Structural Analysis of Polymeric Composite Materials

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

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

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.

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

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

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

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,

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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 1
106 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.

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

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.

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

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-

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

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

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

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.

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

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.

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

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.

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

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.

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

Figure 5 Types of polymer molecular structure.

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

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

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

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

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

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

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

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.

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

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

jC

jF

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

123 85 50 50 85 90 100 150 175 200 215

190 120 60 122 185 195 210 300 350 400 420

Rubbery Rubbery Rubbery Rigid Rigid Rigid Rigid Glassy Glassy Glassy Glassy

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

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.

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



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

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

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)

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

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.

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

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)

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

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

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

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

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

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

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

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.

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

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,

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

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

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

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 n
n, where n is the number of tows between crossover points. A 1
1-bias and a 2
2-bias braided fabric is shown in Fig. 10a and b, respectively. A 1
1 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,

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

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

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

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

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

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.)

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

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-

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

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.

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



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.)

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

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

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

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

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

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,

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

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.

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

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.

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

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.

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

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.

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

REFERENCES 1. 2. 3. 4.

5.

6. 7.

8.

9. 10. 11. 12. 13. 14.

Rodriguez, F. Principles of Polymer Systems; 3rd Ed.; Hemisphere Publ. Co.: New York, 1989; ISBN 0-89116-176-7. Young, R.J.; Lovell, P.A. Introduction to Polymers; 2nd Ed.; Chapman and Hall Publ. Co.: New York, 1991; ISBN 0-89116-176-7. Strong, A.B. Plastics: Materials and Processing; 2nd Ed.; Prentice-Hall: Upper Saddle River, NJ, 2000. ISBN 0-13-021626-7. The Macrogalleria (http://www.psrc.usm.edu/macrog/index.htm), website maintained by the University of Southern Mississippi and devoted to polymeric materials. 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. 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. Cox, B.; Flanagan, G. Handbook of Analytical Methods for Textile Composites. NASA Contractor Report 4750. NASA-Langley Res. Ctr., Hampton, VA, 1997. 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. Kalpakjian, S. Manufacturing Processes for Engineering Materials; 3rd Ed.; Addison-Wesley Longman, Inc.: Menlo Park CA, 1997; ISBN 0-201-82370-5. Schwartz, M.M. Composite Materials Handbook; New York, NY, McGraw-Hill Book Co.: 1983; ISBN 0-07-055743-8. Hyer, M.W. Stress Analysis of Fiber-Reinforced Composite Materials; New York, NY,McGraw-Hill Book Co.: 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. Jones, R.M. Mechanics of Composite Materials; McGraw-Hill Book Co.: New York, NY, 1975; ISBN 0-07-032790-4. Hull, D. An Introduction to Composite Materials; Cambridge University Press: Cambridge, UK, 1981; ISBN 0-521-23991-5.

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

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.

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

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

ð1Þ

The magnitude of the force is given by: qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi jF j ¼ Fx2 þ Fy2 þ F 2z

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.

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

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-

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

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

ð6aÞ

Fy V ¼ cy Vx Fx þ cy Vy Fy þ cy Vz Fz 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

Since iˆ V is a unit vector, then in accordance with Eq. (2): ðcx Vx Þ2 þ ðcxVy Þ2 þ ðcxVz Þ2 ¼ 1

ð8Þ

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

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

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

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

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

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

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 ðsin 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 ðsin 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

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

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 ð0Þðsin hÞ þ ð0Þð0Þ IxW ¼ cos h IyW ¼ cyWxV IxV þ cyWyV IyV þ cyWzV IzV ¼ð0Þðcos hÞ þ ðcos bÞðsin hÞ þ ðsin bÞð0Þ IyW ¼ cos b sin h IzW ¼ czWxV IxV þ czWyV IyV þ czWzV IzV ¼ð0Þðcos hÞ þ ðsin bÞðsin 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 ðcos 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 ðsin 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

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

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): 8 9 2 38 9 cxWx cxWy cxWz < Fx = < FxW = F W ¼ 4 cyWx cyWy cyWz 5 Fy : y ; : ; czWx czWy czWz FzW Fz 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 8 9 8 9 ðcos hÞFx þ ðsin hÞFy < FxW = < = FyW ¼ ðcos b sin hÞFx þ ðcos b cos hÞFy þ ðsin bÞFz : ; : ; ðsin b sin hÞFx þ ðsin b cos hÞFy þ ðcos bÞFz FzW Part (c). Using the specified numerical values and the results of Part (b): 9 8 9 8 ðcos 20B Þð1000 NÞ þ ðsin 20B Þ200 N = < FxW = < FyW ¼ ðcos 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 ðsin 60B cos 20B Þð200 NÞ þ ðcos 60B Þð600 NÞ 8 9 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

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

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.

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

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.

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

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.

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

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

cxV y ¼ sin h

c xV z ¼ 0

cyV x ¼ sin h czV x ¼ 0

cyV y ¼ cos h czV y ¼ 0

cy V z ¼ 0 czV z ¼ 1

Applying Eqs. (6a)–(6c), we have: 8 9 2 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 ; : ; : ; czV x czV y czV z 0 0 1 FzV Fz 0 8 9 < ðcos hÞR = ¼ ðsin 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

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

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),

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

Figure 6 A solid 3-D body in equilibrium.

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

that is, 1 psi=1 lbf/in2. Conversion factors between the two systems of measurement are 1 psi=6895 Pa, or equivalently, 1 Pa=0.1450
103 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

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

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

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

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.

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

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

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

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: 2 3 2 3 rxx sxy sxz rxx sxy sxz 4 syx ryy syz 5 ¼ 4 sxy ryy syz 5 ð10Þ szx szy rzz sxz syz 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) 3 2 rxx sxz syz 5 ! 4 rxy rzz rxz

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.

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

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

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

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  ¼ ½ci V j ½rij ½ci V j T

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

ð12aÞ

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

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

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 czV 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-

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

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

ð15aÞ

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

ð15bÞ

Third stress invariant ¼ W ¼ rxx ryy rzz  rxx r2yz  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

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

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 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 czWy ¼ sin ð35B Þcos ð20B Þ ¼ 0:5390 czWz ¼ cos ð35B Þ ¼ 0:8192 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

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

rxW zW ¼ ð0:9397Þð0:1962Þð50 ksiÞ þ ð0:9397Þð0:5390Þð10 ksiÞ þð0:9397Þð0:8192Þð15 ksiÞ þ ð0:3420Þð0:1962Þð10 ksiÞ þð0:3420Þð0:5390Þð25 ksiÞ þ ð0:3420Þð0:8192Þð30 ksiÞ þð0Þð0:1962Þð15ksiÞþð0Þð0:5390Þð30ksiÞþð0Þð0:8192Þð5 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 ¼40:2802 0:7698 0:57365410 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  5Þ ksi H ¼ 70 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.

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

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:  1 rii rjj  rij rij ¼ rxx ryy þ rxx rzz þ ryy rzz 2    r2xy þ r2xz þ r2yz n U ¼ ð50Þð25Þ þ ð50Þð5Þ þ ð25Þð5Þ  ½ð10Þ2 þ ð15Þ2 o þ ð30Þ2  ðksiÞ2 U¼

U ¼ 350ðksiÞ2 xU–yU–zU coordinate system: U¼

 1 r W Wr W W  ri Wj Wri Wj W 2 ii jj

  U ¼ rxWxWryWyW þ rxWxWrzWzW þ ryWyWrzWzW  r2xWyW þ r2xWzW þ r2yWzW

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

n U ¼ ð40:65Þð43:08Þ þ ð40:65Þð13:72Þ þ ð43:08Þð13:72Þ o  ½ð1:113Þ2 þ ð28:95Þ2 þ ð10:60Þ2  ðksiÞ2 U ¼ 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 ½ð50Þð25Þð5Þ  ð50Þð30Þ2  ð25Þð15Þ2  ð5Þð10Þ2 þ 2ð10Þð15Þð30Þ ðksiÞ3 W ¼ 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 ½ð40:65Þð43:08Þð13:72Þ  ð40:65Þð10:60Þ2  ð43:08Þ
ð28:95Þ2  ð13:72Þð1:113Þ2 þ 2ð1:113Þð28:95Þ
ð10:60Þ ðksiÞ3 W ¼ 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

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

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  ryy Þcpn y  syz cpn z ¼ 0

ð16bÞ

sxz cpn x  syz cpn y þ ðrpn  rzz Þcpn z ¼ 0

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-

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

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  W ¼ 0

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).

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

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 2H  9HU þ 27W b¼ 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 ð18Þ rp1 ; rp2 ; rp3 ¼  > 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

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

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: 2 3 2 3 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 3ð350Þ  ð70Þ2 ¼ 1983 a¼ 3 3  1  3 2H  9HU þ 27W b¼ 27 i 1 h 2ð70Þ3  9ð70Þð350Þ þ 27ð65375Þ ¼ 31801 ¼ 27 vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi u u 3 2 3 3 t 48134 b b a ð48134Þ2 ð1983Þ3 þ A¼  þ þ ¼  þ 2 2 4 27 4 27 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi p ¼ 3 15900 þ ið6010Þ

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

We find that constant A equals the cube root of the complex number: z ¼ 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 ð15900Þ2 þ ð6010Þ2 ¼ 16998   6010 1 u ¼ tan ¼ tan1 ð0:37799Þ ¼ 0:3614 radðorÞ2:780 rad 15900 Since in this case: a ¼ 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.

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

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  sxz cpn z ¼ 0 sxy cpn x þ ðrpn  ryy Þcpn y  syz cpn z ¼ 0 ðcpn x Þ2 þ ðcpn y Þ2 þ ðcpn z Þ2 ¼ 1 Direction cosines for rp1. The three independent equations become: ð54:21  50Þcp1 x þ 10cp1 y  15cp1 z ¼ 0 10cp1 x þ ð54:21  25Þcp1 y  30cp1 z ¼ 0 ðcp1 x Þ2 þ ðcp1 y Þ2 þ ðcp1 z Þ2 ¼ 1 Solving simultaneously, we obtain: cp1 x ¼ 0:9726

cp1 y ¼ 0:1666

cp1 z ¼ 0:1620

Direction cosines for rp2. The three independent equations become: ð43:51  50Þcp2 x þ 10cp2 y  15cp2 z ¼ 0 10cp2 x þ ð43:51  25Þcp2 y  30cp2 z ¼ 0 ðcp2 x Þ2 þ ðcp2 y Þ2 þ ðcp2 z Þ2 ¼ 1 Solving simultaneously, we obtain: cp2 x ¼ 0:05466

cp2 y ¼ 0:8416

cp2 z ¼ 0:5738

Direction cosines for rp3. The three independent equations become: ð27:72  50Þcp3 x þ 10cp3 y  15cp3 z ¼ 0 10cp3 x þ ð27:72  25Þcp3 y  30cp3 z ¼ 0 ðcp3 x Þ2 þ ðcp3 y Þ2 þ ðcp3 z Þ2 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

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

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.

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

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: 2 3 2 32 3 c xV x c xV y c xV z rxV xV rxV yV rxV zV rxx rxy rxz 6r V V r V V r V V 7 6c V 6 7 cyV y cyV z 7 4 yx 54 ryx ryy ryz 5 yy yz 5 ¼ 4 yx rzV xV rzV yV rzV zV czV x czV y czV 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

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

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 ð90B Þ ¼ 0 cyV x ¼ cos ð90B þ hÞ ¼ sin ðhÞ cyV y ¼ cos 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 yy y z 5 ¼ 4 sin h cos h 0 54 syx szV xV szV yV rzV zV 0 0 1 0 0 0 2 3 cos h sin h 0 6 7
4 sin h cos h 0 5 0

0

1

Completing the matrix multiplication indicated results in: 2

rxV xV 4 s yV x V 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 2 ¼6 4cosh sinhrxx þ cosh sinhryy þ ðcos h  sin hÞsxy

sin2 hrxx þ cos2 hryy  2cosh sinhsxy

3 0 7 07 5

0

0

cosh sinhrxx þ cosh sinhryy þ ðcos2 h  sin2 hÞsxy

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  2cos ðhÞsin ðhÞsxy

ð19Þ

sxV yV ¼ cos ðhÞsin ðhÞrxx þ cos ðhÞsin ðhÞryy þ ½cos 2 ðhÞ  sin 2 ðhÞsxy

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

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

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

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 cos ðhÞ 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 ½T 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

ð22Þ

Third stress invariant ¼ W ¼ 0 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.

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

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  r 2 rxx þ ryy xx yy 2 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: h ¼ 25B

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

Figure 15 Plane stress elements associated with Example Problem 5.

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

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

3

8 9 < 70 = 7 sin2 ð25B Þ cos2 ð25B Þ 2cosð25B Þsinð25B Þ 7 5: 15 ; 50 2 B B B B 2 B B cosð25 Þsinð25 Þ cosð25 Þsinð25 Þ cos ð25 Þ  sin ð25 Þ 9 8 9 38 8 9 2 0:8214 0:1786 0:7660 > 70 > 98:5 > > > > > > > < rxV xV > = 6 < = < = 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 cos2 ð25B Þ

sin2 ð25B Þ

2cosð25B Þsinð25B Þ

The rotated stress element is shown in Fig. 15(b). The principal stresses are found through application of Eq. (26): sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi   70 þ 15 70  15 2 rp1 ; rp2 ¼ þð50Þ2 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 2ð50Þ 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

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

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

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

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%.

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

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
106 meter=meter ¼ 1micrometer=meter ¼ 1 Am=m ¼ 1 Ain=in 1
106 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

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

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

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

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: 2 3 2 3 exx exy exz exx exy exz 4 eyx eyy eyz 5 ¼ 4 exy eyy eyz 5 ð34Þ ezx ezy ezz exz eyz 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 2 3 2 3 exx ðcxy =2Þ ðcxz =2Þ exx exy exz 6 7 6 eyy ðcyz =2Þ 7 4 exy eyy eyz 5 ¼ 4 ðcxy =2Þ 5 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

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

ð36aÞ

Alternatively, using matrix notation, the strain tensor transforms according to: ½ei V j V  ¼ ½ci V j ½eij ½ci V j T 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 czV 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 ð37cÞ Third strain invariant ¼ We ¼ eii ejj ekk  3eii ejk ejk þ 2eij ejk eki 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

ð38bÞ

Third strain invariant ¼ We ¼ exx eyy ezz  exx e2yz  eyy e2xz  ezz e2xy þ 2exy exz eyz

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

ð38cÞ

Example Problem 7 Given. A state of strain referenced to an x–y–z coordinate is known to be: 3 2 3 2 1000 Am=m 500 Arad 250 Arad exx exy exz 7 6 7 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

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

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 2 3 1000 500 250 6 7
4 500 1500 750 5 2

250 0:9397

6
4 0:3420 0

750

2000

0:2802 0:7698 0:5736

0:1962

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.

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

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

 1 eiUiU ejUjU  eiUjU eiUjU 2

  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 g ðAm=mÞ2 Ue ¼ 5:625
106 ðAm=mÞ2 ¼ 5:625
106 ðm=mÞ2

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

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 2ð500Þð250Þð750Þ ðAm=mÞ3 W ¼ 2:031
109 ðAm=mÞ3 ¼ 2:031
109 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 2ð727Þð91Þð625Þ ðAm=mÞ3 W ¼ 2:031
109 ðAm=mÞ3 ¼ 2:031
109 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,

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

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 y  exz cpn z ¼ 0

ð39aÞ

exy cpn x þ ðepn  eyy Þcpn y  eyz cpn z ¼ 0

ð39bÞ

exz cpn x  eyz cpn y þ ðepn  ezz Þcpn z ¼ 0

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     ðepn  eyy Þ eyz  ¼ 0  exy     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 A¼  þ 2 4 27 sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3 b b2 a 3 B¼   þ 2 4 27

a¼

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

The three principal strains [i.e., the three roots of Eq. (40)] are then given by: 8 > > > > > > > <

He AþBþ 3   p ffiffiffiffiffiffi ffi AþB AB He þ 3 þ  ep1 ; ep2 ; ep3 ¼ > 3 2 2 > >   > p ffiffiffiffiffiffi ffi > AþB AB He > > :  3 þ 2 2 3

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

exx

6 4 eyx ezx

exy eyy ezy

exz

3

2

1000 Am=m

7 6 eyz 5 ¼ 4 500 Arad ezz 250 Arad

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.

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

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
106 ðm=mÞ2 W ¼ 2:031
109 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
106 Þepn  ð2:031
109 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
106 Þ  ð0:004500Þ2 a¼ 3 3 ¼ 1:125
106  1  3 2He  9He Ue þ 27We b¼ 27 1 h 2ð0:004500Þ3  9ð0:004500Þð5:625
106 Þ b¼ 27 þ27ð2:031
109 Þ ¼ 3:435
1010 sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3 b b2 a3 þ A¼  þ 2 4 27 vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi u u 3 10 t 3:435
10 ð3:435
1010 Þ2 ð1:125
106 Þ3 þ þ ¼ 2 4 27 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3 A ¼ 1:718
1010 þ ið1:524
1010 Þ A ¼ 594:5
106 þ ið146:7
106 Þ sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3 b b2 a 3 þ B¼   2 4 27 vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi u u 3 3:435
1010 t ð3:435
1010 Þ2 ð1:125
106 Þ3 ¼  þ 2 4 27

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

B¼

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3 1:718
1010  ið1:524
1010 Þ

B ¼ 594:5
106  ið146:7
106 Þ 8 H 2689Am=m > > AþBþ > > > >  3  > < A þ B pffiffiffiffiffiffiffi A  B H þ 3  þ ep1 ; ep2 ; ep3 ¼ 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 y  exz cpn z ¼ 0 exy cpn x þ ðepn  eyy Þcpn y  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 y  250cp1 z ¼ 0 500cp1 x þ ð2689  1500Þcp1 y  750cp1 z ¼ 0 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 y  250cp2 z ¼ 0 500cp2 x þ ð1160  1500Þcp2 y  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

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

Direction cosines for ep3. The three independent equations become: ð651  1000Þcp3 x  500cp3 y  250cp3 z ¼ 0 500cp3 x þ ð651  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

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

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 3
3 array: 9 8 9 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: 2 3 2 32 3 e xV xV e xV y V e xV z V c xV x c xV y c xV z exx exy exz 6e V V e V V e V V 7 6c V 6 7 cyV y cyV z 7 4 yx 54 eyx eyy eyz 5 yy yz 5 ¼ 4 yx czV 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 ð90B Þ ¼ 0

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

cyV x ¼ cos ð90B þ hÞ ¼ sin ðhÞ cyV y ¼ cos 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: 2 3 2 32 3 cos h sin h 0 exx exy 0 exV xV exV yV exV zV 6e V V e V V e V V 7 6 76 7 eyy 0 5 4 yx yy y z 5 ¼ 4 sin h cos h 0 54 eyx 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

e xV xV

6 6 e yV x V 4 e z V xV

e x V yV e yV yV e z V yV

exV zV

3

7 e yV z V 7 5 e zV zV

2

3 cos2 hexx þ sin2 heyy þ 2 cos h sin hexy cos h sin hexx þ cos h sin heyy þ ðcos2 h  sin2 hÞexy 0 2 2 ¼4cos h sin hexx þ cos h sin heyy þ ðcos2 h  sin hÞexy sin hexx þ cos2 heyy  2 cos h sin hexy 05 0 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  2 cosðhÞsinðhÞexy

ð42Þ

exV yV ¼ cosðhÞsinðhÞexx þ cosðhÞsinðhÞeyy þ ½cos ðhÞ þ sin ðhÞexy 2

2

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

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

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 ðhÞ 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

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

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: 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: sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi  e  e 2  c 2 exx þ eyy xx yy xy ð47Þ F ep1 ; ep2 ¼ þ 2 2 2 e2pn  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 2 exx  eyy 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).

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

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

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

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): 8 9 2 3 exV xV > > cos2 ð25B Þ sin2 ð25B Þ 2 cosð25B Þsinð25B Þ > > <e = 6 7 yV yV ¼6 cos2 ð25B Þ 2cosð25B Þsinð25B Þ 7 sin2 ð25B Þ 4 5 > 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 hpe ¼ arctan ¼ 29:5B 2 500 þ 1000

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

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

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

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

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

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 ¼ þ ByBz Bz2 By2

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

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 B3 v ¼  2 BxByBz BxBz Bx Bz B3 u B2 cxz B3 w  2 ¼ BxByBz BxBy Bx 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 ¼ þ  2 ð50dÞ ByBz Bx Bz By 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

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

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.

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

(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 3 cos w cos h  sin w sin b sin h cos w sinh þ sin w sin b cos h sin w cos b ¼4 cos b sin h cos b cos h sin b 5 sin w cos h þ cos w sin b sin h sin w sin h  cos w sin b cos h 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.

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

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

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

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

3

1500 Am=m

7 750 Arad 5

750 Arad

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 500 Arad exz 7 6 eyz 5 ¼ 4 500 Arad 1500 Am=m ezz 250 Arad 750 Arad

250 Arad 750 Arad

3 7 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 strain components has been reversed.): 2

exx 6 4 eyx ezx

exy eyy ezy

3 2 1000 Am=m 500 Arad exz 7 6 eyz 5 ¼ 4 500 Arad 1500 Am=m ezz 250 Arad 750 Arad

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

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.

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

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

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

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

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

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

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

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.

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

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.

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

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.

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

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.

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

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

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

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

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

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

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

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

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

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

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

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 cxz ¼ e rxx 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 vyx u e ðorÞ exx ¼  ryy yy Eyy vyz ezz vyz u e ðorÞ ezz ¼  ryy yy Eyy cxy gyy;xy gyy;xy u e ðorÞ cxy ¼ e ryy 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

ð6Þ

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 vzy u e ðorÞ eyy ¼  rzz 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 gzz;yz u e ðorÞ cyz ¼ e rzz zz zz

ð7Þ

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

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.

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

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

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

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 exx ¼ sxy eyy ¼ sxy ezz ¼ sxy ð12Þ 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 cyz ¼ sxz 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 cxy ¼ syz Gyz lyz;xz cxz ¼ syz Gyz gyz;xx exx ¼ syz Gyz gyz;yy eyy ¼ syz Gyz gyz;zz ezz ¼ syz Gyz cyz ¼

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

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

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

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 e33 e22 g22;12 ¼ g22;13 ¼ g22;23 ¼ 0 v23 u

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

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.

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

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.

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



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.

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

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

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

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).

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

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.

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

expected, as DT is increased, the magnitude of all strain components also increases. For modest changes in temperature (say, DT<200 jC), the change in temperature is linearly related to the resulting thermal strain components. That is, the slopes of the six strain vs. DT curves shown in Fig. 10 are constant at relatively low levels of DT. At high levels of DT, the slopes of the curves typically increase. For polymeric composites, the temperature at which the curves become nonlinear is related to the glass transition temperature Tg of the polymeric matrix (the glass transition temperature of a polymer is discussed in Sec. 2 of Chap. 1). Most composite structures are designed to operate at temperatures below the Tg; hence, we will focus our attention on the linear range shown in Fig. 10. Because a change in temperature causes six strains to develop for anisotropic composites, six coefficients of thermal expansion must be defined: eTxx DT cTxy axy u DT

axx u

ayy u

eTyy DT

cT axz u xz DT

eTzz DT cTyz ayz u DT azz u

ð21Þ

Because we have limited our discussion to the linear range shown in Fig. 10, the coefficients of thermal expansion defined by Eq. (21) equal the slopes of the corresponding strain vs. DT curves within the linear range. These properties will henceforth be called linear coefficients of thermal expansion. In SI units, they are usually reported in terms of Am/m jC or Arad/jC (for normal or shear strains, respectively). In English units, they are usually reported in terms of Ain./in. jF or Arad/jF. A CTE can be converted from SI units to English units by multiplying by the factor 5/9. For example, a CTE of 15 Am/m jC equals 8.3 Ain./in. jF. Equation (21) can be easily rearranged and written in matrix form as follows: 3 3 2 T 2 exx cTxy cTxz axx axy axz 7 7 6 6 7 7 6 T 6 6 cyx eTyy cTyz 7 ¼ ðDTÞ6 axy ayy ayz 7 ð22Þ 7 7 6 6 5 5 4 4 cTzx cTzy eTzz axz ayz azz The reader should note that the strains caused by a change in temperature can be transformed from one coordinate system to another, in exactly the same way that mechanically induced strains are transformed. In particular, any of the strain transformation equations reviewed in Chap. 2 (e.g., Eq. (36), Eq. (41), Eq. (43), Eq. (44), or Eq. (47)) can be used to transform thermally induced strains from one coordinate system to another.

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

3.1 Specialization to Orthotropic and Transversely Isotropic Composites We will now apply these general definitions to the case of a unidirectional composite referenced to the principal material coordinate system. It will be seen later that CTEs of an anisotropic composite (i.e., a composite referenced to a nonprincipal material coordinate system) can always be related to those measured in the 1–2–3 coordinate system. Although unidirectional composites are used in the following discussion, a equivalent discussion applies to woven or braided composites, when referenced to the principal material coordinate system. A unidirectional composite subjected to a uniform change in temperature DT and referenced to the 1–2–3 coordinate system is shown in Fig. 11. Because the fibers are distributed symmetrically with respect to the 1 - , 2-, and 3-coordinate axes, a change in temperature does not cause a shear strain to develop. Hence, in the principal material coordinate system, Eq. (21) becomes: a11 ¼

eT11 DT

a22 ¼

eT22 DT

a33 ¼

eT33 DT

a12 ¼ a13 ¼ a23 ¼ 0

ð23Þ

As before, if the fibers are distributed uniformly throughout the 2–3 plane, then a22=a33 and the composite will be transversely isotropic. If the fibers are not uniformly distributed, then a22 p a33, and the composite will be orthotropic. Woven or braided composites are always orthotropic because

Figure 11 Deformations induced in a unidirectional composite by a change in temperature DT (deformations are shown greatly exaggerated for clarity).

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

the distribution of fibers in the 2-direction differs substantially from that in the 3-direction. Experimental methods of measuring thermal expansion coefficients for polymeric composites have not been standardized by organizations such as the ASTM. In practice, the in-plane thermal expansion coefficients exhibited by polymeric composites are most commonly determined through the use of resistance foil strain gages. Strain gage manufacturers often provide recommendations for measuring thermal expansion coefficients using strain gages (e.g., see Ref. 4). Typical CTEs for several common polymeric composites are included in Table 3. 4 MATERIAL PROPERTIES RELATING MOISTURE CONTENT TO STRAIN It is possible for water molecules to diffuse into (or out of ) the overall molecular structure of polymeric materials. In other words, the moisture content of polymeric-based materials, including composites, slowly varies as the relative humidity of the surrounding atmosphere varies. The moisture content of a polymer is usually expressed as a percentage by weight, and typically ranges from f0% to as high as f5%. From a structural point of view, the effects of a change in moisture content are analogous to those caused by a change in temperature. For example, a plot of strains as a function of moisture content would resemble Fig. 10, except that DM would be plotted along the horizontal axis rather than DT. Hence, if an unconstrained anisotropic composite is subjected to a uniform change in moisture content DM, then six components of strain will M M M M M M be induced: exx , eyy , ezz , cxy , cxz , and cyz . The superscript ‘‘M’’ has been used to indicate that these strains are caused solely by a change in moisture content. Strains induced by a change in moisture content are sometimes referred to as hygroscopic strains, and can be just as large as or larger than those associated with a change in temperature. In this text, it will be assumed that strain is linearly related to changes in moisture content. Properties that relate strains to changes in moisture content will be called linear coefficients of moisture expansion, abbreviated as ‘‘CMEs,’’ and will be denoted using the symbol b. Because a change in moisture content causes six strains to develop for anisotropic composites, six CMEs must be defined: bxx u bxy u

eM xx DM cM xy DM

byy u

eM yy DM

cM bxz u xz DM

bzz u byz u

eM zz DM cM yz DM

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

ð24Þ

The units of the CMEs are typically Am/m/%M or Arad/%M. Equation (24) can be easily rearranged and written in matrix form as follows: 2

eM xx

6 6 M 6 cyx 6 4 cM zx

cM xy

cM xz

eM yy

cM yz

cM zy

3

2

bxx

7 6 7 6 7 ¼ ðDMÞ6 bxy 7 6 5 4 M bxz ezz

bxy byy byz

bxz

3

7 7 byz 7 7 5 bzz

ð25Þ

A comparison between Eqs. (24) and (21), or between Eqs. (25) and (23) will reinforce the fact that strains induced by a change in moisture content are analogous (in a mathematical sense) to those caused by a change in temperature. 4.1 Specialization to Orthotropic and Transversely Isotropic Composites In the principal material coordinate system, there is no coupling between a change in moisture content and shear strains. Hence, in the principal material coordinate system, Eq. (24) becomes: b11 u

eM 11 DM

b22 u

eM 22 DM

b33 u

eM 33 DM

ð26Þ

b12 ¼ b13 ¼ b23 ¼ 0 If the composite is based on a unidirectional fabric and fibers are distributed uniformly throughout the 2–3 plane, then b22 = b33 and the composite will be transversely isotropic. If the fibers are not uniformly distributed, or if the composite is based on woven or braided fabrics, then b22 p b33 and the composite will be orthotropic. Recommended experimental methods of measuring linear coefficients of moisture expansion are described in the ASTM standard D5229 [5]. Typical CMEs for several common polymeric composites are included in Table 3.

5 MATERIAL PROPERTIES RELATING STRESS (OR STRAIN) TO FAILURE Material properties that relate stress or strain to failure are measured during either a uniaxial test or a pure shear test. These properties are referred to collectively as material strengths. Before we begin our discussion of material strengths for composite materials, let us briefly review the definitions of

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

material strengths customarily used by engineers to report the failure response of conventional structural materials such as metals or metal alloys. Three idealized plots of the axial strain exx measured during a uniaxial test are shown in Fig. 12a–c. These figures are similar to Fig. 4, except that now: (a) only one of the strain components caused by rxx has been plotted (specifically, only exx has been plotted), and (b) data have been included at high nonlinear stress levels, up to and including the stress level at which the specimen fractures into two (or more) pieces. Materials that exhibit a stress– strain response similar to Fig. 12a are called brittle materials. As indicated, for a brittle material, the stress–strain curve is nearly linear at all stress levels, up to and including the final fracture stress. In fact, a perfectly brittle material exhibits no nonlinear behavior at all; stress is linearly related to strain at all levels, up to final fracture. In contrast, Fig. 12b and c shows the stress–strain behavior for ductile materials. The characteristic feature that distinguishes the two material types is that a ductile material exhibits a far larger region of nonlinear behavior prior to failure than does a brittle material. A second distinction is that the maximum strain a ductile material can withstand prior

Figure 12 Idealized plots of the axial strain qxx caused by application of a tensile uniaxial stress jxx. (a) Stress-strain plot for a brittle material. (b) Stress-strain plot for a modestly ductile material. (c) Stress-strain plot for a highly ductile material.

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

Figure 12 Continued.

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

to final fracture is (in general) far higher than that of a brittle material. The toughness of a material is defined as the area under the entire stress–strain curve. Hence, brittle materials typically possess a low toughness when compared with ductile materials. Four common measures of material strengths (illustrated in Fig. 12c) in terms of stress are the proportional limit rp the percent offset yield strength ry, the ultimate strength ru, and the fracture strength rf. Although the axial strain that corresponds to each of these stress levels can also be defined as a measure of strength, typically only the percent offset yield strain ey and the fracture strain ef are customarily used in this fashion. Many materials exhibit distinctly different strengths in tension vs. compression. Hence, in this text, a superscript ‘‘T’’ or ‘‘C’’ will be used to indicate whether a given strength has been measured in tension or compression, respectively. For example, because Fig. 12 represents stress–strain curves measured in tension, the superscript associated with all measures of strength that appear in these figures includes the character ‘‘T.’’ The proportional limit rxxpT is defined as the maximum stress level at which stress is linearly related to strain. However, for many materials, the deviation from linearity is so gradual that it is difficult to precisely identify the proportional limit based on experimental measurements. In these cases, the stress level at which significant nonlinear behavior begins is defined based on the ‘‘offset method.’’ A line parallel to the initial (linear) region of the stress–stain curve is drawn on the stress–strain curve, but is offset along the strain axis by some standard amount. The stress level at which this offset line intersects the experimental stress–strain curve is defined as the percent offset yield strength ryT xx . A strain offset of either 0.001 or 0.002 m/m (0.1% or 0.2% strain, respectively) is most commonly used, and hence yield strengths measured in this manner are reported as the 0.1% offset yield strength or the 0.2% offset yield strength, respectively. The ultimate strength ruT xx is defined as the maximum stress the material can withstand. The fracture stress rfT xx is defined as the stress that exists at final fracture. Note from Fig. 12 that a given material may not exhibit all four types of material strength. For example, for modestly ductile materials (Fig. 12b), fracture usually occurs at the maximum stress level, and hence for this type of material, there is no distinction between the ultimate strength and the fracture strength. For brittle materials (Fig. 12a), there may be little or no nonlinear behavior prior to fracture, and hence a yield strength is not defined and only the fracture strength (or ultimate strength) would be reported. All of the above definitions may be applied to the data collected during a pure shear test as well. An idealized plot of the shear strain cxy measured during a pure shear test of a modestly ductile material is shown in Fig. 13. This figure is similar to Fig. 5, except that now: (a) only one of the strain

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

Figure 13 Idealized plot of the shear strain cxy caused by application of pure fP fP shear stress sxy, including the shear fracture stress and strains sxy and cxy , and yP yP the shear yield stress and strain sxy and cxy .

components (cxy) caused by sxy has been plotted, and (b) data have been included at high nonlinear stress levels, including the shear stress level at which the specimen fractures into two (or more) pieces. Because anisotropic materials may exhibit different strengths in different directions, strength must be measured in three orthogonal directions; the x-, y-, and z-directions. It is possible for anisotropic materials, including many composites, to exhibit brittle behavior in one direction (say, the x-direction) but ductile behavior in other directions (the y-direction and/or z-direction). 5.1 Specialization to Unidirectional Composites Strengths measured for composites are referenced to the principal material 1– 2–3 coordinate system. In the case of unidirectional composites, strengths are measured parallel to the fibers (i.e., parallel to the 1-axis) and transverse to the fibers (parallel to the 2- and 3-axes). As would be anticipated, the strength of a composite in the 1-direction is determined primarily by the strength of the

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

fibers, whereas strengths in the 2- and 3-directions are determined primarily by the strength of the matrix. Most fibers used in polymeric composites, such as graphite or glass fibers, are nearly perfectly brittle materials. Therefore, most unidirectional polymeric composites exhibit a brittle failure response in the 1-direction. Prior to about 1980, most commercially available polymeric composites were based on relatively brittle thermoset matrices such as epoxies. Consequently, composite material systems available prior to about 1980 exhibited a brittle response in the 2- and 3-directions as well, and exhibited a low overall toughness. Substantial research efforts to develop tougher composite material systems were conducted throughout the 1970s and 1980s, and new-generation composite material systems with substantially increased toughness are now commercially available. This increase in toughness was accomplished primarily by replacing the brittle polymeric matrices with tougher (i.e., more ductile) polymers or polymer blends. Two approaches were used. In the first approach, the toughness of inherently brittle thermosets (such as epoxies) was increased through the addition of a second ductile rubber phase. This resulted in a class of toughened matrices called ‘‘rubber-toughened’’ epoxies. The second approach was to replace the inherently brittle thermoset matrix with a more ductile thermoplastic polymer.* Examples of this latter approach include the use polyketone, polyamide, or polyimide polymers as matrix materials. From a purely structural standpoint, the net result of these developments is that most new composite material systems exhibit a ‘‘modestly ductile’’ response to failure in the 2- and 3-directions, qualitatively similar to Fig. 12b. Consequently, new-generation composites exhibit a much-improved toughness relative to older systems. Methods to characterize the strengths of composite for purposes of structural design have not been widely standardized within the composites industry. Therefore, in this text, composite strengths will be described using terminology similar to that traditionally used with metals and metal alloys. A r11 vs. e11 curve measured for most unidirectional polymeric composites resembles Fig. 12a. That is, in the fiber direction, unidirectional composites exhibit nearly perfectly brittle behavior, although very modest nonlinear behavior may begin to occur at stress levels approaching final fracture. Due to the limited nonlinear response, it is not appropriate to define a yield stress in the 1-direction. Throughout this text, it will be assumed that a uniaxial stress in the fiber direction r11 causes failure due to fracture. That is, at failure, the

* The difference between a thermoset and a thermoplastic polymer is discussed in Sec. 2.4 of Chap. 1.

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

specimen breaks into two or more pieces. The stress that causes fracture is fT fC or r11 , depending on whether r11 is tensile or compressive, denoted r11 fT fC respectively. Similarly, the strain at failure is denoted e11 or e22 . Because very little nonlinear behavior occurs prior to failure, it is possible to relate the fracture stress and strain using Young’s modulus. That is, for most composites, the measured fracture stress and measured fracture strain in the 1direction may be related (to within engineering accuracies) using Young’s modulus: fT rfT 11 i ðE11 Þðe11 Þ

fC rfC 11 i ðE11 Þðe11 Þ

On the other hand, for most modern unidirectional polymeric composites, both the r22 vs. e22 and the r33 vs. e33 curves resemble Fig. 12b. That is, the response measured in the 2- and 3-directions is usually ‘‘modestly ductile.’’ In this case then, ‘‘failure’’ may be defined either on the basis of a tensile yield stress/strain (i.e., on the basis of yielding), or a tensile fracture stress/strain (i.e., on the basis of fracture). In this text, the stress and strain values present at the onset of yielding yT yT yT yC yC yC yC will be denoted (ryT 22 ; e22 ; r33 ; e33 Þ or ðr22 ; e22 ; r33 ; e33 Þ; where the superscript ‘‘T’’ or ‘‘C’’ has once again been used to indicate whether the yield stress/ strain is measured in tension or compression, respectively. Note that it is possible to relate the tensile yield stress and strain (to within engineering accuracy) using Young’s modulus: yT ryT 22 i ðE22 Þðe22 Þ

yC ryC 22 i ðE22 Þðe22 Þ

yT ryT 33 i ðE33 Þðe33 Þ

yC ryC 33 i ðE33 Þðe33 Þ

fT fT , e22 , The stress and strain values present at fracture are denoted (r22 fT fC fC fC fC e33 ) or ðr22 ; e22 ; r33 ; e33 Þ: Because most modern composites exhibit a modestly ductile response in the 2- and 3-directions, the fracture stress is less than the value that would be calculated using a failure strain and Young’s modulus: fT , r33

fT rfT 22 < ðE22 Þðe22 Þ

fC rfC 22 < ðE22 Þðe22 Þ

fT rfT 33 < ðE33 Þðe33 Þ

fC rfC 33 < ðE33 Þðe33 Þ

Shear stress–strain curves (i.e., a plot of s12 vs. c12, s13 vs. c13, or s23 vs. c23) for unidirectional composites typically exhibit a shape somewhere between Fig. 12b and c. That is, the shear response is usually more ductile than that measured for normal stress in the 2-direction, and in some cases may be considered to be ‘‘highly ductile.’’ In the principal material coordinate

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

system (e.g., in the 1–2 coordinate system), the shear response is insensitive to the algebraic sign of the shear stress. That is, in the 1–2 coordinate system, an identical s12 vs. c12 curve will be measured regardless of whether s12 is positive or negative.* As before, a ‘‘shear failure’’ may be defined on the basis of a yield shear stress/strain or a shear fracture stress/strain, depending on application. In this text, the shear stress and strain at yielding will be denoted y y y y y y s12 and c12 (or s13 and c13 , or s23 and c23 ), whereas the shear stress and strain at f f f f f f and c13 , or s23 and c23 ). As before, fracture will be denoted s12 and c12 (or s13 the yield shear stress/strain may be related using the shear modulus, but the shear fracture stress/strain may not: sy12 iðG12 Þðcy12 Þ

sy13 iðG13 Þðcy13 Þ

sy23 iðG23 Þðcy23 Þ

sf12 < ðG12 Þðcf12 Þ

sf13 < ðG13 Þðcf13 Þ

sf23 < ðG23 Þðcf23 Þ

The preceding discussion is for orthotropic composites. Following the discussion presented in Sec. 2, if the composite is transversely isotropic, then the following strengths are related as indicated: yT ryT 22 ¼ r33

yC ryC 22 ¼ r33

fT rfT 22 ¼ r33

fC rfC 22 ¼ r33

sy12 ¼ sy13

sf12 ¼ sf13

Recommended methods of measuring in-plane composite strengths are included in most of the ASTM test standards previously listed in Table 1. Additional ASTM testing standards related specifically to failure of polymeric composites are listed in Table 2. Due to the thin platelike nature of composites, it is difficult to apply well-defined out-of-plane stresses, or to measure the resulting out-of-plane strains. Hence, in practice, the yield and fracture stresses associated with the out-of-plane 3-direction are measured infrequently. Typical yield and fracture strengths measured in the 1–2 plane for three common polymeric composites at room temperatures are included 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. In fact, due to ongoing research and development activities, the failure

* However, in a general nonprincipal material coordinate system, the shear strength is sensitive to the algebraic sign of the shear stress. This important point will be further discussed in Secs. 5 and 6 of Chap. 5.

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

Table 2

ASTM Test Standards Related to Failure of Polymeric Composites (see also Table 1) Designation D3479 D5766 D2344 D5528

E1922 D2290

Title Standard Test Method for Tension–Tension Fatigue of Polymer Matrix Composite Materials Standard Test Method for Open Hole Tensile Strength of Polymer Matrix Composite Laminates Standard Test Method for Apparent Interlaminar Shear Strength of Parallel Fiber Composites by Short-Beam Method Standard Test Method for Mode I Interlaminar Fracture Toughness of Unidirectional Fiber-Reinforced Polymer Matrix Composites Standard Test Method for Translaminar Fracture Toughness of Laminated Polymer Matrix Composite Materials Standard Test Method for Apparent Tensile Strength of Ring or Tubular Plastics and Reinforced Plastics by Split Disk Method

strengths of composites are improved more or less continuously. Therefore, the properties shown in Table 3 may not reflect those of currently available materials. Note that the failure strengths listed in Table 3 represent values measured at room temperatures. This will become an important factor in later chapters. In particular, in Chap. 7, we will consider methods of predicting failure of general composite laminates with multiple fiber angles. As will be seen, failure predictions for a multiangle composite laminate depend (in part) on the difference between the temperature at which the laminate is consolidated (this temperature is called the ‘‘cure temperature’’ throughout this text) and the ‘‘service temperature’’ at which the failure prediction is desired. For example, if a composite laminate is cured at a temperature of 175jC and then cooled to room temperature (20jC), the laminate has experienced a temperature change of 155jC during cooldown. Failure predictions for the laminate are then based on the temperature difference of 155jC, and on failure strengths measured at room temperature. If, instead, a composite is cured at 175jC and then cooled to some other service temperature, say, 100jC, then the laminate has experienced a temperature change of 75jC and failure predictions for the laminate are based on failure strengths measured at 100jC. It is pertinent to point out that the matrix-dominated tensile strengths exhibited by polymeric composites are often lower than the failure strength of the polymeric matrix alone. For example, the tensile strength of a

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

Table 3 Property E11 E22 v12 G12 fT r11 fC r11 yT r22 yC r22 fT r22 fC r22 y s12 f s12 a11 a22 b11 b22

Nominal Material Properties for Common Unidirectional Composites Glass/epoxy

Kevlar/epoxy

Graphite/epoxy

55 GPa (8.0 Msi) 16 GPa (2.3 Msi) 0.28 7.6 GPa (1.1 Msi) 1050 MPa (150 ksi) 690 MPa (100 ksi) 45 MPa (5.8 ksi) 120 MPa (16 ksi) 55 MPa (7.0 ksi) 140 MPa (20 ksi) 40 MPa (4.4 ksi) 70 MPa (10 ksi) 6.7 Am/m jC (3.7 Ain./in. jF) 25 Am/m jC (14 Ain./in. jF) 100 Am/m %M (100 Ain./in. %M) 1200 Am/m %M (1200 Ain./in. %M)

100 GPa (15 Msi) 6 GPa (0.90 Msi) 0.33 2.1 GPa (0.30 Msi) 1380 MPa (200 ksi) 280 MPa (40 ksi) 35 MPa (2.9 ksi) 105 MPa (15 ksi) 45 MPa (4.3 ksi) 140 Msi (20 ksi) 40 MPa (4.0 ksi) 60 MPa (9 ksi) 3.6 Am/m jC (2.0 Ain./in. jF) 58 Am/m jC (32 Ain./in. jF) 175 Am/m %M (175 Ain./in. %M) 1700 Am/m %M (1700 Ain./in. %M)

170 GPa (25 Msi) 10 GPa (1.5 Msi) 0.30 13 GPa (1.9 Msi) 1500 MPa (218 ksi) 1200 MPa (175 ksi) 50 MPa (7.25 ksi) 100 MPa (14.5 ksi) 70 MPa (10 ksi) 130 MPa (18.8 ksi) 75 MPa (10.9 ksi) 130 MPa (22 ksi) 0.9 Am/m jC (0.5 Ain./in. jF) 27 Am/m jC (15 Ain./in. jF) 50 Am/m %M (50 Ain./in. %M) 1200 Am/m %M (1200 Ain./in. %M)

nonreinforced bulk epoxy is commonly about 70 MPa (10 ksi), whereas from Table 3, we see that graphite–epoxy typically possesses a matrix-dominated tensile strength on the order of 50 MPa (7.25 ksi). Even more pronounced is the reduction in tensile strain at fracture: for a nonreinforced bulk epoxy, the tensile strain at fracture commonly ranges from about 1% to 5% (10,000– 50,000 Am/m), whereas for graphite–epoxy, the matrix-dominated tensile fT strain at fracture (e22 ) rarely exceeds about 0.7% (7000 Am/m). The relatively low matrix-dominated strengths exhibited by polymeric composites can be explained on the basis of micromechanics analyses [6–9]. Briefly, two factors lead to low matrix-dominated tensile strengths. The first is thermal stresses induced at the microlevel during cooldown from cure temperatures. Recall that the thermal expansion coefficient of most high-performance fibers is very low and, in fact, is often slightly negative. For example, thermal expansion coefficients of glass, Kevlar, and graphite fibers are about 5, 2, and 0.5 Am/m jC, respectively (see Sec. 3 of Chap. 1). In contrast, the thermal expansion coefficient of polymers is quite high and usually exceeds 30 Am/m jC. Consequently, during cooldown from cure temperatures, the matrix is restrained from thermal contraction by the fibers, leading to self-equilibrating

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

tensile stresses in the matrix and compressive stresses in the fibers. The second factor is the stress-concentrating effect of the fibers. Because most advanced composites are produced with a fiber volume fraction of about 0.65, the tensile stresses induced within the matrix surrounding a fiber are not dictated strictly by the CTE mismatch between matrix and fiber but are also influenced by the presence neighboring fibers. Together, these two factors give rise to thermal stresses at the microlevel that are generally tensile in the matrix and compressive in the fibers. The magnitude of the thermal stresses induced in the fiber is very low relative to the fiber strength. However, the magnitude of the tensile stresses induced in the matrix represents a substantial fraction of the tensile strength of the matrix alone. Numerical micromechanics analyses based on the finite element method have shown that thermal matrix stresses can often be 50–60% of the bulk matrix tensile strength [11–14]. Hence, these thermal stresses are responsible for the low matrix-dominated tensile strengths and tensile strain at fractures exhibited by composites. They also explain in a qualitative sense why the magnitudes of matrix-dominated compressive strengths are invariably higher than matrixyC yT fC fT dominated tensile strengths (i.e., r22 >r22 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

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

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

ð28aÞ

rf ¼ ef Ef ¼ e11 Ef

ð28bÞ

rm ¼ em Em ¼ e11 Em

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

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

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.

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

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  Vf Þ A A

where Vf is the volume fraction of fibers and Vm = (1Vf) 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

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

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 WVf vf e11 DWm ¼ 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 ½Vf vf þ ðVm Þvm e11 W

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

e22 ¼ V f vf þ V m vm e11

Noting as before that Vm=(1Vf), 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

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

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=(1Vf), and we obtain: E22 ¼

Ef Em Ef  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.

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

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 f m Gf þ Gm

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

Gf Gm Gf  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,

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

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 0:25

vyz ¼ 0:60

Gxy ¼ 60 GPa

Gxz ¼ 75 GPa

Gyz ¼ 50 GPa

gxx;xy ¼ 0:30

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 0:15

lxy;xz ¼ 0:10

lxy;yz ¼ 0:05

gxz;yz ¼ 0:10

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

(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 0:25

v23 ¼ 0:60

G12 ¼ 60 GPa

G13 ¼ 75 GPa

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’’ 5
5 mm square. In addition, ‘‘perfect’’ 5
5 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.

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

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

0 61 146 279 433 587 930

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

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.

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

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 1
1 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;

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

American Society for Testing and Materials: West Conshohocken, PA, USA (may also be accessed via the ASTM website at: http://www.astm.org/). 6. Hyer, M.W. Stress Analysis of Fiber-Reinforced Composite Materials; McGrawHill Book Co.: New York, NY; 1998; ISBN 0-07-016700-1. 7. Herakovich, C.T. Mechanics of Fibrous Composites; John Wiley and Sons: New York, NY, 1998; ISBN 0-471-10636-4. 8. Hull, D. An Introduction to Composite Materials; Cambridge University Press: Cambridge, Great Britain, 1981; ISBN 0-521-23991-5. 9. Gibson, R.F. Principles of Composite Material Mechanics; McGraw-Hill Inc.: New York, NY, 1994; ISBN0-07-023451-5.

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

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

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

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 ð1Þ rij ¼ 4 ryx ryy ryz 5 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 3 2 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

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

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 16

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 > > > e2 > S21 > > > 6 > > = 6 <e > 6 S31 3 ¼6 6S > > e > 6 41 > 4> > > 6 > > > 4 S51 > e5 > > > > ; : > e6 S61

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

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

itself be symmetrical. That is, all terms in symmetrical off-diagonal positions must be equal: S21 ¼ S12 S31 ¼ S13

S32 ¼ S23

S41 ¼ S14

S42 ¼ S24

S43 ¼ S34

S51 ¼ S15

S52 ¼ S25

S53 ¼ S35

S54 ¼ S45

S61 ¼ S16

S62 ¼ S26

S63 ¼ S36

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 2 9 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 ¼6 ð9Þ 7 r > cyz > > yz > > > > 6 S41 S42 S43 S44 S45 S46 7> > > > > > 6 > > 7> > > c > > > > 4 S51 S52 S53 S54 S55 S56 5> > > > > xz > > rxz > > > ; ; : : 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 causes ðfrom Eq: 3:14Þ exx ¼ syz 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

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

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

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

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

ð10bÞ



 cyz ¼

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

ð10cÞ



cxz

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

ð10dÞ



cxy

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

ð10eÞ

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

ð10f Þ

Equation (10a)–(10f) can be assembled in matrix form:            38 9 2  9 8 gyz;xx gxy;xx gxz;xx myx 1 mzx > > > > > > > r e xx > > > 7> > 6 Exx > > xx > > E E G G G > > > > yy zz yz xz xy 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 xx yy zz yz xz xy > > > > 7 6 > > > >             > > > > 7 > 6 mxz > > > g g g m 1 > > > > yz yz;zz xy;zz 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 > > 6 > > cyz > > 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 E G G 7 yy zz yz xz xy > > > > > > > > 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

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

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

ð13Þ lyz;xz lxz;yz ¼ Gyz Gxz lyz;xy lxy;yz ¼ Gyz Gxy

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

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

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

where i; j ¼ 16

ð14Þ

Cij is called the ‘‘stiffness matrix.’’* The stiffness matrix is the mathematical inverse of the compliance matrix Cij=Sij1. 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 > > > > 5> > > e5 > > : : ; ; 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.

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



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 > > = <

2 

 m21 6 E22 6 6    6 m21 1 6 6 E E22 6 11 6    6 m23 6 m13 6 6 E11 E22 6 ¼6 > > > 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

1 G13

0

0

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

2

S11

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

S12

S13

0

0

S22

S23

0

0

S23

S33

0

0

0

0

S44

0

0

0

0

S55

0

0

0

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 0

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

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 9 8 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 7> 0 0 C44 > > > > > > > > 7 6 > 4 > >s > > > c > 0 5> 0 0 0 0 C55 > > > 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

S13 S23  S12 S33 S S12 S13  S11 S23 ¼ S 1 ¼ S55

S12 S33  S13 S22 S S11 S22  S212 ¼ S 1 ¼ S66 ð20Þ

C11 ¼

C12 ¼

C13 ¼

C22

C23

C33

C44

C55

C66

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

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: ðE22  m223 E33 ÞE211 X ðm12 m23 þ m13 ÞE11 E22 E33 ¼ X ðm23 E11 þ m12 m13 E22 ÞE22 E33 ¼ X

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

C11 ¼

C12 ¼

C13

C22

C23

C44 ¼ G23

C55 ¼ G13

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 6 E11 7> > E22 E22 > > 6 7 > > > 6 m   1   m  7> > > > 6 7> 21 32 > > > 0 0 0 7> r 6 > 22 > > > 7> 6 E11 > E22 E22 > > 7> 6 >      > > > > 7 6 m12 m 1 > > 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 7 0 0 0 > > > > > > > 7> G22 > 6 > > > > > > > 7 6 > > > >   > > > > 7> 6 > > > 1 > 6 > > > 7 > > > c s 13 > 0 7> 0 0 0 0 > > 13 > 6 > > > > > > > > G12 7 6 > > > > > > > > 7 6   > > > > > > > > 5 4 > > >c > > 1 > > ; : 12 ; : s12 > 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 <e > 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  S23 Þ 0 0

0 0 0 0 S66 0

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

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 C12 ¼ C22 ¼ X Xð1 þ m23 Þ X 2 E22 ðm23 E11 þ m12 E22 Þ C22  C23 E22 C44 ¼ ¼ C66 ¼ C12 ¼ Xð1 þ m23 Þ 2 2ð1 þ m23 Þ

C11 ¼ C23

ð26Þ where: X ¼ E11 ð1  m23 Þ  2m212 E22

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

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 9 8 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 E11 170 GPa 1012 Pa 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 E11 170 GPa 1012 Pa

In this case, Eq. (17) becomes: 9 2 8 5:88 1:76 e11 > > > > > > > > 6 1:76 100:0 e > > 22 > > = 6 < 6 2:06 40:0 e33 ¼6 6 0 > > 6 0 > c23 > > > > > 4 0 0 c13 > > > > ; : 0 0 c12

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

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

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

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

8 8 9 9 350 > 1966 Am=m > > > > > > > > > > > 35 > > > 2284 Am=m > > > > > > > > > < = < = 15 246 Am=m 6
ð10 PaÞ ¼ 30 > 3750 Arad > > > > > > > > > > > > > > 10 1000 Arad > > > > > > > > > ; : ; : 25 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

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

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 12:0 266 1 667 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 9 8 8 9 50 > 1966 Ain:=in: > > > > > > > > > > > > > 5 > 2203 Ain:=in: > > > > > > > > > < = < = 2 596 Ain:=in: 3
ð10 psiÞ ¼ 4 > 3200 Arad > > > > > > > > > > > > > > 750 Arad > 1:5 > > > > > > > > : ; : ; 1750 Arad 3:5 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: 2 2 X ¼ E11 E22  m212 E2 22  m13 E22 E33  m23 E11 E33  2m12 m13 m23 E22 E33

X ¼ fð170Þð10Þ  ð0:30Þ2 ð10Þ2  ð0:35Þ2 ð10Þð8Þ  ð0:40Þ2 ð170Þð8Þ 2ð0:30Þð0:35Þð0:40Þð10Þð8ÞgðGPaÞ2 X ¼ 1457ðGPaÞ2

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

ðE22  m223 E33 ÞE211 fð10 GPaÞ  ð0:40Þ2 ð8 GPaÞgð170 GPaÞ2 ¼ X 1457ðGPaÞ2 ¼ 172:98 GPa

C11 ¼

C12 ¼ ¼

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  ð0:35Þ2 ð8 GPaÞgð10 GPaÞ ¼ X 1457ðGPaÞ2 ¼ 11:602 GPa

C22 ¼

C23 ¼ ¼

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

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

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

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

3 0 0 7 7 0 7 7ðGPaÞ 0 7 7 0 5 13:0

8 9 8 9 1500 Am=m > 245:5 MPa > > > > > > > > > > > > > 2000 Am=m > 19:78 MPa > > > > > > > > > < = < = 1000 Am=m 10:29 MPa
¼ 2500 Arad > > > 20:0 MPa > > > > > > > > > > > > 500 Arad 5:00 MPa > > > > > > > > > ; : : ; 2000 Arad 26:0 MPa Example Problem 4 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  2m212 E22 X ¼ ð25 MsiÞð1  0:40Þ  2ð0:30Þ2 ð1:5 MsiÞ ¼ 14:73 Msi C11 ¼

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

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

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 0:5357 Msi ¼ 2ð1 þ m23 Þ 2ð1 þ 0:40Þ

C66 ¼ G12 ¼ 2:0 Msi Applying Eq. (24), the stresses are: 8 9 2 25:46 0:7637 0:7637 0 0 r11 > > > > > > > > 6 0:7637 1:809 0:07372 0 0 r > > 22 > > < = 6 6 0:7637 0:7372 1:809 0 0 r33 ¼6 6 0 0 0:5357 0 > > s23 > > 6 0 > > > > 4 0 0 0 0 2:0 s13 > > > > ; : 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: 2

eTxx 6 eT 4 yx eTzx

eTxy eTyy eTzy

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

axy ayy

3 axz ayz 7 5

azy

azz

ðrepeatedÞð3:22Þ

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

eM xx

6 cM 4 yx cM zx

cM xy

cM xz

3

2

bxx

eM yy

6 7 cM yz 5 ¼ ðDMÞ4 bxy

cM zy

eM zz

bxz

bxy

bxz

3

byy

byz 7 5

byz

bzz

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

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 9 8 T 9 8 M9 8 > a11 > e11 > e11 > b11 > > > > > > > > > > > > > > > > T > > > > > > > > > > > > > M> > > > > > > > > a e e b > > > > > > > > 22 22 22 22 > > > > > > > > > > > > > > > > > > > > > > > = < eT = < eM = 33 33 33 33 ðandÞ ð28Þ ¼ DT ¼ DT > > > > > cT23 > cM 0 > 0 > > > > > > > > 23 > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > >0 > > > cT13 > > > > cM 0 > > > > > > > > 13 > > > > > > > > > > : : T ; : M; ; ; : > 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 M9 8 b11 > e11 > > > > > > > > > > > > eM > > > > > > > b > > > > 22 22 > > > > > > > > > > > < eM = = 22 33 ¼ DT > > > cM 0 > > > > 23 > > > > > > > > > > > > > > > > M > > c13 > > > 0 > > > > > > > > ; : M; : c12 0

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

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

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 > > > > > <e > =

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 8 9 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=Sij1, as discussed in Sec. 1.

