Composite Materials - Second Edition

Composite Materials A CRC title part of the Taylor Francis imprint a member of the Taylor Francis Group the academic division of TF Informa plc S E C O N D E D I T I O N Boca Raton London New York M E C H A N I C S OF Autar K Kaw 2006 by Taylor Francis Group LLC The cover illustration is an artists rendition of fiber geometries crosssectional views and crack propagation paths in a composite material The author gratefully acknowledges and gives his heartfelt thanks to his longtime friend Dr Suneet Bahl for drawing the cover illustration Published in 2006 by CRC Press Taylor Francis Group 6000 Broken Sound Parkway NW Suite 300 Boca Raton FL 334872742 2006 by Taylor Francis Group LLC CRC Press is an imprint of Taylor Francis Group No claim to original US Government works Printed in the United States of America on acidfree paper 10 9 8 7 6 5 4 3 2 1 International Standard Book Number10 0849313430 Hardcover International Standard Book Number13 9780849313431 Hardcover Library of Congress Card Number 2005049974 This book contains information obtained from authentic and highly regarded sources Reprinted material is quoted with permission and sources are indicated A wide variety of references are listed Reasonable efforts have been made to publish reliable data and information but the author and the publisher cannot assume responsibility for the validity of all materials or for the consequences of their use No part of this book may be reprinted reproduced transmitted or utilized in any form by any electronic mechanical or other means now known or hereafter invented including photocopying microfilming and recording or in any information storage or retrieval system without written permission from the publishers For permission to photocopy or use material electronically from this work please access wwwcopyrightcom httpwwwcopyrightcom or contact the Copyright Clearance Center Inc CCC 222 Rosewood Drive Danvers MA 01923 9787508400 CCC is a notforprofit organization that provides licenses and registration for a variety of users For organizations that have been granted a photocopy license by the CCC a separate system of payment has been arranged Trademark Notice Product or corporate names may be trademarks or registered trademarks and are used only for identification and explanation without intent to infringe Library of Congress CataloginginPublication Data Kaw Autar K Mechanics of composite materials Autar K Kaw2nd ed p cm Mechanical engineering v 29 Includes bibliographical references and index ISBN 0849313430 alk paper 1 Composite materialsMechanical properties I Title II Mechanical engineering series Boca Raton Fla v 29 TA4189C6K39 2005 6201183dc22 2005049974 Visit the Taylor Francis Web site at httpwwwtaylorandfranciscom and the CRC Press Web site at httpwwwcrcpresscom Taylor Francis Group is the Academic Division of Informa plc 1343Disclfm Page 1 Monday September 26 2005 118 PM 2006 by Taylor Francis Group LLC 1343SeriesPage 92805 1029 AM Page 1 Mechanical Engineering Series Frank Kreith Series Editor Published Titles Distributed Generation The Power Paradigm for the New Millennium AnneMarie Borbely Jan F Kreider Elastoplasticity Theory Vlado A Lubarda Energy Audit of Building Systems An Engineering Approach Moncef Krarti Engineering Experimentation Euan Somerscales Entropy Generation Minimization Adrian Bejan The Finite Element Method Using MATLAB 2nd Edition Young W Kwon Hyochoong Bang Fluid Power Circuits and Controls Fundamentals and Applications John S Cundiff Fundamentals of Environmental Discharge Modeling Lorin R Davis Heat Transfer in Single and Multiphase Systems Greg F Naterer Introductory Finite Element Method Chandrakant S Desai Tribikram Kundu Intelligent Transportation Systems New Principles and Architectures Sumit Ghosh Tony Lee Mathematical Physical Modeling of Materials Processing Operations Olusegun Johnson Ilegbusi Manabu Iguchi Walter E Wahnsiedler Mechanics of Composite Materials 2nd Edition Autar K Kaw Mechanics of Fatigue Vladimir V Bolotin Mechanics of Solids and Shells Theories and Approximations Gerald Wempner Demosthenes Talaslidis Mechanism Design Enumeration of Kinematic Structures According to Function LungWen Tsai Multiphase Flow Handbook Clayton T Crowe Nonlinear Analysis of Structures M Sathyamoorthy Optomechatronics Fusion of Optical and Mechatronic Engineering Hyungsuck Cho Practical Inverse Analysis in Engineering David M Trujillo Henry R Busby Pressure Vessels Design and Practice Somnath Chattopadhyay Principles of Solid Mechanics Rowland Richards Jr Thermodynamics for Engineers KauFui Wong Vibration and Shock Handbook Clarence W de Silva Viscoelastic Solids Roderic S Lakes 2006 by Taylor Francis Group LLC Dedication To Sherrie Candace Angelie Chuni Sushma Neha and Trance and in memory of my father Radha Krishen Kaw who gave me the love of teaching movies and music necessarily in that order There is nothing noble about being superior to another man the true nobility lies in being superior to your previous self Upanishads 1343bookfm Page v Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Preface to the Second Edition The first edition of this book was published in 1997 and I am grateful for the response and comments I have received about the book and the accom panying PROMAL software The changes in the book are mainly a result of comments received from students who used this book in a course or as a selfstudy In this edition I have added a separate chapter on symmetric and unsym metric laminated beams All the other chapters have been updated while maintaining the flow of the content Key terms and a summary have been added at the end of each chapter Multiplechoice questions to reinforce the learning from each chapter have been added and are available at the textbook Website httpwwwengusfedukawpromalbookhtml Specifically in Chapter 1 new applications of composite materials have been accommodated With the ubiquitous presence of the Web I have anno tated articles videos and Websites at the textbook Website In Chapter 2 we have added more examples and derivations have been added The appen dix on matrix algebra has been extended because several engineering depart ments no longer teach a separate course in matrix algebra If the reader needs more background knowledge of this subject he or she can download a free ebook on matrix algebra at httpnumericalmethodsengusfedu click on matrix algebra In Chapter 3 derivations are given for the elasticity model of finding the four elastic constants Two more examples can be found in Chapter 5 design of a pressure vessel and a drive shaft The PROMAL program has been updated to include elasticity models in Chapter 3 PROMAL and the accompanying software are available to the eligible buyers of the textbook only at the textbook Website see the About the Software section The software and the manual will be con tinually updated 1343bookfm Page vii Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Preface to the First Edition Composites are becoming an essential part of todays materials because they offer advantages such as low weight corrosion resistance high fatigue strength faster assembly etc Composites are used as materials ranging from making aircraft structures to golf clubs electronic packaging to medical equipment and space vehicles to home building Composites are generating curiosity and interest in students all over the world They are seeing every day applications of composite materials in the commercial market and job opportunities are also increasing in this field The technology transfer initia tive of the US government is opening new and largescale opportunities for use of advanced composite materials Many engineering colleges are offering courses in composite materials as undergraduate technical electives and as graduatelevel courses In addition as part of their continuing education and retraining many practicing engi neers are participating in workshops and taking short courses in composite materials The objective of this book is to introduce a senior undergraduate or graduatelevel student to the mechanical behavior of composites Cover ing all aspects of the mechanical behavior of composites is impossible to do in one book also many aspects require knowledge of advanced graduate study topics such as elasticity fracture mechanics and plates and shells theory Thus this book emphasizes an overview of composites followed by basic mechanical behavior of composites Only then will a student form a necessary foundation for further study of topics such as impact fatigue fracture mechanics creep buckling and vibrations etc I think that these topics are important and the interested student has many wellwritten texts available to follow for that This book breaks some traditional rules followed in other textbooks on composites For example in the first chapter composites are introduced in a questionanswer format These questions were raised through my own thought process when I first took a course in composites and then by my students at the University of South Florida Tampa Also this is the first textbook in its field that includes a professional software package In addi tion the book has a format of successful undergraduate books such as short sections adequate illustrations exercise sets with objective questions and numerical problems reviews wherever necessary simple language and many examples Chapter 1 introduces basic ideas about composites including why com posites are becoming important in todays market Other topics in Chapter 1 include types of fibers and matrices manufacturing applications recy cling and basic definitions used in the mechanics of composites In Chapter 1343bookfm Page ix Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 2 I start with a review of basic topics of stress strain elastic moduli and strain energy Then I discuss the mechanical behavior of a single lamina including concepts about stressstrain relationship for a lamina stiffness and strength of a lamina and the stressstrain response due to temperature and moisture change In Chapter 3 I develop equations for mechanical properties of a lamina such as stiffness strength and coefficients of thermal and mois ture expansion from individual properties of the constituents long contin uous fibers and matrix of composites I introduce experimental characterization of the mechanical properties of a lamina at appropriate places in Chapter 3 Chapter 4 is an extension of Chapter 2 in which the macromechanics of a single lamina are extended to the macromechanics of a laminate I develop stressstrain equations for a laminate based on indi vidual properties of the laminae that make it I also discuss stiffness and strength of a laminate and effects of temperature and moisture on residual stresses in a laminate In Chapter 5 special cases of laminates used in the market are introduced I develop procedures for analyzing the failure and design of laminated composites Other mechanical design issues such as fatigue environmental effects and impact are introduced A separate chapter for using the userfriendly software PROMAL is included for supplementing the understanding of Chapter 2 through Chap ter 5 Students using PROMAL can instantly conduct pragmatic parametric studies compare failure theories and have the information available in tables and graphs instantaneously The availability of computer laboratories across the nation allows the instructor to use PROMAL as a teaching tool Many questions asked by the student can be answered instantly PROMAL is more than a black box because it shows intermediate results as well At the end of the course it will allow students to design laminated composite structures in the class room The computer program still maintains the students need to think about the various inputs to the program to get an optimum design You will find this book and software very interesting I welcome your comments suggestions and thoughts about the book and the software at email promalengusfedu and URL httpwwwengusfedukaw promalbookhtml 1343bookfm Page x Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Acknowledgments I acknowledge all the students who have taken the course on composite materials at the University of South Florida since I first taught it in the spring of 1988 Since then their questions and wish lists have dynamically changed the content of the course I would like to thank my talented students Steven Jourdenais Brian Shanberg Franc Urso Gary Willenbring and Paula Bond for their help with building the PROMAL software PROMAL has been a continuous project since 1988 I thank my dear friend Suneet Bahl who designed yet another unique illustration for the cover for this book His contribution has been inspira tional I thank J Ye J Meyers M Toma A Prasad R Rodriguez K Gan gakhedkar C Khoe P Chalasani and S Johnson for drawing the illustrations proofreading and checking the examples in the text Special thanks go again to R Rodriguez who painstakingly developed the solutions manual for the book using MATHCAD software I would like to thank Sue Britten for helping me in typing the manuscript especially the equations and the endless loop of revisions and changes Her effort was very critical in finishing the project on time I want to thank all the companies that not only sent promotional literature but also made an additional effort to send photographs videos slides design examples etc Individual companies whose information has been used in the book are acknowledged for each citation A sabbatical granted by the University of South Florida in the fall of 2002 was critical in completing this project I thank Professor L Carlsson of Florida Atlantic University who provided the raw data for some of the figures from his book Experimental Characterization of Advanced Composite Materials I thank Dr RY Kim of the University of Dayton Research Institute for providing stressstrain data and photographs for several figures in this book I want to thank Dr GP Tandon of UDRI for several discussions and references on developing the elasticity models for the elastic moduli of unidirectional composites I thank my wife Sherrie and our two children Candace and Angelie for their support and encouragement during this long project In their own way our children have taught me how to be a good teacher I would like to acknowl edge my parents who gave me the opportunities to reach my goals and did that at a great personal sacrifice I am grateful to my father who was a role model for my professional career and taught me many things about being a complete teacher 1343bookfm Page xi Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC I thank Cindy Carelli and Michael Slaughter senior editors of Taylor Francis and their staff for their support and encouragement I want to thank Elizabeth Spangenberger Helena Redshaw Jessica Vakili Naomi Lynch Jonathan Pennell and their staffs for keeping me updated throughout the production process and giving personal attention to many details including design layout equation editing etc of the final product I have to thank the authors of Getting Your Book Published Sage Publica tions for helping me understand the mechanics of publication and how to create a winwin situation for all the involved parties in this endeavor I would recommend their book to any educator who is planning to write a textbook 1343bookfm Page xii Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC About the Author Autar K Kaw is a professor of mechanical engineering at the University of South Florida Tampa Professor Kaw obtained his BE Hons degree in mechanical engineering from Birla Institute of Technology and Science India in 1981 He received his PhD degree in 1987 and MS degree in 1984 both in engineering mechanics from Clemson University South Carolina He joined the faculty of the University of South Florida in 1987 He has also been a maintenance engineer 1982 for FordEscorts Tractors India and a summer faculty fellow 1992 and visiting scientist 1991 at Wright Patterson Air Force Base Professor Kaws main scholarly interests are in the fracture mechanics of composite materials and development of instructional software for engineer ing education His research has been funded by the National Science Foun dation Air Force Office of Scientific Research Florida Department of Transportation Research and Development Laboratories Wright Patterson Air Force Base and Montgomery Tank Lines He is a fellow of the American Society of Mechanical Engineers ASME and a member of the American Society of Engineering Education ASEE He has written more than 35 journal papers and developed several software instructional programs for courses such as Mechanics of Composites and Numerical Methods Professor Kaw has received the Florida Professor of the Year Award from the Council for Advancement and Support of Education CASE and Car negie Foundation for Advancement of Teaching CFAT 2004 Archie Hig don Mechanics Educator Award from the American Society of Engineering Education ASEE 2003 Southeastern Section American Society of Engi neering Education ASEE Outstanding Contributions in Research Award 1996 State of Florida Teaching Incentive Program Award 1994 and 1997 American Society of Engineering Education ASEE New Mechanics Edu cator Award 1992 and Society of Automotive Engineers SAE Ralph Teetor Award 1991 At the University of South Florida he has been awarded the Jerome Krivanek Distinguished Teacher Award 1999 Univer sity Outstanding Undergraduate Teaching Award 1990 and 1996 Faculty Honor Guard 1990 and the College of Engineering Teaching Excellence Award 1990 and 1995 1343bookfm Page xiii Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC About the Software Where can I download PROMAL You can download PROMAL at httpwwwengusfedukawpromal bookhtml In addition to the restrictions for use given later in this section only textbook buyers are authorized to download the software What is PROMAL PROMAL is professionally developed software accompanying this book Taylor Francis Group has been given the rights free of charge by the author to supplement this book with this software PROMAL has five main programs 1 Matrix algebra Throughout the course of Mechanics of Composite Mate rials the most used mathematical procedures are based on linear algebra This feature allows the student to multiply matrices invert square matrices and find the solution to a set of simultaneous linear equations Many students have programmable calculators and access to tools such as MATHCAD to do such manipulations and we have included this program only for convenience This program allows the student to concentrate on the fundamentals of the course as opposed to spending time on lengthy matrix manipulations 2 Lamina properties database In this program the properties of uni directional laminae can be added deleted updated and saved This is useful because these properties can then be loaded into other parts of the program without repeated inputs 3 Macromechanical analysis of a lamina Using the properties of unidi rectional laminae saved in the previously described database one can find the stiffness and compliance matrices transformed stiffness and compliance matrices engineering constants strength ratios based on four major failure theories and coefficients of thermal and moisture expansion of angle laminae These results are then pre sented in textual tabular and graphical forms 4 Micromechanics analysis of a lamina Using individual elastic moduli coefficients of thermal and moisture expansion and specific gravity of fiber and matrix one can find the elastic moduli and coefficients of thermal and moisture expansion of a unidirectional lamina Again the results are available in textual tabular and graphical forms 1343bookfm Page xv Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 5 Macromechanics of a laminate Using the properties of the lamina from the database one can analyze laminated structures These laminates may be hybrid and unsymmetric The output includes finding stiff ness and compliance matrices global and local strains and strength ratios in response to mechanical thermal and moisture loads This program is used for design of laminated structures such as plates and thin pressure vessels at the end of the course Who is permitted to use PROMAL PROMAL is designed and permitted to be used only as a theoreticaledu cational tool it can be used by A university instructor using PROMAL for teaching a formal university level course in mechanics of composite materials A university student using PROMAL to learn about mechanics of composites while enrolled in a formal universitylevel course in mechanics of composite materials A continuing education student using PROMAL to learn about mechanics of composites while enrolled in a formal universitylevel course in mechanics of composite materials A selfstudy student who has successfully passed a formal univer sitylevel course in strength of materials and is using PROMAL while studying the mechanics of composites using a textbook on mechanics of composites If you or your use of PROMAL does not fall into one of these four cate gories you are not permitted to use the PROMAL software What is the license agreement to use the software Software License Grant of License PROMAL is designed and permitted to be used only as a theoreticaleducational tool Also for using the PROMAL software the definition of You in this agreement should fall into one of four categories 1 University instructor using PROMAL for teaching a formal univer sitylevel course in mechanics of composite materials 2 University student using PROMAL to learn about mechanics of composites while enrolled in a formal universitylevel course in mechanics of composite materials 1343bookfm Page xvi Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 3 Continuing education student using PROMAL to learn about mechanics of composites while enrolled in a formal universitylevel course in mechanics of composite materials 4 Selfstudy student who has successfully passed a formal university level course in strength of materials and is using PROMAL while studying the mechanics of composites using a textbook on mechan ics of composites If you or your use of PROMAL does not fall into one of the above four categories you are not permitted to buy or use the PROMAL software Autar K Kaw and Taylor Francis Group hereby grant you and you accept a nonexclusive and nontransferable license to use the PROMAL software on the following terms and conditions only you have been granted an Individual Software License and you may use the Licensed Program on a single personal computer for your own personal use Copyright The software is owned by Autar K Kaw and is pro tected by United States copyright laws A backup copy may be made but all such backup copies are subject to the terms and conditions of this agreement Other Restrictions You may not make or distribute unautho rized copies of the Licensed Program create by decompilation or otherwise the source code of the PROMAL software or use copy modify or transfer the PROMAL software in whole or in part except as expressly permitted by this Agreement If you transfer possession of any copy or modification of the PROMAL software to any third party your license is automatically terminated Such termination shall be in addition to and not in lieu of any equitable civil or other remedies available to Autar K Kaw and Taylor Francis Group You acknowledge that all rights including without limitation copyrights patents and trade secrets in the PROMAL software including without limitation the structure sequence organization flow logic source code object code and all means and forms of operation of the Licensed Program are the sole and exclusive prop erty of Autar K Kaw By accepting this Agreement you do not become the owner of the PROMAL software but you do have the right to use it in accordance with the provision of this Agreement You agree to protect the PROMAL software from unauthorized use reproduction or distribution You further acknowledge that the PROMAL software contains valuable trade secrets and confidential information belonging to Autar K Kaw You may not disclose any component of the PROMAL software whether or not in machine readable form except as expressly provided in this Agreement 1343bookfm Page xvii Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Term This License Agreement is effective until terminated This Agreement will also terminate upon the conditions discussed else where in this Agreement or if you fail to comply with any term or condition of this Agreement Upon such termination you agree to destroy the PROMAL software and any copies made of the PRO MAL software Limited Warranty This limited warranty is in lieu of all other warranties expressed or implied including without limitation any warranties or mer chantability or fitness for a particular purpose The licensed program is furnished on an as is basis and without warranty as to the performance or results you may obtain using the licensed program The entire risk as to the results or performance and the cost of all necessary servicing repair or correction of the PROMAL software is assumed by you In no event will Autar K Kaw or Taylor Francis Group be liable to you for any damages whatsoever including without limitation lost profits lost savings or other incidental or consequential dam ages arising out of the use or inability to use the PROMAL software even if Autar K Kaw or Taylor Francis Group has been advised of the possibility of such damages You should not build design or analyze any actual structure or component using the results from the PROMAL software This limited warranty gives you specific legal rights You may have others by operation of law that vary from state to state If any of the provisions of this agreement are invalid under any applicable statute or rule of law they are to that extent deemed omitted This agreement represents the entire agreement between us and supersedes any proposals or prior agreements oral or written and any other communication between us relating to the subject matter of this agreement This agreement will be governed and construed as if wholly entered into and performed within the state of Florida You acknowledge that you have read this agreement and agree to be bound by its terms and conditions Is there any technical support for the software The program is userfriendly and you should not need technical support However technical support is available only through email and is free for registered users for 30 days from the day of purchase of this book Before using technical support check with your instructor and study the manual and the home page for PROMAL at httpwwwengusfedukaw 1343bookfm Page xviii Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC promalbookhtml At this home page you can also download upgraded pro malexe files Send your questions comments and suggestions for future versions by email to promalengusfedu I will attempt to include your feedback in the next version of PROMAL How do I register the software Register by sending an email to promalengusfedu with registration in the subject line and the body with name universitycontinuing education affiliation postal address email address telephone number and how you obtained a copy of the software ie purchase of book personal copy site license continuing education course OR Register by mailing a post card with name universitycontinuing educa tion affiliation address and email address telephone number and how you obtained a copy of the software ie purchase of book personal copy site license continuing education course to Professor Autar K Kaw ENB 118 Mechanical Engineering Department University of South Florida Tampa FL 336205350 What are the requirements of running the program The program will generally run on any IBMPC compatible computer with Microsoft Windows 98 or later 128 MB of available memory and a hard disk with 50 MB available and Microsoft mouse Can I purchase a copy of PROMAL separately Check the book Website for the latest purchase information for singlecopy sales course licenses and continuing education course prices 1343bookfm Page xix Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Contents 1 Introduction to Composite Materials 1 Chapter Objectives1 11 Introduction 1 12 Classification16 121 Polymer Matrix Composites19 122 Metal Matrix Composites 40 123 Ceramic Matrix Composites45 124 CarbonCarbon Composites 46 13 Recycling FiberReinforced Composites 50 14 Mechanics Terminology51 15 Summary54 Key Terms 54 Exercise Set 55 References57 General References 58 Video References59 2 Macromechanical Analysis of a Lamina 61 Chapter Objectives61 21 Introduction 61 22 Review of Definitions65 221 Stress65 222 Strain 68 223 Elastic Moduli75 224 Strain Energy77 23 Hookes Law for Different Types of Materials 79 231 Anisotropic Material81 232 Monoclinic Material82 233 Orthotropic Material Orthogonally AnisotropicSpecially Orthotropic84 234 Transversely Isotropic Material 87 235 Isotropic Material 88 24 Hookes Law for a TwoDimensional Unidirectional Lamina99 241 Plane Stress Assumption99 242 Reduction of Hookes Law in Three Dimensions to Two Dimensions100 243 Relationship of Compliance and Stiffness Matrix to Engineering Elastic Constants of a Lamina101 25 Hookes Law for a TwoDimensional Angle Lamina109 1343bookfm Page xxi Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 26 Engineering Constants of an Angle Lamina121 27 Invariant Form of Stiffness and Compliance Matrices for an Angle Lamina 132 28 Strength Failure Theories of an Angle Lamina 137 281 Maximum Stress Failure Theory 139 282 Strength Ratio 143 283 Failure Envelopes144 284 Maximum Strain Failure Theory 146 285 TsaiHill Failure Theory149 286 TsaiWu Failure Theory 153 287 Comparison of Experimental Results with Failure Theories158 29 Hygrothermal Stresses and Strains in a Lamina160 291 Hygrothermal StressStrain Relationships for a Unidirectional Lamina163 292 Hygrothermal StressStrain Relationships for an Angle Lamina 164 210 Summary167 Key Terms 167 Exercise Set 168 References174 Appendix A Matrix Algebra175 Key Terms 195 Appendix B Transformation of Stresses and Strains 197 B1 Transformation of Stress 197 B2 Transformation of Strains 199 Key Terms 202 3 Micromechanical Analysis of a Lamina 203 Chapter Objectives203 31 Introduction 203 32 Volume and Mass Fractions Density and Void Content 204 321 Volume Fractions204 322 Mass Fractions 205 323 Density 207 324 Void Content 211 33 Evaluation of the Four Elastic Moduli215 331 Strength of Materials Approach 216 3311 Longitudinal Youngs Modulus218 3312 Transverse Youngs Modulus221 3313 Major Poissons Ratio227 3314 InPlane Shear Modulus 229 332 SemiEmpirical Models 232 3321 Longitudinal Youngs Modulus234 3322 Transverse Youngs Modulus234 1343bookfm Page xxii Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 3323 Major Poissons Ratio236 3324 InPlane Shear Modulus 237 333 Elasticity Approach239 3331 Longitudinal Youngs Modulus241 3332 Major Poissons Ratio249 3333 Transverse Youngs Modulus251 3334 Axial Shear Modulus 256 334 Elastic Moduli of Lamina with Transversely Isotropic Fibers 268 34 Ultimate Strengths of a Unidirectional Lamina 271 341 Longitudinal Tensile Strength271 342 Longitudinal Compressive Strength 277 343 Transverse Tensile Strength284 344 Transverse Compressive Strength 289 345 InPlane Shear Strength291 35 Coefficients of Thermal Expansion296 351 Longitudinal Thermal Expansion Coefficient 297 352 Transverse Thermal Expansion Coefficient 298 36 Coefficients of Moisture Expansion303 37 Summary307 Key Terms 308 Exercise Set 308 References 311 4 Macromechanical Analysis of Laminates 315 Chapter Objectives315 41 Introduction 315 42 Laminate Code316 43 StressStrain Relations for a Laminate 318 431 OneDimensional Isotropic Beam StressStrain Relation 318 432 StrainDisplacement Equations320 433 Strain and Stress in a Laminate325 434 Force and Moment Resultants Related to Midplane Strains and Curvatures326 44 InPlane and Flexural Modulus of a Laminate 340 441 InPlane Engineering Constants of a Laminate 341 442 Flexural Engineering Constants of a Laminate344 45 Hygrothermal Effects in a Laminate 350 451 Hygrothermal Stresses and Strains 350 452 Coefficients of Thermal and Moisture Expansion of Laminates358 453 Warpage of Laminates362 46 Summary363 Key Terms 364 1343bookfm Page xxiii Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Exercise Set 364 References367 5 Failure Analysis and Design of Laminates 369 Chapter Objectives369 51 Introduction 369 52 Special Cases of Laminates370 521 Symmetric Laminates 370 522 CrossPly Laminates 371 523 Angle Ply Laminates 372 524 Antisymmetric Laminates372 525 Balanced Laminate373 526 QuasiIsotropic Laminates373 53 Failure Criterion for a Laminate 380 54 Design of a Laminated Composite 393 55 Other Mechanical Design Issues419 551 Sandwich Composites 419 552 LongTerm Environmental Effects420 553 Interlaminar Stresses421 554 Impact Resistance422 555 Fracture Resistance 423 556 Fatigue Resistance424 56 Summary425 Key Terms 426 Exercise Set 426 References430 6 Bending of Beams 431 Chapter Objectives431 61 Introduction 431 62 Symmetric Beams 433 63 Nonsymmetric Beams444 64 Summary455 Key Terms 455 Exercise Set 456 References457 1343bookfm Page xxiv Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 1 1 Introduction to Composite Materials Chapter Objectives Define a composite enumerate advantages and drawbacks of com posites over monolithic materials and discuss factors that influence mechanical properties of a composite Classify composites introduce common types of fibers and matri ces and manufacturing mechanical properties and applications of composites Discuss recycling of composites Introduce terminology used for studying mechanics of composites 11 Introduction You are no longer to supply the people with straw for making bricks let them go and gather their own straw Exodus 57 Israelites using bricks made of clay and reinforced with straw are an early example of application of composites The individual constituents clay and straw could not serve the function by themselves but did when put together Some believe that the straw was used to keep the clay from cracking but others suggest that it blunted the sharp cracks in the dry clay Historical examples of composites are abundant in the literature Signifi cant examples include the use of reinforcing mud walls in houses with bamboo shoots glued laminated wood by Egyptians 1500 B C and lami nated metals in forging swords A D 1800 In the 20th century modern composites were used in the 1930s when glass fibers reinforced resins Boats 1343bookfm Page 1 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 2 Mechanics of Composite Materials Second Edition and aircraft were built out of these glass composites commonly called fiber glass Since the 1970s application of composites has widely increased due to development of new fibers such as carbon boron and aramids and new composite systems with matrices made of metals and ceramics This chapter gives an overview of composite materials The ques tionanswer style of the chapter is a suitable way to learn the fundamental aspects of this vast subject In each section the questions progressively become more specialized and technical in nature What is a composite A composite is a structural material that consists of two or more combined constituents that are combined at a macroscopic level and are not soluble in each other One constituent is called the reinforcing phase and the one in which it is embedded is called the matrix The reinforcing phase material may be in the form of fibers particles or flakes The matrix phase materials are generally continuous Examples of composite systems include concrete rein forced with steel and epoxy reinforced with graphite fibers etc Give some examples of naturally found composites Examples include wood where the lignin matrix is reinforced with cellu lose fibers and bones in which the bonesalt plates made of calcium and phosphate ions reinforce soft collagen What are advanced composites Advanced composites are composite materials that are traditionally used in the aerospace industries These composites have high performance rein forcements of a thin diameter in a matrix material such as epoxy and alu minum Examples are graphiteepoxy Kevlar epoxy and boron aluminum composites These materials have now found applications in com mercial industries as well Combining two or more materials together to make a composite is more work than just using traditional monolithic metals such as steel and alu minum What are the advantages of using composites over metals Monolithic metals and their alloys cannot always meet the demands of todays advanced technologies Only by combining several materials can one meet the performance requirements For example trusses and benches used in satellites need to be dimensionally stable in space during temperature changes between 256 F 160 C and 200 F 933 C Limitations on coeffi cient of thermal expansion thus are low and may be of the order of 1 Aramids are aromatic compounds of carbon hydrogen oxygen and nitrogen Kevlar is a registered trademark of EI duPont deNemours and Company Inc Wilimington DE Coefficient of thermal expansion is the change in length per unit length of a material when heated through a unit temperature The units are inin F and mm C A typical value for steel is 65 10 6 inin F 117 10 6 mm C 1343bookfm Page 2 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Introduction to Composite Materials 3 10 7 inin F 18 10 7 mm C Monolithic materials cannot meet these requirements this leaves composites such as graphiteepoxy as the only materials to satisfy them In many cases using composites is more efficient For example in the highly competitive airline market one is continuously looking for ways to lower the overall mass of the aircraft without decreasing the stiffness and strength of its components This is possible by replacing conventional metal alloys with composite materials Even if the composite material costs may be higher the reduction in the number of parts in an assembly and the savings in fuel costs make them more profitable Reducing one lbm 0453 kg of mass in a commercial aircraft can save up to 360 gal 1360 l of fuel per year 1 fuel expenses are 25 of the total operating costs of a commercial airline 2 Composites offer several other advantages over conventional materials These may include improved strength stiffness fatigue and impact resis tance thermal conductivity corrosion resistance etc How is the mechanical advantage of composite measured For example the axial deflection u of a prismatic rod under an axial load P is given by 11 where L length of the rod E Youngs modulus of elasticity of the material of the rod Because the mass M of the rod is given by 12 where ρ density of the material of the rod we have Stiffness is defined as the resistance of a material to deflection Strength is defined as the stress at which a material fails Fatigue resistance is the resistance to the lowering of mechanical properties such as strength and stiffness due to cyclic loading such as due to takeoff and landing of a plane vibrating a plate etc Impact resistance is the resistance to damage and to reduction in residual strength to impact loads such as a bird hitting an airplane or a hammer falling on a car body Thermal conductivity is the rate of heat flow across a unit area of a material in a unit time when the temperature gradient is unity in the direction perpendicular to the area Corrosion resistance is the resistance to corrosion such as pitting erosion galvanic etc u PL AE M AL ρ 1343bookfm Page 3 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 4 Mechanics of Composite Materials Second Edition 13 This implies that the lightest beam for specified deflection under a specified load is one with the highest E ρ value Thus to measure the mechanical advantage the E ρ ratio is calculated and is called the specific modulus ratio between the Youngs modulus E and the density ρ of the material The other parameter is called the specific strength and is defined as the ratio between the strength σ ult and the density of the material ρ that is The two ratios are high in composite materials For example the strength of a graphiteepoxy unidirectional composite could be the same as steel but the specific strength is three times that of steel What does this mean to a designer Take the simple case of a rod designed to take a fixed axial load The rod cross section of graphiteepoxy would be same as that of the steel but the mass of graphiteepoxy rod would be one third of the steel rod This reduction in mass translates to reduced material and energy costs Figure 11 shows how composites and fibers rate with other traditional materials in terms of specific strength 3 Note that the unit of specific strength is inches in Figure 11 because specific strength and specific modulus are also defined in some texts as where g is the acceleration due to gravity 322 fts 2 or 981 ms 2 Youngs modulus of an elastic material is the initial slope of the stressstrain curve Density is the mass of a substance per unit volume A unidirectional composite is a composite lamina or rod in which the fibers reinforcing the matrix are oriented in the same direction M PL E 2 4 1 ρ Specific modulus Specific strength E ρ σult ρ Specific modulus Specific strength E g ρ g σult ρ 1343bookfm Page 4 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Introduction to Composite Materials 5 Values of specific modulus and strength are given in Table 11 for typical composite fibers unidirectional composites crossply and quasiisotropic laminated composites and monolithic metals On a first look fibers such as graphite aramid and glass have a specific modulus several times that of metals such as steel and aluminum This gives a false impression about the mechanical advantages of composites because they are made not only of fibers but also of fibers and matrix combined matrices generally have lower modulus and strength than fibers Is the comparison of the specific modulus and specific strength parameters of unidirectional composites to metals now fair The answer is no for two reasons First unidirectional composite structures are acceptable only for carrying simple loads such as uniaxial tension or pure bending In structures with complex requirements of loading and stiffness composite structures including angle plies will be necessary Second the strengths and elastic moduli of unidirectional composites given in Table 11 are those in the direction of the fiber The strength and elastic moduli perpendicular to the fibers are far less FIGURE 11 Specific strength as a function of time of use of materials Source Eager TW Whither advanced materials Adv Mater Processes ASM International June 1991 2529 A unidirectional laminate is a laminate in which all fibers are oriented in the same direction A crossply laminate is a laminate in which the layers of unidirectional lamina are oriented at right angles to each other Quasiisotropic laminate behaves similarly to an isotropic material that is the elastic proper ties are the same in all directions 10 8 6 4 2 0 1400 1500 Wood stone Bronze Cast iron Steel Aluminum Composites Aramid fibers carbon fibers 1600 1700 1800 Year Specific strength 106 in 1900 2000 1343bookfm Page 5 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 6 Mechanics of Composite Materials Second Edition A comparison is now made between popular types of laminates such as crossply and quasiisotropic laminates Figure 12 shows the specific strength plotted as a function of specific modulus for various fibers metals and composites Are specific modulus and specific strength the only mechanical parameters used for measuring the relative advantage of composites over metals No it depends on the application 4 Consider compression of a column where it may fail due to buckling The Euler buckling formula gives the critical load at which a long column buckles as 5 TABLE 11 Specific Modulus and Specific Strength of Typical Fibers Composites and Bulk Metals Material Units Specific gravity a Young s modulus Msi Ultimate strength ksi Specific modulus Msiin 3 lb Specific strength ksiin 3 lb System of Units USCS Graphite fiber Aramid fiber Glass fiber Unidirectional graphiteepoxy Unidirectional glassepoxy Crossply graphiteepoxy Crossply glassepoxy Quasiisotropic graphiteepoxy Quasiisotropic glassepoxy Steel Aluminum 18 14 25 16 18 16 18 16 18 78 26 3335 1798 1233 2625 5598 1392 3420 1010 2750 3000 1000 2998 2000 2248 2176 1540 5410 1280 4010 1060 9400 4000 5129 3555 1365 4541 8609 2408 5259 1747 4229 1065 1065 4610 3959 2489 3764 2368 9359 1968 6937 1630 3336 4258 Material Units Specific gravity Youngs modulus GPa Ultimate strength MPa Specific modulus GPam 3 kg Specific strength MPam 3 kg System of Units SI Graphite fiber Aramid fiber Glass fiber Unidirectional graphiteepoxy Unidirectional glassepoxy Crossply graphiteepoxy Crossply glassepoxy Quasiisotropic graphiteepoxy Quasiisotropic glassepoxy Steel Aluminum 18 14 25 16 18 16 18 16 18 78 26 23000 12400 8500 18100 3860 9598 2358 6964 1896 20684 6895 2067 1379 1550 1500 1062 3730 8825 27648 7308 6481 2758 01278 008857 00340 01131 002144 006000 001310 004353 001053 002652 002652 1148 09850 06200 09377 05900 02331 00490 01728 00406 008309 01061 a Specific gravity of a material is the ratio between its density and the density of water 1343bookfm Page 6 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Introduction to Composite Materials 7 14 where P cr critical buckling load lb or N E Youngs modulus of column lbin 2 or Nm 2 I second moment of area in 4 or m 4 L length of beam in or m If the column has a circular cross section the second moment of area is 15 and the mass of the rod is 16 FIGURE 12 Specific strength as a function of specific modulus for metals fibers and composites 5000 4000 3000 2000 1000 0 0 100 200 Quasiisotropic graphiteepoxy Aluminum Specific modulus Msiin3lb Crossply graphiteepoxy Unidirectional graphiteepoxy Graphite fiber Steel Specific strength Ksiin3lb 300 400 500 600 Pcr EI L π2 2 I d π 4 64 M d L 4 ρ π 2 1343bookfm Page 7 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 8 Mechanics of Composite Materials Second Edition where M mass of the beam lb or kg ρ density of beam lbin 3 or kgm 3 d diameter of beam in or m Because the length L and the load P are constant we find the mass of the beam by substituting Equation 15 and Equation 16 in Equation 14 as 17 This means that the lightest beam for specified stiffness is one with the highest value of E 12 ρ Similarly we can prove that for achieving the minimum deflection in a beam under a load along its length the lightest beam is one with the highest value of E 13 ρ Typical values of these two parameters E 12 ρ and E 13 ρ for typical fibers unidirectional composites crossply and quasiisotropic laminates steel and aluminum are given in Table 12 Comparing these numbers with metals shows composites drawing a better advantage for these two parameters Other mechanical parameters for comparing the perfor mance of composites to metals include resistance to fracture fatigue impact and creep Yes composites have distinct advantages over metals Are there any draw backs or limitations in using them Yes drawbacks and limitations in use of composites include High cost of fabrication of composites is a critical issue For example a part made of graphiteepoxy composite may cost up to 10 to 15 times the material costs A finished graphiteepoxy composite part may cost as much as 300 to 400 per pound 650 to 900 per kilogram Improvements in processing and manufacturing tech niques will lower these costs in the future Already manufacturing techniques such as SMC sheet molding compound and SRIM structural reinforcement injection molding are lowering the cost and production time in manufacturing automobile parts Mechanical characterization of a composite structure is more com plex than that of a metal structure Unlike metals composite mate rials are not isotropic that is their properties are not the same in all directions Therefore they require more material parameters For example a single layer of a graphiteepoxy composite requires nine M L P E cr 2 1 2 1 2 π ρ 1343bookfm Page 8 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Introduction to Composite Materials 9 stiffness and strength constants for conducting mechanical analysis In the case of a monolithic material such as steel one requires only four stiffness and strength constants Such complexity makes struc tural analysis computationally and experimentally more compli cated and intensive In addition evaluation and measurement techniques of some composite properties such as compressive strengths are still being debated Repair of composites is not a simple process compared to that for metals Sometimes critical flaws and cracks in composite structures may go undetected TABLE 12 Specific Modulus Parameters Eρ E12ρ and E13ρ for Typical Materials Material Units Specific gravity Youngs modulus Msi Eρ Msiin3lb E12ρ psi12in3lb E13ρ psi13in3lb System of Units USCS Graphite fiber Kevlar fiber Glass fiber Unidirectional graphiteepoxy Unidirectional glassepoxy Crossply graphiteepoxy Crossply glassepoxy Quasiisotropic graphiteepoxy Quasiisotropic glassepoxy Steel Aluminum 18 14 25 16 18 16 18 16 18 78 26 3335 1798 1233 2625 560 1392 342 1010 275 3000 1000 5128 3555 1365 4541 8609 2408 5259 1747 4229 1065 1065 88806 83836 38878 88636 36384 64545 28438 54980 25501 19437 33666 4950 5180 2558 5141 2730 4162 2317 3740 2154 1103 2294 Material Units Specific gravity Youngs modulus GPa Eρ GPam3kg E12ρ Pam3kg E13ρ Pa13m3kg System of Units SI Graphite fiber Kevlar fiber Glass fiber Unidirectional graphiteepoxy Unidirectional glassepoxy Crossply graphiteepoxy Crossply glassepoxy Quasiisotropic graphiteepoxy Quasiisotropic glassepoxy Steel Aluminum 18 14 25 16 18 16 18 16 18 78 26 23000 12400 8500 18100 3860 9598 2358 6964 1896 20684 6895 01278 008857 0034 01131 002144 0060 00131 004353 001053 002652 002662 2664 2515 1166 2659 1091 1936 8531 1649 7650 583 1010 3404 3562 1759 3535 1878 2862 1593 2571 1481 07582 1577 1343bookfm Page 9 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 10 Mechanics of Composite Materials Second Edition Composites do not have a high combination of strength and fracture toughness compared to metals In Figure 14 a plot is shown for fracture toughness vs yield strength for a 1in 25mm thick mate rial3 Metals show an excellent combination of strength and fracture toughness compared to composites Note The transition areas in Figure 14 will change with change in the thickness of the specimen Composites do not necessarily give higher performance in all the properties used for material selection In Figure 15 six primary material selection parameters strength toughness formability FIGURE 13 A uniformly loaded plate with a crack In a material with a crack the value of the stress intensity factor gives the measure of stresses in the crack tip region For example for an infinite plate with a crack of length 2a under a uniaxial load σ Figure 13 the stress intensity factor is If the stress intensity factor at the crack tip is greater than the critical stress intensity factor of the material the crack will grow The greater the value of the critical stress intensity factor is the tougher the material is The critical stress intensity factor is called the fracture toughness of the material Typical values of fracture toughness are for aluminum and for steel σ σ 2a K a σ π 2366 ksi in 26 MPa m 2548 ksi in 28 MPa m 1343bookfm Page 10 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Introduction to Composite Materials 11 FIGURE 14 Fracture toughness as a function of yield strength for monolithic metals ceramics and metalceramic composites Source Eager TW Whither advanced materials Adv Mater Pro cesses ASM International June 1991 2529 FIGURE 15 Primary material selection parameters for a hypothetical situation for metals ceramics and metalceramic composites Source Eager TW Whither advanced materials Adv Mater Pro cesses ASM International June 1991 2529 Plasticgeneral yielding Kcσy 25 in12 Kcσy 06 in12 Elasticplane strain Ceramics Composites Elasticplasticmixed mode Aluminum Yield strength 103 psi Fracture toughness ksiin 12 Polymers 400 300 200 100 100 200 300 400 500 T it a n i u m St ee l Strength Ceramic Metal Composite Affordability Corrosion resistance Joinability Formability Toughness 1343bookfm Page 11 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 12 Mechanics of Composite Materials Second Edition joinability corrosion resistance and affordability are plotted3 If the values at the circumference are considered as the normalized required property level for a particular application the shaded areas show values provided by ceramics metals and metalceramic com posites Clearly composites show better strength than metals but lower values for other material selection parameters Why are fiber reinforcements of a thin diameter The main reasons for using fibers of thin diameter are the following Actual strength of materials is several magnitudes lower than the theoretical strength This difference is due to the inherent flaws in the material Removing these flaws can increase the strength of the material As the fibers become smaller in diameter the chances of an inherent flaw in the material are reduced A steel plate may have strength of 100 ksi 689 MPa while a wire made from this steel plate can have strength of 600 ksi 4100 MPa Figure 16 shows how the strength of a carbon fiber increases with the decrease in its diameter6 FIGURE 16 Fiber strength as a function of fiber diameter for carbon fibers Reprinted from Lamotte E De and Perry AJ Fibre Sci Technol 3 159 1970 With permission from Elsevier 3 25 2 15 15 75 10 Fiber diameter μm Fiber strength GPa 125 15 1343bookfm Page 12 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Introduction to Composite Materials 13 For higher ductility and toughness and better transfer of loads from the matrix to fiber composites require larger surface area of the fibermatrix interface For the same volume fraction of fibers in a composite the area of the fibermatrix interface is inversely propor tional to the diameter of the fiber and is proved as follows Assume a lamina consisting of N fibers of diameter D The fiber matrix interface area in this lamina is AI N π D L 18 If one replaces the fibers of diameter D by fibers of diameter d then the number of fibers n to keep the fiber volume the same would be 19 Then the fibermatrix interface area in the resulting lamina would be AII n π d L 110 This implies that for a fixed fiber volume in a given volume of composite the area of the fibermatrix interface is inversely pro portional to the diameter of the fiber Fibers able to bend without breaking are required in manufacturing of composite materials especially for woven fabric composites Abil ity to bend increases with a decrease in the fiber diameter and is measured as flexibility Flexibility is defined as the inverse of bend ing stiffness and is proportional to the inverse of the product of the elastic modulus of the fiber and the fourth power of its diameter it can be proved as follows Bending stiffness is the resistance to bending moments According to the Strength of Materials course if a beam is subjected to a pure bending moment M Ductility is the ability of a material to deform without fracturing It is measured by extending a rod until fracture and measuring the initial Ai and final Af crosssectional area Then ductil ity is defined as R 1 AfAi n N D d 2 N D L d π 2 4 Volume of fibers d 1343bookfm Page 13 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 14 Mechanics of Composite Materials Second Edition 111 where v deflection of the centroidal line in or m E Youngs modulus of the beam psi or Pa I second moment of area in4 or m4 x coordinate along the length of beam in or m The bending stiffness then is EI and the flexibility is simply the inverse of EI Because the second moment of area of a cylindrical beam of diameter d is 112 then 113 For a particular material unlike strength the Youngs modulus does not change appreciably as a function of its diameter Therefore the flexibility for a particular material is inversely proportional to the fourth power of the diameter What fiber factors contribute to the mechanical performance of a composite Four fiber factors contribute to the mechanical performance of a composite7 Length The fibers can be long or short Long continuous fibers are easy to orient and process but short fibers cannot be controlled fully for proper orientation Long fibers provide many benefits over short fibers These include impact resistance low shrinkage improved surface finish and dimensional stability However short fibers pro vide low cost are easy to work with and have fast cycle time fab rication procedures Short fibers have fewer flaws and therefore have higher strength Orientation Fibers oriented in one direction give very high stiffness and strength in that direction If the fibers are oriented in more than one direction such as in a mat there will be high stiffness and strength in the directions of the fiber orientations However for the same volume of fibers per unit volume of the composite it cannot match the stiffness and strength of unidirectional composites d v dx M EI 2 2 I πd 4 64 Flexibility Ed 1 4 1343bookfm Page 14 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Introduction to Composite Materials 15 Shape The most common shape of fibers is circular because han dling and manufacturing them is easy Hexagon and square shaped fibers are possible but their advantages of strength and high packing factors do not outweigh the difficulty in handling and processing Material The material of the fiber directly influences the mechanical performance of a composite Fibers are generally expected to have high elastic moduli and strengths This expectation and cost have been key factors in the graphite aramids and glass dominating the fiber market for composites What are the matrix factors that contribute to the mechanical performance of composites Use of fibers by themselves is limited with the exceptions of ropes and cables Therefore fibers are used as reinforcement to matrices The matrix functions include binding the fibers together protecting fibers from the environment shielding from damage due to handling and distributing the load to fibers Although matrices by themselves generally have low mechan ical properties compared to those of fibers the matrix influences many mechanical properties of the composite These properties include transverse modulus and strength shear modulus and strength compressive strength interlaminar shear strength thermal expansion coefficient thermal resis tance and fatigue strength Other than the fiber and the matrix what other factors influence the mechanical performance of a composite Other factors include the fibermatrix interface It determines how well the matrix transfers the load to the fibers Chemical mechanical and reaction bonding may form the interface In most cases more than one type of bonding occurs Chemical bonding is formed between the fiber surface and the matrix Some fibers bond naturally to the matrix and others do not Coupling agents are often added to form a chemical bond The natural roughness or etching of the fiber surface causing inter locking may form a mechanical bond between the fiber and matrix If the thermal expansion coefficient of the matrix is higher than that of the fiber and the manufacturing temperatures are higher than the operating temperatures the matrix will radially shrink more than the fiber This causes the matrix to compress around the fiber Coupling agents are compounds applied to fiber surfaces to improve the bond between the fiber and matrix For example silane finish is applied to glass fibers to increase adhesion with epoxy matrix 1343bookfm Page 15 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 16 Mechanics of Composite Materials Second Edition Reaction bonding occurs when atoms or molecules of the fiber and the matrix diffuse into each other at the interface This interdiffusion often creates a distinct interfacial layer called the interphase with different properties from that of the fiber or the matrix Although this thin interfacial layer helps to form a bond it also forms micro cracks in the fiber These microcracks reduce the strength of the fiber and thus that of the composite Weak or cracked interfaces can cause failure in composites and reduce the properties influenced by the matrix They also allow environmental hazards such as hot gases and moisture to attack the fibers Although a strong bond is a requirement in transferring loads from the matrix to the fiber weak debonding of the fibermatrix interface is used advantageously in ceramic matrix composites Weak interfaces blunt matrix cracks and deflect them along the interface This is the main source of improving toughness of such composites up to five times that of the mono lithic ceramics What is the world market of composites The world market for composites is only 10 109 US dollars as compared to more than 450 109 US dollars for steel The annual growth of composites is at a steady rate of 10 Presently composite shipments are about 3 109 lb annually Figure 17 gives the relative market share of US composite shipments and shows transportation clearly leading in their use Table 13 shows the market share of composites since 1990 12 Classification How are composites classified Composites are classified by the geometry of the reinforcement partic ulate flake and fibers Figure 18 or by the type of matrix polymer metal ceramic and carbon Particulate composites consist of particles immersed in matrices such as alloys and ceramics They are usually isotropic because the par ticles are added randomly Particulate composites have advantages such as improved strength increased operating temperature oxida tion resistance etc Typical examples include use of aluminum par ticles in rubber silicon carbide particles in aluminum and gravel sand and cement to make concrete Flake composites consist of flat reinforcements of matrices Typical flake materials are glass mica aluminum and silver Flake compos 1343bookfm Page 16 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Introduction to Composite Materials 17 FIGURE 17 Approximate shipments of polymerbased composites in 1995 Source Data used in figure published with permission of the SPI Inc httpwwwsocplasorg TABLE 13 US Composites Shipment in 106 lb Including Reinforced Thermoset and Thermoplastic Resin Composites Reinforcements and Fillers Markets 1990 1991 1992 1993 1994 1995 Aircraftaerospacemilitary 39 387 323 254 242 240 Appliancebusiness equipment 153 1352 1432 1475 1607 1665 Construction 468 4200 4830 5300 5969 6269 Consumer products 165 1487 1622 1657 1748 1836 Corrosionresistant equipment 350 3550 3323 3520 3763 3946 Electricalelectronic 241 2311 2600 2749 2993 3151 Marine 375 2750 3044 3193 3635 3751 Transportation 705 6822 7500 8221 9456 9840 Other 79 738 834 893 1018 1066 TOTAL 2575 2360 2551 2726 30431 31764 Source Published with permission of the SPI Inc Appliance business equipment Consumer products Electrical electronics Corrosionresistant equipment Marine Construction Transportation Other Total shipments in 1995 3176 109lb 1441 109 kgs 1343bookfm Page 17 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 18 Mechanics of Composite Materials Second Edition ites provide advantages such as high outofplane flexural modulus higher strength and low cost However flakes cannot be oriented easily and only a limited number of materials are available for use Fiber composites consist of matrices reinforced by short discontin uous or long continuous fibers Fibers are generally anisotropic and examples include carbon and aramids Examples of matrices are resins such as epoxy metals such as aluminum and ceramics such as calciumalumino silicate Continuous fiber composites are emphasized in this book and are further discussed in this chapter by the types of matrices polymer metal ceramic and carbon The fundamental units of continuous fiber matrix composite are unidi rectional or woven fiber laminas Laminas are stacked on top of each other at various angles to form a multidirectional laminate Nanocomposites consist of materials that are of the scale of nanome ters 109 m The accepted range to be classified as a nanocomposite is that one of the constituents is less than 100 nm At this scale the FIGURE 18 Types of composites based on reinforcement shape Out of plane flexural stiffness is the resistance to deflection under bending that is out of the plane such as bending caused by a heavy stone placed on a simply supported plate Anisotropic materials are the opposite of isotropic materials like steel and aluminum they have different properties in different directions For example the Youngs modulus of a piece of wood is higher different in the direction of the grain than in the direction perpendicular to the grain In comparison a piece of steel has the same Youngs modulus in all directions Particulate composites Flake composites Fiber composites 1343bookfm Page 18 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Introduction to Composite Materials 19 properties of materials are different from those of the bulk material Generally advanced composite materials have constituents on the microscale 106 m By having materials at the nanometer scale most of the properties of the resulting composite material are better than the ones at the microscale Not all properties of nanocomposites are better in some cases toughness and impact strength can decrease Applications of nanocomposites include packaging applications for the military in which nanocomposite films show improvement in properties such as elastic modulus and transmission rates for water vapor heat distortion and oxygen8 Body side molding of the 2004 Chevrolet Impala is made of olefin based nanocomposites9 This reduced the weight of the molding by 7 and improved its surface quality General Motors currently uses 540000 lb of nanocomposite materials per year Rubber containing just a few parts per million of metal conducts electricity in harsh conditions just like solid metal Called Metal Rubber it is fabricated molecule by molecule by a process called electrostatic selfassembly Awaited applications of the Metal Rubber include artificial muscles smart clothes flexible wires and circuits for portable electronics10 121 Polymer Matrix Composites What are the most common advanced composites The most common advanced composites are polymer matrix composites PMCs consisting of a polymer eg epoxy polyester urethane reinforced by thin diameter fibers eg graphite aramids boron For example graphite epoxy composites are approximately five times stronger than steel on a weight forweight basis The reasons why they are the most common composites include their low cost high strength and simple manufacturing principles What are the drawbacks of polymer matrix composites The main drawbacks of PMCs include low operating temperatures high coefficients of thermal and moisture expansion and low elastic properties in certain directions What are the typical mechanical properties of some polymer matrix com posites Compare these properties with metals Table 14 gives typical mechanical properties of common polymer matrix composites Some materials such as polymers absorb or deabsorb moisture that results in dimensional changes The coefficient of moisture expansion is the change in length per unit length per unit mass of moisture absorbed per unit mass of the substance 1343bookfm Page 19 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 20 Mechanics of Composite Materials Second Edition Give names of various fibers used in advanced polymer composites The most common fibers used are glass graphite and Kevlar Typical properties of these fibers compared with bulk steel and aluminum are given in Table 15 Give a description of the glass fiber Glass is the most common fiber used in polymer matrix composites Its advantages include its high strength low cost high chemical resistance and good insulating properties The drawbacks include low elastic modulus TABLE 14 Typical Mechanical Properties of Polymer Matrix Composites and Monolithic Materials Property Units Graphite epoxy Glass epoxy Steel Aluminum System of units USCS Specific gravity Youngs modulus Ultimate tensile strength Coefficient of thermal expansion Msi ksi μininF 16 2625 2176 001111 18 5598 1540 4778 78 300 940 65 26 100 400 128 System of units SI Specific gravity Youngs modulus Ultimate tensile strength Coefficient of thermal expansion GPa MPa μmmC 16 1810 1500 002 18 386 1062 86 78 2068 6481 117 26 6895 2758 23 TABLE 15 Typical Mechanical Properties of Fibers Used in Polymer Matrix Composites Property Units Graphite Aramid Glass Steel Aluminum System of units USCS Specific gravity Youngs modulus Ultimate tensile strength Axial coefficient of thermal expansion Msi ksi μininF 18 3335 2998 0722 14 1798 2000 2778 25 1233 2248 2778 78 30 94 65 26 100 400 128 System of units SI Specific gravity Youngs modulus Ultimate tensile strength Axial coefficient of thermal expansion GPa MPa μmmC 18 230 2067 13 14 124 1379 5 25 85 1550 5 78 2068 6481 117 26 6895 2758 23 1343bookfm Page 20 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Introduction to Composite