechanical Testing of Composites, 2nd Ed; Jenkins, C.H., Ed; Bethel, CT: Society for Experimental Mechanics ISBN 088173-284-2, 1998. 4. Rosen, B.W. A simple procedure for experimental determination of the longitudinal shear modulus of unidirectional composites. J Compos Mater 1972, 6, 552.

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Appendix C Tables of Beam Deflections and Slopes

Tables of typical beam deflections and slopes can be found on the following pages.

629

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Table 1

Deflection and Slopes of Cantilever Beams

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Table 1

Continued

Source: Ref. 1.

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Table 2

Deflections and Slopes of Simply Supported Beams

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Table 2

Continued

Source: Ref. 1.

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