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

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 > ; s12 0 0 0 S66 8 9 8 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).

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

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 9 8 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 9 8 8 a11 > b11 > > > > > > > > > > > > > a22 > b > > > > > > > > = = < < 22 > a22 b22 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 0 0 ðC s  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).

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

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 0:25

vyz ¼ 0:60

Gxy ¼ 60 GPa

Gxz ¼ 75 GPa

Gyz ¼ 50 GPa

gxx;xy ¼ 0:30

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 0:15

lxy;xz ¼ 0:10

lxy;yz ¼ 0:05

gxz;yz ¼ 0:10

axx ¼ 5 Am=m jC

ayy ¼ 10 Am=m jC

azz ¼ 20 Am=m  jC

axy ¼ 5 Arad=jC

axz ¼ 15 Arad=jC

ayz ¼ 25 Arad=jC

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

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.

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

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

eyy ¼ 2000 Am=m cxz ¼ 500 Arad

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: 2

r11

6 rij ¼ 4 r21 r31

r12 r22 r32

r13

3

2

75 10 7 r23 5 ¼ 4 10 90 25 30 r33

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.

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

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.

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

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

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

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 2 9 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 > > > > > > > > <e = 6S < = < < = 7 a33 b33 = 0 0 7 r33 33 6 13 S23 S33 0 ¼6 þ DT þ DM 7 > > > > 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: 9 8 9 2 9 8 8 9 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 > > > > > > > > <e = 6S < = = < = < 7 b a 0 0 7 0 33 33 33 6 13 S23 S33 0 ¼6 þ DT þ DM 7 > > 6 0 > 0 > > > > > 0 7 0 0 S44 0 > > > > > c23 > > 6 > > > > 0 > > > 0 > 7> > > > > > > > > > > > > > > > > 7 6 > > > 0 0 0 S55 0 5> > c13 > > 4 0 > 0 > > > > 0 > > > 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 9 2 9 8 8 8 38 b S11 S12 0 > > > > = = = = < a11 > < 11 > < e11 > < 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

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

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 3
3 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 2 9 8 38 Q11 Q12 0 > > = = < r11 > < e11  DTa11  DMb11 > 6 0 7 r22 ¼ 4 Q12 Q22 ð3Þ 5 e22  DTa22  DMb22 > > > > ; ; : : s12 c12 0 0 Q66 The 3
3 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

S12 S22 0

0

31

0 7 5

ð4Þ

S66

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

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

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 e11 þ e22 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

S22 S11 S22  S212

Q12 ¼ Q21 ¼

Q22 ¼

S11 S11 S22  S212

Q66 ¼

S12 S11 S22  S212

1 S66

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

ð10Þ

Alternatively, the elements of the reduced stiffness matrix are related to more familiar engineering constants as follows: Q11 ¼

E211 E11  m212 E22

Q12 ¼ Q21 ¼

Q22 ¼

E11 E22 E11  m212 E22

Q66 ¼ G12

m12 E11 E22 E11  m212 E22

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.

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

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 8 9 6 E 9 78 E11 11 r11 > > 7> < e11 > < = 6 = 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).

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

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: 8 9 2 5:88
1012 1:76
1012 > < e11 > = 6 e22 ¼ 4 1:76
1012 100
1012 > > : ; c12 0 0 9 8 6 > = < 200
10 >
30
106 ðPaÞ > > ; : 50
106

0 0 76:9
1012

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 2 9 8 9 8 8 38 S11 S12 0 > > > > = < e11 > = < r11 > = < 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

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

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
1012 1:76
1012 0 > = < e11 > 6 7 e22 ¼ 4 1:76
1012 100
1012 0 5 > > ; : 12 c12 0 0 76:9
10 8 9 8 9 6 6 > > < 0:9
10 > = < 200
10 > =
30
106 þ ð155Þ þ ð0:5Þ 27
106 > > > > : ; : 6; 50
10 0 9 8 6 > = < 150
10 >
4800
106 > > ; : 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 9 8 > = < 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.

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

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 > = <0> = > < 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

9 38 0 > = < e11  DTa11  DMb11 > 0 7 5 e22  DTa22  DMb22 > > ; : c12 Q66

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

E211 ð170 GPaÞ2 ¼ 170:9 GPa ¼ E11  m212 E22 170 GPa  ð0:30Þ2 ð10 GPaÞ

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

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 10:05 GPa ¼ 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 Þ 150
106 = < 0  ð155Þ0:9
10  ð0:5  
0  ð155Þ 27
106  ð0:5Þ 4800
106 : ; 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=155jCc280jF. 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

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

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

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

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

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

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 > 6 7 e22 ¼ 4 S12 S22 0 5 r22 ð13Þ > > > > ; ; : : s12 0 0 S66 c12 Equation (13) can be rewritten as: 9 2 2 38 e 1 0 0 > S11 S12 < 11 > = 6 7 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 ½S ¼ 4 S12 S22 0 5 ½R 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 ½R ¼ ½S 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 
> > ; ; : c =2 > : c =2 > xy xy

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

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 9 8 9 > > < rxx > = < rxx > =
ryy ¼ ½T  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  ¼ 4 sin2 ðhÞ cos2 ðhÞ 2cosðhÞsinðhÞ 5 cosðhÞsinðhÞ cosðhÞsinðhÞ cos2 ðhÞ  sin2 ðhÞ Inserting these transformation relationships into Eq. (14), we have: 8 9 8 9 e > > < xx > = < rxx > = eyy ½R ½T  ¼ ¼ ½S ½T  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 8 9 e > > = = < xx > < rxx > eyy ¼ ½T 1 ½R 1 ½S ½T  ryy ð16Þ > > > > ; ; : : sxy cxy =2 where:

2

1

6 ½R 1 ¼ 4 0 0 and

2 6 ½T 1 ¼ 4

0 1 0

0

3

7 0 5 1=2

cos2 ðhÞ

sin2 ðhÞ

sin2 ðhÞ

cos2 ðhÞ

cosðhÞsinðhÞ

cosðhÞsinðhÞ

2cosðhÞsinðhÞ

3

7 2cosðhÞsinðhÞ 5

cos2 ðhÞ  sin2 ðhÞ

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

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 > > = = < exx > < rxx > ð17Þ ½R 1 eyy ¼ ½T 1 ½R 1 ½S ½T  ryy > > > > : : ; ; cxy sxy Multiplying both sides of Eq. (17) by the [R] matrix, we arrive at our final result: 9 9 8 8 > > = = < exx > < rxx > eyy ¼ ½R½T 1 ½R1 ½S½T  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 > =  > = < exx > < rxx > eyy ¼ S ryy ð19Þ > > > > ; ; : : cxy sxy where:   S ¼ ½R½T 1 ½R1 ½S½T  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 2 9 8 38 S11 S12 S16 > > = = < exx > < rxx > 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.

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

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  S66 Þcos3 h sin h  ð2S22  2S12  S66 Þ cos h sin3 h

S22 ¼ S11 sin4 h þ ð2S12 þ S66 Þcos2 h sin2 h þ S22 cos4 h S26 ¼ S62 ¼ ð2S11  2S12  S66 Þcos h sin3 h  ð2S22  2S12  S66 Þcos3 h sin h   S66 ¼ 2ð2S11 þ 2S22  4S12  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

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

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 1 eT22 > > > > > > ; ; : T > : > cxy =2 0 2 cos2 ðhÞ sin2 ðhÞ 6 2 ¼6 cos2 ðhÞ 4 sin ðhÞ cosðhÞsinðhÞ cosðhÞsinðhÞ 8 T 9 e > > > = < 11 >
eT22 > > > : T > ; c12 =2

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: 8 9 8 9 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

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

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

ð28Þ

bxy ¼ 2cosðhÞsinðhÞðb11  b22 Þ 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

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

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: 8 9 2 38 exx  DTaxx  DMb 9 rxx > xx > > > Q Q Q > > > > 11 12 16 = = < < 7 e  DTa  DMb 6 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  Q12  2Q66 Þcos3 h sin h ðQ22  Q12  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  Q12  2Q66 Þcos h sin3 h ðQ22  Q12  2Q66 Þcos3 h sin h

  Q66 ¼ ðQ11 þ Q22  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.

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

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
1012 cos4 ð30B Þ

   þ 2 1:76
1012 þ 76:9
1012 cos2 ð30B Þsin2 ð30B Þ   þ 100:0
1012 sin4 ð30B Þ   S11 ¼ 23:32
1012 Pa1 Calculating the remaining elements of the S matrix in similar fashion and applying Eq. (29), we find: 9 2 8 e 23:32
1012 4:327
1012 > < xx > = 6 eyy ¼ 4 4:327
1012 70:38
1012 > ; :c > xy 33:72
1012 47:79
1012 8 9 200
106 >  > < = 1
30
106 ðPaÞ > Pa > : ; 50
106

33:72
1012

3

7 47:79
1012 5 101:3
1012

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-

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

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
106 cos2 ð30jÞ   þ 27
106 sin2 ð30jÞ axx ¼ 6:1 Am=m=o C   ayy ¼ a11 sin2 ðhÞ þ a22 cos2 ðhÞ ¼ 0:9
106 sin2 ð30jÞ   þ 27
106 cos2 ð30jÞ ayy ¼ 20:0 Am=m=o C axy ¼ 2 cosðhÞsinðhÞða11  a22 Þ ¼ 2 cosð30jÞsinð30jÞ  
0:9
106  27
106 axy ¼ 24:2 Arad=o C The moisture expansion coefficients are:   bxx ¼ b11 cos2 ðhÞ þ b22 sin2 ðhÞ ¼ 150
106 cos2 ð30jÞ   þ 4800
106 sin2 ð30jÞ bxx ¼ 1313 Am=m=%M

  byy ¼ b11 sin2 ðhÞ þ b22 cos2 ðhÞ ¼ 150
106 sin2 ð30jÞ   þ 4800
106 cos2 ð30jÞ byy ¼ 3638 Am=m=%M bxy ¼ 2 cosðhÞsinðhÞðb11  b22 Þ ¼ 2 cosð30jÞsinð30jÞ  
150
106  4800
106 bxy ¼ 4027 Arad=%M

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

Hence, Eq. (29) becomes: 9 2 8 23:32
1012 > = < exx > 6 eyy ¼ 4 4:327
1012 > > ; : cxy 33:72
1012

4:327
1012 70:38
1012

33:72
1012

3

7 47:79
1012 5

47:79
1012 101:3
1012 8 9 8 9 6:1
106 > 200
106 > > > = = < < 6 þ ð0:5Þ
30
106 þ ð155Þ 20:0
10 > > > ; ; : : 24:2
106 > 6 50
10 9 8 6 > = < 1313
10 > 3638
106
> > ; : 4027
106

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

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

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 exx  DTaxx  DMbxx 9 > Q16 > = <  7 Q26 5 eyy  DTayy  DMbyy > > ; :  Q66 cxy  DTaxy  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  ð155Þ 6:1
106  ð0:5Þ 1313
106 > > <     =
0  ð155Þ 20:0
106  ð0:5Þ 3638
106 >    > ; : 0  ð155Þ 24:2
106  ð0:5Þ 4027
106 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).

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

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

ð33Þ

S22 ¼ US1  US2 cos 2h þ US3 cos 4h S26 ¼ S62 ¼ US2 sin 2h  2US3 sin 4h S66 ¼ US5  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

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

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 Q Q12 ¼ Q21 ¼ UQ 4  U3 cos 4h 1 Q16 ¼ Q61 ¼ UQ sin 2h þ UQ 3 sin 4h 2 2 Q Q Q22 ¼ UQ 1  U2 cos 2h þ U3 cos 4h 1 Q26 ¼ Q62 ¼ UQ sin 2h  UQ 3 sin 4h 2 2 Q Q66 ¼ UQ 5  U3 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,

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

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 0 5:88
1012 1:76
1012 1 4 S12 S22 0 5 ¼ 4 1:76
1012 100:0
1012 5 0 Pa 0 0 S66 0 0 76:9
1012 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
1012 8 8 i     12 þ 2 1:76
1012 þ 76:9
1012 þ3 100:0
10 ¼ 48:9
1012  1 1 US2 ¼ ðS11  S22 Þ ¼ 5:88
1012  100:0
1012 2 2 ¼ 47:1
1012

1 1h ðS11 þ S22  2S12  S66 Þ ¼ 5:88
1012 þ 100:0
1012 8 8 i   2 1:76
1012  76:9
1012 ¼ 4:06
1012 1 1h US4 ¼ ðS11 þ S22 þ 6S12  S66 Þ ¼ 5:88
1012 þ 100:0
1012 8 8 i   þ6 1:76
1012  76:9
1012 ¼ 2:296
1012 1 1h US5 ¼ ðS11 þ S22  2S12 þ S66 Þ ¼ 5:88
1012 þ 100:0
1012 2 2 i   2 1:76
1012 þ 76:9
1012 ¼ 93:17
1012 US3 ¼

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

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
1012     þ 47:1
1012 cosð60jÞ þ 4:06
1012 cosð120jÞ ¼ 23:3
1012 Pa1 Similarly, using the second equation of Eq. (33) and setting h=30j: S12 ¼ S21 ¼ US4  US3 cos 4h     ¼ 2:296
1012  4:06
1012 cosð120jÞ ¼ 4:33
1012 Pa1 The additional terms within the [S] matrix are calculated using the rest of Eq. (33). A summary of our results is: 2 3 23:3
1012 4:33
1012 33:7
1012   6 7 1 12 ½S  ¼ 6 70:4
1012 47:8
1012 7 4 4:33
10 5 Pa 33:7
1012 47:8
1012 101:3
1012 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

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

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 ¼ ¼ ð38aÞ Exx 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

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

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 cos h þ sin h  þ  cos2 h sin2 h E11 E11 E22 G12 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 ð40bÞ

or equivalently:  gxx;xy ¼

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

ð40cÞ

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

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 2 9 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

ð43aÞ

eyy ¼ S26 sxy

ð43bÞ

cxy ¼ S66 sxy

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

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

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

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

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 DT

axy ¼

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

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

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.

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

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.

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

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.

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

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 Fracture stress in the 1-direction: r11 , 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.

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

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 during the measurement of the fundamental material strength r11 (as

Figure 15 An illustration of what is meant by ‘‘coupling effects’’ of stress on the failure phenomenon.

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

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 at failure, then a coupling effect has ocFig. 8(b) is conducted and r11p r11 curred, 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

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

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 2 3 1 1 7 6 1 r11 r22 4  2 þ  2   2 5 fT r11 ryT ryT 22 33 2

3

ð52Þ

1 1 7 6 1 r11 r33 4  2   2 þ  2 5 fT r11 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: 2 3 ðr11 Þ2 ðr22 Þ2 ðs12 Þ2 1 1 7 6 1  fT 2 þ  2 þ  y 2  r11 r22 4  fT 2 þ  2   2 5 < 1 yT yT yT r11 s12 r11 r22 r22 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 r11 r22  fT 2 þ  2 þ  y 2   fT 2 < 1 r11 s12 r11 ryT 22

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

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 , but the compressive strength in the tensile strength in the 1-direction, r11 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

ð55Þ

þX55 s213 þ X66 s212 þ 2X12 r11 r22 þ 2X13 r11 r33 þ 2X23 r22 r33 < 1 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

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

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:

 fC 2 X1 rfC 11 þ X11 r11 ¼ 1 Solving for X1 and X11, we find: X1 ¼

1 1  fC rfT r 11 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 ryT 22



1 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 ryT 33



1 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

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

r33=s23=s13=s12=0. Substituting these values into the Tsai–Wu criterion and solving for X12 results in: i 1 h ð60Þ X12 ¼ 1  ðX1 þ X2 Þry  ðX11 þ X22 Þðry Þ2 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

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

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 cos2 ð45jÞ 2cosð45jÞsinð45jÞ 5 0 sin2 ð45jÞ > > : ; ; : s12 0 cosð45jÞsinð45jÞ cosð45jÞsinð45jÞ cos2 ð45jÞ  sin2 ð45jÞ or

9 8 8 9 8 9 rxx cos2 ð45jÞ r11 > > > > > rxx =2 > > > > > = > < < < = > = ¼ rxx =2 r22 ¼ 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

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

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 9 8 38 r11 > rxx > cos2 ðhÞ sin2 ðhÞ 2 cosðhÞsinðhÞ > > > > > > = 6 < < = 7 2 2 7 0 r22 ¼ 6 ð h Þ cos ð h Þ 2 cos ð h Þsin ð h Þ sin 5> 4 > > > > > > > ; : : ; 2 2 0 s12 cosðhÞsinðhÞ cosðhÞsinðhÞ cos ðhÞ  sin ðhÞ or equivalently: r11 ¼ rxx cos2 ðhÞ r22 ¼ rxx sin2 ðhÞ

ð64Þ

s12 ¼ rxx cosðhÞsinðhÞ 6.1.1 Maximum Stress Criterion Substituting Eq. (64) into Eq. (50), we obtain: 

rfC rfT 11 11 < rxx < 2 cos ðhÞ cos2 ðhÞ ðandÞ ryC 22 2

sin ðhÞ

< rxx <

ryT 22 sin2 ðhÞ

ð65Þ

ðandÞ jrxx j <

sy12 cosðhÞsinðhÞ

According to the maximum stress criterion, failure will not occur if these five inequalities are satisfied. The predicted tensile and compressive

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

fT fC failure strengths, rxx and rxx , respectively, for a hj off-axis graphite–epoxy laminate are therefore the smallest values returned by the following expressions:*

Tensile strength: rfT xx ¼

rfT 1500 MPa 11 ¼ cos2 ðhÞ cos2 ðhÞ

ð66aÞ

rfT xx ¼

ryT 50 MPa 22 ¼ sin2 ðhÞ sin2 ðhÞ

ð66bÞ

rfT xx

      75 MPa   sy12     ¼ ¼ cosðhÞsinðhÞ   cosðhÞsinðhÞ 

ð66cÞ

Compressive strength: rfC xx ¼

rfC 1200 MPa 11 ¼ cos2 ðhÞ cos2 ðhÞ

ð67aÞ

rfC xx ¼

ryC 100 MPa 22 ¼ 2 sin ðhÞ sin2 ðhÞ

ð67bÞ

rfC xx

      75 MPa   sy12     ¼ ¼ cosðhÞsinðhÞ   cosðhÞsinðhÞ 

ð67cÞ

Note that the superscript fT or fC has been used to denote the failure strength of the unidirectional composite. It should be understood that in this context, ‘‘failure’’ may represent fracture of the fibers or yielding of the matrix. Equations (66a)–(66c) and (67a) (67b) (67c) were used to create the failure envelope for a unidirectional graphite–epoxy laminate shown in Fig. 16. Equations (66a)–(66c) and (67a) (67b) (67c) bound the ‘‘safe’’ region. The reader should note the following: 

The failure envelope shown in Fig. 16 is valid for a uniaxial state of stress only. Specifically, Fig. 16 is valid only if: rxx p 0 ryy ¼ rzz ¼ sxy ¼ sxz ¼ syz ¼ 0 As will be seen later, failure envelopes for other states of stress differ substantially from Fig. 16.

* As before, compressive strength is treated as an algebraically positive number.

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

Figure 16 Failure envelope for a unidirectional graphite–epoxy laminate subjected to a uniaxial stress rxx, based on the maximum stress criterion. 

The mode of failure depends on whether rxx is tensile or compressive, and also on the fiber angle h:

If rxx is tensile, then: Matrix failure is predicted for: 90j
Copyright © 2004 by Marcel Dekker, Inc. All Rights Reserved.

fiber failure. In general, fiber failure will only occur when the composite is tested under carefully controlled laboratory conditions in which the uniaxial stress is aligned with the fiber direction to within a few degrees. Also, note that the fiber angle at which a change from shear failure to matrix failure occurs differs in tension and compression. 6.1.2 Tsai–Hill Criterion According to the Tsai–Hill criterion, failure of an orthotropic composite subjected to plane stress conditions is governed by Eq. (53), whereas failure of a transversely isotropic composite is governed by Eq. (54). For present purposes, we have assumed that the composite is transversely isotropic. Substituting Eq. (64) into Eq. (54), the Tsai–Hill failure criterion predicts that failure will not occur if the following inequality is satisfied: 8 91=2 > < cos2 ðhÞcos2 ðhÞ  sin2 ðhÞ sin4 ðhÞ cos2 ðhÞsin2 ðhÞ > = rxx < þ  2 þ  fT 2  f 2 > > : ; r11 s12 ryT 22 ð68Þ As previously noted, the Tsai–Hill criterion does not automatically account for differences in tensile and compressive strengths. A failure envelope for a unidirectional graphite–epoxy composite will be generated using tensile or compressive strengths, as appropriate. Thus, the tensile strength predicted by the Tsai–Hill criterion is: 91=2 8 > = < cos2 ðhÞcos2 ðhÞ  sin2 ðhÞ sin4 ðhÞ cos2 ðhÞsin2 ðhÞ > rfT ¼ þ þ       xx 2 2 2 > > ; : sf12 rfT ryT 11 22 Similarly, the compressive strength predicted by the Tsai–Hill criterion is: 8 91=2 > < cos2 ðhÞcos2 ðhÞ  sin2 ðhÞ sin4 ðhÞ cos2 ðhÞsin2 ðhÞ > = þ  2 þ rfC  f 2  fC 2 xx ¼ > > : ; s12 r11 ryC 22 Substituting the strength values that have been assumed for graphite–epoxy, we have: ( )1=2   cos2 ðhÞ cos2 ðhÞ  sin2 ðhÞ sin4 ðhÞ cos2 ðhÞsin2 ðhÞ fT rxx ¼ þ þ ð50 MPaÞ2 ð75 MPaÞ2 ð1500 MPaÞ2 ð69aÞ

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

( rfC xx ¼

  cos2 ðhÞ cos2 ðhÞ  sin2 ðhÞ ð1200 MPaÞ2

þ

sin4 ðhÞ ð100 MPaÞ2

þ

cos2 ðhÞsin2 ðhÞ

)1=2

ð75 MPaÞ2 ð69bÞ

A failure envelope based on Eqs. (69a) and (69b) is shown in Fig. 17. As before, it is important to realize that this failure envelope is valid for a uniaxial state of stress only. Failure envelopes obtained using the Tsai–Hill criterion but for other states of stress differ substantially from Fig. 17. 6.1.3 Tsai–Wu Criterion The Tsai–Wu criterion for the case of plane stress is given by Eq. (62). Substituting Eq. (64) into Eq. (62), we obtain:  f 2   rxx X11 cos4 ðhÞ þ X22 sin4 ðhÞ þ cos2 ðhÞ sin2 ðhÞðX66 þ 2X12 Þ   ð70Þ þrfxx X1 cos2 ðhÞ þ X2 sin2 ðhÞ  1 ¼ 0

Figure 17 Failure envelope for a unidirectional graphite–epoxy laminate subjected to a uniaxial stress rxx, based on the Tsai–Hill criterion.

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

The constants that appear in Eq. (70) are calculated using the strength properties that have been assumed for graphite–epoxy and Eqs. (56)–(59), as appropriate: X1 ¼

1 1 1 1 106  ¼  fC ¼ fT 1500 MPa 1200 MPa 6000 Pa r11 r11 1 1 1015 ¼ ¼ fC ð1500 MPaÞð1200 MPaÞ 1800 Pa2 rfT 11 r11

X11 ¼ X2 ¼

1

1 ryC 22

1

X22 ¼

X66



ryT 22

¼

¼

1 1 106  ¼ 50 MPa 100 MPa 100 Pa

1 1015 ¼ ð50 MPaÞð100 MPaÞ 5 Pa2

yC ryT 22 r22  2  2 1 1 1012 ¼ ¼ ¼ y 75 MPa s12 5625 Pa2

As previously discussed, there is no widely accepted technique used to calculate X12. For present purposes, X12 will be calculated in accordance with Eq. (61): 1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 X11 X22 ¼ X12 ¼ 2 2

sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi   15  1015 10 ¼ 1800 Pa2 5 Pa2

pffiffiffiffiffi !  10  15  10 600 Pa2

Substituting these values into Eq. (70), we obtain:  f 2 rxx

(

 1015 cos4 ðhÞ 1800 Pa

2

þ

 15  4 sin ðhÞ 10 5 Pa

2

þ cos2 ðhÞsin2 ðhÞ

"

 15 pffiffiffiffiffi #) 10 10 1012 
5625 Pa2 300 Pa2  þ rfxx

106 cos2 ðhÞ 106 sin2 ðhÞ þ 6000 Pa 100 Pa

ð71Þ 1¼0

Equation (71) is a second-order polynomial in the unknown failure stress, f rxx . For any given fiber angle h, there will be two roots to this equation. The predicted tensile strength equals the algebraically positive root, whereas the

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

predicted compressive strength equals the negative root. For example, for a fiber angle h=30j, Eq. (71) becomes: 

rfxx

  2 4417
1020 2375
1012 f þ r 1¼0 xx Pa Pa2

The two roots of this expression are found to be (125.9
106 Pa, 179.7
106 Pa). Hence, the strengths predicted by the Tsai–Wu criterion for a 30j graphite–epoxy specimen are: rfT xx ¼ 125:9 MPa rfC xx ¼ 179:7 MPa A failure envelope based on the Tsai–Wu criterion for a unidirectional graphite–epoxy laminate subjected to a uniaxial state of stress is shown in Fig. 18. This figure is analogous to those obtained using the maximum stress criterion and the Tsai–Hill criterion (Figs. 16 and 17, respectively). As before,

Figure 18 Failure envelope for a unidirectional graphite–epoxy laminate subjected to a uniaxial stress rxx, based on the Tsai–Wu criterion.

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

it is important to realize that this failure envelope is valid for a uniaxial state of stress only. Failure envelopes based on the Tsai–Wu criterion but for other states of stress differ substantially from Fig. 18. 6.1.4 Comparison The failure envelopes for uniaxial stress obtained on the basis of the three failure criteria considered are compared directly in Fig. 19, and an expanded view of just the first quadrant is presented in Fig. 20. It is apparent that similar predictions are obtained on the basis of all three criteria, although a significant numerical difference occurs at low fiber angles, near the region at which the failure mode shifts from fiber failure to shear matrix failure. However, one should not conclude that the failure criterion described above always leads to similar predictions. In fact, depending on the state of stress considered, the predicted failure envelopes may differ substantially. One stress state that exhibits this effect is the state of pure shear stress, considered in the following subsection.

Figure 19 Comparison of the failure envelopes for a unidirectional graphite– epoxy laminate subjected to a uniaxial stress rxx, obtained using the maximum stress, Tsai–Hill, and Tsai–Wu failure criteria.

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

Figure 20 Comparison of the failure envelopes (first quadrant only) for a unidirectional graphite–epoxy laminate subjected to a uniaxial stress rxx, obtained using the maximum stress, Tsai–Hill, and Tsai–Wu failure criteria.

6.2 Pure Shear Stress States It was mentioned in Sec. 5 of Chap. 3 that the shear strength of composites is sensitive to the algebraic sign of the shear stress when referenced to a nonprincipal material coordinate system. This sensitivity would not be expected based on previous experience with isotropic materials since the shear strength of isotropic materials is not sensitive to algebraic sign. We are now in a position to explain this phenomenon. An off-axis composite ply subjected to a pure shear stress state is shown in Fig. 21. The stresses induced in the 1–2 coordinate system can be determined using Eq. (20) of Chap. 2: 38 9 8 9 2 2 2 r11> ð h Þ sin ð h Þ 2 cos ð h Þsin ð h Þ cos > >0> < = 6 7< = 2 r22 ¼ 6 sin ðhÞ cos2 ðhÞ 2 cosðhÞsinðhÞ 7 4 5> 0 > > : > ; : ; s12 cosðhÞ sinðhÞ cosðhÞ sinðhÞ cos2 ðhÞ  sin2 ðhÞ sxy or equivalently: r11 ¼ 2sxy cosðhÞ sinðhÞ

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

Figure 21 Unidirectional composite subjected to pure shear stress sxy.

r22 ¼ 2sxy cosðhÞ sinðhÞ   s12 ¼ sxy cos2 ðhÞ  sin2 ðhÞ

ð72Þ

6.2.1 Maximum Stress Criterion Substituting Eq. (72) into Eq. (51), we obtain: rfC rfT 11 11 < sxy < 2cosðhÞsinðhÞ 2cosðhÞsinðhÞ

ð73aÞ

(and) ryC 22

2cosðhÞsinðhÞ

< sxy <

ryT 22 2cosðhÞsinðhÞ

ð73bÞ

(and)   sxy  <

sy12 cos2 ðhÞ  sin2 ðhÞ

ð73cÞ

Equations (73a)–(73c) will now be used to generate a failure envelope for a graphite–epoxy laminate subjected to pure shear. With reference to Eqs. (73a)–(73c), the following subtleties in these calculations should be noted: 

Over the range 0j
Copyright © 2004 by Marcel Dekker, Inc. All Rights Reserved.

shear stress sxy will induce a tensile value r11 and a compressive value for r22. However,  Over the range 90j


For 0j
rfT 1500 MPa 11 ¼ 2 cosðhÞ sinðhÞ 2 cosðhÞ sinðhÞ

ð74aÞ

sPxy ¼

ryC 100 MPa 22 ¼ 2 cosðhÞ sinðhÞ 2 cosðhÞ sinðhÞ

ð74bÞ

sPxy ¼

sy12 75 MPa ¼ 2 2 2 cos ðhÞ  sin ðhÞ cos ðhÞ  sin2 ðhÞ

ð74cÞ

For 90j
rfC 1200 MPa 11 ¼ 2 cosðhÞsinðhÞ 2 cosðhÞsinðhÞ

ð74dÞ

sPxy ¼

ryT 50 MPa 22 ¼ 2 cosðhÞsinðhÞ 2 cosðhÞsinðhÞ

ð74eÞ

sPxy ¼

sy12 75 MPa ¼ 2 2 2 cos ðhÞ  sin ðhÞ cos ðhÞ  sin2 ðhÞ

ð74f Þ

Negative shear strengths: 

For 0j
rfC 1200 MPa 11 ¼ 2 cosðhÞsinðhÞ 2 cosðhÞsinðhÞ

ð75aÞ

sN xy ¼

ryT 50 MPa 22 ¼ 2 cosðhÞsinðhÞ 2 cosðhÞsinðhÞ

ð75bÞ

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

sN xy ¼ 

sy12 75 MPa ¼ cos2 ðhÞ  sin2 ðhÞ cos2 ðhÞ  sin2 ðhÞ

ð75cÞ

For 90j
rfT 1500 MPa 11 ¼ 2 cosðhÞsinðhÞ 2 cosðhÞsinðhÞ

ð75dÞ

sN xy ¼

ryC 100 MPa 22 ¼ 2 cosðhÞsinðhÞ 2 cosðhÞsinðhÞ

ð75eÞ

sN xy ¼

sy12 75 MPa ¼ 2 2 2 cos ðhÞ  sin ðhÞ cos ðhÞ  sin2 ðhÞ

ð75f Þ

Equations (74a)–(74f ) and (75a)–(75f ) were used to create the failure envelope shown in Fig. 22. Note the following. 

The failure envelope shown in Fig. 22 is valid for a pure shear stress state only. Failure envelopes for other states of stress differ substantially (for example, compare Figs. 16 and 22, both of which are based on the maximum stress failure criterion).

Figure 22 Failure envelope for a unidirectional graphite–epoxy laminate subjected to a pure shear stress sxy, based on the maximum stress criterion.

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



None of the curves shown in Fig. 22 is associated with fiber failure. The magnitudes of the critical shear stress values returned by Eqs. (74a), (74d), (75a), or (75d) are all large enough that they do not appear in Fig. 22 due to the scale used for the vertical axis. These results indicate that failure of a unidirectional graphite–epoxy laminate subjected to a pure shear stress state will never occur due to fiber failure.  The shear strength of an off-axis unidirectional composite depends on algebraic sign of the shear stress. For example, a 45j specimen is predicted to have a positive shear strength of 100 MPa and a negative shear strength of 50 MPa.  The predicted mode of failure depends on whether sxy is positive or negative as well as on fiber angle h. If sxy is positive, then: Shear failure is predicted for 90j
9 > 1 1 7 cos ðhÞ þ sin ðhÞ = 6 2 <1 s2xy 4 cos2 ðhÞsin2 ðhÞ4  2 þ  2   y 2 5 þ  y 2 > > ; : rfT 2 s12 s12 11 ryT 22 8 > <

2

3

4

4

ð77Þ

Equating the left-hand side to unity and solving for sxy: 91=2 8 > > > > > > > > = < 1 sxy ¼

> cos4 ðhÞ þ sin4 ðhÞ > > > 2 2 1 1 2 > > > >  y 2 ; : 4 cos ðhÞsin ðhÞ ðrfT Þ2 þ ðryT Þ2  2ðsy Þ2 þ 11 12 22 s 12

ð78Þ

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

Recall that the Tsai–Hill criterion does not automatically account for differences in tensile and compressive stresses. Therefore, to predict shear strengths using Eq. (78), the failure strengths used must be selected according to whether r11 and r22 are positive or negative. Positive shear strengths: For 90j
Negative shear strengths: For 90j
A failure envelope based on these failure strengths and Eq. (78) is shown in Fig. 23.

Figure 23 Failure envelope for a unidirectional graphite–epoxy laminate subjected to a pure shear stress sxy, based on the Tsai–Hill criterion.

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

6.2.3 Tsai–Wu Criterion Substituting Eq. (72) into Eq. (62), we obtain: n  h  s2xy X66 cos4 ðhÞ þ sin4 ðhÞ þ 2 cos2 ðhÞsin2 ðhÞ 2X11 þ 2X22 io 4X12  X66 þ 2sxy cosðhÞsinðhÞ½X1  X2  < 1

ð79Þ

Numerical values for constants X1, X2, X11, etc., were calculated in Sec. 6.1.3 (based on strengths assumed for graphite–epoxy). Substituting these values in the left-hand side of Eq. (79) and equating to unity, we obtain: (  1012  4 s2xy cos ðhÞ þ sin4 ðhÞ þ cos2 ðhÞsin2 ðhÞ 2 5625Pa " pffiffiffiffiffi 15  #)

10 10 67
1015 61
106
 sxy cosðhÞsinðhÞ ¼1 3000 150Pa2 75Pa2 ð80Þ Equation (80) is a second-order polynomial in the unknown shear failure f . For a given fiber angle, h, there are two roots to this equation. The stress, sxy fP , equals the algebraically positive root, predicted positive shear strength, sxy fN , equals the negative root. whereas the predicted negative shear strength, sxy For example, for a fiber angle h=30j, Eq. (80) becomes:   2  202:8
1018 88:05
1010 f f sxy 1¼0  sxy Pa Pa2 The two roots of this expression are found to be (95.22
106 Pa, 51.8
106 Pa). Hence, the shear strengths predicted by the Tsai–Wu criterion for a 30j graphite–epoxy laminate are: sfP xy ¼ 95:2 MPa sfN xy ¼ 51:8 MPa A failure envelope based on the Tsai–Wu criterion for a unidirectional graphite–epoxy laminate subjected to a pure shear stress state is shown in Fig. 24. This figure is analogous to those obtained using the maximum stress criterion and the Tsai–Hill criterion (Figs. 22 and 23, respectively). 6.2.4 Comparisons The failure envelopes for pure shear stress obtained on the basis of the three failure criteria considered are compared directly in Fig. 25, and an expanded

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

Figure 24 Failure envelope for a unidirectional graphite–epoxy laminate subjected to a pure shear stress sxy, based on the Tsai–Wu criterion.

view of just the first quadrant is presented in Fig. 26. The difference between predictions obtained using the three failure criteria is more striking in pure shear than was the case in uniaxial stress (e.g., compare Figs. 20 and 26). Predictions on the basis of the Tsai–Hill and Tsai–Wu criteria are similar, although some difference exists. The maximum stress criterion predicts local maximums in shear strength near fiber angles of h=27j and 64j. These maximums are associated with the previously noted change in failure mode, from a shear failure mode to a matrix failure mode (or vice versa). 7 COMPUTER PROGRAMS UNIDIR AND UNIFAIL The results derived in this chapter will be used extensively throughout the remainder of this text. As is already abundantly clear, the calculations associated with any thermomechanical analysis of an anisotropic composite are tedious and time-consuming if performed using a hand calculator. Consequently, most composite analyses are performed with the aid of a digital computer. Two computer programs have been developed to supplement the material presented in this chapter: UNIDIR and UNIFAIL. These programs

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

Figure 25 Comparison of the failure envelopes for a unidirectional graphite– epoxy laminate subjected to pure shear stress sxy, obtained using the Maximum Stress, Tsai–Hill, and Tsai–Wu failure criteria.

can be downloaded at no cost from the following website: http://depts. washington.edu/amtas/computer.html. The analyses that can be performed with the aid of these programs will be discussed in the following subsections. Both programs require the user to provide various numerical values required during the calculations performed. The user must define these values using a consistent set of units. For example, program UNIDIR requires the user to input elastic moduli, thermal expansion coefficients, and moisture expansion coefficients for the composite material system 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 G12 ¼ 13
109 Pa

E22 ¼ 10
109 Pa

a11 ¼ 0:9
106 m=mo C b11 ¼ 150:0
106 m=m%M

v12 ¼ 0:30

a22 ¼ 27:0
106 m=mo C b22 ¼ 4800
106 m=m%M

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

Figure 26 Comparison of the failure envelopes (first quadrant only) for a unidirectional graphite–epoxy laminate subjected to a uniaxial stress rxx, obtained using the maximum stress, Tsai–Hill, and Tsai–Wu failure criteria.

If the analysis requires the user to input numerical values for stresses, then stresses must be input in pascals (not in MPa). A typical value would be rxx=200
106 Pa. If, instead, the analysis requires the user to input numerical values for strains, then strains must be input in m/m (not in Am/m). A typical value would be exx=2000
106 m/m=0.002000 m/m. All temperatures would be input in jC. In contrast, 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 G12 ¼ 1:9
106 psi

E22 ¼ 1:5
106 psi

a11 ¼ 0:5
106 in:=in:o F b11 ¼ 150:0
106 in:=in:%M

v12 ¼ 0:30

a22 ¼ 15
106 in:=in:o F b22 ¼ 4800
106 in:=in:%M

If the analysis requires the user to input numerical values for stresses, then stresses must be input in psi (not in ksi). A typical value would be rxx=30,000 psi. If, instead, the analysis requires the user to input numerical values for

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

strains, then strains must be input in in./in. (not in Ain./in.). A typical value would be exx=2000
106 in./in.=0.002000 in./in. All temperatures would be input in jF. 7.1 Program UNIDIR Program UNIDIR may be used to predict the elastic behavior of unidirectional composites and is based on the material presented in 1 Secs. 2 Secs. 3 Secs. 4. Two different types of analyses may be performed. The program may be used in either of the following: 

Calculating total strains (exx, eyy, cxy) caused by a specified combination of stresses (rxx, ryy, sxy), a uniform temperature change (DT), and a uniform change in moisture content (DM). Calculations performed as a part of Example Problems 1, 2, 4, and 5 are typical of this type of analysis.  Calculating stresses (rxx, ryy, sxy) caused by a specified combination of total strains (exx, eyy, cxy), a uniform temperature change (DT), and a uniform change in moisture content (DM). Calculations performed as a part of Example Problems 3 and 6 are typical of this type of analysis. The program also determines the effective properties of a unidirectional composite based on the definitions described in Sec. 4. An implicit assumption in these calculations is that all material properties (E11, a11, b11, etc.) input by the user correspond to the temperature and moisture content dictated by DT and DM. 7.2 Program UNIFAIL Program UNIFAIL may be used to obtain failure predictions for unidirectional composites based on the maximum stress, Tsai–Hill, or Tsai–Wu failure criteria introduced in Sec. 5. Two different types of analyses may be performed. The program may be used in either of the following: 

Calculating predicted uniaxial and shear strengths of a unidirectional laminate with a specified fiber angle h.  Generating a data file that can subsequently be used to produce failure envelopes for unidirectional composites subjected to several types of plane stress conditions. Note that the program UNIFAIL itself does not create a failure envelope. Rather, the program creates a file (named Envelop.txt) that contains the stress(es) predicted to cause failure of a unidirectional composite as a function of fiber angle, based on the particular failure criterion specified by the user. A

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

failure envelope may then be created using a second software package to ‘‘import’’ the data generated by the program UNIFAIL and then plotting failure stress vs. fiber angle. For example, any of the failure envelopes presented in Sec. 6 may be easily recreated in this way. As before, an implicit fT yT assumption is that failure strengths (r11 , a22 , etc.) input by the user correspond to the values exhibited by the composite at the temperature and moisture content of interest. Also, in the present context, ‘‘failure’’ may represent fracture of the fibers or yielding of the matrix.

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) Computer programs UNIDIR and/or UNIFAIL are referenced in many of the following problems. As described in Sec. 7, these programs can be downloaded from the following website: http://depts.washington.edu/ amtas/computer.html. 1. Calculate the reduced compliance matrix for the materials listed below, first by hand calculation and then using program UNIDIR. Use the material properties listed in Table 3 of Chap. 3. (a) Glass/epoxy. (b) Kevlar/epoxy. (c) Graphite/epoxy. 2. Calculate the reduced stiffness matrix for the materials listed below, first by hand calculation and then using program UNIDIR. Use the material properties listed in Table 3 of Chap. 3. (a) Glass/epoxy. (b) Kevlar/epoxy. (c) Graphite/epoxy. 3. A thin unidirectional glass/epoxy composite laminate is simultaneously subjected to a uniform temperature change DT=175jC, an increase in moisture content DM=0.5%, and the following in-plane stresses: r11 ¼ 350 MPa r22 ¼ 40 MPa s12 ¼ 60 MPa

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

Determine the resulting strains (e11, e22, and c12), first by hand calculation and then using program UNIDIR. Use the material properties listed in Table 3 of Chap. 3. 4. Repeat Problem 3 for a unidirectional Kevlar/epoxy composite laminate. 5. Repeat Problem 3 for a unidirectional graphite/epoxy composite laminate. 6. A thin unidirectional glass/epoxy composite laminate is simultaneously subjected to a uniform temperature change DT=275jF and an unknown plane stress state. The following strains are measured as a result (moisture content remains constant): e11 ¼ 1250 Ain:=in: e22 ¼ 2000 Ain:=in: c12 ¼ 0 Determine the stresses (r11, r22, and s12), first by hand calculation and then using program UNIDIR. Use the material properties listed in Table 3 of Chap. 3. 7. Repeat Problem 6 for a unidirectional Kevlar/epoxy composite laminate. 8. Repeat Problem 6 for a unidirectional graphite/epoxy composite laminate. 9. A square unidirectional glass/epoxy composite laminate with dimensions 1
1 m is clamped between four infinitely rigid walls, as shown in Fig. 27. Material properties are listed in Table 3 of Chap. 3. Initially, the clamped composite is stress-free, but the temperature is subsequently decreased by 100jC. The thermal expansion coefficient of the rigid walls is zero, so the rigid walls do not expand or contract. Calculate the stresses (r11, r22, s12) induced by this change in temperature, first by hand calculation and then using program UNIDIR. (b) Predict whether the composite will fail based on the maximum stress failure criterion. (a)

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

Figure 27 Clamped composite laminate considered in Problems 9, 10, and 11.

10. Repeat Problem 9 for a unidirectional Kevlar/epoxy composite laminate. 11. Repeat Problem 9 for a unidirectional graphite/epoxy composite laminate. 12. A 1
1 m square unidirectional glass/epoxy composite laminate is placed within a cavity defined by four rigid walls, as shown in Fig. 28. An initial gap of 0.050 mm exists between all edges of the ply and the walls. The composite ply adsorbs 1.5% moisture, causing the ply to expand and completely fill the cavity. Temperature remains constant. The rigid walls do not adsorb moisture, and hence do not expand or contract. Assuming the ply does not buckle, calculate the stresses (r11, r22, s12) caused by the change in moisture content. 13. Repeat Problem 12 for a unidirectional Kevlar/epoxy composite laminate. 14. Repeat Problem 12 for a unidirectional graphite/epoxy composite laminate. 15. A perfectly square unidirectional glass/epoxy composite laminate is mounted in a frame consisting of four infinitely rigid frame members,

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

Figure 28 A composite laminate placed within a cavity defined by four rigid walls (considered in Problems 12, 13, and 14).

as shown in Fig. 29(a). The frame members are pinned at each corner. Since the laminate is perfectly square, the angle defined by corners ABC is initially 90j (precisely). A force F is then applied to two diagonal corners, as shown in Fig. 29(b). After F is applied, angle ABC is measured to be 89.50j (precisely). Both temperature and moisture content remain constant. What stresses (r11, r22, s12) are induced in the panel? 16. Repeat Problem 15 for a unidirectional Kevlar/epoxy composite laminate. 17. Repeat Problem 15 for a unidirectional graphite/epoxy composite laminate. 18. Create a plot of the following effective properties for a unidirectional glass/epoxy composite, with fiber angles ranging from 90j to +90j. (a) Exx and Eyy. (b) vxy and vyx.

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

Figure 29 Unidirectional composite laminate considered in Problems 15, 16, and 17.

(c) (d) (e) (f ) (g)

Gxy. gxy,xx and gxy,yy. gxx,xy and gyy,xy. axx, ayy, and axy. bxx, byy, and bxy.

Suggested solution procedure: use program UNIDIR (repeatedly) to calculate the required properties in increments of 5j (that is, calculate for h=0j, 5j, 10j, 15j, 20j, etc.) and then plot these calculations. 19. Repeat Problem 18 for a unidirectional Kevlar/epoxy composite laminate.

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

20. (a)

Using hand calculation, predict the positive and negative shear strengths for a unidirectional 45j glass/epoxy composite laminate based on the maximum stress failure criterion. (b) Use program UNIFAIL to predict the positive and negative shear strengths for a unidirectional 45j glass/epoxy composite laminate based on the maximum stress, Tsai–Hill, and Tsai–Wu failure criteria. Compare these results with your calculations obtained in part (a).

21. Repeat Problem 20 for a unidirectional Kevlar/epoxy composite laminate. 22. Repeat Problem 20 for a unidirectional graphite/epoxy composite laminate. 23. On the same graph, plot failure envelopes for a unidirectional glass/ epoxy composite laminate for the following two conditions: (a) Unidirectional stress: rxx, ryy=sxy=0 (b) Biaxial normal stress: ryy=rxx/10, sxy=0 Use the Tsai–Hill failure criterion. Suggested solution procedure: use program UNIFAIL (twice) to generate data files corresponding to the specified loading conditions, and then plot these data files. 24. Repeat Problem 23 for a unidirectional Kevlar/epoxy composite laminate. 25. Repeat Problem 23 for a unidirectional graphite/epoxy composite laminate. REFERENCES 1.

Reuter, R.C. Concise property transformation relations for an anisotropic lamina. J. Compos. Mater., Vol. 5, April 1971, 270–272. 2. Hill, R. The Mathematical Theory of Plasticity; New York, Oxford University Press, 1998. 3. Tsai, S.W. Strength theories of filamentary structures. In: Fundamental Aspects of Fiber Reinforced Plastic Composites; Schwartz, R.T., Schwartz, H.S., eds; Wiley Interscience: New York, 1968; 3–11. 4. Tsai, S.W.; Wu, E.M. A general theory of strength for anisotropic materials. J. Compos. Mater., Vol. 5, January 1971, 58–80.

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

6 Thermomechanical Behavior of Multiangle Composite Laminates

Thin composite laminates in which the principal material coordinate system of all plies is aligned were considered in Chap. 5. The behavior of laminates, wherein the alignment of the principal material system varies from one ply to the next, will be considered in this chapter. In essence, we will combine the results of Chap. 5 with traditional ‘‘thin-plate theory.’’ This combination will result in an analysis technique commonly known as ‘‘classical lamination theory’’ (CLT). 1 DEFINITION OF A ‘‘THIN PLATE’’ AND ALLOWABLE PLATE LOADINGS A ‘‘thin plate’’ with in-plane dimensions a and b and thickness t is shown schematically in Fig. 1. The plate can be considered ‘‘thin’’ if the plate thickness is less than about one-tenth the in-plane dimensions (i.e., if t
Copyright © 2004 by Marcel Dekker, Inc. All Rights Reserved.

Figure 1 A ‘‘thin plate’’ with in-plane dimensions a and b and thickness t.

state within the plate. Furthermore, in this chapter, we will only consider uniformly distributed loads. That is, we will assume that the loads are constant and uniformly distributed along the edge of the plate. The more general case in which loads vary along the edge of the plate will be considered in Chap. 10. Six types of uniformly distributed loads that give rise to plane stress conditions within the x–y plane are shown in Fig. 2. All load components are shown in an algebraically positive sense. Because the line-of-action of all load vectors shown in Fig. 2 lies within the x–y plane, these load components are often referred to as in-plane loads. First, consider load components Nxx, Nyy, and Nxy. Two subscripts are used to identify these load components. The algebraic sense of each component is interpreted in a manner analogous to that previously used to identify the algebraic sense of individual stress components (discussed in Sec. 2.5). That is, the first subscript indicates the face of the plate a given load acts upon, whereas the second subscript indicates the line of action of the load. A positive load is one that: 

Acts on a positive face and points in a positive coordinate direction, or  Acts on a negative face and points in a negative coordinate direction. The algebraic sense of normal loads Nxx and Nyy (Fig. 2a) is readily apparent and intuitive: A positive (tensile) normal load is one that tends to cause the plate to stretch. The algebraic sense of shear loads Nxy and Nyx (Fig. 2b) is not as immediately apparent, but application of the sign convention just

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

Figure 2 Schematic of allowable plate loadings: (a) in-plane normal forces Nxx and Nyy, (b) in-plane shear forces Nxy and Nyx, (c) bending moment Mxx, (d) bending moment Myy, (e) in-plane torques Mxy and Myx.

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

described will confirm that the shear loads shown in Fig. 2b are indeed positive. That is, two of the shear loads shown are acting on a positive face and point in a positive coordinate direction, whereas two of the shear loads shown act on a negative face and point in a negative coordinate direction. Because individual load components are not allowed to vary spatially (i.e., loads are assumed to be constant and uniformly distributed along each edge of the plate), statical equilibrium requires that the shear loads acting along the xedge and y-edge be orientated tip-to-tip and tail-to-tail, as shown in Fig. 2b. Furthermore, the magnitude of the shear loads must be identical, jNxyj = jNyxj. These requirements are also analogous to those of shear stresses acting on an infinitesimal stress element, as discussed in Sec. 2.5. It is emphasized that Nxx, Nyy, and Nxy are all defined as distributed loads, expressed in units of force/plate length (such as N/m or lbf/in.). The remaining loads shown in Fig. 2 (Mxx, Myy, and Mxy) are bending moments (or torques) distributed along the edge of the plate. Load components Mxx and Myy are uniformly distributed bending moments acting along the x-edge and y-edge of the plate, respectively, as shown in Fig. 2c and d. These loads are shown in an algebraically positive sense. The subscripts assigned to Mxx and Myy may seem puzzling at first because, for example, Mxx represents a bending moment acting about the y-axis. However, Mxx is directly related to the distribution of rxx through the thickness of the plate, as will be shown below. Because Mxx arises due to the distribution of rxx (or vice versa), it is customary to use the same subscripts for both entities. Unfortunately, the convention used to assign an algebraic sign to distributed bending moments varies from one author to the next, and in fact can be arbitrarily chosen. The sign convention used herein is most commonly used in the study of composite plates. An algebraically positive distributed bending moment is defined as one that tends to cause tensile stresses in the positive zface of the plate and compressive stresses in the negative z-face. Referring to Fig. 2c and noting that the positive z-direction is downward as drawn, it is seen that Mxx tends to cause tensile stresses in the positive z-face (i.e., the lower face) of the plate. Hence, Mxx is positive as drawn. A similar observation holds for Myy, as shown in Fig. 2d. Finally, loads Mxy and Myx are defined as uniformly distributed inplane moments (or torques) acting along neighboring edges of the plate, as shown in Fig. 2d. It will be shown below that Mxy and Myx are directly related to in-plane shear stresses sxy and syx, respectively. Because Mxy and Myx arise due to the distribution of sxy and syx, it can be shown that for statical equilibrium to be maintained, jMxyj = jMyxj. An algebraically positive distributed torque is defined as one that tends to cause a positive shear stress in the positive z-face (i.e., the lower face) of the plate. Recall that the units of an applied moment or torque are force–length (such as Nm or lbfin). Because Mxx, Myy, and Mxy all represent uniformly

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

distributed moments acting along the plate edge, they are all expressed in units of force–length/plate length (such as Nm/m or lbfin/in). It is expected that most readers will have considered the behavior of isotropic prismatic beams during earlier studies. It is therefore instructive to contrast the definitions just given for a thin plate, as well as the loads applied thereon, to those encountered in fundamental beam theory. As previously shown in Fig. 1, a ‘‘thin plate’’ is defined as a structure whose thickness, t, is much less than the in-plane dimensions a and b. That is, t<
Figure 3 A prismatic beam with rectangular cross-section (compare with Fig. 1).

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

distributed loads used in plate theory. Because the width of the beam is b, the b b =bNxx and Mxx =bMxx. As two load definitions are related according to Nxx mentioned above, the sign convention used to define an algebraically positive bending moment varies from one author to the next, and can be arbitrarily selected. The bending moment applied to the beam shown in Fig. 3 is considered to be positive according to the sign convention used throughout this text. This corresponds to the convention most commonly used in the study of composite plates. However, according to the sign convention used in many textbooks devoted to fundamental beam theory, the bending moment shown in Fig. 3 would be considered as negative. Hence, the sign convention used to describe bending moments in this and other textbooks devoted to composites differs from the sign convention used in many textbooks devoted to beam theory. The reader must simply be aware of this potential source of confusion, and carefully note which convention has been used when comparing the results described in this text to those developed elsewhere. Let us now return to the topic of thin plates. We wish to relate the external distributed loads applied to the plate to the resulting internal stresses. An edge view of a plate loaded only by distributed load Nxx and moment Mxx is shown in Fig. 4a. A free-body diagram of a section of the plate is shown in Fig. 4b. The free-body diagram has been drawn showing the distributed load Nxx and distributed moment Mxx on the left-hand side, and the resulting internal stress rxx on the right-hand side. The plate is assumed to be in statical equilibrium. Therefore, SF=0, and the force per unit width associated with the distribution of stress rxx through the thickness of the plate must be exactly balanced by the distributed load Nxx. Let the free-body diagram have a width of ‘‘1’’ and consider an incremental strip of height dz. The cross-sectional area dA of this strip is dA=1dz. The incremental force dFxx associated with the stress acting over this thin strip is dFxx=dNxx1=rxxdA=rxxdz. We can now relate the total distributed force Nxx acting on the left-hand side of the freebody diagram to the distribution of rxx acting on the right-hand side by simply ‘‘adding up’’ the forces acting over all incremental strips; that is, we integrate over the thickness of the plate: Z þt=2 Z t=2 Nxx ¼ dNxx ¼ rxx dz ð1aÞ t=2

t=2

In an entirely equivalent manner, we can relate distributed forces Nyy and Nxy to stresses ryy and sxy, respectively: Z t=2 Nyy ¼ ryy dz ð1bÞ Nxy

Z ¼

t=2 t=2

t=2

sxy dz

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

ð1cÞ

Figure 4 Edge view of a thin plate subjected to loads Nxx and Mxx.