Materials 21 poor adhesion to polymers high specific gravity sensitivity to abrasion reduces tensile strength and low fatigue strength Types The main types are Eglass also called fiberglass and Sglass The E in Eglass stands for electrical because it was designed for electrical applications However it is used for many other purposes now such as decorations and structural applications The S in Sglass stands for higher content of silica It retains its strength at high temperatures compared to E glass and has higher fatigue strength It is used mainly for aerospace appli cations Some property differences are given in Table 16 The difference in the properties is due to the compositions of Eglass and Sglass fibers The main elements in the two types of fibers are given in Table 17 Other types available commercially are Cglass C stands for corrosion used in chemical environments such as storage tanks Rglass used in struc tural applications such as construction Dglass dielectric used for applica tions requiring low dielectric constants such as radomes and Aglass appearance used to improve surface appearance Combination types such TABLE 16 Comparison of Properties of EGlass and SGlass Property Units EGlass SGlass System of units USCS Specific gravity Youngs modulus Ultimate tensile strength Coefficient of thermal expansion Msi ksi μininF 254 105 500 28 249 124 665 31 System of units SI Specific gravity Youngs modulus Ultimate tensile strength Coefficient of thermal expansion GPa MPa μmmC 254 7240 3447 504 249 8550 4585 558 TABLE 17 Chemical Composition of EGlass and SGlass Fibers Material Weight EGlass SGlass Silicon oxide Aluminum oxide Calcium oxide Magnesium oxide Boron oxide Others 54 15 17 45 8 15 64 25 001 10 001 08 1343bookfm Page 21 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 22 Mechanics of Composite Materials Second Edition as ECR glass ECR stands for electrical and corrosion resistance and AR glass alkali resistant also exist Manufacturing Glass fibers are made generally by drawing from a melt11 as shown in Figure 19 The melt is formed in a refractory furnace at about 2550F 1400C from a mixture that includes sand limestone and alumina The melt is stirred and maintained at a constant temperature It passes through as many as 250 heated platinum alloy nozzles of about 394 μin 10 μm diameter where it is drawn into filaments of needed size at high speeds of about 361 mih 25 ms These fibers are sprayed with an organic sizing FIGURE 19 Schematic of manufacturing glass fibers and available glass forms From Bishop W in Ad vanced Composites Partridge IK Ed Kluwer Academic Publishers London 1990 Figure 4 p 177 Reproduced with kind permission of Springer Glass feedstock Electrically heated furnace Protective sizing operation water or solvent based Filaments collected together to form a strand Glass filaments Roving Woven roving Untwisted strand Chopping operation Twisting of strand Chopped strand Application of resin binder Chopped strand Mat Weaving Woven Fabric Glass strand wound onto a forming tube and oven dried to remove watersolvent 1343bookfm Page 22 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Introduction to Composite Materials 23 solution before they are drawn The sizing solution is a mixture of binders lubricants and coupling and antistatic agents binders allow filaments to be packed in strands lubricants prevent abrasion of filaments and coupling agents give better adhesion between the inorganic glass fiber and the organic matrix Fibers are then drawn into strands and wound on a forming tube Strands are groups of more than 204 filaments The wound array of strands is then removed and dried in an oven to remove any water or sizing solutions The glass strand can then be converted into several forms as shown in Figure 19 Different forms of various fibers are shown in Figure 110 Give a description of graphite fibers Graphite fibers are very common in highmodulus and highstrength applications such as aircraft components etc The advantages of graphite fibers include high specific strength and modulus low coefficient of thermal expansion and high fatigue strength The drawbacks include high cost low impact resistance and high electrical conductivity Manufacturing Graphite fibers have been available since the late 1800s However only since the early 1960s has the manufacturing of graphite fibers taken off Graphite fibers are generally manufactured from three precursor materials rayon polyacrylonitrile PAN and pitch PAN is the most popular precursor and the process to manufacture graphite fibers from it is given next Figure 111 PAN fibers are first stretched five to ten times their length to improve their mechanical properties and then passed through three heating processes In the first process called stabilization the fiber is passed through a furnace between 392 and 572F 200 and 300C to stabilize its dimensions during the subsequent hightemperature processes In the second process called carbonization it is pyrolized in an inert atmosphere of nitrogen or argon between 1832 and 2732F 1000 and 1500C In the last process called graphitization it is heat treated above 4532F 2500C The graphitization yields a microstructure that is more graphitic than that produced by carbon ization The fibers may also be subjected to tension in the last two heating processes to develop fibers with a higher degree of orientation At the end of this threestep heat treatment process the fibers are surface treated to develop fiber adhesion and increase laminar shear strength when they are used in composite structures They are then collected on a spool Properties Table 18 gives properties of graphite fibers obtained from two different precursors Are carbon and graphite the same No7 they are different Carbon fibers have 93 to 95 carbon content but graphite has more than 99 carbon content Also carbon fibers are produced Pyrolysis is defined as the decomposition of a complex organic substance to one of a simpler structure by means of heat 1343bookfm Page 23 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 24 Mechanics of Composite Materials Second Edition at 2400F 1316C and graphite fibers are typically produced in excess of 3400F 1900C Give a description of the aramid fiber An aramid fiber is an aromatic organic compound made of carbon hydro gen oxygen and nitrogen Its advantages are low density high tensile FIGURE 110 Forms of available fibers Graphic courtesy of MC Gill Corporation httpwww mcgillcorpcom 1343bookfm Page 24 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Introduction to Composite Materials 25 strength low cost and high impact resistance Its drawbacks include low compressive properties and degradation in sunlight Types The two main types of aramid fibers are Kevlar 29 and Kevlar 49 Both types of Kevlar fibers have similar specific strengths but Kevlar 49 has a higher specific stiffness Kevlar 29 is mainly used in bulletproof FIGURE 111 Stages of manufacturing a carbon fiber from PANbased precursors TABLE 18 Mechanical Properties of Two Typical Graphite Fibers Property Units PITCH PAN System of units USCS Specific gravity Youngs modulus Ultimate tensile strength Axial coefficient of thermal expansion Msi ksi μininF 199 55 250 03 178 35 500 07 System of units SI Specific gravity Youngs modulus Ultimate tensile strength Axial coefficient of thermal expansion GPa MPa μmmC 199 3792 1723 054 178 2413 3447 126 Kevlar 29 is a registered trademark of EI duPont deNemours and Company Inc Wilmington DE Kevlar 49 is a registered trademark of EI duPont deNemours and Company Inc Wilmington DE Stretching Offwind creel Surface treatment Windup creel Stabilization 200300C Carbonization 10001500C Graphitization 2500C 1343bookfm Page 25 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 26 Mechanics of Composite Materials Second Edition vests ropes and cables High performance applications in the aircraft indus try use Kevlar 49 Table 19 gives the relative properties of Kevlar 29 and Kevlar 49 Manufacturing The fiber is produced by making a solution of proprietary polymers and strong acids such as sulfuric acid The solution is then extruded into hot cylinders at 392F 200C washed and dried on spools The fiber is then stretched and drawn to increase its strength and stiffness Give names of various polymers used in advanced polymer composites These polymers include epoxy phenolics acrylic urethane and polyamide Why are there so many resin systems in advanced polymer composites Each polymer has its advantages and drawbacks in its use12 Polyesters The advantages are low cost and the ability to be made translucent drawbacks include service temperatures below 170F 77C brittleness and high shrinkage of as much as 8 during curing Phenolics The advantages are low cost and high mechanical strength drawbacks include high void content Epoxies The advantages are high mechanical strength and good adherence to metals and glasses drawbacks are high cost and diffi culty in processing TABLE 19 Properties of Kevlar Fibers Property Units Kevlar 29 Kevlar 49 System of units USCS Specific gravity Youngs modulus Ultimate tensile strength Axial coefficient of thermal expansion Msi ksi μininF 144 9 525 1111 148 19 525 1111 System of units SI Specific gravity Youngs modulus Ultimate tensile strength Axial coefficient of thermal expansion GPa MPa μmmC 144 6205 3620 2 148 1310 3620 2 Shrinkage in resins is found by measuring the density of the resin before and after crosslink ing If ρ is the density before crosslinking and ρ is the density after crosslinking The percent shrinkage is defined as shrinkage ρ ρρ 100 1343bookfm Page 26 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Introduction to Composite Materials 27 As can be seen each of the resin systems has its advantages and drawbacks The use of a particular system depends on the application These consider ations include mechanical strength cost smoke emission temperature excursions etc Figure 112 shows the comparison of five common resins based on smoke emission strength service temperature and cost12 Give a description of epoxy Epoxy resins are the most commonly used resins They are low molecular weight organic liquids containing epoxide groups Epoxide has three mem bers in its ring one oxygen and two carbon atoms The reaction of epichlo rohydrin with phenols or aromatic amines makes most epoxies Hardeners FIGURE 112 Comparison of performance of several common matrices used in polymer matrix composites Graphic courtesy of MC Gill Corporation httpwwwmcgillcorpcom Hardeners are substances that are added to polymers for aiding in curing of composites 10 8 6 4 2 0 Most desirable Phenolic Epoxy Polyester Silicone Polymide Least desirable Smoke emission Maximum strength Service temperature Cost 1343bookfm Page 27 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 28 Mechanics of Composite Materials Second Edition plasticizers and fillers are also added to produce epoxies with a wide range of properties of viscosity impact degradation etc The room temper ature properties of a typical epoxy are given in Table 110 Epoxy is the most common type of matrix material Why Although epoxy is costlier than other polymer matrices it is the most popular PMC matrix More than twothirds of the polymer matrices used in aerospace applications are epoxy based The main reasons why epoxy is the most used polymer matrix material are High strength Low viscosity and low flow rates which allow good wetting of fibers and prevent misalignment of fibers during processing Low volatility during cure Low shrink rates which reduce the tendency of gaining large shear stresses of the bond between epoxy and its reinforcement Available in more than 20 grades to meet specific property and processing requirements Polymers are classified as thermosets and thermoplastics What is the difference between the two Give some examples of both Thermoset polymers are insoluble and infusible after cure because the chains are rigidly joined with strong covalent bonds thermoplastics are formable at high temperatures and pressure because the bonds are weak and TABLE 110 Room Temperature Properties of a Typical Epoxy Property Units Value System of units USCS Specific gravity Youngs modulus Ultimate tensile strength Msi ksi 128 055 120 System of units SI Specific gravity Youngs modulus Ultimate tensile strength GPa MPa 128 3792 8274 Plasticizers are lubricants that improve the toughness flexibility processibility and ductility of polymers This improvement is generally at the expense of lower strength Fillers are ingredients added to enhance properties such as strength surface texture and ultra violet absorption of a polymer and to lower the cost of polymers Typical examples include chopped fabric and wood flour 1343bookfm Page 28 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Introduction to Composite Materials 29 of the van der Waals type Typical examples of thermoset include epoxies polyesters phenolics and polyamide typical examples of thermoplastics include polyethylene polystyrene polyetheretherketone PEEK and polyphenylene sulfide PPS The differences between thermosets and ther moplastics are given in the following table 13 What are prepregs Prepregs are a readymade tape composed of fibers in a polymer matrix Figure 113 They are available in standard widths from 3 to 50 in 76 to 1270 mm Depending on whether the polymer matrix is thermoset or ther moplastic the tape is stored in a refrigerator or at room temperature respec tively One can lay these tapes manually or mechanically at various orientations to make a composite structure Vacuum bagging and curing under high pressures and temperatures may follow Figure 114 shows the schematic of how a prepreg is made 14 A row of fibers is passed through a resin bath The resinimpregnated fibers are then Thermoplastics Thermoset Soften on heating and pressure and thus easy to repair Decompose on heating High strains to failure Low strains to failure Indefinite shelf life Definite shelf life Can be reprocessed Cannot be reprocessed Not tacky and easy to handle Tacky Short cure cycles Long cure cycles Higher fabrication temperature and viscosities have made it difficult to process Lower fabrication temperature Excellent solvent resistance Fair solvent resistance FIGURE 113 Boronepoxy prepreg tape Photo courtesy of Specialty Materials Inc httpwwwspecmaterialscom 1343C001fm Page 29 Wednesday September 28 2005 1029 AM 2006 by Taylor Francis Group LLC 30 Mechanics of Composite Materials Second Edition heated to advance the curing reaction from Astage to the Bstage A release film is now wound over a takeup roll and backed with a release film The release film keeps the prepregs from sticking to each other during storage Give examples of how a polymer matrix composite is manufactured Techniques of manufacturing a polymer matrix composite include filament winding used generally for making pipes and tanks to handle chemicals autoclave forming used to make complex shapes and flat panels for structures in which low void content and high quality are important and resin transfer molding used extensively in the automotive industry because short produc tion runs are necessary Filament winding Fibers are impregnated with a resin by drawing them through an inline resin bath wet winding Figure 116 or prepregs dry winding are wound over a mandrel Wet winding is inexpensive and lets one control the properties of the composite Dry winding is cleaner but more expensive and thus quite uncommon FIGURE 114 Schematic of prepreg manufacturing Reprinted from Mallick PK FiberReinforced Composites Materials Manufacturing and Design Marcel Dekker Inc New York Chap 2 1988 p 62 Courtesy of CRC Press Boca Raton FL Thermosetting resins have three curing stages A B and C Figure 115 Resins are manufactured in the Astage in which the resin may be solid or liquid but is able to flow if heat is applied The Astage is also called the completely uncured stage The Bstage is the middle stage of the reaction of a thermosetting resin used when prepregs are manufactured This stage allows easy processing and handling of composite layers such as graphiteepoxy The Cstage is the final stage in the reaction of a thermosetting resin This stage is accom plished when a composite structure is made out of composite layers Heat and pressure may be applied at the Bstage to cure the resin completely This stage results in irreversible hardening and insolubility Controlled heating elements Metering device Fiber collimator Resin solution Fiber package Backup material release film Takeup roll 1343C001fm Page 30 Wednesday September 28 2005 1029 AM 2006 by Taylor Francis Group LLC Introduction to Composite Materials 31 FIGURE 115 Curing stages of phenolic resins Graphic courtesy of MC Gill Corporation httpwwwmcgillcorpcom Curing stages of phenolic resins OH OH COH COH C O C HO C OH C OH OH OH COH COH C O C C O C HO C C OH OH OH COH OH C C OH H2O OH OH OH OH OH OH OH OH OH H2O A STAGE Low molecular weight linear polymer B STAGE Higher molecular weight partly crosslinked C STAGE Fully crosslinked cured 1343bookfm Page 31 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 32 Mechanics of Composite Materials Second Edition Depending on the desired properties of the product winding patterns such as hoop helical and polar can be developed The product is then cured with or without heat and pressure Depending on the application mandrels are made of wood aluminum steel plaster or salts For example steel mandrels are chosen for manufacturing large quantities of openended cylinders and lowmelting alloys or watersoluble salts are used for closedended cylinders so that one can easily remove the mandrel Autoclave forming This method of manufacturing is used with composites available as prepregs First a peel ply made out of nylon or cellophane coated with Teflon is placed on the mold Teflon is used for easy removal of the part and the peel ply achieves a desired finish that is smooth and wrinkle free Replacing Teflon by mold releasing powders and liquids can also accomplish removal of the part Prepregs of the required number are laid up one ply at a time by automated means or by hand Each ply is pressed to remove any entrapped air and wrinkles The layup is sealed at the edges to form a vacuum seal FIGURE 116 a Filament winding process b filament wound pressure vessel with liner From Chawla KK Composite Materials Science and Engineering SpringerVerlag 1998 Reprinted by per mission of SpringerVerlag Teflon is a registered trademark of EI duPont deNemours and Company Inc Wilmington DE Mold a structure around or in which the composite forms a desired shape Molds are female and male If the composite part is in the mold it is called a female mold if it is made around the mold it is called a male mold See in Figure 117 the male mold that was used in making a humanpowered submarine 1343bookfm Page 32 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Introduction to Composite Materials 33 Now one establishes the bleeder system to get rid of the volatiles and excess resin during the heating and vacuum process that follows later The bleeder system consists of several bleeder sheets made of glass cloth These are placed on the edges and the top of the layup Then vacuum connections are placed over the bleeders and the layup is bagged A partial vacuum is developed to smooth the bag surface The whole assembly is put in an autoclave Figure 118 where heat and pressure are applied with an inert gas such as nitrogen The vacuum system is kept functioning to remove volatiles during the cure cycle and to keep the part conformed to the mold The cure cycle may last more than 5 h Resin transfer molding RTM also called liquid molding A low viscosity resin such as polyester or epoxy resin is injected under low pressure into a closed mold that contains the fiber preform The resin flow is stopped and the part is allowed to cure The cure is done at room temperature or at elevated temperatures The latter is done if the part is to be used for hightemperature FIGURE 117 Humanpowered submarine and its mold Courtesy of Professor GH Besterfield and student section of ASME University of South Florida Tampa 1343bookfm Page 33 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 34 Mechanics of Composite Materials Second Edition application The advantages of RTM are that it is less expensive than hand layup can be automated and does not need refrigerated storage for prepregs Major drawbacks include the capital expense required for having two molds instead of one15 Give typical applications of polymer matrix composites Applications of polymer matrix composites range from tennis racquets to the space shuttle Rather than enumerating only the areas in which polymer based composites are used a few examples have been taken from each industry Emphasis has been placed on why a composite material is the material of choice Aircraft The military aircraft industry has mainly led the use of polymer composites The percentage of structural weight of composites that was less than 2 in F15s in the 1970s has increased to about 30 on the AV8B in the 1990s In both cases the weight reduction over metal parts was more than 20 In commercial airlines the use of composites has been conservative because of safety concerns Use of composites is limited to secondary struc tures such as rudders and elevators made of graphiteepoxy for the Boeing 767 and landing gear doors made of Kevlargraphiteepoxy Composites are also used in panels and floorings of airplanes Some examples of using composites in the primary structure are the allcomposite Lear Fan 2100 plane and the tail fin of the Airbus A310300 In the latter case the tail fin consists of graphiteepoxy and aramid honeycomb It not only reduced the weight of the tail fin by 662 lb 300 kg but also reduced the number of parts from 2000 to 100 Skins of aircraft engine cowls shown in Figure 119 are also made of polymer matrix composites for reducing weight16 With increasing competition in model airplane flying the weight of com posite materials has been reduced Figure 120 shows a World War II model airplane with fuselage made of glassepoxy wings made of balsawood FIGURE 118 Autoclave used for processing polymer matrix composites Photo courtesy of ACP Composites MN httpwwwacpcompositescom 1343bookfm Page 34 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Introduction to Composite Materials 35 FIGURE 119 Aircraft engine cowling Photo provided courtesy of Alliant Techsystems Inc FIGURE 120 Model BF109 WWII German fighter plane using glassepoxymolded fuselage and wing spars of graphiteepoxy Photo courtesy of Russell A Lepré Tampa FL 1343bookfm Page 35 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 36 Mechanics of Composite Materials Second Edition facingsStyrofoam core sandwich construction and wingspars made of graphiteepoxy17 Helicopters and tiltrotors Figure 121 use graphiteepoxy and glass epoxy rotor blades that not only increase the life of blades by more than 100 over metals but also increase the top speeds Space Two factors make composites the material of choice in space appli cations high specific modulus and strength and dimensional stability dur ing large changes in temperature in space Examples include the Graphite epoxyhoneycomb payload bay doors in the space shuttle Figure 122 Weight savings7 over conventional metal alloys translate to higher payloads that cost as much as 1000lb 2208kg Also for the space shuttles graph iteepoxy was chosen primarily for weight savings and for small mechanical and thermal deflections concerning the remote manipulator arm which deploys and retrieves payloads Figure 123 shows a highgain antenna for the space station that uses sandwiches made of graphiteepoxy facings with an aluminum honeycomb core Antenna ribs and struts in satellite systems use graphiteepoxy for their high specific stiffness and its ability to meet the dimensional stability16 requirements due to large temperature excursions in space In June 2004 Paul G Allen and Scaled Composites18 launched the first privately manned vehicle called SpaceshipOne beyond the Earths atmo sphere Figure 124 The spaceship reached a recordbreaking altitude of approximately 62 miles 100 km SpaceshipOne is constructed from graph iteepoxy composite materials a trowelon ablative thermal protection layer19 protects its hotter sections Sporting goods Graphiteepoxy is replacing metals in golf club shafts mainly to decrease the weight and use the saved weight in the head This increase in the head weight has improved driving distances by more than 25 yards 23 m FIGURE 121 The BELL V22 Osprey in combat configuration Courtesy of Bell Helicopter Textron Inc 1343bookfm Page 36 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Introduction to Composite Materials 37 FIGURE 122 Use of composites in the space shuttle Graphic courtesy of MC Gill Corporation httpwwwmcgillcorpcom FIGURE 123 Highgain antenna for space station Photo provided courtesy of Alliant Techsystems Inc Boronaluminum midfuselage truss members High specific strength stiffness Nomex needled felt Protection to 750F Saves 350lbs Over silica insulation Graphiteepoxy OMS pods High strength toweight ratio Boronepoxy reinforced titanium truss members Pressure vessels Fiberglass overwrapping Stowage boxes compartments Fiberglass sand wich face sheet Nomex core Saves 160 lbs over aluminum Graphiteepoxy payload bay doors High strengthto weight ratio Purge vent lines Fiberglass cloth epoxy resin Sleevescryogenic lines Circular knit fiberglasspolyurethane resin Lightweight flexible nonflammable 1343bookfm Page 37 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 38 Mechanics of Composite Materials Second Edition Bicycles use hybrid construction of graphiteepoxy composites wound on an aluminum tubing or chopped Sglass reinforced urethane foam The graphiteepoxy composite increases the specific modulus of the tube and decreases the mass of the frame by 25 Composites also allow frames to consist of one piece which improves fatigue life and avoids stress concen tration found in metallic frames at their joints Bicycle wheels made of carbonpolymide composites offer low weight and better impact resistance than aluminum Tennis and racquetball rackets with graphiteepoxy frames are now com monplace The primary reasons for using composites are that they improve the torsional rigidity of the racquet and reduce risk of elbow injury due to vibration damping Ice hockey sticks are now manufactured out of hybrids such as Kevlarglassepoxy Kevlar is added for durability and stiffness Ski poles made of glasspolyester composites have higher strength flexibility and lower weight than conventional ski poles This reduces stress and impact on upper body joints as the skier plants his poles Medical devices Applications here include the use of glassKevlarepoxy lightweight face masks for epileptic patients Artificial portable lungs are made of graphiteglassepoxy so that a patient can be mobile Xray tables made of graphiteepoxy facing sandwiches are used for their high stiffness light weight and transparency to radiation The latter feature allows the FIGURE 124 First privately manned vehicle SpaceShipOne to go beyond the Earths atmosphere Photo provided courtesy of Scaled Composites httpwwwscaledcom If a loaded machine element has a discontinuity the stresses are different at the discontinuity The ratio between the stresses at the discontinuity and the nominal stress is defined as the stress concentration factor For example in a plate with a small hole the stress concentration factor is three at the edge of the hole Vibration damping is the ability of a material to dissipate energy during vibration Damping of composites is higher than that of conventional metals such as steel and aluminum Damping of composites depends on fiber volume fraction orientation constituent properties and stack ing sequence Damping in composites is measured by calculating the ratio of energy dissipated to the energy stored20 1343bookfm Page 38 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Introduction to Composite Materials 39 patient to stay on one bed for an operation as well as xrays and be subjected to a lower dosage of radiation Marine The application of fiberglass in boats is well known Hybrids of Kevlarglassepoxy are now replacing fiberglass for improved weight sav ings vibration damping and impact resistance Kevlarepoxy by itself would have poor compression properties Housings made of metals such as titanium to protect expensive oceano graphic research instruments during explorations of sea wrecks are cost prohibitive These housings are now made out of glassepoxy and sustain pressures as high as 10 ksi 69 MPa and extremely corrosive conditions Bridges made of polymer composite materials are gaining wide acceptance due to their low weight corrosion resistance longer life cycle and limited earthquake damage Although bridge components made of composites may cost 5lb as opposed to components made of steel reinforced concrete may only cost 030 to 100 per pound the former weighs 80 less than the latter Also by lifetime costs fewer composite bridges need to be built than traditional bridges21 Automotive The fiberglass body of the Corvette comes to mind when considering automotive applications of polymer matrix composites In addi tion the Corvette has glassepoxy composite leaf springs Figure 125 with a fatigue life of more than five times that of steel Composite leaf springs also give a smoother ride than steel leaf springs and give more rapid response to stresses caused by road shock Moreover composite leaf springs offer less chance of catastrophic failure and excellent corrosion resistance22 By weight about 8 of todays automobile parts are made of composites including bumpers body panels and doors However since 1981 the average engine horsepower has increased by 84 while average vehicle weight has increased by more than 20 To overcome the increasing weight but also maintain the safety of modern vehicles some estimate that carbon composite bodies will reduce the weight by 5023 Commercial Fiberreinforced polymers have many other commercial appli cations too Examples include mops with pultruded fiberglass handles Fig FIGURE 125 Rear fiberglass monosprings for Corvettes Photo courtesy of Vette Brakes and Products St Petersburg FL httpwwwvbandpcom 1343bookfm Page 39 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 40 Mechanics of Composite Materials Second Edition ure 126 Some brooms used in pharmaceutical factories have handles that have no joints or seams the surfaces are smooth and sealed This keeps the bacteria from staying and growing To have a handle that also is strong rigid and chemically and heat resistant the material of choice is glassfiber reinforced polypropylene24 Other applications include pressure vessels for applications such as chemical plants Garden tools Figure 12725 can be made lighter than traditional metal tools and thus are suitable for children and people with physically challenged hands Figure 127 shows the Pow ergear Fiskars anvil lopper The handles of the lopper are made of Nyglass composite making it extremely lightweight and durable 122 Metal Matrix Composites What are metal matrix composites Metal matrix composites MMCs as the name implies have a metal matrix Examples of matrices in such composites include aluminum mag nesium and titanium Typical fibers include carbon and silicon carbide Metals are mainly reinforced to increase or decrease their properties to suit the needs of design For example the elastic stiffness and strength of metals can be increased and large coefficients of thermal expansion and thermal FIGURE 126 Fiberglass mop handle Photo courtesy of RTP Company MN 1343bookfm Page 40 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Introduction to Composite Materials 41 and electric conductivities of metals can be reduced by the addition of fibers such as silicon carbide What are the advantages of metal matrix composites Metal matrix composites are mainly used to provide advantages over monolithic metals such as steel and aluminum These advantages include higher specific strength and modulus by reinforcing lowdensity metals such as aluminum and titanium lower coefficients of thermal expansion by reinforcing with fibers with low coefficients of thermal expansion such as graphite and maintaining properties such as strength at high temperatures MMCs have several advantages over polymer matrix composites These include higher elastic properties higher service temperature insensitivity to moisture higher electric and thermal conductivities and better wear fatigue and flaw resistances The drawbacks of MMCs over PMCs include higher processing temperatures and higher densities Do any properties degrade when metals are reinforced with fibers Yes reinforcing metals with fibers may reduce ductility and fracture tough ness26 Ductility of aluminum is 48 and it can decrease to below 10 with FIGURE 127 Strong efficient and lightweight Fiskars Powergear anvil lopper Photo courtesy of Fiskars Brands Inc 1343bookfm Page 41 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 42 Mechanics of Composite Materials Second Edition simple reinforcements of silicon carbide whiskers The fracture toughness of aluminum alloys is 182 to 364 20 to 40 and it reduces by 50 or more when reinforced with silicon fibers What are the typical mechanical properties of some metal matrix compos ites Compare the properties with metals Typical mechanical properties of MMCs are given in Table 111 Show one process of how metal matrix composites are manufactured Fabrication methods for MMCs are varied One method of manufacturing them is diffusion bonding26 Figure 128 which is used in manufacturing boronaluminum composite parts Figure 129 A fiber mat of boron is placed between two thin aluminum foils about 0002 in 005 mm thick A polymer binder or an acrylic adhesive holds the fibers together in the mat Layers of these metal foils are stacked at angles as required by the design The laminate is first heated in a vacuum bag to remove the binder The laminate is then hot pressed with a temperature of about 932F 500C and pressure of about 5 ksi 35 MPa in a die to form the required machine element What are some of the applications of metal matrix composites Metal matrix composites applications are Space The space shuttle uses boronaluminum tubes to support its fuselage frame In addition to decreasing the mass of the space shuttle by more than 320 lb 145 kg boronaluminum also reduced the thermal insulation requirements because of its low thermal con TABLE 111 Typical Mechanical Properties of Metal Matrix Composites Property Units SiC aluminum Graphite aluminum Steel Aluminum System of units USCS Specific gravity Youngs modulus Ultimate tensile strength Coefficient of thermal expansion Msi ksi μininF 26 17 175 69 22 18 65 10 78 30 94 65 26 10 34 128 System of units SI Specific gravity Youngs modulus Ultimate tensile strength Coefficient of thermal expansion GPa MPa μmmC 26 1172 1206 124 22 1241 4482 18 78 2068 6481 117 26 6895 23440 23 ksi in MPa m 1343bookfm Page 42 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Introduction to Composite Materials 43 ductivity The mast of the Hubble Telescope uses carbonreinforced aluminum Military Precision components of missile guidance systems demand dimensional stability that is the geometries of the components cannot change during use27 Metal matrix composites such as SiC aluminum composites satisfy this requirement because they have FIGURE 128 Schematic of diffusion bonding for metal matrix composites Reproduced with permission from Matthews FL and Rawlings RD Composite Materials Engineering and Science Chapman Hall London 1994 Figure 31 p 81 Copyright CRC Press Boca Raton FL Foil Fiber mat Foil Consolidate b d c Stack a Heat and pressure Clean and Trim NDE Secondary Fabrication e f 1343bookfm Page 43 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 44 Mechanics of Composite Materials Second Edition high microyield strength In addition the volume fraction of SiC can be varied to have a coefficient of thermal expansion compatible with other parts of the system assembly Transportation Metal matrix composites are finding use now in auto motive engines that are lighter than their metal counterparts Also because of their high strength and low weight metal matrix com posites are the material of choice for gas turbine engines Figure 130 FIGURE 129 Boronaluminum component made from diffusion bonding Photo courtesy of Specialty Materials Inc httpwwwspecmaterialscom FIGURE 130 Gas turbine engine components made of metal matrix composites Photo courtesy of Specialty Materials Inc httpwwwspecmaterialscom Microyield strength is a major design parameter for elements that are required to be dimen sionally stable It is defined as the stress required to create a plastic residual strain of 1 106 or 1 μm High strength low weight and the ability to perform at high temperatures make metal matrix composites the material of choice for gas turbine engine components 1343bookfm Page 44 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Introduction to Composite Materials 45 123 Ceramic Matrix Composites What are ceramic matrix composites Ceramic matrix composites CMCs have a ceramic matrix such as alumina calcium alumino silicate reinforced by fibers such as carbon or silicon carbide What are the advantages of ceramic matrix composites Advantages of CMCs include high strength hardness high service tem perature limits for ceramics chemical inertness and low density However ceramics by themselves have low fracture toughness Under tensile or impact loading they fail catastrophically Reinforcing ceramics with fibers such as silicon carbide or carbon increases their fracture toughness Table 112 because it causes gradual failure of the composite This combination of a fiber and ceramic matrix makes CMCs more attractive for applications in which high mechanical properties and extreme service temperatures are desired What are the typical mechanical properties of some ceramic matrix com posites Compare them with metals Typical mechanical properties of ceramic matrix composites are given in Table 113 Show one process of how ceramic matrix composites are manufactured One of the most common methods to manufacture ceramic matrix com posites is called the hot pressing method28 Glass fibers in continuous tow are passed through slurry consisting of powdered matrix material solvent such as alcohol and an organic binder Figure 131 The tow is then wound on a drum and dried to form prepreg tapes The prepreg tapes can now be stacked to make a required laminate Heating at about 932F 500C burns out the binder Hot pressing at high temperatures in excess of 1832F 1000C and pressures of 1 to 2 ksi 7 to 14 MPa follows this TABLE 112 Typical Fracture Toughness of Monolithic Materials and Ceramic Matrix Composites Material Fracture toughness MPa Fracture toughness ksi Epoxy Aluminum alloys Silicon carbide SiCAl2O3 SiCSiC 3 35 3 27 30 273 3185 273 246 273 Current service temperatures limits are 750F 400C for polymers 1800F 10000C for metals and their alloys and 2700F 1500C for ceramics m in 1343bookfm Page 45 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 46 Mechanics of Composite Materials Second Edition What are the applications of ceramic matrix composites Ceramic matrix composites are finding increased application in hightem perature areas in which metal and polymer matrix composites cannot be used This is not to say that CMCs are not attractive otherwise especially considering their high strength and modulus and low density Typical appli cations include cutting tool inserts in oxidizing and hightemperature envi ronments Textron Systems Corporation has developed fiberreinforced ceramics with SCS monofilaments for future aircraft engines Figure 132 124 CarbonCarbon Composites What are carboncarbon composites Carboncarbon composites use carbon fibers in a carbon matrix These composites are used in very hightemperature environments of up to 6000F 3315C and are 20 times stronger and 30 lighter than graphite fibers29 What are the advantages of carboncarbon composites Carbon is brittle and flaw sensitive like ceramics Reinforcement of a carbon matrix allows the composite to fail gradually and also gives advan tages such as ability to withstand high temperatures low creep at high temperatures low density good tensile and compressive strengths high fatigue resistance high thermal conductivity and high coefficient of friction Drawbacks include high cost low shear strength and susceptibility to oxi dations at high temperatures Typical properties of carboncarbon compos ites are given in Table 114 TABLE 113 Typical Mechanical Properties of Some Ceramic Matrix Composites Property Units SiCLAS SiCCAS Steel Aluminum System of units USCS Specific gravity Youngs modulus Ultimate tensile strength Coefficient of thermal expansion Msi ksi μininF 21 13 72 2 25 1755 580 25 78 300 940 65 26 100 340 128 System of units SI Specific gravity Youngs modulus Ultimate tensile strength Coefficient of thermal expansion GPa MPa μmmC 21 8963 4964 36 25 121 400 45 78 2068 6481 117 26 6895 2344 23 1343bookfm Page 46 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Introduction to Composite Materials 47 Give a typical method of processing a carboncarbon composite A typical method for manufacturing carboncarbon composites is called lowpressure carbonization29 and is shown in Figure 133 A graphite cloth is taken impregnated by resin such as phenolic pitch and furfuryl ester and laid up in layers It is laid in a mold cured and trimmed The part is then pyrolized converting the phenolic resin to graphite The composite is then impregnated by furfuryl alcohol The process drives off the resin and any volatiles The process is repeated three or four times until the level of porosity is reduced to an acceptable level Each time this process increases FIGURE 131 Schematic of slurry infiltration process for ceramic matrix composites From Chawla KK Science and Business Media from Ceramics Matrix Composites Kluwer Academic Publishers Lon don 1993 Figure 41 p 128 Reproduced with permission of SpringerVerlag Glass impregnated fiber tape Stack of glass impregnated fiber tapes Fiberglass composite Binder burnout 500C Pressure Hot pressing 800925C Glass slurry tank Fibers 1343bookfm Page 47 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 48 Mechanics of Composite Materials Second Edition its modulus and strength Because carboncarbon composites oxidize at temperatures as low as 842F 450C an outer layer of silicon carbide may be deposited30 What are the applications of carboncarbon composites The main uses of carboncarbon composites are the following FIGURE 132 Ceramic matrix composites for high temperature and oxidation resistant application Photo courtesy of Specialty Materials Inc httpwwwspecmaterialscom TABLE 114 Typical Mechanical Properties of CarbonCarbon Matrix Composites Property Units CC Steel Aluminum System of units USCS Specific gravity Youngs modulus Ultimate tensile strength Coefficient of thermal expansion Msi ksi μininF 168 195 5180 111 78 30 94 65 26 10 40 128 System of units SI Specific gravity Youngs modulus Ultimate strength Coefficient of thermal expansion GPa MPa μmmC 168 135 357 20 78 2068 6481 117 26 6895 2344 23 1343bookfm Page 48 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Introduction to Composite Materials 49 Space shuttle nose cones As the shuttle enters Earths atmosphere temperatures as high as 3092F 1700C are experienced Car boncarbon composite is a material of choice for the nose cone because it has the lowest overall weight of all ablative materials high thermal conductivity to prevent surface cracking high specific heat to absorb large heat flux and high thermal shock resistance to low temperatures in space of 238F 150C to 3092F 1700C due to reentry Also the carboncarbon nose remains undamaged and can be reused many times Aircraft brakes The carboncarbon brakes shown in Figure 134 cost 440lb 970kg which is several times more than their metallic counterpart however the high durability two to four times that of steel high specific heat 25 times that of steel low braking dis tances and braking times threequarters that of berylium and large weight savings of up to 990 lb 450 kg on a commercial aircraft such as Airbus A300B2K and A300B4 are attractive29 As mentioned earlier 1 lb 0453 kg weight savings on a fullservice commercial aircraft can translate to fuel savings of about 360 galyear 1360 L year Other advantages include reduced inventory due to longer endurance of carbon brakes Mechanical fasteners Fasteners needed for high temperature applica tions are made of carboncarbon composites because they lose little strength at high temperatures FIGURE 133 Schematic of processing carboncarbon composites Reprinted with permission from Klein AJ Adv Mater Processes 6468 November 1986 ASM International Ablative materials absorb heat through pyrolysis at or near the exposed surfaces Specific heat is the amount of heat required to heat a unit mass of a substance through a unit temperature Processing carboncarbon composites Standard Grphenolic prepreg Layup and cure Step 1 Step 2 Step 3 Resin impregnation Step 4 After 3 impregnations Step 5 Pyrolysis Coating Sealing 1343bookfm Page 49 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 50 Mechanics of Composite Materials Second Edition 13 Recycling FiberReinforced Composites What types of processes are used for recycling of composites The two main processes are called chemical and mechanical processes Why is recycling of composites complex This is because of the many variables in material types thermoset vs thermoplastics long vs short fibers glass vs carbon etc What are the various steps in mechanical recycling of short fiberreinforced composites These are shredding separation washing grinding drying and extrusion Where are mechanically recycled shortfiber composites used The recycled material is available in powder or fiber form Powder form is reused as paste for sheetmolding compounds and the fiber form is used for reinforcement in bulkmolding compounds One cannot use too much of these as replacements because the impact resistance and electrical properties degrade after about 20 content Products from recycled plastics are limited to fences and benches FIGURE 134 Sectioned carboncarbon brake from Airbus A320 From Savage G Science and Business Media from CarbonCarbon Composites Kluwer Academic Publishers London 1993 Figure 92 p 325 Reproduced with kind permission of SpringerVerlag 1343bookfm Page 50 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Introduction to Composite Materials 51 Why is chemical recycling not as popular as mechanical recycling Chemical processing is very costly Processes such as pyrolysis decom posing materials in an oxygenfree atmosphere produce many gases and hydrogenation gives high filler content However General Motors has adapted pyrolysis to recycle composite automobile parts Gases and oils are recovered and the residues are used as fillers in concrete and roof shingles One other problem is the chlorine content The scrap needs to be dehalo genated after separation especially if carbon fibers were used as reinforce ment Glass fibers in recycled composites also pose the problem of low compressive strength of the new material What can one do if the different types of composites cannot be separated Incineration or use as fuel may be the only solution because metals ther mosets and thermoplastics may be mixed and they may be soiled with toxic materials The fuel value of polymer matrix composites is around 5000 BTUlb 11622 kJkg This is about half the value for coal Which chemical process shows the most promise Incineration offers the most promise Its advantages include minimal cost highvolume reduction and no residual material It is also feasible for low scrap volume 14 Mechanics Terminology How is a composite structure analyzed mechanically A composite material consists of two or more constituents thus the anal ysis and design of such materials is different from that for conventional materials such as metals The approach to analyze the mechanical behavior of composite structures is as follows Figure 135 1 Find the average properties of a composite ply from the individual properties of the constituents Properties include stiffness strength thermal and moisture expansion coefficients Note that average properties are derived by considering the ply to be homogeneous At this level one can optimize for the stiffness and strength require ments of a lamina This is called the micromechanics of a lamina Fuel value is the heat transferred when the products of complete combustion of a fuel are cooled to the initial temperature of air and fuel Units of fuel value are Btulbm and Jkg Typical fuel value for lignite coal is 7000 Btulbm 1343bookfm Page 51 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 52 Mechanics of Composite Materials Second Edition 2 Develop the stressstrain relationships for a unidirectionalbidirec tional lamina Loads may be applied along the principal directions of symmetry of the lamina or offaxis Also one develops relation ships for stiffness thermal and moisture expansion coefficients and strengths of angle plies Failure theories of a lamina are based on stresses in the lamina and strength properties of a lamina This is called the macromechanics of a lamina A structure made of composite materials is generally a laminate structure made of various laminas stacked on each other Knowing the macromechan ics of a single lamina one develops the macromechanics of a laminate Stiffness strengths and thermal and moisture expansion coefficients can be FIGURE 135 Schematic of analysis of laminated composites Fiber Matrix Micromechanics of a lamina chapter 3 Macromechanics of a lamina chapter 2 Homogeneous orthotropic layer Macromechanics of a laminate chapter 4 Analysis and design of laminated structures chapter 5 Structural element Laminate 1343bookfm Page 52 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Introduction to Composite Materials 53 found for the whole laminate Laminate failure is based on stresses and application of failure theories to each ply This knowledge of analysis of composites can then eventually form the basis for the mechanical design of structures made of composites Several terms are defined to develop the fundamentals of the mechanical behavior of composites These include the following What is an isotropic body An isotropic material has properties that are the same in all directions For example the Youngs modulus of steel is the same in all directions What is a homogeneous body A homogeneous body has properties that are the same at all points in the body A steel rod is an example of a homogeneous body However if one heats this rod at one end the temperature at various points on the rod would be different Because Youngs modulus of steel varies with temperature one no longer has a homogeneous body The body is still isotropic because the properties at a particular point are still identical in all directions Are composite materials isotropic andor homogeneous Most composite materials are neither isotropic nor homogeneous For example consider epoxy reinforced with long glass fibers If one chooses a location on the glass fiber the properties are different from a location on the epoxy matrix This makes the composite material nonhomogeneous not homogeneous Also the stiffness in the direction parallel to the fibers is higher than in the direction perpendicular to the fibers and thus the prop erties are not independent of the direction This makes the composite mate rial anisotropic not isotropic What is an anisotropic material At a point in an anisotropic material material properties are different in all directions What is a nonhomogeneous body A nonhomogeneous or inhomogeneous body has material properties that are a function of the position on the body What is a lamina A lamina also called a ply or layer is a single flat layer of unidirectional fibers or woven fibers arranged in a matrix What is a laminate A laminate is a stack of plies of composites Each layer can be laid at various orientations and can be made up of different material systems 1343bookfm Page 53 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 54 Mechanics of Composite Materials Second Edition What is a hybrid laminate Hybrid composites contain more than one fiber or one matrix system in a laminate The main four types of hybrid laminates follow Interply hybrid laminates contain plies made of two or more different composite systems Examples include car bumpers made of glass epoxy layers to provide torsional rigidity and graphiteepoxy to give stiffness The combinations also lower the cost of the bumper Intraply hybrid composites consist of two or more different fibers used in the same ply Examples include golf clubs that use graphite and aramid fibers Graphite fibers provide the torsional rigidity and the aramid fibers provide tensile strength and toughness An interplyintraply hybrid consists of plies that have two or more different fibers in the same ply and distinct composite systems in more than one ply Resin hybrid laminates combine two or more resins instead of com bining two or more fibers in a laminate Generally one resin is flexible and the other one is rigid Tests have proven that these resin hybrid laminates can increase shear and work of fracture properties by more than 50 over those of allflexible or allrigid resins31 15 Summary This chapter introduced advanced composite materials and enumerated the advantages and drawbacks of composite materials over monolithic materi als Fiber and matrix factors were discussed to understand their influence on mechanical properties of the composites The classification of the com posites based on the matrix materials polymer metal and ceramics was discussed In addition carboncarbon composites were also examined The manufacturing and mechanical properties and application of compos ites were described Discussion also covered the recycling of composite materials as well as the terminology used in studying the mechanics of composite materials Key Terms Composite Advanced composite materials Specific modulus Specific strength 1343bookfm Page 54 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Introduction to Composite Materials 55 Material selection Fiber factors Matrix factors Classification of composites Polymer matrix composites Resins Prepregs Thermosets Thermoplastics Autoclave Resin transfer molding Metal matrix composites Diffusion bonding Ceramic matrix composites Carboncarbon composites Recycling Isotropic body Anisotropic body Homogeneous body Nonhomogeneous body Lamina Laminate Hybrid laminate Exercise Set 11 What is a composite 12 Why did Israelites reinforce clay with straw 13 Give a brief historical review of composites 14 Give four examples of naturally found composites What are the constituents of these natural composites 15 Airbus A300 saved 300 kg of mass by making the tailfin out of advanced composites Estimate in gallons the amount of fuel saved per year 16 Give the definitions and units of the following in the SI and USCS system of units Coefficient of thermal expansion Coefficient of moisture expansion Thermal conductivity Youngs modulus 1343bookfm Page 55 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 56 Mechanics of Composite Materials Second Edition Ultimate strength Poissons ratio Specific modulus Specific strength Density Specific gravity Ductility Fracture toughness Specific heat 17 Draw the graphs for the ratios E12 ρ vs σ12 ultρ for materials in Table 11 18 A lamina consists of 100 fibers of 10μm diameter The fibers are 10 mm long Find the interfacial area What is the increase in the inter facial area if the diameter of the fiber is reduced to 5 μm and the total volume of fibers is kept constant 19 Compare the flexibility of a 001in diameter steel wire to a 002in diameter aluminum wire The Youngs modulus of steel is 30 Msi and that of aluminum is 10 Msi 110 What are the limitations of modern composites 111 Enumerate six primary material selection parameters that are used in evaluating the use of a particular material 112 How are composites classified 113 Compare the specific modulus specific strength and coefficient of thermal expansion coefficient of a pitch based graphite fiber Kevlar 49 and Sglass 114 Describe one manufacturing method of polymer matrix composites other than those given in Chapter 1 115 Why is epoxy the most popular resin 116 Find ten applications of polymer matrix components other than those given in Chapter 1 117 Give the advantages and drawbacks of metal matrix composites over polymer matrix composites 118 Find three applications of metal matrix composites other than those given in Chapter 1 119 Describe one manufacturing method of metal matrix composites other than given in Chapter 1 120 Find three applications of ceramic matrix composites other than those given in Chapter 1 121 Describe one manufacturing method of ceramic matrix composites other than those given in Chapter 1 1343bookfm Page 56 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Introduction to Composite Materials 57 122 Find three applications of carbon matrix composites other than those given in Chapter 1 123 Describe one manufacturing method of carbon matrix composites other than those given in Chapter 1 124 Give the upper limit of operating temperatures of polymer metal ceramic and carbon matrix composites 125 Define the following Isotropic body Homogeneous body Anisotropic body Nonhomogeneous body Micromechanics Macromechanics Lamina Laminate 126 Give an example of a Homogeneous body that is not isotropic Nonhomogeneous body that is isotropic 127 Do all properties of composites always improve over their individual constituents Give examples 128 How are hybrid composites classified References 1 Mack J Advanced polymer composites Mater Edge 18 January 1988 2 Meetham GW Design considerations for aerospace applications in Handbook of PolymerFiber Composites Jones FR Ed Longman Scientific and Technical Essex England Chap 5 1994 3 Eager TW Whither advanced materials Adv Mater Processes 25 June 1991 4 Ashby MF On engineering properties of materials Acta Metallurgica 37 1273 1989 5 Buchanan GR Mechanics of Materials HRW Inc New York 1988 6 Lamotte E De and Perry AJ Diameter and strainrate dependence of the ultimate tensile strength and Youngs modulus of carbon fiber Fiber Sci Tech nol 3 159 1970 7 Schwartz MM Composite Materials Handbook McGrawHill New York 1984 8 Polymer nanocomposites for packaging applications see httpwwwnatick armymilsoldiermediafactfoodPolyNanohtm last accessed August 31 2004 9 GM GMability advanced technology GM to use nanocomposites on highest volume car see httpwwwgmcomcompanygmabilityadvtech100 newsnanocomposites012704html last accessed August 31 2004 1343bookfm Page 57 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 58 Mechanics of Composite Materials Second Edition 10 Allen L A limber future Popular Sci 36 August 2004 11 Partridge IK Advanced Composites Elsevier Applied Science New York 1989 12 Cooks G Composite resins for the 90s MC Gill Doorway 7 27 Spring 1990 13 Hergenrother PM and Johnston NJ Polymer Mater Sci Eng Proc 59 697 1988 14 Mallick PK FiberReinforced Composites Materials Manufacturing and Design Marcell Dekker Inc New York Chap 2 1988 15 McConnell VP and Stover D Advanced composites performance materials of choice for innovative products HighPerformance Composites Sourcebook 8 November 1995 16 Composite structures Alliant Techsystems Magna UT 17 Lepré RA personal communication 1995 18 Scaled composites see httpwwwscaledcom last accessed September 21 2004 19 How stuff works see httpsciencehowstuffworkscom last accessed Sep tember 21 2004 20 Sun CT and Lu YP Vibration Damping of Structural Elements Prentice Hall Englewood Cliffs NJ 1995 21 Ashley S Bridging the cost gap with composites Mech Eng 118 76 1996 22 Bursel JS Composite Springs Inc technical bulletin October 1990 St Peters burg FL 23 Neil D Our driving conundrum Popular Sci 62 September 2004 24 Clean handling Mech Eng 2627 September 2004 25 Fiskars introduces two new PowerGear pruners press release April 1 2004 see httpwwwfiskarscomenUSpressreleasedonum8res6 last ac cessed September 28 2004 26 Matthews FL and Rawlings RD Composites Materials Engineering and Sci ence SpringerVerlag New York Chap 3 1987 27 Niskanen P and Mohn WR Versatile metalmatrix composites Adv Mat Processes 3 39 1988 28 Chawla KK Ceramic Matrix Composites Chapman Hall London Chap 4 1993 29 Klein AJ Carboncarbon composites Adv Mater Processes 64 November 1986 30 Strife JR and Sheehan JE Ceramic coatings for carboncarbon composites Ceramic Bull 67 369 1988 31 Sheppard LM On the road with composites Adv Mater Processes 36 Decem ber 1986 General References Chung DDL Carbon Fiber Composites ButterworthHeinemann Boston 1994 Gill RM Carbon Fibers in Composite Materials Butterworth and Co London 1972 Geier MH Quality Handbook for Composite Materials Chapman Hall London 1994 Holloway L Polymer Composites for Civil and Structural Engineering Blackie Academic and Professional London 1993 Jones FR Ed Handbook of PolymerFiber Composites Longman Scientific and Tech nical Essex 1994 1343bookfm Page 58 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Introduction to Composite Materials 59 Lubin G 1982 Handbook of Advanced Composites 2nd ed Van Nostrand Reinhold New York Phillips LN Ed Design with Advanced Composite Materials SpringerVerlag New York 1989 Powell PC Engineering with FiberPolymer Laminates Chapman Hall London 1994 Savage G CarbonCarbon Composites Chapman Hall London 1993 Vinson JR and Chou T Composite Materials and Their Use in Structures John Wiley Sons New York 1975 Vinson JR and Sierakowski RL The Behavior of Structures Composed of Composite Materials Martinus Nighoff Publishers Dordrecht 1986 Video References Advanced Composites in Manufacturing Society of Manufacturing Engineering Dear born MI 1986 New Materials Films for the Humanities and Sciences 1989 The Light Stuff Coronet Film and Video Northbrook IL 1988 Tooling for Composites Society of Manufacturing Engineers Dearborn MI 1989 1343bookfm Page 59 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 61 2 Macromechanical Analysis of a Lamina Chapter Objectives Review definitions of stress strain elastic moduli and strain energy Develop stressstrain relationships for different types of materials Develop stressstrain relationships for a unidirectionalbidirec tional lamina Find the engineering constants of a unidirectionalbidirectional lam ina in terms of the stiffness and compliance parameters of the lamina Develop stressstrain relationships elastic moduli strengths and thermal and moisture expansion coefficients of an angle ply based on those of a unidirectionalbidirectional lamina and the angle of the ply 21 Introduction A lamina is a thin layer of a composite material that is generally of a thickness on the order of 0005 in 0125 mm A laminate is constructed by stacking a number of such laminae in the direction of the lamina thickness Figure 21 Mechanical structures made of these laminates such as a leaf spring suspension system in an automobile are subjected to various loads such as bending and twisting The design and analysis of such laminated structures demands knowledge of the stresses and strains in the laminate Also design tools such as failure theories stiffness models and optimization algorithms need the values of these laminate stresses and strains However the building blocks of a laminate are single lamina so under standing the mechanical analysis of a lamina precedes understanding that of a laminate A lamina is unlike an isotropic homogeneous material For example if the lamina is made of isotropic homogeneous fibers and an 1343bookfm Page 61 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 62 Mechanics of Composite Materials Second Edition isotropic homogeneous matrix the stiffness of the lamina varies from point to point depending on whether the point is in the fiber the matrix or the fibermatrix interface Accounting for these variations will make any kind of mechanical modeling of the lamina very complicated For this reason the macromechanical analysis of a lamina is based on average properties and considering the lamina to be homogeneous Methods to find these average properties based on the individual mechanical properties of the fiber and the matrix as well as the content packing geometry and shape of fibers are discussed in Chapter 3 Even with the homogenization of a lamina the mechanical behavior is still different from that of a homogeneous isotropic material For example take a square plate of length and width w and thickness t out of a large isotropic plate of thickness t Figure 22 and conduct the following experiments Case A Subject the square plate to a pure normal load P in direction 1 Measure the normal deformations in directions 1 and 2 δ 1 A and δ 2 A respectively Case B Apply the same pure normal load P as in case A but now in direction 2 Measure the normal deformations in directions 1 and 2 δ 1 B and δ 2 B respectively Note that 21ab However taking a unidirectional square plate Figure 23 of the same dimensions w w t out of a large composite lamina of thickness t and conducting the same case A and B experiments note that the deformations FIGURE 21 Typical laminate made of three laminae Fiber crosssection Matrix material 1A 2B 2A 1B δ δ δ δ 1343bookfm Page 62 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Macromechanical Analysis of a Lamina 63 22ab because the stiffness of the unidirectional lamina in the direction of fibers is much larger than the stiffness in the direction perpendicular to the fibers Thus the mechanical characterization of a unidirectional lamina will require more parameters than it will for an isotropic lamina Also note that if the square plate Figure 24 taken out of the lamina has fibers at an angle to the sides of the square plate the deformations will be different for different angles In fact the square plate would not only have FIGURE 22 Deformation of square plate taken from an isotropic plate under normal loads w 2 w 1 t t w w w Undeformed state Deformed state Undeformed state w δ2A wδ1A p p w δ2B w δ1B w Case A Case B p p Deformed state 1A 2B 2A 1B δ δ δ δ 1343bookfm Page 63 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 64 Mechanics of Composite Materials Second Edition deformations in the normal directions but would also distort This suggests that the mechanical characterization of an angle lamina is further complicated Mechanical characterization of materials generally requires costly and timeconsuming experimentation andor theoretical modeling Therefore the goal is to find the minimum number of parameters required for the mechanical characterization of a lamina Also a composite laminate may be subjected to a temperature change and may absorb moisture during processing and operation These changes in temperature and moisture result in residual stresses and strains in the lam inate The calculation of these stresses and strains in a laminate depends on the response of each lamina to these two environmental parameters In this chapter the stressstrain relationships based on temperature change and moisture content will also be developed for a single lamina The effects of temperature and moisture on a laminate are discussed later in Chapter 4 FIGURE 23 Deformation of a square plate taken from a unidirectional lamina with fibers at zero angle under normal loads w w p p p p w δ2B w δ1B w δ1A w δ2A w w w w 2 1 Fiber cross section Case B Case A Undeformed state Deformed state Deformed state Undeformed state t t 1343bookfm Page 64 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Macromechanical Analysis of a Lamina 65 22 Review of Definitions 221 Stress A mechanical structure takes external forces which act upon a body as surface forces for example bending a stick and body forces for example the weight of a standing vertical telephone pole on itself These forces result in internal forces inside the body Knowledge of the internal forces at all points in the body is essential because these forces need to be less than the strength of the material used in the structure Stress which is defined as the intensity of the load per unit area determines this knowledge because the strengths of a material are intrinsically known in terms of stress Imagine a body Figure 25 in equilibrium under various loads If the body is cut at a crosssection forces will need to be applied on the crosssectional area so that it maintains equilibrium as in the original body At any cross FIGURE 24 Deformation of a square plate taken from a unidirectional lamina with fibers at an angle under normal loads w t t w w p p 2 1 Fiber cross section Undeformed state Deformed state 1343bookfm Page 65 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 66 Mechanics of Composite Materials Second Edition section a force Δ P is acting on an area of Δ A This force vector has a com ponent normal to the surface Δ P n and one parallel to the surface Δ P s The definition of stress then gives 23ab The component of the stress normal to the surface σ n is called the normal stress and the stress parallel to the surface τ s is called the shear stress If one takes a different crosssection through the same point the stress remains