Mechanics Of Composite Structural Elements 2018 [613009]

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Mechanics Of Composite Structural Elements 2018 [613009]

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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 fü r Mechanik
Otto-von-Guericke-Universit ät Magdeburg
Magdeburg, Saxony-AnhaltGermany
Johannes Altenbach
MagdeburgGermanyWolfgang 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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or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilarmethodology now known or hereafter developed.The use of general descriptive names, registered names, trademarks, service marks, etc. in thispublication does not imply, even in the absence of a speci ﬁc statement, that such names are exempt from
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 thisbook are believed to be true and accurate at the date of publication. Neither the publisher nor theauthors or the editors give a warranty, express or implied, with respect to the material contained herein orfor any errors or omissions that may have been made. The publisher remains neutral with regard tojurisdictional claims in published maps and institutional af ﬁliations.
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part of Springer NatureThe 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 de livered at
several universities in Germany and abroad. Throughout the past years the authors
received a lot of suggestions for improvements from student s and colleagues alike.
In addition, the textbook is recommended as the basic readin g 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 cam pus) for third and
fourth year students of the bachelor program in the departme nts of Mechanical Engi-
neering and Civil Engineering. The two weeks’ course includ ed 60 hours of lectures
and tutorials. Finally, the course was concluded with a writ ten exam and a project.
Our special thanks are due to Dr. Christoph Baumann and Sprin ger who provided
personal copies of the ﬁrst edition of the book for the attend ants of the course. As
the result of the discussions with the students the idea was b orn to prepare a second
edition.
By and large the preliminaries of the ﬁrst edition remain unc hanged: the presen-
tation of the mechanics of composite materials is based on th e knowledge of the
ﬁrst and second year of the bachelor program in Engineering M echanics (or in other
countries the courses of General Mechanics and Strength of M aterials). 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 wi th the ﬁrst one:
•some problems are added or clariﬁed (and we hope now better un derstandable),
•Chapter 11 is slightly shortened (some details are no more im portant),
•some details were adopted considering the developments of S pringer’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 hop e you will now ﬁnd
fewer misprints and typos.
We have to acknowledge Dr.-Ing. Heinz K¨ oppe (Otto-von-Gue ricke-Universit¨ at
Magdeburg) and Dipl.-Ing. Christoph Kammer (formaly at Ott o-von-Guericke-
vPreface to the 2ndEdition

vi Preface
Universit¨ at Magdeburg) for ﬁnding a lot of typos. In additi on, we have to thank
Dr. Christoph Baumann (Executive Editor Engineering, Spri nger Nature Singapore)
for the permanent support of the project. We appreciate for a ny 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 1stEdition
Laminate and sandwich structures are typical lightweight e lements with rapidly ex-
panding application in various industrial ﬁelds. In the pas t, these structures were
used primarily in aircraft and aerospace industries. Now, t hey have also found ap-
plication in civil and mechanical engineering, in the autom otive industry, in ship-
building, the sport goods industries, etc. The advantages t hat these materials have
over traditional materials like metals and their alloys are the relatively high speciﬁc
strength properties (the ratio strength to density, etc). I n addition, the laminate and
sandwich structures provide good vibration and noise prote ction, thermal insulation,
etc. There are also disadvantages – for example, composite l aminates are brittle, and
the joining of such elements is not as easy as with classical m aterials. The recycling
of these materials is also problematic, and a viable solutio n is yet to be developed.
Since the application of laminates and sandwiches has been u sed mostly in new
technologies, governmental and independent research orga nizations, as well as big
companies, have spent a lot of money for research. This inclu des the development
of new materials by material scientists, new design concept s by mechanical and
civil engineers as well as new testing procedures and standa rds. The growing de-
mands of the industry for specially educated research and pr acticing engineers and
material scientists have resulted in changes in curricula o f the diploma and master
courses. More and more universities have included special c ourses on laminates and
sandwiches, and training programs have been arranged for po stgraduate studies.
The concept of this textbook was born 10 years ago. At that tim e, the ﬁrst edition
of ”Einf¨ uhrung in die Mechanik der Laminat- und Sandwichtr agwerke”, prepared
by H. Altenbach, J. Altenbach and R. Rikards, was written for German students
only. The purpose of that book consisted the following objec tives:
•to provide a basic understanding of composite materials lik e 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 trea tment of composite
structures and
vii

viii Preface
•to discuss the limitations of analysis and modelling concep ts.
These four items are also included in this textbook. It must b e noted that between
1997 and 2000, there was a common education project sponsore d by the European
Community (coordinator T. Sadowski) with the participatio n of colleagues from
U.K., Belgium, Poland and Germany. One of the main results wa s a new created
course on laminates and sandwiches, and ﬁnally an English te xtbook ”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 v ersions, but has been
signiﬁcantly expanded. It can be characterized by the follo wing items:
•The textbook is written in the style of classical courses of s trength of materials (or
mechanics of materials) and theory of beams, plates and shel ls. In this sense the
course (textbook) can be recommended for master students wi th bachelor degree
and diploma students which have ﬁnished the second year in th e university. In
addition, postgraduates of various levels can ﬁnd a simple i ntroduction to the
analysis and modelling of laminate and sandwich structures .
•In contrast to the traditional courses referred to above, tw o extensions have been
included. Firstly, consideration is given to the linear ela stic material behavior of
both isotropic and anisotropic structural elements. Secon dly, the case of inhomo-
geneous material properties in the thickness direction was also included.
•Composite structures are mostly thin, in which case a dimens ion reduction of the
governing equations is allowed in many applications. Due to this fact, the one-
dimensional equations for beams and the two-dimensional eq uations for plates
and shells are introduced. The presented analytical soluti ons can be related to the
in-plane, out-of-plane and coupled behavior.
•Sandwiches are introduced as a special case of general lamin ates. This results in
signiﬁcant simpliﬁcations because sandwiches with thin or thicker faces can be
modelled and analyzed in the frame of laminate theories of di fferent order and
so a special sandwich theory is not necessary.
•All analysis concepts are introduced for the global structu ral behavior. Local
effects and their analysis must be based on three-dimension al ﬁeld equations
which can usually be solved with the help of numerical method s. It must be
noted that the thermomechanical properties of composites o n polymer matrix at
high temperatures can be essentially different from those a t normal temperatures.
In engineering applications generally three levels of temp erature 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 me chanical properties,
and thus are not included in modelling and analysis. All ther mal and moisture ef-
fects are considered in such a way that the mechanical proper ties can be assumed
unchanged.

Preface ix
•Finite element analysis is only brieﬂy presented. A basic co urse in ﬁnite elements
is necessary for the understanding of this part of the book. I t should be noted
that the ﬁnite element method is general accepted for the num erical analysis of
laminate and sandwich structures. This was the reason to inc lude this item in the
contents of this book.
The textbook is divided into 11 chapters and several appendi ces 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, variat ional methods and the basic
micromechanical models are presented.
•The second part (Chaps. 4–6) can be related to the modelling f rom single laminae
to laminates including sandwiches, the improved theories a nd 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 an d sandwiches. The
modelling of laminated and sandwich plates and shells is res tricted to rectan-
gular plates and circular cylindrical shells. The individu al ﬁber reinforced lam-
inae of laminated structured elements are considered to be h omogeneous and
orthotropic, but the laminate is heterogeneous through the thickness and gener-
ally anisotropic. An equivalent single layer theory using t he classical lamination
theory, and the ﬁrst order shear deformation theory are cons idered. Multilayered
theories or laminate theories of higher order are not discus sed in detail.
•The fourth part (Chap. 10) includes the modelling and analys is of thin-walled
folded plate structures or generalized beams. This topic is not normally consid-
ered in standard textbooks on structural analysis of lamina tes and sandwiches, but
it was included here because it demonstrates the possible ap plication of Vlasov’s
theory of thin-walled beams and semi-membrane shells on lam inated structural
elements.
•Finally, the ﬁfth part (Chap. 11) presents a short introduct ion into the ﬁnite el-
ement procedures and developed ﬁnite classical and general ized beam elements
and ﬁnite plate elements in the frame of classical and ﬁrst or der shear defor-
mation theory. Selected examples demonstrate the possibil ities of ﬁnite element
analysis.
This textbook is written for use not only in engineering curr icula of aerospace, civil
and mechanical engineering, but also in material science an d applied mechanics. In
addition, the book may be useful for practicing engineers, l ectors and researchers in
the mechanics of structures composed of composite material s.
The strongest feature of the book is its use as a textbook. No p rior knowledge of
composite materials and structures is required for the unde rstanding of its content.
It intends to give an in-depth view of the problems considere d and therefore the
number of topics considered is limited. A large number of sol ved problems are
included to assess the knowledge of the presented topics. Th e list of references at
the end of the book focuses on three groups of suggested readi ng:

x Preface
•Firstly, a selection of textbooks and monographs of composi te materials and
structures are listed, which constitute the necessary item s for further reading.
They are selected to reinforce the presented topics and to pr ovide information
on topics not discussed. We hope that our colleagues agree th at the number of
recommended books for a textbook must be limited and we have g iven priority
to newer books available in university libraries.
•Some books on elasticity, continuum mechanics, plates and s hells and FEM are
recommended for further reading, and a deeper understandin g of the mathemati-
cal, mechanical and numerical topics.
•A list of review articles shall enable the reader to become in formed about the
numerous books and proceedings in composite mechanics.
The technical realization of this textbook was possible onl y with the support of
various friends and colleagues. Firstly, we would like to ex press our special thanks
to K. Naumenko and O. Dyogtev for drawing most of the ﬁgures. S econdly, Mrs. B.
Renner and T. Kumar performed many corrections of the Englis h text. At the same
time Mrs. Renner checked the problems and solutions. We rece ived access to the
necessary literature by Mrs. N. Altenbach. Finally, the pro cessing of the text was
done by Mrs. S. Runkel. We would also like to thank Springer Pu blishing for their
service.
Any comments or remarks are welcome and we kindly ask them to b e 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, Micromechan ics
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 Matri ces . 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 . . . . . . . . . . . . . . . . . 4 3
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 Proced ures . . . . . . 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 . . . . . . . . . . . . . . . . . . . . . . . . 6 9
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 Sandwic hes
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 . . . . . . . . . . . . . . . . . . . . . . . . 14 3
4.2.3.2 Antisymmetric Laminates . . . . . . . . . . . . . . . . . . . . . 1 48
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-Sect ions . . 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 . . . . . . . . . . . . . . . . . . . 34 6
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 S tructures
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 Gener al
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, Fini te 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 . . . . . . . . . . 4 52
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 Ordin ary
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 Mechani cs . . . . 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, variat ional methods and the basic mi-
cromechanical models are presented.
The laminates are introduced as layered structures each of t he layers is a ﬁbre-
reinforced material composed of high-modulus, high-stren gth ﬁbers in a polymeric,
metallic, or ceramic matrix material. Examples of ﬁbers use d include graphite, glass,
boron, and silicon carbide, matrix materials are epoxies, p olyamide, aluminium, ti-
tanium, and aluminium. A sandwich is a special class of compo site 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 e quations contain more
than two material parameters. In addition, the transition f rom the general three-
dimensional equations to the special two-dimensional equa tions results in more
complicated constrains. At the same time the introduction o f reduced stiffness and
compliance parameters result in a powerful tool for the anal ysis of laminates.
The variational methods are the base of many numerical solut ion techniques (for
example, the ﬁnite element method). Here only the classical principles and methods
are brieﬂy discussed.
There are many, partly sophisticated micromechanical appr oaches. They are the
base of a better understanding of the local behavior. In the f ocus of this textbook is
the global structural analysis. Thats way the micromechani cal models are presented
only in the elementary form.

Chapter 1
Classiﬁcation of Composite Materials
Fibre reinforced polymer composite systems have become inc reasingly important in
a variety of engineering ﬁelds. Naturally, the rapid growth in the use of composite
materials for structural elements has motivated the extens ion of existing theories in
structural mechanics, therein. The main topics of this text book are
•a short introduction into the linear mechanics of deformabl e solids with an-
isotropic material behavior,
•the mechanical behavior of composite materials as unidirec tional reinforced sin-
gle layers or laminated composite materials, the analysis o f effective moduli,
some basic mechanisms and criteria of failure,
•the modelling of the mechanical behavior of laminates and sa ndwiches, gen-
eral assumptions of various theories, classical laminate t heory (CLT), effect of
stacking of the layers of laminates and the coupling of stret ching, bending and
twisting, ﬁrst order shear deformation theory (FOSDT), an o verview on reﬁned
equivalent single layer plate theories and on multilayered plate modelling,
•modelling and analysis of laminate and sandwich beams, plat es 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, p lates 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 f olded 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 ma terial user as for
classical materials as steel or alloys. Structural enginee ring 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 c omposites and how
they can be classiﬁed, what are the main characteristics and signiﬁcance, micro-
3 © Springer Nature Singapore Pte Ltd. 2018
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 a dvantages 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 thre e categories
•metals,
•ceramics and
•polymers.
It is difﬁcult to give an exact assessment of the advantages a nd disadvantages of
these three basic material classes, because each category c overs whole groups of
materials within which the range of properties is often as br oad as the differences
between the three material classes. But at the simplistic le vel some obvious charac-
teristic properties can be identiﬁed:
•Mostly metals are of medium to high density. They have good th ermal stability
and can be made corrosion-resistant by alloying. Metals hav e useful mechanical
characteristics and it is moderately easy to shape and join. For this reason metals
became the preferred structural engineering material, the y posed less problems
to the designer than either ceramic or polymer materials.
•Ceramic materials have great thermal stability and are resi stant to corrosion,
abrasion and other forms of attack. They are very rigid but mo stly brittle and
can only be shaped with difﬁculty.
•Polymer materials (plastics) are of low density, have good c hemical resistance
but lack thermal stability. They have poor mechanical prope rties, but are eas-
ily fabricated and joined. Their resistance to environment al degradation, e.g. the
photomechanical effects of sunlight, is moderate.
A material is called homogeneous if its properties are the sa me at every point and
therefore independent of the location. Homogeneity is asso ciated with the scale of
modelling or the so-called characteristic volume and the de ﬁnition describes the
average material behavior on a macroscopic level. On a micro scopic level all ma-
terials are more or less homogeneous but depending on the sca le, materials can be
described as homogeneous, quasi-homogeneous or inhomogen eous. 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 anisotro pic. A general anisotropic
material has no planes or axes of material symmetry, but in Se ct. 2.1.3 some special
kinds of material symmetries like orthotropy, transverse i sotropy, 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 ment ioned classes of conven-
tional materials are on the macroscopic level more or less mo nolithic, homogeneous
and isotropic.
The group of materials which can be deﬁned as composite mater ials is extremely
large. Its boundaries depend on deﬁnition. In the most gener al 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 materia l properties derived from
the individual constituents. These properties may have the combined characteristics
of the constituents or they are substantially different. So metimes the material prop-
erties of a composite material may exceed those of the consti tuents. This general
deﬁnition of composites includes natural materials like wo od, traditional structural
materials like concrete, as well as modern synthetic compos ites such as ﬁbre or par-
ticle reinforced plastics which are now an important group o f engineering materials
where low weight in combination with high strength and stiff ness are required in
structural design.
In the more restrictive sense of this textbook a structural c omposite consists of
an assembly of two materials of different nature. In general , one material is dis-
continuous and is called the reinforcement, the other mater ial is mostly less stiff
and weaker. It is continuous and is called the matrix. The pro perties of a composite
material depends on
•The properties of the constituents,
•The geometry of the reinforcements, their distribution, or ientation and concen-
tration usually measured by the volume fraction or ﬁber volu me ratio,
•The nature and quality of the matrix-reinforcement interfa ce.
In a less restricted sense, a structural composite can consi st of two or more phases
on the macroscopic level. The mechanical performance and pr operties of compos-
ite materials are superior to those of their components or co nstituent materials taken
separately. The concentration of the reinforcement phase i s a determining parameter
of the properties of the new material, their distribution de termines the homogeneity
or the heterogeneity on the macroscopic scale. The most impo rtant 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 t wo or more dis-
tinct constituents or phases it must be considered that all c onstituents have to be
present in reasonable proportions that the constituent pha ses have quite different
properties from the properties of the composite material an d that man-made com-
posites are produced by combining the constituents by vario us means. Figure 1.1
demonstrates typical examples of composite materials. Com posites 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
a b
s sss
ss
sstss ssssc d
e f g
♣♣♣♣♣♣♣
♣♣♣♣♣♣♣
♣♣♣♣♣♣♣
♣♣♣♣♣♣♣
♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣
♣♣♣♣♣♣♣♣♣♣
♣♣♣♣♣♣♣♣♣♣
♣♣♣♣♣♣♣♣♣♣
♣♣♣♣♣♣♣♣♣♣
♣♣♣♣♣♣♣♣♣♣
♣♣♣♣♣♣♣♣♣♣
♣♣♣♣♣♣♣♣♣♣
♣♣♣♣♣♣♣♣♣♣
♣♣♣♣♣♣♣♣♣♣
♣♣♣♣♣♣♣♣♣♣
♣♣♣♣♣♣♣♣♣
♣♣♣♣♣♣♣♣♣
♣♣♣♣♣♣♣♣♣
♣♣♣♣♣♣♣♣♣
♣♣♣♣♣♣♣♣♣
♣♣♣♣♣♣♣♣♣
♣♣♣♣♣♣♣♣♣ h
Fig. 1.1 Examples of composite materials with different forms of con stituents 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 lon g ﬁbres are possible. The
arrangement and the orientation of long or short ﬁbres deter mines the mechani-
cal properties of composites and the behavior ranges betwee n a general anisotropy
to a quasi-isotropy. Particulate reinforcements have diff erent 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 major ity of practical applica-
tions particulate reinforced composites are considered to be randomly oriented and
unidirectional
reinforcedbidirectional
reinforcedspatial
reinforcedrandom
orientationpreferred orientationcontinous ﬁbre reinforced
(long ﬁbres)discontinous ﬁbre reinforced
(short ﬁbres)random
orientationpreferred
orientationﬁbre reinforced particle reinforcedComposite
Fig. 1.2 Classiﬁcation of composites

1.1 Deﬁnition and Characteristics 7
the mechanical properties are homogeneous and isotropic (f or mor details Chris-
tensen, 2005; Torquato, 2002). The preferred orientation i n 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 wil l essentially deal with
modelling and analysis of structural elements composed of t his type of composite
material. However, the level of modelling and analysis used in this textbook does not
really differentiate between unidirectional continuous ﬁ bres, oriented short-ﬁbres or
woven ﬁbre composite layers, as long as material characteri stics that deﬁne the layer
response are used. Composite materials can also be classiﬁe d by the nature of their
constituents. According to the nature of the matrix materia l we classify organic,
mineral or metallic matrix composites.
•Organic matrix composites are polymer resins with ﬁllers. T he ﬁbres can be min-
eral (glass, etc.), organic (Kevlar, etc.) or metallic (alu minium, etc.).
•Mineral matrix composites are ceramics with metallic ﬁbres or with metallic or
mineral particles.
•Metallic matrix composites are metals with mineral or metal lic ﬁbres.
Structural composite elements such as ﬁbre reinforced poly mer resins are of par-
ticular interest in this textbook. They can be used only in a l ow 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 m atrix system for com-
posite materials. Typical thermoset matrices include Epoxy ,Polyester ,Polyamide
(Thermoplastics ) and Vinyl Ester , among popular thermoplastics are Polyethylene ,
Polystyrene andPolyether-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 mat rix and a ﬁbre re-
inforcement. The matrix is a polyester or epoxy resin with ﬁl lers. By the addition
of ﬁllers, the characteristics of resins will be improved an d the production costs
reduced. But from the mechanical modelling, a resin-ﬁller s ystem stays as a ho-
mogeneous material and a composite material is a two phase sy stem 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 h igh strength. Their main
drawbacks are the low working temperature, high coefﬁcient s of thermal and mois-
ture expansion and, in certain directions, low elastic prop erties. Most widely used
manufacturing composites are thermosetting resins as unsa turated polyester resins
or epoxy resins. The polyester resins are used as they have lo w production cost.
The second place in composite production is held by epoxy res ins. Although epoxy
is costlier than polyester, approximately ﬁve time higher i n price, it is very popu-
lar in various application ﬁelds. More than two thirds of pol ymer matrices used in
aerospace industries are epoxy based. Polymer matrix compo sites are usually rein-
forced by ﬁbres to improve such mechanical characteristics as stiffness, strength,
etc. Fibres can be made of different materials (glass, carbo n, aramid, etc.). Glass ﬁ-
bres are widely used because their advantages include high s trength, low costs, high

8 1 Classiﬁcation of Composite Materials
chemical resistance, etc., but their elastic modulus is ver y low and also their fatigue
strength. Graphite or carbon ﬁbres have a high modulus and a h igh strength and
are very common in aircraft components. Aramid ﬁbres are usu ally known by the
name of Kevlar, which is a trade name. Summarizing some funct ional requirements
of ﬁbres and matrices in a ﬁbre reinforced polymer matrix com posite
•ﬁbres should have a high modulus of elasticity and a high ulti mate strength,
•ﬁbres should be stable and retain their strength during hand ling and fabrication,
•the variation of the mechanical characteristics of the indi vidual ﬁbres should be
low, their diameters uniform and their arrangement in the ma trix regular,
•matrices have to bind together the ﬁbres and protect their su rfaces 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 th e whole working
period.
The ﬁbre length, their orientation, their shape and their ma terial 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 matr ices by themselves
generally have poorer mechanical properties than compared to ﬁbres, they inﬂuence
many characteristics of the composite such as the transvers e modulus and strength,
shear modulus and strength, thermal resistance and expansi on, etc.
An overview of the material characteristics is given in Sect . 1.4. One of the most
important factors which determines the mechanical behavio r of a composite material
is the proportion of the matrix and the ﬁbres expressed by the ir volume or their
weight fraction. These fractions can be established for a tw o 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 MfandMm(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 relati on 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=ρfVf+ρmVm
V
=ρfvf+ρmvm=ρfvf+ρm(1−vf)(1.1.3)
Starting from the total volume of the composite V=Vf+Vmwe obtain
M
ρ=Mf
ρf+Mm
ρm
and
ρ=1
mf
ρf+mm
ρm(1.1.4)
with
mf=ρf
ρvf,mm=ρm
ρvm
The inverse relation determines
vf=ρ
ρfmf,vm=ρ
ρmmm (1.1.5)
The density ρis determined by Eqs. (1.1.3) or (1.1.4). The equations can b e easily
extended to multi-phase composites.
Mass fractions are easier to measure in material manufactur ing, but volume frac-
tions appear in the theoretical equations for effective mod uli (Sect. 3.1). Therefore,
it is helpful to have simple expressions for shifting from on e fraction to the other.
The quality of a composite material decreases with an increa se 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 ρexpis measured experimentally and
ρtheoris calculated with (1.1.4), the volume fraction of porosity is given by
vpor=ρtheor−ρexp
ρtheor(1.1.6)
1.2 Signiﬁcance and Objectives
Development and applications of composite materials and st ructural 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 requirement s for high performance
materials is not only in aircraft and aerospace structures, but also in the develop-
ment of very powerful experimental equipments and numerica l methods and the
availability of efﬁcient computers. With the development o f composite materials
a new material design is possible that allows an optimal mate rial composition in
connection with the structural design. A useful and correct application of compos-
ite materials requires a close interaction of different eng ineering disciplines such

10 1 Classiﬁcation of Composite Materials
as structural design and analysis, material science, mecha nics of materials, process
engineering, etc. Summarizing the main topics of composite material research and
technology are
•investigation of all characteristics of the constituent an d the composite materials,
•material design and optimization for the given working cond itions,
•development of analytical modelling and solution methods f or determining ma-
terial and structural behavior,
•development of experimental methods for material characte ristics, stress and de-
formation states, failure,
•modelling and analysis of creep, damage and life prediction ,
•development of new and efﬁcient fabrication and recycling p rocedures 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 m aterials such as steel,
alloys, etc. However, to have only material density, stiffn ess and strength in mind
when thinking of composites is a very narrow view of the possi bilities of such ma-
terials as ﬁbre-reinforced plastics because they often may score over conventional
materials as metals not only owing to their mechanical prope rties. Fibre reinforced
plastics are extremely corrosion-resistant and have inter esting electromagnetic prop-
erties. In consequence they are used for chemical plants and for structures which
require non-magnetic materials. Further carbon ﬁbre reinf orced epoxy is used in
medical applications because it is transparent to X-rays.
With applications out of aerospace or aircraft, cost compet itiveness with con-
ventional materials became important. More recently requi rements such as quality
assurance, reproducibility, predictability of the struct ure behavior over its life time,
recycling, etc. became signiﬁcant.
Applications of polymer matrix composites range from the ae rospace industry to
the industry of sports goods. The military aircraft industr y 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 restri ctively and frequently
limited to secondary structural elements. Automotive appl ications, sporting goods,
medical devices and many other commercial applications are examples for the appli-
cation of polymer matrix composites. Also applications in c ivil engineering are now
on the way but it will take some time to achieve wide applicati on of composites in
civil engineering as there are a lot of prescribed condition s to guarantee the reliabil-
ity of structures. But it is clear that over the last decades c onsiderable advances have
been made in the use of composite materials in construction a nd 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 diff erent 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 constit uents, the ﬁbres and ma-
trix. The average characteristics include the elastic modu li, the thermal and mois-
ture expansion coefﬁcients, etc. The micro-mechanics of a l amina does not consider
the internal structure of the constituent elements, but rec ognizes the heterogene-
ity of the ply. The micro-mechanics is based on some simplify ing approximations.
These concern the ﬁbre geometry and packing arrangement, so that the constituent
characteristics together with the volume fractions of the c onstituents yield the av-
erage characteristics of the lamina. Note that the average p roperties are derived by
considering the lamina to be homogeneous. In the frame of thi s textbook only the
micro-mechanics of unidirectional reinforced laminates a re considered (Sect. 3).
The calculated values of the average properties of a lamina p rovide the basis
to predict the macrostructural properties. At the macro-me chanical level, only the
averaged properties of a lamina are considered and the micro structure of the lamina
is ignored. The properties along and perpendicular to the ﬁb re 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 develop ed. 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 stre ngth of an angle ply can
be determined. Failure theories of a lamina are based on stre ngth properties. This
topic is called the macro-mechanics of a single layer or a lam ina (Sect. 4.1).
A laminate is a stack of laminae. Each layer of ﬁbre reinforce ment can have
various orientation and in principle each layer can be made o f different materi-
als. Knowing the macro-mechanics of a lamina, one develops t he macro-mechanics
of the laminate. Average stiffness, ﬂexibility, strength, etc. can be determined for
the whole laminate (Sect. 4.2). The structure and orientati on of the laminae in pre-
scribed sequences to a laminate lead to signiﬁcant advantag es of composite materi-
als when compared to a conventional monolithic material. In general, the mechan-
ical response of laminates is anisotropic. One very importa nt group of laminated
composites are sandwich composites. They consist of two thi n faces (the skins or
sheets) sandwiching a core (Fig. 1.3). The faces are made of h igh strength materials
having good properties under tension such as metals or ﬁbre r einforced laminates
while the core is made of lightweight materials such as foam, resins with special
ﬁllers, called syntactic foam, having good properties unde r compression. Sandwich
composites combine lightness and ﬂexural stiffness. The ma cro-mechanics of sand-
wich composites is considered in Sect. 4.3.
When the micro- and macro-mechanical analysis for laminae a nd laminates are
carried out, the global behavior of laminated composite mat erials is known. The last
step is the modelling on the structure level and to analyze th e global behavior of a
structure made of composite material. By adapting the class ical 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 ana lysis 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 structu res, 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 s heets 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 add ing a core in the
middle increases the resistance. The shear stresses have a m aximum 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 m ore attractive for
some applications than other composite or conventional mat erials.
The most commonly used face materials are aluminium alloys o r ﬁbre reinforced
laminates and most commonly used core materials are balsa wo od, foam and hon-
eycombs (Fig. 1.3). In order to guarantee the advantages of s andwich composites, it
is necessary to ensure that there is perfect bonding between the core and the sheets.
For laminated composites, assumptions are necessary to ena ble the mathematical
modelling. These are an elastic behavior of ﬁbres and matric es, a perfect bonding
between ﬁbres and matrices, a regular ﬁbre arrangement in re gular or repeating ar-
rays, etc.
Summarizing the different length scales of mechanical mode lling structure ele-
ments composed of ﬁbre reinforced composites it must be note d that, independent
of the different possibilities to formulate beam, plate or s hell 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 ﬁb res and the ma-
trix material taking into account the ﬁbre volume fracture a nd the ﬁbre packing
arrangement. The micro-mechanical modelling leads to a cor relation between
constituent properties and average composite properties. In general, simple mix-
ture rules are used in engineering applications (Chap. 3). I f possible, the aver-
age material characteristics of a lamina should be veriﬁed e xperimentally. On
the micro-mechanical level a lamina is considered as a quasi -homogeneous or-
thotropic material.
•Themacroscopic level , where the effective (average) material characteristics o f a
laminate have to be estimated from the average characterist ics of a set of laminae
taking into account their stacking sequence. The macro-mec hanical modelling
leads to a correlation between the known average laminae pro perties and effec-
tive laminate properties. On the macro-mechanical level a l aminate is consid-
ered generally as an equivalent single layer element with a q uasi-homogeneous,
anisotropic material behavior (Chap. 4).
•Thestructural level , where the mechanical response of structural members like
beams, plates, shells etc. have to be analyzed taking into ac count possibilities to
formulate structural theories of different order (Chap. 5) .
In the recent years in the focus of the researchers is an addit ional 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. Partl y 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 conti nuum 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 r equires 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 conc rete, is available much
research and construction experience over many decades, th e codes for structures
composed of conventional materials have been revised conti nuously and so design
engineers pay less attention to material problems because t here is complete docu-
mentation of the material characteristics.
It is quite an another situation for structures made of compo sites. The list of
composite materials is numerous but available standards an d speciﬁcations are very
rare. The properties of each material used for both reinforc ements and matrices of
composites are very much diversiﬁed. The experiences of nea rly all design engineers
in civil or mechanical engineering with composite material s, are insufﬁcient. So it
should be borne in mind that structural design based on compo site materials requires
detailed knowledge about the material properties of the sin gular constituents of the
composite for optimization of the material in the frame of st ructural applications
and also detailed codes for modelling and analysis are neces sary.
The following statements are concentrated on ﬁbre reinforc ed composites with
polymer resins. Material tests of the constituents of compo sites are in many cases a
complicated task and so the material data in the literature a re limited. In engineering
applications the average data for a lamina are often tested t o avoid this problem and
in order to use correct material characteristics in structu ral analysis. But in the area
of material design and selection, it is also important to kno w the properties of all
constituents.
The main properties for the estimation of the material behav ior are
•density ρ,
•Young’s modulus1E,
•ultimate strength σuand
•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 materi al 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 ﬁeld s of light, mechanics (elastic material
parameter, surface tension), energy among others

1.5 Advantages and Limitations 15
speciﬁc moduli E/ρbut in contrast the speciﬁc ultimate stress σu/ρof glass is
signiﬁcantly higher than that of steel and of aluminium allo ys. Table F.2 presents the
mechanical characteristics of selected materials made in t he form of ﬁbres. It should
be borne in mind that the ultimate strength measured for mate rials 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 o f ﬁbres with a very
small diameter of several microns decreases the number of de fects and the values
of ultimate strength increases. Table F.3 gives material pr operties for some selected
matrix materials and core materials of sandwich composites . Tables F.4 and F.5
demonstrate some properties of unidirectional ﬁbre reinfo rced composite materials:
ELis the longitudinal modulus in ﬁbre direction, ETthe transverse modulus, GLTthe
in-plane shear modulus, νLTandνTLare the major and the minor Poisson’s ratio2,
σLu,σTu,σLTuthe ultimate stresses or strengths, αLandαTthe 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 d esign a composite ma-
terial with high modulus and low density or other desired qua lities. The impact of
the costs of the composite material can be low for applicatio ns in the aerospace
industry or high for applications such as in automotive indu stry. 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 compar ed with the possibilities
of conventional materials.
1.5 Advantages and Limitations
The main advantage of polymer matrix composites in comparis on with conventional
materials, such as metals, is their low density. Therefore t wo parameters are com-
monly used to demonstrate the mechanical advantages of comp osites:
1. The speciﬁc modulus E/ρis the Young’s modulus per unit mass or the ratio
between Young’s modulus and density.
2. The speciﬁc strength σu/ρis the tensile strength per unit mass or the ratio be-
tween strength and density
The beneﬁt of the low density becomes apparent when the speci ﬁc modulus and the
speciﬁc strength are considered. The two ratios are high and the higher the speciﬁc
parameters the more weight reduction of structural element s is possible in relation
to special loading conditions. Therefore, even if the stiff ness 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 str ength make composites
2Sim´ eon Denis Poisson (∗21 June 1781 Pithiviers – †25 April 1840 Paris) – mathematici an, ge-
ometer, and physicist, with contributions to mechanics, af ter Poisson an elastic material parameter
is named

16 1 Classiﬁcation of Composite Materials
more attractive. Composite materials are also known to perf orm better under cyclic
loads than metallic materials because of their fatigue resi stance.
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/ρgandσu/ρg,
where gis the acceleration due to the gravity. Furthermore it shoul d be noted that
a single performance indicator is insufﬁcient for the mater ial estimation and that
comparison of the speciﬁc modulus and the speciﬁc strength o f 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 a nd stiffness properties of
ﬁbre-reinforced materials are poor in another important as pect. Their strength de-
pends critically upon the strength of the ﬁbre, matrix inter face and the strength of the
matrix material, if shear stresses are being applied. This l eads to poor shear proper-
ties and this lack of good shear properties is as serious as th e lack of good transverse
properties. For complex structure loadings, unidirection al 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 com posites are usually lam-
inated by using a number of layers. This number of ﬁbre-reinf orced layers can vary
from just a few to several hundred. While generally the major ity 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 transv erse and shear properties.
Additional advantages in the material performances of comp osites are low ther-
mal expansion, high material damping, generally high corro sion resistance and elec-
trical insulation. Composite materials can be reinforced i n any direction and the
structural elements can be optimized by material design or m aterial tailoring.
There are also limitations and drawbacks in the use of compos ite 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 o f some composite
material properties, such as compression or shearing stren gths, are still in discus-
sion.
2. The complexity of material and structural response makes structural modelling
and analysis experimentally and computationally more expe nsive and compli-
cated in comparison to metals or other conventional structu ral materials. There
is also limited experience in the design, calculating and jo ining 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, furthe r the complicated re-
pair technology of composite structures, a lot of recycling problems, etc.
Summarizing, it can be said that the application of composit e materials in struc-
ture design beyond the military and commercial aircraft and aerospace industry and
some special ﬁelds of automotive, sporting goods and medica l devices is still in the

1.6 Problems 17
early stages. But the advancing of technology and experienc e yields an increasing
use of composite structure elements in civil and mechanical engineering and pro-
vides the stimulus to include composite processing, modell ing, 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 t o the mechanical
performance of composites?
4. What are polymer matrix, metal matrix and ceramic matrix c omposites, what are
their main applications?
5. Deﬁne isotropic, anisotropic, homogeneous, nonhomogen eous.
6. Deﬁne lamina, laminate, sandwich. What is micro-mechani cal and macro-
mechanical modelling and analysis?
7. Compare the speciﬁc modulus, speciﬁc strength and coefﬁc ient of thermal ex-
pansion of glas ﬁbre, epoxy resin and steel.
Exercise 1.1. A typical CFK plate (uni-directional reinforced laminae co mposed 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 o n the following data:
•density of the carbon ﬁbres ρf=1,8 g/cm3,
•diameter of the ﬁbres is df=6µm
•ﬁbre volume fraction vf=0.8 and
•density of the epoxy ρm=1,1 g/cm3.
Solution 1.1. Let us assume that the ﬁbres are parallel to the longer plate s ide. In
this case the volume of one ﬁbre V1fis
V1f=πd2
f
4l=π(6µm)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=ρfVf
ρfVf+ρmVm
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 fr action.
References
Altenbach H, Eremeyev V A (2015) On the theories of plates and shells at the
nanoscale. In: Altenbach H, Mikhasev GI (eds) Shell and Memb rane Theories
in Mechanics and Biology: From Macro- to Nanoscale Structur es, Springer Inter-
national Publishing, Cham, pp 25–57
Christensen RM (2005) Mechanics of Composite Materials. Do ver, Mineola (NY)
Cleland AN (2003) Foundations of Nanomechanics. Advanced T exts in Physics,
Springer, Berlin, Heidelberg
Torquato S (2002) Random Heterogeneous Materials – Microst ructure and Macro-
scopic Properties, Interdisciplinary Applied Mathematic s, 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 is otropic.
These assumptions of classical theory of elasticity guaran tee a satisfying quality
of modelling and analysis of structure elements made of conv entional monolithic
materials. Structural analysis of elements composed of com posite materials is based
on the theory of anisotropic elasticity, the elastic proper ties of composites depend
usually on the direction and the deformable solid is anisotr opic. In addition, now the
composite material is not homogeneous at all. It must be assu med that the material
is piecewise homogeneous or quasi-homogeneous.
The governing equations of elastic bodies are nearly the sam e for isotropic and
anisotropic material response. There are equilibrium equa tions, 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 strai ns and displacements. All
these equations are independent of the elastic properties o f the material. Only the
material relations (so-called constitutive equations), w hich describe the relations
between stresses and strains are very different for an isotr opic and an anisotropic
body. This difference in formulating constitutive equatio ns has a great inﬂuence
on the model equations in the frame of the isotropic and the an isotropic 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-cal led initial-boundary
conditions for forces/stresses and/or displacements, yie ld the basic model equations
for linear elastic solids such as differential equations or variational and energy for-
mulations, respectively. All equations for structural ele ments which are given in this
textbook, are founded on these general equations for the the ory of elasticity of linear
elastic anisotropic solids.
19 © Springer Nature Singapore Pte Ltd. 2018
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 Ho oke’s law1, the con-
stitutive equations for anisotropic elastic bodies, and to introduce general relations
for stiffness and strains including transformation rules a nd symmetry relations. The
constitution of a unidirectional composite material and si mpliﬁed approaches for
so-called effective moduli result in an engineering formul ation 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 pass es from all speciﬁc cases
of elastic symmetries to the classical Hooke’s law for an iso tropic body. The only
assumptions are
•all elastic properties are the same in tension and compressi on,
•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-strai n 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 a nisotropy, i.e. the homo-
geneous anisotropic body is characterized by the equivalen ce of parallel directions
passing through different points of the body. Another kind o f anisotropy, which can
be interesting to some applications, e.g. to modelling circ ular 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 propert ies at different points of the
body, the elastic behavior is called curvilinear.
The chapter ends with the derivation of the fundamental equa tions of anisotropic
elasticity and the formulation of variational solution met hods. In Sect. 2.2 the dif-
ferential equations for boundary and initial boundary prob lems are considered. The
classical and generalized variational principles are form ulated 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 mic roscopic scale discon-
tinuous structure of the material. On the macroscopic or phe nomenological scale the
material models are assumed to be continuous and in general h omogeneous. In the
case of ﬁbre reinforced composites, the heterogeneity of th e bulk material is a con-
sequence of the two constituents, the ﬁbres and the matrix, b ut generally there exists
a representative volume element of the material on a charact eristic scale at which the
1Robert Hooke (∗28greg./18jul.July 1635 Freshwater, Isle of Wight – †3 March 1703 London) –
natural philosopher, architect and polymath, ﬁrst constit utive 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 approxim ation. The composite
material can be considered as macroscopically homogeneous and the problem of
designing structural elements made from of composite mater ials can be solved in an
analogous manner as for conventional materials with the hel p 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 reinforcemen ts or reinforced by
stochastically distributed and orientated particles or sh ort ﬁbres, the material behav-
ior of an off-axis forced unidirectional lamina is anisotro pic. In comparison to con-
ventional isotropic materials, the experimental identiﬁc ation of the material param-
eters is much more complicated in the case of anisotropic mat erials. But anisotropic
material behavior also has the advantage of material tailor ing 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 el astic 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 FandT, normal and tangential to the cross-section, respectively .
Assuming a uniform distribution of the forces FandTon the cross-section, the
elementary one-dimensional deﬁnitions for stresses and st rains are given by (2.1.1)
σ=F
A0normal stress σ,
ε=l−l0
l0=△l
l0extensional strain ε,
τ=T
A0shear stress τ,
γ≈tanγ=△u
l0shear strain γ(2.1.1)
The last one deﬁnition is restricted by small strain assumpt ion. This assumption
can be accepted for many composite material applications an d 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 strai ns yield Hooke’s law (2.1.2)

22 2 Linear Anisotropic Materials
F
A
A0
l0 l=l0+△l△l
Fl0△u
TA0
Tγ
Fig. 2.1 Extensional strain εand shear strain γof a body with the length l0and the cross-section
area A0
σ=Eε,E=σ
ε, Eis the elasticity or Young’s modulus ,
τ=Gγ,G=τ
γ, Gis the shear modulus(2.1.2)
For a homogeneous material EandGare parameters with respect to the coordinates.
For the extensionally strained prismatic body (Fig. 2.1) th e phenomenon of con-
traction in a direction normal to the direction of the tensil e loading has to be con-
sidered. The ratio of the contraction, expressed by the tran sverse strain εt, to the
elongation in the loaded direction, expressed by ε, is called Poisson’s ratio ν
εt=−νε,ν=−εt
ε(2.1.3)
For an isotropic bar with an extensional strain ε>0 it follows that εt<02.
Hooke’s law can be written in the inverse form
ε=E−1σ=Sσ (2.1.4)
S=E−1is the inverse modulus of elasticity or the ﬂexibility/comp liance modulus.
For homogeneous material, Sis an elastic parameter.
Consider a tensile loaded prismatic bar composed of differe nt materials
(Fig. 2.2). Since σ=F/Aandσ=Eεthen σA=F=EAεandε= (EA)−1F.
EAis 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 n ot in the focus of the present book.

2.1 Generalized Hooke’s Law 23
l lF F
C1C1
Cn
CnAili
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 arran ged in parallel or in
series. In the ﬁrst case we have
F=n
∑
i=1Fi,A=n
∑
i=1Ai,ε=εi (2.1.5)
Fiare the loading forces on Aiand the strains εiare equal for the total cross-section.
With
F=EAε,Fi=EiAiε,n
∑
i=1Fi=F=n
∑
i=1EiAiε (2.1.6)
follow the coupling equations for the stiffness EiAifor a parallel arrangement
EA=n
∑
i=1EiAi,(EA)−1=1
n
∑
i=1EiAi(2.1.7)
This equal strain treatment is often described as a V oigt3model which is the upper-
bound stiffness.
In the other case, we have
△l=n
∑
i=1△li
andF=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=lε=l(EA)−1F,△li=liεi=li(EiAi)−1F
and
n
∑
i=1△li=/bracketleftigg
n
∑
i=1li(EiAi)−1/bracketrightigg
F (2.1.8)
3Woldemar V oigt (∗2 September 1850 Leipzig – †13 December 1919 G¨ ottingen) – ph ysicist,
worked on crystal physics, thermodynamics and electro-opt ics, the word tensor in its current mean-
ing was introduced by him in 1898, V oigt notation

24 2 Linear Anisotropic Materials
follow the coupling equations for the stiffness EiAiarranged in series
EA=l
n
∑
i=1li(EiAi)−1,(EA)−1=n
∑
i=1li(EiAi)−1
l(2.1.9)
This equal stress treatment is described generally as a Reus s4model which is the
lower bound stiffness. The coupling equations illustrate a ﬁrst clear insight into a
simple calculation of effective stiffness and compliance p arameters for two com-
posite structures.
The three-dimensional state of stress or strain in a continu ous solid is completely
determined by knowing the stress or strain tensor. It is comm on practice to repre-
sent the tensor components acting on the faces of an inﬁnites imal cube with sides
parallel to the reference axes (Fig. 2.3). The sign conventi on 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 an d vice versa. Using the
tensorial notation for the stress tensor σi jand the strain tensor εi jfor the stresses
and the strains we have normal stresses or extensional strai ns respectively for i=j
and shear stresses or shear strains for i/ne}ationslash=j.εi jwith i/ne}ationslash=jare the tensor shear coor-
dinates and 2 εi j=γi j,i/ne}ationslash=jthe engineering shear strains. The ﬁrst subscript of σi j
andεi jindicates 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 εi jfrom the tensor ei jwhich is the tensor of the
relative displacements, ei j=∂ui/∂xj. An application of shear stresses σi jandσji
produces in the i j-plane of the inﬁnitesimal cube (Fig. 2.3) angular rotation s of the
i- and j-directions by ei jandeji. These relative displacements represent a combina-
tion of strain (distorsion) and rigid body rotation with the limiting cases ei j=eji,
eee1eee2eee3
σ11σ12
σ13σ22
σ21σ23σ33
σ32σ31
ε11ε13ε12ε22
ε21ε23ε33
ε32ε31
Fig. 2.3 Stress and strain components on the positive faces of an inﬁn itesimal cube in a set of axis
eee1,eee2,eee3
4Andr´ as (Endre) Reuss (∗1 July 1900 Budapest – †10 May 1968 Budapest) – mechanical eng ineer,
contributions to the theory of plasticity

2.1 Generalized Hooke’s Law 25
j j j
i i iei j ei j
eji
ejiγi j
γi j
ei j=eji=εi j=1
2γi j
ωi j=0ei j=−eji
εi j=γi j=0ei j=2εji=γi j
eji=0a b c
Fig. 2.4 Examples of distorsions and rigid body rotation. aPure shear, bpure rotation, csimple
shear
.
i.e. no rotation, and ei j=−eji, i.e. no distorsion (Fig. 2.4). From the ﬁgure follows
that simple shear is the sum of pure shear and rigid rotation. ei jis positive when
it involves rotating the positive j-direction towards the positive i-direction and vice
versa. Writing the tensor ei jas the sum of symmetric and antisymmetric tensors
ei j=1
2(ei j+eji)+1
2(ei j−eji)=εi j+ωi j (2.1.10)
where εi jis the symmetric strain tensor and ωi jis the antisymmetric rotation ten-
sor. For normal strains, i.e. i=j, there is ei j=εi j, however for i/ne}ationslash=jwe have
γi j=2εi j=ei j+ejiwith the engineering shear strains γi jand the tensorial shear
strains εi j. Careful note should be taken of the factor of two related eng ineering and
tensorial shear strains, γi jis often more convenient for practical use but tensor oper-
ations such as rotations of the axis, Sect. 2.1.2, must be car ried out using the tensor
notation εi j.
The stress and the strain tensors are symmetric tensors of ra nk two. They can be
represented by the matrices
σσσ=
σ11σ12σ13
σ12σ22σ23
σ13σ23σ33
,εεε=
ε11ε12ε13
ε12ε22ε23
ε13ε23ε33
 (2.1.11)
The symmetry of the tensors (2.1.11) reduces the number of un known 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 v alues 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
[σ11σ22σ33σ23≡τ23σ13≡τ13σ12≡τ12]T,
[ε11ε22ε332ε23≡γ232ε13≡γ132ε12≡γ12]T(2.1.12)
The stress and strain states are related by a material law whi ch is deduced from
experimental observations. For a linear elastic anisotrop ic material, the generalized
Hooke’s law relates the stress and the strain tensor
σi j=Ci jklεkl (2.1.13)
Ci jklare the material coefﬁcients and deﬁne the fourth rank elast icity tensor which
in general case contains 81 coordinates. Due to the assumed s ymmetry of σi j=σji
andεi j=εjithe symmetry relations follow for the material tensor
Ci jkl=Cjikl,Ci jkl=Ci jlk (2.1.14)
and reduce the number of coordinates to 36. Introducing a con tracted single-
subscript notation for the stress and strain components and a double-subscript nota-
tion for the elastic parameters, the generalized relation f or stresses and strains can
be written in vector-matrix form
[σi]=[Ci j][εj];i,j=1,2,…, 6;Ci j/ne}ationslash=Cji;i/ne}ationslash=j (2.1.15)
At this stage we have 36 independent material coefﬁcients, b ut a further reduction
of the number of independent values is possible because we ha ve 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
2σi jεi j (2.1.16)
or in a contracted notation
W(εi)=1
2σiεi=1
2Ci jεjεi (2.1.17)
With
∂W
∂εi=σi,∂2W
∂εi∂εj=Ci j,∂2W
∂εj∂εi=Cji
and
∂2W
∂εi∂εj=∂2W
∂εj∂εi
follow the symmetry relations
Ci j=Cji;i,j=1,2,…, 6 (2.1.18)

2.1 Generalized Hooke’s Law 27
and the number of the independent material coefﬁcients is re duced to 21. The gen-
eralized relations for stresses and strains of an anisotrop ic elastic body written again
in a contracted vector-matrix form have a symmetric matrix f orCi j

σ1
σ2
σ3
σ4
σ5
σ6
=
C11C12C13C14C15C16
C22C23C24C25C26
C33C34C35C36
C44C45C46
S Y M C55C56
C66

ε1
ε2
ε3
ε4
ε5
ε6
(2.1.19)
The transformation rules for the contraction of the subscri pts of σi j,εi jandCi jklof
(2.1.13) are given in Tables 2.1 and 2.2.
The elasticity equation (2.1.19) can be written in the inver se form as follows

ε1
ε2
ε3
ε4
ε5
ε6
=
S11S12S13S14S15S16
S22S23S24S25S26
S33S34S35S36
S44S45S46
S Y M S55S56
S66

σ1
σ2
σ3
σ4
σ5
σ6
(2.1.20)
with
[Ci j][Sjk]=[δik]=/braceleftbigg1i=k,
0i/ne}ationslash=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)
σi=Ci jεj, εi=Si jσj; i,j=1,…, 6
σσσ=CCCεεε, εεε=SSSσσσ(2.1.21)
CCC≡[Ci j]is the stiffness matrix and SSS≡[Si j]the compliance or ﬂexibility matrix. Ci j
andSi jare only for homogeneous anisotropic materials constant ma terial parameters
Table 2.1 Transformation of
the tensor coordinates σi jand
εi jto the vector coordinates
σpandεp
σi j σp εi j εp
σ11 σ1 ε11 ε1
σ22 σ2 ε22 ε2
σ33 σ3 ε33 ε3
σ23=τ23σ42ε23=γ23ε4
σ31=τ31σ52ε31=γ31ε5
σ12=τ12σ62ε12=γ12ε6Table 2.2 Transformation
of the tensor coordinates
Ci jklto 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 r eference coordinate
system. A change of the reference system yields a change of th e constant values.
Summarizing the stiffness and the compliance relations, it can be seen that for
a linear elastic anisotropic material 21 material paramete rs have to be measured
experimentally in the general case. But in nearly all engine ering applications there
are material symmetries and the number of material paramete rs can be reduced.
Section 2.1.2 describes some transformation rules for CCCandSSSfollowing from the
change of reference system and the symmetric properties of a nisotropic materials
discussed in Sect. 2.1.3. Furthermore the way that the mater ial parameters Ci jandSi j
are related to the known engineering elastic parameters Ei,Gi jandνi jis considered.
2.1.2 Transformation Rules
If we have a reference system which is characterized by the or thonormal basic unit
vectors eee1,eee2,eee3and another reference system with the vector basis eee′
1,eee′
2,eee′
3. Both
systems are related by a rotation of the coordinate axis (Fig . 2.5), the transformation
rules are
eee′
i=Ri jeeej,eeei=Rjieee′
j,
Ri j≡cos(eee′
i,eeej),Rji≡cos(eeei,eee′
j),i,j=1,2,3 (2.1.22)
These relationships describe a general linear orthogonal c oordinate transformation
and can be expressed in vector-matrix notation
eee′=RRReee,eee=RRR−1eee′=RRRTeee′(2.1.23)
RRRis the transformation or rotation matrix. In the case of an or thogonal set of axes
such as given in Fig. 2.5 the matrix RRRis symmetric and unitary. This means the
determinant of this matrix is unity (Det RRR=|Ri j|=1 and the inverse matrix RRR−1
is identical to the transposed matrix (RRR−1=RRRT). In the special case of a rotation φ
Fig. 2.5 Rotation of a refer-
ence system with the basic
vectors eeeiinto a system with
the basic vectors eee′
ix1x2x3
x1′x2′
x3′
eee1eee′
1eee2eee′
2eee3
eee′
3

2.1 Generalized Hooke’s Law 29
about the direction eee3, the rotation matrix RRRand the inverse matrix RRR−1are
/bracketleftbigg3
Ri j/bracketrightbigg
=
c s 0
−s c0
0 0 1
,/bracketleftbigg3
Ri j/bracketrightbigg−1
=/bracketleftbigg3
Ri j/bracketrightbiggT
=
c−s0
s c 0
0 0 1
, (2.1.24)
and the transformation rules are

eee′
1
eee′
2
eee′
3
=
c s 0
−s c0
0 0 1

eee1
eee2
eee3
,
eee1
eee2
eee3
=
c−s0
s c 0
0 0 1

eee′
1
eee′
2
eee′
3

with c=cosφ,s=sinφ. For rotations ψorθabout the directions eee2oreee1the
rotations matrices [2
Ri j]and[1
Ri j]are
/bracketleftbigg2
Ri j/bracketrightbigg
=
c0−s
0 1 0
s0c
,/bracketleftbigg2
Ri j/bracketrightbigg−1
=/bracketleftbigg2
Ri j/bracketrightbiggT
=
c0s
0 1 0
−s0c
,
/bracketleftbigg1
Ri j/bracketrightbigg
=
1 0 0
0c s
0−s c
,/bracketleftbigg1
Ri j/bracketrightbigg−1
=/bracketleftbigg1
Ri j/bracketrightbiggT
=
1 0 0
0c−s
0s c

with c=cosψor cos θands=sinψor sin θfor rotations about eee2oreee1, respec-
tively.
The transformation rule (2.1.22) can be interpreted as a rul e for vectors or ﬁrst-
rank tensors. The generalization to second-rank tensors yi elds e.g. for the stress
tensor
σ′
i j=RikRjlσkl,σi j=RkiRl jσ′
kl (2.1.25)
For the following reﬂections the transformation rules for t he contracted notation are
necessary. The nine tensor coordinates σi jare shifted to six vector coordinates σp.
The transformations
σ′
p=Tσ
pqσq,σp=/parenleftbig
Tσ
pq/parenrightbig−1σ′
q,p,q=1,…, 6 (2.1.26)
are not tensor transformation rules. The transformation ma trices Tσ
pqand(Tσ
pq)−1
follow by comparison of Eqs. (2.1.25) and (2.1.26). In the sa me manner we can ﬁnd
the transformation rules for the strains
ε′
p=Tε
pqεq,εp=/parenleftbig
Tε
pq/parenrightbig−1ε′
q,p,q=1,…, 6 (2.1.27)
The elements of the transformation matrices [Tσ
pq]and[Tε
pq]are deﬁned in App. B.
Summarizing, the transformation rules for stresses and str ains in a condensed
vector-matrix notation as follows
σσσ′=TTTσσσσ,εεε′=TTTεεεε,σσσ=(TTTσ)−1σσσ′,εεε=(TTTε)−1εεε′(2.1.28)

30 2 Linear Anisotropic Materials
The comparison of
σi j=RkiRl jσ′
klwith σp=(Tσ
pq)−1σ′
q
and
εi j=RkiRl jε′
klwith εp=(Tε
pq)−1ε′
q
yields an important result on the linkage of inverse and tran sposed stress and strain
transformation matrices
(TTTσ)−1=(TTTε)T,(TTTε)−1=(TTTσ)T(2.1.29)
The transformation relations for the stiffness and the comp liance matrices CCCand
SSScan be obtained from the known rules for stresses and strains . With σσσ=CCCεεεand
σσσ′=CCC′εεε′, it follows that
(TTTσ)−1σσσ′=σσσ=CCCεεε=CCC(TTTε)−1εεε′,
σσσ′=TTTσCCC(TTTε)−1εεε′=CCC′εεε′,
TTTσσσσ=σσσ′=CCC′εεε′=CCC′TTTεεεε,
σσσ= (TTTσ)−1CCC′TTTεεεε=CCCεεε,(2.1.30)
respectively.
Considering (2.1.29) the transformation relations for the stiffness matrix are
CCC′=TTTσCCC(TTTσ)T,CCC=(TTTε)TCCC′TTTε(2.1.31)
or in index notation
C′
i j=Tσ
ikTσ
jlCkl,Ci j=Tε
ikTε
jlC′
kl (2.1.32)
The same procedure yields the relations for the compliance m atrix. With
εεε=SSSσσσ,εεε′=SSS′σσσ′(2.1.33)
it follows
(TTTε)−1εεε′=εεε=SSSσσσ=SSS(TTTσ)−1σσσ′,
εεε′=TTTεSSS(TTTσ)−1σσσ′=SSS′σσσ′,
TTTεεεε=εεε′=SSS′σσσ′=SSS′TTTσσσσ,
εεε= (TTTε)−1SSS′TTTσσσσ=SSSσσσ,(2.1.34)
i.e.
εεε′=TTTεSSS(TTTσ)−1σσσ′,εεε=(TTTε)−1SSS′TTTσσσσ (2.1.35)
The comparison leads to the transformation equations for SSSandSSS′
SSS′=TTTεSSS(TTTσ)−1,SSS=(TTTε)−1SSS′TTTσ(2.1.36)
or taking into account (2.1.29)
SSS′=TTTεSSS(TTTε)T,SSS=(TTTσ)TSSS′TTTσ, (2.1.37)

2.1 Generalized Hooke’s Law 31
respectively, in subscript notation
S′
i j=Tε
ikTε
jlSkl,Si j=Tσ
ikTσ
jlS′
kl (2.1.38)
In the special case of a rotation φabout the eee3-direction (Fig. 2.6) the coordinates
of the transformation matrices TTTσandTTTεare given by the (2.1.39) and (2.1.40)
/bracketleftbigg3
Tσ
pq/bracketrightbigg
=
c2s20 0 0 2 cs
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
,/bracketleftbigg3
Tσ
pq/bracketrightbigg−1
=/bracketleftigg
3
Tε
pq/bracketrightiggT
(2.1.39)
/bracketleftigg
3
Tε
pq/bracketrightigg
=
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
−2cs2cs0 0 0 c2−s2
,/bracketleftigg
3
Tε
pq/bracketrightigg−1
=/bracketleftbigg3
Tσ
pq/bracketrightbiggT
(2.1.40)
By all rules following from a rotation of the reference syste m the stresses, strains,
stiffness and compliance parameters in the rotated system a re known. They are sum-
marized in symbolic notation (2.1.41)
σσσ′=TTTσσσσ,εεε′=TTTεεεε,
σσσ=(TTTε)Tσσσ′,εεε=(TTTσ)Tεεε′,
CCC′=TTTσCCC(TTTσ)T,SSS′=TTTεSSS(TTTε)T,
CCC=(TTTε)TCCC′TTTε,SSS=(TTTσ)TSSS′TTTσ(2.1.41)
Fig. 2.6 Rotation about the
eee3-directionx1x2x3,x′
3
x′
1x′
2
eee3,eee′
3
eee′
1eee2eee′
2
eee1
φφ

32 2 Linear Anisotropic Materials
For special cases of a rotation about a direction eeeithe general transformation ma-
trices TTTσandTTTεare substituted byi
TTTσori
TTTε. The case of a rotation about the
eee1-direction yields the coordinates of the transformation ma trices TTTσandTTTεwhich
are given in App. B.
2.1.3 Symmetry Relations of Stiffness and Compliance Matri ces
In the most general case of the three-dimensional generaliz ed Hooke’s law the stiff-
ness and the compliance matrices have 36 non-zero material p arameters Ci jorSi j
but they are each determined by 21 independent parameters. S uch an anisotropic
material is called a triclinic material, it has no geometric symmetry properties. The
experimental tests to determine 21 independent material pa rameters would be difﬁ-
cult to realize in engineering applications. So it is very im portant that the majority
of anisotropic materials has a structure that exhibits one o r more geometric sym-
metries and the number of independent material parameters n eeded to describe the
material behavior can be reduced.
In the general case of 21 independent parameters, there is a c oupling of each
loading component with all strain states and the model equat ions for structure el-
ements would be very complicated. The reduction of the numbe r of independent
material parameters results therefore in a simplifying of t he modelling and analysis
of structure elements composed of composite materials and i mpact 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 i s 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 jandSi j. Mirror
transformation ( x1=x′
1,
x2=x′
2,×3=−x′
3)x1x2x3
eee3
eee1eee2
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 c arried out by a sym-
metry about this plane does not modify the matrices, i.e. tha t the material properties
are identical along any two rays symmetric with respect to th e(x1−x2)plane. The
exploitation of the transformation rules leads to a stiffne ss matrix with the following
structure in the case of monoclinic material behavior
[Ci j]MC=
C11C12C130 0 C16
C12C22C230 0 C26
C13C23C330 0 C36
0 0 0 C44C450
0 0 0 C45C550
C16C26C360 0 C66
(2.1.42)
The compliance matrix has the same structure
[Si j]MC=
S11S12S130 0 S16
S12S22S230 0 S26
S13S23S330 0 S36
0 0 0 S44S450
0 0 0 S45S550
S16S26S360 0 S66
(2.1.43)
The number of non-zero elements Ci jorSi jreduces to twenty, the number of in-
dependent elements to thirteen. The loading-deformation c ouplings are reduced.
Consider for example the stress component σ6≡τ12. There is a coupling with the
extensional strains ε1,ε2,ε3and the shear strain ε6≡γ12but the shear stress σ4or
σ5produces only shear strains.
If an anisotropic material has the plane of elastic symmetry x1−x3then it can be
shown that
[Ci j]MC=
C11C12C130C150
C12C22C230C250
C13C23C330C350
0 0 0 C440C46
C15C25C350C550
0 0 0 C460C66
(2.1.44)
and for the plane of elastic symmetry x2−x3
[Ci j]MC=
C11C12C13C140 0
C12C22C23C240 0
C13C23C33C340 0
C14C24C34C440 0
0 0 0 0 C55C56
0 0 0 0 C56C66
(2.1.45)

34 2 Linear Anisotropic Materials
x1 x1x2 x2x3 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]MChave 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 e xistence 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=
C11C12C130 0 0
C12C22C230 0 0
C13C23C330 0 0
0 0 0 C440 0
0 0 0 0 C550
0 0 0 0 0 C66
(2.1.46)
The compliance matrix has the same structure. An orthotropi c material has 12 non-
zero and 9 independent material parameters. The stress-str ain coupling is the same
as for isotropic material behavior. Normal stresses give ri se 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 i t is invariant with respect
to an arbitrary rotation about a given axis. This material be havior is of special im-
portance in the modelling of ﬁbre-reinforced composite mat erials 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 behavi or lies between isotropic
and orthotropic.
Ifx1is the ﬁbre direction, x2andx3are both rectangular to the ﬁbre direction
and assuming identical material properties in these direct ions is understandable.
The structure of the stiffness matrix of a transversely isot ropic material is given
in (2.1.47)
[Ci j]TI=
C11C12C12 0 0 0
C12C22C23 0 0 0
C12C23C22 0 0 0
0 0 01
2(C22−C23)0 0
0 0 0 0 C550
0 0 0 0 0 C55
(2.1.47)
Ifx2is the ﬁbre direction, x1andx3are both rectangular to the ﬁbre direction and
assuming identical material properties in these direction s then
[Ci j]TI=
C11C12C130 0 0
C12C22C120 0 0
C13C12C110 0 0
0 0 0 C44 0 0
0 0 0 01
2(C11−C13)0
0 0 0 0 0 C44
(2.1.48)
Ifx3is the ﬁbre direction, x1andx2are both rectangular to the ﬁbre direction and
assuming identical material properties in these direction s then
[Ci j]TI=
C11C12C130 0 0
C12C11C130 0 0
C13C13C330 0 0
0 0 0 C440 0
0 0 0 0 C44 0
0 0 0 0 01
2(C11−C12)
(2.1.49)
The compliance matrix has the same structure. For example fo r the case (2.1.47)
[Si j]TI=
S11S12S12 0 0 0
S12S22S23 0 0 0
S12S23S22 0 0 0
0 0 0 2 (S22−S23)0 0
0 0 0 0 S550
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 t he 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 dir ections, i.e. the material
has an inﬁnite number of planes and axes of material symmetry . Most conventional
materials satisfy this behavior approximately on a macrosc opic scale.
The number of independent elasticity parameters is reduced to two and this leads
to the following stiffness matrix in the case of isotropic ma terial behavior
[Ci j]I=
C11C12C120 0 0
C12C11C120 0 0
C12C12C110 0 0
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 of1
2(C11−C12). Tables 2.3 and 2.4 summarize
the stiffness and compliance matrices for all material mode ls described above.
2.1.4 Engineering Parameters
2.1.4.1 Orthotropic Material Behavior
The coordinates Ci jandSi jof the stiffness and compliance matrix are mathematical
symbols relating stresses and strains. For practising engi neers, a clear understand-
ing of each material parameter is necessary and requires a mo re mechanical meaning
by expressing the mathematical symbols in terms of engineer ing parameters such as
moduli Ei,Gi jand Poisson’s ratios νi j. The relationships between mathematical and
engineering parameters are obtained by basic mechanical te sts and imaginary math-
ematical experiments. The basic mechanical tests are the te nsion, 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 st ress and measuring the
strains or vice versa.
It follows that the compliance parameters are directly rela ted 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,G12and Poisson’s ratios νi j,i,j=1,2,3(i/ne}ationslash=j). For orthotropic
materials one can introduce the contracted engineering not ation

2.1 Generalized Hooke’s Law 37
Table 2.3 Three-dimensional stiffness matrices for different mater ial symmetries
Material model Elasticity matrix [Ci j]
Anisotropy:
21 independent
material parameters
C11C12C13C14C15C16
C22C23C24C25C26
C33C34C35C36
C44C45C46
S Y M C55C56
C66

Monoclinic:
13 independent
material parametersSymmetry plane x3=0
C14=C15=C24=C25=C34=C35=C46=C56=0
Symmetry plane x2=0
C14=C16=C24=C26=C34=C36=C45=C56=0
Symmetry plane x1=0
C15=C16=C25=C26=C35=C36=C45=C56=0
Orthotropic:
9 independent
material parameters3 planes of symmetry x1=0,×2=0,×3=0
C14=C15=C16=C24=C25=C26=C34
=C35=C36=C45=C46=C56=0
Transversely isotropic:
5 independent
material parametersPlane 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 other Ci jlike orthotropic
Isotropy:
2 independent
material parametersC11=C22=C33,C12=C13=C23,
C44=C55=C66=1
2(C11−C12)
all other Ci j=0
σ1=E1ε1,σ2=E2ε2,σ3=E3ε3,
σ4=E4ε4,σ5=E5ε5,σ6=E6ε6(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 mate rial symmetries
Material model Compliance matrix [Si j]
Anisotropy:
21 independent
material parameters
S11S12S13S14S15S16
S22S23S24S25S26
S33S34S35S36
S44S45S46
S Y M S55S56
S66

Monoclinic:
13 independent
material parametersSymmetry 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 parameters3 planes of symmetry x1=0,×2=0,×3=0
S14=S15=S16=S24=S25=S26=S34
=S35=S36=S45=S46=S56=0
Transversely isotropic:
5 independent
material parametersPlane 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 jlike orthotropic
Isotropy:
2 independent
material parametersS11=S22=S33,S12=S13=S23,
S44=S55=S66=2(S11−S12)
all other Si j=0
ε1=S11σ1+S12σ2+S13σ3,ε4=S44σ4,
ε2=S12σ1+S22σ2+S23σ3,ε5=S55σ5,
ε3=S13σ1+S23σ2+S33σ3,ε6=S66σ6(2.1.53)
For uniaxial tension in xi-direction, σ1/ne}ationslash=0,σi=0,i=2,…, 6. This reduces (2.1.53)
to
ε1=S11σ1,ε2=S12σ1,ε3=S13σ1,ε4=ε5=ε6=0, (2.1.54)
and the physical tensile tests provides the elastic paramet ersE1,ν12,ν13

2.1 Generalized Hooke’s Law 39
E1=σ1
ε1=1
S11,ν12=−ε2
ε1=−S12E1,ν13=−ε3
ε1=−S13E1 (2.1.55)
Analogous relations resulting from uniaxial tension in x2- and x3-directions and all
Si jare related to the nine measured engineering parameters (3 Y oung’s moduli and
6 Poisson’s ratios) by uniaxial tension tests in three direc tions x1,x2andx3.
From the symmetry of the compliance matrix one can conclude
ν12
E1=ν21
E2,ν23
E2=ν32
E3,ν31
E3=ν13
E1
orνi j
Ei=νji
Ej,νi j
νji=Ei
Ej,i,j=1,2,3(i/ne}ationslash=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) d emonstrate that the nine en-
gineering parameters are not independent parameters and th at in addition to the three
tension tests, three independent shear tests are necessary to ﬁnd the equations
ε4=S44σ4,ε5=S55σ5,ε6=S66σ6,
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 jin (2.1.20) can be substituted by the engineering parameter s

ε1
ε2
ε3
ε4
ε5
ε6
=
1
E1−ν12
E1−ν13
E10 0 0
1
E2−ν23
E20 0 0
1
E30 0 0
1
E40 0
S Y M1
E50
1
E6

σ1
σ2
σ3
σ4
σ5
σ6
(2.1.58)
As seen above, the relations between compliances Si jand engineering parameters
are very simple. This, however, is not the case for the relati ons between the stiffness
and engineering parameters. First we need to invert the comp liance matrix SSSand
to express the stiffness Ci jas 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 [3×3]-matrix must be inverted. The gener al formula is
Ci j=S−1
i j=(−1)i+jUi j
Det[Si j],Det[Si j]=/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleS11S12S13
S12S22S23
S13S23S33/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle, (2.1.60)
where Ui jare the submatrices of SSSto 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 jand engineering parameters given above in
(2.1.58), we obtain
C11=(1−ν23ν32)E1
∆,C12=(ν12+ν13ν32)E2
∆,
C22=(1−ν31ν13)E2
∆,C23=(ν23+ν21ν13)E3
∆,
C33=(1−ν21ν12)E3
∆,C13=(ν13+ν12ν23)E3
∆(2.1.62)
with ∆=1−ν21ν12−ν32ν23−ν13ν31−2ν21ν13ν32. Considering Ei/∆≡Ei,
1/Si≡Eiwe ﬁnally get two subsystems

σ1
σ2
σ3
=
(1−ν23ν32)E1(ν12+ν13ν32)E2(ν13+ν12ν23)E3
(1−ν31ν13)E2(ν23+ν21ν13)E3
SYM (1−ν21ν12)E3

ε1
ε2
ε3
,

σ4
σ5
σ6
=
E40 0
E50
SYM E6

ε4
ε5
ε6
 (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,ν12=ν13,G12=G13,G23=E2
2(1+ν23), (2.1.64)
and we get

2.1 Generalized Hooke’s Law 41
S11=1
E1,S12=S13=−ν12
E1,S44=1
G23=2(1+ν23)
E2,
S22=S33=1
E2,S23=−ν23
E2,S55=S66=1
G12,
C11=(1−ν2
23)E1
∆∗,C12=C13=ν21(1+ν23)E1
∆∗,
C22=C33=(1−ν12ν21)E2
∆∗,C23=(ν23+ν21ν12)E2
∆∗,
C44=G23=E2
2(1+ν23),C55=C66=G12
with ∆∗=(1+ν23)(1−ν23−2ν21ν12).
Similar expressions one gets with the (x1−x2)-plane of isotropy and considering
E1=E2,ν13=ν23,G13=G23,G12=E1
2(1+ν12)(2.1.65)
or with the (x1−x3)-plane of isotropy and
E1=E3,ν12=ν23,G12=G23,G13=E1
2(1+ν13)(2.1.66)
The Young’s modulus and the Poisson’s ratio in the plane of is otropy often will be
designated as ETandνTT.ETcharacterizes elongations or contractions of a trans-
versely isotropic body in the direction of the applied load i n any direction of the
plane of isotropy and νTTcharacterizes contractions or elongations of the body in
the direction perpendicular to the applied load, but parall el to the plane of isotropy.
The shear modulus GTTcharacterizes the material response under shear loading in
the plane of isotropy and takes the form
GTT=ET
2(1+νTT),
i.e. any two of the engineering parameters ET,νTTandGTTcan be used to fully
describe the elastic properties in the plane of isotropy. A t hird independent primary
parameter should be EL. This Young’s modulus characterizes the tension respec-
tively compression response for the direction perpendicul ar to the plane of isotropy.
The fourth primary parameter should be the shear modulus GLTin the planes per-
pendicular to the plane of isotropy. As a ﬁfth primary parame ter can be chosen νLT
orνTL, which characterize the response in the plane of isotropy un der a load in L-
direction or the response in the L-direction under a load in t he plane of isotropy. The
stress-strain relations for an transversely isotropic bod y, if ( x2−x3) is the plane of
isotropy and with the reciprocity relations
νLT
EL=νTL
ET

42 2 Linear Anisotropic Materials
can be used in the following matrix form

ε1
ε2
ε3
ε4
ε5
ε6
=
1
EL−νTL
ET−νTL
ET0 0 0
1
ET−νTT
ET0 0 0
1
ET0 0 0
1
GTT0 0
S Y M1
GLT0
1
GLT

σ1
σ2
σ3
σ4
σ5
σ6
(2.1.67)
with the engineering parameters
E1=EL,E2=E3=ET,E4=G23=GTT=ET
2(1+νTT),
E5=G13=E6=G12=GLT,ν12=ν13=νLT,ν23=νTT,νLT
EL=νTL
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+νTT),ν13=ν23=νTL,ν12=νTT
Notice that in literature the notations of Poisson’s ratios νLTandνTLcan be corre-
spond to the opposite meaning.
2.1.4.3 Isotropic Material Behavior
For an isotropic material behavior in all directions, the nu mber of independent en-
gineering parameters reduces to two
E1=E2=E3=E,ν12=ν23=ν13=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=νE
∆∗∗,C44=C55=C66=G=E
2(1+ν)

2.1 Generalized Hooke’s Law 43
assuming ∆∗∗=(1+ν)(1−2ν).
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 jand engineering parameters are summarized in Table 2.5.
Consider that the elastic properties of an isotropic materi al are determined by two
independent parameters. The elastic parameters Young’s mo dulus Eand Poisson’s
ratio νare generally used because they are determined easily in phy sical tests. But
also the so-called Lam´ e coefﬁcients5λandµ, the shear modulus Gor the bulk
modulus Kcan be used if it is suitable. There are simple relations betw een the
parameters, e.g. as a function of E,ν
λ=Eν
(1+ν)(1−2ν),µ=E
2(1+ν)=G,K=E
3(1−2ν),
ν=λ
2(λ+µ), E=µ(3λ+2µ)
λ+µ, K=λ+2
3µ(2.1.69)
Summarizing the constitutive equations for isotropic, tra nsversely isotropic and
orthotropic materials, which are most important in the engi neering applications of
composite structural mechanics one can ﬁnd that the common f eatures of the re-
lationships between stresses and strains for these materia l 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 c ase 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 indep endent stiffness or
compliances. Therefore we have in comparison with orthotro pic materials to intro-
duce four additional engineering parameters and keeping in mind, that the mono-
clinic case must comprise the orthotropic case, we should no t 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,S36andS46.
The ﬁrst three pair normal strains ε1,ε2,ε3to the shear stress σ6and vice versa
the shear strain ε6to the normal stresses σ1,σ2,σ3. The fourth one couple the shear
strain ε4to the shear stress σ5and vice versa the shear strain ε5to the shear stress σ4.
In a compact notation the strain-stress relations for an ani sotropic 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 (L am´ e’s parameters in elasticity and ﬂuid
mechanics)

44 2 Linear Anisotropic Materials
Table 2.5 Relationships between Si j,Ci jand the engineering parameters for orthotropic, trans-
versely-isotropic and isotropic material
Orthotropic material
S11=E−1
1,S12=S21=−ν12E−1
1,S44=G−1
23=E−1
4
S22=E−1
2,S13=S31=−ν13E−1
1,S55=G−1
13=E−1
5
S33=E−1
3,S23=S32=−ν23E−1
2,S66=G−1
12=E−1
6
νi j=νji(Ei/Ej),Ei=(νi j/νji)Ej i,j=1,2,3
C11=S22S33−S2
23
det[Si j]=(1−ν23ν32)E1
∆,C44=1
S44=G23=E4
C22=S33S11−S2
13
det[Si j]=(1−ν31ν13)E2
∆,C55=1
S55=G31=E5
C33=S11S22−S2
12
det[Si j]=(1−ν21ν12)E3
∆,C66=1
S66=G12=E6
C12=S13S23−S12S33
det[Si j]=(ν12+ν32ν13)E2
∆=(ν21+ν31ν23)E1
∆=C21
C13=S12S23−S13S22
det[Si j]=(ν13+ν12ν23)E3
∆=(ν31+ν21ν32)E1
∆=C31
C23=S12S13−S23S11
det[Si j]=(ν23+ν21ν13)E3
∆=(ν32+ν12ν31)E2
∆=C32
∆=1−ν12ν21−ν23ν32−ν31ν13−2ν21ν13ν32
Transversely-isotropic material
(x2−x3)-plane of isotropy
E1,E2=E3,E5=E6,ν12=ν13,E4=E2
2(1+ν23)
(x1−x2)-plane of isotropy
E1=E2,E3,E4=E5,ν13=ν23,E6=E1
2(1+ν12)
(x1−x3)-plane of isotropy
E1=E3,E2,E4=E6,ν12=ν23,E5=E3
2(1+ν13)
Isotropic material
E1=E2=E3=E,ν12=ν21=ν13=ν31=ν23=ν32=ν,
E4=E5=E6=G=E
2(1+ν)

2.1 Generalized Hooke’s Law 45

ε1
ε2
ε3
ε4
ε5
ε6
=
1
E1−ν21
E2−ν31
E30 0η61
E6
−ν12
E11
E2−ν32
E30 0η62
E6
−ν13
E1−ν23
E21
E30 0η63
E6
0 0 01
E4µ54
E50
0 0 0µ45
E41
E50
η16
E1η26
E2η36
E30 01
E6

σ1
σ2
σ3
σ4
σ5
σ6
(2.1.70)
with the following reciprocal relations
η61
E6=η16
E1,η62
E6=η26
E2,η63
E6=η36
E3,µ54
E5=µ45
E4(2.1.71)
and the compliance components
S16=η61
E6,S26=η62
E6,S36=η63
E6,S45=µ54
E5(2.1.72)
η61,η62andη63are extension-shear coupling coefﬁcients indicating norm al strains
induced by shear stress σ6andη16,η26andη36the shear-extension coupling co-
efﬁcients characterizing shear strain ε6caused by normal stresses. µ45andµ54are
shear-shear coupling coefﬁcients.
The stiffness matrix for the monoclinic material can be foun d as the inverse of
the compliance matrix, but the expressions are unreasonabl e to present in an explicit
form. However, the inverse of a matrix can be easily calculat ed using standard nu-
merical procedures. Also for a generally anisotropic mater ial the compliance can be
formulated with help of eight additional coupling paramete rs but the stiffness matrix
should be calculated numerically.
2.1.5 Two-Dimensional Material Equations
In most structural applications the structural elements ar e simpliﬁed models by
reducing the three-dimensional state of stress and strain a pproximately to a two-
dimensional plane stress or plane strain state. A thin lamin a for instance can be
considered to be under a condition of plane stress with all st ress components in
the out-of-plane direction being approximately zero. The d ifferent 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 unidir ectional long-ﬁbre rein-
forced laminae the most general type of symmetry of the mater ial properties is the

46 2 Linear Anisotropic Materials
x1 x1 x1x2x2 x2x3x3 x3
σ2
σ1σ6σ3
σ2σ4
σ1σ3
σ5a bc
Fig. 2.9 Plane stress state. a(x1−x2)-plane, σ3=σ4=σ5=0,b(x2−x3)-plane, σ1=σ5=
σ6=0,c(x1−x3)-plane, σ2=σ4=σ6=0
monoclinic one the generalized Hooke’s law (2.1.20) is redu ced taking into account
(2.1.43) to
ε1
ε2
ε3
ε4
ε5
ε6
=
S11S12S130 0 S16
S22S230 0 S26
S330 0 S36
S S44S450
Y S550
M S66

σ1
σ2
0
0
0
σ6
(2.1.73)
That means σ3=σ4=σ5=0, and we have three in-plane constitutive equations
ε1=S11σ1+S12σ2+S16σ6
ε2=S12σ1+S22σ2+S26σ6, Si j=Sji
ε6=S16σ1+S26σ2+S66σ6(2.1.74)
and an additional equation for strain ε3inx3-direction
ε3=S13σ1+S23σ2+S36σ6 (2.1.75)
The other strains ε4,ε5are equal to 0 considering the monoclinic material behavior .
If the plane stress assumptions are used to simplify the gene ralized stiffness equa-
tions (2.1.19) taking into account (2.1.42), the result is

σ1
σ2
0
0
0
σ6
=
C11C12C130 0 C16
C22C230 0 C26
C330 0 C36
S C44C450
Y C550
M C66

ε1
ε2
ε3
ε4
ε5
ε6
(2.1.76)
or again three in-plane equations

2.1 Generalized Hooke’s Law 47
σ1=C11ε1+C12ε2+C13ε3+C16ε6,
σ2=C12ε1+C22ε2+C23ε3+C26ε6, Ci j=Cji
σ6=C16ε1+C26ε2+C36ε3+C66ε6(2.1.77)
There are another three equations. At ﬁrst we obtain
σ4=S44ε4+S45ε5=0, σ5=S45ε4+S55ε5=0
Since S44,S45,S55are arbitrary but the stiffness matrix should be positive de ﬁnite it
is obvious that
S44S55−S2
45>0
and from the σ4=σ5=0 condition it follows that ε4,ε5must be equal to 0. Taking
into account the last condition
σ3=C13ε1+C23ε2+C33ε3+C36ε6=0
the strain ε3can be obtained
ε3=−1
C33(C13ε1+C23ε2+C36ε6) (2.1.78)
and eliminated and substituted in Eqs. (2.1.77). After subs tituting ε3using Eq.
(2.1.78) Eqs. (2.1.77) leads to
σi=/parenleftbigg
Ci j−Ci3Cj3
C33/parenrightbigg
εj,i,j=1,2,6, (2.1.79)
respectively
σi=Qi jεj,i,j=1,2,6 (2.1.80)
TheQi jare the reduced stiffness. For the three cases in Fig. 2.9 we o btain
σi=Qi jεj,Qi j=Ci j−Ci3Cj3
C33,i,j=1,2,6,(x1−x2)−plane of symmetry ,
σi=Qi jεj,Qi j=Ci j−Ci1Cj1
C11,i,j=2,3,4,(x2−x3)−plane of symmetry ,
σi=Qi jεj,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 m atrices Si j,Ci jor
Qi jis six. It is very important to note that the elements in the pl ane stress compliance
matrix are simply a subset of the elements from the three-dim ensional compliance
matrix and their numerical values are identical. On the othe r hand, the elements
of the reduced stiffness matrix involve a combination of ele ments from the three-
dimensional stiffness matrix and the numerical values of th eQi jdiffer 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 ε3and the general case leads to
consistent relations for the transverse shear strains ε4andε5.
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 Si j,Ci jorQi jcan be shifted to
the engineering notation. Tables 2.6 and 2.7 summarize the c ompliance 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 ε1,ε2andε6but the four nonzero stress components σ1,σ2,σ3,σ6. Analo-
gous to the plane stress state, here the stress σ3normal to the (x1−x2)plane is not
an independent value and can be eliminated
ε3=S13σ1+S23σ2+S33σ3+S36σ6=0,
σ3=−1
S33(S13σ1+S23σ2+S36σ6)
Therefore in the case of plane strain, the Ci j,i,j=1,2,6 can be taken directly from
the three-dimensional elasticity law and instead of Si jreduced compliances Vi jhave
to be used
Table 2.6 Compliance matrices for various material models, plane str ess state
Material model ε=Sσ ε=Sσ ε=Sσ
Anisotropy: Compliances Si j
6 independentmaterial parameters
ε1
ε2
ε6
=
S11S12S16
S22S26
S66

σ1
σ2
σ6

Orthotropy: S16=S26=0
4 independent material parameters S11=1
E1,S22=1
E2
S12=−ν12
E1=−ν21
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 stre ss state
Material model σ=Qε σ=Qε σ=Qε
Anisotropy: Reduced stiffness Qi j
6 independent material parameters
σ1
σ2
σ6
=
Q11Q12Q16
Q22Q26
Q66

ε1
ε2
ε3

Orthotropy: Q16=Q26=0,Q66=1
S66=G12
4 independent material parameters Q11=S22
∆=E1
1−ν12ν21
Q22=S11
∆=E2
1−ν12ν21
Reference system: on-axis Q12=−S12
∆=ν12E2
1−ν12ν21
∆=S11S22−S2
12
Isotropy: Q16=Q26=0
2 independent material parameters Q11=Q22=E
1−ν2
Reference system: as you like Q12=νE
1−ν2,Q66=E
2(1+ν)=G
σi=Ci jεj,i,j=1,2,6,
εi=Vi jσj,Vi j=Si j−Si3Sj3
S33,i,j=1,2,6
Table 2.8 summarizes for the three-dimensional states and t he plane stress and strain
states the number of non-zero and of independent material pa rameters.
In the two-dimensional equations of anisotropic elasticit y, either reduced stiff-
ness or reduced compliances are introduced into the materia l laws. These equations
are most important in the theory of composite single or multi layered elements, e.g.
of laminae or laminates. The additional transformations ru les which are necessary
in laminae and laminate theories are discussed in more detai l in Chap. 3. Tables 2.6
and 2.7 above shows the relationship between stress and stra in through the compli-
ance[Si j]or the reduced stiffness [Qi j]matrix for the plane stress state and how the
Si jandQi jare related to the engineering parameters. For a unidirecti onal lamina the
engineering parameters are:

50 2 Linear Anisotropic Materials
Table 2.8 Stiffness and compliance parameters for stress and strain e quations σσσ=CCCεεε,
εεε=SSSσσσ,CCC=SSS−1, plane stress state σσσ=QQQεεε,εεε=SSSσσσ,QQQ=SSS−1, plane strain state – σσσ=CCCεεε,
εεε=VVVσσσ,CCC=VVV−1,Qi jandVi jare the reduced stiffness and compliances
Material model Number of non-zero Number of independent
parameters parameters
Three-dimensional Ci j;Si j Ci j;Si j
stress- or i,j=1,…, 6 i,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 Qi j;Si j
(x1−x2)-plane i,j=1,2,6 i,j=1,2,6
Anisotropic 9 6
Orthotropic 5 4
Isotropic 5 2
Plane strain state Ci j;Vi j Ci j;Vi j
(x1−x2)-plane i,j=1,2,6 i,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 th e ﬁbre direction)
ν12major Poisson’s ratio as the ratio of the negative normal str ain in direction 2
to the normal strain in direction 1 only if normal load is appl ied in direction 1
G12in-plane shear modulus for (x1−x2)plane
The four independent engineering elastic parameters are ex perimentally measured
as follows:
•Pure tensile load in direction 1: σ1/ne}ationslash=0,σ2=0,σ6=0
With ε1=S11σ1,ε2=S12σ1,ε6=0 are
E1=σ1
ε1=1
S11,ν12=−ε2
ε1=−S12
S11
•Pure tensile load in direction 2: σ1=0,σ2/ne}ationslash=0,σ6=0
With ε1=S12σ2,ε2=S22σ2,ε6=0 are
E2=σ2
ε2=1
S22,ν21=−ε1
ε2=−S12
S22
ν21is usually called the minor Poisson’s ratio and we have the re ciprocal rela-
tionship ν12/E1=ν21/E2.
•Pure shear stress in the (x1−x2)plane: σ1=σ2=0,σ6/ne}ationslash=0
With ε1=ε2=0,ε6=S66σ6is

2.1 Generalized Hooke’s Law 51
G12=σ6
ε6=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 p arameters: (Q11,Q12,
Q22,Q66),(S11,S12,S22,S66),(E1,E2,ν12,G12).
In Chap. 3 the evaluation of the four engineering elastic par ameters is given ap-
proximately by averaging the ﬁbre-matrix material behavio r. 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 app roach based on the
elementary theory.
2.1.6 Curvilinear Anisotropy
The type of anisotropy considered above was characterized b y the equivalence of
parallel directions passing through different points of th e homogeneous anisotropic
body and we can speak of a rectilinear anisotropy. Another ki nd of anisotropy is the
case, if one chooses a system of curvilinear coordinates in s uch a manner that the
coordinate directions coincide with equivalent direction s of elastic properties at dif-
ferent points of an anisotropic body. The elements of the bod y, which are generated
by three pairs of coordinate surfaces possess identical ela stic properties and we can
speak of a curvilinear anisotropy.
In the frame of this textbook we limit the considerations to c ylindrical 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 cylind rical coordinates x1=r,
x2=θ,x3=zand we have the stress and strain vectors in the contracted si ngle
subscript notation
/bracketleftbigσ1σ2σ3σ4σ5σ6/bracketrightbigT=/bracketleftbigσrσθσzσθzσrzσrθ/bracketrightbigT,
/bracketleftbigε1ε2ε3ε4ε5ε6/bracketrightbigT=/bracketleftbigεrεθεzεθzεrzεrθ/bracketrightbigT(2.1.81)
In the speciﬁc cases of material symmetries the general cons titutive equation
in cylindrical coordinates can be simpliﬁed analogous to th e case of rectilinear
anisotropy.
In the speciﬁc case of an orthotropic cylindrical response t here are three orthog-
onal planes of elastic symmetries. One plane is perpendicul ar 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 sit uation is a transversely
isotropic cylinder or cylindrical tube with the plane of iso tropy ( r−θ). In this case
we obtain, as analog to (2.1.67)

52 2 Linear Anisotropic Materials
z
xr
θz-plane
r−z-plane
θ−z-planez
xθr
Fig. 2.10 Cylindrical orthotropic material symmetry

ε1
ε2
ε3
ε4
ε5
ε6
=
1
ET−νTT
ET−νLT
EL0 0 0
1
ET−νLT
EL0 0 0
1
EL0 0 0
1
GLT0 0
S Y M1
GLT0
1
GTT

σ1
σ2
σ3
σ4
σ5
σ6
, (2.1.82)
where the index T is associated with the coordinate directio nsrandθand the index
L with the coordinate direction zand the reciprocal relations are
νLT
EL=νTL
ET
The stress-strain equations for the orthotropic case of cyl indrical anisotropy are ob-
tained using engineering parameters

2.1 Generalized Hooke’s Law 53

ε1
ε2
ε3
ε4
ε5
ε6
=
1
Er−νθr
Eθ−νzr
Er0 0 0
1
Eθ−νzθ
Er0 0 0
1
Er0 0 0
1
Gθz0 0
S Y M1
Grz0
1
Grθ

σ1
σ2
σ3
σ4
σ5
σ6
(2.1.83)
The indices r,θandzof the engineering parameters are associated with the indic es
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 recipro cal relations
Eiνji=Ejνi j,i,j=r,θ,z
and the Gθz,GrzandGrθmay be written in the more general form E4,E5,E6.
There are two practical situations for a monoclinic materia l 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 substi tuting the subscripts 1,2 and
3 by the engineering parameters to r,θandzand the shear moduli E4,E5andE6
byGθz,GrzandGrθ. 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 woun d cylindrical shells and
we get the strain-stress relations which couple all three no rmal strains to the shear
strain ε4and both shear strains ε5,ε6to both shear stresses σ5,σ6

ε1
ε2
ε3
ε4
ε5
ε6
=
1
E1−ν21
E2−ν31
E3η41
E40 0
1
E2−ν32
E3η42
E40 0
1
E3η43
E40 0
1
E40 0
S Y M1
E5µ65
E61
E6

σ1
σ2
σ3
σ4
σ5
σ6
(2.1.84)
The subscripts 1, 2, 3 of the Young’s moduli and the Poisson’s ratios will be shifted
tor,θ,zand the moduli E4,E5,E6toGθz,Grz,Grθ. There are as above reciprocal
relations for νi j,ηi jandµi j

54 2 Linear Anisotropic Materials
η41
E4=η14
E1,η42
E4=η24
E2,η43
E4=η34
E3,µ65
E6=µ56
E5,
νi j
Ei=νji
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 εand the stress σas functions of A,landF:
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 assumpti ons of the
iso-strain and the iso-stress models.
1. Assumptions
εi=ε,△li=△l,i=1,2,3,A=3
∑
i=1Ai,F=3
∑
i=1Fi
From σ=Eεfollows
σ1=E1ε=70GPa ε,σ2=E2ε=3GPa ε,σ3=E3ε=70GPa ε
With F=σA=EAεandFi=EiAiεfollows
F=3
∑
i=1(EiAi)ε=16,4 GPa εA,E=16,4 GPa,
ε=F
EA=△l
l
yields
△l=Fl
EA=1
16,4Fl
A(GPa)−1
and the solutions are
△l=1
16,4Fl
A(GPa)−1=△l(F,l,A),
ε=△l
l=1
16,4F
A(GPa)−1=ε(F,A),
σi=Eiε=1
16,4EiF
A(GPa)−1=σi(F,A),i=1,2,3

2.1 Generalized Hooke’s Law 55
2. Assumptions
△l=3
∑
i=1△li,Fi=F,εi=△li
li,i=1,2,3
From△l=3
∑
i=1△li=3
∑
i=1liεiandF=EAεfollows
△l=3
∑
i=1/parenleftbiggli
Ei/parenrightbiggF
A=/parenleftbigg0,1
70+0,8
3+0,1
70/parenrightbiggFl
A(GPa)−1,
△l=εl=1
EFl
A=0,2695Fl
A(GPa)−1,E=3,71 GPa
The functions △l,εandσare
△l=0,2695Fl
A(GPa)−1=△l(F,l,A),
ε=0,2695F
A(GPa)−1=ε(F,A),
σ=Eε=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=EA0/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 ρ1,ρ2, the cross-sectional areas A1,A2and
the lengths l1,l2the ratios of the stiffness K1andK2per length and the mass of the
bars areK1
K2=E1A1
E2A2l2
l1,m1
m2=l1A1ρ1
l2A2ρ2
Verify that for l1=l2andm1=m2the ratio K1/K2only depends on the ratio of the
speciﬁc Young’s moduli E1/ρ1andE2/ρ2.
Solution 2.2. Introducing the densities ρ1/ρ2into the stiffness ratio K1/K2yields
K1
K2=E1/ρ1
E2/ρ2m1
m2/parenleftbiggl2
l1/parenrightbigg2
and with m1=m2,l2=l1
K1
K2=E1/ρ1
E2/ρ2

56 2 Linear Anisotropic Materials
Note 2.1. A material with the highest value of E/ρhas the highest tension stiffness.
Exercise 2.3. For a simply supported beam with a single transverse load in t he mid-
dle of the beam we have the following equation
F=48EI
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=48EI/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,ρ1andE2,ρ2but 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/ρ2
1andE2/ρ2
2.
Solution 2.3. Moments of inertia and masses of the two beams are
1. circle cross-sectional: I=πr4/4,m=r2πlρ,
2. square cross-sectional: I=a4/12,m=a2lρ
In case 1. we have
K=48EI
l3=48E
l3πr4
4=48E/ρ2
4l3m2
πl2,
K1
K2=E1/ρ2
1
E2/ρ2
2/parenleftbiggm1
m2/parenrightbigg2/parenleftbiggl2
l1/parenrightbigg5
With l1=l2,m1=m2we obtain
K1
K2=E1/ρ2
1
E2/ρ2
2
In case 2. we have
K=48EI
l3=48E
l3a4
12=4E
l3m2
ρ2l2,
K1
K2=E1/ρ2
1
E2/ρ2
2/parenleftbiggm1
m2/parenrightbigg2/parenleftbiggl2
l1/parenrightbigg5
With l1=l2,m1=m2we obtain
K1
K2=E1/ρ2
1
E2/ρ2
2
Note 2.2. The best material for an optimal bending stiffness of the bea m is that with
the highest value of E/ρ2.
Exercise 2.4. Formulate explicitly the transformation matrices (3
TTTσσσ)−1and(3
TTTεεε)−1
for a rotation about the eee3-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
TTTσσσ)−1=(3
TTTεεε)Tand
(3
TTTεεε)−1=(3
TTTσσσ)T
(3
TTTσσσ)−1=
c2s20 0 0−2cs
s2c20 0 0 2 sc
0 0 1 0 0 0
0 0 0 c s 0
0 0 0−s c 0
cs−cs0 0 0 c2−s2
,
(3
TTTεεε)−1=
c2s20 0 0−cs
s2c20 0 0 sc
0 0 1 0 0 0
0 0 0 c s 0
0 0 0 −s c 0
2cs−2cs0 0 0 c2−s2

Exercise 2.5. Consider the coordinate transformation that corresponds w ith 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 transforma tion 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 forσ
Tpq
R2
11=1,R2
22=1,R2
33=1,
R22R33+R23R32=−1,R11R33+R13R31=−1,R11R22+R12R21=1
and forε
Tpq
R2
11=1,R2
22=1,R2
33=1,
R22R33+R23R22=−1,R11R33+R13R31=−1,R11R22+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 mat erial are
given by
E1=173GPa,E2=33,1GPa,E3=5,17GPa,
E4=3,24GPa,E5=8,27GPa,E6=9,38GPa,
ν12=0,036,ν13=0,25, ν23=0,171
Calculate the stiffness matrix CCCand the compliance matrix SSS.
Solution 2.6. With Table 2.5 we ﬁnd the Si jand the Ci j
S11=E−1
1=5,780 10−3GPa−1,
S12=S21=−ν12E−1
1=−0,208 10−3GPa−1,
S22=E−1
2=30,211 10−3GPa−1,
S13=S31=−ν13E−1
1=−1,445 10−3GPa−1,
S33=E−1
3=193,424 10−3GPa−1,
S23=S32=−ν23E−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−ν12ν21−ν23ν32−ν31ν13−2ν21ν13ν32,
ν21=ν12(E2/E1)=0,0069,ν31=ν13(E3/E1)=0,0075,
ν32=ν23(E3/E2)=0,027,△=0,993,¯Ei=Ei/△,i=1,2,3
C11= (1−ν23ν32)¯E1=173,415GPa,
C22= (1−ν31ν13)¯E2=33,271GPa,
C33= (1−ν12ν21)¯E3=5,205GPa,
C12= (ν12+ν13ν32)¯E2=1,425GPa,
C13= (ν13+ν12ν23)¯E3=1,334GPa,
C23= (ν23+ν21ν13)¯E3=0,899GPa,
C44=E4,C55=E5,C66=E6
With the values for Ci jandSi jthe stiffness matrix CCCand the compliance matrix SSS
can be written

2.2 Fundamental Equations and Variational Solution Proced ures 59
CCC=
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,
SSS=
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 Proced ures
Below we discuss at ﬁrst the fundamental equations of the ani sotropic elasticity for
rectilinear coordinates. The system of equations can be div ided 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 equili brium equations (static
or dynamic) and the kinematic equations (the strain-displa cement 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 b e, the initial conditions
to close the initial-boundary problem. At second, consider ing that closed solutions
are impossible in most of the practical cases approximative solution techniques are
brieﬂy discussed. The main attention will be focussed on var iational 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 t he 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 an d 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 stre ss components. In the
case of dynamic problems the inertia forces are expressed th rough 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 d x1,dx2,dx3:
stress and volume force com-
ponents in x1-direction
x1x2x3
dx1
dx2dx3
σ1+dσ1p1σ6+dσ6σ1σ5+dσ5
σ6
σ5
∂σ1
∂x1+∂σ6
∂x2+∂σ5
∂x3+p1=0,
∂σ6
∂x1+∂σ2
∂x2+∂σ4
∂x3+p2=0,
∂σ5
∂x1+∂σ4
∂x2+∂σ3
∂x3+p3=0,static equations (2.2.1)
∂σ1
∂x1+∂σ6
∂x2+∂σ5
∂x3+p1=ρ∂2u1
∂t2,
∂σ6
∂x1+∂σ2
∂x2+∂σ4
∂x3+p2=ρ∂2u2
∂t2,
∂σ5
∂x1+∂σ4
∂x2+∂σ3
∂x3+p3=ρ∂2u3
∂t2dynamic equations (2.2.2)
The inertial terms in (2.2.2) are dynamic body forces per uni t volume. The den-
sityρfor unidirectional laminae can be calculated e.g. using the rule of mixtures
(Sect. 3.1.1).
•The kinematic equations that are six strain-displacement r elations and six com-
patibility conditions for strains. For linear small deform ation theory, the six
stress-displacement equations couple six unknown strains and three unknown
displacements. Figure 2.12 shows the strains of an inﬁnites imal cube in the
(x1−x2)-plane and we ﬁnd the relations
ε1≡∂u1
∂x1,ε2≡∂u2
∂x2,α=∂u2
∂x1,β=∂u1
∂x2⇒2ε12≡γ12≡∂u1
∂x2+∂u2
∂x1≡ε6
and analogous relations for the ( x2−x3)- and ( x1−x3)-planes yield

2.2 Fundamental Equations and Variational Solution Proced ures 61
x1 x1x2 x2
u1 u1+∂u1
∂x1dx1u2u2+∂u2
∂x2dx2
dx1dx2
A AA′A′CCC′
C′
B BB′
B′∂u1
∂x2dx2
∂u2
∂x1dx1a b
αβ
γ
Fig. 2.12 Strains of the inﬁnitesimal cube shown for the ( x1−x2)-plane. aextensional strains,
bshear strains
ε1=∂u1
∂x1, ε2=∂u2
∂x2, ε3=∂u3
∂x3,
ε4=∂u3
∂x2+∂u2
∂x3,ε5=∂u3
∂x1+∂u1
∂x3,ε6=∂u2
∂x1+∂u1
∂x2(2.2.3)
The displacement ﬁeld of the body corresponding to a given de formation state is
unique, the components of the strain tensor must satisfy the following six com-
patibility conditions
∂2ε1
∂x2
2+∂2ε2
∂x2
1=∂2ε6
∂x1∂x2,∂
∂x3/parenleftbigg∂ε4
∂x1+∂ε5
∂x2−∂ε6
∂x3/parenrightbigg
=2∂2ε3
∂x1∂x2,
∂2ε2
∂x2
3+∂2ε3
∂x2
2=∂2ε4
∂x2∂x3,∂
∂x1/parenleftbigg∂ε5
∂x2+∂ε6
∂x3−∂ε4
∂x1/parenrightbigg
=2∂2ε1
∂x2∂x3,
∂2ε3
∂x2
1+∂2ε1
∂x2
3=∂2ε5
∂x1∂x3,∂
∂x2/parenleftbigg∂ε6
∂x3+∂ε4
∂x1−∂ε5
∂x2/parenrightbigg
=2∂2ε2
∂x3∂x1
In the two-dimensional case the compatibility conditions r educe to a single equa-
tion
∂2ε1
∂x2
2+∂2ε2
∂x2
1−2∂2ε6
∂x1∂x2=0
•The material or constitutive equations which are described in Sect. 2.1 are
εi=Si jσj, σi=Ci jεj (2.2.4)
The generalized Hooke’s law yields six equations relating i n each case six un-
known stress and strain components. The elements of the stif fness matrix CCCand
the compliance matrix SSSare substituted with respect to the symmetry conditions
of the material.

