.co ·cn....Eo~ 0 ... ...._ co Q) N c.. 0....·.E co E... ca>cn -- 0-o -o$ ceo ~c cE ·en _J 00 Q)0>~ ..o .. Design and Optimization of Laminated Composite Materials. New York I Chichester I Weinheim I Brisbane I Singapore I Toronto . Haftka University of Florida Prabhat Hajela Rensselaer Polytechnic Institute @ A Wiley-lnterscience Publication JOHN WILEY & SONS. Zafer Gurdal Virginia Polytechnic Institute and State University Raphael T. INC. paper) I.1 Classification of Composite Materials I 5 1.2. § 1 Introduction 1 Copyright© 1999 by John Wiley & Sons. Danvers. . . p. New York NY 10158-0012. TA418. It is sold with the understanding that the publisher is not engaged in rendering professional services.1.2 Stacking Sequence Optimization I 29 Exercises I 31 References I 32 1.2 Material-Related Design Issues I 21 1. (212) 850-6011.COM. II. fax (212) 850-6008.1 Historical Perspective I 19 1. . Complementation. .2 Rule of Mixtures.2 Stress-Strain Relations I 34 2. . Raphael T. recording. and Interaction I 11 1. 2. . 1956. or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center. .1 2 Mechanics of Laminated Composite Materials . Prabhat. . Raphael T. Published simultaneously in Canada. Hajela. .. ISBN 0-471-25276-X (hardcover : alk. E-Mail:
[email protected] Single Isotropic Layer I 41 v . 3. stored in a retrieval system or transmitted in any form or by any means. fax (978) 750-4744.3. MA 01923. .1. .2.3 Laminate Definition I 16 1. This publication is designed to provide accurate and authoritative information in regard to the subject matter covered. . 605 Third Avenue. . Title. Structural optimization. . .. (978) 750-8400.3. . . III. No part of this publication may be reproduced.4. Includes index.L3G87 1998 620.4 Design Optimization I 24 1. electronic. Requests to the Publisher for permission should be addressed to the Permissions Department. 33 2. Design and optimization of laminated composite materials/Zafer Giirdal. . without either the prior written permission of the Publisher.3 Equilibrium Equations I 36 2. .3 Design of Composite Laminates I 19 1. photocopying. . .1. .2 Properties of Laminated Composites I 9 1. Haftka. Inc. xi This book is printed on acid-free paper. 222 Rosewood Drive.CONTENTS Preface .1. I. If professional advice or other expert assistance is required. Zafer. .2. All rights reserved. . Composite materials.1'18--dc21 98-22855 Printed in the United States of America 10 9 8 7 6 5 4 3 2 I Introduction to Composite Materials I 4 1. John Wiley & Sons. .1 Mathematical Optimization I 24 1. mechanical. . .9. the services of a competent professional person should be sought. scanning or otherwise except as permitted under Section 107 or 108 or the 1976 United States Copyright Act.2 In-Plane Response of Isotropic Layer(s) I 39 2. . . em. .2. .1.1 Material Orthotropy I 9 1.1 Governing Equations for Elastic Medium I 34 2. Prabhat Hajela. . Library of Congress Cataloging-in-Publication Data: Giirdal.1 Plane Stress I 39 2. Laminated materials.2. .4.1 Strain-Displacement Relations I 34 2.2 Fiber-Reinforced Composite Materials I 8 1. Haftka. VI CONTENTS CONTENTS vii 2.2.3 Symmetrically Laminated Layers I 43 2.3 Bending Deformations of Isotropic Layer(s) I 49 2.3.1 Bending Response of a Single Layer I 50 2.3.2 Bending Response of Symmetrically Laminated Layers I 52 2.3.3 Bending-Extension Coupling of Unsymmetrically Laminated Layers I 54 2.4 Orthotropic Layers I 61 2.4.1 Stress-Strain Relations for Orthotropic Layers I 62 2.4.2 Orthotropic Layers Oriented at an Angle I 64 2.4.3 Laminates of Orthotropic Plies I 69 2.4.4 Elastic Properties of Composite Laminates I 73 2.5 Properties of Laminates Made of Sublaminates I 83 Exercises I 87 References I 88 4.1.2 Mathematical Optimization Formulation I 133 4.2 Graphical Solution Procedures I 138 4.2.1 Optimization of Orientations of Layers I 138 4.2.2 Graphical Design of Coefficients of Thermal Expansion I 153 4.2.3 Optimization of Stack Thicknesses I 155 4.3 Dealing with the Discreteness of the Design Problem I 160 Exercises I 166 References I 167 5 Integer Programming . . . . . . . . . 169 3 Hygrothermal Analysis of Laminated Composites . . . 3.1 89 Hygrothermal Behavior of Composite Laminates I 90 3.1.1 Temperature and Moisture Diffusion in Composite Laminates I 91 3.1.2 Hygrothermal Deformations I 94 3.1.3 Residual Stresses I 99 3.1.4 Hygrothermal Laminate Analysis and Hygrothermal Loads I 102 3.1.5 Coefficients of Hygrothermal Laminate Expansion I 105 3.2 Laminate Analysis for Combined Mechanical and HygrothermaL Loads I 108 3.3 Hygrothermal Design Considerations I 118 Exercises I 124 References I 125 5.1 Integer Linear Programming I 170 5.2 In-plane Stiffness Design as a Linear Integer Programming Problem I 172 5.3 Solution of Integer linear Programming Problems I 177 5.3.1 Enumeration I 177 5.3.2 Branch-and-Bound Algorithm I 180 5.4 Genetic Algorithms I 184 5 .4.1 Design Coding I 185 5.4.2 Initial Population I 188 5.4.3 Selection and Fitness I 191 5.4.4 Crossover I 201 5.4.5 Mutation I 206 5.4.6 Permutation, Ply Addition, and Deletion I 208 5.4.7 Computational Cost and Reliability I 209 Exercises I 218 References I 219 6 Failure Criteria for Laminated Composites 6.1 . . . . . . 223 4 Laminate In-Plane Stiffness Design . . . . . . . . . . . 127 4.1 Design Optimization Problem Formulation I 128 4.1.1 Design Formulation of In-Plane Stiffness Problem I 128 Failure Criteria for Isotropic Layers I 226 6.1.1 Maximum Normal Stress Criterion I 226 6.1.2 Maximum Strain Criterion I 227 6.1.3 Maximum Shear Stress (Tresca) Criterion I 229 6.1.4 Distortional Energy (von Mises) Criterion I 230 VIII CONTENTS CONTENTS IX Failure of Fiber-Reinforced Orthotropic Layers I 231 6.2.1 Maximum Stress and Maximum Strain Criteria I 233 6.2.2 Tsai-Hill Criterion I 237 6.2.3 Tsai-Wu Criterion I 241 6.3 Failure of Laminated Composites I 245 6.3.1 Failure under In-Plane Loads I 249 6.3.2 Failure under Bending Loads I 257 Exercises I 258 References I 259 6.2 7 Strength Design of Laminates . . . . . . . . . . . . 261 8.2.1 Linear Problems I 30 l 8.2.2 Changes in Stacking Sequence I 306 8.2.3 Nonlinear Problems I 308 8.3 Flexural Stiffness Design by Integer Linear Programming I 310 8.3.1 Ply-Identity and Stack-Identity Design Variables I 310 8.3.2 Stiffness Design with Fixed Thickness I 313 8.3.3 Buckling Load Maximization with Stiffness and Strength Constraints I 317 8.3.4 Stiffness Design for Minimum Thickness I 322 Exercises I 328 References I 328 Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 331 7.1 Graphical Strength Design I 262 7 .I .1 Design for Specified Laminate Strain Limits I 262 7.1.2 Design of Laminates with Two-Fiber Orientations I 266 7.1.3. Design of Multiple-Ply Laminates with Discrete Fiber Orientations I 271 7.2 Numerical Strength Optimization Using continuous Variables I 274 Strength Design with Thickness Design Variables I 274 7 .2.2 Strength Design with Orientation Angle Design Variables I 286 7.3 Numerical Strength Optimization Using Discrete Variables I 288 Integer Linear Programming for Strength Design I 289 7.3.2 Genetic Algorithms for Strength Design I 292 Exercises I 294 References I 295 8 Laminate Design for Flexural and Combined Response . . . . . . . . . . . . . . . . . . . . . . . . . 297 7.2.1 7.3.1 8.1 Flexural response Equations I 298 8.2 Stiffness Design by Miki's Graphical Procedure I 301 PREFACE With rapid growth of the use of composite materials in many commercial products ranging from sports equipment to high-performance aircraft, literature on composite materials has proliferated. At the time of the publication of this book a simple search of a popular web site for books with the words "composite materials" in their title yielded more than 250 entries. Many of these titles are well-written textbooks on mechanics of composite materials and have been adopted by educational institutions for introductory courses. As the application of composites to commercial products has increased, so has the need for literature that focuses on the design aspects of these materials. However, the number of titles that focus on the mechanics of composites far outnumbers those dealing with design. In particular, books that focus on optimal design of composite materials virtually do not exist. It is the intent of this book to introduce readers to the emerging field of optimal design of laminated composite materials. The first and the foremost reason for writing this book was the desire to acquaint students with the latest techniques in the field. For many years, designers have treated optimization problems involving composite materials with continuous optimization techniques that were ill suited for these problems. The design of a composite laminate stacking sequence generally involves selecting discrete layer thickness and orientation angles-a discrete optimization problem. Researchers in this field have more recently focused on numerical and graphical methods useful for the solution of such problems; this book mirrors that focus. In particular, the book places emphasis on graphical design techniques developed by Professor Miki from Japan. These techniques allow representation of even the most complicated stacking sequences using two parameters bounded by a parabola and provide extremely valuable insight into the multiplicity of solutions available for laminate design problems. Another important motivation for the book was the need to provide condensed coverage that would be of use to the design engineer. Dexi respectively. we recommend that the course cover most of the material in Chapters 1. We have used the material in our respective institutions for a combined senior-level undergraduate and first-year graduate course for several years. the formulation and solution of the in-plane stiffness design· problem is demonstrated. the book may be used in aerospace engineering. The second and third chapters introduce the basic equations and assumptions used in the analysis of laminated composites under mechanical and thermal loads. Chapters 3. For a one-semester course with students who have no previous background in composites or optimization. In addition. The chapter concludes with an introduction to the terminology and formulation of mathematical optimization problems. The authors also appreciate the input and suggestions provided by various individuals. Two formal procedures. 4. Depending on the emphasis of the course. Chapter 4 formulates the in-plane stiffness design as an optimization problem and introduces a simple graphical technique for its so- lution. its natural vibration frequencies. which focus on thermomechanical. respectively. More recently. It is also possible to cover a combination of sections from the remaining chapters as the instructor sees fit. or 8. and mechanical engineering curricula. The first chapter reviews the types of composite materials in use and the terminology established for their description. Applications of composite materials have traditionally originated in weight-critical aerospace structures. Chapter 8 introduces analysis and design for bending requirements. and the buckling response of a simply supported laminate under in-plane loads. Thanks also go to the authors' respective universities for providing the opportunities to teach courses directly relevant to the book's content. 2. The authors wish to express their appreciation for many valuable suggestions from former students in courses that led to the development of this book. Chapter 7 describes the implementation of strength constraints in design optimization based on graphical and mathematical optimization procedures. which is heavily used in the book. namely integer linear programming and genetic algorithms. These include the transverse displacement of a simply supported laminate loaded by transverse loads. Chapters 6 and 7 address strength analysis and design. civil engineering. Researchers in composite materials are likely to benefit from state of the art methods introduced in the book.'"' PREFACE PREFACE XIII sign of composite materials and structures requires both a thorough understanding of the mechanics of laminated composites and familiarity with optimization techniques that enable designers to find practical laminate configurations in an efficient manner. A brief review of the design issues relevant to composite materials is included. They emphasize the computation of elastic properties as functions of variables that can be changed during the design process and the effects of such changes on response quantities such as stresses and strains. strength. Finally. a student wanting to learn about the application of optimization techniques to composite design will need to take a separate course in each subject. The types of composites considered in this book are then identified and their properties are discussed within the context of mechanics. . and parts of Chapter 6. This book has been developed for senior-level undergraduate or early graduate courses in numerical design methods for laminated composite materials. In addition. Commonly used failure criteria for laminated composite materials and their implementation for strength analyses are introduced in Chapter 6. and their fellow faculty members for providing valuable input and stimulating discussions. Japan for introducing us to the graphical representation of laminate optimization problems. engineering science and mechanics. the book has technical material useful for practicing engineers in related fields. may be added. Also provided is a technique to handle the discrete nature of the thickness and orientation design variables. 7. At present. The book combines the study of the mechanics of composite laminates with optimization methods that are most useful for the design of such laminates. Therefore. Special thanks are also due Professor Mitsunori Miki of the Doshisha University. In particular. This is somewhat difficult within the constraints of an undergraduate curriculum. and 5. with special emphasis on laminate design problems. it is probably possible to cover material from one more chapter. or bending design characteristics. suitable for handling discrete optimization problems specific to composite laminate design are introduced in Chapter 5. in . these materials have become popular in civil engineering infrastructure applications (such as bridge and building construction) and mechanical engineering applications (such as mechanisms and lightweight robotic structures). or a space structure. typically related to strength or stiffness. The tasks. changes in the structural dimensions (cross-sectional areas. The measure of goodness of a design depends on the application. Next. the best design often means either the lowest weight (or cost) with limitations on the stiffness (or strength) properties or the maximum stiffness design with prescribed resources and strength limits. member lengths) implemented to improve the performance may yield designs that violate the strength or stiffness requirements. Traditionally. 1 . For example. for reviewing some of the chapters of the book. James H. Walter Dauksher of Boeing. For a given application. stated so simply in two sentences. Professor Ron Kander of Virginia Tech. it is so difficult to satisfy these requirements that the first design ever identified to satisfy them becomes the final design. Sometimes these requirements are difficult to satisfy and may require many iterations by the designer. structural modifications that are likely to improve the performance or reduce the weight or the cost are implemented. He has sponsored research leading to many of the results reported in the textbook. Dr. while resources are measured in terms of weight or cost. In some cases. and his insight into design issues for composite materials enriched our appreciation of the subject.. of NASA Langley Research Center. Therefore. engineers have relied on experience to achieve such designs. Jr. Vasiliev of Moscow State University of Aviation Technology. Zafer Gtirdal Raphael T.'"" PREFACE particular. first a set of essential requirements are identified and designs that satisfy these requirements are obtained. while using the least amount of resources. are often extremely tedious and require a large number of iterations by the designer. be it a vehicle structure.i University. Haftka Prabhat Hajela f. The authors would also like to thank Dr. thicknesses. Starnes. Professor Valery V. I 1 INTRODUCTION Structural designers seek the best possible design. Professor Giinay Anla~+ of Bogazic. a civil engineering structure. most designers limit themselves to aspects of material response that are rather well defined for engineering applications. however. such as steel. is around 22 x 106 lb/in 2 compared to Aluminum's stiffness of 10 x 106 lb/in 2 . see. Therefore. and carry larger static loads than their metallic counterparts. Kirsch (1993). That is. Within the context of optimization. is not a task that can be taken lightly. none of these books covers in detail a subset of the field of structural optimization. respectively. The weight advantage may also be reflected in the performance properties described above by introducing specific stiffness and specific strength. for example. which has a chapter devoted to optimization of laminated composites. most designers limit themselves to configurations known to be manufacturable. Vanderplaats (1998).10 lb/in 3 • In addition to their weight advantage per unit volume. with the exception of the last one. but not all configurations designed to tailor the material properties are manufacturable. Composite materials are also known to perform better under cyclic loads than metallic materials because of their fatigue resistance. It was these advantages that initially stimulated widespread use of composites in structural design. One of the primary reasons for their popularity is their weight advantage. and Haftka and Gtirdal (1993). Formulation of the engineering design optimization. structural optimization.065 lblin 3 .3 2 INTRODUCTION INTRODUCTION Over the past two decades. In order to improve structural performance and meet the requirements of a specific design situation. There are also various composite material response mechanisms that are not fully understood and are still topics for continuing research. The use of mathematical optimization for design relieves the designer of the burden of repeated iterations and transforms the design process into a systematic well-organized activity. Many of the properties of composite materials are unfamiliar to many design engineers who have no formal training in composite materials. There are a number of different ways to manufacture a composite part or a structure. especially for weight-critical applica~ tions such as design of spacecraft and aircraft. the advantages of high specific stiffness and/or specific strength make composites more attractive than alloys. designers obtained a new flexibility through the use of variables that directly change the properties of the material. the weight or the performance becomes the objective function and the variables controlling the structural dimensions (such as thicknesses or cross-sectional areas of members) are design variables that are sized to achieve the best configuration. it is now possible to tailor the properties of the structural material in addition to structural dimensions. the fast pace of advances in composite material manufacturing technology is forging ahead of the present state of established design practices. there is a continued need for developing tools and methodologies for design of composite materials and structures to achieve maximal performance gains. mathematical optimization. see. As will be explained in · this chapter and in a subsequent chapter. and the methods of solution are topics covered in extensive detail in a number of textbooks. and therefore optimal design of structures has acquired a new meaning. has emerged as a powerful tool for structural design. However. However. Arora (1989). even if the stiffness and/or strength performance of a composite material is comparable to that of a conventional alloy. compared to the weight density of Aluminum which is 0. . the weight densities of high-strength Graphite/Epoxy and Glass/Epoxy are 0. Therefore. with the introduction of composite materials into a design problem formulation. Another area which is not fully investigated is the integration of the design and manufacturing aspects of composite structures. defined by dividing the respective property with the material density. namely optimization of structures made of composite materials. Use of composite materials in structural design has gained popularity over the past three decades because of several advantages that these materials offer compared to traditional structural materials. particularly in terms of optimal design. aluminum. the use of composite materials in structural design had significant implications in terms of the design problem formulation. Composite materials such as Graphite/Epoxy and Glass/Epoxy have smaller weight densities compared to metallic materials. structural members made out of composite materials may undergo smaller deformations.056 lb/in3 and 0. thereby limiting the performance gains that can be achieved through the use of composites. Beyond their weight advantage. and various alloys. For example. Stiffness of high strength Graphite/Epoxy. some composites provide better stiffness and strength properties compared to metals as well. Design of a material. for example.. Therefore. Therefore. which deals with either the maximization or minimization of an objective function subject to constraint functions. for example Niu (1992) and Zagainov and Lozino-Lozinski (1996). 4 INTRODUCTION 1.1 INTRODUCTION TO COMPOSITE MATERIALS 5 The application of the methods of mathematical optimization to the design of structures made of composite materials attracted the attention of many researchers. However, the approach has not been fully accepted by practicing engineers, primarily because of the impracticality of many designs obtained via the optimization process. Early design optimization research studies employed mainly extensions of the approaches used for designing structures made of traditional materials and lacked the ability to handle features that are unique to composite materials. For example, composite material design calls for use of design variables that assume values only from a discrete set of prescribed values. Use of traditional optimization techniques, which treat design variables as continuous valued, produces designs with limited practical value. Only recently has research on the design of composite structures started to introduce features that are important to .attain practical designs. This book attempts to address the unique design features associated with a limited class of composite materials, called laminated composite materials, via the application of the methods of mathematical optimization. In subsequent sections of this chapter, we first provide an introduction to the types of composite materials in use, with the intent of familiarizing the reader with general terminology in the materials field. The types of composites t.hat are specifically considered in this book are then identified, and the properties of such composites are discussed within the context of mechanics. A very brief and probably incomplete review of composite material design issues is included to warn designers about the pitfalls associated with tailoring the properties of a composite material. The chapter concludes with a brief discussion of the terminology and formulation of mathematical optimization problems, with a special emphasis on laminate design problems. the resulting composite materials may have the combined characteristics of the constituents or have substantially different properties than the individual constituents. As explained later, sometimes the properties of the composite even exceed those of the constituents. The following classification is given by Schwartz ( 1984). 1.1.1 Classification of Composite Materials One approach to the general classification of composite materials is based on the nature of the constituent materials. The constituents may be either organic or inorganic. The designation of organic refers to materials originating from plants or animals or materials of hydrocarbon origin (natural or synthetic) such as carbon. Inorganic materials are those that cannot be classified as organic matter, for example, metals and minerals. Historically, perhaps the most common structural material of organic nature has been wood. Wood is a composite material because it is composed of two distinct constituents: stiff and strong fibers surrounded by a supporting structure of softer cells. In fact, most modern composite materials imitate wood in that they consist of strong fibers embedded in softer supporting material. The supporting material in such an arrangement is commonly referred to as matrix material. Before getting into a more complete description of the modern composite materials, we continue with general descriptions. For most highly loaded single-material structures, the use of purely organic materials is avoided because of their generally low stiffness and strength properties. They are also sensitive to environmental effects such as temperature and moisture and have low resistance to chemicals and solvents. Despite their disadvantages, organic materials such as polymers are commonly included in composites to bring in certain characteristics that may not be achievable by using only inorganic constituents. For inorganic composites, the constituents may be all metallic, all nonmetallic, or a combination of the two. The most common inorganic materials used for highly loaded structures are metals. Despite their good stiffness and strength properties, metals have high specific weight-a disadvantage in weight-critical applications, such as aircraft and spacecraft. Therefore, metallic composites are used mostly when there is need for specific properties, such as resistance to high temperatures. As a matter of fact, in contrast to single 1.1 INTRODUCTION TO COMPOSITE MATERIALS The most general definition of a composite material is very closely related to the dictionary definition of the word composite, meaning made up of different parts or materials. Composite materials are constructed of two or more materials, commonly referred to as constituents, and have characteristics derived from the individual constituents. Depending on the manner in which the constituents are put together, 8 INTRODUCTION r ! 1.1 INTRODUCTION TO COMPOSITE MATERIALS 7 material structures, which are mostly metallic, most frequently used composite materials are made up of all nonmetallic constituents, which include polymeric constituents as well as inorganic glasslike substances. Polymer matrix materials are made of long chain of organic molecules, generally classified as thermoset or thermoplastic. The two categories have substantially different characteristics as matrix materials. Fibers preimpregnated with thermoset matrix (thermoset prepregs) are soft and malleable. They are tacky and can be draped easily, hence allowing fabrication of t:omplex shapes with ease. They require low processing temperatures but long curing times. They undergo a chemical change (cross-linking of polymer chains) during curing, and the process is irreversible. Thermoplastics, on the other hand, do not require curing and they can be reshaped and reused. However, thermoplastic prepregs are hard and boardlike and lack drape and tack; making them harder to handle. They also require very high processing temperatures to shape them, although processing time can be very short. Thermoplastics also offer high resistance to moisture and can operate at higher temperatures of up to 500°F (260°C) compared to thermoset operating temperatures of about 250°F (121 °C). Typical thermoset matrices include Epoxy, Polyimide, Polyester, and Phenolic materials. Among popular thermoplastics are Polyethylene, Polystyrene, and Polyetheretherketone (PEEK) materials. A more complete discussion of advantages and disadvantages of thermosets and thermoplastics is provided by Niu (1992). A more traditional classification than the one based on the nature of the constituents is derived from their form. Several examples of composite materials with different forms are presented in Fig. 1.1. Particulate composites are generally made up of a randomly dispersed hard particle constituent in a softer matrix. Examples of particulate composites are metal particles in metallic, plastic, or ceramic matrices. A widely used particulate composite is concrete in which gravel is mixed in cement. Flake composites, as the name suggests, are formed by adding thin flakes to the matrix material. Although flake dispersion in the matrix is generally random, the flakes may be made to align with one another, forming a more orderly structure compared to particulate composites. This alignment provides for more uniform properties in the plane of the flakes than with particulate composites. Typical examples of flake Particulate composite Flake composite Fiber reinforced composite Laminated composite Figure 1.1. Composite materials with different forms of constituents. materials are glass, mica, metals, and carbon. The size, shape, and material of the flake to be used depend on the type of application, which also determines the amount of matrix material. The matrix material (either plastics, metals, or epoxy resins) used in a flake composite may make up the bulk of the composite or be in small amounts just sufficient to provide bonding of the flakes. For example, aluminum flakes are sometimes used in molded plastic parts to provide decorative effects. In such a composite, the plastic matrix makes up the bulk of the composite. Fiber-reinforced composites (or fibrous composites) are the most commonly used form of the constituent combinations. The fibers of such composites are generally strong and stiff and therefore serve as the primary load-carrying constituent. The matrix holds the fibers together and serves as an agent to redistribute the loads from a broken fiber to the adjacent fibers in the material when fibers start failing under excessive loads. This property of the matrix constituent contributes to one of the most important characteristics of the fibrous composites, namely, improved strength compared to the individual constituents. This is elaborated upon in Section 1.2.2, which deals with the different mechanisms by which the properties of composites are obtained from those of its constituents. l INTRODUCTION 1.2 PROPERTIES OF LAMINATED COMPOSITES 9 The last form of composite materials is thin layers of material fully bonded together to form so-called composite laminates. The layers of a laminated composite material may be different single materials, such as clad metals that are bonded together or the same material, such as wood put together with different orientations. The layers may be composites themselves, such as fibrous composite layers placed so that different layers have different characteristics. This type of composite is the most commonly encountered laminated composite material used in design of high-performance structures and is the primary subject of this book. The various forms of laminated composite materials are discussed next. .~~~/ Random short fibers Oriented short fibers Plain Fibrous Layers Continuous fibers .7 1.1.2 Fiber-Reinforced Composite Materials An individual layer of a laminated composite material may assume a number of different forms, depending on the arrangement of the fiber constituent (see Fig. 1.2). The layers may be composed of short fibers embedded in a matrix. The short fibers may be distributed at random orientations, or they may be aligned in some manner forming oriented short-fiber composites. A typical example of a random short-fiber composite is fiberglass. Continuous fiber-reinforced composite layers are made up of bundles of small-diameter circular fibers. Typically, the radii of these fibers are in the order of 0.0002 in (0.005 mm), such as the radius of carbon fibers. The largest diameter fibers, such as boron fibers, are of the order of 0.002 in (0.05 mm). Continuous fiber-reinforced composite materials are commercially available in the form of unidirectional tape, with fibers aligned along the length of the tape. The fibers of the tape are preimpregnated with the matrix material, and for this reason the tape is sometimes referred to as a prepreg. As discussed in Section 1.2, this arrangement of aligning the fibers in a given direction ·provides the unique feature of the material properties of fiberreinforced composites. Another form of continuous fiber-reinforced composite layers is the woven fabric type composite, where fiber tows, which are large bundles of fibers (generally 10,000 or more fibers), are woven in two or more directions (see Fig. 1.2). Various fabric types with different weave features producing different material characteristics are available (see, for example, Middleton, 1990). 1.2 PROPERTIES OF LAMINATED COMPOSITES 1.2.1 Material Orthotropy Composite layer properties (such as strength, stiffness, thermal and moisture conductivity, wear and environmental resistance) strongly depend on the form of the reinforcement in the laminate. The directional nature of the fibers in a fiber-reinforced laminate introduces directional dependence to most of those properties. Materials whose properties are independent of direction are called isotropic materials. Conversely, those materials with different properties in different directions are called anisotropic. A special case of anisotropy is the existence of two mutually perpendicular planes of symmetry in material properties. Such materials are referred to as orthotropic. Fibrous composites with either oriented short fibers or continuous fibers are orthotropic in nature. In such composites, properties are defined in 6 (204) 11.59) 0.03 (7.1 along with the fiber volume fraction of the composite. for example.23 0. A quantity. Elastic properties of a unidirectional fiber-reinforced composite layer.27) . is commonly used to measure the amount of fiber in a composite material. "f.2 PROPERTIES OF LAMINATED COMPOSITES 11 the plane of the layer in two directions-the direction along the fibers and the direction perpendicular to the fiber orientation.3) 1. Em. Gm. 1 E and E2._ Complementation. and vm.17) 1. The rule of mixtures employs the volume fraction of the constituents to estimate the properties of the composite.26 Tsai and Hahn. The matrix portion of the composite. 0. are also isotropic.3 loaded along either the fiber direction.&:. E1 .p ~ Vf. the fiber-reinforced material shown in Fig. There are three different mechanisms Jl~Jo.66 .49 (10. Directions that are along the fiber and perpendicular to the fiber directions are commonly referred to as the principal material directions or principal axes of the material.. the fibers. As shown in the next section. XI Ef' Gf' v1 Em. The stiffness of the composite in the fiber direction is much closer to that of the fiber stiffness.6) 1.0 (76) 5. As discussed in the next section.81 (5. 1. A typical example of an orthotropic material property of unidirectional fiber-reinforced composites is the stiffness._o_htain_properties of the composite from those of its constituents. Constituents Graphite/Epoxy Graphite/Epoxy Boron/Epoxy Aramid/Epoxy Glass/Epoxy 106 psi (GPa) 106 psi (GPa) 106 psi (GPa) 26.60 (38. xl' or transverse to the fiber direction. and the stiffness perpendicular to the fiber direction is governed mostly by the properties of the matrix material. (1.. these properties are related to the amount of fiber present in the composite layer. fully describe the mechanical properties of an orthotropic layer in its plane (more detailed discussion of the stiffness properties is given in Section 2.6 0.30 (8. which usually have much higher stiffness than the matrix material. Others have to be measured.1. properties of different natures are determined through different m:ecnanisms. along the fiber and transverse to the fiber directions. ~ '-"-. When the material is loaded along the fiber direction. For example.34 0.80 (5.60 (4. For all practical purposes.5 0. Since the amount of deformation under specified load reflects the stiffness of a material.5) 0. which holds the fibers together.-rj ML C+Urv.3 (181) 20. which must satisfy VI+ Vm = 1.33 (2.4).0 (138) 29. Typical stiffness properties of various unidirectional composite materials systems are given in Table 1. However. One of the ways to estimate composite material properties is summation of the properties of the individual constituents based on their contribution to the overall material volume. referred to as the fiber volume fraction.20 (8. the properties of the composite are no longer isotropic.2. This method is commonly referred to as the rule of mixtures. and Interaction \ ~ ·g . 1980. Four elastic stiffness properties. we have a fiber volume fraction V1 and a matrix volume fraction Vm.04 (7.50) 1.7 0. the deformation is smaller compared to the deformation under a load of the same magnitude applied in a direction perpendicular to the fiber.14) 1/12 vJ 0.28 0. Consider. commonly referred to as the engineering constants.2.1) . x2 . when combined together.3.2 ( \_. and the shear modulus G 12 and Poisson's ratio v in the 12 Table 1.96) 2.IV INTRODUCTION 1. plane of the layer. '" r.30 0. vm Figure 1.68 (18. in the case of a continuous fiber-reinforced composite layer. These are two Young's moduli. is generally isotropic. respectively. 1. the unidirectional composite has different stiffness properties along these two mutually perpendicular directions. Gm.45 0.30) 0.10) 0. Properties of various fiber-reinforced composite layers E1 Material T300/5208 AS4/3501 8(4)/5505 Kevlar49/Ep Scotchply1002 Source: E2 1.0 Rule of Mixtur-es. Some of the properties of a fiber-reinforced composite can be estimated from the properties of its constituents by simple calculations. . However. <51 = om= o.7) p :--.3) unit volume Equation (1.1 The elastic modulus and the weight density of a typical graphite fiber and an epoxy resin are E1 = 230 GPa. Poisson's ratio and the shear modulus for the composite can be obtained in the following form: V12 = v1~+ Vm Vm and _l_=_!i+ Gl2 G! vm em· (1.4.6) 'f m Since the cross-sectional areas are proportional to fiber fractions ~A.2 PROPERTIES OF LAMINATED COMPOSITES 13 Based on the rule of mixtures. (1.5. a property p is estimated from the constituent properties.2. 1. p1 = 17.: unit volume Example 1. the calculations and assumptions may require a more sophisticated model that includes the fiber shape and the amount of fiber in the matrix. we obtain Eq. For the elastic modulus in a direction perpendicular to the fibers. Rule-of-mixtures model of a transverse stiffness of a composite layer.4) becomes oAE1 oA1Et oAmEm L + -y--.4.2).3).5) vm (1. the longitudinal (fiber direction) stiffness property. Assuming the 0 stiffness representation Figure 1. using this rule of mixtures as El =EI~+EmVm. Figure 1. Eq. . p and Pm' as 1 P =P1~+ Pm Vm matrix fiber matrix =P1~+ Pm(l - ~). applied to the flexibility liE.= (A1= (1. Pm = 12. and the total end-deformation o of the composite is identical in the fiber and the matrix.2. Eq. (1.2.2 kN/m 3 .! stiffness representation ' o= B1 +B2 (1.2) i-··-:::1~)------p For example.0 kN/m 3 . (1. see Fig.'.2. where A is the cross-sectional area.45 GPa. Equation (1. Am= VmA). the total deformation at the point of load application in a direction perpendicular to the fiber is the sum of the deflections experienced by the fiber and the matrix. 1.2. (1.6) may be interpreted as the rule of mixtures. The total force causing the deformation is carried partly by the fibers and partly by the matrix P=P1 +Pm. good estimates for the modulus are obtained by modelling the fiber and the matrix as two elastic springs connected in series.12 INTRODUCTION 1.2.3) may be derived with analogy to the calculation of the overall stiffness of two springs connected in parallel. The resulting expression for the transverse modulus (derivation is left to the reader as an exercise) is _!___~ E-E+E. . Using similar arguments.2. (1. .-t. With this model. and Em= 3." !--. The fibers are assumed to be fully bonded to the matrix.4) Assuming that each constituent acts as an axial bar with a forcedisplacement relation o= PU AE. see Fig.-·' .2. EP of the composite may be calculated from the Young's moduli of the constituents E1 and Em.2.2. respectively. Rule-of-mixtures model of a longitudinal stiffness of a composite layer.2.2. 2 -y.5. 3 times that of the matrix material specific stiffness..6 as a function of the fiber volume fraction.1. Note that p is the weight density of the composite system.6. The composite longitudinal specific stiffness (E/p) normalized by the fiber specific stiffness (E/p1 = 13.8 v. Therefore.6 0.5 (GPa).1.2 0. This is demonstrated later in Example 1.0 + 5. (1. Finally.___ 0 0.5226 v1 . E 2. where the maximum value is determined from the geometry of the fiber packing in the unit volume.14 INTRODUCTION 1.45 + 226.935 X 105 2 m. as a function of the fiber volume fraction 0. as the fiber volume fraction increases the density of the composite also increases linearly according to the rule of mixtures equation for the composite density.0 ::f" 5. From Eq. Therefore. in which we formulate a simple design optimization problem. the longitudinal modulus and the weight density are E 1 = Er~+ Em(l.226.4. = !t0.4 (a) 0. plot the longitudinal specific modulus.5 2 = 0. 0 0.1.6 = 3.4.8 0. at the end of this chapter) which allow larger fiber volume fractions. in terms of the fiber volume fraction. E 1.76.52 =o. and the composite density. weight considerations may favor low fiber volume fraction composites as long as stiffness requirements are met. v.. 0. E/p. The magnitude of the longitudinal specific stiffness at that point is about 84% of that of the fiber material. the maximum possible fiber radius is equal to half the unit edge of the volume. we point to the fact that. for example.7854. E1 = p (3.~) vf"._--.J.6) the transverse modulus is E2 = ~Em Figure 1.Vr) E1 230 . since the fiber weight density is larger than that of the matrix.2. and from Eq. vrx = 1t 0.J.0 ~ vf~ v__rax for a unidirectional fiber-reinforced Graphite/Epoxy composite system. 1. The maximum longitudinal and transverse stiffnesses are achieved for the largest allowable fiber volume fraction of vrx = 0. and the transverse specific stiffness (E2 / p) normalized by the matrix spe- . Even larger values of the specific stiffnesses may be possible for other fiber packing models (see.55 Vr The specific stiffnesses are then given as.V1) = 1~.2 V 1 (kN/m ). and the magnitude of the transverse specific stiffness is more than 3.7854. 3 P = p!Vf+ Pm(l.7854.2 (m) ' E2 p ( ) 7..2.4 (b) 0.2).2 __!__-1. Based on the packing geometry provided in Fig. E!Em = 793.4. (1. 1.8 3 0.55 V (GPa) 1 +--.55 Vd X 10 vr 6 12. E 2 /p.45 + 226..17806 v 1 The fiber volume fraction can vary from zero to the maximum value of vrx. and the transverse specific modulus. even though the largest stiffnesses are achieved for large fiber volume fraction composites. cific stiffness (Em!Pm = 276 km) are shown in Fig. 1. Exercise 2. + ( 1 . 2. Normalized composite longitudinal (a) and transverse (b) specific stiffnesses as a function of the fiber volume fraction.37 x 106 m).6 0.2 PROPERTIES OF LAMINATED COMPOSITES 15 packing of the fibers to be represented by the unit volume shown in Fig. p. We first express the specific moduli in the longitudinal and transverse directions.. 16 INTRODUCTION 1.2 PROPERTIES OF LAMINATED COMPOSITES 17 Complementation is another mechanism for the composite material to inherit properties from those of its constituents. Complementation means that one of the constituents contributes a distinct property that is not present in other constituents, and the resulting composite acquires this property. For example, most matrix materials do not conduct electricity. Combining such matrix materials with fibers having good electrical conductivity results in a conductive composite. Also, many polymer matrix composite materials are sensitive to environmental conditions such as moisture. Adding a surface layer that seals the composite against such effects is an effective solution and demonstrates the complementation mechanism. The last mechanism, interaction, is one of the most important mechanisms for obtaining superior composite properties. It is the only mechanism that can cause a composite property to exceed those of the individual constituents. For example, the tensile strength of a glass fiber-reinforced plastic is larger than both the tensile strengths of the plastic matrix and the glass fiber. The explanation of this phenomenon is in the interaction of the fiber and the matrix via load transfer from one to the other. The tensile strength of a fiber is generally governed by defects in the fiber, which are likely to repeat over the length of the fiber. For an unsupported fiber, failure of a single defect is enough to limit the load-carrying capability. On the other hand, for fibers supported in a matrix, upon failure of a fiber due to a defect at one location, the load in the fiber is locally transferred through the matrix to neighboring fibers. Therefore, the load can be increased beyond the first fiber failure load to a point where multiple fiber failures degrade the load-carrying capability of a larger region of the composite and result in an unstable growth of the fiber failures. The stacking sequence lists fiber orientations measured_from a reference axis of the laminate. If the orientation is counterclockwise from the reference direction, it is considered to be positive. The standard stacking sequence lists orientations of the different layers, starting from the top of the laminate to the bottom, in a string separated by slashes. For a laminate with N layers, each made of the same composite material and of the same thickness, t, starting with the top layer with a fiber orientation e" the laminate is represented as [9/9/ · · · ;eN]. ( 1.2.8) 1.2.3 Laminate Definition In the preceding paragraphs we discussed the properties of a single fiber-reinforced composite layer. As stated earlier, the main emphasis of this book is on the design of laminated composite material where the layers of unidirectional fiber-reinforced composites are stacked on top of one another. Such laminates are described according to a standard notation called stacking sequence, which is described in the following. The thickness of each layer, t, in a consolidated form in the laminate is generally provided by the manufacturer's specifications. The total thickness, h, of the laminate is h = tN. Layers oriented at an angle from the reference axes of the laminate are called off-axis layers. When the orientation of a layer coincides with one of the reference axes of the laminate, e = oo ore= 90°, that layer is referred to as an on-axis layer. When several layers with the same orientation are adjacent to one another, it is common to group them together and represent the total number of adjacent layers with the same orientation as a subscript to that particular orientation. For example, the laminate [Oi45 4 / - 45 2] has the top two layers oriented along the reference axis of the laminate, followed by four layers oriented at 45° from the reference axis, followed by two layers of a -45° orientation. When a group of layers is repeated, then the number of repetitions is used as a subscript to the repeating group enclosed in parenthesis. For the following laminate, [Oi(02 /45 2 /90)/0 2], sandwiched between the two layers of oo on the top and two layers of oo at the bottom, the 02 /45 2 /90 group is repeated three times. A laminate is symmetric when the fiber orientations of the bottom half of the laminate are mirror images of the fiber orientations above the mid-plane of the laminate, for example, [-45/30/0/45/45/0/30/45Jr (the subscript T is used after the bracket to indicate that the designation is for the total laminate). Symmetric laminates with an even number of layers are represented by the portion of the stacking sequence above the laminate mid-plane followed by a subscript s after the closing bracket, [-45/30/0/45],. If the layers within the brackets are repeated, the number of repetitions can also be placed after the 18 INTRODUCTION 1.3 DESIGN OF COMPOSITE LAMINATES 19 bracket before the subscript s. That is, the laminate [4510z14510zl 4510z!Oz14510z14510z145Jr is represented in a compact form as [45!02] 3s. In some situations it is desirable to place a negative 9 orientation for every occurrence of the positive 9 layer. Such laminates are referred to as balanced laminates. Pairs of positive and negative orientations do not have to be placed adjacent to each other, but if they are adjacent, the laminate designation may be condensed by putting a plus-minus sign in front of the orientation angle. For example, the [30/ -30/30/ -30h laminate is simply designated as [± 30zJr The [30/30/30/-30h stacking sequence, on the other hand, is designated by [+30]s. In addition to the classifications presented above, several unique cases of laminate stacking sequence definitions are sometimes used. Laminates that have alternating orientations of 0° and goo plies are called cross-ply laminates. In cross-ply laminates, grouping of more than one layer with the same orientation angle is allowed. For example, the symmetric laminate [g02/0J, is a cross-ply laminate although the first two layers near the surfaces of the laminate do not alternate but are of the same orientation angle of goo. Another special case is the angle-ply laminate. All the layers of an angle-ply laminate have the same fiber orientation angle, say 9 = a, with an alternating sign. For example, a 30° symmetric angle-ply laminate could be [30/-30/ 30/-30], (or [±30JJ Finally, a laminate is said to be antisymmetric if the magnitude of the ply orientation angle below the laminate mid-plane is a mirror image of the ply orientations above the mid-plane with signs reversed. For example, [-45/45/ -45/45/ -45/45/ -45/45Jr is an antisymmetric laminate. Antisymmetric laminates with a mixture of off-axis ply orientations (i.e., 9 * oo or goo) are also possible, such as [+301+45/ +451+30h. Note that this last laminate cannot be designated as [+301+45]. because of antisymmetry with respect to the mid-plane (expanded representation of such a symmetric laminate would have been [+301+45/±45/±30h). Antisymmetric laminates always have an even number of equal thickness layers. In the case of cross-ply laminates, antisymmetry implies change of the ply orientation angle from oo to goo or from goo to 0° from one side of the laminate mid-plane to the other. For example, [O/go;o;go]r and [OfgOiOigO]r laminates are antisymmetric cross-ply laminates. 1.3 DESIGN OF COMPOSITE LAMINATES Laminated fibrous composite materials are finding a wide range of applications in structural design, especially for lightweight structures that have stringent stiffness and strength requirements. While they are attractive replacements for metallic materials for many structural applications, the analysis and design of these materials are considerably more complex than those of metallic structures. As mentioned earlier, finding an efficient composite structural design that meets all requirements of a specific application can be achieved not only by sizing the cross-sectional areas and member thicknesses, but also by tailoring of the material properties through selective choice of orientation, number, and stacking sequence of the layers that make up the composite laminate. 1.3.1 Historical Perspective The field of composite design engineering has continued to evolve over the years. In spite of the distinct advantage of composites for tailoring material properties, most early applications of composites were aimed strictly at weight reduction. Metals were replaced with lighter composites with little or no emphasis placed on tailoring the properties. In some instances, designers created quasi-isotropic laminates that largely suppressed the directional properties of the unidirectional layers and made the laminate behave in a manner similar to that of an isotropic material. One such laminate has equal percentages of 0°, +45°, -45°, and goo layers placed symmetrically with respect to the laminate mid-plane, for example [±45/gO/Ol,. Like isotropic materials that have the same elastic properties in every direction, quasiisotropic laminates have elastic properties that are independent of the direction in the plane of the laminate (an in-depth description of quasiisotropic laminates is provided in Chapter 2). Quasi-isotropic laminates were, therefore, a convenient replacement for isotropic materials in weight critical applications; weight savings could be achieved by simply replacing the isotropic material with a similar stiffness laminate that was lighter and probably stronger. For example, the elastic modulus of a [±45/0/gOL T300/5208 Graphite/Epoxy laminate is about E = 70 GPa, which is less than half of the stiffness of the unidirectional material. This stiffness value is comparable to the elastic modulus of Aluminum, 20 INTRODUCTION 1.3 DESIGN OF COMPOSITE LAMINATES 21 which is roughly Eat = 73 GPa. However, the weight density of the Graphite/Epoxy is approximately p = 15 kN/m 3 compared to the weight density of Aluminum, Pat = 26 kN!m\ yielding a specific stiffness of Elp = 4.7 x 106 m for the composite laminate versus E/p = 6 2.8 x 10 m for Aluminum. Therefore, the specific stiffness of the Graphite/Epoxy is 1. 7 times that of the aluminum. Such retrofitting requires a minimal amount of redesign effort and was suggested for aircraft structural components with minimal change in the structural configuration-so-called black aluminum structures. As the number of design engineers with formal training in composite materials increased, tailoring of material properties gained more acceptance. By varying the percentage of layers in the laminate with different orientations, material properties of the entire laminate in different directions can be adjusted to meet the requirements of a design situation. As an example, for a unidirectional material with a fiber direction modulus to transverse modulus ratio of 10, the laminate stiffness ratio E/Ex of an angle-ply laminate [±9], can be varied from 0.1 to 10 by varying the fiber orientation angle 9 from 0° to 90°. Similarly, the bending stiffness of a laminate can be altered by rearranging the relative through-the-thickness location of layers with various orientations. For example, although the in-plane stiffnesses of laminates [Oz190], and [90z10L are identical (same number of layers with same orientations) as will be shown later, the bending stiffnesses of these two laminates are markedly different. The design challenge is, therefore, to find the stacking sequence of a laminate, with different properties in different directions, so as to maximize the utility of the directional nature of the material properties. Despite the weight savings and the material property-tailoring advantages of laminated composites, one factor remains a serious obstacle for widespread use of composite laminates-cost. The cost per pound of a typical high-performance composite material is still much higher than the costs of most structural metals. Therefore, the principal supporter of the composite materials technology so far has been the aerospace industry, where weight savings are often crucial to good performance and can also provide substantial savings in operating costs. In order to make composite structures competitive with metallic counterparts, there is a need to reduce the overall product cost. One area that seems to have substantial potential for savings is the manufacturing of composite parts. Based on advances in various manufac- turing techniques, such as filament winding, thermoforming, and towplacement, it is possible to make net shape components. Many manufacturing processes produce parts that need additional machining for final shape or have limitations on producing the desired shape, requiring manufacture of the component in two or more pieces and subsequent assembly. Net shape parts are largely ready to use in the final product and greatly reduce assembly and manufacturing time. The challenge for composite designers is to incorporate manufacturing considerations into the design process. This is an area in which previous design work has lagged, but is now gaining momentum. 1.3.2 Material-Related Design Issues Even if one does not consider manufacturing related aspects of designing a composite part, there are a number of other issues that makes the design task complex and in some instances intractable. The mechanics of laminated composite materials is generally studied at two distinct levels, commonly referred to as macromechanics and micromechanics. At the macromechanical level the properties of the individual layers are assumed to be known a priori. Macromechanics involves investigation of the interaction of the individual layers of a laminate with one another and their effects on the overall response quantities of the laminate. For example, elastic stiffness properties and the influence of temperature and moisture on the response of laminated composites can be predicted well by macromechanics. Compared to isotropic materials, we need to deal with a more complex model that addresses the material orthotropy and anisotropy and requires more material constants for characterization of the mechanical response of the laminate. However, macromechanics of composite laminates is reasonably well understood in formulating the stiffness analysis of laminates. For a given stacking sequence, the stress-strain relations of a composite laminate (commonly referred to as the constitutive equations) can be derived, and the various coupling mechanisms between the in-plane and out-of-plane deformation modes of a composite laminate can be explained. Thus, the use of macromechanics formulation in designing composite laminates for desired stiffness characteristics is well established. In the first two chapters of this book, the development of macromechanics-based analysis for the stiffness and hygrothermal Use of such analyses along with automated design tools for the in-plane problems are then discussed in Chapters 3-5. on the other hand. Composite properties that must be predicted on the basis of micromechanics.3 DESIGN OF COMPOSITE LAMINATES 23 properties of laminated composites is discussed. the transfer of applied loads from one layer to another creates a complex state of stress between the fibers and the matrix near the layer interface region that is likely to cause failure at the fiber-matrix interface. which are usually lower in magnitude than the primary stresses along the applied load direction. the factors that go into deciding the fiber and matrix materials are not limited to stiffness properties. initiation of these failures is mostly related to microscopic events. at this point we would like to caution the reader about the varied nature of failure modes that can be observed in laminated composite structures. and holes that give rise to critical secondary stresses are ignored. Most engineering structures include local details such as notches. Such through-thethickness failures lead to delaminations at the free edges of the laminate that may affect the in-plane stress state. For example. and thickness variations in the form of ply drop-oft's that give rise to stress concentrations and threedimensional local stresses. holes. and failure of a laminate may involve a complex sequence of failure events in different layers. The failure models discussed in Chapter 6 are suitable for a range of applications. As such. for most engineering applications the elastic stiffness properties of a composite layer (Ep ~' v 12. cost. However. such as the delaminations between the composite layers. such as the throughthe-thickness stresses. that are unique to laminated composite materials. unfortunately. we have shown that the elastic properties of an individual layer can be derived from the elastic properties of the fiber and matrix constituents based on the rule of mixtures. and a proper modeling of those failures can only be made at the micromechanical level. Even though such failures appear at the macroscopic level. there are secondary stresses (i) in the plane of the laminate in a direction transverse to the loading direction generated due to Poisson's effect and (ii) in the through-the-thickness direction of the laminate due to the free edge at hole boundary. fatigue) which are more difficult to include in design optimization. the primary stress field coincides with the direction of loading in the plane of the laminate. Consequently. are not as well understood. and G 12) are determined through tests. in the case of delaminations. The level of accuracy of those equations is. Rather than using the rule of mixtures. for a two-dimensional plate with a circular hole under uniaxial loading. In this book we limit ourselves to only constant thickness laminates without discontinuities under in-plane and bending loads. Earlier in this chapter. Stresses in a laminate generally vary in different layers. Moreover. but involve other considerations (such as response to environmental effects. For most engineering design applications. The common approach taken by designers is to ignore the micromechanical nature of the failures and to limit the analysis to phenomenological failure models that address the specific features of certain classes of commonly observed laminate failures. in some cases this decision is made after experimenting with different fiber and matrix combinations in automated design optimization and evaluating the effects of the different combinations on the overall design. damage tolerance. failure may be predicted by comparing stresses in the layer to the strength of the individual constituents. notches. using a level of detail for analysis that addresses the fiber-matrix interface stresses is unrealistic. Thus.22 INTRODUCTION 1. there are only a handful of cases where the failure of a composite laminate is dictated by the failure of a single layer. For example. Nevertheless. These models are discussed in Chapter 6 along with a short discussion of the nature and modes of failures in laminated composite materials. However. Moreover. the laminate failures are related to a small number of failure modes that are well correlated with the stress and strain levels in the individual layers of the laminate. Composite materials fail due to the failure of either the individual constituents (fiber or matrix) or their interface. Only failure modes that are associated . local details in the form of ply drop-offs. not as good as the laminate constitutive equations. In a single fibrous composite layer under a complex loading. Micromechanics deals with the interaction of the fiber constituent with the matrix material. there is no well-accepted analytical expression for the determination of the strength of a laminated composite material based on the strengths of the individual constituents. Even though the primary loading is in the plane of the laminate. failures in such regions of the laminate may be due to these secondary stresses. In this book. there are failure modes. However. the selection of the fiber and matrix combination that will suit the needs of a design situation is largely a decision made prior to the design optimization. in which case they are called inactive..4. The standard form of the optimization problem is written as where the elements of the vectors xL and xu are the lower and upper bounds on the values of the design variables. note that the constraints are written with a zero on the right-hand side and for a particular choice of the sense of the inequality.3) can be converted into £xu. objective function. The maximization of the goodness is generally performed within some limits that constrain the choice of design. an optimization problem has design variables.4 DESIGN OPTIMIZATION The increased number of design variables is both a blessing and a curse for the designer. an inequality relation in material strains. .4. while inequality constraints can be active or may be satisfied with a margin. To allow for the case that some of the design variables may be discrete or limited otherwise. All satisfied equality constraints are active.4. The possibility of achieving an efficient design that is safe against multiple failure mechanisms. For example. £xu~ £x' (1. gix) ~ i = 1. which are the parameters that are changed during the design process. n8 .. such that h. We use a vector x with n components to describe the design variables. (1. in which the op- .£xu~ 0. we may wish to maximize stiffness for a given laminate thickness rather than minimize thickness subject to stiffness constraints. it is easy to arrange for a zero on the right-hand side of the equalities or inequalities. At a given design point x.4. and the sense can be changed by multiplying the inequality by -1. we just need to change the sign of the objective function. . but the increased number of variables brings the additional burden of selecting those design variables that are important and identifying their values for the best solution of the design problem. an optimization problem has an objective function which measures the goodness or efficiency of the design. That is. = 1. and constraints is as follows. To convert a maximization problem into a minimization problem. xL ~x~xu. Similarly.. Such limits are called constraints.2) 1.Ex~ 0 (1. the notation g(x) for inequality constraints and h(x) for equality constraints.24 INTRODUCTION 1. 1. . Obviously. we use the notation f(x) and for constraints. For the objective function.4..4) and then into £x. Design variables can be continuous or discrete depending on whether they can take values from a continuum or are limited to a number of discrete values. There are more controls to fine-tune the structure to meet design requirements. For example. we use X to denote the domain of these design variables. In many problems we may wish to maximize the objective function.1 Mathematical Optimization In general. j (1. Note that in this standard form we minimize rather than maximize the objective function. A special case of discrete variables are integer variables. Finally.1) 0. An inequality constraint for which the equality condition holds at a particular design point x is called an active constraint. instead of maximizing f(x) we can minimize -f(x). minimize f(x) XE X The set of design points in X that satisfy all the constraints is called the feasible domain. The commonly used notation for design variables. a constraint may be satisfied or violated. 'ne. coupled with the difficulty in selecting the values of a large set of design variables makes mathematical optimization a natural tool for the design of laminated composite structures.4 DESIGN OPTIMIZATION 25 with in-plane stresses in plate-type laminates without discontinuities under in-plane and bending loads are discussed. Formulation and graphical representation of a design problem is shown in the next example.(x) =0. require that E 1 :?: 0. the lowest weight would be achieved by reducing the V1 to zero.2).2.2 ____ j __.45 + 120.2 E1 and £ 2 :?: 0.2.83 V{. formulate the optimization problem to minimize the composite material weight as a function of the fiber volume fraction. f(~) = p = PFJ+ Pm(l - g2 (V ).55 VI (GPa).7a along with a horizontal line that indicates the 20% limit on the modulus.7854 in Example 1.45 GPa. and of the epoxy resin are Em = 3.15 1 _ _ E2 I E or v.7. The longitudinal modulus of the composite should at least be 20% of that of the fiber modulus. respectively. respectively.1 kN/m 3 .2 0.7b. 180.8 v. = 12 + 2. Plot the stiffness constraints to determine the optimal solution for the fiber volume fraction graphically. such that g 1(Vr) EI =0.5 0. Pm = 12.8 124.1 For the fiber packing geometry presented in Fig. E2= E1Em V1Em + (1.83 Example 1.55 V1 ~ 0. The vertical dashed line in the figure indicates the value of the fiber volume fraction at which the longitudinal composite stiffness is equal to 20% of the fiber modulus. 1. 1.0 kN/m3 ..3 0.E f or 21.4 E11Et The constraints on the longitudinal and transverse stiffnesses.4 and fiber and matrix properties of Kevlar and Epoxy provided below. we pose the problem in the following form: minimize f(~) 0 0. (1. The density of the composite to be minimized is the objective function j(~) given in terms of the fiber volume fraction using Eq.18.4.15 E 1• 0. v. 0:::.V1) El 427. However.) = 12 + 2.6 0.1 ~ (kN/m 3). For values of V1 less than approximately 0. and the transverse modulus should be larger than 15% of the composite longitudinal modulus. the longitudinal stiffness constraint is violated.55 V1 (GPa). 1. .180.2. E 1 and £ 2. p1 = 14. Similarly.35.4 (b) 0.4 DESIGN OPTIMIZATION 27 timal solution along with active constraints is identified based on visual inspection of the design space.30.1 V1 Figure 1.120. Since the composite weight density linearly increases with the fiber volume fraction.0. The elastic modulus and the weight density of a typical Kevlar fiber are E1 = 124 GPa.16 < VI _ 10 . an all-epoxy material would not satisfy the requirement on the longitudinal modulus to be greater than 20% of the fiber modulus. A plot of the longitudinal composite modulus normalized by the fiber modulus is shown in Fig. where v_rax is calculated to be 0. 0.29 -0. V1 ~ Vj.______ _ I 0.1 0 I I Using the standard formulation of an optimization. a plot of the transverse composite modulus normalized by the longitudinal composite modulus is shown in Fig. Normalized logitudinal and transverse stiffnesses versus fiber volume fraction.120.V1) = 3.1.26 INTRODUCTION 1. The stiffnesses that are constrained are given by E 1 = EF 1 + Em(I . such as stiffness. Based on the these constraints. tailoring of the material properties is also associated with maximization or minimization of a performance criterion. finding a rounded-off design that does not violate any constraint is often difficult. In either case. It is usually safe to round all thicknesses up. and the vertical dashed line shows the value of the fiber volume fraction at which the transverse composite stiffness is equal to 15% of the longitudinal composite modulus. When ply thicknesses are used as design variables. There are a number of efficient classical algorithms available for the solution of these linear and nonlinear programming problems. if two thickness design variables have values of 3.4.18. and the corresponding solution techniques are referred to as linear programming techniques.18 ::.1) is also called a nonlinear program. When some of the design variables are integer variables. and the desirable response quantities. the usefulness of these algorithms and their efficiency is determined to a large extent by the characteristics of the design space. and stacking sequence of plies that make up the composite laminate.1 and 4. (1. this problem is solved by introducing 0-1 or binary design variables. Most commercially available composite materials come as a unidirectional tape with a fixed thickness. this is an integer problem. The horizontal line in the figure intersects the normalized constraint function once more at a value of the V1 slightly greater than the v_rax which is not a physically acceptable solution. the application of classical mathematical programming algorithms becomes quite cumbersome..7 times the commercially available thickness. the optimal number of layers in the laminate needs to be determined. the reader is advised to refer to one of the textbooks on the subject matter (see. The minimum composite weight is. 0. the feasible range of fiber volume fraction for this problem is approximately 0. number of plies. it may be difficult to write expressions for the objective function and constraints using these variables without the use of multiple "if-then" clauses. The performance criterion may be the weight of the laminate. For a detailed discussion of various design optimization algorithms. Despite the integer nature of the problem. ( 1. they usually control the total thicknesses of contiguous plies of the same orientation. This was partly due to the unavailability of easy-to-use commercial integer programming software and the high computational cost of solving integer programs.. 