Now consider moment Mxx. As before, the plate is assumed to be in statical equilibrium, and hence moments acting about the y-axis must sum to zero: SMy=0. Again consider an incremental strip of height dz, which is located a distance z from the midsurface. The incremental distributed moment dMxx contributed by Nxx acting over the incremental strip is dMxx=dNxxz= rxxzdz. We can obtain the total moment acting on the right-hand side of the free-body diagram by integrating over the thickness of the plate: Z t=2 Mxx ¼ rxx zdz ð2aÞ t=2

In an entirely equivalent manner, we can relate moments Myy and Mxy to stresses ryy and sxy, respectively: Z t=2 Myy ¼ ryy zdz ð2bÞ Mxy

Z ¼

t=2 t=2

t=2

sxy zdz

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

ð2cÞ

Equations (1a) (1b) (1c) (2a) (2b) (2c) show that the uniformly distributed loads and moments applied to the plate edge are directly related to the stresses within the plate. Distributed loads Nxx, Nyy, and Nxy are commonly called stress resultants, and moments Mxx, Myy, and Mxy are commonly called moment resultants. 2 PLATE DEFORMATIONS: THE KIRCHHOFF HYPOTHESIS Let us now consider the deformation of a thin, flat plate. Fig. 5 represents a (magnified) view of the edge of the plate in both the ‘‘initial’’ and ‘‘deformed’’ positions. The positive x-direction is to the right, the positive y-direction is out of the plane of the figure, and the positive z-direction is downward, which is consistent with the original definition shown in Fig. 1. Although we will eventually apply our results to a composite laminate, for the moment, we will not consider the existence of individual plies, and therefore the ply interfaces are not shown in the figure. If the flat plate is loaded and/or is subjected to a change in environment, it will be deformed and (in general) will become curved, as shown in the figure. We will base our analysis on the Kirchhoff hypothesis, which states that a ‘‘straight line that is initially perpendicular to the midplane of the plate remains straight and perpendicular to the midplane of the plate after deformation.’’ For example, let us consider straight line b–o– d. This line is shown in Fig. 5, in the sketch of both the initial and deformed positions of the plate. In accordance with the Kirchhoff hypothesis, line b–o–

Figure 5 Initial and deformed positions of a flat plate.

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

d has been drawn perpendicular to the midplane in both the initial and deformed positions. In effect, we have assumed that out-of-plane shear strains are zero (cxz=cxz=0), which is equivalent to saying that we have assumed that the z-axis is a principal strain axis. We are interested in describing the displacement of an arbitrary point ‘‘c,’’ which is located within the thickness of the plate at some distance zc from the midsurface. Point ‘‘c’’ is shown in Fig. 5, and lies along line b–o–d. Denote the displacement of point ‘‘o’’ in the x-direction and z-direction as distances uo and wo, respectively. From the figure, it can be seen that the distance point ‘‘c’’ has moved in the x-direction, uc, is approximately given by: uc iuo  zc sin a

ð3Þ

where a is the angle formed by the plate midplane and the x-axis in the deformed condition. Equation (3) is approximate because we have ignored any change in plate thickness (i.e., we have ignored any change in distance zc that may have occurred during deformation of the plate). If we now further assume that angle a is small, then we can simplify Eq. (3) using the small-angle approximation, which states that if a is expressed in radians and is less than about 0.1745 rad (about 10j), then: sin aia tan aia cos ai1

ð4Þ

Based on this assumption, Eq. (3) can be written as: uc iuo  zc a

ð5Þ

Now, from Fig. 5, it can be seen that tan a= dw/dx. Applying the small-angle approximation once more, we can say that tan a i a i dw=dx. Substituting this result in Eq. (5), we obtain: dw ð6Þ dx To summarize, we have expressed the displacement in the x-direction of arbitrary point ‘‘c’’ (which we have called distance uc) as a function of the displacement in the x-direction of a point on the plate midsurface (distance uo), the position of point c with respect to the midsurface (length zc), and the slope of the plate midsurface dw/dx. We will also require an expression for the displacement of point ‘‘c’’ in the y-direction. We will denote the displacement of point ‘‘c’’ in the ydirection as vc. Using a procedure that is entirely equivalent to that just described, it can be shown that: uc ¼ u o  z c

vc ¼ vo  zc

dw dy

ð7Þ

Equations (6) and (7) represent the displacements of point ‘‘c’’ in the xdirection and y-direction, respectively, and follow directly from the Kirchhoff

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

hypothesis. We can now determine the infinitesimal in-plane strains at point ‘‘c’’, in accordance with Eq. (49) of Chap. 2: Bu Bv Bv Bu eyy ¼ cxy ¼ þ ð8Þ exx ¼ Bx By Bx By Substituting Eqs. (6) and (7) into Eq. (8), we obtain the following expressions for the strains induced at point ‘‘c’’: Buo B2 w  zc 2 Bx Bx Bvo B2 w  zc 2 ¼ By By

ecxx ¼ ecyy

ccxy ¼

ð9Þ

Buo Bvo B2 w þ  2zc BxBy By Bx

Let: Buo Bx Bv o eoyy ¼ By Buo Bvo coxy ¼ þ By Bx eoxx ¼

B2 w Bx2 B2 w ¼ 2 By

ð10aÞ ð10bÞ ð10cÞ

jxx ¼ 

ð10dÞ

jyy

ð10eÞ

jxy ¼ 2

B2 w BxBy

ð10f Þ

where eoxx ; eoyy ; and coxy are the in-plane strains that exist at the midplane of the plate. The terms jxx, jyy, and jxy are called midplane curvatures, and represent the rate of change of the slope of the midplane of the plane. The reader is likely to have encountered the concept of midplane curvatures during earlier studies of fundamental beam theory. Unfortunately, the algebraic sign convention used to define curvatures varies from one author to the next. The sign convention used throughout this text and defined by Eqs. (10a–10f ) is most commonly used in the study of composite plates. However, in many textbooks devoted to beam theory, curvatures are defined using the opposite sign convention. For example, in beam theory curvature, jxx is often defined as jxx=+B2w/Bx2, rather than jxx=B2w/Bx2 as indicated above. Also, in some textbooks devoted to plate theory, jxy is defined as jxy=+B2w/ BxBy, rather than as jxy=2B2w/BxBy as indicated in Eq. (10f ). These

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

unfortunate deviations from one author to the next have developed over many years, and a universal agreement on algebraic signs or even the fundamental definition of jxy is not likely to occur for the foreseeable future. The reader must simply be aware of these potential sources of confusion, and carefully note that the convention has been used when comparing the results described in this text to those developed elsewhere. Returning now to the discussion of through-thickness strains, note from Fig. 5 that point ‘‘c’’ is located at an arbitrary distance z from the neutral surface. We will therefore discontinue the use of the subscript ‘‘c’’ in Eq. (9). Substituting Eqs. (10a–10f) in Eq. (9), we obtain: exx ¼ eoxx þ zjxx eyy ¼ eoyy þ zjyy cxy ¼

coxy

ð11Þ

þ zjxy

Equation (11) can be conveniently written in matrix form as: 9 8 o 9 9 8 8 exx > exx > jxx > > > > > > > > > > = < = = < < 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;

eyy  1500 Am=m; cxy ¼ 1000 Arad

at z ¼ þt=2 ¼ þ0:5 mm : exx ¼ 250 Am=m; eyy  1100 Am=m; cxy ¼ 800 Arad What midplane strains and curvatures are induced in the plate?

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

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 :

exx ¼ 250 Am=m ¼ eoxx  ð0:0005Þjxx

at z ¼ þt=2 ¼ þ0:0005 m :

exx ¼ 250 Am=m ¼ eoxx 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 1300 Am=m; coxy

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

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

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 9 8 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

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

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: ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi s  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

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

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).

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

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 t=2 ¼ tjxx =2 eyy jz ¼ t=2 ¼ tjyy =2 cxy jz ¼ t=2 ¼ tjxy =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.

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

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

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

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).

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

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

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

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 (znz0), and the thickness of an individual ply k is tk=(zkzk1). For example, the thickness of ply 2 is t2=(z2z1). 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

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

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.

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

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.

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

Sample Problem 2 A ½(0/F30/90)2/45/20s 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; 0 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=z19z0=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  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: 8 9 8 9 < e11 = < exx = eyy e22 ¼ ½Tk : ; : ; 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: 8 9 8 9 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 , aij, and bij) are properties of the ply that exists at position zk, and in particular are

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

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 ½Tz¼zk 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. Solution.

From Sample Problem 1, the midplane strains and curvatures are:

eoxx ¼ 0 Am=m eoyy coxy

jxx ¼ 0:50 rad=m

¼ 1300 Am=m

jyy ¼ 0:40 rad=m

¼ 900 Arad

jxy ¼ 0:20 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

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

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 8 9 8o 9 8 9 8 0 e e 0:50 rad=m > > > j > > > > > = < < xx> = < xx > = < xx> = > < = eyy ¼ eoyy þ z0 jyy ¼ 1300
106 m=m þ ð0:000500 mÞ 0:40 rad=m > > > > > > ; : ; ; ; > > : > : o > : > > : ; cxy cxy z¼z0 jxy 0:20 rad=m 900
106 m=m

j

8 9 8 9 > < 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 9 8 9 8 9 e > e 0:50 rad=m > 0 > j > > > > > < xx > < xx > < xx > < = = > = = < = eyy ¼ eoyy þ z1 jyy ¼ 1300
106 m=m þ ð0:000375 mÞ 0:40 rad=m > > > > > > : : > : > > > ; ; > ; > : ; cxy z¼z1 : coxy ; jxy 900
106 m=m 0:20 rad=m

j

8 9 8 9 > < 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

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

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) 0.500 0.375 0.250 0.125 0.0 0.125 0.250 0.375 0.500

exx (Am/m)

eyy (Am/m)

cxy (Arad)

250 188 125 62 0 62 125 188 250

1500 145 1400 1350 1300 1250 1200 1150 1100

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 8 9 ply 1 2 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

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 0 5 1500 Am=m > > ; : 1 ð1000 AradÞ=2 z ¼ z0

j

9 ply 1 9 ply 1 8 8 > > = = < 250 Am=m > < e11 > e22 ¼ 1500 Am=m > > > > ; ; : : c12 =2 z ¼ z0 1000 Arad z ¼ z0

j

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

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: 8 9 ply 1 2 cos2 h1 > < e11 > = 6 e22 ¼6 sin2 h1 4 > > ; : c12 =2 z ¼ z1 cos h sin h

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 ¼ 4 0 1 0 5 1450 Am=m > > > > ; : ; : 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

sin2 h1

j

2cos h1 sin h1

j

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 sin2 h2 2 cos h2 sin h2 > exx > ply 2 cos2 h2 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 8 9 ply 2 2 38 188 Am=m > ply 2 0:750 0:250 0:866 > > < = < e11 > = 6 7 e22 ¼ 4 0:250 0:750 0:866 5 1450 Am=m > > > > : ; : ; c12 =2 z ¼ z1 0:433 0:433 0:500 ð975 AradÞ=2 z ¼ z1

j

j

j

j

9 ply 2 8 9 ply 2 8 > > = < e11 > < 200 Am=m > = e22 ¼ 1463 Am=m > > > > ; : : ; c12 z ¼ z1 931 Arad z ¼ z1

j

j

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

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

8 9ply 2 2  cos2 ð30jÞ > < e11 > = 6  e22 ¼6  sin2 ð30jÞ 4 > > : ; c12 =2 z ¼ z2 cosð30jÞ sinð30jÞ

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

j

2 cos h2 sin h2

sin2 ð30jÞ cos2 ð30jÞ cosð30jÞ sinð30jÞ

2 cosð30jÞ sinð30jÞ

7 2 cosð30jÞ sinð30jÞ 7 5 cos2 ð30jÞ  sin2 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 ¼ 4 0:250 0:750 0:866 5 1400 Am=m > > > > ; : ; : c12 =2 z ¼ z2 0:433 0:433 0:500 ð950 AradÞ=2 z ¼ z2 8 9 ply 2 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

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

3

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)

e11 (Am/m)

e22 (Am/m)

c12 (Arad)

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

250 188 200 155 1400 1350 691 715 715 738 1250 1200 26 71 188 250

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

Ply 1 Ply 2 Ply 3 Ply 4 Ply 5 Ply 6 Ply 7 Ply 8

Strains are referenced to the 1–2 coordinate system local to individual plies.

For 30j plies: 2   Q 30j

plies

plies

26:06
109

48:3
109

3

27:22
109

7 21:52
109 7 5ðPaÞ

48:3
109

21:52
109

36:05
109

10:05
109

3:016
109

0

170:9
109

0

0

13:00
109

6 9 ¼6 4 26:06
10

For 90j plies: 2   Q 90j

107:6
109

6 9 ¼6 4 3:016
10 0

For 30j plies: 2 107:6
109 6   9 Q 30j plies ¼ 6 4 26:06
10 48:3
109

26:06
109 27:22
109 21:52
109

3 7 7ðPaÞ 5

48:3
109

3

7 21:52
109 7 5ðPaÞ 36:05
109

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

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 ply 1 2 8 9 3 Q11 Q12 Q16 ply 1 > > < rxx > < exx > = = 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

j

38 9 > 250 Am=m > = 7< 7 1500 Am=m 0 5> > ; : 1000 Arad 13:00
109

3:016
109

0

10:05
109 0

9 ply 1 8 9 8 > > = = < rxx > < 38:2 MPa > ¼ 14:3 MPa ryy > > > > ; ; : : sxy z ¼ z0 13 MPa

j

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 ply 1 2 Q 9 8 Q12 Q16 ply 1 8 exx > 11 r > > > xx = = < < 7 6 Q12 Q22 Q26 7 e ryy ¼6 yy 5 4 > > > > ; ; : : sxy z ¼ z1 Q16 Q26 Q66 cxy z ¼ z1

j

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

38 9 188 Am=m > > < = 7 7 1450 Am=m 0 5> > ; : 975 Arad 13:00
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

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 > < 6 7 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

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

8 9 ply > = ryy > ; : sxy >

j

2

2

107:6
109

6 9 ¼6 4 26:06
10 48:3
109

z ¼ z1

26:06
109 27:22
109 21:52
109

9 38 188 Am=m > > > > = < 7 1450 Am=m 21:52
109 7 5> > > 975 Arad > ; 9 : 36:05
10 48:3
109

9 8 9 ply 2 8 > > < rxx > = < 29:5 MPa > = ryy ¼ 13:6 MPa > > > > : ; : ; sxy z ¼ z1 13:0 MPa

j

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)

rxx (MPa)

ryy (MPa)

sxy (MPa)

Q (MPa)

A (MPa)2

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

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

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

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

23.9 13.8 15.7 8.3 242 234 128 132 132 136 218 210 21.4 28.8 47.6 57.8

715 550 567 429 556 651 487 575 575 666 812 884 65.8 175 315 435

Stresses are referenced to the x–y coordinate system.

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

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.

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

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

Ply interface stresses referenced to local 1–2 coordinate systems are summarized in Table 4 and plotted in Fig. 16. Once again, stresses are not linearly distributed through the thickness of the laminate, and instead exhibit a sudden discontinuous change at all ply interface positions. Stress invariants can be used to confirm that the ply stresses referenced to the x–y coordinate system and listed in Table 3 are equivalent to the ply stresses referenced to the 1–2 coordinate system, as listed in Table 4. The concept of ‘‘stress invariants’’ was discussed in Chap. 2. The stress invariants for the case of plane stress are given by Eq. (22) of Chap. 2, repeated here for convenience: First stress invariant ¼ Q ¼ rxx þ ryy Second stress invariant ¼ A ¼ rxx ryy  s2xy

ðrepeatedÞ ð2:22Þ

Third stress invariant ¼ C ¼ 0 For plane stress conditions, the third stress invariant always equals zero, and so C cannot be used to evaluate whether two plane stress states are equivalent. The first and second stress invariants, Q and A, respectively, have been calculated using the ply stress components referenced to both the x–y and 1–2

Table 4 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)

r11 (MPa)

r22 (MPa)

s12 (MPa)

Q (MPa)

A (MPa)2

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

38.2 27.8 29.8 22.2 239 231 120 124 124 128 214 205 8.31 15.9 35.5 46.0

14.3 14.0 14.1 13.9 2.97 3.44 8.08 8.04 8.04 8.00 4.40 4.88 13.1 13.0 12.1 11.8

13.0 12.7 12.1 11.0 12.4 12.0 21.9 20.5 20.5 19.1 11.4 11.0 6.58 5.47 10.7 10.4

23.9 13.8 15.7 8.3 242 234 128 132 132 136 218 210 21.4 28.8 47.6 57.8

715 550 567 429 556 651 487 575 575 666 812 884 65.8 175 315 435

Stresses are referenced to the 1–2 coordinate system.

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

Figure 16 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 1–2 coordinate system.

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

coordinate systems. Values calculated for Q and A are included in the last two columns of both Tables 3 and 4. Identical values are obtained in all cases, indicating the equivalence of the ply stress states described using the two different coordinate systems. 6 CLASSICAL LAMINATION THEORY (CLT) Stress and moment resultants were introduced in Sec. 1. As was discussed, a thin plate subjected to any combination of stress and moment resultants will experience a state of plane stress. Deformations of a thin plate were then considered in Sec. 2. There the Kirchhoff hypothesis was invoked, which allows us to calculate the in-plane strains induced at any location through the thickness of a thin plate. In this section, we will combine the material presented in Secs. 1 and 2, as well as certain material presented in Chap. 5. This will lead to the ability to relate stress and moment resultants to the resulting strains (and hence stresses) induced within a thin composite plate. This combination of analysis tools is commonly known as classical lamination theory (CLT). Stress and moment resultants represent the mechanical loads applied to a laminate. Obviously then, stress and moment resultants will induce strains within the laminate. However, strains may also be induced by environmental factors as well, as discussed in earlier chapters. Of particular importance for polymeric composite laminates are strains due to a change in temperature (DT ) and/or strains due to a change in moisture content (DM ). To simplify our discussion, we will first develop CLT by assuming that constant environmental conditions exist (i.e., we will initially assume DT=DM=0). We will then consider how to account for a change in temperature and/or a change in moisture content. 6.1 Constant Environmental Conditions The stresses rxx induced in a thin composite laminate are related to stress resultant Nxx in accordance with Eq. (1a), repeated here for convenience: Z t=2 Nxx ¼ rxx dz ðrepeatedÞ ð1aÞ t=2

The composite laminate consists of n plies, and the fiber angle may vary from one ply to the next. The stresses in any ply (say, in ply number k) are related to ply strains in accordance with Eq. (30) of Chap. 5, which, for DT=DM=0, becomes: 9 2 9 8 3 8 Q11 Q12 Q16 > > = = < rxx > < exx > 6 7 eyy ¼ 4 Q21 Q22 Q26 5 ð20Þ ryy > > > > ; ; : : cxy k sxy k Q Q Q 61

62

66

k

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

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

Z þ

coxy

Z þ jyy

t=2

t=2 t=2 t=2

t=2

Q16

 k

Z dz þ jxx

t=2

t=2

Q11

 k

ð23Þ

dz



 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 n1

z0 zn1 zn2



dz þ Q11

Z  n

z1 zn

dz

zn1

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

z2

 þeoyy





þ Q12  þcoxy 

þ Q16

n1



n1



Q12

Z   þ Q12 n1  þjxy 

þ Q16

1





Q16

Z

n1



dz þ Q12



dz þ Q16

zn2

Z 

z1

1

z0

zn1

zn2

Z  1

zn1

z1

2

z1

zdz þ Q16

dz þ : : :

z2



z3

dz þ : : :

z2

dz

n

zn1

Z 

Z   zdz þ Q11 3

z2 z1



zn

zn1 Z  z2 2

z1

Z   zdz þ Q12 3

n



zn

zdz þ : : :

z2

z3

zdz þ : : :

z2

zdz zn1 z2

Z 

Z 

z3

zdz

n

zdz þ Q16

zn2

Z   dz þ Q16 3

zn

Z   zdz þ Q12 n



z3

dz

 zdz þ Q12

z0

3



Z   zdz þ Q11 2



Z 

zn1 Z  z2

Z 

zdz þ Q11

z1

dz þ Q12

zn

z0

zn1



z2

n



zn2 Z  z1 1

Z 

 dz þ Q16

z0

zn1

Q11

Z



z0

Z   dz þ Q12 2

zn2 Z  z1

Z



z1

zn1

n1

 þjyy

Q16





þ Q11

1

Z



þjxx 

Q12

Z 

2



zdz þ Q16

z1



zn

zdz

Z  3

z3

zdz þ : : :

z2

ð24Þ

zn1

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  zk1 Þ zk1

or

Z

zk zk1

zdz ¼

1 2 ðz  z2k1 Þ 2 k

Hence, evaluating all integrals that appear in Eq. (24), we obtain: n      Nxx ¼ eoxx Q11 1 ½z1  z0  þ Q11 2 ½z2  z1  þ Q11 3 ½z3  z2  þ : : : o   þ Q11 n ½zn  zn1  n      þeoyy Q12 1 ½z1  z0  þ Q12 2 ½z2  z1  þ Q12 3 ½z3  z2  þ : : :

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

o   þ Q12 n ½zn  zn1  n      þcoxy Q16 1 ½z1  z0  þ Q16 2 ½z2  z1  þ Q16 3 ½z3  z2  þ : : : o   þ Q16 n ½zn  zn1  n            1 þ jxx Q11 1 z21  z20 þ Q11 2 z22  z21 þ Q11 3 z23  z22 2    2 o : z  z2 þ ::þ Q 11 n

n

n1

12 n

n

n1

n            1 þ jyy Q12 1 z21  z20 þ Q12 2 z22  z21 þ Q12 3 z23  z22 2    2 o : z  z2 þ ::þ Q n            1 þ jxy Q16 1 z21  z20 þ Q16 2 z22  z21 þ Q16 3 z23  z22 2    o ð25Þ þ . . . þ Q16 n z2n  z2n1 Equation (25) can be simplified substantially by defining the following terms: n      A11 ¼ Q11 1 ½z1  z0  þ Q11 2 ½z2  z1  þ Q11 3 ½z3  z2  þ : : : o   þ Q11 n ½zn  zn1  n      A12 ¼ Q12 1 ½z1  z0  þ Q12 2 ½z2  z1  þ Q12 3 ½z3  z2  þ : : : o   þ Q12 n ½zn  zn1  n      A16 ¼ Q16 1 ½z1  z0  þ Q16 2 ½z2  z1  þ Q16 3 ½z3  z2  þ : : : o   þ Q16 n ½zn  zn1             1 n Q11 1 z21  z20 þ Q11 2 z22  z21 þ Q11 3 z23  z22 þ : : : 2    o þ Q11 n z2n  z2n1            1 n ¼ Q12 1 z21  z20 þ Q12 2 z22  z21 þ Q12 3 z23  z22 þ : : : 2    o þ Q12 n z2n  z2n1            1 n ¼ Q16 1 z21  z20 þ Q16 2 z22  z21 þ Q16 3 z23  z22 þ : : : 2    o þ Q16 n z2n  z2n1

B11 ¼

B12

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

Nxy ¼

ð26cÞ

A61 eoxx

þ

A62 eoyy

þ

A66 coxy

þ B61 jxx þ B62 jyy þ B66 jxy

where: Aij ¼

n X  Qij k ðzk  zk1 Þ

ð27aÞ

k¼1

Bij ¼

n   2  1X Q z  z2k1 2 k¼1 ij k k

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 2

A61

A62

A66

B11

B12

B16

6 Bij ¼ 4 B21 B61

3

2

A16

A26

A66

B11

B12

B16

B22

7 6 B26 5 ¼ 4 B12

B62

B66

B16

3

B22

7 B26 5

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 

n1



þ Q12

Q12 n1





zdz þ Q11

zn2 Z z1 

Z



 þcoxy

z0 zn1

Z



Q16

1



zdz þ Q12

zn2 Z  z1 z0

zdz

zn1 Z z2  2

Z 

 zdz þ Q16

z2



n

zdz þ Q12

z0

zn1

1



z1 zn

Z 



zdz þ Q12

z1

3



zn

z3

zdz þ : : : :

z2

zdz

n

zn1 Z  z2 2

Z 

z1

Z   zdz þ Q16 3

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

z3 z2

zdz þ : : : :

Z   þ Q16 n1  þjxx





 Q12

Z   þ Q12 n1  þjxy



Q16

Z   þ Q16 n1



zdz þ Q16

Z 

zn2

Q11

Z   þ Q11 n1 þjyy

zn1

Z 

z1

1

1



z dz þ Q11 2

zn2 Z  z1



z dz þ Q12 2

zn2

Z  1

zn1 zn2

z1

Z 

Z   z dz þ Q11 3

z2

2

Z 

2

z1



zn

2

z1

Z   z2 dz þ Q12 3

zn

z2 dz þ : : : :

z2

z dz

zn1 Z  z2

Z   z2 dz þ Q16 n



z3

2

n

z0

z2 dz þ : : : :

z2

z dz

zn1 Z  z2

Z 

 z2 dz þ Q16

z3

2

n

 z2 dz þ Q12

z0

zn1

zn1

z dz þ Q11

z0

zn1

zdz

n



2



zn

2

z1 zn

Z   z2 dz þ Q16 3

z3

z2 dz þ : : : :

z2

z2 dz

ð31Þ

zn1

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  z2k1 2 zk1 or

Z

zk

z2 dz ¼

zk1

 1 3 zk  z3k1 3

Hence, evaluating all integrals, we obtain: Mxx ¼

           1 o n exx Q11 1 z21  z20 þ Q11 2 z22  z21 þ Q11 3 z23  z22 2    2 o z  z2 þ: : : : þ Q n

11 n

n

n1

           1 þ eoyy Q12 1 z21  z20 þ Q12 2 z22  z21 þ Q12 3 z23  z22 2    o : þ : : : þ Q12 n z2n  z2n1            1 n þ coxy Q16 1 z21  z20 þ Q16 2 z22  z21 þ Q16 3 z23  z22 2    o : þ : : : þ Q16 n z2n  z2n1

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

n            1 þ jxx Q11 1 z31  z30 þ Q11 2 z32  z31 þ Q11 3 z33  z32 3    3 o z  z3 þ: : : : þ Q n

11 n

n

n1

           1 þ jyy Q12 1 z31  z30 þ Q12 2 z32  z31 þ Q12 3 z33  z32 3    3 o : þ :::þ Q z  z3 12 n

n

n1

n            1 þ jxy Q16 1 z31  z30 þ Q16 2 z32  z31 þ Q16 3 z33  z32 3    o þ : : : : þ Q16 n z3n  z3n1 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  z3n1            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  z3n1            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  z3n1 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  z3k1 3 k¼1

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 6
6 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 t1 ¼ 0:1875 mm þ 0:125 mm ¼ 0:0625 mm ¼ 0:0000625 m z2 ¼ z1 þ t2 ¼ 0:0625 mm þ 0:125 mm ¼ 0:0625 mm ¼ 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): 2 

Q

3 Q11

 30j ply

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

7 7 Q26 7 5 Q66 26:06
109

48:13
109

3

27:22
109

7 21:52
109 7 5ðPaÞ

21:52
109

36:05
109

For ply 2 (the 0j ply): 3

2 

Q

Q11

 0B ply

Q12

Q16

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

3 7 7ðPaÞ 5

13:00
109

  * 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

3 7 7ðPaÞ 5

13:00
109

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

ðzk  zk1 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  0:0000625Þ A11 ¼ 36:07
106 Pa  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 Q z  z2k1 2 k ¼ 1 11 k k         i 1 h ¼ Q11 1 z21  z20 þ Q11 2 z22  z21 þ Q11 3 z23  z22 2 o n 1 h 107:6
109 ð:0000625Þ2 þ ð0:0001875Þ2 ¼ 2 o

n þ 170:9
109 ð0:0000625Þ2  ð0:0000625Þ2 oi

n þ 10:05
109 ð0:0001875Þ2  ð0:0000625Þ2

B11 ¼ B11 B11

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

B11 ¼ 1:524
103 Pa  m2 The remaining elements of the Bij matrix are found in similar fashion: 2

1:524

6 Bij ¼ 4 0:3601 0:7521

0:3601 2:245 0:3362

0:7521

3

  7 0:3362 5
103 Pa  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  z3k  1 3 k¼1

        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  ð0:0001875Þ3 ¼ 3 o

 n þ 170:9
109 þ ð0:0000625Þ3  ð0:0000625Þ3

D11 ¼ D11

oi

n þ 10:05
109 ð0:0001875Þ3  ð0:0000625Þ3 D11 ¼ 0:2767 Pa  m3 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  3 0:0455 7 5 Pa  m 0:1059

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 ½ABD ¼ 6 6 6 1524 360:1 752:1 6 6 360:1 2245 336:2 4 752:1 336:2 360:1

1524 360:1 752:1 0:2767 0:0620 0:1018

360:1 752:1

3

7 336:2 7 7 7 336:2 360:1 7 7 7 0:0620 0:1018 7 7 2:513 0:0455 7 5 0:0455 0:1059 2245

The [abd] matrix is obtained by inverting the [ABD] matrix, and is found to be: 2

3:757
108 6 6 1:964
109 6 6 6 6 1:038
108 6 ½abd ¼ 6 6 1:440
104 6 6 6 3:905
106 6 4 8:513
105

1:964
109

1:038
108

1:440
104

3:905
106

1:037
107

4:234
108

1:866
105

6:361
104

4:234
108

2:004
107

3:661
104

3:251
104

1:866
105

3:661
104

7:064

6:361
104

3:251
104

4:268
104

1:851
105

3:122
102

3:122
102 4:572

6:429 3:620

8:513
105

7 4:268
104 7 7 7 5 7 1:851
10 7 7 7 7 4:572 7 7 7 3:620 7 5 17:41

Example Problem 5 A [30/0/90]T graphite-epoxy laminate is subjected to the following stress and moment resultants: Nxx ¼ 50 kN=m

Nyy ¼ 10 kN=m

Mxx ¼ 1 N  m=m Myy ¼ 1 N  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

and curvature may be obtained through application of Eq. (38), which becomes: 8 o 9 2 3:757
108 exx > > > > > > 6 > > > > 6 > > o > > 6 1:964
109 > > > eyy > > > 6 > > > > 6 > = 6 1:038
108 < co > 6 xy ¼6 > 6 1:440
104 > > jxx > > > 6 > > > > 6 > > > 6 > > > > > 6 3:905
106 j yy > > > > > 4 > > > ; : jxy 8:513
105




1:964
109

1:038
108

1:440
104

3:905
106

1:037
107

4:234
108

1:866
105

6:361
104

4:234
108

2:004
107

3:661
104

3:251
104

1:866
105

3:661
104

7:064

6:361
104

3:251
104

4:628
10

4

1:851
10

5

3:122
102 4:572

3:122
102 6:429 3:620

8:513
105

7 7 4:628
104 7 7 7 1:851
105 7 7 7 7 4:572 7 7 7 7 3:620 5 17:41

9 8 50
103 > > > > > > > > > > > 3 > > 10
10 > > > > > > > > > > > = < 0 > > > > > > > > > > > > :

> > > > > > > > > > > > ;

1 1 0

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 m1 > > > > > jxx > > > > > > > > > > > > 1 > > > > > 0:096 m j > > > > yy > > > ; > ; : : jxy 1:323 m1 (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: 8 9 > < exx > = eyy > :c > ; xy

j

z ¼ z0

8 8 o 9 8 9 > 2038
106 m=m e > < < xx > = < jxx = > ¼ eoyy þ z0 jyy ¼ 518
106 m=m > > > : ; > : o ; : jxy cxy 55
106 m=m

9 8 < exx = e : yy ; cxy

j

z ¼ z0

8 9 > < 677 Am=m > = ¼ 536 Am=m > > : ; 194 Arad

9 > > =

8 9 14:48 rad=m > > < = þ ð 0:0001875mÞ 0:096 rad=m > > > : ; > ; 1:328 rad=m

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

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)

eyy (Am/m)

cxy (Arad)

677 1133 2943 4753

536 524 512 500

194 28 137 303

0.1875 0.0625 0.0625 0.1875

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 > < rxx > = < exx = 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
106 > 107:6
109 26:06
109 48:13
109 > > = = < rxx > < 7 6 ryy ¼ 4 26:06
109 27:22
109 21:52
109 5 536
106 > > > > ; ; : : sxy z ¼ z0 48:13
109 21:52
109 36:05
109 194
106 9 9 ply 1 8 8 > > = = < 77:5 MPa > < rxx > ryy ¼ 28:1 MPa > > > > ; ; : : 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)

rxx (MPa)

ryy (MPa)

sxy (MPa)

0.1875 0.0625 0.0625 0.0625 0.0625 0.1875

77.5 109.7 192.1 501.5 28.0 46.3

28.1 15.9 1.85 3.73 78.6 71.1

37.1 44.3 0.366 1.78 1.78 3.93

Stresses are referenced to the x–y coordinate system.

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

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  DTayy  DMbxx ð39Þ

 þ Q16 cxy  DTaxy  DMbxy

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

dz

 k

ð40Þ

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  zk1 

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  zk1  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 ¼ DM Q b þ Q b þ Q b ½ z  z  ð42aÞ k k1 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 Q12 axx þ Q22 ayy þ Q26 axy k ½zk  zk1  ð41bÞ NTxy u DT

  Q16 axx þ Q26 ayy þ Q66 axy k ½zk  zk1 

k¼1 n  X

   DT Q11 axx þ Q12 ayy þ Q16 axy k z2k  z2k1 2 k¼1 n     DT X u Q12 axx þ Q22 ayy þ Q26 axy k z2k  z2k1 2 k¼1

MTxx u MTyy

k¼1 n  X

o  Q16 bxx þ Q26 byy þ Q66 bxy k ½zk  zk1 

ð42cÞ

k¼1 n n X

k¼1 n n X

  o DM Q11 bxx þ Q12 byy þ Q16 bxy k z2k  z2k1 2 k¼1 n n   o DM X u Q12 bxx þ Q22 byy þ Q26 bxy k z2k  z2k1 2 k¼1 n n   o DM X u Q16 bxx þ Q26 byy þ Q66 bxy k z2k  z2k1 2 k¼1

MM xx u

MM xy

ð41eÞ

ð42bÞ

NM xy u DM

MM yy

ð41dÞ

n     DT X Q16 axx þ Q26 ayy þ Q66 axy k z2k  z2k1 2 k¼1 n n o X  u DM Q12 bxx þ Q22 byy þ Q26 bxy k ½zk  zk1 

MTxy u NM yy

ð41cÞ

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

ð41f Þ

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): 9 2 9 8 T 9 8 M 9 8 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 > = = < < = 6 = < Nxy > < 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> > > > > > 6 > >Myy > > 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 MT MM 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

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

where, as before:

a

b

b

d



¼

A B

B D

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 Nxx > Nxx > 0> > > > > > > > > > > > > > > > > > > > > > > > > > > > M > > > > > > N N 0 yy > > > > > > yy > > > > > > > > > > > > > > > > > = xy ¼ ¼ > > > > 0> Mxx > MM > > > > > 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

ð5:25Þ

axy ¼ 2cosðhÞsinðhÞða11  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 ð0:9 Am=m B CÞ cos2 ð30B Þ þ ð27 Am=m B CÞ sin2 ð30B Þ

¼ 6:08 Am=m B C aðyy1Þ ¼ ð0:9 Am=m B CÞ sin2 ð30B Þ þ ð27 Am=m B CÞ cos2 ð30B Þ ¼ 20:0 Am=m B C aðxy1Þ ¼ 2cosð30Þ sinð30Þ½ð0:9  27ÞAm=m B C ¼ 24:2 Arad=B C For ply 2 (the 0j ply): 2Þ ¼ ð0:9 Am=m B CÞ cos2 ð0B Þ þ ð27 Am=m B CÞ sin2 ð0B Þ aðxx

¼ 0:9 Am=m B C aðyy2Þ ¼ ð0:9 Am=m B CÞ sin2 ð0B Þ þ ð27 Am=m B CÞ cos2 ð0B Þ ¼ 27:0 Am=m B C aðxy2Þ ¼ 2cosð0B Þ sinð0B Þ½ð0:9  27ÞAm=m B C ¼ 0 Arad=B C

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

For ply 3 (the 90j ply): 3Þ aðxx ¼ ð0:9 Am=m B CÞ cos2 ð90B Þ þ ð27 Am=m B CÞ sin2 ð90B Þ

¼ 27:0 Am=m B C aðyy3Þ ¼ ð0:9 Am=m B CÞ sin2 ð90B Þ þ ð27 Am=m B CÞ cos2 ð90B Þ ¼ 0:9 Am=m B C aðxy3Þ ¼ 2cosð90B Þ sinð90B Þ½ð0:9  27ÞAm=m B C 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  zk1  k¼1

n

  Q11 axx þ Q12 ayy þ Q16 axy 1 ½z1  z0  NTxx ¼ DT    þ Q11 axx þ Q12 ayy þ Q16 axy 2 ½z2  z1   o  þ Q11 axx þ Q12 ayy þ Q16 axy 3 ½z3  z2  n      NTxx ¼ ð155Þ 107:6
109 6:08
106 þ 26:06
109 20:0
106     þ 48:13
109 24:2
106 ð0:0625 þ 0:1875Þ
103        þ 170:9
109 0:9
106 þ 3:016
109 27:0
106      þð0Þð0Þ ð0:0625 þ 0:0625Þ
103 þ 10:05
109 27
106     o þ 3:016
109 0:9
106 þ ð0Þð0Þ ð0:1875  0:0625Þ
103 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 > 3:757
108 > > > > xx > 6 > > > > > 6 1:964
109 > > 6 > eoyy > > > > > > 6 > > > > 6 > = 6 1:038
108 < co > 6 xy ¼6 > > 6 1:440
104 > > j > > 6 xx > > > > 6 > > > > 6 > > > > 6 3:905
106 jyy > > > > 4 > > > > : ; jxy 8:513
105 9 8 4060 > > > > > > > > > > > 7360 > > > > > > > > > > = < 2860 >
> > > > > 0:62 > > > > > > > > > > > > 0:62 > > > > > ; : 0:36

1:964
109

1:038
108

1:440
104

3:905
106

1:037
107

4:234
108

1:866
105

6:361
104

4:234
108

2:004
107

3:661
104

3:251
104

1:866
105

3:661
104

7:064

6:361
104

3:251
104

4:628
104

1:851
105

3:122
102

3:122
102 4:572

6:429 3:620

8:513
105

7 4:628
104 7 7 7 7 1:851
105 7 7 7 7 4:572 7 7 7 7 3:620 5 17:41

9 8 o 9 8 285 Am=m > exx > > > > > > > > > o > > > > > > > > > 1424 Am=m > e > > > > yy > > > > > > > = < o = < 908 Arad > cxy ¼ > > 2:16 m1 > 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  e > = < xx > eyy  > > ; : c xy

9 8 9 8 o 9 8 6 e j > > = = > = < xx > < xx > < 285
10 m=m > ¼ eoyy þ z0 jyy ¼ 1424
106 m=m > > > > ; ; > ; : o > : : cxy jxy 908
106 m=m z ¼ z0 9 8 > < 2:16 rad=m > = þ ð0:0001875 mÞ 10:9 rad=m > > ; : 9:4 rad=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

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 9ply 1 2 9 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  ð155Þð6:08Þ
106 > > > = < 6
½ð3468Þ  ð155Þð20:0Þ
10 > > > > ; : ½ð2672Þ  ð155Þð24:2Þ
106 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)

eyy (Am/m)

cxy (Arad)

120 150 420 690

3468 2100 750 620

2672 1500 320 860

Strains are referenced to the x–y coordinate system.

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

Table 8 Ply Interface Stresses in a [30/0/90]T Graphite-Epoxy Laminate Caused by a Cooldown from 175jC to 20jC Ply number Ply 1 Ply 2 Ply 3

z-coordinate (mm)

rxx (MPa)

ryy (MPa)

sxy (MPa)

0.1875 0.0625 0.0625 0.0625 0.0625 0.1875

53 3.1 43 85 35 37

5.2 0.53 20 33 140 92

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): 2 2 B B bð1Þ xx ¼ ð150 Am=m  %MÞ cos ð30 Þ þ ð4800 Am=m  %MÞ sin ð30 Þ

¼ 1310 Am=m  %M bð1Þ yy

¼ ð150 Am=m  %MÞsin2 ð30B Þ þ ð4800 Am=m  %MÞcos2 ð30B Þ ¼ 3640 Am=m  %M

bð1Þ xy

¼ 2cosð30Þsinð30Þ½ð150  4800ÞAm=m  %M ¼ 4030 Arad=%M

For ply 2 (the 0j ply): 2 B 2 B bð2Þ xx ¼ ð150 Am=m  %MÞ cos ð0 Þ þ ð4800 Am=m  %MÞ sin ð0 Þ

¼ 150 Am=m  %M bð2Þ yy

¼ ð150 Am=m  %MÞ sin2 ð0B Þ þ ð4800 Am=m  %MÞ cos2 ð0B Þ ¼ 4800 Am=m  %M

bð2Þ xy

¼ 2cosð0B Þ sinð0B Þ½ð150  4800ÞAm=m  %M ¼ 0 Arad=%M

For ply 3 (the 90j ply): 2 2 B B bð3Þ xx ¼ ð150 Am=m  %MÞ cos ð90 Þ þ ð4800 Am=m  %MÞ sin ð90 Þ

¼ 4800 Am=m  %M bð3Þ yy

¼ ð150 Am=m  %MÞ sin2 ð90B Þ þ ð4800 Am=m  %MÞ cos2 ð90B Þ ¼ 150 Am=m  %M

bð3Þ xy

¼ 2cosð90B Þsinð90B Þ½ð150  4800Þ Am=m  %M 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  zk1  xx ¼ DM k¼1

n

  NM Q11 bxx þ Q12 byy þ Q16 bxy 1 ½z1  z0  xx ¼ DM    þ Q11 bxx þ Q12 byy þ Q16 bxy 2 ½z2  z1   o  þ Q11 bxx þ Q12 byy þ Q16 bxy 3 ½z3  z2 

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

9 6 NM xx ¼ ðþ0:5Þ ð½ð107:6
10 Þð1312
10 Þ þð26:06
109 Þð3638
106 Þ þ ð48:13
109 Þð4027
106 Þ
½ð0:0625 þ 0:1875Þ
103 Þ þ ð½ð170:9
109 Þð150
106 Þ þð3:016
109 Þð4800
106 Þ þ ð0Þð0Þ½ð0:0625 þ 0:0625Þ
103 Þ þ ð½ð10:05
109 Þð4800
106 Þ þ ð3:016
109 Þ 
ð150
106 Þ þ ð0Þð0Þ½ð0:1875  0:0625Þ
103 Þ 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 = < N = < 233 N=m > xy ¼ 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 3:757
108 1:964
109 exx > > > > > > 6 > > > 6 1:964
109 > > 1:037
107 eoyy > > > > > > 6 > > 8 < o > = 6 6 1:038
10 4:234
108 cxy ¼6 6 4 > > 1:866
105 > jxx > 6 1:440
10 > > > > 6 > > > > 6 6 > > jyy > > 6:361
104 4 3:905
10 > > > > ; : jxy 8:513
105 4:628
104 9 8 4060 þ 8190 > > > > > > > 7360 þ 8460 > > > > > > > > > > = < 2860  233 >
> > 0:62 þ 0:05 > > > > > > > > > > > > 0:62  0:05 > > > > ; : 0:36 þ 0:03

1:038
108

1:440
104

3:905
106

4:234
108

1:866
105

6:361
104

2:004
107

3:661
104

3:251
104

4

3:661
10

4

3:251
10

1:851
105

3:122
102

7:064 2

3:122
10 4:572

6:429 3:620

8:153
105

7 4:628
104 7 7 7 1:851
105 7 7 7 4:572 7 7 7 3:620 5 17:41

* 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.

3

9 8 o 9 8 18 Am=m > exx > > > > > > > > > > > > o > > > > > > 509 Am=m > e > > > > yy > > > > > > > > = = < < 420 Arad o cxy ¼ > > 1:0 m1 > jxx > > > > > > > > > > > > > > > > > 1 > > > > 5:0 m j yy > > > > > > > > ; : ; : jxy 4:4 m1 (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 o 9 8 9 8 exx > jxx > > 18
106 m=m > > > > > > > > < = < < = = > ¼ eoyy þ z0 jyy ¼ 509
106 m=m þ ð0:0001875 mÞ > > > > > > > > > > : ; : o > ; : ; > jxy cxy 420
106 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)

eyy (Am/m)

cxy (Arad)

206 80 44 170

1450 820 190 440

1240 690 150 400

Strains are referenced to the x–y coordinate system.

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

(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 > > > > = < < = 6 7  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 8 9ply 1 107:6
109  r < xx = 6 9 ryy  ¼6 4 26:06
10 : ; sxy z ¼ z0 48:13
109




8 > > <

26:06
109

48:13
109

3

27:22
109

7 21:52
109 7 5

21:52
109

36:05
109

½ð206Þ  ð155Þð6:08Þ  ð0:5Þð1312Þ
106

9 > > =

½ð1450Þ  ð155Þð20:0Þ  ð0:5Þð3638Þ
106 > > > ; : ½ð1240Þ  ð155Þð24:2Þ  ð0:5Þð4027Þ
106 >

9ply 1 8 25 MPa 9 8  > > > > > = = < < rxx >  ryy  ¼ 2:4 MPa > > > ; > : 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)

rxx (MPa)

ryy (MPa)

sxy (MPa)

0.1875 0.0625 0.0625 0.0625 0.0625 0.1875

25 1.4 20 40 16 17

2.2 0.24 9.3 15 65 43

2.2 9.9 9.0 1.9 1.9 5.2

Stressed are referenced to the x–y coordinate system.

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

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: 2

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 > > > > > > > > > > > =

A12

0

0

0

A11

0

0

0

0

0

D11

D12





0

A11  A12 2 0

0

0

D12

D11

0

0

0

0

0

3

8 o 9 8 T9 8 M9 N > >N > > 7> > > > > exx > > > > > > > > 7> > > > > > > > > > > > 7> > > > > T> M> > > > > eo > > 7> > > > > > N N > > > > yy > > > > 7> > > > > > > > > > > > > > > > 7 > > > > > 0 = 7 < co = < 0 = < 0 > 7 xy   7 0 > > > > 7> > > > > > > 0 > > > > > > jxx > 7> 0 > > > > > > > > > > 7> > > > > > > > > > > > 7> 0 > > > > > >j > > > > > 7> 0 yy > > > > 0 > > > > > > >   7> > > > > > > > > > > > > 5 D11  D12 ; : ; ; : : jxy 0 0 2 0 0

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

(46)

where: A11 ¼

Et 1  m2

A12 ¼ mA11 ¼

D11 ¼ D22 ¼

Et3 12ð1  m2 Þ

mEt 1  m2

D12 ¼ mD11 ¼

mEt3 12ð1  m2 Þ

ðD11  D12 Þ Et3 ¼ 2 24ð1 þ mÞ



Eta Etb NT ¼ DT NM ¼ DM ð1  mÞ ð1  mÞ D66 ¼

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

a11

0

0

0

0

2ða11  a12 Þ

0

0

0

0

d11

d12

0

0

d12

d11

0

0

0

0

9 38 Nxx þ NT þ NM > > > > > > 7> > T M> 7> > > 0 N þ N þ N > > yy > 7> > > > 7> = < 0 7 N xy 7 7> > 0 Mxx > 7> > > > 7> > > > > 7> > 0 Myy > 5> > > > > ; : Mxy 2ðd11  d12 Þ 0

where: a11 ¼

1 1 ¼ A11 ð1  m2 Þ Et

d11 ¼ d22 ¼

a12 ¼ ma11

1 12 ¼ D11 ð1  m2 Þ Et3

d66 ¼ 2ðd11  d12 Þ ¼

m m ¼ A11 ð1  m2 Þ Et

d12 ¼ md11 ¼

m 12m ¼ D11 ð1  m2 Þ 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  z2k1 2 k¼1

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 n2 z2n2  z2n3 þ Q11 n1 z2n1  z2n2     þ Q11 n z2n  z2n1 g

Because the laminate is assumed to be symmetrical, it must be that:     Q11 1 ¼ Q11 n     Q11 2 ¼ Q11 n1     Q11 3 ¼ Q11 n2   etc:

ð48Þ

ð49aÞ

Also, due to symmetry, the ply interface positions are located symmetrically about the midplane, and hence (recalling that z0<0 and zn>0): z0 ¼ zn z1 ¼ zn1 z2 ¼ zn2   etc:

ð49bÞ

Together, the relations listed as Eqs. (49a) and (49b) imply that for any symmetrical laminate:         Q11 1 z21  z20 ¼  Q11 n z2n  z2n1         ð50Þ Q11 2 z22  z21 ¼  Q11 n1 z2n1  z2n2    2      Q11 3 z3  z22 ¼  Q11 n2 z2n2  z2n3   etc:

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

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  z2k1 2 k¼1

T In expanded form, Mxx is given by: (    DT  Q11 axx þ Q12 ayy þ Q16 axy 1 z21  z20 MTxx ¼ 2

    þ Q11 axx þ Q12 ayy þ Q16 axy 2 z22  z21     þ Q11 axx þ Q12 ayy þ Q16 axy 3 z23  z22 þ......     þ Q11 axx þ Q12 ayy þ Q16 axy n2 z2n2  z2n3     þ Q11 axx þ Q12 ayy þ Q16 axy n1 z2n1  z2n2 )    2  þ Q11 axx þ Q12 ayy þ Q16 axy n zn  z2n1

ð51Þ

Because the laminate is assumed to be symmetrical, it must be that: 

  2      z  z20 ¼  Q11 axx þ Q12 ayy þ Q16 axy n z2n  z2n1 1 1         Q11 axx þ Q12 ayy þ Q16 axy 2 z22  z21 ¼  Q11 axx þ Q12 ayy þ Q16 axy n1 z2n1  z2n2         Q11 axx þ Q12 ayy þ Q16 axy 3 z23  z22 ¼  Q11 axx þ Q12 ayy þ Q16 axy n2 z2n2  z2n3 Q11 axx þ Q12 ayy þ Q16 axy

(52) T Substituting Eq. (52) into Eq. (51), it is seen that Mxx =0 for a symmetrical 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 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> < = T M 7 0 N þ N þ N xy xy xy 7 7 > d16 7> Mxx > > > > > 7> > > > d26 5> M > > yy > > > > ; : d66 Mxy

(53a)

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 ; ð54aÞ 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

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 > > > > > 0 7 N þ N þ N 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 jh and Q26 jh ¼ Q26 jh 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 jh and bxy jh T ¼ bxy jh 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 > > > 7 > > > Nyy > > 6 A12 A22 > > 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 > > b16 > xx xx > > > > > > 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

8 T 9 8 M9 3 Nxx > Nxx > 0 8 eo 9 > > > > > > xx > > > > > > > > > > > > T > > > > > > 0 7 o > Nyy > NM 7> > > > yy e > > > > > > yy > > > > > 7> > > > > > < o = < 0 = < 0 > = 0 7 7 cxy   7 D16 7> 0 > 0 > > > jxx > > > > > > > > > > > > 7> > > > > > > > > > > > > j D26 7 yy > > > > > > 0 0 > > > > > > 5: ; > > > > > > > j xy : 0 ; : 0 > ; D66 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

8 T 9 8 M9 3 Nxx > Nxx > > B16 8 eo 9 > > > > > > > > > > > xx > > > > > T M > 7 > > > > > > > > > > N N B26 7> o > > > > > > > yy yy eyy > > > > > > > > > > > 7> < = < = < = 7 0 0 0 o 7 cxy   7 > 0 > 0 > D16 7> > > > > > > > jxx > > > > > > > > > > > > 7> > > > > > > > > > > > > jyy > 0 0 D26 5> > > > > ; > > > > : > > > > : : ; ; j xy T M D66 Mxy Mxy

0 0 B26 D12 D22 D26

ð58aÞ and 2 8 o 9 a11 e xx > > 6 > > > > > eo > 6a > > > 6 12 yy > > < o > = 6 0 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

8 9 8 9 3 8 o 9 > NTxx > > NM xx > e > > > > > xx > > > > 7> > > > > > > > > T > M> o > > 0 7> > > > > N N > > > > > > eyy > > yy > > yy > > 7> > > > > > > < < < = = = 7 0 7 co 0 0 xy   7 > > > jxx > D16 7 0 > 0 > > > > > > > > > > > > 7> > > > > > > > > > > > > > > > > > j D26 7 yy > 0 > > 0 > > > > > > > 5: > ; > > > > : ; : ; jxy D66 0 0 ð59aÞ 0

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

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

A66 ¼

1 ðA11  A12 Þ 2

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 ¼ Mxy ¼ Nxy ¼ Mxy ¼ 0 . In this case, Eqs. (44) and (45) laminate, Nxy reduce to: 8 N > > > xx > > > Nyy > > > > > > < Nxy

9 > > > > > > > > > > > > =

2

A11

6 6 A12 6 6 6 6 ¼6 0 > > 6 M > > xx > > 6 B11 > > > > 6 > > > > > > Myy > 6 > 4 B12 > > > > > > : ; B16 Mxy

0

A12 A22



0



B11

B12

B12

B22

B16

B26

D11

D12

B12

A11  A12 2 B16

B22

B26

D12

D22

B26

B66

D16

D26

0

38 o 9 8 T 9 8 M 9 e > > Nxx > > N > B16 > > xx > > xx > > > > > > > > > > 7> > > > > > > eo > > > > > T B26 7> > > > > > M > > > > > > > N N yy 7> > > > yy > yy > > > > > > > > > > > > 7> > o > > > > > < < = = 7< 0 0 = B66 7 cxy   7 > > > 7> j > MTxx > MM > > > xx > > > xx > > > > > D16 7> > > > > > > > > > > > 7> > > > > > > MM > > > > > > MT > > > 7 > > > j D26 5> > > > > > yy yy yy > > > > > > > > > > > > : > ; > : ; : ; 0 0 D66 jxy ð60aÞ

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

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

b11

b12

0

b21

b22

2ða11  a12 Þ b61

b62

a12 a22 0 b21

b61

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 > > > > > > > > > > =

2

A11

6A 6 12 6 6 6 0 ¼6 6 > Mxx > > > 6 0 > > > > 6 > > > > 6 > Myy > > > 4 0 > > > > ; : Mxy 0

0

A12 A11



0

0

0

0

0

0

0

D11

D12



0

A11  A12 2 0

0

0

D12

D22

0

0

D16

D26

0

38 o 9 8 T 9 8 M 9 exx > > Nxx > > Nxx > > > > > > > > > > > > > > > > > > > > > > > o > 0 7 T > M> 7> > > > > e Nyy > Nyy > > > > yy > > > > > > > 7> > > > > > > 7> = > = > = < o > < 0 > < 0 > 7 0 7 cxy   7> > > > > > > jxx > > 0 > > 0 > > > > > > D16 7 > > > > > > 7> > > > > > > > > > > > 7> > > > > > > 0 0 j > > > > > > 5 D26 > yy > > > > > > > > > ; > ; > ; : : : 0 0 jxy D66 0

ð61aÞ

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

0 0

0 0

0 0

0

2ða11  a12 Þ

0

0

0

0

d11

d12

0

0

d12

d22

0

0

d16

d26

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=ABD1. 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

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

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

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

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 eoxx 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

9 38 0 > > > > 7> >N > > yy > > > 7> > > > 7> < 7 0 = 7 d16 7 > > 0 > 7> > > > 7> > > > > 5 0 d26 > > > > ; : 0 d66 0 0 0

Midplane strains induced are therefore: 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

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

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

9 38 0 > > > > > > 7> > > 0 7> > > > > > 7> < 7 Nxy = 7 d16 7 > > 0 > 7> > > > 7> > > > 0 d26 5> > > > > ; : 0 d66 0 0 0

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 Gxy ¼ o ¼  ð69Þ 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 whereas gxy;yy is given by: 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

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

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 mD11 ¼ 12ð1  m2 Þ ðD11  D12 Þ Et3 ¼ ¼ 2 24ð1 þ mÞ

D11 ¼ D22 ¼ D12 D66

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

E t3 xx 

2 12 1  mex xy

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

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 gxx;xy can be defined as ‘‘the midplane shear strain cjxx divided by the midplane 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 eoyy d22 Myy d22

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

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> > > > > 0 > > jyy > > ; : > > : ; 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

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

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 cxy ¼ 6 a16 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

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

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 ½ABD 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: 2

16:0
109 6 6 1:23
109 6 6 6 6 12:0
109 ½abd ¼ 6 6 0 6 6 6 6 0 4 0

1:23
109

12:0
109

0

0

20:0
109

6:0
109

0

0

6:0
109

75:8
109

0 3:51
10

0 1

0

0

0

0

2:92
102

0

1

0

3:92
10

2:92
102 1:408 6:96
101

0

7 7 7 7 7 0 7 7 1 7 3:92
10 7 7 7 6:96
101 7 5 0

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
109 Þ

ex

1 1 ¼ 66:7 GPa ¼ ta22 ð0:000750Þð20
109 Þ

ex

1 1 ¼ 17:6 GPa ¼ ta66 ð0:000750Þð75:8
109 Þ

Exx ¼ Eyy ¼ Gxy ¼

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

3

mex xy mex yx

  a12  1:23
109 ¼ ¼ ¼ 0:077 a11 16:0
109   a12  1:23
109 ¼ ¼ ¼ 0:061 20
109 a22

gex xx;xy ¼

a16 12
109 ¼ ¼ 0:75 a11 16
109

gex yy;xy ¼

a26 6:0
109 ¼ ¼ 0:30 a22 20
109

gex xy;xx ¼

a16 12:0
109 ¼ ¼ 0:16 a66 75:8
109

gex xy;yy ¼

a26 6:0
109 ¼ ¼ 0:079 a66 75:8
109

Effective flexural properties are found to be: E flxx ¼

12 12 ¼ ¼ 81:0 GPa 3 d11 t ð0:351Þð0:000750Þ3

E flyy ¼

12 12 ¼ ¼ 20:2 GPa d22 t3 ð1:408Þð0:000750Þ3

mflxy ¼

d12 2:92
102 ¼ 0:083 ¼ d11 0:351

mflyx ¼

d12 2:92
102 ¼ 0:021 ¼ d22 1:408

gflxx;xy ¼

d16 0:392 ¼ 1:12 ¼ 0:351 d11

gflyy;xy ¼

d26 0:696 ¼ 0:49 ¼ 1:408 d22

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.,

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

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

i 1 h a11 NTxx þ a12 NTyy þ a16 NTxy DT

axx ¼

1  16:0
109 ð52:3Þ  1:23
109 ð94:9Þ ð 1Þ 12:0
109 ð36:9Þ

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

Qij

 k

ðzk  zk1 Þ

k¼1

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

ðrepeatedÞ ð27aÞ

Bij ¼

n    1X Qij k z2k  z2k1 2 k¼1

ðrepeatedÞ ð27bÞ

Dij ¼

n    1X Qij k z3k  z3k1 3 k¼1

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 ¼ ð z  z Þ ¼ t V ¼ cos 2hk ðzk  zk1 Þ VA k k1 0 1 k¼1

VA 2 ¼

n X

k¼1

sin 2hk ðzk  zk1 Þ

VA 3 ¼

k¼1

VA 4 ¼

n X

n X

cos 4hk ðzk  zk1 Þ

ð77Þ

k¼1

sin 4hk ðzk  zk1 Þ ¼ t

k¼1 n  X

n   1X cos 2hk z2k  z2k1 2 k¼1

VB0 ¼

 1 z2  z2k1 ¼ 0 2 k¼1 k

VB2 ¼

n n     1X 1X sin 2hk z2k  z2k1 VB3 ¼ cos 4hk z2k  z2k1 2 k¼1 2 k¼1

VB4 ¼

n   1X sin 4hk z2k  z2k1 2 k¼1

VD 0 ¼

n   1X t3 z3k  z3k1 ¼ 3 k¼1 12

VD 2 ¼

n n     1X 1X sin 2hk z3k  z3k1 VD cos 4hk z3k  z3k1 3 ¼ 3 k¼1 3 k¼1

VB1 ¼

VD 1 ¼

ð78Þ

n   1X cos 2hk z3k  z3k1 3 k¼1

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

ð79Þ

VD 4 ¼

n   1X sin 4hk z3k  z3k1 3 k¼1

Next, consider the steps necessary to calculate the A11 term. In this case, Eq. (27a) becomes: A11 ¼

n X  Qij k ðzk  zk1 Þ

ð80Þ

k¼1

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 ðzk  zk1 Þ 1 þ U2 cos 2h þ U3 cos 4h k

k¼1

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

ðzk  zk1 Þ þ UQ 2

k¼1 n X Q

þ U3

n X

cos 2hk ðzk  zk1 Þ

k¼1

ð82Þ

cos 4hk ðzk  zk1 Þ

k¼1 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: Q A Q A A A22 ¼ UQ 1 V0  U2 V1 þ U3 V3

ð83bÞ

Q A A A12 ¼ A21 ¼ UQ 4 V0  U3 V3

ð83cÞ

A66 ¼

A UQ 5 V0

A16 ¼ A61 A26 ¼ A62

A UQ 3 V3

 1 A ¼ UQ VA þ UQ 3 V4 2 2 2 1 A ¼ UQ VA  UQ 3 V4 2 2 2

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

ð83dÞ ð83eÞ ð83f Þ

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 Q B B B21 ¼ UQ 4 V0  U3 V3 Q B B UQ 5 V0  U3 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: Q D Q D D D11 ¼ UQ 1 V0 þ U2 V1 þ U3 V3

D22 ¼ D12 ¼ D66 ¼

Q D D  UQ 2 V1 þ U3 V3 Q D D D21 ¼ UQ 4 V0  U3 V3 Q D D UQ 5 V0  U3 V3

D26 ¼ D62

ð84f Þ

ð85aÞ

D UQ 1 V0

D16 ¼ D61

ð84eÞ

ð85bÞ ð85cÞ ð85dÞ

1 ¼ UQ VD þ UQ VD 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 ðzk  zk1 Þ þ UQ ½cos 2ðhk  /Þðzk  zk1 Þ ð86Þ 2 k¼1 n X ½cos þUQ 3 k¼1

k¼1

4ðhk  /Þðzk  zk1 Þ

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

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:  Q A A A A22 ¼ UQ 1 V0  U2 V1 cosð2/Þ þ V2 sinð2/Þ ð87bÞ

A  A þUQ 3 V3 cosð4/Þ þ V4 sinð4/Þ  Q A A A A12 ¼ A21 ¼ UQ ð87cÞ 4 V0  U3 V3 cosð4/Þ þ V sinð4/Þ

 Q A A A A66 ¼ UQ ð87dÞ 5 V0  U3 V3 cosð4/Þ þ V4 sinð4/Þ  1 Q A U2 V2 cosð2/Þ  VA 1 sinð2/Þ 2  A A þUQ 3 V4 cosð4/Þ  V3 sinð4/Þ

ð87eÞ

 1 Q A U2 V2 cosð2/Þ  VA 1 sinð2/Þ 2  A A UQ 3 V4 cosð4/Þ  V3 sinð4/Þ

ð87f Þ

A16 ¼ A61 ¼

A26 ¼ A62 ¼

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

 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

 1 Q B U2 V2 cosð2/Þ  V1B sinð2/Þ 2

 þU3Q V4B cosð4/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

D22

D12 D66

ð88aÞ

ð88bÞ ð88cÞ ð88dÞ ð88eÞ

ð88f Þ

ð89aÞ

ð89bÞ ð89cÞ ð89dÞ

 1 Q D U2 V2 cosð2/Þ  V1D sinð2/Þ 2

 þU3Q V4D cosð4/Þ  V3D sinð4/Þ

ð89eÞ

 1 Q D U V2 cosð2/Þ  V1D sinð2/Þ 2 2

 U3Q V4D cosð4/Þ  V3D sinð4/Þ

ð89f Þ

D16 ¼ D61 ¼

D26 ¼ D62 ¼

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

E22 ¼ 10
109 Pa

a11 ¼ 0:9
106 m=m B C

m12 ¼ 0:30

G12 ¼ 13
109 Pa

a11 ¼ 27:0
106 m=m B C

b11 ¼ 150:0
106 m=m  %M b22 ¼ 4800
106 m=m  %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=150
103 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=2000
106 m/m=0.002000 m/m and jxx=0.5 m1. 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

E22 ¼ 1:5
106 psi

a11 ¼ 0:5
106 in=in  B F b11 ¼ 150:0
106 in=in  %M

m12 ¼ 0:30

G12 ¼ 1:9
106 psi

a11 ¼ 15
106 in=in  B F b22 ¼ 4800
106 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
106 in./ in. = 0.002000 in./in. and jxx = 0.01 in1. 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: exx ¼ 1000 in:=in:

eyy ¼ 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.