62 2 Linear Anisotropic Materials
Summarizing all equations, we have 15 independent equation s for 15 unknown com-
ponents of stresses, strains and displacements. In the disp lacement approach, the
stresses and strains are eliminated and a system of three cou pled partial differential
equations for the displacement components are left.
In the static case we have a boundary-value problem, and we ha ve to include
boundary conditions. In the dynamic case the system of parti al differential equations
deﬁnes an initial-boundary-value problem and we have addit ional initial conditions.
A clear symbolic formulation of the fundamental equations i n displacements is
given in vector-matrix notation. With the transposed vecto rsσσσT,εεεTanduuuT
σσσT=[σ1σ2σ3σ4σ5σ6],εεεT=[ε1ε2ε3ε4ε5ε6],uuuT=[u1u2u3](2.2.5)
the transformation and the differential matrices TTTandDDD(nnnis the surface normal
unit vector)
TTT=
n10 0
0n20
0 0 n3
0n3n2
n30n1
n2n10
,DDD=
∂10 0
0∂20
0 0 ∂3
0∂3∂2
∂30∂1
∂2∂10
,ni=cos(nnn,xi)
∂i=∂
∂xi
i=1,2,3(2.2.6)
and the stiffness matrix CCCwe get
DDDTσσσ+ppp=000∈Vstatic equilibrium equations,
DDDTσσσ+ppp=ρ∂2uuu
∂t2∈Vdynamic equilibrium equations,
TTTTσσσ=qqq∈Aqsurface equilibrium equations,
εεε =DDDuuu∈Vkinematic equations,
σσσ =CCCεεε constitutive equations(2.2.7)
Vis the volume and Aqthe surface of the body with surface forces qqq.
Eliminating the stresses and the strains leads to the differ ential equations for the
displacements
Boundary-value problem – elastostatics
DDDTCCCDDDuuu=−ppp∈Vequilibrium for the volume element V,
uuu=uuu∈Auprescribed displacements uuuonAu,
TTTTCCCDDDuuu=qqq∈Aqprescribed surface forces qqqonAq(2.2.8)
Initial boundary-value problem – elastodynamics
DDDTCCCDDDuuu−ρ¨uuu=−ppp∈V equilibrium equation,
uuu=uuu∈Au,TTTTCCCDDDuuu=qqq∈Aq boundary conditions,
uuu(xxx,0)=uuu(xxx,0),˙uuu(xxx,0)=˙uuu(xxx,0)initial conditions(2.2.9)
In the general case of material anisotropic behavior the thr ee-dimensional equa-
tions are very complicated and analytical solutions are onl y possible for some spe-

2.2 Fundamental Equations and Variational Solution Proced ures 63
cial problems. This is independent of the approach to displa cements 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 s tatic hypotheses and the
equations are deduced separately in Chaps. 7 and 8. The simpl iﬁed structural equa-
tions for circular cylindrical shells and thin-walled fold ed structures are given in
Chaps. 9 and 10.
Summarizing the fundamental equations of elasticity we hav e introduced stresses
and displacements as static and kinematic ﬁeld variables. A ﬁeld is said to be stati-
cally admissible if the stresses satisfy equilibrium equat ions (2.2.1) and are in equi-
librium with the surface traction ¯qqqon the body surface Aq, where the traction are
given. A ﬁeld is referred to as kinematically admissible if d isplacements 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 consid ered in principles of vir-
tual work and energy formulations, Sect. 2.2.2. The mutual c orrespondence 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 anisot ropic elasticity may re-
alized as above by a system of partial differential equation s 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 varia tional formulations is in
general twofold. They yield convenient methods for the deri vation 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. T here 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 princi ples formulated for the
total elastic potential energy of the problem or the complem entary potential energy
are very useful in the theory of elasticity and in modelling a nd analysis of structural
elements. The fundamental equations and boundary conditio ns given beforehand
can be derived with the extremal principles and approximate solutions are obtained
by direct variational methods. Both extremal principles fo llow from the principle of
virtual work.
If an elastic body is in equilibrium, the virtual work δWof all acting forces in
moving through a virtual displacement δuuuis zero
δW≡δWa+δWi=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
δWis the total virtual work, δWathe external virtual work of body or volume and
surface forces and δWithe internal virtual work of internal stresses, for the forc es
associated with the stress ﬁeld of a body move the body points through virtual dis-
placements δuuucorresponding to the virtual strain ﬁeld δεεε.
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 di splacements are called
virtual because they are assumed to be inﬁnitesimal while ti me is held constant. The
symbol δis called a variational operator and in the mathematical vie w a virtual dis-
placement is a variation of the displacement function. To us e variational operations
in structural mechanics only the following operations of th eδ-operator are needed
δ/parenleftbiggdnf
dxn/parenrightbigg
=dn
dxn(δf),δ(fn)=n fn−1δfn−1,δ/integraldisplay
fdx=/integraldisplay
δfdx
For a deformable body, the external and the internal work are given in Eqs.
(2.2.11) and (2.2.12), respectively,
δWa=/integraldisplay
VpkδukdV+/integraldisplay
AqqkδukdA, (2.2.11)
δWi=−/integraldisplay
VσkδεkdV (2.2.12)
pkare the components of the actual body force vector pppper unit volume and qk
the components of the actual surface force vector qqq(surface traction per unit area).
Aqdenotes the portion of the boundary on which surface forces a re speciﬁed. σk
andεkare the components of the stress and the strain vector. The ne gative sign in
(2.2.12) indicates that the inner forces oppose the inner vi rtual displacements, e.g.
if the virtual displacement δu1=δε1dx1is subjected an inner force (σ1dx2dx3)the
inner work is (−σ1dx2dx3)δε1dx1. The vectors ppp,qqqanduuuhave three components
but the vectors σσσandεεεhave six components. The double subscript kinpkδukand
qkδukmeans the summarizing on 1 to 3 but in σkδεkon 1 to 6.
The general formulation of the principle of virtual work for a deformable body
δWa+δWi≡δW=0
or /integraldisplay
VpkδukdV+/integraldisplay
AqqkδukdA−/integraldisplay
VσkδεkdV=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 formu lated as follow:
Theorem 2.1 (Principle of virtual work). The sum of virtual work done by the
internal and external forces in arbitrary virtual displace ments satisfying the pre-

2.2 Fundamental Equations and Variational Solution Proced ures 65
scribed geometrical constraints and the strain-displacem ent relations is zero, i.e.
the arbitrary ﬁeld variables δukare kinematically admissible.
An important case is restricted to linear elastic anisotrop ic bodies and is known as
the principle of minimum total potential energy. The extern al virtual work δWais
stored as virtual strain energy δWf=−δWi, i.e. there exists a strain energy density
function
Wf(εεε)=1
2σkεk=1
2Cklεkεl
Assuming conservative elasto-static problems, the princi ple of virtual work takes
the form
δΠ(uuu)≡δΠa(uuu)+δΠ i(εεε)≡0, εεε=εεε(uuu) (2.2.14)
with the total potential energy function Π(uuu)of the elastic body. Πa(uuu)andΠi(εεε)
are the potential functions of the external and the internal forces, respectively,
Πi=Π(εk)=1
2/integraldisplay
VCklεkεldV,
Πa=Πa(uk)=−/integraldisplay
VpkukdV−/integraldisplay
AqqkukdA(2.2.15)
The principle of minimum total potential energy may be state d for linear elastic
bodies with the constraints σσσ=CCCεεε(uuu)as follows:
Theorem 2.2 (Principle of minimum of the total potential ene rgy). Of all the ad-
missible displacement functions satisfying strain-stres s relations and the prescribed
boundary conditions, those that satisfy the equilibrium eq uations 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 mini mum total potential
energy is widely used in solutions to problems of structural mechanics.
The principle of virtual work can be formulated in a compleme ntary statement.
Then virtual forces are introduced and the displacements ar e ﬁxed. In analogy to
(2.2.13) we have the principle of complementary virtual wor k as
δW∗
a+δW∗
i≡δW∗=0
with the complimentary external and internal virtual works
δW∗
a=/integraldisplay
AuukδqkdA,δW∗
i=−/integraldisplay
VεkδσkdV (2.2.16)
7Leonhard Euler (∗15 April 1707 Basel – †7jul./18greg.September 1783 St. Petersburg) – mathe-
matician, physicist, astronomer, logician and engineer, i ntroducing beam theory elements and the
equations of motion
8Joseph-Louis Lagrange, born Giuseppe Lodovico Lagrangia o r Giuseppe Ludovico De la Grange
Tournier, also reported as Giuseppe Luigi Lagrange or Lagra ngia (∗25 January 1736 Turin – †10
April 1813 Paris) – mathematician and astronomer, variatio nal calculations, general mechanics

66 2 Linear Anisotropic Materials
Audenotes the portion of the boundary surface on which displac ements are speci-
ﬁed.
With the complementary stress energy density function
W∗
f(σσσ)=1
2Sklσkσl,δW∗
f(σσσ)=Sklσlδσk=εkδσk
and assuming conservative elasto-static problems, the pri nciple of complementary
work can be formulated as principle of minimum total complem entary energy
δΠ∗=δΠ∗
i+δΠ∗
a
or
δ
/integraldisplay
VW∗(σk)dV−/integraldisplay
AuukqkdA
=0 (2.2.17)
The principle of minimum total complementary energy may be s tated for linear
elastic bodies with constraints εεε=SSSσσσas follows:
Theorem 2.3 (Principle of minimum total complementary ener gy).Of all admis-
sible stress states satisfying equilibrium equations and s tress boundary conditions,
those which are kinematically admissible make the total com plementary 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 o f Castigliano (principle
of virtual forces, force method) correspond to the principl e of minimum potential
energy and complementary energy. The principle of minimum p otential energy is
much more used in solution procedures, because it is usually far easier to formulate
assumptions about functions to represent admissible displ acements as to formulate
admissible stress functions that ensure stresses satisfyi ng 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 obtain ed in its entirety. One
must either satisfy the strain-displacement relations and the displacement bound-
ary conditions exactly and get approximate equilibrium con ditions or vice versa.
Both principles yield the risk to formulate approximate the ories or solution proce-
dures which may be mathematically inconsistent. Reissner’ s variational statement
yields as Euler-Lagrange equations both, the equilibrium e quations and the strain-
displacement relations, and has the advantage that its use w ould yield approximate
theories and solution procedures which satisfy both requir ements to the same de-
gree and would be consistent. Reissner’s variational theor em (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 εεε=CCCσσσwhich satisfy the boundary

2.2 Fundamental Equations and Variational Solution Proced ures 67
conditions, those which also satisfy the equilibrium equat ions and the stress-
displacements relations correspond to a minimum of the func tional ΨRdeﬁned as
ΨR(uuu,σσσ)=/integraldisplay
V[σkεk−Wf(σk)]dV−/integraldisplay
VpkukdV−/integraldisplay
AqqkukdA (2.2.18)
Wf(σk)is the strain energy density function in terms of stresses on ly,ΨRis Reissner’s
functional.
It should be noted that all stress and strain components must be varied while pk
andqkare prescribed functions and therefore ﬁxed. The variation of the functional
ΨR(uuu,σσσ)yields
δΨR=/integraldisplay
V/bracketleftbigg
σkδεk+εkδσk−∂Wf
∂σkδσk/bracketrightbigg
dV−/integraldisplay
VpkδukdV
−/integraldisplay
AqqkδukdA,(2.2.19)
where εkis determined by (2.2.3). δΨR(uuu,σσσ)can be rearranged and we obtain ﬁnally
δΨR=/integraldisplay
V/braceleftbigg/bracketleftbigg
εεε−∂Wf
∂σσσ/bracketrightbigg
δσσσT−/bracketleftbig
DDDTσσσ+ppp/bracketrightbig
δuuuT/bracerightbigg
dV−/integraldisplay
AqqqqδuuuTdA (2.2.20)
Since δσσσandδuuuare arbitrary variations δΨR=0 is satisﬁed only if
εi j=∂Wf(σi j)
∂σi j,∂σi j
∂xj+pi=0 (2.2.21)
Summarizing we have considered two dual energy principles w ithukorσkas admis-
sible functions which have to be varied and one generalized v ariational principle,
where both, ukandσk, 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 d issipative forces
following from external or inner damping effects the total v irtual work has in the
sense of the d’Alambert principle an additional term which r epresents the inertial
forces
δW=−/integraldisplay
Vρ¨ukδukdV−/integraldisplay
VσkδεkdV+/integraldisplay
VpkδukdV+/integraldisplay
AqqkδukdA (2.2.22)
Equation (2.2.22) represents an extension of the principle of virtual work from stat-
ics to dynamics. ρis 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
δt2/integraldisplay
t1(T−Π)dt≡δt2/integraldisplay
t1L(uk)dt=0,T=1
2/integraldisplay
Vρ˙uk˙ukdV (2.2.23)
Π(uuu)is the potential energy given beforehand and T(uuu)is the so-called kinetic
energy. L=T−Πis the Lagrangian function.
Theorem 2.5 (Hamilton’s principle for conservative system s).Of all possible
paths between two points at time interval t 1and t 2along 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(uuu) =1
2/integraldisplay
VσσσεεεTdV−/integraldisplay
VpppuuuTdV−/integraldisplay
AqqqquuuTdA,
δΠ=/integraldisplay
VσσσδεεεTdV−/integraldisplay
VpppδuuuTdV−/integraldisplay
AqqqqδuTdA=0(2.2.24)
Conservative elasto-dynamic problems
L(uuu)=T(uuu)−Π(uuu),T(uuu)=1
2/integraldisplay
Vρ˙uuuT˙uuudV,
δt2/integraldisplay
t1L(uuu)dt=0(2.2.25)
All variations are related to the displacement vector uuu. For the stress and the strain
vector we have to take into consideration that σσσ=σσσ(εεε)=σσσ[εεε(uuu)]and for the time
integrations
δuuu(xxx,t1)=δuuu(xxx,t2)=0 (2.2.26)
For non-conservative systems of elastodynamics, the virtu al work δWincludes an
approximate damping term
−/integraldisplay
Vµ˙uuuTδuuudV
with µas 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 contrib utions to classical mechanics, optics,
and algebra

2.2 Fundamental Equations and Variational Solution Proced ures 69
δW=−/integraldisplay
V(ρ¨uuuT+µ˙uuuT)δuuudV−/integraldisplay
Vδ(σσσTεεε)dV−/integraldisplay
VpppδuuudV
−/integraldisplay
AqqqqδuuudA(2.2.27)
A generalized Hamilton’s principle in conjunction with the Reissner’s variational
statement can be presented as
δ χ(uuu,σσσ)=δt2/integraldisplay
t1[T(uuu)−ψR(uuu,σσσ)]dt=0,
where T(uuu)is the kinetic energy as above, ψR(uuu,σσσ)the Reissner’s functional
(2.2.18).
2.2.3 Variational Methods
The variational principles can be used to obtain, in a mathem atical way, the govern-
ing differential equations and associated boundary condit ions 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 direc t methods, i.e. the approx-
imate solution is obtained directly by applying the same var iational 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 probl em cannot be derived.
Among the approximation methods, Ritz10method 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 theor y and to some extent on
weighted-residual methods.
The Rayleigh-Ritz method is based on variational statement s, e.g. the principle of
minimum total potential energy, which is equivalent to the f undamental differential
equations as well as to the so-called natural or static bound ary conditions including
10Walter Ritz (∗February 22nd, 1878, Sion, Switzerland; † July 7th, 1909, G¨ ottingen) – theoretical
physicist, variational methods
11John 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 i s 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 e lastic potential energy
functional
Π=1
2/integraldisplay
VεεεTCCCεεεdV−/integraldisplay
VpppTuuudV−/integraldisplay
AqqqqTuuudA
=1
2/integraldisplay
V(DDDuuu)TCCCDDDuuudV−/integraldisplay
VpppTuuudV−/integraldisplay
AqqqqTuuudA(2.2.28)
The variations are related to the displacements uuuand the strains εεεwhich have to be
substituted with help of the differential matrix DDD, (2.2.6), by the displacements. The
approximate solution is sought in the form of a ﬁnite linear c ombination. Looking
ﬁrst at a scalar displacement approach, the approximation o f the scalar displacement
function u(x1,x2,x2)is given by the Ritz approximation
˜u(x1,x2,x3)=N
∑
i=1aiϕi(x1,x2,x3)
or
˜u(x1,x2,x3)=N
∑
i=1aiϕi(x1,x2,x3)+ϕ0(x1,x2,x3) (2.2.29)
Theϕiare known functions chosen a priori, named approximation fu nctions 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
˜Πin form of a function of the constants aiandδ˜Π(ai) =0 yields Nstationary
conditions
∂˜Π(ai)
∂ai=0,i=1,2,…, N (2.2.31)
˜Πmay be written as a quadratic form in aiand from Eqs. (2.2.31) follows a system
ofNlinear equations allowing the Nunknown constants aito be determined. In
order to ensure a solution of the system of linear equations a nd a convergence of the
approximate solution to the true solution as the number Nof the aiis increased, the
ϕivalues have to fulﬁll the following requirements:
•ϕ0satisﬁes speciﬁed inhomogeneous geometric boundary condi tions, the so-
called essential conditions of the variational statement a ndϕi,i=1,2,…, Nsat-
isfy the homogeneous form of the geometric boundary conditi ons.
•ϕiare continuous as required in the variational formulation, e.g. they should have
a non-zero contribution to ˜Π.
•ϕiare linearly independent and complete.

2.2 Fundamental Equations and Variational Solution Proced ures 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 pro blems and using vector-
matrix notation it follows
˜uuu(x1,x2,x3)≡
˜u1
˜u2
˜u3
=
aaaT
1ϕϕϕ1
aaaT
2ϕϕϕ2
aaaT
3ϕϕϕ3
≡
ϕϕϕ1ooo ooo
oooϕϕϕ2ooo
ooo oooϕϕϕ3
T
aaa1
aaa2
aaa3
 (2.2.32)
or
˜uuu(x1,x2,x3)=GGGTaaa (2.2.33)
with
GGGT=
ϕϕϕ1ooo ooo
oooϕϕϕ2ooo
ooo oooϕϕϕ3
T
=
ϕϕϕT
1oooToooT
oooTϕϕϕT
2oooT
oooToooTϕϕϕT
3
,aaa=
aaa1
aaa2
aaa3

GGGis the matrix of the approximation functions, ϕϕϕiandoooareN-dimensional vectors
andaaaiareN-dimensional subvectors of the vector aaaof the unknown coordinates.
The application of the Ritz method using the minimum princip le of elastic potential
energy Πhas the following steps:
1. Choose the approximation function ˜uuu=GGGTaaa.
2. Substitute ˜uuuintoΠ
˜Π(˜uuu) =1
2/integraldisplay
V(DDD˜uuu)TCCCDDD˜uuudV−/integraldisplay
VpppT˜uuudV−/integraldisplay
AqqqqT˜uuudA
=1
2aaaTKKKaaa−aaaTfff(2.2.34)
with
KKK=/integraldisplay
V(DDDGGG)TCCC(DDDGGG)dV=/integraldisplay
VBBBTCCCBBBdV,
fff=/integraldisplay
VGGGTpppdV+/integraldisplay
AqGGGTqqqdA
3. Formulate the stationary conditions of ˜Π(aaa)
∂˜Π(aaa)
∂aaa=ooo
i.e. with
∂
∂aaa(aaaTKKKaaa)=2KKKaaa,∂
∂aaa(aaaTfff)=aaa
follows

72 2 Linear Anisotropic Materials
KKKaaa=fff (2.2.35)
KKKis called the stiffness matrix, aaathe vector of unknowns and fffthe force vector.
These notations are used in a generalized sense.
4. Solve the system of linear equations KKKaaa=fff. The vector aaaof unknown coefﬁ-
cients is known.
5. Calculate the approximation solution ˜uuu=aaaTϕϕϕand the ˜εεε=DDD˜uuu,˜σσσ=CCC˜εεε,…
For an increasing number N, the previously computed coefﬁcients of aaaremain un-
changed provided the previously chosen coordinate functio ns 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 carri ed out in an analo-
gous manner and can be summarized as follows. For conservati ve problems we start
with the variational statement (2.2.23). The displacement vector uuuis now a function
ofxxxandtand the aaa-vector a function of t. The stationary condition yields
∂
∂aaa/braceleftbigg1
2aaaT(t)KKKaaa(t)−aaaT(t)fff(t)+¨aaaT(t)MMMaaa(t)/bracerightbigg
=000 (2.2.36)
MMM¨aaa(t)+KKKaaa(t)=fff(t), (2.2.37)
MMM=/integraldisplay
VρGGGTGGGdV,
KKK=/integraldisplay
V(DDDGGG)TCCC(DDDGGG)dV,
fff=/integraldisplay
VGGGTpppdV+/integraldisplay
AqGGGTqqqdA
The matrix GGGdepends on xxx,pppandqqqonxxxandt.MMMis called the mass matrix.
An direct derivation of a damping matrix from the Ritz approx imation analogous
to the KKK- and the MMM-matrix of (2.2.37) is not possible. In most engineering app lica-
tions (2.2.37) has an additional damping term and the dampin g matrix is formulated
approximately as a linear combination of mass- and stiffnes s-matrix (modal damp-
ing)
MMM¨aaa(t)+CCCD˙aaa(t)+KKKaaa(t)=fff(t),CCCD≈αMMM+βKKK (2.2.38)
In the case of the study of free vibrations, we write the time d ependence of aaa(t)in
the form
aaa(t)=ˆaaacos(ωt+ϕ) (2.2.39)
and from (2.2.38) with CCCD=000,fff(t)=000 comes the matrix eigenvalue problem (App.
A.3)
(KKK−ω2MMM)ˆaaa=ooo,det[KKK−ω2MMM]=0 (2.2.40)
ForNcoordinate functions the algebraic equation (2.2.40) yiel dsNeigenfrequen-
cies of the deformable body.

2.2 Fundamental Equations and Variational Solution Proced ures 73
The Rayleigh-Ritz method approximates the continuous defo rmable body by a
ﬁnite number of degree of freedoms, i.e. the approximated sy stem is less ﬂexible
than the actual body. Consequently for the approximated ene rgy potential ˜Π≤Π.
The energy potential converges from below. The approximate displacements satisfy
the equilibrium equations only in the energy sense and not po intwise, unless the
solution converges to the exact solution. The Rayleigh-Rit z 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 hig her 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 sol ution (2.2.29), where
now the constants aiare determined by requiring the residual
RN=A/parenleftigg
N
∑
i=1aiϕi+ϕ0/parenrightigg
−f/ne}ationslash=0 (2.2.42)
be orthogonal to Nlinear independent weight function ψi
/integraldisplay
VRNψidxxx=0,i=1,2,…, N (2.2.43)
ϕ0,ϕishould be linear independent and complete and fulﬁll all bou ndary conditions.
Various known special methods follow from (2.2.43). They di ffer from each other
due to the choice of the weight functions ψi:
•Galerkin’s method12ψi≡ϕi,
•Least-squares method ψi≡A(ϕi),
•Collocation method ψi≡δ(xxx−xxxi)(δ(xxx−xxxi) = 1 if xxx=xxxiotherwise 0)
The Galerkin method is a generalization of the Ritz method, i f 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 equat ions, if the coordinate
functions ϕiin both are the same.
12Boris Grigorjewitsch Galerkin, surname more accurately ro manized as Galyorkin (∗4
Marchgreg./20 Februaryjul.1871 Polozk – 12 July 1945 Leningrad) – mathematician and eng ineer,
contributions to the theory of approximate solutions of par tial differential equations

74 2 Linear Anisotropic Materials
The classical variational methods of Ritz and Galerkin are w idely used to solve
problems of applied elasticity or structural mechanics. Wh en applying the Ritz or
Galerkin method to special problems involving, e.g. a two-d imensional 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=1aiϕi(x1,x2)or ˜u(x1,x2)=N
∑
i=1M
∑
j=1ai jϕ1i(x1)ϕ2j(x2),(2.2.44)
where ϕi(x1,x2)orϕ1i(x1),ϕ2j(x2)are a priori chosen trial functions and the aior
ai jare unknown constants. The approximate solution depends ve ry strongly on the
assumed trial functions.
To overcome the shortcoming of these solution methods Vlaso v13and Kan-
torovich14suggested an approximate solution in the form
˜u(x1,x2)=N
∑
i=1ai(x1)ϕi(x1,x2) (2.2.45)
Theϕiare again a priori chosen trial functions but the ai(x1)are unknown coefﬁcient
functions of one of the independent variables. The conditio nδΠ[˜u(x1,x2)] = 0 or
with d A=dx1dx2 /integraldisplay
ARN(˜u)ϕidA=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 fu nctions ϕias func-
tions of one independent variable, i.e. ϕi=ϕi(x2), since otherwise the system of
ordinary differential equations will have variable coefﬁc ients. The approximate so-
lution ˜ u(x1,x2)tends in regard of the arbitrariness of the assumed trial fun ction
ϕi(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=1ai(x1)ϕi(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
ϕi(x2)≡ai2(x2), i.e.
˜u[I](x1,x2)=N
∑
i=1ai2(x2)ai1(x1),
13Vasily Zakharovich Vlasov (∗11greg./24jul.February 1906 Kareevo – †7 August 1958 Moscow)
– civil engineer
14Leonid Vitaliyevich Kantorovich (∗19 January 1912 St. Petersburg – †7 April 1986 Moscow) –
mathematician and economist, Nobel prize winner in economi cs in 1975

2.2 Fundamental Equations and Variational Solution Proced ures 75
theai1(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 d enoted 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 t he trial function ϕi(x2)
and the iterative procedure converges very rapidly. It can b e 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)ϕ1(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 m ethods is the selection of
coordinate functions, especially for structures with irre gular domains. The limita-
tions of the classical variational methods can be overcome b y numerical methods,
e.g. the ﬁnite element method which is discussed in more deta il in Chap. 11.
2.2.4 Problems
Exercise 2.7. An anisotropic body is subjected to a hydrostatic pressure
σ1=σ2=σ3=−p,σ4=σ5=σ6=0.
1. Calculate the strain state εεε.
2. Calculate the stress state σσσfor a change of the coordinate system obtained by a
rotation TTTσ.
Solution 2.7. The solution can be presented in two parts:
1. The generalized Hooke’s law yields (Eq. 2.1.20)
ε1=−(S11+S12+S13)p,ε4=−(S14+S24+S34)p,
ε2=−(S12+S22+S23)p,ε5=−(S15+S25+S35)p,
ε3=−(S13+S23+S33)p,ε6=−(S16+S26+S36)p
Note 2.3. A hydrostatic pressure in an anisotropic solid yields exten sional 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
σ1=σ2=σ3=σ5=σ6=0,σ4=t.
1. Calculate the strain state εεε.
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
ε1=S14t,ε2=S24t,ε3=S34t,
ε4=S44t,ε5=S45t,ε6=S46t
The anisotropic body has extensional and shear strains in al l coordinate planes.
2. In an isotropic body a pure shear stress state yields only s hearing strains:
S14=S24=S34=0
Exercise 2.9. Consider a prismatic homogeneous anisotropic bar which is ﬁ xed 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, landAare 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 σ1=σ2=σ4=σ5=σ6=0,σ3=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=gρ.
Determine the strain state and the displacements and calcul ate the displacements
in the point (0,0,l).
Solution 2.9. Now one has
1. The generalized Hooke’s law (2.1.20) with
σ1=σ2=σ4=σ5=σ6=0,σ3=F/A=σ/ne}ationslash=0,
gives
ε1=S13σ,ε2=S23σ,ε3=S33σ,
ε4=S34σ,ε5=S35σ,ε6=S36σ
The displacements can be determined by the introduction of t he following equa-
tions
S13σ=u1,1, S23σ=u2,2, S33σ=u3,3,
S34σ=u3,2+u2,3,S35σ=u3,1+u1,3,S36σ=u2,1+u1,2
The equations

2.2 Fundamental Equations and Variational Solution Proced ures 77
u1=σ(S13x1+0.5S36x2),
u2=σ(0.5S36x1+S23x2),
u3=σ(S35x1+S34x2+S33x3)
satisfy the displacement differential equations and the bo undary conditions
which are prescribed at the point x1=x2=x3=0.
Note 2.4. The stress-strain formulae show that an anisotropic tensio n 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 coord inate 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 s train state only.
2. The stress in a bar under its own weight is
σ1=σ2=σ4=σ5=σ6=0,σ3=ρg(l−x3)
The stress-strain displacement equations are
ε1=u1,1=S13ρg(l−x3),ε4=u3,2+u2,3=S34ρg(l−x3),
ε2=u2,2=S23ρg(l−x3),ε5=u3,1+u1,3=S35ρg(l−x3),
ε3=u3,3=S33ρg(l−x3),ε6=u2,1+u1,2=S36ρg(l−x3)
The boundary conditions are identical to case a) and the foll owing displacement
state which satisﬁes all displacement differential equati ons and the conditions at
the point x1=x2=x3=0 can be calculated by integration
u1=ρg[−0.5S35x2
3+S13x1(l−x3)+0.5S36x2(l−x3)],
u2=ρg[−0.5S34x2
3+S23x2(l−x3)+0.5S36x1(l−x3)],
u3=ρg[−0.5S13x2
1+0.5S23x2
2
+0.5S36x1x2+(S34x2+S35x1)l+0.5S33x3(2l−x3)]
Note 2.5. The cross-section does not remain plane, it is deformed to th e shape
of a second-order surface and the bar axis becomes curved. Th e centroid of the
cross-section x3=lis displaced in all three directions
u1(0,0,l) =−0.5ρgS35l2,
u2(0,0,l) =−0.5ρgS34l2,
u3(0,0,l) =0.5ρgS33l2
Exercise 2.10. Show that for a composite beam subjected to a distributed con tinu-
ous load q(x1)the differential equation and the boundary conditions can b e 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 b eam is E1. With the
stress-strain-displacement relations from Bernoulli15beam theory
15Jakob I. Bernoulli (∗27 December 1654jul./6 January 1655greg.Basel – †16 August 1705 Basel)
– mathematician, beam theory

78 2 Linear Anisotropic Materials
σ1=E1ε1,ε1=−x3d2u3
dx2
1
the strain energy function Wfis seen to be
Wf=1
2σ1ε1=1
2E1ε2
1=1
2E1x2
3/parenleftbiggd2u3
dx2
1/parenrightbigg2
and
Πi=l/integraldisplay
0b/2/integraldisplay
−b/2h/2/integraldisplay
−h/2E1
2/parenleftbiggd2u3
dx2
1/parenrightbigg2
x2
3dx3dx2dx1=E1I
2l/integraldisplay
0/parenleftbiggd2u3
dx2
1/parenrightbigg2
dx1
with inertial moment I=bh3/12 for a rectangular cross-section.
In the absence of body forces, the potential function Πaof the external load q(x1)
is
Πa=−l/integraldisplay
0q(x1)u3(x1)dx1
and the total elastic potential energy is
Π(u3)=E1I
2l/integraldisplay
0/parenleftbiggd2u3
dx2
1/parenrightbigg2
dx1−l/integraldisplay
0q(x1)u3(x1)dx1
From the stationary condition δΠ(u3)=0 it follows that
δΠ(u3)=E1I
2l/integraldisplay
0δ/parenleftbiggd2u3
dx2
1/parenrightbigg2
dx1−l/integraldisplay
0q(x1)δu3(x1)dx1=0
There is no variation of E1Iorq(x1), because they are speciﬁed. From the last equa-
tion one gets
δΠ(u3)=E1I
2l/integraldisplay
02d2u3
dx2
1δ/parenleftbiggd2u3
dx2
1/parenrightbigg2
dx1−l/integraldisplay
0q(x1)δu3(x1)dx1=0
The ﬁrst term can be integrated by parts. The ﬁrst integratio n can be performed with
the following substitution
u=d2u3
dx2
1,u′=d3u3
dx3
1,v′=δ/parenleftbiggd2u3
dx2
1/parenrightbigg
,v=δ/parenleftbiggdu3
dx1/parenrightbigg
and yields

2.2 Fundamental Equations and Variational Solution Proced ures 79
δΠ(u3)=E1I

/bracketleftbiggd2u3
dx2
1δ/parenleftbiggdu3
dx1/parenrightbigg/bracketrightbiggl
0−l/integraldisplay
0d3u3
dx3
1δ/parenleftbiggdu3
dx1/parenrightbigg
dx1

−l/integraldisplay
0qδu3dx1=0
The second integration can be performed with the following s ubstitution
u=d3u3
dx3
1,u′=d4u3
dx4
1,v′=δ/parenleftbiggdu3
dx1/parenrightbigg
,v=δu3
and yields
δΠ(u3) =/bracketleftbigg
E1Id2u3
dx2
1δ/parenleftbiggdu3
dx1/parenrightbigg/bracketrightbiggl
0−/bracketleftbigg
E1Id3u3
dx3
1δu3/bracketrightbiggl
0
+l/integraldisplay
0E1Id4u3
dx4
1dx1−l/integraldisplay
0q(x1)δu3dx1=0
Finally we obtain
δΠ(u3) =/bracketleftbigg
E1Id2u3
dx2
1δ/parenleftbiggdu3
dx1/parenrightbigg/bracketrightbiggl
0−/bracketleftbigg
E1Id3u3
dx3
1δu3/bracketrightbiggl
0
+l/integraldisplay
0/bracketleftbigg
E1Id4u3
dx4
1−q(x1)/bracketrightbigg
δu3dx1=0
Since the variations are arbitrary the equation is satisﬁed if
E1I(u3)′′′′=q
and either
E1I(u3)′′=0
oru′
3is speciﬁed and
E1I(u3)′′′=0
oru3is speciﬁed at x1=0 and x1=l.
The beam differential equation is the Euler-Lagrange equat ion 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 th e boundary conditions
include the classical conditions of simply supports, clamp ed and free edges.
Exercise 2.11. The beam of Exercise 2.10 may be moderately thick and the ef-
fects of transverse shear deformation ε5and 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 a dmis-
sible functions for the displacements u1(x1),u3(x1)and for the stresses σ1(x1,x3),
σ3(x1,x3),σ5(x1,x3). For the beam with rectangular cross-section and lateral lo ad-
ingq(x1)follow in the frame of beam theory
u2=0,σ2=σ4=σ6=0
As in the classical beam theory (Bernoulli theory) we assume that the beam cross-
sections undergo a translation and a rotation, the cross-se ctions are assumed to re-
main plane but not normal to the deformed middle surface (Tim oshenko16theory).
Therefore we can assume in the simplest case
u1=x3ψ(x1),u3=w(x1)
and the strain-displacement relations may be written
ε1=∂u1
∂x1=x3ψ′(x1),ε3=∂u3
∂x3=0,ε5=∂u3
∂x1+∂u1
∂x3=w′(x1)+ψ(x1)
For the stresses σ1,σ3andσ5the following functions are assumed
σ1=M
Ix3,I=bh3
12,
σ3=3q
4b/bracketleftigg
x3
h/2+2
3−1
3/parenleftbiggx3
h/2/parenrightbigg2/bracketrightigg
,
σ5=3Q
2A/bracketleftigg
1−/parenleftbiggx3
h/2/parenrightbigg2/bracketrightigg
,A=bh
The assumed functions for σ1andσ5are identical to those of the Bernoulli beam
theory and σ3may be derived from the stress equation of equilibrium in the thick-
ness direction, Eq. (2.2.1), with
σ3(+h/2)=q,σ1(−h/2)=0
The bending moment Mand the shear force resultant Qwill be deﬁned in the usual
manner
M(x1)=+h/2/integraldisplay
−h/2bσ1dx3,Q(x1)=+h/2/integraldisplay
−h/2bσ5dx3
Now Reissner’s functional ΨR(uuu,σσσ), e.g. (2.2.18), takes with the assumption above
the form
16Stepan Prokopovich Timoshenko (∗22 December, 1878 Schpotiwka – † 29 May, 1972
Wuppertal-Elberfeld) – engineer, founder of the modern app lied mechanics

2.2 Fundamental Equations and Variational Solution Proced ures 81
ΨR(uuu,σσσ) =l/integraldisplay
0+h/2/integraldisplay
−h/2/braceleftbigg
σ1ε1+σ5ε5−1
2E/bracketleftbig
σ2
1+σ2
3+2(1+ν)σ2
5/bracketrightbig/bracerightbigg
bdx3dx1
−l/integraldisplay
0qwdx1
=l/integraldisplay
0+h/2/integraldisplay
−h/2/braceleftbigg
σ1×3ψ′(x1)+σ5/bracketleftbig
w′(x1)+ψ(x1)/bracketrightbig
−1
2E/bracketleftbig
σ2
1+σ2
3+2(1+ν)σ2
5/bracketrightbig/bracerightbigg
bdx1−l/integraldisplay
0qwdx1
Substituting MandQand neglecting the term σ3which only depends on qand not
on the basic unknown functions ψ,wrespectively, MandQ, and yields no contribu-
tion to the variation δΨwe obtain
Ψ(w,ψ,M,Q)=l/integraldisplay
0/bracketleftbigg
Mψ′+Q(w′+ψ)−M2
2EI+6νqM
5EA−3Q2
5GA−qw/bracketrightbigg
dx1
δΨR=l/integraldisplay
0/bracketleftbig
Mδψ′+ψ′δM+Q(δw′+δψ)+(w′+ψ)δQ
−M
EIδM+6νq
5EAδM−6Q
5GAδQ−qδw/bracketrightbigg
dx1=0
Integration the terms Mδψ′andQδw′by parts and rearranging the equation
δΨR= [Mδψ+Qδw]l
0+l/integraldisplay
0/braceleftig
[Q+M′]δψ−[Q′−q]δw
+/bracketleftbigg
ψ′−M
EI+6νq
5EA/bracketrightbigg
δM+/bracketleftbigg
ψ+w′−6Q
5GA/bracketrightbigg
δQ/bracerightbigg
dx1=0
The ﬁrst term yields the natural boundary conditions of the v ariational statement:
1. Either M=0 orψmust be prescribed at x1=0,l.
2. Either Q=0 orwmust be prescribed at x1=0,l.
The variations δψ,δw,δMandδQare all arbitrary independent functions of x1and
therefore δΨR=0 only if
−dM(x1)
dx1+Q(x1)=0,dQ(x1)
dx1+q(x1)=0,
dψ(x1)
dx1−M(x1)
EI+6νq(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 w ith 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 mi ddle 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 ψto−w′as assumed in the Bernoulli’s theory. The third term in the
third equation depends on the lateral load qand Poisson’s ratio νand tends to zero
forν→0. This term described the effect of the transverse normal st ressσ3. which
will be vanish if ν=0 as in the classical beam theory.
Substituting the differential equations for ψandwinto the differential relations
forMandQleads to
EI/bracketleftbigg
ψ′′(x1)+ν6
5Q′(x1)
EA/bracketrightbigg
−Q(x1)=0,5
6GA[ψ′(x1)+w′′(x1)]+q(x1)=0
or
EI/bracketleftbigg
ψ′′(x1)−ν6
5q(x1)
EA/bracketrightbigg
−5
6GA[ψ+w′]=0,5
6GA[ψ′(x1)+w′′(x1)]+q(x1)=0
Derivation and rearrangement yield a differential equatio n of 3rd order for ψ(x1)
EIψ′′′(x1)=−q(x1)+ν6
5q′(x1)
EA
With
Q(x1)=dM(x1)
dx1=EI/bracketleftbigg
ψ′′(x1)+ν6
5Q′(x1)
EA/bracketrightbigg
and
Q(x1)=5
6EI
GA[ψ(x1)+w′(x1)]
follows an equation for w′(x1)
w′(x1)=−ψ(x1)+6
5EI
GA/bracketleftbigg
ψ′′(x1)+ν6
5q(x1)
EA/bracketrightbigg
Neglecting with ν=0 the effect of the transverse normal stress σ3we get the Tim-
oshenko’s beam equation
EIψ′′′(x1)=−q(x1),M(x1)=EIψ′(x1),Q(x1)=EIψ′′(x1)
w′(x1)=−ψ(x1)+EIψ′′(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)=−EIw′′(x1),Q(x1)=−EIw′′′(x1)

2.2 Fundamental Equations and Variational Solution Proced ures 83
The derivation of the Timoshenko’s beam equation for lamina ted beams one can
ﬁnd in more detail in Sect. 7.3.
Exercise 2.12. Derive the free vibration equations for the moderately thic k beam,
Exercise 2.11, using Hamilton’s principle in conjunction w ith the Reissner varia-
tional theorem.
Solution 2.12. In order to derive the free vibration equations we apply the g ener-
alized Hamilton’s principle, i.e. the Hamilton’s principl e in conjunction with the
Reissner variational statement
δ χ(uuu,σσσ)=δt2/integraldisplay
t1[T(uuu)−ΨR(uuu,σσσ)]dt=0
T(uuu)is the kinetic energy and ΨR(uuu,σσσ)the Reissner functional of the moderately
thick beam. ΨR(uuu,σσσ)is known from Exercise 2.11 and the kinetic energy for the
beam may be written as
T(uuu) =l/integraldisplay
0h/2/integraldisplay
−h/21
2ρ/bracketleftigg/parenleftbigg∂u
∂t/parenrightbigg2
+/parenleftbigg∂w
∂t/parenrightbigg2/bracketrightigg
bdx3dx1
=l/integraldisplay
01
2ρ/bracketleftigg
I/parenleftbigg∂ψ
∂t/parenrightbigg2
+A/parenleftbigg∂w
∂t/parenrightbigg2/bracketrightigg
dx1
with
I=h/2/integraldisplay
−h/2bx2
3dx3,A=h/2/integraldisplay
−h/2bdx3
and the mass density ρof the beam material. The substitution of T[uuu]above and
Ψ(w,ψ,M,Q)of Exercise 2.11 in the functional χyields
δ χ(uuu,σσσ) =δt2/integraldisplay
t1l/integraldisplay
0/braceleftigg
1
2ρ/bracketleftigg
I/parenleftbigg∂ψ
∂t/parenrightbigg2
+A/parenleftbigg∂w
∂t/parenrightbigg2/bracketrightigg
−/bracketleftbigg
Mψ′+Q(w′+ψ)−M2
2EI+6νqM
5EA−3Q2
5GA−qw/bracketrightbigg/bracerightigg
dtdx1
=t2/integraldisplay
t1l/integraldisplay
0/bracketleftbigg/parenleftbigg
ρI∂2ψ
∂t2δψ+A∂2w
∂t2/parenrightbigg
δw
−Mδψ′−ψ′δM−Q(δw′+δψ)−(w′+ψ)δQ
+M
EIδM−6νq
5EAδM−6Q
5GAδQ−qδw/bracketrightbigg
dtdx1

84 2 Linear Anisotropic Materials
Integration the terms Mδψ′+Qδw′by parts
l/integraldisplay
0Mδψ′dx1=Mδψ/vextendsingle/vextendsingle/vextendsinglel
0−l/integraldisplay
0M′δψdx1,l/integraldisplay
0Qδw′dx1=Qδw/vextendsingle/vextendsingle/vextendsinglel
0−l/integraldisplay
0Q′δwdx1,
rearranging the equation δ χand setting δ χ=0 yield
t2/integraldisplay
t1(Mδψ+Qδw)l
0dt+t2/integraldisplay
t1l/integraldisplay
0/bracketleftbigg/parenleftbigg
ρI∂2ψ
∂t2+Q−M′/parenrightbigg
δψ
−/bracketleftbigg
Q′+qA/parenleftbigg∂2w
∂t2/parenrightbigg/bracketrightbigg
δw+/parenleftbigg
ψ′−M
EI+6νq
5EA/parenrightbigg
δM
+/parenleftbigg
ψ+w′−6Q
5GA/parenrightbigg
δA/bracketrightbigg
dtdx1=0
and the equations for free vibrations follow with q≡0
Q−∂M
∂x1+ρI∂2ψ
∂t2=0,∂Q
∂x1−ρA∂2w
∂t2=0,
∂ψ
∂x1−M
EI=0, ψ+∂w
∂x1−6Q
5GA=0
The underlined terms represent the contribution of rotator y inertia and the effect of
transverse shear deformation.
The system of four equations can be reduced to a system of two e quations for the
unknowns wandψ. Substitution of
Q=EI∂2ψ
∂x2
1−ρI∂2ψ
∂t2,I=bh3
12,A=bh
in the second and fourth equation leads
∂3ψ
∂x3
1−ρ
E∂3ψ
∂t2∂x1−ρA
E∂2w
∂t2=0,ψ+∂w
∂x1−h2
10/bracketleftbiggE
G∂2ψ
∂x2
1−ρ
G∂2ψ
∂t2/bracketrightbigg
=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. Jo urnal of Mathematics
and Physics 29(1-4):90–95

Chapter 3
Effective Material Moduli for Composites
Composite materials have at least two different material co mponents which are
bonded. The material response of a composite is determined b y the material moduli
of all constituents, the volume or mass fractions of the sing le 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 ﬁbr e reinforcement, i.e. the
ﬁbre architecture.
The basic assumptions made in material science approach mod els of ﬁbre rein-
forced composites are:
•The bond between ﬁbres and matrix is perfect.
•The ﬁbres are continuous and parallel aligned in each ply, th ey are packed regu-
larly, i.e. the space between ﬁbres is uniform.
•Fibre and matrix materials are linear elastic, they follow a pproximately 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 h eterogeneity of the material
is neglected and approximately overlayed to a homogeneous m aterial. The most
important composites in structural engineering applicati ons are laminates and sand-
wiches. Each single layer of laminates or sandwich faces is i n general a ﬁbre rein-
forced lamina. For laminates we have therefore two differen t scales of modelling:
•The modelling of the mechanical behavior of a lamina, is call ed the micro-
mechanical or microscopic approach of a composite. The micr o-mechanical
modelling leads to a correlation between constituent prope rties and average ef-
fective composite properties. Most simple mixture rules ar e used in engineering
applications. Whenever possible, the average properties o f a lamina should be
veriﬁed experimentally by the tests described in Sect. 3.1 o r Fig. 3.1.
•The modelling of the global behavior of a laminate constitut ed of several quasi-
homogeneous laminae is called the macroscopic approach of a composite.
Fibre reinforced material is in practice neither monolithi c nor homogeneous, but it
is impossible to incorporate the real material structure in to design and analysis of
85 © Springer Nature Singapore Pte Ltd. 2018
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
✻
❄σL
σLtest
εL,εT
EL=σL
εL,νLT=εT
εL❄✻σT
test
εT
σT
ET=σT
εT✲
✻
❄
✛τ
τtest
γ
GLT=τ
γ
Fig. 3.1 Experimental testing of the mechanical properties of an UD- layer: EL=E1,ET=E2,
GLT=G12,νLT=ν12
composite or any other structural component. Therefore the concept of replacing the
heterogeneous material behavior with an effective materia l which is both homoge-
neous and monolithic, thus characterized by the generalize d Hooke’s law, will be
used in engineering applications. We assume that the local v ariations in stress and
strain state are very small in comparison to macroscopical m easurements of material
behavior.
In the following section some simple approaches to the lamin a properties are
given with help of the mixture rules and simple semi-empiric al consideration. The
more theoretical modelling in Sect. 3.2 has been developed t o establish bounds on
effective properties. The modelling of the average mechani cal 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 ru le of mixtures and the
inverse rule of mixtures is based on the statement that the co mposite property is
the weighted mean of the properties or the inverse propertie s of each constituent
multiplied by its volume fraction. In the ﬁrst case we have th e upper-bound effective
property, in the second – the lower-bound. In composite mech anics these bounds are
related to W. V oigt and A. Reuss. In crystal plasticity the bo unds 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 Cambridg e)
– 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 .) – metall urgist, contribu-
tions to the theory of plasticity

3.1 Elementary Mixture Rules for Fibre-Reinforced Laminae 87
E Young’s modulus
ν Poisson’s ratio
G Shear modulus
σ Stress
ε Strain
V,MV olume, mass
v,mV olume fraction, mass fraction
A Cross-section area
ρ Density
The subscripts f and m refer to ﬁbre and matrix, the subscript s 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 c omposites in terms of
volume fractions is given in Sect. 1.1
ρ=M
V=Mf+Mm
V=ρfVf+ρmVm
V
=ρfvf+ρmvm=ρfvf+ρm(1−vf)(3.1.1)
In literature we also ﬁnd vf≡φfor the ﬁbre volume fraction and we have
ρ=ρfφ+ρm(1−φ) (3.1.2)
In an actual lamina the ﬁbres are randomly distributed over t he 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 typi cal 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 assu med 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 vfmaxabout 0.50 – 0.65. Keep in mind that a lower
ﬁbre volume fraction results in lower laminae strength and s tiffness under tension
in ﬁbre-direction, but a very high ﬁbre volume fraction clos e to the ultimate values
ofvfmay lead to a reduction of the lamina strength under compress ion 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 tensi le or compression load
parallel to the ﬁbres, it can be assumed that the strains of th e ﬁbres, matrix and
composite in the loading direction are the same (Fig. 3.2)
εLf=εLm=εL=∆l
l(3.1.3)
The mechanical model has a parallel arrangement of ﬁbres and matrix (V oigt 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 orFL=σLA=σLfAf+σLmAm (3.1.4)
With Hooke’s law it follows that
σL=ELεL,σLf=ELfεLf,σLm=ELmεLm
or
ELεLA=EfεLfAf+EmεLmAm (3.1.5)
Since the strains of all phases are assumed to be identical (i so-strain condition),
(3.1.5) reduces to
EL=EfAf
A+EmAm
A(3.1.6)
withAf
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)=Efφ+Em(1−φ) (3.1.8)
Equation (3.1.8) is referred to the V oigt 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 FL
✲ ✛l l+∆l✲ ✛
Fig. 3.2 Mechanical model to calculate the effective Young’s modulu sEL

3.1 Elementary Mixture Rules for Fibre-Reinforced Laminae 89
FLf
FL=ELf
ELvf (3.1.9)
Since the ﬁbre stiffness is several times greater than the ma trix 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 serie s of ﬁbre and matrix
(Reuss model, Sect. 2.1.1). The stress resultant FTrespectively the stress σTis equal
for all phases (iso-stress condition)
FT=FTf=FTm,σT=σTf=σTm (3.1.11)
From Fig. 3.3 it follows that
∆b=∆bf+∆bm,εT=∆b
b=∆bf+∆bm
b(3.1.12)
and with
b=vfb+(1−vf)b=bf+bm (3.1.13)
and
εT=∆bf
vfbvfb
b+∆bm
(1−vf)b(1−vf)b
b=vfεTf+(1−vf)εTm (3.1.14)
with
εTf=∆bf
vfb,εTm=∆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 mod ulus ET

90 3 Effective Material Moduli for Composites
σT=ETεT,σTf=ETfεTf,σTm=ETmεTm (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
EmorET=EfEm
(1−vf)Ef+vfEm(3.1.16)
Equation (3.1.16) is referred to Reuss estimate or sometime s 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 (paral lel to the ﬁbres) as
shown in Fig. 3.4. The major Poison’s ratio is deﬁned as the ne gative of the ratio of
the normal strain in the transverse direction to the normal s train in the longitudinal
direction
νLT=−εT
εL(3.1.17)
With
−εT=νLTεL=−∆b
b=−∆bf+∆bm
b=−[vfεTf+(1−vf)εT m],
νf=−εTf
εLf,νm=−εTm
εLm
it follows that
εT=−νLTεL=−vfνfεLf−(1−vf)νmεLm (3.1.18)
The longitudinal strains in the composite, the ﬁbres and the matrix are assumed to be
equal (V oigt model of parallel connection) and the equation for the major Poisson’s
ratio reduces to
✛ ✲✻
❄
✲ ✛b
l✲ ✛ l+∆l❄
✻b−∆b
FL FL
Fig. 3.4 Mechanical model to calculate the major Poisson’s ration νLT

3.1 Elementary Mixture Rules for Fibre-Reinforced Laminae 91
νLT=vfνf+(1−vf)νm=vfνf+vmνm=φνf+(1−φ)νm (3.1.19)
The major Poisson’s ratio νLTobeys the rule of mixture. The minor Poisson’s ratio
νTL=−εL/εTcan be derived with the symmetry condition or reciprocal rel ationship
νTL
ET=νLT
EL,
νTL=νLTET
EL=(vfνf+vmνm)EfEm
(vfEm+vmEf)(vfEf+vmEm)(3.1.20)
The values of Poisson’s ratios for ﬁbres or matrix material r arely differ signiﬁcantly,
so that neither matrix nor ﬁbre characteristics dominate th e major or the minor elas-
tic parameters νLTandνTL.
3.1.5 Effective In-Plane Shear Modulus
Apply a pure shear stress τto 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
γm=τ
Gm,γf=τ
Gf,γ=τ
GLT(3.1.21)
The model is a connection in series (Reuss model) and therefo re
τ=τf=τm,∆=∆f+∆m,∆=ˆbtanγ=γfˆbf+γmˆbm (3.1.22)
and with
ˆb=ˆbf+ˆbm=(vf+vm)ˆb=vfˆb+(1−vf)ˆb (3.1.23)
it follows that
∆f=γfvfˆb,∆m=γm(1−vf)ˆb (3.1.24)

✲ ✛∆
✲✛∆f
✲✛∆m/2✻
❄ˆb ﬁbrematrix
matrix✲✛
γ∆m/2
✠
✛τ
τ ✒✲
ττ
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+φGm(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 isotropi c ﬁbres and matrix
material one gets
Gf=Ef
2(1+νf),Gm=Em
2(1+νm)(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 bet ween ﬁbres and matrix.
There are only two different types of material response: the V oigt or iso-strain model
in which applied strain is the same in both material phases (p arallel response) and
the Reuss or iso-stress model in which the applied stress is t he same in both material
phases (series response).
For an aligned ﬁbre composite the effective material behavi or may be assumed
as transversally isotropic and ﬁve independent effective e ngineering 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+νTT)],
E5=G13=E6=G12=GLT,ν12=ν13=νLT,ν23=νTT,
νLTET=νTLEL
If we make choice of EL,ET,GLT,GTT,νLTas 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,
νLT=vfνf+(1−vf)νm(3.1.27)
The shear modulus GTTcorresponds 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
forGLTandGTTestimation are close to the real situation of shearing loade d ﬁbre
reinforced composites. Therefore the equations for GLTandGTTcannot be expected
as very reliable.
If one considers the approximative predictions for the effe ctive 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 signiﬁc ant 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 isotro pic but the matrix ma-
terial is isotropic. For such cases simple alternative rela tions for the effective engi-
neering moduli of the UD-lamina can be given
EL= vfEf+(1−vf)Em, νLT=vfνLTf+(1−vf)νm,
ET=ETfEm
vfEm+(1−vf)ETf,νT T=νTTfνm
vfνm+(1−vf)νTTf,
GLT=GLTfGm
vfGm+(1−vf)GLTf,GT T=vfGf+(1−vf)Gm(3.1.29)
In Eq. (3.1.29) Em,νm,Gm=Em/2(1+νm)are the isotropic matrix moduli and
Ef,ETf,GLTf,Gf=ETf/2(1+νTTf),νTTf,νLTfare transversally isotropic ﬁbre mod-
uli.Em,νmof the matrix material and Ef,ETf,GLTf,νLTf,νTTforGTTfof 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 dire ction, such as ELandνLT,
are dominated by the ﬁbres. All estimations in this case and e xperimental 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 m ixtures which can
be reduced to the two model connections of V oigt and Reuss, do not agree well with
experimental results. This establishes a need for better mo delling techniques based
on elasticity solutions and variational principle models a nd includes analytical and
numerical solution methods.
Unfortunately, the theoretical models are only available i n the form of compli-
cated mathematical equations and the solution is very limit ed and needs huge effort.
Semi-empirical relationships have been developed to overc ome the difﬁculties with
purely theoretical approaches.

94 3 Effective Material Moduli for Composites
The most useful of those semi-empirical models are those of H alpin and Tsai3
which can be applied over a wide range of elastic properties a nd ﬁbre volume frac-
tions. The Halpin-Tsai relationships have a consistent for m 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, Ha lpin and Tsai proposed
equations that are general and simple in formulation. The mo duli of a unidirectional
composite are given by the following equations
•ELandνLTby the law of mixtures Eqs. (3.1.8), (3.1.19)
•For the other moduli by
M
Mm=1+ξ ηvf
1−ηvf(3.2.1)
Mis the modulus under consideration, e.g. ET,GLT, . . . , ηis a coefﬁcient given
by
η=(Mf/Mm)−1
(Mf/Mm)+ξ(3.2.2)
ξis 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 t he factor ξby comparing
the semi-empirical values with analytical solutions or wit h experimental results.
In addition to the rule of mixtures and the semi-empirical so lution of Halpin and
Tsai there are some solutions available which are based on el asticity models, e.g. for
the model of a cylindrical elementary cell subjected to tens ion. The more compli-
cated formulas for ELandνLTas the formulas given above by the rule of mixtures
yields practically identical values to the simpler formula s and are not useful. But
the elasticity solution for the modulus GLTyields much better results and should be
applied
GLT=GmGf(1+vf)+Gm(1−vf)
Gf(1−vf)+Gm(1+vf)=GmGf(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 r ecommendations
may be possible for an estimation of effective elastic modul i of unidirectional lami-
nae
•ELandνLTshould be estimated by the rule of mixtures
•νT Lfollows from the reciprocal condition
•GLTshould be estimated from (3.2.3) or the Halpin/Tsai formula s (3.2.1) and
(3.2.2)
3Stephen Wei-Lun Tsai (∗6 July 1929, Beijing) – US-American engineer, strength crit eria for
composites, founder of the Journal of Composite Materials

3.3 Problems 95
•ETmay be estimated with help of the Halpin/Tsai formulas. But o nly when reli-
able experimental values of ETandGLTare available for a composite the ξ-factor
can be derived for this case and can be used to predict effecti ve moduli for a range
of ﬁbre volume ratios of the same composite. It is also possib le to look for nu-
merical or analytical solutions for ξbased on elasticity theory. In general, ξmay
vary from zero to inﬁnity, and the Reuss and V oigt models are s pecial cases, e.g.
ET=1+ξ ηvf
1−ηvfEm
forξ=0 and ξ=∞, respectively. In the case of circular cross-sections of th e
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 fra ction
1. the density and the mass fractions of the ﬁbre and matrix,
2. the Young’s moduli E′
1≡ELandE′
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 νLT,νTL,
4. the in-plane shear modulus
The properties of glass and epoxy are taken approximately fr om Tables F.1 and
F.2 as ρglass≡ρf=2,5 gcm−3,νf=0,7,νm=0,3,ρepoxy≡ρm=1,35 gcm−3,
Ef=70 GPa, Em=3,6 GPa.
Solution 3.1. Taking into account the material parameters and the volume f raction
of the ﬁbres one obtains:
1. Using Eq. (1.1.3) the density of the composite is
ρ=ρfvf+ρmvm=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=ρf
ρvf=25
2,1550,7=0,8121, mm=ρm
ρvm=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
ELvf=70
50,080,7=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
νLT=0.2·0.7+0.3·0.3=0.230, νTL=0.23010.71
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 GPa29.17 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 GLTis 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 ELandET
byEL/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=/parenleftbiggvf
Ef1+1−vf
Em/parenrightbigg−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=/parenleftbiggvf
Ef2+1−vf
Em/parenrightbigg−1
=2,3810 Em,
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 areEf=90
GPa, νf=0,2,Gf=35 GPa, Em=3,5 GPa, νm=0,3,Gm=1,3 GPa. The volume
fraction vf=φ=60%. Calculate EL,ET,GLT,νLT,νT Lwith the help of the rule of
mixtures and also GLTwith the improved formula.
Solution 3.3.
EL=Efvf+Em(1−vf)=55,4 GPa, ET=/parenleftbiggvf
Ef+vm
Em/parenrightbigg−1
=8,27 GPa,
νLT=νfvf+νm(1−vf)=0,24, νTL=νLTET
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 GLTvalues 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 vo lume
of glass ﬁbres in the epoxy resin matrix. The material proper ties 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 fract ion one
obtains
1.EL=Efvf+Em(1−vf)=60,52 GPa,
2.FL=σLA=ELεA,Ff=σfA=EfεAf,Fm=σmA=EmεAm
With FL=Ff+Fmit follows that
ELεA=EfεAf+EmεAm⇒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 m a-
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, νf=0,15,νm=0,37. The ﬁbre is
transversally isootropic and transverse Young’s modulus i sETf=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, νLT=0,34.

98 3 Effective Material Moduli for Composites
1. compare the experimental values with predicted values ba sed on rule of mixtures.
2. Using the Halpin-Tsai approximate model for calculating ETandGLT, what value
ofξmust be used in order to obtain moduli that agree with experim ental values?
Solution 3.5. The comparison can be be performed by results from the mixtur e 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,
νLT=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,GLTandνLTare not exactly predicted.
2. The Halpin-Tsai approximation yields with ETf=22 GPa, ξ=2 for ET
ET=1+ξ ηvf
1−ηvfEm,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 ηvf
1−ηvfGm,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 GLTwould
underestimate the predicted value signiﬁcantly
GLT=1+ηvf
1−ηvfGm,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 uncr itically without
experimental measurements.
Exercise 3.6. Let us assume the following material parameters for the shea r 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 νf=0,15,νm=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 Halpi n-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
mm22,5·104Nmm−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 Ha lpin-Tsai at ﬁrst
should be estimated ηwith 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·0,151,2·103Nmm−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
a b cr a
rhr
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(r2π)/4
4r2=π
4=0.785
2. Hexagonal array (Fig. 3.6 b) with a→2r,h→√
3rfollow
Ac=hr,Af=6(1/3)πr2+πr2
and
vu
f=Af
Ac=6(1/3)πr2+πr2)/4
6√
3r2=π
2√
3=0.9069
3. Layer-wise array (Fig. 3.6 c)
vu
f=Af
Ac=r2π
(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 f rom single lami-
nae to laminates including sandwiches, the improved theori es and simplest failure
concepts.
The single layer (lamina) is modelled with the help of the fol lowing assumptions
•linear-elastic isotropic behaviour of the matrix and the ﬁb re materials,
•the ﬁbres are unidirectional oriented and uniformly distri buted
These assumptions result in good stiffness approximations in the longitudinal and
transverse directions. The stiffness under shear is not wel l-approximated.
After the transfer from the local to the global coordinates t he stiffness parame-
ters for the laminate can be estimated. For the classical cas es the effective stiffness
parameters of the laminate is a simple sum up over the laminat e thickness of the
weighted laminae reduced stiffness parameters transferre d into the global coordi-
nate system. Some improved theories are brieﬂy introduced.
The failure concepts are at the moment a research topic chara cterised 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 ma terial. A lamina
or ply is a typical sheet of composite materials, which is gen erally of a thickness
of the order 1 mm. A laminate is constructed by stacking a numb er of laminae in
the direction of the lamina thickness. The layers are usuall y bonded together with
the same matrix material as in the single lamina. A laminate b onded of n(n≥2)
laminae of nearly the same thickness. A sandwich can be deﬁne d 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 r espectively, and of high
strength material for the outer layers, the cover or face she ets. The thickness of the
core is generally much greater than the thickness of the shee ts 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 sandwi ches. However, the lam-
inate elements are single laminae and so understanding the m echanical 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 res ultants are formulated and
basic formulae for stress analysis are derived. These consi derations are expanded to
laminates and sandwiches in Sects. 4.2 and 4.3. The governin g 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 struc tures requires knowl-
edge of the strength and the reliability of composite materi als. Strength failure the-
ories have to be developed in order to compare the actual stre ss states in a material
to a failure criteria. Chapter 6 gives an overview on fractur e modes of laminae. For
laminates, the strength is related to the strength of each in dividual lamina. Various
failure theories are discussed for laminates or sandwiches based on the normal and
shear strengths of unidirectional laminae.
103 © Springer Nature Singapore Pte Ltd. 2018
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 b ased on average ma-
terial properties and by considering the lamina to be homoge neous. The methods
to ﬁnd these average properties based on the individual mech anical values of the
constituents are discussed in Sect. 3.1. Otherwise the mech anical characterization
of laminae can be determined experimentally but it demands s pecial experimental
equipment and is costly and time-consuming. Generally the m odelling goal is to
ﬁnd the minimum of parameters required for the mechanical ch aracterization 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 m ost 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 structu ral level response in a
tractable manner.
•Three of the six components of stress state are generally muc h smaller than the
other three, i.e. the plane stress assumption, which is base d on the manner in
which ﬁbre-reinforced materials are used in such structura l elements as beams,
plates or shells, will be sufﬁcient accurately. With the ass umption that the ( x1−
x2)-plane of the principal co-ordinate system is in-plane str ess state, the in-plane
stress components σ1,σ2,σ6are considered to be much larger in value than the
out-of-plane stress components σ3,σ4,σ5and the last ones are set approximately
to zero.
Using the plane stress assumption it has to be in mind that som e serious inaccuracies
in the mechanical response of laminates can be occurred, Sec t. 4.2.
Therefore, together with the plane stress assumption two ma jor misconceptions
should be avoided:
•The stress components σ3,σ4,σ5equated to zero have to be estimated to their
magnitude and effect. Fibre reinforced material is often ve ry 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 composit e material.
•With assuming σ3is zero does not follow that the associated strain ε3is 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 eng ineering applications:

4.1 Elastic Behavior of Laminae 105
1. Short ﬁbres or particle reinforced components with rando m orientation in the
matrix
The elastic behavior has no preferred direction and is macro scopically quasi-
homogenous and isotropic. The effective elastic moduli are Eandνand the
relations of the in-plane stress components with the in-pla ne strain components
are described (Tables 2.6 and 2.7) by

σ1
σ2
σ6
=
Q11Q120
Q12Q110
0 0 Q66

ε1
ε2
ε6
,

ε1
ε2
ε6
=
S11S120
S12S110
0 0 S66

σ1
σ2
σ6
(4.1.1)
with
Q11=E
1−ν2, S11=1
E,
Q12=Eν
1−ν2, 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 co mposites and is
the main topic of this textbook. The elastic behavior of UD-l aminae depends on
the loading reference coordinate systems. In the on-axis ca se the reference axes
(1′,2′) are identical to the material or principal axes of the lamin a 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 macr oscopically
quasi-homogeneous and orthotropic with four independent m aterial moduli (Ta-
ble 2.6)
E′
1≡EL,E′
2≡ET,E′
6≡G′
12≡GLT,ν′
12≡νLT (4.1.2)
and the in-plane stress-strain relations are
Fig. 4.1 Unidirectional lam-
ina with principal material
axisLandT(on-axis)✻
✲
1′≡L2′≡T
rrrrrrrrrrrrrrh✲✛

106 4 Elastic Behavior of Laminate and Sandwich Composites

σ′
1
σ′
2
σ′
6
=
Q′
11Q′
120
Q′
12Q′
220
0 0 Q′
66

ε′
1
ε′
2
ε′
6
,

ε′
1
ε′
2
ε′
6
=
S′
11S′
120
S′
12S′
220
0 0 S′
66

σ′
1
σ′
2
σ′
6
(4.1.3)
with
Q′
11=E′
1/(1−ν′
12ν′
21), S′
11=1/E′
1,
Q′
22=E′
2/(1−ν′
12ν′
21), S′
22=1/E′
2,
Q′
66=G′
12=E′
6, S′
66=1/G′
12=1/E′
6,
Q′
12=E′
2ν′
12/(1−ν′
12ν′
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 macroscopica lly quasi-homogeneous
and anisotropic. The in-plane stress-strain relations are formulated by fully pop-
ulated matrices with all Qi jandSi jdifferent from zero but the number of inde-
pendent material parameters is still four as in case 2. The tr ansformation rules
are given in detail in Sect. 4.1.2.
A UD-lamina has different stiffness in the direction of the m aterial axes. With
Ef≫Emthe stiffness in the L-direction is ﬁbre dominated and for th e 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 materia l 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≡νLT=ν′
13≡νLT′,
E′
4≡GTT′=E′
2/[2(1+ν′
23)]≡ET/[2(1+νTT′)](4.1.4)
Fig. 4.2 On-axis stress-strain
equations for UD-lamina
(qualitative)✻
✲σ1,σ2,σ6
ε1,ε2,ε6✄✄✄✄✄✄✄✄✄✄✄✄
✦✦✦✦✦✦✦✦✦✦✦✦✦✦✦✦
✥✥✥✥✥✥✥✥✥✥✥✥✥σ1(ε1)
σ2(ε2)
σ6(ε6)✛ ✲σ1 σ1
✻
❄σ2
σ2
✲
✻
✛❄σ6
σ6

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≡νLT,E′
4=E′
2/[2(1+ν′
23)]≡GTT=ET/[2(1+νTT)]. Summarizing, the
stress-strain relations for on-axis loading of UD-laminae in a contracted vector-
matrix notation leads the equations
σσσ′=QQQ′εεε′orσ′
i=Q′
i jε′
j,Q′
i j=Q′
ji
εεε′=SSS′σσσ′orε′
i=S′
i jσ′
j,S′
i j=S′
jii,j=1,2,6 (4.1.5)
The values Q′
i jof the reduced stiffness matrix QQQ′and the S′
i jof the compliance ma-
trixSSS′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). Thes e relations simplify the
problem from a three-dimensional to a two-dimensional or pl ane stress state. Also
the numerical values of the stiffness Q′
i jare 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 als o. 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 a nd shear strains and also
shear stresses applied in the (L-T)-plane do not result in an y normal strains in the
L and T direction. The UD-lamina is therefore also called a sp ecially orthotropic
lamina.
Composite materials are generally processed at high temper ature and then cooled
down to room temperature. For polymeric ﬁbre reinforced com posites 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 resul t 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 st rains 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 long itudinal direction and
transverse the ﬁbre direction of a lamina are not equal since the effective elastic
moduli ELandETand also the thermal and moisture expansion coefﬁcients αth
L,αth
T
andαmo
L,αmo
Trespectively, are different.
The stress-strain relations of a UD-lamina, including temp erature and moisture
differences are given by

ε′
1
ε′
2
ε′
6
=
S′
11S′
120
S′
12S′
220
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. αmo
L,αmo
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′
11Q′
120
Q′
12Q′
220
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 an y 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 tw o 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 h ygrothermal 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
αth
L=αth
fvfEf+αth
m(1−vf)Em
vfEf+(1−vf)Em,
αth
T=αth
fvf(1+νf)+αth
m(1−vf)(1+νm)−[vfνf+(1−vf)νm]αth
L(4.1.9)
If the ﬁbres are not isotropic but have different material re sponse in axial and trans-
verse directions, e.g. in the case of carbon or aramid ﬁbres, the relations for αth
Land
αth
Thave to be changed to
αth
L=αth
LfvfELf+αth
m(1−vf)Em
vfELf+(1−vf)Em,
αth
T= (αth
Tf+νTfαth
Lf)vf+(1+νm)αth
m(1−vf)
−[vfνTf+(1−vf)νm]αth
L(4.1.10)
In most cases the matrix material can be considered isotropi c and therefore the ori-
entation designation L ,T of the matrix material parameters can be dropped.
Discussion and conclusions concerning effective moduli pr esented in Chap. 3 are
valid for effective thermal expansion coefﬁcients too. The simple micro-mechanical
approximations of effective moduli yield proper results fo rαth
Lbut fails to predict
αth
Twith the required accuracy. For practical applications αth
Landαth
Tshould be
normally determined by experimental methods.
Micro-mechanical relations for effective coefﬁcients of m oisture expansion can
be modelled analogously. However, some simpliﬁcation can b e taken into consider-
ation. Usually the ﬁbres, e.g. glass, carbon, boron, etc., d o not absorb moisture that
means αmo
f=0. For isotropic constituents the formulae for αmo
Landαmo
Tare
1Richard Allan Schapery (∗3 March 1935 Duluth, Minnesota, United States) – engineerin g educa-
tor, contributions in the ﬁeld of mechanics of composite mat erials

4.1 Elastic Behavior of Laminae 109
αmo
L=αmo
mEm(1−vf)
vfEf+(1−vf)Em=αmo
mEm(1−vf)
E1,
αmo
T=αmo
m(1−vf)[(1+νm)Efvf+(1−vf)Em−νfvfEm]
vfEf+(1−vf)Em
=αmo
m(1−vf)
E1{(1+νm)Efvf+[(1−vf)−νfvf]Em}(4.1.11)
and for a composite with isotropic matrix but orthotropic ﬁb res the effective moduli
are given by
αmo
L=αmo
mEm(1−vf)
EL,
αmo
T=αmo
m1−vf
EL{(1+νm)ELfvf+[(1−vf)−νLTfvfEm}(4.1.12)
The formulae (4.1.9) – (4.1.12) completes the discussion ab out micro-mechanics
in Chap. 3. Note that for a great ﬁbre volume fraction negativ eαth
Lvalues can be
predicted reﬂecting the dominance of negative values of ﬁbr e expansion αf
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 stra insε3is not the same.
Using the condition σ3=0 follows
ε3=S′
13σ′
1+S′
23σ′
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 longitud inal direction. Laminates
are constituted generally of different layers at different orientations. To study the
elastic behavior of laminates, it is necessary to take a glob al coordinate system for
the whole laminate and to refer the elastic behavior of each l ayer to this reference
system. This is necessary to develop the stress-strain rela tionship for an angle lam-
ina, i.e. an off-axis loaded UD-lamina.
The global and the local material reference systems are give n in Fig. 4.3. We
consider the ply material axes to be rotated away from the glo bal axes by an angle
θ, positive in the counterclockwise direction. This means th at the ( x1,x2)-axes are
at an angle θclockwise from the material axes. Thus transformation rela tions 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≡x12≡x′
2
✟✟✟✟✟✟✟✟
✟✟✟✟✟✟✟✟
✟✟✟✟✟✟✟✟
✟✟✟✟✟✟
✟✟✟✟✟✟✟
✟✟✟
✟✟✟✟✟✟✟✟✟✟✟✟✟✟✟✟✟✟θ✢
needed for the stresses, the strains, the stress-strain equ ations, the stiffness and the
compliance matrices.
The transformation equations for a rotating the reference s ystem ( x′
1,x′
2) or
(x1,x2) counterclockwise or clockwise by an angle θfollow from Sect. 2.1.2 and
are given in Table 4.1. Note the relations for the transforma tion matrices derived in
Sect. 2.1.2
Table 4.1 Transformation rules of the coordinates, displacements, s trains and stresses of a lamina
a) Rotation of the reference systems
/bracketleftbiggx′
1
x′
2/bracketrightbigg
=/bracketleftbiggc s
−s c/bracketrightbigg/bracketleftbiggx1
x2/bracketrightbigg
,/bracketleftbiggx1
x2/bracketrightbigg
=/bracketleftbiggc−s
s c/bracketrightbigg/bracketleftbiggx′
1
x′
2/bracketrightbigg
xxx′=RRRxxx,xxx=RRRTxxx′
b) Transformation of displacements
/bracketleftbiggu′
1
u′
2/bracketrightbigg
=/bracketleftbiggc s
−s c/bracketrightbigg/bracketleftbiggu1
u2/bracketrightbigg
,/bracketleftbiggu1
u2/bracketrightbigg
=/bracketleftbiggc−s
s c/bracketrightbigg/bracketleftbiggu′
1
u′
2/bracketrightbigg
uuu′=RRRuuu,uuu=RRRTuuu′
c) Transformation of strains

ε′
1
ε′
2
ε′
6
=
c2s2sc
s2c2−sc
−2sc2sc c2−s2

ε1
ε2
ε6
,
ε1
ε2
ε6
=
c2s2−sc
s2c2sc
2sc−2sc c2−s2

ε′
1
ε′
2
ε′
6

εεε′=TTTεεεε=/parenleftig
TTTσ′/parenrightigT
εεε,εεε=TTTε′εεε′=(TTTσ)Tεεε′
d) Transformation of stresses

σ′
1
σ′
2
σ′
6
=
c2s22sc
s2c2−2sc
−sc sc c2−s2

σ1
σ2
σ6
,
σ1
σ2
σ6
=
c2s2−2sc
s2c22sc
sc−sc c2−s2

σ′
1
σ′
2
σ′
6

σσσ′=TTTσσσσ=/parenleftig
TTTε′/parenrightigT
σσσ,σσσ=TTTσ′σσσ′=(TTTε)Tσσσ′
with s≡sinθ,c≡cosθ

4.1 Elastic Behavior of Laminae 111
TTTε=/parenleftig
TTTε′/parenrightig−1
=/parenleftig
TTTσ′/parenrightigT
,TTTε′= (TTTε)−1= (TTTσ)T,
TTTσ=/parenleftig
TTTσ′/parenrightig−1
=/parenleftig
TTTε′/parenrightigT
,TTTσ′= (TTTσ)−1= (TTTε)T(4.1.13)
Table 4.2 summarizes the transformation rules for the stres s-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 jand compliance matrix S′
i jin the ref-
erence system ( x′
1,x′
2) to the reduced stiffness matrix Qi jand compliance matrix Si jin the ( x1,x2)-
system
a) Constitutive equations in the ( x1,x2)-reference system

σ1
σ2
σ6
=
Q11Q12Q16
Q12Q22Q26
Q16Q26Q66

ε1
ε2
ε6
,
ε1
ε2
ε6
=
S11S12S16
S12S22S26
S16S26S66

σ1
σ2
σ6

σσσ=QQQεεε,εεε=SSSσσσ,
QQQ=(TTTε)TQQQ′TTTε,SSS=(TTTσ)TSSS′TTTσ,
Qi j=Qji,Q′
i j=Q′
ji,Si j=Sji,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≡sinθ,c≡cosθ

112 4 Elastic Behavior of Laminate and Sandwich Composites
QQQ′=/parenleftig
TTTε′/parenrightigT
QQQTTTε′=TTTσQQQ(TTTσ)T,
SSS′=/parenleftig
TTTσ′/parenrightigT
SSSTTTσ′=TTTεSSS(TTTε)T,
QQQ= (TTTε)TQQQ′TTTε,
SSS= (TTTσ)TSSS′TTTσ(4.1.14)
Starting with the stiffness equation σσσ=QQQεεεand introducing σσσ′=QQQ′εεε′in the
transformation σσσ=TTTσ′σσσ′=/parenleftig
TTTε′/parenrightigT
σσσ′it follows that σσσ=/parenleftig
TTTε′/parenrightigT
QQQ′εεε′and with
εεε′=TTTεεεεthis gives σσσ= (TTTε)TQQQ′TTTεεεε. Comparison of equations σσσ=QQQεεεand
σσσ=(TTTε)TQQQ′TTTεεεεyields

Q11Q12Q16
Q12Q22Q26
Q16Q26Q66
=(TTTε)T
Q′
11Q′
120
Q′
12Q′
220
0 0 Q66
TTTε(4.1.15)
In an analogous way

S11S12S16
S12S22S26
S16S26S66
=(TTTσ)T
S′
11S′
120
S′
12S′
220
0 0 Q66
TTTσ(4.1.16)
can be derived. Note that in (4.1.15) and (4.1.16) the matric es[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 jorSi jare functions of the four independent material
characteristics Q′
i jorS′
i jand the angle θ. The experimental testing is therefore more
simple than for a real anisotropic material with 6 independe nt material values, if the
material axes of the lamina are known.
From the transformation c) in Table 4.2 follows the transfor mation of the engi-
neering parameters EL,ET,GLT,νLTof 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 al l normal and shear
terms of stresses and strains. In Fig. 4.4 these coupling eff ects in an off-axes loaded
UD-lamina are described.
The coupling coefﬁcients
ν12=−ε2
ε1=−S21
S11,ν21=−ε1
ε2=−S12
S22(4.1.17)
are the known Poisson’s ratios and the ratios
ν16=ε6
ε1=S61
S11,ν26=ε6
ε2=S62
S22(4.1.18)
are so called shear coupling values. They are non-dimension al parameters like Pois-
son’s ratio and relate normal stresses to shear strains or sh ear stresses to normal
strains.

