(PDF) Classification of Composite Materials

HolmAltenbach· JohannesAltenbach

WolfgangKissing

Mechanics of

Composite

Structural

Elements

Second Edition

Mechanics of Composite Structural Elements

Holm Altenbach •Johannes Altenbach

Wolfgang Kissing

Mechanics of Composite

Structural Elements

Second Edition

123

Holm Altenbach

Institut für Mechanik

Otto-von-Guericke-Universität Magdeburg

Magdeburg, Saxony-Anhalt

Germany

Johannes Altenbach

Magdeburg

Germany

Wolfgang Kissing

Bad Kleinen, Mecklenburg-Vorpommern

Germany

ISBN 978-981-10-8934-3 ISBN 978-981-10-8935-0 (eBook)

https://doi.org/10.1007/978-981-10-8935-0

Library of Congress Control Number: 2018939885

1st edition: ©Springer-Verlag Berlin Heidelberg 2004

2nd edition: ©Springer Nature Singapore Pte Ltd. 2018

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the relevant protective laws and regulations and therefore free for general use.

The publisher, the authors and the editors are safe to assume that the advice and information in this

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authors or the editors give a warranty, express or implied, with respect to the material contained herein or

for any errors or omissions that may have been made. The publisher remains neutral with regard to

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This Springer imprint is published by the registered company Springer Nature Singapore Pte Ltd.

part of Springer Nature

The registered company address is: 152 Beach Road, #21-01/04 Gateway East, Singapore 189721,

Singapore

This is the second edition of the textbook Mechanics of Composite Structural El-

ements published ﬁrst in 2004. Since that time the course has been delivered at

several universities in Germany and abroad. Throughout the past years the authors

received a lot of suggestions for improvements from students and colleagues alike.

In addition, the textbook is recommended as the basic reading material in a relevant

course at the Otto-von-Guericke-Universit¨at Magdeburg.

In 2016 the ﬁrst author was invited by Prof. Andreas ¨

Ochsner to present a course

with the same title at the Grifﬁth University (Gold Coast campus) for third and

fourth year students of the bachelor program in the departments of Mechanical Engi-

neering and Civil Engineering. The two weeks’ course included 60 hours of lectures

and tutorials. Finally, the course was concluded with a written exam and a project.

Our special thanks are due to Dr. Christoph Baumann and Springer who provided

personal copies of the ﬁrst edition of the book for the attendants of the course. As

the result of the discussions with the students the idea was born to prepare a second

edition.

By and large the preliminaries of the ﬁrst edition remain unchanged: the presen-

tation of the mechanics of composite materials is based on the knowledge of the

ﬁrst and second year of the bachelor program in Engineering Mechanics (or in other

countries the courses of General Mechanics and Strength of Materials). The focus of

the students will be directed to the elementary theory as the starting point of further

advanced courses.

There are some changes in the second edition in comparison with the ﬁrst one:

•some problems are added or clariﬁed (and we hope now better understandable),

•Chapter 11 is slightly shortened (some details are no more important),

•some details were adopted considering the developments of Springer’s templates.

Some references for further reading, but also some original sources are added and

the tables with material data are improved. Of course, we hope you will now ﬁnd

fewer misprints and typos.

We have to acknowledge Dr.-Ing. Heinz K ¨oppe (Otto-von-Guericke-Universit¨at

Magdeburg) and Dipl.-Ing. Christoph Kammer (formaly at Otto-von-Guericke-

v

Preface to the 2nd Edition

vi Preface

Universit¨at Magdeburg) for ﬁnding a lot of typos. In addition, we have to thank

Dr. Christoph Baumann (Executive Editor Engineering, Springer Nature Singapore)

for the permanent support of the project. We appreciate for any comment and sug-

gestion for improvements which should be sent to holm.altenbach@ovgu.de.

Magdeburg and Bad Kleinen, Holm Altenbach

March 2018 Johannes Altenbach

Wolfgang Kissing

Preface to the 1st Edition

Laminate and sandwich structures are typical lightweight elements with rapidly ex-

panding application in various industrial ﬁelds. In the past, these structures were

used primarily in aircraft and aerospace industries. Now, they have also found ap-

plication in civil and mechanical engineering, in the automotive industry, in ship-

building, the sport goods industries, etc. The advantages that these materials have

over traditional materials like metals and their alloys are the relatively high speciﬁc

strength properties (the ratio strength to density, etc). In addition, the laminate and

sandwich structures provide good vibration and noise protection, thermal insulation,

etc. There are also disadvantages - for example, composite laminates are brittle, and

the joining of such elements is not as easy as with classical materials. The recycling

of these materials is also problematic, and a viable solution is yet to be developed.

Since the application of laminates and sandwiches has been used mostly in new

technologies, governmental and independent research organizations, as well as big

companies, have spent a lot of money for research. This includes the development

of new materials by material scientists, new design concepts by mechanical and

civil engineers as well as new testing procedures and standards. The growing de-

mands of the industry for specially educated research and practicing engineers and

material scientists have resulted in changes in curricula of the diploma and master

courses. More and more universities have included special courses on laminates and

sandwiches, and training programs have been arranged for postgraduate studies.

The concept of this textbook was born 10 years ago. At that time, the ﬁrst edition

of ”Einf ¨uhrung in die Mechanik der Laminat- und Sandwichtragwerke”, prepared

by H. Altenbach, J. Altenbach and R. Rikards, was written for German students

only. The purpose of that book consisted the following objectives:

•to provide a basic understanding of composite materials like laminates and sand-

wiches,

•to perform and engineering analysis of structural elements like beams and plates

made from laminates and sandwiches,

•to introduce the ﬁnite element method for the numerical treatment of composite

structures and

vii

viii Preface

•to discuss the limitations of analysis and modelling concepts.

These four items are also included in this textbook. It must be noted that between

1997 and 2000, there was a common education project sponsored by the European

Community (coordinator T. Sadowski) with the participation of colleagues from

U.K., Belgium, Poland and Germany. One of the main results was a new created

course on laminates and sandwiches, and ﬁnally an English textbook ”Structural

Analysis of Laminate and Sandwich Beams and Plates” written by H. Altenbach, J.

Altenbach and W. Kissing.

The present textbook follows the main ideas of its previous versions, but has been

signiﬁcantly expanded. It can be characterized by the following items:

•The textbook is written in the style of classical courses of strength of materials (or

mechanics of materials) and theory of beams, plates and shells. In this sense the

course (textbook) can be recommended for master students with bachelor degree

and diploma students which have ﬁnished the second year in the university. In

addition, postgraduates of various levels can ﬁnd a simple introduction to the

analysis and modelling of laminate and sandwich structures.

•In contrast to the traditional courses referred to above, two extensions have been

included. Firstly, consideration is given to the linear elastic material behavior of

both isotropic and anisotropic structural elements. Secondly, the case of inhomo-

geneous material properties in the thickness direction was also included.

•Composite structures are mostly thin, in which case a dimension reduction of the

governing equations is allowed in many applications. Due to this fact, the one-

dimensional equations for beams and the two-dimensional equations for plates

and shells are introduced. The presented analytical solutions can be related to the

in-plane, out-of-plane and coupled behavior.

•Sandwiches are introduced as a special case of general laminates. This results in

signiﬁcant simpliﬁcations because sandwiches with thin or thicker faces can be

modelled and analyzed in the frame of laminate theories of different order and

so a special sandwich theory is not necessary.

•All analysis concepts are introduced for the global structural behavior. Local

effects and their analysis must be based on three-dimensional ﬁeld equations

which can usually be solved with the help of numerical methods. It must be

noted that the thermomechanical properties of composites on polymer matrix at

high temperatures can be essentially different from those at normal temperatures.

In engineering applications generally three levels of temperature are considered

– normal or room temperature (10◦–30◦C)

– elevated temperatures (30◦–200◦C)

– high temperatures (>200◦C)

High temperatures yield an irreversible variation of the mechanical properties,

and thus are not included in modelling and analysis. All thermal and moisture ef-

fects are considered in such a way that the mechanical properties can be assumed

unchanged.

Preface ix

•Finite element analysis is only brieﬂy presented. A basic course in ﬁnite elements

is necessary for the understanding of this part of the book. It should be noted

that the ﬁnite element method is general accepted for the numerical analysis of

laminate and sandwich structures. This was the reason to include this item in the

contents of this book.

The textbook is divided into 11 chapters and several appendices summarizing the

material properties (for matrix and ﬁber constituents, etc) and some mathematical

formulas:

•In the ﬁrst part (Chaps. 1–3) an introduction into laminates and sandwiches as

structural materials, the anisotropic elasticity, variational methods and the basic

micromechanical models are presented.

•The second part (Chaps. 4–6) can be related to the modelling from single laminae

to laminates including sandwiches, the improved theories and simplest failure

concepts.

•The third part (Chaps. 7–9) discusses structural elements (beams, plates and

shells) and their analysis if they are made from laminates and sandwiches. The

modelling of laminated and sandwich plates and shells is restricted to rectan-

gular plates and circular cylindrical shells. The individual ﬁber reinforced lam-

inae of laminated structured elements are considered to be homogeneous and

orthotropic, but the laminate is heterogeneous through the thickness and gener-

ally anisotropic. An equivalent single layer theory using the classical lamination

theory, and the ﬁrst order shear deformation theory are considered. Multilayered

theories or laminate theories of higher order are not discussed in detail.

•The fourth part (Chap. 10) includes the modelling and analysis of thin-walled

folded plate structures or generalized beams. This topic is not normally consid-

ered in standard textbooks on structural analysis of laminates and sandwiches, but

it was included here because it demonstrates the possible application of Vlasov’s

theory of thin-walled beams and semi-membrane shells on laminated structural

elements.

•Finally, the ﬁfth part (Chap. 11) presents a short introduction into the ﬁnite el-

ement procedures and developed ﬁnite classical and generalized beam elements

and ﬁnite plate elements in the frame of classical and ﬁrst order shear defor-

mation theory. Selected examples demonstrate the possibilities of ﬁnite element

analysis.

This textbook is written for use not only in engineering curricula of aerospace, civil

and mechanical engineering, but also in material science and applied mechanics. In

addition, the book may be useful for practicing engineers, lectors and researchers in

the mechanics of structures composed of composite materials.

The strongest feature of the book is its use as a textbook. No prior knowledge of

composite materials and structures is required for the understanding of its content.

It intends to give an in-depth view of the problems considered and therefore the

number of topics considered is limited. A large number of solved problems are

included to assess the knowledge of the presented topics. The list of references at

the end of the book focuses on three groups of suggested reading:

x Preface

•Firstly, a selection of textbooks and monographs of composite materials and

structures are listed, which constitute the necessary items for further reading.

They are selected to reinforce the presented topics and to provide information

on topics not discussed. We hope that our colleagues agree that the number of

recommended books for a textbook must be limited and we have given priority

to newer books available in university libraries.

•Some books on elasticity, continuum mechanics, plates and shells and FEM are

recommended for further reading, and a deeper understanding of the mathemati-

cal, mechanical and numerical topics.

•A list of review articles shall enable the reader to become informed about the

numerous books and proceedings in composite mechanics.

The technical realization of this textbook was possible only with the support of

various friends and colleagues. Firstly, we would like to express our special thanks

to K. Naumenko and O. Dyogtev for drawing most of the ﬁgures. Secondly, Mrs. B.

Renner and T. Kumar performed many corrections of the English text. At the same

time Mrs. Renner checked the problems and solutions. We received access to the

necessary literature by Mrs. N. Altenbach. Finally, the processing of the text was

done by Mrs. S. Runkel. We would also like to thank Springer Publishing for their

service.

Any comments or remarks are welcome and we kindly ask them to be sent to

holm.altenbach@iw.uni-halle.de.

June 2003

Halle Holm Altenbach

Magdeburg Johannes Altenbach

Wismar Wolfgang Kissing

Contents

Part I Introduction, Anisotropic Elasticity, Micromechanics

1 Classiﬁcation of Composite Materials ............................ 3

1.1 Deﬁnition and Characteristics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.2 Signiﬁcance and Objectives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

1.3 Modelling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

1.4 Material Characteristics of the Constituents . . . . . . . . . . . . . . . . . . . . . 14

1.5 Advantages and Limitations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

1.6 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

References..................................................... 18

2 Linear Anisotropic Materials .................................... 19

2.1 Generalized Hooke’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

2.1.1 Stresses, Strains, Stiffness, and Compliances . . . . . . . . . . . . . 21

2.1.2 Transformation Rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

2.1.3 Symmetry Relations of Stiffness and Compliance Matrices . 32

2.1.3.1 Monoclinic or Monotropic Material Behavior . . . . 32

2.1.3.2 Orthotropic Material Behavior . . . . . . . . . . . . . . . . . 34

2.1.3.3 Transversely Isotropic Material Behavior . . . . . . . . 35

2.1.3.4 Isotropic Material Behavior. . . . . . . . . . . . . . . . . . . . 36

2.1.4 Engineering Parameters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

2.1.4.1 Orthotropic Material Behavior . . . . . . . . . . . . . . . . . 36

2.1.4.2 Transversally-Isotropic Material Behavior . . . . . . . 40

2.1.4.3 Isotropic Material Behavior. . . . . . . . . . . . . . . . . . . . 42

2.1.4.4 Monoclinic Material Behavior . . . . . . . . . . . . . . . . . 43

2.1.5 Two-Dimensional Material Equations . . . . . . . . . . . . . . . . . . . 45

2.1.6 Curvilinear Anisotropy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

2.1.7 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

2.2 Fundamental Equations and Variational Solution Procedures . . . . . . 59

2.2.1 Boundary and Initial-Boundary Value Equations . . . . . . . . . . 59

2.2.2 Principle of Virtual Work and Energy Formulations . . . . . . . . 63

xi

xii Contents

2.2.3 Variational Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69

2.2.3.1 Rayleigh-Ritz Method . . . . . . . . . . . . . . . . . . . . . . . . 69

2.2.3.2 Weighted Residual Methods . . . . . . . . . . . . . . . . . . . 73

2.2.4 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75

References..................................................... 84

3 Effective Material Moduli for Composites . . . . . . . . . . . . . . . . . . . . . . . . . 85

3.1 Elementary Mixture Rules for Fibre-Reinforced Laminae . . . . . . . . . 86

3.1.1 Effective Density . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87

3.1.2 Effective Longitudinal Modulus of Elasticity . . . . . . . . . . . . . 88

3.1.3 Effective Transverse Modulus of Elasticity . . . . . . . . . . . . . . . 89

3.1.4 Effective Poisson’s Ratio . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90

3.1.5 Effective In-Plane Shear Modulus . . . . . . . . . . . . . . . . . . . . . . 91

3.1.6 Discussion on the Elementary Mixture Rules . . . . . . . . . . . . . 92

3.2 Improved Formulas for Effective Moduli of Composites . . . . . . . . . . 93

3.3 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95

Part II Modelling of a Single Laminae, Laminates and Sandwiches

4 Elastic Behavior of Laminate and Sandwich Composites . . . . . . . . . . . . 103

4.1 Elastic Behavior of Laminae . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104

4.1.1 On-Axis Stiffness and Compliances of UD-Laminae . . . . . . . 104

4.1.2 Off-Axis Stiffness and Compliances of UD-Laminae . . . . . . 109

4.1.3 Stress Resultants and Stress Analysis . . . . . . . . . . . . . . . . . . . . 118

4.1.4 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126

4.2 Elastic Behavior of Laminates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131

4.2.1 General Laminates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132

4.2.2 Stress-Strain Relations and Stress Resultants . . . . . . . . . . . . . 135

4.2.3 Laminates with Special Laminae Stacking Sequences . . . . . . 142

4.2.3.1 Symmetric Laminates . . . . . . . . . . . . . . . . . . . . . . . . 143

4.2.3.2 Antisymmetric Laminates . . . . . . . . . . . . . . . . . . . . . 148

4.2.3.3 Stiffness Matrices for Symmetric and

Unsymmetric Laminates in Engineering

Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 150

4.2.4 Stress Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155

4.2.5 Thermal and Hygroscopic Effects . . . . . . . . . . . . . . . . . . . . . . 158

4.2.6 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163

4.3 Elastic Behavior of Sandwiches. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 168

4.3.1 General Assumptions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 169

4.3.2 Stress Resultants and Stress Analysis . . . . . . . . . . . . . . . . . . . . 170

4.3.3 Sandwich Materials with Thick Cover Sheets . . . . . . . . . . . . . 172

4.4 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 174

Contents xiii

5 Classical and Improved Theories .................................177

5.1 General Remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 178

5.2 Classical Laminate Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182

5.3 Shear Deformation Theory for Laminates and Sandwiches . . . . . . . . 188

5.4 Layerwise Theories . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 193

5.5 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 194

References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 200

6 Failure Mechanisms and Criteria ................................201

6.1 Fracture Modes of Laminae . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 202

6.2 Failure Criteria . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 206

6.3 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 219

References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 224

Part III Analysis of Structural Elements

7 Modelling and Analysis of Beams ................................227

7.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 227

7.2 Classical Beam Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 229

7.3 Shear Deformation Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 242

7.4 Sandwich Beams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 248

7.4.1 Stresses and Strains for Symmetrical Cross-Sections . . . . . . . 250

7.4.2 Stresses and Strains for Non-Symmetrical Cross-Sections . . 254

7.4.3 Governing Sandwich Beam Equations . . . . . . . . . . . . . . . . . . . 255

7.5 Hygrothermo-Elastic Effects on Beams . . . . . . . . . . . . . . . . . . . . . . . . 259

7.6 Analytical Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 260

7.7 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 263

8 Modelling and Analysis of Plates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 275

8.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 276

8.2 Classical Laminate Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 277

8.3 Shear Deformation Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 291

8.4 Sandwich Plates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 297

8.5 Hygrothermo-Elastic Effects on Plates . . . . . . . . . . . . . . . . . . . . . . . . . 299

8.6 Analytical Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 302

8.6.1 Classical Laminate Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . 302

8.6.1.1 Plate Strip . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 303

8.6.1.2 Navier Solution. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 308

8.6.1.3 N´adai-L´evy Solution . . . . . . . . . . . . . . . . . . . . . . . . . 312

8.6.2 Shear Deformation Laminate Theory . . . . . . . . . . . . . . . . . . . . 316

8.6.2.1 Plate Strip . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 316

8.6.2.2 Navier Solution. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 320

8.6.2.3 N´adai-L´evy Solution . . . . . . . . . . . . . . . . . . . . . . . . . 322

8.7 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 322

References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 340

xiv Contents

9 Modelling and Analysis of Circular Cylindrical Shells . . . . . . . . . . . . . . 341

9.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 342

9.2 Classical Shell Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 343

9.2.1 General Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 343

9.2.2 Specially Orthotropic Circular Cylindrical Shells

Subjected by Axial Symmetric Loads . . . . . . . . . . . . . . . . . . . 346

9.2.3 Membrane and Semi-Membrane Theories. . . . . . . . . . . . . . . . 350

9.3 Shear Deformation Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 352

9.4 Sandwich Shells . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 360

9.5 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 360

Part IV Modelling and Analysis of Thin-Walled Folded Plate Structures

10 Modelling and Analysis of Thin-walled Folded Structures . . . . . . . . . . 367

10.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 368

10.2 Generalized Beam Models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 371

10.2.1 Basic Assumptions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 372

10.2.2 Potential Energy of the Folded Structure . . . . . . . . . . . . . . . . . 375

10.2.3 Reduction of the Two-dimensional Problem . . . . . . . . . . . . . . 376

10.2.4 Simpliﬁed Structural Models . . . . . . . . . . . . . . . . . . . . . . . . . . 381

10.2.4.1 Structural Model A . . . . . . . . . . . . . . . . . . . . . . . . . . . 381

10.2.4.2 Structural Model B . . . . . . . . . . . . . . . . . . . . . . . . . . . 383

10.2.4.3 Structural Model C . . . . . . . . . . . . . . . . . . . . . . . . . . . 383

10.2.4.4 Structural Model D . . . . . . . . . . . . . . . . . . . . . . . . . . . 384

10.2.4.5 Structural Model E . . . . . . . . . . . . . . . . . . . . . . . . . . . 384

10.2.4.6 Further Special Models by Restrictions of the

Cross-Section Kinematics . . . . . . . . . . . . . . . . . . . . . 385

10.2.5 An Efﬁcient Structure Model for the Analysis of General

Prismatic Beam Shaped Thin-walled Plate Structures . . . . . . 387

10.2.6 Free Eigen-Vibration Analysis, Structural Model A . . . . . . . . 388

10.3 Solution Procedures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 390

10.3.1 Analytical Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 391

10.3.2 Transfer Matrix Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 392

10.4 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 399

References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 406

Part V Finite Classical and Generalized Beam Elements, Finite Plate

Elements

11 Finite Element Analysis .........................................409

11.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 410

11.1.1 FEM Procedure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 410

11.1.2 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 414

11.2 Finite Beam Elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 415

11.2.1 Laminate Truss Elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 416

Contents xv

11.2.2 Laminate Beam Elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 418

11.2.3 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 423

11.3 Finite Plate Elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 425

11.3.1 Classical Laminate Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . 429

11.3.2 Shear Deformation Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . 432

11.4 Generalized Finite Beam Elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . 437

11.4.1 Foundations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 437

11.4.2 Element Deﬁnitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 438

11.4.3 Element Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 440

11.4.4 System Equations and Solution. . . . . . . . . . . . . . . . . . . . . . . . . 444

11.4.5 Equations for the Free Vibration Analysis . . . . . . . . . . . . . . . . 445

11.5 Numerical Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 446

11.5.1 Examples for the Use of Laminated Shell Elements . . . . . . . . 447

11.5.1.1 Cantilever Beam . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 447

11.5.1.2 Laminate Pipe . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 448

11.5.1.3 Sandwich Plate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 451

11.5.1.4 Buckling Analysis of a Laminate Plate . . . . . . . . . . 452

11.5.2 Examples of the Use of Generalized Beam Elements . . . . . . . 456

Part VI Appendices

A Matrix Operations .............................................463

A.1 Deﬁnitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 463

A.2 Special Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 465

A.3 Matrix Algebra and Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 466

B Stress and Strain Transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 471

C Differential Operators for Rectangular Plates . . . . . . . . . . . . . . . . . . . . . 473

C.1 Classical Plate Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 473

C.2 Shear Deformation Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 475

D Differential Operators for Circular Cylindrical Shells . . . . . . . . . . . . . . 477

D.1 Classical Shell Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 477

D.2 Shear Deformation Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 479

E Krylow-Functions as Solution Forms of a Fourth Order Ordinary

Differential Equation ..........................................481

References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 482

F Material’s Properties ...........................................483

References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 483

xvi Contents

G References ....................................................489

G.1 Comprehensive Composite Materiala . . . . . . . . . . . . . . . . . . . . . . . . . . 489

G.2 Selected Textbooks and Monographs on Composite Mechanics . . . . 490

G.3 Supplementary Literature for Further Reading . . . . . . . . . . . . . . . . . . 493

G.4 Selected Review Articles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 494

Index .............................................................497

Part I

Introduction, Anisotropic Elasticity,

Micromechanics

In the ﬁrst part (Chaps. 1–3) an introduction into laminates and sandwiches as

structural materials, the anisotropic elasticity, variational methods and the basic mi-

cromechanical models are presented.

The laminates are introduced as layered structures each of the layers is a ﬁbre-

reinforced material composed of high-modulus, high-strength ﬁbers in a polymeric,

metallic, or ceramic matrix material. Examples of ﬁbers used include graphite, glass,

boron, and silicon carbide, matrix materials are epoxies, polyamide, aluminium, ti-

tanium, and aluminium. A sandwich is a special class of composite materials consist

of two thin but stiff skins and a lightweight but thick core.

The anisotropic elasticity is an extension of the isotropic elasticity. The geomet-

rical relations are assumed to be linear. The constitutive equations contain more

than two material parameters. In addition, the transition from the general three-

dimensional equations to the special two-dimensional equations results in more

complicated constrains. At the same time the introduction of reduced stiffness and

compliance parameters result in a powerful tool for the analysis of laminates.

The variational methods are the base of many numerical solution techniques (for

example, the ﬁnite element method). Here only the classical principles and methods

are brieﬂy discussed.

There are many, partly sophisticated micromechanical approaches. They are the

base of a better understanding of the local behavior. In the focus of this textbook is

the global structural analysis. Thats way the micromechanical models are presented

only in the elementary form.

Chapter 1

Classiﬁcation of Composite Materials

Fibre reinforced polymer composite systems have become increasingly important in

a variety of engineering ﬁelds. Naturally, the rapid growth in the use of composite

materials for structural elements has motivated the extension of existing theories in

structural mechanics, therein. The main topics of this textbook are

•a short introduction into the linear mechanics of deformable solids with an-

isotropic material behavior,

•the mechanical behavior of composite materials as unidirectional reinforced sin-

gle layers or laminated composite materials, the analysis of effective moduli,

some basic mechanisms and criteria of failure,

•the modelling of the mechanical behavior of laminates and sandwiches, gen-

eral assumptions of various theories, classical laminate theory (CLT), effect of

stacking of the layers of laminates and the coupling of stretching, bending and

twisting, ﬁrst order shear deformation theory (FOSDT), an overview on reﬁned

equivalent single layer plate theories and on multilayered plate modelling,

•modelling and analysis of laminate and sandwich beams, plates and shells, prob-

lems of bending, vibration and buckling and

•modelling and analysis of ﬁbre reinforced long thin-walled folded-plate struc-

tural elements.

The textbook concentrates on a simple uniﬁed approach to the basic behavior of

composite materials and the structural analysis of beams, plates and circular cylin-

drical shells made of composite material being a laminate or a sandwich. The in-

troduction into the modelling and analysis of thin-walled folded structural elements

is limited to laminated elements and the CLT. The problems of manufacturing and

recycling of composites will be not discussed, but to use all beneﬁts of the new

young material composite, an engineer has to be more than a material user as for

classical materials as steel or alloys. Structural engineering qualiﬁcation must in-

clude knowledge of material design, manufacturing methods, quality control and

recycling.

In Chap. 1 some basic questions are discussed, e.g. what are composites and how

they can be classiﬁed, what are the main characteristics and signiﬁcance, micro-

3

H. Altenbach et al., Mechanics of Composite Structural

Elements, https://doi.org/10.1007/978-981-10-8935-0_1

4 1 Classiﬁcation of Composite Materials

and macro-modelling, why composites are used, what are the advantages and the

limitations. The App. F contains some values of the material characteristics of the

constituents of composites.

1.1 Deﬁnition and Characteristics

Material science classiﬁes structural materials into three categories

•metals,

•ceramics and

•polymers.

It is difﬁcult to give an exact assessment of the advantages and disadvantages of

these three basic material classes, because each category covers whole groups of

materials within which the range of properties is often as broad as the differences

between the three material classes. But at the simplistic level some obvious charac-

teristic properties can be identiﬁed:

•Mostly metals are of medium to high density. They have good thermal stability

and can be made corrosion-resistant by alloying. Metals have useful mechanical

characteristics and it is moderately easy to shape and join. For this reason metals

became the preferred structural engineering material, they posed less problems

to the designer than either ceramic or polymer materials.

•Ceramic materials have great thermal stability and are resistant to corrosion,

abrasion and other forms of attack. They are very rigid but mostly brittle and

can only be shaped with difﬁculty.

•Polymer materials (plastics) are of low density, have good chemical resistance

but lack thermal stability. They have poor mechanical properties, but are eas-

ily fabricated and joined. Their resistance to environmental degradation, e.g. the

photomechanical effects of sunlight, is moderate.

A material is called homogeneous if its properties are the same at every point and

therefore independent of the location. Homogeneity is associated with the scale of

modelling or the so-called characteristic volume and the deﬁnition describes the

average material behavior on a macroscopic level. On a microscopic level all ma-

terials are more or less homogeneous but depending on the scale, materials can be

described as homogeneous, quasi-homogeneous or inhomogeneous. A material is

inhomogeneous or heterogeneous if its properties depend on location. But in the av-

erage sense of these deﬁnitions a material can be regarded as homogeneous, quasi-

homogeneous or heterogeneous if the scale decreases.

A material is isotropic if its properties are independent of the orientation, they do

not vary with direction. Otherwise the material is anisotropic. A general anisotropic

material has no planes or axes of material symmetry, but in Sect. 2.1.3 some special

kinds of material symmetries like orthotropy, transverse isotropy, etc., are discussed

in detail.

1.1 Deﬁnition and Characteristics 5

Furthermore, a material can depend on several constituents or phases, single

phase materials are called monolithic. The above three mentioned classes of conven-

tional materials are on the macroscopic level more or less monolithic, homogeneous

and isotropic.

The group of materials which can be deﬁned as composite materials is extremely

large. Its boundaries depend on deﬁnition. In the most general deﬁnition we can

consider a composite as any material that is a combination of two or more materi-

als, commonly referred to as constituents, and have material properties derived from

the individual constituents. These properties may have the combined characteristics

of the constituents or they are substantially different. Sometimes the material prop-

erties of a composite material may exceed those of the constituents. This general

deﬁnition of composites includes natural materials like wood, traditional structural

materials like concrete, as well as modern synthetic composites such as ﬁbre or par-

ticle reinforced plastics which are now an important group of engineering materials

where low weight in combination with high strength and stiffness are required in

structural design.

In the more restrictive sense of this textbook a structural composite consists of

an assembly of two materials of different nature. In general, one material is dis-

continuous and is called the reinforcement, the other material is mostly less stiff

and weaker. It is continuous and is called the matrix. The properties of a composite

material depends on

•The properties of the constituents,

•The geometry of the reinforcements, their distribution, orientation and concen-

tration usually measured by the volume fraction or ﬁber volume ratio,

•The nature and quality of the matrix-reinforcement interface.

In a less restricted sense, a structural composite can consist of two or more phases

on the macroscopic level. The mechanical performance and properties of compos-

ite materials are superior to those of their components or constituent materials taken

separately. The concentration of the reinforcement phase is a determining parameter

of the properties of the new material, their distribution determines the homogeneity

or the heterogeneity on the macroscopic scale. The most important aspect of com-

posite materials in which the reinforcement are ﬁbers is the anisotropy caused by the

ﬁber orientation. It is necessary to give special attention to this fundamental charac-

teristic of ﬁbre reinforced composites and the possibility to inﬂuence the anisotropy

by material design for a desired quality.

Summarizing the aspects deﬁning a composite as a mixture of two or more dis-

tinct constituents or phases it must be considered that all constituents have to be

present in reasonable proportions that the constituent phases have quite different

properties from the properties of the composite material and that man-made com-

posites are produced by combining the constituents by various means. Figure 1.1

demonstrates typical examples of composite materials. Composites can be classi-

ﬁed by their form and the distribution of their constituents (Fig. 1.2). The reinforce-

ment constituent can be described as ﬁbrous or particulate. The ﬁbres are continuous

(long ﬁbres) or discontinuous (short ﬁbres). Long ﬁbres are arranged usually in uni-

6 1 Classiﬁcation of Composite Materials

ab

s s ss

s

sts

s s s

s

cd

efg

♣

♣♣

♣

♣♣

♣

♣♣

♣

♣♣

♣♣♣♣♣♣♣

♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣

♣

♣♣♣♣♣♣♣♣♣

♣♣♣♣♣♣♣♣♣h

Fig. 1.1 Examples of composite materials with different forms of constituents and distributions

of the reinforcements. aLaminate with uni- or bidirectional layers, birregular reinforcement with

long ﬁbres, creinforcement with particles, dreinforcement with plate strapped particles, erandom

arrangement of continuous ﬁbres, firregular reinforcement with short ﬁbres, gspatial reinforce-

ment, hreinforcement with surface tissues as mats, woven fabrics, etc.

or bidirectional, but also irregular reinforcements by long ﬁbres are possible. The

arrangement and the orientation of long or short ﬁbres determines the mechani-

cal properties of composites and the behavior ranges between a general anisotropy

to a quasi-isotropy. Particulate reinforcements have different shapes. They may be

spherical, platelet or of any regular or irregular geometry. Their arrangement may be

random or regular with preferred orientations. In the majority of practical applica-

tions particulate reinforced composites are considered to be randomly oriented and

unidirectional

reinforced

bidirectional

reinforced

spatial

reinforced

random

orientation preferred orientation

continous ﬁbre reinforced

(long ﬁbres)

discontinous ﬁbre reinforced

(short ﬁbres)

random

orientation

preferred

orientation

ﬁbre reinforced particle reinforced

Composite

Fig. 1.2 Classiﬁcation of composites

1.1 Deﬁnition and Characteristics 7

the mechanical properties are homogeneous and isotropic (for mor details Chris-

tensen, 2005; Torquato, 2002). The preferred orientation in the case of continuous

ﬁbre composites is unidirectional for each layer or lamina. Fibre reinforced compos-

ites are very important and in consequence this textbook will essentially deal with

modelling and analysis of structural elements composed of this type of composite

material. However, the level of modelling and analysis used in this textbook does not

really differentiate between unidirectional continuous ﬁbres, oriented short-ﬁbres or

woven ﬁbre composite layers, as long as material characteristics that deﬁne the layer

response are used. Composite materials can also be classiﬁed by the nature of their

constituents. According to the nature of the matrix material we classify organic,

mineral or metallic matrix composites.

•Organic matrix composites are polymer resins with ﬁllers. The ﬁbres can be min-

eral (glass, etc.), organic (Kevlar, etc.) or metallic (aluminium, etc.).

•Mineral matrix composites are ceramics with metallic ﬁbres or with metallic or

mineral particles.

•Metallic matrix composites are metals with mineral or metallic ﬁbres.

Structural composite elements such as ﬁbre reinforced polymer resins are of par-

ticular interest in this textbook. They can be used only in a low temperature range

up to 200◦to 300◦C. The two basic classes of resins are thermosets and thermo-

plastics. Thermosetting resins are the most common type of matrix system for com-

posite materials. Typical thermoset matrices include Epoxy,Polyester,Polyamide

(Thermoplastics) and Vinyl Ester, among popular thermoplastics are Polyethylene,

Polystyrene and Polyether-ether-ketone (PEEK) materials. Ceramic based compos-

ites can also be used in a high temperature range up to 1000◦C and metallic matrix

composites in a medium temperature range.

In the following a composite material is constituted by a matrix and a ﬁbre re-

inforcement. The matrix is a polyester or epoxy resin with ﬁllers. By the addition

of ﬁllers, the characteristics of resins will be improved and the production costs

reduced. But from the mechanical modelling, a resin-ﬁller system stays as a ho-

mogeneous material and a composite material is a two phase system made up of a

matrix and a reinforcement.

The most advanced composites are polymer matrix composites. They are charac-

terized by relatively low costs, simple manufacturing and high strength. Their main

drawbacks are the low working temperature, high coefﬁcients of thermal and mois-

ture expansion and, in certain directions, low elastic properties. Most widely used

manufacturing composites are thermosetting resins as unsaturated polyester resins

or epoxy resins. The polyester resins are used as they have low production cost.

The second place in composite production is held by epoxy resins. Although epoxy

is costlier than polyester, approximately ﬁve time higher in price, it is very popu-

lar in various application ﬁelds. More than two thirds of polymer matrices used in

aerospace industries are epoxy based. Polymer matrix composites are usually rein-

forced by ﬁbres to improve such mechanical characteristics as stiffness, strength,

etc. Fibres can be made of different materials (glass, carbon, aramid, etc.). Glass ﬁ-

bres are widely used because their advantages include high strength, low costs, high

8 1 Classiﬁcation of Composite Materials

chemical resistance, etc., but their elastic modulus is very low and also their fatigue

strength. Graphite or carbon ﬁbres have a high modulus and a high strength and

are very common in aircraft components. Aramid ﬁbres are usually known by the

name of Kevlar, which is a trade name. Summarizing some functional requirements

of ﬁbres and matrices in a ﬁbre reinforced polymer matrix composite

•ﬁbres should have a high modulus of elasticity and a high ultimate strength,

•ﬁbres should be stable and retain their strength during handling and fabrication,

•the variation of the mechanical characteristics of the individual ﬁbres should be

low, their diameters uniform and their arrangement in the matrix regular,

•matrices have to bind together the ﬁbres and protect their surfaces from damage,

•matrices have to transfer stress to the ﬁbres by adhesion and/or friction and

•matrices have to be chemically compatible with ﬁbres over the whole working

period.

The ﬁbre length, their orientation, their shape and their material properties are main

factors which contribute to the mechanical performance of a composite. Their vol-

ume fraction usually lies between 0.3 and 0.7. Although matrices by themselves

generally have poorer mechanical properties than compared to ﬁbres, they inﬂuence

many characteristics of the composite such as the transverse modulus and strength,

shear modulus and strength, thermal resistance and expansion, etc.

An overview of the material characteristics is given in Sect. 1.4. One of the most

important factors which determines the mechanical behavior of a composite material

is the proportion of the matrix and the ﬁbres expressed by their volume or their

weight fraction. These fractions can be established for a two phase composite in the

following way. The volume Vof the composite is made from a matrix volume Vm

and a ﬁbre volume Vf(V=Vf+Vm). Then

vf=Vf

V,vm=Vm

V(1.1.1)

with

vf+vm=1,vm=1−vf

are the ﬁbre and the matrix volume fractions. In a similar way the weight or mass

fractions of ﬁbres and matrices can be deﬁned. The mass Mof the composite is

made from Mfand Mm(M=Mf+Mm) and

mf=Mf

M,mm=Mm

M(1.1.2)

with

mf+mm=1,mm=1−mf

are the mass fractions of ﬁbres and matrices. With the relation between volume,

mass and density

ρ

=M/V, we can link the mass and the volume fractions

1.2 Signiﬁcance and Objectives 9

ρ

=M

V=Mf+Mm

V=

ρ

fVf+

ρ

mVm

V

=

ρ

fvf+

ρ

mvm=

ρ

fvf+

ρ

m(1−vf)

(1.1.3)

Starting from the total volume of the composite V=Vf+Vmwe obtain

M

ρ

=Mf

ρ

f

+Mm

ρ

m

and

ρ

=1

mf

ρ

f

+mm

ρ

m

(1.1.4)

with

mf=

ρ

f

ρ

vf,mm=

ρ

m

ρ

vm

The inverse relation determines

vf=

ρ

f

mf,vm=

ρ

m

mm(1.1.5)

The density

ρ

is determined by Eqs. (1.1.3) or (1.1.4). The equations can be easily

extended to multi-phase composites.

Mass fractions are easier to measure in material manufacturing, but volume frac-

tions appear in the theoretical equations for effective moduli (Sect. 3.1). Therefore,

it is helpful to have simple expressions for shifting from one fraction to the other.

The quality of a composite material decreases with an increase in porosity. The

volume of porosity should be less than 5 % for a medium quality and less than 1 %

for a high quality composite. If the density

ρ

exp is measured experimentally and

ρ

theor is calculated with (1.1.4), the volume fraction of porosity is given by

vpor =

ρ

theor −

ρ

exp

ρ

theor

(1.1.6)

1.2 Signiﬁcance and Objectives

Development and applications of composite materials and structural elements com-

posed of composite materials have been very rapid in the last decades. The mo-

tivations for this development are the signiﬁcant progress in material science and

technology of the composite constituents, the requirements for high performance

materials is not only in aircraft and aerospace structures, but also in the develop-

ment of very powerful experimental equipments and numerical methods and the

availability of efﬁcient computers. With the development of composite materials

a new material design is possible that allows an optimal material composition in

connection with the structural design. A useful and correct application of compos-

ite materials requires a close interaction of different engineering disciplines such

10 1 Classiﬁcation of Composite Materials

as structural design and analysis, material science, mechanics of materials, process

engineering, etc. Summarizing the main topics of composite material research and

technology are

•investigation of all characteristics of the constituent and the composite materials,

•material design and optimization for the given working conditions,

•development of analytical modelling and solution methods for determining ma-

terial and structural behavior,

•development of experimental methods for material characteristics, stress and de-

formation states, failure,

•modelling and analysis of creep, damage and life prediction,

•development of new and efﬁcient fabrication and recycling procedures among

others.

Within the scope of this book are the ﬁrst three items.

The most signiﬁcant driving force in the composite research and application was

weight saving in comparison to structures of conventional materials such as steel,

alloys, etc. However, to have only material density, stiffness and strength in mind

when thinking of composites is a very narrow view of the possibilities of such ma-

terials as ﬁbre-reinforced plastics because they often may score over conventional

materials as metals not only owing to their mechanical properties. Fibre reinforced

plastics are extremely corrosion-resistant and have interesting electromagnetic prop-

erties. In consequence they are used for chemical plants and for structures which

require non-magnetic materials. Further carbon ﬁbre reinforced epoxy is used in

medical applications because it is transparent to X-rays.

With applications out of aerospace or aircraft, cost competitiveness with con-

ventional materials became important. More recently requirements such as quality

assurance, reproducibility, predictability of the structure behavior over its life time,

recycling, etc. became signiﬁcant.

Applications of polymer matrix composites range from the aerospace industry to

the industry of sports goods. The military aircraft industry has mainly led the ﬁeld

in the use of polymer composites when compared to commercial airlines which

has used composites, because of safety concerns more restrictively and frequently

limited to secondary structural elements. Automotive applications, sporting goods,

medical devices and many other commercial applications are examples for the appli-

cation of polymer matrix composites. Also applications in civil engineering are now

on the way but it will take some time to achieve wide application of composites in

civil engineering as there are a lot of prescribed conditions to guarantee the reliabil-

ity of structures. But it is clear that over the last decades considerable advances have

been made in the use of composite materials in construction and building industries

and this trend will continue.

1.3 Modelling 11

1.3 Modelling

Composite materials consist of two or more constituents and the modelling, analy-

sis and design of structures composed of composites are different from conventional

materials such as steel. There are three levels of modelling. At the micro-mechanical

level the average properties of a single reinforced layer (a lamina or a ply) have to

be determined from the individual properties of the constituents, the ﬁbres and ma-

trix. The average characteristics include the elastic moduli, the thermal and mois-

ture expansion coefﬁcients, etc. The micro-mechanics of a lamina does not consider

the internal structure of the constituent elements, but recognizes the heterogene-

ity of the ply. The micro-mechanics is based on some simplifying approximations.

These concern the ﬁbre geometry and packing arrangement, so that the constituent

characteristics together with the volume fractions of the constituents yield the av-

erage characteristics of the lamina. Note that the average properties are derived by

considering the lamina to be homogeneous. In the frame of this textbook only the

micro-mechanics of unidirectional reinforced laminates are considered (Sect. 3).

The calculated values of the average properties of a lamina provide the basis

to predict the macrostructural properties. At the macro-mechanical level, only the

averaged properties of a lamina are considered and the microstructure of the lamina

is ignored. The properties along and perpendicular to the ﬁbre direction, these are

the principal directions of a lamina, are recognized and the so-called on-axis stress-

strain relations for a unidirectional lamina can be developed. Loads may be applied

not only on-axis but also off-axis and the relationships for stiffness and ﬂexibility,

for thermal and moisture expansion coefﬁcients and the strength of an angle ply can

be determined. Failure theories of a lamina are based on strength properties. This

topic is called the macro-mechanics of a single layer or a lamina (Sect. 4.1).

A laminate is a stack of laminae. Each layer of ﬁbre reinforcement can have

various orientation and in principle each layer can be made of different materi-

als. Knowing the macro-mechanics of a lamina, one develops the macro-mechanics

of the laminate. Average stiffness, ﬂexibility, strength, etc. can be determined for

the whole laminate (Sect. 4.2). The structure and orientation of the laminae in pre-

scribed sequences to a laminate lead to signiﬁcant advantages of composite materi-

als when compared to a conventional monolithic material. In general, the mechan-

ical response of laminates is anisotropic. One very important group of laminated

composites are sandwich composites. They consist of two thin faces (the skins or

sheets) sandwiching a core (Fig. 1.3). The faces are made of high strength materials

having good properties under tension such as metals or ﬁbre reinforced laminates

while the core is made of lightweight materials such as foam, resins with special

ﬁllers, called syntactic foam, having good properties under compression. Sandwich

composites combine lightness and ﬂexural stiffness. The macro-mechanics of sand-

wich composites is considered in Sect. 4.3.

When the micro- and macro-mechanical analysis for laminae and laminates are

carried out, the global behavior of laminated composite materials is known. The last

step is the modelling on the structure level and to analyze the global behavior of a

structure made of composite material. By adapting the classical tools of structural

12 1 Classiﬁcation of Composite Materials

foam core balsa wood core

foam core with ﬁllers balsa wood core with holes

folded plates core honeycomb core

Fig. 1.3 Sandwich materials with solid and hollow cores

analysis on anisotropic elastic structure elements the analysis of simple structures

as beams or plates may be achieved by analytical methods, but for more general

boundary conditions and/or loading and for complex structures, numerical methods

are used.

The composite structural elements in the restricted view of this textbook are lam-

inated or sandwich composites. The motivation for sandwich composites are two-

fold:

•If a beam is bent, the maximum stresses occur at the top and the bottom surface.

So it makes sense using high strength materials only for the sheets and using low

and lightweight materials in the middle.

1.3 Modelling 13

•The resistance to bending of a rectangular cross-sectional beam is proportional

to the cube of the thickness. Increasing the thickness by adding a core in the

middle increases the resistance. The shear stresses have a maximum in the mid-

dle of a sandwich beam requiring the core to support the shear. This advantage

of weight and bending stiffness makes sandwich composites more attractive for

some applications than other composite or conventional materials.

The most commonly used face materials are aluminium alloys or ﬁbre reinforced

laminates and most commonly used core materials are balsa wood, foam and hon-

eycombs (Fig. 1.3). In order to guarantee the advantages of sandwich composites, it

is necessary to ensure that there is perfect bonding between the core and the sheets.

For laminated composites, assumptions are necessary to enable the mathematical

modelling. These are an elastic behavior of ﬁbres and matrices, a perfect bonding

between ﬁbres and matrices, a regular ﬁbre arrangement in regular or repeating ar-

rays, etc.

Summarizing the different length scales of mechanical modelling structure ele-

ments composed of ﬁbre reinforced composites it must be noted that, independent

of the different possibilities to formulate beam, plate or shell theories (Chaps. 7–9),

three modelling levels must be considered:

•The microscopic level, where the average mechanical characteristics of a lamina

have to be estimated from the known characteristics of the ﬁbres and the ma-

trix material taking into account the ﬁbre volume fracture and the ﬁbre packing

arrangement. The micro-mechanical modelling leads to a correlation between

constituent properties and average composite properties. In general, simple mix-

ture rules are used in engineering applications (Chap. 3). If possible, the aver-

age material characteristics of a lamina should be veriﬁed experimentally. On

the micro-mechanical level a lamina is considered as a quasi-homogeneous or-

thotropic material.

•The macroscopic level, where the effective (average) material characteristics of a

laminate have to be estimated from the average characteristics of a set of laminae

taking into account their stacking sequence. The macro-mechanical modelling

leads to a correlation between the known average laminae properties and effec-

tive laminate properties. On the macro-mechanical level a laminate is consid-

ered generally as an equivalent single layer element with a quasi-homogeneous,

anisotropic material behavior (Chap. 4).

•The structural level, where the mechanical response of structural members like

beams, plates, shells etc. have to be analyzed taking into account possibilities to

formulate structural theories of different order (Chap. 5).

In the recent years in the focus of the researchers is an additional level - the

nanoscale level. There are two reasons for this new direction:

•composites reinforced by nanoparticles and

•nanosize structures.

Both directions are beyond this elementary textbook. Partly new concepts should

be introduced since bulk effects are no more so important and the inﬂuence of sur-

14 1 Classiﬁcation of Composite Materials

face effects is increasing. In this case the classical continuum mechanics approaches

should be extended. More details are presented in Altenbach and Eremeyev (2015);

Cleland (2003).

1.4 Material Characteristics of the Constituents

The optimal design and the analysis of structural elements requires a detailed knowl-

edge of the material properties. They depend on the nature of the constituent mate-

rials but also on manufacturing.

For conventional structure materials such as metals or concrete, is available much

research and construction experience over many decades, the codes for structures

composed of conventional materials have been revised continuously and so design

engineers pay less attention to material problems because there is complete docu-

mentation of the material characteristics.

It is quite an another situation for structures made of composites. The list of

composite materials is numerous but available standards and speciﬁcations are very

rare. The properties of each material used for both reinforcements and matrices of

composites are very much diversiﬁed. The experiences of nearly all design en gineers

in civil or mechanical engineering with composite materials, are insufﬁcient. So it

should be borne in mind that structural design based on composite materials requires

detailed knowledge about the material properties of the singular constituents of the

composite for optimization of the material in the frame of structural applications

and also detailed codes for modelling and analysis are necessary.

The following statements are concentrated on ﬁbre reinforced composites with

polymer resins. Material tests of the constituents of composites are in many cases a

complicated task and so the material data in the literature are limited. In engineering

applications the average data for a lamina are often tested to avoid this problem and

in order to use correct material characteristics in structural analysis. But in the area

of material design and selection, it is also important to know the properties of all

constituents.

The main properties for the estimation of the material behavior are

•density

ρ

,

•Young’s modulus1E,

•ultimate strength

σ

uand

•thermal expansion coefﬁcient

α

.

The material can be made in bulk form or in the form of ﬁbres. To estimate proper-

ties of a material in the form of ﬁbres, the ﬁbre diameter dcan be important.

Table F.1 gives the speciﬁc performances of selected material made in bulk form.

Traditional materials, such as steel, aluminium alloys, or glass have comparable

1Thomas Young (∗13 June 1773 Milverton, Somersetshire - †10 May 1829 London) - polymath

and physician, notable scientiﬁc contributions to the ﬁelds of light, mechanics (elastic material

parameter, surface tension), energy among others

1.5 Advantages and Limitations 15

speciﬁc moduli E/

ρ

but in contrast the speciﬁc ultimate stress

σ

u/

ρ

of glass is

signiﬁcantly higher than that of steel and of aluminium alloys. Table F.2 presents the

mechanical characteristics of selected materials made in the form of ﬁbres. It should

be borne in mind that the ultimate strength measured for materials made in bulk form

is remarkably smaller than the theoretical strengths. This is attributed to defects or

micro-cracks in the material. Making materials in the form of ﬁbres with a very

small diameter of several microns decreases the number of defects and the values

of ultimate strength increases. Table F.3 gives material properties for some selected

matrix materials and core materials of sandwich composites. Tables F.4 and F.5

demonstrate some properties of unidirectional ﬁbre reinforced composite materials:

ELis the longitudinal modulus in ﬁbre direction, ETthe transverse modulus, GLT the

in-plane shear modulus,

ν

LT and

ν

TL are the major and the minor Poisson’s ratio2,

σ

Lu,

σ

Tu,

σ

LTu the ultimate stresses or strengths,

α

Land

α

Tthe longitudinal and the

transverse thermal expansion coefﬁcients.

Summarizing the reported mechanical properties, which are only a small selec-

tion, a large variety of ﬁbres and matrices are available to design a composite ma-

terial with high modulus and low density or other desired qualities. The impact of

the costs of the composite material can be low for applications in the aerospace

industry or high for applications such as in automotive industry. The intended per-

formance of a composite material and the cost factors play an important role and

structural design with composite materials has to be compared with the possibilities

of conventional materials.

1.5 Advantages and Limitations

The main advantage of polymer matrix composites in comparison with conventional

materials, such as metals, is their low density. Therefore two parameters are com-

monly used to demonstrate the mechanical advantages of composites:

1. The speciﬁc modulus E/

ρ

is the Young’s modulus per unit mass or the ratio

between Young’s modulus and density.

2. The speciﬁc strength

σ

u/

ρ

is the tensile strength per unit mass or the ratio be-

tween strength and density

The beneﬁt of the low density becomes apparent when the speciﬁc modulus and the

speciﬁc strength are considered. The two ratios are high and the higher the speciﬁc

parameters the more weight reduction of structural elements is possible in relation

to special loading conditions. Therefore, even if the stiffness and/or the strength

performance of a composite material is comparable to that of a conventional alloy,

the advantages of high speciﬁc stiffness and/or speciﬁc strength make composites

2Sim´eon Denis Poisson (∗21 June 1781 Pithiviers - †25 April 1840 Paris) - mathematician, ge-

ometer, and physicist, with contributions to mechanics, after Poisson an elastic material parameter

is named

16 1 Classiﬁcation of Composite Materials

more attractive. Composite materials are also known to perform better under cyclic

loads than metallic materials because of their fatigue resistance.

The reduction of mass yields reduced space requirements and lower material and

energy costs. The mass reduction is especially important in moving structures. Be-

ware that in some textbooks the speciﬁc values are deﬁned as E/

ρ

gand

σ

u/

ρ

g,

where gis the acceleration due to the gravity. Furthermore it should be noted that

a single performance indicator is insufﬁcient for the material estimation and that

comparison of the speciﬁc modulus and the speciﬁc strength of unidirectional com-

posites to metals gives a false impression. Though the use of ﬁbres leads to large

gains in the properties in ﬁbre direction, the properties in the two perpendicular di-

rections are greatly reduced. Additionally, the strength and stiffness properties of

ﬁbre-reinforced materials are poor in another important aspect. Their strength de-

pends critically upon the strength of the ﬁbre, matrix interface and the strength of the

matrix material, if shear stresses are being applied. This leads to poor shear proper-

ties and this lack of good shear properties is as serious as the lack of good transverse

properties. For complex structure loadings, unidirectional composite structural ele-

ments are not acceptable and so-called angle-ply composite elements are necessary,

i.e. the structural components made of ﬁbre-reinforced composites are usually lam-

inated by using a number of layers. This number of ﬁbre-reinforced layers can vary

from just a few to several hundred. While generally the majority of the layers in the

laminate have their ﬁbres in direction of the main loadings, the other layers have

their ﬁbres oriented speciﬁcally to counter the poor transverse and shear properties.

Additional advantages in the material performances of composites are low ther-

mal expansion, high material damping, generally high corrosion resistance and elec-

trical insulation. Composite materials can be reinforced in any direction and the

structural elements can be optimized by material design or material tailoring.

There are also limitations and drawbacks in the use of composite materials:

1. The mechanical characterization of composite materials is much more complex

than that of monolithic conventional material such as metal. Usually composite

material response is anisotropic. Therefore, the material testing is more compli-

cated, cost and time consuming. The evaluation and testing of some composite

material properties, such as compression or shearing strengths, are still in discus-

sion.

2. The complexity of material and structural response makes structural modelling

and analysis experimentally and computationally more expensive and compli-

cated in comparison to metals or other conventional structural materials. There

is also limited experience in the design, calculating and joining composite struc-

tural elements.

Additional disadvantages are the high cost of fabrication, but improvements in pro-

duction technology will lower the cost more and more, further the complicated re-

pair technology of composite structures, a lot of recycling problems, etc.

Summarizing, it can be said that the application of composite materials in struc-

ture design beyond the military and commercial aircraft and aerospace industry and

some special ﬁelds of automotive, sporting goods and medical devices is still in the

1.6 Problems 17

early stages. But the advancing of technology and experience yields an increasing

use of composite structure elements in civil and mechanical engineering and pro-

vides the stimulus to include composite processing, modelling, design and analysis

in engineering education.

1.6 Problems

1. What is a composite and how are composites classiﬁed?

2. What are the constituents of composites?

3. What are the ﬁbre and the matrix factors which contribute to the mechanical

performance of composites?

4. What are polymer matrix, metal matrix and ceramic matrix composites, what are

their main applications?

5. Deﬁne isotropic, anisotropic, homogeneous, nonhomogeneous.

6. Deﬁne lamina, laminate, sandwich. What is micro-mechanical and macro-

mechanical modelling and analysis?

7. Compare the speciﬁc modulus, speciﬁc strength and coefﬁcient of thermal ex-

pansion of glas ﬁbre, epoxy resin and steel.

Exercise 1.1. A typical CFK plate (uni-directional reinforced laminae composed of

carbon ﬁbres and epoxy matrix) has the size 300 mm ×200 mm ×0.5 mm (length

l×bright b×thickness d). Please show that the ﬁbre volume fraction and the ﬁbre

mass fraction are not the same. The estimate should be based on the following data:

•density of the carbon ﬁbres

ρ

f=1,8 g/cm3,

•diameter of the ﬁbres is df=6

µ

m

•ﬁbre volume fraction vf=0.8 and

•density of the epoxy

ρ

m=1,1 g/cm3.

Solution 1.1. Let us assume that the ﬁbres are parallel to the longer plate side. In

this case the volume of one ﬁbre V1fis

V1f=

π

d2

f

4l=

π

(6

µ

m)2

4·300 mm =8,48 ·10−3mm3

The volume of the plate is

Vplate =300 ·200 ·0,5 mm3=3·104mm3

With the ﬁbre volume fraction vf=0.8 we can calculate the matrix volume fraction

vm

vm=1−vf=1−0.8=0.2

The volume of all ﬁbres is

Vf=vfVplate =0,8·3·104mm3=2,4·104mm3

18 1 Classiﬁcation of Composite Materials

which is equivalent to the following number of ﬁbres nf

nf=Vf

V1f

=2,4·104mm3

8,48 ·10−3mm3=0,28 ·107

The ﬁbre mass fraction can be computed as it follows

mf=Mf

M=Mf

Mf+Mm

=

ρ

fVf

ρ

fVf+

ρ

mVm

With

Vm=Vplate −Vf=3·104mm3−2,4·104mm3=0,6·104mm3

ﬁbre mass fraction is

mf=1,8g/cm3·2,4·104mm3

1,8g/cm3·2,4·104mm3+1,1g/cm3·0,6·104mm3=0,867

This value is slightly grater than the assumed ﬁbre volume fraction.

References

Altenbach H, Eremeyev VA (2015) On the theories of plates and shells at the

nanoscale. In: Altenbach H, Mikhasev GI (eds) Shell and Membrane Theories

in Mechanics and Biology: From Macro- to Nanoscale Structures, Springer Inter-

national Publishing, Cham, pp 25–57

Christensen RM (2005) Mechanics of Composite Materials. Dover, Mineola (NY)

Cleland AN (2003) Foundations of Nanomechanics. Advanced Texts in Physics,

Springer, Berlin, Heidelberg

Torquato S (2002) Random Heterogeneous Materials - Microstructure and Macro-

scopic Properties, Interdisciplinary Applied Mathematics, vol 16. Springer, New

York

Chapter 2

Linear Anisotropic Materials

The classical theory of linear elastic deformable solids is based on the following

restrictions to simplify the modelling and analysis:

•The body is an ideal linear elastic body.

•All strains are small.

•The material of the constituent phases is homogeneous and isotropic.

These assumptions of classical theory of elasticity guarantee a satisfying quality

of modelling and analysis of structure elements made of conventional monolithic

materials. Structural analysis of elements composed of composite materials is based

on the theory of anisotropic elasticity, the elastic properties of composites depend

usually on the direction and the deformable solid is anisotropic. In addition, now the

composite material is not homogeneous at all. It must be assumed that the material

is piecewise homogeneous or quasi-homogeneous.

The governing equations of elastic bodies are nearly the same for isotropic and

anisotropic material response. There are equilibrium equations, which describe the

static or dynamic equilibrium of forces acting on an elastic body. The kinematic

equations describe the strain-displacement relations and the compatibility equations

guarantee a unique solution to the equations relating strains and displacements. All

these equations are independent of the elastic properties of the material. Only the

material relations (so-called constitutive equations), which describe the relations

between stresses and strains are very different for an isotropic and an anisotropic

body. This difference in formulating constitutive equations has a great inﬂuence

on the model equations in the frame of the isotropic and the anisotropic theory of

elasticity. Note that in many cases the material behavior of the constituents can

assumed to be homogeneous and isotropic.

The governing equations, as deﬁned above, including so-called initial-boundary

conditions for forces/stresses and/or displacements, yield the basic model equations

for linear elastic solids such as differential equations or variational and energy for-

mulations, respectively. All equations for structural elements which are given in this

textbook, are founded on these general equations for the theory of elasticity of linear

elastic anisotropic solids.

19

H. Altenbach et al., Mechanics of Composite Structural

Elements, https://doi.org/10.1007/978-981-10-8935-0_2

20 2 Linear Anisotropic Materials

The objective of this chapter is to review the generalized Hooke’s law1, the con-

stitutive equations for anisotropic elastic bodies, and to introduce general relations

for stiffness and strains including transformation rules and symmetry relations. The

constitution of a unidirectional composite material and simpliﬁed approaches for

so-called effective moduli result in an engineering formulation of constitutive equa-

tions for ﬁbre reinforced composites and will be considered in Chap. 3.

The theory of anisotropic elasticity presented in Sect. 2.1 begins with the most

general form of the linear constitutive equations, and passes from all speciﬁc cases

of elastic symmetries to the classical Hooke’s law for an isotropic body. The only

assumptions are

•all elastic properties are the same in tension and compression,

•the stress tensor is symmetric,

•an elastic potential exists and is an invariant with respect to linear orthogonal

coordinate transformation.

In addition to the general three-dimensional stress-strain relationships, the plane

stress and plane strain cases are derived and considered for an anisotropic body and

for all the derived speciﬁc cases of elastic symmetries. The type of anisotropy con-

sidered in Sects. 2.1.1–2.1.5 can be called as rectilinear anisotropy, i.e. the homo-

geneous anisotropic body is characterized by the equivalence of parallel directions

passing through different points of the body. Another kind of anisotropy, which can

be interesting to some applications, e.g. to modelling circular plates or cylindrical

tubes, is considered in a comprehensive formulation in Sect. 2.1.6. If one chooses

a system of curvilinear coordinates in such a manner that the coordinate directions

coincide with equivalent directions of the elastic properties at different points of the

body, the elastic behavior is called curvilinear.

The chapter ends with the derivation of the fundamental equations of anisotropic

elasticity and the formulation of variational solution methods. In Sect. 2.2 the dif-

ferential equations for boundary and initial boundary problems are considered. The

classical and generalized variational principles are formulated and approximate an-

alytical solution methods based on variational principles are discussed.

2.1 Generalized Hooke’s Law

The phenomenological modelling neglects the real on the microscopic scale discon-

tinuous structure of the material. On the macroscopic or phenomenological scale the

material models are assumed to be continuous and in general homogeneous. In the

case of ﬁbre reinforced composites, the heterogeneity of the bulk material is a con-

sequence of the two constituents, the ﬁbres and the matrix, but generally there exists

a representative volume element of the material on a characteristic scale at which the

1Robert Hooke (∗28greg./18jul.July 1635 Freshwater, Isle of Wight - †3 March 1703 London) -

natural philosopher, architect and polymath, ﬁrst constitutive law for elasticity published as an

anagram ceiiinosssttuu which is in Latain ut tensio sic vis and means as the extension, so the force

2.1 Generalized Hooke’s Law 21

properties of the material can be averaged to a good approximation. The composite

material can be considered as macroscopically homogeneous and the problem of

designing structural elements made from of composite materials can be solved in an

analogous manner as for conventional materials with the help of the average mate-

rial properties or the so-called effective moduli. Chapter 3 explains the calculation

of effective moduli in detail.

Unlike metals or polymeric materials without reinforcements or reinforced by

stochastically distributed and orientated particles or short ﬁbres, the material behav-

ior of an off-axis forced unidirectional lamina is anisotropic. In comparison to con-

ventional isotropic materials, the experimental identiﬁcation of the material param-

eters is much more complicated in the case of anisotropic materials. But anisotropic

material behavior also has the advantage of material tailoring to suit the main load-

ing cases.

2.1.1 Stresses, Strains, Stiffness, and Compliances

In preparation for the formulation of the generalized Hooke’s law, a one-dimen-

sional problem will be considered. The deformations of an elastic body can be char-

acterized by displacements or by strains:

•Dilatational or extensional strains

ε

: The body changes only its volume but not

its shape.

•Shear strains

γ

: The body changes only its shape but not its volume.

Figure 2.1 demonstrates extensional and shear strains for a simple prismatic body

loaded by forces Fand T, normal and tangential to the cross-section, respectively.

Assuming a uniform distribution of the forces Fand Ton the cross-section, the

elementary one-dimensional deﬁnitions for stresses and strains are given by (2.1.1)

σ

=F

A0

normal stress

σ

,

ε

=l−l0

l0

=△l

l0

extensional strain

ε

,

τ

=T

A0

shear stress

τ

,

γ

≈tan

γ

=△u

l0

shear strain

γ

(2.1.1)

The last one deﬁnition is restricted by small strain assumption. This assumption

can be accepted for many composite material applications and will be used for both

types of strains.

The material or constitutive equations couple stresses and strains. In linear elas-

ticity the one-to-one transformation of stresses and strains yield Hooke’s law (2.1.2)

22 2 Linear Anisotropic Materials

F

A

A0

l0l=l0+△l

△l

F

l0

△u

TA0

T

γ

Fig. 2.1 Extensional strain

ε

and shear strain

γ

of a body with the length l0and the cross-section

area A0

σ

=E

ε

,E=

σ

ε

,Eis the elasticity or Young’s modulus ,

τ

=G

γ

,G=

τ

γ

,Gis the shear modulus

(2.1.2)

For a homog eneous material Eand Gare parameters with respect to the coordinates.

For the extensionally strained prismatic body (Fig. 2.1) the phenomenon of con-

traction in a direction normal to the direction of the tensile loading has to be con-

sidered. The ratio of the contraction, expressed by the transverse strain

ε

t, to the

elongation in the loaded direction, expressed by

ε

, is called Poisson’s ratio

ν

ε

t=−

νε

,

ν

=−

ε

t

ε

(2.1.3)

For an isotropic bar with an extensional strain

ε

>0 it follows that

ε

t<02.

Hooke’s law can be written in the inverse form

ε

=E−1

σ

=S

σ

(2.1.4)

S=E−1is the inverse modulus of elasticity or the ﬂexibility/compliance modulus.

For homogeneous material, Sis an elastic parameter.

Consider a tensile loaded prismatic bar composed of different materials

(Fig. 2.2). Since

σ

=F/Aand

σ

=E

ε

then

σ

A=F=EA

ε

and

ε

= (EA)−1F.

EA is the tensile stiffness and (EA)−1the tensile ﬂexibility or compliance. The dif-

2The class of auxetic materials that have a negative Poisson’s ratio that means when stretched,

they become thicker perpendicular to the applied force, is not in the focus of the present book.

2.1 Generalized Hooke’s Law 23

l

F

C1

Cn

Ai

li

Fig. 2.2 Tensile bar with stiffness Ci=EiAiarranged in parallel and in series

ferent materials of the prismatic bar in Fig. 2.2 can be arranged in parallel or in

series. In the ﬁrst case we have

F=

n

∑

i=1

Fi,A=

n

∑

i=1

Ai,

ε

=

ε

i(2.1.5)

Fiare the loading forces on Aiand the strains

ε

iare equal for the total cross-section.

With

F=EA

ε

,Fi=EiAi

ε

,

n

∑

i=1

Fi=F=

n

∑

i=1

EiAi

ε

(2.1.6)

follow the coupling equations for the stiffness EiAifor a parallel arrangement

EA =

n

∑

i=1

EiAi,(EA)−1=1

n

∑

i=1

EiAi

(2.1.7)

This equal strain treatment is often described as a Voigt3model which is the upper-

bound stiffness.

In the other case, we have

△l=

n

∑

i=1△li

and F=Fi, the elongation △lof the bar is obtained by addition of the △liof the

different parts of the bar with the lengths liand the tensile force is equal for all

cross-sectional areas. With

△l=l

ε

=l(EA)−1F,△li=li

ε

i=li(EiAi)−1F

and

n

∑

i=1△li="n

∑

i=1

li(EiAi)−1#F(2.1.8)

3Woldemar Voigt (∗2 September 1850 Leipzig - †13 December 1919 G ¨ottingen) - physicist,

worked on crystal physics, thermodynamics and electro-optics, the word tensor in its current mean-

ing was introduced by him in 1898, Voigt notation

24 2 Linear Anisotropic Materials

follow the coupling equations for the stiffness EiAiarranged in series

EA =l

n

∑

i=1

li(EiAi)−1

,(EA)−1=

n

∑

i=1

li(EiAi)−1

l(2.1.9)

This equal stress treatment is described generally as a Reuss4model which is the

lower bound stiffness. The coupling equations illustrate a ﬁrst clear insight into a

simple calculation of effective stiffness and compliance parameters for two com-

posite structures.

The three-dimensional state of stress or strain in a continuous solid is completely

determined by knowing the stress or strain tensor. It is common practice to repre-

sent the tensor components acting on the faces of an inﬁnitesimal cube with sides

parallel to the reference axes (Fig. 2.3). The sign convention is deﬁned in Fig. 2.3.

Positive stresses or strains act on the faces of the cube with an outward vector in

the positive direction of the axis of the reference system and vice versa. Using the

tensorial notation for the stress tensor

σ

i j and the strain tensor

ε

i j for the stresses

and the strains we have normal stresses or extensional strains respectively for i=j

and shear stresses or shear strains for i6=j.

ε

i j with i6=jare the tensor shear coor-

dinates and 2

ε

i j =

γ

i j,i6=jthe engineering shear strains. The ﬁrst subscript of

σ

i j

and

ε

i j indicates the plane xi=const on which the load is acting and the second

subscript denotes the direction of the loading. Care must be taken in distinguish-

ing in literature the strain tensor

ε

i j from the tensor ei j which is the tensor of the

relative displacements, ei j =

∂

ui/

∂

xj. An application of shear stresses

σ

i j and

σ

ji

produces in the i j-plane of the inﬁnitesimal cube (Fig. 2.3) angular rotations of the

i- and j-directions by ei j and eji. These relative displacements represent a combina-

tion of strain (distorsion) and rigid body rotation with the limiting cases ei j =eji,

e

e1

e

e2

e

e3

σ

11

σ

12

σ

13

σ

22

σ

21

σ

23

σ

33

σ

32

σ

31

ε

11

ε

13

ε

12

ε

22

ε

21

ε

23

ε

33

ε

32

ε

31

Fig. 2.3 Stress and strain components on the positive faces of an inﬁnitesimal cube in a set of axis

e

e1,e

e

e2,e

e

e3

4Andr´as (Endre) Reuss (∗1 July 1900 Budapest - †10 May 1968 Budapest) - mechanical engineer,

contributions to the theory of plasticity

2.1 Generalized Hooke’s Law 25

j

jj

i

ii

ei j

eji

γ

i j

γ

i j

ei j =eji =

ε

i j =1

2

γ

i j

ω

i j =0

ei j =−eji

ε

i j =

γ

i j =0

ei j =2

ε

ji =

γ

i j

eji =0

abc

Fig. 2.4 Examples of distorsions and rigid body rotation. aPure shear, bpure rotation, csimple

shear

.

i.e. no rotation, and ei j =−ej i, i.e. no distorsion (Fig. 2.4). From the ﬁgure follows

that simple shear is the sum of pure shear and rigid rotation. ei j is positive when

it involves rotating the positive j-direction towards the positive i-direction and vice

versa. Writing the tensor ei j as the sum of symmetric and antisymmetric tensors

ei j =1

2(ei j +eji) + 1

2(ei j −eji) =

ε

i j +

ω

i j (2.1.10)

where

ε

i j is the symmetric strain tensor and

ω

i j is the antisymmetric rotation ten-

sor. For normal strains, i.e. i=j, there is ei j =

ε

i j, however for i6=jwe have

γ

i j =2

ε

i j =ei j +eji with the engineering shear strains

γ

i j and the tensorial shear

strains

ε

i j. Careful note should be taken of the factor of two related engineering and

tensorial shear strains,

γ

i j is often more convenient for practical use but tensor oper-

ations such as rotations of the axis, Sect. 2.1.2, must be carried out using the tensor

notation

ε

i j.

The stress and the strain tensors are symmetric tensors of rank two. They can be

represented by the matrices

σ

=



σ

11

σ

12

σ

13

σ

12

σ

22

σ

23

σ

13

σ

23

σ

33 

,

ε

=



ε

11

ε

12

ε

13

ε

12

ε

22

ε

23

ε

13

ε

23

ε

33 

(2.1.11)

The symmetry of the tensors (2.1.11) reduces the number of unknown components

for deﬁning these tensors to six components. For this reason, an engineering matrix

notation can be used by replacing the matrix table with nine values by a column

matrix or a vector with six components. The column matrices (stress and strain

vector) are written in Eqs. (2.1.12) in a transposed form

26 2 Linear Anisotropic Materials

[

σ

11

σ

22

σ

33

σ

23 ≡

τ

23

σ

13 ≡

τ

13

σ

12 ≡

τ

12]T,

[

ε

11

ε

22

ε

33 2

ε

23 ≡

γ

23 2

ε

13 ≡

γ

13 2

ε

12 ≡

γ

12]T(2.1.12)

The stress and strain states are related by a material law which is deduced from

experimental observations. For a linear elastic anisotropic material, the generalized

Hooke’s law relates the stress and the strain tensor

σ

i j =Cij kl

ε

kl (2.1.13)

Ci jkl are the material coefﬁcients and deﬁne the fourth rank elasticity tensor which

in general case contains 81 coordinates. Due to the assumed symmetry of

σ

i j =

σ

ji

and

ε

i j =

ε

ji the symmetry relations follow for the material tensor

Ci jkl =Cjikl ,Ci jkl =Cij lk (2.1.14)

and reduce the number of coordinates to 36. Introducing a contracted single-

subscript notation for the stress and strain components and a double-subscript nota-

tion for the elastic parameters, the generalized relation for stresses and strains can

be written in vector-matrix form

[

σ

i] = [Cij ][

ε

j];i,j=1,2,...,6; Ci j 6=Cj i;i6=j(2.1.15)

At this stage we have 36 independent material coefﬁcients, but a further reduction

of the number of independent values is possible because we have assumed the exis-

tence of an elastic potential function.

The elastic strain energy is deﬁned as the energy expended by the action of exter-

nal forces in deforming an elastic body: essentially all the work done during elastic

deformations is stored as elastic energy. The strain energy per unit volume, i.e. the

strain energy density function, is deﬁned as follows

W=1

2

σ

i j

ε

i j (2.1.16)

or in a contracted notation

W(

ε

i) = 1

2

σ

i

ε

i=1

2Ci j

ε

j

ε

i(2.1.17)

With

∂

W

∂ε

i

=

σ

i,

∂

2W

∂ε

i

∂ε

j

=Ci j ,

∂

2W

∂ε

j

∂ε

i

=Cji

and

∂

2W

∂ε

i

∂ε

j

=

∂

2W

∂ε

j

∂ε

i

follow the symmetry relations

Ci j =Cj i;i,j=1,2,...,6 (2.1.18)

2.1 Generalized Hooke’s Law 27

and the number of the independent material coefﬁcients is reduced to 21. The gen-

eralized relations for stresses and strains of an anisotropic elastic body written again

in a contracted vector-matrix form have a symmetric matrix for Ci j







σ

1

σ

2

σ

3

σ

4

σ

5

σ

6







=







C11 C12 C13 C14 C15 C16

C22 C23 C24 C25 C26

C33 C34 C35 C36

C44 C45 C46

S Y M C55 C56

C66













ε

1

ε

2

ε

3

ε

4

ε

5

ε

6







(2.1.19)

The transformation rules for the contraction of the subscripts of

σ

i j,

ε

i j and Ci jkl of

(2.1.13) are given in Tables 2.1 and 2.2.

The elasticity equation (2.1.19) can be written in the inverse form as follows







ε

1

ε

2

ε

3

ε

4

ε

5

ε

6







=







S11 S12 S13 S14 S15 S16

S22 S23 S24 S25 S26

S33 S34 S35 S36

S44 S45 S46

S Y M S55 S56

S66













σ

1

σ

2

σ

3

σ

4

σ

5

σ

6







(2.1.20)

with

[Ci j ][Sjk ] = [

δ

ik] = 1i=k,

0i6=k,i,j,k=1,...,6

In a condensed symbolic or subscript form, Eqs. (2.1.19) and (2.1.20) are (summa-

tion on double subscripts)

σ

i=Ci j

ε

j,

ε

i=Si j

σ

j;i,j=1,...,6

σ

=C

C

ε

,

ε

=S

S

σ

(2.1.21)

C

C≡[Ci j ]is the stiffness matrix and S

S

S≡[Si j]the compliance or ﬂexibility matrix. Ci j

and Si j are only for homogeneou s anisotropic materials constant material parameters

Table 2.1 Transformation of

the tensor coordinates

σ

i j and

ε

i j to the vector coordinates

σ

pand

ε

p

σ

i j

σ

p

ε

i j

ε

p

σ

11

σ

1

ε

11

ε

1

σ

22

σ

2

ε

22

ε

2

σ

33

σ

3

ε

33

ε

3

σ

23 =

τ

23

σ

42

ε

23 =

γ

23

ε

4

σ

31 =

τ

31

σ

52

ε

31 =

γ

31

ε

5

σ

12 =

τ

12

σ

62

ε

12 =

γ

12

ε

6

Table 2.2 Transformation

of the tensor coordinates

Ci jkl to the matrix coordi-

nates Cpq

Ci jkl Cpq

i j : 11, 22, 33 p: 1, 2, 3

23, 31, 12 4, 5, 6

kl : 11, 22, 33 q: 1, 2, 3

23, 31, 12 4, 5, 6

28 2 Linear Anisotropic Materials

with respect to the coordinates. Their values depend on the reference coordinate

system. A change of the reference system yields a change of the constant values.

Summarizing the stiffness and the compliance relations, it can be seen that for

a linear elastic anisotropic material 21 material parameters have to be measured

experimentally in the general case. But in nearly all engineering applications there

are material symmetries and the number of material parameters can be reduced.

Section 2.1.2 describes some transformation rules for C

C

Cand S

S

Sfollowing from the

change of reference system and the symmetric properties of anisotropic materials

discussed in Sect. 2.1.3. Furthermore the way that the material parametersCij and Si j

are related to the known engineering elastic parameters Ei,Gij and

ν

i j is considered.

2.1.2 Transformation Rules

If we have a reference system which is characterized by the orthonormal basic unit

vectors e

e

e1,e

e

e2,e

e

e3and another reference system with the vector basis e

e

e′

1,e

e

e′

2,e

e

e′

3. Both

systems are related by a rotation of the coordinate axis (Fig. 2.5), the transformation

rules are

e

e′

i=Ri je

e

ej,e

e

ei=Rjie

e

e′j,

Ri j ≡cos(e

e

e′

i,e

e

ej),Rji ≡cos(e

e

ei,e

e

e′j),i,j=1,2,3 (2.1.22)

These relationships describe a general linear orthogonal coordinate transformation

and can be expressed in vector-matrix notation

e

e′=R

R

Re

e

e,e

e

e=R

R

R−1e

e

e′=R

R

RTe

e

e′(2.1.23)

R

Ris the transformation or rotation matrix. In the case of an orthogonal set of axes

such as given in Fig. 2.5 the matrix R

R

Ris symmetric and unitary. This means the

determinant of this matrix is unity (Det R

R

R=|Ri j|=1 and the inverse matrix R

R

R−1

is identical to the transposed matrix (R

R

R−1=R

R

RT). In the special case of a rotation

φ

Fig. 2.5 Rotation of a refer-

ence system with the basic

vectors e

e

eiinto a system with

the basic vectors e

e

e′

i

x1

x2

x3

x1′

x2′

x3′

e

e1e

e

e′

1

e

e2

e

e′

2

e

e3

e

e′

3

2.1 Generalized Hooke’s Law 29

about the direction e

e

e3, the rotation matrix R

R

Rand the inverse matrix R

R

R−1are

3

Ri j=



c s 0

−s c 0

0 0 1 

,3

Ri j−1

=3

Ri jT

=



c−s0

s c 0

001

,(2.1.24)

and the transformation rules are





e

e′

1

e

e′

2

e

e′

3

=



c s 0

−s c 0

0 0 1 





e

e1

e

e2

e

e3

,



e

e1

e

e2

e

e3

=



c−s0

s c 0

001





e

e′

1

e

e′

2

e

e′

3



with c=cos

φ

,s=sin

φ

. For rotations

ψ

or

θ

about the directions e

e

e2or e

e

e1the

rotations matrices [2

Ri j]and [1

Ri j]are

2

Ri j=



c0−s

0 1 0

s0c

,2

Ri j−1

=2

Ri jT

=



c0s

0 1 0

−s0c

,

1

Ri j=



100

0c s

0−s c 

,1

Ri j−1

=1

Ri jT

=



1 0 0

0c−s

0s c 



with c=cos

ψ

or cos

θ

and s=sin

ψ

or sin

θ

for rotations about e

e

e2or e

e

e1, respec-

tively.

The transformation rule (2.1.22) can be interpreted as a rule for vectors or ﬁrst-

rank tensors. The generalization to second-rank tensors yields e.g. for the stress

tensor

σ

′

i j =RikRjl

σ

kl ,

σ

i j =RkiRl j

σ

′

kl (2.1.25)

For the following reﬂections the transformation rules for the contracted notation are

necessary. The nine tensor coordinates

σ

i j are shifted to six vector coordinates

σ

p.

The transformations

σ

′

p=T

σ

pq

σ

q,

σ

p=T

σ

pq−1

σ

′

q,p,q=1,...,6 (2.1.26)

are not tensor transformation rules. The transformation matrices T

σ

pq and (T

σ

pq)−1

follow by comparison of Eqs. (2.1.25) and (2.1.26). In the same manner we can ﬁnd

the transformation rules for the strains

ε

′

p=T

ε

pq

ε

q,

ε

p=T

ε

pq−1

ε

′

q,p,q=1,...,6 (2.1.27)

The elements of the transformation matrices [T

σ

pq]and [T

ε

pq]are deﬁned in App. B.

Summarizing, the transformation rules for stresses and strains in a condensed

vector-matrix notation as follows

σ

′=T

T

σσ

σ

,

ε

′=T

T

εε

ε

,

σ

= (T

T

σ

)−1

σ

′,

ε

= (T

T

ε

)−1

ε

′(2.1.28)

30 2 Linear Anisotropic Materials

The comparison of

σ

i j =RkiRl j

σ

′

kl with

σ

p= (T

σ

pq)−1

σ

′

q

and

ε

i j =RkiRl j

ε

′

kl with

ε

p= (T

ε

pq)−1

ε

′

q

yields an important result on the linkage of inverse and transposed stress and strain

transformation matrices

(T

T

σ

)−1= (T

T

ε

)T,(T

T

ε

)−1= (T

T

σ

)T(2.1.29)

The transformation relations for the stiffness and the compliance matrices C

C

Cand

S

Scan be obtained from the known rules for stresses and strains. With

σ

=C

C

ε

and

σ

′=C

C

C′

ε

′, it follows that

(T

T

σ

)−1

σ

′=

σ

=C

C

ε

=C

C

C(T

T

ε

)−1

ε

′,

σ

′=T

T

σ

C

C(T

T

ε

)−1

ε

′=C

C

C′

ε

′,

T

σσ

σ

=

σ

′=C

C

C′

ε

′=C

C

C′T

T

εε

ε

,

σ

= (T

T

σ

)−1C

C

C′T

T

εε

ε

=C

C

ε

,

(2.1.30)

respectively.

Considering (2.1.29) the transformation relations for the stiffness matrix are

C

C′=T

T

σ

C

C(T

T

σ

)T,C

C

C= (T

T

ε

)TC

C

C′T

T

ε

(2.1.31)

or in index notation

C′

i j =T

σ

ik T

σ

jl Ckl ,Ci j =T

ε

ik T

ε

jlC′

kl (2.1.32)

The same procedure yields the relations for the compliance matrix. With

ε

=S

S

σ

,

ε

′=S

S

S′

σ

′(2.1.33)

it follows

(T

T

ε

)−1

ε

′=

ε

=S

S

σ

=S

S

S(T

T

σ

)−1

σ

′,

ε

′=T

T

ε

S

S(T

T

σ

)−1

σ

′=S

S

S′

σ

′,

T

εε

ε

=

ε

′=S

S

S′

σ

′=S

S

S′T

T

σσ

σ

,

ε

= (T

T

ε

)−1S

S

S′T

T

σσ

σ

=S

S

σ

,

(2.1.34)

i.e.

ε

′=T

T

ε

S

S(T

T

σ

)−1

σ

′,

ε

= (T

T

ε

)−1S

S

S′T

T

σσ

σ

(2.1.35)

The comparison leads to the transformation equations for S

S

Sand S

S

S′

S

S′=T

T

ε

S

S(T

T

σ

)−1,S

S

S= (T

T

ε

)−1S

S

S′T

T

σ

(2.1.36)

or taking into account (2.1.29)

S

S′=T

T

ε

S

S(T

T

ε

)T,S

S

S= (T

T

σ

)TS

S

S′T

T

σ

,(2.1.37)

2.1 Generalized Hooke’s Law 31

respectively, in subscript notation

S′

i j =T

ε

ik T

ε

jl Skl ,Si j =T

σ

ik T

σ

jl S′

kl (2.1.38)

In the special case of a rotation

φ

about the e

e

e3-direction (Fig. 2.6) the coordinates

of the transformation matrices T

T

σ

and T

T

ε

are given by the (2.1.39) and (2.1.40)

3

T

σ

pq=







c2s20 0 0 2cs

s2c20 0 0 −2cs

0 0 1 0 0 0

0 0 0 c−s0

0 0 0 s c 0

−cs cs 0 0 0 c2−s2







,3

T

σ

pq−1

="3

T

ε

pq#T

(2.1.39)

"3

T

ε

pq#=







c2s20 0 0 cs

s2c20 0 0 −cs

0 0 1 0 0 0

0 0 0 c−s0

0 0 0 s c 0

−2cs 2cs 0 0 0 c2−s2







,"3

T

ε

pq#−1

=3

T

σ

pqT

(2.1.40)

By all rules following from a rotation of the reference system the stresses, strains,

stiffness and compliance parameters in the rotated system are known. They are sum-

marized in symbolic notation (2.1.41)

σ

′=T

T

σσ

σ

,

ε

′=T

T

εε

ε

,

σ

= (T

T

ε

)T

σ

′,

ε

= (T

T

σ

)T

ε

′,

C

C′=T

T

σ

C

C(T

T

σ

)T,S

S

S′=T

T

ε

S

S(T

T

ε

)T,

C

C= (T

T

ε

)TC

C

C′T

T

ε

,S

S

S= (T

T

σ

)TS

S

S′T

T

σ

(2.1.41)

Fig. 2.6 Rotation about the

e

e3-direction x1

x2

x3,x′

3

x′

1

x′

2

e

e3,e

e

e′

3

e

e′

1

e

e2

e

e′

2

e

e1

φ

32 2 Linear Anisotropic Materials

For special cases of a rotation about a direction e

e

eithe general transformation ma-

trices T

T

σ

and T

T

ε

are substituted by

i

T

σ

or

i

T

ε

. The case of a rotation about the

e

e1-direction yields the coordinates of the transformation matrices T

T

σ

and T

T

ε

which

are given in App. B.

2.1.3 Symmetry Relations of Stiffness and Compliance Matrices

In the most general case of the three-dimensional generalized Hooke’s law the stiff-

ness and the compliance matrices have 36 non-zero material parameters Ci j or Si j

but they are each determined by 21 independent parameters. Such an anisotropic

material is called a triclinic material, it has no geometric symmetry properties. The

experimental tests to determine 21 independent material parameters would be difﬁ-

cult to realize in engineering applications. So it is very important that the majority

of anisotropic materials has a structure that exhibits one or more geometric sym-

metries and the number of independent material parameters needed to describe the

material behavior can be reduced.

In the general case of 21 independent parameters, there is a coupling of each

loading component with all strain states and the model equations for structure el-

ements would be very complicated. The reduction of the number of independent

material parameters results therefore in a simplifying of the modelling and analysis

of structure elements composed of composite materials and impact the engineering

applications. The most important material symmetries are:

2.1.3.1 Monoclinic or Monotropic Material Behavior

A monoclinic material has one symmetry plane (Fig. 2.7). It is assumed that the

Fig. 2.7 Symmetry plane

(x1−x2)of a monoclinic

material. All points of a

body which are symmetric

to this plane have identical

values of Ci j and Si j. Mirror

transformation (x1=x′

1,

x2=x′

2,x3=−x′

3)

x1

x2

x3

e

e3

e

e1e

e

e2

x′

3

2.1 Generalized Hooke’s Law 33

symmetry plane is the (x1−x2)plane. The structure of the stiffness or compliance

matrix must be in that way that a change of a reference system carried out by a sym-

metry about this plane does not modify the matrices, i.e. that the material properties

are identical along any two rays symmetric with respect to the (x1−x2)plane. The

exploitation of the transformation rules leads to a stiffness matrix with the following

structure in the case of monoclinic material behavior

[Ci j]MC =







C11 C12 C13 0 0 C16

C12 C22 C23 0 0 C26

C13 C23 C33 0 0 C36

0 0 0 C44 C45 0

0 0 0 C45 C55 0

C16 C26 C36 0 0 C66







(2.1.42)

The compliance matrix has the same structure

[Si j]MC =







S11 S12 S13 0 0 S16

S12 S22 S23 0 0 S26

S13 S23 S33 0 0 S36

000S44 S45 0

000S45 S55 0

S16 S26 S36 0 0 S66







(2.1.43)

The number of non-zero elements Ci j or Si j reduces to twenty, the number of in-

dependent elements to thirteen. The loading-deformation couplings are reduced.

Consider for example the stress component

σ

6≡

τ

12. There is a coupling with the

extensional strains

ε

1,

ε

2,

ε

3and the shear strain

ε

6≡

γ

12 but the shear stress

σ

4or

σ

5produces only shear strains.

If an anisotropic material has the plane of elastic symmetry x1−x3then it can be

shown that

[Ci j]MC =







C11 C12 C13 0C15 0

C12 C22 C23 0C25 0

C13 C23 C33 0C35 0

0 0 0 C44 0C46

C15 C25 C35 0C55 0

0 0 0 C46 0C66







(2.1.44)

and for the plane of elastic symmetry x2−x3

[Ci j]MC =







C11 C12 C13 C14 0 0

C12 C22 C23 C24 0 0

C13 C23 C33 C34 0 0

C14 C24 C34 C44 0 0

0 0 0 0 C55 C56

0 0 0 0 C56 C66







(2.1.45)

34 2 Linear Anisotropic Materials

x1

x2

x3

a b

Fig. 2.8 Orthotropic material behavior. aSymmetry planes (x1−x2)and (x2−x3),badditional

symmetry plane (x1−x3)

The monoclinic compliance matrices [Si j ]MC have for both cases the same structure

again as the stiffness matrices [Ci j]MC.

2.1.3.2 Orthotropic Material Behavior

An orthotropic material behavior is characterized by three symmetry planes that are

mutually orthogonal (Fig. 2.8). It should be noted that the existence of two orthog-

onal symmetry planes results in the existence of a third. The stiffness matrix of an

orthotropic material has the following structure

[Ci j]O=







C11 C12 C13 0 0 0

C12 C22 C23 0 0 0

C13 C23 C33 0 0 0

0 0 0 C44 0 0

0 0 0 0 C55 0

0 0 0 0 0 C66







(2.1.46)

The compliance matrix has the same structure. An orthotropic material has 12 non-

zero and 9 independent material parameters. The stress-strain coupling is the same

as for isotropic material behavior. Normal stresses give rise to only extensional

strains and shear stresses only shear strains. Orthotropic material behavior is typ-

ical for unidirectional laminae with on-axis loading.

2.1 Generalized Hooke’s Law 35

2.1.3.3 Transversely Isotropic Material Behavior

A material behavior is said to be transversely isotropic if it is invariant with respect

to an arbitrary rotation about a given axis. This material behavior is of special im-

portance in the modelling of ﬁbre-reinforced composite materials with coordinate

axis in the ﬁbre direction and an assumed isotropic behavior in cross-sections or-

thogonal to the ﬁbre direction. This type of material behavior lies between isotropic

and orthotropic.

If x1is the ﬁbre direction, x2and x3are both rectangular to the ﬁbre direction

and assuming identical material properties in these directions is understandable.

The structure of the stiffness matrix of a transversely isotropic material is given

in (2.1.47)

[Ci j]TI =







C11 C12 C12 0 0 0

C12 C22 C23 0 0 0

C12 C23 C22 0 0 0

0001

2(C22 −C23)0 0

0 0 0 0 C55 0

0 0 0 0 0 C55







(2.1.47)

If x2is the ﬁbre direction, x1and x3are both rectangular to the ﬁbre direction and

assuming identical material properties in these directions then

[Ci j]TI =







C11 C12 C13 0 0 0

C12 C22 C12 0 0 0

C13 C12 C11 0 0 0

000C44 0 0

0 0 0 0 1

2(C11 −C13)0

0 0 0 0 0 C44







(2.1.48)

If x3is the ﬁbre direction, x1and x2are both rectangular to the ﬁbre direction and

assuming identical material properties in these directions then

[Ci j]TI =







C11 C12 C13 0 0 0

C12 C11 C13 0 0 0

C13 C13 C33 0 0 0

000C44 0 0

0 0 0 0 C44 0

000001

2(C11 −C12)







(2.1.49)

The compliance matrix has the same structure. For example for the case (2.1.47)

[Si j]TI =







S11 S12 S12 0 0 0

S12 S22 S23 0 0 0

S12 S23 S22 0 0 0

0 0 0 2(S22 −S23)0 0

0 0 0 0 S55 0

0 0 0 0 0 S55







(2.1.50)

36 2 Linear Anisotropic Materials

The number of non-zero elements reduces again to twelve but the independent pa-

rameters are only ﬁve.

2.1.3.4 Isotropic Material Behavior

A material behavior is said to be isotropic if its properties are independent of the

choice of the reference system. There exist no preferred directions, i.e. the material

has an inﬁnite number of planes and axes of material symmetry. Most conventional

materials satisfy this behavior approximately on a macroscopic scale.

The number of independent elasticity parameters is reduced to two and this leads

to the following stiffness matrix in the case of isotropic material behavior

[Ci j]I=







C11 C12 C12 000

C12 C11 C12 000

C12 C12 C11 000

0 0 0 C∗0 0

0 0 0 0 C∗0

0 0 0 0 0 C∗







(2.1.51)

with C∗=1

2(C11 −C12). The compliance matrix has the same structure but with

diagonal terms 2(S11 −S12 )instead of 1

2(C11 −C12). Tables 2.3 and 2.4 summarize

the stiffness and compliance matrices for all material models described above.

2.1.4 Engineering Parameters

2.1.4.1 Orthotropic Material Behavior

The coordinates Cij and Si j of the stiffness and compliance matrix are mathematical

symbols relating stresses and strains. For practising engineers, a clear understand-

ing of each material parameter is necessary and requires a more mechanical meaning

by expressing the mathematical symbols in terms of engineering parameters such as

moduli Ei,Gi j and Poisson’s ratios

ν

i j. The relationships between mathematical and

engineering parameters are obtained by basic mechanical tests and imaginary math-

ematical experiments. The basic mechanical tests are the tension, compression and

torsion test to measure the elongation, the contraction and the torsion of a specimen.

In general, these tests are carried out by imposing a known stress and measuring the

strains or vice versa.

It follows that the compliance parameters are directly related to the engineering

parameters, simpler than those of the stiffness parameters. In the case of orthotropic

materials the 12 engineering parameters are Young’s moduli E1,E2,E3, the shear

moduli G23,G13,G12 and Poisson’s ratios

ν

i j,i,j=1,2,3(i6=j). For orthotropic

materials one can introduce the contracted engineering notation

2.1 Generalized Hooke’s Law 37

Table 2.3 Three-dimensional stiffness matrices for different material symmetries

Material model Elasticity matrix [Ci j ]

Anisotropy:

21 independent

material parameters







C11 C12 C13 C14 C15 C16

C22 C23 C24 C25 C26

C33 C34 C35 C36

C44 C45 C46

S Y M C55 C56

C66







Monoclinic:

13 independent

material parameters

Symmetry planex3=0

C14 =C15 =C24 =C25 =C34 =C35 =C46 =C56 =0

Symmetry planex2=0

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

Symmetry planex1=0

C15 =C16 =C25 =C26 =C35 =C36 =C45 =C56 =0

Orthotropic:

9 independent

material parameters

3 planes of symmetry x1=0,x2=0,x3=0

C14 =C15 =C16 =C24 =C25 =C26 =C34

=C35 =C36 =C45 =C46 =C56 =0

Transversely isotropic:

5 independent

material parameters

Plane of isotropy x3=0

C11 =C22,C23 =C13,C44 =C55,C66 =1

2(C11 −C12)

Plane of isotropy x2=0

C11 =C33,C12 =C23,C44 =C66,C55 =1

2(C33 −C13)

Plane of isotropy x1=0

C22 =C33,C12 =C13,C55 =C66,C44 =1

2(C22 −C23)

all otherCi j like orthotropic

Isotropy:

2 independent

material parameters

C11 =C22 =C33,C12 =C13 =C23,

C44 =C55 =C66 =1

2(C11 −C12)

all otherCi j =0

σ

1=E1

ε

1,

σ

2=E2

ε

2,

σ

3=E3

ε

3,

σ

4=E4

ε

4,

σ

5=E5

ε

5,

σ

6=E6

ε

6

(2.1.52)

with G23 ≡E4,G13 ≡E5,G12 ≡E6.

The generalized Hooke’s law in the form (2.1.19) and (2.1.20) leads, for example,

to the relations

38 2 Linear Anisotropic Materials

Table 2.4 Three-dimensional compliance matrices for different material symmetries

Material model Compliance matrix [Si j ]

Anisotropy:

21 independent

material parameters







S11 S12 S13 S14 S15 S16

S22 S23 S24 S25 S26

S33 S34 S35 S36

S44 S45 S46

S Y M S55 S56

S66







Monoclinic:

13 independent

material parameters

Symmetry plane x3=0

S14 =S15 =S24 =S25 =S34 =S35 =S46 =S56 =0

Symmetry plane x2=0

S14 =S16 =S24 =S26 =S34 =S36 =S45 =S56 =0

Symmetry plane x1=0

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

Orthotropic:

9 independent

material parameters

3 planes of symmetry x1=0,x2=0,x3=0

S14 =S15 =S16 =S24 =S25 =S26 =S34

=S35 =S36 =S45 =S46 =S56 =0

Transversely isotropic:

5 independent

material parameters

Plane of isotropy x3=0

S11 =S22,S23 =S13,S44 =S55 ,S66 =2(S11 −S12)

Plane of isotropy x2=0

S11 =S33,S12 =S23,S44 =S66 ,S55 =2(S33 −S13)

Plane of isotropy x1=0

S22 =S33,S13 =S12,S55 =S66 ,S44 =2(S22 −S23)

all other Si j like orthotropic

Isotropy:

2 independent

material parameters

S11 =S22 =S33,S12 =S13 =S23 ,

S44 =S55 =S66 =2(S11 −S12)

all other Si j =0

ε

1=S11

σ

1+S12

σ

2+S13

σ

3,

ε

4=S44

σ

4,

ε

2=S12

σ

1+S22

σ

2+S23

σ

3,

ε

5=S55

σ

5,

ε

3=S13

σ

1+S23

σ

2+S33

σ

3,

ε

6=S66

σ

6

(2.1.53)

For uniaxial tension in xi-direction,

σ

16=0,

σ

i=0,i=2,...,6. This reduces (2.1.53)

to

ε

1=S11

σ

1,

ε

2=S12

σ

1,

ε

3=S13

σ

1,

ε

4=

ε

5=

ε

6=0,(2.1.54)

and the physical tensile tests provides the elastic parameters E1,

ν

12,

ν

13

2.1 Generalized Hooke’s Law 39

E1=

σ

1

ε

1

=1

S11

,

ν

12 =−

ε

2

ε

1

=−S12E1,

ν

13 =−

ε

3

ε

1

=−S13E1(2.1.55)

Analogous relations resulting from uniaxial tension in x2- and x3-directions and all

Si j are related to the nine measured engineering parameters (3 Young’s moduli and

6 Poisson’s ratios) by uniaxial tension tests in three directions x1,x2and x3.

From the symmetry of the compliance matrix one can conclude

ν

12

E1

=

ν

21

E2

,

ν

23

E2

=

ν

32

E3

,

ν

31

E3

=

ν

13

E1

or

ν

i j

Ei

=

ν

ji

Ej

,

ν

i j

ν

ji

=Ei

Ej

,i,j=1,2,3(i6=j)(2.1.56)

Remember that the ﬁrst and the second subscript in Poisson’s ratios denote stress

and strain directions, respectively. Equations (2.1.56) demonstrate that the nine en-

gineering parameters are n ot independent parameters and that in addition to the three

tension tests, three independent shear tests are necessary to ﬁnd the equations

ε

4=S44

σ

4,

ε

5=S55

σ

5,

ε

6=S66

σ

6,

which yield the relations

S44 =1

G23

=1

E4

,S55 =1

G13

=1

E5

,S66 =1

G12

=1

E6

(2.1.57)

Now all Si j in (2.1.20) can be substituted by the engineering parameters







ε

1

ε

2

ε

3

ε

4

ε

5

ε

6







=







1

E1−

ν

12

E1−

ν

13

E1

0 0 0

1

E2−

ν

23

E2

0 0 0

1

E3

0 0 0

1

E4

0 0

S Y M 1

E5

0

1

E6













σ

1

σ

2

σ

3

σ

4

σ

5

σ

6







(2.1.58)

As seen above, the relations between compliances Si j and engineering parameters

are very simple. This, however, is not the case for the relations between the stiffness

and engineering parameters. First we need to invert the compliance matrix S

S

Sand

to express the stiffness Ci j as a function of the compliances as follows. The shear

relations are uncoupled, and we obtain

C44 =1

S44

=G23,C55 =1

S55

=G13,C66 =1

S66

=G12 (2.1.59)

40 2 Linear Anisotropic Materials

So only a symmetric [3x3]-matrix must be inverted. The general formula is

Ci j =S−1

i j =(−1)i+jUi j

Det[Si j],Det[Si j ] = 

S11 S12 S13

S12 S22 S23

S13 S23 S33 

,(2.1.60)

where Ui j are the submatrices of S

S

Sto the element Si j , and yielding the relations

C11 =S22S33 −S2

23

Det[Si j],C12 =S13S23 −S12S33

Det[Si j],

C22 =S33S11 −S2

13

Det[Si j],C23 =S12S13 −S23S11

Det[Si j],

C33 =S11S22 −S2

12

Det[Si j],C13 =S12S23 −S13S22

Det[Si j]

(2.1.61)

Substituting the relations between Si j and engineering parameters given above in

(2.1.58), we obtain

C11 =(1−

ν

23

ν

32)E1

∆

,C12 =(

ν

12 +

ν

13

ν

32)E2

∆

,

C22 =(1−

ν

31

ν

13)E2

∆

,C23 =(

ν

23 +

ν

21

ν

13)E3

∆

,

C33 =(1−

ν

21

ν

12)E3

∆

,C13 =(

ν

13 +

ν

12

ν

23)E3

∆

(2.1.62)

with

∆

=1−

ν

21

ν

12 −

ν

32

ν

23 −

ν

13

ν

31 −2

ν

21

ν

13

ν

32. Considering Ei/

∆

≡Ei,

1/Si≡Eiwe ﬁnally get two subsystems





σ

1

σ

2

σ

3

=



(1−

ν

23

ν

32)E1(

ν

12 +

ν

13

ν

32)E2(

ν

13 +

ν

12

ν

23)E3

(1−

ν

31

ν

13)E2(

ν

23 +

ν

21

ν

13)E3

SYM (1−

ν

21

ν

12)E3





ε

1

ε

2

ε

3

,





σ

4

σ

5

σ

6

=



E40 0

E50

SYM E6





ε

4

ε

5

ε

6

(2.1.63)

2.1.4.2 Transversally-Isotropic Material Behavior

It should be noted that in the case of transversely isotropic material with the

(x2−x3)-plane of isotropy

E2=E3,

ν

12 =

ν

13,G12 =G13 ,G23 =E2

2(1+

ν

23),(2.1.64)

and we get

2.1 Generalized Hooke’s Law 41

S11 =1

E1

,S12 =S13 =−

ν

12

E1

,S44 =1

G23

=2(1+

ν

23)

E2

,

S22 =S33 =1

E2

,S23 =−

ν

23

E2

,S55 =S66 =1

G12

,

C11 =(1−

ν

2

23)E1

∆

∗,C12 =C13 =

ν

21(1+

ν

23)E1

∆

∗,

C22 =C33 =(1−

ν

12

ν

21)E2

∆

∗,C23 =(

ν

23 +

ν

21

ν

12)E2

∆

∗,

C44 =G23 =E2

2(1+

ν

23),C55 =C66 =G12

with

∆

∗= (1+

ν

23)(1−

ν

23 −2

ν

21

ν

12).

Similar expressions one gets with the (x1−x2)-plane of isotropy and considering

E1=E2,

ν

13 =

ν

23,G13 =G23 ,G12 =E1

2(1+

ν

12)(2.1.65)

or with the (x1−x3)-plane of isotropy and

E1=E3,

ν

12 =

ν

23,G12 =G23 ,G13 =E1

2(1+

ν

13)(2.1.66)

The Young’s modulus and the Poisson’s ratio in the plane of isotropy often will be

designated as ETand

ν

TT.ETcharacterizes elongations or contractions of a trans-

versely isotropic body in the direction of the applied load in any direction of the

plane of isotropy and

ν

TT characterizes contractions or elongations of the body in

the direction perpendicular to the applied load, but parallel to the plane of isotropy.

The shear modulus GTT characterizes the material response under shear loading in

the plane of isotropy and takes the form

GTT =ET

2(1+

ν

TT),

i.e. any two of the engineering parameters ET,

ν

TT and GTT can be used to fully

describe the elastic properties in the plane of isotropy. A third independent primary

parameter should be EL. This Young’s modulus characterizes the tension respec-

tively compression response for the direction perpendicular to the plane of isotropy.

The fourth primary parameter should be the shear modulus GLT in the planes per-

pendicular to the plane of isotropy. As a ﬁfth primary parameter can be chosen

ν

LT

or

ν

TL, which characterize the response in the plane of isotropy under a load in L-

direction or the response in the L-direction under a load in the plane of isotropy. The

stress-strain relations for an transversely isotropic body, if (x2−x3) is the plane of

isotropy and with the reciprocity relations

ν

LT

EL

=

ν

TL

ET

42 2 Linear Anisotropic Materials

can be used in the following matrix form







ε

1

ε

2

ε

3

ε

4

ε

5

ε

6







=







1

EL−

ν

TL

ET−

ν

TL

ET

0 0 0

1

ET−

ν

TT

ET

0 0 0

1

ET

0 0 0

1

GTT

0 0

S Y M 1

GLT

0

1

GLT













σ

1

σ

2

σ

3

σ

4

σ

5

σ

6







(2.1.67)

with the engineering parameters

E1=EL,E2=E3=ET,E4=G23 =GTT =ET

2(1+

ν

TT),

E5=G13 =E6=G12 =GLT,

ν

12 =

ν

13 =

ν

LT,

ν

23 =

ν

TT,

ν

LT

EL

=

ν

TL

ET

With the (x1−x2)-plane of isotropy the engineering parameters are

E1=E2=ET,E3=EL,E4=G23 =E5=G13 =GTL,

E6=G12 =GTT =ET

2(1+

ν

TT),

ν

13 =

ν

23 =

ν

TL,

ν

12 =

ν

TT

Notice that in literature the notations of Poisson’s ratios

ν

LT and

ν

TL can be corre-

spond to the opposite meaning.

2.1.4.3 Isotropic Material Behavior

For an isotropic material behavior in all directions, the number of independent en-

gineering parameters reduces to two

E1=E2=E3=E,

ν

12 =

ν

23 =

ν

13 =

ν

,

G12 =G13 =G23 =G=E

2(1+

ν

)

(2.1.68)

S11 =S22 =S33 =1

E,S12 =S13 =S23 =−

ν

E,

S44 =S55 =S66 =1

G=2(1+

ν

)

E,C11 =C22 =C33 =(1−

ν

)E

∆

∗∗ ,

C12 =C13 =C23 =

ν

E

∆

∗∗ ,C44 =C55 =C66 =G=E

2(1+

ν

)

2.1 Generalized Hooke’s Law 43

assuming

∆

∗∗ = (1+

ν

)(1−2

ν

).

With this, all three-dimensional material laws for various material symmetries

interesting in engineering applications of composites are known. The relations be-

tween Si j,Ci j and engineering parameters are summarized in Table 2.5.

Consider that the elastic properties of an isotropic material are determined by two

independent parameters. The elastic parameters Young’s modulus Eand Poisson’s

ratio

ν

are generally used because they are determined easily in physical tests. But

also the so-called Lam ´e coefﬁcients5

λ

and

µ

, the shear modulus Gor the bulk

modulus Kcan be used if it is suitable. There are simple relations between the

parameters, e.g. as a function of E,

ν

λ

=E

ν

(1+

ν

)(1−2

ν

),

µ

=E

2(1+

ν

)=G,K=E

3(1−2

ν

),

ν

=

λ

2(

λ

+

µ

),E=

µ

(3

λ

+2

µ

)

λ

+

µ

,K=

λ

+2

3

µ

(2.1.69)

Summarizing the constitutive equations for isotropic, transversely isotropic and

orthotropic materials, which are most important in the engineering applications of

composite structural mechanics one can ﬁnd that the common features of the re-

lationships between stresses and strains for these material symmetries are that the

normal stresses are not couplet with shear strains and shear stresses are not coupled

with the normal strains. Each shear stress is only related to the corresponding shear

strain. These features are not retained in the more general case of an monoclinic or

a general anisotropic material.

2.1.4.4 Monoclinic Material Behavior

In the case of monoclinic materials we have 13 mutually independent stiffness or

compliances. Therefore we have in comparison with orthotropic materials to intro-

duce four additional engineering parameters and keeping in mind, that the mono-

clinic case must comprise the orthotropic case, we should not change the engineer-

ing parameters of orthotropic material behavior. Assuming that (x1−x2) is the plane

of elastic symmetry, the additional parameters are related to the compliance matrix

components S16,S26,S36 and S46.

The ﬁrst three pair normal strains

ε

1,

ε

2,

ε

3to the shear stress

σ

6and vice versa

the shear strain

ε

6to the normal stresses

σ

1,

σ

2,

σ

3. The fourth one couple the shear

strain

ε

4to the shear stress

σ

5and vice versa the shear strain

ε

5to the shear stress

σ

4.

In a compact notation the strain-stress relations for an anisotropic material having

one plane of elastic symmetry (x1−x2) are

5Gabriel L´eon Jean Baptiste Lam´e (∗22 July 1795 Tours - †1 May 1870 Paris) - mathematician,

who contributed to the mathematical theory of elasticity (Lam´e’s parameters in elasticity and ﬂuid

mechanics)

44 2 Linear Anisotropic Materials

Table 2.5 Relationships between Si j,Ci j and the engineering parameters for orthotropic, trans-

versely-isotropic and isotropic material

Orthotropic material

S11 =E−1

1,S12 =S21 =−

ν

12E−1

1,S44 =G−1

23 =E−1

4

S22 =E−1

2,S13 =S31 =−

ν

13E−1

1,S55 =G−1

13 =E−1

5

S33 =E−1

3,S23 =S32 =−

ν

23E−1

2,S66 =G−1

12 =E−1

6

ν

i j =

ν

ji(Ei/Ej),Ei= (

ν

i j/

ν

ji)Eji,j=1,2,3

C11 =S22S33 −S2

23

det[Si j]=(1−

ν

23

ν

32)E1

∆

,C44 =1

S44

=G23 =E4

C22 =S33S11 −S2

13

det[Si j]=(1−

ν

31

ν

13)E2

∆

,C55 =1

S55

=G31 =E5

C33 =S11S22 −S2

12

det[Si j]=(1−

ν

21

ν

12)E3

∆

,C66 =1

S66

=G12 =E6

C12 =S13S23 −S12S33

det[Si j]=(

ν

12 +

ν

32

ν

13)E2

∆

=(

ν

21 +

ν

31

ν

23)E1

∆

=C21

C13 =S12S23 −S13S22

det[Si j]=(

ν

13 +

ν

12

ν

23)E3

∆

=(

ν

31 +

ν

21

ν

32)E1

∆

=C31

C23 =S12S13 −S23S11

det[Si j]=(

ν

23 +

ν

21

ν

13)E3

∆

=(

ν

32 +

ν

12

ν

31)E2

∆

=C32

∆

=1−

ν

12

ν

21 −

ν

23

ν

32 −

ν

31

ν

13 −2

ν

21

ν

13

ν

32

Transversely-isotropic material

(x2−x3)-plane of isotropy

E1,E2=E3,E5=E6,

ν

12 =

ν

13,E4=E2

2(1+

ν

23)

(x1−x2)-plane of isotropy

E1=E2,E3,E4=E5,

ν

13 =

ν

23,E6=E1

2(1+

ν

12)

(x1−x3)-plane of isotropy

E1=E3,E2,E4=E6,

ν

12 =

ν

23,E5=E3

2(1+

ν

13)

Isotropic material

E1=E2=E3=E,

ν

12 =

ν

21 =

ν

13 =

ν

31 =

ν

23 =

ν

32 =

ν

,

E4=E5=E6=G=E

2(1+

ν

)

2.1 Generalized Hooke’s Law 45







ε

1

ε

2

ε

3

ε

4

ε

5

ε

6







=







1

E1−

ν

21

E2−

ν

31

E3

0 0

η

61

E6

−

ν

12

E1

1

E2−

ν

32

E3

0 0

η

62

E6

−

ν

13

E1−

ν

23

E2

1

E3

0 0

η

63

E6

0 0 0 1

E4

µ

54

E5

0

0 0 0

µ

45

E4

1

E5

0

η

16

E1

η

26

E2

η

36

E3

0 0 1

E6













σ

1

σ

2

σ

3

σ

4

σ

5

σ

6







(2.1.70)

with the following reciprocal relations

η

61

E6

=

η

16

E1

,

η

62

E6

=

η

26

E2

,

η

63

E6

=

η

36

E3

,

µ

54

E5

=

µ

45

E4

(2.1.71)

and the compliance components

S16 =

η

61

E6

,S26 =

η

62

E6

,S36 =

η

63

E6

,S45 =

µ

54

E5

(2.1.72)

η

61,

η

62 and

η

63 are extension-shear coupling coefﬁcients indicating normal strains

induced by shear stress

σ

6and

η

16,

η

26 and

η

36 the shear-extension coupling co-

efﬁcients characterizing shear strain

ε

6caused by normal stresses.

µ

45 and

µ

54 are

shear-shear coupling coefﬁcients.

The stiffness matrix for the monoclinic material can be found as the inverse of

the compliance matrix, but the expressions are unreasonable to present in an explicit

form. However, the inverse of a matrix can be easily calculated using standard nu-

merical procedures. Also for a generally anisotropic material the compliance can be

formulated with help of eight additional coupling parameters but the stiffness matrix

should be calculated numerically.

2.1.5 Two-Dimensional Material Equations

In most structural applications the structural elements are simpliﬁed models by

reducing the three-dimensional state of stress and strain approximately to a two-

dimensional plane stress or plane strain state. A thin lamina for instance can be

considered to be under a condition of plane stress with all stress components in

the out-of-plane direction being approximately zero. The different conditions for a

plane stress state in the planes (x1−x2),(x2−x3)and (x1−x3)are demonstrated in

Fig. 2.9.

In the following, the plane stress state with respect to the (x1−x2)plane (Fig.

2.9a) is considered. In addition, since in the case of unidirectional long-ﬁbre rein-

forced laminae the most general type of symmetry of the material properties is the

46 2 Linear Anisotropic Materials

x1

x2

x3

σ

2

σ

1

σ

6

σ

3

σ

2

σ

4

σ

1

σ

3

σ

5

abc

Fig. 2.9 Plane stress state. a(x1−x2)-plane,

σ

3=

σ

4=

σ

5=0, b(x2−x3)-plane,

σ

1=

σ

5=

σ

6=0, c(x1−x3)-plane,

σ

2=

σ

4=

σ

6=0

monoclinic one the generalized Hooke’s law (2.1.20) is reduced taking into account

(2.1.43) to







ε

1

ε

2

ε

3

ε

4

ε

5

ε

6







=







S11 S12 S13 0 0 S16

S22 S23 0 0 S26

S33 0 0 S36

SS44 S45 0

YS55 0

MS66













σ

1

σ

2

0

σ

6







(2.1.73)

That means

σ

3=

σ

4=

σ

5=0, and we have three in-plane constitutive equations

ε

1=S11

σ

1+S12

σ

2+S16

σ

6

ε

2=S12

σ

1+S22

σ

2+S26

σ

6,Si j =Sji

ε

6=S16

σ

1+S26

σ

2+S66

σ

6

(2.1.74)

and an additional equation for strain

ε

3in x3-direction

ε

3=S13

σ

1+S23

σ

2+S36

σ

6(2.1.75)

The other strains

ε

4,

ε

5are equal to 0 considering the monoclinic material behavior.

If the plane stress assumptions are used to simplify the generalized stiffness equa-

tions (2.1.19) taking into account (2.1.42), the result is







σ

1

σ

2

0

σ

6







=







C11 C12 C13 0 0 C16

C22 C23 0 0 C26

C33 0 0 C36

SC44 C45 0

YC55 0

MC66













ε

1

ε

2

ε

3

ε

4

ε

5

ε

6







(2.1.76)

or again three in-plane equations

2.1 Generalized Hooke’s Law 47

σ

1=C11

ε

1+C12

ε

2+C13

ε

3+C16

ε

6,

σ

2=C12

ε

1+C22

ε

2+C23

ε

3+C26

ε

6,Ci j =Cj i

σ

6=C16

ε

1+C26

ε

2+C36

ε

3+C66

ε

6

(2.1.77)

There are another three equations. At ﬁrst we obtain

σ

4=S44

ε

4+S45

ε

5=0,

σ

5=S45

ε

4+S55

ε

5=0

Since S44,S45 ,S55 are arbitrary but the stiffness matrix should be positive deﬁnite it

is obvious that

S44S55 −S2

45 >0

and from the

σ

4=

σ

5=0 condition it follows that

ε

4,

ε

5must be equal to 0. Taking

into account the last condition

σ

3=C13

ε

1+C23

ε

2+C33

ε

3+C36

ε

6=0

the strain

ε

3can be obtained

ε

3=−1

C33

(C13

ε

1+C23

ε

2+C36

ε

6)(2.1.78)

and eliminated and substituted in Eqs. (2.1.77). After substituting

ε

3using Eq.

(2.1.78) Eqs. (2.1.77) leads to

σ

i=Ci j −Ci3Cj3

C33 

ε

j,i,j=1,2,6,(2.1.79)

respectively

σ

i=Qi j

ε

j,i,j=1,2,6 (2.1.80)

The Qi j are the reduced stiffness. For the three cases in Fig. 2.9 we obtain

σ

i=Qi j

ε

j,Qi j =Ci j −Ci3Cj3

C33

,i,j=1,2,6,(x1−x2)−plane of symmetry,

σ

i=Qi j

ε

j,Qi j =Ci j −Ci1Cj1

C11

,i,j=2,3,4,(x2−x3)−plane of symmetry,

σ

i=Qi j

ε

j,Qi j =Ci j −Ci2Cj2

C22

,i,j=1,3,5,(x1−x3)−plane of symmetry

The number of unknown independent parameters of each of the matrices Si j,Ci j or

Qi j is six. It is very important to note that the elements in the plane stress compliance

matrix are simply a subset of the elements from the three-dimensional compliance

matrix and their numerical values are identical. On the other hand, the elements

of the reduced stiffness matrix involve a combination of elements from the three-

dimensional stiffness matrix and the numerical values of the Qi j differ from their

counterparts Ci j , i.e. they are actually less than the numerical values for Ci j . In

order to keep consistency with the generalized Hooke’s law, Eq. (2.1.78) should be

48 2 Linear Anisotropic Materials

used when calculating the transverse normal strain

ε

3and the general case leads to

consistent relations for the transverse shear strains

ε

4and

ε

5.

For an orthotropic material behavior under plane stress and on-axis orientation of

the reference system there are four independent parameters and for isotropic behav-

ior there are only two. The mathematical notations Sij ,Ci j or Qi j can be shifted to

the engineering notation. Tables 2.6 and 2.7 summarize the compliance and stiffness

matrices for the plane stress state.

Considering a plane strain state in the (x1−x2)plane we have the three non-zero

strains

ε

1,

ε

2and

ε

6but the four nonzero stress components

σ

1,

σ

2,

σ

3,

σ

6. Analo-

gous to the plane stress state, here the stress

σ

3normal to the (x1−x2)plane is not

an independent value and can be eliminated

ε

3=S13

σ

1+S23

σ

2+S33

σ

3+S36

σ

6=0,

σ

3=−1

S33

(S13

σ

1+S23

σ

2+S36

σ

6)

Therefore in the case of plane strain, the Cij ,i,j=1,2,6 can be taken directly from

the three-dimensional elasticity law and instead of Si j reduced compliances Vi j have

to be used

Table 2.6 Compliance matrices for various material models, plane stress state

Material model

ε

=S

σ

ε

=S

σ

ε

=S

σ

Anisotropy: Compliances Si j

6 independentmaterial parameters 



ε

1

ε

2

ε

6

=



S11 S12 S16

S22 S26

S66 





σ

1

σ

2

σ

6



Orthotropy: S16 =S26 =0

4 independent material parameters S11 =1

E1

,S22 =1

E2

S12 =−

ν

12

E1

=−

ν

21

E2

Reference system: on-axis S66 =1

G12

Isotropy: S16 =S26 =0

2 independent material parameters S11 =S22 =1

E,S12 =−

ν

E,

Reference system: as you like S66 =2(S11 −S12) = 2(1+

ν

)

E=1

G

2.1 Generalized Hooke’s Law 49

Table 2.7 Stiffness matrices for various material models, plane stress state

Material model

σ

=Q

ε

σ

=Q

ε

σ

=Q

ε

Anisotropy: Reduced stiffness Qij

6 independent material parameters 



σ

1

σ

2

σ

6

=



Q11 Q12 Q16

Q22 Q26

Q66 





ε

1

ε

2

ε

3



Orthotropy: Q16 =Q26 =0,Q66 =1

S66

=G12

4 independent material parameters Q11 =S22

∆

=E1

1−

ν

12

ν

21

Q22 =S11

∆

=E2

1−

ν

12

ν

21

Reference system: on-axis Q12 =−S12

∆

=

ν

12E2

1−

ν

12

ν

21

∆

=S11S22 −S2

12

Isotropy: Q16 =Q26 =0

2 independent material parameters Q11 =Q22 =E

1−

ν

2

Reference system: as you like Q12 =

ν

E

1−

ν

2,Q66 =E

2(1+

ν

)=G

σ

i=Ci j

ε

j,i,j=1,2,6,

ε

i=Vij

σ

j,Vij =Si j −Si3Sj3

S33

,i,j=1,2,6

Table 2.8 summarizes for the three-dimensional states and the plane stress and strain

states the number of non-zero and of independent material parameters.

In the two-dimensional equations of anisotropic elasticity, either reduced stiff-

ness or reduced compliances are introduced into the material laws. These equations

are most important in the theory of composite single or multilayered elements, e.g.

of laminae or laminates. The additional transformations rules which are necessary

in laminae and laminate theories are discussed in more detail in Chap. 3. Tables 2.6

and 2.7 above shows the relationship between stress and strain through the compli-

ance [Si j ]or the reduced stiffness [Qi j ]matrix for the plane stress state and how the

Si j and Qi j are related to the engineering parameters. For a unidirectional lamina the

engineering parameters are:

50 2 Linear Anisotropic Materials

Table 2.8 Stiffness and compliance parameters for stress and strain equations

σ

=C

C

ε

,

ε

=S

S

σ

,C

C

C=S

S

S−1, plane stress state

σ

=Q

Q

ε

,

ε

=S

S

σ

,Q

Q

Q=S

S

S−1, plane strain state -

σ

=C

C

ε

,

ε

=V

V

σ

,C

C

C=V

V

V−1,Qi j and Vi j are the reduced stiffness and compliances

Material model Number of non-zero Number of independent

parameters parameters

Three-dimensional Cij ;Si j Ci j;Si j

stress- or i,j=1,...,6i,j=1,...,6

strain state

Anisotropic 36 21

Monotropic 20 13

Orthotropic 12 9

Transversely isotropic 12 5

Isotropic 12 2

Plane stress state Qi j;Si j Qij ;Si j

(x1−x2)-plane i,j=1,2,6i,j=1,2,6

Anisotropic 9 6

Orthotropic 5 4

Isotropic 5 2

Plane strain state Ci j;Vi j Cij ;Vij

(x1−x2)-plane i,j=1,2,6i,j=1,2,6

Anisotropic 9 6

Orthotropic 5 4

Isotropic 5 2

E1longitudinal Young’s modulus in the principal direction 1 (ﬁbre direction)

E2transverse Young’s modulus in direction 2 (orthogonal to the ﬁbre direction)

ν

12 major Poisson’s ratio as the ratio of the negative normal strain in direction 2

to the normal strain in direction 1 only if normal load is applied in direction 1

G12 in-plane shear modulus for (x1−x2)plane

The four independent engineering elastic parameters are experimentally measured

as follows:

•Pure tensile load in direction 1:

σ

16=0,

σ

2=0,

σ

6=0

With

ε

1=S11

σ

1,

ε

2=S12

σ

1,

ε

6=0 are

E1=

σ

1

ε

1

=1

S11

,

ν

12 =−

ε

2

ε

1

=−S12

S11

•Pure tensile load in direction 2:

σ

1=0,

σ

26=0,

σ

6=0

With

ε

1=S12

σ

2,

ε

2=S22

σ

2,

ε

6=0 are

E2=

σ

2

ε

2

=1

S22

,

ν

21 =−

ε

1

ε

2

=−S12

S22

ν

21 is usually called the minor Poisson’s ratio and we have the reciprocal rela-

tionship

ν

12/E1=

ν

21/E2.

•Pure shear stress in the (x1−x2)plane:

σ

1=

σ

2=0,

σ

66=0

With

ε

1=

ε

2=0,

ε

6=S66

σ

6is

2.1 Generalized Hooke’s Law 51

G12 =

σ

6

ε

6

=1

S66

With the help of Tables 2.6 and 2.7, the relating equations of stresses and strains

are given through any of the following combinations of four parameters: (Q11,Q12 ,

Q22,Q66 ),(S11,S12,S22,S66),(E1,E2,

ν

12,G12 ).

In Chap. 3 the evaluation of the four engineering elastic parameters is given ap-

proximately by averaging the ﬁbre-matrix material behavior. There are different ap-

proaches for determining effective elastic moduli, e.g. in a simple way with ele-

mentary mixture rules, with semi-empirical models or an approach based on the

elementary theory.

2.1.6 Curvilinear Anisotropy

The type of anisotropy considered above was characterized by the equivalence of

parallel directions passing through different points of the homogeneous anisotropic

body and we can speak of a rectilinear anisotropy. Another kind of anisotropy is the

case, if one chooses a system of curvilinear coordinates in such a manner that the

coordinate directions coincide with equivalent directions of elastic properties at dif-

ferent points of an anisotropic body. The elements of the body, which are generated

by three pairs of coordinate surfaces possess identical elastic properties and we can

speak of a curvilinear anisotropy.

In the frame of this textbook we limit the considerations to cylindrical anisotropy,

which is also the most common case of this type of anisotropy. The generalized

Hooke’s law equations (2.1.21) are now considered in cylindrical coordinates x1=r,

x2=

θ

,x3=zand we have the stress and strain vectors in the contracted single

subscript notation



σ

1

σ

2

σ

3

σ

4

σ

5

σ

6T=

σ

r

σθσ

z

σθ

z

σ

rz

σ

r

θ

T,



ε

1

ε

2

ε

3

ε

4

ε

5

ε

6T=

ε

r

εθε

z

εθ

z

ε

rz

ε

r

θ

T(2.1.81)

In the speciﬁc cases of material symmetries the general constitutive equation

in cylindrical coordinates can be simpliﬁed analogous to the case of rectilinear

anisotropy.

In the speciﬁc case of an orthotropic cylindrical response there are three orthog-

onal planes of elastic symmetries. One plane is perpendicular to the axis z, another

one is tangential to the surface (

θ

−z) and the third one is a radial plane (Fig. 2.10).

Another case of material symmetry in possible practical situation is a transversely

isotropic cylinder or cylindrical tube with the plane of isotropy (r−

θ

). In this case

we obtain, as analog to (2.1.67)

52 2 Linear Anisotropic Materials

z

x

r

θ

z-plane

r−z-plane

θ

−z-plane

z

x

θ

r

Fig. 2.10 Cylindrical orthotropic material symmetry







ε

1

ε

2

ε

3

ε

4

ε

5

ε

6







=







1

ET−

ν

TT

ET−

ν

LT

EL

0 0 0

1

ET−

ν

LT

EL

0 0 0

1

EL

0 0 0

1

GLT

0 0

S Y M 1

GLT

0

1

GTT













σ

1

σ

2

σ

3

σ

4

σ

5

σ

6







,(2.1.82)

where the index T is associated with the coordinate directions rand

θ

and the index

L with the coordinate direction zand the reciprocal relations are

ν

LT

EL

=

ν

TL

ET

The stress-strain equations for the orthotropic case of cylindrical anisotropy are ob-

tained using engineering parameters

2.1 Generalized Hooke’s Law 53







ε

1

ε

2

ε

3

ε

4

ε

5

ε

6







=







1

Er−

νθ

r

E

θ

−

ν

zr

Er

0 0 0

1

E

θ

−

ν

z

θ

Er

0 0 0

1

Er

0 0 0

1

G

θ

z

0 0

S Y M 1

Grz

0

1

Gr

θ













σ

1

σ

2

σ

3

σ

4

σ

5

σ

6







(2.1.83)

The indices r,

θ

and zof the engineering parameters are associated with the indices

1, 2 and 3 and the strain-stress equations may also be written in a different way by

using the numerical subscripts. Further notice the reciprocal relations

Ei

ν

ji =Ej

ν

i j,i,j=r,

θ

,z

and the G

θ

z,Grz and Gr

θ

may be written in the more general form E4,E5,E6.

There are two practical situations for a monoclinic material behavior. The ﬁrst

case can be one plane of elastic symmetry (r−

θ

) which is rectilinear to the z-axis.

The case is interesting when considering composite discs or circular plates. The

stress-strain equations follow from (2.1.70) after substituting the subscripts 1,2 and

3 by the engineering parameters to r,

θ

and zand the shear moduli E4,E5and E6

by G

θ

z,Grz and Gr

θ

. The second case can be one plane of elastic symmetry (

θ

−z)

as a cylindrical surface with the axis rperpendicular to this surface. This situation

is of practical interest when considering e.g. ﬁlament wound cylindrical shells and

we get the strain-stress relations which couple all three normal strains to the shear

strain

ε

4and both shear strains

ε

5,

ε

6to both shear stresses

σ

5,

σ

6







ε

1

ε

2

ε

3

ε

4

ε

5

ε

6







=







1

E1−

ν

21

E2−

ν

31

E3

η

41

E4

0 0

1

E2−

ν

32

E3

η

42

E4

0 0

1

E3

η

43

E4

0 0

1

E4

0 0

S Y M 1

E5

µ

65

E6

1

E6













σ

1

σ

2

σ

3

σ

4

σ

5

σ

6







(2.1.84)

The subscripts 1, 2, 3 of the Young’s moduli and the Poisson’s ratios will be shifted

to r,

θ

,zand the moduli E4,E5,E6to G

θ

z,Grz,Gr

θ

. There are as above reciprocal

relations for

ν

i j ,

η

i j and

µ

i j

54 2 Linear Anisotropic Materials

η

41

E4

=

η

14

E1

,

η

42

E4

=

η

24

E2

,

η

43

E4

=

η

34

E3

,

µ

65

E6

=

µ

56

E5

,

ν

i j

Ei

=

ν

ji

Ej

,i,j=1,2,3

(2.1.85)

2.1.7 Problems

Exercise 2.1. Calculate for the tensile bar consisting of three parts (Fig. 2.2) the

elongation △l, the strain

ε

and the stress

σ

as functions of A,land F:

1. The stiffness are arranged in parallel:

E1=E3=70 GPa, E2=3 GPa, A1=A3=0,1A,A2=0,8A.

2. The stiffness are arranged in series:

E1=E3=70 GPa, E2=3 GPa, l1=l3=0,1l,l2=0,8l.

Solution 2.1. The solution can be obtained considering the basic assumptions of the

iso-strain and the iso-stress models.

