Finite Element Analysis Method, Verification and Validation (Barna A. Szabó Ivo Babuška)

Finite Element Analysis
WILEY SERIES IN COMPUTATIONAL MECHANICS

Series Advisors:
René de Borst
Perumal Nithiarasu
Tayfun E. Tezduyar
Genki Yagawa
Tarek Zohdi
Fundamentals of the Finite Element Nithiarasu, Lewis and Seetharamu November 2015
Method for Heat and Mass Transfer
Introduction to Computational Contact Konyukhov April 2015
Mechanics: A Geometrical Approach
Extended Finite Element Method: Khoei December 2014
Theory and Applications
Computational Fluid-Structure Bazilevs, Takizawa and Tezduyar January 2013
Interaction: Methods and Applications
Introduction to Finite Strain Theory for Hashiguchi and Yamakawa November 2012
Continuum Elasto-Plasticity
Nonlinear Finite Element Analysis of De Borst, Crisfield, Remmers and Verhoosel August 2012
Solids and Structures, Second Edition
An Introduction to Mathematical Oden November 2011
Modeling: A Course in Mechanics
Computational Mechanics of Discontinua Munjiza, Knight and Rougier November 2011
Introduction to Finite Element Analysis: Szabó and Babuška March 2011
Formulation, Verification and Validation
Finite Element Analysis
Method, Verification and Validation
Second Edition
Barna Szabó
Washington University in St. Louis
Ivo Babuška
The University of Texas at Austin
This second edition first published 2021
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Library of Congress Cataloging-in-Publication Data
Names: Szabó, B. A. (Barna Aladar), 1935- author. | Babuška, Ivo, author.

Title: Finite element analysis : method, verification and validation /
Barna Szabó, Ivo Babuška.
Description: Second edition. | Hoboken, NJ : Wiley, 2021. | Series: Wiley
series in computational mechanics | Includes bibliographical references
and index.
Identifiers: LCCN 2021005389 (print) | LCCN 2021005390 (ebook) | ISBN
9781119426424 (cloth) | ISBN 9781119426387 (adobe pdf) | ISBN
9781119426462 (epub) | ISBN 9781119426479 (obook)
Subjects: LCSH: Finite element method.
Classification: LCC TA347.F5 S98 2021 (print) | LCC TA347.F5 (ebook) |
DDC 620.001/51825–dc23
LC record available at https://lccn.loc.gov/2021005389
LC ebook record available at https://lccn.loc.gov/2021005390
Cover Design: Wiley

Cover Images: © Courtesy of Barna Szabó
Set in 9.5/12.5pt STIXTwoText by SPi Global, Chennai, India
10 9 8 7 6 5 4 3 2 1
v
Contents
Preface to the second edition xi

Preface to the first edition xiii
Preface xv
About the companion website xvii
1 Introduction to the finite element method 1

1.1 An introductory problem 3
1.2 Generalized formulation 6
1.2.1 The exact solution 6
1.2.2 The principle of minimum potential energy 11
1.3 Approximate solutions 12
1.3.1 The standard polynomial space 13
1.3.2 Finite element spaces in one dimension 16
1.3.3 Computation of the coefficient matrices 17
1.3.4 Computation of the right hand side vector 20
1.3.5 Assembly 21
1.3.6 Condensation 24
1.3.7 Enforcement of Dirichlet boundary conditions 24
1.4 Post-solution operations 26
1.4.1 Computation of the quantities of interest 26
1.5 Estimation of error in energy norm 30
1.5.1 Regularity 30
1.5.2 A priori estimation of the rate of convergence 31
1.5.3 A posteriori estimation of error 32
1.5.4 Error in the extracted QoI 36
1.6 The choice of discretization in 1D 38
1.6.1 The exact solution lies in Hk (I), k − 1 > p 38
1.6.2 The exact solution lies in Hk (I), k − 1 ≤ p 39
1.7 Eigenvalue problems 42
1.8 Other finite element methods 46
1.8.1 The mixed method 47
1.8.2 Nitsche’s method 48
2 Boundary value problems 51

2.1 Notation 51
2.2 The scalar elliptic boundary value problem 53
vi Contents
2.2.1 Generalized formulation 53

2.2.2 Continuity 55
2.3 Heat conduction 55
2.3.1 The differential equation 57
2.3.2 Boundary and initial conditions 58
2.3.3 Boundary conditions of convenience 59
2.3.4 Dimensional reduction 61
2.4 Equations of linear elasticity – strong form 67
2.4.1 The Navier equations 70
2.4.2 Boundary and initial conditions 71
2.4.3 Symmetry, antisymmetry and periodicity 72
2.4.4 Dimensional reduction in linear elasticity 73
2.4.5 Incompressible elastic materials 76
2.5 Stokes flow 78
2.6 Generalized formulation of problems of linear elasticity 78
2.6.1 The principle of minimum potential energy 80
2.6.2 The RMS measure of stress 82
2.6.3 The principle of virtual work 83
2.6.4 Uniqueness 84
2.7 Residual stresses 87
2.8 Chapter summary 89
3 Implementation 91
3.1 Standard elements in two dimensions 91
3.2 Standard polynomial spaces 91
3.2.1 Trunk spaces 91
3.2.2 Product spaces 92
3.3 Shape functions 93
3.3.1 Lagrange shape functions 93
3.3.2 Hierarchic shape functions 95
3.4 Mapping functions in two dimensions 97
3.4.1 Isoparametric mapping 97
3.4.2 Mapping by the blending function method 99
3.4.3 Mapping algorithms for high order elements 101
3.5 Finite element spaces in two dimensions 102
3.6 Essential boundary conditions 103
3.7 Elements in three dimensions 103
3.7.1 Mapping functions in three dimensions 105
3.8 Integration and differentiation 106
3.8.1 Volume and area integrals 106
3.8.2 Surface and contour integrals 107
3.8.3 Differentiation 108
3.9 Stiffness matrices and load vectors 109
3.9.1 Stiffness matrices 109
3.9.2 Load vectors 110
Contents vii
3.11 Computation of the solution and its first derivatives 111

3.12 Nodal forces 113
3.12.1 Nodal forces in the h-version 113
3.12.2 Nodal forces in the p-version 115
3.12.3 Nodal forces and stress resultants 117
4 Pre- and postprocessing procedures and verification 119

4.1 Regularity in two and three dimensions 119
4.2 The Laplace equation in two dimensions 120
4.2.1 2D model problem, uEX ∈ Hk (Ω), k − 1 > p 121
4.2.2 2D model problem, uEX ∈ Hk (Ω), k − 1 ≤ p 123
4.2.3 Computation of the flux vector in a given point 126
4.2.4 Computation of the flux intensity factors 128
4.2.5 Material interfaces 131
4.3 The Laplace equation in three dimensions 133
4.4 Planar elasticity 137
4.4.1 Problems of elasticity on an L-shaped domain 137
4.4.2 Crack tip singularities in 2D 139
4.4.3 Forcing functions acting on boundaries 142
4.5 Robustness 143
4.6 Solution verification 148
5 Simulation 155
5.1 Development of a very useful mathematical model 156
5.1.1 The Bernoulli-Euler beam model 156
5.1.2 Historical notes on the Bernoulli-Euler beam model 158
5.2 Finite element modeling and numerical simulation 159
5.2.1 Numerical simulation 159
5.2.2 Finite element modeling 160
5.2.3 Calibration versus tuning 163
5.2.4 Simulation governance 164
5.2.5 Milestones in numerical simulation 165
5.2.6 Example: The Girkmann problem 167
5.2.7 Example: Fastened structural connection 170
5.2.8 Finite element model 176
5.2.9 Example: Coil spring with displacement boundary conditions 180
5.2.10 Example: Coil spring segment 184
6 Calibration, validation and ranking 187

6.1 Fatigue data 187
6.1.1 Equivalent stress 188
6.1.2 Statistical models 189
6.1.3 The effect of notches 190
6.1.4 Formulation of predictors of fatigue life 190
6.2 The predictors of Peterson and Neuber 191
viii Contents
6.2.1 The effect of notches – calibration 193

6.2.2 The effect of notches – validation 195
6.2.3 Updated calibration 197
6.2.4 The fatigue limit 199
6.2.5 Discussion 201
6.3 The predictor Gα 202
6.3.1 Calibration of 𝛽(V, 𝛼) 203
6.3.2 Ranking 204
6.3.3 Comparison of Gα with Peterson′ s revised predictor 205
6.4 Biaxial test data 205
6.4.1 Axial, torsional and combined in-phase loading 206
6.4.2 The domain of calibration 208
6.4.3 Out-of-phase biaxial loading 210
6.5 Management of model development 218
6.5.1 Obstacles to progress 220
7 Beams, plates and shells 223

7.1 Beams 223
7.1.1 The Timoshenko beam 225
7.1.2 The Bernoulli-Euler beam 229
7.2 Plates 234
7.2.1 The Reissner-Mindlin plate 236
7.2.2 The Kirchhoff plate 240
7.2.3 The transverse variation of displacements 243
7.3 Shells 247
7.3.1 Hierarchic thin solid models 249
8 Aspects of multiscale models 255

8.1 Unidirectional fiber-reinforced laminae 255
8.1.1 Determination of material constants 257
8.1.2 The coefficients of thermal expansion 258
8.1.3 Examples 258
8.1.4 Localization 261
8.1.5 Prediction of failure in composite materials 262
8.1.6 Uncertainties 263
8.2 Discussion 264
9 Non-linear models 265

9.1.1 Radiation 265
9.1.2 Nonlinear material properties 266
9.2 Solid mechanics 266
9.2.1 Large strain and rotation 266
9.2.2 Structural stability and stress stiffening 270
Contents ix
9.2.3 Plasticity 275

9.2.4 Mechanical contact 281
Appendix A Definitions 289

A.1 Normed linear spaces, linear functionals and bilinear forms 289
A.1.1 Normed linear spaces 290
A.1.2 Linear forms 290
A.1.3 Bilinear forms 290
A.2 Convergence in the space X 291
A.2.1 The space of continuous functions 291
A.2.2 The space Lp (Ω) 291
A.2.3 Sobolev space of order 1 291
A.2.4 Sobolev spaces of fractional index 292
A.3 The Schwarz inequality for integrals 293
Appendix B Proof of h-convergence 295
Appendix C Convergence in 3D: Empirical results 297
Appendix D Legendre polynomials 301

D.1 Shape functions based on Legendre polynomials 302
Appendix E Numerical quadrature 303

E.1 Gaussian quadrature 303
E.2 Gauss-Lobatto quadrature 304
Appendix F Polynomial mapping functions 307

F.1 Interpolation on surfaces 308
F.1.1 Interpolation on the standard quadrilateral element 309
F.1.2 Interpolation on the standard triangle 309
Appendix G Corner singularities in two-dimensional elasticity 311

G.1 The Airy stress function 311
G.2 Stress-free edges 312
G.2.1 Symmetric eigenfunctions 313
G.2.2 Antisymmetric eigenfunctions 315
G.2.3 The L-shaped domain 315
G.2.4 Corner points 317
Appendix H Computation of stress intensity factors 319

H.1 Singularities at crack tips 319
H.2 The contour integral method 320
H.3 The energy release rate 321
H.3.1 Symmetric (Mode I) loading 322
x Contents
H.3.2 Antisymmetric (Mode II) loading. 323

H.3.3 Combined (Mode I and Mode II) loading. 323
H.3.4 Computation by the stiffness derivative method. 323
Appendix I Fundamentals of data analysis 325

I.1 Statistical foundations 325
I.2 Test data 326
I.3 Statistical models 328
I.4 Ranking 335
I.5 Confidence intervals 335
Appendix J Estimation of fastener forces in structural connections 337
Appendix K Useful algorithms in solid mechanics 341

K.1 The traction vector 341
K.2 Transformation of vectors 342
K.3 Transformation of stresses 343
K.4 Principal stresses 344
K.5 The von Mises stress 344
K.6 Statically equivalent forces and moments 345
K.6.1 Technical formulas for stress 348
Bibliography 351
Index 357
xi
Preface to the second edition
The first edition of this book, published in 1991, focused on the conceptual and algorithmic
development of the finite element method from the perspective of solution verification, that is,
estimation and control of the errors of approximation in terms of the quantities of interest. Since
that time the importance of solution verification became widely recognized. It is a key constituent
of predictive computational science, the branch of computational science concerned with the
prediction of physical events.
Predictive computational science embraces the formulation of mathematical models, definition
of the quantities of interest, code and solution verification, definition of statistical sub-models, cal-
ibration and validation of models, and forecasting physical events with quantified uncertainty. The
second edition covers the main conceptual and algorithmic aspects of predictive computational sci-
ence pertinent to solid mechanics. The formulation and application of design rules for mechanical
and structural components subjected to cyclic loading are used for illustration.
Another objective in writing the first edition was to make some fundamentally important results
of research in the field of applied mathematics accessible to the engineering community. Speaking
generally, engineers and mathematicians view the finite element method very differently. Engi-
neers see the method as a way to construct a numerical problem the solution of which is expected
to provide quantitative information about the response of some physical system, for example a
structural shell, to some form of excitation, such as the application of loads. Their view tends to be
element-oriented: They tend to believe that sufficiently clever formulation of elements can over-
come the various shortcomings of the method.
Mathematicians, on the other hand, view the finite element method as a method for approxi-
mating the exact solution of differential equations cast in a variational form. Mathematicians focus
on a priori and a posteriori error estimation and error control. In the 1970s adaptive procedures
were developed for the construction of sequences of finite element mesh such that the correspond-
ing solutions converged to the exact solution in energy norm at optimal or nearly optimal rates.
An alternative way of achieving convergence in energy norm through increasing the polynomial
degrees of the elements on a fixed mesh was proven in 1981. The possibility of achieving exponential
rates of convergence in energy norm for an important class of problems, that includes the problem
of elasticity, was proven and demonstrated in 1984. This required the construction of sequences of
finite element mesh and optimal assignment of polynomial degrees.
Superconvergent methods of extraction of certain quantities of interest (such as the stress inten-
sity factor) from finite element solutions were developed by 1984.
These developments were fundamentally important milestones in a journey toward the emer-
gence of predictive computational science. Our primary objective in publishing this second edition
xii Preface to the second edition
is to provide engineering analysts and software developers a comprehensive account of the concep-
tual and algorithmic aspects of verification, validation and uncertainty quantification illustrated by
examples.
Quantification of uncertainty involves the application of methods of data analysis. A brief intro-
duction to the fundamentals of data analysis is presented in this second edition.
We recommend this book to students, engineers and analysts who seek Professional Simulation
Engineer (PSE) certification.
We would like to thank Dr. Ricardo Actis for many useful discussions, advice and assistance
provided over many years; Dr. Börje Andersson for providing valuable convergence data relating to
the solution of an interesting model problem of elasticity, and Professor Raul Tempone for guidance
in connection with the application of data analysis procedures.
Barna Szabó and Ivo Babuška

xiii
Preface to the first edition
There are many books on the finite element method today. It seems appropriate therefore to say a
few words about why this book was written. A brief look at the approximately 30-year history of
the finite element method will not only help explain the reasons but will also serve to put the main
points in perspective.
Systematic development of the finite element method for use as an analytical tool in engineering
decision-making processes is usually traced to a paper published in 1956.1 Demand for efficient
and reliable numerical methods has been the key motivating factor for the development of the
finite element method. The demand was greatly amplified by the needs of the space program in the
United States during the 1960s. A great deal was invested into the development of finite element
analysis technology in that period.
Early development of the finite element method was performed entirely by engineers. The names
of Argyris, Clough, Fraeijs de Veubeke, Gallagher, Irons, Martin, Melosh, Pian, and Zienkiewicz
come to mind in this connection.2
Development of the finite element method through the 1960s was based on intuitive reasoning,
analogies with naturally discrete systems, such as structural frames, and numerical experimenta-
tion. The errors of discretization were controlled by uniform or nearly uniform refinement of the
finite element mesh. Mathematical analysis of the finite element method begun in the late 1960s.
Error estimation techniques were investigated during the 1970s. Adaptive mesh refinement proce-
dures designed to reduce the errors of discretization with improved efficiency received a great deal
of attention in this period.3
Numerical experiments conducted in the mid-1970s indicated that the use of polynomials of
progressively increasing degree on a fixed finite element mesh can be much more advantageous
than uniform or nearly uniform mesh refinement.4 To distinguish between reducing errors of dis-
cretization by mesh refinement and the alternative approach, based on increasing the degree of the
polynomial basis functions, the labels h-version and p-version gained currency for the following rea-
son: Usually the symbol h is used to represent the size of the finite elements. Convergence occurs
when the size of the largest element (hmax ) is progressively reduced. Hence the name: h-version.
The polynomial degree of elements is usually denoted by the symbol p. Convergence occurs when
1 Turner MJ, Clough RW, Martin HC and Topp LJ, Stiffness and deflection analysis of complex structures. Journal
of Aeronautical Sciences 23(9), 805–824, 1956.
2 Williamson CF, Jr. A History of the Finite Element Method to the Middle 1960s. Doctoral Dissertation, Boston
University, 1976.
3 Babuška I and Rheinboldt WC. Adaptive approaches and reliability estimations in finite element analysis.
Computer Methods in Applied Mechanics and Engineering, 17/18, 519–540, 1979.
4 Szabó BA and Mehta AK. p-convergent finite element approximations in fracture mechanics. Int. J. Num. Meth.
Engng., 12. 551–560, 1978.
xiv Preface to the first edition
the lowest polynomial degree pmin is progressively increased. Hence the name: p-version. The h-
and p-versions are just special applications of the finite element method which, at least in princi-
ple, allows changing the finite element mesh concurrently with increasing the polynomial degree of
elements. This general approach is usually called the hp-version of the finite element method. The
theoretical basis for the p-version was established by 1981.5 The understanding of how to combine
mesh refinement with p-extensions most effectively was achieved by the mid-1980s.6
The p-version was developed primarily for applications in solid mechanics. A closely related
recent development for applications primarily in fluid mechanics is the spectral element method.7
The 1980s brought another important development: superconvergent methods for the extraction
of engineering data from finite element solutions were developed and demonstrated. At the time of
writing good understanding exists regarding how finite element discretizations should be designed
and how engineering data should be extracted from finite element solutions for optimal reliability
and efficiency.
The knowledge which allows construction of advanced finite element computer programs, that
work efficiently and reliably for very large classes of problems, and all admissible data that char-
acterize those problems, is well established today. It must be borne in mind, however, that finite
element solutions and engineering data extracted from them can be valuable only if they serve the
purposes of making correct engineering decisions.
Our purpose in writing this book is to introduce the finite element method to engineers and engi-
neering students in the context of the engineering decision-making process. Basic engineering and
mathematical concepts are the starting points. Key theoretical results are summarized and illus-
trated by examples. The focus is on the developments in finite element analysis technology during
the 1980s and their impact on reliability, quality assurance procedures in finite element compu-
tations, and performance. The principles that guide the construction of mathematical models are
described and illustrated by examples.
5 Babuška I, Szabó B and Katz IN. The p-version of the finite element method. SIAM J. Numer. Anal., 18, 515–545,
1981.
6 Babuška I. The p- and hp-versions of the finite element method. The state of the art. In: DL Dwoyer, MY Hussaini
and RG Voigt, editors, Finite Elements: Theory and Applications, Springer-Verlag New York, Inc., 1988.
7 Maday Y and Patera AT. Spectral element methods for the incompressible Navier-Stokes equations. In: AK Noor
and JT Oden, editors. State-of-the-Art Surveys on Computational Mechanics, American Society of Mechanical
Engineers, New York, 1989.
xv
Preface
This book, Finite Element Analysis: Method, Verification and Validation, 2nd Edition, is written by
two well-recognized, leading experts on finite element analysis. The first edition, published in 1991,
was a landmark contribution in finite element methods, and has now been updated and expanded
to include the increasingly important topic of error estimation, validation – the process to ascer-
tain that the mathematical/numerical model meets acceptance criteria – and verification – the
process for acceptability of the approximate solution and computed data. The systematic treatment
of formulation, verification and validation procedures is a distinguishing feature of this book and
sets it apart from other texts on finite elements. It encapsulates contemporary research on proper
model selection and control of modeling errors. Unique features of the book are accessibility and
readability for students and researchers alike, providing guidance on modeling, simulation and
implementation issues. It is an essential, self-contained book for any person who wishes to fully
master the finite element method.
xvii
About the companion website
This book is accompanied by a companion website:
www.wiley.com/go/szabo/finite_element_analysis
The website includes solutions manual, PowerPoint slides for instructors, and a link to finite
element software
1
Introduction to the finite element method
This book covers the fundamentals of the finite element method in the context of numerical
simulation with specific reference to the simulation of the response of structural and mechanical
components to mechanical and thermal loads.
We begin with the question: what is the meaning of the term “simulation”? By its dictionary
definition, simulation is the imitative representation of the functioning of one system or process by
means of the functioning of another. For instance, the membrane analogy introduced by Prandtl1
in 1903 made it possible to find the shearing stresses in bars of arbitrary cross-section, loaded by a
twisting moment, through mapping the deflected shape of a thin elastic membrane. In other words,
the distribution and magnitude of shearing stress in a twisted bar can be simulated by the deflected
shape of an elastic membrane.
The membrane analogy exists because two unrelated phenomena can be modeled by the same
partial differential equation. The physical meaning associated with the coefficients of the differen-
tial equation depends on which problem is being solved. However, the solution of one is propor-
tional to the solution of the other: At corresponding points the shearing stress in a bar, subjected
to a twisting moment, is oriented in the direction of the tangent to the contour lines of a deflected
thin membrane and its magnitude is proportional to the slope of the membrane. Furthermore, the
volume enclosed by the deflected membrane is proportional to the twisting moment.
In the pre-computer years the membrane analogy provided practical means for estimating shear-
ing stresses in prismatic bars. This involved cutting the shape of the cross-section out of sheet
metal or a wood panel, covering the hole with a thin elastic membrane, applying pressure to the
membrane and mapping the contours of the deflected membrane. In present-day practice both
problems would be formulated as mathematical problems which would then be solved by a numer-
ical method, most likely by the finite element method.
There are many other useful analogies. For example, the same differential equations simulate
the response of assemblies of mechanical components, such as linear spring-mass-viscous damper
systems and assemblies of electrical components, such as capacitors, inductors and resistors. This
has been exploited by the use of analogue computers. Obviously, it is much easier to build and
manipulate electrical circuitry than mechanical assemblies. In present-day practice both simula-
tion problems would be formulated as mathematical problems which would be solved by a numer-
ical method.
At the heart of simulation of aspects of physical reality is a mathematical problem cast in a gen-
eralized form2 . The solution of the mathematical problem is approximated by a numerical method,
1 Ludwig Prandtl 1875–1953.

2 The generalized form is also called variational form or weak form.
Finite Element Analysis: Method, Verification and Validation, Second Edition. Barna Szabó and Ivo Babuška.
© 2021 John Wiley & Sons, Inc. Published 2021 by John Wiley & Sons, Inc.
Companion Website: www.wiley.com/go/szabo/finite_element_analysis
2 1 Introduction to the finite element method
such as the finite element method, which is the subject of this book. The quantities of interest (QoI)
are extracted from the approximate solution. The errors of approximation in the QoI depend on how
the mathematical problem was discretized3 and how the QoI were extracted from the numerical
solution. When the errors of approximation are larger than what is considered acceptable then the
discretization has to be changed either by an automated adaptive process or by action of the analyst.
Estimation and control of numerical errors are fundamentally important in numerical simula-
tion. Consider, for example, the problem of design certification. Design rules are typically stated in
the form
Fmax ≤ Fall (1.1)
where Fmax > 0 (resp. Fall > 0) is the maximum (resp. allowable) value of a quantity of interest, for
example the first principal stress. Since in numerical simulation only an approximation to Fmax is
available, denoted by Fnum , it is necessary to know the size of the numerical error 𝜏:
|Fmax − Fnum | ≤ 𝜏Fmax . (1.2)
In design and design certification the worst case scenario has to be considered, which is underes-
timation of Fmax , that is,
Fnum = (1 − 𝜏)Fmax . (1.3)
Therefore it has to be shown that
Fnum ≤ (1 − 𝜏)Fall . (1.4)
Without a reliable estimate of the size of the numerical error it is not possible to certify design
and, furthermore, numerical errors penalize design by lowering the allowable value, as indicated
by eq. (1.4). Generally speaking, it is far more economical to ensure that 𝜏 is small than to accept
the consequences of decreased allowable values.
We distinguish between finite element modeling and numerical simulation. As explained in
greater detail in Chapter 5, finite element modeling evolved well before the theoretical basis of
numerical simulation was developed. In finite element modeling a numerical problem is formu-
lated by assembling elements from a library of finite elements that contains intuitively constructed
beam, plate, shell, solid elements of various description. The numerical problem so created may not
correspond to a well defined mathematical problem and therefore a solution may not even exist.
For that reason it is not possible to speak of errors of approximation. Nevertheless, finite element
modeling is widely practiced with success in some cases but with disappointing results in others.
Such practice should be regarded as a practice of art, guided by intuition and experience, rather
than a scientific activity. This is because practitioners of finite element modeling have to balance
two kinds of very large errors: (a) conceptual errors in the formulation and (b) approximation errors
in the numerical solution of an improperly posed mathematical problem.
In numerical simulation, on the other hand, the formulation of mathematical models is treated
separately from their numerical solution. A mathematical model should be understood to be a
precise statement of an idea of physical reality that permits the prediction of the occurrence, or
probability of occurrence, of physical events, given certain data. The intuitive aspects of simulation
are confined to the formulation of mathematical models whereas their numerical solution involves
the application of well established procedures of applied mathematics. Separation of mathematical
3 The term “discretization” refers to processes by which approximating functions are defined. The most widely
used discretizations will be described and illustrated by examples in this and subsequent chapters.
models from their numerical solution makes separate treatment of errors associated with the for-
mulation of mathematical models and their numerical approximation possible. Errors associated
with the formulation of mathematical models are called model form errors. Errors associated with
the numerical solution of mathematical problems are called errors of approximation or errors of
discretization. In the early papers and books on the finite element method no such distinction
was made.
In this chapter we introduce the finite element method as a method by which the exact solution
of a mathematical problem, cast in a generalized form, can be approximated. We also introduce the
relevant mathematical concepts, terminology and notation in the simplest possible setting. Gener-
alization of these concepts to two- and three-dimensional problems will be discussed in subsequent
chapters.
We first consider the formulation of a second order ordinary differential equation without ref-
erence to any physical interpretation. This is to underline that once a mathematical problem was
formulated, the approximation process is independent from why the mathematical problem was
formulated. This important point is often missed by engineering users of legacy finite element codes
because the formulation and approximation of mathematical problems is mixed in finite element
libraries.
We show that the exact solution of the generalized formulation is unique. Approximation of
the exact solution by the finite element method is described and various discretization strategies
are explored. Efficient methods for the computation of QoIs and a posteriori error estimation are
described. This chapter serves as a foundation for subsequent chapters.
We would like to assure engineering students who are not yet familiar with the concepts and
notation of that branch of applied mathematics on which the finite element method is based that
their investment of time and effort to master the contents of this chapter will prove to be highly
rewarding.
1.1 An introductory problem

We introduce the finite element method through approximating the exact solution of the following
second order ordinary differential equation
−(𝜅u′ )′ + cu = f on the closed interval I = [0 ≤ x ≤ 𝓁] (1.5)
with the boundary conditions
u(0) = u(𝓁) = 0 (1.6)
where the prime indicates differentiation with respect to x. It is assumed that 0 < 𝛼 ≤ 𝜅(x) ≤ 𝛽 < ∞
where 𝛼 and 𝛽 are real numbers, 𝜅 ′ < ∞ on I, c ≥ 0 and f = f (x) are defined such that the indicated
operations are meaningful on I. For example, the indicated operations would not be meaningful if
(𝜅u′ )′ , c or f would not be finite in one or more points on the interval 0 ≤ x ≤ 𝓁. The function f is
called a forcing function.
We seek an approximation to u in the form:
∑
n
un = aj 𝜑j (x), 𝜑j (0) = 𝜑j (𝓁) = 0 for all j (1.7)
j=1
where 𝜑j (x) are fixed functions, called basis functions, and aj are the coefficients of the basis func-
tions to be determined. Note that the basis functions satisfy the zero boundary conditions.
Let us find aj such that the integral  defined by

𝓁( )
1
= 𝜅(u′ − u′n )2 + c(u − un )2 dx (1.8)
2 ∫0
is minimum. While there are other plausible criteria for selecting aj , we will see that this criterion
is fundamentally important in the finite element method. Differentiating  with respect to ai and
letting the derivative equal to zero, we have:
𝓁( )
d
= 𝜅(u′ − u′n )𝜑′i + c(u − un )𝜑i dx = 0, i = 1, 2, … , n. (1.9)
dai ∫0
Using the product rule: (𝜅u′ 𝜑i )′ = (𝜅u′ )′ 𝜑i + 𝜅u′ 𝜑′i we write
𝓁 𝓁 ( ′ ′ )
𝜅u′ 𝜑′i dx = (𝜅u 𝜑i ) − (𝜅u′ )′ 𝜑i dx
∫0 ∫0
𝓁
= (𝜅u′ 𝜑i )x=𝓁 − (𝜅u′ 𝜑i )x=0 − (𝜅u′ )′ 𝜑i dx. (1.10)
⏟⏞⏞⏞⏟⏞⏞⏞⏟ ⏟⏞⏞⏞⏟⏞⏞⏞⏟ ∫0
=0 =0
The underbraced terms vanish on account of the boundary conditions, see eq. (1.7). On substituting
this expression into eq. (1.9), we get
𝓁 𝓁
(−(𝜅u′ )′ + cu)𝜑i dx − (𝜅u′n 𝜑′i + cun 𝜑i ) dx = 0
∫0 ⏟⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏟ ∫0
=f (x)
which will be written as

𝓁 𝓁
(𝜅u′n 𝜑′i + cun 𝜑i ) dx = f 𝜑i dx, i = 1, 2, … , n. (1.11)
∫0 ∫0
We define
𝓁 𝓁 𝓁
kij = 𝜅𝜑′i 𝜑′j dx, mij = c𝜑i 𝜑j dx, ri = f 𝜑i dx (1.12)
∫0 ∫0 ∫0
and write eq. (1.11) in the following form
∑
n
(kij + mij )aj = ri , i = 1, 2, … , n (1.13)
j=1
which represents n simultaneous equations in n unknowns. It is usually written in matrix form:
([K] + [M]) {a} = {r}. (1.14)
On solving these equations we find an approximation un to the exact solution u in the sense that
un minimizes the integral .
Example 1.1 Let 𝜅 = 1, c = 1, 𝓁 = 2 and
f = sin(𝜋x∕𝓁) + sin(2𝜋x∕𝓁).
With these data the exact solution of eq. (1.5) is

1 1
u= sin(𝜋x∕𝓁) + sin(2𝜋x∕𝓁).
1 + 𝜋 2 ∕𝓁 2 1 + 4𝜋 2 ∕𝓁 2
0.4
0.3
un
0.2
u
0.1
0
0 0.5 1 1.5 2
x
Figure 1.1 Exact and approximate solutions for the problem in Example 1.1.
We seek an approximation to u in the form:
un = u2 = a1 x(𝓁 − x) + a2 x(𝓁 − x)2 .
On computing the elements of [K], [M] and {r} we get

[ ] [ ] { }
2.6667 2.6667 1.0667 1.0667 1.0320
[K] = [M] = {r} = ⋅
2.6667 4.2667 1.0667 1.2190 1.4191
The solution of this problem is a1 = 0.0556, a2 = 0.2209. These coefficients, together with the basis
functions, define the approximate solution un . The exact and approximate solutions are shown in
Fig. 1.1.
The choice of basis functions

By definition, a set of functions 𝜑j (x), (j = 1, 2, … , n) are linearly independent if
∑
n
aj 𝜑j (x) = 0
j=1
implies that aj = 0 for j = 1, 2, … , n. It is left to the reader to show that if the basis functions are
linearly independent then matrix [M] is invertible.
Given a set of linearly independent functions 𝜑j (x), (j = 1, 2, … , n), the set of functions that can
be written as
∑
n
un = aj 𝜑j (x)
j=1
is called the span and 𝜑j (x) are basis functions of S.

We could have defined other polynomial basis functions, for example;
∑
n
un = ci 𝜓i (x) , 𝜓i (x) = xi (𝓁 − x). (1.15)
i=1
When one set of basis functions {𝜑} = {𝜑1 𝜑2 … 𝜑n }T can be written in terms of another set
{𝜓} = {𝜓1 𝜓2 … 𝜓n }T in the form:
{𝜓} = [B]{𝜑} (1.16)
where [B] is an invertible matrix of constant coefficients then both sets of basis functions are said
to have the same span. The following exercise demonstrates that the approximate solution depends
on the span, not on the choice of basis functions.
Exercise 1.1 Solve the problem of Example 1.1 using the basis functions 𝜑1 = x(𝓁 − x),
𝜑2 = x2 (𝓁 − x) and show that the resulting approximate solution is identical to the approximate
solution obtained in Example 1.1. The span of the basis functions in this exercise and in Example
1.1 is the same: It is the set of polynomials of degree less than or equal to 3 that vanish in the points
x = 0 and x = 𝓁.
Summary of the main points

1. The definition of the integral  by eq. (1.8) made it possible to find an approximation to the exact
solution u of eq. (1.5) without knowing u.
2. A formulation cannot be meaningful unless all indicated operations are defined. In the case of
eq. (1.5) this means that (𝜅u′ )′ and cu are finite on the interval 0 ≤ x ≤ 𝓁. In the case of eq. (1.11)
the integral
𝓁 ( )
𝜅(u′ )2 + cu2 dx
∫0
must be finite which is a much less stringent condition. In other words, eq. (1.8) is meaningful
for a larger set of functions u than eq. (1.5) is. Equation (1.5) is the strong form, whereas eq. (1.11)
is the generalized or weak form of the same differential equation. When the solution of eq. (1.5)
exists then un converges to that solution in the sense that the limit of the integral  is zero.
3. The error e = u − un depends on the span and not on the choice of basis functions.
1.2 Generalized formulation
We have seen in the foregoing discussion that it is possible to approximate the exact solution u of
eq. (1.5) without knowing u when u(0) = u(𝓁) = 0. In this section the formulation is outlined for
other boundary conditions.
The generalized formulation outlined in this section is the most widely implemented formu-
lation; however, it is only one of several possible formulations. It has the properties of stability
and consistency. For a discussion on the requirements of stability and consistency in numerical
approximation we refer to [5].
1.2.1 The exact solution

If eq. (1.5) holds then for an arbitrary function 𝑣 = 𝑣(x), subject only to the restriction that all of the
operations indicated in the following are properly defined, we have
𝓁 ( )
(−𝜅u′ )′ + cu − f 𝑣 dx = 0. (1.17)
∫0
Using the product rule; (𝜅u′ 𝑣)′ = (𝜅u′ )′ 𝑣 + 𝜅u′ 𝑣′ we get
𝓁 𝓁
(−𝜅u′ )′ 𝑣 dx = −(𝜅u′ 𝑣)x=𝓁 + (𝜅u′ 𝑣)x=0 + 𝜅u′ 𝑣′ dx
∫0 ∫0
therefore eq. (1.17) is transformed to:
𝓁 𝓁
(𝜅u′ 𝑣′ + cu𝑣) dx = f 𝑣 dx + (𝜅u′ 𝑣)x=𝓁 − (𝜅u′ 𝑣)x=0 . (1.18)
∫0 ∫0
We introduce the following notation:

𝓁
def
B(u, 𝑣) = (𝜅u′ 𝑣′ + cu𝑣) dx (1.19)
∫0
where B(u, 𝑣) is a bilinear form. A bilinear form has the property that it is linear with respect to
each of its two arguments. The properties of bilinear forms are listed Section A.1.3 of Appendix A.
We define the linear form:
𝓁
def
F(𝑣) = f 𝑣 dx + (𝜅u′ 𝑣)x=𝓁 − (𝜅u′ 𝑣)x=0 . (1.20)
∫0
The forcing function f (x) may be a sum of forcing functions: f (x) = f1 (x) + f2 (x) + …, some or all
of which may be the Dirac delta function4 multiplied by a constant. For example if fk (x) = F0 𝛿(x0 )
then
𝓁 𝓁
fk (x)𝑣 dx = F0 𝛿(x0 )𝑣 dx = F0 𝑣(x0 ). (1.21)
∫0 ∫0
The properties of linear forms are listed in Section A.1.2. Note that F0 𝑣(x0 ) in eq. (1.21) is a linear
form only if 𝑣 is continuous and bounded.
The definitions of B(u, 𝑣) and F(𝑣) are modified depending on the boundary conditions. Before
proceeding further we need the following definitions.
1. The energy norm is defined by

√
def 1
‖u‖E(I) = B(u, u) (1.22)
2
where I represents the open interval I = {x | 0 < x < 𝓁}. This notation should be understood
to mean that x ∈ I if and only if x satisfies the condition to the right of the bar (|). This nota-
tion may be shortened to I = (0, 𝓁), or more generally I = (a, b) where b > a are real numbers.
If the interval includes both boundary points then the interval is a closed interval denoted by
def
I = [0, 𝓁].
We have seen in the introductory example that the error is minimized in energy norm, that
is, ||u − un ||2E(I) , equivalently ||u − un ||E(I) is minimum. The square root is introduced so that
||𝛼u||E(I) = |𝛼|||u||E(I) (where 𝛼 is a constant) holds. This is one of the definitive properties of
norms listed in Section A.1.1.
2. The energy space, denoted by E(I), is the set of all functions u defined on I that satisfy the fol-
lowing condition:
def
E(I) = {u | ||u||E(I) < ∞}. (1.23)
Since infinitely many linearly independent functions satisfy this condition, the energy space is
infinite-dimensional.
̃
3. The trial space, denoted by E(I), is a subspace of E(I). When boundary conditions are prescribed
̃ satisfy those bound-
on u, such as u(0) = û0 and/or u(𝓁) = û𝓁 , then the functions that lie in E(I)
ary conditions. Note that when û0 ≠ 0 and/or û𝓁 ≠ 0 then E(I) ̃ is not a linear space. This is
because the condition stated under item 1 in Section A.1.1 is not satisfied. When u is prescribed
on a boundary then that boundary condition is called an essential boundary condition. If no
essential boundary conditions are prescribed on u then E(I)̃ = E(I).
4 See Definition A.5 in the appendix.

4. The test space, denoted by E0 (I), is a subspace of E(I). When boundary conditions are prescribed
on u, such as u(0) = û0 and/or u(𝓁) = û𝓁 then the functions that lie in E0 (I) are zero in those
boundary points.
If no boundary conditions are prescribed on u then E(I)̃ = E0 (I) = E(I). If u(0) = û0 is prescribed
and u(𝓁) is not known then
def
̃
E(I) = {u | u ∈ E(I), u(0) = û0 } (1.24)
def
E0 (I) = {u | u ∈ E(I), u(0) = 0}. (1.25)
If u(0) is not known and u(𝓁) = û𝓁 is prescribed then
def
̃
E(I) = {u | u ∈ E(I), u(𝓁) = û𝓁 } (1.26)
def
E0 (I) = {u | u ∈ E(I), u(𝓁) = 0}. (1.27)
If u(0) = û0 and u(𝓁) = û𝓁 are prescribed then
def
̃
E(I) = {u | u ∈ E(I), u(0) = û0 , u(𝓁) = û𝓁 } (1.28)
def
E0 (I) = {u | u ∈ E(I), u(0) = 0, u(𝓁) = 0}. (1.29)
We are now in a position to describe the generalized formulation for various boundary conditions
in a concise manner;
1. When u is prescribed on a boundary then the boundary condition is called essential or Dirichlet5
boundary condition. Let us assume that u is prescribed on both boundary points. In this case we
write u = u + u★ where u ∈ E0 (I) is the function to be approximated and u★ ∈ Ẽ is an arbitrary
fixed function that satisfies the boundary conditions. Substituting u + u★ for u in eq. (1.18) we
have:
𝓁 𝓁 𝓁
(𝜅(u★ )′ 𝑣′ + cu★ 𝑣) dx
′
(𝜅u 𝑣′ + cu𝑣) dx = f 𝑣 dx − (1.30)
∫0 ∫0 ∫0
⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟ ⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟
B(u,𝑣) F(𝑣)
and the generalized formulation is stated as follows: “Find u ∈ E0 (I) such that B(u, 𝑣) = F(𝑣)
̃ is independent of the
for all 𝑣 ∈ E0 (I)” where E0 (I) is defined by eq. (1.29). Note that u ∈ E(I)
★
choice of u . Essential boundary conditions are enforced by restriction on the space of admis-
sible functions.
2. When 𝜅u′ = F is prescribed on a boundary then the boundary condition is called Neumann6
boundary condition. Assume that u(0) = û0 and (𝜅u′ )x=𝓁 = F𝓁 are prescribed. In this case
𝓁 𝓁 𝓁
(𝜅(u★ )′ 𝑣′ + cu★ 𝑣) dx
′
(𝜅u 𝑣′ + cu𝑣) dx = f 𝑣 dx + F𝓁 𝑣(𝓁) − (1.31)
∫0 ∫0 ∫0
⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟ ⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟
B(u,𝑣) F(𝑣)
and the generalized formulation is: “Find u ∈ E0 (I) such that B(u, 𝑣) = F(𝑣) for all 𝑣 ∈ E0 (I)”
where E0 (I) is defined by eq. (1.25).
5 Peter Gustav Lejeune Dirichlet 1805–1859.

6 Carl Gottfried Neumann 1832–1925.
An important special case is when c = 0 and (𝜅u′ )x=0 = F0 and (𝜅u′ )x=𝓁 = F𝓁 are prescribed.
In this case:
𝓁 𝓁
𝜅u′ 𝑣′ dx = f 𝑣 dx − F0 𝑣(0) + F𝓁 𝑣(𝓁) (1.32)
∫0 ∫0
⏟⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏟ ⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟
B(u,𝑣) F(𝑣)
and the generalized formulation is “Find u ∈ E(I) such that B(u, 𝑣) = F(𝑣) for all 𝑣 ∈ E(I) where
E(I) is defined by eq. (1.23).” Since the left-hand side is zero for 𝑣 = C (constant) the specified
data must satisfy the condition
𝓁
f dx − F0 + F𝓁 = 0. (1.33)
∫0
3. When (𝜅u′ )x=0 = k0 (u(0) − 𝛿0 ) and/or (𝜅u′ )x=𝓁 = k𝓁 (𝛿𝓁 − u(𝓁)), where k0 > 0, k𝓁 > 0, 𝛿0 and
𝛿𝓁 are given real numbers, is prescribed on a boundary then the boundary condition is
called a Robin7 boundary condition. Assume, for example, that (𝜅u′ )x=0 = k0 (u(0) − 𝛿0 ) and
(𝜅u′ )x=𝓁 = F𝓁 are prescribed. In that case
𝓁 𝓁
(𝜅u′ 𝑣′ + cu𝑣) dx + k0 u(0)𝑣(0) = f 𝑣 dx + F𝓁 𝑣(𝓁) − k0 𝛿0 𝑣(0) (1.34)
∫0 ∫0
⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟ ⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟
B(u,𝑣) F(𝑣)
and the generalized formulation is: “Find u ∈ E(I) such that B(u, 𝑣) = F(𝑣) for all 𝑣 ∈ E(I) where
E(I) is defined by eq. (1.23).”
These boundary conditions may be prescribed in any combination. The Neumann and Robin
boundary conditions are called natural boundary conditions. Natural boundary conditions cannot
be enforced by restriction. This is illustrated in Exercise 1.3.
The generalized formulation is stated as follows: “Find uEX ∈ X such that B(uEX , 𝑣) = F(𝑣) for
all 𝑣 ∈ Y ”. The space X is called the trial space, the space Y is called the test space. We will use
this notation with the understanding that the definitions of X, Y , B(u, 𝑣) and F(𝑣) depend on the
boundary conditions. It is essential for analysts to understand and be able to precisely state the
generalized formulation for any set of boundary conditions.
Under frequently occurring special conditions the mathematical problem can be formulated on a
subdomain and the solution extended to the full domain by symmetry, antisymmetry or periodicity.
The symmetric, antisymmetric and periodic boundary conditions will be discussed in Chapter 2.
Theorem 1.1 The solution of the generalized formulation is unique in the energy space.
The proof is by contradiction: Assume that there are two solutions u1 and u2 in X ⊂ E(I) that
satisfy
B(u1 , 𝑣) = F(𝑣) for all 𝑣 ∈ Y
B(u2 , 𝑣) = F(𝑣) for all 𝑣 ∈ Y .
Using property 1 of bilinear forms stated in the appendix, Section A.1.3, we have
B(u1 − u2 , 𝑣) = 0 for all 𝑣 ∈ Y .
Selecting 𝑣 = u1 − u2 we have B(u1 − u2 , u1 − u2 ) ≡ 2||u1 − u2 ||2E(I) = 0. That is, u1 = u2 in
energy space. Observe that when c = 0 and u1 = u2 + C where C is an arbitrary constant, then
||u1 − u2 ||E(I) = 0.
7 Victor Gustave Robin (1855–1897).


The exact solution of the generalized formulation uEX is called the generalized solution or weak
solution whereas the solution that satisfies equation (1.5) is called the strong solution. The gener-
alized formulation has the following important properties:
1. The exact solution, denoted by uEX , exists for all data that satisfy the conditions 0 < 𝛼 ≤ 𝜅(x) ≤
𝛽 < ∞ where 𝛼 and 𝛽 are real numbers, 0 ≤ c(x) < ∞ and f is such that F(𝑣) satisfies the defini-
tive properties of linear forms listed in Section A.1.2 for all 𝑣 ∈ E(I). Note that 𝜅, c and f can be
discontinuous functions.
2. The exact solution is unique in the energy space, see Theorem 1.1.
3. If the data are sufficiently smooth for the strong solution to exist then the strong and weak
solutions are the same.
4. This formulation makes it possible to find approximations to uEX with arbitrary accuracy. This
will be addressed in detail in subsequent sections.
Exercise 1.2 Assume that u(0) = û0 and (𝜅u′ )x=𝓁 = k𝓁 (𝛿𝓁 − u(𝓁)) are given. State the generalized
formulation.
Exercise 1.3 Consider the sequence of functions un (x) ∈ E(I)

{
−x + (2𝓁∕n + b) for 0 ≤ x ≤ 𝓁∕n
un (x) =
x+b for 𝓁∕n < x ≤ 𝓁
illustrated in Fig. 1.2. Show that un (x) converges to u(x) = x + b in the space E(I) as n → ∞. For the
definition of convergence refer to Section A.2 in the appendix.
This exercise illustrates that restriction imposed on u′ (or higher derivatives of u) at the
boundaries will not impose a restriction on E(I). Therefore natural boundary conditions cannot be
enforced by restriction. Whereas all functions in E(I) are continuous and bounded, the derivatives
do not have to be continuous or bounded.
Exercise 1.4 Show that F(𝑣) defined on E(I) by eq. (1.20) satisfies the properties of linear forms
listed in Section A.1.2 if f is square integrable on I. This is a sufficient but not necessary condition
for F(𝑣) to be a linear form.
Figure 1.2 Exercise 1.3: The function un (x).

u
un 45˚
b + 2l/n
b + l/n
x
0 l/n l
Remark 1.1 F(𝑣) defined on E(I) by eq. (1.20) satisfies the properties of linear forms listed in
Section A.1.2 if the following inequality is satisfied:
𝓁
f 𝑣 dx < ∞ for all 𝑣 ∈ E(I). (1.35)
∫0
1.2.2 The principle of minimum potential energy

Theorem 1.2 The function u ∈ E(I) ̃ that satisfies B(u, 𝑣) = F(𝑣) for all 𝑣 ∈ E0 (I) minimizes
the quadratic functional8 𝜋(u), called the potential energy;
def 1
𝜋(u) = B(u, u) − F(u) (1.36)
2
̃
on the space E(I).
Proof: For any 𝑣 ∈ E0 (I), ||𝑣||E ≠ 0 we have:
1
𝜋(u + 𝑣) = B(u + 𝑣, u + 𝑣) − F(u + 𝑣)
2
1 1
= B(u, u) + B(u, 𝑣) + B(𝑣, 𝑣) − F(u) − F(𝑣)
2 2
1
= 𝜋(u) + B(u, 𝑣) − F(𝑣) + B(𝑣, 𝑣) (1.37)
⏟⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏟ 2
0
where B(𝑣, 𝑣) > 0 unless ||𝑣||E(I) = 0. Therefore any admissible nonzero perturbation of u will
increase 𝜋(u).
This important theorem, called the theorem or principle of minimum potential energy, will be
used in Chapter 7 as our starting point in the formulation of mathematical models for beams, plates
and shells.
Given the potential energy and the space of admissible functions, it is possible to determine
the strong form. This is illustrated by the following example.
Example 1.2 Let us determine the strong form corresponding to the potential energy defined by
𝓁( ( ) ) 𝓁
1 2 1
𝜋(u) = 𝜅 u′ + cu2 dx + k0 u2 (0) − fu dx − k0 𝛿0 u(0) (1.38)
2 ∫0 2 ∫0
̃ = {u | u ∈ E(I), u(𝓁) = û𝓁 }.
with E(I)
Since u minimizes 𝜋(u), any perturbation of u by 𝑣 ∈ E0 (I) will increase 𝜋(u). Therefore 𝜋(u + 𝜖𝑣)
is minimum at 𝜖 = 0 and hence
d𝜋(u + 𝜖𝑣) ||
| = 0. (1.39)
d𝜖 |𝜖=0
Therefore we have
𝓁( ) 𝓁
𝜅u′ 𝑣′ + cu𝑣 dx − f 𝑣 dx + k0 u(0)𝑣(0) − k0 𝛿0 𝑣(0) = 0 (1.40)
∫0 ∫0 ⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟
0
where the last two terms are zero because 𝑣 ∈ E0 (I). Integrating the first term by parts,
𝓁 𝓁
𝜅u′ 𝑣′ dx = 𝜅u′ (𝓁)𝑣(𝓁) − 𝜅u′ (0)𝑣(0) − (𝜅u′ )′ 𝑣 dx
∫0 ⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟ ∫0
0
8 A functional is a real-valued function defined on a space of functions or vectors.

and, substituting this into eq. (1.40), we get

𝓁 ( )
−(𝜅u′ )′ + cu − f 𝑣 dx = 0. (1.41)
∫0
Since this holds for all 𝑣 ∈ E0 (I), the bracketed expression must be zero. In other words, the solution
of the differential equation
−(𝜅u′ )′ + cu = f , (𝜅u′ )x=0 = k0 (u(0) − 𝛿0 ), u(𝓁) = û𝓁 (1.42)
minimizes the potential energy defined by eq. (1.38). This is the strong form of the problem.
Remark 1.2 The procedure in Example 1.2 is used in the calculus of variations for identifying
the differential equation, known as the Euler9 -Lagrange10 equation, the solution of which maxi-
mizes or minimizes a functional. In this example the solution minimizes the potential energy on the
̃
space E(I).
Remark 1.3 Whereas the strain energy is always positive, the potential energy may be positive,
negative or zero.
1.3 Approximate solutions
The trial and test spaces defined in the preceding section are infinite-dimensional, that is, they
span infinitely many linearly independent functions. To find an approximate solution, we construct
finite-dimensional subspaces denoted, respectively, by S ⊂ X, V ⊂ Y and seek the function u ∈ S
that satisfies B(u, 𝑣) = F(𝑣) for all 𝑣 ∈ V. Let us return to the introductory example described in
Section 1.1 and define
∑
n
∑
n
u = un = aj 𝜑j , 𝑣 = 𝑣n = bi 𝜑i
j=1 i=1
where 𝜑i (i = 1, 2, … n) are basis functions. Using the definitions of kij and mij given in eq. (1.12),
we write the bilinear form as
𝓁 ∑ ∑
n n
B(u, 𝑣) ≡ (𝜅u′ 𝑣′ + cu𝑣) dx = (kij + mij )aj bi
∫0 i=1 j=1
= {b}T ([K] + [M]){a}. (1.43)

Similarly,
𝓁 ∑
n
F(𝑣) ≡ f 𝑣 dx = bi ri = {b}T {r} (1.44)
∫0 i=1
where ri is defined in eq. (1.12). Therefore we can write B(u, 𝑣) − F(𝑣) = 0 in the following form:
{b}T (([K] + [M]){a} − {r}) = 0. (1.45)
Since this must hold for any choice of {b}, it follows that
([K] + [M]){a} = {r} (1.46)
9 Leonhard Euler 1707–1783.

10 Joseph-Louis Lagrange 1736–1813.
which is the same system of linear equations we needed to solve when minimizing the integral
, see eq. (1.14). Of course, this is not a coincidence. The solution of the generalized problem:
“Find un ∈ S such that B(un , 𝑣) = F(𝑣) for all 𝑣 ∈ V”, minimizes the error in the energy norm. See
Theorem 1.4.
Theorem 1.3 The error e defined by e = u − un satisfies B(e, 𝑣) = 0 for all 𝑣 ∈ S0 (I). This result
follows directly from
B(u, 𝑣) = F(𝑣) for all 𝑣 ∈ S0 (I)
B(un , 𝑣) = F(𝑣) for all 𝑣 ∈ S0 (I).
Subtracting the second equation from the first we have,
B(u − un , 𝑣) ≡ B(e, 𝑣) = 0 for all 𝑣 ∈ S0 (I). (1.47)
This equation is known as the Galerkin11 orthogonality condition.
Theorem 1.4 If un ∈ S0 (I) satisfies B(un , 𝑣) = F(𝑣) for all 𝑣 ∈ S0 (I) then un minimizes the error
uEX − un in energy norm where uEX is the exact solution:
‖u − u ‖
‖ EX n ‖E(I) = min ||uEX − u||E(I) . (1.48)
u∈S̃
Proof: Let e = u − un and let 𝑣 be an arbitrary function in S0 (I). Then

1 1 1
‖e + 𝑣‖2E(I) ≡ B(e + 𝑣, e + 𝑣) = B(e, e) + B(e, 𝑣) + B(𝑣, 𝑣).
2 2 2
The first term on the right is ||e||2E(I) , the second term is zero on account of Theorem 1.3, the third
term is positive for any 𝑣 ≠ 0 in S0 (I). Therefore ||e||E(I) is minimum.
Theorem 1.4 states that the error depends on the exact solution of the problem uEX and the defi-
̃
nition of the trial space S(I).
The finite element method is a flexible and powerful method for constructing trial spaces. The
basic algorithmic structure of the finite element method is outlined in the following sections.
1.3.1 The standard polynomial space

The standard polynomial space of degree p, denoted by  p (Ist ), is spanned by the monomials
1, 𝜉, 𝜉 2 , … , 𝜉 p defined on the standard element
Ist = {𝜉 | − 1 < 𝜉 < 1}. (1.49)
The choice of basis functions is guided by considerations of implementation, keeping the con-
dition number of the coefficient matrices small, and personal preferences. For the symmetric
positive-definite matrices considered here the condition number C is the largest eigenvalue
divided by the smallest. The number of digits lost in solving a linear problem is roughly equal to
log10 C. Characterizing the condition number as being large or small should be understood in this
context. In the finite element method the condition number depends on the choice of the basis
functions and the mesh.
The standard polynomial basis functions, called shape functions, can be defined in various ways.
We will consider shape functions based on Lagrange polynomials and Legendre12 polynomials.
We will use the same notation for both types of shape function.
11 Boris Grigoryevich Galerkin 1871–1945.

12 Adrien-Marie Legendre 1752–1833.
Lagrange shape functions

Lagrange shape functions of degree p are constructed by partitioning Ist into p sub-intervals. The
length of the sub-intervals is typically 2∕p but the lengths may vary. The node points are 𝜉1 = −1,
𝜉2 = 1 and −1 < 𝜉3 < 𝜉4 < · · · < 𝜉p+1 < 1. The ith shape function is unity in the ith node point and
is zero in the other node points:
∏
p+1
𝜉 − 𝜉k
Ni (𝜉) = , i = 1, 2, … , p + 1, 𝜉 ∈ Ist ⋅ (1.50)
𝜉i − 𝜉k
k=1
k≠i
These shape functions have the following important properties:
{
1 if i = j ∑
p+1
Ni (𝜉j ) = and Ni (𝜉) = 1. (1.51)
0 if i ≠ j i=1
For example, for p = 2 the equally spaced node points are 𝜉1 = −1, 𝜉2 = 1, 𝜉3 = 0. The correspond-
ing Lagrange shape functions are illustrated in Fig. 1.3.
Exercise 1.5 Sketch the Lagrange shape functions for p = 3.
Legendre shape functions

For p = 1 we have
1−𝜉 1+𝜉
N1 = , N2 = ⋅ (1.52)
2 2
For p ≥ 2 we define the shape functions as follows:
√ 𝜉
2i − 3
Ni (𝜉) = P (t) dt i = 3, 4, … , p + 1 (1.53)
2 ∫−1 i−2
1.0
1
N1 = ξ(ξ – 1)
2
–1.0 0 1.0 ξ
1.0
1
N2 = ξ(1 + ξ)
2
–1.0 0 1.0 ξ
1.0
N3 = 1 – ξ 2
–1.0 0 1.0 ξ
Figure 1.3 Lagrange shape functions in one dimension, p = 2.

where Pi (t) are the Legendre polynomials. The definition of Legendre polynomials is given in
Appendix D. These shape functions have the following important properties:
1. Orthogonality. For i, j ≥ 3:
{
+1
dNi dNj 1 if i = j
d𝜉 = (1.54)
∫−1 d𝜉 d𝜉 0 if i ≠ j.
This property follows directly from the orthogonality of Legendre polynomials, see eq. (D.13) in
the appendix.
2. The set of shape functions of degree p is a subset of the set of shape functions of degree p + 1.
Shape functions that have this property are called hierarchic shape functions.
3. These shape functions vanish at the endpoints of Ist : Ni (−1) = Ni (+1) = 0 for i ≥ 3.
The first five hierarchic shape functions are shown in Fig. 1.4. Observe that all roots lie in Ist .
Additional shape functions, up to p = 8, can be found in the appendix, Section D.1.
Exercise 1.6 Show that for the hierarchic shape functions, defined by eq. (1.53), Ni (−1) =
Ni (+1) = 0 for i ≥ 3.
1.0
1–ξ
N1 =
2
–1.0 0 1.0 ξ
1.0
1+ξ
N2 =
2
–1.0 0 1.0 ξ
0
1 3 2
N3 = (ξ – 1)
2 2
5
N4 = ξ(ξ2 – 1)
2√10
1 7
N5 = (5ξ2 – 1)(ξ2 – 1)
8 2
Figure 1.4 Legendre shape functions in one dimension, p = 4.

Exercise 1.7 Show that the hierarchic shape functions defined by eq. (1.53) can be written in the
form:
1 ( )
Ni (𝜉) = √ Pi−1 (𝜉) − Pi−3 (𝜉) i = 3, 4, … (1.55)
2(2i − 3)
Hint: note that Pn (1) = 1 for all n and use equations (D.10) and (D.12) in Appendix D.
1.3.2 Finite element spaces in one dimension

We are now in a position to provide a precise definition of finite element spaces in one dimension.
The domain I = {x | 0 < x < 𝓁} is partitioned into M non-overlapping intervals called finite ele-
ments. A partition, called finite element mesh, is denoted by Δ. Thus M = M(Δ). The boundary
points of the elements are the node points. The coordinates of the node points, sorted in ascending
order, are denoted by xi , (i = 1, 2, … , M + 1) where x1 = 0 and xM+1 = 𝓁. The kth element Ik has the
boundary points xk and xk+1 , that is, Ik = {x | xk < x < xk+1 }.
Various approaches are used for the construction of sequences of finite element mesh. We will
consider four types of mesh design:
1. A mesh is uniform if all elements have the same size. On the interval I = (0, 𝓁) the node points
are located as follows:
xk = (k − 1)𝓁∕M(Δ) for k = 1, 2, 3, … , M(Δ) + 1.
2. A sequence of meshes ΔK (K = 1, 2, …) is quasiuniform if there exist positive constants C1 , C2 ,

independent of K, such that
(K)
𝓁max
C1 ≤ (K)
≤ C2 , K = 1, 2, … (1.56)
𝓁min
(K) (K)
where 𝓁max (resp. 𝓁min ) is the length of the largest (resp. smallest) element in mesh ΔK . In two
and three dimensions 𝓁k is defined as the diameter of the kth element, meaning the diameter
of the smallest circle or sphere that envelopes the element. For example, a sequence of quasi-
uniform meshes would be generated in one dimension if, starting from an arbitrary mesh, the
elements would be successively halved.
3. A mesh is geometrically graded toward the point x = 0 on the interval 0 < x < 𝓁 if the node
points are located as follows:
{
0 for k = 1
xk = (1.57)
qM(Δ)+1−k 𝓁 for k = 2, 3, … , M(Δ) + 1
where 0 < q < 1 is called grading factor or common factor. These are called geometric meshes.
4. A mesh is a radical mesh if on the interval 0 < x < 𝓁 the node points are located by
( )𝜃
k−1
xk = 𝓁, 𝜃 > 1, k = 1, 2, … , M(Δ) + 1. (1.58)
M(Δ)
The question of which of these schemes is to be preferred in a particular application can be

answered on the basis of a priori information concerning the regularity of the exact solution and
aspects of implementation. Practical considerations that should guide the choice of the finite ele-
ment mesh will be discussed in Section 1.5.2.
When the exact solution has one or more terms like |x − x0 |𝛼 , and 𝛼 > 1∕2 is a fractional number,
then the ideal mesh is a geometrically graded mesh and the polynomial degrees are assigned in such
a way that the smallest elements are assigned
√ the lowest polynomial degree, the largest elements the
highest. The optimal grading factor is q = ( 2 − 1)2 ≈ 0.17 which is independent of 𝛼. The assigned
polynomial degrees should increase at a rate of approximately 0.4 [45].
The ideal meshes are radical meshes when the same polynomial degree is assigned to each ele-
ment. The optimal value of 𝜃 depends on p and 𝛼:
p + 1∕2
𝜃= (1.59)
𝛼 − 1∕2 + (n − 1)∕2
where n is the number of spatial dimensions. For a detailed analysis of discretization schemes in
one dimension see reference [45].
The relationship between the kth element of the mesh and the standard element Ist is defined by
the mapping function
1−𝜉 1+𝜉
x = Qk (𝜉) = x + x , 𝜉 ∈ Ist . (1.60)
2 k 2 k+1
A finite element space S is a set of functions characterized by Δ, the assigned polynomial degrees
pk ≥ 1 and the mapping functions Qk (𝜉), k = 1, 2, … , M(Δ). Specifically;
S = S(I, Δ, p, Q) = {u | u ∈ E(I), u(Qk (𝜉)) ∈  pk (Ist ), k = 1, 2, … , M(Δ)} (1.61)
where p and Q represent, respectively, the arrays of the assigned polynomial degrees and the map-
ping functions. This should be understood to mean that u ∈ S if and only if u satisfies the conditions
on the right of the vertical bar (|). The first condition u ∈ E(I) is that u must lie in the energy
space. In one dimension this implies that u must be continuous on I. The expression u(Qk (𝜉)) ∈
 pk (Ist ) indicates that on element Ik the function u(x) is mapped from the standard polynomial
space  pk (Ist ).
The finite element test space, denoted by S0 (I), is defined by the intersection S0 (I) = S(I) ∩ E(I),
that is, u ∈ S0 (I) is zero in those boundary points where essential boundary conditions are pre-
scribed. The number of basis functions that span S0 (I) is called the number of degrees of freedom.
The process by which the number of degrees of freedom is progressively increased by mesh refine-
ment, with the polynomial degree fixed, is called h-extension and its implementation the h-version
of the finite element method. The process by which the number of degrees of freedom is progres-
sively increased by increasing the polynomial degree of elements, while keeping the mesh fixed, is
called p-extension and its implementation the p-version of the finite element method. The process
by which the number of degrees of freedom is progressively increased by concurrently refining the
mesh and increasing the polynomial degrees of elements is called hp-extension and its implemen-
tation the hp-version of the finite element method.
Remark 1.4 It will be explained in Chapter 5 that the separate naming of the h, p and hp versions
is related to the evolution of the finite element method rather than its theoretical foundations.
1.3.3 Computation of the coefficient matrices

The coefficient matrices are computed element by element. The numbering of the coefficients is
based on the numbering of the standard shape functions, the indices range from 1 through pk+1 .
This numbering will have to be reconciled with the requirement that each basis function must be
continuous on I and must have an unique identifying number. This will be discussed separately.
Computation of the stiffness matrix

The first term of the bilinear form in eq. (1.43) is computed as a sum of integrals over the elements
𝓁 ∑
M(Δ) xk+1
𝜅(x)u′n 𝑣′n dx = 𝜅(x)u′n 𝑣′n dx. (1.62)
∫0 ∫xk
k=1
We will be concerned with the evaluation of the integral on the kth element:
(p +1 ) (p +1 )
xk+1 xk+1 ∑ k dNj ∑ k
dNi
𝜅(x)un 𝑣n dx =
′ ′
𝜅(x) aj bi dx.
∫xk ∫xk j=1
dx i=1
dx
The shape functions Ni are defined on the standard domain Ist . Referring to the mapping function
given by eq. (1.60), we have
xk+1 − xk 𝓁
dx = d𝜉 ≡ k d𝜉 (1.63)
2 2
def
where 𝓁k = xk+1 − xk is the length of the kth element. Also,
d d d𝜉 2 d 2 d
= = ≡ ⋅
dx d𝜉 dx xk+1 − xk d𝜉 𝓁k d𝜉
Therefore
(p +1 ) (p +1 )
xk+1
2
+1 ∑ k dNj ∑ k
dNi
𝜅(x)u′n 𝑣′n dx = 𝜅(Qk (𝜉)) aj bi d𝜉.
∫xk 𝓁k ∫−1 j=1
d𝜉 i=1
d𝜉
We define
2
+1
dN dNj
kij(k) = 𝜅(Qk (𝜉)) i d𝜉 (1.64)
𝓁k ∫−1 d𝜉 d𝜉
and write
xk+1 ∑∑
pk +1pk +1
𝜅(x)u′n 𝑣′n dx = kij(k) aj bi ≡ {b}T [K (k) ]{a}. (1.65)
∫xk i=1 j=1
The terms of the stiffness matrix kij(k) depend on the the mapping, the definition of the shape func-
tions and the function 𝜅(x). The matrix [K (k) ] is called the element stiffness matrix. Observe that
kij(k) = kji(k) , that is, [K (k) ] is symmetric. This follows directly from the symmetry of B(u, 𝑣) and the
fact that the same basis functions are used for un and 𝑣n .
In the finite element method the integrals are evaluated by numerical methods. Numerical inte-
gration is discussed in Appendix E. In the important special case when 𝜅(x) = 𝜅k is constant on Ik ,
it is possible to compute [K (k) ] once and for all. This is illustrated by the following example.
Example 1.3 When 𝜅(x) = 𝜅k is constant on Ik and the Legendre shape functions are used then,
with the exception of the first two rows and columns, the element stiffness matrix is perfectly
diagonal:
⎡1∕2 −1∕2 0 0 · · · 0⎤
⎢ ⎥
⎢ 1∕2 0 0 0⎥
2𝜅 ⎢ 1 0 0⎥
[K (k) ] = k ⎢ ⋅
0⎥⎥
(1.66)
𝓁k ⎢ 1
⎢ (sym.) ⋱ ⋮⎥
⎢ 1⎥⎦
⎣
Exercise 1.8 Assume that 𝜅(x) = 𝜅k is constant on Ik . Using the Lagrange shape functions dis-
(k) (k)
played in Fig. 1.3 for p = 2, compute k11 and k13 in terms of 𝜅k and 𝓁k .
Computation of the Gram matrix

The second term of the bilinear form is also computed as a sum of integrals over the elements:
𝓁 ∑
M(Δ) xk+1
c(x)un 𝑣n dx = c(x)un 𝑣n dx. (1.67)
∫0 ∫xk
k=1
We will be concerned with evaluation of the integral

(p +1 ) (p +1 )
xk+1 xk+1 ∑ k ∑k
c(x)un 𝑣n dx = c(x) a j Nj bi Ni dx
∫xk ∫xk j=1 i=1
(p +1 ) (p +1 )
𝓁k +1 ∑k ∑k
= c(Qk (𝜉)) a j Nj bi Ni d𝜉.

2 ∫−1 j=1 i=1
Defining:
𝓁k 1
m(k) = c(Qk (𝜉))Ni Nj d𝜉 (1.68)
ij 2 ∫−1
the following expression is obtained:
xk+1 ∑∑
pk +1pk +1
c(x)un 𝑣n dx = m(k) a b = {b}T [M (k) ]{a} (1.69)
∫xk i=1 j=1
ij j i
where {a} = {a1 a2 … apk +1 }T , {b}T = {b1 b2 … bpk +1 } and
⎡ m(k)
11 m(k)
12
· · · m1,pk +1 ⎤
⎢ m(k) m(k) · · · m2,pk +1 ⎥⎥
(k)
[M ] = ⎢ 21 22 ⋅
⎢ ⋮ ⋱ ⋮ ⎥
⎢m(k) (k)
· · · mpk +1,pk +1 ⎥⎦
⎣ pk +1,1 mpk +1,2
The terms of the coefficient matrix m(k) ij

are computable from the mapping, the definition of the
shape functions and the function c(x). The matrix [M (k) ] is called the element-level Gram matrix13
or the element-level mass matrix. Observe that [M (k) ] is symmetric. In the important special case
where c(x) = ck is constant on Ik it is possible to compute [M (k) ] once and for all. This is illustrated
by the following example.
Example 1.4 When c(x) = ck is constant on Ik and the Legendre shape functions are used then
the element-level Gram matrix is strongly diagonal. For example, for pk = 5 the Gram matrix is:
√ √
⎡2∕3 1∕3 −1∕ 6 1∕3 10 0 0 ⎤
⎢ ⎥
⎢ √ √ ⎥
⎢ 2∕3 −1∕ 6 −1∕3 10 0 0 ⎥
⎢ ⎥
⎢ √ ⎥
[ (k) ] ck 𝓁k ⎢ 2∕5 0 −1∕5 21 0 ⎥
M = ⎢ ⎥ (1.70)
2 ⎢ √ ⎥
⎢ 2∕21 0 −1∕7 45 ⎥
⎢ (sym.) ⎥
⎢ ⎥
⎢ 2∕45 0 ⎥
⎢ ⎥
⎢ ⎥
⎣ 2∕77 ⎦
13 Jörgen Pedersen Gram 1850–1916.

Remark 1.5 For pk ≥ 2 a simple closed form expression can be obtained for the diagonal terms
and the off-diagonal terms. Using eq. (1.55) it can be shown that:
ck 𝓁k 1
+1
m(k) = (Pi−1 (𝜉) − Pi−3 (𝜉))2 d𝜉
ii 2 2(2i − 3) ∫−1
c 𝓁 2
= k k , i≥3 (1.71)
2 (2i − 1)(2i − 5)
and all off-diagonal terms are zero for i ≥ 3, with the exceptions:
ck 𝓁k 1
m(k)
i,i+2
= m(k)
i+2,i
=− √ , i ≥ 3. (1.72)
2 (2i − 1) (2i − 3)(2i + 1)
Remark 1.6 It has been proposed to make the Gram matrix perfectly diagonal by using Lagrange
shape functions of degree p with the node points coincident with the Lobatto points. Therefore
Ni (𝜉j ) = 𝛿ij where 𝛿ij is the Kronecker delta14 . Then, using p + 1 Lobatto points, we get:
ck 𝓁k 1 c 𝓁
m(k) = Ni Nj d𝜉 ≈ k k 𝑤i 𝛿ij
ij 2 ∫ −1 2
where 𝑤i is the weight of the ith Lobatto point. There is an integration error associated with
this term because the integrand is a polynomial of degree 2p. To evaluate this integral exactly
n ≥ (2p + 3)∕2 Lobatto points would be required (see Appendix E), whereas only p + 1 Lobatto
points are used. Throughout this book we will be concerned with errors of approximation that can
be controlled by the design of mesh and the assignment of polynomial degrees. We will assume
that the errors of integration and errors in mapping are negligibly small in comparison with the
errors of discretization.
Exercise 1.9 Assume that c(x) = ck is constant on Ik . Using the Lagrange shape functions of
degree p = 3, with the nodes located in the Lobatto points, compute m(k) 33
numerically using 4
Lobatto points. Determine the relative error of the numerically integrated term. Refer to Remark 1.6
and Appendix E.
Exercise 1.10 Assume that c(x) = ck is constant on Ik . Using the Lagrange shape functions of
degree p = 2, compute m(k) (k)
11 and m13 in terms of ck and 𝓁k .
1.3.4 Computation of the right hand side vector

Computation of the right hand side vector involves evaluation of the functional F(𝑣), usually by
numerical means. In particular, we write:
𝓁 ∑
M(Δ) xk+1
F(𝑣n ) = f (x)𝑣n dx = f (x)𝑣n dx. (1.73)
∫0 ∫xk
k=1
The element-level integral is computed from the definition of 𝑣n on Ik :

(p )
xk+1
𝓁k +1 ∑ k+1
(k)
∑
pk+1
f (x)𝑣n dx = f (Qk (𝜉)) bi Ni d𝜉 = b(k) ri(k) (1.74)
∫xk 2 ∫−1 i=1 i=1
i
14 The definition of 𝛿ij is given by eq. (2.1).

where
def 𝓁k +1
ri(k) = f (Qk (𝜉))Ni (𝜉) d𝜉 (1.75)
2 ∫−1
which is computed from the given data and the shape functions.
Example 1.5 Let us assume that f (x) is a linear function on Ik . In this case f (x) can be written as
1−𝜉 1+𝜉
f (x) = f (xk ) + f (xk+1 ) = f (xk )N1 (𝜉) + f (xk+1 )N2 (𝜉).
2 2
Using the Legendre shape functions we have:
𝓁k +1 2 𝓁 +1
𝓁k ( )
r1(k) = f (xk ) N1 d𝜉 + f (xk+1 ) k N N d𝜉 = 2f (xk ) + f (xk+1 )
2 ∫−1 2 ∫−1 1 2 6
𝓁 +1
𝓁 +1
𝓁k ( )
r2(k) = f (xk ) k N1 N2 d𝜉 + f (xk+1 ) k N 2 d𝜉 = f (xk ) + 2f (xk+1 )
2 ∫−1 2 ∫−1 2 6
𝓁 +1
𝓁 +1
r3(k) = f (xk ) k N1 N3 d𝜉 + f (xk+1 ) k N N d𝜉
2 ∫−1 2 ∫−1 2 3
√
𝓁 3( )
=− k f (xk ) + f (xk+1 ) .
6 2
Exercise 1.11 Assume that f (x) is a linear function on Ik . Using the Legendre shape functions
compute r4(k) and show that ri(k) = 0 for i > 4. Hint: Make use of eq. (1.55).
Exercise 1.12 Let

x − xk
f (x) = fk sin 𝜋, x ∈ Ik
𝓁k
where fk is a constant. Compute r5(k) numerically in terms of fk and 𝓁k using 3, 4 and 5 Gauss points.
See Appendix E. Use the Legendre basis functions.
Exercise 1.13 Assume that f (x) is a linear function on I. Using the Lagrange shape functions for
p = 2, compute r1(k) .
1.3.5 Assembly
Having computed the coefficient matrices and right hand side vectors for each element, it is nec-
essary to form the coefficient matrix and right hand side vector for the entire mesh. This process,
called assembly, executes the summation in equations (1.62), (1.67) and (1.73). The local and global
numbering of variables is reconciled in the assembly process. The algorithm is illustrated by the
following example.
Example 1.6 Consider the three-element mesh shown in Fig. 1.5. The polynomial degrees p1 = 2,
p2 = 1, p3 = 3 are assigned to elements 1, 2, 3 respectively. The basis functions shown in Fig. 1.5
are composed of the mapped Legendre shape functions. For instance, the basis function 𝜑2 (x) is
composed of the mapped shape function N2 from element 1 and the mapped shape function N1
from element 2. This basis function is zero over element 3. Basis function 𝜑6 (x) is the mapped
shape function N3 from element 3. This basis function is zero over elements 1 and 2.
x
k=1 k=2 k=3
x1 = 0 x2 x3 x4 = l
l1 l2 l3
φ1(x)
1
1 2 3 4
φ2(x)
1
φ3(x)
1
φ4(x)
1
φ5(x)
φ6(x)
φ7(x)
Figure 1.5 Typical finite element basis functions in one dimension.
Table 1.1 Local and global numbering in Example 1.6.
Element number
Numbering 1 2 3
local 1 2 3 1 2 1 2 3 4
global 1 2 5 2 3 3 4 6 7
Each basis function is assigned a unique number, called a global number, and this number is
associated with those element numbers and the shape function numbers from which the basis
function is composed. The global and local numbers in this example are indicated in Table 1.1.
We denote c(k)
ij
= kij(k) + m(k)
ij
and, using equations (1.62) and (1.67), write B(un , 𝑣n ) in the following
form:
a1 a2 a5
a2 a3
b1 ⎛c(1) c(1) c(1) ⎞ ( )
B(un , 𝑣n ) = ⎜ 11 12
(1) ⎟
13
+ b2 c(2) c(2)
b2 ⎜c(1) c(1) c23 ⎟ 11 12
⎜
21 22
⎟ b3 c(2) c(2)
b5 ⎝c(1)
31
c(1)
32
c(1)
33 ⎠
21 22
a3 a4 a6 a7
b3 ⎛c(3) c(3) c(3) c(3) ⎞
⎜ 11
(3)
12 13 14 ⎟
+ b4 ⎜c21 c(3)
22
c(3)
23
c(3)
24 ⎟
⎜ (3) ⎟
b6 ⎜c31 c(3)
32
c(3)
33
c(3)
34 ⎟
b7 ⎜c(3) c(3) c(3) c(3) ⎟
⎝ 41 42 43 44 ⎠
where the elements within the brackets are in the local numbering system whereas the coefficients
aj and bi outside of the brackets are in the global system. The superscripts indicate the element num-
bers. The terms multiplied by aj bi are summed to obtain the elements of the assembled coefficient
matrix which will be denoted by cij . For example,
c11 = c(1)
11 , c22 = c(1)
22
+ c(2)
11 , c33 = c(2)
22
+ c(3)
11 , c77 = c(3)
44
.
Assuming that the boundary conditions do not include Dirichlet conditions, the bilinear form can
be written in terms of the 7 × 7 coefficient matrix as:
⎡c11 c12 · · · c17 ⎤ ⎧a1 ⎫

∑ 7 7
∑ ⎢ ⎥⎪ ⎪
⎢ c21 c22 c27 ⎥ ⎪a2 ⎪
B(un , 𝑣n ) =
⋮ ⎥⎨ ⎬
cij aj bi = {b1 b2 · · · b7 }
j=1 i=1
⎢⋮ ⎪⋮⎪
⎢c ⎥
⎣ 71 c72 · · · c77 ⎦ ⎪ a⎪
⎩ 7⎭
≡{b}T [C]{a}. (1.76)
The treatment of Dirichlet conditions will be discussed separately in the next section.
The assembly of the right hand side vector from the element-level right hand side vectors is anal-
ogous to the procedure just described. Referring to eq. (1.73) we write F(𝑣n ) in the following form:
⎧ (3) ⎫
⎧r (1) ⎫ { } ⎪r1(3) ⎪
⎪ 1(1) ⎪ (2)
r1 ⎪r ⎪
F(𝑣n ) ={b1 b2 b5 } ⎨r2 ⎬ + {b2 b3 } (2) + {b3 b4 b6 b7 } ⎨ 2(3) ⎬
⎪r (1) ⎪ r2 ⎪r3(3) ⎪
⎩3 ⎭ ⎪r4 ⎪
⎩ ⎭
⎧ ⎫
⎪r1 ⎪
⎪r ⎪
={b1 b2 · · · b7 } ⎨ 2 ⎬ ≡ {b}T {r}
⎪⋮⎪
⎪r 7 ⎪
⎩ ⎭
where r1 = r1(1) , r2 = r2(1) + r1(2) , r3 = r2(2) + r1(3) , etc.
1.3.6 Condensation
Each element has p − 1 internal basis functions. Those elements of the coefficient matrix which
are associated with the internal basis functions can be eliminated at the element level. This process
is called condensation.
Let us partition the coefficient matrix and right hand side vector of a finite element with p ≥ 2
such that
[ ]{ } { }
C11 C12 a1 r1
=
C21 C22 a2 r2
where the a1 = {a1 a2 }T and a2 = {a3 a4 · · · ap+1 }T . The coefficient matrix is symmetric therefore
C21 = CT12 . Using
a2 = −C−1 −1
22 C21 a1 + C22 r2 (1.77)
we get
( )
C11 − C12 C−1 22 C21 a1 = r1 − C12 C22 r2 .
−1
(1.78)
⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟ ⏟⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏟
Condensed[C] Condensed{r}
The condensed stiffness matrices and load vectors are assembled and the Dirichlet boundary con-
ditions are enforced as described in the following section. Upon solving the assembled system of
equations the coefficients of the internal basis functions are computed from eq. (1.77) for each
element.
1.3.7 Enforcement of Dirichlet boundary conditions

When Dirichlet conditions are specified on either or both boundary points then u ∈ S(I)̃ is split
0 ̃
into two functions; a function u ∈ S (I) and an arbitrary specific function from S(I), denoted by
u★ . We then seek u ∈ S0 (I) such that
𝓁 𝓁 𝓁
(𝜅(u★ )′ 𝑣′ + cu★ 𝑣) dx
′
(𝜅u 𝑣′ + cu𝑣) dx = f 𝑣 dx − (1.79)
∫0 ∫0 ∫0
⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟ ⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟
B(u,𝑣) F(𝑣)
for all 𝑣 ∈ Observe that the solution u = u + u★ is independent of the choice of u★ .

S0 (I).
We denote the global numbers of the basis functions that are unity at x = 0 and x = 𝓁 by K and L
respectively. For instance, in Example 1.6 K = 1 and L = 4. It is advantageous to define u★ in terms
of 𝜑K (x) and 𝜑L (x):
u★ = û0 𝜑K (x) + û𝓁 𝜑L (x) (1.80)
as indicated in Fig. 1.6. When Dirichlet boundary condition is prescribed on only one of the bound-
ary points then this expression is modified to include the term corresponding to that point only.
On substituting eq. (1.80) into eq. (1.79) the second term on the right-hand side of eq. (1.79) can
be written as
𝓁 ∑
Nu
(𝜅(u★ )′ 𝑣′ + cu★ 𝑣) dx = bi (ciK + ciL )
∫0 i=1
where Nu is the number of unconstrained equations, that is, the number of equations prior to
enforcement of the Dirichlet boundary conditions. (For instance, in Example 1.6 Nu = 7.) The coef-
ficients ciK , ciL are elements of the assembled coefficient matrix.
u0
‹
ul
‹
x
x1 = 0 x2 x3 xn–1 xn xn+1 = l
Figure 1.6 Recommended choice of the function u★ in one dimension.
Since 𝑣 ∈ S0 (I), we have bK = bL = 0 and therefore the Kth and Lth rows of matrix [C] are multi-
plied by zero and can be deleted. The Kth and Lth columns of matrix [C] are multiplied by û0 and
û𝓁 respectively, summed and the resulting vector is transferred to the right-hand side. The result-
ing coefficient matrix has the dimension N which is Nu minus the number of Dirichlet boundary
conditions. The number N is called the number of degrees of freedom. It is the maximum number
of linearly independent functions in S0 (I).
Remark 1.7 In order to avoid having to renumber the coefficient matrix once the rows and
columns corresponding to 𝜑K and 𝜑L were eliminated, all elements in the Kth and Lth rows and
columns can be set to zero, with the exception of the diagonal elements, which are set to unity.
The corresponding elements on the right hand side vector are set to û0 and û𝓁 . This is illustrated
by the following example.
Example 1.7 Consider the problem

−u′′ + 4u = 0, u(0) = 1, u(1) = 2
the exact solution of which is
exp(2) − 2 2 − exp(−2)
u= exp(−2x) + exp(2x).
exp(2) − exp(−2) exp(2) − exp(−2)
Using five elements of equal length on the interval I = (0, 1) and p = 1 assigned to each element,
find the finite element solution for this problem.
Referring to equations (1.66) and (1.70), the element-level coefficient matrix for each element is
[ ]
[ (k) ] 79∕15 −73∕15
C = , k = 1, 2, … 5
−73∕15 79∕15
where we used 𝜅k = 1, ck = 4, 𝓁k = 1∕5. The assembled unconstrained coefficient matrix is:
⎡ 79∕15 −73∕15 0 0 0 0⎤
⎢ ⎥
⎢−73∕15 158∕15 −73∕15 0 0 0⎥
⎢ 0 −73∕15 158∕15 −73∕15 0 0⎥
[C] = ⎢
0 0 −73∕15 158∕15 −73∕15 0⎥⋅
⎢ ⎥
⎢ 0 0 0 −73∕15 158∕15 −73∕15⎥
⎢ 0 0 0 0 −73∕15 79∕15⎥⎦
⎣
Upon enforcement of the Dirichlet conditions the system of equations is
⎡ 158∕15 −73∕15 ⎧ ⎫ ⎧ ⎫
0 0⎤ ⎪a2 ⎪ ⎪ 73∕13 ⎪
⎢ ⎥⎪ ⎪ ⎪
−73∕15 158∕15 −73∕15 0⎥ a 3 0 ⎪
[C] = ⎢
⎢ 0 −73∕15 158∕15 −73∕15⎥ ⎨ a ⎬=⎨ 0 ⎬
⎪ 4⎪ ⎪ ⎪
⎢ 0 0 −73∕15 158∕15⎥⎦ ⎪a5 ⎪ ⎪146∕15⎪
⎣ ⎩ ⎭ ⎩ ⎭
alternatively:
⎡1 0 0 0 0 0⎤ ⎧a1 ⎫ ⎧ 1 ⎫
⎢ ⎥⎪ ⎪ ⎪ ⎪
⎢0 158∕15 −73∕15 0 0 0⎥ ⎪a2 ⎪ ⎪ 73∕15 ⎪
⎢0 −73∕15 158∕15 −73∕15 0 0⎥ ⎪a3 ⎪ ⎪ 0 ⎪
[C] = ⎢
⎢ 0 0 −73∕15 158∕15 −73∕15 0⎥⎥ ⎨ ⎬=⎨
⎪a 4 ⎪ ⎪ 0 ⎪
⎬
⎢0 0 0 −73∕15 158∕15 0⎥ ⎪a5 ⎪ ⎪146∕15⎪
⎢0 0 0 0 0 1⎥⎦ ⎪ ⎪ ⎪ ⎪
⎣ ⎩a 6 ⎭ ⎩ 2 ⎭
where the first and sixth equations are placeholders for the boundary conditions a1 = 1, a6 = 2.
The solution is:
{a} = {1.0000 0.8784 0.9012 1.0722 1.4194 2.0000}T .
Exercise 1.14 Solve the problem in Example 1.7 with the boundary conditions u(0) = 1,
u′ (1) = 3.6.
Exercise 1.15 Solve the problem in Example 1.7 with the boundary conditions u′ (0) = −1,
u(1) = 2.
1.4 Post-solution operations

Following assembly of the coefficient matrix and enforcement of the essential boundary conditions
(when applicable) the resulting system of simultaneous equations is solved by one of several meth-
ods designed to exploit the symmetry and sparsity of the coefficient matrix. The solvers are classified
into two broad categories; direct and iterative solvers. Optimal choice of a solver in a particular
application is based on consideration of the size of the problem and the available computational
resources.
At the end of the solution process the finite element solution is available in the form
∑
Nu
uFE = aj 𝜑j (x) (1.81)
j=1
where the indices reference the global numbering and Nu is the number of degrees of freedom plus
the number of Dirichlet conditions.
The basis functions are decomposed into their constituent shape functions and the element-level
solution records are created in the local numbering convention. Therefore the finite element solu-
tion on the kth element is available in the following form:
∑
pk +1
u(k)
FE
= a(k)
j
Nj (𝜉). (1.82)
j=1
1.4.1 Computation of the quantities of interest

The computation of typical engineering quantities of interest (QoI) by direct and indirect methods
is outlined in this section.
Computation of uFE (x𝟎 )

Direct computation of uFE in the point x = x0 involves a search to identify the element Ik in which
point x0 lies and, using the inverse of the mapping function defined by eq. (1.60), the standard
coordinate 𝜉0 ∈ Ist corresponding to x0 is determined:

2x0 − xk − xk+1
𝜉0 = Q−1
k
(x0 ) = (1.83)
xk+1 − xk
and uFE (x0 ) is computed from
∑
pk +1
uFE (x0 ) = a(k)
j
Nj (𝜉0 ). (1.84)
j=1
′
Direct computation of uFE (x𝟎 )
Direct computation of u′FE in the point x0 involves the computation of the corresponding standard
coordinate 𝜉0 ∈ Ist using eq. (1.83) and evaluating the following expression:
( ) ( ) ( )
2 ∑ (k) dNj
pk +1
duFE 2 duFE
= = a (1.85)
dx x=x0 𝓁k d𝜉 𝜉=𝜉0 𝓁k j=1 j d𝜉
𝜉=𝜉0
def
where 𝓁k = xk+1 − xk . The computation of the higher derivatives is analogous.
Remark 1.8 When plotting quantities of interest such as the functions uFE (x) and u′FE (x), the data
for the plotting routine are generated by subdividing the standard element into n intervals of equal
length, n being the desired resolution. The QoIs corresponding to the grid-points are evaluated.
This process does not involve inverse mapping. In node points information is provided from the
two elements that share that node. If the computed QoI is discontinuous then the discontinuity
will be visible at the nodes unless the plotting algorithm automatically averages the QoIs.
′
Indirect computation of uFE (x𝟎 ) in node points
The first derivative in node points can be determined indirectly from the generalized formulation.
For example, to compute the first derivative at node xk from the finite element solution, we select
𝑣 = N1 (Q−1
k
(x)) and use
xk+1 ( ′ ′ ) xk+1 [ ] [ ]
𝜅uFE 𝑣 + cuFE 𝑣 dx = f 𝑣 dx + 𝜅u′FE 𝑣 x=x − 𝜅u′FE 𝑣 x=x . (1.86)
∫xk ∫xk k+1 k
Test functions used in post-solution operations for the computation of a functional are called extrac-
[ ]
tion functions. Here 𝑣 = N1 (Q−1k
(x)) is an extraction function for the functional − 𝜅u′FE x=x . This
k
is because 𝑣(xk ) = 1 and 𝑣(xk+1 ) = 0 and hence
[ ] xk+1 ( ) xk+1
− 𝜅u′FE x=x = 𝜅u′FE 𝑣′ + cuFE 𝑣 dx − f 𝑣 dx
k ∫xk ∫xk
∑
pk +1
= c(k)
1j j
a(k) − r1(k) (1.87)
j=1
where, by definition; c(k)

ij
= kij(k) + m(k)
ij
.
Example 1.8 Let us find u′FE (1) for the problem in Example 1.7 by the direct and indirect meth-
ods. In this case the exact solution is known from which we have u′EX (1) = 3.5978. By direct com-
putation:
( )
2 duFE
u′FE (1) = = 5(a6 − a5 ) = 2.9028 (19.32% error)
𝓁5 d𝜉 𝜉=1
and by indirect computation:

73 79
u′FE (1) = − a5 + a6 = 3.6254 (0.77% error).
15 15
Example 1.9 The following example illustrates that the indirect method can be used for obtain-
ing the QoI efficiently and accurately even when the discretization was very poorly chosen. We will
consider the problem
𝓁 𝓁
u′ 𝑣′ dx = 𝛿(x − x)𝑣 dx = 𝑣(x), u(0) = u(𝓁) = 0
∫0 ∫0
where 𝛿 is the delta function, see Definition A.5 in the appendix. Let us be interested in finding
the approximate value of u′ (0). The data are 𝓁 = 1 and x = 1∕4. We will use one finite element and
p = 2, 3, … This is a poorly chosen discretization because the derivatives of u are discontinuous
in the point x = x, whereas all derivatives of the shape functions are continuous. The proper dis-
cretization would have been to use two or more finite elements with a node point in x = x. Then
the exact solution would be obtained at p = 1.
If we use the Legendre shape functions then the coefficient matrix displayed in eq. (1.66) will
be perfectly diagonal. The first two rows and columns will be zero on account of the boundary
conditions and the diagonal term will be 2. Referring to eq. (1.75) the right hand side vector will be
ri = Ni (𝜉) where 𝜉 = Q−1 (x) = −1∕2.
Therefore the coefficients of the shape functions can be written as ai = ri+2 ∕2 (i = 1, 2, … , p − 1)
where the variables are renumbered through shifting the indices to account for the boundary con-
ditions: a1 = a2 = 0. Hence
1∑
p−1
uFE = N (𝜉)Ni+2 (𝜉)
2 i=1 i+2
and the QoI is:
∑
p−1
dNi+2
u′FE (0) = Ni+2 (𝜉) | .
i=1
d𝜉 𝜉=−1
From the definition of Ni in eq. (1.53) we have
√ √
dNi+2 || 2i + 1 2i + 1
| = Pi (−1) = (−1)i
d𝜉 ||𝜉=−1 2 2
and the QoI can be written as
√
∑
p−1
2i + 1 1∑
p−1
u′FE (0) = Ni+2 (𝜉) (−1)i = (−1)i (Pi+1 (𝜉) − Pi−1 (𝜉))
i=1
2 2 i=1
where we made use of eq. (1.55). The relationships between the polynomial degree ranging from
2 to 100 and the corresponding values of the QoI computed by the direct method are displayed in
Fig. 1.7. It is seen that convergence to the exact value u′EX (0) = 0.75 is very slow.
The indirect method is based on eq. (1.18) which, applied to this example, takes the form
1 1
u′ 𝑣′ dx = 𝛿(x)𝑣 dx + (u′ 𝑣)x=1 − (u′ 𝑣)x=0 .
∫0 ∫0
Selecting 𝑣 = 1 − x and rearranging the terms we get
1
u′ (0) = 𝑣(x) + u′ dx = 𝑣(x) = 0.75
∫0
1.2
1.0
x′FE(0) 0.8 0.75

0.6
0.4
0 20 40 60 80 100
polynomial degree p
Figure 1.7 Example 1.9. Values of u′FE (0) computed by the direct method.
which is the exact solution. The choice 𝑣 = 1 − x was exceptionally fortuitous because it happens
to be the Green’s function (also known as the influence function) for u′ (0). Therefore the extracted
value is independent of the solution u ∈ E0 (I).
Let us choose 𝑣 = 1 − x2 for the extraction function. In this case
1 1
15
u′ (0) = 𝑣(x) − u′ 𝑣′ dx = +2 u′ x dx.
∫0 16 ∫0
Substituting u′FE for u′ :
√
1 ∑
p−1
Ni+2 (𝜉)2i + 1
1
1+𝜉
u′FE x dx = Pi (𝜉) d𝜉
∫0 i=1
2 2 ∫−1 2
√
1∑
p−1 1
2i + 1 3
= Ni+2 (𝜉) P (𝜉)(P0 (𝜉) + P1 (𝜉)) d𝜉 = − ⋅
4 i=1 2 ∫−1 i 32
Taking the orthogonality of the Legendre polynomials (see eq. (D.13)) into account, the sum has to
be evaluated only for p = 2. The extracted value of u′FE (0) for p ≥ 2 is úFE (0) = 0.5156 (31.25% error).
An explanation of why the extraction method is much more efficient than direct computation is
given in Section 1.5.4.
Exercise 1.16 Find u′FE (0) for the problem in Example 1.7 by the direct and indirect methods.
Compute the relative errors.
Exercise 1.17 For the problem in Example 1.9 let 𝑣 = 1 − x3 be the extraction function. Calculate
the extracted value of u′FE (0) for p ≥ 3.
Nodal forces
The vector of nodal forces associated with element k, denoted by {f (k) }, is defined as follows:
(k)
{f (k) } = [K (k) ]{a(k) } − {r } k = 1, 2, … , M(Δ) (1.88)
(k)
where [K (k) ] is the stiffness matrix, {a(k) } is the solution vector and {r } is the load vector corre-
sponding to traction forces, concentrated forces and thermal loads acting on element k.
The sign convention for nodal forces is different from the sign convention for the bar force:
Whereas the bar force is positive when tensile, a nodal force is positive when acting in the direction
of the positive coordinate axis.
Exercise 1.18 Assume that hierarchic basis functions based on Legendre polynomials are used.
Show that when 𝜅 is constant and c = 0 on Ik then
f1(k) + f2(k) = r1(k) + r2(k)
f1(k) 1 2 f2(k)
xk Ik xk+1
Figure 1.8 Exercise 1.18. Notation.
independently of the polynomial degree pk . For sign convention refer to Fig. 1.8. Consider both
thermal and traction loads. This exercise demonstrates that nodal forces are in equilibrium inde-
pendently of the finite element solution. Therefore equilibrium of nodal forces is not an indicator
of the quality of finite element solutions.
1.5 Estimation of error in energy norm

We have seen that the finite element solution minimizes the error in energy norm in the sense of
eq. (1.48). It is natural therefore to use the energy norm as a measure of the error of approximation.
There are two types of error estimators: (a) A priori estimators that establish the asymptotic rate
of convergence of a discretization scheme, given information about the regularity (smoothness) of
the exact solution and (b) a posteriori estimators that provide estimates of the error in energy norm
for the finite element solution of a particular problem.
There is a very substantial body of work in the mathematical literature on the a priori estimation
of the rate of convergence, given a quantitative measure of the regularity of the exact solution and
a sequence of discretizations. The underlying theory is outside of the scope of this book; however,
understanding the main results is important for practitioners of finite element analysis. For details
we refer to [28, 45, 70, 84].
1.5.1 Regularity
Let us consider problems the exact solution of which has the functional form
uEX = x𝛼 𝜑(x), 𝛼 > 1∕2, x ∈ I = (0, 𝓁) (1.89)
where 𝜑(x) is an analytic or piecewise analytic function, see Definition A.1 in the appendix. Our
motivation for considering functions in this form is that this family of functions models the singular
behavior of solutions of linear elliptic boundary value problems near vertices in polygonal and
polyhedral domains. For uEX to be in the energy space, its first derivative must be square integrable
on I. Therefore
𝓁
x2(𝛼−1) dx > 0
∫0
from which it follows that 𝛼 must be greater than 1∕2.
In the following we will see that when 𝛼 is not an integer then the degree of difficulty associated
with approximating uEX by the finite element method is related to the size of (𝛼 − 1∕2) > 0. The
smaller (𝛼 − 1∕2) is, the more difficult it is to approximate uEX .
If α is a fractional number then the measure of regularity used in the mathematical literature is
the maximum number of square integrable derivatives, with the notion of derivative generalized to
fractional numbers. See sections A.2.3 and A.2.4 in the appendix. For our purposes it is sufficient
to remember that if uEX has the functional form of eq. (1.89), and 𝛼 is not an integer, then uEX lies
in the Sobolev space H 𝛼+1∕2−𝜖 (I) where 𝜖 > 0 is arbitrarily small. This means that 𝛼 must be larger
than 1∕2 for the first derivative of uEX to be square integrable. See, for example, [59].
If 𝛼 is an integer then uEX is an analytic or piecewise analytic function and the measure of regu-
larity is the size of the derivatives of uEX . Analogous definitions apply to two and three dimensions.
Remark 1.9 The kth derivative of a function f (x) is a local property of f (x) only when k is an
integer. This is not the case for non-integer derivatives.
1.5.2 A priori estimation of the rate of convergence

Analysts are called upon to choose discretization schemes for particular problems. A sound choice
of discretization is based on a priori information on the regularity of the exact solution. If we know
that the exact solution lies in Sobolev space H k (I) then it is possible to say how fast the error in
energy norm will approach zero as the number of degrees of freedom is increased, given a scheme
by which a sequence of discretizations is generated. Index k can be inferred or estimated from the
input data 𝜅, c and f .
We define
h = max 𝓁j ∕𝓁, j = 1, 2, … M(Δ) (1.90)
j
where 𝓁j is the length of the jth element, 𝓁 is the size of the of the solution domain I = (1, 𝓁). This
is generalized to two and three dimensions where 𝓁 is the diameter of the domain and 𝓁j is the
diameter of the jth element. In this context diameter means the diameter of the smallest circlein
one and two dimensions, or sphere in three dimensions,that contains the element or domain. In
two and three dimensions the solution domain is denoted by Ω.
The a priori estimate of the relative error in energy norm for uEX ∈ H k (Ω), quasiuniform meshes
and polynomial degree p is
⎧ hk−1
⎪ C(k) k−1 ||uEX ||H k (Ω) for k − 1 ≤ p
def ||uEX − uFE ||E(Ω) ⎪ p
(er )E = ≤⎨ (1.91)
||uEX ||E(Ω) ⎪ hp
⎪C(k) k−1 ||uEX ||H p+1 (Ω) for k − 1 > p
⎩ p
where E(Ω) is the energy norm, k is typically a fractional number and C(k) is a positive constant
that depends on k but not on h or p. This inequality gives the upper bound for the asymptotic rate
of convergence of the relative error in energy norm as h → 0 or p → ∞ [22]. This estimate holds
for one, two and three dimensions. For one and two dimensions lower bounds were proven in
[13, 24] and [46] and it was shown that when singularities are located in vertex points then the
rate of convergence of the p-version is twice the rate of convergence of the h-version when both are
expressed in terms of the number of degrees of freedom. It is reasonable to assume that analogous
results can be proven for three dimensions; however, no proofs are available at present.
We will find it convenient to write the relative error in energy norm in the following form
C
(er )E ≤ (1.92)
N𝛽
where N is the number of degrees of freedom and C and 𝛽 are positive constants, 𝛽 is called the alge-
braic rate of convergence. In one dimension N ∝ 1∕h for the h-version and N ∝ p for the p-version.
Therefore for k − 1 < p we have 𝛽 = k − 1. However, for the important special case when the solu-
tion has the functional form of eq. (1.89) or, more generally, has a term like u = |x − x0 |𝜆 and x0 ∈ I
is a nodal point then 𝛽 = 2(k − 1) for the p-version: The rate of p-convergence is twice that of
h-convergence [22, 84].
When the exact solution is an analytic function then uEX ∈ H ∞ (Ω) and the asymptotic rate of
convergence is exponential:
C
(er )E ≤ (1.93)
exp(𝛾N 𝜃 )
where C, 𝛾 and 𝜃 are positive constants, independent of N. In one dimension 𝜃 ≥ 1∕2, in two dimen-
sions 𝜃 ≥ 1∕3, in three dimensions 𝜃 ≥ 1∕5, see [10].
When the exact solution is a piecewise analytic function then eq. (1.93) still holds provided that
the boundary points of analytic functions are nodal points, or more generally, lie on the boundaries
of finite elements.
The relationship between the error e = uEX − uFE measured in energy norm and the error in
potential energy is established by the following theorem.
Theorem 1.5
||e||2E = ||uEX − uFE ||2E(I) = 𝜋(uFE ) − 𝜋(uEX ). (1.94)
Proof: Writing e = uEX − uFE and noting that e ∈ E0 (I), from the definition of 𝜋(uFE ) we have:
1
𝜋(uFE ) =𝜋(uEX − e) = B(uEX − e, uEX − e) − F(uEX − e)
2
1 1
= B(uEX , uEX ) − F(uEX )−B(uEX , e) + F(e) + B(e, e)
2 ⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟ 2
0
=𝜋(uEX ) + ||e||2E(I) .
Remark 1.10 Consider the problem given by eq. (1.5) and assume that 𝜅 and c are constants. In
this case the smoothness of u depends only on the smoothness of f : If f ∈ Ck (I) then u ∈ Ck+2 (I)
for any k ≥ 0. Similarly, if f ∈ H k (I) then u ∈ H k+2 (I) for any k ≥ 0. This is known as the shift
theorem. More generally, the smoothness of u depends on the smoothness of 𝜅, c and F. For a
precise statement and proof of the shift theorem we refer to [21].
Remark 1.11 An introductory discussion on how a priori estimates are obtained under the
assumption that the second derivative of the exact solution is bounded can be found in Appendix B.
1.5.3 A posteriori estimation of error

The goal of finite element computations is to estimate certain quantities of interest (QoIs) such
as, for example, the maximum and minimum values of u or u′ on I = (0, 𝓁). Since finite element
solutions are approximations to an exact solution, it is not sufficient to report the value of a QoI
computed from the finite element solution. It is also necessary to provide an estimate of the rela-
tive error in the QoI, or present evidence that the relative error in the QoI is not greater than an
acceptable value.
In this section we will use the a priori estimates described in Section 1.5.2 to obtain a posteriori
estimates of error in energy norm. It is possible to obtain very accurate estimates for a large class
of problems which includes most problems of practical interest.
Error estimation based on extrapolation

