elasticity --- anisotropy --- laminates solid mechanics
TRANSCRIPT
Pauli Pedersen
E L A S T I C I T Y --- A N I S O T R O P Y ---
L A M I N A T E S
with matrix formulation, finite elements
and an index to matrices
x3= z
x2
x1 σ
1111
σ12
12
σ21
21
σ22
22
–γ
SOLID MECHANICS
TECHNICAL UNIVERSITY OF DENMARK
II
Pauli Pedersen: 1. Contents and Introduction
E L A S T I C I T Y --- A N I S O T R O P Y --- L A M I N A T E S
with matrix formulation, finite elements and an index to matrices
Copyright 1998 by Pauli Pedersen,
but feel free to copy for personal use.
ISBN 87---90416---01---05
III
Pauli Pedersen: 1. Contents and Introduction
FOREWORD
This book originates from notes for a course in elasticity (Pedersen
(1992)) and a translation from Danish is needed because this course is
now taught in English. Extensions from earlier notes on laminates
(Pedersen (1988)) and from a small book on finite elements (Pedersen
(1984)) are naturally included. The influence from recent research on
material optimization (Pedersen (1995a)) will also be seen and finally
the notation is in accordance with the 2 ---contracted notation, advo-
cated in Pedersen (1995b). It is my hope that the reader will find the ef-
forts worth---while, and critics, advices and information about possible
errors will be most welcome (email: [email protected]).
Many students and colleagues have influenced the book and without
mentioning specific names, I am most thankful for their guidance. I
thank Robert Zetterlund for producing many of the figures and a lot of
additional help; I thank Bente Brask Andersen and Betina Christian-
sen for the text---editing with all its revisions. Without their valuable
help the book would not have been finished.
In gratitude of awarm family life, I dedicate this book tomywifeHanne
and my children Niels, Mads and Sine.
January 1997
Pauli Pedersen
IV
Pauli Pedersen: 1. Contents and Introduction
This second version is almost unchanged, but the known errors have
been corrected. Furthermore, the book is now available at the internet:
http://www.fam.dtu.dk/html/pp.html
November 1998
Pauli Pedersen
V
Pauli Pedersen: 1. Contents and Introduction
CONTENTS
page
Foreword III
1. Introduction 1
2. Geometry --- Kinematics 9
2.1 Displacement 9
2.2 Displacement field 10
2.3 Displacement gradients 10
2.4 Stretches and strains in 1---D 11
2.5 Strains from displacement gradients 15
2.6 2---D Strain description 17
2.7 3---D Strain transformation 20
2.8 Dilatation and distortion, deviatoric strains 26
2.9 Compatibility of a strain field 28
2.10 Examples of strain calculations 29
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Pauli Pedersen: 1. Contents and Introduction
3. Statics --- Equilibrium 35
3.1 Force 35
3.2 Moment 37
3.3 Stress concept 38
3.4 Equilibrium 39
3.5 Descriptions of stress state 43
3.6 Hydrostatic stress and deviatoric stress 45
3.7 A classic example 46
4. Physics --- Constitutive Models 53
4.1 Material parameters and modelling 53
4.2 Rotational transformations 56
4.3 Classification of 2---D stiffness for anisotropic elasticity 61
4.4 From 3---D to 2---D constitutive matrix 67
4.5 Energy densities in non---linear elasticity 72
4.6 Measuring the constitutive parameters 79
4.7 Eigenvalues and eigenmodes for the
constitutive matrices 85
4.8 Specific constitutive models 87
5. Stress Concentration Examples 93
5.1 The classic solution for stress concentration around
elliptical holes 93
5.2 Numerical solutions for stresses around elliptical holes 96
5.3 Design with orthotropic materials 100
VII
Pauli Pedersen: 1. Contents and Introduction
6. Stationarity and Extremum Principles of Mechanics 105
6.1 The work equation, an identity 105
6.2 Symbols and definitions 108
6.3 Real stress field and real displacement field 109
6.4 Real stress field and virtual displacement field 111
6.5 Virtual stress field and real displacement field 114
6.6 Minimum total potential 116
6.7 Minimum complementary potential and
two---sided bounds 122
6.8 Table of all principles 124
7. Application of Energy Principles 125
7.1 Elastic energy in straight beams 125
7.2 Simplified beam results 129
7.3 Complementary principles to solve a beam problem 130
7.4 Approximate solution examples 132
7.5 Complementary approximations 135
8. Laminate Analysis 141
8.1 Introduction to laminates 141
8.2 Basic assumptions 142
8.3 Laminate stiffnesses 145
8.4 Special laminate layouts 149
8.5 Advanced laminate models 153
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Pauli Pedersen: 1. Contents and Introduction
9. Bending of Rectangular Orthotropic Plates 155
9.1 Restricted class of problems 155
9.2 Structural equilibrium 157
9.3 Plate differential equations 159
9.4 Boundary conditions 160
9.5 Navier analytical solution 162
9.6 Fourier expansions of loads 164
9.7 Levy analytical solution 168
10. Torsion of Cylindrical Bars 171
10.1 A classic problem of elasticity 171
10.2 Circular cross---sections 172
10.3 Non---circular cross---sections 175
10.4 Analytical solution for hollow elliptic cross---sections 180
10.5 Other analytical solutions 185
10.6 Analytical solution for rectangular solid cross---sections 188
10.7 Thin---walled open cross---sections 195
10.8 Thin---walled closed cross---sections 198
10.9 Examples of thin---walled closed cross---sections 205
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Pauli Pedersen: 1. Contents and Introduction
11. Finite Element Analysis 209
11.1 Contents of this section 209
11.2 Element geometry and nodal positions 210
11.3 Displacement assumptions 213
11.4 Node displacements and the configuration matrix 216
11.5 The strain/displacement matrix and
general equilibrium 221
11.6 Stiffness submatrices and basic matrices 225
11.7 Analytical integration and resulting basic matrices 228
11.8 Equivalent nodal loads and consistent mass matrices 240
11.9 Note on strain evaluation 245
12. An Index to Matrices 249
13. References, List of Symbols and Index 301
X
Pauli Pedersen: 1. Contents and Introduction
1
Pauli Pedersen: 1. Contents and Introduction
1. INTRODUCTION
GOOD We all have our favourite sentences, often written by scientists whose
QUOTATIONS work leaves us with deep respect. Two of my favourite sentences are
Einstein & Infeld:
“Most fundamental ideas of science are essentially simple and may, as a
rule, be expressed in a language comprehensible to everyone”.
and
Lanczos (1949):
“Many of the scientific treatises of today are formulated in a half---mystical
language, as though to impress the reader with the uncomfortable feeling
that he is in the permanent presence of a superman. The present book is
conceived in a humble spirit and is written for humble people. The author
knows from past experience that one outstanding weakness of our present
system of college education is the custom of classing certain fundamental
and apparently simple concepts as “elementary”, and of relegating them to
an age---level at which the student’s mind is not mature enough to grasp
their true meaning. The fruits of this error can be observed daily”.
Writing this book, I shall do my best to live up to the goal of simplicity.
That there is a latent problem when dealing with elasticity and mathe-
matics follows from two other sentences. The first one is rather provok-
ing and will be given without the specific reference, which I cannot
again locate.
2
Pauli Pedersen: 1. Contents and Introduction
“By Mariotte’s time the whole subject of the behaviour of materials and
structures under loads was beginning to be called the science of elasticity
--- for reasons which will become apparent in the next chapter --- and we
shall use this name repeatedly throughout this book. Since the subject
became popular withmathematicians about 150 years ago I am afraid that
a really formidable number of unreadable, incomprehensible books have
been written about elasticity, and generations of students have endured
agonies of boredom in lectures aboutmaterials and structures. In my opin-
ion the mystique and mumbo---jumbo is overdone and often beside the
point”.
and
Arnold (1977):
“The axiomization and algebraization of mathematics, after more than 50
years, has led to the illegibility of such a large number ofmathematical texts
that the threat of complete loss of contact with physics and the natural
sciences has been realized. The author attempts to write in such a way that
this book can be read by not only mathematicians, but also all users of the
theory of differential equations”.
A fifth (and last) sentence by one of my own teachers should also be
stated (translated from Danish)
Meldahl (1960):
“A Technical University should teach the students scientific methods and
therefore not use the very limited time to teach procedures that may be
found in hand---books”.
Meldahl often pointed out the clear distinction between teaching why
and teaching how. Professional engineers should naturally know how,
3
Pauli Pedersen: 1. Contents and Introduction
but with the rapid development of engineering science, understanding
why must have the highest priority.
WORDS OF Now to the title of this book. The book is written for undergraduate
THE TITLE teaching in a one semester course which takes 15---20% of the study
time for the student. Out ofmany possibilities, a limited number of top-
ics have been selected, and one excuse for writing the book is that the
first 10 chapters serve this specific purpose of teaching.
Elasticity means that we restrict ourselves (mainly) to reversible pro-
cesses with a unique solution. A number of classical books like Green
& Zerna (1954), Love (1927(1892)), Lurie (1964), Novozhilov (1961),
Sokolnikoff (1956) have been written on the subject and perhaps some
of the early ones are still the best, like the one by Love. Today elasticity
can be illustrated with solutions directly from commercially available
finite element programs. The importance ofmore complicated “analyt-
ical” methods should therefore be seen in a different light. With this in
mind we shall concentrate on the physics of elasticity rather than on the
mathematics of elasticity.
Anisotropy is normally treated as an extension to isotropy.Of the classi-
cal books we must mention Lekhnitskii (1981). However, most books
on elasticity contain very little about anisotropy.Here we shall take iso-
tropy as a special case of anisotropy and in general we shall strive
towards a general classification of material stiffness. Starting out from
anisotropy offers a number of advantages and does not seem to give our
students additional problems. The extensive use of new, advanced
materials more or less leaves us no choice: we must teach anisotropy
even at the undergraduate level.
4
Pauli Pedersen: 1. Contents and Introduction
Laminates are now used extensively and laminate analysis is an impor-
tant subject within the mechanics of composite materials. With lami-
nates we have the possibility to “designmaterial” which is still an active
area of research.
With matrix formulation means that the traditional tensor notation is
not left out but given in parallel to the matrix notation that dominates
the literature on the finite elementmethod, FEM. The extreme impor-
tance of this methodmust also influence our teaching of the theoretical
basis. Since the appearance of FEM in the mid---fifties our possibilities
to solve field problems have changed in a revolutionary way.
This is also the reason for including a chapter on finite elements in a
presentation which is not the traditional one, but anyhow a presenta-
tion based on many years of actual teaching.
A chapter in the form of an index tomatrices is included, and this index
is intended to be self---explanatory. To a large extent the mathematics
of the book belongs to linear algebra, so there is no need to prove the
results from this theory again, specifically for elasticity.
THE Three aspects of solidmechanics are especially important, i.e.geometry
CHAPTERS --- statics --- physics. We start out with a description of the kinematics,
i.e. the geometrical aspects. Displacements, displacement gradients
KINEMATICS and strains are the involved field quantities, i.e. wemust describe vector
fields.We focus on the fact that strain is a concept and thatmany defini-
tions are possible. Strain is not a measurable physical quantity, but
working extensively with strains, we tend to forget that.
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Pauli Pedersen: 1. Contents and Introduction
EQUILIBRIUM The second chapter is on equilibrium, i.e. the statics of elasticity. Force,
moments and stresses are the involved field quantities, i.e. again vector
fields with the physical quantities being forces. For the concept of stress
we also try to get a physical picture, but we have to remember that stress
is not a directly measurable quantity.
CONSTITUTIVE Then by the physics in geometry---statics---physics wemean the consti---
MODELLING tutive modelling of the relations between strains and stresses. Being a
modelling between two conceptual fields, the constitutive models rely
on the chosen definitions for strains and stresses. In anisotropic mod-
els, the constitutive parameters depend on the directional reference. In
addition to rotational transformations for the strains and stresses, the
rotational transformations for the constitutive parameters are impor-
tant. The goal of this chapter is to treat all possible materials and aim
CLASSIFICATION at a classification, which however is limited to 2---D problems. Non---
linear constitutive modelling is also included.
SOLVED Graphical presentations of solutions obtained by FEM are shown, and
PROBLEMS we focus on the dependence of non---linearity, on the anisotropy and on
modelling of boundary condition. Classical examples of stress con-
centration are put forward, and the influence from shape design is
shown.
BACK TO In chapter six we return to the more abstract theory behind the sta---
THEORY tionarity and extremum principles of mechanics. It is not easy to com-
municate the importance of this theory to our students. An un---tradi-
tional presentation which separates the mathematics from the inter-
pretations is chosen, and the clear distinction between stationarity and
PRINCIPLE OF extremum is put forward. If the principle of virtual work is understood,
VIRTUAL WORK then only little is left.
6
Pauli Pedersen: 1. Contents and Introduction
Examples of application of energy principles hopefully show the practi-
cality of the theory. Furthermore, as we shall see later, FEM is based
on these principles, and this should in fact be motivation enough to
understand the background.
LAMINATES It is a matter of taste whether one calls a laminate a material or a struc-
ture. (In reality any material is a structure.) Anyhow, the growing
importance of laminates and composite materials makes it natural to
include a chapterdescribing the classical (simple) laminate theory.This
chapter is a natural follow---up of the anisotropic constitutive model-
ling in chapter four.
ORTHOTROPIC Bending of rectangular orthotropic plates can be solved at least in a
PLATES semi---analytical way. The differential equation with boundary condi-
tion is derived and solutions with double Fourier expansion are
obtained. In this chapter we see how much can be achieved without
using FEM, but we are naturally restricted by rather simple boundary
conditions. The application of laminates to these plates gives a direct
relation between the two chapters.
TORSION The last subject for the undergraduate course is torsion of cylindrical
bars. Traditionally, this subject belongs to a course in elasticity because
analytical solutions can be obtained and because it demonstrates the
elegant use of the stress function concept. We treat four classes of prac-
tical important problems, classified by the cross---sectional properties
as being circular, non---circular, open thin---walled and closed thin---
walled.