1. Assumptions

ε

i=

ε

,△li=△l,i=1,2,3,A=

3

∑

i=1

Ai,F=

3

∑

i=1

Fi

From

σ

=E

ε

follows

σ

1=E1

ε

=70GPa

ε

,

σ

2=E2

ε

=3GPa

ε

,

σ

3=E3

ε

=70GPa

ε

With F=

σ

A=EA

ε

and Fi=EiAi

ε

follows

F=

3

∑

i=1

(EiAi)

ε

=16,4 GPa

ε

A,E=16,4 GPa,

ε

=F

EA =△l

l

yields

△l=Fl

EA =1

16,4

Fl

A(GPa)−1

and the solutions are

△l=1

16,4

Fl

A(GPa)−1=△l(F,l,A),

ε

=△l

l=1

16,4

F

A(GPa)−1=

ε

(F,A),

σ

i=Ei

ε

=1

16,4

EiF

A(GPa)−1=

σ

i(F,A),i=1,2,3

2.1 Generalized Hooke’s Law 55

2. Assumptions

△l=

3

∑

i=1△li,Fi=F,

ε

i=△li

li

,i=1,2,3

From △l=3

∑

i=1△li=3

∑

i=1

li

ε

iand F=EA

ε

follows

△l=

3

∑

i=1li

EiF

A=0,1

70 +0,8

3+0,1

70 Fl

A(GPa)−1,

△l=

ε

l=1

E

Fl

A=0,2695 Fl

A(GPa)−1,E=3,71 GPa

The functions △l,

ε

and

σ

are

△l=0,2695 Fl

A(GPa)−1=△l(F,l,A),

ε

=0,2695 F

A(GPa)−1=

ε

(F,A),

σ

=E

ε

=F

A=

σ

(F,A)

Exercise 2.2. The relationship between the load Fand the elongation △lof a tensile

bar (Fig. 2.1) is

F=EA0

l0△l=K△l

Eis the Young’s modulus of the material, A0the cross-sectional area of the bar and

l0is the length. The factor K=E A0/l0is the stiffness per length and characterizes

the mechanical performance of the tensile bar. In the case of two different bars

with Young’s moduli E1,E2, densities

ρ

1,

ρ

2, the cross-sectional areas A1,A2and

the lengths l1,l2the ratios of the stiffness K1and K2per length and the mass of the

bars are K1

K2

=E1A1

E2A2

l2

l1

,m1

m2

=l1A1

ρ

1

l2A2

ρ

2

Verify that for l1=l2and m1=m2the ratio K1/K2only depends on the ratio of the

speciﬁc Young’s moduli E1/

ρ

1and E2/

ρ

2.

Solution 2.2. Introducing the densities

ρ

1/

ρ

2into the stiffness ratio K1/K2yields

K1

K2

=E1/

ρ

1

E2/

ρ

2

m1

m2l2

l12

and with m1=m2,l2=l1

K1

K2

=E1/

ρ

1

E2/

ρ

2

56 2 Linear Anisotropic Materials

Note 2.1. A material with the highest value of E/

ρ

has the highest tension stiffness.

Exercise 2.3. For a simply supported beam with a single transverse load in the mid-

dle of the beam we have the following equation

F=48 EI

l3f=K f

Fis the load and fis the deﬂection in the middle of the beam, lis the length of

the beam between the supports and Ithe moment of inertia of the cross-section.

The coefﬁcient K=48E I/l3characterizes the stiffness of the beam. Calculate Kfor

beams with a circle or a square cross-sectional area (radius ror square length a) and

two different materials E1,

ρ

1and E2,

ρ

2but of equal length l, moments of inertia

and masses. Verify that for m1=m2the ratio of the stiffness coefﬁcients K1/K2only

depends on the ratios E1/

ρ

2

1and E2/

ρ

2

2.

Solution 2.3. Moments of inertia and masses of the two beams are

1. circle cross-sectional: I=

π

r4/4,m=r2

π

l

ρ

,

2. square cross-sectional: I=a4/12,m=a2l

ρ

In case 1. we have

K=48EI

l3=48E

l3

π

r4

4=48E/

ρ

2

4l3

m2

π

l2,

K1

K2

=E1/

ρ

2

1

E2/

ρ

2

2m1

m22l2

l15

With l1=l2,m1=m2we obtain

K1

K2

=E1/

ρ

2

1

E2/

ρ

2

In case 2. we have

K=48EI

l3=48E

l3

a4

12 =4E

l3

m2

ρ

2l2,

K1

K2

=E1/

ρ

2

1

E2/

ρ

2

2m1

m22l2

l15

With l1=l2,m1=m2we obtain

K1

K2

=E1/

ρ

2

1

E2/

ρ

2

Note 2.2. The best material for an optimal bending stiffness of the beam is that with

the highest value of E/

ρ

2.

Exercise 2.4. Formulate explicitly the transformation matrices (

3

T

σ

)−1and (

3

T

ε

)−1

for a rotation about the e

e

e3-direction (Fig. 2.6).

2.1 Generalized Hooke’s Law 57

Solution 2.4. With Eqs. (2.1.29), (2.1.39) and (2.1.40) follows (

3

T

σ

)−1= (

3

T

ε

)Tand

(

3

T

ε

)−1= (

3

T

σ

)T

(

3

T

σ

)−1=







c2s20 0 0 −2cs

s2c20 0 0 2sc

0 0 1 0 0 0

0 0 0 c s 0

0 0 0 −s c 0

cs −cs 0 0 0 c2−s2







,

(

3

T

ε

)−1=







c2s2000 −cs

s2c2000 sc

0 0 1 0 0 0

0 0 0 c s 0

0 0 0 −s c 0

2cs −2cs 000c2−s2







Exercise 2.5. Consider the coordinate transformation that corresponds with reﬂec-

tion in the plane x1−x2:x′

1=x1,x′

2=x2,x′

3=−x3. Deﬁne for this case

1. the coordinate transformation matrix [Ri j ]and

2. the stress and strain transformation matrices [

σ

Tpq]and [

ε

Tpq].

Solution 2.5. The solution contains two parts: the coordinate transformation and the

stress/strain transformation.

1. With Ri j =cos(e′

i,ej)(2.1.22) follows

R11 =1,R12 =0,R13 =0,R21 =0,R22 =1,R23 =0,

R31 =0,R32 =0,R33 =−1

and the transformation matrix takes the form

[Ri j] = 



1 0 0

0 1 0

0 0 −1



2. With the help of the transformation matrix App. B we can see that both matrices

are diagonal with the following nonzero elements for

σ

Tpq

R2

11 =1,R2

22 =1,R2

33 =1,

R22R33 +R23R32 =−1,R11R33 +R13R31 =−1,R11 R22 +R12R21 =1

and for

ε

Tpq

R2

11 =1,R2

22 =1,R2

33 =1,

R22R33 +R23R22 =−1,R11R33 +R13R31 =−1,R11 R22 +R12R21 =1

The transformation matrices take the form

58 2 Linear Anisotropic Materials

[

σ

Tpq] =







1 0 0 0 0 0

0 1 0 0 0 0

0 0 1 0 0 0

0 0 0 −1 0 0

0 0 0 0 −1 0

0 0 0 0 0 1







= [

ε

Tpq]

Exercise 2.6. The engineering material parameters for an orthotropic material are

given by

E1=173GPa,E2=33,1GPa,E3=5,17GPa,

E4=3,24GPa,E5=8,27GPa,E6=9,38GPa,

ν

12 =0,036,

ν

13 =0,25,

ν

23 =0,171

Calculate the stiffness matrix C

C

Cand the compliance matrix S

S

S.

Solution 2.6. With Table 2.5 we ﬁnd the Si j and the Ci j

S11 =E−1

1=5,780 10−3GPa−1,

S12 =S21 =−

ν

12E−1

1=−0,208 10−3GPa−1,

S22 =E−1

2=30,211 10−3GPa−1,

S13 =S31 =−

ν

13E−1

1=−1,445 10−3GPa−1,

S33 =E−1

3=193,424 10−3GPa−1,

S23 =S32 =−

ν

23E−1

2=−5,166 10−3GPa−1,

S44 =E−1

4=308,642 10−3GPa−1,

S55 =E−1

5=120,919 10−3GPa−1,

S66 =E−1

6=106,610 10−3GPa−1,

△=1−

ν

12

ν

21 −

ν

23

ν

32 −

ν

31

ν

13 −2

ν

21

ν

13

ν

32,

ν

21 =

ν

12(E2/E1) = 0,0069,

ν

31 =

ν

13(E3/E1) = 0,0075,

ν

32 =

ν

23(E3/E2) = 0,027,△=0,993,¯

Ei=Ei/△,i=1,2,3

C11 = (1−

ν

23

ν

32)¯

E1=173,415GPa,

C22 = (1−

ν

31

ν

13)¯

E2=33,271GPa,

C33 = (1−

ν

12

ν

21)¯

E3=5,205GPa,

C12 = (

ν

12 +

ν

13

ν

32)¯

E2=1,425GPa,

C13 = (

ν

13 +

ν

12

ν

23)¯

E3=1,334GPa,

C23 = (

ν

23 +

ν

21

ν

13)¯

E3=0,899GPa,

C44 =E4,C55 =E5,C66 =E6

With the values for Ci j and Si j the stiffness matrix C

C

Cand the compliance matrix S

S

can be written

2.2 Fundamental Equations and Variational Solution Procedures 59

C

C=







173,415 1,425 1,334 0 0 0

1,425 33,271 0,899 0 0 0

1,334 0,899 5,205 0 0 0

0 0 0 3,24 0 0

0 0 0 0 8,27 0

0 0 0 0 0 9,38







GPa,

S

S=







5,780 −0,208 −1,445 0 0 0

−0,208 30,211 −5,166 0 0 0

−1,445 −5,166 193,424 0 0 0

0 0 0 308,642 0 0

0 0 0 0 120,919 0

0 0 0 0 0 106,610







10−3GPa−1

2.2 Fundamental Equations and Variational Solution Procedures

Below we discuss at ﬁrst the fundamental equations of the anisotropic elasticity for

rectilinear coordinates. The system of equations can be divided into two subsys-

tems: the ﬁrst one is material independent that means we have the same equations

as in the isotropic case. To this subsystem belong the equilibrium equations (static

or dynamic) and the kinematic equations (the strain-displacement equations and the

compatibility conditions). To this subsystem one has to add the constitutive equa-

tions. In addition, we must introduce the boundary and, may be, the initial conditions

to close the initial-boundary problem. At second, considering that closed solutions

are impossible in most of the practical cases approximative solution techniques are

brieﬂy discussed. The main attention will be focussed on variational formulations.

2.2.1 Boundary and Initial-Boundary Value Equations

The fundamental equations of anisotropic elasticity can be formulated and solved

by a displacement, a stress or a mixedapproach. In all cases the starting point are the

following equations:

•The static or dynamic equilibrium equations formulated for an inﬁnitesimal cube

of the anisotropic solid which is subjected to body forces and surface forces char-

acterized by force density per unit surface. In Fig. 2.11 the stress and the volume

force components are shown in the x1-direction. Assuming the symmetry of the

stress tensor, three static equations link six unknown stress components. In the

case of dynamic problems the inertia forces are expressed through displacements,

therefore the equations of motion contain both, six unknown stress and three un-

known displacement components

60 2 Linear Anisotropic Materials

Fig. 2.11 Inﬁnitesimal cube

with lengths dx1,dx2,dx3:

stress and volume force com-

ponents in x1-direction

x1

x2

x3

dx1

dx2

dx3

σ

1+d

σ

1

p1

σ

6+d

σ

6

σ

1

σ

5+d

σ

5

σ

6

σ

5

∂σ

1

∂

x1

+

∂σ

6

∂

x2

+

∂σ

5

∂

x3

+p1=0,

∂σ

6

∂

x1

+

∂σ

2

∂

x2

+

∂σ

4

∂

x3

+p2=0,

∂σ

5

∂

x1

+

∂σ

4

∂

x2

+

∂σ

3

∂

x3

+p3=0,

static equations (2.2.1)

∂σ

1

∂

x1

+

∂σ

6

∂

x2

+

∂σ

5

∂

x3

+p1=

ρ∂

2u1

∂

t2,

∂σ

6

∂

x1

+

∂σ

2

∂

x2

+

∂σ

4

∂

x3

+p2=

ρ∂

2u2

∂

t2,

∂σ

5

∂

x1

+

∂σ

4

∂

x2

+

∂σ

3

∂

x3

+p3=

ρ∂

2u3

∂

t2

dynamic equations (2.2.2)

The inertial terms in (2.2.2) are dynamic body forces per unit volume. The den-

sity

ρ

for unidirectional laminae can be calculated e.g. using the rule of mixtures

(Sect. 3.1.1).

•The kinematic equations that are six strain-displacement relations and six com-

patibility conditions for strains. For linear small deformation theory, the six

stress-displacement equations couple six unknown strains and three unknown

displacements. Figure 2.12 shows the strains of an inﬁnitesimal cube in the

(x1−x2)-plane and we ﬁnd the relations

ε

1≡

∂

u1

∂

x1

,

ε

2≡

∂

u2

∂

x2

,

α

=

∂

u2

∂

x1

,

β

=

∂

u1

∂

x2⇒2

ε

12 ≡

γ

12 ≡

∂

u1

∂

x2

+

∂

u2

∂

x1≡

ε

6

and analogous relations for the (x2−x3)- and (x1−x3)-planes yield

2.2 Fundamental Equations and Variational Solution Procedures 61

x1

x2

u1u1+

∂

u1

∂

x1

dx1

u2

u2+

∂

u2

∂

x2

dx2

dx1

dx2

A

A′

C

C′

BB

B′

∂

u1

∂

x2

dx2

∂

u2

∂

x1

dx1

a b

α

β

γ

Fig. 2.12 Strains of the inﬁnitesimal cube shown for the (x1−x2)-plane. aextensional strains,

bshear strains

ε

1=

∂

u1

∂

x1

,

ε

2=

∂

u2

∂

x2

,

ε

3=

∂

u3

∂

x3

,

ε

4=

∂

u3

∂

x2

+

∂

u2

∂

x3

,

ε

5=

∂

u3

∂

x1

+

∂

u1

∂

x3

,

ε

6=

∂

u2

∂

x1

+

∂

u1

∂

x2

(2.2.3)

The displacement ﬁeld of the body corresponding to a given deformation state is

unique, the components of the strain tensor must satisfy the following six com-

patibility conditions

∂

2

ε

1

∂

x2

2

+

∂

2

ε

2

∂

x2

1

=

∂

2

ε

6

∂

x1

∂

x2

,

∂

x3

∂ε

4

∂

x1

+

∂ε

5

∂

x2−

∂ε

6

∂

x3=2

∂

2

ε

3

∂

x1

∂

x2

,

∂

2

ε

2

∂

x2

3

+

∂

2

ε

3

∂

x2

2

=

∂

2

ε

4

∂

x2

∂

x3

,

∂

x1

∂ε

5

∂

x2

+

∂ε

6

∂

x3−

∂ε

4

∂

x1=2

∂

2

ε

1

∂

x2

∂

x3

,

∂

2

ε

3

∂

x2

1

+

∂

2

ε

1

∂

x2

3

=

∂

2

ε

5

∂

x1

∂

x3

,

∂

x2

∂ε

6

∂

x3

+

∂ε

4

∂

x1−

∂ε

5

∂

x2=2

∂

2

ε

2

∂

x3

∂

x1

In the two-dimensional case the compatibility conditions reduce to a single equa-

tion

∂

2

ε

1

∂

x2

2

+

∂

2

ε

2

∂

x2

1−2

∂

2

ε

6

∂

x1

∂

x2

=0

•The material or constitutive equations which are described in Sect. 2.1 are

ε

i=Si j

σ

j,

σ

i=Ci j

ε

j(2.2.4)

The generalized Hooke’s law yields six equations relating in each case six un-

known stress and strain components. The elements of the stiffness matrix C

C

Cand

the compliance matrix S

S

Sare substituted with respect to the symmetry conditions

of the material.

62 2 Linear Anisotropic Materials

Summarizing all equations, we have 15 independent equations for 15 unknown com-

ponents of stresses, strains and displacements. In the displacement approach, the

stresses and strains are eliminated and a system of three coupled partial differential

equations for the displacement components are left.

In the static case we have a boundary-value problem, and we have to include

boundary conditions. In the dynamic case the system of partial differential equations

deﬁnes an initial-boundary-value problem and we have additional initial conditions.

A clear symbolic formulation of the fundamental equations in displacements is

given in vector-matrix notation. With the transposed vectors

σ

T,

ε

Tand u

u

uT

σ

T= [

σ

1

σ

2

σ

3

σ

4

σ

5

σ

6],

ε

T= [

ε

1

ε

2

ε

3

ε

4

ε

5

ε

6],u

u

uT= [u1u2u3](2.2.5)

the transformation and the differential matrices T

T

Tand D

D

D(n

n

nis the surface normal

unit vector)

T

T=







n10 0

0n20

0 0 n3

0n3n2

n30n1

n2n10







,D

D

D=







∂

10 0

0

∂

20

0 0

∂

3

0

∂

3

∂

2

∂

30

∂

1

∂

2

∂

10







,

ni=cos(n

n

n,xi)

∂

i=

∂

xi

i=1,2,3

(2.2.6)

and the stiffness m atrix C

C

Cwe get

D

DT

σ

+p

p

p=0

0

0∈Vstatic equilibrium equations,

D

DT

σ

+p

p

p=

ρ∂

2u

u

∂

t2∈Vdynamic equilibrium equations,

T

TT

σ

=q

q

q∈Aqsurface equilibrium equations,

ε

=D

D

Du

u

u∈Vkinematic equations,

σ

=C

C

ε

constitutive equations

(2.2.7)

Vis the volume and Aqthe surface of the body with surface forces q

q

q.

Eliminating the stresses and the strains leads to the differential equations for the

displacements

Boundary-value problem - elastostatics

D

DTC

C

CD

D

Du

u

u=−p

p

p∈Vequilibrium for the volume element V,

u

u=u

u

u∈Auprescribed displacements u

u

uon Au,

T

TTC

C

CD

D

Du

u

u=q

q

q∈Aqprescribed surface forces q

q

qon Aq

(2.2.8)

Initial boundary-value problem - elastodynamics

D

DTC

C

CD

D

Du

u

u−

ρ

¨

u

u=−p

p

p∈Vequilibrium equation,

u

u=u

u

u∈Au,T

T

TTC

C

CD

D

Du

u

u=q

q

q∈Aqboundary conditions,

u

u(x

x

x,0) = u

u

u(x

x

x,0),˙

u

u(x

x

x,0) = ˙

u

u(x

x

x,0)initial conditions

(2.2.9)

In the general case of material anisotropic behavior the three-dimensional equa-

tions are very complicated and analytical solutions are only possible for some spe-

2.2 Fundamental Equations and Variational Solution Procedures 63

cial problems. This is independent of the approach to displacements or stresses.

Some elementary examples are formulated in Sect. 2.2.4. The equations for beams

and plates are simpliﬁed with additional kinematic and/or static hypotheses and the

equations are deduced separately in Chaps. 7 and 8. The simpliﬁed structural equa-

tions for circular cylindrical shells and thin-walled folded structures are given in

Chaps. 9 and 10.

Summarizing the fundamental equations of elasticity we have introduced stresses

and displacements as static and kinematic ﬁeld variables. A ﬁeld is said to be stati-

cally admissible if the stresses satisfy equilibrium equations (2.2.1) and are in equi-

librium with the surface traction ¯

q

qon the body surface Aq, where the traction are

given. A ﬁeld is referred to as kinematically admissible if displacements and strains

are restricted by the strain-displacement equations (2.2.3) and the displacement sat-

isﬁes kinematic boundary conditions on the body surface Au, where the displace-

ments are prescribed. Admissible ﬁeld variables are considered in principles of vir-

tual work and energy formulations, Sect. 2.2.2. The mutual correspondence between

static and kinematic ﬁeld variables is established through the constitutive equations

(2.2.4).

2.2.2 Principle of Virtual Work and Energy Formulations

The analytical description of the model equations of anisotropic elasticity may re-

alized as above by a system of partial differential equations but also by integral

statements which are equivalent to the governing equations of Sect. 2.2.1 and based

on energy or variational formulations. The utility of variational formulations is in

general twofold. They yield convenient methods for the derivation of the governing

equations of problems in applied elasticity and provide the mathematical basis for

consistent approximate theories and solution procedures. There are three variational

principles which are used mostly in structural mechanics. There are the principles of

virtual work, the principle of complementary virtual work, Reissner’s6variational

theorem and the related energy principles.

Restricting ourselves to static problems, extremal principles formulated for the

total elastic potential energy of the problem or the complementary potential energy

are very useful in the theory of elasticity and in modelling and analysis of structural

elements. The fundamental equations and boundary conditions given beforehand

can be derived with the extremal principles and approximate solutions are obtained

by direct variational methods. Both extremal principles follow from the principle of

virtual work.

If an elastic body is in equilibrium, the virtual work

δ

Wof all acting forces in

moving through a virtual displacement

δ

u

uis zero

δ

W≡

δ

Wa+

δ

Wi=0 (2.2.10)

6Eric (Max Erich) Reissner (∗January 5th, 1913 in Aachen; † November 1st, 1996 La Jolla, USA)

- engineer, contributions to the theory of beams, plates and shells

64 2 Linear Anisotropic Materials

δ

Wis the total virtual work,

δ

Wathe external virtual work of body or volume and

surface forces and

δ

Withe internal virtual work of internal stresses, for the forces

associated with the stress ﬁeld of a body move the body points through virtual dis-

placements

δ

u

ucorresponding to the virtual strain ﬁeld

δε

ε

.

A displacement is called virtual, if it is inﬁnitesimal, and satisﬁes the geometric

constraints (compatibility with the displacement-strain equations and the boundary

conditions) and all forces are ﬁxed at their values. These displacements are called

virtual because they are assumed to be inﬁnitesimal while time is held constant. The

symbol

δ

is called a variational operator and in the mathematical view a virtual dis-

placement is a variation of the displacement function. To use variational operations

in structural mechanics only the following operations of the

δ

-operator are needed

δ

dnf

dxn=dn

dxn(

δ

f),

δ

(fn) = n f n−1

δ

fn−1,

δ

Zfdx=Z

δ

fdx

For a deformable body, the external and the internal work are given in Eqs.

(2.2.11) and (2.2.12), respectively,

δ

Wa=Z

V

pk

δ

ukdV+Z

Aq

qk

δ

ukdA,(2.2.11)

δ

Wi=−Z

V

σ

k

δε

kdV(2.2.12)

pkare the components of the actual body force vector p

p

pper unit volume and qk

the components of the actual surface force vector q

q

q(surface traction per unit area).

Aqdenotes the portion of the boundary on which surface forces are speciﬁed.

σ

k

and

ε

kare the components of the stress and the strain vector. The negative sign in

(2.2.12) indicates that the inner forces oppose the inner virtual displacements, e.g.

if the virtual displacement

δ

u1=

δε

1dx1is subjected an inner force (

σ

1dx2dx3)the

inner work is (−

σ

1dx2dx3)

δε

1dx1. The vectors p

p

p,q

q

qand u

u

uhave three components

but the vectors

σ

and

ε

have six components. The double subscript kin pk

δ

ukand

qk

δ

ukmeans the summarizing on 1 to 3 but in

σ

k

δε

kon 1 to 6.

The general formulation of the principle of virtual work for a deformable body

δ

Wa+

δ

Wi≡

δ

W=0

or Z

V

pk

δ

ukdV+Z

Aq

qk

δ

ukdA−Z

V

σ

k

δε

kdV=0 (2.2.13)

is independent of the constitutive equations. For the three-dimensional boundary-

value problem of a deformable body the principle can be formulated as follow:

Theorem 2.1 (Principle of virtual work). The sum of virtual work done by the

internal and external forces in arbitrary virtual displacements satisfying the pre-

2.2 Fundamental Equations and Variational Solution Procedures 65

scribed geometrical constraints and the strain-displacement relations is zero, i.e.

the arbitrary ﬁeld variables

δ

ukare kinematically admissible.

An important case is restricted to linear elastic anisotropic bodies and is known as

the principle of minimum total potential energy. The external virtual work

δ

Wais

stored as virtual strain energy

δ

Wf=−

δ

Wi, i.e. there exists a strain energy density

function

Wf(

ε

) = 1

2

σ

k

ε

k=1

2Ckl

ε

k

ε

l

Assuming conservative elasto-static problems, the principle of virtual work takes

the form

δΠ

(u

u

u)≡

δΠ

a(u

u

u) +

δΠ

i(

ε

)≡0,

ε

=

ε

(u

u

u)(2.2.14)

with the total potential energy function

Π

(u

u

u)of the elastic body.

Π

a(u

u

u)and

Π

i(

ε

)

are the potential functions of the external and the internal forces, respectively,

Π

i=

Π

(

ε

k) = 1

2Z

V

Ckl

ε

k

ε

ldV,

Π

a=

Π

a(uk) = −Z

V

pkukdV−Z

Aq

qkukdA

(2.2.15)

The principle of minimum total potential energy may be stated for linear elastic

bodies with the constraints

σ

=C

C

ε

(u

u

u)as follows:

Theorem 2.2 (Principle of minimum of the total potential energy). Of all the ad-

missible displacement functions satisfying strain-stress relations and the prescribed

boundary conditions, those that satisfy the equilibrium equations make the total po-

tential energy an absolute minimum.

The Euler7-Lagrange8equations of the variational problem yield the equilibrium

and mechanical boundary conditions of the problem. The minimum total potential

energy is widely used in solutions to problems of structural mechanics.

The principle of virtual work can be formulated in a complementary statement.

Then virtual forces are introduced and the displacements are ﬁxed. In analogy to

(2.2.13) we have the principle of complementary virtual work as

δ

W∗

a+

δ

W∗

i≡

δ

W∗=0

with the complimentary external and internal virtual works

δ

W∗

a=Z

Au

uk

δ

qkdA,

δ

W∗

i=−Z

V

ε

k

δσ

kdV(2.2.16)

7Leonhard Euler (∗15 April 1707 Basel - †7jul./18greg.September 1783 St. Petersburg) - mathe-

matician, physicist, astronomer, logician and engineer, introducing beam theory elements and the

equations of motion

8Joseph-Louis Lagrange, born Giuseppe Lodovico Lagrangia or Giuseppe Ludovico De la Grange

Tournier, also reported as Giuseppe Luigi Lagrange or Lagrangia (∗25 January 1736 Turin - †10

April 1813 Paris) - mathematician and astronomer, variational calculations, general mechanics

66 2 Linear Anisotropic Materials

Audenotes the portion of the boundary surface on which displacements are speci-

ﬁed.

With the complementary stress energy density function

W∗

f(

σ

) = 1

2Skl

σ

k

σ

l,

δ

W∗

f(

σ

) = Skl

σ

l

δσ

k=

ε

k

δσ

k

and assuming conservative elasto-static problems, the principle of complementary

work can be formulated as principle of minimum total complementary energy

δΠ

∗=

δΠ

∗

i+

δΠ

∗

a

or

δ



Z

V

W∗(

σ

k)dV−Z

Au

ukqkdA

=0 (2.2.17)

The principle of minimum total complementary energy may be stated for linear

elastic bodies with constraints

ε

=S

S

σ

as follows:

Theorem 2.3 (Principle of minimum total complementary energy). Of all admis-

sible stress states satisfying equilibrium equations and stress boundary conditions,

those which are kinematically admissible make the total complementary energy an

absolute minimum.

The Euler-Lagrange equations of the variational statement yield now the compati-

bility equations and the geometrical boundary conditions.

The both well-known principles of structure mechanics, the principle of vir-

tual displacements (displacement method) and the theorem of Castigliano (principle

of virtual forces, force method) correspond to the principle of minimum potential

energy and complementary energy. The principle of minimum potential energy is

much more used in solution procedures, because it is usually far easier to formulate

assumptions about functions to represent admissible displacements as to formulate

admissible stress functions that ensure stresses satisfying mechanical boundary con-

ditions and equilibrium equations. It should be kept in mind that from the two prin-

ciples considered above no approximate theory can be obtained in its entirety. One

must either satisfy the strain-displacement relations and the displacement bound-

ary conditions exactly and get approximate equilibrium conditions or vice versa.

Both principles yield the risk to formulate approximate theories or solution proce-

dures which may be mathematically inconsistent. Reissner’s variational statement

yields as Euler-Lagrange equations both, the equilibrium equations and the strain-

displacement relations, and has the advantage that its use would yield approximate

theories and solution procedures which satisfy both requirements to the same de-

gree and would be consistent. Reissner’s variational theorem (Reissner, 1950) can

be formulated as follows:

Theorem 2.4 (Reissner’s variational theorem, 1950). Of all sets of stress and

displacement functions of an elastic body

ε

=C

C

σ

which satisfy the boundary

2.2 Fundamental Equations and Variational Solution Procedures 67

conditions, those which also satisfy the equilibrium equations and the stress-

displacements relations correspond to a minimum of the functional

Ψ

Rdeﬁned as

Ψ

R(u

u

u,

σ

) = Z

V

[

σ

k

ε

k−Wf(

σ

k)]dV−Z

V

pkukdV−Z

Aq

qkukdA(2.2.18)

Wf(

σ

k)is the strain energy density function in terms of stresses only,

Ψ

Ris Reissner’s

functional.

It should be noted that all stress and strain components must be varied while pk

and qkare prescribed functions and therefore ﬁxed. The variation of the functional

Ψ

R(u

u

u,

σ

)yields

δΨ

R=Z

V

σ

k

δε

k+

ε

k

δσ

k−

∂

Wf

∂σ

k

δσ

kdV−Z

V

pk

δ

ukdV

−Z

Aq

qk

δ

ukdA,

(2.2.19)

where

ε

kis determined by (2.2.3).

δΨ

R(u

u

u,

σ

)can be rearranged and we obtain ﬁnally

δΨ

R=Z

V

ε

−

∂

Wf

∂σ

σ



δσ

σ

T−D

D

DT

σ

+p

p

p

δ

u

uTdV−Z

Aq

q

δ

u

uTdA(2.2.20)

Since

δσ

σ

and

δ

u

uare arbitrary variations

δΨ

R=0 is satisﬁed only if

ε

i j =

∂

Wf(

σ

i j )

∂σ

i j

,

∂σ

i j

∂

xj

+pi=0 (2.2.21)

Summarizing we have considered two dual energy principles with ukor

σ

kas admis-

sible functions which have to be varied and one generalized variational principle,

where both, ukand

σ

k, have to be varied. The considerations are limited to lin-

ear problems of elasto-statics, i.e. the generalized Hooke’s law describes the stress-

strain relations.

Expanding the considerations on dynamic problems without dissipative forces

following from external or inner damping effects the total virtual work has in the

sense of the d’Alambert principle an additional term which represents the inertial

forces

δ

W=−Z

V

ρ

¨uk

δ

ukdV−Z

V

σ

k

δε

kdV+Z

V

pk

δ

ukdV +Z

Aq

qk

δ

ukdA(2.2.22)

Equation (2.2.22) represents an extension of the principle of virtual work from stat-

ics to dynamics.

ρ

is the density of the elastic body.

68 2 Linear Anisotropic Materials

For conservative systems of elasto-dynamics, the Hamilton9principle replaces

the principle of the minimum of the total potential energy

δ

t2

Z

t1

(T−

Π

)dt≡

δ

t2

Z

t1

L(uk)dt=0,T=1

2Z

V

ρ

˙uk˙ukdV(2.2.23)

Π

(u

u

u)is the potential energy given beforehand and T(u

u

u)is the so-called kinetic

energy. L=T−

Π

is the Lagrangian function.

Theorem 2.5 (Hamilton’s principle for conservative systems). Of all possible

paths between two points at time interval t1and t2along which a dynamical sys-

tem may move, the actual path followed by the system is the one which minimizes

the integral of the Lagrangian function.

In the contracted vector-matrix notation we can summarize:

Conservative elasto-static problems

Π

(u

u

u) = 1

2Z

V

σ

σε

ε

TdV−Z

V

p

pu

u

uTdV−Z

Aq

q

qu

u

uTdA,

δΠ

=Z

V

σ

σδε

ε

TdV−Z

V

p

δ

u

uTdV−Z

Aq

q

δ

uTdA=0

(2.2.24)

Conservative elasto-dynamic problems

L(u

u

u) = T(u

u

u)−

Π

(u

u

u),T(u

u

u) = 1

2Z

V

ρ

˙

u

uT˙

u

udV,

δ

t2

Z

t1

L(u

u

u)dt=0

(2.2.25)

All variations are related to the displacement vector u

u

u. For the stress and the strain

vector we have to take into consideration that

σ

=

σ

(

ε

) =

σ

[

ε

(u

u

u)] and for the time

integrations

δ

u

u(x

x

x,t1) =

δ

u

u(x

x

x,t2) = 0 (2.2.26)

For non-conservative systems of elastodynamics, the virtual work

δ

Wincludes an

approximate damping term

−Z

V

µ

˙

u

uT

δ

u

udV

with

µ

as a damping parameter and Eq. (2.2.22) is substituted by

9William Rowan Hamilton (∗4 August 1805 Dublin - †2 September 1865 Dublin) - physicist,

astronomer, and mathematician, who made important contributions to classical mechanics, optics,

and algebra

2.2 Fundamental Equations and Variational Solution Procedures 69

δ

W=−Z

V

(

ρ

¨

u

uT+

µ

˙

u

uT)

δ

u

udV−Z

V

δ

(

σ

T

ε

)dV−Z

V

p

δ

u

udV

−Z

Aq

q

δ

u

udA

(2.2.27)

A generalized Hamilton’s principle in conjunction with the Reissner’s variational

statement can be presented as

δ χ

(u

u

u,

σ

) =

δ

t2

Z

t1

[T(u

u

u)−

ψ

R(u

u

u,

σ

)]dt=0,

where T(u

u

u)is the kinetic energy as above,

ψ

R(u

u

u,

σ

)the Reissner’s functional

(2.2.18).

2.2.3 Variational Methods

The variational principles can be used to obtain, in a mathematical way, the govern-

ing differential equations and associated boundary conditions as the Euler-Lagrange

equations of the variational statement. Now we consider the use of the variational

principles in the solution of the model equations. We seek in the sense of the clas-

sical variational methods, approximate solutions by direct methods, i.e. the approx-

imate solution is obtained directly by applying the same variational statement that

are used to derive the fundamental equations.

2.2.3.1 Rayleigh-Ritz Method

Approximate methods are used when exact solutions to a problem cannot be derived.

Among the approximation methods, Ritz10 method is a very convenient method

based on a variational approach. The variational methods of approximation de-

scribed in this textbook are limited to Rayleigh11-Ritz method for elasto-statics and

elasto-dynamics problems of anisotropic elasticity theory and to some extent on

weighted-residual methods.

The Rayleigh-Ritz method is based on variational statements, e.g. the principle of

minimum total potential energy, which is equivalent to the fundamental differential

equations as well as to the so-called natural or static boundary conditions including

10 Walter Ritz (∗February 22nd, 1878, Sion, Switzerland; † July 7th , 1909, G¨ottingen) - theoretical

physicist, variational methods

11 John William Strutt, 3rd Baron Rayleigh (∗November 12th, 1842, Langford Grove, Maldon,

Essex, UK; † June 30th, 1919, Terling Place, Witham, Essex, UK) - physicist, Nobel Prize in

Physics winner (1904), variational method

70 2 Linear Anisotropic Materials

force boundary conditions . This variational formulation is known as the weak form

of the model equations. The method was proposed as the direct method by Rayleigh

and a generalization was given by Ritz.

The starting point for elasto-static problems is the total elastic potential energy

functional

Π

=1

2Z

V

ε

TC

C

ε

dV−Z

V

p

pTu

u

udV−Z

Aq

q

qTu

u

udA

=1

2Z

V

(D

D

Du

u

u)TC

C

CD

D

Du

u

udV−Z

V

p

pTu

u

udV−Z

Aq

q

qTu

u

udA

(2.2.28)

The variations are related to the displacements u

u

uand the strains

ε

which have to be

substituted with help of the differential matrix D

D

D, (2.2.6), by the displacements. The

approximate solution is sought in the form of a ﬁnite linear combination. Looking

ﬁrst at a scalar displacement approach, the approximation of the scalar displacement

function u(x1,x2,x2)is given by the Ritz approximation

˜u(x1,x2,x3) =

N

∑

i=1

ai

ϕ

i(x1,x2,x3)

or

˜u(x1,x2,x3) =

N

∑

i=1

ai

ϕ

i(x1,x2,x3) +

ϕ

0(x1,x2,x3)(2.2.29)

The

ϕ

iare known functions chosen a priori, named approximation functions or co-

ordinate functions. The aidenote undetermined constants named generalized coor-

dinates. The approximation ˜uhas to make (2.2.28) extremal

˜

Π

(˜u) = ˜

Π

(ai),

δ

˜

Π

(ai) = 0 (2.2.30)

This approximation is characterized by a relative extremum. From (2.2.30) comes

˜

Π

in form of a function of the constants aiand

δ

˜

Π

(ai) = 0 yields Nstationary

conditions

∂

˜

Π

(ai)

∂

ai

=0,i=1,2,...,N(2.2.31)

˜

Π

may be written as a quadratic form in aiand from Eqs. (2.2.31) follows a system

of Nlinear equations allowing the Nunknown constants aito be determined. In

order to ensure a solution of the system of linear equations and a convergence of the

approximate solution to the true solution as the number Nof the aiis increased, the

ϕ

ivalues have to fulﬁll the following requirements:

•

ϕ

0satisﬁes speciﬁed inhomogeneous geometric boundary conditions, the so-

called essential conditions of the variational statement and

ϕ

i,i=1,2,...,Nsat-

isfy the homogeneous form of the geometric boundary conditions.

•

ϕ

iare continuous as required in the variational formulation, e.g. they should have

a non-zero contribution to ˜

Π

.

•

ϕ

iare linearly independent and complete.

2.2 Fundamental Equations and Variational Solution Procedures 71

The completeness property is essential for the convergence of the Ritz approxima-

tion. Polynomial and trigonometric functions are selected examples of complete

systems of functions.

Generalizing the considerations to three-dimensional problems and using vector-

matrix notation it follows

˜

u

u(x1,x2,x3)≡



˜u1

˜u2

˜u3

=





a

aT

1

ϕ

1

a

aT

2

ϕ

2

a

aT

3

ϕ

3





≡



ϕ

1o

o

o o

o

ϕ

2o

o

o o

o

ϕ

3



T



a

a1

a

a2

a

a3

(2.2.32)

or

˜

u

u(x1,x2,x3) = G

G

GTa

a

a(2.2.33)

with

G

GT=



ϕ

1o

o

o o

o

ϕ

2o

o

o o

o

ϕ

3



T

=





ϕ

T

1o

o

oTo

o

oT

o

oT

ϕ

T

2o

o

oT

o

oTo

o

oT

ϕ

T

3





,a

a

a=



a

a1

a

a2

a

a3



G

Gis the matrix of the approximation functions,

ϕ

iand o

o

oare N-dimensional vectors

and a

a

aiare N-dimensional subvectors of the vector a

a

aof the unknown coordinates.

The application of the Ritz method using the minimum principle of elastic potential

energy

Π

has the following steps:

1. Choose the approximation function ˜

u

u=G

G

GTa

a

a.

2. Substitute ˜

u

uinto

Π

˜

Π

(˜

u

u) = 1

2Z

V

(D

D

D˜

u

u)TC

C

CD

D

D˜

u

udV−Z

V

p

pT˜

u

udV−Z

Aq

q

qT˜

u

udA

=1

2a

a

aTK

K

Ka

a

a−a

a

aTf

f

(2.2.34)

with

K

K=Z

V

(D

D

DG

G

G)TC

C

C(D

D

DG

G

G)dV=Z

V

B

BTC

C

CB

B

BdV,

f

f=Z

V

G

GTp

p

pdV+Z

Aq

G

GTq

q

qdA

3. Formulate the stationary conditions of ˜

Π

(a

a

a)

∂

˜

Π

(a

a

a)

∂

a

a=o

o

i.e. with

∂

a

a(a

a

aTK

K

Ka

a

a) = 2K

K

Ka

a

a,

∂

a

a(a

a

aTf

f

f) = a

a

follows

72 2 Linear Anisotropic Materials

K

Ka

a

a=f

f

f(2.2.35)

K

Kis called the stiffness matrix, a

a

athe vector of unknowns and f

f

fthe force vector.

These notations are used in a generalized sense.

4. Solve the system of linear equations K

K

Ka

a

a=f

f

f. The vector a

a

aof unknown coefﬁ-

cients is known.

5. Calculate the approximation solution ˜

u

u=a

a

aT

ϕ

and the ˜

ε

=D

D

D˜

u

u,˜

σ

=C

C

C˜

ε

,...

For an increasing number N, the previously computed coefﬁcients of a

a

aremain un-

changed provided the previously chosen coordinate functions are not changed. Since

the strains are calculated from approximate displacements, strains and stresses are

less accurate than displacements.

The Ritz approximation of elasto-dynamic problems is carried out in an analo-

gous manner and can be summarized as follows. For conservative problems we start

with the variational statement (2.2.23). The displacement vector u

u

uis now a function

of x

x

xand tand the a

a

a-vector a function of t. The stationary condition yields

∂

a

a1

2a

a

aT(t)K

K

Ka

a

a(t)−a

a

aT(t)f

f

f(t) + ¨

a

aT(t)M

M

Ma

a

a(t)=0

0

0 (2.2.36)

M

M¨

a

a(t) + K

K

Ka

a

a(t) = f

f

f(t),(2.2.37)

M

M=Z

V

ρ

G

GTG

G

GdV,

K

K=Z

V

(D

D

DG

G

G)TC

C

C(D

D

DG

G

G)dV,

f

f=Z

V

G

GTp

p

pdV+Z

Aq

G

GTq

q

qdA

The matrix G

G

Gdepends on x

x

x,p

p

pand q

q

qon x

x

xand t.M

M

Mis called the mass matrix.

An direct derivation of a damping matrix from the Ritz approximation analogous

to the K

K

K- and the M

M

M-matrix of (2.2.37) is not possible. In most engineering applica-

tions (2.2.37) has an additional damping term and the damping matrix is formulated

approximately as a linear combination of mass- and stiffness-matrix (modal damp-

ing)

M

M¨

a

a(t) +C

C

CD˙

a

a(t) + K

K

Ka

a

a(t) = f

f

f(t),C

C

CD≈

α

M

M+

β

K

K(2.2.38)

In the case of the study of free vibrations, we write the time dependence of a

a

a(t)in

the form

a

a(t) = ˆ

a

acos(

ω

t+

ϕ

)(2.2.39)

and from (2.2.38) with C

C

CD=0

0

0,f

f

f(t) = 0

0

0 comes the matrix eigenvalue problem (App.

A.3)

(K

K

K−

ω

2M

M

M)ˆ

a

a=o

o

o,det[K

K

K−

ω

2M

M

M] = 0 (2.2.40)

For Ncoordinate functions the algebraic equation (2.2.40) yields Neigenfrequen-

cies of the deformable body.

2.2 Fundamental Equations and Variational Solution Procedures 73

The Rayleigh-Ritz method approximates the continuous deformable body by a

ﬁnite number of degree of freedoms, i.e. the approximated system is less ﬂexible

than the actual body. Consequently for the approximated energy potential ˜

Π

≤

Π

.

The energy potential converges from below. The approximate displacements satisfy

the equilibrium equations only in the energy sense and not pointwise, unless the

solution converges to the exact solution. The Rayleigh-Ritz method can be applied

to all mechanical problems since a virtual statement exists, i.e. a weak form of the

model equations including the natural boundary conditions. If the displacements

are approximate, the approximate eigenfrequencies are higher than the exact, i.e.

˜

ω

≥

ω

.

2.2.3.2 Weighted Residual Methods

Finally some brief remarks on weighted residual methods are given. The fundamen-

tal equation in the displacement approach may be formulated in the form

A(u) = f(2.2.41)

Ais a differential operator. We seek again an approximate solution (2.2.29), where

now the constants aiare determined by requiring the residual

RN=A N

∑

i=1

ai

ϕ

i+

ϕ

0!−f6=0 (2.2.42)

be orthogonal to Nlinear independent weight function

ψ

i

Z

V

RN

ψ

idx

x

x=0,i=1,2,...,N(2.2.43)

ϕ

0,

ϕ

ishould be linear independent and complete and fulﬁll all boundary conditions.

Various known special methods follow from (2.2.43). They differ from each other

due to the choice of the weight functions

ψ

i:

•Galerkin’s method12

ψ

i≡

ϕ

i,

•Least-squares method

ψ

i≡A(

ϕ

i),

•Collocation method

ψ

i≡

δ

(x

x

x−x

x

xi)(

δ

(x

x

x−x

x

xi) = 1 if x

x

x=x

x

xiotherwise 0)

The Galerkin method is a generalization of the Ritz method, if it is not possible to

construct a weak form statement. Otherwise the Galerkin and the the Ritz method for

weak formulations of problems yield the same solution equations, if the coordinate

functions

ϕ

iin both are the same.

12 Boris Grigorjewitsch Galerkin, surname more accurately romanized as Galyorkin (∗4

Marchgreg./20 Februaryjul.1871 Polozk - 12 July 1945 Leningrad) - mathematician and engineer,

contributions to the theory of approximate solutions of partial differential equations

74 2 Linear Anisotropic Materials

The classical variational methods of Ritz and Galerkin are widely used to solve

problems of applied elasticity or structural mechanics. When applying the Ritz or

Galerkin method to special problems involving, e.g. a two-dimensional functional

Π

[u(x1,x2)] or a two-dimensional differential equation A[u(x1,x2)] = f(x1,x2), an

approximative solution is usually assumed in the form

˜u(x1,x2) =

N

∑

i=1

ai

ϕ

i(x1,x2)or ˜u(x1,x2) =

N

∑

i=1

M

∑

j=1

ai j

ϕ

1i(x1)

ϕ

2j(x2),(2.2.44)

where

ϕ

i(x1,x2)or

ϕ

1i(x1),

ϕ

2j(x2)are a priori chosen trial functions and the aior

ai j are unknown constants. The approximate solution depends very strongly on the

assumed trial functions.

To overcome the shortcoming of these solution methods Vlasov13 and Kan-

torovich14 suggested an approximate solution in the form

˜u(x1,x2) =

N

∑

i=1

ai(x1)

ϕ

i(x1,x2)(2.2.45)

The

ϕ

iare again a priori chosen trial functions but the ai(x1)are unknown coefﬁcient

functions of one of the independent variables. The condition

δΠ

[˜u(x1,x2)] = 0 or

with dA=dx1dx2Z

A

RN(˜u)

ϕ

idA=0,i=1,2,...,N

lead to a system of Nordinary differential equations for the unknown functions

ai(x1). Generally it is advisable to choose if possible the trial functions

ϕ

ias func-

tions of one independent variable, i.e.

ϕ

i=

ϕ

i(x2), since otherwise the system of

ordinary differential equations will have variable coefﬁcients. The approximate so-

lution ˜u(x1,x2)tends in regard of the arbitrariness of the assumed trial function

ϕ

i(x2)to a better solution in the x1-direction. The obtained approximative solution

can be further improved in the x2-direction in the following manner. In a ﬁrst step

the assumed approximation

˜u(x1,x2) =

N

∑

i=1

ai(x1)

ϕ

i(x2)(2.2.46)

yields the functions ai(x1)by solving the resulting set of ordinary differential

equations with constant coefﬁcients. In the next step, with ai(x1)≡ai1(x1)and

ϕ

i(x2)≡ai2(x2), i.e.

˜u[I](x1,x2) =

N

∑

i=1

ai2(x2)ai1(x1),

13 Vasily Zakharovich Vlasov (∗11greg./24j ul.February 1906 Kareevo - †7 August 1958 Moscow)

- civil engineer

14 Leonid Vitaliyevich Kantorovich (∗19 January 1912 St. Petersburg - †7 April 1986 Moscow) -

mathematician and economist, Nobel prize winner in economics in 1975

2.2 Fundamental Equations and Variational Solution Procedures 75

the ai1(x1)are the given trial functions and the unknown functions ai2(x2)can be

determined as before the ai1(x1)by solving a set of ordinary differential equations.

After completing the ﬁrst cycle, which yields ˜u[I](x1,x2), the procedure can be con-

tinued iteratively. This iterative solution procedure is denoted in literature as vari-

ational iteration or extended Vlasov-Kantorowich method. The ﬁnal form of the

generated solution is independent of the initial choice of the trial function

ϕ

i(x2)

and the iterative procedure converges very rapidly. It can be demonstrated, that the

iterative generated solutions ˜u(x1,x2)agree very closely with the exact analytical

solutions u(x1,x2)even with a single term approximation

˜u(x1,x2) = a1(x1)

ϕ

1(x2)(2.2.47)

In engineering applications, e.g. for rectangular plates, the single term approxima-

tions yield in general sufﬁcient accuracy.

Summarizing it should be said that the most difﬁcult problem in the application

of the classical variational methods or weighted residual methods is the selection of

coordinate functions, especially for structures with irregular domains. The limita-

tions of the classical variational methods can be overcome by numerical methods,

e.g. the ﬁnite element method which is discussed in more detail in Chap. 11.

2.2.4 Problems

Exercise 2.7. An anisotropic body is subjected to a hydrostatic pressure

σ

1=

σ

2=

σ

3=−p,

σ

4=

σ

5=

σ

6=0.

1. Calculate the strain state

ε

.

2. Calculate the stress state

σ

for a change of the coordinate system obtained by a

rotation T

T

σ

.

Solution 2.7. The solution can be presented in two parts:

1. The generalized Hooke’s law yields (Eq. 2.1.20)

ε

1=−(S11 +S12 +S13)p,

ε

4=−(S14 +S24 +S34)p,

ε

2=−(S12 +S22 +S23)p,

ε

5=−(S15 +S25 +S35)p,

ε

3=−(S13 +S23 +S33)p,

ε

6=−(S16 +S26 +S36)p

Note 2.3. A hydrostatic pressure in an anisotropic solid yields extensional and

shear strains.

2. From (2.1.39) follows

σ

′

1=−p(c2+s2) = −p,

σ

′

4=0,

σ

′

2=−p(s2+c2) = −p,

σ

′

5=0,

σ

′

3=−p,

σ

′

6=−p(−cs +cs) = 0

76 2 Linear Anisotropic Materials

Exercise 2.8. An anisotropic body has a pure shear stress state

σ

1=

σ

2=

σ

3=

σ

5=

σ

6=0,

σ

4=t.

1. Calculate the strain state

ε

.

2. Compare the strain state for the anisotropic case with the isotropic case.

Solution 2.8. The solution can be presented again in two parts:

1. Equations (2.1.20) yield

ε

1=S14t,

ε

2=S24t,

ε

3=S34t,

ε

4=S44t,

ε

5=S45t,

ε

6=S46t

The anisotropic body has extensional and shear strains in all coordinate planes.

2. In an isotropic body a pure shear stress state yields only shearing strains:

S14 =S24 =S34 =0

Exercise 2.9. Consider a prismatic homogeneous anisotropic bar which is ﬁxed at

one end. The origin of the coordinates x1,x2,x3is placed in the centroid of the

ﬁxed section and the x3-axis is directed along the bar axis, land Aare the length

and the cross-section of the undeformed bar. Assume that the bar at the point

x1=x2=x3=0 has no displacement and torsion:

u1=u2=u3=0,u1,3=u2,3=u2,1−u1,2=0

1. A force Facts on the bar on the cross-section x3=land the stress state is de-

termined by

σ

1=

σ

2=

σ

4=

σ

5=

σ

6=0,

σ

3=F/A. Determine the strains, the

displacements and the extension of the axis.

2. The ﬁxed bar is deformed only under its own weight: p1=p2=0, p3=g

ρ

.

Determine the strain state and the displacements and calculate the displacements

in the point (0,0,l).

Solution 2.9. Now one has

1. The generalized Hooke’s law (2.1.20) with

σ

1=

σ

2=

σ

4=

σ

5=

σ

6=0,

σ

3=F/A=

σ

6=0,

gives

ε

1=S13

σ

,

ε

2=S23

σ

,

ε

3=S33

σ

,

ε

4=S34

σ

,

ε

5=S35

σ

,

ε

6=S36

σ

The displacements can be determined by the introduction of the following equa-

tions

S13

σ

=u1,1,S23

σ

=u2,2,S33

σ

=u3,3,

S34

σ

=u3,2+u2,3,S35

σ

=u3,1+u1,3,S36

σ

=u2,1+u1,2

The equations

2.2 Fundamental Equations and Variational Solution Procedures 77

u1=

σ

(S13x1+0.5S36x2),

u2=

σ

(0.5S36x1+S23x2),

u3=

σ

(S35x1+S34x2+S33 x3)

satisfy the displacement differential equations and the boundary conditions

which are prescribed at the point x1=x2=x3=0.

Note 2.4. The stress-strain formulae show that an anisotropic tension bar does not

only lengthens in the force direction x3and contracts in the transverse directions,

but also undergoes shears in all planes parallel to the coordinate planes. The

cross-sections of the bar remain plane. The stress states of an isotropic or an

anisotropic bar are identical, the anisotropy effects the strain state only.

2. The stress in a bar under its own weight is

σ

1=

σ

2=

σ

4=

σ

5=

σ

6=0,

σ

3=

ρ

g(l−x3)

The stress-strain displacement equations are

ε

1=u1,1=S13

ρ

g(l−x3),

ε

4=u3,2+u2,3=S34

ρ

g(l−x3),

ε

2=u2,2=S23

ρ

g(l−x3),

ε

5=u3,1+u1,3=S35

ρ

g(l−x3),

ε

3=u3,3=S33

ρ

g(l−x3),

ε

6=u2,1+u1,2=S36

ρ

g(l−x3)

The boundary conditions are identical to case a) and the following displacement

state which satisﬁes all displacement differential equations and the conditions at

the point x1=x2=x3=0 can be calculated by integration

u1=

ρ

g[−0.5S35x2

3+S13x1(l−x3) + 0.5S36x2(l−x3)],

u2=

ρ

g[−0.5S34x2

3+S23x2(l−x3) + 0.5S36x1(l−x3)],

u3=

ρ

g[−0.5S13x2

1+0.5S23x2

2

+0.5S36x1x2+ (S34x2+S35 x1)l+0.5S33x3(2l−x3)]

Note 2.5. The cross-section does not remain plane, it is deformed to the shape

of a second-order surface and the bar axis becomes curved. The centroid of the

cross-section x3=lis displaced in all three directions

u1(0,0,l) = −0.5

ρ

gS35l2,

u2(0,0,l) = −0.5

ρ

gS34l2,

u3(0,0,l) = 0.5

ρ

gS33l2

Exercise 2.10. Show that for a composite beam subjected to a distributed continu-

ous load q(x1)the differential equation and the boundary conditions can be derived

using the extremal principle of potential energy.

Solution 2.10. The beam is of the length l, width band height h.q(x1)is the lateral

load per unit length. The average elasticity modulus of the beam is E1. With the

stress-strain-displacement relations from Bernoulli15 beam theory

15 Jakob I. Bernoulli (∗27 December 1654jul./6 January 1655greg.Basel - †16 August 1705 Basel)

- mathematician, beam theory

78 2 Linear Anisotropic Materials

σ

1=E1

ε

1,

ε

1=−x3

d2u3

dx2

1

the strain energy function Wfis seen to be

Wf=1

2

σ

1

ε

1=1

2E1

ε

2

1=1

2E1x2

3d2u3

dx2

12

and

Π

i=

l

Z

0

b/2

Z

−b/2

h/2

Z

−h/2

E1

2d2u3

dx2

12

x2

3dx3dx2dx1=E1I

2

l

Z

0d2u3

dx2

12

dx1

with inertial moment I=bh3/12 for a rectangular cross-section.

In the absence of body forces, the potential function

Π

aof the external load q(x1)

is

Π

a=−

l

Z

0

q(x1)u3(x1)dx1

and the total elastic potential energy is

Π

(u3) = E1I

2

l

Z

0d2u3

dx2

12

dx1−

l

Z

0

q(x1)u3(x1)dx1

From the stationary condition

δΠ

(u3) = 0 it follows that

δΠ

(u3) = E1I

2

l

Z

0

δ

d2u3

dx2

12

dx1−

l

Z

0

q(x1)

δ

u3(x1)dx1=0

There is no variation of E1Ior q(x1), because they are speciﬁed. From the last equa-

tion one gets

δΠ

(u3) = E1I

2

l

Z

0

2d2u3

dx2

1

δ

d2u3

dx2

12

dx1−

l

Z

0

q(x1)

δ

u3(x1)dx1=0

The ﬁrst term can be integrated by parts. The ﬁrst integration can be performed with

the following substitution

u=d2u3

dx2

1

,u′=d3u3

dx3

1

,v′=

δ

d2u3

dx2

1,v=

δ

du3

dx1

and yields

2.2 Fundamental Equations and Variational Solution Procedures 79

δΠ

(u3) = E1I



d2u3

dx2

1

δ

du3

dx1l

0−

l

Z

0

d3u3

dx3

1

δ

du3

dx1dx1



−

l

Z

0

q

δ

u3dx1=0

The second integration can be performed with the following substitution

u=d3u3

dx3

1

,u′=d4u3

dx4

1

,v′=

δ

du3

dx1,v=

δ

u3

and yields

δΠ

(u3) = E1Id2u3

dx2

1

δ

du3

dx1l

0−E1Id3u3

dx3

1

δ

u3l

0

+

l

Z

0

E1Id4u3

dx4

1

dx1−

l

Z

0

q(x1)

δ

u3dx1=0

Finally we obtain

δΠ

(u3) = E1Id2u3

dx2

1

δ

du3

dx1l

0−E1Id3u3

dx3

1

δ

u3l

0

+

l

Z

0E1Id4u3

dx4

1−q(x1)

δ

u3dx1=0

Since the variations are arbitrary the equation is satisﬁed if

E1I(u3)′′′′ =q

and either

E1I(u3)′′ =0

or u′

3is speciﬁed and

E1I(u3)′′′ =0

or u3is speciﬁed at x1=0 and x1=l.

The beam differential equation is the Euler-Lagrange equation of the variational

statement

δΠ

=0, u3,u′

3represent essential boundary conditions, and E1Iu′′

3,E1Iu′′′

3

are natural boundary conditions of the problem. Note that the boundary conditions

include the classical conditions of simply supports, clamped and free edges.

Exercise 2.11. The beam of Exercise 2.10 may be moderately thick and the ef-

fects of transverse shear deformation

ε

5and transverse normal stress are taken into

account. Show that the differential equations and boundary condition can be de-

rived using the Reissner’s variational principle. The beam material behavior may be

isotropic.

80 2 Linear Anisotropic Materials

Solution 2.11. To apply Reissner’s variational statement one must assume admis-

sible functions for the displacements u1(x1),u3(x1)and for the stresses

σ

1(x1,x3),

σ

3(x1,x3),

σ

5(x1,x3). For the beam with rectangular cross-section and lateral load-

ing q(x1)follow in the frame of beam theory

u2=0,

σ

2=

σ

4=

σ

6=0

As in the classical beam theory (Bernoulli theory) we assume that the beam cross-

sections undergo a translation and a rotation, the cross-sections are assumed to re-

main plane but not normal to the deformed middle surface (Timoshenko16 theory).

Therefore we can assume in the simplest case

u1=x3

ψ

(x1),u3=w(x1)

and the strain-displacement relations may be written

ε

1=

∂

u1

∂

x1

=x3

ψ

′(x1),

ε

3=

∂

u3

∂

x3

=0,

ε

5=

∂

u3

∂

x1

+

∂

u1

∂

x3

=w′(x1) +

ψ

(x1)

For the stresses

σ

1,

σ

3and

σ

5the following functions are assumed

σ

1=M

Ix3,I=bh3

12 ,

σ

3=3q

4b"x3

h/2+2

3−1

3x3

h/22#,

σ

5=3Q

2A"1−x3

h/22#,A=bh

The assumed functions for

σ

1and

σ

5are identical to those of the Bernoulli beam

theory and

σ

3may be derived from the stress equation of equilibrium in the thick-

ness direction, Eq. (2.2.1), with

σ

3(+h/2) = q,

σ

1(−h/2) = 0

The bending moment Mand the shear force resultant Qwill be deﬁned in the usual

manner

M(x1) =

+h/2

Z

−h/2

b

σ

1dx3,Q(x1) =

+h/2

Z

−h/2

b

σ

5dx3

Now Reissner’s functional

Ψ

R(u

u

u,

σ

), e.g. (2.2.18), takes with the assumption above

the form

16 Stepan Prokopovich Timoshenko (∗22 December, 1878 Schpotiwka - † 29 May, 1972

Wuppertal-Elberfeld) - engineer, founder of the modern applied mechanics

2.2 Fundamental Equations and Variational Solution Procedures 81

Ψ

R(u

u

u,

σ

) =

l

Z

0

+h/2

Z

−h/2

σ

1

ε

1+

σ

5

ε

5−1

2E

σ

2

1+

σ

2

3+2(1+

ν

)

σ

2

5bdx3dx1

−

l

Z

0

qwdx1

=

l

Z

0

+h/2

Z

−h/2

σ

1x3

ψ

′(x1) +

σ

5w′(x1) +

ψ

(x1)

−1

2E

σ

2

1+

σ

2

3+2(1+

ν

)

σ

2

5bdx1−

l

Z

0

qwdx1

Substituting Mand Qand neglecting the term

σ

3which only depends on qand not

on the basic unknown functions

ψ

,wrespectively, Mand Q, and yields no contribu-

tion to the variation

δΨ

we obtain

Ψ

(w,

ψ

,M,Q) =

l

Z

0M

ψ

′+Q(w′+

ψ

)−M2

2EI +6

ν

qM

5EA −3Q2

5GA −qwdx1

δΨ

R=

l

Z

0M

δψ

′+

ψ

′

δ

M+Q(

δ

w′+

δψ

) + (w′+

ψ

)

δ

Q

−M

EI

δ

M+6

ν

q

5EA

δ

M−6Q

5GA

δ

Q−q

δ

wdx1=0

Integration the terms M

δψ

′and Q

δ

w′by parts and rearranging the equation

δΨ

R= [M

δψ

+Q

δ

w]l

0+

l

Z

0n[Q+M′]

δψ

−[Q′−q]

δ

w

+

ψ

′−M

EI +6

ν

q

5EA

δ

M+

ψ

+w′−6Q

5GA 

δ

Qdx1=0

The ﬁrst term yields the natural boundary conditions of the variational statement:

1. Either M=0 or

ψ

must be prescribed at x1=0,l.

2. Either Q=0 or wmust be prescribed at x1=0,l.

The variations

δψ

,

δ

w,

δ

Mand

δ

Qare all arbitrary independent functions of x1and

therefore

δΨ

R=0 only if

−dM(x1)

dx1

+Q(x1) = 0,dQ(x1)

dx1

+q(x1) = 0,

d

ψ

(x1)

dx1−M(x1)

EI +6

ν

q(x1)

5EA =0,dw(x1)

dx1

+

ψ

(x1)−6Q(x1)

5GA =0

82 2 Linear Anisotropic Materials

The both equations for the stress resultants are identical with the equations of the

classical Bernoulli’s theory. The term (w′+

ψ

)in the fourth equation describes the

change in the angle between the beam cross-section and the middle surface during

the deformation. The term (w′+

ψ

)is proportional to the average shear stress Q/A

and so a measure of shear deformation. With GA →∞the shear deformation tends

to zero and

ψ

to −w′as assumed in the Bernoulli’s theory. The third term in the

third equation depends on the lateral load qand Poisson’s ratio

ν

and tends to zero

for

ν

→0. This term described the effect of the transverse normal stress

σ

3. which

will be vanish if

ν

=0 as in the classical beam theory.

Substituting the differential equations for

ψ

and winto the differential relations

for Mand Qleads to

EI 

ψ

′′(x1) +

ν

6

5

Q′(x1)

EA −Q(x1) = 0,5

6GA[

ψ

′(x1) + w′′(x1)] + q(x1) = 0

or

EI 

ψ

′′(x1)−

ν

6

5

q(x1)

EA −5

6GA[

ψ

+w′] = 0,5

6GA[

ψ

′(x1) + w′′(x1)] + q(x1) = 0

Derivation and rearrangement yield a differential equation of 3rd order for

ψ

(x1)

EI

ψ

′′′(x1) = −q(x1) +

ν

6

5

q′(x1)

EA

With

Q(x1) = dM(x1)

dx1

=E I 

ψ

′′(x1) +

ν

6

5

Q′(x1)

EA 

and

Q(x1) = 5

6

EI

GA [

ψ

(x1) + w′(x1)]

follows an equation for w′(x1)

w′(x1) = −

ψ

(x1) + 6

5

EI

GA 

ψ

′′(x1) +

ν

6

5

q(x1)

EA 

Neglecting with

ν

=0 the effect of the transverse normal stress

σ

3we get the Tim-

oshenko’s beam equation

EI

ψ

′′′(x1) = −q(x1),M(x1) = E I

ψ

′(x1),Q(x1) = EI

ψ

′′(x1)

w′(x1) = −

ψ

(x1) + E I

ψ

′′(x1)

ksGA ,ks=5

6

One can note that as G→∞the shear deformation tends to vanish as assumed in

classical beam theory, i.e.

ψ

(x1) = −w′(x1)and the classical beam equations follow

to

EIw′′′′(x1) = q(x1),M(x1) = −E Iw′′(x1),Q(x1) = −EI w′′′(x1)

2.2 Fundamental Equations and Variational Solution Procedures 83

The derivation of the Timoshenko’s beam equation for laminated beams one can

ﬁnd in more detail in Sect. 7.3.

Exercise 2.12. Derive the free vibration equations for the moderately thick beam,

Exercise 2.11, using Hamilton’s principle in conjunction with the Reissner varia-

tional theorem.

Solution 2.12. In order to derive the free vibration equations we apply the gener-

alized Hamilton’s principle, i.e. the Hamilton’s principle in conjunction with the

Reissner variational statement

δ χ

(u

u

u,

σ

) =

δ

t2

Z

t1

[T(u

u

u)−

Ψ

R(u

u

u,

σ

)]dt=0

T(u

u

u)is the kinetic energy and

Ψ

R(u

u

u,

σ

)the Reissner functional of the moderately

thick beam.

Ψ

R(u

u

u,

σ

)is known from Exercise 2.11 and the kinetic energy for the

beam may be written as

T(u

u

u) =

l

Z

0

h/2

Z

−h/2

1

2

ρ

"

∂

u

∂

t2

+

∂

w

∂

t2#bdx3dx1

=

l

Z

0

1

2

ρ

"I

∂ψ

∂

t2

+A

∂

w

∂

t2#dx1

with

I=

h/2

Z

−h/2

bx2

3dx3,A=

h/2

Z

−h/2

bdx3

and the mass density

ρ

of the beam material. The substitution of T[u

u

u]above and

Ψ

(w,

ψ

,M,Q)of Exercise 2.11 in the functional

χ

yields

δ χ

(u

u

u,

σ

) =

δ

t2

Z

t1

l

Z

0(1

2

ρ

"I

∂ψ

∂

t2

+A

∂

w

∂

t2#

−M

ψ

′+Q(w′+

ψ

)−M2

2EI +6

ν

qM

5EA −3Q2

5GA −qw)dtdx1

=

t2

Z

t1

l

Z

0

ρ

I

∂

2

ψ

∂

t2

δψ

+A

∂

2w

∂

t2

δ

w

−M

δψ

′−

ψ

′

δ

M−Q(

δ

w′+

δψ

)−(w′+

ψ

)

δ

Q

+M

EI

δ

M−6

ν

q

5EA

δ

M−6Q

5GA

δ

Q−q

δ

wdtdx1

84 2 Linear Anisotropic Materials

Integration the terms M

δψ

′+Q

δ

w′by parts

l

Z

0

M

δψ

′dx1=M

δψ



l

0−

l

Z

0

M′

δψ

dx1,

l

Z

0

Q

δ

w′dx1=Q

δ

w

l

0−

l

Z

0

Q′

δ

wdx1,

rearranging the equation

δ χ

and setting

δ χ

=0 yield

t2

Z

t1

(M

δψ

+Q

δ

w)l

0dt+

t2

Z

t1

l

Z

0

ρ

I

∂

2

ψ

∂

t2+Q−M′

δψ

−Q′+qA

∂

2w

∂

t2

δ

w+

ψ

′−M

EI +6

ν

q

5EA

δ

M

+

ψ

+w′−6Q

5GA 

δ

Adtdx1=0

and the equations for free vibrations follow with q≡0

Q−

∂

M

∂

x1

+

ρ

I

∂

2

ψ

∂

t2=0,

∂

Q

∂

x1−

ρ

A

∂

2w

∂

t2=0,

∂ψ

∂

x1−M

EI =0,

ψ

+

∂

w

∂

x1−6Q

5GA =0

The underlined terms represent the contribution of rotatory inertia and the effect of

transverse shear deformation.

The system of four equations can be reduced to a system of two equations for the

unknowns wand

ψ

. Substitution of

Q=EI

∂

2

ψ

∂

x2

1−

ρ

I

∂

2

ψ

∂

t2,I=bh3

12 ,A=bh

in the second and fourth equation leads

∂

3

ψ

∂

x3

1−

ρ

E

∂

3

ψ

∂

t2

∂

x1−

ρ

A

E

∂

2w

∂

t2=0,

ψ

+

∂

w

∂

x1−h2

10 E

G

∂

2

ψ

∂

x2

1−

ρ

G

∂

2

ψ

∂

t2=0

The equations for forced or free vibrations are given in Sect. 7.3 in a more general

form.

References

Reissner E (1950) On a variational theorem in elasticity. Journal of Mathematics

and Physics 29(1-4):90–95

Chapter 3

Effective Material Moduli for Composites

Composite materials have at least two different material components which are

bonded. The material response of a composite is determined by the material moduli

of all constituents, the volume or mass fractions of the single constituents in the

composite material, by the quality of their bonding, i.e. of the behavior of the in-

terfaces, and by the arrangement and distribution of the ﬁbre reinforcement, i.e. the

ﬁbre architecture.

The basic assumptions made in material science approach models of ﬁbre rein-

forced composites are:

•The bond between ﬁbres and matrix is perfect.

•The ﬁbres are continuous and parallel aligned in each ply, they are packed regu-

larly, i.e. the space between ﬁbres is uniform.

•Fibre and matrix materials are linear elastic, they follow approximately Hooke’s

law and each elastic modulus is constant.

•The composite is free of voids.

Composite materials are heterogeneous, but in simplifying the analysis of compos-

ite structural elements in engineering applications, the heterogeneity of the material

is neglected and approximately overlayed to a homogeneous material. The most

important composites in structural engineering applications are laminates and sand-

wiches. Each single layer of laminates or sandwich faces is in general a ﬁbre rein-

forced lamina. For laminates we have therefore two different scales of modelling:

•The modelling of the mechanical behavior of a lamina, is called the micro-

mechanical or microscopic approach of a composite. The micro-mechanical

modelling leads to a correlation between constituent properties and average ef-

fective composite properties. Most simple mixture rules are used in engineering

applications. Whenever possible, the average properties of a lamina should be

veriﬁed experimentally by the tests described in Sect. 3.1 or Fig. 3.1.

•The modelling of the global behavior of a laminate constituted of several quasi-

homogeneous laminae is called the macroscopic approach of a composite.

Fibre reinforced material is in practice neither monolithic nor homogeneous, but it

is impossible to incorporate the real material structure into design and analysis of

85

H. Altenbach et al., Mechanics of Composite Structural

Elements, https://doi.org/10.1007/978-981-10-8935-0_3

86 3 Effective Material Moduli for Composites

✻

❄

σ

L

σ

L

test

ε

L,

ε

T

EL=

σ

L

ε

L

,

ν

LT =

ε

T

ε

L

❄

✻

σ

T

test

ε

T

σ

T

ET=

σ

T

ε

T

✲

✻

❄

✛

τ

test

γ

GLT =

τ

γ

Fig. 3.1 Experimental testing of the mechanical properties of an UD-layer: EL=E1,ET=E2,

GLT =G12,

ν

LT =

ν

12

composite or any other structural component. Therefore the concept of replacing the

heterogeneous material behavior with an effective material which is both homoge-

neous and monolithic, thus characterized by the generalized Hooke’s law, will be

used in engineering applications. We assume that the local variations in stress and

strain state are very small in comparison to macroscopical measurements of material

behavior.

In the following section some simple approaches to the lamina properties are

given with help of the mixture rules and simple semi-empirical consideration. The

more theoretical modelling in Sect. 3.2 has been developed to establish bounds on

effective properties. The modelling of the average mechanical characteristics of lam-

inates will be considered in Chap. 4.

3.1 Elementary Mixture Rules for Fibre-Reinforced Laminae

In Sect. 1.1 the formulas for volume fraction, mass fraction and density for ﬁbre

reinforced composites are given by (1.1.1) - (1.1.5). The rule of mixtures and the

inverse rule of mixtures is based on the statement that the composite property is

the weighted mean of the properties or the inverse properties of each constituent

multiplied by its volume fraction. In the ﬁrst case we have the upper-bound effective

property, in the second - the lower-bound. In composite mechanics these bounds are

related to W. Voigt and A. Reuss. In crystal plasticity the bounds were indroduced

by G. Taylor1and O. Sachs2. The notation used is as follows:

1Geoffrey Ingram Taylor (∗7 March 1886 St. John’s Wood, England - †27 June 1975 Cambridge)

- physicist and mathematician, contributions to the theory of plasticity, ﬂuid mechanics and wave

theory

2Oscar Sachs (∗5 April 1896 Moscow - †30 October 1960 Syracus, N.Y.) - metallurgist, contribu-

tions to the theory of plasticity

3.1 Elementary Mixture Rules for Fibre-Reinforced Laminae 87

EYoung’s modulus

ν

Poisson’s ratio

GShear modulus

σ

Stress

ε

Strain

V,MVolume, mass

v,mVolume fraction, mass fraction

ACross-section area

ρ

Density

The subscripts f and m refer to ﬁbre and matrix, the subscripts L ≡1,T≡2 refer to

the principal direction (ﬁbre direction) and transverse to the ﬁbre direction.

3.1.1 Effective Density

The derivation of the effective density of ﬁbre reinforced composites in terms of

volume fractions is given in Sect. 1.1

ρ

=M

V=Mf+Mm

V=

ρ

fVf+

ρ

mVm

V

=

ρ

fvf+

ρ

mvm=

ρ

fvf+

ρ

m(1−vf)

(3.1.1)

In literature we also ﬁnd vf≡

φ

for the ﬁbre volume fraction and we have

ρ

=

ρ

f

φ

+

ρ

m(1−

φ

)(3.1.2)

In an actual lamina the ﬁbres are randomly distributed over the lamina cross-section

and the lamina thickness is about 1 mm and much higher than the ﬁbre diameter

(about 0,01 mm). Because the actual ﬁbre cross-sections and the ﬁbre packing gen-

erally are not known and can hardly be predicted exactly typical idealized regular

ﬁbre arrangements are assumed for modelling and analysis, e.g. a layer-wise, square

or a hexagonal packing, and the ﬁbre cross-sections are assumed to have circular

form. There exists ultimate ﬁbre volume fractions vfmax , which are less than 1 and

depend on the ﬁbre arrangements:

•square or layer-wise ﬁbre packing - vfmax =0.785,

•hexagonal ﬁbre packing - vfmax =0.907

For real UD-laminae we have vfmax about 0.50 - 0.65. Keep in mind that a lower

ﬁbre volume fraction results in lower laminae strength and stiffness under tension

in ﬁbre-direction, but a very high ﬁbre volume fraction close to the ultimate values

of vfmay lead to a reduction of the lamina strength under compression in ﬁbre

direction and under in-plane shear due to the poor bending of the ﬁbres.

88 3 Effective Material Moduli for Composites

3.1.2 Effective Longitudinal Modulus of Elasticity

When an unidirectional lamina is acted upon by either a tensile or compression load

parallel to the ﬁbres, it can be assumed that the strains of the ﬁbres, matrix and

composite in the loading direction are the same (Fig. 3.2)

ε

Lf =

ε

Lm =

ε

L=

∆

l

l(3.1.3)

The mechanical model has a parallel arrangement of ﬁbres and matrix (Voigt model,

Sect. 2.1.1) and the resultant axial force FLof the composite is shared by both ﬁbre

and matrix so that

FL=FLf +FLm or FL=

σ

LA=

σ

LfAf+

σ

LmAm(3.1.4)

With Hooke’s law it follows that

σ

L=EL

ε

L,

σ

Lf =ELf

ε

Lf,

σ

Lm =ELm

ε

Lm

or

EL

ε

LA=Ef

ε

LfAf+Em

ε

LmAm(3.1.5)

Since the strains of all phases are assumed to be identical (iso-strain condition),

(3.1.5) reduces to

EL=Ef

Af

A+Em

Am

A(3.1.6)

with Af

A=Afl

Al =Vf

V=vf,Am

A=Aml

Al =Vm

V=vm(3.1.7)

and the effective modulus ELcan be written as follows

EL=Efvf+Emvm=Efvf+Em(1−vf) = Ef

φ

+Em(1−

φ

)(3.1.8)

Equation (3.1.8) is referred to the Voigt estimate or is more familiarly known as the

rule of mixture. The predicted values of ELare in good agreement with experimental

results. The stiffness in ﬁbre direction is dominated by the ﬁbre modulus. The ratio

of the load taken by the ﬁbre to the load taken by the composite is a measure of the

load shared by the ﬁbre

✛ ✲

FL

✲✛ ll+

∆

l✲✛

Fig. 3.2 Mechanical model to calculate the effective Young’s modulus EL

3.1 Elementary Mixture Rules for Fibre-Reinforced Laminae 89

FLf

FL

=ELf

EL

vf(3.1.9)

Since the ﬁbre stiffness is several times greater than the matrix stiffness, the second

term in (3.1.8) may be neglected

EL≈Efvf(3.1.10)

3.1.3 Effective Transverse Modulus of Elasticity

The mechanical model in Fig. 3.3 has an arrangement in a series of ﬁbre and matrix

(Reuss model, Sect. 2.1.1). The stress resultant FTrespectively the stress

σ

Tis equal

for all phases (iso-stress condition)

FT=FTf =FTm,

σ

T=

σ

Tf =

σ

Tm (3.1.11)

From Fig. 3.3 it follows that

∆

b=

∆

bf+

∆

bm,

ε

T=

∆

b

b=

∆

bf+

∆

bm

b(3.1.12)

and with

b=vfb+ (1−vf)b=bf+bm(3.1.13)

and

ε

T=

∆

bf

vfb

b+

∆

bm

(1−vf)b

b=vf

ε

Tf + (1−vf)

ε

Tm (3.1.14)

with

ε

Tf =

∆

bf

vfb,

ε

Tm =

∆

bm

(1−vf)b

Using Hooke’s law for the ﬁbre, the matrix and the composite

✻

❄

b

❄

✻

b+

∆

b

✻

FT

❄

FT

Fig. 3.3 Mechanical model to calculate the effective transverse modulus ET

90 3 Effective Material Moduli for Composites

σ

T=ET

ε

T,

σ

Tf =ETf

ε

Tf,

σ

Tm =ETm

ε

Tm (3.1.15)

substituting Eqs. (3.1.15) in (3.1.14) and considering (3.1.11) gives the formula of

ET

1

ET

=vf

Ef

+1−vf

Em

=vf

Ef

+vm

Em

or ET=EfEm

(1−vf)Ef+vfEm

(3.1.16)

Equation (3.1.16) is referred to Reuss estimate or sometimes called the inverse rule

of mixtures. The predicted values of ETare seldom in good agreement with exper-

imental results. With Em≪Effollows from (3.1.16) ET≈Em(1−vf)−1, i.e. ETis

dominated by the matrix modulus Em.

3.1.4 Effective Poisson’s Ratio

Assume a composite is loaded in the on-axis direction (parallel to the ﬁbres) as

shown in Fig. 3.4. The major Poison’s 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

ν

LT =−

ε

T

ε

L

(3.1.17)

With

−

ε

T=

ν

LT

ε

L=−

∆

b

b=−

∆

bf+

∆

bm

b=−[vf

ε

Tf + (1−vf)

ε

T m],

ν

f=−

ε

Tf

ε

Lf

,

ν

m=−

ε

Tm

ε

Lm

it follows that

ε

T=−

ν

LT

ε

L=−vf

ν

f

ε

Lf −(1−vf)

ν

m

ε

Lm (3.1.18)

The longitudinal strains in the composite, the ﬁbres and the matrix are assumed to be

equal (Voigt model of parallel connection) and the equation for the major Poisson’s

ratio reduces to

✛ ✲

✻

❄

✲✛

b

l

✲✛ l+

∆

l

❄

✻

b−

∆

b

FLFL

Fig. 3.4 Mechanical model to calculate the major Poisson’s ration

ν

LT

3.1 Elementary Mixture Rules for Fibre-Reinforced Laminae 91

ν

LT =vf

ν

f+ (1−vf)

ν

m=vf

ν

f+vm

ν

m=

φν

f+ (1−

φ

)

ν

m(3.1.19)

The major Poisson’s ratio

ν

LT obeys the rule of mixture. The minor Poisson’s ratio

ν

TL =−

ε

L/

ε

Tcan be derived with the symmetry condition or reciprocal relationship

ν

TL

ET

=

ν

LT

EL

,

ν

TL =

ν

LT

ET

EL

= (vf

ν

f+vm

ν

m)EfEm

(vfEm+vmEf)(vfEf+vmEm)(3.1.20)

The values of Poisson’s ratios for ﬁbres or matrix material rarely differ signiﬁcantly,

so that neither matrix nor ﬁbre characteristics dominate the major or the minor elas-

tic parameters

ν

LT and

ν

TL.

3.1.5 Effective In-Plane Shear Modulus

Apply a pure shear stress

τ

to a lamina as shown in Fig. 3.5. Assuming that the shear

stresses on the ﬁbre and the matrix are the same, but the shear strains are different

γ

m=

τ

Gm

,

γ

f=

τ

Gf

,

γ

=

τ

GLT

(3.1.21)

The model is a connection in series (Reuss model) and therefore

τ

=

τ

f=

τ

m,

∆

=

∆

f+

∆

m,

∆

=ˆ

btan

γ

=

γ

fˆ

bf+

γ

mˆ

bm(3.1.22)

and with

ˆ

b=ˆ

bf+ˆ

bm= (vf+vm)ˆ

b=vfˆ

b+ (1−vf)ˆ

b(3.1.23)

it follows that

∆

f=

γ

fvfˆ

b,

∆

m=

γ

m(1−vf)ˆ

b(3.1.24)









✲✛

∆

✲ ✛

∆

f

✲✛

∆

m/2

✻

❄

ˆ

b

ﬁbre

matrix

✲ ✛

γ

∆

m/2



✠

✛

τ



✒

✲

τ

Fig. 3.5 Mechanical model to calculate the effective in-plane shear modulus GLT

92 3 Effective Material Moduli for Composites

Using Hooke’s law we have

τ

/GLT = (

τ

/Gm)vm+ (

τ

/Gf)vfwhich yields

GLT =GmGf

(1−vf)Gf+vfGm

=GmGf

(1−

φ

)Gf+

φ

Gm

(3.1.25)

or by analogy to (3.1.16)

1

GLT

=vf

Gf

+1−vf

Gm

=vf

Gf

+vm

Gm

which is again a Reuss estimate. Note that assuming isotropic ﬁbres and matrix

material one gets

Gf=Ef

2(1+

ν

f),Gm=Em

2(1+

ν

m)(3.1.26)

3.1.6 Discussion on the Elementary Mixture Rules

Summarizing the rule of mixtures as a simple model to predict effective engineering

moduli it must be kept in mind that there is no interaction between ﬁbres and matrix.

There are only two different types of material response: the Voigt or iso-strain model

in which applied strain is the same in both material phases (parallel response) and

the Reuss or iso-stress model in which the applied stress is the same in both material

phases (series response).

For an aligned ﬁbre composite the effective material behavior may be assumed

as transversally isotropic and ﬁve independent effective engineering moduli have to

be estimated. With x2−x3as the plane of isotropy (Table 2.5) we have

E1=EL,E2=E3=ET,E4=G23 =GTT =ET/[2(1+

ν

TT)],

E5=G13 =E6=G12 =GLT,

ν

12 =

ν

13 =

ν

LT,

ν

23 =

ν

TT,

ν

LTET=

ν

TLEL

If we make choice of EL,ET,GLT,GTT,

ν

LT as the ﬁve independent moduli the rules

of mixture yield

EL=vfEf+ (1−vf)Em,

ET=EfEm

vfEm+ (1−vf)Ef

,

GLT =GfGm

vfGm+ (1−vf)Gf

,

ν

LT =vf

ν

f+ (1−vf)

ν

m

(3.1.27)

The shear modulus GTT corresponds to an iso-shear strain model and is analogous

to the axial tensile modulus case

GTT =vfGf+ (1−vf)Gm(3.1.28)

3.2 Improved Formulas for Effective Moduli of Composites 93

It may be noted that neither the iso-shear stress nor the iso-shear strain condition

for GLT and GTT estimation are close to the real situation of shearing loaded ﬁbre

reinforced composites. Therefore the equations for GLT and GTT cannot be expected

as very reliable.

If one considers the approximative predictions for the effective moduli ELand

ETas a function of the ﬁbre volume fraction vf, i.e., EL=EL(vf),ET=ET(vf),

and the ratio Ef/Emis ﬁxed, it is clear that reinforcing a matrix by ﬁbres mainly

inﬂuences the stiffness in ﬁbre direction (ELis a linear function of vf) and rather

high ﬁbre volume fractions are necessary to obtain a signiﬁcant stiffness increase in

the transverse direction (ETis a non-linear function of vfand rather constant in the

interval 0 <vf<0,5).

Very often ﬁbres material behavior is transversally isotropic but the matrix ma-

terial is isotropic. For such cases simple alternative relations for the effective engi-

neering moduli of the UD-lamina can be given

EL=vfEf+ (1−vf)Em,

ν

LT =vf

ν

LTf + (1−vf)

ν

m,

ET=ETfEm

vfEm+ (1−vf)ETf

,

ν

T T =

ν

TTf

ν

m

vf

ν

m+ (1−vf)

ν

TTf

,

GLT =GLTfGm

vfGm+ (1−vf)GLTf

,GT T =vfGf+ (1−vf)Gm

(3.1.29)

In Eq. (3.1.29) Em,

ν

m,Gm=Em/2(1+

ν

m)are the isotropic matrix moduli and

Ef,ETf,GLTf,Gf=ETf /2(1+

ν

TTf),

ν

TTf,

ν

LTf are transversally isotropic ﬁbre mod-

uli. Em,

ν

mof the matrix material and Ef,ETf,GLTf,

ν

LTf,

ν

TTf or GTTf of the ﬁbre

material can be chosen as the independent moduli.

3.2 Improved Formulas for Effective Moduli of Composites

Effective elastic moduli related to loading in the ﬁbre direction, such as ELand

ν

LT,

are dominated by the ﬁbres. All estimations in this case and experimental results are

very close to the rule of mixtures estimation. But the values obtained for transverse

Young’s modulus and in-plane shear modulus with the rule of mixtures which can

be reduced to the two model connections of Voigt and Reuss, do not agree well with

experimental results. This establishes a need for better modelling techniques based

on elasticity solutions and variational principle models and includes analytical and

numerical solution methods.

Unfortunately, the theoretical models are only available in the form of compli-

cated mathematical equations and the solution is very limited and needs huge effort.

Semi-empirical relationships have been developed to overcome the difﬁculties with

purely theoretical approaches.

94 3 Effective Material Moduli for Composites

The most useful of those semi-empirical models are those of Halpin and Tsai3

which can be applied over a wide range of elastic properties and ﬁbre volume frac-

tions. The Halpin-Tsai relationships have a consistent form for all properties. They

are developed as simple equations by curve ﬁtting to results that are based on the

theory of elasticity.

Starting from results obtained in theoretical analysis, Halpin and Tsai proposed

equations that are general and simple in formulation. The moduli of a unidirectional

composite are given by the following equations

•ELand

ν

LT by the law of mixtures Eqs. (3.1.8), (3.1.19)

•For the other moduli by

M

Mm

=1+

ξ η

vf

1−

η

vf

(3.2.1)

Mis the modulus under consideration, e.g. ET,GLT, ...,

η

is a coefﬁcient given

by

η

=(Mf/Mm)−1

(Mf/Mm) +

ξ

(3.2.2)

ξ

is called the reinforcement factor and depends on

– the geometry of the ﬁbres

– the packing arrangement of the ﬁbres

– the loading conditions.

The main difﬁculty in using (3.2.1) is the determination of the factor

ξ

by comparing

the semi-empirical values with analytical solutions or with experimental results.

In addition to the rule of mixtures and the semi-empirical solution of Halpin and

Tsai there are some solutions available which are based on elasticity models, e.g. for

the model of a cylindrical elementary cell subjected to tension. The more compli-

cated formulas for ELand

ν

LT as the formulas given above by the rule of mixtures

yields practically identical values to the simpler formulas and are not useful. But

the elasticity solution for the modulus GLT yields much better results and should be

applied

GLT =Gm

Gf(1+vf) + Gm(1−vf)

Gf(1−vf) + Gm(1+vf)=Gm

Gf(1+

φ

) + Gm(1−

φ

)

Gf(1−

φ

) + Gm(1+

φ

)(3.2.3)

Summarizing the results of Sects. 3.1 and 3.2 the following recommendations

may be possible for an estimation of effective elastic moduli of unidirectional lami-

nae

•ELand

ν

LT should be estimated by the rule of mixtures

•

ν

T L follows from the reciprocal condition

•GLT should be estimated from (3.2.3) or the Halpin/Tsai formulas (3.2.1) and

(3.2.2)

3Stephen Wei-Lun Tsai (∗6 July 1929, Beijing) - US-American engineer, strength criteria for

composites, founder of the Journal of Composite Materials

3.3 Problems 95

•ETmay be estimated with help of the Halpin/Tsai formulas. But only when reli-

able experimental values of ETand GLT are available for a composite the

ξ

-factor

can be derived for this case and can be used to predict effective moduli for a range

of ﬁbre volume ratios of the same composite. It is also possible to look for nu-

merical or analytical solutions for

ξ

based on elasticity theory. In general,

ξ

may

vary from zero to inﬁnity, and the Reuss and Voigt models are special cases, e.g.

ET=1+

ξ η

vf

1−

η

vf

Em

for

ξ

=0 and

ξ

=∞, respectively. In the case of circular cross-sections of the

ﬁbres

ξ

=2 or

ξ

=1 can be recommended for Halpin-Tsai equation for ETor

GLT. But it is dangerous to use uncritically these values for any given composite.

3.3 Problems

Exercise 3.1. Determine for a glass/epoxy lamina with a 70 % ﬁbre volume fraction

1. the density and the mass fractions of the ﬁbre and matrix,

2. the Young’s moduli E′

1≡ELand E′

2≡ETand determine the ratio of a tensile

load in L-direction taken by the ﬁbres to that of the composite,

3. the major and the minor Poisson’s ratio

ν

LT,

ν

TL,

4. the in-plane shear modulus

The properties of glass and epoxy are taken approximately from Tables F.1 and

F.2 as

ρ

glass ≡

ρ

f=2,5 gcm−3,

ν

f=0,7,

ν

m=0,3,

ρ

epoxy ≡

ρ

m=1,35 gcm−3,

Ef=70 GPa, Em=3,6 GPa.

Solution 3.1. Taking into account the material parameters and the volume fraction

of the ﬁbres one obtains:

1. Using Eq. (1.1.3) the density of the composite is

ρ

=

ρ

fvf+

ρ

mvm=2,5·0,7 gcm−3+1,35 ·0,3 gcm−3=2,155 gcm−3

Using Eq. (1.1.4) the mass fractions are

mf=

ρ

f

ρ

vf=25

2,1550,7=0,8121,mm=

ρ

m

ρ

vm=1,35

2,1550,3=0,1879

Note that the sum of the mass fractions must be 1

mf+mm=0,8121 +0,1879 =1

2. Using Eqs. (3.1.8), (3.1.16) and (3.1.9) we have

EL=70 ·0,7 GPa +3,6·0,3 GPa =50,08 GPa,

96 3 Effective Material Moduli for Composites

1

ET

=0,7

70GPa +0,3

3,6GPa =0,09333 GPa−1,ET=10,71 GPa,

FLf

FL

=ELf

EL

vf=70

50,08 0,7=0,9784

The ratio of the tensile load FLtaken by the ﬁbre is 0,9784.

3. Using Eqs. (3.1.19) and (3.1.20) follows

ν

LT =0.2·0.7+0.3·0.3=0.230,

ν

TL =0.230 10.71

50.08 =0,049

4. Using (3.1.26) and (3.1.25)

Gf=70 GPa

2(1+0.2)=29.17 GPa,Gm=3.6 GPa

2(1+0,3)=1.38 GPa,

GLT =1.38 GPa ·29.17 GPa

29.17 GPa ·0.3+1,38 GPa ·0.7)=4.14 GPa

Using (3.2.3)

GLT =1.38 GPa 29.17 GPa ·1.7+1.38 GPa ·0.3

29.17 GPa ·0.3+1.38 GPa ·1.7=6.22 GPa

Conclusion 3.1. The difference between the both formulae for GLT is signiﬁcant.

The improved formulae should be used.

Exercise 3.2. Two composites have the same matrix materials but different ﬁbre

material. In the ﬁrst case Ef/Em=60, in the second case Ef/Em=30. The ﬁbre

volume fraction for both cases is vf=0,6. Compare the stiffness values ELand ET

by EL/ET,EL/Em,ET/Em.

Solution 3.2. First case

EL1 =Ef1vf+Em(1−vf),Ef1

Em

=60,Ef1 =60Em,

EL1 =36Em+0,4Em=36,4Em,ET1 =vf

Ef1

+1−vf

Em−1

=2,439Em,

EL1

ET1

=14,925,EL1

Em

=36,4,ET1

Em

=2,439

Second case

EL2 =Ef2vf+Em(1−vf),Ef2 =30Em,

EL2 =18Em+0,4Em=18,4Em,ET2 =vf

Ef2

+1−vf

Em−1

=2,3810Em,

EL2

ET2

=7,728,EL2

Em

=18,4,ET2

Em

=2,381

Conclusion 3.2. The different ﬁbre material has a signiﬁcant inﬂuence on the

Young’s moduli in the ﬁbre direction. The transverse moduli are nearly the same.

3.3 Problems 97

Exercise 3.3. For a composite material the properties of the constituents are Ef=90

GPa,

ν

f=0,2,Gf=35 GPa, Em=3,5 GPa,

ν

m=0,3, Gm=1,3 GPa. The volume

fraction vf=

φ

=60%. Calculate EL,ET,GLT,

ν

LT,

ν

T L with the help of the rule of

mixtures and also GLT with the improved formula.

Solution 3.3.

EL=Efvf+Em(1−vf) = 55,4 GPa,ET=vf

Ef

+vm

Em−1

=8,27 GPa,

ν

LT =

ν

fvf+

ν

m(1−vf) = 0,24,

ν

TL =

ν

LTET

EL

=0,0358,

GLT =GmGf

(1−vf)Gf+vfGm

=3,0785 GPa

The improved formula (3.2.3) yields

GLT =Gm

(1+vf)Gf+ (1−vf)Gm

(1−vf)Gf+ (1+vf)Gm

=4,569 GPa

Conclusion 3.3. The difference in the GLT values calculated using the rule of mix-

tures and the improved formula is again signiﬁcant.

Exercise 3.4. A unidirectional glass/epoxy lamina is composed of 70% by volume

of glass ﬁbres in the epoxy resin matrix. The material properties are Ef=85 GPa,

Em=3,4 GPa.

1. Calculate ELusing the rule of mixtures.

2. What fraction of a constant tensile force FLis taken by the ﬁbres and by the

matrix?

Solution 3.4. Withe the assumed material parameters and ﬁbre volume fraction one

obtains

1. EL=Efvf+Em(1−vf) = 60,52 GPa,

2. FL=

σ

LA=EL

ε

A,Ff=

σ

fA=Ef

ε

Af,Fm=

σ

mA=Em

ε

Am

With FL=Ff+Fmit follows that

EL

ε

A=Ef

ε

Af+Em

ε

Am⇒EL=0,7Ef+0,3Em,

60,52 =0,7·85 +3·3,4=59,5+1,02

and therefore FL=60,52 N, Ff=59,5 N, Fm=1,02 N.

Conclusion 3.4. The fractions of a constant tensile load in the ﬁbres and the ma-

trix are: Fibres: 98,31 %, Matrix: 1,69 %

Exercise 3.5. The ﬁbre and the matrix characteristics of a lamina are Ef=220 GPa,

Em=3,3 GPa, Gf=25 GPa, Gm=1,2 GPa,

ν

f=0,15,

ν

m=0,37. The ﬁbre is

transversally isootropic and transverse Young’s modulus is ETf =22 GPa. The ﬁbre

volume fraction is vf=0,56. The experimentally measured effective moduli are

EL=125 GPa, ET=9,1 GPa, GLT =5 GPa,

ν

LT =0,34.

98 3 Effective Material Moduli for Composites

1. compare the experimental values with predicted values based on rule of mixtures.

2. Using the Halpin-Tsai approximate model for calculating ETand GLT, what value

of

ξ

must be used in order to obtain moduli that agree with experimental values?

Solution 3.5. The comparison can be be performed by results from the mixture rules

and improved equations.

1. The rule of mixtures (3.1.27) yields

EL=0,56 ·220 GPa + (1−0,56)·3,3 GPa =124,65 GPa,

ET=220 GPa ·3,3 GPa

0,56 ·3,3 GPa + (1−0,56)·220 GPa =7,36 GPa,

GLT =25 GPa ·1,2 GPa

0,56 ·1,2 GPa + (1−0,56)·25 GPa =2,57 GPa,

ν

LT =0,56 ·0,15 + (1−0,56)·0,37 =0,25

It is seen that the ﬁbre dominated modulus ELis well predicted by the rule of

mixtures, while ET,GLT and

ν

LT are not exactly predicted.

2. The Halpin-Tsai approximation yields with ETf =22 GPa,

ξ

=2 for ET

ET=1+

ξ η

vf

1−

η

vf

Em,

η

=ETf/Em−1

ETf/Em+

ξ

=0,6538,ET=9,02 GPa

With a value

ξ

=2.5 follows for GLT

GLT =1+

ξ η

vf

1−

η

vf

Gm,

η

=Gf/Gm−1

Gf/Gm+

ξ

=0,8339,GLT =4,88 GPa

The predicted value for ETis nearly accurate for

ξ

=2 which is the recom-

mended value in literature but the recommended value of

ξ

=1 for GLT would

underestimate the predicted value signiﬁcantly

GLT =1+

η

vf

1−

η

vf

Gm,

η

=Gf/Gm−1

Gf/Gm+1=0,9084,GLT =3,68 GPa

It can be seen that it is dangerous to accept these values uncritically without

experimental measurements.

Exercise 3.6. Let us assume the following material parameters for the shear mod-

ulus of the ﬁbre and the matrix: Gf=2,5·104Nmm−2,Gm=1,2·103Nmm−2.

The experimental value of the effective shear modulus is Gexp

LT =5·103Nmm−2, the

Poisson’s ratios are

ν

f=0,15,

ν

m=0,37 and the ﬁbre volume fraction is vf=0,56.

For the improvement in the sense of Halpin-Tsai we assume

ξ

=1...2 for a circular

cross-section of the ﬁbres. Compute:

1. the effective shear modulus,

2. the improved effective shear modulus,

3.3 Problems 99

3. the improved effective shear modulus in the sense of Halpin-Tsai

Solution 3.6. The three problems have the following solutions:

1. the effective shear modulus after Eq. (3.1.25)

GLT =1,2·103Nmm−2·2,5·104Nmm−2

(1−0,56)·2,5·104Nmm−2+0,56 ·1,2·103Nmm−2

=2,57 ·103Nmm−2

2. the improved effective shear modulus after Eq. (3.2.3)

GLT =1,2·103N

mm2

2,5·104Nmm−2(1+0,56) + 1,2·103Nmm−2(1−0,56)

2,5·104Nmm−2(1−0,56) + 1,2·103Nmm−2(1+0,56)

=3,68 ·103Nmm−2

3. For the improved effective shear modulus in the sense of Halpin-Tsai at ﬁrst

should be estimated

η

with Eq. (3.2.2)

η

=(2,5·104Nmm−2/1,2·103Nmm−2)−1

(2,5·104Nmm−2/1,2·103Nmm−2) +

ξ

In the case of

ξ

=1 we get

η

=0,9084, if

ξ

=2 we get

η

=0,8686. The shear

modulus itself follows from Eq. (3.2.1)

GLT =1+

ξ η

·0,15

1−

η

·0,15 1,2·103Nmm−2

Finally we get

GLT(

ξ

=1) = 3,68 ·103Nmm−2,GLT(

ξ

=2) = 4,61 ·103Nmm−2

The last result is the closest one to the experimental value. Let us compute the

deviation

δ

=Gexp

LT −Gcomp

LT

Gexp

LT ·100%

The results are

δ

(case a) = 5·103Nmm−2−2,57 ·103Nmm−2

5·103Nmm−2·100% =46,6%

δ

(case b) = 5·103Nmm−2−2,66 ·103Nmm−2

5·103Nmm−2·100% =26,8%

δ

(case c,

ξ

=1) = 5·103Nmm−2−3,68 ·103Nmm−2

5·103Nmm−2·100% =26,3%

δ

(case c,

ξ

=2) = 5·103Nmm−2−4,61 ·103Nmm−2

5·103Nmm−2·100% =7,8%

100 3 Effective Material Moduli for Composites

abc

ra

r

h

r

Fig. 3.6 Fibre arrangements. aSquare array, bhexagonal array, clayer-wise array

The result can be improved if the

ξ

-value will be increased.

Exercise 3.7. Calculate the ultimate ﬁbre volume fractions vu

ffor the following ﬁbre

arrangements:

1. square array,

2. hexagonal array,

3. layer-wise array.

Assume circular ﬁbre cross-sections.

Solution 3.7. For the three ﬁbre arrangements one gets:

1. Square array (Fig. 3.6 a)

vu

f=Af

Ac

=4(r2

π

)/4

4r2=

π

4=0.785

2. Hexagonal array (Fig. 3.6 b) with a→2r,h→√3rfollow

Ac=hr,Af=6(1/3)

π

r2+

π

r2

and

vu

f=Af

Ac

=6(1/3)

π

r2+

π

r2)/4

6√3r2=

π

2√3=0.9069

3. Layer-wise array (Fig. 3.6 c)

vu

f=Af

Ac

=r2

π

(2r)2=

π

4=0.785

Part II

Modelling of a Single Laminae, Laminates

and Sandwiches

The second part (Chaps. 4–6) can be related to the modelling from single lami-

nae to laminates including sandwiches, the improved theories and simplest failure

concepts.

The single layer (lamina) is modelled with the help of the following assumptions

•linear-elastic isotropic behaviour of the matrix and the ﬁbre materials,

•the ﬁbres are unidirectional oriented and uniformly distributed

These assumptions result in good stiffness approximations in the longitudinal and

transverse directions. The stiffness under shear is not well-approximated.

After the transfer from the local to the global coordinates the stiffness parame-

ters for the laminate can be estimated. For the classical cases the effective stiffness

parameters of the laminate is a simple sum up over the laminate thickness of the

weighted laminae reduced stiffness parameters transferred into the global coordi-

nate system. Some improved theories are brieﬂy introduced.

The failure concepts are at the moment a research topic characterised by a large

amount of suggestions for new criteria. With respect to this only some classical

concepts are discussed here.

Chapter 4

Elastic Behavior of Laminate and Sandwich

Composites

A lamina has been deﬁned as a thin single layer of composite material. A lamina

or ply is a typical sheet of composite materials, which is generally of a thickness

of the order 1 mm. A laminate is constructed by stacking a number of laminae in

the direction of the lamina thickness. The layers are usually bonded together with

the same matrix material as in the single lamina. A laminate bonded of n(n≥2)

laminae of nearly the same thickness. A sandwich can be deﬁned as a special case

of a laminate with n=3. Generally, the sandwich is made of a material of low

density for the inner layer, the core or the supporting pith respectively, and of high

strength material for the outer layers, the cover or face sheets. The thickness of the

core is generally much greater than the thickness of the sheets and core and sheets

are bonded to each other at the surfaces.

The design and analysis of structures composed of composite materials demands

knowledge of the stresses and strains in laminates or sandwiches. However, the lam-

inate elements are single laminae and so understanding the mechanical behavior of

a lamina precedes that of a laminate. Section 4.1 introduces elastic behavior of lam-

inae. For in-plane and out-of-plane loading, the stress resultants are formulated and

basic formulae for stress analysis are derived. These considerations are expanded to

laminates and sandwiches in Sects. 4.2 and 4.3. The governing equations of the clas-

sical laminate theory, the shear deformation theory and of a layer-wise theory are

discussed in Chap. 5. A successful design of composite structures requires knowl-

edge of the strength and the reliability of composite materials. Strength failure the-

ories have to be developed in order to compare the actual stress states in a material

to a failure criteria. Chapter 6 gives an overview on fracture modes of laminae. For

laminates, the strength is related to the strength of each individual lamina. Various

failure theories are discussed for laminates or sandwiches based on the normal and

shear strengths of unidirectional laminae.

103

H. Altenbach et al., Mechanics of Composite Structural

Elements, https://doi.org/10.1007/978-981-10-8935-0_4

104 4 Elastic Behavior of Laminate and Sandwich Composites

4.1 Elastic Behavior of Laminae

The macro-mechanical modelling and analysis of a lamina is based on average ma-

terial properties and by considering the lamina to be homogeneous. The methods

to ﬁnd these average properties based on the individual mechanical values of the

constituents are discussed in Sect. 3.1. Otherwise the mechanical characterization

of laminae can be determined experimentally but it demands special experimental

equipment and is costly and time-consuming. Generally the modelling goal is to

ﬁnd the minimum of parameters required for the mechanical characterization of a

lamina.

For the considerations on the elastic behavior of laminae in the following Sects.

4.1.1 - 4.1.3 one has to keep in mind that two assumptions are most important to

model the mechanics of ﬁbre reinforced laminae:

•The properties of the ﬁbres and the matrix can be smeared into an equiva-

lent homogeneous material with orthotropic behavior. This assumption allows

to develop the stress-strain relations and to formulate the response of a ﬁbre-

reinforced lamina sufﬁcient simply to deal with the structural level response in a

tractable manner.

•Three of the six components of stress state are generally much smaller than the

other three, i.e. the plane stress assumption, which is based on the manner in

which ﬁbre-reinforced materials are used in such structural elements as beams,

plates or shells, will be sufﬁcient accurately. With the assumption that the (x1−

x2)-plane of the principal co-ordinate system is in-plane stress state, the in-plane

stress components

σ

1,

σ

2,

σ

6are considered to be much larger in value than the

out-of-plane stress components

σ

3,

σ

4,

σ

5and the last ones are set approximately

to zero.

Using the plane stress assumption it has to be in mind that some serious inaccuracies

in the mechanical response of laminates can be occurred, Sect. 4.2.

Therefore, together with the plane stress assumption two major misconceptions

should be avoided:

•The stress components

σ

3,

σ

4,

σ

5equated to zero have to be estimated to their

magnitude and effect. Fibre reinforced material is often very poor in resisting

stresses transversely to the plane (x1−x2) and therefore out-of-plane stresses

may be small but large enough to cause failure of the composite material.

•With assuming

σ

3is zero does not follow that the associated strain

ε

3is also zero

and ignorable, for the stresses in the (x1−x2)-plane can cause a signiﬁcant strain

response in the x3-direction.

4.1.1 On-Axis Stiffness and Compliances of UD-Laminae

A thin lamina is assumed to be in a plane stress state (Sect. 2.1.5). Three cases of

material behavior of laminae are of special interest for engineering applications:

4.1 Elastic Behavior of Laminae 105

1. Short ﬁbres or particle reinforced components with random orientation in the

matrix

The elastic behavior has no preferred direction and is macroscopically quasi-

homogenous and isotropic. The effective elastic moduli are Eand

ν

and the

relations of the in-plane stress components with the in-plane strain components

are described (Tables 2.6 and 2.7) by





σ

1

σ

2

σ

6

=



Q11 Q12 0

Q12 Q11 0

0 0 Q66 





ε

1

ε

2

ε

6

,





ε

1

ε

2

ε

6

=



S11 S12 0

S12 S11 0

0 0 S66 





σ

1

σ

2

σ

6



(4.1.1)

with

Q11 =E

1−

ν

2,S11 =1

E,

Q12 =E

ν

1−

ν

2,S12 =−

ν

E,

Q66 =G=E

2(1+

ν

),S66 =1

G=2(1+

ν

)

E

2. Long ﬁbres with one unidirectional ﬁbre orientation, so-called unidirectional

laminae or UD-laminae, with loading along the material axis (on-axis case)

This type of material forms the basic conﬁguration of ﬁbre composites and is

the main topic of this textbook. The elastic behavior of UD-laminae depends on

the loading reference coordinate systems. In the on-axis case the reference axes

(1′,2′) are identical to the material or principal axes of the lamina parallel and

transverse to the ﬁbre direction (Fig. 4.1). The 1′-axis is also denoted as L-axis

and the 2′-axis as T-axis (on-axis case). The elastic behavior is macroscopically

quasi-homogeneous and orthotropic with four independent material moduli (Ta-

ble 2.6)

E′

1≡EL,E′

2≡ET,E′

6≡G′

12 ≡GLT,

ν

′

12 ≡

ν

LT (4.1.2)

and the in-plane stress-strain relations are

Fig. 4.1 Unidirectional lam-

ina with principal material

axis Land T(on-axis)

✻

✲

1′≡L

2′≡T

r

h

✲ ✛

106 4 Elastic Behavior of Laminate and Sandwich Composites





σ

′

1

σ

′

2

σ

′

6



=



Q′

11 Q′

12 0

Q′

12 Q′

22 0

0 0 Q′

66







ε

′

1

ε

′

2

ε

′

6



,





ε

′

1

ε

′

2

ε

′

6



=



S′

11 S′

12 0

S′

12 S′

22 0

0 0 S′

66







σ

′

1

σ

′

2

σ

′

6





(4.1.3)

with

Q′

11 =E′

1/(1−

ν

′

12

ν

′

21),S′

11 =1/E′

1,

Q′

22 =E′

2/(1−

ν

′

12

ν

′

21),S′

22 =1/E′

2,

Q′

66 =G′

12 =E′

6,S′

66 =1/G′

12 =1/E′

6,

Q′

12 =E′

2

ν

′

12/(1−

ν

′

12

ν

′

21),S′

12 =−

ν

′

12/E′

1=−

ν

′

21/E′

2

3. UD-laminae with loading along arbitrary axis (x1,x2) different from the material

axis (off-axis case). The elastic behavior is macroscopically quasi-homogeneous

and anisotropic. The in-plane stress-strain relations are formulated by fully pop-

ulated matrices with all Qi j and Si j different from zero but the number of inde-

pendent material parameters is still four as in case 2. The transformation rules

are given in detail in Sect. 4.1.2.

A UD-lamina has different stiffness in the direction of the material axes. With

Ef≫Emthe stiffness in the L-direction is ﬁbre dominated and for the effective

moduli (Sect. 3.1) EL≫ET. Figure 4.2 illustrates qualitatively the on-axis elastic

behavior of the UD-lamina. In thickness direction x3≡T′orthogonal to the (L-T)-

plane a UD-lamina is macro-mechanically quasi-isotropic. The elastic behavior in

the thickness direction is determined by the matrix material and a three-dimensional

model of a single UD-layer yields a transversely-isotropic response with ﬁve inde-

pendent material engineering parameters:

E′

1≡EL,E′

2≡ET=E′

3≡ET′,E′

5=E′

6≡GLT,

ν

′

12 ≡

ν

LT =

ν

′

13 ≡

ν

LT′,

E′

4≡GTT′=E′

2/[2(1+

ν

′

23)] ≡ET/[2(1+

ν

TT′)]

(4.1.4)

Fig. 4.2 On-axis stress-strain

equations for UD-lamina

(qualitative)

✻

✲

σ

1,

σ

2,

σ

6

ε

1,

ε

2,

ε

6

✄✄✄✄✄✄✄✄✄✄✄✄

✦✦✦✦✦✦✦✦✦✦✦✦✦✦✦

✦

✥✥✥✥✥✥✥✥✥✥✥✥

✥

σ

1(

ε

1)

σ

2(

ε

2)

σ

6(

ε

6)

✛ ✲

σ

1

σ

1

✻

❄

σ

2

σ

2

✲

✻

✛

❄

σ

6

σ

6

4.1 Elastic Behavior of Laminae 107

The material behavior in the 2′≡T and 3 =T′directions is equivalent. Therefore,

the notation of the engineering parameters is given by E′

2=E′

3≡ET,E′

5=E′

6≡GLT,

ν

′

12 =

ν

′

13 ≡

ν

LT,E′

4=E′

2/[2(1+

ν

′

23)] ≡GTT =ET/[2(1+

ν

TT)]. Summarizing, the

stress-strain relations for on-axis loading of UD-laminae in a contracted vector-

matrix notation leads the equations

σ

′=Q

Q

Q′

ε

′or

σ

′

i=Q′

i j

ε

′

j,Q′

i j =Q′ji

ε

′=S

S

S′

σ

′or

ε

′

i=S′

i j

σ

′

j,S′

i j =S′ji

i,j=1,2,6 (4.1.5)

The values Q′

i j of the reduced stiffness matrix Q

Q

Q′and the S′

i j of the compliance ma-

trix S

S

S′depend on the effective moduli of the UD-lamina. The term reduced stiffness

is used in relations given by Eqs. (2.1.76) and (4.1.3). These relations simplify the

problem from a three-dimensional to a two-dimensional or plane stress state. Also

the numerical values of the stiffness Q′

i j are actually less than the numerical values

of their respective counterparts C′

i j, see Eq. (2.1.79), of the three-dimensional prob-

lem and therefore the stiffness are reduced in that sense also. For on-axis loading

the elastic behavior is orthotropic and with Q′

16 =Q′

26 =0 and S′

16 =S′

26 =0, there

is as in isotropic materials no coupling of normal stresses and shear strains and also

shear stresses applied in the (L-T)-plane do not result in any normal strains in the

L and T direction. The UD-lamina is therefore also called a specially orthotropic

lamina.

Composite materials are generally processed at high temperature and then cooled

down to room temperature. For polymeric ﬁbre reinforced composites the tempera-

ture difference is in the range of 2000−3000C and due to the different thermal ex-

pansion of the ﬁbres and the matrix, residual stresses result in a UD-lamina and ex-

pansion strains are induced. In addition, polymeric matrix composites can generally

absorb moisture and the moisture change leads to swelling strains and stresses sim-

ilar to these due to thermal expansion. Therefore we speak of hygrothermal stresses

and strains in a lamina. The hygrothermal strains in the longitudinal direction and

transverse the ﬁbre direction of a lamina are not equal since the effective elastic

moduli ELand ETand also the thermal and moisture expansion coefﬁcients

α

th

L,

α

th

T

and

α

mo

L,

α

mo

Trespectively, are different.

The stress-strain relations of a UD-lamina, including temperature and moisture

differences are given by





ε

′

1

ε

′

2

ε

′

6



=



S′

11 S′

12 0

S′

12 S′

22 0

0 0 S′

66







σ

′

1

σ

′

2

σ

′

6



+



ε

′th

1

ε

′th

2

0

+



ε

′mo

1

ε

′mo

2

0

(4.1.6)

with





ε

′th

1

ε

′th

2

0

=



α

′th

1

α

′th

2

0

T,



ε

′mo

1

ε

′mo

2

0

=



α

′mo

1

α

′mo

2

0

M∗(4.1.7)

Tis the temperature change and M∗is weight of moisture absorption per unit weight

of the lamina.

α

mo

L,

α

mo

Tare also called longitudinal and transverse swelling coefﬁ-

108 4 Elastic Behavior of Laminate and Sandwich Composites

cients. Equation (4.1.6) can be inverted to give





σ

′

1

σ

′

2

σ

′

6



=



Q′

11 Q′

12 0

Q′

12 Q′

22 0

0 0 Q′

66







ε

′

1−

ε

′th

1−

ε

′mo

1

ε

′

2−

ε

′th

2−

ε

′mo

2

ε

′

6



(4.1.8)

Note that the temperature and moisture changes do not have any shear strain terms

since no shearing is induced in the material axes. One can see that the hygrothermal

behavior of an unidirectional lamina is characterized by two principal coefﬁcients

of thermal expansion,

α

′th

1,

α

′th

2, and two of moisture expansion,

α

′mo

1,

α

′mo

2. These

coefﬁcients are related to the material properties of ﬁbres and matrix and of the ﬁbre

volume fraction.

Approximate micro-mechanical modelling of the effective hygrothermal coef-

ﬁcients were given by Schapery1and analogous to the micro-mechanical mod-

elling of elastic parameters, Chap. 3, for a ﬁbre reinforced lamina and isotropic

constituents the effective thermal expansion coefﬁcients are

α

th

L=

α

th

fvfEf+

α

th

m(1−vf)Em

vfEf+ (1−vf)Em

,

α

th

T=

α

th

fvf(1+

ν

f) +

α

th

m(1−vf)(1+

ν

m)−[vf

ν

f+ (1−vf)

ν

m]

α

th

L

(4.1.9)

If the ﬁbres are not isotropic but have different material response in axial and trans-

verse directions, e.g. in the case of carbon or aramid ﬁbres, the relations for

α

th

Land

α

th

Thave to be changed to

α

th

L=

α

th

LfvfELf +

α

th

m(1−vf)Em

vfELf + (1−vf)Em

,

α

th

T= (

α

th

Tf +

ν

Tf

α

th

Lf)vf+ (1+

ν

m)

α

th

m(1−vf)

−[vf

ν

Tf + (1−vf)

ν

m]

α

th

L

(4.1.10)

In most cases the matrix material can be considered isotropic and therefore the ori-

entation designation L,T of the matrix material parameters can be dropped.

Discussion and conclusions concerning effective moduli presented in Chap. 3 are

valid for effective thermal expansion coefﬁcients too. The simple micro-mechanical

approximations of effective moduli yield proper results for

α

th

Lbut fails to predict

α

th

Twith the required accuracy. For practical applications

α

th

Land

α

th

Tshould be

normally determined by experimental methods.

Micro-mechanical relations for effective coefﬁcients of moisture expansion can

be modelled analogously. However, some simpliﬁcation can be taken into consider-

ation. Usually the ﬁbres, e.g. glass, carbon, boron, etc., do not absorb moisture that

means

α

mo

f=0. For isotropic constituents the formulae for

α

mo

Land

α

mo

Tare

1Richard Allan Schapery (∗3 March 1935 Duluth, Minnesota, United States) - engineering educa-

tor, contributions in the ﬁeld of mechanics of composite materials

4.1 Elastic Behavior of Laminae 109

α

mo

L=

α

mo

mEm(1−vf)

vfEf+ (1−vf)Em

=

α

mo

mEm(1−vf)

E1

,

α

mo

T=

α

mo

m(1−vf)[(1+

ν

m)Efvf+ (1−vf)Em−

ν

fvfEm]

vfEf+ (1−vf)Em

=

α

mo

m(1−vf)

E1{(1+

ν

m)Efvf+ [(1−vf)−

ν

fvf]Em}

(4.1.11)

and for a composite with isotropic matrix but orthotropic ﬁbres the effective moduli

are given by

α

mo

L=

α

mo

m

Em(1−vf)

EL

,

α

mo

T=

α

mo

m

1−vf

EL{(1+

ν

m)ELfvf+ [(1−vf)−

ν

LTfvfEm}

(4.1.12)

The formulae (4.1.9) - (4.1.12) completes the discussion about micro-mechanics

in Chap. 3. Note that for a great ﬁbre volume fraction negative

α

th

Lvalues can be

predicted reﬂecting the dominance of negative values of ﬁbre expansion

α

f

L, e.g. for

graphite-reinforced material.

Summarizing one has to keep in mind that with the plane stress assumption re-

ferred to the principal material axis L,T the mechanical shear strains and the total

shear strain are identical, i.e.

ε

′th

6=

ε

′mo

6≡0, Eqs. (4.1.6) - (4.1.8). Also the through-

the-thickness total strains

ε

′

4,

ε

′

5are zero and there are no mechanical, thermal or

moisture strains. The conclusion regarding the normal strains

ε

3is not the same.

Using the condition

σ

3=0 follows

ε

3=S′

13

σ

′

1+S′

23

σ

′

2+

α

′th

3T+

α

′mo

3M∗

This equation is the basis for determining the out-of-plane or through-the-thickness

thermal and moisture effects of a laminate.

4.1.2 Off-Axis Stiffness and Compliances of UD-Laminae

A unidirectional lamina has very low stiffness and strength properties in the trans-

verse direction compared with these properties in longitudinal direction. Laminates

are constituted generally of different layers at different orientations. To study the

elastic behavior of laminates, it is necessary to take a global coordinate system for

the whole laminate and to refer the elastic behavior of each layer to this reference

system. This is necessary to develop the stress-strain relationship for an angle lam-

ina, i.e. an off-axis loaded UD-lamina.

The global and the local material reference systems are given in Fig. 4.3. We

consider the ply material axes to be rotated away from the global axes by an angle

θ

, positive in the counterclockwise direction. This means that the (x1,x2)-axes are

at an angle

θ

clockwise from the material axes. Thus transformation relations are

110 4 Elastic Behavior of Laminate and Sandwich Composites

Fig. 4.3 UD-lamina with the

local material principal axis

(1,2)≡(L,T)and the global

reference system (x1,x2)

✻

✲✟✟✟✟✟✟✟✟✟✟✟

✟✯

❆

❆❑

y≡x2

1≡x′

1

x≡x1

2≡x′

2

✟✟✟✟✟✟✟

✟

✟✟✟✟✟✟✟

✟

✟✟✟✟✟✟✟

✟

✟✟✟✟✟

✟

✟✟✟✟

✟

✟✟

✟

✟✟

✟

✟✟

✟

θ

✢

needed for the stresses, the strains, the stress-strain equations, the stiffness and the

compliance matrices.

The transformation equations for a rotating the reference system (x′

1,x′

2) or

(x1,x2) counterclockwise or clockwise by an angle

θ

follow from Sect. 2.1.2 and

are given in Table 4.1. Note the relations for the transformation matrices derived in

Sect. 2.1.2

Table 4.1 Transformation rules of the coordinates, displacements, strains and stresses of a lamina

a) Rotation of the reference systems

x′

1

x′

2=c s

−s c x1

x2,x1

x2=c−s

s c x′

1

x′

2

x

x′=R

R

Rx

x

x,x

x

x=R

R

RTx

x

x′

b) Transformation of displacements

u′

1

u′

2=c s

−s c u1

u2,u1

u2=c−s

s c u′

1

u′

2

u

u′=R

R

Ru

u

u,u

u

u=R

R

RTu

u

u′

c) Transformation of strains





ε

′

1

ε

′

2

ε

′

6



=





c2s2sc

s2c2−sc

−2sc 2sc c2−s2







ε

1

ε

2

ε

6



,



ε

1

ε

2

ε

6



=





c2s2−sc

s2c2sc

2sc −2sc c2−s2







ε

′

1

ε

′

2

ε

′

6





ε

′=T

T

εε

ε

=T

T

σ

′T

ε

,

ε

=T

T

ε

′

ε

′= (T

T

σ

)T

ε

′

d) Transformation of stresses





σ

′

1

σ

′

2

σ

′

6



=





c2s22sc

s2c2−2sc

−sc sc c2−s2







σ

1

σ

2

σ

6



,



σ

1

σ

2

σ

6



=





c2s2−2sc

s2c22sc

sc −sc c2−s2







σ

′

1

σ

′

2

σ

′

6





σ

′=T

T

σσ

σ

=T

T

ε

′T

σ

,

σ

=T

T

σ

′

σ

′= (T

T

ε

)T

σ

′

with s≡sin

θ

,c≡cos

θ

4.1 Elastic Behavior of Laminae 111

T

ε

=T

T

ε

′−1=T

T

σ

′T,T

T

ε

′= (T

T

ε

)−1= (T

T

σ

)T,

T

σ

=T

T

σ

′−1=T

T

ε

′T,T

T

σ

′= (T

T

σ

)−1= (T

T

ε

)T(4.1.13)

Table 4.2 summarizes the transformation rules for the stress-strain relations and for

the values of the stiffness and the compliance matrices. The transformation matrices

of Table 4.2 follow from Sect. 2.1.2. Using the relations (4.1.13) the transformation

rules can be formulated in matrix notation

Table 4.2 Transformation of the reduced stiffness matrix Q′

i j and compliance matrix S′

i j in the ref-

erence system (x′

1,x′

2) to the reduced stiffness matrix Qi j and compliance matrix Si j in the (x1,x2)-

system

a) Constitutive equations in the (x1,x2)-reference system





σ

1

σ

2

σ

6

=



Q11 Q12 Q16

Q12 Q22 Q26

Q16 Q26 Q66 





ε

1

ε

2

ε

6

,



ε

1

ε

2

ε

6

=



S11 S12 S16

S12 S22 S26

S16 S26 S66 





σ

1

σ

2

σ

6



σ

=Q

Q

ε

,

ε

=S

S

σ

,

Q

Q= (T

T

ε

)TQ

Q

Q′T

T

ε

,S

S

S= (T

T

σ

)TS

S

S′T

T

σ

,

Qi j =Qji ,Q′

i j =Q′ji,Si j =Sj i,S′

i j =S′ji

b) Transformation of the reduced stiffnesses







Q11

Q12

Q16

Q22

Q26

Q66







=







c42c2s2s44c2s2

c2s2c4+s4c2s2−4c2s2

c3s−cs(c2−s2)−cs3−2cs(c2−s2)

s42c2s2c44c2s2

cs3cs(c2−s2)−c3s2cs(c2−s2)

c2s2−2c2s2c2s2(c2−s2)2













Q′

11

Q′

12

Q′

22

Q′

66







c) Transformation of the compliances







S11

S12

S16

S22

S26

S66







=







c42c2s2s4c2s2

c2s2c4+s4c2s2−c2s2

2c3s−2cs(c2−s2)−2cs3−cs(c2−s2)

s42c2s2c4c2s2

2cs32cs(c2−s2)−2c3s cs(c2−s2)

4c2s2−8c2s24c2s2(c2−s2)2













S′

11

S′

12

S′

22

S′

66







with s≡sin

θ

,c≡cos

θ

112 4 Elastic Behavior of Laminate and Sandwich Composites

Q

Q′=T

T

ε

′T

Q

QT

T

ε

′=T

T

σ

Q

Q(T

T

σ

)T,

S

S′=T

T

σ

′T

S

ST

T

σ

′=T

T

ε

S

S(T

T

ε

)T,

Q

Q= (T

T

ε

)TQ

Q

Q′T

T

ε

,

S

S= (T

T

σ

)TS

S

S′T

T

σ

(4.1.14)

Starting with the stiffness equation

σ

=Q

Q

ε

and introducing

σ

′=Q

Q

Q′

ε

′in the

transformation

σ

=T

T

σ

′

σ

′=T

T

ε

′T

σ

′it follows that

σ

=T

T

ε

′T

Q

Q′

ε

′and with

ε

′=T

T

εε

ε

this gives

σ

= (T

T

ε

)TQ

Q

Q′T

T

εε

ε

. Comparison of equations

σ

=Q

Q

ε

and

σ

= (T

T

ε

)TQ

Q

Q′T

T

εε

ε

yields





Q11 Q12 Q16

Q12 Q22 Q26

Q16 Q26 Q66 

= (T

T

ε

)T



Q′

11 Q′

12 0

Q′

12 Q′

22 0

0 0 Q66 

T

T

ε

(4.1.15)

In an analogous way





S11 S12 S16

S12 S22 S26

S16 S26 S66 

= (T

T

σ

)T



S′

11 S′

12 0

S′

12 S′

22 0

0 0 Q66 

T

T

σ

(4.1.16)

can be derived. Note that in (4.1.15) and (4.1.16) the matrices [Qi j ]and [Si j ]have

six different elements but the matrices [Q′

i j]and [S′

i j]have only four independent

elements. The elements in Qi j or Si j are functions of the four independent material

characteristics Q′

i j or S′

i j and the angle

θ

. The experimental testing is therefore more

simple than for a real anisotropic material with 6 independent material values, if the

material axes of the lamina are known.

From the transformation c) in Table 4.2 follows the transformation of the engi-

neering parameters EL,ET,GLT,

ν

LT of the UD-lamina in the on-axis-system to the

engineering parameters in the global system (x1,x2). From equation a) in Table 4.2

for an angle lamina it can be seen that there is a coupling of all normal and shear

terms of stresses and strains. In Fig. 4.4 these coupling effects in an off-axes loaded

UD-lamina are described.

The coupling coefﬁcients

ν

12 =−

ε

2

ε

1

=−S21

S11

,

ν

21 =−

ε

1

ε

2

=−S12

S22

(4.1.17)

are the known Poisson’s ratios and the ratios

ν

16 =

ε

6

ε

1

=S61

S11

,

ν

26 =

ε

6

ε

2

=S62

S22

(4.1.18)

are so called shear coupling values. They are non-dimensional parameters like Pois-

son’s ratio and relate normal stresses to shear strains or shear stresses to normal

strains.

4.1 Elastic Behavior of Laminae 113

✛ ✲

✻

❄

❄✻

✲

✛

σ

1

σ

1

σ

2

σ

2

σ

6

σ

6

ε

1

σ

1

=S11 =1

E1

ε

2

σ

1

=S21 =−

ν

12

E1

ε

6

σ

1

=S61 =

ν

16

E1

ε

2

σ

2

=S22 =1

E2

ε

1

σ

2

=S12 =−

ν

21

E2

ε

6

σ

2

=S62 =

ν

26

E2

ε

6

σ

6

=S66 =1

E6

ε

1

σ

6

=S16 =

ν

61

E6

ε

2

σ

6

=S26 =

ν

62

E6

























































Fig. 4.4 Off-axis loaded UD-lamina with one stress component in each case

Hence the strain-stress equation of an angle lamina can be written in terms of

engineering parameters of the off-axis case as





ε

1

ε

2

ε

6

=



1/E1−

ν

21/E2

ν

61/E6

−

ν

12/E11/E2

ν

62/E6

ν

16/E1

ν

26/E21/E6





σ

1

σ

2

σ

6

(4.1.19)

With the compliance engineering parameters

S11 =1

E1

,S12 =−

ν

21

E2

,S21 =−

ν

12

E1

,

S66 =1

E6

,S16 =

ν

61

E6

,S61 =

ν

16

E1

,

S26 =

ν

62

E6

,S62 =

ν

26

E2

,S22 =1

E2

(4.1.20)

it follows from the symmetry considerations of the compliance matrix that

Si j =Sji,i,j=1,2,6; i.e.

ν

12

E2

=

ν

21

E1

,

ν

16

E1

=

ν

61

E6

,

ν

26

E2

=

ν

62

E6

(4.1.21)

but the anisotropic coupling coefﬁcients are

ν

i j 6=

ν

ji,i,j=1,2,6 (4.1.22)

Equation (4.1.19) can be inverted to yield the stress-strain equations in terms of

engineering parameters but these relations would be more complex than (4.1.19).

Using the relationships between engineering parameters and compliances (4.1.20)

in the compliance transformation rule (Table 4.2) we obtain the following transfor-

mations for the engineering parameters of the angle lamina including shear coupling

114 4 Elastic Behavior of Laminate and Sandwich Composites

ratios

ν

16 and

ν

26

1

E1

=1

E′

1

c4+1

G′

12 −2

ν

′

12

E′

1s2c2+1

E′

2

s4,

1

E2

=1

E′

1

s4+1

G′

12 −2

ν

′

12

E′

1s2c2+1

E′

2

c4,

1

G12

=22

E′

1

+2

E′

2

+4

ν

′

12

E′

1−1

G′

12 s2c2+1

G′

12

(c4+s4),(4.1.23)

ν

12

E1

=

ν

21

E2

=

ν

′

12

E′

1

(c4+s4)−1

E′

1

+1

E′

2−1

G′

12 c2s2,

ν

16 =E′

1 2

E′

1

+2

ν

′

12

E′

1−1

G′

12 sc3−2

E′

2

+2

ν

′

12

E′

1−1

G′

12 s3c,

ν

26 =E′

2 2

E′

1

+2

ν

′

12

E′

1−1

G′

12 s3c−2

E′

2

+2

ν

′

12

E′

1−1

G′

12 sc3

Equation (4.1.23) can be also written in the following form

E1=EL

c4+EL

GLT −2

ν

LTs2c2+EL

ET

s4

,

E2=ET

c4+ET

GLT −2

ν

TLs2c2+ET

EL

s4

,

G12 =GLT

c4+s4+22GLT

EL

(1+2

ν

LT) + 2GLT

ET−1s2c2

,

ν

12 =

ν

LT(c4+s4)−1+EL

ET−EL

GLT c2s2

c4+EL

GLT −2

ν

LTc2s2+EL

ET

s4

,

ν

21 =

ν

TL(c4+s4)−1+ET

EL−ET

GLT c2s2

c4+ET

GLT −2

ν

TLc2s2+ET

EL

s4

,

ν

16 =EL 2

EL

+2

ν

LT

EL−1

GLT sc3−2

ET

+2

ν

LT

EL−1

GLT s3c,

ν

26 =ET 2

EL

+2

ν

LT

EL−1

GLT s3c−2

ET

+2

ν

LT

EL−1

GLT sc3

(4.1.24)

The engineering parameters can change rapidly with angel

θ

. This can be inter-

preted as if the ﬁbres are not oriented exactly as intended the values of engineering

parameters are very less or more than expected.