For most practical problems the estimate (1.92) is sufficiently sharp so that the less than or equal
sign (≤) can be replaced by the approximately equal sign (≈) and this a priori estimate can be used
in an a posteriori fashion.
The computed values of the potential energy corresponding to a sequence of finite element spaces
S1 ⊂ S2 ⊂ · · · Sn can be used for estimating the error in energy norm by extrapolation. Sequences of
finite element spaces that have this property are called hierarchic sequences. By Theorem 1.5 and
eq. (1.92) we have:
2
𝜋(uFE ) − 𝜋(uEX ) ≈ (1.95)
N 2𝛽
def
where  = C||uEX ||E(I) . There are three unknowns: 𝜋(uEX ),  and 𝛽. Assume that we have
a sequence of solutions corresponding to the hierarchic sequence of finite element spaces
Si−2 ⊂ Si−1 ⊂ Si . Let us denote the corresponding computed potential energy values by 𝜋i−2 , 𝜋i−1 ,
𝜋i and the degrees of freedom by Ni−2 , Ni−1 , Ni . We will denote the estimate for 𝜋(uEX ) by 𝜋∞ . With
this notation we have:
2
𝜋i − 𝜋∞ ≈ 2𝛽 (1.96)
Np
2
𝜋i−1 − 𝜋∞ ≈ 2𝛽 ⋅ (1.97)
Ni−1
On dividing eq. (1.96) with eq. (1.97) and taking the logarithm we get
𝜋 − 𝜋∞ N
log i ≈ 2𝛽 log i−1 (1.98)
𝜋i−1 − 𝜋∞ Ni
and, repeating with i − 1 substituted for i, it is possible to eliminate 2𝛽 to obtain:
( )
𝜋 i − 𝜋∞ 𝜋i−1 − 𝜋∞ Q
≈ (1.99)
𝜋i−1 − 𝜋∞ 𝜋i−2 − 𝜋∞
where
( )−1
Ni−1 N
Q = log log i−2 ⋅
Ni Ni−1
Equation (1.99) can be solved for 𝜋∞ to obtain an estimate for the exact value of the potential energy.
The relative error in energy norm corresponding to the ith finite element solution in the sequence
is estimated from
( )
𝜋i − 𝜋∞ 1∕2
ei ≈ . (1.100)
|𝜋∞ |
Usually the percent relative error is reported. This estimator has been tested against the known
exact solution of many problems of various smoothness. The results have shown that it works well
for a wide range of problems, including most problems of practical interest; however, it cannot be
guaranteed to work well for all conceivable problems. For example, this method would fail if the
exact solution would happen to be energy-orthogonal to all basis functions associated with (say)
odd values of i.
Remark 1.12 From equation (1.92) we get

log (er )E ≈ log C − 𝛽 log N. (1.101)
On plotting (er )E vs. N on log-log scale a straight line with the slope −𝛽 will be seen for sufficiently
large N. The estimated value of 𝛽, corresponding to the ith solution in the sequence, is denoted by
𝛽i . It is computed from eq. (1.98):
1 log(𝜋i − 𝜋) − log(𝜋i−1 − 𝜋)
𝛽i = ⋅ (1.102)
2 log Ni−1 − log Ni
Examples
The properties of the finite element solution with reference to a family of model problems is dis-
cussed in the following. The problems are stated as follows: Find uFE ∈ S0 (I) such that
𝓁 ( )
𝜅u′FE 𝑣′ + cuFE 𝑣 dx = F(𝑣) for all 𝑣 ∈ S0 (I) (1.103)
∫0
where 𝜅 and c are constants and F(𝑣) is defined such that the exact solution is:
uEX = x𝛼 (𝓁 − x), on I = (0, 𝓁), 𝛼 > 1∕2. (1.104)
As explained in Section 1.5.1, when 𝛼 is not an integer, the case considered in the following, then
this solution lies in the space H 𝛼+1∕2−𝜖 (I). Therefore the asymptotic rate of h-convergence on uni-
form meshes, predicted by eq. (1.92), is 𝛽 = 𝛼 − 1∕2 and the asymptotic rate of p-convergence on a
fixed mesh is 𝛽 = 2𝛼 − 1.
We selected this problem because it is representative of the singular part of the exact solutions of
two-and three-dimensional elliptic boundary value problems.
Referring to Theorem 1.3, we have B(uEX − uFE , 𝑣) = 0 for all 𝑣 ∈ S0 (I) therefore F(𝑣) = B(uEX , 𝑣).
Consequently for the kth element the load vector in the local numbering convention is:
xk+1
ri(k) = (𝜅u′EX 𝜑′i + cuEX 𝜑i ) dx, i = 1, 2, … , pk + 1 (1.105)
∫xk
where by definition 𝜑i (Qk (𝜉)) = Ni (𝜉).
When 1∕2 < 𝛼 < 1 then the first derivative of uEX is infinity in the point x = 0. To avoid having
′
uEX in the integrand, the first term in eq. (1.105) is integrated by parts:
xk+1 ( )x xk+1
𝜅u′EX 𝜑′i dx = 𝜅uEX 𝜑′i xk+1 − 𝜅uEX 𝜑′′i dx.
∫xk k ∫xk
Since 𝜑′′i = 0 for i = 1 and i = 2, we have:
1( ) 1( ) 𝓁 1
r1(k) = − 𝜅uEX x=x + 𝜅uEX x=x + k (cuEX )x=Qk (𝜉) N1 d𝜉
𝓁k k+1 𝓁k k 2 ∫−1
1( ) 1( ) 𝓁 1
r2(k) = 𝜅uEX x=x − 𝜅uEX x=x + k (cuEX )x=Qk (𝜉) N2 d𝜉
𝓁k k+1 𝓁k k 2 ∫−1
and for i ≥ 3 we have:
√ ( 1 )
2i − 3 2 dP
(k)
ri = (𝜅uEX )x=xk+1 − (−1)i (𝜅uEX )x=xk − (𝜅uEX )x=Qk (𝜉) i−2 d𝜉
2 𝓁k ∫−1 d𝜉
𝓁k 1
+ (cuEX )x=Qk (𝜉) Ni d𝜉 (1.106)
2 ∫−1
where Pi−2 (𝜉) is the Legendre polynomial of degree i − 2 and eq. (D.10) was used.
Since the exact solution is known, the exact value of the potential energy can be determined for
any set of values of 𝛼, 𝜅, c and 𝓁. When 𝜅 and c are both constants then
[ ( )
1 𝛼2 (𝛼 + 1)2 2𝛼+1
𝜋(uEX ) = − 𝜅 𝓁 2𝛼−1 − (𝛼 + 1)𝓁 2𝛼 + 𝓁
2 2𝛼 − 1 2𝛼 + 1
( )]
1 1 1
+c 𝓁 2𝛼+1 − 𝓁 2(𝛼+1) + 𝓁 2𝛼+3 . (1.107)
2𝛼 + 1 𝛼+1 2𝛼 + 3
The exact values of the potential energy for the data 𝜅 = 1, c = 50 and 𝓁 = 1 and various values of
𝛼 are shown in Table 1.2.
Table 1.2 Exact values of the potential energy for 𝜅 = 1, c = 50 and 𝓁 = 1.
𝜶 𝝅(uEX ) 𝜶 𝝅(uEX )
0.600 −2.3728354978 1.000 −1.0000000000

0.700 −1.7571858289 1.500 −0.5104166667
0.800 −1.4176885916 2.000 −0.3047619048
0.900 −1.1799028822 3.000 −0.1420634921
When 𝛼 is a fractional number then derivatives higher than 𝛼 will not be finite in x = 0. In the
range 0.5 < 𝛼 < 1 the first derivative in the point x = 0 is infinity. This range of 𝛼 has considerable
practical importance because the exact solutions of two- and three-dimensional problems often
have analogous terms.
When 𝛼 is an integer then all derivatives of uEX are finite. Therefore uEX can be approximated by
Taylor series about any point of the domain I = [0, 𝓁]. It is known that the error term of a Taylor
series truncated at polynomial degree p is bounded by the (p + 1)th derivative of uEX :
p+1
𝓁 p+1 d uEX
max |uFE − uEX | ≤ max | |⋅ (1.108)
(p + 1)! x∈I dxp+1
In the special case when 𝛼 is an integer and pmin ≥ 𝛼 + 1 then uFE = uEX .
Exercise 1.19 Show how eq. (1.106) is obtained from eq. (1.105). Provide details.
Example 1.10 Let us consider model problems in the form of eq. (1.103) with the following
data: 𝓁 = 1, 𝜅 = 1, c = 50 and exact solutions in the form of eq. (1.104) corresponding to 𝛼 = 0.6,
0.7, 0.8, 0.9. We will use a sequence of uniform finite element meshes with M(Δ) = 10, 100, 1000
and pk = p = 2 assigned to all elements. We are interested in the relationship between the esti-
mated and true relative errors. The computed values of the potential energy and their estimated
limit values computed by means of eq. (1.99) are listed in Table 1.3. These are comparable to the
exact values of the potential energy listed in Table 1.2. The estimated limit values of the potential
energy are denoted by 𝜋M(Δ)→∞ .
With the information provided in Tables 1.2 and 1.3 it is possible to compare the estimated and
exact values of the relative error. For example, using eq. (1.100) and
||(uFE )M(Δ) ||2E(I) = |𝜋M(Δ) |
Table 1.3 Example: Computed and estimated values of the potential energy 𝜋. Uniform
mesh refinement, pk = p = 2 for all elements.
M(𝚫) N 𝜶 = 0.6 𝜶 = 0.7 𝜶 = 0.8 𝜶 = 0.9
10 19 −2.17753673 −1.73038992 −1.41382648 −1.17955239

100 199 −2.25079984 −1.74673700 −1.41675042 −1.17984996
1000 1999 −2.29589857 −1.75303348 −1.41745363 −1.17989453
𝜋M(Δ)→∞ −2.37254083 −1.75716094 −1.41768637 −1.17990276

the estimated relative error in energy norm for M(Δ) = 10, 𝛼 = 0.8 is:
√ √
∗
𝜋(uFE ) − 𝜋M(Δ)→∞ −1.41382648 + 1.41768637
(er )E = = = 0.0522
|𝜋M(Δ)→∞ | 1.41768637
or 5.22%. When using the exact value of the potential energy for reference then the relative error is
the same as the estimated relative error to within three digits of accuracy:
√ √
𝜋(uFE ) − 𝜋(uEX ) −1.41382648 + 1.41768859
(er )E = = = 0.0522.
||uEX ||E(I)
2 1.41768859
Exercise 1.20 Compare the estimated and exact values of the relative error in energy norm for
the problem in Example 1.10 for M(Δ) = 100, 𝛼 = 0.7.
Example 1.11 Let us consider once again model problems in the form of eq. (1.103) with the data:
𝓁 = 1, 𝜅 = 1, c = 50 and exact solutions corresponding to 𝛼 = 0.6, 0.7, 0.8, 0.9, see eq. (1.104).
Using a sequence of uniform finite element meshes with M(Δ) = 10, 100, 1000, 10,000 and p = 2
assigned to each element, the results shown in Fig. 1.9 are obtained. The values of 𝛽 were com-
puted by linear regression using eq. (1.101). We observe that 𝛽 = 𝛼 − 1∕2. This is consistent with
the asymptotic estimate given by eq. (1.91).
Example 1.12 Let us consider model problems in the form of eq. (1.103) with the data: 𝓁 = 1,
𝜅 = 1, c = 50 and exact solutions corresponding to 𝛼 = 0.6, 0.7, 0.8, 0.9, see eq. (1.104). Using
a uniform finite element mesh with M(Δ) = 10 and p = 2, 3, 4, 5 assigned to each element,
the results shown in Fig. 1.10 are obtained. The values of 𝛽 were computed by linear regression
using eq. (1.101). We observe that 𝛽 = 2(𝛼 − 1∕2), that is, the rate of convergence is twice that in
Example 1.11. This is consistent with the theoretical results in [22, 84]: The rate of p-convergence
is at least twice the rate of h-convergence when the singular point is a nodal point.
1.5.4 Error in the extracted QoI

In Example 1.9 it was demonstrated that the QoI can be extracted from the finite element solution
efficiently and accurately even when the discretization was very poorly chosen. Let us consider a
quantity of interest Φ(u) and the corresponding extraction function 𝑤 ∈ E(I). The extracted value
of the QoI is
Φ(uFE ) = F(𝑤) − B(uFE , 𝑤) (1.109)
100
Relative error (%)
10
1
α = 0.6, β = 0.1
0.1
α = 0.7, β = 0.2
α = 0.8, β = 0.3 α = 0.9, β = 0.4
0.01
10 102 103 104
N
Figure 1.9 Relative error in energy norm. M(Δ) = 10, 100, 1000, 10000, p = 2.
100
Relative error (%)
10
1
α = 0.6, β = 0.2
α = 0.7, β = 0.4 α = 0.8, β = 0.6 α = 0.9, β = 0.8
0.1
20 25 30 35 40 45 50 55
N
Figure 1.10 Relative error in energy norm. M(Δ) = 10, p = 2, 3, 4, 5.
and the exact value of the QoI is

Φ(uEX ) = F(𝑤) − B(uEX , 𝑤). (1.110)
Subtracting eq. (1.109) from eq. (1.110) we get
Φ(uEX ) − Φ(uFE ) = −B(uEX − uFE , 𝑤). (1.111)
We define a function zEX ∈ E0 (I) such that
B(zEX , 𝑣) = B(𝑤, 𝑣) for all 𝑣 ∈ E0 (I). (1.112)
This operation projects 𝑤 ∈ E(I) onto the space E0 (I). Letting 𝑣 = uEX − uFE we get:
B(zEX , uEX − uFE ) = B(𝑤, uEX − uFE ) for all 𝑣 ∈ E0 (I).
We will write this as
B(uEX − uFE , 𝑤) = B(uEX − uFE , zEX ). (1.113)
Next we define zFE ∈ S0 (I) such that
B(zEX , 𝑣) = B(zFE , 𝑣) for all 𝑣 ∈ S0 (I). (1.114)
This operation projects zEX ∈ E0 (I) onto the space S0 (I). By Galerkin’s orthogonality condition (see
Theorem 1.3) we have
B(uEX − uFE , 𝑣) = 0 for all 𝑣 ∈ S0 (I).
Therefore, letting 𝑣 = zFE , we write eq. (1.113) as
B(uEX − uFE , 𝑤) = B(uEX − uFE , zEX − zFE ) (1.115)
and we can write eq. (1.111) as
Φ(uEX ) − Φ(uFE ) = −B(uEX − uFE , zEX − zFE ). (1.116)
Therefore the error in the extracted data is
|Φ(uEX ) − Φ(uFE )| = |B(uEX − uFE , zEX − zFE )|
≤ 2||uEX − uFE ||E(I) ||zEX − zFE ||E(I) (1.117)
where we used the Schwarz inequality, see Section A.3 in the appendix.
The function zFE made it possible to write the error in the QoI in this form. It does not have to be
computed.
Inequality (1.117) serves to explain why the error in the extracted data can converge to zero faster
than the error in energy norm: If ||zEX − zFE ||E(I) is of comparable magnitude to ||uEX − uFE ||E(I)
then the error in the extracted data is of comparable magnitude to the error in the strain energy,
that is, the error in energy norm squared. But, as seen in Example 1.9, where 𝑤 was much smoother
than uEX , it can be much smaller. In the exceptional case when the extraction function is Green’s
function, the error is zero.
1.6 The choice of discretization in 1D

In an ideal discretization the error (in energy norm) associated with each element would be the
same. This ideal discretization can be approximated by automated adaptive methods in which the
discretization is modified based on feedback information from previously obtained finite element
solutions. Alternatively, based on a general understanding of the relationship between regularity
and discretization, and understanding the strengths and limitations of the software tools available
to them, analysts can formulate very efficient discretization schemes.
1.6.1 The exact solution lies in Hk (I), k − 𝟏 > p

When the solution is smooth then the most efficient finite element discretization scheme is uni-
form mesh and high polynomial degree. However, all implementations of finite element analysis
software have limitations on how high the polynomial degree is allowed to be and therefore it may
not be possible to increase the polynomial degree sufficiently to achieve the desired accuracy. In
such cases the mesh has to be refined. Uniform refinement may not be optimal in all cases, however.
Consider, for example, the following problem:
−𝜖 2 u′′ + cu = f (x), u(0) = ú (𝓁) = 0 (1.118)
where 𝜖 << c, and f is a smooth function. Intuitively, when 𝜖 2 is small then the solution will be close
to u = f ∕c however, because of the boundary condition u(0) = 0, has to be satisfied, the function
u(x) will change sharply over some interval 0 < x < d(𝜖) << 𝓁.
Letting c = 1 and f (x) = 1 the exact solution of this problem is
uEX (x) = 1 − cosh x∕𝜖 + tanh(𝓁∕𝜖) sinh x∕𝜖 (1.119)
which is plotted for various values of 𝜖 on the interval 0 < x∕𝓁 < 0.20 in Fig. 1.11. It is seen that the
gradient at x = 0 rapidly increases with respect to decreasing values of 𝜖.
1.0
ϵ = 0.01
0.8 ϵ = 0.005
0.6
uEX(x)
ϵ = 0.10
0.4
ϵ = 0.05
0.2
0
0 0.05 0.10 0.15 0.20
x/l
Figure 1.11 The solution uEX (x), given by eq. (1.119), in the neighborhood of x = 0 for various values of 𝜖.
This is a simple example of boundary layer problems that arise in models of plates, shells and fluid
flow. Despite the fact that uEX is an analytic function, it may require unrealistically high polynomial
degrees to obtain a close approximation to the solution when 𝜖 is small.
The optimal discretization scheme for problems with boundary layers is discussed in the context
of the hp-version in [85]. The results of analysis indicate that the size of the element at the boundary
is proportional to the product of the polynomial degree p and the parameter 𝜖. Specifically, for the
problem discussed here, the optimal mesh consists of two elements with the node points located at
x1 = 0, x2 = d, x3 = 𝓁, where d = Cp𝜖 with 0 < C < 4∕e.
A practical approach to problems like this is to create an element at the boundary (in higher
dimensions a layer of elements) the size of which is controlled by a parameter. The optimal value
of that parameter is then selected adaptively.
1.6.2 The exact solution lies in Hk (I), k − 𝟏 ≤ p

In this section we consider a special case of the problem stated in eq. (1.103):
𝓁
u′ 𝑣′ dx = F(𝑣), for all 𝑣 ∈ E0 (I) (1.120)
∫0
with the data u(0) = u(𝓁) = 0, 𝓁 = 1 and F(𝑣) defined such that the exact solution is
uEX = x𝛼 (1 − x), 𝛼 > 1∕2, 0<x<1 (1.121)
that is,
𝓁
F(𝑣) = (𝛼x𝛼−1 − (𝛼 + 1)x𝛼 ) 𝑣′ dx. (1.122)
∫0
On integrating by parts, we get the following expression which is better suited for numerical
evaluation:
𝓁
F(𝑣) = − uEX 𝑣′′ dx. (1.123)
∫0
We address the following questions: (a) How does the error in energy norm depend on the param-
eter 𝛼, the mesh Δ and the p-distribution p? and (b) How is this error distributed among the
elements? Understanding these relationships is necessary for making sound choices of discretiza-
tion based on a priori information concerning the regularity of the exact solution.
We compute the potential energy of the difference between the exact solution and its linear inter-
polant for the kth element:
xk+1 ( )
uEX (xk+1 ) − uEX (xk ) 2
(k) 1
𝜋 EX = u′EX − dx.
2 ∫xk xk+1 − xk
The exact solution for 𝛼 = 0.75 and its linear interpolant for M(Δ) = 5, uniform mesh, are shown
in Fig. 1.12.
To obtain the potential energy of the difference between the exact solution and its linear inter-
(k)
polant for the kth element, denoted by 𝜋 FE , we need to solve:
⎡1 0 · · · 0⎤ ⎧ a(k) ⎫ ⎧ (k) ⎫
⎢ ⎥⎪
3 ⎪ ⎪ r3(k) ⎪
⎪
2 ⎢0 1 · · · 0⎥ a4
(k)
⎪ ⎪r ⎪
⎨ ⎬ = ⎨ 4 ⎬⋅ (1.124)
𝓁k ⎢ ⋱ ⎥ ⋮ ⎪ ⎪ ⋮ ⎪
⎢0 0 · · · 1⎥ ⎪ a(k) ⎪ ⎪r (k) ⎪
⎣ ⎦⎪⎩ pk +1 ⎭ ⎩ pk +1 ⎭
0.3
uEX
0.2
0.1
0
0 0.2 0.4 0.6 0.8 1
x
Figure 1.12 The exact solution for 𝛼 = 0.75 and its linear interpolant for M(Δ) = 5, uniform mesh.
The solution is
𝓁
a(k)
i
= k ri(k) , i = 3, 4, … , pk + 1. (1.125)
2
Using eq. (1.106) we get
√ 1( )
(k) 2i − 3 2 dPi−2
ri = − 𝜅 ũ (k) d𝜉, i = 3, 4, … , pk + 1 (1.126)
∫
2 𝓁k −1 EX
x=Qk (𝜉) d𝜉
where ũ (k)
EX
is the difference between uEX and its linear interpolant:
( )
1−𝜉 1+𝜉
ũ (k) = (u )
EX x=Qk (𝜉) − u (x ) + u (x ) (1.127)
EX 2 EX k 2 EX k+1
and compute
1 ∑ (k) (k)
pk +1
(k)
𝜋 FE = − a r .
2 i=3 i i
Referring to Theorem 1.5, the error in energy norm associated with the kth element is
√
(k) (k)
||ek ||E(Ik ) = 𝜋 FE − 𝜋 EX (1.128)
and the relative error in energy norm associated with the kth element is:
||ek ||E(Ik )
(e(k)
r )E = √ ⋅ (1.129)
| (k) |
|𝜋 EX |
| |
The error of approximation over the entire domain is:
(M(Δ) )1∕2
∑ √
||uEX − uFE ||E(I) = ||ek ||E(I )
2
= 𝜋FE − 𝜋EX (1.130)
k
k=1
By Theorem 1.2, the exact value of the potential energy is

1 ( )
1 1 𝛼2 (𝛼 + 1)2
𝜋(uEX ) = − (u′EX )2 dx = − − (𝛼 + 1) + (1.131)
2 ∫0 2 2𝛼 − 1 2𝛼 + 1
and the relative error in energy norm on the entire domain is:
( )
𝜋FE − 𝜋EX 1∕2
(er )E = . (1.132)
|𝜋(uEX )|
Remark 1.13 In estimating the local error we used uFE (xk ) = uEX (xk ). It can be shown that in the
special case of this problem (c = 0) this relationship holds and therefore using the equal sign in
eq. (1.128) is justified. In the general case (c ≠ 0) however, uFE (xk ) ≠ uEX (xk ) and eq. (1.128) will be
an estimate of the local error in the finite element solution. Therefore the equal sign in eq. (1.128)
has to be replaced by the approximately equal (≈) sign and the first equal sign in eq. (1.130) has to
be replaced with the less or equal (≤) sign.
Example 1.13 This example illustrates the distribution of the relative error among the elements
for a fixed mesh and polynomial degree for selected fractional values of 𝛼. Uniform mesh on the
domain (0, 1) with M(Δ) = 5 and pk = 2 for k = 1, 2, … , 5 is used. The exact solution for 𝛼 = 0.75
is shown in Fig. 1.12. The percent relative error in energy norm associated with the kth element,
given by eq. (1.129), is shown in Table 1.4 and the relative error for the entire domain is shown in
the last column.
It is seen that for all values of 𝛼 the maximum error is associated with the first element.
Example 1.14 This example illustrates the distribution of the relative error among the elements
for a fixed mesh and polynomial degree for selected integer values of 𝛼. Uniform mesh on the
domain (0, 1) with M(Δ) = 5 and pk = 2 for k = 1, 2, … , 5 is used. The percent relative error in
energy norm associated with the kth element, given by eq. (1.129), is shown in Table 1.5 and the
relative error for the entire domain is shown in the last column.
The error of approximation for 𝛼 = 1 is zero. This follows directly from Theorem 1.4: The exact
solution is a polynomial of degree 2. Therefore it lies in the finite element space and hence the finite
element solution is the same as the exact solution.
Remark 1.14 In the foregoing discussion it was tacitly assumed that all data computed by numer-
ical integration were accurate and the coefficient matrices of the linear equations were such that
small changes in the right-hand-side vector produce small changes in the solution vector. This
happens when the condition number of the coefficient matrix is reasonably small. In the finite ele-
ment method the condition number depends on the choice of the shape functions, the mapping
functions and the mesh. In one-dimensional setting the mapping is linear and the shape functions
Table 1.4 Example: Element-by-element and total relative errors in energy

norm (percent) for selected fractional values of 𝛼.
Element number
𝜶 1 2 3 4 5 (er )E
1.25 79.49 7.50 2.80 1.63 1.12 4.80

1.15 99.52 4.06 1.63 0.97 0.67 3.92
1.05 29.56 1.24 0.53 0.32 0.22 1.77
0.95 18.89 1.16 0.52 0.32 0.22 2.41
0.85 42.94 3.26 1.52 0.94 0.67 9.84
0.75 60.39 5.14 2.47 1.56 1.11 22.37
0.65 76.07 6.86 3.39 2.16 1.56 42.91
0.55 91.80 8.44 4.28 2.76 2.00 76.22
Table 1.5 Example: Element-by-element and total relative errors in

energy norm (percent) for selected integer values of 𝛼.
Element number
𝜶 1 2 3 4 5 (er )E
1 0 0 0 0 0 0
2 11.00 61.24 15.31 7.02 4.55 2.45
3 20.09 4.69 98.83 16.16 9.24 4.62
4 36.98 8.41 17.24 30.13 14.02 8.00
are energy-orthogonal, therefore round-off errors are not significant. This is not the case in two and
three dimensions, however.
Errors in numerical integration can be particularly damaging. The reader should be mind-
ful of this when applying the concepts and procedures discussed in this chapter to higher
dimensions.
1.7 Eigenvalue problems

The following problem is a prototype of an important class of engineering problems which includes
the undamped vibration of elastic structures:
𝜕2 u
−(𝜅u′ )′ + cu = −𝜇 , x ∈ (0, 𝓁), t ∈ (0, ∞) (1.133)
𝜕t2
where the primes represent differentiation with respect to x. For example, we may think of an elastic
bar of length 𝓁, cross-section A, modulus of elasticity E, in which case 𝜅 ≡ AE > 0 given in units
of Newton (N) or equivalent, the parameter c ≥ 0 is the coefficient of distributed springs (N∕mm2 )
and the parameter 𝜇 > 0 is mass per unit length (kg/m = 10−6 Ns2 ∕mm2 ). The bar is vibrating in its
longitudinal direction.
The boundary conditions are:
u(0, t) = 0, u(𝓁, t) = 0
and the initial conditions are
𝜕u
u(x, 0) = f (x), | = g(x)
𝜕t (x,0)
where f (x) and g(x) are given functions in L2 (I). Here we consider homogeneous Dirichlet boundary
conditions. However, the boundary conditions can be homogeneous Neumann or homogeneous
Robin conditions, or any combination of those.
The generalized form is obtained by multiplying eq. (1.133) by a test function 𝑣 ∈ E0 (I) and inte-
grating by parts:
𝓁 𝓁
𝜕2 u
(𝜅u′ 𝑣′ + cu𝑣) dx = − 𝜇 𝑣 dx. (1.134)
∫0 ∫0 𝜕t2
We now introduce u = U(x)T(t) where U ∈ E0 (I), T ∈ C2 (0, ∞). This is known as separation of
variables. Therefore we get
𝓁 𝓁
𝜕2 T
T (𝜅U ′ 𝑣′ + cU𝑣) dx = − 𝜇U𝑣 dx (1.135)
∫0 𝜕t2 ∫0
which can be written as

𝓁
(𝜅U ′ 𝑣′ + cU𝑣) dx
∫0 1 𝜕2 T
=− = 𝜔2 . (1.136)
𝓁 T 𝜕t2
𝜇U𝑣 dx
∫0
Since the functions on the left are independent of t, the function T depends only on t, both expres-
sions must equal some positive constant denoted by 𝜔2 . That constant has to be positive because
the expression on the left holds for all 𝑣 ∈ E0 (I) and if we select 𝑣 = U then the expression on the
left is positive.
The function T(t) satisfies the ordinary differential equation
𝜕2 T
+ 𝜔2 T = 0 (1.137)
𝜕t2
the solution of which is
T = a cos(𝜔t) + b sin(𝜔t) (1.138)
where 𝜔 is the angular velocity (rad/s). Alternatively 𝜔 is written as 𝜔 = 2𝜋f where f is the fre-
quency (Hz).
To find 𝜔 and U we have to solve the problem
𝓁 𝓁
(𝜅U ′ 𝑣′ + cU𝑣) dx − 𝜔2 𝜇U𝑣 dx = 0 for all 𝑣 ∈ E0 (I) (1.139)
∫0 ∫0
which will be abbreviated as
B(U, 𝑣) − 𝜔2 D(U, 𝑣) = 0 for all 𝑣 ∈ E0 (I). (1.140)
There are infinitely many solutions called eigenpairs (𝜔i , Ui ), i = 1, 2, … , ∞. The set of eigenval-
ues is called the spectrum. If Ui is an eigenfunction and 𝛼 is a real number then 𝛼Ui is also an
eigenfunction. In the following we assume that the eigenfunctions have been normalized so that
𝓁
D(Ui , Ui ) ≡ 𝜇Ui2 dx = 1.
∫0
If the eigenvalues are distinct then the corresponding eigenfunctions are orthogonal: Let (𝜔i , Ui )
and (𝜔j , Uj ) be eigenpairs, i ≠ j. Then from eq. (1.140) we have
B(Ui , Uj ) − 𝜔2i D(Ui , Uj ) = 0

B(Uj , Ui ) − 𝜔2j D(Uj , Ui ) = 0.
Subtracting the second equation from the first we see that if 𝜔i ≠ 𝜔j then Ui and Uj are orthogonal
functions:
𝓁
D(Ui , Uj ) ≡ 𝜇Ui Uj dx = 0 (1.141)
∫0
and hence B(Ui , Uj ) = 0.
Importantly, it can be shown that any function f ∈ E0 (I) can be written as a linear combination
of the eigenfunctions:
‖ ∑ ∞
‖
‖f − ai Ui (x)‖ 2 = 0 (1.142)
‖ ‖L (I)
i=1
where
𝓁
ai = 𝜇f Ui dx. (1.143)
∫0
The Rayleigh15 quotient is defined by
B(u, u)
R(u) = ⋅ (1.144)
D(u, u)
Eigenvalues are usually numbered in ascending order. Following that convention,
𝜔21 ≡ 𝜔2min = min
0
R(u) = R(U1 ) (1.145)
u∈E (I)
that is, the smallest eigenvalue is the minimum of the Rayleigh quotient and the corresponding
eigenfunction is the minimizer of R(u) on E0 (I). This follows directly from eq. (1.140). The kth
eigenvalue minimizes R(u) on the space Ek0 (I)
𝜔2k = min
0
R(u) = R(Uk ) (1.146)
u∈Ek (I)
where
Ek0 (I) = {u | u ∈ E0 (I), B(u, Ui ) = 0, i = 1, 2, … , k − 1}. (1.147)
When the eigenvalues are computed numerically then the minimum of the Rayleigh quotient is
sought on the finite-dimensional space S0 (I). We see from the definition R(u) that the error of
approximation in the natural frequencies will depend on how well the eigenfunctions are approx-
imated in energy norm, in the space S0 (I).
The following example illustrates that in a sequence of numerically computed eigenvalues only
the lower eigenvalues will be approximated well. It is possible, however, at least in principle, to
obtain good approximation for any eigenvalue by suitably enlarging the space S0 (I).
Example 1.15 Let us consider the eigenvalue problem

𝜕2 u 𝜕2 u
𝜅 =𝜇 2, u(0) = u(𝓁) = 0, t ≥ 0. (1.148)
𝜕x 2 𝜕t
This equation models (among other things) the free vibration (natural frequencies and mode
shapes) of a string of length 𝓁 stretched horizontally by the force 𝜅 > 0 (N) under the assumptions
that the displacements are infinitesimal and confined to one plane, the plane of vibration, and the
ends of the string are fixed. The mass per unit length is 𝜇 > 0 (kg/m). We assume that 𝜅 and 𝜇 are
constants. It is left to the reader to verify that the function u defined by
∑
∞
( )
u= ai cos(𝜔i t) + bi sin(𝜔i t) sin(𝜆i x) (1.149)
i=1
where ai , bi are coefficients determined from the initial conditions and

√
𝜋 𝜅
𝜆i = i , 𝜔i = 𝜆i (1.150)
𝓁 𝜇
satisfies eq. (1.148).
If we approximate the eigenfunctions using uniform mesh, p = 2 and plot the ratio (𝜔FE ∕𝜔EX )n
against n∕N, where n is the nth eigenvalue, then we get the curves shown in Fig. 1.13. The curves
show that somewhat more than 20% of the numerically computed eigenvalues will be accurate.
15 John William Strutt, 3rd Baron Rayleigh 1842–1919.

p=2
1.4
(ωFE/ωEX)n
2500 elements
1.2 N = 4999
25 elements
1 N = 49
0 0.2 0.4 0.6 0.8 1

n/N
Figure 1.13 The ratio (𝜔FE ∕𝜔EX )n corresponding to the h version, p = 2.
The higher eigenvalues cannot be well approximated in the space S0 (I). The existence of the jump
seen at n∕N = 0.5 is a feature of numerically approximated eigenvalues by means of standard finite
element spaces using the h-version [2]. The location of the jump depends on the polynomial degree
of elements. There is no jump when p = 1.
If we approximate the eigenfunctions using a uniform mesh consisting of 5 elements, and
increase the polynomial degrees uniformly then we get the curves shown in Fig. 1.14. The curves
show that only about 40% of the numerically computed eigenvalues will be accurate. The error
increases monotonically for the higher eigenvalues and the size of the error is virtually independent
of p.
It is possible to reduce this error by enforcing the continuity of derivatives. Examples are available
in [32]. There is a tradeoff, however: Enforcing continuity of derivatives on the basis functions
reduces the number of degrees of freedom but entails a substantial programming burden because an
adaptive scheme has to be devised for the general case to ensure that the proper degree of continuity
is enforced. If, for example, 𝜇 would be a piecewise constant function then the continuity of the
first and higher derivatives must not be enforced in those points where 𝜇 is discontinuous.
From the perspective of designing a finite element software, it is advantageous to design the
software in such a way that it will work well for a broad class of problems. In the formulation
presented in this chapter C0 continuity is a requirement. Functions that lie in Ck (I) where k > 0
p N
1.4 5 24
10 49
25 124
(ωFE/ωEX)n
100 499
1.2
0 0.2 0.4 0.6 0.8 1

n/N
Figure 1.14 The ratio (𝜔FE ∕𝜔EX )n corresponding to the p version. Uniform mesh, 5 elements.
Table 1.6 Example: p-Convergence of the 24th eigenvalue in

Example 1.16.
p 5 10 15 20
𝜔24 194.296 100.787 98.312 98.312
are also in C0 (I). In other words, the space Ck (I) is embedded in the space C0 (I). Symbolically:
Ck (I) ⊂ C0 (I). The exact eigenfunctions in this example are in C∞ (I).
Example 1.16 Let us consider the problem in Example 1.15 modified so that 𝜇 is a piecewise
constant function defined on a uniform mesh of 5 elements such that 𝜇 = 1 on elements 1, 3 and
5, 𝜇 = 0.2 on elements 2 and 4. In this case the exact eigenfunctions are not smooth and the exact
eigenvalues are not known explicitly.
At p = 5 there are 24 degrees of freedom. Suppose that the 24th eigenvalue is of interest. If we
increase p uniformly then this eigenvalue converges to 98.312. The results of computation are
shown in Table 1.6.
Any eigenvalue can be approximated to an arbitrary degree of precision on a suitably defined
mesh and uniform increase in the degrees of freedom. When 𝜅 and/or 𝜇 are discontinuous func-
tions then the points of discontinuity must be node points.
Observe that the numerically computed eigenvalues converge monotonically from above.
This follows directly from the fact that the eigenfunctions are minimizers of the Rayleigh quotient.
Exercise 1.21 Prove eq. (1.143).
Exercise 1.22 Find the eigenvalues for the problem of Example 1.15 using the generalized for-
mulation and the basis functions 𝜑n (x) = sin(n𝜋x∕𝓁), (n = 1, 2, … , N). Assume that 𝜅 and 𝜇 are
constants and 𝜇∕𝜅 = 1. Let 𝓁 = 10. Explain what makes this choice of basis functions very special.
Hint: Owing to the orthogonality of the basis functions, only hand calculations are involved.
1.8 Other finite element methods

Up to this point we have been concerned with the finite element method based on the generalized
formulation, called the principle of virtual work. There are many other finite element methods. All
finite element methods share the following attributes:
1. Formulation. A bilinear form B(u, 𝑣) is defined on the normed linear spaces X, Y (i.e. u ∈ X,
𝑣 ∈ Y ) and the functional F(𝑣) is defined on Y . The exact solution uEX lies in X and satisfies:
B(uEX , 𝑣) = F(𝑣) for all 𝑣 ∈ Y (1.151)
The normed linear spaces, X, Y , the linear functional F and the bilinear form B satisfy the
respective properties listed in sections A.1.1 and A.1.2.
2. Finite element spaces. The finite-dimensional subspaces Si ⊂ X, Vi ⊂ Y (i = 1, 2, …) are defined
and it is assumed that there are ûi ∈ Si such that the sequence of functions ûi (i = 1, 2, …) con-
verges in the space X to uEX , that is:
||uEX − ûi ||X ≤ 𝜖i 𝜖i → 0 as i → ∞. (1.152)
The functions ûi are not the finite element solutions in general.
1.8 Other finite element methods 47
3. The finite element solution. The finite element solution ui|FE ∈ Si satisfies:
B(ui|FE , 𝑣) = F(𝑣) for all 𝑣 ∈ Vi . (1.153)
4. The stability criterion. The finite element method is said to be stable if
||ui|FE − ûi ||X ≤ C||U − ûi ||X i = 1, 2, … (1.154)
for all possible U ∈ X. The necessary and sufficient condition for a finite element method to be
stable is that for every u ∈ Si there is a 𝑣 ∈ Vi so that
|B(u, 𝑣)| ≥ C ||u||X ||𝑣||Y (1.155)
where C > 0 is a constant, independent of i, or for every 𝑣 ∈ Vi there is a u ∈ Si so that this

inequality holds. This inequality is known as the Babuška-Brezzi condition, usually abbreviated
to “the BB condition”. This condition was formulated by Babuška in 1971 [9] and independently
by Brezzi in 1974 [29].
If the Babuška-Brezzi condition is not satisfied then there will be at least some uEX ∈ X for which
||uEX − ui|FE ||X ↛ 0 as i → ∞ even though there may be uEX ∈ X for which ||uEX − ui|FE ||X → 0
as i → ∞. Examples are presented in [6]. In general it is difficult, or may even be impossible,
to separate those uEX for which the method works well from those for which it does not. The
Babuška-Brezzi condition guarantees that the condition number of the stiffness matrix will not
become too large as i increases.
Remark 1.15 Any implementation of the finite element method must be shown to satisfy the
Babuška-Brezzi condition otherwise there will be some input data for which the method will fail
even though it may work well for other input data. The formulation based on the principle of virtual
work satisfies the Babuška-Brezzi condition.
Exercise 1.23 Show that the finite element method based on the principle of virtual work satisfies
the Babuška-Brezzi condition.
1.8.1 The mixed method

Consider writing eq. (1.5) in the following form:
𝜅u′ − F = 0 (1.156)
−F ′ + cu = f (1.157)
and assume that the boundary conditions are u(0) = u(𝓁) = 0.

In the following we will use the one-dimensional equivalent of the notation introduced in
sections A.2.2 and A.2.3. Multiply eq. (1.156) by G ∈ L2 (I) and eq. (1.157) by 𝑣 ∈ H 1 (I), integrate
by parts and sum the resulting equations to obtain:
𝓁( ) 𝓁( ) 𝓁
du d𝑣
𝜅 G − FG dx + F + cu𝑣 dx = T𝑣dx. (1.158)
∫0 dx ∫0 dx ∫0
We define the bilinear form:
𝓁 ( ) 𝓁 ( ′ )
def
B(u, F; 𝑣, G) = 𝜅u′ G − FG dx + F𝑣 + cu𝑣 dx (1.159)
∫0 ∫0
and the linear form

𝓁
def
F(𝑣) = f 𝑣 dx. (1.160)
∫0
The problem is now stated as follows: Find uEX ∈ H01 (I), FEX ∈ L2 (I) such that
( )
B uEX , FEX ; 𝑣, G = F(𝑣) for all 𝑣 ∈ H01 (I), G ∈ L2 (I). (1.161)
The finite element problem is formulated as follows: Find uFE ∈ S0 (I) where S0 (I) is a subspace
of H01 (I) and FFE ∈ V(I) where V(I) is a subspace of L2 (I) such that
( )
B uFE , FFE ; 𝑣, G = F(𝑣) for all 𝑣 ∈ S0 (I), G ∈ V(I). (1.162)
( ) ( )
We now ask: In what sense will uFE , FFE be close to uEX , FEX ? The answer is that there is a
( )
constant C, independent of the finite element mesh and uEX , FEX , such that
||uEX − uFE ||H 1 (I) + ||FEX − FFE ||L2 (I)
[ ]
≤ C min ||uEX − u||H 1 (I) + min ||FEX − F||L2 (I) (1.163)
provided, however, that S0 (I) and V(I) were properly selected.
For example, let S be the space defined in eq. (1.61) with pk = 1, k = 1, 2 … M(Δ). The space S0 (I)
has the dimension M(Δ) − 1. For V(I) consider three choices:
1. V1 (I) is the set of functions which are constant on each finite element. V1 (I) has the dimension
M(Δ).
2. V2 (I) is the space S defined in (3.11) with pk = 1, k = 1, 2, … , M(Δ) (dimension M(Δ) + 1).
3. V3 (I) is the set of functions which are linear on every element and discontinuous at the nodes
(dimension 2M(Δ)).
For these choices of V(I) the mixed formulation leads to systems of linear equations with 2M(Δ) −
1, 2M(Δ) and 3M(Δ) − 1 unknowns, respectively. In the cases V = V1 and V = V3 , a constant C
exists such that the inequality (1.163) holds for all uEX , and FEX . In the case of V = V2 , however,
such a constant does not exist. This means that no matter how large C is, there exist some uEX ∈
H01 (I) and FEX ∈ L2 (I) and mesh Δ so that the inequality (1.163) is not satisfied. On the other hand,
there will be uEX ∈ H01 (I) and FEX ∈ L2 (I) for which the inequality is satisfied and therefore the
finite element solutions will converge to the underlying exact solution.
1.8.2 Nitsche’s method

Nitsche’s method16 allows the treatment of essential boundary conditions as natural boundary con-
ditions. This has certain advantages in two and three dimensions. An outline of the algorithmic
aspects of the method is presented in the following. For additional details we refer to [51].
Consider the problem:
−u′′ + cu = f (x), x ∈ (0, 𝓁) (1.164)
with the boundary conditions u′ (0) = 0 and u(𝓁) = û𝓁 . However, at x = 𝓁 we substitute the natural
boundary condition:
1
u′ (𝓁) = (û − u(𝓁)) (1.165)
𝜖 𝓁
16 Joachim Nitsche 1926–1996.

1.8 Other finite element methods 49
where 𝜖 is a small positive number, 1∕𝜖 is called penalty parameter. The role of the penalty param-
eter becomes clearly visible if we consider the potential energy
𝓁( )2 𝓁
1 1
Π(u) = (u′ + cu2 ) dx + (u(𝓁) − û𝓁 )2 − f (x)u dx. (1.166)
2 ∫0 2𝜖 ∫0
Letting 𝜖 → 0, the minimizer of the potential energy converges to the solution of the Dirichlet
problem; however, the numerical problem becomes ill-conditioned. Nitsche’s method stabilizes
the numerical problem making it possible to solve it for the full range of boundary conditions,
including 𝜖 = 0.
Stabilization
On multiplying eq. (1.164) by 𝑣 and integrating by parts we get
𝓁 𝓁
−u′ (𝓁)𝑣(𝓁) + (u′ 𝑣′ + cu𝑣) dx = f (x)𝑣 dx. (1.167)
∫0 ∫0
We introduce the stability parameter 𝛾 and multiply eq. (1.165) by 𝑣(𝓁)𝜖∕(𝜖 + 𝛾𝓁) to get
1 ( ′ ) 1
𝜖u (𝓁)𝑣(𝓁) + u(𝓁)𝑣(𝓁) = û 𝑣(𝓁). (1.168)
𝜖 + 𝛾𝓁 𝜖 + 𝛾𝓁 𝓁
Adding eq. (1.167) and eq. (1.168) we get
𝓁
𝛾𝓁 1
(u′ 𝑣′ + cu𝑣) dx − u′ (𝓁)𝑣(𝓁) + u(𝓁)𝑣(𝓁)
∫0 𝜖 + 𝛾𝓁 𝜖 + 𝛾𝓁
𝓁
1
= f (x)𝑣 dx + û 𝑣(𝓁) (1.169)
∫0 𝜖 + 𝛾𝓁 𝓁
and, multiplying eq. (1.165) by 𝑣′ (𝓁)𝜖𝛾𝓁∕(𝜖 + 𝛾𝓁), we have
𝜖𝛾𝓁 ′ 𝛾𝓁 𝛾𝓁
u (𝓁)𝑣′ (𝓁) + u(𝓁)𝑣′ (𝓁) = û 𝑣′ (𝓁). (1.170)
𝜖 + 𝛾𝓁 𝜖 + 𝛾𝓁 𝜖 + 𝛾𝓁 𝓁
Subtracting eq. (1.170) from eq. (1.169) we obtain the generalized formulation:
𝓁
𝛾𝓁 ( ′ )
(u′ 𝑣′ + cu𝑣) dx − u (𝓁)𝑣(𝓁) + u(𝓁)𝑣′ (𝓁)
∫0 𝜖 + 𝛾𝓁
1 𝜖𝛾𝓁 ′
+ u(𝓁)𝑣(𝓁) − u (𝓁)𝑣′ (𝓁)
𝜖 + 𝛾𝓁 𝜖 + 𝛾𝓁
𝓁
1 𝛾𝓁
= f (x)𝑣 dx + û𝓁 𝑣(𝓁) − û 𝑣′ (𝓁). (1.171)
∫0 𝜖 + 𝛾𝓁 𝜖 + 𝛾𝓁 𝓁
Letting 𝜖 = 0 in eq. (1.171) we get the stabilized method proposed by Nitsche [67]:
𝓁 ( ) 1
(u′ 𝑣′ + cu𝑣) dx − u′ (𝓁)𝑣(𝓁) + u(𝓁)𝑣′ (𝓁) + u(𝓁)𝑣(𝓁)
∫0 𝛾𝓁
𝓁
1
= f (x)𝑣 dx + û 𝑣(𝓁) − û𝓁 𝑣′ (𝓁). (1.172)
∫0 𝛾𝓁 𝓁
Numerical example
Letting c = 1, f (x) = 1, 𝓁 = 10 and û𝓁 = 0.25 we construct the numerical problem using one ele-
ment and the hierarchic shape functions defined in Section 1.3.1. By definition:
∑
p+1
∑
p+1
u= aj Nj (𝜉), 𝑣= bi Ni (𝜉) (1.173)
j=1 i=1
Table 1.7 The computed values of u(𝓁).
𝜸 10−3 10−6 10−9 10−12 10−15
u(𝓁) 0.2540348 0.2500004 0.25(0)6 4 0.25(0)9 4 0.25(0)12 4
where p is the polynomial degree. Therefore u(𝓁) = a2 and 𝑣(𝓁) = b2 and, using the Legendre shape
functions, for p = 3 the unconstrained coefficient matrix, without the modifications of Nitsche, is
√ √
⎡1∕2 −1∕2 0 0 ⎤ ⎡2∕3 1∕3 −1∕ 6 1∕3 10 ⎤
⎢ ⎥ ⎢ ⎥
⎢ ⎥ ⎢ √ √ ⎥
⎢ 1∕2 0 0⎥ ⎢ 2∕3 −1∕ 6 −1∕3 10⎥
2 ⎥ + c𝓁 ⎢
[M] = ⎢ ⎥
𝓁⎢ ⎥ 2 ⎢ ⎥
⎢ (sym.) 1 0⎥ ⎢ (sym.) 2∕5 0 ⎥
⎢ ⎥ ⎢ ⎥
⎢ ⎥ ⎢ ⎥
⎣ 1⎦ ⎣ 2∕21 ⎦
Referring to eq. (1.172), the coefficient matrix is modified by the application of Nitsche’s method.
Those modifications in the present case are:
⎡0 1∕𝓁 0 ⎤ 0
⎢ ⎥
⎢ √ √ ⎥
⎢ −1∕𝓁 + 1∕(𝛾𝓁) − 12∕𝓁 − 18∕𝓁 ⎥
[N] = ⎢ ⎥⋅
⎢ ⎥
⎢ (sym.) 0 0 ⎥
⎢ ⎥
⎢ ⎥
⎣ 0 ⎦
The unconstrained right hand side vector without the modifications of Nitsche is:
√
{r} = {𝓁∕2 𝓁∕2 − (𝓁∕2) 2∕3 0}T
and with the modifications of Nitsche it is:
√ √
{rN } = {û𝓁 ∕𝓁 û𝓁 ∕(𝛾𝓁) − û𝓁 ∕𝓁 − û𝓁 12∕𝓁 − û𝓁 18∕𝓁}T .
The numerical results shown in Table 1.7 indicate that the stabilized formulation is remarkably
robust. The notation (0)n indicates that there are n zeros.
51
Boundary value problems
The strong forms of boundary value problems are formulated from first principles with reference
the problems of heat conduction in solid bodies and elasticity. Such mathematical problems appear
in various mathematical models used in numerical simulation of physical systems in structural,
mechanical and aerospace engineering. The generalized (weak) formulations are derived and
examples are presented. Simplifications through dimensional reduction and the use of symmetry,
antisymmetry and periodicity are illustrated by examples.
It is assumed that the reader is familiar with, and has access to, at least one finite element software
product. The numerical solutions presented in this book were obtained with StressCheck1 unless
otherwise noted.
2.1 Notation
The Euclidean space in n dimensions is denoted by ℝn . The Cartesian2 coordinate axes in ℝ3 are
labeled x, y, z (in cylindrical systems r, 𝜃, z) and a vector in ℝn is denoted by u. For example,
u ≡ {ux uy uz } represents a vector in ℝ3 .
The index notation will be introduced gradually, in parallel with the familiar Cartesian notation,
so that readers who are not yet acquainted with this notation can become familiar with it. The basic
rules of index notation are as follows.
1. The Cartesian coordinate axes are labeled x = x1 , y = x2 , z = x3 .
2. In conventional notation the position vector in ℝ3 is x ≡ {x y z}T . In index notation it is simply
xi . A general vector a ≡ {ax ay az } and its transpose is written simply as ai .
3. A free index in ℝn is understood to range from 1 to n.
4. Two free indices represent a matrix. The size of the matrix depends on the range of indices. Thus,
in three dimensions (ℝ3 ):
⎡a11 a12 a13 ⎤ ⎡axx axy axz ⎤
aij ≡ ⎢a21 a22 a23 ⎥ ≡ ⎢ayx ayy ayz ⎥ ⋅
⎢ ⎥ ⎢ ⎥
⎣a31 a32 a33 ⎦ ⎣azx azy azz ⎦
The identity matrix is represented by the Kronecker3 delta 𝛿ij , defined as follows:
{
1 if i = j
𝛿ij = (2.1)
0 if i ≠ j.
1 StressCheck is a trademark of Engineering Software Research and Development, Inc.

2 René Descartes (in Latin: Renatus Cartesius) 1596–1650.
3 Leopold Kronecker 1823–1891.
52 2 Boundary value problems
5. Repeated indices imply summation. For example, the scalar product of two vectors ai and bj is
ai bi ≡ a1 b1 + a2 b2 + a3 b3 . The product of two matrices aij and bij is written as cij = aik bkj .
Definition 2.1 Repeated indices are also called dummy indices. This is because summation
is performed therefore the index designation is immaterial. For example, ai bi ≡ ak bk .
6. In order to represent the cross product in index notation, it is necessary to introduce the permu-
tation symbol eijk . The components of the permutation symbol are defined as follows:
eijk = 0 if the values of i, j, k do not form a permutation of 1, 2, 3

eijk = 1 if the values of i, j, k form an even permutation of 1, 2, 3
eijk = −1 if the values of i, j, k form an odd permutation of 1, 2, 3.
The cross product of vectors aj and bk is written as
ci = eijk aj bk .
Definition 2.2 The permutations (1, 2, 3), (2, 3, 1) and (3, 1, 2) are even permutations. The
permutations (1, 3, 2), (2, 1, 3) and (3, 2, 1) are odd permutations.
7. Indices following a comma represent differentiation with respect to the variables identified by
the indices. For example, if u(xi ) is a scalar function then
𝜕u 𝜕2 u
u,2 ≡ , u,23 ≡ ⋅
𝜕x2 𝜕x2 𝜕x3
The gradient of u is simply u,i .
If ui = ui (xk ) is a vector function in ℝ3 then
𝜕u1 𝜕u2 𝜕u3
ui,i ≡ + +
𝜕x1 𝜕x2 𝜕x3
is the divergence of ui .
8. The transformation rules for Cartesian vectors and tensors are presented in Appendix K.
Example 2.1 The divergence theorem in index notation is:
ui,i dV = ui ni dS (2.2)
∫Ω ∫𝜕Ω
where ui and ui,i are continuous on the domain Ω and its boundary 𝜕Ω, ni is the outward unit
normal vector to the boundary, dV is the differential volume and dS is the differential surface. We
will make use of the divergence theorem in the derivation of generalized formulations.
Exercise 2.1 Write out each term of ci = eijk aj bk where ai = {a1 a2 a3 }T and bi = {b1 b2 b3 }T .
Exercise 2.2 Outline a derivation of the divergence theorem in two dimensions. Hint: Review
the derivation of Green’s theorem and cast it in the form of eq. (2.2).
2.2 The scalar elliptic boundary value problem 53
2.2 The scalar elliptic boundary value problem
The three-dimensional analogue of the model problem introduced in Section 1.1 is the scalar elliptic
boundary value problem
( )
−div [𝜅]grad u + cu = f (x, y, z), (x, y, z) ∈ Ω (2.3)
where
⎡ 𝜅x 𝜅xy 𝜅xz ⎤
[𝜅] = ⎢𝜅yx 𝜅y 𝜅yz ⎥ (2.4)
⎢ ⎥
⎣𝜅zx 𝜅zy 𝜅z ⎦
is a positive-definite matrix4 and c = c(x, y, z) ≥ 0. In index notation eq. (2.3) reads:
−(𝜅ij u,j ),i + cu = f . (2.5)
We will be concerned with the following linear boundary conditions:
1. Dirichlet boundary condition: u = û is prescribed on boundary region 𝜕Ωu . When û = 0 on 𝜕Ωu

then the Dirichlet boundary condition is said to be homogeneous.
2. Neumann boundary condition: The flux vector is defined by
def def
q = −[𝜅]grad u, equivalently; qi = 𝜅ij u,j . (2.6)
def
The normal flux is defined by qn = q ⋅ n ≡ qi ni where n ≡ ni is the unit outward normal to the
boundary. When qn = q̂ n is prescribed on boundary region 𝜕Ωq then the boundary condition
is called a Neumann boundary condition. When q̂ n = 0 on 𝜕Ωq then the Neumann boundary
condition is said to be homogeneous.
3. Robin boundary condition: qn = hR (u − uR ) is given on boundary segment 𝜕ΩR . In this expres-
sion hR > 0 and uR are given functions. When uR = 0 on 𝜕ΩR then the Robin boundary condition
is said to be homogeneous.
4. Boundary conditions of convenience: In many instances the solution domain can be simpli-
fied through taking advantage of symmetry, antisymmetry and/or periodicity. These boundary
conditions are called boundary conditions of convenience.
The boundary segments 𝜕Ωu , 𝜕Ωq , 𝜕ΩR and 𝜕Ωp are non-overlapping and collectively cover the
entire boundary 𝜕Ω. Any of the boundary segments may be empty.
Definition 2.3 The Dirichlet boundary condition is also called essential boundary condition. The
Neumann and Robin conditions area called natural boundary conditions.
2.2.1 Generalized formulation

To obtain the generalized formulation for the scalar elliptic boundary value problem we multiply
eq. (2.5) by a test function 𝑣 and integrate over the domain Ω:
− (𝜅ij u,j ),i 𝑣 dV + cu𝑣 dV = f 𝑣 dV. (2.7)

∫Ω ∫Ω ∫Ω
4 A symmetric matrix [𝜅] of real numbers is positive-definite if xT [𝜅]x > 0 for any x ≠ 𝟎.
This equation must hold for arbitrary 𝑣, provided that the indicated operations are defined. The
first integral can be written as:
(𝜅ij u,j ),i 𝑣 dV = (𝜅ij u,j 𝑣),i dV − 𝜅ij u,j 𝑣,i dV.
∫Ω ∫Ω ∫Ω
Applying the divergence theorem (eq. (2.2)) we have:
(𝜅ij u,j 𝑣),i dV = 𝜅ij u,j ni 𝑣 dS

∫Ω ∫𝜕Ω
where ni is the unit normal vector to the boundary surface. Therefore eq. (2.7) can be written in the
following form:
− 𝜅 u n 𝑣 dS + 𝜅ij u,j 𝑣,i dV + cu𝑣 dV = f 𝑣 dV. (2.8)

∫𝜕Ω ij ,j i ∫Ω ∫Ω ∫Ω
It is customary to write
qi = −𝜅ij u,j and qn = qi ni .
With this notation we have:
𝜅ij u,j 𝑣,i dV + cu𝑣 dV = f 𝑣 dV − qn 𝑣 dS. (2.9)

∫Ω ∫Ω ∫Ω ∫𝜕Ω
This is the generalization of eq. (1.18) to two and three dimensions. As we have seen in Section 1.2,
the specific statement of a generalized formulation depends on the boundary conditions. In the
general case u = û is prescribed on 𝜕Ωu (Dirichlet boundary condition); qn = q̂ n is prescribed on
𝜕Ωq (Neumann boundary condition) and qn = hR (u − uR ) is prescribed on ΩR (Robin boundary
condition), see Section 2.2. We now define the bilinear form as follows:
B(u, 𝑣) = 𝜅ij u,j 𝑣,i dV + cu𝑣 dV + hR u𝑣 dS (2.10)

∫Ω ∫Ω ∫𝜕ΩR
and the linear functional:
F(𝑣) = f 𝑣 dV − qn 𝑣 dS + hR uR 𝑣 dS. (2.11)

∫Ω ∫𝜕Ωq ∫𝜕ΩR
When 𝜕ΩR is empty then the last terms in equations (2.10) and (2.11) are omitted. When Neumann
condition is prescribed on the entire boundary and c = 0 then the data must satisfy the following
condition:
f dV = qn dS. (2.12)
∫Ω ∫𝜕Ω
The space E(Ω) is defined by
def
E(Ω) = {u | B(u, u) < ∞}
and the energy norm

√
def 1
∥u∥E = B(u, u)
2
is associated with E(Ω). The space of admissible functions is defined by:
def
̃
E(Ω) = {u | u ∈ E(Ω), u = û on 𝜕Ωu }.
Here we assume that corresponding to any u = û specified on 𝜕Ωu there is a u★ ∈ E(Ω) such that
̃
u★ = û on 𝜕Ωu . This imposes certain restrictions on û and ensures that E(Ω) is not empty. The space
of test functions is defined by:
def
E0 (Ω) = {u | u ∈ E(Ω), u = 0 on 𝜕Ωu }.
The generalized formulation is now stated as follows: “Find u ∈ E(Ω) ̃ such that B(u, 𝑣) = F(𝑣) for
all 𝑣 ∈ E 0 (Ω)”. A function u that satisfies this condition is called a generalized solution.
The generalized formulation is often stated as a minimization problem. The potential energy
defined by
def1
𝜋(u) = B(u, u) − F(u) (2.13)
2
is formally identical to eq. (1.36) and Theorem 1.2 is applicable: The exact solution of the gener-
̃
alized formulation minimizes the potential energy on the space E(Ω). Alternatively, the potential
energy is defined by
def 1 1 1
Π(u) = 𝜅ij u,i u,j dV + cu2 dV + h (u − uR )2 dS
2 ∫Ω 2 ∫Ω 2 ∫𝜕ΩR R
− fu dV + qn u dS (2.14)
∫Ω ∫𝜕Ωq
so that when u = uR on 𝜕ΩR and ΩR = Ω then Π(u) = 0. Note that 𝜋(u) differs from Π(u) only by a
constant. Therefore the minimizer of 𝜋(u) is the same as the minimizer of Π(u).
Exercise 2.3 Consider the case ΩR = Ω, that is Robin boundary conditions are prescribed on the
entire boundary. Compare the definitions of the potential energy given by equations (2.13) and
(2.14).
Exercise 2.4 Following the proof of Theorem 1.2, show that the generalized formulation mini-
mizes Π(u) given by eq. (2.14).
2.2.2 Continuity
In two and three dimensions u ∈ E(Ω) is not necessarily continuous or bounded. For example, the
function u = log | log r| where r = (x2 + y2 )1∕2 is discontinuous and unbounded in the point r = 0,
yet it lies in E(Ω). This has important implications: Concentrated fluxes are inadmissible data.
Similarly, point constraints are inadmissible except for the enforcement of the uniqueness of the
solution when c = 0 and Neumann conditions are specified on the entire boundary and eq. (2.12)
is satisfied.
Exercise 2.5 Let r = (x2 + y2 )1∕2 and Ω = {r | r ≤ 𝜌0 < 1}. Show that u1 = log r does not lie in
E(Ω) but u2 = log | log r| does. Hint: u2 = log(− log r) when r < 1.
2.3 Heat conduction

Steady state potential flow problems are among the physical phenomena that can be modeled as
scalar elliptic boundary value problems. In this section the formulation of a mathematical problem
that models heat flow by conduction in solid bodies is described.
Mathematical models of heat conduction are based on two fundamental relationships:

the conservation law and Fourier’s law of heat conduction described in the following.
1. The conservation law states that the quantity of heat entering any volume element of the
conducting medium equals the quantity of heat exiting the volume element plus the quantity
of heat retained in the volume element. The heat retained causes a change in temperature in
the volume element which is proportional to the specific heat of the conducting medium c (in
J∕(kg K) units) multiplied by the density 𝜌 (in kg∕m3 units). The temperature will be denoted
by u(x, y, z, t) where t is time.
The heat flow rate across a unit area is represented by a vector quantity called heat flux.
The heat flux is in W∕m2 units, or equivalent, and will be denoted by q = q(x, y, z, t) =
{qx (x, y, z, t) qy (x, y, z, t) qz (x, y, z, t)}T . In addition to heat flux entering and leaving the volume
element, heat may be generated within the volume element, for example from chemical
reactions. The heat generated per unit volume and unit time will be denoted by Q (in W∕m3
units).
Applying the conservation law to the volume element shown in Fig. 2.1, we have:
Δt[qx ΔyΔz − (qx + Δqx )ΔyΔz + qy ΔxΔz − (qy + Δqy )ΔxΔz+
qz ΔxΔy − (qz + Δqz )ΔxΔy + QΔxΔyΔz] = c𝜌ΔuΔxΔyΔz. (2.15)
Assuming that u and q are continuous and differentiable and neglecting terms that go to zero
faster than Δx, Δy, Δz, Δt, we have:
𝜕q 𝜕qy 𝜕qz 𝜕u
Δqx = x Δx, Δqy = Δy, Δqz = Δz, Δu = Δt.
𝜕x 𝜕y 𝜕z 𝜕t
On factoring ΔxΔyΔzΔt the conservation law is obtained:
𝜕qx 𝜕qy 𝜕qz 𝜕u
− − − + Q = c𝜌 ⋅ (2.16)
𝜕x 𝜕y 𝜕z 𝜕t
In index notation:
𝜕u
−qi,i + Q = c𝜌 ⋅ (2.17)
𝜕t
qz + Δqz
qy + Δqy
qx qx + Δqx
Δz
qy
z
y
qz Δy
x Δx
Figure 2.1 Control volume and notation for heat conduction.

2. Fourier’s law states that the heat flux vector is related to the temperature gradient in the
following way:
( )
𝜕u 𝜕u 𝜕u
qx = − kxx + kxy + kxz (2.18)
𝜕x 𝜕y 𝜕z
( )
𝜕u 𝜕u 𝜕u
qy = − kyx + kyy + kyz (2.19)
𝜕x 𝜕y 𝜕z
( )
𝜕u 𝜕u 𝜕u
qz = − kzx + kzy + kzz (2.20)
𝜕x 𝜕y 𝜕z
where the coefficients kxx , kxy , … , kzz are called coefficients of thermal conduction (measured
in W∕(mK) units). It is customary to write kx = kxx , ky = kyy , kz = kzz . The coefficients of thermal
conduction will be assumed to be independent of the temperature u, unless otherwise stated.
Denoting the matrix of coefficients by [K], Fourier’s law of heat conduction can be written as:
q = −[K] grad u. (2.21)
The matrix of coefficients [K] is symmetric and positive-definite. The negative sign indicates
that the direction of heat flow is opposite to the direction of the temperature gradient, that
is, the direction of heat flow is from high to low temperature. In index notation eq. (2.21)
is written as:
qi = −kij u,j . (2.22)
For isotropic materials kij = k𝛿ij .
2.3.1 The differential equation

Combining equations (2.16) through (2.20), we have:
( ) ( )
𝜕 𝜕u 𝜕u 𝜕u 𝜕 𝜕u 𝜕u 𝜕u
kx + kxy + kxz + kyx + ky + kyz +
𝜕x 𝜕x 𝜕y 𝜕z 𝜕y 𝜕x 𝜕y 𝜕z
( )
𝜕 𝜕u 𝜕u 𝜕u 𝜕u
kzx + kzy + kz + Q = c𝜌 (2.23)
𝜕z 𝜕x 𝜕y 𝜕z 𝜕t
which can be written in the following compact form:
( ) 𝜕u
div [K] grad u + Q = c𝜌 (2.24)
𝜕t
or in index notation:
𝜕u
(kij u,j ),i + Q = c𝜌 ⋅ (2.25)
𝜕t
In many practical problems u is independent of time. Such problems are called stationary or
steady state problems. The solution of a stationary problem can be viewed as the solution of some
time-dependent problem, with time-independent boundary conditions, at t = ∞.
In formulating eq. (2.23) we assumed that kij are differentiable functions. In many practical prob-
lems the solution domain is comprised of subdomains Ωi that have different material properties.
In such cases eq. (2.23) is valid on each subdomain. On the boundaries of adjoining subdomains
continuous temperature and flux are prescribed.
To complete the definition of a mathematical model, initial and boundary conditions have to be
specified. This is discussed in the following section.
2.3.2 Boundary and initial conditions

The solution domain will be denoted by Ω and its boundary by 𝜕Ω. We will consider three kinds of
boundary conditions:
1. Prescribed temperature (Dirichlet condition): The temperature u = û is prescribed on boundary

region 𝜕Ωu .
2. Prescribed flux (Neumann condition): The flux vector component normal to the boundary,
denoted by qn , is prescribed on the boundary region 𝜕Ωq . By definition;
def
qn = q ⋅ n ≡ −([K] grad u) ⋅ n ≡ −kij u,j ni (2.26)
where n ≡ ni is the (outward) unit normal to the boundary. The prescribed flux on 𝜕Ωq will be
denoted by q̂ n .
3. Convection (Robin condition): On boundary region 𝜕Ωc the flux vector component qn is pro-
portional to the difference between the temperature of the boundary and the temperature of a
convective medium:
qn = hc (u − uc ), (x, y, z) ∈ 𝜕Ωc (2.27)
where hc is the coefficient of convective heat transfer in W∕(m2 K) units and uc is the (known)
temperature of the convective medium.
The sets 𝜕Ωu , 𝜕Ωq and 𝜕Ωc are non-overlapping and collectively cover the entire boundary. Any
of the sets may be empty.
The boundary conditions may be time-dependent. For time-dependent problems an initial con-
dition has to be prescribed on Ω: u(x, y, z, 0) = U(x, y, z).
It is possible to show that eq. (2.23), subject to the enumerated boundary conditions, has a unique
solution. Stationary problems also have unique solutions, subject to the condition that when flux
is prescribed over the entire boundary 𝜕Ω then the following condition must be satisfied:
Q dV = qn dS. (2.28)
∫Ω ∫𝜕Ω
This is easily seen by integrating
(kij u,j ),i + Q = 0 (2.29)
on Ω and using the divergence theorem, eq. (2.2) and the definition (2.26).
Note that if ui is a solution of eq. (2.29) then ui + C is also a solution, where C is an arbitrary
constant. Therefore the solution is unique up to an arbitrary constant.
In addition to the three types of boundary conditions discussed in this section, radiation may
have to be considered. When two bodies exchange heat by radiation then the flux is proportional to
the difference of the fourth power of their absolute temperatures, therefore radiation is a non-linear
boundary condition. The boundary region subject to radiation, denoted by 𝜕Ωr , may overlap 𝜕Ωc .
Radiation is discussed in Section 9.1.1.
In the following it will be assumed that the coefficients of thermal conduction, the flux prescribed
on Ωq and the coefficient hc prescribed on Ωc are independent of the temperature. This assumption
can be justified on the basis of empirical data in a narrow range of temperatures only.
Exercise 2.6 Discuss the physical interpretation of eq. (2.28).

Exercise 2.7 Show that in cylindrical coordinates r, 𝜃, z the conservation law is of the form:
1 𝜕(rqr ) 1 𝜕q𝜃 𝜕qz 𝜕u
− − − + Q = c𝜌 ⋅ (2.30)
r 𝜕r r 𝜕𝜃 𝜕z 𝜕t
Use two methods: (a) apply the conservation law to an infinitesimal volume element in cylindrical
coordinates and (b) transform eq. (2.16) to cylindrical coordinates.
Exercise 2.8 Show that there are three mutually perpendicular directions (called principal direc-
tions) such that the heat flux is proportional to the (negative) gradient vector. Hint: Consider steady
state heat conduction and let:
[K] grad u = 𝜆 gradu
then show that the principal directions are defined by the normalized eigenvectors.
Remark 2.1 The result of Exercise 2.8 implies that the general form of matrix [K] can be obtained
by rotation from orthotropic material axes.
Exercise 2.9 List all of the physical assumptions incorporated into the mathematical model rep-
resented by eq. (2.23) and the boundary conditions described in this section.
2.3.3 Boundary conditions of convenience

A scalar function is said to be symmetric with respect to a plane of symmetry if in symmetrically
located points the function has equal values. On a plane of symmetry qn = 0. A function is said to be
antisymmetric with respect to a plane of symmetry if in symmetrically located points the function
has equal absolute values but opposite sign. On a plane of antisymmetry u = 0.
In many instances the domain has one or more planes of symmetry, antisymmetry and/or it
may be periodic. For example, the domain shown in Fig. 2.2 has one plane of symmetry; the plane
x = 0, and it is periodic; the shaded subdomain is replicated five times. If the material properties,
source function and boundary conditions are also symmetric, antisymmetric and/or periodic then
it is often advantageous and convenient to formulate the problem on a subdomain and extend the
solution to the entire domain by symmetry, antisymmetry or periodicity.
Figure 2.2 Notation for Example 2.2. y

∂Ω(o)
P+ P–
qn (P+) qn (P–)
θ
+ –
∂Ω p ∂Ω p
x
∂Ω(i)
∂Ω(k) (typ.)
When Ω, [K], Q and the boundary conditions are periodic then a periodic sector of Ω has a
periodic boundary segment pair denoted by 𝜕Ω+p and 𝜕Ω−p . On corresponding points of a periodic
boundary segment pair, P+ ∈ 𝜕Ω+p and P− ∈ 𝜕Ω−p , the boundary conditions are: u(P+ ) = u(P− ) and
q+n = −q−n .
Symmetric, antisymmetric and periodic boundary conditions are illustrated by the following
example.
Example 2.2 Fig, 2.2 is the plan view of a plate-like body of constant thickness t. We assume
that the surfaces z = ±t∕2 are perfectly insulated, the source function is zero and the material is
isotropic. We further assume that on the cylindrical boundary represented by the inner circle (𝜕Ω(i) )
constant temperature u0 is prescribed and on the boundary represented by the outer circle (𝜕Ω(o) )
flux qn (𝜃) is prescribed. On the boundaries of the five circular cut-outs (𝜕Ω(k) , k = 1, 2, … , 5) qn = 0.
Some examples of symmetric, antisymmetric and periodic functions are given here with refer-
ence to Fig. 2.2. Let a, b and c > 0 be given real numbers. The restriction c > 0 is necessary because
the temperature u is in K (Kelvin5 ) units.
1. The functions qn = a + b cos n𝜃 and u0 = c are symmetric with respect to the y axis for n =
2, 4, 6, … This problem can be solved on the half domain with qn = 0 on the boundary coinci-
dent with the y axis.
2. The functions qn = b cos n𝜃 and u0 = 0 are antisymmetric with respect to the y axis for n =
1, 3, 5, … This problem can be solved on the half domain with u = 0 on the boundary coin-
cident with the y axis.
3. The functions qn = a + b cos(5n(𝜃 − 𝜋∕2)) and u0 = c are periodic for n = 1, 2, 3, … This prob-
lem can be solved on a periodic subdomain, for example the shaded sector shown in Fig. 2.2.
Exercise 2.10 Suppose that the problem described in Example 2.2 is modified such that on the
boundary 𝜕Ω(i) flux q(i)
n (𝜃) is prescribed, which is a periodic function. Assuming that on 𝜕Ω
(o)
qn =
(i)
a + b cos(5(𝜃 − 𝜋∕2)), what restriction must be imposed on qn (𝜃)? Hint: Refer to eq. (2.28.)
Numerical treatment of periodic functions

Direct implementation of periodic boundary conditions for arbitrary forcing functions leads to
minimization of the potential energy subject to linear constraints. The constraint conditions are
coupled on periodic boundary segment pairs. However, in the special cases of even and odd loading
functions, symmetric periodic subdomains and symmetric material properties on periodic subdo-
mains, the periodic boundary conditions are not coupled. In fact, the boundary conditions are the
same as the conditions of symmetry and antisymmetry respectively, which makes implementation
much simpler.
A function f (x) defined on the interval −𝓁∕2 < x < 𝓁∕2 is an even function if f (x) = f (−x). It
is an odd function if f (x) = −f (−x). Examples of even functions on −∞ < x < ∞ are cos x, cosh x.
Examples of odd functions are sin x, sinh x. Any function can written as the sum of an even and an
odd function:
f (x) = fe (x) + fo (x) (2.31)
where
1
fe (x) = (f (x) + f (−x)) (2.32)
2
5 William Thomson, Lord Kelvin 1824–1907.

is an even function and

1
fo (x) = (f (x) − f (−x)) (2.33)
2
is an odd function.
Example 2.3 Consider the domain shown in Fig. 2.3. We assume that the source function is zero
and the loading function on the boundary 𝜕Ω(o) is the sum of a symmetric (even) and an antisym-
metric (odd) function:
qn = cos(5(𝜃 − 𝜋∕2)) + sin(5(𝜃 − 𝜋∕2)), 0 ≤ 𝜃 < 2𝜋.
Let us assume that on all other boundary segments qn = 0 and note that qn satisfies eq. (2.28). The
temperature u(x, y) is determined by these boundary conditions up to an arbitrary constant. We will
fix this constant to be 300 K.
The radius of circle 𝜕Ω(o) is ro = 100 mm. The radius of circle 𝜕Ω(i) is ro = 22.5 mm. The radius
of the cut-out circles 𝜕Ω(k) (k = 1, 2, … , 5) is rk = 16.875 mm. The centers of the five cut-out circles
are evenly distributed on a circle of radius rm = 61.25 mm. The thickness is t = 5 mm. The material
is homogeneous and isotropic, the coefficient of thermal conduction is 0.161 W∕(mm2 K).
The solution u on a periodic segment is shown in Fig. 2.3(a). It is the sum of its symmetric part
shown in Fig. 2.3(b) and its antisymmetric part shown in Fig. 2.3(c). For the symmetric part the
boundary condition on the periodic boundary segment pair 𝜕Ω+p , 𝜕Ω−p is qn = 0, for the antisym-
metric part it is u = 0.
2.3.4 Dimensional reduction

In many important practical applications reduction of the number of dimensions is possible with-
out significantly affecting the quantities of interest. In other words, a mathematical model in one
or two dimensions may be an acceptable substitute for the fully three-dimensional model.
Planar problems
Consider a plate-like body shown in Fig. 2.4. The thickness, denoted by tz , will be assumed to be
constant. The mid-surface is the solution domain which will be denoted by Ω. The solution domain
lies in the xy plane. The boundary points of Ω (shown as a dotted line) is denoted by Γ. The unit
outward normal to the boundary is denoted by n.
The heat conduction problem in two dimensions is a special case of the three-dimensional prob-
lem, applicable when the boundary conditions on the surface parallel to the z axis and Q are
500
400
300
200
100
(a) (b) (c)
Figure 2.3 Example 2.3: The solution u of (a) the periodic problem, (b) the symmetric part of the periodic
problem, (c) the antisymmetric part of the periodic problem.
z y Ω Δx
Δy →
t
x →
Γ n
Δs
tz
Figure 2.4 Notation for two-dimensional domains.
independent of z
( ) ( )
𝜕 𝜕u 𝜕u 𝜕 𝜕u 𝜕u 𝜕u
kx + kxy + kyx + ky + Q = c𝜌 (2.34)
𝜕x 𝜕x 𝜕y 𝜕y 𝜕x 𝜕y 𝜕t
where the meaning of Q depends on the boundary conditions prescribed on the top and bottom
surfaces (z = ±tz ∕2) as described in the following.
Eq. (2.34) represents one of two cases:
1. The thickness is large in comparison with the other dimensions and both the material properties
and boundary conditions are independent of z, i.e., u(x, y, z) = u(x, y). This is equivalent to the
case of finite thickness with the top and bottom surfaces (z = ±tz ∕2) perfectly insulated and
Q = Q.
2. The thickness is small in relation to the other dimensions. In this case the two-dimensional
solution is an approximation of the three-dimensional solution. It can be interpreted as the first
term in the expansion of the three-dimensional solution with respect to the z coordinate. The
definition of Q depends on the boundary conditions on the top and bottom surfaces as explained
here.
a) Prescribed flux: Let us denote the heat flux prescribed on the top (resp. bottom) surface by q̂ +n
(resp. q̂ −n ). Note that positive q̂ n is heat flux exiting the body. The amount of heat exiting the
body over a small area ΔA per unit time is (q̂ +n + q̂ −n )ΔA. Dividing by ΔAtz , the heat generated
per unit volume becomes
1
Q = Q − (q̂ +n + q̂ −n ) ⋅ (2.35)
tz
b) Convective heat transfer: Let us denote the coefficient of convective heat transfer at z = tz ∕2
(resp. z = −tz ∕2) by h+c (resp. h−c ) and the corresponding temperature of the convective
medium by u+c (resp. u−c ) then the amount of heat exiting the body over a small area ΔA per
unit time is [h+c (u − u+c ) + h−c (u − u−c )]ΔA. Therefore the heat generated per unit volume is
changed to
1
Q = Q − [h+c (u − u+c ) + h−c (u − u−c )] ⋅ (2.36)
tz
Of course, combinations of these boundary conditions are possible. For example flux may be
prescribed on the top surface and convective boundary conditions may be prescribed on the bottom
surface.
In the two-dimensional formulation u is assumed to be constant through the thickness. There-
fore prescribing a temperature has meaning only if the temperature is the same on the top and
bottom surfaces. Furthermore, the temperature is a continuous function therefore specification of
the temperature on the top and bottom surfaces determines the temperature on the side surfaces
as well. In this case the solution is the prescribed temperature.
E 25.714°
D
R32.0
R30.0
R1.0
5.0° R47.0
C
1.0
A 17.0 B
Figure 2.5 Example 2.4: The solution domain and finite element mesh (mm).
Figure 2.6 Control volume and notation for qy + Δqy

heat conduction in 2D.
qx qx + Δqx
z y
x qy Δy
Δx
Example 2.4 Consider the mathematical problem associated with estimating heat loss per unit
length in a stainless steel pipe with cooling fins. The inner radius of the pipe is 30.0 mm, the outer
radius is 32.0 mm. There are seven equally spaced cooling fins. One-fourteenth of the cross-section
is shown in Fig. 2.5.
The coefficient of thermal conductivity is 0.0236 W/(mmK) and the coefficient of convective heat
transfer is 1.8 × 10−4 W∕(mm2 K). The temperature of the internal surface is 800 K. The external
surface is cooled by convection. The temperature of the convective medium is 300 K.
The solution is a periodic function which is even with respect to segment AB. Therefore the
normal flux is zero on boundary segments AB and DE, i.e., the boundary conditions are symmetry
boundary condition. The estimated heat loss converges to 1816 W/m for the sector shown in Fig. 2.5.
The heat loss for the entire pipe is 14 times this number.
Exercise 2.11 Investigate how the length of the fin influences heat loss. Solve the problem of
Example 2.4 letting dimension AB vary between 7 and 17 mm.
Exercise 2.12 Using the control volume shown in Fig. 2.6, derive eq. (2.34) from first
principles.
Exercise 2.13 Assume that tz = tz (x, y) > 0 is a continuous and differentiable function and the
maximum value of tz is small in comparison with the other dimensions. Assume further that con-
vective heat transfer occurs on the surfaces z = ±tz ∕2 with h+c = h−c = hc and u+c = u−c = uc . Using
a control volume similar to that shown in Fig. 2.6, but accounting for variable thickness, show that
in this case the conservation law for heat conduction in two dimensions is:
𝜕 𝜕 𝜕u
− (tz qx ) − (tz qy ) − 2hc (u − uc ) + Qtz = c𝜌tz ⋅ (2.37)
𝜕x 𝜕y 𝜕t
Remark 2.2 In Exercise 2.13 the thickness tz was assumed to be continuous and differentiable
on Ω. If tz is continuous and differentiable over two or more subdomains of Ω, but discontinuous
on the boundaries of the subdomains, then eq. (2.37) is applicable on each subdomain subject to
the requirement that qn tz is continuous on the boundaries of the subdomains.
Example 2.5 Let the domain

Ω = {x, y | − b < x < b, −c < x < c}
represent the mid-surface of a rectangular plate of constant thickness tz . Assume that the surfaces
(z = ±tz ∕2) are perfectly insulated. On the side surfaces (x = ±b, y = ±c, −t∕2 < z < t∕2) the tem-
perature is constant (u = û 0 ). Let kx = ky = k, kxy = 0 and Q = Q0 where k and Q0 are constants.
The goal is to determine the stationary temperature distribution in the plate. The mathematical
problem is to solve:
( 2 )
𝜕 u 𝜕2 u
k + + Q0 = 0 (2.38)
𝜕x2 𝜕y2
on Ω with the boundary condition u = û 0 .
The solution of this problem can be determined by classical methods6 :
n−1 ( )
Q b2 ∑ (−1) 2
∞
cosh n𝜋y∕2b
u = û 0 + 16 0 3 1 − cos n𝜋x∕2b. (2.39)
k 𝜋 n=1,3,5,… n3 cosh n𝜋c∕2b
This infinite series converges absolutely. It is seen that the classical solution of this seemingly
simple problem is rather complicated and in fact the exact solution can be computed only approx-
imately. However, the truncation error can be made arbitrarily small by computing a sufficiently
large number of terms of the infinite series.
The solution would be far more complicated if (for example) Ω were a general polygonal domain.
Exercise 2.14 Write down the generalized formulation of the problem in Example 2.5 for (a) the
entire domain and (b) taking advantage of symmetry, for one quarter of the domain. Specify B(u, 𝑣),
F(𝑣) and the trial and test spaces for both cases.
Exercise 2.15 Let Ω be a regular pentagon (i.e. a pentagon with equal sides). Taking advantage
of the periodicity of the solution, write down the generalized formulation corresponding to the
problem given by eq. (2.38) on the smallest domain possible. Assume that u = 0 on the boundaries.
Specify the domain, B(u, 𝑣), F(𝑣) and the trial and test spaces.
Exercise 2.16 Assume that the coefficients of thermal conduction kx , kxy , ky are given. Show that
in a Cartesian coordinate system x′ y′ , rotated counterclockwise by the angle 𝛼 relative to the xy
system, the coefficients will be:
[ ] [ ][ ][ ]
kx′ kx′ y′ cos 𝛼 sin 𝛼 kx kxy cos 𝛼 − sin 𝛼
= ⋅
ky′ x′ ky′ − sin 𝛼 cos 𝛼 kyx ky sin 𝛼 cos 𝛼
6 See, for example, Timoshenko, S. and Goodier, J. N., Theory of Elasticity, McGraw-Hill, New York, 2nd ed. 1951,
pp. 275-276.
Hint: A scalar {a}T [K]{a}, where {a} is an arbitrary vector, is invariant under coordinate transfor-
mation by rotation.
Axisymmetric models
Axial symmetry exists when the solution domain can be generated by sweeping a plane figure
around an axis, known as the axis of symmetry, and the material properties and boundary
conditions are axially symmetric. For example, pipes, cylindrical and spherical pressure vessels
are often idealized in this way. In such cases the problem can be formulated in cylindrical
coordinates and, since the solution is independent of the circumferential variable, the number
of dimensions is reduced to two. In the following the z axis will be the axis of symmetry and the
radial (resp. circumferential) coordinates will be denoted by r (resp. 𝜃). Referring to the result of
Exercise 2.7, and letting q𝜃 = 0, the conservation law is
1 𝜕(rqr ) 𝜕qz 𝜕u
− − + Q = c𝜌 ⋅
r 𝜕r 𝜕z 𝜕t
Substituting the axisymmetric form of Fourier’s law:
𝜕u 𝜕u
qr = −kr , qz = −kz
𝜕r 𝜕z
we have the formulation of the axisymmetric heat conduction problem in cylindrical coordinates:
( ) ( )
1 𝜕 𝜕u 𝜕 𝜕u 𝜕u
rkr + kz + Q = c𝜌 ⋅ (2.40)
r 𝜕r 𝜕r 𝜕z 𝜕z 𝜕t
One or more segments of the boundary may lie on the z axis. Implied in the formulation is that
the boundary condition is the zero flux condition on those segments. Therefore it would not be
meaningful to prescribe essential boundary conditions on those boundary segments. To show this,
consider an axisymmetric problem of heat conduction, the solution of which is independent of z.
For simplicity we assume that kr = 1. In this case the problem is essentially one-dimensional:
( )
1 d du
r =0 ri < r < ro .
r dr dr
Assuming that the boundary conditions u(ri ) = û i , u(ro ) = û o are prescribed. the exact solution of
this problem is:
û o − û i û ln ro − û o ln ri
u(r) = ln r + i ⋅ (2.41)
ln ro − ln ri ln ro − ln ri
Consider now the solution in an arbitrary fixed point r = 𝜚 where ri < 𝜚 < ro and let ri → 0:
( )
û o − û i ln 𝜚 û i ln ro ∕ ln ri − û o
limu(𝜚) = lim + = û o .
ri →0 ri →0 ln r ∕ ln r − 1 ln r ln ro ∕ ln ri − 1
o i i
Therefore the solution is independent of û i when ri = 0. It is left to the reader in Exercise 2.17 to
show that du∕dr → 0 as ri → 0, hence the boundary condition at r = 0 is the zero flux boundary
condition.
Exercise 2.17 Refer to the solution given by eq. (2.41). Show that for any 𝜚 > 0
( )
du
→ 0 as ri → 0
dr r=𝜚
independently of û i and û o .
Exercise 2.18 Derive eq. (2.40) by considering a control volume in cylindrical coordinates and
using the assumption that the temperature is independent of the circumferential variable.
Exercise 2.19 Consider the generalized formulation of steady state heat conduction in cylindrical
coordinates in the special case when the solution depends only on the radial variable r:
ro ( ) ( )
du d𝑣 du du
k(r) r dr = rk 𝑣 − rk 𝑣 ⋅
∫ri dr dr dr r=ro dr r=ri
(a) Derive this formulation from eq. (2.40) and (b) apply this formulation to a long pipe of internal
radius ri , external radius ro using the boundary conditions u(ri ) = ûi and
du
qn = −k = hc (u − uc ) at r = ro .
dr
Exercise 2.20 Consider water flowing in a stainless steel pipe. The temperature of the water is
80 ∘ C. The outer surface of the pipe is cooled by air flow. The temperature of the air is 20 ∘ C. The
outer diameter of the pipe is 0.20 m and its wall thickness is 0.01 m. (a) Assuming that convec-
tive heat transfer occurs on both the inner and outer surfaces of the pipe and u is a function of r
only, formulate the mathematical model for stationary heat transfer. (b) Assume that the coefficient
of thermal conduction of stainless steel is 20 W∕mK. Using h(𝑤) c = 750 W∕m2 K for the water and
(a)
hc = 10 W∕m2 K for the air, determine the temperature of the external surface of the pipe and the
rate of heat loss per unit length.
Heat conduction in a bar

One-dimensional models of heat conduction are discussed in this section. It is assumed that (a)
the solution domain is a bar, one end of which is located in x = 0, the other end in x = 𝓁; (b) the
dimensions of the cross-section are small in comparison with 𝓁 and the cross-sectional area A =
A(x) > 0 is a continuous and differentiable function.
If convective heat transfer occurs along the bar, as described in Section 2.3.4, the conservation
law is:
𝜕(Aq) 𝜕u
− − cb (u − ua ) + QA = c𝜌A
𝜕x 𝜕t
where cb = cb (x) is the coefficient of convective heat transfer of the bar (in W∕(mK) units) obtained
from hc by integration:
cb = hc ds
∮
the contour integral taken along the perimeter of the cross-section. Therefore the differential
equation of heat conduction in a bar is:
( )
𝜕 𝜕u 𝜕u
Ak − cb (u − ua ) + QA = c𝜌A ⋅ (2.42)
𝜕x 𝜕x 𝜕t
One of the boundary conditions described in Section 2.3 is prescribed at x = 0 and x = 𝓁.
Example 2.6 Consider stationary heat flow in a partially insulated bar of length 𝓁 and constant
cross-section A. The coefficients k and cb are constant and Q = 0. Therefore eq. (2.42) can be cast
in the following form:
c
u′′ − 𝜆2 (u − ua ) = 0, 𝜆2 = b ⋅
Ak
If the temperature ua is a linear function of x, i.e., ua (x) = a + bx and the boundary conditions are
u(0) = û 0 , q(𝓁) = q̂ 𝓁 then the solution of this problem is:
u = C1 cosh 𝜆x + C2 sinh 𝜆x + a + bx
where:
( )
1 q̂ 𝓁
C1 = û 0 − a, C2 = − + (û 0 − a)𝜆 sinh 𝜆𝓁 + b .
𝜆 cosh 𝜆𝓁 k
Exercise 2.21 Solve the problem described in Example 2.6 using the following boundary condi-
tions: q(0) = q̂ 0 , q(𝓁) = h𝓁 (u(𝓁) − Ua ) where q̂ 0 , h𝓁 , Ua are given data.
Exercise 2.22 A perfectly insulated bar of constant cross-section, length 𝓁, thermal conduction k,
density 𝜌 and specific heat c is subject to the initial condition u(x, 0) = U0 (constant) and the bound-
ary conditions u(0, t) = u(𝓁, t) = 0. Assuming that Q = 0, verify that the solution of this problem
is:
( 2 2 ) ( )
∑ ∞
1 − cos(n𝜋) n𝜋 k x
u = 2U0 exp − 2 t sin n𝜋 ⋅
n=1
n𝜋 𝓁 c𝜌 𝓁
It is sufficient to show that the differential equation, the boundary conditions and the initial con-
dition are satisfied by u = u(x, t).
2.4 Equations of linear elasticity – strong form

Mathematical problems of linear elasticity belong in the category of vector elliptic boundary value
problems. The unknown functions are the components of the displacement vector. In Cartesian
coordinates the displacement vector is:
def
u = ux (x, y, z) ex + uy (x, y, z) ey + uz (x, y, z) ez
≡ {ux (x, y, z) uy (x, y, z) uz (x, y, z)}T
≡ ui (xj ) (2.43)
where ex , ey , ez are the Cartesian basis vectors.
The formulation of problems of linear elasticity is based on three fundamental relationships: the
strain-displacement equations, the stress-strain relationships and the equilibrium equations.
1. Strain-displacement relationships. We introduce the infinitesimal strain-displacement rela-
tionships here. A detailed derivation of these relationships is presented in Section 9.2.1. By
definition, the infinitesimal normal strain components are:
def 𝜕ux def 𝜕uy def 𝜕uz
𝜖x ≡ 𝜖xx = 𝜖y ≡ 𝜖yy = 𝜖z ≡ 𝜖zz = (2.44)
𝜕x 𝜕y 𝜕z
and the shear strain components are:
( )
𝛾xy def 1 𝜕ux 𝜕uy
𝜖xy = 𝜖yx ≡ = +
2 2 𝜕y 𝜕x
( )
𝛾yz def 1 𝜕uy 𝜕uz
𝜖yz = 𝜖zy ≡ = +
2 2 𝜕z 𝜕y
( )
𝛾zx def 1 𝜕uz 𝜕ux
𝜖zx = 𝜖xz ≡ = + (2.45)
2 2 𝜕x 𝜕z
where 𝛾xy , 𝛾yz , 𝛾zx are called the engineering shear strain components. In index notation, the
infinitesimal strain at a point is characterized by the strain tensor
def 1( )
𝜖ij = ui,j + uj,i . (2.46)
2
2. Stress-strain relationships Mechanical stress is defined as force per unit area (N∕m2 ≡ Pa).
Since one Pascal (Pa) is a very small stress, the usual unit of mechanical stress is the Megapascal
(MPa) which can be understood to mean either 106 N∕m2 or 1 N∕mm2 .
The usual notation for stress components is illustrated on an infinitesimal volume element
shown in Fig. 2.7. The indexing rules are as follows: Faces to which the positive x, y, z axes are
normal are called positive faces, the opposite faces are called negative faces. The normal stress
components are denoted by 𝜎, the shear stresses components by 𝜏. The normal stress compo-
nents are assigned one subscript only, since the orientation of the face and the direction of the
stress component are the same. For example, 𝜎x is the stress component acting on the faces to
which the x-axis is normal and the stress component is acting in the positive (resp. negative)
coordinate direction on the positive (resp. negative) face. For the shear stresses, the first index
refers to the coordinate direction of the normal to the face on which the shear stress is acting.
The second index refers to the direction in which the shear stress component is acting.
On a positive (resp. negative) face the positive stress components are oriented in the positive
(resp. negative) coordinate directions. The reason for this is that if we subdivide a solid body
into infinitesimal hexahedral volume elements, similar to the element shown in Fig. 2.7, then
each negative face will be coincident with a positive face. By the action-reaction principle, the
forces acting on those faces must have equal absolute value and opposite sign. In index notation
𝜎11 ≡ 𝜎x , 𝜎12 ≡ 𝜎xy ≡ 𝜏xy , etc.
The mechanical properties of isotropic elastic materials are characterized by the modulus of
elasticity E > 0, Poisson’s ratio7 𝜈 < 1∕2 and the coefficient of thermal expansion 𝛼 > 0. The
𝜎z τyz
τyx
τyx
τzy
τzx 𝜎y
τxz τxy
𝜎x 𝜎x
τxy τxz
𝜎y τzx
z
τyz τzy
y 𝜎z
x
Figure 2.7 Notation for stress components.
7 Simeon Denis Poisson 1781–1840.

stress-strain relationships, known as Hooke’s law8 , are:

1( )
𝜖x = 𝜎 − 𝜈𝜎y − 𝜈𝜎z + 𝛼Δ (2.47)
E x
1 ( )
𝜖y = −𝜈𝜎x + 𝜎y − 𝜈𝜎z + 𝛼Δ (2.48)
E
1( )
𝜖z = −𝜈𝜎x − 𝜈𝜎y + 𝜎z + 𝛼Δ (2.49)
E
2(1 + 𝜈)
𝛾xy ≡ 2𝜖xy = 𝜏xy (2.50)
E
2(1 + 𝜈)
𝛾yz ≡ 2𝜖yz = 𝜏yz (2.51)
E
2(1 + 𝜈)
𝛾zx ≡ 2𝜖zx = 𝜏zx (2.52)
E
where Δ = Δ (x, y, z) is the temperature change with respect to a reference temperature at
which the strain is zero. The strain components represent the total strain; 𝛼Δ is the thermal
strain. Mechanical strain is defined as the total strain minus the thermal strain.
In index notation Hooke’s law can be written as
1+𝜈 𝜈
𝜖ij = 𝜎 − 𝜎 𝛿 + 𝛼Δ 𝛿ij . (2.53)
E ij E kk ij
The inverse is:
𝜎ij = 𝜆𝜖kk 𝛿ij + 2G𝜖ij − (3𝜆 + 2G)𝛼Δ 𝛿ij (2.54)
where 𝜆 and G, called the Lamé constants9 , are defined by
def E𝜈 def E
𝜆= , G= ⋅ (2.55)
(1 + 𝜈)(1 − 2𝜈) 2(1 + 𝜈)
G is also called shear modulus or the modulus of rigidity. Since 𝜆 and G are positive, the admis-
sible range of Poisson’s ratio is −1 < 𝜈 < 1∕2. Typically it is 0 ≤ 𝜈 < 1∕2.
The generalized Hooke’s law states that the components of the stress tensor are linear functions
of the mechanical strain tensor:
𝜎ij = Cijkl (𝜖kl − 𝛼kl Δ ) (2.56)
where Cijkl and 𝛼ij are Cartesian tensors. By symmetry considerations the maximum number of
independent elastic constants that characterize Cijkl is 21. The symmetric tensor 𝛼ij is character-
ized by six independent coefficients of thermal expansion. This is the general form of anisotropy
in linear elasticity.
Remark 2.3 When residual stresses are present in the reference configuration then equation
(2.56) has to be modified. This point is discussed in Section 2.7.
3. Equilibrium Considering the dynamic equilibrium of a volume element, similar to that shown
in Fig. 2.7, except that the edges are of length Δx, Δy, Δz, six equations of equilibrium are writ-
ten: the resultants of the forces and moments must vanish. Assuming that the material is not
loaded by distributed moments (body moments), consideration of moment equilibrium leads to
the conclusion that 𝜏xy = 𝜏yx , 𝜏yz = 𝜏zy , 𝜏zx = 𝜏xz , i.e., the stress tensor is symmetric. Assuming
further that the components of the stress tensor are continuous and differentiable, application of
8 Robert Hooke 1635–1703.

9 Gabriel Lamé 1795–1870.
d’Alembert’s principle10 and consideration of force equilibrium leads to three partial differential
equations:
𝜕𝜎x 𝜕𝜏xy 𝜕𝜏xz 𝜕2 u
+ + + Fx − 𝜚 2x =0 (2.57)
𝜕x 𝜕y 𝜕z 𝜕t
𝜕𝜏xy 𝜕𝜎y 𝜕𝜏yz 𝜕 2 uy

+ + + Fy − 𝜚 =0 (2.58)
𝜕x 𝜕y 𝜕z 𝜕t2
𝜕𝜏xz 𝜕𝜏yz 𝜕𝜎z 𝜕 2 uz

+ + + Fz − 𝜚 =0 (2.59)
𝜕x 𝜕y 𝜕z 𝜕t2
where Fx , Fy , Fz are the components of the body force vector (in N∕m3 units), 𝜚 is the specific
density (in kg∕m3 ≡ Ns2 ∕m4 units). These equations are called the equations of motion. In index
notation:
𝜕 2 ui
𝜎ij,j + Fi = 𝜚 ⋅ (2.60)
𝜕t2
For elastostatic problems the time derivative is zero and the boundary conditions are indepen-
dent of time. This yields the equations of static equilibrium:
𝜎ij,j + Fi = 0. (2.61)
2.4.1 The Navier equations

The equations of motion, called the Navier equations11 , are obtained by substituting eq. (2.54) into
eq. (2.60). In elastodynamics the effects of temperature are usually negligible, hence we will assume
Δ = 0:
𝜕 2 ui
Gui,jj + (𝜆 + G)uj,ji + Fi = 𝜚 ⋅ (2.62)
𝜕t2
In elastostatic problems we have:
Gui,jj + (𝜆 + G)uj,ji + Fi = (3𝜆 + 2G)𝛼(Δ ),i . (2.63)
Exercise 2.23 Derive eq. (2.63) by substituting eq. (2.54) into eq. (2.61). Explain the rules under
which the indices are changed to obtain eq. (2.63). Hint: (𝜖kk 𝛿ij ),j = (uk,k 𝛿ij ),j = uk,ki = uj,ji .
Exercise 2.24 Derive the equilibrium equations from first principles.
Exercise 2.25 In deriving eq. (2.62) and (2.63) it was assumed that 𝜆 and G are constants. For-
mulate the analogous equations assuming that 𝜆 and G are smooth functions of xi ∈ Ω.
Exercise 2.26 Assume that Ω is the union of two or more subdomains and the material prop-
erties are constants on each subdomain but vary from subdomain to subdomain. Formulate the
elastostatic problem for this case.
10 Jean Le Rond d’Alembert 1717–1783.

11 Claude Louis Marie Henri Navier 1785–1836.
2.4.2 Boundary and initial conditions

As in the case of heat conduction, we will consider three kinds of boundary conditions: prescribed
displacements, prescribed tractions and spring boundary conditions. Tractions are forces per
unit area acting on the boundary. Prescribed displacements and tractions are often specified in a
normal-tangent reference frame.
1. Prescribed displacement. One or more components of the displacement vector is prescribed
on all or part of the boundary. This is called a kinematic boundary condition.
2. Prescribed traction. One or more components of the traction vector is prescribed on all or part
of the boundary. The definition of traction vector is given in Appendix K.1.
3. Linear spring. A linear relationship is prescribed between the traction and displacement vector
components. The general form of this relationship is:
Ti = cij (dj − uj ) (2.64)
where Ti is the traction vector, cij is a positive-definite matrix that represents the distributed
spring coefficients; dj is a prescribed function that represents displacement imposed on the
spring and uj is the (unknown) displacement vector function on the boundary. The spring coef-
ficients cij (in N∕m3 units) may be functions of the position xk but are independent of the dis-
placement ui . This is called a “Winkler spring12 ”.
A schematic representation of this boundary condition on an infinitesimal boundary surface
element is shown in Fig. 2.8 under the assumption that cij is a diagonal matrix and therefore
def def def
three spring coefficients c1 = c11 , c2 = c22 , c3 = c33 characterize the elastic properties of the
boundary condition.
Fig. 2.8 should be interpreted to mean that the imposed displacement di will cause a differential
force ΔFi to act on the centroid of the surface element. Suspending the summation rule, the
magnitude of ΔFi is
ΔFi = ci ΔA(di − ui ), i = 1, 2, 3
where ui is the displacement of the surface element. The corresponding traction vector is:
ΔFi
Ti = lim = ci (di − ui ), i = 1, 2, 3.
ΔA→0 ΔA
Figure 2.8 Spring boundary condition. Schematic

representation.
x3 d3
c3Δ A
u3 d2
u2 c2Δ A
c1Δ A u1 x2
d1
ΔA
x1
12 Emil Winkler 1835–1888.

The enumerated boundary conditions may occur in any combination. For example, the displace-
ment vector component u1 , the traction vector component T2 and a linear combination of T3 and
u3 may be prescribed on a boundary segment.
In engineering practice boundary conditions are most conveniently prescribed in the
normal-tangent reference frame. The normal is uniquely defined on smooth surfaces but
the tangential coordinate directions are not. It is necessary to specify the tangential coordinate
directions with respect to the reference frame used. The required coordinate transformations are
discussed in Appendix K.
The boundary conditions are generally time-dependent. For time-dependent problems the initial
conditions, that is, the initial displacement and velocity fields, denoted respectively by U(x, y, z) and
V(x, y, z), have to be prescribed:
( )
𝜕u
u(x, y, z, 0) = U(x, y, z) and = V(x, y, z).
𝜕t (x,y,z,0)
Exercise 2.27 Assume that the following boundary conditions are given in the normal-tangent
reference frame xi′ , x1′ being coincident with the normal: T1′ = c′1 (d′1 − u′1 ); T2′ = T3′ = 0. Using the
transformation xi′ = gij xj , determine the boundary conditions in the xi coordinate system. (See the
Appendix, Section K.2.)
2.4.3 Symmetry, antisymmetry and periodicity

Symmetry and antisymmetry of vectors in two dimensions with respect to the y axis is illustrated
in Fig. 2.9
The definition of symmetry and antisymmetry of vectors in three dimensions is analogous: the
corresponding vector components parallel to a plane of symmetry (resp. antisymmetry) have the
same absolute value and the same (resp. opposite) sign. The corresponding vector components
normal to a plane of symmetry (resp. antisymmetry) have the same absolute value and opposite
(resp. same) sign.
In a plane of symmetry the normal displacement and the shearing traction components are zero.
In a plane of antisymmetry the normal traction is zero and the in-plane components of the dis-
placement vector are zero.
When the solution is periodic on Ω then a periodic sector of Ω has at least one periodic boundary
segment pair denoted by 𝜕Ω+p and 𝜕Ω−p . On corresponding points of a periodic boundary segment
pair, P+ ∈ 𝜕Ω+p and P− ∈ 𝜕Ω−p the normal component of the displacement vector and the periodic
in-plane components of the displacement vector have the same value. The normal component of the
traction vector and the periodic in-plane components of the traction vector have the same absolute
value but opposite sign.
y y
B′ B B′ B
vy → → → →
v u uy vy v u uy
vx A′ A ux vx A′ A ux
x x
(a) symmetry (b) antisymmetry
Figure 2.9 Symmetry and antisymmetry of vectors in two dimensions.

Exercise 2.28 A homogeneous isotropic elastic body with Poisson’s ratio zero occupies the
domain Ω = {x, y, z | |x| < a, |y| < b, |z| < c}. Define tractions on the boundaries of Ω such that
the tractions satisfy equilibrium and the plane z = 0 is a plane of (a) symmetry, (b) antisymmetry.
Exercise 2.29 Consider the domain shown in Fig. 2.2. Assume that Tn = 0 and Tt = 𝜏o where 𝜏o
is constant, is prescribed on 𝜕Ω(o) , Tn = Tt = 0 on the circular boundaries 𝜕Ω(k) and un = ut = 0 on
𝜕Ω(i) . Specify periodic boundary conditions on 𝜕Ω+p and 𝜕Ω−p .
2.4.4 Dimensional reduction in linear elasticity

Owing to the complexity of three-dimensional problems in elasticity, dimensional reduction is
widely used. Various kinds of dimensional reduction are possible in elasticity, such as planar,
axisymmetric, shell, plate, beam and bar models. Each of these model types is sufficiently impor-
tant to have generated a substantial technical literature. In the following models for planar and
axially symmetric problems are discussed. Models for beams, plates and shells will be discussed
separately.
Planar elastostatic models: Notation

We consider a prismatic body of length 𝓁. The material points occupy the domain Ω𝓁 , defined as
follows:
Ω𝓁 = {(x, y, z) | (x, y) ∈ 𝜔, −𝓁∕2 < z < 𝓁∕2, 𝓁 > 0} (2.65)
where 𝜔 ∈ ℝ2 is a bounded domain. The lateral boundary of the body is denoted by
Γ𝓁 = {(x, y, z) | (x, y) ∈ 𝜕𝜔, −𝓁∕2 < z < 𝓁∕2, 𝓁 > 0} (2.66)
and the faces are denoted by
𝛾± = {(x, y, z) | (x, y) ∈ 𝜔, z = ±𝓁∕2}. (2.67)
The notation is shown in Fig. 2.10(a). The diameter of 𝜔 will be denoted by d𝜔 .
The material properties, the volume forces and temperature change acting on Ω𝓁 and tractions
acting on Γ𝓁 will be assumed to be independent of z. Therefore the x, y plane is a plane of symmetry.
The tangential direction will be understood to be with reference to contours parallel to 𝜕𝜔.
∂ω C
ro
γ−
γ+ D
x
y
ω ri
/2 𝛼
z A B x
/2 (a) (b)
Figure 2.10 Notation.