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:

cxy ¼ 750 Arad

Similarly, the strains measured by rosette 2 are: exx ¼ 1000 Ain:=in:

eyy ¼ 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

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

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  8

6 0:2558E  8 6 6 6 0 6 6 0 6 6 6 0 4 0

0:2558E  8 1:763E  8 0 0 0

0

0

0

0

3

7 7 7 7 4:078E  8 0 0 0 7 0 9:029E  2 5:160E  2 2:713E  2 7 7 7 0 5:160E  2 0:6674 9:958E  2 7 5

0

0

0

0

0

0

2:713E  2

9958E  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 8607 N=m NTyy ¼ 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.

(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

Nyy ¼ 7 kN=m

Nxx ¼ 10 N m=m

Myy ¼ Mxy ¼ 0

Nxy ¼ 0

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

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

Nyy ¼ 7 kN=m

Nxx ¼ 10 N m=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.

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

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

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

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

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

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

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

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].

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

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

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

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

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

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].

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

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

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

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.

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

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.

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

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

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

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.

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

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 h

ryy jplies

1 and 3

sxy jplies

1 and 3

rxx jply

ryy jply

n

2

2

i ð1Þ ð2Þ v12  v12 E22 Nxx ¼  hn o n oi2 ð1Þ ð2Þ tp 9E11  E22 v12 þ 2 v12

ð1bÞ

¼0

ð1cÞ

h io ð1Þ ð2Þ ð1Þ 3E11  E22 v12 2v12 þ v12 Nxx ¼  hn o n oi2 ð1Þ ð2Þ tp 9E11  E22 v12 þ 2 v12

ð2aÞ

h i ð1Þ ð2Þ 2 v12  v12 E22 Nxx ¼  hn o n oi2 ð1Þ ð2Þ tp 9E11  E22 v12 þ 2 v12 ¼ 2ryy jplies

ð2bÞ 1 and 3

sxy jply 2 ¼ 0

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.

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

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.

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

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

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

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

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

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

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

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/b<0.5), the Pipes and Pagano results are in excellent agreement with CLT predictions. However, at regions near the free edge (for y/b>0.5), 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<0.70. As y/b ! 1.0, rzz/exx is initially compressive and then becomes tensile, reaching a maximum tensile value of about 2.8 GPa (0.40 Msi) at the free edge. As the free edge is approached, the predicted normalized in-plane shear stress sxy/exx increases very slightly from the value predicted by CLT, reaching a maximum value of about 8.41 GPa (1.22 Msi) at y/ bc0.7 It then decreases to zero at the free edge as it must to satisfy equilibrium at that point. The predicted normalized out-of-plane shear stress sxz/ezz is essentially zero for y/b<0.5. However, it increases very rapidly as y/b!1.0 and, in fact, Pipes and Pagano speculated that sxz is singular at the free edge, although they could not prove this conclusively due to the numerical nature of their study. As will be further discussed below, subsequent analytical studies have shown that free-edge stresses are in fact singular at the free edge. The predicted normalized out-of-plane shear stress syz/ezz is almost nonexistent throughout the entire width of the laminate. Values of syz/ezz less than 0.6 GPa (0.09 Msi) were predicted for y/b>0.75, but then decrease to zero at the free edge (these predictions are barely discernable in Fig. 5).

It should be noted that these are not general results. Rather, these predicted stress distributions are for the specific case of a [45/45]s graphite–

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

epoxy laminate subjected to a remote uniform axial strain exx, based on the ply properties listed above. Different stress distributions would be predicted if a different stacking sequence were involved, if the laminate were subject to different loading conditions, and/or if different material properties were used. Nevertheless, the pioneering study of Pipes and Pagano showed that a threedimensional state of stress generally exists within a narrow region near an unloaded edge of a composite laminate. The width of the region over which appreciable free-edge stresses develop was estimated to be equal to or less than the total thickness of the laminate t. The free-edge shear stress sxz was of particular concern because the numerical analysis indicated that this stress component might be singular near the free edge. The computing power, processing speed, and availability of digital computers were rapidly increasing at about the time the Pipes and Pagano analysis appeared in the literature. The advent of powerful digital computers led to rapid advances in numerical structural analysis methods, particularly in the finite-element technique. Several numerical analyses devoted to free-edge stresses and based on the use of the finite-element technique appeared in the literature following the Pipes and Pagano study (Refs. 11–13) are representative examples. Although predictions based on finite-element methods were in general agreement with the original finite-difference study by Pipes and Pagano, there was growing consensus that a singular stress field exists at the free edge [16–18]. That is, as finite-element analyses became more refined, allowing greater mesh densities to be used to study the free-edge problem, the magnitude of predicted free-edge stresses became larger and larger. The fact that individual stress components (in particular, the values of rzz and sxz) did not converge to a stable value as mesh densities were increased near the free edge is indicative of a singular stress field, although this conclusion could not be proven rigorously on the basis of numerical studies alone. Wang and Choi [20,21] and Zwiers et al. [22] eventually confirmed the singular nature of free-edge stress fields on the basis of analytical studies. The approach taken by Wang and Choi [20,21] is based on the use of complex variable stress potentials, which is an analytical approach originated primarily by Lekhnitski. In contrast, the approach taken by Zwiers et al. [22] is based on the use of complex variable displacement potentials, which is an analytical approach originated primarily by Stroh. Although either approach may be taken, most analytical studies of free-edge stresses (e.g., Refs. 19,20,23–25) have been based on the use of Lekhnitski’s complex stress potentials. These analytical studies have shown conclusively that stresses induced near a free edge are singular in nature and can be written in general form as: rij ¼ fðy; zÞrd ¼

f ðy; zÞ rd

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

ð3Þ

where f ( y,z) represents a generalized function that depends on laminate stacking sequence, material properties, and applied loading, and r represents the distance from the intersection between the interface between two plies and the free edge. It can be shown [16] that the exponent d depends on both material properties as well as the difference in fiber angles between adjacent plies in the laminate. Also, d must be a positive number bounded by 0
Copyright © 2004 by Marcel Dekker, Inc. All Rights Reserved.

Figure 6 Strength of the free-edge stress singularity in [Fh]s graphite–epoxy laminates. (From Ref. 20, with permission from ASME.)

measured at the free-edge using Moire´ interferometry. These authors also considered a [45/45]s graphite–epoxy laminate subjected to axial tensile strain, although the material properties used in this study differed slightly from those in the earlier analyses of Pipes and Pagano and Wang and Choi. They concluded that stress component rzz is compressive at the free edge, in agreement with the Wang and Choi analysis. Hence, it appears that the predicted distribution of the out-of-plane normal stress rzz is particularly sensitive to inclusion of the stress singularity in the problem formulation.

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

Three important points should be emphasized. First, the distribution of free-edge stresses depends on material property and stacking sequence. Because the number of different combinations of materials and stacking sequences is infinitely large, it is not possible to tabulate solutions for freeedge stresses for all laminates that may be encountered in practice; a new analysis is required for each new combination. Second, most analyses that have appeared in the literature have focused on free-edge stresses caused in rectangular test specimens subjected to uniaxial tensile loading. In practice, composite structures are required to support combinations of tensile, compressive, and shear loading, and free-edge stresses may occur at bolt holes, cutouts, or other geometrical discontinuities. Therefore, results of studies that have appeared in the literature may not be directly applicable during the design of a composite structure, even if the same material and stacking sequence are involved. Finally, as discussed by Herakovich et al. [9], the predicted singular nature of free-edge stresses is, in reality, an artifact of available structural analysis methods. That is, all of the analyses described above are based on the assumption of linear–elastic material behavior and, furthermore, all analyses assume that a distinct interface exists between plies. Neither of these assumptions is rigorously valid. A distinct interface does not exist in real composite laminates. Although it is a certainty that high stress levels and stress gradients are induced near a free edge, in reality, high stress levels cause a nonlinear material response, and therefore the magnitude of free-edge stresses is large but finite. This observation is, of course, borne out by experiment because if free-edge stresses were truly singular, then a composite laminate would delaminate immediately upon application of any external load, regardless of the magnitude of the load. 3 PREDICTING LAMINATE FAILURE USING CLT The many factors that must be considered in order to predict the failure of a multiangle composite laminate have been discussed in Secs. 1 and 2. Factors that must be considered at any load level include the existence of thermal and/ or moisture stresses, time-dependent material behavior, and singular (in a mathematical sense) free-edge stresses. As loads are applied and increased, nonlinear behavior (‘‘yielding’’) eventually occurs in one or more plies. As loads are further increased, a multitude of distinct damage ‘‘events’’ (matrix cracking, delaminations, fiber microbuckling, etc.) occurs. It is the integrated effect of all of these factors that governs the final fracture of a composite laminate. Because of these complications, the prediction of fracture of polymeric composites has proven to be a formidable challenge, and methods of predicting fracture under general thermomechanical loading conditions remain an active area of research.

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

Nevertheless, the practicing engineer requires some method of estimating the load-carrying capacity of a composite laminate for purposes of preliminary design. Toward that end, classical lamination theory can be used to estimate when failure of a multiangle laminate will occur. In essence, CLT is used to predict the ply stresses and/or strains that will be induced by a thermomechanical loading of interest. Once ply stresses/strains are known, one of the macromechanics-based failure criteria developed in Sec. 5 of Chap. 5 is used to predict failure of individual plies, ultimately leading to a failure prediction for the laminate as a whole. CLT can be used to predict the first-ply failure of a multiangle laminate with a reasonable degree of accuracy. Recall that, in this text, first-ply failure is meant to imply the initiation of either yielding or fiber fracture in one or more plies. This philosophy results in a conservative estimate of the load-carrying capacity of a composite laminate because (in general) a multiangle laminate can withstand a considerable increase in loading beyond first-ply failure. It is also possible to estimate last-ply failure on the basis of a CLT analysis, although with a substantial decrease in accuracy. Methods of predicting first-ply and last-ply failure using CLT are discussed in Secs. 3.1 and 3.2. 3.1 Predicting First-Ply Failure According to the first-ply failure design philosophy, laminate failure is considered to occur when a given combination of loading, temperature changes, and/or moisture changes causes nonlinear behavior to develop in any ply. The first ply failure stress for the specific case of a [0/45/90/45]s laminate subjected to a uniaxial tensile loading was discussed in Sec. 1 and illustrated schematically in Fig. 1. This same concept can be extended to laminates with different stacking sequences and/or to more general thermomechanical loading conditions. Although several different solution paths may be followed, one approach is illustrated in Fig. 7. The flow diagram shown in this figure is based on the assumption that the laminate of interest is subjected to a uniform change in temperature and/or moisture content as well as some combination of uniform external loads (Nxx, Nyy, Nxy, Mxx, Myy, Mxy). It is further assumed that the relative magnitude of each load component is known. That is, if Nxx=1, then it is assumed that the magnitudes of the remaining loads can be related to Nxx (i.e., Nyy=k1|Nxx|, Nxy=k2|Nxx|, Mxx=k3|Nxx|, Myy=k4|Nxx|, Mxy=k5|Nxx|). The relative magnitudes of each load component (i.e., constants ki) are referred to as ‘‘unit loads’’ in Fig. 7. The calculation steps shown in Fig. 7 can be summarized as follows: 

Step 1. The first step is to define the problem. Ply properties, laminate stacking sequence, unit loads applied to the laminate, stressfree temperature (often assumed to be the cure temperature), service

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

Figure 7 A flow diagram illustrating calculation of the first-ply failure load using CLT.

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













temperature, moisture content, and the failure criterion to be employed must all be specified. Step 2. The [ABD] matrix for the laminate is determined. The reduced stiffness matrix Qij for each material used in the laminate is calculated (using Eq. (11) of Chap. 5), and then the transformed reduced stiffness matrix Qij for each ply is determined (using Eq. (31) of Chap. 5). This allows calculation of the Aij, Bij, and Dij matrices, using Eqs. (20a), (20b), and (27) of Chap. 6, respectively, which are then assembled to form the [ABD] matrix. Step 3. Thermal and moisture stress and moment resultants corresponding to the specified DT and DM are calculated using Eqs. (25) and (28) of Chap. 5, and Eqs. (34) and (35) of Chap. 6. Step 4. Ply stresses and strains in the global x–y coordinate system are determined. This involves calculation of the midplane strains and curvatures using Eq. (37) of Chap. 6, calculation of the ply strains in the x–y coordinate system using Eq. (12) of Chap. 6, and calculation of the ply stresses in the x–y coordinate system using Eq. (30) of Chap. 5. Step 5. If a stress-based failure criterion has been selected for use (such as the maximum stress, Tsai–Hill, or Tsai–Wu failure criterion), then ply stresses are rotated to the local 1–2 coordinate system defined by each ply angle using Eq. (20) of Chap. 2. Alternatively, if a strain-based failure criterion has been selected for use (strain-based failure criteria have not been discussed in this text), the ply strains are rotated to the local 1–2 coordinate system using Eq. (44) of Chap. 2. Steps 6 and 7. The failure criterion is applied to each ply, and the ply ‘‘nearest’’ to failure is identified. The unit loads are then increased, and steps 4 and 5 are repeated iteratively, until the first ply failure condition is reached. Details of the iteration process depend on the failure criterion selected for use. Step 8. Results are outputted once the first-ply failure condition has been identified.

The use of CLT to predict first-ply failure according to the flow diagram shown in Fig. 7 is illustrated in the following Example Problem. Example Problem 1 Predict the first-ply failure load for a [0/30/60]s graphite–epoxy laminate subjected to a uniaxial tensile load Nxx as well as a change in temperature corresponding to cooldown from cure temperatures (175jC) to room temperatures (20jC). Assume no change in moisture content, and that the laminate is

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

stress-free at the cure temperature.* Base the prediction on the maximum stress failure criterion, use properties for graphite–epoxy listed in Table 3 of Chap. 3, and assume that each ply has a thickness of 0.125 mm. Solution. The problem statement implies that: Nyy ¼ Nxy ¼ Mxx ¼ Myy ¼ Mxy ¼ DM ¼ 0 The following properties for graphite–epoxy are taken from Table 3 of Chap. 3: E22 ¼ 10 GPa v12 ¼ 0:30 G12 ¼ 13 GPa E11 ¼ 170 GPa rf11T ¼ 1500 MPa

ryT 22 ¼ 50 MPa

sy12 ¼ 75 MPa

ryC rfC 11 ¼ 1200 MPa 22 ¼ 100 MPa B a11 ¼ 0:9 Am=m  C a22 ¼ 27 Am=m B C Employing CLT calculations discussed in previous chapters, a change in temperature of DT = 155jC (alone) will cause the following stresses in each ply: For the 0j plies (plies 1 and 6), a DT = 155jC will cause: r11 ¼ 55:54 MPa

r22 ¼ 28:36 MPa

s12 ¼ 22:83 MPa

For the 30j plies (plies 2 and 5), a DT=155jC will cause: r11 ¼ 29:59 MPa

r22 ¼ 24:79 MPa

s12 ¼ 0 MPa

For the 60j plies (plies 3 and 4), a DT=155jC will cause: r11 ¼ 55:54 MPa

r22 ¼ 28:36 MPa

s12 ¼ 22:83 MPa

The additional stresses caused by the external load Nxx are superimposed with the initial thermal strains. A CLT analysis gives the following predicted stresses: For the 0j plies (plies 1 and 6): r11 ¼ ð2:750
103 ÞNxx  55:54 MPa r22 ¼ ð0:5193
102 ÞNxx þ 28:36 MPa s12 ¼ ð0:1748
103 ÞNxx þ 22:83 MPa For the 30j plies (plies 2 and 5): r11 ¼ ð1:112
103 ÞNxx þ 29:59 MPa r22 ¼ ð0:1206
103 ÞNxx þ 24:79 MPa s12 ¼ ð0:2646
103 ÞNxx * As discussed in Sec. 5 of Chap. 6, the stress-free temperature is likely to be lower than the cure temperature. For simplicity, it is assumed in this text that the final cure temperature defines the stress-free temperature.

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

For the 60j plies (plies 3 and 4): r11 ¼ ð0:2098
103 ÞNxx  55:54 MPa r22 ¼ ð0:1760
103 ÞNxx þ 28:36 MPa s12 ¼ ð0:8985
102 ÞNxx  22:83 MPa These results can be used to determine the stresses induced by the combined effects of DT = 155jC and any value of Nxx, assuming a linear– elastic response. For example, if Nxx = 100 kN/m, then the stress induced in the fiber direction in plies 1 and 6 is: r11 = (2.750
103)(100
103)55.54 MPa=219.5 MPa. Because the maximum stress failure criterion is used in this sample problem, the value of Nxx necessary to cause stresses to reach critical levels in any ply can be determined directly using the specified failure stresses: For the 0j plies (plies 1 and 6): 1500
106 þ 55:54
106 ¼ 565:7
103 N=m 2:75
103

For fiber failure : N fail xx ¼

For matrix failure : N fail xx ¼

50
106  28:36
106 ¼ 416:7
103 N=m 0:5193
102

75
106  22:83
106 ¼ 559:7
103 N=m 0:1748
103

For shear failure : N fail xx ¼ For the 30j plies (plies 2 and 5):

1500
106  29:59
106 ¼ 1322
103 N=m 1:112
103

For fi ber failure : N fail xx ¼

For matrix failure : N fail xx ¼

50
106  24:79
106 ¼ 209:0
103 N=m 0:1206
103

75
106 ¼ 283:4
103 N=m 0:2646
103

For shear failure : N fail xx ¼ For the 60j plies (plies 3 and 4): For fi ber failure : N fail xx ¼

1200
106 þ 55:54
106 ¼ 5455
103 N=m 0:2098
103

For matrix failure : N fail xx ¼ For shear failure : N fail xx ¼

50
106  28:36
106 ¼ 123:0
103 N=m 0:1760
103

75
106 þ 22:83
106 ¼ 580:6
103 N=m 0:8985
102

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

The first-ply failure load equals the lowest value of Nxx that causes the stress in any ply to reach a critical level. Examination of these results shows that firstply failure is predicted to occur when Nxx=123.0 kN/m due to a matrix yielding failure in the 60j plies (plies 3 and 4). Equivalently, the effective stress at first-ply failure is: rxx ¼

Nxx 123:0 kN=m ¼ 164 MPac24 ksi: ¼ 6ð0:125 mmÞ t

3.2 Predicting Last-Ply Failure Last-ply failure calculated using CLT is based on the so-called ‘‘ply-discount’’ scheme. A flow diagram showing typical steps in the analysis is presented in Fig. 8. The process begins with calculations similar to those described for a first-ply failure analysis. The first step is to define the problem to be considered (e.g., specify materials properties, laminate description, unit loads, temperature and/or moisture changes, and the ply failure criterion to be applied). Standard CLT calculations are then performed to determine the stresses and strains induced in all plies. Unit loads are increased until first-ply failure is predicted, based on the ply failure criterion selected for use. Once a ply is predicted to have failed, the elastic properties of that ply are ‘‘discounted’’ (i.e., changed). The change in elastic properties represents the fact that the ply has either undergone yielding or has fractured. An unresolved question is: In what manner should the change in ply properties be modeled? The traditional approach is based on the assumption that the ply has suffered a matrix crack; therefore, the stiffnesses of a failed ply are reduced in a discrete step. As an example of typical values, the stiffnesses of a failed ply could be related to the intact values according to: Efailed ¼ E11 11 vfailed ¼ v12 12 Efailed ¼ 0:3E22 22 Gfailed ¼ 0:3G12 12 failed In this example, the fiber-dominated properties of the failed ply (E11 and failed v12 ) are not reduced at all from those of an intact ply. In contrast, the failed failed and G12 ) are reduced to 30% of those matrix-dominated properties (E22 of an intact ply. These estimates reflect the fact that matrix fractures occur well before fiber fractures and, in fact, a significant number of fiber fractures do not ordinarily occur until the last-ply failure load is reached. Matrixdominated stiffnesses are not reduced to zero because even a cracked ply will contribute to some extent to the overall laminate stiffness. Although a discrete reduction in ply stiffness has been the traditional approach, it is by no means clear that this technique can be used to model the

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

Figure 8 A flow diagram illustrating calculation of the last-ply failure load using CLT.

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

nonlinear behavior of composites based on new and toughened matrix materials. As discussed in Chap. 3, prior to about 1980, most commercially available high-performance composites were based on relatively brittle polymer matrices. In that case, the assumption that plies fracture before any significant nonlinear (i.e., yielding) behavior occurs may be appropriate. However, the ductility of currently available composites has been greatly enhanced since the 1980s, which means that most systems exhibit a pronounced nonlinear response prior to fracture. In this context then, first-ply failure is defined as the load at which ply yielding first occurs, and the stiffness of yielded plies should be reduced in a smooth manner to represent this phenomenon, rather than by a discrete reduction in ply stiffness. In any event, once a ply or plies is predicted to fail and ply properties are reduced accordingly, all CLT calculations are repeated as indicated in Fig. 8. Note that the [ABD] matrix must be recalculated because the properties of one or more failed plies have been changed. This process is simply repeated until all plies in the laminate are predicted to have failed, at which point the last-ply failure load has been reached. Although last-ply failure analyses have been described routinely in the composites literature, the reader is cautioned that such analyses may be significantly in error. As already discussed, significant errors in prediction may occur at load levels beyond the first-ply failure load because the ductile nature of modern composite material systems is often not properly modeled. A second factor is that as loading is increased, an extensive number of damage events (matrix cracks, delaminations, etc.) eventually occur within the laminate, even for those based on modern toughened matrices, as previously shown in Fig. 1. As damage extends throughout the laminate, the Kirchhoff hypothesis becomes less and less valid. That is, once extensive matrix cracking and/or delaminations have occurred, it is unreasonable to assume that a straight line originally normal to the laminate midplane remains straight and normal, thus invalidating the Kirchhoff hypothesis upon which CLT is based. Once significant damage has occurred, it may not be possible to relate ply stresses and strains to stress and moment resultants using the laminate [ABD] and [abd] matrices. The reader is likely to encounter a last-ply failure analysis during future studies of composite materials and/or structures. The above comments have been included here to urge the reader to view such analyses with a healthy level of skepticism. 4 LAMINATE FIRST-PLY FAILURE ENVELOPES Methods used to determine the first-ply failure load for a laminate subjected to a specific single combination of stress and moment resultants were discussed

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

in Sec. 1 of Chap. 3. Obviously, there are an infinite number of combinations of stress and moment resultants that will cause first-ply failure. Conceptually, those combinations of stress and moment resultants that (collectively) cause first-ply failure define a laminate failure surface. It is difficult to visualize such a surface, however, because there are a total of six stress and moment resultants involved and hence the failure surface involves six ‘‘dimensions.’’ However, a plane that intersects the failure surface can be visualized by considering only two of the six stress resultants. For example, combinations of Nxx and Nyy that cause first-ply failure can be calculated using the approach described in Sec. 1 of Chap. 3, while assuming that the remaining four resultants are zero (Nyy=Mxx=Myy=Mxy=0). A plot of the (Nxx,Nyy) combinations calculated under these assumptions is called a first-ply failure envelope. A typical first-ply failure envelope for a [0/45/90/45]s graphite– epoxy laminate based on the maximum stress failure criterion is shown in Fig. 9. Three curves are shown in the figure. In one case, thermal and moisture stresses and strains are ignored (i.e., this curve was generated by assuming that DT = DM = 0). In a second case, the thermal stresses/strains caused during

Figure 9 First-ply failure envelope for a [0/45/90/45]s graphite–epoxy laminate based on the maximum stress failure criterion and combinations of Nxx and Nyy (Nxy = Mxx = Myy = Mxy = 0).

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

cooldown are included in the analysis (this curve was generated by assuming that DT=155jC, DM=0). This curve illustrates the deleterious effects of thermal stresses and strains on first-ply failure loads, and reinforces the fact that thermal stresses and strains should always be considered during design of a composite structure, especially during consideration of predicted first-ply failure loads. Finally, the third curve includes both thermal effects (based on DT=155jC) as well as a change in moisture content of 1.0% (i.e., DM=1.0). As mentioned earlier, an increase in moisture content often tends to relieve pre-existing thermal stresses. This trend is evident in Fig. 9 because first-ply failure loads for the analysis that include both thermal and moisture effects are slightly higher than the analysis that includes only thermal effects. It should be kept in mind that the first-ply failure load is a conservative design philosophy, and that the failure surfaces shown in Fig. 9 do not represent load combinations that cause laminate fracture. Rather, these curves represent the loads that cause the first instance ply failure on any kind, which in most cases is the onset of yielding in one or more plies. Comparable two-dimensional failure envelopes can be developed based on any two of the six stress and moment resultants. For example, failure envelopes for a [0/45/90/45]s graphite–epoxy laminate based on any combination of (Nxx,Nxy), (Nxx,Mxx), and (Mxx, Myy) are shown in Figures 10–12, respectively. As before, three curves are shown and are based on the maximum stress failure criterion: one in which temperature and moisture effects are neglected, one in which thermal stresses/strains caused during cooldown of the laminate are included, and a third in which both thermal and moisture effects are included. Failure envelopes of distinctly different appearances may be obtained if a different failure criterion is used. For example, failure envelopes obtained using the maximum stress, Tsai–Hill, and Tsai–Wu failure criteria for combinations of Mxx and Myy are compared in Fig. 13. For this loading condition, the Tsai–Wu criterion gives the most conservative first-ply failure envelope, whereas the maximum stress criterion gives the least conservative failure envelope.* Unfortunately, the only way to determine which failure envelope best represents the failure response of a composite laminate is to compare predictions to experimental measurement. In general, for composite laminates based on ductile polymeric matrices, it would be expected that the Tsai–Hill or Tsai–Wu criterion would provide the best prediction, whereas for brittle composites the maximum stress criterion may be best.

* The failure envelope based on the Tsai–Wu criterion was generated by calculating the coupling term X12 using Eq. (61) of Chap. 5. A significantly different failure envelope may be obtained if X12 were determined on the basis of a biaxial test or by some other means (see Sec. 5.3 of Chap. 5).

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

Figure 10 First-ply failure envelope for a [0/45/90/45]s graphite–epoxy laminate based on the maximum stress failure criterion and combinations of Nxx and Nxy (Nyy=Mxx=Myy=Mxy=0).

5 COMPUTER PROGRAM LAMFAIL The computer program LAMFAIL has been developed to supplement the material presented in Secs. 3 and 4 of this chapter. This program can also be downloaded at no cost from the following website: http://depts.washington. edu/amtas/computer.html. Program LAMFAIL can be used to either: 

Perform a first-ply failure analysis, as described in Sec. 1 of Chap. 3 and illustrated in Fig. 7; or  Generate first-ply failure envelopes as discussed in Sec. 4 and illustrated in Figs 9–13. The program prompts the user to input all information necessary to perform the analysis. Properties of up to five different materials may be defined. As previously discussed, an implicit assumption is that moduli and failure strengths inputted by the user correspond to the values exhibited by the composite at the temperature and moisture content of interest. Also, in the present context, ‘‘failure’’ may represent fracture of the fibers (rf11T ; rf11C ) or

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

Figure 11 First-ply failure envelope for a [0/45/90/45]s graphite–epoxy laminate based on the maximum stress failure criterion and combinations of Nxx and Mxx (Nyy = Nxy = Myy = Mxy = 0).

yC y yielding of the matrix (ryT 22 ; r22 ; s12 ). The user specifies the number of plies within the laminate and material type and fiber angle for each ply. The user also selects from the maximum stress, Tsai–Hill, or Tsai–Wu failure criterion.* If a first-ply failure analysis for a particular combination of stress and moment resultants is to be performed, the program prompts the user for ‘‘unit loads’’ (see Fig. 7). Alternatively, if the program is to be used to generate a failure envelope, then the particular pair of resultants to be considered during the analysis is specified by the user (i.e., any two of the six stress resultants Nxx, Nyy, Nxy, Mxx, Myy, Mxy). Note that program LAMFAIL itself does not create a failure envelope. Rather, the program creates a file (named Envelop.txt) that contains the stress resultant pairs predicted to cause first-ply failure of a composite laminate specified by the user. A failure envelope may

* If the Tsai–Wu criterion is selected, then the coupling strength term X12 is calculated according to Eq. (61) of Chap. 5.

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

Figure 12 First-ply failure envelope for a [0/45/90/45]s graphite–epoxy laminate based on combinations of Mxx and Myy (Nxx = Nyy = Nxy = Mxy = 0).

then be created using a second software package to ‘‘import’’ the data generated by program LAMFAIL. The first few lines of the file Envelop.txt created during a typical analysis are shown in Table 1. In this case, the analysis was for a [0/45/90/45]s graphite–expoxy laminate, using properties listed in Table 3 of Chap. 3. The program had been directed to apply the maximum stress failure criterion, and to base the analysis on stress resultant pair Nxx and Nyy. The first few lines of the file represent header information. Notice that the file contains the ply number predicted to fail and the failure type (i.e., fiber, matrix, or shear), in addition to the combinations of Nxx and Nyy predicted to cause failure. Table 1 shows 12 such combinations of Nxx and Nyy. The complete file Envelop.txt contains a total of 2001 pairs. Thus, a firstply failure envelope may be created by importing and plotting the Nxx and Nyy pairs. File Envelop.txt differs slightly if either the Tsai–Hill or Tsai–Wu criterion is selected for use. Specifically, if either of these criteria is selected, then file Envelop.txt will not contain a column with the heading ‘‘failure type’’ because these criteria do not distinguish a particular failure mode in the same sense as does the maximum stress failure criterion. The few 25 lines of the file

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

Figure 13 First-ply failure envelope for a [0/45/90/45]s graphite–epoxy laminate for combinations of Mxx and Myy, based on the maximum stress, Tsai–Hill, and Tsai–Wu criteria (Nxx = Nyy = Nxy = Mxy = 0, DT = 155jC, DM= 1%). Figure 13 A [0/F30]s laminate loaded in biaxial tension, as described in Problem 1.

Envelop.txt created during an analysis for a [0/45/90/45]s graphite–epoxy laminate in which the Tsai–Hill criterion was employed are shown in Table 2.

HOMEWORK PROBLEMS Computer programs CLT and LAMFAIL are used in the following problems. These programs can be downloaded from the following website: http:// depts.washington.edu/amtas/computer.html. 1. Assume that a new room temperature cure graphite–epoxy prepreg system with properties listed in Table 3 of Chap. 3 has been developed. A [0/F30]s laminate is produced using this material, and is therefore initially stress-free and strain-free at room temperature. The room temperature laminate is then subjected to the biaxial tensile loads shown in Fig. 14 (i.e., Nyy = Nxx/2 and Nxy = Mxx = Myy = Mxy = 0). Use the

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

Table 1

First Few Lines of the File Envelop.txt, Created by Program LAMFAIL During an Analysis of a [0/45/90/45]s Graphite–Epoxy Laminate

**Program LAMFAIL*** Laminate failure predictions based on the maximum stress failure criterion Stress-free temperature=175.0j Service temperature=20.0j Change in temperature=155.0j Change in moisture content=1.00% Nxx 0.1590855E+06 0.1590665E+06 0.1590474E+06 0.1590282E+06 0.1590090E+06 0.1589897E+06 0.1589703E+06 0.1589508E+06 0.1589313E+06 0.1589117E+06 0.1588920E+06 0.1588723E+06    

Nyy 0.0000000E+00 0.3187704E+03 0.6387443E+03 0.9599286E+03 0.1282330E+04 0.1605956E+04 0.1930813E+04 0.2256908E+04 0.2584248E+04 0.2912841E+04 0.3242693E+04 0.3573813E+04    

Failed ply 3 3 3 3 3 3 3 3 3 3 3 3    

Failure type Matrix Matrix Matrix Matrix Matrix Matrix Matrix Matrix Matrix Matrix Matrix Matrix    

The maximum stress failure criterion had been selected for use, and stress resultants Nxx and Nyy were considered. The file also contains the number of the ply predicted to fail and the failure type (i.e., fiber, matrix, or shear) for each combination of Nxx and Nyy predicted to cause failure.

following process to determine the tensile loads Nxx and Nyy necessary to cause first-ply failure, according to the maximum stress failure criteria: (a) Use program CLT to determine the stresses (r11,r22,s12) induced in each ply by unit loads Nxx=1, Nyy=1/2 (i.e., k1=0.5). (b) Determine which ply (or plies) is closest to the failure condition dictated by the maximum stress failure criterion, using the failure strengths listed for graphite/epoxy in Table 3 of Chap. 3

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

Table 2 First Few Lines of the File Envelop.txt, Created by Program LAMFAIL During an Analysis of a [0/45/90/45]s Graphite–Epoxy Laminate **Program LAMFAIL*** Laminate failure predictions based on the Tsai–Hill failure criterion Stress-free temperature=175.0j Service temperature=20.0j Change in temperature=155.0j Change in moisture content=1.00% Nxx 0.1565825E+06 0.1565915E+06 0.1566003E+06 0.1566091E+06 0.1566178E+06 0.1566263E+06 0.1566348E+06 0.1566430E+06 0.1566512E+06 0.1566593E+06 0.1566672E+06 0.1566750E+06    

Nyy 0.0000000E+00 0.3138105E+03 0.6289169E+03 0.9453267E+03 0.1263047E+04 0.1582084E+04 0.1902446E+04 0.2224140E+04 0.2547174E+04 0.2871555E+04 0.3197291E+04 0.3524387E+04    

Failed ply 3 3 3 3 3 3 3 3 3 6 3 3    

The Tsai–Hill failure criterion had been selected for use, and stress resultants Nxx and Nyy were considered. The file also contains the number of the ply predicted to fail for each combination of Nxx and Nyy predicted to cause failure.

(c) Calculate the increase in unit loads that will cause first-ply ply failure. (d) Confirm the results of part (c) using program LAMFAIL. 2. Repeat Problem 1 for a room temperature cure glass–epoxy system 3. Repeat Problem 1 for a room temperature cure Kevlar–epoxy system. 4. Repeat Problem 1 for the following load condition: Nyy ¼ Nxx =2

Nxy ¼ 5Nxx

Mxx ¼ Myy ¼ Mxz ¼ 0

5. Repeat Problem 1, except assume that the laminate is produced using a graphite–epoxy system cured at 175jC (350jF). 6. Repeat Problem 1, except assume that the laminate is produced using a glass–epoxy system cured at 175jC (350jF).

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

Figure 14 A [0/F30]s laminate loaded in biaxial tension, as described in Problem 1.

7. Repeat Problem 1, except assume that the laminate is produced using a Kevlar–epoxy system cured at 175jC (350jF). 8. Three different first-ply failure envelopes for a [0/45/90/45]s graphite– epoxy laminate are shown in Fig. 9. The three points listed below lie on these failure envelopes. In each case, use program CLT to determine which ply(ies) is predicted to fail, according to the maximum stress failure criterion: (a) Nxx=327 kN/m, Nyy=0 kN/m (this point lies on the DT=DM=0 curve). (b) Nxx=79.4 kN/m, Nyy=0 kN/m (this point lies on the DT= 155jC, DM=0 curve). (c) Nxx = 159 kN/m, Nyy = 0 kN/m (this point lies on the DT = 155jC, DM = 1% curve). 9. Three different first-ply failure envelopes for a [0/45/90/45]s graphite– epoxy laminate are shown in Fig. 9. The three points listed below lie on these failure envelopes. In each case, use program CLT to determine which ply(ies) is predicted to fail, according to the maximum stress failure criterion: (a) Nxx=0 kN/m, Nyy=32 7kN/m (this point lies on the DT=DM=0 curve). (b) Nxx=0 kN/m, Nyy=79.4 kN/m (this point lies on the DT=155jC, DM=0 curve).

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

(c) Nxx=0 kN/m, Nyy=159 kN/m (this point lies on the DT=155jC, DM=1% curve). 10. Three different first-ply failure envelopes for a [0/45/9045]s graphite– epoxy laminate are shown in Fig. 9. The three points listed below lie on these failure envelopes. In each case, use program CLT to determine which ply(ies) is predicted to fail, according to the maximum stress failure criterion: (a) Nxx=115.2 kN/m, Nyy=212.1 kN/m (this point lies on the DT= DM=0 curve). (b) Nxx=205.2 kN/m, Nyy=91.6 kN/m (this point lies on the DT= 155jC, DM=0 curve). (c) Nxx=172.7 kN/m, Nyy=154.6 kN/m (this point lies on the DT=155jC, DM=1% curve). 11. Use program LAMFAIL to create first-ply envelopes for a [0/F45]s graphite–epoxy laminate subjected to different combinations of Nxx and Myy, based on the maximum stress, Tsai–Hill, and Tsai–Wu failure criteria. Assume that the laminate was cured at 175jC and will be used at room temperatures (20jC) during service, and that it will not experience an increase in moisture content (DM = 0%). Use a plotting format similar to Fig. 13.

REFERENCES 1. 2. 3.

4.

5.

6. 7.

Broek, D. Elementary Engineering Fracture Mechanics; Kluwer Academic Publishers: Hingham, MA, 1986, ISBN 90-247-2580-1. Reifsnider, K.L., Ed. Fatigue of Composite Materials; Elsevier Sci. Publ.: New York, NY, 1991. Selvarathinam, A.S.; Weitsman, Y.J. Shear-lag analysis of transverse cracking and delamination in cross-ply carbon-fibre/epoxy composites under dry, saturated and immersed fatigue conditions. Compos. Sci. Technol. 1999, 59 (14), 2115–2123. Surgeon, M.; Vanswijgenhoven, E.; Wevers, M.; van der Biest, O. Transverse cracking and Poisson’s ratio reduction in cross-ply carbon fibre-reinforced polymers. J. Mater. Sci. 1999, 34 (22), 5513–5517. Joffe, R.; Varna, J. Analytical modeling of stiffness reduction in symmetric and balanced laminates due to cracks in 90j layers. Compos. Sci. Technol. 1999, 59 (11), 1641–1652. Pipes, R.B.; Daniel, I.M. Moire´ analysis of the interlaminar shear edge effect in laminated composites. J. Compos. Mater. 1971, 5, 255. Oplinger, D.W.; Parker, B.S.; Chiang, F.-P. Edge-effect studies in fiber-reinforced laminates. Exp. Mech. 1974, 14 (9), 347–354.

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

8. 9.

10. 11. 12. 13. 14. 15.

16. 17. 18. 19. 20. 21.

22.

23. 24. 25. 26. 27.

Czarnek, R.; Post, D.; Herakovich, C.T. Edge effects in composites by Moire´ interferometry. Exp. Tech. 1983, 7 (1), 18–21. Herakovich, C.T.; Post, D.; Buczek, M.B.; Czarnek, R. Free edge strain concentrations in real composite laminates: experimental–theoretical correlation. J. Appl. Mech. 1985, 52, 787–793. Pipes, R.B.; Pagano, N.J. Interlaminar stresses in composite laminates under uniform axial extension. J. Compos. Mater. 1970, 4, 538–548. Rybicki, E.F. Approximate three-dimensional solutions for symmetric laminates under in-plane loading. J. Compos. Mater. 1971, 5, 354–360. Wang, A.S.D.; Crossman, F.W. Some new results on edge effects in symmetric composite laminates. J. Compos. Mater. 1977, 11, 92–106. Hsu, P.W.; Herakovich, C.T. Edge effects in angle-ply composite laminates. J. Compos. Mater. 1977, 11, 422–428. Herakovich, C.T. On the relationship between engineering properties and delamination of composite materials. J. Compos. Mater. 1981, 15, 336–348. Nailadi, C.L.; Adams, D.O.; Adams, D.F. An experimental and numerical investigation of the free edge problem in composite laminates. Proceedings of the 1998 SEM Spring Conference, Houston, TX, June 1–3, pp. 59–62. Puppo, A.H.; Evensen, H.A. Interlaminar shear in laminated composites under generalized plane stress. J. Compos. Mater. 1970, 4, 204–220. Pagano, N.J.; Pipes, R.B. The influence of stacking sequence on laminate strength. J. Compos. Mater. 1971, 5, 50–55. Pagano, N.J. On the calculation of interlaminar normal stress in composite laminate. J. Compos. Mater. 1974, 8, 65–81. Pagano, N.J. Free-edge stress fields in composite laminates. Int. J. Solids Struct. 1978, 14, 401–406. Wang, S.S.; Choi, I. Boundary-layer effects in composite laminates: Part 1. Freeedge stress singularities. J. Appl. Mech. 1982, 49, 541–548. Wang, S.S.; Choi, I. Boundary-layer effects in composite laminates: Part 2. Freeedge stress solutions and basic characteristics. J. Appl. Mech. 1982, 49, 549– 560. Zwiers, R.I.; Ting, T.C.T.; Spilker, R.L. On the logarithmic singularity of freeedge stress in laminated composites under uniform extension. J. Appl. Mech. 1982, 49, 561–569. Ting, T.C.T. Anisotropic Elasticity Theory and Applications; Oxford University Press: New York, NY, 1996, ISBN 0-19-507447-5. Kassapoglou, C.; Lagace, P.A. An efficient method for the calculation of interlaminar stresses in composite materials. J. Appl. Mech. 1986, 53, 744–750. Kassapoglou, C.; Lagace, P.A. Closed form solutions for the interlaminar stress field in angle-ply and cross-ply laminates. J. Compos. Mater. 1987, 21, 292–308. Becker, W. Closed-form solutions for the free-edge effect in cross-ply laminates. Compos. Struct. 1993, 26 (1–2), 39–45. Stern, M. Families of consistent conforming elements with singular derivative fields. Int. J. Numer. Methods Eng. 1979, 14, 409–421.

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

8 Composite Beams

Composite beams subjected to axial, transverse, or combined loads are discussed in this chapter. Composite beams with rectangular and various thin-walled cross sections are considered. The analysis presented allows the determination of the effective axial and flexural rigidities of a composite beam. Composite beams can be analyzed using the same techniques used to study the behavior of statically determinate or indeterminate isotropic beams, once the effective rigidities are determined. 1 PRELIMINARY DISCUSSION It is expected that readers of this text will have previously studied a subject known as ‘‘mechanics of materials’’ (also called ‘‘strength of materials’’). One of the primary topics considered in mechanics of materials is the structural behavior of a prismatic beam. A prismatic beam is defined as a long, slender, and initially straight structural member whose cross section does not vary along its length. Typical problem objectives are to determine the stresses, strains, and deflections induced in a prismatic beam by various types of external loads. Many excellent textbooks are available to support such studies, a few of which are listed here as Refs. (1–5). In virtually all of these references, it is assumed that the beam is composed of an isotropic material or

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

at most is a beam in which different layers of isotropic materials are bonded or bolted together to form the overall beam.* In contrast, in this chapter, we will consider the behavior of prismatic beams produced using laminated anisotropic composite materials. The traditional pedagogical approach used in textbooks devoted to mechanics of materials, including Refs. (1–5), is to separate the types of loading commonly applied to prismatic beams/bars into three categories: 

Axial loads (i.e., loads acting parallel to the long axis of the beam, which tend to cause the beam to elongate or compress).  Bending loads (i.e., loads acting transverse to the long axis of the beam, which tend to cause the initially straight beam to bend).  Torques (i.e., moments acting about the long axis of the beam, which tend to cause the beam to twist about its long axis). Having made this classification, the stresses, strains, and deflections induced in a prismatic beam by each category of external load are considered in separate sections or chapters. It should be clear that this is an artificial distinction made purely for pedagogical purposes. In actual practice, a prismatic beam or shaft may well be subjected to axial load(s), to bending load(s), and/or to torque(s) simultaneously. In such a case, the stresses, strains, and deflections induced are due to the combined effects of all load components. Composite beams subjected to axial loads and/or bending loads will be discussed in this chapter. Composite beams subjected to torques are not considered in this text. As will be seen, the analysis of composite beams subjected to axial loads and bending loads follows directly from Classical Lamination Theory (CLT). A note of caution regarding free-edge stresses is therefore appropriate. As explained in Chap. 7, free-edge stresses occur because of a mismatch in the effective properties exhibited by adjacent plies in a multiangle composite laminate. A CLT analysis does not account for free-edge stresses, and hence in regions dominated by free-edge stresses, the ply stresses and strains predicted on the basis of CLT are invalid. As a rule of thumb, the region over which significant free-edge stresses occur has a width approximately equal to the laminate thickness. Now, by definition, a ‘‘beam’’ is a structure whose width is similar in magnitude to its thickness (i.e., the dimensions of the beam cross section are much less than the beam length).

* A beam composed of two or more isotropic materials is sometimes called a ‘‘composite’’ beam; in such cases, the use of the term ‘‘composite’’ refers to the fact that the beam is composed of two or more isotropic materials. This is in contrast to the sort of ‘‘composite’’ beam discussed in this text, in which each individual composite ply is anisotropic.

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

Consequently, free-edge effects may significantly influence the ply stresses and strains induced in a composite beam. Despite these concerns, CLT will be used in this chapter to predict the response of composite beams to axial and bending loads. The effects of free-edge stresses will be ignored. The error in predicted response caused by neglecting free-edge stresses is difficult to estimate, in part because the magnitude of free-edge stresses depends on both stacking sequence and elastic properties of the material system used to fabricate the beam. In general, it is expected that free-edge stresses will have little impact on predicted beam stiffness, but may dominate beam failure loads. 2 COMPARING CLASSICAL LAMINATION THEORY TO ISOTROPIC BEAM THEORY Before we begin our discussion of composite beams, it is instructive to compare classical lamination theory (CLT) to results from isotropic beam theory. CLT is based on the behavior of a ‘‘thin plate.’’ A plate with in-plane dimensions a and b and thickness t has been shown previously in Fig. 1 of Chap. 6. The plate can be considered ‘‘thin’’ if the plate thickness is less than about 1/10 the in-plane dimensions, i.e., if t < a/10 and t < b/10. As already discussed in Sec. 1 of Chap. 6, the description of a thin plate can be ‘‘converted’’ to the description of a beam if we let the x-direction define the beam axis and allow the in-plane width of the plate (dimension b) to approach the plate thickness: b c t. In this way, we describe a beam with rectangular cross section bt and length a, as previously shown in Fig. 3 of Chap. 6. Recall that six types of loads are considered in CLT: three stress resultants (Nxx, Nyy, and Nxy) and three moment resultants (Mxx, Myy, and Mxy). For present purposes, we will only consider two of these loads: stress resultant Nxx and moment resultant Mxx. That is, to compare CLT to isotropic beam theory, we will specify that Nyy = Nxy = Myy = Mxy = 0. A further distinction must also be made regarding the manner in which loads are described in CLT vs. beam theory. In CLT, all stress and moment resultants are assumed to be uniformly distributed along the edge of a plate and are described in units corresponding to this assumption. For example, stress resultant Nxx is described in units of N/m (or lbf/in.), and moment resultant Mxx is described in units of N m/m (or lbf in./in.). In contrast, in beam theory, ‘‘point’’ or ‘‘concentrated’’ loads are usually specified. In beam theory, an axial load (corresponding to Nxx) is described in units of N (or lbf ), whereas a bending moment (corresponding to Mxx) is described in units of N m (or lbf in.). In this textbook, the superscript ‘‘b’’ is used to differentiate between loads as defined in beam theory vs. loads as defined in CLT. Thus, an axial load and b b bending moment applied to a beam will be denoted Nxx and Mxx , respec-

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

tively. Since the width of the beam is b, the two load definitions are related b b /b and Mxx = Mxx /b. according to Nxx = Nxx We are now ready to compare CLT with beam theory. From Sec. 6.2 of Chap. 6, the midplane strains and curvatures induced in a general multiangle composite laminate subjected to a combination of mechanical loads, a uniform change in temperature, and a uniform change in moisture content are given by Eq. (45) of Chap. 6, repeated here for convenience: 9 2 9 8 38 Nxx þ NTxx þ NM eoxx > a11 a12 a16 b11 b12 b16 > > 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

0 0

2ða11  a12 Þ 0 0 d11 0 d12 0

0

0 0 0 d12 d11 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): v Et 12v d12 ¼ Et3 24ð1 þ vÞ ¼ 2ðd11  d12 Þ ¼ Et3 1 Et 12 ¼ Et3

a11 ¼ d11 d66

a12 ¼

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

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 Et b b AE E 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

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

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

Mbxx EI

jyy ¼ 

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

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

Upon completing the matrix multiplication indicated above, we find: 8 9 8 b 9 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 axial stress rxx induced in an isotropic beam by Mxx (only) is given by the 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

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

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.

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

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

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

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

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

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,

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

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.

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

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 a
xx of the top flange may be very different than that of the bottom flange. Consequently, if the beam experiences a uniform change in temper-

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

Figure 5 The use of woven or braided fabrics or pre-preg avoids the difficulties shown in Figs. 3 and 4.