4.1 Elastic Behavior of Laminae 113
✛ ✲✻
❄❄✻✲
✛σ1 σ1σ2
σ2 σ6σ6
ε1
σ1=S11=1
E1
ε2
σ1=S21=−ν12
E1
ε6
σ1=S61=ν16
E1ε2
σ2=S22=1
E2
ε1
σ2=S12=−ν21
E2
ε6
σ2=S62=ν26
E2ε6
σ6=S66=1
E6
ε1
σ6=S16=ν61
E6
ε2
σ6=S26=ν62
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 wr itten in terms of
engineering parameters of the off-axis case as

ε1
ε2
ε6
=
1/E1−ν21/E2ν61/E6
−ν12/E11/E2ν62/E6
ν16/E1ν26/E21/E6

σ1
σ2
σ6
 (4.1.19)
With the compliance engineering parameters
S11=1
E1,S12=−ν21
E2,S21=−ν12
E1,
S66=1
E6,S16=ν61
E6, S61=ν16
E1,
S26=ν62
E6,S62=ν26
E2, S22=1
E2(4.1.20)
it follows from the symmetry considerations of the complian ce matrix that
Si j=Sji,i,j=1,2,6; i.e.ν12
E2=ν21
E1,ν16
E1=ν61
E6,ν26
E2=ν62
E6(4.1.21)
but the anisotropic coupling coefﬁcients are
νi j/ne}ationslash=νji,i,j=1,2,6 (4.1.22)
Equation (4.1.19) can be inverted to yield the stress-strai n equations in terms of
engineering parameters but these relations would be more co mplex than (4.1.19).
Using the relationships between engineering parameters an d 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 i ncluding shear coupling

114 4 Elastic Behavior of Laminate and Sandwich Composites
ratios ν16andν26
1
E1=1
E′
1c4+/parenleftbigg1
G′
12−2ν′
12
E′
1/parenrightbigg
s2c2+1
E′
2s4,
1
E2=1
E′
1s4+/parenleftbigg1
G′
12−2ν′
12
E′
1/parenrightbigg
s2c2+1
E′
2c4,
1
G12=2/parenleftbigg2
E′
1+2
E′
2+4ν′
12
E′
1−1
G′
12/parenrightbigg
s2c2+1
G′
12(c4+s4),(4.1.23)
ν12
E1=ν21
E2=/bracketleftbiggν′
12
E′
1(c4+s4)−/parenleftbigg1
E′
1+1
E′
2−1
G′
12/parenrightbigg
c2s2/bracketrightbigg
,
ν16=E′
1/bracketleftbigg/parenleftbigg2
E′
1+2ν′
12
E′
1−1
G′
12/parenrightbigg
sc3−/parenleftbigg2
E′
2+2ν′
12
E′
1−1
G′
12/parenrightbigg
s3c/bracketrightbigg
,
ν26=E′
2/bracketleftbigg/parenleftbigg2
E′
1+2ν′
12
E′
1−1
G′
12/parenrightbigg
s3c−/parenleftbigg2
E′
2+2ν′
12
E′
1−1
G′
12/parenrightbigg
sc3/bracketrightbigg
Equation (4.1.23) can be also written in the following form
E1=EL
c4+/parenleftbiggEL
GLT−2νLT/parenrightbigg
s2c2+EL
ETs4,
E2=ET
c4+/parenleftbiggET
GLT−2νTL/parenrightbigg
s2c2+ET
ELs4,
G12=GLT
c4+s4+2/bracketleftbigg
2GLT
EL(1+2νLT)+2GLT
ET−1/bracketrightbigg
s2c2,
ν12=νLT(c4+s4)−/parenleftbigg
1+EL
ET−EL
GLT/parenrightbigg
c2s2
c4+/parenleftbiggEL
GLT−2νLT/parenrightbigg
c2s2+EL
ETs4,
ν21=νTL(c4+s4)−/parenleftbigg
1+ET
EL−ET
GLT/parenrightbigg
c2s2
c4+/parenleftbiggET
GLT−2νTL/parenrightbigg
c2s2+ET
ELs4,
ν16=EL/bracketleftbigg/parenleftbigg2
EL+2νLT
EL−1
GLT/parenrightbigg
sc3−/parenleftbigg2
ET+2νLT
EL−1
GLT/parenrightbigg
s3c/bracketrightbigg
,
ν26=ET/bracketleftbigg/parenleftbigg2
EL+2νLT
EL−1
GLT/parenrightbigg
s3c−/parenleftbigg2
ET+2νLT
EL−1
GLT/parenrightbigg
sc3/bracketrightbigg(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 th e 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 para meters of a UD-lamina
in off-axis loading can be illustrated by the following step s:
1. Input the basic engineering parameters EL,ET,GLT,νLTreferred to the material
axes of the lamina and obtained by material tests or mathemat ical modelling.
2. Calculate the compliances S′
i jand the reduced stiffness Q′
i j.
3. Application of transformations to obtain the lamina stif fness Qi jand compliances
Si j.
4. Finally calculate the engineering parameters E1,E2,G12,ν12,ν21,ν16,ν26re-
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 t ransversely-isotropic. The
mechanical properties transverse to the ﬁbre direction are provided by weaker ma-
trix material and the effects of transverse shear deformati on may be signiﬁcant. For
such cases, the stress and the strain vector should include a ll six components
σσσT=[σ1σ2σ3σ4σ5σ6],εεεT=[ε1ε2ε3ε4ε5ε6]
For a rotation about the direction eee3(Fig. 2.6) the transformation matrices (2.1.39)
and (2.1.40) are valid and relations for stress and strain ve ctors in the on-axis and
the off-axis reference system are given by
σσσ′=3
TTTσσσσ, εεε′=3
TTTεεεε,3
TTTε′=/parenleftbigg3
TTTσ/parenrightbiggT
=/parenleftbigg3
TTTε/parenrightbigg−1
,
σσσ=/parenleftbigg3
TTTσ/parenrightbigg−1
σσσ′,εεε=/parenleftbigg3
TTTε/parenrightbigg−1
εεε,3
TTTσ′=/parenleftbigg3
TTTε/parenrightbiggT
=/parenleftbigg3
TTTσ/parenrightbigg−1 (4.1.25)
When the stiffness matrix CCC′corresponding to an orthotropic material behavior, see
Eq. (2.1.46), the transformed stiffness matrix CCCmay be written in detail as for mon-
oclinic material behavior, see Eq. (2.1.42)

σ1
σ2
σ3
σ4
σ5
σ6
=
C11C12C130 0 C16
C22C230 0 C26
C330 0 C36
C44C450
S Y M C550
C66

ε1
ε2
ε3
ε4
ε5
ε6
(4.1.26)
TheCi jare the transformed stiffness, i.e. in vector-matrix notat ion
σσσ′=CCC′εεε′
σσσ=CCCεεε,σσσ=3
TTTσ′σσσ′=/parenleftigg
3
TTTε/parenrightiggT
CCC′εεε′=/parenleftigg
3
TTTε/parenrightiggT
CCC′3
TTTεεεε=CCCεεε

116 4 Elastic Behavior of Laminate and Sandwich Composites
we ﬁnally obtain
CCC=/parenleftigg
3
TTTε/parenrightiggT
CCC′3
TTTε, (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 Ci jare not independent material values. They are func-
tions of 9 C′
i jfor 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 jfor 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 εεε′=SSS′σσσ′,εεε=SSSσσσfollow analogously the transformed compliances
SSS=/parenleftbigg3
TTTσ/parenrightbiggT
SSS′3
TTTσ, (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 independ ent material values
for the orthotropic and 5 independent material values for th e transversal-isotropic
case.
The stress-strain relationship for an angle lamina, i.e. an off-axis loaded UD-
lamina, including hygrothermal effects takes the followin g form

ε1
ε2
ε6
=
S11S12S16
S12S22S26
S16S26S66

σ1
σ2
σ6
+
εth
1
εth
2
εth
6
+
εmo
1
εmo
2
εmo
6
, (4.1.31)
where 
εth
1
εth
2
εth
6
=
αth
1
αth
2
αth
6
T,
εmo
1
εmo
2
εmo
6
=
αmo
1
αmo
2
αmo
6
M∗(4.1.32)
with the thermal and moisture expansion coefﬁcients αth
i,αmo
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 coe fﬁcients of both ther-
mal and moisture expansions are pure dilatational in the mat erial coordinate system
(L,T), rotation into the global ( x1,x2) system results in coefﬁcients αth
6,αmo
6. Fur-
thermore if there are no constraints placed on a UD-lamina, n o mechanical strains
will be included in it and therefore no mechanical stresses a re induced. But in lami-
nates, even if there are no constraints on the laminate, the d ifference in thermal and
moisture expansion coefﬁcients of the various laminae of a l aminate induces differ-
ent expansions in each layer and results in residual stresse s. 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
εth
1=ε′th
1c2+ε′th
2s2,
εth
2=ε′th
1s2+ε′th
2c2,
εth
6=2(ε′th
1−ε′th
2)cs,(4.1.34)
follow
αth
1=α′th
1c2+α′th
2s2,
αth
2=α′th
1s2+α′th
2c2,
αth
6=2(α′th
1−α′th
2)cs(4.1.35)
In an anisotropic layer, uniform heating induces not only no rmal strains, but also
shear thermal strains. For a transversal isotropic materia l behavior there is additional
εth
3=αth
3T,αth
3=α′th
3,αth
4=αth
5=0. Because ε3=ε′
3the strain ε3can be obtained
directly from
ε3=αth
3T+αmo
3M∗+S′
13σ′
1+S′
23σ′
2
However, the stresses σ′
1,σ′
2can be written in terms σ1,σ2,σ6referred to the off-
axis coordinate system to obtain an expression for ε3that represents the normal
strain in the x3-direction in terms of the global coordinate system
ε3=εth
3+εmo
3+(S13c2+S23s2)σ1+(S13s2+S23c2)σ2
+2(S13−S23s2)scσ1(4.1.36)
4.1.3 Stress Resultants and Stress Analysis
Sections 4.1.1 and 4.1.2 describe the constitutive equatio ns for UD-laminae in an
on-axis and an off-axis reference system as a relation betwe en stresses and strains.
For each lamina, the stress components can be integrated acr oss their thickness hand
yield stress resultants. Stress resultants can be in-plane forces, transverse forces and
resultant moments. The constitutive equations may then be f ormulated as relations
between mid-plane strain and in-plane forces, transverse s hear strains and transverse
forces and mid-plane curvatures and resultant moments, res pectively.
The in-plane stress resultant force vector, denoted by
NNN=[N1N2N6]T, (4.1.37)
is deﬁned by
NNN=h/2/integraldisplay
−h/2σσσdx3 (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 illustr ated in Fig. 4.5 for constant
in-plane stresses σ1,σ2,σ6across the thickness. In this case we have

4.1 Elastic Behavior of Laminae 119
x1x2x3
N1N6N6N2
h
Fig. 4.5 In-plane force resultants per unit length NNNT=[N1N2N6]
NNN=σσσh (4.1.39)
The reduced stiffness matrix QQQof the lamina has also constant components across
h. The strains of the midplane x3=0 of the lamina are given by
εεε(x1,x2,x3=0)=εεε(x1,x2)
and Eq. (4.1.39) yields
NNN=QQQεεεh=AAAεεε,AAA=QQQh,εεεT=[ε1ε2ε6] (4.1.40)
QQQis the reduced stiffness matrix (Table 4.2 a) and AAAis the off-axis stretching or
extensional stiffness matrix of the lamina. From (4.1.40) i t follows that
εεε=AAA−1NNN=aaaNNN,aaa=AAA−1=SSSh−1(4.1.41)
aaais the off-axis in-plane compliance matrix. AAAandaaaare, like QQQandSSSsymmet-
ric matrices, which have in the general case only non-zero el ements. In the special
cases of on-axis reference systems or isotropic stiffness a nd compliances, respec-
tively, the structure of the matrices is simpliﬁed. AAAis the extensional stiffness and aaa
the extensional compliance matrix expressing the relation ship between the in-plane
stress resultant NNNand the mid-plane strain εεε:

120 4 Elastic Behavior of Laminate and Sandwich Composites
Off-axis extensional stiffness and compliance matrices
AAA=
A11A12A16
A12A22A26
A16A26A66
,aaa=
a11a12a16
a12a22a26
a16a26a66
 (4.1.42)
On-axis extensional stiffness and compliance matrices
AAA=
A11A120
A12A220
0 0 A66
,aaa=
a11a120
a12a220
0 0 a66
 (4.1.43)
If the stresses are not constant across h, resultant moments can be deﬁned
MMM=h/2/integraldisplay
−h/2σσσx3dx3 (4.1.44)
The resultant moment vector is denoted by
MMM=[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
x1x2x3
M6M1M2
M6h
Qs
1Qs
2
Fig. 4.6 Resultant moment vector MMMT=[M1M2M6]and transverse shear resultants QQQsT=[Qs
1Qs
2]

4.1 Elastic Behavior of Laminae 121
twisting strains, which are usually expressed by the relati onship
εεε(x1,x2,x3)=x3κκκ,κκκT=[κ1κ2κ6] (4.1.46)
κκκis the vector of curvature, κ1,κ2correspond to the bending moments M1,M2and
κ6to 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
MMM=QQQκκκh/2/integraldisplay
−h/2×2
3dx3=QQQκκκh3
12=DDDκκκ,κκκ=DDD−1MMM=dddMMM (4.1.47)
DDD=QQQh3/12 is the ﬂexural stiffness matrix and dddthe ﬂexural compliance matrix
expressing the relations between stress couples MMMand the curvatures. For off-axis
and on-axis reference systems the matrices are given by:
Off-axis ﬂexural stiffness and compliance matrices
DDD=
D11D12D16
D12D22D26
D16D26D66
,ddd=
d11d12d16
d12d22d26
d16d26d66
 (4.1.48)
On-axis ﬂexural stiffness and compliance matrices
DDD=
D11D120
D12D220
0 0 D66
,ddd=
d11d120
d12d220
0 0 d66
 (4.1.49)
The transverse shear resultants can be deﬁned in the same way by
QQQs=/bracketleftbigg
Qs
1
Qs
2/bracketrightbigg
=h/2/integraldisplay
−h/2/bracketleftbigg
σs
5
σs
4/bracketrightbigg
dx3 (4.1.50)
QQQsis (like the in-plane resultants) a load vector per unit leng th in the cross section
of the lamina x1=const or x2=const, respectively. The transverse shear resultant
vector QQQsis written with a superscript sto distinguish it from the reduced stiffness
matrix QQQ. When modelling a plane stress state, there are no constitut ive equations
forσ4,σ5and the shearing stresses are calculated with the help of the equilibrium
equations, Eq. (2.2.1), or with help of equilibrium conditi ons of stress resultant,
e.g. Chap. 8. In three-dimensional modelling, including tr ansverse shear strains,
however, constitutive equations for transverse shear resu ltant can be formulated.
For a lamina with resultant forces NNNand moments MMMthe in-plane strain εεεand
the curvature term κκκhave to be combined
εεε(x1,x2,x3)=εεε(x1,x2)+x3κκκ(x1,x2) (4.1.51)

122 4 Elastic Behavior of Laminate and Sandwich Composites
For the stress vector is
σσσ(x1,x2,x3) =QQQ[εεε(x1,x2)+x3κκκ(x1,x2)]
=QQQεεε(x1,x2)+QQQx3κκκ(x1,x2)(4.1.52)
and by integration through the lamina thickness hfollow
NNN=h/2/integraldisplay
−h/2σσσdx3=QQQεεεh+QQQκκκh/2/integraldisplay
−h/2x3dx3=AAAεεε+BBBκκκ,
MMM=h/2/integraldisplay
−h/2σσσx3dx3=QQQεεεh/2/integraldisplay
−h/2x3dx3+QQQκκκh3
12=BBBεεε+DDDκκκ(4.1.53)
The coupling stiffness matrix BBBis zero for a lamina, which is symmetric to the
midplane x3=0, i.e. there are no coupling effects between the NNNandκκκorMMMand
εεε, respectively. In Table 4.3 the constitutive equations of t he lamina resultants are
summarized for a symmetric general angle lamina, for a UD-or thotropic lamina and
for an isotropic layer. In a contracted vector-matrix notat ion, we can formulate the
constitutive equation of a lamina by
/bracketleftbigg
NNN
MMM/bracketrightbigg
=/bracketleftbigg
AAA000
000DDD/bracketrightbigg/bracketleftbigg
εεε
κκκ/bracketrightbigg
, (4.1.54)
where the in-plane stiffness submatrix AAA=QQQhand the plate stiffness submatrix
DDD=QQQh3/12. 000 are zero submatrices. The inverted form of (4.1.54) is impo rtant for
stress analysis
/bracketleftbiggεεε
κκκ/bracketrightbigg
=/bracketleftbiggaaa000
000ddd/bracketrightbigg/bracketleftbiggNNN
MMM/bracketrightbigg
,aaa=AAA−1,
ddd=DDD−1,AAA=QQQh,
DDD=QQQ(h3/12)(4.1.55)
Equation (4.1.52) yields the stress components σi,i=1,2,6
σ1=Q11(ε1+x3κ1)+Q12(ε2+x3κ2)+Q16(ε6+x3κ6) =σ1M+σ1B,
σ2=Q21(ε1+x3κ1)+Q22(ε2+x3κ2)+Q26(ε6+x3κ6) =σ2M+σ2B,
σ6=Q16(ε1+x3κ1)+Q62(ε2+x3κ2)+Q66(ε6+x3κ6) =σ6M+σ6B
σiMare the membrane or in-plane stresses coupled with NiandσiBthe 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 σ4,σ5for 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
σ5(x3)=−x3/integraldisplay
−h/2/parenleftbigg∂σ1
∂x1+∂σ6
∂x2/parenrightbigg
dx3,σ4(x3)=−x3/integraldisplay
−h/2/parenleftbigg∂σ6
∂x1+∂σ2
∂x2/parenrightbigg
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
=
A11A12A160 0 0
A12A22A260 0 0
A16A26A660 0 0
0 0 0 D11D12D16
0 0 0 D12D22D26
0 0 0 D16D26D66

ε1
ε2
ε6
κ1
κ2
κ6
,Ai j=Qi jh,
Di j=Qi jh3
12
Orthotropic single layer or UD-laminae, on-axis

N1
N2
N6
M1
M2
M6
=
A11A120 0 0 0
A12A220 0 0 0
0 0 A660 0 0
0 0 0 D11D120
0 0 0 D12D220
0 0 0 0 0 D66

ε1
ε2
ε6
κ1
κ2
κ6
,Q11=E1
1−ν12ν21,
Q22=E2
1−ν12ν21,
Q12=ν12E2
1−ν12ν21
=ν21E1
1−ν12ν21,
Q66= G12
Isotropic single layer

N1
N2
N6
M1
M2
M6
=
A11A120 0 0 0
A12A110 0 0 0
0 0 A660 0 0
0 0 0 D11D120
0 0 0 D12D110
0 0 0 0 0 D66

ε1
ε2
ε6
κ1
κ2
κ6
,Q11=E
1−ν2,
Q12=νE
1−ν2,
Q66=E
2(1+ν)
and with Eq. (4.1.52) follows
σ5(x3) =−x3/integraldisplay
−h/2/braceleftbigg∂
∂x1[Q11(ε1+x3κ1)+Q12(ε2+x3κ2)+Q16(ε6+x3κ6)]
+∂
∂x2[Q61(ε1+x3κ1)+Q62(ε2+x3κ2)+Q66(ε6+x3κ6)]/bracerightbigg
dx3,
(4.1.57)

124 4 Elastic Behavior of Laminate and Sandwich Composites
✂✂✂✂✂✂✂
✂✂✂✂✂✂✂
❄✻
h0 0 0
σiM σiB σi=σiM+σiBN6
h=σ6MN2
h=σ2MN1
h=σ1M
M6
h3/12×3=σ6BM2
h3/12×3=σ2BM1
h3/12×3=σ1B
σ6=σ6M+σ6Bσ2=σ2M+σ2Bσ1=σ1M+σ1B
Fig. 4.7 In-plane membrane stresses σiM, bending stresses σiBand total stresses σiacross h(qual-
itative)
σ4(x3) =−x3/integraldisplay
−h/2/braceleftbigg∂
∂x1[Q61(ε1+x3κ1)+Q62(ε2+x3κ2)+Q66(ε6+x3κ6)]
+∂
∂x2[Q21(ε1+x3κ1)+Q22(ε2+x3κ2)+Q26(ε6+x3κ6)]/bracerightbigg
dx3
Substituting the midplane strain εεεand the curvature κκκby the resultants NNNandMMM
Eqs. (4.1.57) takes the form
σ5(x3) =−x3/integraldisplay
−h/2/braceleftig
Q11∂
∂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)]/bracerightig
dx3,
(4.1.58)
σ4(x3) =−x3/integraldisplay
−h/2/braceleftig
Q61∂
∂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)]/bracerightig
dx3
The distribution of the transverse shear stresses through t he thickness his obtained
by integration from the bottom surface x3=−h/2 of the lamina to x3
x3/integraldisplay
−h/2QQQdx3=˜AAA(x3) =QQQ(x3+h/2),
x3/integraldisplay
−h/2QQQx3dx3=˜BBB(x3) =QQQ1
2/parenleftbigg
x2
3−h2
4/parenrightbigg (4.1.59)
and we have
˜AAA(−h/2)=˜BBB(−h/2)=000,˜AAA(h/2)≡AAA,˜BBB(h/2)≡BBB
Finally, the transverse shear stress equations (4.1.57) ta ke the form
σ5(x3) =−/bracketleftbigg
˜A11(x3)∂ε1
∂x1+˜A12(x3)∂ε2
∂x1+˜A16(x3)∂ε6
∂x1
+˜B11(x3)∂κ1
∂x1+˜B12(x3)∂κ2
∂x1+˜B16(x3)∂κ6
∂x1
+˜A61(x3)∂ε1
∂x2+˜A62(x3)∂ε2
∂x2+˜A66(x3)∂ε6
∂x2
+˜B61(x3)∂κ1
∂x2+˜B62(x3)∂κ2
∂x2+˜B66(x3)∂κ6
∂x2/bracketrightbigg
,

126 4 Elastic Behavior of Laminate and Sandwich Composites
σ4(x3) =−/bracketleftbigg
˜A61(x3)∂ε1
∂x1+˜A62(x3)∂ε2
∂x1+˜A66(x3)∂ε6
∂x1
+˜B61(x3)∂κ1
∂x1+˜B62(x3)∂κ2
∂x1+˜B66(x3)∂κ6
∂x1
+˜A21(x3)∂ε1
∂x2+˜A22(x3)∂ε2
∂x2+˜A26(x3)∂ε6
∂x2
+˜B21(x3)∂κ1
∂x2+˜B22(x3)∂κ2
∂x2+˜B26(x3)∂κ6
∂x2/bracketrightbigg
or an analogous equation to (4.1.59) by substituting the in- plane strains εεεand the
mid-plane curvatures by the stress resultants NNNandMMM.
The transverse shear stresses σ4,σ5are parabolic functions across the lamina
thickness. If there are no surface edge shear stresses, the c onditions σ5(h/2) =
σ4(h/2)=0 are controlling the performance of the equilibrium equati ons and the
accuracy of the stress analysis.
4.1.4 Problems
Exercise 4.1. For a single layer unidirectional composite, the on-axis el astic behav-
ior is given by EL=140 GPa, ET=9 GPa, νLT=0.3. Calculate the reduced stiffness
matrix QQQ′and the reduced compliance matrix SSS′.
Solution 4.1. EL≡E′
1,ET≡E′
2,νLT=ν′
12. Equation (4.1.3) yields

σ′
1
σ′
2
σ′
6
=
Q′
11Q′
120
Q′
12Q′
220
0 0 Q′
66

ε′
1
ε′
2
ε′
6
,
ε′
1
ε′
2
ε′
6
=
S′
11S′
120
S′
12S′
220
0 0 S′
66

σ′
1
σ′
2
σ′
6
,
Q′
11=E′
1/(1−ν′
12ν′
21), S′
11=1/E′
1,
Q′
22=E′
2/(1−ν′
12ν′
21), S′
22=1/E′
2,
Q′
12=E′
2ν′
12/(1−ν′
12ν′
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
QQQ′=
140,811 2,716 0
2,716 9,052 0
0 0 53 ,846
GPa,
SSS′=
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 th e fol-
lowing properties EL=140 GPa, ET=10 GPa, GLT=6,9 GPa, νLT=0,3 and
θ=450. Calculate the strains ε1,ε2andε6when the panel is loaded by a shear stress
σ6=±τ=±10 MPa.
Solution 4.2. From Fig. 4.4 follows
ε1=±S16σ6,ε2=±S26σ6,ε6=±S66σ6
With EL≡E′
1,ET≡E′
2,GLT≡G′
12andνLT≡ν′
12and 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 ε1=∓0,46·10−3,ε2=ε1,ε6=±1,11·10−3.
Conclusion 4.1. A positive shear load σ6=+τshortens the composite panel in both
directions, a negative shear load σ6=−τwould enlarge the panel in both directions.
Exercise 4.3. In a unidirectional single layer is a strain state ε11=1%=10−2,
ε22=−0.5%=−0.5 10−2,γ12=2%=2 10−2. In the principal directions, the
following engineering parameters of the composite materia l are E′
1=EL=40 GPa,
E′
2=ET=10 GPa, G12=GLT=5 GPa, νLT=0.3. Determine the plane stress state
for the axis (x1,x2)and(x′
1,x′
2)≡(L,T)andθ=450.
Solution 4.3. The stress states σσσ′andσσσare to calculate for a given strain state
ε1=10−2,ε2=−0,5 10−2,ε6=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
−2sc2sc c2−s2

ε1
ε2
ε6

=
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 jand the stresses σ′
iin the on-axis system follow from
(4.1.3)
Q′
11=E′
1/(1−ν′
12ν′
21)=40,9207 GPa, Q′
22=E′
2/(1−ν′
12ν′
21)=10,2302 GPa,
Q′
12=E′
2ν′
12/(1−ν′
12ν′
21)=3,0691 GPa, Q′
66=G′
12=5,0 GPa,

σ′
1
σ′
2
σ′
6
=
Q′
11Q′
120
Q′
12Q′
220
0 0 Q′
66

ε′
1
ε′
2
ε′
6
=
488,5
−38,36
−75,0
10−3GPa
The stresses σiin the off-axis system are calculated with the help of the tra nsforma-
tion rules Table 4.1

σ1
σ2
σ6
=
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′
2andG12/E′
2against the ﬁbre ori-
entation θfor a carbon-epoxy and glass-epoxy lamina using the followi ng 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+/parenleftbigg1
G′
12−2ν′
12
E′
2/parenrightbigg
s2c2+s4
E′
2,
(G12)−1=2/parenleftbigg2
E′
1+2
E′
2+4ν′
12
E′
1−1
G′
12/parenrightbigg
s2c2+1
G′
12(c4+s4)
Now the functions f1(θ)=E1(θ)/E′
2andf2(θ)=G12(θ)/E′
2can be sketched. The
results are shown on Figs. 4.8 and 4.9. Discussion of the func tions f1(θ)andf2(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 maximu m 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-epoxy0
0 10 20 30 40 50 60 70 80 902468101214E1(θ)/E′
2
θ
Fig. 4.9 Variation of
G12(θ)/E′
2against the ﬁbre
orientation for two compos-
ites
carbon-epoxy glass-epoxy0 10 20 30 40 50 60 70 80 900.50.60.70.80.9G12(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 jand the on-axis strains, if the applied on-axis stresses
areσ′
1=2 MPa, σ′
2=−3 MPa, σ′
6=4 MPa,
2. the off-axis compliances Si jand the off-axis and on-axis strains εεε,εεε′ifθ=450
(Fig. 4.3) and the applied off-axis stresses are
σ1=2 MPa, σ2=−3 MPa, σ6=4 MPa,
3. the coefﬁcients of thermal expansion in the off-axis syst em if
αth′
1=9 10−6/0K,αth′
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 10−12Pa−1,
S′
12=−ν′
12/E′
1=−1,667 10−12Pa−1,
S′
22= (E′
2)−1=100 10−12Pa−1,
S′
66= (G′
12)−1=142,86 10−12Pa−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 jcan be calculated (note that c=
cos450=0.7071,s=sin450=0.7071)
S11=61,270 10−12Pa−1,S22=61,271 10−12Pa−1,
S12=−10,160 10−12Pa−1,S66=108,89 10−12Pa−1,
S16=S26=−47,222 10−12Pa−1
Equations (4.1.19) and (4.1.20) yield the strains εi

ε1
ε2
ε6
=
S11S12S16
S12S22S26
S16S26S66

σ1
σ2
σ6

=
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 εεεcan be transformed to the strains εεε′

ε′
1
ε′
2
ε′
6
=
0.5 0.5 0.5
0.5 0.5−0.5
−1 1 0

ε1
ε2
ε6

=
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 αth
ifollow like the strains with
Table 4.1 to

4.2 Elastic Behavior of Laminates 131

αth
1
αth
2
αth
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
15,5
−13,0
10−6/K
Note that in the off-axis system αth
6/ne}ationslash=0.
Exercise 4.6. The micro-mechanical material parameters of a carbon-epox y com-
posite are
ELf=411 GPa, ETf=6.6 GPa, νTLf=0.06,νLTf=0.35,
αth
Lf=−1.2 10−61/K, αth
Tf=27.3 10−61/K,
Em=5.7 GPa, νm=0.316, αm=45 10−61/K, vf=0.5
The experimental tested values of the lamina are
EL=208.6 GPa, ET=6.3 GPa, νLT=0.33,
αth
L=−0.5 10−61/0C,αth
T=29.3 10−61/0C,
Predict the lamina values using the micro-mechanical model ling 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,
νLT=vfνLTf+(1−vf)νm=0.33,
ET=ETfEm/[vfEm+(1−vf)ETf]=6.12 GPa,
αth
L= [αth
LfvfELf+αth
m(1−vf)Em]/EL=−0.57 10−61/K,
αth
T= (αth
Tf+νTfαth
Lf)vf+(1+νm)αth
m(1−vf)−νLTαth
L=4.43 10−61/K
It can be concluded that the simple rules of mixture providin g proper results for lon-
gitudinal material characteristics EL,νLTandαth
L. In this case also ETis predicted
quite well but the formula for αTfails to predict the transverse thermal expansion
coefﬁcient with required accuracy. For engineering applic ations the thermal expan-
sion coefﬁcients should be normally determined by experime ntal methods.
4.2 Elastic Behavior of Laminates
In Sect. 4.1, stress-strain equations were developed for a s ingle 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 jand plate or out-of-plane stiffness Di jwere deﬁned. Assuming symme-
try about the midplane of a lamina in-plane and plate respons es are uncoupled in the
form of a ﬁrst order theory (linear force-displacement rela tions).
The mechanics of laminated composite materials is generall y studied at two dis-
tinct levels, commonly called micromechanics and macromec hanics. In Chap. 3 the
micromechanics was used to study the interaction between th e 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 establ ishes the relationship be-
tween the properties of the constituents and those of the lam ina. All micromechanics
approaches suffer from the problem of measuring the materia l properties of the con-
stituents and generally require correction factors to corr elate with measured lamina
properties. For most engineering design applications an an alysis that addressed to
the micro-mechanical level is unrealistic.
At the macro-mechanical level the properties of the individ ual layers of a lami-
nate are assumed to be known a priori. Macromechanics involv es investigation of the
interaction of the individual layers of a laminate with one a nother 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 ma cromechanical formu-
lations in designing composite laminates for desired mater ial characteristics is well
established. Macromechanics is based on continuum mechani cs, which models each
lamina as homogeneous and orthotropic and ignoring the ﬁbre /matrix interface.
The lamination theory is the mathematical modelling to pred ict the macro-
mechanical behavior of a laminate based on an arbitrary asse mbly of homoge-
neous orthotropic laminae. A two-dimensional modelling is most common, a three-
dimensional theory is very complex and should be limited to s elected problems, e.g.
the analysis of laminates near free edges.
A real structure generally will not consist of a single lamin a. A laminate consist-
ing of more than one lamina bonded together through their thi ckness, for a single
lamina is very thin and several laminae will be required to ta ke 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-la minae it may be an
optimal laminate for unidirectional loading only. One can o vercome this restriction
by making laminates with layers stacked at different ﬁbre an gles corresponding to
complex loading and stiffness requirements. To minimize th e increasing costs and
weights for such approach one have to optimize the laminae an gles. 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 an d analysis of laminates
will be considered. The behavior of a multidirectional lami nate 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-mech anical modelling may
be in the framework of the following assumptions:
•There is a monolithic bonding of all laminae i.e. there is no s lip between laminae
at their interface and no special interface layers are arran ged between the angle
plies.

4.2 Elastic Behavior of Laminates 133
•Each layer is quasi-homogeneous and orthotropic, but the an gle orientations may
be different.
•The strains and displacements are continuous throughout th e laminate. The in-
plane displacements and strains vary linearly through the l aminate thickness.
We will see that the stacking codes of laminates have a great i nﬂuence on the global
mechanical laminate response (Sect. 4.2.3), but there are s ome 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 avo ids 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 or ientations 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 lim it the number of ﬁbre
orientations to a few speciﬁc laminae types, e.g. ﬁbre orien tations of 00,±450
and 900, etc.
Consider a laminate made of nplies shown in Fig. 4.10. Each lamina has a thickness
ofh(k),k=1,2,…, n, and we have
✲✻

✠x3
x1x2Mid-plane x3=0
12n
❄✻❄✻
h
2
h
2x(2)
3
x(1)
3
❄
❄
❄x(0)
3=−h
2✻✻
x(n−1)
3x(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=1h(k)thickness of the laminate,
x(k)
3=−h
2+k
∑
i=1h(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 la minate,
x(k)
3and x(k−1)
3,k=1,2,…, n
are the location coordinates of the top and the bottom surfac e of lamina k. Each
layer of a laminate can be identiﬁed by its location in the lam inate, its material and
its ﬁbre orientation. The layers of the laminate may be symme tric, antisymmetric
or asymmetric to the midplane x3=0.h(k)and the reduced stiffness QQQ(k)may be
different for each lamina, but QQQ(k)is constant for the kth lamina. The following
examples illustrate the laminate code. In Fig. 4.11 the lami nate codes for an un-
symmetric laminate with four layers and a symmetric angle-p ly 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 th e same thickness.
Regular symmetric are those laminates which have an odd numb er of UD-laminae
of equal thicknesses and alternating angle orientations (F ig. 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 or der.
x3=0 x3=0
1234 a θ=00
θ=300
θ=−300
θ=900b θ=00
θ=−450
θ=900
θ=450
θ=450
θ=900
θ=−450
θ=008
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=05
4
3
2
1 θ=−450θ=450θ=900θ=450θ=−450
[−45/45/90/45/−45]≡[−45/45/¯90]S
Fig. 4.12 Regular symmetric angle-ply laminate: orientation of the m idplane θ=900
A general laminate has layers of different orientations θwith−900≤θ≤900.
An angle-ply laminate has ply orientations of θand−θwith 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 wit h identical thickness
and elastic properties but have +θand−θorientations of their principal material
axes with respect to the laminate reference axes. A balanced laminate can be sym-
metric, antisymmetric or asymmetric
[+θ1/−θ1/+θ2/−θ2]S symmetric lay-up,
[θ1/θ2/−θ2/−θ1] antisymmetric lay-up,
[θ1/θ2/−θ1/−θ2] asymmetric lay-up(4.2.2)
Antisymmetric laminates are a special case of balanced lami nates, having the bal-
anced+θand−θpairs of layers symmetrically situated about the middle sur face.
Generally each layer of a laminate can have different ﬁbre an gles, different thick-
nesses and different composite materials. The inﬂuence of t he laminate codes, i.e
the properties and the stacking sequences, on the elastic be havior 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 refer ence 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-pl ane 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

σ1
σ2
σ6
σ4
σ5
(k)
=
Q11Q12Q160 0
Q21Q22Q260 0
Q61Q62Q660 0
0 0 0 Q44Q45
0 0 0 Q54Q55
(k)
ε1
ε2
ε6
ε4
ε5
(k)
(4.2.3)
or in contracted notation
σ(k)
i=Q(k)
i jε(k)
j,i,j=1,2,6,
σ(k)
i=Q(k)
i jε(k)
j,i,j=4,5(4.2.4)
with (see also 2.1.78)
σ(k)
3=0,ε(k)
3=−1
C33(C13ε1+C23ε2+C36ε6) (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)
55are 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 jfrom layer to layer implies the discontinuity
of the stresses when passing from one lamina to another.
From the assumption of macro-mechanical modelling of lamin ates it follows that
εεε(x1,x2,x3)=εεε(x1,x2)+x3κκκ(x1,x2) (4.2.6)
i.e the strains ε1,ε2,ε6vary linearly through the laminate thickness. εεε(x1,x2)is
the vector of the in-plane or membrane strains and x3κκκ(x1,x2)the vector of ﬂex-
ural strains (bending and twisting). κκκ(x1,x2)is the vector of curvature subjected to
bending and twisting. We shall see later (Sect. 5.4) that the re are different curvature
components in the classical and the shear deformation theor y of laminates.
The in-plane stress resultant force vector NNNof a laminate follows by summarizing
the adequate vectors of all laminae
NNN=n
∑
k=1NNN(k),NNNT=[N1N2N6],NNN(k)T=[N(k)
1N(k)
2N(k)
6] (4.2.7)
By analogy it follows that the resultant moment vector is
MMM=n
∑
k=1MMM(k),MMMT=[M1M2M6],MMM(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
QQQs(x1,x2)=n
∑
k=1QQQs(k),QQQsT=[Qs
1Qs
2],QQQs(k)T=[Qs(k)
1Qs(k)
2] (4.2.9)
Equations (4.2.4) and (4.2.6) yield
σσσ(k)=QQQ(k)εεε=QQQ(k)(εεε+x3κκκ) (4.2.10)
and the resultants NNNandMMMfor the laminate are ( k=1,2,…, n)
NNN(k)=/integraldisplay
(h(k))σσσ(k)dx3=σσσ(k)h(k),h(k)=x(k)
3−x(k−1)
3,
NNN=n
∑
k=1σσσ(k)h(k)(4.2.11)
and
MMM(k)=/integraldisplay
(h(k))σσσ(k)x3dx3=σσσ(k)1
2/parenleftig
x(k)2
3−x(k−1)2
3/parenrightig
=σσσ(k)h(k)x(k),
MMM=n
∑
k=1σσσ(k)h(k)x(k)(4.2.12)
with
x(k)=1
2/parenleftig
x(k)
3+x(k−1)
3/parenrightig
For each layer the membrane strains ε1,ε2,ε6, the curvatures κ1,κ2,κ6and the
reduced stiffness Q(k)
11,Q(k)
12,Q(k)
16,Q(k)
22,Q(k)
26,Q(k)
66are constant through each thickness
h(k)and (4.2.11) and (4.2.12) reduces to:
NNN=
n
∑
k=1QQQ(k)x(k)
3/integraldisplay
x(k−1)
3dx3
εεε+
n
∑
k=1QQQ(k)x(k)
3/integraldisplay
x(k−1
3)x3dx3
κκκ
= AAA εεε+ BBB κκκ,
MMM=
n
∑
k=1QQQ(k)x(k)
3/integraldisplay
x(k−1)
3x3dx3
εεε+
n
∑
k=1QQQ(k)x(k)
3/integraldisplay
x(k−1)
3×2
3dx3
κκκ
= BBB εεε+ DDD κκκ(4.2.13)
AAA,BBB,DDDare the extensional, coupling and bending stiffness matric es, respectively.
From (4.2.4) and (4.2.9), the relations for the transverse s hear resultants are

138 4 Elastic Behavior of Laminate and Sandwich Composites
QQQs=n
∑
k=1CCCs(k)x(k)
3/integraldisplay
x(k−1)
3σσσs(k)dx3=
n
∑
k=1CCCs(k)x(k)
3/integraldisplay
x(k−1)
3dx3
γγγs=AAAsγγγs(4.2.14)
with
CCCs=/bracketleftbigg
C44C45
C54C55/bracketrightbigg
,σσσs=/bracketleftbigg
σ4
σ5/bracketrightbigg
,γγγs=/bracketleftbigg
ε4
ε5/bracketrightbigg
,AAAs=/bracketleftbigg
A44A45
A54A55/bracketrightbigg
Equation (4.2.14) is a ﬁrst approach and consists of taking t he transverse shear strain
independent of the coordinate x3.AAAsis the transverse shear stiffness matrix. An
improvement is possible by replacing the transverse shear s tiffness As
i jby(kA)s
i j.
ks
i jare so called shear correction factors (Sect. 5.4). The elem ents of the matrices
AAA,BBB,DDD,AAAsare
Ai j=n
∑
k=1Q(k)
i j/parenleftig
x(k)
3−x(k−1)
3/parenrightig
=n
∑
k=1Q(k)
i jh(k),i,j=1,2,6,
Bi j=1
2n
∑
k=1Q(k)
i j/parenleftbigg
x(k)
32−x(k−1)
32/parenrightbigg
=n
∑
k=1Q(k)
i jx(k)
3h(k),
Di j=1
3n
∑
k=1Q(k)
i j/parenleftbigg
x(k)
33−x(k−1)
33/parenrightbigg
=n
∑
k=1Q(k)
i j/parenleftigg
x(k)
32+h(k)2
12/parenrightigg
h(k),
As
i j=n
∑
k=1C(k)
i j/parenleftig
x(k)
3−x(k−1)
3/parenrightig
=n
∑
k=1C(k)
i jh(k),i,j=4,5(4.2.15)
The constitutive equation for laminates including extensi onal, bending/torsion and
transverse shear strains is the superposition from the so-c alled classical equations
forNNNandMMMand the equation that involves the transverse shear resulta ntQQQs. The
constitutive equation can be written in the following contr acted hypermatrix form

NNN
MMM
QQQs
=
AAA BBB000
BBB DDD000
000 000AAAs

εεε
κκκ
γγγs
 (4.2.16)
The stiffness Q(k)
i jandC(k)
i jin (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 stretc hing and bending/twisting
of a laminate, i.e. in-plane strains result in in-plane resu ltants but also bending
and/or torsion moments and vice versa. Since there are no cou pling effects with

4.2 Elastic Behavior of Laminates 139
the transverse shear strains or shear resultants we conside r the in-plane and ﬂexural
simultaneous equations, (4.2.18), separately

NNN
···
MMM
=
AAA…BBB
. . . .
BBB…DDD

εεε
···
κκκ
,
or

N1
N2
N6
···
M1
M2
M6
=
A11A12A16…B11B12B16
A12A22A26…B12B22B26
A16A26A66…B16B26B66
. . . . . . . . . . . . . . . . . . . . . . . .
B11B12B16…D11D12D16
B12B22B26…D12D22D26
B16B26B66…D16D26D66

ε1
ε2
ε6
···
κ1
κ2
κ6
(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 jfor each lamina kusing the four
elastic moduli, EL,ET,νLT,GLT(4.1.2) and (4.1.3).
•Calculate the values of the transformed reduced stiffness Qi jfor 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 jandDi jfrom (4.2.15).
•Substitute the calculated stiffness and the applied result ant forces and moments
in (4.2.18) and calculate the midplane strains εiand curvatures κi.
•Calculate the global strains εεε(k)in each lamina using (4.2.6) and then the global
stresses σσσ(k)for each lamina kusing (4.2.10).
•Calculate the local strains εεε,(k)and the local stresses σσσ,(k)for each lamina kusing
Table 4.1
The inverted relation (4.2.18) leads to the compliance hype rmatrix for the in-plane
and ﬂexural resultants
/bracketleftbiggεεε
κκκ/bracketrightbigg
=/bracketleftbiggaaa bbb
ccc ddd/bracketrightbigg/bracketleftbiggNNN
MMM/bracketrightbigg
,/bracketleftbiggaaa bbb
ccc ddd/bracketrightbigg
=/bracketleftbiggAAA BBB
CCC DDD/bracketrightbigg−1
(4.2.19)
The compliance submatrices aaa,bbb,ccc,dddfollow from the stiffness submatrices AAA,BBB,DDD.
With
NNN=AAAεεε+BBBκκκ (4.2.20)
it follows that

140 4 Elastic Behavior of Laminate and Sandwich Composites
εεε=AAA−1(NNN−BBBκκκ) (4.2.21)
and using (4.2.13)
MMM=BBBAAA−1NNN−(BBBAAA−1BBB−DDD)κκκ (4.2.22)
The ﬁrst result is a mixed-type constitutive equation

εεε
···
MMM
=
AAA∗…BBB∗
. . . . . .
CCC∗…DDD∗

NNN
···
κκκ
,
AAA∗=AAA−1,BBB∗=−AAA−1BBB,
CCC∗=BBBAAA−1=−BBB∗T,DDD∗=DDD−BBBAAA−1BBB(4.2.23)
With
κκκ=DDD∗−1MMM−DDD∗−1CCC∗NNN (4.2.24)
it follows that
εεε=(AAA∗−BBB∗DDD∗−1CCC∗)NNN+BBB∗DDD∗−1MMM (4.2.25)
and the compliance relation has in contracted notation the f orm

εεε
···
κκκ
=
aaa…bbb
. . . .
ccc…ddd

NNN
···
MMM
,
aaa=AAA∗−BBB∗DDD∗−1CCC∗=AAA∗+BBB∗DDD∗−1BBB∗T,
bbb=BBB∗DDD∗−1,
ccc=−DDD∗−1CCC∗=DDD∗−1BBB∗T=bbbT,
ddd=DDD∗−1(4.2.26)
Equations (4.2.18) and (4.2.26) are inverse relations of th e constitutive equation for
the resultants and the strains of a laminate. The elements of the submatrices AAA,BBB,DDD
andaaa,bbb,ccc,dddare functions of the geometry, the material properties and t he structure
of a laminate and therefore averaged effective elastic lami nate moduli. The subma-
trices AAA,BBB,DDD,aaa,dddare symmetric submatrices. That is not the case for the subma tri-
cesbbbandcccbut with ccc=bbbTthe 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 s ection, all coupling effects
are present. Figure 4.13 illustrates for example the coupli ng states for the resultant
force N1and the resultant moment M1.
In the next Sect. 4.2.3 we shall see that the stacking sequenc e of a laminate in-
ﬂuences the coupling behavior of loaded laminates. In engin eering applications it is
desired to specify the stacking sequence such that a number o f coefﬁcients of the
stiffness matrix will be zero and undesirable couplings bet ween stretching, bending

4.2 Elastic Behavior of Laminates 141

N1
N2
N6
M1
M2
M6
=
A11A12A16B11B12B16
A22A26·B22B26
A66· · B66
D11D12D16
S Y M ·D22D26
· · D66

ε1
ε2
ε6
κ1
κ2
κ6

N1
M1
N1
M1A11 A12 A16
D11 D12 D16
B11 B12 B16
B11 B12 B16strain ε1 strain ε2 shear ε6
curvature κ1 curvature κ2 twisting κ6
curvature κ1 curvature κ2twisting κ6
strain ε1 strain ε2shear ε6
Fig. 4.13 Coupling of strain states: Inﬂuence of the stiffness A1j,D1jandB1j(j=1,2,6)on the
strains εjand the curvature κjof the middle surface of a general laminate loading with N1orM1.
In each case 6 deformation states of N1andM1have 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 perfor mance requirements.
Engineering composite design has continued to evolve over m any years. Most
early applications of composite materials were aimed parti cularly at weight reduc-
tion. Metals were replaced by composites with little or no em phasis placed on tailor-
ing the composite properties. Engineering design created q uasi-isotropic laminates
that largely suppressed the directional material properti es of unidirectional lami-
nae and made the laminate material response similar to that o f 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 mi d-plane. Quasi-isotropic
laminates have elastic properties that are independent of t he direction in the plane
of the laminate, like traditional isotropic engineering ma terials. Therefore, quasi-
isotropic laminates were in the ﬁrst applications of compos ites 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 isot ropic metal with a
similar stiffness laminate that was lighter and probably st ronger. 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 ρand the speciﬁc stiffness E/ρdiffer signiﬁcantly. The speciﬁc stiffness
of graphite/epoxy laminate can be twice that of aluminium. S uch applications of
quasi-isotropic laminates required a minimal amount of red esign effort and there-
fore minimal changes in structural conﬁguration.
By the time the number of design engineers which are trained i n composite ma-
terials increased and the tailoring of material properties gained more acceptance.
To maximize the utility of the non-isotropic nature of lamin ates, the inﬂuence of
the stacking sequence on the structural behavior must be inv estigated in detail and
optimized. Particularly the coupling effects of in-plane a nd 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 eng ineering design of
laminated structures will be introduced. Quite often the de sign of laminates is done
by using laminae that have the same constituents, the same th icknesses, etc. but
have different orientations of their ﬁbre reinforcement di rection with respect to the
global reference system of the laminate and a different stac king sequence of these
layers. In other cases layers with different materials or th icknesses are bonded to a
laminate. The stacking sequence of the layers may result in r educing the coupling
of normal and shear forces, of bending and twisting moments e tc. It can simplify
the mechanical analysis but also gives desired mechanical p erformance. In the fol-
lowing, the mechanical behavior of special symmetric and un symmetric 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 thi ckness of laminae are
the same above and below the midplane, i.e. two symmetric arr anged layers to the
midplane have the same reduced stiffness matrix QQQ(k)≡QQQ(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 jof the coupling submatrix BBBare zero and there are no coupling
relations of stretching and bending
Bi j=1
2n
∑
k=1Q(k)
i j/parenleftbigg
x(k)
32
−x(k−1)
32/parenrightbigg
=1
2n
∑
k=1Q(k)
i j/parenleftig
x(k)
3+x(k−1)
3/parenrightig/parenleftig
x(k)
3−x(k−1)
3/parenrightig
=n
∑
k=1Q(k)
i jx(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 ABDABDABD -matrix of symmetric laminates is uncoupled, i.e. all terms
of the coupling submatrix [Bi j]are zero, see following equation
x3=0n≡1′
1
k′
k❄✻
x(k′)
3
x(k)
3✻x3
Fig. 4.14 Symmetric laminate with identical layers kandk′opposite to the middle surface ( h(k)=
h(k′),QQQ(k)=QQQ(k′))

144 4 Elastic Behavior of Laminate and Sandwich Composites

N1
N2
N6
. .
M1
M2
M6
=
A11A12A16… 0 0 0
A12A22A26… 0 0 0
A16A26A66… 0 0 0
. . . . . . . . . . . . . . . . . . . . . . . .
0 0 0…D11D12D16
0 0 0…D12D22D26
0 0 0…D16D26D66

ε1
ε2
ε6
. .
κ1
κ2
κ6
(4.2.28)
The extensional submatrix AAAand the bending submatrix DDDare in the case of sym-
metric angle ply laminates fully populated and we have in-pl ane normal and shear
strain and out-of-plane bending and torsion couplings. Sin ce the coupling submatrix
BBBis zero the elastic behavior of symmetric laminates is simpl er to analyze than that
of general laminates and symmetric laminates have no tenden cy to warp as a result
of thermal contractions induced during the composite proce ssing. 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=1Q(k)
i jh(k)
=⇒A11=A22,A16=A26=0,
Di j=n
∑
k=1Q(k)
i jh(k)/parenleftigg
x(k)
32+h(k)2
12/parenrightigg
=⇒D11=D22,D16=D26=0,(4.2.29)

4.2 Elastic Behavior of Laminates 145

N1
N2
N6
. .
M1
M2
M6
=
A11A120… 0 0 0
A12A110… 0 0 0
0 0 A66… 0 0 0
. . . . . . . . . . . . . . . . . . . . . . . .
0 0 0…D11D120
0 0 0…D12D110
0 0 0… 0 0 D66

ε1
ε2
ε6
. .
κ1
κ2
κ6
(4.2.30)
This type of symmetric laminates has no stretching-shearin g or bending-torsion
coupling.
•Symmetric cross-ply laminate
A laminate is called a cross-ply laminate or a laminate with s pecially 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)
12ν(k)
21,Q(k)
22≡Q(k′)
22=E(k)
2
1−ν(k)
12ν(k)
21,
Q(k)
12≡Q(k′)
12=ν(k)
21E(k)
1
1−ν(k)
12ν(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 isot ropic layers, but now
A11/ne}ationslash=A22andD11/ne}ationslash=D22, i.e. the laminate has an orthotropic structure

N1
N2
N6
···
M1
M2
M6
=
A11A120… 0 0 0
A12A220… 0 0 0
0 0 A66… 0 0 0
. . . . . . . . . . . . . . . . . . . . . . . .
0 0 0…D11D120
0 0 0…D12D220
0 0 0… 0 0 D66

ε1
ε2
ε6
···
κ1
κ2
κ6
(4.2.32)
Figure 4.15 illustrates examples of symmetric cross-ply la minates. 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 tw isting moments.

146 4 Elastic Behavior of Laminate and Sandwich Composites
S S S
2
13
900
0000
hh
ha
123456
9000090090000900
h(1)h(2)h(3)h(4)=h(3)h(5)=h(2)h(6)=h(1)
1234 00
900
900
00 hhhhb
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 t he same thick-
ness and material where the angles of plies are +θand−θ. An example is the
8 – layer-laminate [±θ1/±θ2/]s. The stiffness coefﬁcients Ai jandDi jwill be
calculated from
Ai j=n
∑
k=1Q(k)
i jh(k),h(k)=h(k′),θ(k)=−θ(k′)
=⇒A16=A26=0,
Di j=1
3n
∑
k=1Q(k)
i j/parenleftbigg
x(k)
33−x(k−1)
33/parenrightbigg
=n
∑
k=1Q(k)
i j/parenleftigg
x(k)
32
+h(k)2
12/parenrightigg
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
=
A11A120… 0 0 0
A12A220… 0 0 0
0 0 A66… 0 0 0
. . . . . . . . . . . . . . . . . . . . . . . .
0 0 0…D11D12D16
0 0 0…D12D22D26
0 0 0…D16D26D66

ε1
ε2
ε6
. .
κ1
κ2
κ6
(4.2.34)
The fact that the in-plane shear coupling stiffness A16andA26are zero is a deﬁn-
ing characteristic of a balanced laminate. In general the be nding/twisting cou-
pling stiffness D16andD26are 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 BBB-matrix are identical to zero and the full (6×6)ABDABDABD – matrix
decouples into two (3×3)matrices, namely
NNN=AAAεεε,MMM=DDDκκκ (4.2.35)
Therefore also the inverse relations degenerates from (6×6)into two(3×3)rela-
tions
εεε=aaaNNN,κκκ=dddMMM (4.2.36)
In these matrix equations aaais the inverse of AAAanddddthe inverse of DDD
a11=A22A66−A2
26
Det[AAA], d11=D22D66−D2
26
Det[DDD],
a12=A26A16−A12A66
Det[AAA],d12=D26D16−D12D66
Det[DDD],
a16=A12A26−A22A16
Det[AAA],d16=D12D26−D22D26
Det[DDD],
a22=A11A66−A2
16
Det[AAA], d22=D11D66−D2
16
Det[DDD],
a26=A12A16−A11A26
Det[AAA],d26=D12D16−D11D26
Det[DDD],
a66=A11A22−A2
12
Det[AAA], d66=D11D22−D2
12
Det[DDD](4.2.37)
with
Det[XXX] =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=νA11, A66=Eh
2(1+ν),
D11=Eh3
12(1−ν2),D12=νD11, D66=Eh3
24(1+ν)
with
h=n
∑
k=1h(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
Theai jare identical to cross-ply layers. The di jare identical to the general sym-
metric case.
4.2.3.2 Antisymmetric Laminates
A laminate is called antisymmetric if the material and thick ness of the laminae are
the same above and below the midplane but the angle orientati ons 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)andx(k′)=−x(k)having the
same thickness h(k)=h(k′)and antisymmetric orientations θ(k)andθ(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 sti ffness 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
=
A11A12 0…B11 0 0
A12A11 0… 0−B110
0 0 A66… 0 0 0
. . . . . . . . . . . . . . . . . . . . . . . . . . .
B11 0 0…D11D12 0
0−B110…D12D11 0
0 0 0… 0 0 D66

ε1
ε2
ε6
. .
κ1
κ2
κ6
(4.2.39)
The constitutive equation (4.2.39) shows that antisymmetr ic cross-ply laminates
only have a tension/bending coupling. It is important to not e that the coupling
coefﬁcient B11approaches zero as the number of plies increases for a consta nt
laminate thickness since it is inversely proportional to th e total number of layers.
•Antisymmetric balanced laminate
Antisymmetric balanced laminates consist of pairs of lamin ae(k)and(k′)at a
distance ¯ x(k)and ¯x(k′)=−¯x(k)with the same material and thickness but orien-
tations θ(k)andθ(k′)=−θ(k). Examples of these laminates are [θ1/−θ1],
[θ1/θ2/−θ2/−θ1], etc. As for all balanced laminates A16=A26=0 and with
Di j=1
3n
∑
k=1Q(k)
i j/parenleftbigg
x(k)
33
−x(k−1)
33/parenrightbigg
(4.2.40)
and /parenleftbigg
x(k)
33−x(k−1)
33/parenrightbigg
=/parenleftbigg
x(k′)
33
−x(k′−1)
33/parenrightbigg
,
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 bendin g/twisting cou-
pling but a coupling of stretching/twisting and bending/sh earing. The constitutive
equation has the following structure

150 4 Elastic Behavior of Laminate and Sandwich Composites

N1
N2
N6
···
M1
M2
M6
=
A11A120… 0 0 B16
A12A220… 0 0 B26
0 0 A66…B16B260
. . . . . . . . . . . . . . . . . . . . . . . .
0 0 B16…D11D120
0 0 B26…D12D220
B16B260… 0 0 D66

ε1
ε2
ε6
···
κ1
κ2
κ6
(4.2.41)
4.2.3.3 Stiffness Matrices for Symmetric and Unsymmetric L aminates in
Engineering Applications
Table 4.4 summarizes the stiffness matrices for symmetric a nd unsymmetric lam-
inates which are used in engineering applications. Symmetr ic laminates avoid the
stretching/bending coupling. But certain applications re quire the use of nonsym-
metric laminates. If possible symmetric balanced laminate s should be used. The
bending and shearing couplings are eliminated and one can sh ow that for symmet-
ric laminates with a constant total thickness hthe values of the bending or ﬂexural
stiffness D16andD26decrease with an increasing number of layers and approach
zero for k−→ ∞. If the stiffness Ai j,Bi jandDi jare calculated, the compliances
ai j,bi j,ci j=bT
i j,di jfollow from (4.2.26) or for symmetric laminates from (4.2.3 7).
The experimental identiﬁcation of the compliances is simpl er than for the stiffness
parameters.
The coupling stiffness Bi jandA16,A26,D16andD26complicate the analysis of
laminates. To minimize coupling effects symmetric balance d laminates should be
created with a ﬁne lamina distribution. Then all Bi jand the A16,A26are identical
to zero and the D16,D26couplings 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 ba lanced lay-up with a ﬁne
lamina interdispersion in order to eliminate in-plane and o ut-of-plane coupling and
the in-plane tension/shearing coupling and to minimize tor sion coupling.
There is furthermore a special class of quasi-isotropic lam inates. The layers of
the laminate can be arranged in such a way that the laminate wi ll behave as an
isotropic layer under in-plane loading. Actually, such lam inates are not isotropic,
because under transverse loading normal to the laminate pla ne and under interlami-
nar shear their behavior is different from real isotropic la yer. That is why one use the
notation quasi-isotropic layer. Because all quasi-isotro pic laminates are symmetric
and balanced the shear coupling coefﬁcients A16,A26are 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 laminate s
Symmetric laminate Unsymmetric laminate
Isotropic layers Balanced laminate
A11A120 0 0 0
A12A110 0 0 0
0 0 A660 0 0
0 0 0 D11D120
0 0 0 D12D110
0 0 0 0 0 D66A11A120B11B12B16
A12A220B12B22B26
0 0 A66B16B26B66
B11B12B16D11D12D16
B12B22B26D12D22D26
B16B26B66D16D26D66
Eq. (4.2.30)
Cross-ply laminate Antimetric balanced laminate
A11A120 0 0 0
A12A220 0 0 0
0 0 A660 0 0
0 0 0 D11D120
0 0 0 D12D220
0 0 0 0 0 D66A11A120 0 0 B16
A12A220 0 0 B26
0 0 A66B16B260
0 0 B16D11D120
0 0 B26D12D220
B16B260 0 0 D66
Eq. (4.2.32) Eq. (4.2.41)
Balanced laminate Cross-ply
A11A120 0 0 0
A12A220 0 0 0
0 0 A660 0 0
0 0 0 D11D12D16
0 0 0 D12D22D26
0 0 0 D16D26D66A11A12 0B11 0 0
A12A11 0 0−B110
0 0 A660 0 0
B11 0 0 D11D12 0
0−B110D12D11 0
0 0 0 0 0 D66
Eq. (4.2.34) Eq. (4.2.39)
Angle-ply laminate Cross-ply
(approximate solution k→∞)
A11A12A160 0 0
A12A22A260 0 0
A16A26A660 0 0
0 0 0 D11D12D16
0 0 0 D12D22D26
0 0 0 D16D26D66A11A120 0 0 0
A12A220 0 0 0
0 0 A660 0 0
0 0 0 D11D120
0 0 0 D12D110
0 0 0 0 0 D66
Eq. (4.2.28)
/bracketleftbigg
0/π
n/2π
n/…/(n−1)π
n/bracketrightbigg
S
or /bracketleftbiggπ
n/2π
n/…/ π/bracketrightbigg
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 s pecial laminae
stacking sequences which are used in laminate design, we hav e considered the fol-
lowing classiﬁcation:
1.General laminates
The stacking sequence, the thickness, the material and the ﬁ bre orientations of
all laminae is quite general. All extensional stiffness Bi jare 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 a n equal distance and
with identical thickness, material and ﬁbre orientation. A ll 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 a n equal distance,
with identical thickness and material, but opposite ﬁbre or ientation. The stiffness
A16,A26,D16andD26are 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 l aminate, 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 t he 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 lam inate, having its
balanced ±pairs of layers symmetrically situated to the middle surfac e.
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 respec t 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/2π
n/…/ π]S
the in-plane stiffness are identical in all directions. Bec ause all these quasi-
isotropic laminates are also balanced we have A11=A22=const,A12=const,
A16=A26=0,Bi j≡0,Di j/ne}ationslash=0
7.Laminates with isotropic layers
If isotropic layers of possible different materials proper ties and thicknesses are
arranged symmetrically to the middle surface the laminate i s 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). aaa,bbb,ccc=bbbT,dddare
the extensional compliance matrix, coupling compliance ma trix and bending com-
pliance matrix, respectively. For a symmetric laminate BBB=000 and it can be shown
thataaa=AAA−1andddd=DDD−1. The in-plane and the ﬂexural compliance matrices aaaand
dddare uncoupled but generally fully populated
εεε=aaaNNN,κκκ=dddMMM (4.2.42)
Equations (4.2.42) lead to effective engineering moduli fo r symmetric laminates.
1. Effective in-plane engineering moduli EN
1,EN
2,GN
12,νN
12:
Substitute N1/ne}ationslash=0,N2=N6=0 inεεε=aaaNNNas

ε1
ε2
ε6
=
a11a12a16
a12a22a26
a16a26a66

N1
0
0
 (4.2.43)
which gives
ε1=a11N1
and the effective longitudinal modulus EN
1is
EN
1≡σ1
ε1=N1/h
a11N1=1
ha11(4.2.44)
In an analogous manner with N1=0,N2/ne}ationslash=0,N6=0 orN1=N2=0,N6/ne}ationslash=0,
the effective transverse modulus EN
2or the effective shear modulus GN
12are
EN
2≡σ2
ε2=N2/h
a22N2=1
ha22, (4.2.45)

154 4 Elastic Behavior of Laminate and Sandwich Composites
GN
12≡σ6
ε6=N6/h
a66N6=1
ha66(4.2.46)
The effective in-plane Poisson’s ratio νN
12can be derived in the following way.
With N1/ne}ationslash=0,N2=N6=0 (4.2.43) yields ε2=a12N1,ε1=a11N1andν12is de-
ﬁned as
νN
12=−ε2
ε1=−a12N1
a11N1=−a12
a11(4.2.47)
The Poisson’s ratio νN
21can be derived directly by substituting N1=N6=0,
N2/ne}ationslash=0 in (4.2.42) and deﬁne νN
21=−ε1/ε2or by using the reciprocal relationship
νN
12/EN
1=νN
21/EN
2. In both cases νN
21is given as
νN
21=−a12
a22(4.2.48)
The effective in-plane engineering moduli can be also formu lated in terms of the
elements of the AAA-matrix
EN
1=A11A22−A2
12
A22h,EN
2=A11A22−A2
12
A11h,GN
12=A66
h,
νN
12=A12
A22,νN
21=A12
A11(4.2.49)
2. Effective ﬂexural engineering moduli EM
1,EM
2,GM
12,νM
12,νM
21:
To deﬁne the effective ﬂexural moduli we start with κκκ=dddMMM. Apply M1/ne}ationslash=0,
M2=0,M6=0 and substitute in the ﬂexural compliance relation to give
κ1=d11M1=M1
EM
1I,I=h3
12
and the effective ﬂexural longitudinal modulus EM
1is
EM
1=12M1
κ1h3=12
h3d11(4.2.50)
Similarly, one can show that the other ﬂexural elastic modul i are given by
EM
2=12
h3d22,νM
12=−d12
d11,GM
12=12
h3d66,νM
21=−d12
d22(4.2.51)
Flexural Poisson’s ratios also have a reciprocal relations hip
νM
12
EM
1=νM
21
EM
2(4.2.52)
In terms of the elements of the DDDmatrix 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,νM
12=D12
D22,νM
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 ef fective 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 functi ons of x3, and the through-
the-thickness stresses, which are calculated approximate ly by integration of the
equilibrium conditions. Taking into account the assumptio ns of macro-mechanical
modelling of laminates the strains ε1,ε2,ε6vary linearly across the thickness of the
laminate
εεε(x1,x2,x3)=εεε(x1,x2)+x3κκκ(x1,x2),h=n
∑
k=1h(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)
εεε′(k)=TTTεεεε(k),x(k−1)
3≤x3≤x(k)(4.2.55)
If the strains are known at any point along the thickness of th e laminate, the stress-
strain relation (Table 4.2) calculates the global stress in each lamina
σσσ(k)=QQQ(k)εεε(k)=QQQ(k)(εεε+x3κκκ),x(k−1)
3≤x3≤x(k)(4.2.56)
By applying the transformation equation for the stress vect or (Table 4.1) the stresses
expressed in the principal material axes can be calculated
σσσ′(k)=TTTσσσσ(k)(4.2.57)
Starting from the strains εεε′(k), the stresses in the kth layer are expressed as follows
σσσ′(k)=QQQ′(k)εεε′(k)(4.2.58)
From (4.2.56), the stresses vary linearly through the thick ness of each lamina and
may jump from lamina to lamina since the reduced stiffness ma trixQQQ(k)changes
from ply to ply since QQQ(k)depends on the material and orientation of the lamina (k).
Figure 4.16 illustrates qualitatively the stress jumps of t he membrane stresses σσσ(k)
M

156 4 Elastic Behavior of Laminate and Sandwich Composites
✻
❄hQ(3)
Q(2)
Q(1)h(1)
h(2)
h(3)❄❄❄✻
✻
✻σ(k)
6M=Q(k)
66ε6σ(k)
2M=Q(k)
22ε2σ(k)
1M=Q(k)
11ε1
σ(k)
6B=Q(k)
66×3κ6σ(k)
2B=Q(k)
22×3κ2σ(k)
1B=Q(k)
11×3κ1
σ(k)
6=σ(k)
6M+σ(k)
6Bσ(k)
2=σ(k)
2M+σ(k)
2Bσ(k)
1=σ(k)
1M+σ(k)
1B
✟✟✟✟✟
✟✟✟✟✟

Fig. 4.16 Qualitatively variation of the in-plane membrane stresses σiM, the bending stresses
σiBand the total stress σithrough 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 NNNand are constant through each lamina
and the bending/torsion stresses σσσ(k)
Bfollowing from the moment resultants MMMand
vary linearly through each ply thickness. The transverse sh ear stresses σ4,σ5follow
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 fo r 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/braceleftigg x(k)
3/integraldisplay
x(k−1)
3/bracketleftbigg
Q(k)
11∂
∂x1(ε1+x3κ1)+Q(k)
12∂
∂x1(ε2+x3κ2)
+Q(k)
16∂
∂x1(ε6+x3κ6)+Q(k)
61∂
∂x2(ε1+x3κ1)
+Q(k)
62∂
∂x2(ε2+x3κ2)+Q(k)
66∂
∂x2(ε6+x3κ6)/bracketrightbigg
dx3/bracerightigg
−x3/integraldisplay
x(m−1)
3/bracketleftbigg
Q(m)
11∂
∂x1(ε1+x3κ1)+Q(m)
12∂
∂x1(ε2+x3κ2)
+Q(m)
16∂
∂x1(ε6+x3κ6)+Q(m)
61∂
∂x2(ε1+x3κ1)
+Q(m)
62∂
∂x2(ε2+x3κ2)+Q(m)
66∂
∂x2(ε6+x3κ6)/bracketrightbigg
dx3, (4.2.59)
σ(m)
4(x3) =−m−1
∑
k=1/braceleftigg x(k)
3/integraldisplay
x(k−1)
3/bracketleftbigg
Q(k)
61∂
∂x1(ε1+x3κ1)+Q(k)
62∂
∂x1(ε2+x3κ2)

4.2 Elastic Behavior of Laminates 157
+Q(k)
66∂
∂x1(ε6+x3κ6)+Q(k)
21∂
∂x2(ε1+x3κ1)
+Q(k)
22∂
∂x2(ε2+x3κ2)dx3+Q(k)
26∂
∂x2(ε6+x3κ6)dx3/bracerightigg
(4.2.60)
−x3/integraldisplay
x(m−1)
3/bracketleftbigg
Q(m
61∂
∂x1(ε1+x3κ1)+Q(m)
62∂
∂x1(ε2+x3κ2)
+Q(m)
66∂
∂x1(ε6+x3κ6)+Q(m)
21∂
∂x2(ε1+x3κ1)
+Q(m)
22∂
∂x2(ε2+x3κ2)+Q(m)
26∂
∂x2(ε6+x3κ6)/bracketrightbigg
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/integraldisplay
x(k−1)
3QQQ(k)dx3=QQQ(k)/parenleftig
x(k)
3−x(k−1)
3/parenrightig
=QQQ(k)h(k),
x(k)
3/integraldisplay
x(k−1)
3QQQ(k)x3dx3=QQQ(k)1
2/parenleftbigg
x(k)
32
−x(k−1)
32/parenrightbigg
=QQQ(k)h(k)x(k)
3=QQQ(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/parenleftig
x3=x(m)
3/parenrightig
σ(m)
4/parenleftig
x3=x(m)
3/parenrightig
=−m
∑
k=1
FFF(k)
1TFFF(k)
6T
FFF(k)
6TFFF(k)
2T
/bracketleftiggηηη1
ηηη2/bracketrightigg
(4.2.62)
with
FFF(k)
1T= [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)],
FFF(k)
2T= [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)],
FFF(k)
6T= [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
ηηηT
1=/bracketleftbigg∂ε1
∂x1∂ε2
∂x1∂ε6
∂x1∂κ1
∂x1∂κ2
∂x1∂κ6
∂x1/bracketrightbigg
,ηηηT
2=/bracketleftbigg∂ε1
∂x2∂ε2
∂x2∂ε6
∂x2∂κ1
∂x2∂κ2
∂x2∂κ6
∂x2/bracketrightbigg
The transverse shear stresses only satisfy the equilibrium conditions but violate the
other fundamental equations of anisotropic elasticity. Th ey 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 str esses 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 matri x of an UD-lamina are
smeared into a single equivalent homogeneous material and t hat the smeared ele-
ments are free of any stresses if temperature is changed. Whe n one considers an
unsmeared material and deals with the individual ﬁbres and t he surrounding matrix,
a temperature change can create signiﬁcant stresses in the ﬁ bre and matrix. When
such selfbalanced stresses are smeared over a volume elemen t, the net result is zero.
However, in a laminate with various laminae of different mat erials 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) αth,αmoare the thermal and moisture
expansion coefﬁcients, Tis the temperature change and M∗the weight of moisture
absorption per unit weight. In the following equations, TandM∗are independent of
thex3-coordinate, i.e. they are constant not only through the thi ckness 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 gradient s inx3-direction are seldom
taken into account and the temperature change Tis then approximately independent
ofx3. Analogous considerations are valid for changes in moistur e.
For a single layer in off-axis coordinates, the hygrotherma l strains and stresses
are given by
εεε=SSSσσσ+αααthT+αααmoM∗,σσσ=QQQ(εεε−αααthT−αααmoM∗) (4.2.63)
or substituting εεε=εεε+x3κκκ
σσσ=QQQ(εεε+x3κκκ−αααthT−αααmoM∗) (4.2.64)
The deﬁnitions of the force and moment resultants NNNandMMM

4.2 Elastic Behavior of Laminates 159
NNN=/integraldisplay
(h)σσσdx3=/integraldisplay
(h)QQQ(εεε+x3κκκ−αααthT−αααmoM∗)dx3,
MMM=/integraldisplay
(h)σσσx3dx3=/integraldisplay
(h)QQQ(εεε+x3κκκ−αααthT−αααmoM∗)x3dx3(4.2.65)
yield the equations
NNN=AAAεεε+BBBκκκ−NNNth−NNNmo,
MMM=BBBεεε+DDDκκκ−MMMth−MMMmo,(4.2.66)
AAA=QQQh,BBB=000,DDD=QQQh3
12
BBB=000 follows from the symmetry of a single layer to its midplane. NNNth,NNNmo,
MMMth,MMMmoare ﬁctitious hygrothermal resultants which are deﬁned in ( 4.2.67). If T
andM∗are independent of x3one can introduce unit thermal and unit moisture
stress resultants ˆNth,ˆMth,ˆNmo,ˆMmo, i.e. resultants per unit temperature or moisture
change
NNNth=/integraldisplay
(h)QQQαααthTdx3=QQQαααthTh=ˆNthT,
MMMth=/integraldisplay
(h)QQQαααthT x3dx3=1
2QQQαααthT h2=ˆMthT,
NNNmo=/integraldisplay
(h)QQQαααmoM∗dx3=QQQαααmoM∗h=ˆNmoM∗,
MMMmo=/integraldisplay
(h)QQQαααmoM∗x3dx3=1
2QQQαααmoM∗h2=ˆMmoM∗.(4.2.67)
NthandNmohave the units of the force resultant, namely N/m, and MthandMmo
the units of the moment resultants, namely Nm/m. The integra l form of the resultant
deﬁnitions makes these deﬁnitions quite general, i.e. if TorM∗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 ˜NNN,˜MMM, equal to the respective sums of
their mechanical and hygrothermal components
˜NNN=NNN+NNNth+NNNmo,˜MMM=MMM+MMMth+MMMmo, (4.2.68)
the extended hygrothermal constitutive equation for a lami na can be written

˜NNN
···
˜MMM
=
AAA… 000
. . . .
000…DDD

εεε
···
κκκ
 (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 for ces and moments are
added to the mechanically applied forces and moments. The in version of (4.2.69)
yields the compliance relation

εεε
···
κκκ
=
aaa… 000
. . . .
000…ddd

˜NNN
···
˜MMM
,aaa=AAA−1,ddd=DDD−1(4.2.70)
The values of the stiffness Ai j,Di jand compliances ai j,di jare the same as for pure
mechanical loading (Table 4.3) and unit stress resultants ˆNth,ˆMth,ˆNmo,ˆMmoare
ˆNth
1= (Q11αth
1+Q12αth
2+Q16αth
6)h,
ˆNth
2= (Q12αth
1+Q22αth
2+Q26αth
6)h,
ˆNth
6= (Q16αth
1+Q26αth
2+Q66αth
6)h,
ˆMth
1= (Q11αth
1+Q12αth
2+Q16αth
6)1
2h2,
ˆMth
2= (Q12αth
1+Q22αth
2+Q26αth
6)1
2h2,
ˆMth
6= (Q16αth
1+Q26αth
2+Q66αth
6)1
2h2(4.2.71)
and analogous for unit moisture stress resultant with αmo
i,i=1,2,6. In the more
general case the integral deﬁnitions have to used.
When a laminate is subjected to mechanical and hygrothermal loading, a lamina k
within the laminate is under a state of stress σσσ(k)and strain εεε(k). The hygrothermoe-
lastic superposition principle shows that the strains εεε(k)in the lamina kare equal
to the sum of the strains produced by the existing stresses an d the free, i.e. unre-
strained, hygrothermal strains and the stresses σσσ(k)follow by inversion
εεε(k)=SSS(k)σσσ(k)+αααth(k)T+αααmo(k)M∗,
σσσ(k)=QQQ(k)(εεε(k)−αααth(k)T−αααmo(k)M∗)
=QQQ(k)(εεε(k)+x3κκκ(k)−αααth(k)T−αααmo(k)M∗)(4.2.72)
When in the lamina kall strains are restrained, then εεε(k)=000 and the hygrothermal
stresses are
σσσ(k)=QQQ(k)/parenleftig
−αααth(k)T−αααmo(k)M∗/parenrightig
(4.2.73)
Integration of the stresses σσσ(k)and the stresses σσσ(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
NNN=n
∑
k=1σσσ(k)h(k)
=
n
∑
k=1QQQ(k)x(k)
3/integraldisplay
x(k−1)
3dx3
εεε+
n
∑
k=1QQQ(k)x(k)
3/integraldisplay
x(k−1)
3x3dx3
κκκ
+
n
∑
k=1QQQ(k)αααth(k)Tx(k)
3/integraldisplay
x(k−1)
3dx3
+
n
∑
k=1QQQ(k)αααmo(k)M∗x(k)
3/integraldisplay
x(k−1)
3dx3
,
MMM=n
∑
k=1σσσ(k)h(k)x(k)
3
=
n
∑
k=1QQQ(k)x(k)
3/integraldisplay
x(k−1)
3x3dx3
εεε+
n
∑
k=1QQQ(k)x(k)
3/integraldisplay
x(k−1)
3×2
3dx3
κκκ
+
n
∑
k=1QQQ(k)αααth(k)Tx(k)
3/integraldisplay
x(k−1)
3x3dx3
+
n
∑
k=1QQQ(k)αααmo(k)M∗x(k)
3/integraldisplay
x(k−1)
3x3dx3
(4.2.74)
With the stiffness matrices AAA,BBBandDDD, Eqs. (4.2.74), can be rewritten in a brief
matrix form
NNN=AAAεεε+BBBκκκ−NNNth−NNNmo,
MMM=BBBεεε+DDDκκκ−MMMth−MMMmo (4.2.75)
The ﬁctitious hygrothermal resultants are given by
NNNth=Tn
∑
k=1QQQ(k)αααth(k)h(k)=Tn
∑
k=1ˆN(k)th,
NNNmo=M∗n
∑
k=1QQQ(k)αααmo(k)h(k)=M∗n
∑
k=1ˆN(k)mo,
MMMth=1
2Tn
∑
k=1QQQ(k)αααth(k)/parenleftbigg
x(k)
32−x(k−1)
32/parenrightbigg
(4.2.76)
=Tn
∑
k=1QQQ(k)αααth(k)x(k)
3h(k)=Tn
∑
k=1ˆM(k)th,
MMMmo=1
2M∗n
∑
k=1QQQ(k)αααmo(k)/parenleftbigg
x(k)
32
−x(k−1)
32/parenrightbigg
=M∗n
∑
k=1QQQ(k)αααmo(k)x(k)
3h(k)=M∗n
∑
k=1ˆM(k)mo

162 4 Elastic Behavior of Laminate and Sandwich Composites
By analogy to the single layer one can introduce total force a nd moment resultants
˜NNNand ˜MMMthe stiffness and compliance equations expanded to the hygr othermal com-
ponents
NNN+NNNth+NNNmo=˜NNN,MMM+MMMth+MMMmo=˜MMM,

˜NNN
···
˜MMM
=
AAA…BBB
. . . .
BBB…DDD

εεε
···
κκκ
,
εεε
···
κκκ
=
aaa…bbb
. . . . .
bbbT…ddd

˜NNN
···
˜MMM
,
aaa=AAA∗−BBB∗DDD∗−1CCC∗,AAA∗=AAA−1,
bbb=BBB∗DDD∗−1, BBB∗=−AAA−1BBB,
ddd=DDD∗−1, DDD∗=DDD−BBBAAA−1BBB(4.2.77)
The coupling effects discussed in Sects. 4.2.2 and 4.2.3 sta y 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 t hermal or moisture force
or moment resultant coefﬁcients may be zero. For temperatur e 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 resultan ts can be
considered, J=th,mo
Symmetric laminates
ˆNJ
1/ne}ationslash=0, ˆMJ
1=0,
ˆNJ
2/ne}ationslash=0, ˆMJ
2=0,
ˆNJ
6/ne}ationslash=0, ˆMJ
6=0
Balanced laminatesˆNJ
1/ne}ationslash=0, ˆMJ
1/ne}ationslash=0,
ˆNJ
2/ne}ationslash=0, ˆMJ
2/ne}ationslash=0,
ˆNJ
6=0, ˆMJ
6/ne}ationslash=0
Symmetric balanced laminates
ˆNJ
1/ne}ationslash=0, ˆMJ
1=0,
ˆNJ
2/ne}ationslash=0, ˆMJ
2=0,
ˆNJ
6=0, ˆMJ
6=0
Cross-ply laminates
ˆNJ
1/ne}ationslash=0, ˆMJ
1/ne}ationslash=0,
ˆNJ
2/ne}ationslash=0, ˆMJ
2/ne}ationslash=0,
ˆNJ
6=0, ˆMJ
6=0

4.2 Elastic Behavior of Laminates 163
Symmetric cross-ply laminates
ˆNJ
1/ne}ationslash=0, ˆMJ
1=0,
ˆNJ
2/ne}ationslash=0, ˆMJ
2=0,
ˆNJ
6=0, ˆMJ
6=0
Summarizing the hygrothermal effects one can see that if bot h mechanical and
hygrothermal loads are applied the mechanical and ﬁctitiou s 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 th e 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 consider ed as
an equivalent homogeneous anisotropic plate in plane stres s state by introducing
average stress σσσ=NNN/handNNN=AAAεεε. Calculate the effective moduli for general
symmetric laminates and for symmetric cross-ply laminates .
Solution 4.7. Equations (4.2.37) yields εεε=aaaNNN,aaa=AAA−1. The components of the
inverse matrix aaaare
a11=(A22A66−A2
26)/∆,a12=(A16A26−A12A66)/∆,
a22=(A11A66−A2
16)/∆,a16=(A12A26−A22A16)/∆,
a26=(A12A16−A11A26)/∆,a66=(A11A22−A2
12)/∆,
∆=Det(Ai j)=A11/vextendsingle/vextendsingle/vextendsingle/vextendsingleA22A26
A26A66/vextendsingle/vextendsingle/vextendsingle/vextendsingle−A12/vextendsingle/vextendsingle/vextendsingle/vextendsingleA12A26
A16A66/vextendsingle/vextendsingle/vextendsingle/vextendsingle+A16/vextendsingle/vextendsingle/vextendsingle/vextendsingleA12A22
A16A26/vextendsingle/vextendsingle/vextendsingle/vextendsingle
The comparison εεε=aaaNNN=haaaσσσwith (4.1.19) leads to
E1=1/ha11,E2=1/ha22,E6=1/ha66,
ν12=−a12/a11,ν21=−a12/a22,ν16=a16/a11,
ν61=a16/a66,ν26=a26/a22,ν62=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 expr essed in
terms of the in-plane stiffness Ai j. With Det (Ai j)=A11A22A66−A2
12A66follow the
effective moduli
E1=(A11A22−A2
12)/hA22,ν12=A12/A22,
E2=(A11A22−A2
12)/hA11,ν21=A12/A11,
G12≡E6=A66/h,ν16=ν61=ν26=ν62=0
Note that these simpliﬁed formulae are not only valid for sym metric 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′
12ν′
21)=140,905 GPa,
Q′
22=E′
2(1−ν′
12ν′
21)=10,065 GPa,
Q′
12=E′
2ν′
12/(1−ν′
12ν′
21)=3,019 GPa,
Q′
66=E′
6=7 GPa
2. Calculation of the reduced stiffness in the laminae (Tabl e 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=1Q(k)
i jh(k)=24
∑
n=1Q(k)
i jh(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,
ν12=ν21=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 σandσ′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
QQQ[00]=
140,9 3,02 0
3,02 10,06 0
0 0 7 ,0
109Pa,
QQQ[900]=
10,06 3,02 0
3,02 140,9 0
0 0 7 ,0
109Pa,
QQQ[±450]=
46,25 32,25±32,71
32,25 46,25±32,71
±32,71±32,71 36,23
109Pa
The axial stiffness matrix AAAis also calculated
AAA=
48,70 14,11 0
14,11 48,70 0
0 0 17 ,29
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/(A11A22−A2
12)=−0,00645 10−6m/N
follows the inverse matrix aaa=AAA−1
aaa=
22,41−6,49 0
−6,49 22,41 0
0 0 57 ,84
10−9m/N
The strains are with (4.2.42)
εεε=aaaNNN=⇒
ε1
ε2
ε6
=aaa
N1
0
0
=
22,41
−6,49
0
10−9(m/N) N1
N1is given in N/m, i.e. εiare dimensionless.
Now the laminar stresses are (Table 4.2)

σ1
σ2
σ6

[00]=QQQ[00]εεε=
3138
2,4
0
N1[N/m2],

σ1
σ2
σ6

[900]=QQQ[900]εεε=
205,8
−846,8
0
N1[N/m2],

166 4 Elastic Behavior of Laminate and Sandwich Composites

σ1
σ2
σ6

[±450]=QQQ[±450]εεε=
827,2
422,6
±520,7
N1[N/m2]
The stresses jump from lamina to lamina. Verify that the resu ltant force N2=0.
The stress components in reference to the principal materia l axes follow with the
transformation rule (Table 4.1)

σ′
1
σ′
2
σ′
6
=
c2s22sc
s2c2−2sc
−cs cs c2−s2

σ1
σ2
σ6
,

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],

σ1
σ2
σ6

[±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 jof the compliance matrix SSS′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 jof the reduced stiffness matrix QQQ′are
Q′
11=E′
1/(1−ν′
12ν′
21)=181,8 109Pa,
Q′
12=ν′
12E′
2/(1−ν′
12ν′
21)=2,897 109Pa,
Q′
22=E′
2/(1−ν′
12ν′
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 obtai ned by inverting the
compliance matrix, i.e. QQQ′=S′S′S′−1. Now the transformed reduced stiffness ma-
trices QQQ[00],QQQ[300],QQQ[−450]have to be calculated with the help of Table 4.2 tak-
ing into account that c=cos00=1,cos300=0,8660,cos(−450) =0,7071 and
s=sin00=0,sin300=0,5,sin(−450)=−0,7071
QQQ[00]=
181,8 2,897 0
2,897 10,35 0
0 0 7 ,17
109Pa,
QQQ[300]=
109,4 32,46 54,19
32,46 23,65 20,05
54,19 20,05 36,74
109Pa,
QQQ[−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 (4.2.1 5) the
extensional stiffness matrix AAAfollows with
Ai j=3
∑
k=1Q(k)
i jh(k)=53
∑
k=1Q(k)
i jmm,
the coupling matrix BBBfollows with
Bi j=3
∑
k=1Q(k)
i jh(k)¯x(k)
3=53
∑
k=1Q(k)
i j¯x(k)
3mm
and the bending stiffness matrix Di jfollows with
Di j=1
33
∑
k=1Q(k)
i jh(k)/parenleftigg
(¯x(k)
3)2+h(k)2
12/parenrightigg
=5
33
∑
k=1Q(k)
i j/parenleftbigg25
12+(¯x(k)
3)2/parenrightbigg
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
andDi jwe 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 jmm2],
Di j=5[(25+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
AAA=
17,39 3,884 0,566
3,884 4,533−1,141
0,566−1,141 4,525
108Pa m,
BBB=
−3,129 0,986−1,072
0,986 1,158−1,072
−1,072−1,072 0,986
106Pa m2,
DDD=
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 h igh strength material
while the core is made of thick and lightweight materials, Se ct. 1.3. The motiva-
tion for sandwich structure elements is twofold. First for b eam or plate bending
the maximum normal stresses occur at the top and the bottom su rface. So it makes
sense using high-strength materials at the top and the botto m and using low and
lightweight strength materials in the middle. The strong an d stiff facings also sup-
port axial forces. Second, the bending resistance for a rect angular cross-sectional
beam or plate is proportional to the cube of the thickness. In creasing the thickness
by adding a core in the middle increases the resistance. The m aximum 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 co mposites attractive in
many applications.
The most commonly used facing materials are aluminium alloy s and ﬁbre re-
inforced plastics. Aluminium has a high speciﬁc modulus, bu t it corrodes without
treatment and can be prone to denting. Therefore ﬁbre reinfo rced laminates, such
as graphite/epoxy or glass/epoxy are becoming more popular as facing materials.
They have high speciﬁc modulus and strength and corrosion re sistance. The ﬁbre
reinforced facing can be unidirectional or woven laminae.
The most commonly used core materials are balsa wood, foam, r esins with spe-
cial ﬁllers and honeycombs (Fig. 1.3). These materials must have high compressive
and shear strength. Honeycombs can be made of plastics, pape r, card-boards, etc.
The strength and stiffness of honeycomb sandwiches depend o n the material and
the cell size and thickness. The following sections conside r 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 laye rs. The thin cover
sheets, i.e. the layers 1 and 3, are laminates of the thicknes sesh(1)for the lower skin
andh(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 sandwic hes are:
1. The thickness of the core is much greater than that of the sk ins,h(2)≫h(1),h(3)
orhc≫hf
2. The strains ε1,ε2,ε6vary linearly through the core thickness hc
εεε=εεε+x3κκκ
3. The sheets only transmit stresses σ1,σ2,σ6and the in-plane strains are uniform
through the thickness of the skins. The transverse shear str esses σ4,σ5are ne-
glected within the skin.
4. The core only transmits transverse shear stresses σ4andσ5, the stresses σ1,σ2
andσ6are neglected.
5. The strain ε3is neglected in the sheets and the core.
With these additional assumptions in the frame of linear ani sotropic elasticity, the
stresses and strains can be formulated.
Strains in the lower and upper sheets:
εεε(l)=εεε∓1
2hcκκκ,ε(l)
i=εi∓1
2hcκi,l=1,3,i=1,2,6 (4.3.1)
The transverse shear strains ε4,ε5are neglected.
Strains in the sandwich core:
εεε(2)=εεεc=εεε+x3κκκ,−hc
2≤x3≤+hc
2(4.3.2)
The transverse shear strains are, in a ﬁrst approach, indepe ndent of the coordinate
x3(4.2.14)
γγγsc=[εc
5εc
4]T(4.3.3)
We shall see in Chap. 5 that in the classical laminate theory a nd the laminate theory
including transverse shear deformations the strain vector εεεis written in an analogous
form, and only the expressions for the curvatures are modiﬁe d.
Stresses in the lower and upper sheets:
In the sheets a plane stress state exists and with assumption 3. the transverse shear
stresses σ4andσ5are neglected. These assumptions imply that for laminated s heets
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 jε(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
σc
1=σc
2=σc
6=0
and the core transmits only the transverse shear stresses
/bracketleftbiggσc
5
σc
4/bracketrightbigg
=/bracketleftbiggC55C45
C45C44/bracketrightbigg/bracketleftbiggε5
ε4/bracketrightbigg
(4.3.5)
or in matrix notation (Eq. 4.2.14)
σσσs=CCCsγγγs(4.3.6)
The coefﬁcients Cc
i jofCCCsare expressed as functions of the coefﬁcients Cc′
i jreferred
to the principal directions by the transformation equation (4.2.17). The coefﬁcients
Cc′
i jin the principal directions are themselves written as funct ions of the shear moduli
of the core (Sect. 2.1, Table 2.5), measured in principal dir ections as follows
Cc′
44=Gc
23,Cc
55=Gc′
13 (4.3.7)
For an isotropic core material a transformation is not requi red.
4.3.2 Stress Resultants and Stress Analysis
The in-plane resultants NNNfor sandwiches are deﬁned by
NNN=−1
2hc/integraldisplay
−(1
2hc+h(1))σσσdx3+1
2hc+h(3)/integraldisplay
1
2hcσσσdx3, (4.3.8)
the moment resultants by

4.3 Elastic Behavior of Sandwiches 171
MMM=−1
2hc/integraldisplay
−(1
2hc+h(1))σσσx3dx3+1
2hc+h(3)/integraldisplay
1
2hcσσσx3dx3 (4.3.9)
and the transverse shear force by
QQQs=1
2hc/integraldisplay
−1
2hcσσσsdx3 (4.3.10)
For the resultants NNNandMMMthe 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 analogo us to (4.2.16) the con-
stitutive equation
NNN
MMM
QQQs
=
AAA BBB000
CCC DDD000
000 000AAAs

εεε
κκκ
γγγs
 (4.3.11)
with the stiffness coefﬁcients
Ai j=A(1)
i j+A(3)
i j,Bi j=1
2hc/parenleftig
A(3)
i j−A(1)
i j/parenrightig
,
Ci j=C(1)
i j+C(3)
i j,Di j=1
2hc/parenleftig
C(3)
i j−C(1)
i j/parenrightig (4.3.12)
and
A(1)
i j=−1
2hc/integraldisplay
−(1
2hc+h1)Q(k)
i jdx3=n1
∑
k=1/integraldisplay
h(k)Q(k)
i jdx3=n1
∑
k=1Q(k)
i jh(k),
A(3)
i j=1
2hc+h3/integraldisplay
1
2hcQ(k)
i jdx3=n2
∑
k=1/integraldisplay
h(k)Q(k)
i jdx3=n2
∑
k=1Q(k)
i jh(k),
C(1)
i j=−1
2hc/integraldisplay
−(1
2hc+h1)Q(k)
i jx3dx3=n1
∑
k=1/integraldisplay
h(k)Q(k)
i jx3dx3=n1
∑
k=1Q(k)
i jh(k)¯x(k)
3,
C(3)
i j=1
2hc+h3/integraldisplay
1
2hcQ(k)
i jx3dx3=n2
∑
k=1/integraldisplay
h(k)Q(k)
i jx3dx3=n2
∑
k=1Q(k)
i jh(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)
n1andn2are the number of layers in the lower and the upper sheet respe ctively and
Cs
i jare the transverse shear moduli of the core. The constitutiv e equations (4.3.11)
for a sandwich composite has a form similar to the constituti ve equation for lami-
nates including transverse shear. It differs only by the ter msCi jinstead of Bi jwhich
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 jand 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, t he coefﬁcients As
i jcan
be corrected by shear correction factors ks
i jand replaced by shear constants (ksAs)i j
to improve the modelling.
In the case of symmetric sandwiches there is no coupling betw een stretching
and bending and the form of the constitutive equation is iden tical to the constitutive
equation for symmetric laminates including transverse she ar.
4.3.3 Sandwich Materials with Thick Cover Sheets
In the case of thick cover sheets it is possible to carry out th e modelling and analysis
with the help of the theory of laminates including transvers e 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. Rest ricting the modelling
to the case of symmetric sandwich composites and to the case w here the core’s
principal direction is in coincidence with the directions o f the reference coordinate
system. The elastic behavior of the composite material is ch aracterized by
•the reduced stiffness parameters Qf
i jfor the face sheets,
•the reduced stiffness parameters Qc
i jand the transverse shear moduli Cc
i jfor the
core
Application of the sandwich theory, Sect. 4.3.2, leads to th e 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 jare 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 sh ear, 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/bracketleftbigg
(hf+hc)2+1
3(hf)2/bracketrightbigg
+1
12Qc
i j(hc)3,i,j=1,2,6(4.3.18)
The shear stiffness coefﬁcients As
i jare 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=2n
∑
k=1Qf(k)
i jhf(k),
DSa
i j=hcn
∑
k=1Qf(k)
i jhf(k)¯x(k)
3,
ALa
i j=2n
∑
k=1Qf(k)
i jhf(k)+hcQc
i j,
DLa
i j=n
∑
k=1Qf(k)
i j/parenleftigg
hf(k)(¯x(k)
3)2+hf(k)3
12/parenrightigg
+Qc
i jhc3
12
The comparison of the analysis based on the sandwich or the la minate theory yields
ALa
i j=ASa
i j/parenleftigg
1+hcQc
i j
2hfQf
i j/parenrightigg
,
DLa
i j=DSa
i j/parenleftigg
1+hf
hchc+(4/3)hf
hc+hf+Qc
i j
6Qf
i j(hc)2
hf(hc+hf)/parenrightigg
,i,j=1,2,6(4.3.20)
AsLa
i j=AsSa
i j/parenleftigg
1+hfCf
i j
2hcCc
i j/parenrightigg
,i,j=4,5 (4.3.21)
Generally the core of the sandwich is less stiff than the cove r 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/parenleftbigg
1+hf
hchc+(4/3)hf
hc+hf/parenrightbigg
(4.3.22)
Equation (4.3.21) stays unchanged.
The bending stiffness Di jare modiﬁed with respect to the theory of sand-
wiches and can be evaluated by the inﬂuence of the sheet thick ness. 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 jor 1,555DSa
i j, a difference of more than 10%, 30% or
50 %.
4.4 Problems
Exercise 4.11. The reduced stiffness Qc
i jandQf
i jof 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 jif 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 differenc e
DLa
i j−DSa
i j
DSa
i j100%
of the stiffness values for DLa
i jandDSa
i jare 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 c over
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 giv en in Fig. 4.17. Cal-
PPPPPPP
✏✏✏✏✏✏✏✲
✲
✲
✛
✛
✛❄
❄
❄
❄✻✲ ✲✻
❄
❄✻
✲ ✛❄
✻
❄
✻✛✻hf
hfx3
x2
hch
bab c
x3 x3
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 s hear 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 resultant sMandQare 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/parenleftbigghc+hf
2/parenrightbigg2
+Ecbhc3
12
EfandEcare the effective Young’s moduli of the face and the core mate rials. The
ﬁrst term presents the local bending stiffness of the faces a bout their own axes, the
third term represents the bending stiffness of the core. Bot h 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
100Ecbhc3/12
Ef[bhf(hc+hf)2]/2<1
100
the ﬁrst and the third term are less than 1% of the second term a nd the bending
stiffness is approximately
D≈Efbhf(hc+hf)2/2
The bending stress distributions through the faces and the c ore are
σf=MEf
Dx3≈±MEf
Dhc+hf
2,σc=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/bracketleftigg
Efhf(hc+hf)
2+Ec
2/parenleftigg
hc2
4−x2
3/parenrightigg/bracketrightigg
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 cons idered constant
τ≈Q
DEfhf(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 ill ustrated. 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 commonl y used structural the-
ories for the modelling and analysis of laminates and sandwi ches is considered,
namely the classical laminate theory and the ﬁrst-order she ar deformation theory.
The classical laminate theory (CLT) and the ﬁrst-order shea r deformation theory
(FSDT) are the most commonly used theories for analyzing lam inated or sand-
wiched beams, plates and shells in engineering application s. 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 limit ed 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 st ructure 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. Furthermo re, the CLT cannot
yield adequate correct through-the-thickness stresses an d 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 str ucture elements, e.g.
the FSDT. CLT and FSDT are so-called equivalent single-laye r theories (ESLT).
Moreover a short overview of so-called discrete-layer or la yerwise theories is given,
which shall overcome the drawbacks of equivalent single lay er theories.
1Gustav Robert Kirchhoff (∗12 March 1824 K¨ onigsberg – †17 October 1887 Berlin) – physi-
cist who contributed to the fundamental understanding of el ectrical circuits, spectroscopy, and the
emission of black-body radiation by heated objects, in addi tion, he formulated a plate theory which
was an extension of the Euler-Bernoulli beam theory
177 © Springer Nature Singapore Pte Ltd. 2018
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 mec hanics illustrates that the
following approaches for the modelling and analysis of beam s and plates composed
of composite materials can be used:
1. So called equivalent single-layer theories: These theor ies are derived from the
three-dimensional elasticity theory by making assumption s concerning the kine-
matics of deformation and/or the stress distribution throu gh the thickness of a
laminate or a sandwich. With the help of these assumptions th e modelling can be
reduced from a 3D-problem to a 2D-problem. In engineering ap plications equiv-
alent single-layer theories are mostly used in the form of th e classical laminate
theory, for very thin laminates, and the ﬁrst order shear def ormation theory, for
thicker laminates and sandwiches.
An equivalent single layer model is developed by assuming co ntinuous displace-
ment and strain functions through the thickness. The stress es jump from ply to
ply and therefore the governing equations are derived in ter ms of thickness aver-
aged resultants. Also second and higher order equivalent si ngle layer theories by
using higher order polynomials in the expansion of the displ acement components
through the thickness of the laminate are developed. Such hi gher order theories
introduce additional unknowns that are often difﬁcult to in terpret in mechanical
terms. The CLT requires C1-continuity of the transverse displacement, i.e. the
displacement and the derivatives must be continuous, unlik e the FSDT requires
C0-continuity only. Higher order theories generally require at least C1-continuity.
2. Three-dimensional elasticity theories such as the tradi tional 3D-formulations of
anisotropic elasticity or the so-called layerwise theorie s: In contrast to the equiv-
alent single-layer theories only the displacement compone nts 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 displacemen t ﬁeld exhibits only
C0-continuity through the thickness directions.
The basic assumption of modelling structural elements in th e 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) + αx3∂w(x1,x2)
∂x1+βx3ψ1(x1,x2)
+γx2
3φ1(x1,x2) + δx3
3χ1(x1,x2),
u2(x1,x2,x3) =v(x1,x2) + αx3∂w(x1,x2)
∂x2+βx3ψ2(x1,x2)
+γx2
3φ2(x1,x2) + δx3
3χ2(x1,x2),
u3(x1,x2,x3) =w(x1,x2) +βx3ψ3(x1,x2) + γx2
3φ3(x1,x2)(5.1.1)