Plane strain
When the boundary conditions on 𝛾+ and 𝛾− are uz = 𝜏zx = 𝜏zy = 0 and on Γ𝓁 normal and tangential
displacements and/or tractions are prescribed that are independent of z, then a plane strain model
may be used. The stress-strain relationships for plane strain models are obtained from Hooke’s law,
see equations (2.47) to (2.52), by letting 𝜖z = 𝛾yz = 𝛾zx = 0 to obtain
⎧ 𝜎 ⎫ ⎡𝜆 + 2G 𝜆 0 ⎤ ⎧ 𝜖x ⎫ ⎧1⎫
⎪ x⎪ ⎢ ⎪ ⎪ ⎪ ⎪
⎨ 𝜎y ⎬ = ⎢ 𝜆 𝜆 + 2G 0 ⎥ ⎨ 𝜖y ⎬ − (3𝜆 + 2G)𝛼Δ ⎨1⎬ ⋅ (2.68)
⎪𝜏xy ⎪ ⎣ 0 ⎥⎪ ⎪ ⎪ ⎪
⎩ ⎭ 0 G ⎦ 𝛾
⎩ xy ⎭ 0
⎩ ⎭
Observe that the boundary conditions prescribed on 𝛾+ and 𝛾− are periodic. Therefore the plane
strain solution is extended by periodic repetition to −∞ < 𝓁 < ∞ independent of what value is
assigned to 𝓁. This implies that uz = 0 and hence 𝜖z = 0. The solution is z-independent, therefore
𝜕ux ∕𝜕z = 𝜕uy ∕𝜕z = 0, hence the shearing strains 𝛾xz and 𝛾yz are zero. The 2D and 3D solutions are
the same in the x y plane. The stress 𝜎z can be computed from the plane strain solution using
𝜎z = −𝜈(𝜎x + 𝜎y ) − E𝛼Δ .
Note that the zero normal displacement condition on 𝛾+ and 𝛾− is an idealization that cannot be
realized in practice.
Plane stress
When the boundary conditions on 𝛾+ and 𝛾− are zero normal and shearing tractions and on Γ𝓁
normal and tangential displacements and/or tractions are prescribed that are independent of z,
then often a plane stress model is used. Letting 𝜎z = 𝜏yz = 𝜏zx = 0, equations (2.47) to (2.52) are
simplified to
⎧𝜎 ⎫ ⎡1 𝜈 0 ⎤⎧𝜖 ⎫ ⎧1⎫
⎪ x⎪ E ⎢𝜈 1 ⎥ ⎪ x⎪
0 ⎥⎨𝜖 ⎬ − E𝛼 Δ ⎪ ⎪
⎨ 𝜎y ⎬ = 2 ⎢ 𝜈 ⎨1⎬ ⋅ (2.69)
⎪𝜏xy ⎪ 1 − 𝜈 ⎢0 0 1 − 𝜈 y
⎥ ⎪𝛾xy ⎪ 1 − ⎪0⎪
⎩ ⎭ ⎣ 2 ⎦⎩ ⎭ ⎩ ⎭
The conditions 𝜎z = 𝜏zx = 𝜏zy = 0 on Ω𝓁 cannot be exactly satisfied by the three-dimensional solu-
tion in the general case, except in the limit when 𝓁 approaches zero. Therefore the solution of a
plane stress model is an approximation of the corresponding 3D model, and the error of approxi-
mation depends on Poisson’s ratio and 𝓁.
Remark 2.4 When the material is homogeneous and isotropic and only tractions are specified
on Γ𝓁 (which must satisfy the equations if equilibrium) and the body force is zero13 , then the prob-
lem is classified as a first fundamental boundary value problem of elasticity [60]. The stress field is
completely determined by the boundary conditions and the compatibility condition. It is indepen-
dent of the material properties E and 𝜈 and therefore independent of whether the model is plane
stress or plane stain. The displacement field can be determined from the stress field up to rigid body
displacements.
Exercise 2.30 Consider the prismatic body that has the annular cross-section ABCD shown in
Fig. 2.10(b). Assume that the faces 𝛾+ and 𝛾− are stress free. The boundary conditions for the other
faces are listed in Table 2.1. Determine or estimate difference between the plane stress and fully
three-dimensional solutions measured (a) in energy norm and (b) the maximum norm of the von
Mises stress using the following data: ro = 200 mm, ri = 175 mm, 𝓁 = 40 mm, 𝛼 = 45∘ , E = 70.0
GPa, 𝜈 = 0.3, qn = −20 MPa, qt = 100 MPa.
13 It is possible to consider a more general case of body forces, but that is not addressed here.
Table 2.1 Exercise 2.30: Boundary conditions
Segment Case 1 Case 2 Case 3
AB symmetry antisymmetry ux = uy = 0
BC Tn = qn ,Tt = 0 Tn = 0,Tt = qt Tn = qn ,Tt = 0
CD symmetry antisymmetry symmetry
2
DA Tn = Tt = 0 Tn = 0,Tt = (ro ∕ri ) qt Tn = Tt = 0
Partial solution: Case 1 is the classical problem of a cylindrical elastic pipe subjected to external
pressure q. The exact stress distribution is known. The domain ABCD represents a 45∘ segment of
the pipe. The symmetry boundary conditions extend the domain by periodic repetition to the entire
cross-section. Clearly, the external pressure satisfies the equations of equilibrium. The difference
between the plane stress and three-dimensional solution is zero in energy norm, independent of
𝓁, and also zero in the maximum norm of the von Mises stress. The maximum von Mises stress is
170.7 MPa.
The Navier equations

The equations of equilibrium are:
𝜕𝜎x 𝜕𝜏xy
+ + Fx = 0 (2.70)
𝜕x 𝜕y
𝜕𝜏yx 𝜕𝜎y
+ + Fy = 0 (2.71)
𝜕x 𝜕y
where Fx , Fy are the components of body force vector F(x, y) (in N∕m3 units).
The Navier equations are obtained by substituting the stress-strain and strain-displacement rela-
tionships into the equilibrium equations. For plane strain;
( ) ( 2 )
𝜕 𝜕ux 𝜕uy 𝜕 ux 𝜕 2 ux E𝛼 𝜕Δ
(𝜆 + G) + +G + = − Fx (2.72)
𝜕x 𝜕x 𝜕y 𝜕x 2 𝜕y 2 1 − 2𝜈 𝜕x
( ) ( 2 )
𝜕 𝜕ux 𝜕uy 𝜕 uy 𝜕 2 uy E𝛼 𝜕Δ
(𝜆 + G) + +G + = − Fy . (2.73)
𝜕y 𝜕x 𝜕y 𝜕x2 𝜕y2 1 − 2𝜈 𝜕y
The boundary conditions are most conveniently written in the normal-tangent (n t) reference
frame illustrated in Fig. 2.4. The relationship between the x y and n t components of displacements
and tractions is established by the rules of vector transformation, described in the Appendix,
Section 2.4.2. The linear boundary conditions listed in Section 2.4.2 are applicable in two
dimensions as well.
Exercise 2.31 Derive eq. (2.69) and eq. (2.68) from Hooke’s law.
Exercise 2.32 Show that for plane stress the Navier equations are:
( ) ( 2 )
E 𝜕 𝜕ux 𝜕uy 𝜕 uy 𝜕 2 uy E𝛼 𝜕Δ
+ +G + = − Fy . (2.74)
2(1 − 𝜈) 𝜕x 𝜕x 𝜕y 𝜕x 2 𝜕y2 1 − 2𝜈 𝜕y
( ) ( 2 )
E 𝜕 𝜕ux 𝜕uy 𝜕 uy 𝜕 2 uy E𝛼 𝜕Δ
+ +G + = − Fy . (2.75)
2(1 − 𝜈) 𝜕y 𝜕x 𝜕y 𝜕x 2 𝜕y2 1 − 2𝜈 𝜕y
Exercise 2.33 Denote the components of the unit normal vector to the boundary by nx and ny .
Show that
Tx = Tn nx − Tt ny
Ty = Tn ny + Tt nx
where Tn (resp. Tt ) is the normal (resp. tangential) component of the traction vector.
Axisymmetric elastostatic models

The radial, circumferential and axial coordinates are denoted by r, 𝜃 and z respectively and the
displacement, stress, strain and traction components are labeled with corresponding subscripts.
The problem is formulated in terms of the displacement vector components ur (r, z) and uz (r, z).
1. The linear strain-displacement relationships in cylindrical coordinates are [105]:
def 𝜕u
𝜖r = r (2.76)
𝜕r
def ur
𝜖𝜃 = (2.77)
r
def 𝜕uz
𝜖z = (2.78)
𝜕z
( )
𝛾rz def 1 𝜕ur 𝜕uz
𝜖rz = = + ⋅ (2.79)
2 2 𝜕z 𝜕r
2. Stress-strain relationships. For isotropic materials the stress-strain relationship is obtained
from eq. (2.54):
⎧ ⎫ ⎡ ⎧ ⎫ ⎧ ⎫
⎪ 𝜎r ⎪ ⎢𝜆 + 2G 𝜆 𝜆 0 ⎤ ⎪ 𝜖r ⎪ ⎪1⎪
⎪ 𝜎𝜃 ⎪ ⎢ 𝜆 ⎥ ⎪ ⎪ E𝛼 ⎪1⎪
𝜆 + 2G 𝜆 0 ⎥ 𝜖𝜃 Δ
⎨𝜎 ⎬ = ⎢ 𝜆 𝜆 𝜆 + 2G 0⎥ ⎨ 𝜖 ⎬ − 1 − 2𝜈 ⎨1⎬ (2.80)
⎪ ⎪ ⎢
z ⎪ ⎪z ⎪ ⎪
⎪𝜏rz ⎪ ⎣ 0 0 0 G⎥⎦ ⎪𝛾rz ⎪ ⎪0⎪
⎩ ⎭ ⎩ ⎭ ⎩ ⎭
3. Equilibrium. The elastostatic equations of equilibrium are [105]:
1 𝜕(r𝜎r ) 𝜕𝜏rz 𝜎𝜃
+ − + Fr = 0 (2.81)
r 𝜕r 𝜕z r
1 𝜕(r𝜏rz ) 𝜕𝜎z
+ + Fz = 0. (2.82)
r 𝜕r 𝜕z
Exercise 2.34 Write down the Navier equations for the axisymmetric model.
2.4.5 Incompressible elastic materials

When 𝜈 → 1∕2 then 𝜆 → ∞ therefore the relationship represented by eq. (2.54) breaks down. Refer-
ring to eq. (2.53), the sum of normal strain components is related to the sum of normal stress
components by:
1 − 2𝜈
𝜖kk = 𝜎kk + 3𝛼Δ . (2.83)
E
def
The sum of normal strain components is called the volumetric strain and will be denoted by 𝜖vol =
𝜖kk . Defining:
def 1
𝜎0 = 𝜎 (2.84)
3 kk
we have
3(1 − 2𝜈)
𝜖vol ≡ 𝜖kk = 𝜎0 + 3𝛼Δ (2.85)
E
where the first term on the right is the mechanical strain, the second term is the thermal strain.
For incompressible materials, that is when 𝜈 = 1∕2, 𝜖vol = 3𝛼Δ is independent of 𝜎0 . Therefore 𝜎0
cannot be computed from the strains in the usual way. Substituting eq. (2.85) into eq. (2.54) and
letting 𝜈 = 1∕2 we have:
2E
𝜎ij = 𝜎0 𝛿ij + (𝜖 − 𝛼Δ 𝛿ij ). (2.86)
3 ij
Substituting into eq. (2.61), and assuming that E and 𝛼 are constant;
2E
(𝜎0 ),i + (𝜖 − 𝛼(Δ ),i ) + Fi = 0.
3 ij,j
Writing
1 1
𝜖ij,j = (u + uj,i ),j = (ui,jj + uj,ij )
2 i,j 2
and interchanging the order of differentiation in the second term, we have:
uj,ij = uj,ji = (uj,j ),i = (𝜖jj ),i = 3𝛼(Δ ),i .
Therefore, for incompressible materials,

1 3
𝜖ij,j = u + 𝛼(Δ ),i .
2 i,jj 2
The problem is to determine ui such that
E( )
(𝜎0 ),i + u + 𝛼(Δ ),i + Fi = 0 (2.87)
3 i,jj
subject to the condition of incompressibility, that is, the condition that volumetric strain can be
caused by temperature change but not by mechanical stress,
ui,i = 3𝛼Δ (2.88)
and the appropriate boundary conditions. If displacement boundary conditions are prescribed over
the entire boundary (𝜕Ωu = 𝜕Ω) then the problem has a solution only if the prescribed displace-
ments are consistent with the incompressibility condition:
ui ni dS = 3 𝛼Δ dV.
∫𝜕Ω ∫Ω
This follows directly from integrating eq. (2.88) and applying the divergence theorem.
Exercise 2.35 An incompressible bar of constant cross-section and length 𝓁 is subjected uniform
temperature change (i.e., Δ is constant). The centroidal axis of the bar is coincident with the x1
axis. The boundary conditions are: u1 (0) = u1 (𝓁) = 0. The body force vector is zero. Explain how
(2.86) and (2.88) are applied in this case to find that 𝜎11 = −E𝛼Δ .
2.5 Stokes flow

The flow of viscous fluids at very low Reynolds numbers14 (Re < 1) is modeled by the Stokes15
equations. There is a close analogy between the Stokes equations and the equations of incompress-
ible elasticity discussed in Section 2.4.5. In fluid mechanics the average compressive normal stress
is the pressure p. The vector ui represents the components of the velocity vector and the shear
modulus of the incompressible elastic solid E∕3 is replaced by the coefficient of dynamic viscosity
𝜇 (measured in Ns∕m2 units):
𝜇ui,jj = p,i − Fi (2.89)
ui,i = 0. (2.90)
Exercise 2.36 Write the Stokes equations in unabridged notation.
Exercise 2.37 Assume that velocities are prescribed over the entire boundary for a Stokes prob-
lem (i.e., 𝜕Ωu = 𝜕Ω). What condition must be satisfied by the prescribed velocities?
Remark 2.5 In his chapter we treated the physical properties, such as coefficients of heat conduc-
tion, the surface coefficient, the modulus of elasticity and Poisson’s ratio as given constants. Readers
should be mindful of the fact that physical properties are empirical data inferred from experimen-
tal observations. Owing to variations in experimental conditions and other factors, these data are
not known precisely and are always subject to restrictions. For example, stress is proportional to
strain up to the proportional limit only, the flux is proportional to the temperature gradient only
within a narrow range of temperatures. In fact, the coefficient of thermal conductivity (k) is typi-
cally temperature-dependent. For example, in the case of AISI 304 stainless steel k changes from
about 15 to about 20 W∕m∘ C in the temperature range of 0 to 400∘ C [18]. For a narrow range of
temperature (say 100 to 200∘ C) the size of the uncertainty in k is about the same as the change in
the mean value. Therefore, using a constant value for k, this range may be “good enough”. Ignoring
temperature-dependence for a much wider range of temperatures may lead to large errors, however.
Taking into account temperature-dependence of the coefficient of thermal conductivity leads
to the formulation of non-linear problems, solutions for which are found by iteratively solving
sequences of linear problems. This will be discussed in Section 9.1.2.
Analysts usually rely on data published in various handbooks. The published data can vary
widely, however. For example, it was reported in [19] that published data on the coefficient of
thermal conductivity of pure iron varies between 71.8 and 80.4 W/(mK).
2.6 Generalized formulation of problems of linear elasticity

Consider the equations of equilibrium (2.60):
𝜎ij,j + Fi = 0. (2.91)
Multiply (2.91) by a test function 𝑣i , and integrate on Ω:
𝜎ij,j 𝑣i dV + Fi 𝑣i dV = 0. (2.92)
∫Ω ∫Ω
14 Osborne Reynolds 1842–1912.

15 George Gabriel Stokes 1819–1903.
Observe that if the equilibrium equation (2.91) is satisfied then eq. (2.92) holds for arbitrary 𝑣i ,
subject to the condition that the indicated operations are defined. We write:
𝜎ij,j 𝑣i dV = (𝜎ij 𝑣i ),j dV − 𝜎ij 𝑣i,j dV

∫Ω ∫Ω ∫Ω
and use the divergence theorem to obtain
𝜎ij,j 𝑣i dV = 𝜎ij nj 𝑣i dS − 𝜎ij 𝑣i,j dV.

∫Ω ∫𝜕Ω ∫Ω
Noting that 𝜎ij nj = Ti (see eq. (K.3) in the Appendix), eq. (2.92) can be written as:
𝜎ij 𝑣i,j dV = Fi 𝑣i dV + Ti 𝑣i dS. (2.93)

∫Ω ∫Ω ∫𝜕Ω
Observe that 𝜎ij 𝑣i,j equals the sum 𝜎11 𝑣1,1 + 𝜎22 𝑣2,2 + 𝜎33 𝑣3,3 plus the sum of pairs like 𝜎12 𝑣1,2 +
𝜎21 𝑣2,1 . Since 𝜎ij = 𝜎ji , this can be written as:
1 1 (𝑣) (𝑣)
𝜎12 𝑣1,2 + 𝜎21 𝑣2,1 = 𝜎12 (𝑣1,2 + 𝑣2,1 ) + 𝜎21 (𝑣2,1 + 𝑣1,2 ) = 𝜎12 𝜖12 + 𝜎21 𝜖21
2 2
where the superscript (𝑣) indicates that these are the infinitesimal strain terms corresponding to
the test function 𝑣i , that is,
def 1
𝜖ij(𝑣) = (𝑣 + 𝑣j,i ).
2 i,j
Therefore:
𝜎ij 𝑣i,j = 𝜎ij 𝜖ij(𝑣)
and eq. (2.93) can be written as:
𝜎ij 𝜖ij(𝑣) dV = Fi 𝑣i dV + Ti 𝑣i dS. (2.94)

∫Ω ∫Ω ∫𝜕Ω
On substituting eq. (2.56) into eq. (2.94) we have:
Cijkl 𝜖ij(𝑣) 𝜖kl dV = Fi 𝑣i dV + Ti 𝑣i dS + Cijkl 𝜖ij(𝑣) 𝛼kl Δ dV. (2.95)

∫Ω ∫Ω ∫𝜕Ω ∫Ω
Let 𝜕Ωu denote the boundary region where ui = û i is prescribed; let 𝜕ΩT denote the boundary
region where Ti = T̂ i is prescribed and let 𝜕Ωs denote the boundary region where Ti = kij (dj − uj )
(i.e., spring boundary condition) is prescribed. Let us define
def
B(u, v) = Cijkl 𝜖ij(𝑣) 𝜖kl dV + kij uj 𝑣i dS (2.96)
∫Ω ∫𝜕Ωs
and
def
F(v) = Fi 𝑣i dV + T̂ i 𝑣i dS + kij dj 𝑣i dS + Cijkl 𝜖ij(𝑣) 𝛼kl Δ dV (2.97)
∫Ω ∫𝜕ΩT ∫𝜕Ωs ∫Ω
where u ≡ ui and v ≡ 𝑣i . In the interest of simplicity we will assume that the material constants E,
𝜈 and 𝛼 are piecewise smooth functions.
The space E(Ω), called the energy space, is defined by
def
E(Ω) = {u | B(u, u) < ∞} (2.98)
and the norm

√
def 1
∥u∥E = B(u, u) (2.99)
2
is associated with E(Ω). The space of admissible functions is defined by:
def
̃
E(Ω) = {ui | ui ∈ E(Ω), ui = û i on 𝜕Ωu }.
Note that this definition imposes a restriction on the prescribed displacement conditions: there has
to be an ui ∈ E(Ω) so that ui = û i on 𝜕Ωu .
The space of test functions is defined by:
def
E0 (Ω) = {ui | ui ∈ E(Ω), ui = 0 on 𝜕Ωu }.
̃
The generalized formulation is stated as follows: “Find u ∈ E(Ω) such that B(u, v) = F(v) for all
0
v ∈ E (Ω)”.
2.6.1 The principle of minimum potential energy

By definition, the potential energy is the functional:
1
def
𝜋(u) = B(u, u) − F(u) (2.100)
2
The principle of minimum potential energy states that the exact solution of the generalized for-
mulation based on the principle of virtual work is the minimizer of the potential energy functional
on the space of admissible functions:
𝜋(uEX ) = min 𝜋(u). (2.101)
̃
u∈E(Ω)
The proof given in Section 1.2.2 is directly applicable to the problem of elasticity.
The definition of the potential energy may be modified by the addition of an arbitrary constant.
Specifically, referring to eq. (2.96) and (2.97), we define the potential energy for the problem of
elasticity as follows:
def 1 1
Π(u) = C (𝜖 − 𝛼ij Δ )(𝜖kl − 𝛼kl Δ ) dV + k (u − di )(uj − dj ) dS
2 ∫Ω ijkl ij 2 ∫𝜕Ωs ij i
− Fi ui dV − Ti ui dS. (2.102)
∫Ω ∫𝜕ΩT
The advantage of this definition, over the definition given by eq. (2.100), is that in the special cases
when a free body is subjected to a temperature change (𝜖ij = 𝛼ij Δ ), or a body with a spring boundary
condition is given a rigid body displacement (ui = di ), then Π(u) = 0, whereas 𝜋(u) ≠ 0.
̃
In the finite element method E(Ω) is replaced by a finite-dimensional subspace S: ̃
Π(uFE ) = min Π(u). (2.103)

̃
u∈S(Ω)
When the sequence of finite element spaces is hierarchic (i.e., S̃ 1 ⊂ S̃ 2 ⊂ · · ·) then the potential
energy converges monotonically.
Exercise 2.38 Compare the definition of 𝜋(u) given by eq. (2.100), with the definition of Π(u)
given by eq. (2.102) and show that the two definitions differ by a constant, defined as follows:
1 1
Π(u) − 𝜋(u) = C 𝛼 𝛼  2 dV + k d d dS.
2 ∫Ω ijkl ij kl Δ 2 ∫𝜕Ωs ij i j
Isotropic elasticity
When the material is isotropic then we can substitute eq. (2.54) into eq. (2.93) to obtain:
( )
𝜆𝜖kk 𝜖ii(𝑣) + 2G𝜖ij 𝜖ij(𝑣) dV = Fi 𝑣i dV + T 𝑣 dS+
∫Ω ∫Ω ∫𝜕Ω i i
E
𝛼 𝜖 (𝑣) dV. (2.104)
∫Ω 1 − 2𝜈 Δ ii
Define the differential operator matrix [D] and the material stiffness matrix [E] as follows:
⎡𝜕 ⎤
⎢ 𝜕x 0 0⎥
⎢ 𝜕 ⎥ ⎡𝜆 + 2G
⎢0 0⎥ 𝜆 𝜆 0 0 0⎤
𝜕y ⎢ ⎥
⎢ 𝜕 ⎥ ⎢ 𝜆 𝜆 + 2G 𝜆 0 0 0⎥
⎢0 0 ⎥ def ⎢
def 𝜕z ⎥ 𝜆 𝜆 𝜆 + 2G 0 0 0⎥
[D] = ⎢ 𝜕 [E] = ⎢ ⋅
0 ⎥⎥
𝜕 (2.105)
⎢ 0⎥ ⎢ 0 0 0 G 0
⎢ 𝜕y 𝜕x ⎥ ⎢ 0 0 0 0 G 0⎥
⎢ 𝜕 𝜕 ⎥ ⎢ 0
⎢0 𝜕z 𝜕y ⎥ ⎣ 0 0 0 0 G⎥⎦
⎢𝜕 𝜕 ⎥
⎢ 0 ⎥
⎣ 𝜕z 𝜕x ⎦
def def
Furthermore, denote u ≡ {u} = {ux uy uz }T and v ≡ {𝑣} = {𝑣x 𝑣y 𝑣z }T . It is left to the reader to
show that eq. (2.104) can be written in the following form:
([D]{𝑣})T [E][D]{u} dV = {𝑣}T {F} dV + {𝑣}T {T} dS

∫Ω ∫Ω ∫𝜕Ω
{ } ⎧1⎫
𝜕𝑣x 𝜕𝑣y 𝜕𝑣z ⎪ ⎪ E𝛼Δ
+
∫Ω 𝜕x 𝜕y 𝜕z ⎨1⎬ 1 − 2𝜈 dV (2.106)
⎪1⎪
⎩ ⎭
def def
where {F} = {Fx Fy Fz }T is the body force vector and {T} = {Tx Ty Tz }T is the traction vector.
Remark 2.6 In the general anisotropic case, represented by eq. (2.95), we have

∫Ω ∫Ω ∫𝜕Ω
+ ([D]{𝑣})T [E]{𝛼}Δ dV (2.107)

∫Ω
where the material stiffness matrix [E] is a symmetric positive definite matrix with 21 independent
def
coefficients and {𝛼} = {𝛼11 𝛼22 𝛼33 2𝛼12 2𝛼23 2𝛼31 }T .
Exercise 2.39 Show that for isotropic elastic materials with 𝜈 ≠ 0 Π(u) can be written in the
following form:
{ ( )2 }
1 1+𝜈 E( )2
Π (u) = 𝜆 𝜖kk − 𝛼Δ + 2G𝜖ij 𝜖ij − 𝛼Δ dV+
2 ∫Ω 𝜈 𝜈
1 ( )( )
kij ui − di uj − dj dS − Fi ui dV − T u dS (2.108)
∫
2 𝜕Ωs ∫Ω ∫𝜕ΩT i i
and verify that when an unconstrained elastic body is subjected to a temperature change Δ (i.e.,
𝜖ij = 𝛼Δ 𝛿ij ) then Π(u) = 0.
2.6.2 The RMS measure of stress

The root-mean-square (RMS) measure of stress is closely related to the energy norm. Using the
notation
def
𝝐 ≡ {𝜖} = [D]{u}, 𝝈 ≡ {𝜎} = [E]{𝜖}, (2.109)
the RMS measure of stress is defined by
( )1∕2
def 1 T
S(𝝈) = {𝜎} {𝜎} dV (2.110)
V ∫Ω
where V is the volume. Therefore we can write
1 1
S2 (𝝈) = ([E]{𝜖})T [E]{𝜖} dV = {𝜖}T [E][E]{𝜖} dV (2.111)
V ∫Ω V ∫Ω
where we used [E]T = [E]. When the boundary conditions do not include spring constraints then
the strain energy is
1
∥u∥2E(Ω) = {𝜖}T [E]{𝜖} dV. (2.112)
2 ∫Ω
Using equations (2.111) and (2.112), we can write
2Λmin 2Λmax
∥u∥2E(Ω) ≤ S2 (𝝈) ≤ ∥ u ∥2E(Ω) (2.113)
V V
where Λmax (resp. Λmin ) is the maximum (resp. minimum) eigenvalue of the matrix [E]. From this
it follows that if uFE converges to uEX in energy norm then S(𝝈 FE ) converges to S(𝝈 EX ) at the same
rate. Equivalently, the error S(𝝈 FE − 𝝈 EX ) is bounded from above and below by the error in energy
norm:
√ √
2Λmin 2Λmax
∥uFE − uEX ∥E(Ω) ≤ S(𝝈 FE − 𝝈 EX ) ≤ ∥uFE − uEX ∥E(Ω) . (2.114)
V V
This result indicates that when Λmax is large then S(𝝈 FE − 𝝈 EX ) can be large even when the error
in energy norm is small. Consider the potential energy expression given by eq. (2.108) and assume
that Δ = 0. The first term of the integrand is 𝜆𝜖kk 2
where 𝜖kk is the infinitesimal volumetric strain.
As 𝜆 → ∞, the minimization of the potential energy results in 𝜖kk → 0. When computing 𝜎kk from
Hooke’s law we have
E
𝜎kk = (3𝜆 + 2G)𝜖kk = 𝜖 (2.115)
1 − 2𝜈 kk
that is, 𝜖kk is multiplied by a large number, which magnifies the error in 𝜖kk . The shear stresses
and the differences of normal stresses are independent of 𝜆 and therefore can be computed directly
from the strains. An indirect method for the computation of 𝜎x + 𝜎y for plane strain problems is
discussed in [98].
Remark 2.7 Using {𝜎} = [E]{𝜖} where [E] is given by eq. (2.105), it is not difficult to see that the
differences of normal stresses and shear stresses can be computed from the numerical simulation
directly. Unlike in the case of the sum of normal stresses, the coefficients do not go to infinity as 𝜈
approaches 1/2.
Exercise 2.40 Find the eigenvalues of matrix [E] for plane strain and plane stress. Partial answer:
For plane strain
E
Λmax = 2(𝜆 + G) ≡ , Λmin = G. (2.116)
(1 − 2𝜈)(1 + 𝜈)
Observe that Λmax → ∞ as 𝜈 → 1∕2.
2.6.3 The principle of virtual work

In many engineering textbooks eq. (2.93) is given a physical interpretation and treated as a fun-
damental principle of continuum mechanics, called the principle of virtual work. In this view, the
test function 𝑣i is understood to be some arbitrary displacement field, imposed by an agent that
is independent of the applied body force Fi and traction Ti . For this reason 𝑣i is called “virtual
displacement”.
The terms on the right-hand side of eq. (2.93) represent work by the body force and the traction
forces acting on the body, collectively called “external forces”, caused by the virtual displacement.
The left-hand side represents virtual work done by the internal stresses. To see this, refer to Fig. 2.11
and assume that vertex A of the infinitesimal hexahedral element, the coordinates of which are xi ,
is subjected to a virtual displacement 𝑣i . Then, since 𝑣i is continuous and differentiable, the face
located at x1 + dx1 will be displaced, relative to point A, by 𝑣i,1 dx1 and the virtual work done by 𝜎11 is:
dW𝜎11 = (𝜎11 dx2 dx3 ) (𝑣1,1 dx1 ) = 𝜎11 𝑣1,1 dV

⏟⏞⏞⏞⏞⏟⏞⏞⏞⏞⏟ ⏟⏞⏟⏞⏟
force displacement
similarly, the virtual work done by 𝜎13 is:
dW𝜎13 = (𝜎13 dx2 dx3 ) (𝑣3,1 dx1 ) = 𝜎13 𝑣3,1 dV

⏟⏞⏞⏞⏞⏟⏞⏞⏞⏞⏟ ⏟⏞⏟⏞⏟
force displacement
etc.
The principle of virtual work states that the virtual work of external forces is equal to the virtual
work of internal stresses. Note that since this result is based on the equilibrium equations (2.91), it
is independent of the material properties and therefore holds for any continuum.
Equation (2.93) is the generic form of the principle of virtual work. Particular statements of the
principle of virtual work depend on the material properties and the boundary conditions.
Exercise 2.41 Starting from eq. (2.16) derive the counterpart of eq. (2.94) for the stationary heat
conduction problem.
dx 2
𝜎31
𝜎13 𝜎13
dx3 𝜎11 𝜎11
v3,1dx1
𝜎31
x3 A dx 2
dx1 v1,1dx1 dx1
x2
x1 (a) (b)
Figure 2.11 Virtual displacements corresponding to (a) 𝜎11 and (b) 𝜎13 .
2.6.4 Uniqueness
The generalized formulation based on the principle of virtual work is unique in the energy space
E(Ω). The proof of uniqueness given by Theorem 1.1 is applicable to the elasticity problem in three
dimensions.
Uniqueness in the energy space does not necessarily mean uniqueness of the displacement field
u. When 𝜕Ωu and 𝜕Ωs are both empty then there are six linearly independent test functions in
E0 (Ω) = E(Ω) for which 𝜖ij(𝑣) = 0 and hence B(u, v) = 0. Three of these functions correspond to rigid
body displacements:
(𝑣)
𝜖11 =0∶ 𝑣(1)
i
=c1 {1 0 0}T
(𝑣)
𝜖22 =0∶ 𝑣(2)
i
=c2 {0 1 0}T
(𝑣)
𝜖33 =0∶ 𝑣(3)
i
=c3 {0 0 1}T
and three correspond to infinitesimal rigid body rotations:
(𝑣)
𝜖12 =0∶ 𝑣(4)
i
=c4 {−x2 x1 0}T rotation about x3
(𝑣)
𝜖23 =0∶ 𝑣(5)
i
=c5 {0 − x3 x2 } T
rotation about x1
(𝑣)
𝜖31 =0∶ 𝑣(6)
i
=c6 {x3 0 − x1 } T
rotation about x2
where c1 , c2 , … , c6 are arbitrary constants. Consequently the body force vector and the surface
tractions must satisfy the following six conditions:
F(v) = 0 ∶ Fi 𝑣(k) dV + Ti 𝑣(k) dS = 0 k = 1, 2, … , 6. (2.117)

∫Ω i ∫𝜕Ω i
The physical interpretation of these conditions is that the body must be in equilibrium, i.e., the
sum of forces and the sum of moments must be zero.
The solution is unique up to rigid body displacements and rotations. In order to ensure unique-
ness of the solution, “rigid body constraints” are imposed, that is, functions that represent rigid
body displacements and rotations are eliminated from the space of test functions.
E0 (Ω) = {𝑣i | 𝑣i ∈ E(Ω), 𝑣(k)
i
= 0, k = 1, 2, … , 6}. (2.118)
The values of the rigid body displacements are arbitrary, therefore the space of admissible functions
is:
̃
E(Ω) = {ui | ui ∈ E(Ω), u(k) = û (k) , k = 1, 2, … , 6}. (2.119)
i i
where û (k)
i
are arbitrary rigid body displacements, usually chosen to be zero.
Figure 2.12 Rigid body constraints. Notation. (C)

U3
X2
C
x3
X3 c
(B)
(A)
U2 (B) U2
U3
(A) B
U1
(A) X1
U3 b
a
A
x1 x2
Rigid body constraints are enforced by setting six displacement components in at least three
non-collinear points to arbitrary values. The usual procedure is as follows: Three non-collinear
points, labeled A, B, C in Fig. 2.12, are selected arbitrarily. A Cartesian coordinate system is asso-
ciated with these points such that points A and B lie on axis X1 , axis X3 is perpendicular to the
plane defined by the points ABC and axis X2 is perpendicular to axes X1 , X3 . The displacement
components U1(A) , U2(A) , U3(A) , U2(B) , U3(B) and U3(C) , shown in Fig. 2.12, are assigned arbitrary values,
typically zero. This will ensure that those modes of displacement that correspond to rigid body
displacements and rotations are removed from the trial space.
For example, denoting the displacement vector components of the origin of the coordinate system
X1 , X2 , X3 , shown in Fig. 2.12, by C1 , C2 , C3 and the infinitesimal rotations about the X1 , X2 and
X2 axes by C4 , C5 , C6 respectively, we can set the displacement vector components in points A, B
and C corresponding to rigid body displacements and rotations to zero:
⎧U1(A) ⎫ ⎡1 0 0 0 0 0 ⎤ ⎧C1 ⎫
⎪ (A) ⎪ ⎢ ⎥⎪ ⎪
⎪U2(A) ⎪ ⎢0 1 0 0 0 −a⎥ ⎪C2 ⎪
⎪U3 ⎪ ⎢0 0 1 0 a 0 ⎥ ⎪C3 ⎪
⎨ (B) ⎬ = ⎢0 1 0 0 0 b ⎥⎥ ⎨ ⎬=0 (2.120)
⎪ U2(B) ⎪ ⎢ ⎪C4 ⎪
⎪ U3 ⎪ ⎢0 0 1 0 −b 0 ⎥ ⎪C5 ⎪
⎪ (C) ⎪ ⎢0 0 1 c 0 0 ⎥⎦ ⎪ ⎪
⎩U ⎭ ⎣
3 ⎩C6 ⎭
where the determinant of the coefficient matrix is c(a + b)2 which is nonzero for any choice of
three non-collinear points. Here we set the displacements equal to zero, which is common practice;
however, arbitrary values could have been assigned. The important point is that the coefficient
matrix in eq. (2.120) must be nonsingular. An additional requirement is that the exact solution in
the points where rigid body constraints are prescribed must be continuous.
Example 2.7 The thin elastic plate-like body with a circular hole, shown in Fig. 2.13, is loaded
by constant normal tractions T0 . The equilibrium conditions (2.117) are obviously satisfied. In this
case there are six rigid body modes and the space of admissible functions is given by eq. (2.119) and
the space of test functions by eq. (2.118). Letting
u(A) (A) (A) (B) (B) (C)

1 = u2 = u3 = u2 = u3 = u3 = 0
the rigid body constraints are enforced.

x2
tz
C
T0 T0
w r0
x1
x3 A B

Figure 2.13 Example 2.7. Notation
x2 x2
tz
D C
plane of line of
C
symmetry symmetry
T0
w r0 r0
F
E b
x1
x3 B A B x1
/2
(a) (b)
In many cases 𝜕Ωu and/or 𝜕Ωs is not empty but the prescribed conditions do not provide suf-
ficient number of constraints to prevent all rigid body displacements and/or rotations. In those
cases it is necessary to provide a sufficient number of rigid body constraints to prevent rigid body
displacement, as illustrated in the following example.
Example 2.8 If the center of the circular hole in the elastic plate-like body of Example 2.7 is
located at x1 = 𝓁∕2 then the solution is symmetric with respect to the plane x1 = 𝓁∕2 and the prob-
lem may be formulated on the half domain shown in Fig. 2.14(a). On a plane of symmetry the
normal displacement component un and the shearing stress components are zero.
Setting un = u1 = 0 on the plane of symmetry prevents displacement in the x1 direction and rota-
tions about the x2 and x3 axes, but does not prevent displacements in the x2 and x3 coordinate
directions, nor rotation about the x1 axis. We may set
u(B)
2
= u(B)
3
= u(C)
3
=0
to prevent rigid body motion allowed by the plane of symmetry.
A three-dimensional problem can be reduced to a planar problem only when there is a plane of
symmetry. The domain of the planar problem lies in this plane of symmetry. Therefore there are at
most three rigid body displacements: two in-plane displacement components and a rotation about
an arbitrary point in the mid-plane.
2.7 Residual stresses 87
For instance, the problem in Example 2.8 can be formulated as a planar problem. In that case
the implied zero normal displacement of the mid-plane prevents displacement in the x3 direction
as well as rotation about the x1 and x2 axes. The line of symmetry shown in Fig. 2.14(b) prevents
displacement in the x1 direction and rotation about the x3 axis. Therefore only one rigid body con-
straint has to be imposed, for example by letting u(A)
2
= 0. Note that the coordinate axes x1 , x2 lie in
the plane x3 = 0.
Exercise 2.42 Refer to Fig. 2.13. Relocate point A to {0 0 0} and point C to {𝓁 𝑤 0}. Assume that
the following constraints are specified:
u(A) (A) (A) (B) (B) (C)

1 = u2 = u3 = u2 = u3 = u3 = 0.
Write down the system of equations analogous to eq. (2.120) and verify that the coefficient matrix
is of full rank.
Remark 2.8 In the foregoing discussion it was assumed that the material is isotropic. If the mate-
rial is not isotropic then symmetry, antisymmetry and periodicity cannot be used in general. An
exception is when the material is orthotropic and the axes of material symmetry are aligned with
the axes of geometric symmetry.
2.7 Residual stresses
Up to this point we tacitly assumed that an elastic body is stress free when the mechanical strain
tensor is zero, see eq. (2.56).
Metal forming, such as forging, rolling and drawing, invariably induce residual stresses in met-
als16 . These residual stresses, present in metal stock, are called bulk residual stresses. In addition,
metal cutting operations, such as turning, drilling and milling, remove thin layers of metal by
shear, leaving behind a layer of residual stresses that rapidly decay with respect to distance from
the surface. The strongly affected boundary layer is typically between 0.2 and 1 mm wide [50].
Shot peening is used for inducing compressive residual stresses to increase durability and to rem-
edy distortions caused by bulk and machining-induced residual stresses. Fastener holes in aircraft
structures are often cold-worked to produce a layer of compressive residual stresses for improved
durability.
In composite materials there is a large difference in the coefficients of thermal expansion of the
fiber and matrix. Residual stresses develop when a part cools after curing.
We denote the reference configuration of an elastic body by Ω0 , its boundary points by 𝜕Ω0 , and
the residual stress field in the reference configuration by 𝜎ij0 . The residual stress field satisfies the
equations of equilibrium and the stress-free boundary conditions:
(0)
𝜎ij,j = 0 in Ω0 and 𝜎ij(0) nj = 0 on 𝜕Ω0 (2.121)
where nj is the unit normal to the boundary. In the presence of residual stresses the generalized
Hooke’s law given by eq. (2.56) is modified to
𝜎ij = 𝜎ij(0) + Cijkl (𝜖kl − 𝛼kl Δ ). (2.122)
16 For example, certain types of aluminum plates are hot-rolled, quenched, over-aged and stretched by the
imposition of 1.5–3.0% strain in the rolling direction.
Γ1 Γ1
Γ2 Γ2 Γ2
Ω0 Ω1 Ω2
The distribution of residual stresses is inferred from measurements of strains in non-destructive

tests and measurements of displacements and strains in destructive and semi-destructive experi-
ments. In destructive and semi-destructive experiments material is removed and the distribution
and magnitude of residual stresses are inferred from displacements and strains caused by the redis-
tribution of residual stresses. Therefore it is necessary to use numerical simulation for the interpre-
tation of physical observations. See, for example, [63, 64].
Let us introduce a cut Γ1 that will produce domain Ω1 and a second cut Γ2 that will produce
domain Ω2 , see Fig. 2.15. Denote the displacement field following the first (resp. second) cut by u(1)
i
(resp. u(2)
i
).
Assuming that the principle of superposition is applicable and the cuts do not cause residual
stresses, we now show that u(2)i
depends on 𝜎ij(0) and Ω2 but not on u(1)
i
or Ω1 Observe that, since the
cuts create free surfaces, u(1)
i
satisfies
B(u(1) , 𝑣i ) = − 𝜎ij(0) nj 𝑣i dS for all 𝑣i ∈ E(Ω1 ). (2.123)

i ∫Γ1
We need to show that
B(u(2) , 𝑣i ) = − 𝜎ij(0) nj 𝑣i dS for all 𝑣i ∈ E(Ω2 ) (2.124)

i ∫Γ2
independently of whether the body was the first cut at Γ2 or first cut at Γ1 and then at Γ2 .
Denote the stress field corresponding to u(1)
i
on Ω1 by sij and let 𝑤i be the displacement field on
Ω2 corresponding to the traction sij acting on Γ2 . Therefore we have
B(𝑤i , 𝑣i ) = sij nj 𝑣i dS for all 𝑣i ∈ E(Ω2 ). (2.125)

∫Γ2
The displacement field corresponding to the second cut is caused by the creation of a free surface
on Γ2 :
( )
B(u(2) − 𝑤i , 𝑣i ) = − 𝜎ij(0) + sij nj 𝑣i dS for all 𝑣i ∈ E(Ω2 ). (2.126)
i ∫Γ2
On adding equations (2.125) and (2.126) we find that u(2) i

satisfies eq. (2.124) which proves the
statement that u(2)
i
depends on 𝜎 (0)
ij
and Ω2 but not on u (1)
i
or Ω1 .
Exercise 2.43 An annular aluminum plate with inner radius rs and outer radius rs + b, constant
thickness da , was joined by shrink fitting to a stainless steel shaft. The configuration is shown in
Fig. 2.16. Denote the mechanical properties of aluminum as follows: modulus of elasticity: Ea , mod-
ulus of rigidity: Ga , mass density: 𝜚a , coefficient of thermal expansion: 𝛼a and the corresponding
mechanical properties of stainless steel as Es , Gs , 𝜚s , 𝛼s . Let 𝓁s = 80 mm, rs = 17.5 mm, da = 15 mm,
b = 150 mm, Ea = 72.0 × 103 MPa, Ga = 28.0 × 103 MPa, 𝜚a = 2800 kg∕m3 , 𝛼a = 23.6 × 10−6 ∕K,
Es = 190 × 103 MPa, Gs = 75.0 × 103 MPa, 𝜚s = 7920 kg∕m3 , 𝛼s = 17.3 × 10−6 ∕K.
Figure 2.16 Notation for Exercise 2.7.

z
da
s
r
rs b
Consider the following conditions: (a) The shaft and the aluminum plate were heated to 220 ∘ C,
the shaft was inserted and then the assembly was cooled to 20 ∘ C. Assume that at 220 ∘ C the clear-
ance between the shaft and the plate was zero and 𝛼a > 𝛼s . (b) The assembly is spinning about the
z axis at an angular velocity 𝜔.
Estimate the value of 𝜔 at which the membrane force in the aluminum plate, Fr , is approximately
zero at r = rs . By definition:
da ∕2
Fr = 𝜎r dz.
∫−da ∕2
Specify 𝜔 in units of cycles per second (Hertz). Note that, in order to have consistent units, kg∕m3
must be converted to Ns2 ∕mm4 .
2.8 Chapter summary
The formulation of mathematical models for linear problems in heat conduction and elasticity was
described in strong and generalized forms.
Mathematical models for heat conduction are based on the conservation law and an empirical
relationship between the derivatives of the temperature u and the flux vector qi . We assumed that
this relationship is linear. In reality, however, the coefficients of thermal conductivity depend on
the value of u as well as the gradient of u. Only within a narrow range of u, and the gradient of u,
should these coefficients be approximated by constants.
Mathematical models for elastic bodies are based on the conservation of momentum (in statics
the equations of equilibrium) and an empirical linear relationship between the stress and strain
tensors. This linear relationship holds for small displacements and strains only. It is important to
bear in mind the limitations of mathematical models imposed by the assumptions incorporated in
the models.
A mathematical model must never be confused with the physical reality that it was conceived to
imitate. This important point will be addressed in greater detail in Chapter 5.
91
Implementation
This chapter is concerned with the algorithmic aspects of the finite element method. The finite
element spaces, standard elements, the corresponding shape functions and mapping functions are
described for two- and three-dimensional formulations. At the end of the solution process the coef-
ficients of the shape functions, the mapping and the material properties are available at the element
level. The quantities of interest are computed from this information either by direct or indirect
methods.
3.1 Standard elements in two dimensions
Two-dimensional finite element meshes are comprised of triangular and quadrilateral elements.
(q)
The standard quadrilateral element will be denoted by Ωst and the standard triangular element by
(t)
Ωst . The definition of standard elements is arbitrary. However, certain conveniences in mapping
and assembly are achieved when the standard elements are defined as shown in Fig. 3.1. Note that
the sides of the elements have the same length (2) as in one-dimension.
3.2 Standard polynomial spaces
The standard polynomial spaces in two and three dimensions are generalizations of the standard
polynomial space  p (Ist ) defined in Section 1.3.1. Whereas the polynomial degree could be charac-
terized by a single number (p) in one dimension, in higher dimensions more than one interpretation
is possible.
3.2.1 Trunk spaces

Trunk spaces are polynomial spaces spanned by the set of monomials 𝜉 i 𝜂 j , i, j = 0, 1, 2, … , p subject
to the restriction i + j = 0, 1, 2, … , p. In the case of quadrilateral elements these are supplemented
by one or two monomials of degree p + 1:
1. Triangles: The dimension of the space is n(p) = (p + 1)(p + 2)∕2. For example, the space  6 (Ω(t)
st )
is spanned by the 28 monomial terms indicated in Fig. 3.2. The space  p (Ω(t) st ) comprises all
polynomials of degree less than or equal to p.
92 3 Implementation
η side 3 η
4 3
3 side 2
side 4 side 2
1.0
side 3
ξ 3
1.0
side 1 side 1
1 1.0 1.0 2 1 1.0 1.0 2 ξ
(a) (b)
Figure 3.1 Standard quadrilateral and triangular elements Ω(q) (t)

st and Ωst .
(q)
S1(Ωst)
1
ξ η
(q)
(t) S4(Ωst)
S6(Ωst) ξ2 ξη η2
ξ3 ξ 2η ξη 2 η3
ξ4 ξ 3η ξ 2η 2 ξη 3 η4
ξ5 ξ 4η ξ 3η 2 ξ 2η 3 ξη 4 η5
ξ6 ξ 5η ξ 4η 2 ξ 3η 3 ξ 2η 4 ξη 5 η6
ξ7 ξ 6η ξ 5η 2 ξ 4η 3 ξ 3η 4 ξ 2η 5 ξη 6 η7
Figure 3.2 Trunk space. Illustration of spanning sets for  1 (Ω(q) 4 (q)
st ),  (Ωst ) and  (Ωst ).
6 (t)
2. Quadrilaterals: Monomials of degree less than or equal to p are supplemented by 𝜉𝜂 for p = 1

(q)
and by 𝜉 p 𝜂, 𝜉𝜂 p for p ≥ 2. For example, the space  4 (Ωst ) is spanned by the 17 monomial terms
(q)
indicated in Fig. 3.2. The dimension of space  p (Ωst ) is
{
4p for p ≤ 3
n(p) = (3.1)
4p + (p − 2)(p − 3)∕2 for p ≥ 4.
3.2.2 Product spaces

In two dimensions product spaces are spanned by the monomials 1, 𝜉, 𝜉 2 , … , 𝜉 p , 1, 𝜂, 𝜂 2 , … , 𝜂 q
and their products. Thus the dimension of product spaces is n(p, q) = (p + 1)(q + 1). Product spaces
(q)
on triangles will be denoted by  p,q (Ω(t)
st ) and on quadrilaterals by  (Ωst ). The spanning set of
p,q
(q)
monomials for the space  4,2 (Ωst ) is illustrated in Fig. 3.3.
p q
1
(q)
S4,2(Ωst)
ξ η
ξ2 ξη η2
ξ3 ξ 2η ξη 2 η3
ξ4 ξ 3η ξ 2η 2 ξη 3 η4
ξ5 ξ 4η ξ 3η 2 ξ 2η 3 ξη 4 η5
ξ6 ξ 5η ξ 4η 2 ξ 3η 3 ξ 2η 4 ξη 5 η6
ξ7 ξ 6η ξ 5η 2 ξ 4η 3 ξ 3η 4 ξ 2η 5 ξη 6 η7
Figure 3.3 Product space. Illustration of spanning set for the space  4,2 (Ω(q)
st ).
3.3 Shape functions
As in the one-dimensional case, we will discuss two types of shape functions: Shape functions
based on Lagrange polynomials and hierarchic shape functions based on the integrals of Legendre
polynomials. We will use the notation Ni (𝜉, 𝜂) (i = 1, 2, … , n) for both. The shape functions in two
dimensions along the sides of each element are the same as the shape functions defined on the
one-dimensional standard element Ist .
3.3.1 Lagrange shape functions

(q)
Elements with shape functions that span  p (Ω(t)st ) and  (Ωst ) with p = 1 and p = 2 are widely
p
used in engineering practice. Each one of the Lagrange shape functions is unity in one of the node
points and zero in the other node points. Therefore the approximating function u can be written as
∑
n
u(𝜉, 𝜂) = ui Ni (𝜉, 𝜂)
i=1
where ui is the value of u in the ith node point.
Quadrilateral elements
(q)
The shape functions of four-node quadrilateral elements span the space  1 (Ωst ):
1
N1 (𝜉, 𝜂) = (1 − 𝜉)(1 − 𝜂) (3.2)
4
1
N2 (𝜉, 𝜂) = (1 + 𝜉)(1 − 𝜂) (3.3)
4
1
N3 (𝜉, 𝜂) = (1 + 𝜉)(1 + 𝜂) (3.4)
4
1
N4 (𝜉, 𝜂) = (1 − 𝜉)(1 + 𝜂) (3.5)
4
94 3 Implementation
(q)
The shape functions of eight-node quadrilateral elements span the space  2 (Ωst ). The shape func-
tions corresponding to the vertex nodes are:
1
N1 (𝜉, 𝜂) = (1 − 𝜉)(1 − 𝜂)(−𝜉 − 𝜂 − 1) (3.6)
4
1
N2 (𝜉, 𝜂) = (1 + 𝜉)(1 − 𝜂)(𝜉 − 𝜂 − 1) (3.7)
4
1
N3 (𝜉, 𝜂) = (1 + 𝜉)(1 + 𝜂)(𝜉 + 𝜂 − 1) (3.8)
4
1
N4 (𝜉, 𝜂) = (1 − 𝜉)(1 + 𝜂)(−𝜉 + 𝜂 − 1) (3.9)
4
and the shape functions corresponding to the mid-side nodes are:
1
N5 (𝜉, 𝜂) = (1 − 𝜉 2 )(1 − 𝜂) (3.10)
2
1
N6 (𝜉, 𝜂) = (1 + 𝜉)(1 − 𝜂 2 ) (3.11)
2
1
N7 (𝜉, 𝜂) = (1 − 𝜉 2 )(1 + 𝜂) (3.12)
2
1
N8 (𝜉, 𝜂) = (1 − 𝜉)(1 − 𝜂 2 ). (3.13)
2
Observe that if we denote the coordinates of the vertices and the midpoints of the sides by (𝜉i , 𝜂i ),
i = 1, 2, … , 8, then Ni (𝜉j , 𝜂j ) = 𝛿ij .
(q)
The shape functions of the nine-node quadrilateral element span  2,2 (Ωst ) (i.e. the product
space). In addition to the four nodes located in the vertices and the four nodes located in the
mid-points of the sides, there is a node in the center of the element. Construction of these shape
functions is left to the reader in the following exercise.
Exercise 3.1 Write down the shape functions for the nine-node quadrilateral element. Sketch the
shape function associated with the node in the center of the element and one of the vertex shape
functions and one of the side shape functions.
Triangular elements
The shape functions for triangular elements are usually written in terms of the triangular coordi-
nates, defined as follows:
( )
1 𝜂
L1 = 1−𝜉− √ (3.14)
2 3
( )
1 𝜂
L2 = 1+𝜉− √ (3.15)
2 3
𝜂
L3 = √ ⋅ (3.16)
3
Note that Li is unity at node i and zero on the side opposite to node i. Also, L1 + L2 + L3 = 1. The
space  1 (Ω(t)
st ) is spanned by the following shape functions:
Ni = Li i = 1, 2, 3. (3.17)
These elements are called “three-node triangles”. For the six-node triangles the shape functions
are:
N1 = L1 (2L1 − 1) (3.18)
N2 = L2 (2L2 − 1) (3.19)
N3 = L3 (2L3 − 1) (3.20)
N4 = 4L1 L2 (3.21)
N5 = 4L2 L3 (3.22)
N6 = 4L3 L1 (3.23)
which span  2 (Ω(t)
st ).
3.3.2 Hierarchic shape functions

Hierarchic shape functions based on the integrals of Legendre polynomials are described for the
nodes, sides and vertices of quadrilateral and triangular elements. The shape functions associated
with the nodes and the sides are the same for the product and trunk spaces. Only the number of
internal shape functions is different.
Quadrilateral elements
The nodal shape functions are the same as those for the four-node quadrilateral, given by
equations (3.2) to (3.5).
The side shape functions are constructed by multiplying the shape functions N3 , N4 , …, defined
for the one-dimensional element (see Fig. 1.4), by linear blending functions. We define:
√ s
def 2k − 1
𝜙k (s) = P (t) dt, k = 2, 3, … (3.24)
2 ∫−1 k−1
Note that the index k represents the polynomial degree. The shape functions of degree p ≥ 2 are
defined for the four sides as follows:
1
side 1: Nk(1) (𝜉, 𝜂) = (1 − 𝜂)𝜙k (𝜉) (3.25)
2
1
side 2: Nk(2) (𝜉, 𝜂) = (1 + 𝜉)𝜙k (𝜂) (3.26)
2
1
side 3: Nk(3) (𝜉, 𝜂) = (1 + 𝜂)𝜙k (−𝜉) (3.27)
2
1
side 4: Nk(4) (𝜉, 𝜂) = (1 − 𝜉)𝜙k (−𝜂) (3.28)
2
where k = 2, 3, … , p. Thus there are 4(p − 1) side shape functions. The argument of 𝜙k is negative
for sides 3 and 4 because the positive orientation of the sides is counterclockwise. This will affect
shape functions of odd degrees only.
The internal shape functions are zero on the sides. For the trunk space there are (p − 2)(p − 3)∕2
internal shape functions (p ≥ 4) constructed from the products of 𝜙k :
Np(k,l) (𝜉, 𝜂) = 𝜙k (𝜉)𝜙l (𝜂) k, l = 2, 3, … , p, k + l = 4, 5, … , p. (3.29)
96 3 Implementation
Vertex/side number
1 2 3 4
Vertex modes
1 1 2 3 4
2 Side modes
5 6 7 8
3
Polynomial degree
9 10 11 12
Internal modes
4
13 14 15 16 17
5
18 19 20 21 22 23
6
24 25 26 27 28 29 30
7 31 32 33 34 35 35 37 38
8 39 40 41 42 43 44 45 46 47
Figure 3.4 Hierarchic shape functions for quadrilateral elements. Trunk space, p = 1 to p = 8.
The shape functions are assigned unique sequential numbers, as shown in Fig. 3.4. For the product
space there are (p − 1)(q − 1) internal shape functions, defined for p, q ≥ 2 by
(k,l)
Npq (𝜉, 𝜂) = 𝜙k (𝜉)𝜙l (𝜂), k = 2, 3, … , p, l = 2, 3, … , q, k + l ≤ p + q. (3.30)
Triangular elements
The nodal shape functions are the same as those for the three-node triangles, given by eq. (3.17).
The side shape functions are constructed as follows: Define:
𝜙 (s)
𝜙̃ k (s) = 4 k 2 k = 2, 3, … (3.31)
1−s
where 𝜙k (s) is the function defined by eq. (3.24). For example,
√
√ √ 7 2
̃ ̃ ̃
𝜙2 (s) = − 6, 𝜙3 (s) = − 10s, 𝜙4 (s) = − (5s − 1), etc.
8
Using
⎧L − L for side 1
⎪ 2 1
s = ⎨L3 − L2 for side 2 (3.32)
⎪L1 − L3 for side 3
⎩
the definition of the side shape functions is
side 1: Nk(1) (L1 , L2 , L3 ) = L1 L2 𝜙̃k (L2 − L1 ) (3.33)
where k = 2, 3, … , p. Thus there are 3(p − 1) side shape functions.
For the trunk space there are (p − 1)(p − 2)∕2 internal shape functions (p ≥ 3) defined as follows:
Np(k,l) = L1 L2 L3 Pk (L2 − L1 )Pl (2L3 − 1) k, l = 0, 1, 2, … , p − 3 (3.36)
where k + l ≤ p − 3 and Pk is the kth Legendre polynomial.
Exercise 3.2 Sketch the shape function N3(2) (L1 , L2 , L3 ).
3.4 Mapping functions in two dimensions
The commonly used mapping procedures for the transformation of the standard element to the
elements of the mesh are outlined in this section.
3.4.1 Isoparametric mapping

The term “isoparametric mapping” is meant to convey the idea that the same shape functions are
used for providing topological descriptions for elements as for the element-level approximations.
If the mapping is of lower (resp. higher) polynomial degree than the approximating functions then
it is said to be subparametric (resp. superparametric).
Isoparametric mapping is based on the Lagrange shape functions described in Section 3.3.1. The
most commonly used isoparametric mapping procedures are the linear and quadratic mappings.
Isoparametric mapping for quadrilateral elements

Linear mapping of quadrilateral elements from the standard quadrilateral element shown in
Fig. 3.1(a) to the kth element is defined by
∑
4
x = Q(k)
x (𝜉, 𝜂) = Ni (𝜉, 𝜂)Xi (3.37)
i=1
∑4
y = Q(k)
y (𝜉, 𝜂) = Ni (𝜉, 𝜂)Yi (3.38)
i=1
where (Xi , Yi ) are the coordinates of vertex i of the kth element numbered in counterclockwise order
and Ni are the shape functions defined by equations (3.2) through (3.5).
Quadratic mapping of quadrilateral elements from the standard quadrilateral element is defined
by:
∑
8
x = Q(k)
x (𝜉, 𝜂) = Ni (𝜉, 𝜂)Xi (3.39)
i=1
∑8
y = Q(k)
y (𝜉, 𝜂) = Ni (𝜉, 𝜂)Yi (3.40)
i=1
where (Xi , Yi ) are the coordinates of the four vertices numbered in counterclockwise order and the
mid-points of the four sides also numbered in counterclockwise order. The side between nodes 1
and 2 is the first side and Ni are the shape functions defined by equations (3.6) through (3.13).
Quadratic isoparametric mapping of a quadrilateral element and typical numbering of the node
points are illustrated in Fig. 3.5(a).
98 3 Implementation
η 7 η
4 3 3
6
6
y 8
ξ 5
2 1
1 5 4
2
x (a) (b) ξ
Figure 3.5 Isoparametric quadrilateral and triangular elements.
Isoparametric mapping for triangular elements

Linear mapping of triangles from the standard triangular element shown in Fig. 3.1(b) to the kth
element is defined by
∑
3
x = Q(k)
x (L1 , L2 , L3 ) = Li Xi (3.41)
i=1
∑3
y = Q(k)
y (L1 , L2 , L3 ) = Li Yi . (3.42)
i=1
Quadratic isoparametric mapping of triangular elements from the standard quadrilateral element
is given by
∑
6
x = Q(k)
x (L1 , L2 , L3 ) = Ni (L1 , L2 , L3 )Xi (3.43)
i=1
∑6
y = Q(k)
y (L1 , L2 , L3 ) = Ni (L1 , L2 , L3 )Yi (3.44)
i=1
where Ni are the shape functions defined by equations (3.18) through (3.23). The mapping of a
triangular element and typical numbering of the node points are illustrated in Fig. 3.5(b).
Remark 3.1 The mapped shape functions are polynomials only in the special cases when the
standard triangle is mapped into a straight-side triangle and when the standard quadrilateral ele-
ment is mapped into a parallelogram. In general the mapped shape functions are not polynomials.
The mapped shape functions are called “pull-back polynomials”. The accuracy of finite element
approximation is governed by the properties of the pull-back polynomials. Depending on exact
solution, approximation by the pull-back polynomials can be better or worse than approximation
by polynomials. In some cases the mapping is designed to improve the approximation.
Exercise 3.3 Show that quadratic parametric mapping applied to straight-side triangular and
quadrilateral elements is identical to linear mapping.
Exercise 3.4 Consider the element shown in Fig. 3.6(b). Note that nodes 4 and 6 are located in
the “quarter point position”, i.e. they are located 1∕4 of the length of sides 1 and 3, respectively, from
vertex 1. Introduce the polar coordinates r, 𝜃 and 𝜌, 𝜙 defined as shown in Fig. 3.6. The triangular
y (x3, y3)
ηt 3
x3/4 3
6 5 6
3 r
θ
ρ 1 5 x
φ
4
1 4 2 ξt
1 1 2
x2/4
(x2, y2)
(a) (b)
Figure 3.6 Notation for (a) the standard triangular element and (b) quarter-point mapping.
coordinates can be written in terms of 𝜌, 𝜙 as follows:

( )
1 1
L1 = 2 − 𝜌 cos 𝜙 − √ 𝜌 sin 𝜙
2 3
( )
1 1
L2 = 𝜌 cos 𝜙 − √ 𝜌 sin 𝜙
2 3
𝜌 sin 𝜙
L3 = √ ⋅
3
√
Show that 𝜌 is proportional to r for any fixed 𝜙 when quadratic parametric mapping is used.
The significance of this is that quarter point mapping modifies the element-level basis functions in
such √
a way that in the neighborhood of vertex 1 the element level basis functions will contain the
term r. Quarter point mapping is frequently used in the solution of fracture mechanics problems.
Remark 3.2 The quarter point element is a special kind of singular element. Singular finite ele-
ments are formulated with the objective to reduce the errors of approximation caused by singular
points. This is done by enlarging the finite element space through the addition of singular functions.
3.4.2 Mapping by the blending function method

To illustrate the method, let us consider a simple case where only one side (side 2) of a quadrilateral
element is curved, as shown in Fig. 3.7. The curve x = x2 (𝜂), y = y2 (𝜂) is given in parametric form
with −1 ≤ 𝜂 ≤ 1. We can now write:
1 1 1
x = (1 − 𝜉)(1 − 𝜂)X1 + (1 + 𝜉)(1 − 𝜂)X2 + (1 + 𝜉)(1 + 𝜂)X3
4 4( 4 )
1 1−𝜂 1+𝜂 1+𝜉
+ (1 − 𝜉)(1 + 𝜂)X4 + x2 (𝜂) − X2 − X3 ⋅ (3.45)
4 2 2 2
Observe that the first four terms in this expression are the linear mapping terms given by
eq. (3.37). The fifth term is the product of two functions: One function, the bracketed expression,
represents the difference between x2 (𝜂) and the x-coordinates of the chord that connects points
(X2 , Y2 ) and (X3 , Y3 ). The other function is the linear blending function (1 + 𝜉)∕2 which is unity
along side 2 and zero along side 4. Therefore we can write:
1 1 1+𝜉
x= (1 − 𝜉)(1 − 𝜂)X1 + (1 − 𝜉)(1 + 𝜂)X4 + x2 (𝜂) ⋅ (3.46)
4 4 2
100 3 Implementation
x = x2(η)
y = y2(η)
3
1–η 1+η
x= X2 + X3
2 2
1–η 1+η
y= Y2 + Y3
y 2 2
4
2
1
x
Figure 3.7 Quadrilateral element with one curved side.
Similarly:
1 1 1+𝜉
y= (1 − 𝜉)(1 − 𝜂)Y1 + (1 − 𝜉)(1 + 𝜂)Y4 + y2 (𝜂) ⋅ (3.47)
4 4 2
In the general case all sides may be curved. We write the curved sides in parametric form:
{
𝜉 on sides 1 and 3
x = xi (s), y = yi (s), −1 ≤ s ≤ +1 where s =
𝜂 on sides 2 and 4
and the subscripts represent the side numbers of the standard element. In this case the mapping
functions are:
1 1 1 1
x = (1 − 𝜂)x1 (𝜉) + (1 + 𝜉)x2 (𝜂) + (1 + 𝜂)x3 (𝜉) + (1 − 𝜉)x4 (𝜂)
2 2 2 2
1 1 1
− (1 − 𝜉)(1 − 𝜂)X1 − (1 + 𝜉)(1 − 𝜂)X2 − (1 + 𝜉)(1 + 𝜂)X3
4 4 4
1
− (1 − 𝜉)(1 + 𝜂)X4 , (3.48)
4
1 1 1 1
y= (1 − 𝜂)y1 (𝜉) + (1 + 𝜉)y2 (𝜂) + (1 + 𝜂)y3 (𝜉) + (1 − 𝜉)y4 (𝜂)
2 2 2 2
1 1 1
− (1 − 𝜉)(1 − 𝜂)Y1 − (1 + 𝜉)(1 − 𝜂)Y2 − (1 + 𝜉)(1 + 𝜂)Y3
4 4 4
1
− (1 − 𝜉)(1 + 𝜂)Y4 . (3.49)
4
The inverse mapping, that is: 𝜉 = Q(k)
𝜉
(x, y), 𝜂 = Q(k)
𝜂 (x, y), cannot be given explicitly in general,
but (𝜉, 𝜂) can be computed very efficiently for any given (x, y) by means of the Newton-Raphson
method or some other iterative procedure.
Exercise 3.5 Refer to Fig. 3.8(a). Show that the mapping of the quadrilateral element by the
blending function method is:
1−𝜉 1+𝜉
x =ri cos(𝜃m + 𝜂𝜃d ) + ro cos(𝜃m + 𝜂𝜃d )
2 2
1−𝜉 1+𝜉
y =ri sin(𝜃m + 𝜂𝜃d ) + ro sin(𝜃m + 𝜂𝜃d )
2 2
where
𝜃1 + 𝜃2 𝜃2 − 𝜃1
𝜃m = , 𝜃d = ⋅
2 2
y y
3
θd
4
3 ro
1
θ2 2 θ1
e x
ri ri
4 θm
2 ro
1 θ1
x
(a) (b)
Figure 3.8 Quadrilateral elements bounded by circular segments.
Exercise 3.6 Refer to Fig. 3.8(b). A quadrilateral element is bounded by two circles. The centers
of the circles are offset as shown. Write down the mapping by the blending function method in
terms of the given parameters. Hint: Using the law of cosines, the radius of the arc between node 2
and node 3 is:
√
r2−3 = −e sin 𝜃 + ro2 − e2 cos2 𝜃
where 𝜃 is the angle measured from the x-axis.
3.4.3 Mapping algorithms for high order elements

High order elements are usually mapped by the blending function method with the bounding
curves approximated by polynomial functions, similar to isoparametric mapping. The reasons for
this are that (a) the boundary curves are generally not available in analytical form and (b) stan-
dard treatment of all bounding curves is preferable from the point of view of implementation. For
example, the boundaries of a domain are typically represented by a collection of splines in com-
puter aided design (CAD) software products. In the interest of generality of implementation, the
bounding curves are interpolated using the Lagrangian basis functions defined by eq. (1.50).
The quality of the approximation depends on the choice of the interpolation points. Specifically,
the interpolation points must be defined so that, for the given polynomial degree of interpolation,
the interpolation function is close to the best possible approximation of the boundary curve by
polynomials in maximum norm. This is realized when the points are chosen so as to minimize
the Lebesque constant. Details are available in Appendix F. In one dimension the abscissas of the
Lobatto points1 are close to the optimal interpolation points. For details we refer to [31].
Exercise 3.7 Refer to Appendix F. Approximate the semicircle of unit radius by polynomials of
degree 5 using (a) the nodal set T15 , (b) the nodal set T25 and (c) six uniformly distributed interpola-
tion points. Compute the maximum relative error in the radius for each case.
Hint: The coordinates of the points on the perimeter of the semi-circle can be written as:
xc = cos(𝜋(1 + 𝜉)∕2), yc = sin(𝜋(1 + 𝜉)∕2), −1 ≤ 𝜉 ≤ 1
1 See Appendix E.
Solution of this problem involves writing a short computer program. The maximum relative error
in the radius in case (a) is 0.058%, in case (b) it is 0.061% and in case (c) it is 1.50%.
Rigid body rotations

In two-dimensional elasticity infinitesimal rigid body rotation is represented by the displacement
vector u = C{−y x}T . Having introduced the mappings x = Q(k) (k)
x (𝜉, 𝜂), y = Qy (𝜉, 𝜂), infinitesimal
rigid body rotations are represented exactly by element k only when Q(k) x (𝜉, 𝜂) ∈  (Ωst ) and
pk
(k)
Qy (𝜉, 𝜂) ∈  k (Ωst ). Iso- and subparametric mappings satisfy this condition, hence rigid body
p
rotation imposed on an element will not induce strains. Superparametric mappings and mapping
by the blending function method when the sides are not polynomials, or are polynomials of degree
higher than pk , do not satisfy this condition. It has been argued that for this reason only iso- and
subparametric mappings should be employed.
This argument is flawed, however. One should view this question in the following light: Errors
are introduced when rigid body rotations are not represented exactly and also by the approxima-
tion of boundary curves. With the blending function method analytic curves, such as circles, are
represented exactly, but the rigid body rotation terms are approximated. With iso- and subparamet-
ric mappings the boundaries are approximated but the rigid body rotation terms are represented
exactly. In either case the errors of approximation will go to zero as the number of degrees of free-
dom is increased whether by mesh refinement or by increasing the polynomial degrees. This is
illustrated by the following exercises.
Exercise 3.8 Refer to Fig. 3.8(a). Let 𝜃1 = 0, 𝜃2 = 60∘ , ri = 1.0 and ro = 2.0. Use E = 200 GPa,
𝜈 = 0.3, plane strain. Impose nodal displacements consistent with rigid body rotation about the
origin: u = C{y − x}. For example let u(1) (1) (2)
x = 0, uy = Cri , uy = Cro where the superscripts indicate
the node numbers and C is the angle of rotation (in radians) about the positive z axis. Let C = 0.1
and compute the maximum equivalent strain2 for p = 1, 2, … , 6. Very rapid convergence to zero
will be seen3 .
Exercise 3.9 Repeat Exercise 3.8 using uniform mesh refinement and p fixed at p = 1 and p = 2.
Plot the maximum equivalent strain vs. the number of degrees of freedom.
Remark 3.3 The constant function is in the finite element space  pk (Ωst ) independent of the
mapping. Therefore rigid body displacements are represented exactly.
3.5 Finite element spaces in two dimensions
Finite element spaces are sets of continuous functions characterized by the finite element mesh Δ,
a polynomial space defined on standard elements and the functions used for mapping the standard
elements into the elements of the mesh. We have seen an example of this in Section 1.3.2 where
finite element spaces in one dimension were described. There the standard element was the interval
2 The equivalent strain is proportional to the root-mean-square of the differences of principal strains and therefore
it is an indicator of the magnitude of shearing strain.
3 Curves are approximated by polynomials of degree 5 in StressCheck. This is a default value that can be changed
by setting a parameter. When the default value is used then the mapping is superparametric for p ≤ 4, isoparametric
at p = 5 and subparametric for p ≥ 6.
I = (−1, 1), the standard space, denoted by  p , was a polynomial space of degree p, and the map-
ping function was the linear function given by eq. (1.63). Finite element spaces in two dimensions
are defined analogously by
def
S = S(Ω, Δ, p, Q) =
{u | u ∈ E(Ω), u(Q(k) (k)
x (𝜉, 𝜂), Qy (𝜉, 𝜂)) ∈  (Ωst ), k = 1, 2, … , M(Δ)}
pk
(3.50)
where p and Q represent, respectively, the arrays of the assigned polynomial degrees and the map-
ping functions. The expression
u(Q(k) (k)
x (𝜉, 𝜂), Qy (𝜉, 𝜂)) ∈  (Ωst )
pk
indicates that the basis functions defined on element Ωk are mapped from the shape functions of a
polynomial space defined on standard triangular and quadrilateral elements.
3.6 Essential boundary conditions

As we have seen in Chapter 1, the essential boundary conditions are enforced by restriction. If
an essential boundary condition can be written as a linear combination of the basis functions then
enforcement is straightforward: The coefficients of the basis functions are known, and enforcement
is treated as in the one-dimensional case.
When the prescribed boundary conditions cannot be written as linear combinations of the basis
functions then it is necessary to approximate the prescribed boundary conditions by the basis func-
tions that are not zero on the element boundaries and then set the coefficients of the basis functions
to the appropriate value.
Let us assume, for example, that side 1 of a quadrilateral element of degree p lies on a boundary
on which a Dirichlet condition uFE = U(s) is to be enforced. Let the parameter values s1 and s2
correspond to nodes 1 and 2. We introduce the transformation
1−𝜉 1+𝜉
s = s(𝜉) = s + s
2 1 2 2
and define
def 1−𝜉 1+𝜉
u(𝜉) = U(s(𝜉)) − U(s1 ) − U(s2 ).
2 2
Using the ordinary least squares method, the function u(𝜉) is approximated by uFE (𝜉) defined by
def ∑
uFE (𝜉) = ai Ni (𝜉, −1)
i∈1 (p)
where 1 (p) is the set of indices of the standard shape functions associated with side 1 that are zero
in the nodal points. The coefficients of the nodal shape functions are U(s1 ) and U(s2 ).
3.7 Elements in three dimensions

Three-dimensional finite element meshes are comprised of hexahedral, tetrahedral, pentahedral
elements, less frequently other types of elements, such as pyramid elements, are used. The standard
hexahedral element, denoted by Ω(h)st , is the set of points
def
Ω(h)
st = {𝜉, 𝜂, 𝜁 | − 1 ≤ 𝜉, 𝜂, 𝜁 ≤ 1}. (3.51)
ζ
ζ 4
4 1 5 6
2
2 3
1
1
1
1
3 1 η
2 1
ξ ξ
1/ 3 η 1
2 3
2/ 3
3
(a) (b)
Figure 3.9 The standard tetrahedral and pentahedral elements Ωst(th) and Ωst(p) .
The standard tetrahedral element, denoted by Ω(th) st , and the standard pentahedral element,
(p)
denoted by Ωst , are shown in Fig. 3.9. Note that the edges of the elements have the length 2.0, as
in one and two dimensions.
The shape functions are analogous to those in one and two dimensions. For example, the
eight-node hexahedral element has vertex shape functions such as:
1
N1 = (1 − 𝜉)(1 − 𝜂)(1 − 𝜁).
8
The 20-node hexahedron is a generalization of the 8-node quadrilateral element to three dimen-
sions. Similarly, the 4-node and 10-node tetrahedra are generalizations of the 3-node and 6-node
triangles.
The hierarchic shape functions are also analogous to the shape functions defined in one and two
dimensions. A detailed description of shape functions for hexahedral elements can be found in
[35]. The shape functions associated with the edges and faces are the same along the edges and on
the faces as in two dimensions. For example, the edge shape function of the pentahedral element,
associated with the edge between nodes 1 and 2, corresponding to p = 2 is:
1−𝜁
N7 = L1 L2 𝜙̃ 2 (L2 − L1 )
2
where 𝜙̃ 2 is defined by eq. (3.31). The face shape function of the pentahedral element, associated
with the face defined by nodes 1, 2 and 3, at p = 3 is:
1−𝜁
N13 = L1 L2 L3 ⋅
2
The internal shape functions of the pentahedral elements are the products of the internal shape
functions defined for the triangular elements in eq. (3.36) and the function 𝜙k (𝜂) defined by
eq. (3.24).
Exercise 3.10 Write down the shape functions N12 for the standard tetrahedral, pentahedral and
hexahedral elements.
3.7.1 Mapping functions in three dimensions

Mesh generators typically produce mapping by linear or quadratic interpolation. The coordinates
of the interpolation points are determined from surface representations the generic form of which
is xi = xi (u, 𝑣) where u, 𝑣 are the surface parameters.
A surface may be composed of one or more patches, each patch parameterized separately. In
other words, the underlying geometry may be piecewise analytic. Various healing procedures are
used for filling gaps that may exist between neighboring surfaces or patches. Element faces may
intersect with more than one patch or surface.
The finite element representation approximates these surfaces by the element-level basis func-
tions. Continuity of the basis functions on element boundaries is enforced. In the h-version of
the finite element method the errors associated with the approximation of surfaces decreases as
the number of elements is increased. This is not the case for the p-version where the number of
elements is fixed. Therefore it is necessary to control the errors associated with mapping indepen-
dently from the number of elements. Extension of the isoparametric mapping procedures to high
order elements, combined with the blending function method, provides a satisfactory solution to
this problem.
The mapping is based on Lagrange polynomials with the collocation points fixed on the standard
element faces subject to the constraints that the mapping of the edges must be common to all ele-
ment types and the interior points selected in such a way that the Lebesque constant is minimum
[31]. The main points are summarized in Appendix F.
Example 3.1 A commercial automatic mesh generator4 created 202 triangular elements on the
spherical surface of unit radius shown in Fig. 3.10. The triangles were mapped by the optimal inter-
polation points for the standard triangle, p = 5. The error in maximum norm over all triangles is
| √ |
|1 − x2 + y2 + z2 | = 4.838 × 10−7 .
| |max
If quadratic isoparametric mapping were used then this error would have been three orders of
magnitude greater (2.875 × 10−4 ).
Figure 3.10 Meshing of a spherical surface with 202

triangular elements.
4 MeshSim®, a product of Simmetrix Inc.

Example 3.2 A spherical shell of radius rs , wall thickness ts = rs ∕100, is subjected to an internal
pressure q. By the membrane theory of shells the principal stress in the shell will be
1 qrs
𝜎1 = 𝜎2 = ⋅ (3.52)
2 ts
Letting p = 2, rs = 1, ts = 0.01, Poisson’s ratio: 𝜈 = 0 and using the same mesh as in Example 3.1, with
(a) optimal interpolation points corresponding to p = 5 and (b) quadratic isoparametric mapping
we get 𝜎1 = 100.0 (four significant digits accuracy) for (a) and 𝜎1 = 102.4 (two significant digits)
for (b).
3.8 Integration and differentiation
In Section 1.3.3, the coefficients of the stiffness matrix, Gram matrix and the right hand side vector
were computed on the standard element. In two and three dimensions the corresponding proce-
dures are analogous; however, with the exception of some important special cases, the mappings
are generally nonlinear.
The mapping functions
x = Q(k)
x (𝜉, 𝜂, 𝜁), y = Q(k)
y (𝜉, 𝜂, 𝜁), z = Q(k)
z (𝜉, 𝜂, 𝜁) (3.53)
map a standard element Ωst onto the kth element Ωk . In the following we will drop the superscript
when it is clear that we refer to the mapping of the kth element. A mapping is said to be proper
when the following three conditions are met: (a) The mapping functions Qx , Qy , Qz are single valued
functions of 𝜉, 𝜂, 𝜁 and possess continuous first derivatives; and (b) the Jacobian determinant |J|
(defined in the next section) is positive in every point of Ωst . The mapping functions used in FEA
must meet these criteria.
3.8.1 Volume and area integrals

Volume integrals on the kth element are computed on the corresponding standard element. The
volume integral of a scalar function F(x, y, z) on Ωk is:
F(x, y, z) dxdydz =  (𝜉, 𝜂, 𝜁)|J| d𝜉d𝜂d𝜁 (3.54)

∫Ωk ∫Ωst
where  (𝜉, 𝜂, 𝜁) = F(Qx (𝜉, 𝜂, 𝜁), Qy (𝜉, 𝜂, 𝜁), Qz (𝜉, 𝜂, 𝜁)) and |J| is the determinant of the Jacobian
matrix5 , called the Jacobian determinant. The Jacobian determinant in eq. (3.54) arises from the
definition of the differential volume: Let us denote the position vector of an arbitrary point P in the
element Ωk by r:
r = xex + yey + zez (3.55)
where ex , ey , ez are the orthogonal basis vectors of a (right-hand) Cartesian coordinate system. By
definition, the differential volume is understood to be the scalar triple product:
( )
𝜕r 𝜕r 𝜕r
dV = dx × dy ⋅ dz = (ex × ey ) ⋅ ez dxdydz = dxdydz. (3.56)
𝜕x 𝜕y 𝜕z
5 Carl Gustav Jacob Jacobi 1804–1851.

3.8 Integration and differentiation 107
Given the change of variables of eq. (3.53), the analogous expression is:
| 𝜕x 𝜕y 𝜕z |
| |
| 𝜕𝜉 𝜕𝜉 𝜕𝜉 |
| |
| |
( ) | |
| |
𝜕r 𝜕r 𝜕r | 𝜕x 𝜕y 𝜕z |
dV = d𝜉 × d𝜂 ⋅ d𝜁 = | | d𝜉d𝜂d𝜁 ≡ |J| d𝜉d𝜂d𝜁. (3.57)
𝜕𝜉 𝜕𝜂 𝜕𝜁 | 𝜕𝜂 𝜕𝜂 𝜕𝜂 |
| |
| |
| |
| 𝜕x 𝜕y 𝜕z |
| |
| 𝜕𝜁 𝜕𝜁 𝜕𝜁 |
| |
From equations (3.56) and (3.57) we get: dxdydz = |J|d𝜉d𝜂d𝜁.
The vectors 𝜕r∕𝜕𝜉, 𝜕r∕𝜕𝜂, 𝜕r∕𝜕𝜁 are a set of right-handed basis vectors, that is, their scalar triple
product yields a positive number. If the Jacobian determinant is negative then the right-handed
coordinate system is transformed into a left-handed one in which case the mapping is improper.
In two dimensions the mapping is
tz
x = Qx (𝜉, 𝜂), y = Qy (𝜉, 𝜂), 𝜁
z = Qz (𝜁) =
2
where tz is the thickness, see Fig. 2.4. When the thickness is constant then the integration in the
transverse (𝜁) direction can be performed explicitly and only an area integral needs to be evaluated:
| 𝜕x 𝜕y |
| |
| 𝜕𝜉 𝜕𝜉 ||
|
dV = tz dA = tz || | d𝜉d𝜂
| (3.58)
| 𝜕x 𝜕y ||
|
| |
| 𝜕𝜂 𝜕𝜂 |
In the finite element method the integrations are performed by numerical quadrature, details
of which are given in Appendix E. The minimum number of quadrature points depends on the
polynomial degree of the shape functions: When the mapping is linear and the material properties
are constant then the coefficients of the stiffness matrix should be exact (up to numerical round-off
errors) otherwise the stiffness matrix may become singular. When the material properties vary, or
the mapping is non-linear, then the number of integration points should be increased so that the
errors in the stiffness coefficients are small. Experience has shown that increasing the number of
integration points in each coordinate direction by two for curved elements is sufficient in most
cases. Similar considerations apply to the load vector.
Exercise 3.11 Show that the Jacobian matrix of straight-side triangular elements, that is, trian-
gular elements mapped by eq. (3.41) and eq. (3.42), is independent of L1 , L2 and L3 and show that
the area of a triangle in terms of its vertex coordinates (Xi , Yi ), i = 1, 2, 3 is:
X1 (Y2 − Y3 ) + X2 (Y3 − Y1 ) + X3 (Y1 − Y2 )
A= ⋅
2
Exercise 3.12 Show that for straight side quadrilaterals the Jacobian determinant is constant only
if the quadrilateral element is a parallelogram.
3.8.2 Surface and contour integrals

Given the mapping functions, each face is parameterized by imposing the appropriate restriction on
the mapping function. For example, if integration is to be performed on the face of a hexahedron
that corresponds to 𝜁 = 1 then, referring to eq. (3.53), the parametric form of the surface
becomes:
x = Qx (𝜉, 𝜂, 1), y = Qy (𝜉, 𝜂, 1), z = Qz (𝜉, 𝜂, 1) (3.59)
and, using the definition of r given in (3.55), the surface integral of a scalar function
F(x, y, z) is:
+1 +1 | 𝜕r 𝜕r |
F(x, y, z) dS =  (𝜉, 𝜂, 1) || × || d𝜉d𝜂
∫ ∫(𝜕Ωk )𝜁 =1 ∫−1 ∫−1 | 𝜕𝜉 𝜕𝜂 |
where  is obtained from F by replacing x, y, z with the mapping functions Qx , Qy . Qz . The treatment
of the other faces is analogous.
In two dimensions the contour integral of a scalar function F(x, y) on the side of a quadrilateral
element corresponding to 𝜂 = 1 is:
+1 | dr |
F(x, y) ds =  (𝜉, 1) || || d𝜉.
∫(𝜕Ωk )𝜂=1 ∫−1 | d𝜉 |
The other sides are treated analogously. The positive sense of the contour integral is counterclock-
wise.
3.8.3 Differentiation
The approximating functions, and hence the solution, are known in terms of the shape functions
defined on standard elements. Therefore differentiation with respect to x, y and z has to be expressed
in terms of differentiation with respect to 𝜉, 𝜂, 𝜁. Using the chain rule, we have:
⎧ 𝜕 ⎫ ⎡ 𝜕x 𝜕y 𝜕z ⎤ 𝜕
⎪ ⎪ ⎢ ⎧ ⎫
⎪ 𝜕𝜉 ⎪ ⎢ 𝜕𝜉 𝜕𝜉 𝜕𝜉 ⎥ ⎪ 𝜕x ⎪
⎪ ⎪ ⎢ ⎥⎪ ⎪
⎪ 𝜕 ⎪ ⎢ 𝜕x ⎥
𝜕y 𝜕z ⎥ ⎪ 𝜕 ⎪
⎨ 𝜕𝜂 ⎬ = ⎢ 𝜕𝜂 ⋅
𝜕𝜂 ⎥ ⎨ 𝜕y ⎬
(3.60)
⎪ ⎪ ⎢ 𝜕𝜂 ⎪ ⎪
⎪ ⎪ ⎢ ⎥⎪ ⎪
⎪ 𝜕 ⎪ ⎢ 𝜕x 𝜕y 𝜕z ⎥⎥ ⎪ 𝜕 ⎪
⎪ 𝜕𝜁 ⎪ ⎣ 𝜕𝜁 ⎩ ⎭
⎩ ⎭ 𝜕𝜁 𝜕𝜁 ⎦ 𝜕z
On multiplying by the inverse of the Jacobian matrix, we have the expression used for computing
the derivatives of the shape functions defined on standard elements:
𝜕x 𝜕y 𝜕z ⎤
−1 ⎧𝜕 ⎫
⎧𝜕 ⎫ ⎡ ⎪ ⎪
𝜕x
⎪ ⎪ ⎢ ⎢ 𝜕𝜉 𝜕𝜉 𝜕𝜉 ⎥ ⎪ 𝜕𝜉 ⎪
⎪ ⎪ ⎢ ⎥ ⎪ ⎪
⎪ 𝜕 ⎪ ⎢ 𝜕x 𝜕y ⎥ ⎪𝜕 ⎪
𝜕z ⎥
⎨ 𝜕y ⎬ = ⎢ 𝜕𝜂 𝜕𝜂 𝜕𝜂 ⎥ ⎨ 𝜕𝜂 ⎬ ⋅ (3.61)
⎪ ⎪ ⎢ ⎥ ⎪ ⎪
⎪ ⎪ ⎢ ⎪ ⎪
⎪ 𝜕 ⎪ ⎢ 𝜕x 𝜕y 𝜕z ⎥⎥ ⎪𝜕 ⎪
⎩ 𝜕z ⎭ ⎣ ⎪ 𝜕𝜁 ⎪
𝜕𝜁 𝜕𝜁 𝜕𝜁 ⎦ ⎩ ⎭
Computation of the first derivatives from a finite element solution in a given point is discussed in
Section 3.11.
3.9 Stiffness matrices and load vectors 109
3.9 Stiffness matrices and load vectors
The algorithms for the computation of stiffness matrices and load vectors for three-dimensional
elasticity are outlined in the following. Their counterparts for two-dimensional elasticity and heat
conduction are analogous. The algorithms are based on eq. (2.107). However, the integrals are eval-
uated element by element:

∫Ωk ∫Ωk ∫𝜕Ωk ∩𝜕ΩT
+ ([D]{𝑣})T [E]{𝛼}Δ dV (3.62)

∫Ωk
where the differential operator matrix [D] and the material stiffness matrix [E] are as defined by
eq. (2.105). The kth element is denoted by Ωk . The second term on the right-hand side represents
the virtual work of tractions acting on boundary segment 𝜕ΩT . This term is present only when one
or more of the boundary surfaces of the element lie on 𝜕ΩT . For the sake of simplicity we assume
that the number of degrees of freedom is the same for all three fields on Ωk . We denote the number
of degrees of freedom per field by n and define the 3 × 3n matrix [N] as follows:
⎡ N 1 N 2 · · · Nn 0 0 · · · 0 0 0 · · · 0 ⎤
[N] = ⎢ 0 0 · · · 0 N1 N2 · · · Nn 0 0 · · · 0 ⎥ .
⎢ ⎥
⎣ 0 0 · · · 0 0 0 · · · 0 N1 N 2 · · · N n ⎦
The jth column of [N], denoted by {Nj }, is the jth shape function vector. We write the trial and test
functions as linear combinations of the shape function vectors:
∑
3n
∑
3n
{u} = aj {Nj } and {𝑣} = bi {Ni }. (3.63)
j=1 i=1
3.9.1 Stiffness matrices

The elements of the stiffness matrix kij can be written in the form:
( )T
kij(k) = [D]{Ni } [E][D]{Nj } dV. (3.64)
∫Ωk
We take advantage of the fact that two elements of {Ni } are zero. The position of the zero elements
depends on the value of the index i. For example, when 1 ≤ i ≤ n then:
⎧𝜕∕𝜕x⎫ ⎡1 0 0⎤
⎪ ⎪ ⎢ ⎥
⎪ 0 ⎪ ⎢ 0 0 0 ⎥ ⎧𝜕∕𝜕x⎫ ⎧𝜕∕𝜕𝜉 ⎫
⎪ 0 ⎪ ⎢ 0 0 0 ⎥⎪ ⎪ −1 ⎪ ⎪
[D]{Ni } = ⎨
𝜕∕𝜕y ⎬ Ni = ⎢ 0 1 0 ⎥ ⎨𝜕∕𝜕y⎬ Ni = [M1 ][Jk ] ⎨𝜕∕𝜕𝜂 ⎬ Ni
⎪ ⎪ ⎢ ⎥ ⎪𝜕∕𝜕z ⎪ ⎪𝜕∕𝜕𝜁 ⎪
⎪ 0 ⎪ ⎢ 0 0 0 ⎥⎩ ⎭ ⎩ ⎭
⎪𝜕∕𝜕z ⎪ ⎢0 0 1⎥
⎩ ⎭ ⎣ ⎦
⏟⏞⏞⏞⏞⏟⏞⏞⏞⏞⏟
[M1 ]
where [Jk ]−1 is the inverse of the Jacobian matrix corresponding to element k, see eq. (3.61), [M1 ]
is a logical matrix. When the index i changes, only [M1 ] has to be replaced. Specifically, when
(n + 1) ≤ i ≤ 2n then [M1 ] is replaced by [M2 ]; when (2n + 1) ≤ i ≤ 3n then [M1 ] is replaced by

[M3 ] which are defined as follows:
⎡ 0 0 0 ⎤ ⎡ 0 0 0 ⎤
⎢ ⎥ ⎢ ⎥
⎢ 0 1 0 ⎥ ⎢ 0 0 0 ⎥
⎢ 0 0 0 ⎥ ⎢ 0 0 1 ⎥
[M2 ] = ⎢
1 0 0 ⎥ [M3 ] = ⎢
0 0 0 ⎥.
⎢ ⎥ ⎢ ⎥
⎢ 0 0 1 ⎥ ⎢ 0 1 0 ⎥
⎢ 0 0 0 ⎥ ⎢ 1 0 0 ⎥
⎣ ⎦ ⎣ ⎦
We define {} = {𝜕∕𝜕𝜉 𝜕∕𝜕𝜂 𝜕∕𝜕𝜁}T and write eq. (3.64) in a form suitable for evaluation by
numerical integration which is described in Appendix E:
( )T
kij(k) = [M𝛼 ][Jk ]−1 {}Ni [E][M𝛽 ][Jk ]−1 {}Nj |Jk | d𝜉d𝜂d𝜁. (3.65)
∫Ωst
The domain of integration is a standard hexahedral, tetrahedral or pentahedral element. The

indices 𝛼 and 𝛽 take on the values 1, 2, 3 depending on the range of the indices i and j. Therefore
(k)
the element stiffness matrix [K (k) ] consists of six blocks [K𝛼𝛽 ]:
⎡ [K (k) ] [K (k) ] [K (k) ] ⎤

⎢ 11 12 13 ⎥
⎥.
(k) (k) (k)
[K ] = ⎢ [K22 ] [K23 ] (3.66)
⎢ sym. [K (k)
] ⎥
⎣ 33 ⎦
Exercise 3.13 Assume that the mapping functions are given for the kth element. For example, in
two dimensions:
x = 𝛼Q(k)
x (𝜉, 𝜂), y = 𝛼Q(k)
y (𝜉, 𝜂)
where 𝛼 > 0 is some real number. Assume further that the element stiffness matrix has been com-
puted for 𝛼 = 1. How will the elements of the stiffness matrix change as functions of 𝛼 in one, two
and three dimensions? Assume that in two dimensions the thickness is independent of 𝛼.
Exercise 3.14 Refer to equations (2.76) to (2.80). Develop an expression for the computation of
the terms of the stiffness matrix, analogous to kij(k) given by eq. (3.65), for axisymmetric elastostatic
models.
3.9.2 Load vectors

The computation of element level load vectors corresponding volume forces, surface tractions and
thermal loading is based on the corresponding terms on the right-hand side of eq. (3.62).
Volume forces
Refer to the first term on the right-hand side of eq. (3.62). The computation of the load vector
corresponding to volume force {F} acting on element k is a straightforward application of eq. (3.54):
ri(k) = {Ni }T {F} |Jk | d𝜉d𝜂d𝜁 i = 1, 2, … , 3n. (3.67)

∫Ωst
3.11 Computation of the solution and its first derivatives 111
Surface tractions
Refer to the second term on the right-hand side of eq. (3.62). Assume that traction vectors are acting
on a hexahedral element on the face 𝜁 = 1. In this case the ith term of the load vector is:
+1 +1 | 𝜕r 𝜕r |
ri(k) = {Ni }T {T} || × || d𝜉d𝜂 (3.68)
∫−1 ∫−1 | 𝜕𝜉 𝜕𝜂 |𝜁 =1
where the range of i is the set of indices of shape functions associated with the face 𝜁 = 1. The other
faces are treated analogously.
Thermal loading
Refer to the third term on the right-hand side of eq. (3.62). The expression for ri(k) corresponding to
thermal loading is:
( )T
ri(k) = [M𝛽 ][Jk ]−1 []{Ni } [E]{𝛼}Δ |Jk | d𝜉d𝜂d𝜁 i = 1, 2, … , 3n (3.69)
∫Ωst
where 𝛽 = 1 when 1 ≤ i ≤ n, 𝛽 = 2 when (n + 1) ≤ i ≤ 2n and 𝛽 = 3 when (2n + 1) ≤ i ≤ 3n. The
matrices [M𝛽 ] and the operator {} are defined in Section 3.9.1.

A finite element space is characterized by a finite element mesh and the polynomial degrees and
mapping functions assigned to the elements of the mesh. The polynomial degrees identify a poly-
nomial space defined on a standard element. The polynomial space is spanned by basis functions,
called shape functions. Two kinds of shape functions, called Lagrange and hierarchic shape func-
tions, were described for quadrilateral and triangular elements. The finite element space is spanned
by the mapped shape functions subject to the continuity requirements described in Section 1.3.2.
Unless the mappings of all elements are polynomial functions of degree equal to or less than
the polynomial degree of elements, rigid body rotation will not be represented exactly by the finite
element solution. Nevertheless, rapid convergence to the correct solution will occur as the finite
element space is progressively enlarged by h-, p- or hp-extension. The mapping functions used in
FEA must be such that the Jacobian determinant is positive in every point within the element.
3.10 Post-solution operations
Following assembly (outlined in Sections 1.3.5) and solution operations, the finite element solution
is stored in the form of data sets that contain the coefficients of the shape functions, the mapping
functions and indices that identify the polynomial space associated with each element.
Some of the data of interest, such as temperature, displacement, flux, strain, stress, can be com-
puted from the finite element solution either by direct or indirect methods while others, such as
stress intensity factors, can be computed by indirect methods only. In this chapter the techniques
used for the computation and verification of engineering data are described.
3.11 Computation of the solution and its first derivatives
If one is interested in the value of the solution in a point (x0 , y0 , z0 ) then the domain has to
be searched to identify the element in which that point lies. Suppose that the point lies in the
kth element. The next step is to find the standard coordinates (𝜉0 , 𝜂0 , 𝜁0 ) from the mapping
functions.
x0 = Q(k)
x (𝜉0 , 𝜂0 , 𝜁0 ), y0 = Q(k)
y (𝜉0 , 𝜂0 , 𝜁0 ), z0 = Q(k)
z (𝜉0 , 𝜂0 , 𝜁0 ). (3.70)
Unless the mapping of the kth element happens to be linear, the inverse of the mapping func-
tion is not known explicitly. Therefore this step involves a root finding procedure, such as the
Newton-Raphson method.
The next step is to look up the parameters that identify the standard space associated with the
element and the computed coefficients of the basis functions. With this information the solution
and its derivatives can be computed. For example, let us assume that the solution is a scalar function
(q)
and the standard space  p,q (Ωst ) is associated with the kth element, denoted by Ωk . Then the finite
element solution in the point (x0 , y0 ) ∈ Ωk is:
∑
n
uFE (x0 , y0 ) = a(k)
i
Ni (𝜉0 , 𝜂0 ) (3.71)
i=1
(q)
where n = (p + 1)(q + 1) is the number of shape functions that span  p,q (Ωst ), Ni (𝜉, 𝜂) are the shape
functions and a(k)
i
are the corresponding coefficients. When the solution is a vector function then
each component of uFE is in the form of eq. (3.71).
Computation of the first derivatives of uFE in the point (x0 , y0 ) involves the computation of the
inverse of the Jacobian matrix in the corresponding point (𝜉0 , 𝜂0 ) and multiplying the derivatives
of the finite element solution with respect to the standard coordinates. Referring to eq. (3.61);
⎧ 𝜕uFE ⎫ ⎡ 𝜕x 𝜕y ⎤−1 ⎧ 𝜕Ni ⎫
⎪ ⎪ ⎢ 𝜕𝜉 ⎪ ⎪
⎪ 𝜕x ⎪ 𝜕𝜉 ⎥ ∑ (k) ⎪ 𝜕𝜉 ⎪
n
⎨ ⎬ =⎢ ⎥ ai ⎨ ⎬ (3.72)
⎪ 𝜕uFE ⎪ ⎢ 𝜕y ⎥⎥ ⎪ 𝜕Ni ⎪
⎪ 𝜕y ⎪ ⎢ 𝜕x i=1
⎪ ⎪
⎩ ⎭ ⎣ 𝜕𝜂 𝜕𝜂 ⎦(𝜉0 ,𝜂0 ) ⎩ 𝜕𝜂 ⎭(𝜉0 ,𝜂0 )
(x0 ,y0 )
where x = Q(k)
x (𝜉, 𝜂), y = Q(k)
y (𝜉, 𝜂). The flux vector (resp. stress tensor) is computed by multiplying
the temperature gradient (resp. the strain tensor) by the thermal conductivity matrix (resp. material
stiffness matrix). The transformation of vectors and tensors is described in Appendix K.
The derivatives of the finite element solution are discontinuous along inter-element boundaries.
Therefore if the point selected for the evaluation of fluxes, stresses, etc. is a node point, or a point
on an inter-element boundary, then the computed value depends on the element selected for the
computation. The degree of discontinuity in the normal and shearing stresses, or the normal flux
component, at inter-element boundaries is an indicator of the quality of the approximation. In
implementations of the h-version the derivatives are typically evaluated in the integration points
and are interpolated over the elements. In graphical displays in the form of contour plots discon-
tinuities of the derivatives at element boundaries are often masked by means of smoothing the
contour lines through averaging.
In the p-version the standard element is subdivided so as to produce a uniform grid, called dis-
play grid, and the solution and its derivatives are evaluated in the grid points. Since the standard
coordinates of the grid points are known, inverse mapping is not involved. The quality of contour
plots depends on the quality of data being displayed and on the fineness of the display grid.
Search for a maximum or minimum value also involves search on a uniform grid defined on the
standard elements. The fineness of the grid, and hence the number of points searched for the mini-
mum or maximum, is controlled by a parameter. In conventional implementations of the h-version
the search grid is typically defined by the integration points or the node points.
Exercise 3.15 Consider two plane elastic elements that have a common edge. Assume that differ-
ent material properties were assigned to the elements. Show that the normal and shearing stresses
corresponding to the exact solution have to be the same along the common edge and hence the
normal and shearing strains will be discontinuous. Hint: Consider equilibrium in the coordinate
system defined in the normal and tangential directions.
3.12 Nodal forces
Recall the definition of nodal forces {f (k) } in Section 1.4.1

(k)
{f (k) } = [K (k) ]{a(k) } − {r } k = 1, 2, … , M(Δ) (3.73)
(k)
where [K (k) ] is the stiffness matrix, {a(k) } is the solution vector and {r } is the load vector corre-
sponding to volume forces and thermal loads acting on element k.
3.12.1 Nodal forces in the h-version

When solving problems of elasticity using finite element analysis based on the h-version, nodal
forces are treated in the same way as concentrated forces are treated in statics. Typical uses of nodal
forces are: (a) isolating some region of interest from a larger structure and treating the isolated
region as if it were a free body held in equilibrium by the nodal forces and (b) computing stress
resultants. The underlying assumption is that nodal forces reliably represent the load path, that
is, the distribution of internal forces in a statically indeterminate structure. This assumption is
usually justified by the argument that nodal forces satisfy the equations of static equilibrium for
any element or group of elements.
The following discussion will show that satisfaction of equilibrium is related to the rank defi-
ciency of unconstrained stiffness matrices. Consequently equilibrium of nodal forces should not
be interpreted as an indicator of the quality of the finite element solution and does not guarantee
that the nodal forces are reliable approximations of the internal forces in a statically indeterminate
structure. On the other hand, nodal forces are useful for the computation of stress resultants.
Let us assume, for example, that Ωk is an 8-node quadrilateral element. The number of degrees
of freedom per field, denoted by n, is 8. The notation is shown on Fig. 3.11.
In expanded notation the elements of the nodal force vector {f (k) } are:
fx(k,i) = fi(k) , fy(k,i) = fn+i

(k)
, i = 1, 2, … , n. (3.74)
Similarly, the elements of the vector {r} are written as follows:

(k,i) (k) (k,i) (k)
rx = ri , ry = r n+i , i = 1, 2, … , n. (3.75)
On adapting the notation used in eq. (3.65) to problems of planar elasticity we write:
( )T ∑2n
(k,i)
fx(k,i) = [M1 ][Jk ]−1 {}Ni [E] {𝜀(k) }a(k) |Jk | d𝜉d𝜂 − r x (3.76)
∫Ω(q) j j
j=1
st
and
( )T ∑2n
(k,i)
fy(k,i) = [M2 ][Jk ]−1 {}Ni [E] {𝜀(k) }a(k) |Jk | d𝜉d𝜂 − r y (3.77)
∫Ω(q) j j
j=1
st
fy(k,4)
fx(k,4)
4
fy(k,7)
fy(k,8) 7 fx(k,7)
8
fx(k,8)
fy(k,1) Ωk fy(k,3)
y
fy(k,5) 3
fx(k,1) 1 fy(k,6) fx(k,3)
fx(k,5) 5 fy(k,2) 6
fx(k,6)
x 2 f (k,2)
x
Figure 3.11 Nodal forces associated with the 8-node quadrilateral element. Notation.
where [Jk ] is the Jacobian matrix and

⎧𝜕 ⎫
⎡1 0⎤ ⎡0 0⎤ ⎪ ⎪
⎪ 𝜕𝜉 ⎪
[M1 ] = ⎢0 0⎥ [M2 ] = ⎢0 1⎥ {} = ⎨ ⎬
⎢ ⎥ ⎢ ⎥ ⎪𝜕 ⎪
⎣0 1⎦ ⎣1 0⎦
⎪ 𝜕𝜂 ⎪
⎩ ⎭
and
{
[M1 ][Jk ]−1 {}Nj for j = 1, 2, … , n
{𝜀(k) } =
j [M2 ][Jk ]−1 {}Nj−n for j = n + 1, n + 2, … , 2n.
The elements of the load vector corresponding to body forces and thermal loads are:
(k,i) ( )T
rx = Ni Fx |Jk | d𝜉d𝜂 + [M1 ][Jk ]−1 {}Ni [E]{𝛼}TΔ |Jk | d𝜉d𝜂 (3.78)
∫ (q)
Ωst ∫ (q)
Ωst
and
(k,i) ( )T
ry = Ni Fy |Jk | d𝜉d𝜂 + [M2 ][Jk ]−1 {}Ni [E]{𝛼}TΔ |Jk | d𝜉d𝜂. (3.79)
∫Ω(q) ∫Ω(q)
st st
The equations of static equilibrium are:

∑
n
fx(k,i) + Fx dxdy = 0 (3.80)
i=1
∫Ωk
∑ (k,i)
n
fy + Fy dxdy = 0 (3.81)
i=1
∫Ωk
and
n ( )
∑
Xi fy(k,i) − Yi fx(k,i) + (xFy − yFx ) dxdy = 0 (3.82)
i=1
∫Ωk
where Xi , Yi are the coordinates of the ith node.