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

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 > yy > > 4 0 > > ; : 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

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

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 ðtbÞE xx AE xx E xx ! ! Nbxx ex ex rxx ex o ¼ vxy ¼ vxy ¼ vxy exx ex ex AE xx E xx

eoxx ¼

ð6aÞ

eoyy

ð6bÞ

coxy

¼

ex gxx;xy

Nbxx ex AE xx

! ¼

ex gxx;xy

rxx ex E xx

! ex eoxx ¼ gxx;xy

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

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

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

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

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.

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

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 > > > > : ; d66 0 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

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

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

b Mxx fl I E xx

jyy ¼  jxy ¼

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 ð11Þ IE xx ¼ 12 d11 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

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

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

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

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-

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

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.

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

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

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: eoxx ¼ eox Vx V ¼ zjxx

ð14Þ

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 > > y Vy V > > 6 = < o a a26 a66 0 cx Vy V ¼ 6 6 16 6 > 0 0 d11 jx VxV > > > 6 0 > > > > 6 > > > > 4 0 jy Vy V > 0 0 d12 > > > ; : 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> > > T M > N þ 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

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

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:

m

þb=2 

þb=2

Mbxx

¼

b=2

Nxx zdz ¼

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

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

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 V<0), these strains will be tensile. These transverse normal strains were neglected during our earlier development and hence are not included in CLT. These strains are not of immediate interest, so they will not be further discussed here. Next, consider the state of stress induced in the beam. Recall from Sec. 9.1 of Chap. 6 that the effective (or nominal) normal stress is given by rx Vx V = rxx = Nxx/t. Substituting this definition into Eq. (15) and rearranging, we have:

jxx ¼

rxx ta11 rxx ¼ ex z zE xx

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

Combining this result with Eq. (16) and solving for effective stress, we have: rxx ¼

Mbxx z I

ð18Þ

Equation (18) is analogous to the flexure formula for isotropic beams [Eq. (3)] and can be used to determine the effective stress at any position z. As would be expected, the effective stress rxx is at a maximum at the outer surfaces of the beam (at z = F t/2) and is zero at the neutral surface (at z = 0). Once the effective stress rxx has been calculated at a point of interest, then the stresses and strains present in individual plies can be determined using the standard CLT analysis methods.

6 EFFECTIVE AXIAL AND FLEXURAL RIGIDITIES FOR THIN-WALLED COMPOSITE BEAMS In this section, the effective axial and flexural rigidities of thin-walled composite beams will be considered. The beam cross section must conform to the symmetries described in Sec. 3. The process used to determine effective rigidities is similar for all cross sections. A composite box-beam will be used to illustrate the process. Referring to Fig. 11, a box-beam is modeled using four symmetric rectangular laminates. The two web laminates are required to have identical stacking sequences, but the two symmetric flange laminates may differ. The thickness of the top and bottom flange laminates is labeled ttf and tbf, respectively. The width of both flanges is labeled b. The two identical web laminates have a thickness of tw and height h. Thus, the overall width and depth of the beam cross section are (b) and (h+ttf+tbf), respectively. To define the axial rigidity, we must determine the axial strains exx b induced when the beam is subjected to an axial load Nxx (only). Further, b Nxx must be applied such that uniform axial strains are induced; that is, the b beam must not bend. This is equivalent to saying that the line of action of Nxx must pass through the centroid of the beam cross section. It is known a priori that the centroid lies along within the x–z plane since we have required symmetry about this plane. However, we have not required symmetry about the x–y plane. Hence, the position of the centroid relative to either outer surface of the beam cross section (that is, coordinates zt and zb shown in Fig. 11) is initially unknown. Note that since the z-axis is positive downwards, in all cases, zt<0 and zb>0. Our first objective is to determine either distance zt or zb. In the following derivation, we will determine distance zb. Note that once zb is known, zt is also known since zt = zb–(h+tbf+ttf).

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

Figure 11 Cross section of a composite box-beam. The positive x-axis is out of the plane of the figure. b Sketches of a composite box-beam subjected to axial load Nxx (only) b are shown in Fig. 12. In this figure, it is assumed that the line of action of Nxx does pass through the centroid. Consequently, a uniform axial strain exx is induced at an arbitrary cross section of the beam, as shown in Fig. 12(a). Since strain is uniform, the stress resultants induced in each segment of the beam can be written as: exx ex Nxx jw ¼ w ¼ tw E xx jw exx ð19aÞ a11 exx ex Nxx jbf ¼ bf ¼ tbf E xx jbf exx ð19bÞ a11 exx ex ð19cÞ Nxx jtf ¼ tf ¼ ttf E xx jtf exx a11

Symbols ‘‘w,’’ ‘‘bf,’’ and ‘‘tf’’ have been used to denote variables associated with the web laminate, bottom flange laminate, and top flange laminate, ex jw represents the effective extensional modulus respectively. For example, E xx of the web laminates.

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

b Figure 12 Side view of a composite box-beam subjected to axial load Nxx . The b line-of-action of Nxx is assumed to pass through the centroid, located distance zb from the bottom surface of the beam. Consequently, (a) uniform axial strains are induced at an arbitrary cross section, and (b) uniform stress resultants are induced in each segment of the beam wall.

The stress resultants given by Eqs. (19a)–(19c) are indicated in the freebody diagram shown in Fig. 12(b). To determine the location of the centroidal axis, we require that the moment associated with the internal stress resultants vanish. Summing moments about the y-axis and equating to zero (SMy=0), we obtain: ð2Þ½ðNxx jw ÞðhÞðzb  tbf  h=2Þ½ðNxx jbf ÞðbÞðzb  tbf =2Þ½ðNxx jtf ÞðbÞ
ðzb  tbf  h  ttf =2Þ ¼ 0 Substituting [Eqs. (19a), (19b), (19c)] and solving for distance zb, we find: 2

3    ex  ex  ex  2A E ð h þ 2t Þ þ t A E þ A E ð2t þ 2h þ t Þ    bf bf bf xx tf xx b tf 1 6 w xx w 7 bf tf  zb ¼ 4 5ð20aÞ ex ex ex    2 2A E  þ A E  þ A E  w

xx

w

bf

xx

bf

tf

xx

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

tf

where Aw = tw h = cross-sectional area of the web laminate. Abf = tbf b = cross-sectional area of the bottom flange laminate. Atf = ttf b = cross-sectional area of the top flange laminate. Noting that the effective extensional Young’s moduli can be written as ex ex ex bf tf E xx Aw ¼ 1=tw aw 11, E xx Abf ¼ 1=tbf a11, and E xx Atf ¼ 1=ttf a11, this result can also be expressed as: (  tf ) tf w bf 1 2habf 11 a11 ðh þ 2tbf Þ þ ba11 tbf a11 þ a11 ð2tbf þ 2h þ ttf Þ ð20bÞ zb ¼ w bf tf tf 2 2habf 11 a11 þ ba11 ða11 þ a11 Þ Equations (20a) and (20b) gives the distance from the centroid to the bottom surface of the composite box-beam. If the bottom and top flange laminates are ex ex identical (i.e., if Abf=Atf, tbf=ttf, and E xx Abf ¼ E xx Atf ), then Eqs. (20a) and (20b) reduces to zb=tbf+h/2, as would be expected. Let us now determine the effective axial rigidity of the composite boxbeam. Since a uniform axial strain (exx) is induced, the stress resultants present in each segment of the beam to the axial strain are given by: exx ex  Nxx jw ¼ w ¼ tw E xx w exx a11  exx ex  Nxx bf ¼ bf ¼ tbf E xx bf exx ð21Þ a 11

 exx ex  Nxx tf ¼ tf ¼ ttf E xx tf exx a11 b Also, the total load applied to the beam as a whole (Nxx ) must equal the sum of the forces acting in each beam segment:

Nbxx ¼ 2ðNxx jw ÞðhÞ þ ðNxx jbf ÞðbÞ þ ðNxx jtf ÞðbÞ Define the effective stress applied to the beam as the force applied to the beam divided by the area of the beam cross section:    2ðNxx w ÞðhÞ þ ðNxx bf ÞðbÞ þ ðNxx tf ÞðbÞ Nbxx rbeam ¼ ¼ ð22Þ 2Aw þ Abf þ Atf Abeam Substituting Eq. (21) into Eq. (22), we find:    2 ex  ex  ex  3 2Aw E xx  þ Abf E xx  þ Atf E xx  w bf tf 5 rbeam ¼ exx 4 2Aw þ Abf þ Atf

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

We can now define the effective Young’s modulus of the beam as:    ex  ex  ex  2Aw E xx  þ Abf E xx  þ Atf E xx  beam w bf tf E xx ¼ 2Aw þ Abf þ Atf

ð23Þ

The effective axial rigidity of the composite box-beam is simply the effective beam Young’s modulus multiplied by the cross-sectional area:    beam ex  ex  ex  AE xx ¼ 2Aw E xx  þ Abf E xx  þ Atf E xx  ð24aÞ w

bf

tf

Alternatively, the effective axial rigidity of the box-beam can be written as: beam

AE xx

¼

2h b b w þ bf þ tf a11 a11 a11

ð24bÞ

If the bottom and top flange laminates are identical (i.e., if Abf=Atf, tbf=ttf, ex ex jbf=E xx jtf), then Eqs. (24a) and (24b) reduce to: and E xx       h b ex  ex  AE beam ¼ 2 Aw E xx  þ Abf E xx  ¼ 2 w þ bf a11 a11 w bf Next, consider the effective flexural rigidity of a composite box-beam. Recall that the flexural rigidity of a beam is defined under a condition of pure bending. A side view of a composite box-beam subjected to pure bending is shown in Fig. 13(a). Since the beam cross section is symmetric about the x–z plane, the long x-axis of the beam must deform into a circular arc with a radius of curvature rxx, as shown. Hence, cross sections of the box-beam 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 encountered in pure bending of a rectangular composite beam with ply orientation parallel to the plane of loading (discussed in Sec. 5.2), as well as in isotropic beams with symmetric rectangular cross sections. Axial normal strains vary linearly with z and are given by: exx ¼ zjxx

ð25Þ

where curvature jxx=1/rxx. We assume that the axial strain induced at any position z is uniform across the width of the web and flange laminates; that is, we assume that exx is not a function of y. Our first objective is to determine the location of the neutral axis. That is, we wish to determine either distance zt or zb, previously shown in Fig. 11. In the following derivation, we will determine distance zb, the distance from the neutral axis to the bottom surface of the beam. Note that this is an arbitrary decision, and a comparable derivation based on distance zt can easily be developed.

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

Figure 13 Side view of a composite box-beam subjected to pure bending. (a) Composite box-beam deformed into a circular arc of radius rxx. (b) Distribution of internal stress resultants NxxAw, NxxAbf, and NxxAtf.

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

First consider the web laminates. We note that for the web laminate, the ply interfaces are parallel to the x–z plane. Also, we have assumed that the strain exx is not a function of y. Hence, Eq. (25) indicates that the midplane strains induced in the web laminates vary with z:  ð26Þ eoxx w ¼ zjxx Since the web laminates  are symmetric, the stress resultant induced at any point in the webs, Nxx w , is given by:    ð27Þ eoxx w ¼ aw 11 Nxx w  Equating Eqs. (26) and (27) and solving for Nxx w , we have:    jxx ex  Nxx w ¼ z w ¼ ztw E xx w jxx a11  This result shows that Nxx w varies linearly with z over the web laminate, as implied in Fig. 13(b). The axial load induced over an incremental strip of width dz is:   dNw xx ¼ Nxx w dz  Hence, the total axial load that supported the web, Nbxx w , is given by:  b  Nxx ¼

m

zb tbf zb htbf

 Nxx w dz ¼

m

zb tbf zb htbf

  ex  tw E xx w jxx zdz

Evaluating this integral, we find:  ex   i tw E xx  jxx h b  w ðzb  tbf Þ2  ðzb  h  tbf Þ2 Nxx  ¼ w 2

ð28Þ

Now consider the bottom flange laminate. Referring to Fig. 11, note that the midplane of the bottom flange is located at z = zb–tbf/2. Substituting this value into Eq. (25), the strain induced at the midplane of the bottom flange laminate is given by:   tbf   ð29Þ eoxx  ¼ zb  jxx bf 2 Since the bottom flange laminate is symmetric, the stress resultant induced in the bottom flange, Nxx bf , is:     Nxx   o  bf bf  exx  ¼ a11 Nxx  ¼ ð30Þ ex  bf bf tbf E xx  bf

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

 Equating Eqs. (29) and (30) and solving for Nxx bf , we find:    tbf  ex   Nxx  ¼ zb  tbf E xx  jxx ð31Þ bf 2 bf  This result gives the value of Nxx bf induced at the midplane of the bottom laminate, shown in Fig. 13(b). The total axial force supported   by the bottom flange, Nbxx bf, is obtained by multiplying stress resultant Nxx bf by the width of the bottom flange, b:     tbf  ex    Abf E xx  jxx ð32Þ Nbxx  ¼ bNxx  ¼ zb  bf bf 2 bf Note that the cross-sectional area of the bottom flange, Abf=btbf, has been used in Eq. (32). Finally, we must determine the total axial force supported by the top flange. From Fig. 11, we see that the midplane of the top flange laminate exists at z=(zb–tbf–h–ttf/2). Based on this value, expressions analogous to Eqs. (28)– (32) can be obtained for the top flange laminate. We find that the total axial load supported by the top flange is given by:    ttf  ex   Nbxx  ¼ zb  tbf  h  Atf E xx  jxx tf 2 tf

ð33Þ

All equations necessary to determine the position of the neutral axis have now been developed. The box-beam is modeled as four rectangular laminates (two web laminates, the bottom flange laminate, and the top flange laminate) and is in pure bending. Hence, the axial loads present in all four laminates must sum to zero:       2Nbxx  þ Nbxx  þ Nbxx  ¼ 0 w

bf

tf

Substituting Eqs. (28), (32), and (33) and solving for zb, we obtain: 2

3    ex  ex  ex  2A E ðh þ 2t Þ þ t A E þ A E ð2t þ 2h þ t Þ    w bf bf bf tf b tf xx xx xx 16 7 w bf tf  zb ¼ 4 5 ð34Þ ex  ex  ex  2 2A E  þ A E  þ A E  w

xx

w

bf

xx

bf

tf

xx

tf

Equation (34) is identical to Eq. (20a). Hence, we conclude that in pure bending, the neutral axis passes through the centroid of the beam cross section. This same conclusion holds true for isotropic beams in pure bending. Now that the position of the neutral axis has been identified, the effective flexural rigidity of the composite box-beam can be determined. First,

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

consider the moment supported by a web laminate. The incremental moment associated with the force dNbxx w located distance z from the neutral axis is given by:        ex    dMbxx  ¼ dNbxx  z ¼ tw E xx  jxx z2 dz w

w

w

The total moment supported by a web laminate is then given by:   Mbxx  ¼ w



m

zb tbf zb htbf

  ex  tw E xx  jxx z2 dz w

Evaluating this integral, we find that the moment supported by a web laminate is given by:  0 1 ex   i tw E xx  jxx h  w A ðzb  tbf Þ3  ðzb  h  tbf Þ3 Mbxx  ¼ @ w 3

ð35Þ

Now consider the moment supported by the bottom flange laminate. Since the midplane of the bottom flange laminate is not located at the neutral axis of the overall beam cross section, the flange laminate will experience both a midplane strain and a midplane curvature. The beam cross section must remain plane and, consequently, the curvature experienced by the bottom flange is identical to that experienced by the beam as a whole: jxx bf = jxx. The moment resultant associated with this curvature is:   fl   bf ¼ t3bf E xx  jxx =12 Mxx  ¼ jxx =d11 bf

bf

This moment resultant contributes to the total moment supported by the bottom flange. In addition, the stress resultant induced Nxxjbf also contributes to the moment supported by the bottom flange and may be of equal or greater importance. The bottom flange laminate is located at z = (zb–tbf/2). Substituting this coordinate into Eq. (25), we find that the midplane strain induced in the bottom flange laminate is   eoxx  ¼ ðzb  tbf =2Þjxx bf

 This midplane strain is related to stress resultant Nxx bf according to:     N eoxx  ¼ abf  xx 11 bf

bf

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

 Equating and solving for Nxx bf , we find:       jxx ex   Nxx  ¼ ðzb  tbf =2Þ bf ¼ ðzb  tbf =2Þ tbf E xx  jxx bf bf a11 This stress resultant acts at a distance (zb–tbf/2) from the neutral axis, and so   is obtained by multiplying by this distance. the moment contributed by N xx bf   Both Nxx bf and Mxx bf are resultants (i.e., loads per unit length) and must be multiplied by the width of the flange to determine the bending moment. Based on the above discussion, the total moment supported by the bottom flange laminate is:  3 2 fl    btbf E xx   ex   bf 5 jxx Mbxx  ¼ 4ðzb  tbf =2Þ2 btbf E xx  þ bf bf 12 This expression can be simplified by noting that the area of the bottom flange is Abf = btbf and, further, that the area moment of inertia of the bottom flange (taken about the local midplane of the bottom flange) is Ibf=btbf3/12:     i h  ex  fl  2 b  Mxx  ¼ ðzb  tbf =2Þ Abf E xx  þ Ibf E xx  jxx bf

bf

ð36Þ

bf

In passing, it might be anticipated that Eq. (36) could be further simplified by expressing the area moment of inertia of the bottom flange laminate about the neutral axis of the beam through application of the parallel axis theorem. It turns out that this is not the case. That is, recall that through the application of the parallel axis theorem, the moment of inertia of the bottom flange laminate about the neutral axis of the beam cross section is given by:   Ibf 

na

¼ ðzb  tbf =2Þ2 Abf þ Ibf

Comparing this expression with Eq. (36), it is seen that the parallel axis theorem cannot be used to advantage because the effective extensional ex  modulus of the bottom flange, Exx bf , differs from the effective flexural fl . modulus of the bottom flange, E xx bf Finally, we consider the moment supported by the top flange. Following an equivalent process, we conclude that the moment support by the top flange laminate is given by:     i h  fl  ex  2 b  Mxx  ¼ ðzb  tbf  h  ttf =2Þ Atf E xx  þ Itf E xx  jxx tf

tf

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

tf

ð37Þ

The total moment applied to the beam, Mob, must equal the sum of the moment components supported by the two web laminates, the bottom flange laminate, and the top flange laminate:       Mbo ¼ 2Mbxx  þ Mbxx  þ Mbxx  w

bf

tf

Also, by definition, the effective flexural rigidity of the box-beam is related to Mob according to: EI ¼

Mbo jxx

Substituting Eqs. (35)–(37), the effective flexural rigidity of a composite boxbeam is given by:  ex  i 2tw E xx  h w ðzb  tbf Þ3  ðzb  h  tbf Þ3 EI ¼ 3  ex  ð38Þ þ ðzb  tbf =2Þ2 Abf E xx  bf   

fl  fl  ex  þðzb  tbf  h  ttf =2Þ2 Atf E xx  þ Ibf E xx  þ Itf E xx  tf

bf

tf

If a box-beam is considered for which the bottom and top flange laminates ex  ex are identical (i.e., if Abf = Atf, tbf = ttf, and E xx bf = E xx tf ), then Eq. (38) reduces to:  2 3 ex    ðtbf þ hÞ2 Abf E xx  ex  fl  bf þ Ibf E xx  5 EI ¼ 2Iw E xx  þ 24 w 4 bf where Iw is the area moment of inertia of a web laminate, taken about the neutral axis of the box-beam. This concludes our analysis of a composite box-beam. To summarize, we have restricted our analysis to composite box-beams that possess certain material and geometric symmetries. Specifically, both the flange and web laminates must be produced using a symmetric stacking sequence, and the beam cross section must be symmetric about the plane of loading, the x–z plane. This implies that the two web laminates must be produced using an identical symmetric stacking sequence, such that the set of fiber angles represented by the two web laminates is symmetric about the x–z plane. The top and bottom flange laminates must also be symmetric, but they need not be identical. Therefore the beam cross section need not be symmetric about the y–z plane. Having defined a beam cross section in this way, then the

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

location of the centroid (which also defines the location of the neutral axis in pure bending) is given by Eqs. (20a) and (20b) or Eq. (34). The effective axial rigidity of the box-beam is given by Eqs. (24a) and (24b), and the effective flexural rigidity is given by Eq. (38). As will be discussed in following sections, once the effective axial and flexural rigidities of the composite box-beam have been determined, then deflection of the beam caused by axial and/or transverse loads can be determined using the methods traditionally employed to study isotropic beams. The general approach described above can also be applied to the other thin-walled cross sections previously discussed in Sec. 3 and shown in Fig. 1. A summary of results for these additional thin-walled cross sections is provided in Tables 1–3. Example Problem 1 Three box-beams are constructed using unidirectional graphite–epoxy prepreg tape with the material properties listed in Table 3 of Chap. 3 and ply thickness of 0.125 mm. The first box-beam is shown in Fig. 14(a). In this case, the web and flange laminates are produced using an identical 8-ply [0/90]2s stacking sequence. The second beam is shown in Fig. 14(b). In this case, additional plies are added to the top flange. Hence, both web laminates and the bottom flange laminate are produced using an 8-ply [0/90]2s stacking sequence, but the top laminate is produced using a 20-ply [(0/90)2s/F45]s stacking sequence. Finally, the third beam is produced using [0/90]2s web laminates and [(0/90)2s/F45]s top and bottom flange laminates. In all three cases, the overall beam width and depth is 30 and 50 mm, respectively. Determine the position of the centroid and the effective axial and flexural rigidities for: (a) The beam shown in Fig. 14(a). (b) The beam shown in Fig. 14(b). (c) The beam shown in Fig. 14(c). Solution Part (a). Since both webs and flanges are produced using an 8-ply stacking sequence, laminate thicknesses are: tw ¼ tbf ¼ ttf ¼ 8ð0:125 mmÞ ¼ 1 mm The length of the bottom and top flange laminates equals the overall width of the beam (30 mm). The length of the web laminates is: h ¼ 50 mm  tbf  ttf ¼ 50mm  2ð1mmÞ ¼ 48mm

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

Table 1

Centroid Location of Some Thin-Walled Composite Beams with Symmetric Web and Flange Laminates

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

Table 2

Effective Axial Rigidities of Some Thin-Walled Composite Beams with Symmetric Web and Flange Laminates

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

Table 3

Effective Flexural Rigidities of Some Thin-Walled Composite Beams with Symmetric Web and Flange Laminates

Figure 14 The composite box-beams considered in Example Problem 1 (not to scale). (a) Composite box-beam with identical stacking sequence in all web and flange laminates. (b) Composite box-beam with additional plies added to the top flange laminate. (c) Composite box-beam with additional plies added to both top and bottom flange laminates.

We can now calculate the area properties of the web and flange laminates: Aw ¼ ðhÞðtw Þ ¼ ð48 mmÞð1 mmÞ ¼ 48
106 m2 Abf ¼ Atf ¼ ðbÞðtbf Þ ¼ ð30 mmÞð1 mmÞ ¼ 30
106 m2 Ibf ¼ Itf ¼

ðbÞðtbf Þ3 ð30 mmÞð1 mmÞ3 ¼ ¼ 2:5
1012 m4 12 12

The effective moduli for the web and flange laminates are calculated using the process described in Sec. 9.1 of Chap. 6 and are found to be:    ex  ex  ex  E xx  ¼ E xx  ¼ E xx  ¼ 90:4 Gpa w bf tf   fl  fl  E xx  ¼ E xx  ¼ 120 Gpa bf

tf

From Table 1, for a composite box-beam, the location of the centroid is given by: 2 3    ex  ex  ex  2A E ðh þ 2t Þ þ t A E þ A E ð2t þ 2h þ t Þ    w bf bf bf tf bf tf xx xx xx 16 7 w bf tf   zb ¼ 4 5 ex  ex  ex  2 2Aw E xx  þ Abf E xx  þ Atf E xx  w

bf

tf

Substituting all known values and completing the indicated calculations, we find: zb ¼ 25 mm This result is as anticipated since the bottom and top flange laminates are identical and the total depth of the beam is 50 mm.

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

From Table 2, for a composite box-beam, the effective axial rigidity is:    ex  ex  ex  ðAE Þ ¼ 2Aw E xx  þ Abf E xx  þ Atf E xx  w

bf

tf

Substituting all known values and completing the indicated calculations, we find: ðAE Þ ¼ 14:1
106 N From Table 3, for a composite box-beam, the effective flexural rigidity is: ðIE Þ ¼

 ex  2tw E xx  h w

ðzb  tbf Þ3  ðzb  h  tbf Þ3

i

3  bf ex  þ ðzb  tbf =2Þ2 Abf E xx þ ðzb  tbf  h  ttf =2Þ2 Atf E xx   

fl  fl  þIbf E xx  þ Itf E xx  bf

tf

tf

Substituting all known values and completing the indicated calculations, we find:   IE ¼ 4:92
103 N  m2 Part (b). The web and bottom flanges are produced using the same 8-ply stacking sequence considered in Part (a), but now the top flange is a 20-ply laminate. Adjusting our previous calculations, we find: tw ¼ tbf ¼ 8ð0:125 mmÞ ¼ 1 mm ttf ¼ 20ð0:125 mmÞ ¼ 2:5 mm h ¼ 50 mm  tbf  ttf ¼ 50 mm  1 mm  2:5 mm ¼ 46:5 mm Aw ¼ ðhÞðtw Þ ¼ ð46:5 mmÞð1 mmÞ ¼ 46:5
106 m2 Abf ¼ ðbÞðtbf Þ ¼ ð30 mmÞð1 mmÞ ¼ 30
106 m2 Atf ¼ ðbÞðttf Þ ¼ ð30 mmÞð2:5 mmÞ ¼ 75
106 m2 Ibf ¼

ðbÞðtbf Þ3 ð30 mmÞð1 mmÞ3 ¼ ¼ 2:5
1012 m4 12 12

ðbÞðttf Þ3 ð30 mmÞð2:5 mmÞ3 ¼ ¼ 39:1
1012 m4 Itf ¼ 12 12   ex  ex  E xx  ¼ E xx  ¼ 90:4 GPa w

bf

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

 fl  E xx  ¼ 120 GPa bf  ex  E xx  ¼ 83:3 GPa tf  fl  E xx  ¼ 94:0 GPa tf

Based on the appropriate expression taken from Table 1, we find that the centroid is located at: zb ¼ 29:4 mm Note that the increased thickness of the top flange laminate has had the effect of moving the centroid 4.4 mm towards the top flange. The new axial and flexural rigidities are: ðAE Þ ¼ 17:4
106 N ðIE Þ ¼ 6:35
103 N  m2 The increased thickness of the top flange laminate has increased the effective axial and flexural rigidities by 23% and 29%, respectively. Part (c). For this cross section, the web laminates are produced using the same 8-ply stacking sequence considered in Part (a), but now, both the bottom and top flanges consist of a 20-ply laminate. We find: tw ¼ 8ð0:125 mmÞ ¼ 1 mm ttf ¼ tbf ¼ 20ð0:125 mmÞ ¼ 2:5 mm h ¼ 50 mm  tbf  ttf ¼ 50 mm  2:5 mm  2:5 mm ¼ 45 mm Aw ¼ ðhÞðtw Þ ¼ ð45 mmÞð1 mmÞ ¼ 45
106 m2 Abf ¼ Atf ¼ ðbÞðtbf Þ ¼ ð30 mmÞð2:5 mmÞ ¼ 75
106 m2 ðbÞðtbf Þ3 ð30 mmÞð2:5 mmÞ3 ¼ ¼ 39:1
1012 m4 Ibf ¼ Itf ¼ 12 12  ex  E xx  ¼ 90:4 GPa w   ex  ex  E xx  ¼ E xx  ¼ 83:3 GPa bf tf   fl  fl  E xx  ¼ E xx  ¼ 94:0 GPa bf

tf

Employing the appropriate expressions from Tables 1–3, we find: zb ¼ 25:0 mm

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

ðAE Þ ¼ 20:6
106 N ðIE Þ ¼ 8:43
103 N m2 Since the top and bottom flanges are identical, the centroid is centered in the overall beam cross section, as was the case in Part (a). There is now a further increase in effective rigidities. Compared to the values calculated in Part (a), the increased thickness of the bottom and top flange laminates has increased the effective axial and flexural rigidities by 46% and 71%, respectively. 7 STATICALLY DETERMINATE AND INDETERMINATE AXIALLY LOADED COMPOSITE BEAMS Deflections of composite beams subjected to axial loads (only) are considered in this section. Beam problems may be divided into two categories: those involving statically determinate beams and those involving statically indeterminate beams. For a statically determinate beam, all unknown reaction forces can be determined through application of the equations of equilibrium. In contrast, for a statically indeterminate beam, the reaction forces are determined through the equations of equilibrium and a consideration of beam deflections. A statically determinate composite beam is shown in Fig. 15(a). As indicated, the beam is clamped at one end and subjected to three known axial loads NBb , NCb , and NDb . In addition, the effective axial rigidity changes discretely at points B and C. In practice, a discrete change in effective axial rigidity is readily accomplished by adding or removing a ply (or plies) from the stacking sequence along the length of the beam. The internal force present in each beam segment can be determined through the use of the free-body diagrams shown in Fig. 15(b)–(d) and the appropriate equation of equilibrium (SFx = 0):  b  Nxx  ¼ NBb þ NCb þ NDb AB  b  Nxx  ¼ NCb þ NDb BC   b Nxx  ¼ NDb CD

The change in length of a beam caused by axial loading is called the elongation, uo. The elongation induced within each segment of the beam shown in Fig. 15 is given by:   uo 

AB

¼

 b  Nxx 

AB

LAB

ðAE xx ÞAB

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

Figure 15 A statically determinate composite beam with changing effective axial rigidity subjected to multiple axial load. (a) Composite Beam. (b) Free-body diagram used to determine reaction force Rxx = NBb+NCb+NDb. (c) Free-body b diagram used to determine internal load in segment BC equals Nxx = NCb+NDb. (d) Free-body diagram used to determine internal load in segment CD equals b = NDb. Nxx

  uo    uo 

BC

CD

¼

¼

 b  Nxx 

BC

LBC

ðAE xx ÞBC  b  Nxx  LCD CD

ðAE xx ÞCD

where LAB, LBC, and LCD represent the length of beam segments AB, BC, and CD, respectively. The elongation of the bar as a whole is simply the sum of the elongation induced over each segment:  b   Li X Nxx i uo ¼ ð39Þ i ðAE xx Þi A statically indeterminate composite beam is shown in Fig. 16. In this case, the beam is clamped at both ends and subjected to two known axial loads

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

Figure 16 A statically indeterminate composite beam subjected to two known intermediate loads. (a) Original structure. (b) Free-body diagram.

NBb and NCb. An unknown reaction force is induced at each end of the bar, Rbxx jA and RbxxjD. Note that both reaction forces have been assumed to act in the +x-direction in Fig. 16. Enforcing the appropriate equation of equilibrium (SFx = 0), we find:     Rbxx  þ NbB þ NbC þ Rbxx  ¼ 0 ð40Þ A

D

b b jA and Rxx jD are unknown, a second independent equation is Since Rxx required in order to solve for the reaction forces. The remaining equations of equilibrium cannot be used to provide the necessary second independent equation since only axial loads are present. The second independent equation is obtained by requiring that the beam satisfy known boundary conditions. That is, since the beam is clamped at both ends, the total elongation of the beam must be zero. By requiring that the total elongation is zero, we develop a second independent equation, known as an equation of compatibility. Treating

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

reaction force Rbxx jD as a redundant force, applying Eq. (39), and equating the total elongation to zero, we find:        b   LAB NBb þ NCb þ Rxx NCb þ Rbxx  LBC   D D þ uo  ¼ AD ðAE xx ÞAB ðAE xx ÞBC ð41Þ    b  Rxx  LCD D þ ¼0 ðAE xx ÞCD Equations (40) and (41) represent two independent equations in two unknowns, Rbxx jA and Rbxx jD. Once these equations are solved to determine the reaction forces, then the axial force induced in any segment of the beam may also be determined. The analysis of statically determinate and indeterminate composite beams will be illustrated in Example Problems 2 and 3, respectively. Example Problem 2 Assume that the composite beam shown in Fig. 15 is a graphite–epoxy boxbeam, produced using a pre-preg system with the material properties listed in Table 3 of Chap. 3 and a ply thickness of 0.125 mm. Over segment AC, the beam has the cross section previously shown in Fig. 14(c). That is, the two web laminates are produced using an 8-ply [0/90]2s stacking sequence, whereas the top and bottom flange laminates are produced using a 20-ply [(0/90)2s/ F 45]s stacking sequence. Over segment CD, the beam has the cross section previously shown in Fig. 14(a). Thus, over this segment, both the web and flange laminates are produced using an 8-ply [0/90]2s stacking sequence. The length of each beam segment is as follows: LAB ¼ 250 mm LBC ¼ 500 mm LCD ¼ 250 mm Thus, the total length of the beam is 1.0 m. Finally, the applied loads are: NBb ¼ 15 kN NCb ¼ 10 kN NDb ¼ 5 kN Determine the total beam elongation produced by this loading condition.

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

Solution. The effective axial rigidities provided by the beam cross sections involved were determined as a part of Example Problem 1 and found to be:     ðAE xx Þ ¼ ðAE xx Þ ¼ 20:6
106 N   ðAE xx Þ

AB

CD

BC

¼ 14:1
106 N

The elongation can be determined through the application of Eq. (39):  b   Li X Nxx i uo ¼ i ðAE xx Þi    b  b  b  Nxx  LBC Nxx  LCD  LAB Nxx BC CD AB uo ¼ þ þ ðAE xx ÞAB ðAE xx ÞBC ðAE xx ÞCD uo ¼

ð30 000NÞð0:25mÞ ð15000NÞð0:50mÞ ð5000NÞð0:25mÞ þ þ 20:6
106 N 20:6
106 N 14:1
106 N

uo ¼ 0:817 mm Example Problem 3 Assume that the composite beam shown in Fig. 16 is a graphite–epoxy boxbeam, produced using a pre-preg system with the material properties listed in Table 3 of Chap. 3 and a ply thickness of 0.125 mm. Over segment AC, the beam has the cross section previously shown in Fig. 14(c). That is, the two web laminates are produced using an 8-ply [0/90]2s stacking sequence, whereas the top and bottom flange laminates are produced using a 20-ply [(0/90)2s/ F 45]s stacking sequence. Over segment CD, the beam has the cross section previously shown in Fig. 14(a). Thus, over this segment, both the web and flange laminates are produced using an 8-ply [0/90]2s stacking sequence. The length of each beam segment is as follows: LAB ¼ 250 mm LBC ¼ 500 mm LCD ¼ 250 mm b Thus, the total length of the beam is 1.0 m. Finally,  loadsbNB = 15 kN and b b   NC = 10 kN. Determine the reaction forces Rxx A and Rxx D .

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

Solution. The reaction forces are determined through solving Eqs. (40) and (41). The effective axial rigidities provided by the beam cross sections involved were determined as a part of Example Problem 1 and found to be:     ðAE xx Þ ¼ ðAE xx Þ ¼ 20:6
106 N AB BC   ðAE xx Þ ¼ 14:1
106 N CD

Using the values provided, Eq. (41) becomes:             25 kN þ Rbxx  ð0:25 mÞ 10 kN þ Rbxx  ð0:5 mÞ Rbxx  ð0:25 mÞ D D D þ þ ¼0 ð20:6
106 NÞ ð20:6
106 NÞ ð14:1
106 NÞ Solving for the redundant force:   Rbxx  ¼ 10:1 kN D

 b  is negative implies that it acts in the direction opposite to The fact that Rxx D that shown in Fig. 16(b).  b  can now be determined through the The unknown reaction force Rxx A application of Eq. (40):       Rbxx  þ NbB þ NbC þ Rbxx  ¼ Rbxx  þ 15 kN þ 10 kN  10:1 kN ¼ 0 A

  Rbxx  ¼ 14:9 kN

D

A

A

 b  Reaction Rxx is also negative, implying that it acts the direction opposite to A that shown in Fig. 16(b).

8 STATICALLY DETERMINATE AND INDETERMINATE TRANSVERSELY LOADED COMPOSITE BEAMS Deflections associated with bending of composite beams are considered in this section. Three types of external loads that lead to beam bending are shown in Fig. 17. A distributed load, q(x), is a load distributed over a length of the beam and is specified in units of either Newtons/meter or pounds-force/inch (N/m or lbf/in.). A point load, P, is a load acting transverse to the axis of the beam and applied at a specific cross section. A point load is usually specified in units of either Newtons or pounds-force (N or lbf ). Finally, a concentrated moment (also called a couple), Mob , is a moment applied at a particular cross section

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

Figure 17 Illustration of external transverse loads that lead to beam bending.

and is specified in units of Newton-meters or pounds-force-in. (N m or lbf in.). As in earlier sections, a superscript ‘‘b’’ has been used to differentiate between a concentrated moment and a moment resultant, Mxx. In the following discussion, the external beam loads shown in Fig. 17 will be referred to as transverse loads. In general, the external transverse forces applied to a beam induce internal forces at all cross sections of a beam. To determine beam deflections, we must determine these internal forces. In order to do so, the manner in which the beam is supported must first be defined. The three most common beam support conditions are illustrated in Fig. 18. A pinned support is one that restrains the beam from deflection in the x- and z-directions, but does not prevent rotation. In contrast, a roller support restrains the beam from deflection in the z-direction, but does not restrain deflections in the xdirection nor does it restrict rotation. Finally, a clamped (or fixed ) support does not allow the beam to rotate or deflect in the x- or z-directions. Some combinations of beam support conditions occur so frequently that they are given special names. A simply supported beam is shown in Fig. 19(a). In this case, one end of the beam is restrained by a pin support, while the other is restrained by a roller support. A cantilever beam is shown in Fig. 19(b) and is defined as a beam clamped at one end and free at the other. A

Figure 18 Common beam support conditions. (a) Pinned support. (b) Roller support. (c) Clamped support.

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

Figure 19 Common combinations of beam support conditions. (a) Simply support beam. (b) Cantilevered beam. (c) Overhanging beam.

third common configuration is an overhanging beam, shown in Fig. 19(c). This condition is similar to a simply supported beam except that the roller support is at an interior cross section, such that one end of the beam is free. Having specified the support conditions and transverse loading applied to a beam, the next step is to determine the reaction forces. In essence, reaction forces are external loads applied to the beam by the beam support(s). Determination of these forces begins by enforcing the six equations of equilibrium. Since we have assumed that all loads act within the x–z plane (the plane of loading), three equations of equilibrium are satisfied automatically: SFy = SMx = SMz = 0. In this section, we do not consider any axial loads; therefore SFx=0 is also satisfied automatically. Consequently, only two equations of equilibrium are available for use. Specifically, to remain in equilibrium, it is required that all forces acting in the z-direction and all moments acting about the y-axis sum to zero: X

Fz ¼ 0

X

My ¼ 0

Beam problems may now be divided into two main categories: statically determinate beams and statically indeterminate beams. For a statically determinate beam, the unknown reaction forces are determined solely through the application of the equations of equilibrium. Since we only have two remaining equilibrium equations, it is apparent that a statically determinate beam can have (at most) two unknown reaction forces. Free-body diagrams for simply supported, cantilevered, and overhanging beams are shown in Fig. 20. All reaction forces are shown in an algebraically positive sense. In each case, only two unknown reaction forces are involved, and hence each of these beams is statically determinate. In contrast, if a beam problem involves three (or more) reaction forces, then the beam is statically indeter-

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

Figure 20 Free-body diagrams for common statically determinate beams. (a) Simply supported beam. (b) Cantilevered beam. (c) Overhanging beam.

minate and the reaction forces are determined through the application of the equations of equilibrium and consideration of beam deflections. An example of a statically indeterminate beam is shown in Fig. 21. In this case, the beam is clamped at one end, supported by a roller support at the other end, and subjected to a known transverse point load P. A total of three unknown (B) b reaction forces exist in this case (R(A) z , Rz , and Mo ), as indicated in the accompanying free-body diagram. The three unknown reaction forces cannot

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

Figure 21 Free-body diagram for a beam statically indeterminate to the first degree.

be determined solely on the basis of equilibrium since only two equations of equilibrium are available for use. The difference between the number of unknown reaction forces and the number of available equations of equilibrium is called the degree of indeterminacy; the beam shown in Fig. 21 is statically indeterminate to the first degree. One way to determine reaction force in this case is to enforce the requirement that the beam z-deflection at the roller support must equal zero, in addition to enforcing the two equations of equilibrium. This leads to third independent equation, a so-called equation of compatibility. Hence, together, the two equations of equilibrium and the equation of compatibility represent three independent equations and can be used to solve for the three unknown reaction forces. In any event, assume for the moment that all reaction forces have been determined for a beam of interest, either through application of the equations of equilibrium or through an appropriate combination of the equations of equilibrium and compatibility. The reaction forces and the applied transverse loads represent external forces applied to the beam, and these induce internal b forces within the beam. Internal forces may consist of a shear force Vxx and b b b bending moment Mxx. In general, Vxz and Mxx both vary along the length of the beam. Note that during the earlier discussion presented in Secs. 5 and 6, we considered composite beams subjected to pure bending—in these earlier b did not vary along the length of sections, the internal bending moment Mxx b = 0. We now the beam. Further, in pure bending, no shear force exists: Vxz wish to determine transverse beam deflections under more general loading b b and Mxx exist. This is often called a state of conditions in which both Vxz b b and Mxx contribute to nonuniform bending. Rigorously speaking, both Vxz b are less transverse beam deflections. However, the deflections due to Vxz b significant than the deflections due to Mxx and can usually be ignored. That is, b we will assume that the contribution of shear force Vxz to the total deflection of composite beams is negligibly small. In essence, we assume Secs. 5 and 6

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

remain valid for the case of nonuniform bending, even though these equations were developed assuming a state of pure bending. This same assumption is made during the study of isotropic beams. The moment-curvature equation for pure bending is: b Mxx ð42Þ IE xx The effective flexural rigidity (IE xx) of composite beams has been developed in previous sections. For rectangular beams with plies orthogonal to the plane of loading, (IE xx ) was developed in Sec. 5.1; for rectangular beams with plies parallel to the plane of loading, (IE xx) was developed in Sec. 5.2; and for thinwalled beams, (IExx) was developed in Sec. 6. Also, for a state of pure bending, the midplane curvature jxx is related to transverse beam deflections according to:

jxx ¼

d2 w ð43Þ dx2 where w = w(x) represents the z-deflection of the composite beam midplane.* Combining Eqs. (42) and (43), we find: jxx ¼ 

d2 w Mb ¼  xx 2 dx IE xx

ð44Þ

Most textbooks devoted to mechanics of materials (Refs. [1–5] for example) describe several techniques to determine transverse beam deflections. These include direct integration of the governing equation [i.e., direct integration of Eq. (44)], the moment-area method, the method of superposition, the techniques based on the use of discontinuity functions, or the use of Castigliono’s Theorem (which is a technique based on energy methods). Two techniques will be briefly discussed here: direct integration of the governing equations and the method of superposition. Direct integration of the governing differential equation is perhaps the most straightforward approach. Once all external forces are known (including reaction forces), then it is a simple matter to express the internal bending b b = Mxx (x). Hence, the beam deflection w(x) moment as a function of x: Mxx is obtained by simply integrating Eq. (44) twice. Constants of integration are determined by enforcing known beam support conditions (i.e., known boundary conditions). The method of superposition is based on the observation that Eq. (44) is a linear differential equation with constant coefficients. Consequently, the

* For pure bending, Eq. (43) is exact and is derived in any text devoted to mechanics of materials, including Refs. [1–5] for example.

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

deflection of a beam subjected to several different load components acting simultaneously can be found by superimposing (i.e., adding together) the deflections caused by each load component acting separately. In practice then, tables of beam deflections caused by simple loading conditions and for commonly encountered beam support conditions are generated, and the deflections caused by a combination of two or more loads are obtained by adding together the deflection caused by each load component. Tables for deflections induced by common individual load components applied to cantilevered and simply supported beams are provided in Appendix C, Tables C.1 and C.2, respectively. Much more extensive tables, which provide solutions for other types of loading and support conditions, are available in engineering handbooks, for example, Ref. [7]. Three example problems will now be presented to illustrate the process of determining composite beam deflections. The deflections of a statically determinate composite beam will be obtained by direct integration of Eq. (44) in Example Problem 4. The method of superposition is then applied to a statically determinate composite beam subjected to a combination of loads in Example Problem 5. Finally, the method of superposition is used to determine the reaction forces and deflections for a statically indeterminate beam in Example Problem 6. Example Problem 4 A cantilevered graphite–epoxy box-beam subjected to a linearly increasing distributed load qðxÞ ¼ ðqo =LÞx is shown in Fig. 22(a): (a)

Obtain analytical expressions giving the reaction forces and transverse beam deflections. (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 qo=200 N/m. Solutions Part (a). A free-body diagram of the beam is shown in Fig. 22(b). Since the beam is clamped, two reaction forces may exist at the left-hand end: Rz and Mob . The first step is to determine the magnitude and algebraic sign of these unknown reaction forces. The beam is statically determinate since only two unknown reaction forces are present. The reaction forces are initially assumed to act in an algebraically positive sense, as shown in Fig. 22(b). To determine Rz and Mob, we first replace the distributed load q(x) with the statically equivalent force qo L/2, located at x = 2L/3. This statically equivalent force is shown in Fig. 22(b). By summing (a) forces in the zdirection, (b) moments about the y-axis, and (c) equating both to zero (SFz =

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

Figure 22 The cantilevered beam considered in Example Problem 4. (a) A cantilevered beam of length L subjected to a linearly increasing distributed load q(x) = (qo/L)x. (b) Free-body diagram used to determined the unknown reaction forces (distributed load qo).

SMy = 0), we obtain analytical expressions giving the reaction forces at the clamped end: Rz ¼

qo L 2

Mbo ¼

qo L2 3

ðaÞ

Note that if qo>0, 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  x2 Þ ðxÞ ¼ Vxz 2L

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

Figure 23 Free-body diagram used to determine internal shear and bending b b and Mxx , respectively, acting at an arbitrary cross section located moment, Vxz 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 ð2L3 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 (x) in Fig. 24. Substituting the above expression for the bending moment Mxx into Eq. (44), we find:

d2 w qo ð2L3 þ 3L2 x  x3 Þ ¼ dx2 6LðIE xx Þ Integrating once results in:   dw qo 3L2 x2 x4 ¼  2L3 x þ þ C1 dx 2 4 6LðIE xx Þ 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  10L2 x þ x3 Þ þ C2 120LðIE xx Þ

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

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  10L2 x þ x3 Þ 120LðIE xx 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

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

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.

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

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  10L2 x þ x3 Þ 120LðIE xx 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 Mo qðxÞ 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

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

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.

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

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  aÞ 6ðIE xx Þ

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

ðBÞ

Rz

¼

Rz x2 ð3L  xÞ 6ðIE xx Þ

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 i 1 Pa2 ð3x  aÞ þ RzðBÞ x2 ð3L  xÞ 6ðIE xx Þ

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 i 1 2 Pa2 ð3L  aÞ þ RðBÞ z ðLÞ ð3L  LÞ ¼ 0 6ðIE xx Þ

which reduces to: 3 2 2RðBÞ z L ¼ Pa ða  3LÞ

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

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 P

ðcÞ

b RðBÞ z L þ Mo ¼ 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

P ða3 þ 2L3  3a2 LÞ 2L3

Pa2 ð3L  aÞ 2L3 Pa Mbo ¼  2 ða2 þ 2L2  3aLÞ 2L RðBÞ z ¼

ðeÞ ðf Þ ðgÞ

Part (b). Based on the specified numerical values, the reaction forces are: RðAÞ ¼ z RðBÞ z ¼ Mbo ¼ 

i ð1000NÞ h 3 3 2 ð0:75mÞ þ 2ð1mÞ  3ð0:75mÞ ð1mÞ ¼ 367N 2ð1mÞ3 ð1000NÞð0:75mÞ2 2ð1mÞ3

½3ð1mÞ  ð0:75mÞ ¼ 633N

i ð1000NÞð0:75mÞ h ð0:75mÞ2 þ 2ð1mÞ2  3ð0:75mÞð1mÞ 2 2ð1mÞ

¼ 117Nm

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

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  xÞ 6ðIE xx Þ

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

ðBÞ

Rz

¼

Rz x2 ð3L  xÞ 6ðIE xx Þ

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 i 1 Px2 ð3a  xÞ þ RzðBÞ x2 ð3L  xÞ 6ðIE xx Þ

ðhÞ

Figure 29 Deflections predicted for the beam considered in Example Problem 6.

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

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

E22 ¼ 10
109 Pa

v12 ¼ 0:30

G12 ¼ 13
10 Pa 9

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

E22 ¼ 1:5
106 psi

v12 ¼ 0:30

G12 ¼ 1:9
10 psi 6

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.

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

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).

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

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).

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

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.

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

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.

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

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

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

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

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

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) Nxy ( y) (x) Mxx ( y) (x) Mxy ( y)

Resultants acting on the positive x-edge: (+x) Nxy ( y) (+x) Mxx ( y) (+x) Mxy ( y)

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

Figure 1 Illustration of varying stress resultants acting on opposite edges of a (+) (y). (b) Variation of stress thin plate. (a) Variation of stress resultants Nxx (+) 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,

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

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 (y) yy (x) N (y) yx (x) M (y) yy (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) of the twisting moment applied to the positive x-face, Mxy ( y), is now (+y) independent of the twisting moment applied to the positive y-face, Myx (x).

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

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

Figure 3 A thin plate subjected to stress resultants and transverse loading that vary along the length and width of the plate.

represent a second type of boundary condition. Types of displacement boundary conditions have already been discussed in Chapter 8. For example, in beam theory it is common to specify that one (or both) end(s) of a beam is (are) ‘‘clamped.’’ If a beam end is clamped then that end of the beam cannot move up/down, and the slope of the beam midplane must equal zero at the clamped end. Hence, when one specifies that a beam end is clamped, one has specified something about the displacement field in the beam, rather than something about the applied loading. Similar displacement boundary conditions can also be defined for thin plates. Hence to define a thin-plate problem of interest we must specify the boundary conditions along each edge of the plate. The specified boundary conditions may involve (a) the loads applied to each edge of the plate, (b) the displacements imposed on each edge of the plate, or (c) some combination thereof. Boundary conditions that are consistent with the assumptions of thin-plate theory will be discussed in detail in Sec. 3. The equations that govern the behavior of thin composite laminates will be developed in the following two sections. First, the equations of equilibrium will be developed in Sec. 2. Boundary conditions that are consistent with thinplate theory will then be developed in Sec. 3. These equations and boundary conditions are quite general and govern the solution to any problem involving a thin plate, including composite laminates. Unfortunately, it is often impossible to obtain exact solutions to the equations of equilibrium that also satisfy the prevailing boundary conditions. Difficulties in obtaining exact solutions occur for either isotropic plates or anisotropic composite plates, and arise from two different sources. First, the boundary conditions encoun-

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

tered in a problem may preclude obtaining an exact solution. For example, an exact solution can be found for a flat rectangular isotropic plate subjected to transverse loading if all four edges are simply supported, but an exact solution cannot be found for an identical flat rectangular isotropic plate subjected to transverse loading if all four edges are clamped. Because boundary conditions are independent of material properties, the difficulties in obtaining exact solutions for certain boundary conditions are encountered for either isotropic or anisotropic composite plates. The second source of difficulty has to do with the anisotropic nature of composites and is not encountered during the study of isotropic plates. Specifically, exact solutions cannot be found for laminates that exhibit the following couplings terms: 

in-plane normal-shear coupling (i.e., laminates for which A16, A26 p 0),  bending-twisting coupling (i.e., laminates for which D16, D26 p 0), and/or  coupling between in-plane loads and out-of-plane twisting (i.e., laminates for which B16, B26 p 0). All of these coupling terms exist for a composite laminate with an arbitrary stacking sequence, and hence an exact solution cannot, in general, be found for composite laminates with an arbitrary stacking sequence. Still another complication associated with environmental factors exists for nonsymmetric laminates. Recall from Chapter 6 that nonsymmetric laminates will bend/warp if subjected to a change in temperature and/or moisture content, because Bij, MijT, MijM p 0 for nonsymmetric laminates. For example, following cure at an elevated temperature a nonsymmetric laminate is likely to be warped at room temperature and may warp further upon exposure to changes in humidity over time. It is for this reason that nonsymmetric laminates are rarely used in practice. An analysis of a nonsymmetric laminate must therefore account for the displacements due to externally applied loads as well as displacements (e.g., warping) due to changes in temperature or moisture content. Because the intention here is to provide a brief introduction to thin plate theory as applied to composites, nonsymmetric laminates will not be considered. In summary then, the equations of equilibrium and boundary conditions that will be developed in the following sections of this Chapter are valid for any symmetric composite laminate. However, exact solutions to these governing equations can only be found for certain combinations of stacking sequences and boundary conditions. Some available exact solutions to the governing equations will be discussed in Chapter 10. Fortunately, techniques are also available that can be used to obtain approximate solutions for those

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

laminates and boundary conditions that do not admit exact solutions. Although approximate, the accuracy of these methods is usually quite good and are suitable for most engineering applications. These approximate solution techniques will be discussed in Chapter 11.

2 THE EQUATIONS OF EQUILIBRIUM FOR SYMMETRIC LAMINATES The externally applied edge loads shown in Fig. 3 induce internal stress and moment resultants. In general, these internal resultants vary throughout the interior regions of the laminate. Ultimately, we wish to relate these varying internal stress and moment resultants to the externally applied loads that induce them. In order to do so we must first develop the equations of equilibrium for a thin plate. In essence, the equations of equilibrium dictate how internal stress and moment resultants may vary over the length and width of the laminate such that static equilibrium is maintained. The equations of equilibrium will be developed in Sec. 2.1 by applying the equations of statics; that is, by requiring that the sum of all forces and moments equal zero. As will be seen, the equations developed in this manner are partial differential equations involving the internal stress and moment resultants, transverse loading q(x,y), and the out-of-plane displacement field, w(x,y). These equations will then be converted to a form more useful for composite studies in Sec. 2.2. This will involve expressing the internal stress and moment resultants in terms of elements of the [ABD] matrix and midplane displacement fields, uo(x,y), vo(x,y), and w(x,y). 2.1 The Equations of Equilibrium Expressed in Terms of Internal Stress and Moment Resultants, Transverse Loading, and Out-of-Plane Displacements Consider a thin laminate subjected to some combination of edge loads and/or edge displacement that induce stress and moment resultants at internal regions of the laminate. The laminate is assumed to be flat prior to application of external edge loads or displacement. In particular, the laminate is assumed to be symmetric and hence Bij =0. Also, because the laminate is symmetric it will not warp prior to the application of external loads if subjected to a change in temperature and/or moisture content (MijT = MijM = 0). An infinitesimal element removed from an internal region of such a laminate is shown in Fig. 4. For clarity, only the midsurface of the element is shown in Fig. 4. The element thickness is actually equal to the total laminate thickness, t. The inplane dimensions are dx and dy. The in-plane normal and shear stress resultants are shown in Fig. 4a, while the moment resultants, out-of-plane

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

Figure 4 Internal stress resultants, moment resultants, and transverse loading acting on an infinitesimal element. (a) In-plane normal and shear stress resultants. (b) Moment resultants, out-of-plane shear stress resultants and transverse loading.

shear resultants, and transverse loading are shown in Fig. 4b (it should be understood that the resultants and transverse loads shown in Fig. 4a,b are applied simultaneously to the element). Once again, it is important to realize that the resultants shown are induced at an interior region represented by the infinitesimal element and do not necessarily equal the external loads applied along the edges of the laminate. A superscript (*) will be used to denote resultants present at interior regions of the laminate. Also, because the element is infinitesimal we do not need to distinguish between shear resultants on adjacent faces, i.e., N*yx =N*xy . All resultants are assumed to vary slightly across the infinitesimal * acting length and width of the element. For example, the stress resultant Nxx on the negative x-face of the element is assumed to vary slightly over distance

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

dx, such that a stress resultant N *xx+(BN *xx/Bx) dx is present on the positive x-face of the element. The equations of equilibrium can be developed using the free-body diagrams shown in Figs. 5 and 6 by applying the standard equations of statics. That is, by requiring that the sum of all forces and moments acting on the infinitesimal element equate to zero: X

Fx ¼

X

Fy ¼

X

Fx ¼

X

Mx ¼

X

My ¼

X

Mz ¼ 0:

Let us first sum forces in the x-direction, based on the free-body diagram shown in Fig. 5. Only those resultants that contribute toward forces in the xdirection are shown in this figure. The element has been deflected out of the original x–y plane by the transverse loading. A view of the x–y plane is shown in Fig. 5a, while a view of the x–z plane is shown in Fig. 5b. In the deflected condition the left and right side of the element form angles a and (a+(Ba/Bx) dx), respectively, with the original x–y plane. Consider the x-directed force associated with Nxx * , acting on the left side of the element. Recall that the units * to a force we must of a stress resultant are (force/length), so to convert Nxx

Figure 5 Free-body diagram of an infinitesimal element, showing only those resultants that contribute toward forces in the x-direction. (a) x-y plane. (b) x-z plane.