5.1 General Remarks 179
A displacement ﬁeld in the form of (5.1.1) satisﬁes the compa tibility conditions for
strains, Sect. 2.2.1, and allows possible cross-sectional warping, transverse shear
deformations and transverse normal deformations to be take n 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 ﬁe ld 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 acc uracy 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 ψ1,ψ2rotations of the
transverse normals referred to the plane x3=0.ψ3has the meaning of extension of
a transverse normal and the remaining functions can be inter preted as warping func-
tions that specify the deformed shape of a straight line perp endicular to the reference
plane of the undeformed structure. In addition, any plate th eory should fulﬁll some
consistency requirements which was ﬁrst time discussed for the simplest case of a
homogeneous isotropic plate in Kienzler (2002) and later ex tended to other cases
by Schneider and Kienzler (2015); Schneider et al (2014). Al so implementations of
higher order theories into ﬁnite element approximations ca nnot be recommended. If
a laminated plate is thick or the 3D stress ﬁeld must be calcul ated in local regions,
a full 3D analysis should be carried out.
The most widely used approach reduces the polynomial functi on of degree three
to a linear or ﬁrst order approximation, which includes the c lassical and the ﬁrst-
order shear deformation theory
u1(x1,x2,x3) =u(x1,x2) + x3ψ1(x1,x2),
u2(x1,x2,x3) =v(x1,x2) + x3ψ2(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
ψ1(x1,x2)=−∂w
∂x1, ψ2(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,ψ1,ψ2.
The strain-displacement equations (2.2.3) give for the ﬁrs t order displacement
approximation a ﬁrst order strain ﬁeld model with transvers e shear
ε1=∂u
∂x1+x3∂ψ1
∂x1,ε2=∂v
∂x2+x3∂ψ2
∂x2,ε3=0,
ε4=∂w
∂x2+ψ2,ε5=∂w
∂x1+ψ1,
ε6=∂u
∂x2+∂v
∂x1+x3/parenleftbigg∂ψ2
∂x1+∂ψ1
∂x2/parenrightbigg(5.1.3)
For the in-plane strains one can write in contracted form
εi(x1,x2,x3)=εi(x1,x2)+x3κi(x1,x2),i=1,2,6,
i.e. the in-plane strains ε1,ε2andε6vary linearly through the thickness h.
The stress-strain relations in on-axis coordinates are
σ′
i=C′
i jε′
j,i,j=1,2,…, 6
From the transformation rule (4.1.27) follow the stiffness coefﬁcients in the off-
axis-coordinates
CCC=3
TTTεT
CCC′3
TTTε
and with (4.1.26) the constitutive equation is

σ1
σ2
σ3
σ4
σ5
σ6
=
C11C12C130 0 C16
C12C22C230 0 C26
C13C23C330 0 C36
0 0 0 C44C450
0 0 0 C45C550
C16C26C360 0 C66

ε1
ε2
ε3
ε4
ε5
ε6
(5.1.4)
Assuming σ3≈0, the stiffness matrix can be rewritten by separating the tr ansverse
shear stresses and strains in analogy to (4.2.3) – (4.2.5)

σ1
σ2
σ6
σ4
σ5
=
Q11Q12Q160 0
Q12Q22Q260 0
Q16Q26Q660 0
0 0 0 C44C45
0 0 0 C45C55

ε1
ε2
ε6
ε4
ε5
(5.1.5)

5.1 General Remarks 181
and from
σ3=C13ε1+C23ε2+C33ε3+C36ε6=0
it follows
ε3=−1
C33(C13ε1+C23ε2+C36ε6)
TheQi jare the reduced stiffness in the off-axis reference system
Qi j=Ci j−Ci3Cj3
C33,i,j=1,2,6,Qi j=Ci j,i,j=4,5
Summarizing, one can say that the ﬁrst order displacement ap proach (5.1.2) includes
the classical and the shear deformation theory for laminate s and sandwiches. In both
cases the in-plane displacements and strains vary linearly through the thickness, but
the explicit expressions for the curvature vector κκκdiffer. 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 equatio ns for the in-plane force
and the moment resultants NNN,MMM. It can be proved that a CLT approach is sufﬁcient
for very thin laminates and it has been used particularly to d etermine the global
response of thin composite structure elements, i.e. deﬂect ions, overall buckling, vi-
bration frequencies, etc. The FSDT approach is sufﬁcient fo r 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 transvers e shear strain. Both ﬁrst
order theories yield a complete understanding of the throug h-the-thickness laminate
response. Transverse normal and shear stresses, however, p lay an important role in
the analysis of beams, plates and shells since they signiﬁca ntly affect characteris-
tic failure modes like, e.g., delamination. The inﬂuence of interlaminar transverse
stresses are therefore taken into account by several failur e criteria. Simple but suf-
ﬁcient accurate methods for determination of the complete s tate 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 t o calculate trans-
verse stress components. In Chap. 11 will be seen that both th e 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 t he 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 equ ations (5.1.2) but makes
additional assumptions:
1. All layers are in a state of plane stress, i.e.
σ3=σ4=σ5=0
2. Normal distances from the middle surface remain constant , i.e. the transverse
normal strain ε3is negligible compared with the in-plane strains ε1,ε2.
3. The transverse shear strains ε4,ε5are 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 th rough 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
ψ1(x1,x2)=−∂w
∂x1,ψ2(x1,x2)=−∂w
∂x2, (5.2.1)
and the displacement approach (5.1.2) and the strain compon ents (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)
ε1=∂u
∂x1−x3∂w2
∂x2
1,ε2=∂v
∂x2−x3∂w2
∂x2
2,ε3=0,
ε4=0,ε5=0,ε6=∂u
∂x2+∂v
∂x1−2×3∂w2
∂x1∂x2(5.2.3)
The condensed form for the in-plane strains can be noted as
εi(x1,x2,x3)=εi(x1,x2)+x3κi,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
ε1=∂u
∂x1,ε2=∂v
∂x2,ε6=∂u
∂x2+∂v
∂x1,
κ1=−∂w2
∂x2
1,κ2=−∂w2
∂x2
2,κ6=−2∂w2
∂x1∂x2
εεεT= [ε1ε2ε6]is the vector of midplane strains (stretching and shearing) and
κκκT= [κ1κ2κ6]the vector of curvature (bending and twisting). For all klayers the
stresses are given in condensed form by
σ(k)
i=Q(k)
i jεi+x3Q(k)
i jκi,i,j=1,2,6 (5.2.4)
and the stiffness equations for the stress resultants follo w from (4.2.13) – (4.2.18).
The classical laminate theory is also called shear rigid the ory, the material equa-
tions yield zero shear stresses σ4,σ5for zero strains ε4,ε5, in the case that the shear
stiffness has ﬁnite values. But the equilibrium conditions yield non-zero stresses
σ4,σ5, if the stresses σ1,σ2andσ6are not all constant. This physical contradiction
will be accepted in the classical theory and the transverse s hear stresses are approx-
imately calculated with the given stresses σ1,σ2,σ6by the equilibrium equations
(4.1.56).
The approximate calculation of transverse shear stresses c an 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 NNN≡000 gives

000
···
MMM
=
AAA…BBB
. . . .
BBB…DDD

εεε
···
κκκ
,
εεε
···
κκκ
=
aaa…bbb
. . . . .
bbbT…ddd

000
···
MMM
 (5.2.5)
that is with Eqs. (4.2.20) – (4.2.26)
εεε=−AAA−1BBBκκκ,MMM=(DDD−BBBAAA−1BBB)κκκ=DDD∗κκκ
εεε=bbbMMM=BBB∗DDD∗−1MMM,κκκ=dddMMM=DDD∗−1MMM (5.2.6)
For symmetric laminates are BBB=000,BBB∗=000,DDD∗=DDDand Eqs. (5.2.6) can be replaced
by
εεε=000,κκκ=DDD−1MMM (5.2.7)
The partial extensional and coupling stiffness ˜AAA(x3),˜BBB(x3), Fig. 5.1, become

184 5 Classical and Improved Theories
❅❅❅
❅
❅ ❅❅❅❅
❅❅❅❅
❅❅❅❅
❅❅❅❅
❅❅❅❅
❅❅❅❅
❅❅❅❅
❅❅❅❅
❅❅❅❅
❅❅❅❅
❅❅❅❅
❅❅❅❅
❅❅❄❄
❄
❄x(m−1)
3x(m−2)
3x(0)
3 x3N
(m−1)
Fig. 5.1 Derivation of partial stiffness ˜AAA(x3)and˜BBB(x3)for the shaded part of the cross-section
˜AAA(x3) =x3/integraldisplay
x(0)
3QQQ(x3)dx3
=m−1
∑
k=1QQQ(k)h(k)+QQQ(m)/parenleftig
x3−x(m−1)
3/parenrightig
,
˜BBB(x3) =x3/integraldisplay
x(0)
3QQQ(x3)x3dx3
=m−1
∑
k=1QQQ(k)s(k)+1
2QQQ(m)/parenleftbigg
x2
3−x(m−1)
32/parenrightbigg
,(5.2.8)
h(k)=x(k)
3−x(k−1)
3,s(k)=h(k)x(k)
3=1
2/parenleftig
x(k)
3+x(k−1)
3/parenrightig/parenleftig
x(k)
3−x(k−1)
3/parenrightig
Outgoing from the equilibrium equations (2.2.1) the shear s tress equations can be
written

5.2 Classical Laminate Theory 185
σ5(x3)=−x3/integraldisplay
x(0)
3/parenleftbigg∂σ1
∂x1+∂σ6
∂x2/parenrightbigg
dx3
=−x3/integraldisplay
x(0)
3/bracketleftbigg∂
∂x1/parenleftig
Q(k)
1jεj+x3Q(k)
1jκj/parenrightig
+∂
∂x2/parenleftig
Q(k)
6jεj+x3Q(k)
6jκj/parenrightig/bracketrightbigg
dx3,
σ4(x3)=−x3/integraldisplay
x(0)
3/parenleftbigg∂σ6
∂x1+∂σ2
∂x2/parenrightbigg
dx3
=−x3/integraldisplay
x(0)
3/bracketleftbigg∂
∂x1/parenleftig
Q(k)
6jεj+x3Q(k)
6jκj/parenrightig
+∂
∂x2/parenleftig
Q(k)
2jεj+x3Q(k)
2jκj/parenrightig/bracketrightbigg
dx3(5.2.9)
or in vector-matrix notation
/bracketleftbiggσ5(x3)
σ4(x3)/bracketrightbigg
=−x3/integraldisplay
x(0)
3/bracketleftbigg1 0 0
0 0 1/bracketrightbigg
Q(k)
1j(εj,x1+x3κj,x1)
Q(k)
2j(εj,x1+x3κj,x1)
Q(k)
6j(εj,x1+x3κj,x1)
dx3
−x3/integraldisplay
x(0)
3/bracketleftbigg
1 0 0
0 0 1/bracketrightbigg
Q(k)
1j(εj,x2+x3κj,x2)
Q(k)
2j(εj,x2+x3κj,x2)
Q(k)
6j(εj,x2+x3κj,x2)
dx3(5.2.10)
with(…),xα=∂…/∂xα,α=1,2,j=1,2,6. Using Eqs. (5.2.6) – (5.2.8)
σσσs(x3)=−BBB1FFF(x3)MMM,x1−BBB2FFF(x3)MMM,x2 (5.2.11)
with
σσσs(x3)=[ σ5σ4]T,MMM,xi=[M1,xiM2,xiM6,xi]T,
FFF(x3)=[ ˜AAA(x3)AAA−1BBB−˜BBB(x3)]DDD∗−1=
F11F12F16
F21F22F26
F61F62F66
(5.2.12)
in the general case if non-symmetrical laminate and
FFF(x3)=˜BBB(x3)DDD−1(5.2.13)
for symmetrical laminates,
BBB1=/bracketleftbigg
1 0 0
0 0 1/bracketrightbigg
,BBB2=/bracketleftbigg
0 0 1
0 1 0/bracketrightbigg
(5.2.14)

186 5 Classical and Improved Theories
are so called Boolean3matrices. Equation (5.2.11) can also be written in componen t
notation.
Equations (5.2.11) and (5.2.16) constitute the straight fo rward equilibrium ap-
proach for transverse shear stresses which only neglects th e inﬂuence of the in-plane
force derivatives NNN,xi, but this is a very minor restriction, since, in most enginee r-
ing applications, the dominating source for transverse she ar stresses are transverse
force resultants. To express the bending moment derivative s by transverse shear
stress resultants it is necessary to assume special selecte d 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) =/integraldisplay
(h)σ5(x3)dx3=n
∑
k=1/integraldisplay
(h)σ(k)
5(x3)dx3,
Qs
2(x2) =/integraldisplay
(h)σ4(x3)dx3=n
∑
k=1/integraldisplay
(h)σ(k)
4(x3)dx3(5.2.16)
Equation (5.2.11) becomes in matrix notation
σσσs(x3)=FFF(x3)QQQs,
σσσs=[σ5(x3)σ4(x3)]T,QQQs=[Qs
1(x1)Qs
2(x2)]T,
FFF=/bracketleftbigg
F11(x3)F62(x3)
F61(x3)F22(x3)/bracketrightbigg (5.2.17)
Summarizing the derivations of transverse shear stresses w e have considered two
cases
1.NNN≡000,MMM=[M1M2M6]T,
2.NNN≡000,MMM=[M1(x1)M2(x2)]T
In case 1. follow Eqs. (5.2.18) and in case 2. Eqs. (5.2.19)
/bracketleftbiggσ5(x3)
σ4(x3)/bracketrightbigg
=/bracketleftbiggF11(x3)F12(x3)F16(x3)
F61(x3)F62(x3)F66(x3)/bracketrightbigg∂
∂x1
M1
M2
M6

+/bracketleftbigg
F61(x3)F62(x3)F66(x3)
F21(x3)F22(x3)F26(x3)/bracketrightbigg∂
∂x2
M1
M2
M6
,(5.2.18)
3George Boole (∗2 November 1815 Lincoln – †8 December 1864 Ballintemp) – math ematician,
educator, philosopher and logician

5.2 Classical Laminate Theory 187
/bracketleftbiggσ5(x3)
σ4(x3)/bracketrightbigg
=/bracketleftbiggF11(x3)F62(x3)
F61(x3)F22(x3)/bracketrightbigg/bracketleftbiggQ1
Q2/bracketrightbigg
,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 applicat ions. In this case
DDD∗=DDD,BBB≡000 and FFF(x3) =−˜BBB(x3)DDD−1. The calculation of the transverse shear
stresses is more simple. The approximate solution for trans verse shear stresses in
the classical laminate theory satisﬁes the equilibrium con dition. The shear stresses
are layerwise parabolic functions and there is no stress jum p at the layer interfaces.
Also in the frame of the classical laminate theory an approxi mate constitutive
equation can be formulated
QQQs=AAAsεεεsor/bracketleftbigg
Qs
1
Qs
2/bracketrightbigg
=/bracketleftbigg
A55A45
A45A44/bracketrightbigg/bracketleftbigg
ε5
ε4/bracketrightbigg
(5.2.20)
Regarding the complementary transverse shear theory formu lated in shear stresses
W∗s
1=1
2/integraldisplay
(h)σσσsT(CCC)−1σσσsdx3 (5.2.21)
and in shear forces
W∗s
2=1
2QQQsT(AAAs)−1QQQs(5.2.22)
The stress vector σσσsis a function of x3only, and therefore the integration is carried
out over x3. In
CCCs=/bracketleftbiggC55C45
C45C44/bracketrightbigg
theCi j,i,j=4,5 are the elastic parameters of the Hooke’s law. In Eq. (5.2.2 1) the
stress can be replaced by the transverse force resultants, E q. (5.2.19). The Qs
ido not
depend on x3and Eq. (5.2.21) yields
W∗s
1=1
2QQQsT
/integraldisplay
(h)FFFT(x3)(CCCs)−1FFF(x3)dx3
QQQs(5.2.23)
FFF(x3)is the reduced elasticity matrix Eq. (5.2.18) and Eq. (5.2.2 3) leads to
W∗s
1=1
2[Qs
1Qs
2]

/integraldisplay
(h)/bracketleftbigg
F11F62
F61F22/bracketrightbiggT/bracketleftbigg
C55C45
C45C44/bracketrightbigg−1/bracketleftbigg
F11F62
F61F22/bracketrightbigg
dx3

/bracketleftbigg
Qs
1
Qs
2/bracketrightbigg
(5.2.24)
With W∗s
1=W∗s
2follows the approximate shear stiffness

188 5 Classical and Improved Theories
AAAs=
/integraldisplay
(h)FFFT(CCCs)−1FFFdx3
−1
(5.2.25)
TheCi jare layerwise constant. The calculation of AAAsdemands an integration over
layerwise deﬁned polynomials of 4th order and can be just sim ple carried out by
programming. For unsymmetrical laminates FFF(x3)is deﬁned by Eq. (5.2.12).
Hygrothermal effects have no inﬂuence on the transverse she ar stresses. In the
classical laminate theory for mechanical and hygrothermal loading as demonstrated
in Sect. 4.2.5, the resultants NNNandMMMmust be substituted by the effective resultants
˜NNNand ˜MMM.
5.3 Shear Deformation Theory for Laminates and Sandwiches
The classical laminate theory allows us to calculate the str esses and strains with
high precision for very thin laminates except in a little ext ended region near the
free edges. The validity of the classical theory has been est ablished by comparing
theoretical results with experimental tests and with more e xact solutions based on
the general equations of the linear anisotropic elasticity theory.
If the width-to-thickness ratio is less about 20, the result s derived from the clas-
sical theory show signiﬁcant differences with the actual me chanical 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 fu rther improvement is
possible by introducing correction factors for the transve rse shear moduli.
The model used now has the same general form, as (5.1.2), for t he displacements,
but contrary to the classical theory, ψ1andψ2are independent functions and a nor-
mal line to the middle surface of the composite remains strai ght under deformation,
however it is not normal to the deformed middle plane. In the s hear deformation the-
ory the actual deformation state is approximated by 5 indepe ndent two-dimensional
functions u,v,w,ψ1,ψ2, in the classical theory by 3 functions u,v,w, respectively.
The strains are deduced from the displacements, (5.1.3). Th e components of the
strains
εεε(x1,x2,x3)=εεε(x1,x2)+x3κκκ(x1,x2),i=1,2,6
again vary linearly through the thickness hand are given by
ε1=∂u
∂x1,ε2=∂v
∂x2,ε6=∂u
∂x2+∂v
∂x1,
κ1=∂ψ1
∂x1,κ2=∂ψ2
∂x2,κ6=∂ψ2
∂x1+∂ψ1
∂x2(5.3.1)

5.3 Shear Deformation Theory for Laminates and Sandwiches 1 89
The components of the vector εεεT=[ε1ε2ε6]are not changed, however the compo-
nents of the curvature vector κκκT=[κ1κ2κ6]are now expressed by the derivatives
of the functions ψ1,ψ2. The stresses in the kth layer can be expressed by

σ1
σ2
σ6
σ4
σ5
(k)
=QQQ(k)
ε1
ε2
ε6
ε4
ε5
(k)
=QQQ(k)
∂u
∂x1+x3∂ψ1
∂x1∂v
∂x2+x3∂ψ2
∂x2∂u
∂x2+∂v
∂x1+x3/parenleftbigg∂ψ2
∂x1+∂ψ1
∂x2/parenrightbigg
∂w
∂x2+ψ2
∂w
∂x1+ψ1
(5.3.2)
The stresses σ1,σ2andσ6are superimposed on the extensional and the ﬂexural
stresses and vary linearly through a layer thickness, the st resses σ4,σ5are, in con-
tradiction to the equilibrium equations, constant through h(k). The strains ε1,ε2,ε6
vary linearly and the strains ε4,ε5constant through the laminate thickness h, i.e. they
vary continuously through the total thickness. Unlike, the corresponding stresses
σ1,σ2,σ6andσ4,σ5vary linearly or remain constant, respectively, through ea ch
layer thickness h(k)only. Therefore is no stress continuity through the laminat e
thickness but stress jumps from ply to ply at their interface s depending on the re-
duced stiffness QQQandQQQs.
With the deﬁnition equations for the stress resultants NNN,MMM,QQQsand the stiffness
coefﬁcients Ai j,Bi j,Di j,As
i jfor laminates (4.2.13) – (4.2.15) or sandwich (4.3.8) –
(4.3.10), (4.3.12) – (4.3.14), respectively, the constitu tive equation can be written in
the condensed hypermatrix form, Eqs. (4.2.16)

NNN
MMM
QQQs
=
AAA BBB000
BBB DDD000
000 000AAAs

εεε
κκκ
γγγs
 (5.3.3)
The stretching, coupling and bending stiffness Ai j,Bi j,Di jstay unchanged in com-
parison to the classical laminate theory. The shear stiffne ss are approximately given
by
As
i j=n
∑
k=1C(k)
i jh(k),i,j=4,5 (5.3.4)
TheC(k)
i jare the constant shear moduli of the kth lamina. These approximated shear
stiffness overestimate the shear stiffness since they are b ased on the assumption of
constant transverse shear strains and also do not satisfy th e transverse shear stresses
vanishing at the top and bottom boundary layers.
The stiffness values can be improved with help of shear corre ction factors (Vla-
choutsis, 1992; Altenbach, 2000; Gruttmann and Wagner, 201 7). In this case the
part of the constitutive equation relating to the resultant sNNN,MMMis not modiﬁed. The

190 5 Classical and Improved Theories
other part relating to transverse shear resultants QQQsis modiﬁed by replacing the
stiffness As
i jby(kA)s
i j. The constants ks
i jare 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/bracketleftigg
1−/parenleftbiggx3
h/2/parenrightbigg2/bracketrightigg
(5.3.5)
and considering that for the kth layer
σ(k)
4=Q(k)
44ε4+Q(k)
45ε5,σ(k)
5=Q(k)
45ε4+Q(k)
55ε5 (5.3.6)
the transverse resultants are:
Q2=n
∑
k=1x(k)
3/integraldisplay
x(k−1)
3σ(k)
4f(x3)dx3
=5
4

n
∑
k=1Q(k)
44ε4x(k)
3/integraldisplay
x(k−1)
3/bracketleftigg
1−/parenleftbiggx3
h/2/parenrightbigg2/bracketrightigg
dx3+n
∑
k=1Q(k)
45ε5x(k)
3/integraldisplay
x(k−1)
3/bracketleftigg
1−/parenleftbiggx3
h/2/parenrightbigg2/bracketrightigg
dx3

,
Q1=n
∑
k=1x(k)
3/integraldisplay
x(k−1)
3σ(k)
5f(x3)dx3
=5
4

n
∑
k=1Q(k)
45ε4x(k)
3/integraldisplay
x(k−1)
3/bracketleftigg
1−/parenleftbiggx3
h/2/parenrightbigg2/bracketrightigg
dx3+n
∑
k=1Q(k)
55ε5x(k)
3/integraldisplay
x(k−1)
3/bracketleftigg
1−/parenleftbiggx3
h/2/parenrightbigg2/bracketrightigg
dx3


The shear stiffness coefﬁcients As
i jof the constitutive equations
Q2=As
44ε4+As
45ε5,Q1=As
45ε4+As
55ε5 (5.3.7)
are calculated by
As
i j=5
4n
∑
k=1Q(k)
i j/bracketleftbigg/parenleftig
x(k)
3−x(k−1)
3/parenrightig
−4
3h2/parenleftbigg
x(k)
33−x(k−1)
33/parenrightbigg/bracketrightbigg
=5
4n
∑
k=1Q(k)
i j/bracketleftigg
h(k)−4
h2h(k)/parenleftigg
h(k)2
12+x(k)
32/parenrightigg/bracketrightigg
,i,j=4,5(5.3.8)

5.3 Shear Deformation Theory for Laminates and Sandwiches 1 91
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/bracketleftbigg
h−4h
h2/parenleftbiggh2
12+0/parenrightbigg/bracketrightbigg
=5
6Gh (5.3.9)
The weighting function (5.3.5) resulting in a shear correct ion factor ksis consistent
with the Reissner theory of shear deformable single layer pl ates (Reissner, 1944)
and slightly differ from Mindlin’s value (Mindlin, 1951).
A second method to determine shear correction factors consi sts of considering
the strain energy per unit area of the composite. Some remark s on this method are
given in Chaps. 7 and 8. However shear correction factors dep end on the special
loading and stacking conditions of a laminate and not the onl y factors is generally
applicable.
A particularly physical foundation to improve the shear sti ffness values AAAsis the
equilibrium approach, Eq. (5.2.25). The sequence of calcul ation steps for determin-
ing improved transverse shear stresses in the frame of the FS DT are analogous to
the CLT and shall be shortly repeated
•ﬁrstly, calculate the improved shear stiffness
AAAs=
/integraldisplay
(h)FFFTCCCs−1FFFdx3
−1
(5.3.10)
•secondly, calculate the resultant transverse shear forces
QQQs=AAAsεεεs(5.3.11)
•thirdly, calculate the improved transverse shear stresses
σσσs=FFFQQQs
AAAs=[Ai j],i,j=5,4,CCCs=[Ci j],i,j=5,4
FFF=/bracketleftbigg
F11F62
F61F22/bracketrightbigg
,QQQs=[Qs
1Qs
2]T,σσσ=[σ5σ4]T,εεεs=[ε5ε4]T(5.3.12)
Relying on the results of calculation improved transverse s hear stresses σσσs, the
transverse normal stress can be evaluated. The following eq uations explain the prin-
cipal way. One starts with solving the equilibrium conditio n for σ3, Eq. (2.2.1)
σ3(x3)=−x3/integraldisplay
x3=0/parenleftbigg∂σ5
∂x1+∂σ4
∂x2/parenrightbigg
dx3+p0 (5.3.13)
p0denotes the transverse load at the starting point of integra tion.
With

192 5 Classical and Improved Theories
FFF(x3)=/bracketleftbiggF11F62
F61F22/bracketrightbigg
=/bracketleftbiggfffT
1
fffT
2/bracketrightbigg
(5.3.14)
we are able to replace the transverse shear stresses in Eq. (5 .3.13) by Eq. (5.3.12)
σ3(x3)=−
x3/integraldisplay
x3=0fffT
1dx3QQQs
,x1+x3/integraldisplay
x3=0fffT
2dx3QQQs
,x2
+p0 (5.3.15)
Only the components of fff1andfff2depend on x3and therefore the derivatives of QQQs
remain unchanged by the integration process. Moreover, Eq. (5.2.12) demonstrates
that only the partial stiffness ˜AAA(x3)and˜BBB(x3)depend on x3, but not the matrices AAA,BBB
andDDD∗. Therefore the integration of FFF(x3)yields
x3/integraldisplay
x3=0FFF(x3)dx3=
x3/integraldisplay
x3=0˜AAA(x3)dx3AAA−1BBB−x3/integraldisplay
x3=0˜BBB(x3)dx3
DDD∗−1=ˆFFF(x3) (5.3.16)
For symmetrical laminates is the coupling matrix BBB≡000 and ˆFFF(x3)can be simpliﬁed
tox3/integraldisplay
x3=0FFF(x3)dx3=−x3/integraldisplay
x3=0˜BBB(x3)dx3DDD−1=ˆFFF(x3) (5.3.17)
Now, Eq. (5.3.15) can be transformed into
σ3(x3)=−/bracketleftig
ˆfffT
1QQQs
,x1+ˆfffT
2dx3QQQs
,x2/bracketrightig
+p0, (5.3.18)
where
ˆfffT
1=[ˆF11ˆF62],ˆfffT
2=[ˆF61ˆF22]
and
Qs
2,×2=(AAAsεεεs),x2
The boundary conditions of vanishing transverse shear stre sses at both surfaces are
fulﬁlled automatically. The boundary conditions for the tr ansverse normal stresses
must be regarded and are taken into account in the integratio n 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 p ublished. The vast ma-
jority falls into the class of plate theories known as displa cement based ones. All
consideration in this textbook are restricted to such theor ies. The term ”higher order
theories” indicates that the displacement distribution ov er 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 r equired to solve the more
complicated equations is not justiﬁed. In addition, the mec hanical interpretation of
the boundary conditions for higher order terms is very difﬁc ult. The most used ESLT

5.4 Layerwise Theories 193
in engineering applications of composite structure elemen ts 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 t o transverse shear than its
homogeneous isotropic counterpart and reduces the range of applicability of CLT.
Increasing in-plane stiffness may alternatively be regard ed as relevant reduction of
its transverse shear strength.
The FSDT yields mostly sufﬁcient accurate results for the di splacements 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 c ause 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-mod elling. Another rel-
ative simple method is to expand the FSDT from ﬁve to six unkno wn 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 sandwich es with thick single lay-
ers. Layerwise displacement approximations provide a more kinematically correct
representation of the displacement functions through the t hickness 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. Other wise so-called full lay-
erwise theories use expansions for all three displacement c omponents. Compared
with equivalent single layer models the partial layerwise m odel provides a more re-
alistic description of the kinematics of composite laminat es and the discrete-layer
behavior of the in-plane components.
Assume a linear displacement approximation (5.1.2) for eac h layer
u(k)
1(x1,x2,x3)=u(k)(x1,x2)+x3ψ(k)
1(x1,x2),
u(k)
2(x1,x2,x3)=v(k)(x1,x2)+x3ψ(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 yie ld 2(n−1)equations and

194 5 Classical and Improved Theories
the equilibrium for the transverse shear stresses yield add itional 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 lay ers 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 l ayerwise 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 l ayerwise theory is
often used in the ﬁnite element method. A comparison of equiv alent single layer
and layerwise theories one can ﬁnd in Reddy (1993).
Summarizing one can say for the class of partial or discrete l ayer-wise models
that all analytical or numerical equations are two-dimensi onal and in comparison to
a real three-dimensional modelling, their modelling and so lution effort, respectively,
is less time and cost consuming. The transverse normal displ acement does not have
a layerwise representation, but compared to the equivalent single layer modelling,
the partial layerwise modelling provides more realistic de scription of the kinematics
of composite laminates or sandwiches by introducing discre te layerwise transverse
shear effects into the assumed displacement ﬁeld.
Discrete layerwise theories that neglect transverse norma l strain are not capa-
ble of accurately determining interlaminar stresses and mo delling localized effects
such as cutouts, free edges, delamination etc. Full or gener alized layerwise theories
include in contrast to the partial layerwise transverse she ar 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 Cha p. 11 the exemplary
consideration of ﬁnite beam and plate elements have been res tricted to CLT and
FSDT.
5.5 Problems
Exercise 5.1. The displacement ﬁeld of a third order laminate (5.1.1) may d eﬁned
byα=−c0,β=1,γ=0,δ=−c1,¯β=¯γ=0.
1. Formulate the displacement equations and recover the dis placement equations
for the classical and the shear deformation laminate theory .
2. Introduce new variables φ1=ψ1−c0∂w/∂x1,φ2=ψ2−c0∂w/∂x2and express
the displacement ﬁeld in terms of φ1andφ2.
3. Substitute the displacements into the linear strain-dis placement relations.
4. Formulate the equations for the transverse shear stresse sσ4,σ5. Find the equa-
tions for c1so that the transverse shear stresses vanish at the top and th e 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/bracketleftbigg
ψ1(x1,x2)−c0∂w(x1,x2)
∂x1/bracketrightbigg
−x3
3c1χ1(x1,x2),
u2(x1,x2,x3) =v(x1,x2)+x3/bracketleftbigg
ψ2(x1,x2)−c0∂w(x1,x2)
∂x2/bracketrightbigg
−x3
3c1χ2(x1,x2),
u3(x1,x2,x3) =w(x1,x2)
Classical laminate theory: c1=0,ψ1=0,ψ2=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)+x3φ1(x1,x2)−c1x3
3χ1(x1,x2),
u2(x1,x2,x3) =v(x1,x2)+x3φ2(x1,x2)−c1x3
3χ2(x1,x2),
u3(x1,x2,x3) =w(x1,x2)
3. Using the strain-displacement equation (2.2.3) and subs titute equations b) we
ﬁnd
ε1=∂u1
∂x1=∂u
∂x1+x3∂φ1
∂x1−x3
3c1∂ χ1
∂x1
=ε0
1+x3εI
1+x3
3εII
1,
ε2=∂u2
∂x2=∂v
∂x2+x3∂φ2
∂x2−x3
3c1∂ χ2
∂x2
=ε0
2+x3εI
2+x3
3εII
2,
ε6=∂u2
∂x1+∂u1
∂x2=∂v
∂x1+∂u
∂x2+x3/parenleftbigg∂φ2
∂x1+∂φ1
∂x2/parenrightbigg
−x3
3c1/parenleftbigg∂ χ2
∂x1+∂ χ1
∂x2/parenrightbigg
=ε0
6+x3εI
6+x3
3εII
6,
ε4=∂u3
∂x2+∂u2
∂x3=∂w
∂x2+φ2−3c1x2
3χ2
=ε0
4+x2
3εII
4,
ε5=∂u3
∂x1+∂u1
∂x3=∂w
∂x1+φ2−3c1x2
3χ1
=ε0
5+x2
3εII
5
Note ε0
i≡εi.
4. The transverse shear stress in the kth layer of a laminate follow with (5.3.5) to
σ(k)
4=Q(k)
44ε4+Q(k)
45ε5=Q(k)
44(ε0
4+x2
3εII
4)+Q(k)
45(ε0
5+x2
3εII
5),
σ(k)
5=Q(k)
45ε4+Q(k)
55ε5=Q(k)
55(ε0
5+x2
3εII
5)+Q(k)
45(ε0
4+x2
3εII
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 ifk=1 orn.

196 5 Classical and Improved Theories
Q(1)
44/parenleftbigg
ε0
4+h2
4εII
4/parenrightbigg
+Q(1)
45/parenleftbigg
ε0
5+h2
4εII
5/parenrightbigg
=0,
Q(n)
44/parenleftbigg
ε0
4+h2
4εII
4/parenrightbigg
+Q(n)
45/parenleftbigg
ε0
5+h2
4εII
5/parenrightbigg
=0,
Q(1)
55/parenleftbigg
ε0
5+h2
4εII
5/parenrightbigg
+Q(1)
45/parenleftbigg
ε0
4+h2
4εII
4/parenrightbigg
=0,
Q(1)
55/parenleftbigg
ε0
5+h2
4εII
5/parenrightbigg
+Q(1)
45/parenleftbigg
ε0
4+h2
4εII
4/parenrightbigg
=0
=⇒ε0
4+h2
4εII
4=0,ε0
5+h2
4εII
5=0
In view of the fact that for c0=1 follows
φ2=ψ2−∂w
∂x2⇒ε4=ψ2−x2
33c1χ2,
ε0
4+h2
4εII
4=0⇒ψ2=h2
43c1χ2
If 3c1=4/h2⇒χ2=ψ2. Analogously follow with 3 c1=4/h2thatχ1=ψ1, i.e
ε0
4+1
3c1εII
4=ψ2−ψ2=0,ε0
5+1
3c1εII
5=ψ1−ψ1=0
The condition 1 /3c1=h2/4, i.e. c1=4/3h2is sufﬁcient to make the transverse
shear stresses σ4andσ5zero 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 sh ear stresses
σ5(x3),σ4(x3)and sketch their distribution across the laminate thicknes shfor
given transverse force resultants Q1=∂M1/∂x1,Q2=∂M2/∂x2andM6≡0.
2. Compare the average shear stiffness with the improved cor rected stiffness values.
Solution 5.2. The solution can be obtained as follows.
1. The reduced stiffness matrix QQQand the shear stiffness matrix CCC≡GGGmust be
calculated for the four layers
00-layers, ν′
21=ν12E′
2/E′
1:

5.5 Problems 197
QQQ[00]≡QQQ′=
E′
1
(1−ν′
12ν′
21)ν′
12E′
2
(1−ν′
12ν′
21)0
ν′
12E′
2
(1−ν′
12ν′
21)E′
2
(1−ν′
12ν′
21)0
0 0 E′
6

=
141,85 2,84 0
2,84 9,46 0
0 0 4 ,3
GPa,
GGG[00]≡GGG′=/bracketleftbiggG′
130
0G′
23/bracketrightbigg
=/bracketleftbigg4,3 0
0 3,2/bracketrightbigg
GPa
900-layers:
QQQ[900]=
9,46 2,84 0
2,84 141,85 0
0 0 4 ,3
GPa,
GGG[900]=/bracketleftbigg
3,2 0
0 4,3/bracketrightbigg
GPa
The bending stiffness matrix follows with (4.2.15)
Di j=4
∑
k=1Q(k)
i j/bracketleftig
(¯x(k)
3)2+(h(k))2/12/bracketrightig
h(k),
DDD=
9,654 0,207 0
0,207 1,379 0
0 0 0 ,314
GPamm3
The corrected ﬂexural stiffness matrix DDD∗(5.2.6) is identical DDDfor symmetric
laminates, i.e. D∗=D, and the FFF(x3)-matrix in (5.2.12) can be simpliﬁed
FFF(x3)=−˜BBB(x3)DDD−1
The inversion of the matrix DDDyields with ∆=22,572 the elements D−1
11of the
inverse matrix DDD−1
D−1
11=D22/∆,D−1
22=D11/∆,D−1
12=D12/∆,D−1
66=(D66)−1
DDD−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 ˜BBB(x3)for the layers of the laminate
can be calculated

198 5 Classical and Improved Theories
˜BBB[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,
˜BBB[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 FFF(x3)=−˜BBB(x3)DDD−1
/bracketleftbiggσ5(x3)
σ4(x3)/bracketrightbigg
=/bracketleftbiggF11F62
F61F22/bracketrightbigg/bracketleftbiggQ1
Q2/bracketrightbigg
,
F[00]=−/bracketleftbigg
6,79 0
0 2,17/bracketrightbigg
x2
3+/bracketleftbigg
1,70 0
0 0,54/bracketrightbigg
mm−1,
F[900]=−/bracketleftbigg0,44 0
0 32,80/bracketrightbigg
x2
3+/bracketleftbigg1,30 0
0 2,46/bracketrightbigg
mm−1
σ5[00](x3) = F11[00](x3)Q1= (−6,79×2
3+1,70)Q1,
σ5[900](x3) =F11[900](x3)Q1= (−0,44×2
3+1,30)Q1,
σ4[00](x3) = F22[00](x3)Q2= (−2,17×2
3+0,54)Q2,
σ4[900](x3) =F22[900](x3)Q2= (−32,80×2
3+2,46)Q2
The distribution of the shear stresses through the laminate thickness his sketched
in Fig. 5.2.
x3 x3
x1x2
0 00.250.250.5 0.5
-0.25-0.25
-0.5 -0.51.30
1.281.28
2.46
0.40.4
σ5(x3)/Q1σ4(x3)/Q2
Fig. 5.2 Distribution of the shear stresses σ5(x3)/Q1andσ4(x3)/Q2across the laminate thickness

5.5 Problems 199
2. A simpliﬁed calculation of the average shear stiffness ¯As
i jyields (4.2.15)
¯As
i j=4
∑
k=1G(k)
i jh(k)=⇒¯AAAs=/bracketleftbigg
3,75 0
0 3,75/bracketrightbigg
GPamm
An improved shear stiffness matrix which include the transv erse shear stress dis-
tribution follows with the help of the complementary strain energy W∗
W∗=1
2/integraldisplay
(h)σσσsTGGG′−1σσσsdx3
=1
2QQQT
/integraldisplay
(h)FFFTGGG′−1FFFdx3
QQQ=1
2QQQTAAAs−1QQQ
With
0,5/integraldisplay
0,25FFFT
[00]GGG′−1FFF[00]dx3=/parenleftbigg/bracketleftbigg
2,000 0
0 0,295/bracketrightbigg
x5
3
−/bracketleftbigg1,667 0
0 0,246/bracketrightbigg
x3
3
+/bracketleftbigg0,625 0
0 0,092/bracketrightbigg
x3/parenrightbigg
GPa−1,
0,25/integraldisplay
0FFFT
[900]GGG′−1FFF[900]dx3=/parenleftbigg/bracketleftbigg0,012 0
0 46,69/bracketrightbigg
x5
3
−/bracketleftbigg0,119 0
0 11,68/bracketrightbigg
x3
3
+/bracketleftbigg
0,529 0
0 1,314/bracketrightbigg
x3/parenrightbigg
GPa−1
follows by the sum up over the four layers the improved matrix AAAs
AAAs=/bracketleftbigg
3,01 0
0 2,54/bracketrightbigg
GPa
The comparison of ¯AAAsandAAAscan be carried out in the form
kkks¯AAAs=AAAs
which yields the shear correction vector
kkks=/bracketleftbigg
0,7718
0,6513/bracketrightbigg

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 la yered plates and
shells. Computational Mechanics 59(1):129–146
Kienzler R (2002) On consistent plate theories. Archive of A pplied Mechanics
72(4):229–247
Mindlin RD (1951) Inﬂuence of rotatory inertia and shear on ﬂ exural motions of
isotropic elastic plates. Trans ASME J Appl Mech 18:31–38
Reddy JN (1993) An evaluation of equivalent-single-layer a nd 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. Mathema tics and Mechanics of
Solids 20(7):871–882
Schneider P, Kienzler R, B¨ ohm M (2014) Modeling of consiste nt second-order plate
theories for anisotropic materials. ZAMM – Journal of Appli ed Mathematics and
Mechanics / Zeitschrift f¨ ur Angewandte Mathematik und Mec hanik 94(1-2):21–
42
Vlachoutsis S (1992) Shear correction factors for plates an d shells. International
Journal for Numerical Methods in Engineering 33(7):1537–1 552

Chapter 6
Failure Mechanisms and Criteria
Failure of structural elements can be deﬁned in a different m anner. As in the case of
buckling, a structural element may be considered failure th ough the material is still
intact, but there are excessive deformations. In Chap. 6 fai lure will be considered to
be the loss of integrity of the composite material itself.
The failure analysis procedures for metallic structures we re well established a
long time ago. In the case of monolithic materials stress con centrations, e.g. around
notches and holes, cause localized failures. For brittle ma terials local failures may
lead to fracture and therefore to a total loss of load-carryi ng capability. For ductile
materials local failure may be in the form of yielding and rem ains localized, i.e., it
is tolerated better than brittle failure. The fail-safe phi losophy has been employed
in the design of metallic structures and is standard in engin eering 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-rei nforced materials is a very
complex topic. While it is important to understand the princ ipal mechanisms of fail-
ure, for many applications it is impossible to detail each st ep of the failure process.
Main causes of failure are design errors, fabrication and pr ocessing 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 w hich 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 inte rnal stresses in adja-
cent laminae may develop a result of external applied loads b ut also by temperature
and moisture changes. Though manufacturing control and mat erial inspection tests
are carried out, structural composites with abnormalities can be produced. The me-
chanical properties of composites may be signiﬁcantly redu ced by high temperature
variations, impact damage, etc. Service anomalies can incl ude improper operation,
faulty maintenance, overloads or environmental incurred d amage.
If structural loadings produce local discontinuities insi de the material we speak
of a crack. Micro-cracking is considered as the nucleation o f micro-cracks at the
microscopic level starting from defects and may cause the in itiation of material
fracture. Macro-cracking is the propagation of a fracture b y the creation of new
201 © Springer Nature Singapore Pte Ltd. 2018
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 m aterials the fraction ini-
tiation is generally well developed before a change in the ma croscopic behavior can
be observed.
If in a laminate macro-cracks occur, it may not be catastroph ic, for it is possible
that some layers fail ﬁrst and the composite continues to tak e more loads until all
laminae fail. Failed laminae may still contribute to the sti ffness and strength of the
laminate. Laminate failure estimations are based on proced ures 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 o f laminates.
In this section the elastic behavior of laminae is primarily discussed from a
macroscopic point of view. But in the case of failure estimat ions 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 t o the ultimate macro-
scopic behavior. For this reason some considerations on mic ro-mechanic failure
mechanisms are made ﬁrst and then failure criteria are discu ssed more in detail.
Summarizing one can say that the ability of failure predicti on is a key aspect in
design of engineering structures. The ﬁrst step is to consid er what is meant by fail-
ure. Material failure of metallic structures is mostly rela ted with material yielding
or rupture, but with composites it is more complex. Therefor e research is ongoing in
developing failure mechanisms and failure criteria for uni directional ﬁbre laminae
and their laminates and in evaluating the accuracy of the fai lure criteria.
6.1 Fracture Modes of Laminae
Composite fracture mechanisms are rather complex because o f 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 reinforce d 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, interlam inar fracture shows the fail-
ure developed between laminae and translaminar fracture is oriented transverse to
the laminate plane. Inter- and intralaminar fractures occu r 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 character ized 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. crazin g.
•The failure is dominated by singularities at the ﬁber-matri x 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 fa ce.
•Crushing failure of the face and core at a support or tensile r espectively shear
failure at fasteners.
The following considerations are restricted to the strengt h of an unidirectional layer
and to the development of reliable criteria for the predicti ng of the failure of lam-
inae and laminates. The failure criteria in engineering app lications are mainly of a
phenomenological character, i.e. analytical approximati ons 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 la yer. In the ﬁbre direc-
tion, as a tensile load is applied, Fig. 6.1a, failure is due t o ﬁbre tensile fracture.
2′=T
1′=L3σL σL
σL>0 σL<0
σT
σT>0σLTσLT
σTσT<0a
b
Fig. 6.1 Fracture modes of a single layer in the case of elementary loa d states. aFibre fracture by
pure tension σL>0 or compression σL<0 (micro-buckling), bMatrix fracture by pure tension
σT>0, pure shearing σLTand pure compression σT<0

204 6 Failure Mechanisms and Criteria
One ﬁbre breaks and the load is transferred through the matri x to the neighboring
ﬁbres which are overloaded and fail too. The failure propaga tes rapidly with small
increasing load. Otherwise a tensile fracture perpendicul ar to the ﬁbres, Fig. 6.1b,
due a combination of different micromechanical failure mec hanisms: tensile failure
of matrix material, tensile failure of ﬁbres across the diam eters, 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 loadi ng 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 ten sile and for compression
tests. If the shear stresses act parallel or transverse to th e ﬁbre orientation there is
✻
❄✻❄✛ ✲ ✲ ✛
✲
✻
❄
✛σL>0 σL>0 σL<0 σL<0
σT>0
σT>0 σT<0σT<0
σLT
σLTmaterial property: σLt material property: σLc
material property: σTc material property: σTt
material property: τSa b
d c
e
Fig. 6.2 Basic strength parameters. aLongitudinal tensile strength σLt,bLongitudinal compres-
sive strength σLc,cTransverse tensile strength σTt,dTransverse compressive strength σTc,eIn-
plane (intralaminar) shear strength τS

6.1 Fracture Modes of Laminae 205
no inﬂuence of the load direction (Fig. 6.3a). Otherwise the positive shear stress
σ6>0 causes tensile in L-direction and compression in T-direct ion and vice versa
forσ6<0 and other strength parameters are standard. The required e xperimental
characterization is relatively simple for the parameters σLtandσTt, but more com-
plicated for the strength parameters σLc,σTcandτS.
In the case of laminates, besides the basic failure mechanis ms for a single
layer, such as ﬁbre fracture, longitudinal and transverse m atrix fraction, ﬁbre-matrix
debonding, etc. described above, another new fracture mode occurs. This mode is
called delamination and consists of separation of layers fr om one another. Through-
the-thickness variation of stresses may be caused even if a l aminate is loaded by
uniform in-plane loads. Generally, the matrix material tha t 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 la minae may cause breaking
of the bond between the layers in mostly localized, small reg ions. However, even if
✲❄✛
✻σ6<0
≡

❅
❅❅

❅❅❅❅❅ ș
❅❅ ❘ ✠
✒

σL=−σ6
σT=σ6✛✻✲
❄σ6>0
≡

❅
❅❅

❅❅❅❅❅ ❘
❅❅ ș ✒
✠

σL=σ6
σT=−σ6✲❄✛
✻σLT<0
≡ ❅
❅❅

❅❅❅❅❅ ș
❅❅ ❘ ✠
✒σ2=−σLT
σ1=σLT✛✻✲
❄σLT>0
≡ ❅
❅❅

❅❅❅❅❅ ❘
❅❅ ș ✒
✠σ2=σLT
σ1=−σLT
ba
✲✻
❅
❅❅ ❘ ✒
11′=L22′=T
✲✻
✒
❅❅❅ ș2′=T2
1′=L
1
Fig. 6.3 In-plane shear. aPositive and negative shear stresses along the principal ma terial 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 in tegrity of a laminate
and can degrade their in-plane load-carrying capability. T herefore, in practical en-
gineering applications it is important to calculate the int erlaminar normal and shear
stresses σ3,σ4andσ5and to check interlaminar failure too.
The deﬁnition of failure may change from case to case and depe nds on the com-
posite material and the kind of loads. For composite materia l, such as UD-laminates,
the end of the elastic domain is associated with the developm ent of micro-cracking.
But in the ﬁrst stage, the initiated cracks do not propagate a nd their development
changes the stiffness of the material very gradually but the degradation is irre-
versible. In the following section failure criteria for lam inae 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 thei r simplest form they are
similar, in principle, to those used for isotropic material s, e.g. maximum stress/strain
and distortional energy theories. The major difference bet ween isotropic materials
and unidirectional ﬁbrous composite materials is the direc tional dependence of the
strength on a macrosopic scale. It is important to realize th at failure criteria are
purely empirical. Their purpose is to deﬁne a failure envelo pe 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 determin ation failure envelopes.
We shall start by considering a single lamina before moving o n to discuss failure
of laminates. Longitudinal tension or pressure, transvers e tension or pressure and
shear are the ﬁve basic modes of failure of a lamina. Generall y the strength in the
principal material axes are regarded as the fundamental par ameters deﬁning failure.
When the lamina is loaded at an angle to the ﬁbres one has to det ermine the stresses
in the principal directions and compare them with the fundam ental strength param-
eters. Failure criteria usually grouped in literature into three different classes: limit
criteria, interactive criteria and hybrid criteria which c ombine selected aspects of
limit and interactive methods. In the following we only disc uss selected criteria of
the ﬁrst two classes.
Failure criteria for homogeneous isotropic materials are w ell established. Macro-
mechanical failure theories for composite materials have b een developed by extend-
ing and adapting isotropic failure theories to account for a nisotropy in stiffness and
strength of the composite. All theories can be expressed as f unctions of the basic
strength parameters referred to the principal material axe s (Fig. 6.2). Some criteria
do not account for interaction of stress components while ot hers do. Some inter-
action criteria require additional strength parameters ob tained 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 lamin ate or sandwich theories
given in Chaps. 4 and 5. A laminate may be assumed to have faile d when the strength
criterion of any one of its laminae is reached (ﬁrst-ply fail ure). 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 relativ e easily. Off-axis criteria
can be obtained by coordinate transformations of stresses o r 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 representati ve and more widely used:
•Maximum stress theory
•Maximum strain theory
•Deviatoric or distorsion strain energy criteria of Tsai-Hi ll1
•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 maxim um stress theory, failure
occurs when at least one stress component along one of the pri ncipal material axes
exceeds the corresponding strength parameter in that direc tion
σL=σLt, σL>0,
σT=σTt, σT>0,
σL=σLc, σL<0,
σT=σTc, σT<0,
|σLT|=τS,(6.2.1)
Note that failure can occur for more than one reason. A layer f ailure does not occur
if
−σLc<σL<σLt,
−σTc<σT<σTt,
−τS<σLT<τS(6.2.2)
For a two-dimensional state of normal stresses, i.e. σL/ne}ationslash=0,σT/ne}ationslash=0,σLT=0, the fail-
ure envelope, Fig. 6.4, takes the form of a rectangle. In the c ase of off-axis tension
or compression of a UD-lamina, Fig. 6.5, the transformed str esses are
σL= σ1cos2θ=σ1c2
σT= σ1sin2θ=σ1s2
σLT=−σ1sinθcosθ=−σ1sc⇒σ1=σL/c2
σ1=σT/s2
σ1=−σLT/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 Gon ville and Caius College, Cambridge,
UK
2Edward Ming-Chi Wu (∗30 September 1938 – †3 June 2009) – US-American engineer

208 6 Failure Mechanisms and Criteria
✻
✲✛ ✲✻
❄σT
σL
σTσL σT
σLσTt
σTcσLcσLt
Fig. 6.4 Failure envelope for UD-lamina under biaxial normal loadin g (max. stress criterion)
✲✻
❆❆❆❆❆❆❆❆❆❆ ❑
✠
✟✟✟✟✟✟✟✟✟✟✟✟✟✟✟✟✟✟ ✯
✟✟✟✟✟✟✟✟✟✟
✟✟✟✟✟
✟✟✟✟✟✟✟✟✟✟✟✟✟ ✟✟✟✟✟✟✟✟
✟✟✟✲ ✛θ
σ1 σ1×2
x1x′
1=xLx′
2=xT
ș
θ
Fig. 6.5 Off-axis unidirectional loading
−σLc<σ1c2<σLt,
−σTc<σ1s2<σTt,
−τS<σ1sc<τS(6.2.4)
The ultimate strength for σ1corresponds to the smallest of the following six values
σ1t=σLt/c2,σ1t=σTt/s2,σ1t=τS/sc,σ1>0,
σ1c=σLc/c2,σ1c=σTc/s2,σ1c=τS/sc,σ1<0(6.2.5)
The failure modes depend on the corresponding ultimate stre ngth σ1u
σ1u=σLt/c2ﬁbre failure,
σ1u=σTt/s2transverse normal stress failure,
σ1u=τS/scin-plane shear failure
In the more general case of off-axis loading, the stress tran sformation rule, Table
4.1, is used

6.2 Failure Criteria 209

σ′
1≡σL
σ′
2≡σT
σ′
6≡σLT
=
c2s22sc
s2c2−2sc
−sc sc c2−s2

σ1
σ2
σ6
 (6.2.6)
Because of the orthotropic symmetry, shear strength is inde pendent of the sign of
σLT(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 theor y may be applicable
for brittle modes of failure of material, e.g. follow from tr ansverse or longitudinal
tension(σL>0,σT>0).
The maximum strain theory is quite similar to the maximum str ess 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 mater ial axes exceeds the
corresponding ultimate strain in that direction
εL=εLt εL>0,
εT=εTt εT>0,
εL=εLc εL<0,
εT=εTc εT<0,
|εLT|=εS(6.2.7)
The lamina failure does not occur if
−εLc<εL<εLt,
−εTc<εT<εTt,
−εS<εLT<εS(6.2.8)
In the case of unidirectional off-axis tension or compressi on (Fig. 6.5), the stress
relations are given by (6.2.3). For the in-plane stress stat e strains in the principal
material axes are 
εL
εT
εLT
=
S′
11S′
120
S′
12S′
220
0 0 S′
66

σL
σT
σLT
 (6.2.9)
By associating (6.2.3) and (6.2.6) and expressing the compl iance parameters S′
i jas
functions of the engineering moduli in the principal direct ions, EL,ET,GLT,νLT,
νT L, it follows that
εL=1
EL(c2−νLTs2)σ1,
εT=1
ET(s2−νTLc2)σ1,
εLT=−1
GLTscσ1(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
andθ=900. The identity of the shear equations is given in both cases. T his implies
that

210 6 Failure Mechanisms and Criteria
εLt=σLt
EL,εLc=−σLc
EL,εTt=σTt
ET,εTc=−σTc
ET,εS=τS
GLT(6.2.11)
and the maximum strain criterion may be rewritten as follows
−σLc<σ1(c2−νLTs2)<σLt,
−σTc<σ1(s2−νTLc2)<σTt,
−τS<σ1sc <τS(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 νLTin the strain criterion. In practice these
terms modify the numerical results slightly. In the special case of a two-dimensional
stress state σL/ne}ationslash=0,σT/ne}ationslash=0,σLT=0, compare Fig. 6.4, the failure envelope takes the
form of a parallelogram for the maximum strain criterion, Fi g. 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 th e principal stress direc-
tions, von Mises3developed a deviatoric or distortional energy criterion fo r isotropic
ductile metals (von Mises, 1913) which can be presented for t he two-dimensional
stress state as
σ2
I+σ2
II−σIσII=σeq
or in a general reference system
σ2
1+σ2
2−σ1σ2+3σ2
6=σeq
σI,σIIare principal stresses, σeqthe equivalent stress. This criterion was extended
in von Mises (1928) and modiﬁed for the case of orthotropic du ctile materials by
Hill (1948)
Aσ2
1+Bσ2
2+Cσ1σ2+Dσ2
6=1 (6.2.13)
✻
✲
σLσT
✭✭✭✭✭✭✭✭✭✭✭✭✭✭✭✭✭✭✭✭✭✭✭
✁
✁
✁
✁
✁✁
✭✭✭✭✭✭✭✭✭✭✭✭✭✭✭✭✭✭✭✭✭✭✭✁
✁
✁
✁
✁✁σTt
σTcσLtσLcσT−νTLσL=σTt
σL−νLTσT=σLt
σT−νTLσL=−σTcσL−νLTσT=−σLc
Fig. 6.6 Failure envelope for UD-lamina under biaxial normal loadin g (max. strain criterion)
3Richard Edler von Mises (∗19 April 1883 Lemberg, Austria-Hungary (now Lviv, Ukraine) –
† 14 July 1953 Boston, Massachusetts) – mathematician who wo rked on solid mechanics, ﬂuid
mechanics, aerodynamics, aeronautics, statistics and pro bability theory, one of founders of the
journal Zeitschrift f¨ ur Angewandte Mathematik und Mechan ik

6.2 Failure Criteria 211
A,B,C,Dare material parameters. Equation (6.2.13) cannot be deﬁne d as distorsion
energy, since in anisotropy distorsion and dilatation ener gies are not separated. The
criterion (6.2.13) was applied to UD-laminae by Tsai and Wu ( 1971)
Aσ2
L+Bσ2
T+CσLσT+Dσ2
LT=1 (6.2.14)
The material parameters A,B,C,Dcan be identiﬁed by tests with acting basic load-
ings
σL=σLU,σT=0,σLT=0⇒A=1
σ2
LU,
σL=0,σT=σTU,σLT=0⇒B=1
σ2
TU,
σL=0,σT=0,σLT=τU⇒D=1
τ2
U(6.2.15)
In dependence on the failure mode, the superscript U must be s ubstituted 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 e qual biaxial normal load-
ingσL=σT/ne}ationslash=0,σLT=0 it can be assumed that the failure follows the maximum
stress criterion, i.e failure will occur when the transvers e stress reaches the trans-
verse strength σTUwhich is much lower than the longitudinal strength σLU. Equa-
tion (6.2.14) yields
/parenleftbiggσL
σLU/parenrightbigg2
+/parenleftbiggσT
σTU/parenrightbigg2
+Cσ2
T=1,σT=σTU=⇒C=−1
σ2
LU(6.2.16)
The Tsai-Hill criterion in the case of plane stress state and on-axis loading may be
written/parenleftbiggσL
σLU/parenrightbigg2
+/parenleftbiggσT
σTU/parenrightbigg2
−σLσT
σ2
LU+/parenleftbiggσLT
τU/parenrightbigg2
=1 (6.2.17)
In the case of tension or compression off the principal mater ial directions, Fig. 6.5,
the Tsai-Hill criterion becomes
/parenleftbiggσ1c2
σLU/parenrightbigg2
+/parenleftbiggσ1s2
σTU/parenrightbigg2
−/parenleftbiggσ1cs
σLU/parenrightbigg2
+/parenleftbiggσ1sc
τU/parenrightbigg2
=1 (6.2.18)
and the strength parameter σ1Uinx1-direction is
/parenleftbigg1
σ1U/parenrightbigg2
=/parenleftbiggc2
σLU/parenrightbigg2
+/parenleftbiggs2
σTU/parenrightbigg2
+/parenleftbigg1
τ2
U−1
σ2
LU/parenrightbigg
c2s2
≈/parenleftbiggc2
σLU/parenrightbigg2
+/parenleftbiggs2
σTU/parenrightbigg2
+/parenleftbiggcs
τ2
U/parenrightbigg2(6.2.19)
The approximated form presumes σLU≫τU.

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 cons iderable interaction
among the strain components and for ductile material the fai lure 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 fail ure surface in stress
space. They took into account only the ﬁrst two terms of the po lynomial criterion
and postulated that fracture of an anisotropic material occ urs when the following
equation is satisﬁed
ai jσi j+ai jklσi jσkl=1 (6.2.20)
or in a contracted notation
aiσi+ai jσiσj=1 (6.2.21)
We are interested in the case of an orthotropic composite mat erial, i.e. a unidirec-
tional lamina, subjected to plane stress state, and the Tsai -Wu criterion may be ex-
pressed as
aLσL+aTσT+aSσS+aLLσ2
L+aTTσ2
T+
aSSσ2
S+2aLTσLσT+2aLSσLσS+2aTSσTσS=1(6.2.22)
Equation (6.2.22) is written in the on-axis system and νLT≡aS.
The linear terms take account the actual differences betwee n composite material
behavior under tension and compression. The term aLTσLσTrepresents independent
interaction among the stresses σLandσTand the remaining quadratic terms describe
an ellipsoid in stress space. Since the strength of a lamina l oaded under pure shear
stress τSin the on-axis system is independent of the sign of the shear s tress, all linear
terms in σSmust vanish
aS=aLS=aTS=0 (6.2.23)
Then the Tsai-Wu criterion for a single layer in on-axis syst em has the form
aLσL+aTσT+aLLσ2
L+aTTσ2
T+aSSσ2
S+2aLTσLσT=1 (6.2.24)
The four quadratic terms in (6.2.24) correspond to the four i ndependent elastic char-
acteristics of orthotropic materials, the linear terms all ow the distinction between
tensile and compressive strength. The coefﬁcients of the qu adratic Tsai-Wu crite-
rion are obtained by applying elementary basic loading cond itions to the lamina
σL=σLt,σT=σS=0
σL=−σLc,σT=σS=0⇒aLσLt+aLLσ2
Lt=1
−aLσLc+aLLσ2
Lc=1⇒aL=1
σLt−1
σLc
aLL=1
σLtσLc
σT=σTt,σL=σS=0
σT=−σTc,σL=σS=0⇒aTσTt+aTTσ2
Tt=1
−aTσTc+aTTσ2
Tc=1⇒aT=1
σTt−1
σTc
aTT=1
σTtσTc

6.2 Failure Criteria 213
σS=τS,σL=σT=0⇒aSSτ2
S=1⇒aSS=1
τ2
S(6.2.25)
The remaining coefﬁcient aLTmust be obtained by biaxial testing
σL=σT=σU,σS=0⇒
(aL+aT)σU+(aLL+aTT+2aLT)σ2
U=1(6.2.26)
σUis the experimentally measured strength under equal biaxia l tensile loading
σL=σT.
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 cri teria lead: The Tsai-Hill
and the Tsai-Wu failure criteria are quadratic interaction criteria which have the
general form
Fi jσiσj+Fiσi=1,i,j=L,T,S (6.2.28)
Fi jandFiare strength parameters and σi,σjthe 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 convention al tensile, compressive
or shear strength terms which can be measured in a convention al experimental test
programme. The strength parameter FLTis 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 th e stress space can be given
in the equation below
σ2
L
σLtσLc+σ2
T
σTtσTc+σ2
LT
τ2
S+2FLTσLσT+/parenleftbigg1
σLt−1
σLc/parenrightbigg
σL+/parenleftbigg1
σTt−1
σTc/parenrightbigg
σT=1
(6.2.29)
Equation (6.2.29) reduces, e.g., for
σLt=σLc,σTt=σTc,FLT=−1
2σ2
Lt
to the Tsai-Hill criterion, for
σLt/ne}ationslash=σLc,σTt/ne}ationslash=σTc,FLT=−1
2σLtσLc
to the Hoffman criterion, and for
σLt/ne}ationslash=σLc,σTt/ne}ationslash=σTc,FLT=−1
2√σLtσLcσTtσTc

214 6 Failure Mechanisms and Criteria
to the Tsai-Wu criterion. Hoffman’s criterion is a simple ge neralization of the Hill
criterion that allows different tensile and compressive st rength parameters (Hoff-
man, 1967).
If one deﬁnes dimensionless stresses as
σ∗
L=√
FLLσL,σ∗
T=√
FTTσT,σ∗
LT=/radicalbig
FSSσLT
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∗
LTσ∗
Lσ∗2
T+F∗
Lσ∗2
L+F∗
Tσ∗2
T=1 (6.2.30)
Note that in the case of isotropic materials with σLt=σLc=σTt=σTcfollow
F∗
L=F∗
T=0. There the principal stress state will have σ∗
LT=σLT=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 lam ina 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 saf e or how much it
can be decreased if the lamina has failed. To overcome this pr oblem, 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. IfR<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 e lastic, for each state of
stress there is a corresponding state of strain and all compo nents of stress and strain
increase by the same proportion.
Summarizing the discussion above, the strength ratio for th e four criteria can be
formulated:
Maximum stress criterion
RLtσ=σLt/σL, σL>0 Strength factor ﬁbre fracture,
RTtσ=σTt/σT, σT>0 Strength factor matrix fracture,
RLcσ=σLc/|σL|, σL<0 Strength factor micro-buckling,
RTcσ=σTc/|σT|, σT<0 Strength factor matrix fracture,
RSσ=τS/|σLT|, Strength factor matrix fracture(6.2.32)

6.2 Failure Criteria 215
Maximum strain criterion
RLtε=εLt/εL, εL>0,
RTtε=εTt/εT, εT>0,
RLcε=εLc/|εL|,εL<0,
RTcε=εTc/|εT|,εT<0,
RSε=εS/|εLT|(6.2.33)
Tsai-Hill-criterion
Only one strength ratio can be introduced
/parenleftbiggRTHσL
σLU/parenrightbigg2
+/parenleftbiggRTHσT
σTU/parenrightbigg2
−RTHσLRTHσT
σ2
LU+/parenleftbiggRTHσLT
τU/parenrightbigg2
=1 (6.2.34)
With the ultimate strength σLU,σTUfor tension and compression the strength ratio
RTHfollows from
1
(RTH)2=/parenleftbiggσL
σLU/parenrightbigg2
+/parenleftbiggσT
σTU/parenrightbigg2
−σLσT
σ2
LU+/parenleftbiggσLT
τU/parenrightbigg2
Tsai-Wu-criterion
The Tsai-Hill and the Tsai-Wu criterion deﬁne only one stren gth ratio RTW
(aLσL+aTσT)RTW+(aLLσ2
L+aTTσ2
T+aSSσ2
S+2aLTσLσT)RTW2=1
or in symbolic notation
ARTW+B(RTW)2=1⇒(RTW)2+A
BRTW=1
B
with the solutions
RTW
1/2=−1
2A
B±/radicalbigg
1
4A2
B2+1
B=1
2B/parenleftig
−A±/radicalbig
A2+4B/parenrightig
RTWmust be positive
RTW=/radicalbig
A2+4B−A/2B (6.2.35)
The procedure for laminate failure estimation on the concep t of ﬁrst ply and last
ply failure is given as follows:
1. Use laminate analysis to ﬁnd the midplane strains and curv atures depending on
the applied mechanical and hygrothermical loads.
2. Calculate the local stresses and strains in each lamina un der 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 g ives the load level of
the failure of the ﬁrst lamina. This load may be called the ﬁrs t 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 pli es. Apply the actual
load level of previous failure.
5. Start again with step 3. to ﬁnd the strength ratios in the un damaged laminae. If
R>1 multiply the applied load by the strength ratio to ﬁnd the lo ad level of the
next ply failure. If R<1, degrade the stiffness and strength characteristics of al l
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 fol lowing 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
QQQ′(k)=
Q′
11Q′
120
Q′
12Q′
220
0 0 Q′
66
≡
QLLQLT0
QLTQTT0
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)
TLand the layer thicknesses h(k)
Transformation of the reduced stiffnesses QQQ′(k)of single layers in on-axis
system to the reduced stiffnesses QQQ(k)of single layers in off-axis system
QQQ(k)=(TTTε′)TQ′Q′Q′(k)TTTε′
Calculate the laminate stiffnesses AAA,BBBandDDD
Ai j=n
∑
k=1Q(k)
i jh(k),Bi j=1
2n
∑
k=1Q(k)
i j/parenleftbigg
x(k)
32−x(k−1)
32/parenrightbigg
=n
∑
k=1Q(k)
i jh(k)x(k)
3,
Di j=1
3n
∑
k=1Q(k)
i j/parenleftbigg
x(k)
33−x(k−1)
33/parenrightbigg
=n
∑
k=1Q(k)
i jh(k)/parenleftbigg
x(k)
32+1
12h(k)2/parenrightbigg
,
x(k)
3=1
2/parenleftig
x(k)
3+x(k−1)
3/parenrightig
Inversion of the matrices AAA,BBBandDDD
Calculate the compliance matrices aaa,bbb,cccanddddof the laminate
aaa=AAA∗−BBB∗DDD∗−1CCC∗,bbb=BBB∗DDD∗−1,ccc=−DDD∗−1CCC∗,ddd=DDD∗−1,
AAA∗=AAA−1,BBB∗=−AAA−1BBB,CCC∗=BBBAAA−1,DDD∗=DDD−BBBAAA−1BBB
Step 2
Calculation of the laminate stress resultants NNNandMMM
by structural analysis of beam or plate structures

6.2 Failure Criteria 217
Step 3.
Calculate the laminate strains εεε=εεε+x3κκκ

εεε
···
κκκ
=
aaa…bbb
. . . .
ccc…ddd

NNN
···
MMM

and the strains for all laminae at lamina interfaces
εεε(k)=εεε0+x(k)
3κκκ,k=0,1,2,…, n
Calculate the stresses for all interface surfaces of single layers
σσσ(k)−=QQQ(k)εεε(k−1),bottom surface of lamina k
σσσ(k)+=QQQ(k)εεεk, top surface of lamina k,k=0,1,2,…, n
Transformation of the interface stresses σσσ(k)−,σσσ(k)+
to the on-axis system of layer kk = 0, 1, 2, . . . , n
Step 4
Failure analysis based on a selected failure criterion in st ress space
Summarizing the strength ratios concept to the general quad ratic interaction criteria
Eq. (6.2.28) we formulate with the maximum values of stresse s
Fi jσ′max
iσ′max
j+Fiσ′max
i=1
Substituting Rσ′applied
i forσ′max
i yield the quadratic equation for the strength ratio R
(Fi jσiσj)R2+(Fiσi)R−1=0
or
aR2+bR−1=0,a=Fi jσiσj,b=Fiσi (6.2.36)
The strength ratio Ris equal to the positive quadratic root
R=−b
2a+/radicaligg/parenleftbiggb
2a/parenrightbigg2
+1
a
As considered above this approach is easy to use because the r esulting ratio provides
a linear scaling factor, i.e.
ifR≤1 failure occurs,
ifR>1, e.g. R=2, the safety factor is 2 and the load can be doubled or the lami nate
thickness reduced by 0.5 before failure occurs.
The same strength ratio can be determined from the equivalen t quadratic criterion
in the strain space. With σσσ=QQQεεεfollows, e.g. with Eqs. (6.2.24) – (6.2.27) the Tsai-
Wu criterion in the strain space as
bLεL+bTεT+bLLε2
L+bTTε2
T+bSSε2
S+2bLTεLεT=1 (6.2.37)

218 6 Failure Mechanisms and Criteria
with
bL=aLQLL+aTQLT,
bT=aTQTT+aLQLT,
bLL=aLLQ2
LL+aTTQ2
LT+2aLTQLLQLT,
bTT=aTTQ2
TT+aLLQ2
LT+2aLTQTTQLT,
bLT=aLLQLLQLT+aTTQTTQLT+aLT(Q2
LT+QLLQTT)(6.2.38)
In the more general form analogous to the strength ratio equa tion is
(Gi jεiεj)R2+(Giεi)R−1=0,
cR2+dR−1=0,
R=−d
c+/radicaligg/parenleftbiggd
2c/parenrightbigg
+1
c(6.2.39)
To determine Rfrom this equivalent quadratic criterion the strain space m ay be
preferred, because laminae strains are either uniform or va ry linearly across each
lamina thickness.
As considered above, the most widely used interlaminar fail ure criteria are the
maximum stress criterion, the maximum strain criterion and the quadratic failure
criteria as a generalization of the von Mises yield criterio n, in particular the Tsai-
Hill and the Tsai-Wu criterion. The interlaminar failure mo des can be ﬁbre breaking,
ﬁbre buckling, ﬁbre pullout, ﬁbre-matrix debonding or matr ix 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 reinfo rced composites.
Interlaminar failure, i.e. failure of the interface betwee n adjacent plies, is a de-
lamination mode. Delamination failure can have different c auses. Weakly bonded
areas impact initial delamination in the inner region of a la minate, whereas delam-
ination along free edges is a result of high interlaminar str esses. 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 experimenta l measured ultimate
stresses or strains for the laminae and realistic or approxi mate assumptions for stiff-
ness degradation of damaged layers. Strength under longitu dinal tensile and com-
pression stresses is usually determined with unidirection al plane specimen, strength
under transverse tension and compression is measured with p lane specimen or cir-
cumferentially reinforced tubes and shear strength is dete rmined in torsion test of
such tubes. Note that compression testing is much more difﬁc ult than tension testing
since there is a tendency of premature failure due to crushin g or buckling.
Summarizing the discussion above on failure analysis one ca n say that for deter-
mination of safety factors of ﬁbre reinforced laminated str uctural elements there is
a strong need for fracture criteria and degradation models w hich are simple enough
for engineering applications but being also in sufﬁcient ag reement with the physical

6.3 Problems 219
reality. In spite of many efforts were made during recent yea rs strength analysis of
laminates is still underdeveloped in comparison to the stre ss and strain analysis.
Essential for recent success in failure analysis was to dist inguish between ﬁbre
failure and inter-ﬁbre failure by separate failure criteri a introduced by Puck4. The
theory and application of Puck’s criterion are detailed des cribed in special literature
(Knops, 2008) and are not considered here. In addition, Chri stensen5has presented
some arguments concerning the best choice of failure criter ia – stress or strain based
(Christensen, 2013). On some actual problems and the state o f the art is reported in
Talreja (2016).
6.3 Problems
Exercise 6.1.
A UD lamina is loaded by biaxial tension σL=13σT,σLT=0. The material is a
glass-ﬁbre epoxy composite with EL=46 GPa, ET=10 GPa, GLT=4,6 GPa,
νLT=0,31. The basic strength parameters are σLt=1400 MPa, σTt=35 MPa,
τS=70 MPa. Compare the maximum stress and the maximum strain cri teria.
Solution 6.1. Maximum stress criterion ( σL<σLt,σT<σTt)
13σT=σL<σLt
σT=σTt<σLt=⇒σT<107,69 MPa
σT<35 MPa
The ultimate stress is determined by the smallest of the two v alues, i.e. failure occurs
by transverse fracture. The stress state is then
σT=35 MPa,σL=13·35=455 MPa <1400 MPa
Maximum strain criterion ( εL<εLt,εT<εTt)
To determine the ultimate strains we assume approximately a linear stress-strain
relation up to fracture. Then follows the ultimate strains
εLt=σLt/EL,εTt=σTt/ET
The strains caused by the biaxial tension state are
εL=SLLσL+SLTσT=1
ELσL−νLT
ELσT=1
EL(σL−νLTσT)<εLt,
εT=SLTσL+STTσT=−νTL
ETσL+1
ETσT=1
ET(σT−νTLσL)<εTt
The maximum strain criterion can be written
4Alfred Puck (∗1927) – engineer and professor, development of a physical-b ased strength criterion
for UD reinforced laminates
5Richard M. Christensen (∗3 July 1932 Idaho Falls, Idaho) – specialist in mechanics of m aterials

220 6 Failure Mechanisms and Criteria
σL−νLTσT<σLt,σT−νTLσL<σTt,νTL=ET
ELνLT
Since σL=13σTfollows
σT<σLt/(13−νLT)=110,32 MPa,
σT<σTt/(1−13νLTET/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
σL=13·110,32=1434,16 MPa,σT=110,32 MPa
The values of both criteria differ signiﬁcantly and the frac ture mode is reversed from
transverse to longitudinal fracture. Because linear elast ic response is assumed to
fail, the criterion can predict strength also in terms of str esses. 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), σ1=3,5 MPa, θ=600. Estimate the state of stress with
the help of the maximum stress, the maximum strain and the Tsa i-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,σLt=1200 MPa, σTt=45 MPa, τS=35 MPa,
εLt=0,033,εTt=0,002,εS=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 ca n be calculated
σ′
1=σ1cos2θ=0,875 MPa <σLt,
σ′
2=σ1sin2θ=2,625 MPa <σTt,
σ′
6=σ1sinθcosθ=−1,515 MPa <τS
The off-axis ultimate tensile strength σ1tis the smallest of the following stresses
σ1=σLt/cos2θ=4800 MPa ,
σ1=σTt/sin2θ=60 MPa,
σ1=τS/sinθcosθ=80,8 MPa
i.e.σ1t=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′
21σ′
2/E′
2=σ′
1/E′
1−ν′
12σ′
2/E′
1,
ε′
2=−ν′
12σ′
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[cos2θ−ν′
12sin2θ]σ1=0,0000047<εLt,
ε′
2=1
E′
2[sin2θ−ν′
12E′
2
E′
1cos2θ]σ1=0,0006<εTt,
ε′
6=1
G′
12sinθcosθσ1=0,0013<εS
All strains are allowed. The composite does not fail.
3. Tsai-Hill criterion
Using (6.2.18) the criterion can be written
/parenleftbiggcos2θ
σLt/parenrightbigg2
+/parenleftbiggsin2θ
σTt/parenrightbigg2
−/parenleftbiggsinθcosθ
σLt/parenrightbigg2
+/parenleftbiggsinθcosθ
τS/parenrightbigg2
<1
σ2
1,
/bracketleftigg/parenleftbigg0,25
1200/parenrightbigg2
+/parenleftbigg0,75
45/parenrightbigg2
−/parenleftbigg0,433
1200/parenrightbigg2
+/parenleftbigg0,433
35/parenrightbigg2/bracketrightigg
MPa−2<0,00043MPa−2,
1
σ2
1=0,0816MPa−2
The left-hand side is smaller than the right-hand side, ther efore the composite
does not fail.
Exercise 6.3. The plane stress state of a UD-lamina is deﬁned by
σ1=2σ,σ2=−3σ,σ6=4σ,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,
σLt=1500MPa ,σLc=1500MPa ,σTt=40MPa,σTc=246MPa,τS=68MPa
The ﬁbre angle is θ=600. Calculate the maximum value for σby 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

2σ
−3σ
4σ
=
1,714
−2,714
−4,165
σ
Using (6.2.2) we ﬁnd the inequalities

222 6 Failure Mechanisms and Criteria
−1500MPa <1,714σ<1500MPa ,
−246MPa <−2,714σ<40MPa,
−68MPa<−4,165σ<68MPa,
=⇒−875,1MPa<σ<875,1MPa,
−14,73MPa<σ<90,64MPa,
−16,33MPa<σ<16,33MPa
The three inequalities are satisﬁed if 0 <σ<16.33 MPa. The maximum stress
state which can be applied before failure is
σ1=32,66 MPa,σ2=48,99 MPa,σ6=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 10−11Pa−1,S′
22=1/E′
2=9,709 10−11Pa−1,S′
66=1/G′
12=
13,95 10−11Pa−1,S′
12=−ν′
12/E′
1=−0,1547 10−11Pa−1

ε′
1
ε′
2
ε′
6
=
S′
11S′
120
S′
12S′
220
0 0 S′
66

σ′
1
σ′
2
σ′
6
=
0,1367
−2,662
−5,809
10−10σ/bracketleftbiggMPa
MPa/bracketrightbigg
Assuming a linear relationship between the stresses and the strains until failure,
we can calculate the ultimate strains in a simple way
εLt=σLt/E′
1=8,287 10−3,εLc=σLc/E′
1=8,287 10−3,
εTt=σTt/E′
2=3,883 10−3,εTc=σTc/E′
2=23,88 10−3,
εS=τS/G′
12=9,483 10−3
and the inequalities (6.2.8) yield
−8,287 10−3<0,1367 10−10σ<8,287 10−3,
−23,88 10−3<−2,662 10−10σ<3,883 10−3,
−9,483 10−3<−5,809 10−10σ<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 t he mode of failure
is shear. For other failure modes there can be signiﬁcant dif ferences, see Exercise
6.1.
3. Tsai-Hill criterion
Using (6.2.17) we have

6.3 Problems 223
/bracketleftigg/parenleftbigg1,714
1500/parenrightbigg2
+/parenleftbigg−2,714
40/parenrightbigg2
−/parenleftbigg1,714
1500/parenrightbigg/parenleftbigg−2,714
1500/parenrightbigg
+/parenleftbigg−4,165
68/parenrightbigg2/bracketrightigg
σ2
1012<1
i.e.σ<10,94.
The Tsai-Hill criterion is an interactive criterion which c annot distinguish the
failure modes. In the form used above it also does not disting uish between com-
pression and tensile strength which can result in an underes timation of the allow-
able loading in compression with other failure criteria. Ge nerally the transverse
tensile strength of a UD-lamina is much less than the transve rse compressive
strength. Therefore the criteria can be modiﬁed. In depende nce of the sign of
theσ′
ithe corresponding tensile or compressive strength is subst ituted. For our
example follows
/bracketleftigg/parenleftbigg1,714
1500/parenrightbigg2
+/parenleftbigg−2,714
246/parenrightbigg2
−/parenleftbigg1,714
1500/parenrightbigg/parenleftbigg−2,714
1500/parenrightbigg
+/parenleftbigg−4,165
68/parenrightbigg2/bracketrightigg
σ2
1012<1
i.e.σ<16,06 MPa.
4. Tsai-Wu criterion
Now (6.2.24) must be applied. The coefﬁcients can be calcula ted
aL=/parenleftbigg1
σLt−1
σLc/parenrightbigg
=0,
aTT=1
σTtσTc=1,0162 10−16Pa−2,
aT=/parenleftbigg1
σTt−1
σTc/parenrightbigg
=2,093 10−8Pa−1,
aSS=1
τ2
Tt=2,1626 10−16Pa−2,
aLL=/parenleftbigg1
σLtσLc/parenrightbigg
=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)(1,714σ)2
+1,0162(10−16)(−2,714σ)2+2,1626(10−16)(−4,165σ)2
+2(−3,360)(10−18)(1,714)(−2,714)σ2<1
The solution of the quadratic equation for σyields 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≡RSσ)
Max. strain criterion: σ=16,33¯σ(¯σ≡RSε)
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≡σare identical with the strength ratios (6.2.32) – (6.2.35).
Remark 6.1. A summary of the examples demonstrates that different failu re criteria
can lead to different results. Unfortunately, there is no on e universal criterion which
works well for all situations of loading and all materials. F or each special class of
problems a careful proof of test data and predicted failure l imits must be conducted
before generalizations can be made. In general, it may be rec ommended that more
than one criterion is used and the results are compared.
References
Christensen RM (2013) The Theory of Materials Failure. Univ ersity Press, Oxford
Gol’denblat II, Kopnov V A (1965) Strength of glass-reinfor ced 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 aniso tropic metals. Pro-
ceedings of the Royal Society of London A: Mathematical, Phy sical and Engi-
neering Sciences 193(1033):281–297
Hoffman O (1967) The brittle strength of orthotropic materi als. Journal of Compos-
ite Materials 1(2):200–206
Knops M (2008) Analysis of Failure in Fiber Polymer Laminate s: The Theory of
Alfred Puck. Springer, Heidelberg
von Mises R (1913) Mechanik des festen K¨ orpers im plastisch en deformablen Zu-
stand. Nachrichten der K¨ oniglichen Gesellschaft der Wiss enschaften G¨ ottingen,
Mathematisch-physikalische Klasse pp 589–592
von Mises R (1928) Mechanik der plastischen form¨ anderung v on kristallen. ZAMM
– Journal of Applied Mathematics and Mechanics / Zeitschrif t f¨ ur Angewandte
Mathematik und Mechanik 8(3):161–185
Talreja R (2016) On failure theories for composite material s. In: Naumenko K,
Aßmus M (eds) Advanced Methods of Continuum Mechanics for Ma terials and
Structures, Springer Singapore, Singapore, pp 379–388
Tsai SW, Wu EM (1971) A general theory of strength for anisotr opic 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 stru ctural elements (beams,
plates and shells) composed of laminates and sandwiches. Th e modelling of lam-
inated and sandwich plates and shells is limited to rectangu lar plates and circular
cylindrical shells. The individual ﬁber reinforced lamina e of laminated structured
elements are considered to be homogeneous and orthotropic, but the laminate is
heterogeneous through the thickness and generally anisotr opic. An equivalent sin-
gle layer theory using the classical lamination theory, and the ﬁrst order shear defor-
mation theory are considered. Multilayered theories or lam inate 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 sig niﬁcance, advantages
and limitations of composite materials and structures and t he material characteris-
tics of the constituents of composite materials were consid ered. Chapter 2 gave a
short introduction to the governing equations of the linear theory of anisotropic ma-
terial behavior. Chapter 3 deﬁned effective material modul i of composites including
elementary mixture rules and improved formulas. Chapter 4 d eveloped in detail the
modelling of the mechanical behavior of laminates and sandw iches in the frame of
classical theories including thermal and hygroscopic effe cts. The constitutive equa-
tions, describing the relationships between stress result ants and in-plane strains and
mid-surface curvatures were developed for unidirectional laminae, laminates and
sandwiches with the assumptions of the classical laminate t heory. Further the calcu-
lation of in-plane and through-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 composi te 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 l ayers, etc. This ba-
sic knowledge can be utilized to develop the modelling and an alysis of structural
elements and structures composed of composite materials.
7.1 Introduction
The analysis of structural elements can be performed by anal ytical and semi-
analytical approaches or by numerical methods. The advanta ge of analytical so-
lutions is their generality allowing the designer to take in to account various design
parameters. Analytical solutions may be either closed form solutions or inﬁnite se-
ries and may be exact solutions of the governing equations or variational approaches.
However, analytical solutions are restricted to the analys is of simple structural el-
ements such as beams, plates and shells with simple geometry . Otherwise numer-
227 © Springer Nature Singapore Pte Ltd. 2018
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 a nalysis. Chapter 7 to
10 describe analytical solutions for one- and two-dimensio nal structural elements.
Chapter 11 gives an insight into numerical solutions based o n the ﬁnite element
method.
In the following sections of Chap. 7 we consider rods, column s 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 tensil e, 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. T he other type of one-
dimensional structural elements, so called plate strips un der cylindrical bending,
are discussed in Chap. 8. The modelling and analysis of gener alized beams based a
thin-walled folded structure are considered in Chap. 10 (ge neralized Vlasov beam
theory).
The elementary or classical beam theory assumes that the tra nsverse shear strains
are negligible and plane cross-sections before bending rem ain plane and normal to
the axis of the beam after bending (Bernoulli-Euler beam the ory, Sect. 7.2). The
assumption of neglecting shear strains is valid if the thick ness his small relative to
the length l(h/l<1/20). In the Bernoulli-Euler beam theory the transverse deﬂe c-
tionu3is 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 differentia l 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 be ams, the results
derived from the Bernoulli-Euler theory can show signiﬁcan t differences with the
actual mechanical behavior, i.e. the deﬂection, stress dis tribution, etc. An improve-
ment is possible by introducing the effect of transverse she ar deformation, i.e. we
apply Timoshenko beam theory (Sect. 7.3). The assumptions o f the classical theory
❅❅❅❅
❅
❅
❅❅❅❅❅❅
❅
❅
❅❅axisb
h
l✲✻
❅❅❅❅❅ ș
x1x3
x2
✻
❄❅
❅❅ ❘❅❅ ș
✛ ✲
Fig. 7.1 Rod/column/beam

7.2 Classical Beam Theory 229
have then to be relaxed in the following way: the transverse n ormals do not remain
perpendicular to the deformed axis of the beam after straini ng. Section 7.4 discuss
some special aspects of sandwich beams.
Laminate or sandwich beams with simple or double symmetric c ross-sections are
most important in engineering applications. The derivatio ns in Sects. 7.2 – 7.5 are
therefore limited to straight beams with simple or double sy mmetric constant cross-
sections which are predominantly rectangular. The bending moments act in a plane
of symmetry. Also cross-sections consisting of partition w alls 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 material s, they design compos-
ite beams made from two or more materials. The design rationa le is quite straight
forward. For bending loading, stiff, strong, heavy or expen sive material must be far
away from the neutral axis at places where its effect will be g reatest. The weaker,
lighter or less expensive material will be placed in the cent ral part of the beam. At
one extreme is a steel-reinforced concrete beam, where weig ht is not a major con-
cern, but strength and cost are. At the other extreme is a sand wich 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-sec tion 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)andE1(x2,x3)=E1(−x2,x3).
With the known equations for the strain ε1and the stress σ1atx1=const
ε1(x3)=ε1+x3κ1,σ1(x2,x3)=ε1E1(x2,x3)+x3κ1E1(x2,x3) (7.2.1)
follow the stress resultants N(x1),M(x1)of a beam
N=ε1/integraldisplay
(A)E1(x2,x3)dA+κ1/integraldisplay
(A)x3E1(x2,x3)dA,
M=ε1/integraldisplay
(A)x3E1(x2,x3)dA+κ1/integraldisplay
(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=aε1+bκ1,a=EmA+(Ef−Em)/integraldisplay
(A)φ(x2,x3)dA,
M=bε1+dκ1,b=(Ef−Em)/integraldisplay
(A)φ(x2,x3)x3dA,
I=/integraldisplay
(A)x2
3dA,d=EmI+(Ef−Em)/integraldisplay
(A)φ(x2,x3)x2
3dA(7.2.4)
The inverse of the stress resultants, (7.2.4), are
ε1=dN−bM
ad−b2,κ1=aM−bN
ad−b2(7.2.5)
and the stress equation (7.2.1) has the form
σ1(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
σf(x3) =dN−bM+(aM−bN)x3
ad−b2Ef,
σm(x3) =dN−bM+(aM−bN)x3
ad−b2Em(7.2.7)
In the case of a double symmetric geometry and ﬁbre volume fra ction function φ,
b=0 and the equations can be simpliﬁed
ε1=N/a,κ1=M/d,
σf(x3)=( N/a+x3M/d)Ef,σm(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)φA=E1A,
b=0,
d=EmI+(Ef−Em)φI=E1I(7.2.9)
and the stress relations for ﬁbre and matrix, (7.2.8), are tr ansformed to
σf(x3)=/parenleftbiggN
A+x3M
I/parenrightbigg/parenleftbiggEf
E1/parenrightbigg
,σm(x3)=/parenleftbiggN
A+x3M
I/parenrightbigg/parenleftbiggEm
E1/parenrightbigg
(7.2.10)
IfEf=EmandE1=E, (7.2.10) becomes the classical stress formula for isotrop ic
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 lo ading. For simplicity,
thermal and hygrothermal effects are ignored. The derivati on 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-sec tion:
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 bendi ng 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
/bracketleftbiggN1
M1/bracketrightbigg
=/bracketleftbiggA11B11
B11D11/bracketrightbigg/bracketleftbiggε1
κ1/bracketrightbigg
(7.2.12)
and from (4.2.14)
Q1=A55ε5 (7.2.13)
If the beam has a midplane symmetry, there is no bending-stre tching coupling, so
thatB11=0 and (7.2.12) becomes
N1=A11ε1,M1=D11κ1 (7.2.14)
Note that in the classical theory, the transverse shear stra in will be ignored, i.e
ε5=0, and there is no constitutive equation for resultant shear forces.
The starting point for derivation of structural equations f or beams is the equilib-
rium equations for stress resultants N,MandQat the undeformed beam element,
Fig. 7.2. The in-plane and transverse stress resultants N1,Q1and the resultant mo-
x3
x1
M M+dMQ Q+dQ
N N+dNq(x1)
n(x1)m(x1)
dx1
Fig. 7.2 Stress resultants N,QandMof 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 clas sical beam theory
(ε1=u′(x1),κ1=−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)−bB11w′′(x1)
M=bB11u′(x1)−bD11w′′(x1)⇐⇒
N(x1)
M(x1)
=
bA11bB11
bB11bD11

u′(x1)
−w′′(x1)