unchanged but the two components of stress normal stress σ n and shear stress τ s will change However it has been proved that a complete definition of stress at a point only needs use of any three mutually orthogonal coordi nate systems such as a Cartesian coordinate system Take the righthand coordinate system xyz Take a crosssection parallel to the yz plane in the body as shown in Figure 26 The force vector Δ P acts FIGURE 25 Stresses on an infinitesimal area on an arbitrary plane Arbitrary plane ΔPs ΔP ΔA ΔPn σn A Pn A lim Δ Δ Δ 0 τs A sP A lim Δ Δ Δ 0 1343bookfm Page 66 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Macromechanical Analysis of a Lamina 67 on an area Δ A The component Δ P x is normal to the surface The force vector Δ P s is parallel to the surface and can be further resolved into components along the y and z axes Δ P y and Δ P z The definition of the various stresses then is 24ac Similarly stresses can be defined for crosssections parallel to the xy and xz planes For defining all these stresses the stress at a point is defined generally by taking an infinitesimal cuboid in a righthand coordinate system FIGURE 26 Forces on an infinitesimal area on the yz plane z ΔPz ΔA Crosssection ΔPyΔP ΔPx y x σx A xP A lim Δ Δ Δ 0 τxy A Py A lim Δ Δ Δ 0 τxz A zP A lim Δ Δ Δ 0 1343bookfm Page 67 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 68 Mechanics of Composite Materials Second Edition and finding the stresses on each of its faces Nine different stresses act at a point in the body as shown in Figure 27 The six shear stresses are related as 25ac The preceding three relations are found by equilibrium of moments of the infinitesimal cube There are thus six independent stresses The stresses σ x σ y and σ z are normal to the surfaces of the cuboid and the stresses τ yz τ zx and τ xy are along the surfaces of the cuboid A tensile normal stress is positive and a compressive normal stress is negative A shear stress is positive if its direction and the direction of the normal to the face on which it is acting are both in positive or negative direction otherwise the shear stress is negative 222 Strain Similar to the need for knowledge of forces inside a body knowing the deformations because of the external forces is also important For example a piston in an internal combustion engine may not develop larger stresses than the failure strengths but its excessive deformation may seize the engine Also finding stresses in a body generally requires finding deformations This is because a stress state at a point has six components but there are only three forceequilibrium equations one in each direction FIGURE 27 Stresses on an infinitesimal cuboid G F σzz σxx σyy τzy τzx τyx τyz τxz τxy H E A B z y x C xy yx τ τ yz zy τ τ zx xz τ τ 1343bookfm Page 68 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Macromechanical Analysis of a Lamina 69 The knowledge of deformations is specified in terms of strains that is the relative change in the size and shape of the body The strain at a point is also defined generally on an infinitesimal cuboid in a righthand coordi nate system Under loads the lengths of the sides of the infinitesimal cuboid change The faces of the cube also get distorted The change in length cor responds to a normal strain and the distortion corresponds to the shearing strain Figure 28 shows the strains on one of the faces ABCD of the cuboid The strains and displacements are related to each other Take the two perpendicular lines AB and AD When the body is loaded the two lines become A B and A D Define the displacements of a point xyz as u uxyz displacement in x direction at point xyz v vxyz displacement in y direction at point xyz w wxyz displacement in z direction at point xyz The normal strain in the x direction ε x is defined as the change of length of line AB per unit length of AB as 26 where FIGURE 28 Normal and shearing strains on an infinitesimal area in the xy plane Q D C B B x C A D y Δy Δx xy θ1 θ2 P A εx AB A B AB AB lim 0 1343bookfm Page 69 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 70 Mechanics of Composite Materials Second Edition 27ab Substituting the preceding expressions of Equation 27 in Equation 26 Using definitions of partial derivatives 28 because for small displacements The normal strain in the ydirection εy is defined as the change in the length of line AD per unit length of AD as 29 where A B A P B P 2 2 x u x x y u x y v x x y Δ Δ Δ 2 v x y 2 AB x Δ u x x y u x y x ε x lim Δ Δ 0 1 2 2 1 Δ Δ Δ x v x x y v x y x 2 1 x u x v x ε 1 2 2 2 1 1 x u x ε u x 1 v x 1 y AD A D AD AD ε lim 0 1343bookfm Page 70 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Macromechanical Analysis of a Lamina 71 AD Δy 210ab Substituting the preceding expressions of Equation 210 in Equation 29 Using definitions of partial derivatives 211 because for small displacements A normal strain is positive if the corresponding length increases a normal strain is negative if the corresponding length decreases The shearing strain in the xy plane γxy is defined as the change in the angle between sides AB and AD from 90 This angular change takes place by the inclining of sides AB and AD The shearing strain is thus defined as 212 A D A Q Q D 2 2 A D y v x y y v x y u x y Δ Δ 2 Δy u x y 2 y y v x y y v x y y ε lim Δ Δ Δ 0 1 2 2 1 u x y y u x y y 2 Δ Δ 1 y v y u y ε 1 2 2 2 1 1 y v y ε u y 1 v y 1 γxy θ θ 1 2 1343bookfm Page 71 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 72 Mechanics of Composite Materials Second Edition where 213ac 214ac Substituting Equation 213 and Equation 214 in Equation 212 215 because for small displacements 1 0 q P B A P AB lim P B v x x y v x y Δ A P u x x y x u x y Δ Δ 2 0 θ Q D A Q AD lim Q D u x y y u x y Δ A Q v x y y y v x y Δ Δ v x x y v x y x u xy x y γ lim Δ Δ Δ Δ 0 0 x x y x u x y x u x y y u x Δ Δ Δ Δ y y v x y y y v x y y Δ Δ Δ Δ v x u x u y u y 1 1 v x u y u x 1 v y 1 1343bookfm Page 72 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Macromechanical Analysis of a Lamina 73 The shearing strain is positive when the angle between the sides AD and AB decreases otherwise the shearing strain is negative The definitions of the remaining normal and shearing strains can be found by noting the change in size and shape of the other sides of the infinitesimal cuboid in Figure 27 as 216ac Example 21 A displacement field in a body is given by u 105x2 6y 7xy v 105yz w 105xy yz2 Find the state of strain at xyz 123 Solution From Equation 28 yz v z w y γ zx w x u z γ z w z ε x u x x x y xz 10 6 7 5 2 10 2 7 5 x z 10 2 1 7 3 5 2 300 10 4 1343bookfm Page 73 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 74 Mechanics of Composite Materials Second Edition From Equation 211 From Equation 216c From Equation 215 y v y y yz 10 5 10 5 z 10 3 5 3 000 10 5 z w z z xy yz 10 5 2 10 5 2 yz 10 2 2 3 5 1 2 10 4 γ xy u y v x y x y xz x yz 10 6 7 10 5 2 5 1343bookfm Page 74 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Macromechanical Analysis of a Lamina 75 From Equation 216a From Equation 216b 223 Elastic Moduli As mentioned in Section 222 three equilibrium equations are insufficient for defining all six stress components at a point For a body that is linearly 10 6 10 0 5 5 6 000 10 5 γ yz v z w y z yz y xy yz 10 10 5 5 2 10 10 5 5 2 y x z 10 2 10 1 3 5 5 2 1 2 10 4 γ zx w x u z x xy yz z x y xz 10 10 6 7 5 2 5 2 10 10 7 5 5 y x 10 2 10 7 1 5 5 9 000 10 5 1343bookfm Page 75 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 76 Mechanics of Composite Materials Second Edition elastic and has small deformations stresses and strains at a point are related through six simultaneous linear equations called Hookes law Note that 15 unknown parameters are at a point six stresses six strains and three displacements Combined with six simultaneous linear equa tions of Hookes law six straindisplacement relations given by Equa tion 28 Equation 211 Equation 215 and Equation 216 and three equilibrium equations give 15 equations for the solution of 15 unknowns1 Because straindisplacement and equilibrium equations are differential equations they are subject to knowing boundary conditions for complete solutions For a linear isotropic material in a threedimensional stress state the Hookes law stressstrain relationships at a point in an xyz orthogonal system Figure 29 in matrix form are 217 FIGURE 29 Cartesian coordinates in a threedimensional body z y x x y z yz zx xy E ε ε ε γ γ γ 1 ν ν ν ν ν ν E E E E E E E E G 0 0 0 1 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 G G 0 0 0 0 0 0 1 x y z yz zx xy σ σ σ τ τ τ 1343bookfm Page 76 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Macromechanical Analysis of a Lamina 77 218 where ν is the Poissons ratio The shear modulus G is a function of two elastic constants E and ν as 219 The 6 6 matrix in Equation 217 is called the compliance matrix S of an isotropic material The 6 6 matrix in Equation 218 obtained by invert ing the compliance matrix in Equation 217 is called the stiffness matrix C of an isotropic material 224 Strain Energy Energy is defined as the capacity to do work In solid deformable elastic bodies under loads the work done by external loads is stored as recoverable strain energy The strain energy stored in the body per unit volume is then defined as 220 Example 22 Consider a bar of crosssection A and length L Figure 210 A uniform tensile load P is applied to the two ends of the rod find the state of stress and strain and strain energy per unit volume of the body Assume that the rod is made of a homogeneous isotropic material of Youngs modulus E x y z yz zx xy E σ σ σ τ τ τ ν 1 1 2 1 1 2 1 1 2 1 0 0 0 ν ν ν ν ν ν ν ν ν E E E E E 1 2 1 1 1 2 1 1 2 1 ν ν ν ν ν ν ν ν 0 0 0 1 2 1 1 2 1 1 1 2 ν ν ν ν ν ν ν ν E E E 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 ν G G G x y z yz zx xy ε ε ε γ γ γ G E 21 ν W x x y y z z xy xy 1 2 σ ε σ ε σ ε τ γ yz yz zx zx τ γ τ γ 1343bookfm Page 77 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 78 Mechanics of Composite Materials Second Edition Solution The stress state at any point is given by 221 If the circular rod is made of an isotropic homogeneous and linearly elastic material then the stressstrain at any point is related as 222 223 The strain energy stored per unit volume in the rod per Equation 220 is FIGURE 210 Cylindrical rod under uniform uniaxial load P P L Crosssection A z y x P x y z yz zx P A σ σ σ τ τ 0 0 0 0 τxy 0 x y z yz zx xy E ε ε ε γ γ γ 1 ν ν ν ν ν ν E E E E E E E E G 0 0 0 1 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 G G 0 0 0 0 0 0 1 P A 0 0 0 0 0 x y z yz P AE P AE P AE ε ε ν ε ν γ 0 zx xy γ γ 0 0 1343bookfm Page 78 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Macromechanical Analysis of a Lamina 79 224 23 Hookes Law for Different Types of Materials The stressstrain relationship for a general material that is not linearly elastic and isotropic is more complicated than Equation 217 and Equation 218 Assuming linear and elastic behavior for a composite is acceptable however assuming it to be isotropic is generally unacceptable Thus the stressstrain relationships follow Hookes law but the constants relating stress and strain are more in number than seen in Equation 217 and Equation 218 The most general stressstrain relationship is given as follows for a threedimen sional body in a 123 orthogonal Cartesian coordinate system 225 where the 6 6 C matrix is called the stiffness matrix The stiffness matrix has 36 constants What happens if one changes the system of coordinates from an orthogonal system 123 to some other orthogonal system 123 Then new stiffness and compliance constants will be required to relate stresses and strains in the new coordinate system 123 However the new stiffness and compli ance matrices in the 123 system will be a function of the stiffness and compliance matrices in the 123 system and the angle between the axes of the 123system and the 123 system W P A P AE P AE 1 2 0 ν P AE 0 0 0 0 0 0 ν 0 1 2 2 2 P A E 1 2 2 σx E 1 2 3 23 31 12 11 12 σ σ σ τ τ τ C CC C C C C C C C C C C C C C 13 14 15 16 21 22 23 24 25 26 31 32 33 C34 C C C C C C C C C C C C C 35 36 41 42 43 44 45 46 51 52 53 54 55 C56 C C C C C C 61 62 63 64 65 66 1 ε 2 3 23 31 12 ε ε γ γ γ 1343bookfm Page 79 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 80 Mechanics of Composite Materials Second Edition Inverting Equation 225 the general strainstress relationship for a three dimensional body in a 123 orthogonal Cartesian coordinate system is 226 In the case of an isotropic material relating the preceding strainstress equation to Equation 217 one finds that the compliance matrix is related directly to engineering constants as 227 and Sij other than in the preceding are zero It can be shown that the 36 constants in Equation 225 actually reduce to 21 constants due to the symmetry of the stiffness matrix C as follows The stressstrain relationship 225 can also be written as 228 where in a contracted notation 229af 1 2 3 23 31 12 11 12 ε ε ε γ γ γ S S S S S S S S S S S S S S S 13 14 15 16 21 22 23 24 25 26 31 32 33 S34 S S S S S S S S S S S S S 35 36 41 42 43 44 45 46 51 52 53 54 55 S56 S S S S S S 61 62 63 64 65 66 1 σ 2 3 23 31 12 σ σ τ τ τ 11 22 33 1 S E S S 12 13 21 23 31 32 S E S S S S S ν 44 55 66 1 S G S S σ ε i ij j j C i 1 6 1 6 4 23 5 31 6 12 σ τ σ τ σ τ 4 23 5 31 6 12 ε γ ε γ ε γ 1343bookfm Page 80 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Macromechanical Analysis of a Lamina 81 The strain energy in the body per unit volume per Equation 220 is expressed as 230 Substituting Hookes law Equation 228 in Equation 230 231 Now by partial differentiation of Equation 231 232 and 233 Because the differentiation does not necessarily need to be in either order 234 Equation 234 can also be proved by realizing that Thus only 21 independent elastic constants are in the general stiffness matrix C of Equation 225 This also implies that only 21 independent constants are in the general compliance matrix S of Equation 226 231 Anisotropic Material The material that has 21 independent elastic constants at a point is called an anisotropic material Once these constants are found for a particular point the stress and strain relationship can be developed at that point Note that W i i i 1 2 1 6 σ ε W Cij j i j i 1 2 1 6 1 6 ε ε W C i j ij ε ε W C j i ji ε ε ij ji C C σ ε i i W 1343bookfm Page 81 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 82 Mechanics of Composite Materials Second Edition these constants can vary from point to point if the material is nonhomoge neous Even if the material is homogeneous or assumed to be one needs to find these 21 elastic constants analytically or experimentally However many natural and synthetic materials do possess material symmetry that is elastic properties are identical in directions of symmetry because symme try is present in the internal structure Fortunately this symmetry reduces the number of the independent elastic constants by zeroing out or relating some of the constants within the 6 6 stiffness C and 6 6 compliance S matrices This simplifies the Hookes law relationships for various types of elastic symmetry 232 Monoclinic Material If in one plane of material symmetry Figure 211 for example direction 3 is normal to the plane of material symmetry then the stiffness matrix reduces to 235 as FIGURE 211 Transformation of coordinate axes for 12 plane of symmetry for a monoclinic material Material symmetry implies that the material and its mirror image about the plane of symmetry are identical 3 1 1 2 2 3 C C C C C C C C C C C C 11 12 13 16 12 22 23 26 13 23 33 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 36 44 45 45 55 16 26 36 66 C C C C C C C C C 1343bookfm Page 82 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Macromechanical Analysis of a Lamina 83 The direction perpendicular to the plane of symmetry is called the principal direction Note that there are 13 independent elastic constants Feldspar is an example of a monoclinic material The compliance matrix correspondingly reduces to 236 Modifying an excellent example2 of demonstrating the meaning of elastic symmetry for a monoclinic material given consider a cubic element of Figure 212 taken out of a monoclinic material in which 3 is the direction perpen dicular to the 12 plane of symmetry Apply a normal stress σ3 to the element Then using the Hookes law Equation 226 and the compliance matrix Equation 236 for the monoclinic material one gets 237af The cube will deform in all directions as determined by the normal strain equations The shear strains in the 23 and 31 plane are zero showing that the element will not change shape in those planes However it will change C C C C C C C C 14 15 24 25 34 35 46 0 0 0 0 0 0 0 56 0 S S S S S S S S S S S S 11 12 13 16 12 22 23 26 13 23 33 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 36 44 45 45 55 16 26 36 66 S S S S S S S S S 1 13 3 ε S σ 2 23 3 ε S σ 3 33 3 ε S σ 23 0 γ 31 0 γ γ σ 12 36 3 S 1343bookfm Page 83 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 84 Mechanics of Composite Materials Second Edition shape in the 12 plane Thus the faces ABEH and CDFG perpendicular to the 3 direction will change from rectangles to parallelograms while the other four faces ABCD BEFC GFEH and AHGD will stay as rectangles This is unlike anisotropic behavior in which all faces will be deformed in shape and also unlike isotropic behavior in which all faces will remain undeformed in shape 233 Orthotropic Material Orthogonally AnisotropicSpecially Orthotropic If a material has three mutually perpendicular planes of material symmetry then the stiffness matrix is given by FIGURE 212 Deformation of a cubic element made of monoclinic material G σ3 σ3 D H A B G F C B A H D E E F 1 2 3 1343bookfm Page 84 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Macromechanical Analysis of a Lamina 85 238 The preceding stiffness matrix can be derived by starting from the stiffness matrix C for the monoclinic material Equation 235 With two more planes of symmetry it gives Three mutually perpendicular planes of material symmetry also imply three mutually perpendicular planes of elastic symmetry Note that nine independent elastic constants are present This is a commonly found material symmetry unlike anisotropic and monoclinic materials Examples of an orthotropic material include a single lamina of continuous fiber composite arranged in a rectangular array Figure 213 a wooden bar and rolled steel The compliance matrix reduces to 239 FIGURE 213 A unidirectional lamina as a monoclinic material with fibers arranged in a rectangular array 3 2 1 C C C C C C C C C C 11 12 13 12 22 23 13 23 33 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 44 55 66 C C C C C C C 16 26 36 45 0 0 0 0 S S S S S S S S S S 11 12 13 12 22 23 13 23 33 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 44 55 66 S S S 1343bookfm Page 85 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 86 Mechanics of Composite Materials Second Edition Demonstrating the meaning of elastic symmetry for an orthotropic mate rial is similar to the approach taken for a monoclinic material Section 232 Consider a cubic element Figure 214 taken out of the orthotropic material where 1 2 and 3 are the principal directions or 12 23 and 31 are the three mutually orthogonal planes of symmetry Apply a normal stress σ3 to the element Then using the Hookes law Equation 226 and the com pliance matrix Equation 239 for the orthotropic material one gets FIGURE 214 Deformation of a cubic element made of orthotropic material G σ3 σ3 D A H F C 3 2 1 E B G D F C E B H A ε σ 1 13 3 S ε σ 2 23 3 S 1343bookfm Page 86 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Macromechanical Analysis of a Lamina 87 240af The cube will deform in all directions as determined by the normal strain equations However the shear strains in all three planes 12 23 and 31 are zero showing that the element will not change shape in those planes Thus the cube will not deform in shape under any normal load applied in the principal directions This is unlike the monoclinic material in which two out of the six faces of the cube changed shape A cube made of isotropic material would not change its shape either however the normal strains ε1 and ε2 will be different in an orthotropic material and identical in an isotropic material 234 Transversely Isotropic Material Consider a plane of material isotropy in one of the planes of an orthotropic body If direction 1 is normal to that plane 23 of isotropy then the stiffness matrix is given by 241 Transverse isotropy results in the following relations Note the five independent elastic constants An example of this is a thin unidirectional lamina in which the fibers are arranged in a square array or ε σ γ γ γ 3 33 3 23 31 12 0 0 0 S C C C C C C C C C C 11 12 12 12 22 23 12 23 22 0 0 0 0 0 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 0 0 0 0 0 22 23 55 55 C C C C C C C C C C C C C 22 33 12 13 55 66 44 22 23 2 1343bookfm Page 87 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 88 Mechanics of Composite Materials Second Edition a hexagonal array One may consider the elastic properties in the two direc tions perpendicular to the fibers to be the same In Figure 215 the fibers are in direction 1 so plane 23 will be considered as the plane of isotropy The compliance matrix reduces to 242 235 Isotropic Material If all planes in an orthotropic body are identical it is an isotropic material then the stiffness matrix is given by 243 Isotropy results in the following additional relationships FIGURE 215 A unidirectional lamina as a transversely isotropic material with fibers arranged in a square array 3 2 1 S S S S S S S S S S 11 12 12 12 22 23 12 23 22 0 0 0 0 0 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 0 0 0 0 0 22 23 55 55 S S S S C C C C C C C C C C 11 12 12 12 11 12 12 12 11 0 0 0 0 0 0 0 0 0 0 0 0 2 0 0 0 0 0 0 2 0 0 0 0 0 0 2 11 12 11 12 11 12 C C C C C C 1343bookfm Page 88 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Macromechanical Analysis of a Lamina 89 This also implies infinite principal planes of symmetry Note the two independent constants This is the most common material symmetry avail able Examples of isotropic bodies include steel iron and aluminum Relat ing Equation 243 to Equation 218 shows that 244ab Note that The compliance matrix reduces to 245 We summarize the number of independent elastic constants for various types of materials C C C C C C C C C 11 22 12 23 66 22 23 11 12 2 2 11 1 1 2 1 C E ν ν ν 12 1 2 1 C E ν ν ν C C 11 12 2 1 2 1 1 2 1 1 2 1 E E ν ν ν ν ν ν E 2 1 ν G S S S S S S S S S S 11 12 12 12 11 12 12 12 11 0 0 0 0 0 0 0 0 0 0 00 0 2 0 0 0 0 0 0 2 0 0 0 0 0 0 2 11 12 11 12 11 1 S S S S S S 2 1343bookfm Page 89 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 90 Mechanics of Composite Materials Second Edition Anisotropic 21 Monoclinic 13 Orthotropic 9 Transversely isotropic 5 Isotropic 2 Example 23 Show the reduction of anisotropic material stressstrain Equation 225 to those of a monoclinic material stressstrain Equation 235 Solution Assume direction 3 is perpendicular to the plane of symmetry Now in the coordinate system 123 Equation 225 with Cij Cji from Equation 234 is 246 Also in the coordinate system 123 Figure 211 247 Because there is a plane of symmetry normal to direction 3 the stresses and strains in the 123 and 123 coordinate systems are related by 248af 1 2 3 23 31 12 11 12 σ σ σ τ τ τ C CC C C C C C C C C C C C C C 13 14 15 16 12 22 23 24 25 26 13 23 33 C34 C C C C C C C C C C C C C 35 36 14 24 34 44 45 46 15 25 35 45 55 C56 C C C C C C 16 26 36 46 56 66 1 ε 2 3 23 31 12 ε ε γ γ γ 1 2 3 2 3 3 1 1 2 σ σ σ τ τ τ C C C C C C C C C C C C 11 12 13 14 15 16 12 22 23 24 25 26 C13 C C C C C C C C C C C C C 23 33 34 35 36 14 24 34 44 45 46 15 25 C35 C C C C C C C C C 45 55 56 16 26 36 46 56 66 1 2 3 2 3 3 1 1 2 ε ε ε γ γ γ 1 1 2 2 3 3 σ σ σ σ σ σ τ τ τ τ τ τ 23 2 3 31 3 1 12 1 2 1343bookfm Page 90 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Macromechanical Analysis of a Lamina 91 249af The terms in the first equation of Equation 246 and Equation 247 can be written as 250ab Substituting Equation 248 and Equation 249 in Equation 250b 251 Subtracting Equation 251 from Equation 250a gives 252 Because γ23 and γ31 are arbitrary 253a Similarly one can show that 254bd Thus only 13 independent elastic constants are present in a monoclinic material Example 24 The stressstrain relation is given in terms of compliance matrix for an orthotropic material in Equation 226 and Equation 239 Rewrite the compliance matrix equations in terms of the nine engineering constants for ε ε ε ε ε ε 1 1 2 2 3 3 γ γ γ γ γ γ 23 2 3 31 3 1 12 1 2 σ ε ε ε γ γ γ 1 11 1 12 2 13 3 14 23 15 31 16 12 C C C C C C σ ε ε ε γ γ 1 11 1 12 2 13 3 1 4 2 3 15 3 1 C C C C C C16 1 2 γ σ ε ε ε γ γ γ 1 11 1 12 2 13 3 14 23 15 31 16 12 C C C C C C 0 2 2 14 23 15 31 C C γ γ C C 14 15 0 C C 24 25 0 C C 34 35 0 C C 46 56 0 1343bookfm Page 91 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 92 Mechanics of Composite Materials Second Edition an orthotropic material What is the stiffness matrix in terms of the engineer ing constants Solution Let us see how the compliance matrix and engineering constants of an orthotropic material are related As shown in Figure 216a apply σ1 0 σ2 0 σ3 0 τ23 0 τ31 0 τ12 0 Then from Equation 226 and Equation 239 ε1 S11σ1 ε2 S12σ1 ε3 S13σ1 FIGURE 216 Application of stresses to find engineering constants of a threedimensional orthotropic body σ1 τ23 τ31 τ12 σ1 a b c e f 3 2 1 d σ2 σ2 σ3 σ3 1343bookfm Page 92 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Macromechanical Analysis of a Lamina 93 γ23 0 γ31 0 γ12 0 The Youngs modulus in direction 1 E1 is defined as 255 The Poissons ratio ν12 is defined as 256 In general terms νij is defined as the ratio of the negative of the normal strain in direction j to the normal strain in direction i when the load is applied in the normal direction i The Poissons ratio ν13 is defined as 257 Similarly as shown in Figure 216b apply σ1 0 σ2 0 σ3 0 τ23 0 τ31 0 τ12 0 Then from Equation 226 and Equation 239 258 259 260 Similarly as shown in Figure 216c apply σ1 0 σ2 0 σ3 0 τ23 0 τ31 0 τ12 0 From Equation 226 and Equation 239 261 E S 1 1 1 11 1 σ ε ν ε ε 12 2 1 12 11 S S ν ε ε 13 3 1 13 11 S S 2 22 1 E S ν21 12 22 S S ν23 23 22 S S E S 3 33 1 1343bookfm Page 93 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 94 Mechanics of Composite Materials Second Edition 262 263 Apply as shown in Figure 216d σ1 0 σ2 0 σ3 0 τ23 0 τ31 0 τ12 0 Then from Equation 226 and Equation 239 ε1 0 ε2 0 ε3 0 γ23 S44τ23 γ31 0 γ12 0 The shear modulus in plane 23 is defined as 264 Similarly as shown in Figure 216e apply σ1 0 σ2 0 σ3 0 τ23 0 τ31 0 τ12 0 Then from Equation 226 and Equation 239 265 Similarly as shown in Figure 216f apply σ1 0 σ2 0 σ3 0 τ23 0 τ31 0 τ12 0 Then from Equation 226 and Equation 239 266 In Equation 255 through Equation 266 12 engineering constants have been defined as follows Three Youngs moduli E1 E2 and E3 one in each material axis ν31 13 33 S S ν32 23 33 S S G S 23 23 23 44 1 τ γ G S 31 55 1 12 66 1 G S 1343bookfm Page 94 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Macromechanical Analysis of a Lamina 95 Six Poissons ratios ν12 ν13 ν21 ν23 ν31 and ν32 two for each plane Three shear moduli G23 G31 and G12 one for each plane However the six Poissons ratios are not independent of each other For example from Equation 255 Equation 256 Equation 258 and Equa tion 259 267 Similarly from Equation 255 Equation 257 Equation 261 and Equa tion 262 268 and from Equation 258 Equation 260 Equation 261 and Equation 263 269 Equation 267 Equation 268 and Equation 269 are called reciprocal Poissons ratio equations These relations reduce the total independent engi neering constants to nine This is the same number as the number of inde pendent constants in the stiffness or the compliance matrix Rewriting the compliance matrix in terms of the engineering constants gives 270 ν ν 12 1 21 2 E E ν ν 13 1 31 3 E E ν ν 23 2 32 3 E E S E E E E E E 1 0 0 0 1 0 0 0 1 12 1 13 1 21 2 2 23 2 ν ν ν ν 31 3 32 3 3 23 31 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 ν ν E E E G G 12 G 1343bookfm Page 95 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 96 Mechanics of Composite Materials Second Edition Inversion of Equation 270 would be the compliance matrix C and is given by 271 where 272 Although nine independent elastic constants are in the compliance matrix S and correspondingly in the stiffness matrix C for orthotropic materials constraints on the values of these constants exist Based on the first law of thermodynamics the stiffness and compliance matrices must be positive definite Thus the diagonal terms of C and S in Equation 271 and Equation 270 respectively need to be positive From the diagonal elements of the compliance matrix S this gives 273 and from the diagonal elements of the stiffness matrix C gives 274 Using the reciprocal relations given by Equation 267 through Equation 269 for and ij 123 we can rewrite the inequalities as follows C E E E E 1 23 32 2 3 21 23 31 2 3 31 21 3 ν ν ν ν ν ν ν ν Δ Δ 2 2 3 21 23 31 2 3 13 31 1 3 32 0 0 0 1 E E E E E E Δ Δ Δ ν ν ν ν ν ν ν 12 31 1 3 31 21 32 2 3 32 12 31 1 3 0 0 0 ν ν ν ν ν ν ν E E E E E E Δ Δ Δ Δ 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 12 21 1 2 23 31 12 ν ν E E G G G Δ 1 2 12 21 23 32 13 31 21 32 13 1 2 ν ν ν ν ν ν ν ν ν E E E3 E1 0 E2 0 E3 0 G12 0 G23 0 G31 0 1 0 23 32 ν ν 1 0 31 13 ν ν 1 0 12 21 ν ν Δ 1 2 0 12 21 23 32 31 13 13 21 32 ν ν ν ν ν ν ν ν ν ν ν ij i ji j E E i j 1343bookfm Page 96 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Macromechanical Analysis of a Lamina 97 For example because then 275a Similarly five other such relationships can be developed to give 275b 275c 275d 275e 275f These restrictions on the elastic moduli are important in optimizing prop erties of a composite because they show that the nine independent properties cannot be varied without influencing the limits of the others 1 0 12 21 ν ν ν ν ν 12 21 1 2 12 1 1 E E ν ν 12 1 2 12 1 E E ν12 1 2 E E ν21 2 1 E E ν32 3 2 E E ν23 2 3 E E ν31 3 1 E E ν13 1 3 E E 1343bookfm Page 97 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 98 Mechanics of Composite Materials Second Edition Example 25 Find the compliance and stiffness matrix for a graphiteepoxy lamina The material properties are given as Solution E GPa 1 181 E GPa 2 10 3 E GPa 3 10 3 ν12 0 28 ν23 0 60 ν13 0 27 G GPa 12 7 17 G GPa 23 3 0 G GPa 31 7 00 S E Pa 11 1 9 12 1 1 1 181 10 5 525 10 S E Pa 22 2 9 11 1 1 1 10 3 10 9 709 10 S E Pa 33 3 9 11 1 1 1 10 3 10 9 709 10 S E Pa 12 12 1 9 12 1 0 28 181 10 1 547 10 ν S E Pa 13 13 1 9 12 1 0 27 181 10 1 492 10 ν S E Pa 23 23 2 9 11 1 0 6 10 3 10 5 825 10 ν S G Pa 44 23 9 10 1 1 1 3 10 3 333 10 S G Pa 55 31 9 10 1 1 1 7 10 1 429 10 1343bookfm Page 98 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Macromechanical Analysis of a Lamina 99 Thus the compliance matrix for the orthotropic lamina is given by The stiffness matrix can be found by inverting the compliance matrix and is given by The preceding stiffness matrix C can also be found directly by using Equa tion 271 24 Hookes Law for a TwoDimensional Unidirectional Lamina 241 Plane Stress Assumption A thin plate is a prismatic member having a small thickness and it is the case for a typical lamina If a plate is thin and there are no outofplane loads it can be considered to be under plane stress Figure 217 If the upper and lower surfaces of the plate are free from external loads then σ3 0 τ31 0 and τ23 0 Because the plate is thin these three stresses within the plate are S G Pa 66 12 9 10 1 1 1 7 17 10 1 395 10 S 5 525 10 1 547 10 1 492 10 0 0 12 12 12 00 1 547 10 9 709 10 5 825 10 0 0 0 1 12 11 11 492 10 5 825 10 9 709 10 0 0 0 0 0 0 3 33 12 11 11 3 10 0 0 0 0 0 0 1 429 10 0 0 0 0 0 0 1 395 10 10 10 10 Pa 1 C S C 1 12 10 0 1850 10 0 7269 10 0 77204 10 0 0 0 0 7269 10 0 1638 10 0 9938 1 10 10 11 00 0 0 0 0 7204 10 0 9938 10 0 1637 10 0 0 0 10 10 10 11 0 0 0 0 3000 10 0 0 0 0 0 0 0 6998 10 0 0 0 0 0 0 0 716 10 10 8 1010 Pa 1343bookfm Page 99 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 100 Mechanics of Composite Materials Second Edition assumed to vary little from the magnitude of stresses at the top and the bottom surfaces Thus they can be assumed to be zero within the plate also A lamina is thin and if no outofplane loads are applied one can assume that it is under plane stress This assumption then reduces the threedimen sional stressstrain equations to twodimensional stressstrain equations 242 Reduction of Hookes Law in Three Dimensions to Two Dimensions A unidirectional lamina falls under the orthotropic material category If the lamina is thin and does not carry any outofplane loads one can assume plane stress conditions for the lamina Therefore taking Equation 226 and Equation 239 and assuming σ3 0 τ23 0 and τ31 0 then 276ab The normal strain ε3 is not an independent strain because it is a function of the other two normal strains ε1 and ε2 Therefore the normal strain ε3 can be omitted from the stressstrain relationship 239 Also the shearing strains γ23 and γ31 can be omitted because they are zero Equation 239 for an orthotropic plane stress problem can then be written as 277 FIGURE 217 Plane stress conditions for a thin plate 2 2 1 3 ε σ σ 3 13 1 23 2 S S γ γ 23 31 0 ε ε γ 1 2 12 11 12 12 22 66 0 0 0 0 S S S S S σ σ τ 1 2 12 1343bookfm Page 100 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Macromechanical Analysis of a Lamina 101 where Sij are the elements of the compliance matrix Note the four indepen dent compliance elements in the matrix Inverting Equation 277 gives the stressstrain relationship as 278 where Qij are the reduced stiffness coefficients which are related to the compliance coefficients as 279ad Note that the elements of the reduced stiffness matrix Qij are not the same as the elements of the stiffness matrix Cij see Exercise 213 243 Relationship of Compliance and Stiffness Matrix to Engineering Elastic Constants of a Lamina Equation 277 and Equation 278 show the relationship of stress and strain through the compliance S and reduced stiffness Q matrices However stress and strains are generally related through engineering elastic constants For a unidirectional lamina these engineering elastics constants are E1 longitudinal Youngs modulus in direction 1 E2 transverse Youngs modulus in direction 2 ν12 major Poissons ratio where the general Poissons ratio νij is defined as the ratio of the negative of the normal strain in direction j to the normal strain in direction i when the only normal load is applied in direction i G12 inplane shear modulus in plane 12 σ σ τ 1 2 12 11 12 12 22 66 0 0 0 0 Q Q Q Q Q ε ε γ 1 2 12 11 22 11 22 12 2 Q S S S S 12 12 11 22 12 2 Q S S S S 22 11 11 22 12 2 Q S S S S 66 66 1 Q S 1343bookfm Page 101 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 102 Mechanics of Composite Materials Second Edition Experimentally the four independent engineering elastic constants are measured as follows and can be related to the four independent elements of the compliance matrix S of Equation 277 Apply a pure tensile load in direction 1 Figure 218a that is 280 Then from Equation 277 FIGURE 218 Application of stresses to find engineering constants of a unidirectional lamina a 2 1 σ1 σ1 τ12 τ12 2 1 c σ2 σ2 2 1 b σ σ τ 1 2 12 0 0 0 1343bookfm Page 102 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Macromechanical Analysis of a Lamina 103 281ac By definition if the only nonzero stress is σ1 as is the case here then 282 283 Apply a pure tensile load in direction 2 Figure 218b that is 284 Then from Equation 277 285ac By definition if the only nonzero stress is σ2 as is the case here then 286 287 The ν21 term is called the minor Poissons ratio From Equation 282 Equation 283 Equation 286 and Equation 287 we have the reciprocal relationship 288 ε σ ε σ γ 1 11 1 2 12 1 12 0 S S E S 1 1 1 11 1 σ ε 12 2 1 12 11 ν ε ε S S σ σ τ 1 2 12 0 0 0 ε σ ε σ γ 1 12 2 2 22 2 12 0 S S E S 2 2 2 22 1 σ ε ν ε ε 21 1 2 12 22 S S ν ν 12 1 21 2 E E 1343bookfm Page 103 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 104 Mechanics of Composite Materials Second Edition Apply a pure shear stress in the plane 12 Figure 218c that is 289 Then from Equation 277 290ac By definition if τ12 is the only nonzero stress as is the case here then 291 Thus we have proved that 292ad Also the stiffness coefficients Qij are related to the engineering constants through Equation 298 and Equation 292 as σ σ τ 1 2 12 0 0 0 and ε1 0 ε2 0 γ τ 12 66 12 S G S 12 12 12 66 1 τ γ S E 11 1 1 S E 12 12 1 ν S E 22 2 1 S G 66 12 1 Q E 11 1 21 12 1 ν ν 1343bookfm Page 104 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Macromechanical Analysis of a Lamina 105 293ad Equation 277 Equation 278 Equation 292 and Equation 293 relate stresses and strains through any of the following combinations of four con stants Q11 Q12 Q22 Q66 or S11 S12 S22 S66 or E1 E2 ν12 G12 The unidirectional lamina is a specially orthotropic lamina because normal stresses applied in the 12 direction do not result in any shearing strains in the 12 plane because Q16 Q26 0 S16 S26 Also the shearing stresses applied in the 12 plane do not result in any normal strains in the 1 and 2 directions because Q16 Q26 0 S16 S26 A woven composite with its weaves perpendicular to each other and short fiber composites with fibers arranged perpendicularly to each other or aligned in one direction also are specially orthotropic Thus any discussion in this chapter or in Chapter 4 Macromechanics of a Laminate is valid for such a lamina as well Mechanical properties of some typical unidirectional lamina are given in Table 21 and Table 22 Example 26 For a graphiteepoxy unidirectional lamina find the following 1 Compliance matrix 2 Minor Poissons ratio 3 Reduced stiffness matrix 4 Strains in the 12 coordinate system if the applied stresses Figure 219 are Use the properties of unidirectional graphiteepoxy lamina from Table 21 Q E 12 12 2 21 12 1 ν ν ν Q E 22 2 21 12 1 ν ν and 66 12 Q G σ σ τ 1 2 12 2 3 4 MPa MPa MPa 1343bookfm Page 105 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 106 Mechanics of Composite Materials Second Edition Solution From Table 21 the engineering elastic constants of the unidirectional graph iteepoxy lamina are 1 Using Equation 292 the compliance matrix elements are TABLE 21 Typical Mechanical Properties of a Unidirectional Lamina SI System of Units Property Symbol Units Glass epoxy Boron epoxy Graphite epoxy Fiber volume fraction Vf 045 050 070 Longitudinal elastic modulus E1 GPa 386 204 181 Transverse elastic modulus E2 GPa 827 1850 1030 Major Poissons ratio ν12 026 023 028 Shear modulus G12 GPa 414 559 717 Ultimate longitudinal tensile strength MPa 1062 1260 1500 Ultimate longitudinal compressive strength MPa 610 2500 1500 Ultimate transverse tensile strength MPa 31 61 40 Ultimate transverse compressive strength MPa 118 202 246 Ultimate inplane shear strength MPa 72 67 68 Longitudinal coefficient of thermal expansion α1 μmmC 86 61 002 Transverse coefficient of thermal expansion α2 μmmC 221 303 225 Longitudinal coefficient of moisture expansion β1 mmkgkg 000 000 000 Transverse coefficient of moisture expansion β2 mmkgkg 060 060 060 Source Tsai SW and Hahn HT Introduction to Composite Materials CRC Press Boca Raton FL Table 17 p 19 Table 71 p 292 Table 83 p 344 Reprinted with permission 1 T ult σ 1 C ult σ 2 T ult σ 2 C ult σ τ12 ult E GPa E GPa G 1 2 12 12 181 10 3 0 28 7 17 ν GPa S Pa 11 9 11 1 1 181 10 0 5525 10 S Pa 12 9 11 1 0 28 181 10 0 1547 10 1343bookfm Page 106 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Macromechanical Analysis of a Lamina 107 2 Using the reciprocal relationship 288 the minor Poissons ratio is 3 Using Equation 293 the reduced stiffness matrix Q elements are TABLE 22 Typical Mechanical Properties of a Unidirectional Lamina USCS System of Units Property Symbol Units Glass epoxy Boron epoxy Graphite epoxy Fiber volume fraction Vf 045 050 070 Longitudinal elastic modulus E1 Msi 560 2959 2625 Transverse elastic modulus E2 Msi 120 2683 149 Major Poissons ratio v12 026 023 028 Shear modulus G12 Msi 060 0811 1040 Ultimate longitudinal tensile strength ksi 15403 18275 21756 Ultimate longitudinal compressive strength ksi 8847 3626 21756 Ultimate transverse tensile strength ksi 4496 8847 5802 Ultimate transverse compressive strength ksi 1712 2930 3568 Ultimate inplane shear strength ksi 1044 9718 9863 Longitudinal coefficient of thermal expansion α1 μininF 4778 3389 00111 Transverse coefficient of thermal expansion α2 μininF 12278 1683 125 Longitudinal coefficient of moisture expansion β1 ininlblb 000 000 000 Transverse coefficient of moisture expansion β2 ininlblb 060 060 060 Source Tsai SW and Hahn HT Introduction to Composite Materials CRC Press Boca Raton FL Table 17 p 19 Table 71 p 292 Table 83 p 344 USCS system used for tables reprinted with permission 1 T ult σ 1 C ult σ 2 T ult σ 2 C ult σ τ12 ult S Pa 22 9 10 1 1 10 3 10 0 9709 10 S Pa 66 9 9 1 1 7 17 10 0 1395 10 ν21 9 9 0 28 181 10 10 3 10 0 01593 1343bookfm Page 107 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 108 Mechanics of Composite Materials Second Edition The reduced stiffness matrix Q could also be obtained by inverting the compliance matrix S of part 1 FIGURE 219 Applied stresses in a unidirectional lamina in Example 26 σ2 3 MPa σ2 σ1 τ12 2 1 τ12 4 MPa σ1 2 MPa Q Pa 11 9 9 181 10 1 0 28 0 01593 181 8 10 Q12 9 0 28 10 3 10 1 0 28 0 01593 2 897 109Pa Q Pa 22 9 9 10 3 10 1 0 28 0 01593 10 35 10 Q Pa 66 9 7 17 10 Q S 1 11 11 0 5525 10 0 1547 10 0 0 1547 10 0 9709 10 0 0 0 0 1395 10 11 10 9 1 1343bookfm Page 108 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Macromechanical Analysis of a Lamina 109 4 Using Equation 277 the strains in the 12 coordinate system are Thus the strains in the local axes are 25 Hookes Law for a TwoDimensional Angle Lamina Generally a laminate does not consist only of unidirectional laminae because of their low stiffness and strength properties in the transverse direction Therefore in most laminates some laminae are placed at an angle It is thus necessary to develop the stressstrain relationship for an angle lamina The coordinate system used for showing an angle lamina is as given in Figure 220 The axes in the 12 coordinate system are called the local axes or the material axes The direction 1 is parallel to the fibers and the direction 2 is perpendicular to the fibers In some literature direction 1 is also called 181 8 10 2 897 10 0 2 897 10 10 35 10 0 0 0 7 9 9 9 9 17 109 Pa ε ε γ 1 2 12 11 11 0 5525 10 0 1547 10 00 0 1547 10 0 9709 10 0 0 0 0 1395 10 11 10 9 2 10 3 10 4 10 15 69 2 6 6 6 94 4 557 9 10 6 ε μ ε μ γ μ 1 2 12 15 69 294 4 557 9 m m m m m m 1343bookfm Page 109 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 110 Mechanics of Composite Materials Second Edition the longitudinal direction L and the direction 2 is called the transverse direction T The axes in the xy coordinate system are called the global axes or the offaxes The angle between the two axes is denoted by an angle θ The stressstrain relationship in the 12 coordinate system has already been established in Section 24 and we are now going to develop the stressstrain equations for the xy coordinate system The global and local stresses in an angle lamina are related to each other through the angle of the lamina θ Appendix B 294 where T is called the transformation matrix and is defined as 295 and FIGURE 220 Local and global axes of an angle lamina 2 1 θ y x σ σ τ σ σ τ x y xy T 1 1 2 12 T c s sc s c sc sc sc c s 1 2 2 2 2 2 2 2 2 1343bookfm Page 110 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Macromechanical Analysis of a Lamina 111 296 297ab Using the stressstrain Equation 278 in the local axes Equation 294 can be written as 298 The global and local strains are also related through the transformation matrix Appendix B 299 which can be rewritten as 2100 where R is the Reuter matrix3 and is defined as 2101 Then substituting Equation 2100 in Equation 298 gives T c s sc s c sc sc sc c s 2 2 2 2 2 2 2 2 c Cos θ s Sin θ σ σ τ ε ε γ x y xy T Q 1 1 2 12 ε ε γ ε ε γ 1 2 12 2 2 T x y xy ε ε γ ε ε γ 1 2 12 1 R T R x y xy R 1 0 0 0 1 0 0 0 2 1343bookfm Page 111 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 112 Mechanics of Composite Materials Second Edition 2102 On carrying the multiplication of the first five matrices on the righthand side of Equation 2102 2103 where are called the elements of the transformed reduced stiffness matrix and are given by 2104af Note that six elements are in the matrix However by looking at Equa tion 2104 it can be seen that they are just functions of the four stiffness elements Q11 Q12 Q22 and Q66 and the angle of the lamina θ Inverting Equation 2103 gives 2105 σ σ τ ε ε x y xy x y T Q R T R 1 1 γ xy σ σ τ x y xy Q Q Q Q Q Q Q Q 11 12 16 12 22 26 16 26 66 Q x y xy ε ε γ Qij Q Q Q c Q s Q Q s c 11 11 4 22 4 12 66 2 2 2 2 Q Q Q Q s c Q c s 12 11 22 66 2 2 12 4 2 4 Q Q s Q c Q Q s c 22 11 4 22 4 12 66 2 2 2 2 Q Q Q Q c s Q Q Q s c 16 11 12 66 3 22 12 66 3 2 2 Q Q Q Q cs Q Q Q c s 26 11 12 66 3 22 12 66 3 2 2 Q Q Q Q Q s c Q s c 66 11 22 12 66 2 2 66 4 4 2 2 Q ε ε γ x y xy S S S S S S S S 11 12 16 12 22 26 16 26 66 S x y xy σ σ σ 1343bookfm Page 112 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Macromechanical Analysis of a Lamina 113 where Sij are the elements of the transformed reduced compliance matrix and are given by 2106af From Equation 277 and Equation 278 for a unidirectional lamina loaded in the material axes directions no coupling occurs between the nor mal and shearing terms of strains and stresses However for an angle lamina from Equation 2103 and Equation 2105 coupling takes place between the normal and shearing terms of strains and stresses If only normal stresses are applied to an angle lamina the shear strains are nonzero if only shearing stresses are applied to an angle lamina the normal strains are nonzero Therefore Equation 2103 and Equation 2105 are stressstrain equations for what is called a generally orthotropic lamina Example 27 Find the following for a 60 angle lamina Figure 221 of graphiteepoxy Use the properties of unidirectional graphiteepoxy lamina from Table 21 1 Transformed compliance matrix 2 Transformed reduced stiffness matrix If the applied stress is σx 2 MPa σy 3 MPa and τxy 4 MPa also find 3 Global strains 4 Local strains 5 Local stresses 6 Principal stresses 7 Maximum shear stress S S c S S s c S s 11 11 4 12 66 2 2 22 4 2 S S s c S S S s c 12 12 4 4 11 22 66 2 2 S S s S S s c S c 22 11 4 12 66 2 2 22 4 2 S S S S sc S S S s c 16 11 12 66 3 22 12 66 3 2 2 2 2 S S S S s c S S S sc 26 11 12 66 3 22 12 66 3 2 2 2 2 66 11 22 12 66 2 2 S 2 2 S 2 S 4 S S s c S 66 s c 4 4 1343bookfm Page 113 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 114 Mechanics of Composite Materials Second Edition 8 Principal strains 9 Maximum shear strain Solution c Cos60 0500 s Sin60 0866 1 From Example 26 Now using Equation 2106a FIGURE 221 Applied stresses to an angle lamina in Example 27 σy 3 MPa σx 2 MPa σy σx τxy τxy 4 MPa 2 1 S Pa 11 11 0 5525 10 1 S Pa 22 10 0 9709 10 1 S Pa 12 11 0 1547 10 1 S Pa 66 9 0 1395 10 1 1343bookfm Page 114 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Macromechanical Analysis of a Lamina 115 Similarly using Equation 2106bf one can evaluate 2 Invert the transformed compliance matrix to obtain the trans formed reduced stiffness matrix S11 11 4 11 0 5525 10 0 500 2 0 1547 10 0 1395 10 0 866 0 5 0 9709 10 0 9 2 2 10 866 0 8053 10 1 4 10 Pa S Pa 12 11 0 7878 10 1 S Pa 16 10 0 3234 10 1 S Pa 22 10 0 3475 10 1 S Pa 26 10 0 4696 10 1 S Pa 66 9 0 1141 10 1 S Q Q 0 8053 10 0 7878 10 0 3234 10 10 11 10 0 7878 10 0 3475 10 0 4696 10 0 32 11 10 10 34 10 0 4696 10 0 1141 10 10 10 9 1 0 2365 10 0 3246 10 0 2005 10 0 3246 10 11 11 11 111 12 11 11 0 1094 10 0 5419 10 0 2005 10 0 5419 10 0 3674 10 11 11 Pa 1343bookfm Page 115 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 116 Mechanics of Composite Materials Second Edition 3 The global strains in the xy plane are given by Equation 2105 as 4 Using transformation Equation 299 the local strains in the lamina are 5 Using transformation Equation 294 the local stresses in the lamina are 6 The principal normal stresses are given by4 2107 ε ε γ x y xy 0 8053 10 0 7878 10 10 11 10 11 10 0 3234 10 0 7878 10 0 3475 10 0 4696 10 0 3234 10 0 4696 10 0 114 10 10 10 1 10 2 10 3 10 4 10 9 6 6 6 0 5534 10 0 3078 10 0 5328 10 4 3 3 ε ε γ 1 2 12 2 0 2500 0 7500 0 8660 0 7 500 0 2500 0 8660 0 4330 0 4330 0 500 0 5534 10 0 3078 10 0 5328 10 2 4 3 3 ε ε γ 1 2 12 4 3 0 1367 10 0 2662 10 0 5809 10 3 σ σ τ 1 2 12 0 2500 0 7500 0 8660 0 7500 0 2500 0 8660 0 4330 0 4330 0 500 2 10 3 10 4 10 6 6 6 0 1714 10 0 2714 10 0 4165 10 7 7 7 Pa σ σ σ σ σ τ maxmin x y x y xy 2 2 2 2 1343bookfm Page 116 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Macromechanical Analysis of a Lamina 117 The value of the angle at which the maximum normal stresses occur is4 2108 Note that the principal normal stresses do not occur along the material axes This should be also evident from the nonzero shear stresses in the local axes 7 The maximum shear stress is given by4 2109 The angle at which the maximum shear stress occurs is4 2110 2 10 3 10 2 2 10 3 10 2 4 10 6 6 6 6 2 6 2 MPa 4 217 5 217 θ τ σ σ p xy x y 1 2 2 1 tan 1 2 2 4 10 2 10 3 10 1 6 6 6 tan 29 000 τ σ σ τ max x y xy 2 2 10 3 10 2 2 2 6 6 2 4 106 2 4 717 MPa θ σ σ τ s x y xy 1 2 2 1 tan 1343bookfm Page 117 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 118 Mechanics of Composite Materials Second Edition 16000 8 The principal strains are given by4 2111 The value of the angle at which the maximum normal strains occur is4 2112 27860 Note that the principal normal strains do not occur along the material axes This should also be clear from the nonzero shear strain in the local axes In addition the axes of principal normal stresses and principal normal strains do not match unlike in isotropic materials 9 The maximum shearing strain is given by4 1 2 2 10 3 10 2 4 10 1 6 6 6 tan ε ε ε ε ε γ maxmin x y x y xy 2 2 2 2 2 0 5534 10 0 3078 10 2 0 5534 10 0 307 4 3 4 8 10 2 0 5328 10 2 3 2 3 2 1 962 10 4 486 10 4 4 θ γ ε ε p xy x y 1 2 1 tan 1 2 0 5328 10 0 5534 10 0 3078 10 1 3 4 3 tan 1343bookfm Page 118 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Macromechanical Analysis of a Lamina 119 2113 The value of the angle at which the maximum shearing strain occurs is4 2114 17140 Example 28 As shown in Figure 222 a 60 angle graphiteepoxy lamina is subjected only to a shear stress τxy 2 MPa in the global axes What would be the value of the strains measured by the strain gage rosette that is what FIGURE 222 Strain gage rosette on an angle lamina γ ε ε γ max x y xy 2 2 0 5534 10 0 3078 10 0 532 10 4 3 2 3 2 6 448 10 4 θ ε ε γ s x y xy 1 2 1 tan 1 2 0 5534 10 0 3078 10 0 5328 10 1 4 3 3 tan 2 1 C A B Fiber Strain gage rosette with gages A B and C y 60 120 120 120 τxy 2 MPa 1343bookfm Page 119 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 120 Mechanics of Composite Materials Second Edition would be the normal strains measured by strain gages A B and C Use the properties of unidirectional graphiteepoxy lamina from Table 21 Solution Per Example 27 the reduced compliance matrix is The global strains in the xy plane are given by Equation 2105 as For a strain gage placed at an angle φ to the xaxis the normal strain recorded by the strain gage is given by Equation B15 in Appendix B For strain gage A φ 0 For strain gage B φ 240 S 0 8053 10 0 7878 10 0 3234 10 0 78 10 11 10 78 10 0 3475 10 0 4696 10 0 3234 1 11 10 10 0 0 4696 10 0 1141 10 10 10 9 1 Pa ε ε γ x y xy 0 8053 10 0 7878 10 10 11 10 11 10 0 3234 10 0 7878 10 0 3475 10 0 4696 10 0 3234 10 0 4696 10 0 114 10 10 10 1 10 0 0 2 10 9 6 6 468 10 9 392 10 2 283 10 5 5 4 ε ε φ ε φ γ φ φ φ x y xy Cos Sin Sin Cos 2 2 εA Cos 6 468 10 0 9 392 10 0 2 5 2 5 2 Sin 283 10 0 0 4 Sin Cos 6 468 10 5 εB 6 468 10 240 9 392 10 240 5 2 5 2 Cos Sin 2 283 10 240 240 4 Sin Cos 1343bookfm Page 120 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Macromechanical Analysis of a Lamina 121 For strain gage C φ 120 26 Engineering Constants of an Angle Lamina The engineering constants for a unidirectional lamina were related to the compliance and stiffness matrices in Section 243 In this section similar techniques are applied to relate the engineering constants of an angle ply to its transformed stiffness and compliance matrices 1 For finding the engineering elastic moduli in direction x Figure 223a apply 2115 Then from Equation 2105 2116ac The elastic moduli in direction x is defined as 2117 1 724 10 4 εC 6 468 10 120 9 392 10 120 5 2 5 2 Cos Sin 2 283 10 120 120 4 Sin Cos 1 083 10 5 σ σ τ x y xy 0 0 0 ε σ x x S 11 ε σ y x S 12 γ σ xy x S 16 E S x x x σ ε 1 11 1343bookfm Page 121 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 122 Mechanics of Composite Materials Second Edition FIGURE 223 Application of stresses to find engineering constants of an angle lamina 2 2 y 1 x 2 y 1 x a b c σx σy σy τxy τxy σx y 1 x 1343bookfm Page 122 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Macromechanical Analysis of a Lamina 123 Also the Poissons ratio νxy is defined as 2118 In an angle lamina unlike in a unidirectional lamina interaction also occurs between the shear strain and the normal stresses This is called shear coupling The shear coupling term that relates the nor mal stress in the xdirection to the shear strain is denoted by mx and is defined as 2119 Note that mx is a nondimensional parameter like the Poissons ratio Later note that the same parameter mx relates the shearing stress in the xy plane to the normal strain in directionx The shear coupling term is particularly important in tensile testing of angle plies For example if an angle lamina is clamped at the two ends it will not allow shearing strain to occur This will result in bending moments and shear forces at the clamped ends5 2 Similarly by applying stresses 2120 as shown in Figure 223b it can be found 2121 2122 2123 The shear coupling term my relates the normal stress σy to the shear strain γxy In the following section 3 note that the same parameter my relates the shear stress τxy in the xy plane to the normal strain εy ν ε ε xy y x S S 12 11 1 1 1 16 1 m E S E x x xy σ γ σ σ τ x y xy 0 0 0 E S y 1 22 νyx S S 12 22 and 1 1 26 1 m S E y 1343bookfm Page 123 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 124 Mechanics of Composite Materials Second Edition From Equation 2117 Equation 2118 Equation 2121 and Equation 2122 the reciprocal relationship is given by 2124 3 Also by applying the stresses 2125 as shown in Figure 223c it is found that 2126 2127 2128 Thus the strainstress Equation 2105 of an angle lamina can also be written in terms of the engineering constants of an angle lamina in matrix form as 2129 The preceding six engineering constants of an angle ply can also be written in terms of the engineering constants of a unidirectional ply using Equation 292 and Equation 2106 in Equation 2117 through Equation 2119 Equation 2121 Equation 2123 and Equation 2128 ν ν yx y xy x E E σ σ τ x y xy 0 0 0 1 1 16 1 m S E x 1 1 26 1 m S E y and G S xy 1 66 ε ε γ ν ν x y xy x xy x x xy x y E E m E E E 1 1 1 m E m E m E G y x y xy x 1 1 1 1 σ σy xy τ 1 11 E S x 1343bookfm Page 124 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Macromechanical Analysis of a Lamina 125 2130 2131 2132 2133 2134 S c S S s c S s 11 4 12 66 2 2 22 4 2 1 1 2 1 1 4 12 12 1 2 2 2 4 E c G E s c E s ν νxy x E S 12 E S s c S S S s c x 12 4 4 11 22 66 2 2 E E s c E E G s c x ν12 1 4 4 1 2 12 2 2 1 1 1 1 22 E S y S s S S c s S c 11 4 12 66 2 2 22 4 2 1 2 1 1 1 4 12 1 12 2 2 2 4 E s E G c s E c ν 1 66 G S xy 2 2 2 4 11 22 12 66 2 2 66 4 4 S S S S s c S s c 2 2 2 4 1 1 1 2 12 1 12 2 2 12 4 4 E E E G s c G s c ν m x S E 16 1 E S S S sc S S S s c 1 11 12 66 3 22 12 66 3 2 2 2 E E E G sc E E G 1 1 12 1 12 3 2 12 1 1 2 2 1 2 2 1 ν ν 2 3 s c 1343bookfm Page 125 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 126 Mechanics of Composite Materials Second Edition 2135 Example 29 Find the engineering constants of a 60 graphiteepoxy lamina Use the properties of a unidirectional graphiteepoxy lamina from Table 21 Solution From Example 27 we have From Equation 2117 m y S E 26 1 E S S S s c S S S sc 1 11 12 66 3 22 12 66 3 2 2 2 2 E E E G s c E E G 1 1 12 1 12 3 2 12 1 1 2 2 1 2 2 1 ν ν 2 3 sc S Pa 11 10 0 8053 10 1 S Pa 12 11 0 7878 10 1 S Pa 16 10 0 3234 10 1 S Pa 22 10 0 3475 10 1 S Pa 26 10 0 4696 10 1 and S Pa 66 9 0 1141 10 1 E GPa x 1 0 8053 10 12 42 10 1343bookfm Page 126 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Macromechanical Analysis of a Lamina 127 From Equation 2118 From Equation 2119 From Equation 2121 From Equation 2123 From Equation 2128 The variations of the six engineering elastic constants are shown as a function of the angle for the preceding graphiteepoxy lamina in Figure 224 through Figure 229 The variations of the Youngs modulus Ex and Ey are inverses of each other As the fiber orientation angle of ply varies from 0 to 90 the value of Ex νxy 0 7878 10 0 8053 10 0 09783 11 10 1 1 0 3234 10 181 10 10 9 mx mx 5 854 E GPa y 1 0 3475 10 28 78 10 1 1 0 4696 10 181 10 10 9 my my 8 499 G GPa xy 1 0 1141 10 8 761 9 1343bookfm Page 127 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 128 Mechanics of Composite Materials Second Edition FIGURE 224 Elastic modulus in directionx as a function of angle of lamina for a graphiteepoxy lamina FIGURE 225 Elastic modulus in directiony as a function of angle of lamina for a graphiteepoxy lamina 200 150 Ex GPa 100 50 0 0 15 30 45 Angle of lamina θ degrees 60 75 90 200 150 100 50 00 15 30 45 Angle of lamina θ degrees 60 75 90 Ey GPa 1343bookfm Page 128 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Macromechanical Analysis of a Lamina 129 FIGURE 226 Poissons ratio νxy as a function of angle of lamina for a graphiteepoxy lamina FIGURE 227 Inplane shear modulus in xyplane as a function of angle of lamina for a graphiteepoxy lamina 03 02 01 0 0 15 30 45 Angle of lamina θ degrees vxy 60 75 90 10 9 8 70 15 30 45 Angle of lamina θ degrees Gxy GPa 60 75 90 1343bookfm Page 129 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 130 Mechanics of Composite Materials Second Edition FIGURE 228 Shear coupling coefficient mx as a function of angle of lamina for a graphiteepoxy lamina FIGURE 229 Shear coupling coefficient my as a function of angle of lamina for a graphiteepoxy lamina 10 8 6 4 2 0 0 15 30 45 60 75 90 Angle of lamina θ degrees mx 10 8 6 4 2 0 0 15 30 45 60 75 90 Angle of lamina θ degrees my 1343bookfm Page 130 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Macromechanical Analysis of a Lamina 131 varies from the value of the longitudinal E1 to the transverse Youngs modulus E2 However the maximum and minimum values of Ex do not necessarily exist for θ 0 and θ 90 respectively for every lamina Consider the case of a metal matrix composite such as a typical SCS 6 Ti6 Al 4V composite The elastic moduli of such a lamina with a 55 fiber volume fraction is E1 272 GPa E2 200 GPa ν12 02770 G12 7733 GPa In Figure 230 the lowest modulus value of Ex is found for θ 63 In fact the angle of 63 at which Ex is minimum is independent of the fiber volume fraction if one uses the mechanics of materials approach Section 331 to evaluate the preceding four elastic moduli of a unidirectional lamina See Exercise 313 In Figure 227 the shear modulus Gxy is maximum for θ 45 and is minimum for 0 and 90 plies The shear modulus Gxy becomes maximum for 45 because the principal stresses for pure shear load on a 45 ply are along the material axis From Equation 2133 the expression for Gxy for a 45 ply is FIGURE 230 Variation of elastic modulus in directionx as a function of angle of lamina for a typical SCS 6Ti6 Al 4V lamina 280 260 240 220 200 180 0 15 30 45 60 75 90 Angle of lamina θ degrees Ex GPa 1343bookfm Page 131 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 132 Mechanics of Composite Materials Second Edition 2136 In Figure 228 and Figure 229 the shear coupling coefficients mx and my are maximum at θ 362 and θ 5378 respectively The values of these coefficients are quite extreme showing that the normalshear coupling terms have a stronger effect than the Poissons effect This phenomenon of shear coupling terms is missing in isotropic materials and unidirectional plies but cannot be ignored in angle plies 27 Invariant Form of Stiffness and Compliance Matrices for an Angle Lamina Equation 2104 and Equation 2106 for the and matrices are not analytically convenient because they do not allow a direct study of the effect of the angle of the lamina on the and matrices The stiffness elements can be written in invariant form as6 2137af where G E E E xy45 1 12 1 2 1 2 ν Q S Q S Q U U U 11 1 2 3 2 4 Cos Cos θ θ Q U U 12 4 3 4 Cos θ Q U U U 22 1 2 3 2 4 Cos Cos θ θ Q U U 16 2 3 2 2 4 Sin Sin θ θ Q U U 26 2 3 2 2 4 Sin Sin θ θ Q U U U 66 1 4 3 1 2 4 Cos θ 1343bookfm Page 132 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Macromechanical Analysis of a Lamina 133 2138ad The terms U1 U2 U3 and U4 are the four invariants and are combinations of the Qij which are invariants as well The transformed reduced compliance matrix can similarly be writ ten as 2139af where U Q Q Q Q 1 11 22 12 66 1 8 3 3 2 4 U Q Q 2 11 22 1 2 U Q Q Q Q 3 11 22 12 66 1 8 2 4 U Q Q Q Q 4 11 22 12 66 1 8 6 4 S S V V V 11 1 2 3 2 4 Cos Cos θ θ S V V 12 4 3 4 Cos θ S V V V 22 1 2 3 2 4 Cos Cos θ θ S V V 16 2 3 2 2 4 Sin Sin θ θ S V V 26 2 3 2 2 4 Sin Sin and θ θ S V V V 66 1 4 3 2 4 4 Cos θ V S S S S 1 11 22 12 66 1 8 3 3 2 V S S 2 11 22 1 2 1343bookfm Page 133 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 134 Mechanics of Composite Materials Second Edition 2140ad The terms V1 V2 V3 and V4 are invariants and are combinations of Sij which are also invariants The main advantage of writing the equations in this form is that one can easily examine the effect of the lamina angle on the reduced stiffness matrix elements Also formulas given by Equation 2137 and Equation 2139 are easier to manipulate for integration differentiation etc The concept is mainly important in deriving the laminate stiffness properties in Chapter 4 The elastic moduli of quasiisotropic laminates that behave like isotropic material are directly given in terms of these invariants Because quasiiso tropic laminates have the minimum stiffness of any laminate these can be used as a comparative measure of the stiffness of other types of laminates7 Example 210 Starting with the expression for from Equation 2104a reduce it to the expression for of Equation 2137a that is Solution Given and substituting V S S S S 3 11 22 12 66 1 8 2 V S S S S 4 11 22 12 66 1 8 6 Q11 Q Q 11 11 4 Cos θ Q Q Q 22 4 12 66 2 2 2 2 Sin Sin Cos θ θ θ 11 Q Q U U U 11 1 2 3 2 4 Cos Cos θ θ Q Q Q Q Q 11 11 4 22 4 12 66 2 2 2 2 Cos Sin Sin Cos θ θ θ θ Cos Cos 2 1 2 2 θ θ Sin Cos 2 1 2 2 θ θ 1343bookfm Page 134 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Macromechanical Analysis of a Lamina 135 we get where Example 211 Evaluate the four compliance and four stiffness invariants for a graphite epoxy angle lamina Use the properties for a unidirectional graphiteepoxy lamina from Table 21 Solution From Example 26 the compliance matrix S elements are Cos Cos and 2 2 1 4 2 θ θ 2 2 Sin Cos Sin θ θ θ Sin Cos 2 2 1 4 2 θ θ Q U U U 11 1 2 3 2 4 Cos Cos θ θ U Q Q Q Q 1 11 22 12 66 1 8 3 3 2 4 U Q Q 2 11 22 1 2 U Q Q Q Q 3 11 22 12 66 1 8 2 4 S Pa 11 11 0 5525 10 1 S Pa 12 11 0 1547 10 1 1343bookfm Page 135 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 136 Mechanics of Composite Materials Second Edition The stiffness matrix Q elements are Using Equation 2138 S Pa 22 10 0 9709 10 1 S Pa 66 9 0 1395 10 1 Q S 1 Q Pa 11 12 0 1818 10 Q Pa 12 10 0 2897 10 Q Pa 22 11 0 1035 10 Q Pa 66 10 0 7170 10 U1 12 11 1 8 3 0 1818 10 3 0 1035 10 2 0 289 77 10 4 0 7171 10 0 7637 10 10 10 11 Pa U P 2 12 11 11 1 2 0 1818 10 0 1035 10 0 8573 10 a U3 12 11 10 1 8 0 1818 10 0 1035 10 2 0 2897 10 4 0 7171 10 0 1971 10 10 11 Pa U4 12 11 10 1 8 0 1818 10 0 1035 10 6 0 2897 10 4 0 7171 10 0 2261 10 10 11 Pa 1343C002fm Page 136 Wednesday September 28 2005 1038 AM 2006 by Taylor Francis Group LLC Macromechanical Analysis of a Lamina 137 Using Equation 2140 28 Strength Failure Theories of an Angle Lamina A successful design of a structure requires efficient and safe use of materials Theories need to be developed to compare the state of stress in a material to failure criteria It should be noted that failure theories are only stated and their application is validated by experiments For a laminate the strength is related to the strength of each individual lamina This allows for a simple and economical method for finding the strength of a laminate Various theories have been developed for studying the failure of an angle lamina The theories are generally based on the normal and shear strengths of a unidirectional lamina An isotropic material such as steel generally has two strength parameters normal strength and shear strength In some cases such as concrete or gray cast iron the normal strengths are different in the tension and compression A simple failure theory for an isotropic material is based on finding the principal normal stresses and the maximum shear stresses These maximum V1 11 11 1 8 3 0 5525 10 3 0 1547 10 2 0 9709 10 0 1395 10 0 5553 10 1 10 9 10 Pa V2 11 11 1 2 0 5525 10 0 1547 10 0 457 8 10 10 1 Pa V3 11 10 1 8 0 5525 10 0 9709 10 2 0 1547 10 11 9 11 0 1395 10 0 4220 10 1 Pa V4 11 10 1 8 0 5525 10 0 9709 10 6 0 1547 10 11 9 11 0 1395 10 0 5767 10 1 Pa 1343bookfm Page 137 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 138 Mechanics of Composite Materials Second Edition stresses if greater than any of the corresponding ultimate strengths indicate failure in the material Example 212 A cylindrical rod made of gray cast iron is subjected to a uniaxial tensile load P Given Crosssectional area of rod 2 in2 Ultimate tensile strength 25 ksi Ultimate compressive strength 95 ksi Ultimate shear strength 35 ksi Modulus of elasticity 10 Msi Find the maximum load P that can be applied using maximum stress failure theory Solution At any location the stress state in the rod is σ P2 From a typical Mohrs circle analysis the maximum principal normal stress is P2 The maximum shear stress is P4 and acts at a crosssection 45 to the plane of maximum normal stress Comparing these maximum stresses to the corresponding ultimate strengths we have and Thus the maximum load is 50000 lb However in a lamina the failure theories are not based on principal normal stresses and maximum shear stresses Rather they are based on the stresses in the material or local axes because a lamina is orthotropic and its properties are different at different angles unlike an isotropic material In the case of a unidirectional lamina there are two material axes one parallel to the fibers and one perpendicular to the fibers Thus there are four normal strength parameters for a unidirectional lamina one for tension and one for compression in each of the two material axes directions The fifth strength parameter is the shear strength of a unidirectional lamina The shear stress whether positive or negative does not have an effect on the reported P or P 2 25 10 50 000 3 lb P or P 4 35 10 140 000 3 lb 1343bookfm Page 138 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Macromechanical Analysis of a Lamina 139 shear strengths of a unidirectional lamina However we will find later that the sign of the shear stress does affect the strength of an angle lamina The five strength parameters of a unidirectional lamina are therefore Unlike the stiffness parameters these strength parameters cannot be trans formed directly for an angle lamina Thus the failure theories are based on first finding the stresses in the local axes and then using these five strength parameters of a unidirectional lamina to find whether a lamina has failed Four common failure theories are discussed here Related concepts of strength ratio and the development of failure envelopes are also discussed 281 Maximum Stress Failure Theory Related to the maximum normal stress theory by Rankine and the maxi mum shearing stress theory by Tresca this theory is similar to those applied to isotropic materials The stresses acting on a lamina are resolved into the normal and shear stresses in the local axes Failure is predicted in a lamina if any of the normal or shear stresses in the local axes of a lamina is equal to or exceeds the corresponding ultimate strengths of the unidirectional lamina Given the stresses or strains in the global axes of a lamina one can find the stresses in the material axes by using Equation 294 The lamina is considered to be failed if 2141ac is violated Note that all five strength parameters are treated as positive numbers and the normal stresses are positive if tensile and negative if compressive Each component of stress is compared with the corresponding strength thus each component of stress does not interact with the others 1 T ult σ Ultimate longitudinal tensile strrength in direction 1 1 C ult σ Ultimate longitudinal compressivee strength in direction 1 2 T ult σ Ultimate transverse tensile strenngth in direction 2 2 C ult σ Ultimate transverse compressive sstrength in direction 2 and τ12 ult Ultimate inplane shear strength in plane 12 σ σ σ 1 1 1 C ult T ult or σ σ σ 2 2 2 C ult T ult or τ τ τ 12 12 12 ult ult 1343bookfm Page 139 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 