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Pauli Pedersen: 1. Contents and Introduction
FINITE A short and rather non---traditional introduction to the element
ELEMENTS analysis for the finite elementmethod is finally presented.To gainmaxi-
mum insight all results are derived analytically. Thus this chapter can
be read also as a supplement to themany good textbooks specifically on
FEM.
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Pauli Pedersen: 1. Contents and Introduction
9
Pauli Pedersen: 2. Geometry --- Kinematics
2. GEOMETRY --- KINEMATICS
displacements, displacement gradients, stretches, strains
2.1 Displacement
NOTATION Displacement will generally be symbolized by the letter v , and already
in this notation there is a source of misunderstanding. Therefore, a few
introductory remarks about the notations v , vi, v and v
T. In fig.
2.1 we show the displacement v , a scalar that gives the displacement
from a reference point P in a given direction. For 1---D problems v
describes the displacement of a specific point (material point).
v
P
x3
x1
x2
v
Fig. 2.1: Illustration of a scalar displacement v and a displacement vector v .
MATERIAL POINT In fig. 2.1 we also illustrate the displacement vector v with two ele---
ments for 2---D problems and with three elements for 3---D problems.
The individual elements of the vector are denoted viwith i= 1, 2 for
2---D and i= 1, 2, 3 for 3---D , i.e.
2–D : vT=
v1v2
(2.1)
3–D : vT=
v1v2v3
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Pauli Pedersen: 2. Geometry --- Kinematics
DIRECTION Thus, with the vector description, we give the direction as well as the
& MAGNITUDE magnitude of the displacement, still related to a specific point. The geo-
metric displacement vector is a physical reality that may be described
by a column matrix v or by a row matrix vT, and these matrices
will differ in different reference coordinate systems but be related by a
transformation matrix [Γ] .
Displacements can be measured.
2.2 Displacement field
MATERIAL FIELD The deformation state of a body is not described by the displacement
of a single point, but by the displacements of all the material points of
the body, i.e. by a field of displacements, here symbolized by
vi= vi(x) (2.2)
where x denotes space, i.e.
2–D : x= x1 , x2(2.3)
3–D : x= x1 , x2 , x3
The relation between material points and space may be chosen as the
state before displacements, but other possibilities exist.
Displacement fields can be measured, for example by means of holo-
graphy.
2.3 Displacement gradients
A pure translation and/or a pure rotation of a body will not result in
deformation of the body. More important than displacements are the
displacement gradients, with short notation denoted vi,j as defined by
11
Pauli Pedersen: 2. Geometry --- Kinematics
vi,j := ∂vi∂xj (2.4)
and thus related to a specific coordinate system x . For a 2---Dproblem,
the four displacement gradients are
v1,1 v1,2 v2,1 v2,2 (2.5)
ACTUAL
GRADIENTS and for 3---D problem, the nine displacement gradients are
v1,1 v1,2 v1,3 v2,1 v2,2 v2,3 v3,1 v3,2 v3,3 (2.6)
It is questionable whether displacement gradients can be measured
directly, but at least they can experimentally be approximated by
vi,j ≈ ∆vi∆xj (2.7)
and measured on a free surface with a strain gauge, as it is normally
called.
2.4 Stretches and strains in 1---D
STRAIN Strain is a concept, introduced to describe deformation of bodies. The
CONCEPT notion of strain refers to a specific point and thus we need a strain field,
to be defined from the field of displacement gradients.
NOT It is essential to note that strain is a conceptwithmany possible defini---
MEASURABLE tions and is thus not ameasurable physical quantity. Tomake this clear
we shall discuss different strain definitions.
12
Pauli Pedersen: 2. Geometry --- Kinematics
P
LL0
Fig. 2.2: Line element through point P with length L0 before deformation and
length L after deformation.
STRETCH The extension ratio λ , often called stretch, is defined by
λ := LL0
(2.8)
and is a quantity with a clear physical interpretation, andwe shall relate
the different strain definitions to this basic quantity.
CAUCHY STRAIN Strains are normally symbolized by the letter to indicate that small
quantities are expected. The most common definition is
C:= ∆L
L0
=L – L
0
L0
= λ – 1 (2.9)
with index C referring to thenameCauchy. This straindefinition is also
called linear strain or engineering strain.
NON ADDITIVE Abasic problemwith this definition is that it is not additive. Thus, if the
deformation is carried through in a sequence of steps, the sum of
Cauchy strains will be different from the total Cauchy strain. Let the
sequence of lengths be
L0< L
1< L
2... < Ln ... < L
N(2.10)
then we see directly that
N
n=1
Ln – Ln–1
Ln–1
≠
LN– L
0
L0
(2.11)
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Pauli Pedersen: 2. Geometry --- Kinematics
LOGARITHMIC An alternative definition of strain is the Hencky strain, also called the
STRAIN logarithmic strain, and is defined by
H:=
L
L0
d= n L
L0
= n(λ)=
(2.12)
n1+ C=
C– 12
C2+ ....
It follows directly from the definition that these strains are additive.We
note that for small strains (<< 1) we get H=
C.
GREEN--- In elasticity theory not restricted to small strains, a strain definition by
LAGRANGE squared lengths is used
STRAIN
G:=
L2 – L20
2L20
=12λ2 – 1= 1
2(λ+ 1)(λ – 1)=
(2.13)12(λ+ 1)
C=
C+
122
C
calledGreen---Lagrange strain. We again note that for small strains we
get G=
C.
To emphasize the fact that strain is a concept we shall mention other
definitions of strain which have been used in the past. The Almansi
strain, which differs from the Green---Lagrange strain only by setting
the change of squared length relative to final length and not initial
length, is defined by
A:=
L2 – L2
0
2L2=
121 – λ
–2 (2.14)
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Pauli Pedersen: 2. Geometry --- Kinematics
The Swaiger strain is similarly related to the Cauchy strain and is
defined by
S:=
L – L0
L= 1 – λ
–1 (2.15)
Lastly, theKuhn strain definition uses changes in third order of length
K:=
L3 – L3
0
3L2
0L=
13λ2 – λ
–1 (2.16)
ASYMPTOTIC All these strain definitions will agree for small strain, i.e. for λ→ 1
EQUIVALENCE (<< 1) we have
C=
H=
G=
A=
S=
K(2.17)
In fig. 2.3 a graphical illustration of these different strain definitions is
given as a function of the stretch λ .
λ
G
K
C
H
S
A
Fig. 2.3: Six different strain measures as a function of stretch (extension ratio).
15
Pauli Pedersen: 2. Geometry --- Kinematics
2.5 Strains from displacement gradients
STRAIN TENSORS Strain tensors are used to describe 2---D and 3---D states of deforma-
tion. These second order tensors are symmetric, so essentially we have
3 components in 2---D and 6 components in 3---D. Restricting ourselves
now to Cauchy strains, denoted by ij , and Green---Lagrange strains,
denoted by ηij , these strains are expressed in displacement gradients
by
ij=12
vi,j+ vj,i (2.18)
and
ηij=12
vi,j+ vj,i+ vk,ivk,j= ij+12vk,ivk,j (2.19)
NORMAL STRAINS Components 11 , 22 , 33η11 , η22 , η33 are termed normal
strains or longitudinal strains and components 12
, 13
, 23
SHEAR STRAINS η12
, η13
, η23
are termed shear strains or tangential strains.
L
A
B
v1, v
2, v
3
dx1+ v
1+ dv
1, v
2+ dv
2, v
3+ dv
3
A B0 , 0 , 0 dx
1, 0 , 0
L0
Fig. 2.4: A line element before and after deformation.
16
Pauli Pedersen: 2. Geometry --- Kinematics
To illustrate the derivation of (2.18), (2.19) we show in fig. 2.4 a line ele-
ment of length dx1, oriented in the x
1---direction of a Cartesian coor-
dinate system before deformation. After deformation the two end
points are displaced as shown and the squared lengths L2
0and L2 can
be calculated:
L2
0= dx2
1
(2.20)
L2= dx2
1+ dv2
1+ 2dx
1dv
1+ dv2
2+ dv2
3
from which follows directly
η11=
L2 – L20
2L20
=
dv1
dx1+
12
dv1
dx1
2
+ dv2
dx1
2
+ dv3dx1
2
(2.21)
in agreement with η11 of (2.19) and with 11 of (2.18) for vi,j→ 0 .
We shall later show that a state of strain can always be described by only
normal strains (the principal ones). Therefore derivations of ηij , ijfor i≠ j will not be illustrated.
CARTESIAN Tobe specific let uswrite out the relations (2.18) and (2.19) for the 2---D
2---D RESULTS case; the Cauchy---strains expressed in displacement gradients are
11= v
1,122= v2,2
(2.22)
12=
12v
1,2+ v
2,1
and the Green---Lagrange strains are
17
Pauli Pedersen: 2. Geometry --- Kinematics
η11=
11+
12v2
1,1+ v2
2,1
η22= 22+12v2
1,2+ v2
2,2 (2.23)
η12=
12+
12v
1,1v1,2+ v
2,1v2,2
It should be noted that strains and displacement gradients are here
described in a specific Cartesian coordinate system. Changing the
coordinate systemwill then be our next subject and in reality we can use
results directly from linear algebra.
2.6 2---D strain description
TENSOR--- The strain state may be described in different notations, which for the
MATRIX--- 2---D case and here with Cauchy strain notation are
VECTOR
strain tensor : ij
xfor i, j= 1, 2
strain matrix : []x= 11
12
12
22x
(2.24)
strain vector : T
x=
11
22
2 12
x
The subscript x is added to indicate the reference to a specific coordi-
nate system x . (The factor of 2 in the vector notation makes rota-
tional transformation much more simple). All these notations (2.24)
will be used, and we shall see later that the stress state can be treated
in a completely similar manner.
18
Pauli Pedersen: 2. Geometry --- Kinematics
2
1
x2
x1
y2
y1
θ
ψ
Fig. 2.5: Three different reference coordinate systems and their mutual rotations.
ROTATIONAL Fig. 2.5 shows three 2D---Cartesian coordinate systems and their
TRANSFORMATION mutual angles. Let the strain state be described in the x---coordinate
system. What is the description in the y---coordinate system? The
answer follows from the matrix index in chapter 12 (Rotational....)
11y=12
11+ 22x+
12
11 – 22xcos 2θ+ 12x sin 2θ
22y=12
11+ 22x– 12
11 – 22x cos 2θ – 12x sin 2θ (2.25)
12y= – 12
11 – 22x sin 2θ+ 12x cos 2θ
ORTHOGONAL We note the π ---periodic nature as a function of θ . From the matrix
TRANSFORMATION index also follows the effective notation with the orthogonal trans-
formation matrix [T] [T]–1= [T]
T
19
Pauli Pedersen: 2. Geometry --- Kinematics
y= [T]x x = [T]Ty
(2.26)
[T] := 12
1+ cos 2θ
1 – cos 2θ
– 2 sin 2θ
1 – cos 2θ
1+ cos 2θ
2 sin 2θ
2 sin 2θ
– 2 sin 2θ
2 cos 2θ
PRINCIPAL For symmetric matrices like the strain matrix, eigenvalues are real and
STRAINS the real eigenvectors are mutually orthogonal. This information gives
directly the following simple description (with index P for principal)
strain matrix : []P= 10 0
2
(2.27)
strain vector : T
P=
120
where the eigenvalues 1,
2are obtained from the determinant
(characteristic polynomium) condition
11 –
12
12
22
– = 0
2 –
11+
22+
1122–
2
12= 0
1
2=
1
2
11+
22 1
2
11–
222+
2
12 (2.28)
cos 2ψ
sin 2ψ
=
=
11
– 22
1–
2
– 212
1–
2
⇒ tan 2ψ=
– 212
11
– 22
20
Pauli Pedersen: 2. Geometry --- Kinematics
with the angle ψ being defined in fig. 2.5.
The principal strains and their directions can be evaluated from the
strain state described in any coordinate system, but here it is shown in
agreement with fig. 2.5, i.e. 11, 22, 12 in (2.28) are known in the
y---coordinate system. The transformation formulas (2.25) also simplify
when principal strains are used as reference andwith 12= 0 in (2.25)
we get
11y=12
1+ 2+ 1
21 – 2
cos 2ψ
22y=12
1+ 2 – 1
21 – 2 cos 2ψ (2.29)
12y= – 12
1 – 2 sin 2ψ
in agreement with (2.28). Note that we have to refer to a specific princi-
pal strain direction; here the 11> 2
direction was chosen but
often the II> II direction is used. The degenerated case of
1= 2 is treated separately because the strain state is then isotropic,
i.e. all directions are principal strain directions with equal principal
strains. The case of 1= –
2correspond to pure shear, because with
ψ= π4 we from (2.29) get 11=
22= 0 and
12=
2=−
1.
2.7 3---D strain transformation
3---D STRAIN The tensor, matrix and vector notations for a 3---D strain state at a
PARAMETERS material point are
21
Pauli Pedersen: 2. Geometry --- Kinematics
ij
xfor i, j= 1, 2, 3
[]x =
11
12
13
,,,
12
22
23
,,,
13
23
33
x
(2.30)
T
x=
11
22
33
2 12
2 13
2 23
x
with index x to indicate that the description refers to a Cartesian
x---coordinate system.
FROM All the results from linear algebra for symmetric, real matrices can now
LINEAR be applied to the strain matrix [] , and we shall start with the proper---
ALGEBRA ties of invariants for similarmatrices (seematrix index). The following
combinations of strains are invariants, i.e. they are the same in all Car-
tesian coordinate systems
STRAIN the trace (1st order norm)
INVARIANTS I1=
11+
22+
33
a 2nd order norm
I2=
1122
– 2
12+
2233
– 2
23
+ 1133
– 2
13 (2.31)
the determinant (3rd order norm)I3= |[]|
the squared Frobenius norm (2nd order norm)
I4= I
12 – 2I
2
and other combinations of the three independent norms. We note that
the length of the strain vector, when definedwith the 2 –factors, is also
invariant
22
Pauli Pedersen: 2. Geometry --- Kinematics
T= I
4(2.32)
EIGENVALUES --- The next property of the strain matrix is the existence of real eigenva---
EIGENVECTORS lueswith real, mutually orthogonal eigenvectors. In terms of the shown
invariants, the third order characteristic polynomium to determine
these eigenvalues is
– 3+ I
12 – I
2 + I
3= 0 (2.33)
and in the coordinate system of principal directions, i.e. the orthogonal
eigenvector directions, the strain matrix []P
simplifies to
PRINCIPAL
STRAINS
[]P=
1
0
0
0
2
0
0
0
3
(2.34)
which means only normal strains. We still have to decide about the
ordering of these principal strains, and normally we choose
3≤
2≤
1(2.35)
EXTREME From eigenvalue theory it also follows that 1
and 3
are the
NORMAL extreme normal strains, i.e.