Differential equations for the displacements
General case
(bA11u′)′′−(bB11w′′)′′=−n′
(bB11u′)′′−(bD11w′′)′′=−q−m′(…)′=d
dx1
Constant stiffness/bracketleftbigg
bA11bB11
bB11bD11/bracketrightbigg/bracketleftbigg
u′′′
−w′′′′/bracketrightbigg
=/bracketleftbigg
−n′
−q−m′/bracketrightbigg
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 loads per unit length and must be mu ltiplied by the
beam width b, i.e. the beam resultants are N=bN1,Q=bQ1,M=bM1.
The differential relations for laminate beams loaded ortho gonally 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)
11ε(k)
1=Q(k)
11(ε1+x3κ1)=Q(k)
11/bracketleftbiggdu(x1)
dx1−x3d2w(x1)
dx2
1/bracketrightbigg
(7.2.15)
or with
ε1=D11N−B11M
A11D11−B2
11,κ1=A11M−B11N
A11D11−B2
11,
one get
σ(k)
1=Q(k)
111
b/parenleftbiggD11N−B11M
A11D11−B2
11+x3A11M−B11N
A11D11−B2
11/parenrightbigg
(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)
11du(x1)
dx1=Q(k)
11N(x1)
bA11,
σ(k)
1B(x1) =Q(k)
11/bracketleftbigg
−x3d2w(x1)
dx2
1/bracketrightbigg
=Q(k)
11x3M(x1)
bD11(7.2.17)
σ(k)
1Mare the layerwise constant stretching or membrane stresses produced by N(x1)
andσ(k)
1Bthe layerwise linear distributed ﬂexural or bending stress es produced by
M(x1). The strain ε1=ε1+x3κ1is 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
effandEM
efffor stretching and ﬂexural loading we can compare the stress
equations of a laminate beam with the stress equation of a sin gle layer beam.
With N/ne}ationslash=0,M=0 Eq. (7.2.12) becomes
/bracketleftbigg
N1
0/bracketrightbigg
=b/bracketleftbigg
A11B11
B11D11/bracketrightbigg/bracketleftbigg
ε1
κ1/bracketrightbigg
(7.2.18)
and with
0=bB11ε1+bD11κ1,κ1=−B11
D11ε1 (7.2.19)
123456

σ1M ε1 σ1B x3κ1✲
x1S✻
❄h✚✚
✚✚☞☞
☞☞
Fig. 7.3 Qualitative distribution of the stresses and strains throu gh 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=bN1andε1
N=bA11ε1+bB11κ1=ε1bA11D11−B2
11
D11,
ε1=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 ε1on the beam axis of a single layer isotropic, homogeneous bea m is
ε1=N/EA. Replacing EbyEN
effgives the strain equation for the laminate beam.
The stresses σ(k)
1in the klayers are then
σ(k)
1=Q(k)
11(ε1+x3κ1)=Q(k)
11ε1/parenleftbigg
1−x3B11
D11/parenrightbigg
=Q(k)
11/parenleftbigg
1−x3B11
D11/parenrightbiggdu
dx1,
σ(k)
1=Q(k)
11
EN
effN
bh/parenleftbigg
1−x3B11
D11/parenrightbigg
=E(k)
1
EN
effN
bh/parenleftbigg
1−x3B11
D11/parenrightbigg(7.2.21)
or for midplane symmetric beams with B11=0
σ(k)
1=E(k)
1
EN
effN
bh=E(k)
1
EN
effN
A,EN
eff=A11
h(7.2.22)
In an analogous manner it follows from (7.2.12) with N=0,M/ne}ationslash=0 that
M=bB11ε1+bD11κ1=κ1bA11D11−B2
11
A11,
κ1=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
κ1=M/EI=12M/bh3,I=bh3/12. Replacing now EbyEM
eff, the stress equations
are

7.2 Classical Beam Theory 235
σ(k)
1=Q(k)
11(ε1+x3κ1)=Q(k)
11κ1/parenleftbigg
−B11
A11+x3/parenrightbigg
=Q(k)
11/parenleftbiggB11
A11−x3/parenrightbiggd2w
dx2
1,
=Q(k)
11
EeffM
I/parenleftbigg
x3−B11
A11/parenrightbigg
=E(k)
1
EeffM
I/parenleftbigg
x3−B11
A11/parenrightbigg(7.2.25)
or with B11=0 for the symmetric case
σ(k)
1=E(k)
1
EM
effM
Ix3,EM
eff=12D11
h3(7.2.26)
If both in-plane and lateral loads occur simultaneously, th e stress in each lamina of
the beam is as for symmetric case
σ(k)
1(x1,x3)=Q(k)
11/parenleftbiggdu
dx1−x3d2w
dx2
1/parenrightbigg
=E(k)
1/parenleftbiggN
EN
effA+M
EM
effIx3/parenrightbigg
, (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, o ne
can say that the equations for u(x1)andw(x1)are identical in form to those of el-
ementary theory for homogeneous, isotropic beams. Hence al l solutions available,
e.g. for deﬂections of isotropic beams under various bounda ry conditions, can be
used by replacing the modulus Ewith EN
efforEM
eff, respectively. The calculation of
the stresses illustrates that constant in-plane layer stre sses produced by Nare propor-
tional to the layer modulus E(k)
1(7.2.22). N/AEN
effis for a cross-section x1=const
a constant value. Analogous are the ﬂexural layer stresses p roportional to E(k)
1×3
(7.2.26). In general, the maximum stress does not occur at th e top or the bottom of
a laminated beam, but the maximum stress location through th e thickness depends
on the lamination scheme.
From the bending moment-curvature relation ( N=0)
M=bD11κ1 (7.2.28)
it follows that
κ1max=Mmax
bD11=−/parenleftbiggd2w
dx2/parenrightbigg
max,
and the maximum stress can be calculated for each lamina
σ(k)
1max=Q(k)
11κ1maxx3=−Q(k)
11/parenleftbiggd2w
dx2/parenrightbigg
maxx3=E(k)
1
EM
effMmax
Ix3 (7.2.29)
σ(k)
1maxmust 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 midp lane 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+dσ(j)
1≈σ(j)
1+dσ(j)
1
dx1dx1
(Fig. 7.4) and no edge shear stresses on the upper and lower fa ces
σ(k)
5dx1−N
∑
j=k+1x(j)
3/integraldisplay
x(j−1)
3/bracketleftigg/parenleftigg
σ(j)
1+dσ(j)
1
dx1dx1/parenrightigg
−σ(j)
1/bracketrightigg
dx3=0 (7.2.30)
or
σ(k)
5=N
∑
j=k+1x(j)
3/integraldisplay
x(j−1)
3dσ(j)
1
dx1dx3=N
∑
j=k+1x(j)
3/integraldisplay
x(j−1)
3E(j)
1
EM
effIdM
dx1x3dx3
With Q=dM/dx1it follows that
σ(k)
5(x1,x3) =Q(x1)
EM
effIN
∑
j=k+1x(j)
3/integraldisplay
x(j−1)
3E(j)
1x3dx3=Q(x1)
EM
effIN
∑
j=k+1E(j)
11
2/parenleftbigg
x(j)
32−x(j−1)
32/parenrightbigg
=Q(x1)
EM
effIN
∑
j=k+1E(j)
1h(j)x(j)
3
(7.2.31)
✲✻
x1x3
✛
σ(k)
5k+1jN
…
…✛
✛
✛
✛
✛σ(k+1)
1σ(j)
1σ(N)
1✲
✲
✲
✲
✲σ(k+1)
1+dσ(k+1)
1σ(j)
1+dσ(j)
1σ(N)
1+dσ(N)
1
✛ ✲dx1✻
x(k)
3✻✻✻✻
x(j−1)
3x(j)
3x(j)
3x(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) yie lds the known parabolic
shear stress distribution through h
σ5(x1,x3)=Q
Ih/2/integraldisplay
x3x3dx3=12Q
bh31
2/parenleftbiggh2
4−x2
3/parenrightbigg
=3Q
2bh/bracketleftbigg
1−4/parenleftigx3
h/parenrightig2/bracketrightbigg
(7.2.32)
With an increasing number of equal thickness layers, the tra nsverse shear stress
distribution (7.2.31) approaches the parabolic function o f 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,i/ne}ationslash=j,i,j=1,2,6. They are summarized for symmet-
ric laminated beams ( N/ne}ationslash=0,M/ne}ationslash=0) in Table 7.2. For symmetric laminated beams
loaded orthogonally to the plane of lamination, the classic al laminate theory yields
identical differential equations for u(x1)andw(x1)with to Bernoulli’s beam theory
of single layer homogeneous isotropic beams, if one substit utesbA11byEA=Ebh
andbD11byEI=Ebh3/12. An equal state is valid for beam vibration and beam
buckling.
The following equations are given without a special derivat ion ( b,A,D11,ρare
constant values):
•Differential equation of ﬂexure (N=0)
d2w(x1)
dx2
1=−M(x1)
bD11,bD11d4w(x1)
dx4
1=q(x1) (7.2.33)
Table 7.2 Stress formulas for symmetric laminated beams, classical t heory
σ(k)
1(x1,x3) =σ(k)
1M(x1)+σ(k)
1B(x1,x3)
=Q(k)
11du(x1)
dx1−Q(k)
11d2w(x1)
dx2
1×3
=Q(k)
11N(x1)
bA11+Q(k)
11M(x1)
bD11x3
=E(k)
1
EN
effN(x1)
A11+E(k)
1
EM
effMx1)
Ix3,
σ(k)
5(x1,x3) =Q(x1)
EM
effIN
∑
j=k+1E(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
bD11d4w(x1,t)
dx4
1+ρAd2w(x1,t)
dt2=q(x1,t),ρ=1
hN
∑
k=1ρ(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(iωt).
•Buckling equation
d2M(x1)
dx2
1+N1(x1)d2w(x1)
dx2
1=0,bD11d4w(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,bD11d4w(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 composit es are stronger shear
deformable than metallic materials and the classical beam t heory 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 potent ial for a ﬂexural beam
(N=0,M/ne}ationslash=0)
Π(w)=1
2l/integraldisplay
0bD11/parenleftbiggd2w(x1)
dx2
1/parenrightbigg2
dx1−l/integraldisplay
0qdx1 (7.2.36)
and the kinetic energy
T(w)=1
2l/integraldisplay
0ρ/parenleftbigg∂2w
∂t/parenrightbigg2
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 qinx3-direction ( N=0,M/ne}ationslash=0,
ε5≈0,νi j≈0) based on the theorem of minimum of total potential energy i s
given in the form
Π[w(x1)] =1
2l/integraldisplay
0bD11/parenleftbiggd2w(x1)
dx2
1/parenrightbigg2
dx1−l/integraldisplay
0qdx1,
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
2l/integraldisplay
0bD11/parenleftbiggd2w(x1)
dx2
1/parenrightbigg2
dx1−1
2l/integraldisplay
0F/parenleftbiggdw(x1)
dx1/parenrightbigg2
dx1,
δΠ[w(x1)] = 0(7.2.39)
The variational formulation for free ﬂexural beam vibratio n (additional to ap-
proaches noted above the rotatory inertia effects are negle cted) can be given by
the Hamilton’s principle
H[w(x1,t)]=t2/integraldisplay
t1L[W(x1,t)]dt,δH[w(x1,t)]= 0 (7.2.40)
The variational formulations can be used for approximate an alytical solution with
the Rayleigh-Ritz procedure or numerical solutions.
In the second case of laminate beams, the loading is in the pla ne of lamination.
We restrict our considerations to symmetric layered beams a nd neglect all Poisson’s
ratio effects. The beam is illustrated in Fig. 7.5. For a symm etric layer stacking there
is no bending-stretching coupling and we have the constitut ive equations
N1=A11ε1,M1=b3
12hA11κ1
or for the beam resultants N=hN1,M=hM1
Fig. 7.5 Laminate beam
loaded in the plane of lamina-
tionq(x1)
x2x3x1
b
h

240 7 Modelling and Analysis of Beams
N=hA11ε1,M=b3
12A11κ1,A11=N
∑
k=1Q(k)
11h(k)(7.2.41)
The differential equations of ﬂexure ( N=0) are
d2w(x1)
dx2
1=−M(x1)
A11b3
12,12A11
b3d4w(x1)
dx4
1=q(x1) (7.2.42)
and the additional equation for the case N/ne}ationslash=0 is
hA11du(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, th e bending differential
equations can be written in the form, given above. The bendin g stiffness is obtained
by combining the results of orthogonal to plane loading and i n-plane loading.
As an example for a box-beam, we consider the beam as shown in F ig. 7.6, which
may be subjected to axial loads in x1-direction, a bending moment with respect to
thex2-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, EI, and
the torsional stiffness, GIt.
The axial force resultant (per unit width) in x1-direction is N1=A11ε1and the
axial load carried by the whole section is then
N(x1)=2NI
1b+2NII
1h=2[(A11)Ib+(A11)IIh]ε1 (7.2.44)
The extensional stiffness for the box cross-section is give n by
Fig. 7.6 Laminated box-beam
with identical top and bottom
panels I and vertical walls IIx2
x3x1
bl
tI
II
h

7.2 Classical Beam Theory 241
(EA)eff=2(A11)Ib+2(A11)IIh (7.2.45)
The box beam is bent in the ( x1−x3) plane, and the moment curvature relation is
M=/bracketleftigg
2(D11)Ib+2(A11)Ib/parenleftbiggh
2/parenrightbigg2
+2(A11)IIh3
12/bracketrightigg
κ1
≈/bracketleftigg
2(A11)Ib/parenleftbiggh
2/parenrightbigg2
+1
6(A11)IIh3/bracketrightigg
κ1(7.2.46)
Since the top and bottom panels are thin relative to the heigh t of the box proﬁle, i.e.
t≪h,(D11)Ican be neglected and the bending stiffness of the box cross-s ection is
(EI)eff≈2(A11)Ib/parenleftbiggh
2/parenrightbigg2
+1
6(A11)IIh3(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
66εI
6,NII
6=AII
66εII
6 (7.2.48)
In the elementary theory of strength of materials the equati on for the angle of twist
of a box-beam is given by
θ=1
2A/contintegraldisplayq(s)
Gtds (7.2.49)
q(s)is the shear ﬂow. In our case, Fig. 7.6, the displacements of t he contours of the
walls of the box beam are denoted by δIandδIIand the angle of twist becomes
θ=δI
(h/2)=δII
(b/2)withδI
l=εI
6,δII
l=εII
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 an alogous. The bending
stiffness is
(EI)eff=/bracketleftigg
2(D11)b+2(A11)b/parenleftbiggh
2/parenrightbigg2
+(A11)h3
12/bracketrightigg
≈A116h2b+h3
12(7.2.53)

242 7 Modelling and Analysis of Beams
Fig. 7.7 I-proﬁle with uni-
form thickness t
bt
h
ifD11≈0. Note that for a one-dimensional thin structural element w hich is sym-
metric with respect to all mid-planes and Poisson’s effect i s neglected, we have the
simple relationships
Q11=E1,A11=E1t,κ1=M
(EI)eff
Summarizing the classical beam equations it must be noted th at 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. ortho tropic axes of material sym-
metry in each ply are not parallel to the beam edges and anisot ropic shear coupling
is displayed.
7.3 Shear Deformation Theory
The structural behavior of many usual beams may be satisfact orily approximated by
the classical Euler-Bernoulli theory. But short and modera tely 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 shortc oming Timoshenko ex-
tended the classical theory by including the effect of trans verse shear deformation.
However, since Timoshenko’s beam theory assumed constant s hear strains through
the thickness ha shear correction factor is required to correct the shear st rain energy.
In this section we study the inﬂuence of transverse shear def ormation upon the
bending of laminated beams. The similarity of elastic behav ior 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 theo ry. Figure 7.8 illus-
trates the cross-section kinematics for the Bernoulli’s an d the Timoshenko’s bending
beam.

7.3 Shear Deformation Theory 243
x1x3
OBA
dx1u(x1)u1(x1)O′B′A′B′′A′′
w′(x1)ψ(x1)
w′(x1)
Bernoulli kinematicsﬂexure curveTimoshenko 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 e quations are identical
with (7.2.12) – (7.2.13), but from Sect. 5.1 the strains of th e Timoshenko’s beam are
ε1=∂u1
∂x1=du
dx1+x3dψ1
dx1,ε2≡0,ε3≡0,
ε5=∂u1
∂x3+∂w
∂x1= ψ1+dw
dx1, ε4≡0,ε6≡0(7.3.1)
i.e. we only have one longitudinal and one shear strain
ε1(x1,x3)=ε1(x1)+x3κ1(x1),ε5(x1,x3)=ψ1(x1)+w′(x1),
ε1(x1)=du
dx1,κ1(x1)=dψ1(x1)
dx1(7.3.2)
When the transverse shear strain are neglected it follows wi thε5≈0 that the rela-
tionship is ψ1(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 N/ne}ationslash=0,M/ne}ationslash=0, the constitutive equations
(stress resultants – strain relations) are given by
N=A11ε1+B11κ1,M=B11ε1+D11κ1,Q=ksA55γs(7.3.3)
with A11=bA11,B11=bB11,D11=bD11,γs=ε5and 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=bn
∑
k=1C(k)
11/parenleftig
x(k)
3−x(k−1
3)/parenrightig
=bn
∑
k=1C(k)
11h(k),
A55=bn
∑
k=1C(k)
55/parenleftig
x(k)
3−x(k−1)
3/parenrightig
=bn
∑
k=1Ck
55h(k),
B11=b1
2n
∑
k=1C(k)
11/parenleftbigg
x(k)
32−x(k−1)
32/parenrightbigg
,
D11=b1
3n
∑
k=1C(k)
11/parenleftbigg
x(k)
33−x(k−1)
33/parenrightbigg
ksis the shear correction factor (Sect. 5.3).
In the static case, the equilibrium equations for the undefo rmed beam element
(Fig. 7.2) yield again for lateral loading q/ne}ationslash=0
dM
dx1−Q=0,dQ
dx1+q=0 (7.3.4)
When considerations are limited to symmetric laminated bea ms the coupling stiff-
ness B11is zero and from (7.3.3) it follows that
M=D11dψ1
dx1,Q=ksA55/parenleftbigg
ψ1+dw
dx1/parenrightbigg
(7.3.5)
Substituting the relations (7.3.5) into (7.3.4) leads to th e differential equations of
ﬂexure
[D11ψ′
1(x1)]′−ksA55[ψ1(x1)+w′(x1)] = 0,
ksA55[ψ1(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 ψ1(x1)
[D11ψ′(x)]′′=−q(x) (7.3.7)
and with
Q=dM
dx1=[D11ψ′
1(x1)]′(7.3.8)
and
Q=ksA55[ψ1(x1)+w′(x1)]

7.3 Shear Deformation Theory 245
follows an equation for w′(x1)
w′(x1)=−ψ1(x1)+[D11ψ′
1(x1)]′
ksA55(7.3.9)
Summarizing the derivations above, the equations for a bent Timoshenko’s beam
are:
[D11ψ′
1(x1)]′′=−q(x1),
M(x1) = D11ψ′
1(x1),
Q(x1) = [ D11ψ′
1(x1)]′,
w′(x1) = −ψ1(x1)+[D11ψ′
1(x1)]′
ksA55(7.3.10)
When the laminated beam problem allows to write the bending m oment 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 determ ine ﬁrst ψ1(x1)and then
w(x1). Otherwise (7.3.6) or (7.3.10) are used to determine w(x1)andψ1(x1).
Integrating the second Eq. (7.3.6) with respect to x1, we obtain
ksA55[w′(x1)+ψ1(x1)]=−/integraldisplay
q(x1)dx1+c1
Substituting the result into the ﬁrst equation of (7.3.6) an d integrating again with
respect to x1yields
D11ψ′
1(x1) =−/integraldisplay/integraldisplay
q(x1)dx1dx1+c1x1+c2,
D11ψ1(x1) =−/integraldisplay/integraldisplay/integraldisplay
q(x1)dx1dx1dx1+c1x2
1
2+c2x1+c3(7.3.11)
Substituting ψ1(x1)andψ′
1(x1)in (7.3.9), considering (7.3.7) and integrating once
more with respect to x1we obtain
w(x1)=1
D11/bracketleftbigg/integraldisplay/integraldisplay/integraldisplay/integraldisplay
q(x1)dx1dx1dx1dx1+c1x3
1
6+c2x2
1
2+c3x1+c4/bracketrightbigg
−1
ksA55/bracketleftbigg/integraldisplay/integraldisplay
q(x1)dx1dx1+c1x1/bracketrightbigg
=wB(x1)+wS(x1)(7.3.12)
The transverse deﬂection consists of two parts. The bending partwB(x1)is the same
as derived in the classical theory. When the transverse stif fness goes to inﬁnity, the
shear deﬂection wS(x1)goes to zero, ψ1(x1)goes to−w′(x1)and the Timoshenko’s
beam theory reduces to the classical Bernoulli’s beam theor y.
The relations for the stresses σ1are the same as in the classical theory. The trans-
verse shear stress can be computed via a constitutive equati on in the Timoshenko
theory

246 7 Modelling and Analysis of Beams
σ(k)
5(x1,x3)=Q(k)
55Q(x1)
ksA55(7.3.13)
The variational formulation for a lateral loaded symmetric laminated beam is given
by
Π(w,ψ1)=Πi+Πa (7.3.14)
with
Πi=1
2l/integraldisplay
0/bracketleftigg
D11/parenleftbiggdψ1
dx1/parenrightbigg2
+ksA55/parenleftbigg
ψ1+dw
dx1/parenrightbigg2/bracketrightigg
dx1,
Πa=−l/integraldisplay
0q(x1)wdx1(7.3.15)
In the more general case of unsymmetric laminated beams and a xial and lateral
loadings we have Π(u,w,ψ1). The Πiexpression can be expanded to
Πi=1
2L/integraldisplay
0/bracketleftigg
A11/parenleftbiggdu
dx1/parenrightbigg2
+2B11du
dx1dψ1
dx1+D11/parenleftbiggdψ1
dx1/parenrightbigg2
+ksA55/parenleftbigg
ψ+dw
dx1/parenrightbigg2/bracketrightigg
dx1
(7.3.16)
andΠahas to include axial and lateral loads.
Since the transverse shear strains are represented as const ant through the lam-
inate thickness, it follows that the transverse stresses wi ll also be constant. In the
elementary beam theory of homogeneous beams, the transvers e shear stress varies
parabolically through the beam thickness and in the classic al laminate theory the
transverse shear stresses vary quadratically through laye r thickness. This discrep-
ancy between the stress state compatible with the equilibri um equations and the
constant stress state of the ﬁrst order shear deformation th eory can be overcome
approximately by introducing a shear correction factor (Se ct. 5.3).
The shear correction factor kscan be computed such that the strain energy W1
due to the classical transverse shear stress equals the stra in energy W2due to the ﬁrst
order shear deformation theory. Consider, for example, a ho mogeneous 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
σ13=τ1=3
2Q
bh/bracketleftigg
1−/parenleftbigg2x3
h/parenrightbigg2/bracketrightigg
,−h
2≤x3≤+h
2(7.3.17)
The transverse stress in the ﬁrst order shear deformation th eory is constant through
the thickness h
σ13=τ2=Q
bh,γ2=Q
ksG(7.3.18)
With W1=W2it follows that

7.3 Shear Deformation Theory 247
1
2/integraldisplay
(A)τ2
1
GdA=1
2/integraldisplay
(A)τ2
2
ksGdA,
3
5Q2
Gbh=1
ksQ2
2Gbh=⇒ks=5
6(7.3.19)
The shear correction factor for a general laminate depends o n lamina properties and
lamina stacking and is given here without a special derivati on by
1
ks=A55bN
∑
k=1x(k)
3/integraldisplay
x(k−1)
3g(k)2(x3)
G(k)dx3, (7.3.20)
g(k)(z)=d11/braceleftigg
−C(k)
11z2
2+k
∑
j=1/bracketleftig
C(j)
11−C(j−1)
11/bracketrightigz(j−1)2
2/bracerightigg
,C(0)
11≡0
d11=1/D11is the beam compliance, C(k)
11=E(k)
1.
Summarizing the beam equations for the ﬁrst order shear defo rmation theory
for symmetrically laminated cross-sections, including vi bration and buckling, the
following relations are valid for constant values of h,b,A,D11,ρ:
•Flexure equations (N=0,M/ne}ationslash=0)
ksA55[ψ1(x1)+w′(x1)]′+q(x1) = 0,
[D11ψ′
1(x1)]′−ksA55[ψ1(x1)+w′(x1)] = 0(7.3.21)
or Eqs. (7.3.10).
•Forced or free vibrations equations
ksA55[ψ1(x1,t)+w′(x1,t)]′−ρ0¨w(x1,t)+q(x1,t) = 0,
[D11ψ′
1(x1,t)]′−ksA55[ψ1(x1,t)+w′(x1,t)]−ρ2¨ψ1(x1,t) =0,(7.3.22)
ρ0=bn
∑
k=1ρ(k)/parenleftig
x(k)
3−x(k−1)
3/parenrightig
,ρ2=bn
∑
k=11
3ρ(k)/parenleftig
x(k)3
3−x(k−1)3
3/parenrightig
The terms involving ρ0andρ2are 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(iωt),ψ1(x1,t)=Ψ1(x1)exp(iωt)
•Buckling equations
ksA55[ψ1(x1)+w′(x1)]′−N(x1)w′′(x1) =0,
[D11ψ′
1(x1)]′−ksA55[ψ1(x1)+w′(x1)] = 0(7.3.23)

248 7 Modelling and Analysis of Beams
or with N(x1)=−F
D11/bracketleftbigg
1−F
ksA55/bracketrightbigg
w′′′′(x1)+Fw′′(x1)=0
The variational formulation for the symmetric bending beam is given by Eq.
(7.3.15).
For vibrations the Lagrange function L(w,ψ1)=T(w,ψ1)−Π(w,ψ1)yields the
Hamilton’s principle
H[w(x1,t),ψ1(x1,t)]=t2/integraldisplay
t1L[w(x1,t),ψ1(x1,t)]dt,
δH[w(x1,t),ψ1(x1,t)]= 0
with
Π[w(x1,t),ψ1(x1,t)] =1
2l/integraldisplay
0[D11ψ′
12+ksA55(ψ1+w′)2]dx1
−l/integraldisplay
0q(x1,t)wdx1,
T[w(x1,t),ψ1(x1,t)] =1
2l/integraldisplay
0[ρ0˙w2+ρ2˙ψ2
1]dx1(7.3.24)
For buckling problems with N(x1)=−Fit follows that
Π[w(x1,t),ψ1(x1,t)] =1
2l/integraldisplay
0[D11ψ′
12+ksA55(ψ1+w′)2]dx1
−1
2l/integraldisplay
0Fw′2dx1(7.3.25)
Equations (7.3.21) to (7.3.25) summarize the bending, buck ling and vibration dif-
ferential and variational statements for laminated beams b ased on the shear defor-
mation theory.
7.4 Sandwich Beams
The similarity of the elastic behavior between symmetric la minates and symmetric
sandwich beams in the ﬁrst order shear deformation theory (S ects. 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 t he ﬂexural and trans-
verse shear stiffness D11andA55the essential difference is at the level of stress
distribution. The model assumptions for sandwich composit es 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 transve rse shear deformation
can be transposed.
In the case of a symmetric sandwich beam with thin cover sheet s we have, for
example, the stiffness values
A11=2Af
11=2n
∑
k=1Q(k)
11h(k),
D11=2hcCf
11=hcn
∑
k=1Q(k)
11h(k)x(k),
As
55=hcGc
13(7.4.1)
nis the number of the face layers.
The coefﬁcient As
55can 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 sheet s. With the inverse
effective shear stiffness G−1
R, given by the Reuss model, and the effective shear
stiffness GV, given by the V oigt model, we have
GV=n
∑
k=1G(k)h(k)
h=A55
bh,G−1
R=n
∑
k=11
G(k)h(k)
h,ks=GR
GV(7.4.2)
The use of sandwich structures is growing very rapidly. Sand wich beams has a
high ratio of ﬂexural stiffness to weight and in comparison t o other beam struc-
tures they have lower lateral deformations, higher bucklin g resistance and higher
natural frequencies. As a result sandwich constructions qu ite often provide a lower
structural weight than other structural elements for a give n set of mechanical and
environmental loads.
The elastic behavior of sandwich beams was modelled by the la minate the-
ory, Sect. 4.3, but it is appropriately to distinguish thin a nd thick sandwich faces.
The differential equations or variational statements desc ribing 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 ord er theory may be
needed.
Because of the continuing popularity of sandwich structure s Sect. 7.4 intends to
recall and summarizes the results of Sects. 4.3 and 5.3 to cov er 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 th e
ﬂexural rigidy is ( D11≡DLa
11,Q11≡E1=E)
bD11=D=b/bracketleftbiggEf(hf)3
6+Efhfd2
2+Ec(hc)3
12/bracketrightbigg
=2Df+Do+Dc(7.4.3)
and both, 2 Dfand Dcare less than 1% of Doifd/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 s tructural sandwich
beam elements, the dominating term in ﬂexural rigidity is th at 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 ﬂexu ral rigidity will be
nearly lost.
The following derivations assume in-plane-, bending- and s hear stiffness for all
layers, i.e. for the faces and the core. Therefore we use the l aminate theory including
transverse shear, Sects. 4.3.3 and 5.3. All calculations ar e ﬁrst restricted to midplane
symmetric beams.
The bending strains vary linearly with x3over the cross-section:
εM
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 con stituent, but there is a
jump in the stresses at the face/core interfaces:
x3
x13
2
1Q
N
ME3=Ef
E2=Ec
E3=Efd=hc+hf
bhf
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)⇒σf=MEf
Dx3
σc=MEc
Dx3(7.4.6)
With Eq. (7.2.12) follows
M=bD11κ1,κ1=M
bD11=h3
12D11M
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
effM
Ix3(k=1,2,3) (7.4.8)
The strains due to in-plane loading are:
εN
1=N
3
∑
k=1Q(k)h(k)=N
3
∑
k=1E(k)h(k)=N
bhh
A11=N
EN
effA(7.4.9)
with
EN
eff=A11
h,A113
∑
k=1Q(k)
11h(k)=3
∑
k=1E(k)h(k),A=bh
εN
1is the strain of the neutral axis. The in-plane stresses foll ow to
σ(k)
1=E(k)
EN
effN
A(k=1,2,3)⇒σf=E(1)
EN
effN
A=E(3)
EN
effN
A
σc=E(2)
EN
effN
A(7.4.10)
The strains and stresses due to in-plane loads and bending ca n be superimposed.
In the same manner as outlined above a general deﬁnition can b e 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 bea ms all derivation can
be given for the upper part of the beam element ( x3≥0). The equilibrium equation in
thex1-direction yield with σ1dσ1≈σ1+(∂σ1/∂x1)dx1and no edge shear stresses
on the upper face

252 7 Modelling and Analysis of Beams
x3
x1
dx1τ(x3)(x(3)
3−x3)
σf
σcσf+dσf
σc+dσcσ5+dσ5
σ5σ1 σ1+dσ1
Fig. 7.10 Sandwich beam element b(x(3)
3−x3)dx1:σ(3)(x3)≡σf(x3),σ(2)(x3)≡σc(x3),
σ5(x3)≡τ(x3)
τ(x3)bdx1−(d+hf)/2/integraldisplay
x3/bracketleftbigg/parenleftbigg
σ1+∂σ1
∂x1dx1/parenrightbigg
−σ1/bracketrightbigg
bdx3=0
⇒τ(x3)=(d+hf)/2/integraldisplay
x3∂σ1
∂x1dx3=0 (7.4.11)
Using the relations dM(x1)/dx1=Q(x1)andσ1=M(E(x3)/D)x3we have
dσ1
dx1=Q(x1)
DE(x3)x3,τ(x3)=Q
bD(d+hf/2)/integraldisplay
x3E(x3)x3bdx3=Q
bDS(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/parenleftbiggh2
4−x2
3/parenrightbigg
;h=(d+hf)
For sandwich beams we have a more generalized deﬁnition for t he ﬁrst moment of
area:
(d+hf)/2/integraldisplay
x3E(x3)x3bdx3=

b/bracketleftbiggEfhfd
2+Ec
2/parenleftbigghc
2−x3/parenrightbigg/parenleftbigghc
2+x3/parenrightbigg/bracketrightbigg
,
|x3|≤hc
2
b/bracketleftbiggEf
2/parenleftbigghc
2+hf−x3/parenrightbigg/parenleftbigghc
2+hf+x3/parenrightbigg/bracketrightbigg
,
hc
2≤|x3|≤hc
2+hf(7.4.13)

7.4 Sandwich Beams 253
The shear stresses for the core and the faces are
τc(x3) =Q
D/bracketleftbiggEfhfd
2+Ec
2/parenleftbigg(hc)2
4−x2
3/parenrightbigg/bracketrightbigg
,
τf(x3) =Q
DEf
2/parenleftbigg(hc)2
4+hchf+(hf)2−x2
3/parenrightbigg (7.4.14)
The maximum shear stress appears at the neutral axes:
τmax=τc(x3=0)=Q
D/parenleftbiggEfhfd
2+Ec(hc)2
8/parenrightbigg
(7.4.15)
The shear stress in the core/face interface is
τc
min≡τf
max=τ/parenleftbigghc
2/parenrightbigg
=Q
D/parenleftbiggEfhfd
2/parenrightbigg
(7.4.16)
There is no jump in the shear stresses at the interfaces and th e 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 diffe rence between τc
maxand
τc
minis less than 1%. As it was outlined in Sect. 4.3, the stress equ ation 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
σf(x3)≈MEf
Do+2Dfx3,
σc(x3)≈0,
τf(x3)≈Q
Do+2DfEf
2/parenleftbigg(hc)2
4+hchf+(hf)2−x2
3/parenrightbigg
,
τc(x3)≈QEfhfd
2(Do+2Df)(7.4.18)
2. The core is weak, Ec≪Ef, and the faces are thin, hf≪hc
σf(x3)≈±M
bhfd,σc(x3)≈0,τf(x3)≈0,τc(x3)≈Q
bd(7.4.19)
This approximation can be formulated as: The faces of the san dwich beam carry
bending moments as constant tensile and compressive stress es 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-Sect ions
In engineering applications also sandwich beams with dissi milar faces are used,
Fig. 7.11. The ﬁrst moment of area is zero when integrated ove r the entire cross-
section and x3is the coordinate from the neutral axes
/integraldisplay
E(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 bec omes
S(x3)=/integraldisplay
E(x3)x3bdx3=/integraldisplay
E(x∗
3+e)bdx∗
3=0,e/integraldisplay
Edx∗
3=−/integraldisplay
Ex∗
3dx∗
3
For the sandwich cross-section, Fig. 7.11, follows
e/parenleftig
E(1)h(1)+E(2)h(2)+E(3)h(3)/parenrightig
=E(1)h(1)/parenleftbigg1
2h(1)+h(2)+1
2h(3)/parenrightbigg
+1
2E(2)h(2)/parenleftig
h(2)+h(3)/parenrightig
and we get an equation for the unknown value e
e=E(1)h(1)/parenleftig
h(1)+2h(2)+h(3)/parenrightig
+E(2)h(2)/parenleftig
h(2)+h(3)/parenrightig
2/parenleftbig
E(1)h(1)+E(2)h(2)+E(3)h(3)/parenrightbig (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)ord−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=/integraltextE(x3)x2
3bdx3yields in the general case
Fig. 7.11 Deﬁnition of the
neutral axis (N.A.) of an
unsymmetrical sandwich:
x3=x∗
3+ex2x3E(3)
E(2)
E(1)dh(3)
bhc=h(2)
h(1)ex∗
3
N.A.

7.4 Sandwich Beams 255
D=1
12/bracketleftig
E(1)(h(1))3+E(2)(h(2))3+E(3)(h(3))3/bracketrightig
+E(1)h(1)(d−e)2+E(3)h(3)e2+E(2)h(2)/bracketleftbigg1
2(h(2)+h(3))−e/bracketrightbigg2 (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 th e usual way
σ1(k)(x3)=ME(k)
Dx3,τ(k)(x3)=Q
bDS(x3) (7.4.26)
For sandwich beams with weak core and thin but dissimilar fac es the stress formulas
are approximately
σ(3)
1≡σf1
1≈MEf1
De=M
bhf1d,
σ(1)
1≡σf2
1≈ −MEf2
D(d−e)=−M
bhf2d,
τ(2)≡τc≈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 r ectangular, 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 rigidi tyG13. The shear
correction factor ksis determined approximately for sandwich beams with thin co ver
sheets with the Reuss formula (7.4.1), or more generally usi ng equivalent shear
strain energy, i.e the potential energy of the applied load e quals 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 a dapt the equations of
this section to the special case of sandwich beams.
The strains ε1andε5≡γare given, (7.3.2), as

256 7 Modelling and Analysis of Beams
ε1=du
dx1+x3dψ
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 0S

u′
ψ′
w′+ψ
 (7.4.29)
For static loading q(x1)/ne}ationslash=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)
IfN≡0 the neutral axes position xN.A.
3is constant along the length of the beam and
is given by
ε1(xN.A.
3)=u′+xN.A.
3ψ′=0,N=Au′+Bψ′=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 e quations for sandwich
beams
DRψ′′(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
DRψ′′′(x1)=−q(x1) (7.4.33)
and with
M′=Q=S(w′+ψ),M=Bu′+DRψ′,M′=Bu′′+DRψ′′
follow
w′(x1)=−ψ(x1)+1
S(Bu′′+DRψ′′)
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 unsymm etrical sandwich beams
DRψ′′(x1)−S[w′(x1)+ψ(x1)] = 0,DR=D−/parenleftbiggB2
A/parenrightbigg
S(w′′(x1)+ψ′(x1)) =−q(x1)
u′(x1) =−B
Aψ′(x1)(7.4.34)

7.4 Sandwich Beams 257
Derivation of the ﬁrst equation and setting in the second equ ation yield one uncou-
pled equation for ψ(x1)
DRψ′′′(x1)=−q(x1) (7.4.35)
The constitutive equations (7.4.4) give the relations
M(x1)=Bu′(x1)+Dψ′(x1),Q(x1)=S[w′(x1)+ψ(x1)] (7.4.36)
and with
M′(x1)=Q(x1)=Bu′′(x1)+Dψ′′(x1),u′(x1)=−B
Aψ′(x1)
follow
w′(x1)=−ψ(x1)+1
S[Bu′′(x1)+DRψ′′(x1)]=−ψ(x1)+DR
Sψ′′(x1)(7.4.37)
Thus we have three uncoupled differential equations:
DRψ′′′(x1) =−q(x1),
w′(x1) =−ψ(x1)+DR
Sψ′′(x1),
u′(x1) =B
Aψ′(x1)(7.4.38)
For symmetrically cross-sections is B≡0,DR≡Dand the differential equations
reduce to
Dψ′′′(x1) =−q(x1),
w′(x1) =−ψ(x1)+D
Sψ′′(x1)),
M(x1) =Dψ′(x1)orψ′(x1)=M(x1)
EM
effI
Q(x1) =Dψ′′(x1)orM′(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 tran sposed with D11=D,
ksA55=S. In dependence of the calculation of the stiffness DandSthe equation
are valid for sandwich beams with thin or thick faces.
The stresses σ1andτcan be calculated with the help of the stress formulas
derived in Sects. 7.4.1 and 7.4.2. For statically determina te structures, M(x1)and
Q(x1)can be calculated with the equilibrium equations and the las t two equations
(7.4.39) can directly used for static calculations.
We consider as an example the cantilever beam, Fig. 7.12, the n:

258 7 Modelling and Analysis of Beams
Fig. 7.12 Symmetrical can-
tilever beam with thin facesx3
x1 hfhf
hc≈d
lF
M(x1)=F(l−x1),Q(x1)=−F,
ψ′(x1)=M(x1)
D=1
DF(l−x1),
ψ(x1)=F
D/parenleftbigg
lx1−1
2×2
1/parenrightbigg
+C1,
ψ(0)=0⇒ C1=0,
w′(x1)=−F
D/parenleftbigg
lx1−1
2×2
1/parenrightbigg
+Q
S=−F
D/parenleftbigg
lx1−1
2×2
1/parenrightbigg
−F
S,
w(x1)=−F
D/parenleftbigg1
2lx2
1−1
6×3
1/parenrightbigg
−Fx1
S+C2,
w(0)=0⇒ C2=0,
w(x1)=−F
D/parenleftbigg1
2lx2
1−1
6×3
1/parenrightbigg
−Fx1
S=wB(x1)+wS(x1),
w(l)=−Fl3
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
σf=±MEf
Dd
2=±M
bdhf=±F(l−x1)
bdhf,τc=Q
bd=−F
bd,σc=τf=0
Consider w(l) =wB(l)+wS(l)it can be seen that the shear deformation strongly
depends on landS. It is important for short and shear weak beams and negligibl e
for slender shear stiff beams.
Summarizing the aspects of sandwich beams it could be demons trated in the
static case that the shear deformation theory for laminated beams is valid for sand-
wich beams, if the stiffness A11,B11andD11of a laminated beam are replaced by
the stiffness A,B,DandS, Eq. (7.4.3), of the sandwich beam. The same conclusion
is valid not only for bending but also for buckling and vibrat ion 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 o f core transverse
deformability is negligible on the bending, vibration and t he overall buckling of
sandwich beams. But in a special case of buckling, called fac e 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 buc kling 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 up on 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 o rder shear deformation
theory, Sect. 7.3. Section 7.4 considered some special aspe cts of sandwich beams
with thin or thick cover sheets and different stiffness of th e core.
In the present section the effects of hygrothermally induce d strains, stresses and
displacements are examined. We assume a moderate hygrother mal loading such that
the mechanical properties remain unchanged for the tempera ture and moisture dif-
ferences considered.
With Eqs. (4.2.63) to (4.2.68) the beam equations (7.2.1) ha ve additional terms
ε1(x3)=ε1+x3κ1=σ1(x2,x3)
E1(x2,x3)
+ [αth(x2,x3)T(x2,x3)+αmo(x2,x3)M∗(x2,x3)],(7.5.1)
σ1(x2,x3) =E1(x2,x3)[ε1+x3κ1
−αth(x2,x3)T(x2,x3)−αmo(x2,x3)M∗(x2,x3)](7.5.2)
αth,αmoare the thermal and moisture expansion coefﬁcients, Tthe temperature
change and M∗the weight of moisture absorption per unit weight. Equation s (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 co mposite beam are
/bracketleftbigg˜N
˜M/bracketrightbigg
=/bracketleftbigg
a b
b d/bracketrightbigg/bracketleftbigg
ε1
κ1/bracketrightbigg
,/bracketleftbigg
ε1
κ1/bracketrightbigg
=/bracketleftbigg
a b
b d/bracketrightbigg−1/bracketleftbigg˜N
˜M/bracketrightbigg
(7.5.4)
The stress formula (7.2.6) yields with Eq. (7.2.5)

260 7 Modelling and Analysis of Beams
σ1(x2,x3) =(dN−bM)+(aM−bN)x3
ad−b2E1(x2,x3)
−E1(x2,x3)[αthT+αmoM∗](7.5.5)
For double symmetric cross-sectional geometry the couplin g 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
σf(x3) = ( ˜N/A+x3˜M/I)(Ef/E1)−Ef(αthT+αmoM∗),
σm(x3) = ( ˜N/A+x3˜M/I)(Em/E1)−Em(αthT+αmoM∗)(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(αthT+αmoM∗) (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 lam inate beam in the frame
of the classical laminate theory and include hygrothermal l oads, (7.2.27) yield
σ(k)
1=E(k)
1˜N
EN
effA+x3˜M
EM
effI−E(k)
1(αth(k)T(k)+αmo(k)M∗(k)) (7.5.8)
The differential equations for deﬂection and midplane disp lacement 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 includin g shear deformation can
be found. The differential equation for a symmetric Timoshe nko’s beam with lateral
loading and hygrothermal effects follows with (7.3.10)
[D11ψ′
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 buck ling 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 the ory and by (7.3.21) to
(7.3.23) for the Timoshenko’s beam theory including transv erse 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)≡bD11wB(x1)
=C1x3
1
6+C2x2
1
2+C3x1+C4+/integraldisplay/integraldisplay/integraldisplay/integraldisplay
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 st iffness 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+ρA
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)=−ρA
bD11W(x1)¨T(t) (7.6.3)
or¨T(t)
T(t)=−ρA
bD11W′′′′(x1)
W(x1)=−ω2(7.6.4)
We get two differential equations
¨T(t)+ωT(t)=0,W′′′′(x1)−ρA
bD11ω2W(x1)=0 (7.6.5)
with the solutions
T(t) = Acosωt+Bsinωt,
W(x1) = C1cosλ
lx1+C2sinλ
lx1+C3coshλ
lx1+C4sinhλ
lx1,
/parenleftbiggλ
l/parenrightbigg4
=ρA
bD11ω2(7.6.6)
The vibration mode is periodic, and ωis called the natural circular frequency. The
mode shapes depend on the boundary conditions of the beam. Co nsider, 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=C2sinλ=0, which implies that

262 7 Modelling and Analysis of Beams
λ=nπ,ω2
n=n4π4
ρAbD11
l4,ωn=/parenleftignπ
l/parenrightig2/radicaligg
bD11
ρA(7.6.7)
For each nthere is a different natural frequency and a different mode s hape. The
lowest natural frequency, corresponding to n=1, is termed the fundamental fre-
quency. If the laminate beam is unsymmetric to the middle sur face, i.e. B11/ne}ationslash=0,
then D11in 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 Timosh enko vibration equa-
tion (7.3.22), involves considerable analytical complica tions. To prove whether
the transverse shear deformation can be important for the na tural frequencies,
we compare the natural frequencies for a simply supported Be rnoulli and Timo-
shenko beam. Using (7.3.22) the boundary conditions for the Timoshenko beam are
w(0,t)=w(l,t)=ψ1(0,t)=ψ1(l,t)=0 and by introducing
w=Asinωtsinnπx1
l,ψ1=Bsinωtcosnπx1
l,
in Eqs. (7.3.22) we can calculate the natural frequencies
ω2
n=n4π4
ρAbD11
l4//parenleftbigg
1+π2D11n2
l2ksA55/parenrightbigg
,ρA≡ρ0 (7.6.8)
i.e.
ωn=ωBernoulli
n/radicaligg
1
1+(n2π2D11)/(l2ksA55)
Transverse shear deformation reduces the values of vibrati on 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 t o (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=π2bD11
l2(Minimum Euler load, Bernoulli beam)
Fcr=π2bD11
l21
1+π2D11
l2ksA55(Minimum buckling load, Timoshenko beam)
(7.6.9)
The buckling loads for clamped beams or beams with more gener al boundary con-
ditions can be calculated analytically analogous to eigenf requencies of vibration
problems.

7.7 Problems 263
For non constant cross-section, stiffness or material para meters there are no exact
analytical solutions. Approximate analytical solutions c an be found with the help of
the Rayleigh-Ritz procedure. In Sect. 7.7 exact and approxi mate 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 strengt h in tension so that the
entire tensile load associated with the bending moment is ca rried by the steel rein-
forcement. Calculate the stresses σm(x3)andσf(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 αhfrom 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 respectiv e
moduli. The stress resultant N(x1)must be zero
σfAf−1
2σm(αh)bαh=0,σm(αh)=σmax
m
With (7.2.1) follows
σf=(h−αh)κ1Ef,σm(αh)=αhκ1Em
i.e.
(h−αh)EfAf−1
2(αh)2bEm=0,
α=EfAf
Embh/parenleftigg
−1+/radicaligg
1+2bh
AfEm
Ef/parenrightigg
or
x3
x1αhb
(h−αh)h
/parenleftbigg
h−αh
3/parenrightbigg
Afσmax
m1
2σmax
mbαh
(σfAf)MAc≡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
σfAf=1
2σmbh2α2
M=(σfAf)/parenleftbigg
h−αh
3/parenrightbigg
=σfAfh/parenleftig
1−α
3/parenrightig
=κ1EfAf(h−αh)/parenleftig
h−α
3/parenrightig
The maximal stress in the concrete is
σm(αh)=−κ1αhEm=MEmαh
EfAf(h−αh)(h−αh/3)
and the reinforcement stress is
σf=κ1(h−αh)Ef=M
Af(h−αh/3)
Exercise 7.2. A symmetric cross-ply laminate beam is shown in Fig. 7.14. Th e 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
x1x3
Lh
3h
3h
3hq(x1)=q0sinπx1
L
Fig. 7.14 Simply supported cross-ply laminated beam [0/90/0]

7.7 Problems 265
Calculate a approximative solution using the Timoshenko be am 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 ¯D11and the shear stiffness ks¯A55follow 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
2L/integraldisplay
0/bracketleftigg
¯D11/parenleftbiggdψ
dx1/parenrightbigg2
+ks¯A55/parenleftbigg
ψ+dw
dx1/parenrightbigg2/bracketrightigg
dx1−L/integraldisplay
0q0sin/parenleftigπx1
L/parenrightig
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/parenleftigπx1
L/parenrightig
,˜ψ(x1)=b1cos/parenleftigπx1
L/parenrightig
and it follows
˜Π(˜w,˜ψ) =1
2L/integraldisplay
0/bracketleftbigg
¯D11/parenleftig
b1π
L/parenrightig2
sin2πx1
L
+ks¯A55/parenleftig
b1cosπx1
L+a1π
Lcosπx1
L/parenrightig2/bracketrightbigg
dx1
−L/integraldisplay
0q0sin/parenleftigπx1
L/parenrightig/parenleftig
a1sinπx1
L/parenrightig
dx1=˜Π(a1,b1)
With (2.2.41) must be ∂˜Π/∂a1=0,∂˜Π/∂b1=0 which yields the two equations
/parenleftbigg¯D11
ks¯A55π
L+L
π/parenrightbigg
b1+ a1=0,
b1+π
La1=q0L
ks¯A55π

266 7 Modelling and Analysis of Beams
and the solution for the unknown constants a1,b1
a1=q0L4
¯D11π4/parenleftbigg
1+¯D11
ks¯A55π2
L2/parenrightbigg
,
b1=−q0L3
¯D11π3
The approximate solutions are now
˜w(x1) =q0L4
¯D11π4/parenleftbigg
1+¯D11
ks¯A55π2
L2/parenrightbigg
sinπx1
L,
˜ψ(x1) =−q0L3
¯D11π3cosπx1
L
Note that
˜wmax=˜w/parenleftbigg
x1=L
2/parenrightbigg
=q0L4
¯D11π4/parenleftbigg
1+¯D11
ks¯A55π2
L2/parenrightbigg
The transverse deﬂection consists of two parts
˜wB(x1) =q0L4
¯D11π4sinπx1
L(bending deﬂections) ,
˜wS(x1) =q0L4
¯D11π4¯D11
ks¯A55π2
L2sinπx1
L
Forks¯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¯A55π2
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/parenleftbigg
x1=L
2/parenrightbigg
=q0L2
π2=12,64 Nm,
˜Qmax=˜Q(x1=0)=q0L
π=52,67 N
The strains ε1follow from Eqs. (7.3.1) or (7.3.2) and (7.3.10)
ε1(x1)=x3ψ′(x1)=x3κ1=x3M(x1)/¯D11
ε1(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/parenleftbigg
x1=L
2,x[3]
3/parenrightbigg
=2,51 10−5,ε(3)
1/parenleftbigg
x1=L
2,x[2]
3/parenrightbigg
=0,84 10−5
The bending stresses σ1(x1,x3)in the 3 layers are for x1=L/2 and x3=x(k)
3
˜σ(3)
1(x(3)
3)=˜ε1(x(3)
3)E′
1=4,327 MPa,
˜σ(3)
1(x(2)
3)=˜ε1(x(2)
3)E′
1=1,448 MPa,
˜σ(2)
1(x(2)
3)=˜ε1(x(2)
3)E′
2=0,579 MPa,
˜σ(2)
1(0)=˜ε1(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 inﬂuen ce 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)sinωt,ψ1(x1,t)=Ψ(x1)sinωt
These functions are substituted in (7.3.22)
ks¯A55[W′′(x1)+Ψ′(x1)]+ρ0ω2W(x1) = 0,
¯D11Ψ′′(x1)−ks¯A55[W′(x1)+Ψ(x1)]+ρ2ω2Ψ(x1) =0
Now we can substitute
ks¯A55Ψ′(x1)=−ρ0ω2W(x1)−ks¯A55W′′(x1),
i.e.
Ψ′(x1)=−ρ0ω2
ks¯A55W(x1)−W′′(x1)
into the derivative of the second equation and we ﬁnd
¯D11W′′′′(x1)+/parenleftbigg¯D11ρ0
ks¯A55+ρ2/parenrightbigg
ω2W′′(x1)
−/parenleftbigg
1−ω2ρ2
ks¯A55/parenrightbigg
ρ0ω2W(x1)=0
or
aW′′′′(x1)+bW′′(x1)−cW(x1)=0
with the coefﬁcients

268 7 Modelling and Analysis of Beams
a=¯D11,b=/parenleftbigg¯D11
ks¯A55+ρ2
ρ0/parenrightbigg
ρ0ω2,c=/parenleftbigg
1−ω2ρ2
ks¯A55/parenrightbigg
ρ0ω2
The linear differential equation of 4th order has constant c oefﬁcients and the general
solutions follow with the solutions λiof the bi-quadratic characteristic equation
aλ4−bλ2−c=0
i.e.(2aλ2−b)2=b2+4ac
λ1−4=±/radicalbigg
1
2a(b±/radicalbig
b2+4ac)
W(x1)=C1sinλ1×1+C2cosλ2×1+C3sinhλ3×1+C4cosh λ4×1
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 C1sinλ1L=0
which implies
λ1n=nπ
L=λn
The bi-quadratic equation can be written alternatively in t erms of ω
Aω4−Bω2+C=0
with
A=ρ2
ks¯A55,B=/bracketleftbigg
1+/parenleftbigg¯D11
ks¯A55+ρ2
ρ0/parenrightbigg
λ2/bracketrightbigg
,C=¯D11
ρ0λ4
i.e. the roots of the equation are
(ω2)1/2=1
2A(B±/radicalbig
B2−4AC)
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
ω2=C
B
with
B=1+¯D11
ks¯A55λ2,C=¯D11
ρ0λ4

7.7 Problems 269
and for ks¯A55→∞follow with ˜B→1 the natural frequency for the Bernoulli beam
model. Substitute λn=nπ/Lfor the simply supported beam we obtain the results:
General case
ω2
n=ks¯A55
2ρ2/bracketleftbigg
1+/parenleftbigg¯D11
ks¯A55+ρ2
ρ0/parenrightbigg/parenleftignπ
L/parenrightig2
−/radicaligg/bracketleftbigg
1+/parenleftbigg¯D11
ks¯A55+ρ2
ρ0/parenrightbigg/parenleftignπ
L/parenrightig2/bracketrightbigg2
−4ρ2
ks¯A55¯D11
ρ0/parenleftignπ
L/parenrightig4

Rotatory inertia neglected ( ρ2≡0)
ω2
n=/parenleftignπ
L/parenrightig4¯D11
ρ0
1−/parenleftignπ
L/parenrightig2¯D11
ks¯A55+¯D11/parenleftignπ
L/parenrightig2

=/parenleftignπ
L/parenrightig4¯D11
ρ0
1
1+¯D11/parenleftignπ
L/parenrightig2
/ks¯A55

Classical beam theory ( ks¯A55→∞,ρ2=0)
ω2
n=/parenleftignπ
L/parenrightig4¯D11
ρ0
Conclusion 7.2. (ωTimoshenko
n )2<(ωBernoulli
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,ρ2/ne}ationslash=0) we have A=0,B=1+λ2ρ2/ρ0,C=λ4¯D11/ρ0,
i.e.
ω2
n=/parenleftignπ
L/parenrightig4¯D11
ρ0
1
1+/parenleftignπ
L/parenrightig2ρ2
ρ0

and we see that also the rotatory inertia decreases the eigen frequencies. All formulas
can be used for computing natural frequencies for all symmet ric laminate and sand-
wich beams. The values for L,ρ0,ρ2,ks,¯D11,¯A55correspond to the special beam
model. Note that the classical laminate theory and the negle cting of rotatory inertia
lead to a overestimation of the natural frequencies.
Exercise 7.4. Calculate the buckling load of a simply supported and a clamp ed sym-
metric laminate or sandwich beam. Compare the results for th e 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/parenleftbigg
1−F
ks¯A55/parenrightbigg
w′′′′(x1)+Fw′′(x1)=0

270 7 Modelling and Analysis of Beams
or
w′′′′(x1)+k2w′′(x1)=0,k2=F
¯D11/parenleftbigg
1−F
ks¯A55/parenrightbigg,Fk2¯D11
1+k2¯D11/ks¯A55
The linear differential equation with constant coefﬁcient s has the characteristic
equation
λ4+k2λ2=0⇒λ2(λ2+k2)=0
with the four roots
λ1/2=0,λ3/4=±ik
and the general solution is
w(x1)=C1sinkx1+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 C1/ne}ationslash=0 follows sin kL=0 implies kL=
nπ,k=nπ/L. Substituting kinto the equation for Fwe obtain
F=/parenleftignπ
L/parenrightig2¯D11
ks¯A55
ks¯A55+/parenleftignπ
L/parenrightig2¯D11

=/parenleftignπ
L/parenrightig2¯D11
1−¯D11/parenleftignπ
L/parenrightig2
/ks¯A55
1+¯D11/parenleftignπ
L/parenrightig2
/ks¯A55

The critical buckling load Fcris given for the minimum ( n=1)
F=/parenleftigπ
L/parenrightig2¯D11
1
1+/parenleftigπ
L/parenrightig2¯D11/ks¯A55

For the classical beam model is ks¯A55→∞and we obtain
F=/parenleftigπ
L/parenrightig2¯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,
¯D11ψ′′(x1)−ks¯A55[w′(x1)+ψ(x1)] = 0
The ﬁrst equation yields
ks¯A55ψ′(x1) =−(ks¯A55−F)w′′(x1),
ks¯A55ψ(x1) =−(ks¯A55−F)w′(x1)+K1
The boundary conditions lead the equations
C2+C4=0,C1sinkL+C2coskL+C3L+C4=0,
−/parenleftbigg
1−F
ks¯A55/parenrightbigg
kC1−C3=0,
−/parenleftbigg
1−F
ks¯A55/parenrightbigg
(kC1coskL−kC2sinkL)−C3=0
Note that
F
k2=¯D11
1+k2¯D11
ks¯A55=/parenleftbigg
1−F
ks¯A55/parenrightbigg
¯D11,
i.e. /parenleftbigg
1−F
ks¯A55/parenrightbigg
=1
1+k2¯D11
ks¯A55,
expressing C4andC3in terms of C1andC2and setting the determinant of the re-
maining homogeneous algebraic equations zero we obtain the buckling equation
2(coskL−1)/parenleftbigg
1+k2¯D11
ks¯A55/parenrightbigg
+kLsinkL=0
With ks¯A55→∞follows the buckling equation for the classical beam
kLsinkL+2cos kL−2=0
Conclusion 7.3. Transverse shear deformation has the effect of decreasing t he
buckling loads, i.e. the classical laminate theory overest imates buckling loads.
The buckling equations can be applied to all symmetric lamin ate 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 for mulation for applied dis-
tributed transverse loading and calculate the Euler differ ential 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 stiffne ss of sandwich beams,
Sect. 7.4, we have
Π(u,w,ψ)=1
2l/integraldisplay
0[Au′2+2Bu′ψ′+Dψ′2+S(w′+ψ)2]dx1−l/integraldisplay
0qwdx1
Taking the variation δΠ=0 one can write the following equation
δΠ=l/integraldisplay
0{Au′δu′+B(u′δψ′+ψ′δu′)+Dψ′δψ′
+S[(w′+ψ)δw′+(w′+ψ)δψ]}dx1−l/integraldisplay
0qδwdx1=0
Integrating by parts, i.e
l/integraldisplay
0f′g′dx=[f′g]l
0−l/integraldisplay
0f′′gdx
yield
l/integraldisplay
0{(Au′′+Bψ′′)δu+[Bu′′+Dψ′′−S(w′+ψ)]δψ+S[(w′′+ψ′)+q]δw}dx1
−[(Au′+Bψ′)δu]l
0−[(Bu′+Dψ′)δψ]l
0−S[(w′+ψ)δw]l
0=0
and the associated differential equations and boundary con ditions are
Au′′+Bψ′′=0,
Bu′′+Dψ′′−S(w′+ψ) =0,
S(w′′+ψ′)+q=0
Putting in u′′=−(B/A)ψ′′into the second equation yield
/parenleftbigg
D−B2
A/parenrightbigg
ψ′′−S(w′+ψ)=0
and we obtain the differential equations (7.4.32)
DRψ′′−S(w′+ψ) =0,
S(w′′+ψ) =−q
The boundary conditions for x1=0,lare

7.7 Problems 273
u=0 or Au′+Bψ′=N=0
ψ=0 or Bu′+Dψ′=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 s impliﬁed: B=
0,DR=D.
Exercise 7.6. A sandwich beam is modelled by the differential equations an d
boundary conditions of Excercise 7.5. Calculate the exact s olution 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+C1x2
1
2+C2x1+C3,
ψ′(0)=0⇒C2=0,
ψ′(l)=0⇒C1=q0l
2DR,
w′(x1) =−ψ(x1)+DR
Sψ′′=q0x3
1
6DR−q0lx2
1
4DR+C3−q0
S/parenleftbigg
x1−l
2/parenrightbigg
,
w(x1) =q0
24DR(x4
1−2lx3
1)+C3x1−q0
2S/parenleftbig
x2
1−lx1/parenrightbig
,
w(l)=0⇒C3=q0l3
24DR,
u′(x1) =−B
Aψ′(x1)=B
Aq0
2DR(x2
1−lx1),
u(x1) =B
Aq0
12DR(2×3
1−3lx1)+C4,
u(0)=0⇒C4=0
Finally we get

274 7 Modelling and Analysis of Beams
ψ(x1) =−q0x3
1
6DR+q0l
2DRx2
1
2+q0l3
24DR,
w(x1) =q0
24DR(x4
1−2lx3
1+l3x1)+q0
2S/parenleftbig
lx1−x2
1/parenrightbig
,
u(x1) =B
Aq0
12DR(2×3
1−3lx1)
For symmetrical beams the solution simpliﬁed with B=0,DR=Dto
w(x1) =q0l4
24D(x4
1−2×3
1+x1)+q0l2
2S(x1−x2
1)
=wB(x1)+wS(x1),
ψ(x1) =q0l3
24D(−1+6×2
1−4×3
1)
u(x1)≡0, x1=x/l

Chapter 8
Modelling and Analysis of Plates
The modelling and analysis of plates constituted of laminat e or sandwich material is
a problem of more complexity than that of beams, considered i n Chap. 7. Generally,
plates are two-dimensional thin structure elements with a p lane 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 restric ted to rectangular plates
including the special case of a plate strip, i.e. a rectangul ar 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 sti ffness, 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 pr oblem. The analysis is
x1,u1x2,u2x3,u3
abh
x1x1
x2x2
x3x3
N1
N2
NnQ1
Q2
QnN6
N6
Nnt
M1
M2
M6M6
Mn
Mnta b
Fig. 8.1 Rectangular plate. aGeometry, bforce resultants N1,N2,N6,Q1,Q2and moment resul-
tants M1,M2,M6.Nn,Nnt,QnandMn,Mntare force and moment resultants for an oblique edge
275 © Springer Nature Singapore Pte Ltd. 2018
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 in troduction to the clas-
sical plate theory and the plate theory including transvers e shear deformations. The
derivations of the principal equations for plates relies up on 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 lay ers. These plates present
couplings of stretching and bending, stretching and twisti ng and bending and twist-
ing and the design engineer has to look for simpliﬁcations.
The ﬁrst and most important simpliﬁcation is to design symme tric laminates
for which no coupling exists between in-plane forces and ﬂex ural moments. The
coupling terms Bi jof the constitutive equations vanish. An additional simpli ﬁca-
tion occurs when no bending-twisting coupling exist, i.e th e terms D16andD26are
zero. As we discussed in Sect. 4.2, in some cases of layer stac king these coupling
terms decrease with an increasing number of layers. Symmetr ic laminates for which
no bending-twisting coupling exists are referred to as spec ially orthotropic lami-
nates. These laminates are considered in detail in this chap ter, because analytical
solutions exist for various loadings and boundary conditio ns. Specially orthotropic
plates are obtained for single layer plates with orthotropi c material behavior or sym-
metric cross-ply laminates. Symmetric balanced laminates with a great number of
layers have approximately a specially orthotropic behavio r. This class of laminates
is greatly simpliﬁed and will be used to gain a basic understa nding 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 cons tant state of trans-
verse shear stresses, but the transverse normal stress is of ten neglected. In the frame-
work of this plate theory, the computation of interlaminar s hear stresses through
constitutive equations is possible, which is simpler than d eriving them through equi-
librium equations.
The most signiﬁcant difference between the classical and ﬁr st-order shear defor-
mation theory is the effect of including transverse shear de formation in the predic-
tion of deﬂections, frequencies or buckling loads. It can be noted that the classical
laminate theory underestimates deﬂections and overestima tes 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 nume rical methods, ap-
plication of analytical methods are much more restricted th an 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 Kirch hoff hypotheses of the
classical plate theory remains valid:
•Transverse normals before deformation remain straight aft er deformation and
rotate such that they remain normal to the middle surface.
•Transverse normals are inextensible, i.e. they have no elon gation.
These assumptions imply that the transverse displacement wis independent of the
thickness coordinate x3, the strains ε3,ε4andε5are zero and the curvatures κiare
given by
[κ1κ2κ6]=/bracketleftbigg
−∂2w
∂x2
1−∂2w
∂x2
2−2∂2w
∂x1∂x2/bracketrightbigg
(8.2.1)
Figure 8.1 shows the plate geometry and the plate stress resu ltants. The equilibrium
equations will be formulated for a plate element d x1dx2(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)
x1x2x3
N1dx2
(N1+dN1)dx2N2dx1 (N2+dN2)dx1N6dx1
(N6+dN6)dx1
dx2 dx2dx1dx1h h
M6dx1
(M6+dM6)dx1M6dx2
(M6+dM6)dx2M1dx2
(M1+dM1)dx2M2dx1(M2+dM2)dx1Q1dx2
(Q1+dQ1)dx2Q2dx1
(Q2+dQ2)dx1p1dx1dx2p2dx1dx2
p3dx1dx2
a b
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 resul tants N1,N2andN6are
uncoupled with the moment resultants M1,M2andM6
∂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 t he static equations for
the undeformed plate element. The further considerations n eglect the in-plane plate
loads p1and p2, i.e. p1=p2=0,p3/ne}ationslash=0. In-plane reactions can be caused by
coupling effects of unsymmetric laminates or sandwich plat es.
Putting the constitutive equations

NNN
···
MMM
=
AAA…BBB
. . . .
BBB…DDD

εεε
···
κκκ
 (8.2.4)
into the equilibrium (8.2.3) and replacing using Eqs. (5.2. 3) the in-plane strains εi
and the curvatures κiby
εεεT= [ε1ε2ε6] =/bracketleftbigg∂u
∂x1∂v
∂x2/parenleftbigg∂u
∂x2+∂v
∂x1/parenrightbigg/bracketrightbigg
,
κκκT= [κ1κ2κ6] =/bracketleftbigg
−∂2w
∂x2
1−∂2w
∂x2
2−2∂2w
∂x1∂x2/bracketrightbigg (8.2.5)
gives the differential equations for general laminated pla tes
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 e quations for the displace-
ments u(x1,x2),v(x1,x2),w(x1,x2). Equation (8.2.6) can be formulated in matrix
form as 
L11L12L13
L21L22L31
L31L32L33

u
v
w
=
0
0
p
,Li j=Lji (8.2.7)
The differential operators are given in App. C.
The differential operators L11,L12andL22are of second order, L13andL23of
third order and L33of fourth order. The homogeneous part of the coupled partial
differential equations (8.2.7) can be reduced to one partia l 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 equatio ns four boundary conditions
must be prescribed for each edge of the plate. The classical b oundary conditions are:
Either
Nnoru,Nntorv,Mnor∂w
∂n,Vn≡Qn+∂Mnt
∂torw(8.2.9)
must be speciﬁed. The subscripts nandtin 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 separ ately to the shear force
resultant Qnand the twisting moment Mntbut 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 respons e variables. One com-
ponent of these pairs involve a force or a moment resultant, t he other a displace-
ment or a rotation. Take into account that in addition to the e dge 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 e dges having elastic con-
straints, are used, e.g. the transverse and/or rotatory pla te conditions
Vn±cTw=0; Mn±cR∂w
∂n=0 (8.2.11)
cTandcRdenote the spring stiffness of the constraints.
In applying the boundary conditions (8.2.9) it is useful to h ave explicit expres-
sions for the stress resultants in a displacement formulati on. 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/parenleftbigg∂u
∂x2+∂v
∂x1/parenrightbigg

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/parenleftbigg∂u
∂x2+∂v
∂x1/parenrightbigg
−B12∂2w
∂x2
1−B22∂2w
∂x2
2−2B26∂2w
∂x1∂x2,
N6=A16∂u
∂x1+A26∂v
∂x2+A66/parenleftbigg∂u
∂x2+∂v
∂x1/parenrightbigg
−B16∂2w
∂x2
1−B26∂2w
∂x2
2−2B66∂2w
∂x1∂x2,
M1=B11∂u
∂x1+B12∂v
∂x2+B16/parenleftbigg∂u
∂x2+∂v
∂x1/parenrightbigg
−D11∂2w
∂x2
1−D12∂2w
∂x2
2−2D16∂2w
∂x1∂x2,
M2=B12∂u
∂x1+B22∂v
∂x2+B26/parenleftbigg∂u
∂x2+∂v
∂x1/parenrightbigg
(8.2.12)
−D12∂2w
∂x2
1−D22∂2w
∂x2
2−2D26∂2w
∂x1∂x2,
M6=B16∂u
∂x1+B26∂v
∂x2+B66/parenleftbigg∂u
∂x2+∂v
∂x1/parenrightbigg
−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
∂2×2−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
∂2×2−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
∂2×2−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 jfor some special cases are given in App. C.
1.Symmetric laminates
Because all coupling stiffness Bi jare 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
L11L120
L12L220
0 0 L33

u
v
w
=
0
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 coupl ed, but with
A16=A26=0,D16=D26=0 the differential operators L11,L22,L33andL12
are reduced. It is no in-plane tension/shearing coupling an d no bending/twisting
coupling.
3.Balanced laminates
For general balanced laminates with A16=A26only the in-plane ten-
sion/shearing coupling is zero, for an antisymmetric balan ced 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

L11L120
L12L220
0 0 L33

u
v
w
=
0
0
p3

with simpliﬁed differential operators L11andL22. 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=−B11or 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 p late.
Summarizing the mathematical structures of the differenti al 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 i s not much simpler
as for a general unsymmetric, unbalanced laminate
•Compared to the general case the mathematical structure of t he symmetric cross-
ply laminate is nearly trivial. A symmetric cross-ply is ort hotropic with respect
to both in-plane and bending behavior, and both are uncouple d.
•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-p lane and
transverse loading.
•For special layer stacking also the force and moment resulta nt Eqs. (8.2.12) are
reduced to more simple equations.
The following developments are restricted to general symme tric 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 Ta ble 8.1 standard bound-
ary conditions are also expressed. The necessary and sufﬁci ent number of boundary
conditions for plates considered here are two at each of the b oundaries. The stan-
dard conditions for the free edge reduce the three static con ditions Mn=0,Qn=0
andMnt=0 to two conditions Mn=0,Vn=0, where Vn=Qn+∂Mnt/∂t=0 is as
discussed above the Kirchhoff effective shear resultant. I n order to avoid mistakes
in the application the equations of Table 8.1, a summary of pl ate stiffness is given.
Table 8.2 contains the plate stiffness for single layer plat es. 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 mois ture changes. In this
case (4.2.63), (4.2.64) must be used instead of (8.2.4) to pu t 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 co nsideration to symmetric
plates. In the case of forced transversal vibration the mome ntum equilibrium equa-
tion (8.2.3) has an additional inertial term
∂2M1
∂x2
1+2∂2M6
∂x1∂x2+∂2M2
∂x2
2=−p3+ρh∂2w
∂t2(8.2.14)
M1,M2,M6,wandp3are functions of x1,x2and the time t,his the total thickness of
the plate and ρthe mass density
h=n
∑
k=1h(k),ρ=1
hn
∑
k=1ρ(k)/parenleftig
x(k)
3−x(k−1)
3/parenrightig
=1
hn
∑
k=1ρ(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 d iscussed 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/ne}ationslash=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
=
D11D12D16
D12D22D26
D16D26D66

−∂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
=
D11D120
D12D220
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
/parenleftbigg
L33+ρh∂
∂t/parenrightbigg
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 jh3
12
Specially orthotropic single layer (on-axis)
D11=Q11h3
12,D12=Q12h3
12,D22=Q22h3
12,D66=Q66h3
12,
Q′
11=/parenleftbiggE′
1
1−ν′
12ν′
21/parenrightbigg
,Q′
22=/parenleftbiggE′
2
1−ν′
12ν′
21/parenrightbigg
,Q′
12=/parenleftbiggν′
12E1
1−ν′
12ν′
21/parenrightbigg
,
Q′
66=G′
12=E′
6
Isotropic single layer
D11=D22=Eh3
12(1−ν2)=D,D12=νD=νEh3
1−ν2,
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−ρh∂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−ρh∂2w
∂t2(8.2.16)
The equation of symmetric laminate plates with isotropic la yers 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 computat ion 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 equ ation 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 resultan ts must be included. For
a coupling of in-plane loads and lateral deﬂection, the equi librium (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=1Q(k)
i j/parenleftbigg
x(k)
33−x(k−1)
33/parenrightbigg
=n
∑
k=1Q(k)
i jh(k)/parenleftigg
x(k)
3+h(k)2
12/parenrightigg
,
x(k)
3=1
2/parenleftig
x(k)
3+x(k−1)
3/parenrightig
, the Q(k)
i jfollow from Table 4.2.
Symmetric balanced laminates
Di j=n
∑
k=1Q(k)
i jh(k)/parenleftigg
x(k)
3+h(k)2
12/parenrightigg
TheQ(k)
i jfollow from Table 4.2.
Symmetric cross-ply laminate (specially orthotropic)
Di j=n
∑
k=1Q(k)
i jh(k)/parenleftigg
x(k)
3+h(k)2
12/parenrightigg
,
D16=D26=0
Q(k)
11=/parenleftbiggE1
1−ν12ν21/parenrightbigg(k)
,Q(k)
22=/parenleftbiggE2
1−ν12ν21/parenrightbigg(k)
,
Q(k)
12=/parenleftbiggν12E1
1−ν12ν21/parenrightbigg(k)
,Q(k)
66=G(k)
12
Symmetric laminate with isotropic layers ( x1-direction equal ﬁbre direction)
Di j=n
∑
k=1Q(k)
i jh(k)/parenleftigg
x(k)
3+h(k)2
12/parenrightigg
,
D16=D26=0,D11=D22,
Q(k)
11=Q(k)
22=/parenleftbiggE
1−ν2/parenrightbigg(k)
,Q(k)
12=/parenleftbiggνE
1−ν2/parenrightbigg(k)
,
Q(k)
66=/parenleftbiggE
2(1+ν)/parenrightbigg(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)
Thespecial case of symmetric laminates with isotropic layers follows from (8.2.19)
with D1=D2. The buckling load is like the natural vibration independen t of the
lateral load and p3is taken to be zero. The classical bifurcation buckling requ ires to
satisfy the governing differential equations derived abov e and the boundary equa-
tions. Both sets of equations are again homogeneous and repr esent an eigenvalue
problem for the buckling modes (eigenvalues) and the mode sh apes (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 equilibri um equations (8.2.3) it is
seen that these equations are identically satisﬁed. Using E q. (4.2.22)
MMM=BBBAAA−1NNN−(BBBAAA−1BBB−DDD)κκκ
and substitute NNNwith help of the Airy‘s stress function and κκκby the derivatives of
wthe third equilibrium equation (8.2.3) yields one coupled p artial differential equa-
tion for Fandw. The necessary second equation yields the in-plane compati bility
condition (Sect. 2.2)
∂2ε1
∂x2
2+∂2ε2
∂x2
1=∂2ε6
∂x1∂x2
together with Eq. (4.2.25) to substitute the strains by the s tress resultants. Suppress-
ing the derivations and restricting to symmetric problems y ield the following in-
plane equations which are summarized in Table 8.4. The stiff ness AAA∗,BBB∗,CCC∗,DDD∗fol-
low with Eq. (4.2.23) as AAA∗=AAA−1,BBB∗=−AAA−1BBB,CCC∗=BBBAAA−1,DDD∗=DDD−BBBAAA−1BBB.
One can see from Table 8.4 that in the general case the mathema tical structure of
the partial differential equation corresponds to an anisot ropic and in the special or-
thotropic case to an orthotropic in-plane behavior of a sing le layer homogeneous
anisotropic or orthotropic plate. A summary of the in-plane stiffness is given in Ta-
ble 8.5. The Q(k)
i jfor angle-ply laminates are calculated in Table 4.2.
Similar to the beam theory the plate equations for ﬂexure, vi bration and buckling
can be given in a variational formulation (Sect. 2.2). This f ormulation provides the
basis for the development of approximate solutions. We rest rict the variational for-
mulation to symmetric laminated plates and to the classical energy principles. From
(2.2.24) it follows with ε3=ε4=ε5≈0 that the elastic potential Πis

8.2 Classical Laminate Theory 287
Table 8.4 In-plane equations, boundary conditions and stress result ants 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
2/integraldisplay
V(σ1ε1+σ2ε2+σ6ε6)dV−/integraldisplay
Ap3(x1,x2)w(x1,x2)dA (8.2.21)
=1
2n
∑
k=1/integraldisplay
Ax(k)
3/integraldisplay
x(k−1)
3(σ(k)
1ε1+σ(k)
2ε2+σ(k)
6ε6)dx3dA−/integraldisplay
Ap3(x1,x2)w(x1,x2)dA
With
εεε(x1,x2,x3)=εεε(x1,x2)+x3κκκ(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=1Q(k)
i j/parenleftig
x(k)
3−x(k−1)
3/parenrightig
=n
∑
k=1Q(k)
i jh(k),i,j=1,2,6
2. Cross-ply laminates
Ai j=n
∑
k=1Q(k)
i jh(k),i,j=1,2,6,A16=A26=0,
Q(k)
11=/parenleftbiggE1
1−ν12ν21/parenrightbigg(k)
,Q(k)
22=/parenleftbiggE2
1−ν12ν21/parenrightbigg(k)
Q(k)
12=/parenleftbiggν12E1
1−ν12ν21/parenrightbigg(k)
,Q(k)
66=G(k)
12
3. Laminates with isotropic layers
Ai j=n
∑
k=1Q(k)
i jh(k),i,j=1,2,6,A16=A26=0,A11=A22
Q(k)
11=Q(k)
22=/parenleftbiggE
1−ν2/parenrightbigg(k)
,Q(k)
12=/parenleftbiggνE
1−ν2/parenrightbigg(k)
,
Q(k)
66=/parenleftbiggE
2(1+ν)/parenrightbigg(k)
=G(k)
4. Single layer
For anisotropic and orthotropic single layers the Ai jfollowed by 1. and 2.
For an isotropic single layer is
A11=A22=A=Eh
1−ν2,A12=νA=νEh
1−ν2,
A66=(1−ν)
2A=Eh
2(1+ν)=G
εεεT=/bracketleftbigg∂u
∂x1∂v
∂x2/parenleftbigg∂u
∂x2+∂v
∂x1/parenrightbigg/bracketrightbigg
,
κκκT=/bracketleftbigg
−∂2w
∂x2
1−∂2w
∂x2
2−2∂2w
∂x1∂x2/bracketrightbigg
and the constitutive equations for the strains and the stres s resultants
σσσ(k)=QQQ(k)(εεε+x3κκκ),
NNN
MMM
=
AAA… 000
. . . .
000…DDD

εεε
κκκ
 (8.2.23)
one obtains the elastic potential for the general case of sym metric plates and for the
special cases of orthotropic or isotropic structure behavi or.