Satisfaction of eq. (3.80) and eq. (3.81) follows from the fact that
∑
n
∑
n
Ni (𝜉, 𝜂) = 1, therefore {} Ni (𝜉, 𝜂) = 0.
i=1 i=1
Satisfaction of eq. (3.82) follows from the mapping functions (3.39) and (3.40) and the fact that
infinitesimal rigid body rotations do not cause strain:
( )T { }
∑n
𝜕 𝜕
−1
[M1 ][Jk ] {} Yi Ni ≡ 0 y = {0 0 1} (3.83)
i=1
𝜕x 𝜕y
( )T { }
∑n
𝜕 𝜕
−1
[M2 ][Jk ] {} Xi Ni ≡ 0 x = {0 0 1}. (3.84)
i=1
𝜕y 𝜕x
(k,i) (k,i)
On substituting equations (3.83) and (3.84) into the expressions for fx(k,i) , fy(k,i) , r x and r y eq. (3.82)
is obtained.
Note that the conditions for static equilibrium are satisfied independently of {a(k) }. Therefore
equilibrium of nodal forces is unrelated to the finite element solution.
Exercise 3.16 Consider the problem of heat conduction in two dimensions. Assume that the
nodal fluxes were computed analogously to eq. (3.73). Define the term which is analogous to {r}
and show that the sum of nodal fluxes plus the integral of the source term over an element is zero.
Use an 8-node quadrilateral element and a 6-node triangular element to illustrate this point.
3.12.2 Nodal forces in the p-version

We have seen in the case of the 8-node quadrilateral element that equilibrium of nodal forces was
related to the facts that the sum of the shape functions is unity and the functions x and y could
be expressed as linear combinations of the shape functions. When the hierarchic shape functions
based on the integrals of Legendre polynomials are used, such as those illustrated in Fig. 3.4, then
the sum of the first four shape functions is unity independently of p. Therefore in the equilibrium
equations (3.80) to (3.82) n = 4.
Example 3.3 A rectangular domain representing an elastic body of constant thickness is sub-
jected to the boundary conditions un = 0, ut = 𝛿 on boundary segments BC and DA shown in
Fig. 3.12(a). The subscripts n and t refer to the normal and tangent directions respectively. Boundary
segments AB and CD are traction-free.
We solve this as a plane stress problem using one element, represented by the shaded part of the
domain, and product spaces. Antisymmetry condition is applied at x = 𝓁∕2. The node numbering
is shown in Fig. 3.12(b). Since only one element is used, the superscript that identifies the
element number is dropped. Letting 𝓁 = 1000 mm, d = 50 mm, b = 20 mm, E = 200 GPA,
𝜈 = 0.295, 𝛿 = 5 mm, the computed values of the nodal forces are shown in Table 3.1. In this
example {r} = 0. Therefore the nodal forces were computed from {f } = [K]{a}. The results shown
in Table 3.1 indicate that equilibrium of nodal forces is satisfied at every p-level, independent of
the accuracy of the finite element solution.
This problem could have been solved on the shaded domain shown in Fig. 3.13 in which case
antisymmetry condition would be prescribed on the x axis. Using the notation shown in Fig. 3.13(b),
the computed nodal forces are shown in Table 3.2 for p = 8. It is seen that the nodal forces once
again satisfy the condition of static equilibrium.
y
D C
x d
A B
/2 /2 b
(a)
fy(4) fy(3)
fx(4) 4 3 fx(3)
fx(1) 1 2 f (2)
x
fy(1) fy(2)
(b)
Figure 3.12 Example 3.3. Notation.
Table 3.1 Example 3.3. Nodal forces (kN). The notation is shown in Fig. 3.12(b). Product spaces.
p fx(1) fx(2) fx(3) fx(4) fy(1) fy(2) fy(3) fy(4)
1 −1971.3 0.000 0.000 1971.3 −98.564 98.564 98.564 −98.564

2 −64.678 0.000 0.000 64.678 −3.234 3.234 3.234 −3.234
3 −50.546 0.000 0.000 50.546 −2.527 2.527 2.527 −2.527
4 −50.190 0.000 0.000 50.190 −2.509 2.509 2.509 −2.509
5 −50.010 0.000 0.000 50.010 −2.500 2.500 2.500 −2.500
6 −49.907 0.000 0.000 49.907 −2.495 2.495 2.495 −2.495
7 −49.843 0.000 0.000 49.843 −2.492 2.492 2.492 −2.492
8 −49.802 0.000 0.000 49.802 −2.490 2.490 2.490 −2.490
y fy(4) fy(3)
fx(4) 4 3 fx(3)
D
x 1 2 fx(2)
A fx(1)
/2 fy(1) fy(2)
(a) (b)
Figure 3.13 Example 3.3. The smallest solution domain.
Exercise 3.17 The nodal forces shown in Table 3.1 (resp. Table 3.2) were computed on the shaded
domain shown in Fig. 3.12(a) (resp. Fig. 3.13(a)). Show that the stress resultants acting on boundary
segments BC and DA are the same. Hint: Sketch and label the values of the nodal forces on the
element shown in Fig. 3.13(b) and its antisymmetric pair.
Table 3.2 Example 3.3. Nodal forces (kN). Solution on the shaded domain shown in Fig. 3.13(a). The
notation is shown in Fig. 3.13(b). Product space.
p fx(1) fx(2) fx(3) fx(4) fy(1) fy(2) fy(3) fy(4)
8 −12.454 −37.353 0.000 49.806 4.922 1.144 1.346 −7.412
3.12.3 Nodal forces and stress resultants

The nodal forces are related to the extraction functions for stress resultants. We illustrate this on
the basis of Example 3.3. We have:
([D]{𝑣})T [E][D]{uFE }b dxdy = (Tx(FE) 𝑣x + Ty(FE) 𝑣y )b ds (3.85)

∫Ω ∫𝜕Ω
where Ω is the domain of the element shown in Fig. 3.12(b) and b is the thickness. If we are inter-
ested in the shear force acting on the side between nodes 2 and 3, denoted by V2,3 , then we select
𝑣x = 0 on Ω and 𝑣y a smooth function of Ω such that 𝑣y = 1 on the side between nodes 2 and 3 and
𝑣y = 0 on the side between nodes 4 and 1. Specifically if we select 𝑣y = N2 (𝜉, 𝜂) + N3 (𝜉, 𝜂) then we
have:
node 3 ∑2n
V2,3 = Ty(FE) b dy = (kn+2,j + kn+3,j )aj = fy(2) + fy(3) (3.86)
∫node 2 j=1
where n is the number of degrees of freedom per field. In Example 3.3 product spaces were used
therefore n = (p + 1)2 .
Exercise 3.18 Solve the problem of Example 3.3 and compute the stress resultants V4,1 and M4,1
on the side between nodes 4 and 1 by direct integration. Keep refining the mesh until satisfac-
tory convergence is observed. Compare the results with those computed from the nodal forces in
Table 3.1. By definition:
+d∕2
M4,1 = Tx yb dy.
∫−d∕2
This exercise shows that integration of stresses is much less efficient than extraction. The presence
of singularities in points A and D influences the accuracy of the numerical integration.
3.13 Chapter summary

In order to meet the requirement of solution verification, it is necessary to show that the errors
in the data of interest do not exceed stated tolerances. In practical problems the exact solution is
typically unknown and it is not possible to determine the errors of approximation with certainty.
It is possible however to show that necessary conditions are satisfied for the errors in the data of
interest to be small.
Error estimation is based on the a priori knowledge that the data of interest corresponding to the
exact solution are finite and independent of the discretization parameter. Therefore a necessary
condition for the error to be small is that the computed data should exhibit convergence to a limit
value as the number of degrees of freedom is increased. An efficient and robust way to achieve
this is to use properly designed meshes and increase the polynomial degree. This will be discussed
further in Chapter 4
119
Pre- and postprocessing procedures and verification
The questions of how to choose an effective finite element discretization scheme, given a set of input
data, and how to extract the quantities of interest from the finite element solution and estimate their
relative errors, are addressed in this chapter.
Preprocessing is concerned with the collection and verification of input data and the formula-
tion of a discretization scheme with the goal to approximate the quantities of interest in an efficient
manner. Based on the problem statement, which includes specification of the domain, the mate-
rial properties, boundary conditions, the quantities of interest and the acceptable error tolerances,
analysts are called upon to take into consideration the technical capabilities of the software tool(s)
available to them and formulate a discretization scheme. This requires an understanding of the
relationship between problem definition and the regularity of the underlying exact solution. For
that reason, this chapter begins with a discussion on the regularity of functions. The main results
of the relevant mathematical theorems are summarized.
Postprocessing is concerned with the extraction of the quantities of interest from the finite ele-
ment solution and estimation of their relative errors. Should it be found that the error tolerances
were exceeded, the discretization has to be revised and a new solution obtained. This chapter also
covers postprocessing procedures for the extraction of flux and stress intensity factors.
4.1 Regularity in two and three dimensions

A rich and elaborate mathematical theory exists on the convergence properties of the finite element
method. The details are beyond the scope of this book; however, an understanding of the meaning
and implications of those theorems is essential in the practice of finite element analysis. The main
point is that the solutions of elliptic boundary value problems on polygonal and polyhedral domains
are typically piecewise analytic functions. Through a judicious combination of local mesh refine-
ment and increase of polynomial degree, it is possible to achieve exponential rates of convergence
and estimate the errors of approximation from the resulting sequence of finite element solutions.
Of particular interest are singular points and curves. Singularities are usually caused by simpli-
fications introduced in the formulation of mathematical problems, which involves the definition
of the solution domain, assignment of material properties and specification of the boundary condi-
tions. In engineering applications it is generally useful to divide the solution domain into domains
of primary and secondary interest. The QoIs are extracted from the domain(s) of primary interest.
Singularities in the domain of secondary interest are nuisances that tend to slow convergence
and may perturb the computed quantities of interest. Singularities may also occur in the domain
of primary interest. For example, in linear elastic fracture mechanics the quantity of interest is the
120 4 Pre- and postprocessing procedures and verification
stress intensity factor. An understanding of why singularities occur and how singularities affect the
numerical solution is essential for the understanding and proper application of the finite element
method.
The regularity of functions is measured by how many derivatives are square integrable. In one
dimension the regularity of the exact solution depends on the smoothness of the coefficients 𝜅 and
c and the forcing function f . In two and three dimensions regularity also depends on the vertex
angles, material properties, boundary conditions and source functions. In three dimensions the
regularity of the solution also depends on the angles at which boundary surfaces intersect.
4.2 The Laplace equation in two dimensions

We consider the solution of the Laplace equation in the neighborhood of a corner point of a
two-dimensional domain, such as point B in Fig. 4.1(a). We assume that the boundary condition
on ΓAB and ΓBC is either u = 0 or 𝜕u∕𝜕n = 0. The Laplace equation in polar coordinates (r, 𝜃) is
𝜕 2 u 1 𝜕u 1 𝜕2 u
Δu ≡ + + = 0. (4.1)
𝜕r 2 r 𝜕r r 2 𝜕𝜃 2
In particular, we are interested in solutions of the form: u = r 𝜆 F(𝜃) with 𝜆 ≠ 0. Such solutions are
typically associated with geometric singularities, boundary conditions and intersections of material
interfaces. Substituting into eq. (4.1), we have:
F ′′ + 𝜆2 F = 0
the general solution of which is
F = a cos 𝜆𝜃 + b sin 𝜆𝜃 (4.2)
where a, b are arbitrary constants. Therefore the solution can be written as:
u = r 𝜆 (a cos 𝜆𝜃 + b sin 𝜆𝜃). (4.3)
Consider, for example, the problem with the boundary condition u = 0 on boundary segments ΓAB
and ΓBC (i.e. 𝜃 = ±𝛼∕2):
a cos 𝜆𝛼∕2 + b sin 𝜆𝛼∕2 = 0
a cos 𝜆𝛼∕2 − b sin 𝜆𝛼∕2 = 0.
y
y
ϱ
C P
𝛼/2 r
ΓBC ΓBC
𝜃
ΓAB B x ΓAB x
𝛼/2 Ωϱ
A Γϱ
(a) (b)
Figure 4.1 Reentrant corner. Notation

Adding and subtracting the two equations we find:

cos 𝜆𝛼∕2 = 0 therefore 𝜆𝛼∕2 = ±(2m − 1)𝜋∕2, m = 1, 2, … (4.4)
sin 𝜆𝛼∕2 = 0 therefore 𝜆𝛼∕2 = ±n𝜋, n = 1, 2, … (4.5)
Note that the values of 𝜆 that satisfy eq. (4.4) and eq. (4.5) can be either positive or negative. How-
ever, the function u, given by eq. (4.3), lies in the energy space only when 𝜆 ≥ 0. Therefore, having
excluded 𝜆 = 0 from consideration, we will be concerned with 𝜆 > 0 only. Denoting
(2m − 1)𝜋
def def 2n𝜋
𝜆(s)
m = , 𝜆n(a) = (4.6)
𝛼 𝛼
where m, n = 1, 2, … the solution can be written in the following form:
∑
∞
(s) ∑
∞
(a)
u= Am r 𝜆m cos 𝜆(s)
m𝜃 + Bn r 𝜆n sin 𝜆(a)
n 𝜃, r ≤ rc (4.7)
m=1 n=1
where rc is the radius of convergence of the infinite series.

Observe that the first sum in eq. (4.7) is a symmetric function with respect to the x axis, the second
is an antisymmetric function. If the solution is symmetric (with respect to the x axis) then Bn = 0,
if it is antisymmetric then Am = 0. Note that when 𝛼 = 𝜋∕k where k = 1, 2, … then the powers of r
are integers and u is an analytic function. For other values of 𝛼 the solution u is not analytic in the
corner point and when 𝛼 > 𝜋 then the first derivative of u with respect to r is infinity in the corner
(s)
point, provided that a1 ≠ 0, and u lies in Sobolev space H 1+𝜆1 −𝜖 (Ω) where 𝜖 > 0 is arbitrarily small.
Exercise 4.1 Consider the solution of Δu = 0 in the neighborhood of corner point B shown in
Fig. 4.1(a), and let u = 0 on ΓBC and the normal derivative 𝜕u∕𝜕n = 0 on ΓAB . Show that u can be
written as
∑∞
(2n − 1)𝜋
u= An r 𝜆n (cos 𝜆n 𝜃 + (−1)n sin 𝜆n 𝜃) where 𝜆n = ⋅
n=1
2𝛼
Hint: The condition: 𝜕u∕𝜕n = 0 on ΓAB is equivalent to 𝜕u∕𝜕𝜃 = 0.
Exercise 4.2 Construct the series expansion analogous to eq. (4.7) for the problem Δu = 0 in the
neighborhood of the corner point B shown in Fig. 4.1(a), given that 𝜕u∕𝜕n = 0 on ΓAB and ΓBC .
Exercise 4.3 Show that u(r, 𝜃), defined by eq. (4.3), is not in the energy space when 𝜆 < 0.
Exercise 4.4 Consider functions u = r 𝜆 F(𝜃, 𝜙) where r, 𝜃, 𝜙 are spherical coordinates centered
on a corner point and F is an analytic function. Show that 𝜕u∕𝜕r is square integrable in three dimen-
sions when 𝜆 > −1∕2.
4.2.1 2D model problem, uEX ∈ Hk (𝛀), k − 1 > p

We consider the L-shaped domain modified by removing a circular sector of radius r0 centered on
the origin. We denote this circular sector by Ω0 . The solution domain is defined by
def
Ω = {(X, Y ) | (X, Y ) ∈ (−1, 1)2 ⧵ [0, 1)2 ⧵ Ω0 }
where the backslash (⧵) is the set operator for subtraction. The solution domain and notation are
shown in Fig. 4.2(a).
Γ3 Y Y
Γ2
Γ4 Γ7 Γ1
X X
r
θ
Γ6
x
Γ5 y
(a) (b)
def
Figure 4.2 The L-shaped domain with a circular cut-out. (a) Notation. (b) The function Q = (qx + qy )FE
corresponding to r0 = 0.05 and p = 8.
The boundary conditions are defined so as to correspond to the exact solution:

uEX = r 2∕3 cos 2𝜃∕3. (4.8)
def
Specifically, the boundary conditions are as follows: u = 0 on Γ1 and Γ2 . On Γq = Γ3 ∪ Γ4 ∪ Γ5 ∪
Γ6 ∪ Γ7 flux qn is specified:
qn = −∇uEX ⋅ nj (4.9)
where nj is the unit normal to the boundary segment Γj (j = 3, 4, 5, 6). On the circular boundary
segment Γ7 eq. (4.9) takes the following form:
𝜕uEX
qn = ⋅
𝜕r
In this problem the singular point is outside of the solution domain hence the exact solution is an
analytic function on the entire domain. The smoothness of the solution increases with increasing
r0 . The exact value of the potential energy is computed from
1
𝜋exact = − q u ds. (4.10)
2 ∫Γq n EX
For r0 = 0.05 we get 𝜋exact = −0.90364617. We examine two discretization schemes in the following:
1. A sequence of uniform meshes characterized by h = 1, 1∕2, 1∕3, … with p = 2 (product space)
assigned to all elements. The a priori estimate for h-convergence on a uniform mesh is given by
eq. (1.91). In this case k − 1 > p and hence
C(k, p)
∥ uEX − uFE ∥E(Ω) ≤ C(k, p)hp ≈ (4.11)
N p∕2
where C(k, p) is a positive constant, independent of h, and we made use of the fact that in two
dimensions N ∝ h−2 . Therefore the asymptotic rate of convergence is 𝛽 = p∕2 = 1. As seen in
Fig. 4.3, this value is approached from above.
2. Uniform mesh of 6 elements with p ranging from 2 to 8 (product space). Since uEX is an analytic
function, the rate of convergence will be exponential. Plotting the relative error in energy norm
vs. N on log-log scale, the absolute value of the slope increases with N. This is visible in Fig. 4.3.
101
h=1
Relative error in energy norm (%)
p=2 uniform mesh

1/2
p = 2, βh = 1
3 1/3
1/4
4 1/5
100 1/6
5 1/7
1/8
6-element mesh 6
1
7
p - extension 1
h - extension 8
10−1
102 103
N
Figure 4.3 The L-shaped domain with a circular cut-out (r0 = 0.05). Comparison of two discretization
schemes.
The function
( )
def 𝜕uFE 𝜕uFE
Q = (qx + qy )FE = − +
𝜕x 𝜕y
corresponding to r0 = 0.05, computed on a 6-element mesh using p = 8 (product space), is shown in
Fig. 4.2(b). The exact solution is an analytic function, hence it can be expanded into a Taylor series
everywhere within the domain and on the boundaries of the domain. The error term depends on
the (p + 1)th derivative of uEX .
4.2.2 2D model problem, uEX ∈ Hk (𝛀), k − 1 ≤ p

We define our second model problem so that the first term of eq. (4.7), defined on the L-shaped
domain shown in Fig. 4.4(a), is the exact solution uEX . This is analogous to the one-dimensional
model problem given by equation (1.5) with the solution uEX = x𝛼 (1 − x) on the interval I = (0, 1).
Γ3 Y Y
(−1,1) (0,1)
Γ2
Γ4 Γ1 (0,0) (1,0)
X X
r
θ
Γ6
x (−1,−1) (1,−1)
Γ5 y
(a) (b)
Figure 4.4 The L-shaped domain. (a) Notation, (b) radically graded 27-element mesh corresponding to the
parameters M = 3 and 𝛾 = 2.
def
We assign the following boundary conditions: u = 0 on Γ1 and Γ2 . On Γq = Γ3 ∪ Γ4 ∪ Γ5 ∪ Γ6 flux
qn is specified consistent with the solution given by eq. (4.8), see eq. (4.9). The exact value of the
potential energy is computed from
1
𝜋exact = − q u ds = −0.9181133309. (4.12)
2 ∫Γq n EX
Except for the vertex located in the origin of the coordinate system, all derivatives of the solution
exist. The solution lies in Sobolev space H 5∕3−𝜖 (Ω), see Section A.2.4. Since the exact value of the
potential energy is known, the relative error in energy norm can be computed for each discretiza-
tion from eq. (1.100) and the realized rates of convergence can be computed from eq. (1.102). We
examine three discretization schemes in the following:
1. A sequence of uniform meshes with p = 2 (product space). The size of the elements is given by
1
hi = , i = 1, 2, … (4.13)
3i
The a priori estimate for h-convergence on a uniform mesh is given by eq. (1.91). In this case
k − 1 < p and hence
C(k, p)
∥ uEX − uFE ∥E(Ω) ≤ C(k, p)hk−1 ≈ (4.14)
N 1∕3
where C(k, p) is a positive constant, independent of h, and we made use of the fact that in two
dimensions N ∝ h−2 . Therefore the asymptotic rate of convergence is 𝛽 = 1∕3.
2. Uniform mesh, h = 1∕3 (27 elements), with p ranging from 2 to 8 (product space). Since the
singular point is a vertex point, the asymptotic rate of convergence of the p-version in energy
norm is twice that of the h-version, therefore 𝛽 = 2∕3. As seen in Fig. 4.5, this is indeed realized.
3. A fixed mesh of 27 elements with radical grading and p ranging from 1 to 8 (product space). The
grading of the radical mesh is controlled by a parameter 𝛾 > 0. The node points on the positive
100
p=1
10 uniform mesh
h = 1/3
1/6 p = 2, β = 1/3
1/12 1/18
2 1/24 1/30
1 3
4 uniform mesh
radical mesh
5 h = 1/3, β = 2/3
M = 3, γ = 4.9
6
0.1 7
8
p - extension
h - extension
0.01
102 103 104
N
Figure 4.5 The L-shaped domain. Comparison of three discretization schemes. Relative error in energy
norm (%).
x axis are
( )
i−1 𝛾
xi = , i = 1, 2, … , M + 1 (4.15)
M
where M is the number of elements that have boundaries on the positive x axis. This partitioning
is extended over the whole domain, as shown in Fig. 4.4(b), where the mesh corresponding to
the parameters M = 3, 𝛾 = 2 is shown. We selected 𝛾 = 4.9 because this value minimizes the
potential energy for the highest polynomial degree allowed by the software, in this case p = 8.
The results for the radical mesh are shown in Table 4.1.
The entry in the last column of Table 4.1 is the effectivity index of the error estimator. The effec-
tivity index, denoted by 𝜃, is defined as the ratio of the estimated relative error to the exact relative
error. The effectivity index can be computed only for those problems for which the exact solution
is known. Observe that the relative errors are slightly overestimated. This occurs because the esti-
mates are based on the assumption that the rate of convergence is algebraic, see eq. (1.92); however,
in the pre-asymptotic range it is stronger than algebraic.
The results for the three discretization schemes considered here are shown in Fig. 4.5. It is
seen that p-extension on a radically graded mesh is much more efficient than h-extension using
a sequence of uniform meshes or p-extension using a fixed uniform mesh. The reason for this is
that for the range of polynomial degrees 1 ≤ p ≤ 8 (supported by the software) radical meshing
overrefines the neighborhood of the singular corner and therefore the error is essentially coming
from the smooth part of the solution where the rate of convergence of p-extensions is exponential.
This is representative of the pre-asymptotic behavior of the p-version when strongly graded meshes
are used. As the p-level is increased, the convergence curve will slow to the asymptotic rate, in this
case 𝛽 = 2∕3. However, at that point the relative error in energy norm will be small.
This example illustrates an important point: in the finite element method the objective in select-
ing a discretization scheme is to achieve the desired accuracy in an efficient fashion. This is best
accomplished when the mesh is laid out in such a way that, for the range of p-values supported by
the software, p-extension is in the pre-asymptotic range.
Table 4.1 L-shaped domain, radical mesh (𝛾 = 4.9), 27 elements, product

space. Estimated and exact relative errors in energy norm.
𝜷 (er )E (%)
p N 𝝅p 𝜽
Est.’d Exact Est.’d Exact
1 33 −0.90505595 − − 11.93 11.93 1.00

2 120 −0.91529501 0.593 0.594 5.54 5.54 1.00
3 261 −0.91698177 0.586 0.587 3.52 3.51 1.00
4 456 −0.91753087 0.593 0.595 2.53 2.52 1.00
5 705 −0.91776880 0.598 0.603 1.95 1.94 1.01
6 1,008 −0.91789036 0.602 0.609 1.57 1.56 1.01
7 1,365 −0.91795963 0.603 0.614 1.31 1.29 1.01
8 1,776 −0.91800230 0.603 0.618 1.11 1.10 1.01
extrapolated: −0.91811642 − 0.667 − 0 −

Dirichlet boundary condition

Let us impose Dirichlet boundary conditions on all boundary segments. Since uEX is a sinusoidal
function and our basis functions are the mapped shape functions, it is necessary to first approximate
uEX by the basis functions on the element boundaries and then enforce the Dirichlet condition on
the coefficients of the basis functions. This procedure was outlined in Section 3.6. We are interested
in the size of the error coming from approximation of the essential boundary conditions.
Remark 4.1 The a posteriori estimator of error described in Section 1.5.3 is based on the assump-
tion that the h- or p-extension is in the asymptotic range, that is, h is sufficiently small or p is
sufficiently large to justify replacement of the ≤ sign with the ≈ sign in eq. (1.91). As we have seen
in this section, the pre-asymptotic rate of convergence is stronger than algebraic.
Exercise 4.5 Explain why the potential energy 𝜋p is negative when Neumann conditions are spec-
ified on Γq but positive when Dirichlet conditions are specified on Γ.
4.2.3 Computation of the flux vector in a given point

The computation flux vectors from the finite element solution in a given point is straightforward.
First, the standard coordinates (𝜉0 , 𝜂0 ) of the given point are determined, then the flux vector is
evaluated in point (𝜉0 , 𝜂0 ) from
{ (FE) } [ ] { }
qx kx kxy −1 𝜕uFE ∕𝜕𝜉
= − [J] (4.16)
qy(FE) kxy ky 𝜕uFE ∕𝜕𝜂
where [J] is the Jacobian matrix defined in Section 3.8.1.

Let us consider, for example the L-shaped domain with a circular cut-out shown in Fig. 4.6(a)
with Dirichlet boundary conditions applied on each boundary segment. The boundary conditions
correspond to the exact solution u = r 2∕3 cos(2𝜃∕3) (in the x y coordinate system).
Table 4.2 L-shaped domain, radical mesh (𝛾 = 4.9), 27 elements, product

space. Estimated and exact relative errors in energy norm.
𝜷 (er )E (%)
p N 𝝅p 𝜽
Est.’d Exact Est.’d Exact
1 16 0.95282287 0.000 19.444 19.444 1.00

2 85 0.91965811 0.932 0.932 4.102 4.102 1.00
3 208 0.91831905 1.127 1.126 1.497 1.497 1.00
4 385 0.91815493 1.300 1.298 0.672 0.673 1.00
5 616 0.91812462 1.394 1.388 0.349 0.351 1.00
6 901 0.91811717 1.442 1.419 0.202 0.204 0.99
7 1240 0.91811490 1.464 1.403 0.126 0.131 0.97
8 1633 0.91811408 1.467 1.330 0.084 0.091 0.93
extrapolated: 0.91811343 0.033 −

100
Relative error (%)

1.0
10
r1 = r0 flux vector
1
energy norm
r2 0.1
1.0
r3
0.01
x y 1 2 3 4 5 6 7
1.0 1.0 Grading exponent (m)
(a) (b)
Figure 4.6 The L-shaped domain with a circular cut-out. (a) 18-element mesh, radical grading in the radial
direction. (b) Relative errors.
The components of exact flux vector in the x, y coordinate system are

2 2
qx = − r −1∕3 cos(𝜃∕3), qy = − r −1∕3 sin(𝜃∕3).
3 3
We define the relative error in the flux vector as
√( )2 ( )2
qx − qx(FE) + qy − qy(FE)
def
efv = 100 −1∕3
(4.17)
2r1 ∕3
where the denominator is the maximum of the absolute value of the flux vector over the domain.
Specifically, we will be interested in the relative error of the flux vector in the point x = r0 , y = 0.
Since this point lies on a line of symmetry, the flux vector is in the direction of the line of symmetry
which is normal to the circular boundary.
We will use the 18-element mesh shown in Fig. 4.6(a) with the radii of the circles defined as
follows:
( )
i−1 m
ri = r1 + (1 − r1 ) , i = 1, 2, 3
3
i.e. radical grading, controlled by the grading exponent m, is used in the radial direction. Letting
r1 = r0 = 0.001 and m = 1, 2, …, p = 8 (product space) we compute the relative errors in the flux
vector in the point (r0 , 0) and the energy norm1 .
The results of computation are shown in Fig. 4.6(b). It is seen that the smallest error in energy
norm occurs at approximately m = 4 and the smallest error in the flux vector (efv ) at approximately
m = 5. At m < 4 the error in energy norm is small, whereas the error efv is large. The reason for this
is that the error in energy norm depends on the square integral of the error in the first derivatives
evaluated over the entire domain, whereas efv is evaluated in a particular point. In this example
the error in the derivatives is large in a small neighborhood of the circular cut-out when m < 4;
however, this error influences the energy norm of the error by a small amount only.
For m < 4 the error in the flux vector is coming mainly from under-refinement of the neighbor-
hood of the small cut-out, the region of primary interest, whereas for m > 5 the error is mainly
from the region of secondary interest.
1 The exact value of the potential energy is known: 𝜋EX = 0.91803479 for r0 = 0.001.
Remark 4.2 It is not difficult to find the value of the grading exponent that yields the minimum
error in energy norm. In general, this is not the grading exponent that is optimal for the quantity
of interest. It is reasonable to expect, however, that the error of approximation in the QoI will not
be far from that minimum. In this example this is the case indeed, as seen in Fig. 4.6(b).
4.2.4 Computation of the flux intensity factors

An algorithm for the computation of the coefficients of the asymptotic expansions from finite
element solutions is described in the following. The algorithm is based on (a) the existence of a
path-independent integral and (b) the orthogonality of the eigenfunctions. The coefficients are
called flux intensity factors.
Consider a two-dimensional domain Ω with boundary Γ. For any two functions in E(Ω) we have:
Δu 𝑣 dxdy = (∇u ⋅ n)𝑣 ds − (∇𝑣 ⋅ n)u ds + Δ𝑣 u dxdy

∫Ω ∮Γ ∮Γ ∫Ω
where we have applied the divergence theorem twice. When both u and 𝑣 satisfy the Laplace
equation, that is Δu = 0 and Δ𝑣 = 0, then this equation becomes:
(∇u ⋅ n)𝑣 ds = (∇𝑣 ⋅ n)u ds. (4.18)

∮Γ ∮Γ
This equation is applicable to Ω and any subdomain of Ω.
Path-independent integral
Now consider a subdomain Ω★ in the neighborhood of a corner point, represented by the shaded
region in Fig. 4.7. Assume that either u = 0 or ∇u ⋅ n = 0 and either 𝑣 = 0 or ∇𝑣 ⋅ n = 0 on Γ★
2
and
Γ★
4
. Then eq. (4.18) becomes:
(∇u ⋅ n)𝑣 ds + (∇u ⋅ n)𝑣 ds = (∇𝑣 ⋅ n)u ds + (∇𝑣 ⋅ n)u ds

∫Γ★ ∫Γ★ ∫Γ★ ∫Γ★
1 3 1 3
which is equivalent to
(∇u ⋅ n)𝑣 ds − (∇𝑣 ⋅ n)u ds = − (∇u ⋅ n)𝑣 ds + (∇𝑣 ⋅ n)u ds.

∫Γ★ ∫Γ★ ∫Γ★ ∫Γ★
1 1 3 3
Observe that integration along Γ★

is clockwise about the corner point whereas integration along
1
Γ★
3
is counterclockwise. Reversing the direction of integration along Γ★
1 so that both integrals
are counterclockwise about the corner point, we find that the two integrals are equal, and
Figure 4.7 Definition of Ω★ .

y

Γ3
Γ4

Γ Γ 1
x

Γ2 Ω
since Ω★ is arbitrary, we may select an arbitrary counterclockwise path Γ★ and the integral
expression
def
IΓ★ = − (∇u ⋅ n)𝑣 ds + (∇𝑣 ⋅ n)u ds (4.19)
∫Γ★ ∫Γ★
will be path-independent.
Orthogonality
Let u = r 𝜆i 𝜙i (𝜃); 𝑣 = r 𝜆j 𝜙j (𝜃), and Γ★ = Γ𝜚 where Γ𝜚 is a circular path of radius 𝜚, centered on the
corner point. Then on Γ𝜚 we have:
( )
𝜕u
∇u ⋅ n = = 𝜆i 𝜚𝜆i −1 𝜙i (𝜃)
𝜕r r=𝜚
( )
𝜕𝑣
∇𝑣 ⋅ n = = 𝜆j 𝜚𝜆j −1 𝜙j (𝜃).
𝜕r r=𝜚
Using ds = 𝜚 d𝜃, eq. (4.19) can be written as
𝛼∕2
IΓ𝜚 = (𝜆j − 𝜆i )𝜚𝜆i +𝜆j 𝜙i (𝜃)𝜙j (𝜃) d𝜃.
∫−𝛼∕2
Since IΓ𝜚 is path-independent, the integral expression must be zero when 𝜆j ≠ ±𝜆i . Note that since
Ω★ does not include the corner point, solutions corresponding to the negative eigenvalues are in
the energy space. In the following we will denote
𝛼∕2 𝛼∕2
def def
Cij = 𝜙i (𝜃)𝜙j (𝜃) d𝜃 and Cij− = 𝜙i (𝜃)𝜙−j (𝜃) d𝜃 (4.20)
∫−𝛼∕2 ∫−𝛼∕2
where 𝜙−j (𝜃) is the eigenfunction corresponding to −𝜆j . The eigenfunctions are orthogonal in the
sense that Cij = 0 when 𝜙j ≠ 𝜙i and 𝜙j ≠ 𝜙−i .
Remark 4.3 In the case of the Laplace operator all eigenvalues are real and simple.
Exercise 4.6 Show that for the asymptotic expression given by eq. (4.7)
{
𝛼∕2 if i = j
Cij =
0 if i ≠ j.
Extraction of Ak
Using the orthogonality property of the eigenfunctions, it is possible to extract approximate values
of the coefficients Ak , k = 1, 2, … from a finite element solution. Let us consider the asymptotic
expansion
∑
∞
uEX = Ai r 𝜆i 𝜙i (𝜃). (4.21)
i=1
Suppose that we are interested in computing Ak . We then define the extraction function 𝑤k as
follows:
def
𝑤k = r −𝜆k 𝜙−k (𝜃)
and evaluate the path-independent integral on a circular path Γ𝜚 centered on the corner point:
IΓ𝜚 (uEX , 𝑤k ) = − (∇uEX ⋅ n)𝑤k ds + (∇𝑤k ⋅ n)uEX ds. (4.22)

∫Γ𝜚 ∫Γ𝜚
It is now left to the reader to show that, utilizing the orthogonality property of the eigenfunctions,
we get:
1
Ak = − − IΓ𝜚 (uEX , 𝑤k ). (4.23)
2Ckk 𝜆k
In the finite element method uEX is replaced by uFE to obtain an approximate value for Ak . This
method, called the contour integral method, is very efficient, as illustrated by the following
example.
Example 4.1 Let us consider the L-shaped domain problem shown in Fig. 4.4 with the bound-
ary conditions u = 0 on Γ1 and Γ2 . On all other boundary segments qn corresponding to the exact
solution
uEX = r 2∕3 cos(2𝜃∕3) (4.24)
is prescribed. Note that this is the leading term of the symmetric part of the expansion given by
eq. (4.7). This example illustrates that if we substitute uFE for uEX in eq. (4.22) then we will get a
good approximation to the exact value of the coefficient A1 which, in this example, was chosen to
be unity.
The extraction function is
𝑤1 (𝜚, 𝜃) = 𝜚−2∕3 cos(2𝜃∕3) (4.25)
and
2
(∇𝑤1 ⋅ n)r = 𝜚 = − 𝜚−5∕3 cos(2𝜃∕3) (4.26)
3
where 𝜚 is the radius of the circle used for extraction.
3𝜋∕4
−
C11 = cos2 (2𝜃∕3) d𝜃 = 3𝜋∕4.
∫−3𝜋∕4
The approximate value of A1 is computed from eq. (4.22) which in this case takes the form
( 3𝜋∕4
1
A1(FE) = − − 𝜚1∕3 (∇uFE ⋅ n) cos(2𝜃∕3) d𝜃 +
2C11 𝜆1 ∫−3𝜋∕4
3𝜋∕4 )
2 −2∕3
𝜚 cos(2𝜃∕3)uFE d𝜃 . (4.27)
3 ∫−3𝜋∕4
We compute uFE and ∇uFE ⋅ n along an arbitrary circular path of radius 𝜚 in integration points and
compute A1(FE) numerically. The integrands are smooth functions therefore the error associated with
numerical integration can be controlled very efficiently. For this example we used a uniform mesh
of 27 elements, h = 1∕3 and an extraction circle of radius 𝜚 = 0.2. The results of computation for
p = 1 to p = 5 (product space) are shown in Table 4.3
It is seen that the relative error in A(FE)

1 decreases much faster than the relative error in energy
norm. For that reason, the contour integral method is said to be superconvergent.
Table 4.3 Example 4.1: The results of extraction by the contour integral method using 𝜚 = 0.2.
polynomial degree (p) 1 2 3 4 5
A(FE)
1 1.0242 1.0075 0.9920 1.0027 0.9987
Rel. error of A(FE)
1 (%) 2.42 0.75 0.80 0.27 0.13
Rel. error in energy norm (%) 11.93 5.54 3.52 2.53 1.95
ϕ1(𝜃) ϕ2(𝜃)
y
ΓAC 1.0 1.0
0.8 ϕ1(𝜃) 0.5
C ΓBC y′
0.6
45° 0
A 0.4
𝜃 ϕ2(𝜃)
0.2 −0.5
ΓAB B x
22.5° x′ 0 −0.1
−0.2
Ω
−200° −100° 0 100°
𝜃 – 22.5°
(a) (b)
Figure 4.8 Example 4.2: The first two normalized eigenfunctions.
Example 4.2 We refer to the results of Exercise 4.1 and construct a model problem on the domain
shown in Fig. 4.8(a) so that the exact solution is a linear combination of the first two terms of the
asymptotic expansion:
uEX = a1 r 𝜆1 (cos 𝜆1 𝜃 − sin 𝜆1 𝜃) + a2 r 𝜆2 (cos 𝜆2 𝜃 + sin 𝜆2 𝜃)
⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟ ⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟
𝜙1 (𝜃) 𝜙2 (𝜃)
where 𝜆1 = 𝜋∕(2𝛼), 𝜆2 = 3𝜋∕(2𝛼). Let 𝛼 = 7𝜋∕4 = 315∘ . The boundary conditions are: 𝜕u∕𝜕n = 0
on ΓAB ; u = 0 on ΓBC , and on ΓAC , i.e., the flux corresponding to uEX is specified:
qn = −a1 𝜆1 r 𝜆1 −1 (cos 𝜆1 𝜃 − sin 𝜆1 𝜃) − a2 𝜆2 r 𝜆2 −1 (cos 𝜆2 𝜃 + sin 𝜆2 𝜃).
We normalize the eigenfunctions as follows. Let 𝜃i be the angle where the absolute value of 𝜙i (𝜃)
is maximum:
def
𝜃i = arg max |𝜙i (𝜃)| where I𝛼 = {𝜃 | − 𝛼∕2 ≤ 𝜃 ≤ +𝛼∕2}.
𝜃∈I𝛼
The normalized eigenfunctions are defined such that the maximum value of 𝜑i (𝜃) on I𝛼 is unity:
def
𝜑i (𝜃) = 𝜙i (𝜃)∕𝜙i (𝜃i ). (4.28)
The functions 𝜑1 (𝜃) and 𝜑2 (𝜃) are shown in Fig. 4.8(b). If we let a1 = 1.0, a2 = 0 then, using a
16-element mesh with one layer of geometrically graded elements around the corner point, trunk
space, the estimated relative error in energy norm is 17.21% at p = 8. Using the method of extraction
described in this section, the computed value of the coefficient of the first normalized eigenfunction
is A1 = 1.348. Its exact value is 1.414, therefore the relative error is 4.66%.
Exercise 4.7 For the problem described in Example 4.2 let a1 = 0 and a2 = 1.0. Determine the
approximate value of the coefficient of the second normalized eigenfunction A2 and estimate the
relative error. The exact value is −1.414.
4.2.5 Material interfaces

A schematic view of the material interface problem in two dimensions is shown in Fig. 4.9. The
shadings represent different materials of constant thermal conductivity.
Figure 4.9 Multi-material interface, notation.

y
Γϱ
C
ΓBC
B
ΓAB x
The intersection of material interfaces with boundaries are singular points. The solution in the
neighborhood of these singular points is characterized by functions that are of the form u = r 𝜆 F(𝜃),
as before. However, here F(𝜃) is a piecewise analytic function.
Since the solution has to satisfy the Laplace equation on each sector, the solution on the kth sector
can be written as
u(k) = r 𝜆 (ak cos(𝜆𝜃) + bk sin(𝜆𝜃)), k = 1, 2, … , n (4.29)
where n is the number of sectors. We assume that homogeneous Dirichlet or Neumann boundary
conditions are prescribed on ΓAB and ΓBC . Imposing continuity on u(k) , and on the flux normal
to the material interface i.e. u(k−1) (r, 𝜃k ) = u(k) (r, 𝜃k ) and qk−1
n (r, 𝜃k ) = −qn (r, 𝜃k ) and enforcing the
k
homogeneous boundary conditions, results in a system of 2n homogeneous equations in ak and bk .

The procedure is illustrated by the following example.
Example 4.3 An aluminum plate with coefficient of thermal conductivity kal = 202 W/(mK) is
bonded to a chrome-nickel steel plate with kcn = 16.3 W/(mK). The edges are offset, forming the
corner shown in Fig. 4.10(a). We assume that the boundaries that lie on the x and y axes are perfectly
insulated.
Four equations are necessary for the enforcement of the homogeneous boundary conditions and
the continuity of the temperature and flux at the material interface:
⎡ − sin(𝜆𝜋∕2) 0 cos(𝜆𝜋∕2) 0 ⎤ ⎧a 1 ⎫
⎢ ⎥⎪ ⎪
⎢ 0 − sin(2𝜆𝜋) 0 cos(2𝜆𝜋) ⎪
⎥⎨ ⎪a 2
sin(𝜆𝜋) ⎥ ⎪b1 ⎬
= 0. (4.30)
⎢ cos(𝜆𝜋) cos(𝜆𝜋) sin(𝜆𝜋) ⎪
⎢−k sin(𝜆𝜋) −k sin(𝜆𝜋) k cos(𝜆𝜋) k cos(𝜆𝜋)⎥ ⎪b ⎪
⎣ al cn al cn ⎦ ⎩ 2⎭
300
y
200
Al
determinant
100
0
x
−100
CrNi steel −200
−300
0 λ1 2 3 4 5 6
λ
(a) (b)
Figure 4.10 Example 4.3. (a) Notation. (b) Value of the determinant of the matrix in eq. (4.30).
These equations have a non-trivial solution only if the determinant of the coefficient matrix is
zero. There are infinitely many eigenvalues. The determinant is plotted as a function of 𝜆 in the
interval 0 < 𝜆 < 5 in Fig. 4.10(b) where the first seven roots are indicated by open circles. The first
eigenvalue is 𝜆1 = 0.5238.
Exercise 4.8 Plot the eigenfunction corresponding to 𝜆1 in Example 4.3. Scale the eigenfunction
so that its maximum value is unity.
Exercise 4.9 Determine the second eigenvalue and the corresponding eigenfunction for the prob-
lem of Example 4.3. Partial answer: 𝜆2 = 1.4762.
Exercise 4.10 Find the first and second eigenvalues for the problem in Example 4.3 modified
so that on the boundaries that lie on the x- and y-axes have zero temperature prescribed. Partial
answer: 𝜆1 = 0.8762.
The Steklov method

It is possible to determine the eigenpairs numerically using a method known as the Steklov
method2 . This method is based on the idea that on a circular boundary of radius 𝜚, shown in
Fig. 4.1(b), the normal derivative of the function u = r 𝜆 F(𝜃) is:
( ) ( )
𝜕u 𝜕u 𝜆
= = 𝜆𝜚𝜆−1 F(𝜃) = u. (4.31)
𝜕n Γ𝜚 𝜕r r=𝜚 𝜚
Consider problems where two or more materials are bonded and the material interfaces3 intersect
the boundary in the origin, see Fig. 4.1(b). Homogeneous natural or essential boundary conditions
are prescribed on ΓAB and ΓBC .
The generalized form of eq. (2.34), subject to the assumptions that Q = 0, steady state conditions
exist, and either u = 0 or 𝜕u∕𝜕n = 0 on ΓAB and ΓBC , is:
𝜆
grad𝑣[𝜅]gradu dxdy = u𝑣 ds for all 𝑣 ∈ E0 (Ω𝜚 ) (4.32)
∫Ω𝜚 𝜚 ∫Γ𝜚
where [𝜅(x, y)] is a positive-definite matrix of material properties which are discontinuous at the
material interfaces4 and Ω𝜚 is defined in Fig. 4.1(b). Equation (4.32) is an eigenvalue problem. The
eigenvalues are real numbers. On solving eq. (4.32) by the finite element method, the eigenfunc-
tions are approximations of F(𝜃) by the (mapped) piecewise polynomial functions. The size of the
numerical characteristic value problem can be reduced to the number of degrees of freedom associ-
ated with the contour Γ𝜚 . For additional discussion on the Steklov method and illustrative examples
we refer to [112].
4.3 The Laplace equation in three dimensions

In this section we consider an extension of the L-shaped domain problem to three dimensions. The
domain, shown in Fig. 4.11, is known as the Fichera cube5 . The Fichera cube is extensively used in
2 Vladimir Andreevich Steklov 1864–1926.

3 The treatment of curved interfaces is not considered here.
4 The elements of the matrix [𝜅] are usually piecewise constant functions.
5 Gaetano Fichera 1922–1996.
Y Y
Z Z
X X
(a) (b)
Figure 4.11 The Fichera domain, 189-element mesh. (a) Uniform mesh: M = 3, 𝛾 = 1, h = 1∕3. (b) Radical
grading with parameters M = 3, 𝛾 = 2.5.
benchmarking various discretization and error control procedures. The domain is defined by
def
Ω = {(X, Y , Z) | (X, Y , Z) ∈ (−1, 1)3 ⧵ [0, 1)3 }. (4.33)
The boundary surfaces lie in the planes X = 0, Y = 0, Z = 0 and X = ±1, Y = ±1, Z = ±1. On the
boundaries X = 0 and Z = 0 homogeneous Dirichlet boundary condition (u = 0) is prescribed, on
boundary Y = 0 homogeneous Neumann condition (qn = 0), on the L-shaped boundaries X = 1,
Y = 1, Z = 1, the Neumann condition qn = 0.1 and on the boundaries X = −1, Y = −1, Z = −1 the
condition qn = −0.5 are prescribed.
In two dimensions it was possible to characterize the regularity of uEX by a single parameter
𝜆. This is not possible in three dimensions, nevertheless, one can expect that the singularities on
the edges coincident with the coordinate axes are similar to the vertex singularity of the L-shaped
domain, the coefficients of the expansion given by eq. (4.7) being functions of the edge coordinate,
and convergence will be characterized by those singularities. See Remark 4.4. Theoretical analysis
of the singularities on domains with piecewise smooth boundaries is addressed in [62].
The exact solution of this problem is not known. The reference value of the potential energy was
estimated by extrapolation, from the results of p-extension on a 189-element radically graded mesh
using the procedure described in Section 1.5.3. The data are listed in Table 4.4.
The grading parameter for the radical mesh was 𝛾 = 6.2. This parameter was chosen because it
minimizes the potential energy at p = 8.
For h-extension, a sequence of uniform mesh was used and the rate of convergence was estimated
with reference to the extrapolated value of the potential energy in Table 4.4. The results are listed
in Table 4.5.
For p-extension, a 189-element uniform mesh was used and the rate of convergence was esti-
mated with reference to the extrapolated value of the potential energy in Table 4.4. The results are
listed in Table 4.6.
The estimated relative error in energy norm vs. degrees of freedom curves are plotted on log-log
scale for h-extension and p-extension on uniform meshes and p-extension on a radical mesh in
Fig. 4.12.
Table 4.4 The Fichera domain. p-convergence,

radical mesh 𝛾 = 6.2, M(Δ) = 189, product space.
p N 𝝅p 𝜷p (er )E (%)
1 288 −3.96897295 − 32.78

2 1,890 −4.34971145 0.423 14.78
3 5,940 −4.41536773 0.492 8.42
4 13,572 −4.43381736 0.533 5.42
5 25,920 −4.44057198 0.562 3.77
6 44,118 −4.44342415 0.565 2.79
7 69,300 −4.44482967 0.577 2.15
8 102,600 −4.44557782 0.577 1.71
extrapolated: −4.44688294 − −
Table 4.5 Fichera domain. h-convergence, uniform mesh refinement,

p = 2 (product space).
h M(𝚫) N 𝝅h kh 𝜷h (er )E (%)
1/3 189 1,890 −4.21669217 0 0 22.75

1/6 1,512 13,572 −4.30253820 1.337 0.118 18.02
1/9 5,103 44,118 −4.33699192 1.336 0.116 15.72
1/12 12,096 102,600 −4.35632002 1.336 0.115 14.27
1/15 23,625 198,090 −4.36893882 1.336 0.114 13.24
1/18 40,824 339,660 −4.37793385 1.336 0.114 12.45
1/21 64,827 536,382 −4.38472583 1.336 0.113 11.82
1/24 96,768 797,328 −4.39006756 1.336 0.113 11.30
1/27 137,781 1,131,570 −4.39439827 1.337 0.113 10.86
1/30 189,000 1,548,180 −4.39799291 1.337 0.113 10.49
1/33 251,559 2,056,230 −4.40103312 1.337 0.113 10.15
1/36 326,592 2,664,792 −4.40364413 1.337 0.113 9.86
1/39 415,233 3,382,938 −4.40591530 1.337 0.113 9.60
reference: −4.44688294 − − −
For the h-version, quasiuniform meshes, we can write eq. (1.91) as

√
𝜋h − 𝜋 ≈ Ch (k, p, uEX )hk−1 (4.34)
where 𝜋h is the potential energy corresponding to the mesh characterized by h, 𝜋 is the (unknown)
exact value of the potential energy, Ch is a constant, independent of h, and k is the index of the
Sobolev space in which uEX lies. Here we used the reference value of 𝜋ref = −4.44688294 (see
Table 4.4) to approximate the exact value of 𝜋.
Table 4.6 The Fichera domain. p-convergence, uniform mesh, M(Δ) = 189, product space.
p N 𝝅p 𝜷p (er )E (%)
1 288 −3.96101944 − 33.05

2 1890 −4.21669217 0.199 22.75
3 5940 −4.30016800 0.197 18.16
4 13572 −4.34150404 0.200 15.39
5 25920 −4.36584861 0.203 13.50
6 44118 −4.38173819 0.205 12.10
7 69300 −4.39284714 0.207 11.02
8 102600 −4.40100653 0.209 10.16
reference: −4.44688294 − −
100
p - extension
h - extension
p=1
Estimated relative error
uniform mesh
in energy norn (%)
h = 1/3 p = 2, βh = 0.113
1/6
1/9
1/15 1/21
2 1/30 1/39
10
3 uniform mesh
4 h = 1/3, βp = 0.209
radical mesh
M = 3, γ = 6.2 5
6
7
8
1
102 103 104 105 106 107
N
Figure 4.12 The Laplace problem on the Fichera domain. Comparison of three discretization schemes.
We estimate the index k from the sequence of finite element solutions of the Fichera problem
shown in Table 4.5 where the column under the heading kh shows the estimated values of k
obtained by the following formula based on eq. (4.34):
1 log(𝜋(hi ) − 𝜋ref ) − log(𝜋(hi−1 ) − 𝜋ref )
kh(i) = 1 + (4.35)
2 log(hi ) − log(hi−1 )
where the index i refers to the ith entry in the table. The results indicate that
lim kh ≈ 1.337. (4.36)
h→0
The rates of convergence computed using the method described in Section 1.5.3 are listed in column
𝛽h . The estimate of 𝛽h is based on eq. (1.92) which can be written as
√ C
𝜋h − 𝜋 ≈ 𝛽 ⋅ (4.37)
N h
In three dimensions we have the proportionality N ∝ h−3 therefore we expect
kh − 1
𝛽h ≈ ≈ 0.113. (4.38)
3
As seen in Table 4.5, this is indeed the case.

Referring to Table 4.6, we see that the estimated rate of p-convergence is 𝛽p ≈ 0.209 and increas-
ing. From this data we may conjecture either that in three dimensions, when all singular arcs are
coincident with element edges, a condition that is usually satisfied in applications of the finite ele-
ment method, 𝛽h < 𝛽p < 2𝛽h , or that 𝛽p → 2𝛽h , as in two dimensions; however, this will occur at
very high values of N. Additional empirical data developed for a problem of elasticity on the Fichera
domain suggests that the second conjecture is more likely to be true. See Appendix C.
Remark 4.4 In Section 4.2 we considered asymptotic expansions of the form given by eq. (4.3).
The coefficients a and b were chosen so as to satisfy the boundary conditions on the intersect-
ing edges. In three dimensions we assume the same functional form along the intersection of two
plane surfaces; however, the coefficients will be functions of the edgewise coordinates. For example,
referring to Fig. 4.11, along the X axis we have
u = r 𝜆 (a(X) cos(𝜆𝜃) + b(X) sin(𝜆𝜃) (4.39)
which has to satisfy the strong form of the Laplace equation, written in cylindrical form;
𝜕 2 u 1 𝜕u 1 𝜕2 u 𝜕2 u
Δu ≡ + + 2 2 + = 0. (4.40)
𝜕r 2 r 𝜕r r 𝜕𝜃 𝜕X 2
This occurs only if a(X) and b(X) are linear functions. If a(X) and b(X) are polynomials of degree
greater than 1 then the functional form of eq. (4.39) has to be augmented by functions, called
shadow functions, defined so that eq. (4.39) is satisfied. Importantly, the shadow functions are
smoother than the corresponding eigenfunctions for the L-shaped domain. Therefore the shadow
functions do not affect the rate of convergence. For details we refer to [112].
4.4 Planar elasticity

The analysis of corner singularities in two-dimensional elasticity is analogous to the way the corner
singularities of the Laplace problem were treated in Section 4.2. However, it is complicated by the
fact that not all eigenvalues are real or simple. Details are given in Appendix G.
4.4.1 Problems of elasticity on an L-shaped domain

The analysis of the corner singularity of an L-shaped domain problem of two-dimensional elasticity,
assuming that the intersecting boundary segments are stress-free, is outlined in Section G.2.3 in the
appendix. The lowest positive eigenvalue is 𝜆1 = 0.54448374. Therefore the exact solution lies in the
Sobolev space H 1+𝜆1 −𝜖 (Ω), hence we expect that the rate of h-convergence on uniform mesh will be
𝛽h = 𝜆1 ∕2 and the rate of p-convergence will be 𝛽p = 𝜆1 .
The stress field corresponding to the first term of the asymptotic expansion is given by eq. (G.28)
and the displacement field is given by eq. (G.29). From this information the exact value of the
potential energy is determined from the integral
1 [ ]
𝜋(uEX ) = − ux (𝜎x nx + 𝜏xy ny ) + 𝜏xy nx + 𝜎y ny ) tz ds
2 ∮
a2 𝓁 2𝜆1 tz
= − 4.15454423 1 (4.41)
E
where nx , ny are the unit normals to the boundary in the x, y coordinate system shown in Fig. 4.4,
𝓁 is a length scale, tz is the thickness (constant) and E is the modulus of elasticity. Plane strain
conditions were assumed with 𝜈 = 0.3.
100
p=1
h = 1/3 uniform mesh
1/6 p = 2, βh = 0.272
10 1/12
1/18
2
1/24 1/30
3
uniform mesh
radical mesh 4 h = 1/3, βp = 0.544
M = 3, γ = 5.0 5
1
6
7
p - extension 8
h - extension
0.1
102 103 104
N
Figure 4.13 A problem of elasticity on an L-shaped domain. Comparison of three discretization schemes.
Plane strain, 𝜈 = 0.3, stress-free boundary conditions on the re-entrant edges.
The results computation for (a) h-extension using a sequence of uniform meshes with h ranging
from 1∕3 to 1∕30, p = 2, product space, (b) p-extension using a 27-element uniform mesh with p
ranging from 1 to 8, product space, and (c) p-extension on a 27-element radical mesh with 𝛾 = 5.0,
product space, are shown in Fig. 4.13. The realized rates of convergence, 𝛽h and 𝛽p are very close to
the theoretical estimates.
The estimated and exact convergence rates and relative errors in energy norm are listed in
Table 4.7 for p-extension on the 27-element radical mesh. The exact values were computed using
the exact value of the potential energy given by eq. (4.41). The estimated values are based on the
extrapolated value of the potential energy. In the last row the error in the extrapolated value is
shown.
Table 4.7 L-shaped domain, free-free boundary conditions, plane strain,

𝜈 = 0.3, radical mesh (𝛾 = 5.0), 27 elements, product space. Estimated and
exact relative errors in energy norm.
𝝅p E 𝜷 (er )E (%)
p N 𝜽
a12 𝓵 2𝝀1 tz Est.’d Exact Est.’d Exact
1 77 −3.87071791 − − 26.137 26.138 1.00

2 263 −4.10911566 0.746 0.746 10.455 10.457 1.00
3 557 −4.14439620 0.999 0.999 4.939 4.942 1.00
4 959 −4.15172126 1.181 1.177 2.600 2.607 1.00
5 1469 −4.15355459 1.241 1.229 1.531 1.543 0.99
6 2087 −4.15412877 1.267 1.236 0.981 1.000 0.98
7 2813 −4.15434288 1.283 1.213 0.669 0.696 0.96
8 3647 −4.15443336 1.283 1.149 0.480 0.517 0.93
extrapolated: −4.15454423 0.192 −

Exercise 4.11 Construct a 27-element mesh for the L-shape domain, similar to the mesh shown
in Fig. 4.4. However, use geometric grading: locate nodal points along the intersecting boundary
segments using eq. (1.57) and extend the mesh to the entire domain. Let q = 0.07. Compare the
relative errors with those in Table 4.7. You will find that the two grading patterns produce similar
results.
4.4.2 Crack tip singularities in 2D

Crack tip singularities have a great practical importance. This is because in damage-tolerant design
it is assumed that small cracks are present at critical locations of a structure at the beginning of its
service life. When the structure is subjected to cyclic loads then the cracks grow. The objective of
design is to ensure that the cracks will not grow by more than a fraction of their critical lengths
between inspection intervals.
The rate of crack growth is correlated with the coefficients of the leading terms of the asymptotic
expansion. This is justified by the idea that the highly nonlinear process of crack growth is driven by
the surrounding elastic stress field, and furthermore, the volume in which the nonlinear processes
take place is sufficiently small so that only the leading terms of the asymptotic expansion have to
be considered. The coefficients of the leading terms are proportional to the stress intensity factors
which are the quantities of interest in linear elastic fracture mechanics (LEFM). This section is
concerned with the extraction of stress intensity factors by the contour integral method, which is
analogous to the method described in Section 4.2.4.
For cracks (𝛼 = 2𝜋) the equations (G.18) and (G.19) in the appendix reduce to one equation:
n
sin 2𝜆𝜋 = 0 therefore 𝜆n = ± , n = 1, 2, 3, …
2
hence all roots are real and simple. The goal is to compute the coefficients of the first terms of the
symmetric (mode I) and antisymmetric (mode II) expansions.
In the engineering literature it is customary write the Cartesian components of the Mode I stress
tensor corresponding in the following form:
( )
K 𝜃 𝜃 3𝜃
𝜎x = √ I cos 1 − sin sin + T + O(r 3∕2 ) (4.42)
2𝜋r 2 2 2
( )
K 𝜃 𝜃 3𝜃
𝜎y = √ I cos 1 + sin sin + O(r 3∕2 ) (4.43)
2𝜋r 2 2 2
KI 𝜃 𝜃 3𝜃
𝜏xy = √ sin cos cos + O(r 3∕2 ) (4.44)
2𝜋r 2 2 2
where −𝜋 ≤ 𝜃 ≤ 𝜋, T is a constant, called the T-stress, see Section H.1. The constant KI is called
the mode I stress intensity factor.
The antisymmetric (Mode II) stress tensor components are usually written in the following form:
( )
K 𝜃 𝜃 3𝜃
𝜎x = − √ II sin 2 + cos cos + O(r 3∕2 ) (4.45)
2𝜋r 2 2 2
K 𝜃 𝜃 3𝜃
𝜎y = √ II sin cos cos + O(r 3∕2 ) (4.46)
2𝜋r 2 2 2
( )
K 𝜃 𝜃 3𝜃
𝜏xy = √ II cos 1 − sin sin + O(r 3∕2 ) (4.47)
2𝜋r 2 2 2
where KII is called the mode II stress intensity factor.
Computation of stress intensity factors

The contour integral method (CIM) for the Navier equations, analogously to the CIM for the
Laplace equation outlined in Section 4.2.4, is based on the existence of a path-independent integral
and the orthogonality of eigenfunctions. Details on the formulation the 2D elasticity problem are
available in [96].
We will consider the Navier equation under the assumption that the volume forces are zero and
the temperature is constant. The path-independent contour integral is:
def
IΓ = (Tx(u) 𝑤x + Ty(u) 𝑤y ) ds − (Tx(w) ux + Ty(w) uy ) ds (4.48)
∫Γ ∫Γ
where the superscript u (resp. w) represents the exact solution (resp. a test function that satisfies
the Navier equation and the homogeneous boundary conditions on the edges that intersect in the
singular point), Γ is an arbitrary contour that begins on one edge and runs in the counterclockwise
direction to the other, as shown in Fig. 4.7. We will assume that Γ is a circular contour, the radius
of which is arbitrary, and we will be interested in the computation of KI , KII and T. Details are
presented in Appendix H.
Under special conditions the stress intensity factors can be determined from the energy release
rate. The procedure for computing stress intensity factors from the energy release rate is described
in Appendix H.
Example 4.4 The plane strain fracture toughness, denoted by KIc , is defined as the value of the
stress intensity factor at which the propagation of cracks becomes rapid and unbounded. It is con-
sidered to be a material property that quantifies the resistance of materials to crack propagation.
Procedures for the measurement of KIc are governed by various standards6 . Compact tension
(CT) specimens and single edge notched bend (SENB) specimens are commonly used. A typical
CT specimen with an initial crack of length a is shown in Fig. 4.14(a). The initial crack is produced
through stable fatigue crack growth.
In this example we consider the following question: Since plane strain cannot be realized in an
experiment, what is the difference between KI of a plane strain model and KI along the crack front
of a CT specimen.
4.0R
60
30
5
A B
12.5 A B
t = 25
a (b)
w = 50
62.5
(a)
Figure 4.14 (a) A typical compact tension test specimen. (b) A typical finite element mesh.
6 See ASTM E399: Standard test method for linear-elastic plane-strain fracture toughness KIc of metallic materials.
See also ASTM E1820, ISO 12737, ISO 12135.
60
3D
KI (MPa mm1/2)
D 55 54.45
plane strain or stress
50
45
A surface
C 40
0 5 10 12.5
B Distance from the plane of symmetry CDE (mm)
(a) (b) (c)
Figure 4.15 One quarter of a compact tension test specimen. (a) Mesh detail, exploded view. (b) 27
element mesh, geometric grading, q = 0.15, (c) computed values of KI for 1 kN applied force and a = 25 mm.
The material is an aluminum alloy with the elastic properties E = 71.7 GPa, 𝜈 = 0.333. Letting
a = 25 mm and using the contour integral method, for an applied force of 1 kN we find that
KI = 54.45 MPa mm1∕2 . This result was confirmed using a 27-element geometrically graded finite
element mesh.
The 27-element mesh was extruded to obtain a three-dimensional representation of the CT spec-
imen, see Fig. 4.15 (b). Since the CT specimen has two planes of symmetry, it is sufficient to use
the quarter model shown in Fig. 4.15(b). Symmetry is prescribed on the rectangular area, the three
vertices A, B, C of which are visible, and on the plane CDE. The cylindrical surface was loaded
by distributed sinusoidal normal tractions, the resultant of which is 1 kN. The other surfaces are
traction-free.
The stress intensity factor KI was computed at 11 equally spaced points along the crack front by
the contour integral method. At each point a sequence of 6 extraction circles were used with the
radii ranging from 0.6 mm (just outside of the innermost elements shown in Fig. 4.15(a)) to 0.8 mm.
The resulting values were extrapolated to zero radius by linear regression. In three dimensions the
finite element solution is affected by the shadow functions (see Remark 4.4). Extrapolation to zero
radius serves to remove the perturbations caused by shadow functions.
The extrapolated values are plotted in Fig. 4.15(c). The maximum value (59.4 MPa mm1∕2 ) occurs
in the plane of symmetry. It is 8.4 % greater than the value obtained by the plane strain model. The
singularity at the point where the crack front intersects the stress-free surface is different from the
crack tip singularity in the case of plane strain. This is because the zero traction conditions have to
be satisfied not only on the faces of the crack but also on the boundary surface.
Remark 4.5 The ASTM Standard E1820-017 provides the following formula for the computation
of KI :
P 2 + a∕𝑤
KI ≈ √ 3∕2
(0.886 + 4.64(a∕𝑤)
t 𝑤 − a∕𝑤)
(1
− 13.32(a∕𝑤)2 + 14.72(a∕𝑤)3 − 5.6(a∕𝑤)4 )
7 Standard Test Method for Measurement of Fracture Toughness.

where P is the applied force. Using the data in Example 4.4 we get KI ≈ 54.64 MPa mm1∕2 which
is a close approximation to the plane strain result. ◽
Remark 4.6 We assumed that the crack front is a straight line. This assumption was necessary in
order to make comparison between the plane strain and 3D models possible. In practice the initial
crack is induced by subjecting the specimen to cyclic loads that produce stable crack growth. This
process results in a curved crack front. In standard tests a test result is rejected if the initial crack
length varies by more than 5% along the crack front. ◽
Exercise 4.12 Assume that the fracture toughness of the material in Example 4.4 is 20 MPa m1∕2 .
Estimate the applied load at which rapid crack propagation is expected to occur when the crack
length is 30 mm.
4.4.3 Forcing functions acting on boundaries

The regularity of the solution is influenced by the forcing function. We consider (a) a concentrated
force and (b) a step function acting on a semi-infinite planar body in the following. The notation is
shown in Fig. 4.16
Concentrated force
In this case the stress function is
Fr
U = − 0 𝜃 sin 𝜃 (4.49)
𝜋
see, for example, [105]. The corresponding stress components are:
1 𝜕U 1 𝜕2 U 2F cos 𝜃
𝜎r = + 2 2 =− 0 (4.50)
r 𝜕r r 𝜕𝜃 𝜋 r
𝜕2 U
𝜎𝜃 = 2 = 0 (4.51)
𝜕r
( )
𝜕 1 𝜕U
𝜏r𝜃 = − =0 (4.52)
𝜕r r 𝜕𝜃
and the displacement components, up to rigid body displacement, with the symmetry constraint
implied on 𝜃 = 0, are:
2F (1 − 𝜈)F0
ur = − 0 cos 𝜃 ln r − 𝜃 sin 𝜃 + C cos 𝜃 (4.53)
𝜋E 𝜋E
y
F0 p0
C
y A x
–𝜃
Tt Tt
Tr B Tr
x (a) (b)
Figure 4.16 (a) Loading by a concentrated force. (b) Loading by a step function.
4.5 Robustness 143
2𝜈F0 2F (1 − 𝜈)F0
u𝜃 = sin 𝜃 + 0 ln r sin 𝜃 + (sin 𝜃 − 𝜃 cos 𝜃) − C sin 𝜃 (4.54)
𝜋E 𝜋E 𝜋E
where C is an arbitrary constant. If we select an arbitrary reference point on the positive x axis,
say x = d, and set ur (d, 0) = 0 then from eq. (4.53) we get C = 2F0 ln d∕(𝜋E). In this case ux (r, 𝜃) is
measured from this point. Note that in the origin ur (0, 𝜃) = ∞ and 𝜕ur ∕𝜕r is not square integrable
on any subdomain that includes the origin. Hence ur is not in the energy space. Concentrated forces
and point constraints are not admissible in two- and three-dimensional elasticity. This point will
be discussed and illustrated by an example in Section 5.2.8.
Step function
The step function is shown in Fig. 4.16(b). The point where the normal traction changes from zero
to −p0 is a singular point. The stress function is:
p r2 ( 1
)
U=− 0 𝜋 + 𝜃 − sin 2𝜃 − 𝜋 ≤ 𝜃 ≤ 0 (4.55)
2𝜋 2
and the stress components are:
1 𝜕U 1 𝜕2 U p ( 1
)
𝜎r = + 2 2 = − 0 𝜋 + 𝜃 + sin 2𝜃 (4.56)
r 𝜕r r 𝜕𝜃 𝜋 2
𝜕2 U p0 ( 1
)
𝜎𝜃 = = − 𝜋 + 𝜃 − sin 2𝜃 (4.57)
𝜕r 2 𝜋 2
( ) p
𝜕 1 𝜕U
𝜏r𝜃 = − = 0 (1 − cos 2𝜃). (4.58)
𝜕r r 𝜕𝜃 2𝜋
In this case the stress components are finite but not single-valued in the singular point. Observe
that along the x axis (𝜃 = 0) 𝜏r𝜃 = 0 in the origin but along the y axis (𝜃 = −𝜋∕2) 𝜏r𝜃 = p0 ∕𝜋. In
the origin it is multi-valued. Similarly, the sum of the normal stresses 𝜎r + 𝜎𝜃 = −2p0 (1 + 𝜃∕𝜋)
ranges from −2p0 to 0 in the origin, depending on the direction from which the origin is
approached.
Exercise 4.13 Refer to Fig. 4.16(b). Let p0 = 1 MPa and the radius of the circle 100 mm. Specify
tractions corresponding to the stresses given by equations (4.56) to (4.58) on the circular boundary.
Plot the sum of normal stresses and report the error in energy norm.
4.5 Robustness
A numerical scheme for the approximation of a parameter-dependent problem is said to be robust

if it is uniformly convergent for all admissible values of the parameter. In this section we consider
robustness with Poisson’s ratio as the parameter.
As Poisson’s ratio 𝜈 approaches 1/2, the rate of h-convergence of low order elements, measured in
energy norm, slows. Furthermore, the error in the first stress invariant, computed from the finite
element solution using Hooke’s law, increases. This is called Poisson ratio locking or volumetric
locking.
To explain why locking occurs, we show that for an arbitrary mesh of regular quadrilateral ele-
ments, p = 1, the number of degrees of freedom is independent of the number of elements when
the material is incompressible and hence convergence cannot occur at all.
Figure 4.17 Poisson ratio locking. Notation.

y
4h
13 14 15 16
3h
9 10 11 12
2h
5 6 7 8
h
1 2 3 4
0
0 h 2h 3h 4h x
Consider the uniform mesh of quadrilateral elements shown in Fig. 4.17. The displacement com-
ponents for the ith element are:
u(i) (i) (i) (i) (i)
x = a1 + a2 x + a3 y + a4 xy (4.59)
u(i)
y = a(i)
5 + a(i)
6
x + a(i)
7 y + a(i)
8
xy. (4.60)
For incompressible elastic materials the volumetric mechanical strain 𝜖kk − 3𝛼Δ is zero, indepen-
dent of the hydrostatic stress 𝜎0 defined by eq. (2.84). Therefore in the case of plane strain the
displacement components satisfy the constraint
𝜕u(i)
x
𝜕u(i)
y
𝜖x(i) + 𝜖y(i) ≡ + = 0. (4.61)
𝜕x 𝜕y
Consequently a(i)
4
= a(i)
8
and a(i) (i)
7 = −a2 and hence
u(i) (i) (i) (i)

x = a1 + a2 x + a3 y (4.62)
u(i)
y = a(i)
5 + a(i)
6
x − a(i)
2
y. (4.63)
Consider now continuity between elements 1 and 2:
a(1) (1) (1) (2) (2) (2)
1 + a2 h + a3 y = a1 + a2 h + a3 y (4.64)
a(1)
5 + a(1)
6
h − a(1)
2
y = a(2)
5 + a(2)
6
h − a(2)
2
y (4.65)
therefore
a(1)
3
= a(2)
3
and a(1)
2
= a(2)
2
which is equivalent to
𝜕u(1)
x 𝜕u(2) 𝜕u(1) 𝜕u(2)
= x and x
= x ⋅
𝜕x 𝜕x 𝜕y 𝜕y
Thus both derivatives of ux are continuous on the interface of elements 1 and 2. The same argument
can be used to show that both derivatives are continuous on the interfaces of elements 2 and 3, 3
and 4, etc. as well as on the interfaces of elements 5 and 6 and so on. Since ux is continuous, it has
to be a linear function over the entire domain, independent of the number of elements.
Similarly, from the condition of continuity between elements 1 and 5, we find
𝜕u(1)
y 𝜕u(5)
x
𝜕u(1)
y 𝜕u(5)
y
= and =
𝜕x 𝜕x 𝜕y 𝜕y
4.5 Robustness 145
and hence both derivatives of uy are continuous on the interfaces of elements 1 and 5, 5 and 9 and
so on. Therefore uy is also a linear function over the entire domain, independent of the number of
elements. Consequently, for the entire domain we have
ux = a 1 + a 2 x + a 3 y (4.66)
uy = a 4 + a 5 x − a 2 y (4.67)
that is, the dimension of the finite element space is 5, no matter how many elements are used. All
other degrees of freedom are needed for satisfying the volumetric and continuity constraints.
Locking does not occur in the p-version because the number of elements and hence the number
of constraint conditions is fixed, independent of the number of degrees of freedom. The elements
can deform while preserving constant volume. However, the computation of normal stresses from
the finite element solution requires special attention. This will be discussed with reference to the
following example, see Example 4.6.
Example 4.5 In this example we consider a classical problem of elasticity, that of a rigid circular
inclusion in an infinite plate, subjected to unidirectional tension at infinity. Plane strain conditions
are assumed. The domain and notation are shown in Fig. 4.18(a). We will use a = 1, b∕a = 5 and
thickness tz = 1. The finite element mesh, consisting of four elements, is shown in Fig. 4.18(b).
The boundary conditions are as follows: Along the circular arc AD both displacement compo-
nents are zero. Along the symmetry lines AB and CD the normal displacement and the shearing
stress are zero. Along the circular arc BC the normal and shearing tractions are given by the strong
form of the classical solution [60]. The displacement components in polar coordinates are:
{ [ ] }
𝜎∞ 2 𝛿 a4
ur = (𝜅 − 1) r + 2 𝛾 a + 𝛽(𝜅 + 1) a + 2 r +
2 2 2 2
cos 2 𝜃 (4.68)
8Gr r2
[ ]
𝜎 2 𝛿 a4
u𝜃 = − ∞ 𝛽 (𝜅 − 1) a2 + 2 r 2 − sin 2 𝜃 (4.69)
8Gr r2
and the stress components are:
[ ( ) ]
𝜎∞ 𝛾a2 2𝛽a2 3𝛿a4
𝜎r = 1− 2 + 1− 2 − 4 cos 2𝜃 (4.70)
2 r r r
[ ( ) ]
𝜎 𝛾a2 3𝛿a4
𝜎𝜃 = ∞ 1 + 2 − 1 − 4 cos 2𝜃 (4.71)
2 r r
( )
𝜎∞ 𝛽a2 3𝛿a4
𝜏r𝜃 = − 1+ 2 + 4 sin 2𝜃 (4.72)
2 r r
y y
C
a 𝜎∞
D r
𝜃
A B x x
2a
b
(a) (b)
Figure 4.18 (a) Rigid circular inclusion in an infinite plate under tension. Notation. (b) Four-element mesh.
where 𝜅, 𝛽, 𝛾, 𝛿 are constants that depend on Poisson’s ratio 𝜈 only. For plane strain:
2 1
𝜅 = 3 − 4𝜈; 𝛽 = − ; 𝛾 = −(1 − 2𝜈); 𝛿 = (4.73)
3 − 4𝜈 3 − 4𝜈
and for plane stress:
3−𝜈 2(1 + 𝜈) 1−𝜈 1+𝜈
𝜅= ; 𝛽=− ; 𝛾=− ; 𝛿= ⋅ (4.74)
1+𝜈 3−𝜈 1+𝜈 3−𝜈
We will use E = 1, 𝜈 = 0.49999. The results demonstrate strong p-convergence in energy norm.
Finite element solutions were obtained by means of StressCheck. The load vector corresponding
to the normal and shearing tractions acting on the circular arc BC was computed by numerical
quadrature using twelve Gauss points per element side. The exact value of the strain energy was
computed for b∕a = 5 as follows:
𝜋∕2 𝜎∞
2 a2 t
1 z
U(u) = (ur 𝜎r + u𝜃 𝜏r𝜃 )b d𝜃 = 7.31883865 ⋅ (4.75)
2 ∫0 E
The computed values of the strain energy for p ranging from 3 to 8 (product space) and the esti-
mated and exact relative errors in energy norm are listed in Table 4.8. It is seen that the slope of the
convergence curve 𝛽 is increasing with p. This is because the exact solution is an analytic function
and hence the rate of p-convergence is exponential. Nevertheless, our a posteriori error estimator,
based on the assumption that the rate of convergence is algebraic (see eq. (1.92)), provides very
reasonable estimates.
Example 4.6 We consider the problem of a rigid circular inclusion in an infinite plate described
in Example 4.5. Plane strain is assumed. The finite element mesh, consisting of four elements, is
shown in Fig. 4.18(b). We are interested in the relative error in the sum of normal stresses 𝜎x + 𝜎y
measured in maximum norm. Referring to equations (4.70), (4.71) and noting that 𝜎x + 𝜎y is a stress
invariant, we define
( ( )2 )
def 2 a
Σ = 𝜎r + 𝜎𝜃 = 𝜎x + 𝜎y = 𝜎∞ 1 + cos 2𝜃
3 − 4𝜈 r
and we define the percent relative error in maximum norm as follows:
def max |Σ − (𝜎x(FE) + 𝜎y(FE) )|
(er )Σ = 100 ⋅ (4.76)
𝜎∞ (1 + 2∕(3 − 4𝜈))
Table 4.8 Convergence of the strain energy (𝜈 = 0.49999). Rigid circular inclusion in an
infinite plate under tension. Four-element mesh, product space, b∕a = 5.
U(u)E Rel. error (%)

p N 𝜷
𝝈∞
2 2
a tz Est.’d Exact
3 72 6.53074253 − 32.81 32.81

4 128 7.29188870 2.93 6.07 6.07
5 200 7.31760970 3.46 1.30 1.30
6 288 7.31873361 3.37 0.38 0.38
7 392 7.31882930 3.86 0.12 0.11
8 512 7.31883783 3.86 0.04 0.03
Exact 7.31883865 − − 0
4.5 Robustness 147
Table 4.9 Example 4.6. Rigid circular inclusion in an

infinite plate under tension. Plane strain. Dependence of
the maximum relative error (er )Σ on Poisson’s ratio.
Four-element mesh, p = 8.
Poisson’s ratio
Space
0.49 0.499 0.4999 0.49999
Product 0.486 2.725 4.988 32.84

Trunk 2.070 8.981 39.68 233.1
The results of computation for p = 8 are listed in Table 4.9. These results show that (er )Σ rapidly
increases as Poisson’s ratio approaches 1∕2, and it increases more rapidly for the trunk space than
for the product space. This is an indication that product space is more robust than the trunk space.
The exact value of the sum of normal stresses and the computed values using the product and
trunk spaces, p = 8, are shown along boundary segment AD in Fig. 4.19 for 𝜈 = 0.49999. For the
trunk space the largest error is in the vicinity of 2 degrees, whereas for the product space the largest
error is at 45 degrees. At 45 degrees the size of the jump is the same for the product and trunk spaces.
This example illustrates that the product space has significantly better convergence properties
than the trunk space for the range of parameters considered. In general, the p-version is more
robust than the h-version.
Exercise 4.14 Solve the rigid inclusion problem described in Example 4.5 for 𝜈 = 0.3, 0.49, 0.499
using b∕a = 5, p = 8, product space. Explain why the error in 𝜏r𝜃 , measured in maximum norm, is
virtually independent of 𝜈.
Exercise 4.15 Repeat the computations in Example 4.6 for plane stress. Explain why the values
of (er )Σ are virtually independent of 𝜈.
10
v = 0.49999
5
σx + σy
−5 exact
product space, p = 8
trunk space, p = 8
−10
0 10 20 30 40 50 60 70 80 90
angle (degrees)
Figure 4.19 Example 4.5. Rigid circular inclusion. The sum of normal stresses at the boundary of the
inclusion. Poisson’s ratio: 0.49999.
Exercise 4.16 Solve the problem of Example 4.5 using a sequence of quasiuniform meshes of 32,
64 and 128 elements, p = 2 (product space) and, using the exact value of the strain energy given
by eq. (4.75), compute the relative error in energy norm for each finite element solution and the
rate of convergence 𝛽 for the sequence of solutions. Is the theoretical rate of convergence (𝛽 = 1)
realized? I not, why not?
Exercise 4.17 Subdivide each of the four elements in Example 4.6 uniformly into 100 elements.
Solve the rigid inclusion problem described in Example 4.5 for 𝜈 = 0.3, 0.49, 0.499 using b∕a = 5,
p = 2, product space. Report the values of (er )Σ . Partial solution: For 𝜈 = 0.49 you will find (er )Σ =
25.34%.
4.6 Solution verification
It is necessary to determine whether the QoIs computed from a finite element solution are suffi-
ciently accurate for the purposes of the analysis. The accuracy of a QoI depends on the accuracy
of the finite element solution, measured in energy norm, and the method by which the QoI is
computed. We have seen a simple illustration of this in Example 1.9. Only direct methods of com-
putation are considered in this section.
Solution verification involves steps to ascertain that (a) the correct input data was used in the
analysis and (b) the quantities of interest are substantially independent of the parameters that char-
acterize the finite element space. This is based on the idea that while we do not know the exact value
of the QoI, we know that it exists, it is unique and it is independent of the discretization. Therefore,
as long as the computed value of a QoI is changing significantly when the number of degrees of
freedom is increased, it cannot be close to its limit value.
This involves obtaining two or more finite element solutions corresponding to a sequence of finite
element spaces and examination of the information generated from the finite element solutions.
The recommended steps are as follows:
1. Display the solution graphically and check whether the solution is reasonable. For example,
plotting the deformed configuration on a 1:1 scale provides information on whether a large error
occurred in specifying the loading and constraint conditions or material properties.
2. Estimate the relative error in energy norm and its rate of convergence. The estimated relative
error in energy norm is a useful indicator of the overall quality of the solution, roughly equiva-
lent to estimating the root-mean-square error in stresses, see Section 2.6.2. The estimated rate of
convergence is an indicator of whether the rate of change of error is consistent with the asymp-
totic rate for the problem class. Substantial deviations from the theoretical rates of convergence
typically indicate errors in the input data or in the finite element mesh. For example, elements
may be highly distorted.
3. Check for the presence of jump discontinuities in stresses and fluxes in regions where the
stresses or fluxes are large. The normal flux and the normal and shearing stress components
at internal element boundaries must be continuous. In regions where the stresses or fluxes are
small, some discontinuity is generally acceptable. Significant jump discontinuities in stresses
and fluxes at element boundaries usually indicates that the mesh is not fine enough or the
polynomial degree is not high enough or the elements are too distorted.
4. Show that the quantities of interest are substantially independent of the mesh and/or the poly-
nomial degree of elements. Estimate the limit value. Report the estimated limit value and the
4.6 Solution verification 149
uniform pressure
a P
P P
b D
F
Rf
Tw
Dw
Te L R0
h symmetry
symmetry
(a) (b)
Figure 4.20 Example 4.7. (a) Notation. (b) Isometric view and finite element mesh. Number of
elements: 26.
percent difference between the estimated limit value and the computed value corresponding to
the highest number of degrees of freedom.
These steps are illustrated by the following example.
Example 4.7 In this example we demonstrate the process of verification. The quantity of interest
is maximum von Mises stress in the fillet region of a shear fitting. The geometric configuration,
notation and boundary conditions are shown in Fig. 4.20. The fitting is loaded by uniform pressure
given by
4F
p=
𝜋(D2𝑤 − D2 )
where F is the applied force, D is the diameter of the hole, D𝑤 is the outer diameter of the area
on which the pressure is acting. The data are as follows: F = 1000 lbs, (4448 N), D = 0.375 in
(9.525 mm), D𝑤 = 0.650 in (16.51 mm), outer radius: R0 = 1.10 in (27.94 mm), wall thickness:
T𝑤 = 0.60 in (15.24 mm), length: L = 3.0 in (76.2 mm), fillet radius: Rf = 0.1 in (2.54 mm), pad
thickness Te = 0.50 in (12.70 mm), offset h = 0.45 in (11.43 mm). Modulus of elasticity: E = 1.05E7
psi (7.24E4 MPa), Poisson’s ratio: 𝜈 = 0.30.
The finite element mesh is shown in Fig. 4.20(b). It consists of 24 hexahedral elements and 2
pentahedral elements. We will perform h- and p-extensions using the trunk space. The initial mesh
for h-extension is the mesh shown in Fig. 4.20(b). We denote the diameter of the largest element in
the initial mesh by h1 . A sequence of meshes was obtained by uniformly subdividing the elements
in the parameter space. This results in hk = h1 ∕k (k = 1, 2, 3, …). The corresponding number of
elements is given by M(Δk ) = 26k3 .
Table 4.10 Example 4.7: Estimated error in energy norm.
p N 𝝅p 𝜷 (er )E
5 2844 −0.41810322 − 3.34

6 4314 −0.41836531 0.99 2.21
7 6263 −0.41848408 1.17 1.43
8 8769 −0.41853071 1.17 0.97
extrapolated: −0.41856984
12.6 +1%
est.’d limit 12.43 h1/9
6
12.4 h1/11
7 p=8 h1/7
−1% 4 5
12.2 h1/5
σvM(P) (ksi)
h1/4 P
12
h1/3
p-version
11.8
h-version
3 h1/2
11.6
11.4 h1, p = 2
103 104
N
Figure 4.21 Example 4.7. Point convergence.
For p-extension the mesh shown in Fig. 4.20(b) was used. The relative error in energy norm at
p = 8 was estimated to be 0.83%. It was found that the maximum von Mises stress occurs in point
P shown in Fig. 4.20. The location parameters are: a = 0.5881 in (14.94 mm) and b = 0.1507 in
(3.83 mm). Point P is located on the intersection between the top surface and the toroidal surface
of the fillet. Therefore the stress is uniaxial in point P.
The values of the von Mises stress corresponding to sequences of finite element solutions
obtained by h- and p- extensions are shown in Fig. 4.21.
As the number of degrees of freedom is increased either by h- or p-extension, the location and
magnitude of the maximum stress change. Here we fixed the location where the maximum was
found and computed sequences of estimated values of the von Mises stress in that location. We see
that convergence is not monotonic either for the h- or p-version.
Remark 4.7 In the foregoing example we excluded from consideration the neighborhood of load
application, based on the following consideration: If the maximum stress would occur in the vicin-
ity of load application then this model would not be suitable for the determination of the maximum
stress. This is because the loading is transferred by mechanical contact which is idealized here as
a uniform pressure distribution. Should the stress distribution in the contact area be of interest, a
different model would have to be formulated8 . Our region of primary interest was the fillet region.
The contact area is a region of secondary interest. ◽
Remark 4.8 Finite element spaces generated by p-extension are hierarchic whereas finite ele-
ment spaces generated by h-extension may or may not be hierarchic. In general, sequences of mesh
created by mesh generators are not hierarchic. ◽
Remark 4.9 In many practical problems the QoIs lie in some small subdomain of the solution
domain. For example, we may be interested in the stresses in the vicinity of a fastener hole in a
large plate. The neighborhood of the fastener hole is the region of primary interest, the rest of the
plate is the region of secondary interest. Errors in the computed data may be caused by insufficient
8 Mechanical contact is described in Section 9.2.4.

Ti
z
ss rm t
a
x ri r
ro
(a) (b)
Figure 4.22 Composite ring. Notation.
discretization in the region of primary or secondary interest, or both. Errors caused by insufficient
discretization of the region of secondary interest are called pollution errors. ◽
The following example highlights errors that can and frequently do occur when the analyst does
not understand the relationship between input data and the regularity of the exact solution.
Example 4.8 A composite ring was made by joining a stainless steel (ss) and a titanium (Ti) ring.
Two analysts were asked to compute the maximum principal stress (𝜎1 )max when the temperature
of the composite ring is increased by 100 C∘ . They were to assume that the two materials are per-
fectly bonded and the assumptions of the linear theory of elasticity are applicable. The geometric
and material parameters are shown in Table 4.11.
Analyst A formulated this as a problem of plane stress. He used two quadrilateral elements on
a 30-degree sector (𝛼 = 30∘ ) and, having performed p-extension (product space), he reported that
the maximum principal stress occurs along the material interface and its value is 22.1 MPa. He
Table 4.11 Geometric and material properties.
Description Symbol Value Units
radius ri 100.0 mm
radius ro 115.0 mm
radius rm 107.5 mm
thickness t 20 mm
modulus of elasticity, ss Es 2.0 × 105 MPa
Poisson’s ratio, ss 𝜈s 0.295 −
coef. of thermal exp. ss 𝛼s 1.17 × 10−5 1∕C∘
modulus of elasticity, Ti Et 1.1 × 105 MPa
Poisson’s ratio, Ti 𝜈t 0.31 −
coef. of thermal exp. Ti 𝛼t 8.64 × 10 −6
1∕C∘
95
p=8
Max. stress (MPa)
90 7
6
A B
85 5
4
80
p=3
75
200 400 600 800 1000 2000
Number of degrees of freedom
Figure 4.23 Divergence of the maximum principal stress. The results were obtained by p-extension on a
16-element geometrically graded mesh. Symmetry boundary conditions were applied on boundary
segment AB.
supported this statement by showing that the error in energy norm converged exponentially and
the maximum normal stress was substantially constant for p = 3, 4, … , 8.
Analyst B formulated this as an axisymmetric problem. He generated a uniform mesh of 800
nine-node quadrilateral elements (that is, p = 2 product space). He reported that the maximum
principal stress occurs at r = rm , z = t∕2 and its value is 52.8 MPa. Relying on his judgment that the
mesh was fine enough, he did not perform solution verification.
Your supervisor asked you to find out why such a large discrepancy exists between the two results.
Having checked the input data, you found no errors. You then solved the axisymmetric problem
using p-extension (product space) on a 16-element, geometrically graded mesh shown in the inset
of Fig. 4.23. On plotting the maximum principal stress vs. the number of degrees of freedom on
semi-log scale you found that it does not converge to a limit value.
Write a succinct memorandum to your supervisor explaining the reasons for the discrepancy.
Solution
Analyst A solved the problem in the plane of symmetry (z = 0) and thus his model does not account
for the singularity caused by the abrupt change in material properties in the points r = rm , z =
±t∕2. The exact solution of this problem is an analytic function. That is why p-convergence was
exponential. Analyst A solved the wrong problem very accurately.
Analyst B solved the correct problem, but with a very large (in fact infinitely large) error in the
computed value of (𝜎1 )max . The exact value of (𝜎1 )max is not finite in the singular point and hence
the absolute error in the reported value of (𝜎1 )max is infinitely large.
You are now faced with the unpleasant problem of having to explain to your supervisor that
asking for the maximum stress under the stated modeling assumptions made no sense.
Discussion
Proper formulation of a mathematical model depends on the quantities of interest (QoI). Analyst
A did not take into consideration that the plane stress model cannot represent the stress raiser at
r = rm , z = ±t∕2 which are singular arcs in 3D, singular points in the axisymmetric formulation.
Therefore the dimensionally reduced model chosen by Analyst A is improper, given the QoI. Had
the QoI been (say) the maximum radial displacement then both models would have been proper.
Analyst A would have reported 0.121 mm, Analyst B would have reported 0.122 mm.
The supervisor was presumably interested in predicting failure (or the probability of failure) and
mistakenly assumed that failure can be correlated with the maximum stress corresponding to the
exact solution of the problem of linear elasticity. However, in this problem the maximum stress is
not a finite number for any temperature change and therefore it cannot be a predictor of failure.
Various theories can be formulated for the prediction of failure events. However, those theories
are subject to the restriction that the driver of failure initiation is a finite number that continuously
depends on the applied load, in this case the temperature change. For example, one could assume
that the generalized stress intensity factor, or the average stress over a volume, is the predictor of
failure initiation. Such predictors would have to be calibrated and tested following a process similar
to the process described in Chapter 6.
155
Simulation
In Chapter 1 simulation was defined as an imitative representation of the functioning of one system
or process by means of the functioning of another. In this chapter we consider the functioning of
mechanical systems and their imitative representation by mathematical models. We are interested
in predicting certain quantities of interest (QoIs), given the geometry, material properties and loads.
The usual QoIs in the simulation of mechanical systems are displacements, stresses, strains, stress
intensity factors, ultimate loads, the probability that a system will survive a given number of load
cycles, etc. These quantities support engineering decisions concerning design, certification and
maintenance.
Simulation begins with a precise statement of an idea of physical reality in the form of a math-
ematical model. The term “mathematical model” should be understood to collectively refer to all
mathematical operations, deterministic or probabilistic, needed for the prediction of the QoIs. The
importance of distinguishing between physical reality and an idea of physical reality was empha-
sized by Wolfgang Pauli1 :
“The layman always means, when he says ‘reality’ that he is speaking of something
self-evidently known; whereas to me it seems the most important and exceedingly difficult task
of our time is to work on the construction of a new idea of reality.”
Implied in this statement is that casting ideas of physical reality into mathematical models has sub-
jective aspects: Various ideas can be proposed for the simulation of the functioning of a particular
system or process. Therefore it is necessary to have objective criteria by which the relative merit
of mathematical models is evaluated. The formulation of mathematical models is described and
illustrated in this chapter.
Whereas Pauli had in mind ideas of reality in the realm of quantum mechanics, his statement
is universally valid. It is applicable to any engineering or scientific project concerned with the for-
mulation of mathematical models for the purpose to generalize data collected from observations of
events. Engineers have the advantage over physicists because it is generally much easier to make
observations and test predictions based on mathematical models in the forefront of engineering
than in the forefront of physics which include phenomena on scales that range from the subatomic
to the astronomical. In either case, the goals are to cast ideas of physical reality into the form of
mathematical models, calibrate the mathematical models and test them in validation experiments
through comparison of predicted outcomes with the outcomes of physical experiments or observa-
tions of physical events.
1 Wolfgang Pauli 1900–1958.
156 5 Simulation
We will view a mathematical model as a transformation of one set of data D into another set F
(the QoI) based on a precisely stated idea of the physical reality I. In short hand,
(D, I) → F (5.1)
where the right arrow represents a mathematical model which (by definition) is the entire process
by which (D, I) is transformed to the quantities of interest.
For example, in the World Wide Failure Exercise [48], investigators sought to predict failure
events in test articles made of composite materials. In that case the failure theory was part of the
mathematical model and F was the set of data needed to predict the outcome of experiments. In
addition, the mathematical models included constitutive laws and multiscale procedures that link
sub-models of macroscopic and microscopic phenomena. When the goal is to predict the probabil-
ity of failure then a statistical sub-model is also part of the mathematical model.
A mathematical model formulated for the prediction of the probability of failure events in high
cycle fatigue will be discussed in Chapter 6.
Remark 5.1 In view of Pauli’s statement, the frequently used term: “physics-based model” does
not have a well defined meaning. Mathematical models are precisely formulated ideas relating to
specific aspects of physical reality. Usually there are several competing models for the prediction
of the same phenomenon. We will discuss objective methods for ranking alternative models in
Chapter 6.
5.1 Development of a very useful mathematical model
To illuminate the process by which ideas of physical reality are cast into the form of a mathematical
model we trace the history of development of a famous and very useful mathematical model known
as the Bernoulli-Euler beam model2 . Students of mechanical and structural engineering usually
learn about this classical model in their second year of study. Interestingly, its development took
188 years and some of the greatest mathematicians of the time contributed to it. We begin with a
brief summary of how the formulation is presented today then outline the major milestones in its
188 years of development.
5.1.1 The Bernoulli-Euler beam model

The formulation of the Bernoulli-Euler beam model for prismatic beams of arbitrary cross-section
can be found in every textbook on the strength of materials. For the sake of simplicity, the for-
mulation is reviewed here for the special case of beams with rectangular cross-section only. The
notation is shown in Fig. 5.1. The origin of the coordinate axes is coincident with the centroid of
the cross-section.
The formulation involves three steps: (a) The mode of deformation is assumed and the distribu-
tion of strain is inferred. (b) A stress-strain relationship is defined. Here we consider linearly elastic
stress-strain relationships only. (c) The equation of equilibrium is written: the bending moment
corresponding to the normal stress acting on a cross-section has to be in equilibrium with the
applied bending moment. The deformation is assumed to be sufficiently small so that the equi-
librium equations can be written with reference to the undeformed configuration.
2 Jacob Bernoulli 1655–1705, Leonhard Euler 1707–1783.

5.1 Development of a very useful mathematical model 157
dθ
y A y
ds
h d B C
x z
Mz Mz D E
A b
(a) Beam segment (b) Section A - A (c) Deformed segment of length ds
Figure 5.1 Rectangular beam in bending. Notation.
Referring to Fig. 5.1, it is assumed that the bending moment causes the cross-sections to rotate
without deformation (plane sections remain plane) hence fibers3 on the convex side lengthen and
fibers on the concave side shorten. Fibers in the xz plane do not deform. The xz plane is called the
neutral plane and its intersection with a cross-section is called the neutral axis. In Fig. 5.1 the z axis
is coincident with the neutral axis.
We denote the radius of curvature by r and the angle subtended by the arc ds from the center of
rotation by d𝜃. The length of arc DE can be written as (r + d)d𝜃. By definition, strain is the change
in length divided by the original length. Since ds lies in the neutral plane, ds is the original length.
Therefore the strain is:
(r + d)d𝜃 − rd𝜃 d y
𝜀= = =− ⋅ (5.2)
rd𝜃 r r
It is assumed that stress is proportional to strain, that is, 𝜎 = E𝜀 where E is the modulus of elasticity
and the stress must be in equilibrium with the bending moment:
h∕2 h∕2
E Mbh3 EI
M=− 𝜎yb dy = y2 b dy = = (5.3)
∫−h∕2 r ∫−h∕2 12 r
where integration is over the cross-section (dA = bdy) and I is the moment of inertia of the
cross-section about the z axis.
We denote the displacement of the neutral axis by u = u(x). The curvature of u is
1 u′′ ′′
r
= ( )3∕2 ≈ u (5.4)
′
1 + (u )2
where the approximate equality (≈) is justified when max (u′ )2 << 1, a condition satisfied in most
mechanical and structural engineering applications. If the beam is loaded by a distributed load q
(given in force/length units) then, using the convention that q is positive in the positive y direction
and M is positive in the sense shown in Fig. 5.1, then the equation of equilibrium is:
M ′′ = q. (5.5)
On combining equations (5.3), (5.4) and (5.5) we get the Bernoulli-Euler beam model:
( )′′
EIu′′ = q. (5.6)
The usual boundary conditions of interest are: simply supported: u = M = 0, fixed or built-in:
u = u′ = 0, free: M = V = 0 where V is called shearing force, it is proportional to u′′′ .
3 In this context a fiber is understood to be the set of points 0 < x < 𝓁 such that y and z are fixed.
158 5 Simulation
5.1.2 Historical notes on the Bernoulli-Euler beam model

The development of the Bernoulli-Euler beam model began with Galileo4 who, in a book published
in 1638, addressed questions relating to how structures resist loads5 . He considered a beam of rect-
angular cross-section, loaded at one end, fixed on the other end and mistakenly assumed that the
resistance of the beam to the applied load is uniformly distributed over the fixed cross-section. As
a result, Galileo overestimated the strength of beams by a factor of 3. Nevertheless he was able
to correctly predict that a beam of rectangular cross-section, bent by moment M about the z-axis
(i.e. Mz = M in Fig. 5.1), will be h∕b times as strong as the same beam bent by M about the y-axis
(My = M).
Hooke6 experimented with beams made of wood and observed that the fibers on the convex side
are extended whereas the fibers on the concave side are compressed. He also observed that forces
and the magnitude of corresponding displacements are in constant ratio. When the components of
stress tensor can be written as linear combinations of the components of the strain tensor then that
relationship is called Hooke’s law.
Jacob Bernoulli investigated the deflected curve of elastic bars. He followed the steps outlined in
Section 5.1.1 however he incorrectly assumed that the strain is zero on the innermost fiber on the
concave side (y = h∕2):
y − h∕2
𝜀= , −h∕2 < y < h∕2
rc
where rc is the radius of the outermost fiber on the convex side. Letting 𝜂 = h∕2 − y, he arrived at
the following formula which is analogous to eq. (5.3) however the moment
h h
Eb Ebh3
M=− 𝜎𝜂 bd𝜂 = 𝜂 2 dy =
∫0 ∫
rc 0 3rc
is 4r∕rc times the correct moment.
Euler used the methods of variational calculus to arrive at the correct functional form:
u′′
M = C( )3∕2
1 + (u′ )2
where, according to Euler, C depends on the elastic properties and, for a beam of rectangular
cross-section, is proportional to bh2 . The correct statement would have been: is proportional to
bh3 , see eq. (5.3). Euler recommended that C should be determined by experimental means. We
now know that C = EI where E is a property of the material, determined by experimental means,
and I a property of the cross-section, see eq. (5.3).
The formulation of the Bernoulli-Euler beam model was completed by Navier7 in 1826 when he
proved that, under the assumptions that the axial force is zero and the material obeys Hooke’s law,
the neutral axis passes through the centroid of the cross-section.
The first engineering use of this model on a large scale was in connection with the design of
the Eiffel8 Tower for the 1889 World’s Fair. Until then bridges and buildings were designed by
precedent. Eiffel demonstrated the practical value of this model.
The foregoing is a greatly abbreviated summary of the history of the development of the
Bernoulli-Euler beam model. For additional information we refer to [102, 106]. This summary
4 Galileo Galilei 1564–1642.