STRAINS
3≤ any ii≤ 1 (2.36)
The important practical problem of transformation of the strain
description []x to the description []y in a rotated Cartesian y---coor-
23
Pauli Pedersen: 2. Geometry --- Kinematics
dinate system, is also a general problem in linear algebra. The orthogo-
nal transformation matrix [Γ] of directional cosines is defined by
[Γ] :=
1
2
3
m1
m2
m3
n1
n2
n3
(2.37)
[Γ]–1= [Γ]
T(i.e. orthogonal)
where 1, 2, 3 are the direction cosines for angles from x1 to
y1, y2, y3 and mi , n i analogously from x2 and x3 , respectively. Then
the transformation formulas are
ROTATIONAL
TRANSFORMATION []x = [Γ]T[]y [Γ]
111213
122223
132333
x
=
1
m1n1
2
m2
n2
3
m3
n3
111213
122223
132333
y
1
2
3
m1
m2
m3
n1
n2
n3
(2.38)
[]y = [Γ] []x [Γ]T
here written out to enable direct interpretation of specific cases.
The normal strain in a given direction, say the x3–direction described
by n1 , n2 , n3 as it follows from (2.38), is
33x= 11y n1 n1+22y n2 n2+ 33y n3 n3
(2.39)
+ 212y n1 n2+ 213y n1 n3+ 223y n2 n3
24
Pauli Pedersen: 2. Geometry --- Kinematics
NORMAL STRAIN which is a specific case of the general result in tensor analysis
IN A DIRECTION= ij ni n j (2.40)
With transformation from principal strains []x = []P (2.38) simpli-
fies to
[]y = [Γ]
1
0
0
0
2
0
0
0
3
[Γ]
T(2.41)
SPECTRAL which gives
DECOMPOSITION []y = 1
T+ 2
mmT+ 3nn
T
= 1
21
12
13
12
22
23
13
23
23
+ 2
m21
m1m2
m1m3
m1m2
m22
m2m3
m1m3
m2m3
m23
+ 3
n21
n1n2
n1n3
n1n2
n22
n2n3
n1n3
n2n3
n23
(2.42)
known as spectral decomposition (see matrix index), which is possible
for all symmetric matrices.
In the strain vector notation the transformation can also be described
by an orthogonal matrix [T] , which is of order 6 by 6 but can conve-
niently be divided into four submatrices of order 3 by 3 . We get
y= [T] x x= [T]T y
[T]=
[T]NN
[T]SN
[T]NS
[T]SS
(2.43)
25
Pauli Pedersen: 2. Geometry --- Kinematics
ORTHOGONAL
TRANSFORMATION
[T]NN=
21
22
23
m2
1
m2
2
m2
3
n21
n22
n23
[T]NS= 2
1m
1
2m
2
3m
3
1n1
2n2
3n3
m1n1
m2n2
m3n3
[T]SN= 2
12
13
23
m1m
2
m1m
3
m2m
3
n1n2
n1n3
n2n3
[T ]SS=
1m
2+
2m
1
1m
3+
3m
1
2m
3+
3m
2
1n2+
2n1
1n3+
3n1
2n3+
3n2
m1n2+m
2n1
m1n3+m
3n1
m2n3+m
3n2
ByMathematica (1992) it is verified that [T]T[T]= [I] (usingGröbner
bases), and at the submatrix level this corresponds to the relations
[T]T
NN[T]
NN+ [T]
T
SN[T]
SN= [I]
[T]T
NN[T]
NS+ [T]
T
SN[T]
SS= [0]
[T]T
SS[T]
SS+ [T]
T
NS[T]
NS= [I]
We note that separating the transformation matrix corresponds to
separating the strain matrix into its normal strains N
and its shear
strains S
T= T
N
T
S=
1122
33
2 12
13
23
(2.44)
With reference to the principal strain description (T
N=
1,
2,
3
and S= 0 ) we again read the spectral decomposition directly
from (2.43)
26
Pauli Pedersen: 2. Geometry --- Kinematics
y= 1
21
22
23
2 122 132 23
+ 2
m21
m22
m23
2 m1m2
2 m1m3
2 m2m3
+ 3
n21
n22
n23
2 n1n2
2 n1n3
2 n2n3
(2.45)
where thedirection cosinesnoware for theprincipal directions, relative
to the y coordinate system.
The specific cases for 2D---problems follow by setting
1= cos θ
2= – sin θ
3= 0
,
,
,
m1= sinθ
m2= cosθ
m3= 0
,
,
,
n1= 0
n2= 0
n3= 1
(2.46)
2.8 Dilatation and distortion, deviatoric strains
Remembering that we are still discussing deformation at a point, i.e. a
concept defined by a limiting process, we shall now divide the deforma-
tion into a pure volumetric change (dilatation) and a change in shape
(distortion)
Deformation= Dilatation+Distortion (2.47)
DILATATION The dilatation is defined by ∆VV for V→ 0 , where V is the vol-
ume at the point in question. In terms of small strains we have
27
Pauli Pedersen: 2. Geometry --- Kinematics
∆VV=
11+
22+
33=
1+
2+
3= I
1(2.48)
and thus the trace of the strain matrix directly gives the dilatation. For
an incompressible material we must therefore have I1= 0 .
DEVIATORIC The deviatoric strains are defined such that they give no dilatation.
STRAINS Normally the symbol e is used, and in tensor notation withKronecker’s
delta δ ij= 1 for i= j and δ ij= 0 for i≠ j wewrite kk= I1
eij= ij – δijkk3 (2.49)
In reality deviatoric strains can be obtained as a projection and in
matrix notation can be described by a projection matrix [P]
([P][P]= [P] and [P]≠ [I])
e= [P] (2.50)
[P] := 13
2 – 1 – 12 – 1
2
symmetric
0003
00003
000003
From this follows e
11+ e
22+ e
33= I
1e= 0 .
DISTORTION The distortion is described directly by the deviatoric strains, that with
I1e= 0 , have principal deviatoric strains determined by
– e3 – I2ee + I
3e= 0 (2.51)
with invariants similar to (2.31).
28
Pauli Pedersen: 2. Geometry --- Kinematics
Defining the mean value of normal strains –
– :=
11+
22+
333 (2.52)
STRAIN we can write the total strain state as
DECOMPOSITION
T= e
T+ –
–– 0 0 0 (2.53)
2.9 Compatibility of a strain field
Not every strain field is compatible with physical reality, i.e. derivable
from a displacement field. Two important cases are always compatible,
i.e. when the strain field is derived from a displacement field and/or
when the strain field is linearly dependent on the coordinates.
If this is not the case it should be verified that the following compatibili-
ty conditions are satisfied
11,23
22,13
33,12
=
=
=
– 23,11+
13,12+
12,13
– 13,22+
12,23+
23,12
– 12,33+
23,13+
13,23COMPATIBILITY
CONDITIONS(2.54)
212,12
213,13
223,23
=
=
=
11,22+
22,11
11,33+
33,11
22,33+
33,22
29
Pauli Pedersen: 2. Geometry --- Kinematics
2.10 Examples of strain calculations
Tobemore familiarwith strain calculation a fewexampleswill be given.
In fig. 2.6 is shown a cross---section of a long, prismatic body of length
L (in figure direction x3). The length of the squared cross---sectional
shape is a . Two sides with a common edge are fixed as shown in the fig-
ure.
x2
d2
d1
x1
x3
a
a
q
Fig. 2.6: Cross---sectional shape and boundary conditions for a long, prismatic body.
Using only two degrees of freedom d1, d
2the following simple dis-
placement field is assumed:
v1=
x1x2
a2d1
v2=
x1x2
a2d2
v3= 0 (2.55)
and we note that the kinematic boundary conditions at x1= 0 and
x2= 0 are satisfied. The Cauchy strain field follows from (2.22)
11= v
1,1=
d1
a2x2
22= v
2,2=
d2
a2x1
(2.56)
12=
12v
1,2+ v
2,1= 1
2a2d
1x1+ d2x2
30
Pauli Pedersen: 2. Geometry --- Kinematics
With v1
and v2 independent of x3 and v3= 0 , direction x3 is a
principal direction with principal strain 3=
33= v
3,3= 0 and
13=
12(v
1,3+ v
3,1)= 0 , 23=
12(v2,3+ v3,2)= 0 .
The two other principal strains are calculated using (2.28)
1
2
= 12a2d
1x2+ d
2x1 d2
1+ d
2
2x2
1+ x
2
2 (2.57)
and we note the trace invariant 11+
22=
1+
2=
(d1x2+ d
2x1)a2 .
Anotherexample is shown in fig. 2.7.Acircular, cylindrical piece of rub-
ber with diameter 2R and thickness h is fixed in a hole in a block of
steel. The rubber is glued to the bottom and the sides of the steel hole.
The steel is modelled without deformation and for the rubber the dis-
placement assumption in the shown Cartesian coordinate system is
v1= v
2= 0 , v
3= – d
1 –
x21+ x2
2
R2x3
h(2.58)
which we see, agrees with the kinematic boundary conditions. The only
degree of freedom is d , the displacement in the negative x3---direction
at x1= x
2= 0 , x
3= h .
31
Pauli Pedersen: 2. Geometry --- Kinematics
x3
x2
x1
p
h
2R
Fig. 2.7: A circular, cylindrical piece of rubber in a stiff block of steel.
The Cauchy strains that follow from (2.58) are
11= v
1,1= 0 22= v2,2= 0
12=
12v
1,2+ v
2,1= 0
13=
12v
1,3+ v
3,1= d
hR2x1x3
(2.59)
23=
12v
2,3+ v
3,2= d
hR2x2x3
33= v3,3=−dh1 −
x21+ x2
2
R2
The strain invariants as defined in (2.31) are
I1= 33
I2= –213+
2
23 (2.60)
I3= 0
and thus the principal strains can be obtained using (2.33)
− 2− 33− 2
13+
2
23= 0 (2.61)
32
Pauli Pedersen: 2. Geometry --- Kinematics
giving
1= 0
(2.62)
2
3
= 1233
2
33
4+
2
13+
2
23
2R
x3
x1
α
x2
β
ωr
x1
v2
–v1
Fig. 2.8: Experimental setup for measuring shear modulus.
A last example is shown in fig. 2.8 which shows an experimental setup
for measuring the shear modulus (to be discussed in section four) of an
isotropic, elasticmaterial, say rubber. The hatched steel parts aremod-
elled as rigid. These two parts are placed with a common x3–axis and
the top of the cone meets the plane circular table at origo of the Carte-
sian coordinate system. The top angle of the cone is (π – 2α) . With ω
33
Pauli Pedersen: 2. Geometry --- Kinematics
being the relative rotationangle (around the x3–axis) between coneand
table, then our displacement assumption for the elastic material is
taken as
v1=
–ωtanα
x2x3
x21+ x2
2
v2=
ω
tanα
x1x3
x21+ x2
2
v3= 0 (2.63)
Note that the boundary conditions are satisfied for x3= 0
(v1= v
2= 0) and for x
3= x2
1+ x2
2 tanα ,where v
1= –ωx
2and
v2= ωx
1. This in fact is the background for the assumption.
From the displacement assumption (2.63) follows the Cauchy strains
with r := x21+ x2
2
11= v
1,1= ω tanαx
1x2x3r
3
22 = v2,2 = – ω tanαx1x2x3r
3
212= v
1,2+ v
2,1= – ω tanαx2
1– x2
2x3r3
(2.64)
213= v
1,3+ v
3,1= – ω tanαx
2r
223= v
2,3+ v
3,2= ω tanαx
1r
33= v
3,3= 0
We note that there is no dilatation (incompressible)
I1=
11+
22+
33= 0 (2.65)
Later, when we have the concept of stress and the models for relations
between stress and strain, we shall return to the three examples that
were discussed in this section.
34
Pauli Pedersen: 2. Geometry --- Kinematics
35
Pauli Pedersen: 3. Statics --- Equilibrium
3. STATICS --- EQUILIBRIUM
forces, moments, pressure, stresses
3.1 Force
NOTATION A force has the dimension N (Newton) and we shall use the notations
Q , Qi, Q for themagnitude, the component in direction x
iand the
force vector, respectively. The individual elements of the vector are
denoted Qiwith i= 1, 2 for 2---D problems and i= 1, 2, 3 for 3---D
problems, i.e.
2D : QT = Q1Q
2
(3.1)
3D : QT = Q1Q
2Q
3
POINT OF In addition to themagnitude and the direction (both given by the vector
ACTION description) we need to specify the point of application of the force, say
a material point P or a coordinate point x .
SPECIFIC Often we give forces as force per length, force per area or force per
FORCES volume, also named line force, area force, volume force, respectively.
Volume force are also called force density, while the more general
name “specific” demands information about the reference, say per
length or per area. Pressure is a particular form of force per area.
36
Pauli Pedersen: 3. Statics --- Equilibrium
FURTHER Traditionally, volume force is denoted by p , pi, p , and line force by
NOTATIONS q , qi, q . Pressure has a given direction by definition and is thus given
alone by its magnitude p ,whichmeans that for this area force we tradi-
tionally use the same symbol as for volume force.
EXTERNAL Fig. 3.1a shows a domain Ω with loads Q , q and p acting from
INTERNAL external sources. For domain Ω these forces are external forces,
although some forces, such as the volume forces, may be acting inside
the domain (resulting from, say gravity). Fig. 3.1b shows two domains
Ω1, Ω
2obtained by separating Ω . The area forces, here symbolised
by σ , are internal forces in Ω , but for the separated bodies Ω1, Ω
2
they are external forces. Thus, the notion of external/internal depends
on our point of view.
a) b)Q
Q
p
p
Ω
Ω2
σ
Ω1
Fig. 3.1: A domain (a body) Ω with external forces Q , q , p and internal forces σ
shown by the separation b).