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

Figure 6 Free-body diagram of an infinitesimal element, showing only those resultants that contribute toward forces in the y-direction: (a) x–y plane; (b) y–z plane.

multiply it by the distance over which it acts, namely, by distance dy. Fig. 5b * is not precisely parallel to the x-axis, due shows that the line-of-action of Nxx to the out-of-plane deflection. The component of the force associated with Nxx * that is parallel to the x-axis is therefore given by: ðN* xx ÞðdyÞcosa We now assume that angle a is small (less than about 10j or 0.1745 rad), such that the small-angle approximation can be applied: cos a c1. With this assumption the component of the force associated with Nxx * that is parallel to the x-axis reduces to: ðN* xx ÞðdyÞ In effect, invoking the small-angle approximation implies that we will ignore the fact that Nxx * is not precisely parallel to the x-axis during the summation of forces in the x-direction. In exactly the same way, the x-directed forces acting on the other three sides of the element are given by: 

x-directed force acting on the positive x-face:   BNxx * dx ðdyÞ Nxx *þ Bx



x-directed force acting on the negative y-face:



x-directed force acting on the positive y-face:   BNxy * dy ðdxÞ *þ Nxy By

ðNxy *ÞðdxÞ

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

Summing the force components acting on all four sides and equating to zero (SFx =0):       BNxy * BNxx * dy dx ¼ 0 ðNxx *Þdy þ Nxx *þ * dx þ Nxy *þ dx dy  Nxy By Bx Simplifying, we obtain: * BNxx * BNxy ¼0 þ By Bx

ð2aÞ

Eq. (2a) is the first equation of equilibrium. It shows that if Nxx * does indeed vary in the x-direction at an interior point in the plate, then N*xy must also vary in the y-direction at a comparable rate, so that static equilibrium is maintained. A second free-body diagram showing only those stress resultants whose line-of-action is parallel to the y-axis is shown Fig. 6. As before, the x–y plane of the element is shown in Fig. 6a while the y–z plane is shown in Fig. 6b. Summing forces in the y-direction, invoking the small angle approximation, and equating to zero (SFy =0), we find:



    BNyy * BNxy * dy dx  Nxy dx dy ¼ 0  Nyy * dx þ Nyy *þ * dy þ Nxy *þ By Bx which simplifies to: BNyy * BNxy * þ ¼0 By Bx

ð2bÞ

Eq. (2b) is the second equation of equilibrium. Let us now consider forces parallel to the z-axis. Due to out-of-plane deflection of the plate midsurface the stress resultants shown in Figs. 5 and 6 have components in the z-direction. For example, from Fig. 5b the component of the force associated with Nxx * that is parallel to the z-axis is given by: *ÞðdyÞsina ðNxx The small angle approximation (where a is expressed in radians) is: sinacac

Bw Bx

* is: Hence, the z-directed component of the force associated with Nxx ðNxx *ÞðdyÞ

Bw Bx

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

In exactly the same way, the z-directed forces acting on the other three sides of the element shown in Fig. 5 are given by: 

z-directed force acting on the positive x-face:    BNxx * Bw B2 w dx þ 2 dx ðdyÞ Nxx *þ Bx Bx Bx



z-directed force acting on the negative y-face:   Bw  Nxy * ðdxÞ Bx



z-directed force acting on the positive y-face:    BNxy * Bw B2 w dy þ dy ðdxÞ Nxy *þ By Bx BxBy

Adding these results, the resultants shown in Fig. 5 result in a component of force in the z-direction given by:    Bw BN*xx Bw B2 w ðdyÞ þ N*xx þ þ 2 dx ðdyÞ dx Bx Bx Bx Bx      Bw BN*xy Bw B2 w ðdxÞ þ N*xy þ þ dy ðdxÞ dy  N*xy Bx Bx BxBy By ðN*xx Þ

When the algebra indicated is completed, two higher-order terms appear: 

BN*xx Bx



 B2 w ðdxÞ2 dy Bx2

 and

BN*xy By



 B2 w dxðdyÞ2 BxBy

Neglecting these higher-order terms, the above expression simplifies to:

BN*xy Bw B2 w BN*xx Bw B2 w þ N*xy þ N*xx 2 þ dxdy Bx BxBy Bx Bx By Bx

ðaÞ

Following an identical procedure, the stress resultants shown in Fig. 6 represent a force in the z-direction given by:

BN*xy Bw B2 w BN*yy Bw B2 w þ N*xy þ dxdy N*yy 2 þ By BxBy By By Bx By

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

ðbÞ

The vertical shear resultants and transverse loading act directly in the zdirection, as previously shown in Fig. 3b. Using a procedure similar to that described above, the sum of these forces in the z-direction is:



  BVyz * BVxz * *dy þ Vxz *þ * dx þ Vyz *þ dy dx dx dy  Vyz ½Vxz By Bx þ ½qðx; yÞdxdy which simplifies to:

* BVxz * BVyz þ qðx; yÞ dxdy þ By Bx

ðcÞ

We are now ready to sum all forces in the z-direction. Adding expressions (a), (b), and (c), equating the resulting sum to zero (SFz =0), and rearranging, there results:   * Bw B2 w B2 w B2 w BNxx * BNxy þ þ * 2 þ Nyy * 2 þ 2Nxy * Nxx By Bx By BxBy Bx Bx   BNyy * BNxy * Bw BVxz * * BVyz þ þ qðx; yÞ ¼ 0 þ þ þ By Bx By By Bx Notice that the terms within the two sets of parenthesis have been previously shown to equal zero, in accordance with Eqs. (2a) and (2b). Hence these terms may be dropped, and our third equation of equilibrium becomes: Nxx *

* B2 w B2 w B2 w BVxz * BVyz þ qðx; yÞ ¼ 0 þ þ þ N * þ 2N * yy xy By Bx2 By2 BxBy Bx

ð2cÞ

Eqs. (2a), (2b), and (2c) represent the requirement that all forces in the x-, y-, and z-directions, respectively, sum to zero. Now consider moment equilibrium about the x-axis. A free-body diagram showing the resultants that contribute to the moment about the x-axis is shown in Fig. 7a. Summing moments and equating to zero (SMx =0), we have:



    BMyy * BMxy * dy dx þ Mxy dx dy Myy * dx  Myy *þ * dy  Mxy *þ By Bx



  BVyz * BVxz * dy dy dxðdyÞ þ Vxz þ Vyz dx dy *þ *þ By Bx 2     dy dy ½Vxz *dy þ ½qðx; yÞdxdy ¼0 2 2

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

Figure 7 Free-body diagrams used to sum moments about the x-, y-, and zaxes. (a) Infinitesimal element used to sum moments about the x-axiz. (b) Infinitesimal element used to sum moments about the y-axis. (c) Infinitesimal element used to sum moments about the z-axis.

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

which simplifies to:   BMyy * BMxy * BVyz * * dx 2 BVxz dxdy dxdy þ Vyz dxðdyÞ þ  * dxdy þ By Bx By Bx 2 qðx; yÞ dxðdyÞ2 ¼ 0
ðdyÞ2 þ 2 The last three terms in this equation contain the higher-order factor (dy)2. Hence dropping the last three terms and simplifying, we obtain: Vyz *¼

BMxy * BMyy * þ Bx By

ð3aÞ

A free-body diagram showing the stress and moment resultants that contribute to the moment about the y-axis is shown in Fig. 7b. Summing moments (SMy = 0), we have:



  BMxy * BMxx * dy dx dx dy þ Mxy ½Mxx *dy  Mxx *þ * dx  Mxy *þ By Bx

 

BVyz * BVxz * dx dy dx dx dyðdxÞ þ Vyz þ Vxz *þ *þ By Bx 2       dy dx  Vyz þ ½qðx; yÞdxdy ¼0 * dx 2 2 Upon completing the algebra indicated, a negligible higher-order term appears (in this case dx2). Neglecting all terms that include this factor and simplifying, we obtain: *¼ Vxz

* BMxx * BMxy þ By Bx

ð3bÞ

Lastly, a free-body diagram showing the stress resultants that contribute to the moment about the z-axis is shown in Fig. 7c. Summing moments (SMz =0), we obtain:       dy BNxx * dy dx dy  Nxx ðNxx *dyÞ *þ 2 Bx 2  

 

BNxy * BNxy * dx dy dx  Nxy dy dx dy þ Nxy *þ *þ Bx By         dx BNyy * dx þ Nyy dy dx *þ * dx  Nyy ¼0 2 2 By

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

Simplifying: 

  * 2 BNxy * BNxx * dxdy2 BNxy dx dy  dxdy2 þ Nxy þ * dx2 dy  dxdy2 Bx By Bx 2


BNyy * dx2 dy ¼0 By 2

All terms in this equation contain a higher-order term (either dx2 or dy2 ). Hence all terms on the left side of the equality are negligible, and the summation of moments about the z-axis provides no new information. In summary, we have developed five useful equations through a summation of forces and moments. The first three, Eqs. (2a)–(2c), were developed by requiring that the sum of forces in the x-, y-, and z-directions sum to zero, while the fourth and fifth equations, Eqs. (3a) and (3b), were developed by requiring that the sum of moments about the x- and y-axes sum to zero. Note that Eqs. (3a) and (3b) relate the out-of-plane shear stress * and Vyz * , to the moment resultants Mxx * , Myy * and Mxy * , and resultants, Vxz that the shear stress resultants also appear in Eq. (2c). This means that only three of the five equations we have developed are independent. The fact that only three equations are independent is fundamentally due to the assumption of plane stress, because this assumption implies there are only three independent stress components. It is customary to substitute Eqs. (3a) and (3b) into Eq. (2c), resulting in: * B2 Mxy * B2 w B2 w B2 w B2 Mxx * B2 Myy þ þ N * þ 2N * þ þ 2 yy xy 2 2 2 2 By BxBy Bx By BxBy Bx þ qðx; yÞ ¼ 0

* Nxx

ð4Þ

Eqs. (2a), (2b), and (4) represent three independent equations of equilibrium for a thin plate subjected to a state of plane stress. If a thin plate is in static equilibrium then the distribution of internal stress and moment resultants induced at any interior region of a rectangular plate must satisfy these equations, which guarantee that SFi =0 and SMi =0. Because these equations will be referenced repeatedly in the following discussion, the three equations of equilibrium are collected here and renumbered for convenience: * BNxx * BNxy ¼0 þ By Bx

ð5aÞ

BNyy * BNxy * þ ¼0 By Bx

ð5bÞ

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

* B2 Mxy * B2 w B2 w B2 w B2 Mxx * B2 Myy þ þ N * þ 2N * þ þ 2 yy xy 2 2 2 2 By BxBy Bx By BxBy Bx þ qðx; yÞ ¼ 0

Nxx *

ð5cÞ

These results have been developed strictly on the basis of static equilibrium. Note that we have not specified what mechanism caused the internal resultants, and so Eqs. (5a)–(5c) are valid regardless of whether the internal resultants were caused by external edge loading, by enforced edge displacements, or by any combination thereof. Also, we have made no assumptions regarding material properties. Therefore Eqs. (5a)–(5c) are valid for any thin plate, including isotropic metallic plates or anisotropic composite plates. We have, however, made one important assumption: we have assumed that the angles a and h, defined by the slope of the deflected midplane and shown in Figs. 5 and 6, are relatively small so that the small angle approximation can be invoked. The third equation of equilibrium, Eq. (5c), can be simplified for cases in which the loads applied to the plate result in a maximum out-of-plane deflection less than about half the laminate thickness: w(x,y)jmax< t/2. In such a case the midplane slopes (i.e., angles a and h) are exceedingly small and * , consequently the z-directed component of the in-plane stress resultants (Nxx Nyy * , and Nxy * ) can be neglected. If the in-plane resultants are ignored during * dy)Bw/Bx =0, for the summation of forces in the z-direction [setting (Nxx example] and the preceding derivation is repeated, the equations of equilibrium simplify as follows: * BNxx * BNxy ¼0 þ By Bx

ð6aÞ

BNyy * BNxy * þ ¼0 By Bx

ð6bÞ

* B2 Mxy * B2 Mxx * B2 Myy þ qðx; yÞ ¼ 0 þ þ2 2 2 By BxBy Bx

ð6cÞ

Eqs. (6a)–(6c) are valid if the maximum out-of-plane displacement is less than about half the plate thickness: w(x,y)jmax< t/2. 2.2 The Equations of Equilibrium Expressed in Terms of the [ABD] Matrix, Transverse Loading, and Midplane Displacement Fields In Chapter 6 we related stress and moment resultants to midplane strains and curvatures, temperature changes, and changes in moisture content. This

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

relationship is summarized by Eq. (44) in Chapter 6. We have since limited our discussion to symmetric laminates (Bij = MTij =MijM) and are now using the superscript ‘‘*’’ to denote internal stress and moment resultants. Consequently, Eq. (44) is modified slightly to become: 

 o  T  M N* e A 0 N N  ð7Þ  ¼ M* j 0 D 0 0 In this equation: 

N* and M* represent internal stress and moment resultants associated with externally applied forces,  NT represents thermal stress resultants associated with a uniform through-thickness change in temperature, DT, and  NM represents moisture stress resultants associated with a uniform through-thickness change in moisture content, DM. Recall that only uniform changes in temperature and/or moisture content are considered in this text. Consequently, NT and NM should be viewed as constants, i.e., they are not functions of x, y, or z. Expanding the first of the six equations represented by Eq. (7), we have: Nxx * ¼ A11 eoxx þ A12 eoyy þ A16 coxy  NTxx  NM xx We will now express the midplane strains and curvatures that appear in this expression in terms of midplane displacement gradients. Before we do so, however, the discussion presented in Sec. 14 of Chapter 2 should be reiterated. In particular, a distinction between finite strains and infinitesimal strains was made at that point. Basically, strains can be considered to be infinitesimal when displacement gradients are small, so that the square of any displacement gradient can be neglected; (Bw/Bx)2 c 0, for example. In this chapter we have already assumed that the slope of the deflected plate midplane is small (i.e., Bw/Bx and Bw/By have already been assumed to be small), which allowed us to apply the small angle approximation. Consequently, we will continue to treat strains as infinitesimal strains. As we will see, this assumption will ultimately lead to the conclusion that in-plane displacement fields uo(x,y) and vo(x,y) (as well as in-plane forces) are independent of the transverse load, q(x,y). Rigorously speaking, this conclusion is incorrect. That is, if a thin plate is subjected to a transverse loading then in-plane displacement fields and/or in-plane forces will change, reflecting a dependence on transverse load. However, if displacement gradients are small, then the changes in-plane displacements or forces are also small and can usually be ignored. In effect, the assumption of infinitesimal strains has eliminated (in a mathematical sense) the coupling between transverse loads and in-plane displacements/

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

forces. Occasionally, this leads to predictions that are nonintuitive. This will be pointed out at appropriate points in following sections (in particular, see Sample Problem 2 in Sec. 4 of Chapter 10). Although the interdependence between out-of-plane and in-plane displacements is ignored throughout most of this text, there is one important exception. This interdependence is included in Chapter 11, during formulation of an approximate analysis technique known as the Ritz method. The interdependence must be included there so as to obtain buckling predictions. In any event, we now assume that strains are infinitesimal and can therefore be expressed in terms of midplane displacement fields in accordance with Eqs. 49(a)–(49f) of Chapter 2: eoxx ¼

Buo Bx

eoyy ¼

Bvo By

coxy ¼

Buo Bvo þ By Bx

jxx ¼

B2 w Bx2

jyy ¼

B2 w By2

jxy ¼ 2

B2 w BxBy

* can be written: Therefore, the internal stress resultant Nxx   Buo Bvo Buo Bvo þ A12 þ A16 þ Nxx * ¼ A11  NTxx  NM xx Bx By By Bx   * is easily obtained: xx Given this result, the quantity BN Bx  2  BNxx * B2 uo B2 vo B uo B 2 vo ¼ A11 þ A16 þ þ A12 Bx Bx2 BxBy BxBy Bx2 Note that this quantity appears as a term in the first equation of equilibrium, Eq. (5a). Also note that neither the thermal nor moisture stress resultant is involved in this result, because as previously noted NT and NM are constants and therefore BNTxx =Bx ¼ BNM xx =Bx ¼ 0. Following an analogous procedure, the second term in Eq. (5a) is found to be:  2  BNxy * B2 uo B 2 vo B uo B 2 vo ¼ A16 þ A26 þ A66 þ By BxBy By2 By2 BxBy Adding these two results, Eq. (5a) can be written in terms of elements of the [ABD] matrix and midplane displacements as follows: A11

B 2 uo B 2 vo B 2 uo B2 vo B2 vo þ 2A16 þ A16 þ ðA12 þ A66 Þ þ A26 2 2 Bx BxBy BxBy Bx By2

B2 uo þA66 ¼0 By2

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

ð8aÞ

Following an identical procedure, the second and third equations of equilibrium [Eqs. (5b) and (5c)] can be written: A16



B 2 uo B2 uo B2 vo B 2 uo B 2 vo þ 2A þ A þ ð A þ A Þ þ A 12 66 26 26 22 Bx2 BxBy BxBy By2 By2

B2 vo þ A66 ¼0 Bx2

ð8bÞ

 2    Buo Bvo Buo Bvo B w T M þ A12 þ A16 þ A11  Nxx  Nxx Bx2 Bx By By Bx

 2    Buo Bvo Buo Bvo B w þ A22 þ A26 þ þ A12  NTyy  NM yy By2 Bx By By Bx

ð8cÞ

 2    Buo Bvo Buo Bvo B w T M þ 2 A16 þ A26 þ A66 þ  Nxy  Nxy BxBy Bx By By Bx

D11

B4 w B4 w B4 w  4D  4D 16 26 Bx4 Bx3 By BxBy3

2ðD12 þ 2D66 Þ

B4 w B4 w  D22 4 þ qðx; yÞ ¼ 0 2 2 Bx By By

Eqs. (8a)–(8c) represent the equations of equilibrium for a symmetric laminate in terms of the [ABD] matrix, transverse loading, and midplane displacement fields. These are valid for any thin symmetric laminate as long as out-of-plane displacements are not excessive. If out-of-plane displacements are very small [i.e., if w(x,y)jmax< t/2] then the equations of equilibrium given by Eqs. (6a)–(6c) are applicable, which become: A11

B 2 uo B 2 vo B 2 uo B 2 vo B 2 vo þ 2A þ A þ ð A þ A Þ þ A 12 66 16 16 26 Bx2 BxBy BxBy Bx2 By2

B2 uo þA66 ¼0 By2 A16

B 2 uo B2 uo B2 vo B 2 uo B 2 vo þ 2A26 þ A26 þ ðA12 þ A66 Þ þ A22 2 2 Bx BxBy BxBy By By2

þ A66

B 2 vo ¼0 Bx2

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

ð9aÞ

ð9bÞ

D11

B4 w B4 w B4 w B4 w  4D  4D  2 ð D þ 2D Þ 16 26 12 66 Bx 4 Bx3 By BxBy3 Bx2 By2

B4 w D22 þ qðx; yÞ ¼ 0 By 4

ð9cÞ

3 BOUNDARY CONDITIONS A summary of the material presented in Secs. 1 and 2 is as follows. We consider a thin rectangular plate of in-plane dimensions a
b, whose edges are parallel to the x- and y- axes. The external edge loads considered in thin plate theory were introduced in Sec. 1. A total of five types of externally applied distributed loads may be present on each of the four edges of a rectangular laminate: one normal stress resultant, one in-plane shear stress resultant, one out-of-plane shear stress resultant, one bending moment resultant, and one twisting moment resultant. Each of these loads is, at most, a function of the coordinate direction tangent to the plate edge. In addition, a transverse loading q(x,y) may be applied to the surface of the laminate. Fig. 3 provides a summary of all externally applied loads considered in thin plate theory. These externally applied edge loads induce a distribution of internal stress and moment resultants at all interior regions of the plate. Generally, internal stress and moment resultants are functions of both x and y and vary throughout the plate. The distribution of internal stress and moment resultants were investigated in Sec. 2.1 by considering free-body diagrams of an infinitesimal element removed from an interior point within the plate. Requiring that the sum of all forces and moments equate to zero resulted in the equations of equilibrium, summarized as Eqs. (5a)–(5c). If the maximum out-of-plane displacement is less than half the laminate thickness, then a simplified version of the equations of equilibrium is applicable, summarized as Eqs. (6a)–(6c). A mathematically equivalent form for a symmetric composite laminate was obtained in Sec. 2.2, where the equations of equilibrium were written in terms of elements of the [ABD] matrix and midplane displacement fields, summarized as Eqs. (8a)–(8c) and Eqs. (9a)– (9c). In this section we will formally define the boundary conditions of the plate. That is, we wish to precisely define what conditions exist along each of the four edges of the rectangular plate. Actually, we have already begun our discussion of boundary conditions, in the sense that the external stress and moment resultants that may be applied along the edges of the laminate were described in Sec. 1. However, in many instances the loads applied along the plate edges are unknown and hence cannot be used as specified boundary

conditions. Rather, the midplane displacements along the edges are known, where displacements of the midplane in the x-, y, and z-directions are denoted uo(x,y), vo(x,y), and w(x,y), respectively, as in earlier chapters. Hence, two categories of boundary conditions can be defined. We can either specify components of the edge displacements, or we can specify components of the edge loads. Boundary conditions involving specified displacements are called geometric (or kinematic) boundary conditions, while boundary conditions involving specified edge loads are called static (or natural) boundary conditions. 3.1 Geometric (Kinematic) Boundary Conditions Geometric boundary conditions are those that dictate some feature of midplane displacements along a plate edge. Each of the four edges of the plate is characterized by a normal and tangential direction. For example, for the positive x-edge shown in Fig. 3 the x-direction is normal to the edge while the y-direction is tangent to the edge. Along this edge we may specify values for displacements in the x-, y-, and z-directions, as well as the slope of the laminate midplane measured in the normal direction. Because this edge is parallel to the y-axis, the displacements imposed along this edge are either constant or, at most, functions of y. That is, we may specify values of uo(+x)( y), vo(+x)( y), w (+x)( y), and Bw(+x)( y)/Bx along the edge x = a. As was the case for edge loads (discussed in Sec. 1), there is no reason to assume that the displacements on opposite edges are identical. Therefore the superscript (+x) has been used to indicate that these displacements are imposed along the positive x-edge of the laminate. Note that by specifying a particular value of the midplane slope at the boundary, (Bw(+x)( y)/Bx), we have specified only one of several possible gradients in the three displacement fields. Initially, one might suspect that, for reasons of mathematical symmetry perhaps, other displacement field gradients should also be specified. For example, perhaps Buo(+x)( y)/By or Bvo(+x)( y)/Bz should be specified along x=a as well. These other gradients need not be considered for two reasons. First, in thin plate theory the boundary conditions are defined for the plate edge at the midplane. By definition the midplane has a thickness of zero. This fundamental definition precludes any midplane boundary conditions with a z-dependency; i.e., the midplane displacement field gradients Buo(+x)(y)/Bz, Bvo(+x)(y)/Bz, or Bw(+x)( y)/Bz are undefined. This leaves us with possible gradients in the xor y-directions. With the exception of (Bw(+x)( y)/Bx), a specified gradient in geometric boundary conditions along a plate edge in the x- or y-directions results in an over-specified problem. For example, having specified displacements in the x-direction along the edge x = a, namely, u o(+x)( y), the value

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

of Buo(+x)( y)/By is also specified as a consequence. A similar consideration of the other possible gradients results in a similar conclusion. Recalling that displacements at any arbitrary interior point of the plate are denoted uo(x,y), vo(x,y), and w(x,y), the geometric boundary conditions that may be specified along each of the four edges of the plate are: For edge x= a:

For edge x = 0: ðþxÞ uo ð yÞ ðþxÞ vo ð yÞ ðþxÞ

u0 ða; yÞ ¼ v0 ða; yÞ ¼ Bwða; yÞ Bw ð yÞ ¼ Bx Bx wða; yÞ ¼ wðþxÞ ð yÞ For edge y= b: u0 ðx; bÞ ¼

ðxÞ

u0 ð0; yÞ ¼ uo ð yÞ ðxÞ v0 ð0; yÞ ¼ vo ð yÞ Bwð0; yÞ BwðxÞ ð yÞ ¼ Bx Bx wð0; yÞ ¼ wðxÞ ð yÞ For edge y =0:

ðþyÞ uo ðxÞ ðþyÞ vo ðxÞ ðþyÞ

v0 ðx; bÞ ¼ Bwðx; bÞ Bw ðxÞ ¼ By By wðx; bÞ ¼ wðþyÞ ðxÞ

ðyÞ

u0 ðx; 0Þ ¼ uo

ðxÞ

ðyÞ vo ðxÞ ðyÞ

v0 ðx; 0Þ ¼ Bwðx; 0Þ Bw ðxÞ ¼ By By wðx; 0Þ ¼ wðyÞ ðxÞ

3.2 Static (Natural) Boundary Conditions We will again use the positive x-edge to demonstrate possible static boundary conditions. Referring to Fig. 3, note that five external stress and moment (+x) (+x) resultants may be specified along this edge: N xx ( y), N (+x) xy ( y), Vxz (+x) (+x) (+x) (+x) (+x) ( y), Mxx ( y), and Mxy ( y). Values for Nxx ( y), Nxy ( y), and Mxx ( y) can be specified independently, without consideration of any other stress (+x) ( y) and or moment results. However, the remaining two resultants, Vxz (+x) Mxy ( y), are not independent and must be considered together. The in(+x) (+x) ( y) and Mxy ( y) was first noted by Kirchhoff terdependence between Vxz in 1850 and occurs because (as far as static equilibrium is concerned) the (+x) ( y) can be replaced by a statically equivalent couple twisting moment Mxy involving two vertical shear resultants. This is illustrated in Fig. 8. The twisting moments acting on two adjacent elements of length dy are shown in Fig. 8a. The element on the right is subjected to a twisting moment (+x) M xy ( y)dy, whereas the element on the left is subjected to a twisting moment ðþxÞ ðþxÞ [Mxy (y)+(dMxy ðyÞ/dy) dy] dy. These twisting moments may be replaced by an equivalent couple involving two vertical shear resultants, as shown in Fig. 8b. The element on the right is subjected to two vertical shear resultants of

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

Figure 8 Representing the twisting moment resultant acting along the edge x = a with statically equivalent vertical shear stress resultants. (a) Twisting moment resultants acting on two adjacent elements of length dy along the edge x = a. (b) Vertical shear resultants acting on adjacent elements along the edge x = a which are statically equivalent to the twisting moments shown in (a).

(+x) magnitude Mxy ( y), acting in the directions shown and separated by distance dy. Similarly, the element on the left is subjected to two vertical ðþxÞ ðþxÞ shear resultants of magnitude [Mxy (y)+(dMxy ( y)/dy)dy], acting in the directions shown and separated by distance dy. Summing the shear force components present at the interface between the two infinitesimal elements, ðþxÞ ðþxÞ ðþxÞ ðþxÞ we obtain {[Mxy ( y)+(dMxy ( y)/dy)dy]Mxy ( y)}=(dMxy ( y)/dy)dy. (+x) Hence the variation of the twist moment M xy ( y) along the edge x =a is ðþxÞ statically equivalent to a vertical shear resultant of magnitude dMxy ( y)/dy. Adding this statically equivalent vertical shear resultant component to the (+x) external vertical shear that is actually applied to the edge, Vxz ( y), we define the ‘‘Kirchhoff’’ shear resultant acting along the edge x= a:

ðyÞ ¼ V ðþxÞ V ðþxKÞ xz xz ðyÞ þ

ðþxÞ

dMxy ðyÞ dy

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

ð10Þ

The superscript ‘‘+xK’’ is used to denote the Kirchhoff shear resultant (sometimes called the ‘‘effective’’ shear resultant) acting along the positive x-edge. Now, the equations of equilibrium (developed in Sec. 2) must be satisfied at all points within the laminate, including the laminate edge. Therefore the distribution of the external vertical shear resultant, V (+x) xz ( y), must satisfy Eq. (3b). Along the edge x=a this equation becomes: V ðþxÞ xz ¼

ðþxÞ

BMxx Bx

ðþxÞ

þ

BMxy By

Recalling that all boundary conditions along the positive x-edge are either ðþxÞ constant or are functions of y only, it must be that dMxx /dx=0. Therefore along the edge x= a: V ðþxÞ ¼ xz

ðþxÞ

BMxy By

ð11Þ

(+x) Eq. (11) shows that the variation of the externally applied shear V xz ( y) along the edge x =a is intimately related to the variation of the externally (+x) applied twist moment Mxy ( y) along this same edge, and that the functional form of these two resultants cannot be specified independently.* Suppose, for example, that the laminate is subjected to a vertical shear resultant that varies (+x) ( y)= Ay+B, where A and linearly over the edge x = a. That is, suppose Vxz B are constants. Under this circumstances, Eq. (11) shows that a twist moment must also be present in order for static equilibrium to be maintained. ðþxÞ Furthermore, the twist moment must vary according to dMxy ( y)/dy= ðþxÞ Ay+B, which implies that Mxy ðyÞ ¼ A=2 ðy2 Þ þ By þ C , where C is a constant of integration. (+x) is constant, i.e., assume As a second case of interest, suppose that V xy (+x) A = 0 and hence that Vxz ( y)= B. In this case the rate of change in the twist ðþxÞ moment must equal dMxy ( y)/dy=B; otherwise, Eq. (11) is not satisfied and static equilibrium is not maintained. This shows that it is not possible to apply constant vertical shear only along x = a; if a constant vertical shear is applied, then a corresponding twist moment that varies over the length of the plate edge must also be present. However, the converse is admissible. That is, if a (+x) = B is applied, then the vertical shear must constant twist moment Mxy

ðþxÞ

ðþxÞ

equal zero: dMxy ( y)/dy=V xz ( y)=0.

* It is noted that a similar interdependence occurs between the shear force and bending moment present in a prismatic beam. That is, from fundamental beam theory, the shear force present at any cross section within a prismatic beam is related to the bending moment according to V = dM/dx.

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

Substituting Eq. (11) into Eq. (10), we find that the effective Kirchhoff shear resultant acting along the edge x=a can be written in two equivalent ways: ðþxKÞ

V xz

ðþxÞ

ðþxÞ

ðyÞ ¼ V xz ðyÞ þ

dMxy ðyÞ dy ð12Þ

or ðþxKÞ

V xz

ðyÞ ¼ 2

ðþxÞ dMxy ðyÞ

dy

Note that the Kirchhoff shear resultant differs from the applied vertical shear (+xK) only if there is variation in the twisting moment. That is, V xz ( y) differs ðþxÞ (+x) from Vxz ( y) only if dMxy ( y)/dy p 0. In effect, the Kirchhoff shear resultant is used to specify both the vertical shear and twist moment acting along the edge of the plate. Recalling that stress and moment resultants at any arbitrary interior point of the plate are denoted Nxx * (x,y), Nyy * (x,y), Nxy * (x,y), Nyx * (x,y), * (x,y), Myy * (x,y), Mxy * (x,y), Myx * (x,y), Vxz * (x,y), and Vyz * (x,y), the static Mxx boundary conditions that may be specified along the edges x= a are as follows: For edge x= a: ðþxÞ

Nxx *ða; yÞ ¼ Nxx ðyÞ *ða; yÞ ¼ NðþxÞ Nxy xy ðyÞ *ða; yÞ ¼ MðþxÞ Mxx xx ðyÞ ðþxÞ

ðþxÞ

BMxy *ða; yÞ BMxx *ða; yÞ dMxy ðyÞ dMxy ðyÞ ¼ V ðþxÞ þ2 ¼2 xz ðyÞ þ By Bx dy dy Static boundary conditions along the edge x= a have been discussed above. Following an analogous procedure, static boundary conditions that may be present along the remaining three edges of the rectangular plate are as follows: For edge x= 0: ðxÞ

Nxx *ð0; yÞ ¼ Nxx ðyÞ ðxÞ

*ð0; yÞ ¼ Nxy ðyÞ Nxy ðxÞ

*ð0; yÞ ¼ Mxx ðyÞ Mxx ðxÞ

ðxÞ

BMxy *ð0; yÞ BMxx *ð0; yÞ dMxy ðyÞ dMxy ðyÞ þ2 ¼2 ¼ V ðxÞ xz ðyÞ þ Bx dy dy By

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

For edge y= b: ðþyÞ

Nyy *ðx; bÞ ¼ Nyy ðxÞ ðþyÞ

Nyx *ðx; bÞ ¼ Nyx ðxÞ ðþyÞ

*ðx; bÞ ¼ Myy ðxÞ Myy ðþyÞ

ðþyÞ

ðyÞ

ðyÞ

BMyy *ðx; bÞ BMyx *ðx; bÞ dM yx ðxÞ dM yx ðxÞ þ2 ¼ V ðþyÞ ¼2 yz ðxÞ þ By Bx dx dy For edge y= 0: ðyÞ

Nyy *ðx; 0Þ ¼ Nyy ðxÞ ðyÞ

*ðx; 0Þ ¼ Nyx ðxÞ Nyx ðyÞ

Myy *ðx; 0Þ ¼ Myy ðxÞ BMyy *ðx; 0Þ BMyx *ðx; 0Þ dM yx ðxÞ dM yx ðxÞ þ2 ¼ V ðyÞ ¼2 yz ðxÞ þ By Bx dx dy 3.3 Combinations of Geometric and Static Boundary Conditions Two fundamental categories of boundary conditions were defined in the preceding subsections: Geometric Boundary Conditions and Static Boundary Conditions. If either a geometric or static boundary condition is specified to equal zero, then that condition is called a homogeneous condition. In contrast, if a condition is required to take on a nonzero value, then the condition is called an inhomogeneous condition. Examples of homogenous and inhomogeneous geometric conditions along the edge x =a are uo(a,y) =0 and uo(a,y)=1 mm, respectively. Examples of homogeneous and inhomogeneous static conditions along the edge x =a are Nxx(a,y)=0 and Nxx(a,y) =1000 N/ m, respectively. Four potential conditions exist in each category for each of the four edges of a rectangular laminate. The possible geometric and static condition for each condition must be viewed as complementary pairs. That is, for each condition we may specify either a geometric requirement or a static requirement, but we cannot specify both. For example, suppose a problem is considered in which it is stipulated that displacements in the x-direction along the edge x =a are zero. That is, we specify that uo(a,y)=0. Note that this requirement is an example of a homogeneous geometric boundary

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

condition. In such a case the geometric boundary condition is specified and is therefore known (i.e., uo(a,y)=0), whereas the corresponding static boundary condition is unknown. In fact, the question becomes, ‘‘what stress resultant N (+x) xx ( y) must exist along the edge x =a, so as to maintain the boundary condition uo(a,y)=0?’’ Conversely, suppose in a different problem it is stipulated that the stress resultant acting normal to the edge x=a must equal some constant value. For example, suppose we specify a boundary condition in which N (+x) xx ( y) =1000 N/m. This is an example of an inhomogeneous static boundary condition, and in this case it is the geometric boundary condition that is unknown. In fact, the question becomes, ‘‘what displacements uo(a,y) exist along the edge x =a, (+x) ( y)=1000 N/m?’’ given that the edge loading is known to be Nxx As stated above, there are four complementary pairs of boundary conditions for each edge of a rectangular plate. For each pair we may specify either a geometric requirement or a static requirement, but we cannot specify both. The conditions that must be specified for each edge of a rectangular plate are summarized below: The following conditions can be specified for the edge x=a: Geometric Condition ðþxÞ

uo ða; yÞ ¼ uo

Static Condition ðþxÞ

ðyÞ

ðORÞ

Nxx *ða; yÞ ¼ Nxx ðyÞ

ðyÞ

ðORÞ

Nxy *ða; yÞ ¼ Nxy ðyÞ

Bwða; yÞ BwðþxÞ ðyÞ ¼ Bx Bx

ðORÞ

Mxx *ða; yÞ ¼ Mxx ðyÞ

wða; yÞ ¼ wðþxÞ ðyÞ

ðORÞ

BMxy *ða; yÞ BMxx *ða; yÞ þ2 By Bx

ðþxÞ

vo ða; yÞ ¼ vo

ðþxÞ

ðþxÞ

¼ VðþxÞ xz ðyÞ þ

ðþxÞ

dMxy ðyÞ dy

ðþxÞ

¼2

dMxy ðyÞ dy

The following conditions can be specified for the edge x =0: Geometric Condition uo ð0; yÞ ¼

ðxÞ uo ðyÞ ðxÞ

vo ð0; yÞ ¼ vo

ðyÞ

Bwð0; yÞ BwðxÞ ðyÞ ¼ Bx Bx

Static Condition ðxÞ

ðORÞ

Nxx *ð0; yÞ ¼ Nxx ðyÞ

ðORÞ

Nxy *ð0; yÞ ¼ Nxy ðyÞ

ðORÞ

Mxx *ð0; yÞ ¼ Mxx ðyÞ

ðxÞ

ðxÞ

wð0; yÞ ¼ wðxÞ ðyÞ

ðORÞ

BMxy *ð0; yÞ BMxx *ð0; yÞ þ2 By Bx ¼ VðxÞ xz ðyÞ þ

ðxÞ

dMxy ðyÞ dy

ðxÞ

¼2

dMxy ðyÞ dy

The following conditions can be specified for the edge y =b: Geometric Condition ðþyÞ

uo ðx; bÞ ¼ uo

Static Condition ðþyÞ

ðxÞ

ðORÞ

Nyx *ðx; bÞ ¼ N yx ðxÞ

ðxÞ

ðORÞ

Nyy *ðx; bÞ ¼ N yy ðxÞ

Bwðx; bÞ BwðþyÞ ðxÞ ¼ By By

ðORÞ

Myy *ðx; bÞ ¼ M yy ðxÞ

wðx; bÞ ¼ wðþyÞ ðxÞ

ðORÞ

BMyy *ðx; bÞ BMyx *ðx; bÞ þ2 By Bx

ðþyÞ

vo ðx; bÞ ¼ vo

ðþyÞ

ðþyÞ

¼ V ðþyÞ yz ðxÞ þ

ðþyÞ

dM yx ðxÞ dx

ðþyÞ

¼2

dM yx ðxÞ dx

The following conditions can be specified for the edge y =0: Geometric Condition ðyÞ

uo ðx; 0Þ ¼ uo

Static Condition ðyÞ

ðxÞ

ðORÞ

Nyx *ðx; 0Þ ¼ N yx ðxÞ

ðxÞ

ðORÞ

Nyy *ðx; 0Þ ¼ N yy ðxÞ

Bwðx; 0Þ BwðyÞ ðxÞ ¼ By By

ðORÞ

Myy *ðx; 0Þ ¼ M yy ðxÞ

wðx; 0Þ ¼ wðyÞ ðxÞ

ðORÞ

BMyy *ðx; 0Þ BMyx *ðx; 0Þ þ2 By Bx

ðyÞ

vo ðx; 0Þ ¼ vo

ðyÞ

ðyÞ

ðyÞ ðxÞ þ ¼ Vyz

ðyÞ

dM yx ðxÞ dx

ðyÞ

¼2

dM yx ðxÞ dx

The number of different combinations that may be defined is enormous. Four conditions must be specified to define the boundary conditions along

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

an edge. Because there are two possibilities in each case, a total of 24 =16 possible combinations exist along each edge. Because there are four edges, a total of 164=65,536 different combinations of boundary conditions can be defined for a thin rectangular plate. However, some combinations are encountered so frequently that they have been given special names. Specifically, free edges, uniformly loaded edges, simply supported edges, and clamped edges are encountered very frequently in practice. These particular edge conditions will be illustrated below by considering the positive x-edge of the plate. 3.3.1 Free Edge A free edge is defined as an edge that is entirely free of external loading. Hence if the edge x =a is a free edge, then the following four homogeneous static boundary conditions must be satisfied: For x=a: Nxx *ða; yÞ ¼ 0 *ða; yÞ ¼ 0 Nxy *ða; yÞ ¼ 0 Mxx BMxy *ða; yÞ BMxx *ða; yÞ ¼0 þ2 By Bx Analogous boundary conditions may be used to specify that any of the remaining three edges are free edges. Each of the internal stress and moments resultants can be expressed in terms of the [ABD] matrix and midplane displacement fields. Recalling that we have limited our discussion to symmetric laminates (Bij =M ijT=M ijM =0) the boundary conditions for a free edge may also be written: For x=a:   Buo ða; yÞ Bvo ða; yÞ Buo ða; yÞ Bvo ða; yÞ T M þ A12 þ A16 þ A11  Nxx ¼0  Nxx Bx By By Bx   Buo ða; yÞ Bvo ða; yÞ Buo ða; yÞ Bvo ða; yÞ T M  Nxy þ A26 þ A66 þ A16  Nxy ¼0 Bx By By Bx B2 wða; yÞ B2 wða; yÞ B2 wða; yÞ  D12  2D16 ¼0 2 2 Bx By BxBy B3 wða; yÞ B3 wða; yÞ B3 wða; yÞ B3 wða; yÞ  ðD  4D þ 4D Þ  2D ¼0 D11 16 12 66 26 Bx3 Bx2 By BxBy2 By3 D11

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

The results listed above are for the edge x =a. Comparable conditions are imposed if any of the other three edges of the plate are free edges. 3.3.2 Simply Supported Edges A simply supported boundary condition is one in which the out-of-plane deflection and bending moment are zero along an edge. Hence the x=a edge of a rectangular plate is simply supported if w(a,y)=0 and Mxx(a,y)=0. Note that these are homogeneous geometric and static boundary conditions, respectively. The corresponding boundary condition in beam theory is a ‘‘pinned support,’’ as discussed in Sec. 8 of Chapter 8. Recall that for the case of a pinned support as defined in beam theory only two requirements are involved (w=Mx =0), whereas in thin plate theory four conditions must be specified along each plate edge. Hence for thin plates four distinct combinations of geometric and static boundary conditions may be classified as a ‘‘simple support.’’ The possible simple-support boundary conditions are often numbered S1 through S4, following a numbering scheme introduced by Almroth (5), and are defined along the edge x=a as follows: For x=a: S1 : wða; yÞ ¼ 0 Mxx *ða; yÞ ¼ 0

ðþxÞ

uo ða; yÞ ¼ uo

ð yÞ

ðþxÞ

ð yÞ

ðþxÞ

ð yÞ

vo ða; yÞ ¼ vo

ðþxÞ *ða; yÞ ¼ 0 Nxx *ða; yÞ ¼ Nxx ð yÞ vo ða; yÞ ¼ vo S2 : wða; yÞ ¼ 0 Mxx

* ða; yÞ ¼ 0 S3 : wða; yÞ ¼ 0 Mxx * ða; yÞ ¼ 0 S4 : wða; yÞ ¼ 0 Mxx

ðþxÞ

uo ða; yÞ ¼ uo

ð yÞ

ðþxÞ Nxy *ða; yÞ ¼ Nxy ð yÞ

ðþxÞ ðþxÞ Nxx * ða; yÞ ¼ Nxx ð yÞ Nxy *ða; yÞ ¼ Nxy ð yÞ

Note that a ‘‘simple support’’ is by definition a mixture of geometric and static boundary conditions. Also, depending on the values specified for ðþxÞ ðþÞ ðþxÞ ðþÞ [uo ( y) or Nxx ( y)] and [vo ( y) or Nxy ( y)], either homogeneous or inhomogeneous boundary conditions may be involved. The conditions involving stress and moment resultants can once again be expressed in terms of elements of the [ABD] matrix and midplane displacement fields. For example, the four conditions that (collectively) define a simple support of type S4 for a symmetric laminate along the edge x=a can be written: S4 simple support, for x=a: wða; yÞ ¼ 0 D11

B2 wða; yÞ B2 wða; yÞ B2 wða; yÞ ¼0 þ D12  2D16 2 2 Bx By BxBy

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

A11

  Buo ða; yÞ Bvo ða; yÞ Buo ða; yÞ Bvo ða; yÞ þA12 þA16 þ Bx By By Bx ðþxÞ

 NTxx NM xx ¼ N xx

  Buo ða; yÞ Bvo ða; yÞ Buo ða; yÞ Bvo ða; yÞ þ A26 þ A66 þ A16 Bx By By Bx ðþxÞ  NTxy  NM xy ¼ Nxy

The other types of simple support can also be expressed in terms of midplane displacements following a comparable procedure. Analogous conditions may be imposed if any of the other three edges of a plate are simply supported. 3.3.3 Clamped Edges A clamped edge (also called a fixed edge) is one in which the out-of-plane deflection and slope are zero along the edge. Hence the x=a edge of a rectangular plate is clamped if two homogenous geometric boundary conditions are satisfied: w(a,y)=0 and Bw(a,y)/Bx=0. Once again, the term ‘‘clamped end’’ is used during the study of beams, as discussed in Sec. 8 of Chapter 8. However, during the study of beams only two geometric requirements are necessary to define a clamped boundary condition, whereas in the case of thin plates four conditions are required. Hence, four distinct combinations can be classified as a ‘‘clamped edge.’’ The possible clamped boundary conditions are often numbered C1 through C4 and are defined as follows: For x=a: C1 : wða; yÞ ¼ 0

Bwða; yÞ ðþxÞ ¼ 0 uo ða; yÞ ¼ uo ðyÞ Bx

C2 : wða; yÞ ¼ 0

Bwða; yÞ ðþÞ ðþxÞ ¼ 0 Nxx *ða; yÞ ¼ Nxx ðyÞ vo ða; yÞ ¼ vo ðyÞ Bx

C3 : wða; yÞ ¼ 0

Bwða; yÞ ðþxÞ ¼ 0 uo ða; yÞ ¼ uo ðyÞ Bx

C4 : wða; yÞ ¼ 0

Bwða; yÞ ðþÞ ðþÞ ¼ 0 Nxx *ða; yÞ ¼ Nxx ðyÞ Nxy *ða; yÞ ¼ Nxy ðyÞ Bx

ðþxÞ

vo ða; yÞ ¼ vo

ðyÞ

ðþÞ

Nxy *ða; yÞ ¼ Nxy ðyÞ

A clamped boundary may or may not involve a mixture of geometric and (+) static conditions, depending on the values specified for [uo(+x)( y) or Nxx ( y)] (+x) (+) and [vo ( y) or Nxy ( y)]. Also, these conditions may be either homoge-

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

neous or inhomogeneous. In those cases in which the internal stress or moment resultants are required to take on specific values at the boundary (e.g., condition C4), the internal stress and moment resultant can be expressed in terms of elements of the [ABD] matrix and midplane displacement fields, as demonstrated above.

4 REPRESENTING ARBITRARY TRANSVERSE LOADS AS A FOURIER SERIES In thin plate theory the distributed transverse loading is allowed to vary in any manner over the x–y plane, i.e., q=q(x,y). However, as will be seen it is particularly easy to obtain a solution if the transverse loading is distributed according to a double sinusoidal variation in x and y:  px   py  qðx; yÞ ¼ qo sin sin ð13Þ a b where qo is a constant and equals the magnitude of the distributed load at the center of the plate, i.e., at x=a/2 and y=b/2. Solutions obtained for this particular distributed transverse loading will be described in Secs. 2 and 3 of Chapter 10. Of course, a double sinusoidal variation is just one of an infinite number of possible transverse loads that may be encountered in practice. However, it can be shown that any arbitrary function q(x,y) can be represented in terms of a double Fourier series: qðx; yÞ ¼

l X l X m¼1 n¼1

qmn sin

 mpx  a

 npy  sin b

ð14Þ

Therefore if a solution is obtained for the simple sinusoidal variation given by Eq. (13), then a solution for an arbitrary transverse loading can be obtained by representing the arbitrary load as the sum of a series of double sinusoidal terms, in accordance with Eq. (14). The constant coefficients qmn that appear in Eq. (14) are determined based on the functional form of q(x,y) and correspond to a particular combination of m and n. To determine the values of qmn, multiply Eq. (14) by the factor sin(nVpy/b)dy, and integrate from 0 to b: Z

b 0

!    Z b X l X l  mpx   npx  nVpy nVpy dy ¼ sin sin dy qðx; yÞsin qmn sin b a b b 0 m¼1 n¼1 

ð15Þ

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

It can be shown that:   Z b  npy  nVpy sin dy ¼ 0; if n p nV sin b b 0   Z b  npy  nVpy b sin sin dy ¼ ; if n ¼ nV b b 2 0 Consequently, upon performing the integration on the right side of Eq. (15), only one term for the range 0
ð17Þ

Eq. (17) allows calculation of the constant coefficients, and hence a given distribution of transverse load q(x,y) can be expressed as the double Fourier series shown in Eq. (14). Eq. (14) involves a summation over an infinite number of terms (as m,n ! l). Of course, it is impossible to use an infinite number of terms; in practice a finite number of terms must be used. As the number of terms used is increased, the series representation given by Eq. (14) resembles the actual load q(x,y) more and more closely. That is, the series representation converges to the actual loading q(x,y) as the number of terms is increased. In practice then, one should use the maximum number of terms that is reasonably possible, to ensure a reasonable series representation of the applied load q(x,y). Three types of commonly encountered transverse loads will be used to illustrate the preceding discussion. First, consider the case of a constant uniform transverse load: qðx; yÞ ¼ qo

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

where qo is the magnitude of the uniformly distributed load. Substituting q(x,y) = qo into Eq. (17), we find the coefficients in the Fourier series expansion associated with specified value of m and n to be: Z Z 4qo a b  mpx   npy  16qo ; m; n ¼ odd integers qmn ¼ sin sin dxdy ¼ 2 ab 0 0 p mn a b qmn ¼

4qo ab

aZ b

Z 0

0

 mpx   npy  sin dxdy ¼ 0; sin a b

m; n ¼ even integers

Hence a constant transverse load q(x,y)=qo can be written: qðx; yÞ ¼

l X l  mpx   npy  16qo X 1 sin sin p2 m ¼ 1 n ¼ 1 mn a b

ðm; n ¼ 1; 3; 5; . . .Þ

ð18Þ

How well Eq. (18) describes a constant transverse loading depends on the number of terms used. A normalized plot of the series representation given by Eq. (18) along the plate centerline y=b/2 is presented in Fig. 9. Curves are shown based on a 9-term expansion (i.e., m,n=1,3,5), a 64-term expansion (m,n=1, 3, 5,. . ., 15), and a 169-term expansion (m,n=1, 3, 5,. . ., 25). The series clearly converges toward a constant loading as the number of terms used is increased, although even with 169 terms the series expansion represents a constant transverse loading in only an approximate sense. As a second example, consider the case of a transverse force P, uniformly distributed over an internal rectangular region of dimensions ai
bi, as shown in Fig. 10. The center of the internal region is located at x=n and y=g, as shown. It is emphasized that P is defined as a force, with units of Newtons or pounds-force, for example. Eq. (17) becomes in this case: qmn ¼

4P abai bi

Z

nþai =2

nai =2

Z

gþbi =2

gbi =2

 mpx   npy  sin sin dxdy a b

Evaluating this integral we find: qmn ¼

     16P mpn npg   mpai  npbi sin sin sin sin p2 mnai bi a b 2a 2b

ð19Þ

Together, Eq. (14) and (19) define the Fourier series expansion of a force P, uniformly distributed over the internal region ai
bi. To illustrate this series expansion, consider the following specific example. Assume that the interior region is centered in the middle of the plate, i.e., let n=a/2, g=b/2. Further,

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

Figure 9 A normalized plot of Eq. (18) along the plate centerline defined by y=b/2.

Figure 10 A rectangular plate with in-plane dimensions a
b, subjected to a force P uniformly distributed over an interior region of dimensions ai
bi. The interior region is centered at the point x=n, y=g.

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

assume that the length and width of the interior region equals one half the length and width of the plate, i.e., let ai =a/2, bi =b/2. For this specific example then, Eq. (19) becomes: qmn ¼

64P p2 mnab

qmn ¼ 0;

sin

ðmp2 Þsinðnp2 Þsinðmp4 Þsinðnp4 Þ;

m; n ¼ odd integers

m; n ¼ even integers

Note that coefficients associated with even integers are zero only because we let n=a/2, g=b/2 in this example. This would not be true if the interior region were not centered on the plate. Substituting this result in Eq. (14), we find: qðx; yÞ ¼

l X l 64P X 1 mp np mp np sin sin sin sin p2 ab m¼1 n¼1 mn 2 2 4 4

ð Þ ð Þ ð Þ ð Þ mpx mpy
sinð sin ; m; n ¼ odd integers a Þ ð b Þ

ð20Þ

As before, how well Eq. (20) describes a load P uniformly distributed over the interior region depends on the number of terms used. A normalized plot of the series representation given by Eq. (20) along the plate centerline y=b/2 is presented in Fig. 11. Curves are shown based on a 9-term expansion (i.e., m,n=1, 3, 5), a 64-term expansion (m,n=1, 3, 5,. . ., 15), and a 169-term

Figure 11 A normalized plot of Eq. (20) along the plate centerline defined by y = b/2.

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

expansion (m,n=1,3,5,. . .,25). The series clearly converges toward a constant loading over the central region of the plate and approaches zero elsewhere. As a third example, consider a concentrated transverse force P, located at x=n and y=g. The Fourier coefficients for this case can be obtained from Eq. (19) by allowing ai !0, bi !0, i.e., the internal area over which load P acts is allowed to shrink to a point. It can be shown that in the limit Eq. (19) becomes: qmn ¼

ð Þ ð Þ

4P mpn mpg sin sin ab a b

ð21Þ

Together, Eqs. (14) and (21) define the Fourier series expansion of a concentrated force P applied at the point x=n and y=g. As a specific example, consider the case in which the concentrated load is applied at the center of the plate, i.e., at n=a/2, g=b/2. In this case Eq. 21 becomes: 8 < 4P ð1Þ½ðmþnÞ=21 ; m; n ¼ odd integers qmn ¼ ab : 0; m; n ¼ even integers Combining this result with Eq. (14), we find that the Fourier series expansion for a concentrated load applied at the center of the plate is given by: qðx; yÞ ¼

l X l 4P X mpx npy sin sin ð1Þ½ðmþnÞ=21 ab m ¼1 n ¼1 a b

ð Þ ð Þ

ð22Þ

where m,n=odd integers. As in the earlier examples, as the number of terms used in the series representation given by Eq. 22 is increased the distribution of q(x,y) resembles a concentrated force P more and more closely. REFERENCES 1. 2. 3. 4. 5.

Timoshenko, S.; Woinowsky-Krieger, S. Theory of Plates and Shells; McGrawHill Book Co.: New York, NY, ISBN 0-07-0647798. Ugural, A.C. Stresses in Plates and Shells; McGraw-Hill Book Co.: New York, NY, ISBN 0-07-065730-0. Whitney, J.M. Structural Analysis of Laminated Anisotropic Plates; Technomic Pub Co.: Lancaster, PA, ISBN 87762-518-2. Turvey, G.J., Marshall, I.H., Eds.; Buckling and Postbuckling of Composite Plates; Chapman and Hall: New York, NY, 1995. Almroth, B.O. Influence of edge conditions on the stability of axially compressed cylindrical shells. AIAA J. 1966, 4 (1), 134–140.

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

10 Some Exact Solutions for Specially Orthotropic Laminates

The equations that govern the behavior of a symmetric composite plate subjected to loads and/or displacements that vary along the edges of the plate were derived in Chap. 9. In this chapter, exact solutions to the governing equations will be obtained for a special class of symmetric laminates called ‘‘specially orthotropic’’ laminates. The solutions presented allow calculation of the deflections caused by the combined effects of in-plane loads, transverse loads, and changes in temperature or moisture content. Buckling caused by in-plane loads and/or temperature changes is also discussed. 1 EQUATIONS OF EQUILIBRIUM FOR A SPECIALLY ORTHOTROPIC LAMINATE The equations of equilibrium for a symmetric composite laminate were presented in Sec. 2.2 of Chap. 9. Boundary conditions consistent with thinplate theory were then discussed in Sec. 3 of Chap. 9. Unfortunately, exact solutions to the equations of equilibrium that satisfy specified boundary conditions often cannot be found. Sometimes, the difficulty in obtaining an exact solution is because of the boundary conditions involved; difficulties of this sort are encountered for both isotropic and anisotropic plates. In other cases, the difficulties are strictly because of the anisotropic nature of composites. 527

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

We have previously restricted our discussion to symmetric stacking sequences, and so Bij ¼ MTij ¼ MM ij = 0 in all cases considered. In addition, in this chapter, we will further restrict our discussion to the so-called specially orthotropic laminates. A specially orthotropic laminate is one for which A16 ¼ A26 ¼ D16 ¼ D26 ¼ NTxy ¼ NM xy = 0. It turns out that this is a very restrictive limitation. Indeed, a review of the stacking sequences presented in Sec. 7 of Chap. 6 will reveal that there are only three stacking sequences that eliminate these coupling stiffnesses and thermal/moisture resultants and can therefore be classified as ‘‘specially orthotropic.’’ These are:   

Unidirectional [0j]n laminates (described in Sec. 7.1 of Chap. 6). Unidirectional [90j]n laminates (described in Sec. 7.1 of Chap. 6). Symmetric cross-ply laminates, e.g., [(0/90)n]s (described in Sec. 7.2 of Chap. 6).