5 Two New Sciences. English translation by H. Crew and A. de Salvio. The Macmillan Company, New York, 1933.
6 Robert Hooke 1635–1703.
7 Claude-Louis Navier 1785–1836.
8 Gustave Eiffel 1832–1923.
is included here to illustrate that casting ideas of physical reality into a mathematical model is a
creative process that involves making assumptions based on insight and experience, attempting
to validate those assumptions through physical experimentation, and confirming, rejecting or
modifying the formulation on the basis of empirical data.
The definition of a mathematical model must always include specification of its range of validity.
In the case of the Bernoulli-Euler beam model the assumption that plane sections remain plane is
valid when the beam is subjected to pure bending but when shearing forces are present then the
shearing deformation may not be negligibly small. This happens when the length to depth ratio is
less than about 10 or when the transverse load has a short wavelength (less than a few times the
depth of the beam). Another limitation is that the stresses must be in the elastic range. For slender
beams the assumption that the deformation is small, and therefore the equilibrium conditions can
be stated with respect to the undeformed configuration, may not be realistic. In such cases it is
necessary to write the equations of equilibrium with reference to the deformed configuration and
a nonlinear mathematical problem has to be solved.
The formulation of mathematical models for slender elastic bodies follows a very different pat-
tern in our time: These models are understood to be dimensionally reduced forms of the general
three-dimensional models of nonlinear continua [4]. The Bernoulli-Euler beam model is a special
case that, notwithstanding its simplicity and limitations, has great practical value.
The formulation of mathematical models in the field of solid mechanics reached a mature state
of development by the mid-twentieth century. Application of these models in engineering practice
was severely limited, however, because the solution of mathematical problems by classical methods
was feasible only in special and highly simplified cases. This limitation was removed when digital
computers became available, opening vast new possibilities in numerical simulation.
5.2 Finite element modeling and numerical simulation

For the reasons that will be discussed in Section 5.2.5, the concept and practice of finite element
modeling evolved more than ten years before the theoretical foundations of numerical simula-
tion were established. The terms “finite element modeling” and “numerical simulation” are often
used interchangeably. Engineers and engineering analysts often refer to numerical simulations but,
more often than not, they have finite element modeling in mind. It is important to clarify the dif-
ference between the two.
5.2.1 Numerical simulation

The main elements of numerical simulation are illustrated schematically in Fig. 5.2 which is a
restatement of eq. (5.1) with the addition that the quantities of interest (F) are solved numerically
and hence we have to rely on a numerical approximation to F which is denoted by Fnum . As we have
seen in Chapter 1, it is not sufficient to compute Fnum , it is also necessary to estimate and control
the error of approximation |F − Fnum |.
The formulation of a mathematical model is a creative process that has room for a wide range of
subjective choices. On the other hand, the solution of the numerical problem and control of error
in the quantities of interest involve the application of algorithmic procedures based on established
theorems of applied mathematics. Once a mathematical model was defined, subjective choices are
confined to the selection of discretization schemes and extraction procedures to the extent that the
available software tools support such choices.
160 5 Simulation
Physical (D, I) Mathematical F Numerical Fnum

reality model solution Prediction
A precise statement of The error | F-Fnum |

an idea of physical reality is controlled
Figure 5.2 The main elements of numerical simulation.
It is essential to recognize the existence of two qualitatively different sources of error: The model
form error incurred when making assumptions in the formulation of a mathematical model and
the numerical error, coming from the approximation of F with Fnum .
The transformation indicated in eq. (5.1) by the right arrow contains coefficients associated with
constitutive equations and models of failure initiation. These coefficients have to be calibrated to
experimental observations. In a calibration processes the mathematical model is assumed to be
correct and its coefficients are determined to match, or nearly match, the outcome of experiments.
If |F − Fnum | is not known to be negligibly small in comparison with the experimental errors then
the coefficients, and hence the predictions, will be polluted by the numerical errors.
Calibrated models are also validated models within the domain and scope of calibration. Vali-
dation experiments performed outside of the domain of calibration provide information about the
size of the domain of calibration. If we allow the domain of calibration to be arbitrarily small then
just about any model can be validated.
Remark 5.2 It is necessary to ensure that the numerical approximation scheme, the discretiza-
tion, meets the conditions of consistency9 and stability10 . The methods described in this book meet
these conditions. For details on consistency and stability we refer to [5].
Remark 5.3 The outcomes of physical experiments are random events, hence calibrated param-
eters are random numbers. When the statistical dispersion of these numbers is small then random-
ness is usually neglected and average values are reported. This is common practice in (for example)
reporting the modulus of elasticity and Poisson’s ratio. However, outcomes can have very large
dispersions as (for example) in fatigue testing of materials. In such cases a statistical sub-model
has to be incorporated in the mathematical model and the QoIs are the predicted probabilities of
outcomes. This is discussed in Chapter 6.
5.2.2 Finite element modeling

Early work on the finite element method was performed entirely by engineers who were familiar
with matrix methods of structural analysis. Their idea was to develop computer codes for the solu-
tion of problems of elasticity and other field problems similar to those that were already in use at
that time for the analysis of structural trusses and frames. They viewed the finite element method
as a “matrix method of structural analysis”.
Idealized trusses are assemblies of bar elements connected at frictionless hinges. An idealized
bar element is assumed to function as a linear spring, loaded by axial tension or compression forces
only. Given the cross-sectional area A, the modulus of elasticity E and the length 𝓁, the relationship
9 Consistency of a discretization refers to a quantitative measure of the extent to which the exact solution satisfies
the discrete problem. This quantitative measure depends on the QoI.
10 Stability of a discretization refers to well-posedness of the discretized problem. A discretization is stable if small
changes in the data produce small changes in the QoI.
Fy(3)
Fy(2)
uy(3) ux(3) Fx(3)
uy(2) 2
ux(2) Fx(2) 3
Fy(1)
uy(1) Fx(1)
Fy(1)
1 u (1) Fy(2)
uy(1) x
uy(2)
1 u (1) F (1)
x x
2 ux(2) Fx(2)
(a) (b)
Figure 5.3 Notation: (a) Truss element, (b) 3-node plane stress or plane strain element.
between the axial force F and the corresponding change in length 𝛿 is F = (AE∕𝓁)𝛿 where the
bracketed term is the spring constant of the bar. A bar element is shown in Fig. 5.3(a). The length
and orientation of a bar element are determined by the coordinates of the hinges, also called nodes,
indicated by the open circles in Fig. 5.3(a).
Letting
{F} = {Fx(1) Fx(2) Fy(1) Fy(2) }T
and
{u} = {u(1) (2) (1) (2) T
x ux uy uy }
the relationship between the nodal force components and the nodal displacement components can
be written as
{F} = [K]{u} (5.7)
where [K] is a singular matrix, called the element stiffness matrix, see eq. (5.9). The relationship
between the nodal forces and the nodal displacements are obtained for the entire truss by writing
the equations of force equilibrium for each node and factoring the nodal displacements. Finally, the
nodal constraints are enforced and the resulting system of linear equations, the coefficient matrix
of which is non-singular, is solved for the nodal displacements.
To extend matrix methods of structural analysis to continuum problems it was necessary to
establish an analogous relationship to eq. (5.7). For truss elements the nodal forces are the stress
resultants. No such interpretation exists for continuum problems, however. For continuum prob-
lems nodal force was interpreted to mean a force-like entity that is energy-equivalent to the strain
energy of the element [113]. For example, for the plane stress or plane strain triangle shown in
Fig. 5.3(b), we have:
{u}T {F} = {u}T [K]{u}
where {u} is the 6 × 1 nodal displacement vector and [K] is a symmetric positive semidefinite 6 × 6
matrix. The strain energy of the element was related to the nodal displacements through the strains
corresponding to the shape functions.
The goal was to formulate element stiffness matrices and load vectors for various applications.
This element-oriented view strongly influenced the development of finite element computer codes
and the intuitive understanding of the finite element method by the engineering community.
162 5 Simulation
(D, i) ? Fnum
Physical Element Numerical
reality library solution Prediction
Intuitive construction The error | ?-Fnum |

of a numerical problem is not defined
Figure 5.4 The main elements of finite element modeling.
The practice of finite element modeling, which is illustrated schematically in Fig. 5.4, is based on
this view.
On comparing Fig. 5.4 with Fig. 5.2 the difference between finite element modeling and numer-
ical simulation should become evident. In finite element modeling a transformation similar to
eq. (5.1) is performed. However, in the place of a precisely formulated idea of physical reality I
in the form of a mathematical model, intuition (i) is employed to construct a numerical problem
through assembling elements from the element library of a finite element software product:
(D, i) → Fnum . (5.8)
Practitioners of finite element modeling tacitly transfer the responsibility for defining the math-
ematical model to the developers of finite element software products. However, the question of
whether a numerical problem corresponds to a properly formulated mathematical model, and the
quantities of interest F are properly defined, cannot be answered without knowing which elements
were chosen and how those elements were formulated and implemented. In general, the details of
implementation are not readily available to users.
The element libraries contain elements identified by the technical theory (e.g. beam, plate, plane
stress, shallow-shell, solid, etc.), the number of nodes, the variational principle employed in the for-
mulation and the integration rule applied (full or reduced). For example, a hexahedral element for
the solution of thermoelastic problems may be designated as a “20-node triquadratic displacement,
trilinear temperature, hybrid, linear pressure, reduced integration” 11 .
Reduced integration means that fewer than the necessary minimum number of integration points
are employed when computing the terms of the stiffness matrix. This topic was addressed in many
papers. The general idea is that elements of low polynomial degree are “too stiff” and they become
less stiff when reduced integration is employed. See, for example, [114]. This is a typical finite mod-
eling argument in that element properties are modified without consideration of the underlying
variational principle. Strang and Fix referred to such practices as “variational crimes” [89].
Subsequently it was realized that reduced integration results in “hourglassing”, a term
that refers to the appearance of spurious deformation modes that have zero strain energy.
Additional numerical schemes had to be invented to counter the effects of hourglassing.
See, for example, [27].
Given that errors associated with finite element modeling cannot be estimated, one can reason-
ably ask whether finite element modeling has any value at all. This question has to be answered in
the affirmative, with certain qualifications, however: Importantly, one must distinguish between
structural and material strength models. At present the only practical way to analyze complicated
engineering structures, such as airframes subjected to static and dynamic loads, automobile bod-
ies under crash conditions and major civil engineering and marine structures, is through finite
element modeling. A finite element model is constructed by an intuitive process guided by expe-
rience. Finite element models of structural components and assemblies of structural components
11 DS Simulia Abaqus/CAE User’s Guide, Abaqus 6.14.

are constructed so as to approximate the stiffness of the components and sub-assemblies. Minor
details such as fillets, rivets and bolts are omitted. The stiffness of the finite element representation
of components and subassemblies are checked against experimental data and the finite element
model is modified (tuned) to match the experimental observations. The goal is to obtain reasonable
estimates of the force distribution among the structural components.
Interestingly, successful applications of finite element models are made possible by an approxi-
mate cancelation of two types of large error: Conceptual errors in the formulation (also known as
variational crimes), and the errors of discretization. The construction of a finite element model is
not a scientific activity but rather an intuition-based, experience-guided artistic undertaking.
Exercise 5.1 Show that for an ideal truss element

⎡ cos2 𝛼 −cos2 𝛼 sin 𝛼 cos 𝛼 − sin 𝛼 cos 𝛼 ⎤
⎢ ⎥
AE ⎢ −cos 𝛼
2 cos2 𝛼 − sin 𝛼 cos 𝛼 sin 𝛼 cos 𝛼 ⎥
[K] = (5.9)
𝓁 ⎢ sin 𝛼 cos 𝛼 − sin 𝛼 cos 𝛼 sin 𝛼
2
−sin 𝛼 ⎥
2
⎢− sin 𝛼 cos 𝛼 sin 𝛼 cos 𝛼 −sin 2
𝛼 sin2 𝛼 ⎥⎦
⎣
where 𝓁 is the length of the truss element and, denoting the coordinates of the nodes by (xi , yi ),
i = 1, 2,
sin 𝛼 = (y2 − y1 )∕𝓁, cos 𝛼 = (x2 − x1 )∕𝓁.
Exercise 5.2 Refer to Exercise 5.1. Show that the nodal forces satisfy the equations of equilibrium
independent of the nodal displacements, no matter what nodal displacements are prescribed.
(a) Fx(1) + Fx(2) = 0, (b) Fy(1) + Fy(2) = 0, (c) Fy(2) cos 𝛼 − Fx(2) sin 𝛼 = 0.
5.2.3 Calibration versus tuning

Mathematical models are calibrated. Finite element models are tuned. The differences between the
two are addressed in this section.
Calibration
In numerical simulation the functional form of a mathematical model is assumed to be correct
for the purpose of calibration. The parameters that represent material properties and boundary
conditions are estimated by matching observed data with what would have been predicted if the
parameters of interest were known. One very simple example is finding the modulus of elasticity
for a material by applying a known force to an elastic bar made of the same material and measuring
the corresponding displacement. Knowing the cross-sectional area and the length of the bar, the
modulus of elasticity can be readily calculated from this information. An example of a far more
challenging calibration problem will be presented in Chapter 6.
Calibrated data have an interval of validity. For example, for elastic moduli that interval is
bounded by zero and the proportional limits in tension and compression. The set of intervals on
which the coefficients are defined, together with the restrictions associated with the assumptions
incorporated in the mathematical model, define the domain of calibration.
In most cases the mathematical problem has to be solved numerically. In such cases it is necessary
to ensure that the errors in the computed quantities of interest are negligibly small in comparison
with experimental errors and uncertainties.
164 5 Simulation
Tuning
In finite element modeling the term “tuning” refers to adjusting the finite element mesh in such
a way as to match experimental observations. Therefore finite element models are interpolators
within a small neighborhood of tuned parameters.
Tuning is widely used in analyzing large structural systems such as automobiles under crash
conditions and structural analysis of airframes. In fact, finite element modeling is the only practical
means available for modeling such large systems at present.
Since validation refers to testing the predictive performance of a mathematical model while
ensuring that the errors of discretization are small, and in finite element modeling the model form
errors are not separated from the errors of discretization, finite element models cannot be validated.
It is possible to match certain observable quantities reasonably well with data obtained from
finite element models and yet have very large errors in other quantities of interest. This will be
demonstrated in Section 5.2.8.
5.2.4 Simulation governance

There are many challenging engineering problems that require the continued development and
improvement of the methods of numerical simulation. For example, as new materials and material
systems are developed, it is necessary to formulate, calibrate and test new constitutive laws and
failure criteria. Taking examples from the field of aerospace engineering, one area of great inter-
est is the development of design rules for fiber-reinforced composite laminates. Rule making for
condition-based maintenance is another important area: Given information concerning the ser-
vice history of a high value asset (such as an airframe), what is the optimal interval for inspection
and repair? What steps should be taken if damage is detected in a principal structural element12 ? To
answer these and similar questions it is necessary to formulate structural and strength models for
the prediction of crack formation and crack propagation rates based on the concepts and methods
of numerical simulation.
Simulation governance is a managerial function concerned with the exercise of command and
control over all aspects of numerical simulation through the establishment of processes for the
systematic improvement of the tools of engineering decision making. This includes: (a) proper
formulation of mathematical models, (b) selection and adoption of the best available numerical
simulation technology, (c) coordination of experimental work with numerical simulation, (d) doc-
umentation and archival of experimental data, (e) application of data and solution verification
procedures, (f) revision of mathematical models in the light of new information collected from
physical experiments and field observations and (g) standardization of design, analysis and certifi-
cation workflows whenever appropriate.
A plan for simulation governance has to be tailored to fit the mission of each organization or
department within an organization: If that mission is to apply established rules of design and certi-
fication then emphasis is on solution verification and standardization. If the mission is to formulate
design rules, or make condition-based maintenance decisions, then verification, validation and
uncertainty quantification must be part of the plan.
12 A principal structural element is an element of an aircraft structure that contributes significantly to the carrying
of flight ground, or pressurization loads and whose integrity is essential in maintaining the overall structural
integrity of the aircraft. See, for example, FAA Advisory Circular No. 25.571-1D.
2 7
1950 1 60 70 3 80 4 5 90 6 2000 10 8
Numerical Error control

Legacy FEA
simulation simple
Error control
FE libraries FE modeling
difficult
Figure 5.5 FEA timeline.
5.2.5 Milestones in numerical simulation

In order to make a clear distinction between finite element modeling and numerical simulation,
and to help understand why and how finite element modeling evolved before the foundations of
numerical simulation were established, a brief review of some of the major milestones in the devel-
opment of the finite element method is presented in the following. These milestones are indicated
by numbered arrows on the finite element analysis (FEA) timeline shown in Fig. 5.5.
1. The first engineering paper on the finite element method was published in 1956 [107]. In the
following year the Soviet Union launched the first satellite, Sputnik, and the space race began.
This brought about substantial investments into engineering and scientific projects in support
of the US space program which funded rapid development of various finite element computer
codes by academic institutions and aerospace companies.
2. NASA issued a request for proposal in 1965 that eventually led to the development of the finite
element software NASTRAN [56], versions of which are still in use today. This marked the begin-
ning of the development of legacy finite element codes. As indicated in the FEA timeline in
Fig. 5.5, the infrastructure of legacy codes was substantially established in the course of the
following 5 to 7 years. Its development was guided by the understanding of the finite element
method that existed in the 1960s. This infrastructure imposes limitations that prevent legacy
FEA codes from keeping pace with the evolution of the finite element method.
3. Systematic exploration of the mathematical foundations of the finite element method began
around 1972 [12]. This is a fundamentally important milestone because mathematicians view
the finite element method very differently from engineers: engineers (generally speaking)
think of the finite element method as a modeling tool that permits assembling various ele-
ments selected from the library of a finite element software product. They believe that the
solution of the corresponding numerical problem approximates the physical response of their
objects of interest such as airframes, turbine disks, pressure vessels, etc. to various loading
conditions.
Mathematicians, on the other hand, view FEM as a method by which approximate solutions can
be obtained to well-defined mathematical problems. For example, the equations of linear elas-
ticity, given a solution domain, material properties, loading and constraint conditions, define
a mathematical problem that has a unique exact solution uEX . The finite element solution,
denoted by uFE , is an approximation to uEX .
166 5 Simulation
Suppose that we are interested in a quantity Φ(uEX ). A key question is this: How close is Φ(uFE )
to Φ(uEX )? Relying on Φ(uFE ) without having some estimate of the size of the error of approxi-
mation in our quantity of interest (QoI) can and often does lead to mistaken conclusions. Finite
element models often do not correspond to a well-defined mathematical problem and therefore
it is not possible to determine what that error is. It is not uncommon in finite element mod-
eling practice that Φ(uEX ) is not a finite number. In such cases reporting Φ(uFE ) is a serious
conceptual error. See, for instance, Example 4.8.
The accuracy of approximation depends on the finite element mesh and the polynomial degree
of the elements. In the early implementations of the finite element method the polynomial
degree of the elements (denoted by p) was fixed at a low value, typically p = 1 or p = 2, and
the error of approximation was controlled by mesh refinement such that the size of the largest
element in the mesh, denoted by h, was reduced. This is known today as the h-version of the
finite element method.
In the mid-1970s research indicated that keeping the finite element mesh fixed and increasing
p, has important advantages [100]. This is known today as the p-version of the finite element
method.
4. In 1981 the p-version was analyzed in [24, 25]. It was proven and demonstrated that for a large
class of problems, which includes two-dimensional linear elasticity, the asymptotic rate of con-
vergence of the p-version in L2 and energy norms is at least twice that of the h-version with
respect to the number of degrees of freedom.
5. It was proven and demonstrated in 1984 that in most problems of engineering interest the
smoothness of the solution is such that when the finite element mesh is properly graded
then the finite element solution converges (in energy norm or L2 norm) exponentially as p
is increased [14, 90]. The h- and p-versions are special cases of the finite element method
where both the mesh and the polynomial degree are important in controlling the error of
approximation. The distinction between the h- and p-versions is rooted in the history of the
development of finite element method rather than in its theoretical foundations. The most
important practical advantage of the p-version is that it makes estimation and control of the
errors of approximation in the quantities of interest simple and efficient.
6. Any mathematical model can be viewed as a special case of a more comprehensive model. There-
fore any mathematical model is a member of a hierarchic sequence. For example, a model based
on the assumptions of the linear theory of elasticity is a special case of a model that accounts for
plastic deformation. Once the solution of a problem of linear elasticity is available, it is possible
to check whether the assumptions incorporated in the model are satisfied. If they are not satis-
fied then the analyst needs to employ a more comprehensive model. The implementation of a
hierarchic modeling framework began in the late 1980s [101]. It provides for seamless transition
from lower to higher models. As indicated in Fig. 5.5, the development and documentation of a
concept demonstrator for hierarchic models was substantially completed by the mid-1990s.
7. The American Society of Mechanical Engineers (ASME) published its first guideline on verifi-
cation and validation (V&V) in computational solid mechanics in 2006. The main point is this:
Whenever engineering decisions are based on the results of numerical simulation, there is an
implied expectation of reliability. Without such expectation it would not be possible to justify the
cost of a simulation project. If simulation produces misleading information then it has a negative
economic value with possibly severe consequences. There are many well documented instances
of expensive repairs, retrofits, project delays and serious safety issues arising from lack of quality
assurance in numerical simulation. Therefore assurance of the quality of the results of numeri-
cal simulation is essential. Specifically, the following quality control steps are recommended by
ASME: (a) code verification, (b) estimation of the errors of approximation in terms of the quan-
tities of interest and (c) validation of the mathematical model through comparison of predicted
outcomes with experimental observations.
8. The concept of simulation governance from the perspective of mechanical and structural engi-
neering was introduced in 2012 [92]. At the 2017 NAFEMS World Congress13 simulation gover-
nance was identified as the first of eight “Big Issues” in the field of numerical simulation.
5.2.6 Example: The Girkmann problem

In order to explore the differences between finite element modeling and numerical simulation in a
specific setting, readers of the bulletin of the International Association for Computational Mechan-
ics, called IACM Expressions, were invited to solve the following problem, known as the Girkmann
problem14 [39], using any finite element analysis software available to them.
A spherical shell of thickness h = 0.06 m, crown radius Rc = 15.00 m is connected to a stiffening
ring at the meridional angle 𝛼 = 2𝜋∕9 (40∘ ). The dimensions of the ring are: a = 0.60 m, b = 0.50
m. The radius of the mid-surface of the spherical shell is Rm = Rc ∕ sin 𝛼. The notation is shown
in Fig. 5.6. The z axis is the axis of rotational symmetry. The shell, made of reinforced concrete, is
assumed to be homogeneous, isotropic and linearly elastic with Young’s modulus E = 20.59 GPa
and Poisson’s ratio 𝜈 = 0.
Consider gravity loading only. The equivalent (homogenized) unit weight of the material is
32.69 kN∕m3 . Assume that uniform normal pressure pAB is acting at the base AB of the stiffening
ring. The resultant of pAB equals the weight of the structure. Assume that the stiffening ring is
weightless. The goals of computation are as follows:
h
Rm
Qα Mα
z
α
Mα Qα
b
Rc
A B
a
Rc
Figure 5.6 The Girkmann problem. Notation.
13 NAFEMS is an International Association for Engineering Modeling, Analysis and Simulation. The acronym
refers to its parent organization which was formed in 1983 in the United Kingdom: National Agency for Finite
Element Methods and Standards.
14 Karl Girkmann 1890–1959.
168 5 Simulation
1. Find the shearing force Q𝛼 in N/m units and the bending moment M𝛼 acting at the junction
between the spherical shell and the stiffening ring in Nm/m units.
2. Determine the location (meridional angle) and the magnitude of the maximum bending
moment in the shell.
3. Verify that the results are accurate to within 5%.
This problem statement appeared in the January 2008 issue if IACM Expressions [76]. This prob-
lem statement differed from the original problem statement by Girkmann in two respects: In [39]
kgf (resp. cm) was used for the unit of force (resp. length), in [76] SI units were used. Also, in [39]
one of the goals of computation was to determine the radial force per unit length between the ring
and the shell. In [76] the goal was to determine the shear force per unit length between the ring and
the shell. The responses received were summarized in a follow-on article published in the January
2009 issue [77] and in greater detail in [99].
Of the 16 solutions received 11 were obtained by finite element modeling methods implemented
in legacy finite element codes, that is, commercial finite element codes that have infrastructures
designed to support finite element modeling. Only two of the respondents who used legacy codes
attempted to perform solution verification. However, the quantities of interest either failed to con-
verge or appeared to have converged to the wrong result. The results are summarized in Table 5.1.
One respondent attempted to demonstrate h-convergence for a shell-solid model on one quarter
of the stiffened shell using six successive uniform mesh refinements. In the sixth refinement 120
million degrees of freedom were used. The sequence of moments corresponding to the six refine-
ments still had not converged but appeared to tend to approximately −205 Nm/m and the shear
force appeared to have converged to approximately 1140 N/m.
Another respondent wrote: “Regarding verification tasks for structural analysis software that has
adequate quality for use in our safety critical profession of structural engineering, the solution of prob-
lems such as the Girkmann problem represents a minuscule fraction of what is necessary to assure
quality.” This statement is obviously true. That is why it is surprising that the answers obtained
with legacy finite element codes had such a large dispersion. As shown in Table 5.1, the reported
Table 5.1 Legacy codes: Summary of results.
Q𝜶 M𝜶 𝝋max Mmax
Element N/m Nm/m degrees Nm/m
axisymmetric solid – 4-node elements 953.7 −10.57 − −

axisymmetric solid – 8-node elements 953.7 −19.67 − −
axisymmetric shell – solid 593.8 −140.12 − −
axisymmetric shell – solid − −78.63 − −
shell – solid 1140.0 −205.00 37.70 215.00
shell – solid 16660.0 17976.6 − −
axisymmetric solid 963.2 −33.73 − −
shell – solid 1015.7 86.30 − 231.09
axisymmetric shell – solid 949.2 −36.62 − −
shell – solid 951.3 −38.35 − −
axisymmetric shell - solid 989.1 −89.11 38.00 238.63
values of the moment at the shell-ring interface ranged between −205 and 17,977 Nm/m. Solu-
tion of the Girkmann problem should be a very short exercise for persons having expertise in finite
element analysis, yet many of the answers were wildly off. Analysts who cannot find a reasonably
accurate solution for the Girkmann problem are not in a position to claim that they can solve much
more complicated problems reliably.
Several respondents were not sure how the shell-ring interface should be treated. One respondent
stated the problem in this way: “Since the (axisymmetric) shell elements have three degrees of freedom
per node, while the (axisymmetric) ring elements have only two, it is not clear how to reconcile this
difference, yet getting this step wrong will give incorrect results that may not be obvious.” This is a
typical finite element modeling dilemma one encounters when different types of elements have to
be joined.
The results obtained by means of numerical simulation codes are listed in Table 5.2. These codes
provide a posteriori error estimation, an essential technical requirement in numerical simulation.
The note “Extraction” in the second entry in Table 5.2 refers to the use of extraction functions for
the computation of the shear force and bending moment at the shell-ring interface [99] from the
finite element solution. This procedure is analogous to the indirect computation of QoI discussed
in Section 1.4.1. The last two entries in Table 5.2 are based on work performed after the completion
of the Girkmann round robin study [66].
An interesting and surprising result of this study was that the bending moment M𝛼 obtained by
Girkmann, which also appeared in translation in [104], differed by a factor of nearly 3 from the veri-
fied solutions obtained with the numerical simulation codes, see Tables 5.2 and 5.3. The reasons for
this were investigated by Pitkäranta et al. [78]. It was found that assumptions made by Girkmann,
namely that the resultant of the distributed membrane force and the resultant of the distributed
axial reaction force pass through the centroid of the footring, were primarily responsible for this
discrepancy. The designation: M-B-RE model15 in Table 5.3 refers a particular adaptation of the
classical membrane theory, with bending theory used for accounting for the boundary layer effects
at the shell-ring interface, and an engineering theory (minimal-energy model) used for modeling
the ring.
Table 5.2 Numerical simulation codes: Summary of results.
Q𝜶 M𝜶 𝝋max Mmax
Model N/m Nm/m degrees Nm/m
axisymmetric solid 934.5 −34.81 − −

axisymmetric solid – Extraction [99] 943.6 −36.81 38.15 255.10
axisymmetric solid 940.9 −36.63 38.20 254.92
thin15 948.4 −37.31 38.20 254.50
axisymmetric solid 940.9 −36.80 38.15 254.80
axisymmetric solid hp version [66] 943.7 −36.79 38.14 254.90
axisymmetric shell-ring h version [66] 942.4 −37.36 38.14 254.10
15 Membrane theory (M), bending theory (B) and energy-based ring theory (RE).
170 5 Simulation
Table 5.3 Solution by classical methods: Summary of results.
Q𝜶 M𝜶
Method N/m Nm/m
Girkmann [39, 104] 1007.4 −110.5

Pitkäranta (M-B-RE model) [78] 944.9 −36.67
Remark 5.4 According to the problem statement, the shell is made of reinforced concrete, yet it
is assumed to be homogeneous and isotropic. While the two statements are contradictory, this is
a commonly used idealization in engineering practice. The inhomogeneity and anisotropy caused
by the reinforcement are neglected in structural analysis but taken into consideration in strength
analysis. In structural analysis problems the QoI are stress resultants, displacements and natural
frequencies. These QoI can be well approximated by homogenized models.
Remark 5.5 In this example we were not concerned with the question of whether the mathe-
matical problem is a realistic model of a spherical shell made of reinforced concrete and supported
by a footring, given the quantities of interest. We were concerned only with the verification of the
numerical solution of this mathematical problem in terms of the QoI.
5.2.7 Example: Fastened structural connection

In this example we analyze the fastened structural connection (lug) shown in Fig. 5.7 with the goal
to determine the shear force imposed by each fastener on the lug and the maximum tensile stress in
the lug in the vicinity of the fastener holes. We will be interested in evaluating the effects of various
modeling assumptions on the quantities of interest. We will consider (a) a model based on the
strength of materials, requiring only simple hand calculations, (b) models based on two assump-
tions concerning the boundary conditions at the fastener holes and (c) a finite element model.
The dimensions (mm) are as follows: a = 75, b = 55, s1 = 32.5, s2 = 40, d1 = 17.5, d2 = 17.5,
rh = 3.2, r1 = 25, r2 = 12.5, the thickness of the lug is t = 6.4 (constant). The lug is made of 2014-T6
aluminum. The elastic properties are: E = 7.52 × 104 MPa, 𝜈 = 0.397. Let F = 12.0 kN, 𝛼 = 35∘ .
d2 s2 Figure 5.7 Lug problem. Notation.
y
d1
1 2
r2 F
s1
rh (typ.) α
3 4
A x
s1
5 6 r1
d1
a b
Model 1: Strength of materials

We estimate the direction and magnitude of the force acting on the plate at each fastener hole by
hand calculation using the model described in Appendix J.
In this model the mathematical problem is simplified by assuming that the lug is a perfectly rigid
body which undergoes displacement and rotation allowed by the elasticity of the fasteners which
is simulated by linear springs. Assuming that the spring rate of each fastener is the same, and using
equations (J.14) and (J.15) (details are given in Appendix J), we find the components of the fastener
forces and determine their direction and magnitude. The results are shown in Table 5.4.
The errors in these data are caused entirely by the modeling assumption that the lug is perfectly
rigid. There is no approximation error in the magnitude and direction of the forces in the fasteners.
The errors are entirely model form errors.
To estimate the maximum stress at the most highly loaded fastener, one can make the modeling
assumption that the fasteners are sufficiently far apart so that the maximum stress is not signifi-
cantly affected by the neighboring fasteners. Therefore a surrogate problem can be solved in which
the maximum stress is computed for two fasteners separated by a large distance and loaded by
smoothly distributed tractions the resultants of which are equal but opposite forces. This approach
is representative of how engineers use simplified models to obtain rough estimates of the QoI.
Specifically, consider the problem shown in Fig. 5.8. This problem is concerned with two fastener
holes of radius rh , the centers of which are located at a distance 2𝓁. The perimeter of the holes is
loaded by sinusoidal normal tractions Tn corresponding to unit forces:
{
(−2∕(rh t𝜋)) cos 𝜃 for |𝜃| ≤ 𝜋∕2
Tn = (5.10)
0 for |𝜃| > 𝜋∕2
where t is the thickness of the plate. Symmetry boundary conditions are prescribed on line segments
AB, CD, DE. The line segments EF and FA are traction free.
Table 5.4 Fastener forces estimated by the method described in Appendix J
Fastener
1 2 3 4 5 6
Force (N) 1675.0 3409.5 1812.3 3479.1 4824.2 5665.0

Angle (deg) 27.6 −64.2 154.7 −118.1 170.8 −147.2
Figure 5.8 Surrogate problem. Notation.

F E
ℓ rh
θ
A B C D
ℓ ℓ
172 5 Simulation
Letting rh = 3.2 mm, t = 6.4 mms and solving the problem for 𝓁∕rh in the range of 10 to 1000, we
find that the maximum value of the principal stress converges to 0.0197 MPa. Therefore the esti-
mated value of the maximum normal stress in the vicinity of the fastener holes is 0.0197 MPa/N.
Hence the estimated maximum stress is 0.0197 × 5665.0 = 111.6 MPa which occurs on the perime-
ter of fastener hole #6.
The use of the surrogate problem introduced a second modeling error. This error can be elimi-
nated by applying sinusoidal normal tractions to the fastener holes which are statically equivalent
to the bearing forces in Table 5.4. The formula for Tn is the same as in eq. (5.10); however, 𝜃 is
measured from the line of action of the resultant force.
The maximum tensile stress in the vicinity of the fasteners is found to be 206.7 MPa, located on
the perimeter of fastener hole #6. On comparing this result with the estimate from the surrogate
problem (111.6 MPa) it is clear that the assumption that the local stress maximum depends only on
the force acting on the fastener hole introduces a substantial modeling error.
Remark 5.6 The surrogate problem is analogous to the dipole problem in electrostatics where
the electrostatic potential field corresponding to a pair of electric charges of equal magnitude but
opposite sign, separated by some distance, is of interest.
Model 2: The fasteners are modeled by linear springs

It is assumed that the normal and shearing tractions at the boundary of each fastener, denoted by
Tn and Tt respectively, are
Tn = −kn un , Tt = 0 (5.11)
where un is the normal displacement and kn is the spring rate, estimated to account for the propping
effect16 of the fastener shank. Assuming plane strain conditions, the spring coefficient, representing
the radial stress to displacement ratio of the fastener shank, is estimated from:
2E
kn = (5.12)
d(1 + 𝜈)(1 − 2𝜈)
where E is the modulus of elasticity, d is the diameter and 𝜈 is Poisson’s ratio of the fastener. We
assume that the fasteners have the same material properties as the lug. The fastener forces and their
directions are shown in Table 5.5.
Since fasteners interact with the lug by mechanical contact, and therefore only compressive trac-
tions can exist between the fasteners and the lug, this model cannot be expected to approximate
the fastener forces well. It serves, however, as a necessary first step toward solving the problem
discussed next.
Table 5.5 Fastener forces estimated by the linear spring model
Fastener
1 2 3 4 5 6
Force (N) 961.9 3426.5 884.8 3538.2 2363.2 8603.4

Angle (deg) 32.8 −39.0 139.0 −137.6 170.0 −153.4
16 The term “propping effect” refers to the resistance of the fastener shank to ovalization of the fastener hole.
Table 5.6 Fastener forces estimated by the nonlinear spring model
Fastener
1 2 3 4 5 6
Force (N) 1228.3 3557.0 1261.0 3593.5 3062.6 7471.1

Angle (deg) 33.1 −49.5 148.2 −132.3 169.3 −152.7
Model 3: The fasteners are modeled by nonlinear springs

This model is the same as Model 2; however, the contact between the fasteners and the lug is rep-
resented by springs that act in compression only:
{
−kn un for un > 0
Tn = (5.13)
0 for un ≤ 0
therefore the problem is nonlinear and must be solved by iteration. In the iterative solution the
first step is to solve Model 2, then continue with cancelation of the tensile spring tractions until a
prescribed tolerance is reached. In solving this problem the tolerance 𝜏 was set at 0.001. This means
that iteration continued until the condition
||ak − ak−1 ||2 ≤ 𝜏||ak ||2 (5.14)
where ||ak ||2 is the Euclidean norm17 of the solution vector at the kth iteration, was reached.
Five iterations were needed. The computed fastener forces are shown in Table 5.6. The maximum
stress in the lug occurs on the perimeter of fastener #6. Its value is 254.9 MPa. These results are not
sensitive to the spring rate assigned to the fasteners. For example, we find that a 10% reduction in
the spring rate causes the maximum principal stress to change by 0.1 N (0.04 %) and the maximum
fastener force by 27 N (0.4 %) only.
Model 4: The three-dimensional contact problem

The previously discussed three models were designed to approximate the solution of the
three-dimensional multibody contact problem discussed in this section. In this model the lug and
fastener shanks are treated as three-dimensional elastic bodies. The fastener shanks are idealized
as cylindrical bodies of radius rh and length 𝓁 centered on each fastener hole. The boundary
conditions at 𝓁 = ±t∕2 are ux = uy = Tz = 0. Frictionless contact condition is prescribed on the
cylindrical surface:
Tn g = 0 (5.15)
where Tn is the normal traction and g is the gap function between the fastener shanks and the lug.
The tangential traction, denoted by Tt , is zero. The initial gap function is assumed to be zero. The
contact problem is solved by iteratively adjusting Tn until eq. (5.15) is satisfied to within a prescribed
tolerance. Denoting the kth increment of Tn by (ΔTn )k , the stopping criterion is
max (ΔTn )k < 𝜏max Tn (5.16)
S S
where S is the cylindrical surface and 𝜏 is the tolerance. For example, letting 𝓁 = t for all fasteners,
𝜏 = 0.01 and using the finite element mesh shown in Fig. 5.9(a) with p = 8, we get the results shown
in Table 5.7.
17 See Definition A.3 in the appendix.

174 5 Simulation
F/2
F
F/2
y
z
x (b)
(a)
Figure 5.9 (a) Three-dimensional contact problem. 50 pentahedral and 108 hexahedral elements.
Hand-mesh. (b) Typical bolted connection detail. Double shear.
Table 5.7 Fastener forces estimated by the multibody contact model
Fastener
1 2 3 4 5 6
Force (N) 1102.8 3395.4 1348.6 3511.7 3117.7 7063.5

Angle (deg) 30.1 −49.4 156.0 −132.6 172.9 −153.0
The fastener forces in Table 5.7 were obtained by numerical integration. The error in equilibrium
was found to be 0.03%. The estimated maximum stress in the lug is (𝜎1 )max = 287.4 MPa which
occurs on the perimeter of fastener hole #6. It was verified by p-extension that the relative error in
the reported results is less than 1%.
Remark 5.7 In the formulation of models 1 through 4 it was assumed that there is no friction
between the fasteners and the lug and between the lug and the plates to which the lug is connected,
as shown in Fig. 5.9(b). This assumption is used in the design of shear joints. To test the predictive
performance of this model in a validation experiment, the structural connection has to be lubri-
cated. Connections can be designed such that forces are transmitted through friction between the
connected plates rather than the shear forces in the fasteners. Such connections are called tension
joints.
Discussion
Mathematical models are formulated with the objective to simulate some aspect of physical reality,
in the present instance the functioning of a fastened structural connection from the perspective of
allowable stress design. The foregoing discussion focused on the definition of four mathematical
models. In the case of Model 1 numerical approximation was not involved. In the cases of Models 2,
3 and 4 the numerical approximation errors were verified to be under 1% in the QoI. Therefore the
differences in the fastener forces shown in Tables 5.4 through 5.7, as well as the differences in the
computed maximum normal stress, are caused by the differences in modeling assumptions.
The credibility of data generated by numerical simulation increases when the number of simpli-
fying assumptions incorporated in a model is decreased. Therefore the results obtained by Model
4 are more credible than the results from the other three models. Note that the complexity of the
model increases as the number of simplifying assumptions is decreased. An important question
associated with the formulation of any mathematical model is: What is the minimum level of com-
plexity that will make the model good enough to serve its intended purpose? This question18 cannot
be answered without testing the effects of modeling assumptions on the QoI. For example, the
maximum stress in the lug, predicted by Model 3 (resp. Model 4), was 254.9 MPa (resp. 288.9 MPa),
11% difference. If this difference is acceptable then Model 3 is good enough for its intended
purpose. We could not have assessed this without actually solving the problem using Model 4,
however.
Up to this point the model of the lug was treated as a deterministic problem assuming that
there is a perfect fit: The holes are exactly the same size as the fastener shanks, there is no gap
or interference between the fastener shanks and the lug. In reality there is a substantial uncer-
tainty concerning the quality of the fit: Even under laboratory conditions, care having been taken
to achieve a sliding fit by carefully reaming the holes, it was not possible to eliminate the effects of
gaps [57]. It is therefore necessary to account for the effects of gaps and interferences. This can be
done by modifying the boundary condition given by eq. (5.13) to:
{
−kn (un + 𝛿n ) for un + 𝛿n > 0
Tn = (5.17)
0 for un + 𝛿n ≤ 0
where 𝛿n is the interference or gap, that is, 𝛿n = (df − dh )∕2 where df is the diameter of the fastener
and dh is the diameter of the hole. Uncertainties in the values of 𝛿n are the dominant uncertainties
in predicting the distribution of fastener forces and hence the maximum stress in the lug.
Example 5.1 In this example we assume that fasteners 1 through 5 are tightly fitted, that is 𝛿n(i) = 0
for i = 1, 2, … 5, but fastener 6 has a 0.025 mm gap; 𝛿n(6) = −0.025 mm. We are interested in the
distribution of the fastener forces and the maximum principal stress computed by the nonlinear
spring model. The results are shown in Table 5.8
On comparing Table 5.8 with Table 5.6, it is seen that the gap at fastener #6 causes a substantial
redistribution of forces. The maximum principal stress in the vicinity of the fasteners is 189.2 MPa.
It occurs on the perimeter of fastener #6. Therefore there is a substantial change in the maximum
stress as well.
The quantities of interest are very sensitive to the quality of fit characterized by 𝛿n(i) . It would be
possible to simulate the response of the structural connection to loads by Monte Carlo simulation.
However, the results will be sensitive to the assumptions concerning the statistical distribution of
𝛿n(i) which would be difficult to validate.
Table 5.8 Example 5.1. Fastener forces estimated by the nonlinear spring model
Fastener
1 2 3 4 5 6
Force (N) 1560.1 4629.5 2099.2 5848.0 5784.9 2732.4

Angle (deg) 36.6 −54.6 150.7 −131.9 172.0 −146.0
18 This question is related to the law of parsimony which is the philosophical principle that if a problem can be
solved by more than one method then one should select the method that involves the fewest assumptions. This
principle is attributed to William of Ockham (c. 1287–1347), an English philosopher.
176 5 Simulation
5.2.8 Finite element model

In finite element modeling practice it is not uncommon to ignore the diameter of the fasteners and
impose displacement constraints on nodes located where the center of each fastener would be. The
fastener forces are then estimated by computing the nodal forces from the finite element solution.
The solution of this problem does not lie in the energy space: If the mesh were progressively
refined then the strain energy, as well as the maximum displacement, would increase indefinitely.
The fastener forces are estimated by summing those nodal force components that correspond to
a constrained vertex over all elements that share that vertex. It will be shown that these fastener
forces satisfy the equations of equilibrium exactly, independent of the finite element solution.
The fastener forces for the 2227-element mesh of six node triangles, shown in Fig. 5.10, are listed
in Table 5.10. Plane stress is assumed. On comparing the entries in Table 5.6 and Table 5.9 it is
seen that the fastener forces are not far apart. This result is a consequence of two large errors
nearly canceling one another. One error is conceptual; point constraints are not allowed in two-
and three-dimensional elasticity. The other error is numerical; the error of approximation in the
fastener forces is large. It is a reasonable conjecture that the fastener forces will converge to the
fastener forces computed for Model 1, see Table 5.4.
In order to demonstrate that the displacements diverge, we compute the average displacement
of the perimeter of the 25 mm diameter hole (radius r2 ) in the direction of the applied force. The
average displacement is defined by
2𝜋
1
dave = (ux cos 𝛼 + uy sin 𝛼) d𝜃. (5.18)
2𝜋 ∫0
where 𝛼 is the angle shown in Fig. 5.7.
Figure 5.10 Finite element mesh consisting of 2227 triangles. The constrained nodes are indicated by
heavy dots.
Table 5.9 Fastener forces estimated by finite element modeling
Fastener
1 2 3 4 5 6
Force (N) 1220.5 3190.9 1175.6 3304.9 3222.8 7760.6

Angle (deg) 29.7 −46.7 146.1 −130.0 171.7 −151.1
0.13
Average displacement (mm)
Model 3 (0.127 mm) converged

0.12 8
6 7
5
Model 4 (0.106 mm) converged 4
0.11 3
2
0.1
Model 2 (0.0928 mm) converged
p=1
0.09
104 105
N
Figure 5.11 Average displacement of the perimeter of the 25 mm diameter hole in the direction of the
applied force vs. degrees of freedom.
A sequence of solutions were obtained by increasing the polynomial degree of the elements uni-
formly from 1 to 8 on the 2227 element mesh shown in Fig. 5.10. The computed values of dave are
plotted against the number of degrees of freedom on a semi-logarithmic graph in Fig. 5.11. Slow
(logarithmic) divergence is observed. The exact value of dave is infinity. In this example the com-
puted value of dave corresponding to p = 2 is 0.105 mm, hence the absolute error is infinity and
the relative error is 100%. However, because the displacements diverge very slowly, credible results
were obtained with this finite element model: The computed values of dave fall between dave pre-
dicted by Model 2 and Model 3, see Fig. 5.11. This is an example of near cancelation of two very
large errors; the conceptual and the numerical errors.
Remark 5.8 A typical engineering quantity of interest is the spring rate of a structural connec-
tion. One possible definition of spring rate is k = F∕dave where F is the applied force indicated in
Fig. 5.7. Another definition of spring rate would be k = F 2 ∕(2U) where U is the strain energy. Since
dave → ∞ and U → ∞ as the number of degrees of freedom is increased, the spring rate predicted by
the finite element model would converge to zero in both cases if the number of degrees of freedom
were progressively increased.
Whereas the displacement and the spring rate of a structural connection could be estimated
reasonably well by finite element modeling, the estimated maximum stress in the vicinity of the
fasteners, computed from the finite element solution, is very sensitive to the discretization and
therefore the probability that a particular discretization would result in near cancelation of the
conceptual and numerical errors is very low.
In the specific instance of the finite element mesh shown in Fig. 5.10, using 3-node and 6-node
plane stress triangles (p = 1 and p = 2 respectively), the computed maximum stress values are
located at fastener #6 in both cases. However, they differ greatly, as shown in Table 5.10. On the
other hand, if we were interested in the strain reading in a point, such as in point A located at
x = −37.0 mm, y = 0 (see Fig. 5.7), denoted by (𝜖x )A , then the result from the finite element model
would be reasonably close to the result from Model 4 and would in fact converge if the size of the
elements were progressively reduced or the polynomial degree increased [20].
The results in Table 5.10 demonstrate that, depending on the quantities of interest, near cance-
lation of the conceptual and numerical errors may or may not occur in finite element modeling.
178 5 Simulation
Table 5.10 Quantities of interest from finite element modeling and Model 3 (ksi).
Element N (𝝈1 )max (𝝐x )A
3-node (p = 1) 2356 184.0 2.72E − 4

6-node (p = 2) 9178 468.6 2.67E − 4
Model 3 254.7 2.87E − 4
Figure 5.12 Nodal forces. Notation.

Fy(J )
fy(K )
J
fx(K) Fx(J )
Element K
Equilibrium of nodal forces

In this section we show that the nodal forces satisfy the equations of equilibrium exactly, indepen-
dent of the finite element solution. Therefore equilibrium of nodal forces should not be interpreted
to mean that the finite element solution is accurate.
We refer to the example of the point-constrained lug shown in Fig. 5.10 and consider the group
of elements that share a constrained node, as shown in Fig. 5.12.
The set of constrained nodes is denoted by  . The set of element numbers that share node J ∈ 
is denoted by J . Uppercase indices refer to global numbering, lowercase indices to local number-
ing. The polynomial degree assigned to element K is denoted by pK . Without loss of generality, we
assume that the local numbering of vertices is such that vertex 1 is coincident with the constrained
node. The inverse of the mapping functions is denoted by
𝜉 = Q(K)
𝜉
(x, y), 𝜂 = Q(K)
𝜂 (x, y)
and define on element K ∈ J

V(K) (K) (K)
x (x, y) = {N1 (Q𝜉 , Q𝜂 ) 0} ,
T
V(K) (K) (K) T
y (x, y) = {0 N1 (Q𝜉 , Q𝜂 )}
where N1 (𝜉, 𝜂) is the shape function associated with vertex 1. By definition, the nodal force compo-
nent fx(K) is
∑
2nK
fx(K) = (K) (K)
k1j aj = B(u(K)
FE
, V(K)
x ), K ∈ J (5.19)
j=1
where kij(K) is an element of the stiffness matrix of finite element K, a(K)

j
is the coefficient of shape
function j in the local numbering convention,
nK = (pK + 1)(pK + 2)∕2
is the number of shape functions.
By definition, the nodal force component Fx(J) is
∑ (K)
Fx(J) = fx = B(u(K)
FE
, v(J)
x ) (5.20)
k∈J
where v(J)
x defined by
∑ (K)
v(J)
x = Vx (5.21)
K∈J
is an extraction function for Fx(J) . It is a function that lies in the finite element space and is zero on
all elements with the exception of elements numbered K ∈ J . Letting wx = {1 0}T we have
( )
∑ (J) 2𝜋
B uFE , wx − vx = r2 t Tx (𝜃)d𝜃 = F cos 𝛼. (5.22)
∫0
J∈
Since wx is rigid body displacement we have B(uFE , wx ) = 0 and, using eq. (5.20), we get the equi-
librium equation
∑ (J)
− Fx = F cos 𝛼. (5.23)
J∈
Analogously letting wy = {0 1}T we get

∑ (K)
Fy(J) = fy = B(u(K)
FE
, v(J)
y ) (5.24)
k∈J
and hence
∑ (J)
− Fy = F sin 𝛼. (5.25)
J∈
To show that the nodal force components satisfy moment equilibrium, we define the rigid body
rotation vector wrot = {−y x}T and write
( )
∑ (J) (J)
2𝜋
B uFE , wrot − (−yJ vx + xJ vy ) = r2 t (−yTx + xTy )d𝜃 (5.26)
∫0
J∈
where xJ , yJ are the coordinates of the Jth node. Since wrot is a rigid body rotation, we have
B(uFE , wrot ) = 0 and, using equations (5.20) and (5.24), we get
∑ 2𝜋
− (−yJ Fx(J) + xJ Fy(J) ) = r2 t (−yTx + xTy ) d𝜃. (5.27)
∫0
J∈
This is the equation of moment equilibrium of the nodal forces and the applied tractions.
Remark 5.9 Whereas the nodal forces depend on the finite element solution, satisfaction of
the three equations of equilibrium (5.23), (5.25) and (5.27) does not. Therefore satisfaction of the
equations of equilibrium by the nodal forces is not an indication of the quality of the finite element
solution.
Remark 5.10 It is possible to show that the nodal forces acting on any subset of elements satisfy
the equations of equilibrium. This is used in engineering practice to isolate parts a structure for
detailed analysis. An isolated part is visualized as a free body subjected to the nodal forces.
Discussion
The point-constrained lug problem does not have an exact solution, nevertheless predictions of
certain data based on finite element approximations of the nonexistent solution can be close to
what would be observed in a physical experiment. One of the reasons for this is that observable
quantities, for example the displacement of any point, strains measured by strain gages, will give a
reasonably close reading to the reading predicted by the finite element model [20].
180 5 Simulation
The second reason is that in some quantities of interest a cancelation of two large errors occur.
This is illustrated by the average displacement of the 25 mm diameter hole in Fig. 5.11. This QoI
cannot be verified because its exact value is infinity. However, divergence with respect to increas-
ing the number of degrees of freedom, whether by h- or p-extension, is very slow and not eas-
ily visible. In the practice of finite element modeling verification of the QoI is rarely performed
and the observed displacement can be close to the displacement predicted by the finite element
model. This will appear to support the (false) conclusion that the finite element model passed a
validation test. Such a conclusion would be false because validation refers to testing the predic-
tive performance of a properly formulated mathematical model. This is possible only if it is first
shown that the errors of numerical approximation of the QoI are small. If the QoI corresponding
to the exact solution is not finite then the model was not properly formulated and hence cannot be
validated.
Since the fastener holes are omitted from the finite element model, this model cannot be used
for estimating the maximum tensile stress in the lug in the vicinity of the fastener holes which,
according to the problem statement considered here, is a QoI.
Exercise 5.3 Explain why B(uFE , wrot ) = 0 where wrot = {−y x}T . Consider two-dimensional
elasticity only.
Exercise 5.4 Write down the three displacement vectors that represent infinitesimal rigid body
rotations about the x, y and z axes in three-dimensional elasticity.
5.2.9 Example: Coil spring with displacement boundary conditions

The centerline of the coil spring shown in Fig. 5.13 is given by
x = rc cos 𝜃 − 𝜋 < 𝜃 < 11𝜋
y = rc sin 𝜃 − 𝜋 < 𝜃 < 11𝜋
Figure 5.13 Coil spring.

⎧0 −𝜋 < 𝜃 ≤ 0
⎪
z = ⎨𝜃 d∕(2𝜋) 0 < 𝜃 ≤ 10𝜋
⎪5d 10𝜋 < 𝜃 < 11𝜋
⎩
where rc = 50.0 mm is the coil radius, d = 25.0 mm is the pitch. The solution domain is such that
any section perpendicular to the centerline is circular with radius r𝑤 = 5.0 mm (the wire radius).
However, the wire is truncated at the ends by the cutting planes z = 0, z = 5d.
The spring is made of AISI 5160 alloy steel, modulus of elasticity: 200 GPa, Poisson’s ratio: 0.285,
yield strength: 285 MPa. Assume that the axial displacement uz at z = 0 is zero and at z = 5d it is
uz = Δ, Δ < 0.
The objectives are to determine the spring rates in N/mm units for Δ = 0 and Δ = −25 mm and
to verify that the errors of approximation in the reported values are not greater than 3%.
The solution is presented in two parts: First, the solution of the linear model is described. By
linear model we understand a problem of linear elasticity: The stress-strain relationship follows
Hooke’s law and the displacement is sufficiently small so that the equilibrium equations can be
written with reference to the undeformed configuration. Second, we solve a geometrically nonlin-
ear model. In this formulation it is assumed that a linear relationship exists between the Cauchy
stress and the Almansi strain tensors. For details we refer to Section 9.2.1. In this formulation
equilibrium is satisfied in the deformed configuration and the effects of stress stiffening (or stress
softening) are taken into account. Solution of the linear problem is the necessary first step toward
solving the nonlinear problem.
Solution of the linear model

We compute the strain energy U(Δ) for an arbitrary fixed value of the displacement Δ imposed
on the spring and use U = kΔ2 ∕2 to determine the spring rate k. Using an automatically gener-
ated mesh of 9,691 tetrahedral isoparametric (10-node) elements and letting the polynomial degree
p range from 2 to 5 (while keeping the mapping functions fixed), we get the results shown in
Table 5.11 where p is the polynomial degree, N is the number of degrees of freedom and the percent
error was estimated by extrapolation using the method described in Section 1.5.3.
Using the extrapolated value of U we get
k = 2U∕Δ2 = 20.83 N/mm.
The estimated relative error in k is less than 0.05%. Note that this error does not account for the
curved surfaces being approximated by piecewise quadratic polynomials. However, given the large
number of 10-node tetrahedra (see Fig. 5.14), the errors of approximation in the surface represen-
tation can be neglected.
Table 5.11 Linear model. Computed values of the strain energy (Δ = −25 mm).
p N U (Nmm) % error
2 58,67 6558.910 0.78

3 173,989 6515.587 0.12
4 385,200 6512.737 0.07
5 721,473 6511.252 0.05
est. limit 6507.872 −

182 5 Simulation
More generally, the stiffness of the spring can be computed by energy methods. These methods
are superconvergent. We assume that the displacements imposed on the boundary at z = 5d can be
written as a linear combination of the displacement in the z direction (Δ) and the rotation about
the x and y axes, denoted by 𝜃x and 𝜃y respectively. The corresponding stress resultants are the axial
force Fz and the moments Mx , My . Their relationship, determined by energy methods19 , is:
⎧ F ⎫ ⎡20.83 0.00 69.40 ⎤ ⎧Δ ⎫

⎪ z⎪ ⎢ ⎪ z⎪
⎨Mx ⎬ = ⎢ 0.00 58073 0.00 ⎥⎥ ⎨ 𝜃x ⎬
⎪My ⎪ ⎣69.40 0.00 58928⎦ ⎪ 𝜃y ⎪
⎩ ⎭ ⎩ ⎭
where the units of the numerical data are N/mm, N or Nmm as appropriate for dimensional con-
sistency. It is seen that for the load case Δ = −25 mm, 𝜃x = 𝜃y = 0 we have Fz = −520.8 N, Mx = 0
and My = −1735 Nmm. Owing to the end conditions, the line of action of the axial force is not coin-
cident with the z−axis but passes through the point (x0 , y0 ) where x0 = −69.40∕20.83 = −3.3 mm
and y0 = 0.
Another method for estimating the spring rate is to compute Fz by integrating the normal stress
on the plane surface at z = 0 or z = 5d and divide Fz by Δ. This method would converge very slowly,
however, because at the edges of the planar surface the stress field is perturbed by singularities. It is
much better to “cut” the spring at an arbitrary value of 𝜃 by a plane perpendicular to its centerline
and compute the resultants of the stress components acting on that plane by numerical integration.
For example, cutting at the mid-point of the spring (𝜃 = 5𝜋), we define a local coordinate axis
such that the x′ , y′ and z′ axes are aligned respectively with the tangent, normal and binormal
of the centerline, as shown in Fig. 5.14. The solution at this cut (as well as at any cut away from
the ends) has sufficient smoothness to permit accurate determination of the stress resultants by
numerical integration.
The computed values of the stress resultants are shown in Table 5.12 for polynomial degrees
ranging from 2 to 5. The results show that the dominant stress resultants are the shear force Fz′ and
the twisting moment Mx′ .
The estimated limit values (corresponding to p → ∞) are the stress resultants computed by the
superconvergent extraction procedure. The small differences between the estimated limit values
and the numerically computed values serve as evidence that the numerically computed values at
Figure 5.14 Solution domain and finite element mesh in

the interval 0 < 𝜃 < 5𝜋.
xʹ
zʹ
yʹ
19 The diagonal terms are computed first from the strain energies corresponding to unit values of Δz , 𝜃x , 𝜃y . Then
the off-diagonal terms are computed from the strain energy values corresponding to pairwise imposition of the unit
displacement components.
Table 5.12 Linear model. Computed stress resultants in the local coordinate system at 𝜃 = 5𝜋
p N Fx ′ Fy ′ Fz′ Mx ′ My ′ Mz′
2 58,785 53.1 34.9 −510.7 −24002.0 253.0 1847.6

3 174,046 −43.5 1.0 −521.7 −24220.8 18.8 1919.7
4 385,332 −40.3 1.7 −519.64 −24217.8 10.9 1923.3
5 721,728 −41.9 0.8 −520.0 −24221.2 11.7 1928.7
est. limit −41.3 0 −519.1 −24228.4 0 1928.0
p = 5 are sufficiently accurate and the errors of approximation are well below the specified tol-
erance of 3%. The development of such evidence, needed for the interpretation of results for the
nonlinear problem, is discussed next. For nonlinear problems the resultants have to be computed
by integration.
On transforming the force vector in Table 5.12 to the global coordinate system we get Fx = Fy = 0,
Fz = −520.7 N from which k ≈ 20.83 N/mm. The quantity of interest computed by two different
methods has the same value to four significant digits.
Solution of the nonlinear model

In the linear model the equations of equilibrium refer to the undeformed configuration. In the
case of this spring the deformed configuration may differ from the original configuration to such
an extent that the difference between the two configurations cannot be neglected.
Since the deformed configuration is not known at the outset, we start with the linear model. We
update the mapping of each element by adding the linear displacement vector to the position vector
corresponding to the original configuration and take into account the nonlinear part of the strain
tensor20 . These steps are repeated until the stopping criterion is satisfied. The stopping criterion is
based on the change in value of the apparent potential energy Πi defined by
1 T
Πi = x [K ]x − xTi ri
2 i i i
where the index i refers to the ith iteration, xi is the solution vector, [Ki ] is the stiffness matrix and ri
is the load vector. Iteration is stopped when the normalized change in Πi , denoted by 𝜏 and defined
by
√
𝜏 = |Πi − Πi−1 |∕|Πi | (5.28)
is less than a specified value 𝜏stop . In this example 𝜏stop = 0.005 was used.
The computed stress resultants for Δ = −25 mm in the global coordinate system are listed in
Table 2. The first group of entries with p ranging from 2 to 5 were computed using the same 9,695
element mesh on which the linear solution was based. The last entry in Table 5.13, set in italic, was
computed using a finer mesh, consisting of 20,907 10-node tetrahedral elements. It is seen that the
quantity of interest (Fz ) converges to three digits accuracy, yielding the estimate for the spring rate:
k = 20.8 at Δ = −25 mm.
20 Depending on the problem, an incremental procedure may have to be used. In the present case it was not
necessary to apply the imposed displacement incrementally.
184 5 Simulation
Table 5.13 Nonlinear model, Δ = −25 mm. Computed stress resultants in the global coordinate system at
𝜃 = 5𝜋
p N Fx Fy Fz Mx My Mz
2 58,785 33.4 −91.5 −505.0 5070.0 502.0 4488.0

3 174,046 1.1 2.0 −521.5 178.9 −1719.1 −108.0
4 385,332 1.7 −1.0 −519.3 325.7 −1578.8 48.3
5 721,728 0.8 0.5 −519.8 253.6 −1646.0 −21.3
5 1,521,598 0.1 −0.1 −520.4 276.0 −1674.4 4.1
Since we have considered geometric nonlinearities only, it is necessary to ascertain that the mate-
rial remains elastic in the range −25 < Δ < 0. On computing the maximum von Mises stress, we
find that its estimated value is 278.8 MPa which is below the yield point of 285 MPa.
In estimating the maximum von Mises stress the boundaries at z = 0 and z = 5d were excluded.
This is because the idealized boundary conditions induce singularities along the curved edges. In a
physical experiment the displacement would have to be imposed by mechanical contact and there-
fore the displacement on the boundaries would not be perfectly constant. These singularities are
artifacts of the modeling assumptions. Solving the contact problem would have introduced other
kinds of local perturbation.
Discussion
Imposing constant normal displacements is one of many possible idealizations of the displacement
boundary condition. Such boundary conditions are called “hard” boundary condition. The “soft”
displacement boundary condition would be to set the integral of the normal displacements to be a
fixed value:
uz dxdy = Δ, uz dxdy = 0
∫𝜕Ω+ ∫𝜕Ω−
and the integral of rotations about the x and y axes to zero:
xuz dxdy = xuz dxdy = yuz dxdy = yuz dxdy = 0

∫𝜕Ω+ ∫𝜕Ω− ∫𝜕Ω+ ∫𝜕Ω−
where 𝜕Ω+ and 𝜕Ω− refer to the planar surfaces at z = 5d and z = 0 respectively. Between the hard
and soft boundary conditions are the “semisoft” boundary conditions corresponding to the nth
moment of the displacement uz (n = 2, 3 …) being set to zero.
The choice of boundary conditions is a modeling decision. Assessment of the sensitivity of the
quantities of interest to the choice of boundary conditions is one of the tasks in formulating a math-
ematical model.
5.2.10 Example: Coil spring segment

In this section we analyze a segment of the coil spring described in Example 5.2.9 assuming that
the spring is compressed by two equal and opposite forces acting along the z axis. The goals are to
estimate the spring rate and the distribution of the von Mises stress.
The spring segment 0 < 𝜃 < 𝜋∕3 is shown in Fig. K.4 (in Appendix K) together with the
18-element mesh used in this example. The imposition of statically equivalent forces and moments
on sections A and B in the local coordinates shown in Fig. K.4 is described in Section K.6. The
distribution of tractions corresponding to the technical formulas for rods is described in Section
K.6.1 and in Example K.4. These tractions satisfy the equations of equilibrium, hence only rigid
body constraints have to be applied.
Solution
The contours of the von Mises stress (MPa) on a 60 degree segment of the coil spring corresponding
to axial force F = 1.0 N are shown in Fig. 5.15. As explained under Remark K.4, the imposition
of tractions through technical formulas induce local perturbations of the solution that decay in
accordance with Saint-Venant’s principle. This is visible in Fig. 5.15.
In order to minimize the error caused by local perturbations of the solution, we estimate the
strain energy of the coil from the strain energy of the set of 6 elements labeled B in Fig. 5.15 that
cover a 20-degree segment of the coil. The results are presented in Table 5.14.
Let us compare the spring rate computed in Section 5.2.9 with the spring rate predicted by the
current model. In this case we use k = F 2 ∕(2U) where F = 1.0 N and U = 5 × 18 × 2.945412E − 4
Nmm to get k = 18.86 N/mm. The difference in spring rates is 9.45%. Since the numerical error
in the computed strain energy is negligibly small in both cases, the difference is caused by the
differences in modeling assumptions.
The solution of this problem by classical methods, based on work by Wahl [109], is available
in engineering handbooks, such as in [111]. Using the notation introduced in Section 5.2.9, the
displacement of a coil spring with n active turns is estimated to be
( ( )2 ( )2 )
4Frc3 n 3 r𝑤 3+𝜈 d
ΔW = 1− + (5.29)
Gr𝑤4 16 rc 2(1 + 𝜈) 2𝜋rc
0.60
0.50
0.40
A 0.30
B
0.20
0.10
C
0
Figure 5.15 Contours of the von Mises stress (MPa) on a 60 degree segment of the coil spring
corresponding to axial force F = 1.0 N.
Table 5.14 Linear model of the coil segment. Computed values of the
strain energy on a 20 degree segment (F = 1.0 N)
p N U (Nmm) % error
5 1152 2.945385E-4 0.00

6 1674 2.945397E-4 0.00
7 2340 2.945407E-4 0.00
8 3168 2.945410E-4 0.00
∞ ∞ 2.945412E-4 −
186 5 Simulation
where F is the axial force and G is the modulus of rigidity. On substituting the parameter values
defined in Section 5.2.9, we find the estimated spring rate to be kW = 19.33 N/mm.
The spring rate predicted by the classical formula is greater than what is predicted by solving
the three-dimensional problem of elasticity. This can be expected because the curved bar model,
on which the classical formula is based, imposes restrictions on the modes of deformation, i.e. the
space E(Ω). Therefore the minimum value of the potential energy for the bar model has to be larger
than for the 3D elasticity model. Consequently the bar model underestimates the strain energy of
the 3D model and therefore overestimates the spring rate, in this case by 2.5%.
The estimated value of the maximum shearing stress is given by the formula [111]:
( ( )2 )
5 r𝑤 7 r𝑤
𝜏max = 𝜏nom 1 + + (5.30)
4 rc 8 rc
where
2Frc
𝜏nom ≡ . (5.31)
𝜋r𝑤
3
On substituting the parameter values defined in Section 5.2.9 we find 𝜏max ∕𝜏nom = 1.134. The
three-dimensional finite element solution converges to 𝜏max ∕𝜏nom = 1.164 a difference of 2.6%.
We see that the classical formulas give very reasonable estimates for the spring stiffness and the
maximum shear stress.
Discussion
The formulation described in this section has the advantage that it is suitable for casting it into the
form of a “smart application”. Smart applications, also called “smart apps”, are expert-designed,
user-friendly software products that allow users, who do not need to be trained analysts, to explore
design options within a parameter space. The formulation described in this section would allow
users to estimate the spring constant, given the parameters of the spring and the elastic properties
of the material, without consideration of the effects of constraints imposed on the ends of the spring.
187
Calibration, validation and ranking
This chapter is concerned with the formulation, calibration, validation and ranking of mathemat-
ical models with reference to a classical problem of mechanical engineering; that of predicting
fatigue failure in metallic machine components and structural elements subjected to alternating
loads in the high cycle regime (> 104 cycles). The mathematical model considered here comprises
three sub-models: (a) a problem of linear elasticity, (b) a predictor of fatigue failure defined on
elastic stress fields and (c) a statistical model.
In fatigue experiments performed on test coupons (that may be notch-free or notched) special
stress fields are imposed on the material. Axial tension-tension or tension-compression tests or
bending tests of rotating round bars are the most common types of fatigue tests. There are various
ways to generalize those special stress fields to arbitrary stress fields. Predictors of fatigue life are
formulated for that purpose. The formulation of predictors is based on intuition and experience.
Various predictors have been and can yet be proposed. The question of how and why one would
select a predictor from among competing predictors is addressed in this chapter.
The following discussion is focused on the calibration and validation of predictors. This involves
the use of data analysis procedures. For readers who are not familiar with those procedures, a brief
overview is provided in Appendix I. A description of the experimental data used in this chapter
and statistical characterization of data collected from fatigue tests of notch-free coupons are also
available in Appendix I.
All numerically computed quantities of interest (QoI) have been verified to ensure that numerical
errors in the QoI are negligible in comparison with the uncertainties associated with the physical
experiments, such as uncertainties in loading conditions, dimensioning tolerances and residual
stresses arising from metal forming operations (such as rolling, drawing and forging) and machin-
ing operations (such as milling, boring and turning).
6.1 Fatigue data
Fatigue data are collected from force- or displacement-controlled tests of smooth and notched
test specimens. Fatigue testing is governed by standards such as ASTM Standards E466 and E606
[7], [8].
The specimens are subjected to a stress field the amplitude of which is a periodic function. The
ratio of the stress components is fixed. Let 𝜎max (resp. 𝜎min ) be the maximum (resp. minimum)
188 6 Calibration, validation and ranking
principal stress. The ratio R = (𝜎min ∕𝜎max ) is called the cycle ratio. The stress amplitude is denoted
by 𝜎a . By definition:
𝜎 − 𝜎min 1−R
𝜎a = max ≡ 𝜎max ⋅ (6.1)
2 2
The mean stress is denoted by 𝜎m . By definition:
𝜎max + 𝜎min 1+R
𝜎m = ≡ 𝜎max ⋅ (6.2)
2 2
Under sinusoidal loading the stress field in a test coupon is
( )
1+R 1−R
𝜎ij (x, t) = 𝜎max (x0 ) + sin(2𝜋ft) Σij (x) (6.3)
2 2
where x is the position vector, 𝜎max (x0 ) is the maximum value of a reference stress in point x0 , t is
time (s), f is the frequency (Hz), Σij (x) is a dimensionless symmetric matrix the elements of which
are the ratios between the stress components and 𝜎max (x0 ). For example, in specimens where the
stress is uniaxial and substantially constant within the test section the elements of Σij (x) are zero
except for Σ11 (x) = 1. Such a specimen is shown in Fig. 6.1. This type of specimen was used in
experiments reported in [40].
In specimens subjected to pure shear 𝜏(t) we have 𝜎max = 𝜏max the elements of Σij (x) are zero,
except for Σ12 (x) = Σ21 (x) = 1.
The test records contain the following information: (a) specimen label, (b) specimen geometry,
(c) the maximum test stress in the test section Smax , (d) the cycle ratio R, (e) the number of cycles at
the end of the test n and (f) notation indicating whether failure occurred outside of the test section
or the test was stopped prior to failure (runout). Typical fatigue test data are shown in Appendix I,
Fig. 1.2.
Remark 6.1 More than one type of load such as shear, bending and axial load may be applied
simultaneously. Cyclic loads may be acting in-phase or out of phase.
6.1.1 Equivalent stress

In many important applications one stress component is so dominant that all other stress compo-
nents are negligible in comparison. The family of predictors
𝜎eq = 𝜎max
1−c
𝜎ac , 0<c<1 (6.4)
where 𝜎eq is called equivalent stress, is widely used in engineering practice for the special case of
uniaxial stress. Substituting eq. (6.1) we get:
R 12.0
Test section
1.000
3.0
4.20 9.60 4.20
Figure 6.1 Notch-free test coupon. The dimensions are in inches. Thickness: 0.090 in. The test section is
within ±1∕2 inches from the minimum cross-section.
6.1 Fatigue data 189
( )
1−R c
𝜎eq = 𝜎max ⋅ (6.5)
2
The predictor with c = 1∕2 is one of the predictors proposed in [88]. We will also use c = 1∕2 in
the following. In fatigue tests 𝜎eq is fixed and the number of cycles (n) at which failure occurs
(i)
is recorded. The set of data (ni , 𝜎eq ), i = 1, 2, … recorded in fatigue tests where the first principal
stress in the test section is either constant or varies with a small gradient is called the S-N data. For
example, S-N data may be collected from rotating round bars, the test section of which is loaded
by a constant bending moment. In that case the principal stress varies linearly between its max-
imum and minimum value and, since the test article is subjected to full stress reversal (R = −1),
𝜎eq = 𝜎max .
Remark 6.2 The definition of equivalent stress is not unique. Consider, for example, the follow-
ing definition of equivalent stress:
(𝛼) ( ) ( 1 − R )c
𝜎eq = 𝛼𝜎max + (1 − 𝛼)𝜎 max , 0≤𝛼≤1 (6.6)
2
where 𝜎 max is the maximum von Mises stress. If the stress is uniaxial then the definitions given
by equations (6.5) and (6.6) are identical for any alpha. However, the two definitions differ for any
other stress condition.
6.1.2 Statistical models

Plotting the values of 𝜎eq at which fatigue failure occurs in notch-free stress coupons subjected to
cyclic loading against log10 n results in a cluster of points that can be well approximated by a curve,
called the S-N curve, see for example Fig. 1.2 in Appendix I.
The S-N curve will be understood to be the median of the statistical distribution of log10 n, given
𝜎eq , and will be denoted by 𝜇(𝜎eq ). Many plausible statistical models can be formulated to represent
the population from which the S-N data were supposedly sampled. We will use one statistical model
for this purpose. This model belongs to the family of random fatigue limit models [73] and is based
on the following assumptions:
1. The statistical distribution of log10 n is normal with mean 𝜇(𝜎eq ) and constant standard
deviation s.
2. The functional form of the mean is:
𝜇(𝜎eq ) = A1 − A2 log10 (𝜎eq − A3 ), 𝜎eq − A3 > 0 (6.7)
where the parameter A3 is called the fatigue limit or endurance limit.
3. The fatigue limit is a random variable and log10 A3 has normal distribution with mean 𝜇f and
standard deviation sf .
Statistical models differ by the assumptions concerning the probability density function of log10 n,
given 𝜎eq , the choice of the functional form 𝜇(𝜎eq ), the definition of 𝜎eq , the functional form of
the standard deviation and, when A3 is treated as a random variable, the functional form of the
probability density function representing the dispersion of A3 and so on.
The parameters of statistical models are estimated by the maximum likelihood method which
is described and illustrated by examples for two functional forms of 𝜇(𝜎eq ) in Appendix I (Section
I.3). Estimation of parameters for statistical models where c is one of the model parameters and
log10 A3 has either normal or smallest extreme value (sev) distribution is described in [17].
Since countless plausible statistical models can be formulated, it is necessary to have a procedure
by which the relative merit of those models can be evaluated with respect to the available data.
Several such methods are described and illustrated by examples in [17]. We will use the Bayes
factor defined in Appendix I for that purpose.
The five parameters corresponding to the random fatigue limit model defined above are displayed
in Table I.5 in the appendix. These parameters maximize the likelihood function for the S-N data
shown in Fig. I.2.
Remark 6.3 It was assumed that the crack initiation process consumed substantially all of the
fatigue lives of the test coupons and once a small crack formed, it propagated very quickly, so that
the number of cycles to failure was substantially the same as the number of cycles at the end of
the initiation process. This assumption was justified by the results of an investigation (not detailed
here) of crack growth rates in a 2024-T3 aluminum sheet assuming that a 0.05 inch (1.3 mm) crack
had formed by an initiation process.
6.1.3 The effect of notches

The effect of notches, fillets, keyways, oil holes and other stress raisers on the fatigue life of mechan-
ical components had been extensively investigated well before computers came to be used for the
solution of stress concentration problems. We mention only the seminal work of Neuber1 [65] and
Peterson2 [74] in this regard and note that a very rich technical literature exists on this important
subject. For a survey we refer to [80].
Designers of machine elements relied on handbooks and design manuals where they could look
up the stress concentration factors associated with various stress raisers. By definition, the stress
concentration factor is the ratio of the maximum stress 𝜎max to a reference stress 𝜎ref :
Kt = 𝜎max ∕𝜎ref . (6.8)
Knowing 𝜎max and R, the equivalent stress can be calculated from eq. (6.5) and, using 𝜎eq , the
expected value of the fatigue life can be estimated from the S-N curve under constant cycle loading.
It was found that this method would significantly underestimate the fatigue limit for notched
specimens when the stress gradient is steep, as in the neighborhood of notch roots. This is illustrated
in Fig. 6.2 where the results of fatigue experiments performed on nine notched specimen types are
displayed. The curve 𝜇(𝜎eq ) is the mean of the random fatigue limit model3 calibrated on the basis
of S-N data obtained for notch-free coupons.
6.1.4 Formulation of predictors of fatigue life

The physical processes that result in the formation of cracks caused by cyclic loading are highly
complex, irreversible processes on the microscopic scale. These processes violate the assumptions
of small strain continuum mechanics which include the assumption that a body can be subdivided
into arbitrarily small volumes, the properties of which remain the same as those of the bulk mate-
rial. Nevertheless, the predictors used for correlating the formation of cracks with the number of
load cycles in high cycle fatigue are usually determined from the solutions of problems of linear
elasticity.
1 Heinz August Paul Neuber 1906–1989.

2 Rudolph Earl Peterson 1901–1982.
3 Since by assumption log10 n is normal, the mean and the median of log10 n are coincident.
120
failure
100 runout
80
σeq (ksi)
60
40
μ(σeq)
20
104 105 106 107
n
Figure 6.2 The results of fatigue experiments performed on nine notched specimen types. See Table I.2 in
Appendix I for details.
This apparent contradiction is resolved through the assumptions that (a) the stresses (or strains)
computed from the solutions of problems based on the linear theory of elasticity represent average
stresses or strains over representative volume elements4 (RVEs) rather than pointwise stresses, (b)
those average stresses can be correlated with crack initiation events through suitably chosen pre-
dictors of fatigue failure and (c) the possibility that small amounts of plastic deformation may occur
is not ruled out. However, it is assumed that any plastic zone is surrounded by elastic material and
therefore the elastic stress field drives the process that causes fatigue failure to occur. Plastic strains
are necessarily small.
Predictors of fatigue life and statistical models have to be formulated for the generalization of
S-N data to multiaxial stresses and conditions where the stresses have steep gradients, such as in
the vicinity of notches. These are phenomenological models that have to be calibrated for each
material.
6.2 The predictors of Peterson and Neuber
The fatigue stress concentration factor Kf was introduced in order to correct for the difference
between predictions based on the maximum equivalent stress and observations of the number
of cycles to failure in notched specimens. By its original definition, Kf is the fatigue limit of a
notch-free specimen divided by the fatigue limit of a notched specimen. However, it will be shown
that Kf can be also used for predicting fatigue life at stress levels higher than the fatigue limit in
the high cycle range:
𝜎max
′
= Kf 𝜎ref . (6.9)
The relationship between Kf and Kt was established by the formula
Kf = 1 + q(Kt − 1) (6.10)
4 By definition, the representative volume element is the smallest volume over which the averaged material
properties are substantially the same as the averaged material properties over the entire volume.
where q is called notch sensitivity factor. Peterson proposed the following empirical formula for q:
1
q= (6.11)
1 + a∕r
were a is a material constant and r is the notch radius. An alternative empirical formula, based on
Neuber’s work on stress concentrations, was employed by Kuhn and Hardrath in [54]:
1 𝜋
qN = √ , f (𝜔) = (6.12)
1 + f (𝜔) A∕r 𝜋−𝜔
where 𝜔 is the flank angle of the notch and A is a material parameter.
Equation (6.9) is an example of an intuitively constructed predictor of fatigue failure caused by
cyclic loading. Neuber and Peterson defined Kf differently. This raises the question: Whose defini-
tion is better? It is possible to show that for a sufficiently small interval of notch radii both work
well but for a large interval neither does. These predictors are based on the following assumptions:
1. The notch sensitivity factor is characterized by a single geometric parameter, the notch radius.
This assumption is applicable to notches and fillets in machine components but is not applicable
to corrosion pits, scratches and other surface defects caused by wear, the kind of damage that
has to be considered when making condition-based maintenance decisions.
2. The parameters a and A are material parameters that must be determined by calibration.
3. The parameters a and A are strongly correlated with the ultimate tensile strength of the material.
Remark 6.4 Reduction of the peak stress using equations (6.10) to (6.12) is closely related to
averaging the normal stress over a material-dependent distance a.
Let us consider, for example, a circular hole of radius r0 in an infinite plate subjected to constant
stress 𝜎∞ acting in the x-direction. Let (r, 𝜃) be polar coordinates with the origin in the center of
the circle with 𝜃 measured from the positive x-axis. Then at 𝜃 = ±𝜋∕2;
( )
2 4
1 r 0 3 r0
𝜎 𝜃 = 𝜎x = 𝜎∞ 1 + + (6.13)
2 r2 2 r4
see, for example, [105]. The maximum stress is at r = r0 : 𝜎max = 3𝜎∞ , that is, Kt = 3. Therefore,
using equations (6.9) through (6.11), we get
( ) ( )
Kt − 1 2
𝜎max = Kf 𝜎∞ = 1 +
′
𝜎∞ = 1 + 𝜎∞ . (6.14)
1 + a∕r0 1 + a∕r0
The average stress over the interval defined by the endpoints (r0 , 𝜋∕2) and (r0 + a, 𝜋∕2) is
( )
( ) 1
r0 +a
𝜎∞ r0 +a 2 4
1 r0 3 r 0
𝜎x ave = 𝜎x (r, 𝜋∕2) dr = 1+ + dr
a ∫r0 a ∫r0 2 r2 2 r4
( )
2 2
1 1 3 1 + a∕r0 + a ∕(3r0 )
= 1+ + 𝜎∞
2 1 + a∕r0 2 (1 + a∕r0 )3
( )
2 2
2 3 a∕r0 + 2a ∕(3r0 )
= 1+ − 𝜎∞
1 + a∕r0 2 (1 + a∕r0 )3
( )
2
= 1+ − O(a∕r0 ) 𝜎∞ . (6.15)
1 + a∕r0
On comparing eq. (6.14) with eq. (6.15) it is seen that the notch sensitivity factors differ by a term
which is of order a∕r0 . When a is much smaller than r0 then this term can be neglected.
6.2.1 The effect of notches – calibration

Calibration of Peterson’s notch sensitivity factor is demonstrated in the following with the aid of the
fatigue test data summarized in Table I.2 in the appendix. We will use test records for six notched
specimen types for estimating parameter a in eq. (6.11). These test records correspond to line items
2 through 7 in Table I.2 and will be called the calibration set. The remaining test records, called the
validation set, will be used for demonstrating the validation process. This will emulate the process
by which test data are collected over time and the calibration and ranking of models are updated
as new data become available. For example, we use data from reports [40] to [44] collected over an
eight-year period. Suppose that we would have formulated a predictor when only half of the data
were available. We would have had to update and possibly revise the predictor in the light of new
information. In industrial and research organizations the management of activities of this kind fall
under the purview of simulation governance.
Not all test records summarized in Table I.2 conform with the assumptions on which Peterson’s
formula is based. For example, one test record for the edge-notched specimen with r = 0.0313 inch
indicates that 𝜎max = 247.8 ksi, 𝜎min = 102.1 ksi and failure occurred at 4500 cycles. The yield
strength of this material is approximately 54 ksi hence the plastic strains are not small (of the order
of 5%). This test is clearly outside of the scope of Peterson’s formula.
The available S-N data is in the range of 8500 < nf < 107 where nf is the number of cycles
at failure. Since we are interested in generalizing these data to notched specimens, we exclude
from consideration notched specimen records that failed below 8500 cycles. The total number of
records available for the six specimen types in the calibration set are shown under the heading
N in Table 6.1. The number of failed specimens and the number of runouts in the retained test
records, called qualified test records, are shown under the headings Nf and Nr respectively.
Using equations (6.8) through (6.10) for each specimen type we have
𝜎max
′ 𝜎eq
′
1 + qk (Kt − 1)
= = , qk = q(rk ) (6.16)
𝜎max 𝜎eq Kt
where 𝜎eq
′ is defined analogously to 𝜎 :
eq
( )
1 − R 1∕2
𝜎eq
′
= 𝜎max
′
. (6.17)
2
We denote the ratio 𝜎eq
′ ∕𝜎 by x and observe that x depends on K and q only. Therefore x
eq k k t k k
is a characterizing parameter for the kth specimen type made of the same material. Given a set of
(i)
independent test records (ni , 𝜎eq ), (i = 1, 2, … , mk ) for the kth notched specimen type we seek xk
Table 6.1 Calibration of Peterson’s material parameter a based on the random fatigue limit model.
k Specimen rk (in) Kt N Nf Nr xk qk ak (in)
2 Open hole 1.5000 2.11 39 25 5 0.7943 0.6091 0.9628

3 Edge notch 0.3175 2.17 42 27 4 0.7601 0.5550 0.2546
4 Fillet notch 0.1736 2.19 32 23 5 0.7159 0.4772 0.1902
5 Edge notch 0.0570 4.43 34 20 5 0.6664 0.5691 0.0431
6 Fillet notch 0.0195 4.83 36 23 4 0.5198 0.3945 0.0299
7 Edge notch 0.0313 5.83 46 22 6 0.5490 0.4557 0.0374
(i)
such that the set of data (ni , xk 𝜎eq ) maximizes the log likelihood function of the random fatigue
limit model defined in Section I.3 in the appendix:
mk [ ( )
∑
LLk (xk |𝜽̂ 3 ) = (i)
(1 − 𝛿i ) ln 𝜙M (𝑤i , xk 𝜎eq )
i=1
( )]
(i)
+ 𝛿i ln 1 − ΦM (𝑤i , xk 𝜎eq ) (6.18)
where 𝜽̂ 3 is the vector of the five statistical parameters that characterize the random fatigue limit
model (see Table I.5) and 𝑤i = log10 ni . The functions 𝜙M and ΦM are the marginal probability den-
sity and marginal cumulative distribution functions defined by equations (I.18) and (I.19) respec-
tively, 𝛿i = 0 if the test resulted in failure, 𝛿i = 1 indicates a runout, that is the specimen did not fail
when the test was stopped at the recorded number of cycles.
Considering qualified test records only, xk is determined such that LLk (xk |𝜽̂ 3 ) is maximum. The
values of xk are listed in Table 6.1 along with the computed values of qk and ak . The formulae for
computing qk and ak are:
( )
xk Kt − 1 1
qk = and ak = r −1 . (6.19)
Kt − 1 qk
Substituting qk for q in eq. (6.10) we find that for the kth specimen type Kf = xk Kt . Therefore Kf can
be used for predicting fatigue life at any stress level in the high cycle range, not just at the fatigue
limit.
Peterson assumed that a is a material constant. However, the results displayed in Table 6.1 indi-
cate otherwise. In fact, the coefficient of variation5 of ak is 141%. Therefore the assumption that the
parameter a is a material constant, independent of the notch radius, does not hold for this mate-
rial. Consequently this assumption, and therefore Peterson′ s formula, has to be rejected on the
basis of the evidence in Table 6.1. The same conclusion applies to Neuber’s definition of the notch
sensitivity factor given by eq. (6.12).
This should be understood to mean that Peterson′ s predictor is not applicable in the interval
of notch radii 0.0195 ≤ rk ≤ 1.50 (inches) used in calibration (see Table 6.1). However, it may be
applicable in a narrow interval of notch radii, say 0.2 ≤ rk ≤ 0.4 (inches) which are commonly used
in machine design. Note, however, that if the domain of parameters is sufficiently small then any
predictor can be calibrated. A calibrated predictor is a validated predictor within the domain of
calibration.
The relationship between log10 a and log10 r is shown in Fig. 6.3 where the open squares represent
the data in Table 6.1. Observe that this relationship is nearly linear within the interval of notch radii
considered here. Using the least squares method we find
log10 a ≈ −0.16641 + 0.83899log10 r, R2 = 0.979 (6.20)
where R2 is the coefficient of determination.
On substituting the eq. (6.20) into eq. (6.11) we get the revised form of Peterson′ s definition of
the notch sensitivity factor for the 24S-T3 aluminum alloy:
1
qrev = , 0.02 < r < 1.5 (6.21)
1 + 0.6817r −0.1610
where 0.02 < r < 1.5 is the domain of calibration. Note that whereas Peterson′ s notch sensitivity
factor has one parameter, the revised formula has two.
5 The coefficient of variation is defined as the ratio of the standard deviation to the absolute value of the mean. It is
a measure of dispersion.
0.5
0
–0.5
r = 0.7600
–1
log10 a
–1.5 r = 0.0710
–2
–2.5 r = 0.0035 R2 = 0.979
–3
–2.5 –2 –1.5 –1 –0.5 0 0.5
log10 r
Figure 6.3 Empirical relationship between the parameters a and r for 24S-T3 (7024-T3) aluminum alloy
calibrated on the basis of the six data points represented by the open squares.
Remark 6.5 The definition of xk is not restricted to the ratio 𝜎eq ′ ∕𝜎 . It can be understood as
eq
a scaling factor for the applied load and hence the entire stress field 𝜎ij = 𝜎ij (x). Therefore it is
applicable to any predictor  = (𝜎ij ) that has the property (xk 𝜎ij ) = xk (𝜎ij ). Another predictor
that has this property will be discussed in Section 6.3.
Remark 6.6 Fitting the data points in Fig. 6.3 by the least squares method is equivalent to assum-
ing that the mean of log10 a is a function of log10 r and the probability density of log10 a is normal
with variance s2 . We can then find the unknown coefficients by maximizing the likelihood function.
To show this let x = log10 r and y = log10 a and write
y = c1 + c2 x + 𝜖, 𝜖 ∼  (0, s2 ).
By assumption, the probability density of 𝜖 is
1 ( )
f = √ exp −𝜖 2 ∕(2s2 ) .
2𝜋s
Let 𝜖i = yi − c1 − c2 xi (i = 1, 2, … , n). Then the likelihood function is
∏
n
1 ( )
L= √ exp −𝜖i2 ∕(2s2 )
i=1 2𝜋s
and the log likelihood function is
( )n
1 ∑
n
1
LL = ln √ − 2 (yi − c1 − c2 xi )2 .
2𝜋s 2s i=1
To find the maximum of LL we differentiate it with respect to c1 and c2 and let the derivatives equal
to zero. We then get the same set of linear equations for c1 , c2 as with the least squares method.
6.2.2 The effect of notches – validation

An ideal predictor would generalize the statistical model calibrated for the notch-free specimens
to the notched specimens. This can never be fully ascertained, however. The best we can do is to
test the hypothesis that the statistical distribution of the data obtained from testing the notched
specimens is the same as the statistical distribution of the S-N data. Specifically, our hypothesis
(i)
is that the probability distribution of the fatigue data (ni , xk 𝜎eq ), k = 1, 2, … , mk , collected from
records of fatigue tests of notched specimens, is consistent with the random fatigue limit model.
The following discussion is concerned with testing Peterson′ s revised predictor against the three
sets of records corresponding to line item 8 through 10 in Table I.2 in Appendix I, the validation
set. The open circles in Fig. 6.3 correspond to those test records.
In order to emulate an ideal validation scenario, we assume that only a test plan is available at this
point and this test plan provides details concerning the type and number of specimens to be tested
and the loading conditions to be applied. Based on this information, the maximum equivalent stress
(𝜎eq )i can be computed for each of the planned tests.
For each specimen type in the validation set we compute xk :
qk (Kt − 1) + 1
xk = where qk = qrev (rk ) (6.22)
Kt
and for each specimen type in the validation set we determine the predicted median from the
inverse of the cumulative distribution function. Specifically, we find 𝑤i ≡ log10 ni such that
(i)
ΦM (𝑤i , xk 𝜎eq ) − 0.5 = 0 (6.23)
where ΦM is the marginal cumulative distribution function of the random fatigue limit model. The
predicted median is the cumulative distribution function of ni = 10𝑤i . When no real root is found or
ni > 108 then a runout is predicted. Additional discussion in available in Appendix I under Remark
I.3. The 0.05 and 0.95 quantiles are computed analogously. We expect the cumulative distribution
function of the outcome of the validation experiments to lie substantially within these quantiles.
Once the outcome of validation experiments is available, we plot and compare the empirical
cumulative distribution function with the predicted median. An example is shown in Fig. 6.4.
Of the three sets of validation experiments the edge notched specimen corresponding to k = 9 is
the most important one because the notch radius r = 0.0035 inch falls well outside of the interval
of notch radii in the calibration set. There are 12 qualified test records for this specimen type. One
runout was predicted, four runouts were recorded.
A Aʹ
1
Edge notch, r = 0.760 in Bʹ
B
0.8
0.6 Empirical CDF

F(n)
0.05 quantile
0.4 0.95 quantile
Predicted median
0.2
0
104 105 106 107 108
n
Figure 6.4 Predicted and empirical cumulative distribution functions for the edge notched specimens with
r = 0.760 inch.
In the calibration process the statistical model and the functional form of the predictor are
assumed to be correct and the parameters are determined such that the predicted and experimental
outcomes are identical. In other words, the predictor is trained to match the calibration data.
Therefore a calibrated predictor interpolates or approximates the calibration data within the
domain of calibration in some sense.
The value of new data outside of the domain of calibration is that the predictive performance of
the model can be tested against that data and the domain of the calibration set can be enlarged.
The value of new data within the domain of calibration is that the dataset on which calibration is
based is enriched.
Edge notched specimen with r = 𝟎.𝟕𝟔𝟎 inch

For the edge notched specimens with r = 0.760 inch the results of the validation experiments are
presented in the form of the empirical cumulative distribution function shown in Fig. 6.4. The
predicted median and the 0.05 and 0.95 quantiles are also shown. The segments labeled A′ − B′
represent the predicted runouts, whereas the segments labeled A − B represent the recorded
runouts. Four runouts were recorded, two runouts were predicted. The empirical CDF lies within
the 0.05 and 0.95 quantiles predicted on the basis of the random fatigue limit model calibrated to
the S-N data of the notch-free specimens, as described in Appendix I.
Remark 6.7 The number of runouts cannot be reliably predicted. A runout is recorded when a
test is stopped prior to the occurrence of failure for any reason. A runout is predicted when eq. (6.23)
does not have a real root or the estimated number of cycles for a given 𝜎eq ′ exceeds 100 million. For
additional discussion see Remark I.3 in the appendix.
Edge notched specimen with r = 𝟎.𝟎𝟎𝟑𝟓 inch

The predicted and empirical cumulative distribution functions for the edge notched specimens
with r = 0.0035 inch are shown in Fig. 6.5. Four runouts were recorded, one runout was predicted.
Conclusion
The realized cumulative distribution functions do not contradict the hypothesis that the notched
fatigue data came from the same population as the S-N data. Therefore we find no reason to reject
Peterson′ s revised predictor.
6.2.3 Updated calibration

Having completed the validation experiments, the validation set is merged with the calibration set
and the calibration process is repeated. This is to take into account all of the available information
concerning the notched fatigue properties of this material.
The data in Table 6.1 is augmented with the calibration data in Table 6.2 and, using the full set
of nine data points, a new empirical relationship between the parameters a and r is obtained. This
relationship is
log10 a = −0.16763 + 0.82291log10 r, R2 = 0.979. (6.24)
Using this relationship, Peterson′ s revised formula is updated to

1
qupd = , 0.003 < r < 1.5. (6.25)
1 + 0.6798r −0.1791
A Aʹ
1
Bʹ
Edge notch, r = 0.0035 in
0.8
B
0.05 quantile
Predicted median
0.6
F(n)
0.4 0.95 quantile
Empirical CDF
0.2
0
104 105 106 107 108
n
Figure 6.5 Predicted and empirical cumulative distribution functions for the edge notched specimens with
r = 0.0035 inch.
Table 6.2 Additional calibration data to augment Table 6.1.
k Specimen rk (in) Kt N Nf Nr xk qk ak (in)
8 Edge notch 0.7600 1.62 31 25 4 0.8164 0.5203 0.7007

9 Edge notch 0.0035 4.48 17 12 4 0.4553 0.2987 0.0082
10 Edge notch 0.0710 4.41 19 12 4 0.6560 0.5551 0.0569
Merging the validation set with the calibration set has resulted in a substantial increase in the
size of the domain of calibration, see eq. (6.21). Using qupd in eq. (6.16), we compute the updated
xk = qupd (rk ) for the qualified test records of nine notched specimen types used for the updated
calibration. The results are shown in Fig. 6.6 where 𝜎eq ′ = x 𝜎 . Compare Fig. 6.6 with Fig. 6.2. The
k eq
median and quantile functions were computed from the random fatigue limit model. On comparing
the two figures it is obvious that 𝜎eq′ is a far better predictor than 𝜎 was.
eq
Remark 6.8 When comparing Fig. 6.6 to Fig. 6.2 it is obvious that 𝜎eq ′ is a much better predictor
than 𝜎eq . However, if we ask the question how much better or worse is 𝜎eq ′ based on q
upd than 𝜎eq
′
based on qrev , simply plotting the two distributions would not provide an answer. A quantitative
measure is needed.
The relative performance of two models is evaluated by the likelihood ratio, in this case Lupd ∕Lrev ,
where L is the likelihood function defined by eq. (I.20) for the random fatigue limit model with the
parameters 𝜽̂ 3 that can be found in Table I.5. On computing the log likelihood function LL defined
by eq. (6.18) we find LLupd = −2534.440 and LLrev = −2537.068. Therefore
Lupd ∕Lrev = exp(LLupd − LLrev ) = 13.8

70 failure
runout
60 0.95 quantile
50
σʹeq (ksi)
40
median
30
20
0.05 quantile
10
104 105 106 107
n
Figure 6.6 The results of fatigue experiments performed on nine notched specimen types. Compare with
Fig. 6.2.
which is a strong indication that the updated model is better than the revised model, given the data
for the nine notched specimen types.
The parameter a = 0.02 is given in [75] for aluminum alloy sheets and bars for use in Peterson′ s
formula, see eq. (6.11). The hypothesis that a is a material constant was rejected on the basis of
the results of calibration summarized in Table 6.1, having observed that the coefficient of varia-
tion was high. We are now in a position to obtain a quantitative measure of the difference between
the predictors based on Peterson′ s original definition of q, denoted by qPet , and qupd . On comput-
ing the log likelihood function corresponding to qPet we find LLPet = −3463.283 hence Lupd ∕LPet =
exp(LLupd − LLPet ) is a very large number indicating that rejection of this model was fully justified.
In reference [37] estimates are given for the parameter A in Neuber’s formula for aluminum
alloys, see eq. (6.12). By interpolation for the ultimate tensile strength of 24S-T3 aluminum (73 ksi),
we find A = 0.018 inch. On computing the log likelihood function corresponding to Neuber’s def-
inition of q, denoted by qNeu we find LLNeu = −2971.704. Therefore the likelihood ratio Lupd ∕LNeu
is a very large number indicating that Neuber’s model should be rejected also.
The q vs. r curves are shown in Fig. 6.7. On comparing Peterson′ s and Neuber’s formulas with
the revised and updated formulas for q, substantial differences are evident, whereas the difference
between the revised and updated formulas is small. Therefore confidence in the updated formula
within the indicated domain of calibration is justified. Observe that the notch sensitivity factor
approaches unity very differently for the Peterson, Neuber and updated Peterson formulas as the
notch radius increases.
6.2.4 The fatigue limit

The notch sensitivity factors of Peterson and Neuber were proposed for estimating the fatigue
limit of notched specimens. In the following we compare the fatigue limit values of the random
fatigue limit model with those computed from the original formulas of Peterson and Neuber. For
reference we will use the mean of the random fatigue limit of the notch-free specimens which
Peterson
1
Notch sensitivity factor q
0.8
Neuber
0.6 qrev
0.4 qupd
0.2
0
0 0.5 1 1.5
Notch radius (inches)
Figure 6.7 Relationship between notch radius and notch sensitivity factor for 24S-T3 aluminum alloy
sheets.
Table 6.3 24S-T3 alloy: Fatigue limit values predicted by the random fatigue limit (RFL), Peterson and
Neuber models.
k Specimen rk Kt xk RFL Peterson Neuber
2 Open hole 1.5000 2.11 0.7962 17.58 21.93 20.93

3 Edge notch 0.3175 2.17 0.7547 16.66 21.37 19.79
4 Fillet notch 0.1736 2.19 0.7381 16.30 20.84 19.16
5 Edge notch 0.0570 4.43 0.5883 12.99 17.64 15.93
6 Fillet notch 0.0195 4.83 0.5408 11.94 13.21 13.50
7 Edge notch 0.0313 5.83 0.5374 11.87 14.95 14.19
8 Edge notch 0.7600 1.62 0.8406 18.56 21.86 20.95
9 Edge notch 0.0035 4.48 0.4937 10.90 7.48 10.18
10 Edge notch 0.0710 4.41 0.5964 13.17 18.33 16.36
is 10𝜇f = 101.344 = 22.08 ksi, see Table I.5. For the notched specimens this number is scaled by xk
defined by eq. (6.22). The results, shown in Table 6.3, indicate that the Peterson and Neuber for-
mulas overestimate the fatigue limit values by a significant margin.
The random fatigue limit model estimates the probability of survival at the mean fatigue limit to
be 50%, see Appendix A, whereas the fatigue limit values computed by the Peterson and Neuber
formulas are deterministic numbers, generally understood to mean that the probability of survival
at stress levels below the fatigue limit is 100%. Therefore those numbers should be smaller, not
larger.
Example 6.1 The fatigue strength of 2024-T3 aluminum alloy is estimated to be 20 ksi (138 MPa)
by the Aluminum Association6 (AA). By this definition fatigue strength is the maximum completely
reversed normal stress that the material is expected to resist at 500 million cycles. This definition
is not consistent with random fatigue limit models which treat fatigue limit and fatigue strength as
6 Source: ASM Material Data Sheet.

random variables. Using the random fatigue limit model characterized by the parameters in Table
I.5, the estimated probability that fatigue failure will occur below 500 million cycles is
Pr (failure|n < 5.0 × 108 , 𝜎eq = 20) = ΦM (𝑤, 𝜎eq ) = 0.1697 (17%)
where ΦM is the marginal cumulative distribution function defined in Appendix I by eq. (I.19). In
other words, the probability of survival at 500 million cycles of fully reversed 20 ksi maximum stress
loading is 83%. Note, however, that 500 million cycles is well outside of the domain of calibration
of the random fatigue limit model described in this chapter and the validity of the model has not
been established for such high number of cycles.
The random fatigue limit model described in this chapter has been calibrated to 10 million cycles
and thus it is possible to answer the following question: “What is the maximum stress at fully
reversed loading such that the probability of survival is 99% at 10 million cycles?” For this we have
to solve the following problem: Find 𝜎eq such that
1 − Pr (failure|n < 107 , 𝜎eq ) = 0.99.
The solution is 𝜎eq = 18.2 ksi. One should bear in mind, however, that prediction of events that have
a low probability of occurrence involves extrapolation even when the event is within the domain of
calibration. This is because statistical models are calibrated against data that have a high probability
of occurrence. At low probability values ΦM (𝑤, 𝜎eq ) is sensitive to the choice of the statistical model
and therefore large model form uncertainties exist. Consequently, validation of statistical models
for the prediction of fatigue limit is not feasible.
6.2.5 Discussion
At this point the reader would be fully justified in expecting a yes or no answer to the question
of whether or not Peterson′ s updated predictor passed the validation test. Disappointing as it may
sound, it is not possible to answer the question in that way. To answer the question yes or no it would
have been necessary to define a validation metric and state a priori the acceptable magnitude of dis-
agreement between the predicted and realized outcomes in terms of the chosen validation metric.
We have not set such a tolerance because any number would have been arbitrary and difficult to
justify.
The purpose of validation is to assess the predictive performance of a mathematical model. The
outcome of a validation experiment is one of the following: (a) the prediction is not consistent with
the observations, hence the model has to be rejected, or (b) the predictions are consistent with the
observations therefore there is no reason to reject the model. A model is validated only within its
domain of calibration. Here we can say with a high degree of confidence that Peterson′ s predic-
tor with the updated notch sensitivity factor qupd has been validated in the domain of calibration
indicated in eq. (6.25).
This claim is not based on some arbitrarily chosen tolerance. Rather, it is in the spirit of Hume7
who wrote: “A wise man proportions his belief to the evidence.” The evidence here is that (a) the
empirical cumulative distribution functions lie substantially within the 0.05 and 0.95 predicted
quantiles as seen (for example) in Fig. 6.4 and (b) upon introduction of new data, which included
data that lie outside of the domain of calibration, predictions based on the formula changed by a
small amount only, indicating that predictions based on qrev are substantially the same as predic-
tions based on qupd . This criterion is discussed in greater detail in [16] where the distance between
7 David Hume 1711–1776.

the cumulative distribution functions associated with the prior and posterior models is assumed to
be an acceptable approximation of the distance between the posterior model and the (unknown)
true model.
Another important objective of validation is to develop information that provides a rational basis
for choosing from among competing models the one that is best suited for the intended use of
the model. Here we used the ratio of likelihoods to show that, given the available data, Peterson′ s
predictor with qupd is better than with qrev and both are far better than Peterson′ s or Neuber’s pre-
dictor with the original definitions for q. Ranking Peterson′ s revised predictor against another kind
of predictor is discussed in Section 6.3.3.
Both Peterson′ s and Neuber’s definitions of the notch sensitivity factor are based on the assump-
tion that notch sensitivity is characterized by a material constant. The results of experiments
reported in [94] show, however, that, at least for 24S-T3 and 75S-T6 aluminum alloys and SAE 4130
steel, this assumption does not hold. Therefore this assumption cannot be valid for all metallic
alloys and may not be valid for any metallic material.
Remark 6.9 Peterson′ s revised notch sensitivity formula has been calibrated for a very special
stress condition only: Since the boundary surface at the notch is stress-free and the stress in the
transverse direction is negligibly small in comparison with the first principal stress, the maximum
stress at the notch is substantially uniaxial. In other words, the domain of calibration is limited to
uniaxial stresses. Generalization to biaxial stresses will be discussed in Section 6.4.
6.3 The predictor G𝛂
A predictor of fatigue failure, proposed in [93], will be discussed in the following. This predictor
is based on the assumption that the onset of fatigue failure can be correlated with the averaged
volume integral of a linear combination of two stress invariants. It is defined as follows:
( ) ( )
1 1 − R 1∕2
G𝛼 (𝜎ij , R) = 𝛼I1 + (1 − 𝛼)𝜎 dV , 0≤𝛼≤1 (6.26)
Vc ∫Ωc 2
where 𝜎ij is the stress tensor field, I1 = 𝜎kk is the first stress invariant, 𝜎 is the von Mises stress,
√ ( )( )
3 1 1
𝜎= 𝜎ij − 𝛿ij 𝜎kk 𝜎ij − 𝛿ij 𝜎kk (6.27)
2 3 3
and Vc is the volume of the domain of integration defined by
Ωc = {x|𝜎1 > 𝛽𝜎max > 0} (6.28)
where 𝜎1 is the first principal stress and 𝜎max > 0 is the maximum macroscopic stress. This is a
generalization of the uniaxial stress to triaxial stress in the sense that G𝛼 is defined for triaxial
stress and, in the special case when constant uniaxial stress 𝜎1 is applied, it has the value of the
equivalent stress
( )
1 − R 1∕2
G𝛼 (𝜎ij , R) = 𝜎1 = 𝜎eq . (6.29)
2
Therefore we will generalize the statistical model defined in Section I.3 by replacing 𝜎eq with G𝛼 in
the definition of the mean:
𝜇(G𝛼 ) = A1 − A2 log10 (G𝛼 − A3 ), G𝛼 − A3 > 0. (6.30)
6.3 The predictor Gα 203
Many generalizations are possible. Predictors are scalars that must be independent of the choice
of the coordinate system. The predictor that has the largest likelihood, given a set of experimen-
tal data, is the one to be preferred. We will compare G𝛼 with Peterson′ s modified predictor using
experimental data for nine notched specimen types.
The predictor G𝛼 has two parameters that have to be determined by calibration. Parameter 𝛼
establishes a convex combination of the first stress invariant I1 and the von Mises stress 𝜎. Param-
eter 𝛽 defines the domain of integration which depends on the material properties and the stress
field in the vicinity of stress raisers and thus it plays a role analogous to Peterson′ s notch sensitivity
factor.
Peterson′ s notch sensitivity factor depends on the notch radius. There are other kinds of stress
raisers, such as scratches and corrosion pits, that cannot be characterized by a notch radius. For this
reason we introduce the highly stressed volume V for characterizing stress raisers. By definition,
{
1 when 𝜎1 (x) > 𝛾𝜎max
V= y(x) dV where y = (6.31)
∫Ω𝜚 0 otherwise.
The domain of integration Ω𝜚 is the neighborhood of a stress raiser. Denoting the location where 𝜎1
is maximum at a stress raiser by x0 we have Ω𝜚 = {x||x − x0 | < 𝜚} where 𝜚 is chosen large enough
to include all points where 𝜎1 (x) > 𝛾𝜎max and small enough to include only one stress raiser or one
group of closely situated stress raisers.
The parameter 𝛾 is independent of the material properties. Its sole purpose is to define the highly
stressed volume V which depends on the stress distribution, and hence the type of loading, but is
independent of the magnitude of the load. The predicted number of fatigue cycles is not sensitive
to the choice of 𝛾. As in reference [93], we will use 𝛾 = 0.85.
6.3.1 Calibration of 𝜷(V, 𝜶)

The calibration of 𝛽 = 𝛽(V, 𝛼) is analogous to the calibration of q discussed in Section 6.2.1 where
xk = 𝜎max
′ ∕𝜎max = 𝜎eq
′ ∕𝜎 was found through maximization of the log likelihood function for each
eq
specimen type in the calibration set. Here we used the nine specimen types listed in line items 2
through 10 in Table I.2 for calibration and, for a fixed 𝛼 and each specimen type, we found 𝛽k by
iteration such that
( )
1 − R 1∕2
G(k)
𝛼 = xk 𝜎max , k = 2, 3, … , 10 (6.32)
2
where the values of xk are the same as those in Sections 6.2.1 and 6.2.3. See Remark 6.5. The 𝛽k
values were fitted by least squares for each 𝛼 using the assumed functional form:
𝛽 k = a1 + a2 log10 Ak + 𝜖, 𝜖 ∼  (0, 𝜎𝜖2 ) (6.33)
where V is the highly stressed volume and 𝜖 is a random variable which was assumed to have
normal distribution with zero mean and standard deviation 𝜎𝜖 . For example, for 𝛼 = 0 we find
a1 = 0.9422, a2 = 0.08184. The calibration curve for 𝛽 is shown in Fig. 6.8. Note that the interval
of the highly stressed volume for which calibration data are available is approximately (1.0E − 8 <
V < 1.0E − 2) in3 .
Using 𝛽 k , we find xk by evaluating
1 ( )
xk = 𝛼I1 + (1 − 𝛼)𝜎 dV (6.34)
𝜎max Vc ∫Ωc
on the domain Ωc = {x|𝜎1 > 𝛽 k 𝜎max > 0}. The numerical results for 𝛼 = 0 are shown in Table 6.4.
0.8
0.6
β β
0.4
0.2
10–8 10–6 10–4 10–2
Highly stressed volume V (in3)
Figure 6.8 The computed 𝛽k values and the 𝛽 function corresponding to 𝛼 = 0.
Table 6.4 24S-T3 aluminum alloy: Computed values of Vk , 𝛽k , 𝛽 k and xk for 𝛼 = 0.
( )
k Specimen Kt Vk in3 xk 𝜷k 𝜷k xk
2 Open Hole 2.11 2.898E-02 0.7943 0.7021 0.7329 0.8163

3 Edge Notch 2.17 4.325E-03 0.7601 0.6826 0.6652 0.7487
4 Fillet Notch 2.19 9.561E-04 0.7159 0.6455 0.6116 0.6951
5 Edge Notch 4.43 1.827E-04 0.6664 0.5742 0.5528 0.6363
6 Fillet Notch 4.83 1.618E-05 0.5198 0.4448 0.4666 0.5501
7 Edge Notch 5.83 5.888E-05 0.5490 0.4487 0.5125 0.5960
8 Edge Notch 1.62 2.875E-02 0.8164 0.7539 0.7326 0.8161
9 Edge Notch 4.48 5.908E-07 0.4553 0.3809 0.3490 0.4325
10 Edge Notch 4.41 3.324E-04 0.6560 0.5645 0.5740 0.6575
Remark 6.10 The values of xk , and hence the computed values of 𝛽k , are influenced by uncer-
tainties in loading conditions, residual stress, variations in surface finish, and tolerances in the
dimensions of the test articles as well as numerical errors. While the presence of systematic errors
cannot be ruled out, the source reports for the experimental data indicate that the investigators
exercised a great deal of care in planning and performing the experiments to avoid such errors. The
relative errors in all numerically computed data have been verified to be less than 1%.
6.3.2 Ranking
Using xk , we evaluate for all specimen types the log likelihood function for a sequence of the model
form parameter 𝛼 = 𝛼j . Specifically, the sequence 𝛼j = 0.2(j − 1), j = 1, 2, … , 6 will be used.
mk [ ( )
∑
LLk (𝛼j |𝜽̂ 3 ) =
(j) (i)
(1 − 𝛿i ) ln 𝜙M (𝑤i , xk 𝜎eq )
i=1
( )]
(i)
+ 𝛿i ln 1 − ΦM (𝑤i , xk 𝜎eq ) , k = 2, 3, … , 10 (6.35)
Table 6.5 Computed log likelihood values LL(j) .
𝜶=0 𝜶 = 0.2 𝜶 = 0.4 𝜶 = 0.6 𝜶 = 0.8 𝜶 = 1.0
−2487.33 −2495.33 −2510.72 −2536.07 −2574.97 −2634.26
where 𝜽̂ 3 is the vector of the five statistical parameters that characterize the random fatigue limit
model (see Table I.5) and 𝑤i = log10 ni . The functions 𝜙M and ΦM are the marginal probability den-
sity and marginal cumulative distribution functions defined by equations (I.18) and (I.19) respec-
tively, 𝛿i = 0 if the test resulted in failure, 𝛿i = 1 indicates runout. The computed values of the log
likelihood function
∑10
LLk (𝛼j |𝜽̂ 3 ), j = 1, 2, … , 6
(j)
LL(j) = (6.36)
k=2
are shown in Table 6.5. It is seen that LL(j) is maximum for 𝛼 = 0. Therefore the model form param-
eter 𝛼 should be set to zero. In the following we will be concerned with this model only.
6.3.3 Comparison of G𝛂 with Peterson′ s revised predictor

Let us compare the predictive performance of G𝛼 with that of the Peterson′ s predictor based on
the revised and updated notch sensitivity index. The basis for comparison is the value of the log
likelihood function which is our validation metric.
The maximum log likelihood for Peterson′ s revised and updated predictor was LLupd = −2534.44,
see Section 6.2.3. This value is substantially lower than the value of the maximum log likelihood
function of the present model with 𝛼 = 0 but is slightly higher than the present model with 𝛼 = 0.6,
see Table 6.5. Based on this evidence we conclude that G𝛼 with 𝛼 = 0 is a better predictor than
Peterson′ s revised predictor, given the data at our disposal and our choice of the statistical model.
Mathematical models (and hence predictors) are ranked on the basis of the log likelihood function,
given the available data.
The values of G𝛼 are plotted against n for the nine notched specimen types considered here and
for 𝛼 = 0 in Fig. 6.9. On comparing Fig. 6.9 with Fig. 6.6 it becomes evident that G𝛼 is preferable to
Peterson′ s revised and updated predictor.
It is not possible to claim that G𝛼 , or any other predictor that may be formulated in the future,
is the best predictor. It is possible to say only, given the data, the statistical model and two predic-
tors, which one is better. When new data become available, or when a new predictor or statistical
model is proposed, then it is possible to evaluate their relative merit objectively on the basis of the
likelihood function following the procedure outlined here.
The formulation of mathematical models is an open-ended problem. There have to be established
policies and procedures for the systematic revision and updating of models when new ideas are
proposed, or when new data become available. The formulation and management of such policies
and procedures are in the domain of simulation governance.
6.4 Biaxial test data

The scope of validation discussed up to this point was confined to substantially uniaxial stress
conditions in the high cycle fatigue regime. In this section we examine whether the predictor G𝛼
70 failure
runout
60
0.95 quantile
50
Gα (ksi)
40
median
30
0.05 quantile
20
10
104 105 106 107
n
Figure 6.9 24S-T3 aluminum alloy: Combined qualified test records for the nine notched specimen types,
𝛼 = 0. The median and quantile functions are those of the random fatigue limit model. Compare with
Fig. 6.6.
(𝛼 = 0), calibrated under uniaxial loading conditions, will pass validation tests under biaxial load-
ing. We will use results of uniaxial and biaxial fatigue tests performed on 2024-T3 aluminum alloy
specimens in the intermediate to high cycle fatigue regime published in [38]. The analysis described
in this section is based on reference [95].
The specimens were fabricated from drawn tubing so as to conform to ASTM standard E22078 .
These specimens feature a 30 mm long cylindrical section with outside diameter of 29 mm, inside
diameter 25.4 mm, wall thickness 1.8 mm. A 3.2 mm diameter cylindrical hole was cut into the
test section by drilling and reaming. The axis of the hole was perpendicular to the test section.
A specimen is shown in Fig. 6.10(a). The surfaces were polished to remove machining marks. The
tests were performed under in-phase, fully reversed (R = −1) axial, torsional and combined loading
conditions with load control in the 0.2 to 7.0 Hz frequency range.
Crack initiation and growth were observed using a 2.0 megapixel digital microscopic camera
capable of 10X to 230X optical zoom. The fatigue crack initiation event was defined as the first
appearance of a 0.2 mm surface crack [38].
6.4.1 Axial, torsional and combined in-phase loading