1993). therefore. most of the early work in design optimization of composite laminates was based on the use of continuous-valued ply thicknesses as design variables. Hence. . Alternatively. become the constraints.28 INTRODUCTION 1. discrete. Arora.4. but this may lead to substantially heavier designs. the variables typically used for design optimization define the fiber orientation. in the case of a design space corresponding to a mixed set of discrete. the requirement on the composite transverse modulus is violated for values of 'j greater than approximately 0. for a large number of design variables. However. the final ply thicknesses can be rounded off to integer multiples of the commercially available ply thickness. For example. 1989. a response quantity can be maximized or minimized subject to a constraint on weight. As will be shown later in Chapter 5. However.22.4.22. which are emerging as strong contenders against the traditional search methods. for example. and continuous design variables.1) where the objective function and constraints are linear functions of x is called a linear program. 'j::. In this case.4 DESIGN OPTIMIZATION 29 The horizontal dashed line indicates the 15% limit on the transverse modulus. and continuous variables is encountered. For example. which is in turn determined by the nature of the objective and constraints functions. when laminate thickness is optimized. These methods belong to a generic category of stochastic search techniques and have been used with some success in problems where the design space may be difficult and where a mix of integer. If such The mathematical statement of an optimization problem defined in Eq. there has been considerable interest in exploring new optimization methods such as genetic algorithms and simulated annealing. integer. 1. In recent years. achieved for "r= 0. The special case of Eq. or Haftka and Gtirdal. It makes sense to round some thicknesses down. Once the optimization is completed. it may be appealing to round off the first one to 3 and the second to 5.2 Stacking Sequence Optimization In the context of optimization. and the mathematical methods for solving the problem are called nonlinear programming techniques. 2. In subsequent chapters. There is a growing interest in the application of integer programming methods to laminate design. relative through-the-thickness placement of the distinct fiber orientations needs to be decided before the thicknesses are determined by the optimizer. (d) Strength of a fiber reinforced composite material cannot be greater than the strength of the individual fiber and matrix constituents. the problem must be formulated with a given stacking sequence. Again.8. That is. Orientations of layers are also occasionally used as design variables. laminate design or the "stacking sequence design" is primarily an integer problem that calls for discrete programming techniques. and therefore mathematical programming based on continuous variables will be avoided unless a design situation specifically calls for the use of such an approach. the designer would be forced to restart the optimization process by limiting that particular orientation to having an upper bound and redefining the stacking sequence so that the ply orientation with the excessive thickness in the previous run appears elsewhere in the laminate. as a function of the fiber radius (0. Therefore. 3.30 INTRODUCTION EXERCISES 31 a design is found.000. (f) The following laminate is a symmetric laminate: [Oi±45/0/ 90i0 /±45/02) T' unit volume Figure 1. which can lead to matrix cracking. which is generally made up of plies with only oo.0 $: r1 $: r[<IX) for a unidirectional fiber-reinforced Graphite/Epoxy composite. p. EXERCISES 1. and ±45° orientations (or occasionally orientations with 15° increments between 0° and 90°). Another problem with using thickness design variables occurs when orientations are limited to a finite set of angles. rather than letting the optimization obtain the best stacking sequence. show that the transverse stiffness. if at the end of the optimization the total ply thickness of a certain fiber orientation comes out to be larger than the thickness of four layers. Answer the following True/False questions: (a) Individual layers of a fully cured laminated fiber-reinforced composite material are referred to as the prepreg. Assuming the packing of the fibers to be represented by the unit volume shown in Fig. Laminates with more than four contiguous layers of the same fiber orientation direction are generally assumed to be not practical because of thermal stresses created during the curing process.8. E/p and E/p. (c) Fiber volume fraction. (b) The stiffness of a composite ply in a direction perpendicular to the fiber is closer to the stiffness of the matrix material than to the stiffness of the fiber. £ 2 .2. for example. it may be a suboptimal design-other designs obtained by rounding off differently will have better objective function values. one of the major difficulties in a realistic design situation is the need for a practical laminate. Still another problem occurs in enforcing constraints on having too many contiguous plies of the same orientations. there are 220 "' 1. Using the stiffness representation models discussed in Section 1.000 possible rounded-off designs.6). and the weight density of the composite. we introduce various approaches that address the discrete nature of the stacking sequence optimization problem. 1. Unit volume . with angles taking any value between oo and 90°. of a fiber-reinforced composite material is given by Eq.2. (1. The primary approach in this book is to employ procedures that lead to practical laminate design. 2. (e) The following laminate is a balanced laminate: [ Oi+4510/ 90i±45!-45Jr. With 20 design variables. plot the longitudinal and transverse specific stiffnesses. Vi' is a measure of the total volume of fibers in a fiber-reinforced composite material. 90°. Then. For these reasons. Chapman & Hall. The objective of this chapter is to discuss how elastic properties are calculated for laminated composite materials from variables that can be changed during the design process. Lancaster. 4. Structural Optimization: Fundamentals and Applications. Elastic properties that form the relation between the stresses and strains are the fundamental quantities that govern the response of the material and. Tsai.. W. (1993). U. D. H. New York. New York). Kluwer Academic Publishers. Inc.32 INTRODUCTION (g) The laminate [0/±45/90/±451s is not a quasi-isotropic laminate. Middleton. Introduction to Optimum Design. Boston. (1980). or buckling loads.. (1998). (1993). therefore. (1989). Composite Materials Handbook. Zagainov. In order to design a structure that will meet requirements on certain response quantities such as displacements. Numerical Optimization Techniques for Engineering Design. M. 1980. I. PA. Third Revised and Expanded Edition. Y.. (h) Bending stiffness matrix of laminated composite materials is a strong function of the sequence of the fiber orientations in the laminate definition. G. Introduction to Composite Materials. Hong Kong Conmilit Press Ltd. M. (1992). Technomic Publishing Co. and Poisson's ratio. In designing structures made up of monolithic materials. for composite materials part of the effort is to design elastic properties of the composite medium. Jones. the structure to be designed. New York. (1996). Hyer. stresses. Composite Materials in Aircraft Structures. CO. all depend on elastic properties such as elastic modulus (Young's modulus). Schwartz. Hong Kong. G. In contrast. McGraw-Hill. and natural frequencies. Elements of Structural Optimization. Longman Scientific and Technical (co-published in the U. N. Composite Materials in Aerospace Design. Colorado Springs. Niu. Composite Aiiframe Structures-Practical Design Information and Data.S. Vinson and Sierakowski. Vanderplaats Research & Development. Tsai and Hahn. 2 MECHANICS OF LAMINATED COMPOSITE MATERIALS REFERENCES Arora. and Giirdal.g. E. Haftka. New York. these properties are normally given and are not included among the set of parameters that the designer varies in order to improve the performance of the structure. buckling loads. 1998. Response quantities such as deformations. McGraw-Hill. M. we need to be able to calculate that response for a specified loading. Vanderplaats. and Lozino-Lozinski. T. with John Wiley & Sons. Write down the shortest possible representation of the following laminates: (a) [+45/ -45/0/0/+45/ -45/0/0/90/0/0/0/0/90/0/0/ -451+451010145/+451r(b) [ +45/-45/0/0/+45/-45/0/0/90/90/0/0/90/90/0/0/90/90/ +45/ -45/0/0/+45/-45Jr. T. S. Inc. The subject of this chapter is covered more extensively in books devoted to the analysis rather than the design of composite materials and structures (e. Kirsch. 1987. shear modulus. and Hahn. C... stresses. H. G. (1984). 33 . (1990). Z. New York. J. S. Springer-Verlag. R. y dv dy dW Ez=a. with respect to the strains. dU dW (2. yields dO'/dEY = dO'/dEx. EY and Ex. respectively.z 0 0 0 0 0 0 0 0 0 0 0 0 Ex 2. and 0' and E represent the vectors of stress (see Fig. y.s c.2 Stress-Strain Relations In addition to the strain-displacement relations.z = cz. The stress-strain relation for a three-dimensional anisotropic linear elastic medium.. and Yy::: Yzx' Yxy denote shear strains.e. ~=~+~.3) Ey Ez Yyz Yzx Yxy 2. which has the same properties in every direction.z ell e.1 GOVERNING EQUATIONS FOR ELASTIC MEDIUM 2. cl2 c.3 = c. 1998). z directions. solution of elasticity problems requires the relations between stresses and strains in the medium. Ez denote the normal strains in the x. and z coordinate axes are denoted as u. Further differentiation of the stresses.4 c. and strain components.2 el2 ell 'tyz I Io I o 0 0 0 0 0 0 c". Cij = Cji' can be proved from the existence of a strain energy density U..3 Cz3 c33 c34 I (2. respectively.3) reduces to the following form. EY.s c. The matrix C is the material stiffness matrix. y.e. 2. where. These displacements along the x.4) or .e.z 2 'tzxl 0 0 ell.1 Strain-Displacement Relations Under loading. (2.z c. We start with the fundamental relations that govern the linear elastic response of an isotropic medium in three dimensions and demonstrate how they are reduced to two-dimensional relations when two or more thin isotropic layers are laminated together.6 c24 Czs Cz6 c34 c3s c36 c44 c4s c46 Czs c3s c4s Css Cs6 c26 c36 c46 cs6 c66 c.6 c.1. an elastic medium undergoes displacements that vary over the domain. upon differentiation of the strain energy density. and Herakovich.. We then extend the treatment to the case of layers with directional properties. respectively.1. from which one can show that c.z 2 I jYyz I IYzx (2. O'J cry I O'z ~ c. Eq. see Boresi (1987). e. also known as Hooke's law. For small deformations the following linear form of the strain displacement relations holds: au Ex= dx' dW dV E =-. For example.2) 'txyJ Io 0 ell. I e.z C22 c23 Cz4 cl4 c.1.34 MECHANICS OF LAMINATI!O COMPOSITE MATERIALS 2. ax and crY.1. In this chapter we present only a relatively brief discussion of the subject matter. there are 21 independent material constants. There are three other stress components 'tzy' 'txz' and 'tyx" but as will be shown in the next section 'tzy='tyz' 'txz='tzx' and 'tyx = 'txy· The symmetry of the material stiffness matrix. ~=~+~.· dV dU ~=~+~. where Ex.-.1 GOVERNING EQUATIONS FOR ELASTIC MEDIUM 35 Ex 1998.z Ey Ez I = I c. is expressed in the following matrix form: O'=CE.1.1. one can show au/dEX = O'x andiJU!dEY= O'y.1).z 2 J lYv (2. because of symmetry.1.1) In the case of a three-dimensional isotropic material. respectively. and w. O'x O'y O'z 'tyz 'tzx 'txy c. v. 1..2.xJ ( d'tyx .x ox+ (y. The rectangular parallelepiped differential element of dimensions &.6) daX a'tyx d'tzx -+ +-=0.4). (2.1. summation of forces in the x direction yields with only two independent material constants.1. A first-order approximation to the stresses at the faces of the differential element is shown in Fig. (2. ( (2.y. and & shown in Fig.1.7) The terms involving ax..2 looking down along the z axis.X az 2 ~xt.dx 2 L\y& + ax+ ax 2 t.::_ ~.1.2.--z~ •-z'tAy x- 2.y 2 aax t:.xJ ( dax t. 2.5) + 'tyx + ay t.1 GOVERNING EQUATIONS FOR ELASTIC MEDIUM 37 ~ Figure 2.. Therefore.'tyx...-::.y&.'tyx .4. 0 acrx t:. az Stress components in a three-dimensional medium. 2.1..dy . MECHANICS OF LAMINATID OOMIIOSITE MATERIALS 2."" 1 ax~ I 'J ~~-· ""r + -ay il<..1..3) and the simplest case of isotropic behavior defined by Eq. there are other special cases of material response between the fully anisotropic case defined by Eq.z to get ~ &J ( &J - The diagonal terms that relate the shear strains to the shear stresses are commonly referred to as the shear modulus G and are obtained in terms of the elastic modulus and Poisson's ratio as c44 = c55 = c66 = ell.('tzx .2 2 Figure 2. 'tyx• and'tz. rft. faces of the element are calculated from the corresponding stresses on those faces.I) and Ciz= (1 + v)(l _ v). 2 vE (2. and Poisson's ratio J.. The stiffness coefficients ell and cl2 are related to modulus of elasticity (elastic modulus or Young's modulus) E.dy 2dtyx ----1----. However. and we divide the rest of the terms by t.2(1 + v) =c. for the sake of clarity we postpone the discussion of special cases of material properties until Section 2. dx ay dz (2.x& .c1z _ 2 E . cr~ ~~x x xy 7PJtcr.. For the sake of clarity.o..3 Equilibrium Equations tzx - For a body to be in equilibrium under the applied loads.l by (1. t-zy \ -=.x ~y t. The net forces acting on the ""' .x& ell = (1 + v)(l - 2J..v)E dax t... ..1 is used to represent the stress state at the center of the element..x I txy + ax.. Az dz z oy + ax 2 dtxy t:. Variation of stresses over a differential element..ax.x + a-rl.y t:. (2. in the absence of body forces..x~y + 'tz..y.ay 2 ( L\yJ t.. Since stresses vary within the material.x cancel. stress components along the z direction are not shown in the figure.1. 2. the actual stresses on the faces of the differential element are different from the ones at the center.X dz 2 t.yJ 2 t.. t..8) Depending on the degree of directionality of the material properties. For the element in Fig. for many of the equations that we will introduce in the following sections the degree of material anisotropy is not of consequence. every differential element in the material needs to be in equilibrium under the internal and external forces acting on it.d'tl. equations of equilibrium for the moments of the stresses about each of the coordinate axes yield 'txy ='tyx' 'txz ='tzx and 'tyz ='tzy• (2.1.2) and 'tyz = Y. if the surface normal is in positive z direction. As we will see later. For example. then from Eq. Eq. l = m = 0 and n = 1.1) l'txy +may+ n'tyz l'txz = Y. Eq. Setting all stress components in the z direction to zero.38 MECHANICS OF LAMINATED COMPOSITE MATERIALS 2.1.10) Note that these equations apply everywhere in the material and are independent of the assumptions for the material behavior or the straindisplacement relations. (2.1.'t2x. the equilibrium equations (2. In special situations where the surface normal is in the same direction as one of the coordinate axes. (2.11) yields (Jz =Z.1. and Z.2. 2.9) O. and therefore.9) vanishes. the last equilibrium equation of Eq. In this section we analyze the in-plane problem where all deformations are in the plane of the layer(s). In case of plane stress.1. One mechanism involves stretching or compression of the layer(s) in their own plane. (2. (2. This holds for two-dimensional plate problems where all loading is limited to the plane of the plate. m. For a thin layer (see Fig. and n (referred to as direction cosines) are cosines of the angles between the direction normal to the boundary and the coordinate axes x. We denote the external forces per unit area in the x.12) Similarly.4). can also be simplified.8) and (2.2 IN-PLANE RESPONSE OF ISOTROPIC LAYER(S) 39 This is the governing partial differential equation of equilibrium in the x direction for the variation of stress components. Y.1.1. (}rxy dx d<JY dy d'tzy dz - From these equations.2.1. The other mechanism involves bending of the layer(s).2 IN-PLANE RESPONSE OF ISOTROPIC LAYER(S) Thin structures made up of thin layer(s) of material can carry loads primarily through two different types of mechanisms.and'tyz stresses are zero at these surfaces. and z. The fact that no normal stress exists in the z direction does not imply that the corresponding . This mechanism is sometimes referred to as the membrane action. it is reasonable to assume that these stress components are zero throughout the layer: (Jz = 0 and 'tyz ='tzx = 0.12) the <J2 . two of the direction cosines vanish.2. and the loads do not have any bending component. respectively.11) + m'tyz + n<Jz = z.1 Plane Stress If there are no forces acting on the z-normal surfaces of the layer(s). 2. the state of stress is called plane stress. the absence of one of the external force components will imply the vanishing of the associated boundary stress component at that point. Similarly. respectively. y. Summation of the forces in the y and z directions produces two other equations of equilibrium. (2. Following the customary notation we use the z axis to denote the direction normal to the layers. (2. 2. + + 0 and d'txz d'tyz dx + dy + d(Jz _ dz - (2. they can be simplified for two-dimensional problems into forms commonly used for plateand shell-type structures. and then the equilibrium conditions for the boundary stresses are lax+ m'txy + n'txz =X. (2.1.9) are simplified. ax. hence. The stresses must also be in equilibrium with the external forces (tractions) acting on the boundaries. y. and the remaining two are given as d<Jx dx + d'tyx _ dy - 0 and d'txy dx + d<Jy _ dy - O. and 'txy' are all in the plane of the layer(s). the three-dimensional form of Hooke's law for an isotropic material.3). and z coordinate directions by X. 'txz =X The remaining stresses. where l. (2. crY.1. E fz=.8) Substituting f 2 from Eq. We will see in Section 2. the plate equilibrium equations and boundary conditions tell us that the in-plane shear stress 'txy and the transverse stress aY are zero everywhere.4) The Sii's in the above equation are referred to as the compliance coefficients. If the stresS-Jstate is known. stress case as 0] . (2. and the axial stress is uniform throughout the layer (see Fig.6) ~if.I 0 1/ 1/ v c. (2.1.vz and E Q66 = G = 2(1 + v) (2.5). (2.2.z. Substituting az = 0 into Eq. we get the stress strain relation for the pi~~. the strain field at every point can be obtained from the inverse of the stress-strain equation of (2. 2.z 1-l/ 0 r~ fy (2.2.z 1/ 1/ Cu---c.2.1 1/ v c. 1-v (2..I -..2.L-Ji' I I I I I I crx In matrix notation Eq.2..2. we obtain a new material stiffness matrix ax} aY { 'txy where -lQll J{fx} Q Q Q 12 11 12 0 0 EY .S12). Two-dimensional plane stress state. For example.2.z 0 0 =I c.Eis.4.2.1. "x = 1.2.5) 0 0 Q66 'Yxy ax=PIA. (2. (2. for a rectangular isotropic layer with unrestrained edges loaded by a uniaxial load P along the x axis.z c.6) are modified to take into account the directio nature of the stiffness properties.2 Single Isotropic Layer A simple in-plane problem may be formulated for a rectangular plate loaded by either uniform edge stresses or uniform edge displacements.3) back into Eq.2 IN-PLANE RESPONSE OF ISOTROPIC LAYER(S) 41 E ~X ~ Q. (2. 7) cry Figure 2.2.40 MECHANICS OF LAMINATED OOMIIOSITE MATERIALS 2. (2.4 ). I E -2(1 + v) I I'Yxy 2.3) where sll s12 S= S~z sll 0 [ 0 s66 =Sa.3.(fx + £).4) and is equal to Substituting the values of C11 andC12 from Eq. we obtain 1/ where the matrix Q is commonly referred to as the reduced material stiffness matrix.2.2.z = 1.lso applicable for orthotropic layers where the definitions of t e Qii's in Eq. (2.:ilr---.vz' vE Q.5). (2.1 that the matrix notation of Eq. ax I a yI 1xyl I ell . (2. With sll = E' 1 sl2 =~ 1-l/ and s66 = 2(Sll. (2. a.z-1-=-C. fz strain fz is zero.4) and collecting the coefficients of the l\ and fY terms together.1.2.5) is generally represented as 0' = QE.9) (2.10) . and solving for in terms of the fx and fY.2. represented by t 0 as shown in Fig. Stress and strain distribution in a single isotropic layer.4 and are computed from Eq. The in-plane deformation assumption together with the plane-stress assumption means that the strains and the stresses are constant throughout the thickness of each layer. Although the stresses -vP Ex= cy= AE. the shear strain Yxy is identically zero everywhere: du Yxy + dx = c. so that the laminated layers deform in unison without experiencing any discontinuity in displacements. it is important to emphasize that the uniformity of the stresses over the domain allowed us to reach the solution discussed above.12) respectively. u(x.(y) = dy and v~-v({E}+c2(x). y = -y0 ). c 1 andc2 are. for example).2. Since the loading is uniaxial tension and the shear stress is zero.8) as p AE' 2. That is. the stresses and displacements are functions of both x andy coordinates. at a given point in the domain.-iffi-ox . 2. At the edges of the laminate. in the most general case. For problems with nonuniform applied loads or domains that . Based on uniformity of the stresses and from symmetry considerations.2. The perfect bond assumption guarantees that strains are constant through the entire thickness of the laminate and.5.4.42 MECHANICS OF LAMINATI!D COMFIOSITE MATERIALS 2.2. Finally. (2. Determination of the displacements and stresses in such cases typically requires solution of the two equilibrium equations as a coupled system. The bending response of two-dimensional layers is discussed in Section 2. the in-plane loads must be applied in such a manner that their resultants are at the mid-plane of the laminate. y = +y0 ) = u(x. u~(:E 'txy )x+c. · the through-the-thickness distribution of the stresses and the strains is still uniform as long as there are no applied loads that create bending of the layer. Nevertheless. a constant stress distribution. the laminate strains could be characterized by the strains in a single layer. respectively (c 1 =Ay + B.distribution ex .12).distribution Figure 2.1) and obtain longitudinal and transverse displacements as. dV (2.3 Symmetrically Laminated Layers When several layers are stacked together. In addition to the layers being in a state of plane stress.2. cause nonuniform internal distribution of stresses (such as plates with holes or openings). properly distributed in-plane loads produce only inplane deformations throughout the laminate thickness.2. c 1 and c2 are constants in this case and can be obtained from boundary conditions on u and v. often by numerical schemes such as finite differences or finite elements.2. a set of additional assumptions is needed in order to derive equations that govern the constitutive behavior of the laminate. each layer is assumed to be bonded perfectly to the adjacent layers. For symmetric laminates. the stacking sequence of the layers (the elastic properties) above the mid-plane is the mirror image of the one below it.2 IN-PLANE RESPONSE OF ISOTROPIC LAVER(S) 43 Ox=!0_/··~ -~P/A /Ox/ l£xl I00-. Such laminates are referred to as symmetric laminates or mid-plane symmetric laminates.3. (2.1. we integrate the first two of Eq. (2. 2. say the mid-plane layer. (2. Since the stresses are constant. the strains are also constant throughout the layer as shown in Fig. Having a constant strain distribution through the thickness does not imply.13) In order to satisfy Eq. however. where A is the cross-sectional area of the layer normal to the loading axis.(y) + c~(x) = 0. Most commonly used laminates have layers with mirror image elastic properties with respect to their mid-plane. linear functions of y and x.11) To obtain displacements of the laminate. (2. therefore. Any offset from the mid-plane would cause bending moments with respect to it and force the laminate to bend. Fig. Such a moment will force the laminate to bend. crx} ~y { xy N1 {ax} N: = L J aY { N zk = t {Nx1 N NY · Nxy (2.A 12 A. MECHANICS OF LAMINATI!C OOMIIOSITE MATERIALS £ 2. for example. 2. = ~k) Eo. see Fig. The possibility of bending is precluded if we limit ourselves to laminates that are symmetric with respect to the laminate mid-plane.2. That is. (2.2.15) ' • •:·>=: Figure 2.)... Equation (2. unless the bending of the laminate is prevented by some external means to keep the laminate in its plane.2. (2.5.18) can be rewritten as Nxl-lAu NY . For an arbitrary stacking sequence. Eq. integration can be replaced by summation over individual layers. and Nxy is the shearing stress resultant.14) will have a net bending moment.5. we get Nx} NY \ Nxy = lN L lQ" Q.2 A22 0 0 Nxy 0 0 A 66 \ J{~} Ey ' 'Y~y (2. layers.7) as (J'(k) Since the stresses are constant within a layer.2 Q.2.zk_.2 Q" k=i 0 J (k) (zk.-j-----. The fact that the stresses vary from one layer to another makes the characterization of the stress-strain behavior of a laminate more difficult. (2.2. This is achieved by through-thethickness integration of the stress components.2. (k) (2.19) where the coefficients of the A matrix are given by .14) dz = (k) L {ax} aY /<. { h/2 Nxy -h/2 'txy =distribution crx = distribution Stress and strain distribution in symmetrically laminated isotropic where Nx and NY are in-plane normal stress resultants. 0 0 Q66 xy where the superscript "o" is used to denote mid-plane strains (recall that strains are constant through the thickness and can be represented by the mid-plane values).2.2.18) . Since the stress resultants are obtained by through-the-thickness integration of stresses.14). -1..2. dividing them by the laminate thickness h yields average laminate stresses..16) are constant within each layer.17) Nxy k=i zk-1 't xy 'txy where k is the layer number. In order to define a simple stress-strain relation we need a quantity that represents the overall effect of a stress component on the laminate response.5) depending on the stiffness of the individual layers (which are assumed to be isotropic in this section) following the stress-strain relation of Eq. This variation is the main reason for the common use of symmetric stacking sequences (or through-the-thickness elastic property distribution) of the layers.. they will vary from one layer to the next (see. With regard to overall stiffness.. (2.2. and zk's are the through-the-thickness locations of the interfaces between the layers. 2. for a laminate of thickness h we define where N is the number of layers.2 IN· PLANE RESPONSE OF ISOTROPIC LAVER(S) 45 (2.p~ Ex X =£0 X ax {ax} N1 N: = J aY dz.) J{£~} E~ '( (2. material2 material! I ~. the stress distribution through the entire laminate thickness given by Eq. the laminate behaves as a single layer with properties that are some averages of the properties of the individual layers. Substituting the stressstrain law.=I (zk-zk_. the subscript (k) denotes quantities in the kth layer. see Section 2.05 ---__L Brass --------------Aluminum Brass ---I I .2 in and brass face sheets of tbr = 0..J 1 .v2 --h eff and 1- v. A66 = L G(k) tk ~I The effective shear modulus is obtained simply as Geff = A66/h.(.Az 12 l A22 J and A. However. tion.24) Figure 2. and Gbr = 5... therefore val= 0.2.6) with unknown effective properties. 2.h _1_ (A 11 A22. ~ * ~ ~ *ci' ax) !Nx) [All 12 A A22 0 01 {E~J E~ .23) _L Equation (2.16)] in the laminate to the strains => V= E 2G .1 The sandwich laminate shown in Fig.339.2.zk-J is the thicknesses of the kth layer.MI:OHANICS OF LAMINATIO OOMIIOSITE MATERIALS N 2.26) (2.=~ (zk.1) ~ 0. Symmetric sandwich laminate of aluminum and brass construc- . the ~ 2 term is identical to A 11 term.6 is made up of an Aluminum core of thickness ta1 = 0.2. Determine the effective elastic properties.2 IN·PLANE RESPONSE OF ISOTROPIC LAYER(S) 47 A. Eeft' Geft' and veff of this laminate.2.6) as G= E 2(1 + v) N=At 0 • (2.22) Once the in-plane stiffness matrix is determined.2. For isotropic laminates considered in this section. by replacing the elastic properties in Eq.4.70 X 106 psi.~ u 22 - N E<kl £.23) is identical in form to the stress-strain relation of Eq.J 1 ~I ~ ~kl E<kl 2 V(k) N N tk.J Q. (2.2. The aluminum and brass have the following properties: Ea1 = lO. (2.60 X 106 psi.2. For more general laminates made of anisotropic layers (such as fiber-reinforced orthotropic layers.0 X 106 psi...19).6. lj(k) k~J (2..2. therefore.Ox 106 psi. for the sake of generality we keep the matrix representation of A as shown in Eq.2 vett= A .19) is written in matrix notation as Example 2.21) and where tk = zk.351 and j/br = 0.:. Ebr = 15.4) these two terms may not be the same. Poisson's ratios for the two materials can be calculated from Eq. (2.2.25) 22 A -A .5) and (2. 0 A66 Y~y (2. (2. (2.2. (2.2. ~I (k) A 12 =£. Ga1 = 3.2.05 in each. (2.5). (2.2. (2. Given the elastic and shear moduli.1.2. (2.tf =h.6)] are l.20) which yield Eett. Eeffand veft' we can write the following two equations in terms of the two unknowns for effective properties: Eeff _Au veffEeff A 12 t-r (br) =0.2. The reduced stiffness matrices for the two materials [see Eqs.2. The matrix A is generally referred to as the extensional material stiffness matrix (or in-plane stiffness matrix) and Eq.zk_ 1) or lj . effective engineering constants of a laminate can be obtained by relating the average stress [see Eq.v2 tk.2" t 1.2. and very close to the weighted average of the Poisson's ratios v's.3).25).008 4.408 4. NY.751 16.751 [ 0 ~ Nx} = h12!Xl BY dz.1 does not limit the strain Ez to being small.2.408 0 0 0 0 3. and the stress components in the z direction are assumed to be negligible so that an approximate state of plane stress prevails. . (2. and Nx.. 5.2 0 0 X r 106 2.3 BENDING DEFORMATIONS OF ISOTROPIC LAYER(S) 49 I 1.1 A. laminated plates are often loaded by forces that bend the laminate. The assumptions invoked in this section. 6 and ing to applied displacements exist).67 X 106 psi.33 _ Veff- _ 1. however.700 X 106 + O O A 66 [ 16.600 From Eq.~08 ~ 0 ] x 10 psi.700 5. Finally.408 4.377 1. andNxy are often used to denote the applied traction resultants.2 + 2 X 0.3 BENDING DEFORMATIONS OF ISOTROPIC LAYER(S) 1.98 . Therefore. the in-plane stiffness matrix can be calculated to be Unfortunately. A second assumption is that the length of that line remains unchanged. note that the definition of the in-plane stress resultants discussed in this section is based on the internal stresses. One assumption is that a straight line perpendicular to the mid-plane before deformation remains straight and perpendicular to the same. the literature does not make any distinction in notation between the internal stress resultants and applied boundary traction resultants. along with the assumption of perfectly bonded layers (with infinitesimally thin nondeformable bonding agents) constitute classical lamination theory or CLT. and the w displacement is constant throughthe-thickness. and Z are applied (or tractions correspond- In addition to in-plane loads. ". the assumption of zero through-the-thickness deformation of the Kirchhoff hypothesis is in- . . 3.05 and 3. The layer(s) are also thin compared to the in-plane dimensions of the layer(s).3467 o 1.408 0 Ql = 4. For bending of laminates we invoke the classical KirchhoffLove assumptions of pure bending.008 11. Note. I .30 X 106 {) . (2.300 Using Eq. r 0... In fact.008 11..977 0 0 l X X 106 lb/in.951 16. as can be observed from Eq..98 _ Geff- X 10 6 =11.382 3. that the plane stress assumption discussed earlier in Section 2. Note that these properties are the weighted average of the elastic moduli E's and G's for the two materials.951 5.2.21). (2. through-the-thickness integration may be used to define applied traction resultants. A.2.1.951 0 0 l [ = X 1 0 0 3 977 1.600 1.27) ] x 10 psi.4.951 Qr = 5.. The assumption of unchanging length of the mid-plane normals of the Kirchhoff hypothesis implies that through-the-thickness deformations in the laminate are zero.98 0..751 5. This important deformation state is best studied by isolating the loads. so that only pure bending deformation of the layer(s) is present without in-plane deformations at the midplane. we then calculate the effective elastic properties E ff e = 1 3. the magnitude of Ez can be comparable to the in-plane strains. the strains in the out-of-plane direction z are neglected. That is.2... jNxy J K Z -h/2 (2..38 _ 3 98 -0.. £\ £\~ 106 psi. Al2 A22 0 0 J= 0.751 16. Y.2. At a boundary point where tractions X.377 3.[ MECHANICS OF LAMINATID COMPOSITE MATERIALS 2. 6 5. 3) aw ax· (2. Since the strains are antisymmetric with respect to the mid-plane of the layer. Since there are no in-plane deformations of the mid-plane. In this case. where the strain within a layer is constant. the dashed lines denote the undeformed configuration and the solid line the deformed one. We seek a quantity that would be representative of the layer stresses.1) From the strain displacement relation of Eq. (2. at any z location is related to the slope by u=-z8x--z_ {::} Yxy ~ 3} z{ xy where { 3} xy =- a2w ax2 a2w I ayl a2w 2-----axay - (2. we get . the mid-plane strains cannot be used to represent the response of the layer in pure bending. (2.0 can be obtained from the rotation of the line segment A'B'C'. -h/2 1 (2. 2.5) and using the strain-curvature relations. B". and C' move after deformations to positions A". The displacement of a point along the x axis.3. In the figure.2. We can get similar expressions for the other two strains and write in vector representation 2.4) Mxy 'txy c z Figure 2. The negative of the second derivative of the out-of-plane displacement is an approximation to the curvature of the layer in the x-z plane and is denoted by Kx.3. the curvatures are also constant through the thickness and are used in a role similar to the mid-plane strains to represent the deformation of a layer.150 MECHANICS OF LAMINAT!C OOMIIOSITE MATERIALS 2. Bending deformations of a single layer. the length of the undeformed line BB' remains unchanged in the deformed positions as shown by BB". (2. Eq.tk3Ex . therefore. Because w is constant through the thickness. But if the point of interest is either below (positive z direction) or above the mid-plane of the layer. but the through-the-thickness integral of the in-plane stresses is zero (the layer is in pure bending).7. the line AA' will shorten. so that points A'.3. the representative quantities for the stresses of a layer are chosen to be moments of the stresses about the mid-plane of the layer.3). Stresses corresponding to these strains are not constant even for a single layer and may be calculated from Eq.distribution crx . the stresses are also antisymmetric.1) we have Ex = au aw ax = -z ar . it is clear that the strain distribution is a linear function of z with Ex(z = 0) = 0 [antisymmetric with respect to the midplane Ex(z) =-Ex( -z)] and. In spite of this inconsistency. 7. indicating compressive strains at z = -t/2. that is.2. Substituting the expressions for the stress components in terms of the strains from Eq.3 BENDING DEFORMATIONS OF ISOTROPIC LAYER(S) 51 consistent with the plane stress assumption. We define moment resultants (sometimes referred to as the stress couples) by x I££1.14). For example. and C". where h is the thickness of the layer. B'. <J/z) =-ax( -z). there will be an elongation or shortening of the line segment along the x axis. (2.1 Bending Response of a Single Layer The cumulative effect of the Kirchhoff-Love assumptions can be best described by studying Fig. so that Ex= zKx. denoted by u.3.3. 2 (2.distribution M 1 x} M: = h/2 J {O' crY z dz.2) In contrast to the in-plane deformation case. the assumptions of classical lamination theory provide a basis for many of the analyses that are published in the engineering literature and have proved to be adequate for most engineering applications.1.3. Strain values at points z :t:. material I -\-~ Ex - distribution ax . we define the matrix that relates the curvatures to the moment resultants as the material flexural stiffness matrix.3 BENDING DEFORMATIONS OF ISOTROPIC LAYER(S) 53 {Ex} {k' x} • EY z dz = [Q J z dz J K'Y h/2 2 'Yxy (2. (2.7) and the straincurvature relations of Eq. the layers of the laminate are perfectly bonded together and deform in a way that straight lines normal to the mid-plane remain straight and normal without changing their length.3) the throughthe-thickness distribution of the bending strains remains linear.3.11) 3 3 )· Q lj(k) ..7). following Eq. Through-the-thickness integration of the stresses to establish the moment resultants involves a summation (2.3). in the case of a laminated medium in bending.3. which are linear in terms of the curvatures.3.3.2 Bending Response of Symmetrically Laminated Layers For multiple layers of isotropic material stacked symmetrically with respect to the mid-plane of the laminate.3) into Eq. 2. 0 D66 Kxy 0 0 12 l hy the moment resultants introduced for a single layer.7) where the coefficients of the flexural matrix D are now defined by 1 D I] . (2. (2. (2.material2 . for the sake of generality we keep the matrix representation of D as shown in Eq.8.2.2 = vDil and D 66 = (1 - v) T· l ! lM kf: lkfxy N =L ~k) f jE: E} z dz N = (L K } ~kl J z dz J{K: . =3 I N (2..! MY Mxl = hl2 Q f rQ Q. .3.3. The stresses can also be represented k=l By substituting the strains from Eq.9) zk {ax} crY z dz. {Mxl lkfy lkfxy = rD D 12 11 D Dzz 0 {Kx} K). The curvature is still independent of the through-the-thickness location in the laminate and can be used as the representative quantity for the laminate bending deformation. l ! lkf: M lkfxy zk = h/2 J {crx} crY z dz = L J N (2. into the definition of the in-plane stress resultants we have material I .3.5) -h/2 1( xy After integration. Bending strains and stresses in laminated layers. Q = Q and. zk 2 (2. -h/2 't xy /r-1 zk-l 't xy (k) (2.3.8) For isotropic laminates considered in this section. These jumps are the result of the change in material properties from one layer to another and are governed by Eq.2..9) yields D" = Dz2 = 12(1. That is. (2.. D. see Fig. (2. Therefore.3. the assumptions described for laminated plates under in-plane loads and the assumptions for a single layer in bending are combined. the moment-curvature relation for the laminate can be rearranged to express it in the matrix form of Eq.6).3. (2. D22 is equal to D11 • However.8.3. (2. there11 22 fore.3. (zk-zk-1 2.z Q2z 11 12 lkfxy -h/2 0 0 0 0 Q66 l MECHANICS OF LAMINATIO COMPOSITE MATERIALS 2.3. stresses not only vary linearly through the thickness but also make discrete jumps across layer interfaces.3. After integration.6) or in matrix notation.10) k=l zk-1 "{ xy (k) k=l zk-1 K xy (2.7) where elements of the D for the single layer isotropic case are given by Eh 3 Dll Substitution of the stress-strain relation of Eq.distribution Figure 2. However.v) 2' D. M=DK. where the in-plane stress resultants are zero. are distributed unsymmetrically with respect to the mid-plane. k=l 1 (2.3.· ..t~-r3--ex .3 BENDING DEFORMATIONS OF ISOTROPIC LAVER(S) p ~ material I Eo +.9. Figure 2. the summation inside the big parenthesis can be shown to be zero (see Exercise 2).14) where the components of the B matrix are given by N Bij = 2 L Qij(k) (z~.2. an unsymmetric laminate with two materials subjected to bending moment M that generate curvatures K.15) The matrix B is called the coupling matrix.7) and the strain-curvature relations Eq.distribution ax. (2. the integrals of stress components. through-the-thickness integrations of the stresses may not vanish and may indicate nonzero in-plane stress resultants. So for this case. but their magnitudes.~.3) into Eq. through-the-thickness distribution of the strains are linear and.13).3 Bending-Extension Coupling of Unsymmetrically Laminated Layers For the symmetric laminates considered so far. because £ 0 = 0.3.3. we obtain l l Nx) NY Nxy = L N Q(k) zk Kx} = [B J z dz J{KY B Kxy 11 12 B B22 0 12 0 0 B66 k=l zk-l 0 l{ Kx} .. Also assume that the in-plane laminate mid-plane strains. -f±-=t.· . Consider first.zLI). L ~k) Jz dzJ1 K: N zk (2. Based on Kirchhoff-Love assumptions..12) ---+ M k=l zk-1 Kxy 3 P X ~ E(l) > E( 2) Since the Qk)'s are symmetric with respect to the mid-plane.· . governed by the stiffness properties of the individual layers. on the other hand. the sequence of layers on one side of the mid-plane is the mirror image of the ones on the other side. see Fig. In-plane loads for such a configuration produce strains that are constant through the thickness and stresses that are constant within each layer but change from one layer to another.16) . In contrast to pure bending of symmetric laminates. (2.. As we will discuss in the following. f {ax} cry k 2k-l 2 dz.material 2 E( ) 2 55 1C } . (k) (2. for example. The moment resultants are zero under in-plane loads. (2. defined to be the stress resultants. (2.. In the case of pure bending deformations. Conversely. JS Kxy (2. The corresponding stresses are also linearly distributed within each layer.· .3.13) k=l 'txy Substituting the stress-strain Eq. are antisymmetric with respect to the midplane. 2.3.. 0 £ are suppressed to be zero by adjusting the load P.14) is represented in matrix notation as N=BK.3.3. through-the-thickness strain and stress distributions are both linear (stresses are piecewise linear) and are distributed antisymmetrically with respect to the laminate mid-plane.· . and Eq. For a more general laminate with N layers.84 MECHANICS OF LAMINATID COMPOSITE MATERIALS jNxy N ) N: . application of bending loads or moments generate in-plane strains as well as bending strains. In this case. the application of in-plane loads produces bending strains as well as in-plane strains.3. (2.· .3. The response of unsymmetric laminates to in-plane loads and pure bending is substantially different. are zero because of the symmetry. = L Q<kJ N J1 E: zk zk-l E} dz k=l Yxy (k) = l 2. Strain and stress distribution in unsymmetric laminates. the in-plane stress resultants are zero. The reason for the latter assumption will be explained later.distribution 2. these in-plane stress resultants are given as I N ) = h/2 N Nx f {cr} ax dz = L -h/2 N:Y 't. Even though through-the-thickness distribution of strains represented by mid-plane strains £ 0 is constant. and stresses within each layer are constant. the unsymmetric profile of the stresses (see Fig.distribution M=BE 0 • (2. For this laminate.2. internal moment resultants are generated. The coupling between the in-plane and out-of-plane response of an unsymmetric laminate has one more element.9 and assume it to be very thin compared to the overall laminate thickness and to have infinitely large in-plane stiffness. Next.3.3. In order to visualize this. Another way of inducing in-plane deformations without laminate curving is to apply in-plane loads with an 0 eccentricity with respect to the laminate mid-plane. [B11 B 12 0 B12 B22 0 0 .22). and the implication of the above equation is that any type of curvature is also associated with in-plane stress resultants. these moment resultants are the result of external actions that keep the laminate in flat equilibrium configurations..-. Eq. the actual strain distribution is a combination of pure bending strains with zero mid-plane value and the finite midplane strain due to the asymmetry of the in-plane stiffnesses of the layers.10. any in-plane stress resultant induced by edge loads will induce curvature. (2.3. This can be m:hieved by placing the laminate between two rigid walls to suppress the curvatures and applying in-plane loading at the laminate midplane. Unsymmetric laminate under in-plane strain state.f aX .16) easily by adding the relation between the in-plane strains and the in-plane stress resultants from Eq. Eq. For more realistic unsymmetric laminates. the net result of these in-plane stress resultants at the edge of the laminate is equal to the applied load P.20) material! -·-------+ __. This can be incorporated into Eq. Therefore. consider a case where we subject the unsymmetric laminate to in-plane strains only by suppressing curvatures. (2. Equation (2. (2. Conversely. Starting from the definition of the moment resultants.3.20) suggests that under pure in-plane deformations.3. bending deformations would cause zero strain at the top surface of the laminate (because of the infinite in-plane stiffness) and finite strains at the laminate mid-plane induced by the curvature. so that it cannot experience any in-plane strains.Oct MECHANICS OF LAMINATID COMF'OSITE MATERIALS 'I :1 BENDING DEFORMATIONS OF ISOTROPIC LAYER(S) 57 Note that the in-plane strains are assumed to be suppressed at this point. . material2 X --8.17) L. Conversely. In effect.10).---1 e material I material2 '----------' e u -r =tl E =£ 0 ---.14)] for pure bending. From inplane equilibrium.3.18) zk-1 £xy Note that the matrix terms in front of the strain vector are the same us the ones obtained earlier [see Eq.3. the relation between the moment resultants and the in-plane strain components is represented as Mx} _ MY Mxy \ or in matrix notation. we use lhc fact that strains are constant to obtain M} N ~k) J {£ M: =L £~ { 0 zk } z dz (k) = lN L k=l ~k) Jz dz {£ Ei · zk } Mxy k=l zk-1 "( xy l 0 (2. consider the top layer in Fig.19) B66 N=BK+At 0 • (2. In a more general case the in-plane strains at the laminate mid-plane will not vanish under pure bending moment.l --.3.l {£~} £~ ' Y~y (2. (2.--.1 0) generates moment resultants about the laminate mid-plane. if external bending moments are applied without in-plane loads. 2. 2.-F Ex - distribution Figure 2. Thus a given £ 0 and zero K can be achieved by applying in-plane loading of N = AE and bending moments M= BE 0 • Note that different eccentricities may be needed for different components of the in-plane loads.. The second set of equations is Eq.089 = 1. that for a given applied bending moment Eq. therefore.2.1. ~ = -0.3.125 27. If in-plane stress resultants are zero.05 in at the bottom.3. (0.125 3 .020 The distances to the layer boundaries are z0 = -0. E0 and K.2.72 2 - 0.125 in.06 1/in.129 1.075' .05 ~r 6.J25)') + 0.21) alone is not enough to solve for the deformation state. but only one set of equations.008 0 16.3..075' .11 0 0 A66 = 6. There are two sets of unknowns. and determine the magnitude of the bending moment required to cause this bending.68 [ (0. (0.J25)') + Q.20) suggests that in-plane deformations would result.1 A two-layer laminate is made up of an aluminum sheet of thickness ta1 = 0.3..17) and. (2.951 0 5. (2.408 4.11 17.008 11.07 5')) = [ 8. (2. the resulting deformation state is a combination of the in-plane and bending deformations governed by 11.006 1/in. then Eq. and Kxy = 0.129 [ ~ 6 ] X 10 lb/in. and the bending-extension stiffness matrix for the laminate is Example 2. the in-plane stiffness matrix for the laminate configuration is calculated to be l l Bll Biz Biz Bzz 0 0 B66J ~ l = k{ Q. (2.21) It is important to note.089 3. Since applied bending moments also cause curvatures as given by Eq.21) can be used to solve for the in-plane strains (or the curvatures).05 [ 5. + 0.72 8. Determine the in-plane strain that the laminate experiences. Using the reduced stiffness matrices Qa1 and Qbr given in Example 2.700 M=BE 0 +DK.2 [ 4. The elastic properties of the aluminum and brass are same as the ones given in Example 2. (2.751 ~ 3. (2.68 ~ ] x 103 lb-in. 0 0 1.3..3. 17.2 in on top and a brass sheet of tbr = 0.3 BENDING DEFORMATIONS OF ISOTROPIC LAYER(S) (2.125 in.0753)) An AI2 12 r0 A22 A 0 Ol = 0.5 l X 10' lb.075 in. and the bending stiffness matrix is DII DI2 Diz Dzz 0 0 D66J ~l = t{ Q.951 0 5. 0 5.72 27. z1 = 0.0.(-o.7)..72 ~ 0 0 9. and z2 = 0.0 1/in. the solution depends also on the in-plane stress resultants. The following example demonstrates a situation in which the in-plane stress resultants are zero.(-o.17) can be used to solve for the curvatures in terms of the in-plane strains (or vice versa).MECHANICS OF LAMINATID COMPOSITE MATERIALS 2.3. however.79 .408 = 0.~51 16. the laminate bends such that Kx = 0. (0.2 Qal + 0.3. After that Eq.600 l ~l X X 59 106 106 3. Under the action of bending moments with no in-plane loads. and the bending deformations are known.1. Such layers are referred to be orthotropic.3. In this case an unsymmetric laminate must bend.11.0 + [ 6. Figure 2. (2. Ei 74.80 MECHANICS OF LAMINATID COMPOSITE MATERIALS 2.72 8. see Fig.21) as K =-D-l Bt 0 • (2. 2. Eqs. However. respectively. Mx} My { Mxy = [27.18 X 10-6 { Yxy 0.21). (2. (2. 2. in the remaining chapters of this book.10 is at the mid-plane without applied moments (M = 0). A11 A. 0 and M and (2. one can solve Eq. the lamina is still orthotropic and is referred to as generally orthotropic.3.3. Since the moment resultants are zero. individual layers then have directional properties with planes of symmetry that are not necessarily oriented along the general x-y axes of the laminated medium.3.68 6. the curvature can be calculated from Eq.06 -D. e. These principal axes correspond to the direction of the fibers and a direction transverse to the fibers and are denoted by subscripts 1 and 2.72 27. the lamina behavior will appear to have the characteristics of an anisotropic material. the moments that generate those can be calculated from Eq. . Given the curvatures of the laminate.3. with respect to the axes of the laminate (an off-axis layer). Planes of material symmetry for an orthotropic layer. we can use Eq. there are two perpendicular planes of symmetry that define two principal axes of material properties. Finally.72 0 0 0 9.72 8.11.0 Once the curvatures and the in-plane strains are known.79 J 10 X 3 I 0.21) have to be solved simultaneously.3.0 J= 1 1 lb-in 1010 258 in 0 Replacing isotropic layers by fiber-reinforced layers affects only the layer stress-strain relations without requiring any changes in the assumptions that form the basis for the laminate constitutive relations. Since the in-plane stress resultants are zero. 0 2.0 = {0} 0 .17) for either in-plane stress resultants given the in-plane strains or in-plane strains for given in-plane stress resultants.5} 74.4 ORTHOTROPIC LAYERS and we can solve this equation to obtain £~} = {-540.17) r 0 0 0 0 A 66 l {E~} Biz E~ + rBll B B ty 12 22 0 0 0 0 B66 ll 0.68 0 0 0 0 5.11 17.2 X 103 6 0. When the principal axes of orthotropy of a lamina are oriented at an angle.11 0] X 10 {-540.22) Knowing the curvatures in terms of the in-plane strains. (2. 2 A12 A22 in the more general case when both N-:~.17) to solve for the values of the in-plane laminate mid-plane strains.5 17.3.06 } -0.4 ORTHOTROPIC LAYERS 61 -:~. 0. However. and the 1-2 axes are the principal axes of orthotropy (sometimes referred to as the principal material directions). (2. 2 Another easy case is when the in-plane loading of Fig. For a unidirectional fiber-reinforced lamina.006 0.52} .006 0. 2./21 ' v 12E 2 1- 1/121/21 c31 = E3 (v.2 63 ~11 1/13 v23 1/32 2. E3 . G23 . £2 £1 £3 £1 £2 £3 (2.1) 't23 =0 and 't31 = 0.7) Q66 .2.and .6) which reduces the stress-strain relations to yl2 Expressions for the Cu's in terms of the orthotropic engineering material constants.1 Stress-Strain Relations for Orthotroplc Layers For an orthotropic lamina.2) 12 = 1 . c33 = E3 (1 - v12 v21) v2.4.3) C44 = G23• Css = G!3 and C66 = G. £ 2 .4. only nine of the constants are independent. respectively. to the strain in the 1 direction.5) v21 =£ v12.113 J.l23 v31) · · cl3 E 1 (v31 + v21 v32) = A . 0"2 0"3 't23 't31 'tl2 o o c. cr3 = 0.v12 v21 = 1 - and where Q66= Gl2' (2.v.1 in the 1-2 principal material plane.4.3 + v.4 ORTHOTROPIC LAYERS v2. Three reciprocity relations provided below without proof may be used to express the Cu's in terms of nine independent material constants: Two of the four independent material constants are the elastic moduli in the 1 and 2 directions. we invoke the plane stress assumptions of Section 2.J. G12 .. £ 1 and£ 2. . and is related to the other elastic properties through the resciprocity relations of Eq.l21 J. (2..4) In addition to the three elastic moduli and three shear moduli. G3 " and vu's are: ell= E 1 (1 . However.4. (2.4. The other two material constants are the shear modulus in the 1-2 plane G 12 and the major Poisson's ratio v12• The major Poisson's ratio is defined as the negative of the ratio of strain in the 2 direction.4..l32- J. there are six Poisson's ratios in the preceding equations giving a total of twelve engineering constants.4..2 c2 2 o o cl3 c33 o o = 0 c23 o o C44 o 0 o o o C55 Cu c.121 + v23) .l12 v21 - V13 v31 . c32 E3 (v13 1. the stress-strain relation in the principal material directions is given by the following set of equations with nine independent constants: .9) .= . I E2 (2. E2 (vl2 where the Qu's are the reduced stiffnesses and are given in terms of four independent engineering material constants in principal material directions as E2 c21 (v12 + v13 v32) Ll c22 = = E2 (1 - l-'13 v31) . 2 C 13 C23 I 0 0 0 0 0 0 0 0 0 0 c66 £1 £2 £3 Y23 Y31 For thin layers with no applied forces acting in the out-of-plane direction. (2.2 v23 V31 .1/12/.l23 v32• (2.4. A v31 + v32) c23 Qll El E2 Q22 Ll 1 . The minor Poisson's ratio is defined as v 21 = -£/£2 under loads along the 2 direction. v. (2.MEOHANICS OF LAMINATIO COMPOSITE MATERIALS 2. under loads acting in the 1 direction only..v23 v32) all [Q Ql2 0"2 = Ql2 Q22 {'t12 ° 0 11 0 0 1 {£1] £2 ' Y12 (2.4.5) cr. E" E2 . £ 1.= . c12 = E 1 (v21 + J..EI V12v21 A Q (2.4.8) A= 1 and J.4.2 v23) A . given by m2 where the various Poisson's ratio V. The transformation of stresses and strains can be accomplished by r-2mn 2mn mn .14) and n =sin e.4.4.1) can be used to solve for the through-the-thickness strain.84 MECHANICS OF LAMINATI!D COMPOSITE MATERIALS :1.4.4.10) In order to convert the strain transformation relation from tensor strain notation to engineering strain notation. COS and the transformation matrix is given by Q12 = (Qll + Q22.4Q66) 2 8 sin2 e + Ql2 (sin 4 e + cos 4 9). the stress-strain relations given in the principal directions of material orthotropy.4. Eq.1 'txyf QRTR 1 { ::J.12) where Exy and E12 are the tensor shear strains which correspond to one-half of the engineering shear strains E=b_ xy 2' {d [~:: ~: ~:: k' ax - Qll Q12 Q16 - -] } (2.4.4. -mn m2-n2 1 (2.4.2). cl3 c23 E3=-cE1-cE2· 33 33 n2 m2 T= r n -mn mn m2-n2 2 2mn .4.4.17) { :~} 't12 = T {::} 'txy and {:~} E12 This transformation produces = T {::].4.4. therefore.11) The transformation matrix for the engineering strains is.4 ORTHOTROPIC LAYERS m2 65 As stated earlier for the isotropic case. m=cose -2mn 1 (2.2 Orthotropic Layers Oriented at an Angle Since orthotropic layers are generally rotated with respect to a reference coordinate system. x-y.4. (2. (2. Eq.10) may be expressed in terms of the engineering properties of the material { Ex} Ey Yxy = R {Ex} Ey ' Exy E3 = v31 + v32l-'21 1v12v2I v32 E1- + V31V12 vl2v21 R= 0 1 0 .15) 1- E2 (2. Exy (2. .16) Substituting the transformation relations for the stresses and engineering strains into Eq. respectively.4. (2.4. 7). (2. (2.13) Qll = Qll 4 COS rlu are related to the Qij by e + 2(Q12 + 2Q66) sin2 e cos2 8 + Q22 sin4 e. (2. From the first equation of (2. n2 m2 Te= n2 2.4.18) = E12=2· Y12 where the transformed reduced stiffnesses (2.4. (2. Yxy Ex (2. the third row of Eq. 0 0 2 [1 001 (2. must be transformed to the reference axes. the plane stress assumption does not preclude the existence of through-the-thickness strains.7).j can be computed from the reciprocity relations of Eq. where By using the definition of the Cy's from Eq. we obtain {::1= .5).6). the transformation matrix is pre.and postmultiplied by a matrix R and the inverse of R. 28) ply with an orientation of 8 is loaded such that only a uniform state of ax = 100 MPa exists.2Ql2. The reduced stiffness matrix for the material is obtained from Eq. (2. I Q26 =2 u2 sin 29Q66 u3 sin 48. . 1 u4 = 8 (Qll + Q22 + 6Q1z. Q 1z.61. G 12 = 7. = U1 + U2 cos 2e + U3 cos 49. Q22 = U1 . U4 = 22.37 + 85.71 cos 48. = (Qll + Q22-'-.4.88 GPa. and Q66 given in terms of the Up U4 .au 4 MECHANICS OF LAMINATED COMFIOSITE MATERIALS 2.21).8). (2.19.17 GPa.19) These equations can be put into a simpler form by using trigonometric identities to replace the terms with powers of sines and cosines of the fiber orientation angle 9 with the sin and cos of multiple angles 29 and 49.18) as Qll Ql2 and the corresponding U values are.4. Both the U's and the Q's are independent of the ply orientation. Ql2 = Q21 = 22.21) Q26 = (Qil Q66 - Ql2. Q22 . Us= 26.61 GPa.4.90 0 0 1GPa 7.2Ql2 + 4Q66). 1 Qzz).4. Tsai and Pagano (1968) defined the following material properties: ul =8 (3Qu + 3Qzz + 2Qlz + 4Q66). (2.2Q66) Sin 9 COS 3 9 + (QI2. 181.4Q66).81 (2.3 GPa. we obtain the terms of the fiber orientation angle 8. Ql6 = ~ Substituting the U's into Eq.1 An off-axis unidirectional Graphite/Epoxy (£ 1 = 181 GPa.Q22 + 2Q66 ) sin 3 9 COS 9. 1 Q= [ 2. Based on this feature.4. Ql6 = (Qll. (2.71 cos 49. 3 = us - U3 cos 49.2Q66) sin2 e cos 2 e + Q66 (sin4 e + cos4 9).4Q66). Us= 8 (QII + Q22. Lhe Q's are dependent on ply orientation for orthotropic materials. the form of Eq.2Ql2.73 cos 29 + 19. (2. E2 = 10.4. u2 sin 29 + u3 sin 48.90 10. v 12 = 0. An advantage of the form of the reduced stiffnesses in Eq.Ql2.73 GPa. Determine the in-plane strains resulting from this loading as a function of the fiber orientation angle.Q22 + 2Q66) sine cos 3 e. = U4 U3 cos 49.20) 2. (2.4.17 ~ which yield a simpler form of the transformed reduced stiffnesses of Eq.2Q66) sin e cos e + (Ql2.U2 cos 2e + U3 cos 49. U1 = 76.71 GPa.35 u2 = 2 (Qil 1 u3 =8 (Qll + Qzz. 1 Example 2. the U's are sometimes called the material invariants.21) is also useful for design optimization.4 ORTHOTROFIIC LAYERS 67 Q22 =Qll sin 9 + 2(Q 1 2 + 2Q66) Sin 2 9 COS 2 9 + Q 22 COS 4 9. and Us which do not depend on the ply orientation. and U3 = 19.4.4.374 GPa. but . As we will see in Chapter 4. (2.21) is the constant parts of {211. U2 = 85. - Q matrix in Qll = 76. strains are obtained from the inverse of Eq. we see that the axial strain Ex increases as the fiber orientation angle increases from oo to 90°.577 + 0.3 and Eqs. and the curvatures K that are constant through the thickness.578 COS 28-0.71 sin 48.553.422 COS 48 -4.22) Yxy Kxy I 0. ~~ O'y ·§ !:": ~ 0. The compressive transverse strain is symmetric with respect to the e = 45° orientation with a peak value of almost 10% of the maximum axial strain obtained for e = 90° at e = 45°. therefore. tension.2.18) symbolically in terms of the fiber orientation angle Ex} {Yxy Ey = {5. in case of bending loads can be superimposed.. (2. lkJ+zk} - Qll - Ql2 - Ql6 E~ Kx (2. In order to show all three strain components on the same figure. (2.2.12 as a function of the fiber orientation angle e.844 sin 48 X 10-ll O'x.71 cos 48.6 0. because of the nonzero Q 16 and Q26 terms.3 Laminates of Orthotropic Plies For the specified stress state of O'x = 100 MPa and cry = 'txy = 0.4. From the figure.11) are obtained by through-the-thickness integration of the stresses in each ply. Ex} = {E~} +z {l(x} . (2. (2.4.3.4 ORTHOTROPIC LAYERS 89 =76.87 sin 28.3.18). the stresses in the kth ply can be expressed in terms of the reduced stiffnesses of that particular ply by substituting Eq. the layer experiences shearing strains at off-axis orientations.8 O'x where the superscript "o" represents the mid-plane strains.4.16).4 0. Q16 = Q61 = 42. Q26 = Q62 = 42. 2. Eq.87 sin 28 + 19. { Y~y Ey E~ ~ (2. Eq.37-85. The strain distribution is. (2.. classical lamination theory (CLT) assumes that the N orthotropic layers described above are perfectly bonded together with an infinitely thin bond line and the in-plane deformations across the bond-line are continuous. .19.23) Figure 2.4.4.3). 8 = 0° and 8 = 90°. 2.3.3. 2. and (2. configurations. Strain versus orientation angle for unidirectional lamina in The stress resultants and moment resultants (stress couples) per unit width of the cross-section acting at a point in the laminate (see Fig. That is.578 sin 28 .1 } [ ~ ~:: ~: ~:. Q66 = 26.73 cos 28 + 19. (2. (2.422 COS 481 -0.MECHANICS OF LAMINATED OOMI'OSITE MATERIALS Q22 2. Therefore.88. The maximum shearing strain is at about e = 36°.22) into the stress-strain relationship. The shearing strain Yxy is zero for the two on-axis The same assumptions used in deriving Eq.4.12. The assumptions of constant through-the-thickness strain distribution E0 in case of in-plane loading and a linear throughthe-thickness variation.22) in Section 2. Using these equations.2 10 20 30 40 50 60 70 80 90 fiber orientation 9 "' ax= JOOMPa kt (jx --.0. normalized values of the three strain components are plotted in Fig.20) in Section 2.19.71 cos 48.71 sin 48.3 are applied when plies with unidirectional properties are bonded together.7).4. However. the transverse strain component is multiplied by -10 and the shear strains by -1. B. k=l N (2.z -h/2 h/2 2 } dz.27) are represented as For example.6 B26 B66 n.4. ~. Starting with the integral form of Eqs. h/2 ~.4.6 B26 B66 and assuming all layers are of the same material.23). z 2 h/2 2 } dz and -h/2 r·) where = rBH B..4. z.29) VI{A.6 D26 D66 r~~ o. (2. A"JrJ Y'/ry r· B.21)] proves to be more useful..1 where the expressions for the V's are given by Aii = L (Qii)<kJ (zk.4.} + U I cos 28 {1.2 A22 A26 A.4.4. O. (2.4. z. o"H KY ~} Kxy + u3 cos 48 {1.. z } dz.B.6 A26 A66 Ey + A. D} ={h. the use of sines and cosines of multiple angles [see Eqs. B 1" D 11 } B"J Ky Kxy = U1 {h.zk_.z. -h/2 I (2.4. xy {A 1" B 1" D 11 } Substituting the stress-strain relations of Eq. (2.B.31) (2.zL). Eqs. }• h/2 Bii =2 ~ (Qii)(kl (z~.4. k=l 1~- (2.26) {A 11 . ·~· •w vvmr""\JCIIII: Ml\1 I: RIALS 2.. -h/2 Z. 0.4. (2. Dii = tL k=l N (Qii)(kJ (z~ - zL). D} =JCOS 48 {1. M=BE 0 +DK. 2 v4{A. Z.4..33) nr~~ = My Mxy A. (2.34) (2. (2.4.. n B.25) 't xy k=l zk-l For some design formulations.2 B22 B26 B.). we have (2.4. the expressions for A 12 and B 11 are given by . for example.4 ORTHOTROPIC LAVERS 71 I N: N) = h/2 -h/2 I {ax} crY dz = (k) L J{ax} crY dz.24) N=AE"+BK.28)-(2.D} =I sin 48 {1. B"j N r} EY + Dl2 D22 D26 Y'/ry D.4.-· .. v2{A.4.30).27) Similar expressions can be found for the other stiffness terms and are summarized in Table 2.30) V3{A. z2 } dz (2. we obtain the following constitutive relations for the laminate: =I Q11 -h/2 h/2 { 1. z.28) h/2 VO{A.2 B22 B26 B.4..D} =I sin 28 {1. z2 } dz.4.32) Nxy 't xy zk-l 't xy i NY Nxy M) M: Mxy = h/2 -h/2 I {ax} crY z dz = (k) L J crY N zk {ax} 't z dz.35) -h/2 In matrix notation. z2 } dz.4. B. -h/2 h/2 Z.DJ = Jcos28 {1.26) and (2. z2 } dz.B. (2. (2. N zk k=l (2. t. the V terms take the following form: N V.27).D) V2{A. (2. A more intricate coupling is due to the B 16 and B 26 terms. D 12 } {A66• B66• D66} 2{A 16.2z~k-i + zL)}' (2.4. Laminates loaded by in-plane loads inducing uniform in-plane Nx and NY stress states will experience twisting curvatures Kxy· The terms B 16 and B26 are often referred to as the extension-twist coupling terms and are caused by the existence of off-axis layers that are not symmetric with respect to the laminate mid-plane.B. then B 16 = B26 = 0 even if the laminate is not symmetric.B.4. B.31) and (2.4.zk_ 1 represent the layer thicknesses and zk =(zk + zk_ 1)12 is the z coordinate of the mid-plane of the kth layer measured from the mid-plane of the laminate.D) = L cos 49(k) {tk. bending. After replacing the integrals with summations and performing the integrations across the individual layer thicknesses. of course. we may be tempted to follow the engineering tradition of associating the stiffness properties of a laminate with quantities that are similar to the classical elastic material properties. Equations (2.26) and (2. t. This is. A.B. B. unless the lamination sequence is symmetric with respect to the laminate mid-plane.i"' tk (z~.2z~k-i + zL)}. B.ik. {A.4. 0) between in-plane and bending deformations of a laminate. k=l N v2{A. The set of integrals in Eq (2. D) =L k=l sin 49(k) { tk.4 Elastic Properties of Composite Laminates h/2 A.D) V4{A. tk (z~.37) The in-plane.u3 v3A = h u4.36) respectively.ik. For example.35) can be replaced by summations since the fiber orientation angle 9 is constant through the thickness of each layer. We saw that there is coupling (B :F.2z~k-I + zL)}. t.'~ MECHANICS OF LAMINATID OOMFIOSITE MATERIALS 2.4.B. However. This is true even in the case of laminates made of isotropic layers. and coupling stiffness matrices derived in the previous section are sufficient to describe fully the elastic response characteristics of a laminate under combined in-plane and bending loads. When restricted to symmetric stacking sequences.4. Clearly.B. k=l N v3{A. from Eqs.4.D) {A 1 ~> Bl!. Dl!} B22• D22} {A 12> B 12 .D) V3{A. B 16 . B. it can be observed that existence of a B 11 term implies coupling between in-plane stress resultant Nx and the laminate curvature along the x axis Kx. UI UI u4 Us 0 0 u2 -u2 0 0 0 0 0 0 0 0 u2 Uz u3 u3 -u3 -u3 0 0 0 0 0 0 2U3 -2U3 where tk =zk.1.ik. D matrices In terms of lamina Invariants Vo{A.4 ORTHOTROPIC LAYERS 73 Table 2. vanishing of the B matrix eliminates coupling between the in-plane and out-of-plane responses of a laminate. In this case. If the laminate is made of isotropic layers. as well as between moment resultant Mx and the mid-plane strain E~.2z~k-l + zL)J.B.u3 cos 49 dz -h/2 h/2 I and h/2 B 11 = u. 2. k=l N v4{A.2 = u4 VOA .32) can be solved independently from one another: one set for the in-plane response and another set for the out-of-plane response. D) =L sin 29(k) { tk. These terms may induce twisting deformations in laminates even if there is no applied twisting action. there are no traditional material properties that can reflect this complex material behavior. tk (z~. not possible for general laminates because of the complex couplings that exist between the different deformation modes. V08 + U2 VIB + U3V38 = U2 I cos 29 {z} dz + U I cos 49 {z} dz. DJ = L cos 29(kJ { tk. D1d 2{A26• B26• D26} {A22. Based on this . 3 -h/2 -h/2 (2. the mid-plane strains are only related to the in-plane stress resultants and the curvatures to the moment resultants (stress couples).4. t. tk (z~.DJ Vl{A. The fiber orientation angle of the layers above the mid-plane can take values of 9 in the range 0° ~ 9 ~ 45°. E 2 = 10. the existence of the A 16 andA 26 terms in the in-plane stiffness matrix yields a coupling behavior termed shear-extension coupling. The following example demonstrates some of the coupling characteristics of unsymmetric laminates. Mx)T = { 1. Starting with the bending stiffness matrix.13.0.2462 cos 29-0.05327 cos(360 + 49).32 sin(360 + 49). A 12 = 114.05327 cos 49 + 0.05327 cos 49 -0.1065-0. G = 7.1231 sin( 180 + 29) . My. The elastic properties of the Glass/Epoxy are E 1 = 181 GPa.74 MECHANICS OF LAMINATED COMPOSITE MATERIALS :J 4 ORTHOTROPIC LAYERS 75 simplification. coupling.1231 sin 29-0. The elements of the in-plane. B 11 = -0.0.13.O}T MN. respectively.6 - 61. B 12 =0. NY. (2.8.55 cos 29 + 13.3 GPa.05327 sin(360 + 49). Nx)T = {0. A 66 = 146. Similarly. but unless the off-axis layers are removed from the laminate they will have a finite value.78 sin 29 + 13. even in the uncoupled case. Balanced laminateshave a layer with a negative 9 orientation for every layer with a positive 9 orientation.55 cos 29 + 13. Their relative magnitude compared to the other bending stiffness terms may be made small by manipulating the stacking sequence (as mentioned in the next paragraph).55 cos(180 + 29) + 13.1 - 13.001 m.17 GPa.35) and Table 2. A 26 = 30.32 cos 49 -61.0.05327 cos(360 + 49). there is nothing simple about the composite laminate deformation characteristics. O.32 cos(360 + 49).4. Example 2.32 cos 49 + 61. The net effect of these terms on the laminate response is the induction of shearing deformation under in-plane normal stress resultants Nx and NY. unlike the bending-twisting terms.78 sin(180 + 29).78 sin(180 + 29) + 13. it may again become tempting to define material properties that mirror the classical material properties of monolithic materials. The positive and the negative parts of the pair do not have to be placed adjacent to each another.32 sin(360 + 49).2 A 16-layer Glass/Epoxy laminate with [(9/(9 + 90h/9)2 / (45/ -45 2/45) 2h stacking sequence is loaded such that the in-plane B 16 =-0.0. Determine the mid-plane strains and curvatures experienced by the laminate for the range of 9 under consideration.4. and the bending stress states are characterized by {Nx. A II= 300. However.1065 + 0. v 12 = 0. There are still terms in the in-plane and bending stiffness matrices that display nonclassical material behavior. The resulting effect is the tendency of the laminate to curl under applied uniform bending moments.32 sin 49 + 30.0.1. Balanced laminates with adjacent layers of plus and minus 9 orientations will have smaller bending-twisting terms compared to ones that have layers with the same orientations grouped together and separated from the layers with negative angles. O. the distance between them has a direct influence on the D 16 and D 26 terms. but can be at separate through-the-thickness locations. These terms exist for all laminates that have layers with off-axis orientations.32 cos(360 + 49).28. .78 sin 29. and the individual layer thickness is 12 0.32 cos 49. Thus.13.13. A 22 = 300.55 cos(180 + 29) + 13.2462 cos(180 + 29). the stiffness terms D 16 and D26 couple the moment resultants Mx and MY with the twisting curvature. 0. A 16 = 30.O}T and {Mx.13.32 cos(360 + 49) (all Au in MN /m). However. these terms may be eliminated q uiteeasily by enforcing a balanced condition of the off-axis layers.32 cos 49 .6+ 61. the terms D 16 and D 26 are commonly referred to as the bending-twisting coupling terms. Although the primary reason for the existence of these terms is the presence of layers with off-axis fiber orientations.0.05327 sin 49 . and bending stiffness matrices for the laminate are computed in terms of the angle 9 from Eq.32 cos(360 + 49). 