Forces can be measured.
37
Pauli Pedersen: 3. Statics --- Equilibrium
3.2 Moment
DIMENSION A moment has the dimension Nm (Newton⋅meter), i.e. the same
dimension as energy. For external moment we shall use the notations
Z ,Zi, Z for magnitude, component and vector. In reality moment
PAIR OF FORCES is a concept that describes the action of a pair of forces, as shown in fig.
3.2.
a) b)
Q QP
a
Z
P× ×
Fig. 3.2: Pair of forces, i.e. two parallel forces of magnitude Q and distance a ,
b) shows the “equivalent” moment Z= Qa acting at point P .
SPECIFIC As for forces we have moments per length, moments per area and
MOMENTS moments per volumewithmoments from rotational inertia as an exam-
ple. We shall use the notations z, zi, z for these specific moments.
The notions of external and internal moments will be used, and again
the classification depends on our point of view. Lastly, it should again
MOMENT be pointed out that moment is merely a calculational concept and we
CONCEPT shall thus see cases (plate theory), in which forces andmoments cannot
be separated.
Moments cannot be measured directly.
38
Pauli Pedersen: 3. Statics --- Equilibrium
3.3 Stress concept
DEFINITION Stress is a concept defined by a limiting process, and the dimensions of
BY LIMIT VALUE stress is force per unit of area, say Nm2 . Firstly, we shall define the
SURFACE surface traction T, Ti , T (again magnitude, component and vec---
TRACTION tor) on a specified surface area ∆A , in terms of the vector components
∆Q
n
∆A
Ti := ∆Qi∆A for ∆A→ 0 (3.2)
SURFACE and related to thematerial point P included in ∆A .The actual surface
ORIENTATION is specified by its normal n , and Ti depends on this surface orienta-
tion.
To describe the surface traction for any surface through a point P we
need information frommore than one surface. This information, called
the stress tensor, stress matrix or stress vector, will here be given a
“physical” interpretation although different definitions are possible,
and stress like strain is merely a concept.
Now, referring to a Cartesian coordinate system, σij is the surface trac-
tion in direction xj for a surface with its normal in direction xi . Thus,
NORMAL σii is a surface traction in the direction of the normal, i.e. perpendicular
STRESS to the surface and is called normal stress or longitudinal stress. A stress
σij for i≠ j is a surface traction, again for a surface with its normal
in direction xi , but now the direction of the traction is in direction xj
39
Pauli Pedersen: 3. Statics --- Equilibrium
SHEAR perpendicular to xiand therefore in the surface plane. This is called
STRESS a shear stress, or tangential stress. The complete analogy to the strain
concept should be noted.
SIGN The definition of sign for the stresses are related to the direction of the
DEFINITION normal for the surface in such a way that normal stresses are positive
in tension and negative in compression. The sign definitions are illus-
trated in fig. 3.3.
From the fact that surfaces change direction and size through deforma-
tion we note that different possibilities for stress definitions exist.
3.4 Equilibrium
Fig. 3.3 shows an infinitesimal box with stress components applied
according to the “physical” definition above. We note that stresses are
positive in the coordinate direction if the surface normal is in the posi-
tive coordinate direction and positive in the negative coordinate direc-
tion if the normal is in the negative coordinate direction.
40
Pauli Pedersen: 3. Statics --- Equilibrium
x3
x1
x2
dx2
dx1
dx3
σ22
σ23
σ21
σ33
σ11+ dσ
11
σ32
σ31
σ13+ dσ
13
σ12+ dσ
12
σ33+ dσ
33
σ32+ dσ
32
σ31+ dσ
31
σ11
σ12
σ13
σ23+ dσ
23
σ22+ dσ
22
σ21+ dσ
21
Fig. 3.3: Infinitesimal volume element showing the sign definition for the stress
components.
Three force equilibriums are directly readable from fig. 3.3. In the x1---
direction, with p1being the volume force component not shown in fig.
3.3, we get
p1dx
1dx
2dx
3+ σ
11,1dx1dx2dx3+ σ21,2dx2dx1dx3+ σ31,3dx3dx1dx2 = 0
i.e. for dx1, dx2, dx3→ (0, 0, 0) we get σ11,1+ σ
21,2+ σ
31,3+ p
1= 0
Note that to find dσ11
we use σ11,1
dx1
with the definition
σ11,1
:= ∂σ11∂x
1.
41
Pauli Pedersen: 3. Statics --- Equilibrium
Similar results for the two other directions can be read, and with tensor
notation we write all these
VOLUME FORCE
EQUILIBRIUM σji,j+ pi≡ 0 (3.3)
indicating by the ≡ symbol that this must hold good everywhere in the
continuum. If there are no volume forces pi= 0 we get σji,j= 0 .
With no external volume moments (as is normally assumed) the stress
components must be in moment equilibrium themselves. For a line
through P (center of the volume element) parallel to the x1–direction
we get from fig. 3.3
σ23dx1dx3
dx2+ σ23,2dx2dx1dx3
dx22 – σ32dx1dx2
dx3 – σ32,3dx3dx1dx2
dx32= 0
i.e. for dx1, dx2, dx3→ (0, 0, 0) we get σ23= σ32
Similar results for the two lines parallel to the x2--- and x3---direction
are obtained, and in tensor notation we write the moment equilibrium
as
MOMENT σij≡ σji (3.4)
EQUILIBRIUM
i.e. the stress tensor is symmetric.
42
Pauli Pedersen: 3. Statics --- Equilibrium
x3
x1
x2
σ33
σ32
σ31
σ21σ22
σ23
σ11
σ12
σ13
T
∆A
n
Fig. 3.4: Infinitesimal tetrahedron element with an oblique surface subjected to the
surface traction T .
In fig. 3.4 we show an oblique plane of area ∆A with surface tractions
Ti and direction cosines of the surface normal ni . For force equilib-
rium we get from fig. 3.4 in direction x1 (area in x2x3 plane is ∆An1
etc.)
T1∆A – σ11∆An
1 – σ21
∆An2 – σ31
∆An3= 0
and for ∆A→ 0 we get T1= σ11n1+ σ21n2+ σ31n3
which generalised in tensor notation for all directions, gives
SURFACE
TRACTION
EQUILIBRIUM Ti≡ σjinj (3.5)
43
Pauli Pedersen: 3. Statics --- Equilibrium
Before moving to the more practical aspects of stress calculations, it
EQUILIBRIUM IN should be pointed out that equilibrium must be satisfied in the
DEFORMED STATE deformed state of the body. Often we can assume the displacements to
be small and give the equilibrium conditions in the undeformed state.
3.5 Descriptions of stress state
ANALOG TO The description of the stress state is completely analogous to the
STRAINS description of the strain state. Thus, in the formulas in section 2.6 (2---D
problems) and section 2.7 (3---D problems), the symbol can just be
changed to σ . We shall here repeat the main results for the 3---D case.
3---D STRESS The tensor,matrix and vector notations for 3---D stress state at amate---
PARAMETERS rial point are
σijx
for i, j= 1, 2, 3
[σ]x =
σ11
σ12
σ13
,,,
σ12
σ22
σ23
,,,
σ13
σ23
σ33
x
(3.6)
σTx =
σ11
, σ22
, σ33
, 2 σ12
, 2 σ13
, 2 σ23
x
with index x to indicate that the description refers to a Cartesian
x---coordinate system.
From linear algebra follows stress invariants
44
Pauli Pedersen: 3. Statics --- Equilibrium
STRESS the trace (1st order norm)
INVARIANTS I1σ= σ
11+ σ
22+ σ
33
a 2nd order norm
I2σ= σ
11σ22– σ2
12+ σ
22σ33– σ2
23
+ σ11σ33– σ2
13 (3.7)
the determinant (3rd order norm)
I3σ= |[σ]|
the squared Frobenius norm
I4σ= I
1σ2 – 2I
2σ= σ
Tσ
PRINCIPAL Eigenvalues σ1,σ
2, σ
3and corresponding eigenvectors are obtained
STRESSES from
– σ3 + I1σσ2 – I
2σσ + I
3σ= 0
(3.8)
σ11– σ
σ12
σ13
σ12
σ22– σ
σ23
σ13
σ23
σ33– σ
1
2
3
=
000
ROTATIONAL Rotational transformations by matrix [Γ] defined in (2.37) are
TRANSFORMATION
[σ]x = [Γ]T[σ]y[Γ]
[σ]y = [Γ] [σ]x[Γ]T
(3.9)
σ= σij n i nj
and description by spectral decomposition is
45
Pauli Pedersen: 3. Statics --- Equilibrium
SPECTRAL [σ]y = σ1T+ σ2
mmT+ σ3nn
T(3.10)
DECOMPOSITION
In vector notation rotational transformation can also be described by
the orthogonal matrix [T] defined in (2.43), giving
σy= [T] σx σx= [T]T σy (3.11)
now with the spectral decomposition written by
σy= σ1
21
22
23
2 122 132 23
+ σ2
m21
m22
m23
2 m1m2
2 m1m3
2 m2m3
+ σ3
n21
n22
n23
2 n1n2
2 n1n3
2 n2n3
(3.12)
3.6 Hydrostatic stress and deviatoric stress
MEAN NORMAL Themean value σ–
of the three normal stress components is also called
STRESS the hydrostatic stress
σ–
=σ
11+ σ
22+ σ
333= σ
1+ σ
2+ σ
33= I
1σ3 (3.13)
and is thus a stress invariant.
DEVIATORIC The deviatoric stress is defined such that its hydrostatic stress is zero.
STRESS Normally, the stress symbol is then changed to s and in analogy with
deviatoric strains (2.49), we have in tensor notation
sij= σij – δ ijσkk3 (3.14)
46
Pauli Pedersen: 3. Statics --- Equilibrium
or in vector notation with projection matrix
s= [P] σ(3.15)
[P] := 13
2 – 1 – 12 – 1
2
symmetric
00
03
00
003
00
0003
and the total stress can then be written
σT= s
T+ σ– σ– σ– 0 0 0 (3.16)
In material models the strength is often assumed to be independent of
the hydrostatic stress, and then a formulation in deviatoric stresses is
useful.
3.7 A classic example
A thick---walled spherical shell is made of isotropic material with
Young’s modulus E and Poisson’s ratio ν . The model is shown in fig.
3.5, placed in a Cartesian coordinate system, with size given by inner
radius a and outer radius b . With point symmetry about origo, the
shell is loaded by an internal pressure p , free at the outer surface and
without volume forces.
47
Pauli Pedersen: 3. Statics --- Equilibrium
x3
x2x
1
ba
p
Fig. 3.5: A sectional cut (through origo) of the spherical shell problem.
Primarily with symmetry consideration and the equilibrium conditions
of this chapter we can solve this problem analytically. This is the reason
for including the example before the constitutive relations between
stress and strain are discussed in detail.
To get symmetric displacements of this symmetrically loaded, symmet-
ric model, the displacement vector vimust be in the direction of the
material point vector xi. With the unknown displacement function
then only a function of the radius r , i.e. v= v(r) , the displacement
field must have the form
DISPLACEMENT vi= v(r) x
ir (3.17)
FIELD
with
r= xix
i = x2
1+ x2
2+ x2
3 ⇒
(3.18)r, i:= ∂r∂x
i= x
ir
48
Pauli Pedersen: 3. Statics --- Equilibrium
The field of displacement gradients follows by differentiation
DISPLACEMENT vi , j= v,rr,j xir+ vδ
ijr – vxir , jr2 or
GRADIENTS (3.19)
v i , j = dvdr
– vr
xixj
r2+ v
r δij
with δij being Kronecker’s delta.
Strains in terms of the Cauchy strain field follow by definition
STRAINS ij=
12vi , j+ vj , i
⇒
ij=dvdr
– vr
xixj
r2+
vr δ ij
as here vi , j= vj , i (3.20)
kk = 11+ 22+ 33=dvdr
– vr+ 3 v
r
Inserting the stress/strain relation termed Hooke’s law (which will be
discussed in detail in the next chapter) we get the stress field
STRESSES σij=
E1+ ν
ij+ν
1 – 2νδijkk
⇒(3.21)
σij=E
1+ νdv
dr–
vrxi
xj
r2+ δij
ν
1 – 2ν+ δ
ij1+ 3ν
1 – 2ν vr
EQUILIBRIUM We can now derive the conditions that follow from equilibriums (3.3),
(3.4) and (3.5). Without volume forces pi≡ 0 the volume force equi-
librium give σij , j≡ 0 . From (3.21) we get
49
Pauli Pedersen: 3. Statics --- Equilibrium
σij , j=
E
1+ νd2v
dr2– dv
dr1r+
vr2r, jxixj
r2+ δ ij
ν
1 – 2ν
(3.22)
+ dvdr
–vrxi , jxj
r2+
xi xj , j
r2–
2xi xj
r3r, j+ δij
1+ ν
1 – 2νdv
dr1r – v
r2r, j
Using
xi,jxj
r2= δ ij
xj
r2;x ixj , j
r2=
3xi
r2;2xixj
r3
xj
r =2xi
r2⇒
xi,jxj
r2+
xixj , j
r2–
2xixj
r3r,j= 1
r2δijx j+ xi
= 2xi
r2and
(3.23)xj
r xixj
r2+ δij
ν
1 – 2ν= xi
r(1 – ν)
(1 – 2ν); δ ijr, j= r, i=
xir
we rewrite (3.22) to
STRESS σij,j=E(1 – ν)
(1+ ν)(1 – 2ν)d2vdr2+ 2 dv
dr1r – 2 v
r2 xir (3.24)
GRADIENTS
We can therefore only have equilibrium σij , j≡ 0 if the displacement
function v= v(r) satisfies the second order differential equation (an
Euler differential equation)
CONDITION
FOR INTERNAL d2vdr2+ 2 dv
dr1r – 2 v
r2≡ 0 (3.25)
EQUILIBRIUM
50
Pauli Pedersen: 3. Statics --- Equilibrium
With A and B being two constants, the solution to (3.25) is
v= Ar+ Br–2 (3.26)
Inserting dvdr= A – 2Br–3 and d2vdr2= 6Br–4 in the differential
equation (3.25) proves (3.26).