The test results are shown in Fig. 6.11 where N is the number of cycles at failure and G𝛼 (𝛼 = 0) is
the predictor defined by eq. (6.26). The solid line represents the median of the random fatigue limit
model characterized by the five parameters in Table I.5.
As previously stated, the purpose of a validation experiment is to test the predictive performance
of a mathematical model. In this case; given the notched specimen described previously, and the
experimental procedure described in [38], the prediction being tested is formulated as follows:
“The probability is x% (say, 90%) that the number of cycles at which failure will occur (Nf ) lies in
8 ASTM E2207 – 15 Standard Practice for Strain-Controlled Axial-Torsional Fatigue Testing with Thin-Walled
Tubular Specimens.
(a) (b)
Figure 6.10 (a) Specimen used in validation experiments. (b) Contours of von Mises stress in torsion.
70
Axial
60 Axial, η > ηlim
Torsion
50 Torsion, η > ηlim
Gα (α = 0)
Torsion runout
40 Axial - torsion (in-phase)
30
20
10
104 105 106
N
Figure 6.11 Outcomes of axial, torsion and combined in-phase fatigue experiments performed on notched
2024-T3 aluminum specimens. Source: reference [38].
the interval N1 ≤ Nf ≤ N2 ”. The limit values N1 and N2 are determined from the survival function
S(N) which, given G𝛼 , is defined by
S(N) = 1 − ΦM (N|G𝛼 ) (6.37)
where ΦM is the marginal cumulative distribution function (CDF) defined by eq. (I.19).
Example 6.2 Let us consider two biaxial fatigue tests reported in [38] with nominal axial stress
of 81 MPa and nominal shear stress of 50 MPa applied in-phase in both tests. The corresponding
value of the predictor G𝛼 (𝛼 = 0) was computed and verified to be G𝛼 = 31.6 ksi (218 MPa). The
survival function is shown in Fig. 6.12.
0.95
1
0.8 Gα = 31.6 ksi

Probability of survival
71.5 %
0.6
0.4
21.6 %
0.2
0
104 105 106
0.05
77,000 671,000
N
Figure 6.12 Survival function corresponding to G𝛼 = 31.6 ksi. Outcomes of combined in-phase
axial-torsion fatigue experiments performed on notched 2024-T3 aluminum specimens.
We determine N1 and N2 from the condition S(N1 ) = 0.05 and S(N2 ) = 0.95 and find
N1 = 671,000, N2 = 77,000. Therefore our model predicts that the probability that failure
will occur in the interval of 77,000 to 671,000 cycles is 90%. The reported experimental values
(Nexp ), shown in Fig. 6.12 are 302,000 and 130,000 cycles. Therefore the outcome of experiments is
within the predicted interval.
An alternative way of reporting the results of validation experiments is to say that the probabilities
of survival, computed from eq. (6.37) by substituting the reported number of cycles at failure Nexp
for N, are 21.6% and 71.5% respectively. Since these probabilities lie in the interval of 5% to 95%, we
consider the predictor G𝛼 (𝛼 = 0) to have passed this validation test.
6.4.2 The domain of calibration

Mathematical models are defined on sets of admissible parameters, specific to the model. Models
are calibrated on a subset of the admissible parameters that define the domain of calibration. The
domain of calibration is an essential attribute of a mathematical model.
The predictor G𝛼 was formulated for the purpose of generalization of the S-N data to notched
specimens in the high cycle regime. The lower limit of what is considered to be high-cycle fatigue
is usually and somewhat arbitrarily set at 10 thousand cycles. In reference [94] fatigue test records
for notched specimens that failed at fewer than 8500 cycles, the lower limit of the number of cycles
for which S-N data are available, were not included in the calibration set.
We introduce a restriction based on the size of the plastic zone relative to the volume of inte-
gration Vc . We define Vyld as the volume where the von Mises stress 𝜎, determined from the linear
solution, is greater than the yield stress 𝜎yld .
def
𝜂 = Vyld ∕Vc . (6.38)
It would be overly restrictive to limit the predictor to purely elastic stress fields because the peak
elastic stress can be very high at notch roots. Therefore such a restriction would severely limit
the applicability of the predictor. It is possible to admit small amounts for plastic deformation,
provided that the plastic zone is sufficiently small so that plastic deformation is controlled by the
surrounding elastic stress field. The effects of plasticity controlled by the elastic stress field are taken
into account through calibration of the parameter 𝛽. Naturally, this leads us to the question: How
large the plastic zone may be? Equivalently, what is the limiting value of 𝜂, denoted by 𝜂 lim ? This
value is a delimiter of the domain of calibration.
Test records that satisfy conditions imposed by the assumptions on which the formulation of
the mathematical model is based are called qualified test records. In the present case the size of
the plastic zone is limited by the condition 𝜂 < 𝜂 lim . The value of 𝜂 lim will be inferred from the
outcome of validation experiments.
The results of experiments, taken from Table 5 of reference [38], and the values of 𝜂 are listed in
Table 6.6 where 𝜎nom and 𝜏nom represent the maximum nominal normal and shearing stresses in
the test section, G𝛼 is the computed value of the predictor, converted to ksi units for consistency
with the calibration records, Nexp is the number of cycles at which failure occurred.
The estimated probabilities of survival to the number of cycles at which failure occurred, com-
puted from eq. (6.37), are listed under the heading “Prob”.
The results in Table 6.6 show that if we limit the domain of the predictor to Nexp > 8500 then 4
out of 13 outcomes fail the validation experiments in the sense that the predicted probabilities of
survival fall outside of the 5% to 95% interval. On closer examination we find, however, that these
specimens had larger amounts of plasticity than the others. For example, if we set 𝜂lim = 0.15 then
two records are disqualified for excessive plastic deformation, the number of qualified records is
11 of which 10 passed the test. This is consistent with the prediction. If we set 𝜂lim < 0.08 then four
records are disqualified and the remaining 7 specimens pass the validation test. In either case, there
is no reason to reject the model.
Since we have access only to a limited number of records of biaxial fatigue tests in the high cycle
regime, we are not in a position to propose a sharp definition for the threshold value of 𝜂. A tentative
value of 10% would appear to be reasonable, however, and consistent with the available data.
Table 6.6 Results of validation experiments performed on notched 2024-T3 aluminum specimens. Source
for the experimental data: Table 5 of Reference [38].
Loading 𝝈nom 𝝉nom G𝜶 Nexp Prob Result 𝜼 Remarks

MPa MPa ksi cycles (%) (%)
Axial 145 0 41.0 9,500 100.0 Fail 17.1 𝜂 > 10%

Axial 130 0 36.7 21,670 99.2 Fail 8.0
Axial 130 0 36.7 31,000 92.1 Pass 8.0
Axial 115 0 32.5 135,450 26.5 Pass 1.6
Axial 115 0 32.5 145,600 22.4 Pass 1.6
Axial 98 0 27.7 735,000 13.5 Pass 0.0
Torsion 0 108 40.3 18,500 97.4 Fail 34.4 𝜂 > 10%
Torsion 0 91 34.0 71,890 61.0 Pass 6.9
Torsion 0 91 34.0 60,140 75.6 Pass 6.0
Torsion 0 76 28.4 215,000 57.6 Pass 0.0
Torsion 0 64 23.9 1,800,000 43.8 Pass 0.0 runout
Combined 81 50 31.6 302,000 21.6 Pass 0.7 in-phase
Combined 81 50 31.6 130,000 71.5 Pass 0.7 in-phase
Based on the evidence developed in the calibration and validation processes, it was asserted in
Section 6.2.5 that Peterson′ s updated predictor was validated in the interval of notch radii 0.003 <
r < 1.5 inches. Here we assert that another predictor, based on integral averages, has been validated
also and, furthermore, it performs better than Peterson′ s updated predictor when tested against the
experimental data summarized in Table I.2. This is based on the observation that the log likelihood
function corresponding to the predictor defined in Section 6.3, evaluated for the entire set of quali-
fied test records, is larger than that corresponding to Peterson′ s updated predictor and has a larger
domain of calibration in the sense that it has been calibrated with reference to the highly stressed
volume and therefore stress raisers do not have to be characterized by notch radii.
These claims of validation and ranking are based on the available experimental data. It is of course
possible that for a different set of data the ranking would change. It is also possible that other kinds
of predictor will be formulated, calibrated and tested that will perform better than the predictor
considered here.
6.4.3 Out-of-phase biaxial loading

The test conditions and validation data for combined out-of-phase axial and torsional loading in
the high cycle range are summarized in Table 6.7 where 𝜎nom and 𝜏nom are the nominal normal and
shearing stresses9 corresponding to the axial and torsional components of the applied load and Nexp
is the number of fully reversed cycles at which failure occurred [38].
The nominal first principal stresses for the axial, torsional and combined loadings are shown in
Fig. 6.13 where we assumed that
(𝜎1 )axl = 𝜎nom sin(2𝜋ft) (6.39)
(𝜎1 )tor = 𝜏nom sin(2𝜋ft + 𝜙) (6.40)
where f is the frequency (Hz), 𝜙 is the phase angle, in the present case 𝜙 = 𝜋∕2. Therefore the
combined value, denoted by (𝜎1 )com , is
(( ) )1∕2
def (𝜎1 )axl (𝜎1 )axl 2
(𝜎1 )com = + 2
+ (𝜎1 )tor . (6.41)
2 2
Accounting for out-of-phase loading requires a modeling assumption in addition to those made
for the simulation of in-phase loading described in [95]. This modeling assumption accounts for
the effects of out-of-phase loading on the notch. Various possibilities exist. The only requirement
Table 6.7 Results of experiments for 90∘ out-of-phase

fully reversed axial and torsional loading.
Exp. 𝝈nom 𝝉nom Nexp

MPa MPa cycles
1 115 71 42760
2 115 71 78100
3 100 58 96800
Source for the data: Table 5 of Reference [38].
9 Nominal stress is the maximum stress in the test section without the notch.
Axial load
120
First principal stress (MPa)
100
80
60
40
20
0
3
Torsion
120
100
80
60
40
20
0
Combined
120
100
80
60
40
20
0
0 0.5 1 1.5 2 2.5 3
number of cycles
Figure 6.13 Nominal stresses in experiments 1 and 2.
is that when the phase angle is zero then the predictor has to be the same as the predictor defined
for in-phase loading. We will discuss two extensions in the following.
Extension A
It is seen in Fig. 6.13 that the maximum value of the axial and combined loading is the same. How-
ever, the material is subjected to high nominal stress over a greater fraction of the cycle under
combined loading. We will assume that the magnifying effects of the out-of-phase loading are pro-
portional to the ratio of the shaded areas in Fig. 6.13. Specifically, we define
1∕2
def
A1 = (𝜎1 )axl dt (6.42)
∫0
1∕2
def
A2 = (𝜎1 )com dt (6.43)
∫0
Table 6.8 Test records used for the calibration of Model B.
# (G𝜶 )axl (G𝜶 )tor Nexp A1 A2

ksi ksi cycles MPa MPa
1 32.5 26.5 42.7E3 36.61 50.52

2 78.1E3
Table 6.9 The outcome of the validation experiment for 90∘ out-of-phase axial and torsional loading.
# (G𝜶 )axl (G𝜶 )tor (G𝜶 )eff N1 N2 Nexp Prob Result

ksi ksi ksi cycles cycles cycles (%)
3 28.3 21.7 33.2 60.4E3 395.3E3 96.8E3 72.6 Pass
and
def A2
(G𝛼 )eff = 𝜅(G𝛼 )axl (6.44)
A1
where 𝜅 is a parameter, independent of A1 and A2 , determined by calibration.
We will use the data in the first two rows in Table 6.7 for calibration, the data in the third row for
validation. The computed values of (G𝛼 )axl , (G𝛼 )tor , A1 and A2 are shown in Table 6.8. On substitut-
ing A1 , A2 and (G𝛼 )axl into eq. (6.44), we find (G𝛼 )eff = 44.8𝜅 ksi.
The objective of calibration is to determine 𝜅 such that the data in the calibration set maxi-
mizes the log likelihood function corresponding to the statistical model. Specifically, letting 𝑤1 =
log10 42.7E3 and 𝑤2 = log10 78.1E3, we have
LL(x) = log(𝜙M (𝑤1 , 44.8x)) + log(𝜙M (𝑤2 , 44.8x)) (6.45)
where 𝜙M is the marginal pdf of the random fatigue limit model given by eq. (I.18) in the appendix.
From this we find
𝜅 = arg max LL(x) = 0.8680. (6.46)
This completes the calibration process. We now examine the predictive performance of (G𝛼 )eff .
Specifically, we test the following prediction: “The probability that the number of cycles Nf at which
failure will occur lies in the interval N1 < Nf < N2 is 90%”. Another way of saying this is that 9 out
of 10 specimens are predicted to fail in the given (load-dependent) interval.
For the test record used in validation, we find A1 = 31.83 MPa, A2 = 42.95 MPa, hence
A2
(G𝛼 )eff = 𝜅(G𝛼 )axl = 33.2 ksi.
A1
Given (G𝛼 )eff , the limits N1 , N2 and the probability of survival to Nexp cycles are computed. The
results are shown in Table 6.9, where the probability of survival to the experimentally observed
number of cycles is shown under the heading “Prob”. Based on these results, we conclude that the
model passes the validation test.
Finally we update the calibration records and recompute 𝜅: The two records used for calibration
are augmented by the record used for validation and eq. (6.45) becomes
LL1 (x) = log(𝜙M (𝑤1 , 44.8x)) + log(𝜙M (𝑤2 , 44.8x)) + log(𝜙M (𝑤3 , 38.2x)) (6.47)
where 𝑤3 = log10 96.8E3 and find the updated value of 𝜅, denoted by 𝜅upd ;
𝜅upd = arg max LL1 (x) = 0.8840. (6.48)
Extension B
We next consider the following extension of in-phase multiaxial loading:
def ( p p )1∕p
(G𝛼,p )eff = (G𝛼 )axl + (G𝛼 )tor , p≥1 (6.49)
where (G𝛼 )axl and (G𝛼 )tor are the values of the predictor G𝛼 corresponding to the axial and torsional
loading, respectively. It is possible to select p in such a way that, for the available out-of-phase biaxial
test records, the likelihood function is maximum. This would calibrate (G𝛼,p )eff . Since we have very
few test records at our disposal, and we are primarily interested in the process of validation and
ranking, we forego that process and let p = 2 and define (G𝛼 )eff ≡ (G𝛼,2 )eff .
The limit values N1 and N2 are determined from the survival function of the statistical model,
see eq. (6.37). The computed values of (G𝛼 )eff and the limits of the predicted interval N1 and N2 ,
corresponding to 5% and 95% probabilities of survival respectively, are listed in Table 6.10. The
column with the heading Nexp shows the probability of survival to the realized number of cycles
calculated from eq. (6.37). Observe that all failures occurred in the predicted intervals, in other
words, there were three successful predictions in three trials.
Remark 6.11 We could have set the survival probabilities to (say) the 98% range. However, the
tail segments of survival functions are very sensitive to the choice of the statistical model and should
not be relied upon in validation experiments. Generally speaking, validation deals with events that
have high probability of occurrence.
Exercise 6.1 Verify the survival probabilities for the ((G𝛼 )eff , Nexp ), entries in Table 6.10.
Ranking
We have formulated two possible extensions for the predictor validated for in-phase loading. Both
predictors passed the validation tests. We now ask: Which one is better? Should we prefer one over
another? Questions such as this have to be addressed in numerical simulation where the intu-
itive process of formulation has to be moderated by objective methods of ranking. To that end, we
evaluate the log likelihood (LL) function for each predictor with reference to the outcome of the
Table 6.10 The outcome of validation experiments for 90∘ out-of-phase axial and torsional loading.
# (G𝜶 )axl (G𝜶 )tor (G𝜶 )eff N1 N2 Nexp Prob Result

ksi ksi ksi cycles cycles cycles (%)
1 32.5 26.5 42.0 22.4E3 80.5E3 42.7E3 45.3 Pass

2 32.5 26.5 42.0 22.4E3 80.5E3 78.1E3 5.7 Pass
3 28.3 21.7 35.6 43.9E3 219.8E3 96.8E3 42.7 Pass
Table 6.11 The computed log likelihood (LL) values for predictors A and B.
Predictor (G𝜶 )eff (ksi) Nexp (cycles) LL
1 2 3 1 2 3
A 39.6 39.6 33.8 42,700 78,100 96,800 −34.33

B 42.0 42.0 35.6 −34.95
validation experiments. The predictor with the greater LL value is the better generalization of the
S-N data to notched specimens subjected to out-of-phase biaxial loading, given the available data.
The updated predictors and the corresponding values of the log likelihood function are listed in
Table 6.11.
The evidence for preferring predictor A to predictor B is based on the likelihood ratio
LA ∕LB = exp(LLA − LLB ) = 1.86 (6.50)
where LA and LB are the likelihood values corresponding to predictors A and B, respectively. If
this ratio is in the interval (1∕3, 3), which is the case here, then the difference is barely worth
mentioning. In other words, the two predictors are in a virtual tie.
Predictive performance
We test the predictive performance of the mathematical model by calculating the number of cycles
N1 (resp. N2 ) corresponding to the 5% (resp. 95%) survival probabilities for each specimen, given
the test conditions. Therefore, if the model is correct then 9 out of 10 specimens will fail in the
predicted interval. We denote the number of outcomes that fall in the predicted interval divided
by the number of tests by 𝜃. Our problem now is to find the most probable value of 𝜃, denoted
by 𝜃0 , and the associated confidence interval, given the experimental data, and prior information
concerning the probability density function of 𝜃.
Bayes′ theorem states:
Pr(D|𝜃)Pr(𝜃)
Pr(𝜃|D) = (6.51)
Pr(D)
where Pr(𝜃|D) is the posterior probability density function, D = (m, n) stands for the available
experimental data in terms of the number of experiments m and the number of successful outcomes
n. The term Pr(𝜃|D) is called the posterior probability density function.
Pr(D|𝜃) is the probability of obtaining the observed data, given 𝜃. The probability of n successes
in m trials is given by the binomial distribution:
m!
Pr(D|𝜃) = 𝜃 n (1 − 𝜃)m−n . (6.52)
n!(m − n)!
Pr(𝜃) is our prior information (the prior) concerning the probability density function (pdf) of 𝜃.
We will use the Beta distribution to estimate this:
Γ(𝛼 + 𝛽) 𝛼−1
Pr(𝜃) = 𝜃 (1 − 𝜃)𝛽−1 (6.53)
Γ(𝛼)Γ(𝛽)
where Γ(⋅) is the Gamma function, 𝛼 > 0 and 𝛽 > 0 are shape parameters. The shape parameters
define the prior probability distribution.
Substituting equations (6.52) and (6.53) into eq. (6.51) we get

Pr(𝜃|D) = C𝜃 n+𝛼−1 (1 − 𝜃)m−n+𝛽−1 (6.54)
where C = 1∕Pr(D) is a normalizing constant that must be chosen such that
1
C 𝜃 n+𝛼−1 (1 − 𝜃)m−n+𝛽−1 d𝜃 = 1. (6.55)
∫0
Noting that C is the normalizing constant for the Beta distribution, we have
Γ(m + 𝛼 + 𝛽)
C= ⋅ (6.56)
Γ(n + 𝛼)Γ(m − n + 𝛽)
When the posterior pdf has the same functional form as the prior, as in the present case, then the
prior is said to be conjugate to the posterior pdf. Using conjugate priors has the advantage that it is
simple to update the posterior pdf.
Selection of the prior

In Bayesian data analysis the selection of the prior pdf is guided by various considerations that fall
into the following broad categories.
1. An uninformative prior, also called objective prior, expresses ignorance concerning the expected
probability distribution of the data. In our case, for example, if we think that a successful out-
come of an experiment is just as likely as an unsuccessful one then, invoking the principle of
indifference, we assume that 𝜃 has uniform distribution, that is Pr(𝜃) ∼ U(0, 1). The principle
of indifference states that if we can enumerate a set of basic, mutually exclusive, probabilities
and have no reason to believe that any one of these is more likely to be true than another, then
we should assign the same probability to all [87]. Uninformative priors may express objective
information. For example, in this case we know that 𝜃 must lie in the interval (0, 1).
2. An informative prior expresses specific, definite information about a variable. For example, we
would have an informative prior if we had access to test records of experiments that existed
before we came into possession of new test records. Those existing test records would provide
specific information relating to the probability distribution of 𝜃.
3. A weakly informative prior expresses partial information about a variable.
Although we had specific, definite prior information about the pdf of 𝜃, detailed in Section 6.4.2
where validation of G𝛼 for in-phase (IP) biaxial loading conditions was considered, we chose an
uninformative prior based on the following consideration: Our goal was to test two extensions of
the definition of G𝛼 formulated to account for out-of phase (OP) biaxial loading conditions. The
prior information was for in-phase loading only. Using anything other than an uninformative prior
would have tended to obscure the relevant information content of the experimental data.
Remark 6.12 Suppose that we would have formulated and tested G𝛼 for out-of phase conditions
first and then tested it under in-phase conditions, then it would have been proper to use the poste-
rior pdf developed from OP data because the IP condition is a special case of the OP condition.
Inferential statistics
For both predictors we had n = 3 successful outcomes in m = 3 experiments. The uniform distri-
bution is equivalent to the Beta distribution with the shape parameters set to 𝛼 = 𝛽 = 1. In this case
C = 4 and the posterior pdf is
Pr(𝜃|D) = 4𝜃 3 . (6.57)
4.0
3.0 posterior
pdf
2.0 prior
1.0 θ2 = θ0
0
0 0. 2 0.4 θ1 0.6 0.8 1.0
θ
Figure 6.14 Posterior pdf corresponding to three successes in three trials.
The best estimate is given by the maximum of the posterior pdf, in the present case the best esti-
mator of 𝜃, denoted by 𝜃0 , is 𝜃0 = 1. It is customary to provide the shortest 95% confidence interval,
also called credible interval. This means that we have to find 𝜃1 and 𝜃2 such that
𝜃2
Pr(𝜃1 < 𝜃0 < 𝜃2 |D) = 4 𝜃 3 d𝜃 ≈ 0.95 (6.58)
∫𝜃1
subject to the condition that 𝜃2 − 𝜃1 is minimum. In the present case we find 𝜃1 = 0.473, 𝜃2 = 1.
The shaded area in Fig. 6.14 represents how much we are justified in believing that 𝜃 lies in the
interval (𝜃1 , 𝜃2 ).
The data tell us that the most likely outcome of the next experiment will be “pass”. Put differ-
ently, the justifiable degree of belief that the next experiment will result in a pass of the validation
experiment is very close to unity, given the available data. This degree of belief (also called plau-
sibility) is modified through the process of conditionalization when new data become available.
Conditionalization will be illustrated in Example 6.3.
Remark 6.13 In this example the pdf is highly asymmetric. While 𝜃0 is the most probable value of
𝜃, the expected value, denoted by 𝜃M , may be more representative because it takes into account the
skewness of the distribution [87]. Its value is: 𝜃M = 0.80. Note that the predicted value of 𝜃 happens
to lie in the middle of the interval (𝜃M , 𝜃0 ).
Validation
Validation criteria are usually stated in terms of a metric, or a similar measure, such as the
Kullback-Leibler divergence, that quantifies the difference between the predicted and realized
pdfs, and an acceptable tolerance [69]. The problem is that setting any particular tolerance would
seem arbitrary and difficult to justify. In validation we are asking the question: “Is there a reason
that would justify rejecting the model?” – In the present case all three experiments passed the
validation test. Clearly, there is no reason to reject this model. Had 𝜃0 been outside of the 95%
confidence interval, we could not trust predictions based on the model and therefore we would
have to consider rejecting it. In such obvious cases the answer is clear. In most cases, however, the
problem will be like this: We performed m experiments, predicted n successes and observed nobs
successes, n ≠ nobs .
In general, a mathematical model is not rejected until and unless a better model is found, in which
case the justification for rejection is based on the relative merit of competing models, measured by
the posterior ratio. When the posterior ratio happens to be close to unity, as in the present instance,
the competing models remain under consideration until new data become available that will serve
to justify a decision one way or another.
5.0
4.0
posterior
3.0
pdf
2.0
prior
1.0 θ2
0
0 0.2 0.4 0.6 θ1 0.8 θ0 1.0
θ
Figure 6.15 Example 2: Prior and posterior pdfs.
Considering that we used an uninformative prior, and only three data points, it is not surprising
that the confidence interval is rather wide. In order to reduce the width of the confidence inter-
val, additional experiments would have to be performed. The length of the confidence interval is
approximately proportional to the square root of the number of experiments [87].
Example 6.3 Suppose, for the sake of illustration, that additional data became available indicat-
ing n = 8 successful outcomes in m = 9 experiments. Let us find the best estimate of 𝜃 and the
confidence interval using the previously obtained posterior pdf for the prior:
Pr(𝜃) ∝ 𝜃 𝛼−1 (1 − 𝜃)𝛽−1
where 𝛼 = 4, 𝛽 = 1. Therefore the new value of C from eq. (6.56) is:
Γ(14)
C= = 156
Γ(11)Γ(3)
and the posterior pdf is
Pr(𝜃|D) = 156 × 𝜃 11 (1 − 𝜃). (6.59)
The best estimate for 𝜃0 , given by the maximum of the posterior pdf, is 𝜃0 = 0.917 and the boundary
points of the shortest 95% confidence interval are found to be 𝜃1 = 0.681, 𝜃2 = 0.995. In view of these
results, our justifiable degree of belief is 95% that the limit value of 𝜃0 lies in this interval.
Remark 6.14 We used the posterior pdf given by eq. (6.57 for the prior in Example 6.3. The result
would have been the same if we used the uniform distribution for the prior and all available records
(n = 11 successes out of m = 12 trials). In other words, the results are independent of the ordering
of the outcomes.
The updated domain of calibration

A mathematical model is a validated model within its domain of calibration. On a sufficiently
small domain of calibration just about any model can be validated. New tests within a domain
of calibration are useful for updating the parameters but cannot be considered as proper validation
experiments. For a validation experiments to be proper, at least one of the admissible parameters
must be outside of the domain of calibration, or the experimental conditions (such as loading
conditions) must be more general than the conditions employed in the calibration experiments.
On successful conclusion of a set of validation experiments, the parameters are updated and the
domain of calibration is revised.
20
m = 160
15
pdf
10
5 m = 10
0
0 0.2 0.4 0.6 0.8 θ0 1
θ
Figure 6.16 Ideal posterior pdfs corresponding to m = 10, 20, 40, 80, 160 ideal data points, 𝜃0 = 0.9.
Table 6.12 Ideal data (𝜃0 = 0.9): Minimum 95% confidence intervals (CI).
m 10 20 40 80 160
𝜃1 0.048 0.044 0.040 0.036 0.033

𝜃2 0.998 0.994 0.990 0.986 0.983
CI 0.361 0.259 0.184 0.131 0.093
The extension of G𝛼 to biaxial in-phase loading was validated with reference to one specimen
type only. The values of the highly stressed volume were within the domain of calibration. For
example, for axial loading: V = 7.28E − 6 in3 , for torsional loading: V = 4.29E − 6 in3 . Whereas
calibration of G𝛼 (𝛽) was under uniaxial loading only, the validation experiments included torsional
and combined torsional and axial loading conditions.
The number of experiments

For the purpose of illustration, let us assume that we have ideal data such that n = m𝜃0 , where
𝜃0 = 0.9 is the ratio of the predicted number of successful outcomes to the number of experiments.
The ideal posterior pdfs are illustrated for m = 10, 20, 40, 80, 160 in Fig. 6.16.
This figure illustrates that the confidence interval converges very slowly. The minimum 95% con-
fidence intervals are listed in Table 6.12 where 𝜃1 and 𝜃2 are the lower and upper limits of the
confidence interval (CI).
The empirical relationship is
CI ≈ 1.123 m−0.4907 . (6.60)
which
√ is consistent with the estimate that the confidence interval is approximately proportional to
1∕ m, see, for example, [87]. The results indicate that, in order to reduce the size of the minimum
95% confidence interval to 0.1, well over 100 experiments would be necessary, which is many times
more than what is usually available in practice.
6.5 Management of model development

While the procedures illustrated in this chapter were specific to the objective to generalize the
results of fatigue tests performed on notch-free and notched coupons under constant cycle uniaxial
Mathematical model
Linear Statistical Formulation

Predictor
Elasticity Model
Calibration S-N data

Calibration
Calibration
for α,β Notched data
Prediction Test conditions Prediction
Validation Validation data Validation
Yes Update notched data

Pass?
and calibration
Disposition
No
Create/update
Reject documentation
Figure 6.17 Schematic representation of the validation process.
loading conditions to variable cycle triaxial conditions, those procedures fit into the generic tasks of
formulation, calibration, prediction, validation and disposition, common to all model development
projects. This is illustrated schematically in Fig. 6.17.
The process is open-ended: The formulation of mathematical models involves making assump-
tions based on intuition, insight and personal preferences. This is a creative process, no boundaries
can be set other than the requirement of logical consistency and the need to recognize that every
assumption imposes some limitation on the model. Those limitations, together with limitations
on the data available for calibration and testing, define the domain of calibration. The domain of
calibration is an essential attribute of a model. New data within the domain of calibration present
an opportunity to update the model parameters, and new data outside of the domain of calibration
present an opportunity to revise the ranking of candidate models and update both the domain of
calibration and the model parameters.
The development of mathematical models typically involves multiple disciplines and cuts across
traditional departmental boundaries in industrial and research organizations. It is necessary for
those organizations to exercise simulation governance, that is, create conditions that facilitate
incremental improvement of simulation practices over time, see Section 5.2.4.
The economic stakes associated with the management of numerical simulation resources can be
very substantial. Properly managed, numerical simulation can be a strategic corporate asset, poorly
managed it can become a major liability. There are many well documented cases of substantial
economic loss attributable to poor management of numerical simulation projects.
The justification for employing a particular mathematical model is based on negation, such as
“we found no reason to reject the model” or “no one proposed a better model”. Conceptually, one
could reject a model based on some fixed criterion. In reality, however, it is very difficult to justify
a particular criterion for rejection. A mathematical model is not rejected until and unless a better
model is found. For example, one would not reject Peterson′ s predictor until a better one, such
as Peterson′ s revised predictor, was found. The justification for preferring the predictor G𝛼 over
Peterson′ s revised predictor is that G𝛼 has been shown to produce better predictions on a larger
domain of calibration. Mathematical models evolve over time through testing, evaluating, compar-
ing, updating and documenting.
6.5.1 Obstacles to progress

We mention two strongly correlated obstacles to the full realization of the potential of numerical
simulation technology in engineering. One is that management has not yet grasped that current
simulation practices in industrial settings are still very far from what is needed for full realization of
the potential of the technology. The second is widespread confusion in the professional community
evidenced by the all too frequent references to fuzzy terminology and empty notions.
An example of fuzzy terminology is “finite element modeling”. As noted in Section 5.2.5, this
term mixes two conceptually and functionally different entities: the mathematical model and its
numerical treatment.
The prime example of empty notions is “physics-based model”. To explain why it is empty, we
quote from a renown theoretical physicist and an influential philosopher of science:
“I take the positivist viewpoint that a physical theory is just a mathematical model and that it is
meaningless to ask whether it corresponds to reality. All that one can ask is that its predictions
should be in agreement with observation.”
Stephen Hawking10
“The empirical basis of objective science has nothing absolute about it. Science does not rest
on solid bedrock. The whole towering structure, the often fantastic and audacious construction
of scientific theories, is built over a swamp. Its foundations are pillars driven from above into
the swamp – not down to any natural ‘given’, ground, but driven just as deeply as is necessary
to support the structure. The reason why we stop driving the pillars deeper into the ground is
not that we have reached solid rock. No, our decision is based on the hope that the pillars will
support the structure.”
Karl Popper11
They are telling us that physical reality cannot be known. Some phenomenological aspects of
reality can be simulated by mathematical means with remarkable precision and success. However,
limitations are always present in the mathematical models used in the foundational as well as in the
applied sciences. Therefore the notion of a “physics-based model” is empty. A mathematical model
is a precise statement of an idea of some aspects of reality and nothing more. The importance of
proper terminology stands long recognized:
10 Stephen William Hawking 1942–2018. “The nature of space and time”. Princeton University Press, 2010 (with
Roger Penrose).
11 Karl Raimund Popper 1902–1994. “The Two Fundamental Problems of the Theory of Knowledge”. Routledge,
2014.
“If names be not correct, language is not in accordance with the truth of things. If language be
not in accordance with the truth of things, affairs cannot be carried on to success.”
Confucius, The Analects – 13
Progress is possible through evolution. The best management can do is to create a friendly envi-
ronment for the evolutionary process to take its course unimpeded. There are many opportunities
for improvement in that area.
223
Beams, plates and shells
Dimensional reduction was discussed in Chapter 2 in connection with planar and axisymmetric
models. Another very important class of dimensionally reduced models is discussed in this chapter.
Our starting point is the generalized formulation of the problem of linear elasticity.
7.1 Beams
In order to present the main points in a simple setting, mathematical models for beams are derived
from the generalized formulation of the problem of two-dimensional elasticity. The formulation of
models for beams in three dimensions is analogous but, of course, more complicated. Referring to
Fig. 7.1, the following assumptions are made:
1. The xy plane is a principal plane, that is, loads applied in the xy plane will not cause displacement
in the direction of the z axis.
2. The x-axis passes through the centroid of the cross-section. The domain of the cross-section is
denoted by 𝜔.
3. The material is elastic and isotropic.
The displacement vector components are written in the following form:
ux = ux|0 (x) + ux|1 (x)y + ux|2 (x)y2 + · · · + ux|m (x)ym (7.1)
uy = uy|0 (x) + uy|1 (x)y + uy|2 (x)y2 + · · · + uy|n (x)yn (7.2)
where the functions ux|k (x), uy|k (x) are called field functions; their multipliers (powers of y) are
called director functions. Writing the displacement components in this form allows us to consider
a hierarchic family of models for beams characterized by the pair of indices (m, n). The highest
member of the hierarchy is the mathematical model based on two-dimensional elasticity which
corresponds to m, n → ∞.
In the following we will denote the exact solution of a hierarchic beam model characterized by
(m,n)
m and n by uEX and the exact solution of the mathematical model based on two-dimensional
(2D)
elasticity by uEX .
224 7 Beams, plates and shells
q(x)
y y c1
F0 Fℓ
x z d
M0 Mℓ
V0 cs Vℓ c2
ℓ b(y)
Any of the boundary conditions described in connection with two-dimensional elasticity may be
specified. However, we will be concerned only with the loadings and constraints typically used in
the analysis of beams. Specifically, the following types of load will be considered:
1. The traction component Tn = Ty acting on the surface y = c1 (see Fig. 7.1):
q(x)
Tn = (7.3)
b(c1 )
where qy is the distributed load (in N/m units).
2. Normal tractions acting on cross-sections written in terms of the axial force F and the bending
moment M. The sign conventions are indicated in Fig. 7.1.
F My
Tn = − (7.4)
A I
where A is the area of the cross-section and I is the moment of inertia of 𝜔 with respect to the
z axis:
A= dydz, I= y2 dydz. (7.5)

∫𝜔 ∫𝜔
3. The shearing tractions Tt (y, z) acting on cross-sections are either constant or defined by a func-
tion such that Tt (c1 , z) = Tt (−c2 , z) = 0 and the resultant is the shear force V:
V VQ(y)
Tt = − or Tt = − (7.6)
A Ib(y)
where Q(y) is the static moment about the z axis of that portion of the cross-section which
extends from y to c1 :
c1
Q(y) = sb(s) ds. (7.7)
∫y
The beam may be supported by an elastic foundation, with foundation modulus cs (x) ≥ 0
(N∕m2 units) and kinematic boundary conditions may be prescribed. We will assume that the
elastic foundation reacts in the y direction only, i.e., the foundation generates a distributed
transverse load qs (x) = −cs (x)uy (x, −c2 ) (N/m units).
We will write the generalized formulations for beam models as applications of the principle of
minimum potential energy. The strain energy is:
𝓁
1 1
U= (𝜎 𝜖 + 𝜎y 𝜖y + 𝜏xy 𝛾xy ) dV + c u2 (x, −c2 ) dx (7.8)
2 ∫Ω x x 2 ∫0 s y
and the potential of external forces is
𝓁 ∑
2
P= quy (x, c1 ) dx + (Tx ux + Ty uy ) dS (7.9)
∫0 i=1
∫𝜔i
7.1 Beams 225
where 𝜔i (i = 1, 2) are the cross-sections at x = 0 and x = 𝓁 respectively. The energy space is defined
in the usual way. The exact solution of a particular model is the minimizer of the potential energy
on the space of admissible functions:
(m,n)
Π(uEX )= min Π(u(m,n) )
̃
u(m,n) ∈E(Ω)
̃
where Π(u) = U(u) − P(u) and E(Ω) is the space of admissible functions.
Remark 7.1 Proper selection of a beam model characterized by the indices (m, n) is
problem-dependent. When the field functions are smooth and the thickness is small then m
and n can be small numbers. The assumptions incorporated into the commonly used beam
models imply that the field functions do not change significantly over distances comparable to the
thickness.
Remark 7.2 The distinction between the notions of mathematical model and its discretization
is blurred by conventions in terminology: It is customary to refer to the various beam, plate
and shell formulations as theories or models. These models are semidiscretizations of the fully
three-dimensional model. Therefore errors that can be attributed to the choice of indices, such as
the indices (m, n) in equations (7.1) and (7.2), and analogous indices defined for plate and shell
models, identify both: a particular semidiscretization of the fully three-dimensional model and
the definition of a beam, plate or shell model.
7.1.1 The Timoshenko beam

Let us consider the simplest model, the model corresponding to the indices m = 1, n = 0. We intro-
duce the notation:
ux|0 (x) = u(x), ux|1 (x) = −𝛽(x), uy|0 (x) = 𝑤(x).
The function 𝛽 represents a positive (i.e., counterclockwise) angle of rotation (with respect to the
z axis). Since a positive angle of rotation multiplied by positive y would result in negative displace-
ment in the x direction, ux|1 = −𝛽. The strain components are as follows:
𝜕ux 𝜕uy 𝜕ux 𝜕uy
𝜖x = = u′ − 𝛽 ′ y, 𝜖y = = 0, 𝛾xy = + = −𝛽 + 𝑤′ (7.10)
𝜕x 𝜕y 𝜕y 𝜕x
where the primes represent differentiation with respect to x. Assuming that the normal stresses 𝜎y
and 𝜎z are negligibly small in comparison with 𝜎x , the stress components are:
𝜎x = E𝜖x = E(u′ − 𝛽 ′ y), 𝜎y = 0, 𝜏xy = G𝛾xy = G(−𝛽 + 𝑤′ ).
Therefore the strain energy is:

𝓁( ) 𝓁
1 1
U= [E(u′ − 𝛽 ′ y)2 + G(−𝛽 + 𝑤′ )2 ] dydz dx + c 𝑤2 dx
2 ∫0 ∫𝜔 2 ∫0 s
where the volume integral was decomposed into an area integral over the cross-section and a line
integral over the length of the beam. Using the notation introduced in eq. (7.5), and noting that
since the y and z axes are centroidal axes, we have
y dydz = 0,
∫𝜔
the strain energy can be written as:

𝓁 𝓁
1 1
U= [EA(u′ )2 + EI(𝛽 ′ )2 + GA(−𝛽 + 𝑤′ )2 ] dx + c 𝑤2 dx. (7.11)
2 ∫0 2 ∫0 s
For the reasons discussed in the following section, the strain energy expression is usually modified
by multiplying the shear term by a factor known as the shear correction factor, denoted by 𝜅. The
modified expression for the strain energy is:
𝓁 𝓁
1 1
U𝜅 = [EA(u′ )2 + EI(𝛽 ′ )2 + 𝜅GA(−𝛽 + 𝑤′ )2 ] dx + c 𝑤2 dx. (7.12)
2 ∫0 2 ∫0 s
The potential of external forces is
𝓁
P= q𝑤 dx + F𝓁 u(𝓁) + M𝓁 𝛽(𝓁) − V𝓁 𝑤(𝓁) − F0 u(0) − M0 𝛽(0) + V0 𝑤(0) (7.13)
∫0
where F, M, V are respectively the axial force, bending moment and sher force, collectively called
stress resultants, which are defined as follows:
F= 𝜎x dydz, M = − 𝜎x y dydz, V = − 𝜏xy dydz. (7.14)

∫𝜔 ∫𝜔 ∫𝜔
The subscripts 0 and 𝓁 refer to the location x = 0 and x = 𝓁, respectively.
The potential energy is defined by:
Π𝜅 = U𝜅 − P. (7.15)
This formulation is known as the Timoshenko beam model1 . Particular applications of the principle
of minimum potential energy depend on the specified loading and boundary conditions.
Shear correction
In this formulation the shear strain is constant on any cross-section (see eq. (7.10)). This is incon-
sistent with the assumption that no shear stresses are applied at the top and bottom surfaces of the
beam, i.e., at y = c1 and y = −c2 , see Fig. 7.1. From equilibrium considerations, it is known that the
shear stress distribution is reasonably well approximated for a wide range of practical problems by
the technical formula
VQ
𝜏xy = −
Ib
where V is the shear force, Q = Q(y) is the function defined by eq. (7.7), I is the moment of inertia
about the z axis and b(y) is the width of the cross section, as shown in Fig. 7.1. The derivation of
this formula can be found in standard texts on strength of materials.
We will consider a rectangular cross-section of depth d and width b in the following. In this case
the shear stress distribution is:
( )
3 V y2
𝜏xy = 1−4 2
2 db d
and the strain energy corresponding to this shear stress in a beam of length Δx is
+d∕2
1 bΔx 1 Δx V 2 6
ΔU𝜏 = 𝜏 2 dy = ⋅
2 G ∫−d∕2 xy 2 G db 5
1 Stephen P. Timoshenko 1878–1972.

7.1 Beams 227
We will adjust the shear modulus G in the (1, 0) beam model so that the strain energy will be the
same:
+d∕2 +d∕2 ( )2
1 bΔx 1 bΔx V 1 Δx V 2
ΔU𝜏(1,0) = 𝜏 2 dy = dy =
2 𝜅G ∫−d∕2 xy 2 𝜅G ∫−d∕2 db 2 𝜅G db
where 𝜅 is the shear correction factor. Letting ΔU𝜏 = ΔU𝜏(1,0) we find that 𝜅 = 5∕6. For this reason
the strain energy expression given by eq. (7.11) is replaced by eq. (7.12). Most commonly 𝜅 = 5∕6
is used, independently of the cross-section, although it is clear from the foregoing discussion that
𝜅 depends on the cross-section.
In static analyses of slender beams the strain energy is typically dominated by the bending term,
hence the solution is not influenced significantly by 𝜅. As the length to depth ratio decreases,
the influence of 𝜅 increases. In vibrating beams the influence of 𝜅 increases with the natural
frequency.
Exercise 7.1 Based on the formulation outlined in Section 1.7, write down the generalized
formulation of the undamped elastic vibration problem for the Timoshenko beam. Assume
ux|0 = 0. Would you expect the natural frequencies computed for the Timoshenko model to be
larger or smaller than the corresponding natural frequencies computed for the beam model (2, 3)?
Why?
Numerical solution
The numerical solution of this problem by the finite element method is very similar to the solution
of the one-dimensional model problem described in Section 1.3. The main difference is that here
we have three field functions u(x), 𝛽(x) and 𝑤(x), which are approximated by the finite element
method. Let us write
M𝛽
∑
Mu
∑ ∑
M𝑤
u= ai Φi (x), 𝛽= bi Φi (x), 𝑤= ci Φi (x)
i=1 i=1 i=1
and denote
a = {a1 a2 … aMu }T , b = {b1 b2 … bM𝛽 }T , c = {c1 c2 … cM𝑤 }T .
The structure of the stiffness matrix becomes visible if we write the strain energy in the following
form:
⎡Ku 0 0 ⎤ ⎧a⎫
1 T T T ⎢ ⎪ ⎪
U𝜅 = {a b c } 0 K𝛽 K𝛽𝑤 ⎥ ⎨b⎬ .
2 ⎢ ⎥
⎣ 0 K𝛽𝑤
T
K𝑤 ⎦ ⎪ c ⎪
⎩ ⎭
Given the assumption that the shearing tractions on the top and bottom surfaces of the beam are
zero and the x axis is coincident with the centroidal axis (see Fig. 7.1), the function u, represent-
ing axial deformation, is not coupled with the rotation 𝛽 and the transverse displacement 𝑤 and
therefore can be solved independently of 𝛽 and 𝑤.
At the element level, assuming constant sectional and material properties, K𝛽 can be constructed
directly from the matrices given by (1.66) and (1.70) and K𝑤 can be constructed from (1.66).
For example, for pk = 2 matrix [K𝛽(k) ] is
⎡ EI + 𝜅GA𝓁k − EI + 𝜅GA𝓁k −
𝜅GA𝓁k
⎤√
⎢ 𝓁k 3 𝓁k 6 ⎥
2 6
⎢ ⎥
⎢ ⎥
(k) ⎢ EI 𝜅GA𝓁k 𝜅GA𝓁k ⎥
[K𝛽 ] = ⎢
𝓁k
+
3
− √ ⎥⋅ (7.16)
⎢ 2 6 ⎥
⎢ (sym) ⎥
⎢ ⎥
⎢ 2EI 𝜅GA𝓁k ⎥
⎣ +
𝓁k 5 ⎦
(k)
The terms of the coupling matrix K𝛽𝑤 are:
+1 dNj
kij(𝛽𝑤) = GA Ni d𝜉.
∫−1 d𝜉
Remark 7.3 A different polynomial degree may be assigned to each field function on each ele-
ment. In the following we will assume that all fields have the same polynomial degree pk on an
element but pk may vary from element to element.
(k)
Exercise 7.2 Write down the terms of the element-level matrices K𝑤 and K𝛽𝑤 for pk = 2.
Example 7.1 Consider a beam of constant cross-section with built-in (fixed) support (i.e.,
ux = uy = 0) at x = 0 and simple support (uy = 0) at x = 𝓁. There is an intermediate support
(uy = 0) at x = 𝓁∕2, as shown in Fig. 7.2(a). The beam is uniformly loaded by a constant dis-
tributed load q = −q0 . The goal is to determine the location and magnitude of the maximum
bending moment. In this case the axial displacement u is zero and the potential energy is defined as
follows:
𝓁 𝓁
1
Π= [EI(𝛽 ′ )2 + 𝜅GA(−𝛽 + 𝑤′ )2 ] dx + q 𝑤 dx. (7.17)
2 ∫0 ∫0 0
q0 = 50.0 kN/m
2.50 m 2.50 m
(a)
MD = 24.2 kNm
A B D
C
1.00 m
MA = –22.5 kNm
MB = –33.3 kNm
(b)
Figure 7.2 (a) Problem definition. (b) Bending moment diagram.

7.1 Beams 229
The spaces of admissible functions are defined as follows:

{ 𝓁 }
̃E𝛽 = 𝛽 ||
def
[(𝛽 ) + 𝛽 ] dx ≤ C < ∞, 𝛽(0) = 0
′ 2 2
| ∫0
{ 𝓁 }
̃E𝑤 = 𝑤 ||
def
(𝑤 ) dx ≤ C < ∞, 𝑤(0) = 0, 𝑤(𝓁∕2) = 0, 𝑤(𝓁) = 0 .
′ 2
| ∫0
The problem is to find 𝛽EX and 𝑤EX by minimization of the potential energy:
Π(𝛽EX , 𝑤EX ) = min

̃
Π(𝛽, 𝑤).
𝛽∈E𝛽
𝑤∈Ẽ 𝑤
Assume that the beam is an S200 × 27 American Standard steel beam2 . The section properties
are: A = 3490 mm, I = 24.0 × 106 mm4 . Let 𝓁 = 5.00 m, E = 200 GPa, 𝜈 = 0.3 and q0 = 50.0 kN/m.
The weight of the beam (27 × 9.81 × 10−3 = 0.265 kN/m) is negligible in relation to the applied load.
The solution was obtained using two finite elements and the shear correction factor 𝜅 = 5∕6. At
p ≥ 4 the exact solution is obtained (up to round-off errors). The results are shown in Fig. 7.2(b).
Shear locking in Timoshenko beams

Let us consider a beam of rectangular cross-section of dimension b × d subject to a distributed load
q = d3 f (x) and assume that the axial force is zero. In this case the potential energy is:
𝓁[ ] 𝓁
1 Ebd3 ′ 2
Π= (𝛽 ) + 𝜅Gbd(−𝛽 + 𝑤′ )2 dx − d3 f (x)𝑤 dx. (7.18)
2 ∫0 12 ∫0
On factoring d3 we have:
𝓁[ ] 𝓁
d3 Eb ′ 2 𝜅Gb
Π= (𝛽 ) + 2 (−𝛽 + 𝑤′ )2 dx − d3 f (x)𝑤 dx. (7.19)
2 ∫0 12 d ∫0
For sufficiently small d values the term 𝜅Gb∕d2 is much larger than Eb∕12. Since the minimum of
Π is sought, this forces 𝛽 → 𝑤′ as d → 0:
𝓁
lim (−𝛽 + 𝑤′ )2 dx = 0. (7.20)
d→0 ∫0
The effect of the constraint 𝛽 = 𝑤′ is that the number of degrees of freedom is reduced. Convergence
can be very slow when low polynomial degrees are used. This is called shear locking. As d → 0, the
solution of the Timoshenko model converges to the solution of the Bernoulli-Euler model which is
described in the following section.
7.1.2 The Bernoulli-Euler beam

The Bernoulli-Euler beam model3 is the limiting case of the Timoshenko model, and all higher
order models, with respect to d → 0. Assuming that u = 0 and cs = 0 and letting 𝛽 = 𝑤′ , for homo-
geneous boundary conditions the potential energy expression (7.15) becomes
𝓁 𝓁
1
Π= EI(𝑤′′ )2 dx − q𝑤 dx. (7.21)
2 ∫0 ∫0
2 In this designation S indicates the cross-section; 200 is the nominal depth in mm, and 27 is the mass per unit
length (kg/m).
3 Leonhard Euler 1707–1783, James Bernoulli 1654–1705.
This is the generalized formulation corresponding to the familiar fourth order ordinary differential
equation that can be found in every introductory text on the strength of materials:
(EI𝑤′′ )′′ = q(x) (7.22)
which is the Bernoulli-Euler beam model. To show this, let 𝑤 ∈ E(I) ̃ be the minimizer of Π and
let 𝑣 ∈ E0 (I) be an arbitrary perturbation of 𝑤. For the sake of simplicity let us assume that the
prescribed displacements, rotations, moments and shear forces are zero. Then,
𝓁 𝓁
1
Π(𝑤 + 𝜀𝑣) = EI(𝑤′′ + 𝜀𝑣′′ )2 dx − q(𝑤 + 𝜀𝑣) dx
2 ∫0 ∫0
will be minimum at 𝜀 = 0:
( )
𝜕Π
= 0.
𝜕𝜀 𝜀=0
Therefore
𝓁 𝓁
EI𝑤′′ 𝑣′′ dx − q𝑣 dx = 0.
∫0 ∫0
On integrating by parts twice, we get:
𝓁
( EI𝑤′′ 𝑣′ )𝓁0 − ((EI𝑤′′ )′ 𝑣)𝓁0 + [(EI𝑤′′ )′′ − q]𝑣 dx = 0.
⏟⏟⏟ ⏟⏟⏟ ∫0
M V
This must hold for any choice of 𝑣 that does not perturb the prescribed essential boundary condi-
tions. Since M = EI𝑤′′ and V = (EI𝑤′′ )′ , the boundary terms vanish and we have the strong form
of the Bernoulli-Euler beam model given by eq. (7.22).
Exercise 7.3 Show that eq. (7.22) can be obtained without the assumption that moments or shear
forces prescribed on the boundaries are zero. For the definition of moments and shear forces refer
to eq. (7.14). Hint: If a non-zero moment and/or shear force is prescribed on a boundary then the
expression for the potential energy given by eq. (7.21) must be modified to account for this term.
Exercise 7.4 Derive the strong form of the Bernoulli-Euler beam model for the case cs ≠ 0.
Exercise 7.5 Assume that u = 0 and cs = 0. Show that the strong form of the Timoshenko
model is:
(EI𝛽 ′ )′ + 𝜅GA(−𝛽 + 𝑤′ ) = 0 (7.23)
[𝜅GA(−𝛽 + 𝑤 )] = −q′ ′
(7.24)
and hence show that in Example 7.1 the exact solution is obtained when p ≥ 4. Hint: The procedure
is analogous to that described in Section 7.1.2.
Numerical solution
For the Bernoulli-Euler beam model the energy space is
𝓁
def |
E(I) = {𝑤 | (𝑤′′ )2 dx ≤ C < ∞}, I = {x | 0 < x < 𝓁}. (7.25)
| ∫0
This implies that 𝑤 ∈ E(I) is continuous and its first derivative is also continuous. Functions
that are continuous up to and including their nth derivatives belong in the space Cn (Ω), see
7.1 Beams 231
1.0
1
N1 = (2 + ξ) (1 – ξ)2
4
–1.0 0 1.0 ξ
1.0
ℓk
N2 = (1 + ξ) (1 – ξ)2
8
45˚
–1.0 0 1.0 ξ
1.0
1
N3 = (2 – ξ) (1 + ξ)2
4
–1.0 0 1.0 ξ
1.0
–1.0 0 45˚ 1.0 ξ

ℓk
N4 = – (1 – ξ) (1 + ξ)2
8
Figure 7.3 The first four C 1 shape functions in one dimension.
Section A.2.1. Until now we have considered functions in C0 (Ω) only. Functions that lie in E(I),
defined by eq. (7.25), must also lie in C1 (I).
The first four shape functions defined on the standard beam element (−1 < 𝜉 < +1) are shown
in Fig. 7.3. Note that N2 and N4 are scaled by the factor 𝓁k ∕2. This is because
d𝑤 2 d𝑤
𝜃= = ⋅
dx 𝓁k d𝜉
To define Ni (𝜉), i ≥ 5, it is convenient to introduce the function 𝜓j (𝜉) which is analogous to 𝜙j (𝜉)
given by eq. (3.24):
√
2j − 3 𝜉 s
𝜓j (𝜉) = P (t) dt ds, j = 4, 5, … (7.26)
2 ∫−1 ∫−1 j−2
where Pj−2 (t) is the Legendre polynomial of degree j − 2. For example,
√
1 5 2
𝜓4 (𝜉) = (𝜉 − 1)2 .
8 2
Then, for i ≥ 5;
Ni (𝜉) = 𝜓i−1 (𝜉).
The stiffness matrix for a Bernoulli-Euler beam element of length 𝓁k , constant EI and p = 5 is:
⎡12 6𝓁k −12 6𝓁k 0 0⎤
⎢ 2 ⎥
⎢ 4𝓁 k
−6𝓁 k 2𝓁k2 0 0⎥
EI ⎢ 12 −6𝓁k 0 0⎥
[K] = 3 ⎢ ⋅
0⎥⎥
(7.27)
𝓁k ⎢ (sym.) 4𝓁k2 0
⎢ 8 0⎥
⎢ 8⎥⎦
⎣
Remark 7.4 The simple model given by eq. (7.22) has been used successfully for the solution of
practical problems for almost 200 years. It is remarkably accurate for the computation of deflec-
tions, rotations, moments and shear forces when the solution does not change substantially over
distances of size d. This is not the case, however, for vibrating beams when the mode shapes have
wavelengths close to d.
Timoshenko proposed high frequency vibrations. Of course, the Timoshenko model has limita-
tions as well. Furthermore, the accuracy of a particular model cannot be ascertained unless it can
be shown that the data of interest are substantially independent of the model characterized by the
indices (m, n), the boundary conditions and other modeling decisions. For these reasons there is a
need for a model hierarchy.
Remark 7.5 The Bernoulli-Euler and Timoshenko beam models are used with the objective to
determine reactions, shear force and bending moment diagrams. Having determined the bending
moment M, the normal stress is computed from the formula:
My
𝜎=− (7.28)
I
where y is a centroidal axis, see Fig. 7.1. For the Bernoulli-Euler model this formula is derived from
𝜎 ≡ 𝜎x = −E𝑤′′ y and M = − 𝜎x y dA = E𝑤′′ y2 dA = EI𝑤′′ .

∫A ∫A
For the Timoshenko model the derivation is analogous. Stresses computed in this way can be very
accurate away from the supports and points where concentrated forces are applied but very inaccu-
rate in the neighborhoods of those points where the assumptions incorporated into these models
do not hold. Nevertheless, engineering design is based on the maximum stress computed from the
formula (7.28) subject to the requirement that ||𝜎||max must be less than an allowable value which
is approximately two-thirds of the yield stress. This makes sense only if we understand that the
7.1 Beams 233
A θA C 2EI B
EI
a a
Figure 7.4 Problem definition for Exercise 7.8.
actual goal is to design beams such that ||M||max is much less than the moment that would cause
extensive plastic deformation.
Exercise 7.6 Determine 𝜓5 (𝜉) and verify k66 in eq. (7.27).
Exercise 7.7 Verify the value of k34 in eq. (7.27).
Exercise 7.8 The beam shown Fig. 7.4 is simply supported on the left and fixed on the right. A
positive rotation 𝜃A is imposed on the simply supported end.
1. Using the Bernoulli-Euler beam model, determine the displacement and rotation of the cen-
troidal axis at point C in terms of 𝜃A and a.
2. Given the exact value of the strain energy:
12 EI 2
UEX = 𝜃
11 a A
what is the moment MA that had to be imposed to cause the 𝜃A rotation of the centroidal axis? Is
the moment positive or negative? (Hint: The work done by the moment equals the strain energy.)
3. If the Timoshenko beam model were used, would the exact value of MA be larger or smaller?
Explain.
4. If the number of degrees of freedom is increased by uniform mesh refinement, does the strain
energy increase, decrease or remain the same? Explain.
Exercise 7.9 The multi-span beam shown in Fig. 7.5 is fixed at the ends and simply supported in
three points. The bending stiffness EI is constant.
1. Taking advantage of the symmetry, state the principle of minimum potential energy and specify
the space of admissible functions for the Bernoulli-Euler beam model.
q0
A B C
a b b a
Figure 7.5 Problem definition for Exercise 7.9.

2. Find an expression for the rotation at support B in terms of the parameters q0 , EI, a and b for
the Bernoulli-Euler beam model.
3. Is it possible to assign polynomial degrees to the elements such that the finite element solution
is the exact solution (up to round-off errors)? Explain.
Exercise 7.10 Based on the formulation outlined in Section 1.7, write down the generalized for-
mulation of the undamped elastic vibration problem for the Bernoulli-Euler beam model.
Exercise 7.11 Model the problem in Example 3.3 as a beam.

(a) Using the Bernoulli-Euler beam model, one element on the interval 0 < x < 𝓁∕2, find the rota-
tion at x = 𝓁∕2 in terms of 𝛿∕𝓁. Hint: Write down the constrained stiffness matrix and load
vector for p = 3. Impose antisymmetry condition at x = 𝓁∕2.
(b) Using the result obtained for part (a), compute the bending moment M0 acting on boundary
at x = 0 by extraction. Hint: Select 𝑣 = −N2 (𝜉) for the extraction function. For the definition of
N2 (𝜉) see Fig. 7.3.
7.2 Plates
The formulation of plate models is analogous to the formulation of beam models. The middle sur-
face of the plate is assumed to lie in the x y plane. The two-dimensional domain occupied by the
middle surface is denoted by Ω and the boundary of Ω is denoted by Γ. The thickness of the plate
is denoted by d and the side surface of the plate is denoted by S, that is: S = Γ × (−d∕2, d∕2). The
displacement vector components are written in the following form:
ux = ux|0 (x, y) + ux|1 (x, y)z + · · · + ux|mx (x, y)zmx
uy = uy|0 (x, y) + uy|1 (x, y)z + · · · + uy|my (x, y)zmy (7.29)
n
uz = uz|0 (x, y) + uz|1 (x, y)z + · · · + uz|n (x, y)z
where ux|0 , ux|1 , uy|0 , etc. are independent field functions. In the following we will refer to a particu-
(m ,m ,n)
lar plate model by the indices (mx , my , n) and denote the corresponding exact solution by uEXx y .
The exact solution of the corresponding problem of three-dimensional elasticity will be denoted by
u(3D)
EX
.
In analyses of plates the stress resultants rather than the stresses are of interest. The stress resul-
tants are the membrane forces:
+d∕2 +d∕2 +d∕2
Fx = 𝜎x dz Fy = 𝜎y dz Fxy = Fyx = 𝜏xy dz (7.30)
∫−d∕2 ∫−d∕2 ∫−d∕2
the transverse shear forces:
+d∕2 +d∕2
Qx = − 𝜏xz dz Qy = − 𝜏yz dz (7.31)
∫−d∕2 ∫−d∕2
and the bending and twisting moments:
+d∕2 +d∕2
Mx = − 𝜎x z dz My = − 𝜎y z dz (7.32)
∫−d∕2 ∫−d∕2
+d∕2
Mxy = −Myx = − 𝜏xy z dz. (7.33)
∫−d∕2
7.2 Plates 235
z Fxy F z
x Qx
Fyx
y y
Qy
Fy
Fy + ΔFy
Fyx + ΔFyx Qy + ΔQy

x x
Fxy + ΔFxy Qx + ΔQx q(x, y)
Fx + ΔFx
(a) (b)
z Mx z Mxy
My y y
Myx
Myx + ΔMyx
My + ΔMy
x x
Mx + ΔMx Mxy + ΔMxy
(c) (d)
Figure 7.6 Sign convention for stress resultants.
Mx , My are called bending moments; Mxy is called twisting moment. The negative sign in the expres-
sions for the shear force and bending moment components is made necessary by the conventions
adopted for the stress resultants shown in Fig. 7.6 and the convention that tensile stresses are pos-
itive. Since 𝜏xy = 𝜏yx , the convention adopted for the twisting moments results in Myx = −Mxy .
The starting point for the formulation of plate models is the principle of virtual work or, equiv-
alently, the principle of minimum potential energy written in terms of the field functions with the
integration performed in the z direction. We will consider a restricted form of the three-dimensional
elasticity problems, using constraints and loads typically used in connection with the analysis of
plates.
The boundary conditions are usually given in terms of the normal-tangent (n, t) system. It is left
to the reader in the following exercises to derive the transformations from the (x, y) to the (n, t)
systems.
Exercise 7.12 Refer to Fig. 7.7(a) and show that

+d∕2
Qn = − 𝜏nz dz = Qx cos 𝛼 + Qy sin 𝛼. (7.34)
∫−d∕2
Exercise 7.13 For the infinitesimal plate element shown in Fig. 7.7(b), subjected to bending and
twisting moments only, show that
Mn = Mx cos2 𝛼 + My sin2 𝛼 + 2Mxy sin 𝛼 cos 𝛼 (7.35)
Figure 7.7 Transformation of stress

t y resultants.
z Qx
Mnt n
α
Qy t Mxy
x
y Mn
Mx
α n
x Qn My
Myx
(a) (b)
Mnt = −(Mx − My ) sin 𝛼 cos 𝛼 + Mxy (cos2 𝛼 − sin2 𝛼). (7.36)

Hint: Use Myx = −Mxy .
Exercise 7.14 Derive equations (7.35) and (7.36) from the definitions
+d∕2 +d∕2
Mn = − 𝜎n z dz Mnt = − 𝜏nt z dz
∫−d∕2 ∫−d∕2
using the transformation given by eq. (K.9) and the formulas (7.32), (7.33).
Remark 7.6 The transformation of moments can be represented by a Mohr circle4 . The maxi-
mum and minimum bending moments, denoted respectively by M1 and M2 , called principal bend-
ing moments, occur at those values of 𝛼 at which Mxy = 0. This can be seen by setting the first
derivative of Mn with respect to 𝛼 equal to zero. The principal moments are:
Mx + My Mx + My
M1 = + R, M2 = −R (7.37)
2 2
where R is the radius of the Mohr circle:
√
( )
Mx − M y 2
R= 2
+ Mxy .
2
7.2.1 The Reissner-Mindlin plate

The plate model (1, 1, 0), known as the Reissner-Mindlin plate5 , is widely used in finite element
analysis. Its formulation is analogous to that of the Timoshenko beam. As in the Timoshenko beam
model, the in-plane displacements are decoupled from the bending and shearing deformations.
For simplicity we will be concerned with the bending and shearing deformations only, i.e., we let
ux|0 = uy|0 = 0. We will use the notation
ux|1 = −𝛽x (x, y), uy|1 = −𝛽y (x, y), uz|0 = 𝑤(x, y)
hence the displacement vector components are of the form:
ux = −𝛽x (x, y)z, uy = −𝛽y (x, y)z, uz = 𝑤(x, y) (7.38)
4 Christian Otto Mohr 1835–1918.

5 Eric Reissner 1913–1996, Raymond David Mindlin 1906–1987.
7.2 Plates 237
and the strain terms are:

𝜕𝛽x 𝜕𝛽y 𝜕𝑤
𝜖x = − z 𝜖y = − z 𝜖z = =0
𝜕x 𝜕y 𝜕z
( )
𝜕𝛽x 𝜕𝛽y 𝜕𝑤 𝜕𝑤
𝛾xy = − + z 𝛾yz = −𝛽y + 𝛾zx = −𝛽x + ⋅
𝜕y 𝜕x 𝜕y 𝜕x
For the stress-strain law the plane stress relationships are used in the x y plane (2.69). In the y z
and z x planes the shear modulus is modified by the shear correction factor 𝜅:
E E
𝜎x = (𝜖 + 𝜈𝜖y ) 𝜎y = (𝜈𝜖 + 𝜖y ) 𝜎z = 0
1 − 𝜈2 x 1 − 𝜈2 x
𝜏xy = G𝛾xy 𝜏yz = 𝜅G𝛾yz 𝜏zx = 𝜅G𝛾zx .
Since the strain component 𝜖z and the stress component 𝜎z are both zero, this choice of stress-strain
law may appear to be contradictory. The justification is that with this material stiffness matrix
the exact solution of the Reissner-Mindlin model will approach the exact solution of the fully
three-dimensional model as the thickness approaches zero:
(110)
||u(3D)
EX
⃗ EX ||E
−u
lim (3D)
= 0. (7.39)
d→0
||⃗uEX ||E
The Reissner-Mindlin model would not have this property, called asymptotic consistency, if the
stress-strain law of three-dimensional elasticity were used.
The strain energy is:
1 1
U𝜅 = (𝜎 𝜖 + 𝜎y 𝜖y + 𝜏xy 𝛾xy + 𝜏yz 𝛾yz + 𝜏zx 𝛾zx ) dxdydz + c 𝑤 dxdy
2 ∫V x x 2 ∫Ω s
where cs (x, y) ≥ 0 is a spring coefficient (in N∕m3 units). On substituting the expressions for stress
and strain and integrating with respect to z, we have:
( ) ( ) ( )
1 𝜕𝛽x 2 𝜕𝛽 𝜕𝛽y 𝜕𝛽y 2 1 − 𝜈 𝜕𝛽x 𝜕𝛽y 2
U𝜅 = D[ + 2𝜈 x + + +
2 ∫Ω 𝜕x 𝜕x 𝜕y 𝜕y 2 𝜕y 𝜕x
( )2 ( )
6𝜅(1 − 𝜈) 𝜕𝑤 6𝜅(1 − 𝜈) 𝜕𝑤 2
+ −𝛽y + + −𝛽x + ] dxdy
d2 𝜕y d2 𝜕x
1
+ c 𝑤 dxdy (7.40)
2 ∫Ω s
where D is the flexural rigidity. With the thickness denoted by tz , the flexural rigidity is:
def Etz3
D= ⋅ (7.41)
12(1 − 𝜈 2 )
The potential of external forces is:
P= q𝑤 dxdy − Qn 𝑤 ds + Mn 𝛽n ds + Mnt 𝛽t ds (7.42)

∫Ω ∮Γ ∮Γ ∮Γ
where the n t z coordinate system, shown in Fig. 7.7(a), is used. Noting that (by definition)
un = −𝛽n z, ut = −𝛽t z and applying the rules of vector transformation we have:
𝛽n = 𝛽x cos 𝛼 + 𝛽y sin 𝛼 (7.43)

𝛽t = −𝛽x sin 𝛼 + 𝛽y cos 𝛼. (7.44)
Particular applications of the principle of minimum potential energy depend on the boundary con-
ditions. The commonly used boundary conditions are:
(a) Fixed: 𝛽n = 𝛽t = 𝑤 = 0
(b) Free: Mn = Mnt = Qn = 0
(c) Simple support can be defined in two different ways for the Reissner-Mindlin plate model:
(i) Soft simple support: 𝑤 = 0, Mn = Mnt = 0
(ii) Hard simple support: 𝑤 = 0, 𝛽t = 0, Mn = 0
(d) Symmetry: 𝛽n = 0, Mnt = 0, Qn = 0
(e) Antisymmetry: Same as hard simple support.
Shear correction for plate models

For the Reissner-Mindlin plate model either the energy or the average mid-surface deflection can
be optimized with respect to the fully three-dimensional model by the choice of the shear correction
factor.
⎧ 5
⎪ 6(1 − 𝜈) for optimal energy
𝜅=⎨ 20 (7.45)
⎪ for optimal displacement.
⎩ 3(8 − 3𝜈)
For the model (1, 1, 1) there is one shear correction factor:
⎧5 for 𝜈 = 0
⎪6 ( √ )
𝜅 = ⎨ 12 − 2𝜈 20𝜈 2 (7.46)
⎪ −1 + 1 + for 𝜈 ≠ 0.
⎩ 𝜈 2 (12 − 2𝜈)2
For all other models (mx , my , n) mx , my ≥ 1, n ≥ 2 the shear correction factor is unity [15].
Example 7.2 Consider the domain of a plate shown in Fig. 7.8(a). The thickness is 2.5 mm. The
material is an aluminum alloy, the elastic properties of which are: E = 71.3 GPa and 𝜈 = 0.33. For
the shear correction factor we use the value which is optimal for energy, see eq. (7.45). The plate is
D y C
8.50 N/mm
25
200
x 0
y –8.50 N/mm
A B z
250 x
(a) (b)
Figure 7.8 Example 7.2: (a) Definition of the domain and 28 element mesh. The dimensions are in
millimeters. (b) Contours of Qx for hard simple supports, p = 8, product space.
7.2 Plates 239
Table 7.1 Example 7.2. Estimated range of

the stress resultants Qx and Qy (N/mm) on
the region of primary interest. Mesh: 112
elements.
Boundary condition Qx Qy
Hard simple support ±8.717 ±8.458

Soft simple support ±8.723 ±8.465
simply supported on boundary segments BC and DA and free on segments AB and CD and on the
circular boundary. It is loaded by a uniformly distributed load q = −2.0 kPa.
The goal is to compute the maximum and minimum values of the stress resultants Qx and Qy
using the Reissner-Mindlin plate model with (a) soft simple supports and (b) hard simple supports.
The region of primary interest is the neighborhood of the hole.
This example highlights some of the important questions that arise when using dimensionally
reduced models. In the present case the questions are: how the choice of soft or hard simple support
affects the quantities of interest, which shear correction factor to select, and how to design the finite
element mesh to capture boundary layer effects. Other than knowing that boundary layer effects
may be significant, these questions cannot be answered a priori. They have to be answered on the
basis of feedback information gathered from finite element solutions.
The finite element mesh, shown in Fig. 7.8(a), was designed with the expectation that boundary
layer effects may be significant. The width of the layer of elements at the boundaries was controlled
by a parameter which was selected so as to minimize the potential energy at the highest polynomial
degree, p = 8. This width was found to be 8.5 mm.
We find the quantities of interest over the elements that cover the region of primary interest,
i.e. we exclude from consideration the elements adjacent to the external boundaries. In order to
realize strong convergence in the quantities of interest, we subdivided each of the 28 elements
shown in Fig. 7.8(a) into four elements to obtain a mesh consisting of 112 elements and performed
p-extension using the product space. The results of computation are shown in Table 7.1. The min-
ima and maxima occur along the perimeter of the hole.
If we sought the maximum of Qx and Qy on the entire domain, rather than on the domain of
primary interest, we would have found that the maximum of Qx diverges in the corner points of
the plate as the number of degrees of freedom is increased, whereas the maximum of Qy would have
converged to the same value as shown in Table 7.1. The corner singularity is an artifact of modeling
assumptions. Since the quantities of interest are not influenced by the singularity in a significant
way, it can be neglected. For the hard simple support the same results as shown in Table 7.1 would
have been obtained.
The contours of Qx are plotted for the hard simple supports case in Fig. 7.8(b). The boundary
layer effects are clearly visible, however the maximum value is not located in the boundary layer.
Exercise 7.15 Using the dimensions, elastic properties and boundary conditions for the plate
described in Example 7.2, find the first three natural frequencies and estimate their relative error.
Assume that the density is 𝜌 = 2780 kg∕m3 (2.78 × 10−9 Ns2 ∕mm4 ). For the shear correction factor
use the value which is optimal for energy.
D C
F
β/2
F β
A B
ℓ B
(a) (b)
Figure 7.9 The rhombic plate problem, Exercise 7.17.
Exercise 7.16 Derive the expression (7.42) from eq. (2.97) assuming that only tractions are
applied on the side surface  = Γ × (−d∕2, d∕2) of the plate. Hint: First show that
Ti ui dS = (Tn un + Tt ut + Tz uz ) dS.
∫𝜕ΩT ∫
Exercise 7.17 The mid-surface of a plate is an equilateral parallelogram (rhombus) character-

ized by the dimension 𝓁 and the angle 𝛽, as shown in Fig. 7.9(a). The plate is of uniform thick-
ness d. The elastic properties are: E = 2.0 × 105 MPa, 𝜈 = 0.3. The plate is uniformly loaded, i.e.,
q = q0 (constant) and simple support conditions are prescribed for all sides. Taking advantage
of the two lines of symmetry, define the solution domain to be the triangle ABE. Let 𝛽 = 𝜋∕6
and 𝓁∕d = 100.
This is one of the benchmark problems that have been used for illustrating the performance
of various plate models and discretization schemes. The challenging aspect of the problem is the
strong singularity at the obtuse corners B and D when simple support is prescribed on all bound-
aries.
Using the Reissner-Mindlin plate model, estimate the displacement 𝑤 and the principal bending
moments M1 and M2 in Point E, assuming (a) hard and (b) soft simple support prescribed for all
sides. Report the displacement and moments in dimensionless form: 𝑤D∕(q0 𝓁 4 ) and Mi ∕(q0 𝓁 2 ),
i = 1, 2. Use the shear correction factor for optimal energy.
7.2.2 The Kirchhoff plate

The Kirchhoff plate model6 is analogous to the Bernoulli-Euler beam model. The formulation is
obtained by letting
𝜕𝑤 𝜕𝑤
𝛽x = , 𝛽y =
𝜕x 𝜕y
6 Gustav Robert Kirchhoff 1824–1887.

7.2 Plates 241
in eq. (7.40). Therefore the strain energy of the Kirchhoff plate model is:
[( )2 ( 2 )2 ( 2 )2 ]
1 𝜕2 𝑤 𝜕2 𝑤 𝜕2 𝑤 𝜕 𝑤 𝜕 𝑤
UK = D + 2𝜈 2 + + 2(1 − 𝜈) dxdy
2 ∫Ω 𝜕x2 𝜕x 𝜕y2 𝜕y2 𝜕x𝜕y
1
+ c 𝑤 dxdy. (7.47)
2 ∫Ω s
The potential of external forces is obtained from (7.42) with the following modification:
𝜕𝑤 𝜕Mnt 𝜕(Mnt 𝑤)
Mnt 𝛽t ds → Mnt ds = − 𝑤 ds + ds
∮Γ ∮Γ 𝜕s ∮Γ 𝜕s ∮Γ 𝜕s
⏟⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏟
0
where we used dt ≡ ds and the assumption that the product Mnt 𝑤 is continuous and differentiable.
Therefore we have:
( )
𝜕𝑤 𝜕Mnt
PK = q𝑤 dxdy + Mn ds − Qn − 𝑤 ds. (7.48)
∫Ω ∮Γ 𝜕n ∮Γ 𝜕s
In the Kirchhoff model simple support has only one interpretation; that of hard simple support. By
definition:
Π(𝑤) = UK (𝑤) − PK (𝑤) (7.49)
and
E(Ω) = {𝑤 | UK (𝑤) ≤ C < ∞}.
̃
Define E(Ω) ⊂ E(Ω) to be the space of functions that satisfy the prescribed kinematic boundary
conditions. The problem is to find:
Π(𝑤EX ) = min Π(𝑤). (7.50)
̃
𝑤∈E(Ω)
This formulation has great theoretical and historical significance (see, for example, [104]). How-
ever, it is not well suited for computer implementation because the basis functions have to be C1
continuous. The difficulties associated with enforcement of C1 continuity are discussed in the fol-
lowing section.
Exercise 7.18 Consider the Kirchhoff plate model and assume homogeneous boundary condi-
tions. Following the procedure of Section 7.1.2, that is, letting
𝜕Π(𝑤 + 𝜖𝑣) |
| =0
𝜕𝜖 |𝜖=0
where Π(𝑤) is defined by eq. (7.49) and 𝑣 is an arbitrary test function in E0 (Ω), show that the
function 𝑤 that minimizes Π satisfies the biharmonic equation:
𝜕4 𝑤 𝜕4 𝑤 𝜕4 𝑤 q
+ 2 + = ⋅ (7.51)
𝜕x4 𝜕x2 𝜕y2 𝜕y4 D
Exercise 7.19 We have seen in Exercise 7.18 that the strong form of Kirchhoff plate model is
the biharmonic equation (7.51). Assume that all sides of the equilateral triangular plate and the
rhombic plate shown in Fig. 7.19 are simply supported. Considering only the symmetric part of the
asymptotic expansion, characterize the smoothness of the homogeneous solution of the Kirchhoff
plate model at each of the singular points. For the rhombic plate let 𝛽 = 𝜋∕6.
D C
β
A B A
ℓ ℓ B
Figure 7.10 Exercise 7.19. Equilateral triangle and rhombic plate. Notation
Hints: (a) The homogeneous biharmonic equation in polar coordinates is given by eq. (G.1).
(b) Letting 𝑤 = r 𝜆+1 F(𝜃), the symmetric eigenfunctions are of the form:
Fi (𝜃) = ai cos(𝜆i − 1)𝜃 + bi cos(𝜆i + 1)𝜃
see eq. (G.13). (c) The zero moment condition on the simply supported edges (Mn = 0) is
equivalent to
( )
1 𝜕2 𝑤
=0
r 2 𝜕𝜃 2 𝜃=±𝛼∕2
where 𝜃 = 0 is the internal bisector of the vertex angle 𝛼 as defined in Fig. 4.1.
Exercise 7.20 The exact solution of the Kirchhoff model for a simply supported and uniformly
loaded equilateral triangular plate is a polynomial of degree 5 (see, for example, [104]). In the center
of the plate the exact values of the displacement 𝑤 and the bending moments are:
𝑤D 1 Mx 1+𝜈
= , = , My = Mx , Mxy = 0
q0 𝓁 4 1728 q0 𝓁 2 72
where D is the plate constant, q0 is the value of the constant distributed load and 𝓁 is the
dimension shown in Fig. 7.19. Compare these results with their counterparts obtained with the
Reissner-Mindlin model for (a) 𝓁∕d = 100, 𝜈 = 0.3 and (b) 𝓁∕d = 10, 𝜈 = 0.3 for hard and soft
simple supports. Investigate the effects of shear correction factors for optimal energy and optimal
displacement. Refer to Section 7.2.1.
Exercise 7.21 Consider a square plate, uniformly loaded, with one side fixed, the opposite side
simply supported, the other two sides free. State the principle of minimum potential energy for this
problem for the Kirchhoff and Reissner-Mindlin models.
Exercise 7.22 Refer to Exercise 7.21. Noting that the Reissner-Mindlin model allows distinction
to be made between hard and soft simple supports whereas the Kirchhoff model does not, estimate
the error of idealization of the Kirchhoff model in relation to the Reissner-Mindlin model through
numerical experiments. Fix the size of the plate and vary the thickness. Compare the estimated
limit values of the strain energy for soft and hard simple supports.
7.2 Plates 243
Enforcement of C𝟏 continuity
The enforcement of C1 continuity in two dimensions requires special consideration. This is because
it is not possible to enforce C1 continuity, and not more than C1 continuity, on polynomial basis
functions. The continuity of the first and higher derivatives is enforced along the sides and at the
vertices the normal derivatives must be continuous also. This means that the first derivatives in four
different directions would have to be continuous. This is not possible unless the second deriva-
tives are also continuous in the vertex points. That, however, causes problems at singular points
where the second derivatives are discontinuous. To overcome this difficulty, composite elements
and elements with rational basis functions have been developed.
It is best to avoid this problem by using plate and shell models that require enforcement of C0
continuity only.
7.2.3 The transverse variation of displacements

In the formulation of plate and shell models the transverse variation of the displacement compo-
nents was represented by polynomials. In this section it is shown that for homogeneous plates of
small thickness the transverse variation of the displacements is indeed best represented by polyno-
mials and for laminated plates by piecewise polynomials.
Consider the infinite strip of unit width shown in Fig. 7.11. Assume that the loading function
q = q(x) is periodic with period L, antisymmetric with respect to the plane y = 0, and satisfies the
equations of equilibrium:
∞ ∞
q dx = xq dx = 0. (7.52)
∫−∞ ∫−∞
Note that periodic loading can be generalized to nonperiodic loading using the Fourier integral
method. We will be interested in the limit process d∕L → 0.
In plate models the transverse load q is referred to the mid-surface. We understand q to be an
antisymmetric distributed load acting on the surfaces Γ+ and Γ− . Let 𝛽 = 2𝜋∕L and assume that
the solution is of the form:
ux = 𝜙(𝛽, y) sin 𝛽x (7.53)
uy = 𝜓(𝛽, y) cos 𝛽x (7.54)
q/2
y
Γ+
d/2
Ω x
d/2
Γ–
L
q/2
Figure 7.11 Infinite strip. Notation.

where 𝜙(𝛽, y) is antisymmetric and 𝜓(𝛽, y) is symmetric with respect to the mid-surface of the strip
(the x-axis). The strain components are:
𝜕ux
𝜖x = = 𝛽𝜙 cos 𝛽x (7.55)
𝜕x
𝜕uy
𝜖y = = 𝜓 ′ cos 𝛽x (7.56)
𝜕y
𝜕u 𝜕uy
𝛾xy = x + = (𝜙′ − 𝛽𝜓) sin 𝛽x (7.57)
𝜕y 𝜕x
where the primes represent differentiation with respect to y. We assume that the material
is orthotropic with the material axes aligned with the x, y and z directions. Therefore the
two-dimensional stress-strain relationship can be written in the form
⎧ 𝜎x ⎫ ⎡E E 0 ⎤ ⎧ 𝜖x ⎫
⎪ ⎪ ⎢ 1 2 ⎪ ⎪
⎨ 𝜎y ⎬ = ⎢E2 E3 0 ⎥⎥ ⎨ 𝜖y ⎬ (7.58)
⎪ ⎪ ⎣ 0 0 E6 ⎦ ⎪ ⎪
⎩𝜏 ⎭
xy ⎩𝛾 ⎭ xy
where E1 , E2 , E3 and E6 may be functions of y. Therefore the stress components are:
𝜎x = (E1 𝛽𝜙 + E2 𝜓 ′ ) cos 𝛽x (7.59)

𝜎y = (E2 𝛽𝜙 + E3 𝜓 ) cos 𝛽x ′
(7.60)
𝜏xy = E6 (𝜙 − 𝛽𝜓) sin 𝛽x.
′
(7.61)
The equilibrium equations with zero body force are

𝜕𝜎x 𝜕𝜏xy
+ =0 (7.62)
𝜕x 𝜕y
𝜕𝜏xy 𝜕𝜎y
+ = 0. (7.63)
𝜕x 𝜕y
On substituting equations (7.59) to (7.61) into the equilibrium equations we get
−E1 𝛽 2 𝜙 − E2 𝛽𝜓 ′ + (E6 𝜙′ )′ − (E6 𝛽𝜓)′ = 0 (7.64)

E6 𝛽𝜙 − E6 𝛽 𝜓 + (E2 𝛽𝜙) + (E3 𝜓 ) = 0.
′ 2 ′ ′ ′
(7.65)
Expanding 𝜙(𝛽, y) and 𝜓(𝛽, y) into a power series with respect to 𝛽:
𝜙(𝛽, y) = 𝜙0 (y) + 𝛽𝜙1 (y) + 𝛽 2 𝜙2 (y) + · · · (7.66)

𝜓(𝛽, y) = 𝜓0 (y) + 𝛽𝜓1 (y) + 𝛽 𝜓2 (y) + · · · 2
(7.67)
and substituting equations (7.66) and (7.66) into equations (7.64) and (7.65) we get:
− E1 (𝛽 2 𝜙0 + 𝛽 3 𝜙1 + 𝛽 4 𝜙2 + · · · ) − E2 (𝛽𝜓0′ + 𝛽 2 𝜓1′ + 𝛽 3 𝜓2′ + · · · )

+ [E6 (𝜙′0 + 𝛽𝜙′1 + 𝛽 2 𝜙′2 + · · · )]′ − [E6 (𝛽𝜓0 + 𝛽 2 𝜓1 + 𝛽 3 𝜓2 + · · · )]′ = 0 (7.68)
E6 (𝛽𝜙′0 + 𝛽 2 𝜙′1 + 𝛽 3 𝜙′2 + · · · ) − E6 (𝛽 𝜓0 + 𝛽 𝜓1 + 𝛽 𝜓2 + · · · )
2 3 4
+ [E2 (𝛽𝜙0 + 𝛽 2 𝜙1 + 𝛽 3 𝜙2 + · · · )]′ + [E3 (𝜓0′ + 𝛽𝜓1′ + 𝛽 2 𝜓2′ + · · · )]′ = 0. (7.69)
Note that these equations hold for any 𝛽.

7.2 Plates 245
Case A: The material properties are independent of y

Letting 𝛽 = 0 and assuming that the material properties are independent of y, from equations (7.68),
(7.69) we have
𝜙′′0 = 0, 𝜓0′′ = 0. (7.70)
Therefore:
𝜙0 (y) = a1 y + a2 , 𝜓0 (y) = b1 y + b0 . (7.71)
Since by assumption 𝜙0 (y) is antisymmetric and 𝜓0 (y) is symmetric:
𝜙0 (y) = a1 y, 𝜓0 (y) = b0 . (7.72)
Differentiating equations (7.68), (7.68) with respect to 𝛽, setting 𝛽 = 0 and using (7.72) the following
equations are obtained:
E2 + E6
𝜙′′1 = 0, 𝜓1′′ = − a1 . (7.73)
E1
Solving for 𝜙1 (y) and dropping the symmetric term; solving for 𝜓1 (y) and dropping the antisym-
metric term, we have:
E2 + E6
𝜙1 = c1 y, 𝜓1 (y) = − a1 y2 + d0 . (7.74)
2E1
Equations (7.72) indicate that the solution should be in the form:
ux (x, y) = ux|1 (x)y (7.75)

2
uy (x, y) = uy|0 (x) + uy|2 (x)y (7.76)
which is the assumed functional form in equations (7.1) and (7.2) modified by the additional
assumptions that ux (x, y) is antisymmetric and uy (x, y) is symmetric and truncating the expressions
at the quadratic term. This choice of the mode of deformation satisfies the equilibrium equations
(7.68) and (7.69) up to the first power of 𝛽. By continuing this process, the equilibrium equations
can be satisfied to an arbitrary power of 𝛽.
Case B: The material properties are symmetric functions of y

If the material properties are not independent of y then setting 𝛽 = 0 in equations (7.68) and (7.69)
we have:
(E6 𝜙′0 )′ = 0, (E1 𝜓0′ )′ = 0. (7.77)
In many practical problems the material properties are symmetric functions of y. For example, the
strip may be made of laminae which are symmetrically arranged with respect to the xz plane, then,
knowing that 𝜙0 (y) is antisymmetric and 𝜓0 (y) is symmetric, we have:
𝜙0 (y) = a1 F0 (y), 𝜓0 = b0 (7.78)
where
y
def 1
F0 (y) = dt. (7.79)
∫0 E6 (t)
Differentiating equations (7.68) and (7.69) with respect to 𝛽 and letting 𝛽 = 0 the following
equations are obtained:
−E2 𝜓0′ + (E6 𝜙′1 )′ − (E6 𝜓0 )′ = 0 (7.80)

E6 𝜙′0 + (E2 𝜙0 ) + ′
(E1 𝜓1′ )′ = 0. (7.81)
Using (7.78), from (7.80) we get
[E6 (𝜙′1 − 𝜓0 )]′ = 0. (7.82)
Solving for 𝜙1 (y) and using the fact that 𝜙1 (y) is antisymmetric:
y
1
𝜙1 (y) = c1 dt + b0 y. (7.83)
∫0 E6 (t)
From equation (7.77) we have E6 𝜙′0 = a1 (constant) therefore equation (7.81) can be written as
a1 + (E2 𝜙0 )′ + (E1 𝜓1′ )′ = 0. (7.84)
Solving for 𝜓1 :
y y y
t 1 E2 (t)
𝜓1 (y) = −a1 dt + d1 dt − a1 F (t) dt + d0 (7.85)
∫0 E1 (t) ∫0 E1 (t) ∫0 E1 (t) 0
since the second term is antisymmetric, d1 = 0. Defining:
y y
t E2 (t)
F1 (y) = dt + F (t) dt (7.86)
∫0 E1 (t) ∫0 E1 (t) 0
we have:
𝜓1 (y) = −a1 F1 (y) + d0 (7.87)
and the solution is of the form:
ux (x, y) = ux|1 (x)y + ux|2 (x)F0 (y) (7.88)

uy (x, y) = uy|0 (x) + uy|2 (x)F1 (y). (7.89)
For additional details and examples we refer to [23].
Remark 7.7 The definition, essential properties and formulation of hierarchic models for lami-
nated plates and shells were first addressed in [1], [23], [91]. If the goal of computation is to estimate
the strength of laminated structural members then it is necessary to identify critical regions in the
macromechanical model and estimate the values of predictors of failure defined at the fiber-matrix
level. Predictors of failure are functionals that have been correlated with failure events through
interpretation of the outcome of physical experiments. Owing to the complexity of the problem,
large aspect ratios, strong boundary layer effects and the requirement of a posteriori error estima-
tion, high order methods, coupled with proper mesh design, are expected to play an increasingly
important role in this field.
Exercise 7.23 Assuming that the material properties are symmetric functions of y, construct ux
and uy so as to satisfy the equilibrium equations (7.68) and (7.69) up to the second power of 𝛽.
7.3 Shells 247
7.3 Shells
The formulation of mathematical models for structural shells is a very large and rather complicated
subject. Only a brief overview of some of the salient points is presented in the following.
A structural shell is characterized by a surface, called mid-surface xi , and the thickness d. Both
are given in terms of two parameters 𝛼1 , 𝛼2 :
xi = xi (𝛼1 , 𝛼2 ), d = d(𝛼1 , 𝛼2 ).
Note that the indices of the parameters 𝛼i take on the values i = 1, 2 whereas the indices of the
spatial coordinates xi range from 1 to 3. Associated with each point of the mid-surface are three
basis vectors. Two of the basis vectors lie in the tangent plane at the point (𝛼1 , 𝛼2 ):
𝜕xi 𝜕xi
b(1) = , b(2) = ⋅
i 𝜕𝛼1 i 𝜕𝛼2
Note that b(1)
i
and b(2)
i
are not necessarily orthogonal. However, in classical treatments of shells, the
parameters 𝛼1 , 𝛼2 are usually chosen so that the basis vectors are orthogonal. The third basis vector
b(3)
i
is the cross product of b(1)
i
and b(2)
i
, therefore it is normal to the tangent plane. These are called
curvilinear basis vectors. The normalized curvilinear basis vectors are denoted by e𝛼 , e𝛽 , en . The
Cartesian unit basis vectors are denoted by ex , ey , ez . A vector u given in terms of the curvilinear
basis vectors is denoted by u(𝛼) , in Cartesian coordinates by u(x) . The transformation is
u(x) = [R]u(𝛼) (7.90)
where the columns of the transformation matrix [R] are the unit vectors e𝛼 , e𝛽 , en .
The classical development of shell models was strongly influenced by the limitations of the meth-
ods available for solving the resulting partial differential equations. The use of curvilinear coor-
dinates allowed the treatment of shells with simple geometric description, such as cylindrical,
spherical and conical shells, by classical methods subject to the condition that the thickness of
the shell is small in relation to its other dimensions. Those limitations no longer exist.
Shells should be viewed as fully three-dimensional solids that may allow a priori restrictions
on the transverse variation of displacements in certain regions, usually not the regions of primary
interest. In virtually all practical applications there are regions where the assumptions incorporated
into shell models do not hold. Examples are the neighborhoods of nozzles, support attachments,
stiffeners, cut-outs and joints where the curvature abruptly changes. From the point of view of
strength analysis those are the usual regions of primary interest. The thickness of shells of engi-
neering interest is rarely very small in relation to the other dimensions.
The hierarchic shell models are generalizations of the hierarchic plate models represented by
eq. (7.29). The displacement vector components are given in the following form:
∑
m𝛼
u𝛼 = u𝛼|i (𝛼, 𝛽)𝜙i (𝜈)
i=0
m𝛽
∑
u𝛽 = u𝛽|i (𝛼, 𝛽)𝜙i (𝜈) (7.91)
i=0
∑ mn
un = un|i (𝛼, 𝛽)𝜙i (𝜈)
i=0
where 𝜙i (𝜈) are called director functions. When the material is isotropic then 𝜙i (𝜈) are polyno-
mials; when the shell is laminated then 𝜙i (𝜈) are piecewise polynomials (see, for example, [1]).
Eq. (7.91) represents a semi-discretization of the problem of fully three-dimensional elasticity, in

the sense that 𝜙i (𝜈) are fixed and thus the problem is reduced from a three-dimensional prob-
lem to a two-dimensional one. A particular shell model is characterized by the set of numbers
(m𝛼 , m𝛽 , mn ).
In the classical treatment of shells the curvilinear basis vectors are retained throughout the anal-
ysis. In the finite element method a generalized formulation is used, most commonly the principle
of virtual work. The algorithmic structure becomes simpler if the displacement components, given
with reference to the curvilinear basis vectors, are transformed to the (global) Cartesian reference
frame by eq. (7.90).
The generic form of the virtual work of internal stresses is given by
+d∕2 ( )T
B(u(𝛼) , v(𝛼) ) = ̃
[D][R]v ̃
(𝛼) [E][D][R]u(𝛼) d𝜈 d𝜔 (7.92)
∫𝜔 ∫−d∕2
where [D]̃ is the differential operator that transforms the displacement vector components given
in terms of the curvilinear coordinates (𝛼, 𝛽, 𝜈) to the Cartesian strain tensor components:
̃ (𝛼) . It is assumed that the material is isotropic. If the material is not isotropic, and its
{𝜖} = [D]u
reference frame differs from the global Cartesian coordinate system, then [E] must be transformed
into the global Cartesian coordinate system.
The generic form of the virtual work of external forces is:
+d∕2 ( )T
F(v(𝛼) ) = [R]F(𝛼) [R]v(𝛼) d𝜈 d𝜔
∫𝜔 ∫−d∕2
+d∕2 ( )T
+ [R]T(𝛼) [R]v(𝛼) d𝜈 ds
∫𝜕𝜔 ∫−d∕2
+d∕2 ( )T
+ ̃
[D][R]v (𝛼) [E]{c}Δ d𝜈 d𝜔 (7.93)
∫𝜔 ∫−d∕2
where F(𝛼) (resp. T(𝛼) ) is the volume force (resp. surface traction vector) given in terms of the curvi-
linear basis, {c} is the vector of coefficients of thermal expansion and Δ is the temperature change.
The Naghdi shell model

The Naghdi shell model7 [61] is analogous to the Reissner-Mindlin plate model: It is assumed that
normals to the mid-surface prior to deformation remain straight lines but not necessarily normals
after deformation. In other words, the kinematic assumptions account for some transverse shearing
deformation. Specifically, the kinematic assumptions are the same as the kinematic assumptions
for the hierarchic shell model (1,1,0):
u𝛼 = u𝛼|0 (𝛼, 𝛽) + u𝛼|1 (𝛼, 𝛽)𝜈

u𝛽 = u𝛽|0 (𝛼, 𝛽) + u𝛽|1 (𝛼, 𝛽)𝜈 (7.94)
un = un|0 (𝛼, 𝛽).
However, the definition of the material stiffness matrix is not the same as for the hierarchic shell
model: Whereas in eq. (7.92) the definition of [E] is that of the three-dimensional stress-strain rela-
tionship, in the Naghdi model [E] is replaced by the stress-strain relationship which incorporates
the assumption that plane stress conditions exist. This is necessary for asymptotic consistency.
7 Paul M. Naghdi 1924–1994.

7.3 Shells 249
The Novozhilov-Koiter shell model

The Novozhilov-Koiter shell model8 [68] is an extension of the Euler-Bernoulli beam model
and the Kirchhoff plate model to shells: It is assumed that normals to the mid-surface prior to
deformation remain normals after deformation. The formulation involves replacement of the
field functions u𝛼 and u𝛽 in Eqn. (7.94) by a linear combination of the first derivatives of un .
Consequently the number of field functions is reduced to three and the second derivatives appear
in Eqn. (7.92), which implies that the space of admissible functions has to have C1 continuity.
From the point of view of computer implementation, the advantages of using only three field
functions are far outweighed by disadvantages of the requirement of C1 continuity and restrictions
on the kinematic boundary conditions. This model is mainly of theoretical and historical interest
today.
7.3.1 Hierarchic thin solid models

An alternative approach to hierarchic semi-discretization is to write the Cartesian components of
the displacement vector in the form
∑
q
ux = ux|i (𝛼, 𝛽)𝜙i (𝜈)
i=0
∑q
uy = uy|i (𝛼, 𝛽)𝜙i (𝜈) (7.95)
i=0
∑q
uz = uz|i (𝛼, 𝛽)𝜙i (𝜈)
i=0
where q is typically a small number. The generic form of the virtual work of internal stresses is
given by the expression
+d∕2 ( )T
B(u(x) , v(x) ) = [D]v(x) [E][D]u(x) d𝜈 d𝜔. (7.96)
∫𝜔 ∫−d∕2
The generic form of the virtual work of external forces is:

+d∕2
F(v(𝛼) ) = F(x) ⋅ v(x) d𝜈 d𝜔 + T(x) ⋅ v(𝛼) dS
∫𝜔 ∫−d∕2 ∫𝜕Ω
+d∕2 ( )T
+ [D]v(x) [E] c𝜏 d𝜈 d𝜔 (7.97)
∫𝜔 ∫−d∕2
where F(x) (resp. T(x) ) is the Cartesian volume force vector (resp. surface traction vector) given in
terms of the curvilinear variables (𝛼, 𝛽, 𝜈) and c𝜏 is the thermal strain. Such models are called
“thin solid” models.
In computer applications anisotropic trunk or anisotropic product spaces are used. The
anisotropic trunk space on the standard hexahedral element Ω(h) st is defined by:
ppq k 𝓁 m
Str (Ω(h) p m p m (h)
st ) = span(𝜉 𝜂 𝜁 , 𝜉 𝜂𝜁 , 𝜉𝜂 𝜁 , (𝜉, 𝜂, 𝜁) ∈ Ωst ,
k, 𝓁 = 0, 1, 2, … , k + 𝓁 ≤ p, m = 0, 1, 2 … , q), p > 1. (7.98)
8 Valentin Valentinovich Novozhilov 1910–1987, Warner Tjardus Koiter 1914–1997.