4.1 Elastic Behavior of Laminae 115

A computational procedure for calculating the elastic parameters of a UD-lamina

in off-axis loading can be illustrated by the following steps:

1. Input the basic engineering parameters EL,ET,GLT ,

ν

LT referred to the material

axes of the lamina and obtained by material tests or mathematical modelling.

2. Calculate the compliances S′

i j and the reduced stiffness Q′

i j.

3. Application of transformations to obtain the lamina stiffness Qi j and compliances

Si j.

4. Finally calculate the engineering parameters E1,E2,G12 ,

ν

12,

ν

21,

ν

16,

ν

26 re-

ferred to the (x1,x2)-system.

Otherwise the engineering parameters referred to the (x1,x2)-coordinate can be cal-

culated directly by Eqs. (4.1.24).

Analogous to the on-axis loaded UD-lamina also the off-axis lamina is in thick-

ness direction x3≡T′orthogonal to the (x1,x2)-plane macro-mechanically quasi-

isotropic and the three-dimensional material behavior is transversely-isotropic. The

mechanical properties transverse to the ﬁbre direction are provided by weaker ma-

trix material and the effects of transverse shear deformation may be signiﬁcant. For

such cases, the stress and the strain vector should include all six components

σ

T= [

σ

1

σ

2

σ

3

σ

4

σ

5

σ

6],

ε

T= [

ε

1

ε

2

ε

3

ε

4

ε

5

ε

6]

For a rotation about the direction e

e

e3(Fig. 2.6) the transformation matrices (2.1.39)

and (2.1.40) are valid and relations for stress and strain vectors in the on-axis and

the off-axis reference system are given by

σ

′=

3

T

σσ

σ

,

ε

′=

3

T

εε

ε

,

3

T

ε

′=3

T

σ

T

=3

T

ε

−1

,

σ

=3

T

σ

−1

σ

′,

ε

=3

T

ε

−1

ε

,

3

T

σ

′=3

T

ε

T

=3

T

σ

−1(4.1.25)

When the stiffness matr ix C

C

C′corresponding to an orthotropic material behavior, see

Eq. (2.1.46) , the transformed stiffness matrix C

C

Cmay be written in detail as for mon-

oclinic material behavior, see Eq. (2.1.42)







σ

1

σ

2

σ

3

σ

4

σ

5

σ

6







=







C11 C12 C13 0 0 C16

C22 C23 0 0 C26

C33 0 0 C36

C44 C45 0

S Y M C55 0

C66













ε

1

ε

2

ε

3

ε

4

ε

5

ε

6







(4.1.26)

The Ci j are the transformed stiffness, i.e. in vector-matrix notation

σ

′=C

C

C′

ε

′

σ

=C

C

ε

,

σ

=

3

T

σ

′

σ

′= 3

T

ε

!T

C

C′

ε

′= 3

T

ε

!T

C

C′3

T

εε

ε

=C

C

ε

116 4 Elastic Behavior of Laminate and Sandwich Composites

we ﬁnally obtain

C

C= 3

T

ε

!T

C

C′3

T

ε

,(4.1.27)

in which the Ci j

C11 =C′

11c4+2C′

12c2s2+C′

22s4+4C′

66c2s2,

C12 =C′

11c2s2+C′

12(c4+s4) + C′

22c2s2−4C′

66c2s2,

C13 =C′

13c2+C′

23s2,C14 =0,C15 =0,

C16 =C′

11c3s−C′

12cs(c2−s2)−C′

22cs3−2C′

66cs(c2−s2),

C22 =C′

11s4+2C′

12c2s2+C′

22c4+4C′

66c2s2,

C23 =C′

13s2+C′

23c2,C24 =0,C25 =0,

C26 =C′

11cs3+C′

12cs(c2−s2)−C′

22c3s+2C′

66cs(c2−s2),

C33 =C′

33,C34 =0,C35 =0,

C36 =C′

13cs −C′

23cs,

C44 =C′

44c2+C′

55s2,

C45 =−C′

44cs +C′

55cs,C46 =0,

C55 =C′

44s2+C′

55c2,C56 =0,

C66 =C′

11c2s2−2C′

12c2s2+C′

22c2s2+C′

66(c2−s2)2(4.1.28)

The 13 non-zero stiffness of Cij are not independent material values. They are func-

tions of 9 C′

i j for a three-dimensional orthotropic material, i.e. of

C′

11,C′

12,C′

13,C′

22,C′

23,C′

33,C′

44,C′

55,C66

and of 5 C′

i j for a transverse-isotropic behavior, i.e. of

C′

11,C′

12,C′

22,C′

23,C′

55

because

C′

13 =C′

12,C′

33 =C′

22,C′

44 =1

2(C′

22 −C′

23),C′

66 =C′

55

With

ε

′=S

S

S′

σ

′,

ε

=S

S

σ

follow analogously the transformed compliances

S

S=3

T

σ

T

S

S′3

T

σ

,(4.1.29)

in which the Si j

4.1 Elastic Behavior of Laminae 117

S11 =S′

11c4+2S′

12c2s2+S′

22s4+S′

66c2s2,

S12 =S′

11c2s2+S′

12(c4+s4) + S′

22c2s2−S′

66c2s2,

S13 =S′

13c2+S′

23s2,S14 =0,S15 =0,

S16 =2S′

11c3s−2S′

12cs(c2−s2)−2S′

22cs3−S′

66cs(c2−s2),

S22 =S′

11s4+2S′

12c2s2+S′

22c4+S′

66c2s2,

S23 =S′

13s2+S′

23c2,S24 =0,S25 =0,

S26 =2S′

11cs3−2S′

12cs(c2−s2)−2S′

22c3s−S′

66cs(c2−s2),

S33 =S′

33,S34 =0,S35 =0,

S36 =2S′

13cs −2S′

23cs,

S44 =S′

44c2+S′

55s2,

S45 =−S′

44cs +S′

55cs,S46 =0,

S55 =S′

44s2+S′

55c2,S56 =0,

S66 =4S′

11c2s2−4S′

12c2s2+2S′

22c2s2+S′

66(s2−c2)2

(4.1.30)

There are again 13 non-zero compliances, but only 9 independent material values

for the orthotropic and 5 independent material values for the transversal-isotropic

case.

The stress-strain relationship for an angle lamina, i.e. an off-axis loaded UD-

lamina, including hygrothermal effects takes the following form





ε

1

ε

2

ε

6

=



S11 S12 S16

S12 S22 S26

S16 S26 S66 





σ

1

σ

2

σ

6

+



ε

th

1

ε

th

2

ε

th

6



+



ε

mo

1

ε

mo

2

ε

mo

6



,(4.1.31)

where 



ε

th

1

ε

th

2

ε

th

6



=



α

th

1

α

th

2

α

th

6



T,



ε

mo

1

ε

mo

2

ε

mo

6



=



α

mo

1

α

mo

2

α

mo

6



M∗(4.1.32)

with the thermal and moisture expansion coefﬁcients

α

th

i,

α

mo

i,i=1,2,6 and the

temperature change Tor the weight of moisture absorption per unit weight M∗,

respectively. It should be remembered that although the coefﬁcients of both ther-

mal and moisture expansions are pure dilatational in the material coordinate system

(L,T), rotation into the global (x1,x2) system results in coefﬁcients

α

th

6,

α

mo

6. Fur-

thermore if there are no constraints placed on a UD-lamina, no mechanical strains

will be included in it and therefore no mechanical stresses are induced. But in lami-

nates, even if there are no constraints on the laminate, the difference in thermal and

moisture expansion coefﬁcients of the various laminae of a laminate induces differ-

ent expansions in each layer and results in residual stresses. This will be explained

fully in Sects. 4.2.4 and 4.2.5. With

ε

′th

1=

α

′th

1T,

ε

′th

2=

α

′th

2T,

ε

′th

6=0,

(4.1.33)

118 4 Elastic Behavior of Laminate and Sandwich Composites

and

ε

th

1=

ε

′th

1c2+

ε

′th

2s2,

ε

th

2=

ε

′th

1s2+

ε

′th

2c2,

ε

th

6=2(

ε

′th

1−

ε

′th

2)cs,

(4.1.34)

follow

α

th

1=

α

′th

1c2+

α

′th

2s2,

α

th

2=

α

′th

1s2+

α

′th

2c2,

α

th

6=2(

α

′th

1−

α

′th

2)cs

(4.1.35)

In an anisotropic layer, uniform heating induces not only normal strains, but also

shear thermal strains. For a transversal isotropic material behavior there is additional

ε

th

3=

α

th

3T,

α

th

3=

α

′th

3,

α

th

4=

α

th

5=0. Because

ε

3=

ε

′

3the strain

ε

3can be obtained

directly from

ε

3=

α

th

3T+

α

mo

3M∗+S′

13

σ

′

1+S′

23

σ

′

2

However, the stresses

σ

′

1,

σ

′

2can be written in terms

σ

1,

σ

2,

σ

6referred to the off-

axis coordinate system to obtain an expression for

ε

3that represents the normal

strain in the x3-direction in terms of the global coordinate system

ε

3=

ε

th

3+

ε

mo

3+ (S13c2+S23s2)

σ

1+ (S13s2+S23c2)

σ

2

+2(S13 −S23s2)sc

σ

1

(4.1.36)

4.1.3 Stress Resultants and Stress Analysis

Sections 4.1.1 and 4.1.2 describe the constitutive equations for UD-laminae in an

on-axis and an off-axis reference system as a relation between stresses and strains.

For each lamina, the stress components can be integrated across their thickness hand

yield stress resultants. Stress resultants can be in-plane forces, transverse forces and

resultant moments. The constitutive equations may then be formulated as relations

between mid-plane strain and in-plane forces, transverse shear strains and transverse

forces and mid-plane curvatures and resultant moments, respectively.

The in-plane stress resultant force vector, denoted by

N

N= [N1N2N6]T,(4.1.37)

is deﬁned by

N

N=

h/2

Z

−h/2

σ

dx3(4.1.38)

The Niare forces per unit length, N1,N2are normal in-plane resultants and N6is

a shear in-plane resultant, respectively. They are illustrated in Fig. 4.5 for constant

in-plane stresses

σ

1,

σ

2,

σ

6across the thickness. In this case we have

4.1 Elastic Behavior of Laminae 119

x1

x2

x3

N1

N6

N2

h

Fig. 4.5 In-plane force resultants per unit length N

N

NT= [N1N2N6]

N

N=

σ

h(4.1.39)

The reduced stiffness matrix Q

Q

Qof the lamina has also constant components across

h. The strains of the midplane x3=0 of the lamina are given by

ε

(x1,x2,x3=0) =

ε

(x1,x2)

and Eq. (4.1.39) yields

N

N=Q

Q

ε

h=A

A

ε

,A

A

A=Q

Q

Qh,

ε

T= [

ε

1

ε

2

ε

6](4.1.40)

Q

Qis the reduced stiffness matrix (Table 4.2 a) and A

A

Ais the off-axis stretching or

extensional stiffness matrix of the lamina. From (4.1.40) it follows that

ε

=A

A

A−1N

N

N=a

a

aN

N

N,a

a

a=A

A

A−1=S

S

Sh−1(4.1.41)

a

ais the off-axis in-plane compliance matrix. A

A

Aand a

a

aare, like Q

Q

Qand S

S

Ssymmet-

ric matrices, which have in the general case only non-zero elements. In the special

cases of on-axis reference systems or isotropic stiffness and compliances, respec-

tively, the structure of the matrices is simpliﬁed. A

A

Ais the extensional stiffness and a

a

the extensional compliance matrix expressing the relationship between the in-plane

stress resultant N

N

Nand the mid-plane strain

ε

:

120 4 Elastic Behavior of Laminate and Sandwich Composites

Off-axis extensional stiffness and compliance matrices

A

A=



A11 A12 A16

A12 A22 A26

A16 A26 A66 

,a

a

a=



a11 a12 a16

a12 a22 a26

a16 a26 a66 

(4.1.42)

On-axis extensional stiffness and compliance matrices

A

A=



A11 A12 0

A12 A22 0

0 0 A66 

,a

a

a=



a11 a12 0

a12 a22 0

0 0 a66 

(4.1.43)

If the stresses are not constant across h, resultant moments can be deﬁned

M

M=

h/2

Z

−h/2

σ

x3dx3(4.1.44)

The resultant moment vector is denoted by

M

M= [M1M2M6]T(4.1.45)

The Miare moments per unit length, M1,M2are bending moments and M6is a

torsional or twisting moment. Figure 4.6 illustrates these moments and a linear stress

distribution across h. The resultant moments yield ﬂexural strains, e.g. bending and

x1

x2

x3

M6

M1

M2

M6

h

Qs

1

Qs

2

Fig. 4.6 Resultant moment vector M

M

MT= [M1M2M6]and transverse shear resultants Q

Q

QsT= [Qs

1Qs

2]

4.1 Elastic Behavior of Laminae 121

twisting strains, which are usually expressed by the relationship

ε

(x1,x2,x3) = x3

κ

,

κ

T= [

κ

1

κ

2

κ

6](4.1.46)

κ

is the vector of curvature,

κ

1,

κ

2correspond to the bending moments M1,M2and

κ

6to the torsion moment M6, respectively. The ﬂexural strains are assumed linear

across h. With Qi j =const across h,i,j=1,2,6 follow

M

M=Q

Q

κ

h/2

Z

−h/2

x2

3dx3=Q

Q

κ

h3

12 =D

D

κ

,

κ

=D

D

D−1M

M

M=d

d

dM

M

M(4.1.47)

D

D=Q

Q

Qh3/12 is the ﬂexural stiffness matrix and d

d

dthe ﬂexural compliance matrix

expressing the relations between stress couples M

M

Mand the curvatures. For off-axis

and on-axis reference systems the matrices are given by:

Off-axis ﬂexural stiffness and compliance matrices

D

D=



D11 D12 D16

D12 D22 D26

D16 D26 D66 

,d

d

d=



d11 d12 d16

d12 d22 d26

d16 d26 d66 

(4.1.48)

On-axis ﬂexural stiffness and compliance matrices

D

D=



D11 D12 0

D12 D22 0

0 0 D66 

,d

d

d=



d11 d12 0

d12 d22 0

0 0 d66 

(4.1.49)

The transverse shear resultants can be deﬁned in the same way by

Q

Qs=Qs

1

Qs

2=

h/2

Z

−h/2

σ

s

5

σ

s

4dx3(4.1.50)

Q

Qsis (like the in-plane resultants) a load vector per unit length in the cross section

of the lamina x1=const or x2=const, respectively. The transverse shear resultant

vector Q

Q

Qsis written with a superscript sto distinguish it from the reduced stiffness

matrix Q

Q

Q. When modelling a plane stress state, there are no constitutive equations

for

σ

4,

σ

5and the shearing stresses are calculated with the help of the equilibrium

equations, Eq. (2.2.1), or with help of equilibrium conditions of stress resultant,

e.g. Chap. 8. In three-dimensional modelling, including transverse shear strains,

however, constitutive equations for transverse shear resultant can be formulated.

For a lamina with resultant forces N

N

Nand moments M

M

Mthe in-plane strain

ε

and

the curvature term

κ

have to be combined

ε

(x1,x2,x3) =

ε

(x1,x2) + x3

κ

(x1,x2)(4.1.51)

122 4 Elastic Behavior of Laminate and Sandwich Composites

For the stress vector is

σ

(x1,x2,x3) = Q

Q

Q[

ε

(x1,x2) + x3

κ

(x1,x2)]

=Q

Q

ε

(x1,x2) + Q

Q

Qx3

κ

(x1,x2)(4.1.52)

and by integration through the lamina thickness hfollow

N

N=

h/2

Z

−h/2

σ

dx3=Q

Q

ε

h+Q

Q

κ

h/2

Z

−h/2

x3dx3=A

A

ε

+B

B

κ

,

M

M=

h/2

Z

−h/2

σ

x3dx3=Q

Q

ε

h/2

Z

−h/2

x3dx3+Q

Q

κ

h3

12 =B

B

ε

+D

D

κ

(4.1.53)

The coupling stiffness matrix B

B

Bis zero for a lamina, which is symmetric to the

midplane x3=0, i.e. there are no coupling effects between the N

N

Nand

κ

or M

M

Mand

ε

, respectively. In Table 4.3 the constitutive equations of the lamina resultants are

summarized for a symmetric general angle lamina, for a UD-orthotropic lamina and

for an isotropic layer. In a contracted vector-matrix notation, we can formulate the

constitutive equation of a lamina by

N

N

M

M=A

A

A0

0

0D

D

D

ε

κ

,(4.1.54)

where the in-plane stiffness submatrix A

A

A=Q

Q

Qh and the plate stiffness submatrix

D

D=Q

Q

Qh3/12. 0

0

0 are zero submatrices. The inverted form of (4.1.54) is important for

stress analysis



ε

κ

=a

a

a0

0

0d

d

dN

N

M

M,a

a

a=A

A

A−1,

d

d=D

D

D−1,

A

A=Q

Q

Qh,

D

D=Q

Q

Q(h3/12)(4.1.55)

Equation (4.1.52) yields the stress components

σ

i,i=1,2,6

σ

1=Q11(

ε

1+x3

κ

1) + Q12(

ε

2+x3

κ

2) + Q16(

ε

6+x3

κ

6) =

σ

1M +

σ

1B,

σ

2=Q21(

ε

1+x3

κ

1) + Q22(

ε

2+x3

κ

2) + Q26(

ε

6+x3

κ

6) =

σ

2M +

σ

2B,

σ

6=Q16(

ε

1+x3

κ

1) + Q62(

ε

2+x3

κ

2) + Q66(

ε

6+x3

κ

6) =

σ

6M +

σ

6B

σ

iMare the membrane or in-plane stresses coupled with Niand

σ

iBthe curvature or

plate stresses coupled with Mi. The membrane stresses are constant and the bending

stresses linear through the lamina thickness (Fig. 4.7).

The transverse shear stresses

σ

4,

σ

5for plane stress state condition are ob-

tained by integration of the equilibrium equations (2.2.1). If the volume forces

p1=p2=0, Eqs. (2.2.1) yield

σ

5(x3) = −

x3

Z

−h/2

∂σ

1

∂

x1

+

∂σ

6

∂

x2dx3,

σ

4(x3) = −

x3

Z

−h/2

∂σ

6

∂

x1

+

∂σ

2

∂

x2dx3(4.1.56)

4.1 Elastic Behavior of Laminae 123

Table 4.3 Stiffness matrices of laminae

Anisotropic single layer or UD-lamina, off-axis







N1

N2

N6

M1

M2

M6







=





A11 A12 A16 000

A12 A22 A26 000

A16 A26 A66 000

0 0 0 D11 D12 D16

0 0 0 D12 D22 D26

0 0 0 D16 D26 D66













ε

1

ε

2

ε

6

κ

1

κ

2

κ

6







,

Ai j =Qi j h,

Di j =Qi j

h3

12

Orthotropic single layer or UD-laminae, on-axis







N1

N2

N6

M1

M2

M6







=





A11 A12 0 0 0 0

A12 A22 0 0 0 0

0 0 A66 000

0 0 0 D11 D12 0

0 0 0 D12 D22 0

0 0 0 0 0 D66













ε

1

ε

2

ε

6

κ

1

κ

2

κ

6







,

Q11 =E1

1−

ν

12

ν

21

,

Q22 =E2

1−

ν

12

ν

21

,

Q12 =

ν

12E2

1−

ν

12

ν

21

=

ν

21E1

1−

ν

12

ν

21

,

Q66 =G12

Isotropic single layer







N1

N2

N6

M1

M2

M6







=





A11 A12 0 0 0 0

A12 A11 0 0 0 0

0 0 A66 000

0 0 0 D11 D12 0

0 0 0 D12 D11 0

0 0 0 0 0 D66













ε

1

ε

2

ε

6

κ

1

κ

2

κ

6







,

Q11 =E

1−

ν

2,

Q12 =

ν

E

1−

ν

2,

Q66 =E

2(1+

ν

)

and with Eq. (4.1.52) follows

σ

5(x3) = −

x3

Z

−h/2

∂

x1

[Q11(

ε

1+x3

κ

1) + Q12(

ε

2+x3

κ

2) + Q16(

ε

6+x3

κ

6)]

+

∂

x2

[Q61(

ε

1+x3

κ

1) + Q62(

ε

2+x3

κ

2) + Q66(

ε

6+x3

κ

6)]dx3,

(4.1.57)

124 4 Elastic Behavior of Laminate and Sandwich Composites

✂✂✂✂✂✂✂

❄

✻

h

0 0 0

σ

iM

σ

iB

σ

i=

σ

iM+

σ

iB

N6

h=

σ

6M

N2

h=

σ

2M

N1

h=

σ

1M

M6

h3/12 x3=

σ

6B

M2

h3/12 x3=

σ

2B

M1

h3/12 x3=

σ

1B

σ

6=

σ

6M +

σ

6B

σ

2=

σ

2M +

σ

2B

σ

1=

σ

1M +

σ

1B

Fig. 4.7 In-plane membrane stresses

σ

iM, bending stresses

σ

iBand total stresses

σ

iacross h(qual-

itative)

σ

4(x3) = −

x3

Z

−h/2

∂

x1

[Q61(

ε

1+x3

κ

1) + Q62(

ε

2+x3

κ

2) + Q66(

ε

6+x3

κ

6)]

+

∂

x2

[Q21(

ε

1+x3

κ

1) + Q22(

ε

2+x3

κ

2) + Q26(

ε

6+x3

κ

6)]dx3

Substituting the midplane strain

ε

and the curvature

κ

by the resultants N

N

Nand M

M

Eqs. (4.1.57) takes the form

σ

5(x3) = −

x3

Z

−h/2nQ11

∂

x1

[a11N1+a12N2+a16N6+x3(d11M1+d12M2+d16M6)]

+Q12

∂

x1

[a21N1+a22N2+a26N6+x3(d21M1+d22M2+d26M6)]

+Q16

∂

x1

[a61N1+a62N2+a66N6+x3(d61M1+d62M2+d66M6)]

+Q61

∂

x2

[a11N1+a12N2+a16N6+x3(d11M1+d12M2+d16M6)]

+Q62

∂

x2

[a21N1+a22N2+a26N6+x3(d21M1+d22M2+d26M6)]

4.1 Elastic Behavior of Laminae 125

+Q66

∂

x2

[a16N1+a26N2+a66N6+x3(d16M1+d26M2+d66M6)]odx3,

(4.1.58)

σ

4(x3) = −

x3

Z

−h/2nQ61

∂

x1

[a11N1+a12N2+a16N6+x3(d11M1+d12M2+d16M6)]

+Q62

∂

x1

[a21N1+a22N2+a26N6+x3(d21M1+d22M2+d26M6)]

+Q66

∂

x1

[a61N1+a62N2+a66N6+x3(d61M1+d62M2+d66M6)]

+Q21

∂

x2

[a11N1+a12N2+a16N6+x3(d11M1+d12M2+d16M6)]

+Q22

∂

x2

[a21N1+a22N2+a26N6+x3(d21M1+d22M2+d26M6)]

+Q26

∂

x2

[a61N1+a62N2+a66N6+x3(d61M1+d62M2+d66M6)]odx3

The distribution of the transverse shear stresses through the thickness his obtained

by integration from the bottom surface x3=−h/2 of the lamina to x3

x3

Z

−h/2

Q

Qdx3=˜

A

A(x3) = Q

Q

Q(x3+h/2),

x3

Z

−h/2

Q

Qx3dx3=˜

B

B(x3) = Q

Q

Q1

2x2

3−h2

4

(4.1.59)

and we have

˜

A

A(−h/2) = ˜

B

B(−h/2) = 0

0

0,˜

A

A(h/2)≡A

A

A,˜

B

B(h/2)≡B

B

Finally, the transverse shear stress equations (4.1.57) take the form

σ

5(x3) =−˜

A11(x3)

∂ε

1

∂

x1

+˜

A12(x3)

∂ε

2

∂

x1

+˜

A16(x3)

∂ε

6

∂

x1

+˜

B11(x3)

∂κ

1

∂

x1

+˜

B12(x3)

∂κ

2

∂

x1

+˜

B16(x3)

∂κ

6

∂

x1

+˜

A61(x3)

∂ε

1

∂

x2

+˜

A62(x3)

∂ε

2

∂

x2

+˜

A66(x3)

∂ε

6

∂

x2

+˜

B61(x3)

∂κ

1

∂

x2

+˜

B62(x3)

∂κ

2

∂

x2

+˜

B66(x3)

∂κ

6

∂

x2,

126 4 Elastic Behavior of Laminate and Sandwich Composites

σ

4(x3) =−˜

A61(x3)

∂ε

1

∂

x1

+˜

A62(x3)

∂ε

2

∂

x1

+˜

A66(x3)

∂ε

6

∂

x1

+˜

B61(x3)

∂κ

1

∂

x1

+˜

B62(x3)

∂κ

2

∂

x1

+˜

B66(x3)

∂κ

6

∂

x1

+˜

A21(x3)

∂ε

1

∂

x2

+˜

A22(x3)

∂ε

2

∂

x2

+˜

A26(x3)

∂ε

6

∂

x2

+˜

B21(x3)

∂κ

1

∂

x2

+˜

B22(x3)

∂κ

2

∂

x2

+˜

B26(x3)

∂κ

6

∂

x2

or an analogous equation to (4.1.59) by substituting the in-plane strains

ε

and the

mid-plane curvatures by the stress resultants N

N

Nand M

M

M.

The transverse shear stresses

σ

4,

σ

5are parabolic functions across the lamina

thickness. If there are no surface edge shear stresses, the conditions

σ

5(h/2) =

σ

4(h/2) = 0 are controlling the performance of the equilibrium equations and the

accuracy of the stress analysis.

4.1.4 Problems

Exercise 4.1. For a single layer unidirectional composite, the on-axis elastic behav-

ior is given by EL=140 GPa, ET=9 GPa,

ν

LT =0.3. Calculate the reduced stiffness

matrix Q

Q

Q′and the reduced compliance matrix S

S

S′.

Solution 4.1. EL≡E′

1,ET≡E′

2,

ν

LT =

ν

′

12. Equation (4.1.3) yields





σ

′

1

σ

′

2

σ

′

6



=



Q′

11 Q′

12 0

Q′

12 Q′

22 0

0 0 Q′

66







ε

′

1

ε

′

2

ε

′

6



,



ε

′

1

ε

′

2

ε

′

6



=



S′

11 S′

12 0

S′

12 S′

22 0

0 0 S′

66







σ

′

1

σ

′

2

σ

′

6



,

Q′

11 =E′

1/(1−

ν

′

12

ν

′

21),S′

11 =1/E′

1,

Q′

22 =E′

2/(1−

ν

′

12

ν

′

21),S′

22 =1/E′

2,

Q′

12 =E′

2

ν

′

12/(1−

ν

′

12

ν

′

21),S′

12 =−

ν

′

12/E′

2,

Q′

66 =G′

12 =E′

2/2(1+

ν

′

12),S′

66 =1/G′

12,

ν

′

21 =

ν

′

12E′

2/E′

1=0,0192,G′

12 =E′

1/2(1+

ν

′

12) = 53,846 GPa,

Q′

11 =140,811 GPa, Q′

22 =9,052 GPa,

Q′

12 =2,716 GPa, Q′

66 =53,846 GPa,

S′

11 =7,143 10−3GPa−1,S′

22 =111,111 10−3GPa−1,

S′

12 =−2,143 10−3GPa−1,S′

66 =18,571 10−3GPa−1,

4.1 Elastic Behavior of Laminae 127

Q

Q′=



140,811 2,716 0

2,716 9,052 0

0 0 53,846 

GPa,

S

S′=



7,143 −2,143 0

−2,143 111,111 0

0 0 18,576 

10−3GPa−1

Exercise 4.2. A composite panel is designed as a single layer lamina with the fol-

lowing properties EL=140 GPa, ET=10 GPa, GLT =6,9 GPa,

ν

LT =0,3 and

θ

=450. Calculate the strains

ε

1,

ε

2and

ε

6when the panel is loaded by a shear stress

σ

6=±

τ

=±10 MPa.

Solution 4.2. From Fig. 4.4 follows

ε

1=±S16

σ

6,

ε

2=±S26

σ

6,

ε

6=±S66

σ

6

With EL≡E′

1,ET≡E′

2,GLT ≡G′

12 and

ν

LT ≡

ν

′

12 and Eq. (4.1.17) is

S′

11 =1/E′

1=7,143 10−3GPa−1,

S′

22 =1/E′

2=100,000 10−3GPa−1,

S′

12 =−

ν

′

12/E′

1=−2,143 10−3GPa−1,

S′

66 =1/G′

12 =144,928 10−3GPa−1

The transformation rule, Table 4.2, yields

S16 =S26 =−0,46 10−1GPa−1,S66 =1,11 10−1GPa−1

The strains are

ε

1=∓0,46 ·10−3,

ε

2=

ε

1,

ε

6=±1,11 ·10−3.

Conclusion 4.1. A positive shear load

σ

6= +

τ

shortens the composite panel in both

directions, a negative shear load

σ

6=−

τ

would enlarge the panel in both directions.

Exercise 4.3. In a unidirectional single layer is a strain state

ε

11 =1% =10−2,

ε

22 =−0.5% =−0.5 10−2,

γ

12 =2% =2 10−2. In the principal directions, the

following engineering parameters of the composite material are E′

1=EL=40 GPa,

E′

2=ET=10 GPa, G12 =GLT =5 GPa,

ν

LT =0.3. Determine the plane stress state

for the axis (x1,x2)and (x′

1,x′

2)≡(L,T)and

θ

=450.

Solution 4.3. The stress states

σ

′and

σ

are to calculate for a given strain state

ε

1=10−2,

ε

2=−0,5 10−2,

ε

6=2 10−2for a UD lamina with E′

1=40 GPa,

E′

2=10 GPa, G′

12 =5 GPa,

ν

′

12 =0,3 and a ﬁbre angle

θ

=450. With Table

4.1 the strains for the off-axis system x1,x2are transferred to the strains for the on-

axis system x′

1,x′

2. Taking into account cos450=sin 450=√2/2=0,707107 we

obtain 



ε

′

1

ε

′

2

ε

′

6



=



c2s2sc

s2c2−sc

−2sc 2sc c2−s2





ε

1

ε

2

ε

6



=



0,5 0,5 0,5

0,5 0,5−0,5

−1 1 0 





10

−5

20 

10−3=



12,5

−7,5

−15 

10−3,

128 4 Elastic Behavior of Laminate and Sandwich Composites

ε

′

1=12,5 10−3,

ε

′

2=−7,5 10−3,

ε

′

6=−15 10−3

and

ν

′

21 =

ν

′

12E′

2

E′

1

=0,075

The reduced stiffness Q′

i j and the stresses

σ

′

iin the on-axis system follow from

(4.1.3)

Q′

11 =E′

1/(1−

ν

′

12

ν

′

21) = 40,9207 GPa, Q′

22 =E′

2/(1−

ν

′

12

ν

′

21) = 10,2302 GPa,

Q′

12 =E′

2

ν

′

12/(1−

ν

′

12

ν

′

21) = 3,0691 GPa, Q′

66 =G′

12 =5,0 GPa,





σ

′

1

σ

′

2

σ

′

6



=



Q′

11 Q′

12 0

Q′

12 Q′

22 0

0 0 Q′

66







ε

′

1

ε

′

2

ε

′

6



=



488,5

−38,36

−75,0

10−3GPa

The stresses

σ

iin the off-axis system are calculated with the help of the transforma-

tion rules Table 4.1





σ

1

σ

2

σ

6

=



c2s2−2sc

s2c22sc

sc −sc c2−s2





σ

′

1

σ

′

2

σ

′

6





=



0,5 0,5−1

0,5 0,5 1

0,5−0,5 0 





488,5

−38,36

−75,0

10−3GPa =



0,300

0,150

0,263 

GPa

Exercise 4.4. Sketch the variation curves E1/E′

2and G12/E′

2against the ﬁbre ori-

entation

θ

for a carbon-epoxy and glass-epoxy lamina using the following material

data:

carbon-epoxy E′

1=140 GPa, E′

2=10 GPa, G′

12 =7 GPa,

ν

′

12 =0,3

glass-epoxy E′

1=43 GPa, E′

2=9 GPa, G′

12 =4,5 GPa,

ν

′

12 =0,27

Discuss the curves.

Solution 4.4. From (4.1.23) follows

(E1)−1=c4

E′

1

+1

G′

12 −2

ν

′

12

E′

2s2c2+s4

E′

2

,

(G12)−1=22

E′

1

+2

E′

2

+4

ν

′

12

E′

1−1

G′

12 s2c2+1

G′

12

(c4+s4)

Now the functions f1(

θ

) = E1(

θ

)/E′

2and f2(

θ

) = G12(

θ

)/E′

2can be sketched. The

results are shown on Figs. 4.8 and 4.9. Discussion of the functions f1(

θ

)and f2(

θ

):

1. The anisotropic ratio E1/E′

2is higher for carbon- than for glass-epoxy.

2. The longitudinal effective modulus E1of the lamina drops sharply as the loading

direction deviates from the ﬁbre direction, especially for-carbon-epoxy.

3. The effective shear modulus of the lamina attains a maximum value at

θ

=450.

4.1 Elastic Behavior of Laminae 129

Fig. 4.8 Variation of

E1(

θ

)/E′

2against the ﬁbre

orientation for two compos-

ites

carbon-epoxy glass-epoxy

0

010 20 30 40 50 60 70 80 90

2

4

6

8

10

12

14

E1(

θ

)/E′

2

θ

Fig. 4.9 Variation of

G12(

θ

)/E′

2against the ﬁbre

orientation for two compos-

ites

carbon-epoxy glass-epoxy

010 20 30 40 50 60 70 80 90

0.5

0.6

0.7

0.8

0.9

G12(

θ

)/E′

2

θ

Exercise 4.5. For a UD-lamina with the elastic properties E′

1=180 GPa,

E′

2=10 GPa, G′

12 =7 GPa,

ν

′

12 =0,3 calculate

1. the on-axis compliances S′

i j and the on-axis strains, if the applied on-axis stresses

are

σ

′

1=2 MPa,

σ

′

2=−3 MPa,

σ

′

6=4 MPa,

2. the off-axis compliances Si j and the off-axis and on-axis strains

ε

,

ε

′if

θ

=450

(Fig. 4.3) and the applied off-axis stresses are

σ

1=2 MPa,

σ

2=−3 MPa,

σ

6=4 MPa,

3. the coefﬁcients of thermal expansion in the off-axis system if

α

th′

1=9 10−6/0K,

α

th′

2=22 10−6/0K.

Solution 4.5. 1. Using (4.1.3) follows

130 4 Elastic Behavior of Laminate and Sandwich Composites

S′

11 = (E′

1)−1=5,556 10−12 Pa−1,

S′

12 =−

ν

′

12/E′

1=−1,667 10−12 Pa−1,

S′

22 = (E′

2)−1=100 10−12 Pa−1,

S′

66 = (G′

12)−1=142,86 10−12 Pa−1,





ε

′

1

ε

′

2

ε

′

6



=



5,556 −1,667 0

−1,667 100 0

0 0 142,86 

10−12Pa−1



2

−3

4

106Pa

=



16,113

−303,334

571,440 

10−6

2. With Table 4.2 the transformed compliances Si j can be calculated (note that c=

cos450=0.7071,s=sin 450=0.7071)

S11 =61,270 10−12 Pa−1,S22 =61,271 10−12 Pa−1,

S12 =−10,160 10−12 Pa−1,S66 =108,89 10−12 Pa−1,

S16 =S26 =−47,222 10−12 Pa−1

Equations (4.1.19) and (4.1.20) yield the strains

ε

i





ε

1

ε

2

ε

6

=



S11 S12 S16

S12 S22 S26

S16 S26 S66 





σ

1

σ

2

σ

6



=



61,270 −10,160 −47,222

−10,160 61,271 −47,222

−47,222 −47,222 108,89 

10−12Pa−1



2

−3

4

106Pa

=

−35,868

−393,01

482,782 

10−6

Using Table 4.1 the strains

ε

can be transformed to the strains

ε

′





ε

′

1

ε

′

2

ε

′

6



=



0.5 0.5 0.5

0.5 0.5−0.5

−1 1 0 





ε

1

ε

2

ε

6



=



0.5 0.5 0.5

0.5 0.5−0.5

−1 1 0 



−35,868

−393,01

482,782 

10−6

=



26,95

−455,83

−357,14 

10−6

3. The transformed thermal expansion coefﬁcients

α

th

ifollow like the strains with

Table 4.1 to

4.2 Elastic Behavior of Laminates 131





α

th

1

α

th

2

α

th

6



=



0.5 0.5−0.5

0.5 0.5 0.5

1−1 0 





α

′th

1

α

′th

2

α

′th

6



=



15,5

−13,0

10−6/K

Note that in the off-axis system

α

th

66=0.

Exercise 4.6. The micro-mechanical material parameters of a carbon-epoxy com-

posite are

ELf =411 GPa, ETf =6.6 GPa,

ν

TLf =0.06,

ν

LTf =0.35,

α

th

Lf =−1.2 10−61/K,

α

th

Tf =27.3 10−61/K,

Em=5.7 GPa,

ν

m=0.316,

α

m=45 10−61/K, vf=0.5

The experimental tested values of the lamina are

EL=208.6 GPa, ET=6.3 GPa,

ν

LT =0.33,

α

th

L=−0.5 10−61/0C,

α

th

T=29.3 10−61/0C,

Predict the lamina values using the micro-mechanical modelling and compare the

calculated and the experimental measured values.

Solution 4.6. Using Eqs. (3.1.27) and (4.1.9)

EL=vfELf + (1−vf)Em=208.35 GPa,

ν

LT =vf

ν

LTf + (1−vf)

ν

m=0.33,

ET=ETfEm/[vfEm+ (1−vf)ETf ] = 6.12 GPa,

α

th

L= [

α

th

LfvfELf +

α

th

m(1−vf)Em]/EL=−0.57 10−61/K,

α

th

T= (

α

th

Tf +

ν

Tf

α

th

Lf)vf+ (1+

ν

m)

α

th

m(1−vf)−

ν

LT

α

th

L=4.43 10−61/K

It can be concluded that the simple rules of mixture providing proper results for lon-

gitudinal material characteristics EL,

ν

LT and

α

th

L. In this case also ETis predicted

quite well but the formula for

α

Tfails to predict the transverse thermal expansion

coefﬁcient with required accuracy. For engineering applications the thermal expan-

sion coefﬁcients should be normally determined by experimental methods.

4.2 Elastic Behavior of Laminates

In Sect. 4.1, stress-strain equations were developed for a single lamina. Mostly im-

portant in engineering applications are isotropic, quasi-isotropic (stochastic distri-

bution of short ﬁbres or particles) and quasi-orthotropic (unidirectional ﬁbre re-

inforced) laminae. Reduced stiffness Qi j , compliances Si j, membrane or in-plane

stiffness Ai j and plate or out-of-plane stiffness Di j were deﬁned. Assuming symme-

try about the midplane of a lamina in-plane and plate responses are uncoupled in the

form of a ﬁrst order theory (linear force-displacement relations).

The mechanics of laminated composite materials is generally studied at two dis-

tinct levels, commonly called micromechanics and macromechanics. In Chap. 3 the

micromechanics was used to study the interaction between the ﬁbres and matrix in a

132 4 Elastic Behavior of Laminate and Sandwich Composites

lamina such that the mechanical behavior of the lamina could be predicted from the

known behavior of the constituents. Micromechanics establishes the relationship be-

tween the properties of the constituents and those of the lamina. All micromechanics

approaches suffer from the problem of measuring the material properties of the con-

stituents and generally require correction factors to correlate with measured lamina

properties. For most engineering design applications an analysis that addressed to

the micro-mechanical level is unrealistic.

At the macro-mechanical level the properties of the individual layers of a lami-

nate are assumed to be known a priori. Macromechanics involves investigation of the

interaction of the individual layers of a laminate with one another and their effects

on the overall response quantities, e.g. elastic stiffness, inﬂuence of temperature and

moisture on the response of laminated composites, etc. Such global response quan-

tities can be predicted well on this level. Thus, the use of macromechanical formu-

lations in designing composite laminates for desired material characteristics is well

established. Macromechanics is based on continuum mechanics, which models each

lamina as homogeneous and orthotropic and ignoring the ﬁbre/matrix interface.

The lamination theory is the mathematical modelling to predict the macro-

mechanical behavior of a laminate based on an arbitrary assembly of homoge-

neous orthotropic laminae. A two-dimensional modelling is most common, a three-

dimensional theory is very complex and should be limited to selected problems, e.g.

the analysis of laminates near free edges.

A real structure generally will not consist of a single lamina. A laminate consist-

ing of more than one lamina bonded together through their thickness, for a single

lamina is very thin and several laminae will be required to take realistic structural

loads. Furthermore the mechanical characteristics of a UD-lamina are very limited

in the transverse direction and by stacking a number of UD-laminae it may be an

optimal laminate for unidirectional loading only. One can overcome this restriction

by making laminates with layers stacked at different ﬁbre angles corresponding to

complex loading and stiffness requirements. To minimize the increasing costs and

weights for such approach one have to optimize the laminae angles. It may be also

useful to stack layers of different composite materials.

4.2.1 General Laminates

In the following section the macro-mechanical modelling and analysis of laminates

will be considered. The behavior of a multidirectional laminate is a function of the

laminae properties, i.e. their elastic moduli, thickness, angle orientations, and the

stacking sequence of the individual layers. The macro-mechanical modelling may

be in the framework of the following assumptions:

•There is a monolithic bonding of all laminae i.e. there is no slip between laminae

at their interface and no special interface layers are arranged between the angle

plies.

4.2 Elastic Behavior of Laminates 133

•Each layer is quasi-ho mogeneous and orthotropic, but the an gle orientations may

be different.

•The strains and displacements are continuous throughout the laminate. The in-

plane displacements and strains vary linearly through the laminate thickness.

We will see that the stacking codes of laminates have a great inﬂuence on the global

mechanical laminate response (Sect. 4.2.3), but there are some rules to guarantee an

optimal global laminate behavior:

•Symmetric laminate stacking yields an uncoupled modelling and analysis of in-

plane and bending/torsion stress-strain relations and avoids distorsions in the pro-

cessing.

•Laminates should be made up of at least three UD-laminae with different ﬁbre

angle orientation.

•The differences of the mechanical properties and the ﬁbre orientations between

two laminae following in the stacking sequence should not be so large that the

so-called interlaminar stresses are small.

•Although it is possible to determine an optimum orientation sequence of lami-

nates for any given load condition, it is more practical from a fabrication stand-

point and from effective experimental lamina testing to limit the number of ﬁbre

orientations to a few speciﬁc laminae types, e.g. ﬁbre orientations of 00,±450

and 900, etc.

Consider a laminate made of nplies shown in Fig. 4.10. Each lamina has a thickness

of h(k),k=1,2,...,n, and we have

✲

✻



✠

x3

x1

x2

Mid-plane x3=0

1

2

n

❄

✻

❄

✻

h

2

h

2

x(2)

3

x(1)

3

❄

❄❄

x(0)

3=−h

2

✻✻

x(n−1)

3

x(n)

3=h

2

.

Fig. 4.10 Laminate made of nsingle layers, coordinate locations

134 4 Elastic Behavior of Laminate and Sandwich Composites

h(k)=x(k)

3−x(k−1)

3,k=1,2,...,nthickness of a lamina,

h=

n

∑

k=1

h(k)thickness of the laminate,

x(k)

3=−h

2+

k

∑

i=1

h(i)distance from the mid-plane,

(4.2.1)

Then

x(n)

3=h

2and x(0)

3=−h

2

are the coordinates of the top and the bottom surface of the laminate,

x(k)

3and x(k−1)

3,k=1,2,...,n

are the location coordinates of the top and the bottom surface of lamina k. Each

layer of a laminate can be identiﬁed by its location in the laminate, its material and

its ﬁbre orientation. The layers of the laminate may be symmetric, antisymmetric

or asymmetric to the midplane x3=0. h(k)and the reduced stiffness Q

Q

Q(k)may be

different for each lamina, but Q

Q

Q(k)is constant for the kth lamina. The following

examples illustrate the laminate code. In Fig. 4.11 the laminate codes for an un-

symmetric laminate with four layers and a symmetric angle-ply laminate with eight

layers are illustrated. A slash sign separates each lamina. The codes in Fig. 4.11

imply that each lamina is made of the same material and is of the same thickness.

Regular symmetric are those laminates which have an odd number of UD-laminae

of equal thicknesses and alternating angle orientations (Fig. 4.12). Since the number

of laminae is odd and symmetry exists at the mid-surface, the 900lamina is denoted

with a bar on the top. The subscript Soutside the code brackets, e.g., in Fig. 4.11 b),

represents that the four plies are repeated in the reverse order.

x3=0x3=0

1

2

3

4

a

θ

=00

θ

=300

θ

=−300

θ

=900

b

θ

=00

θ

=−450

θ

=900

θ

=450

θ

=450

θ

=900

θ

=−450

θ

=00

8

7

6

5

4

3

2

1

[90/−30/30/0][0/−45/90/45/45/90/−45/0]

≡[0/−45/90/45]S

Fig. 4.11 Angle-ply laminates. aunsymmetric 4-layer laminate, bsymmetric 8-layer-laminate

4.2 Elastic Behavior of Laminates 135

x3=0

5

4

3

2

1

θ

=−450

θ

=450

θ

=900

θ

=450

θ

=−450

[−45/45/90/45/−45]≡[−45/45/¯

90]S

Fig. 4.12 Regular symmetric angle-ply laminate: orientation of the midplane

θ

=900

A general laminate has layers of different orientations

θ

with −900≤

θ

≤900.

An angle-ply laminate has ply orientations of

θ

and −

θ

with 00≤

θ

≤900and at

least one lamina has an orientation other than 00or 900. Cross-ply laminates are

those which have only ply orientations of 00and 900.

A laminate is balanced when it consists of pairs of layers with identical thickness

and elastic properties but have +

θ

and −

θ

orientations of their principal material

axes with respect to the laminate reference axes. A balanced laminate can be sym-

metric, antisymmetric or asymmetric

[+

θ

1/−

θ

1/+

θ

2/−

θ

2]Ssymmetric lay-up,

[

θ

1/

θ

2/−

θ

2/−

θ

1]antisymmetric lay-up,

[

θ

1/

θ

2/−

θ

1/−

θ

2]asymmetric lay-up

(4.2.2)

Antisymmetric laminates are a special case of balanced laminates, having the bal-

anced +

θ

and −

θ

pairs of layers symmetrically situated about the middle surface.

Generally each layer of a laminate can have different ﬁbre angles, different thick-

nesses and different composite materials. The inﬂuence of the laminate codes, i.e

the properties and the stacking sequences, on the elastic behavior of laminates will

be considered in Sect. 4.2.3.

4.2.2 Stress-Strain Relations and Stress Resultants

The stiffness matrix of a single lamina referred to the reference system xi,i=1,2,3,

has been formulated in Sect. 4.1.2, Eq. (4.1.26). Extending the assumption of a

plane stress state to laminates with in-plane and out-of-plane loading, the stress-

strain relation (4.1.26) can be rewritten by separating the transverse shear stresses

and strains. The stresses in the kth layer are expressed by means of the reduced

stiffness coefﬁcients Qi j

136 4 Elastic Behavior of Laminate and Sandwich Composites







σ

1

σ

2

σ

6

σ

4

σ

5







(k)

=





Q11 Q12 Q16 0 0

Q21 Q22 Q26 0 0

Q61 Q62 Q66 0 0

0 0 0 Q44 Q45

0 0 0 Q54 Q55







(k)





ε

1

ε

2

ε

6

ε

4

ε

5







(k)

(4.2.3)

or in contracted notation

σ

(k)

i=Q(k)

i j

ε

(k)

j,i,j=1,2,6,

σ

(k)

i=Q(k)

i j

ε

(k)

j,i,j=4,5

(4.2.4)

with (see also 2.1.78)

σ

(k)

3=0,

ε

(k)

3=−1

C33

(C13

ε

1+C23

ε

2+C36

ε

6)(4.2.5)

Q(k)

i j ,i,j=1,2,6 are the reduced stiffness of the kth layer and functions of Q′

i j

(k)and

the ﬁbre orientation angle, the Q(k)

44 ,Q(k)

45 =Q(k)

54 ,Q(k)

55 are identical to the material co-

efﬁcients C(k)

44 ,C(k)

45 =C(k)

54 ,C(k)

55 , which are not reduced by the assumption of a plane

stress state. The discontinuity of Q(k)

i j from layer to layer implies the discontinuity

of the stresses when passing from one lamina to another.

From the assumption of macro-mechanical modelling of laminates it follows that

ε

(x1,x2,x3) =

ε

(x1,x2) + x3

κ

(x1,x2)(4.2.6)

i.e the strains

ε

1,

ε

2,

ε

6vary linearly through the laminate thickness.

ε

(x1,x2)is

the vector of the in-plane or membrane strains and x3

κ

(x1,x2)the vector of ﬂex-

ural strains (bending and twisting).

κ

(x1,x2)is the vector of curvature subjected to

bending and twisting. We shall see later (Sect. 5.4) that there are different curvature

components in the classical and the shear deformation theory of laminates.

The in-plane stress resultant force vector N

N

Nof a laminate follows by summarizing

the adequate vectors of all laminae

N

N=

n

∑

k=1

N

N(k),N

N

NT= [N1N2N6],N

N

N(k)T= [N(k)

1N(k)

2N(k)

6](4.2.7)

By analogy it follows that the resultant moment vector is

M

M=

n

∑

k=1

M

M(k),M

M

MT= [M1M2M6],M

M

M(k)T= [M(k)

1M(k)

2M(k)

6](4.2.8)

The positive directions are corresponding to Figs. 4.5 and 4.6 for a single layer. The

transverse shear resultants given in (4.1.50)

4.2 Elastic Behavior of Laminates 137

Q

Qs(x1,x2) =

n

∑

k=1

Q

Qs(k),Q

Q

QsT= [Qs

1Qs

2],Q

Q

Qs(k)T= [Qs(k)

1Qs(k)

2](4.2.9)

Equations (4.2.4) and (4.2.6) yield

σ

(k)=Q

Q

Q(k)

ε

=Q

Q

Q(k)(

ε

+x3

κ

)(4.2.10)

and the resultants N

N

Nand M

M

Mfor the laminate are (k=1,2,...,n)

N

N(k)=Z

(h(k))

σ

(k)dx3=

σ

(k)h(k),h(k)=x(k)

3−x(k−1)

3,

N

N=

n

∑

k=1

σ

(k)h(k)

(4.2.11)

and

M

M(k)=Z

(h(k))

σ

(k)x3dx3=

σ

(k)1

2x(k)2

3−x(k−1)2

3=

σ

(k)h(k)x(k),

M

M=

n

∑

k=1

σ

(k)h(k)x(k)

(4.2.12)

with

x(k)=1

2x(k)

3+x(k−1)

3

For each layer the membrane strains

ε

1,

ε

2,

ε

6, the curvatures

κ

1,

κ

2,

κ

6and the

reduced stiffness Q(k)

11 ,Q(k)

12 ,Q(k)

16 ,Q(k)

22 ,Q(k)

26 ,Q(k)

66 are constant through each thickness

h(k)and (4.2.11) and (4.2.12) reduces to:

N

N=





n

∑

k=1

Q

Q(k)

x(k)

3

Z

x(k−1)

3

dx3





ε

+





n

∑

k=1

Q

Q(k)

x(k)

3

Z

x(k−1

3)

x3dx3





κ

=A

A

ε

+B

B

κ

,

M

M=





n

∑

k=1

Q

Q(k)

x(k)

3

Z

x(k−1)

3

x3dx3





ε

+





n

∑

k=1

Q

Q(k)

x(k)

3

Z

x(k−1)

3

x2

3dx3





κ

=B

B

ε

+D

D

κ

(4.2.13)

A

A,B

B

B,D

D

Dare the extensional, coupling and bending stiffness matrices, respectively.

From (4.2.4) and (4.2.9), the relations for the transverse shear resultants are

138 4 Elastic Behavior of Laminate and Sandwich Composites

Q

Qs=

n

∑

k=1

C

Cs(k)

x(k)

3

Z

x(k−1)

3

σ

s(k)dx3=





n

∑

k=1

C

Cs(k)

x(k)

3

Z

x(k−1)

3

dx3





γ

s=A

A

As

γ

s(4.2.14)

with

C

Cs=C44 C45

C54 C55 ,

σ

s=

σ

4

σ

5,

γ

s=

ε

4

ε

5,A

A

As=A44 A45

A54 A55 

Equation (4.2.14) is a ﬁrst approach and consists of taking the transverse shear strain

independent of the coordinate x3.A

A

Asis the transverse shear stiffness matrix. An

improvement is possible by replacing the transverse shear stiffness As

i j by (kA)s

i j.

ks

i j are so called shear correction factors (Sect. 5.4). The elements of the matrices

A

A,B

B

B,D

D

D,A

A

Asare

Ai j =

n

∑

k=1

Q(k)

i j x(k)

3−x(k−1)

3=

n

∑

k=1

Q(k)

i j h(k),i,j=1,2,6,

Bi j =1

2

n

∑

k=1

Q(k)

i j x(k)

3

2−x(k−1)

3

2=

n

∑

k=1

Q(k)

i j x(k)

3h(k),

Di j =1

3

n

∑

k=1

Q(k)

i j x(k)

3

3−x(k−1)

3

3=

n

∑

k=1

Q(k)

i j x(k)

3

2+h(k)2

12 !h(k),

As

i j =

n

∑

k=1

C(k)

i j x(k)

3−x(k−1)

3=

n

∑

k=1

C(k)

i j h(k),i,j=4,5

(4.2.15)

The constitutive equation for laminates including extensional, bending/torsion and

transverse shear strains is the superposition from the so-called classical equations

for N

N

Nand M

M

Mand the equation that involves the transverse shear resultant Q

Q

Qs. The

constitutive equation can be written in the following contracted hypermatrix form





N

M

Q

Qs

=



A

A B

B

B0

0

B

B D

D

D0

0

0 0

0

0A

A

As





ε

κ

γ

s

(4.2.16)

The stiffness Q(k)

i j and C(k)

i j in (4.2.15) referred to the laminate’s global reference

coordinate system xi,i=1,2,3, are given in Table 4.2 as functions of the Q′

i j

(k)and

in (4.2.17) as functions of the C′

i j

(k)referred to the material principal directions of

each lamina (k)

C44 =C′

44c2+C′

55s2,

C45 = (C′

55 −C′

44)sc,

C55 =C′

44s2+C′

55c2

(4.2.17)

Equation (4.2.16) illustrates the coupling between stretching and bending/twisting

of a laminate, i.e. in-plane strains result in in-plane resultants but also bending

and/or torsion moments and vice versa. Since there are no coupling effects with

4.2 Elastic Behavior of Laminates 139

the transverse shear strains or shear resultants we consider the in-plane and ﬂexural

simultaneous equations, (4.2.18), separately





N

···

M

M

=





A

A.

.

.B

B

. . . .

B

B.

.

.D

D









ε

···

κ



,

or







N1

N2

N6

···

M1

M2

M6







=







A11 A12 A16

.

.B11 B12 B16

A12 A22 A26

.

.B12 B22 B26

A16 A26 A66

.

.B16 B26 B66

. . . . . . . . . . . . . .. . . . . . . . . .

B11 B12 B16

.

.D11 D12 D16

B12 B22 B26

.

.D12 D22 D26

B16 B26 B66

.

.D16 D26 D66













ε

1

ε

2

ε

6

···

κ

1

κ

2

κ

6







(4.2.18)

The following steps are necessary for analyzing a laminated composite subjected to

forces and moments:

•Calculate the values of the reduced stiffness Q′

i j for each lamina kusing the four

elastic moduli, EL,ET,

ν

LT,GLT (4.1.2) and (4.1.3).

•Calculate the values of the transformed reduced stiffness Qi j for each lamina k

(Table 4.2).

•Knowing the thickness h(k)of each lamina kcalculate the coordinates x(k)

3,x(k−1)

3

to the top and the bottom surface of each ply.

•Calculate all Ai j ,Bi j and Di j from (4.2.15).

•Substitute the calculated stiffness and the applied resultant forces and moments

in (4.2.18) and calculate the midplane strains

ε

iand curvatures

κ

i.

•Calculate the global strains

ε

(k)in each lamina using (4.2.6) and then the global

stresses

σ

(k)for each lamina kusing (4.2.10).

•Calculate the local strains

ε

,(k)and the local stresses

σ

,(k)for each lamina kusing

Table 4.1

The inverted relation (4.2.18) leads to the compliance hypermatrix for the in-plane

and ﬂexural resultants



ε

κ

=a

a

a b

b

c

c d

d

dN

N

M

M,a

a

a b

b

c

c d

d

d=A

A

A B

B

C

C D

D

D−1

(4.2.19)

The compliance submatrices a

a

a,b

b

b,c

c

c,d

d

dfollow from the stiffness submatrices A

A

A,B

B

B,D

D

D.

With

N

N=A

A

ε

+B

B

κ

(4.2.20)

it follows that

140 4 Elastic Behavior of Laminate and Sandwich Composites

ε

=A

A

A−1(N

N

N−B

B

κ

)(4.2.21)

and using (4.2.13)

M

M=B

B

BA

A

A−1N

N

N−(B

B

BA

A

A−1B

B

B−D

D

D)

κ

(4.2.22)

The ﬁrst result is a mixed-type constitutive equation





ε

···

M

M

=





A

A∗.

.

.B

B

B∗

. . . . . .

C

C∗.

.

.D

D

D∗









N

···

κ



,

A

A∗=A

A

A−1,B

B

B∗=−A

A

A−1B

B

B,

C

C∗=B

B

BA

A

A−1=−B

B

B∗T,D

D

D∗=D

D

D−B

B

BA

A

A−1B

B

(4.2.23)

With

κ

=D

D

D∗−1M

M

M−D

D

D∗−1C

C

C∗N

N

N(4.2.24)

it follows that

ε

= (A

A

A∗−B

B

B∗D

D

D∗−1C

C

C∗)N

N

N+B

B

B∗D

D

D∗−1M

M

M(4.2.25)

and the compliance relation has in contracted notation the form





ε

···

κ



=





a

a.

.

.b

b

. . . .

c

c.

.

.d

d









N

···

M

M

,

a

a=A

A

A∗−B

B

B∗D

D

D∗−1C

C

C∗=A

A

A∗+B

B

B∗D

D

D∗−1B

B

B∗T,

b

b=B

B

B∗D

D

D∗−1,

c

c=−D

D

D∗−1C

C

C∗=D

D

D∗−1B

B

B∗T=b

b

bT,

d

d=D

D

D∗−1

(4.2.26)

Equations (4.2.18) and (4.2.26) are inverse relations of the constitutive equation for

the resultants and the strains of a laminate. The elements of the submatrices A

A

A,B

B

B,D

D

and a

a

a,b

b

b,c

c

c,d

d

dare functions of the geometry, the material properties and the structure

of a laminate and therefore averaged effective elastic laminate moduli. The subma-

trices A

A

A,B

B

B,D

D

D,a

a

a,d

d

dare symmetric submatrices. That is not the case for the submatri-

ces b

b

band c

c

cbut with c

c

c=b

b

bTthe compliance hypermatrix is symmetric. The coupling

of different deformation states is a very important quality of the constitutive equa-

tions of laminates. In the general case, considered in this section, all coupling effects

are present. Figure 4.13 illustrates for example the coupling states for the resultant

force N1and the resultant moment M1.

In the next Sect. 4.2.3 we shall see that the stacking sequence of a laminate in-

ﬂuences the coupling behavior of loaded laminates. In engineering applications it is

desired to specify the stacking sequence such that a number of coefﬁcients of the

stiffness matrix will be zero and undesirable couplings between stretching, bending

4.2 Elastic Behavior of Laminates 141







N1

N2

N6

M1

M2

M6







=





A11 A12 A16 B11 B12 B16

A22 A26 ·B22 B26

A66 · · B66

D11 D12 D16

S Y M ·D22 D26

· · D66













ε

1

ε

2

ε

6

κ

1

κ

2

κ

6







N1

M1

N1

M1

A11 A12 A16

D11 D12 D16

B11 B12 B16

strain

ε

1strain

ε

2shear

ε

6

curvature

κ

1curvature

κ

2twisting

κ

6

curvature

κ

1curvature

κ

2twisting

κ

6

strain

ε

1strain

ε

2shear

ε

6

Fig. 4.13 Coupling of strain states: Inﬂuence of the stiffness A1j,D1jand B1j(j=1,2,6)on the

strains

ε

jand the curvature

κ

jof the middle surface of a general laminate loading with N1or M1.

In each case 6 deformation states of N1and M1have to be superposed.

142 4 Elastic Behavior of Laminate and Sandwich Composites

and/or twisting will be avoided. But it is rather difﬁcult to specify an optimum stack-

ing sequence without detailed information about the performance requirements.

Engineering composite design has continued to evolve over many years. Most

early applications of composite materials were aimed particularly at weight reduc-

tion. Metals were replaced by composites with little or no emphasis placed on tailor-

ing the composite properties. Engineering design created quasi-isotropic laminates

that largely suppressed the directional material properties of unidirectional lami-

nae and made the laminate material response similar to that of isotropic materials,

e.g. of metals. We shall see in the following discussion that one of such quasi-

isotropic laminate is given if it has equal percentages of 00,+450,−450and 900

layers placed symmetrically with respect to the laminate mid-plane. Quasi-isotropic

laminates have elastic properties that are independent of the direction in the plane

of the laminate, like traditional isotropic engineering materials. Therefore, quasi-

isotropic laminates were in the ﬁrst applications of composites a convenient replace-

ment for steel or alloys in weight critical applications, e.g. in aerospace industries.

Weight saving could be achieved by simple replacing the isotropic metal with a

similar stiffness laminate that was lighter and probably stronger. If we compare a

graphite/epoxy laminate with an quasi-isotropic stacking sequence of laminae and

aluminium we ﬁnd nearly the same elastic moduli, e.g. E≈70 GPa, but the density

values

ρ

and the speciﬁc stiffness E/

ρ

differ signiﬁcantly. The speciﬁc stiffness

of graphite/epoxy laminate can be twice that of aluminium. Such applications of

quasi-isotropic laminates required a minimal amount of redesign effort and there-

fore minimal changes in structural conﬁguration.

By the time the number of design engineers which are trained in composite ma-

terials increased and the tailoring of material properties gained more acceptance.

To maximize the utility of the non-isotropic nature of laminates, the inﬂuence of

the stacking sequence on the structural behavior must be investigated in detail and

optimized. Particularly the coupling effects of in-plane and out-of-plane responses

affect the effort of laminate structural analysis.

4.2.3 Laminates with Special Laminae Stacking Sequences

Now special cases of laminates which are important in the engineering design of

laminated structures will be introduced. Quite often the design of laminates is done

by using laminae that have the same constituents, the same thicknesses, etc. but

have different orientations of their ﬁbre reinforcement direction with respect to the

global reference system of the laminate and a different stacking sequence of these

layers. In other cases layers with different materials or thicknesses are bonded to a

laminate. The stacking sequence of the layers may result in reducing the coupling

of normal and shear forces, of bending and twisting moments etc. It can simplify

the mechanical analysis but also gives desired mechanical performance. In the fol-

lowing, the mechanical behavior of special symmetric and unsymmetric laminates

are considered.

4.2 Elastic Behavior of Laminates 143

4.2.3.1 Symmetric Laminates

A laminate is called symmetric if the material, angle and thickness of laminae are

the same above and below the midplane, i.e. two symmetric arranged layers to the

midplane have the same reduced stiffness matrix Q

Q

Q(k)≡Q

Q

Q(k′)and the same thickness

h(k)≡h(k′)for opposite coordinates ¯x(k)and ¯x(k′)=−¯x(k)(Fig. 4.14). It follows that

the coefﬁcients Bi j of the coupling submatrix B

B

Bare zero and there are no coupling

relations of stretching and bending

Bi j =1

2

n

∑

k=1

Q(k)

i j x(k)

3

2−x(k−1)

3

2

=1

2

n

∑

k=1

Q(k)

i j x(k)

3+x(k−1)

3x(k)

3−x(k−1)

3

=

n

∑

k=1

Q(k)

i j x(k)

3h(k)=0,i,j=1,2,6

(4.2.27)

With x(k′)

3=−x(k)

3the sum above have two pairs of equal absolute values but oppo-

site signs. Thus the ABD

ABD

ABD-matrix of symmetric laminates is uncoupled, i.e. all terms

of the coupling submatrix [Bi j]are zero, see following equation

x3=0

n≡1′

1



















k′

k

❄

✻

x(k′)

3

x(k)

3

✻

x3

Fig. 4.14 Symmetric laminate with identical layers kand k′opposite to the middle surface (h(k)=

h(k′),Q

Q

Q(k)=Q

Q

Q(k′))

144 4 Elastic Behavior of Laminate and Sandwich Composites







N1

N2

N6

. .

M1

M2

M6







=







A11 A12 A16

.

. 0 0 0

A12 A22 A26

.

. 0 0 0

A16 A26 A66

.

. 0 0 0

........................

0 0 0 .

.

.D11 D12 D16

0 0 0 .

.

.D12 D22 D26

0 0 0 .

.

.D16 D26 D66













ε

1

ε

2

ε

6

. .

κ

1

κ

2

κ

6







(4.2.28)

The extensional submatrix A

A

Aand the bending submatrix D

D

Dare in the case of sym-

metric angle ply laminates fully populated and we have in-plane normal and shear

strain and out-of-plane bending and torsion couplings. Since the coupling submatrix

B

Bis zero the elastic behavior of symmetric laminates is simpler to analyze than that

of general laminates and symmetric laminates have no tendency to warp as a result

of thermal contractions induced during the composite processing. Some important

special cases of symmetric laminates are:

•Symmetric laminate with isotropic layers

Q(k)

11 =Q(k)

22 =Q(k′)

11 =Q(k′)

22 =E(k)

1−

ν

(k),

Q(k)

12 =Q(k′)

12 =

ν

(k)E(k)

1−

ν

(k),

Q(k)

16 =Q(k)

26 =Q(k′)

16 =Q(k′)

26 =0,

Q(k)

66 =Q(k′)

66 =E(k)

2(1+

ν

(k))=G(k),

Ai j =

n

∑

k=1

Q(k)

i j h(k)

=⇒A11 =A22,A16 =A26 =0,

Di j =

n

∑

k=1

Q(k)

i j h(k) x(k)

3

2+h(k)2

12 !

=⇒D11 =D22,D16 =D26 =0,

(4.2.29)

4.2 Elastic Behavior of Laminates 145







N1

N2

N6

. .

M1

M2

M6







=







A11 A12 0.

.

. 0 0 0

A12 A11 0.

.

. 0 0 0

0 0 A66

.

. 0 0 0

. . . . . . . . . . . . . . .. . . . . . . . .

0 0 0 .

.

.D11 D12 0

0 0 0 .

.

.D12 D11 0

0 0 0 .

.

. 0 0 D66













ε

1

ε

2

ε

6

. .

κ

1

κ

2

κ

6







(4.2.30)

This type of symmetric laminates has no stretching-shearing or bending-torsion

coupling.

•Symmetric cross-ply laminate

A laminate is called a cross-ply laminate or a laminate with specially orthotropic

layers if only 00and 900plies were used. The material principal axes and the

global reference axes are identical. If for example for the kth layer the ﬁbre ori-

entation and the x1-direction of the global reference system coincide, we have

Q(k)

11 ≡Q(k′)

11 =E(k)

1

1−

ν

(k)

12

ν

(k)

21

,Q(k)

22 ≡Q(k′)

22 =E(k)

2

1−

ν

(k)

12

ν

(k)

21

,

Q(k)

12 ≡Q(k′)

12 =

ν

(k)

21 E(k)

1

1−

ν

(k)

12

ν

(k)

21

,Q(k)

16 ≡Q(k′)

16 =0,

Q(k)

66 ≡Q(k′)

66 =G(k)

12 ,Q(k)

26 ≡Q(k′)

26 =0

(4.2.31)

and with (4.2.15) A16 =A26 =0,D16 =D26 =0. The stiffness matrix of the

constitutive equation has an adequate structure as for isotropic layers, but now

A11 6=A22 and D11 6=D22, i.e. the laminate has an orthotropic structure







N1

N2

N6

···

M1

M2

M6







=







A11 A12 0.

.

. 0 0 0

A12 A22 0.

.

. 0 0 0

0 0 A66

.

. 0 0 0

. . . . . . . . . . . . . .. . . . . . . . . .

000.

.

.D11 D12 0

000.

.

.D12 D22 0

000.

.

. 0 0 D66













ε

1

ε

2

ε

6

···

κ

1

κ

2

κ

6







(4.2.32)

Figure 4.15 illustrates examples of symmetric cross-ply laminates. With

A16 =A26 =0,D16 =D26 =0 there is uncoupling between the normal and

shear in-plane forces and also between the bending and the twisting moments.

146 4 Elastic Behavior of Laminate and Sandwich Composites

S S S

2

1

3

900

00

h

a

1

2

3

4

5

6

900

00

900

00

900

h(1)

h(2)

h(3)

h(4)=h(3)

h(5)=h(2)

h(6)=h(1)

1

2

3

400

900

00h

h

b

c

Fig. 4.15 Symmetric cross-ply laminate. a3-layer laminate with equal layer thickness, b6-layer

laminate with equal layer thickness in pairs, c4-layer laminate with equal layer thickness

•Symmetric balanced laminate

A laminate is balanced when it consists of pairs of layers of the same thick-

ness and material where the angles of plies are +

θ

and −

θ

. An example is the

8 - layer-laminate [±

θ

1/±

θ

2/]s. The stiffness coefﬁcients Ai j and Di j will be

calculated from

Ai j =n

∑

k=1

Q(k)

i j h(k),h(k)=h(k′),

θ

(k)=−

θ

(k′)

=⇒A16 =A26 =0,

Di j =1

3

n

∑

k=1

Q(k)

i j x(k)

3

3−x(k−1)

3

3

=

n

∑

k=1

Q(k)

i j x(k)

3

2+h(k)2

12 !h(k)

(4.2.33)

with

h(k)=h(k′),

θ

(k)=

θ

(k′),x(k′)

3=−x(k)

3

and the constitutive equation yields







N1

N2

N6

. .

M1

M2

M6







=







A11 A12 0.

.

. 0 0 0

A12 A22 0.

.

. 0 0 0

0 0 A66

.

. 0 0 0

. . . . . . . . . . . . . . .. . . . . . . . .

0 0 0 .

.

.D11 D12 D16

0 0 0 .

.

.D12 D22 D26

0 0 0 .

.

.D16 D26 D66













ε

1

ε

2

ε

6

. .

κ

1

κ

2

κ

6







(4.2.34)

The fact that the in-plane shear coupling stiffness A16 and A26 are zero is a deﬁn-

ing characteristic of a balanced laminate. In general the bending/twisting cou-

pling stiffness D16 and D26 are not zero unless the laminate is antisymmetric.

4.2 Elastic Behavior of Laminates 147

Summarizing the results on symmetric laminates above, it is most important that all

components of the B

B

B-matrix are identical to zero and the full (6×6)ABD

ABD

ABD - matrix

decouples into two (3×3)matrices, namely

N

N=A

A

ε

,M

M

M=D

D

κ

(4.2.35)

Therefore also the inverse relations degenerates from (6×6)into two (3×3)rela-

tions

ε

=a

a

aN

N

N,

κ

=d

d

dM

M

M(4.2.36)

In these matrix equations a

a

ais the inverse of A

A

Aand d

d

dthe inverse of D

D

a11 =A22A66 −A2

26

Det[A

A

A],d11 =D22D66 −D2

26

Det[D

D

D],

a12 =A26A16 −A12 A66

Det[A

A

A],d12 =D26D16 −D12 D66

Det[D

D

D],

a16 =A12A26 −A22 A16

Det[A

A

A],d16 =D12D26 −D22 D26

Det[D

D

D],

a22 =A11A66 −A2

16

Det[A

A

A],d22 =D11D66 −D2

16

Det[D

D

D],

a26 =A12A16 −A11 A26

Det[A

A

A],d26 =D12D16 −D11 D26

Det[D

D

D],

a66 =A11A22 −A2

12

Det[A

A

A],d66 =D11D22 −D2

12

Det[D

D

D]

(4.2.37)

with

Det[X

X

X] = X11(X22X66 −X2

26)−X12(X12X66 −X26X16)

+X16(X12X26 −X22X16);Xi j =Ai j ,Di j

For the special symmetric stacking cases one can ﬁnd

1. Isotropic layers

a11 =a22,a16 =a26 =0,A11 =A22 ,A16 =A26 =0,

d11 =d22,d16 =d26 =0,D11 =D22 ,D16 =D26 =0,

A11 =Eh

1−

ν

,A12 =

ν

A11,A66 =E h

2(1+

ν

),

D11 =Eh3

12(1−

ν

2),D12 =

ν

D11,D66 =E h3

24(1+

ν

)

with

h=

n

∑

k=1

h(k)

148 4 Elastic Behavior of Laminate and Sandwich Composites

2. Cross-play layers

a11 =A22

A11A22 −A2

12

,d11 =D22

D11D22 −D2

12

,

a12 =−A12

A11A22 −A2

12

,d12 =−D12

D11D22 −D2

12

,

a22 =A11

A11A22 −A2

12

,d22 =D11

D11D22 −D2

12

,

a66 =1

A66

,d66 =1

D66

3. Balanced layers

The ai j are identical to cross-ply layers. The di j are identical to the general sym-

metric case.

4.2.3.2 Antisymmetric Laminates

A laminate is called antisymmetric if the material and thickness of the laminae are

the same above and below the midplane but the angle orientations at the same dis-

tance above and below of the midplane are of opposite sign, i.e two symmetric ar-

ranged layers to the midplane with the coordinates ¯x(k)and x(k′)=−x(k)having the

same thickness h(k)=h(k′)and antisymmetric orientations

θ

(k)and

θ

(k′)=−

θ

(k)

(Fig. 4.14).

•Antisymmetric cross-ply laminate

Antisymmetric cross-ply laminates consist of 00and 900laminae arranged in

such a way that for all 00-laminae (k)at a distance ¯x(k)from the midplane there

are 900-laminae (k′)at a distance ¯x(k′)=−¯x(k)and vice versa. By deﬁnition

these laminates have an even number of plies. The reduced stiffness fulﬁlls the

conditions

Q(k)

11 =Q(k′)

22 ,Q(k)

22 =Q(k′)

11 ,Q(k)

12 =Q(k′)

12 ,

Q(k)

16 =Q(k′)

16 =Q(k)

26 =Q(k′)

26 =0,

which yield considering (4.2.15) and the 00and 900layers have the same thick-

ness

A11 =A22,A16 =A26 =0,

B11 =−B22,B12 =B16 =B26 =B66 =0,

D11 =D22,D16 =D26 =0

(4.2.38)

and the constitutive equation has the form

4.2 Elastic Behavior of Laminates 149







N1

N2

N6

. .

M1

M2

M6







=







A11 A12 0.

.

.B11 0 0

A12 A11 0.

.

. 0 −B11 0

0 0 A66

.

. 0 0 0

...........................

B11 0 0 .

.

.D11 D12 0

0−B11 0.

.

.D12 D11 0

0 0 0 .

.

. 0 0 D66













ε

1

ε

2

ε

6

. .

κ

1

κ

2

κ

6







(4.2.39)

The constitutive equation (4.2.39) shows that antisymmetric cross-ply laminates

only have a tension/bending coupling. It is important to note that the coupling

coefﬁcient B11 approaches zero as the number of plies increases for a constant

laminate thickness since it is inversely proportional to the total number of layers.

•Antisymmetric balanced laminate

Antisymmetric balanced laminates consist of pairs of laminae (k)and (k′)at a

distance ¯x(k)and ¯x(k′)=−¯x(k)with the same material and thickness but orien-

tations

θ

(k)and

θ

(k′)=−

θ

(k). Examples of these laminates are [

θ

1/−

θ

1],

[

θ

1/

θ

2/−

θ

2/−

θ

1], etc. As for all balanced laminates A16 =A26 =0 and with

Di j =1

3

n

∑

k=1

Q(k)

i j x(k)

3

3−x(k−1)

3

3(4.2.40)

and x(k)

3

3−x(k−1)

3

3=x(k′)

3

3−x(k′−1)

3

3,

Q(k)

16 =−Q(k′)

16 ,Q(k)

26 =−Q(k′)

26

it follows that

D16 =D26 =0

Note that x(k)

3=−x(k′)

3,h(k)=h(k′)and

Q(k)

11 =Q(k′)

11 ,Q(k)

22 =Q(k′)

22 ,Q(k)

12 =Q(k′)

12 ,

Q(k)

66 =Q(k′)

66 ,Q(k)

16 =−Q(k′)

16 ,Q(k)

26 =−Q(k′)

26

Equation (4.2.15) yields B11 =B22 =B12 =B66 =0. Balanced antisymmetric

laminates have no in-plane shear coupling and also no bending/twisting cou-

pling but a co upling of stretching/twisting and bending/shearing. T he constitutive

equation has the following structure

150 4 Elastic Behavior of Laminate and Sandwich Composites







N1

N2

N6

···

M1

M2

M6







=







A11 A12 0.

.

. 0 0 B16

A12 A22 0.

.

. 0 0 B26

0 0 A66

.

.B16 B26 0

. . . . . . . . . . . . . .. . . . . . . . . .

0 0 B16

.

.D11 D12 0

0 0 B26

.

.D12 D22 0

B16 B26 0.

.

. 0 0 D66













ε

1

ε

2

ε

6

···

κ

1

κ

2

κ

6







(4.2.41)

4.2.3.3 Stiffness Matrices for Symmetric and Unsymmetric Laminates in

Engineering Applications

Table 4.4 summarizes the stiffness matrices for symmetric and unsymmetric lam-

inates which are used in engineering applications. Symmetric laminates avoid the

stretching/bending coupling. But certain applications require the use of nonsym-

metric laminates. If possible symmetric balanced laminates should be used. The

bending and shearing couplings are eliminated and one can show that for symmet-

ric laminates with a constant total thickness hthe values of the bending or ﬂexural

stiffness D16 and D26 decrease with an increasing number of layers and approach

zero for k−→ ∞. If the stiffness Ai j ,Bi j and Di j are calculated, the compliances

ai j,bi j ,ci j =bT

i j,di j follow from (4.2.26) or for symmetric laminates from (4.2.37).

The experimental identiﬁcation of the compliances is simpler than for the stiffness

parameters.

The coupling stiffness Bi j and A16,A26,D16 and D26 complicate the analysis of

laminates. To minimize coupling effects symmetric balanced laminates should be

created with a ﬁne lamina distribution. Then all Bi j and the A16 ,A26 are identical

to zero and the D16 ,D26 couplings are relatively low because of the ﬁne lamina

distribution. Whenever possible it is recommended to limit the number of ﬁbre ori-

entations to a few speciﬁc one, that are 00,±450,900to minimize the processing and

experimental testing effort and to select a symmetric and balanced lay-up with a ﬁne

lamina interdispersion in order to eliminate in-plane and out-of-plane coupling and

the in-plane tension/shearing coupling and to minimize torsion coupling.

There is furthermore a special class of quasi-isotropic laminates. The layers of

the laminate can be arranged in such a way that the laminate will behave as an

isotropic layer under in-plane loading. Actually, such laminates are not isotropic,

because under transverse loading normal to the laminate plane and under interlami-

nar shear their behavior is different from real isotropic layer. That is why one use the

notation quasi-isotropic layer. Because all quasi-isotropic laminates are symmetric

and balanced the shear coupling coefﬁcients A16,A26 are zero. It can be checked in

general any laminate with a lay-up of

4.2 Elastic Behavior of Laminates 151

Table 4.4 Stiffness matrices for symmetric and unsymmetric laminates

Symmetric laminate Unsymmetric laminate

Isotropic layers Balanced laminate

A11 A12 0 0 0 0

A12 A11 0 0 0 0

0 0 A66 000

0 0 0 D11 D12 0

0 0 0 D12 D11 0

0 0 0 0 0 D66

A11 A12 0B11 B12 B16

A12 A22 0B12 B22 B26

0 0 A66 B16 B26 B66

B11 B12 B16 D11 D12 D16

B12 B22 B26 D12 D22 D26

B16 B26 B66 D16 D26 D66

Eq. (4.2.30)

Cross-ply laminate Antimetric balanced laminate

A11 A12 0 0 0 0

A12 A22 0 0 0 0

0 0 A66 000

0 0 0 D11 D12 0

0 0 0 D12 D22 0

0 0 0 0 0 D66

A11 A12 0 0 0 B16

A12 A22 0 0 0 B26

0 0 A66 B16 B26 0

0 0 B16 D11 D12 0

0 0 B26 D12 D22 0

B16 B26 0 0 0 D66

Eq. (4.2.32) Eq. (4.2.41)

Balanced laminate Cross-ply

A11 A12 0 0 0 0

A12 A22 0 0 0 0

0 0 A66 000

0 0 0 D11 D12 D16

0 0 0 D12 D22 D26

0 0 0 D16 D26 D66

A11 A12 0B11 0 0

A12 A11 0 0 −B11 0

0 0 A66 000

B11 0 0 D11 D12 0

0−B11 0D12 D11 0

0 0 0 0 0 D66

Eq. (4.2.34) Eq. (4.2.39)

Angle-ply laminate Cross-ply

(approximate solution k→∞)

A11 A12 A16 000

A12 A22 A26 000

A16 A26 A66 000

0 0 0 D11 D12 D16

0 0 0 D12 D22 D26

0 0 0 D16 D26 D66

A11 A12 0 0 0 0

A12 A22 0 0 0 0

0 0 A66 0 0 0

0 0 0 D11 D12 0

0 0 0 D12 D11 0

0 0 0 0 0 D66

Eq. (4.2.28)

0/

π

n/2

π

n/ . . ./ (n−1)

π

nS

or 

π

n/2

π

n/.../

π

S

152 4 Elastic Behavior of Laminate and Sandwich Composites

is quasi-isotropic for any integer ngreater than 2. The simplest types are laminates

with the following lay-up

[0/60/−60]S,n=3,1200≡ −600

and

[0/+45/−45/90]S,n=4,1350≡ −450

Summarizing the mechanical performance of laminates with special laminae

stacking sequences which are used in laminate design, we have considered the fol-

lowing classiﬁcation:

1. General laminates

The stacking sequence, the thickness, the material and the ﬁbre orientations of

all laminae is quite general. All extensional stiffness Bi j are not zero.

2. Symmetric laminates

For every layer to one side of the laminate reference surface there is a corre-

sponding layer to the other side of the reference surface at an equal distance and

with identical thickness, material and ﬁbre orientation. All coupling stiffness Bi j

are zero.

3. Antisymmetric laminates

For every layer to one side of the laminate reference surface there is a corre-

sponding layer to the other side of the reference surface at an equal distance,

with identical thickness and material, but opposite ﬁbre orientation. The stiffness

A16,A26 ,D16 and D26 are zero.

4. Balanced laminates

For every layer with a speciﬁed thickness, speciﬁc material properties and spe-

ciﬁc ﬁbre orientation there is another layer with identical thickness and material

properties, but opposite ﬁbre orientation anywhere in the laminate, i.e. the corre-

sponding layer with opposite ﬁbre orientation does not have to be on the opposite

side of the reference surface, nor immediately adjacent to the other layer nor any-

where particular. A balanced laminate can be

•General or unsymmetric: A16 =A26 =0

•Symmetric A16 =A26 =0, Bi j =0

•Antisymmetric A16 =A26 =0,D16 =D26 =0

An antisymmetric laminate is a special case of a balanced laminate, having its

balanced ±pairs of layers symmetrically situated to the middle surface.

5. Cross-ply laminates

Every layer of the laminate has its ﬁbers oriented at either 00or 900. Cross-ply

laminates can be

•General or unsymmetric: A16 =A26 =0, B16 =B26 =0, D16 =D26 =0

•Symmetric A16 =A26 =0, D16 =D26 =0, Bi j =0

•Antisymmetric A16 =A26 =0,D16 =D26 =0, B12 =B16 =B26 =B66 =0,

A11 =A22,B11 =B22 ,D11 =D22

4.2 Elastic Behavior of Laminates 153

Symmetric cross-ply laminates are orthotropic with respect to both in-plane and

bending behavior and all coupling stiffness are zero.

6. Quasi-isotropic laminates

For every laminate with a symmetric lay-up of

[0/

π

n/ . . ./ (n−1)

π

n]Sor [0/

π

n/2

π

n/.../

π

]S

the in-plane stiffness are identical in all directions. Because all these quasi-

isotropic laminates are also balanced we have A11 =A22 =const,A12 =const,

A16 =A26 =0, Bi j ≡0,Di j 6=0

7. Laminates with isotropic layers

If isotropic layers of possible different materials properties and thicknesses are

arranged symmetrically to the middle surface the laminate is symmetric isotropic

and we have A11 =A22,D11 =D22 ,A16 =A26 =0,D16 =D26 =0, Bi j =0, i.e.

the mechanical performance is isotropic.

For symmetric laminates the in-plane and ﬂexural moduli can be deﬁned with

help of effective engineering parameters. We start with (4.2.26). a

a

a,b

b

b,c

c

c=b

b

bT,d

d

dare

the extensional compliance matrix, coupling compliance matrix and bending com-

pliance matrix, respectively. For a symmetric laminate B

B

B=0

0

0 and it can be shown

that a

a

a=A

A

A−1and d

d

d=D

D

D−1. The in-plane and the ﬂexural compliance matrices a

a

aand

d

dare uncoupled but generally fully populated

ε

=a

a

aN

N

N,

κ

=d

d

dM

M

M(4.2.42)

Equations (4.2.42) lead to effective engineering moduli for symmetric laminates.

1. Effective in-plane engineering moduli EN

1,EN

2,GN

12,

ν

N

12:

Substitute N16=0,N2=N6=0 in

ε

=a

a

aN

N

Nas





ε

1

ε

2

ε

6

=



a11 a12 a16

a12 a22 a26

a16 a26 a66 





N1

0

0

(4.2.43)

which gives

ε

1=a11N1

and the effective longitudinal modulus EN

1is

EN

1≡

σ

1

ε

1

=N1/h

a11N1

=1

ha11

(4.2.44)

In an analogous manner with N1=0,N26=0,N6=0 or N1=N2=0,N66=0,

the effective transverse modulus EN

2or the effective shear modulus GN

12 are

EN

2≡

σ

2

ε

2

=N2/h

a22N2

=1

ha22

,(4.2.45)

154 4 Elastic Behavior of Laminate and Sandwich Composites

GN

12 ≡

σ

6

ε

6

=N6/h

a66N6

=1

ha66

(4.2.46)

The effective in-plane Poisson’s ratio

ν

N

12 can be derived in the following way.

With N16=0,N2=N6=0 (4.2.43) yields

ε

2=a12N1,

ε

1=a11N1and

ν

12 is de-

ﬁned as

ν

N

12 =−

ε

2

ε

1

=−a12N1

a11N1

=−a12

a11

(4.2.47)

The Poisson’s ratio

ν

N

21 can be derived directly by substituting N1=N6=0,

N26=0 in (4.2.42) and deﬁne

ν

N

21 =−

ε

1/

ε

2or by using the reciprocal relationship

ν

N

12/EN

1=

ν

N

21/EN

2. In both cases

ν

N

21 is given as

ν

N

21 =−a12

a22

(4.2.48)

The effective in-plane engineering moduli can be also formulated in terms of the

elements of the A

A

A-matrix

EN

1=A11A22 −A2

12

A22h,EN

2=A11A22 −A2

12

A11h,GN

12 =A66

h,

ν

N

12 =A12

A22

,

ν

N

21 =A12

A11

(4.2.49)

2. Effective ﬂexural engineering moduli EM

1,EM

2,GM

12,

ν

M

12,

ν

M

21:

To deﬁne the effective ﬂexural moduli we start with

κ

=d

d

dM

M

M. Apply M16=0,

M2=0,M6=0 and substitute in the ﬂexural compliance relation to give

κ

1=d11M1=M1

EM

1I,I=h3

12

and the effective ﬂexural longitudinal modulus EM

1is

EM

1=12M1

κ

1h3=12

h3d11

(4.2.50)

Similarly, one can show that the other ﬂexural elastic moduli are given by

EM

2=12

h3d22

,

ν

M

12 =−d12

d11

,GM

12 =12

h3d66

,

ν

M

21 =−d12

d22

(4.2.51)

Flexural Poisson’s ratios also have a reciprocal relationship

ν

M

12

EM

1

=

ν

M

21

EM

2

(4.2.52)

In terms of the elements of the D

D

Dmatrix we ﬁnd

4.2 Elastic Behavior of Laminates 155

EM

1=12(D11D22 −D2

12)

D22h3,EM

2=12(D11D22 −D2

12)

D11h3,

GM

12 =12D66

h3,

ν

M

12 =D12

D22

,

ν

M

21 =D12

D11

(4.2.53)

Consider unsymmetric laminates, the laminate stiffness or compliance matrices

are not uncoupled and therefore it is not meaningful to use effective engineering

laminate moduli.

4.2.4 Stress Analysis

Laminate stresses may be subdivided into in-plane stresses, which are calculated

below with the classical assumption of linear strain functions of x3, and the through-

the-thickness stresses, which are calculated approximately by integration of the

equilibrium conditions. Taking into account the assumptions of macro-mechanical

modelling of laminates the strains

ε

1,

ε

2,

ε

6vary linearly across the thickness of the

laminate

ε

(x1,x2,x3) =

ε

(x1,x2) + x3

κ

(x1,x2),h=

n

∑

k=1

h(k)(4.2.54)

These global strains can be transformed to the local strains in the principal material

directions of the kth layer through the transformation equations (Table 4.1)

ε

′(k)=T

T

εε

ε

(k),x(k−1)

3≤x3≤x(k)(4.2.55)

If the strains are known at any point along the thickness of the laminate, the stress-

strain relation (Table 4.2) calculates the global stress in each lamina

σ

(k)=Q

Q

Q(k)

ε

(k)=Q

Q

Q(k)(

ε

+x3

κ

),x(k−1)

3≤x3≤x(k)(4.2.56)

By applying the transformation equation for the stress vector (Table 4.1) the stresses

expressed in the principal material axes can be calculated

σ

′(k)=T

T

σσ

σ

(k)(4.2.57)

Starting from the strains

ε

′(k), the stresses in the kth layer are expressed as follows

σ

′(k)=Q

Q

Q′(k)

ε

′(k)(4.2.58)

From (4.2.56), the stresses vary linearly through the thickness of each lamina and

may jump from lamina to lamina since the reduced stiffness matrix Q

Q

Q(k)changes

from ply to ply since Q

Q

Q(k)depends on the material and orientation of the lamina (k).

Figure 4.16 illustrates qualitatively the stress jumps of the membrane stresses

σ

(k)

M

156 4 Elastic Behavior of Laminate and Sandwich Composites

✻

❄

h

Q(3)

Q(2)

Q(1)

h(1)

h(2)

h(3)

❄

✻

σ

(k)

6M =Q(k)

66

ε

6

σ

(k)

2M =Q(k)

22

ε

2

σ

(k)

1M =Q(k)

11

ε

1

σ

(k)

6B =Q(k)

66 x3

κ

6

σ

(k)

2B =Q(k)

22 x3

κ

2

σ

(k)

1B =Q(k)

11 x3

κ

1

σ

(k)

6=

σ

(k)

6M +

σ

(k)

6B

σ

(k)

2=

σ

(k)

2M +

σ

(k)

2B

σ

(k)

1=

σ

(k)

1M +

σ

(k)

1B

✟



✟

✟



Fig. 4.16 Qualitatively variation of the in-plane membrane stresses

σ

iM, the bending stresses

σ

iBand the total stress

σ

ithrough the thickness of the laminate. Assumptions h(1)=h(3),

Q(1)=Q(3)<Q(2),i=1,2,6

which follow from the in-plane resultants N

N

Nand are constant through each lamina

and the bending/torsion stresses

σ

(k)

Bfollowing from the moment resultants M

M

Mand

vary linearly through each ply thickness. The transverse shear stresses

σ

4,

σ

5follow

for a plane stress state assumptions that is in the framework of the classical laminate

theory, Sect. 5.1, not from a constitutive equation but as for the single layer, (4.1.56),

by integration of the equilibrium equations. For any lamina mof the laminate by

analogy to (4.1.57) can be established

σ

(m)

5(x3) = −

m−1

∑

k=1(x(k)

3

Z

x(k−1)

3

Q(k)

11

∂

x1

(

ε

1+x3

κ

1) + Q(k)

12

∂

x1

(

ε

2+x3

κ

2)

+Q(k)

16

∂

x1

(

ε

6+x3

κ

6) + Q(k)

61

∂

x2

(

ε

1+x3

κ

1)

+Q(k)

62

∂

x2

(

ε

2+x3

κ

2) + Q(k)

66

∂

x2

(

ε

6+x3

κ

6)dx3)

−

x3

Z

x(m−1)

3

Q(m)

11

∂

x1

(

ε

1+x3

κ

1) + Q(m)

12

∂

x1

(

ε

2+x3

κ

2)

+Q(m)

16

∂

x1

(

ε

6+x3

κ

6) + Q(m)

61

∂

x2

(

ε

1+x3

κ

1)

+Q(m)

62

∂

x2

(

ε

2+x3

κ

2) + Q(m)

66

∂

x2

(

ε

6+x3

κ

6)dx3,(4.2.59)

σ

(m)

4(x3) = −

m−1

∑

k=1(x(k)

3

Z

x(k−1)

3

Q(k)

61

∂

x1

(

ε

1+x3

κ

1) + Q(k)

62

∂

x1

(

ε

2+x3

κ

2)

4.2 Elastic Behavior of Laminates 157

+Q(k)

66

∂

x1

(

ε

6+x3

κ

6) + Q(k)

21

∂

x2

(

ε

1+x3

κ

1)

+Q(k)

22

∂

x2

(

ε

2+x3

κ

2)dx3+Q(k)

26

∂

x2

(

ε

6+x3

κ

6)dx3)(4.2.60)

−

x3

Z

x(m−1)

3

Q(m

61

∂

x1

(

ε

1+x3

κ

1) + Q(m)

62

∂

x1

(

ε

2+x3

κ

2)

+Q(m)

66

∂

x1

(

ε

6+x3

κ

6) + Q(m)

21

∂

x2

(

ε

1+x3

κ

1)

+Q(m)

22

∂

x2

(

ε

2+x3

κ

2) + Q(m)

26

∂

x2

(

ε

6+x3

κ

6)dx3,

σ

(m)

i(x3=x(m)

3) =

σ

(m+1)

i(x3=x(m)

3),i=4,5;m=1,2,...,n,

x(m+1)

3≤x3≤x(m)

3

With the relationships

x(k)

3

Z

x(k−1)

3

Q

Q(k)dx3=Q

Q

Q(k)x(k)

3−x(k−1)

3=Q

Q

Q(k)h(k),

x(k)

3

Z

x(k−1)

3

Q

Q(k)x3dx3=Q

Q

Q(k)1

2x(k)

3

2−x(k−1)

3

2=Q

Q

Q(k)h(k)x(k)

3=Q

Q

Q(k)s(k)

(4.2.61)

¯x(k)

3is the distance of lamina kfrom the midplane. The shear stresses

σ

(m)

i(x3=x(m)

3),i=4,5, at the top surface of the mth lamina can be formulated by





σ

(m)

5x3=x(m)

3

σ

(m)

4x3=x(m)

3

=−

m

∑

k=1



F

F(k)

1

TF

F

F(k)

6

T

F

F(k)

6

TF

F

F(k)

2

T

"

η

1

η

2#(4.2.62)

with

F

F(k)

1

T= [h(k)Q(k)

(11)h(k)Q(k)

(12)h(k)Q(k)

(16)s(k)Q(k)

(11)s(k)Q(k)

(12)s(k)Q(k)

(16)],

F

F(k)

2

T= [h(k)Q(k)

(21)h(k)Q(k)

(22)h(k)Q(k)

(26)s(k)Q(k)

(21)s(k)Q(k)

(22)s(k)Q(k)

(26)],

F

F(k)

6

T= [h(k)Q(k)

(61)h(k)Q(k)

(62)h(k)Q(k)

(66)s(k)Q(k)

(61)s(k)Q(k)

(62)s(k)Q(k)

(66)]

and

158 4 Elastic Behavior of Laminate and Sandwich Composites

η

T

1=

∂ε

1

∂

x1

∂ε

2

∂

x1

∂ε

6

∂

x1

∂κ

1

∂

x1

∂κ

2

∂

x1

∂κ

6

∂

x1,

η

T

2=

∂ε

1

∂

x2

∂ε

2

∂

x2

∂ε

6

∂

x2

∂κ

1

∂

x2

∂κ

2

∂

x2

∂κ

6

∂

x2

The transverse shear stresses only satisfy the equilibrium conditions but violate the

other fundamental equations of anisotropic elasticity. They vary in a parabolic way

through the thicknesses h(k)of the laminate layers and there is no stress jump if one

crosses the interface between two layers.

4.2.5 Thermal and Hygroscopic Effects

In Sect. 4.1.2 the hygrothermal strains were calculated for unidirectional and angle

ply laminae. As mentioned above, no residual mechanical stresses would develop

in the lamina at the macro-mechanical level, if the lamina is free to expand. Free

thermal strain, e.g., refers to the fact that ﬁbres and matrix of an UD-lamina are

smeared into a single equivalent homogeneous material and that the smeared ele-

ments are free of any stresses if temperature is changed. When one considers an

unsmeared material and deals with the individual ﬁbres and the surrounding matrix,

a temperature change can create signiﬁcant stresses in the ﬁbre and matrix. When

such selfbalanced stresses are smeared over a volume element, the net result is zero.

However, in a laminate with various laminae of different materials and orientations

each individual lamina is not free to deform. This results in residual stresses in the

laminate. As in Eqs. (4.1.31) and (4.1.32)

α

th,

α

mo are the thermal and moisture

expansion coefﬁcients, Tis the temperature change and M∗the weight of moisture

absorption per unit weight. In the following equations, Tand M∗are independent of

the x3-coordinate, i.e. they are constant not only through the thickness h(k)of a sin-

gle layer but through the thickness hof the laminate. Heat transfer in thin laminates,

e.g., is generally quite rapid and, hence, thermal gradients in x3-direction are seldom

taken into account and the temperature change Tis then approximately independent

of x3. Analogous considerations are valid for changes in moisture.

For a single layer in off-axis coordinates, the hygrothermal strains and stresses

are given by

ε

=S

S

σ

+

α

thT+

α

moM∗,

σ

=Q

Q

Q(

ε

−

α

thT−

α

moM∗)(4.2.63)

or substituting

ε

=

ε

+x3

κ

σ

=Q

Q

Q(

ε

+x3

κ

−

α

thT−

α

moM∗)(4.2.64)

The deﬁnitions of the force and moment resultants N

N

Nand M

M

4.2 Elastic Behavior of Laminates 159

N

N=Z

(h)

σ

dx3=Z

(h)

Q

Q(

ε

+x3

κ

−

α

thT−

α

moM∗)dx3,

M

M=Z

(h)

σ

x3dx3=Z

(h)

Q

Q(

ε

+x3

κ

−

α

thT−

α

moM∗)x3dx3

(4.2.65)

yield the equations

N

N=A

A

ε

+B

B

κ

−N

N

Nth −N

N

Nmo,

M

M=B

B

ε

+D

D

κ

−M

M

Mth −M

M

Mmo,(4.2.66)

A

A=Q

Q

Qh,B

B

B=0

0

0,D

D

D=Q

Q

Qh3

12

B

B=0

0

0 follows from the symmetry of a single layer to its midplane. N

N

Nth,N

N

Nmo,

M

Mth,M

M

Mmo are ﬁctitious hygrothermal resultants which are deﬁned in (4.2.67). If T

and M∗are independent of x3one can introduce unit thermal and unit moisture

stress resultants ˆ

Nth,ˆ

Mth,ˆ

Nmo,ˆ

Mmo, i.e. resultants per unit temperature or moisture

change

N

Nth =Z

(h)

Q

α

thTdx3=Q

Q

α

thT h =ˆ

NthT,

M

Mth =Z

(h)

Q

α

thT x3dx3=1

2Q

Q

α

thT h2=ˆ

MthT,

N

Nmo =Z

(h)

Q

α

moM∗dx3=Q

Q

α

moM∗h=ˆ

NmoM∗,

M

Mmo =Z

(h)

Q

α

moM∗x3dx3=1

2Q

Q

α

moM∗h2=ˆ

MmoM∗.

(4.2.67)

Nth and Nmo have the units of the force resultant, namely N/m, and Mth and Mmo

the units of the moment resultants, namely Nm/m. The integral form of the resultant

deﬁnitions makes these deﬁnitions quite general, i.e. if Tor M∗are known func-

tions of x3, the integration can be carried out. But for the temperature changes with

x3and if the material properties change with temperature, the integration can be

complicated, but in general the simple integrated form, Eqs. (4.2.67) can be used.

With the total force and moment resultants ˜

N

N,˜

M

M, equal to the respective sums of

their mechanical and hygrothermal components

˜

N

N=N

N

N+N

N

Nth +N

N

Nmo,˜

M

M=M

M

M+M

M

Mth +M

M

Mmo,(4.2.68)

the extended hygrothermal constitutive equation for a lamina can be written





˜

N

···

˜

M

M

=





A

A.

.

. 0

0

. . . .

0

0.

.

.D

D









ε

···

κ



(4.2.69)

160 4 Elastic Behavior of Laminate and Sandwich Composites

This constitutive equation is identical to that derived for mechanical loading only,

(4.1.54), except for the fact that here the hygrothermal forces and moments are

added to the mechanically applied forces and moments. The inversion of (4.2.69)

yields the compliance relation





ε

···

κ



=





a

a.

.

. 0

0

. . . .

0

0.

.

.d

d









˜

N

···

˜

M

M

,a

a

a=A

A

A−1,d

d

d=D

D

D−1(4.2.70)

The values of the stiffness Ai j ,Di j and compliances ai j ,di j are the same as for pure

mechanical loading (Table 4.3) and unit stress resultants ˆ

Nth,ˆ

Mth,ˆ

Nmo,ˆ

Mmo are

ˆ

Nth

1= (Q11

α

th

1+Q12

α

th

2+Q16

α

th

6)h,

ˆ

Nth

2= (Q12

α

th

1+Q22

α

th

2+Q26

α

th

6)h,

ˆ

Nth

6= (Q16

α

th

1+Q26

α

th

2+Q66

α

th

6)h,

ˆ

Mth

1= (Q11

α

th

1+Q12

α

th

2+Q16

α

th

6)1

2h2,

ˆ

Mth

2= (Q12

α

th

1+Q22

α

th

2+Q26

α

th

6)1

2h2,

ˆ

Mth

6= (Q16

α

th

1+Q26

α

th

2+Q66

α

th

6)1

2h2

(4.2.71)

and analogous for unit moisture stress resultant with

α

mo

i,i=1,2,6. In the more

general case the integral deﬁnitions have to used.

When a laminate is subjected to mech anical and hygrothermal load ing, a lamina k

within the laminate is under a state of stress

σ

(k)and strain

ε

(k). The hygrothermoe-

lastic superposition principle shows that the strains

ε

(k)in the lamina kare equal

to the sum of the strains produced by the existing stresses and the free, i.e. unre-

strained, hygrothermal strains and the stresses

σ

(k)follow by inversion

ε

(k)=S

S

S(k)

σ

(k)+

α

th(k)T+

α

mo(k)M∗,

σ

(k)=Q

Q

Q(k)(

ε

(k)−

α

th(k)T−

α

mo(k)M∗)

=Q

Q

Q(k)(

ε

(k)+x3

κ

(k)−

α

th(k)T−

α

mo(k)M∗)

(4.2.72)

When in the lamina kall strains are restrained, then

ε

(k)=0

0

0 and the hygrothermal

stresses are

σ

(k)=Q

Q

Q(k)−

α

th(k)T−

α

mo(k)M∗(4.2.73)

Integration of the stresses

σ

(k)and the stresses

σ

(k)multiplied by the x3-coordinate

across the thickness h(k)and summation for all laminae gives the force and moment

resultants of the laminate

4.2 Elastic Behavior of Laminates 161

N

N=

n

∑

k=1

σ

(k)h(k)

=





n

∑

k=1

Q

Q(k)

x(k)

3

Z

x(k−1)

3

dx3





ε

+





n

∑

k=1

Q

Q(k)

x(k)

3

Z

x(k−1)

3

x3dx3





κ

+





n

∑

k=1

Q

Q(k)

α

th(k)T

x(k)

3

Z

x(k−1)

3

dx3





+





n

∑

k=1

Q

Q(k)

α

mo(k)M∗

x(k)

3

Z

x(k−1)

3

dx3





,

M

M=

n

∑

k=1

σ

(k)h(k)x(k)

3

=





n

∑

k=1

Q

Q(k)

x(k)

3

Z

x(k−1)

3

x3dx3





ε

+





n

∑

k=1

Q

Q(k)

x(k)

3

Z

x(k−1)

3

x2

3dx3





κ

+





n

∑

k=1

Q

Q(k)

α

th(k)T

x(k)

3

Z

x(k−1)

3

x3dx3





+





n

∑

k=1

Q

Q(k)

α

mo(k)M∗

x(k)

3

Z

x(k−1)

3

x3dx3





(4.2.74)

With the stiffness matrices A

A

A,B

B

Band D

D

D, Eqs. (4.2.74), can be rewritten in a brief

matrix form

N

N=A

A

ε

+B

B

κ

−N

N

Nth −N

N

Nmo,

M

M=B

B

ε

+D

D

κ

−M

M

Mth −M

M

Mmo (4.2.75)

The ﬁctitious hygrothermal resultants are given by

N

Nth =T

n

∑

k=1

Q

Q(k)

α

th(k)h(k)=T

n

∑

k=1

ˆ

N(k)th,

N

Nmo =M∗n

∑

k=1

Q

Q(k)

α

mo(k)h(k)=M∗n

∑

k=1

ˆ

N(k)mo,

M

Mth =1

2T

n

∑

k=1

Q

Q(k)

α

th(k)x(k)

3

2−x(k−1)

3

2(4.2.76)

=T

n

∑

k=1

Q

Q(k)

α

th(k)x(k)

3h(k)=T

n

∑

k=1

ˆ

M(k)th,

M

Mmo =1

2M∗n

∑

k=1

Q

Q(k)

α

mo(k)x(k)

3

2−x(k−1)

3

2

=M∗n

∑

k=1

Q

Q(k)

α

mo(k)x(k)

3h(k)=M∗n

∑

k=1

ˆ

M(k)mo

162 4 Elastic Behavior of Laminate and Sandwich Composites

By analogy to the single layer one can introduce total force and moment resultants

˜

N

Nand ˜

M

Mthe stiffness and compliance equations expanded to the hygrothermal com-

ponents

N

N+N

N

Nth +N

N

Nmo =˜

N

N,M

M

M+M

M

Mth +M

M

Mmo =˜

M

M,





˜

N

···

˜

M

M

=





A

A.

.

.B

B

. . . .

B

B.

.

.D

D









ε

···

κ



,



ε

···

κ



=





a

a.

.

.b

b

.....

b

bT.

.

.d

d









˜

N

···

˜

M

M

,

a

a=A

A

A∗−B

B

B∗D

D

D∗−1C

C

C∗,A

A

A∗=A

A

A−1,

b

b=B

B

B∗D

D

D∗−1,B

B

B∗=−A

A

A−1B

B

B,

d

d=D

D

D∗−1,D

D

D∗=D

D

D−B

B

BA

A

A−1B

B

(4.2.77)

The coupling effects discussed in Sects. 4.2.2 and 4.2.3 stay unchanged and the

stiffness matrices in Table 4.4 can be transferred.

If we can classify a laminate as symmetric, balanced, cross-plied or some combi-

nations of these three laminate stacking types, some of the thermal or moisture force

or moment resultant coefﬁcients may be zero. For temperature or moisture change

that depends on x3, no general statements can be given. However for changes in-

dependent of x3, the following simpliﬁcations for the unit stress resultants can be

considered, J=th,mo

Symmetric laminates

ˆ

NJ

16=0,ˆ

MJ

1=0,

ˆ

NJ

26=0,ˆ

MJ

2=0,

ˆ

NJ

66=0,ˆ

MJ

6=0

Balanced laminates ˆ

NJ

16=0,ˆ

MJ

16=0,

ˆ

NJ

26=0,ˆ

MJ

26=0,

ˆ

NJ

6=0,ˆ

MJ

66=0

Symmetric balanced laminates

ˆ

NJ

16=0,ˆ

MJ

1=0,

ˆ

NJ

26=0,ˆ

MJ

2=0,

ˆ

NJ

6=0,ˆ

MJ

6=0

Cross-ply laminates

ˆ

NJ

16=0,ˆ

MJ

16=0,

ˆ

NJ

26=0,ˆ

MJ

26=0,

ˆ

NJ

6=0,ˆ

MJ

6=0

4.2 Elastic Behavior of Laminates 163

Symmetric cross-ply laminates

ˆ

NJ

16=0,ˆ

MJ

1=0,

ˆ

NJ

26=0,ˆ

MJ

2=0,

ˆ

NJ

6=0,ˆ

MJ

6=0

Summarizing the hygrothermal effects one can see that if both mechanical and

hygrothermal loads are applied the mechanical and ﬁctitious hygrothermal loads can

be added to ﬁnd ply by ply stresses and strains in the laminate, or the mechanical

and hygrothermal loads can be applied separately and then the resulting stresses and

strains of the two problems are added.

4.2.6 Problems

Exercise 4.7. A symmetric laminate under in-plane loading can be considered as

an equivalent homogeneous anisotropic plate in plane stress state by introducing

average stress

σ

=N

N

N/hand N

N

N=A

A

ε

. Calculate the effective moduli for general

symmetric laminates and for symmetric cross-ply laminates.

Solution 4.7. Equations (4.2.37) yields

ε

=a

a

aN

N

N,a

a

a=A

A

A−1. The components of the

inverse matrix a

a

aare

a11 = (A22A66 −A2

26)/

∆

,a12 = (A16A26 −A12 A66)/

∆

,

a22 = (A11A66 −A2

16)/

∆

,a16 = (A12A26 −A22 A16)/

∆

,

a26 = (A12A16 −A11A26)/

∆

,a66 = (A11A22 −A2

12)/

∆

,

∆

=Det(Ai j) = A11 

A22 A26

A26 A66 −A12 

A12 A26

A16 A66 

+A16 

A12 A22

A16 A26 

The comparison

ε

=a

a

aN

N

N=ha

a

σ

with (4.1.19) leads to

E1=1/ha11,E2=1/ha22,E6=1/ha66,

ν

12 =−a12/a11,

ν

21 =−a12/a22,

ν

16 =a16/a11,

ν

61 =a16/a66 ,

ν

26 =a26/a22 ,

ν

62 =a26/a6

These are the effective moduli in the general case. For cross-ply laminates is

A16 =A26 =0 (Eqs. 4.2.31). The effective moduli can be explicitly expressed in

terms of the in-plane stiffness Aij . With Det(Ai j) = A11 A22A66 −A2

12A66 follow the

effective moduli

E1= (A11A22 −A2

12)/hA22,

ν

12 =A12/A22 ,

E2= (A11A22 −A2

12)/hA11,

ν

21 =A12/A11 ,

G12 ≡E6=A66/h,

ν

16 =

ν

61 =

ν

26 =

ν

62 =0

Note that these simpliﬁed formulae are not only valid for symmetric cross-ply lam-

inates (0/90)Sbut also for laminates (±45)S.

164 4 Elastic Behavior of Laminate and Sandwich Composites

Exercise 4.8. Show that a symmetric laminate [±450/00/900]Swith E′

1=140 GPa,

E′

2=10 GPa, E′

6=7 GPa,

ν

′

12 =0,3 has a quasi-isotropic material behavior. The

thicknesses of all plies are onstant h(k)=0,1 mm.

Solution 4.8. The solution will obtained in four steps:

1. Calculation of the on-axis reduced stiffness Q′

i j

With respect to (2.1.56) we obtain

ν

′

12/E′

1=

ν

′

21/E′

2=⇒

ν

′

21 = (

ν

′

12E′

2)/E′

1=0,0214

and ﬁnally from Eqs. (4.1.3) follow

Q′

11 =E′

1(1−

ν

′

12

ν

′

21) = 140,905 GPa,

Q′

22 =E′

2(1−

ν

′

12

ν

′

21) = 10,065 GPa,

Q′

12 =E′

2

ν

′

12/(1−

ν

′

12

ν

′

21) = 3,019 GPa,

Q′

66 =E′

6=7 GPa

2. Calculation of the reduced stiffness in the laminae (Table 4.2)

Qi j[00]≡Q′

i j,

Q11[900]=Q′

22 =10,065 GPa,

Q12[900]=Q′

12 =3,019 GPa,

Q22[900]=Q′

11 =140,905 GPa,

Q66[900]=Q′

66 =7 GPa,

Q16[900]=Q26[900]=0,

Q11[±450]=46,252 GPa,

Q12[±450]=32,252 GPa,

Q22[±450]=46,252 GPa,

Q66[±450]=36,233 GPa,

Q16[±450]=±32,71 GPa,

Q26[±450]=±32,71 GPa

3. Calculation of the axial stiffness Ai j (4.2.15)

Ai j =

8

∑

n=1

Q(k)

i j h(k)=2

4

∑

n=1

Q(k)

i j h(k),

A11 =48,695 106Nm−1=A22 ,A12 =14,108 106Nm−1,

A66 =17,293106Nm−1,A16 =A26 =0

4. Calculation of the effective moduli (example 1)

E1=E2= (A2

11 −A2

12)/hA22 =446,1 GPa, E6=A66 /h=172,9 GPa,

ν

12 =

ν

21 =A12/A22 =0,29

Note that E=2(1+

ν

)G=446,1 GPa, i.e. the isotropy condition is satisﬁed.

Exercise 4.9. Calculate the laminar stresses

σ

and

σ

′in the laminate of previous

example loaded by uniaxial tension N1.

4.2 Elastic Behavior of Laminates 165

Solution 4.9. The following reduced stiffness matrices are calculated

Q

Q[00]=



140,9 3,02 0

3,02 10,06 0

0 0 7,0

109Pa,

Q

Q[900]=



10,06 3,02 0

3,02 140,9 0

0 0 7,0

109Pa,

Q

Q[±450]=



46,25 32,25 ±32,71

32,25 46,25 ±32,71

±32,71 ±32,71 36,23 

109Pa

The axial stiffness matrix A

A

Ais also calculated

A

A=



48,70 14,11 0

14,11 48,70 0

0 0 17,2 9 

106N/m

With

a11 =A22/(A2

11 −A2

12) = 0,02241 10−6m/N,

a22 =A11/(A2

11 −A2

12) = a11 ,

a66 =1/A66 =0,05784 10−6m/N,

a12 =−A12/(A11 A22 −A2

12) = −0,00645 10−6m/N

follows the inverse matrix a

a

a=A

A

A−1

a

a=



22,41 −6,49 0

−6,49 22,41 0

0 0 57,84 

10−9m/N

The strains are with (4.2.42)

ε

=a

a

aN

N

N=⇒



ε

1

ε

2

ε

6

=a

a

a



N1

0

0

=



22,41

−6,49

0

10−9(m/N)N1

N1is given in N/m, i.e.

ε

iare dimensionless.

Now the laminar stresses are (Table 4.2)





σ

1

σ

2

σ

6

[00]

=Q

Q

Q[00]

ε

=



3138

2,4

0

N1[N/m2],





σ

1

σ

2

σ

6

[900]

=Q

Q

Q[900]

ε

=



205,8

−846,8

0

N1[N/m2],

166 4 Elastic Behavior of Laminate and Sandwich Composites





σ

1

σ

2

σ

6

[±450]

=Q

Q

Q[±450]

ε

=



827,2

422,6

±520,7

N1[N/m2]

The stresses jump from lamina to lamina. Verify that the resultant force N2=0.

The stress components in reference to the principal material axes follow with the

transformation rule (Table 4.1)





σ

′

1

σ

′

2

σ

′

6



=



c2s22sc

s2c2−2sc

−cs cs c2−s2





σ

1

σ

2

σ

6

,





σ

′

1

σ

′

2

σ

′

6



[00]

=



3138

2,4

0

N1[N/m2],





σ

′

1

σ

′

2

σ

′

6



[900]

=

−846,8

205,8

0

N1[N/m2],





σ

1

σ

2

σ

6

[±450]

=



1146

104,2

∓202,3

N1[N/m2]

Note 4.1. These stresses are used in failure analysis of a laminate.

Exercise 4.10. A laminate with an unsymmetric layer stacking [−450/300/00]has

three layers of equal thickness h(1)=h(2)=h(3)=5 mm. The mechanical properties

of all UD-laminae are E′

1=181 GPa, E′

2=10,30 GPa, G′

12 =7,17 GPa,

ν

′

12 =0,28

GPa. Determine the laminate stiffness Ai j ,Bi j ,Di j.

Solution 4.10. Using (4.1.3) the elements S′

i j of the compliance matrix S

S

S′are

S′

11 =1/E′

1=0,0055 GPa−1,

S′

12 =−

ν

′

12/E′

1=−0,0015 GPa−1,

S′

22 =1/E′

2=0,0971 GPa−1,

S′

66 =1/G′

12 =0,1395 GPa−1

The minor Poisson’s ratio follows with

ν

′

21 =

ν

′

12E′

2/E′

1=0,01593

Using (4.1.3) the elements Q′

i j of the reduced stiffness matrix Q

Q

Q′are

Q′

11 =E′

1/(1−

ν

′

12

ν

′

21) = 181,8 109Pa,

Q′

12 =

ν

′

12E′

2/(1−

ν

′

12

ν

′

21) = 2,897 109Pa,

Q′

22 =E′

2/(1−

ν

′

12

ν

′

21) = 10,35 109Pa,

Q′

66 =G′

12 =7,17 109Pa

4.2 Elastic Behavior of Laminates 167

Verify that the reduced stiffness matrix could also be obtained by inverting the

compliance matrix, i.e. Q

Q

Q′=S′

S′

S′−1. Now the transformed reduced stiffness ma-

trices Q

Q

Q[00],Q

Q

Q[300],Q

Q

Q[−450]have to be calculated with the help of Table 4.2 tak-

ing into account that c=cos 00=1,cos 300=0,8660,cos(−450) = 0,7071 and

s=sin00=0,sin300=0,5,sin(−450) = −0,7071

Q

Q[00]=



181,8 2,897 0

2,897 10,35 0

0 0 7,17 

109Pa,

Q

Q[300]=



109,4 32,46 54,19

32,46 23,65 20,05

54,19 20,05 36,74 

109Pa,

Q

Q[−450]=



56,66 42,32 −42,87

42,32 56,66 −42,87

−42,87 −42,87 46,59 

109Pa

The location of the lamina surfaces are x(0)

3=−7,5 mm, x(1)

3=−2,5 mm, x(2)

3=2,5

mm, x(3)

3=7,5 mm. The total thickness of the laminate is 15 mm. From (4.2.15) the

extensional stiffness matrix A

A

Afollows with

Ai j =

3

∑

k=1

Q(k)

i j h(k)=5

3

∑

k=1

Q(k)

i j mm,

the coupling matrix B

B

Bfollows with

Bi j =

3

∑

k=1

Q(k)

i j h(k)¯x(k)

3=5

3

∑

k=1

Q(k)

i j ¯x(k)

3mm

and the bending stiffness matrix Di j follows with

Di j =1

3

∑

k=1

Q(k)

i j h(k) (¯x(k)

3)2+h(k)2

12 !=5

3

∑

k=1

Q(k)

i j 25

12 + ( ¯x(k)

3)2mm3

with

¯x(k)

3=1

2(x(k)

3+x(k−1)

3),

i.e. ¯x(1)

3=5 mm, ¯x(2)

3=0 mm, ¯x(3)

3=−5 mm. Summarizing the formulas for Ai j ,Bi j

and Di j we have the equations

Ai j =5[Q(1)

i j +Q(2)

i j +Q(3)

i j ]mm,

Bi j =5[5Q(1)

i j +0Q(2)

i j −5Q(3)

i j mm2],

Di j =5[(2 5 +25/12)Q(1)

i j + (25/12)Q(2)

i j + (25 +25/12)Q(3)

i j ]mm3

168 4 Elastic Behavior of Laminate and Sandwich Composites

and the stiffness matrices follow to

A

A=



17,39 3,884 0,566

3,884 4,533 −1,141

0,566 −1,141 4,525 

108Pa m,

B

B=

−3,129 0,986 −1,072

0,986 1,158 −1,072

−1,072 −1,072 0,986 

106Pa m2,

D

D=



33,43 6,461 −5,240

6,461 9,320 −5,596

−5,240 −5,596 7,663 

103Pa m3

4.3 Elastic Behavior of Sandwiches

One special group of laminated composites used extensively in engineering appli-

cations is sandwich composites. Sandwich panels consist of thin facings, also called

skins or sheets, sandwiching a core. The facings are made of high strength material

while the core is made of thick and lightweight materials, Sect. 1.3. The motiva-

tion for sandwich structure elements is twofold. First for beam or plate bending

the maximum normal stresses occur at the top and the bottom surface. So it makes

sense using high-strength materials at the top and the bottom and using low and

lightweight strength materials in the middle. The strong and stiff facings also sup-

port axial forces. Second, the bending resistance for a rectangular cross-sectional

beam or plate is proportional to the cube of the thickness. Increasing the thickness

by adding a core in the middle increases the resistance. The maximum shear stress

is generally in the middle of the sandwich requiring a core to support shear. The

advantages in weight and bending stiffness make sandwich composites attractive in

many applications.

The most commonly used facing materials are aluminium alloys and ﬁbre re-

inforced plastics. Aluminium has a high speciﬁc modulus, but it corrodes without

treatment and can be prone to denting. Therefore ﬁbre reinforced laminates, such

as graphite/epoxy or glass/epoxy are becoming more popular as facing materials.

They have high speciﬁc modulus and strength and corrosion resistance. The ﬁbre

reinforced facing can be unidirectional or woven laminae.

The most commonly used core materials are balsa wood, foam, resins with spe-

cial ﬁllers and honeycombs (Fig. 1.3). These materials must have high compressive

and shear strength. Honeycombs can be made of plastics, paper, card-boards, etc.

The strength and stiffness of honeycomb sandwiches depend on the material and

the cell size and thickness. The following sections consider the modelling and anal-

ysis of sandwiches with thin and thick cover sheets.

4.3 Elastic Behavior of Sandwiches 169

4.3.1 General Assumptions

A sandwich can be deﬁned as a special laminate with three layers. The thin cover

sheets, i.e. the layers 1 and 3, are laminates of the thicknesses h(1)for the lower skin

and h(3)for the upper skin. The thickness of the core is h(2)≡hc. In a general case

h(1)does not have to be equal to h(3), but in the most important practical case of

symmetric sandwiches h(1)=h(3)≡hf.

The assumptions for macro-mechanical modelling of sandwiches are:

1. The thickness of the core is much greater than that of the skins, h(2)≫h(1),h(3)

or hc≫hf

2. The strains

ε

1,

ε

2,

ε

6vary linearly through the core thickness hc

ε

=

ε

+x3

κ

3. The sheets only transmit stresses

σ

1,

σ

2,

σ

6and the in-plane strains are uniform

through the thickness of the skins. The transverse shear stresses

σ

4,

σ

5are ne-

glected within the skin.

4. The core only transmits transverse shear stresses

σ

4and

σ

5, the stresses

σ

1,

σ

2

and

σ

6are neglected.

5. The strain

ε

3is neglected in the sheets and the core.

With these additional assumptions in the frame of linear anisotropic elasticity, the

stresses and strains can be formulated.

Strains in the lower and upper sheets:

ε

(l)=

ε

∓1

2hc

κ

,

ε

(l)

i=

ε

i∓1

2hc

κ

i,l=1,3,i=1,2,6 (4.3.1)

The transverse shear strains

ε

4,

ε

5are neglected.

Strains in the sandwich core:

ε

(2)=

ε

c=

ε

+x3

κ

,−hc

2≤x3≤+hc

2(4.3.2)

The transverse shear strains are, in a ﬁrst approach, independent of the coordinate

x3(4.2.14)

γ

sc = [

ε

c

5

ε

c

4]T(4.3.3)

We shall see in Chap. 5 that in the classical laminate theory and the laminate theory

including transverse shear deformations the strain vector

ε

is written in an analogous

form, and only the expressions for the curvatures are modiﬁed.

Stresses in the lower and upper sheets:

In the sheets a plane stress state exists and with assumption 3. the transverse shear

stresses

σ

4and

σ

5are neglected. These assumptions imply that for laminated sheets

in all layers of the lower and the upper skins

σ

(k)

4=

σ

(k)

5=0

170 4 Elastic Behavior of Laminate and Sandwich Composites

The other stresses are deduced from the constant strains

ε

(l)

1,

ε

(l)

2,

ε

(l)

6,l=1,3

by the relationships

σ

(k)

i=Q(k)

i j

ε

(l)

j,i,j=1,2,6,l=1,3 (4.3.4)

for the kth layer of the lower (l=1) or the upper (l=3) skin.

Stresses in the sandwich core:

From assumption 4. it follows

σ

c

1=

σ

c

2=

σ

c

6=0

and the core transmits only the transverse shear stresses



σ

c

5

σ

c

4=C55 C45

C45 C44 

ε

5

ε

4(4.3.5)

or in matrix notation (Eq. 4.2.14)

σ

s=C

C

Cs

γ

s(4.3.6)

The coefﬁcients Cc

i j of C

C

Csare expressed as functions of the coefﬁcients Cc′

i j referred

to the principal directions by the transformation equation (4.2.17). The coefﬁcients

Cc′

i j in the principal directions are themselves written as functions of the shear moduli

of the core (Sect. 2.1, Table 2.5), measured in principal directions as follows

Cc′

44 =Gc

23,Cc

55 =Gc′

13 (4.3.7)

For an isotropic core material a transformation is not required.

4.3.2 Stress Resultants and Stress Analysis

The in-plane resultants N

N

Nfor sandwiches are deﬁned by

N

N=

−1

2hc

Z

−(1

2hc+h(1))

σ

dx3+

1

2hc+h(3)

Z

1

2hc

σ

dx3,(4.3.8)

the moment resultants by

4.3 Elastic Behavior of Sandwiches 171

M

M=

−1

2hc

Z

−(1

2hc+h(1))

σ

x3dx3+

1

2hc+h(3)

Z

1

2hc

σ

x3dx3(4.3.9)

and the transverse shear force by

Q

Qs=

1

2hc

Z

−1

2hc

σ

sdx3(4.3.10)

For the resultants N

N

Nand M

M

Mthe integration is carried out over the sheets only and for

the transverse shear force over the core.

By substituting Eqs. (4.3.4) - (4.3.7) for the stresses into the preceding expres-

sions for the force and moment resultants, we obtain analogous to (4.2.16) the con-

stitutive equation





N

M

Q

Qs

=



A

A B

B

B0

0

C

C D

D

D0

0

0 0

0

0A

A

As





ε

κ

γ

s

(4.3.11)

with the stiffness coefﬁcients

Ai j =A(1)

i j +A(3)

i j ,Bi j =1

2hcA(3)

i j −A(1)

i j ,

Ci j =C(1)

i j +C(3)

i j ,Di j =1

2hcC(3)

i j −C(1)

i j 

(4.3.12)

and

A(1)

i j =

−1

2hc

Z

−(1

2hc+h1)

Q(k)

i j dx3=

n1

∑

k=1Z

h(k)

Q(k)

i j dx3=

n1

∑

k=1

Q(k)

i j h(k),

A(3)

i j =

1

2hc+h3

Z

1

2hc

Q(k)

i j dx3=

n2

∑

k=1Z

h(k)

Q(k)

i j dx3=

n2

∑

k=1

Q(k)

i j h(k),

C(1)

i j =

−1

2hc

Z

−(1

2hc+h1)

Q(k)

i j x3dx3=

n1

∑

k=1Z

h(k)

Q(k)

i j x3dx3=

n1

∑

k=1

Q(k)

i j h(k)¯x(k)

3,

C(3)

i j =

1

2hc+h3

Z

1

2hc

Q(k)

i j x3dx3=

n2

∑

k=1Z

h(k)

Q(k)

i j x3dx3=

n2

∑

k=1

Q(k)

i j h(k)¯x(k)

3

(4.3.13)

with x(k)

3=1

2(x(k)

3+x(k−1)

3and

172 4 Elastic Behavior of Laminate and Sandwich Composites

As

i j =hcCs

i j ,i,j=4,5 (4.3.14)

n1and n2are the number of layers in the lower and the upper sheet respectively and

Cs

i j are the transverse shear moduli of the core. The constitutive equations (4.3.11)

for a sandwich composite has a form similar to the constitutive equation for lami-

nates including transverse shear. It differs only by the terms Ci j instead of Bi j which

induce an unsymmetry in the stiffness matrix.

In the case of symmetric sandwiches with identical sheets h(1)=h(3)=hf,

A(1)

i j =A(3)

i j =Af

i j,C(1)

i j =−C(3)

i j =Cf

i j and from this it results that the stiffness

coefﬁcients Eq. (4.3.12) are

Ai j =2Af

i j,Di j =hcCf

i j,Bi j =0,Ci j =0 (4.3.15)

As developed for laminates including shear deformations, the coefﬁcients As

i j can

be corrected by shear correction factors ks

i j and replaced by shear constants (ksAs)i j

to improve the modelling.

In the case of symmetric sandwiches there is no coupling between stretching

and bending and the form of the constitutive equation is identical to the constitutive

equation for symmetric laminates including transverse shear.

4.3.3 Sandwich Materials with Thick Cover Sheets

In the case of thick cover sheets it is possible to carry out the modelling and analysis

with the help of the theory of laminates including transverse shear. Considering the

elastic behavior of sandwich composites we have:

•The stretching behavior is determined by the skins.

•The transverse shear is imposed by the core.

The modelling assumption 1. of Sect. 4.3.1 is not valid. Restricting the modelling

to the case of symmetric sandwich composites and to the case where the core’s

principal direction is in coincidence with the directions of the reference coordinate

system. The elastic behavior of the composite material is characterized by

•the reduced stiffness parameters Qf

i j for the face sheets,

•the reduced stiffness parameters Qc

i j and the transverse shear moduli Cc

i j for the

core

Application of the sandwich theory, Sect. 4.3.2, leads to the following expressions

for the stiffness coefﬁcients of the constitutive equation (upper index Sa), one lamina

ASa

i j =2hfQf

i j,BSa

i j =0,CSa

i j =0,DSa

i j =1

2Qf

i j(hf+hc)hfhc,i,j=1,2,6 (4.3.16)

The shear stiffness coefﬁcients As

i j are in the sandwich theory

AsSa

i j =hcCi j,i,j=4,5,C44 =Gc

23,C55 =Gc

13,C45 =0 (4.3.17)

4.3 Elastic Behavior of Sandwiches 173

Application of the laminate theory including transverse shear, Sect. 4.2.2, (4.2.15)

leads (upper index La) to

ALa

i j =2hfQf

i j +hcQc

i j,BLa

i j =0,

DLa

i j =1

2Qf

i jhf(hf+hc)2+1

3(hf)2+1

12 Qc

i j(hc)3,i,j=1,2,6

(4.3.18)

The shear stiffness coefﬁcients As

i j are now

AsLa

i j =2hfCf

i j +hcCc

i j,i,j=4,5,Cf/c

44 =Gf/c

23 ,Cf/c

55 =Gf/c

13 ,Cf/c

45 =0 (4.3.19)

For symmetric faces with nlaminae Eqs. (4.3.16) - (4.3.19) yield

ASa

i j =2

n

∑

k=1

Qf(k)

i j hf(k),

DSa

i j =hc

n

∑

k=1

Qf(k)

i j hf(k)¯x(k)

3,

ALa

i j =2

n

∑

k=1

Qf(k)

i j hf(k)+hcQc

i j,

DLa

i j =

n

∑

k=1

Qf(k)

i j hf(k)(¯x(k)

3)2+hf(k)3

12 !+Qc

i j

hc3

12

The comparison of the analysis based on the sandwich or the laminate theory yields

ALa

i j =ASa

i j 1+hcQc

i j

2hfQf

i j !,

DLa

i j =DSa

i j 1+hf

hc

hc+ (4/3)hf

hc+hf+Qc

i j

6Qf

i j

(hc)2

hf(hc+hf)!,i,j=1,2,6

(4.3.20)

AsLa

i j =AsSa

i j 1+hfCf

i j

2hcCc

i j !,i,j=4,5 (4.3.21)

Generally the core of the sandwich is less stiff than the cover sheets

Qc

i j ≪Qf

i j

and the relations (4.3.20) can be simpliﬁed

ALa

i j ≈ASa

i j ,DLa

i j ≈DSa

i j 1+hf

hc

hc+ (4/3)hf

hc+hf(4.3.22)

Equation (4.3.21) stays unchanged.