0.32 sin 49 + 30.0. Although the expression for A 26 is not the same as -1 x A 16 above.2708 cos(360 + 48). 2.2708 cos(360 + 48).2708 cos(360 + 48) (all Du in kN.1.06155 + 0.2708 sin(360 + 48). not only the section of the laminate above the mid- D 26 = 0.05327 cos(360 + 48). For 8 = 45°.2462 cos 28 .0.6258 sin( 180 + 28) + 0. D 16 = 0. .2974 cos 48 + 1. B 66 = 0.375 cos 28 + 0.2462 cos(180 + 28). Variation of A1s and A2s as a function of 8 for a [{8/(8 + 90)2/8)2/(45/-452/45)2] r laminate.252 cos(180 + 28) + 0.5° for this laminate. Also. D 22 =6.2708 sin(360 + 48). For the first two values.2708 cos(360 + 48).6873 sin 28-0. the section of the laminate above the mid-plane is the balanced angle-ply (45/-45z145) 2 laminate. 15 10 Note that the three stiffness matrices are fully populated. the curves for A 16 and-A 26 coincide.1065 + 0.05327 sin(360 + 48).4 ORTHOTROPIC LAYERS 77 B 22 =-0.1065 + 0.252 cos(180 + 28) + 0. through trigonometric manipulation it can be shown that for this laminate B n = B 22 and B 11 = -B 12 regardless of the 8 values.13.449. shear-extension coupling terms vanish for the entire laminate.0. and 8 = 45o. D 12 =2.412 + 1.2974 cos 48.2974 cos 48 .13. the section of the laminate above the mid-plane is cross-plied (0/90/0) 2 and (90/0zf90) 2 stacking sequences.6873 sin 28 + 0. For example.2974 cos 48 -0. the only angles that yield a balanced laminate are 8 = 0°. indicating highly coupled deformation modes involving mid-plane strains and curvatures.2974 sin 48 A (MN/m} 25 20 A16 -A26 + 0. For orientation angles that satisfy the balanced condition for the top half of the laminate. respectively. Again.1231 sin 28 + 0.0.1231 sin( 180 + 28) + 0. The variation of the A16 andA 26 terms with the orientation angle 8 is shown in Fig.412.375 cos 28 + 0.0. However.1. Also the extension-twisting coupling terms are B16 = -B 26 • Variation of the magnitudes of the various B terms is shown in Fig. Since there is a 90° difference between the pairs of the layers on the top half.05327 cos 48 + 0. the bottom half of the laminate (45/-45z145) 2 is balanced. D 66 = 3. Note that for 8 = 45°.118-0.0. They vanish for 8 = 0° and 90° angles.14.6258 sin(180 + 28).05327 cos 48 + 0. The maximum values for the shear-extension coupling terms reach a maximum value for 8 = 22.05327 sin 48 . the terms are trigonometric functions of the 8 and some may vanish for certain values of the orientation angle.0. the in-plane 5 5 10 15 20 25 30 35 40 45 e Figure 2. 2. D 11 =6.m). it can be shown through trigonometric manipulation that A 16 =-A 26 • Next we investigate the terms of the B matrix.05327 cos(360 + 48) (all Bu in MN). B 26 =-0.78 MECHANICS OF LAMINATED COMIIOSITE MATERIALS 2.06155 + 0.2974 sin 48 + 0. 8 = 90°. 27 Kxy D16 -53. Figure 2.27) need to be solved.15.0 (2. Variation of 811.6 X 106 E~ l E~) = {0.0 0.27 X 103 Ky [ -92.39) Solution of the six equations for the six unknowns yields E~ =0.27 -53.1 287.27 -92.05 ~Bu 5 10 15 20 25 30 35 40 45 e Figure 2. the stacking sequence is [(30/120z130)zl(45/-45zl 45) 2 Jr. Finally. Y~y Variation of D16 and fk6 as a function of 8 for a [{8/{8 + 90)2/8)2/(45/-452/45)2] r laminate.4. To determine the mid-plane strains and curvatures.27 92. causing all the B terms to vanish. and and D (kN-m) 0.27 92.1 0.14. E~ =0.4.38) -53. For 8 = 30°.78 B (MN) MECHANICS OF LAMINATED COMPOSITE MATERIALS :1. namely.07 -23..733 6.27 53.27] { E~~ 53.3772 Ky [ 0.05095 m/m. but the entire laminate becomes symmetric.6069] X 103 {Kx) + 2. The trigonometric simplifications mentioned earlier are not possible. The constitutive equations become 287. [(45/-45z145) 2 ].2 128.4. or equivalently [(30/-60z130)/(45/-45z145) 2 Jr.0} 0. 822.26) and (2. However.3772 3. there is no orientation angle that will make both terms vanish simultaneously.733 0.4.189 2.402 Kxy = {1.27 53.6069 -0.4 ORTHOTROPIC LAYERS 79 0. coupled equations of the form in Eqs. For space considerations.0 X 106.27 53. 0. the form of the equations will be shown for two numerical values of the orientation angle. 816.066 -D.05095 m/m.2r 0.27 92.15.27 y~y 6. sin 48 and sin ( 180° + 28) terms are slightly different.27 -92.07 -23.15 0.07 [ 23. the bending-twisting coupling terms D 16 andD are 26 shown in Fig.27 53. 8 = 30° and 8 = 45°.0~ 0. The expressions for the D 16 andD terms 26 seem to possess the kind of similarities that exist between the A 16 andA 26 and B 16 andB26 terms.1 23. (2.27] { Kx) + 53.1875 m/m and . 2.2 -23. and 826 as a function of 8 for a [{8/{8 + 90)2/8)2/(45/-452/45)2] r laminate.27 X 103 E~ [ -92.07] 128.27 92.0 xy yo (2. The resulting curves for the laminates as a function of the orientation angle are distinct. =0.07 plane is balanced.27 -53. a closer look reveals that the coefficients of the sin 28 and sin ( 180° + 28) terms. Although the D 26 curve crosses the zero axis. 1231 0.2/m.4 /m.2 Yxy Ey 5 10 15 20 25 30 40 45 e Figure 2.844 0. respectively.686 10 15 20 25 30 35 40 45 e {K} = 110:0 0} K: Kxy Figure 2.018 0. For example.16.3/m. Kxy = -50.0 0. Solution of the six equations for the six unknowns yields Eo x = Eo = yo = 0 •0 y xy and Kx= 233. A single material property value and a single thickness value uniquely identify both the in-plane and the bending stiffness. there can theoretically be infinitely many different bending stiffnesses. which are the crosssectional area A and the second moment of the area /. Strains (m/m) 0.. This is mostly because the utility of such properties is limited. Mid-plane strains as a function of 8 for a [(8/(8 + 90)2/8)2/(45/- 452/45)2] r laminate.41) 452/45)2] r laminate.32/m. the geometric properties of the plate. respectively. the constitutive equations governing the in-plane and bending responses uncouple 273.4.4 141.018 [ 3.772 /m.ou Kx MECHANICS OF LAMINATID OOM. stacking sequence.844 3. Curvatures as a function of 8 for a [(8/{8 + 90)2/8)2/(45/- (2.OSITE MATERIALS 2.0 0.0 200 150 100 50 172. Kxy =-3. The bending stiffness of a laminate is strongly influenced by the relative positioning of the layers with different orientations.17. are conveniently decoupled from the material property.1231 0.1231 X 103 3.4. This is not the case in composite laminates. 0.4 /m.9 [ and 0.17. use of effective engineering material properties may be attempted. For a given total laminate thickness. which corresponds to e = 45°.4 ORTHOTROF'IC LAYERS 81 = 216.0 0. the in-plane and bending stiffnesses of a monolithic plate may simply be expressed by using EA and EI. KY = -104. In this representation.8 l 0 0} X106 {£ Ef = {0 0:0 } (2.16 and 2. For balanced and symmetric laminates where the coupling characteristics are the least pronounced. .40) Yxy 0. 2. even if the properties of the individual layers along their on-axis are identical. given the elastic modulus E of a material.9 141.0 Solution of the mid-plane strains and the curvatures for the complete range of the e is presented in Figs. X 106. KY =-120.4 273. Curvatures ( /m) For the [(45/-45i45) 2].0 s 5. This practice is still not encouraged in composite engineering.1231] 5. There is no simple material property that can be decoupled from the laminate geometry in describing the bending stiffness. which meet the test of true in-plane isotropy. there arc situations in which the representation of the in-plane stiffness properties of a laminate by quantities that emulate traditional elastic stiffness properties may be useful. It can be shown that the new stiffnesses are given by £". the stiffness property will be different and needs to be computed using the transformation relations discussed in Section 2. xy- AJ2 A22 In the preceding paragraphs. they are only referred to as quasi-isotropic because in directions other than the in-plane direction the elastic isotropy disappears.A2 J2 J A22 and pn y = _!_ h (A A 11 22 - An A2 12 J' (2. Also. or [0/90].4... 1 -<1> 2 Gxy Ex Ex ~y Ey J Gxy ~y =£! [ 1/xy Ex (p2 + q2) . however. In a direction other than the two geometric axes. as we will see in later chapters. . Unfortunately. the laminates mentioned above are not the only quasi-isotropic laminates.·--.. A more general condition for laminate quasi-isotropy will be shown in Chapter 4. One final assumption is to enforce A 11 to be lhc same as A22• Such balanced symmetric laminates. Consider. We represent the stiffness matrices (A._1 Ex EY Gxy )p ql ' (2.E p q.42) c. because their stiffness in directions other than the x and y directions may be different.5 PROPERTIES OF LAMINATES MADE OF SUBLAMINATES The stiffness matrices of a laminate made of two or more sublaminates may be derived by using the stiffness matrices of the individual sublaminates directly. £!. such as the [0/±45/90]. 1 n2 n2 ( 1-2 1/xyJ -=::J. Following the approach used in Section 2.4. There are only a small class of laminates. and 190/±30]s laminates. and the superscript "eff' was eliminated from the right-hand sides for clarity.4. for example. Another perspective to this problem will be to imagine that we rotate the entire laminate under consideration. rather than using the Q matrices of the individual plies of the sublaminates. respectively. through an angle -<1> from the x axis and we want to express the new stiffnesses along the x-y coordinates. the method presented here provides understanding of laminate stiffness properties and can be used to construct laminates with certain desired properties hased on the properties of selected sublaminates.(__!_ + __!_ . (2.2.A 12). B.pq Ex EY Gxy Ex ' £t 1 =4 [ -+--+1 1 pq+-(p-q)2.. These laminates will have identical extensional stiffness in every direction and can be shown to have A 66 = 112 (A 11 . the following effective elastic properties may be derived in terms of the values of the in-plane stiffness matrix: ptt X 11 22 =_!_ h (A A .ff =E'/f. [0/±60].4. will have equal stiffnesses E.us PROPERTIES OF LAMINATES MADE OF SUBLAMINATES 83 Nevertheless. G~Y' and ~Y in terms of the stiffnesses described in Eq. 1 2 2 (..3.e_ff xy 1 =hA66 and veff.42). (2.ff G~~ = 2 ( 1 + v~~~ . ~._+!l:_+ FJ Ex E G .43) where p = cos 2 <1> and q = sin 2 <1>.1 2 1/xy -=E.MECHANICS OF LAMINATID OOMFIOSITE MATERIALS .4.. Besides convenience. 2. which reduces lhe shear stiffness to Note the difference between the elastic moduli along the x and the y coordinates of the laminate.44) Although such laminates possess complete in-plane isotropy. which have identical stiffnesses in every direction. a laminate made up of two sublaminates denoted by M and N with nM and nN plies.. X J Y xy X 2. for example I±45].18.2.__+L_+ . we have made a series of assumptions with respect to the laminate construction to simplify the elastic response characteristics. see Fig. which is balanced and symmetric. they will still not behave like isotropic materials. and D) of the two laminates using subscripts M and N and the combined laminate of nM + nN =n plies with subscript MN. 15) and write the difference of the squares of the zk terms as the product of a difference and a sum. . . Stacking sequence of a laminate made of two sublaminates.4) and zN tM (2.)MN= L k=l QiJ(kJ (zk.1) can be divided into S summations. .£. the in-plane stiffness matrix of the new laminate is simply a sum of the in-plane stiffness matrices of the sublaminates. Note that this does not mean that the individual stiffness properties such as Ex and vxy of the new laminate are the sum of the same properties of the sublaminates. then the summation in Eq.. respectively.£..)M+ (AiJ)N. the combined in-plane stiffness matrix is (AiJ)MN= (A. XN zk = Znm Znk =ZNn =n. Rewriting the summations for the individual sublaminates..+l h=n+·+n. If the laminate is made up of S sublaminates. when two or more laminates are put together to make a new laminate. QU<kl (z~.. .5.2.18.5. (2. 2. Qij(kn) tkn 2 + l 2 ~ ~ · · · + . In the case of the two sublaminates M and N of Fig. IJioxM 'M- =n +1 l.£. +I l . + nN =n --. (Aij)MN= L h=I Qij(k) tk + L + L Qij(k) tk.)q' ~ n = .1) • -· ZM 1 Z. and (AiJ)q is the in-plane stiffness matrix of the qth sublaminate... (2. (B.5. ~ Qij(kn) tkn I + .. each with nq plies where q = 1..5. (2. S. nr+nz n+··+n I S Qij(k) tk + .zk-I) (zk + zk-I) h=l h=l 1 n ~- n 1 - S Aij = ~ . + ZN k 'N- !M 2 . We denote the total thickness of sublaminates M and N as tM and tN' respectively. k=l (2. (2..... . the summations can be redefined with respect to the reference axes of the individual sublaminates... ~ Qij(kns'\ fkn ~ S = L (A.5.L Qij(k) t.'i-1 The tk's and Qk's under the summations are independent of the z coordinate and.nk k =nM k =ZMnm t. 2.)MN' we start with the definition of the matrix from Eq. Then the coordinate of the mid-plane of the kth layer measured from the mid-plane of the combined laminate can be expressed in terms of the zMk and zNk as .2) h=n..:zk."'" 84 MECHANICS OF LAMINATI!D COMPOSITE MATERIALS :115 PROPERTIES OF LAMINATES MADE OF SUBLAMINATES 85 Using the definition of the in-plane stiffness matrix from Eq. Then the locations zM and zN of the mid-planes of the sub laminates measured from the mid-plane of the combined laminate are (see Fig. where zk is the z coordinate of the mid-plane of the kth layer measured from the mid-plane of the combined laminate..5.5) (2.20). the in-plane stiffness of the combined laminate is given by n k=l n (A.zL) = 2 L Qij(kl (zk.=~m A ..6) That is.. h=l (2.3) where tknq and Qij(knql are the thickness and transformed reduced stiffness matrix.3..£.zk-I) =L QiJ<kJ tk.. of the kth layer of the qth sublaminate..--- l Figure 2.- I Zk= Zn IN • --!!! 2 X '¥ - where the tk's are the thickness of the kth layer of the combined laminate.5. therefore.l l ' . we have nJ n2 ns For the calculation of the bending-extension coupling stiffness matrix of the combined laminate. n.18) fN zM=-2 (2. (B.)MN= 2 .18. 2' M and N measured from their own mid-planes by We denote the location of the mid-plane of each layer of sublaminates zMk and zNk' respectively. )M+4 (AiJ)N ~~ ~ (2.)N.025/tal = 0.5. by showing that the summation inside the parenthesis of Eq.2/tbr = 0.05/tal = 0.0 X 106 psi.0 = 5. It is interesting to note that besides vanishing BiJ terms from Eq. in which case.5.1/tbr = 0. (2.12) vanish.2 Qij(k) tknN (ZNk + 2 k~I bl nM nM =L bl tN Qij(k) tknM ZMk- 2 L Qij(k) tknM lc=J nN - nN + L bl Qij(k) tknN ZNk +2 tM L bl Qi)(k) tknN· (2. the bending-extension stiffness of the combined laminate does not vanish.70 X 106 psi.5.05] Aluminum and brass have the following properties: Ea1 106 psi.025/tbr = 0. (2. bending-extension coupling. the first and third terms in Eq..5.5.8) are the in-plane stiffness matrices of the sublaminates.5). starting from the definition of the bending stiffness matrix.)M + tM (Bi)N· Compute the A.1/tbr = 0. the bending-twisting terms D 16 andD 26 are also identically zero because the D 16 and D 26 terms of the antisymmetric laminates are equal in magnitude but opposite in sign (the A16 andA 26 terms are already zero because of the balanced condition for the individual sublaminate).1/tal = 0. Show that the bending-extension coupling matrix B is zero for midplane symmetric laminates. (2. ~tMJ ' =L.8) In general. parts of the second and fourth terms of Eq.5.5. By definition.025/tal = 0. Similarly. B.5.5.5.025/tal = 0.05/tbr = 0.7) into Eq. Gbr 106 psi.3.11). (2.2 =zNk + 2 tM - (N for layers of sublaminate M and (2. (2. see Eq. [Hint: the terms (zk + zk_ 1)12 are the locations of the midplane of the individual layers with respect to the midplane of the laminate. (2.10) A practical case of interest is when the individual sublaminates M and N are balanced and symmetric such that (B. tJv ~ = 10.9) Similarly. (2. Gal = 3.60 X x (2. the bending stiffness matrix for the combined laminate can be expressed (see Exercise 4) in terms of the bending. (Di))MN= (Di))M+ (Di))N+4 (A.] .2] (d) [tbr = 0. we obtain (Bij)MN ~ -Qij(k) tknM (tNJ + L.5. ZMk.5. and D. matrices and the effective elastic properties of laminates made up of aluminum and brass layers with the following stacking sequences.025] (c) [tbr = 0.05/tal = 0. (2.2(Ai)M· EXERCISES I.12) Substituting Eq.~ uu MECHANICS OF LAMINATED COMPOSITE MATERIALS - I:XEACISES tM tN 87 zk = zMk .)M = (B.tN (B. and in-plane stiffness matrices of the sublaminates as (DiJ)MN= (Di))M + (DiJ)N + 4 (AiJ)M + 4 (A. 2.11) zk for layers of sublaminate N. (a) [tbr = 0. One such case is that of sublaminate N being antisymmetric to sublaminate M.12). except when the products of the sublaminate thickness 1M and the elements of the in-plane stiffness matrix of sublaminate N ure equal to the products of the sublaminate thickness tN and the elements of the in-plane stiffness matrix of sublaminate M..7) (BiJ)MN= 2 (AiJ)N-2 (Aij)M.025/ta] = 0.1/tbr = 0. All thicknesses are in inches.)N = 0.8) are the bendingextension stiffness matrices of the sublaminates M and N.05] (b) [!br = 0. tM tN (BiJ)MN= (BiJ)M + (BiJ)N + 2 (AiJ)N. Therefore. (2. Ebr = 15. may introduce undesirable effects with changes in moisture content and temperature. S. Martinus Nijhof Publishers. McGraw-Hill. The coupling characteristics. the individual layers will have a tendency to deform differently. Individual ply thicknesses are t = 0. and Pagano N. referred to as the hygrothermal response of the laminate. R. hygrothermal characteristics of the individual layers are directional and depend on the fiber orientation angle. Hyer. Stress Analysis of Fiber-Reinforced Composite Materials. New York. 3 HYGROTHERMAL ANALYSIS OF LAMINATED COMPOSITES 4. and D matrices and the effective elastic properties of the following laminates. W. Halpin. and Sierakowski. Tsai. S. pp... Elasticity in Engineering Mechanics. (1980). John Wiley & Sons. For example.. (1998). Philadelphia. E2 = 13 GPa. and Hahn. Inc. Tsai." In Composite Materials Workshop. Jones. R.10). the 89 . H. J. while providing a favorable design response under applied mechanical loading. REFERENCES Boresi. 2nd Edition. "Invariant Properties of Composite Materials. In Chapter 2 we have seen that the stiffness properties of a composite laminate may be tailored to desired values by arranging the orientation of the layers and the stacking sequence. P. 233-253. W. is brought about because moisture and temperature variations influence the fiber and matrix materials differently. load-adaptive structural systems for aerospace and mechanical engineering applications may be possible through the use of such coupling properties. Technomic Publishing Co. and Chong. Of particular interest are stiffness coupling properties such as extension-twisting and bending-twisting couplings. The latter behavior. Elsevier Publishing Company. L. Complete the steps of derivation of Eq. Tsai.J. Vinson. (1987). G12 = 6.3. W. M. The Behavior of Structures Composed of Composite Materials. Taylor & Francis. Lancaster. (1987).J. (2. R. J. Westport. M. A.5. If these differences are not taken into consideration in establishing the stacking sequence. as they provide additional degrees of freedom in design. MECHANICS OF LAMINATED COMPOSITE MATERIALS Compute the A.013 em. N. Therefore. T.4 GPa. B. C. (1998). Boston. (1998). K.. C.) Technomic Publishing Co. (a) [±45/0/902] 2s (b) [±45/±30/902) 2s (c) [04 /9021zs (d) [(±45/0/902 )/(±45/0/902 ) 2 ] (e) [±45/±60/±15]2s All laminates are made of Graphite/Epoxy with the following properties: E 1 = 128 GPa. W... just like the stiffness properties. Inc. P. PA. (eds. Mechanics of Fibrous Composites. In a laminate under thermal loads. PA. T.88 3. Herakovich. Mechanics of Composite Materials. and v12 = 0. New York. S. Introduction to Composite Materials. and Pagano. New York. (1968). for most practical purposes the cure temperature can be taken to represent a stress-free condition. the laminate is removed from the autoclave and allowed to cool down to room temperature. during the curing process of laminate fabrication. particularly those with resin-based matrix systems. 3. allowing viscous flow to occur in the matrix material with consequent relaxation of the residual stresses. In some situations. Coefficients of thermal expansion Material ( lruphite/Epoxy Boron/Epoxy Uluss/Epoxy ---- a.1 X 10-6 8. the ability to predict hygrothermal deformations may be of significant importance in the design process. At a temperature of about l20°C. Hence. see Table 3. A laminate may consist of several layers at different orientations.1 Temperature and Moisture Diffusion in Composite Laminates Heat conduction through the thickness of a laminate is governed by ·the Fourier equation of heat transfer . for a unidirectional layer. In addition to temperature-related expansions. which begins to flow. the cure temperature may be as high as 380°C. at the elevated temperatures at which the laminate is fabricated.1 HYGROTHERMAL BEHAVIOR OF COMPOSITE LAMINATES In designing the stacking sequence of a composite laminate for anticipated loadings. the coefficients of thermal and moisture-related expansions for fiber and matrix materials are significantly different. the behavior is closer to viscoelastic and the material properties are a function of both time and temperature. Fiber-reinforced composite laminates are generally fabricated at high temperatures. In fiber-reinforced epoxy laminates. For a standard epoxy-based composite system. one must take into consideration the effects of temperature and moisture variations on the behavior of the constituent materials. in any given direction the amount of shrinkage may he different in different layers. slightly heating them. After this curing at elevated temperatures. For example. the temperature is further increased to about 180°C and held at this cure temperature for up to two hours.1. Even though the cure temperature is higher than the temperature at which there are no residual stresses in the system. The fabrication typically involves cutting the prepreg sheets to size. Furthermore. In addition. 3. if one considers a laminate with two layers of the same material but different orientations. In subsequent discussions. As discussed later in this chapter. In actuality. Howl!ver. characteristics of laminates in the presence of combined mechanical and hygrothermal loads are also discussed. Under these circumstances.5 X IQ-6 30. this contributes to the buildup of residual stresses in the laminate.1. The increasing temperature causes a change in the molecular structure of the polymer matrix.1. expand by absorbing moisture from the environment. these differences introduce residual stresses in the system that must be taken into consideration during the design process.02 6.1 HYGROTHERMAL BEHAVIOR OF COMPOSITE LAMINATES 91 laminate may warp as the environmental conditions change.2 X w-6 22. 1998. The process of cross-linking also results in shrinkage of the matrix material.. The present chapter examines the behavior of laminated symmetric and unsymmetric composites under varying temperature and moisture conditions. Table 3. the response of the laminate under thermal or hygral loads is similar to that of a bimetallic plate fabricated from two different isotropic materials. the value of the coefficient of thermal expansion in the fiber direction a 1 is one to three orders of magnitude smaller than the value Uz in a direction perpendicular to the fibers. new cross-linking among the polymer molecules occurs. and the matrix begins to harden. at this temperature. In addition.1 x 10-6 Source: Jones. this chemical shrinkage occurs at elevated temperatures close to the cure temperature. which are necessary to allow for resin curing. and laying them up at the desired orientation angles for the laminate. composite materials. The laminate is then placed in a vacuum bag and heated in an autoclave under pressure.6 x 10-6 X IQ-6 a 2 /°C 22.oc 0. the material behavior is assumed to be lincar elastic with time and temperature-moisture independent material properties. Note that for typical thermoplastic resins. the applied pressure assists in removal of gases and any trapped air. particularly when variations in temperature or humidity are known a priori. with special emphasis on aspects related to the design of such laminates.90 HYGROTHERMAL ANALYSIS OF LAMINATED COMPOSITES :1. Solution of the problem expressed by Eqs. and p is the aggregate density of the composite material. which is analogous to the thermal diffusivity coefficient K/(pc) and is assumed to be constant in the z-direction. in a laminate consisting of layers with various orientations.1) and time t.1. For a unidirectional composite lamina. (3.t) 93 i_ [K az arJ= az P car. Similar to the process of thermal diffusivity. depending on the relative humidity of the atmosphere. e=em.92 HYGROTHERMAL ANALYSIS OF LAMINATED COMPOSITES :t 1 HYGROTHERMAL BEHAVIOR OF COMPOSITE LAMINATES Odeg 90deg C(z. A typical set of initial and boundary conditions for which the above partial differential equation may be solved can be written as where h is the laminate thickness and C.1.1. t) dz.3) is typically provided ir~ terms of a through-the-thickness average specific moisture content. The em is the long-term equilibrium value of the moisture content. at (3. Here em is also a function of material properties that reflect the maximum moisture concentration that can he absorbed in the laminate and is empirically represented as Cm = al(<l>/100)b] where <I> is the relative humidity of the environment. this response under varying humidity conditions is similar to its thermal response-the lamina expands or shrinks to a much greater extent in a direction transverse to the fiber than in the fiber direction (Table 3. K is the thermal conductivity in the thickness direction z. K may be assumed constant with respect to the thickness direction z. T(z. the assumption of constant coefficient of moisture diffusivity in through-the-thickness direction holds for layers with different fiber properties.1.2) Here D is the coefficient of moisture diffusion.1). c is the specific heat of the material. 3. . If all layers of the laminate are of similar material systems and with similar fiber-to-matrix volume ratios.1.4) -h/2 For example. the rate of moisture absorption or desorption is expressed by Pick's equation as D ____ plane of symmetry 90deg Odeg z Figure 3.1. Hence.1. t) is the amount of moisture as a fraction of the dry mass of the composite at a through-the-thickness location z (see Fig. C(z.. (3. -2<z<2.e. for exposures over a long time period the average speci fie moisture concentration is given by e.1. Since the moisture absorption is mostly governed by the resin system. (3. t) is the time dependent through-the-thickness temperature distribution. The resin material in composite systems has an affinity for absorb. (3.1) Here.3) ae = ae 2 2 az at. the different levels of expansion or contraction result in the introduction of residual stresses. rc See Tsai and Hahn (1980). is the initial moisture content.2) and (3. z=(-~·~} t> 0. ( ·• in the composite material defined as h/2 1 e=-.e.' ing moisture or reducing its moisture content. = 1---zexp 8 (-rch22DtJ • em. I C(z.5) e=e.. h h t~ 0. which is a function of the relative humidity of the environment to which the composite is exposed. The term e is also referred to as the specific moisture content. (3. and a and b are constants that reflect material characteristics. Moisture distribution in a laminate. Ef} Ez-Ez t F .Yfz . Note that the expansion in the fiber directions is typically smaller than in a direction transverse to the fibers. the moisture diffusivity coefficient can be assumed constant through the laminate thickness. EM. Eq. assume that this layer also experiences a change in moisture content denoted as 11C. and the superscript F indicates stress-free expansion.4. 3. If the strains of Eq. aggregate density or thermal conductivity) may be ignored in the solution of the heat conduction equation. It is important to emphasize that heat conduction in composite laminates occurs much faster than moisture diffusion. Er where al' ~I' and a 2. respectively. (3.8) Q66 1} = {:~} {~} = {~} {~~} { + 11T+ YI2 yl2 yl2 11C. I.1.7) 0 0 It is important to emphasize again that the stresses may be calculated from Eq. as follows: Consider a single unidirectional layer that is subjected to a temperature change 11T from its stress-free state.6) provides good estimates of moisture content.I. (3.1 HYGROTHERMAL BEHAVIOR OF COMPOSITE LAMINATES H~--~-~- For short time exposures.10) Q66 Yiz.1. Stress-free expansion strains in a unidirectional layer. are expressed as a sum of the free expansion strain.2 Hygrothermal Deformations Also note that no hygrothermal shear deformations are induced in the unidirectional composite layer. such as cryogenic liquid containers. Figure 3.2. Commonly. An exception must be made in the case of very short-period transient loads or environments that have large sustained through-the-thickness thermal gradient along with moisture diffusion.1. (3 .94 HYGROTHERMAL ANALYSIS OF LAMINATED COMPOSITES :1. and mechanically induced strains. 3. the stress components corresponding to the strains would be obtained from the stress-strain relations of a single layer.1.8) if and only if the strains are of mechanical origin.2. (2. (3.ci Hft nh 2' (3.and moisture-related expansion coefficients in the fiber direction and transverse to the fiber direction. overall strains (or total strains) in a layer.9) Stresses in a layer that undergoes both thermal and mechanical strains could then be calculated from cri} = [Qn Ql2 Q12 Q22 {'tl2 0 0 O"z 0 0 1{Ei. an approximate solution presented by Crank (1956). respectively. The resulting free expansion (or stress-free expansion) strains in the lamina can be written as 11 cri} -[Q Q12 {'t12 0 O"z Q Q22 0 12 0 0 1{E1} Ez · yl2 (3. Similarly. = EF +EM. E1.g.1. 7) were of purely mechanical origin. The superscripts FT and FH represent stress-free thermal and hygral effects. where the dark bands indicate the fibers in a unidirectional composite layer. Any changes in the properties of the composite that may be introduced due to variations in moisture content (e. EF =EFT+ EFH.3) of Chapter 2. (3. 95 C-C. These free expansion strains are shown in Fig.1.. ~ 2 are the thermal. Furthermore. --=4 em.1. since equilibrium temperatures are rapidly attained in the laminate. Eq.. the situation is different and somewhat more complicated because each layer with different orientation would have a tendency to expand or contract differently along the axes of the laminate.. To get a clearer understanding of the nature of these residual stresses. therefore.5.3. Figure 3... no expansion of the layer in any direction is allowed) by means of externally applied mechanical strains equal to the hygrothermal strains but opposite in direction.~~L\C ~ . In a general laminate. in the absence of any external constraints.9).tW--1---.4).3. Constraining effect of adjacent layers on 90° layer. then the resulting stresses in the layer would be .5. ~ f I ~ ~~r===~ tr1'W-1---.~----------------------- . • 1 ..1. Free expansion strains in a (45/90/0]Iaminate.. the hygrothermal deformation in each layer would be identical (provided that there is no through-the-thickness variation of temperature or moisture and hygrothermal properties of the layers are the same) and there would be no constraints on layer deformations from adjacent layers. If each layer were allowed to expand freely without the constraining effect of the o deg 90deg I 45deg . individual layers are bonded together. no stresses would be induced in the layers. consider a [0/90Jr laminate shown in Fig. however.. Since in the actual laminate the free expansion strains i i i Figure 3.tW--90degply ---.11) ~12 0 Q66 In the absence of any external constraints. no stresses would be induced in the layers. and the total strain the laminate experiences would have to be determined from the overall equilibrium of the laminate.. Even in the absence of any external constraints.. For example. For a [45/90/0] laminate. for example.4. 3.:\C} .. Again.~---------------------------.. therefore..... -a~ L\T0 (3. . In a composite laminate consisting of several unidirectional layers with the same orientation. Such a constrained deformation would result in the development of residual stresses in the individual layers. A (0/90] cross-ply laminate.} = [Q" { 0 O'z Q12 Qn Q 12 0 0J {-a ~T. the total strains will be equal to the stress-free strains (zero mechanical strains) and.1 HYGAOTHEAMAL BEHAVIOR OF COMPOSITE LAMINATES 97 Stresses would develop in a layer if and only if the hygrothermally induced free deformations were somehow constrained by means of an externally applied mechanical strain field. one can think of each layer deforming against some finite stiffness constraints (as shown in Fig. if the total strains in the layer were completely suppressed to zero (t 1 =0. 3..' Figure 3. the free expansion strains in the x direction (which corresponds to the fiber direction of the 0° layer) are shown in Fig. 3..1. which act like applied mechanical restraints.. Adjacent layers of the laminate provide a constraining influence against free hygrothermal expansion or shrinkage of every layer.tW---1 -M¥-+ a.98 HYGAOTHEAMAL ANALYSIS OF LAMINATED COMPOSITES 3. such stress-free deformations are not possible. (3. the free expansion strains can be written as EFT= (a mz X 1 + a 2nz) t:. (3 .. as shown schematically in Fig. as presented in the following section. the ~·s and t:.1111 HYQROTHERMAL ANALYSIS OF LAMINATED COMPOSITES :1.4. lhc free expansion strains in each layer given by Eq.1 HYQROTHERMAL BEHAVIOR OF COMPOSITE LAMINATES 99 y 0 deg ply produce stresses. In the case of thermal strains only (no hygral strains) and using Eq.7. (3.C are substituted for a's and t:.. as the expansion coefficient in this direction for the former is dominated by the matrix constituent of the lamina.a 2 )mn t:.1. respectively.1.T or rfT =a t:. 7).12) Figure 3.14) 1001( Compression Figure 3. in a direction aligned with the 0° layer and in the opposite state in the 90° direction..3 Residual Stresses For a general laminate made up of layers with different orientations. Even if this laminate is not subjected to an external mechanical load.6. Calculation of the total strains of the laminate in lhis case is not trivial and requires a systematic hygrothermal analysis or the laminate. these equations can be represented as oo goo '-------~~· Tension {~1 'Yxy = {::} axy t:.T.14). (3. which yields EF X 90 deg ply = m2EF + n2EF 1 2' EF y =n2EF + m2EF 1 2' (3. The 90° layer expands much more in the x direction than the oo layer. respectively.. 3.1. the internal reactions that force the layers to remain together can be thought of as mechanical loads that m = cos e and n = sin e. and the superscript FT is replaced by FH in Eq. · where axy is sometimes referred to as the coefficient of thermal shearing. Constrained deformation of the [Q/gQ]Iaminate. Under the constraining effect of the adjacent layer..7.12).13) In matrix notation.T' y y~J = 2( a 1 . where adjacent layer under changed hygrothermal conditions. (3.6.1.1. (2. (3.T' EFT= (a 1n 2 + a 2m 2) t:.2mn E~. 3.1. . Such a laminate would also bend due to the unsymmetric stacking sequence of the laminate.7) can be lransformed to a desired coordinate system x-y following the strain lransformation relation of Eq. In the case of only hygral strains. the expansion would be as shown in Fig. the bonded laminate would expand to some intermediate stage. y~Y = 2mn Ef ... Stress-free expansion of the oo and goo layers.T. 3." . It is clear from these figures that the oo and the 90° layers are in tension and compression.1.T. 16) at a given through-the-thickness location z(k) and the free expansion strains in that layer is the strain that causes stresses. The laminate deformations under consideration are purely hygrothermal in nature. Y < Yw z<k> xy xy l (z) =EN (z)- E~) (z).18) into the stress-strain relation of Chapter 2.4.15) ER} £~ = {£oN} £~N { Y~y <kl X X y~~ + z<kJ {vN} < . discussed in the next paragraph. from the free expansion strains. as they exist in laminates following the curing operation.1. as opposed to its traditional usage to indicate the location of the layer interfaces.1. since they are the cause of residual stresses.4.19) where Qij are the transformed reduced stiffnesses that are defined in terms of the material properties and orientation angles [see Eqs. when layers at different orientations are bonded together.1. These strains are sometimes referred to as residual strains. the laminate will assume some deformed shape that depends on the stacking sequence. the sum of the thermal.IVV HYGROTHERMAL ANALYSIS 011 LAMINATED COMPOSITES 3. (3. Eq. Using the same concept. These hygrothermally induced stresses are sometimes referred to as the residual stresses. (2. of the individual layers. laminate strains that are of nonmechanical origin will be represented by the superscript N and referred to as hygrothermal strains. . EH. EM It is important to note that the strains E~<kl• E~<kl• and 'fxy(k) are the strains in the kth layer induced by the constraining action of the adjacent layers. (3.17) (3. (2. Within the framework of the classical lamination theory discussed in Chapter 2. They are calculated by substituting the strains in Eq.1.19) and (2. Er. therefore.1 HYGROTHERMAL BEHAVIOR OF COMPOSITE LAMINATES 101 In a general laminate. ~ X (3.1.4. The former includes the effect of the restraining action of the adjacent layers on one another as the laminate deforms as a whole. In a symmetric layup. (3. {(J~} a.9). one can calculate stresses induced by the constraining action of the neighboring layers. for unsymmetric layups.1.16) It is important to distinguish the hygrothermal strains in the individual layers. Through-the-thickness distribution of the hygrothermal strains are. given by EN= EoN+ zt(l. (3. (2. such deformations are fully described in terms of mid-plane strains and curvatures represented by Note that (z) indicates through-the-thickness coordinate dependence of the strains. a concept which will be explained later). 'txy w 61 = [Qll Qz1 Q Qzz Q Qz6 Q61 Q62 Q66 12 1 [{E~N} {<11£~} £~ + < .1. the difference between the hygrothermal (total) strains EN of Eq. However. They should not be confused with strains induced in a laminate due to action of mechanical loads acting on the laminate. Even though the stiffness coefficients of most .4. and hygral. changing from one layer to another. The latter ignores this constraining effect and represents the layer strains as if the layers are free to slide over one another. That is.21)].22) based on classical lamination theory. Dependence of the 1:~~> to the z coordinate may arise due to nonuniform through-thethickness distribution of the temperature or moisture. the laminate may also develop curvatures (except when the layup is hygrothermally curvature stable. we define the following strains: EoN} eoN= £~N {YoN xy and KN={~}· < (3. EN. .18) < Y~y <kl which are the total strains and curvatures corresponding to the final shape of the deformed laminate under purely hygrothermal loads. we refrain from using this terminology as it may be confused with overall laminate strains resulting from the application of pure hygrothermal loads (that is.{£F~ c. EF. However. (3. Defining z(k) to be the through-the-thickness location of a point in the kth layer. strains are denoted as EN. Corresponding to these strains.18). Definition of strains that give rise to stresses was made in Eq.1.E. Strain state in a laminate that experiences both in-plane and bending strains is given in Chapter 2 by Eq. it is possible to call the strains associated with thermal or hygral deformation of a laminate "residual strains"). For the remainder of this chapter. there would be in-plane deformations only. aR. 4.1 HYGAOTHEAMAL BEHAVIOR OF COMPOSITE LAMINATES 103 composite materials may be functions of the temperature and moisture.1. there are no external forces or moments acting on the laminate. those integrations yield· vectors of constants that are commonly referred to as the induced hygrothermal or nonmechanical loads. {NN} = { NN] N~ = h/2 J[Q] {EF} E. (2. which leads to the definition of equivalent hygrothermal stress and moment resultants. Finally. see Eqs. Using the definition of the stress and moment resultants. and hending stiffness matrices.20) Rearranging the terms.27) MR] =h/2 {aR} {0} M} J a~ z dz = 0 .102 HYGAOTHEAMAL ANALYSIS OF LAMINATED COMPOSITES :1.21) The computation of the strains and curvatures resulting from purely hygrothermal loading is similar to the computation of strains and curvatures for mechanical loads that create a uniform stress and moments resultants of N. ~.23) -h/2 yF xy 3.25) ~ 0 (3.20) and (3.1. dz yF xy and (3. we know that for general laminates there are residual stresses within the layers. it is important to note that the overall laminate deformations under hygrothermal loads.24) NN xy -h/2 {MN} = M~ = J [Q] h/2 -h/2 NR] =J hl2 {a~} {01 ~ a: dz= o.1. derived in Chapter 2 [see Eqs.C. yF xy (3. except that the stress and moment resultants Substituting the residual stresses of Eq. the stress behavior of the laminate is generally characterized by stress resultants obtained by through-thethickness integration of the layer stresses. are still unknowns at this point. (2.1.4 Hygrothermal Laminate Analysis and Hygrothermal Loads In classical lamination theory.22) and (3.25).1. The terms under the integral above are constant in each layer for a given constant through-the-thickness distribution of !:J.4.28). For a given laminate stacking sequence. which reveals why the terms defined above are called hygrothermal loads: AEoN+ and the vanishing of the moments of the residual stresses by BJt' = NN. there cannot be any stress or moment resultants within the laminate. hence.1.22) oN xy KN xy -h/2 "{F xy [B] EoN} {KN} E~N + [D] K~ {'Y oN xy KN xy h/2 - J [Q] E~ { EF} z dz = 0. However. (3. Eqs. the condition for vanishing stress resultant associated with the residual stresses is given by where A.1.4. B.29). (3. EoN} {KN} J[Q] {EF} [A] E~N + [B] K~ E~ dz = 0. the stress resultants associated with the residual stresses must vanish.1. (2. and D are the extension.19) into Eqs.1.26) (3.4.23) can be written in the following matrix form.24) and (2. z dz.oN and KN.1. this variation is ignored in the present discussion.1. Therefore.T or !:J. { MR xy BE 0 N+ DJt' = MN. For a purely hygrothermal response of a laminate.30)]. (3. -h/2 'tR 'Y 0 (3. and M.21). (3. respectively. (3.1. Determination of the mid-plane strains and curvatures requires completing the)aminate analysis under purely hygrothermal effects.1. { N~y -hn MN] {MN xy {EF} E. {'Y 2 h/ - (3.4.1. bending-extension coupling. we obtain . and (2. 1 HYGROTHERMAL BEHAVIOR OF COMPOSITE LAMINATES 105 are replaced with the equivalent hygrothermal (or nonmechanical) loads.(£oH + H) ~y =11C Y ZJS ' ~xy = !l1C (£~: + z~). 2 K3 = U2 (a1 + a 2) + 2 (U3 + U5 ) (a1 . a calculation of these coefficients is somewhat more complex.22).32) The mid-plane strains and curvatures in these equations are computed from Eq. Such a terminology should be avoided. the coefficients of expansion "at a point" in the laminate are expressed as a x =/1T (£x T + ZKI\ xh 0 1 ay = f1T (£~T + ZK~.5 Coefficients of Hygrothermal Laminate Expansion { toN} KN = [A BJ-I {NN} BD MN· It should be emphasized.1.1. In fact. 3. into Eqs. regardless of the orientation angle. In the most general case.1. For a unidirectional composite laminate. and K3 all depend on the material properties only. For more general laminates.24) and (3. 1 where the constants Kp K2.a (3.31) (3. we have {NT} = {NJ = ~ NI. Complex couplings that exist between the different deformation modes discussed earlier exist under hygrothermal loads. 1 . (3. We first start with the classical definitions of the coefficients of thermal and hygral expansions a= ET I 11T and ~ = EH I !1C. Substituting the definition of the Q matrix terms. + U2 (a 1 . Eq. the strains at a given point in the laminate are expressed in terms of the laminate mid-plane strains and curvatures.7).25) and making use of the free strains of Eqs.1. respectively.1.1.12) and (3. (3.20). it is not too difficult to extend the argument that it is not possible to represent the deformation characteristics of general laminates under hygrothermal loads by defining laminate coefficients.1.4. Therefore. see Eq.1. that the use of letters N and M in hygrothermal loads gives the notion that these quantities are resultants of internal stresses in the laminate.4.30) To a practicing engineer. respectively.a 2). t = t 0 + zK. the thermal loads and moments may be expressed by using the material properties (invariant U's) defined in Chapter 2. the coefficients of moisture. The superscripts T and H denote thermal and hygral effects.K: cos 28 11T z dz -h/2 K sm 28 3 2 cos 28} h12{K1+ K I (3.104 HYGROTHERMAL ANALYSIS 011 LAMINATED COMPOSITES 3. U4 ) (a 1 - = U2 (a1 + a 2 ) + (U1 + 2U3 - a 2 ). t 0 T+ E0 H =toN and KT + KH = KN. (2. the hygrothermal expansion or shrinkage is the same in each layer for a uniform through-the-thickness temperature and moisture distribution.and moisture-related terms. (3. (2.1.4. ). which are in turn computed from I . Finally. (2.28) after differentiating between temperature. involving in-plane and bending deformations.21).4. K 1 = (U1 + U4 ) (a 1 + a 2 ) K2 1 ~x = /lC (£~H + z~). as through-the-thickness integration of the residual stresses actually vanish under pure hygrothermal loading. For such a laminate. Eq.Y JVT} hi2{K1 + K 2 cos 28} K1 cos 28 11T dz -h/2 K 3sm 28 I K~ (3.and thermalrelated expansion for individual layers are often less meaningful than an equivalent coefficient for a general multidirectional composite laminate. 1 axy = !lT (E~~ + ZK~y). Following the discussion in Section 2.1. as they become dependent on layer stiffnesses and the stacking sequence of the laminate.4. though.29) {MT} = { MJ = ~ MTJ Mxy K1 . the coefficients of hygrothermal expansion for the laminate are the same as that of an individual layer. there are special situations in which meaningful definitions may be made. sometimes these terms are erroneously referred to as the thermal stress and moment resultants. Nevertheless. B = 0. 2(a 1 2 a 2 )mn 2 ) MT} {am +an MJ = J[Q] a:n2 +~~2 {Mxy -hn 2(a ~)mn h/2 1- 2 2 ) 11Tzdz=O and (3. we reach the following conditions: If we further assume that the 11T is constant through the thickness of the laminate.1. 2( a. (3. the bending-extension stiffness of a laminate can be eliminated. (3. 2(a 1 . (3. no curvatures would be introduced due to hygrothermal effects.32) cannot be used as parameters that are associated with a laminate as their value is a function of the through-the-thickness coordinate z.1. By imposing mid-plane symmetry conditions for the stacking sequence. in the thermal case.36) -h/2 [3 2)mn Symmetry assumption ensures that all the laminate properties under the integral are mid-plane symmetric. It is also interesting to note that while hygrothermal effects produce no shear strains in a unidirectional layer aligned with the laminate coordinate axes. Eqs. Finally. therefore.a0mn { a m2 2 (3. hygrothermally induced mid-plane strains and curvatures are decoupled and can be solved from NT} Nt {Nxy EoN) { yN 2 2 = hn I [Q] h/2 { a m2 + a n 2 -hn a:n + ~~ 11T dz. For practical applications. In the present section.38) AEoN= NN and D~ = MN.1.1.39) M. Eq.1. in order to achieve zero curvature further assumptions are needed to suppress 1\f".1. Clearly. this dependency on z can be avoided if the hygrothermally induced laminate curvatures are zero. defining the laminate thermal expansion coefficients a through the expression {E 0 N} = {a}I:!T. we can conclude that the laminate coefficients of thermal expansion (and similarly the coefficients of moisture expansion) are ax} a = [Ar' J[Q] { -hn a:y h/2 2 { a 1m + a 1n2 + ~m2 ~n } 2 2 dz.1 HYGROTHERMAL BEHAVIOR OF COMPOSITE LAMINATES 107 {E"T} KT = B D [A B]-l {NT} MT ' {E"H} KH = B D [A BJ' {NH} MH .} {M.1. for laminates with uniform through-the-thickness temperature distribution. .12) and (3.1.34) However. 11T and/or /1C were also mid-plane symmetric.1. = J M~ h/2 [Q] {[3 1m2 + f3 2n 2 ) 13 1n + 132m 2(f3 1 - 2 2 11C z dz = 0. Conditions to accomplish zero curvatures while maintaining some of the stiffness coupling terms are discussed later in Section 3. Therefore. (3. and definitions of the free strains. (3. (3.28) are £~N = [Ar' I -h/2 + ~n 2 ) [Q] a:n 2 + a 2m 11T dz. This would lead to a definition of an equivalent coefficient of expansion for the entire laminate.a 2)mn l (3.1.1. From the definition of hygrothermally induced moments. implying that shear strains are also introduced due to hygrothermal effects. they induce shear expansion coefficients axy or l3xy for general orthotropic laminates.35) { ~xl ~y f3xy = [Ar' h/2 J[QJ -h/2 {[3 1m2 + f3 n2 [3 1n +[32m 2([3 1 -[32 )mn dz.33) For example.37) Therefore.7).39).3. for symmetric laminates with symmetric hygral and/or thermal profiles (of course including the constant through-thethickness profiles).106 HYGROTHERMAL ANALYSIS OF LAMINATED COMPOSITES 3. the stiffness coefficients Qij are in general functions of both the temperature T and the moisture content C.1. (3. the equivalent thermal loads of Eq. and would lead to zero integral if the hygrothermal profiles. (3. the temperature term may be taken out of the inte- . As stated in Eq.1. (3. In this case. we concentrate on a simpler approach for suppressing hygrothermal curvatures. the laminate mid-plane strains from Eq.1.25). these coefficients can be considered simply as the instantaneous values at particular temperature and moisture conditions. (3 .24) are given by Definitions obtained in Eq. 4.1. (2. K3 v2B -h/2 (3.dz = NM -h/2 and J z dz =MM. respectively. NM and MM.2. (2. (3 . (3. Eq. yielding AE 0 M+ B~ + AEoN+ BKN.108 HYGROTHERMAL ANALYSIS OF LAMINATED COMPOSITES ~ :1 LAMINATE ANALYSIS FOR COMBINED MECHANICAL AND HYGROTHERMAL LOADS 109 grals along with the constants Kl' K2 and K3 in thermal loads and moments defined in Eqs. this time in terms of the mid-plane mechanical strains and mechanical curvatures only.5) (3.19). (3.19). For laminates made of layers with the same material properties.1. h/2 h/2 {NT} =~ K 1 VoA + K2 VIA} { Kl VOA . depend solely on the stacking sequence of the laminate and.1. { K3 1 axy (3. <r h/2 (3.41) J o-M dz + J o-R dz = NM -h/2 -h/2 and Jo-M z dz + Jo-R z dz -h/2 = MM.MN = ~.NN = NM and (3. for the sake of derivation we substitute the expression for 1he induced residual stresses from Eq.30). and K 3.2. What is left inside the integrals are the cosines and sines of the orientation angles. The V terms.42) J o-M z dz + BEaN+ DKN.1. on the other hand.1.2.K2 VIA !lT. may be used for designing the laminate stacking sequence.37). Total stresses in a given layer are the sum of these two types of stresses Integration of the mechanical stresses may also be expressed in a fashion similar to Eq.29) and (3.2. 0 Expressing the total laminate mid-plane deformations as a superposition of the mechanical and nonmechanical parts. the coefficients of thermal expansion of a symmetric laminate under uniform thermal loading take the form We have already mentioned in the previous section that through-thethickness integrals of the hygrothermally induced stresses are zero. all the material property information is stored in the constants Kl' K 2. Using definitions of those integrals made in Chapter 2.2 LAMINATE ANALYSIS FOR COMBINED MECHANICAL AND HYGROTHERMAL LOADS For laminates under combined external mechanical and hygrothermal loads. An advantage of this representation is the decoupling of the stacking sequence from the material properties. llowever.19) and rewrite the above equations in the following form: h/2 J o-M dz +AEoN+ h/2 BKN- NN = NM and -h/2 ax} {KI VOA + K2 VIA} ~y = ~ [Ar Kl voA -K2 viA.4) v2A -h/2 3. .MN =MM.4. (3.1. (3. in each layer there are stresses. With these definitions.40) J o. { VIB} h/2 h/2 h/2 (3. therefore.6) BE 0 M + D~ + BE N+ DKN. we now have {MT} =~ K2 -K2 VIB !lT.1. it can be shown that the thermal loads and moments reduce to G (k) =ofk) + G~>" (3.2) K3 v2A -h/2 Since the stresses in the layers have two parts. G~)' resulting from the externally applied mechanical loads. (3.1) Through-the-thickness integration of the layer stresses and their molucnts yields net mechanical stress and moment resultants.1.35).2. in addition to the hygrothermally induced stresses of Eq.3) where the V terms are defined by the summation terms of Eq.2. the thermal loads induced by the temperature difference are obtained as N~ = -78.35 2. In order to include the thermal effects. 'Y~y= 0. X (b) For the goo layers.170 ~ l X 109 £~ = -130 X 22.35 0 0 7. N~ = -78.1. 10.5) ~-27. l -1g5.13) as follows: (a) For the oo layers.8g7 181.170 and due to symmetry of the laminate and the uniformity of the temperature. ~ 0 0 7.170= -130°C that is applied to the composite system.3 0 0 28.£F X X X = -1g5 5 X 10-6' y ER = £ N. J MN/m.8g7 QijOo) = 2.0 (b) For the goo layers.8g7 10.02 X 10-6 = -2. The free expansion strains in each layer corresponding to this change in temperature are obtained from Eq.432 kN /m.56 384.5 X 10-6 = -2g25 X 10-6.1.£F = 2727 X 10-6 y y ' 0 ' =-130 X 0. (a) For the 0° layers.25).18) and the residual stresses are computed by multiplying these strains with the transformed reduced stiffness matrix of the associated layer. and the thermal moment MN were set to zero.8g7 = 2.EF = 2727 X 10-6' y ER =EoN.5 X 10-6 = -2g25 X 10-6.432 kN/m. (3.35 E~ =-130 X 0.1g8 X 10-3 . The nonmechanicallaminate strains corresponding to the above stress resultants may be obtained from Eq.68 l l 0 ~ 0 0 1GPa. 'Y~y= 0. consider the change in temperature dT = 40.65 MPa. The strains that induce the residual stresses in each layer of the laminate are then computed from Eq. £~ = -130 X 22. (3.8g7 10.8 2.1.1.28) as £~N = £~N = -0.EF = -1g5 5 X y y ' X X 10-6' .24) and (3.1. £R X = EoN.3 11. the bending and coupling stiffness matrices are not required.5g Au= 11. { (J~) (J~ "CR xy (O') [181. 0.6 X 10-6.0 0. £~ where the coupling matrix B. (3.8g7 QW0ol = 2.8 0 0 ( ( The thickness of the laminate is 4 X 10-3 m.17oj = 0.112 HYGROTHERMAL ANALYSIS OF LAMINATED COMPOSITES :1 :1 LAMINATE ANALYSIS FOA COMBINED MECHANICAL AND HYGROTHERMAL LOADS 113 Solution The transformed stiffness coefficients for the 0° and goo layers can be written as follows: l81. £R =EoN.8 2.65) 2727 X 10-6 = 27. J GPa Using the definitions of Eqs. (3. 7.6 X 10-6 . and the in-plane stiffness coefficients are obtained as 384.02 X 10-6 =-2. r 0 A Note that because of the symmetry in the laminate. the thermal moments are identically zero leading to zero curvatures. HYGROTHERMAL ANALYSIS OF LAMINATED COMPOSITES :1.. from Fig. the 11C for the 0° ftnd the 90° layers is calculated to be 3. 0. 0. (3. 11T..897 181. (b) 90° layer.01 X 1. the value of the moisture concentration.75 x 10-3 - o.1.8.3 X 3. respectively. 0 0 0. [ 0 0 2.12 + 0.25 X 10-3 - 0.8.1.3 x1?~ 11T. C. N~ = 3.5 1 X { -195.25 X 10-3 + 28 X 10-6 !1T. (3 .5 mm.30 and the stiffness properties Q 11 = 140.5 0. Each layer is 0.0011 0 0.5 0 A= 1. E~ = EfH + E? = 0. The combined thermal. 115 = [ 10. the laminate stiffness and compliance matrices can be computed as follows: 37.75 X 10-3 and 1. Next. • --------------- 90 deg ------~---- 90deg Odeg and the thermal loads of Eq.4 J =EfH +EfT= o.081411T..2.5 37.and moisture-related free expansion strains can be computed for each layer as (a) oo layer.3 x 10--6.65 MPa.0-GPa.2 LAMINATE ANALYSIS FOR COMBINED MECHANICAL AND HYGROTHERMAL LOADS f~L Example 3.0 2727} X 27. Moisture distribution in [0/90]s laminate. = -o. Therefore._ T 1 c Odeg 5.10-3 Ef = EfH + E? = 0.__ -l.8 0 kN/mm. Q22 = 10.0265 0 (kN/mmt'.125 mm thick.3 X 1.0 GPa.25 X 10-3.3 X 10-6 !1T. the laminate strains corresponding to these nonmechanical loads are obtained from Eq.170 ~ ]x Solution 10 9 For the given layer stiffness properties.28): Figure 3.8.01 and~ 2 = 0.0 { J Consider a symmetric cross-ply laminate [0190]s that is immersed in water for a short time so that the moisture variation through the thickness assumes the distribution shown in Fig. Q66 = 5. In this example.oc and ~ = 28 x 10-6/°C in the material axis directions may be assumed for this problem.8 1.8 0 0 7. determine the change in temperature from the temperature at curing.2 I-.1 GPa. 6 E~ = ~H + E~T = 0. The temperature T is assumed to be constant through the laminate depth. Ef l 0.0011 ~ p Q 12 = 3.o1 x 3.0265 -0. at the mid-point of each layer is the average value for that layer and is used in computing the hygroscopic strains.65} 10--6 = -27. 3.59 + 0. Given the hygroscopic expansion coefficients for the material as ~ 1 = 0. and thermal expansion coefficients a.9 GPa. . N/y= 0 Nlm.24) are calculated to be N~ = 3.35 2.897 2. A- = -0.75 X 10-3 + 28 X 10-6 !1T. \ The moisture concentration varies linearly through the thickness of the laminate. 3.. that would produce a zero residual stress cr~ in the 0° layers.080911T. 1 and 3..32 --42..1.87 -42.-.125 + o. and that the laminate is at a uniform operating temperature of 25°C. As in Examples 3.092 + 0 00205 f.T) X 1o--3 .olution 10.oooo415 + o.87 GPa.1 X 10-6 .T) X 10-3 .4 21.. (3.732°C.00205 f. The thickness of each layer group of the laminate is 0.4 37.T 0 X 10- 3 .T} ~ = 0. we need to compute the strains induced by the constraining action of the adjacent layers.(3.( 1. which are sometimes referred to as residual strains. ~ E~ = (0.T 12 y • • ' and setting cr~ = 0 yields a value of the temperature change as f.y axis system as follows: E~ =(cos 45) 2 Ef +(sin 45). The transformed reduced stiffness coefficients for the ±45° layers arc obtained as 56.002068 f.T.87l Q~So) = 42..3 _J Consider an antisymmetric laminate [-45/45/-45/45 4 h made from T300/5208 Graphite/Epoxy.487 X 10-3 . the in-plane.092 + 0. 0 93.116 HYGROTHERMAL ANALYSIS OF LAMINATED COM 1111 LAMINATE ANALYSIS FOR COMIINED MECHANICAL AND HYGROTHERMAL LOADS 117 I E~N} li.6 42..02 X 10-6 1°C and a 2 = 22.6 [A]= 8~6 [ 113. 3 --:7) . [D] = 28.o3 t-.2 84.6 42.1.2.745 X 10- 3 .59 For the laminate layup as described above. with the stiffness properties given in Example 3.T= -10.32 42.002068 f.:so) = 42..T) x 1o-3 = -o. These strains are obtained by subtracting the stress-free expansion strains from the nonmechanical strains as follows [see Eq. Example 3. coupling. The residual stress cr~ in the oo layer is computed as (JR x ~ JMN/m.T) x w-s = o.2 37.2.0 0 In order to compute the value of residual stress. E~ = -3. the free expansion strains in each layer in the fiber direction and transverse to the fibers are obtained as Ef = -3.oooo2595 t-.1 l Nm..87 46.59 56.o. which can be transformed using Eq.oo 1033 .8 [ 0 0 0 0 31.32 56. [ --42.32 56. a 1 = 0.87 42.oooo0237 t-.2 ~ ~ = Q(Oo) £R II X + Q(Oo) ER = 0 002748 + 0 000256 f. [B] = 0 0 [ 0 0 21.079 + 0.T.745 X 10.5 mm.079 + 0. Assume the following values of thermal expansion coefficients in the material axis system: The laminate is at a uniform temperature of 25°C.8 28.18)]: E~ = (0. determine the laminate curvatures.4 kN.12) to the laminate reference x.2. cr~ in the 0° layer. [ 42.87l Q.2.4] 21. = [Ar 1 lN~} .87 GPa. unu flexural stiffness matrices may be computed as follows: 113.87 46.o.6 --42.5 X 10-6 /°C E~ =(sin 45) 2 Ef +(cos 45) 2 Ef = -1. (3.2.028 t-. which implies that each layer is subjected to a temperature change t-.75.T = -155°C. Assuming a laminate cure temperature of 180°C.2 21.1.6 42. 2 Ef = -1. A22 0 A66 B6. Such couplings. the rl!uder should recognize that there would be hygrothermally induced Nlresses and strains in unidirectional laminates as well. may cause warping of the laminate during the cure and cool cycles.2) 0 .1) Note that for such an antisymmetric laminate.8 kN/m.04 X Io-4 -3. only a twisting curvature is induced when the laminate cools from its curing temperature to an operating temperature of 25°C.484 X I o-3. In the following discussions. and there have been several recent studies w. (3. 0 0 Bl6 Du 0 0 0 B26 D21 B6. it is worthwhile to examine specific examples of l'ouplings in composite laminates and their importance in structural dl!sign. If some components of the B matrix are nonzero. 3.here these couplings have been exploited in the design of aerospace structural components. between out-of-plane loads and in-plane deformations. families of unsymmetric composite laminates that provide desired elastic coupling characteristics and maintain shape stability during changes in hygrothermal variations are presented. M~=O.3 HYGROTHERMAL DESIGN CONSIDERATIONS Residual stresses and deformations of laminates induced by changes in hygrothermal conditions constitute a major design consideration. Bending-extension and twisting-extension stiffness coupling terms in unsymmetric laminates. however.2 presented computations of the stresses and Nlrains in the composite material due to moisture and temperature vari111 ions. The laminate strains and curvatures for these thermal loads can be solved from -3.1. may be essential from a design standpoint. As a specific example of the latter.1) B D B]-I {NN} MN Ex Ey = [! ~] Exy Kx Ky Kxy I (3.8 kN /m and W:Y =0. or conversely.1 and 3. B62 0 0 0 0 B62 D12 D22 B16 B26 oI 0 0 D66 (3. In that study. In the absence of such a restriction. N~ =-46.118 HYGROTHERMAL ANALYSIS 01' LAMINATED COMPOSITES :1 :1 HYGROTHERMAL DESIGN CONSIDERATIONS 119 '(~~ =2(sin 45 cos 45) Et'. the moment resultants do not induce in-plane deformations. A review of the literature provides a somewhat misleading conclusion that all general unsymmetric laminates have limited practical use because of their susceptibility to warping. then one can expect couplings between in-plane loads and out-of-plane deformations.624 Sections 3. consider an anti symmetric laminate [<!>I . The thermal loads and moment loads are calculated from Eqs. there is no coupling between the inplane loads and out-of-plane deformations.04 x w-4 = 0 0 0 0. If the components of the B matrix are all zero (as would be true of a symmetric stacking sequence).3.1. Likewise. M~=O and M~=6. as for unsymmetric laminates. A study by Winckler (1986) was the first to explore a class of thermal warp-stable composite laminates.36N.<!>Jr.24) and (3. The stiffness matrix for such a laminate has the following nonzero components: A 11 A 12 0 0 0 0 Az. In order to understand how hygrothermal effects influence deNign decisions.2(sin 45 cos 45) E~· =3. we will assume that all layers url! of the same material. for example.25) as N~ = -46. Let us reexamine the general laminate equation relating stress and moment resultants to strains and curvatures: Nx NY Nxy Mx MY Mxy EoN { (m/m)} =[A KN (m.3. It is a useful exercise to understand the physical mechanism that gives rise to couplings in laminated composites. this can be accomplished by limiting the stacking sequence to be mid-plane symmetric. When the two halves are bonded together as a laminate. the A16 terms of each half contribute to the development of the extension-twist coupling of the combination (Fig. However. Such a laminate would develop a twisting curvature. where the shearing deformation is l'qual in magnitude but opposite in sign. (a) Stress-free and (b) constrained deformation of a [<1>/-<I>]T antisymmetric laminate. an easy way of ensuring hygrothermally curvature-stable laminates is to select a stacking sequence for which Bu = 0. the coupling terms Bu offer a degree of freedom in the design process and have significant merit in the design of aerospace structures. 3. Consider the antisymmetric laminate [<j>J. The large axial centrifugal load in the blade provides a mechanism to induce twist in the blade to a desired spanwise twist distribution if the extension-twist coupling is retained in the composite material.lJa.120 HVGROTHERMAL ANALYSIS OF LAMINATED COMPOSITES :13 HVGROTHERMAL DESIGN CONSIDERATIONS 121 The nonzero B16 and B26 terms represent extension-twisting coupling in the laminate. they are symmetric with respect to their own mid-plane and have no extension-twisting coupling of their own. B22 . is to develop a laminate stacking sequence that preserves mechanical coupling without inducing curvature instability (a terminology used to imply stability of the deformations without twisting under hygrothermal loads). u 1+<1>1n and a [-<j>]n laminate. The optimal twist distribution changes with the rotational speed of the blade. However. To see how this would work.1. \. This is possible if the in-plane hygrothermal shear strain in the laminate is zero [see Eqs.24)]. a nonzero B16 is obtained for the combined laminate. the i\ lh term of the total laminate would be zero. In addition to tension-twist coupling.1. llowever.9b). Hence.<Pnb which has a strong extension-twist l'oupling. we have seen how variations in temperature or moisture conditions induce residual stresses in the laminate. The blade must be appropriately twisted along the span in order to optimize aerodynamic performance. there is an in-plane Nhcar coupling A16 in each half. that is. however. ~Y' under in-plane stress resultants Nx or NY resulting from mechanical loading. A challenge to the design engineer. where the stress and moment resultants due to mechanical loads are replaced by quantities that reflect hygrother. If the only nonzero coupling term is B16. that is. p p p (a) (b) Figure 3. Tension-twisting coupling. This is illustrated in Fig. then for the twisting curvature terms K~Y to be zero.28)]. Although the net resultant of these residual stresses is zero. Note further that the coupling terms /1 11 . one approach would be to develop a laminate for which the components N~ and M~ are exactly zero. the reader is referred to Eq.27). this laminate would also develop a twist with changes in hygrothermal conditions.3.23) and (3. (3. for example. therefore. Provided that the hygrothermal profiles are uniform through the thickness. for all cases other than <j> = oo or 90°. Winckler (1986) presents a class of laminates referred to as HTCC (hygrothermal curvature-stable coupling) laminates that possess the desired properties. If each half of the laminate is considered separately.1. It is clear from this equation that the existence of nonzero Bu terms would indicate deformations involving curvatures under hygrothermal loads.9. and B66 are all zero for this antisymmetrical laminate. (3. can be used in the design of helicopter blades for aeromechanical efficiency.1.1. (3. B 12 . It is possible. that is. In Section 3. to introduce mechanical coupling terms in u laminate without sacrificing hygrothermal curvature stability for temperature or moisture changes.· mal or nonmechanical loads. nonzero B16. to achieve such a state the laminate undergoes deformations that can be quite complex These deformations are computed by using equations that are simila1 to those used in computing the deformations under mechanical loadings [see Eq. . However. (3. This coupling term is of equal magnitude for the two halves. (3. is a finite B 16 which gives rise to the desired tension-twist coupling. a [b d - b] [A BJ B D 1 (3. and twisting curvature Kxy is to be preserved. be formulated as an optimizal ion problem by requiring as an objective a certain strength of coupling us indicated by the magnitude of. That is. Using the partitioned form of the inverted stiffness matrices leads to the mathematical condition. Representing the inverse of the combined stiffness matrices by a partitioned matrix with three 3 x 3 submatrices a.l however.3) ' the second row of matrix equations forK of Eq. exhibits hygrothermal isotropy. When the two halves of the laminate are put together. with an equality constraint on the curvatures as follows: maximize B 16 subject to K = b NN + dM N = 0. there may be cases where we would benefit from having mechanical coupling. narvature stability to prevent warping under changing environmental ronditions. particularly if the condition of antisymmetry in the laminate is not explicitly imposed.(<j> + 90)zl. the in-plane stiffness properties of [0/90]s laminate. As mentioned earlier. It is important to recognize that the selective control of a coupling coefficient such as B 16 can result in undesirable changes in other components of the B matrix.3. the inplane shear coupling for the resultant laminate is zero. rotating entire stacking sequence with respect to the coordinate axes by <1> suits in a change in its in-plane stiffness properties.1.. A number of optimal solutions for such hygrothermally stable laminates obtained through the use of genetic algorithms are presented in Chapter 5. b.·qual to zero. however. Consider. each half of the laminate is symmetric and is hygrothermally curvature stable. Ex= EY.<j>]. unlike mechanical loads. this would require that the coupling wefficient B 16 be nonzero. ropy. moisture and temperature changes are direction independent. the laminate [<j>/90 + <1> ls will have nonzero plane shear-extension coupling terms. but of opposite sign.3 HVGROTHERMAL DESIGN CONSIDERATIONS 123 This observation leads us to define in-plane hygrothermal isotropy and discuss how it differs from in-plane mechanical isotropy. Any adverse effects on ad- .. and d. Other variations of such hygrothermally curvature-stable laminates are possible and are discussed in Winckler ( 1986) and Hajela and Erisman (1993). this laminate has stiffnesses in both x and y directions. It can be shown that any balanced symmetric laminate for which the coefficients of expansion along two mutually perpendicular directions are equal will display in-plane hygrothermal isot. The HTCC laminates discussed here are a candidate solution to this problem. however.1. This design problem can.<j>/. For an arbitr<""'' value of the angle <j>. However.3. One laminate stacking sequence can be written as follows: [<j>/(<j> + 90)zl<j>l. it does not provide for a direct control on the strength of the desired extension-twist coupling term. each half does have nonzero extension-shear strain coupling A 16 when the angle <1> is not oo or 90°. there .4) This laminate is clearly antisymmetric. which are applied specific directions. A 16 and A 26 • The same laminau:: •. (3. therefore. Consider a case where the coupling between in-plane load N. In the context of l'lussical laminate plate theory. A hygrothermally curvature-stable laminate that retains tensiontwist coupling can be assembled from laminates that have · hygrothermal isotropy but are not mechanically isotropic. yet we would prefer hygrothermal Note that constraints on strength and stiffness can be included in this problem in a routine manner. K= bNN + cJMN =0. and the absence of hygrothermal shear strains prevents laminate twisting. In-plane hygrothermal isotropy of each half ensures that there is no induced thermal bending.. However. temperature moisture changes will result in uniform expansion and contraction all directions and the laminate will not experience any shear This is because.122 HVGROTHERMAL ANALYSIS 01' LAMINATED COMPOSITES :1. To derive the condition that causes the curvatures under hygrothermul loads to vanish. for example. As described in Chapter 2. while this approach yields the desired hygrothermal curvature stability.28).28) can be set c. we can start from Eq. say B 16 . Spain. E 2 = 10. Oxford University Press. G 12 = 2. 3. ~2 = 0. ~I= 0.3 m/m.124 HYGROTHERMAL ANALYSIS OF LAMINATED COMPOSITES REFERENCES 125 ditional terms in the B matrix would have to be controlled by adding design constraints in the optimization problem. "{12 = 0. 