The parameter q is fixed, and p > q is increased until convergence is realized. The anisotropic prod-
uct space on Ω(h)
st is defined by:
ppq k 𝓁 m
Spr (Ω(h) (h)
st ) = span(𝜉 𝜂 𝜁 , (𝜉, 𝜂, 𝜁) ∈ Ωst ,
k, 𝓁 = 0, 1, 2, … , p, m = 0, 1, 2 … , q). (7.99)
The definition of anisotropic spaces on the standard pentahedral element is analogous. The
anisotropic spaces with q = 1 are similar but not equivalent to the Reissner-Mindlin plate model or
the Naghdi shell model. The differences are in the number of field functions and the constitutive
laws that have been adjusted for the Reissner-Mindlin plate and the Naghdi shell models to satisfy
the requirement of asymptotic consistency.
Remark 7.8 The advantages of thin solid formulations over shell formulations are that they are
easier to implement and continuity with other bodies, such as stiffeners, are easier to enforce. The
disadvantages are that thin solid formulations cannot be applied to laminated shells unless each
lamina is explicitly modeled and the number of field functions must be the same for each displace-
ment component.
Example 7.3 In this example we compare the solution of the plate problem described in
Example 7.2 with the solution of a thin solid model, q = 1. Our quantity of interest is the
maximum compressive stress on the top of the plate (z = 1.25 mm) in the neighborhood of the
circular hole. We will consider hard simple supports only. We will use the shear correction factor
which is optimal for energy for the Reissner-Mindlin plate and unity for the thin solid model. For
discretization we will use the 28-element mesh shown in Fig. 7.8, p-extension and product space.
Recall that the Reissner-Mindlin model satisfies the condition of asymptotic consistency but it is
not in the hierarchic sequence of thin solid plates, whereas the thin solid plate with q = 1 is in the
hierarchic sequence but does not satisfy the condition of asymptotic consistency.
The contours of the third principal stress (MPa) are shown in Fig. 7.12. The relative error
in the computed stress values were verified to be well under 1%. Therefore the differences in
the results are caused by the differences in the models. In terms of the maximum compres-
sive stress, that difference is 3%. For the Reissner-Mindlin model (resp. the top surface of the
thin solid model, i.e. z = 1.25 mm) the range of the third principal stress is −26.61 < 𝜎1 < 0
0
–6.0
–12.0
–18.0
–24.0
(a) (b)
Figure 7.12 Example 7.3. Contours of the third principal stress (MPa), p = 8 product space.
(a) Reissner-Mindlin model. (b) Top surface of the thin solid model, q = 1.
7.3 Shells 251
Table 7.2 Example 7.4. The first

non-zero natural frequency (Hz)
corresponding to the thin solid models
characterized by q = 1 and q = 3, the
fully 3-dimensional model (3D) and
the Reissner-Mindlin (R-M) plate
model. Product spaces.
p q=1 q=3 3D R-M
5 97.39 90.46 90.46 90.95

6 97.34 90.41 90.41 90.94
7 97.30 90.37 90.37 90.94
8 97.26 90.34 90.33 90.94
∞ 97.12 90.13 90.07 90.94
(resp. −27.40 < 𝜎3 < 4.14). For the thin solid model characterized by q = 3, that range is the same
as for the Reissner-Mindlin model.
The solution of the thin solid model was found to be insensitive to the width of the elements
along the boundaries.
Exercise 7.24 Estimate the maximum bending moment in a plate, modeled as a thin solid, from
the information that (𝜎3 )min = −27.40 MPa. Assume that the membrane forces are zero.
Example 7.4 Using the dimensions, elastic properties and boundary conditions for the plate
described in Example 7.2, let us find the first natural frequency (Hz) predicted by the thin solid
models characterized by q = 1 and q = 3 and compare the results with those computed using the
fully three-dimensional model and the Reissner-Mindlin plate model. We will assume that the den-
sity is the same as in Exercise 7.15: 𝜌 = 2780 kg∕m3 (2.78 × 10−9 Ns2 ∕mm4 ) and use the 28 element
mesh shown in fig. 7.12 and product space.
The results of computation for p ranging from 5 to 8 are listed in Table 7.2. The last row shows the
estimated limit values obtained by extrapolation to p = ∞ by means of the algorithm described in
Section 1.5.3. The computed natural frequencies are monotonically decreasing for each row in the
first three columns. This is because the anisotropic space characterized by q = 1 is a subspace of
the anisotropic space characterized by q = 3 which is a subspace of the isotropic finite element
space. The Reissner-Mindlin (R-M) model is not a member of the hierarchic sequence of thin
solid models because the stress-strain relationship is not consistent with the other models. How-
ever, it has the property of asymptotic consistency. The results indicate that the Reissner-Mindlin
model and the thin solid model with q = 3 are very good approximations to the solution of the fully
three-dimensional model.
Observe that the errors of approximation are negligibly small. Therefore the differences in the
results are caused mainly by differences in modeling assumptions.
Exercise 7.25 The second eigenvalue of the thin solid models and the 3D model in Example 7.4
is comparable to the first eigenvalue of the Reissner-Mindlin model. Explain why.
Table 7.3 Example 7.5. Natural frequencies (Hz) for mode 20

computed using the anisotropic product space prpp3 (Ωst(h) ) and 128
elements.
p N t = 0.01 t = 0.001 t = 0.0001 t = 0.00001
3 13,248 5019.0 1590.7 955.39 942.69

4 23,808 5002.5 1452.3 431.13 198.92
5 37,440 5002.1 1452.1 391.61 133.92
6 54,144 5002.0 1452.0 391.56 110.91
7 73,920 5002.0 1452.0 391.53 103.60
8 96,768 5002.0 1451.9 391.51 101.63
∞ ∞ 5002.0 1451.9 391.42 100.58
Exercise 7.26 If soft simple supports were used in the problem described in Example 7.4, how
many zero eigenvalues would have been found for the thin solid models and the Reissner-Mindlin
model?
Example 7.5 The following example was taken from an ADINA9 technical brief10 in which the
natural frequencies and mode shapes of cylindrical shells are tabulated for various ratios of thick-
ness to radius. The radius of the shell is 0.1 m, the length is 0.4 m, the ends are fixed. The material
properties are E = 2.0 × 1011 Pa, 𝜈 = 0.3, 𝜚 = 7800 kg∕m3 . The same problem was discussed at the
ASME Verification and Validation Symposium in 2012 by T. Yamada et al.11 who presented solu-
tions obtained with MITC4 shell elements [30] on coarse (40 × 20), moderate (80 × 40 and fine
(240 × 120) uniform meshes.
The convergence of the natural frequencies for the 20th mode, computed using the anisotropic
pp3
product space pr (Ω(h)st ) and 128 elements, are shown in Table 7.3. The elements were mapped
using the mean optimal set T25 (6 × 6 collocation points), see Table F.1 in Appendix F. The mode
shapes for t = 0.01 and t = 0.00001 are shown in Fig. 7.13. The contours are proportional to the
normal displacement.
Note that the rate of convergence of the natural frequencies is decreasing as the thickness is
decreased. This is related to the decreasing regularity of the mode shapes, visible in Fig. 7.13.
Exercise 7.27 The mid-surface of a hyperboloidal shell is given by

x2 y2 z2 R2t
2
+ 2 − = 1, −L ≤ z ≤ L, 𝛼2 =
Rt Rt (𝛼L)2 R2c − R2t
where Rt is the throat radius and Rc is the crown radius. Let Rt = 1.0 m, L = 1.0 m and 𝛼 = 1. Denote
the thickness of the shell by d. Assume that the material is elastic with E = 2.0 × 105 MPa, 𝜈 = 0.3.
Assume that a normal pressure p = p0 cos 2𝜃 is acting on the inside surface of the shell where 𝜃 is
9 ADINA is a trademark of ADINA R&D, Inc., Watertown, Massachusetts, USA

10 http://www.adina.com/newsgH53.shtml
11 Verification of shell elements by eigenanalysis of vibration problems.
7.3 Shells 253
(a) (b)
Figure 7.13 Example 7.5. The 20th eigenfunctions for (a) t = 0.01 m and (b) t = 0.00001 m.
the angle measured from the positive x axis as shown in Fig. 7.14. Let p0 = 20.0 kPa. The edge at
z = −L is fixed, i.e., all displacement components are zero, the edge at z = L is free.
1. Construct a solid model of the shell with d defined as a parameter and construct a mesh similar
to that shown in Fig. 7.14.
ppq ppq
2. Using the anisotropic product spaces pr (Ω(h) (h)
st ) and the anisotropic trunk spaces tr (Ωst ),
investigate the rates of convergence in energy norm for d = 0.01 m and d = 0.001 m for
q = 1, 2, 3.
This exercise demonstrates the effects of locking: The relative error is substantially larger for
d = 0.001 m than for d = 0.01 m; however, the estimated asymptotic rates of convergence are close
to 1.0. The strong boundary layer at the fixed edge will be clearly visible when the deformed shape
is plotted.
Figure 7.14 Hyperboloidal shell.
y
θ
x
7.4 Chapter summary
In many cases it is advantageous to use dimensionally reduced models rather than a fully
three-dimensional model. Whether a dimensionally reduced model should be used in a particular
case depends on the goals of computation and the required accuracy. Generally speaking, dimen-
sionally reduced models are well suited for structural analysis when the goals of computation
are to determine structural stiffness, displacements, stress resultants, buckling strength natural
frequencies but are not well suited for strength analysis when the region of primary interest is
the vicinity of support attachments, doublers and external boundaries and the goal is to estimate
strength-related QoIs.
The accuracy of the data of interest depends not only on the discretization used but also on
how well the exact solution of a dimensionally reduced model approximates the exact solution
of the corresponding fully three-dimensional model. The differences between the data of inter-
est determined from the exact solution of a dimensionally reduced model and the corresponding
fully three-dimensional model are model form errors. The hierarchic view of models provides a
conceptual framework for the estimation and control of model form errors.
As the thickness is reduced, the exact solutions of the hierarchic beam, plate and shell mod-
els converge, respectively, to the exact solution of the Bernoulli-Euler beam model, the Kirchhoff
plate model and the Novozhilov-Koiter shell model. These models, unlike the hierarchic models,
require both the displacement functions and their first derivatives to be continuous. Therefore
the exact solutions of the models in the hierarchic family that lie in C0 (Ω) converge to a solution
that lies in C1 (Ω). Unless the polynomial degree is sufficiently high to permit close approximation
of C1 (Ω) continuity, h-convergence will be slow or shear locking may occur. On the other hand,
p-convergence will occur, however entry into the asymptotic range will be at p ≥ 4.
255
Aspects of multiscale models
This chapter is concerned with finding the macromechanical properties of unidirectional

fiber-matrix laminae from the mechanical properties of its constituents. We consider idealized
unidirectional laminae, assuming that the fiber arrangement fits a perfect hexagonal or square
pattern.
8.1 Unidirectional fiber-reinforced laminae

A representative volume element (RVE) is a sample of a heterogeneous material that has the aver-
age physical properties of the heterogeneous material. We are interested in determining the ele-
ments of the macromechanical material stiffness matrix [E] for unidirectional laminae arranged in
hexagonal and rectangular patterns as shown in Fig. 8.1 (a) and (b) respectively.
By definition, {𝜎} = [E]{𝜖} where [E] is a 6 × 6 symmetric, positive-definite matrix and
{𝜎} = {𝜎x 𝜎y 𝜎z 𝜏xy 𝜏yz 𝜏zx }T (8.1)
{𝜖} = {𝜖x 𝜖y 𝜖z 𝛾xy 𝛾yz 𝛾zx }T (8.2)
are the stress and strain vectors respectively. In linear elasticity the elements of [E] are constants,
independent of {𝜖}.
We will be interested in the relationship between the integral average of stress {𝜎} and integral
average of strain {𝜖} with the average taken over the RVE. We will use the notation
{𝜎} = [ERVE ]{𝜖} (8.3)
and determine the macroscopic material stiffness matrix [ERVE ] such that the strain energy for the
heterogeneous RVE equals the strain energy of the homogenized RVE:
1 1
U= {𝜖}T [E]{𝜖} dV = {𝜖}T [ERVE ]{𝜖}V (8.4)
2 ∫V 2
where V is the volume of the RVE. The elements of [ERVE ] must be substantially independent of
how many RVEs are used for their determination.
We formulate an algorithm for the computation of the elements of [ERVE ] by the finite element
method. The algorithm requires determination of the strain energy and the average displacements
on the boundaries of the RVE, both of which converge faster than the energy norm of the solution,
that is superconvergence is realized.
256 8 Aspects of multiscale models
y
a c
z
(a) (b)
x
Figure 8.1 RVEs for unidirectional laminae: (a) hexagonal and (b) rectangular arrangement of fibers.
Owing to the five planes of symmetry of the RVEs shown in Fig. 8.1, the macroscopic material
stiffness matrix is characterized by only six constants as indicated in eq. (8.5).
⎡A e f 0 0 0⎤
⎢ ⎥
⎢e A f 0 0 0⎥
⎢f f B 0 0 0⎥
[ERVE ] = ⎢ ⋅
0 ⎥⎥
(8.5)
⎢0 0 0 C 0
⎢0 0 0 0 D 0⎥
⎢0 0 0 0 0 D⎥⎦
⎣
Therefore the strain energy for the RVE can be written as:
( )
1 2 2 2
U= A(𝜖 x + 𝜖 y ) + B𝜖 z + 2e 𝜖 x 𝜖 y + 2f (𝜖 x 𝜖 z + 𝜖 y 𝜖 z ) dV
2 ∫V
( )
1 2 2 2
+ C𝛾 xy + D(𝛾 yz + 𝛾 zx ) dV (8.6)
2 ∫V
where the overbars indicate integral averages over the volume of the RVE.
We determine the six constants that characterize [ERVE ] by solving six problems in which the
integral averages of the strains are either enforced by imposed displacement boundary conditions
or computed by the finite element method.
Specifically, we denote the surface of the RVE to which the positive x, y and z axis is nor-
mal by X + , Y + and Z + respectively. Similarly we denote the surface of the RVE to which the
negative x, y and z axis is normal by X − , Y − and Z − respectively. The displacement bound-
ary conditions for the six problems are given in Table 8.1 where “sym” indicates symmetry,
that is, the normal displacement is zero, “float” indicates “floating symmetry”. The floating
symmetry condition differs from the symmetry condition in that the normal displacement
is a constant which is determined such that the area integral of the normal stress is zero.
Boundary conditions not shown explicitly in Table 8.1 are homogeneous natural boundary
conditions: For example, for k = 1 we have 𝜏xy = 𝜏xz = 0 on X + . These are periodic boundary
conditions defined in such a way that for each mode of deformation the number of constraints is
minimal.
8.1 Unidirectional fiber-reinforced laminae 257
Table 8.1 Displacement boundary conditions.
k X− X+ Y− Y+ Z− Z+
1 sym ux = a sym float sym float

2 sym float sym float sym uz = c
3 sym ux = a sym uy = b sym float
4 sym ux = a sym float sym uz = c
5 uy = 0 uy = a∕2 ux = 0 ux = b∕2 sym float
6 sym float uz = 0 uz = b∕2 uy = 0 uy = c∕2
The constants A, B, c and f are computed from the equations:

( ) ( ) ( ) ( )
2 2 2
𝜖 x + 𝜖 y A + 𝜖 z B + 2 𝜖 x 𝜖 y k e + 2 𝜖 y 𝜖 z + 𝜖 z 𝜖 x k f = 2Uk ∕V (8.7)
k k
where k = 1, 2, 3, 4 identifies the solution corresponding the boundary conditions shown in

Table 8.1, Uk is the computed strain energy and V is the volume of the RVE.
The constants C and D are computed from
2 2
C = 2U5 ∕(𝛾 xy V), D = 2U6 ∕(𝛾 yz V). (8.8)
8.1.1 Determination of material constants

In the absence of thermal loading the integral average of strain is related to the integral average
of stress by the transformation {𝜖} = [CRVE ]{𝜎}. The matrix [CRVE ] is the macroscopic material
compliance matrix. Obviously [CRVE ] = [ERVE ]−1 .
The compliance matrix [CRVE ] has the same structure as [ERVE ] displayed in eq. (8.5). It is cus-
tomary to write the material constants with reference to the longitudinal (L) and transverse (T)
directions:
𝜈 𝜈
⎡ 1 − T − LT 0 0 0 ⎤
⎢ ET ET EL ⎥
⎢ ⎥
⎢ 𝜈T 1 𝜈LT ⎥
⎢− − 0 0 0 ⎥
⎢ ET ET EL ⎥
⎢ ⎥
⎢ 𝜈TL 𝜈TL 1 ⎥
−
⎢ E − 0 0 0 ⎥
⎢ T E T E L ⎥
[CRVE ] = ⎢ ⎥ (8.9)
⎢ 1 ⎥
⎢ 0 0 0
GT
0 0 ⎥
⎢ ⎥
⎢ ⎥
⎢ 0 0 0 0
1
0 ⎥
⎢ GLT ⎥
⎢ ⎥
⎢ ⎥
⎢ 0 1 ⎥
⎣ 0 0 0 0
GLT ⎦
where EL , ET are the moduli of elasticity, 𝜈T , 𝜈LT , 𝜈TL are the Poison’s ratios and GT , GLT are the
shear moduli. The Poisson ratios 𝜈TL and 𝜈LT are not independent. They must satisfy the following
Table 8.2 Displacement boundary conditions

for the determination of the coefficients of
thermal expansion.
k X− X+ Y− Y+ Z− Z+
7 sym float sym float sym float
condition that guarantees the symmetry of [CRVE ]:

𝜈TL 𝜈
= LT ⋅ (8.10)
ET EL
These material constants can be computed directly from (8.9).
When the elements of [CRVE ] can be computed from five material constants then the material is
said to be transversely isotropic. Specifically, for a transversely isotropic material we have:
ET
GT = ⋅ (8.11)
2(1 + 𝜈T )
Although it is usually assumed that [CRVE ] is transversely isotropic, this assumption is justified only
in the special case of regular hexagonal arrangement of fibers.
8.1.2 The coefficients of thermal expansion

For the computation of the coefficients of thermal expansion we impose the boundary conditions
shown in Table 8.2 and apply an arbitrary constant temperature change  to the RVE. We then
+ +
compute the average displacements on X + , Y + and Z + . Denoting these displacements by ux . uy and
+
uz respectively, the coefficients of thermal expansion in the longitudinal and transverse directions
are computed from
+ + +
𝛼X = ux (a ), 𝛼Y = uy (b ), 𝛼Z ≡ 𝛼L = uz ∕(c ). (8.12)
8.1.3 Examples
In the following examples we consider unidirectional fibers of diameter 7 μm (0.007 mm). The vol-
ume fraction of the fibers Vf is 60%. The fibers are transversely isotropic with the following material
properties: EL = 2.52 × 105 MPa, ET = 1.65 × 104 MPa, 𝜈LT = 0.3, 𝜈T = 0.2, GLT = 4.14 × 104 MPa
where the subscripts L and T refer, respectively, to the longitudinal and transverse directions. The
coefficients of thermal expansion are: 𝛼L = −1.08 × 10−6 ∕∘ C, 𝛼T = 7.2 × 10−6 ∕∘ C. The matrix is
isotropic, its modulus of elasticity is 3.79 × 103 MPa, 𝜈 = 0.3. The coefficient of thermal expansion
is: 𝛼 = 5.4 × 10−5 ∕∘ C.
These material properties were taken from [110]. This will allow comparison with results
obtained by other methods described in the same reference1 . The material properties were
converted to SI units.
All computed quantities of interest were verified by increasing the polynomial degree of elements
from 5 to 8 and estimating their limit values.
1 See Exercise 4.1 on p. 85 and the solutions on p. 272.

Hexagonal pattern
Using the notation for the dimensions in Fig. 8.1 (a) we have: a = b = c = 1.132615 × 10−2 mm.
The algorithm for finding the macroscopic material properties was applied to the RVE shown in
Fig. 8.1(a) and, in order to test for size dependence, also to a domain comprised of 27 RVEs shown
in Fig. 8.2.
The results of computation are summarized in Table 8.3. We note that the number of digits dis-
played in the table would not be justified by the data given in the problem definition. We use four
decimal places only for the purpose of comparing the results. It is seen that the results are virtually
independent of the number of RVEs.
Remark 8.1 Hexagonal arrangements of fibers are usually treated as transversely isotropic mate-
rials, see for example [47], [110]. However, transverse isotropy is realized only in the special case of
regular hexahedral arrangements.
√ Referring to the notation in Fig. 8.1, for the regular hexahedral
arrangement b∕a = 3. In this example, however, b∕a = 1.
Figure 8.2 Solution domain consisting of
27 RVEs (864 finite elements).
Table 8.3 Material properties of the hexagonal

pattern (MPa and 1∕∘ C).
Property one RVE 27 RVEs Diff. (%)
EL 1.5272E+05 1.5272E+05 0.00

ET 7.9143E+03 7.9143E+03 0.00
GLT 5.3708E+03 5.4367E+03 1.23
GT 3.5579E+03 3.5579E+03 0.00
𝜈LT 0.300 0.300 0.00
𝜈T 0.368 0.368 0.00
𝛼L −5.3323E-07 −5.3323E-07 0.00

𝛼T 2.8941E-05 2.8941E-05 0.00
Using eq. (8.11) for the computation of GT from the values of ET and 𝜈T in Table 8.3, we find
GT = 2.8931 × 103 MPa which differs by −18.7% from the computed value. Therefore treating this
as a transversely isotropic material will underestimate the transverse shear modulus by 18.7%. This
is a systematic error that can be easily avoided.
Square pattern
Referring to the notation in Fig. 8.1(b) we have: a = b = c = 8.0088 × 10−3 mm. The algorithm for
finding the macroscopic material properties was applied to the RVE shown in Fig. 8.1(b) and, in
order to test for size dependence, also to a domain comprised of 8 RVEs shown in Fig. 8.3(a). The
periodicity of the solution corresponding to transverse shear is illustrated in Fig. 8.3(b).
To determine the size of the error that would be incurred if this were treated as a transversely
isotropic material, once again we use eq. (8.11) for the computation of GT from the values of ET
and 𝜈T in Table 8.4. We find GT = 3.5583 × 103 MPa which differs by 23% from the value computed
by the present method but is very close to the value reported in Table 8.3.
Region of
interest y
z
(a) x (b)
Figure 8.3 (a) Solution domain consisting of 8 RVEs (128 finite elements). (b) Deformed configuration
under transverse shear.
Table 8.4 Material properties of the square

pattern (MPa and 1∕∘ C).
Property one RVE 8 RVEs Diff. (%)
EL 1.5272E+05 1.5272E+05 0.00

ET 9.0737E+03 9.0785E+03 0.05
GLT 5.4639E+03 5.4698E+03 0.11
GT 2.8932E+03 2.8932E+03 0.00
𝜈LT 0.300 0.300 0.00
𝜈T 2.7513E-01 2.7581E-01 0.24
𝛼L −5.332E-07 −5.332E-07 0.00

𝛼T 2.8941E-05 2.8941E-05 0.00
Table 8.5 Comparison of material properties (psi and 1∕∘ F).
Property Hexagonal Square RM MRM
EL 2.220E+07 2.2201E+07 2.216E+07 2.212E+07

ET 1.151E+06 1.3191E+06 1.02E+06 1.30E+06
GLT 7.808E+05 7.9433E+05 5.0E+05 7.5E+05
GT 5.172E+05 4.2060E+05 4.0E+05 5.2E+05
𝜈LT 0.300 0.300 0.300 0.300
𝜈T 0.368 0.275 0.27 0.26
𝛼L −2.96E-07 −2.96E-07 −2.9E-07 −2.9E-07

𝛼T 1.61E-05 1.61E-05 1.80E-05 1.80E-05
Comparison
In Table 8.5 the computed material properties for one RVE for the hexagonal and square patterns
are presented along with properties computed by the rule of mixtures (RM) and the modified rule
of mixtures (MRM) as reported in [110]. The volume fraction of fibers is 60%. In order to make
direct comparison possible, the data was converted to US customary units. It is seen that both RM
and MRM provide a reasonable approximation for the material properties.
8.1.4 Localization
Localization is concerned with interpretation of the solution of the macroscopic problem on the
scale of RVEs. The main idea is that strain computed from the solution of the macroscopic problem
is virtually constant over one RVE, or a small group of RVEs, and therefore it approximates the
average strain values well. Therefore the displacement components, up to rigid body displacements,
can be written in terms of the average strain values:
ux = 𝜖 x x + 𝛾 xy y∕2 + 𝛾 zx z∕2
uy = 𝛾 xy x∕2 + 𝜖 y y + 𝛾 yz z∕2 (8.13)
uz = 𝛾 zx x∕2 + 𝛾 yz y∕2 + 𝜖 z z.
Assuming that the coordinate system is located in the center of the RVE, or a group of RVEs, such
that −𝓁x ∕2 < x < 𝓁x ∕2, −𝓁y ∕2 < y < 𝓁y ∕2, −𝓁z ∕2 < z < 𝓁z ∕2, the displacement boundary condi-
tions corresponding to equations (8.13) are as shown in Table 8.6. Because the displacement bound-
ary conditions reduce the number of degrees of freedom and introduce local perturbations at the
boundary, it is recommended that at least 8 RVEs be used for solving the local problem.
Remark 8.2 If the exact solution of the macroscopic problem is such that the elements of the
strain tensor are infinity in one or more points then the maximum norm of the computed strain
depends on the discretization and, since the maximum norm of the computed strain is finite, the
error measured in maximum norm is infinitely large. Therefore the goal of computation cannot be
the determination of maximum strain in such cases.
Table 8.6 Displacement boundary conditions for the problem of

localization.
X− X+ Y− Y+ Z− Z+
ux −𝜖 x 𝓁x ∕2 𝜖 x 𝓁x ∕2 −𝛾 xy 𝓁y ∕2 𝛾 xy 𝓁y ∕2 −𝛾 zx 𝓁z ∕2 𝛾 zx 𝓁z ∕2
uy −𝛾 xy 𝓁x ∕2 𝛾 xy 𝓁x ∕2 −𝜖 y 𝓁y ∕2 𝜖 y 𝓁y ∕2 −𝛾 yz 𝓁z ∕2 𝛾 yz 𝓁z ∕2
uz −𝛾 zx 𝓁x ∕2 𝛾 zx 𝓁x ∕2 −𝛾 yz 𝓁y ∕2 𝛾 yz 𝓁y ∕2 −𝜖 z 𝓁z ∕2 𝜖 z 𝓁z ∕2
8.1.5 Prediction of failure in composite materials

In fiber-reinforced composites detectable failure events occur on the scale of one or more RVEs.
While the assumptions of small strain continuum mechanics do not apply to failure events, in
many practically important cases the first appearance of detectable failure can be correlated with
the solution of small strain continuum mechanics problems through the use of suitably defined
predictors. One possible approach is to define integral averages on solution-dependent domains.
This is analogous to the predictor defined in Section 6.3 in connection with fatigue failure in metals.
A similar approach can be applied to composite materials. This will require coordinated experimen-
tal and analytical investigation subject to the procedures of verification, validation and uncertainty
quantification (VVUQ).
Example
In this example we illustrate that the elastic strains are unbounded at fiber terminations, whereas
strains averaged over the matrix phase of an RVE are well defined and converge strongly. We will
consider rectangular fiber arrangement and a domain consisting of 8 RVEs shown in Fig. 8.3. The
dimensions are: 𝓁x = 𝓁y = 𝓁z = 1.6018 × 10−2 mm. The volume fraction is 60% and the elastic prop-
erties of the fiber and matrix are the same as those in Section 8.1.3. The region of interest is that
part of the RVE which is occupied by the matrix, as indicated in Fig. 8.3(a).
The displacement boundary conditions are shown in Table 8.7. Normal displacements consis-
tent with the average strain values 𝜖 x = 3.5 × 10−3 , 𝜖 y = −2.0 × 10−3 , 𝛾 xy = 0 are prescribed on the
boundary surfaces X + and Y + .
Using the finite element mesh consisting of 128 elements shown in Fig. 8.3(a), a sequence of
finite element solutions was obtained corresponding to the polynomial degree p ranging from
1 to 8. For the matrix the equivalent strain is defined as:
√
1 [ ]
𝜖eq = (𝜖 − 𝜖2 )2 + (𝜖2 − 𝜖3 )2 + (𝜖3 − 𝜖1 )2 (8.14)
2(1 + 𝜈)2 1
where 𝜖1 , 𝜖2 and 𝜖3 are the principal strains and 𝜈 the Poisson’s ratio of the matrix. This definition
of equivalent strain satisfies the condition that under uniaxial loading 𝜎eq = E𝜖eq where 𝜎eq is the
von Mises stress and E is the modulus of elasticity. Other definitions of equivalent strain can be
found in the literature.
Table 8.7 Displacement boundary

conditions.
X− X+ Y− Y+ Z− Z+
sym ux sym uy float free

12
p=8
11 7
2
3 5 6
10 4
9
Millistrain
Maximum equivalent strain

8
p=1
7
2 3 4 5 6 7 p=8
6
5 p=1
Equivalent strain averaged over one RVE
4
103 104
Degrees of freedom (N)
Figure 8.4 Example. Equivalent strain in the matrix computed in the region of interest shown in Fig. 8.3(a).
The maximum value and the average value of 𝜖eq in the matrix phase of one RVE were computed.
This is the region of interest indicated in Fig. 8.3. The results of computation are shown in Fig. 8.4.
The results show that max (𝜖eq ) diverges. This would not be visible if p-extension would have been
stopped at p = 6. However, it is seen that the computed data points keep increasing past p = 6. The
divergence of max (𝜖eq ) was confirmed by repeating the process with a refined mesh2 .
The markedly different behaviors of the computed data between low and high N values is typical
for both h- and p-extensions. For large N the sequence of values is in the asymptotic range. For low
N the sequence is in the pre-asymptotic range. Extrapolation to N → ∞ is justified only for high
N values. This raises the obvious question: what is high and what is low? The answer to this is
problem-dependent.
Because max 𝜖eq corresponding to the exact solution is not a finite number, it is not suitable for
use as a predictor of failure. On the other hand, the first derivatives of the exact solution are square
integrable, hence the integral average of 𝜖eq over any volume, in this example over the matrix phase
of an RVE, is a finite number.
The integral average meets the other two requirements as well: small perturbations in D do not
cause large changes in integral averages and critical values of integral averages can be inferred from
observations of the outcomes of coupon experiments.
8.1.6 Uncertainties
The perfectly regular arrangement of fibers analyzed in this chapter is a highly idealized repre-
sentation of reality. As reported in [11], the volume fraction in a plate fabricated under carefully
controlled laboratory conditions ranged from 9.5% to 79.5%. Upon smoothing the data, the aver-
age volume fraction was 59.1% with a range of 43.8% to 69.5%. The volume fraction was lowest
between the plies. The dominant uncertainties are the variations in volume fraction and longitudi-
nal undulations (waviness) of the fibers. Properly calibrated predictors are expected to account for
such variations.
2 These results are not presented here.

8.2 Discussion
An algorithm for finding the macroscopic thermomechanical properties of unidirectional

fiber-matrix composite laminae, given the material properties and volume fractions of the con-
stituents, was outlined in this Chapter. The quantities of interest are computed from the strain
energy values of RVEs corresponding to imposed displacements. It was demonstrated that the
macroscopic material properties are substantially independent of the number of RVEs.
The problem of localization was addressed. The macroscopic strain tensor in a point provides the
information necessary for obtaining solutions on the microscopic scale.
The technical requirements that predictors of damage accumulation and damage propagation
must meet were stated. It was noted that pointwise stresses and strains do not meet those require-
ments; however, many possible definitions of predictors do. One possible definition, the integral
average of the equivalent strain in the matrix phase of an RVE, was discussed.
The evaluation and ranking of predictors has to be performed through a process of model devel-
opment, similar to the process described in Section 6.5. Simulation governance [92] provides a
framework for systematic improvement of predictors over time as new experimental information
becomes available.
265
Non-linear models
The formulation of mathematical models invariably involves making restrictive assumptions such
as neglecting certain geometric features, idealizing the physical properties of the material, idealiz-
ing boundary conditions, neglecting the effects of residual stresses, etc. Therefore any mathematical
model should be understood to be a special case of a more comprehensive model. This is the hier-
archic view of mathematical models.
In order to test whether a restrictive assumption is acceptable for a particular application, it is
necessary to estimate the influence of that assumption on the quantities of interest and, if neces-
sary, revise the model. An exploration of the influence of modeling assumptions on the quantities
of interest is called virtual experimentation. Computational tools designed to support numerical
simulation must have the capability to support virtual experimentation and hence solve nonlinear
problems.
The formulation of nonlinear models is a very large and diverse field. Only a brief introduction is
presented in this chapter, with emphasis on the algorithmic aspects and examples. For additional
discussion and details we refer to other books, such as references [81], [86].
9.1 Heat conduction

Mathematical models of heat conduction often involve radiation heat transfer and the coefficients
of heat conduction are typically functions of the temperature. The formulation of mathematical
models that account for these phenomena is outlined in the following.
9.1.1 Radiation
When two bodies exchange heat by radiation then the flux is proportional to difference of the fourth
power of their absolute temperatures:
qn = 𝜅fs f𝜖 (u4 − u4R ) (9.1)
where u, uR are the absolute temperatures of the radiating bodies, 𝜅 = 5.699 × 10−8 W∕(m2 K4 ) is
the Stefan-Boltzmann constant1 , 0 ≤ fs ≤ 1 is the radiation shape factor and 0 < f𝜖 ≤ 1 is the surface
emissivity which is defined as the relative emissive power of a body compared to that of an ideal
blackbody. The surface emissivity is also equal to the absorption coefficient, defined as that fraction
1 Josef Stefan 1835–1893, Ludwig Boltzmann 1844–1906.
266 9 Non-linear models
of thermal energy incident on a body which is absorbed. In general, surface emissivity is a function
of the temperature2 .
Eq. (9.1) can be viewed as a convective boundary condition where the coefficient of convective
heat transfer depends on the temperature of the radiating bodies. Writing:
qn = 𝜅fs f𝜖 (u4 − u4R ) = 𝜅fs f𝜖 (u2 + u2R )(u + uR )(u − uR )
⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟
hr (u)
we have the form of eq. (2.27) with hc replaced by hr (u). Therefore radiation problems have to be
solved by iteration: first the linear problem is solved, then the coefficient of convective heat transfer
is updated and the solution is repeated. The stopping criterion is based on the size of temperature
change. Usually very few iterations are needed.
9.1.2 Nonlinear material properties

When the conductivity or other material coefficients are functions of the temperature then solu-
tions have to be obtained by iteration. The bilinear and linear forms are split into linear (L) and
nonlinear (NL) parts:
BL (u, 𝑣) + BNL (u, 𝑣) = FL (𝑣) + FNL (u, 𝑣).
To obtain the initial solution u(1) the terms BNL (u, 𝑣) and FNL (u, 𝑣) are neglected. For subsequent
solutions the following problem is solved:
BL (u(k) , 𝑣) = FL (𝑣) + FNL (u(k−1) , 𝑣) − BNL (u(k−1) , 𝑣) k = 2, 3, …
The process is terminated when − u(k) u(k−1) ,
measured in a suitable norm, is sufficiently small.
This process is not guaranteed to converge. If it fails to converge then the forcing function should
be applied in small increments. In most practical problems convergence will occur when the incre-
ments are sufficiently small.
9.2 Solid mechanics

In many practical problems the assumptions that (a) the strains are small, (b) the deformation of the
body is so small that the equilibrium equations written for the deformed configuration would not be
significantly different from the equilibrium equations written for the undeformed configuration, (c)
the stress-strain laws are linear and (d) mechanical contact can be approximated by linear boundary
conditions, such as linear springs, do not hold. In this section the formulation of mathematical
models that account for these phenomena is outlined.
9.2.1 Large strain and rotation

Deformation is characterized by the strain-displacement relationships. Consider an elastic body Ω
in a reference state, as shown in Fig. 9.1. The material points in the reference state are identified
by the position vectors Xi , called Lagrangian coordinates, and in the deformed state by the position
vectors by xi , called Eulerian coordinates. The displacement vector may be written as a function of
Xi in which case the uppercase letter Ui will be used. Alternatively, the displacement vector may
be written as a function of xi , in which case the lowercase letter ui will be used.
2 For example, the surface emissivity of polished stainless steel is 0.22 at 373.15 K and 0.45 at 698.15 K.

Ω′
P′ Q′
y
ds
→
Ω u
P
Y Q
dS
X x
An infinitesimal “fiber” in the undeformed configuration has the length dS and in the deformed
configuration it has the length ds. Let us assume first that the displacement is a function of the
Eulerian coordinates xi . In this case:
dXi = dxi − dui = dxi − ui,j dxj = (𝛿ij − ui,j )dxj .
Therefore:
dS2 = dXk dXk = (𝛿ki − uk,i )dxi (𝛿kj − uk,j )dxj
= (𝛿ij − uj,i − ui,j + uk,i uk,j )dxi dxj
and:
ds2 − dS2 = dxk dxk − dXk dXk
= 𝛿ij dxi dxj − (𝛿ij − uj,i − ui,j + uk,i uk,j )dxi dxj
= (ui,j + uj,i − uk,i uk,j )dxi dxj .
The Almansi strain tensor3 , denoted by eij , is defined by the relationship
ds2 − dS2 = 2eij dxi dxj .
Therefore we have the definition of the Almansi strain tensor:
1
eij = (ui,j + uj,i − uk,i uk,j ). (9.2)
2
In two dimensions, using unabridged notation, this is written as
[( ) ( )]
𝜕ux 1 𝜕ux 2 𝜕uy 2
exx = − +
𝜕x 2 𝜕x 𝜕x
( ) [ ]
1 𝜕ux 𝜕uy 1 𝜕ux 𝜕ux 𝜕uy 𝜕uy
exy = + − + (9.3)
2 𝜕y 𝜕x 2 𝜕x 𝜕y 𝜕x 𝜕y
[( )2 ( )2 ]
𝜕uy 1 𝜕ux 𝜕uy
eyy = − + ⋅
𝜕y 2 𝜕y 𝜕y
When the displacement vector components are written in terms of the Lagrangian coordinates
Xi then xi = Xi + Ui , therefore
dxi = dXi + dUi = dXi + Ui,j dXj = (𝛿ij + Ui,j )dXj
3 Emilio Almansi 1869–1948.

therefore:
ds2 = dxk dxk = (𝛿ki + Uk,i )dXi (𝛿kj + Uk,j )dXj
= (𝛿ij + Uj,i + Ui,j + Uk,i Uk,j )dXi dXj
and:
ds2 − dS2 = dxk dxk − dXk dXk
= (𝛿ij + Uj,i + Ui,j + Uk,i Uk,j )dXi dXj − 𝛿ij dXi dXj
= (Ui,j + Uj,i + Uk,i Uk,j )dXi dXj .
The Green strain tensor, denoted by Eij , is defined by the relationship
ds2 − dS2 = 2Eij dXi dXj .
Therefore we have the definition of the Green strain tensor:

1
Eij = (Ui,j + Uj,i + Uk,i Uk,j ). (9.4)
2
In two dimensions, using unabridged notation, this is written as
[( ) ( )]
𝜕UX 1 𝜕UX 2 𝜕UY 2
EXX = + +
𝜕X 2 𝜕X 𝜕X
( ) [ ]
1 𝜕UX 𝜕UY 1 𝜕UX 𝜕UX 𝜕UY 𝜕UY
EXY = + + + (9.5)
2 𝜕Y 𝜕X 2 𝜕X 𝜕Y 𝜕X 𝜕Y
[( )2 ( )2 ]
𝜕UY 1 𝜕UX 𝜕UY
EYY = + + ⋅
𝜕Y 2 𝜕Y 𝜕Y
When the first derivatives ui,j (resp. Ui,j ) are much smaller than unity then the product terms
uk,i uk,j (resp. Uk,i Uk,j ) are negligible and the strain is called infinitesimal, small or linear strain.
In such cases both the Almansi and Green strain tensors reduce to the definition of infinitesimal
strain:
1
𝜖ij = (ui,j + uj,i ). (9.6)
2
Example 9.1 The following example illustrates the solution of a small-strain large displacement
problem. A 200 mm long (L), 1.0 mm thick (t) and 7.5 mm wide (𝑤) elastic strip is fixed at one end
and loaded at the other end by lineraly distributed normal tractions corresponding to the bending
moment M = EI∕R where R = L∕(2𝜋), I = 𝑤t3 ∕12, E = 2.0 × 105 MPa is the modulus of elasticity
and Poisson’s ratio is zero. According to the Bernoulli-Euler beam model, this moment will bend
the strip into a cylinder of radius R.
The solution, shown in Fig. 9.2, was obtained using a thin solid model, three hexahedral elements
(q)
and anisotropic product space  p,p,q (Ωst ) with p = 8 and q = 1. The stopping criterion is given by
eq. (5.28). In this example 𝜏stop = 0.1 was used.
Remark 9.1 The elastic strip in the foregoing example deforms into a cylinder only when Pois-
son’s ratio is zero. When Poisson’s ratio is not zero then the deformed configuration will be doubly
curved with an anticlastic curvature.
Figure 9.2 Elastic strip bent into a cylinder.
Exercise 9.1 Demonstrate in two dimensions that under arbitrary rigid body rotation all
components of the Almansi strain tensor are zero. Hint: Let the reference configuration of a
two-dimensional body be:
X = R cos 𝜃, Y = R sin 𝜃.
If this body is rotated by angle 𝛼, its new position will be:
x = R cos(𝜃 + 𝛼) y = R sin(𝜃 + 𝛼).
Therefore:
ux = x − X = R cos(𝜃 + 𝛼) − R cos 𝜃
uy = y − Y = R sin(𝜃 + 𝛼) − R sin 𝜃.
Use
𝜕ux 𝜕u 𝜕R 𝜕ux 𝜕𝜃
= x +
𝜕x 𝜕R 𝜕x 𝜕𝜃 𝜕x
= [cos(𝜃 + 𝛼) − cos 𝜃] cos(𝜃 + 𝛼) + [sin(𝜃 + 𝛼) − sin 𝜃] sin(𝜃 + 𝛼)
= 1 − cos 𝛼
etc., to complete the exercise.
Exercise 9.2 Following the procedure indicated in Exercise 9.1, demonstrate that the Green
strain tensor vanishes under rigid body rotation.
Exercise 9.3 Following the procedure indicated in Exercise 9.1, compute the linear strain tensor
defined by eq. (9.6).
Exercise 9.4 Consider a thin wire of length 𝓁 oriented along the x axis. Write expressions for the
Almansi strain ex and the Green strain Ex in terms of an imposed displacement Δ. (a) Show that
ex → 1∕2 as Δ → ∞. (b) Plot the Almansi strain and the Green strain in the range −𝓁 < Δ ≤ 10𝓁.
Hint: Let
1+𝜉 1+𝜉
X= 𝓁, x= (𝓁 + Δ), −1 ≤ 𝜉 ≤ +1.
2 2
9.2.2 Structural stability and stress stiffening

Investigation of buckling and stress-stiffening is generally performed for structures which are
beam-like and slender, or shell-like and thin. Such structures are usually stiffened and typically
there are topological details, loads or boundary conditions for which the assumptions of beam,
plate or shell theories do not hold. For this reason we consider the elastic stability of a fully
three-dimensional body and construct a mathematical model which is not encumbered by the var-
ious restrictions that exist in beam, plate or shell models. From this model various dimensionally
reduced models can be deduced as special cases.
We assume that a three-dimensional elastic body is subjected to an initial stress field 𝜎ij0 which
satisfies the equations of equilibrium of linear elasticity:
𝜎ij,j
0
+ Fi0 = 0 (9.7)
where Fi0 is the body force, and the traction boundary condition:
𝜎ij0 nj = Ti0 on 𝜕ΩT ∪ 𝜕Ωs (9.8)
where Ti0 represents tractions imposed either directly on 𝜕ΩT or through a displacement 𝛿j0 imposed
on a distributed elastic spring on 𝜕Ωs .
The initial stress field may be caused by body forces, surface tractions or temperature, or may
be residual stress caused by manufacturing or cold working processes. The generalized form of
eq. (9.7) is: For all 𝑣i ∈ E0 (Ω)
1
𝜎 0 (𝑣 + 𝑣j,i ) dV = Fi0 𝑣i dV + T 0 𝑣 dS (9.9)
2 ∫Ω ij i,j ∫Ω ∫𝜕ΩT ∪𝜕Ωs i i
where dS represents the differential surface.

Let us assume that the reference configuration is perturbed by a small change in the body force
(F i ); the temperature ( Δ ); the surface tractions (T i ) on 𝜕ΩT ; the displacement imposed on the
∗
distributed spring 𝛿 i on 𝜕Ωs , or imposed displacement (ui ) on boundary segment 𝜕Ωu . The corre-
sponding kinematically admissible displacement field is denoted by ui . It is assumed that the stress
𝜎 ij caused by the perturbation is negligible in relation to the initial stress 𝜎ij0 .
When the reference configuration is not stress-free then the work done by 𝜎ij0 due to the product
terms of the Green strain tensor may not be negligible. Therefore the strain energy is:
1 1
U(ui ) = C 𝜖 𝜖 dV + k u u dS
2 ∫Ω ijkl ij kl 2 ∫𝜕Ωs ij i j
1
+ 𝜎 0 u u dV. (9.10)
2 ∫Ω ij 𝛼,i 𝛼,j
The third integral in eq. (9.10) represents the work done by 𝜎ij0 due to the product terms of the Green
strain tensor. The work done by 𝜎ij0 due to the linear strain terms is exactly canceled by the work
done by Fi0 , Ti0 and 𝛿i0 in the sense of eq. (9.9). The potential energy is then:
Π(ui ) = U(ui )− F i ui dV − T i ui dS − kij 𝛿 j ui dS

∫Ω ∫𝜕ΩT ∫𝜕Ωs
−  Δ Cijkl 𝛼kl ui,j dV (9.11)

∫Ω
where kij is a positive-semidefinite spring rate matrix, and 𝛼kl represents the coefficients of thermal
expansion. For isotropic materials 𝛼kl = 𝛼𝛿kl .
̃
We seek ui ∈ E(Ω) such that Π(ui ) is stationary; that is,
( )
𝜕Π(ui + 𝜀𝑣i )
𝛿Π(ui ) = =0 for all 𝑣i ∈ E0 (Ω). (9.12)
𝜕𝜀 𝜀=0
From this it follows that the principle of virtual work in the presence of an initial stress field is:
̃
Find ui ∈ E(Ω) such that:
Cijkl uk,l 𝑣i,j dV + kij uj 𝑣i dS + 𝜎ij0 u𝛼,j 𝑣𝛼,i dV =

∫Ω ∫𝜕Ωs ∫Ω
F i 𝑣i dV + T i 𝑣i dS + kij 𝛿 j 𝑣i dS +  Δ Cijkl 𝛼kl 𝑣i,j dV (9.13)

for all 𝑣i ∈ E0 (Ω).
The effect of the initial stress 𝜎ij0 depends on its sense: If 𝜎ij0 is predominantly positive (i.e. tensile)
then the stiffness increases. This is called stress stiffening. If, on the other hand, 𝜎ij0 is predominantly
negative, then the stiffness decreases. Of great practical interest is the critical value of the initial
stress at which the stiffness is zero. Define:
𝜎ij0 = 𝜆𝜎ij★ . (9.14)
In stability problems 𝜎ij★ is the pre-buckling stress distribution. In stress stiffening problems 𝜎ij★ is
some reference stress and 𝜆 is some fixed number.
Define the bilinear form B𝜆 (ui , 𝑣i ) by:
B𝜆 (ui , 𝑣i ) = B(ui , 𝑣i ) − 𝜆G(ui , 𝑣i ) (9.15)
where
B(ui , 𝑣i ) = Cijkl ui,j 𝑣k,l dV + kij uj 𝑣i dS

∫Ω ∫𝜕Ωs
G(ui , 𝑣i ) = − 𝜎ij★ u𝛼,i 𝑣𝛼,j dV

∫Ω
F(𝑣i ) = F i 𝑣i dV + T i 𝑣i dS + kij 𝛿 j 𝑣i dS +  Δ Cijkl 𝛼kl 𝑣i,j dV.

̃
The problem is then to find ui ∈ E(Ω) such that
B𝜆 (ui , 𝑣i ) = F(𝑣i ) for all 𝑣i ∈ E0 (Ω). (9.16)
The set of 𝜆 for which eq. (9.16) is uniquely solvable for all F is called the resolvent set. The com-
plement of the resolvent set is the spectrum. In addition to the point spectrum, which consists of
the eigenvalues 𝜆i (i = 1, 2, …), the spectrum may also include values that lie in a continuous spec-
trum. Fortunately, in those problems of engineering interest that require consideration of elastic
stability (the analysis of thin-walled structures) the lowest values of 𝜆 lie in a point spectrum [33].
The effect of stress stiffening is illustrated by considering the free vibration of elastic structures
subjected to initial stress. The mathematical model of free vibration is: Find 𝜔 and ui ∈ E0 (Ω);
ui ≠ 0, such that:
B𝜆 (ui , 𝑣i ) − 𝜔2 𝜌ui 𝑣i dV = 0 (9.17)

∫Ω
where 𝜌 is the specific density of the material and 𝜔 is the natural frequency. The natural frequency
is now a function of 𝜆. If 𝜎ij★ is predominantly compressive (negative) then the structural stiffness
is decreased as 𝜆 > 0 is increased. If 𝜆 is in the point spectrum then there are functions ui which
satisfy eq. (9.17) for 𝜔 = 0. That is, the natural frequency is zero. If 𝜎ij★ is tensile (positive) then the
structural stiffness is increased as 𝜆 is increased. See Exercise 9.8.
Remark 9.2 Using the procedures of variational calculus, the strong form of the equations of
equilibrium is found to be:
𝜎 ij,j + F i + (𝜎kj0 ui,k ),j = 0 (9.18)
where 𝜎 ij = Cijkl (𝜖 kl −  Δ 𝛼kl ). The corresponding natural boundary conditions are:
(𝜎 ij + 𝜎kj0 ui,k )nj = T i on 𝜕ΩT (9.19)

and
(𝜎 ij + 𝜎kj0 ui,k )nj = kij (𝛿 j − uj ) on 𝜕Ωs . (9.20)
Example 9.2 Consider an elastic column of uniform cross-section, area A, moment of inertia I,
length 𝓁 and modulus of elasticity E. The centroidal axis of the column coincides with the x1 axis.
A compressive axial force P is applied, hence 𝜎11
0
= −P∕A. In the classical formulation of elastic
buckling the displacement field is assumed to be
d𝑤
u1 = − x, u2 = 𝑤(x1 ), u3 = 0 (9.21)
dx1 2
where 𝑤 is the transverse displacement. See, for example, [103]. Therefore the only non-zero strain
component is:
d2 𝑤
𝜖11 = x2 (9.22)
dx12
and the term 𝜎ij0 u𝛼,i u𝛼,j can be written as:
⎧( )2 ( )2 ⎫
P ⎪ d2 𝑤 d𝑤 ⎪
𝜎ij0 u𝛼,i u𝛼,j =− ⎨ 2
.
dx1 ⎬
x2 + (9.23)
A ⎪ dx12 ⎪
⎩ ⎭
In this case the strain energy can be written in the following form:
( )2
𝓁( ) 𝓁( )2
1 P d2 𝑤 P d𝑤
U= E− I dx1 − dx1 (9.24)
2 ∫0 A dx12 2 ∫0 dx1
where I is the moment of inertia of the cross-section:
I= x22 dx2 dx3 . (9.25)

∫A
The term P∕A can be neglected in relation to the modulus of elasticity E in (9.24). This
example serves as an illustration of how a dimensionally reduced model is derived from the
three-dimensional formulation.
Example 9.3 The following example illustrates the solution of a buckling problem. An elastic
strip of length L = 200 mm, thickness t = 1.0 mm, width 𝑤 = 7.5 mm is fixed at one end and loaded
at the other end by a constant normal traction Tn , corresponding to 1.0 N compressive force. The
shearing tractions are zero. The modulus of elasticity is E = 2.0 × 105 MPa and Poisson’s ratio is
zero. The goal is to estimate the buckling load factor, that is, the multiplier of the unit applied force
corresponding to the first buckling mode.
The classical formula or the buckling load Fcr is

𝜋 2 EI
Fcr = (9.26)
4L2
where I is the moment of inertia: I = 𝑤t3 ∕12. For the data in this example Fcr = 7.71 N. Therefore
the buckling load factor is 7.71.
Numerical solutions were obtained using a thin solid model, three hexahedral elements (as in
(q)
Example 9.1) and anisotropic product space Sp,p,q (Ωst ) with p ranging from 3 to 8 and q = 1. The
computed value of the buckling load factor was 7.71. Having repeated the computations for q = 3,
the same value was obtained.
Example 9.4 The following example illustrates the effects of stress stiffening on natural frequen-
cies and mode shapes. A 200 mm long, 1.0 mm thick and 7.5 mm wide elastic strip is fixed at one
end and loaded at the other end by normal traction Tn , the shearing tractions are zero. The modulus
of elasticity is E = 2.0 × 105 MPa Poisson’s ratio is 𝜈 = 0.3 and the mass density is 𝜚 = 7.86 × 10−9
Ns2 ∕mm4 (7860 kg∕m3 ).
Solutions were obtained for Tn = 0 and Tn = 150 MPa using a thin solid model, four hexahedral
(q)
elements and anisotropic product space Sp,p,q (Ωst ) with p = 8 and q = 3. The 20th mode shapes for
Tn = 0 and Tn = 150 MPa are shown in Figures 9.3(a) and 9.3(b) respectively. The natural frequency
for Tn = 0 converged to 7019 Hz and for Tn = 150 MPa it converged to 7294 Hz. Az seen in Fig. 9.3
the mode shapes are quite different: one is a torsional mode, the other is a bending mode.
Example 9.5 Consider a cylindrical shell of radius 0.10 m, length 0.40 m, wall thickness 0.001 m.
The material properties are: E = 2.0 × 1011 Pa, 𝜈 = 0.3. The shell is fixed at one end. On the other
end (a) the radial and circumferential displacements are zero and an axial (compressive) displace-
ment Δ is imposed; (b) the axial displacement is zero and tangential displacements are imposed
consistent with rotation 𝜃 about the axis of the shell. The goals of computation are to determine
the critical values of Δ and 𝜃, the corresponding compressive force Fcrit and twisting moment M,
and to verify that the relative errors in Δcrit and 𝜃crit are less than 1%.
(a)
(b)
Figure 9.3 The 20th modes of vibration of the elastic strip in Example 9.4. (a) Without prestress, (b)
prestress of 150 MPa in tension.
Table 9.1 Critical values of the axial displacement Δ (mm)

and the rotation 𝜃 (rad) in Example 9.5.
p N 𝚫crit % Error 𝜽crit % Error
5 11568 2.64 11.81 1.3249 × 10−2 2.16

6 17136 2.44 3.32 1.2983 × 10−2 0.11
−2
7 24240 2.38 0.62 1.2970 × 10 0.01
8 32880 2.37 0.14 1.2969 × 10−2 0.00
∞ ∞ 2.36 − 1.2969 × 10−2 −
(a) (b)
Figure 9.4 The first buckling modes of the cylindrical shell in Example 9.5. (a) Imposed axial
displacement. The contour lines are proportional to the normal displacement. (b) Imposed axial rotation.
The contour lines are proportional to the tangential displacement.
(q)
The computations were performed using 128 quadrilateral thin solid elements  p,p,q (Ωst ) with
p ranging from 5 to 8 (trunk space) and q fixed at q = 3. The results are shown in Table 9.1 and in
Fig. 9.4. The buckling mode shown in Fig. 9.4(a) happens to be an antisymmetric function.
The extrapolated value of the strain energy corresponding to the linear solution at Δ = 0.001 m
is Ulin = 157.65 N m. The strain energy is proportional to Δ2 . Therefore the strain energy at the
critical value of Δ is
Ucrit = Ulin (Δcrit ∕0.001)2 = 157.65(2.36∕1)2 = 878 N m
and the estimated value of the critical force is
Fcrit = (2U∕Δ) = 2 × 878∕0.001 = 1756 kN.
The twisting moment M is computed similarly. In case (b) Ucrit = 102.8 N m and Mcrit = 15.84 kN m.
Remark 9.3 The predicted buckling load should be understood to be an upper bound of the load
carrying capacity of shells. Owing to imperfections and sensitivity to boundary conditions, experi-
mentally obtained buckling loads tend to be substantially lower than the buckling loads predicted
by models such as the model in Example 9.5. Reduction factors of 1∕2 to 1∕4 are used in engineering
practice to account for such uncertainties [49].
Exercise 9.5 Consider the problem of Example 9.5 with the following boundary conditions: The
shell is fixed at one end. On the other end the radial and circumferential displacements are zero and
uniformly distributed axial (compressive) traction is applied the resultant of which is F. Determine
the critical value of F. Explain why it is different from Fcrit in Example 9.5.
Exercise 9.6 Neglecting the term P∕A in (9.24), determine the approximate value of P at which
a column fixed at both ends will buckle. Use one finite element and p = 4. Report the relative
error. Hint: Refer to eq. (7.27) and the definition N5 (𝜉) = 𝜓4 (𝜉) where 𝜓j (𝜉) ( j = 4, 5, …) is given
by eq. (7.26). The exact value of the critical force is 4𝜋 2 EI∕𝓁 2 .
Exercise 9.7 Show that the work done by 𝜎ij0 due to the linear strain terms is exactly canceled by
the work done by Fi0 , Ti0 and 𝛿i0 in the sense of eq. (9.9).
Exercise 9.8 Consider a 50 mm × 50 mm square plate, thickness: 1.0 mm. The material
properties are: E = 6.96 × 104 MPa; 𝜈 = 0.365; 𝜚 = 2.71 × 10−9 Ns2 ∕mm4 . The plate is fixed on
one edge, loaded by a normal traction Tn on the opposite edge and simply supported on the
other edges (soft simple support). Determine the first natural frequency corresponding to Tn =
−50 MPa, Tn = 0 and Tn = 50 MPa. (Partial answer: For Tn = −50 MPa the first natural frequency is
578.6 Hertz.)
Exercise 9.9 If the multi-span beam described in Exercise 7.9 had been pre-stressed by an axial
force such that a constant positive initial stress 𝜎 0 acted on the beam, would the rotation at support
B be larger or smaller than if the beam had not been prestressed? Explain.
9.2.3 Plasticity
The formulation of mathematical models based on the incremental and deformational theories of
plasticity are discussed in the following. In the incremental theory, as the name implies, a relation-
ship between the increment in strain tensor and the corresponding increment in the stress tensor
is defined. In the deformational theory the strain tensor (rather than the increment in the strain
tensor) is related to the stress tensor. It is assumed that the strain components are sufficiently small
to justify small strain representation and the plastic deformation is contained, that is, the plastic
zone is surrounded by an elastic zone. Uncontained plastic flow, as in metal forming processes, is
outside of the scope of the following discussion.
The formulation of plastic deformation is based on three fundamental relationships: (a) yield
criterion, (b) flow rule and (c) hardening rule. We will use the von Mises yield criterion4 and an
associative flow rule known as the Prandtl-Reuss flow rule5 . For a more detailed discussion of this
subject we refer to [86].
Notation
The stress deviator tensor is defined by:
def 1
𝜎̃ ij = 𝜎ij − 𝜎kk 𝛿ij . (9.27)
3
4 Richard von Mises 1883–1953.

5 Ludwig Prandtl 1875–1953, Endre Reuss 1900–1968.
𝜎 𝜎
𝜎 yield 𝜎 yield 𝜎 = H (ε p )
E Es
1
1
ε εp
εp ε
e
Figure 9.5 Typical uniaxial stress-strain curve.
The second invariant of the stress deviator tensor is denoted by J2 and is defined by
1
def
J2 = 𝜎̃ 𝜎̃
2 ij ij
1[ ]
= (𝜎11 − 𝜎22 )2 + (𝜎22 − 𝜎33 )2 + (𝜎33 − 𝜎11 )2 + 6(𝜎12 + 𝜎23 + 𝜎31 ) .
3
In a uniaxial test 𝜎11 is the only non-zero stress component. Therefore J2 = 2𝜎11 2
∕3. The equivalent
√
stress, also called von Mises stress, is J2 scaled such that in the special case of uniaxial stress it is
equal to the uniaxial stress:
√
def 1[ ]
𝜎 = (𝜎11 − 𝜎22 )2 + (𝜎22 − 𝜎33 )2 + (𝜎33 − 𝜎11 )2 + 6(𝜎12 + 𝜎23 + 𝜎31 ) . (9.28)
2
We will interpret uniaxial stress-strain diagrams as a relationship between the equivalent stress and
the equivalent strain. The elastic (resp. plastic) strains will be indicated by the superscript e (resp. p).
The three principal strains are denoted by 𝜖1 , 𝜖2 , 𝜖3 . The equivalent elastic strain is defined by
√
e def 2 √ e
𝜖 = (𝜖1 − 𝜖2e )2 + (𝜖2e − 𝜖3e )2 + (𝜖3e − 𝜖1e )2 (9.29)
2(1 + 𝜈)
where 𝜈 is Poisson’s ratio. The definition of the equivalent plastic strain follows directly from
eq. (9.29) by setting 𝜈 = 1∕2:
√ √
p 2 p p p p p p
𝜖 = (𝜖1 − 𝜖2 )2 + (𝜖2 − 𝜖3 )2 + (𝜖3 − 𝜖1 )2 . (9.30)
3
A typical uniaxial stress-strain curve is shown in Fig. 9.5.
Exercise 9.10 Show that the first derivatives of J2 with respect to 𝜎x , 𝜎y , 𝜎z and 𝜏xy are equal to
𝜎̃ x , 𝜎̃ y 𝜎̃ z and 𝜏̃ xy respectively.
Assumptions
The assumptions on which the formulation of the mathematical problem of plasticity is based are
described in the following.
1. Confined plastic deformation: The strain components are much smaller than unity on the
solution domain and its boundary, and the deformations are small in the sense that equilibrium
equations written for the undeformed configuration are essentially the same as the equilibrium
equations written for the deformed configuration.
2. Decomposition of strain: Assuming that the temperature remains constant, an increment in

the total strain is the sum of the increment in the elastic strain and the increment in the plastic
strain:
p
d𝜖ij = d𝜖ije + d𝜖ij . (9.31)
3. Yield criterion: We define
p p
F(𝜎ij , 𝜖 ) = 𝜎 − H(𝜖 ). (9.32)
When F < 0 then the material is elastic. Plastic deformation may occur only when F = 0. Any
stress state for which F > 0 is inadmissible. This is known as the consistency condition. There-
fore in plastic deformation:
𝜕F p
dF = 0 ∶ d𝜎 − H ′ d𝜖 = 0. (9.33)
𝜕𝜎ij ij
4. Flow rule: The Prandtl-Reuss flow rule states that
p 𝜕F p
d𝜖ij = d𝜖 . (9.34)
𝜕𝜎ij
Example 9.6 In this example we estimate the force-displacement relationship for the shear fit-
ting described in Example 4.7 using the deformation theory of plasticity. The material is 7075-T6
aluminum alloy. The material properties are: Modulus of elasticity: E = 1.05E7 psi (7.24E4 MPa),
Poisson’s ratio: 𝜈 = 0.30. The relationship between the equivalent strain 𝜖 and equivalent stress 𝜎
is given by the Ramberg-Osgood equation:
( )n
𝜎 3 S70E 𝜎
𝜖= + (9.35)
E 7 E S70E
where S70E is the stress corresponding to the point where a line that passes through the origin, and
has the slope of 0.7 times the modulus of elasticity, intersects the uniaxial stress-strain curve. For
the 7075-T6 aluminum alloy S70E = 58.5 ksi (403.3 MPa), n = 15.2.
Incremental stress-strain relationship

An increment in stress is proportional to the elastic strain:
e p
d𝜎ij = Cijkl d𝜖kl = Cijkl (d𝜖kl − d𝜖kl ).
20
15
force (kip)
10
0
0 0.1 0.2
F = 12.0 kips F = 15.0 kips
displacement (in)
Figure 9.6 Example 9.6. The force-displacement relationship and the plastic zone for two loading
conditions. The plastic zone is light grey.
Substituting eq. (9.34) we have:

𝜕F p
d𝜎ij = Cijkl d𝜖kl − Cijkl d𝜖 . (9.36)
𝜕𝜎kl
Using (9.33) we have:
p 𝜕F 𝜕F 𝜕F 𝜕F p
H ′ d𝜖 = d𝜎ij = C d𝜖 − C d𝜖 (9.37)
𝜕𝜎ij 𝜕𝜎ij ijkl kl 𝜕𝜎pq pqrs 𝜕𝜎rs
p
where the dummy indices were suitably renamed. From eq. (9.37) an expression for d𝜖 is obtained:
𝜕F
C
p
𝜕𝜎ij ijkl
d𝜖 = d𝜖kl .
𝜕F 𝜕F
H′ + C
𝜕𝜎pq pqrs 𝜕𝜎rs
On substituting into eq. (9.36) we have the incremental elastic-plastic stress-strain relationship:
⎛ 𝜕F 𝜕F ⎞
⎜ Cijmn C
𝜕𝜎mn 𝜕𝜎u𝑣 u𝑣kl ⎟
d𝜎ij = ⎜Cijkl − ⎟ d𝜖 . (9.38)
⎜ ′ 𝜕F 𝜕F ⎟ kl
⎜ H + Cpqrs
⎝ 𝜕𝜎pq 𝜕𝜎rs ⎟⎠
The bracketed expression in (9.38) is well defined for elastic-perfectly plastic materials (i.e. mate-
rials for which H ′ = 0).
The computations involve the following: Given the current stress 𝜎ij , compute
d𝜎ij = Cijkl d𝜖kl
corresponding to an increment in the applied load. In each integration point compute F(𝜎ij + d𝜎ij ).
If F(𝜎ij + d𝜎ij ) ≤ 0 then nothing further needs to be done. If F(𝜎ij + d𝜎ij ) > 0 then re-compute d𝜎ij
using eq. (9.38). Repeat the process until F(𝜎ij + d𝜎ij ) ≈ 0. The process is started from the linear
solution.
The deformation theory of plasticity

In the deformation theory of plasticity it is assumed that the plastic strain tensor is proportional to
the stress deviator tensor. Referring to Fig. 9.5,
e p 𝜎
𝜖 +𝜖 =
Es
where Es is the secant modulus (0 < Es < E). Since the elastic part of the strain is related to the
stress by Hooke’s law,
e 𝜎
𝜖 =
E
we have
( )
p 1 1
𝜖 = − 𝜎. (9.39)
Es E
In the case of uniaxial stress 𝜎̃ = 2𝜎∕3 and hence eq. (9.39) can be written as:
( )
p 3 1 1
𝜖 = − 𝜎.
̃
2 Es E
By modeling assumption this is generalized to:

( )
p 3 1 1
𝜖ij = − 𝜎̃ ij . (9.40)
2 Es E
For example, in planar problems:
⎧ p⎫ ⎧ ⎫
⎪ 𝜖xp ⎪ 𝜎̃
( )⎪ x⎪
⎪ 𝜖y ⎪ 3 1 1 ⎪ 𝜎̃ y ⎪
⎨ 𝜖 p ⎬ = 2 E − E ⎨ 𝜎̃ ⎬ ⋅ (9.41)
⎪ pz ⎪ s ⎪ z⎪
⎪𝜖xy ⎪ ⎪𝜏̃ xy ⎪
⎩ ⎭ ⎩ ⎭
In the deformation theory of plasticity the elastic-plastic compliance matrix is the matrix [C] that
establishes the relationship between the total strain and stress:
{𝜖} = [C]{𝜎}. (9.42)
The elastic-plastic material stiffness matrix [Eep ] is the inverse of the elastic-plastic compliance
matrix. Using the definition of the stress deviator and the relationship between the plastic strain
and deviatoric stress (9.41), we have:
⎧ 𝜖p ⎫ ( ) ⎡ 2∕3 −1∕3 0⎤ ⎧ 𝜎x ⎫
⎪ xp ⎪ 3 1 1 ⎢ ⎪ ⎪
⎨ 𝜖y ⎬ = 2 E − E ⎢−1∕3 2∕3 0⎥⎥ ⎨ 𝜎y ⎬ .
⎪𝛾xy
p⎪
2⎦ ⎪ ⎪
⎣ 0 ⎩𝜏xy ⎭
s
⎩ ⎭ 0
Using {𝜖} = {𝜖 e } + {𝜖 p }, a relationship is obtained between the total strain components and the
stress tensor:
⎧𝜖 ⎫ ⎛ ⎡ 1 −𝜈 0 ⎤ −1∕2 0⎤⎞ ⎧ 𝜎x ⎫
E − Es ⎡⎢
1
⎪ x⎪ ⎜1 ⎢−𝜈 1 ⎥ ⎪ ⎪
⎨ 𝜖y ⎬ = ⎜ E 0 + −1∕2 1 0⎥⎟ ⎨ 𝜎y ⎬
⎪𝛾xy ⎪ ⎝ ⎢ ⎥ Es E ⎢ ⎥⎟ ⎪ ⎪
⎩ ⎭ ⎣ 0 0 2(1 + 𝜈)⎦ ⎣ 0 0 3⎦⎠ ⎩𝜏xy ⎭
where the bracketed expression is the elastic-plastic material compliance matrix. Since Es < E, it is
invertible.
The solution is obtained by iteration: For each integration point the equivalent stress and strain
are computed. From the stress-strain relationship the secant modulus is computed and the appro-
priate material stiffness matrix [Eep ] is evaluated and a new solution is obtained. The process is
continued until the maximum of the difference between the equivalent stress and the uniaxial
stress, given the equivalent strain, becomes less than a pre-set tolerance (which is typically 1%
or less).
Example 9.7 An interesting benchmark study was performed under a research project entitled
“Adaptive Finite Element Methods in Applied Mechanics” sponsored by the German Research
Foundation6 in the period 1994 to 1999. Nine academic research institutes participated in this
project. Importantly, the project fostered collaboration among researchers from the applied math-
ematics and engineering research communities. At that time many researchers believed that high
order elements cannot be used for solving material nonlinear problems. The results of this investi-
gation served to dispel those beliefs [82].
The following challenge problem was posed: Solve a plane strain problem on a rectangular
domain with a hole in the center and report certain quantities of interest. The domain and
6 Deutsche Forschungsgemeinschaft (DFG).

p0
E D
450
360
270
100
33.3 180
15 90
y C 0
A x B 10
100
(a) (b)
Figure 9.7 Example 9.7: (a) The domain and the initial finite element mesh. (b) The equivalent stress
contours corresponding to p0 = 450 computed on the 54-element mesh at p = 8, product space.
dimensions are shown in Fig. 9.7(a). The boundary segments AB and CD are lines of symmetry.
The boundary segment DE is loaded by uniform normal traction p0 , the shearing traction is
zero. On boundary segments BC and EA the normal and shearing tractions are zero. The values
p0 = 300 and p0 = 450 were specified The investigators were instructed to assume that the material
is isotropic and elastic-perfectly plastic with yield stress 𝜎yield = 450, obeying the deformation
theory of plasticity and the von Mises yield criterion. The shear modulus G = 80193.8 and the bulk
modulus7 K = 164206.0 were given.
We solved the problem using a sequence of quasiuniform meshes with p = 8, product space,
assigned to all elements. The sequence was generated starting from the six element mesh shown
in Fig. 9.7(a) by uniformly dividing the standard quadrilateral element into n2 squares. The mesh
corresponding to n = 3 and the contours of the von Mises stress corresponding to p0 = 450 are
shown in Fig. 9.7(b). Starting from the linear solution, the elastic-plastic material stiffness matrix
[Eep ] was iteratively updated in each integration point until the difference between the equivalent
stress and the uniaxial stress, given the equivalent strain, was less than 0.5%. The computations
were performed with StressCheck.
We computed the following quantities of interest reported in Table 1 of reference [36]: (a) the
integral W defined as
p0 100
W= u (x, 100) dx,
2 ∫x=0 y
(b) the stress component 𝜎y in point B, denoted by 𝜎y(B) and (c) the displacement components uy and
ux in points D and E respectively. The results of computation are shown in Table 9.2. It is seen that
the computed data are virtually independent of the number of degrees of freedom N as the number
of elements (denoted by M) is increased. Also, the data in Table 9.2 match the data reported in [36]
to four significant digits.
E E
7 The shear and bulk moduli are related E and 𝜈 as G = and K = ⋅
2(1 + 𝜈) 3(1 − 2𝜈)
Table 9.2 Example 9.7: The results of computation for

p0 = 300.
M N W 𝝈y(B) uy(D) ux(E)
6 800 2044.983 522.27 0.140327 0.050886

54 7008 2044.986 517.52 0.140327 0.050885
150 19360 2044.984 517.44 0.140327 0.050885
e e p p
Exercise 9.11 Show that in uniaxial tension or compression 𝜖 = 𝜖 1 , 𝜖 = 𝜖 1 and 𝜎 = 𝜎1 .
Exercise 9.12 Derive eq. (9.29) by specializing the root-mean-square of the differences of princi-
e e
pal strains to the one-dimensional case such that 𝜖 = 𝜖 1 .
9.2.4 Mechanical contact

Mathematical models of contact between solid bodies involve non-linear boundary conditions writ-
ten in terms of a gap function g = g(s, t) ≥ 0 where s and t are the parameters of one of the contacting
surfaces: Whenever g = 0, the normal and shearing tractions in corresponding points must have
equal value and opposite sense. When g > 0 then the tractions are zero. The condition g < 0 is not
allowed.
In many practical problems the contacting surfaces are lubricated and therefore, under
quasi-static conditions, the shearing tractions are negligibly small in comparison with the normal
tractions.
We will be concerned with frictionless contact only. Consider two solid bodies Ω1 and Ω2 . The
surfaces are denoted by 𝜕Ω1 and 𝜕Ω2 respectively. We identify a contact zone 𝜕Ωc on one of the
contacting surfaces. The contact zone is a convex subset of one of the contacting surfaces. Contact
is expected to occur within the contact zone. There could be several contact zones and several
bodies can be in contact.
Example 9.8 Let us consider the problem of contact between two elastic bars constrained by
distributed springs. The notation is shown in Fig. 9.8. We assume that the axial stiffness (AE)i and
the spring coefficient ci (i = 1, 2) are constants for each bar and bar 2 is fixed on its right end. The
gap between the bars is g = g0 − U2 + U3 where g0 is the initial gap. The goal is to determine the
elastic spring rate k = F∕U1 as a function of the applied force F.
The differential equations for the bars are:

−(AE)i u′′i + ci ui = 0, i = 1, 2 (9.43)
where the primes indicate differentiation with respect to x. We associate a local coordinate system
with each bar, such that x = 0 at the left end. Therefore U1 = u1 (0), U2 = u1 (𝓁1 ) and U3 = u2 (0).
Figure 9.8 Example: Contact problem in U1 U2

1D. Notation. x x
F Fc Fc
1 g 2
Introducing
√
𝜉 = x∕𝓁i , 𝑣i = ui ∕𝓁i , 𝜆i = ci 𝓁i ∕(AE)i
equations (9.43) are written in dimensionless form:
d2 𝑣i
− + 𝜆2i 𝑣i = 0, 0 ≤ 𝜉 ≤ 1, i = 1, 2. (9.44)
d𝜉 2
The solution is:
𝑣i (x) = ai cosh 𝜆i 𝜉 + bi sinh 𝜆i 𝜉. (9.45)
Referring to the bar on the left, the boundary conditions are:
(AE)1 u′1 (0) = −F, (AE)1 u′1 (𝓁1 ) = −Fc
which correspond to
( ) ( )
d𝑣1 F d𝑣1 Fc
=− , =−
d𝜉 𝜉=0 (AE)1 d𝜉 𝜉=1 (AE)1
therefore the solution is
( )
F cosh 𝜆1 − Fc ∕F
𝑣1 (𝜉) = cosh 𝜆1 𝜉 − sinh 𝜆1 𝜉 . (9.46)
(AE)1 𝜆1 sinh 𝜆1
Given an initial gap g0 > 0, the force needed to close the gap, denoted by F0 , can be computed
from eq. (9.46) by setting 𝑣1 (1) = g0 ∕𝓁1 and Fc = 0. This yields:
F0 = (AE)1 𝜆1 (g0 ∕𝓁1 ) sinh 𝜆1 (9.47)
where we used cosh2 𝜆1 − sinh2 𝜆1 = 1.
Therefore for F ≤ F0 the spring rate is
F F (AE)1 sinh 𝜆1
k= = = 𝜆1 ⋅ (9.48)
U1 𝓁1 𝑣1 (0) 𝓁1 cosh 𝜆1
For any F > F0 a contact force Fc > 0 will develop and the contact condition g = 0 must be satis-
fied. To the bar on the right we apply the boundary conditions
(AE)2 u′2 (0) = −Fc , u2 (𝓁2 ) = 0
which correspond to
( )
d𝑣2 F
=− c , 𝑣2 (1) = 0.
d𝜉 𝜉=0 (AE)2
The solution is:
( )
Fc sinh 𝜆2
𝑣2 (𝜉) = cosh 𝜆2 𝜉 − sinh 𝜆2 𝜉 . (9.49)
(AE)2 𝜆2 cosh 𝜆2
From eq. (9.46) we get
𝓁1 (F − Fc cosh 𝜆1 )
U2 ≡ u1 (𝓁1 ) = 𝓁1 𝑣1 (1) = (9.50)
(AE)1 𝜆1 sinh 𝜆1
and from eq. (9.49) we get
𝓁2 Fc sinh 𝜆2
U3 ≡ u2 (0) = 𝓁2 𝑣2 (0) = ⋅ (9.51)
Using g = g0 − U2 + U3 = 0 we can write Fc = q(F − F0 ) where q is a dimensionless constant that

depends on the parameters that characterize the problem.
( )−1
𝓁2 (AE)1 𝜆1 sinh 𝜆1
q = cosh 𝜆1 + sinh 𝜆2
𝓁1 (AE)2 𝜆2 cosh 𝜆2
𝓁1 F0 𝓁1 F 𝓁1 Fc cosh 𝜆1
g= − +
(AE)1 𝜆1 sinh 𝜆1 (AE)1 𝓁1 sinh 𝜆1 (AE)1 𝜆1 sinh 𝜆1
𝓁2 Fc sinh 𝜆2
+ = 0. (9.52)
In this example it was possible to find Fc in two steps. In the first step the force needed to close
the gap, denoted by F0 , was determined. In the second step the contact force Fc was determined for
F = 𝛼F0 where 𝛼 ≥ 1. In two and three dimensions the problem is far more complicated because
it is necessary to determine the contact surfaces which depend on the contact force, the material
properties and the geometric features of the contacting bodies.
Gap elements in two dimensions

Gap elements are used for the approximation of the gap function g. The gap function measures
the distance between corresponding points in contacting bodies. The condition g > 0 indicates that
corresponding points are not in contact, whereas g < 0 indicate that the contacting bodies intersect.
We will be concerned with frictionless mechanical contact of elastic bodies in two dimensions.
The goal is to find the function Tn such that
gTn = 0, g ≥ 0, Tn(A) = −Tn , Tn(B) = −Tn (9.53)
where Tn(A) (resp. Tn(B) ) represents the normal traction acting on body A (resp. body B).
The mapping of gap elements is analogous to the mapping of quadrilateral elements. We will
assume that the contacting edges correspond to sides 1 and 3. Therefore the mapping of a gap
element is
1 − 𝜂 (A) 1 + 𝜂 (B)
x= f (𝜉) + f (𝜉) (9.54)
2 x 2 x
1 − 𝜂 (A) 1 + 𝜂 (B)
y= f (𝜉) + f (𝜉) (9.55)
2 y 2 y
where ( fx(A) (𝜉), fy(A) (𝜉)) are points on the boundary of body A and, similarly, ( fx(B) (𝜉), fy(B) (𝜉)) are
points on the boundary of body B.
We approximate the functions fx , fy by Lagrange polynomials that have roots in the optimal col-
location points, see Appendix F. For example:
(A) ∑
n
fx(A) (𝜉) ≈ f x (𝜉) = Xi(A) Li (𝜉), Xi(A) = x(𝜉i , −1) (9.56)
i=1
(A) ∑
n
fy(A) (𝜉) ≈ f y (𝜉) = Yi(A) Li (𝜉), Yi(A) = y(𝜉i , −1) (9.57)
i=1
where 𝜉i is the ith optimal collocation point and Li (𝜉) is the corresponding Lagrange interpolating
polynomial.
Example 9.9 Consider a circular (planar) body B of radius R. The coordinates of the center are
x0 = 0, y0 = R. Assume that body B penetrated body A, which is bounded by the line segment y = y0 ,
0.8
3
0.6
boundary of body B: η = 1
0.4
line of symmetry η=0
y
0.2
1
2
0 4 boundary of Body A: η = −1
−0.5 0 0.5 1 1.5 2 2.5 3

x
Figure 9.9 Example 9.9: Gap element with partial penetration.
0 ≤ x ≤ x0 . A gap element, shown in Fig. 9.9, is bounded by a circular segment at 𝜂 = 1 and a

line at 𝜂 = −1. We denote the angle subtended by the circular arc from point (x0 , y0 ) by 𝛼. Then
we have:
( )
Xi(B) = R sin 𝛼(1 + 𝜉i )∕2 (9.58)
( ( ))
Yi(B) = R 1 − cos 𝛼(1 + 𝜉i )∕2 . (9.59)
The gap element characterized by the data R = 5.0, 𝛼 = 𝜋∕6, x0 = 2.5 and y0 = 0.1 is shown in
Fig. 9.9.
Outline of the algorithm

We denote the Jacobian determinant by |J(𝜉, 𝜂)| and for the ith collocation point (i = 1, 2, … , n) we
compute
√
( )2 ( )2
Gi = sign(|J(𝜉i , 0)|) Xi(A) − Xi(B) + Yi(A) − Yi(B) . (9.60)
The gap function is approximated by

∑
n
g(𝜉) = Gi Li (𝜉), −1 ≤ 𝜉 ≤ 1. (9.61)
i=1
Explanation: When partial penetration occurs, as in Fig. 9.9, then |J(𝜉, 𝜂)| < 0 over part of the gap
element. Furthermore, the curve x = x(𝜉, 0), y = y(𝜉, 0), passes through the point (or points) where
sides 1 and 3 intersect and |J(𝜉, 0)| = 0. Note that |J(𝜉, 0)| changes sign in the point of intersection.
To show this, we write the Jacobian determinant for 𝜂 = 0:
( )
(A)
1 dfx dfx(B)
|J(𝜉, 0)| = + (−fy(A) + fy(B) )
4 d𝜉 d𝜉
( (A) )
1 dfy dfy(B)
− + (−fx(A) + fx(B) ) (9.62)
4 d𝜉 d𝜉
and observe that in the point of intersection fx(A) = fx(B) and fy(A) = fy(B) hence |J(𝜉, 0)| = 0.
We assume that the contacting bodies are sufficiently constrained to prevent rigid body displace-
ment. We apply compressive normal tractions to each of the contacting bodies over the boundary
segment where g(𝜉) < 0. Specifically,

Tn = −Cg(𝜉), Tn(A) = −Tn , Tn(B) = −Tn , 𝜉 ∈ {𝜉 | g(𝜉) < 0} (9.63)
where C is an arbitrary positive number. We compute the absolute value of the resultant of the
tractions and denote it by F. We also compute the strain energy values and denote them by U (A)
and U (B) . The estimated (average) spring rates k(A) and k(B) are determined from
F2 F2
k(A) = , k (B)
= ⋅ (9.64)
2U (A) 2U (B)
The condition for closing the gap is:
F F
𝛿 = 𝛿 (A) + 𝛿 (B) = + (9.65)
k(A) k(B)
where 𝛿 is the absolute value of the average of the gap function over the interval where g(𝜉) < 0.
From this we find:
k(A) k(B)
F= 𝛿. (9.66)
k(A) + k(B)
We define the scaling factor s = F∕F and scale Tn such that Tn → s Tn and update the coordinates
of the points in the contact zone by the scaled displacement components. For example,
Xi(A) → Xi(A) + s u(A)
x (𝜉i , −1). (9.67)
The gap function given by eq. (9.61) is re-computed with the updated coordinates and the pro-
cess is repeated. In the second and subsequent steps the tractions applied in the previous step are
incrementally updated. The process is continued until a stopping criterion is satisfied.
This algorithm does not require the element edges in the contact zone to be conforming, nor does
it restrict the number of collocation points, except that the number of collocation points should be
greater than or equal to the number of collocation points used in the mapping of elements.
Example 9.10 A classical contact problem, first solved by Hertz8 , is frictionless contact between
two elastic spheres. The formulas for maximum contact pressure pmax and the radius of the contact
area rc can be found in standard references such as [103]. Specifically, when both spheres have the
same elastic properties and 𝜈 = 0.3, the maximum pressure is
( )1∕3
(r + r )2
1 2
pmax = 0.388 PE2 (9.68)
r12 r22
where P is the compressive force, E is the modulus of elasticity, r1 and r2 are the radii of the con-
tacting spheres. The radius of the contact area is
( )1∕3
Pr1 r2
ac = 1.109 ⋅ (9.69)
E(r1 + r2 )
These equations are based on the assumption that rc << min (r1 , r2 ).
In this example we let r1 = 100 mm, r2 = 25 mm, E = 7.17 × 104 MPa, 𝜈 = 0.3, P = 800 N and
solve it as an axisymmetric problem. Because the contact area, which is the region of primary
interest, is very small in comparison with the radii of the spheres, the starting mesh is graded
in geometric progression toward the contact area using the grading factor 0.15. It is then progres-
sively refined by uniformly subdividing the standard elements to obtain a sequence of quasiuniform
8 Heinrich Rudolf Hertz 1857–1894.

z Tn
A E
r2 500
B
r1 400
300
B
200
100
D 0
C
r
(a) (b)
Figure 9.10 Axisymmetric model of two elastic spheres in frictionless contact. (a) Notation and 300
element mesh. (b) Contours of the von Mises stress in the neighborhood of the contact area.
mesh. One mesh in the sequence, consisting of 250 quadrilateral and 50 triangular elements, is
shown in Fig. 9.10(a). The problem was solved by iteration using the stopping criterion
|p(k) (k−1) (k)

max − pmax | ≤ 0.01|pmax |
where pmax is the maximum contact pressure and k is the iteration counter.
Referring to Fig. 9.10(a), the boundary conditions are as follows: On boundary segments AB
and BC the displacement component ur = 0, the shearing traction is zero. On segment CD the
displacement component uz = 0. The circular arcs DB and BE are traction-free and segment EA
has the normal traction Tn = Tz = −P∕(r22 𝜋), the shearing traction is zero.
The deformed configuration (on 1:1 scale) and the contours of the von Mises stress in the neigh-
borhood of the contact area are shown in Fig. 9.10(b). The maximum occurs in the point r = 0,
z = 99.68 mm.
The computed values of the maximum von Mises stress 𝜎 max (MPa), its location z (mm), the
maximum contact pressure pmax (MPa) and the radius of the contact are rc (mm) are listed for the
quasiuniform sequence of meshes with p = 8 assigned to all elements in Table 9.3. It is seen that the
contact pressure and the maximum von Mises stress are independent of the number of degrees of
freedom to three significant digits. The radius of the contact area was determined to two significant
digits only. This is because the contact pressure oscillates with a small amplitude at the boundary
of the contact area.
Table 9.3 Example 9.10: Computed values of the

quantities of interest with p = 8 assigned to all
elements.
M N 𝝈 max Loc. z pmax rc
108 12985 518.79 99.683 841.9 0.717

300 35721 518.86 99.680 842.3 0.710
588 69721 519.00 99.679 842.2 0.714
Classical 843.7 0.673

Contact pressure (MPa)
500
pmax
1000
0 0.2 0.4 0.6 ac 0.8 1
r (mm)
Figure 9.11 Contact pressure.
The contact pressure is plotted in Fig. 9.11. The variation in the contact pressure with respect to
N is so small that the differences are not visible on this diagram. Note the small oscillation at the
boundary of the contact area. The computed values of the radius of the contact zone (rc ) differ from
the radius (ac ) computed from the classical solution given by eq. (9.69). The reason for this is that
in the classical solution the contacting spheres are approximated by paraboloids of revolution. No
such approximation was necessary in the numerical solution.
Remark 9.4 From the point of view of strength calculations the usual quantities of engineering
interest in contact problems are the maximum shearing stress (for ductile materials) or the maxi-
mum tensile stress (for brittle materials). The maximum shearing stress, or maximum von Mises
stress, is at a small distance from the contact area, as shown in Fig. 9.10(b). The maximum tensile
stress occurs on the spherical boundary outside of the contact area.
9.3 Chapter summary
The formulation of an idea of physical reality starts with simple, usually linear, models with several
restrictive assumptions. The magnitude of the influence of those assumptions on the quantities of
interest have to be estimated and, when necessary, the model has to be revised. Any model should
be viewed as a special case of a more comprehensive model. The objective is to identify the sim-
plest model, given the goals of computation and the corresponding acceptable error tolerances. A
practical way for doing this is to remove limitations and estimate their effects on the quantities
of interest. This is feasible in practice only if the implementation allows seamless transition from
linear to nonlinear models and from one nonlinear model to another.
Analysts must be mindful of the fact that nonlinear models require more information about mate-
rial properties and boundary conditions than linear models. This increases the complexity of the
model as well as the uncertainties associated with the additional information.
289
Appendix A
Definitions
Definition A.1 A function is analytic if and only if its Taylor series about any point x0 ∈ Ω con-
verges to the function in some neighborhood of x0 .
Definition A.2 The real coordinate space of n dimensions, written as ℝn , is the generalization
of the familiar one-, two- and three-dimensional Euclidean space to n dimensions.
Definition A.3 The Euclidean norm of a vector a ∈ ℝn is denoted by ||a||2 and defined as follows:
( n )1∕2
def ∑
‖a‖2 = a2i . (A.1)
i=1
Definition A.4 The terms “supremum” (abbreviated sup) and “infimum” (abbreviated inf) are
generalizations of maximum and minimum, respectively. They are useful for characterizing sets
that do not have a maximum or minimum. For instance, the negative real numbers do not have
a maximum. However, the supremum is uniquely defined: it is zero. The supremum is also called
least upper bound (LUB) and the infimum is also called greatest lower bound (GLB). The terms
“essential supremum” (abbreviated ess sup) and “essential infimum” (abbreviated ess inf) are used
when sets of measure zero are excluded.
Definition A.5 The delta function 𝛿(x), also called Dirac’s delta function1 , has various interpre-
tations. One such interpretation is that it is zero everywhere except in the point x = 0 where it is
undefined. However, its integral is the unit step function. We will understand 𝛿(x) to mean
∞ a+𝜖
𝛿(x − a) f (x) dx = lim 𝛿(x − a) f (x) dx = f (a) (A.2)
∫−∞ 𝜖→0 ∫a−𝜖
where 𝜖 > 0.
A.1 Normed linear spaces, linear functionals and bilinear forms

In the following the definitive properties of normed linear spaces, linear functionals and bilinear
forms are listed. 𝛼 and 𝛽 denote real numbers.
1 Paul Adrien Maurice Dirac 1902–1984.
290 Appendix A Definitions
A.1.1 Normed linear spaces

A normed linear space X is a family of elements u, 𝑣, … which have the following properties:
1. If u ∈ X and 𝑣 ∈ X then (u + 𝑣) ∈ X.
2. If u ∈ X then 𝛼u ∈ X.
3. u+𝑣=𝑣+u
4. u + (𝑣 + 𝑤) = (u + 𝑣) + 𝑤
5. There is an unique element in X, denoted by 0, such that u + 0 = u for any u ∈ X.
6. Associated with every element u ∈ X an unique element −u ∈ X such that u + (−u) = 0.
7. 𝛼(u + 𝑣) = 𝛼u + 𝛼𝑣
8. (𝛼 + 𝛽) u = 𝛼u + 𝛽u
9. 𝛼(𝛽u) = (𝛼𝛽)u
10. 1⋅u=u
11. 0⋅u=0
12. With every u ∈ X we associate a real number ||u||X , called the norm. The norm has the following
properties:
(a) ||u + 𝑣||X ≤ ||u||X + ||𝑣||X . This is called the triangle inequality.
(b) ||𝛼u||X = |𝛼|||u||X .
(c) ||u||X ≥ 0.
(d) ||u||X ≠ 0 if u ≠ 0.
A seminorm has properties (a) to (c) of the norm but lacks property (d). See, for example,
eq. (A.9).
A.1.2 Linear forms

Let X be a normed linear space and F(𝑣) a process which associates with every 𝑣 ∈ X a real number
F(𝑣). F(𝑣) is called a linear form or linear functional on X if it has the following properties:
1. F(𝑣1 + 𝑣2 ) = F(𝑣1 ) + F(𝑣2 )

2. F(𝛼𝑣) = 𝛼F(𝑣)
3. |F(𝑣)| ≤ C||𝑣||X with C independent of 𝑣. The smallest possible value of C is called the norm of F.
A.1.3 Bilinear forms

Let X and Y be normed linear spaces and B(u, 𝑣) a process which associates with every u ∈ X and
𝑣 ∈ Y a real number B(u, 𝑣). B(u, 𝑣) is a bilinear form on X × Y if it has the following properties:
1. B(u1 + u2 , 𝑣) = B(u1 , 𝑣) + B(u2 , 𝑣)

2. B(u, 𝑣1 + 𝑣2 ) = B(u, 𝑣1 ) + B(u, 𝑣2 )
3. B(𝛼u, 𝑣) = 𝛼B(u, 𝑣)
4. B(u, 𝛼𝑣) = 𝛼B(u, 𝑣)
5. |B(u, 𝑣)| ≤ C ||u||X ||𝑣||Y with C independent of u and 𝑣. The smallest possible value of C is called
the norm of B.
The space X is called trial space and functions u ∈ X are called trial functions. The space
Y is called test space and functions 𝑣 ∈ Y are called test functions. B(u, 𝑣) is not necessarily
symmetric.
A.2 Convergence in the space X 291
A.2 Convergence in the space X
A sequence of functions un ∈ X (n = 1, 2, …) converges in the space X to the function u ∈ X if for

every 𝜖 > 0 there is a number n𝜖 such that for any n > n𝜖 the following relationship holds:
||u − un ||X < 𝜖. (A.3)
A.2.1 The space of continuous functions

Let Ω ∈ ℝn be an open bounded domain. The space of continuous functions defined on Ω is denoted
by C0 (Ω). The norm associated with C0 is
‖f ‖C0 (Ω) = sup |f (x)|. (A.4)
x∈Ω
The space Ck (Ω) is the set of functions defined on Ω that have the property that all derivatives, up
to and including the kth derivative, lie in C0 (Ω). The set Ck (Ω) is a proper subset of Ck−1 (Ω) for
k > 0. The set C∞ (Ω) is the set of analytic functions.
A.2.2 The space Lp (𝛀)

The space Lp (Ω) is defined as the set of functions on Ω that satisfy
( )1∕p
‖f ‖Lp (Ω) = |f |p dx <∞ (A.5)
∫Ω
where the integral is the Lebesque2 integral. The space L2 (Ω) is the set of square integrable func-
tions. The space L∞ (Ω) is defined by
‖f (x)‖L∞ (Ω) = ess sup|f (x)| < ∞ (A.6)
x∈Ω
see Definition A.4.
A.2.3 Sobolev space of order 1

Sobolev3 spaces play a central role in the mathematical analysis of the finite element method. For
basic definitions and properties of Sobolev spaces we refer to [55, 59, 84]. The Sobolev space of
order 1 in two dimensions is defined as the set
{ }
def | ‖ 𝜕u ‖2 ‖ 𝜕u ‖2
H (Ω) = u | ‖u‖L2 (Ω) + ‖ ‖
1 2 ‖ ‖ ‖
+‖ ‖ ‖ <∞ (A.7)
| ‖ 𝜕x ‖L2 (Ω) ‖ 𝜕y ‖L2 (Ω)
The Sobolev norm of order one in two dimensions is defined by
( )1∕2
def ‖ 𝜕u ‖2 ‖ 𝜕u ‖2
‖u‖H 1 (Ω) = ‖u‖2L2 (Ω) + ‖ ‖
‖ 𝜕x ‖ 2 + ‖ ‖
‖ ‖ (A.8)
‖ ‖L (Ω) ‖ 𝜕y ‖L2 (Ω)
the first order seminorm is
( )1∕2
def ‖ 𝜕u ‖ ‖ 𝜕u ‖2
2
|u|H 1 (Ω) = ‖ ‖‖ ‖ ‖
+‖ ‖ ‖ (A.9)
‖ 𝜕x ‖L2 (Ω) ‖ 𝜕y ‖L2 (Ω)
2 Henri Léon Lebesgue 1871–1941.

3 Sergei Lvovich Sobolev 1908–1989.
292 Appendix A Definitions
and we define
def
H01 (Ω) = {u | u ∈ H 1 (Ω), u = 0 on 𝜕Ω}. (A.10)
The Sobolev norm and the energy norm are equivalent in the sense that there exist two real numbers
0 < C1 ≤ C2 , independent of u, such that
C1 ‖u‖H 1 (Ω) ≤ ‖u‖E(Ω) ≤ C2 ‖u‖H 1 (Ω) . (A.11)
In the mathematical literature a priori estimates of rates of convergence are given in terms of the
error measured in H 1 norm. Those estimates are valid in the energy norm also.
The standard notation for Sobolev spaces is W k,s (Ω) where the index k is the number of deriva-
tives and index s identifies the associated norm Ls (Ω). The special case s = 2 is a Hilbert space4
denoted by H k (Ω). We have given the definition for the Sobolev space W 1,2 (Ω) ≡ H 1 (Ω).
A.2.4 Sobolev spaces of fractional index

The definition of Sobolev spaces is extended to fractional indices k = m + 𝜇 where m is an integer
and 0 < 𝜇 < 1. Sobolev spaces of fractional index play a fundamentally important role in the theory
of finite element methods. These spaces are used for characterization of the smoothness of func-
tions. The a priori estimates of the rate of convergence (in energy norm) are given in terms of the
indices of these spaces.
A detailed discussion of Sobolev spaces is well outside of the scope of this book, however an intro-
ductory discussion of special cases in one and two dimensions is necessary for the understanding
of theorems pertinent to the convergence properties of the finite element method. For introductory
reading on Sobolev spaces we recommend [34, 59, 84].
In one dimension the function
u = x𝜆 , 𝜆 > 1∕2, x∈I
is in H 1+𝜇 (I) if x𝜆−1−𝜇 is square integrable on I. For I = (0, 1) we have
1
x2𝜆−2−2𝜇 dx < ∞
∫0
therefore
[ ]1
x2𝜆−1−2𝜇
<∞
2𝜆 − 1 − 2𝜇 0
and
2𝜆 − 1 − 2𝜇 > 0.
Selecting the largest value of 𝜇 such that u = x𝜆 is square integrable we have
𝜇 = 𝜆 − 1∕2 − 𝜖
where 𝜖 is an arbitrarily small positive number. For example, u = x2∕3 lies in H 7∕6−𝜖 (I) where
I = (0, 1).
In two dimensions let
u = r 𝜆 𝜙(𝜃), (r, 𝜃) ∈ Ω = (0, 1) × (−𝛼∕2, 𝛼∕2)
4 David Hilbert 1862–1943.

A.3 The Schwarz inequality for integrals 293
where 𝜙(𝜃) is a smooth function and 0 < 𝛼 ≤ 𝜋. The function u is in H 1+𝜇 (Ω) if r 𝜆−1−𝜇 is square
integrable on Ω, that is,
𝛼∕2 1 𝛼∕2 [ 2𝜆−2𝜇 ]1
r
r 2𝜆−2−2𝜇 𝜙2 (𝜃) rdrd𝜃 = 𝜙2 (𝜃) d𝜃 <∞
∫−𝛼∕2 ∫0 ∫−𝛼∕2 2𝜆 − 2𝜇 0
therefore 𝜆 − 𝜇 > 0 and the largest 𝜇 for which this holds is 𝜇 = 𝜆 − 𝜖 where 𝜖 is an arbitrarily small
positive number. For example, the function u = r 2∕3 cos(𝜃) lies in H 5∕3−𝜖 (Ω).
A.3 The Schwarz inequality for integrals
Assume that f (x) ∈ L2 (I) and g(x) ∈ L2 (I) where I = (a, b). Then:
( )1∕2 ( )1∕2
| b | b b
| |
| fg dx | ≤ 2
f dx 2
g dx . (A.12)
|∫a | ∫a ∫a
| |
This is the Schwarz5 inequality for integrals. To prove this we observe that:
b
(f + 𝜆g)2 dx ≥ 0 for any 𝜆 (A.13)
∫a
and therefore:
b b b
f 2 dx + 2𝜆 fgdx + 𝜆2 g2 dx ≥ 0 for any 𝜆. (A.14)
∫a ∫a ∫a
On the left of this inequality is a quadratic expression for 𝜆. To find the roots of this expression we
need only to compute:
√( ) 2
b b b b
− ∫a fgdx ± ∫a fgdx − ∫a g2 dx ∫a f 2 dx
𝜆= b
⋅ (A.15)
∫a g2 dx
Denoting the roots by 𝜆1 and 𝜆2 , eq. (A.14) can be written as:
(𝜆 − 𝜆1 )(𝜆 − 𝜆2 ) ≥ 0
and we observe that the roots cannot be real and simple because then we could select any 𝜆 such
( )( )
that 𝜆1 < 𝜆 < 𝜆2 and we would have 𝜆 − 𝜆1 𝜆 − 𝜆2 < 0. Therefore the radicand must be less
than or equal to zero. This completes the proof.
5 Karl Hermann Amandus Schwarz 1843–1921.

295
Appendix B
Proof of h-convergence
A proof of h-convergence in one dimension, p = 1, subject to the condition that the second deriva-
tive of the exact solution is bounded, is presented here with the objective to provide an introduction,
in the simplest possible setting, to the formulation of a priori error estimates.
Consider the following problem: Find uEX ∈ E0 (I) where I = (0, 𝓁) such that:
𝓁 𝓁
(𝜅u′EX 𝑣′ + cuEX 𝑣) dx = f 𝑣 dx for all 𝑣 ∈ E0 (I) (B.1)
∫0 ∫0
assuming that 𝜅, c and f are such that uEX , u′EX and u′′EX are bounded, continuous functions with
|u′′EX | ≤ C on the interval I, that is, uEX ∈ C2 (I).
Subdivide the domain I into n elements of equal length. The length of each element is: h = 𝓁∕n.
Let un be the linear interpolant of uEX , that is un is a continuous, piecewise linear function such
that:
un (jh) = uEX (jh), j = 0, 1, … , n. (B.2)
We denote Ik = ((k − 1)h, kh) and the error of interpolation on Ik by ek :
def
ek (x) = uEX (x) − un (x), x ∈ Ik , k = 1, 2, … , n. (B.3)
Because ek (x) vanishes at the endpoints of the element, there is a point xk where |ek | is maximal.
′
At this point ek = 0, see Fig. B.1. Since un is linear on Ik , u′′n = 0 therefore:
x x
′ ′′
ek (x) = ek (t) dt = u′′EX (t) dt, x ∈ Ik (B.4)
∫xk ∫xk
and since |u′′EX | ≤ C, we get:
′
|ek (x)| ≤ h C, x ∈ Ik . (B.5)
Let us expand ek into a Taylor series about the point xk and let us assume that xk is located such
that: kh − xk ≤ h∕2. We now have:
′ 1 ′′
ek (kh) = 0 = ek (xk ) + (kh − xk )ek (xk ) + (kh − xk )2 ek (t) (B.6)
2
′
where t is a point between xk and x = kh. Because ek (xk ) = 0:
1 ′′ h2
max |ek (xk )| = |kh − xk |2 |ek (t)| ≤ C. (B.7)
2 8
If xk is closer to (k − 1)h than to kh then we write the Taylor series expression for ek ((k − 1)h) instead
of ek (kh) and obtain the same result. Equation (B.7) is a basic result of interpolation theory. In view
296 Appendix B Proof of h-convergence
Figure B.1 The error ek = uEX − uFE .
_
xk
of (B.4) and (B.7) the strain energy of the error of the interpolant e is:
𝓁( ) kh ( )
1∑
n
1 ′ 2 ′ 2
U(e) = 𝜅(e )2 + ce dx = 𝜅(ek )2 + cek dx
2 ∫0 2 k=1 ∫(k−1)h
( ( 2 )2 )
1 h
≤ nh K1 (Ch) + K2 2
C (B.8)
2 8
where K1 and K2 are constants chosen so that 𝜅(x) ≤ K1 and c(x) ≤ K2 on the interval I. Observe
that K1 , K2 are independent of h. Since nh = 𝓁, there is a constant K such that:
1
U(e) ≤ K𝓁C2 h2 . (B.9)
2
Finally, because the energy norm of the error of the finite element solution e = uEX − uFE is less
than or equal to the energy norm of the error of the interpolant e (see Theorem 1.4) we have:
def √
‖e‖E(Ω) = U(e) ≤ kCh (B.10)
where k depends on 𝜅, c and f but is independent of h. This is in the form of a typical a priori
estimate. A priori estimates indicate how fast the error changes as the discretization is changed,
given information about the smoothness of the exact solution. For example, the smoothness of
the exact solution was characterized by |u′′EX | ≤ C in the foregoing discussion. In general C is not
known a priori.
297
Appendix C
Convergence in 3D: Empirical results
The problem of elasticity was solved on the Fichera domain using the same input data as in refer-
ence [3]. The goal was to compare the empirical rates of convergence of the h- and p-versions in
energy norm, with respect to the number of degrees of freedom, and to estimate the Sobolev index
k of the exact solution. The computations were performed by the finite element analysis software
STRIPE, developed by the Aeronautical Research Institute of Sweden.
Input data
The domain is defined by
def
Ω = {(X, Y , Z) | (X, Y , Z) ∈ (−50, 50)3 ∖[0, 50)3 }. (C.1)
Using the notation shown in Fig. 4.11, on the boundary surfaces that lie in the planes X = 0, Y = 0,
Z = 0 and Y = −50, Z = −50 the tractions are zero: Tx = Ty = Tz = 0. On the boundary surface that
lies in the X = −50 plane Tx = −1, Ty = Tz = 0. On the boundary surfaces that lie in the planes X =
50, Y = 50 and Z = 50 symmetry boundary conditions are prescribed: The normal displacements
and shearing tractions are zero. The modulus of elasticity is E = 1.0E3 and Poisson’s ratio is 0.3.
Reference solution
In order to establish a reference solution, a geometric mesh was constructed. The mesh is charac-
terized by the sequence
{
0 for i = 1
xi = (C.2)
50q m+1−i for i = 2, 3, … , m + 1
which indicates the location of nodes along the x axis. In the same way nodes were located along
the y and z axes. The nodes are in the points of intersection of the planes X = ±xi , Y = ±yj , Z = ±zk ,
(i, j, k = 1, 2, … , m + 1). The number of elements in the mesh is M(Δ) = 7m3 .
The results for m = 8 are shown in Table C.1. It is seen that the potential energy converged to
8 digits precision. The extrapolated value of 𝜋p was used for reference in calculating the rates of
convergence for the p- and h-extensions, denoted by 𝛽p and 𝛽h respectively, in tables C.1 and C.2.
The estimated values of 𝛽, computed by the formula given by eq. (1.102), are shown in Fig. C.1.
Uniform meshes were used for both h- and p-extensions.
The results for h-extension, p = 2, trunk space, are displayed in Table C.2. The parameter m rep-
resents the number of edges along the positive coordinate axes. Therefore h = 50∕m. The values of
kh were computed using the formula (4.35).
298 Appendix C Convergence in 3D: Empirical results
Table C.1 Fichera domain (elasticity). Estimation of the reference value of the potential
energy, p-Convergence, M(Δ) = 3584 (m = 8), geometric mesh q = 1∕7, trunk space.
p N 𝝅p 𝜷p e (%)
3 84,048 −7.1017674042E + 02 − 1.9934

4 153,792 −7.1037322898E + 02 0.9853 1.0992
5 257,520 −7.1044463079E + 02 1.7293 0.4507
6 405,984 −7.1045599535E + 02 1.7004 0.2079
7 609,936 −7.1045850634E + 02 2.0931 0.0887
8 880,128 −7.1045892986E + 02 1.9361 0.0436
9 1,227,312 −7.1045902640E + 02 1.8878 0.0233
10 1,662,240 −7.1045905243E + 02 1.8294 0.0134
11 2,195,664 −7.1045906029E + 02 1.8294 0.0080
extrapolated: −7.1045906487E + 02
0.45 p-ext., M = 189 prod. space

p-ext., M = 189 trunk space
0.4
p-ext., M = 1512 trunk space
estimated values of β
0.35
p-ext., M = 1512 prod. space

0.3
0.25
h-ext., p = 2 prod. space
0.2
h-ext., p = 2 trunk space
0.15
0.1
104 105 106 107
N
Figure C.1 Fichera domain, elasticity. Convergence paths on uniform meshes.
Discussion
The estimation of kh is based on eq. (1.91), whereas the estimation of 𝛽h is based on eq. (1.92). The
two forms are similar and, since in three dimensions N ∝ h−3 , we expect that
kh − 1
⋅ 𝛽h ≈
3
For example, in the entry corresponding to m = 66 in Table C.2, 𝛽h estimated from kh is 0.1838
which is close to the estimate of 0.1847 from eq. (1.92).
Convergence in 3D: Empirical results 299
Table C.2 Fichera domain. h-Convergence, uniform mesh, p = 2 (trunk space).
m N 𝝅h kh 𝜷h (er )E (%)
6 21,132 −7.08307829E + 02 − − 5.5027

9 67,905 −7.09107428E + 02 1.5731 0.1991 4.3617
12 156,960 −7.09483558E + 02 1.5668 0.1946 3.7055
15 301,905 −7.09700398E + 02 1.5633 0.1922 3.2678
18 516,348 −7.09840746E + 02 1.5610 0.1906 2.9501
21 813,897 −7.09938678E + 02 1.5593 0.1895 2.7064
24 1,208,160 −7.10010726E + 02 1.5580 0.1886 2.5121
27 1,712,745 −7.10065854E + 02 1.5570 0.1880 2.3526
30 2,341,260 −7.10109336E + 02 1.5561 0.1874 2.2187
33 3,107,313 −7.10144469E + 02 1.5554 0.1870 2.1043
36 4,024,512 −7.10173423E + 02 1.5548 0.1866 2.0051
39 5,106,465 −7.10197676E + 02 1.5543 0.1863 1.9181
42 6,366,780 −7.10218275E + 02 1.5538 0.1861 1.8410
45 7,819,065 −7.10235977E + 02 1.5534 0.1858 1.7720
48 9,476,928 −7.10251347E + 02 1.5530 0.1856 1.7099
51 11,353,977 −7.10264811E + 02 1.5527 0.1854 1.6535
54 13,463,820 −7.10276699E + 02 1.5524 0.1853 1.6021
57 15,820,065 −7.10287268E + 02 1.5521 0.1851 1.5550
60 18,436,320 −7.10296725E + 02 1.5519 0.1850 1.5116
63 21,326,193 −7.10305233E + 02 1.5517 0.1848 1.4715
66 24,503,292 −7.10312926E + 02 1.5514 0.1847 1.4342
reference value: −7.10459065E + 02
It is known that in two dimensions the limit value of the ratio 𝛽p ∕𝛽h , under conditions that are
normally satisfied in practice, is 2. No analogous theorem exists for three dimensions. Referring to
the apparent limit values of 𝛽 obtained by p- and h-extensions in Fig. C.1, we expect that the same
can be proven for three dimensions as well.
A direct solver was used. It was verified that round-off errors did not influence the computations.
Perturbations caused by round-off affect not fewer than the twelfth digit in the computed potential
energy values.
A surprising finding of this investigation was that entry into the asymptotic range occurs at
very high values of N. As seen in Table C.2, at N = 24.5 million, kh and 𝛽h are still decreasing
monotonically.
301
Appendix D
Legendre polynomials
The Legendre polynomials Pn (𝜉) are solutions of the Legendre differential equation for n = 0, 1, 2, …:
(1 − 𝜉 2 )y′′ − 2𝜉y′ + n(n + 1)y = 0, −1 ≤ 𝜉 ≤ 1. (D.1)
The first eight Legendre polynomials are:
P0 (𝜉) = 1 (D.2)
P1 (𝜉) = 𝜉 (D.3)
1 2
P2 (𝜉) = (3𝜉 − 1) (D.4)
2
1
P3 (𝜉) = (5𝜉 3 − 3𝜉) (D.5)
2
1
P4 (𝜉) = (35𝜉 4 − 30𝜉 2 + 3) (D.6)
8
1
P5 (𝜉) = (63𝜉 5 − 70𝜉 3 + 15𝜉) (D.7)
8
1
P6 (𝜉) = (231𝜉 6 − 315𝜉 4 + 105𝜉 2 − 5) (D.8)
16
1
P7 (𝜉) = (429𝜉 7 − 693𝜉 5 + 315𝜉 3 − 35𝜉). (D.9)
16
Note that Legendre polynomials of even (resp. odd) degree are symmetric (resp. antisymmetric)
functions:
Pn (−𝜉) = (−1)n Pn (𝜉) (D.10)
and Pn (1) = 1 for all n.
Legendre polynomials can be generated from the recursion formula:
(n + 1)Pn+1 (𝜉) = (2n + 1)𝜉Pn (𝜉) − nPn−1 (𝜉), n = 1, 2, … (D.11)
and Legendre polynomials satisfy the following relationship:
′ ′
(2n + 1)Pn (𝜉) = Pn+1 (𝜉) − Pn−1 (𝜉), n = 1, 2, … (D.12)
where the primes represent differentiation with respect to 𝜉. Legendre polynomials satisfy the fol-
lowing orthogonality property:
⎧ 2
+1
⎪ for i = j
Pi (𝜉) Pj (𝜉) d𝜉 = ⎨ 2i + 1 (D.13)
∫−1 ⎪0 for i ≠ j.
⎩
302 Appendix D Legendre polynomials
All roots of Legendre polynomials lie in the interval −1 < 𝜉 < 1. The n roots of Pn (𝜉) are the abscis-
sas 𝜉i for the n-point Gaussian integration:
1 ∑
n
f (𝜉) d𝜉 ≈ 𝑤i f (𝜉i ). (D.14)
∫−1 i=1
Legendre polynomials have the property Pn (1) = 1, Pn (−1) = (−1)n .
D.1 Shape functions based on Legendre polynomials

The first five shape functions based on Legendre polynomials are displayed in Fig. 1.4. Additional
shape functions are listed below.
3
N6 (𝜉) = √ 𝜉(𝜉 2 − 1)(7𝜉 2 − 3) (D.15)
8 √2
1 11 2
N7 (𝜉) = (𝜉 − 1)(21𝜉 4 − 14𝜉 2 + 1) (D.16)
16 2
√
1 13
N8 (𝜉) = 𝜉(𝜉 2 − 1)(33𝜉 4 − 30𝜉 2 + 5) (D.17)
16 2
√
1 15 2
N9 (𝜉) = (𝜉 − 1)(429𝜉 6 − 495𝜉 4 + 135𝜉 2 − 5) (D.18)
128 2
⋮
1
Ni (𝜉) = √ (Pi−1 (𝜉) − Pi−3 (𝜉)). (D.19)
2(2i − 3)
The first eight shape functions are shown in Fig. D.1.
N1 N2
0.5
N8
N6 N7 N4 N5
–0.5
N3
–1
–1 –0.8 –0.6 –0.4 –0.2 0 0.2 0.4 0.6 0.8 1
ξ
Figure D.1 The first eight shape functions based on Legendre polynomials.
303
Appendix E
Numerical quadrature
In the finite element method the terms of the coefficient matrices and right hand side vectors are
computed by numerical quadrature. Most commonly Gaussian1 quadrature is used, in some cases
the Gauss-Lobatto quadrature is used. In one dimension the domain of integration is the standard
element Ist . An integral expression on the standard element is approximated by the sum
+1 ∑
n
f (𝜉) d𝜉 ≈ 𝑤i f (𝜉i ) + Rn (E.1)
∫−1 i=1
where 𝑤i are the weights; 𝜉i are the abscissas and Rn is the error term. The abscissas and weights
are symmetric with respect to the point 𝜉 = 0.
To evaluate an integral on other than the standard domain, the mapping function defined by
eq. (1.60) is used for transforming the domain of integration to the standard domain. For example:
x2 +1
x − x1 1−𝜉 1+𝜉
F(x) dx = F(Q(𝜉)) 2 d𝜉 where Q(𝜉) = x + x⋅
∫x1 ∫−1 2 2 1 2 2
⏟⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏟
f (𝜉)
E.1 Gaussian quadrature

In Gaussian quadrature the abscissa xi is the ith zero of Legendre polynomial Pn . The weights are
computed from:
2
𝑤i = (E.2)
(1 − xi2 )[Pn′ (xi )]2
see, for example Olver et al. [71]. The abscissas and weights for Gaussian quadrature are listed in
Table E.1 up to n = 8. The error term is:
22n+1 (n!)4
Rn = f (2n) (𝜁) − 1 < 𝜁 < +1
(2n + 1)[(2n)!]3
where f (2n) is (2n)th derivative of f . It is seen from the error term that if f (𝜉) is a polynomial of
degree p and Gaussian quadrature is used then the integral will be exact (up to round-off errors)
provided that n ≥ (p + 1)∕2. For example, to integrate a polynomial of degree 5, n = 3 is sufficient.
For other than polynomial functions the rate of convergence depends on how well the integrand
can be approximated by polynomials. It can be shown that if f (𝜉) is a continuous function on Ist
then the sum in eq. (E.1) converges to the true value of the integral.
1 Johann Carl Friedrich Gauss 1777–1855.
304 Appendix E Numerical quadrature
Table E.1 Abscissas and weights for Gaussian quadrature
n ±𝝃i 𝒘i
2 0.57735 02691 89626 1.00000 00000 00000

3 0.00000 00000 00000 0.88888 88888 88889
0.77459 66692 41483 0.55555 55555 55556
4 0.33998 10435 84856 0.65214 51548 62546
0.86113 63115 94053 0.34785 48451 37454
5 0.00000 00000 00000 0.56888 88888 88889
0.53846 93101 05683 0.47862 86704 99366
0.90617 98459 38664 0.23692 68850 56189
6 0.23861 91860 83197 0.46791 39345 72691
0.66120 93864 66265 0.36076 15730 48139
0.93246 95142 03152 0.17132 44923 79170
7 0.00000 00000 00000 0.41795 91836 73469
0.40584 51513 77397 0.38183 00505 05119
0.74153 11855 99394 0.27970 53914 89277
0.94910 79123 42759 0.12948 49661 68870
8 0.18343 46424 95650 0.36268 37833 78362
0.52553 24099 16329 0.31370 66458 77887
0.79666 64774 13627 0.22238 10344 53374
0.96028 98564 97536 0.10122 85362 90376
The integration procedure can be extended directly to the standard quadrilateral element and the
standard hexahedral element. For the standard quadrilateral element:
n𝜉 n𝜂
+1 +1 ∑ ∑
f (𝜉, 𝜂) d𝜉d𝜂 = 𝑤i 𝑤j f (𝜉i , 𝜂j ) (E.3)
∫−1 ∫−1 i=1 j=1
where n𝜉 (resp. n𝜂 ) is the number of quadrature points along the 𝜉 (resp. 𝜂 axis). Usually but not
necessarily n𝜉 = n𝜂 is used. Analogously, for the standard hexahedral element:
n𝜉 n𝜂 n𝜁
+1 +1 +1 ∑ ∑∑
f (𝜉, 𝜂, 𝜁) d𝜉d𝜂d𝜁 = 𝑤i 𝑤j 𝑤k f (𝜉i , 𝜂j , 𝜁k ). (E.4)
∫−1 ∫−1 ∫−1 i=1 j=1 k=1
E.2 Gauss-Lobatto quadrature

In the Gauss-Lobatto quadrature the abscissas are as follows: x1 = −1, xn = 1 and for
i = 2, 3, … , n − 1 the (i − 1)th zero of Pn−1
′
(x) where Pn−1 (x) is the (n − 1)th Legendre polynomial.
The weights are:
⎧ 2
for i = 1 and i = n
⎪ n(n − 1)
𝑤i = ⎨ 2 (E.5)
⎪ for i = 2, 3, … , (n − 1).
⎩ n(n − 1)[P n−1 (xi )]
2
E.2 Gauss-Lobatto quadrature 305
Table E.2 Abscissas and weights for Gauss-Lobatto quadrature
n ±𝝃i 𝒘i
2 1.00000 00000 00000 1.00000 00000 00000

3 0.00000 00000 00000 1.33333 33333 33333
1.00000 00000 00000 0.33333 33333 33333
4 0.44721 35954 99958 0.83333 33333 33333
1.00000 00000 00000 0.16666 66666 66667
5 0.00000 00000 00000 0.71111 11111 11111
0.65465 36707 07977 0.54444 44444 44444
1.00000 00000 00000 0.10000 00000 00000
6 0.28523 15164 80645 0.55485 83770 35486
0.76505 53239 29465 0.37847 49562 97847
1.00000 00000 00000 0.06666 66666 66667
7 0.00000 00000 00000 0.48761 90476 19048
0.46884 87934 70714 0.43174 53812 09863
0.83022 38962 78567 0.27682 60473 61566
1.00000 00000 00000 0.04761 90476 19048
8 0.20929 92179 02479 0.41245 87946 58704
0.59170 01814 33142 0.34112 26924 83504
0.87174 01485 09607 0.21070 42271 43506
1.00000 00000 00000 0.03571 42857 14286
The abscissas and weights for Gauss-Lobatto quadrature are listed in Table E.2 up to n = 8.
The error term is:
−n(n − 1)3 22n−1 [(n − 2)!]4 (2n−2)
Rn = f (𝜁), −1 < 𝜁 < +1
(2n − 1)[(2n − 2)!]3
from which it follows that if f (𝜉) is a polynomial of degree p and Gauss-Lobatto quadrature is used
then the integral will be exact (up to round-off errors) provided that n ≥ (p + 3)∕2.
307
Appendix F
Polynomial mapping functions
The use of isoparametric mapping of elements for polynomial degrees 1 and 2 was introduced in the
1960s. The algorithm for two-dimensional elements was outlined in Section 3.4.1. In this section we
address extension of the idea of isoparametric mapping to polynomial functions of degree greater
than 2 and its combination with the blending function method.
Specifically, we are interested in approximating arbitrary continuous functions such as x2 (𝜂)
and y2 (𝜂) in equations (3.46) and (3.47) by polynomial interpolating functions on the interval
−1 < 𝜂 < 1.
In the interest of simplicity in notation we will seek a polynomial interpolation function of degree
n to f (𝜉) ∈ C0 (Ist ) that is nearly optimal in maximum norm. We denote the polynomial interpolation
functions of degree p by 𝑤p (𝜉) and the set of interpolation points, called the nodal set, by
T = {𝜉1 = −1, 𝜉2 , 𝜉3 , … 𝜉p , 𝜉p+1 = 1}
and the interpolating polynomial is
∑
p+1
𝑤p = f (𝜉k )Nk (𝜉) ≡ T f
k=1
where Nk (𝜉) are the Lagrange shape functions of degree p defined by eq. (1.50) and T is the linear
projection operator that maps continuous functions defined on Ist onto polynomials of degree p
defined on Ist . We define the Lebesque constant 𝜆(T)
∑
p+1
𝜆(T) = max |Ni (𝜉)| (F.1)
𝜉∈Ist
i=1
and note that |T f | ≤ 𝜆(T) ∥ f ∥max .