140 Mechanics of Composite Materials Second Edition Example 213 Find the maximum value of S 0 if a stress of σx 2S σy 3S and τxy 4S is applied to the 60 lamina of graphiteepoxy Use maximum stress failure theory and the properties of a unidirectional graphiteepoxy lamina given in Table 21 Solution Using Equation 294 the stresses in the local axes are From Table 21 the ultimate strengths of a unidirectional graphiteepoxy lamina are 1500 MPa 1500 MPa 40 MPa 246 MPa 68 MPa Then using the inequalities 2141 of the maximum stress failure theory 1500 106 01714 101S 1500 106 246 106 02714 101S 40 106 68 106 04165 101S 68 106 or σ σ τ 1 2 12 0 2500 0 7500 0 8660 0 75 00 0 2500 0 8660 0 4330 0 4330 0 5000 S S S 2 3 4 0 1714 10 0 2714 1 110 0 4165 10 1 1 S σ1 T ult σ1 C ult σ2 T ult σ2 C ult τ12 ult 1343bookfm Page 140 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Macromechanical Analysis of a Lamina 141 8751 106 S 8751 106 1473 106 S 9064 106 1633 106 S 1633 106 All the inequality conditions and S 0 are satisfied if 0 S 1633 MPa The preceding inequalities also show that the angle lamina will fail in shear The maximum stress that can be applied before failure is Example 214 Find the offaxis shear strength of a 60 graphiteepoxy lamina Use the properties of unidirectional graphiteepoxy from Table 21 and apply the maximum stress failure theory Solution The offaxis shear strength of a lamina is defined as the minimum of the magnitude of positive and negative shear stress Figure 231 that can be applied to an angle lamina before failure Assume the following stress state σx 0 σy 0 τxy τ Then using the transformation Equation 294 Using the inequalities 2141 of the maximum stress failure theory we have σ σ τ x y xy MPa MPa MPa 32 66 48 99 65 32 σ σ τ 1 2 12 0 2500 0 7500 0 8660 0 750 0 0 2500 0 8660 0 4330 0 4330 0 5000 0 0 τ σ τ 1 0 866 σ τ 2 0 866 τ τ 12 0 500 1343bookfm Page 141 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 142 Mechanics of Composite Materials Second Edition 1500 0866τ 1500 or 1732 τ 1732 246 0866τ 40 or 4619 τ 2841 68 0500τ 68 or 1360 τ 1360 which shows that τxy 4619 MPa is the largest magnitude of shear stress that can be applied to the 60 graphiteepoxy lamina However the largest positive shear stress that could be applied is τxy 1360 MPa and the largest negative shear stress is τxy 4619 MPa This shows that the maximum magnitude of allowable shear stress in other than the material axes direction depends on the sign of the shear stress This is mainly because the local axes stresses in the direction perpendicular to the fibers are opposite in sign to each other for opposite signs of shear stress σ2 0866τ for positive τxy and σ2 0866τ for negative τxy Because the tensile strength perpendicular to the fiber direction is much lower than the compressive strength perpendicular to the fiber direction the two limiting values of τxy are different FIGURE 231 Positive and negative shear stresses applied to an angle lamina a Positive shear stress τxy τxy 2 y 1 x b Negative shear stress τxy τxy 2 y 1 x 1343bookfm Page 142 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Macromechanical Analysis of a Lamina 143 Table 23 shows the maximum negative and positive values of shear stress that can be applied to different angle plies of graphiteepoxy of Table 21 The minimum magnitude of the two stresses is the shear strength of the angle lamina 282 Strength Ratio In a failure theory such as the maximum stress failure theory of Section 281 it can be determined whether a lamina has failed if any of the inequalities of Equation 2141 are violated However this does not give the information about how much the load can be increased if the lamina is safe or how much the load should be decreased if the lamina has failed The definition of strength ratio SR is helpful here The strength ratio is defined as 2142 The concept of strength ratio is applicable to any failure theory If SR 1 then the lamina is safe and the applied stress can be increased by a factor of SR If SR 1 the lamina is unsafe and the applied stress needs to be reduced by a factor of SR A value of SR 1 implies the failure load Example 215 Assume that one is applying a load of TABLE 23 Effect of Sign of Shear Stress as a Function of Angle of Lamina Angle Degrees Positive ττττxy MPa Negative ττττxy MPa Shear strength MPa 0 6800 S 6800 S 6800 15 7852 S 7852 S 7852 30 1360 S 4619 2T 4619 45 2460 2C 4000 2T 4000 60 1360 S 4619 2T 4619 75 7852 S 7852 S 7852 90 6800 S 6800 S 6800 Note The notation in the parentheses denotes the mode of failure of the angle lamina as follows 1T longitudinal tensile failure 1C longitudinal compressive failure 2T transverse tensile failure 2C transverse compressive failure S shear failure SR Maximum Load Which Can Be Applied Load Applied 1343bookfm Page 143 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 144 Mechanics of Composite Materials Second Edition to a 60 angle lamina of graphiteepoxy Find the strength ratio using the maximum stress failure theory Solution If the strength ratio is R then the maximum stress that can be applied is Following Example 213 for finding the local stresses gives Using the maximum stress failure theory as given by Equation 2141 yields R 1633 Thus the load that can be applied just before failure is Note that all the components of the stress vector must be multiplied by the strength ratio 283 Failure Envelopes A failure envelope is a threedimensional plot of the combinations of the normal and shear stresses that can be applied to an angle lamina just before failure Because drawing three dimensional graphs can be time consuming one may develop failure envelopes for constant shear stress τxy and then use the two normal stresses σx and σy as the two axes Then if the applied stress is within the failure envelope the lamina is safe otherwise it has failed σ σ τ x y xy MPa MPa MPa 2 3 4 σ σ τ x y xy R R R 2 3 4 σ1 1 0 1714 10 R σ2 1 0 2714 10 R τ12 1 0 4165 10 R σ σ τ x y xy MPa MPa 16 33 2 16 33 3 16 3 33 4 Mpa σ σ τ x y xy MPa MPa MPa 32 66 48 99 65 32 1343bookfm Page 144 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Macromechanical Analysis of a Lamina 145 Example 216 Develop a failure envelope for the 60 lamina of graphiteepoxy for a con stant shear stress of τxy 24 MPa Use the properties for the unidirectional graphiteepoxy lamina from Table 21 Solution From Equation 294 the stresses in the local axes for a 60 lamina are given by where σx and σy are also in units of MPa Using the preceding inequalities Various combinations of σx σy can be found to satisfy the preceding inequalities However the objective is to find the points on the failure enve lope These are combinations of σx and σy where one of the three inequalities is just violated and the other two are satisfied Some of the values of σx σy obtained on the failure envelope are given in Table 24 Several methods can be used to obtain the points on the failure envelope for a constant shear stress One way is to fix the value of σx and find the maximum value of σy that can be applied without violating any of the conditions For example for σx 100 MPa from the inequalities we have σ σ σ 1 0 2500 0 7500 20 78 x y MPa σ σ σ 2 0 7500 0 2500 20 78 x y MPa τ σ σ 12 0 4330 0 4330 12 00 x y MPa 1500 0 2500 0 7500 20 78 1500 σ σ x y 246 0 7500 0 2500 20 78 40 σ σ x y 68 0 4330 0 4330 12 00 68 σ σ x y 2061 1939 σy 1201 56 88 σy 29 33 284 80 σy 1343bookfm Page 145 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 146 Mechanics of Composite Materials Second Edition The preceding three inequalities show no allowable value of σy for this value of σx 100 MPa As another example for σx 50 MPa we have from inequalities The preceding three inequalities show two maximum allowable values of the normal stress σy These are σy 9312 MPa and σy 7933 MPa The failure envelope for τxy 24 MPa is shown in Figure 232 284 Maximum Strain Failure Theory This theory is based on the maximum normal strain theory by St Venant and the maximum shear stress theory by Tresca as applied to isotropic materials The strains applied to a lamina are resolved to strains in the local axes Failure is predicted in a lamina if any of the normal or shearing strains in the local axes of a lamina equal or exceed the corresponding ultimate strains of the unidirectional lamina Given the strainsstresses in an angle lamina one can find the strains in the local axes A lamina is considered to be failed if TABLE 24 Typical Values of σx σy on the Failure Envelope for Example 216 σx MPa σy MPa 500 500 500 500 250 250 250 250 931 793 179 135 168 104 160 154 2044 1956 σy 1051 93 12 σy 79 33 234 80 σy ε ε ε 1 1 1 C ult T ult or ε ε ε 2 2 2 C ult T ult or 1343bookfm Page 146 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Macromechanical Analysis of a Lamina 147 2143ac is violated where ultimate longitudinal tensile strain in direction 1 ultimate longitudinal compressive strain in direction 1 ultimate transverse tensile strain in direction 2 ultimate transverse compressive strain in direction 2 ultimate inplane shear strain in plane 12 The ultimate strains can be found directly from the ultimate strength parameters and the elastic moduli assuming the stressstrain response is linear until failure The maximum strain failure theory is similar to the maximum stress failure theory in that no interaction occurs between various components of strain However the two failure theories give different results because the local strains in a lamina include the Poissons ratio effect In fact if the Poissons ratio is zero in the unidirectional lamina the two failure theories will give identical results Example 217 Find the maximum value of S 0 if a stress σx 2S σy 3S and τxy 4S is applied to a 60 graphiteepoxy lamina Use maximum strain failure FIGURE 232 Failure envelopes for constant shear stress using maximum stress failure theory 0 τ 24 MPa 300 200 250 150 100 σx MPa σy MPa 50 0 50 100 200 100 100 200 300 400 γ γ γ 12 12 12 ult ult ε1 T ult ε1 C ult ε2 T ult ε2 C ult γ 12 ult 1343bookfm Page 147 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 148 Mechanics of Composite Materials Second Edition theory Use the properties of the graphiteepoxy unidirectional lamina given in Table 21 Solution In Example 26 the compliance matrix S was obtained and in Example 213 the local stresses for this problem were obtained Then from Equation 277 Assume a linear relationship between all the stresses and strains until failure then the ultimate failure strains are ε ε γ σ σ τ 1 2 12 1 2 12 S 0 5525 10 0 1547 10 0 0 1547 10 0 9 11 11 11 7709 10 0 0 0 0 1395 10 0 1714 10 10 9 1 1 1 0 2714 10 0 4165 10 S 0 1367 10 0 2662 10 0 5809 10 10 9 9 S ε σ 1 1 1 6 9 1500 10 181 10 8 287 1 T ult T ult E 0 3 ε σ 1 1 1 6 9 1500 10 181 10 8 287 1 C ult C ult E 0 3 ε σ 2 2 2 6 9 40 10 10 3 10 3 883 10 T ult T ult E 3 ε σ 2 2 2 6 9 246 10 10 3 10 2 388 1 C ult C ult E 0 2 1343bookfm Page 148 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Macromechanical Analysis of a Lamina 149 The preceding values for the ultimate strains also assume that the com pressive and tensile stiffnesses are identical Using the inequalities 2143 and recognizing that S 0 or which give The maximum value of S before failure is 1633 MPa The same maximum value of S 1633 MPa is also found using maximum stress failure theory There is no difference between the two values because the mode of failure is shear However if the mode of failure were other than shear a difference in the prediction of failure loads would have been present due to the Poissons ratio effect which couples the normal strains and stresses in the local axes Neither the maximum stress failure theory nor the maximum strain failure theory has any coupling among the five possible modes of failure The following theories are based on the interaction failure theory 285 TsaiHill Failure Theory This theory is based on the distortion energy failure theory of VonMises distortional energy yield criterion for isotropic materials as applied to aniso γ τ 12 12 12 6 6 68 10 7 17 10 9 483 1 ult ult G 0 3 8 287 10 0 1367 10 8 287 10 3 10 3 S 2 388 10 0 2662 10 3 883 10 2 9 3 S 9 483 10 0 5809 10 9 483 10 3 9 3 S 606 2 10 606 2 10 6 6 S 14 58 10 89 71 10 6 6 S 16 33 10 16 33 10 6 6 S 0 16 33 S MPa 1343bookfm Page 149 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 150 Mechanics of Composite Materials Second Edition tropic materials Distortion energy is actually a part of the total strain energy in a body The strain energy in a body consists of two parts one due to a change in volume and is called the dilation energy and the second is due to a change in shape and is called the distortion energy It is assumed that failure in the material takes place only when the distortion energy is greater than the failure distortion energy of the material Hill8 adopted the Von Mises distortional energy yield criterion to anisotropic materials Then Tsai7 adapted it to a unidirectional lamina Based on the distortion energy theory he proposed that a lamina has failed if 2144 is violated The components G1 G2 G3 G4 G5 and G6 of the strength criterion depend on the failure strengths and are found as follows 1 Apply to a unidirectional lamina then the lamina will fail Thus Equation 2144 reduces to 2145 2 Apply to a unidirectional lamina then the lamina will fail Thus Equation 2144 reduces to 2146 3 Apply to a unidirectional lamina and assuming that the normal tensile failure strength is same in directions 2 and 3 the lamina will fail Thus Equation 2144 reduces to 2147 4 Apply τ12 τ12ult to a unidirectional lamina then the lamina will fail Thus Equation 2144 reduces to 2148 From Equation 2145 to Equation 2148 G G G G G G G G 2 3 1 2 1 3 2 2 1 2 3 2 3 1 2 2 2 σ σ σ σ σ 2 1 3 σ σ 2 2 2 2 1 1 2 3 4 23 2 5 13 2 6 12 2 G G G G σ σ τ τ τ σ σ 1 1 T ult G G T ult 2 3 1 2 1 σ σ σ 2 2 T ult G G T ult 1 3 2 2 1 σ σ σ 3 2 T ult G G T ult 1 2 2 2 1 σ 2 1 6 12 2 G ult τ 1343bookfm Page 150 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Macromechanical Analysis of a Lamina 151 2149ad Because the unidirectional lamina is assumed to be under plane stress that is σ3 τ31 τ23 0 then Equation 2144 reduces through Equation 2149 to 2150 Given the global stresses in a lamina one can find the local stresses in a lamina and apply the preceding failure theory to determine whether the lamina has failed Example 218 Find the maximum value of S 0 if a stress of σx 2S σy 3S and τxy 4S is applied to a 60 graphiteepoxy lamina Use TsaiHill failure theory Use the unidirectional graphiteepoxy lamina properties given in Table 21 Solution From Example 213 G T ult T ult 1 2 2 1 2 1 2 2 1 σ σ G T ult 2 1 2 1 2 1 σ G T ult 3 1 2 1 2 1 σ G ult 6 12 2 1 2 1 τ σ σ σ σ σ σ σ 1 1 2 1 2 1 2 2 2 T ult T ult T ult ult 2 12 12 2 1 τ τ σ1 1 714 S σ 2 2 714 S 1343bookfm Page 151 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 152 Mechanics of Composite Materials Second Edition Using the TsaiHill failure theory from Equation 2150 1 Unlike the maximum strain and maximum stress failure theories the TsaiHill failure theory considers the interaction among the three unidirectional lamina strength parameters 2 The TsaiHill failure theory does not distinguish between the com pressive and tensile strengths in its equations This can result in underestimation of the maximum loads that can be applied when compared to other failure theories For the load of σx 2 MPa σy 3 MPa and τxy 4 MPa as found in Example 215 Example 217 and Example 218 the strength ratios are given by SR 1094 TsaiHill failure theory SR 1633 maximum stress failure theory SR 1633 maximum strain failure theory TsaiHill failure theory underestimates the failure stress because the trans verse tensile strength of a unidirectional lamina is generally much less than its transverse compressive strength The compressive strengths are not used in the TsaiHill failure theory but it can be modified to use corresponding tensile or compressive strengths in the failure theory as follows 2151 where X1 if σ1 0 if σ1 0 X2 if σ2 0 τ12 4 165 S 1 714 1500 10 1 714 1500 10 6 2 6 S S 2 714 1500 10 2 714 40 10 6 6 2 S S 4 165 68 10 1 6 2 S S 10 94 MPa σ σ σ σ 1 1 2 1 2 2 2 2 X X X Y 2 12 2 1 τ S σ1 T ult σ1 C ult σ1 T ult 1343bookfm Page 152 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Macromechanical Analysis of a Lamina 153 if σ2 0 Y if σ2 0 if σ2 0 S τ12ult For Example 218 the modified TsaiHill failure theory given by Equation 2151 now gives σ 1606 MPa which implies that the strength ratio is SR 1606 modified TsaiHill failure theory This value is closer to the values obtained using maximum stress and maximum strain failure theories 3 The TsaiHill failure theory is a unified theory and thus does not give the mode of failure like the maximum stress and maximum strain failure theories do However one can make a reasonable guess of the failure mode by calculating and τ12τ12ult The maximum of these three values gives the asso ciated mode of failure In the modified TsaiHill failure theory calculate the maximum of σ1X1 σ2Y and τ12S for the associ ated mode of failure 286 TsaiWu Failure Theory This failure theory is based on the total strain energy failure theory of Beltrami TsaiWu9 applied the failure theory to a lamina in plane stress A lamina is considered to be failed if H1σ1 H2σ2 H6τ12 H11 H22 H66 2H12σ1σ2 1 2152 is violated This failure theory is more general than the TsaiHill failure theory because it distinguishes between the compressive and tensile strengths of a lamina σ1 C ult σ2 T ult σ2 C ult 1 714 1500 10 1 714 1500 10 6 2 6 σ σ 2 714 1500 10 2 714 246 10 6 6 2 σ σ 4 165 68 10 1 6 2 σ σ σ 1 1 T ult σ σ 2 2 T ult σ1 2 σ2 2 τ12 2 1343bookfm Page 153 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 154 Mechanics of Composite Materials Second Edition The components H1 H2 H6 H11 H22 and H66 of the failure theory are found using the five strength parameters of a unidirectional lamina as follows 1 Apply to a unidirectional lamina the lam ina will fail Equation 2152 reduces to 2153 2 Apply to a unidirectional lamina the lamina will fail Equation 2152 reduces to 2154 From Equation 2153 and Equation 2154 2155 2156 3 Apply to a unidirectional lamina the lam ina will fail Equation 2152 reduces to 2157 4 Apply to a unidirectional lamina the lamina will fail Equation 2152 reduces to 2158 From Equation 2157 and Equation 2158 2159 2160 σ σ σ τ 1 1 2 12 0 0 T ult H H T ult T ult 1 1 11 1 2 1 σ σ σ σ σ τ 1 1 2 12 0 0 C ult H H C ult C ult 1 1 11 1 2 1 σ σ H T ult C ult 1 1 1 1 1 σ σ H T ult C ult 11 1 1 1 σ σ σ σ σ τ 1 2 2 12 0 0 T ult H H T ult T ult 2 2 22 2 2 1 σ σ σ σ σ τ 1 2 2 12 0 0 C ult H H C ult C ult 2 2 22 2 2 1 σ σ H T ult C ult 2 2 2 1 1 σ σ H T ult C ult 22 2 2 1 σ σ 1343bookfm Page 154 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Macromechanical Analysis of a Lamina 155 5 Apply σ 1 0 σ 2 0 and τ 12 τ 12 ult to a unidirectional lamina it will fail Equation 2152 reduces to 2161 6 Apply σ 1 0 σ 2 0 and τ 12 τ 12 ult to a unidirectional lamina the lamina will fail Equation 2152 reduces to 2162 From Equation 2161 and Equation 2162 2163 2164 The only component of the failure theory that cannot be found directly from the five strength parameters of the unidirectional lamina is H 12 This can be found experimentally by knowing a biaxial stress at which the lamina fails and then substituting the values of σ 1 σ 2 and τ 12 in the Equation 2152 Note that σ 1 and σ 2 need to be nonzero to find H 12 Experimental methods to find H 12 include the following 1 Apply equal tensile loads along the two material axes in a unidirec tional composite If σ x σ y σ τ xy 0 is the load at which the lamina fails then 2165 The solution of Equation 2165 gives 2166 It is not necessary to pick tensile loads in the preceding biaxial test but one may apply any combination of H H ult ult 6 12 66 12 2 1 τ τ H H ult ult 6 12 66 12 2 1 τ τ H6 0 H ult 66 12 2 1 τ H H H H H 1 2 11 22 12 2 2 1 σ σ H H H H H 12 2 1 2 11 22 2 1 2 1 σ σ σ σ σ σ σ 1 2 1343C002fm Page 155 Wednesday September 28 2005 1040 AM 2006 by Taylor Francis Group LLC 156 Mechanics of Composite Materials Second Edition 2167 This will give four different values of H 12 each corresponding to the four tests 2 Take a 45 lamina under uniaxial tension σ x The stress σ x at failure is noted If this stress is σ x σ then using Equation 294 the local stresses at failure are 2168ac Substituting the preceding local stresses in Equation 2152 2169 2170 Some empirical suggestions for finding the value of H 12 include H 12 per TsaiHill failure theory 8 2171ac H 12 per Hoffman criterion 10 H 12 per MisesHencky criterion 11 σ σ σ σ 1 2 σ σ σ σ 1 2 σ σ σ σ 1 2 σ σ σ σ τ σ 1 2 12 2 2 2 H H H H H H 1 2 2 11 22 66 12 2 4 2 1 σ σ H H H H H H 12 2 1 2 11 22 66 2 1 2 σ σ 1 2 1 2 σT ult 1 2 1 1 σ σ T ult C ult 1 2 1 1 1 2 2 σ σ σ σ T ult C ult T ult C ult 1343C002fm Page 156 Wednesday September 28 2005 1040 AM 2006 by Taylor Francis Group LLC Macromechanical Analysis of a Lamina 157 Example 219 Find the maximum value of S 0 if a stress σx 2S σy 3S and τxy 4S are applied to a 60 lamina of graphiteepoxy Use TsaiWu failure theory Use the properties of a unidirectional graphiteepoxy lamina from Table 21 Solution From Example 213 From Equations 2155 2156 2159 2160 2163 and 2164 H1 H2 H6 0 Pa1 H11 H22 H66 Using the MisesHencky criterion for evaluation of H12 Equation 2165c σ1 1 714 S σ2 2 714 S τ12 4 165 S 1 1500 10 1 1500 10 0 6 6 1 Pa 1 40 10 1 246 10 2 093 10 6 6 8 1 Pa 1 1500 10 1500 10 4 4444 10 6 6 19 2 Pa 1 40 10 246 10 1 0162 10 6 6 16 2 Pa 1 68 10 2 1626 10 6 2 16 2 Pa H12 6 6 6 1 2 1 1500 10 1500 10 40 10 246 10 6 18 2 3 360 10 Pa 1343bookfm Page 157 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 158 Mechanics of Composite Materials Second Edition Substituting these values in Equation 2152 we obtain or If one uses the other two empirical criteria for H12 per Equation 2171 this yields Summarizing the four failure theories for the same stress state the value of S obtained is S 1633 maximum stress failure theory S 1633 maximum strain failure theory S 1094 TsaiHill failure theory S 1606 modified TsaiHill failure theory S 2239 TsaiWu failure theory 287 Comparison of Experimental Results with Failure Theories Tsai7 compared the results from various failure theories to some experimen tal results He considered an angle lamina subjected to a uniaxial load in the xdirection σx as shown in Figure 233 The failure stresses were obtained experimentally for tensile and compressive stresses for various angles of the lamina 0 1 714 2 093 10 2 714 8 S S 0 4 165 4 444 10 1 714 19 2 S S 1 0162 10 2 714 2 1626 10 16 2 16 S 4 165 2 S 2 3 360 10 1 714 2 714 1 18 S S S 22 39 MPa S MPa H T ult 22 49 1 2 12 1 2 for σ S MPa for H T ult C ult 22 49 1 2 1 12 1 1 σ σ 1343bookfm Page 158 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Macromechanical Analysis of a Lamina 159 The experimental results can be compared with the four failure theories by finding the stresses in the material axes for an arbitrary stress σx for an angle lamina with an angle θ between the fiber and loading direction as 2172 per Equation 294 The corresponding strains in the material axes are 2173 per Equation 299 Using the preceding local strains and stresses in the four failure theories given by Equation 2141 Equation 2143 Equation 2150 and Equation 2152 one can find the ultimate offaxis load σx that can be applied as a function of the angle of the lamina The following values were used in the failure theories for the unidirectional lamina stiffnesses and strengths FIGURE 233 Offaxis loading in the xdirection in Figure 234 to Figure 237 σx σx 2 1 y x θ σ σ θ 1 2 x Cos σ σ θ 2 2 x Sin τ σ θ θ 12 x Sin Cos ε θ ν θ σ 1 1 2 12 2 1 E x Cos Sin ε θ ν θ σ 2 2 2 21 2 1 E x Sin Cos γ θ θ σ 12 12 1 G x Sin Cos E Msi 1 7 8 1343bookfm Page 159 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 160 Mechanics of Composite Materials Second Edition The comparison for the four failure theories is shown in Figure 234 through Figure 237 Observations from the figures are The difference between the maximum stress and maximum strain failure theories and the experimental results is quite pronounced TsaiHill and TsaiWu failure theories results are in good agreement with experimentally obtained results The variation of the strength of the angle lamina as a function of angle is smooth in the TsaiHill and TsaiWu failure theories but has cusps in the maximum stress and maximum strain failure the ories The cusps correspond to the change in failure modes in the maximum stress and maximum strain failure theories 29 Hygrothermal Stresses and Strains in a Lamina Composite materials are generally processed at high temperatures and then cooled down to room temperatures For polymeric matrix composites this temperature difference is in the range of 200 to 300C for ceramic matrix composites it may be as high as 1000C Due to mismatch of the coefficients of thermal expansion of the fiber and matrix residual stresses result in a lamina when it is cooled down Also the cooling down induces expansional E Msi 2 2 6 ν12 0 25 G Msi 12 1 3 σ1 150 T ult Ksi σ1 150 C ult Ksi σ2 4 T ult Ksi σ2 20 C ult Ksi τ12 6 ult Ksi 1343bookfm Page 160 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Macromechanical Analysis of a Lamina 161 FIGURE 234 Maximum normal tensile stress in the xdirection as a function of angle of lamina using maximum stress failure theory Experimental data reprinted with permission from Introduction to Composite Materials Tsai SW and Hahn HT 1980 CRC Press Boca Raton FL 301 FIGURE 235 Maximum normal tensile stress in the xdirection as a function of angle of lamina using maximum strain failure theory Experimental data reprinted with permission from Introduction to Composite Materials Tsai SW and Hahn HT 1980 CRC Press Boca Raton FL 301 100 Maximum stress failure theory Experimental data points 10 0 15 30 45 Angle of lamina θ degrees σx Ksi 60 75 90 100 10 0 15 30 45 Angle of lamina θ degrees Experimental data points Maximum strain failure theory σx Ksi 60 75 90 1343bookfm Page 161 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 162 Mechanics of Composite Materials Second Edition FIGURE 236 Maximum normal tensile stress in the xdirection as a function of angle of lamina using TsaiHill failure theory Experimental data reprinted with permission from Introduction to Composite Materials Tsai SW and Hahn HT 1980 CRC Press Boca Raton FL 301 FIGURE 237 Maximum normal tensile stress in the xdirection as a function of angle of lamina using TsaiWu failure theory Experimental data reprinted with permission from Introduction to Composite Materials Tsai SW and Hahn HT 1980 CRC Press Boca Raton FL 301 100 TsaiHill failure theory σx Ksi Experimental data points 10 0 15 30 Angle of lamina θ degrees 45 60 75 90 100 10 0 15 30 Angle of lamina θ degrees TsaiWu failure theory Experimental data points σx Ksi 45 60 75 90 1343bookfm Page 162 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Macromechanical Analysis of a Lamina 163 strains in the lamina In addition most polymeric matrix composites can absorb or deabsorb moisture This moisture change leads to swelling strains and stresses similar to those due to thermal expansion Laminates in which laminae are placed at different angles have residual stresses in each lamina due to differing hygrothermal expansion of each lamina The hygrothermal strains are not equal in a lamina in the longitudinal and transverse directions because the elastic constants and the thermal and moisture expansion coef ficients of the fiber and matrix are different In the following sections stressstrain relationships are developed for unidirectional and angle lami nae subjected to hygrothermal loads 291 Hygrothermal StressStrain Relationships for a Unidirectional Lamina For a unidirectional lamina the stressstrain relationship with temperature and moisture difference gives 2174 where the subscripts T and C are used to denote temperature and moisture respectively Note that the temperature and moisture change do not have any shearing strain terms because no shearing strains are induced in the material axes The thermally induced strains are given by 2175 where α1 and α2 are the longitudinal and transverse coefficients of thermal expansion respectively and ΔT is the temperature change The moisture induced strains are given by 2176 where β1 and β2 are the longitudinal and transverse coefficients of moisture respectively and ΔC is the weight of moisture absorption per unit weight of the lamina ε ε γ 1 2 12 11 12 12 22 66 0 0 0 0 S S S S S σ σ τ ε ε ε 1 2 12 1 2 1 0 T T C C ε2 0 ε ε α α 1 2 1 2 0 0 T T T Δ ε ε β β 1 2 1 2 0 0 C C C Δ 1343bookfm Page 163 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 164 Mechanics of Composite Materials Second Edition Equation 2174 can be inverted to give 2177 292 Hygrothermal StressStrain Relationships for an Angle Lamina The stressstrain relationship for an angle lamina takes the following form 2178 where 2179 and 2180 The terms αx αy and αxy are the coefficients of thermal expansion for an angle lamina and are given in terms of the coefficients of thermal expansion for a unidirectional lamina as 2181 σ σ τ 1 2 12 11 12 12 22 66 0 0 0 0 Q Q Q Q Q ε ε ε ε ε ε γ 1 1 1 2 2 2 12 T C T C ε ε γ x y xy S S S S S S S S 11 12 16 12 22 26 16 26 66 S x y xy x T y T x σ σ τ ε ε γ y T x C y C xy C ε ε γ ε ε γ α α α x T y T xy T x y xy T Δ ε ε γ β β β x C y C xy C x y xy C Δ α α α α α x y xy T 2 0 1 1 2 1343bookfm Page 164 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Macromechanical Analysis of a Lamina 165 Similarly βx βy and βxy are the coefficients of moisture expansion for an angle lamina and are given in terms of the coefficients of moisture expansion for a unidirectional lamina as 2182 From Equation 2174 if no constraints are placed on a lamina no mechan ical strains will be induced in it This also implies then that no mechanical stresses are induced However in a laminate even if the laminate has no constraints the difference in the thermalmoisture expansion coefficients of the various layers induces different thermalmoisture expansions in each layer This difference results in residual stresses and will be explained fully in Chapter 4 Example 220 Find the following for a 60 angle lamina of glassepoxy 1 Coefficients of thermal expansion 2 Coefficients of moisture expansion 3 Strains under a temperature change of 100C and a moisture absorption of 002 kgkg Use properties of unidirectional glassepoxy lamina from Table 21 Solution 1 From Table 21 Using Equation 2181 gives β β β β β x y xy T 2 0 1 1 2 α1 6 8 6 10 mm C α2 6 22 1 10 mm C α α α x y xy 2 0 2500 0 7500 0 8660 0 7500 0 2500 0 8660 0 4330 0 4330 0 5000 8 6 10 22 1 10 0 6 6 1343bookfm Page 165 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 166 Mechanics of Composite Materials Second Edition 2 From Table 21 Using Equation 2182 gives 3 Now use Equation 2179 and Equation 2180 to calculate the strains as α α α x y xy 18 73 10 11 98 10 11 6 6 69 10 6 mm C β1 0 mmkgkg β2 0 6 mmkgkg β β β x y xy 2 0 2500 0 7500 0 8660 0 7500 0 2500 0 8660 0 4330 0 4330 0 5000 0 0 0 6 0 β β β x y xy 0 4500 0 1500 0 5196 mmkgkg ε ε γ x y xy 18 73 10 11 98 10 11 6 6 69 10 100 0 4500 0 1500 0 51 6 96 0 02 0 7127 10 0 1802 10 0 9223 10 2 2 2 mm 1343bookfm Page 166 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Macromechanical Analysis of a Lamina 167 210 Summary After reviewing the definitions of stress strain elastic moduli and strain energy we developed the threedimensional stressstrain relationships for different materials These materials range from anisotropic to isotropic The number of independent constants ranges from 21 for anisotropic to 2 for isotropic materials respectively Using plane stress assumptions we reduced the threedimensional problem to a twodimensional problem and devel oped a stressstrain relationship for a unidirectionalbidirectional lamina These relationships were then found for an angle lamina using transforma tion of strains and stresses We introduced failure theories of an angle lamina in terms of strengths of unidirectional lamina Finally we developed stressstrain equations for an angle lamina under thermal and moisture loads In the appendix of this chapter we review matrix algebra and the transformation of stresses and strains Key Terms Mechanical characterization Stress Strain Elastic moduli Strain energy Anisotropic material Monoclinic material Orthotropic material Transversely isotropic material Isotropic material Plane stress Compliance matrix Stiffness matrix Angle ply Engineering constants Invariant stiffness and compliance Failure theories Maximum stress failure theory Maximum strain failure theory TsaiHill theory TsaiWu theory Failure envelopes Hygrothermal stresses Hygrothermal loads 1343bookfm Page 167 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 168 Mechanics of Composite Materials Second Edition Exercise Set 21 Write the number of independent elastic constants for threedimen sional anisotropic monoclinic orthotropic transversely isotropic and isotropic materials 22 The engineering constants for an orthotropic material are found to be Find the stiffness matrix C and the compliance matrix S for the preceding orthotropic material 23 Consider an orthotropic material with the stiffness matrix given by Find 1 The stresses in the principal directions of symmetry if the strains in the principal directions of symmetry at a point in the material are ε1 1 μmm ε2 3 μmm ε3 2 μmm γ23 0 γ31 5 μm m γ12 6 μmm 2 The compliance matrix S 3 The engineering constants E1 E2 E3 ν12 ν23 ν31 G12 G23 G31 4 The strain energy per unit volume at the point where strains are given in part 1 24 Reduce the monoclinic stressstrain relationships to those of an orthotropic material 25 Show the difference between monoclinic and orthotropic materials by applying normal stress in principal directions and shear stress in principal planes one at a time and studying the resulting nonzero and zero strains E Msi E Msi E Msi 1 3 3 12 23 4 3 3 1 0 2 0 ν ν 4 0 6 6 7 2 31 12 23 31 ν G Msi G Msi G Msi C 0 67308 1 8269 1 0577 0 0 0 1 8269 0 67308 1 4423 0 0 0 1 0577 1 4423 0 48077 0 0 0 0 0 0 4 0 0 0 0 0 0 2 0 0 0 0 0 0 1 5 GPa 1343bookfm Page 168 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Macromechanical Analysis of a Lamina 169 26 Write down the compliance matrix of a transversely isotropic mate rial where 23 is the plane of isotropy in terms of the following engineering constants E is the Youngs modulus in the plane of isotropy 23 E is the Youngs modulus in direction 1 that is perpendicular to plane of isotropy 23 ν is the Poissons ratio in the plane of isotropy 23 ν is the Poissons ratio in the 12 plane G is the shear modulus in the 12 plane 27 Find the relationship between the engineering constants of a three dimensional orthotropic material and its compliance matrix 28 What are the values of stiffness matrix elements C11 and C12 in terms of the Youngs modulus and Poissons ratio for an isotropic material 29 Are ν12 and ν21 independent of each other for a unidirectional ortho tropic lamina 210 Find the reduced stiffness Q and the compliance S matrices for a unidirectional lamina of boronepoxy Use the properties of a uni directional boronepoxy lamina from Table 21 211 Find the strains in the 12 coordinate system local axes in a uni directional boronepoxy lamina if the stresses in the 12 coordinate system applied to are σ1 4 MPa σ2 2 MPa and τ12 3 MPa Use the properties of a unidirectional boronepoxy lamina from Table 21 212 Write the reduced stiffness and the compliance matrix for an isotro pic lamina 213 Show that for an orthotropic material Q11 C11 Explain why Also show Q66 C66 Explain why 214 Consider a unidirectional continuous fiber composite Start from σ Q ε and follow the procedure in Section 243 to get E1 ν12 E2 ν21 G12 Q66 215 The reduced stiffness matrix Q is given for a unidirectional lamina is given as follows Q Q Q 11 12 2 22 Q Q 12 22 Q Q Q 22 12 2 11 12 11 Q Q 1343bookfm Page 169 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 170 Mechanics of Composite Materials Second Edition What are the four engineering constants E1 E2 ν12 and G12 of the lamina 216 The stresses in the global axes of a 30 ply are given as σx 4 MPa σy 2 MPa and τxy 3 MPa Find the stresses in the local axes Are the stresses in the local axes independent of elastic moduli Why or why not 217 The strains in the global axes of a 30 ply are given as εx 4 μin in εy 2 μinin and γxy 3 μinin Find the strains in the local axes Are the strains independent of material properties Why or why not 218 Find the transformed reduced stiffness matrix and transformed compliance matrix for a 60 angle lamina of a boronepoxy lam ina Use the properties of a unidirectional boronepoxy lamina from Table 21 219 What is the relationship between the elements of the transformed compliance matrix for a 0 and 90 lamina 220 For a 60 angle lamina of boronepoxy under stresses in global axes as σx 4 MPa σy 2 MPa and τxy 3 MPa and using the properties of a unidirectional boronepoxy lamina from Table 21 find the following 1 Global strains 2 Local stresses and strains 3 Principal normal stresses and principal normal strains 4 Maximum shear stress and maximum shear strain 221 An angle glassepoxy lamina is subjected to a shear stress τxy 04 ksi in the global axes resulting in a shear strain γxy 4683 μinin in the global axes What is the angle of the ply Use the properties of unidirectional glassepoxy lamina from Table 22 222 Find the six engineering constants for a 60 boronepoxy lamina Use the properties of unidirectional boronepoxy lamina from Table 22 223 A bidirectional woven composite ply may yield equal longitudinal and transverse Youngs modulus but is still orthotropic Determine the angles of the ply for which the shear modulus Gxy are maximum and minimum Also find these maximum and minimum values Given E1 69 GPa E2 69 GPa ν12 03 G12 20 GPa Q 5 681 0 3164 0 0 3164 1 217 0 0 0 0 6006 Msi Q S S 1343bookfm Page 170 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Macromechanical Analysis of a Lamina 171 224 A strain gage measures normal strain in a component Experiments12 suggest that errors due to strain gage misalignment are more appre ciable for angle plies of composite materials than isotropic materials 1 Take a graphiteepoxy angle ply of 8 under a uniaxial stress σx 4 Msi Estimate the strain εx as measured by a strain gage aligned in the xdirection Now if the strain gage is misaligned by 3 to the xaxis estimate the measured strain Find the per centage of error due to misalignment Use properties of unidi rectional graphiteepoxy lamina from Table 22 2 Take an aluminum layer under a uniaxial stress σx 4 Msi Estimate the strain εx as measured by a strain gage in the xdirection Now if the strain gage is misaligned by 3 to the xaxis estimate the measured strain Find the percentage of error due to misalignment Assume E 10 Msi v 03 for aluminum 225 A uniaxial load is applied to a 10 ply The linear stressstrain curve along the line of load is related as σx 123εx where the stress is measured in GPa and strain in mm Given E1 180 GPa E2 10 GPa and ν12 025 find the value of 1 shear modulus G12and 2 modulus Ex for a 60 ply 226 The tensile modulus of a 0 90 and 45 graphiteepoxy ply is measured as follows to give E1 2625 Msi E2 1494 Msi Ex 2427 Msi for the 45 ply respectively 1 What is the value Ex for a 30 ply 2 Can you calculate the values of ν12 and G12 from the previous three measured values of elastic moduli 227 Can the value of the modulus Ex of an angle lamina be less than both the longitudinal and transverse Youngs modulus of a unidi rectional lamina 228 Can the value of the modulus Ex of an angle lamina be greater than both the longitudinal and transverse Youngs modulus of a unidi rectional lamina 229 Is the νxy for a lamina maximum for a 45 boronepoxy ply Use properties of unidirectional boronepoxy lamina from Table 22 230 In finding the value of the Youngs modulus Ex for an angle ply lengthtowidth LW ratio of the specimen affects the measured value of Ex The Youngs modulus for a finite lengthtowidth ratio specimen is related to the Youngs modulus Ex for an infinite lengthtowidth ratio specimen by5 Ex 1 E E x x 1 1 ζ 1343bookfm Page 171 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 172 Mechanics of Composite Materials Second Edition where Tabulate the values of ζ for LW 2 8 16 and 64 for a 30 glass epoxy Use properties of unidirectional glassepoxy lamina from Table 22 231 Starting from the expression for the reduced stiffness element derive the expression 232 Initial stressstrain data are given for a uniaxial tensile test of a 45 angle ply Find the inplane shear modulus of the unidirectional lamina G12 Use linear regression analysis for finding slopes of curves If similar data were given for a 35 angle ply would it be sufficient to find the inplane shear modulus of the unidirectional lamina G12 233 Calculate the four stiffness invariants U1 U2 U3 and U4 and the four compliance invariants V1 V2 V3 and V4 for a boronepoxy lamina Use the properties of a unidirectional boronepoxy lamina from Table 22 234 Show that is not a function of the angle of ply 235 Find the offaxis shear strength and mode of failure of a 60 boron epoxy lamina Use the properties of a unidirectional boronepoxy lamina from Table 21 Apply the maximum stress failure maximum strain TsaiHill and TsaiWu failure theories 236 Give one advantage of the maximum stress failure theory over the TsaiWu failure theory σx KPa εx εy 210 413 644 847 1092 01 02 03 04 05 008 016 025 033 042 ζ 1 3 3 2 11 16 2 66 11 2 S S S S L W Q Q Q Q Q s c Q s c 66 11 22 12 66 2 2 66 4 4 2 2 Q U U U 66 1 4 3 1 2 4 Cos θ Q Q Q Q 11 22 12 66 1343bookfm Page 172 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Macromechanical Analysis of a Lamina 173 237 Give one advantage of the TsaiWu failure theory over the maximum stress failure theory 238 Find the maximum biaxial stress σx σ σy σ σ 0 that one can apply to a 60 lamina of graphiteepoxy Use the properties of a unidirectional graphiteepoxy lamina from Table 21 Use maxi mum strain and TsaiWu failure theories 239 Using Mohrs circle show why the maximum shear stress that can be applied to angle laminae differs with the shear stress sign Take a 45 graphiteepoxy lamina as an example Use the properties of a unidirectional graphiteepoxy lamina from Table 21 240 Reduce the TsaiWu failure theory for an isotropic material with equal ultimate tensile and compressive strengths and a shear strength that is 40 of the ultimate tensile strength 241 An offaxis test is used to find the value of H12 for use in the TsaiWu failure theory for a boronepoxy system The five lamina strengths of a unidirectional boronepoxy system are given as follows 188 ksi 361 ksi 9 ksi 45 ksi and τ12ult 10 ksi A 15 specimen fails at a uniaxial load of 33546 ksi Find the value of H12 Does it satisfy the inequality which is a stability criterion for TsaiWu failure theory that says failure surfaces inter cept all stress axes and form a closed geometric surface13 242 Give the units for the coefficient of thermal expansion in the USCS and SI systems 243 Find the freeexpansional strains of a glassepoxy unidirectional lamina under a temperature change of 100C and a moisture absorption of 0002 kgkg Also find the temperature change for which the transverse expansional strains vanish for a moisture absorption of 0002 kgkg Use the properties of a unidirectional glassepoxy lamina from Table 21 244 Find the coefficients of thermal expansion of a 60 glassepoxy lam ina Use the properties of unidirectional glassepoxy lamina from Table 22 245 Give the units for coefficient of moisture expansion in the USCS and SI systems 246 Find the coefficients of moisture expansion of a 60 glassepoxy lamina Use the properties of unidirectional glassepoxy lamina from Table 21 σ1 T ult σ1 C ult σ2 T ult σ2 C ult H H H 12 2 11 22 1343bookfm Page 173 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 174 Mechanics of Composite Materials Second Edition References 1 Timoshenko SP and Goodier JN Theory of Elasticity McGrawHill New York 1970 2 Lekhnitski SG Anisotropic Plates Gordon and Breach Science Publishers New York 1968 3 Reuter RC Jr Concise property transformation relations for an anisotropic lamina J Composite Mater 6 270 1971 4 Buchanan GR Mechanics of Materials HRW Inc New York 1988 5 Halphin JC and Pagano NJ Influence of end constraint in the testing of anisotropic bodies J Composite Mater 2 18 1968 6 Tsai SW and Pagano NJ Composite materials workshop in Progress in Materials Science Series Tsai SW Halftone JC and Pagano NJ Eds Tech nomic Stamford CT 233 1968 7 Tsai SW Strength theories of filamentary structures in Fundamental Aspects of Fiber Reinforced Plastic Composites Schwartz RT and Schwartz HS Eds Wiley Interscience New York 3 1968 8 Hill R The Mathematical Theory of Plasticity Oxford University Press London 1950 9 Tsai SW and Wu EM A general theory of strength for anisotropic materials J Composite Mater 5 58 1971 10 Hoffman O The brittle strength of orthotropic materials J Composite Mater 1 296 1967 11 Tsai SW and Hahn HT Introduction to Composite Materials Technomic Lan caster PA 1980 12 Tuttle ME and Brinson HF Resistancefoil strain gage technology as applied to composite materials Exp Mech 24 54 1985 13 Pipes RB and Cole BW On the offaxis strength test for anisotropic materials in Boron Reinforced Epoxy Systems Hilado CJ Ed Technomic Westport CT 74 1974 14 Chapra SC and Canale RC Numerical Methods for Engineers 2nd ed McGrawHill New York 1988 1343bookfm Page 174 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Macromechanical Analysis of a Lamina 175 Appendix A Matrix Algebra What is a matrix A matrix is a rectangular array of elements The elements can be symbolic expressions andor numbers Matrix A is denoted by Look at the following matrix about the sale of tires given by quarter and make of tires in a Blowoutrus store To determine how many Copper tires were sold in quarter 4 we go along the row Copper and column quarter 4 and find that it is 27 Row i of A has n elements and is and Column j of A has m elements and is Each matrix has rows and columns that define the size of the matrix If a matrix A has m rows and n columns the size of the matrix is denoted by m n The matrix A may also be denoted by Amxn to show that A is a matrix with m rows and n columns Each entry in the matrix is called the entry or element of the matrix and is denoted by aij where i is the row number i 1 2m and j is the column number j 1 2 n of the element The matrix for the tire sales example given earlier could be denoted by the matrix A as This section on matrix algebra is adapted with permission from AK Kaw Introduction to Matrix Algebra Ebook httpnumericalmethodsengusfedu 2004 At the time of printing the complete Ebook can be downloaded free of charge from the given website Quarter 1 Quarter 2 Quarter 3 Quarter 4 Tirestone 25 20 3 2 Michigan 5 10 15 25 Copper 6 16 7 27 A a a a a a a a n n m 11 12 1 21 22 2 1 a a m mn 2 a a a i i in 1 2 a a a j j mj 1 2 1343bookfm Page 175 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 176 Mechanics of Composite Materials Second Edition The size of the matrix is 3 4 because there are three rows and four columns In the preceding A matrix a34 27 What are the special types of matrices Vector A vector is a matrix that has only one row or one column The two types of vectors are row vectors and column vectors Row vector If a matrix has one row it is called a row vector B b1 b2bm and m is the dimension of the row vector Column vector If a matrix has one column it is called a column vector and n is the dimension of the column vector Example A1 Give an example of a row vector Solution B 25 20 3 2 0 is an example of a row vector of dimension 5 Example A2 Give an example of a column vector Solution An example of a column vector of dimension 3 is A 25 20 3 2 5 10 15 25 6 16 7 27 C c cn 1 C 25 5 6 1343bookfm Page 176 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Macromechanical Analysis of a Lamina 177 Submatrix If some rows orand columns of a matrix A are deleted the remaining matrix is called a submatrix of A Example A3 Find some of the submatrices of the matrix Solution Some submatrices of A are Can you find other submatrices of A Square matrix If the number of rows m of a matrix is equal to the number of columns n of the matrix m n it is called a square matrix The entries a11 a22ann are called the diagonal elements of a square matrix Sometimes the diagonal of the matrix is also called the principal or main of the matrix Example A4 Give an example of a square matrix Solution Because it has the same number of rows and columns that is three is a square matrix The diagonal elements of A are a11 25 a22 10 and a33 7 Diagonal matrix A square matrix with all nondiagonal elements equal to zero is called a diagonal matrix that is only the diagonal entries of the square matrix can be nonzero aij 0 i j A 4 6 2 3 1 2 4 6 2 3 1 2 4 6 3 1 4 6 2 4 2 2 A 25 20 3 5 10 15 6 15 7 1343bookfm Page 177 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 178 Mechanics of Composite Materials Second Edition Example A5 Give examples of a diagonal matrix Solution An example of a diagonal matrix is Any or all the diagonal entries of a diagonal matrix can be zero For example the following is also a diagonal matrix Identity matrix A diagonal matrix with all diagonal elements equal to one is called an identity matrix aij 0 i j and aii 1 for all i Example A6 Give an example of an identity matrix Solution An identity matrix is A Zero matrix A matrix whose entries are all zero is called a zero matrix aij 0 for all i and j Example A7 Give examples of a zero matrix 3 0 0 0 2 1 0 0 0 5 A 3 0 0 0 2 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 1343bookfm Page 178 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Macromechanical Analysis of a Lamina 179 Solution Examples of a zero matrix include When are two matrices considered equal Two matrices A and B are equal if The size of A and B is the same number of rows of A is same as the number of rows of B and the number of columns of A is same as number of columns of B and aij bij for all i and j Example A8 What would make equal to A 0 0 0 0 0 0 0 0 0 B 0 0 0 0 0 0 C 0 0 0 0 0 0 0 0 0 0 0 0 D 0 0 0 A 2 3 6 7 B b b 11 22 3 6 1343bookfm Page 179 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 180 Mechanics of Composite Materials Second Edition Solution The two matrices A and B would be equal if b11 2 b22 7 How are two matrices added Two matrices A and B can be added only if they are the same size number of rows of A is same as the number of rows of B and the number of columns of A is same as number of columns of B Then the addition is shown as C A B where cij aij bij for all i and j Example A9 Add the two matrices Solution How are two matrices subtracted Two matrices A and B can be subtracted only if they are the same size number of rows of A is same as the number of rows of B and the number of columns of A is same as number of columns of B The subtraction is given by D A B where dij aij bij for all i and j A 5 2 3 1 2 7 B 6 7 2 3 5 19 C A B 5 2 3 1 2 7 6 7 2 3 5 19 5 6 2 7 3 2 1 3 2 5 7 19 11 9 1 4 7 26 1343bookfm Page 180 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Macromechanical Analysis of a Lamina 181 Example A10 Subtract matrix B from matrix A that is find A B Solution How are two matrices multiplied A matrix A can be multiplied by another matrix B only if the number of columns of A is equal to the number of rows of B to give Cmxn AmxpBpxn If A is an m p matrix and B is a p n matrix then the size of the resulting matrix C is an m n matrix How does one calculate the elements of C matrix for each i 1 2m and j 1 2n A 5 2 3 1 2 7 B 6 7 2 3 5 19 C A B 5 2 3 1 2 7 6 7 2 3 5 19 5 6 2 7 3 2 1 3 2 5 7 19 1 5 5 2 3 12 c a b a b a b a ij ik kj k p i j i j i 1 1 1 2 2 p pj b 1343bookfm Page 181 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 182 Mechanics of Composite Materials Second Edition To put it in simpler terms the ith row and jth column of the C matrix in C AB is calculated by multiplying the ith row of A by the jth column of B that is Example A11 Given find Solution For example the element c12 of the C matrix can be found by multiplying the first row of A by the second column of B c a a a b b b ij i i ip j j pj 1 2 1 2 a b a b a i j i j i 1 1 2 2 p pj b a b ik kj k p 1 A 5 2 3 1 2 7 B 3 2 5 8 9 10 C A B c12 5 2 3 2 8 10 1343bookfm Page 182 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Macromechanical Analysis of a Lamina 183 52 28 310 56 Similarly one can find the other elements of C to give What is a scalar product of a constant and a matrix If A is an n n matrix and k is a real number then the scalar product of k and A is another matrix B where bij kaij Example A12 Let Find 2 A Solution then C 52 56 76 88 A 2 1 3 2 5 1 6 A 2 1 3 2 5 1 6 2 2 2 1 3 2 5 1 6 2 2 1 2 3 2 2 2 A 5 2 1 2 6 4 2 6 4 10 2 12 1343bookfm Page 183 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 184 Mechanics of Composite Materials Second Edition What is a linear combination of matrices If A1 A2Ap are matrices of the same size and k1 k2kp are scalars then is called a linear combination of A1 A2Ap Example A13 If then find Solution What are some of the rules of binary matrix operations Commutative law of addition If A and B are m n matrices then Associate law of addition If A B and C all are m n matrices then k A k A k A p p 1 1 2 2 A A A 1 2 3 5 6 2 3 2 1 2 1 3 2 5 1 6 0 2 2 2 3 3 5 6 A A A 1 2 3 2 0 5 A A A 1 2 3 2 0 5 5 6 2 3 2 1 2 2 1 3 2 5 1 6 0 5 0 2 2 2 3 3 5 6 5 6 2 3 2 1 4 2 6 44 10 2 12 0 1 1 1 1 5 1 75 3 9 2 10 9 5 1 1 5 2 25 10 A B B A 1343bookfm Page 184 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Macromechanical Analysis of a Lamina 185 Associate law of multiplication If A B and C are m n n p and p r size matrices respectively then and the resulting matrix size on both sides is m r Distributive law If A and B are m n size matrices and C and D are n p size matrices then and the resulting matrix size on both sides is m p Example A14 Illustrate the associative law of multiplication of matrices using Solution A B C A B C A B C A B C A C D A C A D A B C A C B C A B C 1 2 3 5 0 2 2 5 9 6 2 1 3 55 B C 2 5 9 6 2 1 3 5 19 27 36 39 A B C 1 2 3 5 0 2 19 27 36 39 91 105 237 276 72 78 1343bookfm Page 185 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 186 Mechanics of Composite Materials Second Edition These illustrate the associate law of multiplication of matrices Is AB BA First both operations AB and BA are only possible if A and B are square matrices of same size Why If AB exists the number of columns of A must be the same as the number of rows of B if BA exists the number of columns of B must be the same as the number of rows of A Even then in general AB BA Example A15 Illustrate whether AB BA for the following matrices Solution A B 1 2 3 5 0 2 2 5 9 6 20 17 51 45 18 12 20 17 51 45 18 12 A B C 2 1 3 5 91 105 237 276 72 78 A B 6 3 2 5 3 2 1 5 A B 6 3 2 5 3 2 1 5 15 27 1 29 1343bookfm Page 186 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Macromechanical Analysis of a Lamina 187 What is the transpose of a matrix Let A be an m n matrix Then B is the transpose of the A if bji aij for all i and j That is the ith row and the jth column element of A is the jth row and ith column element of B Note that B would be an nxm matrix The transpose of A is denoted by AT Example A16 Find the transpose of A Solution The transpose of A is Note that the transpose of a row vector is a column vector and the trans pose of a column vector is a row vector Also note that the transpose of a transpose of a matrix is the matrix that is ATT A Also A BT AT BT cAT cAT What is a symmetric matrix A square matrix A with real elements where aij aji for i 1n and j 1n is called a symmetric matrix This is same as that if A AT then A is a symmetric matrix B A 3 2 1 5 6 3 2 5 14 1 16 28 A B B A 25 20 3 2 5 10 15 25 6 16 7 27 A T 25 5 6 20 10 16 3 15 7 2 25 27 1343bookfm Page 187 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 188 Mechanics of Composite Materials Second Edition Example A17 Give an example of a symmetric matrix Solution A symmetric matrix is because a12 a21 32 a13 a31 6 and a23 a32 8 What is a skewsymmetric matrix A square matrix A with real elements where aij aji for i 1n and j 1n is called a skew symmetric matrix This is same as that if A AT then A is a skew symmetric matrix Example A18 Give an example of a skewsymmetric matrix Solution A skewsymmetric matrix is because a12 a21 1 a13 a31 2 a23 a32 5 Because aii aii only if aii 0 all the diagonal elements of a skewsymmetric matrix must be zero Matrix algebra is used for solving systems of equations Can you illustrate this concept Matrix algebra is used to solve a system of simultaneous linear equations Let us illustrate with an example of three simultaneous linear equations A 21 2 3 2 6 3 2 21 5 8 6 8 9 3 0 1 2 1 0 5 2 5 0 25 5 106 8 a b c 64 8 177 2 a b c 144 12 279 2 a b c 1343bookfm Page 188 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Macromechanical Analysis of a Lamina 189 This set of equations can be rewritten in the matrix form as The preceding equation can be written as a linear combination as follows and further using matrix multiplications gives For a general set of m linear equations and n unknowns can be rewritten in the matrix form as 25 5 64 8 144 12 106 8 1 a b c a b c a b c 77 2 279 2 a b c 25 64 144 5 8 12 1 1 1 106 8 177 2 279 2 25 5 1 64 8 1 144 12 1 106 a b c 8 177 2 279 2 a x a x a x c n n 11 1 22 2 1 1 a x a x a x c n n 21 1 22 2 2 2 a x a x a x c m m mn n m 1 1 2 2 a a a a a a a a a n n m m mn 11 12 1 21 22 2 1 2 x x x c c c n 1 2 1 2 m 1343bookfm Page 189 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 190 Mechanics of Composite Materials Second Edition Denoting the matrices by A X and C the system of equation is A X C where A is called the coefficient matrix C is called the righthand side vector and X is called the solution vector Sometimes A X C systems of equations are written in the augmented form that is Can you divide two matrices because that will help me find the solution vector for a general set of equations given by A X C If ABC is defined it might seem intuitive that A but matrix division is not defined However an inverse of a matrix can be defined for certain types of square matrices The inverse of a square matrix A if exist ing is denoted by A1 such that AA1 I A1A In other words let A be a square matrix If B is another square matrix of the same size so that BA I then B is the inverse of A A is then called invertible or nonsingular If A1 does not exist A is called noninvertible or singular Example A19 Show whether is the inverse of A C a a a c a a a c n n 11 12 1 1 21 22 2 2 a a a c m m mn n 1 2 C B B 3 2 5 3 A 3 2 5 3 1343bookfm Page 190 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Macromechanical Analysis of a Lamina 191 Solution BA I so B is the inverse of A and A is the inverse of B However we can also show that to show that A is the inverse of B Can I use the concept of the inverse of a matrix to find the solution of a set of equations AX C Yes if the number of equations is the same as the number of unknowns the coefficient matrix A is a square matrix Given AX C Then if A1 exists multiplying both sides by A1 A1 AX A1 C IX A1C X A1 C This implies that if we are able to find A1 the solution vector of AX C is simply a multiplication of A1 and the righthand side vector C How do I find the inverse of a matrix If A is an n n matrix then A1 is an n n matrix and according to the definition of inverse of a matrix AA1 I Denoting B A I 3 2 5 3 3 2 5 3 1 0 0 1 A B I 3 2 5 3 3 2 5 3 1 0 0 1 1343bookfm Page 191 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 192 Mechanics of Composite Materials Second Edition Using the definition of matrix multiplication the first column of the A1 matrix can then be found by solving Similarly one can find the other columns of the A1 matrix by changing the righthand side accordingly Example A20 Solve the set of equations A a a a a a a a a n n n n 11 12 1 21 22 2 1 2 ann A a a a a a a n n 1 11 12 1 21 22 2 a a a n n nm 1 2 I 1 0 0 0 1 0 0 1 0 1 a a a a a a a a a n n n n nn 11 12 1 21 22 2 1 2 a a an 11 21 1 1 0 0 25 5 106 8 a b c 1343bookfm Page 192 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Macromechanical Analysis of a Lamina 193 Solution In matrix form the preceding three simultaneous linear equations are writ ten as First we will find the inverse of and then use the definition of inverse to find the coefficients a b c If is the inverse of A then gives three sets of equations 64 8 177 2 a b c 144 12 279 2 a b c 25 5 1 64 8 1 144 12 1 106 a b c 8 177 2 279 2 A 25 5 1 64 8 1 144 12 1 A a a a a a a a a a 1 11 12 13 21 22 23 31 32 33 25 5 1 64 8 1 144 12 1 11 12 13 21 2 a a a a a 2 23 31 32 33 1 0 0 0 1 0 0 0 1 a a a a 25 5 1 64 8 1 144 12 1 11 21 31 a a a 1 0 0 1343bookfm Page 193 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 194 Mechanics of Composite Materials Second Edition Solving the preceding three sets of equations separately gives Therefore Now AX C where 25 5 1 64 8 1 144 12 1 12 22 32 a a a 0 1 0 25 5 1 64 8 1 144 12 1 13 23 33 a a a 0 0 1 a a a 11 21 31 0 04762 0 9524 4 571 a a a 12 22 32 0 08333 1 417 5 000 a a a 13 23 33 0 03571 0 4643 1 429 A 1 0 04762 0 08333 0 03571 0 9524 1 417 0 4643 4 571 5 000 1 429 X a b c 1343bookfm Page 194 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Macromechanical Analysis of a Lamina 195 Using the definition of A1 Computationally and algorithmically more efficient a set of simultaneous linear equations such as those given previously can also be solved by using various numerical techniques These techniques are explained completely in the source httpnumericalmethodsengusfedu of this appendix Some of the common techniques of solving a set of simultaneous linear equations are Matrix inverse method Gaussian elimination method GaussSiedel method LU decomposition method Key Terms Matrix Vector Row vector Column vector Submatrix C 106 8 177 2 279 2 A A X A C X A C 1 1 1 0 04762 0 08333 0 03571 0 9524 1 417 0 4643 4 571 5 000 1 429 106 8 177 2 279 2 a b c 0 2900 19 70 1 050 1343bookfm Page 195 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 196 Mechanics of Composite Materials Second Edition Square matrix Diagonal matrix Identity matrix Zero matrix Equal matrices Addition of matrices Subtraction of matrices Multiplication of matrices Scalar product of matrices Linear combination of matrices Rules of binary matrix operation Transpose of a matrix Symmetric matrix Skew symmetric matrix Inverse of a matrix 1343bookfm Page 196 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Macromechanical Analysis of a Lamina 197 Appendix B Transformation of Stresses and Strains Equation 2100 and Equation 294 give the relationship between stresses strains in the global xy coordinate system and the local 12 coordinate system respectively Note that the transformation is independent of material properties and depends only on the angle between the xaxis and 1axis or the angle through which the coordinate system 12 is rotated anticlockwise B1 Transformation of Stress Consider that σx σy and τxy are the stresses on the rectangular element at a point O in a twodimensional body Figure 238 One now wants to find the values of the stresses σ1 σ2 and τ12 on another rectangular element but at the same point O on the body To do so make a cut at an angle θ normal to direction 1 Now the stresses in the local 12 coordinate system can be related to those in the global xy coordinate system Summing the forces in the direction 1 gives Now and we have B1 σ τ θ σ θ τ θ σ θ 1BC AB AB AC AC xy y xy x Cos Sin Sin Cos 00 σ τ θ σ θ τ θ σ 1 xy y xy x AB BC AB BC AC BC AC B Cos Sin Sin C Cosθ Sinθ AB BC Cos θ AC BC σ τ θ θ σ θ τ θ θ σ θ 1 2 2 xy y xy x Sin Cos Sin Cos Sin Cos σ σ θ σ θ τ θ θ 1 2 2 2 x y xy Cos Sin Sin Cos 1343bookfm Page 197 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 198 Mechanics of Composite Materials Second Edition Similarly summing the forces in direction 2 gives B2 By making a cut at an angle θ normal to direction 2 B3 FIGURE 238 Free body diagrams for transformation of stresses between local and global axes A B C D E x 2 y 1 θ A C D E o σy σy τ12 τ12 τxy τxy σx σ1 σ2 σx σy σ1 σx θ τxy τxy τxy τxy τ σ θ θ σ θ θ τ θ θ 12 2 2 x y xy Sin Cos Sin Cos Cos Sin σ σ θ σ θ τ θ θ 2 2 2 2 x y xy Sin Cos Sin Cos 1343bookfm Page 198 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Macromechanical Analysis of a Lamina 199 In matrix form Equation B1 Equation B2 and Equation B3 relate the local stresses to global stresses as B4 where c Cos θ and s Sin θ The 3 3 matrix in Equation B4 is called the transformation matrix T B5 By inverting B5 B6 This relates the global stresses to local stresses as B7 B2 Transformation of Strains Consider an arbitrary line AB in direction 1 at an angle θ to the xdirection Under loads the line AB deforms to AB By definition of normal strain along AB B8 σ σ τ 1 2 12 2 2 2 2 2 2 2 2 c s sc s c sc sc sc c s σ σ τ x y xy T c s sc s c sc sc sc c s 2 2 2 2 2 2 2 2 T c s sc s c sc sc sc c s 1 2 2 2 2 2 2 2 2 σ σ τ x y xy c s sc s c sc sc sc c 2 2 2 2 2 2 2 s2 1 2 12 σ σ τ ε1 A B AB AB A B AB 1 1343bookfm Page 199 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 200 Mechanics of Composite Materials Second Edition From Figure 239 B9 B10 B11 However from definition of strain B12 B13 Then from Equation B11 through Equation B13 FIGURE 239 Line element for transformation of strains between local and global axes x y B B A A θ Δy Δy Δx Δx 1 1 ε A B AB AB x y 2 2 2 Δ Δ A B x y 2 2 2 Δ Δ Δ Δ Δ x u x x u y y 1 Δ Δ Δ y v x x v y y 1 1343bookfm Page 200 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Macromechanical Analysis of a Lamina 201 Neglecting products and squares of derivatives of strain B14 From Equation B9 Neglecting again the squares of the strains B15 A B u x x u y y v x 2 2 1 Δ Δ Δ Δ x v y y 1 2 A B u x x v y y 2 2 1 2 1 2 Δ Δ 2 2 u y v x x y Δ Δ 1 1 2 2 2 ε A B AB 1 2 1 2 2 2 2 u x x v y y u Δ Δ y v x x y x y Δ Δ Δ Δ 2 2 1 2 1 2 2 2 2 u x x x y v y Δ Δ Δ Δ Δ Δ y x y 2 2 2 2 2 2 u y v x x y x y Δ Δ Δ Δ 1 2 1 2 2 2 2 u x v y u Cos Sin θ θ y v x Sin Cos θ θ 1 1 2 1 2 2 1 2 2 2 ε ε θ ε θ γ θ x y xy Cos Sin Sin Coosθ 1 2 1 2 2 2 1 2 1 2 2 ε ε ε θ ε θ γ θ θ x y xy Cos Sin Sin Cos ε ε θ ε θ γ θ θ 1 2 2 x y xy Cos Sin Sin Cos 1343bookfm Page 201 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 202 Mechanics of Composite Materials Second Edition Similarly one can take an arbitrary line in direction 2 and prove B16 and by taking two straight lines in direction 1 and 2 perpendicular to each other one can prove B17 In matrix form Equation B15 Equation B16 and Equation B17 relate the local strains to global strains B18 where the 3 3 matrix in Equation B18 is the transformation matrix T given in Equation B5 Inverting Equation B18 gives B19 where the 3 3 matrix in Equation B19 is the inverse of the transformation matrix given in Equation B6 Key Terms Transformation of stress Transformation of strain Free body diagram Transformation matrix ε ε θ ε θ γ θ θ 2 2 2 x y xy Sin Cos Sin Cos γ ε θ θ ε θ θ γ θ 12 2 2 2 2 x y xy Sin Cos Sin Cos Cos Sin θ ε ε γ 1 2 12 2 2 2 2 2 2 2 2 c s sc s c sc sc sc c s x y xy 2 2 ε ε γ ε ε γ x y xy c s sc s c sc sc sc c 2 2 2 2 2 2 2 2 2 1 2 12 2 s ε ε γ 1343bookfm Page 202 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 203 3 Micromechanical Analysis of a Lamina Chapter Objectives Develop concepts of volume and weight fraction mass fraction of fiber and matrix density and void fraction in composites Find the nine mechanical and four hygrothermal constants four elastic moduli five strength parameters two coefficients of thermal expansion and two coefficients of moisture expansion of a unidirec tional lamina from the individual properties of the fiber and the matrix fiber volume fraction and fiber packing Discuss the experimental characterization of the nine mechanical and four hygrothermal constants 31 Introduction In Chapter 2 the stressstrain relationships engineering constants and fail ure theories for an angle lamina were developed using four elastic moduli five strength parameters two coefficients of thermal expansion CTE and two coefficients of moisture expansion CME for a unidirectional lamina These 13 parameters can be found experimentally by conducting several tension compression shear and hygrothermal tests on unidirectional lamina laminates However unlike in isotropic materials experimental evaluation of these parameters is quite costly and time consuming because they are functions of several variables the individual constituents of the composite material fiber volume fraction packing geometry processing etc Thus the need and motivation for developing analytical models to find these param eters are very important In this chapter we will develop simple relationships for the these parameters in terms of the stiffnesses strengths coefficients of thermal and moisture expansion of the individual constituents of a compos ite fiber volume fraction packing geometry etc An understanding of this 1343bookfm Page 203 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 204 Mechanics of Composite Materials Second Edition relationship called micromechanics of lamina helps the designer to select the constituents of a composite material for use in a laminated structure Because this text is for a first course in