8.2 Classical Laminate Theory 289
Bending of plates, classical laminate theory :
Angle-ply laminates
Π(w) =1
2/integraldisplay
A/bracketleftbigg
D11/parenleftbigg∂2w
∂x2
1/parenrightbigg
2+D22/parenleftbigg∂2w
∂x2
2/parenrightbigg
2+2D12∂2w
∂x2
1∂2w
∂x2
2+4D66/parenleftbigg∂2w
∂x1∂x2/parenrightbigg
2
+4/parenleftbigg
D16∂2w
∂x2
1+D26∂2w
∂x2
2/parenrightbigg∂2w
∂x1x2/bracketrightbigg
dA−/integraldisplay
Ap3wdA
(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 boundar y conditions or to apply
the direct variational methods of Ritz, Galerkin or Kantoro vich for approximate
solutions.
Vibration of plates, classical laminate theory:
The kinetic energy of a plate is (rotatory energy is neglecte d)
T=1
2/integraldisplay
Aρh/parenleftbigg∂w
∂t/parenrightbigg2
dA, ρ=1
hn
∑
k=1ρ(k)h(k)(8.2.25)
The Hamilton principle for vibrations yields
δH(x1,x2,t)=0
with
H=t2/integraldisplay
t1(T−Π)dt=t2/integraldisplay
t1Ldt (8.2.26)
Buckling of plates, classical laminate theory:
To calculate buckling loads, the in-plane stress resultant sN1,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 Πcan be formu-
lated for angle-ply laminates with bending in-plane forces

290 8 Modelling and Analysis of Plates
Π(w)=1
2/integraldisplay
A/braceleftigg/bracketleftigg
D11/parenleftbigg∂2w
∂x2
1/parenrightbigg2
+D22/parenleftbigg∂2w
∂x2
2/parenrightbigg2
+2D12∂2w
∂x2
1∂2w
∂x2
2
+4D66/parenleftbigg∂2w
∂x1∂x2/parenrightbigg2
+4/parenleftbigg
D16∂2w
∂x2
1+D26∂2w
∂x2
2/parenrightbigg∂2w
∂x1x2/bracketrightigg
−/bracketleftigg
N1/parenleftbigg∂w
∂x1/parenrightbigg2
+N2/parenleftbigg∂w
∂x2/parenrightbigg2
+2N6/parenleftbigg∂w
∂x1∂w
∂x2/parenrightbigg/bracketrightigg
−2p3w/bracerightigg
dA(8.2.27)
The buckling formulation one get with p3≡0. With D16=D26=0 or
D16=D26=0,D11=D22follows the equations for cross-ply laminates plates
and for plates with isotropic layers. The plate stiffness ca n 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 pr inciple and p3/ne}ationslash=0 is
valid to calculate forced vibrations. With p3(x1,x2,t) =0 in vibration equations
orp3(x1,x2) =0 in (8.2.27), we have formulated eigenvalue problems to com pute
natural frequencies or buckling loads.
Summarizing the derivations of governing plate equations i n the frame of classi-
cal laminate theory there are varying degrees of complexity :
•An important simpliﬁcation of the classical two-dimension al plate equations is
the behavior of cylindrical bending. In this case one consid ers a laminated plate
strip with a very high length-to-width ratio. The transvers e 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 a s one-dimensional
problems. In Sect. 8.6 we discuss some applications of cylin drical bending
•In the case of two-dimensional plate equations the ﬁrst degr ee of simpliﬁcation
for plates is to be symmetric. Symmetric laminates can be bro ken into cross-
ply laminates (specially orthotropic plates) with uncoupl ing 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 hom ogeneous orthotropic
plates, symmetric angle-ply laminates of homogeneous anis otropic plates. For
special boundary conditions symmetric cross-ply laminate d rectangular plates
can be solved analytically. The solutions were obtained in t he 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 deformat ions effects. The anal-
ysis and results of the classical laminate theory are sufﬁci ently 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 de formation effects, at
least approximately. The theory of laminate or sandwich pla tes 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 i n one point. The trans-
verse normals do not remain perpendicular to the middle surf ace 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 applicati on undergoes at most a
translation and a rotation. Plate theories based upon this a ssumption are called ﬁrst
order shear deformation theories and are most used in the ana lysis of moderate thick
laminated plates and of sandwich plates. Higher order theor ies which do not require
normals to remain straight are considerably more complicat ed.
Based upon that kinematical assumption of the ﬁrst order she ar deformation the-
ory the displacements of the plate have the form (5.1.2)
u1(x1,x2,x3) =u(x1,x2)+x3ψ1(x1,x2),
u2(x1,x2,x3) =v(x1,x2)+x3ψ2(x1,x2),
u3(x1,x2,x3) =w(x1,x2)(8.3.1)
and with (5.1.3) are the strains
εi(x1,x2,x3)=εi(x1,x2)+x3κi(x1,x2),i=1,2,6,
εεεT=/bracketleftbigg∂u
∂x1∂v
∂x2∂u
∂x2+∂v
∂x1/bracketrightbigg
,κκκT=/bracketleftbigg∂ψ1
∂x1∂ψ2
∂x2∂ψ1
∂x2+∂ψ2
∂x1/bracketrightbigg
,
ε4(x1,x2)=∂w
∂x2+ψ2,ε5(x1,x2)=∂w
∂x1+ψ1(8.3.2)
One can see that a constant state of transverse shear stresse s is accounted for. The
stresses for the kth layer are formulated in (5.3.2) to
σσσ(k)=QQQ(k)εεε(k),σT=[σ1σ2σ6σ4σ5],εεεT=[ε1ε2ε6ε4ε5] (8.3.3)
σ1,σ2,σ6vary linearly and σ4,σ5constant through the thickness hof the plate. With
the stress resultants NNN,MMM,QQQsand stiffness coefﬁcients Ai j,Bi j,Di j,As
i jfor 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 fo rm, (4.2.16). The stiff-
ness coefﬁcients Ai j,Bi j,Di jstay unchanged in comparison to the classical theory
and the As
i jare deﬁned in (5.3.4) and can be improved with the help of shea r correc-
1Raymond David Mindlin (∗17 September 1906 New York – †22 November 1987 Hanover, New
Hempshire) – mechanician, seminal contributions to many br anches of applied mechanics, applied
physics, and engineering sciences

292 8 Modelling and Analysis of Plates
tion factors ks
i jof plates similar to beams (7.3.19) – (7.3.20). The deﬁnitio n of the
positive rotations ψ1,ψ2is 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 cons titutive 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˜L140
˜L21˜L22˜L23˜L240
˜L31˜L32˜L33˜L34˜L35
˜L41˜L42˜L43˜L44˜L45
0 0 ˜L53˜L54˜L55

u
v
ψ1
ψ2
w
=
0
0
0
0
p
(8.3.4)
The differential operators ˜Li jare given in App. C.2 for unsymmetric angle-ply, sym-
metric angle-ply and symmetric cross-ply laminates. Symme tric laminates leading,
additional to (8.3.4), the uncoupled plate equations
/bracketleftbigg˜L11˜L12
˜L21˜L22/bracketrightbigg/bracketleftbiggu
v/bracketrightbigg
=/bracketleftbigg0
0/bracketrightbigg
,
˜L33˜L34˜L35
˜L43˜L44˜L45
˜L53˜L54˜L55

ψ1
ψ2
w
=
0
0
p
 (8.3.5)
Equation (8.3.4) can also formulated in a compact matrix for m
˜LLL˜uuu=˜ppp
˜LLLis a(5×5)matrix and ˜uuu,˜pppare(5×1)matrices.
The governing plate equations including transverse shear d eformations 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 b e prescribed. The
most usual boundary conditions are:
•ﬁxed boundary
w=0,ψn=0,ψt=0
•free boundary
Mn=0,Mnt=0,Qn=0
Fig. 8.3 Positive deﬁnition of
rotations ψix2,vx3,w
x1,u
ψ2ψ1

8.3 Shear Deformation Theory 293
•free edge
Mn=0,ψt=0,w=0
•simply supported boundary
a)w=0,Mn=0,ψt=0 orw=0,∂ψn/∂n=0,ψt=0 (hard hinged support)
b)w=0,Mn=0,Mnt=0
Case b) is more complicated for analytical or semianalytica l solutions. Generally,
boundary conditions require prescribing for each edge one v alue of each of the
following ﬁve pairs: ( uorNn), (vorNnt), (ψnorMn), (ψtorMnt), (worQn).
With ψ1=−∂w/∂x1andψ2=−∂w/∂x2Eq. (8.3.5) can be reduced to the classical
plate equation.
In the following we restrict our development to plates that a re midplane symmet-
ric (Bi j=0), and additional all coupling coefﬁcients (…)16,(…)26,(…)45are zero.
The constitutive equations are then simpliﬁed to
N1=A11ε1+A12ε2,N2=A12ε1+A22ε1,N6=A66ε6,
M1=D11κ1+D12κ2,M2=D12κ1+D22κ2,M6=D66κ6,
Q1=ks
55A55ε5, Q2=ks
44A44ε4(8.3.6)
or in a contracted notation
/bracketleftbiggNNN
MMM/bracketrightbigg
=/bracketleftbiggAAA000
000DDD/bracketrightbigg/bracketleftbiggεεε
κκκ/bracketrightbigg
, QQQs=AAAsεεεs,
NNNT=[N1N2N6],MMMT=[M1M2M6],QQQsT=[Q1Q2],
εεεT=[ε1ε2ε6],κT=[κ1κ2κ6],εεεsT=[ε5ε4],
AAA=
A11A120
A12A220
0 0 A66
,DDD=
D11D120
D12D220
0 0 D66
,AAAs=/bracketleftbigg
ks
55A55 0
0ks
44A44/bracketrightbigg(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 r esultants results in the
following set of governing differential equations for a lam inated composite plate
subjected to a lateral load p3(x1,x2)and including transverse shear deformation
D11∂2ψ1
∂x2
1+(D12+D66)∂2ψ2
∂x1∂x2+D66∂2ψ1
∂x2
2−ks
55A55/parenleftbigg
ψ1+∂w
∂x1/parenrightbigg
=0,
D66∂2ψ2
∂x2
1+(D12+D66)∂2ψ1
∂x1∂x2+D22∂2ψ2
∂x2
2−ks
44A44/parenleftbigg
ψ2+∂w
∂x2/parenrightbigg
=0,
ks
55A55/parenleftbigg∂ψ1
∂x1+∂2w
∂x2
1/parenrightbigg
+ks
44A44/parenleftbigg∂ψ2
∂x2+∂2w
∂x2
2/parenrightbigg
+p3(x1,x2)=0(8.3.8)
Analogous to the classical plate equations the shear deform ation theory can be used
for modelling and analysis of forced vibrations and bucklin g of laminate plates. In

294 8 Modelling and Analysis of Plates
the general case of forced vibrations the displacements u,v,w, the rotations ψ1,ψ2
and the transverse load pin Eq. (8.3.4) are functions of x1,x2andt. In-plane loading
is not considered but in-plane displacements, rotary and co upling inertia terms have
to take into account. Therefore, generalized mass densitie s must be deﬁned
ρ0=n
∑
k=1ρ(k)/parenleftig
x(k)
3−x(k−1)
3/parenrightig
=n
∑
k=1ρ(k)h(k),
ρ1=n
∑
k=1ρ(k)/parenleftbigg
x(k)
32
−x(k−1)
32/parenrightbigg
,
ρ2=n
∑
k=1ρ(k)/parenleftbigg
x(k)
33−x(k−1)
33/parenrightbigg(8.3.9)
Coupling inertia terms ρ1are only contained in unsymmetric plate problems.
If one wishes to determine the natural frequencies of the rec tangular plate con-
sidered above, then in (8.3.8) p3(x1,x2)must be set zero but a term −ρ0∂2w/∂t2
must be added on the right hand side. In addition, because ψ1andψ2are both in-
dependent variables which are independent of the transvers e displacement w, there
will be an oscillatory motion of a line element through the pl ate thickness which re-
sults in rotary inertia terms ρ2∂2ψ1/∂t2andρ2∂2ψ2/∂t2, respectively, on the right
hand side of the ﬁrst two equations of (8.3.8).
The governing equations for the calculation of natural freq uencies of specially
orthotropic plates with A45=0 are
D11∂2ψ1
∂x2
1+(D12+D66)∂2ψ2
∂x1∂x2+D66∂2ψ1
∂x2
2
−ks
55A55/parenleftbigg
ψ1+∂w
∂x1/parenrightbigg
=ρ2∂2ψ1
∂t2,
D66∂2ψ2
∂x2
1+(D12+D66)∂2ψ1
∂x1∂x2+D22∂2ψ2
∂x2
2
−ks
44A44/parenleftbigg
ψ2+∂w
∂x2/parenrightbigg
=ρ2∂2ψ2
∂t2,
ks
55A55/parenleftbigg∂ψ1
∂x2+∂2w
∂x2
1/parenrightbigg
+ks
44A44/parenleftbigg∂ψ2
∂x2+∂2w
∂x2
2/parenrightbigg
=ρ0∂2w
∂t2,(8.3.10)
ρ0=n
∑
k=1ρ(k)(x(k)
3−x(k−1)
3)=n
∑
k=1ρ(k)h(k),
ρ2=1
3n
∑
k=1ρ(k)(x(k)3
3−x(k−1)3
3)
w,ψ1andψ2are functions of x1,x2andt.
In a similar way the governing equations for buckling proble ms can be derived.
In the matrix equations (8.3.4) and (8.3.5) only the differe ntial operator ˜L55is sub-
stituted by

8.3 Shear Deformation Theory 295
˜L55−/parenleftbigg
N1∂2
∂x2
1+2N6∂2
∂x1∂x2+N2∂2
∂x2
2/parenrightbigg
(8.3.11)
For a cross-ply symmetrically laminated plate is with Bi j=0,D16=0,D26=0,
A45=0
D11∂2ψ1
∂x2
1+(D12+D66)∂2ψ2
∂x1∂x2+D66∂2ψ1
∂x2
2−ks
55A55/parenleftbigg
ψ1+∂w
∂x1/parenrightbigg
=0,
D66∂2ψ2
∂x2
1+(D12+D66)∂2ψ1
∂x1∂x2+D22∂2ψ2
∂x2
2−ks
44A44/parenleftbigg
ψ2+∂w
∂x2/parenrightbigg
=0,
ks
55A55/parenleftbigg∂ψ1
∂x2+∂2w
∂x2
1/parenrightbigg
+ks
44A44/parenleftbigg∂ψ2
∂x2+∂2w
∂x2
2/parenrightbigg
=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 Πwe have to consider that in the general case of unsymmetric la minate
plates, including shear deformation, Π=Π(u,v,w,ψ1,ψ2)is a potential function of
ﬁve independent variables and that the strain energy Πihas a membrane, a bending
and a transverse shearing term, i.e.
Πi=1
2/integraldisplay
V(σ1ε1+σ2ε2+σ6ε6+σ5ε5+σ4ε4)dV
=Πm
i+Πb
i+Πs
i(8.3.13)
with
Πm
i=1
2/integraldisplay
V(N1ε1+N2ε2+N6ε6)dA,
Πb
i=1
2/integraldisplay
V(M1κ1+M2κ2+M6κ6)dA,
Πs
i=1
2/integraldisplay
V(Qs
1ε5+Qs
2ε4)dA(8.3.14)
The stress resultants, stiffness and constitutive equatio ns are formulated in Sect. 4.2,
e.g. (4.2.10) – (4.2.17). The elastic potential Πis then given by
Π(u,v,w,ψ1,ψ2) =1
2/integraldisplay
A(εεεTAAAεεε+κκκTBBBεεε+εεεTBBBκκκ+κκκTDDDκκκ
+εεεsTAAAsεεεs)dx1dx2−/integraldisplay
Ap3wdx1dx2(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 dev eloped for plates quite
similar to beams. Approximately one considers a laminate st rip of the width ”1”
orthogonal to the x1-direction and independently another laminate strip ortho gonal
to the x2-direction and calculates the correction factors ks
55andks
44like 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 form ulation for bending
Mindlin’s plates can be simpliﬁed
Π(w,ψ1,ψ2)=1
2/integraldisplay
A(κTDDDκκκ+εεεsTAAAsεs)dx1dx2−/integraldisplay
Ap3wdx1dx2 (8.3.16)
If we restricted the Hamilton’s principle to vibration of sy mmetric plates, the varia-
tional formulation yields
L(w,ψ1,ψ2)=T(w,ψ1,ψ2)−Π(w,ψ1,ψ2),
T(w,ψ1,ψ2)=1
2/integraldisplay
A/bracketleftigg
ρ0/parenleftbigg∂w
∂t/parenrightbigg2
+ρ2/parenleftbigg∂ψ1
∂t/parenrightbigg2
+ρ2/parenleftbigg∂ψ2
∂t/parenrightbigg2/bracketrightigg
dx1dx2(8.3.17)
Πis given by (8.3.16) and ρ0,ρ2by Eqs. (8.3.10), Tis the kinetic energy.
For a symmetric and specially orthotropic Mindlin’s plate a ssuming A45=0 it
follows from (8.3.16) for bending problems that
Π(w,ψ1,ψ2)=1
2/integraldisplay
A/bracketleftigg
D11/parenleftbigg∂ψ1
∂x1/parenrightbigg2
+2D12/parenleftbigg∂ψ1
∂x1∂ψ2
∂x2/parenrightbigg
+D22/parenleftbigg∂ψ2
∂x2/parenrightbigg2
+D66/parenleftbigg∂ψ1
∂x2+∂ψ2
∂x1/parenrightbigg2
+ks
55A55/parenleftbigg
ψ1+∂w
∂x1/parenrightbigg2
+ks
44A44/parenleftbigg
ψ2+∂w
∂x2/parenrightbigg2/bracketrightigg
dx1dx2−/integraldisplay
Ap3wdx1dx2,
(8.3.18)
δΠ(w,ψ1,ψ2)=0
For natural vibration the variational formulation for that plate is
L(w,ψ1,ψ2)=T(w,ψ1,ψ2)−Π(w,ψ1,ψ2),δt2/integraldisplay
t1L(w,ψ1,ψ2)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 ΠaofΠ. Consider a plate with a constant
in-plane force N1it follows

8.4 Sandwich Plates 297
Π(w,ψ1,ψ2)=Πi(w,ψ1,ψ2)−1
2/integraldisplay
N1/parenleftbigg∂w
∂x1/parenrightbigg2
dx1dx2 (8.3.20)
The case of a more general in-plane loading can be transposed from (8.2.27). The
term Πistay unchanged.
8.4 Sandwich Plates
To formulate the governing differential equations or tht va riational statement for
sandwich plates we draw the conclusion from the similarity o f the elastic behavior
between laminates and sandwiches in the ﬁrst order shear def ormation theory that
all results derived above for laminates can be applied to san dwich plates. We restrict
our considerations to symmetric sandwich plates with thin o r thick cover sheets.
Like in the beam theory, there are differences in the express ions for the ﬂexural
stiffness D11,D12,D22,D66and the transverse shear stiffness A55,A44of laminates
and sandwiches (Sects. 4.3.2 and 4.3.3). Furthermore there are essential differences
in the stress distributions. The elastic behavior of sandwi ches and the general model
assumptions are considered in detail in Sect. 4.3. The stiff ness 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=2n
∑
k=1Q(k)
i jh(k),
Di j=hcCf
i j=hcn
∑
k=1Q(k)
i jh(k)¯x(k)
3,¯x(k)
3=1
2/parenleftig
x(k)
3+x(k−1)
3/parenrightig
(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,G23are the
core shear stiffness moduli. Shear correction factors can b e 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 jorAi j=ALa
i j=2hfQf
i j+hcQc
i j,i,j=1,2,6
Di j=DSa
i j=1
2Qf
i j/parenleftig
hf+hc/parenrightig
hfhcor
Di j=DLa
i j=1
2hfQf
i j/bracketleftbigg/parenleftig
hf+hc/parenrightig2
+1
3hf2/bracketrightbigg
+1
12hc3Qc
i j,i,j=1,2,6
(i j)=( 11),(12),(22),(66)
As
i j=As
i jSa=hcCc
i j,As
i j=As
i jLa=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 pla tes 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 wi th thick faces the stiff-
ness ALa
i j,DLa
i j,(i j)=( 11),(22),(66)andALa
i j,(i j) =( 44),(55)should be used. Be-
cause generally Qc
i j≪Qf
i jusually the simpliﬁed stiffness
ALa
i j≈ASa
i j=2Af
i j,DLa
i j≈DSa
i j/parenleftbigg
1+hf
hchc+(4/3)hf
hc+hf/parenrightbigg
yield satisfying results in engineering applications. Thu s 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 continui ng popularity of
sandwich structures was underlined. Sect. 7.4 also recalle d and summarized the
main aspects of modelling and analysis of sandwich structur es. Engineering ap-
plications to sandwich beams were discussed in detail. Keep ing 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 AAA,BBBandDDDof laminated plates are employed.
•Consider the lower face as lamina 1, the core as lamina 2 and th e upper face
as lamina 3 one can include or ignore the effect of the core on t he response to
bending and in-plane loads and the effect of transverse shea r deformation on the
response of the facings.
•The shear deformation theory of laminated plates can be not o nly transposed to
sandwich plates for bending, vibration and buckling induce d 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 equatio ns and variational formula-
tions of Sect. 8.3 stay valid. Some examples for sandwich pla tes 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 signi ﬁcantly the structural
response of ﬁbre-reinforced laminated composites. In Sect s. 8.2 to 8.4 the structural
response of laminated plates as result of mechanical loadin g was considered and
thermal or hygrosgopic loadings were neglected.
This section focuses on hygrothermally induced strains, st resses and displace-
ments of thin or moderate thick laminated plates. We assume a s in Sect. 7.5 mod-
erate hygrothermal loadings such that the mechanical prope rties remain approx-
imately unchanged for the temperature and moisture differe nces considered. Be-
cause the mathematical formulations governing thermal and hygroscopic loadings
are analogous, a uniﬁed derivation is straightforward and w ill be considered in the
frame of the classical laminate theory and the shear deforma tion theory.
The following derivations use the basic equations, Sect. 4. 2.5, on thermal and
hygroscopic effects in individual laminae and in general la minates. The matrix for-
mulations for force and moment resultants, Eq. (4.2.75), ca n be written explicitly
as

N1
N2
N6
M1
M2
M6
=
A11A12A16B11B12B16
A22A26B12B22B26
A66B16B26B66
S D11D12D16
Y D22D26
M D66

ε1
ε2
ε6
κ1
κ2
κ6
−
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=1Q(k)
i j/parenleftig
x(k)
3−x(k−1)
3/parenrightig
,
Bi j=1
2n
∑
k=1Q(k)
i j/parenleftbigg
x(k)
32−x(k−1)
32/parenrightbigg
,
Di j=1
3n
∑
k=1Q(k)
i j/parenleftbigg
x(k)
33
−x(k−1)
33/parenrightbigg(8.5.2)
The thermal and moisture stress resultants NNNth,NNNmo,MMMth,MMMmoare 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 εiand the curvatures κiby the
displacements u,v,w, Eq. (8.2.5), yield the following matrix differential equa tion
for the classical laminate theory
LLLuuu=ppp−∂∂∂/bracketleftbigg
NNN∗
MMM∗/bracketrightbigg
(8.5.3)

300 8 Modelling and Analysis of Plates
LLLuuu=pppis identically to Eq. (8.2.7) with Li jgiven in App. C. NNN∗=NNNth+NNNmo,
MMM∗=MMMth+MMMmoare the hygrothermal stress results and ∂∂∂is a special (3 ×6)
differential matrix
∂∂∂=
−∂
∂x1−∂
∂x20 0 0 0
0−∂
∂x2−∂
∂x10 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. Th e matrix LLLand the dif-
ferential operators Li jare summarized for the most important special laminates in
App. C.
Hygrothermal induced buckling can be modelled as
LuLuLu+∂∂∂/bracketleftbigg
NNN∗
MMM∗/bracketrightbigg
=/parenleftbigg
N1∂2
∂x2
1+2N6∂2
∂x1∂x2+N2∂2
∂x2
2/parenrightbigg
uuu∗(8.5.5)
with uuu∗T=[0 0w]. Prebuckling displacements and stress resultants are dete rmined
by solving Eq. (8.5.5) with NNN≡000. For the corresponding buckling problem, N1,N2
andN6are taken to be the stress resultant functions correspondin g to the prebuckling
state. The buckling loads are found by solving the eigenvalu e problem associated
with (8.5.5), i.e. with NNN∗=000 and MMM∗=000.
Because energy methods are useful to obtain approximate ana lytical solutions
for hygrothermal problems the total potential energy Πis 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
2/integraldisplay
A/braceleftigg
A11/parenleftbigg∂u
∂x1/parenrightbigg2
+2A12/parenleftbigg∂u
∂x1/parenrightbigg/parenleftbigg∂v
∂x2/parenrightbigg
+A22/parenleftbigg∂v
∂x2/parenrightbigg2
+A66/parenleftbigg∂u
∂x2+∂v
∂x1/parenrightbigg2
+D11/parenleftbigg∂2w
∂x2
1/parenrightbigg2
+2D12/parenleftbigg∂2w
∂x2
1/parenrightbigg/parenleftbigg∂2w
∂x2
2/parenrightbigg
+D22/parenleftbigg∂2w
∂x2
2/parenrightbigg2
+4D66/parenleftbigg∂2w
∂x1∂x2/parenrightbigg2
−2N∗
1∂u
∂x1−2N∗
2∂v
∂x2−2N∗
6/parenleftbigg∂u
∂x2+∂v
∂x1/parenrightbigg
+2M∗
1∂2w
∂x2
1+2M∗
2∂2w
∂x2
2−4M∗
6/parenleftbigg∂2w
∂x1∂x2/parenrightbigg
−/bracketleftigg
N1/parenleftbigg∂2w
∂x2
1/parenrightbigg2
+2N6/parenleftbigg∂w
∂x1/parenrightbigg/parenleftbigg∂w
∂x2/parenrightbigg
+N2/parenleftbigg∂2w
∂x2
2/parenrightbigg2/bracketrightigg/bracerightigg
dA(8.5.6)
The classical laminate theory which neglect transverse she ar deformations can lead
to signiﬁcant errors for moderately thick plates and hygrot hermal loadings. Us-

8.5 Hygrothermo-Elastic Effects on Plates 301
ing the shear deformation theory, Sect. 8.3, we can formulat e corresponding to Eq.
(8.3.4).
˜LLL˜uuu=˜ppp−˜∂∂∂/bracketleftbigg
NNN∗
MMM∗/bracketrightbigg
(8.5.7)
The matrix˜∂∂∂is identically to Eq. (8.5.4). For hygrothermal induced buc kling we
have analogous to Eq. (8.5.5)
˜LLL˜uuu+˜∂∂∂/bracketleftbiggNNN∗
MMM∗/bracketrightbigg
=/parenleftbigg
N1∂2
∂x2
1+2N6∂2
∂x1∂x2+N2∂2
∂x2
2/parenrightbigg
uuu∗(8.5.8)
For prebuckling analysis the terms involving N1,N2andN6are 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 hygroth ermal stress resultants
NNN∗andMMM∗. For special laminate stacking the differential operators are summarized
in App. C.
The elastic potential Πis now a function of ﬁve independent functions
u,v,w,ψ1,ψ2. Restricting again to cross-ply laminates the elastic tota l potential Π
can be formulated as
Π=1
2/integraldisplay
A/braceleftigg
A11/parenleftbigg∂u
∂x1/parenrightbigg2
+2A12/parenleftbigg∂u
∂x1∂v
∂x2/parenrightbigg
+A22/parenleftbigg∂v
∂x2/parenrightbigg2
+A66/parenleftbigg∂u
∂x2+∂v
∂x1/parenrightbigg2
+D11/parenleftbigg∂ψ1
∂x1/parenrightbigg2
+2D12/parenleftbigg∂ψ1
∂x1∂ψ2
∂x2/parenrightbigg
+D22/parenleftbigg∂ψ2
∂x2/parenrightbigg2
+D66/parenleftbigg∂ψ1
∂x2+∂ψ2
∂x1/parenrightbigg
+ks
44A44/parenleftbiggdw
dx2+ψ2/parenrightbigg2
+ks
55A55/parenleftbiggdw
dx1+ψ1/parenrightbigg2
(8.5.9)
−2N∗
1∂u
∂x1−2N∗
2∂v
∂x2−2N∗
6/parenleftbigg∂u
∂x2+∂v
∂x1/parenrightbigg
−2M∗
1∂ψ1
∂x1−2M∗
2∂ψ2
∂x2−2M∗
6/parenleftbigg∂ψ1
∂x2+∂ψ2
∂x1/parenrightbigg
−/bracketleftigg
N1/parenleftbigg∂2w
∂x2
1/parenrightbigg2
+2N6/parenleftbigg∂w
∂x1/parenrightbigg/parenleftbigg∂w
∂x2/parenrightbigg
+N2/parenleftbigg∂2w
∂x2
2/parenrightbigg2/bracketrightigg/bracerightigg
dA
Equations (8.5.6) and (8.5.9) are the starting point for sol ving hygrothermal induced
buckling problems e.g. with the Ritz- or Galerkin approxima tion or the ﬁnite ele-
ment method. Analytical solutions are in general not possib le. As considered above,
the force resultants N1,N2andN6have to be calculated in the prebuckling state,
i.e. for NNN≡000 and the calculation force resultants are substituted into Eqs. (8.5.6) or
(8.5.9), respectively, with NNN∗=000,MMM∗=000 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 NNN=000 and substitute an additional term

302 8 Modelling and Analysis of Plates
/integraldisplay
ApwdA
8.6 Analytical Solutions
The analysis of rectangular plates with selected layer stac king and boundary condi-
tions can be carried out analytically in a similar manner to h omogeneous isotropic
and orthotropic plates. The analytical methods of homogene ous 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 stacki ng and analogous bound-
ary conditions. In Sect. 8.6 possibilities of analytical so lutions in the frame of clas-
sical laminate theory and shear deformation theory are demo nstrated for bending,
buckling and vibration problems.
8.6.1 Classical Laminate Theory
There are varying degrees of complexity in laminated plate a nalysis. The least com-
plicated problems are one-dimensional formulations of cyl indrical plate bending.
For cylindrical bending both, symmetric and unsymmetric la minates, are handled in
a unique manner assuming all deformations are one-dimensio nal.
In the case of two-dimensional plate equations the most impo rtant degree of sim-
pliﬁcation is for plates being midplane symmetric because o f their uncoupling in-
plane and out-of-plane response. The mathematical structu re of symmetric angle-ply
plate equations corresponds to homogeneous anisotropic pl ate equations and that of
symmetric cross-ply plate equations to homogeneous orthot ropic plate equations.
To illustrate analytical solutions for rectangular plates in the framework of the clas-
sical laminate theory we restrict our developments to speci ally orthotropic, i.e. to
symmetric cross-ply plates. For this type of laminated plat es the Navier solution
method can be applied to rectangular plates with all four edg es simply supported.
The N´ adai-L´ evy solution (N´ adai, 1925) method can be appl ied to rectangular plates
with two opposite edges have any possible kind of boundary co nditions. For more
general boundary conditions of special orthotropic plates or other symmetric or un-
symmetric rectangular plates approximate analytical solu tions are possible, e.g. us-
ing the Ritz-, the Galerkin- or the Kantorovich methods, Sec t. 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) – e ngineer, 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 d irection with such a lat-
eral load and edge support in this direction that the plate pr oblem may be reduced to
a one-dimensional problem and a quasi-beam solution can be u sed. In the following
we demonstrate analytical solutions for various selected e xamples.
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 ainx1-direction is considered
ﬁnite, the other dimension binx2-direction approximately inﬁnite. The boundary
conditions for the edges x1=0,×1=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. F or 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 wit h the differential equa-
tionbD11w′′′′(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❄❄❄❄❄p3 FM

304 8 Modelling and Analysis of Plates
σ(k)
1(x1,x3) =−Q(k)
11x3d2w
dx2
1,
σ(k)
2(x1,x3) =−Q(k)
12x3d2w
dx2
1,
σ(k)
6(x1,x3) =0
or
σ(k)
1(x1,x3) =−Q(k)
11
D11M1(x1)x3,
σ(k)
2(x1,x3) =−Q(k)
12
D12M2(x1)x3,
σ(k)
6(x1,x3) =0(8.6.2)
and the transverse shear stresses follow from (5.2.19) to
σ4(x1,x3)=F61Q1(x1),σ5(x1,x3)=F11Q1(x1)
and
FFF(x3)=˜BBB(x3)DDD−1
i.e /bracketleftbigg
σ5
σ4/bracketrightbigg
=/bracketleftbigg
F11F62
F61F22/bracketrightbigg/bracketleftbigg
Q1
0/bracketrightbigg
, (8.6.3)
Considering the solutions of the symmetrical laminated pla te strip, we have follow-
ing conclusions:
•The solutions for laminate beams and plate strips are very si milar, but the calcu-
lation of the strip bending stiffness Di jhas to include Poisson’s effects.
•Because of including of Poisson’s effect we have the relatio n
w(x1)strip<w(x1)beam
andM2(x1)/ne}ationslash=0.
•M1(x1)andQ1(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 for ce.
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
andM6=−2D66w′(x1)for specially orthotropic strips.
For unsymmetric laminated plate strips the system of three o ne-dimensional dif-
ferential equations for the displacements u(x1),v(x1)andw(x1)follow with (8.2.6)
as

8.6 Analytical Solutions 305
A11d2u
dx2
1+A16d2v
dx2
1−B11d3w
dx3
1=0
A16d2u
dx2
1+A66d2v
dx2
1−B16d3w
dx3
1=0
D11d4w
dx4
1−B11d3u
dx3
1−B16d3v
dx3
1=p3(8.6.4)
These equations can be uncoupled and analytically solved. T he ﬁrst and the second
equation yield
d2u
dx2
1=B
Ad3w
dx3
1,d2v
dx2
1=C
Ad3w
dx3
1(8.6.5)
with A=A11A66−A2
16,B=A66B11−A16B16,C=A11B16−A16B11. Differentiating
both equations and substituting the results in the third equ ation 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)andw(x1)as
w(x1) =A
D/integraldisplay /integraldisplay /integraldisplay /integraldisplay
p3dx1dx1dx1dx1+C1x3
1
6+C2x2
1
2+C3x1+C4,
u(x1) =B
D/integraldisplay /integraldisplay /integraldisplay
p3dx1dx1dx1+B1x2
1
2+B2x1+B3,
v(x1) =C
D/integraldisplay /integraldisplay /integraldisplay
p3dx1dx1dx1+A1x2
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 f orces and moments resul-
tants
N1=A11du
dx1+A16dv
dx1−B11d2w
dx2
1,
N2=A12du
dx1+A26dv
dx1−B12d2w
dx2
1,

306 8 Modelling and Analysis of Plates
N6=A16du
dx1+A66dv
dx1−B16d2w
dx2
1,
M1=B11du
dx1+B16dv
dx1−D11d2w
dx2
1,
M2=B12du
dx1+B26dv
dx1−D12d2w
dx2
1,
M6=B16du
dx1+B66dv
dx1−D16d2w
dx2
1, (8.6.9)
Q1=B11d2u
dx2
1+B16d2v
dx2
1−D11d3w
dx3
1,
Q2=B16d2u
dx2
1+B66d2v
dx2
1−D16d3w
dx3
1,
V1=B11d2u
dx2
1+B16d2v
dx2
1−D11d3w
dx3
1=Q1,
V2=2B16d2u
dx2
1+2B66d2v
dx2
1−2D16d3w
dx3
1=2Q2
The general symmetric case follows with Bi j=0 and for a symmetric cross-ply strip
areBi j=0 and A16=0,D16=0.
Analytical solutions can also be formulated for vibration a nd buckling of strips
with one-dimensional deformations. The eigen-vibrations of unsymmetrical plate
strips taking account of u(x1,t)=u(x1)eiωt,v(x1,t)=v(x1)eiωt,w(x1,t)=w(x1)eiωt
are mathematically modelled as
A11d2u
dx2
1+A16d2v
dx2
1−B11d3w
dx3
1−ρhω2w=0,
A16d2u
dx2
1+A66d2v
dx2
1−B16d3w
dx3
1−ρhω2v=0,
D11d4w
dx4
1−B11d3u
dx3
1−B16d3w
dx3
1−ρhω2w=0(8.6.10)
u,vandware now functions of x1andt.
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 thi rd Eq. (8.6.10) lead to the
vibration equation
d4w
dx1−A
Dρhω2w=0 (8.6.11)
For a symmetrically laminated cross-ply strip we obtain wit hA/D=1/D11
d4w
dx1−ρhω2
D11w=0,ρh=N
∑
k=1ρ(k)h(k)(8.6.12)

8.6 Analytical Solutions 307
The analytical solutions correspond to the beam solutions i n Sect. 7.6
w(x1)=C1cosλ
ax1+C2sinλ
ax1+C3coshλ
ax1+C4sinhλ
ax1,
/parenleftbiggλ
a/parenrightbigg4
=ρh
D11ω2(symmetric cross-ply strip) ,
/parenleftbiggλ
a/parenrightbigg4
=ρhA
Dω2(general unsymmetric strip)(8.6.13)
For a simply supported strip we have w(0)=w(a)=w′′(0)=w′′(a)=0 and there-
foreC1=C3=C4=0 and C2sin/parenleftig
λ
a/parenrightig
a=0, i.e. with λ=nπfollow
ω2=n4π4
a4ρhD
A,ω=/parenleftignπ
a/parenrightig2/radicaligg
D
ρhA(8.6.14)
Analytical solutions can be calculated for all boundary con ditions of the strip.
In analogous manner analytical solutions follow for the buc kling 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
D11d4w
dx4
1−B11d3u
dx3
1−B16d3v
dx3
1−N1d2w
dx2
1=0 (8.6.15)
and with Eq. (8.6.5) follows
d4w
dx4
1−A
DN1d2w
dx2
1=0,(general case)
d4w
dx4
1−1
D11N1d2w
dx2
1=0,(symmetrical cross-ply case)(8.6.16)
with N1(x1)=−N0. The buckling equations correspond again to the beam equati on
(7.2.35) and can be solved for all boundary conditions of the strip
w(x1)=C1cosλ
ax1+C2sinλ
ax1+C3coshλ
ax1+C4sinhλ
ax1,
/parenleftbiggλ
a/parenrightbigg2
=A
DN0(8.6.17)
For a simply supported strip we have with w(0)=w(a)=w′′(0)=w′′(a)=0
C2sinλ=0,λ=nπ
A nonzero solution is obtained if
N0=n2π2
a2D
A(8.6.18)

308 8 Modelling and Analysis of Plates
Thus the critical buckling load follows with to
N0cr=π2
a2D
A(8.6.19)
Summarizing the developments of analytical solutions for u nsymmetrical laminated
plate strips we have the following conclusions:
•The system of three coupled differential equations for the d isplacements
u(x1),v(x1)andw(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)andv(x1), respectively.
•Analytical solutions for bending of unsymmetrical laminat ed 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 symmet ric case and symmet-
rical cross-ply strips are included as special solutions.
•The derivations of bending equations can be expanded to buck ling and vibration
problems.
•The derivation of analytical solutions for unsymmetrical l aminated 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 al l boundary conditions
p3(x1,x2)
✻
✲x2
x1❄✻
✛ ✲ab Boundary conditions:
w(0,×2)=w(a,x2)=w(x1,0)=w(x1,b)=0
M1(0,×2)=M1(a,x2)=M2(x1,0)=M2(x1,b)=0
Fig. 8.5 Rectangular plate, all edges are simply supported, special ly orthotropic

8.6 Analytical Solutions 309
p3(x1,x2)=∞
∑
r=1∞
∑
s=1prssinαrx1sinβsx2,
prs=4
aba/integraldisplay
0b/integraldisplay
0p3(x1,x2)sinαrx1sinβsx2dx1dx2,
w(x1,x2)=∞
∑
r=1∞
∑
s=1wrssinαrx1sinβsx2(8.6.20)
with αr=rπ/a,βs=sπ/b. The coefﬁcients wrsare 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=1wrs/parenleftbig
D1α4
r+2D3α2
rβ2
s+D2β4
s/parenrightbig
sinαrx1sinβsx2=∞
∑
r=1∞
∑
s=1prssinαrx1sinβsx2
(8.6.21)
and we obtain the coefﬁcients wrs
wrs=prs
D1α4r+2D3α2rβ2s+D2β4s=prs
drs(8.6.22)
The solution becomes
w(x1,x2)=∞
∑
r=1∞
∑
s=1prs
drssinαrx1sinβsx2 (8.6.23)
The load coefﬁcients prsone 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
π2rs,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(D11α2
r+D12β2
s)wrssinαrx1sinβsx2,
M2(x1,x2) =−D12∂2w
∂x2
1−D22∂2w
∂x2
2
=∞
∑
r=1∞
∑
s=1(D12α2
r+D22β2
s)wrssinαrx1sinβsx2,(8.6.25)

310 8 Modelling and Analysis of Plates
M6(x1,x2) =−2D66∂2w
∂x1∂x2=−2D66∞
∑
r=1∞
∑
s=1αrβswrscosαrx1cosβsx2,
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
σσσ(k)(x1,x2,x3)=QQQ(k)x3κκκ=−x3
Q(k)
11Q(k)
120
Q(k)
12Q(k)
220
0 0 Q(k)
66

∂2w/∂x2
1
∂2w/∂x2
2
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=1prs
D1α4r+2D3α2rβ2s+D2β4s
(Q(k)
11α2
r+Q(k)
12β2
s)sinαrx1sinβsx2
(Q(k)
12α2
r+Q(k)
22β2
s)sinαrx1sinβsx2
−2Q(k)
66αrβscosαrx1cosβsx2

(8.6.27)
With the simpliﬁed formula (5.2.19) follows the transverse shear stresses σ(k)
4,σ(k)
5
/bracketleftigg
σ(k)
5(x1,x3)
σ(k)
4(x1,x3)/bracketrightigg
=/bracketleftbiggF11x3F62x3
F61x3F22x3/bracketrightbigg
∞
∑
r=1∞
∑
s=1/bracketleftbigg
(D11α3
r+D12β2
sαr)wrscosαrx1sinβsx2
(D12α2
rβs+D22β3
s)wrssinαrx1cosβsx2/bracketrightbigg (8.6.28)
The Navier solution method can be applied to all simply suppo rted specially or-
thotropic laminated rectangular plates in the same way. For a given lateral load
p3(x1,x2)one can obtain the load coefﬁcients prsby integrating (8.6.20), and by
substituting prsin (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 res ultants and the stresses
and in general with the concentration of lateral loads in par tial 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 antisymmetri c cross-ply laminate
and for symmetric and antisymmetric angle-ply laminates. F or these laminates the
plate equations (8.2.6) are not uncoupled and we have to pres cribe 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=1urscosαrx1sinβsx2,
v(x1,x2) =∞
∑
r=1∞
∑
s=1vrssinαrx1cosβsx2,
w(x1,x2) =∞
∑
r=1∞
∑
s=1wrssinαrx1sinβsx2,
Type 2:
u(x1,x2) =∞
∑
r=1∞
∑
s=1urssinαrx1cosβsx2,
v(x1,x2) =∞
∑
r=1∞
∑
s=1vrscosαrx1sinβsx2,
w(x1,x2) =∞
∑
r=1∞
∑
s=1wrssinαrx1sinβsx2,
αr=πr/a,βs=πs/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,vandwcan be used only for laminates,
whose stiffness A16,A26,B16,B26,D16,D26are 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,D26equal zero, i.e for
symmetric or antisymmetric angle ply laminates.
The Navier solutions can be used for calculating bending, bu ckling 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 rectang ular plates with two op-
posite edges simply supported, a single inﬁnite series meth od can be used. The two
other opposite edges may have arbitrary boundary condition s (Fig. 8.6). N´ adai in-
troduced for isotropic plates the solution of the plate equa tion 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. Wi th the solutions for wh,
suggested by L´ evy, and wp, suggested by N´ adai,
wh(x1,x2)=∞
∑
r=1fr(x2)sinαrx1,wp(x1)=∞
∑
r=1prsinαrx1
D1α4r(8.6.30)
with αr=rπ/aand
p3(x1)=∞
∑
r=1prsinαrx1,pr=2
aa/integraldisplay
0p3(x1)sinαrx1dx1
the boundary conditions for x1=0 and x1=aare satisﬁed.
Substituting (8.6.30) into the plate equation for speciall y orthotropic plates, Table
8.1, it follow for each term fr(x2)a differential equation of 4th order with constant
coefﬁcients
D2d4fr(x2)
dx4
2−2D3α2
rd2fr(x2)
dx2
2+D1α4
rfr(x2)=pr (8.6.31)
or
✻✲x2
x1
❄✻
✛ ✲ab Boundary conditions:
w(0,×2)=w(a,x2)=0
M1(0,×2)=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
2−2D3α2
r
D2d2fr(x2)
dx2
2+D1
D2α4
rfr(x2)=pr
D2(8.6.32)
The homogeneous differential equation, i.e. pr=0, can be solved with
frh(x2)=Crexp(λrαrx2) (8.6.33)
and yields the characteristic equation for the four roots
λ4
r−2D3
D2λ2
r+D1
D2=0=⇒λ2
r=D3
D2±/radicaligg/parenleftbiggD3
D2/parenrightbigg2
−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
λ1/2=±δ1,λ3/4=±δ2,δ1,δ2>0,
frh(x2) =Arcosh δ1αrx2+Brsinhδ1αrx2
+Crcosh δ2αrx2+Drsinhδ2αrx2(8.6.35)
2.(D3/D2)2=(D1/D2): In this case (8.6.34) leads to four real and equal roots
λ1/2=+δ,λ3/4=−δ,δ>0,
frh(x2)=( Ar+Brx2)cosh δαrx2+(Cr+Drx2)sinhδαrx2 (8.6.36)
3.(D3/D2)2<(D1/D2): In this case the roots are complex
λ1/2=δ1±iδ2,λ3/4=−δ1±iδ2,δ1,δ2>0,
frh(x2) = ( Arcosδ2αrx2+Brsinhδ2αrx2)cosh δ1αrx2
+ (Crcosδ1αrx2+Drsinδ1αrx2)sinhδ1αrx2(8.6.37)
For a given plate for which materials and ﬁbre orientations h ave 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 conseque nce 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 lat eral 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=1prsinαrx1
D1α4rsinαrx1 (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 appl ied to eigenvalue
problems. We assume, for instance, that the vibration mode s hapes 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=1wrssinαrx1sinαsx2sinωt (8.6.39)
to represent the expected harmonic oscillation and to satis fy all boundary conditions.
Substituting the expression (8.6.39) into (8.2.16) with p3≡0 yields
[D1α4
r+2D3α2
rα2
s+D2α4
s−ρω2]wrs=0 (8.6.40)
A non-zero value of wrs, i.e. a non-trivial solution, is obtained only if the expres sion
in the brackets is zero, hence we can ﬁnd the equation for the n atural frequencies
ω2
rs=π4
ρh/bracketleftbigg
D1/parenleftigr
a/parenrightig4
+2D3/parenleftigr
a/parenrightig2/parenleftigs
a/parenrightig2
+D2/parenleftigs
a/parenrightig4/bracketrightbigg
(8.6.41)
The fundamental frequency corresponds to r=s=1 and is given by
ω2
11=π4
ρha4/bracketleftbigg
D1+2D3/parenleftiga
b/parenrightig2
+D2/parenleftiga
b/parenrightig4/bracketrightbigg
(8.6.42)
Note that the maximum amplitude wrscannot 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
ω2
rs=krsπ2
a2/radicaligg
D
ρh,krs=/bracketleftbigg
r2+s2/parenleftiga
b/parenrightig2/bracketrightbigg
(8.6.43)
If we consider a buckling problem, e.g. a specially orthotro pic laminated plate sim-
ply supported at all edges with a biaxial compression N1andN2, 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)
π2wrs[D1r4+2D3r2s2γ2+D2s4γ4]=−wrs[N1r2+N2s2γ2]a2(8.6.45)
with γ=a/b. A non-zero solution of the buckling problem ( wrs/ne}ationslash=0) leads to

8.6 Analytical Solutions 315
N1r2+N2s2γ2=−π2
a2[D1r4+2D3r2s2γ2+D2s4γ4] (8.6.46)
We consider the example of uniform compression N1=−NandN2=−κN, where
the boundary force Nis positive. Equation (8.6.46) yields
N=π2(D1r4+2D3r2s2γ2+D2s4γ4)
a2(r2+κs2γ2)
The critical buckling load Ncrcorresponds to the lowest value of N. Ifκ=0 we
have the case of uniaxial compression and the buckling equat ion simpliﬁes to
N=π2
a2r2(D1r4+2D3r2s2γ2+D2s4γ4)
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 Ncris not simple
and depends on the stiffness D1,D2,D3, the length-to-width ratio γ=a/bandr.
However, for a given plate it can be easily determined numeri cally. Summarizing
the discussion of the classical laminate theory applied to l aminate 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 isotr opic Kirchhoff’s plates,
if all or two opposite plate edges are simply supported. Thes e solution methods
can be applied to plate bending, buckling and vibration.
•For more general boundary conditions specially orthotropi c plates may be solved
analytically with the help of the variational approximate s olutions method of
Rayleigh-Ritz or in a more generalized way based on a variati onal method of
Kantorovich.
•Plates with extensional-bending couplings should be solve d numerically, e.g.
with the help of the ﬁnite element method, Chap. 11. Note that in special cases
antisymmetric cross-ply respectively symmetric and antis ymmetric angle-ply
laminates can be analyzed analytically with Navier‘s solut ion method.
In this section we illustrated detailed analytical solutio ns for specially orthotropic
laminates which can predict ”exact” values of deﬂections, n atural frequencies of vi-
bration and critical buckling loads. But even the ”exact” so lutions become approxi-
mate because of the truncation of the inﬁnite series solutio ns or round-off errors in
the solution of nonlinear algebraic equations, etc. Howeve r these solutions help one
to understand, at least qualitatively, the mechanical beha vior of laminates. Many
laminates with certain ﬁbre orientations have decreasing v alues of the coefﬁcients
D16,D26for bending-torsion coupling and they can be analyzed with t he 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 tra nsverse shear deforma-
tions is much more complicated than in the frame of classical laminate theory. Also
for plate analysis including shear deformations the at leas t complicated problem is
cylindrical bending, i.e one-dimensional formulations fo r plate strips.
Unlike to classical plate strips equations only symmetric a nd 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 wi th an inﬁnite length in
thex2-direction and uniformly supported edges x1=0,×1=a, subjected to a load
p3=p(x1). If we restrict the considerations to cross-ply laminated s trips 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,ψ2=0,
u=u(x1),ψ1=ψ1(x1),w=w(x1)from (8.3.4) as
A11d2u
dx2
1+B11d2ψ1
dx2
1=0,
B11d2u
dx2
1+D11d2ψ1
dx2
1−ks
55A55/parenleftbigg
ψ1+dw
dx1/parenrightbigg
=0,
ks
55A55/parenleftbiggdψ1
dx1+d2w
dx2
1/parenrightbigg
+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) = A11du
dx1+B11dψ1
dx1,
N2(x1) = A12du
dx1+B12dψ1
dx1,
N6(x1) = 0,
M1(x1) =B11du
dx1+D11dψ1
dx1,
M2(x1) =B12du
dx1+D12dψ1
dx1,
M6(x1) =0,
Q1(x1) =ks
55A55/parenleftbigg
ψ1+dw
dx1/parenrightbigg
,
Q2(x1) =0(8.6.48)

8.6 Analytical Solutions 317
The three coupled differential equations for u,wandψ1can be reduced to one un-
coupled differential equation for ψ1. The ﬁrst equation yields
d2u
dx2
1=−B11
A11d2ψ1
dx2
1,d3u
dx3
1=−B11
A11d3ψ1
dx3
1(8.6.49)
Differentiating the second equation and substituting the e quation above result in
−B2
11
A11d3ψ1
dx3
1+D11d3ψ1
dx3
1−ks
55A55/parenleftbiggdψ1
dx1+d2w
dx2
1/parenrightbigg
=0
or with/parenleftbigg
D11−B2
11
A11/parenrightbigg
=DR
11
ks
55A55/parenleftbiggdψ1
dx1+d2w
dx2
1/parenrightbigg
=DR
11d3ψ1
dx3
1(8.6.50)
Substituting Eq. (8.6.50) in the third equation (8.6.47) yi eld an uncoupled equation
forψ1(x1)
DR
11d3ψ1
dx3
1=−p3 (8.6.51)
The uncoupled equations for u(x1)andw(x1)follow then as
d2u
dx2
1=−A11
B11d2ψ1
dx2
1,dw
dx1=−ψ1+DR
11
ks
55A55d2ψ1
dx2
1(8.6.52)
The three uncoupled equations can be simple integrated
DR
11ψ1(x1) =−/integraldisplay /integraldisplay /integraldisplay
p3(x1)dx1dx1dx1+C1x2
1
2+C2x1+C3,
w(x1) =1
DR
11/bracketleftigg/integraldisplay /integraldisplay /integraldisplay /integraldisplay
p3(x1)dx1dx1dx1dx1+C1x3
1
6+C2x2
1
2
+C3x1+C4/bracketrightigg
−1
ks
55A55/bracketleftbigg/integraldisplay /integraldisplay
p3(x1)dx1dx1+C1x1/bracketrightbigg
=wB(x1)+wS(x1),
u(x1) = −A11
B111
DR
11/bracketleftbigg/integraldisplay /integraldisplay /integraldisplay
p3(x1)dx1dx1dx1+C1x1+C5/bracketrightbigg(8.6.53)
Thus the general analytical solutions for unsymmetric cros s-ply laminated strips
are calculated. For symmetrical cross-ply laminated strip s the equations yield
DR
11=D11andA11u′′(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
D11d2ψ1
dx2
1−ks
55A55/parenleftbigg
ψ1+dw
dx1/parenrightbigg
=0,
ks
55A55/parenleftbiggdψ1
dx1+d2w
dx2
1/parenrightbigg
+N0d2w
dx1=0(8.6.54)
The equations can be uncoupled. With
/parenleftbiggdψ1
dx1+d2w
dx2
1/parenrightbigg
=D11
ks
55A55d3ψ1
dx3
1,ks
55A55d3ψ1
dx3
1=−ks
55A55d4w
dx4
1−N0d2w
dx1
one obtains analogous to Eq. (7.3.23)
D11/parenleftbigg
1−N0
ks
55A55/parenrightbiggd4w
dx4
1+N0d2w
dx1=0 (8.6.55)
The general solution for the eigenvalue problem (8.6.55) fo llows with
w(x1)=Ceλx1 (8.6.56)
and the characteristic equation
D11/parenleftbigg
1−N0
ks
55A55/parenrightbigg
λ4+N0λ2=0 or D11λ4+k2λ2=0 (8.6.57)
with the solutions
λ1/2=±ik,λ3/4=0
as
w(x1)=C1sinkx1+C2coskx1+C3x1+C4 (8.6.58)
If we assume, e.g. simply supported edges x1(0) = 0,×1(a), follow with
w(0) = w(a) =w′′(0) =w′′(a) =0 the free coefﬁcients C2=C3=C4=0
andC1sinka=0. If C1/ne}ationslash=0 follow with sin ka=0 the solution k=mπ/a=αm
(m=1,2,…)andk2=α2
mand thus
N0
D11/parenleftbigg
1−N0
ks
55A55/parenrightbigg=α2
m,N0=D11ks
55A55α2
m
D11α2m+ks
55A55
The critical buckling load corresponds to the smallest valu e ofN0which is obtained
form=1
Ncr=D11ks
55A55π2
D11π2+ks
55A55a2=π2D11
a21
1+π2D11
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 wer e also considered
in Sect. 7.3. For symmetric cross-ply laminated plate strip s we obtain comparable
equations
D11d2ψ1
dx2
1−ks
55A55/parenleftbigg
ψ1+dw
dx1/parenrightbigg
=ρ2∂2ψ
∂t2,
ks
55A55/parenleftbiggdψ1
dx1+d2w
dx2
1/parenrightbigg
=ρ0∂2w
∂t2(8.6.60)
ρ0andρ2were deﬁned as
ρ0=n
∑
k=1ρ(k)h(k),ρ2=1
3n
∑
k=1ρ(k)/parenleftbigg
x(k)
33
−x(k−1)
33/parenrightbigg
and the terms involving ρ0andρ2are the translatory and the rotatory inertia terms.
ψ1andware functions of x1andtand thus we have partial derivatives. If we assume
again both strip edges simply supported the analytical solu tion follow with
w(x1,t) = C1me−iωmtsinmπx1
a,w(0,t)=w(a,t)=0,
ψ1(x1,t) =C2me−iωmtcosmπx1
a,∂ψ1(0,t)
∂x1=∂ψ1(a,t)
∂x1=0(8.6.61)
Substituting these solution functions into the vibration e quations (8.6.60) follow
/bracketleftbiggD11α2
m+ks
55A55−ρ2ω2
m ks
55A55αm
ks
55A55αm ks
55A55α2
m−ρ0ω2
m/bracketrightbigg/bracketleftbiggC2m
C1m/bracketrightbigg
=/bracketleftbigg0
0/bracketrightbigg
The nontrivial solution of the homogeneous algebraic equat ion yields the eigenfre-
quencies ωm
/vextendsingle/vextendsingle/vextendsingle/vextendsingleD11α2
m+ks
55A55−ρ2ω2
m ks
55A55αm
ks
55A55αm ks
55A55α2
m−ρ0ω2
m/vextendsingle/vextendsingle/vextendsingle/vextendsingle=0 (8.6.62)
or
ρ0ρ2ω4
m−(D11ρ0αm+ks
55A55ρ0+ks
55A55ρ2α2
m)2ω2
m+D11ks
55A55α4
m=0
Aω4
m−Bω2
m+C=0,ω2
m=B
2A±1
2A/radicalbig
B2−4A2C
The general solution for the vibration equations can be form ulated for arbitrary
boundary conditions. For harmonic oscillations we write
w(x1,t)=w(x1)eiωt,ψ1(x1,t)=ψ1(x1)eiωt(8.6.63)
Substituting w(x1,t)and ψ1(x1,t)in the coupled partial differential equations
(8.6.60) yield

320 8 Modelling and Analysis of Plates
D11d2ψ1(x1)
dx2
1−ks
55A55/parenleftbigg
ψ1+dw(x1)
dx1/parenrightbigg
+ρ2ω2ψ(x1) =0,
ks
55A55/parenleftbiggdψ1(x1)
dx1+d2w(x1)
dx2
1/parenrightbigg
+ρ0ω2w(x1) =0(8.6.64)
These both equations can be uncoupled. With
ks
55A55dψ1(x1)
dx1=−ρ0ω2w(x1)−ks
55A55d2w(x1)
dx1
and
D11d3ψ1(x1)
dx3
1−ks
55A55d2w(x1)
dx2
1−ks
55A55dψ1(x1)
dx1−ρ2ω2dψ1(x1)
dx1=0
follow
D11d4w(x1)
dx4
1+/parenleftbiggD11ρ0
ks
55A55+ρ2/parenrightbigg
ω2d2w(x1)
dx2
1−/parenleftbigg
1−ρ2ω2
ks
55A55/parenrightbigg
ρ0ω2w(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)=C1sinλ1×1+C2cosλ2×1+C3sinhλ3×1+C4cosh λ4×1 (8.6.66)
Theλiare the roots of the characteristic algebraic equation of (8 .6.65). The deriva-
tions above demonstrated that for any boundary conditions a n analytical solution
is possible. Unlike to the classical theory we restricted th e considerations in the
frame of the shear deformation theory to cross-ply laminate d strips. Summarizing
the derivations we can draw the following conclusions:
•Cylindrical bending yields simple analytical solutions fo r unsymmetrical and
symmetrical cross-ply laminated plate strips.
•Restricting to symmetrical laminated cross-ply plate stri ps we can obtain ana-
lytical solutions for buckling and vibrations problems, bu t 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 s olutions can be obtain for
symmetric and antisymmetric cross-ply and angle-ply lamin ates 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 th en uncoupled.
Rectangular specially orthotropic plates may be simply sup ported (hard hinged
support) on all four edges.
x1=0,×1=a:w=0,M1=0 respectively∂ψ1
∂x1=0,ψ2=0,
x2=0,×2=b:w=0,M2=0 respectively∂ψ2
∂x2=0,ψ1=0
(8.6.67)
The boundary conditions can be satisﬁed by the following exp ressions:
w(x1,x2) =∞
∑
r=1∞
∑
s=1wrssinαrx1sinβsx2,
ψ1(x1,x2) =∞
∑
r=1∞
∑
s=1ψ1rscosαrx1sinβsx2,αr=rπ
a,βs=sπ
b,
ψ2(x1,x2) =∞
∑
r=1∞
∑
s=1ψ2rssinαrx1cosβsx2(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=1prssinαrx1sinβsx2,
prs=4
aba/integraldisplay
0b/integraldisplay
0p3(x1,x2)sinαrx1sinβsx2dx1dx2(8.6.69)
Now the Navier solution method can be extended to Mindlin’s p lates with all edges
simply supported, but the solution is more complex than for K irchhoff’s plates. Sub-
stituting the expression (8.6.68) and (8.6.69) into the pla te differential equations
(8.3.8) gives
L11L12L13
L12L22L23
L13L23L33

ψ1rs
ψ2rs
wrs
=
0
0
prs
 (8.6.70)
with
L11=D11α2
r+D66β2
s+ks
55A55,L12=(D12+D66)αrβs,L13=ks
55A55αr,
L22=D66α2
r+D22β2
s+ks
44A44,L33=ks
55A55α2
r+ks
44A44β2
s,L33=ks
44A44βs
(8.6.71)
Solving the Eqs. (8.6.65), one obtains
ψ1rs=L12L23−L22L13
Det(Li j)prs,ψ2rs=L12L13−L11L23
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),ψ1(x1,x2),ψ2(x1,x2)are calculated the
curvatures κ1,κ2andκ6may be obtained and the stresses in each lamina follow
from (8.3.3) to
σ1
σ2
σ6
(k)
=x3
Q11Q120
Q12Q220
0 0 Q66
(k)
κ1
κ2
κ6
,
/bracketleftbiggσ5
σ4/bracketrightbigg(k)
=/bracketleftbiggC550
0C44/bracketrightbigg(k)
ψ1+∂w
∂x1
ψ2+∂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 boundar y conditions, respectively,
but the solution procedure is more complicated than in the fr ame of classical plate
theory. We do without detailed considerations and recommen d approximate analyt-
ical solutions or numerical methods to analyze the behavior of general laminated
rectangular plates including shear deformations and suppo rted by any combination
of clamped, hinged or free edges.
Summarizing the discussion of analytical solutions for pla tes including trans-
verse shear deformations one can formulate following concl usion
•Analytical solutions for symmetrical and unsymmetrical la minated plates can be
derived for cylindrical bending, buckling and vibration.
•Navier’s double series solutions can be simple derived for s pecially orthotropic
plates. Navier’s solution method can be also applied to symm etric 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 a nalyze general lami-
nated rectangular plates with general boundary conditions .
•Plates with general geometry or with cut outs etc. should be a nalyzed by numer-
ical methods
8.7 Problems
Exercise 8.1. A plate strip has the width ainx1-direction and is inﬁnitely long in
thex2-direction. The strip is loaded transversely by a uniformly distributed load
p0and simply supported at x1=0,×1=a. Calculate the deﬂection w, the resultant
moments M1,M2,M6and the stresses σ1,σ2,σ6

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 sep arately.
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,
D11d2w
dx2
1=−M1,D12d2w
dx2
1=−M2,M6=0,
D11d4w
dx4
1=p0
The vertical deﬂection w=w(x1)is
w(x1)=1
D11/bracketleftbigg
p0x4
1
24+C1x3
1
6+C2x2
1
2+C3x1+C4/bracketrightbigg
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/bracketleftbigg/parenleftigx1
a/parenrightig4
−2/parenleftigx1
a/parenrightig3
+/parenleftigx1
a/parenrightig/bracketrightbigg
The moment resultants follow as
M1(x1) =−p0a2
2/bracketleftbigg/parenleftigx1
a/parenrightig2
−x1
a/bracketrightbigg
,
M2(x1) =D12
D11M1(x1)=D12
D11p0a2
2/bracketleftbigg/parenleftigx1
a/parenrightig2
−x1
a/bracketrightbigg
,
M6(x1) =0
The strains and stresses at any point can be determined as fol low
κ1=−d2w
dx1=p0a2
2/bracketleftbiggx1
a−/parenleftigx1
a/parenrightig2/bracketrightbigg
,κ2=0,κ6=0,
ε1=x3κ1=x3p0a2
2/bracketleftbiggx1
a−/parenleftigx1
a/parenrightig2/bracketrightbigg
,ε2=0,ε6=0

324 8 Modelling and Analysis of Plates
The stresses in each layer are
00−layers : σ1=σ′
1=x3Q11
D11p0a2
8/bracketleftbiggx1
a−/parenleftigx1
a/parenrightig2/bracketrightbigg
,
σ2=σ′
2=x3Q12
D11p0a2
8/bracketleftbiggx1
a−/parenleftigx1
a/parenrightig2/bracketrightbigg
,
σ6=0,
900−layers : σ2=σ′
1=x3Q11
D11p0a2
8/bracketleftbiggx1
a−/parenleftigx1
a/parenrightig2/bracketrightbigg
,
σ1=σ′
2=x3Q12
D11p0a2
8/bracketleftbiggx1
a−/parenleftigx1
a/parenrightig2/bracketrightbigg
,
σ6=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)
A11d2u
dx2
1−B11d3w
dx3
1=0,A66d2v
dx2
1=0,D11d4w
dx4
1−B11d3u
dx3
1=p3,
N1=A11du
dx1−B11d2w
dx2
1,N2=A12du
dx1−B12d2w
dx2
1,N6=0,
M1=B11du
dx1−D11d2w
dx2
1,M2=B12du
dx1−D12d2w
dx2
1,M6=0
The equilibrium equations for the stress resultants are
dN1
dx1=0,d2M1
dx2
1=−p3
The displacement u(x1)andw(x1)are coupled. Substitute
A11d3u
dx3
1=B11d4w
dx4
1
into the second differential equation yield
/parenleftbigg
D11−B2
11
A11/parenrightbiggd4w
dx4=p3
or with
D11/parenleftbigg
1−B2
11
A11D11/parenrightbigg
=DR
11,p3=p0⇒d4w
dx4
1=p0
DR
11
For the displacement u(x1)follows

8.7 Problems 325
d3u
dx3
1=B11
A11d4w
dx4
1
or with
A11/parenleftbigg
1−B2
11
A11D11/parenrightbigg
=AR
11,d4w
dx4
1=p0
DR
11⇒d3u
dx3
1=B11
D11p0
AR
11
These differential equations can be simple integrated
w(x1) =1
DR
11/bracketleftbiggq0x4
1
24+C1x3
1
6+C2x2
1
2+C3x1+C4/bracketrightbigg
u(x1) =B11
D11AR
11/bracketleftbiggq0x3
1
6+C1x2
1
2+C5x1+C6/bracketrightbigg
Note that with
d2u
dx2
1=B11
A11d3w
dx3
1
in both equations there are equal constants C1. The boundary conditions for w
andM1are 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
11p0a3
12/bracketleftbigg
2/parenleftigx1
a/parenrightig3
−/parenleftigx1
a/parenrightig2/bracketrightbigg
w(x1) =1
DR
11p0a4
24/bracketleftbigg/parenleftigx1
a/parenrightig4
−2/parenleftigx1
a/parenrightig3
+x1
a/bracketrightbigg
The stress and moment resultants follow as
N1(x1)=N6(x1)=0,
N2(x1)=/parenleftbiggA12B11
D11AR
11−B12
DR
11/parenrightbiggp0a2
2/bracketleftbigg/parenleftigx1
a/parenrightig2
−x1
a/bracketrightbigg
,
M1(x1)=−p0a2
2/bracketleftbigg/parenleftigx1
a/parenrightig2
−x1
a/bracketrightbigg
,
M2(x1)=/parenleftbiggB2
11−D12A11
A11/parenrightbiggp0a2
2DR
11/bracketleftbigg/parenleftigx1
a/parenrightig2
−x1
a/bracketrightbigg
,
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. Wit h

326 8 Modelling and Analysis of Plates
1
DR
11=1
D11/parenleftbigg
1−B2
11
A11D11/parenrightbigg>1
D11
the deﬂection of the unsymmetric laminate strip will be grea ter than the deﬂec-
tion of the symmetric laminate. Note that there is no force re sultant N1(x1)in the
unsymmetric case but it is very interesting that there is a fo rce resultant N2as a
function of x1, but N2(0)=N2(a)=0. With
ε1=du
dx1=B11
D11AR
11p0a2
2/bracketleftbigg/parenleftigx1
a/parenrightig2
−x1
a/bracketrightbigg
,ε2=ε6=0,
κ1=−d2
dx2
1=−p0a2
2DR
11/bracketleftbigg/parenleftigx1
a/parenrightig2
−x1
a/bracketrightbigg
, κ2=κ6=0
follow the strains ε1,ε2and the stresses σ1,σ2for 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 q0atx1=a/2. Both edges of the strip are ﬁxed.
Calculate the maximum deﬂection wmaxusing the shear deformation theory.
Solution 8.2. With (8.6.51) and (8.6.52) follow
DR
11≡D11,D11d3ψ1
dx3
1=q0δ/parenleftbigg
x1−1
2a/parenrightbigg
,dw
dx1=−ψ1+D11
ks
55A55d2ψ1
dx2
1
with
δ/parenleftbigg
x1−1
2a/parenrightbigg
=/braceleftbigg0x1/ne}ationslash=a/2
1×1=a/2,/integraldisplay
δ/parenleftbigg
x1−1
2a/parenrightbigg
dx1=/angbracketleftbigg
x1−1
2a/angbracketrightbigg0
<x1−e>is F¨ oppel’s5bracket symbol:
<x1−e>n=/braceleftbigg0 x1<e
(x1−e)nx1>e,
d
dx1<x1−e>n=n<x1−e>n−1,
/integraldisplay
<x1−e>ndx1=1
1+n<x1−e>n+1+C
With (8.6.53) the analytical solutions for ψ1andware given
5August Otto F¨ oppl (∗25 January 1854 Groß-Umstadt – †12 August 1924 Ammerlan) – pr ofessor
of engineering mechanics and graphical statics

8.7 Problems 327
D11ψ1(x1) =1
2q0<x1−1
2a>2+C1x2
1
2+C2x1+C3,
w(x1) =1
D11/bracketleftbigg1
6q0<x1−1
2a>3+C1x3
1
6+C2x2
1
2+C3x1+C4/bracketrightbigg
−1
ks
55A55[q0<x1−1
2a>+C1x1]
ψ1(0)=0 :C3=0,
w(0)=0 :C4=0,
ψ1(a)=0 :1
2q0/parenleftbigg1
2a/parenrightbigg2
+C1/parenleftbigga2
2/parenrightbigg
+C2a=0,
w(a)=0 :1
6q0/parenleftbigg1
2a/parenrightbigg3
+C1/parenleftbigga3
6/parenrightbigg
+C2a2
2−D11
ks
55A55/parenleftig
q0a
2+C1a/parenrightig
=0,
C1=q0
2,C2=−q0a
8,
ψ1(x1) =q0a2
8D11/bracketleftbigg/parenleftigx1
a/parenrightig2
−x1
a/bracketrightbigg
,
w(x1) =−q0a3
48D11/bracketleftbigg
3/parenleftigx1
a/parenrightig2
−4/parenleftigx1
a/parenrightig3/bracketrightbigg
−q0a
2ks
55A55x1
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 a rea=b=1 m,
hf=0,2875 10−3m,hc=24,71 10−3m,Ef=1,42 105MPa, νf=0,3,
Gf=Ef/2(1+νf),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 boundar ies. Calculate the max-
imum ﬂexural displacement wmaxwith the help of a one-term Ritz approximation.
Solution 8.3. The elastic potential Π(w,ψ1,ψ2)of a symmetric and special or-
thotropic Mindlin’s plate is given by (8.3.18). For stiff th in 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=hc/bracketleftig
Qf
i jh(f)¯x(f)
3/bracketrightig
=hchf1
2/parenleftig
hc+hf/parenrightig
Q(f)
i j,
((i j)=( 11),(22),(66),(12))with
Q11=Ef
1−(νf)2=Q22,Q12=νfEf
1−(νf)2,Q66=Gf

328 8 Modelling and Analysis of Plates
and
AS
i j=hcCc
i j=hcGc,(i j)=( 44),(55)
Π(w,ψ1,ψ2) =/integraldisplay
A/braceleftigg
1
2hchf(hc+hf)/bracketleftigg
Q11/parenleftbigg∂ψ1
∂x1/parenrightbigg2
+2Q12/parenleftbigg∂ψ1
∂x1∂ψ2
∂x2/parenrightbigg
+Q22/parenleftbigg∂ψ2
∂x2/parenrightbigg2
+Q66/parenleftbigg∂ψ1
∂x2+∂ψ2
∂x1/parenrightbigg2/bracketrightigg
+kshcGc/bracketleftigg/parenleftbigg
ψ1+∂w
∂x1/parenrightbigg2
+/parenleftbigg
ψ2+∂w
∂x2/parenrightbigg2/bracketrightigg/bracerightigg
dx1dx2
−/integraldisplay
pwdx1dx2
The one-term approximations
w(x1,x2) = a1sin/parenleftigπx1
a/parenrightig
sin/parenleftigπx2
a/parenrightig
,
ψ1(x1,x2) =a2cos/parenleftigπx1
a/parenrightig
sin/parenleftigπx2
a/parenrightig
,
ψ2(x1,x2) =a3sin/parenleftigπx1
a/parenrightig
cos/parenleftigπx2
a/parenrightig
satisfy the boundary conditions. Substituting these appro ximative 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
KKKaaa=qqq
with
aaaT=[a1a2a3],qqqT=[16p/π20 0]
and
KKK=
2hcGcλ2hcGcλ hcGcλ
hcGcλhchf¯xf
3(Q11+Q66)λ2+hcGchchf¯xf
3(Q12+Q66)λ2
hcGcλ hchf¯xf
3(Q12+Q66)λ2hchf¯xf
3(Q22+Q66)λ2+hcGc

with λ=π/a. The solution of the system of three linear equations leads t o
a1=0,0222,a2=a3=−0,046 and the maximum displacement follows to
wmax=w(x1=a/2,×2=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, νm=0.35,νf=0.22,vm=0.4,
vf≡φ=0.6,Gm=Em/2(1+νm) =1.2593 GPa, Gf=Ef/2(1+νf) =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 later al unit load F=1 N
atx1=ξ1,×2=ξ2using 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 cro ss-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,×2)=w(a,x2)=w(x1,0)=w(x1,b)=0,
M1(0,×2)=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=1prs
drssinαrx1sinβsx2
with
drs= [D11α4
r+2(D12+2D66)α2
rβ2
s+D22β4
s],αr=rπ
a,βs=sπ
b,
prs=4F
absinαrξ1sinβsξ2
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=νfvf+νmvm=0,272,ν′
21=ν′
12E′
2/E′
1=0,0328
and (4.1.3)
Q′
11=E′
1/(1−ν′
12ν′
21)=67,97 GPa,
Q′
22=E′
2/(1−ν′
12ν′
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
33
∑
k=1Q(k)
i j/parenleftig
(x(k)
3)3−(x(k−1)
3)3/parenrightig
,
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
π4∞
∑
r=1∞
∑
s=1sinαrξ1sinβsξ2
18492 r4+4650 r2s2+2927 s4sinαrx1sinβsx2
IfF=1 N then w(x1,x2)represents the inﬂuence surface, i.e. the deﬂection at
(x1,x2) due to a unit load at ( ξ1,ξ2). This inﬂuence function w(x1,x2;ξ1,ξ2)is
sometimes called Green’s function of the plate with all boun daries simply sup-
ported. In the more general case of a rectangular plate a/ne}ationslash=bthe Green’s function
is
w(x1,x2;ξ1,ξ2)=F
π4ab∞
∑
r=1∞
∑
s=1sinαrξ1sinβsξ2
drssinαrx1sinβsx2
The Green’s function can be used to calculate the bending sur faces of sim-
ply supported rectangular plates with any transverse loadi ng. With the solution
w(x1,x2)we can calculate the stress resultants M1,M2,M6,Q1,Q2and the stresses
σ1,σ2,σ6,σ5andσ4using (8.6.27) and (8.6.28).
2. Using (8.6.41) the equation for the natural frequencies o f a simply supported
rectangular plate is
ω2
rs=π4
ρh/bracketleftbig
D11α4
r+2(D12+2D66)α2
rβ2
s+D22β4
s/bracketrightbig
with
h=∑
(k)h(k),ρ=1
h∑
(k)ρ(k)/parenleftig
x(k)
3−x(k−1)
3/parenrightig
The fundamental frequency corresponds to r=s=1 and is given by
ω2
11=π4
ρha4/bracketleftbigg
D11+2(D12+2D66)/parenleftiga
b/parenrightig2
+D22/parenleftiga
b/parenrightig4/bracketrightbigg

8.7 Problems 331
Fora=b=1 m and the given material properties we ﬁnd the fundamental n atural
frequency
ω11=1593,5/radicalbig
ρh
Exercise 8.5. Consider a cylindrically orthotropic circular plate with a midplane
symmetric layer stacking under the conditions of axisymmet ric 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,Eθ,νrθ) loaded α)with a radial boundary force
Nr(R) =−Nroandβ)with a body force hpr=hρω2rcaused by spinning the
disk about the axis with an angular velocity ω.
2. Develop the differential equations for transverse loadi ng under the condition of
the ﬁrst order shear deformation theory. Calculate the stre ss resultants for a solid
plate ( R,h,Er,Eθ,νrθ) 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,σ1=σr,σ2=σθ,
σ6=σrθ,ε1=εr,ε2=εθ,ε6=εrθ. For axisymmetric deformations of circular disks
and plates all stresses, strains and displacements are inde pendent of θ, i.e. they are
functions of ralone and σ6=0,ε6=0.
1. For an in-plane loaded cylindrical orthotropic circular disk under the condition
of axisymmetric deformations the equilibrium, constituti ve and geometric
equations are:
•Equilibrium Equations (Fig. 8.7)
With cos(π/2−dθ/2)=sin(dθ/2)≈dθ/2 follow
Fig. 8.7 Disc element
(rdrdθ)hx3
0dθNrpr
Nr+dNrNθdθ
2
Nθdθ
2r+dr
hdr

332 8 Modelling and Analysis of Plates
d(rNr)
dr−Nθ+prr=0
•Constitutive Equations
Nr=A11εr+A12εθ,Nθ=A12εr+A22εθ,Nrθ=0
•Geometric Equations
εr=du
dr,εθ=u
r,γrθ=0
These equations forming the following system of three ordin ary differential equa-
tions
d(rNr)
dr−Nθ=−p0r,
Nr=A11du
dr+A12u
r,Nθ=A12du
dr+A22u
r
involving three unknown quantities Nr,Nθandu. Substituting the stress resul-
tants in the equilibrium equations yield one uncoupled diff erential equation for
u(r)
rd2u
dr2+du
dr−1
rδ2u=−rpr
A11
with δ2=A22/A11or
d2u
dr2+1
rdu
dr−δ2
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 w ith
u(r)=Crλ,λ=±δ
as
u(r)=C1r+δ+C2r−δ
With R1=0,R2=Rwe obtain C2=0,C1=−Nr0/[(A11δ+A12)Rδ−1]and such
Nr(r)=−Nr0/parenleftigr
R/parenrightigδ−1
,Nθ(r)=−Nr0δ/parenleftigr
R/parenrightigδ−1
Conclusion 8.1. Forδ=1 we have an isotropic disk with the well-known solu-
tion Nr=Nθ=−Nr0. For δ>0, i.e. the circumferential stiffness exceeds the
radial stiffness, at r=0 we have Nr=Nθ=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=ρω2rwe obtain the solution of the
inhomogeneous differential equations as
u(r)=C1rδ−1
A11ρω2
9−δ2r3=C1rδ+1
A11ρω2
δ2−9r3
Forδ=1 follow the well-known solution
u(r)=Cr−1−ν2
Eρω2
8r3
2. With Fig. 8.8 we obtain:
•Equilibrium Equations
d(rMr)
dr−Mθ−rQr=0,d(rQr)
dr+rp3=0
•Constitutive Equations
Mr=D11κr+D12κθ,Mθ=D12κr+D22κθ,
Qr=ks
55A55/parenleftbigg
ψr+dw
dr/parenrightbigg
•Geometric Equations
κr=dψr
dr,κθ=ψr
r,γrz=/parenleftbigg
ψr+dw
dr/parenrightbigg
Integrating the second equilibrium equation
Fig. 8.8 Plate element
(rdrdθ)hx3
QrQr+dQr
MrMθ
pz
Mr+dMr
Mθ

334 8 Modelling and Analysis of Plates
Qr(r)=1
r/parenleftbigg
C1−/integraldisplay
p3(r)rdr/parenrightbigg
and substituting Mr,MθandQrin the ﬁrst equilibrium equation yield
rd2ψr
dr2+dψr
dr−1
rδ2
pψr=1
D11/parenleftbigg
C1−/integraldisplay
p3rdr/parenrightbigg
,δ2
p=D22
D11
The general solution has again the form
ψr(r)=C2rδp+C3r−δp+ψp(r)
ψp(r)is the particular solution of the inhomogeneous differenti al equation de-
pending on the form of the loading functions p3(r). The differential equation for
the plate deﬂection w(r)follows with
ksA55dw
dr=Qr−ksA55ψr,
dw
dr=1
ksA55/parenleftbigg
C11
r−1
r/integraldisplay
p3(r)rdr/parenrightbigg
−C2rδp−C3r−δp+ψ0,
w(r) =1
ksA55/parenleftbigg
C1lnr−/integraldisplay1
r/integraldisplay
p3(r)rdrdr/parenrightbigg
−C2rδp+1
δp+1−C3r−δp+1
1−δp+C4−/integraldisplay
ψpdr
For a constant pressure p3(r)=−p0we obtain
w(r) =1
ksA55/parenleftbigg
C1lnr+p0r2
4/parenrightbigg
−C2
1+δpr1+δp−C3
1−δpr1−δp
+C4+C1r2
2D11(δ2p−1)+p0r4
8D11(δ2p−9),
ψr(r) =C2rδp+C3r−δp−C1r
D11(δ2p−1)−p0r3
2D11(δ2p−9)
This general solution is not valid for δp=1 and δp=3 because the particular
solutions ψpfor theses δp-values include terms coinciding with the fundamental
solutions randr3. Therefore, the particular solutions must be determined in an-
other form. For δp=1, i.e. for the isotropic case, one can use ψp=Arlnr+Br3
and for δp=3ψp=Ar+Br3and one obtains the general solutions
δp=1
w(r) =1
ksA55/parenleftbigg
C1lnr+p0r2
4/parenrightbigg
−1
2C2r2−C3lnr
+C4−C1r4
4D11/parenleftbigg
lnr−1
2/parenrightbigg
+p0r4
64D11,
ψr(r) =C2r+C31
r+C1rlnr
2D11+p0r3
16D11

8.7 Problems 335
δp=3
w(r) =1
ksA55/parenleftbigg
C1lnr+p0r2
4/parenrightbigg
−1
4C2r4+C31
2r2
+C4−C1r2
16D11+p0r4
48D11/parenleftbigg
lnr−1
4/parenrightbigg
,
ψr(r) =C2r3+C31
r3−C1rlnr
8D11+p0
12D11r3lnr
The constants C1,C2,C3andC4are determined from the boundary conditions at
the inner and outer plate edge. For solid plates with R1=0,R2=Rthe constants
C1andC3must be zero, otherwise ψrandwtend to inﬁnity at the plate center.
Forδ/ne}ationslash=3 the general solution for solid plates is
w(r) =p0r2
4/bracketleftigg
1
ks
55A55+r2
2D11(δ2p−9)/bracketrightigg
−C2
1+δpr1+δp+C4,
ψr(r) =C2rδp−p0r3
2D11(δ2p−9)
α)Clamped solid circular plate (δp/ne}ationslash=3)
The boundary conditions are ψ1(R)=0,w(R)=0 yield the constants C2andC4
and the solution as
w(r)=−p0
4ksA55(R2−r2)+p0
2D11(δ2p−9)/bracketleftigg
R3−δpr1+δp
1+δp−r4
4+R4(δp−3)
4(1+δp)/bracketrightigg
β)Simply supported solid circular plate (δp/ne}ationslash=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(δ2p−9)/bracketleftigg
(3D11+D12)R3−δp
(δpD11+D12)(1+δp)(r1+δp−R1+δp)+1
4(R4−r4)/bracketrightigg
Note that if the transverse shear deformations are neglecte d we must put
ks
55A55→∞. In the particular case ks
55A55→∞andδp=1 follow the well-known
solutions for the classical theory of isotropic plates, i.e .
α)w(r) =−p0
64D(R2−r2)2=−p0R4
64D/bracketleftig
1−/parenleftigr
R/parenrightig/bracketrightig2
,
β)w(r) =−p0
64D(R2−r2)/parenleftbigg5+ν
1+νR2−r2/parenrightbigg
=−p0R4
64D/bracketleftig
1−/parenleftigr
R/parenrightig/bracketrightig2/bracketleftbigg5+ν
1+ν−/parenleftigr
R/parenrightig2/bracketrightbigg

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 Vlas ov-Kantorovich
method, (2.2.45) – (2.2.47), based on the variation of the po tential energy Π(w).
2. Demonstrate for the special case of a plate clamped at all e dges 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
2a/integraldisplay
−ab/integraldisplay
−b/bracketleftigg
D11/parenleftbigg∂2w
∂x2
1/parenrightbigg2
+D22/parenleftbigg∂2w
∂x2
2/parenrightbigg2
+2D12∂2w
∂x2
1∂2w
∂x2
2+4D66/parenleftbigg∂2w
∂x1∂x2/parenrightbigg2
−2p0/bracketrightigg
dx1dx2
The one-term approximate solution ˜ w(x1,x2) =wi j(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 δΠof the elastic potential energy Π(w)yields
δΠ(w) =1
2a/integraldisplay
−ab/integraldisplay
−b/bracketleftbigg/parenleftbigg
D11∂2w
∂x2
1+D12∂2w
∂x2
2/parenrightbigg
δ/parenleftbigg∂2w
∂x2
1/parenrightbigg
+/parenleftbigg
D12∂2w
∂x2
1+D22∂2w
∂x2
2/parenrightbigg
δ/parenleftbigg∂2w
∂x2
2/parenrightbigg
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✻
✲x1x2
✻❄✻
❄bb
✛ ✲ ✛ ✲a a

8.7 Problems 337
+4D66/parenleftbigg∂2w
∂x1∂x2/parenrightbigg
δ/parenleftbigg∂2w
∂x1∂x2/parenrightbigg
−p0δw/bracketrightbigg
dx1dx2
Substituting ˜ w(x1,x2)=fi(x1)gj(x2)one obtains
δΠ(w) =a/integraldisplay
−a/bracketleftbig
D11A f′′
iδf′′
i+D12B(f′′
iδfi+fiδf′′
i)+D22C fiδfi
+4D66D f′
iδf′
i−¯p0δfi/bracketrightbig
dx1
where
A=b/integraldisplay
−bg2
jdx2,B=b/integraldisplay
−bg′′
jgjdx2,C=b/integraldisplay
−b(g′′
j)2dx2,D=b/integraldisplay
−b(g′
j)2dx2,¯p0=b/integraldisplay
−bp0gjdx2
Integrating Bby parts yield
B=g′
jgj/vextendsingle/vextendsingle/vextendsingleb
−b−b/integraldisplay
−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/integraldisplay
−af′′
iδf′′
idx1
ofδΠ
a/integraldisplay
−af′′
iδf′′
idx1=a/integraldisplay
−af′′
i(δfi)′′dx1=f′′
iδf′
i/vextendsingle/vextendsingle/vextendsinglea
−a−a/integraldisplay
−af′′′
i(δfi)′dx1
=f′′
iδf′
i/vextendsingle/vextendsingle/vextendsinglea
−a−f′′′
iδfi/vextendsingle/vextendsingle/vextendsinglea
−a+a/integraldisplay
−af′′′′
iδfidx1
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,
atx1=±a:[D11A f′′
i(x1)−D12D fi(x1)]δf′
i(x1)=0,
[D11A f′′′
i(x1)−D12D f′
i(x1)+4D66D f′
i(x1)]δfi(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=DD 3
AD1,k4
2=CD 2
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
1ork2
2>k2
1in the form
f(x1)=4
∑
l=1ClΦl(x1)+fp
with fp=¯p0/D2C. The solutions can be simpliﬁed if the problem is symmetric o r
antisymmetric. The constants Clcan be calculated with the boundary conditions
atx1=±a.
2. In the special case of all plate edges are clamped the corre sponding boundary
conditions are
x1=±a:w=0,∂w
∂x1=0,×2=±b:w=0,∂w
∂x2=0
The one-term deﬂection approximation is assumed again in th e form ˜ w(x1,x2)=
wi j(x1,x2)=fi(x1)gj(x2). The Galerkin’s procedure yields
a/integraldisplay
−ab/integraldisplay
−b/parenleftbigg
D1∂4wi j
∂x4
1+2D3∂4wi j
∂x2
1∂x2
2+D2∂4wi j
∂x4
2−p0/parenrightbigg
gjdx2=0
and we obtain
D1
b/integraldisplay
−bg2
jdx2
d4fi
dx4
1+2D3
b/integraldisplay
−bd2gj
dx2
2gjdx2
d2fi
dx2
1+
D2
b/integraldisplay
−bd4gj
dx4
2gjdx2
fi=b/integraldisplay
−bp0gjdx2
Two of the integral coefﬁcients must be integrated by parts

8.7 Problems 339
b/integraldisplay
−bd2gj
dx2
2gjdx2=dgj
dx2gj/vextendsingle/vextendsingle/vextendsingleb
−b−b/integraldisplay
−b/parenleftbiggdgj
dx2/parenrightbigg
dx2,
b/integraldisplay
−bd4gj
dx4
2gjdx2=d3gj
dx3
2gj/vextendsingle/vextendsingle/vextendsingleb
−b−d2gj
dx2
2dgj
dx2/vextendsingle/vextendsingle/vextendsingleb
−b+b/integraldisplay
−b/parenleftbiggd2gj
dx2
2/parenrightbigg2
dx2
The results can be simpliﬁed because for the clamped edges fo llow
gj(±b)=0,dgj
dx2/vextendsingle/vextendsingle/vextendsingle
±b=0
and we obtain the same differential equation as in 1.
D1
b/integraldisplay
−bg2
jdx2
d4fi
dx4
1−2D3
b/integraldisplay
−b/parenleftbiggdgj
dx2/parenrightbigg2
dx2
d2fi
dx2
1+
D2
b/integraldisplay
−bd2gj
dx2
2dx2
fi=b/integraldisplay
−bp0gjdx2
To improve the one-term approximative plate solution we pre sent in a second
step now fi(x1)a priori and obtain in a similar manner a differential equati on for
an unknown function gj(x2)
D2
a/integraldisplay
−af2
idx1
d4gj
dx4
2−2D3
a/integraldisplay
−a/parenleftbiggdfi
dx1/parenrightbigg2
dx2
d2gj
dx2
2+
D1
a/integraldisplay
−a/parenleftbiggd2fi
dx2
2/parenrightbigg2
dx1
gj=a/integraldisplay
−ap0fidx1
In this way we have two ordinary differential equations of th e iterative solution
procedure which can be written
D1Agd4fi
dx4
1−2D3Dgd2fi
dx2
1+D2Cgfi=¯p0g,
D2Afd4gj
dx4
2−2D3Dfd2gj
dx2
2+D1Cfgj=¯p0f
Both equations can be rearranged in the standard form, App. E
d4fi
dx4
1−2k2
1gd2fi
dx2
1+k4
2gfi=kpg,d4gj
dx4
2−2k2
1fd2gj
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 d epend on the
relation between k2
2gandk2
1gork2
2fandk2
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 approximat ion is
w10(x1,x2)=f1(x1)(x2
2−b2)2
and satisfy the boundary conditions w=0,∂w/∂x2=0,×2=±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 ax1cosbx1+C2sinhax1sinbx1+kpg
k4
2g
The constants C1,C2can be calculated with
f1(±a)=0,df1
dx1/vextendsingle/vextendsingle/vextendsingle
±a=0
andw[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 en gineering 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 a spect ratios.
References
N´ adai A (1925) Die elastischen Platten: die Grundlagen und Verfahren zur Berech-
nung ihrer Form¨ anderungen und Spannungen, sowie die Anwen dungen der Theo-
rie der ebenen zweidimensionalen elastischen Systeme auf p raktische Aufgaben.
Springer, Berlin

Chapter 9
Modelling and Analysis of Circular Cylindrical
Shells
In the previous Chaps. 7 and 8 we have considered beams and pla tes, i.e. one- and
two-dimensional structural elements with straight axes an d plane reference surfaces.
Thin-walled laminated or sandwich shells can be also modell ed as two-dimensional
structural elements but with single or double curved refere nce 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 d ifferential geometry
relations.
To give a brief insight into the modelling of shells only the s implest shell ge-
ometry will be selected and the following considerations ar e restricted to circular
cylindrical shells. The modelling and analysis of circular cylindrical shells fabri-
cated from ﬁbre composite material, i.e. its structural the ory, 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 theo ry is capable of accu-
rately predicting the shell behavior. For thick-walled she lls, say R/h<10, a three-
dimensional modelling must be used.
Each single lamina of a ﬁlamentary composite material behav es 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 co ordinates, i.e., the axial
and circumferential directions, the shell is said to be spec ial orthotropic or circum-
ferential cross-ply circular cylindrical shell. Since the often used cylindrical shells
with closely spaced ring and/or stringer stiffeners also ca n 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 ax es 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 bee n not often subjected to
analysis.
In Chap. 9 there are only a short summarizing section on sandw ich shells and
no special section considering hygrothermo-elastic effec ts. Both problems can be
simple retransmitted from the corresponding sections in Ch aps. 7 and 8. Also a spe-
cial discussion of analytical solution methods will be negl ected, because no general
shell problems are considered.
341 © Springer Nature Singapore Pte Ltd. 2018
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 circula r cylindrical shells in the
frame of the classical shell theory and the shell theory incl uding transverse shear
deformations. Figure 9.1 shows a laminated circular cylind rical shell with general
layer stacking, the global coordinates x1=x,x2=s=Rϕ,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 lamina ted 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 axi ally symmetric loads
using the classical shell theory. The mathematically model ling leads in this case to
a
b cx3=z
x2=Rϕx1=xu2u3
u1
h
R
2=x′
21=x′
1
x1x2
θz
h
2h
2z(2)z(n−1)
z(1)
z(0)=−h
2z(n)=h
2
Fig. 9.1 Circular cylindrical shell. aGeometry, global coordinates x1=x,x2=s=Rϕ,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 an d loading will be primary
considered in Chap. 9, because analytical solutions can be d erived. Generally as-
sumed is that each layer having a constant angle of wrap, cons tant volume ratio of
ﬁbre to resin, and the ﬁbre and resin are both isotropic and ho mogeneous within
themselves. The ply material axes, Fig. 9.1 b, will be rotate d 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 equ ations of the classical
shell theory:
•Displacements are small compared to the shell thickness, al l 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 d eformed middle sur-
face, and unchanged in length after deformation.
•All components of translational inertia are included in mod elling 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 th eory 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 negle ct of transverse shear
deformation and transverse normal stress.
In addition we assume that each individual layer is consider ed 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 eac h layer may be of
arbitrary thickness and may be arranged either symmetrical ly or unsymmetrically
with respect to the middle surface.
9.2.1 General Case
The governing differential equations are formulated in ter ms 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, cylindri cal 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
εx=∂u
∂x, εs=∂v
∂s+w
R, εxs=∂u
∂s+∂v
∂x,
κx=−∂2w
∂x2,κs=−∂2w
∂s2+1
R∂v
∂s,κxs=−2∂2w
∂x∂s+1
R∂v
∂x(9.2.2)
The total strains at a arbitrary distance zof the middle surface are
εx=εx+κxz,εs=εs+κsz,εxs=εxs+κxsz
or
εj=εj+κjz,j=(1,2,6)≡(x,s,xs) (9.2.3)
Each individual layer is assumed to be in a state of generaliz ed plane stress, the
Hooke’s law yields
σ(k)
i=Q(k)
i jεj,i,j=(1,2,6) (9.2.4)
and in the general anisotropic case the Qi jmatrix 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/integraldisplay
−h/2σidz,Mi=h/2/integraldisplay
−h/2σizdz,i=(1,2,6)≡(x,s,xs) (9.2.5)
Putting Eq. (9.2.4) into (9.2.5) yields the constitutive eq uations in the known form
/bracketleftbigg
NNN
MMM/bracketrightbigg
=/bracketleftbigg
AAA BBB
BBB DDD/bracketrightbigg/bracketleftbigg
εεε
κκκ/bracketrightbigg
(9.2.6)
with
Ns+∂Ns
∂sdsQs+∂Qs
∂sds
Nsx+∂Nsx
∂xdxNsx+∂Nsx
∂xdxQx+∂Qx
∂xdx
Nx+∂Nx
∂xdx
pspz
px
Nx
QxNxsQsNsNsxds=RdϕdxMsx+∂Msx
∂sds
Ms+∂Ms
∂xdxMx+∂Mx
∂xdxMxs+∂Mxs
∂sds
MxsMxMs
Msxdxds=Rdϕ
Fig. 9.2 Positive directions for stress resultants

9.2 Classical Shell Theory 345
(Ai j,Bi j,Di j)=h/2/integraldisplay
−h/2(1,z,z2)Qi jdz
i.e. for nlaminate layers
Ai j=n
∑
k=1Qi j/parenleftig
z(k)−z(k−1)/parenrightig
,
Bi j=1
2n
∑
k=1Qi j/parenleftig
z(k)2−z(k−1)2/parenrightig
,
Di j=1
3n
∑
k=1Qi j/parenleftig
z(k)3−z(k−1)3/parenrightig
NNNT= [NxNsNxs],MMMT= [MxMsMxs],
εεεT= [εxεsεxs], κκκT= [κxκsκxs](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 tra nsverse shear resul-
tants and one obtains
∂Nx
∂x+∂Nxs
∂s+px =0,
∂Nxs
∂x+∂Ns
∂s+1
R/parenleftbigg∂Ms
∂s+∂Mxs
∂x/parenrightbigg
+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

L11L12L13
L21L22L23
L31L32L33

u
v
w
=−
px
ps
pz
 (9.2.10)
The linear differential operators Li jare 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 cyl indrical 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 include d.
Without detailed derivation on obtains
∂Nx
∂x+∂Nxs
∂s=ρ0∂2u
∂t2,
∂Nxs
∂x+∂Ns
∂s+1
R/parenleftbigg∂Ms
∂s+∂Mxs
∂x/parenrightbigg
=ρ0∂2v
∂t2,
∂2Mx
∂x2+2∂2Mxs
∂x∂s+∂2Ms
∂s2−Ns
R=ρ0∂2w
∂t2(9.2.11)
and Eq. (9.2.10) changes to

L11L12L13
L21L22L23
L31L32L33

u
v
w
=ρ0∂2w
∂t2
u
v
w
 (9.2.12)
with
ρ0=n
∑
k=1z(k)/integraldisplay
z(k−1)ρ(k)
0dz=n
∑
k=1ρ(k)
0h(k)
The stress resultants and the displacement are now function s ofx,sandt.ρ(k)
0is the
mass density of the kth layer, ρ0the mass inertia with respect to the middle surface.
9.2.2 Specially Orthotropic Circular Cylindrical Shells S ubjected
by Axial Symmetric Loads
Now we consider cross-ply laminated circular cylindrical s hells. The laminate stack-
ing may be not middle-surface symmetric, but the ﬁber angles areθ=00orθ=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 ∂/∂sandv,Nxs,Mxszero. For the
loads per unit of the surface area are the following conditio ns 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)
εx=du
dx,εs=w
R,κx=−d2w
dx2,εxs=0,κs=κxs=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(εx+zκx)+Q(k)
12εs=Q(k)
11/parenleftbiggdu
dx−zd2w
dx2/parenrightbigg
+Q(k)
12w
R,
σ(k)
s=Q(k)
12(εx+zκx)+Q(k)
22εs=Q(k)
12/parenleftbiggdu
dx−zd2w
dx2/parenrightbigg
+Q(k)
22w
R,
Q(k)
11=E(k)
x
1−ν(k)
xsν(k)
sx,Q(k)
22=E(k)
s
1−ν(k)
xsν(k)
sx,Q(k)
12=ν(k)
sxE(k)
x
1−ν(k)
xsν(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=A11εx+A12εs+B11κx,
Ns=A12εx+A22εs+B12κx,
Mx=B11εx+B12εs+D11κx,
Ms=B12εx+B22εs+D12κx,(9.2.16)
with εx,εsandκxfrom Eq. (9.2.14).
Putting (9.2.12) and (9.2.14) in the equilibrium equations (9.2.13) one obtains
after a rearrangement
A11d2u
dx2+A12
Rdw
dx−B11d3w
dx3=0,
/bracketleftbiggA11D11−B2
11
A11/bracketrightbiggd4w
dx4+2
R/bracketleftbiggA12B11
A11−B12/bracketrightbiggd2w
dx2
+1
R2/bracketleftbiggA11A22−A2
11
A11/bracketrightbigg
w=pz−A12
A11Nx
R,
and with
DR=A11D11−B2
11
A11,4λ4=1
DRR2A11A22−A2
11
A11(9.2.17)
can ﬁnally be written
d4w
dx4+2
RDR/bracketleftbiggA12B11
A11−B12/bracketrightbiggd2w
dx2+4λ4w=1
DR/parenleftbigg
pz−A12
A11Nx
R/parenrightbigg
(9.2.18)
This is a ordinary differential equation of fourth order wit h 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 E q. (9.2.18) can be
reduced with B11=B12=0,DR=D11as
d4w
dx4+4λ4w=1
D11/parenleftbigg
pz−A12
A11Nx
R/parenrightbigg
,4λ4=1
D11R2A11A22−A2
11
A11(9.2.19)
The inhomogeneous linear differential equation of fourth o rder has constant coefﬁ-
cient and can be analytically solved (App. E)
w(x)=wh(x)+wp(x)
The homogeneous solution wh(x)=Ceαxyields the characteristic equation
α4+4λ4=0
with the conjugate complex roots
α1−4=±λ(1±i),i=√
−1 (9.2.20)
and with e±λx=cosh λx±sinhλx, e±iλx=cosλx±isinλxone obtains the solution
of the homogeneous differential equation as
wh(x) =C1cosh λxcosλx+C2cosh λxsinλx
+C3sinhλxcosλx+C4sinhλxsinλx(9.2.21)
or
wh(x)=e−λx(C1cosλx+C2sinλx)+eλx(C3cosλx+C4sinλx) (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 c ase, i.e.
Mx=−D11d2w
dx2,Qx=dMx
dx=−D11d3w
dx3
the solution can be written as:
w(x) =M0
2λ2D11e−λx(sinλx−cosλx)−Q0
2λ3D11e−λxcosλx
+ML
2λ2D11e−λ(L−x)(sinλ(L−x)−cosλ(L−x))
+QL
2λ3D11e−λ(L−x)cosλ(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,QLatx=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
4λ2D11/bracketleftbigg
p(x)−A12
A11Nx
R/bracketrightbigg
(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. T he trigonomet-
ric terms oscillate between ±1 and are multiplied by exponential terms with a
negative exponent which yields to an exponential decay. If w e set λx=1.5π
orλ(L−x) = 1.5πthen is e−1.5π≈0.009, i.e. the inﬂuence of the boundary
values M0,Q0orML,QLis strong damped to <1%. With 0 ≤x≤1.5π/λor
0≤L−x≤1.5π/λbending boundary layers are deﬁned which depend on the shell
stiffness.
The important point is that at each end of the shell a characte ristic 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,Q0andwpare non-zero,
in the region C-D only ML,QLandwpand in the region B-C only the particular
solution wpis nonzero, i.e. in this region only a membrane solution exis ts. 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
ofu(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
/bracketleftbigg
σx
σs/bracketrightbigg(k)
=/bracketleftbigg
Q11Q12
Q12Q22/bracketrightbigg(k)/bracketleftbigg
εx
εs/bracketrightbigg
+z/bracketleftbigg
Q11Q12
Q12Q22/bracketrightbigg(k)/bracketleftbigg
κx
0/bracketrightbigg
(9.2.25)
A B C DLB LB
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 σxzfollows analogous to the classical beam equations
with (7.2.31).
Summarizing the results of the classical shell equations on e can draw the follow-
ing conclusions:
•The most general case of laminated circular cylindrical she lls is that of arbitrarily
laminated anisotropic layers, i.e. angle-ply layers arbit rarily arranged. The anal-
ysis of these shells is based on approximately analytical me thods using Ritz-,
Galerkin- or Kantorovich method and numerical methods , e.g . FEM.
•Cross-ply laminated shells, i.e shells with orthotropic la yers aligned either ax-
ially or circumferentially and arranged symmetrically wit h 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 symme trical loading the shell
equations reduce in the static case to ordinary differentia l equations of the x-
coordinate and can be solved analytically. If the orthotrop ic layers are arranged
to an unsymmetric laminated cross-ply shell then bending-s tretching, 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 thi ckness and the lay-
ers are arranged symmetrically with respect to the middle su rface, the govern-
ing equations are identical to those of single layer isotrop ic shells. However, if
the isotropic layers are arranged unsymmetrically to the mi ddle surface, there is
a coupling between in-surface, i.e stretching and shear, an d out-of-surface, i.e
bending and twisting, effects.
•Additional to the Kirchhoff’s hypotheses all equations of t he 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 abo ve 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 analyze d in the frame of the so-
called membrane theory. One neglects all moments and transv erse stress resultants,
all stresses are considered approximatively constant thro ugh the shell thickness i.e.
there are no bending stresses and the coupling and bending st iffness are taken to be
zero in the constitutive equations. In some cases it is possi ble to use the membrane
theory for structural analysis of laminated shells. The efﬁ cient structural behavior of
shells based on the shell curvature that yields in wide regio ns 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 sufﬁc ient accurate results in
regions with concentrated loads, boundary constraints or c urvature changes, i.e. in
regions located adjacent to each structural, material or lo ad 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,
εx=∂u
∂x,εs=∂v
∂s+w
R,εxs=∂u
∂s+∂v
∂x=γxs,
Nx=A11εx+A12εs,Ns=A12εx+A22εs,Nxs=A66εxs(9.2.26)
The membrane theory yield three equilibrium conditions to c alculate three unknown
stress resultants, i.e the membrane theory is statically de termined. The membrane
theory is the simplest approach in shell analysis and admit a n approximative an-
alytical solution that is very convenient for a ﬁrst analysi s and design of circular
cylindrical shells.
But the problems which can be solved by the membrane theory ar e unfortunately
limited. To avoid generally to use the more complex bending t heory we can con-
sider a so-called semi-membrane theory of circular cylindr ical shells. The semi-
membrane theory is slightly more complicated than the membr ane theory but more
simpler than the bending theory. The semi-membrane theory w as ﬁrst developed by
Vlasov on the basis of statically and kinematically hypothe ses.
If one intends to construct a semi-membrane theory of compos ite circular cylin-
drical shells bearing in mind the hypotheses underlying the classical single layer
shell theory and the characteristics of the composite struc ture. The semi-membrane
theory for composite circular cylindrical shells introduc es the following assump-
tions:
•The shell wall has no stiffness when bended but in axial direc tion 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.
εs=∂v
∂s+w
R=0
The shear stiffness of composite shells can be small. Theref ore, 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 f ollowing 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,
εx=∂u
∂x,εxs=∂v
∂x+∂u
∂s,εs=∂v
∂s+w
R,(9.2.27)