The surface traction equilibrium at the inner and outer surface gives,
with Ti= 0 at the outer surface,
σijnj= 0 for n j= xjr at r= b (3.27)
and with Ti = – pn i at the inner surface, σijnj = –pn i for nj =
– xjr , ni = – xir , i.e.
σijxj = – p xi at r= a (3.28)
With
σijxjr=E
1+ ν
dvdr
– vrxi
r+
xi
rν
1 – 2ν+ x
i
r1+ 3ν
1 – 2ν vr
(3.29)
σijxjr=
E(1+ ν)(1 – 2ν)
(1 – ν) dvdr+ 2ν vr
xir
the condition (3.27) at the outer surface gives
(1 – ν)A – 2Bb–3+ 2νA+ Bb–3= 0 (3.30)
and the condition (3.28) at the inner surface gives
− E(1+ ν)(1− 2ν)
(1− ν)A− 2Ba−3+ 2νA+ Ba−3
= p (3.31)
51
Pauli Pedersen: 3. Statics --- Equilibrium
SOLUTION Solving the conditions (3.30), (3.31) with respect to the constants A ,
B , we get
A= 1 – 2νE
pa3
b3 – a3; B= 1+ ν
2E
pa3b3
b3 – a3(3.32)
and thereby the final result
(3.33)v=p(1+ ν)
2Ea3
b3 – a3b3r2+ 2 1 – 2ν
1+ ν
r or
with η := ab and := rb written
v2Ebp=
η3
1 − η31+ ν)2+ 2(1 − 2ν)
showing dependence onmaterial E, ν ; on size a , b ; on load p ; and,
lastly, on position r . Then dependence on Poisson’s ratio ν is shown
graphically in fig. 3.6, and we note the limiting case of ν→ 0.5 (incom-
pressibility).Note that dvdν= 0 for b= 3 4 r (r≈ 0.6b) , as seen
in fig. 3.6.Results with both internal and external pressure are obtained
simply by changing the condition (3.30).
52
Pauli Pedersen: 3. Statics --- Equilibrium
ν= – 0.1
0.0
0.3
0.5
0.0
0.3
0.5
ν= – 0.1
ab=
12
ab=
13
r/b
v2Ebp
Fig. 3.6: Displacement functions v= v(r) for a pressurized spherical shell, with Poisson’s ratio ν as parameter
and for two ratios of inner/outer radius.
53
Pauli Pedersen: 4. Physics --- Constitutive Models
4. PHYSICS --- CONSTITUTIVE MODELS
material parameters, anisotropy,
elastic energy, 3---D to 2---D, measuring
4.1 Material parameters and modelling
Material properties like stiffness, compliance and strength can be
described in different notations. To have an unambiguous reference let
us first describe the constitutive relations for 3---D problems by the
MATERIAL fourth---order material stiffness tensor Lijkl defined according to
STIFFNESS
3---D TENSOR σij= Lijkl kl (4.1)
As the index i, j, k and l run through 1, 2 and 3 , we have in (4.1) nine
equations, each with nine terms on the right---hand side. In linear elas-
ticity Lijkl is constant (independent of kl , σij) , but in non---linear
elasticity (still reversible) Lijkl should be interpreted as a “mean
value”, i.e. a secant stiffness tensor. A tangent stiffness tensor would
then be defined by the variational (incremental) equation δσij=
Lijklδkl , and we shall later return to the difference in these descrip-
tions. First let us concentrate on linear elasticity. The inverse relation
MATERIAL to (4.1) is written
COMPLIANCE
3---D TENSOR ij=Mijkl σkl (4.2)
54
Pauli Pedersen: 4. Physics --- Constitutive Models
with thematerial compliance tensor Mijkl , using anew symbolbecause
there is no symbol for inverted tensors.
For 2---D problems the constitutive relations are written
MATERIAL
MODEL
2---D TENSOR σij= Cijkl kl (4.3)
with index i, j, k and l through 1, 2 only, i.e. four equations each with
four terms on the right---hand side. Note that the tensor Lijkl for 3---D
problems and the tensor Cijkl for 2---D problems are given different
symbols. The tensor Lijkl directly describesmaterial behaviour, but the
tensor Cijkl describes both the material behaviour and the modelling
from 3---D to 2---D.
With matrix notation we alternatively write the relation (4.3) by
σy= [C]yy (4.4)
σT := σ11σ22
2 σ12
T:=
1122
2 12
where we have added an index y to point out that the descriptions is
in a Cartesian y coordinate system. In order to avoid misunderstand-
ings we write the specific elements of the [C] matrix using the tensor
elements of (4.3). Furthermore, for later use, we prefer a non---dimen---
CONSTITUTIVE sional representation and write
2---D MATRIX
[C]= C
α1111
α1122
2 α1112
α1122
α2222
2 α2212
2 α1112
2 α2212
2α1212
= C[α] (4.5)
55
Pauli Pedersen: 4. Physics --- Constitutive Models
with C being the dimensional coefficient. Matrix [C] follows directly
from (4.3) with the definitions in (4.4).
The 3---D constitutive matrix will be ordered in different ways depend-
ing on the goal of our analysis. Let us here give it with submatrices
related to a division into normal components and shear components
CONSTITUTIVE σy= [L]y y
3---D MATRIX
σT:= σT
Nσ
T
S= σ
11σ22σ33
2 σ12σ13σ23
T:= T
N
T
S=
1122
33
2 12
13
23
(4.6)
[L]=
[L]NN
[L]T
NS
[L]NS
[L]SS
[L]NN=
L1111
symmetric
L1122
L2222
L1133
L2233
L3333
[L]SS= 2
L1212
symmetric
L1213
L1313
L1223
L1323
L2323
[L]
NS= 2
L1112
L2212
L3312
L1113
L2213
L3313
L1123
L2223
L3323
(Note that different ordering of the shear components is used in the lit-
erature). We see that the general case has here 21 independent consti-
tutive parameters, but most practical materials can be treated with far
fewer parameters, say 9 parameters for the 3---D orthotropic case. In
terms of the [L] matrix, the orthotropic case will have no coupling
between shear and normal components, i.e. [L]NS= [0] . Further-
more there will be no coupling between the three shear components,
i.e. L1213
= L1223
= L1323
= 0 .
56
Pauli Pedersen: 4. Physics --- Constitutive Models
4.2 Rotational transformations
Only isotropic materials have stiffnesses that are independent of the
direction. For anisotropic materials we will derive the necessary trans-
formation formulas. Although we shall mainly concern ourselves with
the 2---D case, let us show the general aspects for 3---D problems.
Substituting in (4.6) the rotational transformation for strain (2.43) and
stress (3.11) we get
[T]σx = [L]y[T]x(4.7)
σx= [T]T[L]y[T]x
For 3---D anisotropic constitutive matrices the rotational transforma-
tions with the orthogonal matrix [T] defined in (2.43) are therefore
given by
3---D STIFFNESS [L]x= [T]T[L]
y[T]
ROTATIONAL (4.8)
TRANSFORMATION [L]y= [T][L]x[T]T
as an alternative to the tensor transformation by
Lijkl
y= aimajnakoalpLmnop
x(4.9)
where the a factors are direction cosines and we have 81 terms on the
right---hand side of (4.9).
For 2---D problems we shall present the detailed results and therefore
introduce the following short notation for the involved trigonometric
functions
57
Pauli Pedersen: 4. Physics --- Constitutive Models
c2 := cos 2θ , c4 := cos 4θ , s2 := sin 2θ , s4 := sin 4θ (4.10)
Then with [T] from (2.26) we have to work out
[C]y =
[T][C]x[T]
Twith [T]= 1
2
1+ c2
1 – c2
– 2 s2
1 – c2
1+ c2
2 s2
2 s2
– 2 s2
2c2
(4.11)
The result can be presented in different forms. Here we prefer themul-
tiple angle approach without powers of trigonometric functions. Fur-
thermore, we give the result in terms of the non---dimensional α com-
ponents in (4.5) and then have
α1111y
α2222y
=
12
α1111+ α2222
x α2c2 – α3
1 – c4 α62s2+ α7s4
α1122y
α1212y
=
α1122x
α1212x
+ α31 – c4 – α7s4 (4.12)
α1112y
α2212y
= – 12α2s2 α3s4+ α6c2 α7c4
with the definition of the practical (not invariant) parameters
α2,α3,α6,α7 by
58
Pauli Pedersen: 4. Physics --- Constitutive Models
α2:=
1
2α
1111– α
2222x
α3:=
1
8α
1111+ α
2222– 2α
1122+ 2α
1212x
(4.13)
α6:= 1
2α
1112+ α
2212x
α7:= 1
2α
1112– α
2212x
A valuable but not well-known alternative to the formulas (4.12) ---
(4.13) is obtained by also representing the [α] matrix in vector form
with 2 factors for the off---diagonal elements. The orthogonal matrix
[R] [R]–1= [R]
T for this transformation is
[R]= 18. (4.14)
3+ 4c2+ c
4
3 – 4c2+ c4
2 – 2c4
2 – 2 c4
– 4s2– 2s
4
– 4s2+ 2s
4
,
,
,
,
,
,
3 – 4c2+ c
4
3+ 4c2+ c4
2 – 2c4
2 – 2 c4
4s2– 2s
4
4s2+ 2s
4
,
,
,
,
,
,
2 – 2c4
2 – 2c4
4+ 4c4
– 2 2 + 2 2 c4
4s4
– 4s4
,
,
,
,
,
,
2 – 2 c4
2 – 2 c4
– 2 2 + 2 2 c4
6+ 2c4
2 2 s4
– 2 2 s4
,
,
,
,
,
,
4s2+ 2s
4
– 4s2+ 2s4
– 4s4
– 2 2 s4
4c2+ 4c
4
4c2– 4c
4
,
,
,
,
,
,
4s2– 2s
4
– 4s2 – 2s4
4s4
2 2 s4
4c2– 4c
4
4c2+ 4c
4
and then with the contracted notation [α]→ α
αT:= α1111 α2222 2α1212 2 α1122 2α1112 2α2212
(4.15)
we can write the transformations simply by
αy= [R]αx αx= [R]Tαy (4.16)
59
Pauli Pedersen: 4. Physics --- Constitutive Models
ORTHOTROPIC For the important case of an orthotropicmaterial, we have α6= α
7=
MATERIAL 0 for the orthotropic directions. The transformation formulas (4.12)
from this direction x simplify to
α1111
y
α2222y
=α1
x α2
xc2+ α3
xc4
α1122y
α1212y
=
α4x
α5x
−
α3xc4 (4.17)
α1112y
α2212y
= −12
α2xs2 α3
xs4
with the definitions of the three additional practical parameters
α1x:= 1
2α1111+ α2222
x– α3
x
α4x:= α1122
x+ α3
x
(4.18)
α5x:= 1
2α
1x−
α4x= α1212
x+ α3
x
traditionally used in laminate theory.
It follows from the α1112yand α2212
ycomponents of (4.17) that it
cannot always be seen directly from the constitutivematrix if amaterial
is orthotropic. The condition for this follows from the general rota-
tional formulas (4.12)
α1112y= α2212
y= 0 for some angle θ (4.19)
60
Pauli Pedersen: 4. Physics --- Constitutive Models
Alternative conditions are
α1112
y+ α2212
y= 0 and α1112
y– α2212
y= 0 (4.20)
which with (4.12) gives
tan(2θ)= 2α6α2x, tan(4θ)= α7α3
x
(4.21)
that with tan(4θ)= 2 tan(2θ)1+ tan2(2θ) is only possible when
CONDITION
FOR α7α2
2– 4α
7α2
6– 4α
6α3α2= 0 (4.22)
ORTHOTROPY
For an isotropic material all directions are principal directions, i.e.
directions orthogonal to symmetry planes. For a 2---D orthotropic
PRINCIPAL material we have two principal (mutually orthogonal) directions, and
MATERIAL have to choose a specific one. In most cases we choose the direction of
DIRECTIONS largest α1111
, i.e. we choose α2≥ 0 .
SPECIFIC For non---orthotropic materials we cannot talk about principal direc---
REFERENCE tion, but it is still convenient to have a specific reference direction. We
DIRECTION again choose the direction of largest α1111
and the solution to this will
be
α1113
= 0⇒ α
6= –α
7(4.23)
(In very special cases there will be more than one solution to (4.23)).
61
Pauli Pedersen: 4. Physics --- Constitutive Models
4.3 Classification of 2---D stiffness for anisotropic elasticity
SYMMETRIC With the goal of dealing with all possiblematerials we shall only restrict
MATRIX the constitutive matrix by the two physical conditions
symmetric matrix
(4.24)positive definite matrix
In terms of the α components this gives the following general condi-
tions
POSITIVE
DEFINITE α1111> 0 α
2222> 0 α
1212> 0
α1111
α2222
– α2
1122> 0 α
2222α1212
– α2
2212> 0 (4.25)
α1111
α1212
– α2
1112> 0 det[α]> 0
POSITIVE It follows that the invariants of the constitutive matrix are all positive,
INVARIANTS
1st order= trace : I1α= α
1111+ α
2222+ 2α
1212> 0
2nd order : I2α= α
1111α2222
– α2
1122+ 2 α
2222α1212
– α2
2212
(4.26)
+ 2α1111
α1212
– α21112
> 0
3rd order= determinant : I3α> 0
The norm of the vector α defined in (4.15) is also positive
62
Pauli Pedersen: 4. Physics --- Constitutive Models
αTα= I
4α= I2
1α– 2I
2α= α
2
1111+ α
2
2222+ 4α2
1212
(4.27)
+ 2α21122+ 4α2
1112+ 4α2
2212> 0
giving
I21α> 2I
2α(4.28)
In a conveniently chosen, specific referencematerial coordinate system
with α1111
≥ α2222
, the conditions (4.25) can be stated more simply
even for the non---orthotropic case with α1113
= 0 as given by (4.23).