The bending stiffness Di j are modiﬁed with respect to the theory of sand-

wiches and can be evaluated by the inﬂuence of the sheet thickness. If for ex-

174 4 Elastic Behavior of Laminate and Sandwich Composites

ample hc=10 mm and the sheet thickness hf=1 mm/ 3 mm or 5 mm we ﬁnd

DLa

i j =1,103DSa

i j /1,326DSa

i j or 1,555DSa

i j , a difference of more than 10%, 30% or

50 %.

4.4 Problems

Exercise 4.11. The reduced stiffness Qc

i j and Qf

i j of a symmetric sandwich satisﬁes

the relation Qc

i j ≪Qf

i j. Evaluate the inﬂuence of the sheet thickness on the bending

stiffness ratio DLa

i j /DSa

i j if the core thickness is constant (hc=10 mm) and the sheet

thickness vary: hf=0.5/1.0/3.0/5.0/8.0/10.0 mm.

Solution 4.11. Using the simpliﬁed formula (4.3.22) the ratio values are

1.051/1.103/1.323/1.555/1.918/2.167 i.e. the difference

DLa

i j −DSa

i j

DSa

i j

100%

of the stiffness values for DLa

i j and DSa

i j are more than 5%/10%/32%/55%/91% or

116%.

Conclusion 4.2. The sandwich formulas of Sect. 4.3.2 should be used for thin cover

sheets only, i.e. hf≪hc.

Exercise 4.12. A sandwich beam has faces of aluminium alloy and a core of

polyurethane foam. The geometry of the cross-section is given in Fig. 4.17. Cal-

PPPPPP

P

✏✏✏✏✏✏

✏

✲

✛

❄

✻

✲✲

✻

❄

✻

✲✛

❄

✻

❄

✻

✛

✻

hf

x3

x2

hch

b

abc

x3x3

Fig. 4.17 Sandwich beam. aGeometry of the cross-section of a sandwich beam, bDistribution

of the bending stress, if the local stiffness of the faces and the bending stiffness of the core are

dropped, cDistribution of the shear stress, if only the core transmit shear stresses

4.4 Problems 175

culate the bending stiffness Dand the distributions of the bending and the shear

stress across the faces and the core, if the stress resultants Mand Qare given.

Solution 4.12. The bending stiffness Dof the sandwich beam is the sum of the

ﬂexural rigidities of the faces and the core

D=2Efbhf3

12 +2Efbhfhc+hf

22

+Ecbhc3

12

Efand Ecare the effective Young’s moduli of the face and the core materials. The

ﬁrst term presents the local bending stiffness of the faces about their own axes, the

third term represents the bending stiffness of the core. Both terms are generally very

small in comparison to the second term. Provided that

(hc+hf)/hf>5,77,[(Efhf)/(Echc)][(hc+hf)/hc]2>100/6

i.e.

Efbhf3/6

Ef[bhf(hc+hf)2]/2<1

100

Ecbhc3/12

Ef[bhf(hc+hf)2]/2<1

100

the ﬁrst and the third term are less than 1% of the second term and the bending

stiffness is approximately

D≈Efbhf(hc+hf)2/2

The bending stress distributions through the faces and the core are

σ

f=MEf

Dx3≈ ±MEf

D

hc+hf

2,

σ

c=MEc

Dx3≈0

The assumptions of the classical beam theory yield the shear stress equation for the

core

τ

=QS(x3)

bI =Q

bD [EfSf(x3) + EcSc(x3)]

≈Q

D"Efhf(hc+hf)

2+Ec

2 hc2

4−x2

3!#

The maximum core shear stress will occur at x3=0. If

Efhf(hc+hf)

Echc2/4>100

the ratio of the maximum core shear stress to the minimum core shear stress is <1%

and the shear stress distribution across the core can be considered constant

τ

≈Q

D

Efhf(hc+hf)

2

and with

176 4 Elastic Behavior of Laminate and Sandwich Composites

D≈Efbhf(hc+hf)2/2

follow

τ

≈Q/b(hc+hf)≈Q/bh. In Fig. 4.17 the distributions of the bending

and shear stresses for sandwich beams with thin faces are illustrated. Note that for

thicker faces the approximate ﬂexural bending rigidity is

D≈Efbhf(hc+hf)/+Efbhf3/12

Chapter 5

Classical and Improved Theories

In this chapter, the theoretical background for two commonly used structural the-

ories for the modelling and analysis of laminates and sandwiches is considered,

namely the classical laminate theory and the ﬁrst-order shear deformation theory.

The classical laminate theory (CLT) and the ﬁrst-order shear deformation theory

(FSDT) are the most commonly used theories for analyzing laminated or sand-

wiched beams, plates and shells in engineering applications. The CLT is an ex-

tension of Kirchhoff’s1classical plate theory for homogeneous isotropic plates to

laminated composite plates with a reasonable high width-to-thickness ratio. For ho-

mogeneous isotropic plates the Kirchhoff ’s theory is limited to thin plates with ratios

of maximum plate deﬂection wto plate thickness h<0.2 and plate thickness/ mini-

mum in-plane dimensions <0.1. Unlike homogeneous isotropic structure elements,

laminated plates or sandwich structures have a higher ratio of in-plane Young’s mod-

uli to the interlaminar shear moduli, i.e. such composite structure elements have a

lower transverse shear stiffness and often have signiﬁcant transverse shear deforma-

tions at lower thickness-to span ratios <0.05. Otherwise the maximum deﬂections

can be considerable larger than predicted by CLT. Furthermore, the CLT cannot

yield adequate correct through-the-thickness stresses and failure estimations. As a

result of these considerations it is appropriate to develop higher-order laminated and

sandwich theories which can be applied to moderate thick structure elements, e.g.

the FSDT. CLT and FSDT are so-called equivalent single-layer theories (ESLT).

Moreover a short overview of so-called discrete-layer or layerwise theories is given,

which shall overcome the drawbacks of equivalent single layer theories.

1Gustav Robert Kirchhoff (∗12 March 1824 K¨onigsberg - †17 October 1887 Berlin) - physi-

cist who contributed to the fundamental understanding of electrical circuits, spectroscopy, and the

emission of black-body radiation by heated objects, in addition, he formulated a plate theory which

was an extension of the Euler-Bernoulli beam theory

177

H. Altenbach et al., Mechanics of Composite Structural

Elements, https://doi.org/10.1007/978-981-10-8935-0_5

178 5 Classical and Improved Theories

5.1 General Remarks

A classiﬁcation of the structural theories in composite mechanics illustrates that the

following approaches for the modelling and analysis of beams and plates composed

of composite materials can be used:

1. So called equivalent single-layer theories: These theories are derived from the

three-dimensional elasticity theory by making assumptions concerning the kine-

matics of deformation and/or the stress distribution through the thickness of a

laminate or a sandwich. With the help of these assumptions the modelling can be

reduced from a 3D-problem to a 2D-problem. In engineering applications equiv-

alent single-layer theories are mostly used in the form of the classical laminate

theory, for very thin laminates, and the ﬁrst order shear deformation theory, for

thicker laminates and sandwiches.

An equivalent single layer model is developed by assuming continuous displace-

ment and strain functions through the thickness. The stresses jump from ply to

ply and therefore the governing equations are derived in terms of thickness aver-

aged resultants. Also second and higher order equivalent single layer theories by

using higher order polynomials in the expansion of the displacement components

through the thickness of the laminate are developed. Such higher order theories

introduce additional unknowns that are often difﬁcult to interpret in mechanical

terms. The CLT requires C1-continuity of the transverse displacement, i.e. the

displacement and the derivatives must be continuous, unlike the FSDT requires

C0-continuity only. Higher order theories generally req uire at least C1-continuity.

2. Three-dimensional elasticity theories such as the traditional 3D-formulations of

anisotropic elasticity or the so-called layerwise theories: In contrast to the equiv-

alent single-layer theories only the displacement components have to be contin-

uous through the thickness of a laminate or a sandwich but the derivatives of

the displacements with respect to the thickness coordinate x3may be discon-

tinuous at the layer interfaces. We say that the displacement ﬁeld exhibits only

C0-continuity through the thickness directions.

The basic assumption of modelling structural elements in the framework of the

anisotropic elasticity is an approximate expression of the displacement components

in the form of polynomials for the thickness coordinate x3. Usually the polynomials

are limited to degree three and can be written in the form

u1(x1,x2,x3) = u(x1,x2) +

α

x3

∂

w(x1,x2)

∂

x1

+

β

x3

ψ

1(x1,x2)

+

γ

x2

3

φ

1(x1,x2) +

δ

x3

3

χ

1(x1,x2),

u2(x1,x2,x3) = v(x1,x2) +

α

x3

∂

w(x1,x2)

∂

x2

+

β

x3

ψ

2(x1,x2)

+

γ

x2

3

φ

2(x1,x2) +

δ

x3

3

χ

2(x1,x2),

u3(x1,x2,x3) = w(x1,x2) +

β

x3

ψ

3(x1,x2) +

γ

x2

3

φ

3(x1,x2)

(5.1.1)

5.1 General Remarks 179

A displacement ﬁeld in the form of (5.1.1) satisﬁes the compatibility conditions for

strains, Sect. 2.2.1, and allows possible cross-sectional warping, transverse shear

deformations and transverse normal deformations to be taken into account. The dis-

placement components of the middle surface are u(x1,x2),v(x1,x2),w(x1,x2). In the

case of dynamic problems the time tmust be introduced in all displacement func-

tions.

The polynomial approach (5.1.1) of the real displacement ﬁeld yields the follow-

ing equivalent single-layer theories

•Classical laminate theories

α

=−1,

β

=

γ

=

δ

=

β

=

γ

=0

•First-order shear deformation theory

α

=0,

β

=1,

γ

=

δ

=

β

=

γ

=0

•Second order laminate theory

α

=0,

β

=1,

γ

=1,

δ

=

β

=

γ

=0

•Third order laminate theory

α

=0,

β

=1,

γ

=1,

δ

=1,

β

=

γ

=0

Theories higher than third order are not used because the accuracy gain is so little

that the effort required to solve the governing equations is not justiﬁed. A third order

theory based on the displacement ﬁeld u1,u2,u3has 11 unknown functions of the

in-plane coordinates x1,x2.u,v,wdenote displacements and

ψ

1,

ψ

2rotations of the

transverse normals referred to the plane x3=0.

ψ

3has the meaning of extension of

a transverse normal and the remaining functions can be interpreted as warping func-

tions that specify the deformed shape of a straight line perpendicular to the reference

plane of the undeformed structure. In addition, any plate theory should fulﬁll some

consistency requirements which was ﬁrst time discussed for the simplest case of a

homogeneous isotropic plate in Kienzler (2002) and later extended to other cases

by Schneider and Kienzler (2015); Schneider et al (2014). Also implementations of

higher order theories into ﬁnite element approximations cannot be recommended. If

a laminated plate is thick or the 3D stress ﬁeld must be calculated in local regions,

a full 3D analysis should be carried out.

The most widely used approach reduces the polynomial function of degree three

to a linear or ﬁrst order approximation, which includes the classical and the ﬁrst-

order shear deformation theory

u1(x1,x2,x3) = u(x1,x2) + x3

ψ

1(x1,x2),

u2(x1,x2,x3) = v(x1,x2) + x3

ψ

2(x1,x2),

u3(x1,x2,x3) = w(x1,x2)

(5.1.2)

180 5 Classical and Improved Theories

The classical approximation can be obtained if

ψ

1(x1,x2) = −

∂

w

∂

x1

,

ψ

2(x1,x2) = −

∂

w

∂

x2

The number of unknown functions reduces to three, that are u,v,w. On the other

hand there are ﬁve independent unknown functions u,v,w,

ψ

1,

ψ

2.

The strain-displacement equations (2.2.3) give for the ﬁrst order displacement

approximation a ﬁrst order strain ﬁeld model with transverse shear

ε

1=

∂

u

∂

x1

+x3

∂ψ

1

∂

x1

,

ε

2=

∂

v

∂

x2

+x3

∂ψ

2

∂

x2

,

ε

3=0,

ε

4=

∂

w

∂

x2

+

ψ

2,

ε

5=

∂

w

∂

x1

+

ψ

1,

ε

6=

∂

u

∂

x2

+

∂

v

∂

x1

+x3

∂ψ

2

∂

x1

+

∂ψ

1

∂

x2

(5.1.3)

For the in-plane strains one can write in contracted form

ε

i(x1,x2,x3) =

ε

i(x1,x2) + x3

κ

i(x1,x2),i=1,2,6,

i.e. the in-plane strains

ε

1,

ε

2and

ε

6vary linearly through the thickness h.

The stress-strain relations in on-axis coordinates are

σ

′

i=C′

i j

ε

′

j,i,j=1,2,...,6

From the transformation rule (4.1.27) follow the stiffness coefﬁcients in the off-

axis-coordinates

C

C=3

T

ε

T

C

C′3

T

ε

and with (4.1.26) the constitutive equation is







σ

1

σ

2

σ

3

σ

4

σ

5

σ

6







=







C11 C12 C13 0 0 C16

C12 C22 C23 0 0 C26

C13 C23 C33 0 0 C36

0 0 0 C44 C45 0

0 0 0 C45 C55 0

C16 C26 C36 0 0 C66













ε

1

ε

2

ε

3

ε

4

ε

5

ε

6







(5.1.4)

Assuming

σ

3≈0, the stiffness matrix can be rewritten by separating the transverse

shear stresses and strains in analogy to (4.2.3) - (4.2.5)







σ

1

σ

2

σ

6

σ

4

σ

5







=





Q11 Q12 Q16 0 0

Q12 Q22 Q26 0 0

Q16 Q26 Q66 0 0

0 0 0 C44 C45

0 0 0 C45 C55













ε

1

ε

2

ε

6

ε

4

ε

5







(5.1.5)

5.1 General Remarks 181

and from

σ

3=C13

ε

1+C23

ε

2+C33

ε

3+C36

ε

6=0

it follows

ε

3=−1

C33

(C13

ε

1+C23

ε

2+C36

ε

6)

The Qi j are the reduced stiffness in the off-axis reference system

Qi j =Ci j −Ci3Cj3

C33

,i,j=1,2,6,Qi j =Cij ,i,j=4,5

Summarizing, one can say that the ﬁrst order displacement approach (5.1.2) includes

the classical and the shear deformation theory for laminates and sandwiches. In both

cases the in-plane displacements and strains vary linearly through the thickness, but

the explicit expressions for the curvature vector

κ

differ. The force and moment re-

sultants can be deﬁned for both theories in the usual way, e.g. (4.2.13), (4.2.14), but

in the classical theory there are only constitutive equations for the in-plane force

and the moment resultants N

N

N,M

M

M. It can be proved that a CLT approach is sufﬁcient

for very thin laminates and it has been used particularly to determine the global

response of thin composite structure elements, i.e. deﬂections, overall buckling, vi-

bration frequencies, etc. The FSDT approach is sufﬁcient for determining in-plane

stresses even if the structure slenderness is not very high.

The CLT neglects all transverse shear and normal effects, i.e. structural deforma-

tion is due entirely to bending and in-plane stretching. The FSDT relaxes the kine-

matic restrictions of CLT by including a constant transverse shear strain. Both ﬁrst

order theories yield a complete understanding of the through-the-thickness laminate

response. Transverse normal and shear stresses, however, play an important role in

the analysis of beams, plates and shells since they signiﬁcantly affect characteris-

tic failure modes like, e.g., delamination. The inﬂuence of interlaminar transverse

stresses are therefore taken into account by several failure criteria. Simple but suf-

ﬁcient accurate methods for determination of the complete state of stress in com-

posite structures are needed to overcome the limitations of the simple ﬁrst order 2D

modelling in the frame of an extended 2D modelling. In Sects. 5.2 and 5.3 a short

description of CLT and FSDT is given including some remarks to calculate trans-

verse stress components. In Chap. 11 will be seen that both the CLT and the FSDT

yield ﬁnite elements with an economical number of degrees of freedom, both have

some drawbacks. CLT-models require C1-continuity which complicates the imple-

mentation in commonly used FEM programs. FSDT-models have the advantage of

requiring only C0-continuity but they can exhibit so-called locking effects if lami-

nates becomes thin. Further details are given in Chap. 11.

182 5 Classical and Improved Theories

5.2 Classical Laminate Theory

The classical laminate theory uses the ﬁrst-order model equations (5.1.2) but makes

additional assumptions:

1. All layers are in a state of plane stress, i.e.

σ

3=

σ

4=

σ

5=0

2. Normal distances from the middle surface remain constant, i.e. the transverse

normal strain

ε

3is negligible compared with the in-plane strains

ε

1,

ε

2.

3. The transverse shear strains

ε

4,

ε

5are negligible. This assumption implies that

straight lines normal to the middle surface remain straight and normal to that

surface after deformation (Bernoulli/Kirchhoff/Love2hypotheses in the theory

of beams, plates and shells).

Further we recall the general assumption of linear laminate theory that each layer

is quasi-homogeneous, the displacements are continuous through the total thickness

h, the displacements are small compared with the thickness hand the constitutive

equations are linear.

From assumptions 2. and 3. it follows from (5.1.3) that

ψ

1(x1,x2) = −

∂

w

∂

x1

,

ψ

2(x1,x2) = −

∂

w

∂

x2

,(5.2.1)

and the displacement approach (5.1.2) and the strain components (5.1.3) are written

by

u1(x1,x2,x3) = u(x1,x2)−x3

∂

w(x1,x2)

∂

x1

,

u2(x1,x2,x3) = v(x1,x2)−x3

∂

w(x1,x2)

∂

x2

,

u3(x1,x2,x3) = w(x1,x2),

(5.2.2)

ε

1=

∂

u

∂

x1−x3

∂

w2

∂

x2

1

,

ε

2=

∂

v

∂

x2−x3

∂

w2

∂

x2

2

,

ε

3=0,

ε

4=0,

ε

5=0,

ε

6=

∂

u

∂

x2

+

∂

v

∂

x1−2x3

∂

w2

∂

x1

∂

x2

(5.2.3)

The condensed form for the in-plane strains can be noted as

ε

i(x1,x2,x3) =

ε

i(x1,x2) + x3

κ

i,i=1,2,6

with

2Augustus Edward Hough Love (∗17 April 1863, Weston-super-Mare – †5 June 1940, Oxford) -

mathematician, mathematical theory of elasticity

5.2 Classical Laminate Theory 183

ε

1=

∂

u

∂

x1

,

ε

2=

∂

v

∂

x2

,

ε

6=

∂

u

∂

x2

+

∂

v

∂

x1

,

κ

1=−

∂

w2

∂

x2

1

,

κ

2=−

∂

w2

∂

x2

2

,

κ

6=−2

∂

w2

∂

x1

∂

x2

ε

T= [

ε

1

ε

2

ε

6]is the vector of midplane strains (stretching and shearing) and

κ

T= [

κ

1

κ

2

κ

6]the vector of curvature (bending and twisting). For all klayers the

stresses are given in condensed form by

σ

(k)

i=Q(k)

i j

ε

i+x3Q(k)

i j

κ

i,i,j=1,2,6 (5.2.4)

and the stiffness equations for the stress resultants follow from (4.2.13) - (4.2.18).

The classical laminate theory is also called shear rigid theory, the material equa-

tions yield zero shear stresses

σ

4,

σ

5for zero strains

ε

4,

ε

5, in the case that the shear

stiffness has ﬁnite values. But the equilibrium conditions yield non-zero stresses

σ

4,

σ

5, if the stresses

σ

1,

σ

2and

σ

6are not all constant. This physical contradiction

will be accepted in the classical theory and the transverse shear stresses are approx-

imately calculated with the given stresses

σ

1,

σ

2,

σ

6by the equilibrium equations

(4.1.56).

The approximate calculation of transverse shear stresses can be simpliﬁed if one

assumes the case of cylindrical bending, i.e. N1=N2=N6≈0,M6≈0. The consti-

tutive equation (4.2.18) or the inverted Eq. (4.2.19) with N

N

N≡0

0

0 gives





0

···

M

M

=





A

A.

.

.B

B

. . . .

B

B.

.

.D

D









ε

···

κ



,



ε

···

κ



=





a

a.

.

.b

b

. . . . .

b

bT.

.

.d

d









0

···

M

M

(5.2.5)

that is with Eqs. (4.2.20) - (4.2.26)

ε

=−A

A

A−1B

B

κ

,M

M

M= (D

D

D−B

B

BA

A

A−1B

B

B)

κ

=D

D

D∗

κ

ε

=b

b

bM

M

M=B

B

B∗D

D

D∗−1M

M

M,

κ

=d

d

dM

M

M=D

D

D∗−1M

M

M(5.2.6)

For symmetric laminates are B

B

B=0

0

0,B

B

B∗=0

0

0,D

D

D∗=D

D

Dand Eqs. (5.2.6) can be replaced

by

ε

=0

0

0,

κ

=D

D

D−1M

M

M(5.2.7)

The partial extensional and coupling stiffness ˜

A

A(x3),˜

B

B(x3), Fig. 5.1, become

184 5 Classical and Improved Theories

❅

❅❅

❅ ❅

❅

❅❅

❅

❄❄

x(m−1)

3x(m−2)

3x(0)

3

x3

N

(m−1)

Fig. 5.1 Derivation of partial stiffness ˜

A

A(x3)and ˜

B

B(x3)for the shaded part of the cross-section

˜

A

A(x3) =

x3

Z

x(0)

3

Q

Q(x3)dx3

=

m−1

∑

k=1

Q

Q(k)h(k)+Q

Q

Q(m)x3−x(m−1)

3,

˜

B

B(x3) =

x3

Z

x(0)

3

Q

Q(x3)x3dx3

=

m−1

∑

k=1

Q

Q(k)s(k)+1

2Q

Q

Q(m)x2

3−x(m−1)

3

2,

(5.2.8)

h(k)=x(k)

3−x(k−1)

3,s(k)=h(k)x(k)

3=1

2x(k)

3+x(k−1)

3x(k)

3−x(k−1)

3

Outgoing from the equilibrium equations (2.2.1) the shear stress equations can be

written

5.2 Classical Laminate Theory 185

σ

5(x3) =−

x3

Z

x(0)

3



∂σ

1

∂

x1

+

∂σ

6

∂

x2dx3

=−

x3

Z

x(0)

3



∂

x1Q(k)

1j

ε

j+x3Q(k)

1j

κ

j+

∂

x2Q(k)

6j

ε

j+x3Q(k)

6j

κ

jdx3,

σ

4(x3) =−

x3

Z

x(0)

3



∂σ

6

∂

x1

+

∂σ

2

∂

x2dx3

=−

x3

Z

x(0)

3



∂

x1Q(k)

6j

ε

j+x3Q(k)

6j

κ

j+

∂

x2Q(k)

2j

ε

j+x3Q(k)

2j

κ

jdx3

(5.2.9)

or in vector-matrix notation



σ

5(x3)

σ

4(x3)=−

x3

Z

x(0)

3

1 0 0

0 0 1 





Q(k)

1j(

ε

j,x1+x3

κ

j,x1)

Q(k)

2j(

ε

j,x1+x3

κ

j,x1)

Q(k)

6j(

ε

j,x1+x3

κ

j,x1)





dx3

−

x3

Z

x(0)

3

1 0 0

0 0 1 





Q(k)

1j(

ε

j,x2+x3

κ

j,x2)

Q(k)

2j(

ε

j,x2+x3

κ

j,x2)

Q(k)

6j(

ε

j,x2+x3

κ

j,x2)





dx3

(5.2.10)

with (...),x

α

=

∂

.../

∂

x

α

,

α

=1,2,j=1,2,6. Using Eqs. (5.2.6) - (5.2.8)

σ

s(x3) = −B

B

B1F

F

F(x3)M

M

M,x1−B

B

B2F

F

F(x3)M

M

M,x2(5.2.11)

with

σ

s(x3) = [

σ

5

σ

4]T,M

M

M,xi= [M1,xiM2,xiM6,xi]T,

F

F(x3) = [ ˜

A

A(x3)A

A

A−1B

B

B−˜

B

B(x3)]D

D

D∗−1=



F11 F12 F16

F21 F22 F26

F61 F62 F66 



(5.2.12)

in the general case if non-symmetrical laminate and

F

F(x3) = ˜

B

B(x3)D

D

D−1(5.2.13)

for symmetrical laminates,

B

B1=1 0 0

0 0 1 ,B

B

B2=0 0 1

0 1 0 (5.2.14)

186 5 Classical and Improved Theories

are so called Boolean3matrices. Equation (5.2.11) can also be written in component

notation.

Equations (5.2.11) and (5.2.16) constitute the straight forward equilibrium ap-

proach for transverse shear stresses which only neglects the inﬂuence of the in-plane

force derivatives N

N

N,xi, but this is a very minor restriction, since, in most engineer-

ing applications, the dominating source for transverse shear stresses are transverse

force resultants. To express the bending moment derivatives by transverse shear

stress resultants it is necessary to assume special selected displacements modes.

If one selects the cylindrical bending around the x1- and the x2-axis one obtains

M6=0,M1,x2=0,M2,x1=0

M1,x1(x1) = Qs

1(x1),M2,x2(x2) = Qs

2(x2)(5.2.15)

with the transverse forces

Qs

1(x1) = Z

(h)

σ

5(x3)dx3=

n

∑

k=1Z

(h)

σ

(k)

5(x3)dx3,

Qs

2(x2) = Z

(h)

σ

4(x3)dx3=

n

∑

k=1Z

(h)

σ

(k)

4(x3)dx3

(5.2.16)

Equation (5.2.11) becomes in matrix notation

σ

s(x3) = F

F

F(x3)Q

Q

Qs,

σ

s= [

σ

5(x3)

σ

4(x3)]T,Q

Q

Qs= [Qs

1(x1)Qs

2(x2)]T,

F

F=F11 (x3)F62 (x3)

F61 (x3)F22 (x3)(5.2.17)

Summarizing the derivations of transverse shear stresses we have considered two

cases

1. N

N

N≡0

0

0,M

M

M= [M1M2M6]T,

2. N

N

N≡0

0

0,M

M

M= [M1(x1)M2(x2)]T

In case 1. follow Eqs. (5.2.18) and in case 2. Eqs. (5.2.19)



σ

5(x3)

σ

4(x3)=F11 (x3)F12 (x3)F16(x3)

F61 (x3)F62 (x3)F66(x3)

∂

x1



M1

M2

M6



+F61 (x3)F62 (x3)F66(x3)

F21 (x3)F22 (x3)F26(x3)

∂

x2



M1

M2

M6

,

(5.2.18)

3George Boole (∗2 November 1815 Lincoln - †8 December 1864 Ballintemp) - mathematician,

educator, philosopher and logician

5.2 Classical Laminate Theory 187



σ

5(x3)

σ

4(x3)=F11 (x3)F62 (x3)

F61 (x3)F22 (x3)Q1

Q2,Qs

1=Qs

1(x1)

Qs

2=Qs

2(x2),

∂

M1(x1)

∂

x1

=Q1,

∂

M2(x2)

∂

x2

=Q2

(5.2.19)

Symmetric laminates are preferred in engineering applications. In this case

D

D∗=D

D

D,B

B

B≡0

0

0 and F

F

F(x3) = −˜

B

B(x3)D

D

D−1. The calculation of the transverse shear

stresses is more simple. The approximate solution for transverse shear stresses in

the classical laminate theory satisﬁes the equilibrium condition. The shear stresses

are layerwise parabolic functions and there is no stress jump at the layer interfaces.

Also in the frame of the classical laminate theory an approximate constitutive

equation can be formulated

Q

Qs=A

A

As

ε

sor Qs

1

Qs

2=A55 A45

A45 A44 

ε

5

ε

4(5.2.20)

Regarding the complementary transverse shear theory formulated in shear stresses

W∗s

1=1

2Z

(h)

σ

sT(C

C

C)−1

σ

sdx3(5.2.21)

and in shear forces

W∗s

2=1

2Q

Q

QsT(A

A

As)−1Q

Q

Qs(5.2.22)

The stress vector

σ

sis a function of x3only, and therefore the integration is carried

out over x3. In

C

Cs=C55 C45

C45 C44 

the Ci j,i,j=4,5 are the elastic parameters of the Hooke’s law. In Eq. (5.2.21) the

stress can be replaced by the transverse force resultants, Eq. (5.2.19). The Qs

ido not

depend on x3and Eq. (5.2.21) yields

W∗s

1=1

2Q

Q

QsT 



Z

(h)

F

FT(x3)(C

C

Cs)−1F

F

F(x3)dx3



Q

Q

Qs(5.2.23)

F

F(x3)is the reduced elasticity matrix Eq. (5.2.18) and Eq. (5.2.23) leads to

W∗s

1=1

2[Qs

1Qs

2]







Z

(h)F11 F62

F61 F22 TC55 C45

C45 C44 −1F11 F62

F61 F22 dx3







Qs

1

Qs

2(5.2.24)

With W∗s

1=W∗s

2follows the approximate shear stiffness

188 5 Classical and Improved Theories

A

As=



Z

(h)

F

FT(C

C

Cs)−1F

F

Fdx3





−1

(5.2.25)

The Ci j are layerwise constant. The calculation of A

A

Asdemands an integration over

layerwise deﬁned polynomials of 4th order and can be just simple carried out by

programming. For unsymmetrical laminates F

F

F(x3)is deﬁned by Eq. (5.2.12).

Hygrothermal effects have no inﬂuence on the transverse shear stresses. In the

classical laminate theory for mechanical and hygrothermal loading as demonstrated

in Sect. 4.2.5, the resultants N

N

Nand M

M

Mmust be substituted by the effective resultants

˜

N

Nand ˜

M

M.

5.3 Shear Deformation Theory for Laminates and Sandwiches

The classical laminate theory allows us to calculate the stresses and strains with

high precision for very thin laminates except in a little extended region near the

free edges. The validity of the classical theory has been established by comparing

theoretical results with experimental tests and with more exact solutions based on

the general equations of the linear anisotropic elasticity theory.

If the width-to-thickness ratio is less about 20, the results derived from the clas-

sical theory show signiﬁcant differences with the actual mechanical behavior and

the modelling must be improved.

A ﬁrst improvement is to include approximately the effect of shear deformation

in the framework of a ﬁrst-order displacement approach. A further improvement is

possible by introducing correction factors for the transverse shear moduli.

The model used now has the same general form, as (5.1.2), for the displacements,

but contrary to the classical theory,

ψ

1and

ψ

2are independent functions and a nor-

mal line to the middle surface of the composite remains straight under deformation,

however it is not normal to the deformed middle plane. In the shear deformation the-

ory the actual deformation state is approximated by 5 independent two-dimensional

functions u,v,w,

ψ

1,

ψ

2, in the classical theory by 3 functions u,v,w, respectively.

The strains are deduced from the displacements, (5.1.3). The components of the

strains

ε

(x1,x2,x3) =

ε

(x1,x2) + x3

κ

(x1,x2),i=1,2,6

again vary linearly through the thickness hand are given by

ε

1=

∂

u

∂

x1

,

ε

2=

∂

v

∂

x2

,

ε

6=

∂

u

∂

x2

+

∂

v

∂

x1

,

κ

1=

∂ψ

1

∂

x1

,

κ

2=

∂ψ

2

∂

x2

,

κ

6=

∂ψ

2

∂

x1

+

∂ψ

1

∂

x2

(5.3.1)

5.3 Shear Deformation Theory for Laminates and Sandwiches 189

The components of the vector

ε

T= [

ε

1

ε

2

ε

6]are not changed, however the compo-

nents of the curvature vector

κ

T= [

κ

1

κ

2

κ

6]are now expressed by the derivatives

of the functions

ψ

1,

ψ

2. The stresses in the kth layer can be expressed by







σ

1

σ

2

σ

6

σ

4

σ

5







(k)

=Q

Q

Q(k)





ε

1

ε

2

ε

6

ε

4

ε

5







(k)

=Q

Q

Q(k)







∂

u

∂

x1

+x3

∂ψ

1

∂

x1

∂

v

∂

x2

+x3

∂ψ

2

∂

x2

∂

u

∂

x2

+

∂

v

∂

x1

+x3

∂ψ

2

∂

x1

+

∂ψ

1

∂

x2

∂

w

∂

x2

+

ψ

2

∂

w

∂

x1

+

ψ

1







(5.3.2)

The stresses

σ

1,

σ

2and

σ

6are superimposed on the extensional and the ﬂexural

stresses and vary linearly through a layer thickness, the stresses

σ

4,

σ

5are, in con-

tradiction to the equilibrium equations, constant through h(k). The strains

ε

1,

ε

2,

ε

6

vary linearly and the strains

ε

4,

ε

5constant through the laminate thickness h, i.e. they

vary continuously through the total thickness. Unlike, the corresponding stresses

σ

1,

σ

2,

σ

6and

σ

4,

σ

5vary linearly or remain constant, respectively, through each

layer thickness h(k)only. Therefore is no stress continuity through the laminate

thickness but stress jumps from ply to ply at their interfaces depending on the re-

duced stiffness Q

Q

Qand Q

Q

Qs.

With the deﬁnition equations for the stress resultants N

N

N,M

M

M,Q

Q

Qsand the stiffness

coefﬁcients Ai j,Bi j ,Di j ,As

i j for laminates (4.2.13) - (4.2.15) or sandwich (4.3.8) -

(4.3.10), (4.3.12) - (4.3.14), respectively, the constitutive equation can be written in

the condensed hypermatrix form, Eqs. (4.2.16)





N

M

Q

Qs

=



A

A B

B

B0

0

B

B D

D

D0

0

0 0

0

0A

A

As





ε

κ

γ

s

(5.3.3)

The stretching, coupling and bending stiffness Ai j ,Bi j ,Di j stay unchanged in com-

parison to the classical laminate theory. The shear stiffness are approximately given

by

As

i j =

n

∑

k=1

C(k)

i j h(k),i,j=4,5 (5.3.4)

The C(k)

i j are the constant shear moduli of the kth lamina. These approximated shear

stiffness overestimate the shear stiffness since they are based on the assumption of

constant transverse shear strains and also do not satisfy the transverse shear stresses

vanishing at the top and bottom boundary layers.

The stiffness values can be improved with help of shear correction factors (Vla-

choutsis, 1992; Altenbach, 2000; Gruttmann and Wagner, 2017). In this case the

part of the constitutive equation relating to the resultants N

N

N,M

M

Mis not modiﬁed. The

190 5 Classical and Improved Theories

other part relating to transverse shear resultants Q

Q

Qsis modiﬁed by replacing the

stiffness As

i j by (kA)s

i j . The constants ks

i j are the shear correction factors. A very

simple approach is to introduce a weighting function f(x3)for the distribution of

the transverse shear stresses through the thickness h.

Assume a parabolic function f(x3)

f(x3) = 5

4"1−x3

h/22#(5.3.5)

and considering that for the kth layer

σ

(k)

4=Q(k)

44

ε

4+Q(k)

45

ε

5,

σ

(k)

5=Q(k)

45

ε

4+Q(k)

55

ε

5(5.3.6)

the transverse resultants are:

Q2=

n

∑

k=1

x(k)

3

Z

x(k−1)

3

σ

(k)

4f(x3)dx3

=5

4









n

∑

k=1

Q(k)

44

ε

4

x(k)

3

Z

x(k−1)

3

"1−x3

h/22#dx3+

n

∑

k=1

Q(k)

45

ε

5

x(k)

3

Z

x(k−1)

3

"1−x3

h/22#dx3











,

Q1=

n

∑

k=1

x(k)

3

Z

x(k−1)

3

σ

(k)

5f(x3)dx3

=5

4









n

∑

k=1

Q(k)

45

ε

4

x(k)

3

Z

x(k−1)

3

"1−x3

h/22#dx3+

n

∑

k=1

Q(k)

55

ε

5

x(k)

3

Z

x(k−1)

3

"1−x3

h/22#dx3









The shear stiffness coefﬁcients As

i j of the constitutive equations

Q2=As

44

ε

4+As

45

ε

5,Q1=As

45

ε

4+As

55

ε

5(5.3.7)

are calculated by

As

i j =5

4

n

∑

k=1

Q(k)

i j x(k)

3−x(k−1)

3−4

3h2x(k)

3

3−x(k−1)

3

3

=5

4

n

∑

k=1

Q(k)

i j "h(k)−4

h2h(k) h(k)2

12 +x(k)

3

2!#,i,j=4,5

(5.3.8)

5.3 Shear Deformation Theory for Laminates and Sandwiches 191

This approach yields for the case of single layer with Q44 =Q55 =G,Q45 =0 a

shear correction factor ks=5/6 for the shear stiffness Gh

As=5

4Gh−4h

h2h2

12 +0=5

6Gh (5.3.9)

The weighting function (5.3.5) resulting in a shear correction factor ksis consistent

with the Reissner theory of shear deformable single layer plates (Reissner, 1944)

and slightly differ from Mindlin’s value (Mindlin, 1951).

A second method to determine shear correction factors consists of considering

the strain energy per unit area of the composite. Some remarks on this method are

given in Chaps. 7 and 8. However shear correction factors depend on the special

loading and stacking conditions of a laminate and not the only factors is generally

applicable.

A particularly physical foundation to improve the shear stiffness values A

A

Asis the

equilibrium approach, Eq. (5.2.25). The sequence of calculation steps for determin-

ing improved transverse shear stresses in the frame of the FSDT are analogous to

the CLT and shall be shortly repeated

•ﬁrstly, calculate the improved shear stiffness

A

As=



Z

(h)

F

FTC

C

Cs−1F

F

Fdx3





−1

(5.3.10)

•secondly, calculate the resultant transverse shear forces

Q

Qs=A

A

As

ε

s(5.3.11)

•thirdly, calculate the improved transverse shear stresses

σ

s=F

F

FQ

Q

Qs

A

As= [Ai j],i,j=5,4,C

C

Cs= [Ci j],i,j=5,4

F

F=F11 F62

F61 F22 ,Q

Q

Qs= [Qs

1Qs

2]T,

σ

= [

σ

5

σ

4]T,

ε

s= [

ε

5

ε

4]T

(5.3.12)

Relying on the results of calculation improved transverse shear stresses

σ

s, the

transverse normal stress can be evaluated. The following equations explain the prin-

cipal way. One starts with solving the equilibrium condition for

σ

3, Eq. (2.2.1)

σ

3(x3) = −

x3

Z

x3=0

∂σ

5

∂

x1

+

∂σ

4

∂

x2dx3+p0(5.3.13)

p0denotes the transverse load at the starting point of integration.

With

192 5 Classical and Improved Theories

F

F(x3) = F11 F62

F61 F22 =f

f

fT

1

f

fT

2(5.3.14)

we are able to replace the transverse shear stresses in Eq. (5.3.13) by Eq. (5.3.12)

σ

3(x3) = −



x3

Z

x3=0

f

fT

1dx3Q

Q

Qs

,x1+

x3

Z

x3=0

f

fT

2dx3Q

Q

Qs

,x2

+p0(5.3.15)

Only the components of f

f

f1and f

f

f2depend on x3and therefore the derivatives of Q

Q

Qs

remain unchanged by the integration process. Moreover, Eq. (5.2.12) demonstrates

that only the partial stiffness ˜

A

A(x3)and ˜

B

B(x3)depend on x3, but not the matrices A

A

A,B

B

and D

D

D∗. Therefore the integration of F

F

F(x3)yields

x3

Z

x3=0

F

F(x3)dx3=



x3

Z

x3=0

˜

A

A(x3)dx3A

A

A−1B

B

B−

x3

Z

x3=0

˜

B

B(x3)dx3

D

D

D∗−1=ˆ

F

F(x3)(5.3.16)

For symmetrical laminates is the coupling matrix B

B

B≡0

0

0 and ˆ

F

F(x3)can be simpliﬁed

to x3

Z

x3=0

F

F(x3)dx3=−

x3

Z

x3=0

˜

B

B(x3)dx3D

D

D−1=ˆ

F

F(x3)(5.3.17)

Now, Eq. (5.3.15) can be transformed into

σ

3(x3) = −hˆ

f

fT

1Q

Q

Qs

,x1+ˆ

f

fT

2dx3Q

Q

Qs

,x2i+p0,(5.3.18)

where

ˆ

f

fT

1= [ ˆ

F11 ˆ

F62 ],ˆ

f

fT

2= [ ˆ

F61 ˆ

F22 ]

and

Qs

2,x2= (A

A

As

ε

s),x2

The boundary conditions of vanishing transverse shear stresses at both surfaces are

fulﬁlled automatically. The boundary conditions for the transverse normal stresses

must be regarded and are taken into account in the integration process.

Summarizing the considerations on single layers or smeared modelling of lam-

inated structures it can be seen that an increasing number of higher order theories

particularly for the analysis of laminated plates has been published. The vast ma-

jority falls into the class of plate theories known as displacement based ones. All

consideration in this textbook are restricted to such theories. The term ”higher order

theories” indicates that the displacement distribution over the thickness is repre-

sented by polynomials of higher than ﬁrst order. In general, a higher approximation

will lead to better results but also requires more expensive computational effort and

the accuracy improvement is often so little that the effort required to solve the more

complicated equations is not justiﬁed. In addition, the mechanical interpretation of

the boundary conditions for higher order terms is very difﬁcult. The most used ESLT

5.4 Layerwise Theories 193

in engineering applications of composite structure elements is the FSDT. The CLT

applications are limited to very thin laminates only, for in comparison to homoge-

neous isotropic plates, the values of the ratio thickness to minimum in-plane dimen-

sion to regard a plate as ”thin” or as ”moderate thick” must be considerably reduced.

Generally, ﬁbre-reinforced material is more susceptible to transverse shear than its

homogeneous isotropic counterpart and reduces the range of applicability of CLT.

Increasing in-plane stiffness may alternatively be regarded as relevant reduction of

its transverse shear strength.

The FSDT yields mostly sufﬁcient accurate results for the displacements and

for the in-plane stresses. However, it may be recalled, as an example, that transverse

shear and transverse normal stresses are main factors that cause delamination failure

of laminates and therefore an accurate determination of the transverse stresses is

needed.

In Sect. 5.3 it was demonstrated that one way to calculate the transverse stresses

is an equilibrium approach in the frame of an extended 2D-modelling. Another rel-

ative simple method is to expand the FSDT from ﬁve to six unknown functions or

degrees of freedom, respectively, by including an x3-dependent term into the poly-

nomial representation of the out-of-plane displacement u3(x1,x2,x3). Several other

possibilities can be found in the literature.

5.4 Layerwise Theories

Layerwise theories are developed for laminates or sandwiches with thick single lay-

ers. Layerwise displacement approximations provide a more kinematically correct

representation of the displacement functions through the thickness including cross-

sectional warping associated with the deformation of thick composite structures.

So-called partial layerwise theories are mostly used which assume layerwise expan-

sions for the in-plane displacement components only. Otherwise so-called full lay-

erwise theories use expansions for all three displacement components. Compared

with equivalent single layer models the partial layerwise model provides a more re-

alistic description of the kinematics of composite laminates and the discrete-layer

behavior of the in-plane components.

Assume a linear displacement approximation (5.1.2) for each layer

u(k)

1(x1,x2,x3) = u(k)(x1,x2) + x3

ψ

(k)

1(x1,x2),

u(k)

2(x1,x2,x3) = v(k)(x1,x2) + x3

ψ

(k)

2(x1,x2),

u(k)

3(x1,x2,x3) = w(x1,x2)

(5.4.1)

with x(k−1)

3≤x3≤x(k)

3;k=1,2,...n. A laminate with nlayers is determined by

(4n+1)unknown functions u(k),v(k),

ψ

(k)

1,

ψ

(k)

2,w;k=1,2,...,n. The continuity

conditions of the displacements at the layer interfaces yield 2(n−1)equations and

194 5 Classical and Improved Theories

the equilibrium for the transverse shear stresses yield additional 2(n−1)equations.

With these 2 ·2(n−1)equations the maximum number of the unknown functions

can be eliminated and we have independent of the number of layers in all cases

(4n+1)−(4n−4) = 5 unknown trial functions. An equivalent single layer model

in the ﬁrst-order shear deformation theory and the partial layerwise model have

the same number of functional degrees of freedom, which are 5. The modelling

of laminates or sandwiches on the assumption of the partial layerwise theory is

often used in the ﬁnite element method. A comparison of equivalent single layer

and layerwise theories one can ﬁnd in Reddy (1993).

Summarizing one can say for the class of partial or discrete layer-wise models

that all analytical or numerical equations are two-dimensional and in comparison to

a real three-dimensional modelling, their modelling and solution effort, respectively,

is less time and cost consuming. The transverse normal displacement does not have

a layerwise representation, but compared to the equivalent single layer modelling,

the partial layerwise modelling provides more realistic description of the kinematics

of composite laminates or sandwiches by introducing discrete layerwise transverse

shear effects into the assumed displacement ﬁeld.

Discrete layerwise theories that neglect transverse normal strain are not capa-

ble of accurately determining interlaminar stresses and modelling localized effects

such as cutouts, free edges, delamination etc. Full or generalized layerwise theories

include in contrast to the partial layerwise transverse shear and transverse normal

stress effects.

Displacement based ﬁnite element models of partial and full layerwise theories

have been developed and can be found in the literature. In Chap. 11 the exemplary

consideration of ﬁnite beam and plate elements have been restricted to CLT and

FSDT.

5.5 Problems

Exercise 5.1. The displacement ﬁeld of a third order laminate (5.1.1) may deﬁned

by

α

=−c0,

β

=1,

γ

=0,

δ

=−c1,¯

β

=¯

γ

=0.

1. Formulate the displacement equations and recover the displacement equations

for the classical and the shear deformation laminate theory.

2. Introduce new variables

φ

1=

ψ

1−c0

∂

w/

∂

x1,

φ

2=

ψ

2−c0

∂

w/

∂

x2and express

the displacement ﬁeld in terms of

φ

1and

φ

2.

3. Substitute the displacements into the linear strain-displacement relations.

4. Formulate the equations for the transverse shear stresses

σ

4,

σ

5. Find the equa-

tions for c1so that the transverse shear stresses vanish at the top and the bottom

of the laminate if c0=1.

Solution 5.1. In the case of a third order displacement ﬁeld one obtains the follow-

ing answers:

1. The starting point is the displacement ﬁeld

5.5 Problems 195

u1(x1,x2,x3) = u(x1,x2) + x3

ψ

1(x1,x2)−c0

∂

w(x1,x2)

∂

x1−x3

3c1

χ

1(x1,x2),

u2(x1,x2,x3) = v(x1,x2) + x3

ψ

2(x1,x2)−c0

∂

w(x1,x2)

∂

x2−x3

3c1

χ

2(x1,x2),

u3(x1,x2,x3) = w(x1,x2)

Classical laminate theory: c1=0,

ψ

1=0,

ψ

2=0,c0=1

First shear deformation theory: c0=c1=0

2. The starting point is now another displacement ﬁeld

u1(x1,x2,x3) = u(x1,x2) + x3

φ

1(x1,x2)−c1x3

3

χ

1(x1,x2),

u2(x1,x2,x3) = v(x1,x2) + x3

φ

2(x1,x2)−c1x3

3

χ

2(x1,x2),

u3(x1,x2,x3) = w(x1,x2)

3. Using the strain-displacement equation (2.2.3) and substitute equations b) we

ﬁnd

ε

1=

∂

u1

∂

x1

=

∂

u

∂

x1

+x3

∂φ

1

∂

x1−x3

3c1

∂ χ

1

∂

x1

=

ε

0

1+x3

ε

I

1+x3

3

ε

II

1,

ε

2=

∂

u2

∂

x2

=

∂

v

∂

x2

+x3

∂φ

2

∂

x2−x3

3c1

∂ χ

2

∂

x2

=

ε

0

2+x3

ε

I

2+x3

3

ε

II

2,

ε

6=

∂

u2

∂

x1

+

∂

u1

∂

x2

=

∂

v

∂

x1

+

∂

u

∂

x2

+x3

∂φ

2

∂

x1

+

∂φ

1

∂

x2−x3

3c1

∂ χ

2

∂

x1

+

∂ χ

1

∂

x2

=

ε

0

6+x3

ε

I

6+x3

3

ε

II

6,

ε

4=

∂

u3

∂

x2

+

∂

u2

∂

x3

=

∂

w

∂

x2

+

φ

2−3c1x2

3

χ

2

=

ε

0

4+x2

3

ε

II

4,

ε

5=

∂

u3

∂

x1

+

∂

u1

∂

x3

=

∂

w

∂

x1

+

φ

2−3c1x2

3

χ

1

=

ε

0

5+x2

3

ε

II

5

Note

ε

0

i≡

ε

i.

4. The transverse shear stress in the kth layer of a laminate follow with (5.3.5) to

σ

(k)

4=Q(k)

44

ε

4+Q(k)

45

ε

5=Q(k)

44 (

ε

0

4+x2

3

ε

II

4) + Q(k)

45 (

ε

0

5+x2

3

ε

II

5),

σ

(k)

5=Q(k)

45

ε

4+Q(k)

55

ε

5=Q(k)

55 (

ε

0

5+x2

3

ε

II

5) + Q(k)

45 (

ε

0

4+x2

3

ε

II

4)

The transverse shear stresses shall vanish at the bottom and the top of the lami-

nate, i.e.

σ

(k)

4(±h/2) =

σ

(k)

5(±h/2) = 0 if k=1 or n.

196 5 Classical and Improved Theories

Q(1)

44 

ε

0

4+h2

4

ε

II

4+Q(1)

45 

ε

0

5+h2

4

ε

II

5=0,

Q(n)

44 

ε

0

4+h2

4

ε

II

4+Q(n)

45 

ε

0

5+h2

4

ε

II

5=0,

Q(1)

55 

ε

0

5+h2

4

ε

II

5+Q(1)

45 

ε

0

4+h2

4

ε

II

4=0,

Q(1)

55 

ε

0

5+h2

4

ε

II

5+Q(1)

45 

ε

0

4+h2

4

ε

II

4=0

=⇒

ε

0

4+h2

4

ε

II

4=0,

ε

0

5+h2

4

ε

II

5=0

In view of the fact that for c0=1 follows

φ

2=

ψ

2−

∂

w

∂

x2⇒

ε

4=

ψ

2−x2

33c1

χ

2,

ε

0

4+h2

4

ε

II

4=0⇒

ψ

2=h2

43c1

χ

2

If 3c1=4/h2⇒

χ

2=

ψ

2. Analogously follow with 3c1=4/h2that

χ

1=

ψ

1, i.e

ε

0

4+1

3c1

ε

II

4=

ψ

2−

ψ

2=0,

ε

0

5+1

3c1

ε

II

5=

ψ

1−

ψ

1=0

The condition 1/3c1=h2/4, i.e. c1=4/3h2is sufﬁcient to make the transverse

shear stresses

σ

4and

σ

5zero at the top and the bottom of the laminate.

Exercise 5.2. A symmetric cross-ply laminate [00/900]Shas the properties h=1

mm, E′

1=141 GPa, E′

2=9,4 GPa, E′

4≡G′

23 =3,2 GPa, E′

5≡G′

13 =E′

6≡G′

12 =4,3

GPa,

ν

′

12 =0,3.

1. Using the simpliﬁed equations (5.2.8) to calculate the shear stresses

σ

5(x3),

σ

4(x3)and sketch their distribution across the laminate thickness hfor

given transverse force resultants Q1=

∂

M1/

∂

x1,Q2=

∂

M2/

∂

x2and M6≡0.

2. Compare the average shear stiffness with the improved corrected stiffness values.

Solution 5.2. The solution can be obtained as follows.

1. The reduced stiffness matrix Q

Q

Qand the shear stiffness matrix C

C

C≡G

G

Gmust be

calculated for the four layers

00-layers,

ν

′

21 =

ν

12E′

2/E′

1:

5.5 Problems 197

Q

Q[00]≡Q

Q

Q′=





E′

1

(1−

ν

′

12

ν

′

21)

ν

′

12E′

2

(1−

ν

′

12

ν

′

21)0

ν

′

12E′

2

(1−

ν

′

12

ν

′

21)

E′

2

(1−

ν

′

12

ν

′

21)0

0 0 E′

6







=



141,85 2,84 0

2,84 9,46 0

0 0 4,3

GPa,

G

G[00]≡G

G

G′=G′

13 0

0G′

23 =4,3 0

0 3,2GPa

900-layers:

Q

Q[900]=



9,46 2,84 0

2,84 141,85 0

0 0 4,3

GPa,

G

G[900]=3,2 0

0 4,3GPa

The bending stiffness matrix follows with (4.2.15)

Di j =

4

∑

k=1

Q(k)

i j h(¯x(k)

3)2+ (h(k))2/12ih(k),

D

D=



9,654 0,207 0

0,207 1,379 0

0 0 0,314 

GPamm3

The corrected ﬂexural stiffness matrix D

D

D∗(5.2.6) is identical D

D

Dfor symmetric

laminates, i.e. D∗=D, and the F

F

F(x3)-matrix in (5.2.12) can be simpliﬁed

F

F(x3) = −˜

B

B(x3)D

D

D−1

The inversion of the matrix D

D

Dyields with

∆

=22,572 the elements D−1

11 of the

inverse matrix D

D

D−1

11 =D22/

∆

,D−1

22 =D11/

∆

,D−1

12 =D12/

∆

,D−1

66 = (D66)−1

D

D−1=



0,104 −0,016 0

−0,016 0,727 0

0 0 3,185 

[GPamm3]−1

Using (5.2.6) the shearing coupling stiffness ˜

B

B(x3)for the layers of the laminate

can be calculated

198 5 Classical and Improved Theories

˜

B

B[00](x3) =



70,93 1,42 0

1,42 4,73 0

0 0 2,30 

GPax2

3−



17,73 0,36 0

0,36 1,18 0

0 0 0,58 

kN,

˜

B

B[900](x3) =



4,73 1,42 0

1,42 70,93 0

0 0 2,30 

GPax2

3−



13,60 0,36 0

0,36 5,32 0

0 0 0,57 

kN

and with F

F

F(x3) = −˜

B

B(x3)D

D

D−1



σ

5(x3)

σ

4(x3)=F11 F62

F61 F22 Q1

Q2,

F[00]=−6,79 0

0 2,17 x2

3+1,70 0

0 0,54 mm−1,

F[900]=−0,44 0

0 32,80 x2

3+1,30 0

0 2,46 mm−1

σ

5[00](x3) = F11[00](x3)Q1= (−6,79x2

3+1,70)Q1,

σ

5[900](x3) = F11[900](x3)Q1= (−0,44x2

3+1,30)Q1,

σ

4[00](x3) = F22[00](x3)Q2= (−2,17x2

3+0,54)Q2,

σ

4[900](x3) = F22[900](x3)Q2= (−32,80x2

3+2,46)Q2

The distribution of the shear stresses through the laminate thickness his sketched

in Fig. 5.2.

x3

x1x2

0

0.25

0.5

-0.25

-0.5

1.30

1.28

2.46

0.4

σ

5(x3)/Q1

σ

4(x3)/Q2

Fig. 5.2 Distribution of the shear stresses

σ

5(x3)/Q1and

σ

4(x3)/Q2across the laminate thickness

5.5 Problems 199

2. A simpliﬁed calculation of the average shear stiffness ¯

As

i j yields (4.2.15)

¯

As

i j =

4

∑

k=1

G(k)

i j h(k)=⇒¯

A

As=3,75 0

0 3,75 GPamm

An improved shear stiffness matrix which include the transverse shear stress dis-

tribution follows with the help of the complementary strain energy W∗

W∗=1

2Z

(h)

σ

sTG

G

G′−1

σ

sdx3

=1

2Q

Q

QT



Z

(h)

F

FTG

G

G′−1F

F

Fdx3



Q

Q

Q=1

2Q

Q

QTA

A

As−1Q

Q

With 0,5

Z

0,25

F

FT

[00]G

G

G′−1F

F

F[00]dx3=2,000 0

0 0,295 x5

3

−1,667 0

0 0,246 x3

3

+0,625 0

0 0,092 x3GPa−1,

0,25

Z

0

F

FT

[900]G

G

G′−1F

F

F[900]dx3=0,012 0

0 46,69 x5

3

−0,119 0

0 11,68 x3

3

+0,529 0

0 1,314 x3GPa−1

follows by the sum up over the four layers the improved matrix A

A

As

A

As=3,01 0

0 2,54 GPa

The comparison of ¯

A

Asand A

A

Ascan be carried out in the form

k

ks¯

A

As=A

A

As

which yields the shear correction vector

k

ks=0,7718

0,6513 

200 5 Classical and Improved Theories

References

Altenbach H (2000) On the determination of transverse shear stiffnesses of or-

thotropic plates. Zeitschrift f¨ur angewandte Mathematik und Physik ZAMP

51(4):629–649

Gruttmann F, Wagner W (2017) Shear correction factors for layered plates and

shells. Computational Mechanics 59(1):129–146

Kienzler R (2002) On consistent plate theories. Archive of Applied Mechanics

72(4):229–247

Mindlin RD (1951) Inﬂuence of rotatory inertia and shear on ﬂexural motions of

isotropic elastic plates. Trans ASME J Appl Mech 18:31–38

Reddy JN (1993) An evaluation of equivalent-single-layer and layerwise theories of

composite laminates. Composite Structures 25(1):21 – 35

Reissner E (1944) On the theory of bending of elastic plates. J Math and Phys

23:184–191

Schneider P, Kienzler R (2015) Comparison of various linear plate theories in the

light of a consistent second-order approximation. Mathematics and Mechanics of

Solids 20(7):871–882

Schneider P, Kienzler R, B ¨ohm M (2014) Modeling of consistent second-order plate

theories for anisotropic materials. ZAMM - Journal of Applied Mathematics and

Mechanics / Zeitschrift f¨ur Angewandte Mathematik und Mechanik 94(1-2):21–

42

Vlachoutsis S (1992) Shear correction factors for plates and shells. International

Journal for Numerical Methods in Engineering 33(7):1537–1552

Chapter 6

Failure Mechanisms and Criteria

Failure of structural elements can be deﬁned in a different manner. As in the case of

buckling, a structural element may be considered failure though the material is still

intact, but there are excessive deformations. In Chap. 6 failure will be considered to

be the loss of integrity of the composite material itself.

The failure analysis procedures for metallic structures were well established a

long time ago. In the case of monolithic materials stress concentrations, e.g. around

notches and holes, cause localized failures. For brittle materials local failures may

lead to fracture and therefore to a total loss of load-carrying capability. For ductile

materials local failure may be in the form of yielding and remains localized, i.e., it

is tolerated better than brittle failure. The fail-safe philosophy has been employed

in the design of metallic structures and is standard in engineering applications. Sim-

ilar procedures for composite materials are not well deﬁned and are the object of

intensive scientiﬁc research up to now. Failure of ﬁbre-reinforced materials is a very

complex topic. While it is important to understand the principal mechanisms of fail-

ure, for many applications it is impossible to detail each step of the failure process.

Main causes of failure are design errors, fabrication and processing errors or unex-

pected service conditions. Design errors can be made in both material and structure.

The stress level carried by each lamina in a laminate depends on the elastic mod-

uli. This may cause large stress gradients between laminae which are oriented at

considerably large angles to each other (e.g. 900). If the stress gradients are close

to a limit value, fracture may occur. Such high levels of internal stresses in adja-

cent laminae may develop a result of external applied loads but also by temperature

and moisture changes. Though manufacturing control and material inspection tests

are carried out, structural composites with abnormalities can be produced. The me-

chanical properties of composites may be signiﬁcantly reduced by high temperature

variations, impact damage, etc. Service anomalies can include improper operation,

faulty maintenance, overloads or environmental incurred damage.

If structural loadings produce local discontinuities inside the material we speak

of a crack. Micro-cracking is considered as the nucleation of micro-cracks at the

microscopic level starting from defects and may cause the initiation of material

fracture. Macro-cracking is the propagation of a fracture by the creation of new

201

H. Altenbach et al., Mechanics of Composite Structural

Elements, https://doi.org/10.1007/978-981-10-8935-0_6

202 6 Failure Mechanisms and Criteria

fracture surfaces at the macroscopic level. For composite materials the fraction ini-

tiation is generally well developed before a change in the macroscopic behavior can

be observed.

If in a laminate macro-cracks occur, it may not be catastrophic, for it is possible

that some layers fail ﬁrst and the composite continues to take more loads until all

laminae fail. Failed laminae may still contribute to the stiffness and strength of the

laminate. Laminate failure estimations are based on procedures for ﬁnding the suc-

cessive loads between the ﬁrst and the last ply failure of the laminate. The failure of

a single layer plays a central function in failure analysis of laminates.

In this section the elastic behavior of laminae is primarily discussed from a

macroscopic point of view. But in the case of failure estimations and strength anal-

ysis of a lamina it is important to understand the underlying failure mechanisms

within the constituents of the composites and their effect to the ultimate macro-

scopic behavior. For this reason some considerations on micro-mechanic failure

mechanisms are made ﬁrst and then failure criteria are discussed more in detail.

Summarizing one can say that the ability of failure prediction is a key aspect in

design of engineering structures. The ﬁrst step is to consider what is meant by fail-

ure. Material failure of metallic structures is mostly related with material yielding

or rupture, but with composites it is more complex. Therefore research is ongoing in

developing failure mechanisms and failure criteria for unidirectional ﬁbre laminae

and their laminates and in evaluating the accuracy of the failure criteria.

6.1 Fracture Modes of Laminae

Composite fracture mechanisms are rather complex because of their anisotropic na-

ture. The failure modes depend on the applied loads and on the distribution of rein-

forcements in the composites. In continuous ﬁbre reinforced composites the types

of fracture may be classiﬁed by these basic forms:

•Intralaminar fracture,

•interlaminar fracture,

•translaminar fracture.

Intralaminar fracture is located inside a lamina, interlaminar fracture shows the fail-

ure developed between laminae and translaminar fracture is oriented transverse to

the laminate plane. Inter- and intralaminar fractures occur in a plane parallel to that

of the ﬁbre reinforcement.

Composite failure is a gradual process. The degradation of a layer results in a

redistribution of stresses in the laminate. It is characterized by different local failure

modes

•The failure is dominated by ﬁber degradation, e.g. rupture, microbuckling, etc.

•The failure is dominated by matrix degradation, e.g. crazing.

•The failure is dominated by singularities at the ﬁber-matrix interface, e.g. crack

propagation, delamination, etc.

6.1 Fracture Modes of Laminae 203

Failure modes of sandwich material may be characterized by

•Tensile failure of the sandwich faces

•Wrinkling failure of the faces due to compressive stresses. Wrinkling is charac-

terized by the eigenmodes of buckling faces.

•Shear failure of core or adhesive failure between core and face.

•Crushing failure of the face and core at a support or tensile respectively shear

failure at fasteners.

The following considerations are restricted to the strength of an unidirectional layer

and to the development of reliable criteria for the predicting of the failure of lam-

inae and laminates. The failure criteria in engineering applications are mainly of a

phenomenological character, i.e. analytical approximations of experimental results,

e.g. by curve ﬁtting.

The fracture of a UD-lamina is the result of the accumulation of various elemen-

tary fracture mechanisms:

•Fibre fracture,

•transverse matrix fracture,

•longitudinal matrix fracture

•fracture of the ﬁbre-matrix interface.

Figure 6.1 illustrates various fracture modes of a single layer. In the ﬁbre direc-

tion, as a tensile load is applied, Fig. 6.1a, failure is due to ﬁbre tensile fracture.

2′=T

1′=L

3

σ

L

σ

L

σ

L>0

σ

L<0

σ

T

σ

T>0

σ

LT

σ

LT

σ

T

σ

T<0

a

b

Fig. 6.1 Fracture modes of a single layer in the case of elementary load states. aFibre fracture by

pure tension

σ

L>0 or compression

σ

L<0 (micro-buckling), bMatrix fracture by pure tension

σ

T>0, pure shearing

σ

LT and pure compression

σ

T<0

204 6 Failure Mechanisms and Criteria

One ﬁbre breaks and the load is transferred through the matrix to the neighboring

ﬁbres which are overloaded and fail too. The failure propagates rapidly with small

increasing load. Otherwise a tensile fracture perpendicular to the ﬁbres, Fig. 6.1b,

due a combination of different micromechanical failure mechanisms: tensile failure

of matrix material, tensile failure of ﬁbres across the diameters, failure of the inter-

face between ﬁbre and matrix. The shear strength, Fig. 6.1b, is limited by the shear

strength of the matrix material, the shear strength between the ﬁbre and the ma-

trix, etc. Figure 6.2 shows the basic strength parameters of a unidirectional lamina

referred to the principal material axes. For in-plane loading of a lamina 5 strength

parameters are necessary, but it is important to have in mind that for composite ma-

terials different strength parameters are measured for tensile and for compression

tests. If the shear stresses act parallel or transverse to the ﬁbre orientation there is

✻

❄

✻

❄

✛ ✲ ✲ ✛

✲

✻

❄

✛

σ

L>0

σ

L>0

σ

L<0

σ

L<0

σ

T>0

σ

T>0

σ

T<0

σ

T<0

σ

LT

σ

LT

material property:

σ

Lt material property:

σ

Lc

material property:

σ

Tc

material property:

σ

Tt

material property:

τ

S

ab

d

c

e

Fig. 6.2 Basic strength parameters. aLongitudinal tensile strength

σ

Lt,bLongitudinal compres-

sive strength

σ

Lc,cTransverse tensile strength

σ

Tt,dTransverse compressive strength

σ

Tc,eIn-

plane (intralaminar) shear strength

τ

S

6.1 Fracture Modes of Laminae 205

no inﬂuence of the load direction (Fig. 6.3a). Otherwise the positive shear stress

σ

6>0 causes tensile in L-direction and compression in T-direction and vice versa

for

σ

6<0 and other strength parameters are standard. The required experimental

characterization is relatively simple for the parameters

σ

Lt and

σ

Tt, but more com-

plicated for the strength parameters

σ

Lc,

σ

Tc and

τ

S.

In the case of laminates, besides the basic failure mechanisms for a single

layer, such as ﬁbre fracture, longitudinal and transverse matrix fraction, ﬁbre-matrix

debonding, etc. described above, another new fracture mode occurs. This mode is

called delamination and consists of separation of layers from one another. Through-

the-thickness variation of stresses may be caused even if a laminate is loaded by

uniform in-plane loads. Generally, the matrix material that holds the laminae of a

laminate together has substantially smaller strength than the in-plane strength of the

layers. Stresses perpendicular to the interface between laminae may cause breaking

of the bond between the layers in mostly localized, small regions. However, even if

✲

❄

✛

✻

σ

6<0

≡













❅❅

❅



❅

❅

❅■

❅

❅❘



✠



✒ 















σ

L=−

σ

6

σ

T=

σ

6

✛

✻

✲

❄

σ

6>0

≡













❅❅

❅



❅

❅

❅❘

❅

❅■



✒



✠ 















σ

L=

σ

6

σ

T=−

σ

6

✲

❄

✛

✻

σ

LT <0

≡

❅❅

❅



❅

❅

❅■

❅

❅❘



✠



✒

σ

2=−

σ

LT

σ

1=

σ

LT

✛

✻

✲

❄

σ

LT >0

≡

❅❅

❅



❅

❅

❅❘

❅

❅■



✒



✠

σ

2=

σ

LT

σ

1=−

σ

LT

b

a

✲

✻

❅❅

❅❘



✒

1

1′=L

2

2′=T

✲

✻



✒

❅

❅■

2′=T

2

1′=L

1

Fig. 6.3 In-plane shear. aPositive and negative shear stresses along the principal material axes,

bPositive and negative shear stresses at 450with the principal material axis

206 6 Failure Mechanisms and Criteria

the size of such delaminations is small they may affect the integrity of a laminate

and can degrade their in-plane load-carrying capability. Therefore, in practical en-

gineering applications it is important to calculate the interlaminar normal and shear

stresses

σ

3,

σ

4and

σ

5and to check interlaminar failure too.

The deﬁnition of failure may change from case to case and depends on the com-

posite material and the kind of loads. For composite material, such as UD-laminates,

the end of the elastic domain is associated with the development of micro-cracking.

But in the ﬁrst stage, the initiated cracks do not propagate and their development

changes the stiffness of the material very gradually but the degradation is irre-

versible. In the following section failure criteria for laminae will be discussed ﬁrst

to allow the designer to have an evaluation of the mechanical strength of laminae.

Secondly, concepts for laminate failure are considered.

6.2 Failure Criteria

Failure criteria for composites are many and varied. In their simplest form they are

similar, in principle, to those used for isotropic materials, e.g. maximum stress/strain

and distortional energy theories. The major difference between isotropic materials

and unidirectional ﬁbrous composite materials is the directional dependence of the

strength on a macrosopic scale. It is important to realize that failure criteria are

purely empirical. Their purpose is to deﬁne a failure envelope by using a minimum

number of test data. Generally, these experimental data are obtained from relatively

simple uniaxial and pure shear tests. Combined stress tests are more difﬁcult to per-

form and should be, if possible, not included in the determination failure envelopes.

We shall start by considering a single lamina before moving on to discuss failure

of laminates. Longitudinal tension or pressure, transverse tension or pressure and

shear are the ﬁve basic modes of failure of a lamina. Generally the strength in the

principal material axes are regarded as the fundamental parameters deﬁning failure.

When the lamina is loaded at an angle to the ﬁbres one has to determine the stresses

in the principal directions and compare them with the fundamental strength param-

eters. Failure criteria usually grouped in literature into three different classes: limit

criteria, interactive criteria and hybrid criteria which combine selected aspects of

limit and interactive methods. In the following we only discuss selected criteria of

the ﬁrst two classes.

Failure criteria for homogeneous isotropic materials are well established. Macro-

mechanical failure theories for composite materials have been developed by extend-

ing and adapting isotropic failure theories to account for anisotropy in stiffness and

strength of the composite. All theories can be expressed as functions of the basic

strength parameters referred to the principal material axes (Fig. 6.2). Some criteria

do not account for interaction of stress components while others do. Some inter-

action criteria require additional strength parameters obtained by more expended

biaxial experimental tests.

6.2 Failure Criteria 207

Laminate failure criteria are applied on a ply-by-ply basis and the load-carrying

capability of the entire composite is predicted by the laminate or sandwich theories

given in Chaps. 4 and 5. A laminate may be assumed to have failed when the strength

criterion of any one of its laminae is reached (ﬁrst-ply failure). However, the failure

of a single layer not necessarily leads to a total fracture of the laminate structure.

Criteria of an on-axis lamina can be determined with relative easily. Off-axis criteria

can be obtained by coordinate transformations of stresses or strains. Based on the

ply-by-ply analysis ﬁrst-ply failure and last-ply failure concepts can be developed.

Failure criteria have been established in the case of a layer. Of all failure criteria

available, the following four are considered representative and more widely used:

•Maximum stress theory

•Maximum strain theory

•Deviatoric or distorsion strain energy criteria of Tsai-Hill1

•Interactive tensor polynomial criterion of Tsai-Wu2

Maximum stress and maximum strain criteria assume no stress interaction while

the other both include full stress interaction. In the maximum stress theory, failure

occurs when at least one stress component along one of the principal material axes

exceeds the corresponding strength parameter in that direction

σ

L=

σ

Lt,

σ

L>0,

σ

T=

σ

Tt,

σ

T>0,

σ

L=

σ

Lc,

σ

L<0,

σ

T=

σ

Tc,

σ

T<0,

|

σ

LT|=

τ

S,

(6.2.1)

Note that failure can occur for more than one reason. A layer failure does not occur

if −

σ

Lc <

σ

L<

σ

Lt,

−

σ

Tc <

σ

T<

σ

Tt,

−

τ

S<

σ

LT <

τ

S

(6.2.2)

For a two-dimensional state of normal stresses, i.e.

σ

L6=0,

σ

T6=0,

σ

LT =0, the fail-

ure envelope, Fig. 6.4, takes the form of a rectangle. In the case of off-axis tension

or compression of a UD-lamina, Fig. 6.5, the transformed stresses are

σ

L=

σ

1cos2

θ

=

σ

1c2

σ

T=

σ

1sin2

θ

=

σ

1s2

σ

LT =−

σ

1sin

θ

cos

θ

=−

σ

1sc ⇒

σ

1=

σ

L/c2

σ

1=

σ

T/s2

σ

1=−

σ

LT/sc

(6.2.3)

and the maximum stress criteria is expressed as follows

1Rodney Hill (∗11 June 1921, Stourton, Leeds - † 2 February 2011, Yorkshire) - applied mathe-

matician and a former Professor of Mechanics of Solids at Gonville and Caius College, Cambridge,

UK

2Edward Ming-Chi Wu (∗30 September 1938 - †3 June 2009) - US-American engineer

208 6 Failure Mechanisms and Criteria

✻

✲

✛ ✲

✻

❄

σ

T

σ

L

σ

T

σ

L

σ

T

σ

L

σ

Tt

σ

Tc

σ

Lc

σ

Lt

Fig. 6.4 Failure envelope for UD-lamina under biaxial normal loading (max. stress criterion)

✲

✻

❆

❆❑

✠

✟✟✟✟✟✟✟✟✟✟✟✟✟✟✟✟✟

✟✯

✟✟✟✟✟✟✟✟✟

✟

✟✟✟✟

✟

✟ ✟✟✟✟✟✟✟

✟

✟✟

✟

✲✛

θ

σ

1

σ

1

x2

x1

x′

1=xL

x′

2=xT

■

θ

Fig. 6.5 Off-axis unidirectional loading

−

σ

Lc <

σ

1c2<

σ

Lt,

−

σ

Tc <

σ

1s2<

σ

Tt,

−

τ

S<

σ

1sc <

τ

S

(6.2.4)

The ultimate strength for

σ

1corresponds to the smallest of the following six values

σ

1t =

σ

Lt/c2,

σ

1t =

σ

Tt/s2,

σ

1t =

τ

S/sc,

σ

1>0,

σ

1c =

σ

Lc/c2,

σ

1c =

σ

Tc/s2,

σ

1c =

τ

S/sc,

σ

1<0(6.2.5)

The failure modes depend on the corresponding ultimate strength

σ

1u

σ

1u =

σ

Lt/c2ﬁbre failure,

σ

1u =

σ

Tt/s2transverse normal stress failure,

σ

1u =

τ

S/sc in-plane shear failure

In the more general case of off-axis loading, the stress transformation rule, Table

4.1, is used

6.2 Failure Criteria 209





σ

′

1≡

σ

L

σ

′

2≡

σ

T

σ

′

6≡

σ

LT 

=



c2s22sc

s2c2−2sc

−sc sc c2−s2





σ

1

σ

2

σ

6

(6.2.6)

Because of the orthotropic symmetry, shear strength is independent of the sign of

σ

LT (Fig. 6.3) and there are ﬁve independent failure modes in the maximum stress

criterion. There is no interaction among the modes although in reality the failure

processes are highly interacting. The maximum stress theory may be applicable

for brittle modes of failure of material, e.g. follow from transverse or longitudinal

tension (

σ

L>0,

σ

T>0).

The maximum strain theory is quite similar to the maximum stress theory. Now

the strains are limited instead of the stresses. Failure of a lamina occurs when at

least one of the strain components along the principal material axes exceeds the

corresponding ultimate strain in that direction

ε

L=

ε

Lt

ε

L>0,

ε

T=

ε

Tt

ε

T>0,

ε

L=

ε

Lc

ε

L<0,

ε

T=

ε

Tc

ε

T<0,

|

ε

LT|=

ε

S

(6.2.7)

The lamina failure does not occur if

−

ε

Lc <

ε

L<

ε

Lt,

−

ε

Tc <

ε

T<

ε

Tt,

−

ε

S<

ε

LT <

ε

S

(6.2.8)

In the case of unidirectional off-axis tension or compression (Fig. 6.5), the stress

relations are given by (6.2.3). For the in-plane stress state strains in the principal

material axes are 



ε

L

ε

T

ε

LT 

=



S′

11 S′

12 0

S′

12 S′

22 0

0 0 S′

66







σ

L

σ

T

σ

LT 

(6.2.9)

By associating (6.2.3) and (6.2.6) and expressing the compliance parameters S′

i j as

functions of the engineering moduli in the principal directions, EL,ET,GLT ,

ν

LT,

ν

T L, it follows that

ε

L=1

EL

(c2−

ν

LTs2)

σ

1,

ε

T=1

ET

(s2−

ν

TLc2)

σ

1,

ε

LT=−1

GLT

sc

σ

1

(6.2.10)

The maximum strain and the maximum stress criteria must lead to identical values

in the cases of longitudinal loading and

θ

=00or transverse unidirectional loading

and

θ

=900. The identity of the shear equations is given in both cases. This implies

that

210 6 Failure Mechanisms and Criteria

ε

Lt =

σ

Lt

EL

,

ε

Lc =−

σ

Lc

EL

,

ε

Tt =

σ

Tt

ET

,

ε

Tc =−

σ

Tc

ET

,

ε

S=

τ

S

GLT

(6.2.11)

and the maximum strain criterion may be rewritten as follows

−

σ

Lc <

σ

1(c2−

ν

LTs2)<

σ

Lt,

−

σ

Tc <

σ

1(s2−

ν

TLc2)<

σ

Tt,

−

τ

S<

σ

1sc <

τ

S

(6.2.12)

By comparing Eqs. (6.2.4) and (6.2.12) we establish that the two criteria differ by

the introduction of the Poisson’s ratio

ν

LT in the strain criterion. In practice these

terms modify the numerical results slightly. In the special case of a two-dimensional

stress state

σ

L6=0,

σ

T6=0,

σ

LT =0, compare Fig. 6.4, the failure envelope takes the

form of a parallelogram for the maximum strain criterion, Fig. 6.6.

One of the ﬁrst interactive criteria applied to anisotropic materials was introduced

by Hill. For a two-dimensional state of stress referred to the principal stress direc-

tions, von Mises3developed a deviatoric or distortional energy criterion for isotropic

ductile metals (von Mises, 1913) which can be presented for the two-dimensional

stress state as

σ

2

I+

σ

2

II −

σ

I

σ

II =

σ

eq

or in a general reference system

σ

2

1+

σ

2

2−

σ

1

σ

2+3

σ

2

6=

σ

eq

σ

I,

σ

II are principal stresses,

σ

eq the equivalent stress. This criterion was extended

in von Mises (1928) and modiﬁed for the case of orthotropic ductile materials by

Hill (1948)

A

σ

2

1+B

σ

2

2+C

σ

1

σ

2+D

σ

2

6=1 (6.2.13)

✻

✲

σ

L

σ

T

✭✭✭✭✭✭✭✭✭✭✭✭✭✭✭✭✭✭✭✭✭✭

✭

✁

✭✭✭✭✭✭✭✭✭✭✭✭✭✭✭✭✭✭✭✭✭✭

✭

✁

σ

Tt

σ

Tc

σ

Lt

σ

Lc

σ

T−

ν

TL

σ

L=

σ

Tt

σ

L−

ν

LT

σ

T=

σ

Lt

σ

T−

ν

TL

σ

L=−

σ

Tc

σ

L−

ν

LT

σ

T=−

σ

Lc

Fig. 6.6 Failure envelope for UD-lamina under biaxial normal loading (max. strain criterion)

3Richard Edler von Mises (∗19 April 1883 Lemberg, Austria-Hungary (now Lviv, Ukraine) -

† 14 July 1953 Boston, Massachusetts) - mathematician who worked on solid mechanics, ﬂuid

mechanics, aerodynamics, aeronautics, statistics and probability theory, one of founders of the

journal Zeitschrift f¨ur Angewandte Mathematik und Mechanik

6.2 Failure Criteria 211

A,B,C,Dare material parameters. Equation (6.2.13) cannot be deﬁned as distorsion

energy, since in anisotropy distorsion and dilatation energies are not separated. The

criterion (6.2.13) was applied to UD-laminae by Tsai and Wu (1971)

A

σ

2

L+B

σ

2

T+C

σ

L

σ

T+D

σ

2

LT =1 (6.2.14)

The material parameters A,B,C,Dcan be identiﬁed by tests with acting basic load-

ings

σ

L=

σ

LU,

σ

T=0,

σ

LT =0⇒A=1

σ

2

LU

,

σ

L=0,

σ

T=

σ

TU,

σ

LT =0⇒B=1

σ

2

TU

,

σ

L=0,

σ

T=0,

σ

LT =

τ

U⇒D=1

τ

2

U

(6.2.15)

In dependence on the failure mode, the superscript U must be substituted by t,c or s

and denotes the ultimate value of stress at failure.

The remaining parameter Cmust be determined by a biaxial test. The C-term

yields the interaction between the normal stresses. Under equal biaxial normal load-

ing

σ

L=

σ

T6=0,

σ

LT =0 it can be assumed that the failure follows the maximum

stress criterion, i.e failure will occur when the transverse stress reaches the trans-

verse strength

σ

TU which is much lower than the longitudinal strength

σ

LU. Equa-

tion (6.2.14) yields



σ

L

σ

LU 2

+

σ

T

σ

TU 2

+C

σ

2

T=1,

σ

T=

σ

TU =⇒C=−1

σ

2

LU

(6.2.16)

The Tsai-Hill criterion in the case of plane stress state and on-axis loading may be

written



σ

L

σ

LU 2

+

σ

T

σ

TU 2

−

σ

L

σ

T

σ

2

LU

+

σ

LT

τ

U2

=1 (6.2.17)

In the case of tension or compression off the principal material directions, Fig. 6.5,

the Tsai-Hill criterion becomes



σ

1c2

σ

LU 2

+

σ

1s2

σ

TU 2

−

σ

1cs

σ

LU 2

+

σ

1sc

τ

U2

=1 (6.2.18)

and the strength parameter

σ

1Uin x1-direction is

1

σ

1U 2

=c2

σ

LU 2

+s2

σ

TU 2

+1

τ

2

U−1

σ

2

LU c2s2

≈c2

σ

LU 2

+s2

σ

TU 2

+cs

τ

2

U2(6.2.19)

The approximated form presumes

σ

LU ≫

τ

U.

212 6 Failure Mechanisms and Criteria

The Tsai-Hill criterion is a single criterion instead of the three subcriteria re-

quired in maximum stress and strain theories. It allows considerable interaction

among the strain components and for ductile material the failure estimation agrees

well with experimental results.

Gol’denblat and Kopnov (1965) proposed a tensor polynomial criterion. Tsai and

Wu modiﬁed this criterion by assuming the existence of a failure surface in stress

space. They took into account only the ﬁrst two terms of the polynomial criterion

and postulated that fracture of an anisotropic material occurs when the following

equation is satisﬁed

ai j

σ

i j +ai jkl

σ

i j

σ

kl =1 (6.2.20)

or in a contracted notation

ai

σ

i+ai j

σ

i

σ

j=1 (6.2.21)

We are interested in the case of an orthotropic composite material, i.e. a unidirec-

tional lamina, subjected to plane stress state, and the Tsai-Wu criterion may be ex-

pressed as

aL

σ

L+aT

σ

T+aS

σ

S+aLL

σ

2

L+aTT

σ

2

T+

aSS

σ

2

S+2aLT

σ

L

σ

T+2aLS

σ

L

σ

S+2aTS

σ

T

σ

S=1(6.2.22)

Equation (6.2.22) is written in the on-axis system and

ν

LT ≡aS.

The linear terms take account the actual differences between composite material

behavior under tension and compression. The term aLT

σ

L

σ

Trepresents independent

interaction among the stresses

σ

Land

σ

Tand the remaining quadratic terms describe

an ellipsoid in stress space. Since the strength of a lamina loaded under pure shear

stress

τ

Sin the on-axis system is independent of the sign of the shear stress, all linear

terms in

σ

Smust vanish

aS=aLS =aTS =0 (6.2.23)

Then the Tsai-Wu criterion for a single layer in on-axis system has the form

aL

σ

L+aT

σ

T+aLL

σ

2

L+aTT

σ

2

T+aSS

σ

2

S+2aLT

σ

L

σ

T=1 (6.2.24)

The four quadratic terms in (6.2.24) correspond to the four independent elastic char-

acteristics of orthotropic materials, the linear terms allow the distinction between

tensile and compressive strength. The coefﬁcients of the quadratic Tsai-Wu crite-

rion are obtained by applying elementary basic loading conditions to the lamina

σ

L=

σ

Lt,

σ

T=

σ

S=0

σ

L=−

σ

Lc,

σ

T=

σ

S=0⇒aL

σ

Lt +aLL

σ

2

Lt =1

−aL

σ

Lc +aLL

σ

2

Lc =1⇒

aL=1

σ

Lt −1

σ

Lc

aLL =1

σ

Lt

σ

Lc

σ

T=

σ

Tt,

σ

L=

σ

S=0

σ

T=−

σ

Tc,

σ

L=

σ

S=0⇒aT

σ

Tt +aTT

σ

2

Tt =1

−aT

σ

Tc +aTT

σ

2

Tc =1⇒

aT=1

σ

Tt −1

σ

Tc

aTT =1

σ

Tt

σ

Tc

6.2 Failure Criteria 213

σ

S=

τ

S,

σ

L=

σ

T=0⇒aSS

τ

2

S=1⇒aSS =1

τ

2

S

(6.2.25)

The remaining coefﬁcient aLT must be obtained by biaxial testing

σ

L=

σ

T=

σ

U,

σ

S=0⇒

(aL+aT)

σ

U+ (aLL +aTT +2aLT)

σ

2

U=1(6.2.26)

σ

Uis the experimentally measured strength under equal biaxial tensile loading

σ

L=

σ

T.

In many cases the interaction coefﬁcient is not critical and is given approxi-

mately. A sufﬁcient approximation is in this case

aLT ≈ −1

2√aLLaTT (6.2.27)

The Tsai-Wu criterion may also be formulated in strain space.

Summarizing the considerations on interactive failure criteria lead: The Tsai-Hill

and the Tsai-Wu failure criteria are quadratic interaction criteria which have the

general form

Fi j

σ

i

σ

j+Fi

σ

i=1,i,j=L,T,S (6.2.28)

Fi j and Fiare strength parameters and

σ

i,

σ

jthe on axis stress components.

For plane stress state six strength parameters FLL,FTT ,FSS ,FLT,FL,FTare re-

quired for implementation of the failure criterion, FLS =FTS =FS=0, see Eq.

(6.2.23). Five of these strength parameters are conventional tensile, compressive

or shear strength terms which can be measured in a conventional experimental test

programme. The strength parameter FLT is more difﬁcult to obtain, since a biaxial

test is necessary and such test is not easy to perform. The two-dimensional repre-

sentation of the general quadratic criterion (6.2.28) in the stress space can be given

in the equation below

σ

2

L

σ

Lt

σ

Lc

+

σ

2

T

σ

Tt

σ

Tc

+

σ

2

LT

τ

2

S

+2FLT

σ

L

σ

T+1

σ

Lt −1

σ

Lc 

σ

L+1

σ

Tt −1

σ

Tc 

σ

T=1

(6.2.29)

Equation (6.2.29) reduces, e.g., for

σ

Lt =

σ

Lc,

σ

Tt =

σ

Tc,FLT =−1

2

σ

2

Lt

to the Tsai-Hill criterion, for

σ

Lt 6=

σ

Lc,

σ

Tt 6=

σ

Tc,FLT =−1

2

σ

Lt

σ

Lc

to the Hoffman criterion, and for

σ

Lt 6=

σ

Lc,

σ

Tt 6=

σ

Tc,FLT =−1

2√

σ

Lt

σ

Lc

σ

Tt

σ

Tc

214 6 Failure Mechanisms and Criteria

to the Tsai-Wu criterion. Hoffman’s criterion is a simple generalization of the Hill

criterion that allows different tensile and compressive strength parameters (Hoff-

man, 1967).

If one deﬁnes dimensionless stresses as

σ

∗

L=√FLL

σ

L,

σ

∗

T=√FTT

σ

T,

σ

∗

LT =pFSS

σ

LT

and normalized strength coefﬁcients as

F∗

L=FL/√FLL,F∗

T=FT/√FT,F∗

LT =FLT/√FLLFTT

Equation (6.2.29) can be rewritten as

σ

∗2

L+

σ

∗2

T+

σ

∗2

LT +2F∗

LT

σ

∗

L

σ

∗2

T+F∗

L

σ

∗2

L+F∗

T

σ

∗2

T=1 (6.2.30)

Note that in the case of isotropic materials with

σ

Lt =

σ

Lc =

σ

Tt =

σ

Tc follow

F∗

L=F∗

T=0. There the principal stress state will have

σ

∗

LT =

σ

LT =0. Equation

(6.2.30) reduces with F∗

LT =−1

2to the known von Mises criterion.

Using the above failure criteria the possibility of a lamina failing can be deter-

mined, for example. In the maximum stress criterion, the lamina failes if any of the

inequalities (6.2.4) are violated. However, the criterion does not give information

about how much the load can be increased by if the lamina is safe or how much it

can be decreased if the lamina has failed. To overcome this problem, strength ratios

are deﬁned as

R=maximum load which can be applied

load applied (6.2.31)

This deﬁnition is applicable to all failure criteria. If R>1, then the lamina is safe

and the applied load can be increased by a factor of R. If R<1 the lamina is unsafe

and the applied load needs to be reduced. A value of R=1 implies the failure load.

The stress ratio factor assumes that the material is linear elastic, for each state of

stress there is a corresponding state of strain and all components of stress and strain

increase by the same proportion.

Summarizing the discussion above, the strength ratio for the four criteria can be

formulated:

Maximum stress criterion

RLt

σ

=

σ

Lt/

σ

L,

σ

L>0 Strength factor ﬁbre fracture,

RTt

σ

=

σ

Tt/

σ

T,

σ

T>0 Strength factor matrix fracture,

RLc

σ

=

σ

Lc/|

σ

L|,

σ

L<0 Strength factor micro-buckling,

RTc

σ

=

σ

Tc/|

σ

T|,

σ

T<0 Strength factor matrix fracture,

RS

σ

=

τ

S/|

σ

LT|,Strength factor matrix fracture

(6.2.32)

6.2 Failure Criteria 215

Maximum strain criterion

RLt

ε

=

ε

Lt/

ε

L,

ε

L>0,

RTt

ε

=

ε

Tt/

ε

T,

ε

T>0,

RLc

ε

=

ε

Lc/|

ε

L|,

ε

L<0,

RTc

ε

=

ε

Tc/|

ε

T|,

ε

T<0,

RS

ε

=

ε

S/|

ε

LT|

(6.2.33)

Tsai-Hill-criterion

Only one strength ratio can be introduced

RTH

σ

L

σ

LU 2

+RTH

σ

T

σ

TU 2

−RTH

σ

LRTH

σ

T

σ

2

LU

+RTH

σ

LT

τ

U2

=1 (6.2.34)

With the ultimate strength

σ

LU,

σ

TU for tension and compression the strength ratio

RTH follows from

1

(RTH)2=

σ

L

σ

LU 2

+

σ

T

σ

TU 2

−

σ

L

σ

T

σ

2

LU

+

σ

LT

τ

U2

Tsai-Wu-criterion

The Tsai-Hill and the Tsai-Wu criterion deﬁne only one strength ratio RTW

(aL

σ

L+aT

σ

T)RTW + (aLL

σ

2

L+aTT

σ

2

T+aSS

σ

2

S+2aLT

σ

L

σ

T)RTW2=1

or in symbolic notation

ARTW +B(RTW)2=1⇒(RTW )2+A

BRTW =1

B

with the solutions

RTW

1/2=−1

2

A

B±r1

4

A2

B2+1

B=1

2B−A±pA2+4B

RTW must be positive

RTW =pA2+4B−A/2B(6.2.35)

The procedure for laminate failure estimation on the concept of ﬁrst ply and last

ply failure is given as follows:

1. Use laminate analysis to ﬁnd the midplane strains and curvatures depending on

the applied mechanical and hygrothermical loads.

2. Calculate the local stresses and strains in each lamina under the assumed load.

3. Use the ply-by-ply stresses and strains in lamina failure theory to ﬁnd the strength

ratios. Multiplying the strength ratio to the applied load gives the load level of

the failure of the ﬁrst lamina. This load may be called the ﬁrst ply failure load.

Using the conservative ﬁrst-ply-failure concept stop here, otherwise go to step 4.

216 6 Failure Mechanisms and Criteria

4. Degrade approximately fully the stiffness of damaged plies. Apply the actual

load level of previous failure.

5. Start again with step 3. to ﬁnd the strength ratios in the undamaged laminae. If

R>1 multiply the applied load by the strength ratio to ﬁnd the load level of the

next ply failure. If R<1, degrade the stiffness and strength characteristics of all

damaged lamina.

6. Repeat the steps above until all plies have failed. That is the last-ply-failure con-

cept.

The laminate failure analysis can be subdivided into the following four parts. The

ﬁrst-ply-failure concept demands only one run through, the last-ply-failure requires

several iterations with degradation of lamina stiffness.

Failure analysis of laminates in stress space:

Step 1

Calculate the stiffnesses

Q

Q′(k)=



Q′

11 Q′

12 0

Q′

12 Q′

22 0

0 0 Q′

66



≡



QLL QLT 0

QLT QTT 0

0 0 QSS 



of all ksingle layers in on-axis system with help of the layer moduli

E(k)

L,E(k)

T,

ν

(k)

TL ,G(k)

TL and the layer thicknesses h(k)

Transformation of the reduced stiffnesses Q

Q

Q′(k)of single layers in on-axis

system to the reduced stiffnesses Q

Q

Q(k)of single layers in off-axis system

Q

Q(k)= (T

T

ε

′)TQ′

Q′

Q′(k)T

T

ε

′

Calculate the laminate stiffnesses A

A

A,B

B

Band D

D

Ai j =

n

∑

k=1

Q(k)

i j h(k),Bi j =1

2

n

∑

k=1

Q(k)

i j x(k)

3

2−x(k−1)

3

2=

n

∑

k=1

Q(k)

i j h(k)x(k)

3,

Di j =1

3

n

∑

k=1

Q(k)

i j x(k)

3

3−x(k−1)

3

3=

n

∑

k=1

Q(k)

i j h(k)x(k)

3

2+1

12h(k)2,

x(k)

3=1

2x(k)

3+x(k−1)

3

Inversion of the matrices A

A

A,B

B

Band D

D

Calculate the compliance matrices a

a

a,b

b

b,c

c

cand d

d

dof the laminate

a

a=A

A

A∗−B

B

B∗D

D

D∗−1C

C

C∗,b

b

b=B

B

B∗D

D

D∗−1,c

c

c=−D

D

D∗−1C

C

C∗,d

d

d=D

D

D∗−1,

A

A∗=A

A

A−1,B

B

B∗=−A

A

A−1B

B

B,C

C

C∗=B

B

BA

A

A−1,D

D

D∗=D

D

D−B

B

BA

A

A−1B

B

Step 2

Calculation of the laminate stress resultants N

N

Nand M

M

by structural analysis of beam or plate structures

6.2 Failure Criteria 217

Step 3.

Calculate the laminate strains

ε

=

ε

+x3

κ





ε

···

κ



=





a

a.

.

.b

b

. . . .

c

c.

.

.d

d









N

···

M

M



and the strains for all laminae at lamina interfaces

ε

(k)=

ε

0+x(k)

3

κ

,k=0,1,2,...,n

Calculate the stresses for all interface surfaces of single layers

σ

(k)−=Q

Q

Q(k)

ε

(k−1),bottom surface of lamina k

σ

(k)+ =Q

Q

Q(k)

ε

k,top surface of lamina k,k=0,1,2,...,n

Transformation of the interface stresses

σ

(k)−,

σ

(k)+

to the on-axis system of layer kk = 0, 1, 2, . . . , n

Step 4

Failure analysis based on a selected failure criterion in stress space

Summarizing the strength ratios concept to the general quadratic interaction criteria

Eq. (6.2.28) we formulate with the maximum values of stresses

Fi j

σ

′max

i

σ

′max

j+Fi

σ

′max

i=1

Substituting R

σ

′applied

ifor

σ

′max

iyield the quadratic equation for the strength ratio R

(Fi j

σ

i

σ

j)R2+ (Fi

σ

i)R−1=0

or

aR2+bR −1=0,a=Fi j

σ

i

σ

j,b=Fi

σ

i(6.2.36)

The strength ratio Ris equal to the positive quadratic root

R=−b

2a+sb

2a2

+1

a

As considered above this approach is easy to use because the resulting ratio provides

a linear scaling factor, i.e.

if R≤1 failure occurs,

if R>1, e.g. R=2, the safety factor is 2 and the load can be doubled or the laminate

thickness reduced by 0.5 before failure occurs.

The same strength ratio can be determined from the equivalent quadratic criterion

in the strain space. With

σ

=Q

Q

ε

follows, e.g. with Eqs. (6.2.24) - (6.2.27) the Tsai-

Wu criterion in the strain space as

bL

ε

L+bT

ε

T+bLL

ε

2

L+bTT

ε

2

T+bSS

ε

2

S+2bLT

ε

L

ε

T=1 (6.2.37)

218 6 Failure Mechanisms and Criteria

with

bL=aLQLL +aTQLT,

bT=aTQTT +aLQLT,

bLL =aLLQ2

LL +aTTQ2

LT +2aLTQLL QLT ,

bTT =aTTQ2

TT +aLLQ2

LT +2aLTQTT QLT ,

bLT =aLLQLLQLT +aTTQTTQLT +aLT(Q2

LT +QLLQTT)

(6.2.38)

In the more general form analogous to the strength ratio equation is

(Gi j

ε

i

ε

j)R2+ (Gi

ε

i)R−1=0,

cR2+dR −1=0,

R=−d

c+sd

2c+1

c

(6.2.39)

To determine Rfrom this equivalent quadratic criterion the strain space may be

preferred, because laminae strains are either uniform or vary linearly across each

lamina thickness.

As considered above, the most widely used interlaminar failure criteria are the

maximum stress criterion, the maximum strain criterion and the quadratic failure

criteria as a generalization of the von Mises yield criterion, in particular the Tsai-

Hill and the Tsai-Wu criterion. The interlaminar failure modes can be ﬁbre breaking,

ﬁbre buckling, ﬁbre pullout, ﬁbre-matrix debonding or matrix cracking. The predic-

tion of the First-Ply Failure with one of the above mentioned criteria is included in

nearly all available analysis tools for layered ﬁbre reinforced composites.

Interlaminar failure, i.e. failure of the interface between adjacent plies, is a de-

lamination mode. Delamination failure can have different causes. Weakly bonded

areas impact initial delamination in the inner region of a laminate, whereas delam-

ination along free edges is a result of high interlaminar stresses. Free edges delam-

ination is one of the most important failure modes in layered composite structures.

Along a free edge a tri-axial stress state is present and must be considered. Free

edge delamination is subject of actual intensive research.

The strength analysis of laminate presupposes experimental measured ultimate

stresses or strains for the laminae and realistic or approximate assumptions for stiff-

ness degradation of damaged layers. Strength under longitudinal tensile and com-

pression stresses is usually determined with unidirectional plane specimen, strength

under transverse tension and compression is measured with plane specimen or cir-

cumferentially reinforced tubes and shear strength is determined in torsion test of

such tubes. Note that compression testing is much more difﬁcult than tension testing

since there is a tendency of premature failure due to crushing or buckling.

Summarizing the discussion above on failure analysis one can say that for deter-

mination of safety factors of ﬁbre reinforced laminated structural elements there is

a strong need for fracture criteria and degradation models which are simple enough

for engineering applications but being also in sufﬁcient agreement with the physical

6.3 Problems 219

reality. In spite of many efforts were made during recent years strength analysis of

laminates is still underdeveloped in comparison to the stress and strain analysis.

Essential for recent success in failure analysis was to distinguish between ﬁbre

failure and inter-ﬁbre failure by separate failure criteria introduced by Puck4. The

theory and application of Puck’s criterion are detailed described in special literature

(Knops, 2008) and are not considered here. In addition, Christensen5has presented

some arguments concerning the best choice of failure criteria - stress or strain based

(Christensen, 2013). On some actual problems and the state of the art is reported in

Talreja (2016).

6.3 Problems

Exercise 6.1.

A UD lamina is loaded by biaxial tension

σ

L=13

σ

T,

σ

LT =0. The material is a

glass-ﬁbre epoxy composite with EL=46 GPa, ET=10 GPa, GLT =4,6 GPa,

ν

LT =0,31. The basic strength parameters are

σ

Lt =1400 MPa,

σ

Tt =35 MPa,

τ

S=70 MPa. Compare the maximum stress and the maximum strain criteria.

Solution 6.1. Maximum stress criterion (

σ

L<

σ

Lt,

σ

T<

σ

Tt)

13

σ

T=

σ

L<

σ

Lt

σ

T=

σ

Tt <

σ

Lt =⇒

σ

T<107,69 MPa

σ

T<35 MPa

The ultimate stress is determined by the smallest of the two values, i.e. failure occurs

by transverse fracture. The stress state is then

σ

T=35 MPa,

σ

L=13 ·35 =455 MPa <1400 MPa

Maximum strain criterion (

ε

L<

ε

Lt,

ε

T<

ε

Tt)

To determine the ultimate strains we assume approximately a linear stress-strain

relation up to fracture. Then follows the ultimate strains

ε

Lt =

σ

Lt/EL,

ε

Tt =

σ

Tt/ET

The strains caused by the biaxial tension state are

ε

L=SLL

σ

L+SLT

σ

T=1

EL

σ

L−

ν

LT

EL

σ

T=1

EL

(

σ

L−

ν

LT

σ

T)<

ε

Lt,

ε

T=SLT

σ

L+STT

σ

T=−

ν

TL

ET

σ

L+1

ET

σ

T=1

ET

(

σ

T−

ν

TL

σ

L)<

ε

Tt

The maximum strain criterion can be written

4Alfred Puck (∗1927) - engineer and professor, development of a physical-based strength criterion

for UD reinforced laminates

5Richard M. Christensen (∗3 July 1932 Idaho Falls, Idaho) - specialist in mechanics of materials

220 6 Failure Mechanisms and Criteria

σ

L−

ν

LT

σ

T<

σ

Lt,

σ

T−

ν

TL

σ

L<

σ

Tt,

ν

TL =ET

EL

ν

LT

Since

σ

L=13

σ

Tfollows

σ

T<

σ

Lt/(13 −

ν

LT) = 110,32 MPa,

σ

T<

σ

Tt/(1−13

ν

LTET/EL) = 282,72 MPa

The ultimate stress is given by the lowest of both values, i.e. failure occurs by

longitudinal fracture and the stress state is then

σ

L=13 ·110,32 =1434,16 MPa,

σ

T=110,32 MPa

The values of both criteria differ signiﬁcantly and the fracture mode is reversed from

transverse to longitudinal fracture. Because linear elastic response is assumed to

fail, the criterion can predict strength also in terms of stresses. In reality the relation

between ultimate stress and strain is more complex.

Exercise 6.2. Consider an off-axis unidirectional tension of a glass ﬁbre/polyster

resin laminate (Fig. 6.5),

σ

1=3,5 MPa,

θ

=600. Estimate the state of stress with

the help of the maximum stress, the maximum strain and the Tsai-Hill failure cri-

terion. The lamina properties are E′

1=30 GPa, E′

2=4 GPa, G′

12 =1,2 GPa,

ν

′

12 =0,28,

ν

′

21 =0,037,

σ

Lt =1200 MPa,

σ

Tt =45 MPa,

τ

S=35 MPa,

ε

Lt =0,033,

ε

Tt =0,002,

ε

S=0,0078.

Solution 6.2. The solution is split with respect to different criteria.

1. Maximum stress criterion

Using (6.2.6) the stresses in the principal material axes can be calculated

σ

′

1=

σ

1cos2

θ

=0,875 MPa <

σ

Lt,

σ

′

2=

σ

1sin2

θ

=2,625 MPa <

σ

Tt,

σ

′

6=

σ

1sin

θ

cos

θ

=−1,515 MPa <

τ

S

The off-axis ultimate tensile strength

σ

1t is the smallest of the following stresses

σ

1=

σ

Lt/cos2

θ

=4800 MPa,

σ

1=

σ

Tt/sin2

θ

=60 MPa,

σ

1=

τ

S/sin

θ

cos

θ

=80,8 MPa

i.e.

σ

1t =60 MPa. All stresses

σ

′

iare allowable, the lamina does not fail.

2. Maximum strain criterion

From the Hooke’s law for orthotropic materials follows

ε

′

1=

σ

′

1/E′

1−

ν

′

21

σ

′

2/E′

2=

σ

′

1/E′

1−

ν

′

12

σ

′

2/E′

1,

ε

′

2=−

ν

′

12

σ

′

1/E′

1+

σ

′

2/E′

2,

ε

′

6=

σ

′

6/E′

6

The transformation for

σ

′

iyields

6.3 Problems 221

ε

′

1=1

E′

1

[cos2

θ

−

ν

′

12 sin2

θ

]

σ

1=0,0000047 <

ε

Lt,

ε

′

2=1

E′

2

[sin2

θ

−

ν

′

12

E′

2

E′

1

cos2

θ

]

σ

1=0,0006 <

ε

Tt,

ε

′

6=1

G′

12

sin

θ

cos

θσ

1=0,0013 <

ε

S

All strains are allowed. The composite does not fail.

3. Tsai-Hill criterion

Using (6.2.18) the criterion can be written

cos2

θ

σ

Lt 2

+sin2

θ

σ

Tt 2

−sin

θ

cos

θ

σ

Lt 2

+sin

θ

cos

θ

τ

S2

<1

σ

2

1

,

"0,25

1200 2

+0,75

45 2

−0,433

1200 2

+0,433

35 2#MPa−2<0,00043MPa−2,

1

σ

2

1

=0,0816MPa−2

The left-hand side is smaller than the right-hand side, therefore the composite

does not fail.

Exercise 6.3. The plane stress state of a UD-lamina is deﬁned by

σ

1=2

σ

,

σ

2=−3

σ

,

σ

6=4

σ

,

σ

>0

The material properties are

E′

1=181GPa,E′

2=10,3GPa,

ν

′

12 =0,28,G′

12 =7,17GPa,

ν

′

21 =0,01593,

σ

Lt =1500MPa,

σ

Lc =1500MPa,

σ

Tt =40MPa,

σ

Tc =246MPa,

τ

S=68MPa

The ﬁbre angle is

θ

=600. Calculate the maximum value for

σ

by using the different

failure criteria.

Solution 6.3. The solution is given for the four cases separately.

1. Maximum stress criterion

Transformation of the stresses from the off-axis to on-axis reference system

yields (Table 4.1)





σ

′

1

σ

′

2

σ

′

6



=



0,250 0,750 0,866

0,750 0,250 −0,866

−0,433 0,433 −0,500 





2

σ

−3

σ

4

σ



=



1,714

−2,714

−4,165 



σ

Using (6.2.2) we ﬁnd the inequalities

222 6 Failure Mechanisms and Criteria

−1500 MPa <1,714

σ

<1500 MPa,

−246 MPa <−2,714

σ

<40 MPa,

−68 MPa <−4,165

σ

<68 MPa,

=⇒−875,1MPa <

σ

<875,1 MPa,

−14,73 MPa <

σ

<90,64 MPa,

−16,33 MPa <

σ

<16,33 MPa

The three inequalities are satisﬁed if 0 <

σ

<16.33 MPa. The maximum stress

state which can be applied before failure is

σ

1=32,66 MPa,

σ

2=48,99 MPa,

σ

6=65,32 MPa

The mode of failure is shear.

2. Maximum strain criterion

Using the transformation rule (4.1.5) for strains

ε

′

ifollows with S′

11 =

1/E′

1=0,5525 10−11 Pa−1,S′

22 =1/E′

2=9,709 10−11 Pa−1,S′

66 =1/G′

12 =

13,95 10−11 Pa−1,S′

12 =−

ν

′

12/E′

1=−0,1547 10−11 Pa−1





ε

′

1

ε

′

2

ε

′

6



=



S′

11 S′

12 0

S′

12 S′

22 0

0 0 S′

66







σ

′

1

σ

′

2

σ

′

6



=



0,1367

−2,662

−5,809 

10−10

σ

MPa

MPa

Assuming a linear relationship between the stresses and the strains until failure,

we can calculate the ultimate strains in a simple way

ε

Lt =

σ

Lt/E′

1=8,287 10−3,

ε

Lc =

σ

Lc/E′

1=8,287 10−3,

ε

Tt =

σ

Tt/E′

2=3,883 10−3,

ε

Tc =

σ

Tc/E′

2=23,88 10−3,

ε

S=

τ

S/G′

12 =9,483 10−3

and the inequalities (6.2.8) yield

−8,287 10−3<0,1367 10−10

σ

<8,287 10−3,

−23,88 10−3<−2,662 10−10

σ

<3,883 10−3,

−9,483 10−3<−5,809 10−10

σ

<9,483 10−3

or

−606,2 106<

σ

<606,2 106,

−14,58 106<

σ

<89,71 106,

−16,33 106<

σ

<16,33 106

The inequalities are satisﬁed if 0 <

σ

<16,33 MPa, i.e. there is the same maxi-

mum value like using the maximum stress criterion, because the mode of failure

is shear. For other failure modes there can be signiﬁcant differences, see Exercise

6.1.

3. Tsai-Hill criterion

Using (6.2.17) we have

6.3 Problems 223

"1,714

1500 2

+−2,714

40 2

−1,714

1500 −2,714

1500 +−4,165

68 2#

σ

2

1012 <1

i.e.

σ

<10,94.

The Tsai-Hill criterion is an interactive criterion which cannot distinguish the

failure modes. In the form used above it also does not distinguish between com-

pression and tensile strength which can result in an underestimation of the allow-

able loading in compression with other failure criteria. Generally the transverse

tensile strength of a UD-lamina is much less than the transverse compressive

strength. Therefore the criteria can be modiﬁed. In dependence of the sign of

the

σ

′

ithe corresponding tensile or compressive strength is substituted. For our

example follows

"1,714

1500 2

+−2,714

246 2

−1,714

1500 −2,714

1500 +−4,165

68 2#

σ

2

1012 <1

i.e.

σ

<16,06 MPa.

4. Tsai-Wu criterion

Now (6.2.24) must be applied. The coefﬁcients can be calculated

aL=1

σ

Lt −1

σ

Lc =0,

aTT =1

σ

Tt

σ

Tc

=1,0162 10−16Pa−2,

aT=1

σ

Tt −1

σ

Tc =2,093 10−8Pa−1,

aSS =1

τ

2

Tt

=2,1626 10−16Pa−2,

aLL =1

σ

Lt

σ

Lc =4,44441 10−19Pa−2,

aLT ≈ −1

2√aLLaTT =−3,360 10−18Pa−2

Substituting the values of the coefﬁcients in the criterion it yields the following

equation

0·(1,714)

σ

+2,093(10−8)(−2,714)

σ

+4,4444(10−19)(1,714

σ

)2

+1,0162(10−16)(−2,714

σ

)2+2,1626(10−16)(−4,165

σ

)2

+2(−3,360)(10−18)(1,714)(−2,714)

σ

2<1

The solution of the quadratic equation for

σ

yields

σ

<22,39 MPa.

224 6 Failure Mechanisms and Criteria

Summarizing the results of the four failure criteria we have

Max. stress criterion:

σ

=16,33 ¯

σ

(¯

σ

≡RS

σ

)

Max. strain criterion:

σ

=16,33 ¯

σ

(¯

σ

≡RS

ε

)

Tsai-Hill criterion:

σ

=10,94 ¯

σ

(¯

σ

≡RTH)

Mod. Tsai-Hill criterion:

σ

=16,06 ¯

σ

(¯

σ

≡RTHm)

Tsai-Wu criterion:

σ

=22,39 ¯

σ

(¯

σ

≡RTW)

The values ¯

σ

≡

σ

are identical with the strength ratios (6.2.32) - (6.2.35).

Remark 6.1. A summary of the examples demonstrates that different failure criteria

can lead to different results. Unfortunately, there is no one universal criterion which

works well for all situations of loading and all materials. For each special class of

problems a careful proof of test data and predicted failure limits must be conducted

before generalizations can be made. In general, it may be recommended that more

than one criterion is used and the results are compared.

References

Christensen RM (2013) The Theory of Materials Failure. University Press, Oxford

Gol’denblat II, Kopnov VA (1965) Strength of glass-reinforced plastics in the com-

plex stress state. Polymer Mechanics 1(2):54–59

Hill R (1948) A theory of the yielding and plastic ﬂow of anisotropic metals. Pro-

ceedings of the Royal Society of London A: Mathematical, Physical and Engi-

neering Sciences 193(1033):281–297

Hoffman O (1967) The brittle strength of orthotropic materials. Journal of Compos-

ite Materials 1(2):200–206

Knops M (2008) Analysis of Failure in Fiber Polymer Laminates: The Theory of

Alfred Puck. Springer, Heidelberg

von Mises R (1913) Mechanik des festen K ¨orpers im plastischen deformablen Zu-

stand. Nachrichten der K ¨oniglichen Gesellschaft der Wissenschaften G ¨ottingen,

Mathematisch-physikalische Klasse pp 589–592

von Mises R (1928) Mechanik der plastischen form¨anderung von kristallen. ZAMM

- Journal of Applied Mathematics and Mechanics / Zeitschrift f¨ur Angewandte

Mathematik und Mechanik 8(3):161–185

Talreja R (2016) On failure theories for composite materials. In: Naumenko K,

Aßmus M (eds) Advanced Methods of Continuum Mechanics for Materials and

Structures, Springer Singapore, Singapore, pp 379–388

Tsai SW, Wu EM (1971) A general theory of strength for anisotropic materials.

Journal of Composite Materials 5(1):58–80

Part III

Analysis of Structural Elements

The third part (Chaps. 7–9) is devoted to the analysis of structural elements (beams,

plates and shells) composed of laminates and sandwiches. The modelling of lam-

inated and sandwich plates and shells is limited to rectangular plates and circular

cylindrical shells. The individual ﬁber reinforced laminae of laminated structured

elements are considered to be homogeneous and orthotropic, but the laminate is

heterogeneous through the thickness and generally anisotropic. An equivalent sin-

gle layer theory using the classical lamination theory, and the ﬁrst order shear defor-

mation theory are considered. Multilayered theories or laminate theories of higher

order are not discussed in detail.

Chapter 7

Modelling and Analysis of Beams

In Chap. 1 the classiﬁcation of composite materials, the signiﬁcance, advantages

and limitations of composite materials and structures and the material characteris-

tics of the constituents of composite materials were considered. Chapter 2 gave a

short introduction to the governing equations of the linear theory of anisotropic ma-

terial behavior. Chapter 3 deﬁned effective material moduli of composites including

elementary mixture rules and improved formulas. Chapter 4 developed in detail the

modelling of the mechanical behavior of laminates and sandwiches in the frame of

classical theories including thermal and hygroscopic effects. The constitutive equa-

tions, describing the relationships between stress resultants and in-plane strains and

mid-surface curvatures were developed for unidirectional laminae, laminates and

sandwiches with the assumptions of the classical laminate theory. Further the calcu-

lation of in-p lane and throug h-the-thickness stresses was considered. Chapter 5 gave

an introduction to classical and reﬁned laminate theories. In Chap. 6 selected failure

mechanisms and criteria were brieﬂy discussed. These parts of the book give the

basic knowledge, how the design engineer can tailor composite materials to obtain

the desired properties by the appropriate choice of the ﬁbre and matrix constituents,

a laminate or a sandwich material, the stacking sequence of layers, etc. This ba-

sic knowledge can be utilized to develop the modelling and analysis of structural

elements and structures composed of composite materials.

7.1 Introduction

The analysis of structural elements can be performed by analytical and semi-

analytical approaches or by numerical methods. The advantage of analytical so-

lutions is their generality allowing the designer to take into account various design

parameters. Analytical solutions may be either closed form solutions or inﬁnite se-

ries and may be exact solutions o f the governing equations or variational appro aches.

However, analytical solutions are restricted to the analysis of simple structural el-

ements such as beams, plates and shells with simple geometry. Otherwise numer-

227

H. Altenbach et al., Mechanics of Composite Structural

Elements, https://doi.org/10.1007/978-981-10-8935-0_7

228 7 Modelling and Analysis of Beams

ical methods have to be applied more general for structural analysis. Chapter 7 to

10 describe analytical solutions for one- and two-dimensional structural elements.

Chapter 11 gives an insight into numerical solutions based on the ﬁnite element

method.

In the following sections of Chap. 7 we consider rods, columns and beams. These

are one-dimensional structural elements with a thickness hand a width bwhich are

small relative to the element length l, i.e. h,b≪l. When this element is loaded by

an axial force only one speaks of a rod if the loading is tensile, and of columns if

the load is compressive. One calls this element a beam when it is acted upon by

lateral loads. In general a combination of lateral and axial loadings is possible and

so we shall speak of beams under lateral and axial loadings. The other type of one-

dimensional structural elements, so called plate strips under cylindrical bending,

are discussed in Chap. 8. The modelling and analysis of generalized beams based a

thin-walled folded structure are considered in Chap. 10 (generalized Vlasov beam

theory).

The elementary or classical beam theory assumes that the transverse shear strains

are negligible and plane cross-sections before bending remain plane and normal to

the axis of the beam after bending (Bernoulli-Euler beam theory, Sect. 7.2). The

assumption of neglecting shear strains is valid if the thickness his small relative to

the length l(h/l<1/20). In the Bernoulli-Euler beam theory the transverse deﬂec-

tion u3is assumed to be independent of coordinates x2,x3of the cross-section (Fig.

7.1), i.e. u3≡w=w(x1). In Sect. 7.2 the governing equations of the classical beam

theory for composite beams are considered. The differential equations and varia-

tional formulations will be developed in detail for bending only, the equations for

vibration and buckling are brieﬂy summarized.

In the case of sandwich beams or moderately thick laminate beams, the results

derived from the Bernoulli-Euler theory can show signiﬁcant differences with the

actual mechanical behavior, i.e. the deﬂection, stress distribution, etc. An improve-

ment is possible by introducing the effect of transverse shear deformation, i.e. we

apply Timoshenko beam theory (Sect. 7.3). The assumptions of the classical theory

❅

❅❅❅

❅

❅❅❅

❅

axis

b

h

l

✲

✻

❅

❅■

x1

x3

x2

✻

❄

❅❅

❅❘

❅

❅■

✛ ✲

Fig. 7.1 Rod/column/beam

7.2 Classical Beam Theory 229

have then to be relaxed in the following way: the transverse normals do not remain

perpendicular to the deformed axis of the beam after straining. Section 7.4 discuss

some special aspects of sandwich beams.

Laminate or sandwich beams with simple or double symmetric cross-sections are

most important in engineering applications. The derivations in Sects. 7.2 - 7.5 are

therefore limited to straight beams with simple or double symmetric constant cross-

sections which are predominantly rectangular. The bending moments act in a plane

of symmetry. Also cross-sections consisting of partition walls in and orthogonal to

the plane of bending, e.g. I- or box beams, are considered.

7.2 Classical Beam Theory

Frequently, as engineers try to optimize the use of materials, they design compos-

ite beams made from two or more materials. The design rationale is quite straight

forward. For bending loading, stiff, strong, heavy or expensive material must be far

away from the neutral axis at places where its effect will be greatest. The weaker,

lighter or less expensive material will be placed in the central part of the beam. At

one extreme is a steel-reinforced concrete beam, where weight is not a major con-

cern, but strength and cost are. At the other extreme is a sandwich beam used e.g. in

an aircraft with ﬁbre-reinforced laminate cover sheets and a foam core. In that case,

stiffness and weight are essential but cost not.

First we consider elementary beam equations: The cross-section area Acan have

various geometries but must be symmetric to the x3-axis. The ﬁbre reinforcement

of the beam is parallel to the x1-axis and the volume fraction is a function of the

cross-sectional coordinates x2,x3, i.e. vf=vf(x2,x3). The symmetry condition yields

vf(x2,x3) = vf(−x2,x3)and E1(x2,x3) = E1(−x2,x3).

With the known equations for the strain

ε

1and the stress

σ

1at x1=const

ε

1(x3) =

ε

1+x3

κ

1,

σ

1(x2,x3) =

ε

1E1(x2,x3) + x3

κ

1E1(x2,x3)(7.2.1)

follow the stress resultants N(x1),M(x1)of a beam

N=

ε

1Z

(A)

E1(x2,x3)dA+

κ

1Z

(A)

x3E1(x2,x3)dA,

M=

ε

1Z

(A)

x3E1(x2,x3)dA+

κ

1Z

(A)

x2

3E1(x2,x3)dA

(7.2.2)

The effective longitudinal modulus of elasticity is (3.1.8)

E1=Efvf+Emvm=Em+

φ

(x2,x3)(Ef−Em)(7.2.3)

and with Ef=const, Em=const,

φ

(x2,x3) =

φ

(−x2,x3)it follows that

230 7 Modelling and Analysis of Beams

N=a

ε

1+b

κ

1,a=EmA+ (Ef−Em)Z

(A)

φ

(x2,x3)dA,

M=b

ε

1+d

κ

1,b= (Ef−Em)Z

(A)

φ

(x2,x3)x3dA,

I=Z

(A)

x2

3dA,d=EmI+ (Ef−Em)Z

(A)

φ

(x2,x3)x2

3dA

(7.2.4)

The inverse of the stress resultants, (7.2.4), are

ε

1=dN −bM

ad −b2,

κ

1=aM −bN

ad −b2(7.2.5)

and the stress equation (7.2.1) has the form

σ

1(x2,x3) = dN −bM + (aM −bN)x3

ad −b2E1(x2,x3)(7.2.6)

Taking into consideration the different moduli Efand Em, the ﬁbre and matrix

stresses are

σ

f(x3) = dN −bM + (aM −bN)x3

ad −b2Ef,

σ

m(x3) = dN −bM + (aM −bN)x3

ad −b2Em

(7.2.7)

In the case of a double symmetric geometry and ﬁbre volume fraction function

φ

,

b=0 and the equations can be simpliﬁed

ε

1=N/a,

κ

1=M/d,

σ

f(x3) = (N/a+x3M/d)Ef,

σ

m(x3) = (N/a+x3M/d)Em(7.2.8)

For a uniform ﬁbre distribution,

φ

(x2,x3) = const, (7.2.3) – (7.2.4) give

a=EmA+ (Ef−Em)

φ

A=E1A,

b=0,

d=EmI+ (Ef−Em)

φ

I=E1I

(7.2.9)

and the stress relations for ﬁbre and matrix, (7.2.8), are transformed to

σ

f(x3) = N

A+x3

M

IEf

E1,

σ

m(x3) = N

A+x3

M

IEm

E1(7.2.10)

If Ef=Emand E1=E, (7.2.10) becomes the classical stress formula for isotropic

beam with axial and lateral loadings

σ

(x3) = N

A+M

Ix3(7.2.11)

7.2 Classical Beam Theory 231

Now we consider laminate beams loaded by axial and lateral loading. For simplicity,

thermal and hygrothermal effects are ignored. The derivation of the beam equations

presume the classical laminate theory (Sects. 4.1 and 4.2). There are two different

cases of simple laminated beams with rectangular cross-section:

1. The beam is loaded orthogonally to the plane of lamination.

2. The beam is loaded in the plane of lamination.

In the ﬁrst case, we start from the constitutive equations (4.2.18).

The beam theory makes the assumption that in the case of bending and stretch-

ing in the (x1−x3)-plane of symmetry, i.e. no unsymmetrical or skew bending,

N2=N6=0, M2=M6=0 and that all Poisson’s effects are neglected.

With these assumptions (4.2.18) is reduced to

N1

M1=A11 B11

B11 D11 

ε

1

κ

1(7.2.12)

and from (4.2.14)

Q1=A55

ε

5(7.2.13)

If the beam has a midplane symmetry, there is no bending-stretching coupling, so

that B11 =0 and (7.2.12) becomes

N1=A11

ε

1,M1=D11

κ

1(7.2.14)

Note that in the classical theory, the transverse shear strain will be ignored, i.e

ε

5=0, and there is no constitutive equation for resultant shear forces.

The starting point for derivation of structural equations for beams is the equilib-

rium equations for stress resultants N,Mand Qat the undeformed beam element,

Fig. 7.2. The in-plane and transverse stress resultants N1,Q1and the resultant mo-

x3

x1

MM+dM

QQ+dQ

NN+dN

q(x1)

n(x1)

m(x1)

dx1

Fig. 7.2 Stress resultants N,Qand Mof the inﬁnite beam element, q(x1),n(x1)are line forces,

m(x1)is a line moment

232 7 Modelling and Analysis of Beams

Table 7.1 Differential relations for laminate beams based on the classical beam theory

(

ε

1=u′(x1),

κ

1=−w′′(x1))

Relations between stress resultants and loading

N′(x1) = −n(x1),Q′(x1) = −q(x1),

M′(x1) = Q(x1)−m(x1),M′′(x1) = −q(x1)−m′(x1)

Relations between stress resultants and strains

N=bA11u′(x1)−bB11 w′′(x1)

M=bB11u′(x1)−bD11 w′′(x1)⇐⇒ 



N(x1)

M(x1)

=



bA11 bB11

bB11 bD11 





u′(x1)

−w′′(x1)



Differential equations for the displacements

General case

(bA11u′)′′ −(bB11 w′′)′′ =−n′

(bB11u′)′′ −(bD11 w′′)′′ =−q−m′(...)′=d

dx1

Constant stiffness

bA11 bB11

bB11 bD11  u′′′

−w′′′′ =−n′

−q−m′

Midplane symmetric laminates (B11 =0)

(bA11u′)′=−n

(bD11w′′)′′ =q+m′

bA11u′′ =−n

bD11w′′′′ =q+m′

Special cases

m′(x1) = 0 :

u′′(x1) = −n(x1)

bA11

,w′′′′(x1) = q(x1)

bD11

N′(x1) = −n(x1),Q′(x1)= −q(x1),

M′(x1) = Q(x1)−m(x1)

m(x1) = 0,n(x1) = 0 :

u′(x1) = −N

bA11

=const,w′′′′(x1) = q(x1)

bD11

Q′(x1) = −q(x1),M′(x1) = Q(x1)

ment M1in (7.2 .12)–(7.2.13) are load s per unit length and must be multiplied by th e

beam width b, i.e. the beam resultants are N=bN1,Q=bQ1,M=bM1.

The differential relations for laminate beams loaded orthogonally to the plane of

lamination are summarized in Table 7.1. Note that when Nis a compressive load,

we have to consider additional stability conditions.

The stresses

σ

(k)

1(x1,x3)in the kth layer are given by

7.2 Classical Beam Theory 233

σ

(k)

1=Q(k)

11

ε

(k)

1=Q(k)

11 (

ε

1+x3

κ

1) = Q(k)

11 du(x1)

dx1−x3

d2w(x1)

dx2

1(7.2.15)

or with

ε

1=D11N−B11M

A11D11 −B2

11

,

κ

1=A11M−B11N

A11D11 −B2

11

,

one get

σ

(k)

1=Q(k)

11

1

bD11N−B11M

A11D11 −B2

11

+x3

A11M−B11N

A11D11 −B2

11 (7.2.16)

In the most usual case of midplane symmetric beams the stress equations (7.2.15),

(7.2.16) can be simpliﬁed to

σ

(k)

1M(x1) = Q(k)

11

du(x1)

dx1

=Q(k)

11

N(x1)

bA11

,

σ

(k)

1B(x1) = Q(k)

11 −x3

d2w(x1)

dx2

1=Q(k)

11 x3

M(x1)

bD11

(7.2.17)

σ

(k)

1Mare the layerwise constant stretching or membrane stresses produced by N(x1)

and

σ

(k)

1Bthe layerwise linear distributed ﬂexural or bending stresses produced by

M(x1). The strain

ε

1=

ε

1+x3

κ

1is continuous and linear through the total beam

thickness h. The stresses

σ

(k)

1are continuous and linear through each single layer

and have stress jumps at the layer interfaces (Fig. 7.3) With the help of effective

moduli EN

eff and EM

eff for stretching and ﬂexural loading we can compare the stress

equations of a laminate beam with the stress equation of a single layer beam.

With N6=0,M=0 Eq. (7.2.12) becomes

N1

0=bA11 B11

B11 D11 

ε

1

κ

1(7.2.18)

and with

0=bB11

ε

1+bD11

κ

1,

κ

1=−B11

D11

ε

1(7.2.19)

1

2

3

4

5

6



σ

1M

ε

1

σ

1B

x3

κ

1

✲

x1

S

✻

❄

h

✚





☞

Fig. 7.3 Qualitative distribution of the stresses and strains through the beam thickness hassuming

Q(1)

11 =Q(6)

11 >Q(3)

11 =Q(4)

11 >Q(2)

11 =Q(5)

11

234 7 Modelling and Analysis of Beams

follows the equations for N=bN1and

ε

1

N=bA11

ε

1+bB11

κ

1=

ε

1bA11D11 −B2

11

D11

,

ε

1=D11

(A11D11 −B2

11)bN=D11h

(A11D11 −B2

11)

N

bh =N

EN

effA(7.2.20)

with

EN

eff =A11D11 −B2

11

D11h,A=bh

The strain

ε

1on the beam axis of a single layer isotropic, homogeneous beam is

ε

1=N/EA. Replacing Eby EN

eff gives the strain equation for the laminate beam.

The stresses

σ

(k)

1in the klayers are then

σ

(k)

1=Q(k)

11 (

ε

1+x3

κ

1) = Q(k)

11

ε

11−x3

B11

D11 

=Q(k)

11 1−x3

B11

D11 du

dx1

,

σ

(k)

1=Q(k)

11

EN

eff

N

bh 1−x3

B11

D11 =E(k)

1

EN

eff

N

bh 1−x3

B11

D11 

(7.2.21)

or for midplane symmetric beams with B11 =0

σ

(k)

1=E(k)

1

EN

eff

N

bh =E(k)

1

EN

eff

N

A,EN

eff =A11

h(7.2.22)

In an analogous manner it follows from (7.2.12) with N=0,M6=0 that

M=bB11

ε

1+bD11

κ

1=

κ

1bA11D11 −B2

11

A11

,

κ

1=A11

(A11D11 −B2

11)bM=A11h3

12(A11D11 −B2

11)

M

bh3

12

=M

EM

effI,(7.2.23)

with

EM

eff =12(A11D11 −B2

11)

A11h3,I=bh3

12 (7.2.24)

For an isotropic homogeneous single layer beam of width band thickness hone get

κ

1=M/EI =12M/bh3,I=bh3/12. Replacing now Eby EM

eff, the stress equations

are

7.2 Classical Beam Theory 235

σ

(k)

1=Q(k)

11 (

ε

1+x3

κ

1) = Q(k)

11

κ

1−B11

A11

+x3

=Q(k)

11 B11

A11 −x3d2w

dx2

1

,

=Q(k)

11

Eeff

M

Ix3−B11

A11 =E(k)

1

Eeff

M

Ix3−B11

A11 

(7.2.25)

or with B11 =0 for the symmetric case

σ

(k)

1=E(k)

1

EM

eff

M

Ix3,EM

eff =12D11

h3(7.2.26)

If both in-plane and lateral loads occur simultaneously, the stress in each lamina of

the beam is as for symmetric case

σ

(k)

1(x1,x3) = Q(k)

11 du

dx1−x3

d2w

dx2

1=E(k)

1N

EN

effA+M

EM

effIx3,(7.2.27)

A=bh,I=bh3/12,EN

eff =A11/h,EM

eff =12D11/h3

Conclusion 7.1. Summarizing the equations for symmetric laminated beams, one

can say that the equations for u(x1)and w(x1)are identical in form to those of el-

ementary theory for homogeneous, isotropic beams. Hence all solutions available,

e.g. for deﬂections of isotropic beams under various boundary conditions, can be

used by replacing the modulus Ewith EN

eff or EM

eff, respectively. The calculation of

the stresses illustrates that constant in-plane layer stresses produced by Nare propor-

tional to the layer modulus E(k)

1(7.2.22). N/AE N

eff is for a cross-section x1=const

a constant value. Analogous are the ﬂexural layer stresses proportional to E(k)

1x3

(7.2.26). In general, the maximum stress does not occur at the top or the bottom of

a laminated beam, but the maximum stress location through the thickness depends

on the lamination scheme.

From the bending moment-curvature relation (N=0)

M=bD11

κ

1(7.2.28)

it follows that

κ

1max =Mmax

bD11

=−d2w

dx2max

,

and the maximum stress can be calculated for each lamina

σ

(k)

1max =Q(k)

11

κ

1maxx3=−Q(k)

11 d2w

dx2max

x3=E(k)

1

EM

eff

Mmax

Ix3(7.2.29)

σ

(k)

1max must be compared with the allowable strength value.