2. London. E 2 = 10. P. The thermal (a) and moisture (~) related expansion coefficients of a high-strength Carbon/Epoxy unidirectional layer in the material axes system are given as follows: a 1 =. and Erisman. Q 11 = 181.02 X 10-6 (m/m)JOC. Consider an unsymmetric composite laminate [Oz1902 ] fabricated from a high-strength Carbon/Epoxy unidirectional layer of thickness 0. (1993).897GPa. A. Determine the total strains in each layer of the laminate if the operating temperature is 20°C and the laminate is subjected to the following stress resultants: Nx = 15 KN/m. The elastic constants for the material are E 1 = 140. Uz = 28.3 x 10--<i (rn/m)/°C. Winckler. (b) Find the residual stresses in the 0° and 90° layers for the operating conditions of part (a).3 GPa.0 X 10-6 (mlmWC. S. Mathematics of Diffusion.897 GPa. A theory for warp-free unsymmetric composite plates and beams with extension-bending coupling.34 GPa.6 m/m. Determine the equivalent laminate coefficients of thermal and moisture related expansion for a symmetric laminate [0/90/45/-45] 5 fabricated using this unidirectional layer. Q33 =7. New York.0 EXERCISES 1.0 m/m.3. Q 12 = 2.0 GPa. The material axes stiffness coefficients of this layer are known to be as follows: E 1 = 140. a 2 = 22.3 GPa. ~2 = 0. Mechanics of Composite Materials. and Hahn. PA. £2 = 0. dissertation. July 7-9. Lancaster.0 GPa.15 mm and with properties given in problem 1. Q22 = 10. G12 = 2. 2nd Edition. REFERENCES Crank. Troy. . S. . Ph.012 X 10-3. T. determine the angle 8 that minimize the steady-state laminate expansion strain £x if it were submerged in an identical cooled fluid. Assume that the stress-free curing temperature is l25°C. R. v12 = 0. 4.. Philadelphia.6 X 10-3. PA.17GPa. Introduction to Composite Materials. (1956).8 GPa. All layers in this laminate are of equal thickness t0 = 0.0. A unidirectional composite layer undergoes the following steady state expansions when immersed in a cooled liquid medium: £1 = 0. Technomic Publishing Co. Tsai. (1998). H. Taylor & Francis. If this layer is used to fabricate an angle-ply laminate [82 /-8 4 /8 2]. J. Zaragosa.and moisture-related expansion and the stiffness coefficients are given as a 1 = -0.125 mm.01 rn/m. J.5%. Consider a Graphite/Epoxy laminate with a stacking sequence of [0/90/45 2L.5 X 10--<i (m/m)/°C. Rensselaer Polytechnic Institute." Proceedings of OPT/'93. (1980). Hajela. NY= 10 KN/m. W. ~~ = 0.897 GPa. M. assuming the cure temperature to be 122°C. "Genetic Search Methods in the Design of Thermally Balanced Composite Laminates.D.3. (1986). (a) Find the stress-free hygrothermal expansion strains ef and £~ at temperature 22°C and a moisture content of 0. The material axes coefficients of thermal.Iones. v12 = 0. There are also many applications where the design of the laminate is primarily governed by in-plane stiffness requirements. Not only is it simpler to perform the analysis. Unlike stiffness. because of the incomplete understanding of the various failure modes of composite laminates. This is the result of the existence of many complex modes of failure that will be discussed in Chapter 6.4 LAMINATE IN-PLANE STIFFNESS DESIGN In Chapter 2 we discussed the use of classical lamination theory (CLT) for calculating the stiffness properties of composite laminates. That is. Another argument for studying in-plane stiffness design problems alone is that they are simpler than most other laminate design problems. strength requirements are often transformed into stiffness requirements. In this chapter we will use the expressions obtained in that chapter for designing laminates with desired in-plane stiffness properties. In many applications the designer is concerned with the strength as well as the stiffness of laminates. Furthermore. the strength of composite laminates is not easy to predict. This is because the 127 . limiting laminate strains as surrogate strength constraint is a common design practice. but in-plane stiffness does not depend on the arrangement or stacking sequence of the layers with various orientations in the laminate. which was developed for designing composite laminates for buckling constraints. As discussed in Chapter 2 (Section 2. which is equivalent to an average stress d. = N/h. without regard to their z location.1) 4. For example.· i ng that induces transverse stress resultant NY.A22 A12 (4. defined for the composite laminate. the stress resultants are typically specified based on the stress analysis of the laminate under applied loads.G . is assumed to produce the following axial and transverse strains: Ex= crJE~ff and Ey = -1/~~Ex' respectively.h _. These elastic constants are defined in a manner analogous to the definition of the elastic moduli H1. and the Ex' EY' Gxy' andvxy shown below will be used to represent the elastic properties of a composite laminate: Ex. and the shearing stress resultant Ntr = haxy leads to a shear strain y:~ = ax/Gxy· Altogether. £ 2.h _ .hEeff Vxy 'y ' X X X __1_ (. The third section shows how the graphical solution technique can be used also for the case where the thicknesses and the angles are limited to discrete values.128 LAMINATE IN·PLANE STIFFNESS DESIGN 4. For the sake of convenience. Similarly.rff. in-plane stiffness design is particularly suitable for introducing the basic concepts and simplest methods of design optimization. comparison of these equations to the constitutive relations expressed in terms of the A matrix.1.1/ yx X ' y y _ (jxy _ Nxy Yxy.4.hPff y. the PASCO program (Stroud and Anderson._ z. To further simplify the formulation. permits limits on the Aij's. One prerequisite for separating the in-plane response from the bending response is a zero bending-extension coupling matrix B.1. we formulate the design in terms specific to the problem at hand. In this section we also review optimization formulation and terminology.1/eff.42) in terms of the elements of the A matrix. G xy and ~~f. the superscript "eff' will be dropped in the remainder of this chapter. for an orthotropic laminate (zero or negligible A 16 and A 26 ). 1981).2) For laminate design problems. G12 and Poisson's ratios v12 and l/21 • That is. Second.) __1_ (N effN) Ey. xy. We seek a laminate that will not exceed given strain limits under these resultants and that will have some desirable stiffness properties. laminates which have a -0 layer for each +0 layer) where A 16 = A 26 = 0.) __1_ (N _ ettN ) Ex.2. Sometimes.1 Design Formulation of In-Plane Stiffness Problem 1 Gxy= h A66 The simplest measures of laminate in-plane stiffness to obtain are the coefficients of the A matrix.yx ax .Eeff ax xy ay . which can be achieved by symmetry of the stacking sequence with respect to the laminate mid-plane. we recast the problem as a mathematical optimization problem. Finally. The thickness of the laminate will either be given.!_ (A11A22. yields the effective elastic constants shown in Eq.Ar2J A 22 and Ey.!_ (AuA22Aj2J ' A 11 4. we consider in this chapter only the design of balanced laminates (that is. such as Eq.4). and 1/ -~.1 DESIGN OPTIMIZATION PROBLEM FORMULATION Finding an optimal design is often handled as a three-step process.hG . however.1. E~ff. (2. As such. under load. First. The first section seeks to define and formulate the stiffness design problems that we want to solve. The material in this chapter is organized into three sections. E~ff. These are frequently used as design requirements in the composite materials literature. a specified value of the stress resultant Nx. ure also used as stiffness measures.23).4. (2. The second section describes graphical solution techniques for optimization of orientation angles (sometimes referred to as ply orientations) and stack thickness treated as continuous variables.Eeff ay. we solve the problem using one or more available optimization methods. the induced strains are characterized by E~tt and v~~t.1 DESIGN OPTIMIZATION PROBLEM FORMULATION 129 in-plane stiffness matrix A simply sums the contributions of the inplane stiffness properties of layers with various orientations. effective engineering elastic constants. this leads to the following form of the constitutive equations: __1_ (. xy xy (4. or we will . Since the expressions for Q 1 p Q 12. El::. (a) we get 1 Ey <E . .89 x 106 psi. G 12 = 0. For this all-zero laminate. (2. (b) iYx)::.000 lb/in.000 Similarly.3. Of course. t <G <cu Gxyxy. We use Eq. the design problem may be formulated as find a laminate to minimize h E~::. of the associated quantities.Y' 18.xy' 0. from Eq.500. Q 12 = 0. Consider the following three laminates as design candidates: (i) [O]n.25 Q22] 6.•~u LAMINATE IN·FILANE STIFFNESS DESIGN . Au= hQij' where Q. Example 4. Ex ::. (ii) For the [±45]ns laminate. (b) we obtain or h ~ 0. Therefore.8) to obtain the Qij's.000 lb/in. Q22 andQ 66 in Eq. by definition. Q66 = 0.j are the constants associated with the 45° layers: A 22 =Au= h Qu(45°) = - = h [0. determine the smallest value of n that will satisfy the specified strain limits.1.6% shear strain. but is the number of sublaminates in half of the laminates for cases (ii) and (iii).4. there is no difference between the contribution of the 45° and -45° layers.1.004 or hEx~ 2.540 in.1.908 x 106 psi._!2) and G.28) _!_o calculate the Aij's. v 12 = 0.93 x 106 psi. it is not necessary to have upper and lower limits for every strain component and effective engineering constant. So that from Eq. Ex::. 4. vlxy<u < Vxy u "xy- where h is the laminate thickness.135 in. The allowable laminate strains are 0.67 x 106 psi. A Graphite/Epoxy laminate (E 1 = 18. That is. (4.e.5 x 106 h ~ 2.006 xy or hGxy ~ 500. E~.005. we will need to have a total of I 08 layers [0] 108 yielding h = 0. (2. Y~Y' E~::. (4..3) or h ~ 0.538 in.5 (Q 12 + 2 Q66) + 0.1 Since the thickness h has to be an integer multiple of the layer thickness of 0. We next use Eqs. Then identify the lightest laminate.~00 ::.000 (4.005 in) is to be designed for two load conditions: (a) Nx = 10.93 X 106 psi.y<£U . Q 11 = 18. E y y- (i) For the first load case we expect that Ex will be larger than Ev' so that from the first equation of Eq. respectively.4. the calculation is more involved. and the superscripts u and l denote upper and lower limits. and (b) Nxy = 3. Q22 = 1.361 x 106 h lb/in.000 lb/in. (2.1) the strain limit may be expressed as Ex= 1 ~~00::.500. (ii) [±45]ns' and (iii) [±45190/0]ns· For each laminate.. Ex= E 1 and Gxy= G 12.000 lb/in. Note that n is the number of layers for case (i).5723 x 106 psi.1. 0.4% normal strain and 0.4.19) involve only squares of cosines and sines. E 2 = 1. and t = 0.1) we get Yxy = 3 < Eu Y' .1 DESIGN OPTIMIZATION FIAOILEM FORMULATION 131 seek the laminate of minimum thickness that satisfies the strain limits and has acceptable stiffness properties.5 x 106 psi.25 Q11 + 0. 0.93 x 106 h ~ 500. X (a) such that For the second load case we have only Yxy' and from the last of Eq.4. 787 in.04 in.1.501 x 106 h lb/in. so that the Nxy load forced the large thickness. The first inequality is more stringent and yields h ~ 0. (2. =4.000 lb/in. (4. Using Eq.4. so we need h = 0. 2 Q 12 .500. mathematicians and numerical analysts have developed a host of methods that are available for solving optimization problems. (4.000 lb/in.5 h [Q12 + "(212(45°)) = 2.36 or [45/-45!9010] 9s.894 X 106 h lb/in.2 Q66) + 0. =0.2) we obtain .1 represents a typical optimization problem for the design of a laminate subject to limits on its in-plane stiffness.4. Note that the all-zero laminate was very flexible in shear. which is the lightest laminate. and the solution could be found easily. h has to be a multiple of 8t = 0.2 Mathematical Optimization Formulation = 0.5 Q 12 ] E)z = (A 11A22 -Ai 2)/A 22 =7.25 h [Q 12 (0°) + "Qn(90°) + {?12 (45°) + (2 12 (-45°)] = 0.25 (Q 11 + Q22 . Due to space limitations.500. A 66 = h Q66 (45°) Gxyh = A66 = 2.000 lb/in.25 h [(!66(0o) + "Q66(90o) + {2"66(45o) + Q66(-45o)] = 0.176 x 106 h lb/in. However. the designer can make use of optimization software packages embodying various methods.552 x 106 h lb/in. So we get h = 0. From Eqs. Since the laminate is made by repeating a stack with four layers and is symmetric.1. Using Eq. 4.1 DESIGN OPTIMIZATION PROBLEM FORMULATION 133 A 12 = h (2 12 (45°) = h [0. h must be an integer multiple of 4t = 0. we can cover only a small fraction of these methods in this text. = 4.132 LAMINATE IN-PLANE STIFFNESS DESIGN 4.5 Q66] = h [0.1.858 x 106 h lb/in.858 x 106 h lb/in.1. the same problem is quite formidable.25 (Q 11 + Q22 - Then from Eqs.1.894 x 106 h lb/in. Then from Eqs.28) we get A11 4.331. A 12 = 0.858 x 106 h ~ 500. and we will return to it later in this chapter. (a) and (b) we obtain 3.325 x 106 h lb/in = A22 . For this example we limited ourselves to three possible stacking sequences.80 in or [45!-45] 4os· (iii) The last laminate is the quasi-isotropic one.894 x 106 h ~ 500.2) we get E)z = (A 11A22 -Ai 2 )/A22 = 3. The situation was reversed for the (45/-45) laminate which was too flexible in the axial direction.02 in.4.537 X A66 106 h lb/in. The objective of the present section is to illustrate how the in-plane stiffness design problem can be formulated in the standard format of Section 1.19) and (2. Once a composite design problem has been formulated in this general format. The quasi-isotropic laminate represents a good compromise for the two load cases.552 x 106 h ~ 2. Example 4.25 h [Q 11 (0°) + Q 11 (90°) + Q 11 (45°) + Q 11 (-45°)] = 0. Since the laminate has repeating stacks with two layers each and is symmetric.000 lb/in.1. Gxyh =A 66 = 4.4 Q 66 ) + 0.25 h [Q 11 + Q22 + 2 Q 11 (45°)] = 8.176 x 106 h ~ 2.5 h [Q66 + Q66(45°)] = 2. (a) and (b) we get 7. Fortunately. The first inequality is more stringent and yields h ~ 0. Without the limitation to a small number of stacking sequences. general terminology and notation most commonly used in mathematical optimization papers and textbooks was introduced earlier in Section 1.4. 2. In the example.1. An optimization problem usually also has constraints that limit the choice of design. If we replace this design variable with a continuous variable. which is the thickness h. so that we need a constraint that will prevent more than one of these design variables to be nonzero.= 0 if it is not. [45/-45]. However. j = 1. ng. The example had another design variable. In the example we have done essentially that. In the example. It is a choice design variable. When the number of choices is small. We would like to use the choice design variable and formulate the problem in the standard format of Section 1.5) This is a cumbersome definition of J. which in most cases eliminates the need for conditional statements. . since the density was constant for all candidate laminates. In Example 4. . This is accomplished easily by the equality constraint.. in the example. it is common for designers to replace integer variables with continuous ones.. ne. we solved three optimization problems with one variable (n). One way to characterize this choice is by an integer variable taking the values 1. For example. (4. That is. #X) $0. we can solve the problem and then round up the optimum value of the variable to the nearest integer value. for the all-zero laminate.1.1. . 4nt. enumeration is not efficient. we solved for the continuous variable h. and the total thickness h was our measure of lightness. xr. or quasi-isotropic laminate.(x) = 0.1.lit4 LAMINATE IN·PI. .. The other design variable cannot be treated this way. If we have a choice with r options. we use f to denote the objective function. and X. as we will soon see. For example. h. x X. X. the thickness can he expressed as f(x) =h = l nt. There are no obvious equality constraints hi in the example hut. one of the design variables was n. but one which is quite acceptable for some optimization software dealing with integer design variables. 2. which in the example represent the limits on I he strains. Such programs may accept user-defined subroutines for calculating lhe objective functions and constraints. whenever this does not introduce large errors. We use g1 to denote the inequality constraints.1 DESIGN OPTIMIZATION PAOILI!M FORMULATION 135 E An optimization problem has an objective function that measures the goodness or efficiency of the design. which was the choice of laminate.4) As noted in Section 1. i = 1.4. which are the parameters that can be changed in the design process. and the user needs to formulate the optimization problem with the objective function and constraints defined in terms of the basic four algebraic operations of addition. the solution was n = I 08.1. Finally. in Example 4. corresponding to the choice of all-zero. respectively. but not a small one. enumeration is a reasonable procedure. which is the number of stacks of the basic repeating sublaminate. the number of stacks n is an integer. multiplication.1: minimize such that j(x). Binary variables are integer variables which take only the values 0 or 1.• . .1 in the standard formulation we have to express the thickness and the strain limits in lcrms of the choice design variables. Optimization methods for problems with continuous design variables are usually much more efficient than methods that have to handle integer variables. we can define r binary design variables Xp . and noninteger values do not have much meaning and cannot be rounded up or down in an obvious way. if i = 3. (4.1 our goal was to design the lightest laminate.1 we had two constraints that represented limits on the strains under two load conditions. For this reason. an optimization problem has design variables. with one design variable corresponding to one option.= 1 if the ith option is exercised. and then rounded it up to get an integer value of n. there is a standard way to convert a choice problem into an equivalent problem involving binary design variables. or 3. 8nt. Fortunately. This is an integer design variable.. and division. In order to formulate the problem of Example 4.4. This approach is called enumeration.. if i = 1' if i = 2. many programs available for solving integer optimization problems do not accept such subroutines. We have skirted the difficulty by solving the problem separately for each one of the choices.ANE STIFFNESS DESIGN 4. the definition of the design variables can require the introduction of such equality constraints. subtraction. Of course. For example. we cannot exercise more than one option. However. Instead of solving one optimization problem with two variables. for more complex problems.1. Thus h was the objective function. 138 LAMINATE IN·PLANE STIFFNESS DESIGN 4 1 DESIGN OPTIMIZATION PAOILEM FORMULATION 137 ~ x.x.500.5 x 1 + 3. (18. we can write the constraint on Gxy as 0. we can easily write any property of the laminate in terms of the corresponding prop· erties of the possible options. Of course. vxy .1) as minimize t xix 1+ 2 x 2 + 4 x 3) =h = n t(x1+ 2x2 + 4x3). and i = 3 to a quasi-isotropic laminate. such that X. = 0 if we do not. the objective function may be written as f(x) Finally. Ex= (18. In that example we found the values of Ex associated with the three laminates to be Ex! = 18. 2.552 x 3 ) Now the limit on Ex may be written as X 106 psi. consider a situation four possible types of laminates.93 x1 + 4.. I i=i A similar procedure can be followed for the strain constraints expressed in terms of the limits on the values of Ex and Gxy obtained in Example 4.5 .5 X 106 psi.n t (x 1 + 0. and 4. 2.1.176 x 2 + 7. The natural choice of design variables for this problem is the number of stacks.2 Formulate the design problem of Example 4.= 1• £.1.1.5 x 106 • Substituting the expression for the thickness h and rearranging.Y.5 x 1 + 3.1.5 x 1 + 3. we can rewrite the constraint as 2.t xix 1+ 0. [±45]ns' and [±45190/0]ns' respectively. Once we have these binary design variables defined. i.5. with Poisson's ratios v~Y' v.1 as a standard optimization problem. For example. of the binary variables represent the number of layers in each of the repeating sublaminates [O]n.552 x 3 ) ~ 0. 1.5 x 2 + 0. hEx~ 2.000 lb/in.25 x 3)(0.176 X 106 psi.5 x 2 + 0. Ex2 = 3. to make sure that we choose exactly one laminate we need to add the constraint x 1 +x2 +x3 = 1.1.5 x 2 + 0. h (18.176 x 2 + 7. Since Ex is independent of the thickness. v~Y' v. we can write it for any of the three laminates as Example 4.5 x 1 + 3. n. In terms of these binary design variables. Then the value of Poisson's ratio of the selected laminate is obtained by specifying the design variables x" x 2. x . v xy + x2 v xy + x3 v xy + x4 v xy The following example demonstrates how we use binary choice vari· abies to formulate Example 4.894 x 3) ~ 0. (1. Following a similar procedure. i = 2 to a (±45°)n laminate. We replace this single variable with three binary variables x" x 2.552 x 3 ) x 106 ~ 2. such that x 1 + x 2 + x 3 = 1.n t (x 1 + 0. _ I 2 3 4 Ex3 = 7.5 .858 x 2 + 2. i = 1 corresponds to all zero laminate. x in 3 4 h Gxy ~ 500.= 1 if we choose the ith laminate.176 x 2 + 7.. .4.25 x 3 )(18.552 x 3 ) ~ 0. defining a general integer variable x4 =n we can formulate the optimization problem in the form of Eq.176 x 2 + 7.552 X 106 psi. that ranges from 1 to 3 and indicates the choice of laminate. and a choice variable. That is.000 lb/in.25 x 3) X where the coefficients.1 in standard format.~ . x3. and X. we start this section with a discussion of the opuuuLauu• of orientation angles. ~ (0. The formulation of laminate design problems in the standard is important for applying sophisticated optimization algorithms or izing standard optimization software. G12 = 6. and v12 = 0.1.0 GPa at 9 = 36°.1 or by simple graphical procedures. while constraints are imposed on other elastic properties. we obtain the moduli and Poisson's ratio as a function of 9 and plot them in Fig. Using Eq. For example.3. Orientation angles are continuous varia ranging from -90° to 90°. subject to the requirement that Gxy is at least 25 GPa and that vxy is not larger than 1.28) for the AJs and Eq.~ 0 z E ft1 10 20 30 40 50 60 70 80 90 (+9/-9) 8 Figure 4. using optimization terminology.1.858 X 2 + 2. such as when we want to design an angle-ply laminate [9/-9]ns· The laminate can also be more complex.4. 4.138 LAMINATE IN·PLANE STIFFNESS 4 :1 GRAPHICAL SOLUTION PROC!OUPIES 139 0. provided that additional layers have fixed orientations.t x/x 1 + 0. Design an angle-ply laminate [9/-9]s with maximum Ex. Elastic properties for Graphite/Epoxy.1. and the optimization can be solved by plotting the pertinent elastic . This is illustrated in the following example. However.894 x 3) 0.2) for the effective clastic constants. '0 . x 4 positive integer. for some problems design problem is simple enough so that it can be solved by inspectwn as was Example 4.. (2. Another way to say that.1. The range for the design variable 9 that satisfies both constraints is called the feasible domain.2.2 GRAPHICAL SOLUTION PROCEDURES The in-plane properties of a composite laminate are determined the orientation angle of the layers and the number of layers used a given orientation angle. From the figure we see that the requirement on Gxy is satisfied in the interval 28° ~ 9 ~ 62° and that the limit on vxy is satisfied for e ~ 21° or e :?: 36°. :> ~ 4. Note that at the optimum one of the constraints is active.4 GPa.:)~ ~ We will consider first laminates with a fixed total thickness.25 x 3) X . while the number of layers is an · variable. The section is devoted to such procedures. that is [(±9/0/90)nL· Since the thickness is fixed. is that the optimum is on the boundary of the feasible domain.5 x 2 + 0.·onstants as a function of 9 and picking the optimum design by inNpcdion of their graphs.1 Consider a laminate made of Graphite/Epoxy with E 1 = 128 GPa.93 X 1 + 4. (4. The objective function in this case could be one of the effective elastic properties.1 Optimization of Orientations of Layers 1:. x 3 ) binary. 4. (xl' x2 . n is not a design variable. Since graphical procedures are easier to apply to variables. In this range the maximum value for the objective function Ex is 36. . simplest layer orientation design problem involves design of a laminate with a single orientation angle. and it is 36° ~ 9 ~ 62°. lixample 4.5. H2 = 13 GPa.2. we may want to design a laminate with some 0° and 90° layers plus some angle-ply stacks with undetermined angle. (4. = 1 0 v. the lamination parameters are v~ A. The major effort of Miki's design procedure is the construction of a lamination parameter diagram which describes the region of allowable combinations of lamination parameters.35) by the total laminate thickness.J vk cos 4 8k• k=l (4. (4. (4.5. laminate. the elements of the A matrix divided by the laminate thickness can be determined from the following equations.2. u4 -v! u3 Us 0 -v3 VI I r) = cos 28 and v.4.• u2 . v. For a laminate of total thickness h.. The key to Miki's procedure is that for a balanced composite laminate the engineering constants of the laminate can be described in terms of two lamination parameters that capture all the relevant in-··· formation associated with the stacking sequence.4) and (4. (4. -1 ~ (V.7).5). which can be obtained from Table 2. and I is the number of different ±8 groups. These two paramewhich are obtained by normalizing the in-plane ters are V~ and components of V 1A and V3A in Eq. The effective engineering constants can then be obtained from the A~'s using Eq. oo or 90°) can be included in the stacking sequence.2. and C correspond to laminates with [0].) ~ 1.4) k=l where I is the number of different ±8 groups.5). For example.6) (4. and v.1: The other expressions in Eq.-0. [±8nL (/ = 1). Two other points that we will consider in some of the examples in this book are point D (0.2. the lamination parameters are given as I I V• 1 = h = L.2. and [90] orientation angles. (A*12)2 = A*II A* 22- A. Indeed. 1983) for the design of laminates with prescribed in-plane stiffness properties.5.e. 4. defined in Eq.. (2. the values of the lamination parameters are always bounded.1. E X Values of all possible combinations of the lamination parameters are.l A..2 described hy Eq.2 GRAF'HICAL SOLUTION I'IIIOOIDUIII!B 141 A more general graphical procedure was introduced by Miki ( 1982. (4.2.2. =2 V? -1. cos2 28 = 112 (cos 48 + 1)..J vk cos 28k VIA ""' and v3A = ""' v3* = h L. therefore. are the orientation-invariant material properties.' liiiU LAMINATE IN·PLANE STIFFNESS DESIGN 4. (2. i=l (4. If the laminate consists of two or more fiber orientations. B. [±45L. stacks of unidirectional layers with principal material axes aligned with the axes of the laminate (i. in which the volume fraction of the layers with ±ei orientation angles is v. = cos 48.1) Using the trigonometric identity.-0.2.2 (4. where the total number of layers in the laminat~ is N = •f'Lf= 1 N. (4. then it can be shown that Eq.7) becomes an inequality v.~2v?-1. Note that points A.2 A~2 A~6 v· U -v. For a laminate with only one fiber orientation angle.2). corresponding to the [±30L laminate and point E ( 0.5).1.5) From Eqs. located along the boundary curve ABC in Fig.2.7) where A~=Aufh and where the U. corresponding to the [±60].= 1.2. The stacking sequences of laminates amenable for treatment by Miki's procedure may be written as [(±ei)NI(±8I_1)N I··· /(±8 1 )N]. In addition to the balanced angle-ply sublaminates. respectively. the two parameters are related as v.2. The procedure is suitable for balanced angle-ply laminates made up of multiple stacks of layers with different orientation angles.. (4.4.8) . LV.20).2) can be modified in a similar manncr to obtain the other effective engineering constants. (V~. (iii) the vertical line V~ = 0.tJ/1 I I v. So we use v 1 to denote the volume fraction of 0° layers and v2 the volume fraction of ±60° layers. (ii) The solution along the line CD is entirely similar to the solution along the line AE. v = (v" v2 . Therefore. The point. Then. The line AC in the lamination diagram constitutes a special case.. That is.4. '"' .2 Consider the origin of the lamination diagram. 4.2. v1). c ~" 3 . 2/3). which is at a distance of 0. v. (4. (4.2.A the volume fraction of 0° layers by v1 and the volume fraction of 90° luyers by v2.4).2: (i) line AE. 0) + 0. This means that the lamination parameters along a straight line connecting two points in the diagram represent laminates with volume fractions proportional to the distance from these points.r)l from Pq has a vector v of volume fractions Example 4. from Eq.6.. . and (iv) the horizontal line (i) Point A represents an all oo laminate and point E an all ±60° laminate. (ii) line CD. 1 v = 0. -1 B Figure 4. 4. v = (1 . from Eq. 0) and v2 = (0. corresponds to [0/902 ]s or to any other cross-ply laminate with the same ratio of oo and 90° layers.9) we have at point S. 0) and point C has v = (0. In most cases a straight line connecting two points on the diagram will pass in the interior. as it lies entirely on the boundary of the diagram.2.9) still holds.4(0. The origin is one-third of the way from E and two-thirds of the way from A. respectively. any point along the line can also be obtained by connecting other points to one another. 1). then. 4. That is. so r = 2/3. The rule expressed by Eq. 1) = (0. This is illustrated in the following example.~'I. From Eq. "''.2 from C.r)vs + rvq. let points Ps and Pq. Denoting ..0 = 0.9) For example.2.. the origin represent a laminate with one-third oo layers and two-thirds ±60° layers.0). The distance between points A and C is I= 2 units. [0/±60]. (4. respectively. v =(113)(1. 1) =(1/3. and vq. point A has v =(1. The allowable region of the lamination parameters is the area bounded by the curve ABC in Fig. 1).. Then the point P lying on the line Ps Pq at a distance rl from Ps and (1 . (4. Lamination parameter diagram of a [±8]s laminate. 1. (4. The volume fraction vectors corresponding to points A and E are v 1 = (1. Point S. Each point of this region is called a lamination point and corresponds to one or more laminates with specific stiffness properties. be characterized by vectors of volume fractions v.2. V~ = = 0. t I>J.) = (0.2.2. However.2 represent laminates made of combinations of 0° and 90° layers only.4) we see that the lamination parameters are a linear function of the volume fractions.6(1. Any point inside the lamination parameter diagram must correspond to laminates with two or more fiber orientations. so that the origin is also the laminate [90/±30]. 0) + (2/3)(0. v. located at a distance I of each other. for example..142 LAMINATE IN·PLANE STIFFNESS DESIGN 4 2 GRAPHICAL SOLUTION PFIOCEOUAEB 143 v. The lamination characteristic of the points inside the lamination diagram may be related to the characteristics of the points along the edges of the diagram. The dimensionality of the volume fraction vectors is determined by the number of distinct fiber orientation angles.=o.. 0. and obtain four laminates that represent this point by observing that it lies along the following four lines in Fig.9). v. has r = IU~/2.2.8 from A and 1. irrespective of the number of different layer orientation angles.. all the points along the line AC in Fig.• .2. and therefore we have multiple laminates corresponding to the same point. such as [±45/0/90L laminate.VI.5) which correspond to a laminate where we have equal parts of 0°. = V 1 COS 48 1 + V2 COS 482. The volume fraction of the first orientation is the ratio of the distance between the origin and the point in the segment CE to the total length of the line. = 0. (4. which corresponds lo [0/±60).5. We can find combinations of two angles by passing a line through the point. 90°. For example.4) becomes v. = 0 has the same quasi-isotropic property. Similarly. since a point is defined by two parameters. where the volume fraction is 113. =0 with the boundary of the diagram between C and B. So we use v1.25.2.5° corresponds to the intersection of the line v. However. vi' 8. 90° and the angle defined by the intersection of the line with the other boundary (ABC). which corresponds to [±30/901. The geometrical procedure clearly indicates that if we want to have only two orientation angles. the point 8 = ±67 . and going through the procedure illustrated in the previous example by using the interior point and the two boundary points. are sufficient for designing laminates for given lamination parameters and hence for prescribed stiffness requirements. =0 corresponds to either 48 = ±90° (8 = ±22. or 48 = ±270° (8 = ±67. (4.5°). Tsai and Pagano also show that any laminate with 11.. =cos48.5°). 0). and 82 for obtaining a particular lamination point. 45°.2. Instead of going through the geometrical procedure described above. and the other side in the range CE.10) as 8. (4. We thus have three parameters. (4.10) where v2 = 1. Point (0. and v3 to denote the volume fraction of 0°. A point in the interior of the lamination diagram may represent a very large number of laminates with some of them including many orientation angles. This attribute of quasi-isotropic laminates is not difficult to prove (Tsai and Pagano.5 COS-I T2. but only within some ranges. = 0 with the boundary of the diagram between A and B. (iv) From Eq.2 indicates that one side of the line must intersect the segment AD of the boundary. 4. 90°. because points along AC are combinations of 0° and 90°.6).. 8 1 and 82. a quasi-isotropic laminate made of two orientation angles can be defined by passing a line through the =0). that line canquasi-isotropic lamination point (Vi= not intersect the boundary of the lamination domain along the top boundary (AC in Fig. 1968). oo. rotated by 22. v. and ±45° layers. 0.5]. these can vary within a range that depends on the lamination point.5 cos-' Tl' where 82 = 0. the geometric procedure for creating a quasiisotropic laminate. the two angles can be obtained by solving Eq.2). Similarly. and the other side the segment EC. For example. 0. 0.9) we get v = (0. 0. It is apparent from the diagram that the volume fraction is minimal when the line begins at point A. we can solve the problem algebraically as follows. v2. and a volume fraction of 2/3. However. a two-ply-orientation quasi-isotropic laminate cannot have an orientation 8 in the region 30° < 8 < 60°. It is easy to check that the laminate in case (iv) is the same laminate as that of case (iii)..2. The point 8 = ±22.2. which was discussed above. For a laminate with two orientation angles. 1) then has a volume fraction vector v 1 = (0.5°. and -45° layers. 1) has equal amounts of 0° and 90° layers. respectively.5° corresponds to the intersection of the horizontal line v.5. The origin lies in the middle between these two laminates. = v.5/±67. so that r = 0. For a given volume fraction in the appropriate range.2. indicated that one side of the line must lie in the segment AD. along the diagram boundary we have v. These laminates are called quasi-isotropic because we can rotate them by any angle without changing their in-plane properties. One of the angles or the volume fraction v1 can be assigned a value arbitrarily. (4. so that it corresponds to a laminate with equal proportions of all four angles such as [±22. Eq. while point B has a volume fraction vector v2 = (0.5. the volume fraction of the first orientation is maximal along the line DE.·- LAMINATE IN·PLANI! STIFFNESS DESIGN 4. Thus. (4. Point B is an all ±45° laminate. 1). finding its intersection with the diagram boundary on the two opposite sides of the point. 4. Similarly. Thus a line through the quasi-isotropic point intersecting AC will correspond to a laminate with three orientation angles.25. V~ = V 1 COS 28 1 + Vz COS 282 and v. Inspection of Fig. and from Eq.2. the laminate in case (ii) is the laminate of case (i) rotated by 90°. this means that only two orientation angles..2 GRAPHICAL SOLUTION PAOCECURES 145 (iii) The point (0. The origin is halfway between these two points.11) . Therefore. the value of the quantity I+ 2V? is equal to 0. For v1 = 0. With a minimum of four layers per angle (to achieve the symmetric and balance requirement) we need to consider only v 1 = 0.2.1 has lamination parameters V~ = 0.1° and 92 = 0.25 0./2 X 0..2. only with the two angles switched.KaKc = --=----K.5 cos.75 X 0 ·9 = 0.V~).09771 v. working with integer number of layers limits the possible choice of lamination parameters.. for v 1 = 0. Kc = 2 v~ + 2 (v 1 .5 and 1.75 because this will give us the same solution as v 1 = 0. (4. relative total thickness of the layers with 9 1 and 92 orientations must remain between 0.25.3896) = 33.2.25 X (0. we get v2 For a quasi-isotropic laminate Eq. For the case of three angles with 9 3 = 0°.3673.3 As we will see later.5172 . symmetric..75 X 0.::.12).63 3 ]. Miki (1982) gives the two angles for a given choice of volume fractions as 1 8 1 =2cos = 0. . We cannot have v1 = 0 because the lamination point is not on the boundary.- v 1 cos 29 1 + v2 cos 292 + v3 and T 1 =-----~~--~-------------------- 2 X 0. Kb = 2 vz(v 3 . ---T2Vz VITI 0 These simple equations let us easily find quasi-isotropic 1 with only two distinct orientation angles for a given volume vp as long as v1 stays within the limits of 113 and 2/3 (T1 and T2 remain between 0 and 1).1 (0.2.5 cos. We do not need to consider v 1 = 0.9000 and T2 = or 0.4.. therefore. = V 1 COS 49 1 + V 2 COS 492 + V 3 .9° and 92 = 0.25 and v 1 = 0.___::. v.Y2v 1 vz(l+V. We first consider a two-angle laminate.v T I I T2- und v1 + v2 + v3 = I. Design a balanced..3896 0.5° Ka = 2 viv 1 + v2). = -0.5. 2. the optimum laminate for Example 4. [±15/±62. is 20-layer quasi-isotropic laminate.9°..JK~.0.63°. It is also easy to show that.V3) v 1 .25 0.9) = 12.25 X 0.1(0..5172.1(0. many combinations of orientation angles and volume fractions produce the same lamination parameters and... For balanced angle-ply laminates with more than two orientations. the same stiffness properties. That is..6448. The smallness of the number indicates that it is close to the boundary of the lamination diagram.2.Vz 8.5 cos.25 2 X 0.0..15) or X where T2 8 -Kb ± .5172) ± --. or (4.l 344 = 0.146 LAMINATE IN·FILANE STIFFNESS DESIGN ~ 2 GRAPHICAL SOLUTION FIFlOC!OUREB 147 TI = 2v 1 V~±.5 cos..5172 . 1 82 =2cos 8..-2 V?) 2vi ' _ v~. For example. -I 0.16) 91 = 0.1(0. Example 4. we have V~ = v.) v 1 + 2 V.::. for example. for laminate to be quasi-isotropic.(1 + v. a so that 91 = 0.09771..1344 -I (V~. it is easy to that we get 9 1 = 15° and 92 = 62.25. For this lamination point. (4.2.6448) = 24.. (4.12) reduces to -+ TI - {f.2V~)v3 .1344) = 41. from Eq. 16-layer laminate with these lamination parameters. Since we have only 16 layers. 9/(±33.5 ~ ~ X 0. The great advantage of this two-step procedure is that the optimization phase has only two unknowns. and hence the same in-plane stiffness properties.Uf + u~ 3 Ui2 VI +2 U4-EJ ' (4.Gxy u3 (4.5 cos. (2.09771 2x0. and so it lends itself to graphical solution.42) and (4.7382) = 21. Example 4. for v 1 = 0.3 demonstrates that once we find the desired laminalion parameters it is possible to find many stacking sequences that lead to these parameters. The design space is the area bounded by the curve ABC in Fig.25 .0..4°.1(0.375.5 cos.2.25) = 0.2. we find a set of stacking sequences that we can use to achieve the desired lamination parameters.= Us.. 2962 so that.5 X (0.1336.5172) = -0.5 + 0.5 we get Tl 2 X 0.V 2Ex v~ + E.51722 = 0. Next we consider the case of three angles with the third one being 0°.5172) .25 (0.2962) = 36.2. Then. The equations needed for such contours. v2 = 0.25(0.148 LAMINATE IN-PLANE STIFFNESS DESIGN 4. as a second step.17) Kb = 2 X 0.( 1 . A very useful aid in the graphical solution of the laminate design problem is to draw contours of constant effective engineering stiffnesses.5 X 0. or 0.4)2].11(±24..E) (4.U1 .2. the symmetric case of v1 one pair of solutions el = 0. From Eq.2.2962 = 0. obtained by combining Eqs. it is impossible to find a laminate that will have the desired lamination parameters. we cannot use them here. U~V? + U2EYV~ + EYU 1 .5 X 0. This indicates that even with v3 = 0.2962 = 0.vxyul + u4 (1 + vxy)U3 (4. These are [±12.25. Since it is impossible to have a smaller volume fraction for the 0° layers. we found three laminates that have the same required V~ and v. Ex contours: V* _ u~v?.5 To summarize the results..0.18) + 2 X 0. [±41.2.1 ). l (±21.KaKc is negative.Vi+ U~ U/2 V 1 + 2 V 4 .9) 3 ].25.5172) ± . with constraints on the effective engineering constants limiting further the set of possible design points (called the feasible domain). we find that the quantity under the radical of the first Eq. and we cannot continue the computation. 7382 = 0.5) 3Js.5172 T2 = 0. K~.2.19) Gxy contours: v.2° and 82 = 0.2)/(±36. Kc = 2 X 0.3673) X 0.5 yields only = 0.16). One possibility is v1 = 0. 7382. v3 =0.5172 .4. (4.25 2 + 2(0.0. as expected.5.25 (25% zero layers). (4.5 2 X 0.V2 X 0.20) .07644.7382 and T2 or 0. This indicates that we can break the design of laminates for in-plane stiffness requirements into two steps.2.5 . vxy contours: For these values.25 EY contours: v. are = 0.2.1(0.16) we have Ka = 2 X 0. 4.5 0.2 GRAPHICAL SOLUTION PROCEDURES 149 Similarly. First we find the optimum set of lamination parameters that will maximize or minimize a selected property while satisfying constraints on other properties. V~ and v.2 X 0. V* 3 vxyu2 v~ . 030 GPa.3673.20) this line is V1 .604 =-0. [±9]. Ez. represents a laminate stiffer in the x direction. G"'. the laminates that produce the maximum Poisson's ratio are [±25]. = 20. A positive v.547 GPa.. respectively.2.2. For example.2 GRAPHICAL SOLUTION PPIOC!OURES 151 Contours of constant laminate effective engineering properties are shown in Fig. The equation of this line. This is demonstrated in the next example. respectively. 4. For design problems where one or more of the effective engineering constants are constrained. II ~ VI • Redo Example 4. with the shear stiffness increasing as v. is .540 GPa.4. The first laminalion parameter.1 to be U 1 = 57. Similarly. represents the shear stiffness of the laminate. stacking sequence.8). In the region above the line Gxy< 25 GPa.3. is seen to represent the relative stiffness of the laminate in the x and y directions. From Eq. (2. the line corresponding to vxy = 1 separates the lamination diagram into two regions with the bottom one corresponding to vxy > 1. The contours shown in Fig. for the Graphite/Epoxy of Example 4.004. The lamination parameter v.. ! ~ v"' VJ • U4 = 17.4 '\ ".004 GPa. represents a laminate stiffer in the y direction. EY. (4.4). Using Eq.2.GP8 Vj ·-. on the other hand.2.20). U5 = 20.604 GPa.19). sr---------~--------~ The line corresponding to Gxv = 25 GPa divides the lamination diagram into two regions (see Fig.GPa TJ00/5208 Graphite/Epoxy and Scotchply 1002 Glass/Epoxy materials.. if no other constraints are specified.GPa E1. and the maximum shear stiffness for [±451. ·~\ ~ • I • \ ' >~ / - v.4. we first find the material invariants U.3. while a negative v. As expected. 4. U3 = 13. Contours of constant effective engineering elastic properties.150 LAMINATE IN·PLANE STIFFNESS DESIGN 4. (2. (4. The figure indicates that. ' '\ \f \ ( -·) . determination of the value of lamination angle [±9]s that maximize the effective Poisson's ratio is not so obvious and is a function of lamina properties via Eqs. appropriate contours can be superimposed lo identify the feasible design space and the lamination point that maximizes (or minimizes) the desired stiffness property.4. the maximum values of the Ex.20) and (2.3 help in correlating laminate propl'rties with the two in-plane lamination parameters. becomes more negative..25 13.2. v. Example 4. and in the region below Gxy > 25 GPa. 4.1 without limiting the laminate to an angleply. and [±31]. maximum Ex and EY are obtained for oo and 90° laminates. from Eq. andGxy are all achieved for lamination points located around the boundary of the design space that require only one lamination angle. U2 = 58.. (a) Figure 4.. laminate. 604 = 82.3673 x 13.604 = 22. (a) and (b) intersect. from Eq.536 GPa.1. A.563 GPa.4.3672. Ex = 82. (3. I .1. It has substantially larger Gxy than required.54 + 0.35).604 = 22.5172 x 58.2.2. v. The previous example. Then. it can be shown that the coefficients of expansion of a balanced symmetric laminate take the following form: . is the unhatched part of the diagram. This is a standard graphical representation of the constraint.547 + 0.5172 x 58.2.. In Chapter 3 the coefficients of thermal expansion of symmetric laminates are derived.4.4704. from Eq. 2 x 13. = 0.1 for an angle-ply laminate. The feasible domain.2.0 Figure 4. the constraint on Poisson's ratio excludes a large portion of the boundary of the lamination diagram near the optimum.03v. where the A 16 and A26 terms also vanish..A . The lamination parameters which correspond to the intersection point are v.547-0. Figure 4.2 Graphical Design of Coefficients of Thermal Expansion When restricted to in-plane respose. = -0.2.2.03-0. Eq. I I ' l 'i '! ~ '>J "\\'k'\1 ~~~l'\l'\ 1~':/ vi"\~''" ' . { 2 (4.. 1 = 57. the expression can further be simplified as a. I I ~ . which is the region where both requirements are satisfied.21) Gxy = 25 GPa ~. deformation of laminated composites under thermal loads may be represented by coefficients of therfor the laminate derived in terms of the properties mal expansion.537 22.2 GRAPHICAL SOLUTION PROCEDURES 153 v.3673 x 13. 1328v. (4.563 X 22. V3 20 35 50 80 95 110 4.03-0.3)..1). By imposing the balanced laminate condition. A..3673 X 13.2.604 (b) A~ 2 = 17.537 22. v. (2. in terms of the in-plane V terms defined in Eq.4.547+ 17 ·54 = 2 .3.-57. As can be seen from Figure 4. on the individual layers.537 GPa. The optimum angle-ply design in Example 4. 58. This value of Ex is 67% larger than the value obtained in Example 4.42).A ~ K. (4. 2 = 57. Example 4.536 2 = 60 0 GP · a. point 0.-t. Note that the constraint boundaries are shaded on the side of the boundary where the constraint is violated.4 also shows some contours of Ex. At this point.. v• 1 axl J Kt VoA + K V1 A} ~J ~ ~ [AJ-' ( ' v. Using the definition of the remaining A terms in terms of the V's and their normalized form V*'s and performing the matrix inversion and the multiplication.5172.4 which is quite far from the optimum design. provided several laminates that realize the optimum lamination parameters of point 0.2. These contours indicate that the maximum value of Ex is realized at the top right corner of the feasible domain where the two lines of Eqs. Lamination parameter diagram for Example 4. corresponds to point A in Figure 4. This is because an angle-ply design has to stay on the boundary of the lamination diagram. vxy = 1..152 LAMINATE IN·PLANE STIFFNESS DESIGN 4.4. at the expense of the objective Ex. [36/-36] 5 . That is. there is more freedom in establishing thermal isotropy. The elastic properties of the material are in the U terms. we can derive the following equation for thermal isotropy: v~ (K2 r I 1 v'1 Figure 4. Compared to the condition of mechanical isotropy that is limited to a single point. However.31).5.. and K 3 . 4.U3 (U4 +U 1 )-V?U~+(Ui-U~)) + K1 (U 1 - U4 ) ' (4. U3 (U4 + U1) . is one of the characteristics of an isotropic layer. on those lines.24) 4.2. at the origin. that the only way to make the material dependent part vanish is when a 1 = a 2. (4. Such contour lines can be used to determine laminate designs that satisfy the desired thermal characteristics. which are located along the v.K 1 U 2 + K 2 U4 ) .2. Finally. For example. defined by combinations of the V~ and v.25) There are infinitely many laminates on Miki's in-plane lamination diagram.2.2. A more interesting result of Eq. see Eq.. but to the total thickness of a stack of contiguous layers .K 1 U2 + K 2 U4 ) = o. The coefficients of thermal expansion of the individual layers (which are assumed to be the same in each layer) are captured in Kp K2 . = 0.22) a= y V~ (K2 U 1 -K1 U2 +K2 U4 ) + V~ 2 K 2 U2 -2 K 1 U3 v. k=l (4.u~)) '\ \ .24) is the condition that v~ =I vk cos 2ek = o. which Beside orientation angles.2. Graphite/Epoxy laminates with zero coefficients of thermal expansion.3 Optimization of Stack Thicknesses All the terms inside the parentheses are material dependent. whereas the stacking sequence is determined by the V~ term.23)..1: I ""'~ I (4.. the contour lines corresponding to zero values of ax andaY are shown in Fig.22) and (4. there are infinitely many laminates. line between the [0190]ns and [±45]ns laminates. v~ = v.2. U 1 .5 for a Graphite/Epoxy material. The equations shown above may be used to draw the contour lines corresponding to specified values of the coefficients of thermal expansion of laminates on the in-plane lamination diagram.2." The term generally referred not to the actual layer thickness. ax= 2(2V.U4) 2 (2 v. Early work mostly disregarded the discrete nature of this variable and replaced it with a variable called "ply thickness.1.154 LAMINATE IN·PLANE STIFFNESS DESIGN 4 2 GRAPHICAL SOLUTION PFIOCEOUFIES 155 _ V~ (K2 U 1 .V~ 2 K 2 U 2 + 2 K1 U3 v. For the material system used in this figure. (4. the two lines do not cross.2.23) Just as in the case of elastic properties.v? u~ +cui. that have a zero value of either ax or ay. the above equation completely isolates the stacking sequence dependent terms from the material properties... Obviously. Such a laminate would have no deformation in any in-plane direction under thermal loads. one of the ways to satisfy the condition is to get the terms inside the parantheses to vanish.-K 1 (U 1 . no laminate has both zero ax and zero ay. (4. (3. the concept of thermal isotropy introduced in Chapter 3 requires that the coefficient of thermal expansion be the same along two mutually perpendicular directions. By imposing the condition ax= a_v in Eqs. we can optimize the number of layers that we have in any given orientation. It can be shown (see Exercise 6). 1. (4. . Ni. Such a procedure is not guaranteed to result in an optimal solution. we express the in-plane stiffnesses in terms of the design variables. (4. goo.2.t. ±45°. i=l (4. the way Eq.28) Note that the xk represent the total stack thickness associated with a given orientation. and the optimum design is identified. A typical design problem is to seek the lightest laminate that satisfies design requirements on in-plane stiffnesses. A 16 and A 26 are zero. Au= L (Qu\k)xk.. We define the stack-thickness design variables to be the total thickness for each orientation. namely. (4. .s.1.4. However.. In most applications. With the in-plane stiffness expressed as a simple linear function of the design variables. Fig.1.2. .27) Next. This procedure is demonstrated in the following example. objective function contours are drawn in the feasible domain. (4.500.lCU:I LAMINATE IN·F'LANE STIFFNESS DESIGN 4.2.2. it is usually easy to obtain expressions for constraints on engineering stiffnesses and write them in the standard form of Eq.26) defined the thickness design variables... We define the design variables as in Eq. the objective function f(x) can be taken to be the total thickness n It is usually easy to determine on which side of this boundary the constraint is satisfied. This optimum design usually corresponds to noninteger values of the N.2. graphical solutions are based on tracing the constraint boundaries. it is easy to use.1). and then the feasible domain may be defined (as was done in the lamination diagram.1. f(9 1)2NI Js.1 ).4). . When the laminate thickness is large.. so that the candidate laminate is [±45N/ON/gO]s. oo and 9. Furthermore.000 lb/in. where h is the total laminate thickness. = ' 2tN. Example 4. x. and n is the number of layer groups with distinct orientation angle. x 1 = 0.2.28) we can write the in-plane stiffnesses as h Ex ~ 2. and goo.2 GRAPHICAL SOLUTION F'AOCEOURES n 157 with the same orientation. if 9. and we can treat the laminate as if it were made of [(9n)2Nn1(9n_ 1) 2Nn-1 f . Eq.28) applies to design variables associated with either angle-ply stacks or oo and goo stacks. Assume that we have determined that we need only two goo layers. 4. and the objective function is . The other four terms in the A matrix are independent of the sign of the angle (recall that the lamination parameters are defined in terms of cosines). In this case. k=l (4. Then. if9. We seek the lightest laminate that satisfies the two design constraints of Example 4. Next.02 N 1 and x 2 = 0. (2.=goo. So the final problem is to find a nearby integer solution. Since the laminate is balanced.. ng. Therefore.01 N 2 .5 Consider the design problem of Example 4. only inequality constraints are used. . It is then possible to look for the optimum design of the actual (discrete) problem in the vicinity of the continuous optimum by rounding up the stack thickness to nearest integer multiple of the layer thickness (as was done in Example 4. Since this terminology is confusing. using Eq. .26) where the factors of 2 and 4 are due to the symmetry and balance requirements. As in the case of ply-orientation variables. the contributions of the layers with orientation 9 and with orientation -9 are the same. it may be advantageous to treat the stack thicknesses as continuous design variables and find the optimum design of the continuous problem. we will use the term "stack thickness" to refer to such total thickness.t. and it allows graphical solution technique.26). that is. that is. f(x) =LX.4.2. We consider again the design of a symmetric and balanced laminate with multiple angle-ply stacks [(±9JN/(±9n_1)Nn_/ · · 1(±9 1 )NJ~· Now the orientation angles are fixed.000 lb/in and h Gxy ~ 500. where i = 1..2. (1. the curves gix) =0. and the n unknowns are the numbers of layers of each specified orientation. In this section we demonstrate the use of such a procedure. .1. .=0o or 9. for a laminate with orientation angles limited to 0°. n. (4. j = 1. especially when the number of stack variables is only one or two.2g) {4t N. we get A 11 = 1... Au = 0.4. For example..2.. we need to express the constraints in terms of the design variables. ~~ ..1..............000 A 22 ... f = 0.000 or 2.. The constraint on h Gxy is simpler.A 11A 22 + A 12 ~ 0... ~~ + 57.........93 X 104 + 4.... Design space for Example 4..1 layers.2.....8 Xz) ~ 0..... with the objective function....23 X 2 ) 2 ~~ ~~ ~~ X ~~ 2 .361 104 + 6. From the definition of the design variables we see that this corresponds to N 1 = 3..0 I + x 1 + x 2• Next....04 ..21 in thick laminate is 38% thinner than the best laminate that we obtained in Example 4.A 66 ~ 0... From Eq.. using Eq..2055 in or 41.1.. 4 ..42) to write the constraint on hEx as AIIA2222 Constraints boundaries can be plotted by using software packages such as Mathematica.67 X 104 + 6.935 and N2 = 11. .... which corresponds to [(±45)i0 12 /90J_. Using the values of the Qu' s obtained in Example 4..08 X 0....500..... . + 250( 18. Substituting for A 66 and dividing by 104 .... (2... (2.93 106 x 2 • Next...8 x 2 ) + (0.... ~~ .908 X = 18.5723 A 66 X 104 + 4.. ~~ ..... and for g 2 the solution of a linear equation.908 X 106 x 2 ..5723 + 450.42) we see that it may be written as 500.501 X 106 x 1 + 0.... off = 0..361 X 106 x 1 + 18.......5. As a first step we write the Au's in terms of the design variables..2 GRAPHICAL SOLUTION PIIIOCEOURES 159 j{x) = h =0. the optimum design may involve ~-=---= 2:: Aiz 2. we get g 1(xi' x 2 ) =-(1.. I x 1 + 190.. f = 0.. we cannot be sure that the process will yield the optimum design... (4..... 106 X 1 + 1... 2 Substituting the expressions of the AiJ's into this constraint and dividing by 108 . we get gz(xi' x 2) 0 0 0......3 .1168 in..0787 in and x 2 = 0. Solving these two equations we obtain x 1 = 0..67 + 636. A 22 X A 12 = 0..16 0.67 X 106 x 2 .1 X 1 .....1 x 1 + I 867 x 2 ) (18..2 =49. Usually when we use this type of rounding up to obtain a laminate with an integer number of layers from the optimum continuous-thickness solution.1..6 shows the constraint boundaries and the contours of the objective function.. It is clear that the optimum is realized at the intersection of the constraints..1 X1 + 190...4.5723 X X 106 x 2 ...000 . Figure 4... This 42-ply 0....908 + 636.. In absence of such a capability...68. For g 1 this requires the solution of a quadratic equation...... we use Eq.01 Q u(90°) + x 1 Q u( 45°) + x 2 Qu...8 x 1 - 93 x 2 ~ 0.. Figure 4.. = 0...... where g 1 = g2 = 0... ......500.....67 + 636...28).......1.10" LAMINATE IN·IILANE STIFFNESS DESIGN 4.12 1 0..... The obvious integer solution is obtained by rounding up both n urn bers to N 1 = 4 and N 2 = 12.858 X 106 X 1 + 0.07-485......6. -~ 0. the total thickness........ we can plot the constraint boundaries by selecting values of x 1 and solving for the corresponding values of x 2 by setting gi = 0. an extension of Miki's lamination diagram method.373 0. while the other.160 LAMINATE IN·PLANE STIFFNESS DESIGN 4. and that the optimum is [(±45)/0 21. This is demonstrated in the following example.615 0. Table 4. Here we will introduce only two simple methods.1 Ex (GPa) 24. in many problems rounding up does not produce the optimum design. In general. in most applications. Candidate laminates for Example 4.78 27. So. This is the price we pay for limiting ourselves to the discrete set of Table 4. usually 0°.81 26.77 52. One aspect of the discrete nature of the problem is that only a finite number of laminates need to be considered as candidates for the design.00 20. for laminate Solve the problem in Example 4. Unfortunately. In such cases we may want to use methods devised to solve the discrete optimization directly. methods that are developed by a branch of optimization theory called integer programming. However.81 26.1 for a 16-layer balanced and symmetric laminate made of 0°.4.2.1 4. There are twenty-five combinations of orientation angles that satisfy the conditions of symmetry and balance. the number of possible designs is not very large.90 21.87 Gxy Stacking Sequence [(±45)z/904ls [(±45)z/0/903ls [(±45)z/Ozf902]. exhaustive enumeration of all possible designs is a reasonable solution strategy.3 DEALING WITH THE DISCRETENESS OF THE DESIGN PROBLEM In the last example we were fortunate that the optimum discrete solution (corresponding to integer number of layers) was obtained by simply rounding up the thicknesses obtained from the continuous optimum.3. which is less than the continuous optimum.61 0.2. The continuous optimum is a lower bound for the real optimum which has the additional restriction of integer number of layers. which is applicable to general discrete optimization problems. (GPa) l/xy 20. The second reason is preference by industry for laminates with a few fixed orientation angles.3.91 75. In fact. with Ex= 48. The constraint on vxy is not critical for any of the laminates. It is seen that there are four laminates that satisfy the constraint on Gxy being at least 25 GPa.672 0.00 20. [(±45)zl03/90)s [(±45)z/04L [(±45h/90zL [(±45)/0/90Js [(±45)/02]. for in-plane stiffness design problems. Some of the general methods of integer programming will be discussed in Chapter 5 in more detail. Therefore. Occasionally. the stacking sequence does not change the in-plane stiffness properties. Therefore.. here we can tell immediately that the solution we obtained cannot be improved upon.305 0. and 90° layers.9 GPa. we are restricted to a small set of prescribed orientation angles.3 DEALING WITH THE OISCFI!TENESS OF THE DESIGN PROBLEM 161 rounding down one of the ply thicknesses and rounding up of another. and 90°. Note that the value of Ex is 19 percent lower than the continuous optimum of Ex= 60. often the optimum discrete design is not even in the general neighborhood of the continuous optimum.1 lists laminates with at least 50% ±45° layers in the order of increasing percentages of these layers. ±45°. as long as the total number of layers of each orientation remains the same.709 . [(±45)4]. Therefore.480 0. The first is associated with the fact that we must have an integer number of layers. However. the optimum design is not even close to the rounded up continuous solution. Example 4. The first method is referred to as exhaustive enumeration. Since the requirement on Gxy is difficult to satisfy. along with the Ex' Gxy' and vxy for each laminate. ±45°. it is impossible for the optimum design to be lighter than the continuous optimum.07 48. only laminates with a substantial percentage of ±45° layers should be considered.203 0. design problems that number can be very large because the stacking sequence of the laminate (the position in the laminate of the various layers) may be as important as the number of layers of a given orientation.00 20.98 38.00 26.408 0.20 64.243 0.1. was devised specifically for the in-plane stiffness design problem.16 39. There are two reasons for the discrete nature of the laminate design problem.0 GPa found in Example 4. This preference is based on ease of manufacturing a laminate with commonly used angles and on the availability of test data for such laminates. The reason for this is that a lighter laminate will have 40 layers for a thickness of 0.2.81 33.00 20. 1) to eliminate v 1• The resulting expression v. (4.162 LAMINATE IN·PLANE STIFFNESS DESIGN 4. [0] [±15] ~r--IT' 1 v~ -1 [±45] Figure 4. 4. Therefore. respectively. 4. and VI + Vz + V3 = 1.1) and then use the first relation of Eq.3. (4.1 for an angle-ply laminate.1) where vi' v2 .3. for laminates made up of layers with 0°. By using Eq.8. For v3 = 0.3.~ . From the definition of the in-plane lamination parameters. (4. we find equations of lines that are parallel to the line passing through A and C. For other permissible values of the v3 between 0 and 1. When the laminate is thick. Eq. vertices of the polygons are placed at those locations on the perimeter of the diagram that correspond to the selected angles. for example. respectively. \. Consider.3. Eq. a procedure similar to the one in Example 4.3.1) and (4. 7.2. (4. laminates made of only el = 0°.8a.1 .2).3. Similarly. and 90° =V1 - V3 and v.2.. by eliminating v2 and v3 from the Eqs.2. exhaustive enumeration may become cumbersome. we have V~ vj [90)•. and 83 = 90° layers. for v3 = 0 (laminates with no 90° layers). the laminate is an angle-ply laminate with one fiber orientation. !!:·· f l v. and we may want to consider the use of Miki's lamination parameter diagrams.2) C 190 4 llt [0/90 3 ]J0 2 /90 2 ]!0 3 /901. .3) corresponds to the line connecting points A and B in Fig. (4. A D A . laminates with 25% 90° layers.8a and 4... Such lines are shown in Fig.4). (4. angles. If the design point is on the perimeter of the diagram.2 can be used to determine the possible laminates. I 10 4 1. = 2V~ + 4v3 - 1 is the parametric equation of a straight line in the in-plane lamination parameter diagram. which was determined without limiting ourselves to a discrete set of angles. e2 = ±45°. For points inside the diagram.3 DEALING WITH THE OISCAETENESS OF THE DESIGN PROBLEM 163 angles.25. For example. They showed that the feasible region for laminates with fixed orientation angles is a polygon with vertices located on the envelope of the lamination parameter diagram. we obtain other lines that are parallel to the line AB.. and v3 are the fraction of the layers with 0°.8b. Miki and Sugiyama (1991) showed that the diagrams can be used for designing laminates with predetermined orientation angles. ±45°. In-plane lamination diagram for angle-ply laminates with 15° incre- passes through the origin parallel to AB. all angle-ply laminates with orientations limited to integer multiples of 15° are shown in Fig. 4. (4. However. Ex is still 36% higher than the value that we found in Example 4.3. . .3. In-plane lamination diagram for laminates with specified orientation .. = V 1 - V2 + V 3. (4. the line (a) (b) Figure 4. given a set of permissible integer orientation angles. ments. and 90° orientations. ±45°.7. As expected.1.2) we eliminate v2 from the second relation of Eq. . In the vertical direction. tA (a) (b) Figure 4. For example. in addition to the vertices. which represents the solution for part (a) of this Example. which corresponds to variation in the number of ±45° layers. ±45.1 for a 16-layer balanced and symmetric laminate made of (a) 0°.. the total number of ±45° layers has to be even. with four different lamination angles.3. 4. Consider. but are combinations of the three available angles.. the point [0/45 2].. so that it can correspond to 1±60/02 Js and along the line BD. If such a line terminates at another discrete design point at the opposite end of the polygon. and goo. As we saw in Example 4. while the point [0/45 3] is not a possible balanced laminate.2.. For this case.. For example. point A corresponds to the laminate 0~6 . ±45° and goo layers and (b) 0°. However. we can divide the sides of the polygon into I/2 intervals instead of I intervals. From the nodes we obtained along the edges. Consider first Fig. the design space is a triangle with vertices at A. ±60°. ±60°. for example. In Exercise 8. The number of lamination points along the edges and interior points of the polygons corresponding to laminates with combinations of two or more orientation angles is determined by the total number of layers in the laminate.l equally spaced design points along the edges and along the internal lines that join two vertices.2. the design space is a trapezoid.2.2. To avoid the problem of obtaining points that do not correspond to a balanced laminate..2 Use the lamination diagram to solve the problem in Example 4. there are five equally spaced design points with fiber orientations varying incrementally from one vertex to another as shown in Fig. =0. we can have only four divisions because of the balanced laminate requirement. the reader is asked to provide all possible labels for the intersection points shown in Fig. The labels in Fig.. To solve this problem we draw the feasible domain on the diagram as in Example 4. ±30°. . and goo layers are marked. then it is easy to label the design that would be in the interior by looking at the designs at the two end points. We denote the total number of layers as N and I N/2. the origin.8a include only 45° layers. Note that the design points inside the triangular region also follow an incremental pattern.4 and add the discrete points that correspond to the possible laminates. along the line AB. With N = 12 and I= 6. The number of layers is 16.. the origin lies along the line AC. 4.2. 4. ±60°. and C as shown in the figure. so that it also corresponds to I±30/g02J. . and as noted in the previous Ex Ex= 48.. so we can have eight divisions along the line AC that corresponds to cross-ply laminates (laminates with only 0° and goo layers). as both correspond to the same lamination parameters. when the laminate has to be balanced. and the point immediately to its left corresponds to (0/gO)s.8b. and goo orientations. the sides of the trapezoid were divided into only three intervals. ±30°. For laminates with 0. where some of the vertices corresponding to a 12-ply balanced symmetric laminate made of oo. and we note that the parallel lines indicate the existence of additional points along the line AD beside the two interior points that we started with. = Example 4. we also draw lines parallel to the lines that join vertices. B. ±45°. the point closest to A is (±45/06 )s· Altogether there are 25 possible laminates.164 LAMINATE IN·PLANE STIFFNESS DESIGN 4. can be replaced by [0/±45].8a.. and goo layers. we can have multiple definitions of some of the points in the diagram. ±30°. All the intersection points correspond to balanced symmetric laminates. ±60°.n. For example. 4.ga. c' 1. Then. For orientation angles of 0°. ·""· ·"· ·"· I· D t v. and goo layers. This is illustrated in Fig. and goo angles (triangular design space). For example.3. 4. ±30°.. so that it corresponds also to [0/±30/±60/gO].9 A Jlt ' .3 DEALING WITH THE OISOAETENESS OF THE DESIGN PROBLEM 185 orientations and 0°. GPal ~ ~'A 1 " " . but each 45° layer can be replaced by a -45° layer.8b. However. we obtain I . for an eight-ply (total) laminate with 0°. we can show that it also lies along the line v. In-plane lamination diagram for Example 4.9. . Part I." Proceedings of the AIAAIASME/ASCE/ AHS/ ASC 32th Structures.the feasible domain." Trans. which corresponds to [(±45)/0 2]. 3.) ICCM-IV. we get v. 1725-1731.1 for a [8/-8/0°). Miki.75 cos(4 X 45°) + 0. only four of them are in . = -0. 2. Tokyo.75 cos(4 x 30°) which can be satisfied only when a 1 = a 2• 7. K.25 cos(4 X 0°) = 0.1.. 51-55. The labeling of all the points in the diagram is left as an exercise to the reader (Exercise 8). An examination of the contours of Ex in Example 4. REFERENCES EXERCISES 1. laminate. ±45°.75 cos(2 X 45°) + 0. 2.75 cos(2 x 30°) + 0. Repeat Example 4. (1981 ). which is not simply a rotated version of one of the quasi-isotropic laminates described in the text.3673.1. =0. + 0. and Umekawa.9a. only four are in the feasible domain. Hayashi. "Optimum Design of Laminated Composite Plates Using Lamination Parameters.4. Next we turn our attention to part (b) and Fig. For the problem of Example 4. Because of the availability of test data. with an Ex= 48.1. Miki.5. (1982).25 cos(2 X 0°) = 0. JSCM 9. S.8b. and the optimum is [(±30)/±60]s. (eds.2.1). then repeat for decreasing values of h until the feasible domain vanishes. Capability and Analytical Foundations. find the lightest symmetric and balanced laminate that is made only of 0°. Use Miki's graphical design procedure to obtain the lightest laminate for Example 4. Vol. 4. Baltimore. Therefore. [Hint: Start with h = 0." NASA TM 80181.9 GPa (see Table 4.34 in and draw the feasible domain on the lamination diagram. "Material Design of Composite Laminates with Required InPlane Elastic Properties.. BC.9b.IVV LAMINATE IN•PLANE STIFFNESS DESIGN REFERENCES 167 example.5172. (1991). MD. M. Find a quasi-isotropic laminate that includes ±20° layers. Miki. 4. Of all possible lamination points. 275-283. and CD are divided into only four subintervals. v. M. tells us that the optimum point is at the top right region in the feasible domain. so the segments AB. By using the definitions of the K's and the U's.2.E22 Vf2 ' v. it is desirable to work with laminates having stacking angles that are 45° or 90° apart. 2. (1) 6.25 cos(2 x 60°) = 0. and Sugiyama. Design such a laminate that has the lamination parameters v~ = 0. "A Graphical Method for Designing Fibrous Laminated Composites with Required In-plane Stiffness." Progress in Science and Engineering of Composites.2. show that this condition can further be reduced to = 0.. 8.2. Repeat Example 4. Y. M. Provide all the possible labels for all intersection points in (a) Figure 4.24)] is the condition that (K2 U 1 -K1 U2 +K2 U4 )=0. T. S. and Anderson. Kawata. W. (4. Stroud. and 90° layers. the two laminates have the same A matrix and the same effective engineering constants.25. and Materials Conference. and for (al-~) E2 E II 22 =0 (2) Ell . 