ISOTROPIC For the isotropic case with only two parameters α1111
, α1122
= TWO α2222
= α1111
PARAMETER (4.29)
α1212
= α1111
− α1122
2
We get the condition
|α1122
|< α1111
(4.30)
For the orthotropic case with four parameters α1111
, α2222
, α1122
,
α1212
we get the conditions
ORTHOTROPIC
= FOUR 0< α2222
α2222
≤ α1111
, 0< α1212
PARAMETERS (4.31)
|α1122
|< α1111
α2222
α1111
α2222
≤ 12
α1111
+ α2222
And, lastly, for the non---orthotropic case with five parameters α1111
,
α2222
, α1122
, α1212
, α1123
, the conditions are
63
Pauli Pedersen: 4. Physics --- Constitutive Models
NON---ORTHOTROPIC
= FIVE 0< α2222
α2222≤ α
1111 , 0< α
1212
PARAMETERS (4.32)
|α1122
|< α1111
α2222
α1111
α2222
< 12
α1111
+ α2222
|α2212
|< α2222
α1212
–α2
1122α1212
α1111
While the parameter α2gives information about the “level” of aniso-
tropy (for isotropy α2= 0) , the parameter α
3gives information
about the relative shear stiffness as it follows from the definition (4.13)
α3:= α
1111+ α
2222– 2α
1122 – 4α
12128 (4.33)
The term α1111+ α
2222– 2α
1122 is positive as stated in (4.32), so for
low value of α1212
we have α3> 0 . For the isotropic case we get
α3= 0 and for high value of α
1212(not restricted) we get α
3< 0 .
We accordingly classify as follows
LOW/HIGH α3≥ 0 : Model of low shear stiffness
SHEAR (4.34)
STIFFNESS α3< 0 : Model of high shear stiffness
As an example, by angle design, laminate models can give α3> 0 as
well as α3< 0 .
A classification by the inverted constitutive matrix (the compliance
matrix) is also important. Restricting ourselves to orthotropic descrip-
tion we get
64
Pauli Pedersen: 4. Physics --- Constitutive Models
α1111
α1122
0
α1122
α2222
0
0
0
2α1212
–1
=
α2222
– α1122
0
– α1122
α1111
0
0
0
α1111
α2222
– α2
1122
2α1212
1α1111
α2222
– α2
1122
(4.35)
andbasedon thiswecandefine β parameters, similar to the α parame-
ters
β2=
12
α2222
– α1111
(4.36)
β3=
α
2222+ α
1111+ 2α
1122−
α1111
α2222− α
2
1122
α1212
The sign of β3then give rise to a further classification
LOW/HIGH β3≥ 0 : Model of low shear compliance
SHEAR
COMPLIANCE β3< 0 : Model of high shear compliance
This later classification is not just an alternativeway of giving the classi-
fication (4.34). We shall show this in terms of the engineering moduli,
to be discussed further in section 4.6. With EL, E
Tbeing the two
length moduli, GLT
the shear modulus and νLT
a Poissons ratio then
EL> ν2
LTET> 0 , G
LT> 0 give the positive definite condition.
Then, for increasing value of GLT
, we get the following classification:
65
Pauli Pedersen: 4. Physics --- Constitutive Models
Low shear stiffness, high shear compliance α3> 0 , β
3< 0 for
GLT<
ELET
EL+ 1+ 2ν
LTE
T
(4.37)
Low shear stiffness, low shear compliance α3> 0 , β
3> 0 for
ELET
EL+ 1+ 2ν
LTE
T
< GLT<
ELE
L+ E
T1 – 2ν
LT
4EL– ν
2
LTET
(4.38)
and, lastly, high shear stiffness, low shear compliance (α3< 0 ,
β3> 0) for
ELE
L+ E
T1 – 2ν
LT
4EL– ν
2
LTET
< GLT
(4.39)
The detailed analysis that gives the results (4.37)---(4.39) can be found
in Cheng & Pedersen (1997).
In table 4.1 we show schematically the classification of 2---D stiffness
models.
66
Pauli Pedersen: 4. Physics --- Constitutive Models
Any reference axis A conveniently chosen, specific reference axis
[α] symmetric=
α1111
α1122
2 α1112
α1122
α2222
2 α2212
2 α1112
2 α2212
2α1212
⇒ 6 parameters[α] positive definite
ANISOTROPIC
α7α2
2– 4α7α
2
6– 4α6α3α2
= 0
Practical parameters :
α2:= 1
2(α
1111– α
2222)
α3:= 1
8((α
1111+ α
2222)
–2(α1122+ 2α
1212))
α6:= 1
2(α
1112+ α
2212)
α7:= 1
2(α
1112– α
2212)
[α]=
α1111
α1122
0
α1122
α2222
2 α2212
0
2 α2212
2α1212
⇒ 5 parameters (+ direction of axis)
NON--
ORTHOTROPIC
≠ 0
positive
definite
0< α2222
; 0< α1212
|α1122|< α1111α2222
|α2212|< α2222α1212 –α
2
1122α
1212
α1111
axis from α1111> α
2222and α
1112= 0,
i.e. α2> 0 and α
6= – α7
[α]=
α1111
α1122
0
α1122
α2222
0
0
0
2α1212
⇒ 4 parameters (+ direction of axis)
axis from α1111≥ α
2222and
α1112= α
2212= 0,
i.e. α2≥ 0 and α
6= α7= 0
ORTHOTROPICpositive
definite
0< α2222
; 0< α1212
|α1122
|< α1111
α2222
< 12
(α1111+ α
2222)
α2,α
3 α3
HIGH SHEARMODULUS
≠ 0, 0 < 0
LOW SHEARMODULUS
≥ 0
ISOTROPIC
OTHERS
[α]=
α1111
α1122
0
α1122
α1111
0
0
0
(α1111
– α1122
)⇒ 2 parameters
positive definite |α1122
|<α1111
= 0, 0PLANE STRESS, isotropic
PLANE STRAIN, isotropic
Table 4.1: Classification of 2---D stiffness models.
67
Pauli Pedersen: 4. Physics --- Constitutive Models
4.4 From 3---D to 2---D constitutive matrix
The 3---D constitutive matrix in (4.6) is ordered by division into normal
and shear components. Here we need a division ordered into plane
components (index P) , transverse shear components (index T) and the
out---of---plane normal component (index W) .
σ= [L]
σT:= σT
Pσ
T
TσW= σ
11σ22
2 σ12
2 σ13
2 σ23
σ33
T:= T
P
T
TW=
1122
2 12
2 13
2 23
33
[L]=
[L]PP
[L]T
PT
LT
PW
[L]PT
[L]TT
LT
TW
LPW
LTW
LWW
(4.40)
[L]PP=
L1111
symmetric
L1122
L2222
2 L1112
2 L2212
2L1212
[L]
PT=
2 L1113
2 L2213
2L1213
2 L1123
2 L2223
2L1223
LT
PW= L
1133L2233
2 L3312
[L]TT=
2L1313
symmetric
2L1323
2L2323
LT
TW= 2 L
33132 L
3323 L
WW= L
3333
From the shown submatrix description we read the three matrix equa-
tions
68
Pauli Pedersen: 4. Physics --- Constitutive Models
SUBMATRIX
EQUATIONS σP= [L]
PP
P+ [L]
PT
T+ L
PWW
(4.41)
σT= [L]
T
PT
P+ [L]
TT
T+ L
TWW
(4.42)
σW= L
T
PW
P+ L
T
TW
T+ L
WWW
(4.43)
which is the basis for the derived results.
In the literature we find different definitions for models like “plane
stress” and “plane strain”, so we shall primarily discuss these. Let the
model “plane strain” be defined by
PLANE
STRAIN 33= 0 ,
13=
23= 0 (4.44)
1st MODEL
i.e. a principal strain direction with no strain. With T= 0 and
W= 0 follows from (4.41) the plane constitutive model
σP= [C]
Pwith [C]= [L]
PP(4.45)
i.e. directly a part of the 3---D constitutive matrix. The out---of---plane
stresses σ33
and σ13, σ
23arenot necessarily zero, andwith
Pdeter-
mined, they can be found from (4.42), (4.43)
σT= [L]
T
PT
P, σ
W= L
T
PW
P(4.46)
In the theory of plasticity we see “plane strain” defined by
PLANE
STRAIN 33= 0 , σ
13= σ
23= 0 (4.47)
2nd MODEL
and we shall analyse when this definition agrees with the definition
(4.44). The condition of [L]PT= [0] follows directly from (4.46) and
69
Pauli Pedersen: 4. Physics --- Constitutive Models
this is the case in material orthotropic directions. More generally from
(4.47) and (4.42) follows
T= – [L]
–1
TT[L]
T
PT
P(4.48)
which, inserted in (4.41), gives the result
σP= [C]
Pwith [C]= [L]
PP– [L]
PT[L]
–1
TT[L]
T
PT(4.49)
and then, from (4.43), the resulting σW
component
σW= LT
PW– L
T
TW[L]
–1
TT[L]
T
PT
P(4.50)
The first “plane stress” model is defined by
PLANE
STRESS σ33= 0 ,
13=
23= 0 (4.51)
1st MODEL
i.e. a mixed assumption as in (4.47). From (4.43) follows with (4.51)
W= – L
T
PW
PL
WW(4.52)
which, inserted in (4.41), gives the result
σP= [C]
Pwith [C]= [L]
PP– L
PWL
T
PWL
WW(4.53)
and the unknown transverse shear stresses can then be evaluated from
(4.42)
σT= [L]T
PT– L
TWL
T
PWL
WW
P(4.54)
70
Pauli Pedersen: 4. Physics --- Constitutive Models
The more common definition of a “plane stress” model is
PLANE
STRESS σ33= 0 , σ
13= σ
23= 0 (4.55)
2nd MODEL
i.e. a principal stress direction with no stress. This can only be identical
to the definition (4.51) when, according to (4.54), we have
[L]T
PT– L
TWL
T
PWLWW= [0] (4.56)
For orthotropic directions we have [L]PT= [0] and L
TW= 0 so
for this case (4.56) is satisfied, but generally the twomodelswill givedif-
ferent results. By simple matrix multiplications we can show that when
the constitutive matrix can be decomposed into a dyadic product
[L] = XXT , (4.56) will be satisfied.
After this more general discussion let us list the practical results, at
least valid for orthotropic directions. From (4.40), (4.44), (4.45) and
(4.47) we have with a plane strain assumption
PLANE
STRAIN 33= 0 ,
13=
23= 0 , σ
13= σ
23= 0
ORTHOTROPIC (4.57)
[C]=
L1111
symmetric
L1122
L2222
2 L1112
2 L2212
2L1212
with Lijkl being the 3---D parameters.
Similarly, from (4.40), (4.51), (4.53) and (4.55), we have with a plane
stress assumption
71
Pauli Pedersen: 4. Physics --- Constitutive Models
PLANE
STRESS σ33= 0 , σ
13= σ
23= 0 ,
13=
23= 0
ORTHOTROPIC (4.58)
C =
L1111 – L2
1133L3333
symmetric
L1122 – L1133L2233L3333
L2222 – L2
2233L3333
2 L1112 – L1133L3312L3333
2 L2212 – L2233L3312L3333
2L1212 – L2
3312L3333
again with Lijkl being the 3---D parameters.
To illustrate the use of (4.57) and (4.58) we insert the parameters of an
isotropic material with Young’s modulus E and Poisson’s ratio ν and
get from
L1111= L2222= L3333=E(1 – ν)
(1+ ν)(1 – 2ν)ISOTROPIC
CASEL1122= L1133= L2233=
ν
1 – νL1111 (4.59)
L1212= L1313= L2323=E
2(1+ ν)
(remaining components are zero)
for the plane strain case
[C]=E(1 – ν)
(1+ ν)(1 – 2ν)
1
symmetric
ν
1 – ν
1
0
0
1 – ν
1 – ν
(4.60)
And for the plane stress case
[C]= E1 – ν
2
1 ν
100
1 – ν
(4.61)
72
Pauli Pedersen: 4. Physics --- Constitutive Models
We end the section by concluding that other 2---D models can easily by
derived with the help of the submatrix description (4.40).
4.5 Energy densities in non---linear elasticity
We shall in this section restrict the analysis to power law non---linear
elasticity, and although the general case of 3---D anisotropic behaviour
will be treated, we shall at first deal with 1---D problems.
ENERGY With , d being strain and differential strain and σ , dσ being the
DEFINITIONS conjugated stress and differential stress, strain energy density (per vol-
ume) is defined by
1–D STRAIN
ENERGY DENSITYdu := σd ; u=
0
σ ~d ~
(4.62)
showing symbolically the difference between the integration variable
~ and the final strain . The stress energy density uC , also named
complementary energy density, is defined by
1–D STRESS
ENERGY DENSITYduC := dσ ; uC =
σ
0
σ~dσ~ (4.63)
73
Pauli Pedersen: 4. Physics --- Constitutive Models
u
uC
σ
Fig.4.1:Graphical illustrationofstrainenergydensity u andstressenergydensity uC .
and the results of these definitions are illustrated in fig. 4.1. It follows
from algebra that
d(σ)= σd+ dσ (4.64)
or, graphically from fig. 4.1, that independent of the functions σ=
σ() , = (σ) we have
COMPLEMENTARY
ENERGY u+ uC = σ (4.65)
DENSITIES
also explaining the naming of complementary energies.
The model of power law non---linear elasticity for 1---D problems and
in terms of σ= σ() is
σ= Ep or σ= E||p–1 (4.66)
with the constants E , p being material parameters. The drastic influ-
ence of the power p is shown in fig. 4.2.
74
Pauli Pedersen: 4. Physics --- Constitutive Models
p= σE
p= 0.8
p= 0.6
p= 0.4
p= 0.2
0 0.02 0.04 0.06 0.08 0.1
Fig. 4.2: Stress---strain curves for different values of the power p .