236 7 Modelling and Analysis of Beams

The calculation of the transverse shear stress

σ

(k)

5(x1,x3)is analogous as in the

elementary beam theory. We restrict our calculation to midplane symmetric beams

and therefore all derivations can be given for the upper part of the beam element

(x3≥0). The equilibrium equations in the x1-direction lead with

σ

(j)

1+d

σ

(j)

1≈

σ

(j)

1+d

σ

(j)

1

dx1

(Fig. 7.4) and no edge shear stresses on the upper and lower faces

σ

(k)

5dx1−

N

∑

j=k+1

x(j)

3

Z

x(j−1)

3

"

σ

(j)

1+d

σ

(j)

1

dx1

dx1!−

σ

(j)

1#dx3=0 (7.2.30)

or

σ

(k)

5=

N

∑

j=k+1

x(j)

3

Z

x(j−1)

3

d

σ

(j)

1

dx1

dx3=

N

∑

j=k+1

x(j)

3

Z

x(j−1)

3

E(j)

1

EM

effI

dM

dx1

x3dx3

With Q=dM/dx1it follows that

σ

(k)

5(x1,x3) = Q(x1)

EM

effI

N

∑

j=k+1

x(j)

3

Z

x(j−1)

3

E(j)

1x3dx3=Q(x1)

EM

effI

N

∑

j=k+1

E(j)

1

2x(j)

3

2−x(j−1)

3

2

=Q(x1)

EM

effI

N

∑

j=k+1

E(j)

1h(j)x(j)

3

(7.2.31)

✲

✻

x1

x3

✛

σ

(k)

5

k+1

j

N

.

✛

σ

(k+1)

1

σ

(j)

1

σ

(N)

1✲

✲

σ

(k+1)

1+d

σ

(k+1)

1

σ

(j)

1+d

σ

(j)

1

σ

(N)

1+d

σ

(N)

1

✛ ✲

dx1

✻

x(k)

3

✻

x(j−1)

3

x(j)

3

x(j)

3

x(N)

3

Fig. 7.4 Beam element b(x(N)

3−x(k)

3)dx1with ﬂexural normal stresses

σ

(j)

1and the interlaminar

stress

σ

(k)

5

7.2 Classical Beam Theory 237

For a single layer homogeneous, isotropic beam (7.2.31) yields the known parabolic

shear stress distribution through h

σ

5(x1,x3) = Q

I

h/2

Z

x3

x3dx3=12Q

bh3

1

2h2

4−x2

3=3Q

2bh 1−4x3

h2(7.2.32)

With an increasing number of equal thickness layers, the transverse shear stress

distribution (7.2.31) approaches the parabolic function of the single layer beam.

All stress equations presume that the Poisson’s effects can be completely ne-

glected, i.e. Q(k)

i j =Di j =0,i6=j,i,j=1,2,6. They are summarized for symmet-

ric laminated beams (N6=0,M6=0) in Table 7.2. For symmetric laminated beams

loaded orthogonally to the plane of lamination, the classical laminate theory yields

identical differential equations for u(x1)and w(x1)with to Bernoulli’s beam theory

of single layer homogeneous isotropic beams, if one substitutes bA11 by E A =Ebh

and bD11 by EI =Ebh3/12. An equal state is valid for beam vibration and beam

buckling.

The following equations are given without a special derivation (b,A,D11,

ρ

are

constant values):

•Differential equation of ﬂexure (N=0)

d2w(x1)

dx2

1

=−M(x1)

bD11

,bD11

d4w(x1)

dx4

1

=q(x1)(7.2.33)

Table 7.2 Stress formulas for symmetric laminated beams, classical theory

σ

(k)

1(x1,x3) =

σ

(k)

1M(x1) +

σ

(k)

1B(x1,x3)

=Q(k)

11

du(x1)

dx1−Q(k)

11

d2w(x1)

dx2

1

x3

=Q(k)

11

N(x1)

bA11

+Q(k)

11

M(x1)

bD11

x3

=E(k)

1

EN

eff

N(x1)

A11

+E(k)

1

EM

eff

Mx1)

Ix3,

σ

(k)

5(x1,x3) = Q(x1)

EM

effI

N

∑

j=k+1

E(j)

1h(j)¯x(j)

3,

A=bh,I=bh3/12,

EN

eff =A11/h,EM

eff =12D11/h3,¯x(j)=1

2(x(j)

3+x(j−1)

3)

238 7 Modelling and Analysis of Beams

•Forced or free vibrations

bD11

d4w(x1,t)

dx4

1

+

ρ

Ad2w(x1,t)

dt2=q(x1,t),

ρ

=1

h

N

∑

k=1

ρ

(k)h(k)(7.2.34)

Rotational inertia terms are neglected. For free vibration with q=0 the solution

is assumed periodic: w(x1,t) = W(x1)exp(i

ω

t).

•Buckling equation

d2M(x1)

dx2

1

+N1(x1)d2w(x1)

dx2

1

=0,bD11

d4w(x1)

dx4

1−N1(x1)d2w(x1)

dx2

1

=0

(7.2.35)

or with N(x1) = −F

d2M(x1)

dx2

1−Fd2w(x1)

dx2

1

=0,bD11

d4w(x1)

dx4

1

+Fd2w(x1)

dx2

1

=0

All solutions of the elementary beam theory for single layer isotropic beams can

transferred to laminate beams. Note that laminate composites are stronger shear

deformable than metallic materials and the classical beam theory is only acceptable

when the ratio l/h>20.

The equations for ﬂexure, vibration and buckling can also be given in a vari-

ational formulation (Sect. 2.2.2). With the elastic potential for a ﬂexural beam

(N=0,M6=0)

Π

(w) = 1

2

l

Z

0

bD11 d2w(x1)

dx2

12

dx1−

l

Z

0

qdx1(7.2.36)

and the kinetic energy

T(w) = 1

2

l

Z

0

ρ



∂

2w

∂

t2

dx1(7.2.37)

the Lagrange function is given by L(w) = T(w)−

Π

(w)(Sect. 2.2.2).

The variational formulation for a symmetric laminated beam without bending-

stretching coupling subjected to a lateral load qin x3-direction (N=0,M6=0,

ε

5≈0,

ν

i j ≈0) based on the theorem of minimum of total potential energy is

given in the form

Π

[w(x1)] = 1

2

l

Z

0

bD11 d2w(x1)

dx2

12

dx1−

l

Z

0

qdx1,

δΠ

[w(x1)] = 0

(7.2.38)

The variational formulation for the buckling of a symmetric laminate beam with

N(x1) = −Fis

7.2 Classical Beam Theory 239

Π

[w(x1)] = 1

2

l

Z

0

bD11 d2w(x1)

dx2

12

dx1−1

2

l

Z

0

Fdw(x1)

dx12

dx1,

δΠ

[w(x1)] = 0

(7.2.39)

The variational formulation for free ﬂexural beam vibration (additional to ap-

proaches noted above the rotatory inertia effects are neglected) can be given by

the Hamilton’s principle

H[w(x1,t)] =

t2

Z

t1

L[W(x1,t)]dt,

δ

H[w(x1,t)] = 0 (7.2.40)

The variational formulations can be used for approximate analytical solution with

the Rayleigh-Ritz procedure or numerical solutions.

In the second case of laminate beams, the loading is in the plane of lamination.

We restrict our considerations to symmetric layered beams and neglect all Poisson’s

ratio effects. The beam is illustrated in Fig. 7.5. For a symmetric layer stacking there

is no bending-stretching coupling and we have the constitutive equations

N1=A11

ε

1,M1=b3

12hA11

κ

1

or for the beam resultants N=hN1,M=hM1

Fig. 7.5 Laminate beam

loaded in the plane of lamina-

tion

q(x1)

x2

x3

x1

b

h

240 7 Modelling and Analysis of Beams

N=hA11

ε

1,M=b3

12 A11

κ

1,A11 =

N

∑

k=1

Q(k)

11 h(k)(7.2.41)

The differential equations of ﬂexure (N=0) are

d2w(x1)

dx2

1

=−M(x1)

A11

b3

12 ,12A11

b3

d4w(x1)

dx4

1

=q(x1)(7.2.42)

and the additional equation for the case N6=0 is

hA11

du(x1)

dx1

=N(x1)(7.2.43)

The calculation of stresses is analogous to case 1 of layered beams.

When beam proﬁles consist of partition-walls in the plane of loading and orthog-

onal to the plane of loading, e.g. I-proﬁles or box-beams, the bending differential

equations can be written in the form, given above. The bending stiffness is obtained

by combining the results of orthogonal to plane loading and in-plane loading.

As an example for a box-beam, we consider the beam as shown in Fig. 7.6, which

may be subjected to axial loads in x1-direction, a bending moment with respect to

the x2-axis and a twisting moment with respect to the x1-axis. For an isotropic beam

the stiffness needed are the extensional stiffness, EA, the ﬂexural stiffness, E I, and

the torsional stiffness, GIt.

The axial force resultant (per unit width) in x1-direction is N1=A11

ε

1and the

axial load carried by the whole section is then

N(x1) = 2NI

1b+2NII

1h=2[(A11)Ib+ (A11)I I h]

ε

1(7.2.44)

The extensional stiffness for the box cross-section is given by

Fig. 7.6 Laminated box-beam

with identical top and bottom

panels I and vertical walls II

x2

x3

x1

b

l

t

I

II

h

7.2 Classical Beam Theory 241

(EA)eff =2(A11)Ib+2(A11)I I h(7.2.45)

The box beam is bent in the (x1−x3) plane, and the moment curvature relation is

M="2(D11)Ib+2(A11)Ibh

22

+2(A11)II

h3

12 #

κ

1

≈"2(A11)Ibh

22

+1

6(A11)II h3#

κ

1

(7.2.46)

Since the top and bottom panels are thin relative to the height of the box proﬁle, i.e.

t≪h,(D11)Ican be neglected and the bending stiffness of the box cross-section is

(EI )eff ≈2(A11)Ibh

22

+1

6(A11)II h3(7.2.47)

If the box-beam is acted by a torsional moment MTthis is equivalent to the moment

of the shear ﬂows with respect to the x1-axis and we have

MT=2NI

6b(h/2) + 2NII

6h(b/2),NI

6=AI

66

ε

I

6,NII

6=AII

66

ε

II

6(7.2.48)

In the elementary theory of strength of materials the equation for the angle of twist

of a box-beam is given by

θ

=1

2AIq(s)

Gt ds(7.2.49)

q(s)is the shear ﬂow. In our case, Fig. 7.6, the displacements of the contours of the

walls of the box beam are denoted by

δ

Iand

δ

II and the angle of twist becomes

θ

=

δ

I

(h/2)=

δ

II

(b/2)with

δ

I

l=

ε

I

6,

δ

II

l=

ε

II

6(7.2.50)

From (7.2.48) - (7.2.50) we have

MT=bh

l[(AI

66)h+ (AII

66)b]

θ

(7.2.51)

and the torsional stiffness of the cross-section is

(GIt)eff =bh

l[(AI

66)h+ (AII

66)b](7.2.52)

For the I-proﬁle in Fig. 7.7 the calculation for bending is analogous. The bending

stiffness is

(EI )eff ="2(D11)b+2(A11)bh

22

+ (A11)h3

12 #

≈A11

6h2b+h3

12

(7.2.53)

242 7 Modelling and Analysis of Beams

Fig. 7.7 I-proﬁle with uni-

form thickness t

b

t

h

if D11 ≈0. Note that for a one-dimensional thin structural element which is sym-

metric with respect to all mid-planes and Poisson’s effect is neglected, we have the

simple relationships

Q11 =E1,A11 =E1t,

κ

1=M

(EI )eff

Summarizing the classical beam equations it must be noted that the effect of Pois-

son’s ratio is negligible only if the length-to-with ratio l/bis large (l≫b), otherwise

the structure behavior is more like a plate strip than a beam (Sect. 8.2). This is of

particular importance for angle-ply laminates, i.e. orthotropic axes of material sym-

metry in each ply are not parallel to the beam edges and anisotropic shear coupling

is displayed.

7.3 Shear Deformation Theory

The structural behavior of many usual beams may be satisfactorily approximated by

the classical Euler-Bernoulli theory. But short and moderately thick beams or lami-

nated composite beams which l/hratios are not rather large cannot be well treated

in the frame of the classical theory. To overcome this shortcoming Timoshenko ex-

tended the classical theory by including the effect of transverse shear deformation.

However, since Timoshenko’s beam theory assumed constant shear strains through

the thickness ha shear correction factor is required to correct the shear strain energy.

In this section we study the inﬂuence of transverse shear deformation upon the

bending of laminated beams. The similarity of elastic behavior of laminate and sand-

wich beams with transverse shear effects included allows us generally to transpose

the results from laminate to sandwich beams. When applied to beams, the ﬁrst order

shear deformation theory is known as Timoshenko’s beam theory. Figure 7.8 illus-

trates the cross-section kinematics for the Bernoulli’s and the Timoshenko’sbending

beam.

7.3 Shear Deformation Theory 243

x1

x3

O

B

A

dx1

u(x1)

u1(x1)

O′

B′

A′B′′

A′′

w′(x1)

ψ

(x1)

w′(x1)

Bernoulli kinematics

ﬂexure curve

Timoshenko kinematics

w(x1)

x2

Fig. 7.8 Kinematics of a bent Timoshenko- and Bernoulli-beam in the (x1−x3) plane

When all Poisson’s effects are neglected the constitutive equations are identical

with (7.2.12) - (7.2.13), but from Sect. 5.1 the strains of the Timoshenko’s beam are

ε

1=

∂

u1

∂

x1

=du

dx1

+x3

d

ψ

1

dx1

,

ε

2≡0,

ε

3≡0,

ε

5=

∂

u1

∂

x3

+

∂

w

∂

x1

=

ψ

1+dw

dx1

,

ε

4≡0,

ε

6≡0

(7.3.1)

i.e. we only have one longitudinal and one shear strain

ε

1(x1,x3) =

ε

1(x1) + x3

κ

1(x1),

ε

5(x1,x3) =

ψ

1(x1) + w′(x1),

ε

1(x1) = du

dx1

,

κ

1(x1) = d

ψ

1(x1)

dx1

(7.3.2)

When the transverse shear strain are neglected it follows with

ε

5≈0 that the rela-

tionship is

ψ

1(x1) = −w′(x1)and that is the Bernoulli’s kinematics.

244 7 Modelling and Analysis of Beams

In the general case of an unsymmetric laminated Timoshenko’s beam loaded

orthogonally to the laminated plane and N6=0,M6=0, the constitutive equations

(stress resultants - strain relations) are given by

N=A11

ε

1+B11

κ

1,M=B11

ε

1+D11

κ

1,Q=ksA55

γ

s(7.3.3)

with A11 =bA11,B11 =bB11 ,D11 =bD11,

γ

s=

ε

5and the stiffness equations are

from (4.2.15) and Q(k)

11 ≡C(k)

11 =E(k)

1,Q(k)

55 ≡C(k)

55 =G(k)

13

A11 =b

n

∑

k=1

C(k)

11 x(k)

3−x(k−1

3)=b

n

∑

k=1

C(k)

11 h(k),

A55 =b

n

∑

k=1

C(k)

55 x(k)

3−x(k−1)

3=b

n

∑

k=1

Ck

55h(k),

B11 =b1

2

n

∑

k=1

C(k)

11 x(k)

3

2−x(k−1)

3

2,

D11 =b1

3

n

∑

k=1

C(k)

11 x(k)

3

3−x(k−1)

3

3

ksis the shear correction factor (Sect. 5.3).

In the static case, the equilibrium equations for the undeformed beam element

(Fig. 7.2) yield again for lateral loading q6=0

dM

dx1−Q=0,dQ

dx1

+q=0 (7.3.4)

When considerations are limited to symmetric laminated beams the coupling stiff-

ness B11 is zero and from (7.3.3) it follows that

M=D11

d

ψ

1

dx1

,Q=ksA55 

ψ

1+dw

dx1(7.3.5)

Substituting the relations (7.3.5) into (7.3.4) leads to the differential equations of

ﬂexure

[D11

ψ

′

1(x1)]′−ksA55[

ψ

1(x1) + w′(x1)] = 0,

ksA55 [

ψ

1(x1) + w′(x1)]′+q(x1) = 0(7.3.6)

Derivation of the ﬁrst equation of (7.3.6) and setting in the second equation yields a

differential equation of 3rd order for

ψ

1(x1)

[D11

ψ

′(x)]′′ =−q(x)(7.3.7)

and with

Q=dM

dx1

= [D11

ψ

′

1(x1)]′(7.3.8)

and

Q=ksA55[

ψ

1(x1) + w′(x1)]

7.3 Shear Deformation Theory 245

follows an equation for w′(x1)

w′(x1) = −

ψ

1(x1) + [D11

ψ

′

1(x1)]′

ksA55

(7.3.9)

Summarizing the derivations above, the equations for a bent Timoshenko’s beam

are: [D11

ψ

′

1(x1)]′′ =−q(x1),

M(x1) = D11

ψ

′

1(x1),

Q(x1) = [D11

ψ

′

1(x1)]′,

w′(x1) = −

ψ

1(x1) + [D11

ψ

′

1(x1)]′

ksA55

(7.3.10)

When the laminated beam problem allows to write the bending moment Mand the

transverse force Qin terms of the known applied lateral loads q, like in statically

determined beam problems, (7.3.5) can be utilized to determine ﬁrst

ψ

1(x1)and then

w(x1). Otherwise (7.3.6) or (7.3.10) are used to determine w(x1)and

ψ

1(x1).

Integrating the second Eq. (7.3.6) with respect to x1, we obtain

ksA55[w′(x1) +

ψ

1(x1)] = −Zq(x1)dx1+c1

Substituting the result into the ﬁrst equation of (7.3.6) and integrating again with

respect to x1yields

D11

ψ

′

1(x1) = −ZZ q(x1)dx1dx1+c1x1+c2,

D11

ψ

1(x1) = −ZZZ q(x1)dx1dx1dx1+c1

x2

1

2+c2x1+c3

(7.3.11)

Substituting

ψ

1(x1)and

ψ

′

1(x1)in (7.3.9), considering (7.3.7) and integrating once

more with respect to x1we obtain

w(x1) = 1

D11 ZZZZ q(x1)dx1dx1dx1dx1+c1

x3

1

6+c2

x2

1

2+c3x1+c4

−1

ksA55 ZZ q(x1)dx1dx1+c1x1

=wB(x1) + wS(x1)

(7.3.12)

The transverse deﬂection consists of two parts. The bending part wB(x1)is the same

as derived in the classical theory. When the transverse stiffness goes to inﬁnity, the

shear deﬂection wS(x1)goes to zero,

ψ

1(x1)goes to −w′(x1)and the Timoshenko’s

beam theory reduces to the classical Bernoulli’s beam theory.

The relations for the stresses

σ

1are the same as in the classical theory. The trans-

verse shear stress can be computed via a constitutive equation in the Timoshenko

theory

246 7 Modelling and Analysis of Beams

σ

(k)

5(x1,x3) = Q(k)

55

Q(x1)

ksA55

(7.3.13)

The variational formulation for a lateral loaded symmetric laminated beam is given

by

Π

(w,

ψ

1) =

Π

i+

Π

a(7.3.14)

with

Π

i=1

2

l

Z

0"D11d

ψ

1

dx12

+ksA55

ψ

1+dw

dx12#dx1,

Π

a=−

l

Z

0

q(x1)wdx1

(7.3.15)

In the more general case of unsymmetric laminated beams and axial and lateral

loadings we have

Π

(u,w,

ψ

1). The

Π

iexpression can be expanded to

Π

i=1

2

L

Z

0"A11 du

dx12

+2B11

du

dx1

d

ψ

1

dx1

+D11 d

ψ

1

dx12

+ksA55 

ψ

+dw

dx12#dx1

(7.3.16)

and

Π

ahas to include axial and lateral loads.

Since the transverse shear strains are represented as constant through the lam-

inate thickness, it follows that the transverse stresses will also be constant. In the

elementary beam theory of homogeneous beams, the transverse shear stress varies

parabolically through the beam thickness and in the classical laminate theory the

transverse shear stresses vary quadratically through layer thickness. This discrep-

ancy between the stress state compatible with the equilibrium equations and the

constant stress state of the ﬁrst order shear deformation theory can be overcome

approximately by introducing a shear correction factor (Sect. 5.3).

The shear correction factor kscan be computed such that the strain energy W1

due to the classical transverse shear stress equals the strain energyW2due to the ﬁrst

order shear deformation theory. Consider, for example, a homogeneous beam with

a rectangular cross-section A=bh. The classical shear stress distribution following

from the course of elementary strength of materials is given by

σ

13 =

τ

1=3

2

Q

bh "1−2x3

h2#,−h

2≤x3≤+h

2(7.3.17)

The transverse stress in the ﬁrst order shear deformation theory is constant through

the thickness h

σ

13 =

τ

2=Q

bh ,

γ

2=Q

ksG(7.3.18)

With W1=W2it follows that

7.3 Shear Deformation Theory 247

1

2Z

(A)

τ

2

1

GdA=1

2Z

(A)

τ

2

ksGdA,

3

5

Q2

Gbh =1

ks

Q2

2Gbh =⇒ks=5

6(7.3.19)

The shear correction factor for a general laminate depends on lamina properties and

lamina stacking and is given here without a special derivation by

1

ks=A55b

N

∑

k=1

x(k)

3

Z

x(k−1)

3

g(k)2(x3)

G(k)dx3,(7.3.20)

g(k)(z) = d11 (−C(k)

11

z2

2+

k

∑

j=1hC(j)

11 −C(j−1)

11 iz(j−1)2

2),C(0)

11 ≡0

d11 =1/D11 is the beam compliance, C(k)

11 =E(k)

1.

Summarizing the beam equations for the ﬁrst order shear deformation theory

for symmetrically laminated cross-sections, including vibration and buckling, the

following relations are valid for constant values of h,b,A,D11 ,

ρ

:

•Flexure equations (N=0,M6=0)

ksA55 [

ψ

1(x1) + w′(x1)]′+q(x1) = 0,

[D11

ψ

′

1(x1)]′−ksA55[

ψ

1(x1) + w′(x1)] = 0(7.3.21)

or Eqs. (7.3.10).

•Forced or free vibrations equations

ksA55 [

ψ

1(x1,t) + w′(x1,t)]′−

ρ

0¨w(x1,t) + q(x1,t) = 0,

[D11

ψ

′

1(x1,t)]′−ksA55[

ψ

1(x1,t) + w′(x1,t)]−

ρ

2¨

ψ

1(x1,t) = 0,(7.3.22)

ρ

0=b

n

∑

k=1

ρ

(k)x(k)

3−x(k−1)

3,

ρ

2=b

n

∑

k=1

1

3

ρ

(k)x(k)3

3−x(k−1)3

3

The terms involving

ρ

0and

ρ

2are called translatory or rotatory inertia terms. For

free vibrations we assume that the transverse load qis zero and the motion is

periodic:

w(x1,t) = W(x1)exp(i

ω

t),

ψ

1(x1,t) =

Ψ

1(x1)exp(i

ω

t)

•Buckling equations

ksA55 [

ψ

1(x1) + w′(x1)]′−N(x1)w′′(x1) = 0,

[D11

ψ

′

1(x1)]′−ksA55[

ψ

1(x1) + w′(x1)] = 0(7.3.23)

248 7 Modelling and Analysis of Beams

or with N(x1) = −F

D11 1−F

ksA55 w′′′′(x1) + F w′′(x1) = 0

The variational formulation for the symmetric bending beam is given by Eq.

(7.3.15).

For vibrations the Lagrange function L(w,

ψ

1) = T(w,

ψ

1)−

Π

(w,

ψ

1)yields the

Hamilton’s principle

H[w(x1,t),

ψ

1(x1,t)] =

t2

Z

t1

L[w(x1,t),

ψ

1(x1,t)]dt,

δ

H[w(x1,t),

ψ

1(x1,t)] = 0

with

Π

[w(x1,t),

ψ

1(x1,t)] = 1

2

l

Z

0

[D11

ψ

′

1

2+ksA55(

ψ

1+w′)2]dx1

−

l

Z

0

q(x1,t)wdx1,

T[w(x1,t),

ψ

1(x1,t)] = 1

2

l

Z

0

[

ρ

0˙w2+

ρ

2˙

ψ

2

1]dx1

(7.3.24)

For buckling problems with N(x1) = −Fit follows that

Π

[w(x1,t),

ψ

1(x1,t)] = 1

2

l

Z

0

[D11

ψ

′

1

2+ksA55(

ψ

1+w′)2]dx1

−1

2

l

Z

0

Fw′2dx1

(7.3.25)

Equations (7.3.21) to (7.3.25) summarize the bending, buckling and vibration dif-

ferential and variational statements for laminated beams based on the shear defor-

mation theory.

7.4 Sandwich Beams

The similarity of the elastic behavior between symmetric laminates and symmetric

sandwich beams in the ﬁrst order shear deformation theory (Sects. 4.3 and 5.3) al-

lows us to transpose the results derived above to the bending of sandwich beams.

7.4 Sandwich Beams 249

In addition to the differences between the expressions for the ﬂexural and trans-

verse shear stiffness D11 and A55 the essential difference is at the level of stress

distribution. The model assumptions for sandwich composites with thin and thick

cover sheets are considered in detail in Sects. 4.3.1 to 4.3.3. There one can ﬁnd the

stiffness values A11,D11,A55 . With these values, all differential and variational for-

mulation of the theory of laminated beams including transverse shear deformation

can be transposed.

In the case of a symmetric sandwich beam with thin cover sheets we have, for

example, the stiffness values

A11 =2Af

11 =2

n

∑

k=1

Q(k)

11 h(k),

D11 =2hcCf

11 =hcn

∑

k=1

Q(k)

11 h(k)x(k),

As

55 =hcGc

13

(7.4.1)

nis the number of the face layers.

The coefﬁcient As

55 can be corrected by a shear correction factor ks. In addition

to the calculation of ks, derived for a laminated beam an approximate formula was

developed by Reuss, for sandwich beams with thin cover sheets. With the inverse

effective shear stiffness G−1

R, given by the Reuss model, and the effective shear

stiffness GV, given by the Voigt model, we have

GV=

n

∑

k=1

G(k)h(k)

h=A55

bh ,G−1

R=

n

∑

k=1

1

G(k)

h(k)

h,ks=GR

GV

(7.4.2)

The use of sandwich structures is growing very rapidly. Sandwich beams has a

high ratio of ﬂexural stiffness to weight and in comparison to other beam struc-

tures they have lower lateral deformations, higher buckling resistance and higher

natural frequencies. As a result sandwich constructions quite often provide a lower

structural weight than other structural elements for a given set of mechanical and

environmental loads.

The elastic behavior of sandwich beams was modelled by the laminate the-

ory, Sect. 4.3, but it is appropriately to distinguish thin and thick sandwich faces.

The differential equations or variational statements describing the structural behav-

ior of sandwich beams generally based in the ﬁrst order shear deformation theory

Sect. 5.3, and only if very ﬂexible cores are used a higher order theory may be

needed.

Because of the continuing popularity of sandwich structures Sect. 7.4 intends to

recall and summarizes the results of Sects. 4.3 and 5.3 to cover some of the most

important aspects of sandwich beam applications.

250 7 Modelling and Analysis of Beams

7.4.1 Stresses and Strains for Symmetrical Cross-Sections

Figure 7.9 shows a sandwich beam with a symmetrical lay up, i.e. the faces have the

same thickness hfand are of the same material. As derived in Sect. 4.3 and 4.4 the

ﬂexural rigidy is (D11 ≡DLa

11 ,Q11 ≡E1=E)

bD11 =D=bEf(hf)3

6+Efhfd2

2+Ec(hc)3

12 =2Df+Do+Dc(7.4.3)

and both, 2Dfand Dcare less than 1% of Doif d/hf>5,77 and

(6Efhfd2)/Ec(hc)3>100. Thus, for a sandwich with thin faces hf≪hcand a weak

core, Ec≪Ef, the ﬂexural rigidity is approximately

D≈Do=bEfhfd2

2(7.4.4)

It can be noted that in most engineering applications using structural sandwich

beam elements, the dominating term in ﬂexural rigidity is that of the faces bending

about the neutral axes of the beam, i.e. the dominating part Doof the total rigidity

Doriginating from a direct tension-compression of the cover sheets. But is there

no monolithic bonding between the faces and the core the ﬂexural rigidity will be

nearly lost.

The following derivations assume in-plane-, bending- and shear stiffness for all

layers, i.e. for the faces and the core. Therefore we use the laminate theory including

transverse shear, Sects. 4.3.3 and 5.3. All calculations are ﬁrst restricted to midplane

symmetric beams.

The bending strains vary linearly with x3over the cross-section:

ε

M

1=M

Dx3(7.4.5)

Unlike the bending strains, which vary linearly with x3over the whole cross-section,

the bending stresses vary linearly within each material constituent, but there is a

jump in the stresses at the face/core interfaces:

x3

x1

3

2

1

Q

N

M

E3=Ef

E2=Ec

E3=Ef

d=hc+hf

b

hf

hc

hf

Fig. 7.9 Symmetrical sandwich beam: N=bN1,Q=bQ1,M=bM1are the beam stress resultants

7.4 Sandwich Beams 251

σ

(k)

1=ME(k)

Dx3(k=1,2,3)⇒

σ

f=MEf

Dx3

σ

c=MEc

Dx3

(7.4.6)

With Eq. (7.2.12) follows

M=bD11

κ

1,

κ

1=M

bD11

=h3

12D11

M

bh3/12 =M

EM

effI,(7.4.7)

with

Eeff =12

h3D11,I=bh3

12

and the stress equations can be written as

σ

(k)

1=E(k)

EM

eff

M

Ix3(k=1,2,3)(7.4.8)

The strains due to in-plane loading are:

ε

N

1=N

3

∑

k=1

Q(k)h(k)

=N

3

∑

k=1

E(k)h(k)

=N

bh

h

A11

=N

EN

effA(7.4.9)

with

EN

eff =A11

h,A11

3

∑

k=1

Q(k)

11 h(k)=

3

∑

k=1

E(k)h(k),A=bh

ε

N

1is the strain of the neutral axis. The in-plane stresses follow to

σ

(k)

1=E(k)

EN

eff

N

A(k=1,2,3)⇒

σ

f=E(1)

EN

eff

N

A=E(3)

EN

eff

N

A

σ

c=E(2)

EN

eff

N

A

(7.4.10)

The strains and stresses due to in-plane loads and bending can be superimposed.

In the same manner as outlined above a general deﬁnition can be found for shear

strains and shear stresses. Consider the beam element b(x(3)

3−x3)dx1, Fig. 7.10. The

upper edge of the sandwich, i.e. x3= (d+hf)/2, Fig. 7.9, is stress free and we have

τ

[(d+hf)/2] = 0.

Since we restrict on calculations to midplane symmetric beams all derivation can

be given for the upper part of the beam element (x3≥0). The equilibrium equation in

the x1-direction yield with

σ

1d

σ

1≈

σ

1+ (

∂σ

1/

∂

x1)dx1and no edge shear stresses

on the upper face

252 7 Modelling and Analysis of Beams

x3

x1

dx1

τ

(x3)

(x(3)

3−x3)

σ

f

σ

c

σ

f+d

σ

f

σ

c+d

σ

c

σ

5+d

σ

5

σ

5

σ

1

σ

1+d

σ

1

Fig. 7.10 Sandwich beam element b(x(3)

3−x3)dx1:

σ

(3)(x3)≡

σ

f(x3),

σ

(2)(x3)≡

σ

c(x3),

σ

5(x3)≡

τ

(x3)

τ

(x3)bdx1−

(d+hf)/2

Z

x3

σ

1+

∂σ

1

∂

x1

dx1−

σ

1bdx3=0

⇒

τ

(x3) =

(d+hf)/2

Z

x3

∂σ

1

∂

x1

dx3=0 (7.4.11)

Using the relations dM(x1)/dx1=Q(x1)and

σ

1=M(E(x3)/D)x3we have

d

σ

1

dx1

=Q(x1)

DE(x3)x3,

τ

(x3) = Q

bD

(d+hf/2)

Z

x3

E(x3)x3bdx3=Q

bD S(x3)(7.4.12)

S(x3)is the ﬁrst moment of the area (x(3)

3−x3)b. For a single layer homogeneous

and isotropic beam we have the well-known formula

S(x3) = b

2h2

4−x2

3;h= (d+hf)

For sandwich beams we have a more generalized deﬁnition for the ﬁrst moment of

area:

(d+hf)/2

Z

x3

E(x3)x3bdx3=











bEfhfd

2+Ec

2hc

2−x3hc

2+x3,

|x3| ≤ hc

2

bEf

2hc

2+hf−x3hc

2+hf+x3,

hc

2≤|x3| ≤ hc

2+hf

(7.4.13)

7.4 Sandwich Beams 253

The shear stresses for the core and the faces are

τ

c(x3) = Q

DEfhfd

2+Ec

2(hc)2

4−x2

3,

τ

f(x3) = Q

D

Ef

2(hc)2

4+hchf+ (hf)2−x2

3(7.4.14)

The maximum shear stress appears at the neutral axes:

τ

max =

τ

c(x3=0) = Q

DEfhfd

2+Ec(hc)2

8(7.4.15)

The shear stress in the core/face interface is

τ

c

min ≡

τ

f

max =

τ

hc

2=Q

DEfhfd

2(7.4.16)

There is no jump in the shear stresses at the interfaces and the shear stresses are zero

at the outer ﬁbres of the faces. If we have

4Efhfd

Ec(hc)2>100 (7.4.17)

the shear stresses in the core are nearly constant. The difference between

τ

c

max and

τ

c

min is less than 1%. As it was outlined in Sect. 4.3, the stress equation in sandwich

beams very often can be simpliﬁed.

Summarizing the stress estimations due to bending and shear for symmetrical

sandwich beams we have the following equations:

1. The core is weak, Ec≪Ef, but the faces can be thick

σ

f(x3)≈MEf

Do+2Dfx3,

σ

c(x3)≈0,

τ

f(x3)≈Q

Do+2Df

Ef

2(hc)2

4+hchf+ (hf)2−x2

3,

τ

c(x3)≈QEfhfd

2(Do+2Df)

(7.4.18)

2. The core is weak, Ec≪Ef, and the faces are thin, hf≪hc

σ

f(x3)≈ ± M

bhfd,

σ

c(x3)≈0,

τ

f(x3)≈0,

τ

c(x3)≈Q

bd (7.4.19)

This approximation can be formulated as: The faces of the sandwich beam carry

bending moments as constant tensile and compressive stresses and the core car-

ries the transverse forces as constant shear stresses.

254 7 Modelling and Analysis of Beams

7.4.2 Stresses and Strains for Non-Symmetrical Cross-Sections

In engineering applications also sandwich beams with dissimilar faces are used,

Fig. 7.11. The ﬁrst moment of area is zero when integrated over the entire cross-

section and x3is the coordinate from the neutral axes

ZE(x3)x3bdx3=0 (7.4.20)

The location of neutral axis is unknown. With the coordinate transformation

x∗

3=x3−efrom a known axis of the cross-section the equation above becomes

S(x3) = ZE(x3)x3bdx3=ZE(x∗

3+e)bdx∗

3=0,eZEdx∗

3=−ZEx∗

3dx∗

3

For the sandwich cross-section, Fig. 7.11, follows

eE(1)h(1)+E(2)h(2)+E(3)h(3)

=E(1)h(1)1

2h(1)+h(2)+1

2h(3)+1

2E(2)h(2)h(2)+h(3)

and we get an equation for the unknown value e

e=

E(1)h(1)h(1)+2h(2)+h(3)+E(2)h(2)h(2)+h(3)

2E(1)h(1)+E(2)h(2)+E(3)h(3)(7.4.21)

If the core is weak, E(2)≪(E(1),E(3))we have approximately

e=E(1)h(1)d

E(1)h(1)+E(3)h(3)or d−e=E(3)h(3)d

E(1)h(1)+E(3)h(3),(7.4.22)

where d=1

2h(1)+h(2)+1

2h(3).

The bending stiffness D=RE(x3)x2

3bdx3yields in the general case

Fig. 7.11 Deﬁnition of the

neutral axis (N.A.) of an

unsymmetrical sandwich:

x3=x∗

3+e

x2

x3

E(3)

E(2)

E(1)

d

h(3)

b

hc=h(2)

h(1)

e

x∗

3

N.A.

7.4 Sandwich Beams 255

D=1

12 hE(1)(h(1))3+E(2)(h(2))3+E(3)(h(3))3i

+E(1)h(1)(d−e)2+E(3)h(3)e2+E(2)h(2)1

2(h(2)+h(3))−e2(7.4.23)

and can be simpliﬁed for E(2)≪(E(1),E(3))but thick faces as

D≈E(1)(h(1))3

12 +E(3)(h(3))3

12 +E(1)h(1)E(3)h(3)d2

E(1)h(1)+E(3)h(3)(7.4.24)

For thin faces the ﬁrst two terms vanish

D≈Do=E(1)h(1)E(3)h(3)d2

E(1)h(1)+E(3)h(3)(7.4.25)

Now the bending and shearing stresses can be calculated in the usual way

σ

1(k)(x3) = ME(k)

Dx3,

τ

(k)(x3) = Q

bD S(x3)(7.4.26)

For sandwich beams with weak core and thin but dissimilar faces the stress formulas

are approximately

σ

(3)

1≡

σ

f1

1≈MEf1

De=M

bhf1d,

σ

(1)

1≡

σ

f2

1≈ −MEf2

D(d−e) = −M

bhf2d,

τ

(2)≡

τ

c≈Q

bd ,

τ

(3)=

τ

(1)≈0

(7.4.27)

7.4.3 Governing Sandwich Beam Equations

The following derivations assumed, as generally in Chap. 7, straight beams with at

least single symmetric constant cross-sections which are rectangular, i.e we consider

single core sandwich beams. The faces can be thin or thick and symmetrical or non-

symmetrical. The bending moments and axial forces act in the plane of symmetry

(x1−x3). The inﬂuence of transverse shear deformation is included, because the

core of sandwich beams has a low transverse modulus of rigidity G13. The shear

correction factor ksis determined approximately for sandwich beams with thin cover

sheets with the Reuss formula (7.4.1), or more generally using equivalent shear

strain energy, i.e the potential energy of the applied load equals the strain energy

of the beam to account for the nonuniform shear distribution through the thickness.

The shear deformation theory (Sect. 7.3) is valid and we can adapt the equations of

this section to the special case of sandwich beams.

The strains

ε

1and

ε

5≡

γ

are given, (7.3.2), as

256 7 Modelling and Analysis of Beams

ε

1=du

dx1

+x3

d

ψ

dx1

,

γ

=dw

dx1

+

ψ

With

A11 =A,B11 =B,D11 =D,ksA55 =S(7.4.28)

the constitutive equations (7.2.12), (7.2.13) yield





N

M

Q

=



A B 0

B D 0

0 0 S





u′

ψ

′

w′+

ψ



(7.4.29)

For static loading q(x1)6=0,n(x1)≡0 the equilibrium equations are as in the clas-

sical beam theory, Table 7.1,

N′=0,Q′+q=0,M′−Q=0 (7.4.30)

If N≡0 the neutral axes position xN.A.

3is constant along the length of the beam and

is given by

ε

1(xN.A.

3) = u′+xN.A.

3

ψ

′=0,N=Au′+B

ψ

′=0

xN.A.

3=−u′

ψ

′=B

A(7.4.31)

Thus, if the stiffness A,B,D,Sare constant, the substitution of Eq. (7.4.29) into

(7.4.30) yields the following two governing differential equations for sandwich

beams

DR

ψ

′′(x1)−S[w′(x1) +

ψ

(x1)] = 0,

S[w′′(x1) +

ψ

′(x1)] = −q(x1)(7.4.32)

with DR=D−(B2/A). Derivation of the ﬁrst equation and setting in the second

equation yield

DR

ψ

′′′(x1) = −q(x1)(7.4.33)

and with

M′=Q=S(w′+

ψ

),M=Bu′+DR

ψ

′,M′=Bu′′ +DR

ψ

′′

follow

w′(x1) = −

ψ

(x1) + 1

S(Bu′′ +DR

ψ

′′)

For symmetrical cross-sections is B=0,DR=D. In the general case, if all stiff-

ness are constant and unequal zero the substitution of (7.4.4) into (7.4.5) yields the

governing simultaneous differential equations for unsymmetrical sandwich beams

DR

ψ

′′(x1)−S[w′(x1) +

ψ

(x1)] = 0,DR=D−B2

A

S(w′′(x1) +

ψ

′(x1)) = −q(x1)

u′(x1) = −B

A

ψ

′(x1)

(7.4.34)

7.4 Sandwich Beams 257

Derivation of the ﬁrst equation and setting in the second equation yield one uncou-

pled equation for

ψ

(x1)

DR

ψ

′′′(x1) = −q(x1)(7.4.35)

The constitutive equations (7.4.4) give the relations

M(x1) = Bu′(x1) + D

ψ

′(x1),Q(x1) = S[w′(x1) +

ψ

(x1)] (7.4.36)

and with

M′(x1) = Q(x1) = Bu′′(x1) + D

ψ

′′(x1),u′(x1) = −B

A

ψ

′(x1)

follow

w′(x1) = −

ψ

(x1) + 1

S[Bu′′(x1) + DR

ψ

′′(x1)] = −

ψ

(x1) + DR

S

ψ

′′(x1)(7.4.37)

Thus we have three uncoupled differential equations:

DR

ψ

′′′(x1) = −q(x1),

w′(x1) = −

ψ

(x1) + DR

S

ψ

′′(x1),

u′(x1) = B

A

ψ

′(x1)

(7.4.38)

For symmetrically cross-sections is B≡0, DR≡Dand the differential equations

reduce to

D

ψ

′′′(x1) = −q(x1),

w′(x1) = −

ψ

(x1) + D

S

ψ

′′(x1)),

M(x1) = D

ψ

′(x1)or

ψ

′(x1) = M(x1)

EM

effI

Q(x1) = D

ψ

′′(x1)or M′(x1) = S(w′(x1) +

ψ

(x1))

(7.4.39)

The equations (7.4.39) correspond to the equations (7.3.10) of the laminated beam

and the analytical solutions (7.3.11), (7.3.12) can be transposed with D11 =D,

ksA55 =S. In dependence of the calculation of the stiffness Dand Sthe equation

are valid for sandwich beams with thin or thick faces.

The stresses

σ

1and

τ

can be calculated with the help of the stress formulas

derived in Sects. 7.4.1 and 7.4.2. For statically determinate structures, M(x1)and

Q(x1)can be calculated with the equilibrium equations and the last two equations

(7.4.39) can directly used for static calculations.

We consider as an example the cantilever beam, Fig. 7.12, then:

258 7 Modelling and Analysis of Beams

Fig. 7.12 Symmetrical can-

tilever beam with thin faces

x3

x1

hf

hc≈d

l

F

M(x1) = F(l−x1),Q(x1) = −F,

ψ

′(x1) = M(x1)

D=1

DF(l−x1),

ψ

(x1) = F

Dlx1−1

2x2

1+C1,

ψ

(0) = 0⇒C1=0,

w′(x1) = −F

Dlx1−1

2x2

1+Q

S=−F

Dlx1−1

2x2

1−F

S,

w(x1) = −F

D1

2lx2

1−1

6x3

1−Fx1

S+C2,

w(0) = 0⇒C2=0,

w(x1) = −F

D1

2lx2

1−1

6x3

1−Fx1

S=wB(x1) + wS(x1),

w(l) = −Fl 3

3D−Fl

S

Assume thin faces and weak core, i.e

hf≪hc,Ec≪Ef

we have

D=Do=bEfhfd2

2,S=ksGcbd

and the stresses are

σ

f=±MEf

D

d

2=±M

bdhf=±F(l−x1)

bdhf,

τ

c=Q

bd =−F

bd ,

σ

c=

τ

f=0

Consider w(l) = wB(l) + wS(l)it can be seen that the shear deformation strongly

depends on land S. It is important for short and shear weak beams and negligible

for slender shear stiff beams.

Summarizing the aspects of sandwich beams it could be demonstrated in the

static case that the shear deformation theory for laminated beams is valid for sand-

wich beams, if the stiffness A11 ,B11 and D11 of a laminated beam are replaced by

the stiffness A,B,Dand S, Eq. (7.4.3), of the sandwich beam. The same conclusion

is valid not only for bending but also for buckling and vibration and for differential

7.5 Hygrothermo-Elastic Effects on Beams 259

and variational formulations. In this way all formulas (7.3.19) to (7.3.23) can easy

transposed to symmetrically sandwich beams.

In the considerations above we have assumed that the effect of core transverse

deformability is negligible on the bending, vibration and the overall buckling of

sandwich beams. But in a special case of buckling, called face wrinkling the trans-

verse normal stiffness of the core has an important inﬂuence. Wrinkling is a form

of local instability of thin faces associated with short buckling waves. This phe-

nomenon was not discussed here.

7.5 Hygrothermo-Elastic Effects on Beams

In Sects. 7.2 and 7.4 the effect of mechanical loads acting upon ﬁbre reinforced

beams with E1(x2,x3) = E1(−x2,x3)and laminated or sandwich beams was consid-

ered. The considerations for laminated beams as derived are valid in the framework

of the classical laminate theory, Sect. 7.2, and of the ﬁrst order shear deformation

theory, Sect. 7.3. Section 7.4 considered some special aspects of sandwich beams

with thin or thick cover sheets and different stiffness of the core.

In the present section the effects of hygrothermally induced strains, stresses and

displacements are examined. We assume a moderate hygrothermal loading such that

the mechanical properties remain unchanged for the temperature and moisture dif-

ferences considered.

With Eqs. (4.2.63) to (4.2.68) the beam equations (7.2.1) have additional terms

ε

1(x3) =

ε

1+x3

κ

1=

σ

1(x2,x3)

E1(x2,x3)

+ [

α

th(x2,x3)T(x2,x3) +

α

mo(x2,x3)M∗(x2,x3)],

(7.5.1)

σ

1(x2,x3) = E1(x2,x3)[

ε

1+x3

κ

1

−

α

th(x2,x3)T(x2,x3)−

α

mo(x2,x3)M∗(x2,x3)] (7.5.2)

α

th,

α

mo are the thermal and moisture expansion coefﬁcients, Tthe temperature

change and M∗the weight of moisture absorption per unit weight. Equations (7.2.4)

have now additional terms Nth ,Nmo,Mth ,Mmo, the so-called ﬁctitious hygrothermal

resultants, (4.2.67), and with

˜

N=N+Nth +Nmo,˜

M=M+Mth +Mmo (7.5.3)

the extended hygrothermal constitutive equation for the composite beam are

˜

N

˜

M=a b

b d 

ε

1

κ

1,

ε

1

κ

1=a b

b d −1˜

N

˜

M(7.5.4)

The stress formula (7.2.6) yields with Eq. (7.2.5)

260 7 Modelling and Analysis of Beams

σ

1(x2,x3) = (dN −bM) + (aM −bN)x3

ad −b2E1(x2,x3)

−E1(x2,x3)[

α

thT+

α

moM∗]

(7.5.5)

For double symmetric cross-sectional geometry the coupling coefﬁcient is zero and

the stress equation can be simpliﬁed.

For uniform ﬁbre distribution, i.e.

φ

=const, (7.2.9) follow for a,b,dand the

stress relations for ﬁbres and matrices material are

σ

f(x3) = ( ˜

N/A+x3˜

M/I)(Ef/E1)−Ef(

α

thT+

α

moM∗),

σ

m(x3) = ( ˜

N/A+x3˜

M/I)(Em/E1)−Em(

α

thT+

α

moM∗)(7.5.6)

With Ef=Em=E1=Ecomes the stress equation for isotropic beams with mechan-

ical and hygrothermal loadings

σ

(x3) = ˜

N

A+x3

˜

M

I−E(

α

thT+

α

moM∗)(7.5.7)

For laminate or sandwich beams the developments are similar. All problems are

linear and the principle of superposition is valid and can be used to calculate the

hygrothermal effects. Consider for example a symmetric laminate beam in the frame

of the classical laminate theory and include hygrothermal loads, (7.2.27) yield

σ

(k)

1=E(k)

1

˜

N

EN

effA+x3

˜

M

EM

effI−E(k)

1(

α

th(k)T(k)+

α

mo(k)M∗(k))(7.5.8)

The differential equations for deﬂection and midplane displacement of a symmetric

laminated beam are

[D11w′′(x1)]′′ =q(x1)−Mth(x1)′′ −Mmo(x1)′′,

[A11u′x1)]′=−n(x1) + Nth (x1)′−Nmo(x1)′(7.5.9)

In an analogous manner the differential equations including shear deformation can

be found. The differential equation for a symmetric Timoshenko’s beam with lateral

loading and hygrothermal effects follows with (7.3.10)

[D11

ψ

′

1(x1)]′′ =−q(x1) + Mth(x1)′′ +Mmo(x1)′′ (7.5.10)

The relation for the layer stresses

σ

(k)

1are identical to the classical theory. The trans-

verse shear stresses

σ

(k)

5are not changed by hygrothermal effects.

7.6 Analytical Solutions

The differential equations for bending, vibration and buckling of symmetric lam-

inated beams loaded orthogonally to the plane of lamination are summarized by

7.6 Analytical Solutions 261

(7.2.33) to (7.2.35) for the classical Bernoulli’s beam theory and by (7.3.21) to

(7.3.23) for the Timoshenko’s beam theory including transverse shear deformation.

All stiffness and material parameters are constant values.

The simplest problem is the analysis of bending. The general solution of the

differential equation of 4th order (7.2.33) for any load q(x1)is given by

bD11w(x1)≡bD11 wB(x1)

=C1

x3

1

6+C2

x2

1

2+C3x1+C4+ZZZZ q(x1)dx1dx1dx1dx1

(7.6.1)

The general solution of the Timoshenko’s beam is given by (7.3.12) in the form

w(x1) = wB(x1) + wS(x1). The correction term wS(x1)describes the inﬂuence of the

shear deformation and it decreases with increasing shear stiffness ksA55.

The free vibration of Bernoulli’s beams is modelled by (7.2.34), rotatory inertia

terms are neglected. The partial differential equation

∂

4w(x1,t)

∂

x4

1

+

ρ

A

bD11

∂

2w(x1,t)

∂

t2=0 (7.6.2)

can be separated with w(x1,t) = W(x1)T(t)and yields

W′′′′(x1)T(t) = −

ρ

A

bD11

W(x1)¨

T(t)(7.6.3)

or ¨

T(t)

T(t)=−

ρ

A

bD11

W′′′′(x1)

W(x1)=−

ω

2(7.6.4)

We get two differential equations

¨

T(t) +

ω

T(t) = 0,W′′′′(x1)−

ρ

A

bD11

ω

2W(x1) = 0 (7.6.5)

with the solutions

T(t) = Acos

ω

t+Bsin

ω

t,

W(x1) = C1cos

λ

lx1+C2sin

λ

lx1+C3cosh

λ

lx1+C4sinh

λ

lx1,



λ

l4

=

ρ

A

bD11

ω

2

(7.6.6)

The vibration mode is periodic, and

ω

is called the natural circular frequency. The

mode shapes depend on the boundary conditions of the beam. Consider, for exam-

ple, a simply supported beam, we have W(0) = W′′(0) = W(l) = W′′(l) = 0 and

therefore C1=C3=C4=0 and C2sin(

λ

/l)l=C2sin

λ

=0, which implies that

262 7 Modelling and Analysis of Beams

λ

=n

π

,

ω

2

n=n4

π

4

ρ

A

bD11

l4,

ω

n=n

π

l2sbD11

ρ

A(7.6.7)

For each nthere is a different natural frequency and a different mode shape. The

lowest natural frequency, corresponding to n=1, is termed the fundamental fre-

quency. If the laminate beam is unsymmetric to the middle surface, i.e. B11 6=0,

then D11 in Eq. (7.6.7) can be approximately replaced by (A11D11 −B2

11)/A11, the

so called reduced or apparent ﬂexural stiffness.

Including shear deformation effects, i.e. using the Timoshenko vibration equa-

tion (7.3.22), involves considerable analytical complications. To prove whether

the transverse shear deformation can be important for the natural frequencies,

we compare the natural frequencies for a simply supported Bernoulli and Timo-

shenko beam. Using (7.3.22) the boundary conditions for the Timoshenko beam are

w(0,t) = w(l,t) =

ψ

1(0,t) =

ψ

1(l,t) = 0 and by introducing

w=Asin

ω

tsin n

π

x1

l,

ψ

1=Bsin

ω

tcos n

π

x1

l,

in Eqs. (7.3.22) we can calculate the natural frequencies

ω

2

n=n4

π

4

ρ

A

bD11

l4/1+

π

2D11n2

l2ksA55 ,

ρ

A≡

ρ

0(7.6.8)

i.e.

ω

n=

ω

Bernoulli

ns1

1+ (n2

π

2D11)/(l2ksA55 )

Transverse shear deformation reduces the values of vibration frequencies. As in

the case of static bending the inﬂuence of shear on the values of vibration frequen-

cies depends on the ratio E1/G13 ≡E1/E5and the ratio l/h, i.e the span length

between the supports to the total thickness of the laminate. For more general bound-

ary conditions we can develop a mode shape function similar to (7.6.6).

In an analogous way, one can show that the buckling loads for a simply supported

Bernoulli and Timoshenko beam with a compression load Ffollow from (7.2.35)

and (7.3.23) and are

Fcr =

π

2bD11

l2(Minimum Euler load, Bernoulli beam)

Fcr =

π

2bD11

l2

1

1+

π

2D11

l2ksA55

(Minimum buckling load, Timoshenko beam)

(7.6.9)

The buckling loads for clamped beams or beams with more general boundary con-

ditions can be calculated analytically analogous to eigenfrequencies of vibration

problems.

7.7 Problems 263

For non constant cross-section, stiffness or material parameters there are no exact

analytical solutions. Approximate analytical solutions can be found with the help of

the Rayleigh-Ritz procedure. In Sect. 7.7 exact and approximate analytical solution

procedures are illustrated for selected beam problems.

7.7 Problems

Exercise 7.1. A reinforced concrete beam is loaded by a bending moment M

(Fig. 7.13). It is assumed that the concrete has zero strength in tension so that the

entire tensile load associated with the bending moment is carried by the steel rein-

forcement. Calculate the stresses

σ

m(x3)and

σ

f(x3)in the concrete part (Am,Em)

and the steel reinforcements (Af,Ef).

Solution 7.1. The neutral axis x1of the beam is in an unknown distance

α

hfrom the

top, Ac≡Amis the effective area of the concrete above the x1-axis. The strains will

vary linearly from the x1-axis and the stresses will equal strain times the respective

moduli. The stress resultant N(x1)must be zero

σ

fAf−1

2

σ

m(

α

h)b

α

h=0,

σ

m(

α

h) =

σ

max

m

With (7.2.1) follows

σ

f= (h−

α

h)

κ

1Ef,

σ

m(

α

h) =

α

h

κ

1Em

i.e.

(h−

α

h)EfAf−1

2(

α

h)2bEm=0,

α

=EfAf

Embh −1+s1+2bh

Af

Em

Ef!

or

x3

x1

α

h

b

(h−

α

h)

h

h−

α

h

3

Af

σ

max

m

1

2

σ

max

mb

α

h

(

σ

fAf)

M

Ac≡Am

Fig. 7.13 Reinforced concrete beam loaded by pure bending

264 7 Modelling and Analysis of Beams

α

=1

m(−1+√1+2m),m=Embh

EfAf

Now the bending moment is with

σ

fAf=1

2

σ

mbh2

α

2

M= (

σ

fAf)h−

α

h

3=

σ

fAfh1−

α

3

=

κ

1EfAf(h−

α

h)h−

α

3

The maximal stress in the concrete is

σ

m(

α

h) = −

κ

1

α

hEm=MEm

α

h

EfAf(h−

α

h)(h−

α

h/3)

and the reinforcement stress is

σ

f=

κ

1(h−

α

h)Ef=M

Af(h−

α

h/3)

Exercise 7.2. A symmetric cross-ply laminate beam is shown in Fig. 7.14. The ma-

terial properties and the geometry are deﬁned by

E′

1=17,24 104MPa,E′

2=0,6895 104MPa,

G′

12 =G′

13 =0,3448 104MPa,G′

23 =0,1379 104MPa,

ν

′

12 =0,25,

L=240 mm,b=10 mm,h(1)=h(2)=h(3)=8 mm,h=24 mm,

q0=0,6895 N/mm

x1

x3

L

h

3

h

3

h

3

h

q(x1) = q0sin

π

x1

L

Fig. 7.14 Simply supported cross-ply laminated beam [0/90/0]

7.7 Problems 265

Calculate a approximative solution using the Timoshenko beam model and two one-

term Ritz procedures.

Solution 7.2. Let us introduce the cross-section geometry

x[0]

3=−12 mm,x[1]

3=−4 mm,x[2]

3=4 mm,x[3]

3=12 mm

The shear correction factor can be calculated with Eq. (7.3.20) to ks=0,569. The

bending stiffness ¯

D11 and the shear stiffness ks¯

A55 follow with Eq. (7.3.3)

¯

D11 =b

3[E1((−4)3−(−12)3) + E2(43−(−4)3) + E1(123−43)]

=1,92 109Nmm2,

ks¯

A55 =ksb[G128+G238+G128] = 3,76 105N

The variational formulation for a lateral loaded symmetric laminate beam is given

with (7.3.15)

Π

(w,

ψ

) = 1

2

L

Z

0"¯

D11 d

ψ

dx12

+ks¯

A55 

ψ

+dw

dx12#dx1−

L

Z

0

q0sin

π

x1

Lwdx1

The essential boundary conditions are

w(x1=0) = 0,w(x1=L) = 0,

ψ

′(x1=0) = 0,

ψ

′(x1=L) = 0

The approximate functions are

˜w(x1) = a1sin 

π

x1

L,˜

ψ

(x1) = b1cos 

π

x1

L

and it follows

˜

Π

(˜w,˜

ψ

) = 1

2

L

Z

0¯

D11 b1

π

L2

sin2

π

x1

L

+ks¯

A55 b1cos

π

x1

L+a1

π

Lcos

π

x1

L2dx1

−

L

Z

0

q0sin 

π

x1

La1sin

π

x1

Ldx1=˜

Π

(a1,b1)

With (2.2.41) must be

∂

˜

Π

/

∂

a1=0,

∂

˜

Π

/

∂

b1=0 which yields the two equations

¯

D11

ks¯

A55

π

L+L

π

b1+a1=0,

b1+

π

La1=q0L

ks¯

A55

π

266 7 Modelling and Analysis of Beams

and the solution for the unknown constants a1,b1

a1=q0L4

¯

D11

π

41+¯

D11

ks¯

A55

π

2

L2,

b1=−q0L3

¯

D11

π

3

The approximate solutions are now

˜w(x1) = q0L4

¯

D11

π

41+¯

D11

ks¯

A55

π

2

L2sin

π

x1

L,

˜

ψ

(x1) = −q0L3

¯

D11

π

3cos

π

x1

L

Note that

˜wmax =˜wx1=L

2=q0L4

¯

D11

π

41+¯

D11

ks¯

A55

π

2

L2

The transverse deﬂection consists of two parts

˜wB(x1) = q0L4

¯

D11

π

4sin

π

x1

L(bending deﬂections),

˜wS(x1) = q0L4

¯

D11

π

4

¯

D11

ks¯

A55

π

2

L2sin

π

x1

L

For ks¯

A55 →∞follows ˜wS→0, i.e. ˜wB(x1)is the solution of the Bernoulli beam

model and we found

wTimoshenko =k1wBernoulli

with

k1=1+¯

D11

ks¯

A55

π

2

L2=1,875

For the laminate beam with h/L=1/10 the Bernoulli model cannot be accepted, the

relative error for the maximum value of the deﬂection is 46,7 %. Equations (7.3.10)

lead

˜

Mmax =˜

Mx1=L

2=q0L2

π

2=12,64 Nm,

˜

Qmax =˜

Q(x1=0) = q0L

π

=52,67 N

The strains

ε

1follow from Eqs. (7.3.1) or (7.3.2) and (7.3.10)

ε

1(x1) = x3

ψ

′(x1) = x3

κ

1=x3M(x1)/¯

D11

ε

1(x1)is linear distributed across hand we calculate the following values for the

cross-section x1=L/2

7.7 Problems 267

˜

ε

(3)

1x1=L

2,x[3]

3=2,51 10−5,

ε

(3)

1x1=L

2,x[2]

3=0,84 10−5

The bending stresses

σ

1(x1,x3)in the 3 layers are for x1=L/2 and x3=x(k)

3

˜

σ

(3)

1(x(3)

3) = ˜

ε

1(x(3)

3)E′

1=4,327 MPa,

˜

σ

(3)

1(x(2)

3) = ˜

ε

1(x(2)

3)E′

1=1,448 MPa,

˜

σ

(2)

1(x(2)

3) = ˜

ε

1(x(2)

3)E′

2=0,579 MPa,

˜

σ

(2)

1(0) = ˜

ε

1(0)E′

2=0 MPa

Exercise 7.3. Find the analytical solution for the natural vibrations of a simply sup-

ported symmetric laminate or sandwich beam. Test the inﬂuence of the transverse

shear deformation and the rotatory inertia upon the natural frequencies.

Solution 7.3. Starting point are the (7.3.22) with q(x1,t)≡0 and the boundary con-

ditions

w(0,t) = w(l,t) = 0,

ψ

′(0,t) =

ψ

′(l,t) = 0

For a simply supported beam we can assume the periodic motion in the form

w(x1,t) = W(x1)sin

ω

t,

ψ

1(x1,t) =

Ψ

(x1)sin

ω

t

These functions are substituted in (7.3.22)

ks¯

A55[W′′(x1) +

Ψ

′(x1)] +

ρ

0

ω

2W(x1) = 0,

¯

D11

Ψ

′′(x1)−ks¯

A55[W′(x1) +

Ψ

(x1)] +

ρ

2

ω

2

Ψ

(x1) = 0

Now we can substitute

ks¯

A55

Ψ

′(x1) = −

ρ

0

ω

2W(x1)−ks¯

A55W′′(x1),

i.e.

Ψ

′(x1) = −

ρ

0

ω

2

ks¯

A55

W(x1)−W′′(x1)

into the derivative of the second equation and we ﬁnd

¯

D11W′′′′(x1) + ¯

D11

ρ

0

ks¯

A55

+

ρ

2

ω

2W′′(x1)

−1−

ω

2

ρ

2

ks¯

A55 

ρ

0

ω

2W(x1) = 0

or

aW ′′′′(x1) + bW′′(x1)−cW(x1) = 0

with the coefﬁcients

268 7 Modelling and Analysis of Beams

a=¯

D11,b=¯

D11

ks¯

A55

+

ρ

2

ρ

0

ρ

0

ω

2,c=1−

ω

2

ρ

2

ks¯

A55 

ρ

0

ω

2

The linear differential equation of 4th order has constant coefﬁcients and the general

solutions follow with the solutions

λ

iof the bi-quadratic characteristic equation

a

λ

4−b

λ

2−c=0

i.e. (2a

λ

2−b)2=b2+4ac

λ

1−4=±r1

2a(b±pb2+4ac)

W(x1) = C1sin

λ

1x1+C2cos

λ

2x1+C3sinh

λ

3x1+C4cosh

λ

4x1

For a simply supported beam the boundary conditions are

W(0) = 0,W(L) = 0,

Ψ

′(0) = 0,

Ψ

′(L) = 0

or the equivalent equations

W(0) = 0,W(L) = 0,W′′(0) = 0,W′′(L) = 0

The boundary conditions lead to the result C2=C3=C4=0 and C1sin

λ

1L=0

which implies

λ

1n=n

π

L=

λ

n

The bi-quadratic equation can be written alternatively in terms of

ω

A

ω

4−B

ω

2+C=0

with

A=

ρ

2

ks¯

A55

,B=1+¯

D11

ks¯

A55

+

ρ

2

ρ

0

λ

2,C=¯

D11

ρ

0

λ

4

i.e. the roots of the equation are

(

ω

2)1/2=1

2A(B±pB2−4AC)

It can be shown that B2−4AC >0. Therefore the frequency given by −√B2−4AC

is the smaller of the two roots.

When the rotatory inertia is neglected follows A≡0 and the frequency is given

by

ω

2=C

B

with

B=1+¯

D11

ks¯

A55

λ

2,C=¯

D11

ρ

0

λ

4

7.7 Problems 269

and for ks¯

A55 →∞follow with ˜

B→1 the natural frequency for the Bernoulli beam

model. Substitute

λ

n=n

π

/Lfor the simply supported beam we obtain the results:

General case

ω

2

n=ks¯

A55

2

ρ

21+¯

D11

ks¯

A55

+

ρ

2

ρ

0n

π

L2

−s1+¯

D11

ks¯

A55

+

ρ

2

ρ

0n

π

L22

−4

ρ

2

ks¯

A55

¯

D11

ρ

0n

π

L4



Rotatory inertia neglected (

ρ

2≡0)

ω

2

n=n

π

L4¯

D11

ρ

0



1−n

π

L2¯

D11

ks¯

A55 +¯

D11 n

π

L2





=n

π

L4¯

D11

ρ

0





1

1+¯

D11 n

π

L2/ks¯

A55







Classical beam theory (ks¯

A55 →∞,

ρ

2=0)

ω

2

n=n

π

L4¯

D11

ρ

0

Conclusion 7.2. (

ω

Timoshenko

n)2<(

ω

Bernoulli

n)2, i.e. the shear deformation de-

creases the frequencies of natural vibration. In the case of classical beam theory with

rotatory inertia (ks¯

A55 →0,

ρ

26=0) we have A=0,B=1+

λ

2

ρ

2/

ρ

0,C=

λ

4¯

D11/

ρ

0,

i.e.

ω

2

n=n

π

L4¯

D11

ρ

0





1

1+n

π

L2

ρ

2

ρ

0







and we see that also the rotatory inertia decreases the eigenfrequencies. All formulas

can be used for computing natural frequencies for all symmetric laminate and sand-

wich beams. The values for L,

ρ

0,

ρ

2,ks,¯

D11,¯

A55 correspond to the special beam

model. Note that the classical laminate theory and the neglecting of rotatory inertia

lead to a overestimation of the natural frequencies.

Exercise 7.4. Calculate the buckling load of a simply supported and a clamped sym-

metric laminate or sandwich beam. Compare the results for the classical beam the-

ory and the beam theory including shear deformation.

Solution 7.4. Staring point are Eqs. (7.3.23) with N(x1) = −F, i.e.

¯

D11 1−F

ks¯

A55 w′′′′(x1) + F w′′(x1) = 0

270 7 Modelling and Analysis of Beams

or

w′′′′(x1) + k2w′′(x1) = 0,k2=F

¯

D11 1−F

ks¯

A55 ,Fk2¯

D11

1+k2¯

D11/ks¯

A55

The linear differential equation with constant coefﬁcients has the characteristic

equation

λ

4+k2

λ

2=0⇒

λ

2(

λ

2+k2) = 0

with the four roots

λ

1/2=0,

λ

3/4=±ik

and the general solution is

w(x1) = C1sin kx1+C2coskx1+C3x1+C4

1. Simply supported beam

Boundary conditions are w(0) = w(L) = 0,w′′(0) = w′′(L) = 0, which leads the

constants C2=C3=C4=0 and for C16=0 follows sin kL =0 implies kL =

n

π

,k=n

π

/L. Substituting kinto the equation for Fwe obtain

F=n

π

L2¯

D11 





ks¯

A55

ks¯

A55 +n

π

L2¯

D11







=n

π

L2¯

D11 



1−

¯

D11 n

π

L2/ks¯

A55

1+¯

D11 n

π

L2/ks¯

A55







The critical buckling load Fcr is given for the minimum (n=1)

F=

π

L2¯

D11 





1

1+

π

L2¯

D11/ks¯

A55







For the classical beam model is ks¯

A55 →∞and we obtain

F=

π

L2¯

D11

2. At both ends ﬁxed beam (clamped beam)

Now we have the boundary conditions

w(0) = w(L) = 0,

ψ

(0) =

ψ

(L) = 0

From (7.3.23) follows

7.7 Problems 271

ks¯

A55[w′′(x1) +

ψ

′(x1)] −Fw′′(x1) = 0,

¯

D11

ψ

′′(x1)−ks¯

A55[w′(x1) +

ψ

(x1)] = 0

The ﬁrst equation yields

ks¯

A55

ψ

′(x1) = −(ks¯

A55 −F)w′′(x1),

ks¯

A55

ψ

(x1) = −(ks¯

A55 −F)w′(x1) + K1

The boundary conditions lead the equations

C2+C4=0,C1sin k L +C2cos k L +C3L+C4=0,

−1−F

ks¯

A55 kC1−C3=0,

−1−F

ks¯

A55 (kC1coskL −kC2sinkL)−C3=0

Note that

F

k2=¯

D11

1+k2¯

D11

ks¯

A55

=1−F

ks¯

A55 ¯

D11,

i.e. 1−F

ks¯

A55 =1

1+k2¯

D11

ks¯

A55

,

expressing C4and C3in terms of C1and C2and setting the determinant of the re-

maining homogeneous algebraic equations zero we obtain the buckling equation

2(coskL −1)1+k2¯

D11

ks¯

A55 +kL sin kL =0

With ks¯

A55 →∞follows the buckling equation for the classical beam

kL sin kL +2 coskL −2=0

Conclusion 7.3. Transverse shear deformation has the effect of decreasing the

buckling loads, i.e. the classical laminate theory overestimates buckling loads.

The buckling equations can be applied to all symmetric laminate and sandwich

beams if the corresponded material and stiffness values are calculated and sub-

stituted.

Exercise 7.5. A sandwich beam is modelled by the laminated beam version and

the shear deformation theory. Consider the variational formulation for applied dis-

tributed transverse loading and calculate the Euler differential equation and the

boundary conditions.

272 7 Modelling and Analysis of Beams

Solution 7.5. The elastic potential

Π

(u,w,

ψ

)for unsymmetrical laminated beams

is given by Eq. (7.3.16). Using the notations for the stiffness of sandwich beams,

Sect. 7.4, we have

Π

(u,w,

ψ

) = 1

2

l

Z

0

[Au′2+2Bu′

ψ

′+D

ψ

′2+S(w′+

ψ

)2]dx1−

l

Z

0

qwdx1

Taking the variation

δΠ

=0 one can write the following equation

δΠ

=

l

Z

0{Au′

δ

u′+B(u′

δψ

′+

ψ

′

δ

u′) + D

ψ

′

δψ

′

+S[(w′+

ψ

)

δ

w′+ (w′+

ψ

)

δψ

]}dx1−

l

Z

0

q

δ

wdx1=0

Integrating by parts, i.e

l

Z

0

f′g′dx= [ f′g]l

0−

l

Z

0

f′′gdx

yield

l

Z

0{(Au′′ +B

ψ

′′)

δ

u+ [Bu′′ +D

ψ

′′ −S(w′+

ψ

)]

δψ

+S[(w′′ +

ψ

′) + q]

δ

w}dx1

−[(Au′+B

ψ

′)

δ

u]l

0−[(Bu′+D

ψ

′)

δψ

]l

0−S[(w′+

ψ

)

δ

w]l

0=0

and the associated differential equations and boundary conditions are

Au′′ +B

ψ

′′ =0,

Bu′′ +D

ψ

′′ −S(w′+

ψ

) = 0,

S(w′′ +

ψ

′) + q=0

Putting in u′′ =−(B/A)

ψ

′′ into the second equation yield

D−B2

A

ψ

′′ −S(w′+

ψ

) = 0

and we obtain the differential equations (7.4.32)

DR

ψ

′′ −S(w′+

ψ

) = 0,

S(w′′ +

ψ

) = −q

The boundary conditions for x1=0,lare

7.7 Problems 273

u=0 or Au′+B

ψ

′=N=0

ψ

=0 or Bu′+D

ψ

′=M=0

w=0 or S(w′+

ψ

) = Q=0

u,

ψ

,wrepresent the essential and N,M,Qthe natural boundary conditions of the

problem. For symmetric sandwich beams the equations can be simpliﬁed: B=

0,DR=D.

Exercise 7.6. A sandwich beam is modelled by the differential equations and

boundary conditions of Excercise 7.5. Calculate the exact solution for a simply sup-

ported beam with q(x1) = q0,N(x1) = 0.

Solution 7.6. The boundary conditions are:

w(0) = w(l) = 0,M(0) = M(l) = 0

Using the equations (7.4.38)

ψ

′′′(x1) = −q0

DR

,

ψ

′′(x1) = −q0x1

DR

+C1,

ψ

′(x1) = −q0x2

1

2DR

+C1x1+C2,

ψ

(x1) = −q0x3

1

6DR

+C1

x2

1

2+C2x1+C3,

ψ

′(0) = 0⇒C2=0,

ψ

′(l) = 0⇒C1=q0l

2DR

,

w′(x1) = −

ψ

(x1) + DR

S

ψ

′′ =q0x3

1

6DR−q0lx2

1

4DR

+C3−q0

Sx1−l

2,

w(x1) = q0

24DR

(x4

1−2lx3

1) + C3x1−q0

2Sx2

1−lx1,

w(l) = 0⇒C3=q0l3

24DR

,

u′(x1) = −B

A

ψ

′(x1) = B

A

q0

2DR

(x2

1−lx1),

u(x1) = B

A

q0

12DR

(2x3

1−3lx1) + C4,

u(0) = 0⇒C4=0

Finally we get

274 7 Modelling and Analysis of Beams

ψ

(x1) = −q0x3

1

6DR

+q0l

2DR

x2

1

2+q0l3

24DR

,

w(x1) = q0

24DR

(x4

1−2lx3

1+l3x1) + q0

2Slx1−x2

1,

u(x1) = B

A

q0

12DR

(2x3

1−3lx1)

For symmetrical beams the solution simpliﬁed with B=0,DR=Dto

w(x1) = q0l4

24D(x4

1−2x3

1+x1) + q0l2

2S(x1−x2

1)

=wB(x1) + wS(x1),

ψ

(x1) = q0l3

24D(−1+6x2

1−4x3

1)

u(x1)≡0,x1=x/l

Chapter 8

Modelling and Analysis of Plates

The modelling and analysis of plates constituted of laminate or sandwich material is

a problem of more complexity than that of beams, considered in Chap. 7. Generally,

plates are two-dimensional thin structure elements with a plane middle surface. The

thickness his small relatively to the two other dimensions a,b(Fig. 8.1).

In Chap. 8 all derivatives are as a matter of priority restricted to rectangular plates

including the special case of a plate strip, i.e. a rectangular plate element which

is very long, for instance in the x2-direction and has ﬁnite dimension in the x1-

direction. When the transverse plate loading, the plate stiffness, and the boundary

conditions for the plate edges x1=const are independent of the coordinate x2, the

plate strip modelling can be reduced to a one-dimensional problem. The analysis is

x1,u1

x2,u2

x3,u3

a

b

h

x1

x2

x3

N1

N2

Nn

Q1

Q2

Qn

N6

Nnt

M1

M2

M6

Mn

Mnt

ab

Fig. 8.1 Rectangular plate. aGeometry, bforce resultants N1,N2,N6,Q1,Q2and moment resul-

tants M1,M2,M6.Nn,Nnt ,Qnand Mn,Mnt are force and moment resultants for an oblique edge

275

H. Altenbach et al., Mechanics of Composite Structural

Elements, https://doi.org/10.1007/978-981-10-8935-0_8

276 8 Modelling and Analysis of Plates

nearly the same as in the beam theory.Chapter 8 gives a ﬁrst introduction to the clas-

sical plate theory and the plate theory including transverse shear deformations. The

derivations of the principal equations for plates relies upon the basic considerations

of Chap. 5.

8.1 Introduction

In the theory of plate bending the most complex problem is the modelling and anal-

ysis of laminate plates with an arbitrary stacking of the layers. These plates present

couplings of stretching and bending, stretching and twisting and bending and twist-

ing and the design engineer has to look for simpliﬁcations.

The ﬁrst and most important simpliﬁcation is to design symmetric laminates

for which no coupling exists between in-plane forces and ﬂexural moments. The

coupling terms Bi j of the constitutive equations vanish. An additional simpliﬁca-

tion occurs when no bending-twisting coupling exist, i.e the terms D16 and D26 are

zero. As we discussed in Sect. 4.2, in some cases of layer stacking these coupling

terms decrease with an increasing number of layers. Symmetric laminates for which

no bending-twisting coupling exists are referred to as specially orthotropic lami-

nates. These laminates are considered in detail in this chapter, because analytical

solutions exist for various loadings and boundary conditions. Specially orthotropic

plates are obtained for single layer plates with orthotropic material behavior or sym-

metric cross-ply laminates. Symmetric balanced laminates with a great number of

layers have approximately a specially orthotropic behavior. This class of laminates

is greatly simpliﬁed and will be used to gain a basic understanding of laminate plate

response. Like in Chap. 7 for beams, we consider the plates in the framework of the

classical and the ﬁrst order shear deformation theory. For a better understanding the

assumptions of both plate theories given in Sects. 5.1 and 5.2 are reviewed.

The ﬁrst order shear deformation theory accounted for a constant state of trans-

verse shear stresses, but the transverse normal stress is often neglected. In the frame-

work of this plate theory, the computation of interlaminar shear stresses through

constitutive equations is possible, which is simpler than deriving them through equi-

librium equations.

The most signiﬁcant difference between the classical and ﬁrst-order shear defor-

mation theory is the effect of including transverse shear deformation in the predic-

tion of deﬂections, frequencies or buckling loads. It can be noted that the classical

laminate theory underestimates deﬂections and overestimates frequencies as well as

buckling loads when the plate side-to-thickness ratio is of the order 20 or less. For

this reason it is necessary to include shear deformation for moderately thick plates.

In general, moderately thick plates must be computed by numerical methods, ap-

plication of analytical methods are much more restricted than in the classical plate

theory.

8.2 Classical Laminate Theory 277

8.2 Classical Laminate Theory

In the classical laminate theory one presumes that the Kirchhoff hypotheses of the

classical plate theory remains valid:

•Transverse normals before deformation remain straight after deformation and

rotate such that they remain normal to the middle surface.

•Transverse normals are inextensible, i.e. they have no elongation.

These assumptions imply that the transverse displacement wis independent of the

thickness coordinate x3, the strains

ε

3,

ε

4and

ε

5are zero and the curvatures

κ

iare

given by

[

κ

1

κ

2

κ

6] = −

∂

2w

∂

x2

1−

∂

2w

∂

x2

2−2

∂

2w

∂

x1

∂

x2(8.2.1)

Figure 8.1 shows the plate geometry and the plate stress resultants. The equilibrium

equations will be formulated for a plate element dx1dx2(Fig. 8.2) and yield three

force and two moments equations

ւ

∂

N1

∂

x1

+

∂

N6

∂

x2

=−p1,

→

∂

N6

∂

x1

+

∂

N2

∂

x2

=−p2,

↑

∂

Q1

∂

x1

+

∂

Q2

∂

x2

=−p3,

→

∂

M1

∂

x1

+

∂

M6

∂

x2

=Q1,

ւ

∂

M6

∂

x1

+

∂

M2

∂

x2

=Q2

(8.2.2)

x1

x2

x3

N1dx2

(N1+dN1)dx2

N2dx1(N2+dN2)dx1

N6dx1

(N6+dN6)dx1

dx2

dx1

h

M6dx1

(M6+dM6)dx1

M6dx2

(M6+dM6)dx2

M1dx2

(M1+dM1)dx2

M2dx1

(M2+dM2)dx1

Q1dx2

(Q1+dQ1)dx2

Q2dx1

(Q2+dQ2)dx1

p1dx1dx2

p2dx1dx2

p3dx1dx2

ab

Fig. 8.2 Stress resultants applied to a plate element

278 8 Modelling and Analysis of Plates

The transverse shear force resultants Q1,Q2can be eliminated and the ﬁve equations

(8.2.2) reduce to three equations. The in-plane force resultants N1,N2and N6are

uncoupled with the moment resultants M1,M2and M6

∂

N1

∂

x1

+

∂

N6

∂

x2

=−p1,

∂

N6

∂

x1

+

∂

N2

∂

x2

=−p2,

∂

2M1

∂

x2

1

+2

∂

2M6

∂

x1

∂

x2

+

∂

2M2

∂

x2

2

=−p3

(8.2.3)

The equations are independent of material laws and present the static equations for

the undeformed plate element. The further considerations neglect the in-plane plate

loads p1and p2, i.e. p1=p2=0, p36=0. In-plane reactions can be caused by

coupling effects of unsymmetric laminates or sandwich plates.

Putting the constitutive equations





N

···

M

M

=





A

A.

.

.B

B

. . . .

B

B.

.

.D

D









ε

···

κ



(8.2.4)

into the equilibrium (8.2.3) and replacing using Eqs. (5.2.3) the in-plane strains

ε

i

and the curvatures

κ

iby

ε

T= [

ε

1

ε

2

ε

6] = 

∂

u

∂

x1

∂

v

∂

x2

∂

u

∂

x2

+

∂

v

∂

x1,

κ

T= [

κ

1

κ

2

κ

6] = −

∂

2w

∂

x2

1−

∂

2w

∂

x2

2−2

∂

2w

∂

x1

∂

x2(8.2.5)

gives the differential equations for general laminated plates

A11

∂

2u

∂

x2

1

+2A16

∂

2u

∂

x1

∂

x2

+A66

∂

2u

∂

x2

2

+A16

∂

2v

∂

x2

1

+ (A12 +A66)

∂

2v

∂

x1

∂

x2

+A26

∂

2v

∂

x2

2−B11

∂

3w

∂

x3

1−3B16

∂

3w

∂

x2

1

∂

x2−(B12 +2B66)

∂

3w

∂

x1

∂

x2

2−B26

∂

3w

∂

x3

2

=0,

A16

∂

2u

∂

x2

1

+ (A12 +A66)

∂

2u

∂

x1

∂

x2

+A26

∂

2u

∂

x2

2

+A66

∂

2v

∂

x2

1

+2A26

∂

2v

∂

x1

∂

x2

+A22

∂

2v

∂

x2

2−B16

∂

3w

∂

x3

1−(B12 +2B66)

∂

3w

∂

x2

1

∂

x2−3B26

∂

3w

∂

x1

∂

x2

2−B22

∂

3w

∂

x3

2

=0,

D11

∂

4w

∂

x4

1

+4D16

∂

4w

∂

x3

1

∂

x2

+2(D12 +2D66)

∂

4w

∂

x2

1

∂

x2

2

+4D26

∂

4w

∂

x1

∂

x3

2

+D22

∂

4w

∂

x4

2−B11

∂

3u

∂

x3

1−3B16

∂

3u

∂

x2

1

∂

x2−(B12 +2B66)

∂

3u

∂

x1

∂

x2

2−B26

∂

3u

∂

x3

2

−B16

∂

3v

∂

x3

1−(B12 +2B66)

∂

3v

∂

x2

1

∂

x2−3B26

∂

3v

∂

x1

∂

x2

2−B22

∂

3v

∂

x3

2

=p3

(8.2.6)

8.2 Classical Laminate Theory 279

Equations (8.2.6) are three coupled partial differential equations for the displace-

ments u(x1,x2),v(x1,x2),w(x1,x2). Equation (8.2.6) can be formulated in matrix

form as 



L11 L12 L13

L21 L22 L31

L31 L32 L33 





u

v

w

=



0

p

,Li j =Lji (8.2.7)

The differential operators are given in App. C.

The differential operators L11 ,L12 and L22 are of second order, L13 and L23 of

third order and L33 of fourth order. The homogeneous part of the coupled partial

differential equations (8.2.7) can be reduced to one partial equation of eight order

[(L11L22 −L2

12)L33 −(L11L2

23 −2L12L13L23 +L2

13L22 )]w=0 (8.2.8)

Consistent with the eight order set of differential equations four boundary conditions

must be prescribed for each edge of the plate. The classical boundary conditions are:

Either

Nnor u,Nnt or v,Mnor

∂

w

∂

n,Vn≡Qn+

∂

Mnt

∂

tor w(8.2.9)

must be speciﬁed. The subscripts nand tin the boundary conditions above denote

the coordinates normal and tangential to the boundary. It is well known that in the

classical plate theory the boundary cannot responded separately to the shear force

resultant Qnand the twisting moment Mnt but only to the effective or Kirchhoff

shear force resultant

Vn≡Qn+

∂

Mnt

∂

t(8.2.10)

Equations (8.2.9) may be used to represent any form of simple edge conditions, e.g.

clamped, simply supported and free.

The boundary conditions (8.2.9) represent pairs of response variables. One com-

ponent of these pairs involve a force or a moment resultant, the other a displace-

ment or a rotation. Take into account that in addition to the edge conditions it can

be necessary to fulﬁl the point corner conditions, e.g. for a free corner. Sometimes

more general boundary conditions, which are applicable to edges having elastic con-

straints, are used, e.g. the transverse and/or rotatory plate conditions

Vn±cTw=0; Mn±cR

∂

w

∂

n=0 (8.2.11)

cTand cRdenote the spring stiffness of the constraints.

In applying the boundary conditions (8.2.9) it is useful to have explicit expres-

sions for the stress resultants in a displacement formulation. According to Eqs.

(8.2.5) and (8.2.4) the stress resultants can be written as

N1=A11

∂

u

∂

x1

+A12

∂

v

∂

x2

+A16 

∂

u

∂

x2

+

∂

v

∂

x1

280 8 Modelling and Analysis of Plates

−B11

∂

2w

∂

x2

1−B12

∂

2w

∂

x2

2−2B16

∂

2w

∂

x1

∂

x2

,

N2=A12

∂

u

∂

x1

+A22

∂

v

∂

x2

+A26 

∂

u

∂

x2

+

∂

v

∂

x1

−B12

∂

2w

∂

x2

1−B22

∂

2w

∂

x2

2−2B26

∂

2w

∂

x1

∂

x2

,

N6=A16

∂

u

∂

x1

+A26

∂

v

∂

x2

+A66 

∂

u

∂

x2

+

∂

v

∂

x1

−B16

∂

2w

∂

x2

1−B26

∂

2w

∂

x2

2−2B66

∂

2w

∂

x1

∂

x2

,

M1=B11

∂

u

∂

x1

+B12

∂

v

∂

x2

+B16 

∂

u

∂

x2

+

∂

v

∂

x1

−D11

∂

2w

∂

x2

1−D12

∂

2w

∂

x2

2−2D16

∂

2w

∂

x1

∂

x2

,

M2=B12

∂

u

∂

x1

+B22

∂

v

∂

x2

+B26 

∂

u

∂

x2

+

∂

v

∂

x1(8.2.12)

−D12

∂

2w

∂

x2

1−D22

∂

2w

∂

x2

2−2D26

∂

2w

∂

x1

∂

x2

,

M6=B16

∂

u

∂

x1

+B26

∂

v

∂

x2

+B66 

∂

u

∂

x2

+

∂

v

∂

x1

−D16

∂

2w

∂

x2

1−D26

∂

2w

∂

x2

2−2D66

∂

2w

∂

x1

∂

x2

,

Q1=B11

∂

2u

∂

x2

1

+2B16

∂

2u

∂

x1

∂

x2

+B66

∂

2u

∂

x2

2

+B16

∂

2v

∂

x2

1

+ (B12 +B66)

∂

2v

∂

x1

∂

x2

+B26

∂

2v

∂

2x2−D11

∂

3w

∂

x3

1−3D16

∂

3w

∂

x2

1

∂

x2−(D12 +2D66)

∂

3w

∂

x1

∂

x2

2−D26

∂

3w

∂

x3

2

,

Q2=B16

∂

2u

∂

x2

1

+ (B12 +B66)

∂

2u

∂

x1

∂

x2

+B26

∂

2u

∂

x2

2

+B66

∂

2v

∂

x2

1

+2B26

∂

2v

∂

x1

∂

x2

+B22

∂

2v

∂

2x2−D16

∂

3w

∂

x3

1−(D12 +2D66)

∂

3w

∂

x2

1

∂

x2−3D26

∂

3w

∂

x1

∂

x2

2−D22

∂

3w

∂

x3

2

,

V1=B11

∂

2u

∂

x2

1

+3B16

∂

2u

∂

x1

∂

x2

+2B66

∂

2u

∂

x2

2

+B16

∂

2v

∂

x2

1

+ (B12 +2B66)

∂

2v

∂

x1

∂

x2

+2B26

∂

2v

∂

2x2−D11

∂

3w

∂

x3

1−4D16

∂

3w

∂

x2

1

∂

x2−(D12 +D66)

∂

3w

∂

x1

∂

x2

2−2D26

∂

3w

∂

x3

2

,

8.2 Classical Laminate Theory 281

V2=2B16

∂

2u

∂

x2

1

+ (B12 +2B66)

∂

2u

∂

x1

∂

x2

+B26

∂

2u

∂

x2

2

+2B66

∂

2v

∂

x2

1

+3B26

∂

2v

∂

x1

∂

x2

+B22

∂

2v

∂

x2

2−2D16

∂

3w

∂

x3

1−(D12 +4D66)

∂

3w

∂

x2

1

∂

x2−4D26

∂

3w

∂

x1

∂

x2

2−D22

∂

3w

∂

x3

2

The coupled system of three partial differential equations (8.2.6) or (8.2.7), respec-

tively, can be simpliﬁed for special layer stacking, Sect. 4.2.3. The differential op-

erators Li j for some special cases are given in App. C.

1. Symmetric laminates

Because all coupling stiffness Bi j are zero the in-plane and the out-of-plane dis-

placement response are uncoupled. With L13 =L31 =0,L23 =L32 =0 Eq. (8.2.7)

simpliﬁes to





L11 L12 0

L12 L22 0

0 0 L33 





u

v

w

=



0

p3

(8.2.13)

The plate equation reduces to L33w=p3and corresponds to the plate equation

of an anisotropic homogeneous plate.

2. Antisymmetric laminates

The in-plane and the transverse part of Eq. (8.2.7) are coupled, but with

A16 =A26 =0,D16 =D26 =0 the differential operators L11 ,L22,L33 and L12

are reduced. It is no in-plane tension/shearing coupling and no bending/twisting

coupling.

3. Balanced laminates

For general balanced laminates with A16 =A26 only the in-plane ten-

sion/shearing coupling is zero, for an antisymmetric balanced laminate we have

A16 =A26 =0,D16 =D26 =0 and for symmetric balanced laminates follow

A16 =A26 =0, Bi j =0. The last case yields the equations





L11 L12 0

L12 L22 0

0 0 L33 





u

v

w

=



0

p3



with simpliﬁed differential operators L11 and L22. Only the in-plane equations

correspond to an orthotropic stiffness behavior.

4. Cross-ply laminates

The stacking can be unsymmetrical, i.e. A16 =A26 =0,D16 =D26 =0,

B16 =B26 =0, antisymmetrical, i.e. A16 =A26 =0,D16 =D26 =0,

B12 =B16 =B26 =B66 =0, B22 =−B11 or symmetrical with A16 =A26 =0,

D16 =D26 =0,Bi j =0. Cross-ply laminates have an orthotropic response to

both in-plane and bending and no in-plane/bending coupling. The plate equation

L33w=p3corresponds to the equation of an homogeneous orthotropic plate.

Summarizing the mathematical structures of the differential equations in depen-

dence on the layer stacking the following conclusions can be drawn:

282 8 Modelling and Analysis of Plates

•The mathematical structure of a general balanced laminate is not much simpler

as for a general unsymmetric, unbalanced laminate

•Compared to the general case the mathematical structure of the symmetric cross-

ply laminate is nearly trivial. A symmetric cross-ply is orthotropic with respect

to both in-plane and bending behavior, and both are uncoupled.

•The most simple mathematical structure yields the laminate with symmetrical

arranged isotropic layers. With A11 =A22,A16 =A26 =0,Bi j =0, D11 =D22,

D16 =D26 =0, it corresponds to a single layer isotropic plate with in-plane and

transverse loading.

•For special layer stacking also the force and moment resultant Eqs. (8.2.12) are

reduced to more simple equations.

The following developments are restricted to general symmetric plates and plates

with specially orthotropic behavior. The equations will be signiﬁcant simpliﬁed, for

example in the general case all Bi j =0 and for specially orthotropic plates there are

additional D16 =D26 =0. The in-plane and the ﬂexural equations are uncoupled. Ta-

ble 8.1 summarizes the most important plate equations. In Table 8.1 standard bound-

ary conditions are also expressed. The necessary and sufﬁcient number of boundary

conditions for plates considered here are two at each of the boundaries. The stan-

dard conditions for the free edge reduce the three static conditions Mn=0, Qn=0

and Mnt =0 to two conditions Mn=0,Vn=0, where Vn=Qn+

∂

Mnt/

∂

t=0 is as

discussed above the Kirchhoff effective shear resultant. In order to avoid mistakes

in the application the equations of Table 8.1, a summary of plate stiffness is given.

Table 8.2 contains the plate stiffness for single layer plates. The plate stiffness for

symmetric laminates are given in Table 8.3. In all equations the hygrothermal effects

are neglected, but it is no problem to include thermal or moisture changes. In this

case (4.2.63), (4.2.64) must be used instead of (8.2.4) to put into the equilibrium

equations. This will be considered in Sect. 8.5.

The classical laminate theory can be used also for modelling and analysis of vi-

bration and buckling of laminated plates. We restrict the consideration to symmetric

plates. In the case of forced transversal vibration the momentum equilibrium equa-

tion (8.2.3) has an additional inertial term

∂

2M1

∂

x2

1

+2

∂

2M6

∂

x1

∂

x2

+

∂

2M2

∂

x2

2

=−p3+

ρ

h

∂

2w

∂

t2(8.2.14)

M1,M2,M6,wand p3are functions of x1,x2and the time t,his the total thickness of

the plate and

ρ

the mass density

h=

n

∑

k=1

h(k),

ρ

=1

h

n

∑

k=1

ρ

(k)x(k)

3−x(k−1)

3=1

h

n

∑

k=1

ρ

(k)h(k)(8.2.15)

The rotatory inertia is neglected. The Eqs. (8.2.14), (8.2.4) and (8.2.5) yield the plate

equations for force vibration. For the both layer stacking discussed above we obtain:

8.2 Classical Laminate Theory 283

Table 8.1 Plate equation, boundary conditions and stress resultants of symmetric laminates

1. General case: Bi j =0,Di j 6=0i,j=1,2,6

D11

∂

4w

∂

x4

1

+4D16

∂

4w

∂

x3

1

∂

x2

+2(D12 +2D66)

∂

4w

∂

x2

1

∂

x2

2

+4D26

∂

4w

∂

x1

∂

x3

2

+D22

∂

4w

∂

x4

2

=p3

2. Specially orthotropic laminates: Bi j =0,D16 =D26 =0

D11

∂

4w

∂

x4

1

+2(D12 +2D66)

∂

4w

∂

x2

1

∂

x2

2

+D22

∂

4w

∂

x4

2

=p3

or with D11 =D1,D22 =D2,D12 +2D66 =D3

D1

∂

4w

∂

x4

1

+2D3

∂

4w

∂

x2

1

∂

x2

2

+D2

∂

4w

∂

x4

2

=p3

3. Laminates with isotropic layers: D11 =D22 =D1,(D12 +2D66 ) = D3

D1

∂

4w

∂

x4

1

+2D3

∂

4w

∂

x2

1

∂

x2

2

+D1

∂

4w

∂

x4

2

=p3

Typical boundary conditions: 1. Simply supported edge: w=0,Mn=0

2. Clamped edge: w=0,

∂

w/

∂

n=0

3. Free edge: Mn=0,Vn=Qn+

∂

Mnt/

∂

t=0

Stress resultants:

1. General case





M1

M2

M6

=



D11 D12 D16

D12 D22 D26

D16 D26 D66 



−

∂

2w/

∂

x2

1

−

∂

2w/

∂

x2

2

−2

∂

2w/

∂

x1

∂

x2

,

Q1=

∂

M1

∂

x1

+

∂

M6

∂

x2

Q2=

∂

M6

∂

x1

+

∂

M2

∂

x2

2. Specially orthotropic





M1

M2

M6

=



D11 D12 0

D12 D22 0

0 0 D66 



−

∂

2w/

∂

x2

1

−

∂

2w/

∂

x2

2

−2

∂

2w/

∂

x1

∂

x2

,

Q1=

∂

M1

∂

x1

+

∂

M6

∂

x2

Q2=

∂

M6

∂

x1

+

∂

M2

∂

x2

3. Isotropic layers (like 2. with D11 =D22)

1. General case of symmetric plates

L33 +

ρ

h

∂

tw=p3

or explicitly

284 8 Modelling and Analysis of Plates

Table 8.2 Plate stiffness for single layer

Anisotropic single layer

Di j =Q(k)

i j

h3

12

Specially orthotropic single layer (on-axis)

D11 =Q11

h3

12 ,D12 =Q12

h3

12 ,D22 =Q22

h3

12 ,D66 =Q66

h3

12 ,

Q′

11 =E′

1

1−

ν

′

12

ν

′

21 ,Q′

22 =E′

2

1−

ν

′

12

ν

′

21 ,Q′

12 =

ν

′

12E1

1−

ν

′

12

ν

′

21 ,

Q′

66 =G′

12 =E′

6

Isotropic single layer

D11 =D22 =Eh3

12(1−

ν

2)=D,D12 =

ν

D=

ν

Eh3

1−

ν

2,

D66 =1−

ν

2D=Eh3

24(1+

ν

),

D11

∂

4w

∂

x4

1

+4D16

∂

4w

∂

x3

1

∂

x2

+2(D12 +2D66)

∂

4w

∂

x2

1

∂

x2

2

+4D26

∂

4w

∂

x1

∂

x3

2

+D22

∂

4w

∂

x4

2

=p3−

ρ

h

∂

2w

∂

t2

with w=w(x1,x2,t),p= (x1,x2,t).

2. Specially orthotropic plates

D1

∂

4w

∂

x4

1

+2D3

∂

4w

∂

x2

1

∂

x2

2

+D2

∂

4w

∂

x4

2

=p3−

ρ

h

∂

2w

∂

t2(8.2.16)

The equation of symmetric laminate plates with isotropic layers follows from

(8.2.16) with D1=D2, the plate stiffness are taken from Table 8.2 (single layer

plates) or Table 8.3 (laminates). In the case of the computation of natural or eigen-

vibrations, the forcing function p3(x1,x2,t)is taken to be zero and the time depen-

dent motion is a harmonic oscillation. The differential equation is homogeneous,

leading an eigenvalue problem for the eigenvalues (natural frequencies) and the

eigenfunctions (mode shapes).

To predict the buckling for plates, in-plane force resultants must be included. For

a coupling of in-plane loads and lateral deﬂection, the equilibrium (8.2.2) will be

formulated for the deformed plate element with p1=p2=p3=0 and modiﬁed to

8.2 Classical Laminate Theory 285

Table 8.3 Plate stiffness for symmetric laminates

Symmetric angle ply laminate

Di j =

n

∑

k=1

Q(k)

i j x(k)

3

3−x(k−1)

3

3=

n

∑

k=1

Q(k)

i j h(k) x(k)

3+h(k)2

12 !,

x(k)

3=1

2x(k)

3+x(k−1)

3, the Q(k)

i j follow from Table 4.2.

Symmetric balanced laminates

Di j =

n

∑

k=1

Q(k)

i j h(k) x(k)

3+h(k)2

12 !

The Q(k)

i j follow from Table 4.2.

Symmetric cross-ply laminate (specially orthotropic)

Di j =

n

∑

k=1

Q(k)

i j h(k) x(k)

3+h(k)2

12 !,

D16 =D26 =0

Q(k)

11 =E1

1−

ν

12

ν

21 (k)

,Q(k)

22 =E2

1−

ν

12

ν

21 (k)

,

Q(k)

12 =

ν

12E1

1−

ν

12

ν

21 (k)

,Q(k)

66 =G(k)

12

Symmetric laminate with isotropic layers (x1-direction equal ﬁbre direction)

Di j =

n

∑

k=1

Q(k)

i j h(k) x(k)

3+h(k)2

12 !,

D16 =D26 =0,D11 =D22,

Q(k)

11 =Q(k)

22 =E

1−

ν

2(k)

,Q(k)

12 =

ν

E

1−

ν

2(k)

,

Q(k)

66 =E

2(1+

ν

)(k)

∂

2M1

∂

x2

1

+2

∂

2M6

∂

x1

∂

x2

+

∂

2M2

∂

x2

2

=N1

∂

2w

∂

x2

1

+N2

∂

2w

∂

x2

2

+2N6

∂

2w

∂

x1

∂

x2

,

∂

N1

∂

x1

+

∂

N6

∂

x2

=0,

∂

N6

∂

x1

+

∂

N2

∂

x2

=0

(8.2.17)

In the general case of a symmetric laminate, the plate equation can be expressed by

D11

∂

4w

∂

x4

1

+4D16

∂

4w

∂

x3

1

∂

x2

+2(D12 +2D66)

∂

4w

∂

x2

1

∂

x2

2

+4D26

∂

4w

∂

x1

∂

x3

2

+D22

∂

4w

∂

x4

2

=N1

∂

2w

∂

x2

1

+N2

∂

2w

∂

x2

2

+2N6

∂

2w

∂

x1

∂

x2

(8.2.18)

286 8 Modelling and Analysis of Plates

and for specially orthotropic laminates

D1

∂

4w

∂

x4

1

+2D3

∂

4w

∂

x2

1

∂

x2

2

+D2

∂

4w

∂

x4

2

=N1

∂

2w

∂

x2

1

+N2

∂

2w

∂

x2

2

+2N6

∂

2w

∂

x1

∂

x2

(8.2.19)

The special case of symmetric laminates with isotropic layers follows from (8.2.19)

with D1=D2. The buckling load is like the natural vibration independent of the

lateral load and p3is taken to be zero. The classical bifurcation buckling requires to

satisfy the governing differential equations derived above and the boundary equa-

tions. Both sets of equations are again homogeneous and represent an eigenvalue

problem for the buckling modes (eigenvalues) and the mode shapes (eigenfunc-

tions).

To calculate the in-plane stress resultants N1,N2,N6it is usually convenient to

represent they by the Airy stress function F(x1,x2)

N1=

∂

2F

∂

x2

2

,N2=

∂

2F

∂

x2

1

,N6=−

∂

2F

∂

x1

∂

x2

(8.2.20)

If Eqs. (8.2.19) are substituted into the ﬁrst two equilibrium equations (8.2.3) it is

seen that these equations are identically satisﬁed. Using Eq. (4.2.22)

M

M=B

B

BA

A

A−1N

N

N−(B

B

BA

A

A−1B

B

B−D

D

D)

κ

and substitute N

N

Nwith help of the Airy‘s stress function and

κ

by the derivatives of

wthe third equilibrium equation (8.2.3) yields one coupled partial differential equa-

tion for Fand w. The necessary second equation yields the in-plane compatibility

condition (Sect. 2.2)

∂

2

ε

1

∂

x2

2

+

∂

2

ε

2

∂

x2

1

=

∂

2

ε

6

∂

x1

∂

x2

together with Eq. (4.2.25) to substitute the strains by the stress resultants. Suppress-

ing the derivations and restricting to symmetric problems yield the following in-

plane equations which are summarized in Table 8.4. The stiffness A

A

A∗,B

B

B∗,C

C

C∗,D

D

D∗fol-

low with Eq. (4.2.23) as A

A

A∗=A

A

A−1,B

B

B∗=−A

A

A−1B

B

B,C

C

C∗=B

B

BA

A

A−1,D

D

D∗=D

D

D−B

B

BA

A

A−1B

B

B.

One can see from Table 8.4 that in the general case the mathematical structure of

the partial differential equation corresponds to an anisotropic and in the special or-

thotropic case to an orthotropic in-plane behavior of a single layer homogeneous

anisotropic or orthotropic plate. A summary of the in-plane stiffness is given in Ta-

ble 8.5. The Q(k)

i j for angle-ply laminates are calculated in Table 4.2.

Similar to the beam theory the plate equations for ﬂexure, vibration and buckling

can be given in a variational formulation (Sect. 2.2). This formulation provides the

basis for the development of approximate solutions. We restrict the variational for-

mulation to symmetric laminated plates and to the classical energy principles. From

(2.2.24) it follows with

ε

3=

ε

4=

ε

5≈0 that the elastic potential

Π

is

8.2 Classical Laminate Theory 287

Table 8.4 In-plane equations, boundary conditions and stress resultants for symmetric laminates

1. Angle-ply laminates

B∗

i j =0,C∗

i j =0,i,j=1,2,6

A∗

22

∂

4F

∂

x4

1−2A∗

26

∂

4F

∂

x3

1

∂

x2

+(2A∗

12 +A∗

66)

∂

4F

∂

x2

1

∂

x2

2−2A∗

16

∂

4F

∂

x1

∂

x3

2

+A∗

11

∂

4F

∂

x4

2

=0

2. Cross-ply laminates

B∗

i j =0,A∗

16 =A∗

26 =0,

A∗

22

∂

4F

∂

x4

1

+ (2A∗

12 +A∗

66)

∂

4F

∂

x2

1

∂

x2

2

+A∗

11

∂

4F

∂

x4

2

=0

or with A∗

11 =A∗

1,A∗

22 =A∗

2,(2A∗

12 +A∗

66) = A∗

3,

A∗

2

∂

4F

∂

x4

1

+2A∗

3

∂

4F

∂

x2

1

∂

x2

2

+A∗

1

∂

4F

∂

x4

2

=0

3. Laminates with isotropic layers

A∗

1=A∗

2=A∗

3=1,

∂

4F

∂

x4

1

+2

∂

4F

∂

x2

1

∂

x2

2

+

∂

4F

∂

x4

2

=0

Typical boundary conditions

Edge x1=const

∂

2F

∂

x2

2

=N1(x1=const,x2),−

∂

2F

∂

x1

∂

x2

=N6(x1=const,x2),

For an unloaded edge follow N1=0,N6=0

Stress resultants

N1=

∂

2F

∂

x2

2

,N2=

∂

2F

∂

x2

1

,N6=−

∂

2F

∂

x1

∂

x2

Π

=1

2Z

V

(

σ

1

ε

1+

σ

2

ε

2+

σ

6

ε

6)dV−Z

A

p3(x1,x2)w(x1,x2)dA(8.2.21)

=1

2

n

∑

k=1Z

A

x(k)

3

Z

x(k−1)

3

(

σ

(k)

1

ε

1+

σ

(k)

2

ε

2+

σ

(k)

6

ε

6)dx3dA−Z

A

p3(x1,x2)w(x1,x2)dA

With

ε

(x1,x2,x3) =

ε

(x1,x2) + x3

κ

(x1,x2)(8.2.22)

288 8 Modelling and Analysis of Plates

Table 8.5 In-plane stiffness for symmetric laminates

1. Angle-ply laminates

Ai j =

n

∑

k=1

Q(k)

i j x(k)

3−x(k−1)

3=

n

∑

k=1

Q(k)

i j h(k),i,j=1,2,6

2. Cross-ply laminates

Ai j =

n

∑

k=1

Q(k)

i j h(k),i,j=1,2,6,A16 =A26 =0,

Q(k)

11 =E1

1−

ν

12

ν

21 (k)

,Q(k)

22 =E2

1−

ν

12

ν

21 (k)

Q(k)

12 =

ν

12E1

1−

ν

12

ν

21 (k)

,Q(k)

66 =G(k)

12

3. Laminates with isotropic layers

Ai j =

n

∑

k=1

Q(k)

i j h(k),i,j=1,2,6,A16 =A26 =0,A11 =A22

Q(k)

11 =Q(k)

22 =E

1−

ν

2(k)

,Q(k)

12 =

ν

E

1−

ν

2(k)

,

Q(k)

66 =E

2(1+

ν

)(k)

=G(k)

4. Single layer

For anisotropic and orthotropic single layers the Ai j followed by 1. and 2.

For an isotropic single layer is

A11 =A22 =A=Eh

1−

ν

2,A12 =

ν

A=

ν

Eh

1−

ν

2,

A66 =(1−

ν

)

2A=Eh

2(1+

ν

)=G

ε

T=

∂

u

∂

x1

∂

v

∂

x2

∂

u

∂

x2

+

∂

v

∂

x1,

κ

T=−

∂

2w

∂

x2

1−

∂

2w

∂

x2

2−2

∂

2w

∂

x1

∂

x2

and the constitutive equations for the strains and the stress resultants

σ

(k)=Q

Q

Q(k)(

ε

+x3

κ

),



N

M

M

=





A

A.

.

. 0

0

. . . .

0

0.

.

.D

D









ε

κ



(8.2.23)

one obtains the elastic potential for the general case of symmetric plates and for the

special cases of orthotropic or isotropic structure behavior.

8.2 Classical Laminate Theory 289

Bending of plates, classical laminate theory

:

Angle-ply laminates

Π

(w) = 1

2Z

AD11 

∂

2w

∂

x2

12+D22 

∂

2w

∂

x2

22+2D12

∂

2w

∂

x2

1

∂

2w

∂

x2

2

+4D66 

∂

2w

∂

x1

∂

x22

+4D16

∂

2w

∂

x2

1

+D26

∂

2w

∂

x2

2

∂

2w

∂

x1x2dA−Z

A

p3wdA

(8.2.24)

Cross-ply laminates

D16 =D26 =0

Laminates with isotropic layers

D16 =D26 =0,D11 =D22

The principle of minimum of the total potential yields

δΠ

[w(x1,x2)] = 0

as the basis to derive the differential equation and boundary conditions or to apply

the direct variational methods of Ritz, Galerkin or Kantorovich for approximate

solutions.

Vibration of plates, classical laminate theory:

The kinetic energy of a plate is (rotatory energy is neglected)

T=1

2Z

A

ρ

h

∂

w

∂

t2

dA,

ρ

=1

h

n

∑

k=1

ρ

(k)h(k)(8.2.25)

The Hamilton principle for vibrations yields

δ

H(x1,x2,t) = 0

with

H=

t2

Z

t1

(T−

Π

)dt =

t2

Z

t1

Ldt(8.2.26)

Buckling of plates, classical laminate theory:

To calculate buckling loads, the in-plane stress resultants N1,N2,N6must be in-

cluded into the potential

Π

. These in-plane stress resultants are computed in a ﬁrst

step or are known a priori. With the known N1,N2,N6the potential

Π

can be formu-

lated for angle-ply laminates with bending in-plane forces

290 8 Modelling and Analysis of Plates

Π

(w) = 1

2Z

A("D11 

∂

2w

∂

x2

12

+D22 

∂

2w

∂

x2

22

+2D12

∂

2w

∂

x2

1

∂

2w

∂

x2

2

+4D66 

∂

2w

∂

x1

∂

x22

+4D16

∂

2w

∂

x2

1

+D26

∂

2w

∂

x2

2

∂

2w

∂

x1x2#

−"N1

∂

w

∂

x12

+N2

∂

w

∂

x22

+2N6

∂

w

∂

x1

∂

w

∂

x2#−2p3w

)dA

(8.2.27)

The buckling formulation one get with p3≡0. With D16 =D26 =0 or

D16 =D26 =0, D11 =D22 follows the equations for cross-ply laminates plates

and for plates with isotropic layers. The plate stiffness can be taken from Tables 8.2

or 8.3.

The variational principle

δΠ

=0 applied to (8.2.24) and (8.2.27) yield solutions

for bending and bending with in-plane forces. Hamilton’s principle and p36=0 is

valid to calculate forced vibrations. With p3(x1,x2,t) = 0 in vibration equations

or p3(x1,x2) = 0 in (8.2.27), we have formulated eigenvalue problems to compute

natural frequencies or buckling loads.

Summarizing the derivations of governing plate equations in the frame of classi-

cal laminate theory there are varying degrees of complexity:

•An important simpliﬁcation of the classical two-dimensional plate equations is

the behavior of cylindrical bending. In this case one considers a laminated plate

strip with a very high length-to-width ratio. The transverse load and all displace-

ments are functions of only x1and all derivatives with respect to x2are zero.

The laminated beams, Chap. 7, and the laminated strips under cylindrical bend-

ing are the two cases of laminated plates that can be treated as one-dimensional

problems. In Sect. 8.6 we discuss some applications of cylindrical bending

•In the case of two-dimensional plate equations the ﬁrst degree of simpliﬁcation

for plates is to be symmetric. Symmetric laminates can be broken into cross-

ply laminates (specially orthotropic plates) with uncoupling in-plane and bend-

ing response (Bi j =0) and vanishing bending-twisting terms (D16 =D26 =0)

and angle-ply laminates (only Bi j =0). The governing equations of symmetric

cross-ply laminates have the mathematical structure of homogeneous orthotropic

plates, symmetric angle-ply laminates of homogeneous anisotropic plates. For

special boundary conditions symmetric cross-ply laminated rectangular plates

can be solved analytically. The solutions were obtained in the same manner as

for homogeneous isotropic plates, Sect. 8.6.

•Laminates with all coupling effects are more complicated to analyze. Generally,

approximate analytical or numerical methods are used.

8.3 Shear Deformation Theory 291

8.3 Shear Deformation Theory

In Sect. 8.2 we have neglected the transverse shear deformations effects. The anal-

ysis and results of the classical laminate theory are sufﬁciently accurate for thin

plates, i.e. a/h,b/h>20. Such plates are often used in civil engineering. For mod-

erately thick plates we have to take into account the shear deformation effects, at

least approximately. The theory of laminate or sandwich plates corresponds then

with the Reissner or Mindlin1plate theory. In the Reissner-Mindlin theory the as-

sumptions of the Kirchhoff’s plate theory are relaxed only in one point. The trans-

verse normals do not remain perpendicular to the middle surface after deformation,

i.e. a linear element extending through the thickness of the plate and perpendicular

to the mid-surface prior to loading, upon the load application undergoes at most a

translation and a rotation. Plate theories based upon this assumption are called ﬁrst

order shear deformation theories and are most used in the analysis of moderate thick

laminated plates and of sandwich plates. Higher order theories which do not require

normals to remain straight are considerably more complicated.

Based upon that kinematical assumption of the ﬁrst order shear deformation the-

ory the displacements of the plate have the form (5.1.2)

u1(x1,x2,x3) = u(x1,x2) + x3

ψ

1(x1,x2),

u2(x1,x2,x3) = v(x1,x2) + x3

ψ

2(x1,x2),

u3(x1,x2,x3) = w(x1,x2)

(8.3.1)

and with (5.1.3) are the strains

ε

i(x1,x2,x3) =

ε

i(x1,x2) + x3

κ

i(x1,x2),i=1,2,6,

ε

T=

∂

u

∂

x1

∂

v

∂

x2

∂

u

∂

x2

+

∂

v

∂

x1,

κ

T=

∂ψ

1

∂

x1

∂ψ

2

∂

x2

∂ψ

1

∂

x2

+

∂ψ

2

∂

x1,

ε

4(x1,x2) =

∂

w

∂

x2

+

ψ

2,

ε

5(x1,x2) =

∂

w

∂

x1

+

ψ

1

(8.3.2)

One can see that a constant state of transverse shear stresses is accounted for. The

stresses for the kth layer are formulated in (5.3.2) to

σ

(k)=Q

Q

Q(k)

ε

(k),

σ

T= [

σ

1

σ

2

σ

6

σ

4

σ

5],

ε

T= [

ε

1

ε

2

ε

6

ε

4

ε

5](8.3.3)

σ

1,

σ

2,

σ

6vary linearly and

σ

4,

σ

5constant through the thickness hof the plate. With

the stress resultants N

N

N,M

M

M,Q

Q

Qsand stiffness coefﬁcients Ai j,Bi j ,Di j ,As

i j for laminates

or sandwiches given in Eqs. (4.2.13) - (4.2.15) or (4.3.8) - (4.3.22), respectively, the

constitutive equation can be formulated in a hypermatrix form, (4.2.16). The stiff-

ness coefﬁcients Ai j ,Bi j,Di j stay unchanged in comparison to the classical theory

and the As

i j are deﬁned in (5.3.4) and can be improved with the help of shear correc-

1Raymond David Mindlin (∗17 September 1906 New York - †22 November 1987 Hanover, New

Hempshire) - mechanician, seminal contributions to many branches of applied mechanics, applied

physics, and engineering sciences

292 8 Modelling and Analysis of Plates

tion factors ks

i j of plates similar to beams (7.3.19) – (7.3.20). The deﬁnition of the

positive rotations

ψ

1,

ψ

2is illustrated in Fig. 8.3. The equilibrium equations (8.2.2)

- (8.2.3) stay unchanged.

Substituting the kinematic relations (5.3.1) into the constitutive equations (5.3.3)

and then these equations into the ﬁve equilibrium equations (8.2.2) one obtains the

governing plate equations for the shear deformation theory in a matrix form as







˜

L11 ˜

L12 ˜

L13 ˜

L14 0

˜

L21 ˜

L22 ˜

L23 ˜

L24 0

˜

L31 ˜

L32 ˜

L33 ˜

L34 ˜

L35

˜

L41 ˜

L42 ˜

L43 ˜

L44 ˜

L45

0 0 ˜

L53 ˜

L54 ˜

L55













u

v

ψ

1

ψ

2

w







=





0

p







(8.3.4)

The differential operators ˜

Li j are given in App. C.2 for unsymmetric angle-ply, sym-

metric angle-ply and symmetric cross-ply laminates. Symmetric laminates leading,

additional to (8.3.4), the uncoupled plate equations

˜

L11 ˜

L12

˜

L21 ˜

L22 u

v=0

0,



˜

L33 ˜

L34 ˜

L35

˜

L43 ˜

L44 ˜

L45

˜

L53 ˜

L54 ˜

L55 





ψ

1

ψ

2

w

=



0

p

(8.3.5)

Equation (8.3.4) can also formulated in a compact matrix form

˜

L

L˜

u

u=˜

p

˜

L

Lis a (5×5)matrix and ˜

u

u,˜

p

pare (5×1)matrices.

The governing plate equations including transverse shear deformations are a set

of three coupled partial equations of second order, i.e. the problem is of sixth order

an for each edge of the plate three boundary conditions must be prescribed. The

most usual boundary conditions are:

•ﬁxed boundary

w=0,

ψ

n=0,

ψ

t=0

•free boundary

Mn=0,Mnt =0,Qn=0

Fig. 8.3 Positive deﬁnition of

rotations

ψ

i

x2,v

x3,w

x1,u

ψ

2

ψ

1

8.3 Shear Deformation Theory 293

•free edge

Mn=0,

ψ

t=0,w=0

•simply supported boundary

a) w=0, Mn=0,

ψ

t=0 or w=0,

∂ψ

n/

∂

n=0,

ψ

t=0 (hard hinged support)

b) w=0, Mn=0, Mnt =0

Case b) is more complicated for analytical or semianalytical solutions. Generally,

boundary conditions require prescribing for each edge one value of each of the

following ﬁve pairs: (uor Nn), (vor Nnt), (

ψ

nor Mn), (

ψ

tor Mnt), (wor Qn).

With

ψ

1=−

∂

w/

∂

x1and

ψ

2=−

∂

w/

∂

x2Eq. (8.3.5) can be reduced to the classical

plate equation.

In the following we restrict our development to plates that are midplane symmet-

ric (Bi j =0), and additional all coupling coefﬁcients (...)16,(...)26,(...)45 are zero.

The constitutive equations are then simpliﬁed to

N1=A11

ε

1+A12

ε

2,N2=A12

ε

1+A22

ε

1,N6=A66

ε

6,

M1=D11

κ

1+D12

κ

2,M2=D12

κ

1+D22

κ

2,M6=D66

κ

6,

Q1=ks

55A55

ε

5,Q2=ks

44A44

ε

4

(8.3.6)

or in a contracted notation

N

N

M

M=A

A

A0

0

0D

D

D

ε

κ

,Q

Q

Qs=A

A

As

ε

s,

N

NT= [N1N2N6],M

M

MT= [M1M2M6],Q

Q

QsT= [Q1Q2],

ε

T= [

ε

1

ε

2

ε

6],

κ

T= [

κ

1

κ

2

κ

6],

ε

sT= [

ε

5

ε

4],

A

A=



A11 A12 0

A12 A22 0

0 0 A66 

,D

D

D=



D11 D12 0

D12 D22 0

0 0 D66 

,A

A

As=ks

55A55 0

0ks

44A44 

(8.3.7)

Substituting the constitutive equations for M1,M2,M6,Q1,Q2into the three equilib-

rium equations (8.2.2) of the moments and transverse force resultants results in the

following set of governing differential equations for a laminated composite plate

subjected to a lateral load p3(x1,x2)and including transverse shear deformation

D11

∂

2

ψ

1

∂

x2

1

+(D12 +D66)

∂

2

ψ

2

∂

x1

∂

x2

+D66

∂

2

ψ

1

∂

x2

2−ks

55A55 

ψ

1+

∂

w

∂

x1=0,

D66

∂

2

ψ

2

∂

x2

1

+(D12 +D66)

∂

2

ψ

1

∂

x1

∂

x2

+D22

∂

2

ψ

2

∂

x2

2−ks

44A44 

ψ

2+

∂

w

∂

x2=0,

ks

55A55 

∂ψ

1

∂

x1

+

∂

2w

∂

x2

1+ks

44A44 

∂ψ

2

∂

x2

+

∂

2w

∂

x2

2+p3(x1,x2) = 0

(8.3.8)

Analogous to the classical plate equations the shear deformation theory can be used

for modelling and analysis of forced vibrations and buckling of laminate plates. In

294 8 Modelling and Analysis of Plates

the general case of forced vibrations the displacements u,v,w, the rotations

ψ

1,

ψ

2

and the transverse load pin Eq. (8.3.4) are functions of x1,x2and t. In-plane loading

is not considered but in-plane displacements, rotary and coupling inertia terms have

to take into account. Therefore, generalized mass densities must be deﬁned

ρ

0=

n

∑

k=1

ρ

(k)x(k)

3−x(k−1)

3=

n

∑

k=1

ρ

(k)h(k),

ρ

1=

n

∑

k=1

ρ

(k)x(k)

3

2−x(k−1)

3

2,

ρ

2=

n

∑

k=1

ρ

(k)x(k)

3

3−x(k−1)

3

3

(8.3.9)

Coupling inertia terms

ρ

1are only contained in unsymmetric plate problems.

If one wishes to determine the natural frequencies of the rectangular plate con-

sidered above, then in (8.3.8) p3(x1,x2)must be set zero but a term −

ρ

0

∂

2w/

∂

t2

must be added on the right hand side. In addition, because

ψ

1and

ψ

2are both in-

dependent variables which are independent of the transverse displacement w, there

will be an oscillatory motion of a line element through the plate thickness which re-

sults in rotary inertia terms

ρ

2

∂

2

ψ

1/

∂

t2and

ρ

2

∂

2

ψ

2/

∂

t2, respectively, on the right

hand side of the ﬁrst two equations of (8.3.8).

The governing equations for the calculation of natural frequencies of specially

orthotropic plates with A45 =0 are

D11

∂

2

ψ

1

∂

x2

1

+ (D12 +D66)

∂

2

ψ

2

∂

x1

∂

x2

+D66

∂

2

ψ

1

∂

x2

2

−ks

55A55 

ψ

1+

∂

w

∂

x1=

ρ

2

∂

2

ψ

1

∂

t2,

D66

∂

2

ψ

2

∂

x2

1

+ (D12 +D66)

∂

2

ψ

1

∂

x1

∂

x2

+D22

∂

2

ψ

2

∂

x2

2

−ks

44A44 

ψ

2+

∂

w

∂

x2=

ρ

2

∂

2

ψ

2

∂

t2,

ks

55A55 

∂ψ

1

∂

x2

+

∂

2w

∂

x2

1+ks

44A44 

∂ψ

2

∂

x2

+

∂

2w

∂

x2

2=

ρ

0

∂

2w

∂

t2,

(8.3.10)

ρ

0=

n

∑

k=1

ρ

(k)(x(k)

3−x(k−1)

3) =

n

∑

k=1

ρ

(k)h(k),

ρ

2=1

3

n

∑

k=1

ρ

(k)(x(k)3

3−x(k−1)3

3)

w,

ψ

1and

ψ

2are functions of x1,x2and t.

In a similar way the governing equations for buckling problems can be derived.

In the matrix equations (8.3.4) and (8.3.5) only the differential operator ˜

L55 is sub-

stituted by

8.3 Shear Deformation Theory 295

˜

L55 −N1

∂

2

∂

x2

1

+2N6

∂

2

∂

x1

∂

x2

+N2

∂

2

∂

x2

2(8.3.11)

For a cross-ply symmetrically laminated plate is with Bi j =0,D16 =0, D26 =0,

A45 =0

D11

∂

2

ψ

1

∂

x2

1

+ (D12 +D66)

∂

2

ψ

2

∂

x1

∂

x2

+D66

∂

2

ψ

1

∂

x2

2−ks

55A55 

ψ

1+

∂

w

∂

x1=0,

D66

∂

2

ψ

2

∂

x2

1

+ (D12 +D66)

∂

2

ψ

1

∂

x1

∂

x2

+D22

∂

2

ψ

2

∂

x2

2−ks

44A44 

ψ

2+

∂

w

∂

x2=0,

ks

55A55 

∂ψ

1

∂

x2

+

∂

2w

∂

x2

1+ks

44A44 

∂ψ

2

∂

x2

+

∂

2w

∂

x2

2

=N1

∂

2w

∂

x2

1

+2N6

∂

2w

∂

x1

∂

x2

+N2

∂

2w

∂

x2

2

(8.3.12)

The variational formulation of laminated plates including shear deformation may

be based for example upon the principle of minimum potential energy for static

problems and the Hamilton’s principle for dynamic problems. Formulating the elas-

tic potential

Π

we have to consider that in the general case of unsymmetric laminate

plates, including shear deformation,

Π

=

Π

(u,v,w,

ψ

1,

ψ

2)is a potential function of

ﬁve independent variables and that the strain energy

Π

ihas a membrane, a bending

and a transverse shearing term, i.e.

Π

i=1

2Z

V

(

σ

1

ε

1+

σ

2

ε

2+

σ

6

ε

6+

σ

5

ε

5+

σ

4

ε

4)dV

=

Π

m

i+

Π

b

i+

Π

s

i

(8.3.13)

with

Π

m

i=1

2Z

V

(N1

ε

1+N2

ε

2+N6

ε

6)dA,

Π

b

i=1

2Z

V

(M1

κ

1+M2

κ

2+M6

κ

6)dA,

Π

s

i=1

2Z

V

(Qs

1

ε

5+Qs

2

ε

4)dA

(8.3.14)

The stress resultants, stiffness and constitutive equations are formulated in Sect. 4.2,

e.g. (4.2.10) - (4.2.17). The elastic potential

Π

is then given by

Π

(u,v,w,

ψ

1,

ψ

2) = 1

2Z

A

(

ε

TA

A

ε

+

κ

TB

B

ε

+

ε

TB

B

κ

+

κ

TD

D

κ

+

ε

sTA

A

As

ε

s)dx1dx2−Z

A

p3wdx1dx2

(8.3.15)

296 8 Modelling and Analysis of Plates

In (8.3.15) the in-plane loads p1,p2are not included and must be added in gen-

eral loading cases. Shear correction coefﬁcients can be developed for plates quite

similar to beams. Approximately one considers a laminate strip of the width ”1”

orthogonal to the x1-direction and independently another laminate strip orthogonal

to the x2-direction and calculates the correction factors ks

55 and ks

44 like in Chap. 7

for beams. Sometimes the shear correction factors were used approximately equal

to homogeneous plates, i.e, ks

44 =ks

45 =ks

55 =ks=5/6.

Mostly we have symmetric laminates and the variational formulation for bending

Mindlin’s plates can be simpliﬁed

Π

(w,

ψ

1,

ψ

2) = 1

2Z

A

(

κ

TD

D

κ

+

ε

sTA

A

As

ε

s)dx1dx2−Z

A

p3wdx1dx2(8.3.16)

If we restricted the Hamilton’s principle to vibration of symmetric plates, the varia-

tional formulation yields

L(w,

ψ

1,

ψ

2) = T(w,

ψ

1,

ψ

2)−

Π

(w,

ψ

1,

ψ

2),

T(w,

ψ

1,

ψ

2) = 1

2Z

A"

ρ

0

∂

w

∂

t2

+

ρ

2

∂ψ

1

∂

t2

+

ρ

2

∂ψ

2

∂

t2#dx1dx2

(8.3.17)

Π

is given by (8.3.16) and

ρ

0,

ρ

2by Eqs. (8.3.10), Tis the kinetic energy.

For a symmetric and specially orthotropic Mindlin’s plate assuming A45 =0 it

follows from (8.3.16) for bending problems that

Π

(w,

ψ

1,

ψ

2) = 1

2Z

A"D11 

∂ψ

1

∂

x12

+2D12 

∂ψ

1

∂

x1

∂ψ

2

∂

x2+D22 

∂ψ

2

∂

x22

+D66 

∂ψ

1

∂

x2

+

∂ψ

2

∂

x12

+ks

55A55 

ψ

1+

∂

w

∂

x12

+ks

44A44 

ψ

2+

∂

w

∂

x22#dx1dx2−Z

A

p3wdx1dx2,

(8.3.18)

δΠ

(w,

ψ

1,

ψ

2) = 0

For natural vibration the variational formulation for that plate is

L(w,

ψ

1,

ψ

2) = T(w,

ψ

1,

ψ

2)−

Π

(w,

ψ

1,

ψ

2),

δ

t2

Z

t1

L(w,

ψ

1,

ψ

2)dt=0 (8.3.19)

To calculate buckling loads the in-plane stress resultants must be, like in the Kirch-

hoff’s plate theory, included into the part

Π

aof

Π

. Consider a plate with a constant

in-plane force N1it follows

8.4 Sandwich Plates 297

Π

(w,

ψ

1,

ψ

2) =

Π

i(w,

ψ

1,

ψ

2)−1

2ZN1

∂

w

∂

x12

dx1dx2(8.3.20)

The case of a more general in-plane loading can be transposed from (8.2.27). The

term

Π

istay unchanged.

8.4 Sandwich Plates

To formulate the governing differential equations or tht variational statement for

sandwich plates we draw the conclusion from the similarity of the elastic behavior

between laminates and sandwiches in the ﬁrst order shear deformation theory that

all results derived above for laminates can be applied to sandwich plates. We restrict

our considerations to symmetric sandwich plates with thin or thick cover sheets.

Like in the beam theory, there are differences in the expressions for the ﬂexural

stiffness D11,D12 ,D22,D66 and the transverse shear stiffness A55,A44 of laminates

and sandwiches (Sects. 4.3.2 and 4.3.3). Furthermore there are essential differences

in the stress distributions. The elastic behavior of sandwiches and the general model

assumptions are considered in detail in Sect. 4.3. The stiffness relations for sand-

wiches with thin and thick skins are also given there:

•Symmetric sandwiches with thin cover sheets (4.3.12) - (4.3.14)

Ai j =2Af

i j =2

n

∑

k=1

Q(k)

i j h(k),

Di j =hcCf

i j =hcn

∑

k=1

Q(k)

i j h(k)¯x(k)

3,¯x(k)

3=1

2x(k)

3+x(k−1)

3

(i j) = (11),(12),(22),(66)

As

i j =hcCs

i j,(i j ) = (44),(55),Cs

44 =Gs

23,Cs

55 =Gs

13

(8.4.1)

hcis the thickness of the core, nis the number of faces layers and G13 ,G23 are the

core shear stiffness moduli. Shear correction factors can be calculated similarly

to the beams approximately with the help of (7.4.2).

298 8 Modelling and Analysis of Plates

•Symmetric sandwiches with thick cover sheets (4.3.16) - (4.3.17) of one lamina

Ai j =ASa

i j =2hfQf

i j or Ai j =ALa

i j =2hfQf

i j +hcQc

i j,i,j=1,2,6

Di j =DSa

i j =1

2Qf

i j hf+hchfhcor

Di j =DLa

i j =1

2hfQf

i j hf+hc2+1

3hf2+1

12 hc3Qc

i j,i,j=1,2,6

(i j) = (11),(12),(22),(66)

As

i j =As

i j

Sa =hcCc

i j,As

i j =As

i j

La =2hfCf

i j +hcCc

i j,(i j ) = (44),(55)

Cf

44 =Gf

23,Cc

44 =Gc

23,Cf

55 =Gf

13,Cc

55 =Gc

13

(8.4.2)

With these stiffness values for the two types of sandwich plates the differential equa-

tions (8.3.8), (8.3.10) or the variational formulations (8.3.18) - (8.3.20) of the theory

of laminate plates including transverse shear deformation can be transposed to sand-

wich plates.

Equation (4.3.22) demonstrated that for sandwich plates with thick faces the stiff-

ness ALa

i j ,DLa

i j ,(i j) = (11),(22),(66)and ALa

i j ,(i j) = (44),(55)should be used. Be-

cause generally Qc

i j ≪Qf

i j usually the simpliﬁed stiffness

ALa

i j ≈ASa

i j =2Af

i j,DLa

i j ≈DSa

i j 1+hf

hc

hc+ (4/3)hf

hc+hf

yield satisfying results in engineering applications. Thus is valid for isotropic-facing

sandwich plates and for sandwich plates having orthotropic composite material fac-

ings (cross-ply laminates).