5. Indeed for V~ we get for both points V~ for example [(±8/±(8 + 45°)1.25 cos(4 x 60°) = -0.] One of the conditions for achieving a thermally isotropic laminate [see Eq.1. M.5 with a symmetric and balanced laminate made of oo and ±30° layers. Note that this point has exactly the same lamination parameters as the solution of part (a). "PASCO: Structural Panel Analysis and Sizing Code. pp. This time we have two angles that have to be balanced. Structural Dynamics. J. :~I . (1983). (b) Figure 4. J. has general applicability beyond in-plane stiffness design. One method. The second method. can be formulated as linear integer programming problems. S. Furthermore. and Pagano N. Section 5. algorithms that deal with nonlinear integer programming problems must be used. For problems that cannot be formulated as linear integer programming problems. The objective of the present chapter is to describe methods that are more generally applicable to the design of laminated composites. but for most problems it is too computationally expensive. Such algorithms have not reached the level of efficiency and reliability achieved for linear problems. of the type discussed in Chapter 4.3 describes the branch-and-bound algorithm. Miki's lamination parameter diagram.168 LAMINATE IN·F'LANE STIFFNESS DESIGN Tsai.. and efficient algorithms that are guaranteed to find the global optimum are available. N. the choice of algorithm is highly problem dependent. Pagano. This class of problems has been studied extensively by developers of optimization methods. which is the most popular general algorithm for linear integer programming. Section 5.2 shows how in-plane stiffness design problems. Section 5. but cannot handle many problems of interest. Tsai. Westport. Halpin. enumeration. W..." In Composite Materials Workshop. C. "Invariant Properties of Composite Materials. S. (eds.1 describes the formulation of linear integer programming problems.J. can be generalized to a few other problems. 5 INTEGER PROGRAMMING In the previous chapter we illustrated the use of two simple integer programming techniques for the design of laminates subject to inplane stiffness requirements. ( 1968). W. We 169 . J.) Technomic Publishing Co. 233-253. 1 INTEGER LINEAR PROGRAMMING An optimization problem is called a linear programming (LP) problem if the objective function and the constraint functions are linear functions of the design variables. . . an unrestricted design variable X. Additionally. The general LP problem is minimize f(x) = cTx = cTx < such that Ax ~ b. If a variable is to have an unrestricted sign. where xu and xL are the vectors of upper and lower bounds for the design variables.xu . YJ. genetic algorithms are fairly easy to implement and to tailor to the particular laminate design problem at hand. is replaced by two new variables as x. A is an m x n matrix of constraint coefficients. Note that the signs of the design variables are restricted to being positive in the standard form.1 INTEGER LINEAR PROGRAMMING 171 chose to concentrate on a single class of algorithms. the problem is referred to as mixed integer linear programming (MILP) problem. X. The standard form of an LP problem is minimize f(x) where Cis greater than the absolute value of the lowest possible value of the unrestricted variable.:::: XL. therefore.1. However. are often written separately.1. This standard form is then solved by a variety of solution techniques. If all the variables are so restricted. In the case of LP problems encountered in laminate design. (5. Section 5. for example. If the designer has a good idea of the lowest possible value of the variable that can assume negative values.:::: 0 integer.3) such that A.1) X.:::: 0. but depending on relative magnitudes of the two. For example. This more general form of LP can be converted to the standard form by a change of variables.2) where c is an n x I vector of constant coefficients. the problem is called an integer linear programming (ILP) problem or pure ILP problem. A disadvantage of the approach just described is the doubling of the number of design variables. Such upper and lower bounds on the design variables are often treated differently than constraints that involve more than one design variable.x + A 2y = b. + C. x::. x.1. such that Ax= b.170 INTEGER PROGRAMMINQ ti. some or all of the variables of a LP problem are restricted to take only integer values. The bounds. Some commercially available LP codes are also restricted to handle positive-valued design variables. If some of the design variables are allowed to be continuous.os =x. and its standard form is minimize f(x) =cJx + ciy (5. The reader is referred to.4 provides an introduction to genetic algorithms and their application to composite laminate design. x. with which we have had extensive and good experience in the design of composite laminates.:::o.. and b is an m x I vector of constants. genetic algorithms. Arora ( 1989) for more information on solution of LP problems. then a more practical approach is to define the new variable by adding a positive constant C to the variable with an unrestricted sign.1 =x+-x-:I I where both x7 and xj are limited to being positive. most notably the Simplex method. the outcome may yield a positive or a negative value for the variable X. 5. (5. then the standard procedure is to replace it with the difference of two positive variables. More general LP problems allow design variables of unrestricted signs and general inequalities. frequently the values of the design variables are restricted not to exceed or drop below values that ure dictated by either availability of the resources or physical limitations. · · ·+COS 49N12 ).2.2. For example. For example. the elements of the in-plane stiffness matrix A are also linear functions of the number of layers. A 11 If several of the individual layers are made up of the same orientation angles. (5.2. it is demonstrated in the following that for laminates with prescribed fiber orientations.1). (5.U2 v. ..2 IN· PLANE STIFFNESS DESIGN AS A LINEAR INTEGER PROGRAMMING PROBLEM 173 It is also common to have problems where design variables are used to indicate a yes-or-no type decision-making situation. =h (no. (5. A quick review of the in-plane stiffness problem reveals that certain in-plane laminate properties are indeed linear functions of the design variables. Most of the remaining discussion in this chapter assumes to be pure ILP. A 66 = h( U5 .• i = 1. say nP layers with SP and nq layers with Sq. (5. A 22 = h( U 1 .35) we see that the elements of the in-plane stiffness matrix A. but the in-plane stiffness coefficients are still linear.2. and engineering for many years.2nf+ nn). (5. =~(COS 49 1 +COS 492 +COS 49 3 + · · ·+COS 29N12 ).4) cancels the h in Eq. = ~ (np cos 491' + nq cos 48q).1 and Eq: (2.2. from Table 2.3) 5.2. Many pwe. Therefore. Furthermore. the cosine terms are constants and.uuu" for the solution of ILP problems permit the user to specify that of the variables are binary. . in a laminate stacking sequence design. + U 3v.). the lamination parameters are not linear functions of the number of layers. Restricting ourselves to balanced symmetric laminates. because the h in the denominators in Eq. therefore. therefore. The in-plane lamination parameters are given by 29 3 + · · · + VN/2 COS 29N12 ). As a result. ±45°. it is possible to express the in-plane lamination parameters as linear functions of the number of layers of a given orientation.5) :Ill .4) by expanding the summation over each layer in one-half of the laminate: V~ = 2(vl COS 29 1 + V2 COS 292 + V 3 COS For a constant thickness laminate with specified values of the fiber orientations 91' and 9q. the presence of a layer with a specified fiber orientation its absence) can be represented by a binary variable.4. v~ =~(no -nn) and 2t v. For example. and 90° layers. consider a balanced symmetric laminate which contains only oo. to total laminate thickness. then Eq. are equal to the ratio of the thickness of a layer. v. efficient software for solving LP problems is readily available and.. N12.). there is a substantial incentive to formulate laminate design problems as LP problems. Such problems are referred to as zero-one or binary ILP problems.). A 12 = h(U4 . =2(Vl COS 491 + V 2 COS 48 2 + V3 COS 49 3 + · · · + VN/ 2 COS 49N12 ). =~(COS 29 1 +COS 292 +COS 29 3 + v. Hence. the volume fractions of individual layers I'.2. the in-plane stiffness parameters are linear functions of the number of layers nP and nq. economics.4) = h(U1 + U2 v.172 INTEGER PROGRAMMING 5. v. in the common case when all layers have the same thickness t. h.2. (4..2) In this representation.3) takes the following form: v~ = ~ (np cos 291' + nq cos 28q). we rewrite Eq. t. (5. are all linear functions of the in-plane lamination parameters V~ and v.).U3v.U3 v. (5. as we will demonstrate in Chapter 8. When the laminate thickness is not constant. v. + U3 v.2 IN-PLANE STIFFNESS DESIGN AS A LINEAR INTEGER PROGRAMMING PROBLEM Linear programming (LP) problems have been important in management. 34.U2 )]no. The weight of the laminate is proportional to the total laminate thickness h which we have seen is a linear function of the number layers. formulate the problem of determining the number of 0°. the objective function is one of the stiffness properties. the laminate has constant thickness of 2t(n0 + 2n1 + nn) = h = 2 X 10-3 m. U3 = 8.[2U4 + 2U3 . £ 2 =5.54 GPa. and t=0. In the weight minimit. The properties of individual layers are E 1 =76 GPa.v~v (2U1 .v~v (U1 + U3 + U2)] ~ o.2.9) v.U3 (no.Uina. A 12 = 2t[(no + 2n1 + nn) U4 . Constraints are placed on the stiffnesses and occasionally on the maximum or minimum percentages of layers of given orientation. and n1 is the number of ±45° layer groups in half of the laminate.'s to be U 1 = 32.Uino. we calculate the U. Additional stiffness and layer percentage constraints may also be present.2 IN·PLANE STIFFNESS DESIGN AS A LINEAR INTEGER PROGRAMMING PROBLEM 175 where no is the number of oo layers.4.42) and (5.2U3 )]n1 -[U4 - U 3 .2U3nr+ (U3 .2n1 + nn)]. we minimize the weight of the laminate.5 GPa. we can formulate as LP any in-plane design problem that involves the weight and linear expressions in the elements of the A matrix. (2. which is equivalent to minimizing the thickness when the laminate is made from a single material. and 90° layers to maximize the stiffness coefficient A 11 while keeping the shear stiffness Gxy of the laminate larger than 12 GPa and A 22 larger than 20 MN/m.174 INTEGER PROGRAMMING 5. (5. A 11 = 2t[(no + 2nr+ nn) U1 + (U2 + U3) no. The weight minimization problem is left as an exercise (see Exercise l ).125x 10-3 m.2. The thickness of the laminate may be written as h = 2t(no + 2nf+ nn).2. if we require that the laminate has a Poisson's ratio larger than a certain value vxy ~ v~Y' using Eqs.44 GPa.U2) no. nn is the number of 90° layers.5) we see that the are no longer linear in the numbers stiffness parameters V~ and layers of distinct orientation angles. the coefficients of stiffness matrix are still linear functions of the number of layers. stituting the in-plane stiffness parameters into Eq.7) the coefficients of the stiffness matrix A are + (U3 .U2 )nn]. Example 5.2. However. (5. or the thickness. The following example demonstrates a stiffness maximization problem. A 22 = 2t[(no + 2n1 + nn) U1 + (U3 . (5. The effective engineering elastic constants are not linear functions of the numbers of layers. In-plane laminate design problems can usually be classified as weight minimization or stiffness maximization. Using the material properties specified. the laminate shear stiffness Gxv and Poisson's ratio vxy' can be used to formulate linear constraints.2. Thus. and so from Eq. For this problem. For example. but two of them. while the weight.ation problem.1 For a 16-layer balanced symmetric laminate. This equation may be rearranged in the standard form of an inequality constraint as .2. G 12 =2. ±45°. U2 = 35.1) yields following expressions for stiffness coefficients: -1 U4 - U3 . In the stiffness maximization problem. or the total number of layers is specified.2U3nr+ (U3 + U2 )nn].55 GPa. A66 = 2t[(no + 2n1 + nn) U5 .3 GPa.652 GPa.2n1 + nn)].v~v (U 1 + U3 . and U5 = 10.7) we can impose the following linear constraint: (n 0 + 2n1 + nn)U4 .95 GPa. Substituting this expression into Eq. (5.U2)n 0 - 2U3nr+ (U3 + U2)nn].2n1 + nn) ~ v~v [(n 0 + 2n1 + nn) U1 U4 = 10. v 12 =0. However.1..2. For many laminate design problems. n1. we can vary no from zero to 8 .95.326 n1 + 11. Then for each value of n1 .4.724 nn. and use what is left for nn. the shear stiffness is Gxy = 10.1.90. which corresponds to all ±45° laminates.11.3.176 INTEGER PROGRAMMINQ .05 n 0 - 4. n0 . no + 2 nf + nn = 8. even if the number of possible designs is in the thousands. nn E 0.050. 1. these operations change the value of the objective function at the optimum. The results are summarized in Table 5.. obtained by assuming them to be continuous.2n1 .05 nn.326 n1 .1. and that change must be reversed in order to obtain the optimum of the original problem. ] . For problems with 11 integer design variables there are 2n possible rounded-off designs.724 no+ 4. it may be more reasonable to use methods specifically developed for MILP problems. nn = 0 which corresponds to a [Oz1±453 ]. Note that the constant term in the A 11 is ignored for the objective function because most linear programming codes do not allow constants in their input.88. Also note that the constraint equation (a) is added to the problem to insure that the total number of layers in the laminate adds up to 16.05 no. for a problem with 10 design variables there are 1024 possibilities. and the problem of choosing the best one is formidable for large n. For example. 5.3.2.082 n0 + 2.082 no + 2.000.6.3.05 nn $44. 8..326n. 2. .326 n1 . the design variable n0 and nn can take any integer value between 1 and 8.1. with the condition of having exactly 16 layers given as no+ 2nf+ nn = 8. Furthermore. Increasing the number of design variables to 20 would increase the number of possibilities to more than 1.163 n.163 nn. A 22 = 64. for some problems the optimum design may not even be one of these rounded-off designs. nn) subject to = 11.4).724 no. Adding a constant to the objective function or multiplying it with a positive constant does not change the optimal values of the design variables. Therefore. instead of using continuous optimization methods followed by rounding. !13 SOLUTION OF INTEQER LINEAR PROGRAMMING PROBLEMS 177 A 11 = 64. The solution of the problem is no= 2. n1= 3. . A 66 = 21.326 n1 - 6. 724 nn 6.6.1 by enumerating all possible 16-ply laminates. but the largest integer number n1 can assume is 4.163 n0 + 4. laminate.2.2. For problems with a small number of design variables. as we can have only five possible values (0.1 Solve Example 5.163 n1 .4.082 nn 2:: 1.082 nn GPa. Based on this equation.1 Enumeration n1 E 0.000. none of the rounded-off designs may be feasible. all in MN/m.3. The possible laminates are [0no 1±45nr190nn ]5 . while for others. -1. to the nearest acceptable integer value. . Example 5. enumeration of all possible solutions is a reasonable alternative to rounding. It is convenient to enumerate on the basis of the number nf of ±45° pairs. Thus.2.88.1. The optimization problem is then formulated as maximize f(no..88 + 11. the cost of an analysis is low enough so that enumeration is a reasonable procedure.4.3 SOLUTION OF INTEGER LINEAR PROGRAMMING PROBLEMS A common approach to solving MILP problems is to round off the optimum values of the variables. 5. and these are marked by the letter "F" (for feasible design).28 135.95 10.00 56.11 29.18 64.29 2.88 47.95 15.53 91.29 2.08 91.62 10.95 10.62 6.45 47.98 55.85 109. Branching from this solution are two paths corresponding to the two alternatives for the first variable.73 135.62 6.88 82.3.29 2.41 46.65 I 00.58 A 22 (MN/m) 153.58 Gxy (GPa) 2. i = 1. we have used the condition that the total number of layers is 16 to reduce the number of combinations of no.29 2. 5.31 73. [Osls [±45/906]..95 10.62 6. respectively (Fig. [±45/0i902L [±45/0sf90L [±45/06]. [0/9051.33 74.64 28.96 117.29 6. Example 5. .31 109.86 11. Enumeration tree for binary ILP problem with constraint h= r.09 28. The example also introduces some of the terminology that will be useful for the most important algorithm for ILP discussed later in this chapter.1.33 47.43 82.65 64.00 47.43 38. Laminate [90sls [0/907L [Oz/906 ].73 99.23 38.98 20. n1 . the branch-and-bound technique. 5..t1 iXi= 5. t\. Among these the [±45/02].96 82.3 SOLUTION OF INTEGER LINEAR PROGRAMMING PROBLEMS 179 Enumeration of all possible 16 balanced symmetric laminates A 11 (MN/m) 11.45 56.64 64. [±45zi904L [±45z101903L [±45z102/902L [±45 2/0/901 [±45 2/04].21 100.41 82. . [Oi904].76 73. [07/901.178 Table 5.23 74. design has the highest A 11 • As can be seen from this example.29 2.85 29.62 6.2 Consider the binary ILP problem of choosing a combination of five variables such that the following summation is satisfied: h= I i=l ixi=5. 5) with a function value of h = 0.11 64.29 2.76 37. A systematic way of taking advantage of constraints to reduce the number of cases that need to be inspected by enumeration is called an enumeration tree.28 F 19.53 55. .29 2.29 2..08 126. A decision tree representing the progression of the solution of this problem is composed of nodes and branches that represent the solutions and the combinations of variables that lead to those solutions.1.18 99. The top node of the tree (leftmost in Fig.1) corresponds to a solution for which all the variables are turned off (xi = 0.21 37. [±45/0/90L [±45/021 [±454).62 6. [±45/0/9051 [±45/02/9041.95 10.28 20.51 153. [Osf903L [OJ902].27 F 15.1).62 6. [±45/0/903].09 126. andnn that we needed to check.60F pic from Garfinkel and Nemhauser ( 1972) demonstrates the use of a constraint in an enumeration tree.51 117. An exam- Figure 5. The branch with Only four of the laminates satisfy the constraints that A 22 be at least 20 MN/m and that Gxy be at least 12 GPa.27 F 15. [±453/90zl.86 46. 8. One of the problems. Next.3 and xk+I =Xk+I = 2. namely 25 = 32. an enumeration tree can yield substantial savings. 5. the problem is solved. "' "' ' ' '' ' ' Figure 5. (5. since imposing conditions that require noninteger-valued variables to take integer values can only cause the objective function to increase. needed to identify all possible solutions. One of these possibilities is to have no feasible solution for the new problem. Tomlin. The original algorithm. In this case. and is indicated by a vertical line. The other three vertices are said to be live and can be branched further by considering the alternatives for the remaining variables in a sequential manner until either the created nodes are fathomed or the branches arrive at feasible solutions to the problem. after considering 19 possible combinations of variables. These two problems actually do branch the feasible design space of the LP-1 into two segments. LP-3.. the number of analyses required can be reduced. 3 .2 Branch-and-Bound Algorithm The concept behind the enumeration technique forms the basis for this powerful algorithm suitable for MILP problems (Lawler and Wood. The node arrived at by taking x 1 = x 2 = 1 has h = 3 and is terminated because further branching would mean adding a number that would cause h to exceed its required value of 5. and the other. This initial prob- The second step of the algorithm is to branch from the node into two new LP problems by adding a new constraint to the LP-1 that involves only one of the noninteger variables.1. For the purpose of illustration. Another possibility is to reach an all-integer feasible solution (see LP-3 of Fig. but the value of the objective function will provide an upper bound for the optimum solution because it is feasible.3). Illustration of branch-and-bound decision tree for MILP problems. Iem is labeled as LP-1 and is placed in the top node of the enumeration tree as shown in Fig. Such a vertex is said to be fathomed. we identified three feasible solutions. these two nodes are branched by considering the on and off alternatives for the second variable. will have a constraint that xk is larger than the smallest integer larger than x!. For a laminate design problem in which trials with different combinations of variables could require lengthy analyses. 1970). . 1966. 5. Consider a hypothetical example of a MILP problem of the type of Eq. The other branch is the same as the initial solution and can be branched further.2. We denote that upper bound as fu and update it when we find other integer solutions with lower values than the current fu· Once an upper bound has been established. developed by Land and Doig (1960).3 SOLUTION OF INTEGER LINEAR PROGRAMMING PROBLEMS xk+l S: 181 x 1 = 1 has h = I and tolerates turning additional variables on without running the risk of exceeding the required function value of 5. so that nodes that result in designs with objective functions outside the bounds can be fathomed. xk and xk+l' violate the integer requirement with xk 5. which are marked by an asterisk. will require the value of the branched variable xk to be less than or equal to the largest integer smaller than x!. The first step of the algorithm is to solve the LP problem obtained from the MILP problem by assuming the variables to be continuous valued. This is a 40% reduction in the total number of possible trials. . and thereby. there is no need to continue. In that case the new node will be fathomed. If there are no solutions with an =x! = 4. . any node that has an LP solution with a larger value of the objective function will be fathomed. There are several possibilities for the solutions of these two new problems. it is assumed that only two variables. and only those solutions that have the potential of producing an objective function between JL andfu will be pursued.2). the node will again be fathomed. LP-2. For the present example. If all the x variables for the resulting solution have integer values.2. say xk.3. Suppose several of the variables assume noninteger values and the objective function value is f 1• The f 1 value will form a lower bound JL =f 1for the MILP.180 INTEGER PROGRAMMINCJ !1. relies on calculating upper and lower bounds on the objective function. typically provide only a local optimum. Example 5. The next . Branch-and-bound is only one of the algorithms for the solution of ILP or MILP problems. the efficiency of traditional random search techniques is poor for problems that have a large number of variables. the branch with n1 S: 2 does not have a solution because again we cannot satisfy the constraint on Gxy' and the branch with n1 ~ 3 yields [0/±45 3]. An= 94. 5. Linear problems have a big advantage over nonlinear ones in that we can count on getting the global optimum for the problem. with All= 94 GPa.3. A methodical way of dealing with multiple minima for discrete optimization problems is to use random search techniques that sample the design space for a global minimum without the use of derivatives. Live nodes are then again by considering one of the remaining noninteger values.00 no solution Figure 5.3. 5. and their implementation calls for the use of a random selection process guided by probabilistic decisions. and since we do not have any more live nodes. for example.182 b. then computational savings can be obtained because of the elimination branches that cannot generate solutions lower than the upper bound A rule of thumb for choosing the noninteger variable to be is to take the variable with the largest fraction. Now that no and nn are integers. Algorithms for nonlinear problems. However. we choose.2. they may still include noninteger-valued variables (LP-2 Fig. then the node with}~ is an optimum solution. With no having the larger fraction. If a selected node branching variable leads to an upper bound close to the objective tion of the LP-1 early in the enumeration scheme. and suiting solutions are analyzed until all the nodes are fathomed. using software for solving continuous linear programming problems.3.2). it is incorporated into many commercially available computer programs (see. There are a number of other techniques that are capable of handling general discrete-valued problems (see.25 GPa. have emerged more recently as tools ideally suited for optimization problems where a global minimum is sought.3 Solve Example 5. among all the live nodes. 1989). simulated annealing (Laarhoven. on the other hand.0 GPa. it is the optimum solution. with A11 = 87. with A11 = 74. 1989). Both algorithms mimic naturally observed phenomena. the problem that has the smallest value of the objective function. we branch on n0 • The branch with the additional constraint of no~ 4 does not have a solution because the constraint on Gxy cannot be satisfied. Johnson and Powell.3 SOLUTION OF INTEGER LINEAR F'ROGRAMMING PROBLEMS 183 objective function smaller thanfu. For the selection the node to be branched. and are labeled as live nodes.1 by the branch-and-bound method. The performance of the branch-and-bound algorithm depends ily on the choice of the noninteger variable to be used for and the selection of the node to be branched. The first node shows the solution with a noninteger number of layers [0 3 _5 /±45 2 _24 ]. The decision tree for this problem is shown in Fig. these two algorithms are also powerful tools for problems with discrete-valued design variables. Some of these algorithms are good not only for ILP problems but also for nonlinear programming problems with integer variables. for example. If there are other solutions with an objective function than fu. because of its simplicity. However. Kovacs. Branch-and-bound decision tree for laminate design problem.. Two algorithms.3. 5. node is most likely to generate a feasible design with a tighter bound. 1978 and Schrage. 1980). 1987) and genetic algorithms (Goldberg. we can branch only on nf' As shown in Fig. The branch with an additional constraint of noS: 3 yields the solution [0/±45 2 _5 ]. This is an integer solution. In addition to being able to locate near-global solutions. and mutation) and a fourth operator. 1993. For example.4 GENETIC ALGORITHMS Genetic algorithms use techniques derived from biology and rely on the application of Darwin's principle of survival of the fittest. this is quite simple. Pan. This aspect of the genetic algorithms is responsible for increased chances of obtaining global or near-global optima. mimic the mechanics of natural genetics for artificial systems based on operations that are the counterparts of the natural ones (even called by the same names). (5. rather than converging on a single point early in the process. For example. consider the minimization problem minimize f(x). crossover. Le Riche and Haftka. we reduce the risk of converging to a local minimum. a population of biological creatures is allowed to evolve over gene~" tions. Genetic algorithms have been applied to optimal structural design only recently. developed by Holland (1975). reproduction. the design variable vector can be a binary string represented as . Unlike many search algorithms that move from one point to another in the design variable space. For historical reasons. these operations involve simple. 1993.184 INTEGER PROGRAMMING !1. Hajela and Erisman. 1993. In biological populations these are stored in chromosomal strings. 5.1 Design Coding Application of the operators of the genetic algorithm to a search problem first requires the representation of the possible combinations of the variables in terms of bit strings that are counterparts of the chromosomes in biological genetics. crossover. These operations may appear like a completely random search of extremum in parameter space based on function values only. since the outcome of the search is random. genetic algorithms work with a population of strings.1) where the variables are integer and can assume values in the range {15?x1.. and Kogiso et al. random exchanges of location of numbers in strings that represent the design variables. because individuals carrying them more chances to breed. Here we discuss the genetic representation of a minimization problem and focus on the mechanics of three commonly used genetic operations (namely. and the reader is referred to Goldberg (1989) for discussion of their theoretical properties. Rao. The length of the bit string used to represent each variable is dictated in this case by the range of the variables. in particular some that have re· cently been applied to the design of composite laminates. However. genetic algorithms have been experimentally proved to be robust. By keeping many solution points that may have the potential of being close to minima (local or global) in the pool during the search process. Applications of the algorithm to design with composite materials are recent (see Callahan and Weeks 1992.4 GENETIC ALGORITHMS 185 section describes genetic algorithms.4. x4 }. For integer design variables. and {3?x3 ?0}. individual characteristics that are useful for survival tend to passed on to future generations. and Venkayya (1991) addressed the optimal selection of discrete actuator locations in actively controlled structures via genetic algorithms. As will be described in the following paragraphs. recently proposed for buckling optimization of composite plates. The first application of the algorithm to a structural design was presented by Goldberg and Samtani (1986).x4 ?0}. 5. x = {xp x 2 . More recently. repeated genetic optimization will often yield different good designs. x 3 . Also. binary numbers were predominantly used for representing design information as bit strings.4. easy-to-program. 1994 ). This aspect can be very useful to the designer when the design space includes many local optima of comparable performance.. the concept of parallelism is even more basic to genetic algorithms in that evolutionary selection can improve in parallel many different characteristics of the design. and inversion of the somal strings. Working on a population of designs also suggests the possibility of implementation on parallel computers. Nagendra et al. The mechanics of natural is based on operations that result in structured yet randomized change of genetic information (hence biological traits) between chromosomal strings of the reproducing parents and consists of duction. who solved a well known 10-bar truss weight minimization problem. Hajela (1990) used genetic search for several structural design problems for which the design space is known to be either nonconvex or disjoint. Genetic algorithms. However. permutation (sometimes called gene swap). occasional mutation. {7?x2 ?0}. 536 strings of 8 digits. Then we can represent the symmetric and balanced laminate [(±8L)NI(±8L_1)N I . for example. For designing the stacking sequence of a laminate. X4 X3 (5. 2. which actually produces increments of 0.00 1. with a very large alphabet it is more difficult to ensure that every possible gene is present in the initial population. where the gene looks like the stacking sequence. i = 1. x. but only 4 8 = 65.. Of course. -45°. The plus or minus pair of off-axis angles are normally placed adjacent to one another in order to keep the values of the D 16 and D 26 terms small.00087 in the value of the variable. -X2 = 5..01 $xi::. the smallest number of digits is m = 11. For example.. instead of the required value of 0. This reduction in the length of the string can significantly improve the speed of the genetic search. We define stack variables. as will be described later. and 90° layers with fixed thickness t can be represented by a string containing the digits 1. we will need a constraint requiring that N 1 + N 2 + N 3 = 12. This allows us to reduce further the length of the string for balanced symmetric laminates. . xu.g.. laminates are further assumed to be balanced. and can vary between 0 and 12. This reduction in the length of the string. For example. For problems where the design variables are continuous within a range :i: ::. if {0.---I0I X2 187 I I I 0 1I . 2. 1993).. 45°.xf::.2) where binary representations of the individual variables are strung head-to-tail. ±45°. as long as we take precautions with the crossover and mutation operators. does not permit us to represent laminates containing stacks with an odd number of 0° or 90o layers. and.81} and the required precision is 0. The integers N" N 2. xi::. requiring a -45° layer for every 45° layer. binary digits needed for an appropriate representation can be calculated from ' I I 2m~ ((xf. limit ourselves to only symmetric laminates) we can simply represent the same laminate by [2 3 1 1 1 1 4 4]. xf} the precision needed for the final value is _xincr. N 3 ].3 x 109 16-digit strings that correspond to distinct designs. which has some theoretical advantages. respectively. /(±8 1)NL as a string with the L integers Ni. in this example. For example. defines the number of bits needed. The size of the subintervals. ±45. If we make use of the symmetry condition (that is. we typically employ a code assigning an integer digit to each possible orientation angle. and we need only the number of layers in each given orientation. In many problems we can use the actual values of the design variables in the string. Michalewicz 1992. however. This natural representation reduces the effort of coding and decoding. ::. For a laminate made of only o. If a variable is defined in a range {.. respectively. x3 = 3. For example. Variants of genetic algorithms that employ real numbers and are applied to function optimization have been dubbed evolutionary algorithms by the European school. • . the stacking sequence [±45/0i90 2L may be represented as [45 0 0 90]. then the number m of. which determines the of the solution obtained by the genetic algorithm.4 GENETIC ALGORITHMS 0 I I 0 X1 --. the variables are x 1 = 6. The use of real variables in strings has enjoyed more popularity in Europe than in the United States (see. Back and Schwefel. 3. represent the number of stacks of ±30°. and 3 representation. studies comparing binary to other representations tend to find superior performance by nonbinary representations (e. the common practice is to divide the intervals into a finite number of subintervals so as to discretize the design variables. or' with a longer string containing the binary representation of these integers. For most applications. However. 1. ±45°. the stacking sequence of a laminate that can only admit oo.4. Instead we can I. the in-plane design of a 48-ply balanced and symmetric laminate with ±30°.xf)/xincr + 1). The 16-ply laminate [±45/0i902L is then represented by the genetic code [2 3 1 1 1 1 4 4 4 4 1 1 1 1 3 2]. but it requires more specialized programming tailored to the specific set of orientations used..186 INTEGER PROGRAMMING 5. Furuya and Haftka 1993)...001. For in-plane laminate design problems. which represent two adjacent layers with the same orientation. x4 = 11. based on mimicking evolution.. the stacking sequence is often unimportant. and N 3 . Binary variables correspond to the smallest possible alphabet. Therefore. and ±60° stacks can be represented by a string [N" N 2 . The European school often uses the term "evolutionary strategies" to describe methods. such as genetic algorithms. L.. and 4.. i1 ~1 . the [±45/0i902L laminate is represented by [2 1 1 3]. and 9~ layers we use the 1. The conversion of variables to binary numbers introduces a degree of complexity that is not always necessary. There are 4 16 = 4. For example.. It is also possible to use natural coding. and ±60° layers. the design of a thin laminate that can be represented by a short string will require a 1"'11 . 5. As will be seen later. and we use a coding with one number for each or each two-layer stack. so that each digit in the string can take integer values in the range (0. This can present a problem when the ber of layers in the laminate is unknown. So with the representation we have a much larger design space. and these voids will have to be pushed out. but we seek thinner laminates satisfy these constraints. so that the size of the initial population of designs determines the size of the population in all future generations as well. with N3 = 12-N 1 -N2. and the great majority of the designs generated by the genetic on the way to the optimum will violate the constrai N1 + N 2 + N 3 = 12. for the number of strings. Then we repeat it again to generate the requisite n5 laminates (see Fig. while the design space associated with [Nl' N2 ] only 169 possible designs. we have a smaller design space. and then transform it to a choice of an integer between zero and four by taking the value LsrJ. Thus the numbers are not truly random. The design space associated with [Nl' N2 .2 Initial Population We can use information on the design to bias the initial population makeup. 45°. by [0 0 1 4 2 3]. will then represent the laminate [0 5 ). For example.. because of constraint on N 3 . Such random number generators typically require an input "seed" the first time they are called. the design of a structure composed of several laminates will require an even larger population size. with a constraint that N 3 ~ 0. Le Riche and Haftka (1993) found that the optimum population size was H. the optimal size of the population increases with problem size. The seed determines the sequence of random numbers that will be generated by repeatedly calling the function or subroutine. with zero corresponding to an absent layer.4 GENETIC ALGORITHMS 189 have a string composed only of [NpN2]. with this approach we need to take the precaution of not having O's appear anywhere except for the leftmost positions in the string. Similarly.4). 1 0 4 2 0 3 => 0 0 1 4 2 3 => [0/90/45/-45]. for use in genetic algorithms the distinction is not important.4. for the design of 48-ply balanced symmetric laminates ( 12-digit string) subject to buckling and strength constraints. In general. Subroutines or functions that generate random numbers distributed uniformly be(ween 0 and 1 are widely available. satisfy the constraint N 3 ~ 0). Similarly. N 3] has 13 3 = 2197 possible designs.4.188 INTEGER PROGRAMMING 5. smaller population size than the design of a thicker laminate that must he represented by a longer string. Using RAND. we first generate a random number r between 0 and 1. with the zero in the leftmost position corresponding to the absent layer. 5. consider the case of the stacking sequence design of a laminate made of 0°. We can then repeat this process for each layer to generate the entire laminate. We will denote that size by ns. resulting in sl'ower progress towards the optimum• With the second representation. It is easy to check that. This second representation is much more ef"' ficient. That is. For the design of a simple wing made up of 18 different laminates.. we will represent [0/90/45/-45]. Genetic algorithms generally work with a population of strings are all of the same length.0. where LxJ indicate the integer part of a real number x (floor function). Limiting ourselves to symmetric lamina this means that we can use 6-digit strings to code the laminate. for example. In the following we will assume that random numbers are generated by a function RAND(ISEED) and we will denote by r the output of the function.4) 5. and 90° layers. as in the case of thic11. there are only 91 possible designs. A singular optimum is an isolated optimum that represents a discontinuity in design space. For example. assume that we found that all the constraints be satisfied with a 12-ply laminate. Choosing the population size is at the present a matter of trial and error. One case where we need such bias is when we encounter a singular optimum associated with 0° layers. (5. 1994). but the Most genetic algorithms work with a fixed-size population. and we use them extensively in genetic algorithms. by [0 1 1 1 1 1]. and are often called pseudorandom numbers.4). The initial population is typically generated at random. However. -45°. We can solve this problem by making length of our strings correspond to an upper limit on the thickness the laminate and adding another digit to represent an absent layer stack. We can get a singular optimum in the design of composite laminates when the objective function favors designs with ±45° or 90° layers. more than half of the designs generated by the genetic algorithm be possible designs (that is. Subroutine init of Fig. For example.m:::s:s minimization. the best population size was about 50 (Le Riche.4 will produce zeroes or "voids" inside the laminate. 1 ). a 48-ply laminate under uniaxial tension.4 GENETIC ALGORITHMS 191 subroutine init(lamin. with each digit being an integer between 0 and nmax. . for example. However. if gmax> 0. otherwise (5. Consider.190 INTEGER PROGRAMMING 5. and we want to add a penalty to the objective function to help the search move into the feasible design. Consider the standard formulation of an optimization problem..4. If we start with a design with a small number of 0° layers. For unconstrained problems.3 Selection and Fitness loading is primarily in the 0° direction. v-Eigma). Eq. we apply a series of genetic operators that produce a new generation and replace the initial population with the new one.ns do 100 j=1. and Haftka. as explained in the following. Fortran subroutine for generating initial population of ns strings of length L. With genetic algorithms we have to use a similar strategy. If we reduce the number of 0° layers by two to (±45 9 /0 2 )s. j = 1.5) r 1111 1 ' 11!1 . the fitness of a design can be defined as the objective function (for maximization problems) or a constant minus the objective (for minimization problems). Several methods exist for handling constraints.g. such as (±45 8/0 4 )s. If gmax is positive the design is infeasible. The design margin of safety is then defined by the most critical constraint gmax= max/g). and this is a hotly debated topic (e. Le Riche. If the load is high enough. 1995). For constrained problems the fitness also must consider constraint violations or constraint margins. . We can define an augmented objective function j* as J = fJ + p gmax + D. However. where for simplicity we omit the equality constraints: minimize f(x).nmax) integer lamin(L. x ~ E X such that g/x) 0. We assume that the constraints are normalized so that a constraint value of -0.. while a constraint value of 0. The following is one simple way for handling constraints.L. ( 1. the strains in the zero fibers will be higher than the strains in the 45° or the -45° fibers (see Exercise 2 at the end of this chapter). The first of these operations is selection. Knopf-Lenoir. 5.L lamin(j. a similar problem with a singular optimum occurs with discrete thickness designs.. If gmax is negative we have a feasible design with a positive margin of safety.1 corresponds to a 10% margin. This phenomenon was reported by Schmit and Farshi (1973) for continuous layer thickness design variables. When continuous optimization techniques are used. the laminate will first fail due to breakage of the 0° fibers. where we want to maximize Gxy subject to constraints on strains in the layers. Genetic algorithms therefore need to define fitness and use it in a procedure that selects pairs of parent designs that will be used to create child designs for future generations. ng.ns) do 100 i= 1. if we eliminate the last oo layer and get an all ±45° laminate.4.4.1 corresponds to a 10% deficiency. this difficulty has to be countered by considering designs that have oo layers and also designs that do not have 0° layers. the situation will worsen as the strains in the oo layers will increase. The all ±45° optimum is called singular because it cannot be obtained by gradual reduction of the oo layers.4.i)=(nmax+ 1)*rand(iseed) 100 continue return end Figure 5. seeding the initial population with some designs without any oo layers. The selection process in genetic algorithms mimics biology in giving more fit designs a higher chance to breed and pass their genes to future generations. Once an initial population is generated.ns. Kogiso et al. (1994) have shown that such seeding helps in getting optimal no-zero-layer designs and does not appear to have adverse effects even when the optimum is not singular. The optimum design will usually be an all ±45° laminate. the strains in the zero direction will not matter because we will not have any fibers in this direction. and we want to reduce the objective function as a bonus for that margin. 71.15. gi = 0. (5. consider a situation ' where we require the laminate to be balanced. from j* as <l>=c-J*. but with different constraint margins. Such double change has very little probability of happening with the standard genetic operators described later. This example illustrates a case where the easiest transition between two feasible designs proceeds through an intermediate infeasible design. Example 5. (a) Determine the largest value of the bonus parameter £ that may be used for this set of designs without causing the GA (ge- . should be chosen as a small infeasibility penalty that applies even for very small constraint violations. we have designs that are only slightly infeasible. resulting in an unbalanced laminate. it may be desirable to periodically update c as well. r=7.69. For example.6) where c is a constant chosen large enough so that <1> is always positive.1 Consider the following optimization problem. we would have to use very large values of p to ensure that these designs did not have the best augmented objective. the unbalanced laminate will have such low fitness value that it will have very little chance of being selected for reproduction. g1 = -0.192 INTEGER PROGRAMMING 5. 3 and p that penalize infeasible design. but is quite common. which could be maximized or minimized. so that these conditions will be met. For example. The bonus parameter £ is needed when the optimum design is not unique. 3. gi = 0. gi = -0. all having the same thickness. we often can obtain several laminates. A population of four designs is calculated to have the following objective function and constraint function values: Design 1: Design 2: Design 3: Design 4: l =6. When we design laminates for minimum thickness we often encounter such nonuniqueness because the thickness can be only a multiple of the basic layer thickness. = -0. rather than the concept of an objective function.4.79. So to achieve a transition from the pair of [0/90] layers to the [±45] stack. so that we want to find the optimum design with the highest safety margin.4. The second parameter. but not so large as to almost obliterate the relative differences between different designs. It should typically be a very small percentage of the average value of the objective function. However.50. we do not want to use a very large penalty parameter because the design process often benefits from shortcuts taken via infeasible design. If we penalize constraint violation too heavily. the bonus parameter £ needs to be small enough so that no design has a lower j* than the optimum design. f 3 = 6. <j>. The first. We can have intermediate designs where only one layer is replaced.4. gi = -0. 3. maximize f(x) such that g 1(x) ~ 0 and gz(x) ~ 0. Occasionally. Genetic algorithms traditionally use the concept of fitness. and having to rely on a double change will slow down convergence. Therefore. This situation is not .4 GENETIC ALGOFIITHMS 193 In Eq. but we have a coding that permits unbalanced designs. However.28. we can define the fitness. a 0° layer is replaced by a 45° layer. Without the first penalty parameter. p. This situation is the reason for using two penalty parameters instead of one. To accommodate the minimization problem. we must have two changes in the genetic string occur simultaneously. but which have very good objective functions. (5. penalizes substantial constraint violations. gi = -0. The penalty parameter p needs to be chosen large enough so that all infeasible designs have a higher value of f* than the optimum design. Thus. no matter how large the constraint margin is.49. g~ f2 = 7.o2. When large changes in the magnitude of f occur during the optimization. large values for penalty parameters are often found to slow down the convergence of the genetic algorithm. Now suppose we have a design that needs to replace a pair of [0/90] layers with a stack of [±45] layers.5) there are two parameters.11.16. which is higher for the better individuals.51.30. because the value of 8max for these designs is very small.· unique to balance constraints. The bonus parameter ensures that the selection process favors the design with the highest margin of safety. g~ = 0. 69+ 0. the smallest penalty parameter that can be used without causing the GA to pick an infeasible design as the best design of the population is the larger of the two values.51 .69 + E 0. If we use a penalty parameter of p = 1. In the preceding example we picked the permissible values of the penalty and bonus parameters by inspecting four possible designs.. the fitness of the best feasible design with the specified bonus parameter of E = 0. other designs may exist that will require more stringent values for these parameters. the largest value of the bonus parameter should be limited such that the fitness <)> 3 is smaller than the fitness <)> 1• Therefore. If these parameters are inappropriate we will find that the genetic algorithm will yield infeasible designs.69. with ns designs having fitness values of <j>i. Note that this is a maximization problem so we add the bonus to the objective function [as opposed to the function of Eq. we may want to start with fairly substantial bonus parameters.4 GENETIC ALGORITHMS 195 netic algorithm) to pick an inferior design as the best design of the population. For part (b).88.11 and <)> 3 = 6. ns. Typically.p 0. we penalize those designs that have constraint violation. (5. Therefore.30-0.07. we will add the bonus only to the objective functions of Design 1 and Design 3.5.. Adding a bonus to these constraints for the two designs will change the fitness function of the respective designs. we need to try to estimate reasonable values for these parameters based on our knowledge of the problem. Of course.76. p = 2. For Design 1. Once we define the fitness of all the designs in the population.02-0. Therefore. is unacceptable. where . In general.88. and adding too large a bonus may make the fitness <1> of the Design 3 better than that of Design 1. (b) If the bonus parameter is E = 0. 0. namely.4. 0. That is.2. and then we can increase the penalty parameters. designs that violate one or more of the constraints have better objective function values than designs that satisfy all constraints. with each design being assigned to a sector of the roulette wheel with an area proportional to its fitness. we consider only designs that satisfy both constraints.11) < (6. determine the smallest value of the penalty parameter p that can be used without causing the GA to pick an infeasible design as the best design of the population.01.5) from which we subtract the bonus]. = 6.71. For the fourth design.p 0. .49) :::::} E::. as is the case in this example. the ith designs gets a fraction ri of the wheel. Similarly.16 > 6. For part (a) of this problem. This is not an acceptable situation. If the penalty parameter is not large enough.6.194 INTEGER PROGRAMMING !5. therefore.2 is <)> 1 = 6.11 = 6. the fitness function of the design with a constraint violation may be better than the fitness of the best design which satisfies all the constraints. If we find that some of the designs with high constraint margins produce fitnesses larger than fitnesses for designs that have a better value of actual objective function but smaller constraint margins. In this example.07. a common procedure for selecting parent designs is to simulate a biased roulette wheel.0.01. g~ = 0. 11 is the most critical and for Design 3. the fitness of Design 2 becomes <)> 2 = 6. However. E(0. which is larger than the fitness of the best feasible design and. It pays to begin with relatively small penalty parameters. The bonus is calculated on the basis of the most critical constraint.51. Clearly.28 > 6. <)> 1 For the second design.71 :::::} p < 1. gl = 0. i = 1. we can reduce the bonus parameter. .51 is the most critical.2 and the penalty parameter o = 0. the value of the penalty parameter that will cause its fitness to become larger than <)> 1 is 7. the value of the penalty parameter that will cause <)> 3 to become larger than <)> 1 is 7.71 :::::} p < 2.01. Design 2 and Design 4.49 + E 0. Design 1 has a better objective function than Design 3. Design 3 has a constraint margin that is larger than that of Design 1. 45 + o.0. 0. 0.15}. {R0 .6 with the integers on the slices indicating the location of the design in the initial fitness list (also note that the angles associated with the R. The locations of the R 1.ns if (place .35. r 4 . subroutine select(R. = L rj' j=l i = 1' .6120.1473. 0.9217. we place the random number around the circumference going counter clockwise from R0 = 0 and find the interval in which the random number falls. r 6 } = {0.1560. That is. 5. and the wheel for this generation of designs is shown in Fig.3471}.4.35 + o. 6 are marked along the circumference of the wheel. Fortran subroutine for selecting two parents by simulating a roulette In order to implement the selection process using the roulette wheel.65. ' ns. <j> 5 .1357. Therefore.9). 0.5156.0 and 1.2 <l>j I ' j=l On the computer the roulette wheel can be implemented by generating a random number r between zero and one.38 + o.0. RU-1)) go to 300 parent(i)=j-1 go to 400 300 continue parent(i)=ns 400 continue return end Figure 5. were generated: r = {0.. 0. <1> 4 .. R4 . l = Example 5.1357.7123. 0. 0. 0.0.65 + o.45. 0.6o + o. the sum of all fitnesses yields n.. 0. The procedure is coded as subroutine select in Fig. R3.2 place=rand(iseed) do 300 j=2. we chose a parent design using Eq.gt. .8) For a generation of six designs. 0. In order to select parents for crossover. R5. the following fitnesses were calculated: {<1>1' <1> 2 . . 0.4. Construct the roulette wheel and determine the individuals selected for the crossover operation. r 3. <j> 3 .5 and demonstrated in the following example. For each value of the random number r. and R.4. 0.196 INTEGER PROGRAMMING <I>. 0.9419. 0.2825 < R 2 . <1> 6 } = {0.0.2519. 0. (5.8).2825.. and then selecting the ith design if R1_ 1 :::. Rl' R2. 5. R6 } = {0.2825 we have R1 < 0.38.3683.5.'s are obtained by multiplying them by 360°). For the given list of fitnesses.4. six random numbers between 0. (5.1744. r<R1. For the first random number r = 0.60.58.4. (5.15 = 2. 0.ns. 0.parent) integer parent(2) dimension R(ns) do 400 i=1. 0. the fractions of the wheel occupied by the different designs are given by { rl' r 2 . where R 0 = 0. wheel. 1.4 GENETIC ALGORITHMS 197 r. we create another list of numbers following Eq. n 5. r 5.7675.0581 }.2326.0}. (5. . i = 1.9) :L j=l <1>j = o. 1626. In that case. 2}. 0. (a) for the standard fitness-based selection. 0. $6 } = {0. R0 < 0.i + 1) (5. For the same set of random numbers used in Example 5. {0. we have R 3 < 0. $5 .61.2 for a fitness list of {<I>P $2. 1.0}.6.7a. we have Defining fitness on the basis of the numerical value of the objective function or the augmented function carries the disadvantage that towards the end of the optimization. 4. R5 <0.60. For this reason. 5.1653. Roulette wheel for example 5. another popular strategy is to define fitness based on the relative rank of the designs. Note that none of the designs is selected twice as parents. and progress slows down.2825 INTEGER PROGRAMMING 15. 0. and R4 < 0. the selection pressure in favor of the better design becomes small. The list of summation of fractions of the area of the wheel is {R 1. r5 . 6. 0.62. r3 . .7123 < R4 .4. R 2 . the fractions of the wheel occupied by the different designs are given by {rp r2.6721.9217 Example 5.10) In this case the portion of the roulette wheel assigned to the design. 5.1680.4. 0. R4 . and R2 <0.2825 < R2 .1734}.r4 . r 6 } r=0.4.7123 Figure 5. and the wheel for this generation of designs is shown in Fig. 4.1545. 4.(ns + 1)ns r = 0.1560 i + l.8266.4.7).9217 < R5 . = R1 < 0.= 199 n~- r = 0.1560 < R2 . $3 .6120<R4 .6120 < R4 R 1 < 0.64}. which represent a typical population towards the end of a genetic optimization when the difference between the designs becomes small. (5.2.65.7123 < R 5 . R 5 . 0. (5. Design 2 is selected. 0. 0. Eq.1 therefore. the ith-ranked design is reassigned a fitness value of R 1 < 0. yielding the following parents: {2. R3 . 2. 0.4 GENETIC ALGORITHMS <I>. 0.5068.3471 < R2 • The parent designs selected for the next generations are {2. For the remaining random numbers.4. 5.4. (a) In this case of fitness-based selection.. Use the same six random numbers used in the example.3 Repeat Example 5. R4 < 0. 3}.57. when the differences between competing designs become small. R 3 <0. 0. becomes rRJ 2(ns.3471<R3. 1. = {0.11) .1762.1560 < R 1. 0. $4. 0.4.1680.198 r = 0. (b) for the rank-based selection process. R6 } R3 < 0.9217<R6.3306. 0. Once pairs of parents are selected.57}. 1}. R4 .7b. a large penalty multiplier causes the fitness function to become negative.4 Crossover r = 0. Designs 4 and 5 of the original population with low fitnesses are not selected. 5.9524.1560 r = 0. 1. begins by generating a random integer k between 1 and L .1905.7.2} designs of the ranked population as parents. 6}. R2. 1. The fractions of the wheel occupied by the rank-based designs are given by Eq.3471<R 2 . 0.3. 0. for the same set of random numbers used in the previous example. 0.9217<R5 . This feature becomes especially handy in cases where. 3. On the other hand.1429.7143. 2. r4 . 1. (5. Another advantage of the rank-based selection procedure is its capability to handle negative values of the fitness function. R3 . The list of summation of fractions of the areas of the wheel is r Designs 1 and 3 in the original population.2.61. the singlepoint crossover.8572. 5}. 0. 0. {0. which are among the designs with the highest fitnesses.7123 l:z) based onfitnesses 3 r = 0. we have R0 < 0.3.2381. 0. R 5 . for designs that violate the constraint. R 0 < 0.62.65. 0.0952. R2 <0.200 INTEGER PROGRAMMING 5. q>.4 GENETIC ALGORITHMS 201 (b) Reordering the designs according to ranking we have {0.4.1. 5. and the wheel for this generation of designs is shown in Fig. 6. This number defines a cutoff point in each of the two strings and separates each into two substrings. r 6 } = These random numbers select the {1. {R 1. Note that the numbers in the list above are ordered from best to poorest rather than by the original order of the designs.7123 'b) based on ranking Figure 5. The original designs corresponding to these parents are {3. 0.60.4. 3. 4.1560 RJ 5. 0. 2.6120<R3 and R1 <0. = 0. 4.1560 < Ri' The new rank-based fitnesses for these designs are {6.2857 corresponds to the best design.5238.2825 < Rp R2 < 0. r5 . Roulette wheel for Example 5.7123 < R3 .2857. or Design 3 of the original list. 2. Denoting.0476}. where L is the string length. 0. for convenience.2825 r = 0. 1. 0. So r 1 = 0.2857. where the corresponding designs from the original list are {3.4.64. the two parts as left (initial part) and right (final part) substrings. we splice together the left part of the string of . as well as the combination of negative and positive ones. the mating of the pair also involves a random process called crossover. 0. 0. The simplest crossover.11) as {rp r2 . 0.0} and. are each selected twice as parents. r3 . 1. R4 <0. R6 } = {0.5. (5. represented to an accuracy of 0. Assume the length variable to have an expected range of (0. we use real numbers to represent the variables.8. However. but because of the mixing of the strings the crossover becomes a more random process and the performance of the algorithm might degrade (De Jong. on the other hand. One or both of the child designs are then selected for the next generation. Implementation of crossover begins with the generation of a single random number r uniformly distributed between 0 and 1. One solution to this problem is to use the previously discussed two-layer stacks. The two possible child designs are child 1: 0 0 1 2 1 2 3 4 1 [0/45/0/45/-45/90/0].4. [0/45/-45/90/0Jv. instead of getting child designs that have some intermediate value between the two parent designs. are represented by binary strings. Note that even though both parent laminates are balanced. which reduces the effectiveness of the crossover operator for combining existing traits from parent designs into child designs. In fact.05.15) The two possible child designs are child 1: child 2: 000000 111111 l = 0. the . for example. When integer or real variables. where r is our (0. Consider. Consider.4. Booker ( 1987) showed that. crossover may generate ~hild designs that do not bear any resemblance to the parent designs.2. and a crossover point k = 5: parent 1: parent 2: 0 0 1 2 1113 1 1 1 0 0 0 0 1112 3 4 1 [0/45/0/-45/03). it contributes to the exploration of new design alternatives.2) in. An exception to this is the two-point crossover.4.05 (b + 1).14) The following crossover scenario. 1) uniformly distributed random variable. one can expect that in some situations the additional exploratory power of the binary crossover operator is beneficial. The processes of selection and crossover are repeated until there are n5 child designs. (5. The binary number has a range from 0 to 26 .12) ~rossed. However. In this kind of representation. (5.4.8 and 2. 3.16) l = 3. so that we calculate l as l =0. neither child is. we run into the opposite problem.. instead of letting the end of the string be the second point that defines the string segment to be 1=2.. then one of the parents is cloned into the next generation.4. a process that is usually handled by the mutation operator described in the next section. If. Crossover is typically implemented with some probability p c· If crossover is not indicated.1)r + 1]. two symmetric laminates coded in a string of length L = 9.13) choosing the endpoint randomly improves the performance of the algorithm. The decisions on which parent to clone or which child design to select are not important. It is probably preferable to let the crossover operator control recombination of existing traits and the mutation operator handle exploration of new traits.202 INTEGER PROGRAMMING 5. 1975). as the parents were selected at random.4 GENETIC ALGORITHMS 203 one parent with the right part of the string of the other parent.7. Multiple point crossovers in which information between the two parents is swapped among more string segments is also possible. child2: 0 0 0 0 1 3 1 1 1 [0/-45/0J. (5. (5.05. The random integer needed for defining the cutoff point can be generated as [(L . This indicates the disadvantage of coding that distinguishes between +8 and -8 in balanced laminates. we obtain designs that are more extreme than either parent.1 = 63. That is. One can check that most of the designs obtained by crossover from these two parents will have this property. a rectangular panel with a length variable l. between two panels of lengths 0.05 in by a 6-digit binary number b. the one-point crossover can be viewed as a special case of the two-point crossover in which one end of the string is the second crossover point.45. such as number of layers or panel dimensions. for example.4. If the number is less than p c crossover will be performed by generating another random number for the cutoff point. Another solution is discussed in Section 5.45 can take place: parent 1: parent 2: 0 0111 1 1 1 11110000 l = 0. 204 INTEGER PROGRAMMING !5.4 GENETIC ALGORITHMS 205 real value of each of the variables is assigned to a single location in. the genetic string. With binary representation of these real numbe1 crossover can create child designs that have intermediate values of tho. parent genes. However, with each real number represented by a gene, crossover can only reproduce values present in the initial lation. This situation may be acceptable, since the mutation opcwLul,~ described in the next section, can create new values. Another solution to this problem is to replace the ordinary crossover with an crossover. For example, if we have a string of length L with or real variables, we generate the crossover cutoff point as a real dom number w = rL between 0 and L. Then we take the first variables from one parent, the last L - L w J- 1 from the second and we average the LwJ + 1 variable between the two parents ac'-ulum)i to the fraction w- LwJ. That is, if in location LwJ + 1 the first p has the value x 1 and the second parent x 2, then the child design have and then rounded down to an integer. If the cutoff involved one of the real numbers in the string, the last step of rounding would have not been required. The process is coded in subroutine cross for integer variables, as shown in Fig. 5.8. r = (w- LwJ) x 1 + (1- w + LwJ) r. Consider, for example, a problem where we design a balanced, metric laminated plate with the design variables as the length a width b of the plate, and the numbers of stacks of o;, ±45°, and in half of the thickness of the plate, denoted as N 0 , Nw andN90, respectively. The genetic string will be [a, b, N 0 , N 45 , N 9 oJ. The crossover scenario will correspond to a 10.1 x 7.3 [O]z0 plate crossed, with a 12.5 x 6.3 [Oi±45/904 ], plate with a cutoff point of 2.35: parent 1: parent 2: The (only) child design is 10.1 7.3 50 0, 12.5 6.3 2 1 2. child 10.1 7.3 3 1 2. (5.4.19) That is, the child design is a 10.1 x 7.3 [Oi±451904 ]s plate. Note that with the cutoff point being 2.35, the first two numbers in the string were taken from parent 1, the last two from parent 2, while the third number was calculated as N 0 = (2.35 - 2)5 + (1 - 2.35 + 2)2 = 3.05 subroutine cross(parent,chldlm,prntlm,ns, pc,L) integer parent(2),prntlm(L,ns ),chldlm(L) id 1=parent(l) id2=parent(2) clone=rand(iseed) if (clone .gt. pc) then do 100 i=1,L chldlm(i)=prntlm(i,id 1) 100 continue return else endif rcut=L *rand(iseed) icut=int(rcut) if (icut .eq. 0) go to 300 do 200 i= 1,icut chldlm(i)=prntlm(i,id1) 200 continue 300 continue chldlm(icut+ 1)=(rcut-icut)*prntlm(icut+ 1,id 1) $+(1 +icut-rcut)*prntlm(icut+ 1,id2) if (icut .eq. L-1) return do 400 i=icut+2,L chldlm(i)=prntlm(i,id2) 400 continue return end Figure 5.8. Fortran subroutine for averaging crossover with integer variables in a string of length L. Crossover is performed with probability pc, otherwise the first parent is cloned. (5.4.20) 206 !1.4 GENETIC ALGORITHMS 207 5.4.5 Mutation Mutation performs the valuable task of preventing premature loss important genetic information by introduction of occasional random alteration of a string. Inferior designs may have some good traits can get lost in the gene pool when these designs are not selected parents. Additionally, mutation is needed when integer or real c is used because in most cases there is low probability that all possible genes are represented in the initial population. Consider, example an initial population of 20 designs, each with 6 genes, each gene can take a value from 1 to 10. Let us calculate the ability that no member of the initial population will have a value 10 for one of the 6 genes. The probability that any individual does not have the value 10 in the first gene is 0.9. The that none of the designs have that value is 0.9 20 = 0.121. That is, probability that at least one design will have a gene with a value 10 is 0.879. The probability that for each one of the 6 genes we find one design (not necessarily the same for different genes) with value of 10 for that gene is 0.879 6 = 0.46. That is, there is more a 50% chance that the value 10 is missing from at least one gene at least one location in the string) for every member of the population. Because this is the highest value for the gene, crossover cannot create this gene, only mutation can. Mutation is implemented by changing, at random, the value of digit in the string with small probability. Based on small rates of currence in biological systems and on numerical experiments, the of the mutation operation on the performance of a genetic is considered to be a secondary effect. Goldberg (1989) suggested rate of mutation Pm of one in one-thousand bit operations. H when integer coding is used, higher mutation rates are required. experience with stacking sequence design suggests a rate of about in a hundred, with smaller populations requiring higher mutation to preserve diversity in the population. Mutation is implemented by generating a random number between 0 and 1 and applying a mutation if its value is smaller than Pm· Another random number can then be used to select a replacement for the digit being mutated. When we use 0 to denote absent layers, a mutation may create a void inside the laminate, so that we have to pack the laminate, as described in Section 5.4.2, after the mutation operation. When real design variables are coded as binary numbers, there are situations where small changes in the design cannot be achieved by mutations. For example, consider the case when we have an integer design variable varying from 0 to 31, coded as a 5-digit binary number. The string [01111] corresponds to 15 and the string [10000] corresponds to 16. Even though the two designs are close, it is extremely unlikely that a mutation will change one string into another. This can occasionally result in slower progress for the algorithm. For example, if the optimum design corresponds to 16 and 15 is a close second best, we can run into situations where most of the members of the population have a value of 15 at some stage of the optimization. Making the final improvement from 15 to 16 can then take a very long time. To counteract this problem, one can use another binary coding, called the gray code, which does not have the abrupt change in digits caused by small changes in the value of the number. Alternatively, we can add another mutation operator, called local mutation, applied to the original variables rather than the coded ones. This local mutation operator is also useful when we do not use binary numbers but use the actual values of the variables in the string. Subroutine locmut, shown in Fig. 5.9, performs such a mutation. The size of the mutation is dictated by the parameter range. subroutine locmut(lamin,L,pm,range) dimension lamin(L) do 100 i=1,L dodont=rand( iseed) if (dodont .gt. pm) go to 100 mute=range*rand(iseed) if (dodont .lt. 0.5) then lamin(i)=lamin(i)-mute if (lamin(i) .lt. 0) lamin(i)=O else lamin(i)=lamin(i)+mute endif I 00 continue return end Figure 5.9. Fortran subroutine for local mutation that perturbs integer variables by the ±range with probability pm. 208 INTEGER PROGRAMMING 5.4 GENETIC ALGORITHMS 209 2 3 1 4 4 1 2 1 1 3, The mutation coded in subroutine locmut in Fig. 5.9 is based on a uniform distribution. When real variables are used in the string, the mutation is often a normally distributed random number with a zero mean and a specified standard deviation (see Back and Schwefel, 1993). before ply-swap: after ply -swap: 2 3 2 4 4 11 1 1 3, (5.4.21) 5.4.6 Permutation, Ply Addition, and Deletion For stacking sequence design, several specialized mutation operators have been suggested in the literature. One class of operators is permutation operators, which change the stacking sequence without changing the composition of the laminate. These operators change the bending stiffness of the laminate without changing its in-plane properties. They are particularly useful for laminate problems with constraints on both in-plane and bending properties, as explained in what follows. In problems of stacking sequence design, there are many laminates with the same in-plane properties and different bending properties. For example, the laminates [0/902 ]s, [90/0/90],, and [90/0], have the same in-plane stiffnesses but different bending stiffnesses. In fact, any permutation of the stacking sequence preserves the in-plane properties. In particular, if we consider the optimum stacking sequence for a combined in-plane and bending problem, there may be many laminates with identical layer composition, but with different (nonoptimal) stacking sequences. Since there are many possible laminates, the genetic algorithm typically finds laminates with optimal composition of orientation angles early in the search, and the rest of the search is devoted to shuffling the stacking sequence to produce the desired bending properties. Achieving a permutation in stacking sequence by a standard mutation operator is difficult because it calls for two simultaneous mutations, which have a low probability of occurring. For this reason, various permutation operators have been proposed for stacking sequence design (e.g., Le Riche and Haftka 1993, 1995; Marcelin, Trompette, and Dornberger, 1995; Nagendra et al., 1996). The simplest and possibly the most effective permutation operator is the ply-swap, where two layers are interchanged. For example, for a symmetric balanced laminate [45/-45/0/90/0/45/0/-45], with positions 3 and 7 chosen for a ply-swap, we have (0°, 45°, -45°, 90° coded as 1, 2, 3, 4, respectively): leading to [45/-45/45/90/0/-45], laminate. Occasionally, the number of plies of each orientation is fixed, and the laminate design requires only obtaining the best stacking sequence with the given orientations. This is equivalent to finding the best permutation of a baseline laminate. Specialized codings and crossover have been developed for permutation problems arising in scheduling (e.g., Michalewicz, 1992). These can be adapted to composite optimization, and have proved to be highly efficient (Liu, Haftka, and Akgi.in, 1998). When the thickness of a laminate is not specified, we use empty layers or stacks with special codes (e.g., zero) to allow for variable thickness. The standard mutation operator in this case controls the processes of adding layers, deleting layers, and changing their orientation in a collective manner. These operators are referred to as the ply-addition, ply-deletion, and ply-orientation mutations, respectively. Assigning different probabilities to the three different forms of the mutation operator may provide performance improvements. For example, if a single mutation probability is used, we may change the laminate thickness more frequently than we need. The optimum laminate thickness is usually achieved early in the genetic search, so that the probability of adding or subtracting layers should be low compared to the probability of changing the orientation. Le Riche and Haftka (1995) showed the benefit of using separate ply-addition, ply-deletion, and ply-orientation mutations. Nagendra et al. (1995) showed similar gains and also suggested separate ply-swap operators for interchanging layers in the same laminate and between different laminates in the components of a stiffened panel (e.g., between the skin and the stiffener laminates). 