352 9 Modelling and Analysis of Circular Cylindrical Shells
Nx=A11εx,Nxs=A66εxs,Ms=D22κs,Qs=ks
44A44/parenleftbigg
ψs+∂w
∂s/parenrightbigg
The system (9.2.27) can be reduced. For the circular cylindr ical shell the unknown
functions and loads can be represented with trigonometric s eries and after some
manipulations we obtained one uncoupled ordinary differen tial 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 sh ell theory is only sufﬁ-
ciently accurate for thin shells. For moderately thick shel ls we have to take, at least
approximately, the transverse shear deformation effects i nto account. The Kirch-
hoff’s hypotheses are again relaxed in one point: the transv erse normals do not
remain perpendicular to the middle-surface after deformat ion, but a line element
through the shell thickness perpendicular to the middle-su rface prior loading, un-
dergoes at most a translation and rotation upon the load appl ications, no stretching
or curvature.
The following considerations are restricted to axial symme trical problems of
symmetrical laminated cross-ply circular cylindrical she lls including transverse
shear deformation. We start with a variational formulation including the trapeze
effect, i.e. Love’s ﬁrst approximation is not valid. For axi al 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
εxs=0,κs=0,κxs=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 di splacements
ux(x,z) =u(x)+zψx(x),
us(x,z) =0,
uz(x,z) =w(x)(9.3.2)
The strain-displacement relations are
εx=du
dx+zdψx
dx,
εs=w
R+z,
εxz=ψx+dw
dx(9.3.3)

9.3 Shear Deformation Theory 353
and the stresses in the kth layer of the shell are

σx
σs
σxz
(k)
=
Q11Q120
Q12Q220
0 0 Q55
(k)
du
dx+zdψx
dx
w
r+z
ψx+dw
dx
(9.3.4)
For a special orthotropic shell the Qi jare
Q11=Ex
1−νxsνsx,Q22=Es
1−νxsνsx,
Q12=νxsEx
1−νxsνsx,Q55=Gxz(9.3.5)
The stress resultant forces and couples are deﬁned as
Nx=n
∑
k=1/integraldisplay
h(k)σ(k)
x/parenleftig
1+z
R/parenrightig
dz, Ns=n
∑
k=1/integraldisplay
h(k)σ(k)
sdz,
Mx=n
∑
k=1/integraldisplay
h(k)σ(k)
xz/parenleftig
1+z
R/parenrightig
dz,Ms=n
∑
k=1/integraldisplay
h(k)σ(k)
szdz,
Qx=n
∑
k=1/integraldisplay
h(k)σ(k)
xz/parenleftig
1+z
R/parenrightig
dz,(9.3.6)
and one obtains with (9.3.4) and (9.3.6)
Nx=A11du
dx+A12w
R+1
RD11dψx
dx,
Ns=A12du
dx+T22w,
Mx=D11/parenleftbigg1
Rdu
dx+dψx
dx/parenrightbigg
,
Ms=D12dψx
dx+˜T22w,
Qx=A55/parenleftbigg
ψx+dw
dx/parenrightbigg
orQx=ks
55A55/parenleftbigg
ψx+dw
dx/parenrightbigg(9.3.7)
The stiffness coefﬁcients are

354 9 Modelling and Analysis of Circular Cylindrical Shells
(Ai j,Di j) =n
∑
k=1/integraldisplay
h(k)Q(k)
i j(1,z2)dz,(i j)=( 11),(12),
T22=n
∑
k=1/integraldisplay
h(k)Q(k)
22dz
R+z=n
∑
k=1Q(k)
22/bracketleftig
ln/parenleftig
1+z
R/parenrightig/bracketrightigh/2
−h/2
≈1
RA22+1
R3D22,
˜T22=n
∑
k=1/integraldisplay
h(k)Q(k)
22zdz
R+z=n
∑
k=1Q(k)
22R/bracketleftigz
R−ln/parenleftig
1+z
R/parenrightig/bracketrightigh/2
−h/2
≈1
R2D22(9.3.8)
andks
55is the shear correction factor.
The variational formulation for the axial symmetrically ci rcular cylindrical shell
with symmetrically laminated θ=00andθ=900laminae and coincided principal
material and structural axes is given as
Π(u,w,ψx) = Πi−Πa,
Πi(u,w,ψx) =1
2h/2/integraldisplay
−h/2π/integraldisplay
0L/integraldisplay
0(σxεx+σsεs+σxzεxz)dx(R+z)dϕdz
=1
2n
∑
k=1/integraldisplay
h(k)π/integraldisplay
0L/integraldisplay
0/parenleftig
Q(k)
11ε2
x
+2Q(k)
12εxεs+Q(k)
22ε2
s+Q(k)
55ε2
xz/parenrightig
(R+z)dxdϕdz(9.3.9)
Using Eq. (9.3.3) one obtains
Πi=1
22π/integraldisplay
0L/integraldisplay
0/braceleftigg
R/bracketleftigg
A11/parenleftbiggdu
dx/parenrightbigg2
+D11/parenleftbiggdψx
dx/parenrightbigg2
+2D11du
dxdψx
dx
+2A12wdu
dx+T22w2+Rks
55A55/bracketleftigg
ψ2
x+2ψxdw
dx+/parenleftbiggdw
dx/parenrightbigg2/bracketrightigg/bracerightigg
dxdϕ,
Πa=2π/integraldisplay
0L/integraldisplay
0pz(x)wRdxdϕ(9.3.10)
Equation (9.3.10) can be used for solving shell problems by t he variational meth-
ods of Ritz, Galerkin or Kantorovich. It can be also used to de rive the differential
equations and boundary conditions but this will be done late r on the direct way.
Hamilton’s principle is formulated to solve vibration prob lems. The potential
energy function Πis given with (9.3.10) but all displacements are now functio ns 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=1T(k)=1
2n
∑
k=1/integraldisplay
h(k)2π/integraldisplay
0L/integraldisplay
0ρ(k)/bracketleftigg/parenleftbigg∂u
∂t/parenrightbigg2
+2z∂u
∂t∂ψx
∂t
+z2/parenleftbigg∂ψx
∂t/parenrightbigg2
+/parenleftbigg∂w
∂t/parenrightbigg2/bracketrightigg
Rdϕdxdz
=1
22π/integraldisplay
0L/integraldisplay
0/braceleftigg
Rρ0/bracketleftigg/parenleftbigg∂u
∂t/parenrightbigg2
+/parenleftbigg∂w
∂t/parenrightbigg2/bracketrightigg
+2Rρ1∂u
∂t∂ψx
∂t+Rρ2/parenleftbigg∂ψx
∂t/parenrightbigg2/bracerightigg
dϕdx(9.3.11)
In Eq. (9.3.11) ρ(k)is the mass density of the kth layer, ρ0andρ2are the mass
and the moment of inertia with respect to the middle surface p er unit area and ρ1
represents the coupling between extensional and rotationa l motions. ρ1does not
appear in equations for homogeneous shells.
Now with the Lagrange function L(u,w,ψx) =T(u,w,ψx)−Π(u,w,ψx)the
Hamilton’s principle is obtained as
δt2/integraldisplay
t1L(u,w,ψx)dt=0 (9.3.12)
The direct derivation of the differential equations for sym metrical cross-ply circular
cylindrical shells follow using the constitutive, kinemat ics and equilibrium equa-
tions. The stiffness matrix is deﬁned as
/bracketleftbiggNNN
MMM/bracketrightbigg
=/bracketleftbiggAAA000
000DDD/bracketrightbigg/bracketleftbiggεεε
κκκ/bracketrightbigg
,QQQs=AAAsεεεs,
NNNT= [NxNsNxs],MMMT=[MxMsMxs],QQQsT=[QxQs],
εεεT=/bracketleftbigg∂u
∂x/parenleftbigg∂v
∂s+w
R/parenrightbigg /parenleftbigg∂u
∂s+∂v
∂x/parenrightbigg/bracketrightbigg
,
κκκT=/bracketleftbigg∂ψx
∂x∂ψs
∂s/parenleftbigg∂ψx
∂s+∂ψs
∂x/parenrightbigg/bracketrightbigg
,
εεεsT=/bracketleftbigg/parenleftbigg
ψx+∂w
∂x/parenrightbigg /parenleftbigg
ψs+∂w
∂s−v
R/parenrightbigg/bracketrightbigg(9.3.13)
For symmetric cross-ply shells all Bi jare zero and also the Ai j,Di jwith
(i j) = ( 16),(26)and(45). Such we have
AAA=
A11A120
A12A220
0 0 A66
,DDD=
D11D120
D12D220
0 0 A66
,AAAs=/bracketleftbiggks
55A55 0
0ks
44A44/bracketrightbigg

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+ρ0∂2u
∂t2+ρ1∂2ψx
∂t2,
∂Nxs
∂x+∂Ns
∂s+Qs
R=−ps+ρ0∂2v
∂t2+ρ1∂2ψs
∂t2,
∂Qx
∂x+∂Qs
∂s−Ns
R=−pz+ρ0∂2w
∂t2,
∂Mx
∂x+∂Mxs
∂s−Qx=ρ2∂2ψx
∂t2+ρ1∂2u
∂t2,
∂Mxs
∂x+∂Ms
∂s−Qs=ρ2∂2ψs
∂t2+ρ1∂2v
∂t2(9.3.14)
ρ0,ρ1,ρ2are like in (9.3.11) generalized mass density and are deﬁned in (8.3.9).
Putting the constitutive equations (9.3.12) in the equilib rium equations (9.3.14) the
equations can be manipulated in similar manner to those of th e classical theory and
one obtains the simultaneous system of differential equati ons

˜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
ψx
ψs
w
=−
px
ps
0
0
pz
+
ρ00ρ10 0
0ρ00ρ10
ρ10ρ20 0
0ρ10ρ20
0 0 0 0 ρ0
∂2
∂t2
u
v
ψx
ψs
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,ψs=0,
the spatial dependence can be written as products of two trig onometric functions
and the complete form of vibrations can be taken as
u(x,ϕ,t) =∞
∑
r=1∞
∑
s=1Urseiωrstcosαmxcosnϕ,
v(x,ϕ,t) =∞
∑
r=1∞
∑
s=1Vrseiωrstsinαmxsinnϕ,
w(x,ϕ,t) =∞
∑
r=1∞
∑
s=1Wrseiωrstsinαmxcosnϕ, (9.3.16)
ψx(x,ϕ,t) =∞
∑
r=1∞
∑
s=1Ψrseiωrstcosαmxcosnϕ,

9.3 Shear Deformation Theory 357
ψs(x,ϕ,t) =∞
∑
r=1∞
∑
s=1˜Ψrseiωrstsinαmxsinnϕ,
where Urs,Vrs,Wrs,Ψrs,˜Ψrsdenote amplitudes, αm=mπ/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 fo r a particular pair ( m,n)
gives the frequency and amplitude ratio corresponding to th ese wave numbers. For
arbitrary boundary conditions the Ritz’ or Galerkin’s meth od can be recommended
to obtain the characteristic equations for solving the eige nvalue problem. Then the
natural frequencies and the mode shapes can be calculated. T he solution process is
manageable, but involved.
If one restricts the problem to statics and to axially symmet rical 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
εx=du
dx,εs=w
R,κx=dψx
dx,εxz=ψx+dw
dx(9.3.18)
•Constitutive equations
Nx=A11εx+A12εs,
Ns=A12εx+A22εs,
Mx=D11dψx
dx,
Qx=ks
55A55εxz(9.3.19)
All derivatives ∂/∂s(…)andv,εxs,κs,εsz,Nxs,Mxs,Qsare zero. The stress resultant-
displacement relations follow as
Nx=A11du
dx+A12w
R,
Ns=A12du
dx+A22w
R,
Mx=D11dψx
dx,
Qx=ks
55A55/parenleftbigg
ψx+dw
dx/parenrightbigg(9.3.20)
With d Nx/dx=0 we have Nx=const=N0and one obtains
du
dx=1
A11/parenleftig
N0−A12w
R/parenrightig
(9.3.21)
The equilibrium equation (9.3.17) for Qxyields

358 9 Modelling and Analysis of Circular Cylindrical Shells
d
dx/parenleftbigg
ψx+dw
dx/parenrightbigg
=1
ks
55A55/parenleftbiggNs
R−pz/parenrightbigg
=1
ks
55A55/bracketleftbigg1
R/parenleftbigg
A12du
dx+A22w
R/parenrightbigg
−pz/bracketrightbigg (9.3.22)
and the equilibrium equation d Mx/dx−Qx=0
D11d2ψx
dx2−ks
55A55/parenleftbigg
ψx+dw
dx/parenrightbigg
=0 (9.3.23)
After some manipulations follow with (9.3.22), (9.3.23) tw o differential equations
forψxandwas
dψx
dx=−d2w
dx2+1
ks
55A55/bracketleftbigg1
R/parenleftbiggA12
A11N0−A2
12−A11A22
A11w
R/parenrightbigg
−pz/bracketrightbigg
,
dw
dx=D11
ks
55A55d2ψx
dx2−ψx(9.3.24)
Differentiating the second equation (9.3.24) and eliminat ingψxthe ﬁrst equation
leads to one uncoupled differential equation of fourth orde r for w(x)
d4w
dx4−1
ks
55A551
R2A11A22−A2
12
A11d2w
dx2+1
R2A11A22−A2
12
D11A11w
=1
D11/parenleftbigg
−A12
A11N0
R+pz/parenrightbigg
−1
ks
55A55d2pz
dx2,
dψx
dx=−d2w
dx2+1
ks
55A55A11A22−A2
12
R2A11w+1
ks
55A55/parenleftbiggA12
A11N0
R−pz/parenrightbigg(9.3.25)
With ks
55A55→∞Eqs. (9.3.25) simplify to the corresponding equations of th e clas-
sical shell theory (9.2.19).
The governing equation of the axisymmetric problem for or ci rcular cylindrical
shell in the frame of the shear deformation theory can be writ ten as
d4w
dx4−2k1d2w
dx2+k4
2w=kp (9.3.26)
with
k2
1=1
ks
55A551
RA11A22−A2
12
A11,k4
2=1
ks
55A551
R2A11A22−A2
12
D11A11,
kp=1
D11/parenleftbigg
−A12
A11N0
R+pz/parenrightbigg
−1
D11ks
55A55d2pz
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)=Ceαxyield the characteristic equation
α4−2k2
1α2+k4
2=0 (9.3.28)
with the roots
α1−4=±/radicalbigg
k2
1±/radicalig
k4
1−k4
1
which can be conjugate complex, real or two double roots depe nding on the relations
of the constants k1andk2.
The general solution can be written as (App. E)
w(x)=4
∑
i=1CiΦi(x)+wp(x)
The functions Φi(x)are given in different forms depending on the roots of the cha r-
acteristic equation (9.3.28). The roots and the functions Φi(x)are summarized in
App. E. The most often used solution form in engineering appl ication is given for
k2
2>k2
1.
For short shells with edges affecting one another, the Φi(x)involving the hy-
perbolic functions are convenient. If there are symmetry co nditions to the middle
cross-section x=L/2 the solution can be simpliﬁed, for we have Φ3=Φ4=0.
For long shells with ends not affecting one another applying theΦi(x)that involve
exponential functions.
Analogous to the classical shell solution for long shells a b ending-layer solution
can be applied. Only inside the bending-layer region with th e characteristic length
LBthe homogeneous 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 theo ry one can say
•If one restricts the consideration to symmetrical cross-pl y circular cylindrical
shells subjected to axially symmetric loadings the modelli ng and analysis is most
simpliﬁed and correspond to the classical shell theory.
•In more general cases including static loading and vibratio n and not neglecting
the trapeze effect the variational formulation is recommen ded and approximative
analytical or numerical solutions should be applied.
•Circular cylindrical shells are one of the most used thin-wa lled structures of con-
ventional or composite material. Such shells are used as res ervoirs, pressure ves-
sels, chemical containers, pipes, aircraft and ship elemen ts. This is the reason for
a long and intensive study to model and analyze circular cyli ndrical 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 w eight then constructions
with other materials. But there is not only weight saving int eresting, but in several
engineering applications the core material of a sandwich co nstruction can be also
used as thermal insulator or sound absorber. Therefore one c an ﬁnd numerous lit-
erature on modelling and analysis for sandwich shells subje cted static, dynamic or
environmental loads .
But as written in Sects. 7.4 and 8.4 sandwich constructions a re, simply consid-
ered, laminated constructions involving three laminae: th e lower face, the core, and
the upper face. And by doing so, one can employ all methods of m odelling and
analysis of laminated structural elements.
It was discussed in detail in Sect. 8.4 that, considering san dwich structural ele-
ments, we have to keep in mind the assumptions on the elastic b ehavior of sand-
wiches. Such there are differences in the expressions for th e ﬂexural bending and
transverse shear stiffness in comparison with laminated ci rcular cylindrical shells
and essential differences in the stress distribution over t he 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-pl ane moduli of elasticity
to the transverse shear moduli is high and transverse shear d eformations are mostly
included in its structural modelling. For this reason, the ﬁ rst order shear deformation
theory of laminated shells is used in priority for sandwich s hells. But for thin-walled
sandwich shells with a higher shear stiffness approximatel y the classical sandwich
theory can be used.
The correspondence between laminated and sandwich shells i s for vibration or
buckling problems limited and only using for overall buckli ng 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
11and the thicknesses hf1,hf3. The shell has an orthotropic
material behavior and the material principal axes shall coi ncide 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 di stributed load is pz=pz(x).

9.5 Problems 361
Formulate the differential equation using a perturbation c onstant 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 cl assical
shell theory one can use the differential equation (9.2.18)
d4w
dx4+2
RDR/bracketleftbiggA12B11
A11−B12/bracketrightbiggd2w
dx2+4λ4w=1
DR/parenleftbigg
pz−A12
A11Nx
R/parenrightbigg
The stiffness parameter are calculated for sandwiches with thin faces, Sect. 4.3.2,
h≈hc
A11=Qf1
11hf1+Qf3
11hf3=Qf1
11hf1/parenleftigg
1+Qf3
11hf3
Qf1
11hf1/parenrightigg
,
D11=/parenleftbiggh
2/parenrightbigg2
Qf1
11hf1+/parenleftbiggh
2/parenrightbigg2
Qf3
11hf3=/parenleftbiggh
2/parenrightbigg2
Qf1
11hf1/parenleftigg
1+Qf3
11hf3
Qf1
11hf1/parenrightigg
,
B11=−h
2Qf1
11hf1+h
2Qf3
11hf3=h
2Qf1
11hf1/parenleftigg
−1+Qf3
11hf3
Qf1
11hf1/parenrightigg
B12andD12can be calculated analogous. A asymmetry constant can be deﬁ ned 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 ηis for any sandwich
construction given as
−1<η<+1
The differential equation (9.2.18) can be written with the c onstant ηas
d4w
dx4+2
RDR/bracketleftbiggA12√D11√A11−√A11B12√D11
B11/bracketrightbigg
ηd2w
dx2+4λ4w=1
DR/parenleftbigg
pz−A12
A11Nx
R/parenrightbigg
Since|η|<1 one can ﬁnd w(x)in the form of a perturbation solution
w(x)=∞
∑
n=0wn(x)ηn
Forn=0 follow
d4w0
dx4−4λ4w0=−1
DR/parenleftbigg
pz−A12
A11Nx
R/parenrightbigg
and for n≥1

362 9 Modelling and Analysis of Circular Cylindrical Shells
d4wn
dx4+4λ4wn=−2
RDR/bracketleftbiggA12√D11√A11−√A11B12√D11
B11/bracketrightbigg
ηd2wn−1
dx2
The left hand side corresponds to the middle-surface symmet ric shell with axially
symmetric loading. The right hand side corresponds to the se cond derivation of the
previously obtained w-solution.
Conclusion 9.1. The perturbation solution yield the solution of the differe ntial equa-
tion as a successive set of solutions of axially symmetric pr oblems of which many
solutions are available. The perturbation solution conver ges to the exact solution. In
many engineering applications w(x)=wo(w)+ηw1(x)will be sufﬁcient accurate.
Exercise 9.2. A symmetrical cross-ply circular cylindrical shell is load ed 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 MLandQL
w(x)=M0
2λ2D11e−λx(sinλx−cosλx)−Q0
2λ3D11cosλx
The condition of no radius changing yields
w(x=0)=0⇒−M0
2λ2D11−Q0
2λ3D11=0⇒M0
Q0=−1
λ
Exercise 9.3. For a long ﬂuid container, Fig. 9.4 determinate the displace ment w(x)
and the stress resultants Ns(x)andMx(x). The container has a symmetrical cross-ply
layer stacking and can be analyzed in the frame of the classic al 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 ex ponent
Fig. 9.4 Long ﬂuid container,
L>LBLp(x)R R
x

9.5 Problems 363
wh(x)=e−λx(C1sinλx+C2cosλx)
The particular solution wp(x)follow with
p(x)=p0/parenleftig
1−x
L/parenrightig
and Eq. (9.2.24) as
wp(x)=p0
4λ4D11/parenleftig
1−x
L/parenrightig
The boundary constraints are
w(0)=0⇒C2=−p0
4λ4D11,dw(0)
dx=0⇒C1=−p0
4λ4D11/parenleftig
1−x
L/parenrightig
and we obtain the solutions
w(x)=p0
4λ4D11/braceleftig
1−x
L−/bracketleftig
cosλx+/parenleftig
1−x
L/parenrightig
sinλx/bracketrightig
e−λx/bracerightig
In addition,
Ns=A11εx+A22εs=A11du
dx+A22w
R
i.e. with εx=0
Ns=A22
Rw(x),Mx=D11κx=−D11d2w
dx2,Qx=dMx
dx=−D11d3w
dx3
Exercise 9.4. Consider a cantilever circular cylindrical shell, Fig. 9.5 . The normal
and shear forces NxandNxsas 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
xL
FVFH
MBMTϕs=Rϕ
R Nxs
Fig. 9.5 Tension, bending and torsion of a cantilever circular cylin drical shell

364 9 Modelling and Analysis of Circular Cylindrical Shells
Ns=Rpz,∂Nxs
∂x=−ps,∂Nx
∂x=−px−∂Nxs
∂s
andpz=ps=px=0 yield
Ns=0,Nxs=const,Nx=const
The distributions of FH,FV,MBandMTover the cross-section contour x=Lcan be
represented as
Nx(x=L) =1
2πR/parenleftbigg
FH+2MB
Rcosϕ/parenrightbigg
,
Nrs(x=L) =1
2πR/parenleftbiggMT
R+2FVsinϕ/parenrightbigg
and yield the reduced forces FV,FMand moments MB,MT
2π/integraldisplay
0Nx(x=L)Rdϕ=FH,
2π/integraldisplay
0Nxs(x=L)sinϕRdϕ=1
πRπ/integraldisplay
02FVsin2ϕRdϕ=FV,
4π/2/integraldisplay
0Nx(x=L)Rcosϕdϕ=4
πMBπ/2/integraldisplay
0cos2ϕdϕ=MB,
2π/2/integraldisplay
0RNxs(x=L)Rdϕ=1
Rππ/2/integraldisplay
0MTRdϕ=MT
The equilibrium equations yield
Nx(x) =FH
2πR−[MB+FV(L−x)]cosϕ
πR,
Ns(x) =0,
Nsx=MT
2πR12+FV
πRsinϕ

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

The fourth part (Chap. 10) includes the modelling and analys is of thin-walled folded
plate structures or generalized beams. This topic is not nor mally considered in stan-
dard textbooks on structural analysis of laminates and sand wiches, but it is included
here because it demonstrates the possible application of Vl asov’s theory of thin-
walled beams and semi-membrane shells on laminated structu ral elements.

Chapter 10
Modelling and Analysis of Thin-walled Folded
Structures
The analysis of real structures always is based on a structur al and mathematical
modelling. It is indispensable for obtaining realistic res ults that the structural model
represents sufﬁciently accurate the characteristic struc ture behavior.
Generally the structural modelling can be divided into thre e structure levels
•Three-dimensional modelling . It means structural elements, their dimensions in
all three directions are of the same order, we have no prefera ble direction.
•Two-dimensional modelling . One dimension of a structural element is signiﬁcant
smaller in comparison with the other both, so that we can rega rd it as a quasi
two-dimensional element. We have to distinguish plane and c urved 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 ele ment as quasi one-
dimensional. We call it rod, column,bar, beam or arch and can distinguish straight
and curved forms also.
The attachment of structural elements to one of these classe s is not well deﬁned
rather it must be seen in correlation with the given problem.
Many practical problems, e.g., in mechanical or civil engin eering 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 f ourth modelling class
was developed, the modelling class of thin-walled beams and so-called beam shaped
shells including also folded plate structures. In this four th modelling class it is typ-
ical that we have structures with a signiﬁcant larger dimens ion in one direction
(the length) in comparison with the dimensions in transvers e 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 fram e of the
Bernoulli’s and Timoshenko’s beam theory which cannot appl ied generally to thin-
walled beams. The modelling of two-dimensional laminate st ructures as plates and
shells was the subject of the Chaps. 8 and 9. In the present Cha p. 10 the investi-
367 © Springer Nature Singapore Pte Ltd. 2018
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 Structu res
gation of beams with thin-walled cross-sections and beam sh aped shells especially
folded structures is carried out. Chapter 10 starting in Sec t. 10.1 from a short recall
of the classical beam models. In Sect. 10.2 a generalized bea m model for prismatic
thin-walled folded plate structures is introduced, includ ing 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 distin guish their global and local
structural behavior. Global bending, vibration or bucklin g is the response of the
whole structure to external loading and is formulated in a gl obal coordinate system.
A typical example for global structure behavior is the deﬂec tion of a ship hull on
the waves. But the deﬂections and stresses in a special domai n of the ship e.g. in the
region of structure loading or deck openings or the vibratio ns or buckling of single
deck plates represent typical local effects.
A necessary condition for a global analysis is that the geome try 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-dimensio nal structure.
Of course there are interactions between global and local ef fects, 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 f or local considerations
by using special local co-ordinate systems. The reactions o f local to global effects
whereas are neglected.
From this point of view the global analysis of thin-walled be ams and beam
shaped shells can be done approximately by describing them a s 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 thr ee fundamental
hypotheses:
•There are no deformations of the cross-sectional contour.
•The cross-section is plane also in case of deformed structur es.
•The cross-section remain orthogonally to the deformed syst em axis.
As a result of bending without torsion we have normal stresse sσand strains εonly
in longitudinal direction. Shear deformations are neglect ed. The shear stresses τ
caused by the transverse stress resultants are calculated w ith 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 c ross-section it is supposed
that the bending stresses σband the shear stresses τqare 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 gi ven by Saint-
Venant1for considering the torsional stress. Under torsional stre sses the cross-
sections have out-of-plane warping, but assuming that thes e are the same in all cross-
sections and they are not constrained we have no resulting lo ngitudinal 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 stresse s is based on a closed shear
ﬂow in the cross-section. For closed thin-walled cross-sec tions 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 approximatel y. The ﬁrst two basic hy-
potheses of the Bernoulli’s model are remained. The plane cr oss-section stays plane
in this case but is not orthogonally to the system axes in the d eformed structure. For
the torsional stress also the relationships of Saint-Venan t are used.
Rather soon the disadvantages of this both classical beam mo dels were evident
for modelling and analysis of general thin-walled beam shap ed structures. Espe-
cially structures with open cross-section have the endeavo r for warping, and be-
cause the warping generally is not the same in all cross-sect ions, there are additional
normal stresses, so-called warping normal stresses and the y lead to warping shear
stresses too. Therefore the torsional moment must be divide d into two parts, the
Saint-Venant part and the second part caused by the warping s hear stresses.
Very fundamental and general works on this problem were done by Vlasov. Be-
cause his publications are given in Russian language they st ayed 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 Germ an 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 pa pers.
The Vlasov’s beam model for thin-walled beams with open cros s-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 mi d-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 bea ms with closed cross-
sections leads to nonsatisfying results, because the inﬂue nces of the cross-sectional
contour deformations and of the mid-plane shear strains in t he 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 long itudinal bending mo-
ments and the torsional moments in the plates of folded struc tures with closed cross-
sections are neglected. By this way we have in longitudinal d irection only membrane
stresses and in transversal direction a mixture of membrane and bending stresses.
This two-dimensional structural model can be reduced to a on e-dimensional one
by taking into account the Kantorovich relationships in for m 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 engin eer

370 10 Modelling and Analysis of Thin-walled Folded Structu res
functions. One of them describes a given deformation state o f the cross-section,
considered as a plane frame structure and the other is an unkn own 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). Startin g 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-mome nt shell model and all
classical and generalized linear beam models could be deriv ed by neglecting special
terms in the elastic energy potential function or by assumin g special conditions for
the contour deformation states. In Sect. 10.2 the derivatio n of generalized folded
structural model is given for anisotropic plates, e.g. off- axis loaded laminates. The
derivations are restricted to prismatic systems with strai ght system axes only.
Summarizing one can conclude from the above discussion ther e are several rea-
sons why for thin-walled structures must be given special co nsideration in design
and analysis. In thin-walled beams the shear stresses and st rains are relatively much
larger than those in beams with solid, e.g. rectangular, cro ss-sections. The assump-
tions of Bernoulli’s or Timoshenko’s beam theory can be viol ated e.g. by so-called
shear lag effects, which result in a non-constant distribut ion of normal bending
stresses which are different from that predicted by the Bern oulli hypotheses for
beams carrying only bending loads. When twisting also occur s warping effects, e.g.
warping normal and shear stresses, have to add to those arisi ng from bending loads.
The warping of the cross-section is deﬁned as its out-of-pla ne distorsion in the direc-
tion of the beam axis and violated the Bernoulli’s hypothese s and the Timoshenko’s
hypotheses too.
Because of their obvious advantages ﬁbre reinforced lamina ted 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 aer ospace, naval or automotive
industry. In addition to the known advantages of high streng th 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-s tacking sequence, can be
successfully exploited to enhance the response characteri stics of aerospace or naval
vehicles.
In order to be able to determine the behavior of these composi te beam structures,
consistent mechanical theories and analytical tools are re quired. So a Vlasov type
theory for ﬁber-reinforced beams with thin-walled open cro ss-sections made from
mid-plane symmetric ﬁber reinforced laminates was develop ed but in the last 30
years many improved or simpliﬁed theories were published.
Because primary or secondary structural conﬁgurations suc h as aircraft wings,
helicopter rotor blades, robot arms, bridges and other stru ctural elements in civil en-
gineering can be idealized as thin- or thick-walled beams, e specially as box beams,
beam models appropriate for both thin- and thick-walled geo metries which include
the coupled stiffness effects of general angle-ply laminat es, transverse shear defor-
mation of the cross-section and the beam walls, primary and s econdary warping,
etc. were developed. But nearly all governing equations of t hin- 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 defor m in their own
planes.
This assumption implies that the normal strain εsin the contour direction is small
compared to the normal strain εzparallel to the beam axis. This is particular valid
for thin-walled open cross-sections, for thin-walled clos ed cross-section with stiff-
eners (transverse sheets) and as the wall thickness of close d cross-sections increase.
Chapter 10 focuses the considerations to a more general mode l of composite thin-
walled beams which may be include the classical Vlasov assum ptions 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 str ucture shall be deﬁned as
a prismatic thin-walled structure which can be formed by fol ding a ﬂat rectangu-
lar plate or joining thin plate strips along lines parallel t o their length. Figure 10.1
demonstrates thin-walled structures of the type deﬁned abo ve. The plate strips can
be laminates.
10.2 Generalized Beam Models
Section 10.2 deﬁnes the outline of modelling beam shaped, th in-walled prismatic
folded plate structures with open, one or multi-cell closed or mixed open-closed
cross-sections. The considerations are limited to global s tructural response. Assum-
ing the classical laminate theory for all laminated plate st rips of the beam shaped
structure the elastic energy potential function is formula ted. 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-dimension al functional is re-
duced to an approximate one-dimensional one. A priori ﬁxed g eneralized coordi-

372 10 Modelling and Analysis of Thin-walled Folded Structu res
nate functions describing the cross-section kinematics ar e introduced. Generalized
displacement functions which depend on the system coordina texonly are the inde-
pendent functions of the reduced variational statement whi ch leads to a system of
matrix differential equations, the Euler equations of the v ariational 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 gener alized coordinate func-
tions, i.e. the cross-section kinematics. All results are d iscussed under the viewpoint
of a sufﬁcient general structural model for engineering app lications. A general struc-
tural model is recommended which includes all above noted fo rms of cross-sections
and enables to formulate efﬁcient numerical solution proce dures.
10.2.1 Basic Assumptions
A prismatic system is considered, its dimensions are signiﬁ cant 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 w ell 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 th e validity of the Kirchhoff
hypotheses is supposed or we use the classical laminate theo ry only. All constants
of each strip are constant in x-direction. For the displacements we can write
n1,w1s1,v1x,u1
n2,w2s2,v2x,u2
ni,wi si,vix,uix
yz
ti
dil
Fig. 10.2 Thin-walled folded structure geometry and co-ordinate sys tems

10.2 Generalized Beam Models 373
ui=ui(x,si),vi=vi(x,si),wi=wi(x,si) (10.2.1)
uiandviare 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 boundarie s 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 e ach strip we can use the
constitutive relationship given as
/bracketleftbiggNNN
MMM/bracketrightbigg
i=/bracketleftbiggAAA BBB
BBB DDD/bracketrightbigg
i/bracketleftbiggεεε
κκκ/bracketrightbigg
i(10.2.4)
or
Nxi
Nsi
Nxsi
Mxi
Msi
Mxsi
=
A11iA12iA16iB11iB12iB16i
A12iA22iA26iB12iB22iB26i
A16iA26iA66iB16iB26iB66i
B11iB12iB16iD11iD12iD16i
B12iB22iB26iD12iD22iD26i
B16iB26iB66iD16iD26iD66i

εxi
εsi
εxsi
κxi
κsi
κxsi

The following steps are necessary for calculating the eleme nts of the matrices
AAA,BBB,DDDfor the ith strip:
•Calculate the reduced stiffness matrix QQQ′for each lamina (k)of the strip (i)by
using the four elastic moduli EL,ET,νLT,GLT, Eqs. (4.1.2) and (4.1.3).
•Calculate the values of the transformed reduced stiffness m atrix QQQfor 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 σ1,σ2,σ6here are
σxi,σsi,σxsi. 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/integraldisplay
−ti/2σxidni,Nsi=ti/2/integraldisplay
−ti/2σsidni,Nxsi=ti/2/integraldisplay
−ti/2σxsidni,
(10.2.5)

374 10 Modelling and Analysis of Thin-walled Folded Structu res
Mxi=ti/2/integraldisplay
−ti/2σxinidni,Msi=ti/2/integraldisplay
−ti/2σsinidni,Mxsi=ti/2/integraldisplay
−ti/2σxsinidni
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 NsniandNxnifollow with the help
of the equilibrium conditions for a strip element.
In the same way here we have the following deﬁnitions for the e lements of the
deformation vector [ε1ε2ε6κ1κ2κ6]T≡[εxiεsiεxsiκxiκsiκxsi]Twith
εxi=∂ui
∂x=u′
i, εsi=∂vi
∂si=v•
i,
εxsi=∂ui
∂si+∂vi
∂x=u•
i+v′
i, κxi=−∂2wi
∂x2=−w′′
i, (10.2.6)
Fig. 10.3 Loads and resultant
forces and moments in the ith
stripnix
siNsiNxsiNxi
Nxsi
NsiNxsipxi
psi
qsi|x=0
qxi|x=0
x
si
niMxsiNsniMsiNxniMxsi
Mxi
Msi
Mxsi
Nsni
qni|x=0pni

10.2 Generalized Beam Models 375
κsi=−∂2wi
∂s2
i=−w••
i, κxsi=−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 obt ained by summarizing
the energy of all the nstrips
Π=1
2∑
(i)l/integraldisplay
0di/integraldisplay
0/bracketleftbig
NNNTMMMT/bracketrightbig
i/bracketleftbiggεεε
κκκ/bracketrightbigg
idsidx−Wa (10.2.7)
With equation (10.2.4) the vectors of the resultant forces a nd moments can be ex-
pressed and we obtain
Π=1
2∑
(i)l/integraldisplay
0di/integraldisplay
0/bracketleftbigg
εεε
κκκ/bracketrightbiggT
i/bracketleftbigg
AAA BBB
BBB DDD/bracketrightbigg
i/bracketleftbigg
εεε
κκκ/bracketrightbigg
idsidx−Wa (10.2.8)
The external work of the loads is also the sum of all the nstrips
Wa=∑
i/braceleftigg
1
2l/integraldisplay
0di/integraldisplay
02(pxiui+psivi+pniwi)dsidx
+di/integraldisplay
0/bracketleftbigg
(qxiui+qsivi+qniwi)/vextendsingle/vextendsingle/vextendsingle/vextendsingle
x=0+ (qxiui+qsivi+qniwi)/vextendsingle/vextendsingle/vextendsingle/vextendsingle
x=l/bracketrightbigg
dsi/bracerightigg (10.2.9)
After some steps considering the Eqs. (10.2.4) and (10.2.6) Eq. (10.2.9) leads to
Π=∑
(i)/braceleftigg
1
2l/integraldisplay
0di/integraldisplay
0/bracketleftbig
A11iu′2
i+2A12iu′
iv•
i+2A16iu′
i(u•
i+v′
i)
+A22iv•
i2+2A26iv•
i(u•
i+v′
i)+A66i/parenleftbig
u•
i+v′
i/parenrightbig2
−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••
i2+4D26iw••
iw′
i•+4D66iw′
i•2
−2(pxiui+psivi+pniwi)/bracketrightbig
dsidx

376 10 Modelling and Analysis of Thin-walled Folded Structu res
−di/integraldisplay
0[(qxiui+qsivi+qniwi)|x=0+(qxiui+qsivi+qniwi)|x=l]dsi/bracerightigg
10.2.3 Reduction of the Two-dimensional Problem
Equation (10.2.10) represents the complete folded structu re 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 exceptio n of some very simple
cases. Therefore here we will take another way. As the main ob ject of this section
we will ﬁnd approximate solutions by reducing the two-dimen sional problem to an
one-dimensional one taking into account the so-called Kant orovich 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)ϕi j(si) =UUUTϕϕϕ=ϕϕϕTUUU,
vi(x,si) =∑
(k)Vk(x)ψik(si) =VVVTψψψ=ψψψTVVV,
wi(x,si) =∑
(k)Vk(x)ξik(si) = VVVTξξξ=ξξξTVVV(10.2.11)
Here the ϕi j(si),ψik(si),ξik(si)are a priori given trial functions of the co-ordinates
siandUj(x),Vk(x)unknown coefﬁcient functions of the longitudinal co-ordin atex.
Vlasov deﬁned the ϕi j(si),ψik(si),ξik(si)as the generalized co-ordinate functions
and the Uj(x),Vk(x)as the generalized displacement functions. Of course it is v ery
important for the quality of the approximate solution, what kind and which number
of generalized co-ordinates ϕi j(si),ψik(si),ξik(si)are used.
Now we consider a closed thin-walled cross-section, e.g. th e 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 ϕi j.n∗is the number of
parallel strip edge lines of cross-section.
•The strains εsi(si)can be neglected, i.e. εsi≈0. The trial functions ψik(si)are
then constant functions in all strips and we have n∗∗linear independent ψik(si)
andξik(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 un it displacement states
in longitudinal direction (ϕ)and in transversal directions (ψ,ξ). Usually however
generalized co-ordinate functions are used, which allow me chanical interpretations.
In Fig. 10.4, e.g., the generalized co-ordinate functions f or a one-cellular rectangu-
lar cross-section are shown. ϕ1characterizes the longitudinal displacement of the

10.2 Generalized Beam Models 377
y
zxw1u
v1
u
w2v2 uv3
w3uw4v4
d1d4
d3
d2t1
t2t3t4
1
aaa
aa
b bbbbc
ϕ1 ϕ2 ϕ3 ϕ4
ψ1 ψ2ψ3 ψ4
1 1
1111 1 1
ξ1 ξ2ξ3 ξ4d1=d3=dS
d2=d4=dG
a=dS/2,b=dG/2
c=dSdG/4
Fig. 10.4 Generalized coordinate functions of an one-cellular recta ngular cross-section

378 10 Modelling and Analysis of Thin-walled Folded Structu res
whole cross-section, ϕ2andϕ3its rotations about the global y- and z-axes. ϕ1,ϕ2,
ϕ3represent the plane cross-section displacements, while ϕ4shows its warping.
ψ2andψ3characterize the plan cross-section displacements in z- and y-direction
andψ1the rotation of the rigid cross-section about the system axi sx.ψ4deﬁnes a
cross-sectional contour deformation, e.g. a distorsion. T he generalized co-ordinate
functions ξ1,ξ2,ξ3,ξ4represent displacements of the strips corresponding to ψ1,
ψ2,ψ3,ψ4. For the example of a box-girder cross-section there is n∗=4,m∗=4
andn∗∗=8−4=4.
In the following more general derivations the strains εsiwill be included, we will
take into account more complicated forms of warping functio ns and therefore there
are no restrictions for the number of generalized co-ordina te functions. After the
input of Eq. (10.2.11) into (10.2.10) and with the deﬁnition of the 28 matrices
ˆAAA1=∑
(i)di/integraldisplay
0A11iϕϕϕϕϕϕTdsi,ˆAAA2=∑
(i)di/integraldisplay
0A16iϕϕϕ•ϕϕϕTdsi,
ˆAAA3=∑
(i)di/integraldisplay
0A66iϕϕϕ•ϕϕϕ•Tdsi,ˆAAA4=∑
(i)di/integraldisplay
0A66iψψψψψψTdsi,
ˆAAA5=∑
(i)di/integraldisplay
0A26iψψψ•ψψψTdsi,ˆAAA6=∑
(i)di/integraldisplay
0A22iψψψ•ψψψ•Tdsi,
ˆAAA7=∑
(i)di/integraldisplay
0D11iξξξξξξTdsi,ˆAAA8=∑
(i)di/integraldisplay
0D16iξξξ•ξξξTdsi,
ˆAAA9=∑
(i)di/integraldisplay
0D66iξξξ•ξξξ•Tdsi,ˆAAA10=∑
(i)di/integraldisplay
0D12iξξξ••ξξξTdsi,
ˆAAA11=∑
(i)di/integraldisplay
0D26iξξξ••ξξξ•Tdsi,ˆAAA12=∑
(i)di/integraldisplay
0D22iξξξ••ξξξ••Tdsi,
ˆAAA13=∑
(i)di/integraldisplay
0A16iϕϕϕψψψTdsi,ˆAAA14=∑
(i)di/integraldisplay
0A12iϕϕϕψψψ•Tdsi, (10.2.12)
ˆAAA15=∑
(i)di/integraldisplay
0A66iϕϕϕ•ψψψTdsiˆAAA16=∑
(i)di/integraldisplay
0A26iϕϕϕ•ψψψ•Tdsi,
ˆAAA17=∑
(i)di/integraldisplay
0B11iϕϕϕξξξTdsi,ˆAAA18=∑
(i)di/integraldisplay
0B16iϕϕϕξξξ•Tdsi,

10.2 Generalized Beam Models 379
ˆAAA19=∑
(i)di/integraldisplay
0B12iϕϕϕξξξ••Tdsi,ˆAAA20=∑
(i)di/integraldisplay
0B16iϕϕϕ•ξξξTdsi,
ˆAAA21=∑
(i)di/integraldisplay
0B66iϕϕϕ•ξξξ•Tdsi,ˆAAA22=∑
(i)di/integraldisplay
0B26iϕϕϕ•ξξξ••Tdsi,
ˆAAA23=∑
(i)di/integraldisplay
0B16iψψψξξξTdsi,ˆAAA24=∑
(i)di/integraldisplay
0B66iψψψξξξ•Tdsi,
ˆAAA25=∑
(i)di/integraldisplay
0B26iψψψξξξ••Tdsi,ˆAAA26=∑
(i)di/integraldisplay
0B12iψψψ•ξξξTdsi,
ˆAAA27=∑
(i)di/integraldisplay
0B26iψψψ•ξξξ•Tdsi,ˆAAA28=∑
(i)di/integraldisplay
0B22iψψψ•ξξξ••Tdsi
and the load vectors
fffx=∑
(i)di/integraldisplay
0pxiϕϕϕdsi,rrrx=∑
(i)di/integraldisplay
0qxiϕϕϕdsi,
fffs=∑
(i)di/integraldisplay
0psiψψψdsi,rrrs=∑
(i)di/integraldisplay
0qsiψψψdsi,
fffn=∑
(i)di/integraldisplay
0pniξξξdsi,rrrn=∑
(i)di/integraldisplay
0qniξξξdsi(10.2.13)
the potential energy in matrix form is obtained as follows
Π=1
2l/integraldisplay
0/bracketleftbigg
UUU′TˆAAA1UUU′+VVVTˆAAA6VVV+UUUTˆAAA3UUU+2UUUTˆAAA15VVV′
+VVV′TˆAAA4VVV′+VVV′′TˆAAA7VVV′′+VVVTˆAAA12VVV+4VVV′TˆAAA9VVV′
+2UUU′TˆAAA14VVV+2UUUTˆAAA2UUU′+2UUU′TˆAAA13VVV′−2UUU′TˆAAA17VVV′′
−2UUU′TˆAAA19VVV−4UUU′TˆAAA18VVV′+2UUU′TˆAAA16VVV+2VVVTˆAAA5VVV′
−2VVVTˆAAA26VVV′′−2VVVTˆAAA28VVV−4VVVTˆAAA27VVV′−2UUUTˆAAA20VVV′′
−2VVV′TˆAAA23VVV′′−2UUUTˆAAA22VVV−2VVV′TˆAAA25VVV−4UUUTˆAAA21VVV′
−4VVV′TˆAAA24VVV′+2VVVTˆAAA10VVV′′+4VVV′TˆAAA8VVV′′+4VVVTˆAAA11VVV′
−2(UUUTfffx+VVVTfffs+VVVTfffn)/bracketrightbigg
dx
−(UUUTrrrx+VVVTrrrs+VVVTrrrn)|x=0−(UUUTrrrx+VVVTrrrs+VVVTrrrn)|x=l(10.2.14)

380 10 Modelling and Analysis of Thin-walled Folded Structu res
The variation of the potential energy function (10.2.14) an d using
∂Π
∂UUU−d
dx/parenleftbigg∂Π
∂UUU′/parenrightbigg
=000,∂Π
∂VVV−d
dx/parenleftbigg∂Π
∂VVV′/parenrightbigg
+d2
dx2/parenleftbigg∂Π
∂VVV′′/parenrightbigg
=000,
δUUUT/bracketleftbigg∂Π
∂UUU′/bracketrightbigg
x=0,l=0,δVVVT/bracketleftbigg∂Π
∂VVV′−d
dx/parenleftbigg∂Π
∂VVV′′/parenrightbigg/bracketrightbigg
x=0,l=0,
δVVV′T/bracketleftbigg∂Π
∂VVV′′/bracketrightbigg
x=0,l=0(10.2.15)
leads to a system of matrix differential equations and matri x boundary conditions of
the complete thin-walled folded plate structure
−ˆAAA1UUU′′+(ˆAAA2−ˆAAAT
2)UUU′+ˆAAA3UUU
+ˆAAA17VVV′′′−(ˆAAA13−2ˆAAA18+ˆAAA20)VVV′′
+(−ˆAAA14+ˆAAA15+ˆAAA19−2ˆAAA21)VVV′+(ˆAAA16−ˆAAA22)VVV =fffx
−ˆAAAT
17UUU′′′−(ˆAAAT
13−2ˆAAAT
18+ˆAAAT
20)UUU′′
+(ˆAAAT
14−ˆAAAT
15−ˆAAAT
19+2ˆAAAT
21)UUU′+(ˆAAAT
16−ˆAAAT
22)UUU
+ˆAAA7VVV′′′′+(2ˆAAAT
8−2ˆAAA8+ˆAAA23−AAAT
23)VVV′′′
−(ˆAAA4+4ˆAAA9−ˆAAA10−ˆAAAT
10−4ˆAAA24+ˆAAA26+ˆAAAT
26)VVV′′
+(ˆAAA5−ˆAAAT
5+2ˆAAA11−2ˆAAAT
11+ˆAAA25−ˆAAAT
25−2ˆAAA27+2ˆAAAT
27)VVV′
+(ˆAAA6+ˆAAA12−2ˆAAA28)VVV =fffs+fffn(10.2.16)
δUUUT[ˆAAA1UUU′+ˆAAAT
2UUU−ˆAAA17VVV′′+(ˆAAA13−2ˆAAA18)VVV′
+(ˆAAA14−ˆAAA19)VVV±rrrx]x=0,l =0
δVVVT[ˆAAAT
17UUU′′+(ˆAAAT
13−2ˆAAAT
18+ˆAAAT
20)UUU′+(ˆAAAT
15−2ˆAAAT
21)UUU
−ˆAAA7VVV′′′+(2ˆAAA8−2ˆAAAT
8−ˆAAA23+ˆAAAT
23)VVV′′
+(ˆAAA4+4ˆAAA9−ˆAAAT
10−4ˆAAA24+ˆAAAT
26)VVV′
+(ˆAAAT
5+2ˆAAAT
11−ˆAAA25−2ˆAAAT
27)VVV±rrrs±rrrn]x=0,l=0
δVVV′T[−ˆAAAT
17UUU′−ˆAAAT
20UUU
+ˆAAA7VVV′′+(2ˆAAAT
8−ˆAAAT
23)VVV′+(ˆAAAT
10−ˆAAAT
26)VVV]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 o f derivation simpliﬁed
models are usual:
•Neglecting of special terms in the potential energy functio n 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 κxi,
•the transversal strains εsi,
•the shear deformations of the mid-planes εxsiand
•the torsional curvatures κxsi
in the strips. But not all possibilities for simpliﬁed model s shall be taken into ac-
count. We will be restricted the considerations to:
A a structure model with neglected longitudinal curvatures κxionly,
B a structure model with neglected longitudinal curvatures κxiand neglected tor-
sional curvatures κxsi,
C a structure model with neglected longitudinal curvatures κxiand neglected
transversal strains εsi,
D a structure model with neglected longitudinal curvatures κxi, neglected transver-
sal strains εsiand torsional curvatures κxsi, and
E a structure model with neglected longitudinal curvatures κxi, neglected transver-
sal strains εsiand neglected shear strain εxsiof the mid-planes of the strips.
In Fig. 10.5 is given an overview on the development of struct ural simpliﬁed models.
10.2.4.1 Structural Model A
The starting point is the potential energy equation (10.2.1 0), in which all terms
containing w′′
ihave to vanish. Together with (10.2.11) and (10.2.12) we ﬁnd that in
this case
ˆAAA7=000,ˆAAA8=000,ˆAAA10=000,ˆAAA17=000,ˆAAA20=000,ˆAAA23=000,ˆAAA26=000
The matrix differential equations (10.2.16) and the bounda ry conditions (10.2.17)
change then into

382 10 Modelling and Analysis of Thin-walled Folded Structu res
κsi/ne}ationslash=0,εsi/ne}ationslash=0,εxsi/ne}ationslash=0,κxsi/ne}ationslash=0
κxi=0
εsi/ne}ationslash=0 εsi=0
εxsi/ne}ationslash=0 εxsi=0 εxsi/ne}ationslash=0 εxsi=0
κxsi/ne}ationslash=0 κxsi=0 κxsi/ne}ationslash=0 κxsi=0 κxsi/ne}ationslash=0
A B C D E
Fig. 10.5 Overview to the derivation of usual simpliﬁed models for thi n-walled folded plate struc-
tures
−ˆAAA1UUU′′+(ˆAAA2−ˆAAAT
2)UUU′+ˆAAA3UUU−(ˆAAA13−2ˆAAA18)VVV′′
+(−ˆAAA14+ˆAAA15+ˆAAA19−2ˆAAA21)VVV′+(ˆAAA16−ˆAAA22)VVV =fffx,
−(ˆAAAT
13−2ˆAAAT
18)UUU′′+(ˆAAAT
14−ˆAAAT
15−ˆAAAT
19+2ˆAAAT
21)UUU′
+(ˆAAAT
16−ˆAAAT
22)UUU−(ˆAAA4+4ˆAAA9−4ˆAAA24)VVV′′
+(ˆAAA5−ˆAAAT
5+2ˆAAA11−2ˆAAAT
11+ˆAAA25−ˆAAAT
25−2ˆAAA27+2ˆAAAT
27)VVV′
+(ˆAAA6+ˆAAA12−2ˆAAA28)VVV =fffs+fffn,(10.2.18)
δUUUT/bracketleftig
ˆAAA1UUU′+ˆAAAT
2UUU+(ˆAAA13−2ˆAAA18)VVV′
+(ˆAAA14−ˆAAA19)VVV±rrrx/bracketrightig
x=0,l=0,
δVVVT/bracketleftig
(ˆAAAT
13−2ˆAAAT
18)UUU′+(ˆAAAT
15−2ˆAAAT
21)UUU
+(ˆAAA4+4ˆAAA9−4ˆAAA24)VVV′
+(ˆAAAT
5+2ˆAAAT
11−ˆAAA25−2ˆAAAT
27)VVV±rrrs±rrrn/bracketrightig
x=0,l=0(10.2.19)

10.2 Generalized Beam Models 383
10.2.4.2 Structural Model B
Here the longitudinal curvatures κxiand the torsional curvatures κxsiare 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 th e ma-
trices ˆAAA9,ˆAAA11,ˆAAA18,ˆAAA21,ˆAAA24,ˆAAA27are null-matrices. This leads to the following matrix
differential equations and the corresponding boundary con ditions:
−ˆAAA1UUU′′+(ˆAAA2−ˆAAAT
2)UUU′+ˆAAA3UUU−ˆAAA13VVV′′
+(−ˆAAA14+ˆAAA15+ˆAAA19)VVV′+(ˆAAA16−ˆAAA22)VVV =fffx,
−ˆAAAT
13UUU′′+(ˆAAAT
14−ˆAAAT
15−ˆAAAT
19)UUU′+(ˆAAAT
16−ˆAAAT
22)UUU−ˆAAA4VVV′′
+(ˆAAA5−ˆAAAT
5+ˆAAA25−ˆAAAT
25)VVV′+(ˆAAA6+ˆAAA12−ˆAAA28)VVV =fffs+fffn,(10.2.20)
δUUUT/bracketleftig
ˆAAA1UUU′+ˆAAAT
2UUU+ˆAAA13VVV′+(ˆAAA14−ˆAAA19)VVV±rrrx/bracketrightig
x=0,l=0,
δVVVT/bracketleftig
ˆAAAT
13UUU′+ˆAAAT
15UUU+ˆAAA4VVV′+(ˆAAAT
5−ˆAAA25)VVV±rrrs±rrrn/bracketrightig
x=0,l=0(10.2.21)
10.2.4.3 Structural Model C
In this structure model the longitudinal curvatures κxiand the strains εsiare ne-
glected. Therefore in this case in the potential energy func tion (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 struct ure model A here the
matrices ˆAAA5,ˆAAA6,ˆAAA14,ˆAAA16,ˆAAA27,ˆAAA28are null-matrices and in this way we obtain the
following matrix differential equations and the correspon ding boundary conditions:
−ˆAAA1UUU′′+(ˆAAA2−ˆAAAT
2)UUU′+ˆAAA3UUU−(ˆAAA13−2ˆAAA18)VVV′′
+(ˆAAA15+ˆAAA19−2ˆAAA21)VVV′−ˆAAA22VVV =fffx,
−(ˆAAAT
13−2ˆAAAT
18)UUU′′+(−ˆAAAT
15−ˆAAAT
19+2ˆAAAT
21)UUU′−ˆAAAT
22UUU
−(ˆAAA4+4ˆAAA9−4ˆAAA24)VVV′′
+(2ˆAAA11−2ˆAAAT
11+ˆAAA25−ˆAAAT
25)VVV′+ˆAAA12VVV =fffs+fffn,(10.2.22)
δUUUT/bracketleftig
ˆAAA1UUU′+ˆAAAT
2UUU+(ˆAAA13−2ˆAAA18)VVV′−ˆAAA19VVV±rrrx/bracketrightig
x=0,l=0,
δVVVT/bracketleftig
(ˆAAAT
13−2ˆAAAT
18)UUU′+(ˆAAAT
15−2ˆAAAT
21)UUU
+(ˆAAA4+4ˆAAA9−4ˆAAA24)VVV′+(2ˆAAAT
11−ˆAAA25)VVV±rrrs±rrrn/bracketrightig
x=0,l=0(10.2.23)

384 10 Modelling and Analysis of Thin-walled Folded Structu res
10.2.4.4 Structural Model D
This structure model neglects the longitudinal curvatures κxi, the strains εsiand
the torsional curvatures κxsi. 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 (10.2.1 2) that
additionally to the structure model C the matrices ˆAAA9,ˆAAA11,ˆAAA18,ˆAAA21,ˆAAA24are null-
matrices. We obtain the matrix differential equations and t he boundary conditions
in the following form:
−ˆAAA1UUU′′+(ˆAAA2−ˆAAAT
2)UUU′+ˆAAA3UUU
−ˆAAA13VVV′′+(ˆAAA15+ˆAAA19)VVV′−ˆAAA22VVV=fffx,
−ˆAAAT
13UUU′′+(−ˆAAAT
15−ˆAAAT
19)UUU′−ˆAAAT
22UUU
−ˆAAA4VVV′′+(ˆAAA25−ˆAAAT
25)VVV′+ˆAAA12VVV=fffs+fffn,(10.2.24)
δUUUT/bracketleftig
ˆAAA1UUU′+ˆAAAT
2UUU+ˆAAA13VVV′−ˆAAA19VVV±rrrx/bracketrightig
x=0,l=0,
δVVVT/bracketleftig
ˆAAAT
13UUU′+ˆAAAT
15UUU+ˆAAA4VVV′−ˆAAA25VVV±rrrs±rrrn/bracketrightig
x=0,l=0(10.2.25)
10.2.4.5 Structural Model E
Now the longitudinal curvatures κxi, the transversal strains εsiand the shear strains
εxsiof the mid-planes shall be neglected. Therefore in the poten tial energy function
all the terms containing w′′
iandv•
ivanish again. The neglecting of the shear strains
of the mid-planes leads with Eq. (10.2.6) to
εxsi=∂ui
∂si+∂vi
∂x=u•
i+v′
i=0
and we can see that the generalized displacement functions UUU(x)andVVV(x)and also
the generalized co-ordinate functions ϕϕϕ(si)andψψψ(si)are no more independent from
each other
u•
i=−v′
i,UUUTϕϕϕ•=−VVV′Tψψψ,ϕϕϕ•=ψψψ,UUU=−VVV′,UUU′=−VVV′′,UUU′′=−VVV′′′(10.2.26)
Therefore the potential energy function must be reformulat ed before the variation
ofUUUandVVV. 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
2l/integraldisplay
0/bracketleftbigg
VVV′′TˆAAA1VVV′′+VVVTˆAAA12VVV+2VVV′′TˆAAA19VVV+4VVV′′TˆAAA18VVV′
+4VVV′TˆAAA9VVV′+4VVVTˆAAA11VVV′−2(UUUTfffx+VVVTfffs+VVVTfffn)/bracketrightbigg
dx
−(UUUTrrrx+VVVTrrrs+VVVTrrrn)|x=0−(UUUTrrrx+VVVTrrrs+VVVTrrrn)|x=l(10.2.27)
The variation of the potential energy, see also (10.2.15), l eads to the matrix differ-
ential equations and the boundary conditions for the struct ural model E
ˆAAA1VVV′′′′+2(ˆAAA18−ˆAAAT
18)VVV′′′+(−4ˆAAA9+ˆAAA19+ˆAAAT
19)VVV′′
+2(ˆAAA11−ˆAAAT
11)VVV′+ˆAAA12VVV =fff′
x+fffs+fffn,(10.2.28)
δVVVT/bracketleftig
−ˆAAA1VVV′′′+2(−ˆAAA18+ˆAAAT
18)VVV′′+(4ˆAAA9−ˆAAA19)VVV′
+2ˆAAAT
11VVV+fffx±rrrx±rrrn/bracketrightbig
x=0,l=0,
δVVV′T/bracketleftig
ˆAAA1VVV′′+2ˆAAA18VVV′+ˆAAA19VVV∓rrrx/bracketrightig
x=0,l=0(10.2.29)
10.2.4.6 Further Special Models by Restrictions of the Cros s-Section
Kinematics
All the ﬁve given simpliﬁed structure models include the neg lecting of the longitu-
dinal curvatures κxiin the strips. Because in the case of a beam shaped thin-walle d
structure the inﬂuence κxion the deformation state and the stresses of the whole
structure can be seen as very small, its neglecting is vindic ated 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 decreasi ng of the order of deriva-
tions of the generalized displacement functions UUUandVVVin the potential energy.
This is an important effect for practical solution strategi es of the model equations.
The structure models A and C can be used for the analysis of thi n-walled beam
shaped structures with open or closed cross-sections. The d ifference exists only in
the including or neglecting of the strains εsiin 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 vali d only for structures
with closed cross-sections, because there the torsional cu rvatures 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 th e displacements and
the stress state of the structure how in opposite to the case o f a closed cross-section.
Further for each of the ﬁve considered models we can develop m odel variants
restricting the cross-section kinematics by selection of s pecial sets of generalized
co-ordinate functions ϕϕϕ,ψψψ,ξξξ. 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 Structu res
ofψψψ,ξξξ-co-ordinates is limited to n∗∗, see Sect. 10.2.3. Restricting in this model ad-
ditionally the ϕϕϕ-co-ordinates to n∗, the semi-moment shell theory for an anisotropic
behavior of the strips is obtained. The Eqs. (10.2.24) and (1 0.2.25) stay unchanged.
A symmetric stacking sequence in all the strips leads for thi s model to a further
simpliﬁcation, because then we have no coupling between str etching and bending
in the strips, all the elements of the coupling matrix BBBvanish and therefore all the
matrices ˆAAA17−ˆAAA28are null-matrices. In this special case the following matri x dif-
ferential equations and boundary conditions are valid
−ˆAAA1UUU′′+(ˆAAA2−ˆAAAT
2)UUU′+ˆAAA3UUU−ˆAAA13VVV′′+ˆAAA15VVV′=fffx,
−ˆAAAT
13UUU′′−ˆAAAT
15UUU′−ˆAAA4VVV′′+ˆAAA12VVV =fffs+fffn,(10.2.30)
δUUUT/bracketleftig
ˆAAA1UUU′+ˆAAAT
2UUU+ˆAAA13VVV′±rrrx/bracketrightbig
x=0,l=0,
δVVVT/bracketleftbigˆAAAT
13UUU′+ˆAAAT
15UUU+ˆAAA4VVV′±rrrs±rrrn/bracketrightig
x=0,l=0(10.2.31)
If we have symmetric cross-ply laminates in all the plate str ips 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 t heBBB-matrix the elements
A16=A26=0,D16=D26=0. With these the matrices ˆAAA2,ˆAAA13, are null-matrices,
Eq. (10.2.12). In this case the Eqs. (10.2.30) and (10.2.31) lead to
−ˆAAA1UUU′′+ˆAAA3UUU+ˆAAA15VVV′=fffx,
−ˆAAA15UUU′−ˆAAA4VVV′′+ˆAAA12VVV=fffs+fffn,(10.2.32)
δUUUT/bracketleftig
ˆAAA1UUU′±rrrx/bracketrightig
x=0,l=0,
δVVVT/bracketleftig
ˆAAAT
15UUU+ˆAAA4VVV′±rrrs±rrrn/bracketrightig
x=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 matric esˆAAAconsider the
anisotropic behavior of the strips.
Otherwise if starting from the structure model E then the gen eralized co-ordinate
functions ψψψ,ξξξare restricted to three co-ordinates, representing the rig id cross-
section, i.e. ( ψ1,ξ1), the rotation of the cross-section about the x-axis, ( ψ2,ξ2) and
(ψ3,ξ3) the displacements in the global y- and z-direction. If we additionally assume
only four ϕ-functions, then the ﬁrst three representing the plane cros s-section, i.e.
ϕ1the displacement in x-direction, ϕ2,ϕ3the rotations about the y- and z-axes and
ϕ4is a linear warping function, the so-called unit warping fun ction according to
the sectorial areas law. Therefore we have a structural mode l similar the classical
Vlasov beam model. The difference is only that in the classic al Vlasov beam model
isotropic material behavior is assumed and here the anisotr opic 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)=ϕy=w′(x)rotation of the plane cross-section about the y-axis,
U3(x)=ϕz=v′(x)rotation of the plane cross-section about the z-axis,
U4(x)=ωθ′(x) the warping with ωas 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 warp ing of the
cross-section and use only the functions ϕ1,ϕ2,ϕ3representing 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-wal led cross-sections
and anisotropic material behavior. In a similar way a specia lized Timoshenko beam
model can be obtained, if we restrict the generalized co-ord inate functions ϕito 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 f or structures with
cross-sections consisting of single thin plate strips with out any rule of their arrange-
ment in the cross-section. The curvatures κxiare generally neglected. The special-
ized beam equations described above cannot be compared dire ctly with the beam
equations in Chap. 7 because the derivation there is not rest ricted to thin-walled
folded plate cross-sections.
10.2.5 An Efﬁcient Structure Model for the Analysis of Gener al
Prismatic Beam Shaped Thin-walled Plate Structures
Because in the following only beam shaped thin-walled struc tures are analyzed the
neglecting of the inﬂuence of the longitudinal curvatures κxiin the single plate
strips is vindicated. How it was mentioned above, we have in t his 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 procedur es.
The selected structure model shall enable the analysis of th in-walled structures
with open, closed and mixed open/closed cross-sections. Be cause the inﬂuence of
κxsiis small only for closed cross-sections but the inﬂuence of εxsican be neglected
for open cross-sections only, the selected structure model for general cross-sections
have to include the torsional curvatures and the shear strai ns of the mid-planes.
The strains εsihave in the most cases only a small inﬂuence and could be ne-
glected generally. But we shall see in Chap. 11 that includin gεsiin the model equa-
tions leads an effective way to deﬁne the shape functions for special ﬁnite elements
and therefore also the εsiare included in the selected structure model.

388 10 Modelling and Analysis of Thin-walled Folded Structu res
Summarizing the above discussion the structure model A is se lected as an uni-
versal model for the modelling and analysis of beam shaped th in-walled plate struc-
tures. An extension of the equations to eigen-vibration pro blems 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 anal ysis also shall be restricted
to global vibration response. Local vibrations, e.g. vibra tion of single plates, are
excluded. A structure model neglecting the longitudinal cu rvatures cannot describe
local plate strip vibrations. Further only free undamped vi brations are considered.
The starting point is the potential energy function, Eq. (10 .2.10), but all terms
including κxiare neglected. With the potential energy Π(ui,vi,wi)
Π=∑
(i)1
2l/integraldisplay
0di/integraldisplay
0/bracketleftbigg
A11iu′2
i+2A12iu′
iv•
i+2A16iu′
i(u•
i+v′
i)
+A22iv•
i2+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••
i2+4D26iw••
iw′
i•+4D66iw′
i•2/bracketrightbigg
dsidx(10.2.34)
and the kinetic energy T(ui,vi,wi)
T(uuu)=1
2∑
(i)l/integraldisplay
0di/integraldisplay
0ρiti/bracketleftigg/parenleftbigg∂ui
∂t/parenrightbigg2
+/parenleftbigg∂vi
∂t/parenrightbigg2
+/parenleftbigg∂wi
∂t/parenrightbigg2/bracketrightigg
dsidx, (10.2.35)
where ρiis the average density of the ith plate strip
ρi=1
tin
∑
k=1ρ(k)
it(k)
i (10.2.36)
Because we have thin plate strips only, rotational terms of t he kinetic energy can be
neglected.
The reduction of the two-dimensional problem is carried out again with the gen-
eralized co-ordinate functions ϕϕϕ(si),ψψψ(si),ξξξ(si), but we must remark that the gener-
alized displacement functions UUU,VVVare time-dependent and therefore they are writ-
ten in the following with a tilde. The reduction relationshi ps are

10.2 Generalized Beam Models 389
ui(x,si,t) =∑
(j)˜Uj(x,t)ϕi j(si) = ˜UUUTϕϕϕ=ϕϕϕT˜UUU,
vi(x,si,t) =∑
(k)˜Vk(x,t)ψik(si) = ˜VVVTψψψ=ψψψT˜VVV,
wi(x,si,t) =∑
(k)˜Vk(x,t)ξik(si) = ˜VVVTξξξ=ξξξT˜VVV(10.2.37)
Additionally to the ˆAAAmatrices equation (10.2.12) the following matrices are deﬁ ned:
ˆBBB1=∑
(i)di/integraldisplay
0ρitiϕϕϕϕϕϕTdsi,ˆBBB2=∑
(i)di/integraldisplay
0ρitiψψψψψψTdsi,ˆBBB3=∑
(i)di/integraldisplay
0ρitiξξξξξξTdsi(10.2.38)
and we obtain the so-called Lagrange function L=T−Π
L=∑
(i)1
2l/integraldisplay
0/bracketleftbigg
˙˜UUUTˆBBB1˙˜UUU+˙˜VVVTˆBBB2˙˜VVV+˙˜VVVTˆBBB3˙˜VVV
−(˜UUU′TˆAAA1˜UUU′+2˜UUU′TˆAAA14˜VVV+2˜UUUTˆAAA2˜UUU′+2˜UUU′TˆAAA13˜VVV′
+˜VVVTˆAAA6˜VVV+2˜UUUTˆAAA16˜VVV+2˜VVVTˆAAA5˜VVV′+˜UUUTˆAAA3˜UUU
+2˜UUUTˆAAA15˜VVV′+˜VVV′TˆAAA4˜VVV′−2˜UUU′TˆAAA19˜VVV−4˜UUU′TˆAAA18˜VVV′
−2˜VVVTˆAAA28˜VVV−4˜VVVTˆAAA27˜VVV′−2˜UUUTˆAAA22˜VVV−2˜VVV′TˆAAA25˜VVV
−4˜UUUTˆAAA21˜VVV′−4˜VVV′TˆAAA24˜VVV′
+˜VVVTˆAAA12˜VVV+4˜VVVTˆAAA11˜VVV′+4˜VVV′TˆAAA9˜VVV′)/bracketrightbigg
dx(10.2.39)
The time derivations of the generalized displacement funct ions are written with the
point symbol
˙˜UUU=∂˜UUU
∂t,˙˜VVV=∂˜VVV
∂t(10.2.40)
The Hamilton principle yields the variational statement
δt2/integraldisplay
t1Ldt=t2/integraldisplay
t1δLdt=0,L=L(x,t,˜UUU,˜VVV,˜UUU′,˜VVV′,˙˜UUU,˙˜VVV) (10.2.41)
and we obtain two differential equations
∂L
∂˜UUU−d
dx/parenleftbigg∂L
∂˜UUU′/parenrightbigg
−d
dt/parenleftbigg∂L
∂˙˜UUU/parenrightbigg
=000,
∂L
∂˜VVV−d
dx/parenleftbigg∂L
∂˜VVV′/parenrightbigg
−d
dt/parenleftbigg∂L
∂˙˜VVV/parenrightbigg
=000(10.2.42)

390 10 Modelling and Analysis of Thin-walled Folded Structu res
If further harmonic relationships for the generalized disp lacement functions are as-
sumed
˜UUU(x,t)=UUU(x)sinω0t,˜VVV(x,t)=VVV(x)sinω0t (10.2.43)
and we obtain after some steps the following matrix differen tial equations:
−ˆAAA1UUU′′+(ˆAAA2−ˆAAAT
2)UUU′+(ˆAAA3−ω2
0ˆBBB1)UUU−(ˆAAA13−2ˆAAA18)VVV′′
+(−ˆAAA14+ˆAAA15+ˆAAA19−2ˆAAA21)VVV′+(ˆAAA16−ˆAAA22)VVV =000,
−(ˆAAAT
13−2ˆAAAT
18)UUU′′+(ˆAAAT
14−ˆAAAT
15−ˆAAAT
19+2ˆAAAT
21)UUU′
+(ˆAAAT
16−ˆAAAT
22)UUU−(ˆAAA4+4ˆAAA9−4ˆAAA24)VVV′′
+(ˆAAA5−ˆAAAT
5+2ˆAAA11−2ˆAAAT
11+ˆAAA25−ˆAAAT
25−2ˆAAA27+2ˆAAAT
27)VVV′
+(ˆAAA6+ˆAAA12−2ˆAAA28−ω2
0ˆBBB2−ω2
0ˆBBB3)VVV =000,(10.2.44)
δUUUT[ˆAAA1UUU′+ˆAAAT
2UUU+(ˆAAA13−2ˆAAA18)VVV′+(ˆAAA14−ˆAAA19)VVV]x=0,l=0
δVVVT[(ˆAAAT
13−2ˆAAAT
18)UUU′+(ˆAAAT
15−2ˆAAAT
21)UUU+(ˆAAA4+4ˆAAA9−4ˆAAA24)VVV′
+(ˆAAAT
5+2ˆAAAT
11−ˆAAA25−2ˆAAAT
27)VVV]x=0,l =0(10.2.45)
With these equations given above the global free vibration a nalysis of prismatic
beam shaped thin-walled plate structures can be done sufﬁci ent exactly.
10.3 Solution Procedures
Two general kinds of solution procedures may be taken into ac count
•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 differenti al equations is carried
out. In the other case, the variational statement of the prob lem is, e.g., solved by
the Ritz or Galerkin method, and in general, the procedures y ield in an approximate
analytical series solution.
Numerical solution procedures essentially consist of meth ods outgoing from the
differential equation or from the corresponding variation al problem. The numeri-
cal solutions of differential equations may include such me thods as ﬁnite difference
methods, Runge3-Kutta4methods and transfer matrix methods. The main represen-
tative for the second way is the ﬁnite element method (FEM). A fter a few remarks
3Carl David Tolm´ e Runge (∗30 August 1856 Bremen – †3 January 1927 G¨ ottingen) – mathema ti-
cian and physicist
4Martin Wilhelm Kutta (∗3 November 1867 Pitschen – †25 December 1944 F¨ urstenfeldbr uck) –
mathematician

10.3 Solution Procedures 391
in Sect. 10.3.1 about analytic solution possibilities for t he here considered prob-
lems, the numerical solution procedure using the transfer m atrix 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 thi n-walled structures are
discussed in Chap. 11.
10.3.1 Analytical Solutions
For the generalized beam models given in Sect. 10.2 only for s impliﬁed special
cases analytical solutions are possible. If we use, for exam ple, the structure model
D in connection with a symmetric cross-ply stacking in all pl ates, what means
that the differential equations are from the same type as in c ase of the isotropic
semi-moment shell theory of Vlasov, analytical solutions c an be developed for spe-
cial cross-sections geometry. It is very useful to choose or thogonal generalized co-
ordinate functions ϕϕϕ,ψψψ,ξξξ, because it yields the possibility of decomposition of the
system of differential equations into some uncoupled parti al systems. For example,
the generalized co-ordinate functions ϕϕϕin Fig. 10.4 are completely orthogonal and
in this way the matrix ˆAAA1is a diagonal matrix. Therefore some couplings between
the single differential equations vanish.
A suitable method for construction an exact solution is the K rylow5method or
the so-called method of initial parameters. The ﬁrst step fo r the application of this
method is to convert the system of differential equations in to an equivalent differen-
tial equation of n-th order.
L[y(x)]=n
∑
ν=0aνy(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 initi al 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/integraldisplay
0Kn(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 Structu res
yp(x)=/vextenddo ble/vextenddo ble/vextenddo ble/vextenddo ble
x>x0x/integraldisplay
x0Kn(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)
+/vextenddo ble/vextenddo ble/vextenddo ble/vextenddo ble
x>x0x/integraldisplay
x0Kn(x−t)r0(t)dt+/vextenddo ble/vextenddo ble/vextenddo ble/vextenddo ble
x>x1x/integraldisplay
x1Kn(x−t)r1(t)dt(10.3.5)
+/vextenddo ble/vextenddo ble/vextenddo ble/vextenddo ble
x>x2x/integraldisplay
x2Kn(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 genera l 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 a re the starting point of a
numerical solution by transfer matrix method. At ﬁrst the sy stem of higher order dif-
ferential equations has to transfer into a system of differe ntial equations of ﬁrst order
using the natural boundary conditions as deﬁnitions of gene ralized cross-sectional
forces.
For sake of simplicity this solution method shall be demonst rated for the structure
model D and for a symmetric cross-ply stacking in all plates o f the structure. Then
the following system of differential equations and boundar y conditions are valid,
see also Sect. 10.2.4.6 and Eqs. (10.2.32) and (10.2.33)
ˆAAA1UUU′′−ˆAAA3UUU−ˆAAA15VVV′+fffx=000,
ˆAAAT
15UUU′+ˆAAA4VVV′′−ˆAAA12VVV+fffs+fffn=000,(10.3.6)
δUUUT/bracketleftig
ˆAAA1UUU′±rrrx/bracketrightig
x=0,l=0,
δVVVT/bracketleftig
ˆAAAT
15UUU+ˆAAA4VVV′±rrrs±rrrn/bracketrightig
x=0,l=0(10.3.7)
Equations (10.3.7) leads to the deﬁnitions of the generaliz ed cross-sectional forces,
i.e. generalized longitudinal forces and transverse force s.

10.3 Solution Procedures 393
PPP=ˆAAA1UUU′, (10.3.8)
QQQ=ˆAAAT
15UUU+ˆAAA4VVV′, (10.3.9)
It can be shown that we have with Eqs. (10.3.8), (10.3.9) real ly the deﬁnitions of
generalized forces
PPP=∑
(i)di/integraldisplay
0ϕϕϕσxitidsi=∑
(i)di/integraldisplay
0ϕϕϕNxidsi,
QQQ=∑
(i)di/integraldisplay
0ψψψτxsitidsi=∑
(i)di/integraldisplay
0ψψψNxsidsi
In the here considered structure model we have only membrane stresses σxiandτxsi
because the longitudinal curvatures and longitudinal bend ing 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=A11iεxi=A11iu′
i=A11iϕϕϕTUUU′,
Nxsi=A66iεxsi=A66i(u•
i+v′
i) =A66i(ϕϕϕ•TUUU+ψψψTVVV′)
Considering (10.2.12) we obtain again the deﬁnitions of the generalized forces given
in (10.3.8) and (10.3.9)
PPP=∑
(i)di/integraldisplay
0A11iϕϕϕϕϕϕTdsiUUU′=ˆAAA1UUU′,
QQQ=∑
(i)di/integraldisplay
0A66i(ψψψϕϕϕ•TUUU+ψψψψψψTVVV′)dsi=ˆAAAT
15UUU+ˆAAA4VVV′
The inversion of the Eqs. (10.3.8) and (10.3.9) leads to
UUU′=ˆAAA−1
1PPP, (10.3.10)
VVV′=−ˆAAA−1
4ˆAAAT
15UUU+ˆAAA−1
4QQQ (10.3.11)
With the ﬁrst derivatives of Eqs. (10.3.8) and (10.3.9) and a fter the input of (10.3.10)
and (10.3.11) into (10.3.6) we obtain the following system o f differential equations
of ﬁrst order
UUU′=ˆAAA−1
1PPP,
VVV′=−ˆAAA−1
4ˆAAAT
15UUU+ˆAAA−1
4QQQ,
PPP′= (ˆAAA3−ˆAAA15ˆAAA−1
4ˆAAAT
15)UUU+ˆAAA15ˆAAA−1
4QQQ−fffx,
QQQ′=ˆAAA12VVV−fffs−fffn(10.3.12)
respectively written in matrix notation

394 10 Modelling and Analysis of Thin-walled Folded Structu res

UUU
VVV
PPP
QQQ
1
′
=
000 0 00ˆAAA−1
1 000 ooo
−ˆAAA−1
4ˆAAAT
15 000 000 ˆAAA−1
4 ooo
(ˆAAA3−ˆAAA15ˆAAA−1
4ˆAAAT
15)000 000ˆAAA15ˆAAA−1
4−fffx
000 ˆAAA12000 0 00−fffs−fffn
oooToooToooToooT0

UUU
VVV
PPP
QQQ
1

yyy′=BBByyy (10.3.13)
BBBis the system matrix and yyythe so-called state vector containing all the generalized
displacement functions UUUandVVVand all the generalized forces PPPandQQQ. 000 and oooin
theBBB-matrix are null matrices and vectors.
The next step is a discretization of the one-dimensional pro blem, see Fig. 10.6.
Between the state vectors at the point j+1 and the point jwe have generally the
relationship
yyyj+1=WWWjyyyj (10.3.14)
where WWWjis 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 equat ions is
yyy′(x)=BBByyy(x),
yyy(x)=yyy(x0)eB(x−x0)
∆j✲✛ x=x0 x=xN
0 N j−1 j j+1
Fig. 10.6 Discretization of the one-dimensional structure

10.3 Solution Procedures 395
and we ﬁnd that eBBB(x−x0)can be deﬁned as the transfer matrix from the point x0tox.
Therefore the transfer matrix between two points xjandxj+1generally is obtained
as
WWWj=eBBB(xj+1−xj)(10.3.15)
The numerical calculation of transfer matrices can be carri ed out by series develop-
ment of the exponential function
WWWj=III+△jBBB+△2
j
2!BBB2+△3
j
3!BBB3+… (10.3.16)
and also by using a Runge-Kutta method
WWWj=III+△j
6(MMM1j+2MMM2j+2MMM3j+MMM4j), (10.3.17)
MMM1j=BBB(xj),
MMM2j=BBB(xj+1
2△j)(III+1
2△jMMM1j),
MMM3j=BBB(xj+1
2△j)(III+1
2△jMMM2j),
MMM4j=BBB(xj+△j)(III+△jMMM3j)
In both equations IIIare unit matrices of the same rank as the system matrix. The
boundary conditions of the problem can be expressed by a matr ix equation
yyy0=AAAxxx∗(10.3.18)
Here AAAis the so-called start matrix containing the boundary condi tions at x=x0
andxxx∗is the vector of the unknown boundary values there. In the las t 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 c orresponding row
the unknown value is associated with an element of the unknow n vector xxx∗. For
example, in Eq. (10.3.19) a start matrix is shown in case of a f ree structure end, it
means all the displacements are unknown and all forces are gi ven.

396 10 Modelling and Analysis of Thin-walled Folded Structu res

U1
…
Um
V1
…
Vn
P1
…
Pm
Q1
…
Qn
1

0=
1 0
… 0
1 0
1 0
… 0
1 0
P10
…
Pm0
Q10
…
Qn0
1

x∗
1
…
…
…
…
x∗
m+n
1
,
yyy0= AAA0 xxx∗(10.3.19)
Now the multiplications with the transfer matrices can be ca rried out over all sec-
tions(xj,xj+1)until x=xN. With the equation
SSSyyyN=000 (10.3.20)
the boundary conditions are formulated at x=xN.SSSis the so-called end matrix con-
taining in its last column the negative values of the given di splacements or forces. A
unit in an other column of each row yields the association to a n element of the state
vector yyyN. Equation (10.3.21) shows the end matrix for a clamped end, w here all
displacements are given. This matrix equation leads to a sys tem of linear equations
for the unknowns in the vector xxx∗

1 …−U1N
… … …
1 …−UmN
1…−V1N
… … …
1…−VnN

U1
…
Um
V1
…
Vn
P1
…
Pm
Q1
…
Qn
1

N=
0
…
0
0
…
0
(10.3.21)
The real parts of the eigenvalues of the system matrix BBBlead to numerical instable
solutions especially for long beam structures. From the mec hanical point of view it
means that the inﬂuence of the boundary conditions of both st ructure ends to each
other are very low and with this we have a nearly singular syst em of linear equa-

10.3 Solution Procedures 397
tions. For the consolidation of this problem intermediate c hanges of the unknowns
are carried out, by formulation of a new start matrix AAAat such an intermediate point.
Usually the generalized displacements are chosen as the new unknowns. The fol-
lowing equations show the general procedure schedule
yyy0=AAA0xxx∗
0
yyy1=WWW0yyy0=WWW0AAA0xxx∗
0
…
yyyi=WWWi−1WWWi−2…WWW0AAA0xxx∗
0=FFFixxx∗
0
yyyi=AAA1xxx∗
1 ﬁrst change of unknowns
yyyi+1=WWWiyyyi=WWWiAAA1xxx∗
1
…
yyyj=FFFjxxx∗
l−1
yyyj=AAAlxxx∗
l lth change of unknowns
yyyj+1=WWWjAAAlxxx∗
l
…
yyyk=FFFkxxx∗
n−1
yyyk=AAAnxxx∗
n nth change of unknowns
yyyk+1=WWWkAAAnxxx∗
n
… (10.3.22)
yyyN=FFFNxxx∗
n
SSSyyyN=000system of linear equations for the solution of the unknowns x∗
n
The multiplications of the state vector yyy0with 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 FFFi. A new unknown vector xxx∗
1is deﬁned by the new
start matrix AAA1and this procedure is repeated at the following intermediat e points.
General for the l-th intermediate change of unknowns the Eq. (10.3.23) is cur rent
FFFjxxx∗
l−1=AAAlxxx∗
l (10.3.23)
With a segmentation of the state vector yyyjinto the sub-vectors yyyvfor the displace-
ments and yyykfor the forces

398 10 Modelling and Analysis of Thin-walled Folded Structu res
yyyj=
UUU
VVV
PPP
QQQ
1

j=
yyyv
yyyk
1

j, (10.3.24)
we obtain a separated form of Eq. (10.3.23)

FFF1jfff1j
FFF2jfff2j
oooT1
/bracketleftbigg˜xxx∗
l−1
1/bracketrightbigg
=
AAA1laaa1l
AAA2laaa2l
oooT1
/bracketleftbigg˜xxx∗
l
1/bracketrightbigg
(10.3.25)
With the assumption that the displacements are the new unkno wns we ﬁnd that the
sub-matrix AAA1lis a unit matrix and the sub-vector aaa1lis a null vector
AAA1l=III,aaa1l=ooo (10.3.26)
This leads to
FFF1j˜xxx∗
l−1+fff1j=˜xxx∗
l, (10.3.27)
˜xxx∗
l−1=FFF−1
1j(˜xxx∗
l−fff1j),xxx∗
l−1=/bracketleftbiggFFF−1
1j(˜xxx∗
l−fff1j)
1/bracketrightbigg
(10.3.28)
and than the second equation of (10.3.25) yields the structu re of the new start matrix
FFF2jFFF−1
1j(˜xxx∗
l−fff1j)+fff2j=AAA2l˜xxx∗
l+aaa2l,
AAA2l=FFF2jFFF−1
1j,aaa2l=fff2j−FFF2jFFF−1
1jfff1j,(10.3.29)
AAAl=
III o oo
FFF2jFFF−1
1jfff2j−FFF2jFFF−1
1jfff1j
oooT1
 (10.3.30)
At such an intermediate change point it is also possible to co nsider the introduction
of concentrated generalized forces or the disposition of su pports with given general-
ized displacements. Than the new start matrix must be modiﬁe d additionally, in the
ﬁrst case by a modiﬁcation of the sub-vector aaa2land in the second case by consid-
eration of the jump behavior of the forces at this point. But m ore details about this
shall not be given here.
With the end matrix SSSand the end state vector yyyNthe relationship SSSyyyNyields
a system of linear equations for the last unknown vector xxx∗
nand after this all the
unknown vectors can be calculated by repeatedly using Eq. (1 0.3.28).
The transfer matrix method with intermediate changes of the unknown state vec-
tors yields in contrast to the classical transfer method a nu merical stable procedure
also for long beam structures. From the mechanical point of v iew 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 anal ysis of eigen-
vibrations. There we have a modiﬁed system matrix BBBcontaining frequency de-
pendent terms
yyy′=BBB(ω0)yyy (10.3.31)
Therefore, the transfer matrices can be calculated only wit h assumed values for the
frequencies. The end matrix leads to a homogenous system of l inear equations. Its
coefﬁcient determinant must be zero. The assumed frequenci es are to vary until this
condition is fulﬁlled sufﬁciently.
The transfer matrix method with numerical stabilization wa s 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 al l plates an analogous
mathematical model structure as isotropic semi-moment she ll structures. Therefore,
the procedure can be simply transferred to such laminated th in-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 a nd their implemen-
tation in a FEM-program system is more generally and more efﬁ ciently. 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 h ere
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 the y 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 structu re model D shall be used
and because we have cross-ply stacking, the Eqs. (10.2.32) a re valid

400 10 Modelling and Analysis of Thin-walled Folded Structu res
ˆAAA1UUU′′−ˆAAA3UUU−ˆAAA15VVV′+fffx=000,
ˆAAAT
15UUU′+ˆAAA4VVV′′−ˆAAA12VVV+fffs+fffn=000
Solution 10.1. At ﬁrst we have to calculate the matrices ˆAAA1,ˆAAA3,ˆAAA4,ˆAAA12and ˆAAA15,
their deﬁnitions are given in Eq. (10.2.12)
ˆAAA1=∑
(i)di/integraldisplay
0A11iϕϕϕϕϕϕTdsi,
ˆAAA3=∑
(i)di/integraldisplay
0A66iϕϕϕ•ϕϕϕ•Tdsi,
ˆAAA4=∑
(i)di/integraldisplay
0A66iψψψψψψTdsi,
ˆAAA12=∑
(i)di/integraldisplay
0D22iξξξ••ξξξ••Tdsi,
ˆAAA15=∑
(i)di/integraldisplay
0A66iϕϕϕ•ψψψTdsi,
The co-ordinate functions ϕϕϕ,ψψψ,ξξξare also shown in Fig. 10.4. For solving the inte-
grals to obtain the ˆAAA-matrices, the functions ϕϕϕ,ψψψ,ξξξmust be written as functions of
the co-ordinates siof each strip. In accordance with Fig. 10.4 we ﬁnd
ϕ1(si)=+ 1,i=1,2,3,4; ϕ•
1(si)=0,i=1,2,3,4;
ϕ2(s1)=−dS
2/bracketleftbigg
1−2/parenleftbiggs1
dS/parenrightbigg/bracketrightbigg
;ϕ•
2(si)=ψ2(si),i=1,2,3,4;
ϕ2(s2)=dS
2;
ϕ2(s3)=+dS
2/bracketleftbigg
1−2/parenleftbiggs3
dS/parenrightbigg/bracketrightbigg
;
ϕ2(s4)=−dS
2;
ϕ3(s1)=dG
2;ϕ•
3(si)=ψ3(si),i=1,2,3,4;
ϕ3(s2)=+dG
2/bracketleftbigg
1−2/parenleftbiggs2
dG/parenrightbigg/bracketrightbigg
;
ϕ3(s3)=−dG
2;
ϕ3(s4)=−dG
2/bracketleftbigg
1−2/parenleftbiggs4
dG/parenrightbigg/bracketrightbigg
;

10.4 Problems 401
ϕ4(s1)=+dSdG
4/bracketleftbigg
1−2/parenleftbiggs1
dS/parenrightbigg/bracketrightbigg
;ϕ•
4(si)=ψ4(si),i=1,2,3,4;
ϕ4(s2)=−dSdG
4/bracketleftbigg
1−2/parenleftbiggs2
dG/parenrightbigg/bracketrightbigg
;
ϕ4(s3)=+dSdG
4/bracketleftbigg
1−2/parenleftbiggs3
dS/parenrightbigg/bracketrightbigg
;
ϕ4(s4)=−dSdG
4/bracketleftbigg
1−2/parenleftbiggs4
dG/parenrightbigg/bracketrightbigg
;
ψ1(s1)=−dG
2; ψ2(s1)=+ 1;
ψ1(s2)=−dS
2; ψ2(s2)=0;
ψ1(s3)=−dG
2; ψ2(s3)=−1;
ψ1(s4)=−dS
2; ψ2(s4)=0;
ψ3(s1)=0; ψ4(s1)=−dG
2;
ψ3(s2)=−1; ψ4(s2)=+dS
2;
ψ3(s3)=0; ψ4(s3)=−dG
2;
ψ3(s4)=+ 1; ψ4(s4)=+dS
2;
ξ1(s1)=−dS
2/bracketleftbigg
1−2/parenleftbiggs1
dS/parenrightbigg/bracketrightbigg
;ξ••
1(si)=0,i=1,2,3,4;
ξ1(s2)=−dG
2/bracketleftbigg
1−2/parenleftbiggs2
dG/parenrightbigg/bracketrightbigg
;
ξ1(s3)=−dS
2/bracketleftbigg
1−2/parenleftbiggs3
dS/parenrightbigg/bracketrightbigg
;

402 10 Modelling and Analysis of Thin-walled Folded Structu res
ξ1(s4)=−dG
2/bracketleftbigg
1−2/parenleftbiggs4
dG/parenrightbigg/bracketrightbigg
;
ξ2(s1)=0; ξ••
2(si)=0,i=1,2,3,4;
ξ2(s2)=+ 1;
ξ2(s3)=0;
ξ2(s4)=−1;
ξ3(s1)=+ 1; ξ••
3(si)=0,i=1,2,3,4;
ξ3(s2)=0;
ξ3(s3)=−1;
ξ3(s4)=0;
Some additional considerations are necessary to determine the functions ξ4(si). The
generalized co-ordinate function ξ4is corresponding to ψ4and represents therefore
a double antisymmetric deﬂection state of the cross-sectio n. The cross-section is
double symmetric in its geometry and in the elastic behavior . Therefore we must
have an antisymmetric function ξ4(si)in each strip. It means that the following
conditions are valid
ξ4(si=0) = ξi0,ξ4(si=di) =−ξi0,
ξ•
4(si=0) = α0,ξ•
4(si=di) = α0,
ξ••
4(si=0) =κsi0,ξ••
4(si=di) =−κsi0
Supposing a polynomial function of the third order, we can wr ite
ξ••
4(si) =κsi0/parenleftbigg
1−2si
di/parenrightbigg
,
ξ•
4(si) = κsi0di/bracketleftigg/parenleftbiggsi
di/parenrightbigg
−/parenleftbiggsi
di/parenrightbigg2/bracketrightigg
+α0,
ξ4(si) =κsi0d2
i
6/bracketleftigg
3/parenleftbiggsi
di/parenrightbigg2
−2/parenleftbiggsi
di/parenrightbigg3/bracketrightigg
+α0disi
di+ξi0
The condition ξ4(si=di)=−ξi0leads to
α0di=−/parenleftbiggκsi0d2
i
6+2ξi0/parenrightbigg
and than we obtain
ξ4(si)=κsi0d2
i
6/bracketleftigg
3/parenleftbiggsi
di/parenrightbigg2
−2/parenleftbiggsi
di/parenrightbigg3
−si
di/bracketrightigg
+ξi0/bracketleftbigg
1−2/parenleftbiggsi
di/parenrightbigg/bracketrightbigg
With the antisymmetric properties mentioned above we ﬁnd

10.4 Problems 403
κs10=κs30=κS0,
κs20=κs40=κG0
The continuity of the rotation angles at the corners and the e quilibrium 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),
D22SκS0=−D22GκG0
lead to the unknown curvatures
κS0=−12D22G
dGD22S+dSD22G,κG0=+12D22S
dGD22S+dSD22G
and we obtain
ξ4(s1) =−2D22Gd2
S
dGD22S+dSD22G/bracketleftigg
3/parenleftbiggs1
dS/parenrightbigg2
−2/parenleftbiggs1
dS/parenrightbigg3
−s1
dS/bracketrightigg
+dS
2/bracketleftbigg
1−2s1
dS/bracketrightbigg
,
ξ4(s2) =2D22Sd2
G
dGD22S+dSD22G/bracketleftigg
3/parenleftbiggs2
dG/parenrightbigg2
−2/parenleftbiggs2
dG/parenrightbigg3
−s2
dG/bracketrightigg
−dG
2/bracketleftbigg
1−2s2
dG/bracketrightbigg
,
ξ4(s3) =−2D22Gd2
S
dGD22S+dSD22G/bracketleftigg
3/parenleftbiggs3
dS/parenrightbigg2
−2/parenleftbiggs3
dS/parenrightbigg3
−s3
dS/bracketrightigg
+dS
2/bracketleftbigg
1−2s3
dS/bracketrightbigg
,
ξ4(s4) =2D22Sd2
G
dGD22S+dSD22G/bracketleftigg
3/parenleftbiggs4
dG/parenrightbigg2
−2/parenleftbiggs4
dG/parenrightbigg3
−s4
dG/bracketrightigg
−dG
2/bracketleftbigg
1−2s4
dG/bracketrightbigg
,
ξ••
4(s1) =−12D22G
dGD22S+dSD22G/bracketleftbigg
1−2s1
dS/bracketrightbigg
,
ξ••
4(s2) = +12D22S
dGD22S+dSD22G/bracketleftbigg
1−2s2
dG/bracketrightbigg
,
ξ••
4(s3) =−12D22G
dGD22S+dSD22G/bracketleftbigg
1−2s3
dS/bracketrightbigg
,
ξ••
4(s4) = +12D22S
dGD22S+dSD22G/bracketleftbigg
1−2s4
dG/bracketrightbigg

404 10 Modelling and Analysis of Thin-walled Folded Structu res
Now all elements of the matrices can be calculated. Here only the calculation of the
element ˆA122of the matrix ˆAAA1shall be derived in a detailed manner
ˆA111=4
∑
i=1di/integraldisplay
0A11iϕ1(si)ϕ1(si)dsi=2(A11SdS+A11GdG),
ˆA112=4
∑
i=1di/integraldisplay
0A11iϕ1(si)ϕ2(si)dsi=0,
ˆA113=4
∑
i=1di/integraldisplay
0A11iϕ1(si)ϕ3(si)dsi=0,
ˆA114=4
∑
i=1di/integraldisplay
0A11iϕ1(si)ϕ4(si)dsi=0,
ˆA122=4
∑
i=1di/integraldisplay
0A11iϕ2(si)ϕ2(si)dsi
=A11Sd2
S
4dS/integraldisplay
0/parenleftbigg
1−2s1
dS/parenrightbigg2
ds1+A11Gd2
S
4dG/integraldisplay
0ds2
+A11Sd2
S
4dS/integraldisplay
0/parenleftbigg
1−2s3
dS/parenrightbigg2
ds3+A11Gd2
S
4dG/integraldisplay
0ds4
=A11Sd3
S
12+A11Gd2
S
4dG+A11Sd3
S
12+A11Gd2
S
4dG
=d2
S
6(A11SdS+3A11GdG),
ˆA123=4
∑
i=1di/integraldisplay
0A11iϕ2(si)ϕ3(si)dsi=0,
ˆA124=4
∑
i=1di/integraldisplay
0A11iϕ2(si)ϕ4(si)dsi=0,
ˆA133=4
∑
i=1di/integraldisplay
0A11iϕ3(si)ϕ3(si)dsi=d2
G
6(3A11SdS+A11GdG),
ˆA134=4
∑
i=1di/integraldisplay
0A11iϕ3(si)ϕ4(si)dsi=0,
ˆA144=4
∑
i=1di/integraldisplay
0A11iϕ4(si)ϕ4(si)dsi=d2
Sd2
G
24(A11SdS+A11GdG)

10.4 Problems 405
With all elements ˆA1i jwe obtain the matrix ˆAAA1to
ˆAAA1=
A1110 0 0
0A1220 0
0 0 A1330
0 0 0 A144

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 ϕiare orthogonal to each
other and therefore the matrix ˆAAA1is a diagonal matrix
In the same way the other ˆAAAimatrices are obtained
ˆAAA3=
0 0 0 0
0 2A66SdS 0 0
0 0 2 A66GdG 0
0 0 0dSdG
2(A66SdG+3A66GdS)
,
ˆAAA4=
A 0 0 0
0 2A66SdS 0 0
0 0 2 A66GdG0
B 0 0 A
,
ˆAAA12=
0 0 0 0
0 0 0 0
0 0 0 0
dSdG
2(A66SdG−A66GdS)0 096D22GD22S
dGD22S+dSD22G
,
ˆAAA15=
0 0 0 0
0 2A66SdS 0 0
0 0 2 A66GdG0
B 0 0 A

with
A=dSdG
2(A66SdG+A66GdS),B=dSdG
2(A66SdG−A66GdS)
Now the system of differential equations can be developed wi th the help of Eq.
(10.2.32).