composite materials details will be explained only for the simple models based on the mechanics of materials approach and the semiempirical approach Results from other methods based on advanced topics such as elasticity are also explained for completeness As mentioned in Chapter 2 a unidirectional lamina is not homogeneous However one can assume the lamina to be homogeneous by focusing on the average response of the lamina to mechanical and hygrothermal loads Figure 31 The lamina is simply looked at as a material whose properties are different in various directions but not different from one location to another Also the chapter focuses on a unidirectional continuous fiberreinforced lamina This is because it forms the basic building block of a composite structure which is generally made of several unidirectional laminae placed at various angles The modeling in the evaluation of the parameters is dis cussed first This is followed by examples and experimental methods for finding these parameters 32 Volume and Mass Fractions Density and Void Content Before modeling the 13 parameters of a unidirectional composite we intro duce the concept of relative fraction of fibers by volume This concept is critical because theoretical formulas for finding the stiffness strength and hygrothermal properties of a unidirectional lamina are a function of fiber volume fraction Measurements of the constituents are generally based on their mass so fiber mass fractions must also be defined Moreover defining the density of a composite also becomes necessary because its value is used in the experimental determination of fiber volume and void fractions of a composite Also the value of density is used in the definition of specific modulus and specific strength in Chapter 1 321 Volume Fractions Consider a composite consisting of fiber and matrix Take the following symbol notations FIGURE 31 A nonhomogeneous lamina with fibers and matrix approximated as a homogeneous lamina Nonhomogeneous lamina Homogeneous lamina 1343bookfm Page 204 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Micromechanical Analysis of a Lamina 205 v cfm volume of composite fiber and matrix respectively ρ c fm density of composite fiber and matrix respectively Now define the fiber volume fraction V f and the matrix volume fraction V m as and 31a b Note that the sum of volume fractions is from Equation 31 as 322 Mass Fractions Consider a composite consisting of fiber and matrix and take the following symbol notation w cfm mass of composite fiber and matrix respectively The mass fraction weight fraction of the fibers W f and the matrix W m are defined as 32a b Note that the sum of mass fractions is V v v f f c V v v m m c V V f m 1 v v v f m c W w w f f c and W w w m m c W W f m 1 1343bookfm Page 205 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 206 Mechanics of Composite Materials Second Edition from Equation 32 as From the definition of the density of a single material 33ac Substituting Equation 33 in Equation 32 the mass fractions and vol ume fractions are related as 34a b in terms of the fiber and matrix volume fractions In terms of individual constituent properties the mass fractions and volume fractions are related by 35a b One should always state the basis of calculating the fiber content of a composite It is given in terms of mass or volume Based on Equation 34 it is evident that volume and mass fractions are not equal and that the mismatch between the mass and volume fractions increases as the ratio between the density of fiber and matrix differs from one f m c w w w w r v w r v w r v c c c f f f m m m and f f c f W V ρ ρ and m m c m W V ρ ρ f f m f m f m f W V V V ρ ρ ρ ρ W V V V m f m m m m 1 1 ρ ρ 1343bookfm Page 206 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Micromechanical Analysis of a Lamina 207 323 Density The derivation of the density of the composite in terms of volume fractions is found as follows The mass of composite w c is the sum of the mass of the fibers w f and the mass of the matrix w m as 36 Substituting Equation 33 in Equation 36 yields and 37 Using the definitions of fiber and matrix volume fractions from Equation 31 38 Now consider that the volume of a composite v c is the sum of the volumes of the fiber v f and matrix v m 39 The density of the composite in terms of mass fractions can be found as 310 Example 31 A glassepoxy lamina consists of a 70 fiber volume fraction Use proper ties of glass and epoxy from Table 31 and Table 32 respectively to deter mine the Table 31 and Table 32 give the typical properties of common fibers and matrices in the SI sys tem of units respectively Note that fibers such as graphite and aramids are transversely isotro pic but matrices are generally isotropic The typical properties of common fibers and matrices are again given in Table 33 and Table 34 respectively in the USCS system of units w w w c f m ρ ρ ρ c c f f m m v v v ρ ρ ρ c f f c m m c v v v v c f f m m V V ρ ρ ρ v v v c f m 1 W W c f f m m ρ ρ ρ 1343bookfm Page 207 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 208 Mechanics of Composite Materials Second Edition 1 Density of lamina 2 Mass fractions of the glass and epoxy 3 Volume of composite lamina if the mass of the lamina is 4 kg 4 Volume and mass of glass and epoxy in part 3 Solution 1 From Table 31 the density of the fiber is TABLE 31 Typical Properties of Fibers SI System of Units Property Units Graphite Glass Aramid Axial modulus Transverse modulus Axial Poissons ratio Transverse Poissons ratio Axial shear modulus Axial coefficient of thermal expansion Transverse coefficient of thermal expansion Axial tensile strength Axial compressive strength Transverse tensile strength Transverse compressive strength Shear strength Specific gravity GPa GPa GPa μ mm C μ mm C MPa MPa MPa MPa MPa 230 22 030 035 22 13 70 2067 1999 77 42 36 18 85 85 020 020 3542 5 5 1550 1550 1550 1550 35 25 124 8 036 037 3 50 41 1379 276 7 7 21 14 TABLE 32 Typical Properties of Matrices SI System of Units Property Units Epoxy Aluminum Polyamide Axial modulus Transverse modulus Axial Poissons ratio Transverse Poissons ratio Axial shear modulus Coefficient of thermal expansion Coefficient of moisture expansion Axial tensile strength Axial compressive strength Transverse tensile strength Transverse compressive strength Shear strength Specific gravity GPa GPa GPa μ mm C mmkgkg MPa MPa MPa MPa MPa 34 34 030 030 1308 63 033 72 102 72 102 34 12 71 71 030 030 27 23 000 276 276 276 276 138 27 35 35 035 035 13 90 033 54 108 54 108 54 12 f 3 2500 kg m ρ 1343bookfm Page 208 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Micromechanical Analysis of a Lamina 209 From Table 32 the density of the matrix is Using Equation 38 the density of the composite is 2 Using Equation 34 the fiber and matrix mass fractions are TABLE 33 Typical Properties of Fibers USCS System of Units Property Units Graphite Glass Aramid Axial modulus Transverse modulus Axial Poissons ratio Transverse Poissons ratio Axial shear modulus Axial coefficient of thermal expansion Transverse coefficient of thermal expansion Axial tensile strength Axial compressive strength Transverse tensile strength Transverse compressive strength Shear strength Specific gravity Msi Msi Msi μ inin F μ inin F ksi ksi ksi ksi ksi 3335 319 030 035 319 07222 3889 2997 2898 1116 609 522 18 1233 1233 020 020 5136 2778 2778 2248 2248 2248 2248 508 25 1798 116 036 037 0435 2778 2278 2000 4002 1015 1015 3045 14 TABLE 34 Typical Properties of Matrices USCS System of Units Property Units Epoxy Aluminum Polyamide Axial modulus Transverse modulus Axial Poissons ratio Transverse Poissons ratio Axial shear modulus Coefficient of thermal expansion Coefficient of moisture expansion Axial tensile strength Axial compressive strength Transverse tensile strength Transverse compressive strength Shear strength Specific gravity Msi Msi Msi μ inin F ininlblb ksi ksi ksi ksi ksi 0493 0493 030 030 01897 35 033 1044 1479 1044 1479 493 12 1030 1030 030 030 3915 1278 000 4002 4002 4002 4002 2001 27 05075 05075 035 035 01885 50 033 783 1566 783 1566 783 12 ρm kg m 1200 3 ρc kg m 2500 0 7 1200 0 3 2110 3 1343bookfm Page 209 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 210 Mechanics of Composite Materials Second Edition Note that the sum of the mass fractions 3 The volume of composite is 4 The volume of the fiber is The volume of the matrix is Wf 2500 2110 0 3 0 8294 Wm 1200 2110 0 3 0 1706 W W f m 0 8294 0 1706 1 000 v w c c c ρ 4 2110 1 896 10 3 3 m v V v f f c 0 7 1 896 10 3 1 327 10 3 3 m v V v m m c 0301896 10 3 1343bookfm Page 210 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Micromechanical Analysis of a Lamina 211 The mass of the fiber is The mass of the matrix is 06826 kg 324 Void Content During the manufacture of a composite voids are introduced in the com posite as shown in Figure 32 This causes the theoretical density of the composite to be higher than the actual density Also the void content of a FIGURE 32 Photomicrographs of crosssection of a lamina with voids 0 5688 10 3 3 m w v f f f ρ 2500 1 327 10 3 3 318 kg w v m m m ρ 1200 0 5688 10 3 1343bookfm Page 211 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 212 Mechanics of Composite Materials Second Edition composite is detrimental to its mechanical properties These detriments include lower Shear stiffness and strength Compressive strengths Transverse tensile strengths Fatigue resistance Moisture resistance A decrease of 2 to 10 in the preceding matrixdominated properties gen erally takes place with every 1 increase in the void content 1 For composites with a certain volume of voids V v the volume fraction of voids V v is defined as 311 Then the total volume of a composite v c with voids is given by 312 By definition of the experimental density ρ ce of a composite the actual volume of the composite is 313 and by the definition of the theoretical density ρ ct of the composite the theoretical volume of the composite is 314 Then substituting the preceding expressions 313 and 314 in Equation 312 The volume of void is given by V v v v v c v v v v c f m v v w c c ce ρ v v w f m c ct ρ w w v c ce c ct v ρ ρ 1343bookfm Page 212 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Micromechanical Analysis of a Lamina 213 315 Substituting Equation 313 and Equation 315 in Equation 311 the volume fraction of the voids is 316 Example 32 A graphiteepoxy cuboid specimen with voids has dimensions of a b c and its mass is Mc After it is put it into a mixture of sulfuric acid and hydrogen peroxide the remaining graphite fibers have a mass Mf From independent tests the densities of graphite and epoxy are ρf and ρm respectively Find the volume fraction of the voids in terms of a b c Mf Mc ρf and ρm Solution The total volume of the composite vc is the sum total of the volume of fiber vf matrix vm and voids vv 317 From the definition of density 318a 318b The specimen is a cuboid so the volume of the composite is 319 Substituting Equation 318 and Equation 319 in Equation 317 gives v w v c ce ct ce ct ρ ρ ρ ρ V v v v v c ct ce ct ρ ρ ρ v v v v c f m v v M f f ρf v M M m c f m ρ v c abc 1343bookfm Page 213 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 214 Mechanics of Composite Materials Second Edition and the volume fraction of voids then is 320 Alternative Solution The preceding problem can also be solved by using Equation 316 The theoretical density of the composite is 321 where Vf is the theoretical fiber volume fraction given as 322 The experimental density of the composite is 323 Substituting Equation 321 through Equation 323 in the definition of void volume fractions given by Equation 316 324 Experimental determination the fiber volume fractions of the constituents of a composite are found generally by the burn or the acid digestion tests These tests involve taking a sample of composite and weighing it Then the density abc M M M v f f c f m v ρ ρ V v abc abc M M M v v f f c f m 1 1 ρ ρ ρ ρ ρ ct f f m f V V 1 V volume of fibers volume of fibers volume f of matrix V M M M M f f f f f c f m ρ ρ ρ ρce Mc abc V abc M M M v f f c f m 1 1 ρ ρ 1343bookfm Page 214 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Micromechanical Analysis of a Lamina 215 of the specimen is found by the liquid displacement method in which the sample is weighed in air and then in water The density of the composite is given by 325 where wc weight of composite wi weight of composite when immersed in water ρw density of water 1000 kgm3 or 624 lbft3 For specimens that float in water a sinker is attached The density of the composite is then found by 326 where wc weight of composite ws weight of sinker when immersed in water ww weight of sinker and specimen when immersed in water The sample is then dissolved in an acid solution or burned2 Glassbased composites are burned and carbon and aramidbased composites are digested in solutions Carbon and aramidbased composites cannot be burned because carbon oxidizes in air above 300C 572F and the aramid fiber can decompose at high temperatures Epoxybased composites can be digested by nitric acid or a hot mixture of ethylene glycol and potassium hydroxide polyamide and phenolic resinbased composites use mixtures of sulfuric acid and hydrogen peroxide When digestion or burning is complete the remaining fibers are washed and dried several times and then weighed The fiber and matrix weight fractions can be found using Equation 32 The densities of the fiber and the matrix are known thus one can use Equation 34 to determine the volume fraction of the constituents of the composite and Equation 38 to calculate the theoretical density of the composite 33 Evaluation of the Four Elastic Moduli As shown in Section 243 there are four elastic moduli of a unidirectional lamina ρ ρ c c c i w w w w ρ ρ c c c s w w w w w w 1343bookfm Page 215 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 216 Mechanics of Composite Materials Second Edition Longitudinal Youngs modulus E1 Transverse Youngs modulus E2 Major Poissons ratio ν12 Inplane shear modulus G12 Three approaches for determining the four elastic moduli are discussed next 331 Strength of Materials Approach From a unidirectional lamina take a representative volume element that consists of the fiber surrounded by the matrix Figure 33 This representa tive volume element RVE can be further represented as rectangular blocks The fiber matrix and the composite are assumed to be of the same width h but of thicknesses tf tm and tc respectively The area of the fiber is given by 327a The area of the matrix is given by 327b and the area of the composite is given by 327c The two areas are chosen in the proportion of their volume fractions so that the fiber volume fraction is defined as 328a and the matrix fiber volume fraction Vm is A representative volume element RVE of a material is the smallest part of the material that represents the material as a whole It could be otherwise intractable to account for the distribu tion of the constituents of the material A t h f f A t h m m A t h c c V A A t t f f c f c 1343bookfm Page 216 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Micromechanical Analysis of a Lamina 217 328b The following assumptions are made in the strength of materials approach model The bond between fibers and matrix is perfect The elastic moduli diameters and space between fibers are uniform The fibers are continuous and parallel FIGURE 33 Representative volume element of a unidirectional lamina 3 Lc tm2 tr tm2 tc tc h h 2 1 V A A t t V m m c m c f 1 1343bookfm Page 217 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 218 Mechanics of Composite Materials Second Edition The fibers and matrix follow Hookes law linearly elastic The fibers possess uniform strength The composite is free of voids 3311 Longitudinal Youngs Modulus From Figure 34 under a uniaxial load Fc on the composite RVE the load is shared by the fiber Ff and the matrix Fm so that 329 The loads taken by the fiber the matrix and the composite can be written in terms of the stresses in these components and crosssectional areas of these components as 330a 330b 330c where σcfm stress in composite fiber and matrix respectively Acfm area of composite fiber and matrix respectively Assuming that the fibers matrix and composite follow Hookes law and that the fibers and the matrix are isotropic the stressstrain relationship for each component and the composite is FIGURE 34 A longitudinal stress applied to the representative volume element to calculate the longitudinal Youngs modulus for a unidirectional lamina tm2 tm2 tf tc σc σc h F F F c f m F A c c c σ F A f f f σ F A m m m σ 1343bookfm Page 218 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Micromechanical Analysis of a Lamina 219 331a 331b and 331c where εcfm strains in composite fiber and matrix respectively E1fm elastic moduli of composite fiber and matrix respectively Substituting Equation 330 and Equation 331 in Equation 329 yields 332 The strains in the composite fiber and matrix are equal εc εf εm then from Equation 332 333 Using Equation 328 for definitions of volume fractions 334 Equation 334 gives the longitudinal Youngs modulus as a weighted mean of the fiber and matrix modulus It is also called the rule of mixtures The ratio of the load taken by the fibers Ff to the load taken by the composite Fc is a measure of the load shared by the fibers From Equation 330 and Equation 331 335 In Figure 35 the ratio of the load carried by the fibers to the load taken by the composite is plotted as a function of fibertomatrix Youngs moduli ratio EfEm for the constant fiber volume fraction Vf It shows that as the fiber to matrix moduli ratio increases the load taken by the fiber increases tre mendously σ ε c E c 1 σ ε f f f E σ ε m m m E E A E A E A c c f f f m m m 1ε ε ε E E A A E A A f f c m m c 1 E E V E V f f m m 1 F F E E V f c f f 1 1343bookfm Page 219 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 220 Mechanics of Composite Materials Second Edition Example 33 Find the longitudinal elastic modulus of a unidirectional glassepoxy lamina with a 70 fiber volume fraction Use the properties of glass and epoxy from Table 31 and Table 32 respectively Also find the ratio of the load taken by the fibers to that of the composite Solution From Table 31 the Youngs modulus of the fiber is Ef 85 GPa From Table 32 the Youngs modulus of the matrix is Em 34 GPa Using Equation 334 the longitudinal elastic modulus of the unidirectional lamina is Using Equation 335 the ratio of the load taken by the fibers to that of the composite is FIGURE 35 Fraction of load of composite carried by fibers as a function of fiber volume fraction for constant fiber to matrix moduli ratio 1 08 06 04 020 20 40 60 80 100 Fiber to matrix moduli ratio EfEm Vf 02 Vf 04 Vf 06 Vf 08 Fiber to composite load ratio FfFc E GPa 1 85 0 7 3 4 0 3 60 52 1343bookfm Page 220 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Micromechanical Analysis of a Lamina 221 Figure 36 shows the linear relationship between the longitudinal Youngs modulus of a unidirectional lamina and fiber volume fraction for a typical graphiteepoxy composite per Equation 334 It also shows that Equation 334 predicts results that are close to the experimental data points3 3312 Transverse Youngs Modulus Assume now that as shown in Figure 37 the composite is stressed in the transverse direction The fibers and matrix are again represented by rectan gular blocks as shown The fiber the matrix and composite stresses are equal Thus 336 where σcfm stress in composite fiber and matrix respectively Now the transverse extension in the composite Δc is the sum of the trans verse extension in the fiber Δf and that is the matrix Δm FIGURE 36 Longitudinal Youngs modulus as function of fiber volume fraction and comparison with experimental data points for a typical glasspolyester lamina Experimental data points repro duced with permission of ASM International 70 60 50 40 30 20 10 0 03 04 Fiber volume fraction Vf Experimental data points Longitudinal Youngs modulus E1 GPa 05 06 F F f c 85 60 52 0 7 0 9831 σ σ σ c f m 1343bookfm Page 221 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 222 Mechanics of Composite Materials Second Edition 337 Now by the definition of normal strain 338a 338b and 338c where tcfm thickness of the composite fiber and matrix respectively εcfm normal transverse strain in the composite fiber and matrix respectively Also by using Hookes law for the fiber matrix and composite the normal strains in the composite fiber and matrix are 339a 339b and 339c FIGURE 37 A transverse stress applied to a representative volume element used to calculate transverse Youngs modulus of a unidirectional lamina tm2 tf tc σc σc h tm2 c f m Δ Δ Δ Δc c t c ε Δ f f f t ε Δm m m t ε ε σ c c E 2 ε σ f f Ef ε σ m m Em 1343bookfm Page 222 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Micromechanical Analysis of a Lamina 223 Substituting Equation 338 and Equation 339 in Equation 337 and using Equation 336 gives 340 Because the thickness fractions are the same as the volume fractions as the other two dimensions are equal for the fiber and the matrix see Equa tion 328 341 Equation 341 is based on the weighted mean of the compliance of the fiber and the matrix Example 34 Find the transverse Youngs modulus of a glassepoxy lamina with a fiber volume fraction of 70 Use the properties of glass and epoxy from Table 31 and Table 32 respectively Solution From Table 31 the Youngs modulus of the fiber is Ef 85 GPa From Table 32 the Youngs modulus of the matrix is Em 34 GPa Using Equation 341 the transverse Youngs modulus E2 is Figure 38 plots the transverse Youngs modulus as a function of fiber volume fraction for constant fibertomatrix elastic moduli ratio EfEm For metal and ceramic matrix composites the fiber and matrix elastic moduli are of the same order For example for a SiCaluminum metal matrix composite EfEm 4 and for a SiCCAS ceramic matrix composite EfEm 2 The transverse Youngs modulus of the composite in such cases changes more smoothly as a function of the fiber volume fraction 1 1 1 E2 E t t E t t f f c m m c 1 E2 V E V E f f m m 1 0 7 85 0 3 3 4 10 37 2 2 E E GPa 1343bookfm Page 223 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 224 Mechanics of Composite Materials Second Edition For polymeric composites the fibertomatrix moduli ratio is very high For example for a glassepoxy polymer matrix composite EfEm 25 The transverse Youngs modulus of the composite in such cases changes appre ciably only for large fiber volume fractions Figure 38 shows that for high EfEm ratios the contribution of the fiber modulus only increases substantially for a fiber volume fraction greater than 80 These fiber volume fractions are not practical and in many cases are physically impossible due to the geometry of fiber packing Figure 39 shows various possibilities of fiber packing Note that the ratio of the diameter d to fiber spacing s ds varies with geometrical packing For circular fibers with square array packing Figure 39a 342a This gives a maximum fiber volume fraction of 7854 as s d For circular fibers with hexagonal array packing Figure 39b 342b FIGURE 38 Transverse Youngs modulus as a function of fiber volume fraction for constant fiber to matrix moduli ratio 40 EfEm1 EfEm5 EfEm 25 EfEm 125 30 20 10 0 0 02 04 06 Fiber volume fraction Vf Transverse Youngs modulus ratio E2Em 08 1 d s Vf 4 1 2 π d s Vf 2 3 1 2 π 1343bookfm Page 224 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Micromechanical Analysis of a Lamina 225 This gives a maximum fiber volume fraction of 9069 because s d These maximum fiber volume fractions are not practical to use because the fibers touch each other and thus have surfaces where the matrix cannot wet out the fibers In Figure 310 the transverse Youngs modulus is plotted as a function of fiber volume fraction using Equation 341 for a typical boronepoxy lamina Also given are the experimental data points4 In Figure 310 the experimental and analytical results are not as close to each other as they are for the longitudinal Youngs modulus in Figure 36 FIGURE 39 Fiber to fiber spacing in a square packing geometry and b hexagonal packing geometry s s d d b a 1343bookfm Page 225 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 226 Mechanics of Composite Materials Second Edition FIGURE 310 Theoretical values of transverse Youngs modulus as a function of fiber volume fraction for a BoronEpoxy unidirectional lamina Ef 414 GPa νf 02 Em 414 GPa νm 035 and comparison with experimental values Figure b zooms figure a for fiber volume fraction between 045 and 075 Experimental data from Hashin Z NASA tech rep contract no NAS1 8818 November 1970 40 30 20 10 02 04 06 Fiber volume fraction Vf Transverse Youngs modulus E2 GPa Experimental data points Mechanics of materials approach 08 1 0 0 40 30 20 10 0 045 055 Experimental data points Mechanics of materials approach Fiber volume fraction Vf Transverse Youngs modulus E2 GPa 065 075 1343bookfm Page 226 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Micromechanical Analysis of a Lamina 227 3313 Major Poissons Ratio The major Poissons ratio is defined as the negative of the ratio of the normal strain in the transverse direction to the normal strain in the longitudinal direction when a normal load is applied in the longitudinal direction Assume a composite is loaded in the direction parallel to the fibers as shown in Figure 311 The fibers and matrix are again represented by rectangular blocks The deformations in the transverse direction of the composite is the sum of the transverse deformations of the fiber and the matrix as 343 Using the definition of normal strains 344a 344b FIGURE 311 A longitudinal stress applied to a representative volume element to calculate Poissons ratio of unidirectional lamina σ1 σ1 h a b tr tc tm2 tm2 tcδc T tfδf T tf Lc tc tm2 tm2 δc T δ f T δm T δ δ δ c T f T m T ε δ f T f T ft ε δ m T m T tm 1343bookfm Page 227 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 228 Mechanics of Composite Materials Second Edition and 344c where εcfm transverse strains in composite fiber and matrix respectively Substituting Equation 344 in Equation 343 345 The Poissons ratios for the fiber matrix and composite respectively are 346a 346b and 346c Substituting in Equation 345 347 where v12fm Poissons ratio of composite fiber and matrix respectively longitudinal strains of composite fiber and matrix respec tively However the strains in the composite fiber and matrix are assumed to be the equal in the longitudinal direction which from Equation 347 gives 348 ε δ c T c T ct t t t c c T f f T m m T ε ε ε ν ε ε f f T f L ν ε ε m m T m L ν ε ε 12 c T c L t t t c c L f f f L m m m L ν ε ν ε ν ε 12 ν12 f m εc f m L ε ε ε c L f L m L t t t c f f m m ν ν ν 12 ν ν ν 12 f f c m m c t t t t 1343bookfm Page 228 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Micromechanical Analysis of a Lamina 229 Because the thickness fractions are the same as the volume fractions per Equation 328 349 Example 35 Find the major and minor Poissons ratio of a glassepoxy lamina with a 70 fiber volume fraction Use the properties of glass and epoxy from Table 31 and Table 32 respectively Solution From Table 31 the Poissons ratio of the fiber is νf 02 From Table 32 the Poissons ratio of the matrix is νm 03 Using Equation 349 the major Poissons ratio is From Example 33 the longitudinal Youngs modulus is E1 6052 GPa and from Example 34 the transverse Youngs modulus is E2 1037 GPa Then the minor Poissons ratio from Equation 283 is 3314 InPlane Shear Modulus Apply a pure shear stress τc to a lamina as shown in Figure 312 The fibers and matrix are represented by rectangular blocks as shown The resulting ν ν ν 12 f f m m V V ν12 0 2 0 7 0 3 0 3 0 230 ν ν 21 12 2 1 0 230 10 37 60 52 0 03941 E E 1343bookfm Page 229 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 230 Mechanics of Composite Materials Second Edition shear deformations of the composite δc the fiber δf and the matrix δm are related by 350 From the definition of shear strains 351a 351b and 351c where γcfm shearing strains in the composite fiber and matrix respec tively tcfm thickness of the composite fiber and matrix respectively From Hookes law for the fiber the matrix and the composite 352a 352b and FIGURE 312 An inplane shear stress applied to a representative volume element for finding inplane shear modulus of a unidirectional lamina tm2 tm2 tr tc h τc τc c f m δ δ δ δ γ c c ct δ γ f f ft δ γ m m m t γ τ c c G 12 γ τ f f Gf 1343bookfm Page 230 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Micromechanical Analysis of a Lamina 231 352c where G12fm shear moduli of composite fiber and matrix respectively From Equation 350 through Equation 352 353 The shear stresses in the fiber matrix and composite are assumed to be equal τc τf τm giving 354 Because the thickness fractions are equal to the volume fractions per Equation 328 355 Example 36 Find the inplane shear modulus of a glassepoxy lamina with a 70 fiber volume fraction Use properties of glass and epoxy from Table 31 and Table 32 respectively Solution The glass fibers and the epoxy matrix have isotropic properties From Table 31 the Youngs modulus of the fiber is Ef 85 GPa and the Poissons ratio of the fiber is νf 02 The shear modulus of the fiber γ τ m m Gm τ τ τ c c f f f m m m G t G t G t 12 1 1 1 G12 G t t G t t f f c m m c 1 G12 V G V G f f m m G E GPa f f f 2 1 85 2 1 0 2 35 42 ν 1343bookfm Page 231 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 232 Mechanics of Composite Materials Second Edition From Table 32 the Youngs modulus of the matrix is Em 34 GPa and the Poissons ratio of the fiber is νm 03 The shear modulus of the matrix is From Equation 355 the inplane shear modulus of the unidirectional lamina is Figure 313a and Figure 313b show the analytical values from Equation 355 of the inplane shear modulus as a function of fiber volume fraction for a typical glassepoxy lamina Experimental values4 are also plotted in the same figure 332 SemiEmpirical Models The values obtained for transverse Youngs modulus and inplane shear modulus through Equation 341 and Equation 355 respectively do not agree well with the experimental results shown in Figure 310 and Figure 313 This establishes a need for better modeling techniques These tech niques include numerical methods such as finite element and finite differ ence and boundary element methods elasticity solution and variational principal models5 Unfortunately these models are available only as compli cated equations or in graphical form Due to these difficulties semiempirical models have been developed for design purposes The most useful of these models include those of Halphin and Tsai6 because they can be used over a wide range of elastic properties and fiber volume fractions Halphin and Tsai6 developed their models as simple equations by curve fitting to results that are based on elasticity The equations are semiempirical in nature because involved parameters in the curve fitting carry physical meaning G E GPa m m m 2 1 3 40 2 1 0 3 1 308 ν 1 0 70 35 42 0 30 1 308 4 014 12 12 G G GPa 1343bookfm Page 232 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Micromechanical Analysis of a Lamina 233 FIGURE 313 Theoretical values of inplane shear modulus as a function of fiber volume fraction and com parison with experimental values for a unidirectional glassepoxy lamina Gf 3019 GPa Gm 183 GPa Figure b zooms figure a for fiber volume fraction between 045 and 075 Experimental data from Hashin Z NASA tech rep contract No NAS18818 November 1970 30 20 10 0 0 02 04 06 Fiber volume fraction Vf Inplane shear modulus GPa Experimental data points Mechanics of materials approach 08 1 30 20 10 0 045 055 Mechanics of materials approach Experimental data points Fiber volume fraction Vf Inplane shear modulus GPa 065 075 1343bookfm Page 233 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 234 Mechanics of Composite Materials Second Edition 3321 Longitudinal Youngs Modulus The HalphinTsai equation for the longitudinal Youngs modulus E1 is the same as that obtained through the strength of materials approach that is 356 3322 Transverse Youngs Modulus The transverse Youngs modulus E2 is given by6 357 where 358 The term ξ is called the reinforcing factor and depends on the following Fiber geometry Packing geometry Loading conditions Halphin and Tsai6 obtained the value of the reinforcing factor ξ by com paring Equation 357 and Equation 358 to the solutions obtained from the elasticity solutions For example for a fiber geometry of circular fibers in a packing geometry of a square array ξ 2 For a rectangular fiber cross section of length a and width b in a hexagonal array ξ 2ab where b is in the direction of loading6 The concept of direction of loading is illustrated in Figure 314 Example 37 Find the transverse Youngs modulus for a glassepoxy lamina with a 70 fiber volume fraction Use the properties for glass and epoxy from Table 31 and Table 32 respectively Use HalphinTsai equations for a circular fiber in a square array packing geometry Solution Because the fibers are circular and packed in a square array the reinforcing factor ξ 2 From Table 31 the Youngs modulus of the fiber is Ef 85 GPa E E V E V f f m m 1 E E V V m f f 2 1 1 ξη η η ξ E E E E f m f m 1 1343bookfm Page 234 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Micromechanical Analysis of a Lamina 235 From Table 32 the Youngs modulus of the matrix is Em 34 GPa From Equation 358 From Equation 357 the transverse Youngs modulus of the unidirectional lamina is For the same problem from Example 34 this value of E2 was found to be 1037 GPa by the mechanics of materials approach Figure 315a and Figure 315b show the transverse Youngs modulus as a function of fiber volume fraction for a typical boronepoxy composite The HalphinTsai equations 357 and the mechanics of materials approach Equation 341 curves are shown and compared to experimental data points As mentioned previously the parameters ξ and η have a physical meaning For example FIGURE 314 Concept of direction of loading for calculation of transverse Youngs modulus by HalphinTsai equations σ2 σ2 a b η 85 3 4 1 85 3 4 2 0 8889 E E 2 2 3 4 1 2 0 8889 0 7 1 0 8889 0 7 20 20 GPa 1343bookfm Page 235 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 236 Mechanics of Composite Materials Second Edition EfEm 1 implies η 0 homogeneous medium EfEm implies η 1 rigid inclusions EfEm 0 implies voids 3323 Major Poissons Ratio The HalphinTsai equation for the major Poissons ratio ν12 is the same as that obtained using the strength of materials approach that is FIGURE 315 Theoretical values of transverse Youngs modulus as a function of fiber volume fraction and comparison with experimental values for boronepoxy unidirectional lamina Ef 414 GPa νf 02 Em 414 GPa νm 035 Figure b zooms figure a for fiber volume fraction between 045 and 075 Experimental data from Hashin Z NASA tech rep contract no NAS18818 November 1970 0 5 10 15 20 25 30 35 40 0 02 04 06 08 1 Transverse Youngs modulus E2 GPa Fiber volume fraction Vf HaphinTsai equation Experimental data points Mechanics of materials approach 0 10 20 30 40 045 055 065 075 Fiber volume fraction Vf Transverse Youngs modulus E2 GPa HaphinTsai equation Experimental data Points Mechanics of materials approach η ξ 1 1343bookfm Page 236 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Micromechanical Analysis of a Lamina 237 359 3324 InPlane Shear Modulus The HalphinTsai6 equation for the inplane shear modulus G12 is 360 where 361 The value of the reinforcing factor ξ depends on fiber geometry packing geometry and loading conditions For example for circular fibers in a square array ξ 1 For a rectangular fiber crosssectional area of length a and width b in a hexagonal array where a is the direction of loading The concept of the direction of loading7 is given in Figure 316 The value of ξ 1 for circular fibers in a square array gives reasonable results only for fiber volume fractions of up to 05 For example for a typical glassepoxy lamina with a fiber volume fraction of 075 the value of in plane shear modulus using the HalphinTsai equation with ξ 1 is 30 lower than that given by elasticity solutions Hewitt and Malherbe8 sug gested choosing a function 362 FIGURE 316 Concept of direction of loading to calculate inplane shear modulus by HalphinTsai equations τ12 τ12 a b ν ν ν 12 f f m m V V G G V V m f f 12 1 1 ξη η η ξ G G G G f m f m 1 ξ 3 log e a b ξ 1 40 10 Vf 1343bookfm Page 237 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 238 Mechanics of Composite Materials Second Edition Example 38 Using HalphinTsai equations find the shear modulus of a glassepoxy composite with a 70 fiber volume fraction Use the properties of glass and epoxy from Table 31 and Table 32 respectively Assume that the fibers are circular and are packed in a square array Also get the value of the shear modulus by using Hewitt and Malherbes8 formula for the reinforc ing factor Solution For HalphinTsais equations with circular fibers in a square array the rein forcing factor ξ 1 From Example 36 the shear modulus of the fiber is Gf 3542 GPa and the shear modulus of the matrix is Gm 1308 GPa From Equation 361 From Equation 360 the inplane shear modulus is For the same problem the value of G12 4013 GPa was found by the mechanics of materials approach in Example 35 Because the volume fraction is greater than 50 Hewitt and Mahelbre8 suggested a reinforcing factor Equation 362 Then from Equation 361 η 35 42 1 308 1 35 42 1 308 1 0 9288 G G 12 1 308 1 1 0 9288 0 7 1 0 9288 0 7 12 6 169 GPa ξ 1 40 1 40 0 7 2 130 10 10 Vf 1343bookfm Page 238 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Micromechanical Analysis of a Lamina 239 From Equation 360 the inplane shear modulus is Figure 317a and Figure 317b show the inplane shear modulus as a func tion of fiber volume fraction for a typical glassepoxy composite The Hal phinTsai equation 360 and the mechanics of materials approach Equation 355 are shown and compared to the experimental4 data points 333 Elasticity Approach In addition to the strength of materials and semiempirical equation approaches expressions for the elastic moduli based on elasticity are also available Elasticity accounts for equilibrium of forces compatibility and Hookes law relationships in three dimensions the strength of materials approach may not satisfy compatibility andor account for Hookes law in three dimensions and semiempirical approaches are just as the name implies partly empirical The elasticity models described here are called composite cylinder assem blage CCA models4912 In a CCA model one assumes the fibers are circular in crosssection spread in a periodic arrangement and continuous as shown in Figure 318 Then the composite can be considered to be made of repeating elements called the representative volume elements RVEs The RVE is considered to represent the composite and respond the same as the whole composite does The RVE consists of a composite cylinder made of a single inner solid cylinder fiber bonded to an outer hollow cylinder matrix as shown in Figure 319 The radius of the fiber a and the outer radius of the matrix b are related to the fiber volume fraction Vf as 363 Appropriate boundary conditions are applied to this composite cylinder based on the elastic moduli being evaluated η 35 42 1 308 1 35 42 1 308 2 130 0 8928 G12 1 308 1 2 130 0 8928 0 7 1 0 8928 0 7 8 130 G GPa V a b f 2 2 1343bookfm Page 239 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 240 Mechanics of Composite Materials Second Edition FIGURE 317 Theoretical values of inplane shear modulus as a function of fiber volume fraction compared with experimental values for unidirectional glassepoxy lamina Gf 3019 GPa Gm 183 GPa Figure b zooms figure a for fiber volume fraction between 045 and 075 Experimental data from Hashin Z NASA tech rep contract No NAS18818 November 1970 FIGURE 318 Periodic arrangement of fibers in a crosssection of a lamina 0 10 20 30 0 02 04 06 08 1 Fiber volume fraction Vf Inplane shear modulus G12 GPa Experimental data points Mechanics of materials approach HalphinTsai equation Fiber volume fraction Vf Inplane shear modulus G12 GPa Experimental data points Mechanics of materials approach HalphinTsai equation 10 0 20 30 055 045 065 075 1343bookfm Page 240 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Micromechanical Analysis of a Lamina 241 3331 Longitudinal Youngs Modulus To find the elastic moduli along the fibers we will apply an axial load P in direction 1 Figure 319 The axial stress σ1 in direction 1 then is 364 Now in terms of Hookes law 365 where E1 longitudinal Youngs modulus 1 axial strain in direction 1 Thus from Equation 364 and Equation 365 we have 366 FIGURE 319 Composite cylinder assemblage CCA model used for predicting elastic moduli of unidirec tional composites Fiber Matrix 2 θ 3 r 2 Fiber Matrix b a 1 z σ π 1 2 P b σ1 1 1 E E P b 1 1 2 π E P b 1 2 1 π 1343bookfm Page 241 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 242 Mechanics of Composite Materials Second Edition To find E1 in terms of elastic moduli of the fiber and the matrix and the geometrical parameters such as fiber volume fraction we need to relate the axial load P and the axial strain 1 in these terms Assuming the response of a cylinder is axisymmetric the equilibrium equation in the radial direction is given by13 367 where σr radial stress σθ hoop stress The normal stressnormal strain relationships in polar coordinates rθz for an isotropic material with Youngs modulus E and Poissons ratio ν are given by 368 The shear stresses and shear strains are zero in the rθz coordinate system for axisymmetric response The strain displacement equations for axisymmetric response are 369a 369b 369c where d dr r r r σ σ σθ 0 σ σ σ ν ν ν ν ν θ r z E E 1 1 2 1 1 2 1 ν ν ν ν ν ν ν ν ν E E E 1 2 1 1 2 1 1 1 2 1 1 2 1 1 2 1 1 2 1 ν ν ν ν ν ν ν ν ν ν E E E E 1 1 2 1 ν ν ν r z θ r du dr θ u r z dw dz 1343bookfm Page 242 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Micromechanical Analysis of a Lamina 243 u displacement in radial direction w displacement in axial direction Substituting the straindisplacement equations 369ac in the stressstrain equations 368 and noting that z 1 everywhere gives 370 which is rewritten for simplicity as 371 where the constants of the stiffness matrix are 372a 372b Substituting Equation 371 in the equilibrium equation 367 gives 373 The solution to the linear ordinary differential equation is found by assum ing that σ σ σ ν ν ν ν ν θ r z E E 1 1 2 1 1 2 1 ν ν ν ν ν ν ν ν ν E E E 1 2 1 1 2 1 1 1 2 1 1 2 1 1 2 1 1 2 1 ν ν ν ν ν ν ν ν ν ν E E E E d 1 1 2 1 ν ν ν u dr u r 1 σ σ σ θ r z C C C C C C C C C 11 12 12 12 11 12 12 12 11 1 du dr u r C E 11 1 1 2 1 ν ν ν C E 12 1 2 1 ν ν ν d u dr r du dr u r 2 2 2 1 0 1343bookfm Page 243 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 244 Mechanics of Composite Materials Second Edition 374 Substituting Equation 374 in Equation 373 gives 375 The preceding expression 375 requires that An 0 n except for n 1 and n 1 376 Therefore the form of the radial displacement is 377 To keep the terminology simple assume that the form of the radial dis placement with different names for the constants 378 The preceding equations are valid for a cylinder with an axisymmetric response Thus the radial displacement uf and um in the fiber and matrix cylinders respectively can be assumed of the form u A r n n n n n A r r nA r r A r n n n n n n n n 1 1 1 2 1 2 0 n n n n A r n n n 1 1 0 2 n A r n n n 2 2 1 0 n n A r n n n 1 1 0 2 u A r A r 1 1 u Ar B r 1343bookfm Page 244 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Micromechanical Analysis of a Lamina 245 379 380 However because the fiber is a solid cylinder and the radial displacement uf is finite Bf 0 otherwise the radial displacement of the fiber uf would be infinite Thus 381 382 Differentiating Equation 381 and Equation 382 gives 383a 383b Using Equation 383a and Equation 383b in Equation 370 the stressstrain relationships for the fiber are 384 where the stiffness constants of the fiber are 385 u A r B r r a f f f 0 u A r B r a r b m m m u A r r a f f 0 u A r B r a r b m m m du dr A f f du dr A B r m m m 2 σ σ σ θ r f f z f f f f f f C C C C C C 11 12 12 12 11 12 f f f f f f C C C A A 12 12 11 1 C E f f f f f 11 1 1 2 1 ν ν ν C E f f f f f 12 1 2 1 ν ν ν 1343bookfm Page 245 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 246 Mechanics of Composite Materials Second Edition and the stressstrain relationships for the matrix are 386 where the stiffness constants of the matrix are 387a 387b How do we now solve for the unknown constants Af Am Bm and ε1 The following four boundary and interface conditions will allow us to do that 1 The radial displacement is continuous at the interface r a 388 Then from Equation 381 and Equation 382 389 2 The radial stress is continuous at r a 390 Then from Equation 384 and Equation 386 391 σ σ σ θ r m m z m m m m m m C C C C C C 11 12 12 12 11 12 m m m m m m m m C C C A B r A B r 12 12 11 2 2 1 C E m m m m m 11 1 1 2 1 ν ν ν C E m m m m m 12 1 2 1 ν ν ν u r a u r a f m A a A a B a f m m σ σ r f r m r a r a C A C A C C A B a C f f f f f m m m m 11 12 1 11 2 12 12 A B a C m m m 2 12 1 1343bookfm Page 246 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Micromechanical Analysis of a Lamina 247 3 Because the surface at r b is traction free the radial stress on the outside of matrix r b is zero 392 Then Equation 384 gives 393 4 The overall axial load over the fibermatrix crosssectional area in direction 1 is the applied load P then 394 Because the axial normal stress σz is independent of θ 395 Now 396 Then from Equation 384 and Equation 386 397 σr m r b 0 C A B b C A B b C m m m m m m m 11 2 12 2 12 1 0 σz A dA P σ θ π z b rdrd P 0 2 0 σ π z b rdr P 2 0 σ σ z z f r a 0 σz m a r b C A C A C rdr C A B r f f f f f a m m m 12 12 11 1 0 12 2 2 π C A B r C rdr m m m m 12 2 11 1 2π P a b 1343bookfm Page 247 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 248 Mechanics of Composite Materials Second Edition Solving Equation 389 Equation 391 Equation 393 and Equa tion 397 we get the solution to Af Am Bm and ε1 Using the resulting solution for 1 and using Equation 366 398 Although the preceding expression can be written in a compact form by using definitions of shear and bulk modulus of the material we avoid doing so because results given in Equation 398 can now be found symbolically by computational systems such as Maple14 Note that the first two terms of Equation 398 represent the mechanics of materials approach result given by Equation 334 Example 39 Find the longitudinal Youngs modulus for a glassepoxy lamina with a 70 fiber volume fraction Use the properties for glass and epoxy from Table 31 and Table 32 respectively Use equations obtained using the elasticity model Solution From Table 31 the Youngs modulus of fiber is Ef 85 GPa the Poissons ratio of the fiber is νf 02 From Table 32 the Youngs modulus of matrix is Em 34 GPa Bulk modulus of an elastic body is defined as the slope of the applied hydrostatic pressure vs volume dilation curve Hydrostatic stress is defined as σxx σyy σzz p τxy 0 τyz 0 τzx 0 and volume dilation Dv is defined as the sum of resulting normal strains Dv εx εy εz The bulk modulus K is used for finding volume changes in a given body subjected to hydrostatic pressure E P b 1 2 1 π E V E V f f m f 1 2 1 2 1 2 2 E E V V E V V V m f f f m f f m f m f m f ν ν ν ν ν E V V V m f f f f f f f 1 2 2 2 2 ν ν ν ν 1343bookfm Page 248 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Micromechanical Analysis of a Lamina 249 and the Poissons ratio of the matrix is νm 03 Using Equation 398 the longitudinal Youngs modulus For the same problem the longitudinal Youngs modulus was found to be 6052 GPa from the mechanics of materials approach as well as the Hal phinTsai equations 3332 Major Poissons Ratio In Section 3331 we solved the problems of an axially loaded cylinder This same problem can be used to determine the axial Poissons ratio ν12 because of the definition of major Poissons ratio as 399 when a body is only under an axial load in direction 1 From the definition of radial strain from Equation 369a that at r b 3100 the major Poissons ratio is 3101 Using Equation 3101 E1 9 9 9 85 10 0 7 3 4 10 1 0 7 2 3 4 10 85 10 0 7 0 2 0 3 1 0 7 85 10 2 9 2 9 0 3 0 7 0 3 0 7 0 3 0 7 1 3 4 10 2 9 2 2 1 2 0 7 0 2 0 2 0 7 0 2 2 0 2 0 7 60 53 10 60 53 9 Pa GPa ν12 1 r r m r b u b b ν12 1 u r b b m 1343bookfm Page 249 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 250 Mechanics of Composite Materials Second Edition 3102 Using the solution obtained in Section 3331 for Am Bm and 1 by solving Equation 389 Equation 391 Equation 393 and Equation 397 we get 3103 Although the preceding expression can be written in a compact form by using definitions of shear and bulk modulus of the material we avoid doing so because results given in Equation 3103 can be found symbolically by computational systems such as Maple14 Note that the first two terms of Equation 3103 are the same as the mechanics of materials approach result given by Equation 334 Example 310 Find the major Poissons ratio for a glassepoxy lamina with a 70 fiber volume fraction Use the properties for glass and epoxy from Table 31 and Table 32 respectively Use equations obtained using the elasticity model Solution Using Equation 3103 the major Poissons ratio is For the same problem the major Poissons ratio was found to be 02300 from the mechanics of materials approach as well as the HalphinTsai equations ν12 2 1 A B b m m ν ν ν ν ν ν ν 12 2 2 f f m m f m f m f m m f f m V V V V E E E E E E V V V E m f m f m f m m f f f f ν ν ν ν ν ν 2 2 1 2 2 2 2 V V V E f f f f f f m ν ν ν 2 1 2 ν12 0 2 0 7 0 3 0 3 0 7 0 3 0 2 0 3 2 85 10 0 3 0 3 85 10 85 10 3 9 2 9 9 4 10 3 4 10 0 2 2 3 4 10 0 2 9 9 9 2 2 0 3 2 0 7 0 3 0 3 0 7 1 00 7 85 10 2 0 3 0 7 0 2 2 0 9 2 7 0 2 0 7 0 2 1 3 4 10 0 2 9 2238 1343bookfm Page 250 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Micromechanical Analysis of a Lamina 251 3333 Transverse Youngs Modulus The CCA model only gives lower and upper bounds of the transverse Youngs modulus of the composite However for the sake of completeness we will summarize the result from a threephase model This model Figure 320 however yields an exact solution12 for the transverse shear modulus G23 However the transverse Youngs modulus can be found as follows Assuming that the resulting composite properties are transversely isotropic a valid assumption for hexagonally arranged fibers 23 plane is isotropic 3104 where ν23 transverse Poissons ratio The transverse Poissons ratio ν23 is given by15 3105 where 3106 The bulk modulus K of the composite under longitudinal plane strain is FIGURE 320 Threephase model of a composite Fiber Matrix Equivalent homogenous medium E G 2 23 23 2 1 ν ν23 23 23 K mG K mG m K E 1 4 12 2 1 ν 1343bookfm Page 251 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 252 Mechanics of Composite Materials Second Edition 3107 The bulk modulus Kf of the fiber under longitudinal plane strain is 3108 The bulk modulus Km of the matrix under longitudinal plane strain is 3109 To derive the solution for G23 for use in Equation 3104 is out of scope of this book however for the sake of completeness the final solution is given next Based on the threephase model Figure 320 where the fiber is sur rounded by matrix which is then surrounded by a homogeneous material equivalent to the composite the transverse shear modulus G23 is given by the acceptable solution of the quadratic equation12 3110 where K K K G V K K G V K G V K G m f m m f m m f f m m m m Vf K E f f f f 2 1 1 2 ν ν K E m m m m 2 1 1 2 ν ν A G G B G G C m m 23 2 23 2 0 A V V G G G G f f f m f m f 3 1 1 2 η G G G G V V f m m f m f m m f f f m η η η η η η 3 G G G G f m f m m 1 1 η B V V G G G G f f f m f m f 3 1 1 2 η 1 2 1 1 1 η η m f m f m f m f G G G G V G Gm f η 2 2 1 1 3 G G V V G G f m m f f f m f m η η η G G G G V f m f f m m f f η η η 3 1343bookfm Page 252 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Micromechanical Analysis of a Lamina 253 3111 3112 Then using Equation 3104 through Equation 3109 we get the transverse Youngs modulus E2 Example 311 Find the transverse Youngs modulus for a glassepoxy lamina with a 70 fiber volume fraction Use the properties for glass and epoxy from Table 31 and Table 32 respectively Use equations obtained using the elasticity model Solution From Equation 3112 From Equation 3108 and Equation 3109 C V V G G G G f f f m f m f 3 1 1 2 η η η m f m f m f f m f f G G G G V G G G G 1 1 m m f Vf η η 3 η ν m m 3 4 η ν f f 3 4 ηf 3 4 0 2 2 2 ηm 3 4 0 3 1 8 Kf 85 10 2 1 0 2 1 2 0 2 9 59 03 109 Pa 1343bookfm Page 253 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 254 Mechanics of Composite Materials Second Edition From Equation 3107 The three constants of the quadratic Equation 3110 are given by Equation 3111 as Km 3 4 10 2 1 0 3 1 2 0 3 9 3 269 109 Pa K 3 269 10 59 03 10 1 308 10 0 3 59 0 9 9 9 3 10 3 269 10 1 308 10 0 7 5 9 9 9 9 03 10 1 308 10 0 3 3 269 10 1 308 9 9 9 10 9 0 7 11 66 109 Pa A 3 0 7 1 0 7 35 42 10 1 308 10 1 3 2 9 9 5 42 10 1 308 10 2 2 9 9 35 42 10 1 308 10 1 8 2 2 1 8 35 42 10 1 9 9 9 308 10 1 8 2 2 0 7 9 3 0 7 1 8 35 42 10 1 308 10 1 35 42 9 9 10 1 308 10 1 8 1 9 9 476 0 B 3 0 7 1 0 7 35 42 10 1 308 10 1 2 9 9 35 42 10 1 308 10 2 2 9 9 1 2 1 8 35 42 10 1 308 10 35 42 10 1 3 9 9 9 08 10 1 0 7 1 9 1343bookfm Page 254 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Micromechanical Analysis of a Lamina 255 Substituting values of A B and C in Equation 3110 gives G23 5926 109 Pa 1953 109 Pa Thus the acceptable solution is From Equation 3106 From Equation 3105 1 8 1 35 42 10 1 308 10 2 2 2 35 42 9 9 10 1 308 10 1 8 2 2 0 7 0 9 9 3 7 2 1 8 1 35 42 10 1 308 10 1 35 42 9 9 10 1 308 10 1 8 35 42 10 1 308 10 1 8 2 9 9 9 9 2 0 7 723 0 3 C 3 0 7 1 0 7 35 42 10 1 308 10 1 3 2 9 9 5 42 10 1 308 10 2 2 1 8 35 42 10 1 9 9 9 308 10 35 42 10 1 308 10 1 0 7 1 9 9 9 35 42 10 1 308 10 2 2 35 42 10 1 3 9 9 9 08 10 1 8 2 2 0 7 3222 9 3 476 0 1 308 10 2 723 0 1 308 23 9 2 23 G G 10 3222 0 278 4 10 1106 10 9 18 23 2 9 G GG23 3222 0 G Pa 23 9 5 926 10 m 1 4 11 66 10 0 2238 60 53 10 9 2 9 1 039 1343bookfm Page 255 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 256 Mechanics of Composite Materials Second Edition From Equation 3104 1551 GPa For the same problem the transverse Youngs modulus was found to be 1037 GPa from the mechanics of materials approach and 2020 GPa from the HalphinTsai equations Figure 321a and Figure 321b show the transverse Youngs modulus as a function of fiber volume fraction for a typical boronepoxy unidirectional lamina The elasticity equation 3104 HalphinTsai equation 360 and the mechanics of materials approach Equation 355 are shown and compared to the experimental data points 3334 Axial Shear Modulus To find the axial shear modulus G12 of a unidirectional composite we consider the same concentric cylinder model Figure 319 Consider a long fiber of radius a and shear modulus Gf surrounded by a long concentric cylinder of matrix of outer radius b and shear modulus Gm The composite cylinder Figure 319 is subjected to a shear strain in the 12 plane Following the derivation41216 the normal displacements in the 1 2 3 direction for the fiber or matrix are assumed of the following form 3113a b c where is the applied shear strain to the boundary ν23 9 9 9 11 66 10 1 039 5 926 10 11 66 10 1 0 39 5 926 109 0 3089 E2 9 2 1 0 3089 5 926 10 15 51 109 Pa γ 12 0 u x F x x 1 12 0 2 2 3 2 γ u x 2 12 0 1 2 γ u3 0 γ 12 0 1343bookfm Page 256 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Micromechanical Analysis of a Lamina 257 The preceding assumption of the form of the displacements is based on a semiinverse method17 that is beyond the scope of this book Individual expressions for displacement of the fiber and matrix will be shown later in the derivation From the straindisplacement13 equations and the expressions for the dis placement field in Equation 3113a b c FIGURE 321 Theoretical values of transverse Youngs modulus as a function of fiber volume fraction and comparison with experimental values for boronepoxy unidirectional lamina Ef 414 GPa νf 02 Em 414 GPa νm 035 Figure b zooms figure a for fiber volume fraction between 045 and 075 Experimental data from Hashin Z NASA tech rep contract No NAS18818 November 1970 0 10 20 30 40 0 02 04 06 08 1 Fiber volume fraction Vf Transverse Youngs modulus E2 GPa HalphinTsai equation Experimental data points Mechanics of materials approach Elasticity solution Fiber volume fraction Vf Transverse Youngs modulus E2 GPa HalphinTsai equation Experimental data points Mechanics of materials approach Elasticity solution 0 10 20 30 40 045 055 065 075 1343bookfm Page 257 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 258 Mechanics of Composite Materials Second Edition 3114 af Because all normal strains in the 1 2 and 3 directions are zero all the normal stresses in 1 2 3 directions are also zero Also τ23 0 because γ23 0 Using Equation 3114e and Equation 3114f the only possible nonzero stresses are 11 1 1 u x 0 22 2 2 u x 0 33 3 3 u x 0 γ 23 2 3 3 2 u x u x 0 γ 12 1 2 2 1 u x u x F x2 γ 31 1 3 3 1 u x u x F x3 1343bookfm Page 258 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Micromechanical Analysis of a Lamina 259 3115a 3115b where G is the shear modulus of the material The equilibrium condition derived from the fact that the sum of the forces in direction 1 is zero gives13 3116 With Equation 3115a and Equation 3115b and σ1 0 the equilibrium equation 3116 reduces it to 3117 Converting Equation 3117 to polar coordinates needs the following 3118 3119 give 3120a 3120b From Equation 3118 Equation 3119 and Equation 3120a b τ γ 12 12 G G F x2 τ γ 13 13 G G F x3 σ τ τ 1 1 12 2 13 3 0 x x x 2 2 2 2 3 2 0 F x F x x 2 rCos θ x 3 rSin θ r x x 2 2 2 3 2 θ tan 1 3 2 x x 1343bookfm Page 259 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 260 Mechanics of Composite Materials Second Edition 3121a 3121b 3121c 2 2 2 2 r r x x r x x r 2 2 Cosθ 2 2 3 3 r r x x r x x r 3 3 Sinθ θ x x x x x 2 3 2 2 3 2 2 1 1 x x x 3 2 2 3 2 rSin r θ 2 Sin r θ θ x x x x 3 3 2 2 2 1 1 1 x x x 2 2 2 3 2 1343bookfm Page 260 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Micromechanical Analysis of a Lamina 261 3121d Now using the chain rule for derivatives 3122 and using Equation 3121a and Equation 3121c 3123 Repeating a similar chain of rule of derivatives on Equation 3122 3124a Similarly 3124b Substituting Equation 3124a and Equation 3124b in Equation 3117 yields 3125 The solution of Equation 3125 is given by 3126 only after noting that the complete solution to Equation 3125 is of the form rCos r θ Cos r θ F x F r r x F x 2 2 2 θ θ F x Cos F r Sin r F 2 θ θ θ 2 2 2 2 2 2 2 2 2 2 1 1 F x Cos F r Sin r F r r F θ θ θ 2 1 Sin Cos r r F θ θ θ 2 3 2 2 2 2 2 2 2 2 1 1 F x Sin F r Cos r F r r F θ θ θ 2 1 Sin Cos r r F θ θ θ 2 2 2 2 2 1 1 0 F r r F r r F θ F r Ar B r Cos θ θ 1343bookfm Page 261 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 262 Mechanics of Composite Materials Second Edition 3127 but that the surface r b of the composite cylinder is only subjected to displacements 3128 3129 3130 Thus the function Frθ of Equation 3126 for the fiber Ff and matrix Fm is given by 3131 3132 How do we find A1 B1 A2 and B2 The following boundary and interface conditions are applied to find these four unknowns 1 The axial displacements of the fiber u1f and the matrix u1m at the interface r a are continuous 3133 F r A A r B r C Sin n D Cos n n n n n n n n θ θ 0 1 θ u r b x m r b 1 12 0 2 2 γ γ θ 12 0 2 bCos u r b x m r b 2 12 0 1 2 γ γ θ 12 0 2 bSin u r b 3m 0 F r A r B r Cos f θ θ 1 1 F r A r B r Cos m θ θ 2 2 u r a u r a f m 1 1 1343bookfm Page 262 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Micromechanical Analysis of a Lamina 263 Now from Equation 3113a 3134 At r a 3135 Similarly from Equations 3313a 3318 and 3131 3136 Equating Equation 3135 and Equation 3136 per Equation 3133 gives 3137 2 The displacement of the fiber u1f is given by Equations 3313a 3318 and 3131 as 3138 Because r 0 is a point on the fiber and displacement in the fiber is finite 3139 3 The shear stress in the fiber τ1rf and that in the matrix τ1rm are con tinuous at the interface r a 3140 u x F x x f f 1 12 0 2 2 3 2 γ γ θ θ θ 12 0 2 rCos F rCos rSin f u r a aCos A a B a Cos 1f 12 0 1 1 2 γ θ θ u r a aCos A a B a Cos 1m 12 0 2 2 2 γ θ θ A a B a A a B a 1 1 2 2 u rCos A r B r Cos 1f 12 0 1 1 2 γ θ θ B1 0 τ τ 1 1 r f r m r a r a 1343bookfm Page 263 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 264 Mechanics of Composite Materials Second Edition First we need to derive an expression for τ1r from transforming stresses between 1r and 13 coordinates 3141 Using Equation 3115a b in Equation 3141 3142 3143 Substituting Equation 3121a and Equation 3121b in Equation 3143 gives 3144 Thus in the fiber from Equation 3131 3145 and in the matrix from Equation 3132 τ θ τ θ τ 1 12 13 r Cos Sin τ θ θ 1 2 3 r Cos G F x Sin G F x G Cos F x Sin F x θ θ 2 3 τ1 2 2 3 3 r G x r F x x r F x τ1r G F r τ1r f f f G F r G A B r Cos f 1 1 2 θ τ1r m m m G F r 1343bookfm Page 264 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Micromechanical Analysis of a Lamina 265 3146 Equating Equation 3145 and Equation 3146 at r a per Equation 3140 gives 3147 4 The displacement due to the applied shear strain of at the bound ary r b of the composite cylinder is given by 3148a 3148b Based on Equation 3113a and Equation 3132 3149 From Equation 3148b and Equation 3149 we get 3150 Solving the three simultaneous equations Equation 3137 Equation 3147 and Equation 3150 to find A1 A2 and B2 B1 0 from Equa tion 3139 we get G A B r Cos m 2 2 2 θ G A B a G A B a f m 1 1 2 2 2 2 γ 12 0 u r b x m r b 1 12 0 2 2 γ γ θ 12 0 2 bCos u r b x F x x m m r b 1 12 0 2 2 3 2 γ γ θ θ 12 0 2 2 2 bCos A b B b Cos A b B b b 2 2 12 0 γ 1343bookfm Page 265 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 266 Mechanics of Composite Materials Second Edition 3151 3152 3153 where from Equation 363 the fiber volume fraction Vf is substi tuted for The shear modulus G12 can be now be found as 3154 where because based on Equation 3115a 3155 Using Equation 3121a and Equation 3121b 3156 Using Equation 3131 and Equation 3132 in Equation 3156 gives A G G V G V m m f f f 1 12 0 2 1 1 γ A G G G V G V f m m f f f 2 12 0 1 1 γ B a G G G V G V m f m f f f 2 2 12 0 1 1 γ a b 2 2 G m r b 12 12 12 0 τ γ τ12 m r b r b shear stress at τ12 2 m m m G F x G F r r x F x m m m 2 2 θ θ τ θ θ θ 12m m m m G F r Cos F Sin r 1343bookfm Page 266 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Micromechanical Analysis of a Lamina 267 3157 At r b θ 0 3158 Substituting values of A2 and B2 from Equation 3152 and Equation 3153 respectively in Equation 3158 yields 3159 and the shear modulus G12 can be found as This gives 3160 Example 312 Find the shear modulus G12 for a glassepoxy composite with 70 fiber volume fraction Use the properties for glass and epoxy from Table 31 and Table 32 respectively Use the equations obtained using the elasticity model Solution From Example 36 Gf 3542 GPa and Gm 1308 GPa Using Equation 3160 the inplane shear modulus is τ θ θ 12 2 2 2 2 2 m Gm A B r Cos Cos A r B r Sin Sin r θ θ G A B r Cos A B r Sin m 2 2 2 2 2 2 2 2 θ θ τ θ 12 0 2 2 2 m r b Gm A B b τ θ 12 0 1 1 1 m r b m f f m f f f m G G V G V G V G 1 12 0 Vf γ G m r b 12 12 0 12 0 τ γ θ G G G V G V G V G V m f f m f f f m f 12 1 1 1 1 1343bookfm Page 267 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 268 Mechanics of Composite Materials Second Edition For the same problem the shear modulus G12 is found to be 4014 GPa from the mechanics of materials approach and 6169 GPa from the HalphinTsai equations Figure 322a and Figure 322b show the inplane shear modulus as a func tion of fiber volume fraction for a typical glassepoxy unidirectional lamina The elasticity equation 3160 HalphinTsai equation 360 and the mechan ics of materials approach Equation 355 are shown and compared to the experimental data points A comparison of the elastic moduli from the mechanics of materials approach the HalphinTsai equations and elasticity models Example 33 through Example 311 is given in Table 35 334 Elastic Moduli of Lamina with Transversely Isotropic Fibers Glass aramids and graphite are the three most common types of fibers used in composites among these aramids and graphite are transversely isotropic From the definition of transversely isotropic materials in Chapter 2 such fibers have five elastic moduli If L represents the longitudinal direction along the length of the fiber and T represents the plane of isotropy Figure 323 perpendicular to the longi tudinal direction the five elastic moduli of the transversely isotropic fiber are EfL longitudinal Youngs modulus EfT Youngs modulus in plane of isotropy νfL Poissons ratio characterizing the contraction in the plane of isot ropy when longitudinal tension is applied νfT Poissons ratio characterizing the contraction in the longitudinal direction when tension is applied in the plane of isotropy GfT inplane shear modulus in the plane perpendicular to the plane of isotropy The elastic moduli using strength of materials approach for lamina with transversely isotropic fibers18 are G12 9 9 9 1 308 10 35 42 10 1 0 7 1 308 10 11 0 7 35 42 10 1 0 7 1 308 10 1 0 7 9 9 6 169 10 6 169 9 Pa GPa 1343bookfm Page 268 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Micromechanical Analysis of a Lamina 269 FIGURE 322 Theoretical values of inplane shear modulus as a function of fiber volume fraction compared with experimental values for unidirectional glassepoxy lamina Gf 3019 GPa Gm 183 GPa Figure b zooms figure a for fiber volume fraction between 045 and 075 Experimental data from Hashin Z NASA tech rep contract No NAS18818 November 1970 TABLE 35 Comparison of Predicted Elastic Moduli Method E1 GPa E2 GPa ν12 G12 GPa Mechanics of materials HalphinTsai Elasticity 6052 6052 6053 1037 2020 1551 02300 02300 02238 4014 6169 6169a a The HalphinTsai equations and the elasticity model equations give the same value for the shear modulus Can you show that this is not a coincidence 0 10 20 30 0 02 04 06 08 1 Fiber volume fraction Vf Inplane shear modulus G12 GPa Experimental data points Mechanics of materials approach HalphinTsai equation Elasticity solution Fiber volume fraction Vf Inplane shear modulus G12 GPa Experimental data points Mechanics of materials approach HalphinTsai equation Elasticity solution 10 0 20 30 055 045 065 075 1343bookfm Page 269 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 270 Mechanics of Composite Materials Second Edition and 3161ad The preceding expressions are similar to those of a lamina with isotropic fibers The only difference is that appropriate transverse or longitudinal properties of the fiber are used In composites such as carboncarbon the matrix is also transversely isotropic In that case the preceding equations cannot be used and are given elsewhere1519 FIGURE 323 Longitudinal and transverse direction in a transversely isotropic fiber L T E E V E V fL f m m 1 1 E2 V E V E f fT m m ν ν ν 12 fT f m m V V 1 G12 V G V G f fT m m 1343bookfm Page 270 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Micromechanical Analysis of a Lamina 271 34 Ultimate Strengths of a Unidirectional Lamina As shown in Chapter 2 one needs to know five ultimate strength parameters for a unidirectional lamina Longitudinal tensile strength Longitudinal compressive strength Transverse tensile strength Transverse compressive strength Inplane shear strength τ12ult In this section we will see whether and how these parameters can be found from the individual properties of the fiber and matrix by using the mechanics of materials approach The strength parameters for a unidirectional lamina are much harder to predict than the stiffnesses because the strengths are more sensitive to the material and geometric nonhomogeneities fibermatrix interface fabrication process and environment For example a weak inter face between the fiber and matrix may result in premature failure of the composite under a transverse tensile load but may increase its longitudinal tensile strength For these reasons of sensitivity some theoretical and empir ical models are available for some of the strength parameters Eventually the experimental evaluation of these strengths becomes important because it is direct and reliable These experimental techniques are also discussed in this section 341 Longitudinal Tensile Strength A simple mechanics of materials approach model is presented Figure 324 Assume that Fiber and matrix are isotropic homogeneous and linearly elastic until failure The failure strain for the matrix is higher than for the fiber which is the case for polymeric matrix composites For example glass fibers fail at strains of 3 to 5 but an epoxy fails at strains of 9 to 10 Now if σfult ultimate tensile strength of fiber Ef Youngs modulus of fiber σmult ultimate tensile strength of matrix Em Youngs modulus of matrix σ1 T ult σ1 C ult σ2 T ult σ2 C ult 1343bookfm Page 271 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 272 Mechanics of Composite Materials Second Edition then the ultimate failure strain of the fiber is 3162 and the ultimate failure strain of the matrix is 3163 Because the fibers carry most of the load in polymeric matrix composites it is assumed that when the fibers fail at the strain of εfult the whole composite fails Thus the composite tensile strength is given by 3164 Once the fibers have