In Sect. 4.3 generally and in Sect. 7.4 for beams the continuing popularity of

sandwich structures was underlined. Sect. 7.4 also recalled and summarized the

main aspects of modelling and analysis of sandwich structures. Engineering ap-

plications to sandwich beams were discussed in detail. Keeping this in mind, the

derivations to sandwich plates can be restricted here to few conclusions:

•Most sandwich structures can be modelled and analyzed using the shear defor-

mation theory for laminated plates.

•Generally, the stiffness matrices A

A

A,B

B

Band D

D

Dof laminated plates are employed.

•Consider the lower face as lamina 1, the core as lamina 2 and the upper face

as lamina 3 one can include or ignore the effect of the core on the response to

bending and in-plane loads and the effect of transverse shear deformation on the

response of the facings.

•The shear deformation theory of laminated plates can be not only transposed to

sandwich plates for bending, vibration and buckling induced by mechanical loads

but include also other loading, e.g. hydrothermal effects.

With the special sandwich stiffness including or ignore in-plane, bending and trans-

verse shear deformation response all differential equations and variational formula-

tions of Sect. 8.3 stay valid. Some examples for sandwich plates are considered in

Sect. 8.7.

8.5 Hygrothermo-Elastic Effects on Plates 299

8.5 Hygrothermo-Elastic Effects on Plates

Elevated temperature and absorbed moisture can alter signiﬁcantly the structural

response of ﬁbre-reinforced laminated composites. In Sects. 8.2 to 8.4 the structural

response of laminated plates as result of mechanical loading was considered and

thermal or hygrosgopic loadings were neglected.

This section focuses on hygrothermally induced strains, stresses and displace-

ments of thin or moderate thick laminated plates. We assume as in Sect. 7.5 mod-

erate hygrothermal loadings such that the mechanical properties remain approx-

imately unchanged for the temperature and moisture differences considered. Be-

cause the mathematical formulations governing thermal and hygroscopic loadings

are analogous, a uniﬁed derivation is straightforward and will be considered in the

frame of the classical laminate theory and the shear deformation theory.

The following derivations use the basic equations, Sect. 4.2.5, on thermal and

hygroscopic effects in individual laminae and in general laminates. The matrix for-

mulations for force and moment resultants, Eq. (4.2.75), can be written explicitly

as







N1

N2

N6

M1

M2

M6







=







A11 A12 A16 B11 B12 B16

A22 A26 B12 B22 B26

A66 B16 B26 B66

SD11 D12 D16

YD22 D26

MD66













ε

1

ε

2

ε

6

κ

1

κ

2

κ

6







−







Nth

1

Nth

2

Nth

6

Mth

1

Mth

2

Mth

6







−







Nmo

1

Nmo

2

Nmo

6

Mmo

1

Mmo

2

Mmo

6







(8.5.1)

with the known matrix elements, Eq. (4.2.15),

Ai j =

n

∑

k=1

Q(k)

i j x(k)

3−x(k−1)

3,

Bi j =1

2

n

∑

k=1

Q(k)

i j x(k)

3

2−x(k−1)

3

2,

Di j =1

3

n

∑

k=1

Q(k)

i j x(k)

3

3−x(k−1)

3

3

(8.5.2)

The thermal and moisture stress resultants N

N

Nth,N

N

Nmo,M

M

Mth,M

M

Mmo are resultants per unit

temperature or moisture change, Eqs. (4.2.67).

Substituting the hygrothermal constitutive equation (8.5.1) into the equilibrium

equations (8.2.3) and replacing the in-plane strains

ε

iand the curvatures

κ

iby the

displacements u,v,w, Eq. (8.2.5), yield the following matrix differential equation

for the classical laminate theory

L

Lu

u

u=p

p

p−

∂

N

N

N∗

M

M∗(8.5.3)

300 8 Modelling and Analysis of Plates

L

Lu

u

u=p

p

pis identically to Eq. (8.2.7) with Li j given in App. C. N

N

N∗=N

N

Nth +N

N

Nmo,

M

M∗=M

M

Mth +M

M

Mmo are the hygrothermal stress results and

∂

is a special (3 ×6)

differential matrix

∂

=





−

∂

x1−

∂

x2

0 0 0 0

0−

∂

x2−

∂

x1

0 0 0

0 0 0 −

∂

2

∂

x2

1−

∂

2

∂

x2

2−2

∂

2

∂

x1

∂

x2







(8.5.4)

For selected layer stacking Eq. (8.5.3) can be simpliﬁed. The matrix L

L

Land the dif-

ferential operators Li j are summarized for the most important special laminates in

App. C.

Hygrothermal induced buckling can be modelled as

Lu

Lu +

∂

N

N

N∗

M

M∗=N1

∂

2

∂

x2

1

+2N6

∂

2

∂

x1

∂

x2

+N2

∂

2

∂

x2

2u

u

u∗(8.5.5)

with u

u

u∗T= [0 0 w]. Prebuckling displacements and stress resultants are determined

by solving Eq. (8.5.5) with N

N

N≡0

0

0. For the corresponding buckling problem, N1,N2

and N6are taken to be the stress resultant functions corresponding to the prebuckling

state. The buckling loads are found by solving the eigenvalue problem associated

with (8.5.5), i.e. with N

N

N∗=0

0

0 and M

M

M∗=0

0

0.

Because energy methods are useful to obtain approximate analytical solutions

for hygrothermal problems the total potential energy

Π

is formulated. Restricting

to symmetrical problems with A16 =A26 =0 and D16 =D26 =0, i.e to cross-ply

laminates, we have

Π

(u,v,w) = 1

2Z

A(A11 

∂

u

∂

x12

+2A12 

∂

u

∂

x1

∂

v

∂

x2+A22 

∂

v

∂

x22

+A66 

∂

u

∂

x2

+

∂

v

∂

x12

+D11

∂

2w

∂

x2

12

+2D12

∂

2w

∂

x2

1

∂

2w

∂

x2

2

+D22 

∂

2w

∂

x2

22

+4D66 

∂

2w

∂

x1

∂

x22

−2N∗

1

∂

u

∂

x1−2N∗

2

∂

v

∂

x2−2N∗

6

∂

u

∂

x2

+

∂

v

∂

x1

+2M∗

1

∂

2w

∂

x2

1

+2M∗

2

∂

2w

∂

x2

2−4M∗

6

∂

2w

∂

x1

∂

x2

−"N1

∂

2w

∂

x2

12

+2N6

∂

w

∂

x1

∂

w

∂

x2+N2

∂

2w

∂

x2

22#)dA

(8.5.6)

The classical laminate theory which neglect transverse shear deformations can lead

to signiﬁcant errors for moderately thick plates and hygrothermal loadings. Us-

8.5 Hygrothermo-Elastic Effects on Plates 301

ing the shear deformation theory, Sect. 8.3, we can formulate corresponding to Eq.

(8.3.4).

˜

L

L˜

u

u=˜

p

p−˜

∂

N

N

N∗

M

M∗(8.5.7)

The matrix ˜

∂

is identically to Eq. (8.5.4). For hygrothermal induced buckling we

have analogous to Eq. (8.5.5)

˜

L

L˜

u

u+˜

∂

N

N

N∗

M

M∗=N1

∂

2

∂

x2

1

+2N6

∂

2

∂

x1

∂

x2

+N2

∂

2

∂

x2

2u

u

u∗(8.5.8)

For prebuckling analysis the terms involving N1,N2and N6are ignored. Then buck-

ling loads are calculated by substituting the values N1,N2,N6determined for the

prebuckling state into Eq. (8.5.8) dropping now the hygrothermal stress resultants

N

N∗and M

M

M∗. For special laminate stacking the differential operators are summarized

in App. C.

The elastic potential

Π

is now a function of ﬁve independent functions

u,v,w,

ψ

1,

ψ

2. Restricting again to cross-ply laminates the elastic total potential

Π

can be formulated as

Π

=1

2Z

A(A11 

∂

u

∂

x12

+2A12 

∂

u

∂

x1

∂

v

∂

x2+A22 

∂

v

∂

x22

+A66 

∂

u

∂

x2

+

∂

v

∂

x12

+D11 

∂ψ

1

∂

x12

+2D12 

∂ψ

1

∂

x1

∂ψ

2

∂

x2+D22 

∂ψ

2

∂

x22

+D66 

∂ψ

1

∂

x2

+

∂ψ

2

∂

x1

+ks

44A44 dw

dx2

+

ψ

22

+ks

55A55 dw

dx1

+

ψ

12

(8.5.9)

−2N∗

1

∂

u

∂

x1−2N∗

2

∂

v

∂

x2−2N∗

6

∂

u

∂

x2

+

∂

v

∂

x1

−2M∗

1

∂ψ

1

∂

x1−2M∗

2

∂ψ

2

∂

x2−2M∗

6

∂ψ

1

∂

x2

+

∂ψ

2

∂

x1

−"N1

∂

2w

∂

x2

12

+2N6

∂

w

∂

x1

∂

w

∂

x2+N2

∂

2w

∂

x2

22#)dA

Equations (8.5.6) and (8.5.9) are the starting point for solving hygrothermal induced

buckling problems e.g. with the Ritz- or Galerkin approximation or the ﬁnite ele-

ment method. Analytical solutions are in general not possible. As considered above,

the force resultants N1,N2and N6have to be calculated in the prebuckling state,

i.e. for N

N

N≡0

0

0 and the calculation force resultants are substituted into Eqs. (8.5.6) or

(8.5.9), respectively, with N

N

N∗=0

0

0,M

M

M∗=0

0

0 to calculate the buckling loads. If there are

transverse loads p, Eqs. (8.5.3) or (8.5.7) the bending problem follows from (8.5.6)

or (8.5.7) by setting N

N

N=0

0

0 and substitute an additional term

302 8 Modelling and Analysis of Plates

Z

A

pwdA

8.6 Analytical Solutions

The analysis of rectangular plates with selected layer stacking and boundary condi-

tions can be carried out analytically in a similar manner to homogeneous isotropic

and orthotropic plates. The analytical methods of homogeneous isotropic plates, e.g.

the double series solutions of Navier2or the single series solutions of N´adai3-L´evy4

can be applied to laminated plates with special layer stacking and analogous bound-

ary conditions. In Sect. 8.6 possibilities of analytical solutions in the frame of clas-

sical laminate theory and shear deformation theory are demonstrated for bending,

buckling and vibration problems.

8.6.1 Classical Laminate Theory

There are varying degrees of complexity in laminated plate analysis. The least com-

plicated problems are one-dimensional formulations of cylindrical plate bending.

For cylindrical bending both, symmetric and unsymmetric laminates, are handled in

a unique manner assuming all deformations are one-dimensional.

In the case of two-dimensional plate equations the most important degree of sim-

pliﬁcation is for plates being midplane symmetric because of their uncoupling in-

plane and out-of-plane response. The mathematical structure of symmetric angle-ply

plate equations corresponds to homogeneous anisotropic plate equations and that of

symmetric cross-ply plate equations to homogeneous orthotropic plate equations.

To illustrate analytical solutions for rectangular plates in the framework of the clas-

sical laminate theory we restrict our developments to specially orthotropic, i.e. to

symmetric cross-ply plates. For this type of laminated plates the Navier solution

method can be applied to rectangular plates with all four edges simply supported.

The N´adai-L ´evy solution (N ´adai, 1925) method can be applied to rectangular plates

with two opposite edges have any possible kind of boundary conditions. For more

general boundary conditions of special orthotropic plates or other symmetric or un-

symmetric rectangular plates approximate analytical solutions are possible, e.g. us-

ing the Ritz-, the Galerkin- or the Kantorovich methods, Sect. 2.2.3, or numerical

methods are applied, Chap. 11.

2Claude Louis Marie Henri Navier (∗10 February 1785 Dijon - †21 August 1836 Paris) - engineer

and physicist

3´

Arp´ad N´adai (∗3 April 1883 Budapest - †18 July 1963 Pittsburgh) - professor of mechanics,

contributions to the plate theory and theory of plasticity

4Maurice L´evy (∗28 February 1838 Ribeauvill´e - †30 September 1910 Paris) - engineer, total

strain theory

8.6 Analytical Solutions 303

As considered above the simplest problem of plate bending is the so-called cylin-

drical bending for a plate strip i.e. a very long plate in one direction with such a lat-

eral load and edge support in this direction that the plate problem may be reduced to

a one-dimensional problem and a quasi-beam solution can be used. In the following

we demonstrate analytical solutions for various selected examples.

8.6.1.1 Plate Strip

The model ”plate strip” (Fig. 8.4) describes approximately the behavior of a rect-

angular plate with a/b≪1. The plate dimension ain x1-direction is considered

ﬁnite, the other dimension bin x2-direction approximately inﬁnite. The boundary

conditions for the edges x1=0,x1=amay be quite general, but independent of x2

and the lateral load is p3=p3(x1). All derivatives with respect to x2are zero and

the plate equation reduces to a one-dimensional equation. For symmetric laminated

strips Eqs. (8.2.6) and (8.2.9) reduce to

D11w′′′′(x1) = p3(x1),

M1(x1) = −D11w′′(x1),

M2(x1) = −D12w′′(x1),

M6(x1) = −D16w′′(x1)(general case),

M6(x1) = 0 (specially orthotropic case,D16 =0),

Q1(x1) = M′

1(x1) = −D11w′′′(x1),

Q2(x1) = M′

6(x1) = −D16w′′′(x1)(general case),

Q2(x1) = 0 (specially orthotropic case)

(8.6.1)

When one compares the differential equation of the strip with the differential equa-

tion bD11w′′′′(x1) = q(x1)of a beam it can be stated that all solutions of the beam

equation can be used for the strip.

For the normal stresses in the layer kthe equations are

Fig. 8.4 Plate strip

✲

✻

x2

x1

✻ ✻

a

❄❄ ❄❄❄

p3FM

304 8 Modelling and Analysis of Plates

σ

(k)

1(x1,x3) = −Q(k)

11 x3

d2w

dx2

1

,

σ

(k)

2(x1,x3) = −Q(k)

12 x3

d2w

dx2

1

,

σ

(k)

6(x1,x3) = 0

or

σ

(k)

1(x1,x3) = −Q(k)

11

D11

M1(x1)x3,

σ

(k)

2(x1,x3) = −Q(k)

12

D12

M2(x1)x3,

σ

(k)

6(x1,x3) = 0

(8.6.2)

and the transverse shear stresses follow from (5.2.19) to

σ

4(x1,x3) = F61 Q1(x1),

σ

5(x1,x3) = F11 Q1(x1)

and

F

F(x3) = ˜

B

B(x3)D

D

D−1

i.e 

σ

5

σ

4=F11 F62

F61 F22 Q1

0,(8.6.3)

Considering the solutions of the symmetrical laminated plate strip, we have follow-

ing conclusions:

•The solutions for laminate beams and plate strips are very similar, but the calcu-

lation of the strip bending stiffness Di j has to include Poisson’s effects.

•Because of including of Poisson’s effect we have the relation

w(x1)strip <w(x1)beam

and M2(x1)6=0.

•M1(x1)and Q1(x1)of the strip and the beam are identical. If M6(x1) = 0 then

V1≡Q1, i.e there is no special effective Kirchhoff transverse force.

The solutions for plate strips with cylindrical bending can be transposed to

lateral loads p3(x1,x2) = x2p(x1). From w(x1,x2) = x2w(x1)it follows that

x2w′′′′(x1) = x2p(x1)/D11 . The displacement w(x1)and the stress resultants

M1(x1),Q1(x)of the plate strip with the lateral load p(x1)have to be multiplied

by the coordinate x2to get the solution for the lateral load x2p(x1). Note that in

contrast to the case above, here M6=−2D66w′(x1)−D16w′′(x1)in the general case

and M6=−2D66w′(x1)for specially orthotropic strips.

For unsymmetric laminated plate strips the system of three one-dimensional dif-

ferential equations for the displacements u(x1),v(x1)and w(x1)follow with (8.2.6)

as

8.6 Analytical Solutions 305

A11

d2u

dx2

1

+A16

d2v

dx2

1−B11

d3w

dx3

1

=0

A16

d2u

dx2

1

+A66

d2v

dx2

1−B16

d3w

dx3

1

=0

D11

d4w

dx4

1−B11

d3u

dx3

1−B16

d3v

dx3

1

=p3

(8.6.4)

These equations can be uncoupled and analytically solved. The ﬁrst and the second

equation yield

d2u

dx2

1

=B

A

d3w

dx3

1

,d2v

dx2

1

=C

A

d3w

dx3

1

(8.6.5)

with A=A11A66 −A2

16,B=A66B11 −A16B16,C=A11 B16 −A16B11. Differentiating

both equations and substituting the results in the third equation of (8.6.4) we obtain

one differential equation of fourth order in w(x1)

d4w

dx4

1

=A

Dp3,D=D11A−B11B−B16C(8.6.6)

Equation (8.6.6) can be integrated to obtain w(x1)and than follow with (8.6.5)

d3u

dx3

1

=B

Dp3,d3v

dx3

1

=C

Dp3(8.6.7)

For a transverse load p3=p3(x1)we obtain the analytical solutions for the displace-

ments u(x1),v(x1)and w(x1)as

w(x1) = A

DZZZ Z p3dx1dx1dx1dx1+C1

x3

1

6+C2

x2

1

2+C3x1+C4,

u(x1) = B

DZZZp3dx1dx1dx1+B1

x2

1

2+B2x1+B3,

v(x1) = C

DZZZp3dx1dx1dx1+A1

x2

1

2+A2x1+B3

(8.6.8)

With Eqs. (8.6.5) follows A1=B1=C1and we have eight boundary conditions to

calculate 8 unknown constants, e.g. for clamped supports

u(0) = u(a) = v(0) = v(a) = w(b) = w(a) = 0,w′(0) = w′(a) = 0,

Eqs. (8.2.12) yield the one-dimensional equations for the forces and moments resul-

tants

N1=A11

du

dx1

+A16

dv

dx1−B11

d2w

dx2

1

,

N2=A12

du

dx1

+A26

dv

dx1−B12

d2w

dx2

1

,

306 8 Modelling and Analysis of Plates

N6=A16

du

dx1

+A66

dv

dx1−B16

d2w

dx2

1

,

M1=B11

du

dx1

+B16

dv

dx1−D11

d2w

dx2

1

,

M2=B12

du

dx1

+B26

dv

dx1−D12

d2w

dx2

1

,

M6=B16

du

dx1

+B66

dv

dx1−D16

d2w

dx2

1

,(8.6.9)

Q1=B11

d2u

dx2

1

+B16

d2v

dx2

1−D11

d3w

dx3

1

,

Q2=B16

d2u

dx2

1

+B66

d2v

dx2

1−D16

d3w

dx3

1

,

V1=B11

d2u

dx2

1

+B16

d2v

dx2

1−D11

d3w

dx3

1

=Q1,

V2=2B16

d2u

dx2

1

+2B66

d2v

dx2

1−2D16

d3w

dx3

1

=2Q2

The general symmetric case follows with Bi j =0 and for a symmetric cross-ply strip

are Bi j =0 and A16 =0,D16 =0.

Analytical solutions can also be formulated for vibration and buckling of strips

with one-dimensional deformations. The eigen-vibrations of unsymmetrical plate

strips taking account of u(x1,t) = u(x1)ei

ω

t,v(x1,t) = v(x1)ei

ω

t,w(x1,t) = w(x1)ei

ω

t

are mathematically modelled as

A11

d2u

dx2

1

+A16

d2v

dx2

1−B11

d3w

dx3

1−

ρ

h

ω

2w=0,

A16

d2u

dx2

1

+A66

d2v

dx2

1−B16

d3w

dx3

1−

ρ

h

ω

2v=0,

D11

d4w

dx4

1−B11

d3u

dx3

1−B16

d3w

dx3

1−

ρ

h

ω

2w=0

(8.6.10)

u,vand ware now functions of x1and t.

If the in-plane inertia effects are neglected the Eqs. (8.6.5) are valid. Differenti-

ating these equations and substituting the result in the third Eq. (8.6.10) lead to the

vibration equation

d4w

dx1−A

D

ρ

h

ω

2w=0 (8.6.11)

For a symmetrically laminated cross-ply strip we obtain with A/D=1/D11

d4w

dx1−

ρ

h

ω

2

D11

w=0,

ρ

h=

N

∑

k=1

ρ

(k)h(k)(8.6.12)

8.6 Analytical Solutions 307

The analytical solutions correspond to the beam solutions in Sect. 7.6

w(x1) = C1cos

λ

ax1+C2sin

λ

ax1+C3cosh

λ

ax1+C4sinh

λ

ax1,



λ

a4

=

ρ

h

D11

ω

2(symmetric cross-ply strip),



λ

a4

=

ρ

hA

D

ω

2(general unsymmetric strip)

(8.6.13)

For a simply supported strip we have w(0) = w(a) = w′′(0) = w′′(a) = 0 and there-

fore C1=C3=C4=0 and C2sin

λ

aa=0, i.e. with

λ

=n

π

follow

ω

2=n4

π

4

a4

ρ

h

D

A,

ω

=n

π

a2sD

ρ

hA (8.6.14)

Analytical solutions can be calculated for all boundary conditions of the strip.

In analogous manner analytical solutions follow for the buckling behavior of

strips which are subjected to an initial compressive load N1=−N0. The third equa-

tion of (8.6.4) is formulated with p3=0 as

D11

d4w

dx4

1−B11

d3u

dx3

1−B16

d3v

dx3

1−N1

d2w

dx2

1

=0 (8.6.15)

and with Eq. (8.6.5) follows

d4w

dx4

1−A

DN1

d2w

dx2

1

=0,(general case)

d4w

dx4

1−1

D11

N1

d2w

dx2

1

=0,(symmetrical cross-ply case)

(8.6.16)

with N1(x1) = −N0. The buckling equations correspond again to the beam equation

(7.2.35) and can be solved for all boundary conditions of the strip

w(x1) = C1cos

λ

ax1+C2sin

λ

ax1+C3cosh

λ

ax1+C4sinh

λ

ax1,



λ

a2

=A

DN0

(8.6.17)

For a simply supported strip we have with w(0) = w(a) = w′′(0) = w′′(a) = 0

C2sin

λ

=0,

λ

=n

π

A nonzero solution is obtained if

N0=n2

π

2

a2

D

A(8.6.18)

308 8 Modelling and Analysis of Plates

Thus the critical buckling load follows with to

N0cr =

π

2

a2

D

A(8.6.19)

Summarizing the developments of analytical solutions for unsymmetrical laminated

plate strips we have the following conclusions:

•The system of three coupled differential equations for the displacements

u(x1),v(x1)and w(x1)can be uncoupled and reduced to one differential equa-

tion of fourth order for w(x1)and two differential equations of third order for

u(x1)and v(x1), respectively.

•Analytical solutions for bending of unsymmetrical laminated plate strips can be

simple derived for all possible boundary conditions. In the general case all stress

resultants (8.6.4) are not equal to zero. The general symmetric case and symmet-

rical cross-ply strips are included as special solutions.

•The derivations of bending equations can be expanded to buckling and vibration

problems.

•The derivation of analytical solutions for unsymmetrical laminated strips can,

like for the symmetrical case, expanded to lateral loads p3(x1,x2) = x2p(x1).

8.6.1.2 Navier Solution

Figure 8.5 shows a specially orthotropic rectangular plate simply supported at all

edges with arbitrary lateral load p3(x1,x2). In the Navier solution one expands the

deﬂection w(x1,x2)and the applied lateral load p(x1,x2), respectively, into double

inﬁnite Fourier sine series because that series satisﬁes all boundary conditions

p3(x1,x2)

✻

✲

x2

x1

❄

✻

✛ ✲

a

bBoundary conditions:

w(0,x2) = w(a,x2) = w(x1,0) = w(x1,b) = 0

M1(0,x2) = M1(a,x2) = M2(x1,0) = M2(x1,b) = 0

Fig. 8.5 Rectangular plate, all edges are simply supported, specially orthotropic

8.6 Analytical Solutions 309

p3(x1,x2) =

∞

∑

r=1

∞

∑

s=1

prs sin

α

rx1sin

β

sx2,

prs =4

ab

a

Z

0

b

Z

0

p3(x1,x2)sin

α

rx1sin

β

sx2dx1dx2,

w(x1,x2) =

∞

∑

r=1

∞

∑

s=1

wrs sin

α

rx1sin

β

sx2

(8.6.20)

with

α

r=r

π

/a,

β

s=s

π

/b. The coefﬁcients wrs are to be determined such that the

plate equation (Table 8.1) is satisﬁed.

Substituting Eqs. (8.6.20) into the plate equation yields

∞

∑

r=1

∞

∑

s=1

wrs D1

α

4

r+2D3

α

2

r

β

2

s+D2

β

4

ssin

α

rx1sin

β

sx2=

∞

∑

r=1

∞

∑

s=1

prs sin

α

rx1sin

β

sx2

(8.6.21)

and we obtain the coefﬁcients wrs

wrs =prs

D1

α

4

r+2D3

α

2

r

β

2

s+D2

β

4

s

=prs

drs

(8.6.22)

The solution becomes

w(x1,x2) =

∞

∑

r=1

∞

∑

s=1

prs

drs

sin

α

rx1sin

β

sx2(8.6.23)

The load coefﬁcients prs one obtains by integrating (8.6.20) for the given lateral

loading p3(x1,x2). For a uniform distributed load p3(x1,x2) = p=const we obtain,

for instance,

prs =16p

π

2rs ,r,s=1,3,5,... (8.6.24)

From Table 8.1, the equations for the moment resultants are:

M1(x1,x2) = −D11

∂

2w

∂

x2

1−D12

∂

2w

∂

x2

2

=

∞

∑

r=1

∞

∑

s=1

(D11

α

2

r+D12

β

2

s)wrs sin

α

rx1sin

β

sx2,

M2(x1,x2) = −D12

∂

2w

∂

x2

1−D22

∂

2w

∂

x2

2

=

∞

∑

r=1

∞

∑

s=1

(D12

α

2

r+D22

β

2

s)wrs sin

α

rx1sin

β

sx2,

(8.6.25)

310 8 Modelling and Analysis of Plates

M6(x1,x2) = −2D66

∂

2w

∂

x1

∂

x2

=−2D66

∞

∑

r=1

∞

∑

s=1

α

r

β

swrs cos

α

rx1cos

β

sx2,

Q1(x1,x2) =

∂

M1

∂

x1

+

∂

M6

∂

x2

,V1(x1,x2) = Q1+

∂

M6

∂

x2

,

Q2(x1,x2) =

∂

M6

∂

x1

+

∂

M2

∂

x2

,V1(x1,x2) = Q2+

∂

M6

∂

x2

,

and with the stress relation for the klayers

σ

(k)(x1,x2,x3) = Q

Q

Q(k)x3

κ

=−x3





Q(k)

11 Q(k)

12 0

Q(k)

12 Q(k)

22 0

0 0 Q(k)

66









∂

2w/

∂

x2

1

∂

2w/

∂

x2

2

∂

2w/

∂

x1

∂

x2

(8.6.26)

one obtains the solutions for the in-plane stresses

σ

(k)

1,

σ

(k)

2,

σ

(k)

6







σ

(k)

1

σ

(k)

2

σ

(k)

6







=

∞

∑

r=1

∞

∑

s=1

prs

D1

α

4

r+2D3

α

2

r

β

2

s+D2

β

4

s







(Q(k)

11

α

2

r+Q(k)

12

β

2

s)sin

α

rx1sin

β

sx2

(Q(k)

12

α

2

r+Q(k)

22

β

2

s)sin

α

rx1sin

β

sx2

−2Q(k)

66

α

r

β

scos

α

rx1cos

β

sx2







(8.6.27)

With the simpliﬁed formula (5.2.19) follows the transverse shear stresses

σ

(k)

4,

σ

(k)

5

"

σ

(k)

5(x1,x3)

σ

(k)

4(x1,x3)#=F11 x3F62 x3

F61 x3F22 x3

∞

∑

r=1

∞

∑

s=1(D11

α

3

r+D12

β

2

s

α

r)wrs cos

α

rx1sin

β

sx2

(D12

α

2

r

β

s+D22

β

3

s)wrs sin

α

rx1cos

β

sx2(8.6.28)

The Navier solution method can be applied to all simply supported specially or-

thotropic laminated rectangular plates in the same way. For a given lateral load

p3(x1,x2)one can obtain the load coefﬁcients prs by integrating (8.6.20), and by

substituting prs in (8.6.3) follows the wrs. Some conclusions can be drawn from the

application of the Navier solution:

•The solution convergence is rapid for the lateral deﬂection w(x1,x2)and uniform

loaded plates. The convergence decreases for the stress resultants and the stresses

and in general with the concentration of lateral loads in partial regions.

•The solution convergence is more rapid for the stresses

σ

(k)

1in the ﬁbre direction

but is not as rapid in calculating

σ

(k)

2.

The Navier solutions can be also developed for antisymmetric cross-ply laminate

and for symmetric and antisymmetric angle-ply laminates. For these laminates the

plate equations (8.2.6) are not uncoupled and we have to prescribe in-plane and

out-of-plane boundary conditions. It is easy to review that the Naviers double series

solutions Type 1 and Type 2, i.e.

8.6 Analytical Solutions 311

Type 1:

u(x1,x2) =

∞

∑

r=1

∞

∑

s=1

urs cos

α

rx1sin

β

sx2,

v(x1,x2) =

∞

∑

r=1

∞

∑

s=1

vrs sin

α

rx1cos

β

sx2,

w(x1,x2) =

∞

∑

r=1

∞

∑

s=1

wrs sin

α

rx1sin

β

sx2,

Type 2:

u(x1,x2) =

∞

∑

r=1

∞

∑

s=1

urs sin

α

rx1cos

β

sx2,

v(x1,x2) =

∞

∑

r=1

∞

∑

s=1

vrs cos

α

rx1sin

β

sx2,

w(x1,x2) =

∞

∑

r=1

∞

∑

s=1

wrs sin

α

rx1sin

β

sx2,

α

r=

π

r/a,

β

s=

π

s/asatisfy the following alternative boundary conditions for se-

lected laminated plates:

•Simply supported boundary conditions, Type 1

x1=0 and x1=a

w=0,M1=0,v=0,N1=0

x2=0 and x2=b

w=0,M2=0,u=0,N2=0

The Naviers double series Type 1 for u,vand wcan be used only for laminates,

whose stiffness A16,A26 ,B16,B26,D16,D26 are zero, i.e for symmetric or anti-

symmetric cross-ply laminates

•Simple supported boundary conditions, Type 2

x1=0 and x1=a

w=0,M1=0,u=0,N6=0

x2=0 and x2=b

w=0,M2=0,v=0,N6=0

The Navier double series solution Type 2 for u,v,wcan be used only for laminate

stacking sequences with A16,A26,B11 ,B12,B22,B66,D16,D26 equal zero, i.e for

symmetric or antisymmetric angle ply laminates.

The Navier solutions can be used for calculating bending, buckling and vibration.

For buckling the edge shear force N6and, respectively, for vibration the in-plane

inertia terms must be necessarily zero.

312 8 Modelling and Analysis of Plates

8.6.1.3 N´

adai-L´

evy Solution

For computing the bending of specially orthotropic rectangular plates with two op-

posite edges simply supported, a single inﬁnite series method can be used. The two

other opposite edges may have arbitrary boundary conditions (Fig. 8.6). N´adai in-

troduced for isotropic plates the solution of the plate equation in the form

w(x1,x2) = wp(x1) + wh(x1,x2),p3=p3(x1),(8.6.29)

where wp(x1)represents the deﬂection of a plate strip and wh(x1,x2)is the solution

of the homogeneous plate equation ( p3=0). whmust be chosen such that w(x1,x2)

in (8.6.29) satisfy all boundary conditions of the plate. With the solutions for wh,

suggested by L´evy, and wp, suggested by N ´adai,

wh(x1,x2) =

∞

∑

r=1

fr(x2)sin

α

rx1,wp(x1) =

∞

∑

r=1

prsin

α

rx1

D1

α

4

r

(8.6.30)

with

α

r=r

π

/aand

p3(x1) =

∞

∑

r=1

prsin

α

rx1,pr=2

a

Z

0

p3(x1)sin

α

rx1dx1

the boundary conditions for x1=0 and x1=aare satisﬁed.

Substituting (8.6.30) into the plate equation for specially orthotropic plates, Table

8.1, it follow for each term fr(x2)a differential equation of 4th order with constant

coefﬁcients

D2

d4fr(x2)

dx4

2−2D3

α

2

r

d2fr(x2)

dx2

2

+D1

α

4

rfr(x2) = pr(8.6.31)

or

✻

✲

x2

x1

❄

✻

✛ ✲

a

bBoundary conditions:

w(0,x2) = w(a,x2) = 0

M1(0,x2) = M1(a,x2) = 0

For the edges x2=±b/2 may be

arbitrary b.c.

Fig. 8.6 Rectangular specially orthotropic rectangular plate with two opposite edges simply sup-

ported

8.6 Analytical Solutions 313

d4fr(x2)

dx4

2−2D3

α

2

r

D2

d2fr(x2)

dx2

2

+D1

D2

α

4

rfr(x2) = pr

D2

(8.6.32)

The homogeneous differential equation, i.e. pr=0, can be solved with

frh(x2) = Crexp(

λ

r

α

rx2)(8.6.33)

and yields the characteristic equation for the four roots

λ

4

r−2D3

D2

λ

2

r+D1

D2

=0=⇒

λ

2

r=D3

D2±sD3

D22

−D1

D2

(8.6.34)

In the case of isotropic plates it follows with D1=D2=D3=Dthere are repeated

roots ±1.

For specially orthotropic laminated plates the form of frh(x2)depends on the

character of the roots of the algebraic equation of 4th order. There are three different

sets of roots:

1. (D3/D2)2>(D1/D2): In this case (8.6.34) leads to four real and different roots

λ

1/2=±

δ

1,

λ

3/4=±

δ

2,

δ

1,

δ

2>0,

frh(x2) = Arcosh

δ

1

α

rx2+Brsinh

δ

1

α

rx2

+Crcosh

δ

2

α

rx2+Drsinh

δ

2

α

rx2

(8.6.35)

2. (D3/D2)2= (D1/D2): In this case (8.6.34) leads to four real and equal roots

λ

1/2= +

δ

,

λ

3/4=−

δ

,

δ

>0,

frh(x2) = (Ar+Brx2)cosh

δα

rx2+ (Cr+Drx2)sinh

δα

rx2(8.6.36)

3. (D3/D2)2<(D1/D2): In this case the roots are complex

λ

1/2=

δ

1±i

δ

2,

λ

3/4=−

δ

1±i

δ

2,

δ

1,

δ

2>0,

frh(x2) = (Arcos

δ

2

α

rx2+Brsinh

δ

2

α

rx2)cosh

δ

1

α

rx2

+ (Crcos

δ

1

α

rx2+Drsin

δ

1

α

rx2)sinh

δ

1

α

rx2

(8.6.37)

For a given plate for which materials and ﬁbre orientations have been speciﬁed only,

one of the three cases exists. However in the design problem, trying to ﬁnd the best

variant, more than one case may be involved with the consequence of determin-

ing not just four constants Ar,Br,Cr,Dr, but eight or all twelve to calculate which

construction is optimal for the design.

Concerning the particular solution, it is noted that the lateral load may be at most

linear in x2too, i.e p3(x1,x2) = p3(x1)q(x2)with qat most linear in x2. The solution

wpin (8.6.29) is then replaced by

wp(x1,x2) = q(x2)

∞

∑

r=1

prsin

α

rx1

D1

α

4

r

sin

α

rx1(8.6.38)

314 8 Modelling and Analysis of Plates

With the solution w(x1,x2) = wh(x1,x2) + wp(x1,x2)the stress resultants and

stresses can be calculated in the usual way.

The Navier and N´adai-L´evy solution method can be also applied to eigenvalue

problems. We assume, for instance, that the vibration mode shapes of a laminated

plate with specially orthotropic behavior, which is simply supported at all four

edges, is identical to an isotropic plate. We choose

w(x1,x2,t) =

∞

∑

r=1

∞

∑

s=1

wrs sin

α

rx1sin

α

sx2sin

ω

t(8.6.39)

to represent the expec ted harmonic oscillation and to satisfy all boundary cond itions.

Substituting the expression (8.6.39) into (8.2.16) with p3≡0 yields

[D1

α

4

r+2D3

α

2

r

α

2

s+D2

α

4

s−

ρω

2]wrs =0 (8.6.40)

A non-zero value of wrs, i.e. a non-trivial solution, is obtained only if the expression

in the brackets is zero, hence we can ﬁnd the equation for the natural frequencies

ω

2

rs =

π

4

ρ

hD1r

a4+2D3r

a2s

a2+D2s

a4(8.6.41)

The fundamental frequency corresponds to r=s=1 and is given by

ω

2

11 =

π

4

ρ

ha4D1+2D3a

b2+D2a

b4(8.6.42)

Note that the maximum amplitude wrs cannot be determined, only the vibration

mode shapes are given by (8.6.39). In the case of an isotropic plate the natural

frequencies are with D1=D2=D3=D

ω

2

rs =krs

π

2

a2sD

ρ

h,krs =r2+s2a

b2(8.6.43)

If we consider a buckling problem, e.g. a specially orthotropic laminated plate sim-

ply supported at all edges with a biaxial compression N1and N2, it follows from

(8.2.19) that

D1

∂

4w

∂

x4

1

+2D3

∂

4w

∂

x2

1

∂

x2

2

+D2

∂

4w

∂

x4

2

=N1

∂

2w

∂

x2

1

+N2

∂

2w

∂

x2

2

(8.6.44)

The Navier solution method yields with (8.6.39)

π

2wrs[D1r4+2D3r2s2

γ

2+D2s4

γ

4] = −wrs[N1r2+N2s2

γ

2]a2(8.6.45)

with

γ

=a/b. A non-zero solution of the buckling problem (wrs 6=0) leads to

8.6 Analytical Solutions 315

N1r2+N2s2

γ

2=−

π

2

a2[D1r4+2D3r2s2

γ

2+D2s4

γ

4](8.6.46)

We consider the example of uniform compression N1=−Nand N2=−

κ

N, where

the boundary force Nis positive. Equation (8.6.46) yields

N=

π

2(D1r4+2D3r2s2

γ

2+D2s4

γ

4)

a2(r2+

κ

s2

γ

2)

The critical buckling load Ncr corresponds to the lowest value of N. If

κ

=0 we

have the case of uniaxial compression and the buckling equation simpliﬁes to

N=

π

2

a2r2(D1r4+2D3r2s2

γ

2+D2s4

γ

4)

For a given r, the smallest value of Nis obtained for s=1, because sappears only

in the numerator. To determine which rprovides the smallest value Ncr is not simple

and depends on the stiffness D1,D2,D3, the length-to-width ratio

γ

=a/band r.

However, for a given plate it can be easily determined numerically. Summarizing

the discussion of the classical laminate theory applied to laminate plates we can

formulate the following conclusions:

•Specially orthotropic laminate plates can be analyzed with the help of the Navier

solution or the N´adai-L´evy solution of the theory of isotropic Kirchhoff’s plates,

if all or two opposite plate edges are simply supported. These solution methods

can be applied to plate bending, buckling and vibration.

•For more general boundary conditions specially orthotropic plates may be solved

analytically with the help of the variational approximate solutions method of

Rayleigh-Ritz or in a more generalized way based on a variational method of

Kantorovich.

•Plates with extensional-bending couplings should be solved numerically, e.g.

with the help of the ﬁnite element method, Chap. 11. Note that in special cases

antisymmetric cross-ply respectively symmetric and antisymmetric angle-ply

laminates can be analyzed analytically with Navier‘s solution method.

In this section we illustrated detailed analytical solutions for specially orthotropic

laminates which can predict ”exact” values of deﬂections, natural frequencies of vi-

bration and critical buckling loads. But even the ”exact” solutions become approxi-

mate because of the truncation of the inﬁnite series solutions or round-off errors in

the solution of nonlinear algebraic equations, etc. However these solutions help one

to understand, at least qualitatively, the mechanical behavior of laminates. Many

laminates with certain ﬁbre orientations have decreasing values of the coefﬁcients

D16,D26 for bending-torsion coupling and they can be analyzed with the help of the

solution methods for specially orthotropic plates.

316 8 Modelling and Analysis of Plates

8.6.2 Shear Deformation Laminate Theory

The analysis of laminated rectangular plates including transverse shear deforma-

tions is much more complicated than in the frame of classical laminate theory. Also

for plate analysis including shear deformations the at least complicated problem is

cylindrical bending, i.e one-dimensional formulations for plate strips.

Unlike to classical plate strips equations only symmetric and unsymmetric cross-

ply laminates can be handled in a unique manner. In the case of two-dimensional

plate equations we restrict the developments of analytical solutions for bending,

buckling and vibrations analogous to Eqs. (8.3.6) - (8.3.8) to midplane symmetric

cross-ply plates with all Bi j =0 and additional A16 =A26 =D16 =D26 =0,A45 =0.

8.6.2.1 Plate Strip

Consider ﬁrst the cylindrical bending for the plate strip with an inﬁnite length in

the x2-direction and uniformly supported edges x1=0,x1=a, subjected to a load

p3=p(x1). If we restrict the considerations to cross-ply laminated strips the gov-

erning strip equations follow with A16 =A26 =0,B16 =B26 =0, D16 =D26 =0,

A45 =0 and result in a cylindrical deﬂected middle surface with v=0,

ψ

2=0,

u=u(x1),

ψ

1=

ψ

1(x1),w=w(x1)from (8.3.4) as

A11

d2u

dx2

1

+B11

d2

ψ

1

dx2

1

=0,

B11

d2u

dx2

1

+D11

d2

ψ

1

dx2

1−ks

55A55 

ψ

1+dw

dx1=0,

ks

55A55 d

ψ

1

dx1

+d2w

dx2

1+p3(x1) = 0

(8.6.47)

The stress resultants Ni(x)1,Mi(x1),i=1,2,6 and Qj,j=1,2 are with (8.3.2) and

(8.3.6)

N1(x1) = A11

du

dx1

+B11

d

ψ

1

dx1

,

N2(x1) = A12

du

dx1

+B12

d

ψ

1

dx1

,

N6(x1) = 0,

M1(x1) = B11

du

dx1

+D11

d

ψ

1

dx1

,

M2(x1) = B12

du

dx1

+D12

d

ψ

1

dx1

,

M6(x1) = 0,

Q1(x1) = ks

55A55 

ψ

1+dw

dx1,

Q2(x1) = 0

(8.6.48)

8.6 Analytical Solutions 317

The three coupled differential equations for u,wand

ψ

1can be reduced to one un-

coupled differential equation for

ψ

1. The ﬁrst equation yields

d2u

dx2

1

=−B11

A11

d2

ψ

1

dx2

1

,d3u

dx3

1

=−B11

A11

d3

ψ

1

dx3

1

(8.6.49)

Differentiating the second equation and substituting the equation above result in

−B2

11

A11

d3

ψ

1

dx3

1

+D11

d3

ψ

1

dx3

1−ks

55A55 d

ψ

1

dx1

+d2w

dx2

1=0

or with D11 −B2

11

A11 =DR

11

ks

55A55 d

ψ

1

dx1

+d2w

dx2

1=DR

11

d3

ψ

1

dx3

1

(8.6.50)

Substituting Eq. (8.6.50) in the third equation (8.6.47) yield an uncoupled equation

for

ψ

1(x1)

DR

11

d3

ψ

1

dx3

1

=−p3(8.6.51)

The uncoupled equations for u(x1)and w(x1)follow then as

d2u

dx2

1

=−A11

B11

d2

ψ

1

dx2

1

,dw

dx1

=−

ψ

1+DR

11

ks

55A55

d2

ψ

1

dx2

1

(8.6.52)

The three uncoupled equations can be simple integrated

DR

11

ψ

1(x1) = −ZZZp3(x1)dx1dx1dx1+C1

x2

1

2+C2x1+C3,

w(x1) = 1

DR

11 "ZZZ Z p3(x1)dx1dx1dx1dx1+C1

x3

1

6+C2

x2

1

2

+C3x1+C4#−1

ks

55A55 ZZ p3(x1)dx1dx1+C1x1

=wB(x1) + wS(x1),

u(x1) = −A11

B11

1

DR

11 ZZZp3(x1)dx1dx1dx1+C1x1+C5

(8.6.53)

Thus the general analytical solutions for unsymmetric cross-ply laminated strips

are calculated. For symmetrical cross-ply laminated strips the equations yield

DR

11 =D11 and A11u′′(x1) = 0. Restricting to symmetrical cross-ply laminated

strips analytical solutions for buckling or vibrations can be developed analogous to

Timoshenko’s beams or to the classical strip problems.

For a buckling load N1(x1) = −N0follow with p3=0

318 8 Modelling and Analysis of Plates

D11

d2

ψ

1

dx2

1−ks

55A55 

ψ

1+dw

dx1=0,

ks

55A55 d

ψ

1

dx1

+d2w

dx2

1+N0

d2w

dx1

=0

(8.6.54)

The equations can be uncoupled. With

d

ψ

1

dx1

+d2w

dx2

1=D11

ks

55A55

d3

ψ

1

dx3

1

,ks

55A55

d3

ψ

1

dx3

1

=−ks

55A55

d4w

dx4

1−N0

d2w

dx1

one obtains analogous to Eq. (7.3.23)

D11 1−N0

ks

55A55 d4w

dx4

1

+N0

d2w

dx1

=0 (8.6.55)

The general solution for the eigenvalue problem (8.6.55) follows with

w(x1) = Ce

λ

x1(8.6.56)

and the characteristic equation

D11 1−N0

ks

55A55 

λ

4+N0

λ

2=0 or D11

λ

4+k2

λ

2=0 (8.6.57)

with the solutions

λ

1/2=±ik,

λ

3/4=0

as

w(x1) = C1sin kx1+C2coskx1+C3x1+C4(8.6.58)

If we assume, e.g. simply supported edges x1(0) = 0,x1(a), follow with

w(0) = w(a) = w′′(0) = w′′(a) = 0 the free coefﬁcients C2=C3=C4=0

and C1sinka =0. If C16=0 follow with sin ka =0 the solution k=m

π

/a=

α

m

(m=1,2,...)and k2=

α

2

mand thus

N0

D11 1−N0

ks

55A55 =

α

2

m,N0=D11ks

55A55

α

2

m

D11

α

2

m+ks

55A55

The critical buckling load corresponds to the smallest value of N0which is obtained

for m=1

Ncr =D11ks

55A55

π

2

D11

π

2+ks

55A55a2=

π

2D11

a2

1

1+

π

2D11

a2ks

55A55

(8.6.59)

It can be seen that analogous to the Timoshenko’s beam, Sect. 7.3, the including of

shear deformations decreases the buckling loads.

8.6 Analytical Solutions 319

The free vibrations equations of the Timoshenko’s beams were also considered

in Sect. 7.3. For symmetric cross-ply laminated plate strips we obtain comparable

equations

D11

d2

ψ

1

dx2

1−ks

55A55 

ψ

1+dw

dx1=

ρ

2

∂

2

ψ

∂

t2,

ks

55A55 d

ψ

1

dx1

+d2w

dx2

1=

ρ

0

∂

2w

∂

t2

(8.6.60)

ρ

0and

ρ

2were deﬁned as

ρ

0=

n

∑

k=1

ρ

(k)h(k),

ρ

2=1

3

n

∑

k=1

ρ

(k)x(k)

3

3−x(k−1)

3

3

and the terms involving

ρ

0and

ρ

2are the translatory and the rotatory inertia terms.

ψ

1and ware functions of x1and tand thus we have partial derivatives. If we assume

again both strip edges simply supported the analytical solution follow with

w(x1,t) = C1me−i

ω

mtsin m

π

x1

a,w(0,t) = w(a,t) = 0,

ψ

1(x1,t) = C2me−i

ω

mtcos m

π

x1

a,

∂ψ

1(0,t)

∂

x1

=

∂ψ

1(a,t)

∂

x1

=0

(8.6.61)

Substituting these solution functions into the vibration equations (8.6.60) follow

D11

α

2

m+ks

55A55 −

ρ

2

ω

2

mks

55A55

α

m

ks

55A55

α

mks

55A55

α

2

m−

ρ

0

ω

2

mC2m

C1m=0

0

The nontrivial solution of the homogeneous algebraic equation yields the eigenfre-

quencies

ω

m



D11

α

2

m+ks

55A55 −

ρ

2

ω

2

mks

55A55

α

m

ks

55A55

α

mks

55A55

α

2

m−

ρ

0

ω

2

m

=0 (8.6.62)

or

ρ

0

ρ

2

ω

4

m−(D11

ρ

0

α

m+ks

55A55

ρ

0+ks

55A55

ρ

2

α

2

m)2

ω

2

m+D11ks

55A55

α

4

m=0

A

ω

4

m−B

ω

2

m+C=0,

ω

2

m=B

2A±1

2ApB2−4A2C

The general solution for the vibration equations can be formulated for arbitrary

boundary conditions. For harmonic oscillations we write

w(x1,t) = w(x1)ei

ω

t,

ψ

1(x1,t) =

ψ

1(x1)ei

ω

t(8.6.63)

Substituting w(x1,t)and

ψ

1(x1,t)in the coupled partial differential equations

(8.6.60) yield

320 8 Modelling and Analysis of Plates

D11

d2

ψ

1(x1)

dx2

1−ks

55A55 

ψ

1+dw(x1)

dx1+

ρ

2

ω

2

ψ

(x1) = 0,

ks

55A55 d

ψ

1(x1)

dx1

+d2w(x1)

dx2

1+

ρ

0

ω

2w(x1) = 0

(8.6.64)

These both equations can be uncoupled. With

ks

55A55

d

ψ

1(x1)

dx1

=−

ρ

0

ω

2w(x1)−ks

55A55

d2w(x1)

dx1

and

D11

d3

ψ

1(x1)

dx3

1−ks

55A55

d2w(x1)

dx2

1−ks

55A55

d

ψ

1(x1)

dx1−

ρ

2

ω

2d

ψ

1(x1)

dx1

=0

follow

D11

d4w(x1)

dx4

1

+D11

ρ

0

ks

55A55

+

ρ

2

ω

2d2w(x1)

dx2

1−1−

ρ

2

ω

2

ks

55A55 

ρ

0

ω

2w(x1) = 0

or

ad4w(x1)

dx4

1

+bd2w(x1)

dx2

1−cw(x1) = 0 (8.6.65)

The general solution can be derived as

w(x1) = C1sin

λ

1x1+C2cos

λ

2x1+C3sinh

λ

3x1+C4cosh

λ

4x1(8.6.66)

The

λ

iare the roots of the characteristic algebraic equation of (8.6.65). The deriva-

tions above demonstrated that for any boundary conditions an analytical solution

is possible. Unlike to the classical theory we restricted the considerations in the

frame of the shear deformation theory to cross-ply laminated strips. Summarizing

the derivations we can draw the following conclusions:

•Cylindrical bending yields simple analytical solutions for unsymmetrical and

symmetrical cross-ply laminated plate strips.

•Restricting to symmetrical laminated cross-ply plate strips we can obtain ana-

lytical solutions for buckling and vibrations problems, but for general boundary

conditions the analytical solution can be with difﬁculty.

8.6.2.2 Navier Solution

Navier’s double series solution can be used also in the frame of the shear deforma-

tions plate theory. Analogous to Sect. 8.6.1 double series solutions can be obtain for

symmetric and antisymmetric cross-ply and angle-ply laminates with special types

of simply supported boundary conditions. In the interest of brevity the discussion

8.6 Analytical Solutions 321

is limited here to symmetrical laminated cross-ply plates, i.e. specially orthotropic

plates. The in-plane and out-of-plane displacements are then uncoupled.

Rectangular specially orthotropic plates may be simply supported (hard hinged

support) on all four edges.

x1=0,x1=a:w=0,M1=0 respectively

∂ψ

1

∂

x1

=0,

ψ

2=0,

x2=0,x2=b:w=0,M2=0 respectively

∂ψ

2

∂

x2

=0,

ψ

1=0

(8.6.67)

The boundary conditions can be satisﬁed by the following expressions:

w(x1,x2) =

∞

∑

r=1

∞

∑

s=1

wrs sin

α

rx1sin

β

sx2,

ψ

1(x1,x2) =

∞

∑

r=1

∞

∑

s=1

ψ

1rs cos

α

rx1sin

β

sx2,

α

r=r

π

a,

β

s=s

π

b,

ψ

2(x1,x2) =

∞

∑

r=1

∞

∑

s=1

ψ

2rs sin

α

rx1cos

β

sx2

(8.6.68)

The mechanical loading p3(x1,x2)can be also expanded in double Fourier sine se-

ries

p3(x1,x2) =

∞

∑

r=1

∞

∑

s=1

prs sin

α

rx1sin

β

sx2,

prs =4

ab

a

Z

0

b

Z

0

p3(x1,x2)sin

α

rx1sin

β

sx2dx1dx2

(8.6.69)

Now the Navier solution method can be extended to Mindlin’s plates with all edges

simply supported, but the solution is more complex than for Kirchhoff’s plates. Sub-

stituting the expression (8.6.68) and (8.6.69) into the plate differential equations

(8.3.8) gives





L11 L12 L13

L12 L22 L23

L13 L23 L33 





ψ

1rs

ψ

2rs

wrs 

=



0

prs 

(8.6.70)

with

L11 =D11

α

2

r+D66

β

2

s+ks

55A55,L12 = (D12 +D66)

α

r

β

s,L13 =ks

55A55

α

r,

L22 =D66

α

2

r+D22

β

2

s+ks

44A44,L33 =ks

55A55

α

2

r+ks

44A44

β

2

s,L33 =ks

44A44

β

s

(8.6.71)

Solving the Eqs. (8.6.65), one obtains

ψ

1rs =L12L23 −L22 L13

Det(Li j )prs ,

ψ

2rs =L12L13 −L11 L23

Det(Li j)prs,wrs =L11L22 −L2

12

Det(Li j)prs

(8.6.72)

Det(Li j)is the determinant of the matrix in (8.6.65).

322 8 Modelling and Analysis of Plates

If the three kinematic values w(x1,x2),

ψ

1(x1,x2),

ψ

2(x1,x2)are calculated the

curvatures

κ

1,

κ

2and

κ

6may be obtained and the stresses in each lamina follow

from (8.3.3) to





σ

1

σ

2

σ

6



(k)

=x3



Q11 Q12 0

Q12 Q22 0

0 0 Q66 



(k)



κ

1

κ

2

κ

6

,



σ

5

σ

4(k)

=C55 0

0C44 (k)





ψ

1+

∂

w

∂

x1

ψ

2+

∂

w

∂

x2







(8.6.73)

In a analogous manner natural vibrations and buckling loads can be calculated for

rectangular plates with all edges hard hinged supported.

8.6.2.3 N´

adai-L´

evy Solution

The N´adai-L´evy solution method can also be used to develop analytical solutions for

rectangular plates with special layer stacking and boundary conditions, respectively,

but the solution procedure is more complicated than in the frame of classical plate

theory. We do without detailed considerations and recommend approximate analyt-

ical solutions or numerical methods to analyze the behavior of general laminated

rectangular plates including shear deformations and supported by any combination

of clamped, hinged or free edges.

Summarizing the discussion of analytical solutions for plates including trans-

verse shear deformations one can formulate following conclusion

•Analytical solutions for symmetrical and unsymmetrical laminated plates can be

derived for cylindrical bending, buckling and vibration.

•Navier’s double series solutions can be simple derived for specially orthotropic

plates. Navier’s solution method can be also applied to symmetric or antisym-

metric cross-ply and angle-ply laminates, but the solution time needed is rather

high.

•Ritz’s, Galerkin’s or Kantorovich’s methods are suited to analyze general lami-

nated rectangular plates with general boundary conditions.

•Plates with general geometry or with cut outs etc. should be analyzed by numer-

ical methods

8.7 Problems

Exercise 8.1. A plate strip has the width ain x1-direction and is inﬁnitely long in

the x2-direction. The strip is loaded transversely by a uniformly distributed load

p0and simply supported at x1=0,x1=a. Calculate the deﬂection w, the resultant

moments M1,M2,M6and the stresses

σ

1,

σ

2,

σ

6

8.7 Problems 323

1. for a symmetrical four layer plate [0/90/90/0],

2. for a unsymmetrical four layer plate [0/0/90/90]

Solution 8.1. The solutions are presented for both stacking sequences separately.

1. The plate strip is a symmetric cross-ply laminate, i.e. Bi j =0, D16 =D26 =0.

The governing differential equations are

d2M1

dx2

1

=−p0,dM1

dx1

=Q1,

D11

d2w

dx2

1

=−M1,D12

d2w

dx2

1

=−M2,M6=0,

D11

d4w

dx4

1

=p0

The vertical deﬂection w=w(x1)is

w(x1) = 1

D11 p0

x4

1

24 +C1

x3

1

6+C2

x2

1

2+C3x1+C4

Satisfying the boundary conditions

w(0) = 0,w(a) = 0,M1(0) = 0,M1(a) = 0

yield the unknown constants C1-C4

C1=−q0a

2,C2=0,C3=q0a3

24 ,C4=0

and as result the complete solution for the deﬂection w(x1)

w(x1) = p0a4

24D11 x1

a4−2x1

a3+x1

a

The moment resultants follow as

M1(x1) = −p0a2

2x1

a2−x1

a,

M2(x1) = D12

D11

M1(x1) = D12

D11

p0a2

2x1

a2−x1

a,

M6(x1) = 0

The strains and stresses at any point can be determined as follow

κ

1=−d2w

dx1

=p0a2

2x1

a−x1

a2,

κ

2=0,

κ

6=0,

ε

1=x3

κ

1=x3

p0a2

2x1

a−x1

a2,

ε

2=0,

ε

6=0

324 8 Modelling and Analysis of Plates

The stresses in each layer are

00−layers :

σ

1=

σ

′

1=x3

Q11

D11

p0a2

8x1

a−x1

a2,

σ

2=

σ

′

2=x3

Q12

D11

p0a2

8x1

a−x1

a2,

σ

6=0,

900−layers :

σ

2=

σ

′

1=x3

Q11

D11

p0a2

8x1

a−x1

a2,

σ

1=

σ

′

2=x3

Q12

D11

p0a2

8x1

a−x1

a2,

σ

6=0

2. The plate strip is an unsymmetric cross-ply laminate, i.e. A16 =A26 =0,B16 =

B26 =0,D16 =D26 =0. The governing equations follow from Eqs. (8.2.6) and

(8.2.12)

A11

d2u

dx2

1−B11

d3w

dx3

1

=0,A66

d2v

dx2

1

=0,D11

d4w

dx4

1−B11

d3u

dx3

1

=p3,

N1=A11

du

dx1−B11

d2w

dx2

1

,N2=A12

du

dx1−B12

d2w

dx2

1

,N6=0,

M1=B11

du

dx1−D11

d2w

dx2

1

,M2=B12

du

dx1−D12

d2w

dx2

1

,M6=0

The equilibrium equations for the stress resultants are

dN1

dx1

=0,d2M1

dx2

1

=−p3

The displacement u(x1)and w(x1)are coupled. Substitute

A11

d3u

dx3

1

=B11

d4w

dx4

1

into the second differential equation yield

D11 −B2

11

A11 d4w

dx4

=p3

or with

D11 1−B2

11

A11D11 =DR

11,p3=p0⇒d4w

dx4

1

=p0

DR

11

For the displacement u(x1)follows

8.7 Problems 325

d3u

dx3

1

=B11

A11

d4w

dx4

1

or with

A11 1−B2

11

A11D11 =AR

11,d4w

dx4

1

=p0

DR

11 ⇒d3u

dx3

1

=B11

D11

p0

AR

11

These differential equations can be simple integrated

w(x1) = 1

DR

11 q0x4

1

24 +C1

x3

1

6+C2

x2

1

2+C3x1+C4

u(x1) = B11

D11AR

11 q0x3

1

6+C1

x2

1

2+C5x1+C6

Note that with

d2u

dx2

1

=B11

A11

d3w

dx3

1

in both equations there are equal constants C1. The boundary conditions for w

and M1are identically to case 1.

The in-plane boundary conditions are formulated for a ﬁxed-free support, i.e.

u(0) = 0,N1(a) = 0. The boundary conditions lead to the six unknown constants

C1-C6and the solution functions are

u(x1) = B11

D11AR

11

p0a3

12 2x1

a3−x1

a2

w(x1) = 1

DR

11

p0a4

24 x1

a4−2x1

a3+x1

a

The stress and moment resultants follow as

N1(x1) = N6(x1) = 0,

N2(x1) = A12B11

D11AR

11 −B12

DR

11 p0a2

2x1

a2−x1

a,

M1(x1) = −p0a2

2x1

a2−x1

a,

M2(x1) = B2

11 −D12A11

A11 p0a2

2DR

11 x1

a2−x1

a,

M6(x1) = 0

It is interesting to compare the results of case 1. and case 2. The forms of w(x1)

are for the two cases identical except for the magnitude. With

326 8 Modelling and Analysis of Plates

1

DR

11

=1

D11 1−B2

11

A11D11 >1

D11

the deﬂection of the unsymmetric laminate strip will be greater than the deﬂec-

tion of the symmetric laminate. Note that there is no force resultant N1(x1)in the

unsymmetric case but it is very interesting that there is a force resultant N2as a

function of x1, but N2(0) = N2(a) = 0. With

ε

1=du

dx1

=B11

D11AR

11

p0a2

2x1

a2−x1

a,

ε

2=

ε

6=0,

κ

1=−d2

dx2

1

=−p0a2

2DR

11 x1

a2−x1

a,

κ

2=

κ

6=0

follow the strains

ε

1,

ε

2and the stresses

σ

1,

σ

2for the 00and 900-layers in a

similar manner like case 1. With B11 =B12 =0 case 2. yields the symmetrical

case 1.

Exercise 8.2. A plate strip of the width awith a symmetrical cross-ply stacking is

subjected a downward line load q0at x1=a/2. Both edges of the strip are ﬁxed.

Calculate the maximum deﬂection wmax using the shear deformation theory.

Solution 8.2. With (8.6.51) and (8.6.52) follow

DR

11 ≡D11,D11

d3

ψ

1

dx3

1

=q0

δ

x1−1

2a,dw

dx1

=−

ψ

1+D11

ks

55A55

d2

ψ

1

dx2

1

with

δ

x1−1

2a=0x16=a/2

1x1=a/2,Z

δ

x1−1

2adx1=x1−1

2a0

<x1−e>is F¨oppel’s5bracket symbol:

<x1−e>n=0x1<e

(x1−e)nx1>e,

d

dx1

<x1−e>n=n<x1−e>n−1,

Z<x1−e>ndx1=1

1+n<x1−e>n+1+C

With (8.6.53) the analytical solutions for

ψ

1and ware given

5August Otto F¨oppl (∗25 January 1854 Groß-Umstadt – †12 August 1924 Ammerlan) - professor

of engineering mechanics and graphical statics

8.7 Problems 327

D11

ψ

1(x1) = 1

2q0<x1−1

2a>2+C1

x2

1

2+C2x1+C3,

w(x1) = 1

D11 1

6q0<x1−1

2a>3+C1

x3

1

6+C2

x2

1

2+C3x1+C4

−1

ks

55A55

[q0<x1−1

2a>+C1x1]

ψ

1(0) = 0 : C3=0,

w(0) = 0 : C4=0,

ψ

1(a) = 0 : 1

2q01

2a2

+C1a2

2+C2a=0,

w(a) = 0 : 1

6q01

2a3

+C1a3

6+C2

a2

2−D11

ks

55A55 q0

a

2+C1a=0,

C1=q0

2,C2=−q0a

8,

ψ

1(x1) = q0a2

8D11 x1

a2−x1

a,

w(x1) = −q0a3

48D11 3x1

a2−4x1

a3−q0a

2ks

55A55

x1

a=wB+wS,

wmax =q0a3

192D11

+q0a

4ks

55A55

The classical plate theory yields with ks

55A55 →∞the known value

wmax =q0a3

192D11

Exercise 8.3 (Bending of a quadratic sandwich plate). A quadratic sandwich

plate has a symmetric cross-section. The plate properties are a=b=1 m,

hf=0,2875 10−3m, hc=24,71 10−3m, Ef=1,42 105MPa,

ν

f=0,3,

Gf=Ef/2(1+

ν

f),Gc=22 MPa. The cover sheet and the core material are

isotropic, hf≪hc. The transverse uniform distributed load is p=0,05 MPa. The

boundary conditions are hard hinged support for all boundaries. Calculate the max-

imum ﬂexural displacement wmax with the help of a one-term Ritz approximation.

Solution 8.3. The elastic potential

Π

(w,

ψ

1,

ψ

2)of a symmetric and special or-

thotropic Mindlin’s plate is given by (8.3.18). For stiff thin cover sheets and a core

which transmits only transverse shear stresses the bending and shear stiffness for

isotropic face and core materials are (8.4.1)

Di j =hcC(f)

i j =hchQf

i jh(f)¯x(f)

3i=hchf1

2hc+hfQ(f)

i j ,

((i j) = (11),(22),(66),(12)) with

Q11 =Ef

1−(

ν

f)2=Q22,Q12 =

ν

fEf

1−(

ν

f)2,Q66 =Gf

328 8 Modelling and Analysis of Plates

and

AS

i j =hcCc

i j =hcGc,(i j) = (44),(55)

Π

(w,

ψ

1,

ψ

2) = Z

A(1

2hchf(hc+hf)"Q11 

∂ψ

1

∂

x12

+2Q12 

∂ψ

1

∂

x1

∂ψ

2

∂

x2

+Q22 

∂ψ

2

∂

x22

+Q66 

∂ψ

1

∂

x2

+

∂ψ

2

∂

x12#

+kshcGc"

ψ

1+

∂

w

∂

x12

+

ψ

2+

∂

w

∂

x22#)dx1dx2

−Zpwdx1dx2

The one-term approximations

w(x1,x2) = a1sin 

π

x1

asin 

π

x2

a,

ψ

1(x1,x2) = a2cos

π

x1

asin

π

x2

a,

ψ

2(x1,x2) = a3sin

π

x1

acos

π

x2

a

satisfy the boundary conditions. Substituting these approximative functions into

Π

follow ˜

Π

=˜

Π

(a1,a2,a3)and the conditions for a minimum of

Π

, i.e.

∂

˜

Π

/

∂

a1=0,

i=1,2,3 yield the equations for the undetermined coefﬁcients ai

K

Ka

a

a=q

q

with

a

aT= [a1a2a3],q

q

qT= [16p/

π

20 0]

and

K

K=





2hcGc

λ

2hcGc

λ

hcGc

λ

hcGc

λ

hchf¯xf

3(Q11 +Q66)

λ

2+hcGchchf¯xf

3(Q12 +Q66)

λ

2

hcGc

λ

hchf¯xf

3(Q12 +Q66)

λ

2hchf¯xf

3(Q22 +Q66)

λ

2+hcGc







with

λ

=

π

/a. The solution of the system of three linear equations leads to

a1=0,0222,a2=a3=−0,046 and the maximum displacement follows to

wmax =w(x1=a/2,x2=a/2) = a1=2,22 cm.

Exercise 8.4. A simply supported laminate plate [00/900/00]has the following ma-

terial properties: Em=3.4 GPa, Ef=110 GPa,

ν

m=0.35,

ν

f=0.22, vm=0.4,

vf≡

φ

=0.6, Gm=Em/2(1+

ν

m) = 1.2593 GPa, Gf=Ef/2(1+

ν

f) = 45.0820

GPa, h(1)=h(2)=h(3)=5 mm, a=b=1 m.

8.7 Problems 329

1. Formulate the equation for the bending surface for a lateral unit load F=1 N

at x1=

ξ

1,x2=

ξ

2using the classical laminate theory.

2. Formulate the equation for the natural frequencies of the laminate plate using the

classical plate theory and neglecting the rotatory inertia.

Solution 8.4. The solutions for both cases can be presented as follows.

1. The stacking sequence of the layers yields a symmetric cross-ply plate which is

specially orthotropic (Table 8.1) Bi j =0,D16 =D26 =0

D11

∂

4w

∂

x4

1

+2(D12 +2D66)

∂

4w

∂

x2

1

∂

x2

2

+D22

∂

4w

∂

x2

2

=p3(x1,x2)

The boundary conditions are (Fig. 8.5)

w(0,x2) = w(a,x2) = w(x1,0) = w(x1,b) = 0,

M1(0,x2) = M1(a,x2) = M2(x1,0) = M2(x1,b) = 0

The Navier’s double inﬁnite series solution (8.6.21) - (8.6.23) leads to

w(x1,x2) =

∞

∑

r=1

∞

∑

s=1

prs

drs

sin

α

rx1sin

β

sx2

with

drs = [D11

α

4

r+2(D12 +2D66)

α

2

r

β

2

s+D22

β

4

s],

α

r=r

π

a,

β

s=s

π

b,

prs =4F

ab sin

α

r

ξ

1sin

β

s

ξ

2

With (section 2.2.1)

E′

1=Efvf+Emvm=67,36 GPa,E′

2=EfEm

Efvm+Emvf

=8,12 GPa,

G′

12 =GfGf

Gfvm+Gmvf

=3,0217 GPa

ν

′

12 =

ν

fvf+

ν

mvm=0,272,

ν

′

21 =

ν

′

12E′

2/E′

1=0,0328

and (4.1.3)

Q′

11 =E′

1/(1−

ν

′

12

ν

′

21) = 67,97 GPa,

Q′

22 =E′

2/(1−

ν

′

12

ν

′

21) = 8,194 GPa,

Q′

12 =Q′

21 =

ν

′

12Q′

22 =2,229 GPa,

Q′

66 =G′

12 =3,02 GPa

follow (4.2.15) the stiffness

330 8 Modelling and Analysis of Plates

Di j =1

3

∑

k=1

Q(k)

i j (x(k)

3)3−(x(k−1)

3)3,

Q(1)

i j =Q(3)

i j =Q[00]

i j =Q′

i j,Q(2)

i j =Q[900]

i j ,Q(2)

11 =Q(1)

22 ,

Q(2)

22 =Q(1)

11 ,Q(2)

66 =Q(1)

66 ,

x(0)

3=−7,5 mm,x(1)

3=−2,5 mm,

x(2)

3=2,5 mm,x(3)

3=7,5 mm,

D11 =18492 Nm,D22 =2927 Nm,

D12 =D21 =627 Nm,D66 =849 Nm

The equation for the bending surface is

w(x1,x2) = Fa2

π

4

∞

∑

r=1

∞

∑

s=1

sin

α

r

ξ

1sin

β

s

ξ

2

18492r4+4650r2s2+2927s4sin

α

rx1sin

β

sx2

If F=1 N then w(x1,x2)represents the inﬂuence surface, i.e. the deﬂection at

(x1,x2) due to a unit load at (

ξ

1,

ξ

2). This inﬂuence function w(x1,x2;

ξ

1,

ξ

2)is

sometimes called Green’s function of the plate with all boundaries simply sup-

ported. In the more general case of a rectangular plate a6=bthe Green’s function

is

w(x1,x2;

ξ

1,

ξ

2) = F

π

4ab

∞

∑

r=1

∞

∑

s=1

sin

α

r

ξ

1sin

β

s

ξ

2

drs

sin

α

rx1sin

β

sx2

The Green’s function can be used to calculate the bending surfaces of sim-

ply supported rectangular plates with any transverse loading. With the solution

w(x1,x2)we can calculate the stress resultants M1,M2,M6,Q1,Q2and the stresses

σ

1,

σ

2,

σ

6,

σ

5and

σ

4using (8.6.27) and (8.6.28).

2. Using (8.6.41) the equation for the natural frequencies of a simply supported

rectangular plate is

ω

2

rs =

π

4

ρ

hD11

α

4

r+2(D12 +2D66)

α

2

r

β

2

s+D22

β

4

s

with

h=∑

(k)

h(k),

ρ

=1

h∑

(k)

ρ

(k)x(k)

3−x(k−1)

3

The fundamental frequency corresponds to r=s=1 and is given by

ω

2

11 =

π

4

ρ

ha4D11 +2(D12 +2D66)a

b2+D22 a

b4

8.7 Problems 331

For a=b=1 m and the given material properties we ﬁnd the fundamental natural

frequency

ω

11 =1593,5

p

ρ

h

Exercise 8.5. Consider a cylindrically orthotropic circular plate with a midplane

symmetric layer stacking under the conditions of axisymmetric loading and dis-

placements.

1. Develop the differential equations for in-plane loading. Calculate the stress re-

sultants for a solid disk (R,h,Er,E

θ

,

ν

r

θ

) loaded

α

)with a radial boundary force

Nr(R) = −Nro and

β

)with a body force h pr=h

ρω

2rcaused by spinning the

disk about the axis with an angular velocity

ω

.

2. Develop the differential equations for transverse loading under the condition of

the ﬁrst order shear deformation theory. Calculate the stress resultants for a solid

plate (R,h,Er,E

θ

,

ν

r

θ

) loaded by a uniform constant pressure p3(r)≡ −p0and

α

)clamped, respectively,

β

)simply supported at the boundary r=R.

Solution 8.5. With Sect. 2.1.6 we obtain x1=xr,x2=

θ

,x3=z,

σ

1=

σ

r,

σ

2=

σθ

,

σ

6=

σ

r

θ

,

ε

1=

ε

r,

ε

2=

εθ

,

ε

6=

ε

r

θ

. For axisymmetric deformations of circular disks

and plates all stresses, strains and displacements are independent of

θ

, i.e. they are

functions of ralone and

σ

6=0,

ε

6=0.

1. For an in-plane loaded cylindrical orthotropic circular disk under the condition

of axisymmetric deformations the equilibrium, constitutive and geometric

equations are:

•Equilibrium Equations (Fig. 8.7)

With cos(

π

/2−d

θ

/2) = sin(d

θ

/2)≈d

θ

/2 follow

Fig. 8.7 Disc element

(rdrd

θ

)h

x3

0

d

θ

Nr

pr

Nr+dNr

N

θ

d

θ

2

N

θ

d

θ

2

r+dr

h

dr

332 8 Modelling and Analysis of Plates

d(rNr)

dr −N

θ

+prr=0

•Constitutive Equations

Nr=A11

ε

r+A12

εθ

,N

θ

=A12

ε

r+A22

εθ

,Nr

θ

=0

•Geometric Equations

ε

r=du

dr,

εθ

=u

r,

γ

r

θ

=0

These equations forming the following system of three ordinary differential equa-

tions d(rNr)

dr −N

θ

=−p0r,

Nr=A11

du

dr+A12

u

r,N

θ

=A12

du

dr+A22

u

r

involving three unknown quantities Nr,N

θ

and u. Substituting the stress resul-

tants in the equilibrium equations yield one uncoupled differential equation for

u(r)

rd2u

dr2+du

dr−1

r

δ

2u=−rpr

A11

with

δ

2=A22/A11 or

d2u

dr2+1

r

du

dr−

δ

2

r2u=−pr

A11

α

)Radial boundary force

−pr

A11

=0,Nr(R) = −Nr0,R1=0,R2=R

The general solution of the differential equations follow with

u(r) = Cr

λ

,

λ

=±

δ

as

u(r) = C1r+

δ

+C2r−

δ

With R1=0,R2=Rwe obtain C2=0,C1=−Nr0/[(A11

δ

+A12)R

δ

−1]and such

Nr(r) = −Nr0r

R

δ

−1,N

θ

(r) = −Nr0

δ

r

R

δ

−1

Conclusion 8.1. For

δ

=1 we have an isotropic disk with the well-known solu-

tion Nr=N

θ

=−Nr0. For

δ

>0, i.e. the circumferential stiffness exceeds the

radial stiffness, at r=0 we have Nr=N

θ

=0, otherwise for

δ

<0, i.e. the ra-

8.7 Problems 333

dial stiffness exceeds the circumferential, at r=0 we have inﬁnitely high stress

resultants or stresses, respectively.

β

)Body force caused by rotation With pr=

ρω

2rwe obtain the solution of the

inhomogeneous differential equations as

u(r) = C1r

δ

−1

A11

ρω

2

9−

δ

2r3=C1r

δ

+1

A11

ρω

2

δ

2−9r3

For

δ

=1 follow the well-known solution

u(r) = Cr −1−

ν

2

E

ρω

2

8r3

2. With Fig. 8.8 we obtain:

•Equilibrium Equations

d(rMr)

dr−M

θ

−rQr=0,d(rQr)

dr+rp3=0

•Constitutive Equations

Mr=D11

κ

r+D12

κθ

,M

θ

=D12

κ

r+D22

κθ

,

Qr=ks

55A55 

ψ

r+dw

dr

•Geometric Equations

κ

r=d

ψ

r

dr,

κθ

=

ψ

r

r,

γ

rz =

ψ

r+dw

dr

Integrating the second equilibrium equation

Fig. 8.8 Plate element

(rdrd

θ

)h

x3

Qr

Qr+dQr

Mr

M

θ

pz

Mr+dMr

M

θ

334 8 Modelling and Analysis of Plates

Qr(r) = 1

rC1−Zp3(r)rdr

and substituting Mr,M

θ

and Qrin the ﬁrst equilibrium equation yield

rd2

ψ

r

dr2+d

ψ

r

dr−1

r

δ

2

p

ψ

r=1

D11 C1−Zp3rdr,

δ

2

p=D22

D11

The general solution has again the form

ψ

r(r) = C2r

δ

p+C3r−

δ

p+

ψ

p(r)

ψ

p(r)is the particular solution of the inhomogeneous differential equation de-

pending on the form of the loading functions p3(r). The differential equation for

the plate deﬂection w(r)follows with

ksA55

dw

dr=Qr−ksA55

ψ

r,

dw

dr=1

ksA55 C1

1

r−1

rZp3(r)rdr−C2r

δ

p−C3r−

δ

p+

ψ

0,

w(r) = 1

ksA55 C1ln r−Z1

rZp3(r)rdrdr

−C2

r

δ

p+1

δ

p+1−C3

r−

δ

p+1

1−

δ

p

+C4−Z

ψ

pdr

For a constant pressure p3(r) = −p0we obtain

w(r) = 1

ksA55 C1ln r+p0r2

4−C2

1+

δ

p

r1+

δ

p−C3

1−

δ

p

r1−

δ

p

+C4+C1r2

2D11(

δ

2

p−1)+p0r4

8D11(

δ

2

p−9),

ψ

r(r) = C2r

δ

p+C3r−

δ

p−C1r

D11(

δ

2

p−1)−p0r3

2D11(

δ

2

p−9)

This general solution is not valid for

δ

p=1 and

δ

p=3 because the particular

solutions

ψ

pfor theses

δ

p-values include terms coinciding with the fundamental

solutions rand r3. Therefore, the particular solutions must be determined in an-

other form. For

δ

p=1, i.e. for the isotropic case, one can use

ψ

p=Ar ln r+Br3

and for

δ

p=3

ψ

p=Ar +Br3and one obtains the general solutions

δ

p=1

w(r) = 1

ksA55 C1ln r+p0r2

4−1

2C2r2−C3lnr

+C4−C1r4

4D11 lnr−1

2+p0r4

64D11

,

ψ

r(r) = C2r+C3

1

r+C1

rln r

2D11

+p0r3

16D11

8.7 Problems 335

δ

p=3

w(r) = 1

ksA55 C1ln r+p0r2

4−1

4C2r4+C3

1

2r2

+C4−C1

r2

16D11

+p0r4

48D11 ln r−1

4,

ψ

r(r) = C2r3+C3

1

r3−C1

rlnr

8D11

+p0

12D11

r3lnr

The constants C1,C2,C3and C4are determined from the boundary conditions at

the inner and outer plate edge. For solid plates with R1=0,R2=Rthe constants

C1and C3must be zero, otherwise

ψ

rand wtend to inﬁnity at the plate center.

For

δ

6=3 the general solution for solid plates is

w(r) = p0r2

4"1

ks

55A55

+r2

2D11(

δ

2

p−9)#−C2

1+

δ

p

r1+

δ

p+C4,

ψ

r(r) = C2r

δ

p−p0r3

2D11(

δ

2

p−9)

α

)Clamped solid circular plate (

δ

p6=3)

The boundary conditions are

ψ

1(R) = 0,w(R) = 0 yield the constants C2and C4

and the solution as

w(r) = −p0

4ksA55

(R2−r2) + p0

2D11(

δ

2

p−9)"R3−

δ

pr1+

δ

p

1+

δ

p−r4

4+R4(

δ

p−3)

4(1+

δ

p)#

β

)Simply supported solid circular plate (

δ

p6=3)

We take now the boundary conditions w(R) = 0 and Mr(R) = 0 and have the

solution

w(r) = −p0

4ksA55

(R2−r2)2

+p0

2D11(

δ

2

p−9)"(3D11 +D12)R3−

δ

p

(

δ

pD11 +D12)(1+

δ

p)(r1+

δ

p−R1+

δ

p) + 1

4(R4−r4)#

Note that if the transverse shear deformations are neglected we must put

ks

55A55 →∞. In the particular case ks

55A55 →∞and

δ

p=1 follow the well-known

solutions for the classical theory of isotropic plates, i.e.

α

)w(r) = −p0

64D(R2−r2)2=−p0R4

64Dh1−r

Ri2,

β

)w(r) = −p0

64D(R2−r2)5+

ν

1+

ν

R2−r2

=−p0R4

64Dh1−r

Ri25+

ν

1+

ν

−r

R2

336 8 Modelling and Analysis of Plates

Exercise 8.6. A rectangular uniformly loaded symmetric cross-ply plate, Fig. 8.9, is

clamped at the edges x2=±band can be arbitrary supported at the edges x1=±a.

The deﬂection w(x1,x2)may be represented in separated-variables form w(x1,x2) =

wi j(x1,x2) = fi(x1)gj(x2).

1. Formulate one-term approximate solutions using the Vlasov-Kantorovich

method, (2.2.45) - (2.2.47), based on the variation of the potential energy

Π

(w).

2. Demonstrate for the special case of a plate clamped at all edges the extended

Kantorovich method using the Galerkin’s equations.

Solution 8.6. The differential equation and the elastic potential energy can be for-

mulated, Table 8.1 and Eq. (8.2.24),

D1

∂

4w

∂

x4

1

+2D3

∂

4w

∂

x2

1

∂

x2

2

+D2

∂

4w

∂

x4

2

=p0,

with D1=D11,D2=D22,D3=D12 +2D66,pz=p0

Π

(w) = 1

2

a

Z

−a

b

Z

−b"D11 

∂

2w

∂

x2

12

+D22 

∂

2w

∂

x2

22

+2D12

∂

2w

∂

x2

1

∂

2w

∂

x2

2

+4D66 

∂

2w

∂

x1

∂

x22

−2p0#dx1dx2

The one-term approximate solution ˜w(x1,x2) = wij (x1,x2) = fi(x1)gj(x2)has an

unknown function fi(x1)and a priori chosen trial function gj(x2), which satisfy at

least the geometric boundary conditions at x2=±b.

1. The variation

δΠ

of the elastic potential energy

Π

(w)yields

δΠ

(w) = 1

2

a

Z

−a

b

Z

−bD11

∂

2w

∂

x2

1

+D12

∂

2w

∂

x2

2

δ



∂

2w

∂

x2

1

+D12

∂

2w

∂

x2

1

+D22

∂

2w

∂

x2

2

δ



∂

2w

∂

x2

2

Fig. 8.9 Rectangular uni-

formly loaded plate, cross-ply

symmetrically laminated,

clamped at the longitudinal

edges x2=±band arbitrary

boundary conditions at the

edges x1=±a

✻

✲

x1

x2

✻

❄

✻

❄

b

✛ ✲✛ ✲ aa

8.7 Problems 337

+4D66 

∂

2w

∂

x1

∂

x2

δ



∂

2w

∂

x1

∂

x2−p0

δ

wdx1dx2

Substituting ˜w(x1,x2) = fi(x1)gj(x2)one obtains

δΠ

(w) =

a

Z

−aD11A f ′′

i

δ

f′′

i+D12B(f′′

i

δ

fi+fi

δ

f′′

i) + D22C fi

δ

fi

+4D66D f ′

i

δ

f′

i−¯p0

δ

fidx1

where

A=

b

Z

−b

g2

jdx2,B=

b

Z

−b

g′′

jgjdx2,C=

b

Z

−b

(g′′

j)2dx2,D=

b

Z

−b

(g′j)2dx2,¯p0=

b

Z

−b

p0gjdx2

Integrating Bby parts yield

B=g′jgj

b

−b−

b

Z

−b

(g′j)2dx2=−D

because gj(±b) = 0 for plates with clamped or simply supported edges x2=±b.

Now we integrated by parts the term

a

Z

−a

f′′

i

δ

f′′

idx1

of

δΠ

a

Z

−a

f′′

i

δ

f′′

idx1=

a

Z

−a

f′′

i(

δ

fi)′′dx1=f′′

i

δ

f′

i

a

−a−

a

Z

−a

f′′′

i(

δ

fi)′dx1

=f′′

i

δ

f′

i

a

−a−f′′′

i

δ

fi

a

−a+

a

Z

−a

f′′′′

i

δ

fidx1

and the condition

δΠ

=0 yields the ordinary differential equations and the nat-

ural boundary conditions for fi(x1)

D1A f ′′′′

i(x1)−2D3D f ′′

i(x1) + D2C fi(x1) = ¯p0,

at x1=±a:[D11A f ′′

i(x1)−D12D fi(x1)]

δ

f′

i(x1) = 0,

[D11A f ′′′

i(x1)−D12D f ′

i(x1) + 4D66D f ′

i(x1)]

δ

fi(x1) = 0

If a plate edge is clamped, we have f=0,f′=0, if it is simply supported, we

have f=0,f′′ =0 and if it is free, we have

338 8 Modelling and Analysis of Plates

(D11A f ′′

i+D12B fi) = 0,(D11A f ′′′

i+D12B f ′

i−4D66B f ′

i) = 0

The differential equation for f(x1)can be written in the form

f′′′′(x1)−2k2

1f′′(x1) + k4

2f(x1) = kp

with

k2

1=DD3

AD1

,k4

2=CD2

AD1

,kp=¯p0

AD1

The solutions of the differential equation are given in App. E in dependence on

k2

2<k2

1,k2

2=k2

1or k2

2>k2

1in the form

f(x1) =

4

∑

l=1

Cl

Φ

l(x1) + fp

with fp=¯p0/D2C. The solutions can be simpliﬁed if the problem is symmetric or

antisymmetric. The constants Clcan be calculated with the boundary conditions

at x1=±a.

2. In the special case of all plate edges are clamped the corresponding boundary

conditions are

x1=±a:w=0,

∂

w

∂

x1

=0,x2=±b:w=0,

∂

w

∂

x2

=0

The one-term deﬂection approximation is assumed again in the form ˜w(x1,x2) =

wi j(x1,x2) = fi(x1)gj(x2). The Galerkin’s procedure yields

a

Z

−a

b

Z

−bD1

∂

4wi j

∂

x4

1

+2D3

∂

4wi j

∂

x2

1

∂

x2

2

+D2

∂

4wi j

∂

x4

2−p0gjdx2=0

and we obtain

D1



b

Z

−b

g2

jdx2



d4fi

dx4

1

+2D3



b

Z

−b

d2gj

dx2

2

gjdx2



d2fi

dx2

1

+

D2



b

Z

−b

d4gj

dx4

2

gjdx2

fi=

b

Z

−b

p0gjdx2

Two of the integral coefﬁcients must be integrated by parts

8.7 Problems 339

b

Z

−b

d2gj

dx2

2

gjdx2=dgj

dx2

gj

b

−b−

b

Z

−bdgj

dx2dx2,

b

Z

−b

d4gj

dx4

2

gjdx2=d3gj

dx3

2

gj

b

−b−d2gj

dx2

2

dgj

dx2

b

−b+

b

Z

−bd2gj

dx2

22

dx2

The results can be simpliﬁed because for the clamped edges follow

gj(±b) = 0,dgj

dx2±b=0

and we obtain the same differential equation as in 1.

D1



b

Z

−b

g2

jdx2



d4fi

dx4

1−2D3



b

Z

−bdgj

dx22

dx2



d2fi

dx2

1

+

D2



b

Z

−b

d2gj

dx2

2

dx2

fi=

b

Z

−b

p0gjdx2

To improve the one-term approximative plate solution we present in a second

step now fi(x1)a priori and obtain in a similar manner a differential equation for

an unknown function gj(x2)

D2



a

Z

−a

f2

idx1



d4gj

dx4

2−2D3



a

Z

−adfi

dx12

dx2



d2gj

dx2

2

+

D1



a

Z

−ad2fi

dx2

22

dx1

gj=

a

Z

−a

p0fidx1

In this way we have two ordinary differential equations of the iterative solution

procedure which can be written

D1Ag

d4fi

dx4

1−2D3Dg

d2fi

dx2

1

+D2Cgfi=¯p0g,

D2Af

d4gj

dx4

2−2D3Df

d2gj

dx2

2

+D1Cfgj=¯p0f

Both equations can be rearranged in the standard form, App. E

d4fi

dx4

1−2k2

1g

d2fi

dx2

1

+k4

2gfi=kpg,d4gj

dx4

2−2k2

1f

d2gj

dx2

2

+k4

2fgj=kp f

with

340 8 Modelling and Analysis of Plates

k2

1g=DgD3

AgD1

,k4

2g=CgD2

AgD1

,kpg =¯p0g

AgD1

,k2

1f=DfD3

AfD1

,k4

2f=CfD2

AfD1

,kp f =¯p0f

AfD1

The solutions of both equations are summarized in App. E and depend on the

relation between k2

2gand k2

1gor k2

2fand k2

1f, respectively.

The iterations start by choosing the ﬁrst approximation as

w[1]

10 =f1(x1)g1(x2),w[2]

21 =f2(x1)g1(x2),w[3]

22 =f2(x1)g2(x2),...

In the special case under consideration the ﬁrst approximation is

w10(x1,x2) = f1(x1)(x2

2−b2)2

and satisfy the boundary conditions w=0,

∂

w/

∂

x2=0,x2=±b. For a number

of widely used composite material we have k2

2>k2

1. Because the problem is

symmetric we have then the simpliﬁed solution

f1(x1) = C1cosh ax1cos bx1+C2sinhax1sinbx1+kpg

k4

2g

The constants C1,C2can be calculated with

f1(±a) = 0,df1

dx1±a=0

and w[0]

10 (x1,x2)is determined. Now one can start the next step

w[1]

11 (x1,x2) = f1(x1)g1(x2)

with the function f1(x1)as the a priori trial function. The iteration steps can be

repeated until the convergence is satisfying. In the most engineering applications

w[0]

11 (x1,x2) = f1(x1)g1(x2)

can be used as satisfying closed analytical solution, i.e. w[1]

11 (x1,x2)is suitable for

engineering analysis of deﬂection and stresses in a clamped rectangular special

orthotropic plate with uniform lateral load and different aspect ratios.

References

N´adai A (1925) Die elastischen Platten: die Grundlagen und Verfahren zur Berech-

nung ihrer Form ¨ander ungen und Spannungen, sowie die Anwendu ngen der Theo-

rie der ebenen zweidimensionalen elastischen Systeme auf praktische Aufgaben.

Springer, Berlin

Chapter 9

Modelling and Analysis of Circular Cylindrical

Shells

In the previous Chaps. 7 and 8 we have considered beams and plates, i.e. one- and

two-dimensional structural elements with straight axes and plane reference surfaces.

Thin-walled laminated or sandwich shells can be also modelled as two-dimensional

structural elements but with single or double curved reference surfaces. To cover

shells of general shape a special book is necessary, because a general treatment

of shells of any geometry demands a detailed application of differential geometry

relations.

To give a brief insight into the modelling of shells only the simplest shell ge-

ometry will be selected and the following considerations are restricted to circular

cylindrical shells. The modelling and analysis of circular cylindrical shells fabri-

cated from ﬁbre composite material, i.e. its structural theory, depends on the ra-

dius/thickness ratio R/h. For thin-walled shells, i.e. for R/h≫1 (R/h>10), either

the classical or the ﬁrst order shear deformation shell theory is capable of accu-

rately predicting the shell behavior. For thick-walled shells, say R/h<10, a three-

dimensional modelling must be used.

Each single lamina of a ﬁlamentary composite material behaves again macro-

scopically as if it were a homogeneous orthotropic material. If the material axes of

all laminae are lined up with the shell-surface principal coordinates, i.e., the axial

and circumferential directions, the shell is said to be special orthotropic or circum-

ferential cross-ply circular cylindrical shell. Since the often used cylindrical shells

with closely spaced ring and/or stringer stiffeners also can be approximated by con-

sidering them to be specially orthotropic, a greater number of analysis have been

carried out for such shell type. If the material-symmetry axes are not lined up with

the shell principal axes, the shell is said to be anisotropic, but since there is no struc-

tural advantage for shells constructed in this way it has been not often subjected to

analysis.

In Chap. 9 there are only a short summarizing section on sandwich shells and

no special section considering hygrothermo-elastic effects. Both problems can be

simple retransmitted from the corresponding sections in Chaps. 7 and 8. Also a spe-

cial discussion of analytical solution methods will be neglected, because no general

shell problems are considered.

341

H. Altenbach et al., Mechanics of Composite Structural

Elements, https://doi.org/10.1007/978-981-10-8935-0_9

342 9 Modelling and Analysis of Circular Cylindrical Shells

9.1 Introduction

Chapter 9 gives a short introduction to the theory of circular cylindrical shells in the

frame of the classical shell theory and the shell theory including transverse shear

deformations. Figure 9.1 shows a laminated circular cylindrical shell with general

layer stacking, the global coordinates x1=x,x2=s=R

ϕ

,x3=z, and the principal

material coordinates 1 =x′

1,2=x′

2. In the theory of circular cylindrical shells the

most complex problem is the modelling and analysis of laminated shells with an

arbitrary stacking of the layers and arbitrary loading. The at least complex problem

is a mid-plane symmetric cross-ply laminated shell with axially symmetric loads

using the classical shell theory. The mathematically modelling leads in this case to

a

bc

x3=z

x2=R

ϕ

x1=x

u2

u3

u1

h

R

2=x′

2

1=x′

1

x1

x2

θ

z

h

2

h

2z(2)

z(n−1)

z(1)

z(0)=−h

2

z(n)=h

2

Fig. 9.1 Circular cylindrical shell. aGeometry, global coordinates x1=x,x2=s=R

ϕ

,bshell

middle surface, principal material coordinates x′

1=1,x′

2=2, ﬁbre angle

θ

,claminate structure, n

layers, layer coordinates z(k), layer thickness h(k)=z(k)−z(k−1)

9.2 Classical Shell Theory 343

an ordinary differential equation. This type of stacking and loading will be primary

considered in Chap. 9, because analytical solutions can be derived. Generally as-

sumed is that each layer having a constant angle of wrap, constant volume ratio of

ﬁbre to resin, and the ﬁbre and resin are both isotropic and homogeneous within

themselves. The ply material axes, Fig. 9.1 b, will be rotated away from the global

axes by an angle

θ

, positive in the counterclockwise direction.

9.2 Classical Shell Theory

The following hypotheses are the basis to derivative the equations of the classical

shell theory:

•Displacements are small compared to the shell thickness, all strain-displacement

relations may be assumed to be linear.

•The Kirchhoff hypothesis is applicable, i.e. line elements normal to the middle

surface before deformation remain straight, normal to the deformed middle sur-

face, and unchanged in length after deformation.

•All components of translational inertia are included in modelling vibration prob-

lems, but all components of rotatory inertia are neglected.

•The ratio of the shell thickness h to the radius R of the middle surface is small as

compared with unity and Love’s ﬁrst-approximation shell theory is used which

deﬁne a thin or classical shell theory: h/R≪1 and all terms 1 +(z/R)≈1. It can

be shown that this relationship is consistent with the neglect of transverse shear

deformation and transverse normal stress.

In addition we assume that each individual layer is considered to behave macro-

scopically as a homogeneous, anisotropic, linear-elastic material, that all layers are

assumed to be bonded together with a perfect bond and that each layer may be of

arbitrary thickness and may be arranged either symmetrically or unsymmetrically

with respect to the middle surface.

9.2.1 General Case

The governing differential equations are formulated in terms of the three middle-

surface displacement components (u1≡ux,u2≡us,u3≡uz)

u1(x1,x2,0) = u(x,s),u2(x1,x2,0) = v(x,s),u3(x1,x2,0) = w(x,s)(9.2.1)

The strain displacement relations for a circular, cylindrical shell of any material,

neglecting the effects of transverse shear deformation and using Love’s ﬁrst approx-

imation are given by

344 9 Modelling and Analysis of Circular Cylindrical Shells

ε

x=

∂

u

∂

x,

ε

s=

∂

v

∂

s+w

R,

ε

xs =

∂

u

∂

s+

∂

v

∂

x,

κ

x=−

∂

2w

∂

x2,

κ

s=−

∂

2w

∂

s2+1

R

∂

v

∂

s,

κ

xs =−2

∂

2w

∂

x

∂

s+1

R

∂

v

∂

x

(9.2.2)

The total strains at a arbitrary distance zof the middle surface are

ε

x=

ε

x+

κ

xz,

ε

s=

ε

s+

κ

sz,

ε

xs =

ε

xs +

κ

xsz

or

ε

j=

ε

j+

κ

jz,j= (1,2,6)≡(x,s,xs)(9.2.3)

Each individual layer is assumed to be in a state of generalized plane stress, the

Hooke’s law yields

σ

(k)

i=Q(k)

i j

ε

j,i,j= (1,2,6)(9.2.4)

and in the general anisotropic case the Qi j matrix is full populated (Table 4.2).

Using again the Love’s ﬁrst approximation 1 + (z/R)≈1), i.e. neglecting the

difference in the areas above and below the middle surface z=0, the force and

moment resultants, Fig. 9.2, are deﬁned analogous to plates

Ni=

h/2

Z

−h/2

σ

idz,Mi=

h/2

Z

−h/2

σ

izdz,i= (1,2,6)≡(x,s,xs)(9.2.5)

Putting Eq. (9.2.4) into (9.2.5) yields the constitutive equations in the known form

N

N

M

M=A

A

A B

B

B D

D

D

ε

κ

(9.2.6)

with

Ns+

∂

Ns

∂

sds

Qs+

∂

Qs

∂

sds

Nsx +

∂

Nsx

∂

xdx

Nsx +

∂

Nsx

∂

xdx

Qx+

∂

Qx

∂

xdx

Nx+

∂

Nx

∂

xdx

ps

pz

px

Nx

Qx

Nxs

QsNs

Nsx

ds=Rd

ϕ

dx

Msx +

∂

Msx

∂

sds

Ms+

∂

Ms

∂

xdx

Mx+

∂

Mx

∂

xdx

Mxs +

∂

Mxs

∂

sds

Mxs

Mx

Ms

Msx

dx

ds=Rd

ϕ

Fig. 9.2 Positive directions for stress resultants

9.2 Classical Shell Theory 345

(Ai j,Bi j ,Di j ) =

h/2

Z

−h/2

(1,z,z2)Qi jdz

i.e. for nlaminate layers

Ai j =

n

∑

k=1

Qi j z(k)−z(k−1),

Bi j =1

2

n

∑

k=1

Qi j z(k)2−z(k−1)2,

Di j =1

3

n

∑

k=1

Qi j z(k)3−z(k−1)3

N

NT= [NxNsNxs ],M

M

MT= [MxMsMxs],

ε

T= [

ε

x

ε

s

ε

xs],

κ

T= [

κ

x

κ

s

κ

xs](9.2.7)

The equilibrium equations follow with Fig. 9.2 as

∂

Nx

∂

x+

∂

Nxs

∂

s+px=0,

∂

Mx

∂

x+

∂

Mxs

∂

s−Qx=0,

∂

Nxs

∂

x+

∂

Ns

∂

s+Qs

R+ps=0,

∂

Mxs

∂

x+

∂

Ms

∂

s−Qs=0,

∂

Qx

∂

x+

∂

Qs

∂

s−Ns

R+pz=0

(9.2.8)

The moment equations (9.2.8) can be used to eliminate the transverse shear resul-

tants and one obtains

∂

Nx

∂

x+

∂

Nxs

∂

s+px=0,

∂

Nxs

∂

x+

∂

Ns

∂

s+1

R

∂

Ms

∂

s+

∂

Mxs

∂

x+ps=0,

∂

2Mx

∂

x2+2

∂

2Mxs

∂

x

∂

s+

∂

2Ms

∂

s2−Ns

R+pz=0

(9.2.9)

Substituting Eqs. (9.2.6) into (9.2.9) yields a set of three coupled partial differential

equations for the three displacements u,v,w, which can be written in matrix form





L11 L12 L13

L21 L22 L23

L31 L32 L33 





u

v

w

=−



px

ps

pz

(9.2.10)

The linear differential operators Li j are deﬁned in App. D. For symmetrically ar-

ranged layers the differential operators can be simpliﬁed, but the matrix (9.2.10)

stay full populated (App. D).

If we consider natural vibrations of laminated circular cylindrical shells in

Eqs. (9.2.9) and (9.2.10) the distributed loads px,ps,pzare taken zero, i.e.

346 9 Modelling and Analysis of Circular Cylindrical Shells

px=ps=pz=0, but all components of translatory inertia must be included.

Without detailed derivation on obtains

∂

Nx

∂

x+

∂

Nxs

∂

s=

ρ

0

∂

2u

∂

t2,

∂

Nxs

∂

x+

∂

Ns

∂

s+1

R

∂

Ms

∂

s+

∂

Mxs

∂

x=

ρ

0

∂

2v

∂

t2,

∂

2Mx

∂

x2+2

∂

2Mxs

∂

x

∂

s+

∂

2Ms

∂

s2−Ns

R=

ρ

0

∂

2w

∂

t2

(9.2.11)

and Eq. (9.2.10) changes to





L11 L12 L13

L21 L22 L23

L31 L32 L33 





u

v

w

=

ρ

0

∂

2w

∂

t2



u

v

w

(9.2.12)

with

ρ

0=

n

∑

k=1

z(k)

Z

z(k−1)

ρ

(k)

0dz=

n

∑

k=1

ρ

(k)

0h(k)

The stress resultants and the displacement are now functions of x,sand t.

ρ

(k)

0is the

mass density of the kth layer,

ρ

0the mass inertia with respect to the middle surface.

9.2.2 Specially Orthotropic Circular Cylindrical Shells Subjected

by Axial Symmetric Loads

Now we consider cross-ply laminated circular cylindrical shells. The laminate stack-

ing may be not middle-surface symmetric, but the ﬁber angles are

θ

=00or

θ

=900

and the principal material axes 1′−2′−3 coincide with the structural axes x,s,z, i.e.

the stiffness A16 =A26 =0,D16 =D26 =0. In the case of axial symmetry loading

and deformations there are both, all derivations

∂

/

∂

sand v,Nxs ,Mxs zero. For the

loads per unit of the surface area are the following conditions valid

px=0,ps=0,pz=pz(x)

The equilibrium equations (9.2.8) reduce to

dNx

dx=0,dQx

dx−Ns

R+pz=0,dMx

dx−Qx=0

or eliminating Qx

dNx

dx=0,d2Mx

dx2−Ns

R=−pz(9.2.13)

9.2 Classical Shell Theory 347

The strain-displacement relations follow from (9.2.2)

ε

x=du

dx,

ε

s=w

R,

κ

x=−d2w

dx2,

ε

xs =0,

κ

s=

κ

xs =0 (9.2.14)

and the stresses from (9.2.4) and (9.2.14) with Q16 =Q26 =Q66 =0

σ

(k)

x=Q(k)

11 (

ε

x+z

κ

x) + Q(k)

12

ε

s=Q(k)

11 du

dx−zd2w

dx2+Q(k)

12

w

R,

σ

(k)

s=Q(k)

12 (

ε

x+z

κ

x) + Q(k)

22

ε

s=Q(k)

12 du

dx−zd2w

dx2+Q(k)

22

w

R,

Q(k)

11 =E(k)

x

1−

ν

(k)

xs

ν

(k)

sx

,Q(k)

22 =E(k)

s

1−

ν

(k)

xs

ν

(k)

sx

,Q(k)

12 =

ν

(k)

sx E(k)

x

1−

ν

(k)

xs

ν

(k)

sx

,

ν

(k)

xs

E(k)

x

=

ν

(k)

sx

E(k)

s

(9.2.15)

The constitutive equations (9.2.6) can be written as follow

Nx=A11

ε

x+A12

ε

s+B11

κ

x,

Ns=A12

ε

x+A22

ε

s+B12

κ

x,

Mx=B11

ε

x+B12

ε

s+D11

κ

x,

Ms=B12

ε

x+B22

ε

s+D12

κ

x,

(9.2.16)

with

ε

x,

ε

sand

κ

xfrom Eq. (9.2.14).

Putting (9.2.12) and (9.2.14) in the equilibrium equations (9.2.13) one obtains

after a rearrangement

A11

d2u

dx2+A12

R

dw

dx−B11

d3w

dx3=0,

A11D11 −B2

11

A11 d4w

dx4+2

RA12B11

A11 −B12d2w

dx2

+1

R2A11A22 −A2

11

A11 w=pz−A12

A11

Nx

R,

and with

DR=A11D11 −B2

11

A11

,4

λ

4=1

DRR2

A11A22 −A2

11

A11

(9.2.17)

can ﬁnally be written

d4w

dx4+2

RDRA12B11

A11 −B12d2w

dx2+4

λ

4w=1

DRpz−A12

A11

Nx

R(9.2.18)

This is a ordinary differential equation of fourth order with constant coefﬁcients and

can be solved by standard methods .

348 9 Modelling and Analysis of Circular Cylindrical Shells

For the most important case of a symmetrical layer stacking Eq. (9.2.18) can be

reduced with B11 =B12 =0,DR=D11 as

d4w

dx4+4

λ

4w=1

D11 pz−A12

A11

Nx

R,4

λ

4=1

D11R2

A11A22 −A2

11

A11

(9.2.19)

The inhomogeneous linear differential equation of fourth order has constant coefﬁ-

cient and can be analytically solved (App. E)

w(x) = wh(x) + wp(x)

The homogeneous solution wh(x) = Ce

α

xyields the characteristic equation

α

4+4

λ

4=0

with the conjugate complex roots

α

1−4=±

λ

(1±i),i=√−1 (9.2.20)

and with e±

λ

x=cosh

λ

x±sinh

λ

x, e±i

λ

x=cos

λ

x±isin

λ

xone obtains the solution

of the homogeneous differential equation as

wh(x) = C1cosh

λ

xcos

λ

x+C2cosh

λ

xsin

λ

x

+C3sinh

λ

xcos

λ

x+C4sinh

λ

xsin

λ

x(9.2.21)

or

wh(x) = e−

λ

x(C1cos

λ

x+C2sin

λ

x) + e

λ

x(C3cos

λ

x+C4sin

λ

x)(9.2.22)

The particular solution wp(x)of the inhomogeneous equation depends on the load-

ing term.

In solving (9.2.19), another solution form may be utilized, the so-called bending-

layer solution. Note the Eqs. (9.2.16) for the symmetrical case, i.e.

Mx=−D11

d2w

dx2,Qx=dMx

dx=−D11

d3w

dx3

the solution can be written as:

w(x) = M0

2

λ

2D11

e−

λ

x(sin

λ

x−cos

λ

x)−Q0

2

λ

3D11

e−

λ

xcos

λ

x

+ML

2

λ

2D11

e−

λ

(L−x)(sin

λ

(L−x)−cos

λ

(L−x))

+QL

2

λ

3D11

e−

λ

(L−x)cos

λ

(L−x) + wp(x)

(9.2.23)

Instead of the general constants Ci,i=1,2,3,4 the resultant stress moments M0,ML

and resultant stress forces Q0,QLat x=0 respectively x=Lare used as integration

constants. To determine the wp(x)solution one have to consider Nxin Eq. (9.2.19)

9.2 Classical Shell Theory 349

as a constant value following by boundary condition and p(x)have to be restricted

to cases where d4p(x)/dx4=0, what is almost true from view point of practical

applications. It can be easy seen that

wp(x) = 1

4

λ

2D11 p(x)−A12

A11

Nx

R(9.2.24)

is a solution of the inhomogeneous differential equation (9.2.19).

The advantage of the solution from (9.2.23) is easily seen. The trigonomet-

ric terms oscillate between ±1 and are multiplied by exponential terms with a

negative exponent which yields to an exponential decay. If we set

λ

x=1.5

π

or

λ

(L−x) = 1.5

π

then is e−1.5

π

≈0.009, i.e. the inﬂuence of the boundary

values M0,Q0or ML,QLis strong damped to <1%. With 0 ≤x≤1.5

π

/

λ

or

0≤L−x≤1.5

π

/

λ

bending boundary layers are deﬁned which depend on the shell

stiffness.

The important point is that at each end of the shell a characteristic length LB

can be calculated and the M0- and Q0-terms approach zero at the distance x>LB

from x=0 while the ML- and QL-terms approach zero at the same distance LB

from x=L. In the boundary layer region bending stresses induced from M0,Q0a

ML,QLare superimposed to membrane stresses induced from pz. Looking at a long

shell, Fig. 9.3, with L>LBin the region A-B only M0,Q0and wpare non-zero,

in the region C-D only ML,QLand wpand in the region B-C only the particular

solution wpis nonzero, i.e. in this region only a membrane solution exists. With the

calculated w(x)the ﬁrst differential equation (9.2.17) with B11 =0 can be solved

and yields the displacement function u(x). It should be noted that only some terms

of u(x)decay away from the boundary edges.

In the case of axially symmetric loading and deformation the bending stresses in

each lamina are given by



σ

x

σ

s(k)

=Q11 Q12

Q12 Q22 (k)

ε

x

ε

s+zQ11 Q12

Q12 Q22 (k)

κ

x

0(9.2.25)

ABCD

LBLB

L

Fig. 9.3 Long circular cylindrical shell: Bending boundary regions (A-B) and (C-D), membrane

region (B-C)

350 9 Modelling and Analysis of Circular Cylindrical Shells

The transverse shear stress

σ

xz follows analogous to the classical beam equations

with (7.2.31).

Summarizing the results of the classical shell equations one can draw the follow-

ing conclusions:

•The most general case of laminated circular cylindrical shells is that of arbitrarily

laminated anisotropic layers, i.e. angle-ply layers arbitrarily arranged. The anal-

ysis of these shells is based on approximately analytical methods using Ritz-,

Galerkin- or Kantorovich method and numerical methods , e.g. FEM.

•Cross-ply laminated shells, i.e shells with orthotropic layers aligned either ax-

ially or circumferentially and arranged symmetrically with respect to the shell

middle surface have governing shell equations which are the same as those for

a single-layer specially orthotropic shell. For axis symmetrical loading the shell

equations reduce in the static case to ordinary differential equations of the x-

coordinate and can be solved analytically. If the orthotropic layers are arranged

to an unsymmetric laminated cross-ply shell then bending-stretching, coupling is

induced and the governing equations are more complex.

•When circular cylindrical shells are laminated of more than one isotropic layer

with each layer having different elastic properties and thickness and the lay-

ers are arranged symmetrically with respect to the middle surface, the govern-

ing equations are identical to those of single layer isotropic shells. However, if

the isotropic layers are arranged unsymmetrically to the middle surface, there is

a coupling between in-surface, i.e stretching and shear, and out-of-surface, i.e

bending and twisting, effects.

•Additional to the Kirchhoff’s hypotheses all equations of the classical shell the-

ory assumed Love’s ﬁrst approximation, i.e. the ratio h/Ris so small compared

to 1 that the difference in the areas of shell wall element above and below the

middle surface can be neglected.

9.2.3 Membrane and Semi-Membrane Theories

Thin-walled singe layer shells of revolution can be analyzed in the frame of the so-

called membrane theory. One neglects all moments and transverse stress resultants,

all stresses are considered approximatively constant through the shell thickness i.e.

there are no bending stresses and the coupling and bending stiffness are taken to be

zero in the constitutive equations. In some cases it is possible to use the membrane

theory for structural analysis of laminated shells. The efﬁcient structural behavior of

shells based on the shell curvature that yields in wide regions of shells of revolution

approximately a membrane response upon loading as the basic state of stresses and

strains. The membrane theory is not capable to predict sufﬁcient accurate results in

regions with concentrated loads, boundary constraints or curvature changes, i.e. in

regions located adjacent to each structural, material or load discontinuity. Restrict-

ing the consideration again to circular cylindrical shells with unsymmetric cross-ply

stacking we arrive the following equations

9.2 Classical Shell Theory 351

Mx=Ms=Mxs =0,Qx=Qs=0,

∂

Nx

∂

s+

∂

Nxs

∂

s=−px,

∂

Nxs

∂

s+

∂

Ns

∂

s=−ps,Ns=Rpz,

ε

x=

∂

u

∂

x,

ε

s=

∂

v

∂

s+w

R,

ε

xs =

∂

u

∂

s+

∂

v

∂

x=

γ

xs,

Nx=A11

ε

x+A12

ε

s,Ns=A12

ε

x+A22

ε

s,Nxs =A66

ε

xs

(9.2.26)

The membrane theory yield three equilibrium conditions to calculate three unknown

stress resultants, i.e the membrane theory is statically determined. The membrane

theory is the simplest approach in shell analysis and admit an approximative an-

alytical solution that is very convenient for a ﬁrst analysis and design of circular

cylindrical shells.

But the problems which can be solved by the membrane theory are unfortunately

limited. To avoid generally to use the more complex bending theory we can con-

sider a so-called semi-membrane theory of circular cylindrical shells. The semi-

membrane theory is slightly more complicated than the membrane theory but more

simpler than the bending theory. The semi-membrane theory was ﬁrst developed by

Vlasov on the basis of statically and kinematically hypotheses.

If one intends to construct a semi-membrane theory of composite circular cylin-

drical shells bearing in mind the hypotheses underlying the classical single layer

shell theory and the characteristics of the composite structure. The semi-membrane

theory for composite circular cylindrical shells introduces the following assump-

tions:

•The shell wall has no stiffness when bended but in axial direction and when

twisted, i.e. D11 =D66 =0,B11 =B66 =0.

•The Poisson’s effect is neglected, i.e. A12 =0,B12 =0,D12 =0.

•With the assumptions above follow Mx=Mxs =0,Qx=0.

•The cross-section contour is inextensible, i.e.

ε

s=

∂

v

∂

s+w

R=0

The shear stiffness of composite shells can be small. Therefore, the assumption

of the classical single layer semi-membrane theory that the shear stiffness is in-

ﬁnitely large, is not used.

Taking into account the assumptions above, one obtains the following set of eleven

equations for eleven unknown functions.

∂

Nx

∂

x+

∂

Nxs

∂

s=0,

∂

Nxs

∂

x+

∂

Ns

∂

s+Qs

R+ps=0,

∂

Ms

∂

s−Qs=0,

∂

Qs

∂

s−Ns

R−pz=0,

ε

x=

∂

u

∂

x,

ε

xs =

∂

v

∂

x+

∂

u

∂

s,

ε

s=

∂

v

∂

s+w

R,(9.2.27)

352 9 Modelling and Analysis of Circular Cylindrical Shells

Nx=A11

ε

x,Nxs =A66

ε

xs,Ms=D22

κ

s,Qs=ks

44A44 

ψ

s+

∂

w

∂

s

The system (9.2.27) can be reduced. For the circular cylindrical shell the unknown

functions and loads can be represented with trigonometric series and after some

manipulations we obtained one uncoupled ordinary differential equation of fourth

order for wn(x),n=0,1,2,.... The detailed derivation of the governing solutions

shall not be considered.

9.3 Shear Deformation Theory

Analogous to plates, considered in Chap. 8, the classical shell theory is only sufﬁ-

ciently accurate for thin shells. For moderately thick shells we have to take, at least

approximately, the transverse shear deformation effects into account. The Kirch-

hoff’s hypotheses are again relaxed in one point: the transverse normals do not

remain perpendicular to the middle-surface after deformation, but a line element

through the shell thickness perpendicular to the middle-surface prior loading, un-

dergoes at most a translation and rotation upon the load applications, no stretching

or curvature.

The following considerations are restricted to axial symmetrical problems of

symmetrical laminated cross-ply circular cylindrical shells including transverse

shear deformation. We start with a variational formulation including the trapeze

effect, i.e. Love’s ﬁrst approximation is not valid. For axial symmetrical problems

we have the following simpliﬁcations of the shell equations:

All derivatives

∂

/

∂

s(...)are zero and for the strains, stress resultants and loads we

assume

ε

xs =0,

κ

s=0,

κ

xs =0,

Nxs =0,Mxs =0,

ps=0,px=0,pz=pz(x)

(9.3.1)

The kinematical assumptions yield with (5.1.2) the shell displacements

ux(x,z) = u(x) + z

ψ

x(x),

us(x,z) = 0,

uz(x,z) = w(x)

(9.3.2)

The strain-displacement relations are

ε

x=du

dx+zd

ψ

x

dx,

ε

s=w

R+z,

ε

xz =

ψ

x+dw

dx

(9.3.3)

9.3 Shear Deformation Theory 353

and the stresses in the kth layer of the shell are





σ

x

σ

s

σ

xz 



(k)

=



Q11 Q12 0

Q12 Q22 0

0 0 Q55 



(k)





du

dx+zd

ψ

x

dx

w

r+z

ψ

x+dw

dx







(9.3.4)

For a special orthotropic shell the Qi j are

Q11 =Ex

1−

ν

xs

ν

sx

,Q22 =Es

1−

ν

xs

ν

sx

,

Q12 =

ν

xsEx

1−

ν

xs

ν

sx

,Q55 =Gxz

(9.3.5)

The stress resultant forces and couples are deﬁned as

Nx=

n

∑

k=1Z

h(k)

σ

(k)

x1+z

Rdz,Ns=

n

∑

k=1Z

h(k)

σ

(k)

sdz,

Mx=

n

∑

k=1Z

h(k)

σ

(k)

xz1+z

Rdz,Ms=

n

∑

k=1Z

h(k)

σ

(k)

szdz,

Qx=

n

∑

k=1Z

h(k)

σ

(k)

xz 1+z

Rdz,

(9.3.6)

and one obtains with (9.3.4) and (9.3.6)

Nx=A11

du

dx+A12

w

R+1

RD11

d

ψ

x

dx,

Ns=A12

du

dx+T22w,

Mx=D11 1

R

du

dx+d

ψ

x

dx,

Ms=D12

d

ψ

x

dx+˜

T22w,

Qx=A55 

ψ

x+dw

dxor Qx=ks

55A55 

ψ

x+dw

dx

(9.3.7)

The stiffness coefﬁcients are

354 9 Modelling and Analysis of Circular Cylindrical Shells

(Ai j,Di j ) =

n

∑

k=1Z

h(k)

Q(k)

i j (1,z2)dz,(i j) = (11),(12),

T22 =

n

∑

k=1Z

h(k)

Q(k)

22

dz

R+z=

n

∑

k=1

Q(k)

22 hln1+z

Rih/2

−h/2

≈1

RA22 +1

R3D22,

˜

T22 =

n

∑

k=1Z

h(k)

Q(k)

22

zdz

R+z=

n

∑

k=1

Q(k)

22 Rhz

R−ln1+z

Rih/2

−h/2

≈1

R2D22

(9.3.8)

and ks

55 is the shear correction factor.

The variational formulation for the axial symmetrically circular cylindrical shell

with symmetrically laminated

θ

=00and

θ

=900laminae and coincided principal

material and structural axes is given as

Π

(u,w,

ψ

x) =

Π

i−

Π

a,

Π

i(u,w,

ψ

x) = 1

2

h/2

Z

−h/2

π

Z

0

L

Z

0

(

σ

x

ε

x+

σ

s

ε

s+

σ

xz

ε

xz)dx(R+z)d

ϕ

dz

=1

2

n

∑

k=1Z

h(k)

π

Z

0

L

Z

0Q(k)

11

ε

2

x

+2Q(k)

12

ε

x

ε

s+Q(k)

22

ε

2

s+Q(k)

55

ε

2

xz(R+z)dxd

ϕ

dz

(9.3.9)

Using Eq. (9.3.3) one obtains

Π

i=1

2

π

Z

0

L

Z

0(R"A11 du

dx2

+D11 d

ψ

x

dx2

+2D11

du

dx

d

ψ

x

dx

+2A12wdu

dx+T22w2+Rks

55A55"

ψ

2

x+2

ψ

x

dw

dx+dw

dx2#)dxd

ϕ

,

Π

a=

2

π

Z

0

L

Z

0

pz(x)wRdxd

ϕ

(9.3.10)

Equation (9.3.10) can be used for solving shell problems by the variational meth-

ods of Ritz, Galerkin or Kantorovich. It can be also used to derive the differential

equations and boundary conditions but this will be done later on the direct way.

Hamilton’s principle is formulated to solve vibration problems. The potential

energy function

Π

is given with (9.3.10) but all displacements are now functions of

xand the time t. If we analyze natural vibrations the transverse load pzis taken zero.

The kinetic energy follows as

9.3 Shear Deformation Theory 355

T=

n

∑

k=1

T(k)=1

2

n

∑

k=1Z

h(k)

2

π

Z

0

L

Z

0

ρ

(k)"

∂

u

∂

t2

+2z

∂

u

∂

t

∂ψ

x

∂

t

+z2

∂ψ

x

∂

t2

+

∂

w

∂

t2#Rd

ϕ

dxdz

=1

2

π

Z

0

L

Z

0(R

ρ

0"

∂

u

∂

t2

+

∂

w

∂

t2#

+2R

ρ

1

∂

u

∂

t

∂ψ

x

∂

t+R

ρ

2

∂ψ

x

∂

t2)d

ϕ

dx

(9.3.11)

In Eq. (9.3.11)

ρ

(k)is the mass density of the kth layer,

ρ

0and

ρ

2are the mass

and the moment of inertia with respect to the middle surface per unit area and

ρ

1

represents the coupling between extensional and rotational motions.

ρ

1does not

appear in equations for homogeneous shells.

Now with the Lagrange function L(u,w,

ψ

x) = T(u,w,

ψ

x)−

Π

(u,w,

ψ

x)the

Hamilton’s principle is obtained as

δ

t2

Z

t1

L(u,w,

ψ

x)dt=0 (9.3.12)

The direct derivation of the differential equations for symmetrical cross-ply circular

cylindrical shells follow using the constitutive, kinematics and equilibrium equa-

tions. The stiffness matrix is deﬁned as

N

N

M

M=A

A

A0

0

0D

D

D

ε

κ

,Q

Q

Qs=A

A

As

ε

s,

N

NT= [NxNsNxs ],M

M

MT= [MxMsMxs],Q

Q

QsT= [QxQs],

ε

T=

∂

u

∂

x

∂

v

∂

s+w

R

∂

u

∂

s+

∂

v

∂

x,

κ

T=

∂ψ

x

∂

x

∂ψ

s

∂

s

∂ψ

x

∂

s+

∂ψ

s

∂

x,

ε

sT=

ψ

x+

∂

w

∂

x

ψ

s+

∂

w

∂

s−v

R

(9.3.13)

For symmetric cross-ply shells all Bi j are zero and also the Ai j ,Di j with

(i j) = (16),(26)and (45). Such we have

A

A=



A11 A12 0

A12 A22 0

0 0 A66 

,D

D

D=



D11 D12 0

D12 D22 0

0 0 A66 

,A

A

As=ks

55A55 0

0ks

44A44 

356 9 Modelling and Analysis of Circular Cylindrical Shells

The static equilibrium equations are identical with (9.2.8). For vibration analysis

inertia terms have to be added and one can formulate

∂

Nx

∂

x+

∂

Nxs

∂

s=−px+

ρ

0

∂

2u

∂

t2+

ρ

1

∂

2

ψ

x

∂

t2,

∂

Nxs

∂

x+

∂

Ns

∂

s+Qs

R=−ps+

ρ

0

∂

2v

∂

t2+

ρ

1

∂

2

ψ

s

∂

t2,

∂

Qx

∂

x+

∂

Qs

∂

s−Ns

R=−pz+

ρ

0

∂

2w

∂

t2,

∂

Mx

∂

x+

∂

Mxs

∂

s−Qx=

ρ

2

∂

2

ψ

x

∂

t2+

ρ

1

∂

2u

∂

t2,

∂

Mxs

∂

x+

∂

Ms

∂

s−Qs=

ρ

2

∂

2

ψ

s

∂

t2+

ρ

1

∂

2v

∂

t2

(9.3.14)

ρ

0,

ρ

1,

ρ

2are like in (9.3.11) generalized mass density and are deﬁned in (8.3.9).

Putting the constitutive equations (9.3.12) in the equilibrium equations (9.3.14) the

equations can be manipulated in similar manner to those of the classical theory and

one obtains the simultaneous system of differential equations







˜

L11 ˜

L12 ˜

L13 ˜

L14 ˜

L15

˜

L21 ˜

L22 ˜

L23 ˜

L24 ˜

L25

˜

L31 ˜

L32 ˜

L33 ˜

L34 ˜

L35

˜

L41 ˜

L42 ˜

L43 ˜

L44 ˜

L45

˜

L51 ˜

L52 ˜

L53 ˜

L54 ˜

L55













u

v

ψ

x

ψ

s

w







=−





px

ps

0

pz







+





ρ

00

ρ

10 0

0

ρ

00

ρ

10

ρ

10

ρ

20 0

0

ρ

10

ρ

20

0 0 0 0

ρ

0







∂

2

∂

t2







u

v

ψ

x

ψ

s

w







(9.3.15)

The linear differential operators are deﬁned in App. D.2.

For free vibrations the loads px,ps,pzare zero and the shell will perform sim-

ple harmonic oscillations with the circular frequency

ω

. Corresponding to simple

supported conditions on both ends of the cylinder, i.e

Nx=0,v=0,w=0,Mx=0,

ψ

s=0,

the spatial dependence can be written as products of two trigonometric functions

and the complete form of vibrations can be taken as

u(x,

ϕ

,t) =

∞

∑

r=1

∞

∑

s=1

Ursei

ω

rstcos

α

mxcosn

ϕ

,

v(x,

ϕ

,t) =

∞

∑

r=1

∞

∑

s=1

Vrsei

ω

rstsin

α

mxsinn

ϕ

,

w(x,

ϕ

,t) =

∞

∑

r=1

∞

∑

s=1

Wrsei

ω

rstsin

α

mxcos n

ϕ

,(9.3.16)

ψ

x(x,

ϕ

,t) =

∞

∑

r=1

∞

∑

s=1

Ψ

rsei

ω

rstcos

α

mxcosn

ϕ

,

9.3 Shear Deformation Theory 357

ψ

s(x,

ϕ

,t) =

∞

∑

r=1

∞

∑

s=1

˜

Ψ

rsei

ω

rstsin

α

mxsinn

ϕ

,

where Urs,Vrs,Wrs,

Ψ

rs,˜

Ψ

rs denote amplitudes,

α

m=m

π

/l,m,nare the longitudinal

and the circumferential wave numbers. Substituting Eqs. (9.3.16) into (9.3.15) re-

sults in a homogeneous algebraic system and its solutions for a particular pair (m,n)

gives the frequency and amplitude ratio corresponding to these wave numbers. For

arbitrary boundary conditions the Ritz’ or Galerkin’s method can be recommended

to obtain the characteristic equations for solving the eigenvalue problem. Then the

natural frequencies and the mode shapes can be calculated. The solution process is

manageable, but involved.

If one restricts the problem to statics and to axially symmetrical loading

pz=pz(x)the Eqs. (9.3.13) - (9.3.15) can be simpliﬁed:

•Equilibrium equations

dNx

dx=0,dMx

dx−Qx=0,dQx

dx−Ns

R+pz=0 (9.3.17)

•Strain-displacement equations

ε

x=du

dx,

ε

s=w

R,

κ

x=d

ψ

x

dx,

ε

xz =

ψ

x+dw

dx(9.3.18)

•Constitutive equations

Nx=A11

ε

x+A12

ε

s,

Ns=A12

ε

x+A22

ε

s,

Mx=D11

d

ψ

x

dx,

Qx=ks

55A55

ε

xz

(9.3.19)

All derivatives

∂

/

∂

s(...)and v,

ε

xs,

κ

s,

ε

sz,Nxs ,Mxs,Qsare zero. The stress resultant-

displacement relations follow as

Nx=A11

du

dx+A12

w

R,

Ns=A12

du

dx+A22

w

R,

Mx=D11

d

ψ

x

dx,

Qx=ks

55A55 

ψ

x+dw

dx

(9.3.20)

With dNx/dx=0 we have Nx=const =N0and one obtains

du

dx=1

A11 N0−A12

w

R(9.3.21)

The equilibrium equation (9.3.17) for Qxyields

358 9 Modelling and Analysis of Circular Cylindrical Shells

d

dx

ψ

x+dw

dx=1

ks

55A55 Ns

R−pz

=1

ks

55A55 1

RA12

du

dx+A22

w

R−pz(9.3.22)

and the equilibrium equation dMx/dx−Qx=0

D11

d2

ψ

x

dx2−ks

55A55 

ψ

x+dw

dx=0 (9.3.23)

After some manipulations follow with (9.3.22), (9.3.23) two differential equations

for

ψ

xand was

d

ψ

x

dx=−d2w

dx2+1

ks

55A55 1

RA12

A11

N0−A2

12 −A11A22

A11

w

R−pz,

dw

dx=D11

ks

55A55

d2

ψ

x

dx2−

ψ

x

(9.3.24)

Differentiating the second equation (9.3.24) and eliminating

ψ

xthe ﬁrst equation

leads to one uncoupled differential equation of fourth order for w(x)

d4w

dx4−1

ks

55A55

1

R2

A11A22 −A2

12

A11

d2w

dx2+1

R2

A11A22 −A2

12

D11A11

w

=1

D11 −A12

A11

N0

R+pz−1

ks

55A55

d2pz

dx2,

d

ψ

x

dx=−d2w

dx2+1

ks

55A55

A11A22 −A2

12

R2A11

w+1

ks

55A55 A12

A11

N0

R−pz

(9.3.25)

With ks

55A55 →∞Eqs. (9.3.25) simplify to the corresponding equations of the clas-

sical shell theory (9.2.19).

The governing equation of the axisymmetric problem for or circular cylindrical

shell in the frame of the shear deformation theory can be written as

d4w

dx4−2k1

d2w

dx2+k4

2w=kp(9.3.26)

with

k2

1=1

ks

55A55

1

R

A11A22 −A2

12

A11

,k4

2=1

ks

55A55

1

R2

A11A22 −A2

12

D11A11

,

kp=1

D11 −A12

A11

N0

R+pz−1

D11ks

55A55

d2pz

dx2

The differential equation can be analytical solved

w(x) = wh(x) + wp(x)

9.3 Shear Deformation Theory 359

The particular solution wp(x)has e.g. with d2pz/dx2=0, the form

wp(x) = kp

k4

2

=R2pz−RA12N0

A11A22 −A2

12

(9.3.27)

The homogeneous solution wh(x) = Ce

α

xyield the characteristic equation

α

4−2k2

1

α

2+k4

2=0 (9.3.28)

with the roots

α

1−4=±rk2

1±qk4

1−k4

1

which can be conjugate complex, real or two double roots depending on the relations

of the constants k1and k2.

The general solution can be written as (App. E)

w(x) =

4

∑

i=1

Ci

Φ

i(x) + wp(x)

The functions

Φ

i(x)are given in different forms depending on the roots of the char-

acteristic equation (9.3.28). The roots and the functions

Φ

i(x)are summarized in

App. E. The most often used solution form in engineering application is given for

k2

2>k2

1.

For short shells with edges affecting one another, the

Φ

i(x)involving the hy-

perbolic functions are convenient. If there are symmetry conditions to the middle

cross-section x=L/2 the solution can be simpliﬁed, for we have

Φ

3=

Φ

4=0.

For long shells with ends not affecting one another applying the

Φ

i(x)that involve

exponential functions.

Analogous to the classical shell solution for long shells a bending-layer solution

can be applied. Only inside the bending-layer region with the characteristic length

LBthe homogen eous part whand the particular part wpof the general solution whave

to superimposed. Outside the bending-layer region, i.e for x>LBor (L−x)>LB

only wpcharacterizes the shell behavior.

Summarizing the results of the shear deformation shell theory one can say

•If one restricts the consideration to symmetrical cross-ply circular cylindrical

shells subjected to axially symmetric loadings the modelling and analysis is most

simpliﬁed and correspond to the classical shell theory.

•In more general cases including static loading and vibration and not neglecting

the trapeze effect the variational formulation is recommended and approximative

analytical or numerical solutions should be applied.

•Circular cylindrical shells are one of the most used thin-walled structures of con-

ventional or composite material. Such shells are used as reservoirs, pressure ves-

sels, chemical containers, pipes, aircraft and ship elements. This is the reason for

a long and intensive study to model and analyze circular cylindrical shells and as

result efﬁcient theories and solutions methods are given in literature.