5.4.7 Computational Cost and Reliability Genetic algorithms tend to be costly, requiring thousands of analyses. The cost depends on the size of the design space and the way constraints hinder movement in the design space. Therefore, to keep the computational cost manageable, it is useful to try to keep down the 90°. ±45°. with the understanding that the outermost 2 in the string: corresponds to 45°.145. for the 48-ply laminate. The digit 2 represents either a 45° or a -45° layer.4 Use genetic optimization to find the thinnest balanced symmetric Graphite/Epoxy laminate made of 0°.000 lb/in.8 x 1014 • As noted in the Section on crossover (Section 5.4 GENETIC ALGORITHMS 211 size of the design space and avoid formulating constraints in such a way that they will greatly constrain the movement in the design space. The constraint reflecting the balanced laminate requirement is a good example of this principle. We write the constraints in a normalized form as . and violations are less common as a result of crossover.23) For example. we may a bit on performance. but only the number of layers in each orientation. it can be shown that the standard deviation 0'0 of the apparent reliability is 0'0 =---/ r(l .1% of the optimum.4.210 INTEGER PROGRAMMING 5. we can simply run the algorithm several times and take the best answer. Consider for example. Reliability can be increased by increasing the number of generations that the algorithm is run and by increasing the number of strings n. some of the performance gains. is f = 2(no + n9o) + 4(n3o + n4s + n6o)· The two constraints are a limit of 0. and 902 .006 on the shear strain Yxy under a load of Nxy = 3. With only three possible digits the of the design space shrinks to 3 12 = 531.7.4.000 lb/in and a limit of 0. each execution of the algorithm can result in a different answer. The symmetry allows modeling of only half of the laminate. Le Riche and Haftka (1993) defined practical reliability as the probability that the optimum is "practically equal" to the global optimum. Example 5. Also.004 on the axial strain Ex under a load of Nx = 10. with a standard deviation of 0. n 30. The size of the design space is now 324 = 2. ±60°. which is the objective function. With the symmetry and the balance condition. they defined the term "practically equal" to mean within 0. That is. with 1 representing 0° and 3 representing 90° layers. and 90° layers that satisfies the strain constraints of Example 4. However.441. If the algorithm is run m times and is successful in finding the global optimum ms of these runs. for example. This point is illustrated in the following example. the apparent reliability may be quite different from the true reliability r. multiple runs may also produce multiple solutions that could be useful to the designer. duce the string length to 12. and the bal constraint is violated for many laminates generated by crossover mutation. then the apparent reliability ra is ra= ms/m. because we exclude laminates with an odd ber of layers of the same orientation in a stack.1. ±30°. now we can never have more than a single-layer violation of the balance condition. the child laminate is often balanced. If we use stacks of two layers each 0 2 . (5. and so on. and ±45° orientations. (5. if two-parent laminates are balanced. which is still 1000 times smaller than the original design space. n 60 . The computational cost of genetic algorithms also depends on our requirements in terms of reliability.r)/m. and n90 . Of course. so that the string length is 24.1. we have a very large design space. n45 . we automatically satisfy the balance constraint and. the design of a 48-ply symmetric and balanced laminate to be made of 0°.• Alternatively. 2. if we have 7 successful runs out of 10. For the buckling and strength problems that they solved. we can still salvage. In fact. Since this is an in-plane problem. the apparent reliability is 0. With four possible digits to rep·\ resent the four orientations. It is important to realize that because of the random aspects of the algorithm. the total number of layers in the laminate. and 3 to represent the possible layers.22) However. with a bit of statistics. the next 2 corresponds to -45°.1.4.8 x lOll. if m is small. We can use.4). If we have a where this consideration is important enough. All of these strategies cause an increase in computational cost. ±45. the size of the design space is 4 24 2. We therefore select the design variables as n 0 .4. The reliability of the genetic algorithm is usually established by testing with problems with known solutions. High reliability means that almost always the answer is close enough to the global optimum. the digits 1. we do not need to design the stacking sequence. 049. 5.9.4. However. then rq(n) = 1. the probability that we will not find the optimum in two runs is 0. if we find that optimization runs that last 100 generations give us 40% reliability. For the genetic optimization we use a string of (n 0 . g 2 = -0. we settled on a mutation rate.e lmax (gp g2)1. [0 10/±30/±45/±60]. the difference is so minute that it is much smaller than the error associated with predicting the strains using classical lamination theory (or any other theory). and the probability that we will not find the optimum in three runs is 0. that is. respectively.6%. That is.95 with standard deviation of 0. To determine the magnitude of p we note that in Example 4. Example 5.1 the total number of layers varied between 68 and 158. if max (gp g2 ) > 0. g 2 = -0. and (0 10/±30/±45)s. On the other hand. n60 .62 = 36%.4.1 $ 0.63 = 21. For any of these choices. f* ={ f. n45 ..9. For example. if we denote the reliability achieved by q runs each with n generations as r/n).4 demonstrates that multiple genetic runs can produce near-equivalent designs.8 and the local mutation of Fig. The optimizations yielded four designs with practically identical performance.0006. with nmax = 20. This last design is slightly superior in that it has a minimum margin of 0. After some experimentation. To ensure feasibility we chose p = 300. with an average of 160.24) I . However. The last design has g 1 = -0. thus presenting a choice that could be decided on the basis of other considerations. For a population size of 80 and 50 generations. a mutation range of ±5.4. 5. we ran the optimization 20 times and monitored the reliability of the process.[1. Additionally. To determine the population size and number of generations. two runs give us a reliability of 64% and three runs a reliability of 78. say 10%. It is worthwhile to note again that composite stiffness design problems often have multiple optima and that genetic algorithms help the designer find all or many of these optima. close to zero. As can be seen from the equation for the objective function. These were [0 12/±45 4 ]s. n90 ).1 (for a discussion of the selection of the bonus and penalty parameters. (5. and the first three have g 1 = -0.4.0001.006 ~-1$0. In general.004 . will increase the strains by 10% and so will result in gmax becoming about 0.0. it is important to build up intuition for these parameters from simple problems. For the initial population we used subroutine init of Fig. and an augmented objective function of ! + p max (gp gz). [09/±30/±45/90]. e g2 = 0.4.4. Reducing the number of layers by. To provide a bonus for feasible designs with constraint margin. We need the penalty associated with this violation to result in an increment to the augmented objective of at least 10% of that objective.22) and (5. 5.0365. We need to compare these numbers with the reliability achieved in a single run of 200 generations and 300 generations. depending on the choice of orientation angles.212 INTEGER PROGRAMMING 5.r 1(n)V (5. we need to choose e to be small enough so that even a large constraint margin will not overpower a difference in thickness. and a crossover probability.3 and Example 5.23). refer to Section 5. we want it to be large enough to reward constraint margins for designs with the same thickness. This means a p of between 68 and 158.1). Pm = 0. Pc = 0. it is often computationally advantageous to make multiple runs rather than a single long genetic optimization.06% instead of the 0. one constraint is critical. this corresponds to a reliability of 0. 19 of the 20 optimizations gave the same final objective. Using Eqs. for more expensive problems we may not be able to engage in such extensive experimentation.01% obtained for the other three.4 GENETIC ALGORITHMS 213 g1 X .. then the probability that we will not find the optimum in one run is 60%.4. For this we chose e = 0. All four designs have 40 layers.15.4%.1. We used the averaging crossover of Fig. otherwise. this choice permits the initial population to have up to 320 layers. n30 . Of course.0006.1. The couplings as well as the tip displacement and twist are given in terms of the inverse of the (A. Then the results could serve as guidelines for choosing the number of runs for similar but more computationally expensive cases. The constraints require the tip displacement to be less than 0. In addition to constraints on maximum deflection and twist at the tip. The problem formulation allows . with the corresponding locations of submatrices in the inverse denoted as au.3)].p respectively [see Eq. single runs are most efficient when the desired reliability is below about 80%. s (Type I) and [TJITJ + 60/TJ + 120]n s (Type II)..005 in. For higher reliabilities.dl1PL3 di6Mxf} 3 + 2 . The beam is to be designed to maximize extensiontwisting and bending-extension coupling.. multiple runs are more efficient. '1r~ . the objective function is (c 11b11 1+ c2 1b161). an additional constraint was imposed in this problem that limited the number of layers in the laminate to a maximum of 24. The tip displacement wtip is given as wtip- Figure 5. on how we want to emphasize the two coupling terms.. h. v12 = 0. Any additional symmetric layer groups with equally' spaced angles can also be added to the laminate without disturbing the hygrothermal curvature stability.304. £ 2 = 1.1 in and the tip twist to be less than 0.3.10.. The tip twist <!>tip is given as <!>tip= di6PL2 ~6MxyL --4.214 INTEGER PROGRAMMING 5. with constraints on stiffness in flexure and twist under the applied loads.. Cantilevered beam with a tip load and a moment. antisymmetric laminate [8/(8 + 90) 6 /8/-8/-(8 + 90)6 /-8 3] provides one solution that would meet the desired objective (see Chapter 3). t = 0.+ 2 Assuming that we want to maximize a linear combination of the two coupling coefficients b11 and b 16 . The coefficients c 1 and c2 are chosen depending A hygrothermally curvature-stable. and are defined as follows: E 1 = 20 X 106 psi.. However. The hygrothermal curvature stability can also be realized in more general unsymmetric composite laminates that are obtained by stacking two symmetric layer groups [8/8 + 90]. Finding the optimum value of 8 is a one-dimensional optimization problem and can be solved by plotting the objective function and constraints as a function of 8. (3. Most of the discussion in the section on genetic algorithm focused on integer codings.12 in T ~LOin ce=p 4 Mxy Section A-A Example 5.4. Le Riche and Haftka (1995) found that for designing the stacking sequence of unstiffened panels for strength and buckling constraints.65 x 106 psi.- ~ _l_ 0. binary codings are also popular. and we close the chapter with an example of an application of a genetic algorithm based on binary coding. D) matrix. Material properties are assumed to be those corresponding to HT-S/4617 Graphite/Epoxy.p and d. .3 X 106 psi..4 GENETIC ALGORITHMS 215 p ~A This equation can be used to compare the relative efficiency of multiple runs for simple cases for which we can afford to make a large number of runs.5 The cantilever composite beam shown in Figure 5.10 is subjected to a transverse load P and a twisting moment Mxy· Use a genetic algorithm to design a hygrothermally curvature-stable laminate for the beam.01 rad. G 12 = 0. B. ~A L=lOin .. and 200.5. Mutation with a probability of 0.(c 11b11 1+ c 21b161) such that n 1p 1 + n2p 2 :::::.544 . [ (I 10 3 d PL + d 16M _11_3_ 2xy L 2 1) 1]: : :. respectively. respectively.. and similar solutions were identified in each case. For three distinct choices of weighting coefficients c 1 and c2 . 2 2 3 2 2 Pz 10-6 10-6 2 3 4 5 2 3 2 3 3 4 3 2 3 3 2 2 3 2 2 -16.544 25. The first test case required the maximization of an equally weighted sum of the b11 and b16 coefficients.2-5. this is not a serious concern. which could assume the values of 2 or 3.743 -17. It is possible to have both layer groups be of the same type.25 and c 1 = c2 = 0.031 18.000 function evaluations were permitted. p 1 and p 2 • That is. As seen in Tables 5. When the tension-twist coupling Table 5.2. The population size was selected as 200.8. however. The optimization problem for finding the optimum laminate with the angles e and 11 was formulated as minimize 6. and the coupling coefficients b 11 and b 16 were 122. and the number of layers in each layer group. the two excess binary string combinations were assigned to 0° and 90°. The variables n 1 and n2 were both obtained as unity. because of a large number of possible choices of e and 11 to select from. as the thinner laminate would maximize the compliance coefficients..5.02 per string was used.6 e 126 174 126 126 126 11 30 168 168 168 168 p. For the case where c 1 = 0.3. Summary of optimization results for weighting coefficients c1 = 0. two digits each for n 1 and n2. similar results were obtained.218 INTEGER PROGRAMMING 5. The orientation angles e and 11 were varied discretely between 0° and 174° in increments of 6°. The choice of layer group is assigned to two variables p 1 and p2 .924 -26.544 -26. a total of 10. These results are summarized in Tables 5. the basic orientation angles for these groups were 8 1 = 108° and 82 = 150°. - [ 100 W":L' + ~.4 GENETIC ALGORITHMS 217 for the two-layer groups to be assigned to the two available positions on the laminate. 150. p 1 = 2 if we select the first layer group to go with n 1. a single-digit string is sufficient to represent each of these variables.98 X 10-6 in/lb-in. In the second test case. in that the search is biased in favor of the overrepresented variables.LI )- 1] s 0 The design variables for this problem are the basic orientation angles. which was set to 10 x 10-4 . n 1 and n2.4.2 and 5. 0.210 -26. It is important to indicate that there is an inherent danger in representing certain variables by more strings than others.44 X 10-6 in/lb-in and 436. five simulations of genetic search were performed. The best design obtained for this case was that in each of the two available slots. nl nz b. A 16-digit binary string is required to represent all design variables for this prob- lem (ten digits for e and 11. and p 1 = 3 if we select the second layer group to go with n 1• The constant 6. 24. with similar or different basic orientation angles. Hence. and one digit each for p 1 andp 2).031 18. this is to be expected.839 -26.5 Run ID Group Type b. c2 = 0. e and 11. the solutions corresponded to a use of both Type I and Type II groups in establishing an optimal laminate for the given design constraints. The variables n 1 and n2 varied between 1 and 4 and required a two-digit binary string to uniquely represent the four possibilities. A two-point crossover was used with a probability of 0. the number of layer groups.031 18. c2 = 0. Two variations of the above problem were considered.~. This experiment was repeated for population sizes of 100. is used to convert the maximization problem into a minimization. the bending and twist displacement constraints were activated in addition to the maximum thickness requirement.75. In this problem. The 30 possible values of e were each represented by one of the unique combinations available from a five-digit binary string. layer group I was selected. with a constraint only on the total allowable thickness of the beam. In each case. "Improving Search in Genetic Algorithms. Formulate the problem as a linear integer programming problem and use optimization software (e. Summary of optimization results for weighting coefficients c1 = 0.25 = = Run ID Group Type bll ni bi6 9 36 0 126 0 168 1l 168 48 168 48 42 PI Pz nz I0-6 10-6 2 3 4 5 2 3 2 3 3 3 2 3 2 2 3 2 3 2 2 2 3 2 3 3 -26.031 18. and the stiffness term A 22 is greater than 30 MN/m. H. Use the following material constants: U 1 = 57.1758 -26. G 12 = 2. Table 5. "An Overview of Evolutionary Algorithms for Parameter Optimization. 61-73. E 2 = 5. (1993). e:z = 0.218 INTEGER PROGRAMMING REFERENCES 219 Table 5. CA.5 GPa..545 -36.9194 0..545 -36. It is also worthwhile to note that the presence of the bending and twisting stiffness constraints increases the laminate thickness by introducing n 1 and n 2 values different from unity. U 2 = 58. the .6 GPa. 4. Los Altos. design 16-.-P.696 7.34. This makes the all-±45° design singular in that it cannot be reached by gradually reducing the number of oo layers.125 mm. the Type I group is mechanically orthotropic and is the predominant source of the b 16 coupling coefficient.1758 28. (1989).25.4. Repeat Example 5. S.9194 0. Use the following material properties: E 1 = 76 GPa. 1-23.75). ±45°. v 12 = 0.545 -36. (1987). pp. Summary of optimization results for weighting coefficients c1 0. the Type I group was used exclusively as shown in Table 5.. c2 0.. the maximum strain in any fiber direction increases and then decreases.9194 is emphasized to a greater degree than bending extension (c 1 = 0.545 2 3 4 5 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 3 3 3 0.g. J.25. McGraw-Hill. T. This is easily explained by the observation that the Type II group is mechanically isotropic and would not contribute to the extension-twist coupling coefficient. Solve problem 1 by genetic optimization. Davis. 2. 5. Design a minimum thickness laminate with 0°.544 28." Evolutionary Computation 1(1).705 -6. Booker.696 7. 9 144 36 36 36 144 1l 126 54 54 54 126 PI Pz nz IQ-6 10-6 -36.4 using binary coding of the design variables.0 GPa. Inc. Microsoft Excel) to solve it. Introduction to Optimum Design..4.5 GPa.031 18.7722 shear modulus is less than 16 GPa. REFERENCES Arora. L.3 GPa.). Back. and t = 0.75 Run ID Group Type bii ni bi6 For a 40-ply Graphite/Epoxy laminate under unit axial load. and 90° layers such that the laminate's Poisson's ratio is greater than 0. L. c 2 = 0. U4 = 17.9194 0. Study the effects of the probability of mutation and the magnitude of the penalty multiplier on the progress and final results of the design optimization. show that as you decrease the number of zeros in the laminate." In Genetic Algorithms and Simulated Annealing.. 3. Note that the total laminate thickness is not expected to be larger than 20 layers.544 28. Using genetic optimization. and 32-ply balanced symmetric laminates to maximize the Poisson's ratio of the laminate subjected to the constraint that EY > 40 GPa. New York. and U5 = 20. EXERCISES 1.6. U3 = 13.5 GPa. Morgan Kaufmann Publishers. 24-.75.0 GPa.545 -36.9194 0.3. In contrast. .4. and Schwefel. (ed. from (±45 8/0 4 ) 5 to (±45/0 2 ) 5 and finally to (±45 10 ). February. pp. The Scientific Press. (1996). D.A. (1986). Marcelin. Redwood City. and Samtani. 76-9381. (1989). R. 536-542.220 INTEGER PROGRAMMING REFERENCES 221 Callahan. Vol. 2418-2436. J.. T. Lawler. Ph.. and Nemhauser. G. Berlin. (1994).. H. Springer Verlag. and Haftka. L.. Reidel Puslishing. J. (1973). 7. T. pp. R." AIAA J.. Z. Nagendra." In Genetic Algorithms: Proceedings of 6th International Conference on Genetic Algorithms (ICGA95) Eshelman. (1989). (1998). 951-956.-S. (1995). Long Beach. Integer Programming." Proceedings of the Ninth Conference on Electronic Computation. P. 5140B. and Aarts. 519-536. S. (1995). T. K. (1978). San Diego..." Proceedings of OPT/'93. "Optimization of Laminate Stacking Sequence for Buckling Load Maximization by Genetic Algorithm. van. A. E. Akademiai Kiad6. (1993). "Genetic Algorithms for Placing Actuators on Space Structures. Dordrecht. (1970). Holland." AIAA/ASMEIAHS/ASCE/ASC 33rd Structures.. Combinatorial Methods of Discrete Programming..497-520. AlsoAIAA Journal Vol. De Jong. R. Fifth International Conference on Genetic Algorithms. B. Part 2. "Optimum Design of Composite Laminates Using Genetic Algorithms.. Simulated Annealing: Theory and Applications. Tomlin. Jestin. Part 4.. Schmit. H. A..).. (1994). "Branch-and-Bound Methods-A Survey. S. Pan. Inc. L. R. J. T. E. (1991). S. Hajela. Structural Dynamics and Materials Conference.. and Gtirdal." In Integer and Nonlinear Programming. 143. Adaptation of Natural and Artificial Systems. R. (ed. R. E. July 17-22." Proceedings. Integer. 543-555.. Structural Dynamics and Materials Conference. Fourth Edition. Gtirdal Z." Proceedings." In Design and Implementation of Optimization Software. J. "Improved Genetic Algorithm for Minimum Thickness Composite Laminate Design. MA. L. B. Greenberg. M.. The Netherlands. July. Nato Advanced Study Institutes Series. 699-719. D. and Machine Learning. 1141-1152. Optimization.D. (1975). Nagendra. S.). and Nagendra. and Wood. "Optimum Laminate Design for Strength and Stiffness.P. AIAA/ASME!ASCE! AHS/ASC 39th Structures.. Doctoral Dissertation. "Genetic Search-An Approach to the Nonconvex Optimization Problem. Rao. P. (1966). D. Johnson.. J. and Akgtin. (1987). and Weeks. pp.." Int. Kovacs. and Powell." Structural Optimization 9(3/4). (1960). 5(2). . 95-117. L. M. E.. Zaragosa. 29(6). Optimization of composite structures by genetic algorithms. and Quadratic Programming with LINDO. R. (1993). R. pp. Hajela. Le Riche. Laarhoven. ASCE.. and Haftka. H. Goldberg. "Design of a Blade Stiffened Composite Panel by a Genetic Algorithm. 236-244. 161. University Microfilms No. Land. Meth.-L. Le Riche. Le Riche. "Integer Programming Codes. Mathematical Methods of Operations Research Series. J. J. Goldberg. (ed. Le Riche. P. TX. University of Michigan Press. MI. L. Virginia Polytechnic Institute and State University. Num. and Erisman. Haftka R. R. Engrg." Operations Research 14. "Engineering Optimization via Genetic Algorithm. 471-482. C. 437-450. L. "Design of Composite Laminates by a Genetic Algorithm with Memory. Trompette. N. Structural Dynamics and Materials Conference. April20-23. Michalewicz.. "An Automatic Method for Solving Discrete Programming Problems.. Sifthoff & Noordhoff. T. M. CA. Genetic Algorithms in Search. "Optimization of Composite Beam Structures using a Genetic Algorithm. Knopf-Lenoir. Ann Arbor. (1980). "Permutation Genetic Algorithm for Stacking Sequence Optimization. 28(7). IL. G. University of Michigan." Econometrica 28. Dissertation. Watson. AIAAIASME/ASCE/ AHS!ASC 34th Structures. Abadie.). (1995)... (ed. S. H. and Domberger. A. E. April19-21... Urbana.. and Venkayya. Urbina. Furuya. S. T. 31(5). (1993). 942-943. J..A. "A Segregated Genetic Algorithm for Constrained Structural Optimization. E. (1993). V. Spain. R. D." Computers and Structures 58(3). B. Kogiso. and Haftka. E.. Reading. "Optimal Placement of Actuators in Actively Controlled Structures Using Genetic Algorithms. 2. Addison-Wesley Publishing Co. "Improved Genetic Algorithm for the Design of Stiffened Composite Panels. Z. CA. (1972). 225-240. B.." Composites Engineering 2(3).. John Wiley & Sons. Gtirdal. "Branch-and-Bound Methods for Integer and Non-convex Programming. Linear. Haftka. Inc. 1205-1210. (1992). New York. Analysis of the behavior of a class of genetic adaptive systems. (1975). Garfinkel. A. T. R." AIAA J. R. Liu.. P. Z. L. T. 149-160. K. Haftka. D. T. Haftka. and Haftka. pp. L. S. New York. Schrage. "Genetic Search Methods in the Design of Thermally Balanced Composite Laminates. Genetic Algorithms+ Data Structures= Evolution Programs. T. Elsevier Publishing Co." Composite Engineering. July 7-9." Mechanics of Composite Materials and Structures 1(1). Dissertation Abstracts International36(10). G. (1992). R. T. and Watson. and Doig. pp. CA. Dallas. (1990). L." Proceedings. and Farshi. Budapest. In the case of monolithic isotropic materials. . unlike stresses and strains. there is no sound analytical foundation for the determination of strength.The fundamental stress-strain relations for laminated composite -----~ materials are developed in Chapter 2. along with equilibrium equations and appropriate boundary conditions for a specific configuration. These relations.6 FAILURE CRITERIA FOR LAMINATED COMPOSITES Prediction of failure of structural components is ysually (!ccomplisl:u~g l?~lJ_Il~tiQIJ£ oLthe stresses or strain_§__tQ __f!!at~rial strength . local failures generally lead to complete fracture and total 223 . ' ~. First. stress concentrations (such as areas around notches and holes) at which one or more stress components assume large values compared to nominal stresses in the material are likely to cause localized failures.:. can be used to determine the stresses or strains under prescribed loads. on the other hand. Determination of the strength of a laminated composite material. The next difficulty is caused by through-the-thickness variation of stresses within the laminate even if the laminate is loaded by uniform in-plane loads. For brittle materials. This is true even for monolithic isotropic materials. where the strength properties are determined empirically based on tension and compression coupon test data. presents several difficulties. Variation of stresses within a structure is nothing new for stress analysts. Under certain loading. but the strength in a direction perpendicular to the fiber is controlled by the matrix material that holds the fibers together and is usually an order of magnitude lower than the fiber direction strength. failures progress from one layer to others as the load is increased.. ··------. local failure is in the form of yielding that remains localized and. In this chapter. For some applications first failure of any layer is not acceptable because it degrades the strength and stiffness of the laminate.. application of loads along the fiber direction may induce failure transverse to the fibers because of the stresses developed by Poisson's effect. · · .based __on first faL __ _ is commonly referred to as the first-ply failure cxiterion. In such a case. on the other hand. Prediction of delamination failures requires evaluation of interlaminar normal 0'2 and shear Txz and 'tyz stresses.. However. Stresses calculated in the principal material axes would have components both along the fiber direction and transverse to the · fiber direction. we provide a brief introduction to the topic of strength. Just as the elastic properties change as a function of layer orientation. in this book we will concentrate on the in-plane stresses and failures resulting from them. It is also possible that. . large stresses in a direction perpendicular to the interface between the layers can be generated. We know that in-plane elastic responses of such layers demonstrate coupling between the shearing and normal deformations. . be carried safely. The majority of failure discussion in this chapter is limited to plane stress problems. it is possible that as the loading is increased stresses in one or more of the layers of a laminate would reach their limiting strength earlier than other layers. Finally. with an emphasis on the strength of laminated composite materials. Therefore. leading to what is commonly referred to as progressive failule. a load applied to a laminate along the fiber direction may . but are at an angle to it. Yet. therefore.: ~pplied perpendicularly to the fibers. It is. the definition tl of failure for the laminate becomes fuzzy. A more comprehensive coverage of the subject can be obtained in Tsai (1968). These stresses are mostly localized and cause breaking of the bond between the layers within a small region. . matrix normal strength. · . even if the size of delaminations is small. in some other applications the laminate is considered to be unfailed as long as it carries at least a portion of the first-ply failure load stably (that is. Using unidirectional fibrous composite laminates with orthotropic properties brings in additional complexities to failure prediction that we do not encounter with isotropic layers. tolerated better than brittle failure. therefore.224 FAILURE CRITERIA FOR LAMINATED COMPOSITES FAILURE CRITERIA FOR LAMINATED COMPOSITES 225 loss of load-carrying capability. Another difficulty that arises from the lamination of layers together is the occurrence of damage in the form of delaminations. Calculation of these stresses is complicated and usually requires expensive analyses. there is an additional level of complication that is the result of stacking together several layers of material with different orientations and properties. Generally. For ductile materials. Hashin (1980). For the sake of simplicity. whereas the same load may cause failure if it is . determination of whether the failure of such a configuration is controlled by the fiber strength. at the macroscopic level the strength of unidirectional layers is direction dependent.. for a lamina restrained in a direction transverse to the fibers.1 "strength evaluation becomes even more complicated if the fibers are neither perpendicular to or along a given uniaxial in-plane stress state.. Since the stresses and the strengths may be different in different layers.n:nal!~~ unidirectional composite materials behave in a brittle manner . without unstable failure of the remaining layers). and Soni (1983). however.. they may affect the integrity of the laminate and can degrade the in-plane loadcarrying capability of a laminate.. Wu (1974). The strength of a layer of unidirectional material in the fiber direction is generally high. we include a brief discussion of the application of failure theories to bending of laminated composites. At the end of the chapter.. in practical engineering applications it is important to check interlaminar failures and take appropriate measure against such failures. In the case of laminated materials. Failure prediction_. or a combination of all of them is difficult. therefore. as well as a shear stress component.· -----::. Most high-p~ffQ. we discuss the failure of laminates with emphasis on fiber-reinforced laminates. matrix shearing strength. We then extend the discussion to fiber-reinforced orthotropic layers. However. likely that the laminate would suffer damage in the form of local (and most probably brittle) failures in those layers before it fails completely and ceases to carry any loads": That is. the matrix material that holds the individual layers of the laminate together has substantially smaller strength than the in-plane strength of the layers. We first discuss some of the basic failure theories that are commonly used for isotropic materials. . failure is usually associated yielding if the material is ductile or fracture if it is brittle. The failure envelope in the strain space is defined by the following simple inequalities: -cue < c.. Sue .1. Failure envelopes in the principal stress space.1 by solid lines..'' "' -"':' . 1 .. crY. Limiting ourselves to thin layers under plane stress.The following are the most frequently found failure criteria for · isotropic materials.___ a:un:um stram crzterron I Sy. 6. the failure strengths in tension and compression are generally assumed to be the same and are measured by the yield strength of the material. This assumption leads to the following set of inequalities used for failure description: -SY < cr1 < SY and -SY < cr2 < SY. and 'txY' the maximum normal stress corresponds to one of the principal stresses and are given by /> \ I ' - - . The failure envelope in this case is defined by The maximum strain failure criterion is similar to the maximum stress criterion in that the material fails when the maximum principal strain at a point exceeds the failure strain of the same material in a simple tension or compression test.' "' . 6. together with the other failure criteria described in the following subsections. 6.1).1.1 Maximum Normal Stress Criterion Distortion energy criterion I This criterion assumes that failure occurs when the maximum normal stress in the plane of the thin layer exceeds the material strength. Therefore. ( 2 + 1 xy-sue< cr.226 FAILURE CRITERIA FOR LAMINATED COMPOSITES 6. the square failure envelope is generally offset with respect to the principal stress axes (this offset is not shown in Fig. 6. \ M . < Eut and . . Maximum shear stress criterion I . .' ..1. The failure envelope in strain space is shown by solid lines in Fig. The normal stress failure criteria is more suitable for brittle materials than the ductile materials. (6.' . A graphical representation of the maximum normal stress criterion is shown in the stress space in Fig. / . respectively.''/ .2) where SY is the yield strength of the material.3) where sue and sut are the ultimate compressive and the ultimate tensile strengths. ~.1 FAILURE CRITERIA FOFIISOTFIOPIC LAVERS 227 6..-. For most materials the ultimate compressive strength is many times greater than the tensile strength.2 Maximum Strain Criterion 2 For the case of isotropic ductile layers.~ uc .. anq 'txy. "' .4) where cue and Eut are the ultimate compressive and the ultimate tensile strains. A stress state corresponding to any point outside the boundaries will cause failure.' 1.'~ . 1 1 11111 ... crY. For brittle materials. respectively..1.1 FAILURE CRITERIA FOR ISOTROPIC LAYERS SY' Sutl cr2 . the yield strength in Eq.'/ Maximum normal stress criterion S S Y' cr.1.' ..... < sut and -sue< (J2 <SUI' (6. and therefore the boundaries are referred to as the failure envelope.'I cr.2 = ax+a 2 Y ± a -ay X 2 ) Figure 6. For a given stress state ax. .. which has different values in tension and compression. 6. in the most general case of combined loading we deal with only in-plane stresses ax. For monolithic isotropic materials. (6.Euc < £2 < cut' (6. A stress state that yields a combination of the principal stresses cr1 and <J2 and remains inside the square is safe.1.1.2) is replaced by ultimate strength.2.. 6.8). However.10) s ut E and Eue =E' sue (6.11) The maximum strain failure envelope is shown in the stress space in Fig.cr21< SY.6) and the failure envelope is defined by -sue< crll/0"2 Similar expressions for t 23 and t 31 yield equations that involve the differences cr2 . Therefore.II 1 1111 I .1.2. (6.. for an isotropic material.1. (6. the maximum shear stress in the plane is cr 1 vcr2 E E and cr E2= l-£· vcri (6. for a material that has a yield strength of SY. Eq.8) For a general state of plane stress. identical to the maximum stress !1111 . 6.. based on an additional stipulation that the failure strains are obtained from a test configuration where the stress state was a uniform uniaxial normal stress only. the maximum shear stress at yield is tmax = S/2. Ssy' is given by It is also possible to represent the maximum strain failure envelope in the stress space. hence to prevent yield we must also satisfy the maximum normal stress criterion <SUI and -sue< 0"2.1 FAILURE CRITEFIIA FOR ISOTROPIC LAVERS 229 02 Maxim um strain criteri'on Eut / Eut 01 stress criterion. This criterion is mostly used to predict yielding in ductile isotropic materials.1.1) the minimum and maximum principal stresses have the same sign.1.1.1 by long dashed lines.1. 2 or lcr 1 .. the shear stress criterion is. the lines for the maximum strain criterion are skewed due to Poisson's ratio in the above equations.7) lcrll < sy and lcr21 < sy.1. or the yield strength in shear. The maximum shear stress in a tension or compression test specimen develops at an orientation 45 degrees from the axis of the specimen and its value is half the value of the stress in the axial direction.cry ( 2 ] 2 2 (6.crl' respectively. (6. Using the first two equations of the strain-stress relation. for plane stress cr3 is zero. Failure stresses corresponding to the failure strains can also be obtained from the same equations.228 FAILURE CRITERIA FOR LAMINATED COMPOSITES 6. 6._..9) +t. therefore. Compared to the maximum normal For the first and third quadrants of the principal stress plane (see Fig. we have £=. 2.!!!. Yielding is prevented if the following conditions are met: lt 12 1< s.cr3 and cr3 .. (2. Failure envelopes in the principal strain space.5) cr 1 .vcrl < sut• (6. the principal strains are expressed in terms of the principal stresses as E =---1 ssy =s. Yielding is predicted when the maximum shear stress at a point exceeds the shear stress in a tension or compression test specimen at yield.2.3 Maximum Shear Stress (Tresca) Criterion Figure 6.cr2 =± tl2 = 2 ax. Therefore.1. . either in tension or compression. surface treatment 6. (6.1.13). Use of micromechanical models is required to determine a reliable estimate of the strength (see. principal stress directions coincide with the longitudinal axis of the specimen. Strengths in the fiber and transverse to the fiber directions are typically determined using unidirectional test specimens (sometimes referred to as test coupons) that have fibers running along and transverse to the specimen axis. respectively. (6. it deals with the distortion strain energy density. 6.1. As mentioned in the introduction to this chapter. For loads that are primarily parallel to the fibers.1. principal stresses that are oriented at an angle from the principal material axes may be generated by using off-axis specimens which have fibers oriented at an angle from the specimen axis. That is. failure is controlled by the failure of the much weaker matrix material. As with the maximum shear stress failure criterion. microscopic defects such as voids.l ud = ~ ua. The off-axis strength is influenced to a large extent by such factors as fiber volume fraction. In a test environment. However. the principal stresses in layers of a laminate are commonly at an angle to the principal axes of the layers. off-axis strength prediction is difficult even if micromechanical models are used because of a number of manufacturing-related factors that are difficult to incorporate into analytical models. Strength evaluation for such a specimen as a function of the off-axis angle is not simple because it is not obvious whether the failure is controlled by the fibers or the matrix.14) its boundary is shown in Fig.1. independent of the direction of principal stresses.cr 1 cr2+a~< s. (6.1.12) to be smaller than its failure value in Eq. For off-axis specimens loaded along the axis of the specimen.2 FAILURE OF FIBER·REINFORCED ORTHOTROPIC LAVERS 231 criterion as predicted by Eq.6 of Beer and Johnston. loaded by axial loads. However. (6. composite layers are much stronger in the fiber direction than in the direction perpendicular to the fibers.1 0) is more restrictive than the maximum normal stress criterion as shown by the short dashed line in Fig. ratio of fiber to matrix strength. however.12) (see Section 10. the calculation of the magnitudes of principal stresses is sufficient to predict failure of isotropic materials based on the failure criteria discussed in the previous section. (6. the magnitudes are compared with the limiting material strength values to avoid failure. The distortion energy criterion is based on the assumption that yielding occurs when the distortion energy density at a point in the material exceeds the distortion energy density of a tension or compression test specimen at yield. for example.13) For plane stress problems the safe region of the stress space can be described by setting the Ud in Eq.11 ). 1992). Under general loading conditions. cr2= cr3= 0 and cr. the material strength is generally governed by the failure of the fibers. The major difference between the isotropic materials and unidirectional fibrous composite materials is the directional dependence of the strength on a macroscopic scale. 1974a).1 by the long and short dashed lines. The distortion strain energy is the part of the strain energy associated with shape change and it is given as 1 + j.cr.1. 6. 6. This inequality yields the following quadratic expression: crf. For loads transverse to the fibers. Ud3E Y 1+v (6. Chamis.?J (6. = SY at failure.1. the maximum shear stress criterion predicted by Eq.230 FAILURE CRITERIA FOR LAMINATED COMPOSITES 6. for example.1. in the second and fourth quadrants.. the distortion energy criterion is generally used for yielding of ductile materials.2 FAILURE OF FIBER-REINFORCED ORTHOTROPIC LAYERS Given a two-dimensional stress state. fiber matrix interface strength. and fiber matrix interfacial debonds.cr3) 2+ (cr3.- crz) 2+ (crz. --~--sz.4 Distortional Energy (von Mises) Criterion The distortion energy theory is different from the other failure criteria in that instead of working directly with stresses or strains. Therefore. For a simple tension or compression test.1. which influences the adhesion between the fibers and the matrix. because of the difficulties described in the previous paragraphs the strength of the homogeneous fiber matrix system is defined only in the fiber and transverse to the fiber directions. + + Despite the fact that the fibrous layers are made of distinct phases that control their strengths. For example. Chamis (1974b) Typical failure mechanisms under compressive loads along the fiber direction are shown in Fig.4.3. the stresses that are used for comparison with the strengths are the stresses along the principal material axes. in contrast to isotropic materials in which we have generally three distinct strength quantities (tension. the stresses in the principal material axes are represented as 0'1 and 0'2 . (a) Fiber microbuckling-extensional mode.232 FAILURE CRITERIA FOR LAMINATED COMPOSITES 6. The basic premise in predicting the failure of fiber-reinforced layers using maximum stress and maximum strain criteria is the same as the one used for isotropic materials. 6. (c) kink band formation. not the principal stresses that are used for the case of isotropic materials.2. These are the stresses along the fiber and transverse to the fiber directions. {b) fiber microbuckling-shear mode. fibrous composites typically have five distinct strength properties: tension and compression. Strength along the principal material directions is influenced by micromechanical events as well. for strength determination the layer is usually assumed to be homogeneous. It is also important to note that. Recall that the same assumption is used in determination of the elastic properties of the material. Failure is predicted when the maximum stresses in the medium along the fiber or transverse to the fiber directions exceed the respective strengths of the tension or compression test specimens with fibers placed along the axis of the specimen or transverse to it. An in-depth review of micromechanical theories of compressive failure is provided by Schultheisz and Waas ( 1996). and shear). 6. the failure mechanism depends on the sense of the stresses. In the remainder of this chapter. Of course the effects on strength of micromechanical features that bring into consideration the interaction of the matrix and the fiber constituents are not only apparent for the off-axis strength.3. in contrast to the elastic properties. two of each (one in the fiber direction and another one in a direction perpendicular to the fiber) and shear strength. Compressive failure modes in unidirectional layers. is known to have a strong effect on off-axis strength. which are the principal axes of material orthotropy. The failure envelope under normal stresses is defined by .1 Maximum Stress and Maximum Strain Criteria (a) + + + + (b) (c) (d) Figure 6. as well as the degree of adhesion achieved between the fiber and the matrix. Prediction of the failure of laminates made of composite layers with different orientations is discussed in Section 6. Strength of a composite layer in any other direction is evaluated based on various failure criteria presented in the following subsections.3. Failure stresses of a composite specimen in tension and compression are quite different from each other and both are strongly influenced by the properties of the fiber and matrix materials. However. compression. Note that since the strengths are specified only along the fiber and transverse to the fiber directions. Discussion of the factors that affect the failure mode and differences between the failure characteristics is beyond the scope of this book. which can uniquely be calculated in any direction (see Section 2. Depending on a variety of factors any one of those modes might dominate the failure. (d) shear failure on a 45 degree plane. in addition to failure loads that are quite different in the fiber and transverse to the fiber directions.2 FAILURE OF FIBER·REINFORCEC ORTHOTROPIC LAYERS 233 of fibers.2). 2) a1 > -Xc. in which case 't 12 = S. Xc = 2500 MPa.05. ax= ~/sin 2 8 or ax= S/sin e cos e. by substituting m = cos e and n = sin e. fibrous composite layers are also vulnerable to failures caused by shearing stresses. O'Y. With 0'/0'x denoted as r.14). For a given stress state at a point (ax. From 3° up to approximately 42°. the strength is controlled by lillil 1111'. = E. In addition to failure initiated by normal stresses.0. cipal material axis oriented at an angle e from the x-axis (called the axial direction for the rest of this example) loaded by (i) pure axial tension.0'2 <0. (c) Example 6. (b) fort" 2 E2 > 0.2 FAILURE OF FIBEFI·FIEINFORCED ORTHOTROPIC LAYERS 235 a.2. we have 'txy = 0. (c) are plotted in Fig. and 't_x).125 x 10-3 m) with the prin- Tensile strengths obtained from Eq.23. (2. cr 2 = (n 2 + rm 2 )ax• 't 12 xt c.3) where S is the ultimate in-plane shear strength of a specimen under pure shear loading. For very small values of the fiber orientation (up to about 3°) the tensile strength along the fiber direction controls the failure. a. e : :.n 'txy 2 l (a) (6. forcr 1. using Eqs. ly12 1< Y~2 = G s 12 • ax= X/cos 2 e. It is also possible that. As noted earlier. and a 2 are the stresses along the fiber and·· transverse to the fiber directions.1 A unidirectional Boron/Epoxy layer (£1 = 204 GPa. 6. From Eq. (6. < X1.2. the layer can fail because of shearing stress.4a as a function of the fiber orientation e. the stresses in the principal material direction are expressed as 2 2 0'1} a {'t 2 12 = T {O'x} O'Y = [ m n 'txy n2 -2mn 2mn {O'x} m crY . v 12 = 0. and the failure occurs when 0' 1 or 0'2 reaches X1 or Y1. and S = 67 MPa. Typical strength properties of the Boron/Epoxy are X1 = 1260 MPa.59 GPa. 90°. the matrix equation (a) can be rewritten as cr 1 = (m 2 + rn 2 )ax.234 FAILURE CRITERIA FOFI LAMINATED COMPOSITES 6.2. Yc = 202 MPa. for some off-axis orientations. magnitude of the compressive stress allowables and are positive numbers).4.' E2 yt < E~ = E = (r-1) mnO'x.15.5 GPa. E = 18.12) and (2. 2 G12 = 5. a2 < Y1 a2 > -Yc for a 1. 2 2 -mn mn m .4. not the principal stresses that are used for the case of isotropic materials. respectively. Since the strength of fibrous layers can be quite different in tension and compression. (i) For pure axial tension we have av = 0 and r = 0. and t = 0.1) (6. (b). the subscripts t and c are used to distinguish the two strengths (Xc and Yc represent the. Determine the strength of this layer as a function of the fiber orientation angle in the range 0° :::. (ii) pure axial compression. Y1 = 61 MPa. and 't 12 is the shear stress in the principal material direction (not the maximum shear stress). respectively. a 2 > 0. IIIII I . The maximum stress criterion for shearing stresses takes the form l't 12 1 < S. where X and Y represent the ultimate strengths along and transverse to the fiber directions. E2 < 0.2. and (iii) combined biaxial tensile loading with r=O"/ax=0. The maximum strain criterion similarly states that failure when one of the following inequalities is violated: E1 < For the three cases considered in this example. threeeq uations that control the maximum applied stress ax that can be allowed are E I XC > -£1=--E' I E2 >- E~ =- Yc E 2 for E1. (deg) 75 90 (a) r = crJcrx = 0.15 (jx = s l(r. 30 = (sin2 0 + r cos 2 0)' e. (deg) (b) compressive loading 6.2 FAILURE OF FIBER·REINFOACED ORTHOTROPIC LAYERS 237 2500 ~ 2000 fiber failure ~ 2000 ~ transverse matrix failure ~ '0~ 1500 Strength values as a function of the orientation are plotted in Fig. Compared to the pure tension case. the three equations that govern the applied stress at failure are 2000 1750 1500 'fiber "'-. (deg) 45 60 75 90 15 30 45 60 9.05.5b for the two values of the stress ratio r. Tensile and compressive strengths of unidirectional Boron/Epoxy.--failure ~ '0~ 1250 1000 750 shear failure transverse matrix failure ~ 1250 (j X xt = --::---'---2 2 ( COS 0+r Sin 0)' 250 yt (jx I 15 :--:: :-:. (c) are replaced by Xc and Yc. (iii) In the biaxial tension case with a specified stress ratio cr/crx = r. The compressive strength is shown in Fig. I~ 1!:1 1111111 . the matrix strength controls the failure. Figure 6. There are various quadratic failure criteria for composite layers. adding a small transverse tension precludes fiber failure for small fiber orientations. Quadratic failure criteria similar to the von Mises criterion can account for the interaction of the stresses. in the case of r = 0. 6. the strength is controlled by transverse tensile failure. Biaxial strength of unidirectional Boron/Epoxy. For orientation angles larger than 42°. Experimental observations for both isotropic materials and orthotropic.15.4.. the strengths X1 and Y1 in Eq. 60 75 90 15 (a) tensile loading 90 15 30 45 9. Hoffman. the increased strengths in compression (both in the fiber and transverse to the fiber directions) cause the material to be more vulnerable to shear strength failure. shear failure.5. The shear strength is independent of the sign.1) sin 0 cos 01 .236 3000 fiber failure FAILURE CRITERIA FOR LAMINATED COMPOSITES 3000 6. respectively. Even for 0 = oo.2 Tsai-Hill Criterion While the application of maximum stress and maximum strain criteria is straightforward.. strength is controlled by shear failure for a fiber orientation range of approximately 5° ~ 0 ~ 40°. Compared to the tensile case. transverse matrix tensile failure is the dominant mode for the entire range of the fiber orientation angle. However._failure 2000 1750 1500 fiber . materials show that such interactions can affect the failure of material. (ii) In the case of pure compression. 6. these criteria fail to represent interactions of different stress components in failure mechanisms.. which controls the failure in the range of about 1. fiber-reinforced.5 to 71°. The most frequently used ones are the Tsai-Hill. For r = 0.-::5.5a and 6.05 (b) r = crjcrx = 0. and Tsai- Figure 6.2.4b. . cr 1 + [ Y. respectively. (6.9) for the coefficients produces F= 2x.cr3)2 + H(cr3 .7) reduces to what is generally referred to as the Tsai-Hill criterion: 1 2 1 2 I 1 2 x2 cri + y2 0'2. uniaxial tests in the 2 and 3 directions yield (F +G) Y~ =I and (G+H)Y~=l. 6. In- . at failure (cr 1=X1.2. therefore. (6. with its compressive counterpart. which is referred to as the Hoffman criterion: (JI (JI(J2 (J2 1 t 1 ) c 2 . 1968 and Wu.cr2)2 + G(cr2 .2.rY t + X t )S n Similarly. for each normal stress direction we use results from two experiments. Hill (1948) proposed an extension of the von Mises yield to anisotropic materials with equal strengths in tension and compression.Y.2 FAILURE OF FIBER·REINFORCED ORTHOTROPIC LAYERS 239 Wu criteria (see Tsai. 1974).7) is obtained from a shear test with a pure 't 12 stress causing failure at S. t + [(r 2 . Example 6.cr2)2 + G(cr 2 . respectively.rYt2 + Y t2 )SZm4 + (r 2 Y 2 t .s.6b for tensile and compressive stresses. one for compression and one for tension.2. the stress ratio r = 0. For a three-dimensional stress state. If one of the normal stress components. and Y1 are replaced by Xc and Yc.12) we evaluate the coefficients by matching the criterion to results obtained under simple load conditions as described for the Tsai-Hill criterion.· 1 1 I G= y2.13) c x. Evaluation of the coefficients (see Exercise 2 at the end of the chapter) yields the following failure criterion.r. Hill's criterion yields.11).8) and (6. cr2 = cr3 ='tu = 0).2r + l)(x.s 2 )]m2n 2 } cr. and S. where coefficients for the stresses are related to the yield strengths different directions. respectively. Starting with the following quadratic form for the failure criterion: (Jx ~ --. the safe level of stress is given by where the final coefficient N in Eq. In this case. since the failure stress in the 2 and 3 directions are both equal Yr The solution of Eqs.238 FAILURE CRITERIA FOR LAMINATED COMPOSITES 6.2.1 into the Tsai-Hill failure criterion. Hill's criterion is given F(cr 1 . Tsai (1968) extended Hill's criterion to composites by relating those coefficients to the longitudinal.2 Repeat Example 6.2. t t For the pure axial tension (i) and pure axial compression (ii).2..6a and 6.2.2.X X + YY 2 + [X .r.· I Under the plane stress assumption. For a tension test specimen with fibers running along the length of the men.1 to determine the strength of Boron/Epoxy layers using the Tsai-Hill criterion as a function of the fiber orientation angle 0° ~ e ~ 90°.cr3 ) 2 + H(cr3 . Strength as a function of orientation is plotted in Fig.y ) cr2 + Sl t (6. Eq. is compressive then we replace the corresponding strength quantity. X. A generalization of the Hill's criterion that allows for different tensile and compressive strengths is offered by Hoffman (1967). (b) of Example 6. and shear failure strengths. (6. X.2.x2 crlcr2 + s2 't12 < 1. (6. = 1. in the case of compression.2.2X 2 t t and H= 2X.2. we obtain the following single equation for the maximum stress ax that can be applied to the layer: 2 2 2 4 {(r 2X t2 . ~ x. (F+H) F(cr 1.cr 1) 2 + L'r:~ 3 + Mti3 + Nr:i2 < 1. Substituting Eq. (6. Eq.r.Yt2S 2) cos 82 sine"' where. cr 1 or cr2. X1 or Y1.X XX tc tc tc 1 1 't12 2 < 1.cr 1 ? + Pcr 1 + Qcr 2 + Ra3 + L't~ 3 + M'ti 3 + N'ti 2 < 1. verse.IX 2S2 t sin S4 + Y2 S2 t XtYrSt cos 84 + (X t2 Yt2 . (6. In this subsection introduce the Tsai-Hill and Hoffman criteria. .2.. (deg) 60 / I \. eluded in the figure as dashed lines are the stress limits from Fig.7.2.3 Tsai-Wu Criterion A more general form of the failure criterion for orthotropic materials under plane stress assumption is expressed as F 11 af + F22 a~ + F66 -cf2 + 2F 1 p1 -c 12 a 2 + 2F16a 1 + 2Fz6az'tu + F.?> ____ ..{---"' Tsai-Hill 15 (a) 30 . Failure predictions are plotted in Fig. Tsai-Hill predicts the failure to be at about ax= 893 MPa._~. 6. Axial strength of Boron/Epoxy layer based on Tsai-Hill criterion. the strengths predicted by the Tsai-Hill and maximum stress criteria are identical for the oo and 90° layers. The strength predicted by Tsai-Hill is almost identical to the ones predicted by the maximum stress criteria used in Example 6. - 200f 90 ~~\' / I /._ '~ ' -~ transverse I matrix 1 failure I t:> ~~ I ' 1 1 o..\ y \ shear failure ~ 800 1 d" 600 400 200 1 \\ '.. For a 0° layer..240 FAILURE CRITERIA FOR LAMINATED COMPOSITES \_ 3000rl I I I I I I I 6. the largest difference between the Tsai-Hill and a maximum stress criterion prediction is for the fiber-orientation angle of about 3°. respectively. F2e Fl' F2.. (6.2.. J ' transverse matrix failure I I ---=----"' 1 I 1 1 1 . where two of the maximum stress criteria intersect. Use of an interactive failure criterion. F 16 .15) .05. it was shown in Example 6.-/-. For cases of uniaxial tension or compression. heightens the sensitivity of the failure load to the existence of biaxial stress state with small transverse stress.1 except in the neighborhood of the fiber-orientation angles 0. (deg) 45 60 75 90 r = O"/O"x = 0. if the failure stresses of a specimen loaded along the fiber direction in tension and compression are X1 and Xc.05 a. F 22 . r = 0. the failure stress predicted by the Tsai-Hill criterion is substantially lower than the one predicted by the maximum stress criterion based on shear failure. Figure 6.05 and r = 0. where maximum stress criterion caused by fiber tensile failure and matrix shear failure is predicted at a tensile stress level of ax= 1264 MPa. then we get F11 X~ + F 1 X1 = 1. the failure of the layer is due to transverse matrix failure..2. jibe r 1200 1 1 failure 1 I I I 0 \ ~ 8o t:> ~ 1000 \ 1 1 shear 1 1 • /failure •.15. and F6 requires using the results of experiments on unidirectional fiber-reinforced specimens under simple load conditions..' matrix 1 failure 1 - -~~'-.6. In the case of biaxial loading.. however. "'"'-._ ""-.::::. an almost 30% lower strength than the maximum stress criterion. (deg) (b) compressive loading 30 a...---45 9. For the same fiber orientation.1 that for r =a/ax= 0. For example. see Fig. \ transverse.al + Fzaz + F6-c12 < 1.7a and 6..1500 ~ ~ ~2ooo \ \ 200 .. Biaxial tensile strength of Boron/Epoxy layer based on Tsai-Hill criterion. 241 1400r:-~ 1200~1 I I I I I I ""'fiber failure I 2500~I "' 1 'V!_ber ~-~ 1 \ I I I I failure 1 1 1 1 : I 1200t-l .fiber \ failure 1400 I / 1-11.. 6.4 for comparison. ~ • t:>~ woo~'' II shear Jailure ~ 800 I I I I 1 1 ~ 1000 I I I I I I I II \ ---. existence of transverse stresses can cause transverse matrix failure even for the 0° layer._-___. 6. 6. (6. Fw F 12 . For example.6. 45 15 30 (b) I/ failure ~-r transverse matrix I I I I 30 (a) 75 90 tensile loading 45 60 75 9. considered in the previous example... For example.15 60 75 90 Figure 6. such as the Tsai-Hill.7b for two levels of stress ratio.•~ I\ '7. (deg) r =O"/O"x =0.14) Evaluation of the strength coefficients F 11 .2 FAILURE OF FII!R·RI!INFORCED ORTHOTROPIC LAYERS 1400r 1 .2. JFzz' Fz F*--~ 12.2.27) We next consider pure positive and negative shear tests where the specimens fail at a stress level of 't 12 = ±S.17) The last two equations are almost identical except for the sign change in front of the F 16 term.+ [1 1) az<l.21) Solution of these equations leads to the determination of two more coefficients.z+2F.2F.2. which are given as F 6 =0 and 1 F66= S2' (6. with X1 = Xc and Y1 = Yc reduces to Tsai-Hill's criterion. is the one that reflects the effect of interaction of the two normal stresses on failure: 1 a..28) and normalized strength coefficients as Fli?S~omb + F66S~omb + 2F.+yy 2 1 az+s2't.2. F 12 =.15) and (6. namely F 16 and F26 .2.6rS~omb + F.112x.15) and (6.2.242 FAILURE CRITERIA FOR LAMINATED COMPOSITES 6.z.= .. since the value of scomb is independent of the shear direction Equations (6. (6.2.p. which ignores the interaction of the two normal stresses. (6..2.JFzzaz. For example.2..14) we have There are a number of different approaches one can devise to establish a failure criterion from this equation. F 66S 2 + F 6S = 1 and F 66S 2 . F.23) F~ = . (6. (6.16) is obtained by considering tension and compression tests in a direction transverse to the fiber direction. Subtracting Eq..22) In order to determine the coefficients that reflect the interaction of the normal and shearing stresses. X-X y--yc c t t (6. For example.2.JF66't. Therefore.F 2 Yc (6.rScomb = 1. These equations are F 22 Y.16) are a system of two equations in two unknowns. (6. solution of which yields F =__!_1 X x 1 1 F11r S~omb + F66S~omb. + F 2 Y1 = 1. For example.2. The criterion generally referred to as the Tsai-Wu criterion is a case where Eq.2. (6.'IF. the specimen fails at 't 12 =±Scomb' then from Eq.1.2.24) from Eq. F 12 is assumed to have a nonzero value and the value is generally selected to make the criterion agree with one of the simple criteria discussed earlier. Narayanaswami and Adelman (1977) suggested F 12 = 0.az+ 2 1 2 [1 XX t and their solution is 1 1 Fz =y--yc t and 1 Fzz = ytyc· (6.F6 S = 1. (6. F 12. F22 Y~.2.2. Usually.26) = 1. (6. 2 (6.JFII' F. (6.2.IF22 F12 (6. we need experiments that will generate combined stress states involving a 1 and -r 12 and a 2 and 't 12.112X1Xc to Hoffman's criterion.24) and c Fll = Xtxc· 1 (6. 't~z = .18) Fz6 = 0. however.16) and.23) yields F 16 =0. (6.20) c t c 1) a. and F 12 =.F 1Xc= 1.6rS~omb +FirS comb= 1.2. (6. for the experiment under consideration.2. (6.19) The only coefficient left to be determined.2 FAILURE OF FIBER·REINFORCED ORTHOTROPIC LAYERS 243 F 11 X~.2. Another set of equations similar to Eqs..2.29) . If. = .2.14) as explained in the following.27) is made to resemble the von Mises criterion of Eq. consider an experiment with l-r 121> 0 and a 1 > 0 with a 2 = 0 and a/'t 12 = ±r.. We define nondimensional stresses as a~= ~a" a.2.25) A similar reasoning for an experiment that involves a combination of a 2 and 't 12 leads to the determination of (6..2. 27) as 1400 /200 2 2 2+ 2F* cr*cr* + F*cr* + F*cr* < 1 cr* I + cr* 2 + 't* 12 12 I 2 I I 2 2 · (6.3 FAILURE OF LAMINATED COMPOSITES There are two roots of this equation. Compared to the Tsai-Hill. Laminate strength prediction is carried out by evaluating the stress state within each layer based on the classical lamination theory . Eq.ax= 1. the stress ratio.3 Repeat parts (i) and (ii) of Example 6. Eq. (a) of Example 6. see Exercise 3 at the end of the chapter. xtxc + ytyc cri cr~ + 'tiz_ 2 s crlcr2 .Vxtxcytyc +(_!_ _ _!_Jcr 1 +(_!_ _ _!_Jcr <1.Txy loads is left as an exercise. ~ 800 J Tsai-Hill J' Tsai-Wu 400 (6. Strength of Boron/Epoxy layer based on Tsai-Wu criterion. (6. we can proceed with the strength prediction of a laminate.2.31) 200 200 or F12= 1 30 45 60 75 90 /5 30 45 60 75 90 2.3 FAILURE OF LAMINATED COMPOSITES 245 1400 1200 /000 and rewrite Eq. Substituting Eq.33). It is expected.30) ~ Note that in the case of isotropic materials with X1 =Xc = Y1 = Yc. r==m=2=n=z=J cr2 . we have F~ = F.2. = 0.2. xt XC yt yc 2 (6.244 FAILURE CRITERIA FOR LAMINATED COMPOSITES 6.2. Also the principal stress state will have 't~ 2 = 't = 0.32) 9.2.2. however.2. e : :. 12 Therefore.30) reduces to the von Mises criterion if we chose 1 F~2 =-2. but the differences are relatively minor.8b for tensile and compressive uniaxial loadings along with the strengths predicted by the Tsai-Hill criterion. and the other one corresponds to the compressive stresses. that the biggest difference between the two criteria would be for biaxial loads.1 into the Tsai-Wu failure criterion.33) Example 6. (6. 90°' using Tsai-Wu criterion F~ 2 =.~For the uniaxial loading. (deg) (a) tensile loading 9. A comparison of the Tsai-Hill and Tsai-Wu criteria presented in the previous example is for uniaxial tensile and compressive loads. 0° :::. (deg) (b) compressive loading For this choice the form of the Tsai-Wu criterion is Figure 6. (6. we have the following quadratic equation to solve for the maximum allowable stress level: m4 n4 m2 n2 ( xx+yy+----:51 c t to the tensile failure of the layer.2.YxC xt YC yt X c t m2 n2 m2 n2J + (x-+y--x--y.8.YXIXCYIYC .cry and ax. (6.1 to determine the uniaxial tensile and compressive strengths of a Boron/Epoxy layer as a function of the fiber-orientation angle. prediction of the Tsai-Wu criterion is slightly more conservative for tensile failure and slightly less conservative for compressive failure. Strength values as a function of orientation are plotted in Fig.8a and 6.2. 6. The root with a plus sign in front of the terms under a radical is the stress that corresponds Once the strength of a single layer under a combined state of stress is established. r = 0'/0'x is zero.2. Comparison of the two criteria under combined ax. t t c c 6. and 90° layers).246 FAILURE CRITERIA FOR LAMINATED COMPOSITES 6. for example. raising the stresses in them. such failures are in the form of local yielding of the layer upon which the loads around the region of local failure are gradually transferred to the adjacent layers without a catastrophic failure. the increased stresses in the other layers may not cause failure and overall laminate integrity may be maintained. A highly schematic representation of the load versus deflection diagram of a laminate with such a failure is shown in Fig. The interlaminar strength or the interlaminar fracture toughness (which is a measure of the resistance of a material to crack growth) of laminates may not be sufficient to resist rapid crack growth between the layers. This is generally referred to as first-ply failure. for a Graphite/Epoxy laminate E/£2 = 15. however. the constituent materials for most composite laminates are brittle in nature and do not tolerate local failures well. since the low stiffness matrix carries only a small fraction of the applied loads. therefore. for a cross-ply laminate [0/90]. the material is usually capable of sustaining nearly constant loads in a plastic state as the loading is increased. which . depending on the testing conditions). may or may not be able to carry loads beyond failure initiation. Although the layers oriented along the load axis carry a major portion of the applied loads. However. For example. which are susceptible to experiencing damage due to external impact. almost 94% of the loads will be carried by the oo layers. The transverse tensile failure of the layer appears in the form of matrix cracks between the fibers. because of its low strength. Although the stiffness properties of the laminate as a whole may change noticeably. the slope of the initial load path changes. By making the ±45° layers the outermost layers. demonstrating the stiffness reduction. possible loss of load-carrying capability is minimized. the important issue is to establish what constitutes laminate failure. on the other hand. In a laminate. In a laminate made of a mix of various fiber orientation angles and loaded along the x-coordinate axis. As mentioned in the introduction to this chapter. Although yielding implies failure initiation. Therefore. The second factor is the large difference of stiffness and strength between the two principal material (fiber and transverse to the fiber) directions in a layer. This is especially true for laminates tested under load-controlled rather than displacement-controlled loading. 6.3 FAILURE OF LAMINATED COMPOSITES 247 described in Chapter 2 and implementing one of the criteria described in Section 6. First. as there will be a sudden jump in the stresses in the remaining layers.4 with 50% oo and 50% 90° layers. their strength properties along the fiber direction are also substantially higher than their transverse strength. Upon crushing. It is. Under compressive loads. A fiber-reinforced laminate. This is precisely the reason for placing the load-carrying 0° layers in the interior of laminates for damage tolerance considerations. failure of a loadcarrying layer may dictate the failure of the laminate. some of the layers are likely to reach their limiting stresses before the rest of the layers and fail first. common that the lightly loaded layers may experience the first failures in a laminate. which further destabilizes the laminate. depending on the nature of the first failure and to some extent on the sign of the stresses. In a quasi-isotropic laminate (with 0°. Under these circumstances. failure of a loadcarrying layer may in. owing to their large stiffness carry most of the applied loads.9 (in an actual test. However. the kinks in the plot are also associated with discrete jumps in load or deflection or both. 0° layers (if there are any). the stiffness of the layer will be locally reduced and the other layers in the laminate will carry a larger portion of the applied loads. For example. they carry 58% of the applied loads. Note that upon failure.2. There are two factors that contribute to this behavior. loaded in compression along the x axis.duce through-the-thickness imperfections and damage between the layers of a laminate that can create local threedimensional stress states. This behavior is further complicated because of the layered nature of composite laminates with weak interlaminar normal strength. transverse cracking of a layer may simply result in internal redistribution of the stresses within the laminate so that still larger loads can be sustained. This may also be true in case of transverse matrix tensile failure of a layer placed at a 90° orientation from the major load axis. helps release local stress concentrations around the failure initiation ·. ±45°. the 90° layer might carry less than 6% of the applied loads. stresses in layers with different orientations or different material properties are generally different. Lack of local yielding.