For conveniencewe shall in the following assume > 0 and it then fol-
lows that the secant modulus Es and the tangent modulus Et will be
Es :=σ
= Ep–1 ; Et :=
dσd= pEp–1 (4.67)
Inserting (4.66) in (4.62) we obtain the strain energy density for this
material model
u= E
0
p~ d ~
=
1p+ 1
Ep+1 (4.68)
and from (4.65) with (4.66)
u+ uC = Ep+1 (4.69)
75
Pauli Pedersen: 4. Physics --- Constitutive Models
i.e. from (4.68) the stress energy density (written in strain)
uC =p
p+ 1E
p+1 (4.70)
RATIO OF Note that uC = uC(σ) but in (4.70) has been substituted for σ , and
DENSITIES we find the important relation between the energy densities
(4.71)uC = p u
These important results we can also derive in terms of stresses
= σE
1p
= σEn= σ
EσE
n–1 with n= 1p (4.72)
which, inserted in (4.63), gives
uC = 1Enσ
0
σ~
n
dσ~
=
1En
1n+ 1
σn+1 (4.73)
an alternative expression for (4.70).
Extension to 2---D and 3---D problems is not trivial, and the definitions
of effective strain e and of effective stress σe must be chosen cor-
rectly. In matrix formulation the differential strain energy density du
is
du= σTd ; u=
0
du (4.74)
and, in analogy with (4.66), the constitutive secant modulus is
76
Pauli Pedersen: 4. Physics --- Constitutive Models
2–D OR 3–D
SECANTσ= e
0p–1E0
[α]= E p–1e
[α] (4.75)
MODULUS
MATRIXi.e. [C]= E
p–1e
[α] and E= E0p–1
0
with the non---dimensional [α] matrix defined as in (4.5) for 2---D
problems and E being the dimensional constant coefficient. The refer-
ence strain is 0 and the corresponding reference modulus is E0 . It
REFERENCE follows from (4.75) that at the reference strain 0 , the scalar secant
STRAIN AND modulus is E0 independent of the power p . The fact that the matrix
MODULUS [α] is also constant means that the anisotropic nature is unchanged;
only the stiffness magnitude changes through the term p–1e
.
Inserting (4.75) in (4.74), with [α] being symmetric, we get
du = Ep–1e
T[α]d (4.76)
and the question is now when the matrix product T[α]d can be
integrated andhow e should be defined?From the energy related def-
inition
EFFECTIVE
STRAIN 2e :=
T[α] (4.77)
follows with [α] constant
2ede = 2T[α]d (4.78)
and thus inserted in (4.76)
du= E pe de (4.79)
which again gives the 1---D result
77
Pauli Pedersen: 4. Physics --- Constitutive Models
ENERGY
DENSITIES u= 1p+ 1
E p+1e (4.80)
now just in terms of the effective strain measure as defined in (4.77).
From (4.75) and (4.77) also follows
u+ uC = σT= E
p+1e
; uC =p
p+ 1E
p+1e
(4.81)
and thus again the result (4.71).
The constitutive tangent matrix is obtained from (4.75)
dσ= p – 1ep–2
E[α]de+ ep–1
E[α]d (4.82)
and inserting (4.78), we get
2–D OR 3–D
TANGENT
MODULUS MATRIX
dσ= ep–1
E[α][I] – 1 – p
2e
T[α]d (4.83)
i.e. considerably more complicated than for the 1---D case.
From (4.75) the formulation in terms of effective stress σe is
2---D OR 3---D
SECANT = σn–1e
1En
[α]–1σ with σe = Epe and n= 1p (4.84)
COMPLIANCE
MATRIX
78
Pauli Pedersen: 4. Physics --- Constitutive Models
and the effective stress (energy related definition) is
EFFECTIVE
STRESS σ2e := σ
T[α]–1σ (4.85)
From this definition we could alternatively derive the results (4.80),
(4.81), but expressed in stresses, as
uC = 1En
1n+ 1
σn+1e (4.86)
identical to the 1---D result (4.73).
The reference stress in formulation (4.84) is the modulus E . Let us
assume that at a given stress level σe= σ0the constitutive secant com-
pliance matrix is given by
σe=σ
0=
1E0
[α]–1σ (4.87)
REFERENCE Then, according to (4.84),
STRESS
σn–10
1En =
1E0
⇒ E= E1n0
σ(n–1)n0
= Ep
0σ1–p
0(4.88)
which, togetherwith the definition of E= E01–p
0in (4.75) leads to the
relation
σ0= E00 (4.89)
for the reference values, in contrast to the relation (4.84) for the effec-
tive values.
79
Pauli Pedersen: 4. Physics --- Constitutive Models
4.6 Measuring the constitutive parameters
The constitutive parameters in the 3---D [L] matrix or in the 2---D [C]
matrixmust beobtained experimentally for agivenmaterial/model.Let
PLANE us assume a plane, orthotropic case; thematrix [C] of order three then
ORTHOTROPY has four parameters, which we can determine by three experiments.
The three load cases A , B and C correspond in the orthotropic direc-
tions to σAT= σ1, 0, 0 , σ
BT= σ0, 1, 0 and σ
CT=
σ0, 0, 1 and we measure the corresponding elongations (contrac-
ENGINEERING tions); from these we can find the Cauchy strains which by definition
MODULI give the engineering moduli EL, E
T, G
LTand the Poisson’s ratio
νLT
.
THREE σAT= σ1 0 0 ⇒
AT= σ 1
EL
– νLT
EL
0EXPERIMENTS
σBT= σ0 1 0 ⇒
BT = σ– ν
TL
ET
1ET
0 (4.90)
σCT= σ0 0 1 ⇒
CT = σ0 0 1
2GLT
(G
LTdefined by σ
12= G
LT2
12in pure shear), i.e. we aremeasuring
the compliance matrix [C]–1 . From symmetry follows
νLTE
L= ν
TLE
T(4.91)
and we therefore have
[C]–1= 1
EL
1
– νLT
0
– νLT
ELE
T
0
0
0
EL2G
LT
(4.92)
80
Pauli Pedersen: 4. Physics --- Constitutive Models
Inverting (4.92), we get the modulus matrix
[C]=EL
α0
1
νLT
ET
EL
0
νLT
ET
EL
ET
EL
0
0
0
2GLT
EL
α0
(4.93)
α0:= 1 – ν2
LTETE
L
The practical parameters α1, α
2, α
3, α
4, α
5defined in (4.13) and
(4.18) can be given in the engineering parameters by
α1= 1
83+ 3+ 2ν
LTE
TE
L+ 4α
0G
LTE
L
α2= 1
21 – E
TE
L
α3= 1
81+ 1 – 2ν
LTE
TE
L– 4α
0G
LTE
L (4.94)
α4= 1
81+ 1+ 6ν
LTE
TE
L– 4α
0G
LTE
L
α5= 1
2α
1– α
4 , α
0= 1 – ν2
LTETE
L
as follows directly from the definitions. Table 4.2. shows data for actual
materials (laminate plies)
81
Pauli Pedersen: 4. Physics --- Constitutive Models
Materials GPa (Giga Pascal) Practical non---dimensional parameters
EL ET GLTνLT α
0α1
α2
α3
α4 α
5
181.0 10.30 7.17 0.28 0.9955 0.4200 0.4716 0.1084 0.1244 0.1478
Graphite/Epoxy 138.0 8.96 7.10 0.30 0.9942 0.4298 0.4675 0.1027 0.1222 0.1538p p y
207.0 5.17 2.59 0.25 0.9964 0.3922 0.4875 0.1203 0.1266 0.1328
204.0 18.50 5.59 0.23 0.9952 0.4278 0.4547 0.1175 0.1384 0.1447
Boron/Epoxy 207.0 20.70 6.90 0.30 0.9910 0.4365 0.4500 0.1135 0.1435 0.1465p y
213.7 23.44 5.17 0.28 0.9914 0.4358 0.4452 0.1190 0.1498 0.1430
Aramid/Epoxy 76.0 5.50 2.30 0.34 0.9916 0.4233 0.4638 0.1129 0.1375 0.1429
Glass/Epoxy38.6 8.27 4.14 0.26 0.9855 0.5221 0.3929 0.0850 0.1407 0.1907
Glass/Epoxy53.8 17.90 8.96 0.25 0.9792 0.6021 0.3336 0.0643 0.1474 0.2273
ISOTROPIC E E E2(1+ ν )
ν 1 – ν2 1 0 0 ν (1 – ν)2
Table 4.2: Actual material non---dimensional parameters, calculated from Tsai and Hahn (1980).
(Note! α1+ α2+ α3= 1)
PLANE When using constitutive parameters for a specific calculation we have
STRESS OR to decide themodel for analysis. Parallel to this we have to decide the
PLANE model for experiments. Is plane stress or plane strain dominant in our
STRAIN experimental model? The answer is possibly somewhere in between. In
reality, the model for analysis cannot be totally separated from the
model for experiments although we are often forced to do so. Anyhow,
there is a clear lack of experimental data compared to the very refined
models used in analysis.
STRAIN Fig. 4.3 shows a model of a gauge that can measure the strains in three
GAUGES directions a , b, c given relative to the x
1---direction. A sketch
MEASUREMENTS of the three load cases A , B and C is also shown. We shall show the
necessary analysis to obtain the modulus matrix [C] . Relations
between the normal strains in gauge directions and the 11
, 22
, 12
strains in the x---coordinate system are taken from (2.25) or (2.40)
82
Pauli Pedersen: 4. Physics --- Constitutive Models
Fig. 4.3: Gauge geometry and sketch of load cases.
a
b
c
=
cos2a
cos2
b
cos2c
sin2a
sin2
b
sin2c
sin 2a 2
sin 2b 2
sin 2c 2
11
22
2 12
(4.95)
~= [F] by definitions
Thus, the strains which give the first column of the compliance matrix
[C]–1
from the load case σAT= σ1 0 0 is 1
σ[F]
–1~A , and simi-
larly for load cases σB= σ0 1 0 and σ
C= σ0 0 1 . In all, we
get
[C]−1=
1σ[F]−1
A
~ ~B
C
~ (4.96)
and then directly in terms of the measured gauge normal strains
[C]= σA
~ B
~ C
~ −1
[F] (4.97)
83
Pauli Pedersen: 4. Physics --- Constitutive Models
IDENTIFICATION Traditional laboratory tests are designed to give the desired quantities
BY DYNAMIC in themost direct way. Forexample, determinationofmaterial param---
MEASUREMENTS eters of constitutive laws leads to special design of tests samples and
choice of applied load in an attempt to obtain homogeneous stress---
strain fields. These idealised tests are not easy toperform, however, and
especially for composite systems, problems often arise. Furthermore,
the tests have a local natureat thepoint of the strain gauge, for example,
so for more general material information a series of tests is required.
The identification approach is an experimental strategy from the oppo-
site point of view. The experiment is made as simple as possible to give
reliable results. On the other hand, this often leads to complex non---
homogeneous stress---strain fields where the desired quantities are not
among the directly measured quantities. This causes a complicated
interpretation, but nowadays this is comfortably taken care of by com-
puter calculations.
Although our aim is to find static material data, we decide to measure
eigenfrequencies of a free rectangular plate because excellent agree-
ment between measured and calculated eigenfrequencies can be
obtained. A structural eigenfrequency is an integrated quantity and we
thus obtain material quantities that are valid in the mean for the entire
structure.
Fig. 4.4: Scheme of the experimental set---up.
84
Pauli Pedersen: 4. Physics --- Constitutive Models
The interpretation of the measurement can be formulated as an opti-
mization problem inwhichweminimizean error functional that expres-
ses the difference between model analysis response and experimental
response. The optimization is solved by an iterative procedure based on
analytical sensitivity analyses.
Eigenfrequency measurements are fast and simple to perform using
modern equipment, and there is nothing to prevent them from being
performed under different conditions. This means that the identifica-
tion method is suitable for the study of the moduli dependence due to,
say, moisture or temperature.
For the 2---D orthotropic case the important non---dimensional param-
eters are α2(level of anisotropy), α
3(relative shear stiffness) and α
4
(indirect effects), with definitions given in (4.94). In the identification
approach we primarily work with these quantities and then, when a
solution is obtained by means of the inverse formulas get the ratios of
the engineering parameters:
ETE
L= 1 – 2α
2; ν
LT= α
4– α
31 – 2α
2
(4.98)
GLTE
L= 1 – α
2– 3α
3– α
42α
0; α
0= 1 – ν2
LTETE
L
The dimensional constant of the constitutive matrix is determined by
the absolute quantities involved, i.e. plate dimensions, plate mass den-
sity and the absolute values of the measured frequencies.
A short introduction to the approach is given in Pedersen & Frederik-
sen (1992) and the thesis by Frederiksen (1992) in addition to recent
papers on more advanced techniques, Frederiksen (1996, 1997a,
1997b) gives all the necessary details about this experimental/numeri-
cal method.
85
Pauli Pedersen: 4. Physics --- Constitutive Models
4.7 Eigenvalues and eigenmodes for the constitutive matrices
The 2---D constitutive matrices, as stated in table 4.1, are positive defi-
nite, i.e. having only positive eigenvalues. With the eigenvalues and
eigenmodes we have essentially all information about a symmetric
constitutive matrix, and we shall therefore give more details about
eigenvalues and corresponding eigenmodes.
3---D For 3---D problems with a constitutive matrix [L] of order six we can
ISOTROPIC at least give the analytical solution for the isotropic case with only two
CASE parameters E , ν
[L]= Eν
∧
1
symmetric
ν
~
1
ν
~
ν
~
1
0
00
1 – ν
~
0
00
0
1 – ν
~
0
00
0
0
1 – ν
~
ν
∧
ν
~
:=
:=
1 – ν
(1+ ν)(1 – 2ν)
ν
1 – ν
(4.99)
with the six eigenvalues λiand corresponding eigenvectors being
λ1= 2G := E
(1+ ν),
1T = 0 0 0 1 0 0
λ2= 2G ,
2T = 0 0 0 0 1 0
λ3= 2G ,
3T = 0 0 0 0 0 1
(4.100)
λ4= 2G ,
4T = –1 0 1 0 0 0
λ5= 2G ,
5T = –1 1 0 0 0 0
λ6= 3K := E
(1 – 2ν),
6T = 1 1 1 0 0 0
86
Pauli Pedersen: 4. Physics --- Constitutive Models
that is a five---multiple eigenvalue with pure shear and the last eigenva-
lue with pure dilatation, thereby introducing the parameter K , termed
the bulk modulus. Similar simple results are found for 2---D isotropic
cases, and they will be a special case of a more general case, i.e. the ort-
hotropic case.