360 9 Modelling and Analysis of Circular Cylindrical Shells

9.4 Sandwich Shells

Sandwich shells are widely used in many industrial branches because sandwich con-

structions often results in designs with lower structural weight then constructions

with other materials. But there is not only weight saving interesting, but in several

engineering applications the core material of a sandwich construction can be also

used as thermal insulator or sound absorber. Therefore one can ﬁnd numerous lit-

erature on modelling and analysis for sandwich shells subjected static, dynamic or

environmental loads .

But as written in Sects. 7.4 and 8.4 sandwich constructions are, simply consid-

ered, laminated constructions involving three laminae: the lower face, the core, and

the upper face. And by doing so, one can employ all methods of modelling and

analysis of laminated structural elements.

It was discussed in detail in Sect. 8.4 that, considering sandwich structural ele-

ments, we have to keep in mind the assumptions on the elastic behavior of sand-

wiches. Such there are differences in the expressions for the ﬂexural bending and

transverse shear stiffness in comparison with laminated circular cylindrical shells

and essential differences in the stress distribution over the thickness of the shell

wall. The stiffness parameter for sandwich shells depend on the modelling of sand-

wiches having thin or thicker faces, in the same manner as for plates, Eqs. (8.4.1)

and (8.4.2).

For sandwich constructions generally the ratio of the in-plane moduli of elasticity

to the transverse shear moduli is high and transverse shear deformations are mostly

included in its structural modelling. For this reason, the ﬁrst order shear deformation

theory of laminated shells is used in priority for sandwich shells. But for thin-walled

sandwich shells with a higher shear stiffness approximately the classical sandwich

theory can be used.

The correspondence between laminated and sandwich shells is for vibration or

buckling problems limited and only using for overall buckling and vibration. There

are some special local problems like face wrinkling and core shear instability in

buckling or the face must be additional considered as a shell of elastic foundation

on the core and also shear mode vibration can occur where each face is vibrating

out of phase with the other face. These problems are detailed discussed in a number

of special papers and can not considered in this book.

9.5 Problems

Exercise 9.1. A circular cylindrical sandwich shell has two unequal faces with the

reduced stiffness Qf1

11,Qf3

11 and the thicknesses hf1,hf3. The shell has an orthotropic

material behavior and the material principal axes shall coincide with the structural

axes x,s. The core with the thickness hcdoes not contribute signiﬁcantly to the

extensional and the ﬂexural shell stiffness. The lateral distributed load is pz=pz(x).

9.5 Problems 361

Formulate the differential equation using a perturbation constant to characterize the

asymmetry of the sandwich and ﬁnd the perturbation solution way.

Solution 9.1. The shell problem is axially symmetric. In the frame of the classical

shell theory one can use the differential equation (9.2.18)

d4w

dx4+2

RDRA12B11

A11 −B12d2w

dx2+4

λ

4w=1

DRpz−A12

A11

Nx

R

The stiffness parameter are calculated for sandwiches with thin faces, Sect. 4.3.2,

h≈hc

A11 =Qf1

11hf1+Qf3

11hf3=Qf1

11hf1 1+Qf3

11hf3

Qf1

11hf1!,

D11 =h

22

Qf1

11hf1+h

22

Qf3

11hf3=h

22

Qf1

11hf1 1+Qf3

11hf3

Qf1

11hf1!,

B11 =−h

2Qf1

11hf1+h

2Qf3

11hf3=h

2Qf1

11hf1 −1+Qf3

11hf3

Qf1

11hf1!

B12 and D12 can be calculated analogous. A asymmetry constant can be deﬁned as

η

=B11

√D11A11

=−1+Qf3

11hf3

Qf1

11hf1

1+Qf3

11hf3

Qf1

11hf1

For a symmetric sandwich wall is B11 =0 and so

η

=0. For an inﬁnite stiffness of

face 1 follows

η

→−1 and of face 3

η

→+1, i.e. the constant

η

is for any sandwich

construction given as

−1<

η

<+1

The differential equation (9.2.18) can be written with the constant

η

as

d4w

dx4+2

RDRA12√D11

√A11 −√A11B12 √D11

B11 

η

d2w

dx2+4

λ

4w=1

DRpz−A12

A11

Nx

R

Since |

η

|<1 one can ﬁnd w(x)in the form of a perturbation solution

w(x) =

∞

∑

n=0

wn(x)

η

n

For n=0 follow

d4w0

dx4−4

λ

4w0=−1

DRpz−A12

A11

Nx

R

and for n≥1

362 9 Modelling and Analysis of Circular Cylindrical Shells

d4wn

dx4+4

λ

4wn=−2

RDRA12√D11

√A11 −√A11B12 √D11

B11 

η

d2wn−1

dx2

The left hand side corresponds to the middle-surface symmetric shell with axially

symmetric loading. The right hand side corresponds to the second derivation of the

previously obtained w-solution.

Conclusion 9.1. The perturbation solution yield the solution of the differential equa-

tion as a successive set of solutions of axially symmetric problems of which many

solutions are available. The perturbation solution converges to the exact solution. In

many engineering applications w(x) = wo(w) +

η

w1(x)will be sufﬁcient accurate.

Exercise 9.2. A symmetrical cross-ply circular cylindrical shell is loaded at the

boundary x=0 by an axially symmetric line pressure Q0and line moment M0.

Calculate the ratio M0/Q0that the boundary shell radius does not change if the shell

is very long.

Solution 9.2. We use the solution (9.2.23) with wp=0 and neglect for the long shell

the inﬂuence of MLand QL

w(x) = M0

2

λ

2D11

e−

λ

x(sin

λ

x−cos

λ

x)−Q0

2

λ

3D11

cos

λ

x

The condition of no radius changing yields

w(x=0) = 0⇒ − M0

2

λ

2D11 −Q0

2

λ

3D11

=0⇒M0

Q0

=−1

λ

Exercise 9.3. For a long ﬂuid container, Fig. 9.4 determinate the displacement w(x)

and the stress resultants Ns(x)and Mx(x). The container has a symmetrical cross-ply

layer stacking and can be analyzed in the frame of the classical laminate theory.

Solution 9.3. For a long circular cylindrical shell the solution for wh(x), Eq.

(9.2.22), can be reduced to the ﬁrst term with the negative exponent

Fig. 9.4 Long ﬂuid container,

L>LB

L

p(x)

R R

x

9.5 Problems 363

wh(x) = e−

λ

x(C1sin

λ

x+C2cos

λ

x)

The particular solution wp(x)follow with

p(x) = p01−x

L

and Eq. (9.2.24) as

wp(x) = p0

4

λ

4D11 1−x

L

The boundary constraints are

w(0) = 0⇒C2=−p0

4

λ

4D11

,dw(0)

dx=0⇒C1=−p0

4

λ

4D11 1−x

L

and we obtain the solutions

w(x) = p0

4

λ

4D11 n1−x

L−hcos

λ

x+1−x

Lsin

λ

xie−

λ

xo

In addition,

Ns=A11

ε

x+A22

ε

s=A11

du

dx+A22

w

R

i.e. with

ε

x=0

Ns=A22

Rw(x),Mx=D11

κ

x=−D11

d2w

dx2,Qx=dMx

dx=−D11

d3w

dx3

Exercise 9.4. Consider a cantilever circular cylindrical shell, Fig. 9.5. The normal

and shear forces Nxand Nxs as are distributed along the contour of the cross-section

x=Lthat they can reduced to the axial force FH, the transverse force FV, the bending

moment MBand the torsion moment MT. Calculate the resultant membrane stress

forces with the membrane theory.

Solution 9.4. With (9.2.26) we have the following equations

x

L

FV

FH

MB

MT

ϕ

s=R

ϕ

RNxs

Fig. 9.5 Tension, bending and torsion of a cantilever circular cylindrical shell

364 9 Modelling and Analysis of Circular Cylindrical Shells

Ns=Rpz,

∂

Nxs

∂

x=−ps,

∂

Nx

∂

x=−px−

∂

Nxs

∂

s

and pz=ps=px=0 yield

Ns=0,Nxs =const,Nx=const

The distributions of FH,FV,MBand MTover the cross-section contour x=Lcan be

represented as

Nx(x=L) = 1

2

π

RFH+2MB

Rcos

ϕ

,

Nrs(x=L) = 1

2

π

RMT

R+2FVsin

ϕ



and yield the reduced forces FV,FMand moments MB,MT

2

π

Z

0

Nx(x=L)Rd

ϕ

=FH,

2

π

Z

0

Nxs(x=L)sin

ϕ

Rd

ϕ

=1

π

R

π

Z

0

2F

Vsin2

ϕ

Rd

ϕ

=F

V,

4

π

/2

Z

0

Nx(x=L)Rcos

ϕ

d

ϕ

=4

π

MB

π

/2

Z

0

cos2

ϕ

d

ϕ

=MB,

2

π

/2

Z

0

RNxs(x=L)Rd

ϕ

=1

R

π

/2

Z

0

MTRd

ϕ

=MT

The equilibrium equations yield

Nx(x) = FH

2

π

R−[MB+FV(L−x)] cos

ϕ

π

R,

Ns(x) = 0,

Nsx =MT

2

π

R12 +FV

π

Rsin

ϕ

Part IV

Modelling and Analysis of Thin-Walled

Folded Plate Structures

The fourth part (Chap. 10) includes the modelling and analysis of thin-walled folded

plate structures or generalized beams. This topic is not normally considered in stan-

dard textbooks on structural analysis of laminates and sandwiches, but it is included

here because it demonstrates the possible application of Vlasov’s theory of thin-

walled beams and semi-membrane shells on laminated structural elements.

Chapter 10

Modelling and Analysis of Thin-walled Folded

Structures

The analysis of real structures always is based on a structural and mathematical

modelling. It is indispensable for obtaining realistic results that the structural model

represents sufﬁciently accurate the characteristic structure behavior.

Generally the structural modelling can be divided into three structure levels

•Three-dimensional modelling. It means structural elements, their dimensions in

all three directions are of the same order, we have no preferable direction.

•Two-dimensional modelling. One dimension of a structural element is signiﬁcant

smaller in comparison with the other both, so that we can regard it as a quasi

two-dimensional element. We have to distinguish plane and curved elements e.g.

discs, plates and shells.

•One-dimensional modelling. Here we have two dimensions (the cross-section)

in the same order and the third one (the length) is signiﬁcant larger in com-

parison with them, so that we can regard such a structural element as quasi one-

dimensional. We call it ro d, column,bar, beam or arch and can d istinguish straight

and curved forms also.

The attachment of structural elements to one of these classes is not well deﬁned

rather it must be seen in correlation with the given problem.

Many practical problems, e.g., in mechanical or civil engineering lead to the

modelling and analysis of complex structures containing so-called thin-walled ele-

ments. As a result of the consideration of such structures a fourth modelling class

was developed, the modelling class of thin-walled beams and so-called beam shaped

shells including also folded plate structures. In this fourth modelling class it is typ-

ical that we have structures with a signiﬁcant larger dimension in one direction

(the length) in comparison with the dimensions in transverse directions (the cross-

section) and moreover a signiﬁcant smaller thickness of the walls in comparison

with the transverse dimensions.

In Chap. 7 the modelling of laminate beams is given in the frame of the

Bernoulli’s and Timoshenko’s beam theory which cannot applied generally to thin-

walled beams. The modelling of two-dimensional laminate structures as plates and

shells was the subject of the Chaps. 8 and 9. In the present Chap. 10 the investi-

367

H. Altenbach et al., Mechanics of Composite Structural

Elements, https://doi.org/10.1007/978-981-10-8935-0_10

368 10 Modelling and Analysis of Thin-walled Folded Structures

gation of beams with thin-walled cross-sections and beam shaped shells especially

folded structures is carried out. Chapter 10 starting in Sect. 10.1 from a short recall

of the classical beam models. In Sect. 10.2 a generalized beam model for prismatic

thin-walled folded plate structures is introduced, including all known beam models.

Section 10.3 discusses some solution procedures and in Sect. 10.4 selected problems

are demonstrated.

10.1 Introduction

Analyzing thin-walled structures it can be useful to distinguish their global and local

structural behavior. Global bending, vibration or buckling is the response of the

whole structure to external loading and is formulated in a global coordinate system.

A typical example for global structure behavior is the deﬂection of a ship hull on

the waves. But the deﬂections and stresses in a special domain of the ship e.g. in the

region of structure loading or deck openings or the vibrations or buckling of single

deck plates represent typical local effects.

A necessary condition for a global analysis is that the geometry of the structure

allows its description in a global co-ordinate system, i.e. the thin-walled structure is

sufﬁcient long how it is given in case of a quasi one-dimensional structure.

Of course there are interactions between global and local effects, and in the most

cases these interactions are nonlinear. Usually the global analysis is taken as the

basic analysis and its results are the boundary conditions for local considerations

by using special local co-ordinate systems. The reactions of local to global effects

whereas are neglected.

From this point of view the global analysis of thin-walled beams and beam

shaped shells can be done approximately by describing them as one-dimensional

structures with one-dimensional model equations. For such problems the classical

beam model of J. Bernoulli was used. This model is based on three fundamental

hypotheses:

•There are no deformations of the cross-sectional contour.

•The cross-section is plane also in case of deformed structures.

•The cross-section remain orthogonally to the deformed system axis.

As a result of bending without torsion we have normal stresses

σ

and strains

ε

only

in longitudinal direction. Shear deformations are neglected. The shear stresses

τ

caused by the transverse stress resultants are calculated with the help of the equilib-

rium equations, but they are kinematically incompatible.

The Bernoulli’s beam model can be used for beams with compact and sufﬁcient

stiff thin-walled cross-sections. In case of thin-walled cross-section it is supposed

that the bending stresses

σ

band the shear stresses

τ

qare distributed constantly over

the thickness t. If we have closed thin-walled cross-sections a statically indetermi-

nate shear ﬂow must be considered.

10.1 Introduction 369

A very important supplement to Bernoulli’s beam model was given by Saint-

Venant1for considering the torsional stress. Under torsional stresses the cross-

sections have out-of-plane warping, but assuming that these are the same in all cross-

sections and they are not constrained we have no resulting longitudinal strains and

normal stresses. In this way we have also no additional shear stresses. The distribu-

tion of the so-called Saint-Venant torsional shear stresses is based on a closed shear

ﬂow in the cross-section. For closed thin-walled cross-sections the well-known ele-

mentary formulae of Bredt2can be used.

The Timoshenko’s beam model is an extension of the Bernoulli’s beam model.

It enables to consider the shear deformations approximately. The ﬁrst two basic hy-

potheses of the Bernoulli’s model are remained. The plane cross-section stays plane

in this case but is not orthogonally to the system axes in the deformed structure. For

the torsional stress also the relationships of Saint-Venant are used.

Rather soon the disadvantages of this both classical beam models were evident

for modelling and analysis of general thin-walled beam shaped structures. Espe-

cially structures with open cross-section have the endeavor for warping, and be-

cause the warping generally is not the same in all cross-sections, there are additional

normal stresses, so-called warping normal stresses and they lead to warping shear

stresses too. Therefore the torsional moment must be divided into two parts, the

Saint-Venant part and the second part caused by the warping shear stresses.

Very fundamental and general works on this problem were done by Vlasov. Be-

cause his publications are given in Russian language they stayed unknown in west-

ern countries for a long time. In 1958 a translation of Vlasov’s book ”General Shell

Theory and its Application in Technical Sciences” into German language was pub-

lished (Wlassow, 1958) and some years later his book on thin-walled elastic beams

was published in English (Vlasov, 1961). By Vlasov a general and systematic ter-

minology was founded, which is used now in the most present papers.

The Vlasov’s beam model for thin-walled beams with open cross-sections is

based on the assumption of a rigid cross-section contour too, but the warping ef-

fects are considered. Neglecting the shear strains of the mid-planes of the walls the

warping of the beam cross-section are given by the so-called law of sectorial areas.

The application of this Vlasov beam model to thin-walled beams with closed cross-

sections leads to nonsatisfying results, because the inﬂuences of the cross-sectional

contour deformations and of the mid-plane shear strains in the walls are signiﬁcant

in such cases.

Therefore a further special structural model was developed by Vlasov in form

of the so-called semi-moment shell theory, in which the longitudinal bending mo-

ments and the torsional moments in the plates of folded structures with closed cross-

sections are neglected. By this way we have in longitudinal direction only membrane

stresses and in transversal direction a mixture of membrane and bending stresses.

This two-dimensional structural model can be reduced to a one-dimensional one

by taking into account the Kantorovich relationships in form of products of two

1Adh´emar Jean Claude Barr´e de Saint-Venant (∗23 August 1797 Villiers-en-Bi`ere – †6 January

1886 Saint-Ouen) - mechanician and mathematician

2Rudolf Bredt (∗17 April 1842 Barmen - †18 May 1900 Wetter) - mechanical engineer

370 10 Modelling and Analysis of Thin-walled Folded Structures

functions. One of them describes a given deformation state of the cross-section,

considered as a plane frame structure and the other is an unknown function of the

longitudinal co-ordinate.

In 1994 the authors of this textbook published a monograph on thin-walled folded

plate structures in German (Altenbach et al, 1994). Starting from a general struc-

tural model for isotropic structures also a short outlook to anisotropic structures was

given. The general model equations including the semi-moment shell model and all

classical and generalized linear beam models could be derived by neglecting special

terms in the elastic energy potential function or by assuming special conditions for

the contour deformation states. In Sect. 10.2 the derivation of generalized folded

structural model is given for anisotropic plates, e.g. off-axis loaded laminates. The

derivations are restricted to prismatic systems with straight system axes only.

Summarizing one can conclude from the above discussion there are several rea-

sons why for thin-walled structures must be given special consideration in design

and analysis. In thin-walled beams the shear stresses and strains are relatively much

larger than those in beams with solid, e.g. rectangular, cross-sections. The assump-

tions of Bernoulli’s or Timoshenko’s beam theory can be violated e.g. by so-called

shear lag effects, which result in a non-constant distribution of normal bending

stresses which are different from that predicted by the Bernoulli hypotheses for

beams carrying only bending loads. When twisting also occurs warping effects, e.g.

warping normal and shear stresses, have to add to those arising from bending loads.

The warping of the cross-section is deﬁned as its out-of-plane distorsion in the direc-

tion of the beam axis and violated the Bernoulli’s hypotheses and the Timoshenko’s

hypotheses too.

Because of their obvious advantages ﬁbre reinforced laminated composite beam

structures are likely to play an increasing role in design of the present and, espe-

cially, of future constructions in the aeronautical and aerospace, naval or automotive

industry. In addition to the known advantages of high strength or high stiffness to

weight ratio, the various elastic and structural couplings, which are the result of the

directional nature of composite materials and of laminae-stacking sequence, can be

successfully exploited to enhance the response characteristics of aerospace or naval

vehicles.

In order to be able to determine the behavior of these composite beam structures,

consistent mechanical theories and analytical tools are required. So a Vlasov type

theory for ﬁber-reinforced beams with thin-walled open cross-sections made from

mid-plane symmetric ﬁber reinforced laminates was developed but in the last 30

years many improved or simpliﬁed theories were published.

Because primary or secondary structural conﬁgurations such as aircraft wings,

helicopter rotor blades, robot arms, bridges and other structural elements in civil en-

gineering can be idealized as thin- or thick-walled beams, especially as box beams,

beam models appropriate for both thin- and thick-walled geometries which include

the coupled stiffness effects of general angle-ply laminates, transverse shear defor-

mation of the cross-section and the beam walls, primary and secondary warping,

etc. were developed. But nearly all governing equations of thin- and thick-walled

composite beams adopt the basic Vlasov assumption:

10.2 Generalized Beam Models 371

Fig. 10.1 Thin-walled prismatic folded plate structures with open or closed cross-sections

The contours of the original beam cross-section do not deform in their own

planes.

This assumption implies that the normal strain

ε

sin the contour direction is small

compared to the normal strain

ε

zparallel to the beam axis. This is particular valid

for thin-walled open cross-sections, for thin-walled closed cross-section with stiff-

eners (transverse sheets) and as the wall thickness of closed cross-sections increase.

Chapter 10 focuses the considerations to a more general model of composite thin-

walled beams which may be include the classical Vlasov assumptions or may be

relax these assumptions, e.g. by including the possibility of a deformation of the

cross-section in its own plane, etc.

In the following a special generalized class of thin-walled structures is consid-

ered, so-called folded plate structures. A folded plate structure shall be deﬁned as

a prismatic thin-walled structure which can be formed by folding a ﬂat rectangu-

lar plate or joining thin plate strips along lines parallel to their length. Figure 10.1

demonstrates thin-walled structures of the type deﬁned above. The plate strips can

be laminates.

10.2 Generalized Beam Models

Section 10.2 deﬁnes the outline of modelling beam shaped, thin-walled prismatic

folded plate structures with open, one or multi-cell closed or mixed open-closed

cross-sections. The considerations are limited to global structural response. Assum-

ing the classical laminate theory for all laminated plate strips of the beam shaped

structure the elastic energy potential function is formulated. The energy potential

is a two-dimensional functional of the coordinate xof the structure axis and the

cross-section contour coordinate s.

Following the way of Vlasov-Kantorovich the two-dimensional functional is re-

duced to an approximate one-dimensional one. A priori ﬁxed generalized coordi-

372 10 Modelling and Analysis of Thin-walled Folded Structures

nate functions describing the cross-section kinematics are introduced. Generalized

displacement functions which depend on the system coordinate xonly are the inde-

pendent functions of the reduced variational statement which leads to a system of

matrix differential equations, the Euler equations of the variational statement, and

to the possible boundary equations.

The general structural model can be simpliﬁed by neglecting selected terms in the

energy formulation or by restricting the number of the generalized coordinate func-

tions, i.e. the cross-section kinematics. All results are discussed under the viewpoint

of a sufﬁcient general structural model for engineering applications. A general struc-

tural model is recommended which includes all above noted forms of cross-sections

and enables to formulate efﬁcient numerical solution procedures.

10.2.1 Basic Assumptions

A prismatic system is considered, its dimensions are signiﬁcant larger in one di-

rection (the length) in comparison with these in transverse directions. The system

consists of nplane thin-walled strip elements; it means their thickness is signiﬁcant

smaller than the strip width, i.e. ti≪di. Rigid connections of the plate strips along

their length lines are supposed. Closed cross-sections as well as open cross-sections

and combined forms are possible. In Fig. 10.2 a general thin-walled folded structure

is shown. There is a global co-ordinate system x,y,zwith any position. In each strip

we have a local co-ordinate system x,si,ni, the displacements are ui,vi,wi. We re-

strict our considerations to prismatic structures only and neglect the transverse shear

strains in the strips normal to their mid-planes, it means the validity of the Kirchhoff

hypotheses is supposed or we use the classical laminate theory only. All constants

of each strip are constant in x-direction. For the displacements we can write

n1,w1

s1,v1

x,u1

n2,w2

s2,v2

x,u2

ni,wisi,vi

x,ui

x

y

z

ti

di

l

Fig. 10.2 Thin-walled folded structure geometry and co-ordinate systems

10.2 Generalized Beam Models 373

ui=ui(x,si),vi=vi(x,si),wi=wi(x,si)(10.2.1)

uiand viare the displacements in the mid-plane and wiis the deﬂection normal to

the mid-plane of the ith strip. As loads are considered:

•surface forces, distributed on the unit of the mid-plane

pxi=pxi(x,si),psi=psi(x,si),pni=pni(x,si)(10.2.2)

•line forces, distributed on the length unit of the boundaries of the structure

qxi|x=0=qxi(0,si),qsi|x=0=qsi(0,si),qni|x=0=qni(0,si)

qxi|x=l=qxi(l,si),qsi|x=l=qsi(l,si),qni|x=l=qni(l,si)(10.2.3)

If a linear anisotropic material behavior is supposed, for each strip we can use the

constitutive relationship given as

N

N

M

Mi

=A

A

A B

B

B D

D

Di

ε

κ

i

(10.2.4)

or







Nxi

Nsi

Nxsi

Mxi

Msi

Mxsi







=







A11iA12iA16iB11iB12iB16i

A12iA22iA26iB12iB22iB26i

A16iA26iA66iB16iB26iB66i

B11iB12iB16iD11iD12iD16i

B12iB22iB26iD12iD22iD26i

B16iB26iB66iD16iD26iD66i













ε

xi

ε

si

ε

xsi

κ

xi

κ

si

κ

xsi







The following steps are necessary for calculating the elements of the matrices

A

A,B

B

B,D

D

Dfor the ith strip:

•Calculate the reduced stiffness matrix Q

Q

Q′for each lamina (k)of the strip (i)by

using the four elastic moduli EL,ET,

ν

LT,GLT, Eqs. (4.1.2) and (4.1.3).

•Calculate the values of the transformed reduced stiffness matrix Q

Q

Qfor each lam-

ina (k)of the strip (i)(Table 4.2).

•Considering the stacking structure, it means, considering the positions of all lam-

inae in the ith strip calculate the matrix elements Akli,Bkli,Dkli, (4.2.15).

It must be noted that the co-ordinates x1,x2,x3used in Sect. 4.1.3 are corresponding

to the coordinates x,si,niin the present chapter and the stresses

σ

1,

σ

2,

σ

6here are

σ

xi,

σ

si,

σ

xsi. For the force and moment resultants also the corresponding notations

Nxi,Nsi,Nxsi,Mxi,Msi,Mxsiare used and we have to take here:

Nxi=

ti/2

Z

−ti/2

σ

xidni,Nsi=

ti/2

Z

−ti/2

σ

sidni,Nxsi=

ti/2

Z

−ti/2

σ

xsidni,

(10.2.5)

374 10 Modelling and Analysis of Thin-walled Folded Structures

Mxi=

ti/2

Z

−ti/2

σ

xinidni,Msi=

ti/2

Z

−ti/2

σ

sinidni,Mxsi=

ti/2

Z

−ti/2

σ

xsinidni

In Fig. 10.3 the orientations of the loads, see Eqs. (10.2.2) and (10.2.3), and the

resultant forces and moments in the ith wall are shown. In the frame of the classical

laminate theory the transverse force resultants Nsniand Nxnifollow with the help

of the equilibrium conditions for a strip element.

In the same way here we have the following deﬁnitions for the elements of the

deformation vector [

ε

1

ε

2

ε

6

κ

1

κ

2

κ

6]T≡[

ε

xi

ε

si

ε

xsi

κ

xi

κ

si

κ

xsi]Twith

ε

xi=

∂

ui

∂

x=u′

i,

ε

si=

∂

vi

∂

si

=v•

i,

ε

xsi=

∂

ui

∂

si

+

∂

vi

∂

x=u•

i+v′

i,

κ

xi=−

∂

2wi

∂

x2=−w′′

i,(10.2.6)

Fig. 10.3 Loads and resultant

forces and moments in the ith

strip

ni

x

si

Nsi

Nxsi

Nxi

Nxsi

Nsi

Nxsi

pxi

psi

qsi|x=0

qxi|x=0

x

si

ni

Mxsi

Nsni

Msi

Nxni

Mxsi

Mxi

Msi

Mxsi

Nsni

qni|x=0

pni

10.2 Generalized Beam Models 375

κ

si=−

∂

2wi

∂

s2

i

=−w••

i,

κ

xsi=−2

∂

2wi

∂

x

∂

si

=−2w′

i•

10.2.2 Potential Energy of the Folded Structure

The potential energy of the whole folded structure can be obtained by summarizing

the energy of all the nstrips

Π

=1

2∑

(i)

l

Z

0

di

Z

0N

N

NTM

M

MTi

ε

κ

i

dsidx−Wa(10.2.7)

With equation (10.2.4) the vectors of the resultant forces and moments can be ex-

pressed and we obtain

Π

=1

2∑

(i)

l

Z

0

di

Z

0

ε

κ

T

iA

A

A B

B

B D

D

Di

ε

κ

i

dsidx−Wa(10.2.8)

The external work of the loads is also the sum of all the nstrips

Wa=∑

i(1

2

l

Z

0

di

Z

0

2(pxiui+psivi+pniwi)dsidx

+

di

Z

0(qxiui+qsivi+qniwi)x=0

+ (qxiui+qsivi+qniwi)x=ldsi)(10.2.9)

After some steps considering the Eqs. (10.2.4) and (10.2.6) Eq. (10.2.9) leads to

Π

=∑

(i)(1

2

l

Z

0

di

Z

0A11iu′2

i+2A12iu′

iv•

i+2A16iu′

i(u•

i+v′

i)

+A22iv•

i

2+2A26iv•

i(u•

i+v′

i) + A66iu•

i+v′

i2

−2B11iu′

iw′′

i−2B12iu′

iw••

i−2B12iv•

iw′′

i

−4B16iu′

iw′

i•−2B16i(u•

i+v′

i)w′′

i−2B22iv•

iw••

i−4B26iv•

iw′

i•

−2B26i(u•

i+v′

i)w••

i−4B66i(u•

i+v′

i)w′

i•

+D11iw′′2

i+2D12iw′′

iw••

i+4D16iw′′

iw′

i•(10.2.10)

+D22iw••

i

2+4D26iw••

iw′

i•+4D66iw′

i•2

−2(pxiui+psivi+pniwi)dsidx

376 10 Modelling and Analysis of Thin-walled Folded Structures

−

di

Z

0

[(qxiui+qsivi+qniwi)|x=0+ (qxiui+qsivi+qniwi)|x=l]dsi)

10.2.3 Reduction of the Two-dimensional Problem

Equation (10.2.10) represents the complete folded structure model, because it con-

tains all the energy terms of the membrane stress state and of the bending/torsional

stress state under the validity of the Kirchhoff hypotheses. An analytical solution of

this model equations is really impossible with the exception of some very simple

cases. Therefore here we will take another way. As the main object of this section

we will ﬁnd approximate solutions by reducing the two-dimensional problem to an

one-dimensional one taking into account the so-called Kantorovich separation rela-

tionships (Sect. 2.2).

For the displacements ui,vi,wiin the ith strip we write the approximative series

solutions

ui(x,si) = ∑

(j)

Uj(x)

ϕ

i j (si) = U

U

UT

ϕ

=

ϕ

TU

U

U,

vi(x,si) = ∑

(k)

Vk(x)

ψ

ik(si) = V

V

VT

ψ

=

ψ

TV

V

V,

wi(x,si) = ∑

(k)

Vk(x)

ξ

ik(si) = V

V

VT

ξ

=

ξ

TV

V

(10.2.11)

Here the

ϕ

i j(si),

ψ

ik(si),

ξ

ik(si)are a priori given trial functions of the co-ordinates

siand Uj(x),Vk(x)unknown coefﬁcient functions of the longitudinal co-ordinate x.

Vlasov deﬁned the

ϕ

i j(si),

ψ

ik(si),

ξ

ik(si)as the generalized co-ordinate functions

and the Uj(x),Vk(x)as the generalized displacement functions. Of course it is very

important for the quality of the approximate solution, what kind and which number

of generalized co-ordinates

ϕ

i j (si),

ψ

ik(si),

ξ

ik(si)are used.

Now we consider a closed thin-walled cross-section, e.g. the cross-section of a

box-girder, and follow the Vlasov’s hypotheses:

•The out-of-plane displacements ui(si)are approximately linear functions of si. In

this case there are n∗linear independent trial functions

ϕ

i j.n∗is the number of

parallel strip edge lines of cross-section.

•The strains

ε

si(si)can be neglected, i.e.

ε

si≈0. The trial functions

ψ

ik(si)are

then constant functions in all strips and we have n∗∗ linear independent

ψ

ik(si)

and

ξ

ik(si)with n∗∗ =2n∗−m∗.m∗is the number of strips of the thin-walled

structure and n∗is deﬁned above.

The generalized co-ordinate functions can be obtained as unit displacement states

in longitudinal direction (

ϕ

)and in transversal directions (

ψ

,

ξ

). Usually however

generalized co-ordinate functions are used, which allow mechanical interpretations.

In Fig. 10.4, e.g., the generalized co-ordinate functions for a one-cellular rectangu-

lar cross-section are shown.

ϕ

1characterizes the longitudinal displacement of the

10.2 Generalized Beam Models 377

y

z

x

w1u

v1

u

w2

v2u

v3

w3

u

w4

v4

d1

d4

d3

d2

t1

t2

t3

t4

1

a

b

bc

ϕ

1

ϕ

2

ϕ

3

ϕ

4

ψ

1

ψ

2

ψ

3

ψ

4

1

11

ξ

1

ξ

2

ξ

3

ξ

4

d1=d3=dS

d2=d4=dG

a=dS/2,b=dG/2

c=dSdG/4

Fig. 10.4 Generalized coordinate functions of an one-cellular rectangular cross-section

378 10 Modelling and Analysis of Thin-walled Folded Structures

whole cross-section,

ϕ

2and

ϕ

3its rotations about the global y- and z-axes.

ϕ

1,

ϕ

2,

ϕ

3represent the plane cross-section displacements, while

ϕ

4shows its warping.

ψ

2and

ψ

3characterize the plan cross-section displacements in z- and y-direction

and

ψ

1the rotation of the rigid cross-section about the system axis x.

ψ

4deﬁnes a

cross-sectional contour deformation, e.g. a distorsion. The generalized co-ordinate

functions

ξ

1,

ξ

2,

ξ

3,

ξ

4represent displacements of the strips corresponding to

ψ

1,

ψ

2,

ψ

3,

ψ

4. For the example of a box-girder cross-section there is n∗=4, m∗=4

and n∗∗ =8−4=4.

In the following more general derivations the strains

ε

siwill be included, we will

take into account more complicated forms of warping functions and therefore there

are no restrictions for the number of generalized co-ordinate functions. After the

input of Eq. (10.2.11) into (10.2.10) and with the deﬁnition of the 28 matrices

ˆ

A

A1=∑

(i)

di

Z

0

A11i

ϕ

ϕϕ

ϕ

Tdsi,ˆ

A

A2=∑

(i)

di

Z

0

A16i

ϕ

•

ϕ

Tdsi,

ˆ

A

A3=∑

(i)

di

Z

0

A66i

ϕ

•

ϕ

•Tdsi,ˆ

A

A4=∑

(i)

di

Z

0

A66i

ψ

ψψ

ψ

Tdsi,

ˆ

A

A5=∑

(i)

di

Z

0

A26i

ψ

•

ψ

Tdsi,ˆ

A

A6=∑

(i)

di

Z

0

A22i

ψ

•

ψ

•Tdsi,

ˆ

A

A7=∑

(i)

di

Z

0

D11i

ξ

ξξ

ξ

Tdsi,ˆ

A

A8=∑

(i)

di

Z

0

D16i

ξ

•

ξ

Tdsi,

ˆ

A

A9=∑

(i)

di

Z

0

D66i

ξ

•

ξ

•Tdsi,ˆ

A

A10 =∑

(i)

di

Z

0

D12i

ξ

••

ξ

Tdsi,

ˆ

A

A11 =∑

(i)

di

Z

0

D26i

ξ

••

ξ

•Tdsi,ˆ

A

A12 =∑

(i)

di

Z

0

D22i

ξ

••

ξ

••Tdsi,

ˆ

A

A13 =∑

(i)

di

Z

0

A16i

ϕ

ϕψ

ψ

Tdsi,ˆ

A

A14 =∑

(i)

di

Z

0

A12i

ϕ

ϕψ

ψ

•Tdsi,(10.2.12)

ˆ

A

A15 =∑

(i)

di

Z

0

A66i

ϕ

•

ψ

Tdsiˆ

A

A16 =∑

(i)

di

Z

0

A26i

ϕ

•

ψ

•Tdsi,

ˆ

A

A17 =∑

(i)

di

Z

0

B11i

ϕ

ϕξ

ξ

Tdsi,ˆ

A

A18 =∑

(i)

di

Z

0

B16i

ϕ

ϕξ

ξ

•Tdsi,

10.2 Generalized Beam Models 379

ˆ

A

A19 =∑

(i)

di

Z

0

B12i

ϕ

ϕξ

ξ

••Tdsi,ˆ

A

A20 =∑

(i)

di

Z

0

B16i

ϕ

•

ξ

Tdsi,

ˆ

A

A21 =∑

(i)

di

Z

0

B66i

ϕ

•

ξ

•Tdsi,ˆ

A

A22 =∑

(i)

di

Z

0

B26i

ϕ

•

ξ

••Tdsi,

ˆ

A

A23 =∑

(i)

di

Z

0

B16i

ψ

ψξ

ξ

Tdsi,ˆ

A

A24 =∑

(i)

di

Z

0

B66i

ψ

ψξ

ξ

•Tdsi,

ˆ

A

A25 =∑

(i)

di

Z

0

B26i

ψ

ψξ

ξ

••Tdsi,ˆ

A

A26 =∑

(i)

di

Z

0

B12i

ψ

•

ξ

Tdsi,

ˆ

A

A27 =∑

(i)

di

Z

0

B26i

ψ

•

ξ

•Tdsi,ˆ

A

A28 =∑

(i)

di

Z

0

B22i

ψ

•

ξ

••Tdsi

and the load vectors

f

fx=∑

(i)

di

Z

0

pxi

ϕ

dsi,r

r

rx=∑

(i)

di

Z

0

qxi

ϕ

dsi,

f

fs=∑

(i)

di

Z

0

psi

ψ

dsi,r

r

rs=∑

(i)

di

Z

0

qsi

ψ

dsi,

f

fn=∑

(i)

di

Z

0

pni

ξ

dsi,r

r

rn=∑

(i)

di

Z

0

qni

ξ

dsi

(10.2.13)

the potential energy in matrix form is obtained as follows

Π

=1

2

l

Z

0U

U

U′Tˆ

A

A1U

U

U′+V

V

VTˆ

A

A6V

V

V+U

U

UTˆ

A

A3U

U

U+2U

U

UTˆ

A

A15V

V

V′

+V

V

V′Tˆ

A

A4V

V

V′+V

V

V′′Tˆ

A

A7V

V

V′′ +V

V

VTˆ

A

A12V

V

V+4V

V

V′Tˆ

A

A9V

V

V′

+2U

U

U′Tˆ

A

A14V

V

V+2U

U

UTˆ

A

A2U

U

U′+2U

U

U′Tˆ

A

A13V

V

V′−2U

U

U′Tˆ

A

A17V

V

V′′

−2U

U

U′Tˆ

A

A19V

V

V−4U

U

U′Tˆ

A

A18V

V

V′+2U

U

U′Tˆ

A

A16V

V

V+2V

V

VTˆ

A

A5V

V

V′

−2V

V

VTˆ

A

A26V

V

V′′ −2V

V

VTˆ

A

A28V

V

V−4V

V

VTˆ

A

A27V

V

V′−2U

U

UTˆ

A

A20V

V

V′′

−2V

V

V′Tˆ

A

A23V

V

V′′ −2U

U

UTˆ

A

A22V

V

V−2V

V

V′Tˆ

A

A25V

V

V−4U

U

UTˆ

A

A21V

V

V′

−4V

V

V′Tˆ

A

A24V

V

V′+2V

V

VTˆ

A

A10V

V

V′′ +4V

V

V′Tˆ

A

A8V

V

V′′ +4V

V

VTˆ

A

A11V

V

V′

−2(U

U

UTf

f

fx+V

V

VTf

f

fs+V

V

VTf

f

fn)dx

−(U

U

UTr

r

rx+V

V

VTr

r

rs+V

V

VTr

r

rn)|x=0−(U

U

UTr

r

rx+V

V

VTr

r

rs+V

V

VTr

r

rn)|x=l

(10.2.14)

380 10 Modelling and Analysis of Thin-walled Folded Structures

The variation of the potential energy function (10.2.14) and using

∂Π

∂

U

U−d

dx

∂Π

∂

U

U′=0

0

0,

∂Π

∂

V

V−d

dx

∂Π

∂

V

V′+d2

dx2

∂Π

∂

V

V′′ =0

0

0,

δ

U

UT

∂Π

∂

U

U′x=0,l

=0,

δ

V

VT

∂Π

∂

V

V′−d

dx

∂Π

∂

V

V′′ x=0,l

=0,

δ

V

V′T

∂Π

∂

V

V′′ x=0,l

=0

(10.2.15)

leads to a system of matrix differential equations and matrix boundary conditions of

the complete thin-walled folded plate structure

−ˆ

A

A1U

U

U′′ + ( ˆ

A

A2−ˆ

A

AT

2)U

U

U′+ˆ

A

A3U

U

+ˆ

A

A17V

V

V′′′ −(ˆ

A

A13 −2ˆ

A

A18 +ˆ

A

A20)V

V

V′′

+(−ˆ

A

A14 +ˆ

A

A15 +ˆ

A

A19 −2ˆ

A

A21)V

V

V′+ ( ˆ

A

A16 −ˆ

A

A22)V

V

V=f

f

fx

−ˆ

A

AT

17U

U

U′′′ −(ˆ

A

AT

13 −2ˆ

A

AT

18 +ˆ

A

AT

20)U

U

U′′

+( ˆ

A

AT

14 −ˆ

A

AT

15 −ˆ

A

AT

19 +2ˆ

A

AT

21)U

U

U′+ ( ˆ

A

AT

16 −ˆ

A

AT

22)U

U

+ˆ

A

A7V

V

V′′′′ + (2ˆ

A

AT

8−2ˆ

A

A8+ˆ

A

A23 −A

A

AT

23)V

V

V′′′

−(ˆ

A

A4+4ˆ

A

A9−ˆ

A

A10 −ˆ

A

AT

10 −4ˆ

A

A24 +ˆ

A

A26 +ˆ

A

AT

26)V

V

V′′

+( ˆ

A

A5−ˆ

A

AT

5+2ˆ

A

A11 −2ˆ

A

AT

11 +ˆ

A

A25 −ˆ

A

AT

25 −2ˆ

A

A27 +2ˆ

A

AT

27)V

V

V′

+( ˆ

A

A6+ˆ

A

A12 −2ˆ

A

A28)V

V

V=f

f

fs+f

f

fn

(10.2.16)

δ

U

UT[ˆ

A

A1U

U

U′+ˆ

A

AT

2U

U

U−ˆ

A

A17V

V

V′′ + ( ˆ

A

A13 −2ˆ

A

A18)V

V

V′

+( ˆ

A

A14 −ˆ

A

A19)V

V

V±r

r

rx]x=0,l=0

δ

V

VT[ˆ

A

AT

17U

U

U′′ + ( ˆ

A

AT

13 −2ˆ

A

AT

18 +ˆ

A

AT

20)U

U

U′+ ( ˆ

A

AT

15 −2ˆ

A

AT

21)U

U

−ˆ

A

A7V

V

V′′′ + (2ˆ

A

A8−2ˆ

A

AT

8−ˆ

A

A23 +ˆ

A

AT

23)V

V

V′′

+( ˆ

A

A4+4ˆ

A

A9−ˆ

A

AT

10 −4ˆ

A

A24 +ˆ

A

AT

26)V

V

V′

+( ˆ

A

AT

5+2ˆ

A

AT

11 −ˆ

A

A25 −2ˆ

A

AT

27)V

V

V±r

r

rs±r

r

rn]x=0,l=0

δ

V

V′T[−ˆ

A

AT

17U

U

U′−ˆ

A

AT

20U

U

+ˆ

A

A7V

V

V′′ + (2ˆ

A

AT

8−ˆ

A

AT

23)V

V

V′+ ( ˆ

A

AT

10 −ˆ

A

AT

26)V

V

V]x=0,l=0

(10.2.17)

In Eqs. (10.2.17) the upper sign (+) is valid for the boundary x=0 of the structure

and the lower one (−) for the boundary x=l. This convention is also valid for all

following simpliﬁed models.

10.2 Generalized Beam Models 381

10.2.4 Simpliﬁed Structural Models

Starting from the complete folded structure model two ways of derivation simpliﬁed

models are usual:

•Neglecting of special terms in the potential energy function of the complete

folded plate structure.

•Restrictions of the cross-section kinematics by selection of special generalized

co-ordinate functions.

For the ﬁrst way we will consider the energy terms caused by

•the longitudinal curvatures

κ

xi,

•the transversal strains

ε

si,

•the shear deformations of the mid-planes

ε

xsiand

•the torsional curvatures

κ

xsi

in the strips. But not all possibilities for simpliﬁed models shall be taken into ac-

count. We will be restricted the considerations to:

A a structure model with neglected longitudinal curvatures

κ

xionly,

B a structure model with neglected longitudinal curvatures

κ

xiand neglected tor-

sional curvatures

κ

xsi,

C a structure model with neglected longitudinal curvatures

κ

xiand neglected

transversal strains

ε

si,

D a structure model with neglected longitudinal curvatures

κ

xi, neglected transver-

sal strains

ε

siand torsional curvatures

κ

xsi, and

E a structure model with neglected longitudinal curvatures

κ

xi, neglected transver-

sal strains

ε

siand neglected shear strain

ε

xsiof the mid-planes of the strips.

In Fig. 10.5 is g iven an overview on the development of structural simpliﬁed models.

10.2.4.1 Structural Model A

The starting point is the potential energy equation (10.2.10), in which all terms

containing w′′

ihave to vanish. Together with (10.2.11) and (10.2.12) we ﬁnd that in

this case

ˆ

A

A7=0

0

0,ˆ

A

A8=0

0

0,ˆ

A

A10 =0

0

0,ˆ

A

A17 =0

0

0,ˆ

A

A20 =0

0

0,ˆ

A

A23 =0

0

0,ˆ

A

A26 =0

0

The matrix differential equations (10.2.16) and the boundary conditions (10.2.17)

change then into

382 10 Modelling and Analysis of Thin-walled Folded Structures

κ

si6=0,

ε

si6=0,

ε

xsi6=0,

κ

xsi6=0

κ

xi=0

ε

si6=0

ε

si=0

ε

xsi6=0

ε

xsi=0

ε

xsi6=0

ε

xsi=0

κ

xsi6=0

κ

xsi=0

κ

xsi6=0

κ

xsi=0

κ

xsi6=0

A B C D E

Fig. 10.5 Overview to the derivation of usual simpliﬁed models for thin-walled folded plate struc-

tures

−ˆ

A

A1U

U

U′′ + ( ˆ

A

A2−ˆ

A

AT

2)U

U

U′+ˆ

A

A3U

U

U−(ˆ

A

A13 −2ˆ

A

A18)V

V

V′′

+(−ˆ

A

A14 +ˆ

A

A15 +ˆ

A

A19 −2ˆ

A

A21)V

V

V′+ ( ˆ

A

A16 −ˆ

A

A22)V

V

V=f

f

fx,

−(ˆ

A

AT

13 −2ˆ

A

AT

18)U

U

U′′ + ( ˆ

A

AT

14 −ˆ

A

AT

15 −ˆ

A

AT

19 +2ˆ

A

AT

21)U

U

U′

+( ˆ

A

AT

16 −ˆ

A

AT

22)U

U

U−(ˆ

A

A4+4ˆ

A

A9−4ˆ

A

A24)V

V

V′′

+( ˆ

A

A5−ˆ

A

AT

5+2ˆ

A

A11 −2ˆ

A

AT

11 +ˆ

A

A25 −ˆ

A

AT

25 −2ˆ

A

A27 +2ˆ

A

AT

27)V

V

V′

+( ˆ

A

A6+ˆ

A

A12 −2ˆ

A

A28)V

V

V=f

f

fs+f

f

fn,

(10.2.18)

δ

U

UThˆ

A

A1U

U

U′+ˆ

A

AT

2U

U

U+ ( ˆ

A

A13 −2ˆ

A

A18)V

V

V′

+( ˆ

A

A14 −ˆ

A

A19)V

V

V±r

r

rxix=0,l=0,

δ

V

VTh(ˆ

A

AT

13 −2ˆ

A

AT

18)U

U

U′+ ( ˆ

A

AT

15 −2ˆ

A

AT

21)U

U

+( ˆ

A

A4+4ˆ

A

A9−4ˆ

A

A24)V

V

V′

+( ˆ

A

AT

5+2ˆ

A

AT

11 −ˆ

A

A25 −2ˆ

A

AT

27)V

V

V±r

r

rs±r

r

rnix=0,l=0

(10.2.19)

10.2 Generalized Beam Models 383

10.2.4.2 Structural Model B

Here the longitudinal curvatures

κ

xiand the torsional curvatures

κ

xsiare neglected

and therefore in the potential energy additionally to w′′

i≈0 in model A all terms

containing w′

i•have to vanish. Additionally to the case of model A now also the ma-

trices ˆ

A

A9,ˆ

A

A11,ˆ

A

A18,ˆ

A

A21,ˆ

A

A24,ˆ

A

A27 are null-matrices. This leads to the following matrix

differential equations and the corresponding boundary conditions:

−ˆ

A

A1U

U

U′′ + ( ˆ

A

A2−ˆ

A

AT

2)U

U

U′+ˆ

A

A3U

U

U−ˆ

A

A13V

V

V′′

+(−ˆ

A

A14 +ˆ

A

A15 +ˆ

A

A19)V

V

V′+ ( ˆ

A

A16 −ˆ

A

A22)V

V

V=f

f

fx,

−ˆ

A

AT

13U

U

U′′ + ( ˆ

A

AT

14 −ˆ

A

AT

15 −ˆ

A

AT

19)U

U

U′+ ( ˆ

A

AT

16 −ˆ

A

AT

22)U

U

U−ˆ

A

A4V

V

V′′

+( ˆ

A

A5−ˆ

A

AT

5+ˆ

A

A25 −ˆ

A

AT

25)V

V

V′+ ( ˆ

A

A6+ˆ

A

A12 −ˆ

A

A28)V

V

V=f

f

fs+f

f

fn,

(10.2.20)

δ

U

UThˆ

A

A1U

U

U′+ˆ

A

AT

2U

U

U+ˆ

A

A13V

V

V′+ ( ˆ

A

A14 −ˆ

A

A19)V

V

V±r

r

rxix=0,l=0,

δ

V

VThˆ

A

AT

13U

U

U′+ˆ

A

AT

15U

U

U+ˆ

A

A4V

V

V′+ ( ˆ

A

AT

5−ˆ

A

A25)V

V

V±r

r

rs±r

r

rnix=0,l=0

(10.2.21)

10.2.4.3 Structural Model C

In this structure model the longitudinal curvatures

κ

xiand the strains

ε

siare ne-

glected. Therefore in this case in the potential energy function (10.2.10) all terms

containing w′′

iand v•

ihave to vanish. Considering the equations (10.2.11) and

(10.2.12) we ﬁnd, that additionally to the case of the structure model A here the

matrices ˆ

A

A5,ˆ

A

A6,ˆ

A

A14,ˆ

A

A16,ˆ

A

A27,ˆ

A

A28 are null-matrices and in this way we obtain the

following matrix differential equations and the corresponding boundary conditions:

−ˆ

A

A1U

U

U′′ + ( ˆ

A

A2−ˆ

A

AT

2)U

U

U′+ˆ

A

A3U

U

U−(ˆ

A

A13 −2ˆ

A

A18)V

V

V′′

+( ˆ

A

A15 +ˆ

A

A19 −2ˆ

A

A21)V

V

V′−ˆ

A

A22V

V

V=f

f

fx,

−(ˆ

A

AT

13 −2ˆ

A

AT

18)U

U

U′′ + (−ˆ

A

AT

15 −ˆ

A

AT

19 +2ˆ

A

AT

21)U

U

U′−ˆ

A

AT

22U

U

−(ˆ

A

A4+4ˆ

A

A9−4ˆ

A

A24)V

V

V′′

+(2ˆ

A

A11 −2ˆ

A

AT

11 +ˆ

A

A25 −ˆ

A

AT

25)V

V

V′+ˆ

A

A12V

V

V=f

f

fs+f

f

fn,

(10.2.22)

δ

U

UThˆ

A

A1U

U

U′+ˆ

A

AT

2U

U

U+ ( ˆ

A

A13 −2ˆ

A

A18)V

V

V′−ˆ

A

A19V

V

V±r

r

rxix=0,l=0,

δ

V

VTh(ˆ

A

AT

13 −2ˆ

A

AT

18)U

U

U′+ ( ˆ

A

AT

15 −2ˆ

A

AT

21)U

U

+( ˆ

A

A4+4ˆ

A

A9−4ˆ

A

A24)V

V

V′+ (2ˆ

A

AT

11 −ˆ

A

A25)V

V

V±r

r

rs±r

r

rnix=0,l=0

(10.2.23)

384 10 Modelling and Analysis of Thin-walled Folded Structures

10.2.4.4 Structural Model D

This structure model neglects the longitudinal curvatures

κ

xi, the strains

ε

siand

the torsional curvatures

κ

xsi. In the potential energy function all terms containing

w′′

i,v•

i,w′

i•have to vanish and we ﬁnd together with (10.2.11) and (10.2.12) that

additionally to the structure model C the matrices ˆ

A

A9,ˆ

A

A11,ˆ

A

A18,ˆ

A

A21,ˆ

A

A24 are null-

matrices. We obtain the matrix differential equations and the boundary conditions

in the following form:

−ˆ

A

A1U

U

U′′ + ( ˆ

A

A2−ˆ

A

AT

2)U

U

U′+ˆ

A

A3U

U

−ˆ

A

A13V

V

V′′ + ( ˆ

A

A15 +ˆ

A

A19)V

V

V′−ˆ

A

A22V

V

V=f

f

fx,

−ˆ

A

AT

13U

U

U′′ + (−ˆ

A

AT

15 −ˆ

A

AT

19)U

U

U′−ˆ

A

AT

22U

U

−ˆ

A

A4V

V

V′′ + ( ˆ

A

A25 −ˆ

A

AT

25)V

V

V′+ˆ

A

A12V

V

V=f

f

fs+f

f

fn,

(10.2.24)

δ

U

UThˆ

A

A1U

U

U′+ˆ

A

AT

2U

U

U+ˆ

A

A13V

V

V′−ˆ

A

A19V

V

V±r

r

rxix=0,l=0,

δ

V

VThˆ

A

AT

13U

U

U′+ˆ

A

AT

15U

U

U+ˆ

A

A4V

V

V′−ˆ

A

A25V

V

V±r

r

rs±r

r

rnix=0,l=0(10.2.25)

10.2.4.5 Structural Model E

Now the longitudinal curvatures

κ

xi, the transversal strains

ε

siand the shear strains

ε

xsiof the mid-planes shall be neglected. Therefore in the potential energy function

all the terms containing w′′

iand v•

ivanish again. The neglecting of the shear strains

of the mid-planes leads with Eq. (10.2.6) to

ε

xsi=

∂

ui

∂

si

+

∂

vi

∂

x=u•

i+v′

i=0

and we can see that the generalized displacement functionsU

U

U(x)and V

V

V(x)and also

the generalized co-ordinate functions

ϕ

(si)and

ψ

(si)are no more independent from

each other

u•

i=−v′

i,U

U

UT

ϕ

•=−V

V

V′T

ψ

,

ϕ

•=

ψ

,U

U

U=−V

V

V′,U

U

U′=−V

V

V′′,U

U

U′′ =−V

V

V′′′ (10.2.26)

Therefore the potential energy function must be reformulated before the variation

of U

U

Uand V

V

V. Considering the vanishing terms w′′

i,v•

iand (10.2.26) we obtain the

potential energy function in the following form:

10.2 Generalized Beam Models 385

Π

=1

2

l

Z

0V

V

V′′Tˆ

A

A1V

V

V′′ +V

V

VTˆ

A

A12V

V

V+2V

V

V′′Tˆ

A

A19V

V

V+4V

V

V′′Tˆ

A

A18V

V

V′

+4V

V

V′Tˆ

A

A9V

V

V′+4V

V

VTˆ

A

A11V

V

V′−2(U

U

UTf

f

fx+V

V

VTf

f

fs+V

V

VTf

f

fn)dx

−(U

U

UTr

r

rx+V

V

VTr

r

rs+V

V

VTr

r

rn)|x=0−(U

U

UTr

r

rx+V

V

VTr

r

rs+V

V

VTr

r

rn)|x=l

(10.2.27)

The variation of the potential energy, see also (10.2.15), leads to the matrix differ-

ential equations and the boundary conditions for the structural model E

ˆ

A

A1V

V

V′′′′ +2(ˆ

A

A18 −ˆ

A

AT

18)V

V

V′′′ + (−4ˆ

A

A9+ˆ

A

A19 +ˆ

A

AT

19)V

V

V′′

+2(ˆ

A

A11 −ˆ

A

AT

11)V

V

V′+ˆ

A

A12V

V

V=f

f

f′

x+f

f

fs+f

f

fn,

(10.2.28)

δ

V

VTh−ˆ

A

A1V

V

V′′′ +2(−ˆ

A

A18 +ˆ

A

AT

18)V

V

V′′ + (4ˆ

A

A9−ˆ

A

A19)V

V

V′

+2ˆ

A

AT

11V

V

V+f

f

fx±r

r

rx±r

r

rnx=0,l=0,

δ

V

V′Thˆ

A

A1V

V

V′′ +2ˆ

A

A18V

V

V′+ˆ

A

A19V

V

V∓r

r

rxix=0,l=0

(10.2.29)

10.2.4.6 Further Special Models by Restrictions of the Cross-Section

Kinematics

All the ﬁve given simpliﬁed structure models include the neglecting of the longitu-

dinal curvatures

κ

xiin the strips. Because in the case of a beam shaped thin-walled

structure the inﬂuence

κ

xion the deformation state and the stresses of the whole

structure can be seen as very small, its neglecting is vindicated here. The main ad-

vantage of the given ﬁve simpliﬁed structure models however is that by neglecting

the longitudinal curvatures in the strips we have a decreasing of the order of deriva-

tions of the generalized displacement functions U

U

Uand V

V

Vin the potential energy.

This is an important effect for practical solution strategies of the model equations.

The structure models A and C can be used for the analysis of thin-walled beam

shaped structures with open or closed cross-sections. The difference exists only in

the including or neglecting of the strains

ε

siin the strips. Usually they can be ne-

glected, if we have not temperature loading or concentrated transversal stiffeners in

the analyzed structure. The structure models B and D are valid only for structures

with closed cross-sections, because there the torsional curvatures and the torsional

moments Mxsiare very small. The use of the structure model E is vindicated only

for beam shaped structures with open cross-sections. There the shear strains of the

mid-planes of the plate strips have only small inﬂuence on the displacements and

the stress state of the structure how in opposite to the case of a closed cross-section.

Further for each of the ﬁve considered models we can develop model variants

restricting the cross-section kinematics by selection of special sets of generalized

co-ordinate functions

ϕ

,

ψ

,

ξ

. For example, in the structure model B the number of

generalized co-ordinate functions is unlimited. In model D in contrast the number

386 10 Modelling and Analysis of Thin-walled Folded Structures

of

ψ

,

ξ

-co-ordinates is limited to n∗∗, see Sect. 10.2.3. Restricting in this model ad-

ditionally the

ϕ

-co-ordinates to n∗, the semi-moment shell theory for an anisotropic

behavior of the strips is obtained. The Eqs. (10.2.24) and (10.2.25) stay unchanged.

A symmetric stacking sequence in all the strips leads for this model to a further

simpliﬁcation, because then we have no coupling between stretching and bending

in the strips, all the elements of the coupling matrix B

B

Bvanish and therefore all the

matrices ˆ

A

A17 −ˆ

A

A28 are null-matrices. In this special case the following matrix dif-

ferential equations and boundary conditions are valid

−ˆ

A

A1U

U

U′′ + ( ˆ

A

A2−ˆ

A

AT

2)U

U

U′+ˆ

A

A3U

U

U−ˆ

A

A13V

V

V′′ +ˆ

A

A15V

V

V′=f

f

fx,

−ˆ

A

AT

13U

U

U′′ −ˆ

A

AT

15U

U

U′−ˆ

A

A4V

V

V′′ +ˆ

A

A12V

V

V=f

f

fs+f

f

fn,

(10.2.30)

δ

U

UThˆ

A

A1U

U

U′+ˆ

A

AT

2U

U

U+ˆ

A

A13V

V

V′±r

r

rxx=0,l=0,

δ

V

VTˆ

A

AT

13U

U

U′+ˆ

A

AT

15U

U

U+ˆ

A

A4V

V

V′±r

r

rs±r

r

rnix=0,l=0(10.2.31)

If we have symmetric cross-ply laminates in all the plate strips and one of the main

axes of them is identical with the global x-axis, there is no stretching/shearing or

bending/twisting coupling and therefore additionally to the B

B

B-matrix the elements

A16 =A26 =0,D16 =D26 =0. With these the matrices ˆ

A

A2,ˆ

A

A13, are null-matrices,

Eq. (10.2.12). In this case the Eqs. (10.2.30) and (10.2.31) lead to

−ˆ

A

A1U

U

U′′ +ˆ

A

A3U

U

U+ˆ

A

A15V

V

V′=f

f

fx,

−ˆ

A

A15U

U

U′−ˆ

A

A4V

V

V′′ +ˆ

A

A12V

V

V=f

f

fs+f

f

fn,(10.2.32)

δ

U

UThˆ

A

A1U

U

U′±r

r

rxix=0,l=0,

δ

V

VThˆ

A

AT

15U

U

U+ˆ

A

A4V

V

V′±r

r

rs±r

r

rnix=0,l=0(10.2.33)

and we ﬁnd that we have equations of the same type how in case of the classical

semi-moment shell theory of Vlasov. Here however the matrices ˆ

A

Aconsider the

anisotropic behavior of the strips.

Otherwise if starting from the structure model E then the generalized co-ordinate

functions

ψ

,

ξ

are restricted to three co-ordinates, representing the rigid cross-

section, i.e. (

ψ

1,

ξ

1), the rotation of the cross-section about the x-axis, (

ψ

2,

ξ

2) and

(

ψ

3,

ξ

3) the displacements in the global y- and z-direction. If we additionally assume

only four

ϕ

-functions, then the ﬁrst three representing the plane cross-section, i.e.

ϕ

1the displacement in x-direction,

ϕ

2,

ϕ

3the rotations about the y- and z-axes and

ϕ

4is a linear warping function, the so-called unit warping function according to

the sectorial areas law. Therefore we have a structural model similar the classical

Vlasov beam model. The difference is only that in the classical Vlasov beam model

isotropic material behavior is assumed and here the anisotropic behavior of the plate

strips is considered. Note that for comparison of these both models usually the fol-

lowing correlations should be taken into account:

10.2 Generalized Beam Models 387

U1(x) = u(x)displacements of the plane cross-section in x-direction,

U2(x) =

ϕ

y=w′(x)rotation of the plane cross-section about the y-axis,

U3(x) =

ϕ

z=v′(x)rotation of the plane cross-section about the z-axis,

U4(x) =

ωθ

′(x)the warping with

ω

as the unit warping function,

V1(x) =

θ

(x)rotation of the rigid cross-section about the x-axis,

V2(x) = v(x)displacement of the rigid cross-section in y-direction,

V3(x) = w(x)displacement of the rigid cross-section in z-direction

More details about the equations shall not be given here.

If we in this anisotropic Vlasov beam model suppress the warping of the

cross-section and use only the functions

ϕ

1,

ϕ

2,

ϕ

3representing the plane cross-

section kinematics and do not take into account the torsion, we obtain a special-

ized Bernoulli beam model for laminated beams with thin-walled cross-sections

and anisotropic material behavior. In a similar way a specialized Timoshenko beam

model can be obtained, if we restrict the generalized co-ordinate functions

ϕ

ito the

three functions for the plane cross-section kinematics and take into account only

the two

ψ

-co-ordinates for the displacements in y- and z-direction, here however

starting from the structural model D.

The both above discussed quasi beam models are specialized for structures with

cross-sections consisting of single thin plate strips without any rule of their arrange-

ment in the cross-section. The curvatures

κ

xiare generally neglected. The special-

ized beam equations described above cannot be compared directly with the beam

equations in Chap. 7 because the derivation there is not restricted to thin-walled

folded plate cross-sections.

10.2.5 An Efﬁcient Structure Model for the Analysis of General

Prismatic Beam Shaped Thin-walled Plate Structures

Because in the following only beam shaped thin-walled structures are analyzed the

neglecting of the inﬂuence of the longitudinal curvatures

κ

xiin the single plate

strips is vindicated. How it was mentioned above, we have in this case a decreasing

of the order of derivations of the generalized displacement functions in the potential

energy, and this is very important for the solution procedures.

The selected structure model shall enable the analysis of thin-walled structures

with open, closed and mixed open/closed cross-sections. Because the inﬂuence of

κ

xsiis small only for closed cross-sections but the inﬂuence of

ε

xsican be neglected

for open cross-sections only, the selected structure model for general cross-sections

have to include the torsional curvatures and the shear strains of the mid-planes.

The strains

ε

sihave in the most cases only a small inﬂuence and could be ne-

glected generally. But we shall see in Chap. 11 that including

ε

siin the model equa-

tions leads an effective way to deﬁne the shape functions for special ﬁnite elements

and therefore also the

ε

siare included in the selected structure model.

388 10 Modelling and Analysis of Thin-walled Folded Structures

Summarizing the above discussion the structure model A is selected as an uni-

versal model for the modelling and analysis of beam shaped thin-walled plate struc-

tures. An extension of the equations to eigen-vibration problems is given in Sect.

10.2.6.

10.2.6 Free Eigen-Vibration Analysis, Structural Model A

Analogous to the static analysis, the eigen-vibration analysis also shall be restricted

to global vibration response. Local vibrations, e.g. vibration of single plates, are

excluded. A structure model neglecting the longitudinal curvatures cannot describe

local plate strip vibrations. Further only free undamped vibrations are considered.

The starting point is the potential energy function, Eq. (10.2.10), but all terms

including

κ

xiare neglected. With the potential energy

Π

(ui,vi,wi)

Π

=∑

(i)

1

2

l

Z

0

di

Z

0A11iu′2

i+2A12iu′

iv•

i+2A16iu′

i(u•

i+v′

i)

+A22iv•

i

2+2A26iv•

i(u•

i+v′

i) + A66i(u•

i+v′

i)2

−2B12iu′

iw••

i−4B16iu′

iw′

i•−2B22iv•

iw••

i−4B26iv•

iw′

i•

−2B26i(u•

i+v′

i)w••

i−4B66i(u•

i+v′

i)w′

i•

+D22iw••

i

2+4D26iw••

iw′

i•+4D66iw′

i•2dsidx

(10.2.34)

and the kinetic energy T(ui,vi,wi)

T(u

u

u) = 1

2∑

(i)

l

Z

0

di

Z

0

ρ

iti"

∂

ui

∂

t2

+

∂

vi

∂

t2

+

∂

wi

∂

t2#dsidx,(10.2.35)

where

ρ

iis the average density of the ith plate strip

ρ

i=1

ti

n

∑

k=1

ρ

(k)

it(k)

i(10.2.36)

Because we have thin plate strips only, rotational terms of the kinetic energy can be

neglected.

The reduction of the two-dimensional problem is carried out again with the gen-

eralized co-ordinate functions

ϕ

(si),

ψ

(si),

ξ

(si), but we must remark that the gener-

alized displacement functions U

U

U,V

V

Vare time-dependent and therefore they are writ-

ten in the following with a tilde. The reduction relationships are

10.2 Generalized Beam Models 389

ui(x,si,t) = ∑

(j)

˜

Uj(x,t)

ϕ

i j (si) = ˜

U

UT

ϕ

=

ϕ

T˜

U

U,

vi(x,si,t) = ∑

(k)

˜

Vk(x,t)

ψ

ik(si) = ˜

V

VT

ψ

=

ψ

T˜

V

V,

wi(x,si,t) = ∑

(k)

˜

Vk(x,t)

ξ

ik(si) = ˜

V

VT

ξ

=

ξ

T˜

V

(10.2.37)

Additionally to the ˆ

A

Amatrices equation (10.2.12) the following matrices are deﬁned:

ˆ

B

B1=∑

(i)

di

Z

0

ρ

iti

ϕ

ϕϕ

ϕ

Tdsi,ˆ

B

B2=∑

(i)

di

Z

0

ρ

iti

ψ

ψψ

ψ

Tdsi,ˆ

B

B3=∑

(i)

di

Z

0

ρ

iti

ξ

ξξ

ξ

Tdsi(10.2.38)

and we obtain the so-called Lagrange function L=T−

Π

L=∑

(i)

1

2

l

Z

0˙

˜

U

UTˆ

B

B1˙

˜

U

U+˙

˜

V

VTˆ

B

B2˙

˜

V

V+˙

˜

V

VTˆ

B

B3˙

˜

V

−(˜

U

U′Tˆ

A

A1˜

U

U′+2˜

U

U′Tˆ

A

A14 ˜

V

V+2˜

U

UTˆ

A

A2˜

U

U′+2˜

U

U′Tˆ

A

A13 ˜

V

V′

+˜

V

VTˆ

A

A6˜

V

V+2˜

U

UTˆ

A

A16 ˜

V

V+2˜

V

VTˆ

A

A5˜

V

V′+˜

U

UTˆ

A

A3˜

U

+2˜

U

UTˆ

A

A15 ˜

V

V′+˜

V

V′Tˆ

A

A4˜

V

V′−2˜

U

U′Tˆ

A

A19 ˜

V

V−4˜

U

U′Tˆ

A

A18 ˜

V

V′

−2˜

V

VTˆ

A

A28 ˜

V

V−4˜

V

VTˆ

A

A27 ˜

V

V′−2˜

U

UTˆ

A

A22 ˜

V

V−2˜

V

V′Tˆ

A

A25 ˜

V

−4˜

U

UTˆ

A

A21 ˜

V

V′−4˜

V

V′Tˆ

A

A24 ˜

V

V′

+˜

V

VTˆ

A

A12 ˜

V

V+4˜

V

VTˆ

A

A11 ˜

V

V′+4˜

V

V′Tˆ

A

A9˜

V

V′)dx

(10.2.39)

The time derivations of the generalized displacement functions are written with the

point symbol

˙

˜

U

U=

∂

˜

U

∂

t,˙

˜

V

V=

∂

˜

V

∂

t(10.2.40)

The Hamilton principle yields the variational statement

δ

t2

Z

t1

Ldt=

t2

Z

t1

δ

Ldt=0,L=L(x,t,˜

U

U,˜

V

V,˜

U

U′,˜

V

V′,˙

˜

U

U,˙

˜

V

V)(10.2.41)

and we obtain two differential equations

∂

L

∂

˜

U

U−d

dx

∂

L

∂

˜

U

U′−d

dt

∂

L

∂

˙

˜

U

U=0

0

0,

∂

L

∂

˜

V

V−d

dx

∂

L

∂

˜

V

V′−d

dt

∂

L

∂

˙

˜

V

V=0

0

(10.2.42)

390 10 Modelling and Analysis of Thin-walled Folded Structures

If further harmonic relationships for the generalized displacement functions are as-

sumed

˜

U

U(x,t) = U

U

U(x)sin

ω

0t,˜

V

V(x,t) = V

V

V(x)sin

ω

0t(10.2.43)

and we obtain after some steps the following matrix differential equations:

−ˆ

A

A1U

U

U′′ + ( ˆ

A

A2−ˆ

A

AT

2)U

U

U′+ ( ˆ

A

A3−

ω

2

0ˆ

B

B1)U

U

U−(ˆ

A

A13 −2ˆ

A

A18)V

V

V′′

+(−ˆ

A

A14 +ˆ

A

A15 +ˆ

A

A19 −2ˆ

A

A21)V

V

V′+ ( ˆ

A

A16 −ˆ

A

A22)V

V

V=0

0

0,

−(ˆ

A

AT

13 −2ˆ

A

AT

18)U

U

U′′ + ( ˆ

A

AT

14 −ˆ

A

AT

15 −ˆ

A

AT

19 +2ˆ

A

AT

21)U

U

U′

+( ˆ

A

AT

16 −ˆ

A

AT

22)U

U

U−(ˆ

A

A4+4ˆ

A

A9−4ˆ

A

A24)V

V

V′′

+( ˆ

A

A5−ˆ

A

AT

5+2ˆ

A

A11 −2ˆ

A

AT

11 +ˆ

A

A25 −ˆ

A

AT

25 −2ˆ

A

A27 +2ˆ

A

AT

27)V

V

V′

+( ˆ

A

A6+ˆ

A

A12 −2ˆ

A

A28 −

ω

2

0ˆ

B

B2−

ω

2

0ˆ

B

B3)V

V

V=0

0

0,

(10.2.44)

δ

U

UT[ˆ

A

A1U

U

U′+ˆ

A

AT

2U

U

U+ ( ˆ

A

A13 −2ˆ

A

A18)V

V

V′+ ( ˆ

A

A14 −ˆ

A

A19)V

V

V]x=0,l=0

δ

V

VT[( ˆ

A

AT

13 −2ˆ

A

AT

18)U

U

U′+ ( ˆ

A

AT

15 −2ˆ

A

AT

21)U

U

U+ ( ˆ

A

A4+4ˆ

A

A9−4ˆ

A

A24)V

V

V′

+( ˆ

A

AT

5+2ˆ

A

AT

11 −ˆ

A

A25 −2ˆ

A

AT

27)V

V

V]x=0,l=0

(10.2.45)

With these equations given above the global free vibration analysis of prismatic

beam shaped thin-walled plate structures can be done sufﬁcient exactly.

10.3 Solution Procedures

Two general kinds of solution procedures may be taken into account

•analytic solutions and

•numerical solutions

The consideration below distinguish exact and approximate analytical solution pro-

cedures. In the ﬁrst case an exact solution of the differential equations is carried

out. In the other case, the variational statement of the problem is, e.g., solved by

the Ritz or Galerkin method, and in general, the procedures yield in an approximate

analytical series solution.

Numerical solution procedures essentially consist of methods outgoing from the

differential equation or from the corresponding variational problem. The numeri-

cal solutions of differential equations may include such methods as ﬁnite difference

methods, Runge3-Kutta4methods and transfer matrix methods. The main represen-

tative for the second way is the ﬁnite element method (FEM). After a few remarks

3Carl David Tolm´e Runge (∗30 August 1856 Bremen – †3 January 1927 G¨ottingen) - mathemati-

cian and physicist

4Martin Wilhelm Kutta (∗3 November 1867 Pitschen – †25 December 1944 F¨urstenfeldbruck) -

mathematician

10.3 Solution Procedures 391

in Sect. 10.3.1 about analytic solution possibilities for the here considered prob-

lems, the numerical solution procedure using the transfer matrix method is consid-

ered in detail in Sect. 10.3.2. The application of the FEM and the development of

special one-dimensional ﬁnite elements for beam shaped thin-walled structures are

discussed in Chap. 11.

10.3.1 Analytical Solutions

For the generalized beam models given in Sect. 10.2 only for simpliﬁed special

cases analytical solutions are possible. If we use, for example, the structure model

D in connection with a symmetric cross-ply stacking in all plates, what means

that the differential equations are from the same type as in case of the isotropic

semi-moment shell theory of Vlasov, analytical solutions can be developed for spe-

cial cross-sections geometry. It is very useful to choose orthogonal generalized co-

ordinate functions

ϕ

,

ψ

,

ξ

, because it yields the possibility of decomposition of the

system of differential equations into some uncoupled partial systems. For example,

the generalized co-ordinate functions

ϕ

in Fig. 10.4 are completely orthogonal and

in this way the matrix ˆ

A

A1is a diagonal matrix. Therefore some couplings between

the single differential equations vanish.

A suitable method for construction an exact solution is the Krylow5method or

the so-called method of initial parameters. The ﬁrst step for the application of this

method is to convert the system of differential equations into an equivalent differen-

tial equation of n-th order.

L[y(x)] =

n

∑

ν

=0

a

ν

y(

ν

)(x) = r(x)(10.3.1)

Its homogeneous solution shall be written as

yh(x) = y(0)K1(x) + y′(0)K2(x) + ...+y(n−1)(0)Kn(x)(10.3.2)

The free constants of the solution are expressed by the initial constants, i.e. the

function y(x)and its derivatives till the (n−1)th order at x=0. A particular solution

can be obtained with

yp(x) =

x

Z

0

Kn(x−t)r(t)dt(10.3.3)

or in case that r0(x)is not deﬁned for x≥0 but for x≥x0

5Aleksei Nikolajewitsch Krylow (∗3 August 1863jul./15 August 1863greg.Wisjaga - †26 October

1945 Leningrad) - naval engineer, applied mathematician

392 10 Modelling and Analysis of Thin-walled Folded Structures

yp(x) = 



x>x0

x

Z

x0

Kn(x−t)r0(t)dt(10.3.4)

Equation (10.3.4) is a quasi closed analytical solution for the differential equation

of the structure model D and different functions ri(x)for respectively x>xi,i=

0,1,...,n

y(x) = y(0)K1(x) + y′(0)K2(x) + ...+y(n−1)(0)Kn(x)

+



x>x0

x

Z

x0

Kn(x−t)r0(t)dt+



x>x1

x

Z

x1

Kn(x−t)r1(t)dt(10.3.5)

+



x>x2

x

Z

x2

Kn(x−t)r2(t)dt+...

Complete closed analytical solutions for isotropic double symmetric thin-walled

box-girders and general loads one can ﬁnd in (Altenbach et al, 1994). Also analytical

solution for a two-cellular box-girder including shear lag effects is given there. But

in the majority of engineering applications refer to general laminated thin-walled

structures, an analytical solution has to be ruled out.

10.3.2 Transfer Matrix Method

The differential equations and their boundary conditions are the starting point of a

numerical solution by transfer matrix method. At ﬁrst the system of higher order dif-

ferential equations has to transfer into a system of differential equations of ﬁrst order

using the natural boundary conditions as deﬁnitions of generalized cross-sectional

forces.

For sake of simplicity this solution method shall be demonstrated for the structure

model D and for a symmetric cross-ply stacking in all plates of the structure. Then

the following system of differential equations and boundary conditions are valid,

see also Sect. 10.2.4.6 and Eqs. (10.2.32) and (10.2.33)

ˆ

A

A1U

U

U′′ −ˆ

A

A3U

U

U−ˆ

A

A15V

V

V′+f

f

fx=0

0

0,

ˆ

A

AT

15U

U

U′+ˆ

A

A4V

V

V′′ −ˆ

A

A12V

V

V+f

f

fs+f

f

fn=0

0

0,(10.3.6)

δ

U

UThˆ

A

A1U

U

U′±r

r

rxix=0,l=0,

δ

V

VThˆ

A

AT

15U

U

U+ˆ

A

A4V

V

V′±r

r

rs±r

r

rnix=0,l=0(10.3.7)

Equations (10.3.7) leads to the deﬁnitions of the generalized cross-sectional forces,

i.e. generalized longitudinal forces and transverse forces.

10.3 Solution Procedures 393

P

P=ˆ

A

A1U

U

U′,(10.3.8)

Q

Q=ˆ

A

AT

15U

U

U+ˆ

A

A4V

V

V′,(10.3.9)

It can be shown that we have with Eqs. (10.3.8), (10.3.9) really the deﬁnitions of

generalized forces

P

P=∑

(i)

di

Z

0

ϕ

ϕσ

xitidsi=∑

(i)

di

Z

0

ϕ

Nxidsi,

Q

Q=∑

(i)

di

Z

0

ψ

ψτ

xsitidsi=∑

(i)

di

Z

0

ψ

Nxsidsi

In the here considered structure model we have only membrane stresses

σ

xiand

τ

xsi

because the longitudinal curvatures and longitudinal bending moments are neglected

in all plates. Additional with cross-ply stacking are A16 =A26 =0. Therefore and

with Eqs. (10.2.4), (10.2.6), (10.2.11) we can write

Nxi=A11i

ε

xi=A11iu′

i=A11i

ϕ

TU

U

U′,

Nxsi=A66i

ε

xsi=A66i(u•

i+v′

i) = A66i(

ϕ

•TU

U

U+

ψ

TV

V

V′)

Considering (10.2.12) we obtain again the deﬁnitions of the generalized forces given

in (10.3.8) and (10.3.9)

P

P=∑

(i)

di

Z

0

A11i

ϕ

ϕϕ

ϕ

TdsiU

U

U′=ˆ

A

A1U

U

U′,

Q

Q=∑

(i)

di

Z

0

A66i(

ψ

ψϕ

ϕ

•TU

U

U+

ψ

ψψ

ψ

TV

V

V′)dsi=ˆ

A

AT

15U

U

U+ˆ

A

A4V

V

V′

The inversion of the Eqs. (10.3.8) and (10.3.9) leads to

U

U′=ˆ

A

A−1

1P

P

P,(10.3.10)

V

V′=−ˆ

A

A−1

4ˆ

A

AT

15U

U

U+ˆ

A

A−1

4Q

Q

Q(10.3.11)

With the ﬁrst derivativesof Eqs. (10.3.8) and (10.3.9) and after the input of (10.3.10)

and (10.3.11) into (10.3.6) we obtain the following system of differential equations

of ﬁrst order

U

U′=ˆ

A

A−1

1P

P

P,

V

V′=−ˆ

A

A−1

4ˆ

A

AT

15U

U

U+ˆ

A

A−1

4Q

Q

Q,

P

P′= ( ˆ

A

A3−ˆ

A

A15 ˆ

A

A−1

4ˆ

A

AT

15)U

U

U+ˆ

A

A15 ˆ

A

A−1

4Q

Q

Q−f

f

fx,

Q

Q′=ˆ

A

A12V

V

V−f

f

fs−f

f

fn

(10.3.12)

respectively written in matrix notation

394 10 Modelling and Analysis of Thin-walled Folded Structures







U

V

P

Q

1







′

=





0

0 0

0

0ˆ

A

A−1

10

0

0o

o

−ˆ

A

A−1

4ˆ

A

AT

15 0

0

0 0

0

0ˆ

A

A−1

4o

o

(ˆ

A

A3−ˆ

A

A15 ˆ

A

A−1

4ˆ

A

AT

15)0

0

0 0

0

0ˆ

A

A15 ˆ

A

A−1

4−f

f

fx

0

0ˆ

A

A12 0

0

0 0

0

0−f

f

fs−f

f

fn

o

oTo

o

oTo

o

oTo

o

oT0













U

V

P

Q

1







y

y′=B

B

By

y

y(10.3.13)

B

Bis the system matrix and y

y

ythe so-called state vector containing all the generalized

displacement functions U

U

Uand V

V

Vand all the generalized forces P

P

Pand Q

Q

Q. 0

0

0 and o

o

oin

the B

B

B-matrix are null matrices and vectors.

The next step is a discretization of the one-dimensional problem, see Fig. 10.6.

Between the state vectors at the point j+1 and the point jwe have generally the

relationship

y

yj+1=W

W

Wjy

y

yj(10.3.14)

where W

W

Wjis the transfer matrix for the structure section j−(j+1).

A ﬁrst order differential equation

y′(x) = b y(x),b=const

has the solution

y(x) = Cebx

and with

y(x0) = Cebx0→C=y(x0)e−bx0,

we obtain

y(x) = y(x0)eb(x−x0)

In the same way the solution of the matrix differential equations is

y

y′(x) = B

B

By

y

y(x),

y

y(x) = y

y

y(x0)eB(x−x0)

∆

j

✲✛

x=x0x=xN

0N

j−1jj+1

Fig. 10.6 Discretization of the one-dimensional structure

10.3 Solution Procedures 395

and we ﬁnd that eB

B

B(x−x0)can be deﬁned as the transfer matrix from the point x0to x.

Therefore the transfer matrix between two points xjand xj+1generally is obtained

as

W

Wj=eB

B

B(xj+1−xj)(10.3.15)

The numerical calculation of transfer matrices can be carried out by series develop-

ment of the exponential function

W

Wj=I

I

I+△jB

B

B+△2

j

2! B

B

B2+△3

j

3! B

B

B3+... (10.3.16)

and also by using a Runge-Kutta method

W

Wj=I

I

I+△j

6(M

M

M1j+2M

M

M2j+2M

M

M3j+M

M

M4j),(10.3.17)

M

M1j=B

B

B(xj),

M

M2j=B

B

B(xj+1

2△j)(I

I

I+1

2△jM

M

M1j),

M

M3j=B

B

B(xj+1

2△j)(I

I

I+1

2△jM

M

M2j),

M

M4j=B

B

B(xj+△j)(I

I

I+△jM

M

M3j)

In both equations I

I

Iare unit matrices of the same rank as the system matrix. The

boundary conditions of the problem can be expressed by a matrix equation

y

y0=A

A

Ax

x

x∗(10.3.18)

Here A

A

Ais the so-called start matrix containing the boundary conditions at x=x0

and x

x

x∗is the vector of the unknown boundary values there. In the last column of the

start matrix the known boundary values are included. For the unknown boundary

values the last column elements are zero, and by a unit in the corresponding row

the unknown value is associated with an element of the unknown vector x

x

x∗. For

example, in Eq. (10.3.19) a start matrix is shown in case of a free structure end, it

means all the displacements are unknown and all forces are given.

396 10 Modelling and Analysis of Thin-walled Folded Structures







U1

...

Um

V1

...

Vn

P

1

...

P

m

Q1

...

Qn

1





0

=







1 0

... 0

1 0

... 0

1 0

P

10

...

P

m0

Q10

...

Qn0

1













x∗

1

...

x∗

m+n

1







,

y

y0=A

A

A0x

x

x∗

(10.3.19)

Now the multiplications with the transfer matrices can be carried out over all sec-

tions (xj,xj+1)until x=xN. With the equation

S

Sy

y

yN=0

0

0 (10.3.20)

the boundary conditions are formulated at x=xN.S

S

Sis the so-called end matrix con-

taining in its last column the negative values of the given displacements or forces. A

unit in an other column of each row yields the association to an element of the state

vector y

y

yN. Equation (10.3.21) shows the end matrix for a clamped end, where all

displacements are given. This matrix equation leads to a system of linear equations

for the unknowns in the vector x

x

x∗







1... −U1N

... ... ...

1... −UmN

1... −V1N

... ... ...

1... −VnN













U1

...

Um

V1

...

Vn

P

1

...

P

m

Q1

...

Qn

1





N

=







0

...

0

...

0







(10.3.21)

The real parts of the eigenvalues of the system matrix B

B

Blead to numerical instable

solutions especially for long beam structures. From the mechanical point of view it

means that the inﬂuence of the boundary conditions of both structure ends to each

other are very low and with this we have a nearly singular system of linear equa-

10.3 Solution Procedures 397

tions. For the consolidation of this problem intermediate changes of the unknowns

are carried out, by formulation of a new start matrix A

A

Aat such an intermediate point.

Usually the generalized displacements are chosen as the new unknowns. The fol-

lowing equations show the general procedure schedule

y

y0=A

A

A0x

x

x∗

0

y

y1=W

W

W0y

y

y0=W

W

W0A

A

A0x

x

x∗

0

...

y

yi=W

W

Wi−1W

W

Wi−2...W

W

W0A

A

A0x

x

x∗

0=F

F

Fix

x

x∗

0

y

yi=A

A

A1x

x

x∗

1ﬁrst change of unknowns

y

yi+1=W

W

Wiy

y

yi=W

W

WiA

A

A1x

x

x∗

1

...

y

yj=F

F

Fjx

x

x∗

l−1

y

yj=A

A

Alx

x

x∗

llth change of unknowns

y

yj+1=W

W

WjA

A

Alx

x

x∗

l

...

y

yk=F

F

Fkx

x

x∗

n−1

y

yk=A

A

Anx

x

x∗

nnth change of unknowns

y

yk+1=W

W

WkA

A

Anx

x

x∗

n

... (10.3.22)

y

yN=F

F

FNx

x

x∗

n

S

Sy

y

yN=0

0

0 system of linear equations for the solution of the unknowns x∗

n

The multiplications of the state vector y

y

y0with transfer matrices are carried out until

the ﬁrst intermediate change of unknowns. The product of the transfer matrices and

the start matrix makes the matrix F

F

Fi. A new unknown vector x

x

x∗

1is deﬁned by the new

start matrix A

A

A1and this procedure is repeated at the following intermediate points.

General for the l-th intermediate change of unknowns the Eq. (10.3.23) is current

F

Fjx

x

x∗

l−1=A

A

Alx

x

x∗

l(10.3.23)

With a segmentation of the state vector y

y

yjinto the sub-vectors y

y

yvfor the displace-

ments and y

y

ykfor the forces

398 10 Modelling and Analysis of Thin-walled Folded Structures

y

yj=





U

V

P

Q

1





j

=



y

yv

y

yk

1

j

,(10.3.24)

we obtain a separated form of Eq. (10.3.23)





F

F1jf

f

f1j

F

F2jf

f

f2j

o

oT1

˜

x

x∗

l−1

1=



A

A1la

a

a1l

A

A2la

a

a2l

o

oT1

˜

x

x∗

l

1(10.3.25)

With the assumption that the displacements are the new unknowns we ﬁnd that the

sub-matrix A

A

A1lis a unit matrix and the sub-vector a

a

a1lis a null vector

A

A1l=I

I

I,a

a

a1l=o

o

o(10.3.26)

This leads to

F

F1j˜

x

x∗

l−1+f

f

f1j=˜

x

x∗

l,(10.3.27)

˜

x

x∗

l−1=F

F

F−1

1j(˜

x

x∗

l−f

f

f1j),x

x

x∗

l−1=F

F

F−1

1j(˜

x

x∗

l−f

f

f1j)

1(10.3.28)

and than the second equation of (10.3.25) yields the structure of the new start matrix

F

F2jF

F

F−1

1j(˜

x

x∗

l−f

f

f1j) + f

f

f2j=A

A

A2l˜

x

x∗

l+a

a

a2l,

A

A2l=F

F

F2jF

F

F−1

1j,a

a

a2l=f

f

f2j−F

F

F2jF

F

F−1

1jf

f

f1j,(10.3.29)

A

Al=



I

I o

o

F

F2jF

F

F−1

1jf

f

f2j−F

F

F2jF

F

F−1

1jf

f

f1j

o

oT1

(10.3.30)

At such an intermediate change point it is also possible to consider the introduction

of concentrated generalized forces or the disposition of supports with given general-

ized displacements. Than the new start matrix must be modiﬁed additionally, in the

ﬁrst case by a modiﬁcation of the sub-vector a

a

a2land in the second case by consid-

eration of the jump behavior of the forces at this point. But more details about this

shall not be given here.

With the end matrix S

S

Sand the end state vector y

y

yNthe relationship S

S

Sy

y

yNyields

a system of linear equations for the last unknown vector x

x

x∗

nand after this all the

unknown vectors can be calculated by repeatedly using Eq. (10.3.28).

The transfer matrix method with intermediate changes of the unknown state vec-

tors yields in contrast to the classical transfer method a numerical stable procedure

also for long beam structures. From the mechanical point of view correspond each

intermediate change x=x∗

la substitution of the structure section 0 ≤x≤x∗

lby

generalized elastic springs.

10.4 Problems 399

The transfer matrix procedure is also applicable to the analysis of eigen-

vibrations. There we have a modiﬁed system matrix B

B

Bcontaining frequency de-

pendent terms

y

y′=B

B

B(

ω

0)y

y

y(10.3.31)

Therefore, the transfer matrices can be calculated only with assumed values for the

frequencies. The end matrix leads to a homogenous system of linear equations. Its

coefﬁcient determinant must be zero. The assumed frequencies are to vary until this

condition is fulﬁlled sufﬁciently.

The transfer matrix method with numerical stabilization was applied successfully

to several isotropic thin-walled box-beam structures. The structure model D con-

sidered above has for a symmetrical cross-ply stacking of all plates an analogous

mathematical model structure as isotropic semi-moment shell structures. Therefore,

the procedure can be simply transferred to such laminated thin-walled beam struc-

tures. An application to other structure models, Sect. 10.2, is in principle possible

but rather expansive and not efﬁcient.

The development and application of special ﬁnite elements and their implemen-

tation in a FEM-program system is more generally and more efﬁciently. FEM will

be discussed in detail in Chap. 11.

10.4 Problems

Exercise 10.1. Establish the system of differential equations for the box-girder with

a rectangular cross-section, which is shown in Fig. 10.4. It shall be supposed that its

dimensions are symmetric to both axes and therefore we have here

t1=t3=tS,

t2=t4=tG,

d1=d3=dS,

d2=d4=dG,

Further we have a cross-ply stacking in all the plate strips. The stiffness of both

horizontally arranged strips (index G) are the same, but they are different from the

stiffness of the vertically arranged strips (index S), what means that

A11 1 =A11 3 =A11 S ,

A11 2 =A11 4 =A11 G ,

A66 1 =A66 3 =A66 S ,

A66 2 =A66 4 =A66 G ,

D22 1 =D22 3 =D22 S,

D22 2 =D22 4 =D22 G

For the calculation of this box girder the simpliﬁed structure model D shall be used

and because we have cross-ply stacking, the Eqs. (10.2.32) are valid

400 10 Modelling and Analysis of Thin-walled Folded Structures

ˆ

A

A1U

U

U′′ −ˆ

A

A3U

U

U−ˆ

A

A15V

V

V′+f

f

fx=0

0

0,

ˆ

A

AT

15U

U

U′+ˆ

A

A4V

V

V′′ −ˆ

A

A12V

V

V+f

f

fs+f

f

fn=0

0

Solution 10.1. At ﬁrst we have to calculate the matrices ˆ

A

A1,ˆ

A

A3,ˆ

A

A4,ˆ

A

A12 and ˆ

A

A15,

their deﬁnitions are given in Eq. (10.2.12)

ˆ

A

A1=∑

(i)

di

Z

0

A11 i

ϕ

ϕϕ

ϕ

Tdsi,

ˆ

A

A3=∑

(i)

di

Z

0

A66 i

ϕ

•

ϕ

•Tdsi,

ˆ

A

A4=∑

(i)

di

Z

0

A66 i

ψ

ψψ

ψ

Tdsi,

ˆ

A

A12 =∑

(i)

di

Z

0

D22 i

ξ

••

ξ

••Tdsi,

ˆ

A

A15 =∑

(i)

di

Z

0

A66 i

ϕ

•

ψ

Tdsi,

The co-ordinate functions

ϕ

,

ψ

,

ξ

are also shown in Fig. 10.4. For solving the inte-

grals to obtain the ˆ

A

A-matrices, the functions

ϕ

,

ψ

,

ξ

must be written as functions of

the co-ordinates siof each strip. In accordance with Fig. 10.4 we ﬁnd

ϕ

1(si) = +1,i=1,2,3,4;

ϕ

•

1(si) = 0,i=1,2,3,4;

ϕ

2(s1) = −dS

21−2s1

dS;

ϕ

•

2(si) =

ψ

2(si),i=1,2,3,4;

ϕ

2(s2) = dS

2;

ϕ

2(s3) = + dS

21−2s3

dS;

ϕ

2(s4) = −dS

2;

ϕ

3(s1) = dG

2;

ϕ

•

3(si) =

ψ

3(si),i=1,2,3,4;

ϕ

3(s2) = + dG

21−2s2

dG;

ϕ

3(s3) = −dG

2;

ϕ

3(s4) = −dG

21−2s4

dG;

10.4 Problems 401

ϕ

4(s1) = + dSdG

41−2s1

dS;

ϕ

•

4(si) =

ψ

4(si),i=1,2,3,4;

ϕ

4(s2) = −dSdG

41−2s2

dG;

ϕ

4(s3) = + dSdG

41−2s3

dS;

ϕ

4(s4) = −dSdG

41−2s4

dG;

ψ

1(s1) = −dG

2;

ψ

2(s1) = +1;

ψ

1(s2) = −dS

2;

ψ

2(s2) = 0;

ψ

1(s3) = −dG

2;

ψ

2(s3) = −1;

ψ

1(s4) = −dS

2;

ψ

2(s4) = 0;

ψ

3(s1) = 0;

ψ

4(s1) = −dG

2;

ψ

3(s2) = −1;

ψ

4(s2) = + dS

2;

ψ

3(s3) = 0;

ψ

4(s3) = −dG

2;

ψ

3(s4) = +1;

ψ

4(s4) = + dS

2;

ξ

1(s1) = −dS

21−2s1

dS;

ξ

••

1(si) = 0,i=1,2,3,4;

ξ

1(s2) = −dG

21−2s2

dG;

ξ

1(s3) = −dS

21−2s3

dS;

402 10 Modelling and Analysis of Thin-walled Folded Structures

ξ

1(s4) = −dG

21−2s4

dG;

ξ

2(s1) = 0;

ξ

••

2(si) = 0,i=1,2,3,4;

ξ

2(s2) = +1;

ξ

2(s3) = 0;

ξ

2(s4) = −1;

ξ

3(s1) = +1;

ξ

••

3(si) = 0,i=1,2,3,4;

ξ

3(s2) = 0;

ξ

3(s3) = −1;

ξ

3(s4) = 0;

Some additional considerations are necessary to determine the functions

ξ

4(si). The

generalized co-ordinate function

ξ

4is corresponding to

ψ

4and represents therefore

a double antisymmetric deﬂection state of the cross-section. The cross-section is

double symmetric in its geometry and in the elastic behavior. Therefore we must

have an antisymmetric function

ξ

4(si)in each strip. It means that the following

conditions are valid

ξ

4(si=0) =

ξ

i0,

ξ

4(si=di) = −

ξ

i0,

ξ

•

4(si=0) =

α

0,

ξ

•

4(si=di) =

α

0,

ξ

••

4(si=0) =

κ

si0,

ξ

••

4(si=di) = −

κ

si0

Supposing a polynomial function of the third order, we can write

ξ

••

4(si) =

κ

si01−2si

di,

ξ

•

4(si) =

κ

si0di"si

di−si

di2#+

α

0,

ξ

4(si) =

κ

si0d2

i

6"3si

di2

−2si

di3#+

α

0di

si

di

+

ξ

i0

The condition

ξ

4(si=di) = −

ξ

i0leads to

α

0di=−

κ

si0d2

i

6+2

ξ

i0

and than we obtain

ξ

4(si) =

κ

si0d2

i

6"3si

di2

−2si

di3

−si

di#+

ξ

i01−2si

di

With the antisymmetric properties mentioned above we ﬁnd

10.4 Problems 403

κ

s10=

κ

s30=

κ

S0,

κ

s20=

κ

s40=

κ

G0

The continuity of the rotation angles at the corners and the equilibrium equation

ξ

•

4(s1=dS) =

ξ

•

4(s2=0),

ξ

•

4(s2=dG) =

ξ

•

4(s3=0),

ξ

•

4(s3=dS) =

ξ

•

4(s4=0),

ξ

•

4(s4=dG) =

ξ

•

4(s1=0),

D22S

κ

S0 =−D22G

κ

G0

lead to the unknown curvatures

κ

S0 =−12D22G

dGD22S +dSD22G

,

κ

G0 = + 12D22S

dGD22S +dSD22G

and we obtain

ξ

4(s1) = −2D22Gd2

S

dGD22S +dSD22G "3s1

dS2

−2s1

dS3

−s1

dS#

+dS

21−2s1

dS,

ξ

4(s2) = 2D22Sd2

G

dGD22S +dSD22G "3s2

dG2

−2s2

dG3

−s2

dG#

−dG

21−2s2

dG,

ξ

4(s3) = −2D22Gd2

S

dGD22S +dSD22G "3s3

dS2

−2s3

dS3

−s3

dS#

+dS

21−2s3

dS,

ξ

4(s4) = 2D22Sd2

G

dGD22S +dSD22G "3s4

dG2

−2s4

dG3

−s4

dG#

−dG

21−2s4

dG,

ξ

••

4(s1) = −12D22G

dGD22S +dSD22G 1−2s1

dS,

ξ

••

4(s2) = + 12D22S

dGD22S +dSD22G 1−2s2

dG,

ξ

••

4(s3) = −12D22G

dGD22S +dSD22G 1−2s3

dS,

ξ

••

4(s4) = + 12D22S

dGD22S +dSD22G 1−2s4

dG

404 10 Modelling and Analysis of Thin-walled Folded Structures

Now all elements of the matrices can be calculated. Here only the calculation of the

element ˆ

A122 of the matrix ˆ

A

A1shall be derived in a detailed manner

ˆ

A111 =

4

∑

i=1

di

Z

0

A11i

ϕ

1(si)

ϕ

1(si)dsi=2(A11SdS+A11GdG),

ˆ

A112 =

4

∑

i=1

di

Z

0

A11i

ϕ

1(si)

ϕ

2(si)dsi=0,

ˆ

A113 =

4

∑

i=1

di

Z

0

A11i

ϕ

1(si)

ϕ

3(si)dsi=0,

ˆ

A114 =

4

∑

i=1

di

Z

0

A11i

ϕ

1(si)

ϕ

4(si)dsi=0,

ˆ

A122 =

4

∑

i=1

di

Z

0

A11i

ϕ

2(si)

ϕ

2(si)dsi

=A11S

d2

S

4

dS

Z

01−2s1

dS2

ds1+A11G

d2

S

4

dG

Z

0

ds2

+A11S

d2

S

4

dS

Z

01−2s3

dS2

ds3+A11G

d2

S

4

dG

Z

0

ds4

=A11S

d3

S

12 +A11G

d2

S

4dG+A11S

d3

S

12 +A11G

d2

S

4dG

=d2

S

6(A11SdS+3A11GdG),

ˆ

A123 =

4

∑

i=1

di

Z

0

A11i

ϕ

2(si)

ϕ

3(si)dsi=0,

ˆ

A124 =

4

∑

i=1

di

Z

0

A11i

ϕ

2(si)

ϕ

4(si)dsi=0,

ˆ

A133 =

4

∑

i=1

di

Z

0

A11i

ϕ

3(si)

ϕ

3(si)dsi=d2

G

6(3A11SdS+A11GdG),

ˆ

A134 =

4

∑

i=1

di

Z

0

A11i

ϕ

3(si)

ϕ

4(si)dsi=0,

ˆ

A144 =

4

∑

i=1

di

Z

0

A11i

ϕ

4(si)

ϕ

4(si)dsi=d2

Sd2

G

24 (A11SdS+A11GdG)

10.4 Problems 405

With all elements ˆ

A1i j we obtain the matrix ˆ

A

A1to

ˆ

A

A1=





A111 0 0 0

0A122 0 0

0 0 A133 0

0 0 0 A14 4







with

A111 =2(A11SdS+A11GdG),

A122 =d2

S

6(A11SdS+3A11GdG),

A133 =d2

G

6(3A11SdS+3A11GdG),

A144 =d2

Sd2

G

24 (A11SdS+3A11GdG)

One can see that the generalized co-ordinate functions

ϕ

iare orthogonal to each

other and therefore the matrix ˆ

A

A1is a diagonal matrix

In the same way the other ˆ

A

Aimatrices are obtained

ˆ

A

A3=





0 0 0 0

0 2A66SdS0 0

002A66GdG0

0 0 0 dSdG

2(A66SdG+3A66GdS)







,

ˆ

A

A4=





A0 0 0

0 2A66SdS0 0

0 0 2A66GdG0

B0 0 A







,

ˆ

A

A12 =





0 0 0 0

dSdG

2(A66SdG−A66GdS)0 0 96D22GD22S

dGD22S+dSD22G







,

ˆ

A

A15 =





0 0 0 0

0 2A66SdS0 0

0 0 2A66GdG0

B0 0 A







with

A=dSdG

2(A66SdG+A66GdS),B=dSdG

2(A66SdG−A66GdS)

Now the system of differential equations can be developed with the help of Eq.

(10.2.32).

406 10 Modelling and Analysis of Thin-walled Folded Structures

2(A11SdS+A11GdG)U′′

1=−fx1,

d2

S

6(A11SdS+3A11GdG)U′′

2−2A66SdS(U2+V′

2) = −fx2,

2A66SdS(U′

2+V′′

2) = −(fs2+fn2),

d2

G

6(3A11SdS+A11GdG)U′′

3−2A66GdG(U3+V′

3) = −fx3,

2A66GdG(U′

3+V′′

3) = −(fs3+fn3),

d2

Sd2

G

24 (A11SdS+A11GdG)U′′

4

−dSdG

2(A66SdG+A66GdS)(U4+V′

4)

−dSdG

2(A66SdG−A66GdS)V′

1=−fx4,

dSdG

2(A66SdG−A66GdS)(U′

4+V′′

4)

+dSdG

2(A66SdG+A66GdS)V′′

1=−(fs1+fn1),

dSdG

2(A66SdG+A66GdS)(U′

4+V′′

4)

+dSdG

2(A66SdG−A66GdS)V′′

1

−96D22GD22S

dGD22S+dSD22G

V4=−(fs4+fn4)

We can see, that the system of differential equations is divided into four decou-

pled partial systems. The ﬁrst equation describes the longitudinal displacement, the

second and the third partial systems represent the bending about the global y- and

z-axes and the fourth - the torsion, the warping and the contour deformation of the

cross-section. An analytic solution of the fourth partial system is more difﬁcult like

the solutions of the ﬁrst three partial systems but it is possible too. The analytical

solution of an analogous system for an isotropic box girder is given in detail by

Vlasov and by the authors of this book, see also the remarks in 10.3.1.

References

Altenbach J, Kissing W, Altenbach H (1994) D¨unnwandige Stab- und Stabschalen-

tragwerke. Vieweg-Verlag, Brauschweig/Wiesbaden

Vlasov VZ (1961) Thin-walled elastic beams. National Science Foundation and De-

partment of Commerce, Arlington, VI

Wlassow WS (1958) Allgemeine Schalentheorie und ihre Anwendung in der Tech-

nik. Akademie-Verlag, Berlin

Part V

Finite Classical and Generalized Beam

Elements, Finite Plate Elements

The ﬁfth part (Chap. 11) presents a short introduction into the ﬁnite element proce-

dures and developed ﬁnite classical and generalized beam elements and ﬁnite plate

elements in the frame of classical and ﬁrst order shear deformation theory. Selected

examples demonstrate the possibilities of ﬁnite element analysis.

Chapter 11

Finite Element Analysis

The Finite Element Method (FEM) is one of the most effective methods for the nu-

merical solution of ﬁeld problems formulated in partial differential equations. The

basic idea of the FEM is a discretization of the continuous structure into substruc-

tures. This is equivalent to replacing a domain having an inﬁnite number of degrees

of freedom by a system having a ﬁnite number of degrees of freedom. The actual

continuum or structure is represented as an assembly of subdivisions called ﬁnite

elements. These elements are considered to be interconnected at speciﬁed joints

which are called nodes. The discretization is deﬁned by the so-called ﬁnite element

mesh made up of elements and nodes.

We assume one-dimensional elements, when one dimension is very large in com-

parison with the others, e.g. truss or beam elements, two-dimensional elements,

when one dimension is very small in comparison with the others, e.g. plate or shell

elements, and volume elements. From the mechanical point of view the nodes are

coupling points of the elements, where the displacements of the coupled elements

are compatible. On the other hand from the mathematical point of view the nodes are

the basic points for the approximate functions of the displacements inside a ﬁnite el-

ement and so at these nodes the displacements are compatible. It must be noted here

that all considerations are restricted to the displacement method. The force method

or hybrid methods are not considered in this book.

An important characteristic of the discretization of a structure is the number of

degrees of freedom. To every node, a number of degrees of freedom will be assigned.

These are nodal constants which usually (but not necessarily) have a mechanical or

more general physical meaning. The number of degrees of freedom per element is

deﬁned by the product of the number of nodes per element and degrees of freedom

per node. The number of degrees of freedom in the structure is the product of the

number of nodes and the number of degrees of freedom per node.

Chapter 11 contains an introduction to the general procedure of ﬁnite element

analysis in a condensed form (Sect. 11.1). For more detailed information see the

vast amount of literature. In Sects. 11.2 and 11.3 the development of ﬁnite beam

elements and ﬁnite plate elements for the analysis of laminate structures is given.

Section 11.4 contains the development of generalized ﬁnite beam elements based on

409

H. Altenbach et al., Mechanics of Composite Structural

Elements, https://doi.org/10.1007/978-981-10-8935-0_11

410 11 Finite Element Analysis

a generalized structure model for beam shaped thin-walled folded structures given

in Sect. 10.2. In Sect. 11.5 the results of some numerical applications show the

inﬂuences of chosen parameters on the behavior of laminate structures.

11.1 Introduction

The principle of the total minimum potential energy and the Hamilton‘s principle

are given in Sect. 2.2.2 in connection with analytical variational approaches, they are

also the theoretical basis of the FEM solutions of elastostatic and of dynamic prob-

lems. In this way we have variational problems. For such problems the Ritz method

may be used as a so-called direct solution method (see Sect. 2.2.3). In the classical

Ritz method the approximation functions are deﬁned for the whole structure, and

so for complex geometries it is difﬁcult to realize the requirements of satisfying

the boundary conditions and of the linear independence and completeness of these

functions.

One way to overcome these difﬁculties is by the discretization of the structure

into a number of substructures, if possible of the same kind (ﬁnite elements). Then

the approximation functions can be deﬁned for the elements only and they must sat-

isfy the conditions of geometrical compatibility at the element boundaries. Because

it is usual to deﬁne different types of ﬁnite elements, we have special types of ap-

proximation functions for each element type. Here the approximation functions are

denoted Ni, the so-called shape functions. They are arranged in a matrix N

N

N, the ma-

trix of the shape functions of the particular element type. The following introduction

to the FEM procedure is given in a general but condensed form and illustrates that

the step-by-step ﬁnite element procedure can be stated as follows:

•Discretization of the structure,

•Selection of a suitable element displacement model,

•Derivation of element stiffness matrices and load vectors,

•Assembly of element equations to obtain the system equations,

•Calculation of the system equations for the unknown nodal displacements,

•Computation of element strains and stresses

11.1.1 FEM Procedure

The starting point for elastostatic problems is the total potential energy given in Eq.

(2.2.28). In accordance with the Ritz method the approximation

˜

u

u(x

x

x) = N

N

N(x

x

x)v

v

v(11.1.1)

is used for the displacement ﬁeld vector u

u

u. Here N

N

Nis the matrix of the shape func-

tions, they are functions of the position vector x

x

x, and v

v

vis the element displacement

11.1 Introduction 411

vector. The matrix N

N

Nhas the same number of rows as the displacement vector u

u

has components and the same number of columns as the element displacement vec-

tor v

v

vhas components. If the element has nKE nodes and the degree of freedom

for each node is nF, the element displacement vector v

v

vcontains nKE subvectors

v

vi,i=1,...,nKE with nFcomponents in each, and so v

v

vhas nKE nFcomponents. The

number of components of the displacement ﬁeld vector u

u

uis nu. Then the structure

of the matrix N

N

Nis generally

N

N= [N1I

I

InuN2I

I

Inu...N˜nI

I

Inu],˜n=nKE nF

nu

(11.1.2)

with I

I

Inuas unit matrices of the size (nu,nu). Therefore the size of N

N

Nis generally

(nu,nKE nF). In dependence on the kind of continuity at the element boundaries, the

so-called C(0)- or C(1)-continuity, see below, two cases can be distinguished. In the

case of C(0)-continuity nFequals nuand therefore ˜nis equal nKE, we have only nKE

shape functions Ni, whereas we can have up to nKE nFshape functions in the case of

C(1)-continuity.

For the stresses and the strains we obtain from (11.1.1)

σ

(x

x

x) = C

C

ε

(x

x

x) = C

C

CD

D

DN

N

N(x

x

x)v

v

v,

ε

(x

x

x) = D

D

Du

u

u(x

x

x) = D

D

DN

N

N(x

x

x)v

v

v=B

B

B(x

x

x)v

v

v(11.1.3)

With the approximation (11.1.1) the total potential energy is a function of all the

nodal displacement components arranged in the element displacement vector v

v

v, e.g.

Π

=

Π

(v

v

v). The variation of the total potential energy

δΠ

=

δ

v

vT



Z

V

B

BTC

C

CB

B

Bv

v

vdV−Z

V

N

NTp

p

pdV−Z

Aq

N

NTq

q

qdA



(11.1.4)

leads with

δΠ

=0 to

δ

v

vT(K

K

Kv

v

v−f

f

fp−f

f

fq) = 0 (11.1.5)

K

Kis the symmetric stiffness matrix with the size (nK E nF,nKE nF)

K

K=Z

V

B

BTC

C

CB

B

BdV(11.1.6)

and f

f

fpand f

f

fqare the vectors of the volume forces and the surface forces

f

fp=Z

V

N

NTp

p

pdV,f

f

fq=Z

Aq

N

NTq

q

qdA(11.1.7)

If the components of

δ

v

vare independent of each other, we obtain from (11.1.5) a

system of linear equations

K

Kv

v

v=f

f

f,f

f

f=f

f

fp+f

f

fq(11.1.8)

412 11 Finite Element Analysis

For elastodynamic problems, we have to consider that forces and displacements are

also dependent on time and the Hamilton‘s principle is the starting point for the

FEM procedure. Assuming again the independence of the components of

δ

v

vthe

matrix equation is

M

M¨

v

v(t) + K

K

Kv

v

v(t) = f

f

f(t)(11.1.9)

for elastic systems without damping effects. M

M

Mis symmetric mass matrix

M

M=Z

V

ρ

N

NTN

N

NdV(11.1.10)

and f

f

f(t)the vector of the time dependent nodal forces. Assuming the damping pro-

portional to the relative velocities, an additional term C

C

CD˙

v

v(t)can be supplemented

formally in Eq. (11.1.9)

M

M¨

v

v(t) +C

C

CD˙

v

v(t) + K

K

Kv

v

v(t) = f

f

f(t),(11.1.11)

where C

C

CDis the damping matrix. C

C

CDhas the same size as the matrices K

K

Kand M

M

and usually it is formulated approximately as a linear combination of K

K

Kand M

M

M. The

factors

α

and

β

can be chosen to give the correct damping at two frequencies

C

CD≈

α

M

M+

β

K

K(11.1.12)

In selecting the shape functions N

N

Ni(x

x

x)it must be remembered that these functions

must be continuous up to the (n−1)th derivative, if we have derivatives of the nth

order in the variational problem, i.e. in the total potential energy or in the Hamil-

ton’s function. In this case only the results of FEM approximations converge to the

real solutions by increasing the number of elements. For more-dimensional ﬁnite

elements in this way it is to realize that the displacements are compatible up to the

(n−1)th derivative at the boundaries of adjacent elements, if they are compatible at

the nodes.

In plane stress or plane strain problems and in general three-dimensional prob-

lems the vector u

u

ucontains displacements only (no rotations) and the differential

operator D

D

Dis of the 1st order. In this way we must only satisfy the displacements

compatibility at the element boundaries that means the so-called C(0)-continuity.

By using beam or plate models especially of the classical Bernoulli beam model

or the classical Kirchhoff plate model, the rotation angles are expressed by deriva-

tives of the displacements of the midline or the midplane and the differential op-

erator D

D

Dis of the second order. Then we have to satisfy the compatibility of dis-

placements and rotations at the element boundaries. In such cases we speak about a

C(1)-continuity and ﬁnding the shape functions Niis more difﬁcult.

Because we have no differential operator in connection with the mass matrix M

M

M,

it would be possible to use other, more simple functions N∗

ifor it. In such a case

the mass matrix would have another population, e.g. a diagonal matrix structure is

possible. Then we speak about a so-called condensed mass matrix, otherwise we

have a consistent mass matrix. By using the condensed mass matrix we have less

11.1 Introduction 413

computational expense than by using the consistent mass matrix, but a decreasing

convergence to the real results is possible.

All equations considered above are only valid for a single element and strictly

they should have an additional index E. For example, we have the inner element

energy

UE=1

2v

v

vT

EZ

VE

B

BTC

C

CB

B

BdVv

v

vE=1

2v

v

vT

EK

K

KEv

v

vE(11.1.13)

with the element stiffness matrix

K

KE=Z

VE

B

BTC

C

CB

B

BdV(11.1.14)

Since the energy is a scalar quantity, the potential energy of the whole structure can

be obtained by summing up the energies of the single elements. Previously a system

displacement vector containing the displacements of all nodes of the whole system

must be deﬁned. By a so-called coincidence matrix L

L

LEthe correct position of each

single element is determ ined. L

L

LEis a Boolean matrix of the size (nKE nF,nKnF) with

nKas the number of nodes of the whole structure.

The element displacement vector v

v

vEis positioned into the system displacement

vector v

v

vby the equation

v

vE=L

L

LEv

v

v(11.1.15)

and we obtain the system equation by summing up over all elements

∑

i

L

LT

iEK

K

KiEL

L

LiE !v

v

v="∑

i

L

LiE (f

f

fiE p +f

f

fiEq )#

K

Kv

v

v=f

f

(11.1.16)

The system stiffness matrix is also symmetric, but it is a singular matrix, if the

system is not ﬁxed kinematically, i.e., we have no boundary conditions constraining

the rigid body motion. After consideration of the boundary conditions of the whole

system, K

K

Kbecomes a positive deﬁnite matrix and the system equation can be solved.

Then with the known displacements v

v

vthe stresses and deformations are calculated

using the element equations (11.1.1) and (11.1.3).

For elastodynamic problems, the system stiffness matrix and the system mass

matrix are obtained in the same manner and we have the system equation

M

M¨

v

v(t) +C

C

CD˙

v

v(t) + K

K

Kv

v

v(t) = f

f

f(t)(11.1.17)

For investigation of the eigen-frequencies of a system without damping harmonic

vibrations are assumed and with

v

v(t) = ˆ

v

vcos(

ω

t+

ϕ

)(11.1.18)

and C

C

CD=0

0

0,f

f

f(t) = o

o

othe matrix eigen-value problem follows

414 11 Finite Element Analysis

(K

K

K−

ω

2M

M

M)v

v

v=o

o

o(11.1.19)

and the eigen-frequencies and the eigen-vectors characterizing the mode shapes can

be calculated.

11.1.2 Problems

Exercise 11.1. A plane beam problem is given. The beam is divided into three plane

two-node beam elements. The number of nodal degrees of freedom is three (u,w,

ϕ

):

1. What size are the element stiffness matrix and the system stiffness matrix before

the consideration of the boundary conditions?

2. Show the coincidence matrix L

L

L2of the second element lying between the nodes

2 and 3!

3. Show the population of the system stiffness matrix and the boundary conditions,

if the beam is ﬁxed at node 1 (cantilever beam)! Do the same as in the previous

case but consider that the beam is simply supported (node 1 is constrained for the

deﬂections uand wand node 4 only for the deﬂection w)!

Solution 11.1. For the plane beam problem one gets

1. With nKE =2 and nF=3 the element stiffness matrix has the size (6,6). Be-

cause we have 4 nodes (nK=4) the size of the system stiffness matrix before the

consideration of the boundary conditions is (12,12).

2. The coincidence matrices in this case have the size (6,12). Because it must be

v

v2=L

L

L2v

v

we obtain the coincidence matrix for the element Nr. 2

L

L2=







0 0 0 1 0 0 0 0 0 0 0 0

0 0 0 0 1 0 0 0 0 0 0 0

0 0 0 0 0 1 0 0 0 0 0 0

0 0 0 0 0 0 1 0 0 0 0 0

0 0 0 0 0 0 0 1 0 0 0 0

0 0 0 0 0 0 0 0 1 0 0 0







3. The system stiffness matrix without consideration of the boundary conditions is

deﬁned by

K

K=

4

∑

i=1

L

LT

iK

K

KiL

L

Li

In this case we obtain the following population of the matrix K

K

11.2 Finite Beam Elements 415

u1w1

ϕ

1u2w2

ϕ

2u3w3

ϕ

3u4w4

ϕ

4







[(v)] [(v)] [(v)] [(v)] [(v)] [(v)] [(0)] [(0)] [(0)] [(0)] [(0)] [(0)]

[(v)] [(v)] (v) (v) (v) (v) (0) (0) (0) (0) (0) (0)

[(v)] [(v)] (v)v+x v +x v +x x x x 0[(0)] 0

[(0)] [(0)] (0)x x x x +z x +z x +z z [z]z

[(0)] [(0)] (0)0 0 0 z z z z [z]z

[(0)] [(0)] [(0)] [(0)] [(0)] [(0)] [z] [z] [z] [z] [z] [z]

[(0)] [(0)] [(0)] 0 0 0 z z z z [z]z







u1

w1

ϕ

1

u2

w2

ϕ

2

u3

w3

ϕ

3

u4

w4

ϕ

4

v- components of the stiffness matrix of element No. 1, x- components of the

stiffness matrix of element No. 2, z- components of the stiffness matrix of ele-

ment No. 3.

Considering the boundary conditions for a cantilever beam clamped at node 1

(u1=0,w1=0,

ϕ

1=0) we have to cancel the ﬁrst three rows and the ﬁrst three

columns in the obtained matrix - characterized by brackets (...). If we have a

simply supported beam with u1=0,w1=0 and w4=0, the ﬁrst two rows and

columns and the row and the column No. 11 must be deleted - characterized by

square brackets [...].

11.2 Finite Beam Elements

A beam is a quasi one-dimensional structure, the dimensions of the cross-section

of it are very small in comparison to its length. The connection of the centers of

the cross-sectional areas is called the midline of the beam. We distinguish between

straight beams and beams with an in-plane or spatial curved midline, respectively.

Here we consider beams with a straight midline only.

Generally such a beam can be loaded by tension/compression, one- or two-axial

bending and torsion. Especially with respect to the use of laminate beams the fol-

lowing investigations are restricted to tension/compression and one-axial bending.

For two-axial bending and torsion, laminate beams are not so predestined.

Laminate beams consist of UD-laminae mostly have a rectangular cross-section

of the dimension b(width) and h(hight) and very often the laminae are arranged

symmetrically to the midline. We will assume this special case for the following

development of ﬁnite laminate beam elements. In this way we have no coupling of

tension and bending and we can divide our considerations into the development of

laminate elements for tension/compression, so-called laminate truss elements, and

laminate beam elements for bending only.

416 11 Finite Element Analysis

11.2.1 Laminate Truss Elements

The laminate truss element is a very simple element. It is assumed to be a straight

structure of the length lwith a constant cross-sectional area A. The nodal degree of

freedom is one - the displacement uin axial direction (Fig. 11.1). In the potential

energy we have only the ﬁrst derivative and so we can use a two-node truss element

with linear shape functions Ni(x1)and Nj(x1), which satisﬁes C(0)-continuity

u(x1) = N

N

Nv

v

vE,v

v

vT

E= [uiuj],N

N

N= [Ni(x1)Nj(x1)] (11.2.1)

The two shape functions (see also Fig. 11.2) are

Ni(x1) = 1−x1

l,Nj(x1) = x1

l(11.2.2)

With the stress resultant

N(x1) = Z

A

σ

dA=A11

ε

1(x1) = A11

du

dx1

,A11 =b

n

∑

k=1

C(k)

11 hk(11.2.3)

and the longitudinal load per length n(x1)the total potential energy can be written

as

Π

(u) = 1

2

l

Z

0

A11u′2dx1−

l

Z

0

n(x1)udx1(11.2.4)

and for the element stiffness matrix we obtain

Fig. 11.1 Laminate truss

element l

x1

n(x1)

uiuj

x1

l

x1

l

0

1.0

0.5

NiNj

Fig. 11.2 Shape functions of the two-node truss element

11.2 Finite Beam Elements 417

K

KE=A11

l

Z

0

N

N′TN

N

N′dx1=A11

l1−1

−1 1 (11.2.5)

The element force vector is deﬁned as

f

fnE =

l

Z

0

N

NTn(x1)dx1

If we assume that n(x1)is a linear function with niand njas the intensities at the

nodes

n(x1) = N

N

Nni

nj

then

f

fnE =

l

Z

0

N

NTN

N

Ndx1ni

nj=l

62 1

1 2 ni

nj(11.2.6)

In case of nodal forces f

f

fPE = [FiFj]T, the vector f

f

fPE must be added to the vector

f

fnE

f

fE=f

f

fnE +f

f

fPE (11.2.7)

The system equation can be obtained in dependence on the structure of the whole

system, deﬁned by a coincidence matrix together with the transformation of all el-

ement equations into a global coordinate system. Considering the boundary con-

ditions, the system equation can be solved and with the known displacements the

stresses can be calculated for each element.

For vibration analysis, the element mass matrix (11.1.10) has to be used

M

ME=Z

V

ρ

N

NTN

N

NdV=

l

Z

0

ρ

N

NTN

N

Ndx1,

ρ

=1

h

n

∑

k=1

ρ

(k)h(k)

All parts of the cross-section have the same translation uand the corresponding

acceleration ¨umultiplied by the distributed mass produces a distributed axial inertia

force. Instead of handling the distributed mass directly, we generate ﬁctitious nodal

masses contained in the consistent mass matrix

M

ME=

ρ

Al

62 1

1 2 (11.2.8)

With the system equation, obtained in the same manner as for elastostatic problems,

the eigen-frequencies and mode shapes can be calculated.

418 11 Finite Element Analysis

11.2.2 Laminate Beam Elements

For the analysis of laminate beams in this book two theories are considered, the

classical laminate theory and the shear deformation theory. The classical laminate

theory is based on the Bernoulli beam model and the shear deformation theory on

the Timoshenko beam model. The Bernoulli beam model neglects the shear strains

in the bending plane and so it seems to be less realistic for the calculation of laminate

beams. Therefore it is better to use the Timoshenko beam model, which includes the

shear strains in a simple form (Chap. 7).

In the following discussion, only the shear deformation theory is used and we

assume a simple rectangular cross-section with a symmetric arrangement of the UD-

laminae. This means that we have no coupling of tension and bending. The main

advantage of the shear deformation theory in comparison with the Bernoulli theory

is that the cross-sectional rotation angle

ψ

is independent of the displacement wand

therefore the differential operator D

D

Din the strain energy is of the 1st order. In this

way we can use elements with C(0)-continuity, and a two-node element with linear

shape functions is possible. The nodal degrees of freedom are 2 (w,

ψ

). In Fig. 11.3

such a two-node beam element is shown. The element displacement vector is

v

vT

E= [wi

ψ

iwj

ψ

j](11.2.9)

For the displacement vector u

u

uthe approximation (11.2.11) is used

u

u(x1) = w(x1)

ψ

(x1)=N

N

Nv

v

vE,(11.2.10)

where the matrix of the shape functions is

N

N=Ni(x1)1 0

0 1 Nj(x1)1 0

0 1  (11.2.11)

with the shape functions (11.2.2), see also Fig. 11.2.

A better element accuracy can be expected, if we consider a three-node element,

as shown in Fig. 11.4. Then the element displacement vector is

v

vT

E= [wi

ψ

iwj

ψ

jwk

ψ

k](11.2.12)

and for the matrix N

N

Nwe obtain

Fig. 11.3 Two-node beam

element l

x1

x3

wiwj

ψ

i

ψ

j

11.2 Finite Beam Elements 419

Fig. 11.4 Three-node beam

element

l/2

x1

x3

wiwjwk

ψ

k

ψ

i

ψ

j

N

N=Ni(x1)1 0

0 1 Nj(x1)1 0

0 1 Nk(x1)1 0

0 1  (11.2.13)

with the shape functions

Ni(x1) = 1−3x1

l+2x2

1

l2,Nj(x1) = 4x1

l−4x2

1

l2,Nk(x1) = −x1

l+2x2

1

l2,(11.2.14)

which are shown in Fig. 11.5. A further increase in the element accuracy can be

achieved with a four-node element, see Fig. 11.6. Here the element displacement

vector and the matrix of the shape functions are

v

vT

E= [wi

ψ

iwj

ψ

jwk

ψ

kwl

ψ

l](11.2.15)

x1

l

x1

l

x1

l

0

1.0

0.5

-0.125

Ni

Nk

Nj

Fig. 11.5 Shape functions of the three-node element

420 11 Finite Element Analysis

Fig. 11.6 Four-node beam

element

x1

x3,w

l

wiwjwkwl

l/3l/3l/3

ψ

i

ψ

j

ψ

k

ψ

l

N

N=Ni(x1)1 0

0 1Nj(x1)1 0

0 1Nk(x1)1 0

0 1Nl(x1)1 0

0 1 (11.2.16)

with the shape functions

Ni=1−11

2

x1

l+9x1

l2−9

2x1

l3,Nj=9x1

l−45

2x1

l2+27

2x1

l3,

Nk=−9

2

x1

l+18 x1

l2−27

2x1

l3,Nl=x1

l−9

2x1

l2+9

2x1

l3

(11.2.17)

which are shown in Fig. 11.7. The three types of beam elements given above show

the possibility of using elements of different accuracy. Of course, using the element

with higher number of nodes means that less elements and a more coarse mesh

can be used, but the calculations of the element stiffness matrices will be more

computationally expensive.

The further relationships are developed formally independent of the chosen num-

ber of element nodes. The element stiffness matrix is obtained with (11.1.6)

K

KE=

l

Z

0

B

BTC

C

CB

B

Bdx1=

l

Z

0

N

NTD

D

DTC

C

CD

D

DN

N

Ndx1(11.2.18)

Here

D

D=





0d

dx1

d

dx1

0



,C

C

C=D11 0

0ksA55 (11.2.19)

with the stiffness

D11 =b

3

n

∑

k=1

C(k)

11 x(k)3

3−x(k−1)3

3,A55 =b

n

∑

k=1

C(k)

55 h(k)(11.2.20)

and the shear correction factor ksgiven in (7.3.20).

For calculation of the element force vector, we assume that the element is loaded

by a distributed transverse load per length q(x1)and we can write the external work

as

WE=

l

Z

0

q(x1)w(x1)dx1=v

v

vT

Ef

f

fE

11.2 Finite Beam Elements 421

and with

w(x1) = [w

ψ

]1

0=u

u

uTR

R

R=v

v

vT

EN

N

NTR

R

R(11.2.21)

the element force vector f

f

fEis obtained

f

fE=

l

Z

0

N

NTR

R

Rq(x1)dx1(11.2.22)

If single nodal forces or moments are acting, they must be added.

The system equation can be obtained in dependence on the structure of the whole

system deﬁned by a coincidence matrix together with the transformation of all el-

ement equations into a global coordinate system. After considering the boundary

conditions, the system equation can be solved. After this the stress resultants are

obtained for all elements

0.00.0

0.20.2

0.40.4

0.60.6

0.80.8

1.01.0

x1

l

x1

l

x1

l

x1

l

−0.2

0.0

0.2

0.4

0.6

0.8

1.0

−0.4

1.2

NiNj

NkNl

i

j

k

l

Fig. 11.7 Shape functions of the four-node element

422 11 Finite Element Analysis

σ

E=M

QE

=C

C

CD

D

DN

N

Nv

v

vE(11.2.23)

For elastodynamic problems the mass matrix must be calculated. The Timoshenko

beam model includes in the general case axial, transversal and rotational inertia

forces and moments. So it must be noted that the laminae of the beam have different

velocities in x1-direction

TE=1

2

l

Z

0

b/2

Z

−b/2

x(n)

3

Z

x(0)

3

ρ

(˙u2+˙w2)dx1dx2dx3,˙u=˙u0+x3˙

ψ

(11.2.24)

After integration with respect to dx2and dx3follow

TE=1

2

l

Z

0

ρ

0(˙u2

0+˙w2) + 2

ρ

1˙u0˙

ψ

+

ρ

2˙

ψ

2dx1(11.2.25)

with the so-called generalized densities

ρ

0=b

n

∑

k=1

ρ

(k)x(k)

3−x(k−1)

3,

ρ

1=b1

2

n

∑

k=1

ρ

(k)x(k)

3

2−x(k−1)

3

2,

ρ

2=b1

3

n

∑

k=1

ρ

(k)x(k)

3

3−x(k−1)

3

3

(11.2.26)

ρ

(k)is the density of the kth lamina.

Because we assumed a symmetric arrangement of the laminae in the cross-

section it follows that

ρ

1=0,˙u0=0

and therefore

TE=1

2

l

Z

0

(

ρ

0˙w2+

ρ

2˙

ψ

2)dx1=1

2

l

Z

0

˙

u

uTR

R

R0˙

u

udx1(11.2.27)

with the matrix R

R

R0

R

R0=

ρ

00

0

ρ

2(11.2.28)

Using (11.1.1)

TE=1

2

l

Z

0

˙

v

vT

EN

N

NTR

R

R0N

N

N˙

v

vEdx1

the element mass matrix M

M

MEis obtained

11.2 Finite Beam Elements 423

M

ME=

l

Z

0

N

NTR

R

R0N

N

Ndx1(11.2.29)

The system equation is established in the same manner as for elastostatic problems,

and with the assumption of harmonic vibrations, the eigen-frequencies and the mode

shapes can be calculated.

11.2.3 Problems

Exercise 11.2. Let us assume a two-node beam element.

1. Calculate the element stiffness matrix for a two-node beam element by analytical

integration!

2. Calculate the element force vector for a two-node beam element, loaded by a

linear distributed transverse load per length q(x1). The intensities at the nodes

are qiand qj!