broken can the composite take more load The stress that the matrix can take alone is given by σmult 1 Vf Only if this stress is greater than Equation 3164 is it possible for the composite to take more load The volume fraction of fibers for which this is possible is called the minimum fiber volume fraction Vfminimum and is FIGURE 324 Stressstrain curve for a unidirectional composite under uniaxial tensile load along fibers Fiber Composite Matrix σfult σmult εfult εmult Stress σ Strain ε ε σ f ult f ult Ef ε σ m ult m ult Em σ σ ε 1 1 T ult f ult f f ult m f V E V σ1 T ult 1343bookfm Page 272 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Micromechanical Analysis of a Lamina 273 3165 It is also possible that by adding fibers to the matrix the composite will have lower ultimate tensile strength than the matrix In that case the fiber volume fraction for which this is possible is called the critical fiber volume fraction Vfcritical and is 3166 Example 313 Find the ultimate tensile strength for a glassepoxy lamina with a 70 fiber volume fraction Use the properties for glass and epoxy from Table 31 and Table 32 respectively Also find the minimum and critical fiber volume fractions Solution From Table 31 Ef 85 GPa σfult 1550 MPa Thus From Table 32 Em 34 GPa σmult 72 MPa σ σ m ult f minimum f ult f minimum V V 1 ε f ult m f minimum E V 1 V E E f minimum m ult m f ult f ult m f σ ε σ ε ult m ult σ σ σ ε m ult f ult f critical f ult m f V E V 1 critical V E E f critical m ult m f ult f ult m f σ ε σ ε ult ε f ult 1550 10 85 10 0 1823 10 6 9 1 1343bookfm Page 273 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 274 Mechanics of Composite Materials Second Edition Thus Applying Equation 3164 the ultimate longitudinal tensile strength is Applying Equation 3165 the minimum fiber volume fraction is This implies that if the fiber volume fraction is less than 06422 the matrix can take more loading after all the fibers break Applying Equation 3166 the critical fiber volume fraction is This implies that if the fiber volume fraction were less than 06732 the composite longitudinal tensile strength would be less than that of the matrix Experimental evaluation The general test method recommended for tensile strength is the ASTM test method for tensile properties of fiberresin com posites D3039 Figure 325 A tensile test geometry Figure 326 to find εm ult 72 10 3 4 10 0 2117 10 6 9 1 σ1 6 1 1550 10 0 7 0 1823 10 3 4 T ult 10 1 0 7 1104 9 MPa Vf miminum 72 10 3 4 10 0 1823 10 15 6 9 1 50 10 3 4 10 0 1823 10 72 10 0 642 6 9 1 6 2 10 0 6422 2 Vf critical 72 10 3 4 10 0 1823 10 1 6 9 1 5550 10 3 4 10 0 1823 10 0 6732 10 6 9 1 2 0 6732 1343bookfm Page 274 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Micromechanical Analysis of a Lamina 275 the longitudinal tensile strength consists of six to eight 0 plies that are 125 mm 12 in wide and 229 mm 10 in long The specimen is mounted with strain gages in the longitudinal and transverse directions Tensile stresses are applied on the specimen at a rate of about 05 to 1 mmmin 002 to 004 inmin A total of 40 to 50 data points for stress and strain is taken until a specimen fails The stress in the longitudinal direction is plotted as a function FIGURE 325 Tensile coupon mounted in the test frame for finding the tensile strengths of a unidirectional lamina Photo courtesy of Dr RY Kim University of Dayton Research Institute Dayton OH FIGURE 326 Geometry of a longitudinal tensile strength specimen Tab thickness 32 mm Tab length 38 mm T t 60 W Gage length 153 mm Length 229 mm 1343bookfm Page 275 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 276 Mechanics of Composite Materials Second Edition of longitudinal strain as shown in Figure 327 The data are reduced using linear regression The longitudinal Youngs modulus is the initial slope of the σ1 vs ε1 curve From Figure 327 the following values are obtained Discussion Failure of a unidirectional ply under a longitudinal tensile load takes place with 1 Brittle fracture of fibers 2 Brittle fracture of fibers with pullout 3 Fiber pullout with fibermatrix debonding The three failure modes are shown in Figure 328 The mode of failure depends on the fibermatrix bond strength and fiber volume fraction20 For low fiber volume fractions 0 Vf 040 a typical glassepoxy composite FIGURE 327 Stressstrain curve for a 08 laminate under a longitudinal tensile load Data courtesy of Dr RY Kim University of Dayton Research Institute Dayton OH 3000 2000 1000 0 0 05 Longitudinal tensile strain Longitudinal tensile stress MPa 1 15 2 E GPa 1 187 5 σ1 2896 T ult MPa ε1 1 560 T ult 1343bookfm Page 276 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Micromechanical Analysis of a Lamina 277 exhibits a mode 1 type failure For intermediate fiber volume fractions 04 Vf 065 mode 2 type failure occurs For high fiber volume fractions Vf 065 it exhibits mode 3 type of failure 342 Longitudinal Compressive Strength The model used for calculating the longitudinal tensile strength for a unidi rectional lamina cannot also be used for its longitudinal compressive strength because the failure modes are different Three typical failure modes are shown in Figure 329 Fracture of matrix andor fibermatrix bond due to tensile strains in the matrix andor bond Microbuckling of fibers in shear or extensional mode Shear failure of fibers Ultimate tensile strains in matrix failure mode A mechanics of materials approach model based on the failure of the composite in the transverse direction due to transverse tensile strains is given next20 Assuming that one is applying a longitudinal compressive stress of magnitude σ1 then the magnitude of longitudinal compressive strain is given by 3167 Because the major Poissons ratio is ν12 the transverse strain is tensile and is given by 3168 FIGURE 328 Modes of failure of unidirectional lamina under a longitudinal tensile load a b c ε σ 1 1 1 E ε ν σ 2 12 1 1 E 1343bookfm Page 277 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 278 Mechanics of Composite Materials Second Edition Using maximum strain failure theory if the transverse strain exceeds the ultimate transverse tensile strain the lamina is considered to have failed in the transverse direction Thus 3169 The value of the longitudinal modulus E1 and the major Poissons ratio v12 can be found from Equation 334 and Equation 349 respectively However for the value of one can use the empirical formula FIGURE 329 Modes of failure of a unidirectional lamina under a longitudinal compressive load Fiber microbuckling extensional mode Fiber microbuckling shear mode Transverse tensile failure of matrix Shear failure a b c d ε2 T ult σ ε ν 1 1 2 12 c ult T ult E ε2 T ult 1343bookfm Page 278 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Micromechanical Analysis of a Lamina 279 3170 or the mechanics of materials formula 3171 where ultimate tensile strain of the matrix d diameter of the fibers s centertocenter spacing between the fibers Equation 3170 and Equation 3171 will be discussed later in Section 343 Shearextensional fiber microbuckling failure mode local buckling models for calculating longitudinal compressive strengths have been developed2122 Because these results are based on advanced topics only the final expressions are given 3172 where 3173a and 3173b Note that the extensional mode buckling stress is higher than the shear mode buckling stress for most cases Extensional mode buckling is prev alent only in low fiber volume fraction composites Shear stress failure of fibers mode A unidirectional composite may fail due to direct shear failure of fibers In this case the rule of mixtures gives the shear strength of the unidirectional composite as 3174 ε ε 2 1 3 1 T ult m T ult Vf ε ε 2 1 1 T ult m T ult m f d s E E εm T ult min σ1 1 2 c ult c S Sc S V V E E V E E V c f f m f f m f f 1 2 1 3 1 S G V c m f 2 1 Sc 1 Sc 2 τ τ τ 12 ult f ult f m ult m V V 1343bookfm Page 279 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 280 Mechanics of Composite Materials Second Edition where τfult ultimate shear strength of the fiber τmult ultimate shear strength of the matrix The maximum shear stress in a lamina under a longitudinal compressive load is at 45 to the loading axis Thus 3175 Three models based on each of the preceding failure modes were dis cussed to find the magnitude of the ultimate longitudinal compressive strength One may caution that these models are not found to match the experimental results as is partially evident in the comparison of experimen tal and predicted values23 of longitudinal compressive strength given in Table 36 Comparison with other equations 3169 and 3175 is not avail able because the properties of constituents are not given in the reference23 although fiber buckling is the most probable mode of failure in advanced polymer matrix composites Several factors may contribute to this discrepancy including Irregular spacing of fibers causing premature failure in matrixrich areas Less than perfect bonding between the fibers and the matrix Poor alignment of fibers Not accounting for Poissons ratio mismatch between the fiber and the matrix Not accounting for the transversely isotropic nature of fibers such as aramids and graphite In addition there is controversy concerning the techniques used in measur ing compressive strengths TABLE 36 Comparison of Experimental and Predicted Values of Longitudinal Compressive Strength of Unidirectional Laminaea Material Experimental strength Equation 378a MPa Equation 378b MPa Glasspolyester Type I carbonepoxy Kevlar 49epoxy 6001000 700900 240290 8700 22800 13200 2200 2900 2900 a Vf 050 Source Hull D Introduction to Composite Materials Cambridge University Press 1981 Table 72 Reprinted with the permission of Cambridge University Press σ1 c σ1 2 c σ τ τ 1 2 c ult f ult f m ult m V V 1343bookfm Page 280 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Micromechanical Analysis of a Lamina 281 Example 314 Find the longitudinal compressive strength of a glassepoxy lamina with a 70 fiber volume fraction Use the properties of glass and epoxy from Table 31 and Table 32 respectively Assume that fibers are circular and are in a square array Solution From Table 31 the Youngs modulus for the fiber is Ef 85 GPa and the Poissons ratio of the fiber is νf 020 The ultimate tensile strength of the fiber is σfult 1550 MPa and the ultimate shear strength of the fiber is τfult 35 MPa From Table 32 the Youngs modulus of the matrix is Em 34 GPa and the Poissons ratio of the matrix is νm 030 The ultimate normal strength of the matrix is σmult 72 MPa and the ultimate shear strength of the matrix is τmult 34 MPa From Example 33 the longitudinal Youngs modulus of the unidirectional lamina is E1 6052 GPa From Example 35 the major Poissons ratio of the unidirectional lamina is 1343bookfm Page 281 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 282 Mechanics of Composite Materials Second Edition ν12 023 Using Equation 342a the fiber diameter to fiber spacing ratio is The ultimate tensile strain of the matrix is Using the transverse ultimate tensile strain failure mode formula 376 From the empirical Equation 3170 Using the lesser of these two values of ultimate transverse tensile strain and Equation 3169 Using shearextensional fiber microbuckling failure mode formulas 3173a d s 4 0 7 0 9441 1 2 π εm ult 72 10 3 40 10 0 2117 10 6 9 1 ε2 1 9 9 0 2117 10 0 9441 3 4 10 85 10 1 T ult 1 0 1983 10 2 ε2 1 1 3 0 2117 10 1 0 7 0 2373 10 T ult 2 ε2 T ult σ1 9 2 60 52 10 0 1983 10 0 23 521 8 C ult MPa SC 1 9 9 2 0 7 1 0 7 3 4 10 85 10 0 7 3 4 10 85 10 3 1 0 7 21349 9 9 MPa 1343bookfm Page 282 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Micromechanical Analysis of a Lamina 283 From Example 36 the shear modulus of the matrix is Using Equation 3173b Thus from Equation 3172 the ultimate longitudinal compressive strength is Using shear stress failure of fibers mode the ultimate longitudinal com pressive strength from Equation 3175 is Taking the minimum value of the preceding the ultimate longitudinal compressive strength is predicted as Experimental evaluation The compressive strength of a lamina has been found by several different methods A highly recommended method is the IITRI Illinois Institute of Technology Research Institute compression test24 Figure 330 shows the ASTM D3410 Celanese IITRI fixture mounted in a test frame A specimen Figure 331 consists generally of 16 to 20 plies of 0 lamina that are 64 mm 14 in wide and 127 mm 5 in long Strain gages are mounted in the longitudinal direction on both faces of the specimen to check for parallelism of the edges and ends The specimen is compressed at a rate of 05 to 1 mmmin 002 to 004 inmin A total of 40 to 50 data points for stress and strain are taken until the specimen fails The stress in the longitudinal direction is plotted as a function of longitudinal strain and is shown for a typical graphiteepoxy lamina in Figure 332 The data are reduced using linear regression and the modulus is the initial slope of the stressstrain curve From Figure 332 the following values are obtained G m 1 308 GPa S MPa C 2 9 1 308 10 1 0 7 4360 min σ1 21349 4360 4360 C ult MPa σ1 6 6 2 35 10 0 7 34 10 0 3 6 C ult 9 4 MPa σ1 69 4 C ult MPa 1343bookfm Page 283 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 284 Mechanics of Composite Materials Second Edition 343 Transverse Tensile Strength A mechanics of materials approach model for finding the transverse tensile strength of a unidirectional lamina is given next25 Assumptions used in the model include A perfect fibermatrix bond Uniform spacing of fibers FIGURE 330 IITRI fixture mounted in a test frame for finding the compressive strengths of a lamina Data reprinted with permission from Experimental Characterization of Advanced Composites Carlsson LA and Pipes RB Technomic Publishing Co Inc 1987 p 76 Copyright CRC Press Boca Raton FL E GPa c 1 199 σ1 1908 c ult MPa and ε1 0 9550 c ult 1343bookfm Page 284 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Micromechanical Analysis of a Lamina 285 The fiber and matrix follow Hookes law There are no residual stresses Assume a plane model of a composite as shown by the shaded portion in Figure 333 In this case s distance between center of fibers d diameter of fibers The transverse deformations of the fiber δf the matrix δm and the com posite δc are related by 3176 FIGURE 331 Geometry of a longitudinal compressive strength specimen Data reprinted with permission from Experimental Characterization of Advanced Composites Carlsson LA and Pipes RB Tech nomic Publishing Co Inc 1987 p 76 Copyright CRC Press Boca Raton FL w L1 L2 L1 mm 1271 127 15 127 01 or 64 01 L2 mm w mm Strain gage Specimen dimensions δ δ δ c f m 1343bookfm Page 285 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 286 Mechanics of Composite Materials Second Edition FIGURE 332 Stressstrain curve for a 024 graphiteepoxy laminate under a longitudinal compressive load Data courtesy of Dr RY Kim University of Dayton Research Institute Dayton OH FIGURE 333 Representative volume element to calculate transverse tensile strength of a unidirectional lamina 2000 1500 1000 500 0 0 02 04 Longitudinal compressive strain Longitudinal compressive stress MPa 06 08 1 Matrix Fiber σ2 σ2 d d s Fiber Matrix Matrix 1343bookfm Page 286 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Micromechanical Analysis of a Lamina 287 Now by the definition of strain the deformations are related to the trans verse strains 3177a 3177b 3177c where εcfm the transverse strain in the composite fiber and matrix respectively Substituting the expressions in Equation 382 in Equation 381 we get 3178 Now under transverse loading one assumes that the stresses in the fiber and matrix are equal see derivation of transverse Youngs modulus in Sec tion 3212 Then the strains in the fiber and matrix are related through Hookes law as 3179 Substituting the expression for the transverse strain in the fiber εf in Equation 3178 the transverse strain in the composite 3180 If one assumes that the transverse failure of the lamina is due to the failure of the matrix then the ultimate transverse failure strain is 3181 where ultimate tensile failure strain of the matrix The ultimate transverse tensile strength is then given by 3182 where is given by Equation 3181 The preceding expression assumes that the fiber is perfectly bonded to the matrix If the adhesion δ ε c s c δ ε f d f δ ε m m s d ε ε ε c f m d s d s 1 E E f f m m ε ε ε ε c m f m d s E E d s 1 ε ε 2 1 T ult m f m T ult d s E E d s εm T ult σ ε 2 2 2 T ult T ult E ε2 T ult 1343bookfm Page 287 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 288 Mechanics of Composite Materials Second Edition between the fiber and matrix is poor the composite transverse strength will be further reduced Example 315 Find the ultimate transverse tensile strength for a unidirectional glassepoxy lamina with a 70 fiber volume fraction Use properties of glass and epoxy from Table 31 and Table 32 respectively Assume that the fibers are circular and arranged in a square array Solution From Example 314 the ultimate transverse tensile strain of the lamina is the lower estimate from using Equation 3170 and Equation 3171 From Example 34 the transverse Youngs modulus of the lamina is E2 1037 GPa Using Equation 3182 the ultimate transverse tensile strength of the lamina is Experimental evaluation The procedure for finding the transverse tensile strength is the same as for finding the longitudinal tensile strength Only the specimen dimensions differ The standard width of the specimen is 254 mm 1 in and 8 to 16 plies are used This is mainly done to increase the amount of load required to break the specimen Figure 334 shows the typical stressstrain curve for a 90 graphitepeek laminate From Figure 334 the following data are obtained E2 9963 GPa Discussion Predicting transverse tensile strength is quite complicated Under a transverse tensile load factors other than the individual properties of the fiber and matrix are important These include the bond strength between the fiber and the matrix the presence of voids and the presence ε2 2 0 1983 10 T ult σ2 9 2 10 37 10 0 1983 10 20 56 T ult MPa σ2 53 28 T ult MPa ε2 0 5355 T ult 1343bookfm Page 288 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Micromechanical Analysis of a Lamina 289 of residual stresses due to thermal expansion mismatch between the fiber and matrix Possible modes of failure under transverse tensile stress include matrix tensile failure accompanied by fiber matrix debonding and or fiber splitting 344 Transverse Compressive Strength Equation 3182 which was developed for evaluating transverse tensile strength can be used to find the transverse compressive strengths of a lamina The actual compressive strength is again lower due to imperfect fibermatrix interfacial bond and longitudinal fiber splitting Using com pressive parameters in Equation 3182 3183 where 3184 ultimate compressive failure strain of matrix FIGURE 334 Stressstrain curve for a 9016 graphiteepoxy laminate under a transverse tensile load Data courtesy of Dr RY Kim University of Dayton Research Institute Dayton OH 60 40 20 0 0 02 04 Transverse tensile strain Transverse tensile stress MPa 06 σ ε 2 2 2 C ult C ult E ε ε 2 1 C ult m f m C ult d s E E d s εm C ult 1343bookfm Page 289 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 290 Mechanics of Composite Materials Second Edition Example 316 Find the ultimate transverse compressive strength of a glassepoxy lamina with 70 fiber volume fraction Use the properties of glass and epoxy from Table 31 and Table 32 respectively Assume that the fibers are circular and packed in a square array Solution From Table 31 the Youngs modulus of the fiber is Ef 85 GPa From Table 32 the Youngs modulus of the matrix is Em 34 GPa and the ultimate compressive strength of the matrix is From Example 34 the transverse Youngs modulus is E2 1037 GPa From Example 314 the fiber diameter to fiber spacing ratio is The ultimate compressive strain of the matrix is From Equation 3184 the ultimate transverse compressive strain of the lamina is and from Equation 3183 the ultimate transverse compressive strength is Experimental evaluation The procedure for finding the transverse compres sive strength is the same as that for finding the longitudinal compressive strength The only difference is in the specimen dimensions The width of the specimen is 127 mm 12 in and 30 to 40 plies are used in the test σm C ult MPa 102 d s 0 9441 εm C 102 10 3 4 10 0 03 6 9 ε2 9 9 0 9441 3 4 10 85 10 1 0 9441 C 0 03 0 2810 10 2 σ2 9 2 10 37 10 0 2810 10 29 14 C ult MPa 1343bookfm Page 290 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Micromechanical Analysis of a Lamina 291 Figure 335 shows the typical stressstrain curve for a 90 graphiteepoxy laminate From Figure 335 the following data are obtained26 Discussion Methods for predicting transverse compressive strength are also not yet satisfactory Several modes of failure possible under a transverse compressive stress include matrix compressive failure matrix shear failure and matrix shear failure with fibermatrix debonding andor fiber crushing 345 InPlane Shear Strength The procedure for finding the ultimate shear strength for a unidirectional lamina using a mechanics of materials approach follows that described in Section 343 Assume that one is applying a shear stress of magnitude τ12 FIGURE 335 Stressstrain curve for a 9040 graphiteepoxy laminate under a transverse compressive load perpendicular to the fibers Data reprinted with permission from Experimental Characterization of Advanced Composites Carlsson LA and Pipes RB Technomic Publishing Co Inc 1987 p 79 240 200 160 120 80 40 05 1 15 2 Transverse compressive strain Transverse compressive stress MPa 25 3 35 0 0 E GPa c 2 93 σ2 198 c ult MPa ε2 2 7 c ult 1343bookfm Page 291 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 292 Mechanics of Composite Materials Second Edition and then that the shearing deformation in the representative element is given by the sum of the deformations in the fiber and matrix 3185 By definition of shearing strain 3186a 3186b and 3186c where γ12cfm the inplane shearing strains in the composite fiber and matrix respectively Substituting the Equation 3186ac in Equation 3185 3187 Now under shearing stress loading one assumes that the shear stress in the fiber and matrix are equal see derivation of shear modulus in Section 3314 Then the shearing strains in the fiber and matrix are related as 3188 Substituting the expression for γ12f from Equation 3188 in Equation 3187 3189 If one assumes that the shear failure is due to failure of the matrix then 3190 where γ12m ult ultimate shearing strain of the matrix Δ Δ Δ c f m Δc c s γ 12 Δ f f d γ 12 Δm m s d γ 12 γ γ γ 12 12 12 1 c f m d s d s γ γ 12 12 m m f f G G γ γ 12 12 1 c m f m d s G G d s γ γ 12 12 1 ult m f mult d s G G d s 1343bookfm Page 292 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Micromechanical Analysis of a Lamina 293 The ultimate shear strength is then given by 3191 Example 317 Find the ultimate shear strength for a glassepoxy lamina with 70 fiber volume fraction Use properties for glass and epoxy from Table 31 and Table 32 respectively Assume that the fibers are circular and arranged in a square array Solution From Example 36 the shear modulus of the fiber is Gf 3542 GPa the shear modulus of the matrix is Gm 1308 GPa and the inplane shear modulus of the lamina is G12 4014 GPa From Example 314 the fiber diameter to fiber spacing ratio is From Table 32 the ultimate shear strength of the matrix is τ12mult 34 MPa Then the ultimate shearing strain of the matrix is τ γ 12 12 12 12 1 ult ult m f G G d s G G d s γ 12 mult d s 0 9441 γ 12 6 9 1 34 10 1 308 10 0 2599 10 mult 1343bookfm Page 293 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 294 Mechanics of Composite Materials Second Edition Using Equation 3191 the ultimate inplane shear strength of the unidi rectional lamina is Experimental determination One of the most recommended methods27 for calculating the inplane shear strength is the 452S laminated tensile cou pon Figure 336 A 452S laminate is an eightply laminate with 45 45454545454545 distribution of plies on top of each other FIGURE 336 Schematic of a 452S laminate shear test See Section 42 of Chapter 4 for an explanation on laminate codes σx σx 45 45 1 2 3 45 45 Straingage rosette τ12 9 9 4 014 10 0 9441 1 308 10 35 42 10 ult 9 1 1 0 9441 0 2599 10 9 469 MPa 1343bookfm Page 294 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Micromechanical Analysis of a Lamina 295 An axial stress σx is applied to the eightply laminate the axial strain εx and transverse strain εy are measured If the laminate fails at a load of σxult the ultimate shear strength of a unidirectional lamina is given by 3192 and the ultimate shear strain of a unidirectional lamina is 3193 An eightply 452S laminate is used for several reasons First according to maximum stress and strain failure theories of Chapter 2 each lamina fails in the shear mode and at the same load The stress at which it fails is simply twice the shear strength of a unidirectional lamina and is independent of the other mechanical properties of the lamina as reflected in Equation 3192 Second the shear strain is measured simply by strain gages in two perpendicular directions and does not require the values of elastic constants of the lamina Equation 3192 and Equation 3193 can be derived using concepts from Chapter 4 and Chapter 5 The inplane shear strength is simply half of the maximum uniaxial stress that can be applied to the laminate The initial slope of the τ12 vs γ12 curve gives the shear modulus G12 A total of 40 to 50 points are taken for the stress and strains until the specimen fails From Figure 337 the following values are obtained for a typical graphiteepoxy lamina Discussion The prediction of the ultimate shear strength is complex Similar parameters such as weak interfaces the presence of voids and Poissons ratio mismatch make modeling quite complex Theoretical methods for obtaining the strength parameters also include statistical and advanced methods Statistical methods include accounting for variations in fiber strength fibermatrix adhesion voids fiber spacing fiber diameter alignment of fibers etc Advanced methods use elasticity finite element methods boundary element methods finite difference methods etc τ σ 12 2 ult x ult γ ε ε 12 ult x ult y ult G GPa 12 5 566 τ12 87 57 MPa γ 12 2 619 ult 1343bookfm Page 295 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 296 Mechanics of Composite Materials Second Edition 35 Coefficients of Thermal Expansion When a body undergoes a temperature change its dimensions relative to its original dimensions change in proportion to the temperature change The coefficient of thermal expansion is defined as the change in the linear dimen sion of a body per unit length per unit change of temperature For a unidirectional lamina the dimensions changes differ in the two directions 1 and 2 Thus the two coefficients of thermal expansion are defined as α1 linear coefficient of thermal expansion in direction 1 mmC in inF α2 linear coefficient of thermal expansion in direction 2 mmC in inF The following are the expressions developed for the two thermal expan sion coefficients using the thermoelastic extremum principle28 3194 FIGURE 337 Shear stressshear strain curve obtained from a 452S graphiteepoxy laminate under a tensile load Data courtesy of Dr RY Kim University of Dayton Research Institute Dayton OH 100 75 50 25 0 0 1 2 Shear strain Shear stress MPa 3 α α α 1 1 1 E E V E V f f f m m m 1343bookfm Page 296 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Micromechanical Analysis of a Lamina 297 3195 where αf and αm are the coefficients of thermal expansion for the fiber and the matrix respectively 351 Longitudinal Thermal Expansion Coefficient As an example Equation 3194 can be derived using the mechanics of materials approach29 Consider the expansion of a unidirectional lamina in the longitudinal direction under a temperature change of ΔT If only the temperature ΔT is applied the unidirectional lamina has zero overall load F1 in the longitudinal direction Then 3196 3197 where Acfm the crosssectional area of composite fiber and matrix respectively σ1fm the stress in composite fiber and matrix respectively Although the overall load in the longitudinal direction 1 is zero stresses are caused in the fiber and the matrix by the thermal expansion mismatch between the fiber and the matrix These stresses are 3198a and 3198b Substituting Equation 3198a and Equation 3198b in Equation 3197 and realizing that the strains in the fiber and matrix are equal εf εm ε1 3199 For free expansion in the composite in the longitudinal direction 1 the longitudinal strain is 3200 α ν α ν α α ν 2 1 12 1 1 f f f m m m V V F A A A c f f m m 1 1 0 σ σ σ σ σ f f m m V V 0 σ ε α f f f f E T Δ σ ε α m m m m E T Δ ε α α f f f f m m m f f m m E V E V E V E V T Δ ε α 1 1 ΔT 1343bookfm Page 297 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 298 Mechanics of Composite Materials Second Edition Because the strains in the fiber and composite are also equal ε1 εf from Equation 3199 and Equation 3200 Using Equation 334 for the definition of longitudinal Youngs modulus 3201 The longitudinal coefficient of thermal expansion can be rewritten as 3202 which shows that it also follows the rule of mixtures based on the weighted mean of αEE1 of the constituents 352 Transverse Thermal Expansion Coefficient Due to temperature change ΔT assume that the compatibility condition that the strain in the fiber and matrix is equal in direction 1 that is 3203 Now the stress in the fiber in the longitudinal direction 1 is 3204 and the stress in the matrix in longitudinal direction 1 is 3205 α α α 1 f f f m m m f f m m E V E V E V E V α α α 1 1 1 E E V E V f f f m m m α α α 1 1 1 f f f m m m E E V E E V m f 1 σ α α f f f f f f E E E T 1 1 1 1 Δ σ α α m m m m m m E E E T 1 1 1 1 Δ 1343bookfm Page 298 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Micromechanical Analysis of a Lamina 299 The strains in the fiber and matrix in the transverse direction 2 are given by using Hookes law 3206 3207 The transverse strain of the composite is given by the rule of mixtures as 3208 Substituting Equation 3206 and Equation 3207 in Equation 3208 3209 and because 3210 we get 3211 Substituting 3212 in the preceding equation it can be rewritten as 3213 f f f f f T E 2 1 α ν σ Δ m m m m m T E 2 1 α ν σ Δ 2 2 2 f f m m V V 2 1 α ν α α f f f f f f T E T E V Δ Δ α ν α α m m m m m m T E T E V Δ Δ 1 2 2 α ΔT α α ν α α α ν α α 2 1 1 f f f f m m m m V V ν ν ν 12 f f m m V V α ν α ν α α ν 2 1 12 1 1 f f f m m m V V 1343bookfm Page 299 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 300 Mechanics of Composite Materials Second Edition Example 318 Find the coefficients of thermal expansion for a glassepoxy lamina with 70 fiber volume fraction Use the properties of glass and epoxy from Table 31 and Table 32 respectively Solution From Table 31 the Youngs modulus of the fiber is Ef 85 GPa and the Poissons ratio of the fiber is νf 02 The coefficient of thermal expansion of the fiber is αf 5 106 mmC From Table 32 the Youngs modulus of the matrix is Em 34 GPa the Poissons ratio of the matrix is νm 03 and the coefficient of thermal expansion of the matrix is αm 63 106 mmC From Example 33 the longitudinal Youngs modulus of the unidirectional lamina is E1 6052 GPa From Example 35 the major Poissons ratio of the unidirectional lamina is ν12 02300 Now substituting the preceding values in Equation 3194 and Equation 3195 the coefficients of thermal expansion are 1343bookfm Page 300 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Micromechanical Analysis of a Lamina 301 In Figure 338 the two coefficients of thermal expansion of glassepoxy are plotted as a function of fiber volume fraction It should be noted that the longitudinal thermal expansion coefficient is lower than the transverse thermal expansion coefficient in polymeric matrix composites Also in some cases the thermal expansion coefficient of the fibers is negative and it is thus possible for a lamina to have zero thermal expansions in the fiber directions This property is widely used in the man ufacturing of antennas doors etc when dimensional stability in the pres ence of wide temperature fluctuations is desired Experimental determinations The linear coefficients of thermal expansion are determined experimentally by measuring the dimensional changes in a FIGURE 338 Longitudinal and transverse coefficients of thermal expansion as a function of fiber volume fraction for a glassepoxy unidirectional lamina Properties of glass and epoxy from Table 31 and Table 32 α1 9 6 9 6 1 60 52 10 5 10 85 10 0 7 63 10 3 4 10 0 3 5 978 10 9 6 m m C α2 6 6 1 0 2 5 0 10 0 7 1 0 3 63 10 00 3 5 978 10 0 23 27 40 10 6 6 m m C 30 20 10 0 0 02 04 Fiber volume fraction Vf Coefficients of thermal expansion μmmC 06 08 1 α2 α1 1343bookfm Page 301 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 302 Mechanics of Composite Materials Second Edition lamina that is free of external stresses A test specimen is made of a 50 50 mm 2 in 2 in eightply laminated unidirectional composite Figure 339 Two strain gages are placed perpendicular to each other on the specimen A temperature sensor is also placed The specimen is put in an oven and the temperature is slowly increased Strain and temperature measurements are taken and plotted as a function of each other as given in Figure 340 The data are reduced using linear regression The slope of the two straintem perature curves directly gives the coefficient of thermal expansion From Figure 340 the following values are obtained for a typical graphite epoxy laminate26 α1 13 106 mmC α2 339 106 mmC FIGURE 339 Unidirectional graphiteepoxy specimen with strain gages and temperature sensors for finding coefficients of thermal expansion Reprinted with permission from Experimental Characterization of Advanced Composites Carlsson LA and Pipes RB Technomic Publishing Co Inc 1987 p 98 Copyright CRC Press Boca Raton FL 1343bookfm Page 302 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Micromechanical Analysis of a Lamina 303 36 Coefficients of Moisture Expansion When a body absorbs water as is the case for resins in polymeric matrix composites it expands The change in dimensions of the body are measured by the coefficient of moisture expansion defined as the change in the linear dimension of a body per unit length per unit change in weight of moisture content per unit weight of the body Similar to the coefficients of thermal expansion there are two coefficients of moisture expansion one in the lon gitudinal direction 1 and the other in the transverse direction 2 β1 linear coefficient of moisture expansion in direction 1 mmkg kg ininlblb β2 linear coefficient of moisture expansion in direction 2 mmkg kg ininlblb The following are the expressions for the two coefficients of moisture expansion30 FIGURE 340 Induced strain as a function of temperature to find the coefficients of thermal expansion of a unidirectional graphiteepoxy laminate Reprinted with permission from Experimental Charac terization of Advanced Composites Carlsson LA and Pipes RB Technomic Publishing Co Inc 1987 p 102 Copyright CRC Press Boca Raton FL 12 1 08 06 ε2 ε1 04 02 0 0220 60 100 140 Temperature C Induced strain 180 220 1343bookfm Page 303 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 304 Mechanics of Composite Materials Second Edition 3214 3215 where ΔCf the moisture concentration in the fiber kgkg lblb ΔCm the moisture concentration in the matrix kgkg lblb βf the coefficient of moisture expansion of the fiber mmkg kg ininlblb βm the coefficient of moisture expansion of the matrix mm kgkg ininlblb Note that unlike the coefficients of thermal expansion the content of moisture enters into the formula because the moisture absorption capacity in each constituent can be different However in most polymeric matrix composites fibers do not absorb or deabsorb moisture so the expressions for coefficients of moisture expansion do become independent of moisture contents Substituting ΔCf 0 in Equation 3214 and Equation 3215 3216 3217 Further simplification for composites such as graphiteepoxy with high fibertomatrix moduli ratio EfEm and no moisture absorption by fibers leads to 3218 3219 Similar to the derivation for the longitudinal coefficient of thermal expan sion in Section 35 Equation 3214 can be derived using the mechanics of materials approach Consider the expansion of a unidirectional lamina in the longitudinal direction because of change in moisture content in the compos ite The overall load in the composite F1 is zero that is β β β ρ ρ ρ 1 1 f f f f m m m m f f f m m m c C V E C V E E C V C V Δ Δ Δ Δ β ν β ν β ρ ρ 2 1 1 V C V C V C V f f f f m m m m m m m f f Δ Δ Δ ΔC f c ρ β ν 1 12 β ρ ρ β 1 1 E E m c m m β ν ρ ρ β β ν 2 1 12 1 m c m m β1 0 and β ν ρ ρ β 2 1 m c m m 1343bookfm Page 304 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Micromechanical Analysis of a Lamina 305 3220 where Acfm the crosssectional areas of the fiber matrix and composite respectively σ1fm the stresses in the fiber matrix and composite respectively The stresses in the fiber and matrix caused by moisture are 3221 3222 Substituting Equation 3221 and 3222 in Equation 3220 and knowing that the strains in the fiber and matrix are equal εf εm 3223 For free expansion of the composite in the longitudinal direction the longitudinal strain is 3224 where ΔCc the moisture concentration in composite Because the strains in the fiber and the matrix are equal 3225 Equation 3225 can be simplified by relating the moisture concentration in the composite ΔCc to the moisture concentration in the fiber ΔCf and the matrix ΔCm The moisture content in the composite is the sum of the moisture contents in the fiber and the matrix 3226 where wcfm the mass of composite fiber and matrix respectively Thus F A A A c f f m m 1 1 0 σ σ σ and σ σ f f m m V V 0 σ ε β f f f f f E C Δ σ ε β m m m m m E C Δ ε β β f f f f f m m m m f f m m C V E C V E E V E V Δ Δ ε β 1 1 ΔCc β β β 1 f f f f m m m m f f m m c C V E C V E E V E V C Δ Δ Δ Δ Δ Δ C w C w C w c c f f m m 1343bookfm Page 305 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 306 Mechanics of Composite Materials Second Edition 3227 where Wfm the mass fractions of the fiber and matrix respectively Substituting Equation 3227 in Equation 3225 3228 Using Equation 34 and Equation 334 one can rewrite Equation 3228 in terms of fiber volume fractions and the longitudinal Youngs modulus as 3229 Example 319 Find the two coefficients of moisture expansion for a glassepoxy lamina with 70 fiber volume fraction Use properties for glass and epoxy from Table 31 and Table 32 respectively Assume that glass does not absorb moisture Solution From Table 31 the density of the fiber is ρf 2500 kgm3 From Table 32 the density of the matrix is ρm 1200 kgm3 the swelling coefficient of the matrix is βm 033 mmkgkg and the Youngs modulus of the matrix is Em 34 GPa The Poissons ratio of the matrix is νm 03 Δ Δ Δ C C W C W c f f m m β β β 1 f f f f m m m m f f m m f f m C V E C V E E V E V C W C Δ Δ Δ Δ Wm β β β ρ ρ ρ 1 1 f f f f m m m m f f f m m m c C V E C V E E C V C V Δ Δ Δ Δ 1343bookfm Page 306 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Micromechanical Analysis of a Lamina 307 From Example 31 the density of the composite is ρc 2110 kgm3 From Example 33 the longitudinal Youngs modulus of the lamina is E1 6052 GPa From Example 35 the major Poissons ratio is ν12 0230 Thus the longitudinal coefficient of moisture expansion from Equation 3216 is and the transverse coefficient of moisture expansion from Equation 3217 is Experimental determination A specimen is placed in water and the moisture expansion strain is measured in the longitudinal and transverse directions Because moisture attacks strain gage adhesives micrometers are used to find the swelling strains 37 Summary After developing the concepts of fiber volume and weight fractions we developed equations for density and void content We found the four elastic moduli constants of a unidirectional lamina using three analytical approaches strength of materials HalphinTsai and elasticity Analytical models and experimental techniques for the five strength parameters the two coefficients of thermal expansion and the two coefficients of moisture expansion for a unidirectional lamina were discussed β1 9 9 3 4 10 60 52 10 2110 1200 0 33 0 3260 10 1 m m kg kg β2 1 1 0 3 2110 1200 0 33 0 3260 10 0 23 00 0 7468 m m kg kg 1343bookfm Page 307 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 308 Mechanics of Composite Materials Second Edition Key Terms Volume fraction Weight mass fraction Density Void volume fraction Void content Elastic moduli Array packing HalphinTsai equations Elasticity models Transversely isotropic fibers Strength ASTM standards Failure modes IITRI compression test Shear test Coefficient of thermal expansion Coefficient of moisture expansion Exercise Set 31 The weight fraction of glass in a glassepoxy composite is 08 If the specific gravity of glass and epoxy is 25 and 12 respectively find the 1 Fiber and matrix volume fractions 2 Density of the composite 32 A hybrid lamina uses glass and graphite fibers in a matrix of epoxy for its construction The fiber volume fractions of glass and graphite are 40 and 20 respectively The specific gravity of glass graphite and epoxy is 26 18 and 12 respectively Find 1 Mass fractions 2 Density of the composite 33 The acid digestion test left 2595 g of fiber from a composite specimen weighing 3697 g The composite specimen weighs 1636 g in water If the specific gravity of the fiber and matrix is 25 and 12 respec tively find the 1 Theoretical volume fraction of fiber and matrix 2 Theoretical density of composite 1343bookfm Page 308 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Micromechanical Analysis of a Lamina 309 3 Experimental density 4 Weight fraction of fiber and matrix 5 Void fraction 34 A resin hybrid lamina is made by reinforcing graphite fibers in two matrices resin A and resin B The fiber weight fraction is 40 for resin A and resin B the weight fraction is 30 each If the specific gravity of graphite resin A and resin B is 12 26 and 17 respec tively find 1 Fiber volume fraction 2 Density of composite 35 Find the elastic moduli of a glassepoxy unidirectional lamina with 40 fiber volume fraction Use the properties of glass and epoxy from Table 33 and Table 34 respectively 36 Show that if the fibers are much stiffer than the matrix that is Gf Gm 37 Assume that fibers in a unidirectional lamina are circularly shaped and in a square array Calculate the ratio of fiber diameter to fiber centertocenter spacing ratio in terms of the fiber volume fraction 38 Circular graphite fibers of 10 μm diameter are packed in a hexagonal array in an epoxy matrix The fiber weight fraction is 50 Find the fibertofiber spacing between the centers of the fibers The density of graphite fibers is 1800 kgm3 and epoxy is 1200 kgm3 39 Find the elastic moduli for problem 35 using HalphinTsai equa tions Assume that the fibers are circularly shaped and are in a square array Compare your results with those of problem 35 310 A unidirectional glassepoxy lamina with a fiber volume fraction of 70 is replaced by a graphiteepoxy lamina with the same longitu dinal Youngs modulus Find the fiber volume fraction required in the graphiteepoxy lamina Use properties of glass graphite and epoxy from Table 31 and Table 32 311 Sometimes the properties of a fiber are determined from the measured properties of a unidirectional lamina As an example find the exper imentally determined value of the Poissons ratio of an isotropic fiber from the following measured properties of a unidirectional lamina 1 Major Poissons ratio of composite 027 2 Poissons ratio of the matrix 035 3 Fiber volume fraction 065 G G V m f 12 1 1343bookfm Page 309 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 310 Mechanics of Composite Materials Second Edition 312 Using elasticity model equations find the elastic moduli of a glass epoxy unidirectional lamina with 40 fiber volume fraction Use the properties of glass and epoxy from Table 33 and Table 34 respec tively Compare your results with those obtained by using the strength of materials approach and the HalphinTsai approach Assume that the fibers are circularly shaped and are in a square array for the HalphinTsai approach 313 A measure of degree of orthotropy of a material is given by the ratio of the longitudinal to transverse Youngs modulus Given the prop erties of glass graphite and epoxy from Table 31 and Table 32 and using the mechanics of materials approach to find the longitudinal and transverse Youngs modulus find the fiber volume fraction at which the degree of orthotropy is maximum for graphiteepoxy and glassepoxy unidirectional laminae 314 What are three common modes of failure of a unidirectional com posite subjected to longitudinal tensile load 315 Do high fiber volume fractions increase the transverse strength of a unidirectional lamina 316 Find the five strength parameters of a unidirectional glassepoxy lamina with 40 fiber volume fraction Use the properties of glass and epoxy from Table 33 and Table 34 317 A rod is designed to carry a uniaxial tensile load of 1400 N with a factor of safety of two The designer has two options for the mate rials steel or 66 fiber volume fraction graphiteepoxy Use the properties of graphite and epoxy from Table 31 and Table 32 Assume the following properties for steel Youngs modulus of steel 210 GPa Poissons ratio of steel 03 Tensile strength of steel 450 MPa Specific gravity of steel 78 The cost of graphiteepoxy is five times that of steel by weight List your material of choice if the criterion depends on just 1 Mass 2 Cost 318 Find the coefficients of thermal expansion for a 60 unidirectional glassepoxy lamina with a 60 fiber volume fraction Use properties of glass and epoxy from Table 33 and Table 34 respectively 319 If one plots the transverse coefficient of thermal expansion α2 as a function of fiber volume fraction Vf for a unidirectional glassepoxy lamina α2 αm for a certain fiber volume fraction Find this range of fiber volume fraction Use properties of glass and epoxy from Table 31 and Table 32 respectively 1343bookfm Page 310 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Micromechanical Analysis of a Lamina 311 320 Find the fiber volume fraction for which the unidirectional glass epoxy lamina transverse thermal expansion coefficient is a maxi mum Use properties of glass and epoxy from Table 31 and Table 32 respectively 321 Prove31 that it is possible to have the transverse coefficient of thermal expansion of a unidirectional lamina greater than the coefficient of thermal expansion of the matrix α2 αm only if 322 The coefficient of thermal expansion perpendicular to the fibers of a unidirectional glassepoxy lamina is given as 28 μmmC Use the properties of glass and epoxy from Table 33 and Table 34 to find the coefficient of thermal expansion of the unidirectional glass epoxy lamina in the direction parallel to the fibers 323 There are large excursions of temperature in space and thus com posites with zero or near zero thermal expansion coefficients are attractive Find the volume fraction of the graphite fibers for which the thermal expansion coefficient is zero in the longitudinal direction of a graphiteepoxy unidirectional lamina Use all the properties of graphite and epoxy from Table 31 and Table 32 respectively but assume that the longitudinal coefficient of thermal expansion of graphite fiber is 13 106 mmC 324 Find the coefficients of moisture expansion of a glassepoxy lamina with 40 fiber volume fraction Use the properties of glass and epoxy from Table 31 and Table 32 respectively 325 Assume a 60 fiber volume fraction glassepoxy lamina of cuboid dimensions 25 cm along the fibers 10 cm 0125 mm Epoxy absorbs water as much as 8 of its weight Use the properties of glass and epoxy from Table 31 and Table 32 respectively and find 1 Maximum mass of water that the specimen can absorb 2 Change in volume of the lamina if the maximum possible water is absorbed Assume that the coefficient of moisture expansion through the thick ness is the same as the coefficient of moisture expansion in the transverse direction and that the glass fibers absorb no moisture References 1 Judd NCW and Wright WW Voids and their effects on the mechanical properties of composites an appraisal SAMPE J 10 14 1978 E E E E f m f m f m f m 1 1 1 ν ν ν ν or 1343bookfm Page 311 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 312 Mechanics of Composite Materials Second Edition 2 Geier MH Quality Handbook for Composite Materials Chapman Hall Lon don 1994 3 Adams RD Damping properties analysis of composites in Engineered Mate rials Handbook vol 1 ASM International Metals Park OH 1987 4 Hashin Z Theory of fiber reinforced materials NASA tech rep contract no NAS18818 November 1970 5 Chamis LC and Sendeckyj GP Critique on theories predicting thermoelastic properties of fibrous composites J Composite Mater 2 332 1968 6 Halphin JC and Tsai SW Effect of environment factors on composite mate rials Air Force tech rep AFMLTR67423 June 1969 7 Foye RL An evaluation of various engineering estimates of the transverse properties of unidirectional composite SAMPE 10 31 1966 8 Hewitt RL and Malherbe MDDe An approximation for the longitudinal shear modulus of continuous fiber composites J Composite Mater 4 280 1970 9 Hashin Z and Rosen BW 1964 The elastic moduli of fiber reinforced mate rials ASME J Appl Mech 31 223 1964 10 Hashin Z Analysis of composite materials a survey ASME J Appl Mech 50 481 1983 11 Knott TW and Herakovich CT Effect of fiber orthotropy on effective com posite properties J Composite Mater 25 732 1991 12 Christensen RM Solutions for effective shear properties in three phase sphere and cylinder models J Mech Phys Solids 27 315 1979 13 Timoshenko SP and Goodier JN Theory of Elasticity McGrawHill New York 1970 14 Maple 90 Advancing mathematics See httpwwwmaplesoftcom 15 Hashin Z Analysis of properties of fiber composites with anisotropic constit uents ASME J Appl Mech 46 543 1979 16 Hyer MW Stress Analysis of FiberReinforced Materials WCB McGrawHill New York 1998 17 Hashin Z Theory of Fiber Reinforced Materials NASA CR1974 1972 18 Whitney JM and Riley MB Elastic properties of fiber reinforced composite materials AIAA J 4 1537 1966 19 Hill R Theory of mechanical properties of fiberstrengthened materials I Elastic behavior J Mech Phys Solids 12 199 1964 20 Agrawal BD and Broutman LJ Analysis and Performance of Fiber Composites John Wiley Sons New York 1990 21 Dow NF and Rosen BW Evaluations of filament reinforced composites for aerospace structural applications NASA CR207 April 1965 22 Schuerch H Prediction of compressive strength in uniaxial boron fiber metal matrix composites AIAA J 4 102 1966 23 Hull D An Introduction to Composite Materials Cambridge University Press 1981 24 Hofer KE Rao N and Larsen D Development of engineering data on mechanical properties of advanced composite materials Air Force tech rep AFMLTR72205 part 1 September 1972 25 Kies JA Maximum strains in the resin of fiberglass composites NRL rep No 5752 AD274560 1962 26 Carlsson LA and Pipes RB Experimental Characterization of Advanced Com posite Materials Technomic Publishing Company Inc Lancaster PA 1996 1343bookfm Page 312 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Micromechanical Analysis of a Lamina 313 27 Rosen BW A simple procedure for experimental determination of the longi tudinal shear modulus of unidirectional composites J Composite Mater 21 552 1972 28 Schepery RA Thermal expansion coefficients of composite materials based on energy principles J Composite Mater 2 380 1968 29 Greszak LB Thermoelastic properties of filamentary composites presented at AIAA 6th Structures and Materials Conference April 1965 30 Tsai SW and Hahn HT Introduction to Composite Materials Technomic Pub lishing Company Inc Lancaster PA 1980 31 Kaw AK On using a symbolic manipulator in mechanics of composites ASEE Computers Educ J 3 61 1993 1343bookfm Page 313 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 315 4 Macromechanical Analysis of Laminates Chapter Objectives Understand the code for laminate stacking sequence Develop relationships of mechanical and hygrothermal loads applied to a laminate to strains and stresses in each lamina Find the elastic moduli of laminate based on the elastic moduli of individual laminae and the stacking sequence Find the coefficients of thermal and moisture expansion of a laminate based on elastic moduli coefficients of thermal and moisture expan sion of individual laminae and stacking sequence 41 Introduction In Chapter 2 stressstrain equations were developed for a single lamina A real structure however will not consist of a single lamina but a laminate consisting of more than one lamina bonded together through their thickness Why First lamina thicknesses are on the order of 0005 in 0125 mm implying that several laminae will be required to take realistic loads a typical glassepoxy lamina will fail at about only 750 lb per inch 131350 Nm width of a normal load along the fibers Second the mechanical properties of a typical unidirectional lamina are severely limited in the transverse direction If one stacks several unidirectional layers this may be an optimum laminate for unidirectional loads However for complex loading and stiff ness requirements this would not be desirable This problem can be over come by making a laminate with layers stacked at different angles for given loading and stiffness requirements This approach increases the cost and weight of the laminate and thus it is necessary to optimize the ply angles Moreover layers of different composite material systems may be used to develop a more optimum laminate 1343bookfm Page 315 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 316 Mechanics of Composite Materials Second Edition Similar to what was done in Chapter 2 the macromechanical analysis will be developed for a laminate Based on applied inplane loads of extension shear bending and torsion stresses and strains will be found in the local and global axes of each ply Stiffnesses of whole laminates will also be calculated Because laminates can also be subjected to hygrothermal loads of temperature change and moisture absorption during processing and ser vicing stresses and strains in each ply will also be calculated due to these loads Intuitively one can see that the strengths stiffnesses and hygrother mal properties of a laminate will depend on Elastic moduli Stacking position Thickness Angle of orientation Coefficients of thermal expansion Coefficients of moisture expansion of each lamina 42 Laminate Code A laminate is made of a group of single layers bonded to each other Each layer can be identified by its location in the laminate its material and its angle of orientation with a reference axis Figure 41 Each lamina is repre sented by the angle of ply and separated from other plies by a slash sign The first ply is the top ply of the laminate Special notations are used for FIGURE 41 Schematic of a laminate Fiber direction x θ z y 1343bookfm Page 316 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Macromechanical Analysis of Laminates 317 symmetric laminates laminates with adjacent lamina of the same orientation or of opposite angles and hybrid laminates The following examples illus trate the laminate code 045906030 denotes the code for the above laminate It consists of five plies each of which has a different angle to the reference x axis A slash separates each lamina The code also implies that each ply is made of the same material and is of the same thickness Sometimes 045906030 T may also denote this laminate where the subscript T stands for a total laminate 04590 2 600 denotes the laminate above which consists of six plies Because two 90 plies are adjacent to each other 90 2 denotes them where the subscript 2 is the number of adjacent plies of the same angle 04560 s denotes the laminate above consisting of six plies The plies above the midplane are of the same orientation material and thickness as the plies below the midplane so this is a symmetric laminate The top three plies are written in the code and the subscript s outside the brackets repre sents that the three plies are repeated in the reverse order 045 s denotes this laminate which consists of five plies The num ber of plies is odd and symmetry exists at the midsurface therefore the 60 ply is denoted with a bar on the top 0 45 90 60 30 0 45 90 90 60 0 0 45 60 60 45 0 0 45 60 45 0 60 1343bookfm Page 317 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 318 Mechanics of Composite Materials Second Edition 0 Gr 45 B s denotes the above laminate It consists of six plies the 0 plies are made of graphiteepoxy and the 45 angle plies are made of boron epoxy Note the symmetry of the laminate Also the 45 notation indicates that the 0 ply should be followed by a 45 angle ply and then by a 45 angle ply A notation of 45 would indicate the 45 angle ply is followed by a 45 angle ply 43 StressStrain Relations for a Laminate 431 OneDimensional Isotropic Beam StressStrain Relation Consider a prismatic beam of crosssection A Figure 42a under a simple load P the normal stress at any crosssection is given by 41 The corresponding normal strain for a linearly elastic isotropic beam is 42 where E is the Youngs modulus of the beam Note the assumption that the normal stress and strain are uniform and constant in the beam and are dependent on the load P being applied at the centroid of the cross section Now consider the same prismatic beam in a pure bending moment M Figure 42b The beam is assumed to be initially straight and the applied loads pass through a plane of symmetry to avoid twisting Based on the elementary strength of material assumptions The transverse shear is neglected Crosssections retain their original shape The yz plane before and after bending stays the same and normal to the x axis Graphiteepoxy 0 Boronepoxy 45 Boronepoxy 45 Boronepoxy 45 Boronepoxy 45 Graphiteepoxy 0 x P A σ x P AE ε 1343bookfm Page 318 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Macromechanical Analysis of Laminates 319 Then at a distance z from the centroidal line 43 where ρ is the radius of curvature of the beam If the material is linearly elastic and isotropic 44 and FIGURE 42 A beam under a axial load b bending moment and c combined axial and bending moment P M M M P P M P x z x x x z z a b c εxx z ρ σxx Ez ρ 1343bookfm Page 319 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 320 Mechanics of Composite Materials Second Edition 45 where and is defined as the second moment of area and M is the overall bending moment Now if the same beam is under the influence of axial load P and bending moment M Figure 42c then 46 47 where ε 0 is the strain at z 0 that is the centroid line of the beam and κ curvature of the beam This shows that under a combined uniaxial and bending load the strain varies linearly through the thickness of the beam Introducing the stressstrain relations of a laminate in this manner has been important because it forms a clear basis for developing similar relationships for a laminate in the next section There the straindisplacement equations similar to Equation 47 will be developed in two dimensions 432 StrainDisplacement Equations In the previous section the axial strain in a beam was related to the midplane strain and curvature of the beam under a uniaxial load and bending In this section similar relationships will be developed for a plate under inplane loads such as shear and axial forces and bending and twisting moments Figure 43 The classical lamination theory is used to develop these rela tionships The following assumptions are made in the classical lamination theory to develop the relationships 1 Each lamina is orthotropic Each lamina is homogeneous A line straight and perpendicular to the middle surface remains straight and perpendicular to the middle surface during deformation γ xz γ yz 0 xx Mz I σ I z dA A 2 xx 1 AE P z EI M ε ε ε ρ xx z 0 1 ε ε κ xx z 0 1343bookfm Page 320 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Macromechanical Analysis of Laminates 321 The laminate is thin and is loaded only in its plane plane stress σ z τ xz τ yz 0 Displacements are continuous and small throughout the laminate u v w h where h is the laminate thickness Each lamina is elastic No slip occurs between the lamina interfaces Consider a side view of a plate in the Cartesian xyz coordinate system as shown in Figure 44 The origin of the plate is at the midplane of the plate that is z 0 Assume u 0 v 0 and w 0 to be displacements in the x y and z directions respectively at the midplane and u v and w are the displacements at any point in the x y and z directions respectively At any point other than the midplane the two displacements in the xy plane will depend on the axial location of the point and the slope of the laminate midplane with the x and y directions For example as shown in Figure 44 FIGURE 43 Resultant forces and moments on a laminate FIGURE 44 Figure showing the relationship between displacements through the thickness of a plate to midplane displacements and curvatures x x y a b Nx Mx Mxy Myx My Nxy Nyx Ny y z z Midplane Crosssection before loading Crosssection after loading h2 h2 z z x α α zα A A uo wo 1343bookfm Page 321 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 322 Mechanics of Composite Materials Second Edition 48 where 49 Thus the displacement u in the x direction is 410 Similarly taking a crosssection in the yz plane would give the displace ment in the y direction as 411 Now from the definitions of the three strains Equation 216 in the xy plane and Equation 410 and Equation 411 412a 412b and 412c u u z 0 α α w x 0 u u z w x 0 0 v v z w y o 0 εx o u x u x z w x 2 0 2 εy o v y v y z w y 2 0 2 γ xy u y v x u y v x z w x y 0 0 2 0 2 1343bookfm Page 322 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Macromechanical Analysis of Laminates 323 The straindisplacement equations 412a to 412c can be written in matrix form as 413 The two arrays on the righthand sides of Equation 413 are the definitions of the midplane strains 414 and the midplane curvatures 415 respectively Therefore the laminate strains can be written as 416 ε ε γ x y xy u x v y u y v x 0 0 0 0 z w x w y w 2 0 2 2 0 2 2 0 2 x y ε ε γ x y xy u x v y u y v 0 0 0 0 0 0 0 x κ κ κ x y xy w x w y w 2 0 2 2 0 2 2 2 0 x y ε ε γ ε ε γ x y xy x y xy 0 0 0 z x y xy κ κ κ 1343bookfm Page 323 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 324 Mechanics of Composite Materials Second Edition Equation 416 shows the linear relationship of the strains in a laminate to the curvatures of the laminate It also indicates that the strains are independent of the x and y coordinates Also note the similarity between Equation 416 and Equation 47 which was developed for the one dimensional beam Example 41 A 0010 in thick laminate is subjected to inplane loads The midplane strains and curvatures are given as follows and Find the global strains at the top surface of the laminate Solution Because the value of z in Equation 416 is measured from the midplane z 0005 in at the top surface of the laminate The global strains at the top surface from Equation 416 are ε ε γ x y xy 0 0 0 2751 1331 1125 μ in in κ κ κ x y xy 1 965 0 2385 1 773 in in ε ε γ x y xy 2751 1331 1125 10 0 005 1 965 0 2385 1 773 6 7074 2524 7740 μ in in 1343bookfm Page 324 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Macromechanical Analysis of Laminates 325 433 Strain and Stress in a Laminate If the strains are known at any point along the thickness of the laminate the stressstrain Equation 2103 calculates the global stresses in each lamina 417 The reduced transformed stiffness matrix corresponds to that of the ply located at the point along the thickness of the laminate Substituting Equation 416 into Equation 417 418 From Equation 418 the stresses vary linearly only through the thickness of each lamina Figure 45 The stresses however may jump from lamina to lamina because the transformed reducedstiffness matrix changes from ply to ply because depends on the material and orientation of the ply These global stresses can then be transformed to local stresses through the transformation Equation 294 Local strains can be transformed to global strains through Equation 299 The local stresses and strains can then be used in the failure criteria discussed in Chapter 2 to find when a laminate fails The only question remaining in the macromechanical analysis of a laminate now is how to find the midplane strains and curvatures if the FIGURE 45 Strain and stress variation through the thickness of the laminate σ σ τ x y xy Q Q Q Q Q Q Q Q 11 12 16 12 22 26 16 26 66 Q x y xy ε ε γ Q σ σ τ x y xy Q Q Q Q Q Q Q Q 11 12 16 12 22 26 16 26 66 0 0 0 11 Q z Q Q x y xy ε ε γ 12 16 12 22 26 16 26 66 Q Q Q Q Q Q Q x y xy κ κ κ Q Q z MidPlane Strain variation Stress variation Laminate 1343bookfm Page 325 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 326 Mechanics of Composite Materials Second Edition loads applied to the laminate are known This question is answered in the next section 434 Force and Moment Resultants Related to Midplane Strains and Curvatures The midplane strains and plate curvatures in Equation 416 are the unknowns for finding the lamina strains and stresses However Equation 418 gives the stresses in each lamina in terms of these unknowns The stresses in each lamina can be integrated through the laminate thickness to give resultant forces and moments or applied forces and moments The forces and moments applied to a laminate will be known so the midplane strains and plate curvatures can then be found This relationship between the applied loads and the midplane strains and curvatures is developed in this section Consider a laminate made of n plies shown in Figure 46 Each ply has a thickness of tk Then the thickness of the laminate h is 419 Then the location of the midplane is h2 from the top or the bottom surface of the laminate The zcoordinate of each ply k surface top and bottom is given by Ply 1 FIGURE 46 Coordinate locations of plies in a laminate h0 1 2 3 h2 h2 z Midplane hn hn1 hk hk1 k1 k1 tk n k h1 h2 h3 h tk k n 1 h h top surface 0 2 1343bookfm Page 326 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Macromechanical Analysis of Laminates 327 Ply k k 2 3n 2 n 1 Ply n 420 Integrating the global stresses in each lamina gives the resultant forces per unit length in the xy plane through the laminate thickness as 421a 421b 421c where h2 is the half thickness of the laminate h h t bottom surface 1 1 2 h h t top surface k k 1 1 1 2 h h t bottom surface k k 2 1 h h t top surface n n 1 2 h h bottom surface n 2 N dz x x h h σ 2 2 N dz y y h h σ 2 2 N dz xy xy h h τ 2 2 1343bookfm Page 327 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 328 Mechanics of Composite Materials Second Edition Similarly integrating the global stresses in each lamina gives the result ing moments per unit length in the xy plane through the laminate thick ness as 422a 422b 422c where Nx Ny normal force per unit length Nxy shear force per unit length Mx My bending moments per unit length Mxy twisting moments per unit length The resulting force and moment in the laminate are written in matrix form per Equation 421 and Equation 422 as 423a 423b which gives M zdz x x h h σ 2 2 M zdz y y h h σ 2 2 M zdz xy xy h h τ 2 2 x y xy h h x y xy N N N 2 2 σ σ τ dz x y xy h h x y xy M M M 2 2 σ σ τ zdz 1343bookfm Page 328 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Macromechanical Analysis of Laminates 329 424a 424b Substituting Equation 418 in Equation 424 the resultant forces and moments can be written in terms of the midplane strains and curvatures 425a 425b In Equation 425a and Equation 425b the midplane strains and plate curvatures are independent of the zcoordinate Also the transformed reduced stiffness matrix is constant for each ply Therefore Equation 425 can be rewritten as N N N dz x y xy x y xy k σ σ τ h h k n k k 1 1 M M M zdz x y xy x y xy k σ σ τ h h k n k k 1 1 N N N Q Q Q Q Q Q Q Q x y xy 11 12 16 12 22 26 16 26 66 0 0 0 Q dz k x y xy hk ε ε γ 1 1 h k n k Q Q Q Q Q Q Q Q Q k x y 11 12 16 12 22 26 16 26 66 κ κ κxy h h k n zdz k k 1 1 M M M Q Q Q Q Q Q Q Q x y xy 11 12 16 12 22 26 16 26 66 0 0 0 Q zdz k x y xy hk ε ε γ 1 1 h k n k Q Q Q Q Q Q Q Q Q k x y 11 12 16 12 22 26 16 26 66 κ κ κxy h h k n z dz k k 2 1 1 Q k 1343bookfm Page 329 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 330 Mechanics of Composite Materials Second Edition 426a 426b Knowing that and substituting in Equation 426 gives N N N Q Q Q Q Q Q Q Q x y xy 11 12 16 12 22 26 16 26 66 1 0 1 Q dz k n k h h x k k ε ε γ y xy 0 0 Q Q Q Q Q Q Q Q Q k k n 11 12 16 12 22 26 16 26 66 1 zdz h h x y xy k k 1 κ κ κ M M M Q Q Q Q Q Q Q Q x y xy 11 12 16 12 22 26 16 26 66 1 1 Q zdz k k n h h x k k ε0 0 0 ε γ y xy Q Q Q Q Q Q Q Q Q k k n 11 12 16 12 22 26 16 26 66 1 z dz h h x y xy k k 2 1 κ κ κ dz h h h h k k k k 1 1 zdz h h k k h h k k 1 2 2 1 2 1 z dz h h h h k k k k 2 3 1 3 1 1 3 1343bookfm Page 330 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Macromechanical Analysis of Laminates 331 427a 427b where 428a 428b 428c The A B and D matrices are called the extensional coupling and bending stiffness matrices respectively Combining Equation 427a and