It is mentioned in passing that the definition of a ‘‘specially orthotropic laminate’’ is inconsistently used in the literature, and is sometimes relaxed from the definition just described. Specifically, the definition of a specially orthotropic laminate may be relaxed if the laminate is symmetric and is subjected to in-plane loading only. In problems of this type, the coupling stiffnesses D16 and D26 play no role. Therefore researchers studying the behavior of symmetric composites subjected to in-plane loads and displacements (only) often specify that a specially orthotropic laminate is one in which A16 ¼ A26 ¼ NTxy ¼ NM xy = 0, but place no restriction on the values of D16 or D26. For example, Lekhnitski [1] and Savin [2] have obtained solutions that allow calculation of the in-plane stresses induced in regions near elliptical, rectangular, or triangular holes in anisotropic composite plates, and these solutions are valid if the laminate is symmetric and subjected to in-plane loads only. The solutions by Lekhnitski and Savin are said to be applicable to specially orthotropic laminates, but in fact only require that A16 ¼ A26 ¼ NTxy ¼ NM xy = 0. Hence both symmetric balanced laminates and symmetric angle-ply laminates (see Sec. 7 of Chap. 6) are specially orthotropic under this relaxed definition. Nevertheless, in this chapter, we consider situations in which the composite laminate is subjected to both in-plane and out-of-plane loading, and the therefore the bending stiffnesses of the laminate plays an important role. Therefore we must maintain the more restrictive definition and stipulate that a specially orthotropic laminate must have A16 ¼ A26 ¼ D16 ¼ D26 ¼ NTxy ¼ NM xy = 0. For a specially orthotropic laminate, the equations of equilibrium for a symmetric laminate [i.e., Eq. (8) of Chap. 9] are simplified further still, and become: A11

@ 2 uo @ 2 vo @ 2 uo þ A66 þ ðA12 þ A66 Þ ¼0 2 @x @x@y @y2

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

ð1aÞ

ðA12 þ A66 Þ

@ 2 uo @ 2 vo @ 2 uo þ A22 þ A ¼0 66 @x@y @y2 @x2



 2  @uo @vo @ w þ A12  NTxx  NM A11 xx @x2 @x @y

 2  @uo @vo @ w þ A22  NTyy  NM þ A12 yy @y2 @x @y    2  @uo @vo @ w þ þ 2 A66 @x@y @y @x D11

ð1bÞ

ð1cÞ

@4w @4w @4w  2ðD12 þ 2D66 Þ 2 2  D22 4 þ qðx; yÞ ¼ 0 4 @x @x @y @y

In the following sections, we will first obtain solutions for plate deflections under relatively simple loading conditions. It will then be seen that solutions for simple loading condition form the basis for predicting displacements caused by complex transverse loads. There are, in essence, two techniques that can be used to obtain exact solutions for problems involving arbitrary transverse loads. Both methods place limitations on the type of boundary conditions that can be considered. The first approach was developed by Navier in about 1820, and is known as the Navier solution [3]. The Navier solution is applicable if all four edges of the specially orthotropic rectangular plate are simply supported. The Navier solution applied to the case of a simply supported specially orthotropic panel subjected to arbitrary transverse loading will be described in Sec. 6. The second exact solution technique was developed by Levy in 1899, and is known as the Levy solution [3]. In this case, two opposite edges of the plate must be simply supported, while the boundary conditions on the remaining two edges are arbitrary and may be clamped or free, for example. The Levy solution technique will not be discussed in this textbook, and the interested reader is referred to Refs. 1–5 for a discussion of this method. A final preliminary comment is that thin composite plates (or in fact any thin-walled structure) are prone to buckling if subjected to compressive inplane loads and/or high shear loads. Buckling of specially orthotropic laminates is discussed in Secs. 7 and 8, respectively.

2 IN-PLANE DISPLACEMENT FIELDS IN SPECIALLY ORTHOTROPIC LAMINATES The overall goal in this chapter is to predict deflections induced in specially orthotropic composite plates that are simply supported along all four edges.

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

As discussed in Sec. 3 of Chap. 9, four types of simple supports may be defined for a thin plate. For the edge x=a, these are: * ða; yÞ ¼ 0 S1: wða; yÞ ¼ 0 Mxx S2: wða; yÞ ¼ 0 M* ða; yÞ ¼ 0

uo ða; yÞ ¼ uðþxÞ ðyÞ o ðþxÞ * N ða; yÞ ¼ N ðyÞ

* ða; yÞ ¼ 0 S3: wða; yÞ ¼ 0 Mxx

uo ða; yÞ ¼ uðþxÞ ðyÞ o

vo ða; yÞ ¼ vðþxÞ ðyÞ o N* ða; yÞ ¼ NðþxÞ ðyÞ

* ða; yÞ ¼ 0 S4: wða; yÞ ¼ 0 Mxx

* ða; yÞ ¼ NðþxÞ ðyÞ Nxx xx

* ða; yÞ ¼ NðþxÞ ðyÞ Nxy xy

xx

xx

xx

vo ða; yÞ ¼ vðþxÞ ðyÞ o

xy

xy

Hence the definition of a simply supported edge requires that either in-plane displacements, in-plane stress resultants, or some combination thereof must be specified along the edge, in addition to the requirement that out-of-plane displacements and bending moments vanish along the edge. The following sequence of events is assumed to occur during fabrication and assembly of a simply supported composite plate. We assume that the laminate is cured at an elevated temperature, and that the laminate is stressand strain-free at the cure temperature.* Following cure, the laminate is cooled to room temperature, and therefore midplane displacements (as well as thermal stress resultants and associated ply strains and stresses) are induced during cooldown to room temperature. The laminate is then trimmed to the desired dimensions a
b and assembled in a surrounding structure that provides simple supports along all four edges. According to this scenario then, midplane displacements have already been induced within the laminate prior to assembly in the simple supports. Whether additional in-plane displacements subsequently occur, due to application of q(x,y), a further change in temperature, and/or a change in moisture content, depends on the type of simple supports involved. That is, the development of additional in-plane displacements depends on whether simple supports of type S1, S2, S3, or S4 are imposed along each edge of the plate. All problems considered here will be based on the following. First, we assume that opposite edges of the plate are subjected to the same type of simple support. For example, if the edge x = 0 is subjected to the type S1 simple support, then by assumption the edge x = a is also subjected to type S1 supports. Second, we assume that stress resultants Nxx, Nyy, and/or Nxy applied to opposite edges of the plate (if any) are identical and uniformly distributed along the edge. This loading condition is precisely equivalent to

*As mentioned in Section 6.2 of Chap. 6, in practice the stress- and strain-free temperature is often 20–50jC below the final cure temperature. This complication has been ignored throughout this text, and it is assumed that the laminate is stress- and strain-free at the cure temperature.

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

that assumed during the development of CLT in Chap. 6. Therefore we can develop general expressions giving the in-plane displacement fields caused by any combination of Nij, DT, and/or DM using a standard CLT analysis. We will then specialize these general expressions to correspond to type S1, S2, S3, or S4 simple supports. To begin this process, we relate midplane strains and curvatures to stress resultants using Eq. (45) of Chap. 6. For a symmetric specially orthotropic laminate subjected to uniform in-plane stress resultants, a uniform change in temperature, and/or a uniform change in moisture content, Eq. (45) of Chap. 6 becomes: 8 o 9 2 9 38 exx > Nxx þ NTxx þ NM a11 a12 0 0 0 0 > > xx > > > > > > > > 6 a12 a22 0 > eoyy > > Nyy þ NT þ NM > > 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 d 0 0 0 d > > > > j 12 22 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 Þþ a ðN þ N þ N Þ uo ðx; yÞ ¼ a11 ðNxx þNTxx þNM 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

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

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 Þy

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

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

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

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

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

temperature changes. Therefore type S1 boundary conditions for all four edges of the plate become: For x ¼ 0

For x ¼ a :

wð0; yÞ ¼ 0

wða; yÞ ¼ 0

@2w @2w þ D12 2 ¼ 0 2 @x @y ðxÞ uo ð0; yÞ ¼ uo ðyÞ ¼ 0   ðxÞ 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 :

* ð0; yÞ ¼ D Mxx 11

wðx; 0Þ ¼ 0

* ða; yÞ ¼ D Mxx 11

wðx; bÞ ¼ 0

2 2 @ w @ w * ðx; bÞ ¼ D @ w þ D @ w ¼ 0 þ D ¼ 0 M 12 11 12 yy @x2 @y2 @x2 @y2 ðyÞ ðþxÞ c uo ðx; 0Þ ¼ uo ðxÞ ¼ ðaxx DT Þx uo ðx; bÞ ¼ uo ðxÞ ¼ ðaxx DT c Þx   ðyÞ ðþxÞ 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

2

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.

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

Following this process we find: c¼

qo  

c  1   D 2 D ð DT  DT Þ 1  11 22 p4 4 þ 2 2 ðD12 þ 2D66 Þ þ 4 þ A11 axx þ A12 ayy þ 2 A12 axx þ A22 ayy 2 2 a b a b p 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 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

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.

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

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 ½ABD ¼ 6 6 0 6 6 4 0 0

4:52
106

0

0

0

95:6
106

0

0

0

0 0 0 0

19:5
106 0 0 0

0 40:1 0:848 0

0

3

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  jC

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

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 < 155jC Note that the temperatures to be considered are all at or below room temperature. Because the effective thermal expansion coefficients are algebraically positive, if the laminate were not constrained by the S1 simple supports, it would tend to contract as temperature is lowered. Because it is in fact constrained by the simple supports, the in-plane stress resultants that develop as temperature is lowered tend to be tensile. Therefore thermal buckling is not of concern. Substituting all known values, Eq. (8) becomes:  px   py  324

wðx; yÞ ¼ sin sin ðmetersÞ 88:6ðDT c  DTÞ 0:30 0:15 p4 278 þ p2 This expression can be used to calculate the out-of-plane displacement induced at any point (x,y) over the surface of the plate. The maximum out-of-

Figure 2 Maximum out-of-plane displacement for the graphite–epoxy plate considered in Sample Problem 1 as a function of temperature.

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

plane displacement occurs at the center of the plate, i.e., at x = a/2 = 0.15 m and y = b/2 = 0.075 m. Because DT c = 20jC–175jC = 155jC, and temperatures ranging from 50jC < T < 20jC are to be considered, the quantity (DT cDT ) ranges from 70jC z (DT cDT ) z 0jC in this problem. A plot of maximum out-of-plane displacement as a function of temperature is shown in Fig. 2. At room temperature (20jC), a maximum deflection of 12 mm is predicted. As would be expected, the plate becomes stiffer as the temperature is decreased, in the sense that out-of-plane displacements are decreased because of the in-plane tensile loads that develop as temperature is decreased. At the lowest temperature considered (50jC), a maximum deflection of 3.7 mm is predicted. 4 SPECIALLY ORTHOTROPIC LAMINATES SUBJECT TO SIMPLE SUPPORTS OF TYPE S4 In this section, we consider the specially orthotropic plate shown in Fig. 3. As in the preceding section, the plate is rectangular with thickness t and in-plane dimensions a
b. However, we now assume that each edge of the plate is subject to simple supports of type S4, rather than type S1. Hence displacement

Figure 3 Thin rectangular plate of thickness t and in-plane dimensions a
b, py subjected to a transverse load q(x,y) = qosin(px a )sin( b ). All four edges of the plate are subject to simple supports of type S4 (compare with Figure 1).

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

fields along the plate edges are not required to take on any specified value. Rather, we assume that a uniform normal stress resultant is applied to each edge, as shown in Fig. 3. Note that shear loading is not considered: Nxy = Nyx = 0. The plate is also subjected to a uniform change in temperature, DT, and a transverse loading that varies over the x–y plane according to:  px   py  qðx; yÞ ¼ qo sin sin ð9Þ a b We will not consider any change in moisture content, i.e., DM = 0. Also, we will not consider the possibility of buckling in this section, although buckling is a possibility if either of the in-plane normal stress resultants is compressive. Buckling under type S4 simple supports will be discussed in Sec. 9. The boundary conditions that define type S4 simple support were discussed in Sec. 3.3.2 of Chap. 9. Because the stress resultants applied along each edge are uniform and constant, we can write: * ð0; yÞ ¼ N* ða; yÞ ¼ N Nxx xx xx * ðx; bÞ ¼ N N*yy ðx; 0Þ ¼ Nyy yy * ða; yÞ ¼ N* ðx; 0Þ ¼ N* ðx; bÞ ¼ 0 N*xy ð0; yÞ ¼ Nxy yx yx Because we have limited consideration to symmetric specially orthotropic T M laminates, A16 ¼ A26 ¼ D16 ¼ D26 ¼ Bij ¼ NTxy ¼ NM xy ¼ Mij ¼ Mij = 0. Therefore the boundary conditions can be written as: For x = 0,a: wð0; yÞ ¼ wða; yÞ ¼ 0 * ð0; yÞ ¼ Mo ða; yÞ ¼ D Mxx 11 xx

ð10aÞ @2w @2w þ D ¼0 12 @x2 @y2

* ð0; yÞ ¼ N* ða; yÞ ¼ A @uo þ A @vo  NT ¼ N Nxx 11 12 xx xx xx @x @y   @uo @vo * ð0; yÞ ¼ N* ða; yÞ ¼ A þ Nxy ¼0 66 xy @y @x

ð10bÞ ð10cÞ ð10dÞ

For y = 0,b: wðx; 0Þ ¼ wðx; bÞ ¼ 0

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

ð11aÞ

2 2 * ðx; 0Þ ¼ M* ðx; bÞ ¼ D @ w þ D @ w ¼ 0 Myy 12 22 yy @x2 @y2

ð11bÞ

* ðx; 0Þ ¼ N* ðx; bÞ ¼ A @uo þ A @vo  NT ¼ N Nyy 12 22 yy yy yy @x @y   @uo @vo * * þ Nyx ðx; 0Þ ¼ Nyx ðx; bÞ ¼ A66 ¼0 @y @x

ð11cÞ ð11dÞ

Using Eqs. (3a) and (3b) for the assumed conditions (i.e., Nxy = DM = 0), the in-plane displacements fields are:   uo ðx; yÞ ¼ a11 Nxx þ a12 Nyy þ DT axx x ð12aÞ   ð12bÞ vo ðx; yÞ ¼ a12 Nxx þ a22 Nyy þ DT ayy y Let us confirm that these equations satisfy the appropriate boundary conditions. Substituting Eqs. (12a) and (12b) into boundary condition (10c) and rearranging, we find that the following expression must be satisfied: Nxx ðA11 a11 þ A12 a12 Þ þ Nyy ðA11 a12 þ A12 a22 Þ þ DTðA11 axx þ A12 ayy Þ  NTxx ¼ Nxx Because the laminate is specially orthotropic, 2 3 A11 A12 0 ½A ¼ 4 A12 A22 0 5 0 0 A66 2 A22 A12 6 ðA A  A2 Þ ðA A  A2 Þ 11 22 6 11 22 12 12 6 A12 A11 6 1 ½a ¼ ½A ¼ 6 6 ðA11 A22  A212 Þ ðA11 A22  A212 Þ 6 4 0 0 Therefore by direct substitution we find: ðA11 a11 þ A12 a12 Þ ¼ 1 ðA11 a12 þ A12 a22 Þ ¼ 0 Also, from Eq. (73b) of Chap. 6, we find (for A16 = 0): NTxx ¼ DTðA11 axx þ A12 ayy Þ

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

3 0 7 7 7 7 0 7 7 7 1 5 A66

Hence the boundary condition represented by Eq. (10c) is satisfied identically by the in-plane displacement fields. A similar process can be used to confirm that the boundary condition given by Eq. (11c) is also satisfied. Both Eqs. (10d) and (11d) are satisfied as well, because @uo/@y = @vo/@x = 0. Now consider out-of-plane displacements w(x,y). Guided by the functional form of the transverse pressure [i.e., Eq. (9)], we once again assume the out-of-plane displacement field is given by:  px   py  sin ð13Þ wðx; yÞ ¼ c sin a b where c is an unknown constant. Substituting this assumed form into boundary conditions [Eqs. (10a), (10b), (11a), and (11b)] 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. Substituting Eqs. (12a), (12b), and (13) into the third equation of equilibrium, Eq. (1c), and solving for constant c, we find: c¼

qo

ð14aÞ 1 2 1 1 1 4 p 4 D11 þ 2 2 ðD12 þ 2D66 Þ þ 4 D22 þ 2 2 Nxx þ 2 2 Nyy a ab b p a pb

Using the definition of the plate aspect ratio, R = a/b, this result can also be written as: c¼



qo R 4 b4

p4 D11 þ 2R2 ðD12 þ2D66 Þ þ R4 D22 þ

 a2  Nxx þ Nyy R2 2 p



ð14bÞ

The predicted out-of-plane deflection is obtained by combining either Eq. (14a) or (14b) with Eq. (13). Using Eq. (14b) for example, we have: wðx; yÞ ¼

qo R4 b4 sinðpx=aÞsinðpy=bÞ

 a2  2 4 2 4 p D11 þ 2R ðD12 þ 2D66 Þ þ R D22 þ 2 Nxx þ Nyy R p

ð15Þ

Equations (12a), (12b), and (15) give the predicted displacement fields induced in the plate and represent the solution to this problem. To summarize, we have found the displacement fields induced in a symmetric specially orthotropic type S4 simply supported plate subjected to a sinusoidally varying transverse load, a uniform change in temperature DT, and uniform stress resultants Nxx and Nyy. A typical application of this solution is discussed in Sample Problem 2. It should be kept in mind that we

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

have not considered the possibility of buckling in this section. The solution we have obtained is not valid if Nxx and/or Nyy are compressive and have magnitudes large enough to cause buckling. Buckling under type S4 simple supports will be discussed in Sec. 9. Sample Problem 2 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 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 in-plane tensile loading (i.e., Nxx = Nyy) and transverse pressure given by q(x,y) = 40 {sin [(px)/a]} {sin[(py)/b]} (kPa). Temperature remains constant and no change in moisture content occurs (DM = 0). (a)

Plot the out-of-plane displacements induced along the centerline defined by y = 0.075 m, if in-plane loads Nxx = Nyy = 50 kN/m are applied. (b) Plot the maximum out-of-plane displacement as a function of inplane loads, over the range 0<(Nxx = Nyy)<70 kN/m. (c) Compare the maximum out-of-plane displacement caused by the specified transverse load at room temperature if the plate is subject to (i) type S1 simple supports (as discussed in Sec. 3), and (ii) type S4 simple supports, with Nxx = Nyy = 0 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. A [(02/90)2]s graphite–epoxy laminate was also considered in Sample Problem 1, and numerical values for the [ABD] matrix are listed there. As before, the 12-ply laminate has a total thickness t = 1.5 mm and aspect ratio R = a/b = 2.0. Using these laminate stiffnesses, dimensions, and the specified transverse loading, Eq. (15) becomes: 3 2 7  px   py  6 324 7sin   sin ðmetersÞ wðx; yÞ6 4  5 0:090  0:3 0:15 4 p 278 þ 2 Nxx þ 4Nyy p This expression can be used to calculate the out-of-plane displacement induced at any point (x,y) over the surface of the plate.

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

Part (a). Using Nxx ¼ Nyy = 50 kN/m and y = 0.075 m, out-of-plane displacements are:  px  ðmetersÞ wðx; yÞ ¼ ½1:30
103 sin 0:3 A plot of these displacements over the length of the plate (0<x<0.3m is shown in Fig. 4(a). Displacements are zero at the edges defined by x = 0, 0.3 m, as dictated by the specified boundary conditions. As would be expected because of symmetry, out-of-plane displacement is maximum at the center of the plate, and equals 1.3 mm for the loading considered. Part (b). The maximum out-of-plane displacement occurs at the center of the plate, i.e., at x = a/2 = 0.15 m and y = b/2 = 0.075 m, and at this point the maximum out-of-plane displacement is given by: 3 2 6 wjmax ¼ 6 4

 p4

7 324  7 ðmetersÞ  5 0:090  278 þ 2 Nxx þ 4Nyy p

A plot of maximum out-of-plane displacement as a function of in-plane tensile loads is shown in Fig. 4b. As would be intuitively expected, the plate is stiffened by the application of in-plane loading. That is, the maximum out-ofplane displacement is decreased as in-plane tensile loads are increased. A maximum deflection of 12 mm occurs when Nxx = Nyy = 0, whereas the maximum deflection is reduced to 0.96 mm when Nxx = Nyy = 70 kN/m. Part (c). This same panel and transverse loading was considered in Sample Problem 1, except type S1 simple supports were assumed. Thus in Sample Problem 1 in-plane displacements were fixed and were not allowed to change when the transverse load was applied. In contrast, type S4 simple supports are assumed in this problem; in-plane stress resultants are specified rather than in-plane displacements. Referring to the results presented in these two sample problems, we find that identical deflections are predicted, despite the differences in boundary conditions. That is, a maximum deflection of 12 mm is predicted at room temperature for type S1 condition, and an identical 12 mm deflection is predicted if Nxx = Nyy = 0 for type S4 conditions. This result may seem nonintuitive and (rigorously speaking) is incorrect. That is, for type S4 boundary conditions, a transverse loading will cause a change in inplane displacements. Therefore one might anticipate that the out-of-plane displacement for type S4 conditions would be increased, relative to type S1 conditions. However, the relative increase is very small if displacement gradients are small. Thus the relative increase in out-of-plane displacements

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

Figure 4 Out-of-plane displacements induced in the graphite–epoxy plate considered in Sample Problem 2. (a) Out-of-plane displacements induced along centerline y = 0.075 m when Nxx = Nyy = 50 kN/m, (b) Maximum out-of-plane displacements as a function of in-plane loads Nxx = Nyy.

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

for type S4 conditions is not predicted because we have based our analysis on infinitesimal strains. The consequences of the infinitesimal strain assumption were alluded to in Sec. 2.2 of Chap. 9. It was noted there that this assumption ultimately leads to the conclusion that in-plane displacement fields uo(x,y) and vo(x,y) are independent of the transverse load, q(x,y). The comparison between the results of Sample Problems 1 and 2 presented here is an illustration of this independence. Of course, results for the two different boundary conditions are identical because we have considered the case in which Nxx = Nyy = 0. If Nxx and/or Nyy p 0, then the transverse displacements for a plate supported by type S4 simple supports is quite different from that of a plate supported by type S1 supports. 5 SPECIALLY ORTHOTROPIC LAMINATES WITH TWO SIMPLY SUPPORTED EDGES OF TYPE S1 AND TWO EDGES OF TYPE S2 In this section, we consider the specially orthotropic plate shown in Fig. 5. As in the preceding section, the plate is rectangular with thickness t and in-plane

Figure 5 Thin rectangular plate of thickness t and in-plane dimensions a x b, subjected to a transverse load q(x,y) = qo {sin[(px)/a]}{sin[(py)/b]}. Edges x = 0,a are subject to simple supports of type S2, whereas edges y = 0,b are subject to simple support of type S1 (compare with Figs. 1 and 3).

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

dimensions a
b. We now assume that the two edges x = 0,a are subjected to simple supports of type S2. That is, we specify that these two edges are subjected to known stress resultants Nxx and known displacements vo. In contrast, the two edges y = 0,b are subject to simple supports of type S1, which means that we require these edges to maintain known in-plane displacements uo and vo. The plate is also subjected to a uniform change in temperature, DT, and a transverse loading that varies over the x–y plane according to:  px  py  qðx; yÞ ¼ qo sin ð16Þ a b We will not consider a change in moisture content, i.e., DM = 0. Also, we will not consider the possibility of buckling in this section, although buckling is a possibility if Nxx is compressive or if the change in temperature DT tends to cause the laminate to expand. As in earlier sections, we assume the laminate was cured at an elevated temperature and cooled to room temperature prior to assembly in a surrounding structure that provides simple supports. Consider the midplane strains vo(x,y) induced during cooldown. Using Eq. (3b), this can be written (with Nxx = Nyy = Nxy = DM = 0): vo ðx; yÞ ¼ ðDT c ayy Þy

ð17Þ

We can use this expression to specify known values of vo along all four edges of the plate. Because a known stress resultant Nxx is applied to the two edges x = 0 and x = a, all quantities necessary to define the type S2 simple support along these two edges are known. To define the type S1 simple supports along the edges y = 0,b, we must specify known values of uo. Toward that end, equate Eqs. (3b) and (17):   1 ða66 Nxy Þx þ a12 Nxx þ a22 Nyy þ DT ayy y ¼ ðDT c ayy Þy 2 This relation must be satisfied along the edges y = 0,b. Substituting y = 0, it is seen that Nxy must vanish (Nxy = 0). Substituting y = b, and solving for Nyy, we obtain: Nyy ¼

ayy ðDT c  DTÞ  a12 Nxx a22

ð18Þ

Equation (18) gives the stress resultant Nyy that must be provided by the simple supports along the y = 0,b so as to maintain the stipulated value of vo represented by Eq. (17). Note that if temperature does not change from room temperature (i.e., if DT = DT c ), and if no stress resultant Nxx is applied (i.e., if Nxx = 0), then Nyy = 0, as would be expected.

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

We can now calculate the displacement uo implied by these conditions, using Eq. (3b):  

ayy ðDT c  DTÞ  a12 Nxx uo ðx; yÞ ¼ a11 Nxx þ a12 þ DT axx x a22 This expression can be written as:

 a212 a12 c ayy ðDT  DTÞ þ axx DT x uo ðx; yÞ ¼ Nxx a11  þ a22 a22

ð19Þ

Equations (17) and (19) can be used to specify the displacement boundary conditions along all four edge of the plate. A summary of all boundary conditions associated with the problem considered in this section is: For x = 0,a wð0; yÞ ¼ wða; yÞ ¼ 0

ð20aÞ

2 2 * ð0; yÞ ¼ M* ða; yÞ ¼ D @ w þ D @ w ¼ 0 Mxx 11 12 xx @x2 @y2

ð20bÞ

* ð0; yÞ ¼ N* ða; yÞ ¼ A @uo þ A @vo  NT ¼ N Nxx 11 12 xx xx xx @x @y

ð20cÞ

vo ð0; yÞ ¼ vo ða; yÞ ¼ ðDT c ayy Þy

ð20dÞ

For y = 0,b: wðx; 0Þ ¼ wðx; bÞ ¼ 0

ð21aÞ

* ðx; 0Þ ¼ M* ðx; bÞ ¼ D @ w þ D @ w ¼ 0 Mxx 12 22 yy @x2 @y2 2

2

uo ðx; 0Þ ¼ uo ðx; bÞ

 a212 a12 c ¼ Nxx a11  ayy ðDT  DTÞ þ axx DT x þ a22 a22 vo ðx; 0Þ ¼ 0

vo ðx; bÞ ¼ ðDT c ayy Þb

ð21bÞ

ð21cÞ ð21dÞ

Now consider out-of-plane displacements w(x,y). Guided by the functional form of the transverse pressure [i.e., Eq. (16)], we once again assume the out-of-plane displacement field is given by:  px   py  wðx; yÞc sin sin ð22Þ a b where c is an unknown constant. Substituting this assumed form into boundary conditions Eqs. (20a), (20b), (21a), and (21b) will reveal that they

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

are identically satisfied for any value of c. Hence the value of constant c must be determined by enforcing the third equation of equilibrium. Substituting Eqs. (19), (17), and (22) into the third equation of equilibrium, Eq. (1c), and solving for constant c, we find: c¼



D11 2ðD12 þ 2D66 Þ D22 Nxx p4 4 þ þ 4 þ 2 a b p a2 b2



qo

 ayy ðA12 A22  A212 Þ 1 A12 c  ðDT  DT Þ þ a2 A11 b2 p2 b2 A11

ð23Þ Using the definition of the plate aspect ratio, R = a/b, this result can also be written as: c¼

qo R4 b4  

 R b 1 A12 ayy ðA12 A22  A212 Þ p4 D11 þ 2R2 ðD12 þ 2D66 Þ þ R4 D22 þ 2 Nxx 2  ðDT c  DT Þ þ p A11 R A11

4 2

ð24Þ The predicted out-of-plane deflection is obtained by combining either Eq. (23) or (24) with Eq. (22). Using Eq. (24), for example, we have: wðx; yÞ ¼

qo R4 b4 sinðpx=aÞsinðpy=bÞ  

 ayy ðA12 A22  A212 Þ R4 b2 1 A12 Nxx 2  ðDT c  DT Þ þ p4 D11 þ 2R2 ðD12 þ 2D66 Þ þ R4 D22 þ 2 R p A11 A11

ð25Þ

Equations (17), (19), and (25) give the predicted displacement fields induced in the plate and represent the solution to this problem. To summarize, we have found the displacement fields induced in a symmetric specially orthotropic plate subjected to a sinusoidally varying transverse load and a uniform change in temperature DT. The two edges x = 0,a are subject to simple supports of type S2, whereas the two edges y = 0,b are subject to simple supports of type S1. A typical application of this solution is discussed in Sample Problem 3. It should be kept in mind that the possibility of buckling has not been considered. Buckling is a possibility if Nxx is compressive or if the change in temperature DT tends to cause the laminate to expand. Sample Problem 3A [(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

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

type S2 simple supports along the two edges x = 0 and x = a, and type S1 simple supports along the two edges y = 0 and y = b. 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 in-plane tensile loading Nxx along x = 0,a and a transverse pressure given by q(x,y) = 40 {sin[(px)/a]}{sin[(py)/b]} (kPa). (a ) Plot the maximum out-of-plane displacement as a function of tensile load over the range 0
3 324

 p4 278 þ

0:099 82:5 Nxx þ 2 ðDTc  DTÞ p2 p

7  px   py  7 5sin 0:3 sin 0:15 ðmetersÞ

This expression can be used to calculate the out-of-plane displacement induced at any point (x,y) over the surface of the plate. Part (a). The maximum out-of-plane displacement occurs at the center of the plate, i.e., at x = a/2 = 0.15 m and y = b/2 = 0.075. Because the plate remains at room temperature, DT = DT c, under these conditions the maximum out-of-plane displacement is given by: wjmax ¼

324   ðmetersÞ 0:099 4 p 278 þ 2 Nxx p

A plot of maximum out-of-plane displacement as a function of Nxx is shown in Fig. 6(a). As would be intuitively expected, the plate is stiffened as Nxx is

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

Figure 6 Out-of-plane displacements induced in the graphite–epoxy plate considered in Sample Problem 3. (a) Maximum out-of-plane displacements at room temperature as a function of in-plane load Nxx. (b) Maximum out-of-plane displacements as a function of temperature (Nxx = 50 kN/m).

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

increased. A maximum deflection of 12 mm occurs when Nxx = 0, whereas the maximum deflection is reduced to 3.4 mm when Nxx = 70 kN/m. Part (b). As before, the maximum out-of-plane displacement occurs at x = a/2 = 0.15 m and y = b/2 = 0.075. Because a constant in-plane tensile load Nxx = 50 kN/m is applied, the maximum out-of-plane displacement is given by: wjmax ¼

324

ðmetersÞ 82:5 p4 781 þ 2 ðDT c  DTÞ p

A plot of maximum out-of-plane displacement as a function of temperature is shown in Fig. 6(b). At room temperature (20jC), a maximum deflection of 4.3 mm is predicted. As would be expected, the plate becomes stiffer as the temperature is decreased. Out-of-plane displacements are decreased because of the in-plane tensile load Nyy that develops as temperature is decreased. At the lowest temperature considered (50jC), a maximum deflection of 2.4 mm is predicted. 6 THE NAVIER SOLUTION APPLIED TO A SPECIALLY ORTHOTROPIC LAMINATE SUBJECT TO SIMPLE SUPPORTS OF TYPE S4 In Sec. 4, we developed the solution for a specially orthotropic laminate subjected to homogenous simple supports of type S4 along all four edges, uniform in-plane loads Nxx and Nyy, a uniform temperature change DT, and a sinusoidal transverse loading given by:  px   py  qðx; yÞ ¼ qo sin sin a b In-plane and transverse displacements caused by this thermomechanical loading are given by Eqs. (12a), (12b), and (15). Now recall from Sec. 4 of Chap. 9 that any transverse loading may be represented in terms of the double-Fourier series given by Eq. (14) of Chap. 9, repeated here for convenience: l X l  mpx   npy  X qðx; yÞ ¼ qmn sin sin ð9:14Þ a b m¼1 n¼1 Thus any transverse loading can be viewed as the sum of a large number of sinusoidal load components. The displacements caused by an arbitrary transverse loading applied to a plate with simple supports of type S4 can therefore be obtained using the same approach as that used in Sec. 4. The only

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

differences are that the transverse load is given by Eq. 14 of Chap. 9 and the transverse deflections w(x,y) are assumed to be of the form: l X l  mpx   npy  X sin ð26Þ cmn sin wðx; yÞ ¼ a b m¼1 n¼1 where cmn are unknown constants to be determined using the equations of equilibrium. The assumed form and solution for in-plane displacements uo(x,y) and vo(x,y) remain unchanged and are given by Eqs. (12a) and (12b). Substituting Eqs. (26) and Eq. (14) of Chap. 9 into the third equation of equilibrium, Eq. (1c), and equating coefficients (following the same procedures as used in Sec. 4), we obtain: qmn a4 b4

"

cmn ¼

4

2

 4

p4 D11 ðmbÞ þ 2ðD12 þ 2D66 ÞðmnabÞ þ D22 ðnaÞ þ

ab p

2 n

2

Nxx ðmbÞ þ Nyy ðnaÞ

2

o

#

ð27aÞ Using the definition of the plate aspect ratio, R=a/b, this result can also be written as: cmn ¼

qmn R4 b4



p4 D11 m4 þ 2ðD12 þ 2D66 ÞðmnRÞ2 þ D22 ðnRÞ4 þ

o

a2 n 2 2 N m þ N ðnRÞ xx yy p2 ð27bÞ

Substituting this result into Eq. (26) completes the solution to the problem. To summarize, the midplane displacements induced in a symmetric specially orthotropic laminate subjected to an arbitrary transverse loading given by Eq. (15) of Chap. 9, constant and uniform in-plane loads Nxx and Nyy, a uniform temperature change DT, and homogenous simple supports of type S4 along all four edges are given by: 

 A22 Nxx  A12 Nyy ð28aÞ uo ðx; yÞ ¼ þ axx DT x A11 A22  A212  vo ðx; yÞ ¼

wðx; yÞ ¼

A11 Nyy  A12 Nxx A11 A22  A212



þ ayy DT y

ð28bÞ

l X l R4 b4 X qmn sinðmpx=aÞsinðnpy=bÞ o

p4 m¼1 n¼1 a2 n D11 m4 þ 2ðD12 þ 2D66 ÞðmnRÞ2 þ D22 ðnRÞ4 þ 2 Nxx m2 þ Nyy ðnRÞ2 p

ð28cÞ

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

The method of using a double-Fourier series expansion to represent the transverse loading and out-of-plane displacements was first suggested by Navier in about 1820, and is known as the Navier solution. An application of this technique is illustrated in Sample Problem 4. The solution presented here is for a plate subjected to type S4 simple supports, but the same approach may be used to study simply supported plates with type S1 supports (as in Sec. 3), or plates with mixed simple supports (as in Sec. 5). As previously mentioned, a second technique known as the Levy solution [1–5] can also be used to obtain exact solutions for symmetric specially orthotropic laminates. While the Navier solution requires that all four edges be simply supported, the Levy solution only requires that two opposite edges of the plate are simply supported; the boundary conditions on the remaining two edges are arbitrary and may be clamped or free, for example. The Levy solution technique will not be discussed in this textbook, and the interested reader is referred to the references cited for a discussion of this approach. Sample Problem 4 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 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 in-plane tensile loading Nxx = Nyy = 50 kN/m and uniform transverse loading q(x,y) = 100 kPa. The temperature remains constant and no change in moisture content occurs (DM = 0). Plot the out-of-plane displacements induced along the centerline defined by y = 0.075 m. 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. A [(02/90)2]s graphite–epoxy laminate was also considered in Sample Problem 1, and numerical values for the [ABD] matrix are listed there. As before, the 12-ply laminate has a total thickness t = 1.5 mm and aspect ratio R = a/b = 2.0. The double-Fourier series expansion of a uniform transverse loading was discussed in Sec. 4 of Chap. 9. The coefficients in the Fourier series expansion were found to be: qmn ¼

16qo ; p2 mn

m; n ¼ odd integers

Combining these coefficients with Eq. (28c) allows prediction of out-of-plane displacements. A plot of these displacements along the centerline of the plate

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

Figure 7 Out-of-plane displacements induced in the graphite–epoxy plate considered in Sample Problem 4, along centerline y = 0.075 m.

defined by y = b/2 = 0.075 m is shown in Fig. 7. Curves are shown based on a 1-term expansion (i.e., m,n = 1), a 4-term expansion (m,n = 1,3), and a 9term expansion (m,n = 1,3,5). As would be expected due to symmetry, out-ofplane displacement is maximum at the center of the plate. The solution rapidly converges. The maximum displacement predicted on the basis of a 1-, 4-, and 9-term expansion equals 5.27, 4.71, and 4.78 mm, respectively. If 100 terms were used (m,n = 1,3,. . .,19), the maximum predicted displacement is 4.77 mm. 7 BUCKLING OF RECTANGULAR SPECIALLY ORTHOTROPIC LAMINATES SUBJECT TO SIMPLE SUPPORTS OF TYPE S4 This section is devoted to the phenomenon known as plate ‘‘buckling.’’ A brief discussion of what the term ‘‘buckling’’ means in the context of a thin plate is in order. Consider an initially flat symmetric laminate subjected to ðxÞ ðþxÞ ðyÞ constant and uniform boundary edge loads Nxx ¼ Nxx = Nxx and Nyy = ðþy Þ Nyy = Nyy (shear resultants are assumed zero: Nxy = Nyx = 0). As we have seen in preceding sections, coupling stiffnesses Bij = 0 for all symmetric laminates. Hence, according to our earlier analyses, we would not expect these edge loads to cause out-of-plane displacements because in-plane loads and out-of-plane displacements are (apparently) uncoupled for a symmetric laminate.

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

This statement is always true if both Nxx and Nyy are tensile; a flat symmetric laminate subjected to tensile edge loads Nxx and/or Nyy will remain flat regardless of the magnitude of load (unless, of course, the loads are high enough to cause fracture). However, suppose that either Nxx or Nyy (or both) are compressive. If the magnitude of the compressive load(s) is (are) relatively low, then the initially flat laminate remains flat; that is, only in-plane displacements uo(x,y) and vo(x,y) are induced and out-of-plane displacement remain zero: w(x,y)=0. However, as the compressive load(s) is (are) increased, the laminate may exhibit a sudden out-of-plane displacement, w(x,y) p 0. This is the phenomenon is known as ‘‘buckling.’’ The load level at which out-of-plane displacements initially occur is called the critical buckling load. In general, a thin plate does not collapse at the critical buckling load, and can support a further increase in load. However, out-of-plane displacements rapidly increase as in-plane loads are increased beyond the initial buckling load, signaling imminent structural failure. The out-of-plane displacements that occur at the onset of buckling exhibit a characteristic pattern. This characteristic pattern is called the buckling mode. A typical buckling mode is illustrated in Fig. 8. A threedimensional view of the buckling mode is shown in Fig. 8(a), while a 2-D view (which can be thought of as a ‘‘topographical map’’ of out-of-plane displacements) is shown in Fig. 8(b). In the discussion to follow, predicted buckling modes will be illustrated in a form similar to Fig. 8(b). The buckling mode exhibited by a given laminate depends on the stiffness of the laminate, plate aspect ratio, applied loading, and the boundary conditions. A particular buckling mode is described in terms of the number of points of relative maximum out-of-plane displacements in the x and y directions. For the buckling mode shown in Fig. 8, there are two points of relative maximum displacement in the x direction, whereas there is only one point of relative maximum displacement in the y direction. Hence the pattern shown in Fig. 8 is called ‘‘mode [2,1].’’ In this section, we consider buckling of an initially flat rectangular specially orthotropic laminate subjected to simple supports of type S4 and ðxÞ ðþxÞ ðyÞ ðþyÞ uniform edge loads Nxx ¼ Nxx ¼ Nxx and Nyy ¼ Nyy = Nyy, at least one of which is compressive. If the laminate was cured at an elevated temperature and then cooled to room temperature, then pre-existing thermal stress resultants associated with cooldown, NTxx and NTyy , are present. It is assumed that no transverse loading is applied and that no change moisture content occurs: q(x,y) = DM = 0. Because the applied edge loads are constant and uniform, prior to buckling the internal stress resultants at all points within the laminate are also constant and uniform, and are equal to the external edge loads. Hence, prior

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

Figure 8 A typical mode [2,1] buckling mode, caused in a rectangular plate by a uniaxial compressive loading Nxx.

to buckling, the internal stress resultants at all points within the laminate are uniform and given by: @uo @vo ðþxÞ þ A12  NTxx ¼ NðxÞ N*xx ðx; yÞ ¼ A11 xx ¼ Nxx ¼ Nxx @x @y * ðx; yÞ ¼ A @uo þ A @vo  NT ¼ NðyÞ ¼ NðþyÞ ¼ N Nyy 12 22 yy yy yy yy @x @y   @uo @vo * ðx; yÞ ¼ A Nxy þ ¼0 66 @y @x

ð29aÞ ð29bÞ ð29cÞ

It is emphasized that Eqs. (29a), (29b), and (19c) are valid prior to buckling. After buckling has occurred and significant out-of-plane displacements have

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

developed, the internal stress and moment resultants may no longer be constant and uniform, and may not equal the applied edge loads. The remaining type S4 simple support boundary conditions are: Along x = 0: wðx; yÞ ¼ 0 * ð0; yÞ ¼ D Mxx 11

ð30aÞ @ 2 wð0; yÞ @ 2 wð0; yÞ þ D12 ¼0 2 @x @y2

ð30bÞ

Along x = a: wða; yÞ ¼ 0 * ða; yÞ ¼ D Mxx 11

ð31aÞ @ wða; yÞ @ wða; yÞ þ D12 ¼0 @x2 @y2 2

2

ð31bÞ

Along y = 0: wðx; 0Þ ¼ 0 * ðx; 0Þ ¼ D Myy 12

ð32aÞ @ 2 wðx; 0Þ @ 2 wðx; 0Þ þ D22 ¼0 2 @x @y2

ð32bÞ

Along y = b: wðx; bÞ ¼ 0 * ðx; bÞ ¼ D Myy 12

ð33aÞ @ wðx; bÞ @ wðx; bÞ þ D22 ¼0 2 @x @y2 2

2

ð33bÞ

The equation of equilibrium governing out-of-plane displacements, Eq. (1c), is [with q(x,y) = DM = 0]:

 2  @uo @vo @ w þ A12  NTxx @x2 @x @y

 2  @uo @vo @ w T þ A22  Nyy þ A12 @y2 @x @y    2  @uo @vo @ w @4w þ  D11 4 þ 2 A66 @x@y @x @y @x

A11

 2ðD12 þ 2D66 Þ

ð34Þ

@4w @4w  D ¼0 22 @x2 @y2 @y4

Equations (29a), (29b), and (29c) are valid up to the onset of buckling; that is, we assume that internal stress resultants are constant and uniform as the

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

buckling phenomenon is approached. Hence substituting Eqs. (29a), (29b), and (29c) into Eq. (34), we have: @4w @4w @4w þ 2ðD þ 2D Þ þ D 12 66 22 @x4 @x2 @y2 @y4  2   2  @ w @ w ¼ Nxx þ Nyy @x2 @y2

D11

ð35Þ

Notice that thermal stress and moment resultants do not appear in Eq. (35). This reveals that the buckling response of symmetric specially orthotropic laminates is independent of temperature, when subject to simple supports of type S4. As will be discussed in Sec. 8, if the laminate were subject to boundary conditions that constrain in-plane displacements, e.g., simple supports of type S1, then buckling may be caused by a temperature change. We now assume out-of-plane displacements are given by:  mpx   npy  wðx; yÞ ¼ c sin sin ð36Þ a b where m and n are positive integers and c is an unknown constant representing the maximum out-of-plane deflection. It is easy to show that Eq. (36) satisfies the boundary conditions, Eqs. (30a), (30b), (31a), (31b), (32a), (32b), and (33a), (33b), so the next step is to evaluate the conditions under which the equation of equilibrium governing out-of-plane deflections is satisfied. Substituting Eq. (36) into Eq. (35) and simplifying, we find that in order for equilibrium to be maintained, it is required that:  mp 4  mp 2  np 2  np 4 D11 þ2ðD12 þ 2D66 Þ þD22 a a b b  mp 2  np 2 ¼ Nxx Nyy ð37Þ a n The unknown constant c has canceled out and does not appear in the requirement for equilibrium, Eq. (37). Referring to Eq. (36), note that integers m and n dictate the spatial variation of the out-of-plane displacement field, i.e., m and n define how w(x,y) varies with x and y, and consequently define the buckling mode. Constant c represents the magnitude of w(x,y). Hence while we are able to determine m and n and thus predict the buckling mode on the basis of Eq. (37), we cannot predict the magnitude of out-of-plane displacements. Although Eq. (37) is valid for any combination of Nxx and Nyy, it is convenient to rearrange Eq. (37) for three different loading conditions. As explained above, Nxx and/or Nyy must be compressive to cause buckling. Assume for the moment that a constant transverse tension Nyy z 0 is applied,

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

in which Nxx case must be compressive to cause buckling. Solving Eq. (37) for Nxx, we find: " p2 Nxx ¼ 2 2 2 D11 ðmbÞ4 þ 2ðD12 þ 2D66 ÞðmnabÞ2 m a b ð38Þ  2 2 # na b þ D22 ðnaÞ4 þ Nyy p Using the definition of the plate aspect ratio, R = a/b, this result can also be written as: " p2 Nxx ¼ D11 m4 þ 2ðD12 þ 2D66 ðmnRÞ2 ðmaÞ2 ð39Þ   # naR 2 4 þ D22 ðnRÞ þ Nyy p Hence, given some value for Nyy z 0 and assumed integer values for m and n, either Eq. (38) or (39) can be used to calculate a corresponding value for Nxx. Because m and n can be any combination of positive integers, there are an infinite number of values for Nxx that satisfy Eq. (38) or (39). The critical c , corresponds to the particular combination of m buckling load, denoted Nxx and n that leads to the value of Nxx with lowest magnitude. The combination of m and n that correspond to this lowest load define the predicted critical buckling mode. The following observations are based on inspection of Eq. (38) or (39). Because we have assumed for the moment that Nyy z 0, all variables that appear on the right side of the equality sign are algebraically positive. Hence c c must be algebraically negative, i.e., the critical buckling load Nxx is Nxx predicted to be compressive, as would be expected. Secondly, the minimum c magnitude of Nxx will always correspond to n = 1 because Nyy z 0. Finally, note that if the constant transverse load Nyy is increased, then the magnitude c is increased. That is, a transverse tension will cause an increase in the of Nxx critical buckling load. As a second loading condition of interest, let us now assume that a constant Nxx z 0 is applied, and that a compressive load in the y direction causes buckling. Solving Eq. (37) for Nyy, we find: " p2 Nyy ¼ 2 4 2 D11 ðmbÞ4 þ 2ðD12 þ 2D66 ÞðmnabÞ2 n ab ð40Þ   # mab 2 4 þ D22 ðnaÞ þ Nxx p

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

Using the plate aspect ratio, R = a/b, this result can be written as: " p2 Nyy ¼ D11 m4 þ 2ðD12 þ 2D66 ÞðmnRÞ2 2 ðnbR2 Þ #  ma 2 4 þ D22 ðnRÞ þ Nxx p

ð41Þ

As before, m and n can be any combination of positive integers, and therefore there are an infinite number of values for Nyy that satisfy Eq. (40) or (41). The c corresponds to the particular combination of m critical buckling load Nyy and n that leads to the value of Nyy with lowest magnitude, and the combination of m and n that correspond to this load define the predicted buckling mode. Because it has been assumed that Nxx z 0, m = 1 in all cases, and if the constant tensile load Nxx is increased, the magnitude of the critical c will increase. buckling load Nyy Finally, consider a third loading condition in which buckling is caused by the simultaneous increase in both Nxx and Nyy. Further, assume the two loads are linearly related. That is, assume Nyy = kNxx, where k is a known constant. Substituting this relation into Eq. (37) and solving for Nxx, we find: h i 4 2 4 p2 D11 ðmbÞ þ 2ðD12 þ 2D66 ÞðmnabÞ þ D22 ðnaÞ h i Nxx ¼  ð42Þ ðabÞ2 ðmbÞ2 þ kðnaÞ2 As before, m and n can be any combination of positive integers, and therefore there are an infinite number of values for Nxx that satisfy Eq. (42). The critical c c buckling condition, defined by the two simultaneous loads, Nxx and Nyy = c kNxx , corresponds to the particular combination of m and n that leads to the value of Nxx (and Nyy) with lowest magnitude. The combination of m and n that correspond to this load condition defines the predicted buckling mode. Note that if k < 0, then Nxx and Nyy are of opposite algebraic signs. This implies that there are two distinct buckling load conditions, one in which (Nxx < 0, Nyy > 0), and a second in which (Nxx > 0, Nyy < 0). Numerical examples illustrating buckling predictions for symmetric specially orthotropic simply supported laminates will be discussed in Sample Problems 5 and 6. Two important concluding comments are made in passing. First, from the preceding discussion, the reader may have inferred that buckling is only caused by Nxx, by Nyy, or by some combination thereof. This is not the case. Other forms of loading may cause buckling, e.g., a shear load Nxy. In this chapter, only buckling caused by Nxx and/or Nyy is considered. The reader interested in buckling caused by other types of loading is referred to Refs. 4 and 5.

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

Second, recall that the analysis presented above does not allow prediction of the magnitude of out-of-plane displacements induced at buckling. Buckling of thin plates is rarely catastrophic, and in general finite out-ofplane displacements occur at and above the critical buckling load. That is, a thin plate does not collapse once the buckling load is reached and can support a further increase in load, albeit at a reduced level of stiffness. Methods to predict the magnitude of out-of-plane displacement for load levels at or above the initial buckling load are known as ‘‘post-buckling’’ analyses, and are not discussed in this text. Sample Problem 5 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 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). c and mode for this laminate, if (a) Predict the critical buckling load Nxx 0 V Nyy V 400 kN/m. c (b) Predict the critical buckling load Nyy and mode for this laminate, if 0 V Nxx V 400 kN/m.

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. A [(02/90)2]s graphite–epoxy laminate was also considered in Sample Problem 1, and numerical values for the [ABD] matrix are listed there. As before, the 12-ply laminate has a total thickness t = 1.5 mm and aspect ratio R = a/b = 2.0. Part (a). this case:

Nxx

It is noted that n = 1, because Nyy z 0. Equation (39) becomes in "  2 # p2 0:6 4 2 ¼ 40:1m þ 65:34m þ 172:8 þ Nyy p 0:09m2

A plot of the critical buckling load for 0 V Nyy V 400 kN/m is presented in c is increased as Nyy is increased. A change in Fig. 9(a). As expected, Nxx buckling mode also occurs as Nyy is increased. The plate buckles in mode [2,1] over the range 0 V Nyy < 35 kN/m, in mode [3,1] over the range 35 kN/ m V Nyy < 150 kN/m, and in mode [4,1] over the range 150 kN/m V Nyy < 400 kN/m. These buckling modes are illustrated in Fig. 9(b,c,d), respectively.

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

Figure 9 Buckling response of the [(02/90)2]s graphite–epoxy plate considered in Sample Problem 5.

Part (b). this case:

It is noted that m = 1, because Nxx z 0. Equation (41) becomes in

Nyy ¼



p2 0:09 4 2 40:1 þ 65:34n þ 172:8ðnÞ þ N xx p2 0:36n2

A plot of the critical buckling load for 0 V Nxx V 400 kN/m is presented in c Fig. 10(a). As expected, Nyy is increased as Nxx is increased. A change in

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

Figure 10 Buckling response of the [(02/90)2]s graphite–epoxy plate considered in Sample Problem 5.

buckling mode also occurs as Nxx is increased. The plate buckles in mode [1,1] over the range 0 V Nxx < 72 kN/m, and in mode [1,2] over the range 72 kN/m V Nxx < 400 kN/m. These buckling modes are illustrated in Fig. 10(b,c), respectively. c Note that the magnitudes of Nyy calculated in part (b) are far lower than c those calculated for Nxx in part (a). This pronounced difference is largely because of the stacking sequence involved. For the [(02/90)2]s laminate under

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

consideration, 8 of 12 plies are 0j plies (i.e., plies with fibers parallel to the x axis). Hence the resistance to buckling because of a compressive load Nxx is far higher than resistance to buckling because of a compressive Nyy. Sample Problem 6 A structure is being designed that will involve a [(02/90)2]s graphite–epoxy laminate with a width (in the y direction) of 150 mm. The length of the panel (in the x direction) has not yet been established, and could be anywhere from 150 to 750 mm. During service, the panel will be subjected to a compressive load Nxx (only), and simple supports of type S4 along all four edges. Buckling is therefore of concern. Predict the buckling load and mode for the panel, for any panel length ranging from 150 to 750 mm. Solution. A [(02/90)2]s graphite–epoxy laminate was also considered in Sample Problem 1, and numerical values for the [ABD] matrix are listed there. As before, the 12-ply laminate has a total thickness t = 1.5 mm. Buckling loads and modes will be predicted using Eq. (39). According to the problem statement, b = 0.15 m, and 0.15 m < a < 0.75 m. The plate aspect

Figure 11 Predicted buckling loads and modes as a function of aspect ratio for the [(02/90)2]s graphite–epoxy panel considered in Sample Problem 6.