• points. although the oo layers make up only 25% of the total thickness. For laminates with ductile layers (layers made of conventional ductile metals such as aluminum and steel). matrix crushing failure is the likely local failure in that layer. which are oriented along the x axis. As stated in the . can cause unstable brittle fracture leading to catastrophic failure of a laminate. Even in those situations where one may be willing to accept local failures in the form of transverse matrix cracking. and stresses in layers with the same orientations are identical. Given the strains in the principal material directions for each layer. depending upon the direction of the applied loading. Maximum Stress Failure Criterion. (k) (6. For specified values of laminate stress resultants (Nx. the implementation of this failure criterion calls for the evaluation of strains along the principal material axes for each fiber orientation.4)-(6. stresses in the kth layer are calculated from {'t" crl} cr2 12 (k) -lQu . An example of this kind of a failure will be discussed in Section 6. introduction. and Nxy) or average laminate stresses (<h. but also on the stacking sequence of the laminate.1 Failure under In-Plane Loads For plane stress. These residual strains may either help or worsen the load-carrying capacity of a ply. We then use Eqs.Ql2 0 Ql2 Q22 0 0 0 Q66 j{£1} £2 . implementation of failure criteria is straightforward.lNxl :Y = hA. (2. it is often assumed that initial failure of a layer implies laminate failure. Layer-by-layer application of either the maximum stress or maximum strain failure criteria with only X1 and Xc (ignoring the ~and Yc strengths) would serve for the purpose. Also such local failures are likely to create weak links in the laminate that may initiate future failures.3 FAILUAE OF LAMINATED COMPOSITES 249 6.4.lcrxl ~Y = A*.9. with h being the laminate thickness (see Chapter 2). This is referred to as first-ply failure criterion. layer-by-layer application of either a strain-based or a stress-based criterion requires calculations to be performed only for layers with different properties or fiber orientations.3.lcrxl ~Y { £~} 1 1 1 . The laminate strains are calculated from £~ = A. Therefore. Deflection Figure 6. and 1:xy). In some cases it is not desirable to have local damage.3. With the laminate strains known. it is important that any residual stresses or strains that may exist in a composite laminate are accounted for during the strength analysis. (6.2.1) 1. 0 £ {'Y12 1 f2 } (k) = T~k) {£ £~ } = YxY l n2 m2 m2 n2 mn -mn 1 {£~} £ 0 (6. As will be discussed briefly in Section 6.2) -2mn 2mn m2 . the number of failures will keep increasing progressively (therefore referred to as progressive failure) until the failure of an additional layer will cause failure of all the remaining layers. As explained in Chapter 3. Finally.16). Ny. cry. (6.y xy xy xy where A* = A/h.1. even in the form of transverse matrix cracks.3. The first-ply failure criterion provides a conservative estimate of the laminate failure load since first failure in the form of matrix cracks may not lead to laminate failure.3) "{12 . the residual stress levels introduced during manufacturing depend not only on the operating temperature and humidity. prediction of failure under bending loads and deformations is more involved.6) for each layer to check for failure. Progressive failure of a composite laminate.248 Load FAILURE CRITERIA FOR LAMINATED COMPOSITES 6.n=sin9k. Strains are constant through the thickness.3. Therefore. strains for each layer in the principal material system can be calculated using the engineering strain transformation matrix of Eq.3. that changes laminate elastic properties. a first-ply failure criterion might still be useful. Maximum Strain Failure Criterion.2.2.n2 (k) YxY ' where m=cos9k.3. 34. For the (Ozl±45)s Kevlar/Epoxy laminate. r E~45 ol = 3. strains in the ±45° layer in the principal material directions are calculated from Eq. crx + 80. f~ = 18. f~= 3.42. However.n2)Q66 where all average stresses are in GPa and the strains are in millistrains (10.08028 0.30 Example 6. In this case.3 FAILUFI! OF LAMINATED COMPOSITES 251 Substituting Eqs.216 10. Draw the strength envelope for the laminate in the crx-crY plane based on maximum strain criterion and determine the maximum value of Nx that can be applied.28 cry.302 45 = crx + 31. we have £ 1 = £ 2 and.250 FAILURE CRITERIA FOR LAMINATED COMPOSITES 6.1) we have f (Oo) 1 -Ex- {~:} K(k) = K(k) { 12 (k) Yxy :J} =K<kl ~~:]· A •-t 23 .Q 12)] m2QI2 + n2Q22 n2QI2 + m2Q22 mn(Q 12 .0 0.12 cry.85 cry= 18.538 0. [ 0.0 0.09 ax+ 80.1 A (Ozl±45)s Kevlar/Epoxy laminate is loaded by biaxial loads with N/Nx = 0. E 2 = 5.3.3. Following is a set of typical elastic and strength properties for Kevlar/Epoxy: E 1 = 76.85 [ 50.45 ax . -16. '¥12 . and y·.02345 -{).09.125 X 10.30 GPa. Similarly..78 X 10-3 . 2 = 14. due to the lack of applied shear. (6.4) 91 · 31 'txy' - y where the matrix K is given by m2QII + n2QI2 n2Q.72 cry+ 45. 85 cry. (6.09 X 3 3 10. f~ = 2.5.72 cry= -3.18. For the 45° layers. and -16.42 X 10-3. The failure envelope for the 0° layer is given by the four lines (no shear failure since iJ~ol = 0) 23.0 10. the failure envelope would also normally be defined by six equations.0 A= 10.952 and l These four lines constituting the failure envelope for the 0° layer are shown in Fig. 3 J. therefore.1) and (6.64.112 = 0. the normal strain component with the smaller limiting value controls the failure. the in-plane stiffness matrix is Equating the strains to their appropriate limiting values specified in the problem statement produces the equations for the failure envelopes. G 12 = 2.2) as (6.3 m/m).85 23.64 X 10-3.538 14. 6.00 GPa. .85 cry= -3.302 crx + 31.45.I + m2QI2 mn(Q 11 .16.85 ax+ 80.651xy' crx + 97.1 = -0.3.72 cry= 2.50 GPa.65 ::rxy' crx + 31.0 GPa 0.45 crx.16 . failure is controlled by transverse tension failure 3. and t = 0.45 crx-16.2) into the above equation and rearranging we have so that from Eq.3. (6.5) E) ol = 3. f~ = 9.28 cry= -9 .3. y)~o) = -40.3.0 ] 1 A*. (6.72 cry.668 0.3.'Yxy(Oo) - Eioo) = Ey = -16.Q22 ) • -2mnQ66 2mnQ66 (m2.28 cry= 2.m.0 MN/m 0.18 X 10.01685 0.18.302 crx + 31.01685 0..1 Oa.09131 longitudinal compressive failure 3.302 The following example demonstrates the use of the maximum strain criterion for a laminate under combined loads.0 0. 5 (N/Nx = 0. which mark the load limits.10.85 ax+ 80. At point A.28 cry= 2.78.09 which yields. Altogether -161 X 103 N/m ~ Nx ~ 93. .6 x 103 N/m. Transverse tensile failure of the 0° fibers governs the failure at point B: -16. ~ . Failure envelopes for the oo and 45o layers. ~-c::_""G'"' Figure 6.6 X 103 N/m.252 (a) 0° layers FAILURE CRITERIA FOR LAMINATED COMPOSITES 6. Ox(GPa) 1 = .16=-0.5) shows the specified combination of the biaxial loads. -0.4 0.85 + 80.28 X 0. longitudinal compression 3. Keeping the same 0'/0'x ratio.4 t compression (b) +45° or -45° layer Oy (GPa) 0. failure is due to longitudinal compressive strain in the 45° layer. Failure envelope for the laminate. Point A corresponds to compressive failure of the 45° layer in the fiber direction.. The failure envelope is drawn in Fig.3 to the failure of the 0° layer due to transverse tension.~/ . or (3.3 = 0.18/23.161GPa or Nx=hax=-161xl0 3 N/m.09. This line intersects the failure envelope at two points. Superposition of the failure envelopes for each layer gives the failure envelope for the laminate shown in Fig.78.12 cry= 14.18._.e~ Oy (GPa) 0.10b. -40. .302 ax+ 31.30 ax+ 97. Point B corresponds ax= 2.0936 GPa or Nx = h ax= 93. crx=-3.:S(]y.09/19.30 ax+ 97.12 cry= -14.302 + 31. 6.18 and (-16.5 ax. 6.11.72 X 0.3 FAILURE 0111' LAMINATI!D COMPOSITES 253 0".11.5) ax= -3. The thin line marked 0'/0'x = 0.5) ax= 2. Values of the maximum average axial stress ax that can be applied are determined by picking the appropriate equation corresponding to either point A or point B and substituting cry= 0.72 cry= -3. Figure 6. the safe region of the stress plane will be any point located between A and B. A and B.4 . and shear failure -40. s = 68 MPa. both the fiber and matrix properties are degraded. transverse. the longitudinal. who proposed two types of material property degradation. In such instances. where the approximations described in the preceding discussion are used.691 0 0. The laminate is subjected to a tensile uniaxial stress resultant Nx in the direction of the fibers of the oo layer. v12 =0.00828. E 2 =10.17GPa. the fiber direction will continue to carry loads as before. A more refined version of a progressive failure model has recently been developed by Tsai (1992). The matrix mode of degradation assumes formation of microcracks in the matrix material under transverse tension and involves a set of empirical constants. if the applied load produces a tensile load that further opens the cracks in the matrix material. As an example. and continues to be. These are the fiber and matrix modes of degradations. if the ply failure is due to matrix cracking.0 8. and G12 for that particular ply be set to zero. For the T300/5208 Graphite/Epoxy system. In the context of classical lamination theory. In this case.0 ] 0. It is important to emphasize that these strains are somewhat meaningless and should not be the basis of any post-failure analysis.3GPa. G 12 =7.691 288. the laminate strains can be obtained in terms of Nx from the following equations: \ Nx~ r288. E 1 =181GPa.2 0. These failed plies are assumed to occupy the same geometric location in the laminate as prior to failure. at which the first ply fails. a simplistic and conservative approach can be adopted. a prediction of the additional load capacity of the laminate is based on modeling of the laminate's stiffnesses subsequent to failure of a single ply or a group of plies. it is assumed to have lost the ability to carry any shear loads.2 Consider a symmetric cross-ply laminate with a layup sequence of [90/0i90 3Jr where each ply is 0. The fiber degradation model also uses an empirical fiber degradation factor. Determine the value of Nx. Laminate strength analysis is illustrated with the help of the following example problem. leading to changes in every stiffness property of the layer (£" E 2 .2 0 = 8. A more detailed description of the model is provided in Section 9 of Tsai (1992). and the laminate analysis is used to predict finite strains and curvatures in these plies. Solution The maximum allowable strains for fracture under tension loading are first computed as follows: El r xt = E = 0.51 {Ex1 MN Ey ---. failure of the fibers in such an approach would suggest that each of the constants El' E2. and a straightforward approximation for this case is more difficult to obtain. The reader should note. and the ultimate loadcarrying capacity of the laminate N:rlt· Use the maximum strain criterion for the calculations.254 FAILURE CRITERIA FOR LAMINATED COMPOSITES 6. However. G 12. Example 6. which are based on micromechanics equations. an active area of research.00388. yt = 40 MPa.0 21. and G12 • Such approximations are derived from a more detailed understanding of the failure mechanism of a particular ply. In order to define the constraints for preliminary design. E" of the failed ply. and v12 ) through mi- cromechanics relations.28. Such modeling was. A less conservative approximation used in the modeling of reduced effective stiffness of laminates is based on a selective reduction of the elastic constants E" E2. 1 t E2 yt = E = 0. Also. yc = 246 MPa. and shear stress to failure and the stiffness components are given as follows: xt =XC= 1500 MPa. the stress levels in these plies will be vanishingly small quantities because of the choice of elastic constants for the failed plies. there would be a need to modify the elastic constant. that the failed matrix material can be responsible for introducing local fiber instabilities.0 0.-· Yxy (b) .25 x 10-3 m thick. however.3 FAILUFIE OF LAMINATED COMPOSITES 255 Progressive Failure of Laminates. the elastic constant E 2 can also be assumed to be zero. where the elastic constants for failed plies are set to very low numerical values. particularly if a compressive load is applied in the fiber direction.3. 2 (a) For the in-plane loading as defined and the laminate in-plane stiffness computed from the given ply characteristics. Finally. as they do in the case of plane stress.345 4.0 ] {0.0 in this case.3KN/m. matrix failure in the 90° layers is most likely to be a series of cracks in the matrix material that run parallel to the fiber direction. Therefore. this layer would fail when Ex= E~ = 0. which implies an incremental strain ~Ex= 0. the incremental load necessary to reach these strain levels is computed from the following equations: 6.00441 0. To determine if the laminate has any residual load-carrying capacity. a new in-plane stiffness matrix must be evaluated. and since Ex corresponds to the transverse direction strain for the 90° layer.3 =o. and is somewhat higher than that obtained with the more conservative approximation.2 8.76 (c) (6.256 FAILURE CRITERIA FOR LAMINATED COMPOSITES 6. For isotropic monolithic plates in bending. A conservative approximation is to assume that all stiffness contributions from the 90° layers can be neglected.0 ] 0. for a cross-ply laminate.691 288.691 8. The implication of the linear variation of the strains is that layers with the same properties and orientations will not reach their limiting stresses or strains simultaneously.3.8) For this assumption. the strains vary linearly through the thickness of the laminate.345 287.7 KN/m.7) N~it=Nr+mx= 1306.0 0. it is possible to determine the order of ply failure by inspection.6 KN/m.117 x 10-4 from the second row. Hence. and hence one chooses E 1 = 181 GPa and E 2 = G 12 = 0.3.0 [ 0.0 0. The ultimate load is obtained as N:." The shear strain component 'Yxy = 0.3.00388.0 for this layer to obtain a modified laminate in-plane stiffness matrix as follows: 272.2 Failure under Bending Loads /1NXJ j 0 0 = [288. Once again.76 ~N · (6. The layer stiffness in the fiber directions still contributes to the laminate stiffness. Substituting the value of Ex to the right-hand side of Eq.345 4. If a less conservative approximation for the failure load is desired.7KN/m where the superscript "fpf' denotes "first-ply failure. out of two layers with the same orientations located on the same side of .0 10.51 'Yxy (d) (6. Under combined loads it is possible that.0 &Y MN_ L\ m 21. through-the-thickness location of failures will be more localized in the case of bending-related failures.0044.0 0. (b). and hence.02388 and this confirms that fiber failure in tension will precede matrix failure for the 0° layers.3 FAILUAI! OF LAMINATED COMPOSITES 257 23 In the cross-ply laminate considered in this example.0 m · 10. The remaining oo plies now carry all of the applied load and would fail when the strain Ex in the direction of the applied loading reaches a value E1 1 = 0.7 4. failure occurs near the surface where the strains are highest. one can selectively reduce the stiffness coefficients of the failed 90° layers on the basis of the mode of failure.7 4. The ultimate load is thus obtained as Note that the compressive strain in the transverse direction for the oo layers at this load is EY = -0. then using the first row of Eq. we first solve for the transverse strain EY = 0. which yields the following modified in-plane stiffness matrix: 272. the allowable strains in compression in the transverse direction are obtained as Under bending loads.00828.0 0.4 KN/m. the degree of conservatism in first-ply failure criterion will be increased.3. the incremental load is computed as Wx = 1199. as 11Nx = 1194.2 0.345 15.00134. (b) we can determine the load corresponding to failure as ~pf = 111. For composite laminates this is not necessarily the case. Note that Ex is the larger of the two extensional strains.6) 246 x wE~ = 10.52 [ 0.0] MN 0.0 0.0 0.Jt = ~pf + 11Nx = 1311. 1 for Glass/Epoxy with the following properties. (b) Determine the maximum axial average tensile stress that the laminate can carry if it is loaded such that ajax= 0.2. Yc = 17. For the unidirectional Boron/Epoxy layer of Example 6. Xc = 88. 93-151.2. 4. if the laminate is under loads that generate constant shear stresses through the thickness but linearly varying normal stresses in each layer. H. Strength properties: X 1 = 154 x 103 psi. the one closer to the mid-plane may fail earlier than the one closer to the surface. L. using the TsaiHill and Tsai-Wu failure criteria. Mechanics of Materials. Broutman. Determine the strength of this layer based on the Tsai-Hill failure criterion as a function of the fiber orientation in the range oo : : . strength criterion needs to be checked on both sides of the laminate because of the different allowables in tension and compression. Yc= 118 MPa. 0.2. New York. J. An off-axis unidirectional Glass/Epoxy layer (£ 1 = 18.59GPa. Jr.3. C. using (a) Maximum stress criterion (b) Tsai-Hill criterion and plot and compare the results. Vol. Draw the strength envelope for the laminate in the ax-ay plane based on Tsai-Hill criterion. C. and N for Hoffman's strength criterion of Eq. P. Furthermore. and (b) s = :rx/ax= 0. 6. New York.1 and 6. to be on the safe side.23. R. pp. .60 x 106 psi. For the unidirectional Boron/Epoxy layer of Examples 6. The unidirectional Boron/Epoxy layer of Example 6. Chamis. Q.5GPa. Y1 = 4. and S = 10. Xc = 610 MPa.50 X 103 psi. 6. G12 = 4. Yet the same shear stress may be high enough in the layer closer to the laminate mid-plane to cause failure.and t = 0. and plot the strength for r = 0. H. v 12 = 0.4 x 103 psi. Typical strength properties of Boron/Epoxy layers are X 1 = 1260 MPa. Broutman. Elastic properties: £ 1 = 5. and Krock. J. Y1 =31 MPa. A Boron/Epoxy laminate (see properties in Exercise 1) with (0/±45/902)zs stacking sequence is loaded by biaxial loads. Xc = 2500 MPa. "Micromechanics Strength Theories. oo < e < 90°. G 12 =5. (eds. G.1 and Example 6.2. Therefore. L. even if the laminate is symmetric. Y1 = 61 MPa. Yc = 202 MPa. Plot the failure envelope and show the load limits under biaxial load a/ax= 0. M. 2.45) with its principal ma2 terial axis rotated from the laminate X axis by 8 degrees is loaded in shear resulting in a pure shear stress state 'txy· Determine the maximum value of the shear stress 'txy that can be applied to this layer as a function of the fiber-orientation angle. EXERCISES 1. r::::. Chamis.E2 =18. e : : . H. McGraw-Hill. The strength properties of the unidirectional Glass/Epoxy layer are X 1 = 1062 M P a. R. F. and t = 0.3.13). Derive the strength coefficients F. L.2.2. (6.7 GPa.258 FAILURE CRITERIA FOR LAMINATED COMPOSITES REFERENCES 259 the mid-plane of the laminate. 5. evaluation of the strength criteria in every layer is advisable.2.) Academic Press.05 and r = 0. E. R.ax under combined ax-1:xy stresses in the range 0 ::::.5 X 103 psi. (eds.125 x 1o-3 m) is to be loaded by combined axial and shear stresses with :rx/ax = r. and Johnston. P. 0." In Composite Materials. plot the failure envelopes for e = 0°' 45°' and 90° fiber orientations as a function of (a) r= ajax under combined ax-ay stresses and (b) s = :rx.20 x 106 psi.14 GPa.60 x 106 psi. (1992).15." In Composite Materials. New York. 31-77. 90°. C. (a) Determine the maximum value of pure axial tensile and compressive average stresses that the laminate can carry without failure.15 using the Tsai-Hill and Tsai-Wu failure criteria.26.6 GPa. and S = 72 MPa. and S = 67 MPa.5. and Krock. plot the failure envelopes as a function of the fiber orientation angle e under (a) combined stresses of r=a/ax= 0.1 X 103 psi.25.1 in. R. v 12 =0. G 12 =0. Second Edition. "Mechanics of Load Transfer at the Interface. pp. £ 2 = 1. 3. C. For example. (1974a).25 and 0 ::::. the existence of the larger normal stresses close to the laminate surface may dissipate the effect of shear stress in those layers preventing shear failure. Vol. (1974b). and v 12 = 0.1 (E 1 = 204GPa. REFERENCES Beer.) Academic Press. s ::::. £ = 8. 5. respectively. M. S. S. (1983). 1-42. which treat the design variables as continuously varying quantities.2 discusses the general aspects of the use of traditional mathematical optimization techniques. 0. W." Journal of Applied Mechanics 47. L. (1996). Narayanaswami. 329-334. 5.." Journal of Composite Materials 11(4). R. pp. (1967). Broutman. Soni." In Composite Materials Vol. R. for problems with strength constraints. (1977). 191-247. C. R. Z.) Academic Press. M. J. 366-377. "The Brittle Strength of Orthotropic Materials. T. "Compressive Failure of Composites. 261 Jl I. Tsai. Tsai. R." Journal of Reinforced Composites and Plastics 2(1). "A Comparative Study of Failure Envelopes in Composite Laminates." Prog. and Waas. "A Theory of the Yielding and Plastic Flow of Anisotropic Metals. 3-11. liill IIIII 1 1 111 1 . Schwartz. (1980). Hoffman. Wu. introduced earlier for in-plane stiffness design. and Adelman. The first section of this chapter extends the use of Miki's graphical technique. R. in order to provide an insight into the design of laminates with strength constraints. "Evaluation of the Tensor Polynomial and Hoffman Strength Theories for Composite Materials. S.7-9-7-11. However. 200-297." Proceedings of the Royal Society A(l93) 281-297. H. we limit the discussion of this chapter to the in-plane response of laminates. E. However. New York. Think Composites. (1992). Laminates with fiber-orientation variables that are either continuous or discrete valued are considered. 34-42. "Strength Theories of Filamentary Structures.M. ( 1968). 7 STRENGTH DESIGN OF LAMINATES The simple nature of the in-plane stiffness design problems considered in Chapter 4 lends itself to graphical approaches or other approaches that do not require computerized solution techniques. A. W. "Failure Criteria for Unidirectional Composites. Schultheisz. (eds." In Fundamental Aspects of Fiber Reinforced Plastic Composites. and Krock. Aerospace Sci. Hill. 32. to designing laminates with strength constraints (Miki 1986. pp. (1974). Theory of Composites Design. Such traditional optimization techniques commonly require derivatives of the problem constraints as well as objective function. "Strength and Fracture of Composites. Dayton Ohio. R. Part I: Testing and Micromechanical Theories. This is particularly true for problems involving combined bending and in-plane response of laminates. Section 7. the complex nature of the strength criteria discussed in the previous chapter rarely permits us to incorporate strength constraints into design problems without the aid of computer-based techniques. ( 1948).) Wiley Interscience. Miki and Sugiyama." Journal of Composite Materials 1(2). S.280 FAILURE CRITERIA FOR LAMINATED COMPOSITES Hashin. 1991). H. 9-8-9-14. pp. and Schwartz H. New York.7-137-15. (eds. examples of a more versatile approach for nonlinear discrete optimization problems. k=l (7 .1 GRAPHICAL STRENGTH DESIGN A* A* 4 A In this section we use Miki's graphical procedure to design laminates with stress or strain constraints.U~V? + 2U/U 1 + U4 ) v. justify retaining this section. for the purpose of laminate strain calculations. From the definition of the matrix in Table 2. we make use of in-plane lamination parameters to express laminate strains in the general laminate coordinate system and ply level strains in the principal material directions for graphical representation on Miki's in-plane lamination diagram.1. V~ and v..1.1.1 GRAPHICAL STRENGTH DESIGN 263 Therefore.) ax.3) Although it is not common practice. Inverting the from Eq. equations developed in this section form the basis for the discussions in Sections 7 . starting with the stress-strain relation for a laminate defined in Chapter 6.1 Design for Specified Laminate Strain Limits v~ =I vk cos 2ek k=l and v.3. A* = U U 1 U U1 4 [ 0 o lu2 o ol l u3 o1 + o -U o ~ + -U 2 3 -u3 U3 0 0 U5 0 0 0 0 ~ v. defined in Section 4. Finally. the matrix for a balanced laminate is given as 7. derivation of the derivatives of the strength constraints with respect to the thickness variables is demonstrated.1) . A linearization scheme for forming integer linear programming is introduced.2. However.. namely genetic algorithms introduced in Chapter 5. which requires use of discrete variables within the context of strength design.1. The convenience of this approach is because calculation of the laminate strains does not require explicit knowledge about the orientation of the individual plies in the laminate. the easiest application of strength constraints is at the laminate level where we may limit the magnitude of strains along the laminate x-y axes. For a balanced symmetric laminate made up of I ply groups with orientations ±ei. strains are calculated from Eq. (2. xy (7 .1. we revisit the basic relations. In order to demonstrate this aspect. ' .1): hxy j:j} = A*-~~~:]. Clearly.5) U~)..262 STRENGTH DESIGN OF LAMINATES 7. (7. =I vk cos 4ek. hence. .1.2 and 7 ..1.1) we obtain A* E~ = (U -U )-U V 1 +2U3(U 1 +U4 )V 3 * 2 4 1 (U 1 .1. all the information about the stacking sequence of the laminate is contained in the two in-plane lamination parameters. and V~ and v. (7.2) and substituting into Eq. (6.) ax+ (U 1 + U2 V~ + U3 v.20). ax.3 and. and txy at a point in the laminate. Given a set of specified stress resultants or average stress loading cry.4. Equations needed for calculation of stresses and strains in the layers of a laminate were derived in Chapter 2. In Section 7. are the in-plane lamination parameters.1. All the information about the laminate is contained in the in-plane lamination parameters. where vi is the fraction of the total thickness occupied by the ply group ±ei.) (Ui- cry (7 . where i = 1.(U4 .2) where the U/s are the material invariants defined by Eq. where = A/h.U2 V~ + U3 v. is discussed.4) o _ EY- -(U4 .3 treatment of stacking sequence design.) 2 2 2 •2 cry ' (7.U3 v. . the in-plane lamination parameters are given as 7. is demonstrated.1.U3 v. I.1. In the following. The reader should be warned that this approach is more of a habit inherited from designing conventional monolithic materials and should be avoided for laminated composites.1. -u3 l (7. U 4) Ex+ 1 .U 2Ey I Uicrx +cry).6) can be used to find the regions of the in-plane lamination diagram that correspond to laminates satisfying the strain limits for a specified load combination.4. = 0 establishes this point as feasible and helps us decide the feasible side of each constraint. and there is no guarantee that a selected laminate will not fail because of excessive strains in the principal material direction of the individual layers. for theE~ constraint.4 )-(7 .V* 1 . U3 = 19. The in-plane lamination parameters of a laminate that has the largest axial stiffness correspond to point A on the diagram.1. ' (7. Contours of the boundaries of these regions are obtained by solving Eqs.1.3 GPa.-txy + U5ya 3xy These are plotted in Fig.6108.004.9). (2. U4 = 22.1. = -0. That is. U 2 = 85. Design a laminate that will withstand an average combined stress of ax= 500 MPa.17 GPa. =0. for the Y'zy constraint. then Eqs.1. E~ = 0.3785 + 26.U 4) Ey2cry I .3682GPa.0115 V~ + 35. and txy = 100 MPa. E.2.008.2UiU1 + U4) E. without exceeding laminate strains of E~ = 0.U 2Ex 1 Ulax +a). V3=--~--~~--------~--~~~--~~- * 2 a U .1. Example 7. E~. From Eq.20) we calculate U1 = 76.1.6089 v~ + 3..264 (} -----. are prescribed.008. A check of the strains at v~ = v.4 )-(7 .7296 for the E~ constraint. the laminate level strain limits must be used with extreme caution (if not completely avoided) since they do not represent a type of formal failure criterion we presented in the preceding chapter. (7 . andy~.7) v.6) for v.6805 V? v.1.7325 GPa.9) Vj However. G 12 =7. there are infinitely many laminates with various ply orientation combinations that correspond to the values of the in-plane lamination parameters inside the feasible region.1.28.+ (U 2 a U2crx . and U5 = 26.1 GRAPHICAL STRENGTH DESIGN 265 t Yxy. cry= 250 MPa. Graphical representation of laminate strain limits. in terms of the v. 7 . (7. E2 =10.1 The material properties of a Graphite/Epoxy material are given as E 1 = 181 GPa.7)-(7.1. Substituting the U/s into Eqs.8804 GPa. = -16. and 1/12 =0.2U/U1 + U4 ) E~ (7. with the strains set to their limiting values The laminate should also have the largest possible stiffness in the x direction. The following example demonstrates the use of the in-plane lamination diagram to design laminates with constraints on laminate strains.6) If limits on the laminate strains. uya 3 xy (7.6074 GPa. (7 .1.U v·· 3 3 (7.:2: 5 STRENGTH DESIGN OF LAMINATES 7.U .1. Figure 7.8) V: .1.5787 v? v. and y~ 2 = 0. bounds for the regions of the lamination diagram that satisfy the strain limits are obtained: v· 3 2 2 aV* 2 U4cryUlcrx .1. .7104 GPa.V* 2 aV* 2 U4crxUlay+ (U 2 I . For the ±70.4. Eq. the ply strains are same as the laminate strains. (7 . Eq.v" we can solve for the cosines of the ply orientation angles in terms of the in-plane lamination parameters. yielding v2 = 0.1 GRAPHICAL STRENGTH DESIGN 267 mny" xy' 0 0 n 2)yxy' which is the intersection of the shear strain and the transverse strain limits. and y12 = -5. E2 = 1. the very same quantity that we are trying to design.07° layer.2 Design of Laminates with Two-Fiber Orientations With the laminate strains determined. (6.) 2v2 Yxy' o 7.2). E~. There- . Ek 2 2 = n2E" x + m E" y 0 X = -2mnE + 2mnEy + (m 2 Yk 12 (7 .3. (6.1 T1) El 2 Ex+ 2 Ey + 2 o 2v 1 and T2 =~+ + v. For ply orientations other than 0°. cos 29 2 = T2.1.3.6). strains in the principal material directions are same as the laminate strains.10) can be solved for the fraction of the laminate with oo layers and the second fiber orientation.4.4)-(7.1 0) can be written as e.T) I E + sin(cos. n = sin 9k.1623 X IQ. In the next section we discuss the implementation of a graphical design procedure for a special class of laminates with ply-level strain constraints. These equations are given in Eq. The simplest general case is a laminate with only two different stacks of fiber orientations 9 1 and 92 • With these two-fiber orientation angles.4) [see Section 2.3292 and 92 = ±70.3). we need to express m and n in terms of the lamination parameters v~ and v.10) where m = cos 9k..4. we need to know the orientations of the individual layers. and y~ 2 ) to be same as the specified laminate strain limits. From Eq._l+T1 0 l-T1 o sin(cos.2. using basic trigonometric identities. (7.3. Therefore the laminate is safe against individual ply failures.16) for more detail]. stresses can be calculated by multiplying the strains with the Q matrix.2. assuming the ply strain limits (E~. (6. In order to show graphically the plylevel strains on the lamination diagram. in order to design laminates subject to stress or strain constraints in the principal material directions of individual layers. strains in the ±70. (7.5276 X IQ.1. (6.3.1. the oo layer of the laminate passes the test automatically.12)(2.1. it is not guaranteed that the individual layers will satisfy strain limits in the principal material directions. We start with the definitions of the lamination parameters of Eq.07 layer are E 1 = 6.1 T1) E2 2 Ex+ 2 EY2 .1.T) 1 Ey + T l yo xy' (7.1.11) For given volume fractions v 1 and v2 = 1 . As stated earlier.9424 x I0-3 . For oo layers.2) the strains in the principal material directions of the kth layer are Ek I e. it is possible to generate values of the in-plane lamination parameters that cover the entire design space of Miki's diagram and are given by ~ (7 . = v 1 cos 49 1 + v2 cos 492 .2 and equations (2. calculation of the ply level strains requires the use of transformation for layers with fiber orientations other than 0° plies.1. with coordinates of~= 0.J. (4.3.11): cos 29 1 = T 1 and where T~=~± v2(1- = v 1 cos 29 1 + v 2 cos 292 and v. and the laminate strains are obtained in terms of the in-plane stiffness parameters and the applied average stresses in Eqs. X Yxy' 1 1 0 0 r12 = -sin(cos.14) = m2Eo + n2Eo + mnyo x y xy' Similar expressions can be written for the layers of the second orientation. we can calculate the strain in the principal material directions from Eq.1.7296.266 STRENGTH DESIGN OF LAMINATES 7. = 0.1.1.13) Then. the ply strains in the principal material directions need to be calculated from Eq. If the ply orientation is oo degrees._l-T1 0 l+T1 o sin(cos. As demonstrated in Example 7 .07°.4180 and v. (4. regardless of the orientation of the ply. In this case.12) 2~2 + v. For this problem. If we assume that the laminate has two different fiber orientations with one of them being 0° layers. v 1(1.2~ 2 (7.f.3.1. Having the strains in the principal material direction.2) and the stresses from Eq.) (7 .1. 1. and v.. E1 = 0.5.3210 x 10 21 -7.6513) v.700 v? -7.6513) (14.sign for T2) are used to generate the in-plane lamination diagram for a laminate with v1 = 0.13).0. The laminate should also have the largest possible stiffness in the x direction.2246 v. without exceeding ply strains of E~ = 8.2866 X 10 19 V.1. + 1.1433 X 10 19 V~ + 1.sign for T2 ) or the second set (with .1.1. and the shear stiffness should be larger than 20 GPa.284). (7 .3. Example 7. plane. the largest stiffness in the x direction is obtained at the upper right corner (point A) of the feasible design space where the transverse strain in the plies with the first orientation (second of the strian equations shown above) and the shear stiffness constraints are active. the laminate strains are calculated as £0=--------~--------~~--------~~ 3.28.3638 2 ) +vi (4.3501 x 1021 V? + 3.4 (v 2 = 0. For example. A plot of the feasible design space for the second sign of T and T2 is shown in Fig.17. Using the given material properties and Eqs.4. Contours of laminate stiffness Ex are also shown on Fig. E~ =3. -5.7613) + T1 (Vi (12. either the first set of six equations (say. E 1 = 181 GPa.674 (14. In this case the in-plane stiffness parameters are . ~z- at v. + 6.10 V)-268.2. replaced by T2. 7 . (7. For a specified value of the volume fraction v1 (v2 = 1.4299 Vi)~ (7. The design space generated for this case is shown in Fig.48 X 10-3. The in-plane stiffness parameters that correspond to point A are v~ Substituting these strains into Eq.3501 vi'.7884 X 10 18 + 2..2532 X 10 19 -4.1. x 5.3501 X 1021 V? + 3. which corresponds to an all 0° laminate.700 Vi 2 -7.3. E2 +vi (4. cry= 250 MPa.6.2.642)(197. These lines divide the in-plane stiffness diagram into feasible regions where the strains are below the limiting ply-level strains and into infeasible regions where the strain limits are exceeded.4783 X 10 19 V.6).2 Using the graphical procedure and plies with the following material properties.. 10 v. (7 . 7.8771 v..17. It was shown in Section 4. which is also defined in Eq. and the optimal design is marked 1 as point B. Equations for the strains in the principal material direction of the layers with the second fiber orientation 82 are same as the ones above except that the T 1 is Note. Because of the constraints. we can now show the limits on stresses and strains in the individual layers on the in-plane stiffness diagram. however.3210 X 1021 -7.1 GRAPHICAL.4783 X 10 19 V.8034 V)-10.3210) 10 2 ' where T 1 is defined in Eq.10.· 7..80) (2.8771 v.80) = 2 -5.1 that the laminate with the largest stiffness is on the upper right corner of the in-plane lamination diagram.4744.5. The objective is to find the laminate with the largest stiffness in the x direction.sign for T1 and + sign for T2 ) can be used to plot the lines corresponding to the limiting values of the strains in the V~-v. E 2 = 10.6).674 v 3.268 STRENGTH DESIGN OF LAMINATES 7.8034 v.14 )..642) (197.3638 2 ) and the values of the T 1 and T2 corresponding to these in-plane stiffness parameters are T 1 = 1.0 indicating that no real value of the orientation angle 8 1 exists for this case.4 )-(7 .2246 v.1. and ~xy = 0. = 0. limiting strains for both ply orientations (+ sign for T 1 and .v1). design a laminate that will stand an average combined stress of ax= 500 MPa. ply level strains in the layers with el orientations are a. The next example demonstrates the use of the lamination diagram with a strain-based failure criterion implemented for each layer of a two-angle laminate.395. that the T1 value is larger than 1. (7 .284). + sign for T 1 and . G12 = 7.' Eo= y l'~y = 0.277 a.8275 (V 3'.9017 x 1021 v.2 along with the shear stiffness constraint. STRENGTH DESIGN 288 fore.7613) + T1 (Vi (12.9017 X 1021 v. and 1'~2 = 9.1.17 GPa. resulting in a total of 12 equations for the strains.88 X 10-3.13).29 X 10-3.349.5. 7 . 5. + 6.1936.8275 (V 3 .1.4.0.3 GPa. and v 12 = 0.268.9017 and T2 = 0. Because of the presence of the ± sign in the equation for T1 and T2 there are two sets of equations for each ply orietnation. 397. If the design space bounded by the limiting strain curves does not include a discrete point. 4. Graphical representation of the laminate strain limits. y· . Figure 7. It is also possible to draw the limiting values of the ply-level strains for other combinations of v1 and v • Although 2 some of these additional cases may yield no feasible regions in the in-plane stiffness diagram. . v.4. and v. The following example demonstrates the use of the procedure. since the ply orientation angles are known. The fiber orientation angles for the optimal laminate are el = 59.1 GRAPHICAL STRENGTH DESIGN 271 Vj ·. . .4842 and T2 =0. I I 1 v.2. ' v. {.157°. we can assume one or more of the prescribed angles to be absent from the laminate.3. = 0. Fig. = 0. In this case. I .3 Design of Multiple-Ply Laminates with Discrete Fiber Orientations The use of the graphical procedure is most suitable for designing laminates with discrete fiber orientations. 7.1. We also have to make sure that we do not consider discrete points associated with the omitted fiber orientation.48° and e2 = 5. Figure 7. we can draw the failure envelopes of plies with the specified angles in the in-plane lamination diagram.270 Vj STRENGTH DESIGN OF LAMINATES 7. The results presented here correspond to a selected value of v1 = 0. ·1·\·11' . Active constraints for the laminate are the transverse strain in the layer with the second fiber orientation angle and the shear strain limit. thereby increasing the area of the design space. I . and the values of the T 1 and T2 corresponding to these in-plane stiffness parameters are T1 =-0. Graphical representation of the laminate strain limits. 1 -1 . \"l~ .349.9838. in general there will be other laminates that will satisfy the requirements of the problem.8) on this in-plane lamination diagram and identify points with various ply orientation combinations that fall within the appropriate feasible region of the diagram.3. We then superimpose the polygons for the laminates with prescribed angles (see Section 4.. make up the boundaries of the region with feasible laminate stacking sequences.942v.0. Subsequently.35 0 2.87 42.66 42.V~ 2 + 3902.0 can be generated using graphical tools. 5. E 2 = 10.87 56. 35ov.942v. *2 + 3.59 1 Stresses in the 90° layers have the same equations.87 GPa.' 0 1 Yxy-268. only.897 10.376 + 8. Using the given material properties in Eqs. and S = 68 MPa. 3 71. 42. E~ = 5321.4. Y1 = 40 MPa. ±45°.1534. and 90° layers for the present problem are shown in Fig.3 Using the graphical procedure and layers with the following material properties: E 1 = 181 GPa. there is a triangular pattern of acceptable discrete points.573V~ + 3. . and V 12 = 0. Because of the discrete nature of the problem. cr(O') _ Y - (O'J = 'txy With laminate strains calculated.70 MPa. x-y. The laminate should also have the largest possible elastic modulus in the x direction. as selection of the applicable Tsai-Hill constraint boundaries depends on the orientation angles in a given laminate.20 + 63.1. ±45°.8. One has to be careful in choosing the best design. and 90° layers. cry= 100 MPa. As discussed in Chapter 4.897 10. Equations for the stresses in the 45° layers are slightly lengthier and for brevity are not shown.8 2. For example.1.35 0 0 r81.170 1 Q<+4S"l = r56. For example. Some of those points for 16layer laminates are also shown in the figure. v. and 90° layers to calculate the stresses in those layers in the laminate coordinates.4 )-(7 .1V. +45°. Parametric plots of such equations corresponding to the right-hand side value of 1. Having the stresses in the laminate coordinates. E~ = 5321.902 V * MPa. V~ 2 + 3902. These curves.1. the sign of the Q16 andQ26 terms is reversed. stress transformation equations can be used to calculate the stresses in the principal material directions.321 -7. 7.573v. 5. v~ + 728.376. Strength properties of the layers are X1 = 1500 MPa. and 1xy = 100 MPa.1 GRAPHICAL STRENGTH DESIGN 273 Example 7.688.66 42. not all points on the diagram are acceptable. for the oo layer we have ~ O') _ - 993.8-197. ±45°.6). and 90° layers are Q0'l = 2. the principal material direction stresses can be substituted into the Tsai-Hill criterion.32 42.350V 12 + 3.17 GPa.21 v.272 STRENGTH DESIGN OF LAMINATES 7. -7350. except the ax and crY are interchanged. for the present problem. G12 = 7. design a 16-layer symmetric laminate with discrete ply orientations of 0°. for problems with oo.8 0 GPa. we can next use the equations for the transformed stiffness matrices for the 0°. ±45°. The transformed reduced stiffness matrices for the 0°.87 46. superimposed on Miki's in-plane lamination diagram. which is now in terms of the V~ and v. + 3.897 0 181. 7 . and 90° which will stand an average combined stress of ax= 100 MPa. • * MPa. without violating the Tsai-Hill failure criterion.321 -7. 0 7. the laminate mid-plane strains are calculated as 5. Multiplying the strains with the appropriate transformed reduced stiffness matrices.170 1 90 Q< 'l = r2. the most restrictive Tsai-Hill criterion is the one for the -45° layer.' 5. (7 .902V3 71. 2. we find the stresses.1.897 0 0 GPa.28. Also shown in the figure are the contours of the elastic modulus of the laminate in the x direction.971 v.86V~ + 52. -7350. v.32 42. Tsai-Hill constraints for the 0°. For the -45° layer.1 v.3 GPa. 274 V* 3 STRENGTH DESIGN OF LAMINATES 7. At this point. The laminate is considered to be under the action of combined uniform in-plane stress resultants. requiring robust algorithms for full convergence to an optimal solution. referred to as the thickness variables. the design of a composite laminate is a combinatorial problem with discrete layer thicknesses and orien- subject to gkj= (PJkl Elk+ QYl E2k + RYl 'Y12k). A 66 ~A6~. their convergence behavior is slow compared to the thickness variables. If the laminate to be designed is to have a balanced stacking sequence. A 22 ~A~~. were commonly used as design variables. Graphical representation of Tsai-Hill constraints for discrete angles. therefore. only the thicknesses. . the best laminate is at point A.3) . then not all the points shown on the diagram are possible. If. (7. which was explicitly expressed in terms of the thickness design variables. = 1. k = 1.1) Pk tk 7.2 NUMERICAL STRENGTH OPTIMIZATION USING CONTINUOUS VARIABLES 275 Figure 7. (7. are used as design variables. Once the design is optimized using continuous variables.2. By inspection. and most past work on laminate optimization is based on the use of such variables.2. the Tsai-Hill constraints clearly eliminate the regions of the lamination diagram that does not include 45° layers (the horizontal line at v.. I. orientation variables influence only the strength constraints and. there are two possibilities.2 NUMERICAL STRENGTH OPTIMIZATION USING CONTINUOUS VARIABLES As emphasized in earlier chapters. Nx. then we should not be using that boundary. If the laminate being designed does not include a -45° layer. I = N/2. the total thicknesses of contiguous layers with the same orientation. tk. tation angles. we relax the balanced condition and design a laminate with only +45° layers. on the other hand.1 Strength Design with Thickness Design Variables Schmit and Farshi (1973) were among the first to consider the optimal design of a symmetric balanced laminate with fixed orientation angles. which has a higher stiffness than the balanced design.2) All ~Af{. then we can ignore the Tsai-Hill failure boundary for the -45° layer and choose the [Osf90/+45 2 ]s laminate at point B.. For the present loading. 7. of one-half of the total number of layers. with orientations taking any value between 0° and 90°. which corresponds to [Oi90i±45]s. However. Orientation angles were also occasionally used as design variables.1 ~ 0. because most optimization problems with strength constraints were formulated to minimize the laminate weight.4. NY. Because of the laminate symmetry.2. In weight minimization problems. .0). and Nxy· The optimization problem is formulated in the following form: W= minimize L2 k=i (7. the final thicknesses (or orientations) can be rounded off to integer multiples of the commercially available layer thickness (or conventional orientation angles).2. in part. For example. The less frequent use of ply orientation angles was. the majority of the commercially available optimization software is for continuous-valued design variables. 11) yl2 ' .7) at.276 STRENGTH DESIGN OF LAMINATES 7.2.2. E 0 =(Ex..2. at. we have ack _I d£o aco dyo =mz_x +n2~+mn~ dt.2.8) ' In matrix notation. Also specified are the limits on the in-plane stiffness terms with the following allowable values·. of the kth layer.Ec' 4 • 2 (7. (7.2. Qyl. Differentiating Eq. (t) = g . These derivatives are related to the derivatives of the laminate mid-plane strains with respect to the ith layer thickness..-1 R<kl. (2.6) results in the following values of the coefficients: For tensile normal stresses.2. 0 at.5) into a sequence of linear programs..4. andy12k are the strains in the principal material direction in the kth layer. 0 at. ' aco ac" dyo =nz_x +m2~-mn-=2 1 £1 t' QCkl = 0 I ' R<kl = 0 I (7.2.. at. therefore.2.2. Aan Aa 11 andAan II' 22• 66' For a simple maximum strain criterion.0 Q4 .13) For compressive normal stresses. and d£ 12klat.2. which puts bounds on the values of the strains in the principal material directions. and the Ew £ 2k. these equations take the form (k)_O P4 ' (kl. y12kf. The coefficient R is the inverse of the shear failure strain written twice (once for positive shear and once for negative shear). see Haftka and Giirdal (1993).9) ~ = Te(k) I a~:(k) dEo dt.2) is a nonlinear function of the thickness variables and.. p<kl =I Schmit and Farshi transformed the nonlinear programming problem described in Eqs. The strength constraint of Eq. Nxy?' the derivative of the laminate strains with respect to the thickness variables can be deter- .2. . the failure envelope has six facets with P and Q defined as the inverse of the normal failure strains in the longitudinal and transverse directions to the fibers. a£ at. (7. d£ 2/dt. .) [ p\kl ____!! + Q<kl ~ + RCkl Y12k . a£ at.. = 1.2.1)-(7.4) j p<kl=O 6 ' Q<kl=O 6 ' R<kl=6 -1 (7.. and Te(k) is the engineering strain transformation matrix for the kth layer defined by Eq. a procedure commonly referred to as sequential linear programming. For example.14) For shear stresses.5) where pk and tk are the density and the thickness. For a specified in-plane loading N = (Nx.2.2. I. for k = 1. are the derivatives of the principal material direction strains in the kth layer with respect to the thickness of the ith layer. J. (7 .1. p<kl = -=!_ 3 ~=-2mn-x +2mn-Y +(m 2 -n 2 )~. respectively. . once in tension and once compression. respectively.16). O. are coefficients that define the jth boundary of a failure envelope for each layer (k) in the strain space.12) where d£ 1/dt.2. Yxyf. . ] df. (k)_O P5 ' <kl = 0 Q5 ' R<kl __1_ 5 - (7. (7. ' I (7.. I 01 } df. the maximum strain criteria of Eqs..2. ai.10) Yi2 ' where E(k) = (Ew £ 2k. (7. (7. 1 at. £Y. Pjkl. ' ac ac a J i=] l l l (7. ack _2 P<kl-O 2 ' = _!_ Q<kl 2 t' Ez R<kl = 0• 2 (7.t . is linearized as (t. at. c' £1 QCkl = 0 3 ' R<kl = 0 3 (7. ' ayo at.6) ' at. Ryl. (6.4)-(6.(t ) + ~ g kjL I kj 0 £. NY.1 0).2 NUMERICAL STRENGTH OPTIMIZATION USING CONTINUOUS VARIABLES 277 and -tk:::. . at. With the provided elastic stiffness properties.06).28).4 For the specified stress resultants. Substituting this relation into Eq.3 0.00535.15) Rearranging. and Nxy = 0.2. The laminate is under loading that induces Nx = 4000 lb/in. Note that the 2 in front of the variable x stems from the fact that the laminate is mid-plane symmetric and the x is related to the thickness of the total oo layers in the laminate.278 STRENGTH DESIGN OF LAMINATES 7. we can determine the derivatives of the mid-plane strains needed in equation Eq.02) + 7.2 NUMERICAL STRENGTH OPTIMIZATION USING CONTINUOUS VARIABLES 279 mined by differentiating the in-plane portion (zero B matrix) of the stress-strain relation of Eq. U3 = 1.3 608.02.2. (7.02) + 2.0 kip/in. o (i)t · (7. G12 =0. £~ = £~ = 0.508 (2x. As defined in Eq.979 Msi.4.2. x ::. the laminate mid-plane strains corresponding to the nominal design are . (7. Therefore.'s are computed: U1 = 7. to obtain dN dA d£ 0 . Eo. Design of a laminate with 0°.0.3.16) The maximum strain failure limits for the material are ci = £~ = 0.t 0 +A-=0.= . (7.31).=Q(i)' I (7. linear approximations to the strain constraints can be constructed using Eq.005 in. NY= 1000 lb/in. the derivative of the A matrix with respect to the ith thickness variable is simply equal to the transformed reduced stiffnesses of the ith layer. Example 7. Properties of the layers correspond to that of AS4/3501-6 Carbon/Epoxy material: £ 1 =18.14) as ~=-A:1 o 1Q dt. N =At". the derivative of the mid-plane strain with respect to the thickness variables may be written as dto =-A-1 ()A dt. The in-plane stiffness matrix for the nominal design is 269.15).9 0.0. the in-plane stiffness matrix A is a linear function of the thickness variables.1 = -1.973 (2x. [0/±45/902 ].02.17) For a balanced symmetric laminate of [Ox /±45/90 2 ]s.and y~ 2 = 0.0 206.[0/±45 2 /902]..18) A 22 = -6. (2.06). dt. ±45°.414 Msi. U4 = 2.979 (2x + 0. (7.2.46 Msi.. (2.81 Msi. A 66 Once the derivatives of the strains in the fiber and transverse to the fiber directions are calculated from Eq.14).0 ] A= 184. • dt. (7.4. [ 0. The linear approximation is to be performed for a nominal point x = 2t.481 Msi. Formulate the linear approximations for the strains in the principal material directions in each layer with respect to the change in the thickness of the oo layer.12) at any step of the sequential linearizations. dt.06).0 0. The thickness of the individual layers is t = 0. U2 = 8.2.973 (2x.0. aA _ a.0.7 184. which corresponds to the initial quasi-isotropic laminate. .414 (2x + 0. A 12 = -1.02) + 2. the following U.3 Msi. (7. £ 2 =1.2.02) + 7.973 Msi. 0. where x = t ·:X. the terms of the A matrix can be computed in terms of the thickness-related variable x as A 11 = 10.783 Msi.2. dt. and plot the strains in the individual layers as the thickness of the 0° layers change in the range 0 ::.783 (2x + 0.2.06).0115. and V 12 = 0.0. and 90° layers is to be started from an initial design that corresponds to a 16-ply quasi-isotropic laminate.979 (2x + 0. and U5 = 2.2.45 (2x. 5 0.0362 0.0 0.5] 0. -1 1 0 Te(-45o) = r 0. 0 ::::. x ::::.4.638 ) 0. and plotted as shown in Figs.015 0. 4t.6. (2. The filled circles in the figures correspond to the strains for the quasi-isotropic laminate.301 ) 0.280 1.5217 1. t(o') = -0. -0.492 0 0.337 Strain These computations may be performed for each value of the thickness variable.638 ) %. j 0.0362 %.2 NUMERICAL STRENGTH OPTIMIZATION USING CONTINUOUS VARIABLES 281 0. the magnitude of the shear strain y12 is the same in the +45° and -45° layer with only the sign reversing between the two layers.5 and 7.5 0.015 Strains in the individual layers can be computed by using the engineering strain transformation matrix Eq.0 t('45 o) = j oo and goo layers.01 (45°) (-45°) (-45°) £2 £1 £2 0. 0.301 %.0 j Strain 0.02 -0.5.5217 1 r 0. and there is no applied shear stress resultant Nxy' there is no shear strain in the 0° layer. 7.o l STRENGTH DESIGN OF LAMINATES 7. Eq.724 =A.5 0.02 -0. 0 0 1 001 Te(goo) 1 o =o 1 0 0 r 0 0 -1 l .01 0.5] 0.5 -0.015 0.6.015 Figure 7.5 -0.5 0. 0. Principal material direction strains in the 45° and -45° layers. Since the laminate is balanced and remains balanced as the thickness of the oo layers change.5 .5 . 0. which is given by the reduced transformed stiffness matrix of the layer being changed. we need the derivative of the laminate mid-plane strains with respect to the thickness variable.16): 1 Te(Oo) = r 0 1 0 . 1 -1 0 0. Clearly.5 0.01 -0.005 ~----~--~~==~~==------~--====~x Te(45o) = r 0.005 -0. Also. We first compute the derivative of the in-plane stiffness matrix with respect to the thickness variable.N= --D. In order to construct the linear approximations to the principal material direction strains.005 0.0 o. as the thickness of the 0° layers decrease the strains in principal material directions in all the layers increase.01 0.724 t 0 -0.005 -r--------~--------~--------~--------~x 0. w-3 = -0.01 0.005 Figure 7. Principal material direction strains in the Therefore. .0 {4} 1 x 4.5 0.02 0. 724 -0.1146 .2t) 0.0 j Using these derivatives. EC45o)(x) dA dX l18. the approximate longitudinal strain in the oo layer e)0l indicates that the thickness of the layer may be reduced to zero without failure. £~.0 0.638! -0. 0.2585 90° layer.0 4. as expected.724 0.2t) -0. however.81 l = E(goo/X = 2t) +(X.4412 o.5217 0.2t) dE(oo) a.4024) 0. however. which produces ~= dX dE ~-0. this would indicate a constraint violation.4024 0. Note.1439 .--(). is much lower than the fiber direction failure strain. 0. that the variable x used in the laminate stacking sequence description controls the thickness of the 0° layers on both sides of the laminate simultaneously.0 0.-• Etlo) = l 0. (7.2 NUMERICAL STRENGTH OPTIMIZATION USING CONTINUOUS VARIABLES 283 (7 . 0. In this particular problem.638 ) ~-0.1146 .003.0362 ' 0.0 2 =.t(O'') X= 2t) + (x.0 ~= dX dE {-0. that the strain to failure in the transverse to the fiber direction.17).2t) a~.ol Msi = Q(0°) = 0.8 shows the shear strains in the principal material direction in the +45° layer along with the corresponding shear failure strain.1146 .7 and 7. In fact.282 STRENGTH DESIGN OF LAMINATES 7. reaches its critical level at around x = 0. -0.0 I dE -~-0.0 0. 0. Although the approximations are reasonably good within the immediate vicinity of the nominal point.0 0.1146 -0. 7. (7. In particular.0 E(9 oo) = 1 -0. EC90o)(x) dE dX 0 [ 1.638 + (x.0 0.. if the first-ply failure criterion is used in designing the laminate.492 Jr 18.8.0 0. for the derivative of the mid-plane strain. their accuracy degrades.301 + (x.4412 0.2585 ~= dX dE ! 0.14).0 Two examples of the plot of the linear strain approximations along with the actual strains are shown in Figs.2.43 0. Therefore.4412 1.81 - = t(45°/x = 2t) + (x.0 0. The actual strain.1439} 0. The magnitude of the shear strain .471 0.1439) -0.0 1.4024) dX . 7 actual and approximate longitudinal strain in the oo layer and the transverse strain in the 90° layer are shown along with the corresponding failure strains.0 0. only the thickness of the 0° layers is being changed.013 to prevent the violation. L ( ) _ ( E(0°) X . 7.4024) -0. 0 0.0362 + (x.4024 .2t) -0.0 Derivatives of the principal material direction strains in the individual layers are found from the derivatives of the mid-plane strains using Eq.0.337 0. the linear approximations for the principal material direction strains in the individual layers are oo layer.18) by 2: E( 45 ol = 1 0 301) ~-0.0 0.1146! 0. on the other hand.0 0.0362) 0. Figure 7. Therefore. In Fig.2.4412 0. we multiply Eq. especially for small values of the thickness variable x. 45° layer. Therefore.1439 .43 0.2.2t) d~:Oo).o)' Note.5217 1. The derivative term computed above is the derivative of the A matrix with respect to change in the single layer. transverse strain in the 90° layer for the nominal quasi-isotropic laminate exceeds the failure strain. 0.471 0. and the variable x would be pushed to a higher value of about 0. who presented the results of an optimization study using conventional laminates with 0°. they have demonstrated that it does make a difference whether we initially select a [0/±45/90].0025 ~~------~--------~------~~--~~~x in the +45° layer for the nominal design is smaller than the failure strain and would not be critical at this particular design point. Linear shear strain approximation in the 45° layer.284 strain STAENQTH DESIGN OF LAMINATES 7.015 0.0175 -0. the optimization procedure can resize the remaining layers to achieve a weight lower than the one achieved before.8.01 -----------~~--------s - Figure 7. This is best exemplified by Schmit and Farshi (1973 ). Although derivative-based optimization methods are quite powerful search procedures. Compared to the second laminate..005 0.015 -0. For example.0075 0. it is quite possible that once a layer with a thickness different from its lower bound is removed.01 -0. This can make the design procedure difficult. . laminate even though at the end of the design iterations the thickness of the 90° plies of the first laminate vanishes. one has to be careful in the case of composite laminate design problems with continuous layer thickness design variables.0075 -0.0025 -0.005 -0. ±45°. One of the difficulties with such problems is the choice of the initial laminate stacking sequence. even though the thickness of the layer is vanishingly small.1. That is. In order to achieve a true optimal solution. and the removal of such layers is not an option. ef ----------- 0. laminate or a [0/±45].02 0.7.005 -0. strain 0. which are generally different than zero. Multiple load conditions also tend to produce designs where ply removal may not be possible. therefore.02 Figure 7. for a laminate under uniaxial stress and limits on shear stiffness. are imposed.01 0. Therefore.01 0.02 0. The final design of the first laminate has a critical strength constraint for the 90° layer. [0/±45]. Linear strain approximations in the oo and 90° layers.0125 0.0175 0.015 0.0125 -0. as this may influence the final design optimization result. and 90° orientation angles under various combinations of in-plane normal and shear loads. lower limits. the designer has to repeat the optimization process with different laminate definitions.2 NUMERICAL STRENGTH OPTIMIZATION USING CONTINUOUS VARIABLES 28!5 0. for most practical applications the presence of plies with fibers running in prescribed directions (such as fibers transverse to the load direction) is desirable. However. the fact that a layer assumes a value different from its lower bound may not mean that the particular layer is essential for the optimal design. Results of these laminates obtained from Schmit and Farshi ( 1973) are summarized in Table 7 . it is about 9% thicker due to an additional 0° layer required for the first laminate to prevent violation of the strength constraint in the 90° layer. especially by removing the layer(s) that converge to their lower bounds. However. because of the need to try all possible combinations of preselected angles.005 0. the orientation angles of the layers as well as their thicknesses may be used as design variables.000000 0. and for large shear loading without the longitudinal load.52 0 4 6 6 0 [0/±45Js 1 where b0 is defined solely in terms of the stiffness and strength properties in the principal material directions. NY. ~12 = b2 y 2 hy o• l 2 3 4 oo +45° -45° 90° L... 0..2 NUMERICAL STRENGTH OPTIMIZATION USING CONTINUOUS VARIABLES 287 Table 7. Nxy This approximate failure envelope is given by E2 x Final Design % ..034194 0.034194 0. An alternative to the weight objective function minimization is the maximization of the laminate strength as demonstrated by Park (1982) and Massard (1984). Minimum weight laminates with stiffness constraints loaded In axial compression Layup and Layer Number [01±45!90]. the optimum angle reached 45° for NY= NJ2.129124 0. As the transverse load NY increased. the optimal angle increased and reached a value of about 73 o for NY= Nxy· The continuous laminate proved to .032281 0. As mentioned earlier..52 35. 0. The optimum angle for the [-e/0/+e].018793 0. [-e/90/+eJ. In order to demonstrate the use of ply orientations as design variables.032281 0. covering a range from -e 0 to eo orientations. this failure envelope is in the strain space and does not include interactive strain terms such as ExEy or Ex"fxy The objective function to be minimized is defined as 2 2 I 2 f = Ex + Ey + 2 '¥xy' oo +45° +450 2 3 21.!.2. Only balanced symmetric laminates are considered by Park.059438 28. Indeed. therefore.012555 0. five of which were the following conventional layups: [-e/+eJ. Orientation Angle Initial Design Final Design (deg) t. t. laminate depended on the magnitude of the transverse load NY. some optimization algorithms may experience convergence difficulties. and was equal to 60° for NY= Nx.064890 0. the stronger the laminate against first-ply failure.44 39. Number of Plies (rounded) + E2 + . many design codes can treat both as design variables. however. A quadratic first-ply failure criterion based on an approximate failure envelope in the strain space is used by Park for laminates under which represents the square of the norm of the strain vector. Similarly. Results by Park showed that under combined loading the best laminate. t.102583 0.023048 0.286 STRENGTH DESIGN OF LAMINATES 7.032281 0. type.032281 0.023441 0. and six different laminates were studied.023441 0. One key feature of this approximate strain failure envelope is that it applies to laminate strains and does not require ply-level strain calculations. 7. the objective function is independent of the orientation of the layers and. (in) t1 (in) various in-plane loading conditions Nx.023048 0. and was equal to oo for NY= 0.034194 L. [-e/0/+eJ. the larger the loads that can be applied to the laminate before the failure envelope is violated. for optimization problems formulated as minimum weight designs. [-e/0/90/+e]s.96 35. The smaller the objective function value.1. according to the first-ply failure criterion. for the [-e/-45/+45/+e]s laminate (with shear loading and no axial loading). and [-e/-45/+45/+e]s. the best was the [-e/-45/ +45/+SJ.2 Strength Design with Orientation Angle Design Variables In order to find a laminate stacking sequence that is best suited to a specified loading. we concentrate in this section on examples with orientation variables only. As the transverse load was increased. therefore.12 39. the optimal angle was 45° for NY= 0. laminate.44 3 6 6 Source: Schmit and Farshi (1973). The sixth laminate was called a continuous laminate and was assumed to have a fiber orientation changing linearly from the top surface to the mid-plane of the laminate. Note that compared to the quadratic failure criteria discussed in Chapter 6. for large longitudinal loading without shear was the [-e/0/+eJ. such as the Tsai-Hill criterion or the Tsai-Wu criterion. (7. l (7. but such intuition may not always lead to optimal designs when working with composite materials. constraint linearization is one of the useful techniques in handling difficult nonlinear functions in powerful linear optimization algorithms. which may be used either as a design constraint or the objective function. or apply linearization schemes to those nonlinear response quantities so that integer linear programming may be used. a different form of linearization may be more appropriate. which involves discrete values of both the thickness and the orientation angle variables. In Example 7 . and equating the derivative of the failure index of Eq. crY. the inplane stiffness parameters capture all the information about the thicknesses and orientations of the layers of a laminate. For a set of prescribed orientation angles.3 NUMERICAL STRENGTH OPTIMIZATION USING DISCRETE VARIABLES Practical stacking sequence design of a composite laminate requires the use of the methods of integer programming discussed in Chapter 5. The same approach may also be used to form linear approximations with respect to the orientation of the individual layers in a laminate.2.3. The strength of a composite laminate.2) with respect to the fiber orientation to zero.1 Integer Linear Programming for Strength Design != cr~J . optimal placement of the fibers is not along the principal stress directions. As introduced in Chapter 2. (x "Y + ('s 2 2 2 )(2 1• used for the strength prediction of a unidirectional composite (Tsai. ±45°.6(at 0 aAij " [Aij. In this section a linearization that can be used to cast the nonlinear strain quantities in terms of the elements of the in-plane stiffness matrix is presented. A strength constraint linearized in terms of the in-plane stiffnesses. powerful integer linear programming techniques may be used.288 STRENGTH DESIGN OF LAMINATES 7. For nonlinear problems. however.11): treated accordingly.2. The advantage of linearization with respect to the in-plane stiffness terms is their form which lends itself to integer linear programming for the laminate. 7.1) . Consider. can be used directly in an integer linear programming problem. The linear approximation to the mid-plane strains of a laminate is given by E~(x) = E (X) + ~ 0 1. for example.7) shows the expressions for the in-plane stiffnesses for a balanced symmetric laminate made of only oo. the in-plane stiffness parameters can easily be turned into linear expressions in terms of the number of layers of the prescribed orientations. 7. In the following sections. the designer has to either use a nonlinear programming approach that can handle discrete variables. For example. For those design problems that can be cast in the form of linear programming. implementation of the genetic algorithm to the design of laminates with strength constraints is demonstrated. The quantities X. The above results were intuitively appealing in that the fibers were mostly placed in a direction parallel to the applied loads. but depends on the values of the strength quantities as well as the applied stresses. However. first a linearization scheme which is suitable for the stacking sequence design of laminates is presented. 1968). therefore.1 one such linearization is demonstrated for constructing strain approximations with respect to a thickness design variable. (6. 'Cxy and the fiber orientation 9. such as genetic algorithms.2. and S is the shear strength of a ply.Au(xo)].(cricrzJ + (crzJ tiz) : :. and 90° layers.3 NUMERICAL STRENGTH OPTIMIZATION USING DISCRETE VARIABLES 289 have the best overall performance under combined longitudinal and shear loadings with a range ±65° for NY= 0. Yare the normal strengths in directions parallel and transverse to the fibers. Brandmaier (1970) showed that if the transverse normal strength Y is less than the shear strength S. Eq. (5 . for stacking sequence design. Next. the Tsai-Hill criterion of Eq.2. is clearly a nonlinear function of both the thickness and orientation variables and must be Traditionally.2.3. This can be demonstrated (see Exercise 1) by expressing the stresses in the principal material directions in terms of the applied stresses ax. 3.3.-+-=.7) Also note that some of the terms which are zero in the x-y coordinate system are not zero in the principal material directions. For example. for a balanced laminate.] = _ oE~ + 0 ) oE.16).3. further manipulation of the derivative terms is possible. ..A12Ny £X- with the rest of the derivative terms. A 16 =A 26 = 0. we need to compute the values of the derivatives of the mid-plane strains with respect to the in-plane stiffness terms at the nominal design point. such as CJE~/CJA 66 . respectively. the mid-plane strains are computed from o.290 STRENGTH DESIGN OF LAMINATES 7.6) CJAll 0 CJE\450) = CJE~450J =-(All. Derivatives of the strains in the principal material directions of the individual layers can be computed using the engineering strain transformation relation of Eq.11) CJE~ _ 2 A 12 E~ ~ CJA 12 H . ±45°. CJAz2-.A~6' (7.14) CJA2z OE CJA 12 0 _ CJA22 2H - 2 A 12 E. The expressions for the derivative terms can be obtained from the in-plane constitutive relations connecting mid-plane strains to the stress resultants.2) A ' where H = A 11 A 22 .3.3. For specified discrete angles.5) iJE(450J CJE(450J CJA11 CJAu _1_=_2_= (A -A ) 12 22 £0 x' (7. + ~ H H' (7.12) CJAu Further. (2.3) CJE~45o) 1 [ dE~ CJ£o CJy~y] . (7. Example 8. CJE(450J N CJE _ All E. For example.Alz) Eo. at the nominal design point X0 • In order to set up the linearization. the strain derivatives in the 45° layer are computed by using {J£(45°) I H ' o_ A11Ny-Al2Nx EyH .) and £ 0 (xJ are the values of the in-plane stiffness terms and the mid-plane laminate strains.3.Ai2.2 ()Au CJAu .3. the derivatives with respect to the in-plane stiffness terms can be shown to be 0 0 OEX A 22 £X CJA11 =-~.2. it can be shown that (7.=L CJAu . and y: = Nxy xy 66 (7.3..4. 1 --=--2.8) CJA66 xy A66 (7.3.9) An example of the use of linearized strain constrains in buckling load maximization design of laminated composites is discussed in Section 8. For example. being zero. Y (7..3.CJAu ' (7.3.H' CJE~ A 11 E~ Nx --=--+CJA22 H H' (7.3.3.10) (7. H _ Nx H' (7.o] 1 [ CJ£o x CJ£o y v f xy CJAij =2 CJAu + CJAu + CJAu ' (7.. Therefore. ~.3.3.4) iJy\4.3..3 NUMERICAL STRENGTH OPTIMIZATION USING DISCRETE VARIABLES 291 where Au(x. and 90°. CJAu.A22Nx.15) (7. with the discrete orientation angles limited to 0°.13) (JAil CJE" 2H = _ A 22 E.H ' CJy~Y Nxy CJA66 =.3. 1.1' ly12. lyl21 ~ y~2 = 0. The objective function <1> was the number of plies.1. A.89 Msi. The stacking sequence is required to have no more than four contiguous layers of the same orientation. E is a bonus parameter for designs with a margin. E =0.. designers often introduce constraints that require the design of the stacking sequence.E(Acr. The problem was simplified by assuming that the layers came in stacks of 02 .1 E~ Y'h] ' Example 7. The strain constraint is expressed in terms of a single parameter. However.1 Design the lightest symmetric and balanced Graphite/Epoxy laminate made of 0°. Using A. Then the strain constraint is expressed as 1. <i>= !!!!:. able except for very thin laminates. because the minimum function is not smooth. we often require that there be no more than four contiguous plies of the same orientation. and 90° layers to carry loads of Nx =-10. That is.5. Material properties are E 1 = 18. 7. in order to be on the safe side. Use the maximum strain criterion. Because the calculation of ply strains or stresses takes very little computer time. ±45.3.015. and the reliability as a function of the number of generations is shown in Fig. { mm .0. otherwise. so that a laminate can be coded using a four-letter alphabet including the three stacks and an empty stack. E~ = 0. 