Equation 427b gives six simultaneous linear equations and six unknowns as 429 The extensional stiffness matrix A relates the resultant inplane forces to the inplane strains and the bending stiffness matrix D relates the resultant bending moments to the plate curvatures The coupling stiffness N N N A A A A A A A A x y xy 11 12 16 12 22 26 16 26 66 0 0 0 11 1 A B B x y xy ε ε γ 2 16 12 22 26 16 26 66 B B B B B B B x y xy κ κ κ M M M B B B B B B B B x y xy 11 12 16 12 22 26 16 26 66 0 0 0 11 1 B D D x y xy ε ε γ 2 16 12 22 26 16 26 66 D D D D D D D x y xy κ κ κ A Q h h i j ij ij k k k k n 1 1 1 2 6 1 2 6 B Q h h i j ij ij k k k k n 1 2 1 2 6 1 2 1 2 1 22 6 D Q h h i j ij ij k k k k n 1 3 1 2 6 1 3 1 3 1 22 6 N N N M M M A A A x y xy x y xy 11 12 166 11 12 16 12 22 26 12 22 26 16 26 66 16 2 B B B A A A B B B A A A B B 6 66 11 12 16 11 12 16 12 22 26 12 22 26 1 B B B B D D D B B B D D D B 6 26 66 16 26 66 0 0 B B D D D x y ε ε γ κ κ κ xy x y xy 0 1343bookfm Page 331 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 332 Mechanics of Composite Materials Second Edition matrix B couples the force and moment terms to the midplane strains and midplane curvatures The following are the steps for analyzing a laminated composite subjected to the applied forces and moments 1 Find the value of the reduced stiffness matrix Q for each ply using its four elastic moduli E1 E2 ν12 and G12 in Equation 293 2 Find the value of the transformed reduced stiffness matrix for each ply using the Q matrix calculated in step 1 and the angle of the ply in Equation 2104 or in Equation 2137 and Equation 2138 3 Knowing the thickness tk of each ply find the coordinate of the top and bottom surface hi i 1 n of each ply using Equation 420 4 Use the matrices from step 2 and the location of each ply from step 3 to find the three stiffness matrices A B and D from Equation 428 5 Substitute the stiffness matrix values found in step 4 and the applied forces and moments in Equation 429 6 Solve the six simultaneous equations 429 to find the midplane strains and curvatures 7 Now that the location of each ply is known find the global strains in each ply using Equation 416 8 For finding the global stresses use the stressstrain Equation 2103 9 For finding the local strains use the transformation Equation 299 10 For finding the local stresses use the transformation Equation 294 Example 42 Find the three stiffness matrices A B and D for a threeply 03045 graphiteepoxy laminate as shown in Figure 47 Use the unidirectional FIGURE 47 Thickness and coordinate locations of the threeply laminate in Example 42 and Example 43 Q Q z 75 mm 5 mm 5 mm 5 mm z 0 30 45 z 25 mm z 25 mm z 75 mm 1343bookfm Page 332 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Macromechanical Analysis of Laminates 333 properties from Table 21 of graphiteepoxy Assume that each lamina has a thickness of 5 mm Solution From Example 26 the reduced stiffness matrix for the 0 graphiteepoxy ply is From Equation 2104 the transformed reduced stiffness matrix for each of the three plies is The total thickness of the laminate is h 00053 0015 m The midplane is 00075 m from the top and the bottom of the laminate Thus using Equation 420 the locations of the ply surfaces are h0 00075 m h1 00025 m h2 00025 m h3 00075 m From Equation 428a the extensional stiffness matrix A is Q 181 8 2 897 0 2 897 10 35 0 0 0 7 17 109 Pa Q Q 0 181 8 2 897 0 2 897 10 35 0 0 0 7 17 109 Pa Q 30 109 4 32 46 54 19 32 46 23 65 20 05 54 19 20 05 36 74 109 Pa Q 45 56 66 42 32 42 87 42 32 56 66 42 87 42 87 42 87 46 59 109 Pa 1343bookfm Page 333 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 334 Mechanics of Composite Materials Second Edition From Equation 428b the coupling stiffness matrix B is A Q h h ij ij k k k k 1 1 3 A 181 8 2 897 0 2 897 10 35 0 0 0 7 17 10 0 0025 0 0075 9 109 4 32 46 54 19 32 46 23 65 20 05 54 19 20 05 36 74 10 0 0025 0 0025 9 56 66 42 32 42 87 42 32 56 66 42 87 42 87 42 887 46 59 10 0 0075 0 0025 9 A 1 739 10 3 884 10 5 663 10 3 884 10 4 5 9 8 7 8 33 10 1 141 10 5 663 10 1 141 10 4 525 1 8 8 7 8 008 Pa m B Q h h ij ij k k k k 1 2 2 1 2 1 3 B 1 2 181 8 2 897 0 2 897 10 35 0 0 0 7 17 10 0 0025 0 0075 9 2 2 1 2 109 4 32 46 54 19 32 46 23 65 20 05 54 19 20 05 36 74 10 0 0025 0 0025 9 2 2 1343bookfm Page 334 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Macromechanical Analysis of Laminates 335 From Equation 428c the bending stiffness matrix D is Example 43 A 03045 graphiteepoxy laminate is subjected to a load of Nx Ny 1000 Nm Using the properties of unidirectional graphiteepoxy from Table 21 and assuming that each lamina is 5 mm thick find 1 2 56 66 42 32 42 87 42 32 56 66 42 87 42 87 4 22 87 46 59 10 0 0075 0 0025 9 2 2 B 3 129 10 9 855 10 1 972 10 9 855 10 1 6 5 6 5 158 10 1 972 10 1 072 10 1 072 10 9 85 6 6 6 6 5 105 2 Pa m D Q h h ij ij k k k k 1 3 3 1 3 1 3 D 1 3 181 8 2 897 0 2 897 10 35 0 0 0 7 17 10 0 0025 0 0075 9 3 3 1 3 109 4 32 46 54 19 32 46 23 65 20 05 54 19 20 05 36 74 10 0 0025 0 0025 9 3 3 1 3 56 66 42 32 42 87 42 32 56 66 42 87 42 87 4 22 87 46 59 10 0 0075 0 0025 9 3 3 D 3 343 10 6 461 10 5 240 10 6 461 10 9 4 3 3 3 320 10 5 596 10 5 240 10 5 596 10 7 663 3 3 3 3 103 Pa m3 1343bookfm Page 335 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 336 Mechanics of Composite Materials Second Edition 1 Midplane strains and curvatures 2 Global and local stresses on top surface of 30 ply 3 Percentage of load Nx taken by each ply Solution 1 From Example 42 the three stiffness matrices A B and D are Because the applied load is Nx Ny 1000 Nm the midplane strains and curvatures can be found by solving the following set of six simultaneous linear equations Equation 429 A 1 739 10 3 884 10 5 663 10 3 884 10 4 5 9 8 7 8 33 10 1 141 10 5 663 10 1 141 10 4 525 1 8 8 7 8 008 Pa m B 3 129 10 9 855 10 1 072 10 9 855 10 1 6 5 6 5 158 10 1 072 10 1 072 10 1 072 10 9 85 6 6 6 6 5 105 2 Pa m D 3 343 10 6 461 10 5 240 10 6 461 10 9 4 3 3 3 320 10 5 596 10 5 240 10 5 596 10 7 663 3 3 3 3 103 Pa m3 x y xy x y xy N N N M M M 1 739 10 3 884 10 5 663 10 3 129 10 9 855 10 9 8 7 6 5 11 072 10 3 884 10 4 533 10 1 141 10 9 855 6 8 8 8 10 1 158 10 1 072 10 5 663 10 1 141 10 5 6 6 7 8 4 525 10 1 072 10 1 072 10 9 855 10 3 1 8 6 6 5 29 10 9 855 10 1 072 10 3 343 10 6 461 10 6 5 6 4 3 3 5 6 6 5 240 10 9 855 10 1 158 10 1 072 10 6 4 661 10 9 320 10 5 596 10 1 072 10 1 072 3 3 3 6 110 9 855 10 5 240 10 5 596 10 7 663 10 6 5 3 3 3 ε ε γ κ κ κ x y xy x y xy 0 0 0 1343bookfm Page 336 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Macromechanical Analysis of Laminates 337 This gives 2 The strains and stresses at the top surface of the 30 ply are found as follows First the top surface of the 30 ply is located at z h1 00025 m From Equation 416 Using the stressstrain Equation 2103 for an angle ply The local strains and local stress as in the 30 ply at the top surface are found using transformation Equation 299 as ε ε γ κ κ κ x y xy x y xy 0 0 0 3 1 23 10 3 492 10 7 598 10 7 6 7 m m 2 971 10 3 285 10 4 101 10 5 4 4 1m ε ε γ x y xy top 30 7 3 123 10 3 492 110 7 598 10 0 0025 2 971 1 6 7 00 3 285 10 4 101 10 5 4 4 2 380 10 4 313 10 1 785 10 7 6 6 m m σ σ τ x y xy top 30 109 4 32 46 54 19 3 2 46 23 65 20 05 54 19 20 05 36 74 10 9 7 6 6 2 380 10 4 313 10 1 785 10 6 930 10 7 391 10 3 381 10 4 4 4 Pa 1343bookfm Page 337 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 338 Mechanics of Composite Materials Second Edition and transformation Equation 294 as The values of global and local strains and stresses at the top middle and bottom surfaces of each ply are shown in Table 41 to Table 44 3 The portion of the load Nx taken by each ply can be calculated by integrating the stress σxx through the thickness of each ply How ever because the stress varies linearly through each ply the portion of the load Nx taken is simply the product of the stress σxx at the middle of each ply see Table 42 and the thickness of the ply TABLE 41 Global Strains mm in Example 43 Ply no Position εx εy γxy 1 0 Top Middle Bottom 8944 108 1637 107 2380 107 5955 106 5134 106 4313 106 3836 106 2811 106 1785 106 2 30 Top Middle Bottom 2380 107 3123 107 3866 107 4313 106 3492 106 2670 106 1785 106 7598 107 2655 107 3 45 Top Middle Bottom 3866 107 4609 107 5352 107 2670 106 1849 106 1028 106 2655 107 1291 106 2316 106 ε ε γ 1 2 12 2 0 7500 0 2500 0 8660 0 250 0 0 7500 0 8660 4 330 0 4330 0 5000 2 380 10 4 313 10 1 785 10 2 7 6 6 ε ε γ 1 2 12 7 6 4 837 10 4 067 10 2 636 10 6 m m σ σ τ 1 2 12 0 7500 0 2500 0 8660 0 2500 0 7500 0 8660 0 4330 0 4330 0 5000 6 930 10 7 391 10 3 381 10 4 4 4 9 973 10 4 348 10 1 890 10 4 4 4 Pa 1343bookfm Page 338 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Macromechanical Analysis of Laminates 339 Portion of load Nx taken by 0 ply 4464 1045 103 2232 Nm Portion of load Nx taken by 30 ply 1063 1055 103 5315 Nm Portion of load Nx taken by 45 ply 4903 1045 103 2452 Nm TABLE 42 Global Stresses Pa in Example 43 Ply no Position σx σy τxy 1 0 Top Middle Bottom 3351 104 4464 104 5577 104 6188 104 5359 104 4531 104 2750 104 2015 104 1280 104 2 30 Top Middle Bottom 6930 104 1063 105 1434 105 7391 104 7747 104 8102 104 3381 104 5903 104 8426 104 3 45 Top Middle Bottom 1235 105 4903 104 2547 104 1563 105 6894 104 1840 104 1187 105 3888 104 4091 104 TABLE 43 Local Strains mm in Example 43 Ply no Position ε1 ε2 γ12 1 0 Top Middle Bottom 8944 108 1637 107 2380 107 5955 106 5134 106 4313 106 3836 106 2811 106 1785 106 2 30 Top Middle Bottom 4837 107 7781 107 1073 106 4067 106 3026 106 1985 106 2636 106 2374 106 2111 106 3 45 Top Middle Bottom 1396 106 5096 107 3766 107 1661 106 1800 106 1940 106 2284 106 1388 106 4928 107 TABLE 44 Local Stresses Pa in Example 43 Ply no Position σ1 σ2 σ12 1 0 Top Middle Bottom 3351 104 4464 104 5577 104 6188 104 5359 104 4531 104 2750 104 2015 104 1280 104 2 30 Top Middle Bottom 9973 104 1502 105 2007 105 4348 104 3356 104 2364 104 1890 104 1702 104 1513 04 3 45 Top Middle Bottom 2586 105 9786 104 6285 104 2123 104 2010 104 1898 104 1638 104 9954 103 3533 103 1343bookfm Page 339 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 340 Mechanics of Composite Materials Second Edition The sum total of the loads shared by each ply is 1000 Nm 2232 5315 2452 which is the applied load in the xdirection Nx Percentage of load Nx taken by 0 ply Percentage of load Nx taken by 30 ply Percentage of load Nx taken by 45 ply 44 InPlane and Flexural Modulus of a Laminate Laminate engineering constants are another way of defining laminate stiff nesses Showing Equation 429 in short notation 430 where 223 2 1000 100 22 32 531 5 1000 100 53 15 245 2 1000 100 24 52 N M A B B D ε κ 0 N N N N x y xy M M M M x y xy 1343bookfm Page 340 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Macromechanical Analysis of Laminates 341 Inverting Equation 430 gives 431 where 432a and 432b The A B and D matrices are called the extensional compliance matrix coupling compliance matrix and bending compliance matrix respectively 441 InPlane Engineering Constants of a Laminate For a symmetric laminate B 0 and it can be shown that A A1 and D D1 Then from Equation 431 433 ε ε ε γ 0 0 0 0 x y xy κ κ κ κ x y xy ε κ 0 A B C D N M A B C D A B B D 1 C B T ε ε γ x y xy A A A A A 0 0 0 11 12 16 12 22 A A A A N N N x y xy 26 16 26 66 1343bookfm Page 341 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 342 Mechanics of Composite Materials Second Edition The preceding equations allow us to define effective inplane moduli in terms of the extensional compliance matrix A as follows2 Effective inplane longitudinal modulus Ex Apply the load Nx 0 Ny 0 Nxy 0 and then substitute in Equation 433 as 434 which gives Now the effective inplane longitudinal modulus Ex is 435 Effective inplane transverse modulus Ey Apply the load Nx 0 Ny 0 Nxy 0 and then substitute in Equation 433 as 436 which gives The effective inplane transverse modulus Ey is 437 Effective inplane shear modulus Gxy ε ε γ x y xy A A A A A 0 0 0 11 12 16 12 22 A A A A Nx 26 16 26 66 0 0 εx A Nx 0 11 E N h A N hA x x x x x σ ε0 11 11 1 ε ε γ x y xy A A A A A 0 0 0 11 12 16 12 22 A A A A Ny 26 16 26 66 0 0 εy A Ny 0 22 E N h A N hA y y y y y σ ε0 22 22 1 1343bookfm Page 342 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Macromechanical Analysis of Laminates 343 Apply Nx 0 Ny 0 Nxy 0 and then substitute in Equation 433 as 438 which gives The effective inplane shear modulus Gxy is 439 Effective inplane Poissons ratio νxy From the derivation for the effective longitudinal Youngs modulus Ex where the load applied is Nx 0 Ny 0 Nxy 0 from Equation 434 we have 440 441 The effective Poissons ratio νxy is then defined as 442 Effective inplane Poissons ratio νyx From the derivation for the effective transverse Youngs modulus Ey where the load applied is Nx 0 Ny 0 Nxy 0 from Equation 436 we have 443 444 The effective Poissons ratio νyx is then defined as ε ε γ x y xy A A A A A 0 0 0 11 12 16 12 22 A A A A Nxy 26 16 26 66 0 0 γ xy A Nxy 0 66 G N h A N hA xy xy xy xy xy τ γ 0 66 66 1 εy A Nx 0 12 εx A Nx 0 11 ν ε ε xy y x x x A N A N A A 0 0 12 11 12 11 εx A Ny 0 12 εy A Ny 0 22 1343bookfm Page 343 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 344 Mechanics of Composite Materials Second Edition 445 Note here that a reciprocal relationship exists between the two effective Poissons ratios νxy and νyx From Equation 435 and Equation 442 446a From Equation 437 and Equation 445 446b From Equation 446a and Equation 446b 447 442 Flexural Engineering Constants of a Laminate Also for a symmetric laminate the coupling matrix B 0 then from Equation 431 448 ν ε ε yx x y y y A A N N A A 0 0 12 22 12 22 νxy Ex A A hA A h 12 11 11 12 νyx Ey A A hA A h 12 22 22 12 ν ν xy x yx y E E κ κ κ x y xy D D D D D D 11 12 16 12 22 26 D D D M M M x y xy 16 26 66 1343bookfm Page 344 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Macromechanical Analysis of Laminates 345 Equation 448 allows us to define effective flexural moduli in terms of the bending compliance matrix D as follows2 Apply Mx 0 My 0 Mxy 0 and then substitute in Equation 448 as 449 which gives 450 The effective flexural longitudinal modulus is 451 Similarly one can show that the other flexural elastic moduli are given by 452 453 454 455 Flexural Poissons ratios and also have a reciprocal relationship as given by 456 κ κ κ x y xy D D D D D D 11 12 16 12 22 26 D D D Mx 16 26 66 0 0 κx D Mx 11 Ex f E M h h D x f x x 12 12 3 3 11 κ E h D y f 12 3 22 G h D xy f 12 3 66 νxy f D D 12 11 νyx f D D 12 22 νxy f νyx f ν ν xy f x f yx f y f E E 1343bookfm Page 345 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 346 Mechanics of Composite Materials Second Edition In unsymmetric laminates the stressstrain relationships in Equation 429 are not uncoupled between force and moment terms Therefore in those cases the effective inplane stiffness constants and flexural stiffness constants are not meaningful Example 44 Find the inplane and flexural stiffness constants for a threeply graphiteepoxy laminate Use the unidirectional properties of graphite epoxy from Table 21 Each lamina is 5 mm thick Solution From Example 42 the transformed reduced stiffness matrix is Then from Equation 2104 the transformed reduced stiffness matrix is The total thickness of the laminate is h 0005 3 0015 m The midplane is 00075 m from the top and bottom surfaces of the laminate Thus h0 00075 m h1 00025 m h2 00025 m h3 00075 m From Equation 428a 0 90 s Q 0 181 8 2 897 0 2 897 10 35 0 0 0 7 17 109 Pa Q 90 10 35 2 897 0 2 897 181 8 0 0 0 7 17 109 Pa A Q h h ij ij k k k k 1 1 3 1343bookfm Page 346 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Macromechanical Analysis of Laminates 347 which gives Inverting the extensional stiffness matrix A we get the extensional com pliance matrix as The inplane engineering constants are found as follows From Equation 435 from Equation 437 A 181 8 2 897 0 2 897 10 35 0 0 0 7 17 10 0 0025 0 0075 9 10 35 2 897 0 2 897 181 8 0 0 0 7 17 109 0 0025 0 0025 181 8 2 897 0 2 897 10 35 0 0 0 7 17 109 0 0075 0 0025 A 1 870 10 4 345 10 0 4 345 10 1 013 10 0 0 9 7 7 9 0 1 076 108 Pa m A 5 353 10 2 297 10 0 2 297 10 9 10 11 11 886 10 0 0 0 9 298 10 1 10 9 Pa m E hA GPa x 1 1 0 015 5 353 10 124 5 11 10 E hA GPa y 1 1 0 015 9 886 10 67 43 22 10 1343bookfm Page 347 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 348 Mechanics of Composite Materials Second Edition From Equation 439 from Equation 442 and from Equation 445 Note that the reciprocal relationship of the Poissons ratios can be verified with the preceding values From Equation 428c G hA GPa xy 1 1 0 015 9 289 10 7 17 66 9 νxy A A 12 11 11 10 2 297 10 5 353 10 0 04 292 νyx A A 12 22 11 10 2 297 10 9 886 10 0 02 323 ν ν xy x yx y E E D Q h h ij ij k k k k 1 3 3 1 3 1 3 D 1 3 181 8 2 897 0 2 897 10 35 0 0 0 7 17 10 0 0025 0 0075 9 3 3 1 3 10 35 2 897 0 2 897 181 8 0 0 0 7 17 1 0 0 0025 0 0025 9 3 3 1 3 181 8 2 897 0 2 897 10 35 0 0 0 7 17 1 0 0 0075 0 0025 9 3 3 1343bookfm Page 348 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Macromechanical Analysis of Laminates 349 which gives Inverting the bending stiffness matrix D we get The flexural engineering constants are found as follows From Equation 451 and from Equation 452 From Equation 453 from Equation 454 and from Equation 455 D 4 925 10 8 148 10 0 8 148 10 4 696 10 0 0 4 2 2 3 0 2 017 103 3 Pa m D 2 032 10 3 526 10 0 3 526 10 2 136 5 6 6 10 0 0 0 4 959 10 1 4 4 3 Pa m E h D G x f 12 12 0 015 2 032 10 175 0 3 11 3 5 Pa E h D G y f 12 12 0 015 2 136 10 16 65 3 22 3 4 Pa G h D G xy f 12 12 0 015 4 959 10 7 17 3 66 3 4 Pa νxy f D D 12 11 6 5 3 526 10 2 032 10 0 1 735 νyx f D D 12 22 6 4 3 526 10 2 136 10 0 016 51 1343bookfm Page 349 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 350 Mechanics of Composite Materials Second Edition The reciprocal relationship of the Poissons ratios can be verified with the preceding values Also note that in the preceding example of a crossply laminate the inplane shear moduli and the flexural shear moduli are the same 45 Hygrothermal Effects in a Laminate In Section 29 the hygrothermal strains were calculated for an angle and unidirectional lamina subjected to a temperature change ΔT and moisture content change ΔC As mentioned if the lamina is free to expand no residual mechanical stresses would develop in the lamina at the macromechanical level However in a laminate with various plies of different angles or mate rials each individual lamina is not free to deform This results in residual stresses in the laminate3 451 Hygrothermal Stresses and Strains Sources of hygrothermal loads include cooling down from processing tem peratures operating temperatures different from processing temperatures and humid environments such as in aircraft flying at high altitudes Each ply in a laminate gets stressed by the deformation differences of adjacent lamina Only the strains in excess of or less than the hygrothermal strains in the unrestricted lamina produce the residual stresses These strain differ ences are called mechanical strains and the stresses caused by them are called mechanical stresses The mechanical strains induced by hygrothermal loads alone are 457 where the superscript M represents the mechanical strains T stands for the free expansion thermal strain and C refers to the free expansion mois ture strains xy f x f yx f y f E E ν ν ε ε γ ε ε γ x M y M xy M x y xy ε ε γ ε ε γ x T y T xy T x C x C xy C 1343bookfm Page 350 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Macromechanical Analysis of Laminates 351 Using stressstrain Equation 2103 the hygrothermal stresses in a lamina are then given by 458 where TC stands for combined thermal and moisture effects Hygrothermal stresses induce zero resultant forces and moments in the laminate and thus in the nply laminate shown in Figure 46 459 460 From Equation 458 to Equation 460 461a and 461b Substituting Equation 457 and Equation 416 in Equation 461 gives σ σ γ x TC y TC xy TC Q Q Q Q Q Q 11 12 16 16 26 26 16 26 66 Q Q Q x M y M xy M ε ε γ σ σ τ σ σ x TC y TC xy TC h h x TC dz 2 2 0 y TC xy TC k h h k n dz k k τ 1 1 σ σ τ σ σ τ x TC y TC xy TC x TC y TC xy T zdz 0 C k h h k n h h zdz k k 1 1 2 2 Q Q Q Q Q Q Q Q Q x M y M 11 12 16 12 22 26 16 26 66 ε ε γ xy M h h k n dz k k 1 1 0 Q Q Q Q Q Q Q Q Q x M y M 11 12 16 12 22 26 16 26 66 ε ε γ xy M h h k n zdz k k 1 1 0 1343bookfm Page 351 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 352 Mechanics of Composite Materials Second Edition 462 463 The four arrays on the righthand sides of Equation 462 and Equation 463 are given by 464 465 466 467 The loads in Equation 464 to Equation 467 are called fictitious hygro thermal loads and are known One can calculate the midplane strains and curvatures by combining Equation 462 and Equation 463 A A A A A A A A A x o y 11 12 16 12 22 26 16 26 66 0 ε ε γ xy B B B B B B B B B 0 11 12 16 12 22 26 16 26 66 κ κ κ x y xy x T y T xy T N N N N N N x C y C xy C B B B B B B B B B x o y o 11 12 16 12 22 26 16 26 66 ε ε γ xy o D D D D D D D D D 11 12 16 12 22 26 16 26 666 κ κ κ x y xy x T y T xy T M M M M M M x C y C xy T N N N N T Q Q Q Q Q T x T y T xy T Δ 11 12 16 12 22 26 16 26 66 Q Q Q Q k x y xy α α α k k k k n h h 1 1 M M M M T Q Q Q Q T x T y T xy T 1 2 11 12 16 1 Δ 2 22 26 16 26 66 Q Q Q Q Q k x y xy α α α k k k k n h h 2 1 2 1 N N N N C Q Q Q Q Q C x C y C xy C Δ 11 12 16 12 22 26 16 26 66 Q Q Q Q k x y xy β β β k k k k n h h 1 1 M M M M C Q Q Q Q C x C y C xy C 1 2 11 12 16 1 Δ 2 22 26 16 26 66 Q Q Q Q Q k x y xy β β β k k k k n h h 2 1 2 1 1343bookfm Page 352 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Macromechanical Analysis of Laminates 353 468 Using Equation 416 469 one can calculate the global strains in any ply of the laminate These global strains are the actual strains in the laminate However it is the difference between the actual strains and the free expansion strains which results in mechanical stresses The mechanical strains in the kth ply are given by Equa tion 457 as 470 The mechanical stresses in the kth ply are then calculated by 471 The fictitious hygrothermal loads represent the loads in Equation 464 to Equation 467 which one can apply mechanically to induce the same stresses and strains as by the hygrothermal load Thus if both mechanical and hygrothermal loads are applied one can add the mechanical loads to the fictitious hygrothermal loads to find the plybyply stresses and strains in the laminate or separately apply the mechanical and hygrothermal loads and then add the resulting stresses and strains from the solution of the two problems Example 45 Calculate the residual stresses at the bottom surface of the 90 ply in a two ply 090 graphiteepoxy laminate subjected to a temperature change of N M N M A B B D T T C C ε κ 0 ε ε γ ε ε γ x y xy x y xy 0 0 0 z x y xy κ κ κ ε ε γ ε ε γ x M y M xy M k x y xy k x T y T xy T x C y C xy C ε ε γ ε ε γ k σ σ τ x y xy k Q Q Q Q Q Q Q Q 11 12 16 12 22 26 16 26 Q66 k x M y M xy M k ε ε γ 1343bookfm Page 353 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 354 Mechanics of Composite Materials Second Edition 75C Use the unidirectional properties of a graphiteepoxy lamina from Table 21 Each lamina is 5 mm thick Solution From Table 21 the coefficients of thermal expansion for a 0 graphiteepoxy ply are From Equation 2175 the transformed coefficients of thermal expansion are and From Example 44 the reduced transformed stiffness matrices are According to Equation 464 the fictitious thermal forces are α α α 1 2 12 7 4 0 200 10 0 225 10 0 m m C α α α x y xy 0 7 4 0 200 10 0 225 10 0 m m C α α α x y xy 90 4 7 0 225 10 0 200 10 0 m m C Q 0 181 8 2 897 0 2 897 10 35 0 0 0 7 17 GPa Q 90 10 35 2 897 0 2 897 181 8 0 0 0 7 17 GPa 1343bookfm Page 354 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Macromechanical Analysis of Laminates 355 According to Equation 465 the fictitious thermal moments are Using Equation 428 the stiffness matrices A B and D are calcu lated as N N N x T y T xy T 75 181 8 2 897 0 2 897 10 35 0 0 0 7 17 10 0 200 10 0 9 7 225 10 0 0 000 0 005 75 4 10 35 2 897 0 2 897 181 8 0 0 0 7 17 109 0 225 10 0 200 10 0 0 005 0 0 4 7 00 1 131 10 1 131 10 0 5 5 Pa m M M M x T y T xy T 1 2 75 181 8 2 897 0 2 897 10 35 0 0 0 7 17 10 0 200 10 0 9 7 225 10 0 0 000 0 005 1 2 4 2 2 75 10 35 2 897 0 2 897 10 35 0 0 0 7 17 10 0 225 10 0 200 10 0 0 0 9 4 7 05 0 000 1 538 10 1 538 10 0 2 2 2 2 Pa m A 9 608 10 2 897 10 0 2 897 10 9 608 10 0 0 8 7 7 8 0 7 170 107 Pa m 1343bookfm Page 355 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 356 Mechanics of Composite Materials Second Edition These stiffness matrices A B and D are to be used in Equation 468 to give the midplane strains and curvatures This gives B Pa 2 143 10 0 0 0 2 143 10 0 0 0 0 6 6 m2 D 8 007 10 2 414 10 0 2 414 10 8 007 10 0 0 3 2 2 3 0 5 975 102 3 Pa m N M A B B D T T ε κ 0 1 131 10 1 131 10 0 1 538 10 1 538 10 0 5 5 2 2 9 608 10 2 897 10 0 2 1 8 7 443 10 0 0 2 897 10 9 608 10 0 0 2 143 10 0 0 0 7 6 7 8 6 170 10 0 0 0 2 143 10 0 0 8 007 10 2 414 10 0 0 7 6 3 2 2 143 10 0 2 414 10 8 007 10 0 0 0 0 0 0 5 975 10 6 2 3 2 0 0 0 ε ε γ κ κ κ x y xy x y xy ε ε γ κ κ κ x y xy x y xy 0 0 0 3 907 10 3 907 10 0 1 276 10 1 276 10 0 4 4 1 1 m m m 1 1343bookfm Page 356 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Macromechanical Analysis of Laminates 357 Equation 416 gives the actual strains at the bottom surface h2 0005 of the 90 ply as The mechanical strains result in the residual stresses Thus if one subtracts the strains that would have been caused by free expansion from the actual strains one can calculate the mechanical strains Equation 2179 gives the free expansion thermal strains in the 90 ply From Equation 457 the mechanical strains at the bottom of the 90 ply are thus ε ε γ x y xy 90 4 3 907 10 3 907 10 4 1 1 0 0 005 1 276 10 1 276 10 0 1 029 10 2 475 10 0 3 4 m m ε ε γ x T y T xy T 0 225 10 0 200 10 4 7 0 75 0 16875 10 0 1500 10 0 2 5 m m ε ε γ x M y M xy M 1 029 10 2 475 10 3 4 2 5 0 0 16875 10 0 1500 10 0 0 6585 10 0 2490 10 0 3 3 1343bookfm Page 357 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 358 Mechanics of Composite Materials Second Edition The stressstrain Equation 471 gives the mechanical stresses at the bot tom surface of the 90 ply as The global strains and stresses in all the plies of the laminate are given in Table 45 and Table 46 respectively 452 Coefficients of Thermal and Moisture Expansion of Laminates The concept of finding coefficients of thermal and moisture expansion of laminates is again well suited only for symmetric laminates because in this case the coupling stiffness matrix B 0 and no bending occurs under hygrothermal loads TABLE 45 Global Strains for Example 43 Ply no Position εx εy γxy 1 0 Top Middle Bottom 2475 104 7160 105 3907 104 1029 103 7098 104 3907 104 00 00 00 2 90 Top Middle Bottom 3907 104 7098 104 1029 103 3907 104 7160 105 2475 104 00 00 00 TABLE 46 Global Stresses Pa for Example 43 Ply no Position σy σy τxy 1 0 Top Middle Bottom 4718 107 9912 106 6701 107 7535 106 9912 106 1229 107 00 00 00 2 90 Top Middle Bottom 1229 107 9912 106 7535 106 6701 107 9912 106 4718 107 00 00 00 σ σ τ x y xy 90 10 35 2 897 0 2 897 181 88 0 0 0 7 17 10 0 6585 10 0 2490 1 9 3 0 0 3 7 535 10 4 718 10 0 6 7 Pa 1343bookfm Page 358 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Macromechanical Analysis of Laminates 359 The coefficients of thermal expansion are defined as the change in length per unit length per unit change of temperature Three coefficients of thermal expansion one in direction x αx and the others in direction y αy and in the plane xy αxy are defined for a laminate Assuming ΔT 1 and C 0 472 where NT is the resultant thermal force given by Equation 464 corre sponding to ΔT 1 and ΔC 0 Similarly assuming ΔT 0 and ΔC 1 the moisture expansion coefficients can be defined as 473 where NC is the resultant moisture force given by Equation 466 corre sponding to ΔT 0 and ΔC 1 Example 46 Find the coefficients of thermal and moisture expansion of a graph iteepoxy laminate Use the properties of a unidirectional graphiteepoxy lamina from Table 21 Solution From Example 44 the extensional compliance matrix is Corresponding to a temperature change of ΔT 1C the fictitious thermal forces are α α α ε ε γ x y xy x y xy 0 0 0 ΔC T A A A A A A A A A 0 1 11 12 16 12 22 26 16 26 6 Δ 6 N N N x T y T xy T β β β ε ε γ x y xy x y xy 0 0 0 ΔT C A A A A A A A A A 0 1 11 12 16 12 22 26 16 26 6 Δ 6 N N N x C y C xy C 0 90 s A 5 353 10 2 297 10 0 2 297 10 9 10 11 11 886 10 0 0 0 9 298 10 1 10 9 Pa m 1343bookfm Page 359 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 360 Mechanics of Composite Materials Second Edition Equation 472 and the extensional compliance matrix from Example 44 then give the coefficients of thermal expansion for the laminate Corresponding to a moisture content of ΔC 1 kgkg the fictitious mois ture forces are N N N T Q Q Q Q Q Q x T y T xy T Δ 11 12 16 12 22 26 Q Q Q h k x y xy k 16 26 66 α α α k k k h 1 1 3 1 181 8 2 897 0 2 897 10 35 0 0 0 7 17 10 0 200 10 0 225 10 0 0 002 9 7 4 5 0 0075 1 10 35 2 897 0 2 897 181 35 0 0 0 7 17 10 0 225 10 0 200 10 0 0 00 9 4 7 25 0 0025 1 181 8 2 897 0 2 897 10 35 0 0 0 7 17 10 0 200 10 0 225 10 0 0 007 9 7 4 5 0 0025 1 852 10 2 673 10 0 3 3 Pa m α α α x y xy 5 353 10 2 297 10 10 11 0 2 297 10 9 886 10 0 0 0 9 298 10 11 10 9 1 852 10 2 673 10 0 3 3 9 303 10 2 600 10 0 7 6 m m C 1343bookfm Page 360 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Macromechanical Analysis of Laminates 361 Equation 473 gives the coefficients of moisture expansion for the lami nate as N N N C Q Q Q Q Q Q x C y C xy C Δ 11 12 16 12 22 26 Q Q Q h k x y xy k 16 26 66 β β β hk k n 1 1 1 181 8 2 897 0 2 897 10 35 0 0 0 7 17 10 0 0 6 0 0 0025 0 0075 9 1 10 35 2 897 0 2 897 181 35 0 0 0 7 17 10 0 6 0 0 0 0025 0 0025 9 1 181 8 2 897 0 2 897 10 35 0 0 0 7 17 10 0 0 6 0 0 0075 0 0025 9 4 842 10 7 077 10 0 7 7 Pa m β β β x y xy 5 353 10 2 297 10 10 11 0 2 297 10 9 886 10 0 0 0 9 298 10 11 10 9 4 842 10 7 077 10 0 7 7 2 430 10 6 885 10 0 2 2 m m kg kg 1343bookfm Page 361 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 362 Mechanics of Composite Materials Second Edition 453 Warpage of Laminates In laminates that are not symmetric a temperature difference results in out ofplane deformations This deformation is also called warpage4 and is cal culated by integrating the curvaturedisplacement Equation 415 474a 474b 474c From the integration of Equation 474 the outofplane deflection w can be derived Integrating Equation 474a 475 where f1y and f2y are unknown functions Substituting Equation 475 in Equation 474c 476 This gives 477 where C1 is an unknown constant of integration From Equation 475 and Equation 477 478 Substituting Equation 478 in Equation 474b 479 κx w x 2 2 κ y w y 2 2 κxy w x y 2 2 w x f y x f y x κ 2 1 2 2 κxy w x y df y dy 2 2 2 1 f y y C xy 1 1 2 κ w x xy C x f y x xy κ κ 2 1 2 2 2 κ y w y d f y dy 2 2 2 2 2 1343bookfm Page 362 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Macromechanical Analysis of Laminates 363 This gives 480 Substituting Equation 480 in Equation 478 481 The terms C1x C2y C3 are simply rigid body motion terms and one can relate the warpage to be 482 Example 47 Find the warpage in 090 graphiteepoxy laminate under a temperature change of 75C Use the properties of graphiteepoxy from Table 21 Solution From Example 45 the midplane curvatures of the laminate are given by Thus the warpage at any point xy on the plane from Equation 482 is w 06383 101 x2 06383 101 y2 Note that this warpage is calculated relative to the point xy 00 46 Summary In this chapter we introduced the laminate code for laminate stacking sequence Then we developed the theory for the elastic response of a lam inate subjected to mechanical loads such as inplane loads and bending moments and thermal and hygrothermal loads This theory allowed us to calculate plybyply global as well as local stresses and strains in each ply f y y C y C y 2 2 2 3 2 κ w x y xy C x C y C x y xy 1 2 2 2 1 2 3 κ κ κ w x y xy x y xy 1 2 2 2 κ κ κ κ κ κ x y xy 1 276 10 1 276 10 0 1 1 1m 1343bookfm Page 363 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 364 Mechanics of Composite Materials Second Edition and also to calculate the effective inplane and bending flexural moduli for a laminate Key Terms Laminate code Classical lamination theory Midplane strains and stresses Plybyply strains and stresses Inplane and flexural modulus Hygrothermal stresses in a laminate Warpage Exercise Set 41 Condense the following expanded laminate codes 1 0454590 2 0454545450 3 0906060900 4 04560450 5 4545454545454545 42 Expand the following laminate codes 1 4545S 2 4545290S 3 4503S 4 45302 5 45302 43 A laminate of 0015 in thickness under a complex load gives the following midplane strains and curvatures x y xy x y xy 0 0 0 2 1 γ κ κ κ 0 3 10 4 10 1 2 10 1 5 10 2 6 10 6 6 6 4 4 4 in in in 1 1343bookfm Page 364 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Macromechanical Analysis of Laminates 365 Find the global strains at the top middle and bottom surface of the laminate 44 Do global strains vary linearly through the thickness of a laminate 45 Do global stresses vary linearly through the thickness of a laminate 46 The global strains at the top surface of a 04560s laminate are given as and the midplane strains in this laminate are given as What are the midplane curvatures in this laminate if each ply is 0005 in thick 47 The global stresses in a threeply laminate are given at the top and bottom surface of each ply Each ply is 0005 in thick Find the resultant forces and moments on the laminate if it has a top cross section of 4 in 4 in Ply no σxx psi Top Bottom 1 2 3 3547 104 9267 103 7201 103 2983 103 1658 104 2435 104 Ply no σyy psi Top Bottom 1 2 3 2425 104 1638 104 3155 103 7087 103 9432 103 3553 104 Ply no τxy psi Top Bottom 1 2 3 2946 104 1299 104 5703 103 5564 103 1317 104 2954 104 x y xy γ 1 686 10 6 500 10 2 1 8 8 43 10 7 ε ε γ x y xy 0 0 0 6 4 8 388 10 4 762 10 3 129 10 3 1343bookfm Page 365 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 366 Mechanics of Composite Materials Second Edition 48 Find the three stiffness matrices A B and D for a 06060 glassepoxy laminate Use the properties of glassepoxy unidirec tional lamina from Table 22 and assume the lamina thickness to be 0005 in Also find the mass of the laminate if the top surface area of the laminate is 5 in 7 in Use densities of glass and epoxy from Table 33 and Table 34 respectively 49 Give expressions for the stiffness matrices A B and D for an isotropic material in terms of its thickness t Youngs modulus E and Poissons ratio ν 410 Show that for a symmetric laminate the coupling stiffness matrix is equal to zero 411 A beam is made of two bonded isotropic strips as shown in the Figure 48 The two strips are of equal thickness Find the stiffness matrices A B and D 412 Rewrite the expressions for the stiffness matrices A B and D in terms of the transformed reduced stiffness matrix elements thick ness of each ply and the location of the middle of each ply with respect to the midplane of the laminate This is called the parallel axis theorem for the laminate stiffness matrices 413 Find the local stresses at the top of the 60 ply in a 06060 graphiteepoxy laminate subjected to a bending moment of Mx 50 Nmm Use the properties of a unidirectional graphiteepoxy lam ina from Table 21 and assume the lamina thickness to be 0125 mm What is the percentage of the bending moment load taken by each of the three plies 414 Find the forces and moments required in a 06060 graphite epoxy laminate to result in bending curvature of κx 01 in1 and κy 01 in1 Use the properties of a unidirectional graphiteepoxy lamina from Table 22 and assume the lamina thickness to be 0005 in 415 Find the extensional and flexural engineering elastic moduli of a 4545s graphiteepoxy laminate Verify the reciprocal relation ships for the Poissons ratios Use the properties of a unidirectional graphiteepoxy lamina from Table 21 FIGURE 48 Laminate made of two isotropic plies Strip 1 E1 V1 Strip 2 E2 V2 1343bookfm Page 366 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Macromechanical Analysis of Laminates 367 416 Find the residual stresses at the top of the 60 ply in a 06060 graphiteepoxy laminate subjected to a temperature change of 150F Each lamina is 0005 in thick use the properties of a unidi rectional graphiteepoxy lamina from Table 22 417 For a 045s glassepoxy laminate find the coefficients of the thermal expansion Use the properties of a unidirectional glass epoxy lamina from Table 21 Assume thickness of each lamina as 0125 mm Also find the change in the volume of the laminate if the crosssectional area is 100 mm 50 mm and the temperature change is 100C 418 Find the coefficients of moisture expansion of a 045s graphite epoxy laminate The properties of a unidirectional graphiteepoxy lamina are given in Table 21 Assume thickness of each lamina as 0125 mm 419 Find the local stresses at the middle of the 30 ply in a 3045 glass epoxy laminate that is subjected to the following mechanical and hygrothermal loads Nx 108 lbin ΔT 100F ΔC 5 Use the properties of a unidirectional glassepoxy lamina given in Table 22 The thickness of each lamina is 0005 m 420 Find the difference between the vertical deflection through the thickness at the center and the four corners of a 060 graphite epoxy cuboid laminate The thickness of each ply is 0005 in and the top surface dimensions of the laminate are 20 in 10 in The laminate is subjected to a temperature change of 75F Use the properties of a unidirectional graphiteepoxy lamina given in Table 22 References 1 Ashton JE Halphin JD and Petit PH Primer on Composite Materials Anal ysis Technomic Publishing Company West Port CT 1969 2 Soni SR A digital algorithm for composite laminate analysis Fortran AFWALTR814073 WPAFB report 1983 3 Hahn HT Residual stresses in polymer matrix composite laminates J Com posite Mater 10 266 1976 4 Zewi IG Daniel IM and Gotro JT Residual stresses and warpage in wovenglassepoxy laminates Exp Mech 27 44 1987 1343bookfm Page 367 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 369 5 Failure Analysis and Design of Laminates Chapter Objectives Understand the significance of stiffness and hygrothermal and mechanical response of special cases of laminates Establish the failure criteria for laminates based on failure of indi vidual lamina in a laminate Design laminated structures such as plates thin pressure vessels and drive shafts subjected to inplane and hygrothermal loads Introduce other mechanical design issues in laminated composites 51 Introduction The design of a laminated composite structure such as a flat floor panel or a pressure vessel starts with the building block of laminae in which fiber and matrix are combined in a manufacturing process such as filament wind ing or prepregs The material of the fiber and matrix processing factors such as packing arrangements and fiber volume fraction determine the stiffness strength and hygrothermal response of a single lamina These properties can be found by using the properties of the individual constituents of the lamina or by experiments as explained in Chapter 3 Then the laminate can have variations in material systems and in stacking sequence of plies to tailor a composite for a particular application In Chapter 4 we developed analysis to find the stresses and strains in a laminate under inplane and hygrothermal loads In this chapter we will first use that analysis and failure theories studied in Chapter 2 to predict failure in a laminate Then the fundamentals learned in Chapter 4 and the failure analysis discussed in this chapter will be used to design structures using laminated composites 1343bookfm Page 369 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 370 Mechanics of Composite Materials Second Edition First special cases of laminates that are important in the design of laminated structures will be introduced Then the failure criterion analysis will be shown for a laminate Eventually we will be designing laminates mainly on the basis of optimizing for cost weight strength and stiffness Other mechanical design issues are briefly introduced at the end of the chapter 52 Special Cases of Laminates Based on angle material and thickness of plies the symmetry or antisym metry of a laminate may zero out some elements of the three stiffness matri ces A B and D These are important to study because they may result in reducing or zeroing out the coupling of forces and bending moments normal and shear forces or bending and twisting moments This not only simplifies the mechanical analysis of composites but also gives desired mechanical performance For example as already shown in Chapter 4 the analysis of a symmetric laminate is simplified due to the zero coupling matrix B Mechanically symmetric laminates result in no warpage in a flat panel due to temperature changes in processing 521 Symmetric Laminates A laminate is called symmetric if the material angle and thickness of plies are the same above and below the midplane An example of symmetric laminates is For symmetric laminates from the definition of B matrix it can be proved that B 0 Thus Equation 429 can be decoupled to give 51a 0 30 60 30 0 0 30 60 s N N N A A A A A A A A x y xy 11 12 16 12 22 26 16 26 66 0 0 0 A x y xy ε ε γ 1343bookfm Page 370 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Failure Analysis and Design of Laminates 371 51b This shows that the force and moment terms are uncoupled Thus if a laminate is subjected only to forces it will have zero midplane curvatures Similarly if it is subjected only to moments it will have zero midplane strains The uncoupling between extension and bending in symmetric laminates makes analyzing such laminates simpler It also prevents a laminate from twisting due to thermal loads such as cooling down from processing temper atures and temperature fluctuations during use such as in a space shuttle etc 522 CrossPly Laminates A laminate is called a crossply laminate also called laminates with specially orthotropic layers if only 0 and 90 plies were used to make a laminate An example of a cross ply laminate is a 090 2 090 laminate For crossply laminates A 16 0 A 26 0 B 16 0 B 26 0 D 16 0 and D 26 0 thus Equation 429 can be written as 52 In these cases uncoupling occurs between the normal and shear forces as well as between the bending and twisting moments If a crossply lami nate is also symmetric then in addition to the preceding uncoupling the coupling matrix B 0 and no coupling takes place between the force and moment terms 0 90 90 0 90 M M M D D D D D D D D x y xy 11 12 16 12 22 26 16 26 66 D x y xy κ κ κ N N N M M M A A B x y xy x y xy 11 12 0 11 12 12 22 12 22 66 66 11 12 11 1 0 0 0 0 0 0 0 0 B A A B B A B B B D D 2 12 22 12 22 66 66 0 0 0 0 0 0 0 B B D D B D ε ε γ κ κ κ x y xy x y xy 0 0 0 1343bookfm Page 371 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 372 Mechanics of Composite Materials Second Edition 523 Angle Ply Laminates A laminate is called an angle ply laminate if it has plies of the same material and thickness and only oriented at θ and θ directions An example of an angle ply laminate is 40404040 If a laminate has an even number of plies then A 16 A 26 0 However if the number of plies is odd and it consists of alternating θ and θ plies then it is symmetric giving B 0 and A 16 A 26 D 16 and D 26 also become small as the number of layers increases for the same laminate thickness This behavior is similar to the symmetric crossply laminates However these angle ply laminates have higher shear stiffness and shear strength properties than crossply laminates 524 Antisymmetric Laminates A laminate is called antisymmetric if the material and thickness of the plies are the same above and below the midplane but the ply orientations at the same distance above and below the midplane are negative of each other An example of an antisymmetric laminate is From Equation 428a and Equation 428c the coupling terms of the extensional stiffness matrix A 16 A 26 0 and the coupling terms of the bending stiffness matrix D 16 D 26 0 53 40 40 40 40 45 60 60 45 N N N M M M A A B x y xy x y xy 11 12 0 11 12 16 12 22 12 22 26 66 16 26 66 11 0 0 0 B B A A B B B A B B B B B22 16 11 12 12 22 26 12 22 16 26 66 66 0 0 0 0 B D D B B B D D B B B D ε ε γ κ κ κ x y xy x y xy 0 0 0 1343bookfm Page 372 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Failure Analysis and Design of Laminates 373 525 Balanced Laminate A laminate is balanced if layers at angles other than 0 and 90 occur only as plus and minus pairs of θ and θ The plus and minus pairs do not need to be adjacent to each other but the thickness and material of the plus and minus pairs need to be the same Here the terms A 16 A 26 0 An example of a balanced laminate is 304030303040 From Equation 428a 54 526 QuasiIsotropic Laminates For a plate of isotropic material with Youngs modulus E Poissons ratio ν and thickness h the three stiffness matrices are 55 56 30 40 30 30 30 40 N N N M M M A A B x y xy x y xy 11 12 0 11 12 16 12 22 12 22 26 26 16 26 66 11 0 0 0 B B A A B B B A B B B B B12 16 11 12 16 12 22 26 12 22 26 16 26 66 B D D D B B B D D D B B B D16 26 66 0 0 0 D D x y xy x y ε ε γ κ κ κxy A E E E E E 1 1 0 1 1 0 0 0 2 1 2 2 2 2 ν ν ν ν ν ν ν h B 0 0 0 0 0 0 0 0 0 1343bookfm Page 373 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 374 Mechanics of Composite Materials Second Edition 57 A laminate is called quasiisotropic if its extensional stiffness matrix A behaves like that of an isotropic material This implies not only that A 11 A 22 A 16 A 26 0 and but also that these stiffnesses are indepen dent of the angle of rotation of the laminate The reason for calling such a laminate quasiisotropic and not isotropic is that the other stiffness matrices B and D may not behave like isotropic materials Examples of quasi isotropic laminates include 0 60 0 4590 s and 036723672 Example 51 A 0 60 graphiteepoxy laminate is quasiisotropic Find the three stiffness matrices A B and D and show that 1 2 B 0 unlike isotropic materials 3 D matrix is unlike isotropic materials Use properties of unidirectional graphiteepoxy lamina from Table 21 Each lamina has a thickness of 5 mm Solution From Example 26 the reduced stiffness matrix Q for the 0 graphiteepoxy lamina is From Equation 2104 the transformed reduced stiffness matrices for the three plies are D E E E E 12 1 12 1 0 12 1 12 1 2 2 2 2 ν ν ν ν ν ν 0 0 0 24 1 3 E h ν A A A 66 11 12 2 A A A A A A A 11 22 16 26 66 11 12 0 2 Q 181 8 2 897 0 2 897 10 35 0 0 0 7 17 109 Pa 1343bookfm Page 374 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Failure Analysis and Design of Laminates 375 The total thickness of the laminate is h 00053 0015 m The midplane is 00075 m from the top and bottom of the laminate Thus using Equation 420 h 0 00075 m h 1 00025 m h 2 00025 m h 3 00075 m Using Equation 428a to Equation 428c one can now calculate the stiffness matrices A B and D respectively as shown in Example 42 Q 0 181 8 2 897 0 2 897 10 35 0 0 0 7 17 109 Pa Q 60 23 65 32 46 20 05 32 46 109 4 54 19 20 05 54 19 36 74 109 Pa Q 60 23 65 32 46 20 05 32 46 109 4 54 19 20 05 54 19 36 74 109 Pa A 1 146 0 3391 0 0 3391 1 146 0 0 0 0 4032 109 Pa m B 3 954 0 7391 0 5013 0 7391 2 476 1 355 0 5013 1 355 0 7391 106 2 Pa m 1343bookfm Page 375 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 376 Mechanics of Composite Materials Second Edition 1 From the extensional stiffness matrix A 04032 109 Pam A66 This behavior is similar to that of an isotropic material However a quasiisotropic laminate should give the same A matrix if a constant angle is added to each of the layers of the laminate For example adding 30 to each ply angle of the 060 laminate gives a 309030 laminate which has the same A matrix as the 0 60 laminate 2 Unlike isotropic materials the coupling stiffness matrix B of the 060 laminate is nonzero 3 In an isotropic material and In this example unlike isotropic materials D11 D22 because D11 2807 103 Pam3 D 28 07 5 126 2 507 5 126 17 35 6 774 2 507 6 774 6 328 103 3 Pa m A A Pa m 11 22 9 1 146 10 A A 16 26 0 A A 11 12 9 2 1 146 0 3391 2 10 D D 11 22 D D 16 26 0 D D D 66 11 12 2 1343bookfm Page 376 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Failure Analysis and Design of Laminates 377 D22 1735 103 Pam3 D16 0 D26 0 as D16 2507 103 Pam3 D26 6774 103 Pam3 because 1147 103 Pam3 D66 6328 103 Pam3 One can make a quasiisotropic laminate by having a laminate with N N 3 lamina of the same material and thickness where each lamina is oriented at an angle of 180N between each other For example a threeply laminate will require the laminae to be oriented at 1803 60 to each other Thus 06060 3090 30 and 457515 are all quasiisotropic laminates One can make the preceding combinations symmetric or repeated to give quasiisotropic laminates such as 060s 060s and 0602s laminates The symmetry of the laminates zeros out the coupling matrix B and makes its behavior closer not same to that of an isotropic material Example 52 Show that the extensional stiffness matrix for a general Nply quasiisotropic laminate is given by 58 D D D 11 12 66 2 D D 11 12 3 3 2 28 07 10 5 126 10 2 A U U U U U U h 1 4 4 1 1 4 0 0 0 0 2 1343bookfm Page 377 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 378 Mechanics of Composite Materials Second Edition where U1 and U4 are the stiffness invariants given by Equation 2132 and h is the thickness of the laminate Also find the inplane engineering stiffness constants of the laminate Solution From Equation 2131a for a general angle ply with angle θ U1 U2 Cos2θ U3 Cos4θ 59 For the kth ply of the quasiisotropic laminate with an angle θk U1 U2 Cos2θk U3 Cos4θk 510 where From Equation 428a 511 where tk thickness of kth lamina Because the thickness of the laminate is h and all laminae are of the same thickness 512 and substituting Equation 510 in Equation 511 513 Q11 Q k 11 θ π θ π θ π θ π θ π 1 2 1 2 1 N N k N N N k N N A t Q k k k N 11 11 1 t h N k N k 1 2 A h N U U U hU U k k k N 11 1 2 3 1 1 2 4 Cos Cos θ θ 2 3 1 1 2 4 h N U h N k k N k k N Cos Cos θ θ 1343bookfm Page 378 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Failure Analysis and Design of Laminates 379 Using the following identity1 514 Then 515a 515b Thus 516a Similarly it can be shown that 516b 516c 516d Therefore 517 Cos Sin Sin kx N x x k N 1 1 2 2 2 1 2 Cos 2 0 1 1 θk k N for N Cos 4 0 3 1 θk k N for N A U h 11 1 A U h 12 4 A U h 22 1 A U U h 66 1 4 2 A U U U U U U h 1 4 4 1 1 4 0 0 0 0 2 1343bookfm Page 379 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 380 Mechanics of Composite Materials Second Edition Because Equation 515b is valid only for N 3 this proves that one needs at least three plies to make a quasiisotropic laminate For a symmetric quasiisotropic laminate the extensional compliance matrix is given by 518 From the definitions of engineering constants given in Equations 435 437 439 442 and 445 and using Equation 518 the elastic moduli of the laminate are independent of the angle of the lamina and are given by 519a 519b 519c 53 Failure Criterion for a Laminate A laminate will fail under increasing mechanical and thermal loads The laminate failure however may not be catastrophic It is possible that some layer fails first and that the composite continues to take more loads until all the plies fail Failed plies may still contribute to the stiffness and strength of the laminate The degradation of the stiffness and strength properties of each failed lamina depends on the philosophy followed by the user When a ply fails it may have cracks parallel to the fibers This ply is still capable of taking load parallel to the fibers Here the cracked ply can be replaced by a hypothetical ply that has no transverse A h U U U U U U U U U U U 1 0 1 1 2 4 2 4 1 2 4 2 4 1 2 4 2 1 1 2 U U U 4 2 1 4 0 0 0 2 E E E A h U U U x y iso 1 11 1 2 4 2 1 G G A h U U xy iso 1 2 66 1 4 ν ν ν xy yx iso A A U U 12 22 4 1 1343bookfm Page 380 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Failure Analysis and Design of Laminates 381 stiffness transverse tensile strength and shear strength The longi tudinal modulus and strength remain unchanged When a ply fails fully discount the ply and replace the ply of near zero stiffness and strength Near zero values avoid singularities in stiffness and compliance matrices The procedure for finding the successive loads between first ply failure and last ply failure given next follows the fully discounted method 1 Given the mechanical loads apply loads in the same ratio as the applied loads However apply the actual temperature change and moisture content 2 Use laminate analysis to find the midplane strains and curvatures 3 Find the local stresses and strains in each ply under the assumed load 4 Use the plybyply stresses and strains in ply failure theories dis cussed in Section 28 to find the strength ratio Multiplying the strength ratio to the applied load gives the load level of the failure of the first ply This load is called the first ply failure load 5 Degrade fully the stiffness of damaged ply or plies Apply the actual load level of previous failure 6 Go to step 2 to find the strength ratio in the undamaged plies If the strength ratio is more than one multiply the strength ratio to the applied load to give the load level of the next ply failure and go to step 2 If the strength ratio is less than one degrade the stiffness and strength properties of all the damaged plies and go to step 5 7 Repeat the preceding steps until all the plies in the laminate have failed The load at which all the plies in the laminate have failed is called the last ply failure The procedure for partial discounting of fibers is more complicated The noninteractive maximum stress and maximum strain failure criteria are used to find the mode of failure Based on the mode of failure the appropriate elastic moduli and strengths are partially or fully discounted Example 53 Find the plybyply failure loads for a graphiteepoxy laminate Assume the thickness of each ply is 5 mm and use properties of unidirec tional graphiteepoxy lamina from Table 21 The only load applied is a tensile normal load in the xdirection that is the direction parallel to the fibers in the 0 ply 0 90 s 1343bookfm Page 381 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 382 Mechanics of Composite Materials Second Edition Solution Because the laminate is symmetric and the load applied is a normal load only the extensional stiffness matrix is required From Example 44 the extensional compliance matrix is which from Equation 51a gives the midplane strains for symmetric lam inates subjected to Nx 1 Nm as The midplane curvatures are zero because the laminate is symmetric and no bending and no twisting loads are applied The global strains in the top 0 ply at the top surface can be found as follows using Equation 416 Using Equation 2103 one can find the global stresses at the top surface of the top 0 ply as A 5 353 10 2 297 10 0 2 297 10 9 10 11 11 886 10 0 0 0 9 298 10 1 10 9 Pa m ε ε γ x y xy 0 0 0 10 5 353 10 2 297 10 11 0 ε ε γ x y xy 5 353 10 2 297 10 10 11 0 0 0075 0 0 0 5 353 10 2 297 10 0 10 11 σ σ τ x y xy top 0 181 8 2 897 0 2 897 1 00 35 0 0 0 7 17 10 5 353 10 2 29 9 10 7 10 0 11 1343bookfm Page 382 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Failure Analysis and Design of Laminates 383 Using the transformation Equation 294 the local stresses at the top surface of the top 0 ply are All the local stresses and strains in the laminate are summarized in Table 51 and Table 52 TABLE 51 Local Stresses Pa in Example 53 Ply no Position σ1 σ2 τ12 1 0 Top Middle Bottom 9726 101 9726 101 9726 101 1313 100 1313 100 1313 100 00 00 00 2 90 Top Middle Bottom 2626 100 2626 100 2626 100 5472 100 5472 100 5472 100 00 00 00 3 0 Top Middle Bottom 9726 101 9726 101 9726 101 1313 100 1313 100 1313 100 00 00 00 TABLE 52 Local Strains in Example 53 Ply no Position ε1 ε2 τ12 1 0 Top Middle Bottom 5353 1010 5353 1010 5353 1010 2297 1011 2297 1011 2297 1011 00 00 00 2 90 Top Middle Bottom 2297 1011 2297 1011 2297 1011 5353 1010 5353 1010 5353 1010 00 00 00 3 0 Top Middle Bottom 5353 1010 5353 1010 5353 1010 2297 1011 2297 1011 2297 1011 00 00 00 9 726 10 1 313 0 1 Pa σ σ τ 1 2 12 0 1 0 0 0 1 0 0 0 1 top 9 726 10 1 313 10 0 9 726 10 1 1 0 1 313 0 Pa 1343bookfm Page 383 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 384 Mechanics of Composite Materials Second Edition The TsaiWu failure theory applied to the top surface of the top 0 ply is applied as follows The local stresses are σ1 9726 101 Pa σ2 1313 Pa τ12 0 Using the parameters H1 H2 H6 H11 H22 H66 and H12 from Example 219 the TsaiWu failure theory Equation 2152 gives the strength ratio as 0 9726 101 SR 2093 108 1313 SR 0 0 44444 1019 9726 1012SR2 10162 1016 13132SR2 21626 1016 02 23360 1018 9726 101 1313SR21 SR 1339 107 The maximum strain failure theory can also be applied to the top surface of the top 0 ply as follows The local strains are Then according to maximum strain failure theory Equation 2143 the strength ratio is given by SR min 1500 106181 1095353 1010 246 106103 1092297 1011 1548 107 The strength ratios for all the plies in the laminate are summarized in Table 53 using the maximum strain and TsaiWu failure theories The symbols in TABLE 53 Strength Ratios in Example 53 Ply no Position Maximum strain TsaiWu 1 0 Top Middle Bottom 1548 107 1T 1548 107 1T 1548 107 1T 1339 107 1339 107 1339 107 2 90 Top Middle Bottom 7254 106 2T 7254 106 2T 7254 106 2T 7277 106 7277 106 7277 106 3 0 Top Middle Bottom 1548 107 1T 1548 107 1T 1548 107 1T 1339 107 1339 107 1339 107 ε ε γ 1 2 12 10 11 5 353 10 2 297 10 0 000 1343bookfm Page 384 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Failure Analysis and Design of Laminates 385 the parentheses in the maximum strain failure theory column denote the mode of failure and are explained at the bottom of Table 23 From Table 53 and using the TsaiWu theory the minimum strength ratio is found for the 90 ply This strength ratio gives the maximum value of the allowable normal load as and the maximum value of the allowable normal stress as where h thickness of the laminate The normal strain in the xdirection at this load is Now degrading the 90 ply completely involves assuming zero stiffnesses and strengths of the 90 lamina Complete degradation of a ply does not allow further failure of that ply For the undamaged plies the lami nate has two reduced stiffness matrices as and for the damaged ply Using Equation 428a the extensional stiffness matrix N N m x 7 277 106 N h Pa x 7 277 10 0 015 0 4851 10 6 9 εx 0 10 5 353 10 7 277 1 first ply failure 0 3 895 10 6 3 0 90 s Q 181 8 2 897 0 2 897 10 35 0 0 0 7 17 GPa Q GPa 0 0 0 0 0 0 0 0 0 1343bookfm Page 385 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 386 Mechanics of Composite Materials Second Edition Inverting the new extensional stiffness matrix A the new extensional compliance matrix is which gives midplane strains subjected to Nx 1 Nm by Equation 51a as A Q h h ij ij k k k k 1 1 3 A 181 8 2 897 0 2 897 10 35 0 0 0 7 17 10 0 005 9 0 0 0 0 0 0 0 0 0 10 9 0 005 181 8 2 897 0 2 897 10 35 0 0 0 7 17 109 0 005 A 181 8 2 897 0 2 897 10 35 0 0 0 7 17 107 Pa m A 5 525 10 1 547 10 0 1 547 10 9 10 10 10 709 10 0 0 0 1 395 10 1 9 8 Pa m ε ε γ x y xy 0 0 10 5 525 10 1 547 10 10 10 9 8 0 1 547 10 9 709 10 0 0 0 1 395 10 1 0 0 1343bookfm Page 386 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Failure Analysis and Design of Laminates 387 These strains are close to those obtained before the ply failure only because the 90 ply takes a small percentage of the load out of the normal load in the xdirection The local stresses in each layer are found using earlier techniques given in this example and are shown in Table 54 The strength ratios in each layer are also found using methods given in this example and are shown in Table 55 From Table 56 and using TsaiWu failure theory the minimum strength ratio is found in both the 0 plies This strength ratio gives the maximum value of the normal load as TABLE 54 Local Stresses after First Ply Failure in Example 53 Ply no Position σ1 σ2 τ12 1 0 Top Middle Bottom 10000 102 10000 102 10000 102 00 00 00 00 00 00 2 90 Top Middle Bottom 3 0 Top Middle Bottom 10000 102 10000 102 10000 102 00 00 00 00 00 00 TABLE 55 Local Strains after First Ply Failure in Example 53 Ply no Position ε1 ε2 γ12 1 0 Top Middle Bottom 525 1010 5525 1010 5525 1010 1547 1010 1547 1010 1547 1010 00 00 00 2 90 Top Middle Bottom 3 0 Top Middle Bottom 5525 1010 5525 1010 5525 1010 1547 1010 1547 1010 1547 1010 00 00 00 ε ε γ x y xy 0 0 0 10 5 525 10 1 547 10 10 0 N N m x 1 5 107 1343bookfm Page 387 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 388 Mechanics of Composite Materials Second Edition and the maximum value of the allowable normal stress as where h is the thickness of the laminate The normal strain in the xdirection at this load is The preceding load is also the last ply failure LPF because none of the layers is left undamaged Plotting the stress vs strain curve for the laminate until last ply failure shows that the curve will consist of two linear curves each ending at each ply failure The slope of the two lines will be the Youngs modulus in x direction for the undamaged laminate and for the FPF laminate that is using Equation 435 until first ply failure and TABLE 56 Strength Ratios after First Ply Failure in Example 53 Ply no Position Max strain TsaiWu 1 0 Top Middle Bottom 15000 107 1T 15000 1071T 15000 1071T 15000 107 15000 107 15000 107 2 90 Top Middle Bottom 3 0 Top Middle Bottom 15000 1071T 15000 1071T 15000 1071T 15000 107 15000 107 15000 107 N h Pa x 1 5 10 0 015 1 0 10 7 9 εx o last ply failure 5 525 10 1 5 10 10 7 8 288 10 3 E GPa x 1 0 015 5 353 10 124 5 10 1343bookfm Page 388 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Failure Analysis and Design of Laminates 389 after first ply failure and until last ply failure Figure 51 Example 54 Repeat Example 53 for the first ply failure and use TsaiWu failure theory now with an additional thermal load a temperature change of 75C Solution The laminate is symmetric and the load applied is a normal load and a temperature change Thus only the extensional stiffness matrix is needed From Example 53 FIGURE 51 Stressstrain curve showing plybyply failure of a laminated composite 0 300 600 900 1200 1500 0 0005 001 Normal strain εx Normal stress Nxh MPa First ply failure Last ply failure E N h N h x x x last ply failure first play ffailure last play failure first p ε ε x o x o lay failure 0 1 10 0 4851 10 8 288 10 10 9 3 3 895 10 117 2 3 GPa A 5 353 10 2 297 10 0 2 297 10 9 10 11 11 886 10 0 0 0 9 298 10 1 10 9 Pa m 1343bookfm Page 389 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 390 Mechanics of Composite Materials Second Edition Corresponding to a temperature change of 75C the mechanical stresses can be found as follows The fictitious thermal forces given by Equation 464 are Because the laminate is symmetric the fictitious thermal moments are zero This also then gives only midplane strains in the laminate without any plate curvatures The midplane strain due to the thermal load is given by The laminate is symmetric and no bending or torsional moments are applied therefore the global strains in the laminate are the same as the midplane strains The free expansional thermal strains in the top 0 ply are N N N x T y T xy T 75 181 8 2 897 0 2 897 10 35 0 0 0 7 17 10 0 200 10 0 22 9 7 5 10 0 0 0025 0 0075 4 75 10 35 2 897 0 2 897 181 8 0 0 0 7 17 10 0 225 10 0 200 10 0 0 0 9 4 7 025 0 0025 75 181 8 2 897 0 2 897 10 35 0 0 0 7 17 10 0 200 10 0 225 10 0 0 0 9 7 4 075 0 0025 1 389 10 2 004 10 0 5 5 Pa m ε ε γ x y xy 0 0 0 10 5 353 10 2 297 10 11 11 10 9 0 2 297 10 9 886 10 0 0 0 9 298 10 1 389 10 2 004 10 0 5 5 0 6977 10 0 1950 10 0 4 3 1343bookfm Page 390 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Failure Analysis and Design of Laminates 391 From Equation 470 the global mechanical strain at the top surface of the top 0 ply is From Equation 2103 the global mechanical stresses at the top of the top 0 ply are Now if the mechanical loads were given the resulting mechanical stresses could then be added to the previous stresses due to the temperature difference x y xy T 0 α α α Δ 0 200 10 0 225 10 0 75 7 4 0 1500 10 0 16875 10 0 5 2 0 6977 10 0 1950 10 0 0 1500 4 3 10 0 16875 10 0 0 6827 10 5 2 4 0 14925 10 0 2 σ σ τ x y xy 181 8 2 897 0 2 897 10 35 0 0 0 7 17 10 0 6827 10 0 14925 10 9 4 2 6 7 0 8 088 10 1 524 10 0 Pa 1343bookfm Page 391 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 392 Mechanics of Composite Materials Second Edition Then the failure criteria could be used to find out whether the ply has failed However we are asked to find out the mechanical load that could be applied in the presence of the temperature difference This can be done as follows The stress at the top of the 0 ply per Example 53 for a unit load Nx 1 Nm is If the unknown load is Nx then the overall stress at the top surface of the top 0 ply is Now the failure theories can be applied to find the value of Nx Using transformation equation 294 the local stresses at the top surface of the top 0 ply are Using the parameters H1 H2 H6 H11 H22 H66 and H12 from Example 219 the TsaiWu failure criterion Equation 2146 is 0 8088 106 9726 101 Nx 2093 1081524 107 1313 100 Nx00 44444 1019 8088 106 9726 101 Nx2 10162 1016 1524 107 1313 100 Nx2 21626 1016 02 23360 1018 8088 106 9726 101 Nx 1524 107 1313 100 Nx 1 σ σ τ x y xy 9 726 10 1 313 10 0 0 1 0 Pa σ σ τ x y xy Nx 8 088 10 9 726 10 1 6 1 524 10 1 313 10 0 7 0 N Pa x σ σ τ 1 2 12 6 1 8 088 10 9 726 10 1 5 Nx 24 10 1 313 10 0 7 0 N Pa x 1343bookfm Page 392 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Failure Analysis and Design of Laminates 393 As can be seen this results in a quadratic polynomial in the lefthand side of the strength criteria that is This gives two roots for which the inequality is satisfied for Nx 1100 107 and Nx 1695 107 Because the load Nx is tensile Nx 1100 107 is the valid solution Similarly the values of strength ratios for all the plies in the laminate are found and summarized in Table 57 Using the lowest value of strength ratio of 4279 106 gives Nx 4279 106 Nm as the load at which the first ply failure would take place Compare this with the value of Nx 7277 106 in Example 53 in which no temperature change was applied 54 Design of a Laminated Composite Because we have developed the laminated plate theory for composites sub jected to inplane mechanical loads temperature and moisture the designs in this chapter are also limited to such loads and simple shapes Factors not covered in this section include stability outof plane loads and fracture impact and fatigue resistance interlaminar strength damping characteris tics vibration control and complex shapes These factors are introduced briefly in Section 55 Design of laminated composites includes constraints on optimizing and constraining factors such as Cost Mass as related to aerospace and automotive industry to reduce energy cost TABLE 57 Strength Ratios of Example 54 Ply no Position TsaiWu 1 0 Top Middle Bottom 1100 107 1100 107 1100 107 2 90 Top Middle Bottom 4279 106 4279 106 4279 106 3 0 Top Middle Bottom 1100 107 1100 107 1100 107 3 521 10 2 096 10 0 6566 0 15 2 8 N N x x 1343bookfm Page 393 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 394 Mechanics of Composite Materials Second Edition Stiffness to limit deformations as related to aircraft skins to avoid buckling Thermal and moisture expansion coefficients as related to space antennas to maintain dimensional stability These factors are similar to those used with designing with monolithic mate rials thus the main issue with designing with composites as opposed to monolithic materials involves understanding the orthotropic nature of com posite plies The possibility of different fibermatrix systems combined with the vari ables such as fiber volume fraction first dictate the properties of a lamina Then laminae can be placed at angles and at particular distances from the midplane in the laminate The material systems and the stacking sequence then determine the stresses and strains in the laminate The failure of the composite may be based on the first ply failure FPF or the last ply failure LPF Although one may think that all plies failing at the same time is an ideal laminate others may argue that differences between the two give time for detection and repair or replacement of the part Laminate selection is a computationally intensive and repetitive task due to the many possibilities of fibermatrix combinations material systems and stacking sequence Computer programs have made these calculations easy and the reader is directed to use the PROMAL2 program included in this book or any other equivalent program of choice to fully appreciate designing with composites A more scientific approach to optimization of laminated composites is out of scope of this book and the reader is referred to Gurdal et al3 Example 55 1 An electronic device uses an aluminum plate of crosssection 4 in 4 in to take a pure bending moment of 13000 lbin The factor of safety is 2 Using the properties of aluminum given in Table 34 find the thickness of the plate 2 The designer wants at least to halve the thickness of the plate to make room for additional hardware on the electronic device The choices include unidirectional laminates of graphiteepoxy glass epoxy or their combination hybrid laminates The ply thickness is 0125 mm 00049213 in Design a plate with the lowest cost if the manufacturing cost per ply of graphiteepoxy and glassepoxy is ten and four units respectively