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

ratio therefore varies over 1 V R V 5. Because transverse loading is zero (Nyy = 0), n = 1 and Eq. (39) becomes: i p2 h Nxx ¼ 40:1m4 þ 16:34ðmRÞ2 þ 10:8R4 2 ðmaÞ A plot of the predicted critical buckling load over the specified range in aspect ratio is presented in Fig. 11. The buckling mode is predicted to increase as the aspect ratio increases: mode [1,1] is predicted over the range 1
Copyright © 2004 by Marcel Dekker, Inc. All Rights Reserved.

where axx and ayy are the effective thermal expansion coefficients of the laminate and DT c is the change in temperature from cure to room temperature. Therefore boundary conditions along all four edges are: For x ¼ 0 :

For x ¼ a :

wð0; yÞ ¼ 0

wða; yÞ ¼ 0

* ð0; yÞ ¼ 0 Mxx

* ða; yÞ ¼ 0 Mxx

ðxÞ

uo ð0; yÞ ¼ uo

ðxÞ

vo ð0; yÞ ¼ vo

ðþxÞ

ðyÞ ¼ ðaxx DT c Þa

ðþxÞ

ðyÞ ¼ ðayy DT c Þy

ðyÞ0

uo ða; yÞ ¼ uo

ðyÞ ¼ ðayy DT c Þy

vo ð0; yÞ ¼ vo

For y ¼ 0 :

For y ¼ b :

wðx; 0Þ ¼ 0

wðx; bÞ ¼ 0

* ðx; 0Þ ¼ 0 Myy

* ðx; bÞ ¼ 0 Myy

ðyÞ

uo ðx; 0Þ ¼ uo

ðyÞ

vo ðx; 0Þ ¼ vo

ðþyÞ

ðxÞ ¼ ðaxx DT c Þa

ðþyÞ

ðyÞ ¼ ðayy DT c Þb

ðxÞ ¼ ðaxx DT c Þx

uo ðx; bÞ ¼ uo

ðxÞ ¼ 0

vo ðx; bÞ ¼ vo

After assembly, we assume the laminate is subjected to a uniform change in temperature. The change in temperature is referenced to the strain-free temperature (assumed to be the cure temperature) and is represented by DT = (current temperature)  (cure temperature). If the laminate was not simply supported and instead was free to expand or contract, then a uniform change in temperature would simply cause a change in midplane displacements and no external edge loads would result. Because the laminate is instead subject to simple supports of type S1, no changes in midplane displacements are allowed to occur and external edge loads develop as temperature changes. Depending on the magnitude of the temperature change, these thermally induced edge loads may cause the laminate to buckle. Although the edge loads are thermally induced, they are external mechanical loads nevertheless and consequently the analysis presented in Sec. 7 is still applicable. The onset of thermal buckling occurs if Eq. (37) is satisfied, repeated here for convenience: D11

 mp 4 a

¼ Nxx

þ2ðD12 þ 2D66 Þ

 mp 2 a

Nyy

 mp 2  np 2

a  np 2

b

þD22

 np 4 b

b

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

In the present case, loads Nxx and Nyy are caused by a deviation from room temperature, and are given by: Nxx ¼ ðDT c  DT ÞðA11 axx þ A12 ayy Þ

ð43aÞ

Nyy ¼ ðDT c  DT ÞðA12 axx þ A22 ayy Þ

ð43bÞ

Substituting Eqs. (43a) and (43b) into Eq. (37), solving for the temperature difference, and using the definition of the plate aspect ratio, R = a/b, we obtain: 3 2 p2 6 D11 m4 þ 2ðD12 þ 2D66 ÞðmnRÞ2 þ D22 ðnRÞ4 7 5 4  m 2 b2 R4 2 n ðA12 a¯ xx þ A22 a¯ yy Þ þ ðA11 a¯ xx þ A12 a¯ yy Þ R It is convenient to express the temperature differences involved as:

ðDT c DT Þ ¼

ð44aÞ

DTc  DT ¼ TRT  T where TRT=room temperature (i.e., the temperature at which the laminate was mounted in the simple support fixture); T = current temperature.With this change in notation we have: 3 2 T¼

p2 6 D11 m4 þ 2ðD12 þ 2D66 ÞðmnRÞ2 þ D22 ðnRÞ4 7 5TRT 4  m 2 b2 R4 2 n ðA12 a¯ xx þ A22 a¯ yy Þ þ ðA11 a¯ xx þ A12 a¯ yy Þ R

ð44bÞ

Equations (44a) and (44b) are entirely equivalent, and both are based on the assumption that the simply supported laminate is assembled at room temperature, TRT. In practice, it is conceptually simplest to use Eq. (44b). By specifying a particular laminate and room temperature, all variables on the right side of the equality sign in Eq. (44b) are known, except for variables m and n. Because m and n can be any combination of positive integers, there are an infinite number of values of temperature T that will satisfy Eq. (44b). The temperature at which thermal buckling will occur, denoted Tbk, corresponds to the particular combination of m and n that leads to the value of T of lowest magnitude. The combination of m and n that correspond to this lowest temperature define the predicted critical thermal buckling mode. Now, based on the results of Sample Problem 6, one might anticipate that the thermal buckling mode exhibited would vary as a function of aspect ratio. That is, based on earlier analyses, one would expect that the values of m and n that correspond to the critical thermal buckling load would vary with aspect ratio. Equations (44a) and (44b) allow for such dependence. However, if physically reasonable material properties are used during the evaluation of Eqs. (44a) and (44b), then numerical experiments show that m = n = 1 in all

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

cases. Thus a thermal buckling mode [1,1] is always predicted when realistic material properties are used (at least for the simply supported boundary conditions considered in this section), regardless of the aspect ratio or stacking sequence considered. Assuming that this observation always holds true and that m = n = 1, Eq. (44b) reduces to: 3 2 D11 þ 2ðD12 þ 2D66 ÞR2 þ D22 R4 7 5 þ TRT 1 ðA12 axx þ A22 ayy Þ þ 2 ðA11 axx þ A22 ayy Þ R This latter result shows that the thermal buckling temperature decreases with an increase in aspect ratio. In the limit (i.e., as R!l), the thermal buckling temperature becomes: Tbk ¼

p2 6 4 b2 R 4

Tbk ¼

p2 D22 þ TRT b2 ðA12 axx þ A22 ayy Þ

Sample Problem 7 Two [02/90)2]s graphite–epoxy laminates are cured at 175jC and then cooled to room temperature (20jC). After cooling, one laminate is trimmed to inplane dimensions of 300
150 mm, whereas the second is trimmed to in-plane dimensions of 3000
150 mm. Both laminates are then mounted in assemblies that provide type S1 simple supports along all four edges, and subjected to a uniform increase in temperature. Determine the temperature at which each plate will buckle. 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. A [(02/90)2]s graphite–epoxy laminate was also considered in Sample Problem 1, and numerical values for the [ABD] matrix and effective thermal expansion coefficients are listed there. The aspect ratios involved in this problem are: R ¼ 300 mm=150 mm ¼ 2 and R ¼ 3000 mm=150 mm ¼ 20: The temperature at which thermal buckling is predicted to occur is obtained through application of Eq. (44b). It is found that both laminates are predicted to buckle in mode [1,1]. For the laminate with aspect R = 2, thermal buckling is predicted to occur when temperature is raised to 51jC, whereas for the laminate with aspect ratio R = 20, thermal buckling is predicted at a temperature of 40jC.

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

9 COMPUTER PROGRAM SPORTHO The computer program SPORTHO has been developed to supplement the material presented in this chapter. This program can also be downloaded at no cost from the following website: http://depts.washington.edu/amtas/ computer.html. Program SPORTHO is applicable to symmetric specially orthotropic laminates. The program can be used to calculate the transverse deflections according to the analyses presented in Secs. 3 through 6, to calculate buckling loads according to the analysis presented in Sec. 7, or to calculate the temperature at which thermal buckling will occur, according to the analysis presented in Sec. 8. The user is prompted to input all information necessary to perform these calculations. Properties of up to five different materials may be defined. Numerical values must be defined using a consistent set of units in all cases. 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

E22 ¼ 10
109 Pa

v12 ¼ 0:30

G12 ¼ 13
109 Pa

Because 1 Pa = 1 N/m2, all other lengths must be input in meters. For example, ply thickness 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). Inplane plate dimensions must also be input in meters. 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

In this case, all lengths would be input in inches. REFERENCES 1. 2. 3. 4. 5.

Lekhnitskii, S.G. Anisotropic Plates, translated by S. W. Tsai and T. Cheron, Taylor and Francis Books Ltd, London, UK, ISBN 0-677-20670-4, 1968. Savin, G.N. Stress Concentration Around Holes; New York: Pergamon Press, 1961. Timoshenko, S.; Woinowsky-Krieger, S. Theory of Plates and Shells; New York, NY: McGraw-Hill Book Co (ISBN 0-07-0647798), 1987. Whitney, J.M. Structural Analysis of Laminated Anisotropic Plates; Lancaster, PA: Technomic Pub Co (ISBN 87762-518-2). Buckling and Postbuckling of Composite Plates. Turvey, G.J.; Marshall, I.H., Eds; Chapman and Hall: New York, NY, 1995.

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

11 Some Approximate Solutions for Symmetric Laminates

Exact solutions for many symmetric composite laminates commonly used in practice cannot be found. In this chapter, methods of obtaining approximate numerical solutions for such laminates are discussed. The approximate solutions presented allow prediction of the deflections caused by the combined effects of in-plane loads, transverse loads, and changes in temperature or moisture content. Buckling caused by in-plane loads and/or temperature changes is also discussed. 1 PRELIMINARY DISCUSSION The equations of equilibrium for a thin symmetric composite laminate were derived in Sec. 2 of Chap. 9 based on a summation of forces and moments. Boundary conditions consistent with thin-plate theory were then discussed in Sec. 3 of Chap. 9. It turns out that exact solutions to these equations and boundary conditions can only be obtained if A16 = A26 = D16 = D26 = N Txy = N M xy = 0; that is, exact solutions are only available for specially orthotropic laminates. A few exact solutions for simply supported and symmetric specially orthotropic laminates were presented in Chap. 10. Unfortunately, many stacking sequences widely used in practice are not specially orthotropic. For example, symmetric quasi-isotropic laminates are not specially orthotropic because D16,D26 p 0 for this stacking sequence. 571

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

Hence, the solutions presented in Chap. 10 are not rigorously valid for many laminates encountered in practice. Fortunately, approximate numerical solutions are available that are suitable for use with any laminate including symmetric quasi-isotropic laminates. A brief introduction to these approximate solutions techniques will be presented in this chapter. Readers interested in a more detailed discussion are referred to Refs. (1,2). It is appropriate to clarify the distinction between ‘‘exact’’ and ‘‘approximate’’ solutions in the present context. In particular, confusion may arise in that exact solutions often involve a series with an infinite number of terms (as in the Navier solutions presented in Sec. 6 of Chap. 10, for example). A numerical evaluation of these solutions will contain some level of error because a finite number of terms must obviously be used during any practical application. However, the source of error in these cases arises from an inability to describe the applied loading in an exact analytical sense, rather than the solution itself. Furthermore, in a practical application, the number of terms used can be increased to obtain numerical results that satisfy all boundary conditions and the equations of equilibrium to any desired number of significant figures. The distinction between an exact and approximate solution does not refer to the need to use a finite number of terms during a numerical evaluation. Rather, the distinction refers to the rigor with which the boundary conditions are satisfied. Recall from Sec. 3 of Chap. 9 that it is necessary to specify four boundary conditions along each edge of a thin plate. Furthermore, recall that any boundary condition can be classified as either a geometric boundary condition or a static (also called a ‘‘natural’’) boundary condition. Geometric boundary conditions are those that dictate some feature of midplane displacements along a plate edge, whereas static boundary conditions are those that dictate some external load applied along a plate edge. Many common edge conditions consist of a combination of geometric and static boundary conditions. For example, a simply supported edge is one in which out-of-plane deflection is zero along the edge (a geometric condition) and the bending moment is zero along the edge (a static condition); the remaining two conditions may be either geometric or static. Consequently there are four ‘‘types’’ of simply supported edges (see Sec. 3.3.2 of Chap. 9). Now, during derivation of an exact solution all four boundary conditions along each edge of the plate are accounted for during derivation of the solution and are satisfied exactly. For example, in the solutions for simply supported specially orthotropic plates discussed in Chap. 10 all geometric and static boundary conditions are satisfied exactly by the solutions presented. In contrast, in an approximate solution, only the geometric boundary conditions are specified and enforced directly. Static boundary conditions are not en-

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

forced directly. Therefore, solutions obtained using approximate analysis techniques may, or may not, satisfy the static boundary conditions. The approximate solutions to be discussed in this chapter are based on the principle of minimum potential energy. Solution techniques that follow from this principle are known as energy methods. Consider a solid body subjected to some external loading. Briefly stated, the principle of minimum potential energy states that for all possible displacement fields that satisfy given boundary conditions, the displacement field that actually occurs is one in which the potential energy of the solid body assumes a minimum value. This statement is exact. In fact, it is possible to rederive the equations of equilibrium and associated boundary conditions (and, hence, all of the results presented in Chaps. 9 and 10) based on this principle. Energy methods will not be used to rederive the equations of equilibrium and associated boundary conditions in this textbook; the reader interested in this derivation is referred to the text by Whitney (1). Rather, the reason energy methods are of interest here is that they form the basis for approximate numerical solution techniques that are based on the principle of minimum potential energy. The two most common approximate numerical techniques are the Ritz method and the Galerkin method. The Ritz method will be developed and applied to several problems in this chapter. The Galerkin method will not be discussed, and the reader interested in this method is referred to Refs. (1,2). The essential elements of the Ritz method are as follows. The potential energy of a solid body is denoted P. The mathematical form of P depends on details of the problem under consideration and will be discussed in later sections. At this point, suffice it to say that the potential energy P of a solid body can be calculated if (a) the elastic properties of the body are known, (b) the external forces applied to the body are known, and (c) the resulting displacement fields induced in the body are known. During a typical structural analysis, the elastic properties and the external forces are known, but the displacement fields are not (in fact, the objective during a structural analysis is often to determine the displacement fields induced by some specified loading). During application of the Ritz method it is assumed, in effect, that the functional form of the displacement fields are known. For general thin-plate problems the midplane displacement fields are assumed to be of the form: uo ðx; yÞ ¼

M1 X N1 X

amn Umn ðx; yÞ

ð1aÞ

m ¼ 1n ¼ 1

vo ðx; yÞ ¼

M2 X N2 X

bmn Vmn ðx; yÞ

m ¼ 1n ¼ 1

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

ð1bÞ

wðx; yÞ ¼

M3 X N3 X

cmn Wmn ðx; yÞ

ð1cÞ

m ¼ 1n ¼ 1

where amn, bmn, and cmn are unknown constants; and Umn, Vmn, and Wmn are known functions that vary over x and y. In general, the number of terms used to describe each displacement field may differ (e.g., M1 does not necessarily equal M2 or M3). Also, the number of terms used to describe the variation in x and y may differ (e.g., M1 does not necessarily equal N1). Having assumed the functional form for the displacement fields as represented by Eqs. (1a)–(1c), then the potential energy P of an elastic plate subjected to specified external loading can be calculated. Note that the values of constants amn, bmn, and cmn in Eqs. (1a)–(1c) effectively define the magnitudes of the displacement fields. Hence, as constants amn, bmn and/or cmn are increased or decreased (and assuming that elastic properties and external forces remain constant), the potential energy of the body is increased or decreased accordingly. The principle of minimum potential energy states that the displacement field actually adopted by a solid body is one in which the potential energy is a minimum. This condition is therefore defined by the following criteria:  BP m ¼ 1; 2; . . . ; M1 ¼0 ð2aÞ n ¼ 1; 2; . . . ; N1 Bamn  BP m ¼ 1; 2; . . . ; M2 ¼0 ð2bÞ n ¼ 1; 2; . . . ; N2 Bbmn  BP m ¼ 1; 2; . . . ; M3 ¼0 ð2cÞ n ¼ 1; 2; . . . ; N3 Bcmn Equations (2a)–(2c) lead to (M1
N1)+(M2
N2)+(M3
N3) equations that must be satisfied simultaneously. Hence, by solving these equations for constants amn, bmn, and cmn, the magnitudes of the displacement fields given by Eqs. (1a)–(1c) that correspond to the minimum potential energy are known and the problem is solved. The validity of the solution obtained hinges on whether Eqs. (1a)–(1c) adequately represent the displacement fields actually induced in the structure. It is seen therefore that solutions obtained using the Ritz method are based on the functions Umn, Vmn, and Wmn that appear in Eqs. (1a)–(1c). These functions are more-or-less arbitrarily selected, but must posses two important characteristics: (a) they must be continuous and differentiable to at least the second order, and (b) they must satisfy the geometric boundary conditions. This latter characteristic is the source of the approximate nature of the Ritz analysis. That is, the mathematical forms of functions Umn(x,y), Vmn(x,y), and Wmn(x,y) are selected to satisfy the prevailing geometric boundary conditions, but the static boundary conditions are not considered

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

during this selection. Hence, these functions may, or may not, satisfy the prevailing static boundary conditions. If the functions Umn(x,y), Vmn(x,y), and Wmn(x,y) do not satisfy both geometric and static boundary conditions, then the solution obtained using the Ritz approach will be approximate. On the other hand, if the forms selected for Umn, Vmn, and Wmn do happen to satisfy both geometric and static boundary conditions, then the solution obtained using the Ritz method is exact. The challenge of identifying functional forms for Umn, Vmn, and Wmn that satisfy both geometric and static boundary conditions will be illustrated for a simply supported plate. Functions Umn, Vmn, and Wmn are usually selected to be separable functions of x and y. That is, function Wmn, for example, is usually selected to be of the form: Wmn ¼ Xm ðxÞYn ðyÞ For simply supported plates, it is common to assume that Xm and Yn are sinusoidal functions of x and y:     mpx npy Yn ðyÞ ¼ sin ð3Þ Xm ðxÞ ¼ sin a b where a and b are the length and width of a rectangular composite plate, respectively. Thus, based on this selection, Eq. (1c) becomes:     M3 X N3 X mpx npy wðx; yÞ ¼ cmn sin sin ð4Þ a b m ¼ 1n ¼ 1 These functional forms for Xm and Yn are appropriate for simply supported plates because they satisfy the geometric boundary conditions, regardless of the magnitudes of cmn. That is, Eq. (4) gives w(x,y) = 0 for x = 0,a and y = 0, b for all m,n. They do not necessarily satisfy the static boundary conditions for a simply supported plate, however. Recall from Sec. 3 of Chap. 9 that the bending moment must vanish along a simply supported edge, resulting in the following static boundary condition along the two edges x = 0, a (a comparable condition must be satisfied along edges y=0,b): D11

B2 w B2 w B2 w ¼ M*xx ¼ 0 þ D12 2  2D16 2 Bx By BxBy

Upon substituting the assumed displacement field [Eq. (4)] into this static boundary condition, the first two terms lead to: " # M3 X N3  mp 2  mpx   npy  X B2 w D11 2 ¼ D11 cmn sin sin ¼ 0; Bx a a b m ¼ 1n ¼ 1 for x ¼ 0; a

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

B2 w D12 2 ¼ D12 By

"

M3 N3 X X m ¼ 1n ¼ 1

 cmn

np b

2

  # mpx npy sin ¼ 0; sin a b 

for x ¼ 0; a However, the third term leads to: "       # M3 N3 X X B2 w mp np mpx npy ¼ 2D16 2D16 cmn cos cos BxBy a b a b m ¼ 1n ¼ 1 Evaluating this third term along x=0, a:   M3 N3 X B2 w 2p2 D16 X npy ¼ ; for x ¼ 0; a ðmnÞ cmn cos 2D16 BxBy b ab m ¼ 1 n ¼ 1 Hence, the static boundary condition associated with simply supported edges x = 0,a is satisfied only if:  npy  ¼ 0; for 0 V y V b ðmnÞcmn cos b m ¼ 1n ¼ 1 M3 X N3 X

ð5Þ

If only a single term is used to describe the displacement field (i.e., if M3= N3=1), then the static boundary condition is clearly not satisfied even approximately because:  py  c11 cos p 0; for all 0 V y V b b However, if the number of terms used is increased (i.e., as M3 and/or N3 are increased), then Eq. (5) may be satisfied more and more exactly, through proper selection of the values of constants cmn. For example, if four terms are used (i.e., if M3=N3=2), then the static boundary condition along edges x=0,a is satisfied if:      py   py  2py 2py c11 cos þ 2c12 cos þ 2c21 cos þ 4c22 cos ¼ 0; b b b b for all 0 V y V b It is now possible to satisfy the static boundary condition along edges x = 0, a exactly by setting c21 = c11/2 and c22 = c12/2. Of course, selecting constants that satisfy these requirements may not lead to the displacement field that represents the state of minimum potential energy because constants c11, c12, c21, and c22 must also satisfy the static boundary conditions along the edges y = 0, b. Still, it is apparent that by increasing the number of terms used to describe the displacement field, it is possible to satisfy the static boundary conditions along all four edges more and more exactly.

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

In general then, the validity of a solution based on the Ritz approach is increased as the number of terms is increased. Assuming functions Xm and Yn are selected to satisfy the geometric boundary conditions, then a Ritz analysis will converge toward the exact solution as M3 and/or N3 are increased. It is also appropriate to note that the difficulty in obtaining an exact solution is due to the D16 term. If D16 = 0 (i.e., if the laminate were specially orthotropic), then the geometric and static boundary conditions associated with the simply supported edges x = 0,a would be satisfied exactly by the assumed sinusoidal form for Xm and Yn. A relatively general description of how the Ritz approach is applied has been presented in the preceding paragraphs. We will now specialize this approach for application to the specific problems discussed in this chapter. First, as in previous chapters, we limit our discussion to rectangular symmetric laminates, simply supported along all four edges. Second, for all problems considered herein, the in-plane displacement fields uo(x,y) and vo(x,y) that exist prior to application of q(x,y) can be deduced based on a CLT analysis (as discussed in the next section) and are known a priori. Hence, for present purposes, there is no need to express uo(x,y) or vo(x,y) in terms of the double series as listed as Eqs. (1a,b) nor to determine the magnitudes of inplane displacement fields using Eqs. (2a,b). For the problems considered herein, only w(x,y) is unknown. The out-of-plane displacement field will be expressed using the double series listed as Eq. (1c), and the magnitude of outof-plane displacements will be determined through application of Eq. (2c). Functions Xm and Yn will be assumed to be sinusoidal functions as listed as Eq. (3); thus, the assumed form for w(x,y) is given by Eq. (4). To simplify nomenclature, let M3!M and N3!N; thus, w(x,y) will be written:     M N X X mpx npy wðx; yÞ ¼ cmn sin sin ð6Þ a b m ¼ 1n ¼ 1 Although not discussed in this text, solutions are also available based on alternate (nonsinusoidal) function forms for Xm and Yn. Alternate forms include polynomials in x and y: Xm ðxÞ ¼ ðx2  axÞ2 xm1 Yn ðyÞ ¼ ðy2  ayÞ2 yn1 forms involving other trigonometric functions:





2mpx Xm ðxÞ ¼ 1  cos a  

2npy Yn ðyÞ ¼ 1  cos b

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

or forms involving so-called beam functions:         km x km x km x km x  cm cosh þ sin  sinh Xm ðxÞ ¼ cm cos a a a a 

kn x Yn ðxÞ ¼ cn cos b





kn x  cn cosh b





kn x þ sin b





kn x  sinh b



These alternate forms may be used to model other boundary conditions such as a clamped edge. The reader interested application of these alternate forms is referred to Refs. [1,2,4].

2 IN-PLANE DISPLACEMENT FIELDS In this section, we will derive expressions that give the midplane displacement fields induced in a symmetric laminate for specific loading conditions. The derivation is based on the following sequence of events. First, assume the laminate is cured at an elevated temperature and that the laminate is stressand strain-free at the cure temperature.* Following cure, the laminate is cooled to room temperature. The laminate is then trimmed to the desired final rectangular dimensions a
b. Finally, the laminate is subjected to constant and uniformly distributed stress resultants (Nxx, Nyy, and/or Nxy) along all four edges and/or a further change in temperature. Although we will not consider the possibility of a uniform change in moisture content directly, from previous discussion, it should be clear that changes in moisture content could be accounted for (in a mathematical sense) using the same techniques used to model uniform changes in temperature. The sequence of events described above are precisely those assumed during the development of CLT in Chap. 6. Because the externally applied stress resultants are uniformly distributed along each edge, the stress resultants induced at all interior points are equal to the edge loads: * ð0; yÞ ¼ Nxx * ða; yÞ ¼ Nxx Nxx * ðx; bÞ ¼ Nyy N*yy ðx; 0Þ ¼ Nyy * ða; yÞ ¼ Nyx * ðx; 0Þ ¼ Nyx * ðx; bÞ ¼ Nxy N*xy ð0; yÞ ¼ Nxy

* As mentioned in Sec. 6.2 of Chap. 6, in practice, the stress- and strain-free temperature is often 20–50jC below the final cure temperature. This complication has been ignored throughout this text, and it is assumed that the laminate is stress- and strain-free at the cure temperature.

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

For a symmetric laminate subjected to uniform in-plane stress resultants and a change in temperature, Eq. (45) of Chap. 6 becomes: 38 9 8 0 9 2 a11 a12 a16 0 0 0 > exx > Nxx þ NTxx > > > > > > > 6 > > > 7> > > > > > > > 7 6 > > > > > > > 0 > T 7 6 > > > > a a 0 0 0 a e N þ N > > > > 12 22 26 yy yy yy 7> 6 > > > > > > > 7 6 > > > > > > > > > 6 > > 0 > > 7> > > T > > > 7 6 0 0 7< Nxy þ Nxy > = < cxy = 6 a16 a26 a66 0 7 ¼6 7> > 6 > > > > > 6 0 0 0 d11 d12 d16 7 jxx > > > > 0 > > > > 7> 6 > > > > > > > > 7 6 > 6 > > > > > > > 7 > > > > > > > > 7 6 > > > > 0 0 d12 d22 d26 7> jyy > 6 0 0 > > > > > > > > > 5> 4 > > > > > > > > : ; : ; 0 0 0 d16 d26 d66 jxy 0 As in earlier chapters, we assume infinitesimal strain levels, and therefore midplane strains are related to midplane displacements according to Eq. (10) of Chap. 6: e0xx ¼

Buo Bx

e0yy ¼

Bvo By

c0xy ¼

Buo Bvo þ By Bx

Consequently, midplane displacement fields are related to the stress and thermal resultants as follows: Buo ¼ a11 ðNxx þ NTxx Þ þ a12 ðNyy þ NTyy Þ þ a16 ðNxy þ NTxy Þ Bx Bvo ¼ a12 ðNxx þ NTxx Þ þ a22 ðNyy þ NTyy Þ þ a26 ðNxy þ NTxy Þ By Buo Bvo þ ¼ a16 ðNxx þ NTxx Þ þ a26 ðNyy þ NTyy Þ þ a66 ðNxy þ NTxy Þ By Bx

ð7aÞ ð7bÞ ð7cÞ

Integrating Eq. (7a) with respect to x, we find: h      i uo ðx; yÞ ¼ a11 Nxx þ NTxx þ a12 Nyy þ NTyy þ a16 Nxt þ NTxy 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. (7b) with respect to y, we find: h      i vo ðx; yÞ ¼ a12 Nxx þ NTxx þ a22 Nyy þ NTyy þ a26 Nxy þ NTxy y þ f2 ðxÞ þ k2 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

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

are zero at the origin (i.e., at x = y = 0) and, consequently, k1 = k2 = 0. Substituting these expressions for uo(x,y) and vo(x,y) into Eq. (7c), we find: Buo Bvo Bf1 Bf2 þ ¼ þ ¼ a16 ðNxx þ NTxx Þ þ a26 ðNyy þ NTyy Þ By Bx By Bx þ a66 ðNxy þ NTxy Þ Because all terms on the right side of the equality are known constants, it follows that f1 and f2 must be at most linear function of y and x, respectively: f1 ðyÞ ¼ k3 y f2 ðxÞ ¼ k4 x Hence, we can write: k3 þ k4 ¼ a16 ðNxx þ NTxx Þ þ a26 ðNyy þ NTyy Þ þ a66 ðNxy þ NTxy Þ Because all quantities that appear in this relation are constants, k3 and k4 can take on any value as long as they sum to the expression on the right side of the equality. To determine particular values that satisfy this expression, 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 (3):   1 Buo Bvo  xxy ¼ 2 By Bx Requiring that xxy=0 leads to:     i 1h  a16 Nxx þ NTxx þ a26 Nyy þ NTyy þ a66 Nxy þ NTxy k3 ¼ k4 ¼ 2 Combining the preceding results, we conclude that the in-plane midplane displacements induced in a symmetric composite panel by the combination of uniform in-plane stress resultants and a change in temperature are given by: h      i uo ðx; yÞ ¼ a11 Nxx þ NTxx þ a12 Nyy þNTyy þ a16 Nxy þ NTxy x þ

vo ðx; yÞ ¼

    i 1h  a16 Nxx þ NTxx þ a26 Nyy þ NTyy þ a66 Nxy þ NTxy y 2

    i 1h  a16 Nxx þ NTxx þ a26 Nyy þ NTyy þ a66 Nxy þ NTxy x 2 h      i þ a12 Nxx þ NTxx þ a22 Nyy þ NTyy þ a26 Nxy þ NTxy y

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

These expressions can be simplified through the use of the effective thermal expansion coefficients of the laminate, defined by Eqs. (73a) of Chap. 6 and repeated here for convenience: 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 ayy ¼ ð6:73aÞ DT i 1 h a16 NTxx þ a26 NTyy þ a66 NTxy axy ¼ DT Hence, we see that the midplane displacement fields can be written: h i uo ðx; yÞ ¼ a11 Nxx þ a12 Nyy þ a16 Nxy þ DT axx x þ

i 1h a16 Nxx þ a26 Nyy þ a66 Nxy þ DT axy y 2

i 1h a16 Nxx þ a26 Nyy þ a66 Nxy þ DT axy x vo ðx; yÞ ¼ 2 h i þ a12 Nxx þ a22 Nyy þ a26 Nxy þ DT ayy y

ð7Þ

Note that Eqs. (7a)–(7c) are valid only if both stress resultants and temperature changes are uniform. In the following sections, we will use these in-plane displacement fields to obtain solutions based on the Ritz method for simply supported composite plates. Recall from Sec. 3 of Chap. 9 that four distinct combinations of geometric and static boundary conditions, numbered S1 through S4, can be defined as a ‘‘simple support.’’ The distinction between the different types of simple supports has to do with the boundary condition assumed for the inplane displacement fields. For example, to define a simple support of type S1, one specifies known values of in-plane displacements, whereas to define a simple support of type S4, one specifies known values of in-plane stress resultants. For the problems considered here, we are able to calculate the midplane displacement field induced by a specified combination of edge loads (or vice versa). This is possible because we have assumed all stress resultants applied at the edge of the plate are constant and uniform. Of course, if the stress resultants were not uniform but rather varied along the plate edges, then it would be more difficult (and, in most cases, impossible) to determine associated in-plane displacement fields. Because we have limited discussion to cases in which stress resultants applied to the edges are constant and uniform, the midplane displacement

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

fields given by Eqs. (7a)–(7c) can be used to obtain solutions for any of the four types of simple-support boundary conditions, S1 through S4. Only one type of simple support will be considered in this chapter because this discussion is intended to be a brief introduction to the Ritz method. Specifically, type S4 simple supports will be assumed throughout the remainder of this chapter. Thus, solutions will be obtained using the Ritz method for problems in which uniform in-plane stress resultants Nxx, Nyy, and/or Nxy are applied to the edge of the plate. Although not discussed herein, similar analyses can be performed assuming type S1, S2 or S3 simple supports, or combinations thereof.

3 POTENTIAL ENERGY IN A THIN COMPOSITE PLATE In this section, we will develop the equations necessary to calculate the potential energy of a thin elastic plate subjected to a combination of loads and uniform temperature changes. The possibility of a uniform change in moisture content will not be considered although from previous discussion, it should be clear that changes in moisture content can be accounted for (in a mathematical sense) using the same techniques used to model uniform changes in temperature. Type S4 boundary conditions are assumed for all four edges of the plate. Two energy terms will be encountered in the following discussion. First, we will consider the work done when a transverse load q(x,y) is applied to a thin plate. This energy term is denoted W. Recall that the fundamental definition of ‘‘work’’ is force multiplied by the distance through which it travels. Also, recall that the transverse load applied to a plate q(x,y) has been defined using units of force/area. The product ( q)(dx)(dy) represents the transverse force provided by q(x,y) acting over an infinitesimal element of area (dx)(dy). The distance through which this force travels equals the out-of-plane deflection of the plate at that point, w(x,y). Therefore, the total work done by the transverse load acting over the entire surface of the plate is given by: ZZ W¼ ½qðx; yÞ½wðx; yÞdxdy ð8Þ Secondly, we will consider the strain energy within the plate in the deformed condition. This energy term is denoted U. A general expression giving the strain energy within a linear elastic solid body subjected to an arbitrary state of stress is: ZZZ 1 U¼ ðrxx exx þ ryy eyy þ rzz ezz þ syz cyz þ sxz cxz þ sxy cxy Þdxdydz 2

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

Note that evaluation of strain energy involves integration over the entire volume of the body. In our case, we wish to calculate the strain energy within a thin symmetric composite laminate subjected to a state of plane stress. We will therefore specialize the above general expression of strain energy for present purposes. First, because plane stress is assumed (rzz = syz = sxz = 0), we can immediately discard terms involving these stress components: ZZZ 1 ðrxx exx þ ryy eyy þ sxy cxy Þdxdydz ð9Þ U¼ 2 Next, the stresses in ply k of the laminate are given by Eqs. (30) of Chap. 5, which become (for DM = 0): 8 9 2 9 3 8 < rxx = Q11 Q12 Q16 < exx  DTaxx = eyy  DTayy ryy ¼ 4 Q12 Q22 Q26 5 : ; : ; rxy k Q16 Q26 Q66 k cxy  DTaxy k Substituting these expressions for ply stresses into Eq. (9) and rearranging, we find: U¼

ZZ Z h 1 ðkÞ ðkÞ ðkÞ ðkÞ ðkÞ ðkÞ Q11 e2xx þ 2Q12 exx eyy þ 2Q16 exx cxy þ 2Q26 eyy cxy þ Q22 e2yy þ Q66 c2xy 2

  ðkÞ ðkÞ ðkÞ ðkÞ ðkÞ DT aðkÞ Q þ a Q þ a Q xx 11 yy 12 xy 16 exx   ðkÞ ðkÞ ðkÞ ðkÞ ðkÞ DT aðkÞ Q þ a Q þ a Q xx 12 yy 22 xy 26 eyy   i ðkÞ ðkÞ ðkÞ ðkÞ ðkÞ DT aðkÞ Q þ a Q þ a Q xx 16 yy 26 xy 66 cxy dxdydz

ð10Þ

We now invoke the Kirchhoff hypothesis, which allows us to relate ply strains at any through-thickness position z to midplane strains and curvatures, in accordance with Eqs. (12)of Chap. 6, repeated here for convenience: exx ¼ e0xx þ zjxx eyy ¼ e0yy þ zjyy cxy ¼ c0xy þ zjxy Substitution of Eqs. (12) of Chap. 6 into Eq. (10) results in: U¼

ZZZ n h i  2 1 Q11 e0xx þ 2ze0xx jxx þ z2 j2xx 2

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

h i þ2Q12 e0xx e0yy þ ze0xx jyy þ ze0yy jxx þ z2 jxx jyy h i þ2Q16 e0xx c0xy þ ze0xx jxy þ zc0xy jxx þ z2 jxx jxy h i þ2Q26 e0yy c0xy þ ze0yy jxy þ zc0xy jyy þ z2 jyy jxy h 2 i þQ22 e0yy þ2ze0yy jyy þ z2 j2yy h 2 i þQ66 c0xy þ2zc0xy jxy þ z2 j2xy h i  DT axx Q11 þ ayy Q12 þ axy Q16 e0xx þ zjxx h i  DT axx Q12 þ ayy Q22 þ axy Q26 e0yy þ zjyy i o h DT axx Q16 þ ayy Q26 þ axy Q66 c0xy þ zjxy dxdydz

ð11Þ

Next, integrate Eq. (11) over the thickness of the laminate, i.e., over the range t/2 V z V t/2. During this process, a number of integrals will be encountered that were previously evaluated in Chap. 6. A few specific examples are:

Z

t=2

Q11 dz; which after integration becomes A11

t=2

Z

t=2

Q11 zdz; which after integration becomes B11

t=2

Z

t=2

Q11 z2 dz; which after integration becomes D11

t=2

Z

t=2



DT axx Q11 þ ayy Q12 þ axy Q16 dz;

t=2

which after integration becomes NTxx

Z t=2 DT axx Q11 þ ayy Q12 þ axy Q16 zdz; t=2

which after integration becomes MTxx Hence, after integration Eq. (11) can be written:

ZZ   2    2 1 U¼ A11 e0xx þ 2A12 e0xx e0yy þ A22 e0yy þ 2 A16 e0xx þ A26 e0yy c0xy 2

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

  þ A66 c2xy þ 2B11 e0xx jxx þ 2B12 e0xx jyy þ e0yy jxx     þ 2B16 e0xx jxy þ c0xy jxx þ 2B26 e0yy jxy þ c0xy jyy þ 2B22 e0yy jyy þ 2B66 c0xy jxy þ D11 j2xx þ 2D12 jxx jyy   þ 2 D16 jxx þ D26 jyy jxy þ D22 j2yy þ D66 j2xy  NTxx e0xx  NTyy e0yy  NTxy c0xy  MTxx jxx  MTyy jyy  MTxy jxy dxdy

ð12Þ

T = Because we have limited discussion to symmetric laminates, Bij = Mxx T T Myy = Mxy = 0 in all cases considered. Equations (12) therefore simplifies to: ZZ   2    2 1 A11 e0xx þ 2A12 e0xx e0yy þ A22 e0yy þ2 A16 e0xx þ A26 e0yy c0xy U¼ 2   þ A66 c2xy þ D11 j2xx þ 2D12 jxx jyy þ 2 D16 jxx þ D26 jyy jxy ð13Þ þ D22 j2yy þ D66 j2xy  NTxx e0xx  NTyy e0yy  NTxy c0xy dxdy

We have now developed expressions for the two energy terms necessary for our purposes: the work W done by a transverse load applied to a laminate [Eq. (8)] and the strain energy U within a deformed laminate [Eq. (13)]. We wish to form an appropriate combination of these terms so as to represent the total potential energy of a symmetric composite laminate. An itemized conceptual description of how U and W are combined is presented below. Mathematical implementation of these concepts for the particular class of problems considered in this text is then discussed in separate subsections. The reader is urged to carefully consider the follow conceptual description before considering the mathematical formulation that follows. Step 1. We assume that in-plane stress resultants Nij are applied to the laminate first, before the application of any other load(s) that cause bending. Our first step is therefore to calculate the strain energy within a laminate subjected to Nij only even if other loads are involved in the problem under consideration. We will label this component of strain energy UI. Calculation of UI is straightforward because we have limited our discussion to symmetric laminates. That is, for a symmetric laminate, stress resultants Nij are solely responsible for the development of midplane strains and do not cause curvatures to develop. If we had included nonsymmetric laminates in our analysis, then Bij p 0. If this were the case, then Nij would also contribute to curvatures; furthermore, both Mij and q(x,y) would contribute to

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

midplane strains. These coupling effects would greatly complicate our analysis. Because symmetry has been assumed, the laminate remains flat following application of Nij. Therefore, the only loads that contribute to midplane strains are stress resultants Nij. The strain energy associated with Nij only (i.e., the strain energy component UI) can therefore be obtained from Eq. (13) simply by setting jxx=jyy= jxy=0. Ultimately our expression for UI will involve in-plane displacements u(x,y) and v(x,y), but will not involve out-of-plane displacements w(x,y). Step 2. Next, we calculate strain energy associated with midplane curvatures only. That is, we ignore any preexisting midplane strains and base our calculation of strain energy based solely on curvatures jxx, jyy, and jxy. We will label this component of the strain energy UII. Because the laminate is symmetric, midplane curvatures are caused by the combined effects of the transverse load q(x,y) and/or bending moment resultants Mij, but are independent of stress resultants Nij. Strain energy component UII can be obtained from Eq. (13) simply by setting e0xx =e0yy =e0xy =0. Ultimately, our expression for UII will involve out-of-plane displacements w(x,y), but will not involve in-plane displacements u(x,y) or v(x,y). Step 3. As described above, in step 1, we calculate the strain energy UI associated with stress resultants Nij prior to the application of any load(s) that cause the laminate to bend. Of course, when transverse loads and/or bending moments are subsequently applied, the laminate will bend. Now, once bending has occurred, then calculation of strain energy as performed in step 1 is incomplete. That is, as the laminate begins to bend, the in-plane strains that exist prior to bending will change, resulting in a change in the strain energy associated with stress resultants Therefore, in step 3, we calculate the change in strain energy caused by bending and associated with in-plane stress resultants. We will label this component of the strain energy UIII. We assume that the stress resultants Nij that exist prior to bending remain constant during bending. That is, we assume that only inplane strains change as bending occurs. The change in in-plane strains will be related to out-of-plane displacements w(x,y). Hence, our expression for UIII will involve w(x,y), but will not involve in-plane displacements u(x,y) or v(x,y). Step 4. In step 4, we calculate the work W done by the transverse load q(x,y), in accordance with Eq. (6). The mathematical form of our expression for W will obviously depend on the nature of the transverse load. Thus, for example, the mathematical form of W for a uniform transverse load, q(x,y)=qo, will differ from the mathematical form of W if the transverse load varies sinusoidally: q(x,y)= qosin(px/a)sin(py/b).

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

Step 5. Finally, in step 5, we form the desired expression for the total potential energy P in the composite plate. The total potential energy is given by: P = UI+UII+UIIIW. 3.1 Evaluation of Strain Energy Component UI Strain energy component UI is obtained from Eq. (11) by setting jxx=jyy= jxy=0: ZZ   2    2 1 A11 e0xx þ 2A12 e0xx e0yy þ A22 e0yy þ 2 A16 e0xx þ A22 e0yy c0xy UI ¼ 2 2 T 0 T 0 T 0 þ A66 cxy  Nxx exx  Nyy eyy  Nxy cxy dxdy As in earlier chapters, we assume infinitesimal strain levels and therefore midplane strains are related to midplane displacements according to Eq. (10) of Chap. 6: e0xx ¼

Buo Bx

e0yy ¼

Bvo By

c0xy ¼

Buo Bvo þ By Bx

With this substitution, we have:       ZZ (  1 Buo 2 Buo Bvo Bvo 2 UI ¼ A11 þ 2A12 þ A22 2 Bx Bx By By      Buo Bvo Buo Bvo Buo Bvo 2 þ A26 þ þ þ 2 A16 þ A66 ð14Þ Bx By By Bx By Bx      ) Buo Bvo Buo Bvo T T T þ  Nxx  Nyy  Nxy dxdy Bx By By Bx Equation (14) gives the strain energy component UI for any in-plane displacement fields uo(x,y) and vo(x,y). We will now integrate this expression using the displacement fields induced by uniform stress resultants and a uniform change in temperature, as given by Eqs. (7). To avoid a very lengthy expression, we make the following change in notation: h i C1 ¼ a11 Nxx þ a12 Nyy þ a16 Nxy þ DT axx h i C2 ¼ a12 Nxx þ a22 Nyy þ a26 Nxy þ DT ayy ð15Þ C3 ¼

i 1h a16 Nxx þ a26 Nyy þ a66 Nxy þ DT axy 2

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

With this change in notation, Eqs. (7) can be written: uo ðx; yÞ ¼ C1 x þ C3 y vo ðx; yÞ ¼ C3 x þ C2 y Substituting these expressions into Eq. (14), we have: Z Z 1 b an A11 ðC1 Þ2 þ 2A12 ðC1 ÞðC2 Þ þ A22 ðC2 Þ2 UI ¼ 2 0 0 þ 2ðA16 C1 þ A26 C2 Þð2C3 Þ þ A66 ð2C3 Þ2  NTxx ðC1 Þ o  NTyy ðC2 Þ  NTxy ð2C3 Þ dxdy Evaluating this definite integral, we find: h UI ¼ A11 ðC1 Þ2 þ 2A12 ðC1 ÞðC2 Þ þ A22 ðC2 Þ2   þ 4 A16 C1 C3 þ A26 C2 C3 þA66 C23 i  NTxx C1  NTyy C2  2NTxy C3 ðabÞ Substituting Eqs. (73b) of Chap. 6, which give the thermal stress resultants in terms of elements of the [A] matrix, and Eq. (5), we find that UI can be written:   ð16Þ UI ¼ C1 Nxx þ C2 Nyy þ 2C3 Nxy ðabÞ 3.2 Evaluation of Strain Energy Component UII As previously discussed, for symmetric laminates, strain energy component UII can be obtained from Eq. (13) by setting e0xx ¼ e0yy ¼ e0xy =0: ZZ n 1 UII ¼ D11 j2xx þ 2D12 jxx jyy þ 2ðD16 jxx þ D26 jyy Þjxy þ D22 j2yy 2 o þ D66 j2xy dxdy Midplane curvatures are related to out-of-plane displacements according to Eq. (10) of Chap. 6: jxx ¼ 

B2 w Bx2

jyy ¼ 

B2 w By2

jxy ¼ 2

B2 w BxBy

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

With this substitution, we have:  2  2  Z Z (  2 2 1 B w B w B w D11 þ2D UII ¼ 12 2 2 2 Bx Bx By2  2   2  2 B w Bw B w þ 4 D16 þ D26 2 2 Bx By BxBy  2 2 )  2 2 Bw B w þ D22 dxdy þ4D66 By2 BxBy

ð17Þ

We will now integrate Eq. (17) for the class of problems considered in this text. In all problems, we consider simply supported plates and assume outof-plane displacements are given by Eq. (6), repeated here for convenience:     M N X X mpx npy wðx; yÞ ¼ cmn sin sin a b m ¼ 1n ¼ 1 Because Eq. (6) will be used in all problems considered, we will integrate Eq. (17) based on this displacement field. The following derivatives appear in Eq. (17):       M M X X B2 w mp 2 mpx npy ¼ cmn sin sin Bx2 a a b m ¼ 1 n ¼ 1  2     M N X X B2 w np mpx npy ¼  c sin sin mn By2 b a b m ¼ 1 n ¼ 1        M N X X B2 w mp np mpx npy ¼ cmn cos cos BxBy m ¼ 1 n ¼ 1 a b a b Consider the first term under the integral sign in Eq. (17). Upon substituting the expression for (B2w/Bx2) listed above, this term becomes:  2 2 Z Z 1 b a B w D11 dxdy 2 0 0 Bx2 " 2    #  Z Z M N X X 1 b a mp mpx npy D11  cmn sin sin ¼ 2 0 0 a a b m ¼ 1 n ¼ 1 " #  2     M N X X ip ipx jpy
 cij sin sin dxdy a a b i ¼ 1 j ¼ 1

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

The evaluation of this integral is greatly simplified by noting the following identities:

Z

a 0

Z

b 0

    mpx ipx sin sin dx ¼ 0; if m p i a a     npy ipy sin sin dy ¼ 0; if n p j b b

Hence, during integration of this term, we need only consider: 1 2

Z bZ



a

D11 0

0

Z Z 1 b ¼ 2 0

B2 w Bx2

a

D11 0

2 dxdy

M M X X

"

 c2mn

m ¼ 1n ¼ 1

mp a

4

 2

sin

 #  mpx 2 npy sin dxdy a b

After integration and evaluation, this term becomes:

Z Z 1 b 2 0



a

D11 0

B2 w Bx2

2 dxdy ¼

M N X X p4 b D m4 c2mn 11 8a3 m ¼ 1 n ¼ 1

The following terms also appear in Eq. (17) and are evaluated in a similar manner:

Z Z 1 b 2 0 Z Z 1 b 2 0 Z Z 1 b 2 0



a

2D12 0



a

D22 0

4D66 0

B2 w By2



a

B2 w Bx2



 M N X X B2 w p4 D m2 n2 c2mn dxdy ¼ 12 By2 4ab m ¼ 1 n ¼ 1

2

B2 w BxBy

dxdy ¼

M N X X p4 a D n4 c2mn 22 8b3 m ¼ 1 n ¼ 1

2 dxdy ¼

M N X X p4 D66 m2 n2 c2mn 2ab m ¼ 1 n ¼ 1

The remaining terms in Eq. (17) involve D16 and D26. Upon substituting the appropriate derivatives, the first of these becomes:  2  2  Z Z 1 b a B w B w 4D16 dxdy 2 2 0 0 Bx BxBy

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

"    #   Z Z M N X X 1 b a mp 2 mpx npy ¼ 4D16  cmn sin sin 2 0 0 a a b m ¼ 1 n ¼ 1 "       # M N X X ip jp ipx jpy dxdy
cij cos cos a b a b i ¼ 1 j ¼ 1 This integration does not simplify as readily as those considered above. We make use of the following identities:

Z

a 0

Z

b 0

8 0; if ðm þ iÞ is even >    < mpx ipx cos dx ¼ sin 2ma > a a : ; if ðm þ iÞ is odd p½m2  i2  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  

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 m2 nij cmn cij ¼ 2 D16 ðMIÞðNJ Þ ½m2  i 2 ½n2  j 2  a m ¼ 1n ¼ 1i ¼ 1j ¼ 1 where: ðMIÞ ¼ ½ð1Þm ð1Þi  1 ðNJ Þ ¼ ½ð1Þn ð1Þ j  1 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

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

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 cmn cij ¼ 2 D26 ðMIÞðNJ Þ b ½m2  i 2 ½n2  j 2  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 ðD D þ 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 dx ¼ dx 1 þ 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 n þ n þ ... ð1 þ nÞ1=2 ¼ 1 þ n  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: e0;b yy ¼

 2 1 Bw 2 By

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

ð19bÞ

 c0;b xy

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 which this force moves as bending develops equals e0;b xx dx. The incremental strain energy associated with this force and caused by bending is therefore dUIII=(Nxxdy)(e0;b xx 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 e0;b ð20Þ xx þ Nyy eyy 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  NTxx Bx By By Bx Similarly,  Nyy ¼ A12  Nxy ¼ A16

Buo Bx Buo Bx



 þ A22



 þ A26

Bvo By Bvo By



 þ A26



 þ A66

Buo Bvo þ By Bx Buo Bvo þ By Bx

  NTyy   NTxy

Substituting these expressions as well as Eqs. (19a,b,c) into Eq. (20), we obtain:      Z Z   1 Buo Bvo Buo Bvo þ UIII ¼ A11 þ A12 þ A16  NTxx 2 Bx By By Bx

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

        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  NTxx

N X

 cmn

m ¼ 1n ¼ 1



    !2 mp mpx npy cos sin a a b

n o þ A12 C1 þ A22 C2 þ 2A26 C3  NTyy


M N X X

 cmn

m ¼ 1n ¼ 1

    !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 sin cos dxdy 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  NTxx a

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



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 Nxx þ Nyy þ 2m2 ncmn Nxy cmn 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 2 Because we have incorporated the term e0;b xx = 1/2(Bw/Bx) 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

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

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 sin dxdy ð23Þ 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

    mpx npy sin cmn sin a b ¼ 1

M N X X m ¼ 1n

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.

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

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 ð1Þ  1 ð1Þ  1 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



M N  4 X X p 2 bm4 2m2 n2 an4 D þ þ 2D Þ þ D cmn ðD 11 12 66 22 8 a3 ab b3 m ¼ 1n ¼ 1 2p2 mncmn

i ¼ 1j ¼ 1 M N  2 X X p m ¼ 1n

( 

M X N  X cij



ij ðm 2



i 2 Þðn 2



j 2Þ

  m2 n2 D16 þ 2 D26 ðMI ÞðNJ Þ 2 a b

g



m2 b n2 a Nxx þ Nyy a b 8 ¼ 1 

M N X X  j þ2m2 ncmn Nxy ðMI ÞðNJ Þ cij ðm2  i 2 Þðn2  j 2 Þ i ¼ 1j ¼ 1 c2mn

M N  X X abqo cmn m ¼ 1n ¼ 1

p2 mn

ð1Þ  1

ð25Þ

)

m

g

n

ð1Þ  1

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

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

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

F2 ¼

F8 ¼ 2Nxy

p4 a D22 8b3 2p2 F6 ¼ 2 D16 a abqo F9 ¼ p2 F3 ¼

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:  2 4 2 2 4 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







F9 ð1Þm  1 ð1Þn  1 mn ( M X N X cij þ cmn 2 2 ðm  i Þðn2  j 2 Þ i ¼ 1j ¼ 1 

) 3 3 2
F6 ðm nijÞ þ F7 ðmn ijÞ þ F8 ðm njÞ ðMIÞðNJÞ þ cmn

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

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

ð27Þ

40 c11 c22 ð2F6 þ 2F7 þ F8 Þ 9 40  c12 c21 ð2F6 þ 2F7 þ F8 Þ  4c11 F9 9

þ

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  4F9 ¼ 0 Bc11 9 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

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

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 300
150 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 0 7 0 0 50:2
10 7 6 ½ABD 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

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

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).

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

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.

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

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 300
150 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

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

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.

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

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 sin sin dxdy W ¼ qo sin cmn sin 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 dy ¼ b sin > 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 D þ þ 2D Þ þ D cmn ðD 11 12 66 22 8 a3 ab b3 m ¼ 1n ¼ 1   M X N X ij 2p2 mncmn cij 2  i 2 ½n 2  j 2  m ½ i ¼ 1j ¼ 1 




m2 n2 ðMI ÞðNJ Þ D þ D 16 26 a2 b2

ð28Þ

g



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 2 cij ðMI ÞðNJ Þ þ 2m ncmn Nxy ðm2  i 2 Þðn2  j 2 Þ i ¼ 1j ¼ 1

 qo abc11  4

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

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 a b 2b 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 p2 ai bi m¼1 n¼1 mn a b 2a 2b

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  i 2 ½n2  j 2  ½ 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

i ¼ 1j ¼ 1

( 

M X N  X cij



j ðMI ÞðNJ Þ ðm2  i 2 Þðn2  j 2 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

M X N  4 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  i 2 ½n 2  j 2  m ½ i ¼1j ¼ 1  2 

m n2
2 D16 þ 2 D26 ðMI ÞðNJ Þ a b 2p2 mncmn

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

(  P

þ2m2 ncmn Nxy

M X N  X cij i ¼1j ¼ 1

M X N X

 cmn sin

m ¼1n ¼1

g



j ðMI ÞðNJ Þ ðm2  i 2 Þðn2  j 2 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

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

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: " M X N 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 F6 ðm3 nijÞ þ F7 ðmn3 ijÞ ð31Þ 2  i 2 Þðn2  j 2 Þ ðm i ¼ 1j ¼ 1

)# þF8 ðm2 njÞ ðMI ÞðNJ Þ 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.

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

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

2

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 ð2F6 þ 2F7 Þ 0 0 6 2ðF1 þ F2 þ F3 þ F4 Þ 7 9 6 7 6 7 40 6 7 ð2F þ 4F þ 16F þ F Þ  þ 2F Þ 0 0 2ðF 1 2 3 4 6 7 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 0 0 8ð4F1 þ 4F2 þ 4F3 þ F4 Þ ð2F6 þ 2F7 Þ 9 9 8 9 8 0> c11 > > > > > > > > > > > > > > > > > > > > > > = <0> = > < c12 > ¼
> > >0> > > > c > > > > > > > > 21 > > > > > > > > > ; > : ; : > c22 0

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

0

40 ð2F6 þ 2F7 Þ 9

2ðF1 þ 4F2 þ 16F3 Þ

40  ð2F6 þ 2F7 Þ 9

0

2ð16F1 þ 4F2 þ F3 Þ

0

0

8ð4F1 þ 4F2 þ 4F3 Þ



40 ð2F6 þ 2F7 Þ 9 0

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

3 7 7 7 7 7 7 7 7 7 7 5

8 9 9 8 98 2 0 0 0 >> c11 > c11 > > > > > > > > > > > > > > > > > > > > > > > > > > = < = < 0 2 0 0 c12 = 12
¼ F4 >> > > > > > > > > > > > > > > c c21 > 0 0 8 0 > 21 > > > > > > > > > > > > : ;: ; ; : c22 c22 0 0 0 8 98 8 9 ð32Þ > > p2 b c11 > > > > > > > > 0 0 0 > > > > > > > > 4a > > >c > > > => < < 12 > = 2 p b ¼ Nxx 0 0 0 > > > c21 > a > > > > > > > > > > > > 2 > > > > > > > > p b > : > ; ; : 0 0 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

3 40 ð2F 2ðF þ F þ F Þ 0 0 þ 2F Þ 1 2 3 6 7 6 7 9 6 7 6 7 40 6 7  ð2F6 þ 2F7 Þ 0 0 2ðF1 þ 4F2 þ 16F3 Þ 6 7 9 6 7 6 7 40 6 7 0 0  ð2F6 þ 2F7 þ F8 Þ 2ð16F1 þ 4F2 þ F3 Þ 6 7 9 6 7 4 40 5 ð2F6 þ 2F7 Þ 0 0 8ð4F1 þ 4F2 þ 4F3 Þ 9 9 8 98 c 9 8 c11 > 80 >> 11 > > > > > > > > >> > > > 0 0 0  > > > > > > > > > > > 9 > > > > > > > > > > > > > > > > > > > 80 c c 12 12 > > > > = = => < < < 0 0 > 0 9 ¼ Nxy ð33Þ
80 >> > > > > > > > c > c21 > 0 0 > > > > > > > > > > 21 > > > 0 > > > > > > 9 > > > > > > > > > > > > > >> > > > > > > > :  80 0 ; ; : 0 0 ;: c22 c22 9

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

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

E22 ¼ 10
109 Pa

v12 ¼ 0:30

G12 ¼ 13
10 Pa 9

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 500
300 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.

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

2. 3. 4. 5.

Turvey, G.J.; Marshall, I.H., Eds. Buckling and Postbuckling of Composite Plates. Chapman and Hall: New York, NY, 1995. Fung, Y.C. A First Course in Continuum Mechanics. Prentice-Hall: Englewood Cliffs, NJ, 1969. Timoshenko, S.; Woinowsky-Krieger. Theory of Plates and Shells. McGraw-Hill Book Co.: New York, NY, 1987. ISBN 0-07-0647798. 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.

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

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

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

u ¼ tan1 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 tan1(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).

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

(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 tan1( ) function corresponds to the proper quadrant, as determined by the algebraic signs of h and v.

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

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

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

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.

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

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.

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

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

cos ðhÞ þ E11 4



1  1 2v12 sin4 ðhÞ  cos2 ðhÞsin2 ðhÞ þ G12 E22 E11

Solving this equation for G12, we obtain: G12 ¼

E11 E22  Exx



Exx E11 E22 cos2 ðhÞsin2 ðhÞ  ðB:1Þ E22 cos4 ðhÞ þ 2v12 E22 cos2 ðhÞsin2 ðhÞ  E11 sin4 ðhÞ

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

Exx E11 E22 4E11 E22  Exx ½E22 þ 2v12 E22  E11 

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 rxx =2 c12 Aþ45j ¼ ðexx  eyy Þ In contrast, the shear stress and strain induced in the 45j plies are given by: s12 A45j ¼ rxx =2 c12 A45j ¼ ðexx  eyy Þ

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

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.

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

Appendix C Tables of Beam Deflections and Slopes

Tables of typical beam deflections and slopes can be found on the following pages.

629

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

Table 1

Deflection and Slopes of Cantilever Beams

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

Table 1

Continued

Source: Ref. 1.

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

Table 2

Deflections and Slopes of Simply Supported Beams

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

Table 2

Continued

Source: Ref. 1.

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