406 10 Modelling and Analysis of Thin-walled Folded Structu res
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+dSD22GV4 =−(fs4+fn4)
We can see, that the system of differential equations is divi ded into four decou-
pled partial systems. The ﬁrst equation describes the longi tudinal displacement, the
second and the third partial systems represent the bending a bout the global y- and
z-axes and the fourth – the torsion, the warping and the contou r deformation of the
cross-section. An analytic solution of the fourth partial s ystem is more difﬁcult like
the solutions of the ﬁrst three partial systems but it is poss ible too. The analytical
solution of an analogous system for an isotropic box girder i s 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 Sta b- und Stabschalen-
tragwerke. Vieweg-Verlag, Brauschweig/Wiesbaden
Vlasov VZ (1961) Thin-walled elastic beams. National Scien ce Foundation and De-
partment of Commerce, Arlington, VI
Wlassow WS (1958) Allgemeine Schalentheorie und ihre Anwen dung 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 t he ﬁnite element proce-
dures and developed ﬁnite classical and generalized beam el ements and ﬁnite plate
elements in the frame of classical and ﬁrst order shear defor mation theory. Selected
examples demonstrate the possibilities of ﬁnite element an alysis.

Chapter 11
Finite Element Analysis
The Finite Element Method (FEM) is one of the most effective m ethods for the nu-
merical solution of ﬁeld problems formulated in partial dif ferential equations. The
basic idea of the FEM is a discretization of the continuous st ructure into substruc-
tures. This is equivalent to replacing a domain having an inﬁ nite number of degrees
of freedom by a system having a ﬁnite number of degrees of free dom. The actual
continuum or structure is represented as an assembly of subd ivisions called ﬁnite
elements. These elements are considered to be interconnect ed at speciﬁed joints
which are called nodes. The discretization is deﬁned by the s o-called ﬁnite element
mesh made up of elements and nodes.
We assume one-dimensional elements, when one dimension is v ery large in com-
parison with the others, e.g. truss or beam elements, two-di mensional elements,
when one dimension is very small in comparison with the other s, 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 poin t of view the nodes are
the basic points for the approximate functions of the displa cements 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 stru cture is the number of
degrees of freedom. To every node, a number of degrees of free dom will be assigned.
These are nodal constants which usually (but not necessaril y) have a mechanical or
more general physical meaning. The number of degrees of free dom per element is
deﬁned by the product of the number of nodes per element and de grees 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 procedur e 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 develo pment of ﬁnite beam
elements and ﬁnite plate elements for the analysis of lamina te structures is given.
Section 11.4 contains the development of generalized ﬁnite beam elements based on
409 © Springer Nature Singapore Pte Ltd. 2018
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 f olded structures given
in Sect. 10.2. In Sect. 11.5 the results of some numerical app lications show the
inﬂuences of chosen parameters on the behavior of laminate s tructures.
11.1 Introduction
The principle of the total minimum potential energy and the H amilton‘s principle
are given in Sect. 2.2.2 in connection with analytical varia tional approaches, they are
also the theoretical basis of the FEM solutions of elastosta tic and of dynamic prob-
lems. In this way we have variational problems. For such prob lems 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 w hole structure, and
so for complex geometries it is difﬁcult to realize the requi rements of satisfying
the boundary conditions and of the linear independence and c ompleteness of these
functions.
One way to overcome these difﬁculties is by the discretizati on 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 o nly and they must sat-
isfy the conditions of geometrical compatibility at the ele ment boundaries. Because
it is usual to deﬁne different types of ﬁnite elements, we hav e special types of ap-
proximation functions for each element type. Here the appro ximation functions are
denoted Ni, the so-called shape functions. They are arranged in a matri xNNN, the ma-
trix of the shape functions of the particular element type. T he following introduction
to the FEM procedure is given in a general but condensed form a nd illustrates that
the step-by-step ﬁnite element procedure can be stated as fo llows:
•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 equation s,
•Calculation of the system equations for the unknown nodal di splacements,
•Computation of element strains and stresses
11.1.1 FEM Procedure
The starting point for elastostatic problems is the total po tential energy given in Eq.
(2.2.28). In accordance with the Ritz method the approximat ion
˜uuu(xxx)=NNN(xxx)vvv (11.1.1)
is used for the displacement ﬁeld vector uuu. Here NNNis the matrix of the shape func-
tions, they are functions of the position vector xxx, and vvvis the element displacement

11.1 Introduction 411
vector. The matrix NNNhas the same number of rows as the displacement vector uuu
has components and the same number of columns as the element d isplacement vec-
torvvvhas components. If the element has nKEnodes and the degree of freedom
for each node is nF, the element displacement vector vvvcontains nKEsubvectors
vvvi,i=1,…, nKEwith nFcomponents in each, and so vvvhasnKEnFcomponents. The
number of components of the displacement ﬁeld vector uuuisnu. Then the structure
of the matrix NNNis generally
NNN=[N1IIInuN2IIInu…N˜nIIInu],˜n=nKEnF
nu(11.1.2)
with IIInuas unit matrices of the size (nu,nu). Therefore the size of NNNis generally
(nu,nKEnF). In dependence on the kind of continuity at the element boun daries, the
so-called C(0)- orC(1)-continuity, see below, two cases can be distinguished. In t he
case of C(0)-continuity nFequals nuand therefore ˜ nis equal nKE, we have only nKE
shape functions Ni, whereas we can have up to nKEnFshape functions in the case of
C(1)-continuity.
For the stresses and the strains we obtain from (11.1.1)
σσσ(xxx) =CCCεεε(xxx)=CCCDDDNNN(xxx)vvv,
εεε(xxx) = DDDuuu(xxx)=DDDNNN(xxx)vvv=BBB(xxx)vvv(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 disp lacement vector vvv, e.g.
Π=Π(vvv). The variation of the total potential energy
δΠ=δvvvT
/integraldisplay
VBBBTCCCBBBvvvdV−/integraldisplay
VNNNTpppdV−/integraldisplay
AqNNNTqqqdA
 (11.1.4)
leads with δΠ=0 to
δvvvT(KKKvvv−fffp−fffq)=0 (11.1.5)
KKKis the symmetric stiffness matrix with the size ( nKEnF,nKEnF)
KKK=/integraldisplay
VBBBTCCCBBBdV (11.1.6)
andfffpandfffqare the vectors of the volume forces and the surface forces
fffp=/integraldisplay
VNNNTpppdV,fffq=/integraldisplay
AqNNNTqqqdA (11.1.7)
If the components of δvvvare independent of each other, we obtain from (11.1.5) a
system of linear equations
KKKvvv=fff,fff=fffp+fffq (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 st arting point for the
FEM procedure. Assuming again the independence of the compo nents of δvvvthe
matrix equation is
MMM¨vvv(t)+KKKvvv(t)=fff(t) (11.1.9)
for elastic systems without damping effects. MMMis symmetric mass matrix
MMM=/integraldisplay
VρNNNTNNNdV (11.1.10)
andfff(t)the vector of the time dependent nodal forces. Assuming the d amping pro-
portional to the relative velocities, an additional term CCCD˙vvv(t)can be supplemented
formally in Eq. (11.1.9)
MMM¨vvv(t)+CCCD˙vvv(t)+KKKvvv(t)=fff(t), (11.1.11)
where CCCDis the damping matrix. CCCDhas the same size as the matrices KKKandMMM
and usually it is formulated approximately as a linear combi nation of KKKandMMM. The
factors αandβcan be chosen to give the correct damping at two frequencies
CCCD≈αMMM+βKKK (11.1.12)
In selecting the shape functions NNNi(xxx)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 potentia l energy or in the Hamil-
ton’s function. In this case only the results of FEM approxim ations converge to the
real solutions by increasing the number of elements. For mor e-dimensional ﬁnite
elements in this way it is to realize that the displacements a re compatible up to the
(n−1)th derivative at the boundaries of adjacent elements, if t hey are compatible at
the nodes.
In plane stress or plane strain problems and in general three -dimensional prob-
lems the vector uuucontains displacements only (no rotations) and the differe ntial
operator DDDis of the 1st order. In this way we must only satisfy the displa cements
compatibility at the element boundaries that means the so-c alled C(0)-continuity.
By using beam or plate models especially of the classical Ber noulli 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 DDDis of the second order. Then we have to satisfy the compatibil ity 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 MMM,
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 diagon al matrix structure is
possible. Then we speak about a so-called condensed mass mat rix, otherwise we
have a consistent mass matrix. By using the condensed mass ma trix we have less

11.1 Introduction 413
computational expense than by using the consistent mass mat rix, but a decreasing
convergence to the real results is possible.
All equations considered above are only valid for a single el ement and strictly
they should have an additional index E. For example, we have the inner element
energy
UE=1
2vvvT
E/integraldisplay
VEBBBTCCCBBBdVvvvE=1
2vvvT
EKKKEvvvE (11.1.13)
with the element stiffness matrix
KKKE=/integraldisplay
VEBBBTCCCBBBdV (11.1.14)
Since the energy is a scalar quantity, the potential energy o f 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 no des of the whole system
must be deﬁned. By a so-called coincidence matrix LLLEthe correct position of each
single element is determined. LLLEis a Boolean matrix of the size ( nKEnF,nKnF) with
nKas the number of nodes of the whole structure.
The element displacement vector vvvEis positioned into the system displacement
vector vvvby the equation
vvvE=LLLEvvv (11.1.15)
and we obtain the system equation by summing up over all eleme nts
/parenleftigg
∑
iLLLT
iEKKKiELLLiE/parenrightigg
vvv=/bracketleftigg
∑
iLLLiE(fffiE p+fffiEq)/bracketrightigg
KKKvvv=fff(11.1.16)
The system stiffness matrix is also symmetric, but it is a sin gular matrix, if the
system is not ﬁxed kinematically, i.e., we have no boundary c onditions constraining
the rigid body motion. After consideration of the boundary c onditions of the whole
system, KKKbecomes a positive deﬁnite matrix and the system equation ca n be solved.
Then with the known displacements vvvthe stresses and deformations are calculated
using the element equations (11.1.1) and (11.1.3).
For elastodynamic problems, the system stiffness matrix an d the system mass
matrix are obtained in the same manner and we have the system e quation
MMM¨vvv(t)+CCCD˙vvv(t)+KKKvvv(t)=fff(t) (11.1.17)
For investigation of the eigen-frequencies of a system with out damping harmonic
vibrations are assumed and with
vvv(t)=ˆvvvcos(ωt+ϕ) (11.1.18)
andCCCD=000,fff(t)=ooothe matrix eigen-value problem follows

414 11 Finite Element Analysis
(KKK−ω2MMM)vvv=ooo (11.1.19)
and the eigen-frequencies and the eigen-vectors character izing 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 p lane
two-node beam elements. The number of nodal degrees of freed om is three ( u,w,ϕ):
1. What size are the element stiffness matrix and the system s tiffness matrix before
the consideration of the boundary conditions?
2. Show the coincidence matrix LLL2of 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 i n the previous
case but consider that the beam is simply supported (node 1 is constrained for the
deﬂections uandwand 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
vvv2=LLL2vvv
we obtain the coincidence matrix for the element Nr. 2
LLL2=
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
KKK=4
∑
i=1LLLT
iKKKiLLLi
In this case we obtain the following population of the matrix KKK

11.2 Finite Beam Elements 415
u1w1ϕ1u2w2ϕ2u3w3ϕ3u4w4ϕ4 
[(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) (v) (v) (0) (0) (0) (0) (0) (0)
[(v)] [(v)] (v)v+x v+x v+x x x x 0[(0)]0
[(v)] [(v)] (v)v+x v+x v+x x x x 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)x x x x +z x+z x+z z[z]z
[(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
ϕ1
u2
w2
ϕ2
u3
w3
ϕ3
u4
w4
ϕ4
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 c lamped at node 1
(u1=0,w1=0,ϕ1=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 – char acterized 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 connecti on 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 m idline, 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 la minate beams the fol-
lowing investigations are restricted to tension/compress ion 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 rectangu lar 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 ca se for the following
development of ﬁnite laminate beam elements. In this way we h ave no coupling of
tension and bending and we can divide our considerations int o the development of
laminate elements for tension/compression, so-called lam inate 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 ass umed 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)andNj(x1), which satisﬁes C(0)-continuity
u(x1)=NNNvvvE,vvvT
E=[uiuj],NNN=[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)=/integraldisplay
AσdA=A11ε1(x1)=A11du
dx1,A11=bn
∑
k=1C(k)
11hk (11.2.3)
and the longitudinal load per length n(x1)the total potential energy can be written
as
Π(u)=1
2l/integraldisplay
0A11u′2dx1−l/integraldisplay
0n(x1)udx1 (11.2.4)
and for the element stiffness matrix we obtain
Fig. 11.1 Laminate truss
elementlx1n(x1)
ui uj
x1
lx1
l
00
001.0
1.01.0
1.00.5 0.5NiNj
Fig. 11.2 Shape functions of the two-node truss element

11.2 Finite Beam Elements 417
KKKE=A11l/integraldisplay
0NNN′TNNN′dx1=A11
l/bracketleftbigg
1−1
−1 1/bracketrightbigg
(11.2.5)
The element force vector is deﬁned as
fffnE=l/integraldisplay
0NNNTn(x1)dx1
If we assume that n(x1)is a linear function with niandnjas the intensities at the
nodes
n(x1)=NNN/bracketleftbigg
ni
nj/bracketrightbigg
then
fffnE=l/integraldisplay
0NNNTNNNdx1/bracketleftbiggni
nj/bracketrightbigg
=l
6/bracketleftbigg2 1
1 2/bracketrightbigg/bracketleftbiggni
nj/bracketrightbigg
(11.2.6)
In case of nodal forces fffPE=[FiFj]T, the vector fffPEmust be added to the vector
fffnE
fffE=fffnE+fffPE (11.2.7)
The system equation can be obtained in dependence on the stru cture of the whole
system, deﬁned by a coincidence matrix together with the tra nsformation of all el-
ement equations into a global coordinate system. Consideri ng 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) h as to be used
MMME=/integraldisplay
VρNNNTNNNdV=l/integraldisplay
0ρNNNTNNNdx1,ρ=1
hn
∑
k=1ρ(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 a xial inertia
force. Instead of handling the distributed mass directly, w e generate ﬁctitious nodal
masses contained in the consistent mass matrix
MMME=ρAl
6/bracketleftbigg2 1
1 2/bracketrightbigg
(11.2.8)
With the system equation, obtained in the same manner as for e lastostatic 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 a re considered, the
classical laminate theory and the shear deformation theory . The classical laminate
theory is based on the Bernoulli beam model and the shear defo rmation theory on
the Timoshenko beam model. The Bernoulli beam model neglect s the shear strains
in the bending plane and so it seems to be less realistic for th e calculation of laminate
beams. Therefore it is better to use the Timoshenko beam mode l, which includes the
shear strains in a simple form (Chap. 7).
In the following discussion, only the shear deformation the ory 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 be nding. The main
advantage of the shear deformation theory in comparison wit h the Bernoulli theory
is that the cross-sectional rotation angle ψis independent of the displacement wand
therefore the differential operator DDDin 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 ar e 2 (w,ψ). In Fig. 11.3
such a two-node beam element is shown. The element displacem ent vector is
vvvT
E=[wiψiwjψj] (11.2.9)
For the displacement vector uuuthe approximation (11.2.11) is used
uuu(x1)=/bracketleftbiggw(x1)
ψ(x1)/bracketrightbigg
=NNNvvvE, (11.2.10)
where the matrix of the shape functions is
NNN=/bracketleftbigg
Ni(x1)/bracketleftbigg1 0
0 1/bracketrightbigg
Nj(x1)/bracketleftbigg1 0
0 1/bracketrightbigg/bracketrightbigg
(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 th ree-node element,
as shown in Fig. 11.4. Then the element displacement vector i s
vvvT
E=[wiψiwjψjwkψk] (11.2.12)
and for the matrix NNNwe obtain
Fig. 11.3 Two-node beam
elementlx1x3
wi wj
ψi ψj

11.2 Finite Beam Elements 419
Fig. 11.4 Three-node beam
element
l/2 l/2x1x3
wiwj wk
ψk ψi ψj
NNN=/bracketleftbigg
Ni(x1)/bracketleftbigg1 0
0 1/bracketrightbigg
Nj(x1)/bracketleftbigg1 0
0 1/bracketrightbigg
Nk(x1)/bracketleftbigg1 0
0 1/bracketrightbigg/bracketrightbigg
(11.2.13)
with the shape functions
Ni(x1)=1−3×1
l+2×2
1
l2,Nj(x1)=4×1
l−4×2
1
l2,Nk(x1)=−x1
l+2×2
1
l2,(11.2.14)
which are shown in Fig. 11.5. A further increase in the elemen t accuracy can be
achieved with a four-node element, see Fig. 11.6. Here the el ement displacement
vector and the matrix of the shape functions are
vvvT
E=[wiψiwjψjwkψkwlψl] (11.2.15)
x1
lx1
lx1
l
0000
00
1.0
1.01.0
1.01.0
1.0
0.50.5 0.5
-0.125-0.125Ni
NkNj
Fig. 11.5 Shape functions of the three-node element

420 11 Finite Element Analysis
Fig. 11.6 Four-node beam
element
x1x3,w
lwi wj wk wl
l/3 l/3 l/3
ψi ψj ψk ψl
NNN=/bracketleftbigg
Ni(x1)/bracketleftbigg1 0
0 1/bracketrightbigg
Nj(x1)/bracketleftbigg1 0
0 1/bracketrightbigg
Nk(x1)/bracketleftbigg1 0
0 1/bracketrightbigg
Nl(x1)/bracketleftbigg1 0
0 1/bracketrightbigg/bracketrightbigg
(11.2.16)
with the shape functions
Ni=1−11
2×1
l+9/parenleftigx1
l/parenrightig2
−9
2/parenleftigx1
l/parenrightig3
,Nj=9×1
l−45
2/parenleftigx1
l/parenrightig2
+27
2/parenleftigx1
l/parenrightig3
,
Nk=−9
2×1
l+18/parenleftigx1
l/parenrightig2
−27
2/parenleftigx1
l/parenrightig3
,Nl=x1
l−9
2/parenleftigx1
l/parenrightig2
+9
2/parenleftigx1
l/parenrightig3
(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 mor e coarse mesh
can be used, but the calculations of the element stiffness ma trices will be more
computationally expensive.
The further relationships are developed formally independ ent of the chosen num-
ber of element nodes. The element stiffness matrix is obtain ed with (11.1.6)
KKKE=l/integraldisplay
0BBBTCCCBBBdx1=l/integraldisplay
0NNNTDDDTCCCDDDNNNdx1 (11.2.18)
Here
DDD=
0d
dx1d
dx10
,CCC=/bracketleftbiggD11 0
0ksA55/bracketrightbigg
(11.2.19)
with the stiffness
D11=b
3n
∑
k=1C(k)
11/parenleftig
x(k)3
3−x(k−1)3
3/parenrightig
,A55=bn
∑
k=1C(k)
55h(k)(11.2.20)
and the shear correction factor ksgiven in (7.3.20).
For calculation of the element force vector, we assume that t he element is loaded
by a distributed transverse load per length q(x1)and we can write the external work
as
WE=l/integraldisplay
0q(x1)w(x1)dx1=vvvT
EfffE

11.2 Finite Beam Elements 421
and with
w(x1)=[ wψ]/bracketleftbigg1
0/bracketrightbigg
=uuuTRRR=vvvT
ENNNTRRR (11.2.21)
the element force vector fffEis obtained
fffE=l/integraldisplay
0NNNTRRRq(x1)dx1 (11.2.22)
If single nodal forces or moments are acting, they must be add ed.
The system equation can be obtained in dependence on the stru cture of the whole
system deﬁned by a coincidence matrix together with the tran sformation of all el-
ement equations into a global coordinate system. After cons idering the boundary
conditions, the system equation can be solved. After this th e stress resultants are
obtained for all elements
0.0 0.00.0 0.0
0.2 0.20.2 0.2
0.4 0.40.4 0.4
0.6 0.60.6 0.6
0.8 0.80.8 0.8
1.0 1.01.0 1.0
x1
lx1
lx1
lx1
l
−0.2−0.2
−0.2−0.2
0.00.0
0.00.0
0.20.2
0.20.2
0.40.4
0.40.4
0.60.6
0.60.6
0.80.8
0.80.8
1.01.0
1.01.0
−0.4−0.4
1.21.2Ni Nj
Nk Nl
ii
ii
jj
jj
kk
kk
ll
ll
Fig. 11.7 Shape functions of the four-node element

422 11 Finite Element Analysis
σσσE=/bracketleftbiggM
Q/bracketrightbigg
E=CCCDDDNNNvvvE (11.2.23)
For elastodynamic problems the mass matrix must be calculat ed. The Timoshenko
beam model includes in the general case axial, transversal a nd rotational inertia
forces and moments. So it must be noted that the laminae of the beam have different
velocities in x1-direction
TE=1
2l/integraldisplay
0b/2/integraldisplay
−b/2x(n)
3/integraldisplay
x(0)
3ρ(˙u2+˙w2)dx1dx2dx3,˙u=˙u0+x3˙ψ (11.2.24)
After integration with respect to d x2and d x3follow
TE=1
2l/integraldisplay
0/bracketleftbig
ρ0(˙u2
0+˙w2)+2ρ1˙u0˙ψ+ρ2˙ψ2/bracketrightbig
dx1 (11.2.25)
with the so-called generalized densities
ρ0=bn
∑
k=1ρ(k)/parenleftig
x(k)
3−x(k−1)
3/parenrightig
,
ρ1=b1
2n
∑
k=1ρ(k)/parenleftbigg
x(k)
32−x(k−1)
32/parenrightbigg
,
ρ2=b1
3n
∑
k=1ρ(k)/parenleftbigg
x(k)
33
−x(k−1)
33/parenrightbigg(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
ρ1=0,˙u0=0
and therefore
TE=1
2l/integraldisplay
0(ρ0˙w2+ρ2˙ψ2)dx1=1
2l/integraldisplay
0˙uuuTRRR0˙uuudx1 (11.2.27)
with the matrix RRR0
RRR0=/bracketleftbiggρ00
0ρ2/bracketrightbigg
(11.2.28)
Using (11.1.1)
TE=1
2l/integraldisplay
0˙vvvT
ENNNTRRR0NNN˙vvvEdx1
the element mass matrix MMMEis obtained

11.2 Finite Beam Elements 423
MMME=l/integraldisplay
0NNNTRRR0NNNdx1 (11.2.29)
The system equation is established in the same manner as for e lastostatic problems,
and with the assumption of harmonic vibrations, the eigen-f requencies 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 bea m element by analytical
integration!
2. Calculate the element force vector for a two-node beam ele ment, loaded by a
linear distributed transverse load per length q(x1). The intensities at the nodes
areqiandqj!
3. Calculate the element mass matrix for a two-node beam elem ent!
Solution 11.2. The three solutions are;
1. In the case of a two-node beam element the matrix of the shap e functions is
NNN=/bracketleftbigg
Ni(x1)/bracketleftbigg1 0
0 1/bracketrightbigg
Nj(x1)/bracketleftbigg1 0
0 1/bracketrightbigg/bracketrightbigg
with Ni(x1)=1−(x1/l),Nj(x1)=x1/l. The element stiffness matrix is deﬁned
by (11.2.18)
KKK=l/integraldisplay
0NNNTDDDTCDNCDNCDN dx1,
where in (11.2.19) are given
DDD=
0d
dx1d
dx10
,CCC=/bracketleftbigg¯D11 0
0ks¯A55/bracketrightbigg
with ¯D11andks¯A55in according to (11.2.20). After execution the matrix opera –
tions we obtain for the stiffness matrix

424 11 Finite Element Analysis
KKKE=l/integraldisplay
0
ks¯A55/parenleftbiggdNi
dx1/parenrightbigg2
0 ks¯A55dNi
dx1dNj
dx10
0 ¯D11/parenleftbiggdNi
dx1/parenrightbigg2
0 ¯D11dNi
dx1dNj
dx1
ks¯A55dNi
dx1dNj
dx10 ks¯A55/parenleftbiggdNj
dx1/parenrightbigg2
0
0 ¯D11dNi
dx1dNj
dx10 ¯D11/parenleftbiggdNj
dx1/parenrightbigg2
dx1
and ﬁnally
KKKE=1
l
ks¯A55 0−ks¯A55 0
0 ¯D11 0−¯D11
−ks¯A55 0 ks¯A55 0
0−¯D11 0 ¯D11

2. The element force vector to calculate with respect to (11. 2.22)
fffE=l/integraldisplay
0NNNTRRRq(x1)dx1with RRR=/bracketleftbigg
1
0/bracketrightbigg
For the loading function q(x1)we can write
q(x1)=[ Ni(x1)Nj(x1)]/bracketleftbiggqi
qj/bracketrightbigg
and then we ﬁnd
fffE=l/integraldisplay
0
Ni(x1)2Ni(x1)Nj(x1)
0 0
Ni(x1)Nj(x1)Nj(x1)2
0 0
/bracketleftbigg
qi
qj/bracketrightbigg
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)
MMME=l/integraldisplay
0NNNTRRR0NNNdx1,RRR0=/bracketleftbiggρ00
0ρ2/bracketrightbigg
with the generalized densities ρ0andρ2(11.2.26).
Inserting the matrix of the shape functions given above and e xecuting the matrix
operations we obtain
l/integraldisplay
0
Ni(x1)2ρ0 0 Ni(x1)Nj(x1)ρ0 0
0 Ni(x1)2ρ2 0 Ni(x1)Nj(x1)ρ2
Ni(x1)Nj(x1)ρ0 0 Nj(x1)2ρ0 0
0 Ni(x1)Nj(x1)ρ2 0 Nj(x1)2ρ2
dx1

11.3 Finite Plate Elements 425
and after integration the element mass matrix is in this case
MMME=l
6
2ρ00ρ00
0 2ρ20ρ2
ρ00 2ρ00
0ρ20 2ρ2

11.3 Finite Plate Elements
Plates are two-dimensional structures that means that one d imension, the thickness,
is very small in comparison to the others and in the unloaded s tate they are plane.
Usually, the midplane between the top and the bottom plate su rfaces 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 fo llowing 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 th e single layer as
quasi-homogeneous and orthotropic. In each layer we can hav e different materials,
different thicknesses and especially different angle orie ntations of the ﬁbres. The
whole plate is assumed to be a continuous structure. The stac king sequence of the
single layers has a great inﬂuence on the deformation behavi or of the plate. Plates
can be loaded by distributed and concentrated loads in all di rections, so called in-
plane and out of plane loading. In a special case of laminate p lates, 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 dis tinguishes between
the classical laminate theory and the shear deformation the ory like the modelling
of beams. The plate model based on the classical laminate the ory usually is called
Kirchhoff plate with its main assumption that points lying o n a line orthogonal to
the midplane before deformation are lying on such a normal li ne after deformation.
This assumption is an extended Bernoulli hypothesis of the b eam model to two-
dimensional structures.
The application of the classical laminate theory should be r estricted 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 c alled the Mindlin plate
model. The following development of ﬁnite laminate plate el ements will be carried
out for both models. Here we will be restricted to symmetric l aminate plates in both
cases, it means that we have no coupling of membrane and bendi ng/twisting states
and we will consider bending only.
In both cases we consider a triangular ﬁnite plate element. T he approximation
of complicated geometric forms, especially of curved bound aries, can be done
easily with triangular elements. Usually special coordina tes are used for triangu-
lar elements. The triangle is deﬁned by the coordinates of th e three corner points
Pi(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△=/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle1x1y1
1x2y2
1x3y3/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle,A1=/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle1x y
1x2y2
1x3y3/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle,
A2=/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle1x1y1
1x y
1x3y3/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle,A3=/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle1x1y1
1x2y2
1x y/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle(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 coordinat es
Fig. 11.8 Natural triangle
coordinatesxy
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 s ides.
Because the shape functions Niused for the approximation of the deformation
ﬁeld in a triangular plate element are usually written as fun ctions of the natural
element coordinates, it is necessary to ﬁnd relationships f or the derivatives of the
shape functions with respect to the global cartesian coordi nates. At ﬁrst the deriva-
tives of the shape functions Niare given by the natural triangle coordinates L1and
L2.L3depends from L1andL2, 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
=JJJ
∂Ni
∂x
∂Ni
∂y
(11.3.7)
JJJis the Jacobi matrix of the coordinate transformation
JJJ=
∂x
∂L1∂y
∂L1
∂x
∂L2∂y
∂L2
=/bracketleftbiggcj−bj
−cibi/bracketrightbigg
(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
DetJJJ=cjbi−bjci=∆ (11.3.10)
Fig. 11.9 Natural triangle
coordinates of special pointsxy
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
JJJ−1=1
∆/bracketleftbiggbibj
cicj/bracketrightbigg
(11.3.11)
and then we obtain for the derivatives of the shape functions Niwith respect to the
cartesian coordinates xandy

∂Ni
∂x
∂Ni
∂y
=JJJ−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=/parenleftbigg∂x
∂L1/parenrightbigg2∂2Ni
∂x2+2∂x
∂L1∂y
∂L1∂2Ni
∂x∂y+/parenleftbigg∂y
∂L1/parenrightbigg2∂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 ma trix form
JJJ∗
∂2Ni
∂x2
∂2Ni
∂x∂y
∂2Ni
∂y2
=
∂2Ni
∂L2
1
∂2Ni
∂L1∂L2
∂2Ni
∂L2
2
(11.3.14)
where JJJ∗is a modiﬁed or extended Jacobi matrix
JJJ∗=
/parenleftbigg∂x
∂L1/parenrightbigg2
2∂x
∂L1∂y
∂L1/parenleftbigg∂y
∂L1/parenrightbigg2
∂x
∂L1∂x
∂L2∂x
∂L2∂y
∂L1+∂x
∂L1∂y
∂L2∂y
∂L1∂y
∂L2/parenleftbigg∂x
∂L2/parenrightbigg2
2∂x
∂L2∂y
∂L2/parenleftbigg∂y
∂L2/parenrightbigg2
(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
=JJJ∗−1
∂2Ni
∂L2
1
∂2Ni
∂L1∂L2
∂2Ni
∂L2
2
(11.3.16)
Of course, by consequently using the natural triangle coord inates it follows that the
integrands in the energy terms are functions of these coordi nates. Therefore we have
to consider for the variables of integration the relationsh ip
dA=dxdy=DetJJJdL1dL2=∆dL1dL2 (11.3.17)
In Sects. 11.3.1 and 11.3.2 the development of triangular ﬁn ite 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 symmet ric laminate plate, see
also (8.2.24)
Π(w) =1
2/integraldisplay
A/bracketleftigg
D11/parenleftbigg∂2w
∂x2/parenrightbigg2
+D22/parenleftbigg∂2w
∂y2/parenrightbigg2
+2D12∂2w
∂x2∂2w
∂y2+4D66/parenleftbigg∂2w
∂x∂y/parenrightbigg2
+4/parenleftbigg
D16∂2w
∂x2+D26∂2w
∂y2/parenrightbigg∂2w
∂x∂y/bracketrightigg
dA−/integraldisplay
ApzwdA(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 gr eat number of very thin
layers, we use the following simpliﬁed strain energy approx imately

430 11 Finite Element Analysis
Π(w) =1
2/integraldisplay
A/bracketleftigg
D11/parenleftbigg∂2w
∂x2/parenrightbigg2
+D22/parenleftbigg∂2w
∂y2/parenrightbigg2
+2D12∂2w
∂x2∂2w
∂y2+4D66/parenleftbigg∂2w
∂x∂y/parenrightbigg2
−2pzw/bracketrightigg
dA(11.3.19)
The total potential energy of the classical plate model cont ains 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 d irection 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 w e 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 trian gular 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 derivat ives in normal direction
only.
It is a disadvantage when using this element in a general prog ram system that we
have a different number of degrees of freedom at the nodes. Th erefore an elimina-
tion, a so-called static condensation of the nodal constant s of the mid-side nodes,
can be done and then we have only 18 degrees of freedom for the e lement. 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 ﬁn ite element is given by
a special 5th order polynomial, it contains however a comple te 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
elementxy
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)=NNN(L1,L2)vvv (11.3.21)
with NNNas the matrix of the 18 shape functions (here it has only one ro w) and the el-
ement displacement vector vvvincluding 18 components. For the differential operator
DDDOPmust be written
DDDOP=/bracketleftbigg∂2
∂x2∂2
∂y22∂2
∂x∂y/bracketrightbiggT
(11.3.22)
and after this, see also Eq. (11.1.3), the matrix BBBleads to
BBB=DDDOPNNN (11.3.23)
Since the shape functions are functions of the natural trian gle coordinates L1andL2,
for the derivatives by the cartesian coordinates we have to t ake into consideration
(11.3.16). The element stiffness matrix follows according to (11.1.6)
KKKE=/integraldisplay
AEBBBTDBDBDBdA
and with the substitution of the integration variable Eq. (1 1.3.17)
KKKE=1/integraldisplay
01−L1/integraldisplay
0BBBTDBDBDB△dL2dL1 (11.3.24)
Here DDDis the matrix of the plate stiffness, the coupling of bending and twisting is
neglected ( D16=D26=0)
DDD=
D11D120
D12D220
0 0 D66

According to (11.1.7) we obtain the element force vector
fffE=/integraldisplay
AENNNTpdA=1/integraldisplay
01−L1/integraldisplay
0NNNTp△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 mas s matrix must be
calculated. According to (11.1.10), the element mass matri x reduces to
MMME=/integraldisplay
VEρNNNTNNNdV=1/integraldisplay
01−L1/integraldisplay
0ρNNNTNNNh△dL2dL1 (11.3.26)

432 11 Finite Element Analysis
with ρas an average density
ρ=1
hn
∑
k=1ρ(k)h(k)(11.3.27)
Note that the classical laminate theory does not consider th e rotary kinetic energy.
The integrations in the (11.3.24) for the element stiffness matrix KKKE, (11.3.25)
for the element force vector fffEand (11.3.26) for the element mass matrix must
be carried out numerically. Only the force vector fffEcan 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 orde r 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 shea r deformation theory,
considers the shear deformation in a simpliﬁed form. In the M indlin plate model
the Kirchhoff’s hypotheses are relaxed. Transverse normal s to the midplane do not
remain perpendicular to the middle surface after deformati on. 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,ψ1,ψ2) =1
2/integraldisplay
A/bracketleftigg
D11/parenleftbigg∂ψ1
∂x/parenrightbigg2
+2D12/parenleftbigg∂ψ1
∂x∂ψ2
∂y/parenrightbigg
+D22/parenleftbigg∂ψ2
∂y/parenrightbigg2
+D66/parenleftbigg∂ψ1
∂y+∂ψ2
∂x/parenrightbigg2
+ks
55A55/parenleftbigg
ψ1+∂w
∂x/parenrightbigg2
+ks
44A44/parenleftbigg
ψ2+∂w
∂y/parenrightbigg2/bracketrightigg
dxdy−/integraldisplay
Apzwdxdy
δΠ(w,ψ1,ψ2) =0
(11.3.28)
or written in matrix form
Π(w,ψ1,ψ2)=1
2/integraldisplay
A(κκκTDDDκκκ+εεεsTAAAsεεεs)dxdy−/integraldisplay
Ap3wdxdy (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
DDD=
D11D120
D12D220
0 0 D66
,AAAs=/bracketleftbigg
ks
55A55 0
0ks
44A44/bracketrightbigg
(11.3.30)
The stiffness are given in detail in (4.2.15) and for the shea r correction factor see
Sect. 8.3. Note that we have, in the elastic potential three i ndependent deformation
components (the deﬂection wand the rotations ψ1andψ2), so the displacement ﬁeld
vector uuuhas three components ( nu=3), see also (11.1.2).
For the curvatures and the shear strains we have
κκκ=DDDbuuu,εεεs=DDDsuuu (11.3.31)
where DDDbandDDDsare the matrices of the differential operators
DDDb=
0∂
∂x0
0 0∂
∂y
0∂
∂y∂
∂x
,DDDs=
∂
∂x1 0
∂
∂y0 1
(11.3.32)
The most important property of the elastic potential howeve r is that it contains ﬁrst
derivatives only. Therefore, we have to guarantee only C(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 s ix-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 ha ve the nodal and the
element displacement vectors
vvvT
i=[wiψxiψyi],vvvT
E=[vvvT
ivvvT
jvvvT
kvvvT
lvvvT
mvvvT
n] (11.3.33)
and according to (11.1.2) with nu=nF,˜n=nKEthe matrix of the shape functions is
given by
NNN=[NiIII3NjIII3NkIII3NlIII3NmIII3NnIII3], (11.3.34)
Fig. 11.11 Six-node ﬁnite
plate element with nodal
degrees of freedomxyz
ψ1iψ2iwi
il jmk
n

434 11 Finite Element Analysis
where III3are 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 expr essed by
κκκ=DDDbuuu=DDDbNNNvvvE=BBBbvvvE,BBBb=DDDbNNN
εεεs=DDDsuuu=DDDsNNNvvvE=BBBsvvvE,BBBs=DDDsNNN(11.3.36)
and consideration of (11.3.12) leads to the element stiffne ss matrix, see also (11.1.6)
consisting of two parts
KKKE=KKKb
E+KKKs
E,KKKb
E=/integraldisplay
AEBBBbTDDDBBBbdxdy,KKKs
E=/integraldisplay
AEBBBsTAAAsBBBsdxdy (11.3.37)
Because the shape functions in NNNare functions of the natural triangle co-ordinates,
the integration variables must be substituted by (11.3.17) , and then we ﬁnd
KKKb
E=1/integraldisplay
01−L1/integraldisplay
0BBBbTDDDBBBb∆dL2dL1,KKKs
E=1/integraldisplay
01−L1/integraldisplay
0BBBsTAAAsBBBs∆dL2dL1 (11.3.38)
To obtain the element force vector fffEa load vector qqqmust be deﬁned with the same
number of components as the displacement ﬁeld vector uuu. Because only surface
loading p(x,y)is considered here, it leads to
qqqT=[p0 0]
and then the element force vector is
fffE=/integraldisplay
AENNNTqqqdxdy,fffE=1/integraldisplay
01−L1/integraldisplay
0NNNTqqq∆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 integr ation formulae of the
same order as used for shape polynomials, here of the second o rder. It must be done
in this manner for the ﬁrst part KKKb
Eof the stiffness matrix, for the second part of KKKs
E
a lower order can be used. Such a different kind of integratio n for the two parts of
the stiffness matrix is called selective integration.
For dynamic analysis the element mass matrix MMMEmust 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
2/integraldisplay
VEρ˙uuuT˙uuudV=1
2/integraldisplay
AEh
2/integraldisplay
−h
2ρ(˙w2+˙ψ2
1+˙ψ2
2)dzdA (11.3.40)
If the so called generalized densities are used
ρ0=n
∑
k=1ρ(k)[z(k)−z(k−1)]=n
∑
k=1ρ(k)h(k),
ρ1=1
2n
∑
k=1ρ(k)[z(k)2−z(k−1)2],
ρ2=1
3n
∑
k=1ρ(k)[z(k)3−z(k−1)3](11.3.41)
and it is noted here that ρ1=0, because we have assumed symmetric laminates only,
then for the kinetic energy we obtain
TE=1
2/integraldisplay
AE˙vvvTRRR0˙vvvdA (11.3.42)
RRR0is a matrix of the generalized densities
RRR0=
ρ00 0
0ρ20
0 0 ρ2
 (11.3.43)
Using the approximation for the displacement ﬁeld vector ac cording to Eq. (11.1.1)
we obtain
TE=1
2vvvT
E/integraldisplay
AENNNTRRR0NNNdAvvvE (11.3.44)
and the element mass matrix is
MMME=/integraldisplay
AENNNTRRR0NNNdA,MMME=1/integraldisplay
01−L1/integraldisplay
0NNNTRRR0NNN∆dL1dL2 (11.3.45)
with substitution of the integration variables.
The ﬁnite laminate plate element developed above is called P L18, where the
number 18 gives the degrees of freedom of all element nodes. T his element can be
used only for laminate plates with laminae arranged symmetr ically to the midplane,
where we have no coupling of membrane and bending/twisting s tates and we have
no in-plane loading.

436 11 Finite Element Analysis
In many cases we have nonsymmetric laminates and we have a cou pling 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 freed om 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
vvvT
i=[uiviwiψxiψyi],vvvT
E=[vvvT
ivvvT
jvvvT
kvvvT
lvvvT
mvvvT
n] (11.3.46)
The structure of the matrix of the shape functions NNNis in this more general case
NNN=[NiIII5NjIII5NkIII5NlIII5NmIII5NnIII5] (11.3.47)
with III5as unit matrices of the size (5,5), the shape functions remai n unchanged.
The total potential energy for this case is, see also (8.3.15 ),
Π(u,v,w,ψ1,ψ2) =1
2/integraldisplay
A(εεεTAAAεεε+κκκTBBBεεε+εεεTBBBκκκ+κκκTDDDκκκ
+εεεsTAAAsεεεs)dx1dx2−/integraldisplay
Apzwdxdy(11.3.48)
and we have to take into consideration the membrane stiffnes s matrix AAAand the
coupling matrix BBBadditionally, the element stiffness matrix consists of fou r parts
KKKE=KKKm
E+KKKmb
E+KKKb
E+KKKs
E (11.3.49)
representing the membrane state ( KKKm
E), the coupling of membrane and bending states
(Kmb
E), the bending state ( KKKb
E) and the transverse shear state ( KKKs
E).
The general form for the element force vector (11.3.39) is un changed, it must be
noted that the loading vector qqqhere has another structure containing loads in three
directions
qqqT=[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 RRR0is
RRR0=
ρ00 0 ρ10
0ρ00 0 ρ1
0 0 ρ00 0
ρ10 0 ρ20
0ρ10 0 ρ2
(11.3.51)
The remarks about the realization of the integrations remai ns unchanged here. Of
course they are all more complicated for this element. Such a n extended element
would be called PL30, because the degree of freedom of all nod al displacements is
30. Further details about this extended element are not give n here.

11.4 Generalized Finite Beam Elements 437
11.4 Generalized Finite Beam Elements
In civil engineering and also in mechanical engineering a sp ecial kind of struc-
tures are used very often structures consisting of thin-wal led elements with sig-
niﬁcant larger dimensions in one direction (length) in comp arison with the dimen-
sions in the transverse direction. They are called beam shap ed 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 t he reduction of the
two-dimensional problem to an one-dimensional one followi ng the way of Vlasov-
Kantorovich. This folded structure model contains all the e nergy 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 s tructure 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 κxi, the shear strains εxsi, the torsional curvatures κxsior the transversal
strains εsiof the strips. Because the inﬂuence of the longitudinal curv atures κxiof
the single strips to the deformation state and the stress sta te of the whole structure
is very small for beam shaped structures, they are neglected generally. The shear
strains εxsiof the strips can be neglected for structures with open cross -sections,
but not in the case of closed cross-sections. In opposite to t his the torsional curva-
tures κxsican be neglected for closed cross-sections, but not for open cross-sections.
Therefore, because we had in mind to ﬁnd a generalized beam mo del as well the
shear strains as the torsional curvatures are considered. A lthough 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-o rdinate functions for a
general cross-section systematically. Therefore as a gene ralized structure model for
beam shaped thin-walled folded plate structures the struct ure model A (see Sects.
10.2.4 and 10.2.5) is chosen, in which only the longitudinal curvatures κxiof 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 st rips the longitudinal curva-
tures κxiare neglected all terms containing w′′
ihave to vanish. It leads together with
equations (10.2.11), (10.2.12) and with ˆAAA7=000,ˆAAA8=000,ˆAAA10=000,ˆAAA17=000,ˆAAA20=000,
ˆAAA23=000,ˆAAA26=000 to a simpliﬁcation of the potential energy equation (10.2. 10)

438 11 Finite Element Analysis
Π=1
2l/integraldisplay
0/bracketleftbigg
UUU′TˆAAA1UUU′+VVVTˆAAA6VVV+UUUTˆAAA3UUU+2UUUTˆAAA25VVV′
+VVV′TˆAAA4VVV′+VVVTˆAAA12VVV+4VVV′TˆAAA9VVV′
+2UUU′TˆAAA14VVV+2UUUTˆAAA2UUU′+2UUU′TˆAAA13VVV′
−2UUU′TˆAAA19VVV−4UUU′TˆAAA18VVV′+2UUU′TˆAAA16VVV
+2VVVTˆAAA5VVV′−2VVVTˆAAA28VVV−4VVVTˆAAA27VVV′
−2UUUTˆAAA22VVV−2VVV′TˆAAA25VVV−4UUUTˆAAA21VVV′
−4VVV′TˆAAA24VVV′+4VVVTˆAAA11VVV′−2(UUUTfffx+VVVTfffs+VVVTfffn)/bracketrightbigg
dx
−(UUUTrrrx+VVVTrrrs+VVVTrrrn)|x=0−(UUUTrrrx+VVVTrrrs+VVVTrrrn)|x=l(11.4.1)
We can see that the one-dimensional energy function contain s 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 t he potential energy
only a C(0)continuity is to satisfy at the element boundaries and there fore it would
be possible to use a two-node element with linear shape funct ions. 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 l ike (11.2.15)
N1(x)=1−3x
l+2×2
l2,N2(x)=4x
l−4×2
l2,N3(x)=−x
l+2×2
l2(11.4.2)
They are shown in Fig. 11.5.
Because a generalized ﬁnite beam element with a general cros s-section shall be
developed at ﬁrst we must ﬁnd a rule to deﬁne the cross-sectio n topology. We will
use for it the proﬁle node concept. For this we will see the mid lines 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 mid dle of each strip there
Fig. 11.12 Three-node gener-
alized beam element1 2 3
x
l

11.4 Generalized Finite Beam Elements 439
are additional proﬁle nodes, they are called secondary proﬁ le nodes. Figure 11.13
shows an example for it. The topology of the thin-walled cros s-section is described
sufﬁciently by the co-ordinates of the main proﬁle nodes. Ad ditionally the stiffness
parameters of each strip must be given. The connections of th e strips in the main
proﬁle nodes are supposed as rigid.
For the generation of the generalized deﬂection co-ordinat e functions ϕ,ψ,ξis
assumed that a main proﬁle node has four degrees of freedom, t he 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 no des lead linear
generalized co-ordinate functions ϕ,ψand cubic functions ξbetween the adjacent
nodes. For an increasing the accuracy the activation of the s econdary proﬁle node
degrees of freedom is optional, they are shown in Fig. 11.15. In this case ϕandψare
quadratic and ξpolynomials of 4th and 5th order between the adjacent main no des.
Therefore a more complex deformation kinematics of the cros s-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 c o-ordinate functions
only in the adjacent intervals of the proﬁle line
Fig. 11.13 Description of a
general cross-section1 1
22
33 4 4
5
5 66
7 78zyy1 s2 s4 s5s6
s3 s7s8
– 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 axia l parallel arranged strips
are shown. Figure 11.17 gives the supplements for slanting a rranged strips.
11.4.3 Element Equations
In the case of non-activated degrees of freedom of the second ary proﬁle nodes we
have a degree of freedom of an element node of four times the nu mber of main
proﬁle nodes (4 nMPN) and the element displacement vector consists of 12 nMPN
components
vvvT=[vvv1vvv2vvv3],vvvj=/bracketleftbigg¯uuuj
¯vvvj/bracketrightbigg
(11.4.3)
¯uuuj,¯vvvjcontain the values of the generalized displacement functio ns at the node j.
The displacement vector uuu(x)contains here the generalized displacement functions
UUU(x)andVVV(x)and in accordance with Eq. (11.1.1) we obtain for the interpo lation
uuu(x)=/bracketleftbiggUUU(x)
VVV(x)/bracketrightbigg
=NvNvNv (11.4.4)
The matrix of the shape functions for the chosen three-node e lement is
NNN=[N1(x)III N2(x)III N3(x)III], (11.4.5)
where IIIare the unit matrices of the size 4 nMPN and with this the matrix NNNhas
the format (4 nMPN, 12nMPN). 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 11 11 10.25
0.25
1/16
−1/8dϕ1=1−s
dϕ2=s
dϕ3=s
d−/parenleftigs
d/parenrightig2
ψ1=1−s
d
ξ1=0ψ2=s
d
ξ2=0ψ3=s
d−/parenleftigs
d/parenrightig2
ξ3=0
ξ4=1−3/parenleftigs
d/parenrightig2
+2/parenleftigs
d/parenrightig3
ψ4=0ξ5=3/parenleftigs
d/parenrightig2
−2/parenleftigs
d/parenrightig3
ψ5=0ξ8=/parenleftigs
d/parenrightig2
−2/parenleftigs
d/parenrightig3
+/parenleftigs
d/parenrightig4
ψ8=0
ξ6=s/bracketleftbigg
1−2s
d+/parenleftigs
d/parenrightig2/bracketrightbigg
ψ6=0ξ7=s/bracketleftbigg
−s
d+/parenleftigs
d/parenrightig2/bracketrightbigg
ψ7=0ξ9=/parenleftigs
d/parenrightig2
−4/parenleftigs
d/parenrightig3
+5/parenleftigs
d/parenrightig4
−2/parenleftigs
d/parenrightig5
ψ4=0
Fig. 11.16 Generalized co-ordinate functions for axial parallel arra nged strips

442 11 Finite Element Analysis
1
23x
zy
n,wx,u
s,vα
ψ1=/parenleftig
1−s
d/parenrightig
cosα
ξ1=−/bracketleftbigg
1−3/parenleftigs
d/parenrightig2
+2/parenleftigs
d/parenrightig3/bracketrightbigg
sinα
ψ2=s
dcosα
ξ2=−/bracketleftbigg
3/parenleftigs
d/parenrightig2
−2/parenleftigs
d/parenrightig3/bracketrightbigg
sinα
ψ4=/parenleftig
1−s
d/parenrightig
sinα
ξ4=/bracketleftbigg
1−3/parenleftigs
d/parenrightig2
+2/parenleftigs
d/parenrightig3/bracketrightbigg
cosα
ψ5=s
dsinα
ξ5=/bracketleftbigg
3/parenleftigs
d/parenrightig2
−2/parenleftigs
d/parenrightig3/bracketrightbigg
cosα
Fig. 11.17 Supplements for slanting arranged strips

11.4 Generalized Finite Beam Elements 443
displacement functions we have to write
UUU(x)=LLLT
10uuu(x)=LLLT
10NvNvNv,VVV(x)=LLLT
01uuu(x)=LLLT
01NvNvNv (11.4.6)
with the matrices
LLLT
10=[III000],LLLT
01=[000III] (11.4.7)
In the ﬁrst case ( LLL10)IIIis a unit matrix of the size nMNP and the null matrix has the
format ( nMPN, 3nMPN), in the second case ( LLL01)IIIis a unit matrix of the size 3 nMPN
and the null matrix has the format (3 nMPN,nMPN).
Of course in the case of activated degrees of freedom of the se condary pro-
ﬁle nodes all the dimensions given above are increased corre spondingly. Inserting
the generalized displacement functions (11.4.6) into the p otential energy equation
(11.4.1) we obtain
Π=1
2vvvTKvKvKv−fffTvvv (11.4.8)
The condition
∂Π
∂vvv=000 (11.4.9)
leads to the element equation
KvKvKv=fff (11.4.10)
with the symmetric element stiffness matrix
KKK=
KKK11KKK12KKK13
KKKT
12KKK22KKK23
KKKT
13KKKT
23KKK33
 (11.4.11)
For the sub-matrices Kmnwe ﬁnd the general equation
Kmn=3
∑
h=1
ˆAAA1hImn1
+ˆAAA2hImn2
+ˆAAAT
2hImn3
+ˆAAA3hImn4(ˆAAA13h−2ˆAAA18h)Imn1
+(ˆAAA15h−2ˆAAA21hImn2
+(ˆAAA14h−ˆAAA19h)Imn3
+(ˆAAA16h−ˆAAA22h)Imn4
(ˆAAAT
13h−2ˆAAAT
18h)Imn1
+(ˆAAAT
15h−2ˆAAAT
21h)Imn2
+(ˆAAAT
14h−ˆAAAT
19h)Imn3
+(ˆAAAT
16h−ˆAAAT
22h)Imn4(ˆAAA4h+ˆAAA9h−2ˆAAA24h−2ˆAAAT
24h)Imn1
+(ˆAAA5h+2ˆAAA11h−ˆAAAT
25h−2ˆAAA27h)Imn2
+(ˆAAAT
5h+2ˆAAA11h−ˆAAA25h−2ˆAAAT
27h)Imn3
+(ˆAAA6h+ˆAAA12h−ˆAAA28h−ˆAAAT
28h)Imn4

(11.4.12)
with

444 11 Finite Element Analysis
Imn1=l/integraldisplay
0NhN′
mN′
ndx,Imn2=l/integraldisplay
0NhNmN′
ndx,
Imn3=l/integraldisplay
0NhN′
mNndx,Imn4=l/integraldisplay
0NhNmNndx
To include approximately slight non-prismatic structures the matrices of the stiff-
ness parameters ˆAAAi, see Eq. (10.2.12), are interpolated in the element in the sa me
manner as the displacements
ˆAAAi=3
∑
h=1ˆAAAihNh (11.4.13)
ˆAAAihare the matrices at the nodes h=1,2,3.
The element force vector is obtained as
fff=
fff1
fff2
fff3
 (11.4.14)
with the sub-vectors
fffm=3
∑
h=1
fffxhl/integraldisplay
0NhNmdx
(((fsh+fnh)l/integraldisplay
0NhNmdx
(11.4.15)
Here fffxh,fffsh,fffnhare the generalized load vectors, see Eq. (10.2.13), at the n odes
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 bou ndary conditions of the
whole structure are to consider, otherwise this matrix is si ngular, if the structure
is not ﬁxed kinematically. The solution of the system equati ons lead to the nodal
displacements and with them the strains and curvatures in th e single strips of each
element can be calculated, see Eqs. (10.2.6) and (10.2.11),

11.4 Generalized Finite Beam Elements 445
εx(x,si) =3
∑
h=1N′
h¯uuuT
hϕϕϕ, εs(x,si) =3
∑
h=1Nh¯vvvT
hψψψ•,
εxs(x,si) =3
∑
h=1(Nh¯uuuT
hϕϕϕ•+N′
hvvvT
hψψψ),
κs(x,si) =−3
∑
h=1Nh¯vvvT
hξξξ••, κxs(x,si) =−23
∑
h=1N′
h¯vvvT
hξξξ•(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

εx
εs+nkκs
εxs+nkκxs
 (11.4.17)
These stress resultants are related on the strip co-ordinat e axes xandsi. Therefore, it
is necessary to transform them into the material co-ordinat e system of the kth lamina
(for the transformation relationship see Table 4.1)

NLk
NTk
NLTk
=
cos2αk sin2αk 2sin αkcosαk
sin2αk cos2αk−2sin αkcosαk
−sinαkcosαksinαkcosαkcos2αk−sin2αk

Nxk
Nsk
Nxsk
(11.4.18)
Than the stresses of the kth lamina are obtained
σLk=NLk
tk,σTk=NTk
tk,τLTk=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

εLk
εTk
εLTk
=QQQ′
k−1
NLk
NTk
NLTk
 (11.4.20)
There QQQ′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 h armonic vibrations for
the considered generalized beam element to the element equa tion
(KKK−ω2MMM)vvv=000 (11.4.21)

446 11 Finite Element Analysis
KKKis the element stiffness matrix, see Eqs. (11.4.11) and (11. 4.12), and MMMis the ele-
ment mass matrix. The element mass matrix is obtained with th e matrices ˆBBB1,ˆBBB2,ˆBBB3,
see Eqs. (10.2.38)
MMM=
MMM11MMM12MMM13
MMMT
12MMM22MMM23
MMMT
13MMMT
23MMM33
 (11.4.22)
with
MMMmn=3
∑
h=1/bracketleftbiggˆBBB1hImn4 000
000(ˆBBB2h+ˆBBB3h)Imn4/bracketrightbigg
(11.4.23)
There the ˆBBBmatrices are also interpolated in the element by using the sh ape func-
tions. In this way slight non-prismatic structures are cons iderable 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 matr ix. After consider-
ation the boundary conditions the eigen-value problem can b e 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 FE M program packages
is increasing. In universal FEM program systems we have gene rally the possibility
to consider anisotropic material properties, e.g. in the pr ogram system COSMOS/M
we can use volume elements with general anisotropic materia l 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 n o spe-
cial laminate plate elements, the laminate shell elements a re used also for the anal-
ysis of laminate and sandwich plates. Perhaps, because of th e higher signiﬁcance
of two-dimensional laminate structures in comparison with beam shaped structures
laminate beam elements are missing in nearly all universal F EM program systems.
The generalized beam elements, Sect. 11.4, are e.g. impleme nted only in the FEM
program system COSAR.
For the following numerical examples the program system COS MOS/M is used.
In COSMOS/M a three node and a four node thin laminate shell el ement are
available (SHELL3L; SHELL4L). Each node has 6 degrees of fre edom. The element
can consist of up to 50 layers. Each layer can have different m aterial parameters,
different thicknesses and especially different angles of ﬁ bre directions. We have no
restrictions in the stacking structure, symmetric, antisy mmetric and nonsymmetric
structures are possible. The four nodes of the SHELL4L eleme nt must not arranged
in-plane. By the program in such case a separation is done int o two or four triangular
partial elements. Further there is a SHELL9L element availa ble. It has additional

11.5 Numerical Results 447
nodes at the middles of the four boundaries and in the middle o f 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 lamina te shell element
SHELL4L shall be demonstrated. At ﬁrst a thin-walled beam sh aped laminate struc-
ture with L-cross-section under a concentrated force loadi ng is considered, and the
second example is a thin-walled laminate pipe under torsion al loading. In both cases
the inﬂuence of the ﬁbre angles in the layers is tested. The us e of the laminate shell
element for the static and dynamic analysis of a sandwich pla te is shown in the
third example. A buckling analysis of a laminate plate is dem onstrated 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,νxy,νyx,Gxy. It is loaded by a concentrated force F, see Fig. 11.18.
The material parameters are
LF
+α+α
−α
33
4
30040010
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,
νxy=0.30,νyx=0.021
The ﬁbre angle αshall 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. 1 1.20, 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 ( σx) and
perpendicular to it ( σy) 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,
σlay1,max,x= 7,487 kN/cm2,σlay1,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 11.2 2. The results show
that for such a beam shaped structure the main stresses are ly ing 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 materia l parameters Ex,Ey,
νxy,νyx,Gxyis ﬁxed at left end and loaded by a torsional moment, see Fig. 1 1.23.
The material parameters are the same as in the previous examp le:
Ex=1,53 104kN/cm2,Ey=1,09 103kN/cm2,Gxy=560 kN/cm2,
νxy=0,30,νyx=0,021
The ﬁbre angle αshall 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)xy
z
Fig. 11.20 Cantilever beam
deformed shapexy
z

11.5 Numerical Results 449
vy vzFibre AngleDisplacements/cm
0◦ 10◦20◦ 30◦40◦0.511.522.533.544.5
Fig. 11.21 Displacements of the corner point at the free edge
σxσyFibre AngleStresses kN/cm2
0◦ 10◦20◦ 30◦40◦0246810
Fig. 11.22 Maximal stresses at the bottom of layer No. 1

450 11 Finite Element Analysis
LMt
+α
−α 33
D
6
Fig. 11.23 Laminate pipe: geometry, cross-section and stacking seque nce ( Mt=1200 kNcm,
L=2 m, D=200 mm)
ters and properties into COSMOS/M the FEM model can be illust rated (Fig. 11.24).
From the results of the analysis only the twisting angle of th e free edge shall
be considered here. For this we have to list the results for th e displacements in y-
direction of two nodes at the free edge, lying in opposite to e ach other, e.g. the nodes
255 and 663 in our FE model. The twisting angle is calculated b y
ϕ=(vy,255−vy,663)/D
Carrying out the analysis for all ﬁbre angles we obtain the re sults, given in Fig.
11.25. The diagram demonstrates the well known fact that in c ase of pure shear
loading the main normal stresses are lying in a direction wit h 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 ﬁbr e angels yield the greatest
shear rigidity.
Fig. 11.24 FE-model of Lam-
inate Pipe in COSMOS/M
(800 elements, 816 nodes)xy
z

11.5 Numerical Results 451
Fibre AngleTwisting Angle
0◦ 10◦20◦30◦40◦12468
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 boun daries and simply
supported at one of the long boundaries. The cover sheets con sist of an aluminium
alloy and the core of foam of polyurethan. The material param eters are:
AlZnMgCu0.5F450:
ρ=2.7·103kg/m3,E=7.0·1010N/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 p ressure loading p,
a vibration analysis will be performed is asked. We have to ca lculate the 4 lowest
eigenfrequencies and the mode shapes, respectively. Note t hat we use in this exam-
ple only the basic units of the SI-system, so we avoid the calc ulation of correction
factors for the obtained eigenfrequencies.
The FE-model is given in Fig. 11.27. The static analysis lead s the displacements
and stresses. We consider only the stresses of the bottom of t he lower cover sheet
(layer 3, top). The Figs. 11.28 and 11.29 show the plots of str ess distributions for the
ﬂexural stresses σxandσz, Fig. 11.30 the distribution of the von Mises equivalent
stress. The lowest 4 eigenfrequencies and their 4 mode shape s are shown in the
following Fig. 11.31. The static and frequency computation s conﬁrm the successful
application of the SHELL4L element for a sandwich plate.

452 11 Finite Element Analysis
6.0 m4.0 m p=1200 N/m2
aluminium alloy
polyurethan foam24 30 mm
Fig. 11.26 Sandwich plate
Fig. 11.27 FE-model of Sandwich Plate in COSMOS/M (600 elements, 651 no des)
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
xy
z
2.3717E+0071.4703E+0075.6887E+006-3.3255E+006-1.2340E+007-2.1354E+007-3.0368E+007-3.9382E+007-4.8396E+007σx
Fig. 11.28 Stresses in x-direction for the bottom of the lower cover sheet
xy
z
1.1711E+0078.1902E+0064.6695E+0061.1488E+006-2.3719E+006-5.8926E+006-9.4133E+006-1.2934E+007-1.6455E+007σz
Fig. 11.29 Stresses in z-direction for the bottom of the lower cover sheet

454 11 Finite Element Analysis
xy
z
4.4196E+0073.9218E+0073.4240E+0072.9262E+0072.4284E+0071.9306E+0071.4328E+0079.3496E+0064.3715E+006von Mises
Fig. 11.30 V on 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,
νxy=0.30,νyx=0.021
For the stacking structure two cases shall be considered, a s ymmetric (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 calculate s a buckling factor νB
to multiply the unit loading for obtaining the buckling load .
The FE-model created in COSMOS/M by the input of all properti es and param-
eters is shown in Fig. 11.33. The calculation for α=30◦leads to a buckling factor
νB=1,647 and to the buckling mode shown in Fig. 11.34. In the same ma nner the
calculations for the other ﬁbre angels and for the antisymme tric laminate were per-
formed. The results for the buckling factors are shown in a di agram in Fig. 11.35.
The buckling modes are symmetric to the symmetric axis in loa ding 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. Th e buckling modes for
the antisymmetric laminate are very similar but not identic al to the buckling modes
1.5 m1.0 m
+α+α
+α+α
−α−α
−α−α4×25case 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 an gle near 45◦leads to
the highest buckling load for a quadratic plate. It shall be n oted that the antimetric
stacking sequence of the laminate improved the buckling sta bility.
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 o f the general purpose
FEM-program system COSAR. The real handling of the FEM-proc edures are not
given here, but two simple examples shall demonstrate the po ssibilities of these el-
ements for global static or dynamic structure analysis.
Figure 11.36 shows thin-walled cantilever beams with open o r closed cross-
sections and different loadings. All these beam structure m odels have equal length,
hight and width and also the total thicknesses of all laminat e strips are equal, inde-
pendent of the number of the layers.

11.5 Numerical Results 457
n, symm. n, antim.Fibre AngleBuckling Value
0 10 20 30 40 50 60 70 80 9011.21.41.61.822.2
Fig. 11.35 Results of the Buckling Analysis
The stacking structure may be symmetric or antisymmetric. F igure 11.37 shows
the two considered variants: case A with three laminae and sy mmetric 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, νLT=0.30,
ET=10900 N/mm2, νTL=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 proﬁ le nodes for both
cross-sections but three secondary proﬁle nodes for the ope n and four for the closed
cross-section. The numerical analysis shall demonstrate t he inﬂuence of the stacking
structure. Figure 11.39 illustrates the relative changes o f 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 o f the vertical
deﬂections wBin comparison to the wAvalues in the case of symmetric stacking.
Generally, only two degrees of freedom of secondary proﬁle n odes were activated.
In a separate analysis the inﬂuence of a higher degrees of fre edom in the sec-
ondary proﬁle nodes was considered. As a result it can be reco mmended that for
antisymmetric layer structures and for open proﬁles more th an two degrees of free-

458 11 Finite Element Analysis
F=260 kN
FF
x xxxx
y yyyy
z
zzzz
400 400800 8005000
Fig. 11.36 Cantilever beams, geometry and loading

11.5 Numerical Results 459
a bss
n n−α−α+α
+α+α
Fig. 11.37 Stacking structure of the laminates. aSymmetric sandwich, btwo-layer
4 44
33 3 3
2
22 2
1 1 11
Fig. 11.38 Cross-sections with main proﬁle nodes ( •) and secondary proﬁle nodes ( ×)
dom should be activated. Ignoring the activation of seconda ry proﬁle node degrees
of freedom leads to nonrealistic structure stiffness. The s tructure model is to stiff
and therefore the deﬂections are to small.
The second example concerned the eigen-vibration analysis . For the closed cross-
section with symmetric layer stacking the inﬂuences of the d egree of freedom of
secondary proﬁle nodes and of the variation of the ﬁbre angle s were considered. As
a result it can be stated that the inﬂuence of higher degrees o f freedom of the sec-
ondary proﬁle nodes is negligible but the inﬂuence of the ﬁbr e angles is signiﬁcant.
Figure 11.40 illustrates the inﬂuence of the ﬁbre angle vari ations on the eigenfre-
quencies of the beam, which can be used for structure optimiz ation.
Summarizing Sect. 11.5 one have to say that only a small selec tion of one- and
two-dimensional ﬁnite elements was considered. Many ﬁnite plate and shell ele-
ments were developed using equivalent single layer theorie s for laminated struc-
tures but also multi-layered theories are used. Recent revi ew articles give a detailed
overview on the development, implementation and testing of different ﬁnite lami-
nate and sandwich elements.

460 11 Finite Element Analysis
/G30/G32/G34/G36/G38/G31 /G30/G31 /G32/G31 /G34
/G30 /GB0 /G31 /G30 /GB0 /G32 /G30 /GB0 /G33 /G30 /GB0 /G34 /G30 /GB0 /G35 /G30 /GB0 /G36 /G30 /GB0 /G37 /G30 /GB0 /G38 /G30 /GB0 /G39 /G30 /GB0∆w
%
Fig. 11.39 Relative changes (wB−wA)/wA=∆w(α)of the vertical deﬂections wAandwB
/G30/G32 /G30/G34 /G30/G36 /G30/G38 /G30/G31 /G30 /G30/G31 /G32 /G30/G31 /G34 /G30
/G30 /GB0 /G31 /G30 /GB0 /G32 /G30 /GB0 /G33 /G30 /GB0 /G34 /G30 /GB0 /G35 /G30 /GB0 /G36 /G30 /GB0 /G37 /G30 /GB0 /G38 /G30 /GB0 /G39 /G30 /GB0Hz
1. EF
2. EF
3. EF
4. EF
Fig. 11.40 Inﬂuence of the ﬁbre angle on the ﬁrst four eigen-frequencie s of the cantilever box-
beam

Part VI
Appendices

This part is focussed on some basics of mathematics and mecha nics like matrix
operations (App. A), stress and strain transformations (Ap p. B), differential opera-
tors for rectangular plates (App. C) and differential opera tors for circular cylindrical
shells (App. D). In addition, the Krylow functions as soluti on 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 temperat ure, time, etc. Last but
not least there are given some references for further readin g (App. G).

Appendix A
Matrix Operations
The following short review of the basic matrix deﬁnitions an d operations will pro-
vide a quick reference and ensure that the particular use of v ector-matrix notations
in this textbook is correct understood.
A.1 Deﬁnitions
1.Rectangular matrix
AAA=
a11a12··· ··· a1n
a21a22··· ··· 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 jis called the (i,j)-element of AAA. There
are two important special cases
aaa=
a1
…
am
=[ai]
is am×1 matrix or column vector, while
aaaT=/bracketleftbig
a1···an/bracketrightbig
=[ai]T
is a 1×nmatrix or row vector.
With aaarespectively aaaTa matrix AAAcan be written
463 © Springer Nature Singapore Pte Ltd. 2018
H. Altenbach et al., Mechanics of Composite Structural
Elements , https://doi.org/10.1007/978-981-10-8935-0

464 A Matrix Operations
AAA=/bracketleftbigaaa1···aaan/bracketrightbig
,aaaj=
a1j
…
am j
,j=1,…, n
or
AAA=
aaaT
1
···
aaaT
m
,aaaT
j=/bracketleftbigaj1···ajn/bracketrightbig
,j=1,…, m
Ifn=mthe matrix is square of the order n×n. For a square matrix the elements
ai jwith i=jdeﬁne the principal matrix diagonal and are located on it.
2.Determinant of a square matrix A AA
|AAA|=/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsinglea11a12··· ··· a1n
a21a22··· ··· a2n
……………
an1an2··· ··· ann/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle=|ai j|=detAAA
The determinant of a matrix AAAwith elements ai jis given by
|AAA|=a11M11−a12M12+a13M13−…(−1)1+na1nM1n,
where the minor Mi jis the determinant of the matrix |AAA|with missing row iand
column j. Note the following properties of determinants:
•Interchanging two rows or two columns changing the sign of |AAA|.
•If all elements in a row or a column of AAAare zero then |AAA|=0.
•Multiplication by a constant factor cof all elements in a row or column of AAA
multiplies |AAA|byc.
•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 landmthen the determinant
must be zero.
3.Regular matrix
A square matrix AAAis regular if |AAA|/ne}ationslash=0.
4.Singular matrix
A square matrix AAAis singular if |AAA|=0.
5.Trace of a matrix
The trace of a square matrix AAAis the sum of all elements of the principal diagonal,
i.e.
trAAA=m
∑
k=1akk
6.Rank of a matrix
The rank rk (AAA)of am×nmatrix AAAis the largest value of rfor which there exist a

A.2 Special Matrices 465
r×rsubmatrix of AAAthat is non-singular. Submatrices are smaller arrays of k×k
elements ai jof the matrix AAA, i.e. k≤mifm≤nork≤nifn≤m.
A.2 Special Matrices
In the following the δi jdenotes the Kronecker symbol
δi j=/braceleftbigg0i/ne}ationslash=j
1i=j
1.Null matrix 000
All elements ai jof am×nmatrix are identically equal zero
ai j≡0,i=1,…, m,j=1,…, n
2.Diagonal matrix D DD=diag[aii]=diag[ai jδi j]
A diagonal matrix is a square matrix in which all elements are zero except those
on the principal diagonal
ai j=0,i/ne}ationslash=j,ai j/ne}ationslash=0,i=j
3.Unit matrix I II=[δi j]
A unit or identity matrix is a special case of the diagonal mat rix for which ai j=1
when i=jandai j=0 when i/ne}ationslash=j.
4.Transpose A AATof a matrix A AA
The transpose of a matrix AAAis found by interchanging rows and columns. If
AAA= [ai j]follow AAAT= [aT
i j]with aT
i j=aji. A transposed matrix is denoted by a
superscript T. Note (AAAT)T=AAA
5.Symmetric matrix A AAS
A square matrix AAAis said to be symmetric if for all i/ne}ationslash=j ai j=aji, i.e. AAA=AAAT. A
symmetric matrix is denoted by a superscript S.
6.Skew-symmetric matrix A AAA
A square matrix AAAis said to be skew-symmetric if all principal diagonal eleme nts
are equal zero and for all i/ne}ationslash=j ai j=−aji, i.e. AAA=−AAAT. A skew-symmetric
matrix is denoted by a superscript A.
7. Any matrix can be decomposed in a symmetric and a skew-symm etric part in a
unique manner
AAA=AAAS+AAAA
Proof. Since
AAAS=1
2(AAA+AAAT)
and
AAAA=1
2(AAA−AAAT)

466 A Matrix Operations
the sum AAAS+AAAAis equal to AAA.
A.3 Matrix Algebra and Analysis
1.Addition and subtraction
Am×nmatrix AAAcan be added or subtracted to a m×nmatrix BBBto form a m×n
matrix CCC
AAA±BBB=CCC,ai j±bi j=ci j,i=1,…, m,j=1,…, n
Note AAA+BBB=BBB+AAA,AAA−BBB=−(BBB−AAA)=−BBB+AAA,(AAA±BBB)T=AAAT±BBBT.
2.Multiplication
•Multiplication the matrix AAAby a scalar αinvolves the multiplication of all
elements of the matrix by the scalar
αAAA=AAAα=[αai j],
(α±β)AAA=αAAA±βAAA,
α(AAA±BBB)=αAAA±αBBB
•The product of a (1×n)matrix (row vector aaaT) and a(n×1)matrix (column
vector bbb) forms a (1×1)matrix, i.e. a scalar α
aaaTbbb=bbbTaaa=α,α=n
∑
k=1akbk
•The product of a (m×n)matrix AAAand a(n×1)column vector bbbforms a
(m×1)column vector ccc
AbAbAb=ccc,ci=n
∑
j=1ai jbj=ai1b1+ai2b2+…+ainbn,i=1,2,…, m
The forgoing product is only possible if the number of column s ofAAAis equal
the number of rows of bbb.
Note A.1. b bbTAAAT=cccT
•IfAAAis a(m×n)matrix and BBBa(p×q)matrix the product ABABAB=CCCexists if
n=p, in which case CCCis a(m×q)matrix. For n=pthe matrix AAAandBBBare
said to be conformable for multiplication. The elements of t he matrix CCCare
ci j=n=p
∑
k=1aikbk j,i=1,2,…, m,j=1,2,…, q
Note A.2. AB ABAB/ne}ationslash=BABABA,AAA(BBB±CCC)=ABABAB±ACACAC,(ABABAB)T=BBBTAAAT

A.3 Matrix Algebra and Analysis 467
(l×m)
AAA(m×n)
BBB(n×p)
CCC=(l×p)
DDD
3.Inversion and division
AIAIAI=IAIAIA=AAA,AAA−1AAA=AAAAAA−1=III,/parenleftbig
AAA−1/parenrightbig−1=AAA
The matrix inversion is based on the existence of a n×nunit matrix IIIand a square
n×nmatrix AAA.AAA−1is the inverse of AAAwith respect to the matrix multiplication
AAAAAA−1=AAA−1AAA=III. IfAAA−1exist, the matrix AAAis invertible or regular, otherwise
non-invertible or singular. Matrix division is not deﬁned.
Note A.3. (ABABAB)−1=BBB−1AAA−1,(ABCABCABC)−1=CCC−1BBB−1AAA−1,…
•Cofactor matrix
With the minors Mi jintroduced above to deﬁne the determinant |AAA|of a matrix
AAAa so-called cofactor matrix AAAc=[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 AAAis the transpose of the cofactor
matrix
adjAAA=(AAAc)T
Note A.4. Because symmetric matrices possess symmetric cofactor mat rices
the adjoint of a symmetric matrix is the cofactor matrix itse lf
adjAAAS=/parenleftbig
AAAS/parenrightbigc
It can be shown that
AAA(adjAAA)=|AAA|III
i.e.
AAA(adjAAA)
|AAA|=III=AAAAAA−1⇒AAA−1=adjAAA
|AAA|
Inverse matrices have some important properties
/parenleftbig
AAAT/parenrightbig−1=/parenleftbig
AAA−1/parenrightbigT
and if AAA=AAAT
AAA−1=/parenleftbig
AAA−1/parenrightbigT
i.e. the inverse matrix of a symmetric matrix AAAis also symmetric.
Note A.5. Symmetric matrices posses symmetric transposes, symmetri c cofac-
tors, symmetric adjoints and symmetric inverses.

468 A Matrix Operations
4.Powers and roots of square matrices
Ifn×nmatrix AAAis conformable with itself for multiplication, one may deﬁn e its
powers
AAAn=AAAAAA…AAA,
and for symmetric positive semideﬁnite matrices
AAA1
n=n√
AAA,AAA−n=/parenleftbig
AAA−1/parenrightbign
and if AAAis regular
(AAAm)n=AAAmn,AAAmAAAn=AAAm+n,
5.Matrix eigenvalue problems
The standard eigenvalue problem of a quadratic n×nmatrix AAAis of the form:
ﬁnd(λ,xxx)with/ne}ationslash=/ne}ationslash=/ne}ationslash=000 such that
AAAxxx=λxxxor(AAA−λIII)xxx=000
KKK=[AAA−λIII]is called the characteristic matrix of AAA, detKKK=0 is called the char-
acteristic determinant or equation of AAA. The characteristic determinant produces
a characteristic polynomial with powers of λup to λnand therefore when it set
equal zero having nroots which are called the eigenvalues. If the characterist ic
equation has ndistinct roots, the polynomial can be factorized in the form
(λ−λ1)(λ−λ2)…(λ−λn)=0
If we put λ=0 in the characteristic equation we get
detAAA=λ1λ2…λn
Inserting any root λiinto the standard eigenvalue equation leads to
[AAA−λiIII]xxxi=000,i=1,2,…, n
xxxiare the eigendirections (eigenvectors) which can be comput ed from the last
equation considering the orthogonality condition. A nontr ivial solution exists if
and only if
det[AAA−λiIII]=0
Note A.6. If we have the 3×3 symmetric matrix the eigendirection xxxican be com-
puted for each λifrom the polynomial of third order. Three different solutio ns
are possible:
•all solutions λiare distinct – three orthogonal eigendirections can be comp uted
(but their magnitudes are arbitrary),
•one double solution and one distinct solutions – only one eig endirection can
be computed (its magnitudes is arbitrary), and
•all solutions are identically – no eigendirections can be co mputed.

A.3 Matrix Algebra and Analysis 469
Anyway, the orthogonality condition xxxT·xxx=1 should be taken into account.
The general eigenvalue problem is given in the form
AAAxxx=λBBBxxx
which can be premultiplied by BBB−1to produce the standard form
BBB−1AAAxxx=BBB−1λBBBxxx=(BBB−1AAA)xxx=λIIIxxx=λxxx
resulting in
(BBB−1AAA−λIII)xxx=000
Note A.7. In the case of non-symmetric matrix AAAthe eigenvalue can be complex.
6.Differentiating and integrating
•To differentiate a matrix one differentiates each matrix el ement ai jin the con-
ventual manner.
•To integrate a matrix one integrates each matrix element ai jin the conventual
manner. For deﬁnite integrals, each term is evaluated for th e limits of integra-
tion.
7.Partitioning of matrices
A useful operation with matrices is partitioning into subma trices. These subma-
trices may be treated as elements of the parent matrix and man ipulated by the
standard matrix rules reviewed above. The partitioning is u sually indicated by
dashed partitioning lines entirely through the matrix
MMM=[mi j]=
AAA…BBB
··· ··· ···
CCC…DDD

For a m×nmatrix MMMwe may have submatrices AAA(i×j),BBB(i×p),CCC(k×j),
DDD(k×p)with i+k=m,j+p=n, i.e.
MMMm×n=
AAAi×j…BBBi×(n−j)
··· ··· ···
CCC(m−i)×j…DDD(m−i)×(n−j)
,

AAA…BBB
··· ··· ···
CCC…DDD
±
EEE…FFF
··· ··· ···
GGG…HHH
=
AAA±EEE…BBB±FFF
··· ··· ···
CCC±GGG…DDD±HHH
,

470 A Matrix Operations

AAA…BBB
··· ··· ···
CCC…DDD

EEE…FFF
··· ··· ···
GGG…HHH
=
AEAEAE+BGBGBG…AFAFAF+BHBHBH
··· ··· ···
CECECE+DGDGDG…CFCFCF+DHDHDH

The multiplications are only deﬁned if the correspondent ma trices are con-
formable for multiplication

AAA…BBB
··· ··· ···
CCC…DDD
T
=
AAAT…CCCT
··· ··· ···
BBBT…DDDT

If the matrix
MMM=
AAA…BBB
··· ··· ···
CCC…DDD

is symmetric ( MMM=MMMT), it follows AAA=AAAT,DDD=DDDT,BBB=CCCT,CCC=BBBT

Appendix B
Stress and Strain Transformations
Stress and strain transformations under general orthogona l coordinate transforma-
tioneee′=RRReeeore′
i=Ri jej:
1.σ′
p=Tσ
pqσq. The matrix [Tσ
pq]is deﬁned by

R2
11 R2
12 R2
13 2R12R13 2R11R13 2R11R12
R2
21 R2
22 R2
23 2R22R23 2R21R23 2R21R22
R2
31 R2
32 R2
33 2R32R33 2R31R33 2R31R32
R21R31R22R32R23R33R22R33+R23R32R21R33+R23R31R21R32+R22R31
R11R31R12R32R13R33R12R33+R13R32R11R33+R13R31R11R32+R12R31
R11R21R12R22R13R23R12R23+R13R22R11R23+R13R21R11R22+R12R21

2.ε′
p=Tε
pqεq. The matrix [Tε
pq]is deﬁned by

R2
11 R2
12 R2
13 R12R13 R11R13 R11R12
R2
21 R2
22 R2
23 R22R23 R21R23 R21R22
R2
31 R2
32 R2
33 R32R33 R31R33 R31R32
2R21R312R22R322R23R33R22R33+R23R32R21R33+R23R31R21R32+R22R31
2R11R312R12R322R13R33R12R33+R13R32R11R33+R13R31R11R32+R12R31
2R11R212R12R222R13R23R12R23+R13R22R11R23+R13R21R11R22+R12R21

3. Rotation about the eee1-direction, Fig. B.1:
[Ri j]=
1 0 0
0c s
0s c
,eee′=1
RRR eee
471 © Springer Nature Singapore Pte Ltd. 2018
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′
1x2x3
x′
2 x′
3
eee1,eee′
1eee2eee3
eee′
2 eee′
3Φ
Φ
Fig. B.1 Rotation about the eee1-direction
[1
Tσ
pq]=
1 0 0 0 0 0
0c2s22cs0 0
0s2c2−2cs0 0
0−cs cs c2−s20 0
0 0 0 0 c−s
0 0 0 0 s c
,σσσ′=1
TTTσσσσ
[1
Tε
pq]=
1 0 0 0 0 0
0c2s2cs 0 0
0s2c2−cs0 0
0−2cs2cs c2−s20 0
0 0 0 0 c−s
0 0 0 0 s c
,εεε′=1
TTTεεεε

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

L11L12L13
L22L23
sym L33

u
v
w
=
0
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=−/bracketleftbigg
B11∂3
∂x3
1+3B16∂3
∂x2
1∂x2+(B12+2B66)∂3
∂x1∂x2
2+B26∂3
∂x3
2/bracketrightbigg
,
473 © Springer Nature Singapore Pte Ltd. 2018
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=−/bracketleftbigg
B22∂3
∂x3
2+3B26∂3
∂x1∂x2
2+(B12+2B66)∂3
∂x2
1∂x2+B16∂3
∂x3
1/bracketrightbigg
2. General symmetric laminates

L11L120
L220
sym L33

u
v
w
=
0
0
p

Bi j=0, i.e. L13=L31=0,L23=L32=0.L11,L22,L33,L12as above in 1.
3. Balanced symmetric laminates

L11L120
L220
sym L33

u
v
w
=
0
0
p

In addition to 2. both A16andA26are zero, i.e. L13=L31=0,L23=L32=0 and
L11,L22andL33simplify with A16=A26=0.
4. Cross-ply symmetric laminates

L11L120
L220
sym L33

u
v
w
=
0
0
p

In addition to 3. both D16andD26are zero, i.e. L33simpliﬁes.
5. Balanced unsymmetric laminates

L11L12L13
L22L23
sym L33

u
v
w
=
0
0
p

With A16=A26=0 only the operators L11,L22andL12of 1. can be simpliﬁed.
6. Cross-ply unsymmetric laminates

L11L12L13
L22L23
sym L33

u
v
w
=
0
0
p

In addition to 5., D16,D26,B16andB26are 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˜L140
˜L22˜L23˜L240
˜L33˜L34˜L35
˜L44˜L45
S Y M ˜L55

u
v
ψ1
ψ2
w
=
0
0
0
0
p

with ˜L11=L11,˜L22=L22,˜L12=L12(theLi jcan 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=−/parenleftbigg
ks
55A55∂2
∂x2
1+ks
45A45∂2
∂x1∂x2+ks
44A44∂2
∂x2
2/parenrightbigg
,
˜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=−/parenleftbigg
ks
55A55∂
∂x1+ks
45A45∂
∂x2/parenrightbigg
,
˜L45=˜L54=−/parenleftbigg
ks
45A45∂
∂x1+ks
44A44∂
∂x2/parenrightbigg
with ks
45=/radicalbigks
44ks
55.
2. General symmetric laminates
Bi j=0, i.e. ˜L13=˜L31,˜L14=˜L41,˜L23=˜L32and˜L24=˜L42are zero
/bracketleftbigg˜L11˜L12
˜L12˜L22/bracketrightbigg/bracketleftbiggu
v/bracketrightbigg
=0

˜L33˜L34˜L35
˜L34˜L44˜L45
˜L53˜L54˜L55

ψ1
ψ1
w
=
0
0
p


476 C Differential Operators for Rectangular Plates
3. Cross-ply symmetric laminates
In addition to 2. both D16,D26andA16,A26,A45are 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

L11L12L13
L22L23
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 © Springer Nature Singapore Pte Ltd. 2018
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 Shel ls
+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
AllBi 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 cyl indrical shells there is
a coupling of the in-plane and out-of-plane displacements a nd 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 lam inates
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
ψ1
ψ2
w
=−
px
ps
0
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=/parenleftbigg
B121
R−As
55/parenrightbigg∂
∂x+/parenleftbigg
B261
R−ks
45A45/parenrightbigg∂
∂s,
˜L44=D66∂2
∂x2+2D26∂2
∂x∂s+D22∂2
∂s2−ks
44A44,
˜L45=/parenleftbigg
B261
R−ks
45A45/parenrightbigg∂
∂x+/parenleftbigg
B221
R−ks
44A44/parenrightbigg∂
∂s,
˜L55=A55∂2
∂x2+2ks
45A45∂2
∂x∂s−A22)1
R2

480 D Differential Operators for Circular Cylindrical Shel ls
and
˜L51=−˜L15,˜L52=−˜L25,˜L53=−˜L35,˜L54=−˜L45,
ks
45=/radicalig
ks
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 differ ential equation
w′′′′−k2
1w′′+k4
2w=0
can be presented in the form of so-called Krylow functions (F ilonenko-Boroditsch,
1952):
1.k2
2>k2
1
α1=−α2=a+ib, α3=−α4=a−ib,
a=/radicalbigg
1
2(k2
2+k2
1), b=/radicalbigg
1
2(k2
2−k2
1)
and the solutions are
Φ1=cosh axcosbx, Φ2=sinhaxsinbx,
Φ3=cosh axsinbx, Φ4=sinhaxcosbx
or
Φ1=e−axcosbx, Φ2=e−axsinbx,
Φ3=eaxsinbx, Φ4=eaxcosbx
2.k2
2<k2
1
α1=−α2=a, α3=−α4=b,
a=/radicalbigg
k2
1−/radicalig
k4
1−k4
2, b=/radicalbigg
k2
1+/radicalig
k4
1−k4
2
and the solutions are
Φ1=cosh ax, Φ2=cosh bx,
Φ3=sinhax, Φ4=sinhbx
or
Φ1=e−ax, Φ2=e−bx,
Φ3=eax, Φ4=ebx
481 © Springer Nature Singapore Pte Ltd. 2018
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 Or dinary Differential Equation
3.k2
2=k2
1
α1=α2=a,α3=α4=−a
and the solutions are
Φ1=cosh ax, Φ2=xsinhax,
Φ3=sinhax, Φ4=xcosh ax
or
Φ1=e−ax, Φ2=xe−ax,
Φ3=eax, Φ4=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. Th e 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 approxi mate relations are used
(e.g. 1 kgf ≈10 N).
With respect to the quick changes in application composite m aterials all material
data are only examples showing the main tendencies. Every ye ar new materials are
developed and for the material data one have to contact direc tly the companies.
References
Czichos H, Hennecke M (eds) (2012) H¨ utte – das Ingenieurwis sen, 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 Composit e Materials. McGraw-
Hill, Boston et al.
Vasiliev V , Morozov E (2001) Mechanics and Analysis of Compo site Materials.
Elsevier, Amsterdam
483 © Springer Nature Singapore Pte Ltd. 2018
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 temp erature (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 σu σu/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

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