Use the properties of unidirectional graphiteepoxy and glassepoxy laminae from Table 22 3 Did your choice of the laminate composite design decrease the mass If so by how much 1343bookfm Page 394 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Failure Analysis and Design of Laminates 395 Solution 1 The maximum normal stress in a plate under bending is given by 520 where M bending moment lbin t thickness of plate in I second moment of area in4 For a rectangular crosssection the second moment of area is 521 where b width of plate in Using the given factor of safety Fs 2 and given b 4 in the thickness of the plate using the maximum stress criterion is 522 where σult 4002 Ksi from Table 34 2 Now the designer wants to replace the 09872 in thick aluminum plate by a plate of maximum thickness of 04936 in half that of aluminum made of laminated composites The bending moment per unit width is σ M t I 2 I bt 3 12 t MF b s ult 6 σ t in 6 13000 2 4 40 02 10 0 9872 3 M lb in in xx 13 000 4 3 250 1343bookfm Page 395 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 396 Mechanics of Composite Materials Second Edition Using the factor of safety of two the plate is designed to take a bending moment per unit width of The simplest choices are to replace the aluminum plate by an all graphiteepoxy laminate or an all glassepoxy laminate Using the procedure described in Example 53 or using the PROMAL2 pro gram the strength ratio for using a single 0 ply for the previous load for glassepoxy ply is SR 5494 105 The bending moment per unit width is inversely proportional to the square of the thickness of the plate so the minimum number of plies required would be 135 plies This gives the thickness of the allglassepoxy laminate as tGlEp 135 00049213 in 06643 in The thickness of an allglassepoxy laminate is more than 04935 in and is thus not acceptable Similarly for an all graphiteepoxy laminate made of only 0 plies the minimum number of plies required is NGrEp 87 plies This gives the thickness of the plate as tGrEp 87 00049213 04282 in M lb in in xx 3 250 2 6 500 NGl Ep 1 5 494 10 5 1343bookfm Page 396 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Failure Analysis and Design of Laminates 397 The thickness of an allgraphiteepoxy laminate is less than 04936 in and is acceptable Even if an allgraphiteepoxy laminate is acceptable because graphiteepoxy is 25 times more costly than glassepoxy one would suggest the use of a hybrid laminate The question that arises now concerns the sequence in which the unidirectional plies should be stacked In a plate under a bending moment the magnitude of ply stresses is maximum on the top and bottom face Because the longitudinal tensile and compressive strengths are larger in the graphiteepoxy lamina than in a glassepoxy lamina one would put the former as the facing material and the latter in the core The maximum number of plies allowed in the hybrid laminate is 100 plies Several combinations of 100ply symmetric hybrid laminates of the form are now subjected to the applied bending moment Minimum strength ratios in each laminate stacking sequence are found Only if the strength ratios are greater than one that is the laminate is safe is the cost of the stacking sequence determined A summary of these results is given in Table 58 From Table 58 an acceptable hybrid laminate with the lowest cost is case VI TABLE 58 Cost of Various GlassEpoxyGraphiteEpoxy Hybrid Laminates Case Number of plies Minimum SR Cost Glassepoxy m Graphiteepoxy 2n I II III IV V VI VII 0 20 60 80 70 68 66 87 80 40 20 30 32 34 1023 1342 1127 08032 09836 1014 1043 870 880 640 592 604 Nmax Maximum Allowable Thickness Thickness of each ply 0 4936 0 0049213 n Gr m Gl n Gr 0 0 0 16 68 16 0 0 0 Gr Gl Gr 1343bookfm Page 397 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 398 Mechanics of Composite Materials Second Edition 3 The volume of the aluminum plate is VAl 4 4 09871 157936 in3 The mass of the aluminum plate is specific gravity 27 from Table 32 MAl VAl ρAl 15793 27 36127 102 1540 lbm The volume of the glassepoxy in the hybrid laminate is VGlEp 4 4 00049213 68 5354 in3 The volume of graphiteepoxy in the hybrid laminate is VGrEp 4 4 00049213 32 2520 in3 Using the specific gravities of glass graphite and epoxy given in Table 31 and Table 32 and considering that the density of water is 36127 102 lbmin3 ρGl 25 36127 102 09032 101 lbmin3 ρGr 18 36127 102 06503 101 lbmin3 ρEp 12 36127 102 04335 101 lbmin3 The fiber volume fraction is given in Table 21 and substituting in Equation 38 the density of glassepoxy and graphiteepoxy lam inae is ρGlEp 09032 101 045 04335 101 055 06449 101 lbmin3 1343bookfm Page 398 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Failure Analysis and Design of Laminates 399 ρGrEp 06503 101 070 04335 101 030 05853 101 lbmin3 The mass of the hybrid laminate then is Mh 5354 06449 101 252005853 101 04928 lbm The percentage savings using the composite laminate over alumi num is 68 This example dictated the use of unidirectional laminates How will the design change if multiple loads are present Examples of mul tiple loads include a leaf spring subjected to bending moment as well as torsion or a thin pressure vessel subjected to an internal pressure to yield a biaxial state of stress In such cases one may have a choice not only of material systems and their combination but also of orientation of plies Combinations of angle plies can be infinite so attention may be focused on angle plies of 0 90 45 and 45 and their combinations This reduces the possibilities to a finite number for a limited number of material systems however but the number of combinations can still be quite large to handle Example 56 An electronic device uses an aluminum plate of 1in thickness and a top crosssectional area of 4 in 4 in to take a pure bending moment The designer wants to replace the aluminum plate with graphiteepoxy unidi rectional laminate The ply thickness of graphiteepoxy is 0125 mm 00049213 in 1 Use the properties of aluminum and unidirectional graphiteepoxy as given in Table 34 and Table 22 respectively to design a plate of graphiteepoxy with the same bending stiffness in the needed direc tion of load as that of the aluminum beam 2 Does the laminate design decrease the mass If so by how much 1540 04928 1540 100 1343bookfm Page 399 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 400 Mechanics of Composite Materials Second Edition Solution 1 The bending stiffness Eb of the aluminum plate is given by Eb EI 523 E where E Youngs modulus of aluminum b width of beam h thickness of beam Eb 103 106 3433 106 lbin2 To find the thickness of a graphiteepoxy laminate with unidirec tional plies and the same flexural rigidity let us look at the bending stiffness of a laminate of thickness h Eb ExI Ex where Ex Youngs modulus in direction of fibers Because Ex E1 2625 Msi for a 0 ply from Table 22 3433 106 2625 106 giving h 0732 in Thus a 1in thick aluminum beam can be replaced with a graphite epoxy laminate of 0732 in thickness Note that although the Youngs modulus of graphite epoxy is approximately 25 times that of aluminum the thickness of aluminum plate is approximately only 1 12 bh3 1 2 4 1 3 1 12 3 bh 1 2 4 h3 1343bookfm Page 400 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Failure Analysis and Design of Laminates 401 14 times that of the graphite epoxy of laminate because the bending stiffness of a beam is proportional to the cube of the thickness Thus the lightest beam for such bending would be influenced by the cube root of the Youngs moduli From the thickness of 0732 in of the laminate and a thickness of 00049312 in of the lamina the number of 0 graphiteepoxy plies needed is The resulting graphiteepoxy laminate then is 0149 2 The volume of the aluminum plate VAl is VAl 4 4 10 16 in3 The mass of the aluminum plate is specific gravity 27 from Table 32 density of water is 36127 102 lbmin3 MAl VAl ρAl 16 27 36127 102 1561 lbm The volume of a 0149 graphiteepoxy laminate is VGrEp 4 4 00049213 149 1173 in3 The density of a graphiteepoxy from Example 55 is ρGrEp 05853 101 The mass of the graphiteepoxy laminate beam is MGrEp 05853 101 1173 06866 lbm n 0 732 0 0049213 149 lbm in3 1343bookfm Page 401 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 402 Mechanics of Composite Materials Second Edition Therefore the percentage saving in using graphiteepoxy composite laminate over aluminum is 56 Example 57 A 6ftlong cylindrical pressure vessel Figure 52 with an inner diameter of 35 in is subjected to an internal gauge pressure of 150 psi The vessel operates at room temperature and curing residual stresses are neglected The cost of a graphiteepoxy lamina is 250 unitslbm and cost of a glassepoxy lamina is 50 unitslbm The following are other specifications of the design 1 Only 0 45 45 60 60 and 90 plies can be used 2 Only symmetric laminates can be used 3 Only graphiteepoxy and glassepoxy laminae as given in Table 22 are available but hybrid laminates made of these two laminae are allowed The thickness of each lamina is 0005 in FIGURE 52 Fiber composite pressure vessel Note that the thickness of each lamina is given as 0005 in and is not 0125 mm 00004921 in as given in the material database of the PROMAL program Material properties for two new materials need to be entered in the database M M M Al Gr Ep Al 1 561 0 6866 1 561 100 z Sectional front view Side view pd 2t σy pd 4t σx x y 1343bookfm Page 402 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Failure Analysis and Design of Laminates 403 4 Calculate specific gravities of the laminae using Table 33 and Table 34 and fiber volume fractions given in Table 22 5 Neglect the end effects and the mass and cost of ends of the pressure vessel in your design 6 Use TsaiWu failure criterion for calculating strength ratios 7 Use a factor of safety of 195 Design for ply orientation stacking sequence number of plies and ply material and give separate designs laminate code including materials based on each of the following design criteria 1 Minimum mass 2 Minimum cost 3 Both minimum mass and minimum cost You may be unable to minimize mass and cost simultaneously that is the design of the pressure vessel for the minimum mass may not be same as for the minimum cost In that case give equal weight to cost and mass and use this as your optimization function 524 where A mass of composite laminate B mass of composite laminate if design was based only on minimum mass C cost of composite laminate D cost of composite laminate if design was based only on min imum cost Solution LOADING For thinwalled cylindrical pressure vessels the circumferential stress or hoop stress σy and the longitudinal or axial stress σx is given by 525a 525b where F A B C D σx pr t 2 σy pr t 1343bookfm Page 403 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 404 Mechanics of Composite Materials Second Edition p internal gage pressure psi r radius of cylinder in t thickness of cylinder in For our case we have giving For the forces per unit length 526a 526b p psi r in 150 35 2 17 5 σ σ x y t t t 150 17 5 2 1 3125 10 150 17 5 3 2 625 103 t N t t t lb in x x σ 1 3125 10 1 3125 10 3 3 N t t t lb in y y σ 2 625 10 2 625 10 3 3 1343C005fm Page 404 Wednesday September 28 2005 1041 AM 2006 by Taylor Francis Group LLC Failure Analysis and Design of Laminates 405 MASS OF EACH PLY The mass of a graphite epoxy ply is where VGrEp volume of a graphite epoxy ply in3 ρGrEp density of a graphiteepoxy ply lbmin3 VGrEp πLdtp where L length of the cylinder in d diameter of the cylinder in tp thickness of graphiteepoxy ply in Because L 6 ft d 35 in and tp 0005 in VGrEp π6 12350005 39584 in3 The density of a graphiteepoxy lamina is From Table 22 the fiber volume fraction Vf of the graphite epoxy is 07 Thus The matrix volume fraction Vm then is The specific gravity of graphite and epoxy is given in Table 33 and Table 34 respectively as sGr 18 and sEp 12 given that the density of water is 36127 102 lbmin3 m V Gr Ep Gr Ep Gr Ep ρ ρ ρ ρ Gr Ep Gr f Ep m V V Vf 0 7 V V m f 1 1 0 7 0 3 1343bookfm Page 405 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 406 Mechanics of Composite Materials Second Edition Therefore the mass of a graphiteepoxy lamina is The mass of a glassepoxy ply is where VGlEp volume of glassepoxy in3 ρGlEp density of glassepoxy lbmin3 VGlEp πLdtp 39584 in3 The density of a glassepoxy lamina is From Table 22 the fiber volume fraction Vf of the glassepoxy is 045 thus The matrix volume fraction Vm then is ρGr Ep 1 8 3 6127 10 0 7 1 2 3 6127 2 10 0 3 5 8526 10 2 2 3 lbm in m V Gr Ep Gr Ep Gr Ep ρ 39 584 5 8526 10 2 2 3167 lbm m V Gl Ep Gl Ep Gl Ep ρ ρ ρ ρ Gl Ep Gl f Ep m V V Vf 0 45 V V m f 1 1 0 45 0 55 1343bookfm Page 406 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Failure Analysis and Design of Laminates 407 The specific gravity of glass and epoxy is given in Table 33 and Table 34 respectively as and given that the density of water is 36127 102 lbmin3 Therefore the mass of glassepoxy lamina is COST OF EACH PLY The cost of a graphiteepoxy ply is where mGrEp mass of graphiteepoxy ply cGrEp unit cost of graphiteepoxy ply Because the cost of a graphiteepoxy ply is s s Gl Ep 2 5 1 2 ρG Ep 2 5 3 6127 10 0 45 1 2 3 612 2 7 10 0 55 6 4487 10 2 2 3 lbm in m V Gl Ep Gl Ep Gl Ep ρ 39 584 6 4487 10 2 2 5526 lbm C m c Gr Ep Gr Ep Gr Ep m 23167 lbm Gr Ep c units lbm Gr Ep 250 CGrEp 2 3167 250 579 17 units 1343bookfm Page 407 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 408 Mechanics of Composite Materials Second Edition Similarly the cost of a glassepoxy ply is Because mGlEp 25526 lbm and the cost of a glassepoxy ply is CGlEp 2552650 12763 units 1 To find the design for minimum mass consider a composite laminate made of graphiteepoxy with 0902s We simply choose this lam inate as Ny 2Nx and thus choose two 90 plies for every 0 ply For this laminate from PROMAL we get the minimum strength ratio as SR 06649 Because the required factor of safety is 195 we need 18 plies 09023s is a possible choice because it gives a strength ratio of 1995 However is this laminate with the minimum mass Choosing some other choices such as laminates with 60 laminae a graphite epoxy 604s laminate gives an SR 1192 and that is lower than the required SR of 195 A 09023s laminate made of glassepoxy gives a strength ratio of SR 05192 and that is also lower than the needed strength ratio of 195 Other combinations tried used more than 18 plies A sum mary of possible combinations is shown in Table 59 C m c Gl Ep Gl Ep Gl Ep c units lbm Gl Ep 50 1 95 0 6649 6 1343bookfm Page 408 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Failure Analysis and Design of Laminates 409 Thus one can say that the laminate for minimum mass is the first stacking sequence in Table 59 Number of plies 18 Material of plies graphiteepoxy Stacking sequence 09023s Mass of laminate 18 23167 41700 lbm Cost of laminate 41700 250 10425 units 2 To find the design for minimum cost we found in part 1 that the 09023s graphiteepoxy laminate is safe but the same stacking sequence for glassepoxy gives a SR 05192 Therefore we may need four times more plies of glassepoxy to keep it safe to obtain a factor of safety of 195 If so would it be cheaper than the 09023s graphiteepoxy laminate Yes it would because a glassepoxy costs 12763 units per ply as opposed to 57917 units per ply for graphite epoxy Choosing 090212s glassepoxy laminate gives SR 2077 Are there other combinations that give an SR 195 but use less than the 72 plies used in 090212s Stacking sequences of 60 plies such as 904510s and 6015s were tried and were acceptable designs The results from some of the stacking sequences are summarized in Table 510 Therefore we can say that the laminate for minimum cost is as follows TABLE 59 Mass and Cost of Possible Stacking Sequences for Minimum Mass Stacking sequence No plies Minimum strength ratio Mass lbm Cost units 09023s Graphiteepoxy 18 1995 41700 10425 604s Graphiteepoxy 16 1192 09023s Glassepoxy 18 05192 605s Graphiteepoxy 20 1490 452603s Graphiteepoxy 20 2332 46334 11583 TABLE 510 Mass and Cost of Possible Stacking Sequences for Minimum Cost Stacking sequence No plies Minimum strength ratio Mass lbm Cost units 09023s Graphiteepoxy 18 1995 41700 10425 452603s Graphiteepoxy 20 2291 46334 11583 090212s Glassepoxy 72 2077 18379 9189 904510s Glassepoxy 60 1992 15316 7658 6015s Glassepoxy 60 2033 15316 7658 1343bookfm Page 409 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 410 Mechanics of Composite Materials Second Edition Number of plies 60 Material of plies glassepoxy Stacking sequence 6015s Mass of laminate 60 25526 15316 lbm Cost of laminate 15316 50 7658 units 3 Now how do we find the laminate that minimizes cost and mass We know that the solutions to part 1 and 2 are different Thus we need to look at other combinations However before doing so let us find the minimizing function for parts 1 and 2 The mini mizing function is given as where A mass of composite laminate B mass of composite laminate if design was based only on minimum mass C cost of composite laminate D cost of composite laminate if design was based only on min imum cost From part 1 B 41700 lbm and from part 2 D 7658 units then the minimizing function is for the 09023s graphiteepoxy laminate obtained in part 1 for the 6015s glassepoxy laminate obtained in part 2 Therefore the question is whether a laminate that has an optimiz ing function value of less than 2361 can be found If not the answer is the same as the laminate in part 1 Table 511 gives the summary of some of the laminates that were tried to find minimum value of F The third stacking sequence in Table 511 is the one in which in the 09023s graphiteepoxy laminate of part 1 six of the graphite F A B C D F 41 700 41 700 10425 7658 2 361 F 153 16 41 700 7658 7658 4 673 1343bookfm Page 410 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Failure Analysis and Design of Laminates 411 epoxy plies of 0902 sublaminate group are substituted with 24 glassepoxy plies of the 0902 sublaminate group Thus it seems that 09023s graphiteepoxy laminate is the answer to part 3 Although more combinations should have been attempted to come to a definite conclusion it is left to the reader to try other hybrid combinations using the PROMAL program Example 58 Drive shafts Figure 53 in cars are generally made of steel An automobile manufacturer is seriously thinking of changing the material to a composite material The reasons for changing the material to composite materials are that composites 1 Reduce the weight of the drive shaft and thus reduce energy consumption 2 Are fatigue resistant and thus have a long life 3 Are noncorrosive and thus reduce maintenance costs and increase life of the drive shaft 4 Allow single piece manufacturing and thus reduce manufacturing cost The design constraints are as follows 1 Based on the engine overload torque of 140 Nm the drive shaft needs to withstand a torque of 550 Nm TABLE 511 Optimizing Function Values for Different Stacking Sequences Stacking sequence Mass lbm Cost Minimum strength ratio F 09023s graphiteepoxy part a 41700 10425 1995 2361 6018s glassepoxy part b 15316 7658 20768 4672 0GrEp902 GrEp0GrEp902 GrEp0GlEp 902 GlEp0GlEp902 GlEp0GlEp902GlEp0GlEp 902 GlEps 89063 10013 2012 3443 FIGURE 53 Fiber composite drive shaft 100 mm 148 cm 1343bookfm Page 411 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 412 Mechanics of Composite Materials Second Edition 2 The shaft needs to withstand torsional buckling 3 The shaft has a minimum bending natural frequency of at least 80 Hz 4 Outside radius of drive shaft 50 mm 5 Length of drive shaft 148 cm 6 Factor of safety 3 7 Only 0 45 45 60 60 and 90 plies can be used For steel use the following properties Youngs modulus E 210 GPa Poissons ratio ν 03 Density of steel ρ 7800 kgm3 Ultimate shear strength τult 80 MPa For the composite use properties of glassepoxy from Table 21 and Table 31 and assume that ply thickness is 0125 mm Design the drive shaft using 1 Steel 2 Glassepoxy Solution 1 STEEL DESIGN Torsional strength The primary load in the drive shaft is torsion The max imum shear stress τmax in the drive shaft is at the outer radius ro and is given as 527 where T maximum torque applied in drive shaft Nm ro outer radius of shaft m J polar moment of area m4 Because the ultimate shear strength of steel is 80 MPa and the safety factor used is 3 using Equation 527 gives τmax Tr J o 80 10 3 550 0 050 2 0 050 0 0486 6 4 4 π r r i i 3 m 1343bookfm Page 412 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Failure Analysis and Design of Laminates 413 Therefore the thickness of the steel shaft is t r o r i 0050 004863 1368 mm Torsional buckling This requirement asks that the applied torsion be less than the critical torsional buckling moment For a thin hollow cylinder made of isotropic materials the critical buckling torsion T b is given by 4 528 where r m mean radius of the shaft m t wall thickness of the drive shaft m E Youngs modulus Pa Using the thickness t 1368 mm calculated in criterion 1 and the mean radius The value of critical torsional buckling moment is larger than the applied torque of 550 Nm Natural frequency The lowest natural frequency for a rotating shaft is given by 5 529 T r t E t r b m m 2 0 272 2 2 3 π r r r m m o i 2 0 050 0 04863 2 0 049315 Tb 2 0 049315 0 001368 0 272 210 10 0 2 9 001368 0 049315 109442 3 2 N m f EI mL n π 2 4 1343C005fm Page 413 Wednesday September 28 2005 1042 AM 2006 by Taylor Francis Group LLC 414 Mechanics of Composite Materials Second Edition where g acceleration due to gravity ms2 E Youngs modulus of elasticity Pa I second moment of area m4 m mass per unit length kgm L length of drive shaft m Now the second moment of area I is The mass per unit length of the shaft is m π ro 2 ri 2 ρ π 00502 0048632 7800 3307 kgm Therefore This value is greater than the minimum desired natural frequency of 80 Hz Thus the steel design of a hollow shaft of outer radius 50 mm and thickness t 1368 mm is an acceptable design 2 COMPOSITE MATERIALS DESIGN Torsional strength Assuming that the drive shaft is a thin hollow cylinder an element in the cylinder can be assumed to be a flat laminate The only nonzero load on this element is the shear force Nxy If the average shear stress is τxyaverage the applied torque then is T shear stress area moment arm I r r o i π π 4 4 0 050 0 04863 5 162 10 4 4 4 4 7 4 m fn π 2 210 10 5 162 10 3 307 1 48 12 9 7 4 9 8 Hz 1343bookfm Page 414 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Failure Analysis and Design of Laminates 415 530 The shear force per unit width is given by Nxy Because t ro ri then 531 To find approximately how many layers of glassepoxy may be needed to resist the shear load choose a fourply 45s laminate Inputting a value of Nxy 35014 Nm into the PROMAL program the minimum strength ratio obtained using TsaiWu theory is 1261 A strength ratio of at least 3 is needed so the number of plies is increased proportionately as The next laminate chosen is 45245s laminate A minimum strength ratio of 358 is obtained so it is an acceptable design based on torsional strength criterion Torsional buckling An orthotropic thin hollow cylinder will buckle torsion ally if the applied torque is greater than the critical torsional buckling load given by4 532 T r r r xy average o i m τ π 2 2 τxy averaget r r r m o i 2 N T r N m xy m 2 550 2 0 050 35 014 2 2 π π 3 1 261 4 10 T r t E E t r c m x y m 2 0 272 2 3 1 4 3 2 π 1343bookfm Page 415 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 416 Mechanics of Composite Materials Second Edition From PROMAL the longitudinal Youngs moduli Ex and the transverse Youngs moduli Ey of the 45245s glassepoxy laminate based on proper ties from Table 21 are Ex 1251 GPa Ey 1251 GPa Because lamina thickness is 0125 mm the thickness of the tenply 452 45s laminate t is t 10 0125 125 mm The mean radius rm is Therefore This is less than the applied torque of 550 Nm Thus the 45245s laminate would torsionally buckle Per the formula the torsional buckling is proportional to Ey 34 and Ex 14 Because the modulus in the ydirection is more effective in increasing the critical torsional buckling load it will be necessary to substitute by or add 90 plies Natural frequency Although the 45245s laminate is inadequate per the torsional buckling criterion let us still find the minimum natural frequency of the drive shaft which is given by5 r r t mm m o 2 50 1 25 2 49 375 Tc 2 0 049375 0 00125 0 272 12 51 10 2 π 9 9 3 1 4 2 3 12 51 10 0 00125 0 049375 262 N m 1343bookfm Page 416 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Failure Analysis and Design of Laminates 417 533 Now The mass per unit length of the beam is Thus Because the minimum bending natural frequency is required to be 80 Hz this requirement is also not met by the 45245s laminate The minimum natural frequency can be increased by increasing the value of Ex because the natural frequency fn is proportional to To achieve this 0 plies can be added or substituted f E I mL n x π 2 4 E GPa I r r x i 12 51 4 4 0 050 0 04875 0 4 4 4 π π 4 7 4 4 728 10 m m r r L L r r o i o i π ρ π ρ 2 2 2 2 2 2 0 05 0 04875 1785 0 6922 kg m fn π 2 12 51 10 4 728 10 0 6922 1 48 66 9 7 4 3 Hz Ex 1343bookfm Page 417 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 418 Mechanics of Composite Materials Second Edition From the three criteria we see that 45 plies increase the torsional strength 90 plies increase the critical torsional buckling load and the 0 plies increase the natural frequency of the drive shaft Therefore having 45 90 and 0 plies may be the key to an optimum design In Table 512 several other combinations have been evaluated to find an acceptable design The last stacking sequence 4590302s is a 12ply laminate and meets the three requirements of torsional strength critical torsional buckling load and minimum natural frequency MASS SAVINGS The savings in the mass of the drive shaft are calculated as follows Mass of steel drive shaft π ro 2 ri 2 L ρ π 00502 0048632 148 7800 4894 kg The thickness t of the 4590302s glassepoxy shaft is The inner radius of the 4590302s glassepoxy shaft then is TABLE 512 Acceptable and Nonacceptable Designs of Drive Shaft Based on Three Criteria of Torsional Strength Critical Torsional Buckling Load and Minimum Natural Frequency Laminate stacking sequence No plies Minimum strength ratio Critical torsional buckling load Nm Ex GPa Ey GPa Minimum natural frequency Hz Acceptable design 04524590s 14 3982 7978 1644 1644 756 No 0245290s 14 3248 8288 2016 1616 837 Yes 045290s 12 3006 5641 1707 1707 772 No 0452s 10 2764 2912 1786 1276 792 No 4590302s 12 4127 7635 1944 2447 824 Yes Design constraints 3 550 80 Note Numbers given in bold italics to show the reason for unacceptable designs t mm 12 0 125 1 5 1343bookfm Page 418 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Failure Analysis and Design of Laminates 419 ri ro t 005 00015 00485 m Mass of 4590302s glassepoxy shaft is π ro 2 ri 2 L ρ π 0052 004852 148 1758 1226 kg Percentage mass saving over steel is Would an 11ply 45904 s glassepoxy laminate meet all the require ments 55 Other Mechanical Design Issues 551 Sandwich Composites One group of laminated composites used extensively is sandwich compos ites Sandwich panels consist of thin facings also called skin sandwiching a core The facings are made of highstrength material such as steel and composites such as graphiteepoxy the core is made of thick and lightweight materials such as foam cardboard plywood etc Figure 54 The motivation in doing this is twofold First if a plate or beam is bent the maximum stresses occur at the top and bottom surfaces Therefore it makes sense to use highstrength materials only at the top and bottom and low and lightweight strength materials in the middle The strong and stiff facings also support axial forces Second the resistance to bending of a rectangular crosssectional beamplate is proportional to the cube of the thickness Thus increasing the thickness by adding a core in the middle increases this resistance Note that the shear forces are maximum in the 4 894 1 226 4 894 100 75 90 1343bookfm Page 419 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 420 Mechanics of Composite Materials Second Edition middle of the sandwich panel thus requiring the core to support shear This advantage in weight and bending stiffness makes sandwich panels more attractive than other materials Sandwich panels are evaluated based on strength safety weight durability corrosion resistance dent and puncture resistance weatherability and cost6 The most commonly used facing materials are aluminum alloys and fiber reinforced plastics Aluminum has high specific modulus but it corrodes without treatment and is prone to denting Fiberreinforced plastics such as graphiteepoxy and glassepoxy are becoming popular as facing materials because of their high specific modulus and strength and corrosion resistance Fiberreinforced plastics may be unidirectional or woven laminae The most commonly used core materials are balsa wood foam and hon eycombs Balsa wood has high compressive strength 1500 psi good fatigue life and high shear strength 200 psi Foams are lowdensity polymers such as polyuretherane phenolic and polystyrene Honeycombs are made of plastic paper cardboard etc The strength and stiffness of honeycomb depend on the material and its cell size and thickness Adhesives join the facing and core materials and thus are critical in the overall integrity of the sandwich panel Adhesives come in forms of film paste and liquid Common examples include vinyl phenolic modified epoxy and urethane 552 LongTerm Environmental Effects Section 45 has already discussed the effects caused by temperature and moisture such as residual stresses and strains What effect do these and FIGURE 54 Fiberglass facings with a Nomex7 honeycomb core Picture Courtesy of MC Gill Corporation httpwwwmcgillcorpcom 1343bookfm Page 420 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Failure Analysis and Design of Laminates 421 other environmental factors such as corrosive atmospheres and temperatures and humidity variations have over the long term on composites These elements may lessen the adhesion of the fibermatrix interface such as between glass and epoxy Epoxy matrices soften at high temperatures affect ing properties dominated by the matrix such as transverse and inplane shear stiffness and strength and flexural strength For example Quinn7 found that a glassepoxy composite rod absorbed as much as 04 of water over 150 days of immersion The effect of this moisture absorption on flexural modulus is shown in Figure 55 553 Interlaminar Stresses Due to the mismatch of elastic moduli and angle between the layers of a laminated composite interlaminar stresses are developed between the lay ers These stresses which are normal and shear can be high enough to cause edge delamination between the layers810 Delamination eventually limits the life of the laminated structure Delamination can be further caused due to nonoptimum curing and introduction of foreign bodies in the structure11 In Figure 56 theoretical interlaminar shear and normal stresses are plotted as a function of normalized distance zero at the center line and one at the free edge from the center line of a 45s graphiteepoxy laminate The interlaminar stresses given are for the bottom surface of the top ply of the laminate and are found by using equations of elasticity9 Away from the edges these stresses are the same as predicted by the classical lamination theory discussed in Chapter 4 However near the edges the normal shear stress τxy decreases to zero and the outofplane shear stress τxz becomes infinite not shown The classical lamination theory and elasticity results give different results because the former violates equilibrium and boundary conditions at the interface For example for a simple state of stress on the 45s laminate the classical lamination theory predicts nonzero values for the stresses σxx σyy and τxy for each ply This is not true at the edges where σy and τxy are actually zero because they are free boundaries Interlaminar stresses pose a challenge to the designer and there are some ways to counter their effects Pagano and Pipes9 found theoretically that keeping the angle symmetry and number of plies the same but changing the stacking sequence influences the interlaminar stresses The key to chang ing the stacking sequence is to decrease the interlaminar shear stresses with out increasing the tensile if any interlaminar normal stresses For example a laminate stacking sequence of 3090s produces tensile interlaminar nor mal stresses under a uniaxial tensile load however if the sequence is just changed to 9030s it produces compressive interlaminar normal stresses This makes the latter sequence less prone to delamination Other techniques to improve tolerance to delamination include using toughened resin systems12 and interleaved systems in which a discrete layer of resin with high toughness and strain to failure is added on top of a layer1314 1343bookfm Page 421 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 422 Mechanics of Composite Materials Second Edition 554 Impact Resistance The resistance to impact of laminated composites is important in applica tions such as a bullet hitting a military aircraft structure or even the contact of a composite leaf spring in a car to runaway stones on a gravel road The resistance to impact depends on several factors of the laminate such as the FIGURE 55 Moisture absorption as a function of time and its effect on flexural modulus of a glasspolyester composite rod Reprinted from Quinn JA in Design with Advanced Composite Materials Phil lips LN Ed 1990 Figure 310 p 91 and Figure 311 p 92 SpringerVerlag Heidelberg 06 04 02 0 Moisture absorption 150 100 50 0 Number of days 50 48 47 45 44 0 01 02 03 04 Flexural modulus GPa Moisture absorption 1343bookfm Page 422 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Failure Analysis and Design of Laminates 423 material system interlaminar strengths stacking sequence and nature of the impact such as velocity mass and size of the impacting object Impact reduces strengths of the laminate and also initiates delamination in com posites Delamination becomes more problematic because many times visual inspection cannot find it Solutions for increasing impact resistance and residual impact strength have included toughened epoxies and inter leaved laminates In the former epoxies are toughened by liquid rubber and in the latter case a discrete toughened layer is added to the laminae at selected places 555 Fracture Resistance When a crack develops in an isotropic material the stresses at the crack tip are infinite The intensity of these infinite stresses is called the stress intensity factor If the stress intensity factor is greater than the critical stress intensity factor for that material the crack is considered to grow catastrophically Another parameter called the strain energy release rate is also used in determining fracture resistance This is the rate of the energy release as the crack grows If this rate is greater than the critical strain energy release rate of the material the crack will grow catastrophically The strain energy release rate and stress intensity factor are related to each other in isotropic materials In composites the mechanics of fracture is not as simple First cracks can grow in the form of fiber breaks matrix breaks debonding between fiber and matrix and debonding between layers Second no single critical stress FIGURE 56 Normal and shear stresses at the interface of bottom surface of top ply in a fourply laminate Reprinted from Pagano NJ and Soni SR in Interlaminar Response of Composite Materials Pagano NJ Ed 1989 p 9 Elsevier Science New York with kind permission from authors 50 40 30 20 10 0 Stress KPa 0 025 05 075 1 Normalized distance from laminate center Laminate center line Laminate free edge σx τxy σy 1343bookfm Page 423 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 424 Mechanics of Composite Materials Second Edition intensity factors and strain energy release rates can determine the fracture mechanics process Fiber breaks may occur because of the brittle nature of fibers Some fibers may break because statistically some fibers are weaker than others and thus fail at low strains The matrix may then break because of high strains caused by the fiber breaks In ceramic matrix composites the matrix failure strain is lower than that of the fiber Therefore matrix breaks precede fiber breaks In fact fiber breaks are seen to occur only close to the ultimate failure of the composite Also matrix breaks may keep occurring parallel to the crack length When a fiber or matrix breaks the crack does not grow in a selfsimilar fashion It may grow along the interface that blunts the crack and improves the fracture resistance of the composites or it may grow into the next con stituent resulting in uncontrolled failure The competition between whether a crack grows along the interface or jumps to the adjoining constituent depends on the material properties of the fiber matrix and the interface as well as the fiber volume fraction Fracture mechanics in composites is still an open field because there are several mechanisms of failure and developing uniform criteria for the mate rials looks quite impossible 556 Fatigue Resistance Structures over time are subjected to repeated cyclic loading such as the fluctuating loads on an aircraft wing This cyclic loading weakens the mate rial and gives it a finite life For example a composite helicopter blade may have a service life of 10000 hours Fatigue data for composite materials are collected using several different data such as plotting the peak stress applied during the loading as a function of the number of cycles The allowable peak stress decreases as the number of cycles to failure is increased The peak stress is compared to the static strength of the composite structure If these peak stresses are comparably larger than the allowable ultimate strength of the composite fatigue does not influence the design of the composite structure This is the case in graph iteepoxy composites in which the allowable ultimate strength is low due to its low impact resistance Other factors that influence the fatigue properties are the laminate stacking sequence fiber and matrix properties fiber volume fraction interfacial bond ing etc For example for quasiisotropic laminates SN curves are quite different from those of unidirectional laminates In this case the 90 plies develop transverse cracks which influence the elastic moduli and strength of the laminate Although the influence is limited because 90 plies do not contribute to the static stiffness and strength in the first place the stress concentrations caused by these cracks may lead to damage in the 0 plies Other damage modes include fiber and matrix breaks interfacial and inter laminar debonding etc Laminate stacking sequence influences the onset of edge delamination For example Foye and Baker15 conducted tensile fatigue 1343bookfm Page 424 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Failure Analysis and Design of Laminates 425 testing of boronepoxy laminates and found the dependence of fatigue life on stacking sequence A 4515s laminate had a higher fatigue life than a 1545s laminate Figure 57 Both laminates have the same number and angle of plies and only the stacking sequence has been changed Loading factors such as tension andor compression temperature mois ture and frequency of loading also determine the fatigue behavior of com posites For example for compressive fatigue loading or tensioncompressive fatigue loading carbonepoxy composites have very low peak strains because compression can cause layer buckling etc In such cases the dom inance of fiber effects is not present but matrix fibermatrix interfaces and the layers play a more important role Nonmechanical issues are also important in design of composite struc tures These include fire resistance smoke emission lightning strikes elec trical and thermal conductivity recycling potential electromagnetic interference etc 56 Summary In this chapter we introduced the special case of laminates and their effect on the stiffness matrices and response to external loads We established FIGURE 57 Comparison of residual strength as a function of number of cycles for two laminates Reprinted from Pagano NJ and Soni SR in Interlaminar Response of Composite Materials Pagano NJ Ed 1989 p 12 Elsevier Science New York with kind permission from authors 80 70 60 50 40 30 104 105 Cycles Residual strength Ksi 106 45451515S 15154545S 107 1343bookfm Page 425 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 426 Mechanics of Composite Materials Second Edition failure criteria for laminates using the plybyply failure theory Examples of designing laminated structures such as plates thin pressure vessels and drive shafts were given Other mechanical design issues such as environ mental effects interlaminar stresses impact resistance fracture resistance and fatigue resistance were discussed Key Terms Special laminates Crossply laminates Angle ply laminates Antisymmetric laminates Balanced laminates Quasiisotropic laminates Failures of laminates Design of laminates Sandwich composites Environmental effects Interlaminar stresses Impact resistance Fracture resistance Fatigue resistance Exercise Set 51 Classify the following laminates 30454530 30303030 303030 45303045 09009009090 0909090900 0183654729018365472 52 Write an example of laminate code for the following Symmetric laminate Antisymmetric laminate Symmetric crossply laminate 1343bookfm Page 426 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Failure Analysis and Design of Laminates 427 Symmetric angleply laminate Balanced angleply laminate 53 Give an example of a laminate with zero coupling stiffness matrix B 54 Is a nonzero B matrix attributed to the orthotropy of layers 55 Is a nonzero B matrix attributed to the unsymmetrical stacking of laminae in a laminate 56 Show numerically that a 090 laminate is not a quasiisotropic laminate Use the properties of unidirectional glassepoxy lamina from Table 22 57 Does a symmetric quasiisotropic laminate have A B and D stiffness matrices like that of an isotropic material 58 Are 06060 and 606060 quasiisotropic laminates 59 Are midplane strains andor midplane curvatures always zero for symmetric laminates 510 Find 1 the extensional stiffness matrix and 2 the extensional elastic moduli of the following graphiteepoxy laminate 018365472 9018365472s Use properties of unidirectional graphite epoxy lamina from Table 21 511 Show that A12 U4h for a quasiisotropic laminate 512 A 090s laminate made of glassepoxy is subjected to an axial load Nx Use properties of unidirectional glassepoxy lamina from Table 22 and assume that each layer is 0005 in thick 1 Use the maximum stress failure theory to find the first and last ply failure of the laminate 2 Draw the stressstrain curve for the laminate till the last ply failure 513 Using TsaiWu theory find the plybyply failure of a 4545s graphiteepoxy laminate under a pure bending moment Mx Use properties of unidirectional graphiteepoxy lamina from Table 21 and assume each layer is 0125 mm thick 514 Repeat the preceding exercise in the presence of a temperature change of ΔT 150F and a moisture content of ΔC 04 515 Develop a comparison table to show the elastic moduli Ex Ey νxy and Gxy and the tensile strengths in x and y directions shear strength in the xy plane of the two laminates and glass epoxy laminate Use properties of unidirectional glassepoxy lamina from Table 22 and assume failure based on first ply failure FPF 516 Find the angle in θns graphiteepoxy sublaminate for maximum value of each of the elastic moduli 1 Ex 0 90 s 45 45 s 1343bookfm Page 427 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 428 Mechanics of Composite Materials Second Edition 2 Ey 3 Gxy Use properties of unidirectional graphiteepoxy lamina from Table 21 517 The bending stiffness of a laminate does not decrease substantially by replacing some of the plies at the midplane 1 Find the percentage decrease in the longitudinal bending mod ulus of a 08 glassepoxy laminate if four of the plies closest to the midplane are replaced by a core of negligible stiffness 2 What is the percentage decrease in the longitudinal bending modulus of a 0904545s glassepoxy laminate if four of the plies closest to the midplane are replaced by a core of negligible stiffness Use properties of unidirectional glassepoxy lamina from Table 21 518 A designer uses a 08 glassepoxy laminate to manufacture a rotat ing blade The inplane longitudinal modulus is adequate but the inplane shear modulus is not A suggestion is to replace the 08 glassepoxy laminate by a 452s graphiteepoxy laminate Use the properties of unidirectional glassepoxy lamina and unidirectional graphiteepoxy lamina from Table 22 to find 1 Whether the longitudinal modulus increases or decreases and by how much 2 Percentage increase or decrease in the inplane shear modulus with the replacement 519 Design a symmetric graphiteepoxy crossply sublaminate such that the thermal expansion coefficient in the xdirection is zero Use the properties of unidirectional graphiteepoxy laminate from Table 21 however assume that the longitudinal coefficient of thermal expan sion is 03 106 mmC 520 1 Find the coefficient of thermal expansion of a symmetric quasi isotropic graphiteepoxy laminate 2 If you were able to change the longitudinal Youngs modulus of the unidirectional graphiteepoxy lamina without affecting the value of other properties what value would you choose to get zero thermal expansion coefficient for the quasiisotropic laminate Use the properties of unidirectional graphiteepoxy lamina given in Table 21 except choose the longitudinal coefficient of thermal expansion as 03 106 mmC 521 Certain laminated structures such as thin walled hollow drive shafts are designed for maximum shear stiffness Find the angle θ for a symmetric θns graphiteepoxy laminate such that the in 1343bookfm Page 428 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Failure Analysis and Design of Laminates 429 plane shear stiffness is a maximum Use the properties of unidirec tional graphiteepoxy lamina from Table 22 522 A thinwalled pressure vessel is manufactured by a filament winding method using glassepoxy prepregs Find the optimum angles θ if the pressure vessel is made of θns sublaminate with 1 Spherical construction for maximum strength 2 Cylindrical construction for maximum strength 3 Cylindrical construction for no change in the internal diameter Apply TsaiWu failure theory Use properties of unidirectional glassepoxy lamina from Table 22 523 A cylindrical pressure vessel with flat ends of length 6 ft and inner diameter of 35 in is subjected to an internal gauge pressure of 150 psi Neglect the end effects and the mass of ends of the pressure vessel in your design Take the factor of safety as 195 1 Design the radial thickness of the pressure vessel using steel For steel assume that the Youngs modulus is 30 Msi Poissons ratio is 03 specific gravity of steel is 78 and the ultimate normal tensile and compressive strength is 36 ksi 2 Find the axial elongation of the steel pressure vessel designed in part 1 assuming plane stress conditions 3 Find whether graphiteepoxy would be a better material to use for minimizing mass if in addition to resisting the applied pres sure the axial elongation of the pressure vessel does not exceed that of the steel pressure vessel The vessel operates at room temperature and curing residual stresses are neglected for sim plification The following are other specifications of the design Only 0 45 45 60 60 and 90 plies can be used The thickness of each lamina is 0005 in Use specific gravities of the laminae from Example 56 Use TsaiWu failure criterion for calculating strength ratios 524 Revisit the design problem of the drive shaft in Example 58 Use graphiteepoxy laminate with ply properties given in Table 21 to design the drive shaft 1 If minimizing mass is still an issue would a graphiteepoxy laminate be a better choice than glassepoxy 2 If cost is the only issue is glassepoxy laminate steel or graph iteepoxy the best choice Assume total manufacturing cost of graphiteepoxy is five times that of glassepoxy on a perunit mass basis and that the glassepoxy and steel cost the same on a perunitmass basis 1343bookfm Page 429 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 430 Mechanics of Composite Materials Second Edition References 1 Gradshetyn IS and Ryzhik IM Table of Integrals Series and Products Aca demic Press New York 1980 2 PROMAL for Windows software Mechanical Engineering Department Uni versity of South Florida Tampa 1996 3 Gurdal Z Haftka RT and Hajela P Design and Optimization of Laminated Composite Materials John Wiley Sons New York 1999 4 James ML Smith GM Wolford JC and Whaley PW Vibration of Mechan ical and Structural Systems Harper and Row New York 1989 5 Column Research Committee of Japan Eds Handbook of Structural Stability Tokyo Corona Publishing 1971 6 Sandwich panel review Part IIV MC Gill Doorway 28 1991 7 Quinn JA Properties of thermoset polymer composites and design of pultru sions in Design with Advanced Composite Materials Philips LN Ed Spring erVerlag New York 1989 Chap 3 8 Pipes RB and Pagano NJ Interlaminar stresses in composite laminates under uniform axial extension J Composite Mater 4 538 1970 9 Pagano NJ and Pipes RB The influence of stacking sequence on laminate strengths Int J Mech Sci 15 679 1973 10 Wang SS Edge delamination in angleply composite laminates AIAA J 21 256 1984 11 Sela N and Ishai O Interlaminar fracture toughness and toughening of laminated composite materials a review Composites 20 423 1989 12 Williams JG OBrien TK and Chapman III AJ Comparison of toughened composite laminates using NASA standard damage tolerance tests NASA CP 2321 ACEE Composite Structure Technology Conference Seattle WA August 1984 13 Chen SF and Jeng BZ Fracture behavior of interleaved fiberresin compos ites Composites Sci Technol 41 77 1991 14 Kaw AK and Goree JG Effect of Interleaves on fracture of laminated com posites Part II ASME J Appl Mech 57 175 1990 15 Foye RL and Baker DJ Design of orthotropic laminates 11th Annual AIAA Conference on Structures Structural Dynamics and Materials Denver CO April 1970 1343bookfm Page 430 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 431 6 Bending of Beams Chapter Objectives Develop formulas to find the deflection and stresses in a beam made of composite materials Develop formulas for symmetric beams that are narrow or wide Develop formulas for nonsymmetric beams that are narrow or wide 61 Introduction To study mechanics of beams made of laminated composite materials we need to review the beam analysis of isotropic materials Several concepts applied to beams made of isotropic materials will help in understanding beams made of composite materials We are limiting our study to beams with transverse loading or applied moments The bending stress in an isotropic beam Figure 61 and Figure 62 under an applied bending moment M is given by 12 61 where z distance from the centroid I second moment of area The bending deflections w are given by solving the differential equation σ Mz I 1343bookfm Page 431 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 432 Mechanics of Composite Materials Second Edition 62 where E Youngs modulus of the beam material The term of is defined as the curvature 63 FIGURE 61 Bending of a beam FIGURE 62 Curvature of a bended beam M M z x O z Neutral axis ρ EI d w dx M 2 2 d w dx 2 2 κx w x 2 2 1343bookfm Page 432 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Bending of Beams 433 giving 64 The formula for the bending stress is only valid for an isotropic material because it assumes that the elastic moduli is uniform in the beam In the case of laminated materials elastic moduli vary from layer to layer 62 Symmetric Beams To keep the introduction simple we will discuss beams that are symmetric and have a rectangular crosssection 3 Figure 63 Because the beam is sym metric the loads and moments are decoupled in Equation 429 to give 65 or 66 Now if bending is only taking place in the x direction then FIGURE 63 Laminated beam showing the midplane and the neutral axis zc Midplane z Neutral axis EI M x κ M M M D x y xy x y xy κ κ κ κ κ κ x y xy x y xy D M M M 1 1343bookfm Page 433 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 434 Mechanics of Composite Materials Second Edition 67 that is 68a 68b 68c where are the elements of the D 1 matrix as given in Equation 428c Because defining midplane curvatures Equation 415 69 the midplane deflection w 0 is not independent of y However if we have a narrow beam that is the length to width ratio Lb is sufficiently high we can assume that w 0 w 0 x only 610 Writing in the form similar to Equation 62 for isotropic beams My 0 Mxy 0 κ κ κ x y xy x D M 1 0 0 κx D Mx 11 κy D Mx 12 κxy D Mx 16 Dij κx w x 2 0 2 κy w y 2 0 2 κxy w x y 2 2 0 κx x d w dx D M 2 0 2 11 1343bookfm Page 434 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Bending of Beams 435 611 where b width of beam E x effective bending modulus of beam I second moment of area with respect to the xy plane From Equation 68a and 611 we get 612 Also 613 614 To find the strains we have from Equation 416 615a 615b 615c These global strains can be transformed to the local strains in each ply using Equation 295 616 The local stresses in each ply are obtained using Equation 273 as d w dx M b E I x x 2 0 2 E h D x 12 3 11 I bh 3 12 M M b x x x zκ y y zκ γ κ xy z xy 1 2 12 1 γ γ k x y xy R T R k 1343bookfm Page 435 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 436 Mechanics of Composite Materials Second Edition 617 The global stresses in each ply are then obtained using Equation 289 as 618 Example 61 A simply supported laminated composite beam of length 01 m and width 5 mm Figure 64 made of graphiteepoxy has the following layup of 0 903030 s A uniform load of 200 Nm is applied on the beam What is the maximum deflection of the beam Find the local stresses at the top of the third ply 30 from the top Assume that each ply is 0125 mm thick and the properties of unidirectional graphiteepoxy are as given in Table 21 Solution The shear and bending moment diagrams for the beam are given in Figure 65 The bending moment is maximum at the center of the beam and is given by 619 where q load intensity Nm L length of the beam m FIGURE 64 Uniformly loaded simply supported beam σ σ τ γ 1 2 12 1 2 12 k k Q σ σ τ σ σ τ x y xy k T 1 1 2 12 k M qL 2 8 5 mm q 200 Nm 01 m 1343bookfm Page 436 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Bending of Beams 437 The maximum bending moment then is 025 Nm Without showing the calculations because they are shown in detail in Chapter 4 see Example 42 we get FIGURE 65 Shear a and bending moment b diagrams of a simply supported beam q 200 Nm b a 0 0 20 20 005 01 025 0 0 005 01 025 M 200 0 1 8 2 D 1 015 10 5 494 10 4 234 10 5 494 1 1 1 1 00 5 243 10 1 567 10 4 234 10 1 567 1 1 0 1 1 0 9 055 10 1 1 3 Pa m 1343bookfm Page 437 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 438 Mechanics of Composite Materials Second Edition To find the maximum deflection of the beam δ we use the isotropic beam formula 620 Now in Equation 612 Thus From Equation 613 D 1 1 3 2 1 009 10 9 209 10 4 557 10 9 20 9 10 1 926 10 2 901 10 4 557 10 2 901 3 1 2 2 10 1 131 10 1 2 0 3 Pa m δ 5 384 qL4 E I x h 8 0 125 10 3 0 001 m D Pa m 11 1 3 1 009 10 1 E h D x 12 12 0 001 1 009 10 1 189 10 3 11 3 1 11Pa I bh m 3 3 3 13 4 12 5 10 0 001 12 4 167 10 1343bookfm Page 438 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Bending of Beams 439 Therefore from Equation 620 The maximum curvature is at the middle of the beam and is given by The global strains Equation 615 at the top of the third ply 30 are δ 5 200 0 1 384 1 189 10 4 167 10 4 11 1 3 5 256 10 3 m 5 256 mm χ χ χ x y xy D D D qL 11 12 16 2 8b 1 009 10 9 209 10 4 557 10 2 1 3 2 00 0 1 8 0 005 2 1 009 10 9 209 10 4 557 10 5 1 3 2 0 5 045 0 4605 2 279 1 m x y xy x y xy z γ κ κ κ 0 00025 5 045 0 4605 2 279 1343bookfm Page 439 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 440 Mechanics of Composite Materials Second Edition The global stresses Equation 618 at the top of the third ply 30 then are Example 62 In Example 61 the widthtoheight ratio in the crosssection of the beam is b h 51 5 This may be considered as a narrowbeam crosssection If the b h ratio were large the crosssection may be considered to be wide beam What are the results of Example 61 if one considers the beam to be a wide beam Solution In the case of wide beams we consider Then from Equation 65 1 261 10 1 151 10 5 696 10 3 4 4 m m σ σ τ γ x y xy x y xy Q 1 094 10 3 246 10 5 419 10 3 246 10 2 11 10 10 10 365 10 2 005 10 5 419 10 2 005 10 10 10 10 10 3 674 10 1 261 10 1 151 10 10 3 4 5 698 10 4 1 034 10 2 680 10 4 511 10 8 7 7 Pa κ κ y xy 0 M M M D D D D D D D D D x y xy 11 12 16 12 22 26 16 26 666 0 0 κx 1343bookfm Page 440 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Bending of Beams 441 giving 621 622 Thus from Equation 69a Equation 611 and Equation 621 and from Equation 620 The relative difference in the value of deflection between the assumption of a wide and narrow beam is M D x x 11κ M M b D b x x 11κ E D h x 12 12 1 015 10 0 001 1 218 10 11 3 1 3 11Pa δ 5 200 0 1 384 1 218 10 4 167 10 4 11 1 3 3 5 131 10 5 131 m mm a narrow wide narrow δ δ δ 100 5 256 5 131 5 256 100 2 357 1343bookfm Page 441 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 442 Mechanics of Composite Materials Second Edition Because there is only 2357 difference in the maximum deflection does this mean that the assumption of wide beams influences the stresses only by a similar amount From Equation 621 Because κy 0 κxy 0 The global strains Equation 615 at the top of the third ply 30 are The global stresses Equation 618 at the top of the third ply 30 then are κx Mx D 11 50 1 015 101 4 926 1 m κ κ κ x y xy m 4 926 0 0 1 x y xy x y xy z γ κ κ κ 0 00025 4 926 0 0 1 232 10 0 0 3 m m 1343bookfm Page 442 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Bending of Beams 443 The relative differences in the stresses obtained using wide and narrow beam assumptions are σ σ τ γ x y xy x y xy Q 1 094 10 3 246 10 5 419 10 3 246 10 2 11 10 10 10 365 10 2 005 10 5 419 10 2 005 10 10 10 10 10 3 674 10 1 232 10 0 0 10 3 1 348 10 3 999 10 6 676 10 8 7 7 Pa a x narrow x wide x narrow σx σ σ σ 100 1 034 10 1 348 10 1 034 10 8 8 8 30 37 a y narrow y wide y narrow σy σ σ σ 100 2 680 10 3 999 10 2 680 10 100 7 7 7 49 22 1343bookfm Page 443 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 444 Mechanics of Composite Materials Second Edition 63 Nonsymmetric Beams In the case of nonsymmetric beams the loads and moment are not decou pled The relationship given by Equation 429 is or Assuming that the preceding 6 6 inverse matrix is denoted by J that is 623 then 624 a xy narrow xy wide xy narrow τxy τ τ τ 100 4 511 10 6 676 10 4 511 10 100 7 7 7 48 00 N M A B B D 0 κ 0 1 κ A B B D N M A B B D J 1 x y xy x y xy J J 0 0 0 11 1 γ κ κ κ 2 13 14 15 16 21 22 23 24 25 26 31 32 33 3 J J J J J J J J J J J J J J 4 35 36 41 42 43 44 45 46 51 52 53 54 55 5 J J J J J J J J J J J J J J 6 61 62 63 64 65 66 J J J J J J N N x y xy x y xy N M M M 1343bookfm Page 444 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Bending of Beams 445 If bending is taking place only in the xdirection then Mx is the only nonzero component giving 625 The strain distribution in the beam then from Equation 416 is 626a 626b 626c Because the beam is unsymmetric the neutral axis does not coincide with the midplane The location of the neutral axis zn is where x 0 From Equation 626a giving 627 x x J M 0 14 y x J M 0 24 γ xy x J M 0 34 κx x J M 44 κy y J M 54 κxy xy J M 64 x x x z 0 κ y y y z 0 κ γ γ κ xy xy z xy 0 0 0 x n x z κ J M z J M x n x 14 44 z J J n 14 44 1343bookfm Page 445 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 446 Mechanics of Composite Materials Second Edition Because from Equation 415 the deflection w0 is not independent of y However if we have a narrow beam that is the lengthtowidth ratio Lb is sufficiently high we can assume that w0 w0x only 628 writing in the form 629 where b width of beam Ex effective bending modulus of beam I second moment of area with respect to the xyplane From Equation 628 and Equation 629 we get 630 Also κx w x 2 0 2 κy w y 2 0 2 κxy w x y 2 2 0 κx x d w dx J M 2 0 2 44 d w dx M b E I x x 2 0 2 E h J x 12 3 44 I bh 3 12 M M b x 1343bookfm Page 446 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Bending of Beams 447 To find the strains we have from Equation 416 631a 631b 631c These global strains can be transformed to the local strains in each ply using Equation 295 632 The local stresses in each ply are obtained using Equation 273 as 633 The global stresses in each ply are then obtained using Equation 289 as 634 Example 63 A simply supported laminated composite beam Figure 64 of length 01 m and width 5 mm made of graphiteepoxy has the following layup 090 30302 A uniform load of 200 Nm is applied on the beam What is the maximum deflection of the beam Find the local stresses at the top of the third ply 30 from the top Assume that each ply is 0125 mm thick and the properties of unidirectional graphiteepoxy are as given in Table 21 Solution The stiffness matrix found by using Equation 428 and Equation 429 is x x o x zκ y y o y zκ γ γ κ xy xy z xy 0 1 2 12 1 γ γ k x y xy R T R k σ σ τ γ 1 2 12 1 2 12 k k Q σ σ τ σ σ τ x y xy k T 1 1 2 12 k 1343bookfm Page 447 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 448 Mechanics of Composite Materials Second Edition The inverse of the matrix is Now in Equation 630 From Equation 613 1 027 10 1 768 10 3 497 10 1 848 10 1 84 8 7 10 3 8 10 1 694 10 1 768 10 5 986 10 2 608 10 3 3 7 7 9 11 848 10 1 848 10 6 267 10 3 497 10 2 6 3 3 2 10 08 10 2 195 10 1 694 10 6 267 10 1 848 10 9 7 3 2 3 3 3 3 1 848 10 1 848 10 1 694 10 9 231 1 473 4 2 334 10 1 848 10 1 848 10 6 267 10 1 473 4 1 3 3 2 3319 1 567 10 1 694 10 6 267 10 1 848 10 4 1 3 2 3 234 10 1 567 10 1 829 1 1 1 068 10 3 409 10 7 009 10 4 298 10 8 9 10 6 7 241 10 9 809 10 3 409 10 1 829 10 6 6 9 8 44 042 10 6 097 10 1 142 10 3 083 10 10 6 5 6 10 10 8 7 009 10 4 042 10 5 035 10 6 339 10 6 6 5 6 3 460 10 4 989 10 4 298 10 6 097 1 00 6 339 10 1 194 10 4 335 10 1 940 6 6 1 2 110 7 241 10 1 142 10 3 460 10 4 355 2 6 5 6 10 2 551 10 5 480 10 9 809 10 3 08 2 1 3 6 33 10 4 989 10 1 940 10 5 480 10 6 1 6 5 2 3 223 10 1 h 8 0 125 10 3 0 001 m J Pa m 44 1 3 1 194 10 1 E h J x 12 12 0 001 1 194 10 1 005 10 3 44 3 1 1 1 Pa I bh 3 12 1343bookfm Page 448 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Bending of Beams 449 Thus from Equation 620 The maximum bending moment occurs at the middle of the beam and is given by Calculating the midplane strains and curvature from Equation 624 gives 5 10 0 001 12 3 3 4 167 10 13 4 m δ 5 200 0 1 384 1 005 10 4 167 10 4 11 1 3 3 6 219 10 6 219 m mm M qL N m max 2 2 8 200 0 1 8 0 25 M M b N m m x max max 0 25 0 005 50 1343bookfm Page 449 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 450 Mechanics of Composite Materials Second Edition giving The global strains Equation 631 at the top of the third ply 30 are x y xy x y xy 0 0 0 1 0 γ κ κ κ 68 10 3 409 10 7 009 10 4 298 10 7 8 9 10 6 2241 10 9 809 10 3 409 10 1 829 10 4 6 6 9 8 0042 10 6 097 10 1 142 10 3 083 10 7 10 6 5 6 009 10 4 042 10 5 035 10 6 339 10 10 10 8 6 3 460 10 4 989 10 4 298 10 6 097 10 6 5 6 6 6 1 2 6 339 10 1 194 10 4 335 10 1 940 10 2 6 5 6 7 241 10 1 142 10 3 460 10 4 355 1 00 2 551 10 5 480 10 9 809 10 3 083 2 1 3 6 110 4 989 10 1 940 10 5 480 10 6 123 6 5 2 3 10 0 0 0 50 0 0 1 x y xy 0 0 0 4 4 2 149 10 3 048 10 γ 3 169 10 4 κ κ κ x y xy 5 970 2 178 9 700 10 1 x y xy x y xy x z γ γ κ κ 0 0 0 y xy κ 2 149 10 3 048 10 3 169 10 4 4 4 0 00025 5 970 2 178 9 700 10 1 1 278 10 2 397 10 7 431 10 3 4 5 m m 1343bookfm Page 450 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Bending of Beams 451 The global stresses Equation 634 at the top of the third ply 30 are Example 64 In Example 63 the widthtoheight ratio in the crosssection of the beam is bh 51 5 This may be considered as a narrowbeam crosssection If the bh ratio were large the crosssection may be considered to be wide beam What are the results of Example 63 if one considers the beam to be a wide beam Solution In the case of the wide beams we consider Then from Equation 624 635 we get σ σ τ γ x y xy x y xy Q 1 094 10 3 246 10 5 419 10 3 246 10 2 11 10 10 10 365 10 2 005 10 5 419 10 2 005 10 10 10 10 10 3 674 10 1 278 10 2 397 10 7 10 3 4 431 10 5 1 280 10 3 431 10 6 170 10 8 7 7 Pa κy 0 x y xy x xy J J 0 0 0 11 12 0 γ κ κ J J J J J J J J J J J J J J 13 14 15 16 21 22 23 24 25 26 31 32 33 34 J J J J J J J J J J J J J J 35 36 41 42 43 44 45 46 51 52 53 54 55 56 J J J J J J M 61 62 63 64 65 66 0 0 0 x y M 0 1343bookfm Page 451 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 452 Mechanics of Composite Materials Second Edition 636a 636b 636c 636d 636e 636f To find the neutral axis x 0 we use Equation 636a and Equation 636e to give 637 638 From Equation 69a Equation 611 and Equation 638 Thus from Equation 620 we get x x y J M J M 0 14 15 y x y J M J M 0 24 25 γ xy x y J M J M 0 34 35 κx x y J M J M 44 45 0 54 55 J M J M x y κxy x y J M J M 64 65 z J J J J J J J J n 14 55 15 54 44 55 45 54 M bM b J J J J J beam x x 55 44 55 54 45 κ E h J J J J J x 12 12 0 001 2 551 1 3 55 44 55 45 54 3 00 1 194 10 2 551 10 4 355 10 1 1 1 2 4 355 10 1 071 10 2 11 Pa δ 5 200 0 1 384 1 072 10 4 167 10 4 11 1 3 1343bookfm Page 452 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Bending of Beams 453 From Example 63 the maximum bendings moment per unit width is From Equation 636e From Equation 635 5 830 10 5 830 3 m mm M N m m x max 50 M J J M y x max 54 55 4 355 10 2 551 10 50 2 1 8 497 N m m x y xy x y 0 0 0 0 1 068 1 γ κ κ 00 3 409 10 7 009 10 4 298 10 7 241 8 9 10 6 110 9 809 10 3 409 10 1 829 10 4 042 6 6 9 8 110 6 097 10 1 142 10 3 083 10 7 009 10 6 5 6 10 4 042 10 5 035 10 6 339 10 3 4 10 10 8 6 660 10 4 989 10 4 298 10 6 097 10 6 6 5 6 6 3339 10 1 194 10 4 335 10 1 940 10 7 6 1 2 2 241 10 1 142 10 3 460 10 4 355 10 2 6 5 6 2 551 10 5 480 10 9 809 10 3 083 10 1 3 6 6 4 989 10 1 940 10 5 480 10 6 123 10 5 2 3 1 0 0 0 50 8 497 0 1 534 10 2 078 10 3 463 10 4 4 4 5 602 0 1 017 1343bookfm Page 453 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 454 Mechanics of Composite Materials Second Edition The global strains Equation 615 at the top of the third ply 30 are The global stresses Equation 618 at the top of the third ply 30 are The relative differences in the stresses obtained using wide and narrow beam assumptions are x y xy x y xy γ γ 0 0 0 z x y xy κ κ κ 1 534 10 2 078 10 4 4 3 463 10 0 00025 5 602 0 1 0 4 17 1 247 10 2 078 10 9 221 3 4 10 5 m m σ σ τ γ x y xy x y xy Q 1 094 10 3 246 10 5 419 10 3 246 10 11 10 10 1 0 10 10 10 2 365 10 2 005 10 5 419 10 2 005 10 10 10 3 3 674 10 1 247 10 2 078 10 4 3 8 9 221 10 1 382 10 4 354 1 00 6 833 10 7 7 a a x narrow x wide x narrow σx σ σ σ 100 1343bookfm Page 454 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Bending of Beams 455 64 Summary In this chapter we reviewed the bending of isotropic beams and then extended the knowledge to study stresses and deflection in laminated com posite beams The beams could be symmetric or unsymmetric and wide or narrow crosssectioned Differences in the deflection and stress are calculated between the results of a wide and a narrow beam Key Terms Bending stress Second moment of area 1 280 10 1 382 10 1 280 10 100 7 97 8 8 8 a y narrow y wide y narrow σx σ σ σ 100 3 431 10 4 354 10 3 431 10 100 26 90 7 7 7 a xy narrow xy wide xy narrow τxy τ τ τ 100 6 170 10 6 836 10 6 170 10 100 10 79 7 7 7 1343bookfm Page 455 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC 456 Mechanics of Composite Materials Second Edition Symmetric beams Wide beams Narrow beams Unsymmetric beams Exercise Set 61 A simply supported laminated composite beam Figure 66 made of glassepoxy is 75 mm long and has the layup of 302s A uniform load is applied on the beam that is 5 mm in width Assume each ply is 0125 mm thick and the properties of glassepoxy are from Table 21 1 What is the maximum deflection of the beam 2 Find the local stresses at the top of the laminate 62 A simply supported laminated composite beam Figure 66 made of glassepoxy is 75 mm long and has the layup of 304 A uniform load is applied on the beam that is 5 mm in width Assume each ply is 0125 mm thick and the properties of glassepoxy are from Table 21 1 What is the maximum deflection of the beam 2 Find the local stresses at the top of the laminate 63 Calculate the bending stiffness of a narrow beam crossply laminate 0902s Now compare it by using the average modulus of the lam inate Assume that each ply is 0125 mm thick and the properties of glassepoxy are from Table 21 FIGURE 66 Uniformly loaded simply supported beam 5 mm q 100 Nm 0075 m 1343bookfm Page 456 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC Bending of Beams 457 References 1 Buchanan GR Mechanics of Materials HRW Inc New York 1988 2 Ugural AC and Fenster SK Advanced Strength and Applied Elasticity 3rd ed Prentice Hall Englewood Cliffs NJ 1995 3 Swanson SR Introduction to Design and Analysis with Advanced Composite Ma terials Prentice Hall Englewood Cliffs NJ 1997 1343bookfm Page 457 Tuesday September 27 2005 1153 AM 2006 by Taylor Francis Group LLC