1 . for genetic algorithms it is a convenient way to represent the constraints. Based on previous experience. the population size was set at 8. It can be seen that about 300 generations (2400 analyses) are . E~ = 0. A.A. cr with a safety factor off= 1. G 12 = 0. which depend strongly on the stacking sequence.292 STFIENQTH DESIGN OF LAMINATES 7. The maximum number of stacks was taken to be 20. where E denotes 2 an empty stack.3 NUMEFIICAL STRENGTH OPTIMIZATION USING DISCRETE VARIABLES 293 7. ±45 o. This simplification is reason- where p is a penalty parameter for violations of contiguity. n.5.cr' which is the load multiplier at which one of the strain limits will be exceeded.005 in.5 Msi.0. For example. In the application here. for example. or 902 . and the probability of a stack swap set at 1. deleting. and Nxy = 1.cr ~ 0.3. and () is a penalty parameter. + 0. where i is taken over all the layers. with penalties for violating the strength and contiguity constraints and a bonus for constraint margin. The following example illustrates the use of a genetic algorithm for design subject to strength constraints. 1£21: : :. Lumping all the strain constraints together in a single one does not work well for derivative-based optimization algorithms. However. the probability of crossover at 1. the same example will be used in Chapter 8 with additional buckling constraints. [. the objective function is given as p[n. where the use of odd numbers of 0° or 90° plies can result in substantial relative reduction of the optimal thickness. for bending loads the stacking sequence itself needs to be designed. Even when there are no bending loads.000 lb/in. =-mm cr J i IE!il' 1~. the laminate [(±45)i0i90 L is coded as [E E 45 0 0 90]. v12 = 0. in order to minimize matrix cracking problems.029.000 lb/in. NY= -2.2 Genetic Algorithms for Strength Design Genetic algorithms are a natural tool for strength design of composite laminates.E2 .9.008.1)]. if Acr ~ 1.A.000 lb/in.3. With this coding. and () = 0. E 2 = 1. and t = 0. or adding a stack all set at 0. the large number of analyses required for genetic optimization is easily affordable. For in-plane loads and symmetric laminates.93 Msi. the probabilities of mutating. The optimization was repeated 100 times. with IE 11:::::. the design variables can be the number of layers of each orientation. and 'txy• show that the stationary values of the Tsai-Hill function REFERENCES Brandmaier.4. N. with the fiber strain constraint in the oo layers being the most critical constraint. Reinf. in each case: E 1 = 134 GPa. Kluwer Academic Publishers. Mxy= 0. (1984). "Optimum Design of Laminated Composite Plates Subject to Axial Compression.cos 28 = b. Reliability as a function of number of generations for Example 7.irdal. Figure 7. 1. E. M.6.10 GPa. 422-425. K. 0." J.. Japan-U. and MY= aA." Composites' 86: Recent Advances in Japan and the United States. Massard.6.0.. 100 150 Generations 200 250 300 Use a genetic algorithm to design a maximum strength 16-ply balanced symmetric laminate with 0°. (b) Mx =A. 300-345. H. Boston. CCM-III. where a= and 2'txy _ 1 and b= ay+O'x 1-a? cry. and Kobayashi. G 12 = 5. ±45°.2aA-.1 2. ax. Proc.9~------~---------r--------.28. R. Kawata. J~(i)-(i~)+(i)+(~) are achieved for a cos 28 + sin 28 = 0 .2. Composite Materials 4.4. "Computer Sizing of Composite Laminates for Strength. Y are the normal strengths in parallel and transverse directions to the fibers and S is the shear strength. and t = 0. (1986). yc = 200 MPa.1. where a= 0.90 GPa.. Elements of Structural Optimization Third Revised and Expanded Edition.S.3. and 90° plies under combined loadings of (a) Mx=A-andMxy=aA. Use Hoffman's failure criteria for the strength maximization with the following material properties and determine the value of A. (1993). T. where a = 0. 0. 0. Plastics and Composites 3. EXERCISES 1. 0.125 mm. Miki. ( 1970). crY. yt = 80 MPa. A. S. S = 160 MPa. (eds. 0.ax ~2. "Optimum Filament Orientation Criteria. E 2 = 8. v12 = 0. needed for about 80% reliability.2. The optimum design obtained was [±45/90i0i±4510i±45L.. Haftka.a}. and Gi. where X.---------r--------. XC= 1100 MPa. Tokyo. 673-680." J. xt = 2130 MPa. T. For a unidirectional laminate under uniform applied stresses. 0.9.1. REFERENCES 295 and a sin 28. z.2' crY a=XIY and ~=XIS. Umekawa. pp.).294 STRE!NQTH DESIGN OF LAMINATES 0. 3-11. MD. flexural stiffness response often depends not only on the thickness of the laminate but also on its other dimensions. Layers that are located on the outside of the laminate contribute much more to the flexural stiffness of the laminate than identical layers that are located near the center plane.. and Sugiyama. pp. J. Meth. "Optimum Design of Laminated Fibre Composite Plates. Unlike in-plane response." J. These methods have been applied to problems of laminate design for in-plane response. Baltimore. the order of the layers with different orientations is as important as their total thickness." In Fundamental Aspects of Fiber Reinforced Plastic. 275-283. S. (eds. For in-plane response. Z. (1968). "Stacking Sequence Optimization of Simply Supported Laminates with Stability and Strain Constraints. (1973)." AIM J. Wiley Interscience. R. (1992). S. H. The present chapter is concerned with the application of these methods to flexural response and combined in-plane and flexural response. (1982). "Strength Theories of Filamentary Structures. R. Haftka. Structural Dynamics. 623-640. For flexural response. Nagendra.. J. In this chapter we limit ourselves to the flexural response of rectangular plates. Engng. Num. and Farshi. 2132-2137. Matis. B. Y. and Giirdal. Schmit. We start this chapter with a summary of the equations governing some of the important response characteristics that 297 . T. and Materials Conference. 11.STPII!NQTH DESIGN OF LAMINATES Miki. ( 1991 ). L. Schwartz. "Optimum Design of Laminated Composite Plates Using Lamination Parameters. A. only the total thickness of all layers of a given orientation matters. S. T. W. 8 LAMINATE DESIGN FOR FLEXURAL AND COMBINED RESPONSE The previous chapters have laid the foundation for graphical and analytical methods of laminate design. The major difference between design for in-plane response and flexural response is in the importance of the stacking sequence. and Schwartz.341-355. 16. "An Optimal Design of Simple Symmetric Laminates Under the First Ply Failure Criterion. Part I. 30(8). W. Camp. New York." Proceedings of the AIAAIASME! ASCE!AHS!ASC 32th Structures." Int..). Tsai.. Park. M. 1) yields a4 q 0 sin(nx/a) sin(ny/b) 1t 4 (8. so that the contribution to D and D 26 of the layers with positive angles is balanced ex16 actly by layers of negative angles.1 FLEXURAL RESPONSE EQUATIONS terms D and D26 . a .1. . Except for very thin laminates. the pressure distribution is expanded in a double Fourier series. For a simply supported plate under sinusoidally varying pressure.2) )114· il w ()4w ()4w ()4w ()4 w Dtt ()x4 + 4Dt6 ()x3()y + 2(Dt2 + 2D66) ()x2()y2 + 4D26 ()x()y3 + D22 ()l =q. We will use this criterion in the more general setting of flexural response.1) The bending-twisting coupling terms D 16 and D 26 in the equation complicate its solution.1. y) = q0 sm-..1. However. The relative magnitude of D 16 and D 26 can be assessed by the following two nondimensional terms D16 The equations governing the out-of-plane displacement w of a symmetric and balanced laminate subjected to a pressure loading q (see Fig. Furthermore. Laminate under transverse loading. q(x. for a uniform pressure distribution of magnitude q0 . the bending-twisting coupling that they represent is sometimes undesirable.5) m=1. As an example. These include the out-of-plane displacement of a simply supported laminate loaded by transverse loads and its natural vibration frequencies and the buckling response of a simply supported laminate under in-plane loads. These equations are limited to the special case of balanced symmetric laminates.__L .L .1 FLEXURAL RESPONSE EQUATIONS 299 are commonly used in design.- r•!fr X ' '' '' ' (see Kedward and Whitney. =I . where . 8.X ..'' 16q0 1t +z Figure 8. making it impossible to obtain closed form solutions that are useful for simple design problems.1.L wmnsm--sm-b. .ny (8.1) are ()4 y= (D3 D 11 22 )114 and 0= D3 D ( 22 D26 11 (8.1. (8.. The source of the D 16 and D 26 terms are the off-axis layers that have nonzero Q 16 and Q26 terms. (8. (8.nx. because the contribution of the positive and negative 16 angle plies also depends on their distance from the laminate center plane. sm b' .1. For more general out-of-plane loadings. m1t.4).298 LAMINATE DESIGN FOR FLEXURAL AND COMBINED RESPONSE 8. 1990).3. For balanced laminates the contributions of the positive angle plies cancels the contribution of the negative angle plies to the in-plane shear-extension coupling terms A 16 and A 26 • The same does not hold for the bending-twisting coupling stiffness q(x. this can be accomplished by simply stacking a +8 layer always next to a -8 layer. 8.1.2.y) Nemeth (1985) suggests that the effects of D 16 and D 26 for buckling are small for y and 0 below 0.3) the solution of Eq. the transverse displacement field is obtained as w=--6- /.3.. nny .4) 4 w= [D 11 + 2(D 12 + 2D66) (alb) + D 22 (alb) ] · 2 a ------------.. and the solution for each term is obtained in a form similar to Eq. we will assume that these terms are made negligible by proper arrangement of layers in the laminate.1. (8. 8) where m.1 Linear Problems For a symmetric laminate we define flexural lamination parameters as * 12Vm Wt= h3 =""' £. (8. we replace q in Eq. 1986) and will be solved using integer linear programming in the next section. For the transverse vibrations of a laminated plate. If the plate is loaded such that Nx = -Wxo and NY = -ANyo' then the critical value of 'A corresponding to a buckling load with m half waves in the x direction and n half waves in the y direction is 'Acr' and is determined as ~ s. 1.9) For a simply supported plate with no shear load.10) The buckling load multiplier is obtained by finding the lowest value of \r for all combinations of m and n.35). (4. Applications to the maximization of the buckling load of a laminate are somewhat cumbersome with Miki's graphical procedure (see Miki. and we will discuss the use of this diagram for maximizing flexural stiffness under lateral loads and for maximizing the fundamental frequency of a laminated plate.. For a simply supported plate. the critical values of m and n are small. any balanced symmetric angle-ply laminate with multiple orientations can be represented as a point in a region bounded by 2 W*3 . k=l l sk cos 29k and w. NY. the vibration modes are sinusoidal in both the x and y directions and the natural vibration frequencies are given as ffimn= -~ "--D 11 (mla) 4 + 2(D 12 + 2D66 ) (m/a) 2 (nlb) 2 + D2z(nlb)4 . (:) 4 2 + 2(D. When the plate is loaded by in-plane loads that introduce uniform in-plane stress resultants Nx.2. 'I ph 2 (8. and Nxy' the equation of equilibrium in the direction normal to the plate (neglecting D 16 and D26 ) is d4w d4w d4w d2w d2w d2w Dll dx4 + 2(DI2 + 2D66) axZal + Dzz dy4 =Nx dx2 + 2Nxy dxdy +NY al· 8. 3 (8.2 STIFFNESS DESIGN BY MIKI'S GRAPHICAL PROCEDURE Miki (1985) showed that a diagram analogous to the in-plane lamination diagram discussed earlier can be constructed for designing laminates for flexural response. Unless the plate has a very high aspect ratio (such as an alb of 3 or larger). +2D") (::) + D22 ~) 4 r (8. (2.300 LAMINATE DESIGN FOR FLEXURAL AND COMBINED RESPONSE 8.1_ (2z.2. k=l I (8. and t denotes time. q=-ph a2w ar' (8.=(2z. .2. or extreme ratios of Therefore.1.7) where p denotes the mass density of the plate.1.8) is obtained for the flexural lamination parameters "~ cr ' (m 2 4 2(D 1t [D 11 (mla) 2D66 )(m/a) 2 (n/b) 2 D 22 (n/b) 4] 12 n)---~~------~~--=------------=----(m/a) 2 Nxo (n/b) 2 Nyo + + + w.- _ 12V3b = h3 L sk cos 49k.2.2.1.4. and (8. (8.. The fundamental frequency is associated with m = n = 1 (see Vinson and Sierakowski. w~:::.1.1.2) Miki (1985) shows that a relation of the same form as Eq.. 1987).4) ..1) with the inertia load 8. This diagram is referred to as the flexural lamination diagram. Nxy is zero and the buckling mode is sinusoidal. large D 11 /D 22 such as 3 or more).2 STIFFNESS DESIGN BY MIKI'S GRAPHICAL PROCEDURE 301 w= ~ ~n [o. ~ 2W~ 2 -1. (8.1) where Vm and V3n are given in Eq. I is the number of different ±9 groups. respectively.16) the Du's (in particular..3) - + (8.2.._ h) h 1 J.2W* 1 -1 -1::. n are the number of half waves in the x and y directions. = s 1 COS 28 1 + s2 COS 282. [±8]ns• with D22 = h3) (U 1(12 (h3) W~U2 + w. lj . For convenience in dealing with the frequencies.s 1• If we specify s 1 . However. Indeed. t = 0. we may get flexural lamination parameters inside the boundary.005 in.u3 ). (8.12pa (Omn. (~ .2.8) as • * 2 2 Q_ * na . In both cases we want to maximize the stiffness parameter S given by S = D.L with various combinations of n 1 and n2 • For desired values of the flexural lamination parameters w~ and w. we can usually find a large number of laminates with two or more distinct orientation angles. where Tl= 82 = 0.5 cos-l T.2. : w. and contours of constant S are straight lines on Miki's flexural diagram. S is a linear function of W~ and w. ± 'V4siW?.93 x 106 psi. we rewrite Eq. Example 8.5) and the points on the straight portion of the boundary corresponding to w.2 = 12 (U4- w. w..1) becomes D.1 r~ (8. Eq.6) where s2 = 1 .2.10) D 66 = (~. v 12 = 0.u3). that correspond to the interior of the flexural lamination diagram.* 4 = rt4h2 -Dum + 2(Dl2 + 2D66)m n b + Dz2 b ' 2s 1 w.4 in/sec 2 to obtain the mass density. and therefore to an [±8]ns angle-ply laminate.8) Consider now the design for minimum displacement under sinusoidal load. + W.8) with m = n = 1. (8.057 lb/in 3 .5 COS-I T2.2. Note that the weight density has to be divided by g = 386. This means that the laminate with the highest fundamental frequency corresponds to a boundary point on Miki's parabola. = 1 are cross-ply laminates [On 190n. (8. where 12DiJ D*. = cos 48 (8. = s 1 COS 48 1 + Sz COS 482. G 12 = 0. Eq. (b) Design the plate so that its fundamental frequency is 60Hz.. w. (8. + w.) (U 5 - w. The material properties are £ 1= 18.9) However. (8. the two angles can be obtained as 81 = 0.1.2. or the design for maximum fundamental frequency. = u~ }u. Assume that this second frequency corresponds to two half waves in the longer dimension.3.08 in.). Eq. if we specify constraints on several frequencies. + w. and the second frequency is 120Hz. E 2 = 1. (8. (8.2 2s 1 (2W?.u.7) Consider the design of a 16-layer 20 x 15 in laminated Graphite/Epoxy plate. The total thickness of the laminate is 0.2 STIFFNESS DESIGN BY MIKI'S GRAPHICAL PROCEDURE 303 Points on the parabolic boundaries correspond to designs with unique angle-ply laminates with only one lamination angle.s2) sl T = W* 1 -s 1T 1 2 s2 (8..=-h3 ..1 we have amn 42 ()2 ()4 (a) D.u. Therefore.s2w.2.5 x 106 psi.1.302 LAMINATE DESIGN FOR FLEXURAL AND COMBINED RESPONSE 8. = cos 28 and w. the Dij's are linear functions of W~ and w.2. and p = 0.4).89 X 106 psi.) 2 )4 (~J+ D.1. (a) Design the plate to maximize its fundamental vibration frequency.2. from Table 2. For a two-angle (/ = 2) laminate.u3 ). + 2(D. respectively). equal to the slope of the parabola w. are a 2. = 2 w. The maximum value of the frequency can be obtained graphically. \ = 12(0.0 Hz. for a precise value we note that.n ~ I''{ \<Ill 11 \.8).58 X 108.1197 and w.) x 108 • To maximize the fundamental frequency ro11 we need to maximize a.2 STIFFNESS DESIGN BY MIKI'S GRAPHICAL PROCEDURE 305 We first calculate the material constants U1 from Eq. '.2. or I . (a) for a 11 and Uz 1 . 6 U2 = 8.304 LAMINATE DESIGN FOR FLEXURAL AND COMBINED RESPONSE 8.. This. we get a 11 = (6.0762W~.8109W~. Substituting these expressions into the equations for the D~ and then to Eq. So the optimum design is a [±55. II 1t4 X 0.1. U3 = 1.37°.057 /386. However. There are many possible laminates. (b) which corresponds to a frequency of ro11 = 419. (a).) X 107.3821 6 X 106 psi.20).08 2 =2.. -1 . = -0.2 as the line for ro11 = 60 Hz and the line for ro21 = 120 Hz. As can be seen from Fig.0226 X 107. U5 = 2. Next.37]8s laminate. 8.5 Hz.4. we need to find a laminate that will have the flexural parameters calculated in the preceding. The fiber-orientation angle at the optimum can be found as 8 = 0. this gives us two possibilities..082 a2. which are contours of equal a. and calculating w. Fundamental vibration frequency on flexural lamination diagram..057 /386. (2. • wj O.4238. = 0.2708W.0. to 66.2708 Solving for W~ =4w. 8 1 and 8 2. W~opt = -0. does not indicate that the frequency is completely insensitive to the angle. That is.opt w. t: I rI .1 w~opt = 55.7490. (2. The reader can check that replacing the calculated angle with a more conventional angle of 60° would reduce the fundamental frequency by only a fraction of a percent. = 8. a"' and ro"' we get a 11 U4 = 2. For a 16-layer symmetric and balanced laminate.1." which is a linear function of W~ and Contours of equal fundamental frequency. 10 psi.5366 X 10 psi. 8.5 cos.2.724 Hz. W? - 0. [(±8 1)/(±82) 21. we consider the problem of designing a laminate with the first two frequencies of 60 Hz and 120 Hz (377 rad/sec and 754 rad/sec. The desired design is found at the intersection of these two lines.I~ 1~ 1 Iii I Iii I .4) X 204 X 7542 1t4 x 0. at the maximum.4617W.2 rad/sec.457 X 107. The corresponding values of a 11 and aw from Eq.9643 X 106 psi. an all-zero laminate has a frequency of only 43. (b) we get two straight lines. Equating these values to the expressions in Eq.3542. or 66.8109 1. for example. shown in Fig. 12 =120Hz a = 12(0.3253 X -1.4) X 204 X 3772 = 6 . Finally. with the aid of Eq." are straight lines on Miki's diagram as shown in Figure 8.1202. the line a. = constant is tangent to the parabola so that its slope is 1. for illustration we will consider a laminate with only two angles. at w~ = -0. however.2.7790 + 1.n =60Hz Figure 8.4. to obtain U1= 8. = (2.8943 X 106 psi. 62°.306 LAMINATE DESIGN FOR FLEXURAL AND COMBINED RESPONSE 8.578125. then z = 0.2.3 and stack 't + 1 in the laminate on the right. using Eq. a change in stacking sequence of a laminate requires corresponding changes in the thickness fraction of the layers that are permuted. Often it is desirable to obtain all of these designs rather than just the optimum. 2 ---layer 1 T2 = -0. these are small enough that all six buckling loads are within 2% of one another. we may have many designs that are very close in performance to the optimum one.2 Changes in Stacking Sequence Figure 8.7451.6) 3]s.2.2 STIFFNESS DESIGN BY MIKI'S GRAPHICAL PROCEDURE r-- 307 ~-. Shin et al. r-1 - .11) Exercise 3).J Then.7500 or T 1 = -0.7) we calculate the fiber orientation angles to be 91 = 20. and [±85.·J-(~·J 2 (8. (1989) have shown that. If we select the second laminate.0/ (±29.375h. z~ = z*.J-r. so that we always get a physically realizable result provided that the layer thicknesses are allowed to vary continuously. However.12) or ( . this condition is 3 3 + z~_ 1• (8. Furthermore. given a laminate with more than one orientation angle. Ply-permutation with constant flexural stiffness. T2 = 0.1) and (8. The figure shows the bottom half of a laminate [(±e1)NI(±91_ 1)N I· · · /(±9 )N L 1 with I different orientation angles.2. the stacking order of the layers with different orientation angles can be permuted arbitrarily without changing the Du's of the laminate.3)31. layert layer t+1 ]T ~J z•3 3 ZH 1 . We have thus proved that we can permute any two adjacent stacks. when we solve a stacking sequence design problem.2 under uniaxial compression.3.2. 8. (8.2) gives us s1 =0.-1 layer 1 s 2 = 0. All six laminates have the same Du's and therefore the same buckling load. Thus. lie on the boundary of Miki's parabola. such that W~ and w. the composition of a laminate with given flexural stiffness is not unique. However. these have been converted here to numbers of layers for a laminate with a total of 50 layers. For the angle associated with stack 't in the laminate on the left in Fig. layer I layer I - 8.96°. Substituting these values into Eq. 92 = 29.8) we obtain T 1 = 0. As can be seen from Example 8. The rounding introduces variations in the flexural properties. The process of permuting the layers is illustrated in Fig.2.2. (1989).·J~r.3. In the figure. z1 =0. For exam- It is easy to check that we get the same condition for the other permuted stack. layer t+l r z. z* always lies between z~_ 1 and zH 1 (see . (8.5113.2.0.2) we see that one can satisfy this condition if the value of si associated with each angle does not change.. we consider permu'ting the order of the angle in the 'tth stack with that of the adjacent ('t + 1)th stack. (1989) are in terms of volume fractions. We assume that all the other stacks remain unchanged. Table 8.2. remain unchanged. (8. (8. 92 = 69. From Eqs.z~ . The results reported by Shin et al. J ~· =-_! z1 +I layert =-_j That is.5h. z0 =0. The noninteger number of layers ni was obtained by maximizing the buckling load of a Graphite/Epoxy plate with an aspect ratio of 1.421875and 1 [±9 1 /(±92)3].. The numbers in parentheses correspond to rounding of the exact number of layers for the purpose of obtaining an integer number of layers for a 50-ply laminate.1. except when W~ anctw. 8. Obviously we can change the stacking sequence in any order by doing a sequence of binary permutations.9/(±69.. and Eq. the two laminates are [±20.2. and we want to find a new boundary between the two stacks.9845.1 shows results obtained for six laminates by Shin et al.92°.33° or 91 = 84. (8. .6). As long as we can calculate the value of the response for pairs of W~ and w. calculates first the Dij's using Eq. discussed in Chapter 5.1. /On/±45n.4. W is given as Figure 8.915 0.2 Design the plate of Example 8.1.. (a).2.. Optimum designs with equivalent D matrix Stacking Sequence n..)a w= ""' £. m1t . plane as shown in Figure 8. [On.ls [90n.1 we could get linear equations of the frequency contours and plot them on Miki's flexure diagram. wmnsm2sm 2..1±45n.f90n/ ± 45n)s [On.f0n. that satisfies the equation w =c. (8. we consider the design of a plate for minimum displacement. (a) values of m and n from 1 to 19 were used. it is not necessary that we have a closed-form equation for the contour in order to be able to plot it. Since the value of the applied pressure % does not influence the choice of optimum plate.. n1t £.ls 0.1 for minimum displacement at its center under uniform pressure.2. and with 100 terms the series has converged to six significant figures. for given values of W~ and w. can be used to yield such multiple optimal and near-optimal designs.2.10) and then calculates w using Eq.840 1. Example 8. In Eq.847 17. we can generate contours numerically and plot them.585 7. the designer may want to sacrifice a percent or two of performance if one of the near-optimal laminates has some desirable properties that were not included in the formulation of the original design problem. /90n/On)s [90n.2.265 (nz)" n3 10.119 17.2.3.848 3.610 4.785 3. where w mn is given by Eq. . nz 3.. For that purpose. .f90n)s [±45n. Our objective is to find the values of W~ and w.1.5) for the displacement.120 1. The contours for a given value W = c were obtained by varying W~ and solving for the value of w.500 3. which yields a value of 8. (n. Most of the contribution was due to the first term in the series. This approach is useful when the contour equations are nonlinear. /On/90. ple.)s [±45n.915 2. The figure indicates that the optimum is obtained on the boundary of the parabola that corresponds to an angle-ply laminate. we minimize W = (1t6 /16)(w I q 0 ) instead of w.2 STIFFNESS DESIGN BY MIKI'S GRAPHICAL PROCEDURE 309 (a) 308 LAMINATE DESIGN FOR FLEXURAL AND COMBINED RESPONSE - Table8.823 11.2.848 11.2.3. Contours of equal w were plotted in the W~-w.823 10.119 (n3)a (10) (18) (18) (12) (12) (10) m=l... ""' . 1v.785 2.SO. However. a subroutine was written that. The solution was performed with subroutine NEQNF of the IMSL library. At the center of the plate (x = a/2. for a total of 100 terms.! ±45n.8. that minimize winside the Miki's parabola. Flexural lamination diagram for Example 8.4. n=l. Genetic algorithms. As an example of this approach. (8. The optimal fiber orientation is calculated to be 8* = 56..3 Nonlinear Problems V3 In Example 8. y = b/2).500 (1) (1) (3) (3) (3) (3) (4) (3) (1) (7) (5) (2) "Layer numbers are rounded such that each laminate has an integer number of layers with a total of 50 layers. We use Eq. both the in-plane integrals VoA• VIA' and v3A and the flexural integrals Von• VID. + n. -h/2 k=l h/2 N/2 n (8.Jf. we will need a constraint that will prevent more than one of these variables from being equal to 1 simultaneously. the change in critical buckling mode with changes in lamination sequence [that is.f ~ . Haftka and Walsh (1992).(k-1) ](o. 2 t N/2 Von=3 k=l 3 LPk N/2 [( zk ] t _ ('~') ~ 2. Of course.8)]. For this reason. 45°.4) fr 2t vm =. A variable o. It is because the first term in Eq. will be described by f ~ == fi =o3 == o4 = 1. yielding a value of w == 45.. and ±45° layers.nk). and V3n [see Eq. we treat stiffness and strength design problems with discrete angles by the integer programming techniques described in Chapter 5.1.).ff.') ]~ ~ 3 3 N/2 2 . Furthermore.3) 8.4.A =Jcos 29 dz = 2 t L (ok. and/~. 90°. n. (a) dominates the series.. for buckling design. the flexural stiffness matrix D is not.'I [k'.1 Ply-Identity and Stack-Identity Design Variables For the flexural response. n.05 in 3/lb.' I k=l [k' _ (k -l)'](o. ] N/2 3 k-1 (8. As we will see later. or is equal to 1 if there is a 0°. . rather than in accordance with the convention that lists the layers from the outside in.3 - 3 ~ p. . Therefore. The closeness of the optimal angles for the maximum frequency and minimum deflection laminates is no coincidence.).3. As in Example 8. -h/2 k=l (8.1 0)] creates further complications.5) . 8.1. (2. the matrix D can be made a linear function of the design variables if we use zero-one ply-identity design variables. the integrals Von• Vm.310 LAMINATE DESIGN FOR FLEXURAL AND COMBINED RESPONSE 8. . with the other twelve variables associated with the laminate being zero.3. -h/2 k=l (8.3. N/2. where i = 1 . in the ith layer. see Miki ( 1986). +fl +J. and V3n are expressed as 3 3 While the in-plane stiffness matrix A is a linear function of the number of layers of a given orientation. Thus. if the laminate is made up of 0°. While Miki's method is applicable to such problems. 90°. the stacking sequence can be defined in terms of a set of four ply-orientation identity variables. [ ( t) _[zk..2. the laminate [0/±45].. (8. that are zero-one integer variables.when using a somewhat similar but more conventional laminate of [±60°] 4s is minimal. or -45° layer.n.1) v. it is convenient to number the layers so that the first one is nearest to the plane of symmetry of the laminate. Vw and V3A are given in terms of the ply-identity variables and the thickness of a single layer t is given as h/2 N/2 VOA = Jdz = 2 t L (ok + nk + + f'k). respectively.3 FLEXURAL STIFFNESS DESIGN BY INTEGER LINEAR PROGRAMMING 311 w == 44.35)] are linear functions of the design variables.89 in 3/lb. (8. However. o.3 FLEXURAL STIFFNESS DESIGN BY INTEGER LINEAR PROGRAMMING In many applications.1.f 'k). For example. the loss of performance . In this section we focus on problems that can be formulated as linear integer programming problems so that they are readily solved by commonly available software. . coo ze. (8. and this term is proportional to the square of the frequency [see Eq. the points on the diagram corresponding to the discrete set of angles are not as simple to obtain as in the case of in-plane stiffness design.3.. The integrals V0A. the change in m and n in Eq.2) h/2 N/2 v3A = Jcos 49 dz = 2 t L (ok + nk .3. the layer orientation angles that we can use are limited to a discrete set. ~ith these zero-one ply-identity variables.3. the A and D matrices are linear functions of these variables. LPk cos 4ek k=l 3 ( 4 t J . with ! 1 = o 2 = 1 and o 1 = n 1 = f 2 = n 2 = 0. o.3.6) VoA = 4 t L (ok + nk + fk). The use of stacks constrains the solution. (8.3.312 LAMINATE DESIGN FOR FLEXURAL AND COMBINED RESPONSE 3 8.9) where ff andfT do not appear in the expression for V1A and Vm since the cosine of 90° is equal to zero.2 When layers empty dition Stiffness Design with Fixed Thickness the total thickness of the laminate and hence the number of are given. Therefore.. For example. The variable pk in Eq. k=l (8. ±e. k=l N/4 (8.1) ](ok + nk + fk).12) . the laminate [0i±45L will now be described by 6 design variables instead of 16. so that Eqs. k=l (8. One strategy that may be used to reduce that number is to replace the ply-identity design variables with stack-identity design variables that control a stack of two or more layers instead of a single layer. With two-layer stacks. (8. 90°.4.3. we use two-layer stacks also for 0° and 90° layers. the stacks are ±45°.3.3.(k.fk). empty layers will be used to reduce the laminate thickness during optimization. The use of stack variables as just described simplifies somewhat the equations for the in-plane and flexural integrals. It is also convenient to have all stacks the same thickness.. (8. The most natural stack for balanced laminates is a stack of two layers.3. [k . 8.. However.3. Constraints must be applied during the optimization to ensure that pk can be zero only for the outermost layers. and n. and ±45° layers.4)-(8.('~· J 3 l (8. The ply-identity design variables have to satisfy the conthat for each layer at least one of the variables is equal to 1 .3 FLEXURAL STIFFNESS DESIGN BY INTEGER LINEAR PROGRAMMING N/4 313 2t N/2 [ v3D = -3.3...6) is unity if the kth layer is occupied and is zero if it is empty: Pk = 0 k + nk +if+ IT v3A = 4 t L (ok + nk.3 £.1)3](ok.£.16t3 ~ k3 3 V3D.3.(k.6) become (8. 1.. V1A = 4 t L (ok.8) = 2 t IWN/2 k=l N/4 (k-l?](ok+nk- n -JT).nk). For a laminate made up of 0°. 3 k=l N/4 As we will demonstrate in Section 8.11) VID = 1~t3 N/4 L W.3. k=l (8. respectively. The use of ply-identity design variables has the disadvantage that it creates a very large number of design variables.3. especially for laminates with large number of layers. k=l N/4 (8. and so we may obtain a suboptimal design... we need only N/4 sets of stack-identity design variables and each set has fewer variables because positive and negative angles are lumped in a single stack. and the corresponding stack-identity design variables associated with the ith stack are denoted by o.7) and 16t3 ~ 3 3 Von= .[ -(k-1)](ok+nk-jk).10) -:. the difference for the case of two-layer stacks is usually small. we do not have to worry about the possibility of layers.3.. and 90.f.3.nk).13) The simplification associated with the use of stacks is not without a price. 105 D~ 2 = 2.1 with the added condition that the orientation angles are multiples of 45°.0 - D ~ 2 = 0..n3) + 37(o4 . 90°.12) and (8. (8. .0. and (c). (8. Eqs. 0 .13097 106 -0.30692 X l05 106 .n4 ) .n 1 + 7(o2 .. With a 16-ply laminate.3253 D.1). 0 3Vm * t =64 W 1 16 3 and v3n.2.n 1 + 7(o 2 . f 1 + n 1 + 7(o 2 - j 2 + n2) + 19(o3 - f 3 + n3) (a) + 37(o4 .j 2+ n2) + 1~0 3 . .j 3+ n4).3. k = 1. 2 such that o 1 . ok.8943 X 106 -0.30692 X l0 V..3.3. D~ 1 = 8. .2.+n.1 Solve Example 8. 5 10 V~ 0 + 0.30692 X V. (8. i = 1.30692 X 10 6 To demonstrate how we use the stack-identity design variables.3. For our problem.314 LAMINATE DESIGN FOR FLEXURAL AND COMBINED RESPONSE 8. 8.V~n = 0.v. Instead of V 10 and V30 . given by Eq. 01+!1 +n1 = 1. instead of the fundamental frequency ro11 • For the plate considered in the example..5366 x 106 -0.+.n2) + 19(o3 . If we use two-layer stacks.13097 106 .'s in Example 8.. o1 - all= D~~ + [ ~ )cn~2 + 2D~6) Dij=J.D and w.J· 12Dij 3 2 6 +[ 1 i )n. X l05V. 0 . Example 8. .30692 X 5 - D~ 1 = 0. N = 16. (c) Using Eqs.D = 01.14) Next. 0 106 V~ 0 + 0.30692 X 10 V. v.13097 X 10 6 V~ 0 + 0. D ~6 = 0. X X l0 5 V.13097 X 106 V~ 0 + 0.13) give o3 +!3 + n3 = 1.0 .2.fk.= 1.3253 X 106 + 0. the corresponding condition is Dij's using the values for the 5 o. this condition is o. For our laminate. we maximize the frequency parameter a 11 .j 1 + nl + 7(02.0.15) (b) 1.3253 8. so that we have N/4 = 4 sets of stack-identity design variables.8943 X where 106 + 0. alb = 4/3 so that our objective function is X X V.0 .30692 X V. the following mixed integer linear programming problem can be formulated: maximize 81D~ 1 + 288(D~ 2 + 2D~ 6 ) + 256D. (8. see Eq.j3 + n3) + 37(04. we write Eq.3.3253 2. (a) in that example.2. made up of and ±45° layers.. .3 FLEXURAL STIFFNESS DESIGN BY INTEGER LINEAR PROGRAMMING 315 I and the others are equal to 0.0 - D. As in Example 8. Then add the condition that the laminate is stiff enough in the x direction by requiring A 11 to be at least 50% of its value for an all-zero laminate. (b). ..1: (8. we consider an example of frequency maximization.D = Q.0.2 = 0. (a).n2) + 19(o3 .5366 2.2. 2 = 8.j3 + n4). where we have used h = 16t to establish the relation between v~D and w~ and between v.+n.n4).16t 3 * _ 3 v3o = 64 w.n 3) + 37(o4 .10) for the U.+Jf+f~= oo. . i=l.f. V ~D = o 1 .30692 x 1osv. o2 + n2 + !2 = 1.. it is convenient to work with Vm= * X 106 . 4.1.N/4.2. D~ 6 = 2. and nk. N/2. we replace v.~Dm and v. = 0..1. The solution obtained was o 1 = f 2 =f 3 = f 4 = 1.3.3 FLEXURAL STIFFNESS DESIGN BY INTEGER LINEAR PROGRAMMING 317 +!4 + n4 = I. = 0.A-h + n. A.1. = V3A/h.336 x 106 or 8. which corresponds to an all ±45° laminate. For an all-zero laminate we have o 1 = o 2 = o 3 = o = 1.1.25 h (0 1 + 0 2 + 0 3 + 0 4 . n). we use Table 2.9643 v. Note that the addition of a zero stack near the centerline changes A 11 substantially. with the 4 other stack-identity variables being zero.*= min Acr (m.1 to write A 11 To solve the constrained problem. The LINDO program. The all ±45° laminate is rather soft with an All of only about a third of an all-zero laminate.10) that correspond to the lowest buckling load are not known a priori and depend on the design.* is not a linear function of the ply-identity or stack-identity design variables. + n2 + n3 + n4). ~ 9.25 h (o. These additional constraints were added to the original problem and solved with LINDO. + U3 v. we need to add the constraint of Eq. (8. for the present problem. n) is a linear function.10) we get.n) is given by Eq. This problem was solved with the LINDO program (Schrage.. even though \r (m. being of indefinite sign. using h VIA= = I6t.D by v. and then the COnStraint On All becomes A 11 1h = u. The fundamental frequency for this laminate is 65.n3 .Dp. with the other stack-identity design variables being zero. It is convenient to USe v.fz. so that V A = V A = h.1). Fortunately. However. 4V. To circumvent this difficulty. the variables v.2. That is.n =0. m. The reason for the difficulty is that the wave numbers m and n in Eq. if we use the . (8. assumes that all variables are positive. o1 +o2 +o3 + o4 - n1 - n2 . the standard approach is to replace each variable of an indefinite sign with the difference of two positive variables (see Section 5. (8.Dp. 0. 1989).672x 106 h lb/in.3. we have a standard approach for resolving this difficulty.011.377 Hz. Therefore.* as an additional design variable to the problem. That is. We want to maximize A.v.J.3821 (8. (d) along with two equality constraints that come from the definition of v.086 Hz.. + U2 v..h + U2 v.0 can be negative. (d) where Acr(m.9) and (8.n4) and v3A 8.10).3 Buckling Load Maximization with Stiffness and Strength Constraints Optimization to maximize the buckling load of a laminated plate is more complicated than optimization to maximize the fundamental frequency.n. For example.16) v. = V1A/h and v. + 1. the laminate is [(±45)/0 2 ]. but has very little influence on the Dij's (because the contribution of a layer to the flexural terms depends on the square of its distance from the centerline). and v.3. which is about 2% lower than the optimal design with unrestricted angles found in Example 8. + 02 + 03 + 04. The fundamental frequency ro11 of the laminate was reduced by another half a percent to 65. To formulate the condition that All is at least 50% of that of an all-zero laminate.Dm· The solution obtained by LINDO is / 1 = J. v. - n2. 1 3 and All= h(U1 + U2 + U3 ) = 18. Again.n4 - 4V.w From Eqs. However.3. like many other linear and integer linear programming codes. It consists of adding A. = f 3 = f 4 = I.316 04 LAMINATE DESIGN FOR FLEXURAL AND COMBINED RESPONSE 8.* given as A. by 0 v. o1 + o 2 + o 3 + o 4 - / 1 -/ 2- f3 . had to be replaced by the difference of positive variables.n3. 0 and lf.f4 + n 1 + n 2 + n 3 + n 4 - = u.A + U3 V. ~ 1. and v. by adding the zero layer near the centerline we are able to change the inplane stiffness with minimal effect on the fundamental frequency. = 1. The linearization is performed in terms of the elements of the in-plane stiffness matrix.. Once the laminate mid-plane strains are expressed as a linear function of the design variables.. 0 (8.20) can now be added to the problem formulation of Eq.. The stress resultants used in the buckling load equation. Equation (8. which are already linear in the ply.J.* Ei ~fan. = 1. In the case of zero-one ply-identity variables.N/4.3.3.17) also allows us to incorporate the stiffness constraints discussed in the previous section. and can be solved by using software such as LINDO.. By moving /.fall< 2fall A .318 LAMINATE DESIGN FOR FLEXURAL AND COMBINED RESPONSE 8.3. (8.* is used as one of the design variables in the formulation described in this section.3. are simply initial reference values. A. (8. stress constraints can be obtained from the principal material direction strains by multiplying them with the transformed reduced stiffness matrices of the appropriate layers.18). is still incomplete. . the maximization of the buckling load will be formulated as (Nagendra et al.17) for designing laminates that have maximum buckling load and are strength-failure resistant. constraints on principal material direction strains may be formed by using the strain transformation matrices. oi + ni +J. N/4.. + .. then the constraint is expressed as.18) i = 1. . (8. 1992) find to maximize o.and stack-identity variables. In using sequential linear programming.. The actual values of the stresses are dependent on the buckling load factor. n1. it is also possible to incorporate strength constraints into the present formulation. and expanding 1/A.. n). The minimization over m and n is performed by checking for all values of m between 1 and m1. The magnitude of the strains and stresses in the laminate depend on the magnitude of the applied loads.* ~ Acr (m.. which are mostly used to specify their ratio with Since Ei is a function of the design variables and A. . It was demonstrated in Chapter 7 that strength constraints expressed in terms of laminate mid-plane strains may be approximated by linearizing them. (8. such that /. So the optimization problem of Eq... A final implementation note is for the method of solution. This requires addition of six more constraints to the problem at each step of the sequential integer linear programming solutions. . a straight use of integer linear programming is not advisable as the quality of the approximation may deteriorate away from the nominal point. A* m respect to one another (say. ni. . i= 1. Since the formulation involves a local approximation for the strength constraint..* Aof . (8. 0 The linear strain constraint of Eq. .1...3. m1 . N/ Nx = 0.3 FLEXURAL STIFFNESS DESIGN BY INTEGER LINEAR PROGRAMMING 319 two-layer stack design variables corresponding to the 90.10) for \r is linear in the Dij's.20) where A is the buckling load factor for the nominal design.Ao). The incorporation of the strength constraints into the buckling maximization problem.3. The formulation presented in Eq. 1 (-1] 1 (8. if the value of a strain component corresponding to the reference values of the in-plane stress resultant is Ei for the ith constraint. and its limiting value is Ea11. and ±45° layers. the above constraint is again nonlinear.10). n = 1.* in a linear Taylor's series.17) /.18) can be written as. The standard procedure recommended in this case is the use of sequential linear programming..5). (8. Eq..Ao + A~ (A*.. as we would prefer the panel not to fail due to strength before the buckling load.3.19) A. and all values of n between 1 and np where m1 and n1 are selected to be large enough to include all possible critical buckling modes. An alternative approach is to apply bounds on the extensional stiffnesses Aij' expressed in terms of the ply-identity variables. (7 . however. which are linear functions of the stack-identity design variables.3.1.. This is simply accomplished by expanding the elements of the in-plane stiffness matrix in terms of the ply-identity. imposing move limits on the design variables is not practical.1 ).*. _!__ /. (8. . . If desired. (8.* and oi.* to the right-hand side of Eq. the constraint equation of Eq.3.3 .. or stack-identity variables.17) is an integer linear programming problem. . move limits on design variables are generally invoked so that designs generated based on approximate constraints remain in or near the feasible design space. is reached. That is. Moreover.3. see Eq.. (8. the quantity being formulated for maximization. N/Nx = 0._j.::t ~0 g-~ to :.~n=0.2~-ca e. Ol Ql Ol Ql Ol Ql .:::. u::: .:::.89 X 106 psi. ~ ~ ~ ~.. and 90° orientation angles.0 :::l "' ~ 0 ~ "' E :::l E ·x "' 0 O>LOroo 0 ~ LO c ~~lil - . 8. The laminate is to be designed for two combinations of in-plane stress resultant ratios... lO Ill c.93 x 106 psi. "'~ Ill c.£ '-' c: v"":"ii5E ~8l § ~ om o .2.3.25 the failure load factor decreased from 13441.cz ::J .:-o ~85-ffia Ol ~ -~ ctl 'E iii c 0 (.. .:OL!) +IT"""g'c.~ "9"9"9 0"' 0> "d' Example 8.. Design results for the 48-ply laminates under two different combination of biaxial loads.3. a>"":2E LO co ·ca ::s va>-E 8ro~o 8 -~ COl Q5 -o c ctl Q) .c g . v12 = 0. are presented in Fig.1 :::JO ~ e_ ~ 321 . (8.@ tiic Cii ~ cOl a~~ E ~~~E 'OJ~ +I -.. ±45°... The first one represents the laminate for maximum buckling load obtained by sequential use of integer linear programming with no constraints. "C. ~(")-~0 ex) tri l!! :::J C) ci g oti o ". and t = 0.) c_J == ti ::J ::J ."''Cil ~ c .2 A a = 20-in long rectangular Graphite/Epoxy laminate with an aspect ratio of alb = 4 is to be designed to maximize its buckling load under combined longitudinal and transverse loads.25 and N/Nx = 0. E va>-E lncO~::s Ol '(ij Q) c ~~-iij·a ou.£·. The failure strains in the principal material directions are E~n=0. The reference value used for the longitudinal stress resultant is Nxo = 1.21) The results of a design study formulated using the approach described in this section are summarized in the following example taken from Nagendra et al. G 12 = 0.c . ci 1i _o:g o o ~ 0> "' LO 0)0') II .§::s ~~u . Since the method used involves local approximations.5 schematically.c "' ~ -~ 0 11 LO ~ USc (/) to "' '@ tlc c Ol "' c C> c iii ~ Q~~o 'N 5 . For each of the biaxial load combinations considered.25 and N/Nx = 0.A l l!j< 0' - i.. Compared to the design without strength failure constraint. Designs with a higher confidence of being globally optimum can be generated by using the genetic search algorithm discussed in Chapter 5.2 .~ 0 .0%) because of the inclusion of the strain constraint. The second one is the maximum buckling load design with the strain constraint.q.!: . Inclusion of four stacks of oo layers was responsible ~11 C\1 (j) (j) ~ 0 N -~ Ul C "' c en "' c .and 1:~=0.c ou. 011~0 +I C\1. the final design may be a locally optimal design. The elastic properties for the layers are £ 1 = 18..3.5. The laminate is to have a 48-layer balanced symmetric layup with oo.j=I.Q 'OJ Ol(!J 0 c 0') == -"' 1.:::.0 lb/in so that the computed value of the A at any step corresponds to the buckling value.. C.320 At-Au$0 LAMINATE DESIGN FOR FLEXURAL AND COMBINED RESPONSE and A lj .005 in.85 to 12622. £ 2 = 1.!: 32 Ol (. C.:-o "0 LOII'iilO o 'OJ l{) 11_0"00 'N co 0> roz Ol- - ~ == 't\ c-' ~-0~0 "' co ~ ~~ 'N ~ ~ ...015."' ·- c Ol Q) 1! _g. the last design is generated using the genetic algorithm discussed earlier and verified to be actually the global optimum design for each load case..5 x 106 psi. N/Nx = 0. Finally.029. for NY= 0..0 +I ro . Vl CQl ctl ~ 0N E ~.008.5 with strain constraints in the individual layers.::JO "' .) Cll - .~ "' "' "' .f: 0 Ole.44 (by about 6...§ C\J 'Nu:. ~ -~(9 .cz . three stacking sequences are shown in Fig. (1992).Ql 0~ t! 8"rJ O>~::JE '§ Cll £ C1ro~O ~C\i~E ~~~ oC\1.c .!: ·..5. 8. This means that we need the next higher available thickness.25) 8. . will be a 20-ply laminate. so is the (i . If we do not have some additional requirements that will help us choose between all the possible laminates. the dependence on h.23) which require that if the ith stack is occupied. That is.46 MN/m. . (8.N/4. (8. However.1. Although the total number of stacks with different orientations was identical between the global optimal genetic algorithm design and the integer linear programming design. is approximately proportional to the cube of the thickness (it will be exactly proportional to the thickness if we can preserve the value of the lamination parameters W~ and w.3. being linear in the Du's. therefore. the case of minimizing the thickness of a laminated plate subject to the requirement that it does not buckle under an Nx = 0. which.3. which require every layer to be occupied. For the load ratio of NY= 0. What we have to do is to select N as an upper bound for the number of layers (for example by finding a design that satisfies all the constraints). the square of romn in Eq. but instead replace this equation by oi+ni+. i=l (8.h5. with two-layer stack design variables.1. For design subject to constraints on frequencies. the load factor was fractionally smaller. we can expect many 20-ply laminates to be feasible. i= 1. was globally optimum. The thickness minimization problem is more complicated because we do not know the value of the number of layers. Assume. with a 20-ply lami3 nate we can expect a maximum buckling load of about 0. Therefore.22) thus allowing all the stack-identity variables to assume zero value for a given layer..3. The buckling load of the laminate.8) is no longer a linear function of these design variables.3 PL!XUFIAL STIFFNESS DESIGN BY INTEGER LINEAR PROGRAMMING 323 for the decrease of the buckling load factor.25 = 1.N/4.4 Stiffness Design for Minimum Thickness Instead of maximizing the frequency or buckling load of a laminate of given thickness. Because of The left side of Eq... we can replace a constraint on the frequency such as (!)mn ~ (l)req (8. (8. furthermore.h_" i=2. (8. . Consider. The design obtained from the sequential integer linear programming approach was identical to the genetic algorithm design and.3. (8. we also have to be careful in formulating the constraints so as to preserve linearity of the constraint in term of stack-identity design variables. we should at least try to obtain the 20-ply laminate with the highest buckling load.1)th. we often need to solve the problem of minimizing the thickness of a laminate subject to a limit on the frequency or buckling load. Compared to the genetic algorithm design. which was verified to be the global optimum stacking sequence. Such a laminate will satisfy the constraint with the highest margin.3.3. (8.3. a slight variation of the locations of the 0° and the 90° stacks was responsible for the small difference of the buckling load factor. Another problem associated with minimum thickness design has to do with nonuniqueness of the solution. N.322 LAMINATE DESIGN FOR FLEXURAL AND COMBINED RESPONSE 8. for example. we may prefer the design that satisfies the constraints with the largest margin. we may have multiple designs of the same thickness that satisfy all the constraints. We can no longer use equations such as Eq.25) is a linear function of the Du's [see Eq.3.15). This can be achieved by subtracting from the objective function (thickness) a small constant times the buckling load.5. the design without the strain constraint violated the shear strength in the 45° layers by 7%. h ro~eq =4 t (!)~ L (oi +h + n)..).5. With such a large margin between the maximum buckling load and the required buckling load. The following example demonstrates the use of this technique. the optimum buckling load is 0. .75 MN/m. This equation has to be supplemented with oi+ni+f.4). (8. to use in equations such as Eq.8 MN/m.24) by N/4 h ro~n?.8)] and therefore of the stack-identity design variables. In such a circumstance. These latter conditions eliminate the possibility of ending with designs with empty stacks in the interior of the laminate. 1. that if we optimize a 16-ply laminate for maximum buckling load. .75 x 1.oi-l +ni-l +. ) = 1.. (~) 4 l (a) ro0 =50 Hz = 314. D 66 =(~.1. this translates to 20 layers. We calculate the material constants Ui from Eqs.25) may be written as __ z h-4. we need h = 0.z = }u Using Eq. from Eq. and defining ~ = roj 1 h. (8.20) to obtain U 1 = 8. we choose an all-zero laminate. D.057/386. v12 = 0.D 11 = 0.3253 X Then. For a symmetric and balanced laminate. For convenience. + U + U 2 3) = 1.093 in or 18. (8. (8. we use Table 2.7500 x 1Q4 h3 lb-in. o.. Next. + V.4. or with stack design variables to N/4 = 5 sets of stack design variables. 0. With =:a: [D.3 Design a minimum thickness laminate for a 20 x 15 in Graphite/Epoxy plate so that its fundamental frequency is at least 50 Hz. = 1. 106 V0n + 8.8) we get = 537. t = 0.4) as o 1 + n 1 + f 1 + 7(o2 + n 2 + f 2) + 19(o3 + n 3 + f 3) + 37(o4 + n 4 + ! 4) + 61(os + ns + fs). 98696h.3821 106 Vm + 1. }u.1. we write the equation for V0n from Eq.4.6 layers. the optimization problem is formulated as minimize h 4 - U 3) = 4. 6 . U4 = 2. so that W~ = w.3.127 _ [ D + (32) (D +2D )+ (256) D 22] ~98.16 rad/s.8) as h roj. we specialize Eq.9643 X 106 V3n. We rewrite Eq.U C011 such that = 7. and p = 0. The optimum laminate is guaranteed to be no thicker. (~.8) and (2. U 3 = 1. (2. + Oz + n2 + !2 + o3 + n3 + f. + !. Eq.25) to our case. (8. G 12 = 0. (8. with similar equations for V 1n and V3n. Us= 2.7689 X 104 h3 lb-in.324 LAMINATE DESIGN FOR FLEXURAL AND COMBINED RESPONSE 8.10).5366 X 106 psi. o 1 + n 1 + f 1 + 7(o2 + n2 + f 2) + 19(o3 + n3 + ! 3) + 37(o4 + n4 + f 4 ) + 61(o5 + n 5 + f 5 ) .8943 X 106 psi.93 x 106 psi.4 into Eq. (a) and equations of the type of (b) and (c).5 x 10 V0n = 0.74 h Hz.5 x 106 psi. we calculate = U 1V0n + U2 Vm + U 3 V 30 X D 11 = (~.3. U 2 = 8.+ 2(D.89 x 106 psi.057 lb/in 3 .3.9643 X 106 psi. ~~ 3) Substituting these and p = 0.5 x 106 Von= 0.5896 X)(}' h' lb-in.3253 X Next. We can do that by selecting a single orientation angle and determining how thick this angle-ply laminate needs to be in order to satisfy the 50 Hz requirement. So. + n. The material properties are £ 1 = 18. (c) D.1.1. + o4 + n4 + !4 + Os + n 5 + f 5 . As a first step.3.2.}us.05095 [81D 11 + 288(D 12 + 2D66 ) + 256D 22] . + 2D (~) 00) 2 + D.50h = 0. for a frequency of 50 Hz.5560 x 106 h3 lb-in.005 in. WI 66 12 11 1 81 9 Next.V. we need to estimate an upper bound on thickness of the laminate by finding one laminate that satisfies that 50 Hz requirement.~ = 0.3821 X 106 psi.3.1 to express D 11 as D 11 or 8. = u~ }u. (b) 106 psi. (8. £ 2 = 1.3 FLEXURAL STIFFNESS DESIGN BY INTEGER LINEAR PROGRAMMING 325 Example 8.696h. The first was associated with the scaling of the variables. and II= 1o-5 ~. 8.3.326 LAMINATE DESIGN FOR FLEXURAL AND COMBINED RESPONSE 8.3821 x 6 106 V 10 + 1. +/.9643 ~ 1.4 Hz.. the formulation takes the form find i o.3. The second problem ables L (o..29) vlD 1.9643 2.n 3) + 37(o4 . i = 0. m = 1. n. .D 11 = o. laminate with a frequency of 55. n./ 4) + 61(o5 + n5 . mf' n = 1.28) such that Acr (m. o5 + n5 + fs ~ o4 + n4 + f4. but it is small enough so that increasing h by adding another layer increases the objective function even though it increases the frequency.n5) .0 Hz. to minimize When we attempted to solve this optimization problem with the LINDO program (Schrage.1. (8. the objective function was changed to h. Since V 10 and V3n can take negative values. .. + n. 'nf. n) 'A*~ ~'A*. (8. and ~ = roi 1 h was of the order of 10 5 • LINDO warned about possible ill conditioning associated with such scale differences.f.26) 0 3 + n3 +!3 ~ 0 2 + nz + fz. the Du's were of the order of 10. N/4. are binary integers.3253 X V3np ..n 2) + 19(o3 .1. corresponding to a [90/06 ].. and').V!Dm' V3n := / ) 3 + 37(o4 + n4 . This time the solution from LINDO was / 1 =f 2 =f 1 =f 4 =I or [±45]4s laminate with a fundamental frequency of 65.3. m.8. 1989).... X + nl +!1 + nz + fz 02 ~ 01 + nl + f1.f. o1 + n 1 / 1 + 7(o2 + n2 - / 2) + 19(o3 + n 3 - was associated.fs).3253 X 106 V0n. . = 1..E'A* i=1 (8. The second term gives designs with higher frequencies a lower objective function. (8.3. To check whether we can obtain a solution with the same thickness but higher frequency.D 22 = o.0. 106 V3n.V3Dm· 10 V0n + 8. . n) ~ 1. . two problems surfaced.n 1+ 7(o2 .*. The problem was solved by using scaled vari= 1Cf ViD.5 X 106 V3D = 0.5366 X 10 Von. 2. The solution obtained was o 1 = o2 = o3 = n4 = 1..1. For the problem of thickness minimization of a laminate capable of sustaining a specified loading Nx and NY without buckling.3.. If we want to get the minimum thickness laminate with the largest margin.27) o. .3821 X 106 V 10 + 1. the formulation is similar. as in Example 8..0001fj. N/4 i = 1.9643 x 106 V3n . .9643 x 106 V3n. .8943 01 6 X 106 V3n. The integrals such as V0n were of the order of 105 . they had to be replaced by the difference of two positive variables V!D := V!Dp .1.n4) + 61 (o 5 . The buckling constraint will take the form of min Acr (m. 8. 04 + n4 +!4 ~ 0 3 + n3 + f3.5 x 106 V10 = o..1. with the assumption by LINDO that all variables are positive.D 66 = 0..). 5. 1' 3..D12 = 0.n X 106 V0n.3 FLEXURAL STIFFNESS DESIGN BY INTEGER LINEAR PROGRAMMING 327 o 1. 328 LAMINATE DESIGN FOR FLEXURAL AND COMBINED RESPONSE REFERENCES 329 oi + ni +!.2. Loaded Symmetric Composite Plates. (eds. and Gtirdal. Philadelphia. R. User's Manual for LINDO.~oi-l +ni-l +. R. 348-369. L. Japan-U. pp.. and Gillespie. R. and Kobayashi. (eds.) American Society for Testing and Materials. ASTM STP 864. M. T. CCM-III. R. "Importance of Anisotropy on Buckling of Compression.. S. 4. pp. T.) Proc. M. Martin us Nijhoff Publishers. Redwood City. Fourth Edition. (8. "Design of Laminated Fibrous Composite Plates with Required Flexural Thickness. Prove that the solution of Eq.2. M." Recent Advances in Composites in the United States and Japan. "Stacking Sequence Optimization of Simply Supported Laminates with Stability and Strain Constraints." AIAA J.. 5.1 is to be designed so that the fundamental frequency is at least 10 Hz.1 for maximum buckling load under biaxial loading with N/Nx = 0. Vinson. Miki. Delaware Composites Design Encyclopedia. and the second frequency is as high as possible. Scientific Press. 1831-1835.001) used to make the optimization procedure favor laminates with high buckling load over laminates of the same thickness with lower buckling load. T.. The Behavior of Structures Composed of Composite Materials. 814819. S.3. J. Kedward.30) and oi + ni +!. Watson..25._ 1. E is a small number (of the order of 0. 2132-2137.01 or 0." Composites '86: Recent Advances in Japan and the United . and Sierakowski. Lancaster. Kawata. 387-400. Y. 2. Nemeth. 24(11). 3. REFERENCES Haftka. Use Miki's diagram to find the stacking sequence of the optimum laminate. Nagendra. 30(8). M." AIAA J. (1986). (1992). (1985). and Taya." J. T. ~ 1. T. (1990). J.S. (1989). K. "Stacking Sequence Optimization for Buckling of Laminated Plates by Integer Programming. CA. Design Studies. ..2.) Technomic Publishing Co. (8. Haftka.f.2. and Plaut. R. J. W.. R. Umekawa.12) satisfies z~_ 1 ~z* ~zH 1 • States. Shin. (8. H. Composite Materials 23.3. Haftka. (eds. "Optimum Design of Laminated Composite Plates Subject to Axial Compression.31) Here." AIAA J. The plate of Example 8. (1989). Vol. Design the plate of Example 8. Z.. i = 1. 673-680. (1992). Repeat Example 8.2 for a 15 in square plate. 30(3). A.3. Carlsson. Schrage. Find the optimum design for Example 8. (1986). L. K. P. PA. and Walsh. . R. and Whitney J. Dordrecht. S. A. (1987). Tokyo. "Design of Laminated Plates for Maximum Buckling Load. L. EXERCISES 1. N/4.. Vinson. 5.3 without the constraint of using two-layer stack design variables. M. Miki. J. L. L. 106 Bending-twisting coupling. 180. 246 Constituent properties. 54. 293-294 Boron fibers. 207. 301 Balanced symmetric laminate. 328 Buckling load maximization. 55 Bending -extension stiffness. 83. 49. 324 Angle-ply stacks. 94. 18. 7 fibrous. 8 Boron/Epoxy. 8 Compressive failure. 252. 4-5. 129. 298 Carbon fibers. 58. 232. 74. 156 Anisotropic materials. 50. 135 Binary string. 238 ultimate. 91. 232. 68. 119-120 Aramid/Epoxy. 6 short fiber. 214 Binary design variables. 256 Compressive strength. 43 Constitutive equations. 254 Coefficient of thermal expansion. 227 Constituent materials. 317 Buckling mode. 103. 12 Constitutive behavior. 70. 6 laminated. 90. 287 Bending of laminates. 189. 179. 91 Average laminate stresses. 245. 49. 87. 4 Composite strength. 162. 47. 138. 246 Angle-ply laminate. 327 Buckling load. 7 flake. 11. 301. 8 particulate. 204 Balanced laminate. 310. 242 Compliance coefficients. 125. 298 Binary coding. 300. 195. 74. 298. 128. 128. 18. 11 Branch-and-bound.INDEX Active constraint. 153 Combined stresses. 309. 86. 2. 61. 208 Bending-extension coupling. 87. 127. 172 Bonus parameter. 169. 212. 292. 45 Averaging crossover. 11 AS4/3501 Graphite/Epoxy. 318. 102. 192-193. 79. 318 Buckling of a laminate. 33. 128. 223. 87. 185. 214 Bending-extension coupling matrix. 246 Brittle materials. 20. 234. 233 Composites fiber reinforced. 210. 203 Binary variables. 244 B( 4)/5505. 172. 103. 226. 236. 74. 300. 8 Carbon/Epoxy. 20. 52 Bending stiffness matrix. 85. 9 Antisymmetric laminate. 89. 20. 290 331 . 220 Brittle fracture. 214. 146. 11. 182. 18. 260 Compressive strain. 11 AS4/350 l-6 Carbon/Epoxy. 90. 41 Composite materials. 78. 118-119. 279 Autoclave. 25 Aluminum. 77. 246 Buckling constraint. 21. 124 AS4/3501-6. 69. 279 Classical lamination theory. 184 selection. 134 buckling. 274 discrete. 201. 171. 120 Curvature-stable laminates. 47 Elastic modulus Aluminum. 46-47. 190. 303 Flexural stiffness. 16 Engineering strain transformation. 224. 183 Enumeration tree. 199. 128. 184. 25. 174. 46 Graphite/Epoxy. 301. 320 coding. 65. 254. 174 layer percentage. 28. 189. 322 inequality. 189 Genetic operators crossover. 256. 169. 184 mutation. 191. 226 normal stresses. 226. 227. 231 Fiberglass. 73. 206 practical reliability. 8 Filament winding. 23 Derivatives of constraints. 215. 201 gene swap. 89 hygrothermal.'. 82 Effective shear modulus. 237. 134. 204 multiple point. 182 strength. 230 Ductile materials. 189. 143. 242. 174 Effective engineering stiffnesses. 249. 89. 175 normalized. 247. 202 probability. 246. 232 Fiber radius. 226 Fracture toughness. 232. 30. 54. 171 continuous. 225 Fiber failure. 298 twisting-entension. 74. 24.·:. 241. 262. 155 Displacement minimization. 28. 134. 224. 69. 161 shear stiffness. I 05 Cross-linking. 261. 210 ranking of designs. 4. 209 roulette wheel. 310 integer. 252. 5. 191 initial population. 210. 230 first-ply. 122. 226 maximum shear stress. 286 ply-identity. 229 maximum normal stress. 149 Fiber direction. ::~~ . 46 Fabrication of laminates. 58. 11. 291 ' · . 237 Tresca. 60. 298 extension-shear. 4. 135 frequency. 207. 270 stiffness. 122 extension-twisting. 193. 211 strength. 308 Distortional energy (von Mises) criterion. 249 orthotropic. 249 maximum stress. 238. 26. 78. 214 Curvatures. 36 Evolutionary strategies. 275 Discrete design variables. 255. 120. 208 strain. 134. 214 elastic properties. 244. 26 Equality constraints. 138. 202 single-point. 191 twisting. 56. 118 Coupling response. 138 engineering stiffnesses. 102 INDEX INDEX 333 Feasible domain. 25. 160. 285 Hoffman. 287 first-ply failure. 248 quadratic. 81. 317 Generally orthotropic. 232 Failure model. 100. 179. 185 notation. 244. 260 distortion energy. 118-119. 229 maximum strain. 225 Delaminations. 278 Engineering strains.·:. 139. 6 Flexural lamination diagram. I 0. 314 Fundamental frequency. 28. 24 orientation. 73. 229 Tsai-Hill. 238. 224. 174 shear. 195. 16 Epoxy. 149. 90 Failure brittle materials. 261. 178 Environmental effects. 134 binary. 218 Extensional stiffness matrix. 18. 73. 233 Electrical conductivity. 248. 246 Frequency constraints. 261 Continuous optimization. 61 Genetic algorithms. 287 Flake composites. 183. 188 mutation. 237-238. 183. 149 Effective Poisson's ratio. 218 shear-extension. 29. 261 Constraint functions. 85. 225. 312. 73-74. 214 bending-twisting. 186 Extension-shear coupling. 216. 274. 21 First-ply failure. 261. 240.:. 184 inversion. 301-303. 232 Feasible design. 224. 250. 24. 233 Failure mechanisms. 209. 100. 246. 156. 79. 322 Frequency maximization. 81. 317. 189. 123. 103. 111. 191 notation. 213. 239 isotropic. 234. 241 progressive failure. 24 Poisson's ratio. 97. 302 Crossover averaging. 224 Failure criteria. 7 Fiber strength. 140. 29 stack-identity. 135. 120. 247 Delamination damage. 223-224. 64 Enumeration. 310. 24-25. 258. 157 equality. 201 Cure temperature. 177. 89.. 19 Elastic properties. 198 reliability. 51-52. 24. 313 rounding off. 2. 231 Fiber tows/bundles. 181. 191 . 197. 243 Failure envelope. 310 Discrete variables. 135 Equilibrium equations. 217. 135. 150 Effective elastic properties of a laminate. 89. 310 Fracture. 10. 134. 177 Coupling bending-extension. 230. 289 Constraint derivatives. 91 Cross-ply laminate. 230 Effective engineering properties.. 19 effective.. 214. 203. 16 Fiber microbuckling. 225. ~. 137. 323 thickness. 301 Flexural stiffness matrix. 272. 226-227. 120. 193. 287 von Mises. 195 seeding. 22. 248. 8 Fiber-reinforced composites. 184 reproduction. 24.332 Constraint shear stiffness. 238. 228. 211. 24. 122 Extension-twisting coupling. 74. 24. 223-224 ductile materials. 165. 91 Curvature. 24. 187. 195 Constraints. 183-185. 236. 23 Failure stress. 8 Fiber volume fraction. 128. 170 Constraint margin. 103 hygrothermal. 285. 191. 172 bounds. 287 Tsai-Wu. 226. 60 Damage tolerance. 206 permutation. 161. 205. 261 Design variables. 153. 123. 184. 289. 87.. 292. 4. 185 fitness.i( l Engineering strain transformation matrix. 301 Flexural lamination parameters. 214. 74. 61. 89 Hygrothermal strains. 144. 211. 119 hygrothermal. 28. average. 178. 102 Hygrothermal response. 20. 5 Inorganic materials. 225 Matrix strength. 303. 191. 250 Kevlar49/Ep. 77. 89 Moisture diffusion coefficient. 265. 118-119 Laminate buckling. 301. 206 Natural vibration frequency. 323. 192. 150. 81 Optimization. 317. 111. 134. 225 Isotropic hygrothermal. 176. 73-74. 107. 103 Hygrothermal moment resultants. 308 Minimum displacement. 206 probability. 269. 49 Layer volume fraction. 232 Micromechanics. 288 flexural stiffness design. 160. 18. 226 Inequality constraints. 194. 69. 200. 273 Graphite/Epoxy. 271. 182. 171 Moisture. 127-128. 223 Local optimum. 67. 171 sequential. 154. 229 Maximum strain criterion. 21 Moment resultants. 162. orthotropic. 248 Mass density. 262. 264. 129 Kevlar. 18. 30. 174 Inorganic composites. 202. 258-259 Scotchply-1002. 146. 162. 247. 275 Laminated composites.111. 198. 146. 2. 22 Laminate stresses. symmetric. 307. 231 INDEX y INDEX 335 orthotropic materials. 5 Orientation angle. 141 Lamination parameters. 74. 183 Graphical optimization. 301 flexural. 172. 142 quasi-isotropic. 99 Hygrothermal curvatures. 102 Hygrothermally curvature-stable laminates. 61 Orthotropic laminate. 100 Normal strain. 188. 129 quasi-isotropic.143. 216. 187. 102 Hygrothermal stress resultants. 166. 156. 184. 140. 148. in-plane. 235. 210. 145 Lamination theory. 262 In-plane stiffness design. 303. 154. 9 Mathematical optimization problem. 231 strength. 315. I 02 Hygrothermal isotropy. 169. 69 Miki's lamination diagram. 43. 83 In-plane lamination parameters. 231-232 On-axis layer. 120. 122 Hygrothermally stable laminates. 262 Lamination parameter diagram. 233 Maximum stress failure criteria. 204. 215 T300/5208. 144. 220. 24 Off-axis layer. 231 Matrix shear strength. 292. II. 128 Matrix constituent. 134-135. 122. 11 HT-S/4617. 227. 319. 190.334 specialized operators. 327 Integer programming. 19. 174. 8 Laminates-quasi-isotropic. 298 specimen. 10. 124. 170. 122 Hygrothermal loads. 130. 24-25. 122 in-plane. 267. 151. 214. 11. 233 Maximum strain failure criteria. 262-263. 180. 172. 91. 28 Nonmechanical strains. 183 Longitudinal modulus. 326 Linear integer programming. 314 Maximum fundamental frequency. 29 Integer variables. 286 Orthotropic. 120. 140. 21 Manufacturing induced stresses. 54. 74. 172 Local failures. 26 Kevlar/Epoxy. 298 Laminate failure. 169-170. 34 Objective function. 259. 255. 18. 109. 173 LINDO. 206. 138. 192 Mixed integer linear programming. 165. 107. 132 Lamination diagram. 87. 301 in-plane. 62 Materials anisotropic. 132. 139. 106 unsymmetric. 119 balanced. 316. 157. 134 Interlaminar stresses. 140. 319 zero-one. 300 Material constants. 98 Matrix material. 72 Orientation design variables. 208. 300 Nonlinear programming. 275. 18. 317 Maximum frequency. 5 Integer design variables. 5. 67. 249 Maximum stress criterion. 49-50. 27 Macromechanics. 289 notation. 164. 158. 53. 57. 45 Laminate weight. 74. 54 Laminate angle-ply. 160. 310 flexure. 100. 262 Lamination point. 261 Graphical strength design. 142. 152. 212. 254 Laminate in-plane stiffness design. 24. 261-262. 214-215 fabrication. 172. 83 Isotropic materials. 226 Maximum shear stress (Tresca) criterion. 247 symmetric. 151 Global optimum. 278. 310-311. 17. 120. 288 Integer programming problem. 322 Minimum thickness laminates. 100. 10. 17. 276 . 92 Moisture effects. 172. 302 curvature-stable. 135. 249 Microbuckling. 324 AS4/3501. 9 orthotropic. 224. 146. 169 flexural. 91. 287 coefficient of thermal expansion. 321. 264 Laminate strength. 267. 162. orthotropic materials. 309 antisymmetric. 140-141. 103. 255 Hoffman criterion. 153 cross-ply. 196. 68. 214 In-plane isotropy. 313. 21-22 Mid-plane strains. 231 Maximum buckling load. 298 balanced. 171 mixed. 208 Glass/Epoxy. 2. 151. 116. 17-18. 183 Mutation. 39 Hygral strains. progressive. 168 Laminate shear stiffness. 310 In-plane stresses. 51. 218. 34. 90 hygrothermally stable. 298. 160. 288 integer. 125. 138. 302-303 Maximum normal stress criterion. 185 Integer linear programming. 9. 310 Linear programming. 83. 102. 102 Multiple minima. 186. 325. 133 Organic materials. 136. 239 Hooke's law. 11 Kirchhoff-Love assumptions. 151 Laminate strain limits. 120 orthotropic. 17. 141. 155. 173. 302 Minimum thickness design. 19.247. 176. 97. 261 In-plane stiffness matrix. 320. 130. 9 isotropic. 73. 287 genetic coding. 195 Permutation of stacking sequence. 10 I shear. 123. 174. 313 Ply-level strains. 35 transformed reduced. 237. 208. 64 Thermal bending. 283 Strain constraints. 69. 128. I 02 nonmechanical. 54 coupling. 225. 51. 164. 73 Twisting-entension coupling. 291 Strain transformation matrix. 35 Sequential linear programming. 243-244 Von Mises failure criterion. 22 Strength properties. 58. 136. 278 reduced. 74. 234 Unbalanced laminates. 233. 224. 233 Stress concentrations. 39. 91 through-the-thickness. 244 Tsai-Wu failure criterion. 10 Transverse failure. 10 I. 299 Simulated annealing. 254 Quadratic failure criteria. 238. 11. 237. 146. . 156 Stack-identity design variables. 36 effective. 8. 54. 248 Roulette wheel. 262 Tow-placement. 300 Volume fraction. 10. 227 Tensor strains. 277. engineering. 223 Stress couples. 8 Unsymmetric laminates. 100 normal. 34 residual. 73. 8 Simplex method. 287 Twisting curvature. 23 Symmetric laminates. 6 Thickness design variables. 97. 227 Strain transformation. 272. 183 Ranking of fitnesses. 292. 142. 318 hygrothermal. 91. 47 Shear stiffness. 219 Practical reliability. 19. 119. 144. 241.. 237 Quasi-isotropic. 145 Von Mises criterion. 206 Progressive failure. 275. 297 Stacking sequence design. 74. 132. . 298 Transverse modulus. 278. 83. 306 Plane stress. 40 Stiffness matrix transformed reduced. 275. 74 Vibration frequency. 298 Short fiber composites. 43. 16 ultimate. 118 Weight density. 225 Orthotropic material constrants. 249 Principal material directions. 229. 10. 230 Strain space. 312. 13. 278 Transformed reduced stiffness matrix. 6 Thermoset composites. 238 Tsai-Hill failure criterion. 198 Reduced effective stiffness. 20 Graphite/Epoxy. 189 Specific heat. 56. 121 Shear modulus.. 46. 34 Strength constraint. 53. 10. 249 Ply orientations. 246. 310 in-plane. 45. 319 Shear coupling. 9 Orthotropic properties. 226. 173.336 Orthotropic layers. . 175 Wood. 153. 33. 60. 41. 171 Simply supported plate. 214. 195 Rule of mixtures. 120 Twisting deformations. 20 Weight minimization. 175 Stochastic search. 247 Transverse direction. 229 Poisson's ratio maximization. 289 Strength of a laminate. 92 Thermal effects. 21 Transformation matrix. 128 Ply-identity variables. 249 Strain-displacement relation. 289 Strength design. 61. 174 Shear strain. 17. 101. 28 Transverse normal strength. 118-119 bending-extension coupling. 137 Strain energy density. 246 Young's modulus. 20. 29 Strain hygral. 224 Poisson's ratio. 65. 248 Transverse loads. 288 Stacking sequence permutation. 74. 267 Poisson's effect. 122 Thermal diffusivity. 118 Ultimate strains. 248 Progressive failure of laminates. 287 Tsai-Wu criterion. 41 three-dimensional. 65. 192 Unidirectional tape. 210 Prepreg. 249. I 01 Residual stresses. 106. 300. 118 Warping. 228 Principal stresses. 21 Thermoplastic composites. 229 Yielding. 83. 227. 279 Random search. 64. 10. 20 Graphite/Epoxy. 226. 34 Shear strength. 288 Tresca failure criterion. 282 Reliability of genetic algorithms. 275 Thickness variable. 278. 17. 64 Stresses residual. 47 Reduced transformed stiffness matrix. 2 Aluminum. 5 Woven fabric composite. 229. 202 genetic representation. 84. 263. I 0. 55 extensional. 254 Reduced stiffness matrix. 90 Principal material axes. 46 flexural. 92 Specific moisture content. 248. 65. 261. 36. 317. 306 Stiffness matrix bending-extension. 6 PEEK. 20 Specific strength. 209 Residual strains. 34 thermal. 318 Transverse cracking. 231 Probability of mutation. 89 Thermal isotropy. I 02 Stress space. 288 Shear-extension coupling. 154 Thermal strains. 49. 228 Stress transformation. 8 Yield strength. 231. 64 f. 291 Principal strains. 29. 92-93 Specific stiffness. 99 Strain approximation. 219 Singular optimum. 183. 113 Stiffness maximization. 227 Ultimate strength. 12. 113. 84. 2. 2 Stack thickness. 323 Stacking sequence. 99 Thermoforming. 6 Penalty parameters. 129. 290. 69 Stress resultants. 318 two-dimensional. 247 Quasi-isotropic laminate. 226. 64. 103. 229 Tsai-Hill criterion. 187 Tensile strength. ' )· I f~. 30. 230 Warp-stable laminates. 224 Out-of-plane displacement. 298 Particulate composites. 62 Orthotropic materials. 102. 22 INDEX INDEX 337 engineering strain. 99 hygrothermal.