Let 𝑤★
p be the best polynomial approximation of f in maximum norm. Then
|f − T f | = |f − 𝑤★ ★ ★ ★
p + 𝑤p − T f | ≤ |f − 𝑤p | + |𝑤p − T f | (F.2)
where we used the triangle inequality. Noting that 𝑤★

p can be written as 𝑤★
p =  T 𝑤★
p , we have
𝑤★ ★
p − T f = T (𝑤p − f )
and eq. (F.2) can be written as

∥ f − T f ∥max ≤∥ f − 𝑤★ ★
p ∥max + ∥ T (f − 𝑤p )∥max (F.3)
from which it follows that the Lebesque constant bounds the interpolation error:
∥ f − T f ∥max ≤ (1 + 𝜆(T)) ∥ f − 𝑤★
p ∥max (F.4)
308 Appendix F Polynomial mapping functions
1.8
1.6722
1.6
yp 1.4
1.2
1
–1 –0.8 –0.6 –0.4 –0.2 0 0.2 0.4 0.6 0.8 1
ξ
Figure F.1 The function yp on the optimal nodal set for p = 5.
that is, the interpolation polynomial is at most a factor (1 + 𝜆(T)) worse than the best possible
polynomial approximation of degree p. The set of interpolation points of degree p that minimizes
p
the Lebesque constant will be denoted by T1 . This set is also called the optimal nodal set or the set
of optimal collocation points.
Example F.1 The optimal nodal set for p = 5 is:

T15 = {−1 − 0.734127 − 0.268907 0.268907 0.734127 1}. (F.5)
The function
∑
p+1
yp = |Ni (𝜉)|
i=1
is plotted for the optimal nodal set in Fig. F.1. The maximum of yp is the Lebesque constant:
(yp )max = 𝜆(T) = 1.6722.
p
Extension of the optimal interpolation set T1 to two dimensions is technically difficult. For this
reason another optimal set, known as the mean optimal set, is used. The mean optimal set in one
p
dimension, denoted by T2 , minimizes
1∑
p+1
𝜂2 = |Ni (𝜉)|2 d𝜉. (F.6)
∫−1 i=1
p p
The optimal sets T1 and T2 are listed in Table F.1.
F.1 Interpolation on surfaces
In three-dimensional problems the boundary surfaces are covered by triangles and/or rectangles
that are mapped from the standard quadrilateral and triangular elements. In this section the opti-
mal interpolation points are described for the standard quadrilateral and triangular elements.
F.1 Interpolation on surfaces 309
Table F.1 Coordinates of one-dimensional optimal sets T1p and T2p from
reference [31]. Only the positive interior coordinates are listed.
p T1p T2p
3 0.4177913013559897 0.4306648
4 0.6209113046899123 0.6363260
5 0.2689070447719729 0.2765187
0.7341266671891752 0.7485748
6 0.4461215299911067 0.4568660
0.8034402382691066 0.8161267
7 0.1992877299056662 0.2040623
0.5674306027472533 0.5790145
0.8488719610366557 0.8598070
8 0.3477879716116667 0.3551496
0.6535334790799030 0.6649023
0.8802308527184540 0.8896327
F.1.1 Interpolation on the standard quadrilateral element

On the standard quadrilateral element the coordinates of the interpolation points are the tensor
p
product of the one-dimensional interpolation set T2 .
F.1.2 Interpolation on the standard triangle

Certain restrictions apply to the definition of optimal sets for the standard triangle. These
restrictions are made necessary by the requirement that mapped surfaces have to be continuous.
Therefore the interpolation points on the edges of the standard triangle must have the same
locations as on the standard quadrilateral element. Therefore on each edge of the standard triangle
p
the interpolation points are the same as the one-dimensional T2 interpolation set. Only the interior
points are determined by minimizing the function
n
∑p
𝜂Δ2 = |Ni (𝜉, 𝜂)|2 d𝜉d𝜂 (F.7)

∫Ω(t) i=1
st
where np = (p + 1)(p + 2)∕2. Furthermore, since the standard triangle has three-fold symmetry, the
interpolation points also have three-fold symmetry. The triangular coordinates of the six interior
interpolation points of the optimal nodal set for p = 5 are listed in Table F.2. Note that the coordi-
nates in the second and third lines are even permutations of the coordinates in the first line, indi-
cating three-fold symmetry. Lines 4 to 6 are also even permutations of the triangular coordinates.
310 Appendix F Polynomial mapping functions
Figure F.2 Optimal interpolation points for the

standard triangle, p = 5.
1.5
1
η
0.5
0
–1 –0.5 0 0.5 1
ξ
Table F.2 Triangular coordinates of the interior interpolation points for p = 5.
L1 L2 L3
0.152754 0.152754 0.694493
0.694493 0.152754 0.152754
0.152754 0.694493 0.152754
0.416888 0.416888 0.166225
0.166225 0.416888 0.416888
0.416888 0.166225 0.416888

311
Appendix G
Corner singularities in two-dimensional elasticity
The following discussion of corner singularities in two-dimensional elasticity is analogous to the

discussion of corner singularities in Section 4.2.2. The corner problem in two-dimensional elasticity
is complicated by the fact that the eigenvalues may be complex numbers.
G.1 The Airy stress function

It is assumed in the following that the material is isotropic and elastic, hence the material properties
are characterized by the two material constants E and 𝜈, and the volume forces are zero. We examine
the solution in the neighborhood of corner points when the intersecting edges are stress free. The
treatment of other cases is analogous.
The stress fields in planar elasticity can be derived from the Airy stress function1 denoted by
U(r, 𝜃). The Airy stress function satisfies the biharmonic equation:
( 2 )( 2 )
𝜕 1 𝜕 1 𝜕2 𝜕 1 𝜕 1 𝜕2
+ + + + U = 0. (G.1)
𝜕r 2 r 𝜕r r 2 𝜕𝜃 2 𝜕r 2 r 𝜕r r 2 𝜕𝜃 2
The components of the stress tensor in polar coordinates are related to U by the following formulas
(see, for example, [105]):
( )
1 𝜕U 1 𝜕2 U 𝜕2 U 𝜕 1 𝜕U
𝜎r = + 2 2 , 𝜎𝜃 = , 𝜏r𝜃 = − ⋅ (G.2)
r 𝜕r r 𝜕𝜃 𝜕r 2 𝜕r r 𝜕𝜃
The Cartesian components of the stress tensor are:
𝜕2 U 𝜕2 U 𝜕2 U
𝜎x = , 𝜎y = , 𝜏xy = − ⋅ (G.3)
𝜕y2 𝜕x2 𝜕x𝜕y
The stress function can be written in complex variable form2 :
U = ℜ(z𝜑(z) + 𝜒(z)) (G.4)
where ℜ(⋅) means the real part of the expression in the bracket, 𝜑(z) and 𝜒(z), called complex poten-
tials, are analytic functions of the complex variable z. The overbar indicates the complex conjugate.
The stresses components in polar coordinates are related to 𝜑(z) and 𝜒(z) as follows:
( )
𝜎r + 𝜎𝜃 = 2 𝜑′ (z) + 𝜑′ (z) = 4ℜ(𝜑′ (z)) (G.5)
−1 ( )
𝜎𝜃 − 𝜎r + 2i𝜏r𝜃 = 2zz z𝜑′′ (z) + 𝜒 ′′ (z) . (G.6)
1 Sir George Biddell Airy 1801–1892.

2 This formula is attributed to Edouard Jean-Baptiste Goursat 1858–1936.
312 Appendix G Corner singularities in two-dimensional elasticity
The stresses components in Cartesian coordinates are related to 𝜑(z) and 𝜒(z) as follows:
( )
𝜎x + 𝜎y = 2 𝜑′ (z) + 𝜑′ (z) = 4ℜ(𝜑′ (z)) (G.7)
( )
𝜎y − 𝜎x + 2i𝜏xy = 2 z𝜑′′ (z) + 𝜒 ′′ (z) . (G.8)
The components of the displacement vector in polar coordinates (up to rigid body displacement
and rotation) are related to 𝜑(z) and 𝜒(z) as follows:
( )
1∕2
2G(ur + iu𝜃 ) = z−1∕2 z 𝜅𝜑(z) − z 𝜑′ (z) − 𝜒 ′ (z) (G.9)
and in Cartesian coordinates:
2G(ux + iuy ) = 𝜅𝜑(z) − z 𝜑′ (z) − 𝜒 ′ (z) (G.10)
where 𝜅 is defined as
{3−𝜈
def for plane stress
𝜅 = 1+𝜈 (G.11)
3 − 4𝜈 for plane strain
This is known as the Kolosov-Muskhelishvili method3 . Details and derivations are available in
books on elasticity, such as [60, 105].
We will be interested in solutions corresponding to
𝜑(z) = (a1 − ia2 )z𝜆 , 𝜒(z) = (a3 − ia4 )z𝜆+1 , 𝜆 ≥ 0, 𝜆 ≠ 1 (G.12)
where ai (i = 1, 2, 3, 4) and 𝜆 are real numbers. The corresponding stress function is:
U = r 𝜆+1 (a1 cos(𝜆 − 1)𝜃 + a2 sin(𝜆 − 1)𝜃 + a3 cos(𝜆 + 1)𝜃 + a4 sin(𝜆 + 1)𝜃). (G.13)
In the case of 𝜆 = 1,
𝜑(z) = a1 z − ia2 z log z, 𝜒(z) = (a3 − ia4 )z2 (G.14)
the corresponding stress function is:
U = r 2 (a1 + a2 𝜃 + a3 cos 2𝜃 + a4 sin 2𝜃). (G.15)
G.2 Stress-free edges

We refer to Fig. 4.1 and assume that the boundary segments ΓAB and ΓBC are stress free, that is,
𝜎𝜃 = 𝜏r𝜃 = 0 at 𝜃 = ±𝛼∕2. Using equations (G.13) and (G.2) we find:
𝜎𝜃 = r 𝜆−1 𝜆(𝜆 + 1)[a1 cos(𝜆 − 1)𝜃 + a2 sin(𝜆 − 1)𝜃
+ a3 cos(𝜆 + 1)𝜃 + a4 sin(𝜆 + 1)𝜃]
𝜏r𝜃 = r 𝜆−1 𝜆(𝜆 − 1)[a1 sin(𝜆 − 1)𝜃 − a2 cos(𝜆 − 1)𝜃)]
+ r 𝜆−1 𝜆(𝜆 + 1)[a3 sin(𝜆 + 1)𝜃 − a4 cos(𝜆 + 1)𝜃].
On setting 𝜎𝜃 = 𝜏r𝜃 = 0 at 𝜃 = ±𝛼∕2, following straightforward algebraic manipulation, we have:
⎡ cos(𝜆 − 1) 𝛼 𝛼
cos(𝜆 + 1) ⎤
⎢ 2 2 ⎥ ⎧a ⎫
⎢ ⎥ ⎪ 1⎪
⎢ 𝛼 𝛼⎥⎨ ⎬ = 0 (G.16)
⎢−Λ sin(𝜆 − 1) 2 sin(𝜆 + 1) 2 ⎥ ⎪a3 ⎪
⎢ ⎥⎩ ⎭
⎣ ⎦
3 Gury Vasilievich Kolosov 1867–1936; Nikolai Ivanovich Muskhelishvili 1891–1976.

and
𝛼 𝛼
⎡ sin(𝜆 − 1) sin(𝜆 + 1) ⎤ ⎧a ⎫
⎢ 2 2 ⎥ ⎪ 2⎪
⎢ ⎥⎨ ⎬ = 0 (G.17)
⎢−Λ cos(𝜆 − 1) 𝛼 cos(𝜆 + 1) 𝛼 ⎥ ⎪a4 ⎪
⎣ 2 2⎦⎩ ⎭
where:
def 1−𝜆
Λ= ⋅
1+𝜆
Note that a1 and a3 (resp. a2 and a4 ) are coefficients of symmetric (resp. antisymmetric) functions
in eq. (G.13). Therefore, analogously to eq. (4.7), U(r, 𝜃) can be written in terms of sums of sym-
metric and antisymmetric functions. The symmetric functions associated with the eigenvalues of
eq. (G.16) are called Mode I eigenfunctions and the antisymmetric functions associated with the
eigenvalues of eq. (G.17) are called Mode II eigenfunctions.
Nontrivial solutions exist if either the determinant of eq. (G.16) or the determinant of eq. (G.17)
vanishes. This will occur if either
𝛼 𝛼 𝛼 𝛼
cos(𝜆 − 1) sin(𝜆 + 1) + Λ sin(𝜆 − 1) cos(𝜆 + 1) = 0
2 2 2 2
which can be simplified to
sin 𝜆 𝛼 + 𝜆 sin 𝛼 = 0, 𝜆 ≠ 0, ±1 (G.18)
or
𝛼 𝛼 𝛼 𝛼
sin(𝜆 − 1) cos(𝜆 + 1) + Λ cos(𝜆 − 1) sin(𝜆 + 1) = 0
2 2 2 2
which can be simplified to:
sin 𝜆 𝛼 − 𝜆 sin 𝛼 = 0, 𝜆 ≠ 0, ±1. (G.19)
We denote
sin 𝜆𝛼
def def sin 𝛼
Q(𝜆𝛼) = and q(𝛼) = (G.20)
𝜆𝛼 𝛼
and discuss the eigenvalues corresponding to the symmetric and antisymmetric eigenfunctions
separately in the following. For a detailed treatment of this subject we refer to [53].
Example G.1 Let us find the first eigenvalue for 𝛼 = 3𝜋∕2 (270∘ ) from eq. (G.18). In this case we
def
need to find t = 𝜆𝛼 such that
def sin t 2
f (t) = − = 0.
t 3𝜋
Using a root finding routine, we get t = 2.56581916 and 𝜆 = 0.54448374.
G.2.1 Symmetric eigenfunctions

Equation (G.18) can be written as
Q(𝜆𝛼) + q(𝛼) = 0. (G.21)
The function Q(𝜆𝛼) is plotted on the interval 0 < 𝜆𝛼 < 4𝜋 in Fig. G.1. The problem is to find the
roots of eq. (G.21) for a given 𝛼.
1.0
0.8
0.6
0.4
Q(λα) D
0.2 B
E
0
–0.2 C
A
–0.4
0 π/2 π 3π/2 2π 5π/2 3π 7π/2 4π
λα
Figure G.1 The function Q(𝜆𝛼) on the interval 0 < 𝜆𝛼 < 4𝜋.
Observe the following:
1. In the interval 0 < 𝛼 < 𝛼A where 𝛼A = 2.553591 (146.31∘ ) the line −q(𝛼) has no points in com-
mon with Q(𝜆𝛼), therefore there are no real roots.
2. At 𝛼 = 𝛼A the line −q(𝛼A ) is tangent to Q(𝜆𝛼) in point A. Therefore there is a double root at this
angle. There are double roots at 𝛼 = 𝛼B(1) = 3.625739 (207.74∘ ), 𝛼 = 𝛼B(2) = 5.499379 (315.09∘ ) and
also at 𝛼 = 𝛼C = 2.875839 (164.77∘ ).
3. In the interval 𝛼A < 𝛼 < 𝛼B there are at least two real and simple roots.
4. At 𝛼 = 𝜋 and 𝛼 = 2𝜋 there are infinitely many real roots. Furthermore, at 𝛼 = 𝜋 all roots are
integers.
5. Point D corresponds to 𝛼 = 3𝜋∕2 (270∘ ) where 𝜆𝛼 = 2.565819, hence 𝜆 = 0.544484. See Example
G.1. There are no other symmetric real roots at this angle.
6. Point E corresponds to 𝛼 = 2𝜋 where 𝜆𝛼 = 𝜋, hence 𝜆 = 1∕2. Point E also corresponds to 𝛼 = 𝜋
which is a special case, discussed next.
In formulating eq. (G.21) we excluded 𝜆 = 1 from consideration. When 𝜆 = 1 then U is given by

eq. (G.15). Considering the symmetric terms only and using eq. (G.2), we have
𝜎𝜃 = 2(a1 + a3 cos 2𝜃), 𝜏r𝜃 = 2a3 sin 2𝜃.
Letting 𝜎𝜃 (±𝛼∕2) = 𝜏r𝜃 (±(𝛼∕2) = 0, we find that non-trivial solution exists only if
[ ]
1 cos 𝛼
det = 0.
0 sin 𝛼
Therefore 𝛼 = n𝜋, (n = 1, 2, … ). Point E in Fig. G.1 represents 𝛼 = 𝜋.
Remark G.1 To find the complex roots of eq. (G.21) we write 𝜆 = 𝜉 + i𝜂. Therefore eq. (G.21)
becomes
sin(𝜉𝛼 + i𝜂𝛼)
= q(𝛼) (G.22)
𝜉𝛼 + i𝜂𝛼
which is equivalent to the following system of two equations:

sin 𝜉𝛼 cosh 𝜂𝛼 = 𝜉q(𝛼) (G.23)
cos 𝜉𝛼 sinh 𝜂𝛼 = 𝜂q(𝛼). (G.24)

For details we refer to [108].
G.2.2 Antisymmetric eigenfunctions

Equation (G.19) can be written as
Q(𝜆𝛼) − q(𝛼) = 0. (G.25)
Note that 𝜆 = 1 trivially satisfies eq. (G.25) for all 𝛼 and recall that we have excluded 𝜆 = 1 from
consideration when we formulated eq. (G.25). There are no real roots in the interval 0 < 𝛼 < 𝛼B
where 𝛼B = 2.777068 (159.11∘ ). There are at least two real roots in the interval 𝛼B < 𝛼 < 𝛼C where
𝛼C(1) = 3.463416 (198.44∘ ); 𝛼C(2) = 5.732235 (328.43∘ ). As in the case of Mode I, there are infinitely
many real roots at 𝛼 = 𝜋 and 𝛼 = 2𝜋 and at 𝛼 = 𝜋 all roots are integers. There is only one real root
in the interval 𝛼C < 𝛼 < 𝛼A where 𝛼A = 4.493409 (257.45∘ ).
The angle 𝛼A is a special angle, which corresponds to 𝜆 = 1. To show this, we consider the anti-
symmetric terms in eq. (G.15). Using eq. (G.2), we have
𝜎𝜃 = 2(a2 𝜃 + a4 sin 2𝜃), 𝜏r𝜃 = −(a2 + 2a4 cos 2𝜃).
Letting 𝜎𝜃 (±𝛼∕2) = 𝜏r𝜃 (±𝛼∕2) = 0, we find that non-trivial solution exists when
[ ]
𝛼∕2 sin 𝛼
det = 0.
1 2 cos 𝛼
Therefore 𝛼 = tan 𝛼. In the interval 0 < 𝛼 < 2𝜋 there is one root; 𝛼 = 𝛼A = 4.493409 (257.45∘ ). This
corresponds to point A in Fig. G.1.
G.2.3 The L-shaped domain

The L-shaped domain problem (𝛼 = 3𝜋∕2) described here is a widely used benchmark problem
in finite element analysis. It is representative of a geometric detail that frequently occurs in finite
element analysis, typically because fillets are omitted for convenience. The resulting singularity
may influence the accuracy of the quantities of interest.
It is assumed that the edges that intersect in the singular point are stress-free. In Example G.1
it was found that the lowest positive eigenvalue of eq. (G.18) is 𝜆1 = 0.544483737. The Airy stress
function corresponding to 𝜆1 can be written as:
U = a1 r 𝜆1 +1 (cos(𝜆1 − 1)𝜃 + Q1 cos(𝜆1 + 1)𝜃) (G.26)
where a1 is an arbitrary real number and Q1 = 0.543075579. Equation (G.26) is equivalent to
U = a1 ℜ(zz𝜆1 + Q1 z𝜆1 +1 ). (G.27)
Using eq. (G.27) and eq. (G.3) it is possible to show that the stress components corresponding to 𝜆1
are:
[( ) ]
𝜎x = a1 𝜆1 r 𝜆1 −1 2 − Q1 (𝜆1 + 1) cos(𝜆1 − 1) 𝜃 − (𝜆1 − 1) cos(𝜆1 − 3) 𝜃
[( ) ]
𝜎y = a1 𝜆1 r 𝜆1 −1 2 + Q1 (𝜆1 + 1) cos(𝜆1 − 1) 𝜃 + (𝜆1 − 1) cos(𝜆1 − 3) 𝜃 (G.28)
[ ]
𝜏xy = a1 𝜆1 r 𝜆1 −1 (𝜆1 − 1) sin(𝜆1 − 3) 𝜃 + Q1 (𝜆1 + 1) sin(𝜆1 − 1) 𝜃 .
Using eq. (G.10) and eq. (G.27) it is possible to show that the displacement components correspond-
ing to 𝜆1 , up to rigid body displacement and rotation terms, are:
a [( ) ]
ux = 2G1 r 𝜆1 𝜅 − Q1 (𝜆1 + 1) cos 𝜆1 𝜃 − 𝜆1 cos(𝜆1 − 2) 𝜃
a [( ) ] (G.29)
uy = 2G1 r 𝜆1 𝜅 + Q1 (𝜆1 + 1) sin 𝜆1 𝜃 + 𝜆1 sin(𝜆1 − 2) 𝜃
where 𝜅 is defined by eq. (G.11).
Complex eigenvalues
In planar elasticity, in contrast to the Laplace equation, 𝜆 can be complex and it can be either a
simple or a multiple root. If 𝜆 is complex then its conjugate is also a root. In the case of multiple
roots special treatment is necessary which is not discussed here. We refer to [72] for details.
Consider the Airy stress function in the form: U = r 𝜆+1 F(𝜃). If 𝜆 is complex we write 𝜆 = 𝜉 + i𝜂
and F = f + ig. Therefore
i𝜂
U = r (𝜉+1+i𝜂) (f + ig) = r (𝜉+1) e(ln r ) (f + ig).
Writing
i𝜂 )
e(ln r = e(i𝜂 ln r) = cos(𝜂 ln r) + i sin(𝜂 ln r)
we have:
U = r 𝜉+1 [(f cos(𝜂 ln r) − 𝜂 sin(𝜂 ln r)) + i(f sin(𝜂 ln r) − 𝜂 cos(𝜂 ln r))]. (G.30)
Both the real and imaginary parts of U are solutions of the biharmonic equation (G.1). Since ln r →
−∞ as r → 0, the sinusoidal terms oscillate with a wavelength approaching zero.
Remark G.2 Note that the number of square integrable derivatives, and hence the rate of con-
vergence, is independent of the imaginary part.
Table G.1 Lowest positive values of ℜ(𝜆) for three types of homogeneous boundary
conditions prescribed on intersecting straight boundary segments.
𝜶 free-free fixed-free fixed-fixed

𝕽(𝝀(s)
1
) 𝕽(𝝀(a)
1
) 𝕽(𝝀1 ) 𝕽(𝝀(s)
1
) 𝕽(𝝀(a)
1
)
45∘ 5.39053 9.56271 1.30434 5.57328 2.60831

90∘ 2.73959 4.80825 0.75835 2.82579 1.49046
135∘ 1.88537 3.24281 0.69339 1.57323 1.16088
180∘ 1.00000 2.00000 0.50000 1.00000 1.00000
225∘ 0.67358 1.30209 0.40594 0.73554 0.87723
270∘ 0.54448 0.90853 0.34032 0.60404 0.74446
315∘ 0.50501 0.65970 0.28784 0.53793 0.60945
360∘ 0.50000 0.50000 0.25000 0.50000 0.50000
G.2.4 Corner points

The lowest values of ℜ(𝜆) associated with the points of intersection of straight boundary segments
in two-dimensional elasticity are listed in Table G.1 for three types of homogeneous boundary con-
ditions prescribed on the intersecting boundary segments. The angle intersection 𝛼 has the same
sense as in Fig. 4.1.
1. Free-free conditions: On both edges 𝜎𝜃 = 𝜏r𝜃 = 0. In this case the eigenvalues are independent
of Poisson’s ratio 𝜈.
2. Fixed-free condition (plane stress, 𝜈 = 0.3): on one edge ur = u𝜃 = 0, on the other edge 𝜎𝜃 =
𝜏r𝜃 = 0
3. Fixed-fixed condition (plane stress, 𝜈 = 0.3): ur = u𝜃 = 0 on both edges.
In the free-free and fixed-fixed cases the eigenfunctions are either symmetric or antisymmet-
ric. These are indicated in Table G.1 with the superscripts s and a, respectively. At 𝛼 = 180∘ the
eigenvalues are integers.
319
Appendix H
Computation of stress intensity factors
The goal of computation in linear elastic fracture mechanics is to estimate stress intensity factors.
The contour integral method and the energy release rate method, outlined in the following, are
widely used for this purpose.
H.1 Singularities at crack tips

At a crack tip 𝛼 = 2𝜋, see equations (G.16) and (G.17). The Airy stress function corresponding to
the symmetric part of the asymptotic expansion is:
∑
∞
Us = ai ℜ(zz𝜆i + Qi z𝜆i +1 ), 𝜆i = i∕2 (H.1)
i=1
where Qi = (2 − i)∕(2 + i) when i is odd and Qi = −1 when i is even. The first two terms of this
asymptotic expansion play a central role in linear elastic fracture mechanics. The stress components
are obtained from eq. (G.3). For example:
( ) ( )
𝜕 2 Us 𝜕 𝜕Us 𝜕z 𝜕Us 𝜕z 𝜕 𝜕Us 𝜕Us
𝜎x = = + =i −
𝜕y2 𝜕y 𝜕z 𝜕y 𝜕z 𝜕y 𝜕y 𝜕z 𝜕z
( ) ( )
𝜕 𝜕U 𝜕U 𝜕 𝜕U 𝜕U
= − s + s
+ s
− s
(H.2)
𝜕z 𝜕z 𝜕z 𝜕z 𝜕z 𝜕z
therefore:
𝜎x = a1 𝜆1 ℜ(2z𝜆1 −1 − (𝜆1 − 1)zz𝜆1 −2 − Q1 (𝜆1 + 1)z𝜆1 −1 )
[( ) ]
= a1 𝜆1 r 𝜆1 −1 2 − Q1 (𝜆1 + 1) cos(𝜆1 − 1) 𝜃 − (𝜆1 − 1) cos(𝜆1 − 3) 𝜃
a (3 1
)
= √1 cos(𝜃∕2) + cos(5𝜃∕2) . (H.3)
r 4 4
√
Letting a1 = KI ∕ 2𝜋 and using trigonometric identities the functional form used in linear elastic
fracture mechanics can be obtained:
( )
K 𝜃 𝜃 3𝜃
𝜎x = √ I cos 1 − sin sin (H.4)
2𝜋r 2 2 2
see eq. (4.42). It is left to the reader to show that eq. (H.3) and eq. (H.4) are equivalent.
The stress components corresponding to the second term of eq. (H.1) are: 𝜎x = 4a2 , 𝜎y = 𝜏xy = 0.
The stress 𝜎x = 4a2 (constant) is called the T-stress and denoted by T. This is the second term in
eq. (4.42). The other stress components are obtained analogously.
320 Appendix H Computation of stress intensity factors
The Airy stress function corresponding to the antisymmetric part of the asymptotic expansion is:
∑
∞
Ua = bi ℑ(zz𝜆i + i z𝜆i +1 ), 𝜆i = i∕2 (H.5)
i=1
where ℑ(⋅) represents the imaginary part of (⋅) and i = −1 when i is odd, i = (2 − i)∕(2 + i) when
i is even. This can be verified by comparing eq. (H.5) with the antisymmetric terms in eq. (G.13).
The stress components corresponding to the second term of eq. (H.5) are: 𝜎x = 𝜎y = 𝜏xy = 0.
Using equations (H.1), (H.5) and (G.2), (G.3) the stress distribution in the neighborhood of a crack
tip can be determined up to the coefficients ai , bi (i = 1, 2, … , ∞). Procedures for the determination
of these coefficients from finite element solutions are described in Sections H.2 and H.3.
H.2 The contour integral method
The contour integral method was described in connection with the Laplace equation in
Section 4.2.4. In this section the contour integral method for the computation of the Mode I stress
intensity factor in planar elasticity is outlined.
The Airy stress function corresponding to the symmetric part of the asymptotic expansion is given
by eq. (H.1). The extraction function for the coefficient of the first term, denoted by w, corresponds
to the first negative eigenvalue 𝜆1 = −1∕2. Therefore 𝜑(z) = z−1∕2 and 𝜒(z) = 3z1∕2 in eq. (G.4) and
the Airy stress function is
U = Cℜ(zz−1∕2 + 3z1∕2 ) (H.6)
where C is an arbitrary real number in units of MPa m1∕2 . The stress components 𝜎x(w) , 𝜎y(w) , 𝜏xy
(w)
are determined from equations (G.7) and (G.8):

C
𝜎x(w) = − r −3∕2 (cos(3𝜃∕2) + 3 cos(7𝜃∕2)) (H.7)
4
C
𝜎y(w) = − r −3∕2 (7 cos(3𝜃∕2) − 3 cos(7𝜃∕2)) (H.8)
4
(w) 3C −3∕2
𝜏xy = r (sin(3𝜃∕2) − sin(7𝜃∕2)) (H.9)
4
and the traction vector components are computed from
Tx(w) = 𝜎x(w) cos 𝜃 + 𝜏xy

(w)
sin 𝜃 (H.10)
Ty(w) = 𝜏xy
(w)
cos 𝜃 + 𝜎y(w) sin 𝜃. (H.11)
The components of w are determined from eq. (G.10):

[( ) ]
r −1∕2 3 1
𝑤x = C 𝜅− cos(𝜃∕2) + cos(5𝜃∕2) (H.12)
2G 2 2
[( ) ]
r −1∕2 3 1
𝑤y = −C 𝜅+ sin(𝜃∕2) − sin(5𝜃∕2) (H.13)
2G 2 2
where G is the shear modulus. The path-independent integral, evaluated on a circle of radius r, is:
𝜋 𝜋
def
IΓ (u, w) = (Tx(u) 𝑤x + Ty(u) 𝑤y )r d𝜃 − (Tx(w) ux + Ty(w) uy )r d𝜃. (H.14)
∫−𝜋 ∫−𝜋
Table H.1 Values of the function F(𝜅).
𝝂 plane stress plane strain
0 −4𝜋 −4𝜋
0.1 −11.4240 −11.3097
0.2 −10.4720 −10.0531
0.3 −9.66644 −8.79646
0.4 −8.97598 −7.53982
0.5 −8.37758 −6.28319
This
√ integral is analogous to eq. (4.19). The stress intensity factor KI is defined by convention to be
a1 2𝜋 where a1 is the coefficient of the first term in eq. (H.1). Using the orthogonality property of
eigenfunctions we get
a1 C
IΓ (u, w) = F(𝜅) (H.15)
G
where F(𝜅) is defined by
def
F(𝜅) = GIΓ (u1 ∕a1 , w∕C). (H.16)
The function u1 is the displacement field corresponding to the first term in eq. (H.1). It is deter-
mined using eq. (G.10):
a1 r 1∕2 [( 1
)
1
]
u(1)
x = 𝜅− cos(𝜃∕2) − cos(3𝜃∕2) (H.17)
2G 2 2
a1 r 1∕2 [( 1
)
1
]
u(1)
y = 𝜅 + sin(𝜃∕2) − sin(3𝜃∕2) . (H.18)
2G 2 2
The√corresponding stress field is given by equations (4.42) to (4.44) with a1 substituted for
KI ∕ 2𝜋. Note that F(𝜅) is dimensionless and can be computed explicitly. Values of F(𝜅) are listed
in Table H.1. √
From eq. (H.15), using KI = a1 2𝜋, we get
√ √
2𝜋G 2𝜋G
KI = I (u, w) = I (u, w∕C). (H.19)
CF(𝜅) Γ F(𝜅) Γ
Note that since IΓ (u, w∕C) has the dimension m1∕2 , the dimension of KI is MPa m1∕2 . Note further
that since C is arbitrary, it can be chosen to be 1 MPa m3∕2 . To obtain an approximation for KI , we
substitute the finite element solution uFE for u. The derivation of the extraction function for KII is
analogous.
H.3 The energy release rate

The relationship between the stress intensity factors and the energy release rate  is derived in the
following for a plane elastic body of thickness tz . We assume that the body force and thermal load
are zero. By definition,
def 𝜕Π
=− (H.20)
𝜕a
where Π is the potential energy defined by eq. (2.102) and a is the crack length.
H.3.1 Symmetric (Mode I) loading

Consider the state of stress at the crack tip. In the coordinate system centered on the crack tip, as
shown in Fig. H.1, neglecting terms of higher order,
K
𝜎y = √ I 0 < x. (H.21)
2𝜋x
The displacement of the crack face is
(1 + 𝜈)(𝜅 + 1) KI √
uy = √ −x x≤0 (H.22)
E 2𝜋
where 𝜅 is defined by eq. (G.11).
Assume now that the crack length increases by a small amount Δa, as shown in Fig. H.1. In this
case
(1 + 𝜈)(𝜅 + 1) KI √
uy = √ Δa − x 0 ≤ x ≤ Δa. (H.23)
E 2𝜋
The work required to return the crack to its original length, that is to close the length increment
Δa, is:
2
√
Δa
1 (1 + 𝜈)(𝜅 + 1) KI tz Δa
Δa − x
ΔW = 2 𝜎 u t dx = dx. (H.24)
∫0 2 y y z E 2𝜋 ∫0 x
⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟
Δa𝜋∕2
Hence:
(1 + 𝜈)(𝜅 + 1) 2
ΔW = KI tz Δa. (H.25)
4E
In order to restore the crack to its initial length, energy equal to ΔW had to be imparted to
the elastic body. This is the energy expended in crack growth, called Griffith’s surface energy1 .
y
x′ = x – Δa
uy
Δa
Figure H.1 Notation.
1 Alan Arnold Griffith 1893–1963.

The potential energy had to decrease by the same amount when the crack increment occurred.
Hence:
ΔΠ 𝜕Π (1 + 𝜈)(𝜅 + 1) 2
 = − lim =− = KI tz . (H.26)
Δa→0 Δa 𝜕a 4E
H.3.2 Antisymmetric (Mode II) loading

When the loading is purely antisymmetric then the relationship between the energy release rate and
the stress intensity factor is analogous to the symmetric case, however instead of equations (H.21)
and (H.22) we have:
K
𝜏xy = √ II 0<x (H.27)
2𝜋x
and
(1 + 𝜈)(𝜅 + 1) KII tz √
ux = √ −x x ≤ 0. (H.28)
E 2𝜋
The derivation of the relationship between  and KII , under the assumption of perfectly antisym-
metric loading, is left to the reader.
H.3.3 Combined (Mode I and Mode II) loading

In view of the fact that the solutions corresponding to Mode I and Mode II loadings are energy
orthogonal, we have:
Π(uI + uII ) = Π(uI ) + Π(uII ) (H.29)
where uI and uII are the Mode I and Mode II solutions respectively. Therefore in the case of
combined loading we have:
(1 + 𝜈)(𝜅 + 1) 2
= (KI + KII2 ) tz . (H.30)
4E
Consequently the stress intensity factors are related to  as follows:
⎧
⎪ E for plane stress
2 2 ⎪ t
(KI + KII ) = ⎨ z (H.31)
⎪ E
⎪ (1 − 𝜈 2 )t for plane strain.
⎩ z
H.3.4 Computation by the stiffness derivative method

In the following the vector of coefficients of the basis functions computed by the finite element
method is denoted by x which is a function of the crack length a. Let us assume that we have
computed x = x(a) for a problem of linear elastic fracture mechanics. The potential energy is
1 T
Π(a) = x Kx − xT r
2
where K = K(a) is the stiffness matrix and r = r(a) is the load vector. Following crack extension,
the potential energy is:
1
Π(a + Δa) = (xT + ΔxT ) (K + ΔK) (x + Δx) − (xT + ΔxT )(r + Δr)
2
1 1
= Π(a) + ΔxT (Kx − r) + ΔxT KΔx + xT ΔKx+
⏟⏞⏟⏞⏟ 2 2
this is zero
1
x ΔKΔx + ΔxT ΔKΔx − xT Δr − ΔxT Δr.
T
2
Therefore:
Π(a + Δa) − Π(a) 1 𝜕K 𝜕r
 = − lim = − xT x + xT . (H.32)
Δa→0 Δa 2 𝜕a 𝜕a
In finite element computations 𝜕K∕𝜕a and 𝜕r∕𝜕a are approximated by finite differences:
𝜕K K(a + Δa) − K(a − Δa)
≈
𝜕a 2Δa
𝜕r r(a + Δa) − r(a − Δa)
≈ ⋅
𝜕a 2Δa
This involves re-computation of the stiffness matrices for those elements only which have a vertex
on the crack tip. In most cases 𝜕r∕𝜕a is either zero or negligibly small.
325
Appendix I
Fundamentals of data analysis
“Without data, you’re just another person with an opinion.” W. E. Deming1 .
Parametric data analysis, also called statistical inference, is a process by which the statistical
attributes of a large population are estimated from observed data which are viewed as samples
taken from the large population.
The fundamental concepts and algorithmic procedures of parametric data analysis needed for
understanding validation and ranking of mathematical models are introduced. It is assumed that
the reader is already familiar with the basic terminology and concepts of statistics, such as those that
can be found in the first few chapters in introductory texts: random variables, statistical indepen-
dence, correlation, mean, variance, commonly used probability density functions and conditional
probability.
I.1 Statistical foundations

Frequentist probability or frequentism is one interpretation of probability; it defines an event’s
probability as the limit of its relative frequency in a large number of trials. This interpretation
supports the statistical needs of experimental scientists and pollsters; probabilities can be found
in principle by a repeatable objective process and are thus devoid of subjectivity.
Bayesian probability is another interpretation of probability: instead of frequency or propensity
of some phenomenon, probability is interpreted as reasonable expectation representing a state
of knowledge, or as quantification of a personal belief. Validation and ranking of mathematical
models involves arguments based on Bayesian probability. In the following discussion probability
should be understood in this sense.
Data analysis is largely based on three basic theorems of statistics, known as the product rule,
Bayes’ theorem and marginalization. These theorems are stated in this section and their appli-
cations in data analysis are illustrated in Section I.3. For detailed introductory treatment of data
analysis we recommend references [87] and [52].
The product rule

The product rule states that the probability of X and Y both being true is equal to the probability
of X being true, given Y , times the probability of Y being true.
Pr (X, Y ) = Pr (X | Y ) Pr (Y ) = Pr (Y | X) Pr (X). (I.1)
1 William Edwards Deming 1900–1993.
326 Appendix I Fundamentals of data analysis
The second equality follows from the fact that the ordering of X and Y is immaterial.
Bayes’ theorem
Bayes’ theorem follows directly from the product rule:
Pr (Y | X) Pr (X)
Pr (X | Y ) = ⋅ (I.2)
Pr (Y )
The importance of this theorem becomes obvious if we replace X with M, representing the statistical
model, and Y with D, representing the available data:
Pr (D | M) Pr (M)
Pr (M | D) = ⋅ (I.3)
Pr (D)
Thus Bayes’ theorem relates the probability that a statistical model is the true representation of the
(unknown) statistical properties of the population from which data D was sampled, to the proba-
bility that we would have observed data D if the model were true.
The term Pr (M | D) is called posterior probability. It represents our degree of belief in the validity
of model M in the light of the data D. The term Pr (D | M) is the likelihood function. The term Pr (M)
is called prior probability which represents our degree of belief in the validity of model M before the
current data is analyzed and the term Pr (D) is called marginal likelihood or evidence. It is a scaling
factor needed to ensure that the integral of the right hand side over the entire range of parameters
is unity.
Marginalization
Marginalization is used in data analysis to eliminate parameters that enter into the analysis but are
not of primary interest. Those parameters are called nuisance parameters. The probability that X is
true equals the probability that X and Y is true integrated over the entire range of possible values
for Y :
∞
Pr (X) = Pr (X, Y ) dY . (I.4)
∫−∞
In this instance Y is the nuisance parameter.
I.2 Test data

The examples presented in Chapter 6 are based on fatigue test records for 24S-T3 aluminum alloy2
specimens extracted from references [40–44].
Surface finish is known to be of major importance in determining fatigue strength. In
order to avoid introducing scratches and residual stresses, the surfaces of the specimens were
electropolished.
The average static properties of the material are listed in Table I.1. The heading “Grain (L)” (resp.
“Cross (T)”) indicates that the specimen was cut such that the long dimension was oriented in the
direction of the grain (resp. across the grain).
The fatigue tests were conducted at 1100 cycles per minute (18.3 Hz). The estimated precision
of setting and maintaining loads was approximately ±3 % for tension-tension tests and ±5 % for
tension-compression tests.
2 The current designation of this alloy is 2024-T3.

I.2 Test data 327
Table I.1 Average static properties of 24S-T3 (2024-T3) aluminum from reference [40].
Property Grain (L) Cross (T)
Modulus of elasticity in compression 10,650 ksi 10,450 ksi

Elongation in 2 inches, percent 18.2 % 18.3 %
Yield strength (0.2 % offset) in tension 54.0 ksi 50.0 ksi
Yield strength (0.2 % offset) in compression 44.5 ksi 50.5 ksi
Ultimate strength in tension 73.0 ksi 71.0 ksi
The test records contain the following information: (a) Specimen dimensions in inches3 , (b) spec-
imen label, (c) the maximum stress in the test section or notch root Smax in psi units4 , (d) the cycle
ratio R, (e) the number of cycles at the end of the test N and (f) notes indicating remarkable obser-
vations relating to the test, such as whether failure occurred outside of the test section or the test
was stopped prior to failure (run-out). Two of the notched test specimens are shown in Fig. I.1.
The test data are summarized in Table I.2. For the notch-free specimens the range of 𝜎max was
(24, 58) ksi, the range of 𝜎min was (−50, 29) ksi.
In the column labeled (Kt )act the actual stress concentration factors, defined as the ratio of maxi-
mum stress to the average stress in the test section, are shown. In the column labeled A, the highly
stressed area is shown. This area, multiplied by the thickness, approximates the highly stressed
volume defined in eq. (6.31) for 𝛾 = 0.85. Errors in the computed values of A have been verified to
be not greater than 1%.
In the last column the specimen count is listed with and without run-outs. The number of spec-
imens without counting run-outs is in brackets. This count does not include specimens that were
tested at stress levels higher than 58.0 ksi and specimens that were disregarded by the investigators
because of flaws, buckling or failure in the grips.
r = 0.3175
1.500 0.375 2.25
8.1825 8.1825
r = 0.0195 arbitrary large radius
1.500 2.25
6.0 5.0 6.0
Figure I.1 Notched test specimens of type #3 and #6.
3 1 inch = 25.40 mm.

4 pounds per square inch; 1 psi = 6.895 kPa.
Table I.2 Summary of test records extracted from NACA & NASA technical reports for 24S-T3 (2024-T3)
aluminum specimens.
k Specimen Ref. rk (in) (Kt )act A (in2 ) Count
1 Notch free [40] 12.0 1.00 test section 53 (44)

2 Open hole [41] 1.5000 2.11 2.898 E−02 39 (34)
3 Edge notch [41] 0.3175 2.17 4.325 E−03 42 (38)
4 Fillet [41] 0.1736 2.19 9.561 E−04 32 (27)
5 Edge notch [41] 0.0570 4.43 1.827 E−04 34 (29)
6 Fillet [41] 0.0195 4.83 1.618 E−05 36 (32)
7 Edge notch [42] 0.0313 5.83 5.888 E−05 46 (40)
8 Edge notch [43] 0.7600 1.62 2.875 E−02 31 (27)
9 Edge notch [44] 0.0035 4.48 5.908 E−07 17 (13)
10 Edge notch [44] 0.0710 4.41 3.324 E−04 19 (15)
I.3 Statistical models
In this section we formulate three statistical models that could reasonably be expected to represent
the population from which the data collected from constant cycle fatigue experiments of metal
specimens was sampled and estimate their parameters. Evaluation of the relative merit of these
models is discussed in Section I.4.
The available information comprises a detailed description of the testing apparatus, the geomet-
ric configuration and surface finish of the test specimens, and a set of records consisting of the
number of cycles at which failure occurred or the test was stopped, the maximum and minimum
load values and notes concerning unusual events, such as failure outside of the test section, buck-
ling and so on. A representative set of test data, called S-N data, taken from [40], is shown in Fig. I.2.
There are 53 data points of which 9 are runouts, that is, the test was stopped before failure occurred.
The statistical term for runouts is “right-censored data”.
50 Failure
Equivalent stress (ksi)
Runout
40
30
20
104 105 106 107

n
Figure I.2 S-N Data for 24S-T3 (2024-T3) aluminum alloy.

In these tests the equivalent stress 𝜎eq was controlled. Various definitions of equivalent stress are
possible, see Sections 6.1.1, 6.3. The number of cycles to failure N is assumed to be an independent
and identically distributed random variable. In this context “independent” should be understood
to mean that the occurrence of one outcome does not affect the probability of occurrence of other
outcomes. This assumption is justified noting that each outcome corresponds to a different test
specimen. The term “identically distributed” refers to an assumption that there exists a probability
distribution and each recorded number of cycles ni is a particular realization of that probability
distribution. A statistical model is a precise statement of the assumed statistical distribution of N,
given 𝜎eq . In other words, a statistical model is a statement of an idea of what is expected to happen,
whereas the test results tell us what in fact did happen.
For example, we may assume that W = log10 N is normally distributed with mean 𝜇 = 𝜇(𝜎eq ) and
standard deviation s = s(𝜎eq ). Conventionally, upper case letters denote random variables, the cor-
responding lower case letters denote their realizations. Therefore the probability density function
of W can be written as
( )
1 (𝑤 − 𝜇)2
Pr (𝑤 | 𝜇, s) = √ exp − . (I.5)
2𝜋 s 2s2
which has the property
( )
1
∞
(𝑤 − 𝜇)2
√ exp − d𝑤 = 1. (I.6)
2𝜋 s ∫−∞ 2s2
Introducing the change of variables 𝑤 = log10 n = ln n∕ ln(10) we have d𝑤 = dn∕(n ln(10)) and
hence
( )
1 (log10 n − 𝜇)2
Pr (log10 n | 𝜇, s) = √ exp − (I.7)
2𝜋 s n ln(10) 2s2
which also has the essential property of probability density functions that their integral over the
domain of definition is unity:
( )
1
∞
(log10 n − 𝜇)2
√ exp − dn = 1.
2𝜋 s n ln(10) ∫0 2s2
The cumulative distribution function of the standard normal distribution is:
( ) ( ( ))
log10 n − 𝜇 1 log10 n − 𝜇
Φ = 1 + erf (I.8)
s 2 s
where erf is the error function.
Various plausible assumptions can be made concerning the functional forms of 𝜇(𝜎eq ) and s(𝜎eq ).
We will consider two functional forms for 𝜇(𝜎eq ): The bilinear form
{
log10 N0 − m1 (𝜎eq − S0 ) for 𝜎eq ≥ S0
𝜇1 (𝜎eq ) = (I.9)
log10 N0 + m2 (S0 − 𝜎eq ) for 𝜎eq < S0
and the logarithmic form
{
A1 − A2 log10 (𝜎eq − A3 ) for 𝜎eq − A3 > 0
𝜇2 (𝜎eq ) = (I.10)
∞ for 𝜎eq − A3 ≤ 0
where 𝜎eq , S0 and A3 are in ksi units5 . Since the argument of the log function has to be dimension-
less, the argument should be understood to have been scaled by 1 ksi. The subscript on 𝜇 will be
dropped when there is no possibility of confusion.
5 kilo-pounds per square inch. 1 ksi = 6.895 MPa.

This functional form implies that there is a material property, called fatigue limit, denoted by
A3 in eq. (I.10). Fatigue failure can occur only for 𝜎eq > A3 . For 𝜎eq ≤ A3 the fatigue life is infinity.
We will refer to this as the fatigue limit model. In reference [73] it was proposed that A3 should be
treated as a random variable. We will consider that possibility as well and will refer to that model
as the random fatigue limit (RFL) model.
We assume, for the sake of simplicity, that the standard deviation s is constant. Therefore the
bilinear model is characterized by the five parameters 𝜽1 = {S0 N0 m1 m2 s}T and the assumption
that log10 N is normally distributed.
The fatigue limit model is characterized by the four parameters 𝜽2 = {A1 A2 A3 s}T and the ran-
dom fatigue limit model is characterized by the five parameters 𝜽3 = {A1 A2 s 𝜇f sf }T where 𝜇f and
sf are the mean and standard deviation of log10 A3 respectively.
Statistical models are calibrated against the available data by maximizing the likelihood function.
By definition, the likelihood function is the joint probability density of a random sample. For the
bilinear and fatigue limit models being considered here the likelihood function is
( )1−𝛿i ( ( ))𝛿i
∏ n (i) 2
exp(−(log10 ni − 𝜇(𝜎eq )) ∕2s2 ) (i)
log10 ni − 𝜇(𝜎eq )
L= √ 1−Φ (I.11)
i=1 2𝜋sni ln(10) s
where 𝛿i = 0 if the test specimen failed at ni cycles, 𝛿i = 1 if a runout was recorded, that is, the test
was stopped at ni cycles before failure occurred. The runout data carry the information that the
specimen survived ni cycles. Maximizing L with respect to the parameters of the statistical model
yields the combination of model parameters that maximize the probability of observations actually
obtained. In other words, calibration is based on the idea that, assuming the functional form of a
statistical model to be correct, the set of model parameters that would have produced the calibration
data with the greatest probability is the best estimate for the model parameters. We will denote the
parameters that maximize the likelihood function by 𝜽̂ k .
Since logarithm is a monotonically increasing function, the parameters that maximize L also
maximize LL ≡ ln L. Therefore LL, called the log likelihood function, can be used in place of the
likelihood function when computing the statistical parameters. Solving for the statistical param-
eters involves solving a multivariate nonlinear problem which is usually easier to do when the
function being maximized is the log-likelihood function rather than the likelihood function. The
log likelihood function corresponding to eq. (I.11) is
( )
∑ n (i) 2
(log10 ni − 𝜇(𝜎eq )) (√ )
LL = − (1 − 𝛿i ) + ln 2𝜋sni ln(10)
i=1
2s2
( ( ))
∑ n
log10 ni − 𝜇(𝜎eq (i)
)
+ 𝛿i ln 1 − Φ . (I.12)
i=1
s
The bilinear model

The estimated parameters for the bilinear model and the maximum value of LL are shown in
Table I.3. The parameter S0 is in ksi units, the parameters m1 , m2 are in ksi−1 units, the stan-
dard deviation s is dimensionless. The function 𝜇(𝜎eq ) characterized by the parameters in Table I.3,
known as the S-N curve, is shown in Fig. I.3.
Remark I.1 The parameters were computed by means of the Matlab function mle (maximum
likelihood estimate) using several seed values to ascertain, with a high degree of confidence, that
the parameters correspond to the global maximum of the likelihood function.
Table I.3 Estimated parameters 𝜽̂ 1 for the bilinear model.
Model S0 log10 N0 m1 m2 s LL(𝜽̂ 1 )
Bilinear 31.305 5.26352 0.0596 0.1779 0.4583 −601.931
m1
50
40
m2 S0
30
1
20
104 105 N0 106 107

n
Figure I.3 Bilinear S-N curve for 24S-T3 aluminum alloy.
Table I.4 Estimated parameters 𝜽̂ 2 for the fatigue limit model.
Model A1 A2 A3 s LL(𝜽̂ 2 )
Fatigue limit 9.019 3.198 17.856 0.4662 −602.112
The fatigue limit model

The estimated parameters for the fatigue limit model are shown in Table I.4. The parameter A3 is
in ksi units, the other parameters are dimensionless. The S-N curve is shown in Fig. I.4.
Remark I.2 Fatigue limit and endurance limit are expressions used to describe a property of
materials: the amplitude of cyclic stress at full reversal (R = −1) that can be applied to the material
without causing fatigue failure at any number of cycles. Fatigue strength is the value of 𝜎eq at a
fixed number of cycles (usually n = 5 × 108 cycles) read from the S-N curve: 𝜎eq = 𝜇 −1 (log10 n).
Ferrous alloys and titanium alloys appear to have fatigue limits in the megacycle range. However,
as noted in [26], for some low carbon steels the difference between the fatigue strength at 106 and
109 cycles is less than 50 MPa, whereas for other steels this difference ranges from 50 to 200 MPa.
Non-ferrous metals, such as aluminium and copper alloys, do not have fatigue limits. Neverthe-
less, fatigue limit models are being used for non-ferrous metals, see (for example) the MMPDS
Handbook [58].
Since fatigue limit is a material property, it is a random number that has to be inferred from
a statistical model. The estimated mean value of the fatigue limit or fatigue strength tends to be
sensitive to the choice of the statistical model. See Example 6.1.
50
40
30
20
104 105 106 107

n
Figure I.4 MMPDS Fatigue limit model, S-N curve for 24S-T3 aluminum alloy.
The random fatigue limit model

We introduce the notation 𝑣 = log10 A3 and assume that 𝑣 is normally distributed with mean 𝜇f
and standard deviation sf . We invoke two fundamental theorems in statistics; the product rule and
marginalization. The product rule takes the form:
Pr (𝑤, 𝑣 | 𝜎eq ) = Pr (𝑤 | 𝜎eq , 𝑣) Pr (𝑣 | 𝜎eq ) (I.13)
where 𝑤 = log10 n as defined previously. The equation of marginalization is

log10 𝜎eq
Pr (𝑤 | 𝜎eq ) = Pr (𝑤, 𝑣 | 𝜎eq ) d𝑣 (I.14)
∫−∞
where the upper limit of the integral is determined from the condition that 𝜎eq − A3 > 0. Therefore:
log10 𝜎eq
Pr (𝑤 | 𝜎eq ) = Pr (𝑤 | 𝜎eq , 𝑣) Pr (𝑣 | 𝜎eq ) d𝑣. (I.15)
∫−∞
Since by hypothesis 𝑤 is normally distributed with mean 𝜇 and standard deviation s, the first
term of the integrand in eq. (I.15) is the probability density function:
exp(−(𝑤 − 𝜇(𝜎eq , 10𝑣 ))2 ∕(2s2 ))
Pr (𝑤 | 𝜎eq , 𝑣) = 𝜙(𝑤, 𝜎eq , 𝑣) = √ ⋅ (I.16)
s 2𝜋
The second term of the integrand follows from the assumption that 𝑣 is normally distributed.
Therefore the probability density of 𝑣 is:
exp(−(𝑣 − 𝜇f )2 ∕(2s2f ))
Pr (𝑣 | 𝜎eq ) = f (𝑣) = √ , 𝑣 < log10 (𝜎eq ) (I.17)
sf 2𝜋
The marginal probability density function of 𝑤, given 𝜎eq , is:
log10 𝜎eq
Pr (𝑤 | 𝜎eq ) = 𝜙M (𝑤, 𝜎eq ) = 𝜙(𝑤, 10𝑣 )f (𝑣) d𝑣 (I.18)
∫−∞
and the marginal cumulative distribution function (CDF) of 𝑤, given 𝜎eq , is:
( ( ))
1
log10 𝜎eq 𝑤 − 𝜇(𝜎eq , 10𝑣 )
ΦM (𝑤, 𝜎eq ) = 1 + erf √ f (𝑣) d𝑣. (I.19)
2 ∫−∞ s 2
Table I.5 Estimated parameters 𝜽̂ 3 for the random fatigue limit model.
Model A1 A2 s 𝝁f sf LL(𝜽̂ 3 )
RFL 7.191 1.991 0.1255 1.3438 0.0488 −576.734
(i)
Given a set of independent observations (𝑤i , 𝜎eq ), (i = 1, 2, … , n) the likelihood function is
∏
n
(i) 1−𝛿i (i) 𝛿i
L(𝜽) = [𝜙M (𝑤i , 𝜎eq )] [1 − ΦM (𝑤i , 𝜎eq )] (I.20)
i=1
where 𝛿i = 0 if the test resulted in failure, 𝛿i = 1 if the test was stopped before failure occurred.
The maximum likelihood estimate of 𝜽, denoted by 𝜽̂ 3 , maximizes L(𝜽), or equivalently, the cor-
responding log likelihood function. The estimated parameters for the random fatigue limit model
are shown in Table I.5. The S-N curve is shown in Fig. I.5.
Another way to visualize the relationship between the random fatigue limit model and the S-N
data is to display the empirical cumulative distribution function of the S-N data and compare that
distribution with the median corresponding to the random fatigue limit model.
Given 𝜎eq , the p-quantile corresponding to the statistical model can be computed from the inverse
of the cumulative distribution function. Specifically, we find 𝑤i = log10 ni such that
Q(p) ≡ ΦM (𝑤i , 𝜎eq ) − p = 0, 0<p<1 (I.21)
where ΦM is the marginal cumulative distribution function defined by eq. (I.19). Letting p = 0.5
we get the predicted median. The results are shown in Fig. I.6. There were nine runouts recorded
which are indicated by open circles.
A runout is predicted when the function Q(p) = 0 does not have a real root at p = 0.5 or the
predicted number of cycles is greater than 100 million. See Remark I.3.
A recorded runout indicates the number of cycles at which a test was stopped and the specimen
did not fail. Therefore a predicted runout is not necessarily a prediction of a recorded runout. Nine
runouts were recorded for the S-N data, four runouts are predicted by the random fatigue limit
model.
50 failure
runout
40
30
20
104 105 106 107

n
Figure I.5 Random fatigue limit model: S-N data for 24S-T3 aluminum alloy.
0.8
0.05 quantile Runout (9)
0.6
F(n)
0.4
0.95 quantile
median
0.2
Empirical CDF
0
104 105 106 107
n
Figure I.6 Empirical CDF of the S-N data compared with the median predicted by the random fatigue limit
model.
Remark I.3 The probability that failure will occur at n < 10𝑤 cycles, given 𝜎eq , is ΦM (𝑤, 𝜎eq ), see
eq. (I.19). Letting 𝑤 → ∞ we find:
log10 𝜎eq
Pr (failure | n < ∞, 𝜎eq ) = f (𝑣) d𝑣 (I.22)
∫−∞
therefore
log10 𝜎eq
Pr (no failure | n < ∞, 𝜎eq ) = 1 − f (𝑣) d𝑣. (I.23)
∫−∞
A feature of the random fatigue limit model is that there is a non-zero probability that failure will
not occur at any stress level. At high stress levels this probability is negligibly small. When 𝜎eq is
equal to the mean fatigue limit then this probability is 50%.
For example, the estimated mean of the random fatigue limit of 24S-T3 aluminum is 10𝜇f = 22.07
ksi. The marginal cumulative distribution functions are shown in Fig. I.7 for various values of 𝜎eq .
1
σeq = 50 0.866
Pr(failure ∣ n < ∞, σeq)
0.8
25
ΦM(w,σeq)
0.6 30
0.500
0.4 22.07
40
0.2 σeq = 20 0.190
0
104 105 106 107 108
n = 10w
Figure I.7 Marginal cumulative distribution functions of the random fatigue limit model for 24S-T3
aluminum. The stress values are in ksi units.
I.5 Confidence intervals 335
The limit values for 𝜎eq = 25, 22.07, 20 ksi with respect to n → ∞ are shown on the right. It is
seen that in the neighborhood of 𝜎eq = 10𝜇f a small change in 𝜎eq results in a large change in the
predicted value of n. Therefore the random fatigue limit model is not a reliable predictor of n when
𝜎eq is close to 10𝜇f .
Remark I.4 The integrals in equations (I.18) and (I.19) have to be evaluated numerically.
Remark I.5 The assumption that f (𝑣) has the functional form of eq. (I.17) is one of many pos-
sible assumptions. In reference [17] f (𝑣) was assumed to have normal and smallest extreme value
distributions.
I.4 Ranking
Of necessity, several assumptions have to be made in the formulation of statistical models. These
assumptions are based on prior experience, insight and intuition. It is possible to objectively rank
models based on their predictive performance, measured by the likelihood function, with reference
to the available data. Since the available data tend to increase over time and new statistical models
may be proposed, it is necessary to establish a process for systematic revision and updating models
over time. In industrial and research organizations this falls under the administration of simulation
governance and management [92]. Ranking, based on Bayes’ theorem, is discussed in this section.
The Bayes factor

Ranking statistical models can be based on assessment of the relative merit of model Mi with respect
to model Mj as quantified by the Bayes factor BFij which is defined by
def Pr (D | Mi )
BFij = ⋅ (I.24)
Pr (D | Mj )
The term Pr (D) cancels and, unless we have some reason to assign to the ratio of prior probabilities
Pr (Mi )∕ Pr (Mj ) a number other than unity, we have
Pr (Mi | D) Pr (D | Mi )
=
Pr (Mj | D) Pr (D | Mj )
L(D | 𝜽i )
= = exp(LL(D | 𝜽i ) − LL(D | 𝜽j )). (I.25)
L(D | 𝜽j )
Let us assign the index 1, 2, 3 to the bilinear, fatigue limit and random fatigue limit models respec-
tively. From the log likelihood values in tables I3, I4 and I5 we find:
BF31 ≈ 8.8 × 1010 , BF32 ≈ 1.1 × 1011
that is, the random fatigue limit model is very strongly preferred over the other two models.
I.5 Confidence intervals

Confidence intervals can be estimated by means of the profile likelihood function. Let 𝜃k be a
parameter for which we are interested in computing the confidence interval and 𝜽k the vector of all
1
Profile likelihood
0.75
0.5
0.25 1.241 1.388

0
1.2 1.25 1.3 1.35 1.4 1.45
μf
Figure I.8 Profile likelihood and estimated confidence interval for 𝜇f .
other parameters associated with a statistical model. In other words, 𝜽k comprises all parameters
of the statistical model except 𝜃k . The profile likelihood function for 𝜃k is defined by
L(𝜃k , 𝜽k )
R(𝜃k ) = max (I.26)
𝜽k ̂
L(𝜽)
where 𝜽̂ is the set of parameters that maximizes the likelihood function. Wilks’ theorem6 states
that, as the sample size approaches infinity, the statistic −2 ln R(𝜃k ) will approach the chi-squared
distribution with one degree of freedom provided that the statistical model is correct. Therefore the
confidence interval can be estimated from the formula
−2 ln(R(𝜃k )) ≤ 𝜒1;1−𝛼
2
(I.27)
where is the 100(1 − 𝛼) percentile of the chi-square distribution with one degree of free-
𝜒1;1−𝛼
2
dom. For example, referring to the parameters of the random fatigue limit model in Table I.5, the
estimated 95% confidence interval for 𝜇f is (1.241, 1.388). The profile likelihood function and the
confidence interval for 𝜇f are shown in Fig. I.8.
6 Samuel Stanley Wilks 1906–1964.

337
Appendix J
Estimation of fastener forces in structural connections
A method for estimating forces in individual fasteners in a fastener group is described in the follow-
ing. Discussion of this method, used in traditional structural engineering practice, can be found in
textbooks on the mechanics of materials, see for example [79]. It is an example of a highly simplified
mathematical model applied to a rather complicated problem.
The simplified model is based on the assumption that a fastened structural component behaves as
a perfectly rigid body. The fasteners behave as linearly elastic springs. Therefore the displacement
and rotation of the fastened component is determined by the displacements of the linear springs.
Consider the solution domain of a rigid planar body of constant thickness t. The domain Ω is
bounded by the contour Γ. Inside Γ are n circular holes of radius 𝜚k , denoted by Ω(k)
𝜚 , k = 1, 2, … , n.
The solution domain and notation are shown in Fig. J.1. Formally,
def
Ω = Ω − Ω(1) (2) (n)
𝜚 ∪ Ω𝜚 ∪ · · · ∪ Ω𝜚 .
The boundary of Ω(k) (k)

𝜚 is denoted by Γ𝜚 . For simplicity we will use 𝜚k = 𝜚 in the following.
Tractions Tx , Ty are prescribed on Γ. The forces Fx , Fy and the moment M are defined such that
the equations of equilibrium are satisfied:
Fx + Tx t ds = 0 (J.1)
∫Γ
Fy + Ty t ds = 0 (J.2)
∫Γ
( )
M+ −Tx y + Ty x t ds = 0 (J.3)
∫Γ
where Fx , Fy and M are the resultants of forces imposed by the fasteners on the rigid planar body.
The goal is to estimate the forces fx(k) , fy(k) in a fastened structural connection. We will denote the
spring rate of the kth fastener by 𝜅k and the displacement vector components in the center of the
hole by u(k) (k) (k) (k) (k) (k)
x and uy . Therefore fx = 𝜅k ux and fy = 𝜅k uy . The origin of the coordinate system is
in the center of rotation which is defined such that the following equations are satisfied:
∑
n
∑
n
𝜅k xk = 0, 𝜅k yk = 0. (J.4)
k=1 k=1
Denoting the displacement vector components of the center of rotation by u(c) (c)
x and uy and the
rotation by 𝜃 (c) , we have:
u(k) (c) (c)
x = ux − 𝜃 rk sin 𝛼k = ux − 𝜃 yk
(c) (c)
(J.5)
u(k)
y = u(c)
y + 𝜃 rk cos 𝛼k =
(c)
u(c)
y + 𝜃 xk .
(c)
(J.6)
338 Appendix J Estimation of fastener forces in structural connections
y
–
Γ
Fy fy(k) Γϱ(k)
yk
M rk fx(k)
∝k
Fx xk x
–
Ω
Figure J.1 Notation.
Therefore:
fx(k) = 𝜅k (u(c)
x − 𝜃 yk )
(c)
(J.7)
fy(k) = 𝜅k (u(c)
y + 𝜃 xk ).
(c)
(J.8)
The resultants can be written in terms of the fastener forces:

∑
n
∑
n
Fx = fx(k) = u(c)
x 𝜅k (J.9)
k=1 k=1
∑n
∑n
Fy = fy(k) = u(c)
y 𝜅k (J.10)
k=1 k=1
∑n
∑
n
M= (fy(k) xk − fx(k) yk ) = 𝜃 (c) 𝜅k (xk2 + y2k ) (J.11)
k=1 k=1
where we used eq. (J.4). Writing the displacements of the center of rotation in terms of the stress
resultants:
Fx Fy M
u(c)
x = , u(c)
y = , 𝜃 (c) =
∑n
∑ n
∑
n
𝜅k 𝜅k 𝜅k (xk2 + y2k )
k=1 k=1 k=1
equations (J.7) and (J.8) can be written for the jth fastener as:
(j)
𝜅j 𝜅j yj
fx = Fx − M (J.12)
∑n
∑ n
𝜅k 𝜅k (xk2 + y2k )
k=1 k=1
(j)
𝜅j 𝜅j xj
fy = Fy + M. (J.13)
∑n
∑n
𝜅k 𝜅k (xk2 + y2k )
k=1 k=1
Estimation of fastener forces in structural connections 339
In the special case when 𝜅j has the same value for all j we have:
(j) Fx yj
fx = − n M (J.14)
n ∑ 2
(xk + y2k )
k=1
(j)
Fy xj
fy = + M. (J.15)
n ∑
n
(xk2 + y2k )
k=1
Remark J.1 Note that no discretization is necessary, therefore the errors in predictions based on
this model are due entirely to model form errors.
341
Appendix K
Useful algorithms in solid mechanics
Some of the frequently used algorithms in finite element analysis applied to problems in solid
mechanics are summarized. The basic properties of the stress tensor and traction vector and their
transformation are reviewed. The equations of static equilibrium of forces and moments are stated.
K.1 The traction vector
Let us assume that the state of stress at a point is known and let us determine the components of
the traction vector Tx , Ty , Tz acting on the inclined face of the infinitesimal tetrahedral element
shown in Fig. K.1. By definition, the traction vector, also known as the stress vector, is
ΔF
T = lim
ΔA→0 ΔA
where ΔF is the differential force, acting on the inclined face of the tetrahedron, the area of which
is ΔA. This force is in equilibrium with the resultants of the stresses acting on the other three faces
of the tetrahedron.
We first show that the unit normal to the inclined plane, denoted by n, has the components
ΔAx ΔAy ΔAz
nx = , ny = , nz = (K.1)
ΔA ΔA ΔA
where ΔAx (resp. ΔAy , ΔAz ) is the area of that face of the tetrahedron to which the x axis (resp. y,
z axis) is normal, see Fig. K.1(a). Consider the cross product:
c = a × b = (Δyey − Δxex ) × (Δzez − Δyey )
= ΔyΔzex + ΔxΔzey + ΔxΔyez = 2ΔAx ex + 2ΔAy ey + 2ΔAz ez
where ex , ey , ez are the unit basis vectors. The vector c is normal to the inclined plane and its
absolute value is 2ΔA. Therefore n = c∕|c| = c∕2ΔA and the components of the unit normal are
as given by eq. (K.1).
The equations of equilibrium are:
∑
ΔFx = 0 ∶ Tx ΔA − 𝜎x ΔAx − 𝜏yx ΔAy − 𝜏zx ΔAz = 0
∑
ΔFy = 0 ∶ Ty ΔA − 𝜏xy ΔAx − 𝜎y ΔAy − 𝜏zy ΔAz = 0
∑
ΔFz = 0 ∶ Tz ΔA − 𝜏xz ΔAx − 𝜏yz ΔAy − 𝜎z ΔAz = 0.
342 Appendix K Useful algorithms in solid mechanics
→ z
ez
Tz
→
c
τxy
ΔAy
Δz ΔA τyz σx
→
b τyx Ty
→
ΔAx ey
τxz
σy
τzx
τzy y
ΔAz
→ → Tx σz
ex a Δy
x
Δx
(a) (b)
Figure K.1 Notation.
On dividing by ΔA, and making use of the fact that the stress tensor is symmetric, we have:
⎧T ⎫ ⎡ 𝜎 𝜏 𝜏 ⎤ ⎧ n ⎫
⎪ x ⎪ ⎢ x xy xz ⎥ ⎪ x ⎪
⎨Ty ⎬ = ⎢𝜏yx 𝜎y 𝜏yz ⎥ ⎨ny ⎬ ⋅ (K.2)
⎪Tz ⎪ ⎣𝜏zx 𝜏zy 𝜎z ⎦ ⎪nz ⎪
⎩ ⎭ ⎩ ⎭
In index notation this can be written as:
Ti = 𝜎ij nj . (K.3)
K.2 Transformation of vectors
Consider a Cartesian coordinate system xi′ rotated relative to the xi system as shown in Fig. K.2.
Let 𝛼ij be the angle between the axis xi′ and the axis xj and let
gij = cos 𝛼ij . (K.4)
In other words, the ith row of gij is the unit vector in the direction of axis xi′
in the unprimed system.
Therefore if ai is an arbitrary vector in the unprimed system then the same vector in the primed
system is:
a′i = gij aj . (K.5)
Conversely, the jth column of gij is the unit vector in the direction of axis xj in the primed system.
Therefore if a′r is an arbitrary vector in the primed system then the same vector in the unprimed
system is:
ar = gkr a′k . (K.6)
Given the definition of gij and the orthogonality of the coordinate systems, gij multiplied by its
transpose must be the identity matrix:
gri grj = gis gjs = 𝛿ij . (K.7)
K.3 Transformation of stresses 343
z x3
z′ x′3
y′ x′2
y x2
x x1
x′
x′1
Figure K.2 Coordinate transformation. Notation.
In other words gij is an orthogonal matrix. This can be proven formally as follows:
ai = gri a′r
a′s = gsj aj
and
a′r = 𝛿rs a′s = 𝛿rs gsj aj .
Therefore,
ai ≡ 𝛿ij aj = gri 𝛿rs gsj aj

⏟⏟⏟
gri grj
and, for an arbitrary vector aj we have:
(𝛿ij − gri grj )aj = 0.
Consequently the bracketed term must be zero. This completes the proof.
K.3 Transformation of stresses
Referring to the definition of traction vectors given in Section K.1, we have:
Ti′ = 𝜎ik′ n′k = 𝜎ik′ gks ns
and applying the transformation rule (K.5):
Ti′ = gir Tr = gir 𝜎rs ns .
Therefore we have:
(gks 𝜎ik′ − gir 𝜎rs )ns = 0.
Since this equation holds for arbitrary ns , the bracketed term must vanish. Consequently:
gks 𝜎ik′ = gir 𝜎rs

multiplying by gjs ,
gjs gks 𝜎ik′ = gir gjs 𝜎rs
⏟⏟⏟
𝛿jk
and using eq. (K.7), we have the transformation rule for stresses:
𝜎ij′ = gir gjs 𝜎rs . (K.8)
Remark K.1 Denoting the stress tensor by [𝜎] and gij by [g], eq. (K.8) is the symmetric matrix
triple product
[𝜎 ′ ] = [g][𝜎][g]T . (K.9)
K.4 Principal stresses
It is possible to find a plane such that the traction vector, acting on that plane, is normal to the
plane (i.e., the shearing components are zero):
𝜎ij nj = Tni ≡ T𝛿ij nj
where T is the magnitude of the traction vector. Therefore:
(𝜎ij − T𝛿ij )nj = 0. (K.10)
This is a eigenvalue problem. Since 𝜎ij is symmetric, all eigenvalues are real. The eigenvalues are
called principal stresses and the eigenvectors define the directions of the principal stresses. Since
the eigenvectors are mutually orthogonal, in every point there is an orthogonal coordinate system
in which the stress state is characterized by normal stresses only. This coordinate system is uniquely
defined only when all eigenvalues are simple.
It follows from eq. (K.10) that the principal stresses are the roots of the following characteristic
equation:
T 3 − I1 T 2 − I2 T − I3 = 0 (K.11)
where
1
I1 = 𝜎kk , (𝜎 𝜎 − 𝜎ii 𝜎jj ), I3 = det (𝜎ij ).
I2 = (K.12)
2 ij ij
The principal stresses are denoted by a single index: 𝜎1 , 𝜎2 , 𝜎3 and ordered such that 𝜎1 ≥ 𝜎2 ≥ 𝜎3 .
The principal stresses characterize the state of stress in a point.
The principal stresses do not depend on the coordinate system in which the stress components are
given. Therefore the coefficients I1 , I2 and I3 are invariant with respect to rotation of the coordinate
system. These coefficients are called the first, second and third stress invariant respectively.
K.5 The von Mises stress

The stress deviator tensor is defined by
1
𝜎̂ ij = 𝜎ij − 𝛿ij 𝜎kk . (K.13)
3
It has the property that its first invariant is zero. The second invariant of the stress deviator tensor,
denoted by J2 , is particularly important in materials science and engineering because, according to
a widely used phenomenological model proposed by von Mises1 , ductile materials begin to yield
when J2 reaches a critical value. This is known as the von Mises yield criterion or the distortion
energy criterion. Since 𝜎̂ kk = 0, the second invariant is
1
J2 = 𝜎̂ 𝜎̂ . (K.14)
2 ij ij
Writing J2 in terms of the principal stresses we have:
( )
1 1 1 1
J2 = (𝜎1 − 𝜎kk )2 + (𝜎2 − 𝜎kk )2 + (𝜎3 − 𝜎kk )2
2 3 3 3
1( )
= (𝜎1 − 𝜎2 ) + (𝜎2 − 𝜎3 ) + (𝜎3 − 𝜎1 ) .
2 2 2
(K.15)
6
According to the von Mises yield criterion at the onset of yielding in ductile materials J2 has a
critical value which is a material property found though uniaxial tensile tests of specimens, such as
the specimen shown in Fig. 6.1. At the onset of yielding 𝜎11 = 𝜎yld , 𝜎22 = 𝜎33 = 0 in the test section.
Therefore the critical value of J2 is
1 2
J2crit = 𝜎 .
3 yld
The von Mises stress, denoted by 𝜎, is defined by
√
(𝜎1 − 𝜎2 )2 + (𝜎2 − 𝜎3 )2 + (𝜎3 − 𝜎1 )2
𝜎= ⋅ (K.16)
2
At the onset of yielding 𝜎 = 𝜎yld . The von Mises stress is one possible generalization of the uniaxial
yield stress to arbitrary stress conditions.
Example K.1 Let us predict the onset of yielding in pure shear, that is, 𝜎12 = 𝜎21 = 𝜏, all other
components of the stress tensor are zero. We will use eq. (K.16) to find the value of 𝜏 at which the
material begins to yield. The principal stresses are: 𝜎1 = 𝜏, 𝜎2 = 0, 𝜎3 = −𝜏. Therefore
√
𝜎 = 3𝜏yld = 𝜎yld
√
hence 𝜏yld = 𝜎yld ∕ 3.
K.6 Statically equivalent forces and moments
Suppose that a force Fi and moment Mi act on a rigid body. The line of action of Fi passes through
point P. Force Fi and moment Mi are statically equivalent to force Fi′ whose line of action passes
through point P′ and moment Mi′ if the following condition is satisfied:
Fi′ = Fi (K.17)
Mi′ = Mi − eijk rj Fk (K.18)
where rj is the position vector of point P′ with reference to point P as indicated in Fig. K.3. The
moment vector components are indicated by double arrows.
1 Richard Edler von Mises 1883–1953.

x′3
M′3
x3
F′3
M3 P′
rj F′2
F′1 M′2 x′2
F3
P x′1 M′1
F1 x2
F2 M2
M1
x1
Figure K.3 Statically equivalent forces and moments. Notation.
Statically equivalent force-moment pairs are used in assigning traction boundary conditions such
that the conditions of equilibrium are satisfied.
Example K.2 Consider a segment of a coil spring shown in Fig. 5.13. The centerline is given by
d
x1 = x = rc cos 𝜃, x2 = y = rc sin 𝜃, 𝜃, 0 < 𝜃 < 𝜋∕3.
x3 = z = (K.19)
2𝜋
The spring is compressed by two equal and opposite forces F acting along the z-axis. We are inter-
ested in finding the components of the statically equivalent forces on cross-sections A and B in the
local coordinate systems indicated in Fig. K.4 where the coordinate axes xA , yA , zA and xB , yB , zB are
coincident respectively with the tangent, normal and binormal of the centerline at cross-sections
A and B.
The solution involves two steps: First, the statically equivalent forces and moments acting on
cross-sections A and B are determined in local coordinate systems the axes of which are parallel to
the x, y, z system. For this we use equations (K.17), (K.18). Second, the the force components are
transformed to the local coordinate systems using eq. (K.5).
Step 1: The forces and moments acting on cross-sections A and B, in coordinate systems the
origins of which are in the points of intersection of the centerline with the cross-sections and the
axes are parallel with the global coordinate axes, in conventional notation, are:
FA = F{0 0 1}T
MA = −{rc 0 0}T × {0 0 F}T = rc F{0 1 0}T
FB = F{0 0 − 1}T
MB = −{rc cos(𝜋∕3) rc sin(𝜋∕3) d∕6}T × {0 0 − F}T
= rc F{sin(𝜋∕3) − cos(𝜋∕3) 0}T .
yB
xB
B
zB
yA
A
zA
z
y x
Figure K.4 Example K.2: Segment of a coil spring.
Step 2: To transform the forces and moments to the local coordinate systems we need the trans-
formation matrices gijA and gijB , defined by eq. (K.4). We denote the tangent, normal and binormal
unit vectors by ti , ni and bi respectively. By definition, bi is the cross product of ti and ni , that is,
bi = eijk tj nk . Using eq. (K.19) and defining 𝛽 = d∕(2𝜋rc ), these unit vectors are:
1
ti = √ {− sin 𝜃 cos 𝜃 𝛽} (K.20)
1 + 𝛽2
ni = {− cos 𝜃 − sin 𝜃 0} (K.21)
1
bi = √ {𝛽 sin 𝜃 − 𝛽 cos 𝜃 1}. (K.22)
1 + 𝛽2
Therefore the transformation matrix gij is
⎡ √ − sin 𝜃 √
cos 𝜃 𝛽⎤
1 ⎢− 1 + 𝛽 2 cos 𝜃 − 1 + 𝛽 2 sin 𝜃 0 ⎥ ⋅
gij = √ (K.23)
1 + 𝛽 2 ⎢⎣ 𝛽 sin 𝜃 −𝛽 cos 𝜃 1⎦
⎥
In this example cross-section A is located at 𝜃 = 0. Therefore the local force and moment vectors
denoted by FAL and MAL are:
⎡ √0 1 𝛽⎤ ⎧0⎫ ⎧𝛽 ⎫
1⎢ ⎪ ⎪ F ⎪ ⎪
FAL = √ − 1 + 𝛽 2 0 0⎥ ⎨ 0 ⎬ = √ ⎨0⎬
1 + 𝛽 2 ⎢⎣ 0
⎥⎪ ⎪
−𝛽 1 ⎦ ⎩F ⎭ 1 + 𝛽2 ⎪1⎪
⎩ ⎭
⎡ √0 1 𝛽⎤ ⎧ 0 ⎫ ⎧1⎫
MAL = √
1⎢− 1 + 𝛽 2 0 0 ⎥ ⎪⎨
⎪
rc F ⎬ = √
rc F ⎪ ⎪
⎨ 0 ⎬⋅
1 + 𝛽 2 ⎢⎣ ⎥
−𝛽 1 ⎦ ⎪ ⎪ 1 + 𝛽2 ⎪−𝛽 ⎪
0 ⎩ 0 ⎭ ⎩ ⎭
y y y
Ωy
y2
Ωz
z2 z1 x
z z x z
x y1
Ω
(a) (b) (c)
Figure K.5 Notation.
Similarly, for cross-section B located at 𝜃 = 𝜋∕3, we have:

√
⎡− 3∕2 1∕2 𝛽⎤ ⎧ 0 ⎫ ⎧𝛽 ⎫
B 1 ⎢ √ √ ⎥⎪ ⎪ −F ⎪ ⎪
FL = √ ⎢− 1 + 𝛽 ∕2 − 3(1 + 𝛽 )∕2 0 ⎥ ⎨ 0 ⎬ = √
2 2
⎨0⎬
1 + 𝛽 2 ⎢√ ⎥⎪ ⎪ 1 + 𝛽2 ⎪ ⎪
⎣ 3𝛽∕2
√
−𝛽∕2 1 ⎦ ⎩−F ⎭
√ ⎩1⎭
⎡ − 3∕2 1∕2 𝛽 ⎤ ⎧rc F 3∕2⎫
B 1 ⎢ √ √ ⎥⎪ ⎪
ML = √ ⎢− √1 + 𝛽 2 ∕2 − 3(1 + 𝛽 2 )∕2 0 ⎥ ⎨ −rc F∕2 ⎬
1 + 𝛽2 ⎢ ⎥⎪ ⎪
⎣ 3𝛽∕2 −𝛽∕2 1⎦ ⎩ 0 ⎭
⎧1⎫
−r F ⎪ ⎪
= √ c ⎨ 0 ⎬⋅
1 + 𝛽2 ⎪ ⎪
⎩−𝛽 ⎭
Remark K.2 Note that in the foregoing example FBL = −FAL and MBL = −MAL . This is because in
this special case the transformation of the force and moment vectors to the local system is indepen-
dent of 𝜃.
K.6.1 Technical formulas for stress

The term “rod” will be used as a generic name for any slender body such as arch, bar, beam, col-
umn, spring and shaft in the following. In the formulation of mathematical models for rods it is
often necessary to apply traction boundary conditions that have known resultants. The question
is: Given the stress resultants, what surface tractions should be prescribed? The answer to this
question is not unique; however, in the technical theory of rods there are commonly used relation-
ships between stress distributions corresponding to moments and forces acting on a cross-section.
These distributions are based on assumptions concerning the mode of deformation of straight rods
supplemented by equilibrium considerations. A brief summary is presented in the following. For
detailed discussion we refer to introductory texts on the strength of materials.
The origin of the local coordinate system is in the centroid of the cross-section of a rod, as indi-
cated in Fig. K.5. The locus of the centroids is the axis of the rod. In the large majority of problems
of practical interest the axis of the rod is a continuous and differentiable function. We assume this
to be the case in the following. For curved rods the tangent, normal and binormal unit vectors
associated with the axis of the rod define the local coordinate system. The notation is shown in
Fig. K.5.
Normal traction
The stress 𝜎x is assumed to be a linear function over a cross-section. Therefore it can be written as
𝜎x = a0 + a1 y + a2 z. (K.24)
By definition, the axial force Fx and the moments acting about the y and z axes, denoted respectively
by My and Mz are:
Fx = 𝜎x dydz, My = 𝜎x z dydz, Mz = − 𝜎x y dydz. (K.25)

∫Ω ∫Ω ∫Ω
On substituting eq. (K.24) into eq. (K.25) we get the following three equations:
⎡A 0 0 ⎤ ⎧a0 ⎫ ⎧ Fx ⎫
⎢0 I I ⎥ ⎪ ⎪ ⎪ ⎪
yz ⎨a1 ⎬ = ⎨−Mz ⎬ (K.26)
⎢ z
⎥⎪ ⎪ ⎪
⎣ 0 Iyz Iy ⎦ ⎩a2 ⎭ ⎩ My ⎪⎭
where A is the area of the cross-section and
Iy = z2 dydz, Iz = y2 dydz, Iyz = yz dydz (K.27)

∫Ω ∫Ω ∫Ω
are the moments of inertia of the cross-section. The zero entries in the coefficient matrix occur
because the y and z axes are centroidal axes and hence the first moments of the area are zero. On
solving eq. (K.26) we get the normal traction denoted by Tx :
Fx Mz Iy + Iyz My My Iz + Iyz Mz
Tx = − 2
y+ 2
z. (K.28)
A Iy Iz − Iyz Iy Iz − Iyz
Remark K.3 For straight rods the y and z axes are usually chosen such that Iyz = 0. When a
cross-section has one or more axis of symmetry then Iyz = 0.
Shearing tractions
The formulas for shearing tractions corresponding to the shearing forces Fy and Fz are developed
from considerations of equilibrium of straight rods. Details of the formulation can be found in any
introductory text on the strength of materials. The formulas are:
Fy Qz (y) Fz Qy (z)
Ty(F) = , Tz(F) = (K.29)
Iz tz Iy ty
where Qz (y) and Qy (z) are the first moments of the shaded areas identified are Ωy and Ωz in
figures K.5(b) and K.5(c) about the z and y axis respectively, tz = z2 − z1 and ty = y2 − y1 .
Example K.3 For a circular cross-section of radius r𝑤

r𝑤 √
2 2
Qz = 2 y r𝑤 2
− y2 dy = (r𝑤 − y2 )3∕2
∫y 3
√
2 4 ∕4. Therefore
and tz = 2 r𝑤 − y2 , Iz = r𝑤
2 2 2 2
4 r𝑤 − y 4 r𝑤 − z
Ty(F) = Fy and similarly Tz(F) = Fz ⋅ (K.30)
3 r𝑤 4
𝜋 3 r𝑤 4
𝜋
Finding the shearing tractions corresponding to the twisting moment Mx for general
cross-sections involves the solution of Poisson’s equation, usually by numerical means. However,
in the important special case of straight rods with circular cross-section a closed form solution
exists, the formulation of which is also available in introductory texts. In polar coordinates (r 𝜃),
the origin being coincident with the center of the circular cross-section, 𝜏r = 0 and
Mx r
𝜏𝜃 = (K.31)
J
where 𝜏𝜃 is normal to the radius, J = r𝑤4
𝜋∕2 is the polar moment of inertia and rc is the radius of
the circular cross-section. The shearing tractions corresponding to Mx are
2Mx z 2Mx y
Ty(M) = − , Tz(M) = ⋅ (K.32)
4
r𝑤 𝜋 4
r𝑤 𝜋
Remark K.4 The formulas (K.28), (K.30) and (K.32) are based on assumptions that are exactly
satisfied for straight rods of constant cross-section under special loading conditions only. These
formulas are useful in the general case because the corresponding tractions are in equilibrium
with the applied forces and moments and therefore the resultants of the difference between the
exact tractions and the tractions given by these formulas is zero. Therefore the error in the imposed
tractions causes local perturbations in the solution which decay in accordance with Saint-Venants’s
principle2 .
Example K.4 For the problem described in Example K.2 the tractions acting on section A in the
local coordinate system indicated in Fig. K.4 are:
FxA MzA y MyA z
Tx(A) = − +
A I I
A
Fy Qz MxA z
Ty(A) = −
Itz J
FzA Qy MxA y
Tz(A) = +
Ity J
where
F𝛽 F
FxA = √ , FyA = 0, FzA = √
1+ 𝛽2 1 + 𝛽2
rc F r F𝛽
MxA = √ , MyA = 0, MzA = − √ c
1 + 𝛽2 1 + 𝛽2
r 4𝜋 r 4𝜋 Q r 2 − y2 Qy 2 − z2
r𝑤
𝜋, I = 𝑤 , J = 𝑤 , = 𝑤
z
2
A = r𝑤 , = ⋅
4 2 tz 3 ty 3
In view of Remark K.2, the tractions acting on Section B in the local coordinate system are:
Tx(B) = −Tx(A) , Ty(B) = −Ty(A) , Tz(B) = −Tz(A) .
2 Adhémar Jean Claude Barré de Saint-Venant 1797–1886.

351
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357
Index
a flux 58
adaptive methods 38 homogeneous 53
ADINA 252 kinematic 71
admissible functions 80, 84 natural 9, 10, 53
Airy stress function 311 Neumann 8, 53, 54
Almansi strain 181 of convenience 53
analytic function 31, 289 periodic 9, 60, 72
angular velocity 43 radiation 265
antisymmetry 121 Robin 9, 53, 54
assembly 21 symmetric 9
asymptotic consistency 237, 250 symmetry 59, 72
temperature 58
b traction 71
Babŭska-Brezzi condition 47 Winkler spring 71
basis function 3 boundary layer 253
basis vectors buckling load factor 273
curvilinear 247
Bayes factor 190, 335 c
Bayes’ theorem 325, 326 calculus of variations 12
beam models 223 calibration 160, 187, 330
Bernoulli-Euler beam 229 calibration set 193
Timoshenko beam 226 Cauchy stress 181
bending moment 235 censored data 328
Bernoulli-Euler beam model 156, 268 coefficient
biharmonic equation 241, 311 of convective heat transfer 58
bilinear form 7, 290 of dynamic viscosity 78
binormal 347 of thermal conduction 57, 61, 64
boundary condition 58 of thermal conductivity 112
antisymmetric 9 of thermal expansion 68
antisymmetry 59, 72 spring 71
convection 58 coefficient of determination 194
Dirichlet 8, 53, 54 coefficient of variation 194
displacement 71 coil spring 346
essential 8, 53, 103 complex potentials 311
358 Index
compliance matrix director functions 223, 247

elastic-plastic 279 Dirichlet, see boundary condition 53
condensation 24 discretization 2, 39
condition number 13, 41 display grid 112
confidence interval 335 divergence theorem 52, 54, 79
conservation law 56 domain
consistency 160 of calibration 194
consistency condition 277
constant e
Stefan-Boltzmann 265 effectivity index 125
constraint eigenfunction
rigid body 84 Mode I, 313
contact 281 Mode II, 313
contact zone 281 eigenpair 43
continuity 45, 55 eigenvalue 313
C1 243 element
contour integral 108 eight-node quadrilateral 94
contour integral method 130 four-node quadrilateral 93
contour plot 112 nine-node quadrilateral 94
convergence 291 six-node triangle 95
monotonic 81 three-node triangle 95
radius of 121 element stiffness matrix 161
coordinate space 289 emissivity 265
coordinates endurance limit 189, 331
Eulerian 266 energy norm 7, 54, 292
Lagrangian 266 energy release rate 140, 321
triangular 94 energy space 7
cross product 347 equations of motion 70
cumulative distribution function equations of static equilibrium 70
329 equilibrium
marginal 207 dynamic 69
cycle ratio 188 equivalence of norms 292
equivalent stress 188, 276
d error
d’Alembert’s principle 70 a priori estimate 296
data analysis model form 3
parametric 187, 325 numerical 2
degrees of freedom 17, 25 of approximation 3
derivative pollution 151
fractional 30 error estimation 33
design rules 2 asymptotic 31
diameter of an element 16 essential infimum 289
differential volume 106 essential supremum 289
differentiation 108 Euler-Lagrange equation 12
dimension 51 evidence 326
dimensional reduction 61 extension
dipole problem 172 p-version 151, 174
Index 359
extraction 2 h
extraction function 27, 29, 117, 129, 179 h-extension 17
h-version 17
f hierarchic spaces 150
fastener 337 highly stressed volume 203
fatigue limit 189, 191, 199, 330, Hilbert space 292
331 Hooke’s law 69, 158
fatigue strength 200, 331 generalized 69
field functions 223
hourglassing 162
finite element 16
hp-extension 17
hexahedral 104
hp-version 17
pentahedral 104
standard quadrilateral 91
i
standard triangular 91
incompressibility
tetrahedral 104
condition of 77
finite element mesh 16
index notation 51
finite element method 3
infimum 289
h-version 166
initial condition 58, 72
p-version 166
integration 106
finite element model 2
interpolation 307
finite element space
isoparametric mapping, see mapping 97
anisotropic 252
hierarchic 33
flank angle 192
j
Jacobian
flux
determinant 106
heat 56, 59
vector 53 matrix 106, 112
flux intensity factor 128 inverse of 108
forcing function 3, 142
Fourier’s law 56 k
free body 179 Kronecker delta 51
frequency 43
function l
even 60 L-shaped domain 315
odd 60 Lagrange polynomial 13
functional Lamé constants 69
quadratic 11 Laplace equation 120
law of parsimony 175
g Lebesque constant 105, 307
Galerkin orthogonality 13 legacy codes 168
gap elements 283 Legendre polynomial 13, 97, 301, 303
generalized form 1 likelihood function 190, 330
generalized solution 10, 55 likelihood ratio 198
uniqueness of 84 linear form 7, 290
Girkmann problem 167 linear independence 5
global number 23 linear space 290
grading factor 16 Lobatto point 20
360 Index
locking 143, 253 n

shear 229 NASTRAN 165
log likelihood 330 Navier equations 70
Neumann, see boundary condition 53
m neutral axis 157
mapping neutral plane 157
by blending functions 99, 307 Newton-Raphson method 100, 112
high order elements 101 Nitsche’s method 48
improper 107 nodal displacement 161
inverse 100, 112 nodal force 29, 113, 161
isoparametric 97, 101 nodal set 307
linear isoparametric 97 node point 16
proper 106 norm 290
quadratic isoparametric 97 L2 166
subparametric 97 energy 166
superparametric 97 notch sensitivity 192, 194
mapping function 17 notch sensitivity factor 202
marginal likelihood 326 nuisance parameter 326
marginalization 326, 332 numerical error 2
material stiffness matrix 81 numerical simulation 2
mathematical model 2, 155
hierarchic models 223 p
range of validity 159 p-convergence 31
Matlab 330 p-extension 17
matrix p-version 17
Gram 19 path-independent integral 129
mass 19 permutation symbol 52
stiffness 18 physical reality 155
maximum likelihood 189 physics-based model 156
maximum norm 307 piecewise polynomials 243
mean optimal set 252, 308 plane strain 74
median 196 plane stress 237
membrane analogy 1 plasticity 275
membrane force 234 plates
mesh Kirchhoff plate 240
geometric 16 Reissner-Mindlin plate 236
quasiuniform 16, 280 Poisson’s equation 349
radical 16 Poisson’s ratio 68
uniform 16 polar moment of inertia 350
Mode I eigenfunctions 313 pollution error, see error 151
model form error 171 posterior probability 326
model hierarchy 265 potential energy 55, 80
modulus of elasticity 68, 157 sign of 12
modulus of rigidity, see shear modulus potential flow 55
69 predicted median 333
Mohr circle 236 predictor of fatigue life 187
moment of inertia 349 pressure 78
Index 361
principal direction 59 resolvent set 271

principal moments 236 Reynolds number 78
principal stress 344 rigid body
principal structural element 164 displacement 84
principle of minimum potential rotation 84, 102, 269
energy Robin, see boundary condition 53
application to beams 224 robustness 143
application to plates 238 rod 348
principle of virtual work 83 runout 188, 328, 330
prior probability 326
probability s
Bayesian 325 S-N curve 189, 330
frequentist 325 S-N data 189, 328
product rule 325, 332 Saint-Venants’s principle 185
product space 92, 95 scalar product 52
anisotropic 250 Schwarz inequality 37, 293
profile likelihood 335 scope
propping effect 172 of calibration 202
pull-back polynomials 98 of validation 205
secant modulus 278
q seminorm 290
QoI, see quantity of interest 2 separation of variables 42
quadrature 107, 303 shadow function 137, 141
Gauss-Lobatto 303, 304 shape function 13, 161
Gaussian 303 hierarchic 15
qualified test record 193 shape function vector 109
quantile 333 shape functions
quantity of interest 2, 26, 155, 170, 171, 187, 3-dimensional 104
265 Lagrange 93
quarter point element 99 shear correction factor 226, 227, 237, 238
quasiuniform, see mesh 16 shear force 234
shear modulus 69
r shearing tractions 349
radiation 58 shell
random fatigue limit 189, 330 hierarchic models 247
ranking 187 hyperboloidal 252
rate of convergence 148 structural 247
algebraic 31 shell model
exponential 32 Naghdi 248
Rayleigh quotient 44 Novozhilov-Koiter 249
reduced integration 162 shift theorem 32
reduction factor 274 simulation 1, 155
region of primary interest 150, 239 simulation governance 164, 167, 193, 205,
region of secondary interest 150 219
regularity 16, 30 smart application 186
representative volume element 191, 255 Sobolev space 291
residual stress 87 solution verification 168
362 Index
solver stress intensity factor 139, 321

direct 26 generalized 153
iterative 26 stress invariant 202
space stress stiffening 181, 271
energy 79 stress vector 341
Euclidean 51 StressCheck 51, 280
finite element 17, 102 superconvergence 130, 182
infinite-dimensional 7 supremum 289
span 5, 12 surface finish 326
spectrum 43, 271 surface integral 108
continuous 271 survival function 207
point 271 symmetry 121
spring rate 177 axial 65
square integrable function 291
stability 47, 160 t
stability parameter 49 T-stress 139, 319
standard element 103 Taylor series 35
statical equivalence 345 technical formulas 185
stationary problems 57 temperature
statistical inference 325 absolute 58, 265
steady state problems 57 test function 80, 290
Steklov method 133 test section 188
stiffness matrix 18 test space 8, 9, 290
elastic-plastic 279 test stress 188
Stokes equations 78 thin solid model 249, 268, 273
strain 157 traction 71
Almansi 267 traction vector 341
enginering shear 68 trial function 290
equivalent elastic 276 trial space 7, 9, 290
equivalent plastic 276 trunk space 91, 95
Green 268 anisotropic 249
infinitesimal 67 truss 160
mechanical 69 tuning 164
normal 67 twisting moment 235
shear 67
thermal 69 u
total 69 uncertainty 78
volumetric 76 aleatory 163
stress uniqueness 9
transformation of 344
resultant 226 v
stress concentration factor 190 validation 166, 187, 195
fatigue 191 validation metric 205
stress deviator tensor 344 validation scenario 196
Index 363
validation set 193, 196 virtual experimentation 265

variational crime 162 virtual work 83
vector viscosity 78
transformation of 342 von Mises stress 184, 189, 202, 276
verification 148, 166 von Mises yield criterion 345
data 119
solution 152 w
virtual displacement 83 Winkler spring, see boundary condition 71
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What topics are covered in the book?

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What are some of the key concepts discussed?

What are some of the key concepts discussed?

Finite Element Analysis

WILEY SERIES IN COMPUTATIONAL MECHANICS

Method, Veriﬁcation and Validation

Limit of Liability/Disclaimer of Warranty

Library of Congress Cataloging-in-Publication Data

Names: Szabó, B. A. (Barna Aladar), 1935- author. | Babuška, Ivo, author.

Cover Design: Wiley

Set in 9.5/12.5pt STIXTwoText by SPi Global, Chennai, India

Preface to the second edition xi

1 Introduction to the ﬁnite element method 1

2 Boundary value problems 51

2.2.1 Generalized formulation 53

3.11 Computation of the solution and its first derivatives 111

4 Pre- and postprocessing procedures and veriﬁcation 119

6 Calibration, validation and ranking 187

6.2.1 The effect of notches – calibration 193

7 Beams, plates and shells 223

8 Aspects of multiscale models 255

9 Non-linear models 265

9.2.3 Plasticity 275

Appendix A Deﬁnitions 289

Appendix B Proof of h-convergence 295

Appendix C Convergence in 3D: Empirical results 297

Appendix D Legendre polynomials 301

Appendix E Numerical quadrature 303

Appendix F Polynomial mapping functions 307

Appendix G Corner singularities in two-dimensional elasticity 311

Appendix H Computation of stress intensity factors 319

H.3.2 Antisymmetric (Mode II) loading. 323

Appendix I Fundamentals of data analysis 325

Appendix J Estimation of fastener forces in structural connections 337

Appendix K Useful algorithms in solid mechanics 341

Preface to the second edition

Barna Szabó and Ivo Babuška

Preface to the ﬁrst edition

About the companion website

This book is accompanied by a companion website:

Introduction to the ﬁnite element method

1 Ludwig Prandtl 1875–1953.

Fmax ≤ Fall (1.1)

|Fmax − Fnum | ≤ 𝜏Fmax . (1.2)

Fnum = (1 − 𝜏)Fmax . (1.3)

Therefore it has to be shown that

Fnum ≤ (1 − 𝜏)Fall . (1.4)

1.1 An introductory problem

Let us find aj such that the integral  defined by

which will be written as

which represents n simultaneous equations in n unknowns. It is usually written in matrix form:

([K] + [M]) {a} = {r}. (1.14)

Example 1.1 Let 𝜅 = 1, c = 1, 𝓁 = 2 and

With these data the exact solution of eq. (1.5) is

We seek an approximation to u in the form:

un = u2 = a1 x(𝓁 − x) + a2 x(𝓁 − x)2 .

On computing the elements of [K], [M] and {r} we get

The choice of basis functions