3. Calculate the element mass matrix for a two-node beam element!

Solution 11.2. The three solutions are;

1. In the case of a two-node beam element the matrix of the shape functions is

N

N=Ni(x1)1 0

0 1 Nj(x1)1 0

0 1 

with Ni(x1) = 1−(x1/l),Nj(x1) = x1/l. The element stiffness matrix is deﬁned

by (11.2.18)

K

K=

l

Z

0

N

NTD

D

DTCDN

CDN

CDNdx1,

where in (11.2.19) are given

D

D=





0d

dx1

d

dx1

0



,C

C

C=¯

D11 0

0ks¯

A55 

with ¯

D11 and ks¯

A55 in according to (11.2.20). After execution the matrix opera-

tions we obtain for the stiffness matrix

424 11 Finite Element Analysis

K

KE=

l

Z

0







ks¯

A55 dNi

dx12

0ks¯

A55

dNi

dx1

dNj

dx1

0

0¯

D11 dNi

dx12

0¯

D11

dNi

dx1

dNj

dx1

ks¯

A55

dNi

dx1

dNj

dx1

0ks¯

A55 dNj

dx12

0

0¯

D11

dNi

dx1

dNj

dx1

0¯

D11 dNj

dx12







dx1

and ﬁnally

K

KE=1

l





ks¯

A55 0−ks¯

A55 0

0¯

D11 0−¯

D11

−ks¯

A55 0ks¯

A55 0

0−¯

D11 0¯

D11







2. The element force vector to calculate with respect to (11.2.22)

f

fE=

l

Z

0

N

NTR

R

Rq(x1)dx1with R

R

R=1

0

For the loading function q(x1)we can write

q(x1) = [Ni(x1)Nj(x1)]qi

qj

and then we ﬁnd

f

fE=

l

Z

0







Ni(x1)2Ni(x1)Nj(x1)

0 0

Ni(x1)Nj(x1)Nj(x1)2

0 0





qi

qjdx1=l

6





2qi+qj

0

qi+2qj

0







3. The element mass matrix for such a two-node beam element we ﬁnd in according

to (11.2.29)

M

ME=

l

Z

0

N

NTR

R

R0N

N

Ndx1,R

R

R0=

ρ

00

0

ρ

2

with the generalized densities

ρ

0and

ρ

2(11.2.26).

Inserting the matrix of the shape functions given above and executing the matrix

operations we obtain

l

Z

0







Ni(x1)2

ρ

00Ni(x1)Nj(x1)

ρ

00

0Ni(x1)2

ρ

20Ni(x1)Nj(x1)

ρ

2

Ni(x1)Nj(x1)

ρ

00Nj(x1)2

ρ

00

0Ni(x1)Nj(x1)

ρ

20Nj(x1)2

ρ

2







dx1

11.3 Finite Plate Elements 425

and after integration the element mass matrix is in this case

M

ME=l

6





2

ρ

00

ρ

00

0 2

ρ

20

ρ

2

ρ

00 2

ρ

00

0

ρ

20 2

ρ

2







11.3 Finite Plate Elements

Plates are two-dimensional structures that means that one dimension, the thickness,

is very small in comparison to the others and in the unloaded state they are plane.

Usually, the midplane between the top and the bottom plate surfaces is deﬁned as

the reference plane and is taken as the plane of x−y. The z-direction corresponds

to the thickness direction. To avoid double indexes in the following relationships in

this section we will use the coordinates x,y,zinstead of x1,x2,x3. Laminate plates

consist of a number of bonded single layers. We assume that the single layer as

quasi-homogeneous and orthotropic. In each layer we can have different materials,

different thicknesses and especially different angle orientations of the ﬁbres. The

whole plate is assumed to be a continuous structure. The stacking sequence of the

single layers has a great inﬂuence on the deformation behavior of the plate. Plates

can be loaded by distributed and concentrated loads in all directions, so called in-

plane and out of plane loading. In a special case of laminate plates, if we have an

arrangement of the single layers symmetric to the midplane, the in-plane and out of

plane states are decoupled.

In Chap. 8 the modelling of laminate plates is given and it distinguishes between

the classical laminate theory and the shear deformation theory like the modelling

of beams. The plate model based on the classical laminate theory usually is called

Kirchhoff plate with its main assumption that points lying on a line orthogonal to

the midplane before deformation are lying on such a normal line after deformation.

This assumption is an extended Bernoulli hypothesis of the beam model to two-

dimensional structures.

The application of the classical laminate theory should be restricted to the anal-

ysis of very thin plates only. For moderate thick plates it is better to use the shear

deformation theory. The plate model based on this theory is called the Mindlin plate

model. The following development of ﬁnite laminate plate elements will be carried

out for both models. Here we will be restricted to symmetric laminate plates in both

cases, it means that we have no coupling of membrane and bending/twisting states

and we will consider bending only.

In both cases we consider a triangular ﬁnite plate element. The approximation

of complicated geometric forms, especially of curved boundaries, can be done

easily with triangular elements. Usually special coordinates are used for triangu-

lar elements. The triangle is deﬁned by the coordinates of the three corner points

P

i(xi,yi),i=1,2,3. A point P(x,y)within the triangle is also deﬁned by the natural

426 11 Finite Element Analysis

triangle coordinates L1,L2,L3,P(L1,L2,L3), see Fig. 11.8. There

L1=A1

A△

,L2=A2

A△

,L3=A3

A△

(11.3.1)

with the triangle area A△and the partial areas A1,A2,A3,A△=A1+A2+A3. There-

fore

L1+L2+L3=1 (11.3.2)

The areas A1,A2,A3,A△can be expressed by determinants

A△=

1x1y1

1x2y2

1x3y3

,A1=

1x y

1x2y2

1x3y3

,

A2=

1x1y1

1x y

1x3y3

,A3=

1x1y1

1x2y2

1x y 

(11.3.3)

and for the coordinates L1,L2,L3of the point P(x,y)

L1=1

2A△

[(x2y3−x3y2) + (y2−y3)x+ (x3−x2)y],

L2=1

2A△

[(x3y1−x1y3) + (y3−y1)x+ (x1−x3)y],

L3=1

2A△

[(x1y2−x2y1) + (y1−y2)x+ (x2−x1)y]

(11.3.4)

and the coordinates x,ycan be expressed by

x=x1L1+x2L2+x3L3,y=y1L1+y2L2+y3L3(11.3.5)

Considering Eq. (11.3.2) we obtain for cartesian coordinates

Fig. 11.8 Natural triangle

coordinates x

y

A1

A2

A3

1(i)

2(j)

3(k)

P

11.3 Finite Plate Elements 427

x=L1(x1−x3) + L2(x2−x3) + x3,y=L1(y1−y3) + L2(y2−y3) + y3(11.3.6)

In Fig. 11.9 the natural triangle coordinates L1,L2,L3are illustrated for some special

points: the corner points and the points in the middle of the sides.

Because the shape functions Niused for the approximation of the deformation

ﬁeld in a triangular plate element are usually written as functions of the natural

element coordinates, it is necessary to ﬁnd relationships for the derivatives of the

shape functions with respect to the global cartesian coordinates. At ﬁrst the deriva-

tives of the shape functions Niare given by the natural triangle coordinates L1and

L2.L3depends from L1and L2, see (11.3.2). So we consider only two independent

coordinates. Here we have







∂

Ni

∂

L1

∂

Ni

∂

L2







=





∂

x

∂

L1

∂

y

∂

L1

∂

x

∂

L2

∂

y

∂

L2













∂

Ni

∂

x

∂

Ni

∂

y







=J

J

J





∂

Ni

∂

x

∂

Ni

∂

y







(11.3.7)

J

Jis the Jacobi matrix of the coordinate transformation

J

J=





∂

x

∂

L1

∂

y

∂

L1

∂

x

∂

L2

∂

y

∂

L2







=cj−bj

−cibi(11.3.8)

and the expressions bi,bj,ci,cjare

bi=y2−y3,bj=y3−y1,ci=x3−x2,cj=x1−x3(11.3.9)

With

DetJ

J

J=cjbi−bjci=

∆

(11.3.10)

Fig. 11.9 Natural triangle

coordinates of special points x

y

1(i)

(1; 0; 0)2(j)

(0; 1; 0)

3(k)(0; 0; 1)

6(n)

(0.5; 0; 0.5)5(m)

(0; 0.5; 0.5)

4(l)

(0.5; 0.5; 0)

428 11 Finite Element Analysis

it follows that

J

J−1=1

∆

bibj

cicj(11.3.11)

and then we obtain for the derivatives of the shape functions Niwith respect to the

cartesian coordinates xand y







∂

Ni

∂

x

∂

Ni

∂

y







=J

J

J−1





∂

Ni

∂

L1

∂

Ni

∂

L2







(11.3.12)

In case of the classical laminate theory, the second partial derivatives

∂

2Ni

∂

L2

1

,

∂

2Ni

∂

L1L2

,

∂

2Ni

∂

L2

2

are also required. For this we must put the result for

∂

Ni/

∂

L1instead of Niinto the

ﬁrst row of Eq. (11.3.7), and we obtain

∂

2Ni

∂

L2

1

=

∂

x

∂

L12

∂

2Ni

∂

x2+2

∂

x

∂

L1

∂

y

∂

L1

∂

2Ni

∂

x

∂

y+

∂

y

∂

L12

∂

2Ni

∂

y2(11.3.13)

In the same manner we can do so with

∂

Ni/

∂

L2and the second row of Eq. (11.3.7)

and for the mixed second partially derivative with

∂

Ni/

∂

L2and the ﬁrst row or vice

versa. The three relationships obtained can be written in matrix form

J

J∗







∂

2Ni

∂

x2

∂

2Ni

∂

x

∂

y

∂

2Ni

∂

y2







=







∂

2Ni

∂

L2

1

∂

2Ni

∂

L1

∂

L2

∂

2Ni

∂

L2

2







(11.3.14)

where J

J

J∗is a modiﬁed or extended Jacobi matrix

J

J∗=









∂

x

∂

L12

2

∂

x

∂

L1

∂

y

∂

L1

∂

y

∂

L12

∂

x

∂

L1

∂

x

∂

L2

∂

x

∂

L2

∂

y

∂

L1

+

∂

x

∂

L1

∂

y

∂

L2

∂

y

∂

L1

∂

y

∂

L2



∂

x

∂

L22

2

∂

x

∂

L2

∂

y

∂

L2

∂

y

∂

L22







(11.3.15)

11.3 Finite Plate Elements 429

Now the second partial derivatives of the shape functions by the cartesian coordi-

nates can be calculated







∂

2Ni

∂

x2

∂

2Ni

∂

x

∂

y

∂

2Ni

∂

y2







=J

J

J∗−1







∂

2Ni

∂

L2

1

∂

2Ni

∂

L1

∂

L2

∂

2Ni

∂

L2

2







(11.3.16)

Of course, by consequently using the natural triangle coordinates it follows that the

integrands in the energy terms are functions of these coordinates. Therefore we have

to consider for the variables of integration the relationship

dA=dxdy=DetJ

J

JdL1dL2=

∆

dL1dL2(11.3.17)

In Sects. 11.3.1 and 11.3.2 the development of triangular ﬁnite plate elements will

be shown in a condensed way for the classical laminate theory and for the shear

deformation theory, respectively.

11.3.1 Classical Laminate Theory

The starting point is the total potential energy of an symmetric laminate plate, see

also (8.2.24)

Π

(w) = 1

2Z

A"D11 

∂

2w

∂

x22

+D22 

∂

2w

∂

y22

+2D12

∂

2w

∂

x2

∂

2w

∂

y2+4D66 

∂

2w

∂

x

∂

y2

+4D16

∂

2w

∂

x2+D26

∂

2w

∂

y2

∂

2w

∂

x

∂

y#dA−Z

A

pzwdA

(11.3.18)

with the stiffness Di j,i,j=1,2,6, see Table 8.3. The strain energy simpliﬁes the

couplings, if we assume special orthotropic laminates (e.g. cross-ply-laminates).

We have no bending-twisting coupling, i.e. D16 =D26 =0. Supposing in other cases

these coupling terms as very small, especially if we have a great number of very thin

layers, we use the following simpliﬁed strain energy approximately

430 11 Finite Element Analysis

Π

(w) = 1

2Z

A"D11 

∂

2w

∂

x22

+D22 

∂

2w

∂

y22

+2D12

∂

2w

∂

x2

∂

2w

∂

y2+4D66 

∂

2w

∂

x

∂

y2

−2pzw#dA

(11.3.19)

The total potential energy of the classical plate model contains second derivatives

and so we have to realize C(1)-continuity at the element boundaries. This means,

continuity of the deﬂections and the derivatives in normal direction to the bound-

aries. It must be noted that we do not have C(1)-continuity, if the ﬁrst derivatives

at the corner points of adjacent elements are equal because we have to guarantee

the continuity of the derivatives in the normal direction at all boundary points of

adjacent elements.

It can be shown that we have to use a polynomial with minimum of 18 coefﬁ-

cients, and because we want to have a complete polynomial, we choose a polynomial

of ﬁfth order with 21 coefﬁcients. Therefore we deﬁne a triangular ﬁnite plate ele-

ment with 6 nodes as shown in Fig. 11.10. At the corner nodes 1,2,3(i,j,k)we have

6 degrees of freedom, the deﬂection, the ﬁrst derivatives in both directions and the

three curvatures, but at the mid-side nodes the ﬁrst derivatives in normal direction

only.

It is a disadvantage when using this element in a general program system that we

have a different number of degrees of freedom at the nodes. Therefore an elimina-

tion, a so-called static condensation of the nodal constants of the mid-side nodes,

can be done and then we have only 18 degrees of freedom for the element. The el-

ement is converted into a three-node element, the nodes 4,5,6(l,m,n)vanish. The

polynomial approximation of the displacement ﬁeld in the ﬁnite element is given by

a special 5th order polynomial, it contains however a complete polynomial of 4th

order. In this way we obtain 18 shape functions Ni(L1,L2,L3),i=1,2,...,18 which

are not illustrated here. Because the coordinates L1,L2,L3are not independent, see

(11.3.2), L3usually is eliminated by

L3=1−L1−L2(11.3.20)

Fig. 11.10 Six-node plate

element x

y

1(i)

2(j)

3(k)

6(n)5(m)

4(l)

11.3 Finite Plate Elements 431

According to (11.1.1) we have the approximation

w(x,y) = N

N

N(L1,L2)v

v

v(11.3.21)

with N

N

Nas the matrix of the 18 shape functions (here it has only one row) and the el-

ement displacement vector v

v

vincluding 18 components. For the differential operator

D

DOP must be written

D

DOP =

∂

2

∂

x2

∂

2

∂

y22

∂

2

∂

x

∂

yT

(11.3.22)

and after this, see also Eq. (11.1.3), the matrix B

B

Bleads to

B

B=D

D

DOPN

N

N(11.3.23)

Since the shape functions are functions of the natural triangle coordinates L1and L2,

for the derivatives by the cartesian coordinates we have to take into consideration

(11.3.16). The element stiffness matrix follows according to (11.1.6)

K

KE=Z

AE

B

BTDB

DB

DBdA

and with the substitution of the integration variable Eq. (11.3.17)

K

KE=

1

Z

0

1−L1

Z

0

B

BTDB

DB

DB△dL2dL1(11.3.24)

Here D

D

Dis the matrix of the plate stiffness, the coupling of bending and twisting is

neglected (D16 =D26 =0)

D

D=



D11 D12 0

D12 D22 0

0 0 D66 



According to (11.1.7) we obtain the element force vector

f

fE=Z

AE

N

NTpdA=

1

Z

0

1−L1

Z

0

N

NTp△dL2dL1(11.3.25)

where p(x,y) = p(L1,L2)is the element surface load.

For the ﬂexural vibration analysis of plates the element mass matrix must be

calculated. According to (11.1.10), the element mass matrix reduces to

M

ME=Z

VE

ρ

N

NTN

N

NdV=

1

Z

0

1−L1

Z

0

ρ

N

NTN

N

Nh△dL2dL1(11.3.26)

432 11 Finite Element Analysis

with

ρ

as an average density

ρ

=1

h

n

∑

k=1

ρ

(k)h(k)(11.3.27)

Note that the classical laminate theory does not consider the rotary kinetic energy.

The integrations in the (11.3.24) for the element stiffness matrix K

K

KE, (11.3.25)

for the element force vector f

f

fEand (11.3.26) for the element mass matrix must

be carried out numerically. Only the force vector f

f

fEcan be calculated analytically,

if we have a constant surface loading p(x,y) = const. For the numerical solutions

it is recommended that integration formulae of the same order are used like the

polynomials for the shape functions, in this case of the ﬁfth order.

11.3.2 Shear Deformation Theory

The Mindlin plate model, which is based on the ﬁrst order shear deformation theory,

considers the shear deformation in a simpliﬁed form. In the Mindlin plate model

the Kirchhoff’s hypotheses are relaxed. Transverse normals to the midplane do not

remain perpendicular to the middle surface after deformation. In Sect. 8.3 the basic

equations are given for this plate model.

Here the starting point is the total potential energy, and if we restrict ourselves to

symmetric and special orthotropic laminates, we have

Π

(w,

ψ

1,

ψ

2) = 1

2Z

A"D11 

∂ψ

1

∂

x2

+2D12 

∂ψ

1

∂

x

∂ψ

2

∂

y+D22 

∂ψ

2

∂

y2

+D66 

∂ψ

1

∂

y+

∂ψ

2

∂

x2

+ks

55A55 

ψ

1+

∂

w

∂

x2

+ks

44A44 

ψ

2+

∂

w

∂

y2#dxdy−Z

A

pzwdxdy

δΠ

(w,

ψ

1,

ψ

2) = 0

(11.3.28)

or written in matrix form

Π

(w,

ψ

1,

ψ

2) = 1

2Z

A

(

κ

TD

D

κ

+

ε

sTA

A

As

ε

s)dxdy−Z

A

p3wdxdy(11.3.29)

The matrices of the plate stiffness for this case (D16 =D26 =0) and the shear stiff-

ness with A45 =0 are, see also (8.3.7),

11.3 Finite Plate Elements 433

D

D=



D11 D12 0

D12 D22 0

0 0 D66 

,A

A

As=ks

55A55 0

0ks

44A44 (11.3.30)

The stiffness are given in detail in (4.2.15) and for the shear correction factor see

Sect. 8.3. Note that we have, in the elastic potential three independent deformation

components (the deﬂection wand the rotations

ψ

1and

ψ

2), so the displacement ﬁeld

vector u

u

uhas three components (nu=3), see also (11.1.2).

For the curvatures and the shear strains we have

κ

=D

D

Dbu

u

u,

ε

s=D

D

Dsu

u

u(11.3.31)

where D

D

Dband D

D

Dsare the matrices of the differential operators

D

Db=







0

∂

x0

0 0

∂

y

0

∂

y

∂

x







,D

D

Ds=





∂

x1 0

∂

y0 1







(11.3.32)

The most important property of the elastic potential however is that it contains ﬁrst

derivatives only. Therefore, we have to guarantee onlyC(0)-continuity at the element

boundaries and it will be possible to take a three-node ﬁnite element with linear

shape functions, but it shall be not done here.

Due to the better approximation properties we will choose a six-node element

with polynomials of the second order as shape functions. The six-node element with

its nodal degrees of freedom is shown in Fig. 11.11. Then we have the nodal and the

element displacement vectors

v

vT

i= [wi

ψ

xi

ψ

yi],v

v

vT

E= [v

v

vT

iv

v

vT

jv

v

vT

kv

v

vT

lv

v

vT

mv

v

vT

n](11.3.33)

and according to (11.1.2) with nu=nF,˜n=nKEthe matrix of the shape functions is

given by

N

N= [NiI

I

I3NjI

I

I3NkI

I

I3NlI

I

I3NmI

I

I3NnI

I

I3],(11.3.34)

Fig. 11.11 Six-node ﬁnite

plate element with nodal

degrees of freedom x

y

z

ψ

1i

ψ

2i

wi

ilj

m

k

n

434 11 Finite Element Analysis

where I

I

I3are unit matrices of the size (3,3). The shape functions are

Ni= (2L1−1)L1,Nj= (2L2−1)L2,Nk= (2L3−1)L3,

Nl=4L1L2,Nm=4L2L3,Nn=4L1L3

(11.3.35)

They are functions of the natural triangle co-ordinates L1,L2,L3, see (11.3.1) -

(11.3.4).

The curvatures and the shear strains in (11.3.29) can be expressed by

κ

=D

D

Dbu

u

u=D

D

DbN

N

Nv

v

vE=B

B

Bbv

v

vE,B

B

Bb=D

D

DbN

N

ε

s=D

D

Dsu

u

u=D

D

DsN

N

Nv

v

vE=B

B

Bsv

v

vE,B

B

Bs=D

D

DsN

N

N(11.3.36)

and consideration of (11.3.12) leads to the element stiffness matrix, see also (11.1.6)

consisting of two parts

K

KE=K

K

Kb

E+K

K

Ks

E,K

K

Kb

E=Z

AE

B

BbTD

D

DB

B

Bbdxdy,K

K

Ks

E=Z

AE

B

BsTA

A

AsB

B

Bsdxdy(11.3.37)

Because the shape functions in N

N

Nare functions of the natural triangle co-ordinates,

the integration variables must be substituted by (11.3.17), and then we ﬁnd

K

Kb

E=

1

Z

0

1−L1

Z

0

B

BbTD

D

DB

B

Bb

∆

dL2dL1,K

K

Ks

E=

1

Z

0

1−L1

Z

0

B

BsTA

A

AsB

B

Bs

∆

dL2dL1(11.3.38)

To obtain the element force vector f

f

fEa load vector q

q

qmust be deﬁned with the same

number of components as the displacement ﬁeld vector u

u

u. Because only surface

loading p(x,y)is considered here, it leads to

q

qT= [p0 0]

and then the element force vector is

f

fE=Z

AE

N

NTq

q

qdxdy,f

f

fE=

1

Z

0

1−L1

Z

0

N

NTq

q

∆

dL2dL1(11.3.39)

with the substitution of integration variables.

The integrations in (11.3.39) can be done analytically only in the case of constant

surface loading p=const. In the other cases it must be calculated numerically. For

the numerical integration it is recommended to apply integration formulae of the

same order as used for shape polynomials, here of the second order. It must be done

in this manner for the ﬁrst part K

K

Kb

Eof the stiffness matrix, for the second part of K

K

Ks

E

a lower order can be used. Such a different kind of integration for the two parts of

the stiffness matrix is called selective integration.

For dynamic analysis the element mass matrix M

M

MEmust also be calculated. For

the shear deformation theory the rotatory kinetic energy is usually taken into con-

11.3 Finite Plate Elements 435

sideration. The kinetic energy of an element is then

TE=1

2Z

VE

ρ

˙

u

uT˙

u

udV=1

2Z

AE

h

2

Z

−h

2

ρ

(˙w2+˙

ψ

2

1+˙

ψ

2

2)dzdA(11.3.40)

If the so called generalized densities are used

ρ

0=

n

∑

k=1

ρ

(k)[z(k)−z(k−1)] =

n

∑

k=1

ρ

(k)h(k),

ρ

1=1

2

n

∑

k=1

ρ

(k)[z(k)2−z(k−1)2],

ρ

2=1

3

n

∑

k=1

ρ

(k)[z(k)3−z(k−1)3]

(11.3.41)

and it is noted here that

ρ

1=0, because we have assumed symmetric laminates only,

then for the kinetic energy we obtain

TE=1

2Z

AE

˙

v

vTR

R

R0˙

v

vdA(11.3.42)

R

R0is a matrix of the generalized densities

R

R0=



ρ

00 0

0

ρ

20

0 0

ρ

2

(11.3.43)

Using the approximation for the displacement ﬁeld vector according to Eq. (11.1.1)

we obtain

TE=1

2v

v

vT

EZ

AE

N

NTR

R

R0N

N

NdAv

v

vE(11.3.44)

and the element mass matrix is

M

ME=Z

AE

N

NTR

R

R0N

N

NdA,M

M

ME=

1

Z

0

1−L1

Z

0

N

NTR

R

R0N

N

∆

dL1dL2(11.3.45)

with substitution of the integration variables.

The ﬁnite laminate plate element developed above is called PL18, where the

number 18 gives the degrees of freedom of all element nodes. This element can be

used only for laminate plates with laminae arranged symmetrically to the midplane,

where we have no coupling of membrane and bending/twisting states and we have

no in-plane loading.

436 11 Finite Element Analysis

In many cases we have nonsymmetric laminates and we have a coupling of mem-

brane and bending/twisting states or there are in-plane and out-of-plane loadings.

Then an element is necessary where the nodal degrees of freedom also include the

deﬂections in x- and y-direction u,v. For such an element, assuming six nodes again,

the nodal and the element displacement vectors are

v

vT

i= [uiviwi

ψ

xi

ψ

yi],v

v

vT

E= [v

v

vT

iv

v

vT

jv

v

vT

kv

v

vT

lv

v

vT

mv

v

vT

n](11.3.46)

The structure of the matrix of the shape functions N

N

Nis in this more general case

N

N= [NiI

I

I5NjI

I

I5NkI

I

I5NlI

I

I5NmI

I

I5NnI

I

I5](11.3.47)

with I

I

I5as unit matrices of the size (5,5), the shape functions remain unchanged.

The total potential energy for this case is, see also (8.3.15),

Π

(u,v,w,

ψ

1,

ψ

2) = 1

2Z

A

(

ε

TA

A

ε

+

κ

TB

B

ε

+

ε

TB

B

κ

+

κ

TD

D

κ

+

ε

sTA

A

As

ε

s)dx1dx2−Z

A

pzwdxdy

(11.3.48)

and we have to take into consideration the membrane stiffness matrix A

A

Aand the

coupling matrix B

B

Badditionally, the element stiffness matrix consists of four parts

K

KE=K

K

Km

E+K

K

Kmb

E+K

K

Kb

E+K

K

Ks

E(11.3.49)

representing the membrane state (K

K

Km

E), the coupling of membrane and bending states

(Kmb

E), the bending state (K

K

Kb

E) and the transverse shear state (K

K

Ks

E).

The general form for the element force vector (11.3.39) is unchanged, it must be

noted that the loading vector q

q

qhere has another structure containing loads in three

directions

q

qT= [pxpypz0 0](11.3.50)

the general form for the element mass matrix is the same as in (11.3.45), but here

the matrix of the generalized densities R

R

R0is

R

R0=





ρ

00 0

ρ

10

0

ρ

00 0

ρ

1

0 0

ρ

00 0

ρ

10 0

ρ

20

0

ρ

10 0

ρ

2







(11.3.51)

The remarks about the realization of the integrations remains unchanged here. Of

course they are all more complicated for this element. Such an extended element

would be called PL30, because the degree of freedom of all nodal displacements is

30. Further details about this extended element are not given here.

11.4 Generalized Finite Beam Elements 437

11.4 Generalized Finite Beam Elements

In civil engineering and also in mechanical engineering a special kind of struc-

tures are used very often structures consisting of thin-walled elements with sig-

niﬁcant larger dimensions in one direction (length) in comparison with the dimen-

sions in the transverse direction. They are called beam shaped shell structures. Beam

shaped shell structures include folded plate structures as the most important class

of such structures. In Chap. 10 the modelling of folded plate structures was con-

sidered and there a generalized beam model was developed by the reduction of the

two-dimensional problem to an one-dimensional one following the way of Vlasov-

Kantorovich. This folded structure model contains all the energy terms of the mem-

brane stress state and of the bending/twisting stress state under the validity of the

Kirchhoff hypotheses. Outgoing from this complete folded structure model some

simpliﬁed structure models were developed (see Sect. 10.2.4) by neglecting of se-

lected energy terms in the potential function e.g. the terms caused by the longitudinal

curvatures

κ

xi, the shear strains

ε

xsi, the torsional curvatures

κ

xsior the transversal

strains

ε

siof the strips. Because the inﬂuence of the longitudinal curvatures

κ

xiof

the single strips to the deformation state and the stress state of the whole structure

is very small for beam shaped structures, they are neglected generally. The shear

strains

ε

xsiof the strips can be neglected for structures with open cross-sections,

but not in the case of closed cross-sections. In opposite to this the torsional curva-

tures

κ

xsican be neglected for closed cross-sections, but not for open cross-sections.

Therefore, because we had in mind to ﬁnd a generalized beam model as well the

shear strains as the torsional curvatures are considered. Although the inﬂuence of

the transversal strains in most cases is very small, they are considered too, because

with this we have a possibility to deﬁne the generalized co-ordinate functions for a

general cross-section systematically. Therefore as a generalized structure model for

beam shaped thin-walled folded plate structures the structure model A (see Sects.

10.2.4 and 10.2.5) is chosen, in which only the longitudinal curvatures

κ

xiof the

strips are neglected.

11.4.1 Foundations

The starting point for the development of generalized ﬁnite beam elements is the

potential energy, see equation (10.2.10). Because in all strips the longitudinal curva-

tures

κ

xiare neglected all terms containing w′′

ihave to vanish. It leads together with

equations (10.2.11), (10.2.12) and with ˆ

A

A7=0

0

0, ˆ

A

A8=0

0

0, ˆ

A

A10 =0

0

0, ˆ

A

A17 =0

0

0, ˆ

A

A20 =0

0

0,

ˆ

A

A23 =0

0

0, ˆ

A

A26 =0

0

0 to a simpliﬁcation of the potential energy equation (10.2.10)

438 11 Finite Element Analysis

Π

=1

2

l

Z

0U

U

U′Tˆ

A

A1U

U

U′+V

V

VTˆ

A

A6V

V

V+U

U

UTˆ

A

A3U

U

U+2U

U

UTˆ

A

A25V

V

V′

+V

V

V′Tˆ

A

A4V

V

V′+V

V

VTˆ

A

A12V

V

V+4V

V

V′Tˆ

A

A9V

V

V′

+2U

U

U′Tˆ

A

A14V

V

V+2U

U

UTˆ

A

A2U

U

U′+2U

U

U′Tˆ

A

A13V

V

V′

−2U

U

U′Tˆ

A

A19V

V

V−4U

U

U′Tˆ

A

A18V

V

V′+2U

U

U′Tˆ

A

A16V

V

+2V

V

VTˆ

A

A5V

V

V′−2V

V

VTˆ

A

A28V

V

V−4V

V

VTˆ

A

A27V

V

V′

−2U

U

UTˆ

A

A22V

V

V−2V

V

V′Tˆ

A

A25V

V

V−4U

U

UTˆ

A

A21V

V

V′

−4V

V

V′Tˆ

A

A24V

V

V′+4V

V

VTˆ

A

A11V

V

V′−2(U

U

UTf

f

fx+V

V

VTf

f

fs+V

V

VTf

f

fn)dx

−(U

U

UTr

r

rx+V

V

VTr

r

rs+V

V

VTr

r

rn)|x=0−(U

U

UTr

r

rx+V

V

VTr

r

rs+V

V

VTr

r

rn)|x=l

(11.4.1)

We can see that the one-dimensional energy function contains only derivatives of

the ﬁrst order.

11.4.2 Element Deﬁnitions

Outgoing from Eq. (11.4.1) a one-dimensional ﬁnite element can be deﬁned. Be-

cause we have no higher derivatives than of the ﬁrst order in the potential energy

only a C(0)continuity is to satisfy at the element boundaries and therefore it would

be possible to use a two-node element with linear shape functions. To have a better

accuracy here we will take a three-node element using second order polynomials as

shape functions, Fig. 11.12. The shape functions are again like (11.2.15)

N1(x) = 1−3x

l+2x2

l2,N2(x) = 4x

l−4x2

l2,N3(x) = −x

l+2x2

l2(11.4.2)

They are shown in Fig. 11.5.

Because a generalized ﬁnite beam element with a general cross-section shall be

developed at ﬁrst we must ﬁnd a rule to deﬁne the cross-section topology. We will

use for it the proﬁle node concept. For this we will see the midlines of all strips as the

cross-sections proﬁle line. The start- and the endpoints of each strip on this proﬁle

line are deﬁned as the so-called main proﬁle nodes. In the middle of each strip there

Fig. 11.12 Three-node gener-

alized beam element

123

x

l

11.4 Generalized Finite Beam Elements 439

are additional proﬁle nodes, they are called secondary proﬁle nodes. Figure 11.13

shows an example for it. The topology of the thin-walled cross-section is described

sufﬁciently by the co-ordinates of the main proﬁle nodes. Additionally the stiffness

parameters of each strip must be given. The connections of the strips in the main

proﬁle nodes are supposed as rigid.

For the generation of the generalized deﬂection co-ordinate functions

ϕ

,

ψ

,

ξ

is

assumed that a main proﬁle node has four degrees of freedom, the displacements

in the directions of the global co-ordinate axes x,y,zand the rotation about the

global xaxis, see Fig. 11.14. The displacements of the main proﬁle nodes lead linear

generalized co-ordinate functions

ϕ

,

ψ

and cubic functions

ξ

between the adjacent

nodes. For an increasing the accuracy the activation of the secondary proﬁle node

degrees of freedom is optional, they are shown in Fig. 11.15. In this case

ϕ

and

ψ

are

quadratic and

ξ

polynomials of 4th and 5th order between the adjacent main nodes.

Therefore a more complex deformation kinematics of the cross-section is consid-

erable. The generalized co-ordinate functions for any thin-walled cross-section are

here deﬁned as follows:

1. Main node displacements or rotations result in non-zero co-ordinate functions

only in the adjacent intervals of the proﬁle line

Fig. 11.13 Description of a

general cross-section

11

2

3

344

5

56

6

7 7

8

z

y

y1s2s4s5s6

s3s7

s8

- main proﬁle nodes (MPN)

- secondary proﬁle nodes SPN)

deﬂection in x-direction deﬂection in y-direction

deﬂection in z-direction rotation about x-axis

Fig. 11.14 Main proﬁle node degrees of freedom

440 11 Finite Element Analysis

second order v-deﬂection second order u-deﬂection

fourth order w-deﬂection ﬁfth order w-deﬂection

Fig. 11.15 Secondary proﬁle node degrees of freedom

2. Secondary node displacements or rotations result in non-zero co-ordinate func-

tions only in the interval between the adjacent main nodes.

In Fig. 11.16 the generalized coordinate functions for axial parallel arranged strips

are shown. Figure 11.17 gives the supplements for slanting arranged strips.

11.4.3 Element Equations

In the case of non-activated degrees of freedom of the secondary proﬁle nodes we

have a degree of freedom of an element node of four times the number of main

proﬁle nodes (4 nMP N ) and the element displacement vector consists of 12 nMPN

components

v

vT= [v

v

v1v

v

v2v

v

v3],v

v

vj=¯

u

uj

¯

v

vj(11.4.3)

¯

u

uj,¯

v

vjcontain the values of the generalized displacement functions at the node j.

The displacement vector u

u

u(x)contains here the generalized displacement functions

U

U(x)and V

V

V(x)and in accordance with Eq. (11.1.1) we obtain for the interpolation

u

u(x) = U

U

U(x)

V

V(x)=Nv

Nv

Nv (11.4.4)

The matrix of the shape functions for the chosen three-node element is

N

N= [N1(x)I

I

I N2(x)I

I

I N3(x)I

I

I],(11.4.5)

where I

I

Iare the unit matrices of the size 4 nMPN and with this the matrix N

N

Nhas

the format (4 nMPN , 12 nM PN ). Following the equation (11.4.4) for the generalized

11.4 Generalized Finite Beam Elements 441

1 3 2

x,u

n,ws,v

d

main proﬁle node functions secondary proﬁle node functions

u-deﬂections

v-deﬂections

w-deﬂections

1

11

1

11 0.25

0.25

1/16

−1/8d

ϕ

1=1−s

d

ϕ

2=s

d

ϕ

3=s

d−s

d2

ψ

1=1−s

d

ξ

1=0

ψ

2=s

d

ξ

2=0

ψ

3=s

d−s

d2

ξ

3=0

ξ

4=1−3s

d2+2s

d3

ψ

4=0

ξ

5=3s

d2−2s

d3

ψ

5=0

ξ

8=s

d2−2s

d3+s

d4

ψ

8=0

ξ

6=s1−2s

d+s

d2

ψ

6=0

ξ

7=s−s

d+s

d2

ψ

7=0

ξ

9=s

d2−4s

d3

+5s

d4−2s

d5

ψ

4=0

Fig. 11.16 Generalized co-ordinate functions for axial parallel arranged strips

442 11 Finite Element Analysis

1

2

3

x

z

y

n,w

x,u

s,v

α

ψ

1=1−s

dcos

α

ξ

1=−1−3s

d2+2s

d3sin

α

ψ

2=s

dcos

α

ξ

2=−3s

d2−2s

d3sin

α

ψ

4=1−s

dsin

α

ξ

4=1−3s

d2+2s

d3cos

α

ψ

5=s

dsin

α

ξ

5=3s

d2−2s

d3cos

α

Fig. 11.17 Supplements for slanting arranged strips

11.4 Generalized Finite Beam Elements 443

displacement functions we have to write

U

U(x) = L

L

LT

10u

u

u(x) = L

L

LT

10Nv

Nv

Nv,V

V

V(x) = L

L

LT

01u

u

u(x) = L

L

LT

01Nv

Nv

Nv (11.4.6)

with the matrices

L

LT

10 = [I

I

I0

0

0],L

L

LT

01 = [0

0

0I

I

I](11.4.7)

In the ﬁrst case (L

L

L10)I

I

Iis a unit matrix of the size nMNP and the null matrix has the

format (nMPN , 3 nM PN ), in the second case (L

L

L01)I

I

Iis a unit matrix of the size 3 nMPN

and the null matrix has the format (3 nM PN ,nMPN ).

Of course in the case of activated degrees of freedom of the secondary pro-

ﬁle nodes all the dimensions given above are increased correspondingly. Inserting

the generalized displacement functions (11.4.6) into the potential energy equation

(11.4.1) we obtain

Π

=1

2v

v

vTKv

Kv

Kv −f

f

fTv

v

v(11.4.8)

The condition

∂Π

∂

v

v=0

0

0 (11.4.9)

leads to the element equation

Kv

Kv =f

f

f(11.4.10)

with the symmetric element stiffness matrix

K

K=





K

K11 K

K

K12 K

K

K13

K

KT

12 K

K

K22 K

K

K23

K

KT

13 K

K

KT

23 K

K

K33





(11.4.11)

For the sub-matrices Kmn we ﬁnd the general equation

Kmn =

3

∑

h=1







ˆ

A

A1hImn1

+ˆ

A

A2hImn2

+ˆ

A

AT

2hImn3

+ˆ

A

A3hImn4

(ˆ

A

A13h−2ˆ

A

A18h)Imn1

+( ˆ

A

A15h−2ˆ

A

A21hImn2

+( ˆ

A

A14h−ˆ

A

A19h)Imn3

+( ˆ

A

A16h−ˆ

A

A22h)Imn4

(ˆ

A

AT

13h−2ˆ

A

AT

18h)Imn1

+( ˆ

A

AT

15h−2ˆ

A

AT

21h)Imn2

+( ˆ

A

AT

14h−ˆ

A

AT

19h)Imn3

+( ˆ

A

AT

16h−ˆ

A

AT

22h)Imn4

(ˆ

A

A4h+ˆ

A

A9h−2ˆ

A

A24h−2ˆ

A

AT

24h)Imn1

+( ˆ

A

A5h+2ˆ

A

A11h−ˆ

A

AT

25h−2ˆ

A

A27h)Imn2

+( ˆ

A

AT

5h+2ˆ

A

A11h−ˆ

A

A25h−2ˆ

A

AT

27h)Imn3

+( ˆ

A

A6h+ˆ

A

A12h−ˆ

A

A28h−ˆ

A

AT

28h)Imn4







(11.4.12)

with

444 11 Finite Element Analysis

Imn1=

l

Z

0

NhN′

mN′

ndx,Imn2=

l

Z

0

NhNmN′

ndx,

Imn3=

l

Z

0

NhN′

mNndx,Imn4=

l

Z

0

NhNmNndx

To include approximately slight non-prismatic structures the matrices of the stiff-

ness parameters ˆ

A

Ai, see Eq. (10.2.12), are interpolated in the element in the same

manner as the displacements

ˆ

A

Ai=

3

∑

h=1

ˆ

A

AihNh(11.4.13)

ˆ

A

Aih are the matrices at the nodes h=1,2,3.

The element force vector is obtained as

f

f=



f

f1

f

f2

f

f3

(11.4.14)

with the sub-vectors

f

fm=

3

∑

h=1







f

fxh

l

Z

0

NhNmdx

(

(fsh +fnh)

l

Z

0

NhNmdx







(11.4.15)

Here f

f

fxh,f

f

fsh,f

f

fnh are the generalized load vectors, see Eq. (10.2.13), at the nodes

h=1,2,3.

11.4.4 System Equations and Solution

The system equations can be obtained by using the Eqs. (11.1.15) and (11.1.16) with

the coincidence matrices, determining the position of each element in the whole

structure. In the so founded system stiffness matrix the boundary conditions of the

whole structure are to consider, otherwise this matrix is singular, if the structure

is not ﬁxed kinematically. The solution of the system equations lead to the nodal

displacements and with them the strains and curvatures in the single strips of each

element can be calculated, see Eqs. (10.2.6) and (10.2.11),

11.4 Generalized Finite Beam Elements 445

ε

x(x,si) =

3

∑

h=1

N′

h¯

u

uT

h

ϕ

,

ε

s(x,si) =

3

∑

h=1

Nh¯

v

vT

h

ψ

•,

ε

xs(x,si) =

3

∑

h=1

(Nh¯

u

uT

h

ϕ

•+N′

hv

v

vT

h

ψ

),

κ

s(x,si) = −

3

∑

h=1

Nh¯

v

vT

h

ξ

••,

κ

xs(x,si) = −2

3

∑

h=1

N′

h¯

v

vT

h

ξ

•

(11.4.16)

Now we can obtain the stress resultants in the kth lamina, which has the distance nk

from the mid plane of the strip





Nxk

Nsk

Nxsk 

=



A11kA12kA16k

A12kA22kA26k

A16kA26kA66k





ε

x

ε

s+nk

κ

s

ε

xs +nk

κ

xs 

(11.4.17)

These stress resultants are related on the strip co-ordinate axes xand si. Therefore, it

is necessary to transform them into the material co-ordinate system of the kth lamina

(for the transformation relationship see Table 4.1)





NLk

NTk

NLTk

=



cos2

α

ksin2

α

k2sin

α

kcos

α

k

sin2

α

kcos2

α

k−2sin

α

kcos

α

k

−sin

α

kcos

α

ksin

α

kcos

α

kcos2

α

k−sin2

α

k





Nxk

Nsk

Nxsk 

(11.4.18)

Than the stresses of the kth lamina are obtained

σ

Lk=NLk

tk

,

σ

Tk=NTk

tk

,

τ

LTk=NLTk

tk

(11.4.19)

In some cases the strains in the kth lamina related to the material co-ordinate system

are important for the failure assessment of the lamina. Then they can be calculated

with help of the following matrix equation





ε

Lk

ε

Tk

ε

LTk

=Q

Q

Q′

k−1



NLk

NTk

NLTk

(11.4.20)

There Q

Q

Q′is the reduced stiffness matrix of the kth lamina.

11.4.5 Equations for the Free Vibration Analysis

The variation statement given by the Hamilton’s principle, see Eq. (10.2.41) leads

with the Lagrange function (10.2.39) and the assumption of harmonic vibrations for

the considered generalized beam element to the element equation

(K

K

K−

ω

2M

M

M)v

v

v=0

0

0 (11.4.21)

446 11 Finite Element Analysis

K

Kis the element stiffness matrix, see Eqs. (11.4.11) and (11.4.12), and M

M

Mis the ele-

ment mass matrix. The element mass matrix is obtained with the matrices ˆ

B

B1,ˆ

B

B2,ˆ

B

B3,

see Eqs. (10.2.38)

M

M=





M

M11 M

M

M12 M

M

M13

M

MT

12 M

M

M22 M

M

M23

M

MT

13 M

M

MT

23 M

M

M33





(11.4.22)

with

M

Mmn =

3

∑

h=1ˆ

B

B1hImn40

0

0(ˆ

B

B2h+ˆ

B

B3h)Imn4(11.4.23)

There the ˆ

B

Bmatrices are also interpolated in the element by using the shape func-

tions. In this way slight non-prismatic structures are considerable too. The system

equations can be developed in a similar way as it was done for a static analysis. Here

we have to ﬁnd a system stiffness matrix and a system mass matrix. After consider-

ation the boundary conditions the eigen-value problem can be solved and the mode

shapes can be estimated.

11.5 Numerical Results

Additional to a great number of special FEM programs general purpose FEM pro-

gram systems are available. The signiﬁcance of universal FEM program packages

is increasing. In universal FEM program systems we have generally the possibility

to consider anisotropic material properties, e.g. in the program system COSMOS/M

we can use volume elements with general anisotropic material behavior and plane

stress elements can have orthotropic properties.

Laminate shell elements are available e.g. in the universal FEM program systems

ANSYS, NASTRAN or COSMOS/M. In many program systems we have no spe-

cial laminate plate elements, the laminate shell elements are used also for the anal-

ysis of laminate and sandwich plates. Perhaps, because of the higher signiﬁcance

of two-dimensional laminate structures in comparison with beam shaped structures

laminate beam elements are missing in nearly all universal FEM program systems.

The generalized beam elements, Sect. 11.4, are e.g. implemented only in the FEM

program system COSAR.

For the following numerical examples the program system COSMOS/M is used.

In COSMOS/M a three node and a four node thin laminate shell element are

available (SHELL3L; SHELL4L). Each node has 6 degrees of freedom. The element

can consist of up to 50 layers. Each layer can have different material parameters,

different thicknesses and especially different angles of ﬁbre directions. We have no

restrictions in the stacking structure, symmetric, antisymmetric and nonsymmetric

structures are possible. The four nodes of the SHELL4L element must not arranged

in-plane. By the program in such case a separation is done into two or four triangular

partial elements. Further there is a SHELL9L element available. It has additional

11.5 Numerical Results 447

nodes at the middles of the four boundaries and in the middle of the element. For

the following examples only the element SHELL4L is used.

11.5.1 Examples for the Use of Laminated Shell Elements

By the following four examples the application of the laminate shell element

SHELL4L shall be demonstrated. At ﬁrst a thin-walled beam shaped laminate struc-

ture with L-cross-section under a concentrated force loading is considered, and the

second example is a thin-walled laminate pipe under torsional loading. In both cases

the inﬂuence of the ﬁbre angles in the layers is tested. The use of the laminate shell

element for the static and dynamic analysis of a sandwich plate is shown in the

third example. A buckling analysis of a laminate plate is demonstrated by the fourth

example. In all 4 cases a selection of results is given.

11.5.1.1 Cantilever Beam

A cantilever beam with L-cross-section consists of 3 layers with the given material

parameters Ex,Ey,

ν

xy,

ν

yx,Gxy . It is loaded by a concentrated force F, see Fig. 11.18.

The material parameters are

L

F

+

α

+

α

−

α

3

4

300

400

10

Fig. 11.18 Cantilever beam: cross-section and stacking structure (F= 4.5 kN, L= 4 m, all other

geometrical values in mm)

448 11 Finite Element Analysis

Ex=1.53 ·104kN/cm2,Ey=1.09 ·103kN/cm2,Gxy =560kN/cm2,

ν

xy =0.30,

ν

yx =0.021

The ﬁbre angle

α

shall be varied:

α

=0◦,10◦,20◦,30◦,40◦.

The FEM model after the input of all properties into COSMOS/M is illustrated

in Fig. 11.19. The computing yields a lot of results. In Fig. 11.20, e.g., is shown

the deformed shape for a ﬁbre angle of

α

=30◦. Here should be selected only

the displacements of the corner node at the free edge (node No. 306 in our FE

model) in y- and z-direction and the maximal stresses in ﬁbre direction (

σ

x) and

perpendicular to it (

σ

y) for the left side of the vertical part of the cross-section (layer

No. 1, bottom):

v306,y= -2,204 cm, v306,z= -1,805 cm,

σ

lay1,max,x= 7,487 kN/cm2,

σ

lay1,max,y= 0,824 kN/cm2

Similar the displacements and stresses for the ﬁbre angles

α

=0◦,10◦,20◦,40◦are

calculated, and the results are shown in Figs. 11.21 and 11.22. The results show

that for such a beam shaped structure the main stresses are lying in the longitudinal

direction and therefore the ﬁbre angle 0◦leads to the most effective solution.

11.5.1.2 Laminate Pipe

A laminate pipe consisting of 2 layers with the given material parameters Ex,Ey,

ν

xy,

ν

yx,Gxy is ﬁxed at left end and loaded by a torsional moment, see Fig. 11.23.

The material parameters are the same as in the previous example:

Ex=1,53 104kN/cm2,Ey=1,09 103kN/cm2,Gxy =560 kN/cm2,

ν

xy =0,30,

ν

yx =0,021

The ﬁbre angle

α

shall be varied:

α

=0◦,15◦,30◦,45◦. After the input of all parame-

Fig. 11.19 FE-model of can-

tilever beam in COSMOS/M

(650 elements, 714 nodes)

x

y

z

Fig. 11.20 Cantilever beam

deformed shape

x

y

z

11.5 Numerical Results 449

vyvz

Fibre Angle

Displacements/cm

0◦10◦20◦30◦40◦

0.5

1

1.5

2

2.5

3

3.5

4

4.5

Fig. 11.21 Displacements of the corner point at the free edge

σ

x

σ

y

Fibre Angle

Stresses kN/cm2

0◦10◦20◦30◦40◦

0

2

4

6

8

10

Fig. 11.22 Maximal stresses at the bottom of layer No. 1

450 11 Finite Element Analysis

L

Mt

+

α

−

α

3

D

6

Fig. 11.23 Laminate pipe: geometry, cross-section and stacking sequence (Mt=1200 kNcm,

L=2 m, D=200 mm)

ters and properties into COSMOS/M the FEM model can be illustrated (Fig. 11.24).

From the results of the analysis only the twisting angle of the free edge shall

be considered here. For this we have to list the results for the displacements in y-

direction of two nodes at the free edge, lying in opposite to each other, e.g. the nodes

255 and 663 in our FE model. The twisting angle is calculated by

ϕ

= (vy,255 −vy,663 )/D

Carrying out the analysis for all ﬁbre angles we obtain the results, given in Fig.

11.25. The diagram demonstrates the well known fact that in case of pure shear

loading the main normal stresses are lying in a direction with an angle of 45◦to the

shear stresses. Therefore here the ﬁbre angels of +45◦/−45◦to the longitudinal

axis are the most effective arrangements, because these ﬁbre angels yield the greatest

shear rigidity.

Fig. 11.24 FE-model of Lam-

inate Pipe in COSMOS/M

(800 elements, 816 nodes)

x

y

z

11.5 Numerical Results 451

Fibre Angle

Twisting Angle

0◦10◦20◦30◦40◦

1

2

4

6

8

Fig. 11.25 Twisting angle of the free edge

11.5.1.3 Sandwich Plate

The sandwich plate (Fig. 11.26) is clamped at both short boundaries and simply

supported at one of the long boundaries. The cover sheets consist of an aluminium

alloy and the core of foam of polyurethan. The material parameters are:

AlZnMgCu0.5F450:

ρ

=2.7·103kg/m3,E=7.0·1010 N/m2,

ν

=0.34

polyurethan foam:

ρ

= 150 kg/m3,E=4.2·107N/m2,

ν

=0.30

Additional to a stress analysis of the plate under constant pressure loading p,

a vibration analysis will be performed is asked. We have to calculate the 4 lowest

eigenfrequencies and the mode shapes, respectively. Note that we use in this exam-

ple only the basic units of the SI-system, so we avoid the calculation of correction

factors for the obtained eigenfrequencies.

The FE-model is given in Fig. 11.27. The static analysis leads the displacements

and stresses. We consider only the stresses of the bottom of the lower cover sheet

(layer 3, top). The Figs. 11.28 and 11.29 show the plots of stress distributions for the

ﬂexural stresses

σ

xand

σ

z, Fig. 11.30 the distribution of the von Mises equivalent

stress. The lowest 4 eigenfrequencies and their 4 mode shapes are shown in the

following Fig. 11.31. The static and frequency computations conﬁrm the successful

application of the SHELL4L element for a sandwich plate.

452 11 Finite Element Analysis

6.0 m

4.0 m

p=1200 N/m2

aluminium alloy

polyurethan foam

24 30 mm

Fig. 11.26 Sandwich plate

Fig. 11.27 FE-model of Sandwich Plate in COSMOS/M (600 elements, 651 nodes)

11.5.1.4 Buckling Analysis of a Laminate Plate

For a rectangular laminate plate consisting of 4 layers with the given material pa-

rameters a buckling analysis shall be carried out. The plate is simply supported at

11.5 Numerical Results 453

x

y

z

2.3717E+007

1.4703E+007

5.6887E+006

-3.3255E+006

-1.2340E+007

-2.1354E+007

-3.0368E+007

-3.9382E+007

-4.8396E+007

σ

x

Fig. 11.28 Stresses in x-direction for the bottom of the lower cover sheet

x

y

z

1.1711E+007

8.1902E+006

4.6695E+006

1.1488E+006

-2.3719E+006

-5.8926E+006

-9.4133E+006

-1.2934E+007

-1.6455E+007

σ

z

Fig. 11.29 Stresses in z-direction for the bottom of the lower cover sheet

454 11 Finite Element Analysis

x

y

z

4.4196E+007

3.9218E+007

3.4240E+007

2.9262E+007

2.4284E+007

1.9306E+007

1.4328E+007

9.3496E+006

4.3715E+006

von Mises

Fig. 11.30 Von Mises stress for the bottom of the lower cover sheet

Fig. 11.31 Mode shapes for the lowest four eigenfrequencies: f1=5,926 Hz (top-left),

f2=12,438 Hz (top-right), f3=13,561 Hz (bottom-left), f4=19,397 Hz (bottom-right)

11.5 Numerical Results 455

all boundaries and loaded by a uniaxial uniform load, see Fig. 11.32. Material pa-

rameters are again the same as in the previous examples

Ex= 1.53 104kN/cm2,Ey= 1.09 103kN/cm2,Gxy = 560 kN/cm2,

ν

xy =0.30,

ν

yx =0.021

For the stacking structure two cases shall be considered, a symmetric (case I) and

a antisymmetric (case II) laminate structure (Fig. 11.32). The ﬁbre angle is to vary:

α

=0◦,15◦,30◦,45◦,60◦,75◦,90◦. For the buckling analysis in COSMOS/M a unit

pressure loading must be created, and the program calculates a buckling factor

ν

B

to multiply the unit loading for obtaining the buckling load.

The FE-model created in COSMOS/M by the input of all properties and param-

eters is shown in Fig. 11.33. The calculation for

α

=30◦leads to a buckling factor

ν

B=1,647 and to the buckling mode shown in Fig. 11.34. In the same manner the

calculations for the other ﬁbre angels and for the antisymmetric laminate were per-

formed. The results for the buckling factors are shown in a diagram in Fig. 11.35.

The buckling modes are symmetric to the symmetric axis in loading direction. For

the symmetric laminates the buckling modes for

α

=0◦,15◦,30◦are nearly the

same, see Fig. 11.34. For ﬁbre angles 45◦,60◦,75◦,90◦the buckling modes have

different shapes, they are shown in the following ﬁgures. The buckling modes for

the antisymmetric laminate are very similar but not identical to the buckling modes

1.5 m

1.0 m

+

α

+

α

+

α

+

α

−

α

−

α

−

α

−

α

4×25

case I case II

Fig. 11.32 Rectangular laminate plate

Fig. 11.33 FE-model of the

laminate plate in COSMOS/M

(600 elements, 651 nodes)

456 11 Finite Element Analysis

Fig. 11.34 Buckling modes for symmetric laminates

α

=30◦(top-left),

α

=45◦(top-right),

α

=

60◦(middle-left),

α

=75◦(middle-right),

α

=90◦(bottom)

of the symmetric laminates. They are not given here. A ﬁbre angle near 45◦leads to

the highest buckling load for a quadratic plate. It shall be noted that the antimetric

stacking sequence of the laminate improved the buckling stability.

11.5.2 Examples of the Use of Generalized Beam Elements

Generalized ﬁnite elements for the analysis of thin-walled beam shaped plate struc-

tures, Sect. 11.4, were implemented and tested in the frame of the general purpose

FEM-program system COSAR. The real handling of the FEM-procedures are not

given here, but two simple examples shall demonstrate the possibilities of these el-

ements for global static or dynamic structure analysis.

Figure 11.36 shows thin-walled cantilever beams with open or closed cross-

sections and different loadings. All these beam structure models have equal length,

hight and width and also the total thicknesses of all laminate strips are equal, inde-

pendent of the number of the layers.

11.5 Numerical Results 457

n, symm. n, antim.

Fibre Angle

Buckling Value

010 20 30 40 50 60 70 80 90

1

1.2

1.4

1.6

1.8

2

2.2

Fig. 11.35 Results of the Buckling Analysis

The stacking structure may be symmetric or antisymmetric. Figure 11.37 shows

the two considered variants: case A with three laminae and symmetric stacking and

case B with two laminae and antisymmetric stacking. The ﬁbre reinforced material

is characterized again by the following effective moduli

EL=153000 N/mm2,

ν

LT =0.30,

ET=10900 N/mm2,

ν

TL =0.021,

GLT =5600 N/mm2,

ρ

=2 g/cm3

The ﬁbre angles shall be varied.

Figure 11.38 shows the proﬁle nodes. There are four main proﬁle nodes for both

cross-sections but three secondary proﬁle nodes for the open and four for the closed

cross-section. The numerical analysis shall demonstrate the inﬂuence of the stacking

structure. Figure 11.39 illustrates the relative changes of the cantilever beam in the

loaded point, if the symmetric stacking structure (wA) is change to the antisymmetric

one (wB). The antisymmetric layer stacking leads to higher values of the vertical

deﬂections wBin comparison to the wAvalues in the case of symmetric stacking.

Generally, only two degrees of freedom of secondary proﬁle nodes were activated.

In a separate analysis the inﬂuence of a higher degrees of freedom in the sec-

ondary proﬁle nodes was considered. As a result it can be recommended that for

antisymmetric layer structures and for open proﬁles more than two degrees of free-

458 11 Finite Element Analysis

F=260 kN

F

x

y

z

400 400

800

5000

Fig. 11.36 Cantilever beams, geometry and loading

11.5 Numerical Results 459

ab

ss

nn

−

α

−

α

+

α

+

α

+

α

Fig. 11.37 Stacking structure of the laminates. aSymmetric sandwich, btwo-layer

44

4

3

333

2

22

111

1

Fig. 11.38 Cross-sections with main proﬁle nodes (•) and secondary proﬁle nodes (×)

dom should be activated. Ignoring the activation of secondary proﬁle node degrees

of freedom leads to nonrealistic structure stiffness. The structure model is to stiff

and therefore the deﬂections are to small.

The second example concern ed the eigen-vibration analysis. For the closed cross-

section with symmetric layer stacking the inﬂuences of the degree of freedom of

secondary proﬁle nodes and of the variation of the ﬁbre angles were considered. As

a result it can be stated that the inﬂuence of higher degrees of freedom of the sec-

ondary proﬁle nodes is negligible but the inﬂuence of the ﬁbre angles is signiﬁcant.

Figure 11.40 illustrates the inﬂuence of the ﬁbre angle variations on the eigenfre-

quencies of the beam, which can be used for structure optimization.

Summarizing Sect. 11.5 one have to say that only a small selection of one- and

two-dimensional ﬁnite elements was considered. Many ﬁnite plate and shell ele-

ments were developed using equivalent single layer theories for laminated struc-

tures but also multi-layered theories are used. Recent review articles give a detailed

overview on the development, implementation and testing of different ﬁnite lami-

nate and sandwich elements.

460 11 Finite Element Analysis

0

2

4

6

8

1 0

1 2

1 4

0 ° 1 0 ° 2 0 ° 3 0 ° 4 0 ° 5 0 ° 6 0 ° 7 0 ° 8 0 ° 9 0 °

∆

w

%

Fig. 11.39 Relative changes (wB−wA)/wA=

∆

w(

α

)of the vertical deﬂections wAand wB

0

2 0

4 0

6 0

8 0

1 0 0

1 2 0

1 4 0

0 ° 1 0 ° 2 0 ° 3 0 ° 4 0 ° 5 0 ° 6 0 ° 7 0 ° 8 0 ° 9 0 °

Hz

1. EF

2. EF

3. EF

4. EF

Fig. 11.40 Inﬂuence of the ﬁbre angle on the ﬁrst four eigen-frequencies of the cantilever box-

beam

Part VI

Appendices

This part is focussed on some basics of mathematics and mechanics like matrix

operations (App. A), stress and strain transformations (App. B), differential opera-

tors for rectangular plates (App. C) and differential operators for circular cylindrical

shells (App. D). In addition, the Krylow functions as solution forms of a special

fourth order ordinary differential equation are discussed (App. E) and some mate-

rial’s properties are given in App. F. In the last one section like always in this book

material or constitutive parameters are used as usual. Note that any material param-

eter is a parameter since there are dependencies on temperature, time, etc. Last but

not least there are given some references for further reading (App. G).

Appendix A

Matrix Operations

The following short review of the basic matrix deﬁnitions and operations will pro-

vide a quick reference and ensure that the particular use of vector-matrix notations

in this textbook is correct understood.

A.1 Deﬁnitions

1. Rectangular matrix

A

A=





a11 a12 ······ a1n

a21 a22 ······ a2n

.

..

.

..

.

..

.

..

.

am1am2······ amn







= [ai j]

Rectangular matrix with i=1,2,...,mrows and j=1,2,...,ncolumns, is a

rectangular-ordered array of quantities with mrows and ncolumns. m×nor

often (m,n)is the order of the matrix, ai j is called the (i,j)-element of A

A

A. There

are two important special cases

a

a=





a1

.

am





= [ai]

is a m×1 matrix or column vector, while

a

aT=a1···an= [ai]T

is a 1 ×nmatrix or row vector.

With a

a

arespectively a

a

aTa matrix A

A

Acan be written

463

H. Altenbach et al., Mechanics of Composite Structural

Elements, https://doi.org/10.1007/978-981-10-8935-0

464 A Matrix Operations

A

A=a

a

a1···a

a

an,a

a

aj=





a1j

.

am j





,j=1,...,n

or

A

A=



a

aT

1

···

a

aT

m



,a

a

aT

j=aj1···ajn ,j=1,...,m

If n=mthe matrix is square of the order n×n. For a square matrix the elements

ai j with i=jdeﬁne the principal matrix diagonal and are located on it.

2. Determinant of a square matrix A

A

|A

A

A|=

a11 a12 ······a1n

a21 a22 ······a2n

.

..

.

..

.

..

.

..

.

an1an2······ann



=|ai j|=detA

A

The determinant of a matrix A

A

Awith elements ai j is given by

|A

A

A|=a11M11 −a12M12 +a13M13 −...(−1)1+na1nM1n,

where the minor Mi j is the determinant of the matrix |A

A

A|with missing row iand

column j. Note the following properties of determinants:

•Interchanging two rows or two columns changing the sign of |A

A

A|.

•If all elements in a row or a column of A

A

Aare zero then |A

A

A|=0.

•Multiplication by a constant factor cof all elements in a row or column of A

A

multiplies |A

A

A|by c.

•Adding a constant multiple of row or column kto row or column ldoes not

change the determinant.

•If one row kis a linear combination of the rows land mthen the determinant

must be zero.

3. Regular matrix

A square matrix A

A

Ais regular if |A

A

A| 6=0.

4. Singular matrix

A square matrix A

A

Ais singular if |A

A

A|=0.

5. Trace of a matrix

The trace of a square matrix A

A

Ais the sum of all elements of the principal diagonal,

i.e.

trA

A

A=

m

∑

k=1

akk

6. Rank of a matrix

The rank rk(A

A

A)of a m×nmatrix A

A

Ais the largest value of rfor which there exist a

A.2 Special Matrices 465

r×rsubmatrix of A

A

Athat is non-singular. Submatrices are smaller arrays of k×k

elements ai j of the matrix A

A

A, i.e. k≤mif m≤nor k≤nif n≤m.

A.2 Special Matrices

In the following the

δ

i j denotes the Kronecker symbol

δ

i j =0i6=j

1i=j

1. Null matrix 0

0

All elements ai j of a m×nmatrix are identically equal zero

ai j ≡0,i=1,...,m,j=1,...,n

2. Diagonal matrix D

D

D=diag[aii] = diag[ai j

δ

i j]

A diagonal matrix is a square matrix in which all elements are zero except those

on the principal diagonal

ai j =0,i6=j,ai j 6=0,i=j

3. Unit matrix I

I

I= [

δ

i j]

A unit or identity matrix is a special case of the diagonal matrix for which aij =1

when i=jand ai j =0 when i6=j.

4. Transpose A

A

ATof a matrix A

A

The transpose of a matrix A

A

Ais found by interchanging rows and columns. If

A

A= [ai j]follow A

A

AT= [aT

i j]with aT

i j =aji . A transposed matrix is denoted by a

superscript T. Note (A

A

AT)T=A

A

5. Symmetric matrix A

A

AS

A square matrix A

A

Ais said to be symmetric if for all i6=j ai j =aji , i.e. A

A

A=A

A

AT. A

symmetric matrix is denoted by a superscript S.

6. Skew-symmetric matrix A

A

AA

A square matrix A

A

Ais said to be skew-symmetric if all principal diagonal elements

are equal zero and for all i6=j aij =−aji , i.e. A

A

A=−A

A

AT. A skew-symmetric

matrix is denoted by a superscript A.

7. Any matrix can be decomposed in a symmetric and a skew-symmetric part in a

unique manner

A

A=A

A

AS+A

A

AA

Proof. Since

A

AS=1

2(A

A

A+A

A

AT)

and

A

AA=1

2(A

A

A−A

A

AT)

466 A Matrix Operations

the sum A

A

AS+A

A

AAis equal to A

A

A.

A.3 Matrix Algebra and Analysis

1. Addition and subtraction

Am×nmatrix A

A

Acan be added or subtracted to a m×nmatrix B

B

Bto form a m×n

matrix C

C

A

A±B

B

B=C

C

C,ai j ±bi j =ci j,i=1,...,m,j=1,...,n

Note A

A

A+B

B

B=B

B

B+A

A

A,A

A

A−B

B

B=−(B

B

B−A

A

A) = −B

B

B+A

A

A,(A

A

A±B

B

B)T=A

A

AT±B

B

BT.

2. Multiplication

•Multiplication the matrix A

A

Aby a scalar

α

involves the multiplication of all

elements of the matrix by the scalar

α

A

A=A

A

α

= [

α

ai j],

(

α

±

β

)A

A

A=

α

A

A±

β

A

A,

α

(A

A

A±B

B

B) =

α

A

A±

α

B

•The product of a (1×n)matrix (row vector a

a

aT) and a (n×1)matrix (column

vector b

b

b) forms a (1×1)matrix, i.e. a scalar

α

a

aTb

b

b=b

b

bTa

a

a=

α

,

α

=

n

∑

k=1

akbk

•The product of a (m×n)matrix A

A

Aand a (n×1)column vector b

b

bforms a

(m×1)column vector c

c

Ab

Ab =c

c

c,ci=

n

∑

j=1

ai jbj=ai1b1+ai2b2+...+ainbn,i=1,2,...,m

The forgoing product is only possible if the number of columns of A

A

Ais equal

the number of rows of b

b

b.

Note A.1. b

b

bTA

A

AT=c

c

cT

•If A

A

Ais a (m×n)matrix and B

B

Ba(p×q)matrix the product AB

AB

AB =C

C

Cexists if

n=p, in which case C

C

Cis a (m×q)matrix. For n=pthe matrix A

A

Aand B

B

Bare

said to be confor mable for multiplication. The elements of the matrix C

C

Care

ci j =

n=p

∑

k=1

aikbk j ,i=1,2,...,m,j=1,2,...,q

Note A.2. AB

AB

AB 6=BA

BA

BA,A

A

A(B

B

B±C

C

C) = AB

AB

AB ±AC

AC

AC,(AB

AB

AB)T=B

B

BTA

A

AT

A.3 Matrix Algebra and Analysis 467

(l×m)

A

(m×n)

B

(n×p)

C

C=

(l×p)

D

3. Inversion and division

AI

AI =IA

IA

IA =A

A

A,A

A

A−1A

A

A=A

A

AA

A

A−1=I

I

I,A

A

A−1−1=A

A

The matrix inversion is based on the existence of a n×nunit matrix I

I

Iand a square

n×nmatrix A

A

A.A

A

A−1is the inverse of A

A

Awith respect to the matrix multiplication

A

AA

A

A−1=A

A

A−1A

A

A=I

I

I. If A

A

A−1exist, the matrix A

A

Ais invertible or regular, otherwise

non-invertible or singular. Matrix division is not deﬁned.

Note A.3. (AB

AB

AB)−1=B

B

B−1A

A

A−1,(ABC

ABC

ABC)−1=C

C

C−1B

B

B−1A

A

A−1,...

•Cofactor matrix

With the minors Mi j introduced above to deﬁne the determinant |A

A

A|of a matrix

A

Aa so-called cofactor matrix A

A

Ac= [Ai j]can be deﬁned, where

Ai j = (−1)i+jMi j

The cofactor matrix is denoted by the superscript c.

•Adjoint or adjugate matrix

The adjoint matrix of the square matrix A

A

Ais the transpose of the cofactor

matrix

adjA

A

A= (A

A

Ac)T

Note A.4. Because symmetric matrices possess symmetric cofactor matrices

the adjoint of a symmetric matrix is the cofactor matrix itself

adjA

A

AS=A

A

ASc

It can be shown that

A

A(adjA

A

A) = |A

A

A|I

I

i.e. A

A

A(adjA

A

A)

|A

A

A|=I

I

I=A

A

AA

A

A−1⇒A

A

A−1=adjA

A

|A

A

A|

Inverse matrices have some important properties

A

A

AT−1=A

A

A−1T

and if A

A

A=A

A

AT

A

A−1=A

A

A−1T

i.e. the inverse matrix of a symmetric matrix A

A

Ais also symmetric.

Note A.5. Symmetric matrices posses symmetric transposes, symmetric cofac-

tors, symmetric adjoints and symmetric inverses.

468 A Matrix Operations

4. Powers and roots of square matrices

If n×nmatrix A

A

Ais conformable with itself for multiplication, one may deﬁne its

powers

A

An=A

A

AA

A

A...A

A

A,

and for symmetric positive semideﬁnite matrices

A

A1

n=n

√A

A

A,A

A

A−n=A

A

A−1n

and if A

A

Ais regular

(A

A

Am)n=A

A

Amn,A

A

AmA

A

An=A

A

Am+n,

5. Matrix eigenvalue problems

The standard eigenvalue problem of a quadratic n×nmatrix A

A

Ais of the form:

ﬁnd (

λ

,x

x

x)with 6=

6=

6=0

0

0 such that

A

Ax

x

x=

λ

x

xor (A

A

A−

λ

I

I)x

x

x=0

0

K

K= [A

A

A−

λ

I

I]is called the characteristic matrix of A

A

A, detK

K

K=0 is called the char-

acteristic determinant or equation of A

A

A. The characteristic determinant produces

a characteristic polynomial with powers of

λ

up to

λ

nand therefore when it set

equal zero having nroots which are called the eigenvalues. If the characteristic

equation has ndistinct roots, the polynomial can be factorized in the form

(

λ

−

λ

1)(

λ

−

λ

2)...(

λ

−

λ

n) = 0

If we put

λ

=0 in the characteristic equation we get

detA

A

A=

λ

1

λ

2...

λ

n

Inserting any root

λ

iinto the standard eigenvalue equation leads to

[A

A

A−

λ

iI

I

I]x

x

xi=0

0

0,i=1,2,...,n

x

xiare the eigendirections (eigenvectors) which can be computed from the last

equation considering the orthogonality condition. A nontrivial solution exists if

and only if

det[A

A

A−

λ

iI

I

I] = 0

Note A.6. If we have the 3x3 symmetric matrix the eigendirection x

x

xican be com-

puted for each

λ

ifrom the polynomial of third order. Three different solutions

are possible:

•all solutions

λ

iare distinct - three orthogonal eigendirections can be computed

(but their magnitudes are arbitrary),

•one double solution and one distinct solutions - only one eigendirection can

be computed (its magnitudes is arbitrary), and

•all solutions are identically - no eigendirections can be computed.

A.3 Matrix Algebra and Analysis 469

Anyway, the orthogonality condition x

x

xT·x

x

x=1 should be taken into account.

The general eigenvalue problem is given in the form

A

Ax

x

x=

λ

B

Bx

x

which can be premultiplied by B

B

B−1to produce the standard form

B

B−1A

A

Ax

x

x=B

B

B−1

λ

B

Bx

x

x= (B

B

B−1A

A

A)x

x

x=

λ

I

Ix

x

x=

λ

x

resulting in

(B

B

B−1A

A

A−

λ

I

I)x

x

x=0

0

Note A.7. In the case of non-symmetric matrix A

A

Athe eigenvalue can be complex.

6. Differentiating and integrating

•To differentiate a matrix one differentiates each matrix element ai j in the con-

ventual manner.

•To integrate a matrix one integrates each matrix element ai j in the conventual

manner. For deﬁnite integrals, each term is evaluated for the limits of integra-

tion.

7. Partitioning of matrices

A useful operation with matrices is partitioning into submatrices. These subma-

trices may be treated as elements of the parent matrix and manipulated by the

standard matrix rules reviewed above. The partitioning is usually indicated by

dashed partitioning lines entirely through the matrix

M

M= [mi j] = 





A

A.

.

.B

B

·········

C

C.

.

.D

D







For a m×nmatrix M

M

Mwe may have submatrices A

A

A(i×j),B

B

B(i×p),C

C

C(k×j),

D

D(k×p)with i+k=m,j+p=n, i.e.

M

Mm×n=





A

Ai×j

.

.B

B

Bi×(n−j)

··· ··· ···

C

C(m−i)×j

.

.D

D

D(m−i)×(n−j)







,







A

A.

.

.B

B

·········

C

C.

.

.D

D





±





E

E.

.

.F

F

·········

G

G.

.

.H

H







=





A

A±E

E

E.

.

.B

B

B±F

F

··· ··· ···

C

C±G

G

G.

.

.D

D

D±H

H







,

470 A Matrix Operations







A

A.

.

.B

B

·········

C

C.

.

.D

D











E

E.

.

.F

F

·········

G

G.

.

.H

H







=





AE

AE +BG

BG

BG .

.

.AF

AF

AF +BH

BH

··· ··· ···

CE

CE +DG

DG

DG .

.

.CF

CF

CF +DH

DH







The multiplications are only deﬁned if the correspondent matrices are con-

formable for multiplication







A

A.

.

.B

B

·········

C

C.

.

.D

D







T

=





A

AT.

.

.C

C

CT

··· ··· ···

B

BT.

.

.D

D

DT







If the matrix

M

M=





A

A.

.

.B

B

·········

C

C.

.

.D

D







is symmetric (M

M

M=M

M

MT), it follows A

A

A=A

A

AT,D

D

D=D

D

DT,B

B

B=C

C

CT,C

C

C=B

B

BT

Appendix B

Stress and Strain Transformations

Stress and strain transformations under general orthogonal coordinate transforma-

tion e

e

e′=R

R

Re

e

eor e′

i=Ri jej:

1.

σ

′

p=T

σ

pq

σ

q. The matrix [T

σ

pq]is deﬁned by







R2

11 R2

12 R2

13 2R12R13 2R11 R13 2R11R12

R2

21 R2

22 R2

23 2R22R23 2R21 R23 2R21R22

R2

31 R2

32 R2

33 2R32R33 2R31 R33 2R31R32

R21R31 R22 R32 R23R33 R22R33 +R23 R32 R21R33 +R23R31 R21R32 +R22 R31

R11R31 R12 R32 R13R33 R12R33 +R13 R32 R11R33 +R13R31 R11R32 +R12 R31

R11R21 R12 R22 R13R23 R12R23 +R13 R22 R11R23 +R13R21 R11R22 +R12 R21







2.

ε

′

p=T

ε

pq

ε

q. The matrix [T

ε

pq]is deﬁned by







R2

11 R2

12 R2

13 R12R13 R11 R13 R11R12

R2

21 R2

22 R2

23 R22R23 R21 R23 R21R22

R2

31 R2

32 R2

33 R32R33 R31 R33 R31R32

2R21R31 2R22 R32 2R23R33 R22 R33+R23 R32 R21R33 +R23R31 R21R32 +R22 R31

2R11R31 2R12 R32 2R13R33 R12 R33+R13 R32 R11R33 +R13R31 R11R32 +R12 R31

2R11R21 2R12 R22 2R13R23 R12 R23+R13 R22 R11R23 +R13R21 R11R22 +R12 R21







3. Rotation about the e

e

e1-direction, Fig. B.1:

[Ri j] = 



1 0 0

0c s

0s c 

,e

e

e′=1

R

R e

e

471

H. Altenbach et al., Mechanics of Composite Structural

Elements, https://doi.org/10.1007/978-981-10-8935-0

472 B Stress and Strain Transformations

x1,x′

1

x2

x3

x′

2

x′

3

e

e1,e

e

e′

1

e

e2

e

e3

e

e′

2

e

e′

3

Φ

Fig. B.1 Rotation about the e

e

e1-direction

[

1

T

σ

pq] =







1 0 0 0 0 0

0c2s22cs 0 0

0s2c2−2cs 0 0

0−cs cs c2−s20 0

0 0 0 0 c−s

0 0 0 0 s c







,

σ

′=

1

T

σσ

σ

[

1

T

ε

pq] =







1 0 0 0 0 0

0c2s2cs 0 0

0s2c2−cs 0 0

0−2cs 2cs c2−s20 0

0 0 0 0 c−s

0 0 0 0 s c







,

ε

′=

1

T

εε

ε

Appendix C

Differential Operators for Rectangular Plates

Below two cases will be discussed

•the classical plate theory and

•the shear deformation theory.

C.1 Classical Plate Theory

1. General unsymmetric laminates





L11 L12 L13

L22 L23

sym L33 





u

v

w

=



0

p

,

L11 =A11

∂

2

∂

x2

1

+2A16

∂

2

∂

x1

∂

x2

+A66

∂

2

∂

x2

2

,

L22 =A22

∂

2

∂

x2

2

+2A26

∂

2

∂

x1

∂

x2

+A66

∂

2

∂

x2

1

,

L33 =D11

∂

4

∂

x4

1

+4D16

∂

4

∂

x3

1

∂

x2

+2(D16 +2D66)

∂

4

∂

x2

1

∂

x2

2

+4D26

∂

4

∂

x1

∂

x3

2

+D22

∂

4

∂

x4

2

,

L12 =A16

∂

2

∂

x2

1

+ (A12 +A66)

∂

2

∂

x1

∂

x2

+A26

∂

2

∂

x2

2

,

L13 =−B11

∂

3

∂

x3

1

+3B16

∂

3

∂

x2

1

∂

x2

+ (B12 +2B66)

∂

3

∂

x1

∂

x2

2

+B26

∂

3

∂

x3

2,

473

H. Altenbach et al., Mechanics of Composite Structural

Elements, https://doi.org/10.1007/978-981-10-8935-0

474 C Differential Operators for Rectangular Plates

L23 =−B22

∂

3

∂

x3

2

+3B26

∂

3

∂

x1

∂

x2

2

+ (B12 +2B66)

∂

3

∂

x2

1

∂

x2

+B16

∂

3

∂

x3

1

2. General symmetric laminates





L11 L12 0

L22 0

sym L33 





u

v

w

=



0

p



Bi j =0, i.e. L13 =L31 =0,L23 =L32 =0. L11,L22 ,L33,L12 as above in 1.

3. Balanced symmetric laminates





L11 L12 0

L22 0

sym L33 





u

v

w

=



0

p



In addition to 2. both A16 and A26 are zero, i.e. L13 =L31 =0, L23 =L32 =0 and

L11,L22 and L33 simplify with A16 =A26 =0.

4. Cross-ply symmetric laminates





L11 L12 0

L22 0

sym L33 





u

v

w

=



0

p



In addition to 3. both D16 and D26 are zero, i.e. L33 simpliﬁes.

5. Balanced unsymmetric laminates





L11 L12 L13

L22 L23

sym L33 





u

v

w

=



0

p



With A16 =A26 =0 only the operators L11 ,L22 and L12 of 1. can be simpliﬁed.

6. Cross-ply unsymmetric laminates





L11 L12 L13

L22 L23

sym L33 





u

v

w

=



0

p



In addition to 5., D16,D26 ,B16 and B26 are zero, i.e. all operators of 1. can be

simpliﬁed.

C.2 Shear Deformation Theory 475

C.2 Shear Deformation Theory

1. General unsymmetrical laminates







˜

L11 ˜

L12 ˜

L13 ˜

L14 0

˜

L22 ˜

L23 ˜

L24 0

˜

L33 ˜

L34 ˜

L35

˜

L44 ˜

L45

S Y M ˜

L55













u

v

ψ

1

ψ

2

w







=





0

p







with ˜

L11 =L11,˜

L22 =L22,˜

L12 =L12 (the Li j can be taken from Appendix C.1)

and

˜

L33 =D11

∂

2

∂

x2

1

+2D16

∂

2

∂

x1

∂

x2

+D66

∂

2

∂

x2

2−ks

55A55,

˜

L44 =D66

∂

2

∂

x2

1

+2D26

∂

2

∂

x1

∂

x2

+D22

∂

2

∂

x2

2−ks

44A44,

˜

L55 =−ks

55A55

∂

2

∂

x2

1

+ks

45A45

∂

2

∂

x1

∂

x2

+ks

44A44

∂

2

∂

x2

2,

˜

L13 =˜

L31 =B11

∂

2

∂

x2

1

+2B26

∂

2

∂

x1

∂

x2

+B66

∂

2

∂

x2

2

,

˜

L14 =˜

L41 =˜

L23 =˜

L32 =B16

∂

2

∂

x2

1

+ (B12 +B66)

∂

2

∂

x1

∂

x2

+B26

∂

2

∂

x2

2

,

˜

L24 =˜

L42 =B66

∂

2

∂

x2

1

+2B26

∂

2

∂

x1

∂

x2

+B22

∂

2

∂

x2

2

,

˜

L34 =˜

L43 =D16

∂

2

∂

x2

1

+ (D12 +D66)

∂

2

∂

x1

∂

x2

+D26

∂

2

∂

x2

2

,

˜

L35 =˜

L53 =−ks

55A55

∂

x1

+ks

45A45

∂

x2,

˜

L45 =˜

L54 =−ks

45A45

∂

x1

+ks

44A44

∂

x2

with ks

45 =pks

44ks

55.

2. General symmetric laminates

Bi j =0, i.e. ˜

L13 =˜

L31,˜

L14 =˜

L41,˜

L23 =˜

L32 and ˜

L24 =˜

L42 are zero

˜

L11 ˜

L12

˜

L12 ˜

L22 u

v=0





˜

L33 ˜

L34 ˜

L35

˜

L34 ˜

L44 ˜

L45

˜

L53 ˜

L54 ˜

L55 





ψ

1

ψ

1

w

=



0

p



476 C Differential Operators for Rectangular Plates

3. Cross-ply symmetric laminates

In addition to 2. both D16 ,D26 and A16,A26 ,A45 are zero.

Appendix D

Differential Operators for Circular Cylindrical

Shells

Below two cases will be considered

•the classical case and

•the ﬁrst order shear deformation theory.

D.1 Classical Shell Theory

1. General unsymmetrical laminates





L11 L12 L13

L22 L23

SYM L33 





u

v

w

=−



px

ps

pz



L11 =A11

∂

2

∂

x2+2A16

∂

2

∂

x

∂

s+A66

∂

2

∂

s2,

L12 = (A16 +R−1B16)

∂

2

∂

x2

1

+ (A12 +R−1B12 +A66 +R−1B66)

∂

2

∂

x

∂

s

+ (A26 +R−1B26)

∂

2

∂

s2,

L13 =R−1A16

∂

x+R−1A26

∂

s−B11

∂

3

∂

x3−3B16

∂

3

∂

x2

∂

s

−(B12 +B66)

∂

3

∂

x

∂

s2−B26

∂

3

∂

s3,

L22 = (A66 +2R−1B66 +R−2D66)

∂

2

∂

x2

477

H. Altenbach et al., Mechanics of Composite Structural

Elements, https://doi.org/10.1007/978-981-10-8935-0

478 D Differential Operators for Circular Cylindrical Shells

+2(A26 +2R−1B26 +2R−2D26)

∂

2

∂

x

∂

s

+ (A22 +2R−1B22 +R−2D22)

∂

2

∂

s2,

L23 =R−1(A26 +R−1B26)

∂

x+R−1(A22 +R−1B22)

∂

s

−(B16 +R−1D16)

∂

3

∂

x3−[B12 +2B66 +R−1(D12 +2D66)]

∂

3

∂

x2

∂

s

−3(B26 +R−1D26)

∂

3

∂

x

∂

s2−(B22 +R−1D22)

∂

3

∂

s3,

L33 =R−2(A22 +R−1B22) + 2R−1B12

∂

2

∂

x2+4R−1B26

∂

2

∂

x

∂

s+2R−1B22

∂

2

∂

s2

−D11

∂

4

∂

x4−4D16

∂

4

∂

x3

∂

s−2(D12 +2D66)

∂

4

∂

x2

∂

s2

−4D26

∂

4

∂

x

∂

s3−D22

∂

4

∂

s4

2. General symmetrical laminates

All Bi j =0, but the matrix [Li j ]is full populated, i.e. all [Li j ]are nonequal zero.

Note that for general symmetrically laminated circular cylindrical shells there is

a coupling of the in-plane and out-of-plane displacements and stress resultants.

3. Cross-ply symmetrical laminates

Bi j =0,A16 =A26 =0,D16 =D26 =0

4. Cross-ply antisymmetrical laminates

B22 =−B11,all other Bi j =0,A16 =A26 =0,D16 =D26 =0

5. Axisymmetric deformations of symmetrical cross-ply laminates

Additional to 3. all derivative

∂

/

∂

sand the displacement vare taken zero and

yield

L12 =L13 =L23 ≡0

D.2 Shear Deformation Theory 479

D.2 Shear Deformation Theory

1. General unsymmetrical laminates







˜

L11 ˜

L12 ˜

L13 ˜

L14 ˜

L15

˜

L21 ˜

L22 ˜

L23 ˜

L24 ˜

L25

˜

L31 ˜

L32 ˜

L33 ˜

L34 ˜

L35

˜

L41 ˜

L42 ˜

L43 ˜

L44 ˜

L45

˜

L51 ˜

L52 ˜

L53 ˜

L54 ˜

L55













u

v

ψ

1

ψ

2

w







=−





px

ps

0

pz







,

˜

L11 =A11

∂

2

∂

x2+2A16

∂

2

∂

x

∂

s+A66

∂

2

∂

s2,

˜

L12 =A16

∂

2

∂

x2+ (A12 +A66)

∂

2

∂

x

∂

s+A26

∂

2

∂

s2,

˜

L13 =B11

∂

2

∂

x2+2B16

∂

2

∂

x

∂

s+B66

∂

2

∂

s2,

˜

L14 =B16

∂

2

∂

x2+ (B12 +B66)

∂

2

∂

x

∂

s+B26

∂

2

∂

s2,

˜

L15 =1

RA12

∂

x+1

RA26

∂

s,

˜

L22 =A66

∂

2

∂

x2+2A26

∂

2

∂

x

∂

s+A22

∂

2

∂

s2−1

R2ks

44A44 ,

˜

L23 =B16

∂

2

∂

x2+ (B16 +B66)

∂

2

∂

x

∂

s+B26

∂

2

∂

s2+1

Rks

45A45 ,

˜

L24 =B66

∂

2

∂

x2+2B26

∂

2

∂

x

∂

s+B22

∂

2

∂

s2+1

Rks

44A44,

˜

L25 = (A12 +ks

55A55)1

R

∂

x+ (A26 +ks

45A45)1

R

∂

s,

˜

L33 =D11

∂

2

∂

x2+2D16

∂

2

∂

x

∂

s+D66

∂

2

∂

s2−ks

55A55

˜

L34 =D16

∂

2

∂

x2+ (D12 +D66)

∂

2

∂

x

∂

s+D26

∂

2

∂

s2−ks

45A45,

˜

L35 =B12

1

R−As

55

∂

x+B26

1

R−ks

45A45

∂

s,

˜

L44 =D66

∂

2

∂

x2+2D26

∂

2

∂

x

∂

s+D22

∂

2

∂

s2−ks

44A44,

˜

L45 =B26

1

R−ks

45A45

∂

x+B22

1

R−ks

44A44

∂

s,

˜

L55 =A55

∂

2

∂

x2+2ks

45A45

∂

2

∂

x

∂

s−A22)1

R2

480 D Differential Operators for Circular Cylindrical Shells

and

˜

L51 =−˜

L15,˜

L52 =−˜

L25,˜

L53 =−˜

L35,˜

L54 =−˜

L45,

ks

45 =qks

44ks

55

2. Cross-ply symmetrical laminates

Bi j =0,i,j=1,2,6,

A16 =A26 =A45 =0,D16 =D26 =0

3. Cross-ply antisymmetrical laminates

B22 =−B11,all other Bi j =0,i,j=1,2,6,

A16 =A26 =A45 =0,D16 =D26 =0

Appendix E

Krylow-Functions as Solution Forms of a Fourth

Order Ordinary Differential Equation

The solutions of the following fourth order ordinary differential equation

w′′′′ −k2

1w′′ +k4

2w=0

can be presented in the form of so-called Krylow functions (Filonenko-Boroditsch,

1952):

1. k2

2>k2

1

α

1=−

α

2=a+ib,

α

3=−

α

4=a−ib,

a=r1

2(k2

2+k2

1),b=r1

2(k2

2−k2

1)

and the solutions are

Φ

1=cosh ax cosbx,

Φ

2=sinhaxsinbx,

Φ

3=cosh ax sinbx,

Φ

4=sinhaxcosbx

or

Φ

1=e−ax cosbx,

Φ

2=e−ax sinbx,

Φ

3=eax sinbx,

Φ

4=eax cosbx

2. k2

2<k2

1

α

1=−

α

2=a,

α

3=−

α

4=b,

a=rk2

1−qk4

1−k4

2,b=rk2

1+qk4

1−k4

2

and the solutions are

Φ

1=cosh ax,

Φ

2=coshbx,

Φ

3=sinhax,

Φ

4=sinhbx

or

Φ

1=e−ax,

Φ

2=e−bx,

Φ

3=eax,

Φ

4=ebx

481

H. Altenbach et al., Mechanics of Composite Structural

Elements, https://doi.org/10.1007/978-981-10-8935-0

482 E Krylow-Functions as Solution Forms of a Fourth Order Ordinary Differential Equation

3. k2

2=k2

1

α

1=

α

2=a,

α

3=

α

4=−a

and the solutions are

Φ

1=coshax,

Φ

2=xsinhax,

Φ

3=sinhax,

Φ

4=xcoshax

or

Φ

1=e−ax,

Φ

2=xe−ax,

Φ

3=eax,

Φ

4=xeax

References

Filonenko-Boroditsch MM (1952) Festigkeitslehre, vol II. Verlag Technik, Berlin

Appendix F

Material’s Properties

Below material properties for classical materials, for the constituents of various

composites and for unidirectional layers are presented. The information about the

properties were taken from different sources (see the Handbooks, Textbooks and

Monographs at the end of this appendix).

Note that the presentation of material data in a unique way is not so easy due to

the incompleteness of material data in the original sources. This means that there are

some empty places in the above following tables. The authors of this textbook were

unable to ﬁll out these places. Another problem is connected with the different unit

systems in the original sources. For recalculation approximate relations are used

(e.g. 1 kgf ≈10 N).

With respect to the quick changes in application composite materials all material

data are only examples showing the main tendencies. Every year new materials are

developed and for the material data one have to contact directly the companies.

References

Czichos H, Hennecke M (eds) (2012) H ¨utte - das Ingenieurwissen, 34th edn.

Springer, Berlin, Heidelberg

Grote KH, Feldhusen J (eds) (2014) Dubbel - Taschenbuch f¨ur den Maschinenbau,

24th edn. Springer Vieweg, Berlin, Heidelberg

Hyer M (1998) Stress Analysis of Fiber-Reinforced Composite Materials. McGraw-

Hill, Boston et al.

Vasiliev V, Morozov E (2001) Mechanics and Analysis of Composite Materials.

Elsevier, Amsterdam

483

H. Altenbach et al., Mechanics of Composite Structural

Elements, https://doi.org/10.1007/978-981-10-8935-0

484 F Material’s Properties

Table F.1 Material properties of conventional materials at room temperature (bulk form), after

Grote and Feldhusen (2014)] and Czichos and Hennecke (2012)

Density Young’s Maximum Ultimate Maximum Coefﬁcient

modulus speciﬁc strength speciﬁc of thermal

modulus strength expansion

ρ

E E/

ρ σ

u

σ

u/

ρ α

(g/cm3) (GPa) (MNm/kg) (MPa) (kNm/kg) (10−6/◦K)

Steel 7.8-7.85 180-210 27 340-2100 270 13

Gray cast iron 7.1-7.4 64-181 25 140-490 69 9-12

Aluminium 2.7-2.85 69-72 27 140-620 230 23

Titanium 4.4-4.5 110 25 1000-1200 273 11

Magnesium 1.8 40 22 260 144 26

Beryllium 1.8-1.85 300-320 173 620-700 389

Nickel 8.9 200 22 400-500 56 13

Zirconium 6.5 100 15 390 60 5.9

Tantalum 16.6 180 11 275 17 6.5

Tungsten 19.3 350 18 1100-4100 212 6.5

Glass 2.5 70 28 700-2100 840 3.5-5.5

References 485

Table F.2 Material properties of ﬁbre materials, after Hyer (1998)

Density Young’s Maximum Ultimate Maximum Coefﬁcient Diameter

modulus speciﬁc strength speciﬁc of thermal

modulus strength expansion

- ﬁbre

direction

ρ

E E/

ρ σ

u

σ

u/

ρ α

d

(g/cm3) (GPa) (MNm/kg) (MPa) (kNm/kg) (10−6/◦K (

µ

m)

E-Glass 2.54 72.4 29 3450 1358 5 8-14

C-Glass 2.49 68.9 28 3160 1269 7.2

S-Glass 2.49 85.5 34 4600 1847 5.6 10

Carbon

Intermediate 1.78-1.82 228-276 155 2410-2930 1646 -0.1- -0.5 8-9

modulus

High modulus 1.67-1.9 331-400 240 2070-2900 1736 -1- -4 5-7

High strength 1.85 240 130 3500 1892 -1- -4 5-7

Polymeric ﬁbres

Kevlar-29 1.44 62 43 2760 1917 -2 12

Aramid (Kevlar-49) 1.48 131 89 2800-3792 2562 -2 12

Spectra 900 0.97 117 121 2580 2660 38

Boron 2.63 385 146 2800 1065 4 100-140

Boron Carbide 2.5 480 192 2100-2500 1000 50

Boron Nitride 1.9 90 47 1400 737 7

Titanium Carbide 4.9 450 92 1500 306 280

486 F Material’s Properties

Table F.3 Material properties of matrix and core materials, after Hyer (1998)

Density Young’s Shear Young’s Ultimate Coefﬁcient

modulus modulus modulus strength of thermal

(tension) (compression) expansion

ρ

EtG Ec

σ

u

α

(g/cm3) (GPa) (GPa) (GPa) (MPa) (10−6/◦K)

Thermosetting polymers

Polyester 1.2-1.3 3-4.2 0.7-2 90-250 40-90 80-150

Vinyl ester 1.15 3-4 127 65-90 80-150

Bismaleimide 1.32 3.6 1.8 200 48-78 49

Polyimide 1.43-1.89 3.1-4.9 70-120 90

Epoxy 1.1-1.6 3-6 1.1-1.2 100-200 30-100 45-80

Thermoplastic polymers

PEEK 1.32 3.6 1.38 140 92-100 47

PPS 1.34 2.5 70-75 54-100

Polysulfone 1.24 2.5 70-75 56-100

Polypropylene 0.9 1-1.4 0.38-0.54 25-38 110

Nylon 1.14 1.4-2.8 0.54-1.08 34 60-75 90

Polcarbonate 1.06-1.2 2.2-2.4 86 45-70 70

Ceramics

Borosilicate glass 2.3 64 26.4 100 3

Balsa wood 0.1-0.19 2-6 8-18

Polystyrene 0.03-0.07 0.02-0.03 0.25-1.25

References 487

Table F.4 Material properties of selected unidirectional composites

E-Glass/ S-Glass/ Kevlar/ Boron/ Carbon

epoxy epoxy epoxy epoxy epoxy

Fibre volume fraction vf0.55 0.50 0.6 0.5 0.63

Density (g/cm3)2.1 2.0 1.38 2.03 1.58

Longitudinal modulus EL(GPa) 39 43 87 201 142

Transverse modulus ET(GPa) 8.6 8.9 5.5 21.7 10.3

In-plane shear modulus GLT (GPa) 3.8 4.5 2.2 5.4 7.2

Major Poisson’s ratio

ν

LT 0.28 0.27 0.34 0.17 0.27

Minor Poisson’s ratio

ν

TL 0.06 0.06 0.02 0.02 0.02

Longitudinal ultimate stress

σ

Lu (MPa) 1080 1280 1280 1380 2280

Transverse ultimate stress

σ

Tu (MPa) 39 49 30 56 57

In-plane ultimate shear stress

σ

LTu (MPa) 89 69 49 62 71

Longitudinal thermal

expansion coefﬁcients

α

L(10−6/◦K) 7 5 -2 6.1 -0.9

Transverse thermal

expansion coefﬁcient

α

T(10−6/◦K) 21 26 60 30 27

488 F Material’s Properties

Table F.5 Typical properties of unidirectional composites as functions of the ﬁbre volume fraction,

after Vasiliev and Morozov (2001)

Glass/ Carbon/ Carbon/ Aramid/ Boron/ Boron/ Carbon/ Al2O3/

epoxy epoxy PEEK epoxy epoxy aluminium carbon aluminium

Fibre volume

fraction vf0.65 0.62 0.61 0.6 0.5 0.5 0.6 0.6

Density

ρ

(g/cm3) 2.1 1.55 1.6 1.32 2.1 2.65 1.75 3.45

Longitudinal

modulus EL(GPa) 60 140 140 95 210 260 170 260

Transverse

modulus ET(GPa) 13 11 10 5.1 19 140 19 150

In-plane shear

modulus GLT (GPa) 3.4 5.5 5.1 1.8 4.8 60 9 60

Major Poisson’s

ratio

ν

LT 0.3 0.27 0.3 0.34 0.21 0.3 0.3 0.24

Longitudinal ultimate

tensile stress

σ

t

Lu (GPa) 1.8 2 2.1 2.5 1.3 1.3 0.34 0.7

Longitudinal ultimate

compressive stress

σ

c

Lu (GPa) 0.65 1.2 1.2 0.3 2 2 0.18 3.4

Transverse ultimate

tensile stress

σ

t

Tu (GPa) 0.04 0.05 0.075 0.03 0.07 0.14 0.007 0.19

Transverse ultimate

compressive stress

σ

c

Tu (GPa) 0.09 0.17 0.25 0.13 0.3 0.3 0.05 0.4

In-plane ultimate

shear stress

σ

LTu (GPa) 0.05 0.07 0.16 0.03 0.08 0.09 0.03 0.12

Appendix G

References

G.1 Comprehensive Composite Materiala

1. Editors-in-chief Kelly, A. and Zweben, C.: Comprehensive Composite Materials.

Pergamon, Oxford, 2000.

•Vol. 1: Fiber Reinforcements and General Theory of Composites (ed. by T.-W.

Chou)

•Vol. 2: Polymer Matrix Composites (ed. by R. Talreja & J.-A. E. M˚anson)

•Vol. 3: Metal Matrix Composites (ed. by T. W. Clyne)

•Vol. 4: Carbon/Carbon, Cement, and Ceramic Matrix Composites (ed. R. War-

ren)

•Vol. 5: Test Methods, Nondestructive Evaluation, and Smart Materials (ed. by

L. Carlsson, R.L. Crane & K. Uchino)

•Vol. 6: Design and Applications (ed. by M.G. Bader, K.T. Kedward & Y.

Sawada)

2. Editors-in-chief Zweben, C. and Beaumont, P.: Comprehensive Composite Ma-

terials II. Elsevier, 2017.

•Vol. 1: Reinforcements and General Theories of Composites (ed. by E.E. Gd-

outos)

•Vol. 2: Polymer Matrix Composites: Fundamentals (ed. by R. Talreja)

•Vol. 3: Polymer Matrix Composites: Manufacture and Applications (ed. by A.

Poursartip)

•Vol. 4: Metal Matrix Composites (ed. by T.W. Clyne)

•Vol. 5: Ceramic and Carbon Matrix Composites (ed. by M.B. Ruggles-Wrenn)

•Vol. 6: Nanocomposites and Multifunctional Materials (ed. by T. Peijs and

E.T. Thostenson)

•Vol. 7: Testing, Nondestructive Evaluation and Structural Health Monitoring

(ed by R. Crane)

•Vol. 8: Design and Analysis of Composite Structures (ed. by A. Johnson and

C. Soutis)

489

H. Altenbach et al., Mechanics of Composite Structural

Elements, https://doi.org/10.1007/978-981-10-8935-0

490 G References

G.2 Selected Textbooks and Monographs on Composite

Mechanics

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of Fiber Composites. John Wiley & Sons, Hoboken (NJ), 3. ed., 2006.

2. Altenbach, H., J. Altenbach, R. Rikards: Einf ¨

uhrung in die Mechanik der

Laminat- und Sandwichtragwerke - Modellierung und Berechnung von Balken

und Platten aus Verbundwerkstoffen. Dt. Verl. f¨ur Grundstofﬁndustrie, Stuttgart,

1996.

3. Altenbach, H., J. Altenbach, W. Kissing: Structural Analysis of Laminate and

Sandwich Beams and Plates. Lubelskie Towarzystwo Naukove, Lublin, 2001.

4. Altenbach, H., W. Becker (Eds): Modern Trends in Composite Laminates Me-

chanics. CISM Courses and Lectures. Springer, Wien, New York, 2003.

5. Ashbee K.H.G.: Fundamental Principles of Reinforced Composites. Technomic,

Lancaster et al., 2. ed., 1993.

6. Becker, W.: Beitr¨

age zur analytischen Behandlung ebener Laminate. Habil.-

Schrift, TH Darmstadt 1993.

7. Bergmann, H.W.: Konstruktionsgrundlagen f¨

ur Faserverbundbauteile. Springer,

Berlin u.a., 1992.

8. Berthelot, J.-M.: Composite Materials. Mechanical Behaviour and Structure

Analysis. Springer, New York et al., 1999.

9. Bogdanovich, A.E., C.M. Pastore: Mechanics of Textile and Laminated Compos-

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10. Buhl, H. (Ed.): Advanced Aerospace Materials. Materials Research and Engi-

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11. Carlsson, L.A., D.F. Adams, D.F., R.B. Pipes: Experimental Characterization of

Advanced Composite Materials. CRC Press, Boca Raton, 4rd edition, 2014.

12. Chawla, K.K.: Composite Materials. Science and Engineering. Springer, New

York, 2012.

13. Chung, D.D.L.: Composite Materials: Functional Materials for Modern Tech-

nologies. Springer, London, 2003.

14. Chung, D.D.L.: Composite Materials: Science and Applications. Springer, Lon-

don, 2010.

15. Daniel, I.M., O. Ishai: Engineering Mechanics of Composite Materials. Oxford

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16. Davies, J.M. (Ed.): Lightweight Sandwich Construction. Blackwell Science,

Oxford et al., 2001.

17. Decolon, C.: Analysis of Composite Structures. HPS, London, 2002.

18. Delhaes, P. (Ed.): Fibres and Composites. Taylor & Francis, London, 2003.

19. Dimitrienko, Yu.I.: Thermomechanics of Composites under High Temperature.

Springer Netherlands, Dordrecht, 2016.

20. Ehrenstein, G.W.: Faserverbund-Kunststoffe. Hanser, M¨unchen, Wien, 2nd ed.,

2006.

G.2 Selected Textbooks and Monographs on Composite Mechanics 491

21. Friedrich, K. (Ed.): Application of Fracture Mechanics to Composite Materials.

Bd. 6 Composite Material Series Elsevier, Amsterdam, 1989.

22. Gay, D.: Composite Materials: Design and Applications. CRC Press, Boca

Raton, 3rd. ed., 2014.

23. Geier, M.H.: Quality Handbook for Composite Materials. Chapman & Hall,

London et al., 1994.

24. Gibson, R.F.: Principles of Composite Material Mechanics. CRC Press, Boca

Raton, 4th ed., 2016.

25. Gibson, R.F.: Dynamic Mechanical Behavior of Composite Materials and Struc-

tures. CRC Press, Boca Raton, 2002.

26. G ¨urtal, Z., Haftka, R.T., Hajela, P.: Design and Optimization of Laminated Com-

posite Materials. John Willy & Sons Inc. New-York, 1999.

27. Harper, C.A. (Ed.): Handbook of Plastics, Elastomers, and Composites.

McGraw-Hill, New York et al., 4th edition, 2002.

28. Harris, B.: Engineering Composite Materials. IOM Communications Ltd., Lon-

don, 2nd edition, 1999.

29. Hoa, S.V.: Analysis for Design of Fiber Reinforced Plastic Vessels and Pipes.

Technomic, Lancaster, Basel, 1991.

30. Hoa, S.V., Wei Fang: Hybrid Finite Element Method for Stress Analysis of Lam-

inated Composites. Kluwer Academic Publishers. Dordrecht, 1998.

31. Hollaway, L.: Polymer Composites for Civil and Structural Engineering. Blackie

Academic & Professional, London et al., 1993.

32. Hult, J., F.G. Rammersdorfer (Eds): Engineering Mechanics of Fibre Rein-

forced Polymers and Composite Structures. CISM Courses and Lectures No.

348. Springer, Wien, New York, 1994.

33. Hyer, M.W.: Stress Analysis of Fibre-Reinforced Composite Materials. DEStech

Publ., Lancaster (PA), Uptadet ed., 2009.

34. Hull, D., T.W. Clyne: An Introduction to Composite Materials. Cambridge Uni-

versity Press. 2nd ed., 2003.

35. Jones, R.M.: Mechanics of Composite Materials. Taylor & Francis, London,

1999.

36. Kachanov, L.M.: Delamination Buckling of Composite Materials. Mechanics of

Elastic Stability, Vol. 14. Kluwer, Dordrecht, Boston, London, 1988.

37. Kalamkarov, A.L.: Composite and Reinforced Elements of Construction. Wiley

& Sons, Chichester et al., 1992.

38. Kalamkarov, A.L., A.G. Kolpakov: Analysis, Design and Optimization of Com-

posite Structures. Wiley & Sons, Chichester et al., 1997.

39. Kaw, A.K.: Mechanics of Composite Materials. CRC Press, Boca Rotan, New

York, 2nd ed., 2006.

40. Kim, D.-H.: Composite Structures for Civil and Architectural Engineering.

E&FN SPON, London et al., 1995.

41. Kollar, L.P., G.S. Springer: Mechanics of Composite Structures. Cambridge

University Press, Cambridge, 2003.

42. Matthews, F.L., R.D. Rawlings: Composite Materials: Engineering and Science.

Woodhead Publishing, Cambridge, 1999.

492 G References

43. Matthews, F.L., Davies, G.A.O., Hitching, D., Soutis, C.: Finite Element Mod-

elling of Composite Materials and Structures. CRC Press. Woodhead Publishing

Limited, Cambridge, 2000.

44. McCullough, R.L.: Micromechanical Materials Modelling. Delaware Compos-

ites Design Encyclopedia, Vol. 2. Technomic, Lancaster, Basel, 1991.

45. Michaeli, W., D. Huybrechts, M. Wegener: Dimensionieren mit Faserverbund-

kunststoffen: Einf¨

urung und praktische Hilfen. Hanser, M ¨unchen, Wien, 1995.

46. Milton, G. W.: The Theory of Composites. Cambridge University Press, Cam-

bridge, 2002.

47. Mittelstedt, C., W. Becker: Strukturmechanik ebener Laminate. TU Darmstadt,

Darmstadt, 2016.

48. Moser, K.: Faser-Kunststoff-Verbund: Entwurfs- und Berechnungsgrundlagen.

VDI-Verlag, D ¨usseldorf, 1992.

49. Newaz, G.M. (Ed.): Delamination in Advanced Composites. Technomic, Lan-

caster, 1991.

50. Nethercot, D.A.: Composite Construction. Spon Press, New York, 2003.

51. Ochoa, O.O., J.N. Reddy: Finite Element Analysis of Composite Laminates.

Solid Mechanics and its Applications, Vol. 7. Kluwer, Dordrecht, Boston, Lon-

don, 1992.

52. Plantema, F.J.: Sandwich Constructions. John Wiley & Sons, New York, 1966.

53. Powell, P.C.: Engineering with Fibre-polymer Laminates. Chapman & Hall,

London et al., 1994.

54. Puck, A.: Festigkeitsanalyse an Faser-Matrix-Laminaten: Realistische Bruchkri-

terien und Degradationsmodelle. Hanser, M”unchen, 1996.

55. Reddy, J.N.: Mechanics of Laminated Composite Plates - Theory and Analysis.

CRC Press, Boca Rotan et al., 1997.

56. Reddy, J.N.: Mechanics of Laminated Composite Plates and Shells: Theory and

Analysis. CRC Press, Boca Rotan et al., 2004.

57. Reddy, J.N., A. Miravete: Practical Analysis of Composite Laminates. CRC

Press, Boca Rotan et al., 1995.

58. Rohwer, K.: Modelle und Methoden zur Berechnung von Laminaten aus uni-

direktionalen Faserverbunden. Fortschritt-Berichte VDI: Reihe 1 Konstruktion-

stechnik, Maschinenelemente Nr. 264. VDI-Verlag, D¨usseldorf, 1996.

59. Schulte, K., B. Fiedler:: Structure and Properties of Composite Materials.

TUHH-Technologie GmbH, Hamburg, 2. Auﬂ., 2005.

60. Sih, G.C., A. Carpinteri, G. Surace (Eds): Advanced Technology for Design

and Fabrication of Composite Materials and Structures. Engng. Appl. of Fract.

Mech., Vol.14. Kluwer Academic Publ., Dordrecht, Boston, London, 1995.

61. Sih, G.C., A.M. Skudra (Eds): Failure Mechanics of Composites. Handbook of

Composites, Bd. 3. North-Holland, Amsterdam, New York, Oxford, 1985.

62. Talreja, R. (Ed.): Damage Mechanics of Composite Materials. Composite Ma-

terials Series, Vol. 9. Elsevier, Amsterdam et al., 1994.

63. Tarnopol’ski, J.M., T. Kincis: Test Methods for Composites. Van Nostrand Rein-

hold, New York, 1985.

64. Tsai, S.W.: Composites Design. Think Composites, Dayton, Paris, Tokyo, 1988.

G.3 Supplementary Literature for Further Reading 493

65. Turvey, G.J., I.H. Marshall (Eds.): Buckling and Postbuckling of Composite

Plates. Chapman & Hall, London, 1995.

66. Vasiliev, V.V., Jones, R.M. (Engl. Ed. Editor): Mechanics of Composite Struc-

tures. Taylor & Francis, Washington, 1993.

67. Vasiliev, V.V., Morozov, E.V.: Mechanics and Analysis of Composite Materials.

Elsevier, London, 2001.

68. Vasiliev, V.V., Morozov, E.V.: Advanced Mechanics of Composite Materials and

Structural Elements. Elsevier, London, 3rd ed., 2013.

69. Vinson, J.R.: The Behavior of Shells Composed of Isotropic and Composite

Materials. Solid Mechanics and its Applications, Vol. 18. Kluwer, Dordrecht,

Boston, London, 1993.

70. Vinson, J.R., R.L. Sierakowski: The Behavior of Structures Composed of Com-

posite Materials. Springer, Dordrecht et al., 3rd ed., 2008.

71. Whitney, J.M.: Structural Analysis of Laminated Anisotropic Plates. Technomic

Publishing Co. Inc., Lancaster, 1987.

72. Ye, J.: Laminated Composite Plates and Shells: 3D Modelling. Springer, London

et al., 2003.

73. Zweben, C., H.T. Hahn, T.-W. Chou: Mechanical Behavior and Properties of

Composite Materials. Delaware Composites Design Encyclopedia, Bd. 1. Tech-

nomic, Lancaster, Basel, 1989.

G.3 Supplementary Literature for Further Reading

1. Altenbach, H., J. Altenbach, K. Naumenko: Ebene Fl¨

achentragwerke - Grund-

lagen der Modellierung und Berechnung von Scheiben und Platten. Springer

Vieweg, Berlin, Heidelberg, New York, 2. Auﬂ., 2016.

2. Altenbach, H.: Kontinuumsmechanik - Einf¨

uhrung in die materialunabh¨

angigen

und materialabh ¨

angigen Gleichungen. Springer Vieweg, Berlin, Heidelberg, 3.

Auﬂ., 2015.

3. Altenbach, J., W. Kissing, H. Altenbach: D¨

unnwandige Stab- und Stabschalen-

tragwerke. Vieweg-Verlag, Braunschweig/Wiesbaden, 1994.

4. Altenbach, H., J. Altenbach, A. Zolochevsky: Erweiterte Deformationsmodelle

und Versagenskriterien der Werkstoffmechanik. Deutscher Verlag f ¨ur Grund-

stofﬁndustrie, Stuttgart, 1995.

5. Ambarcumyan, S.A.: Theory of Anisotropic Plates: Strength, Stability, and Vi-

brations. Hemisphere Publishing, Washington, 1991.

6. Betten, J.: Kontinuumsmechanik. Springer-Verlag. Berlin, Heidelberg, New York,

2nd edition, 2001.

7. Betten, J. Finite Elemente f¨

ur Ingenieure - Grundlagen, Matrixmethoden, Kon-

tinuum. Springer-Verlag, Berlin, Heidelberg, New York, 2nd edition, 2003.

8. Haupt, P.: Continuum Mechanics and Theory of Materials. Springer-Verlag.

Berlin, Heidelberg, New York, 2nd edition, 2002

494 G References

9. Lekhnitskii, S.G.: Anisotropic Plates. Gordon and Breach Science Publishers,

London, 3rd print, 1987.

10. Lekhnitskij, S.G.: Theory of Elasticity of an Anisotropic Body. Mir Publishers,

Moscow, 1981.

11. Lewi ´nski, T., J.J. Telega: Plates, Laminates and Shells Asymptotic Analysis and

Homogenization. World Scientiﬁc, Singapore, 2000.

12. Wlassow, V.S.: Allgemeine Schalentheorie und ihre Anwendung in der Technik.

Akademie-Verlag, Berlin, 1958.

13. Zienkiewicz, O.C., R.L. Taylor: The Finite Element Method, Vol. 2: Solid Me-

chanics. McGraw Hill, Oxford, 5th ed., 2000.

G.4 Selected Review Articles

1. Altenbach, H.: Modellierung des Deformationsverhaltens mehrschichtiger Fl ¨

a-

chentragwerke - ein ¨

Uberblick zu Forschungsrichtungen und -tendenzen. Wiss.

Ztschr. TH Magdeburg 32(4): 86 – 94, 1988.

2. Bert, C.W.A. A critical evaluation of new plate theories applied to laminated

composites. Comp. Struc. 2, 329 – 347, 1984.

3. Carrera, E., L. Demasi, M. Manganello: Assessment of plate elements on bendi-

jng and vibration of composite structures. Mech. of Adv. Mat. and Struct. 9, 333

– 357, 2002.

4. Christenson, R.M.: A survey of and evaluation methodology for Fiber Composite

Material Failure Theories. In ”Mechanics for a New Millennium”, Eds H. Aref

and J.W. Philips, 25 – 40, 1998.

5. Ha, K.H.: Finite element analysis of sandwich construction: a critical review.

Sandwich Construction 1, 69 – 85, 1989.

6. Failure criteria in ﬁbre-reinforced polymer composites. Special Issue of Com-

posites Science and Technology 58, 1998.

7. Hashin, Z.: Analysis of composite materials - A Survey. Trans. ASME. J. Appl.

Mech. 50: 481 – 505, 1983.

8. Hohe, J., W. Becker: Effective stress-strain relations for two-dimensional cel-

lular sandwich cores: Homogenization, material models, and properties. Appl.

Mech. Rev. 54: 61 – 87, 2001.

9. Irschik, H.: On vibration of layered beams and plates. ZAMM 73 (4-5), T34 –

T45, 1993.

10. Leissa, W.W.: A review of laminated composite plate buckling. Appl. Mech. Rev.

40 (5), 575 – 590, 1987.

11. Lui, M., P. Habip: A survey of modern developments in the analysis of sandwich

structures. Appl. Mech. Rev. 18(2), 93 – 98, 1965.

12. Mallikarguwa, T. Kant: A critical review and some results of recently developed

reﬁned theories of ﬁber-reinforced laminated composites and sandwiches. Comp.

Structures 23, 293 – 312, 1993.

G.4 Selected Review Articles 495

13. Nahas, M.N.: Survey of failure and post-failure theories of laminated ﬁbre-

reinforced composites. J. Composites Technology & Research 8: 138 – 153,

1986.

14. Naumenko, K., J. Altenbach, H. Altenbach, V.K. Naumenko: Closed and ap-

proximate analytical solutions for rectangular Mindlin plates. Acta Mechanica

147: 153 – 172, 2001.

15. Noor, A.K., W.S. Burton: Assessment of shear deformation theories for multilay-

ered composite plates. Appl. Mech. Rev. 41(1): 1 – 13, 1989.

16. Noor, A.K., W.S. Burton, C.W. Bert: Computational models for sandwich panels

and shells. Appl. Mech. Rev. 49(3), 155 – 199, 1996.

17. Noor, A.K., W.S. Burton, J.U. Peters: Assessment of computational models for

multilayered composite cylinders. Int. J. Solids Structures, 27 (10), 1269 – 1286,

1991.

18. Vinson, J.R.: Sandwich structures. Appl. Mech. Rev. 54(3), 201 – 214, 2001.

19. Reddy, J.N.: A review of reﬁned theories of laminated composite plates. Shock

Vibr. Dig. 22: 3 – 17, 1990.

20. Reddy, J.N.: An evaluation of equivalent-single-layer and layer theories of com-

posite laminates. Composite Structures 25, 21 – 35, 1993.

21. Reddy, J.N., Robbins Jr., D.H. Theories and computational models for composite

laminates. Appl. Mech. Rev. 47 (6) 147 – 169, 1994.

22. Rohwer, K. Computational models for laminated composites. Z. Flugwiss. Wel-

traumforsch. 17, 323 – 330, 1993.

Index

Airy stress function, 290

analytical solution, 306

anisotropy

curvilinear, 51

cylindrical, 51

rectilinear, 51

approximation function, 414

auxetic material, 22

basic modes of failure of a lamina, 210

beam, 419

beam element, 413

laminate, 450

beam equations, 251

elementary, 233

beam resultants, 236

beam shaped shell, 371

beam shaped thin-walled folded structure, 414

beam theory

Bernoulli-Euler, 232

classical, 232

elementary, 232

generalized

Vlasov , 232

Timoshenko, 232, 246

bending

cylindrical, 294

bending stress, 126

bending-layer solution, 352

Bernoulli

Jakob I., 77

Bernoulli beam, 416

Bernoulli beam model, 422

Saint-Venant supplement, 373

specialized, 391

Boole

George, 190

boundary condition, 296

boundary conditions, 283

Bredt

Rudolf, 373

buckling, 286, 288, 297, 304, 305, 310, 318,

321

buckling equation, 242, 251

bulk modulus, 43

Christensen

Richard M., 223

classical beam model, 372

classical laminate theory, 181, 281, 286, 293,

303, 306, 376, 422, 432

closed thin-walled cross-section, 380

CLT, 181

coincidence matrix, 417

collocation method, 73

column, 232

compatibility conditions, 60

compliance hypermatrix, 143

compliance matrix, 27, 30, 32, 36

bending, 157

coupling, 157

extensional, 157

ﬂexural, 125

monoclinic material, 33

off-axis extensional, 124

off-axis ﬂexural, 125

off-axis in-plane, 123

on-axis extensional, 124

on-axis ﬂexural, 125

orthotropic material, 34

transversely isotropic material, 35

compliance modulus, 22

compliance submatrix, 143

concentrated transversal stiffener, 389

497

H. Altenbach et al., Mechanics of Composite Structural

Elements, https://doi.org/10.1007/978-981-10-8935-0

498 Index

constitutive equations, 62

constitutive equations of the lamina resultants,

126

continuity

element boundaries, 415

contraction, 22

coordinate functions, 70

coupling inertia term, 298

curvature, 281, 282

curvatures

longitudinal, 385

torsional, 385

cylindrical bending, 306, 320

d’Alambert principle, 67

damping matrix, 416

delamination, 209

density, 392

differential equation of ﬂexure, 241

differential matrix, 62

discretization of the structure, 414

displacement

virtual, 64

displacement method, 66, 413

distorsion strain energy criteria of Tsai-Hill,

211

distortional energy criterion, 214

effective compliance, 24

effective hygrothermal coefﬁcients, 112

effective moduli, 21

effective stiffness, 24

effective thermal expansion coefﬁcients, 112

eigen-vibration, 403

eigenvalue problem, 318, 322

elastic parameters, 43

elasticity modulus, 22

elasticity tensor, 26

end matrix, 400

energy principles, 63

engineering parameters, 36

equilibrium equations

dynamic, 59, 62

plate, 281

static, 59, 62

surface, 62

equivalent single layer theory

higher order, 182

equivalent single-layer theory, 181, 183

ESLT, 181

Euler

Leonhard, 65

extended Vlasov-Kantorowich method, 75

extensional strains, 21, 24

external virtual work, 64

extremal principles, 63

F¨oppl

August Otto, 330

face wrinkling, 263

ﬁnite element, 413

beam, 413

generalized, 413

one-dimensional, 442

plate, 413

laminate, 439

triangular, 429, 434

ﬁnite element procedure, 414

ﬁrst order shear deformation theory, 246, 251,

280, 295, 301, 436

ﬁrst-order shear deformation theory, 181, 183

ﬁrst-ply failure, 211

ﬁxed boundary, 296

ﬂexibility matrix, 27

ﬂexibility modulus, 22

ﬂexure equations, 251

folded plate structure, 371, 375

force method, 66, 413

force resultant, 348, 377

forced vibration, 242, 251

forces

in-plane, 122

transverse, 122

four-node element, 423

fracture

interlaminar, 206

intralaminar, 206

translaminar, 206

fracture modes, 207

free boundary, 296

free edge, 297

free vibration, 242, 251

FSDT, 181

Galerkin

Boris Grigorjewitsch, 73

Galerkin’s method, 73

general three-dimensional problem, 416

generalized beam, 441

generalized co-ordinate function, 380, 392

generalized coordinates, 70

generalized densities

matrix, 439, 440

generalized density, 426

generalized displacement function, 392

generalized mass density, 298

Hamilton

Index 499

William Rowan, 68

Hamilton principle, 68, 299, 393

Hamilton’s principle, 449

Hamilton‘s principle, 414, 416

harmonic oscillation, 318

higher order theory, 295

Hill

Rodney, 211

Hoffman criterion, 217

Hooke

Robert, 20

Hooke’s law, 21

generalized, 26

inverse form, 22

hybrid criteria, 210

hybrid method, 413

hygrothermal loading, 263

hypotheses

Bernoulli, 186

Kirchhoff, 186

Love, 186

in-plane strain, 282

in-plane stress resultant force vector, 122

inner energy

element, 417

interactive criteria, 210, 214

interactive tensor polynomial criterion of

Tsai-Wu, 211

internal virtual work, 64

iso-strain condition, 90

iso-stress condition, 91

Jacobi matrix, 431

Kantorovich

Leonid Vitaliyevich, 74

Kantorovich separation relationships, 380

kinematic equations, 60, 62

kinetic energy, 392

rotatory, 438

Kirchhoff

Gustav Robert, 181

Kirchhoff hypotheses, 281, 376

Kirchhoff plate, 416, 429

Kirchhoff shear force resultant, 283

Krylov method, 395

Krylow

Aleksei Nikolajewitsch, 395

Krylow functions, 485

Kutta

Martin Wilhelm, 394

L´evy

Maurice, 306

Lagrange

Joseph-Louis, 65

Lagrange function, 393

Lagrangian function, 68

Lam´e

Gabriel L´eon Jean Baptiste, 43

Lam´e coefﬁcients, 43

lamina, 11

laminae

UD-laminae, 109

unidirectional, 109

laminate, 11

angle-ply, 139, 293, 294

antisymmetric, 138, 156, 285

balanced, 153

cross-ply, 152

asymmetric, 138

balanced, 139, 156, 285

beam element, 419

cross-ply, 139, 156, 285, 293, 294

general, 139, 156

isotropic layers, 293

nonsymmetric, 440

quasi-isotropic, 157

special orthotropic, 436

specially orthotropic, 280

symmetric, 138, 156, 285, 436

angle-ply, 138

balanced, 150

cross-ply, 149

general case, 289

regular, 138

special case with isotropic layers, 290

with isotropic layers, 148

truss element, 419

unsymmetric, 138

with isotropic layers, 157

laminate beam

symmetric, 252

laminate code, 138

laminate plate

arbitrary stacking, 280

symmetric, 288

laminate theory

classical, 183

second order, 183

third order, 183

laminates

specially orthotropic, 290

last-ply failure, 211

layer, 11

Layerwise theory, 197

layerwise theory, 182

500 Index

least-squares method, 73

limit criteria, 210

Love

Augustus Edward Hough, 186

Love’s ﬁrst-approximation shell theory, 347

macro-mechanical level, 11

macro-mechanical modelling, 108

macroscopic approach, 87

macroscopic level, 13

mass matrix

condensed, 416

consistent, 416

element, 439, 450

symmetric, 416

material

anisotropic, 4

arranged in parallel, 23

arranged in series, 23

ceramics, 4

composite

advantages, 15

classiﬁcation, 5

examples, 5

limitations, 16

matrix, 7

reinforcement, 5

heterogeneous, 4

homogeneous, 4

inhomogeneous, 4

isotropic, 4, 36, 42

metals, 4

monoclinic, 32, 43

monolithic, 5

orthotropic, 34

polymers, 4

transversely isotropic, 35, 40

triclinic, 32

maximum strain theory, 213

maximum stress theory, 211

membrane stress, 126

membrane theory, 354

mesh, 413

method of initial parameters, 395

micro-mechanical level, 11

microscopic approach, 87

microscopic level, 13

mid-plane strain, 123

Mindlin

Raymond David, 295

Mindlin plate, 300, 325, 429, 436

Mindlin plate theory, 295

Mises

Richard Edler von, 214

modelling

one-dimensional, 371

three-dimensional, 371

two-dimensional, 371

moment

resultant, 122, 124

moment resultant, 348, 377

multidirectional laminate, 136

N´adai

´

Arp´ad, 306

N´adai-L´evy solution, 306, 316, 326

nanoscale level, 13

Navier

Claude Louis Marie Henri, 306

Navier solution, 306, 312, 324

nodal force

vector

time dependent, 416

node, 413

normal stress, 21

normal stresses, 24

number of degrees of freedom, 413

off-axis case, 110

off-axis loaded UD-lamina, 113

off-axis stretching, 123

on-axis case, 109

one-dimensional element, 413

one-dimensional structure, 419

optimal global laminate behavior, 137

plane beam problem, 418

plane strain problem, 416

plane strain state, 48

plane stress problem, 416

plane stress state, 45

plate, 279, 429

bending, 293

buckling, 293

moderately thick, 295

specially orthotropic, 288, 298

symmetric

general case, 287

vibration, 293

plate element, 413

laminate, 450

plate strip, 232, 307, 320

symmetrical laminated, 308

unsymmetric laminated, 308

plate theory

classical

Kirchhoff, 181

ply, 11

Index 501

Poisson

Sim´eon Denis, 15

Poisson effect, 308

Poisson’s ratio, 22, 43

major, 92

minor, 93

potential energy, 383, 392, 420, 436, 440

folded structure, 379

single element, 417

whole structure, 417

prebuckling, 304

principle of complementary virtual work, 63,

65

principle of minimum potential energy, 299

principle of minimum total complementary

energy, 66

principle of minimum total potential energy, 65

principle of the total minimum potential

energy, 414

principle of virtual displacements, 66

principle of virtual forces, 66

principle of virtual work, 63, 64

prismatic structure, 376

proﬁle node

main, 442

proﬁle node concept, 442

proﬁle nodes

secondary, 443

Puck

Alfred, 223

Rayleigh, 69

Rayleigh-Ritz method, 69

reduced compliances, 48

reduced stiffness, 47

Reissner

Eric (Max Erich), 63

Reissner plate theory, 295

Reissner theory, 195

Reissner’s functional, 67

Reissner’s variational theorem, 63, 66

resultant

in-plane force, 282, 288

moment, 282

transverse shear force, 282

resultant moment vector, 124

Reuss

Andr´as (Endre), 24, 88

Reuss estimate, 92

Reuss model, 24, 91, 93, 94

Ritz

Walter, 69

Ritz approximation, 70, 72

Ritz method, 69, 414

rod, 232

rotation matrix, 28

rotational term, 392

rotatory inertia, 251, 286, 323

rule of mixture, 90

rule of mixtures, 88, 94, 96

inverse, 88, 92

Runge

Carl David Tolm ´e, 394

Sachs

Oscar, 88

Saint-Venant

Adh´emar Jean Claude Barr´e de, 373

sandwich, 11

symmetric

thick cover sheets, 302

thin cover sheets, 301

sandwich beam

dissimilar faces, 258

symmetric, 252

sandwich composites, 172

sandwich plate, 301

Schapery

Richard Allan, 112

selective integration, 438

semi-empirical solution of Halpin and Tsai, 96

semi-membrane theory, 355

shape function, 414, 415, 423, 424, 431, 434,

442

linear, 420

shear correction coefﬁcient, 300

shear correction factor, 142, 195, 246, 251

shear deformation theory, 306

shear deformations of the mid-planes, 385

shear lag effect, 374, 396

shear modulus, 22, 43

shear rigid theory, 187

shear strains, 21

engineering, 25

tensorial, 25

shear stress, 21

shell, 345

circular cylindrical, 345

circumferential cross-ply, 345

special orthotropic, 345

classical theory, 345, 346

hypotheses, 347

ﬁrst order shear deformation theory, 345

moderately thick, 356

thin-walled, 345

transverse shear deformations, 346

shell element, 413

laminate, 450

502 Index

simply supported boundary, 297

six-node element, 437

stacking

symmetric, 390

stacking codes of laminates, 137

stiffness

plate, 288

reduced, 111

shear, 194

stiffness matrix, 27, 30, 32, 36

bending, 141

coupling, 141

element, 417, 450

extensional, 123, 141

ﬂexural, 125

isotropic material, 36

monoclinic material, 33

off-axis extensional, 124

off-axis ﬂexural, 125

on-axis extensional, 124

on-axis ﬂexural, 125

orthotropic material, 34

reduced, 377

symmetric, 415

transverse shear, 142

transversely isotropic material, 35

stiffness submatrix, 143

strain tensor, 24, 25

strain vector, 25

strain-displacement relations, 60

stress resultant, 283, 299, 449

plate, 281

stress resultants, 233

stress tensor, 24, 25

stress vector, 25

structural behavior

global, 372

local, 372

structural level, 13

Strutt

John William, 69

submatrix

bending, 148

coupling, 147

extensional, 148

surface force

vector, 415

Taylor

Geoffrey Ingram, 88

tensile compliance, 22

tensile ﬂexibility, 22

tensile stiffness, 22

theorem of Castigliano, 66

thin-walled beam, 371

three-node element, 422, 437, 442

Timoshenko

Stepan Prokopovich, 80

Timoshenko beam model, 373, 422

specialized, 391

total virtual work, 64

transfer matrix, 398

transfer matrix method, 396

transformation matrix, 28, 29, 62

translatory inertia, 251, 323

transversal strains, 385

transverse force resultant, 378

transverse shear, 376

transverse shear deformation, 246

transverse shear resultant, 125

transverse shear stress, 295

trial function, 380

triangle co-ordinates

natural, 438

triangle coordinates

natural, 430

truss element, 413

Tsai

Stephen Wei-Lun, 96, 211

Tsai-Hill criterion, 215

Tsai-Wu criterion, 216

two-dimensional element, 413

two-dimensional structural element, 345

two-dimensional structure, 429

two-node element, 442

beam, 422

truss, 420

unknown coefﬁcient function, 380

variational formulation

axial symmetrically circular cylindrical

shell, 358

variational iteration method, 75

variational operations, 64

vector of curvature, 125

vibration, 286, 310, 318, 421

forced, 297

forced transversal, 286

free, 323, 392

Vlasov

Vasily Zakharovich, 74

Vlasov beam model, 373, 391

Vlasov hypotheses, 380

Voigt

Woldemar, 23, 88

Voigt estimate, 90

Voigt model, 23, 90, 92, 94

Index 503

volume element, 413

volume force

vector, 415

warping, 373

weak form of the model equations, 70

weighted residual methods, 73

weighted-residual methods, 69

Wu

Edward Ming-Chi, 211

Young

Thomas, 14

Young’s modulus, 22, 43

Classification of Composite Materials

Abstract

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