2---D For 2---D orthotropicmodels the constitutivematrix [C] relative to the
ORTHOTROPIC orthotropic directions can be written
CASE
[C]=EL
α0
1
α4– α
3
0
α4– α
3
1 – 2α2
0
0
0
1 – α2– 3α
3– α
4
(4.101)
using the parameters defined in (4.94). The three eigenvalues and the
corresponding eigenvectors for this case are
λ1=
EL
α0
(1 – α2)+ α
2
2+ (α
4– α
3)2 with
1T =
1 ,
– α2+ α2
2+ (α
4– α
3)2
α4– α
3
, 0
λ2=
EL
α0
(1 – α2) – α2
2+ (α
4– α
3)2 with
(4.102)
2T =
1 ,
– α2– α2
2+ (α
4– α
3)2
α4– α
3
, 0
λ3=
EL
α0
(1 – α2) – 3α
3– α
4 with
3T = 0 0 1
87
Pauli Pedersen: 4. Physics --- Constitutive Models
2---D For the 2---D isotropic case we have α2= α
3= 0 and thus
ISOTROPIC
CASE λ1=
Eα0
(1+ α4)
1T= 1 1 0
λ2=
Eα0
(1 – α4)
2T = 1 –1 0 (4.103)
λ3=
Eα0
(1 – α4)
3T = 0 0 1
i.e. one dilatation case and two pure shear cases.
4.8 Specific constitutive models
Metals often create crystals with cubic symmetry, either cubic surface
centered as aluminium and copper or cubic space centered as iron and
nickel. These crystals are illustrated in fig. 4.5.
Fig. 4.5: Surface centered and space centered crystal models.
88
Pauli Pedersen: 4. Physics --- Constitutive Models
CUBIC For an elastic crystal with cubic symmetry the constitutive matrix in the
SYMMETRY Cartesian coordinate system aligned with the cube faces can be written
[L]x=
21+ λ
λ
λ
0
0
0
λ
21+ λ
λ
0
0
0
λ
λ
21+ λ
0
0
0
0
0
0
22
0
0
0
0
0
0
22
0
0
0
0
0
0
22
x
(4.104)
This material model has three parameters and is a most simple ortho-
tropic case with isotropy only for 2=
1. Comparing the isotropic
case of (4.104) with (4.99), we find
λ= E ν(1+ ν)(1 – 2ν)
(4.105)
= 2=
1= G= E 1
2(1+ ν)
Parameters λ, are named Lame parameters. Now back to the more
general case (4.104) we shall determine the corresponding compliance
matrix [L]–1
xand get
[L]–1
x=
m1
m2
m2
0
0
0
m2
m1
m2
0
0
0
m2
m2
m1
0
0
0
0
0
0
1(22)
0
0
0
0
0
0
1(22)
0
0
0
0
0
0
1(22)
x
(4.106)
89
Pauli Pedersen: 4. Physics --- Constitutive Models
with m1=
1+ λ
1(2
1+ 3λ)
m2=
– λ21(2
1+ 3λ)
Let us analyse how such a material reacts to pure normal stress in three
different directions. First we subject the material to unidirectional
stress σ in the x1---direction (direction cosines 1 0 0) . For this case
the spectral decomposition (3.12) with σ1= σ and σ
2= σ
3= 0
gives the following stress vector
σT
x= σ1 0 0 0 0 0 (4.107)
and then from = [L]–1
xσ with (4.106) the resulting strain vector
T
x= σm
1m
2m
20 0 0 (4.108)
The modulus of elasticity E100= E
100(
1, λ,
2) for this test, defined
by σ11= E
10011
, will therefore be
E100
MODULUS E100= 1m
1=
1(2
1+ 3λ)(
1+ λ) (4.109)
Next we apply unidirectional stress σ in a direction given by
1 1 0 2 and the spectral decomposition then gives
σT
x= σ1 1 0 2 0 02
(4.110)
T
x= σ
2m
1+m
2m
1+m
2m
2+m
22 (2
2) 0 0
The normal strain in direction 1 1 0 2 will from (2.40) be
= 12(
11+
22+ 2
12)= σ
42(m
1+m
2)+ 1
2 (4.111)
90
Pauli Pedersen: 4. Physics --- Constitutive Models
and the modulus of elasticity E110
from σ= E therefore is
E110
MODULUS E110= 4 2
1+ λ
1(2
1+ 3λ)
+ 12
–1 (4.112)
Finally we use direction 1 1 1 3 for the unidirectional stress σ
and for this test case get
σx= σ1 1 1 2 2 2 3
(4.113)
x= σ
3m
1+ 2m
2m
1+ 2m
2m
1+ 2m
22 (2
2) 2 (2
2) 2 (2
2)
Then the normal strain in direction 1 1 1 3 will be
= 13(
11+
22+
33+ 2
12+ 2
13+ 2
23)
(4.114)
= σ9
3m1+ 6m
2+ 3
2
with resulting modulus of elasticity in this direction
E111
MODULUS E111= 3 1
(21+ 3λ)
+ 12
–1 (4.115)
Inserting the isotropic case (4.105) in (4.109), (4.112) and (4.115) we
will find as expected E100= E
110= E
111= E .
TRANSVERSAL Another important case is named transversal isotropy. We use this
ISOTROPY case for a drawn wire and also as a model for wood.
If the drawn direction (the growth direction) is chosen as the x3---direc-
tion then the constitutive matrix with five parameters λ1, λ
2, λ
3,
1,
2
is
91
Pauli Pedersen: 4. Physics --- Constitutive Models
[L]x=
21+ λ
1
λ1
λ2
0
0
0
λ1
21+ λ
1
λ2
0
0
0
λ2
λ2
λ3
0
0
0
0
0
0
21
0
0
0
0
0
0
22
0
0
0
0
0
0
22
x
(4.116)
The corresponding compliance matrix [L]–1
xis
[L]–1
x=
m1
m2
m3
0
0
0
m2
m1
m3
0
0
0
m3
m3
m4
0
0
0
0
0
0
1(21)
0
0
0
0
0
0
1(22)
0
0
0
0
0
0
1(22)
x
(4.117)
With
D := 41(
1+ λ
1)λ
3− 4
1λ22
m1:= 2(
1+ λ
1)λ
3– λ2
2D= 12
1
m2:= λ2
2– λ
1λ3D
m3:= – 2
1λ2D
m4:= 4
1
1+ λ
1D
With unidirectional stress in the isotropic plane, say the x1---direction
we, using x= [L]–1
xσx , get
92
Pauli Pedersen: 4. Physics --- Constitutive Models
σx= σ1 0 0 0 0 0
(4.118)
x= σm1 m2 m3 0 0 0
and thereby the modulus of elasticity in the isotropic plane
Eisotropic plane = 1m1= 21 (4.119)
With unidirectional stress in the direction perpendicular to the iso-
tropic plane, i.e. the x3---direction we have
σx= σ0 0 1 0 0 0
(4.120)
x= σm3 m3 m4 0 0 0
and the resulting modulus of elasticity
E⊥isotropic plane= 1m4=
2(1+ λ1)λ3− 2λ22
2(1+ λ1)(4.121)
93
Pauli Pedersen: 5. Stress Concentration Examples
5. STRESS CONCENTRATION EXAMPLES
effects of non---linearity, effects of anisotropy
5.1 The classic solution for stress concentration around elliptical
holes
Fig. 5.1 shows anelliptical holewith half---axes a , b in the x1, x
2direc-
tions, respectively. Two angles are involved in the problem, firstly the
LOAD angle from the x1---direction to the direction of the load. For the load
DIRECTION a uniaxial stress state is assumed and the stress level far away from the
hole is σ∞
. The other angle θ specifies the point of the elliptical
boundary at which we determine the actual tangential stress σθ, which
a
b
σθ()
θ
σ∞
x2
x1
Fig. 5.1: Elliptical hole in an infinite plane domain, subjected to stress far away
from the hole.
94
Pauli Pedersen: 5. Stress Concentration Examples
is a principal stress because the stress normal to and at the elliptical
boundary is zero.
The coordinates for the point at which the tangential stress is given are
x1= a cos t , x
2= b sin t (5.1)
where t is the traditional parameter for describing an ellipse. The for-
mula for the stress can be found in such classical works as Savin (1961)
UNIAXIAL with reference back toMuskhelishvili (1934) or even in handbooks like
STRESS STATE Hedner ed. (1992). The tangential stress is
σθ()= σ
∞
1 – m2+ 2mcos(2) – 2 cos2(θ+ )
1+m2 – 2mcos(2θ)(5.2)
with m := (a – b)(a+ b)
For a circular hole with m = 0 we get σθ= σ
∞1 – 2 cos(2θ) , i.e.
σθmax
= 3σ∞
. This means a stress concentration factor of 3 and a
stress variation from – σ∞
to 3σ∞
along the circular boundary.
For the case of biaxial stress state as shown in fig. 5.2 we find the result
by directly adding two cases = 0 and = π2 of (5.2) and get
σθ=
11+m2 – 2m cos(2θ)
(σ1+ σ
2)(1 – m2)+ (σ
1− σ
2)2m− cos(2θ) (5.3)
CIRCULAR and for a circular hole we get, with m= 0 ,
HOLE
σθ(a= b)= σ
1+ σ
2–
σ1– σ
22 cos(2θ) (5.4)
95
Pauli Pedersen: 5. Stress Concentration Examples
From this last result we see that constant tangential stress is only pos---
STRESS sible for the biaxial stress state of σ1= σ
2= σ , which gives
CONCENTRATION
FACTOR σθa= b, σ
1= σ
2= σ
= 2σ (5.5)
and we may talk about a stress concentration factor of 2 for this most
simple case. In the more general case of (5.4) the maximum tangential
stress at the circular boundary as found from dσθdθ= 0 at θ= 0 ,
π2 is
3σ2− σ
1≤ σ
θ≤ 3σ
1− σ
2
3σ1− σ
2≤ σ
θ≤ 3σ
2− σ
1
for
for
σ1> σ
2
σ2> σ
1
(5.6)
Let us go back to the elliptical hole and for the result (5.3) find the
condition for a constant tangential stress, i.e. dσθdθ= 0 for all θ .
We find after some algebra
ab= σ1σ
2⇒ σ
θconstant (5.7)
MINIMUM a result frequently used as a test example in optimal shape design for
STRESS minimum stress concentration. For this case the constant stress will be
CONCENTRATION
σθab=
σ1
σ2
= σ1+ σ
2(5.8)
Now all these classical results are based on assumptions of plane, linear
elasticity, of infinite domains, and of isotropicmaterial. In the next sec-
tion we shall use some numerical examples to study the influence of
these assumptions.
96
Pauli Pedersen: 5. Stress Concentration Examples
5.2 Numerical solutions for stresses around elliptical holes
The essential influence from a finite width of a strip with a circular hole
is shown in Savin (1961) based on early results using series expansions,
i.e. before the finite elementmethodwas developed. Today such results
can be obtained with almost any finite element program, and graphical
illustrations of stress state, strain state, energy density state give direct
access to the result. With the reference model shown in fig. 5.2 we shall
discuss such results and extend the problem to include anisotropic and
even non---linear elasticity.
σ1=
32σ2
σ2
σ1
x1
x2
a= 32b
a
b
Fig. 5.2: Plane reference model for the study of stress/strain concentration around
an elliptical hole in a finite domain.
The first results will illustrate the influence of the ratio of the hole size
to the finite domain size. With isotropy and linear elasticity and
σ1= 3 , σ
2= 2 , ab= 32 , we for infinite domain from (5.8)
97
Pauli Pedersen: 5. Stress Concentration Examples
Fig. 5.3: Larger principal stress fields resulting from σ1= 3 , σ
2= 2 , and “opti-
mal” hole shape of ab = 32 . Only the hole size in this finite domain is changed.
should have σθ= 5 and constant. For the small hole model we get
(σθ)max = 5.14 and for the larger hole model we get (σ
θ)max = 7.45 ,
in both cases rather constant along the elliptical boundary. In fig. 5.3 the
total stress fields are illustrated, showing principal stress directions and
a hatching proportional to σ2 (squared to focus on the stress con-
centrations). In the rest of this chapter we shall use this directional
hatching technique to illustrate stress fields, strain fields and energy
density field when dealing with anisotropic materials.
The general conclusion from the results in fig. 5.3 is in good agreement
with the early results shown in Savin (1961). It tells us that the analytical
results for infinite domains should be used with great caution, because
the ratio of hole size to domain size has an influence which should not
be neglected. For the larger hole the tangential stress along the ellipti-
cal boundary varies from 4.7 to 7.5 , while for the smaller hole the vari-
ation is only 5.05 to 5.14 .
98
Pauli Pedersen: 5. Stress Concentration Examples
p = 0.75
p = 0.50 p = 0.25
σmax = 5.14
p = 1.00
σmax = 4.69
σmax = 4.32 σmax = 3.94
Fig. 5.4: Stress field for solutions with non--- linear elasticity.
Let us then for the problem with the smaller hole study the influence
from possible material non---linearity, using a power law model as
described in (4.75). Concentrating on the field close to the hole bound-
ary we show first, in fig. 5.4, the stress results for p = 1.0 , 0.75 , 0.50
and 0.25 . (The material difference for these cases is illustrated in fig.
4.2).
99
Pauli Pedersen: 5. Stress Concentration Examples
p = 1.00 p = 0.75
p = 0.50 p = 0.25
max= .0026 max= .0028
max= .0033 max= .0051
Fig. 5.5: Strain fields for solutions with non--- linear elasticity.
The main effect of the non---linearity on the stress field is a releasing
effect. Themaximumprincipal stress for the four cases is σmax = 5.14 ,
4.69 , 4.32 , 3.94 , respectively, and the fields also clearly showmoreuni-
formity for increasing non---linearity (power p decreasing).
100
Pauli Pedersen: 5. Stress Concentration Examples
The strain fields and the energy density fields show the opposite effect,
which is increasing localisation with increasing non---linearity. Fig. 5.5
shows principal strain fields with hatching proportional to strain inten-
sity. The maximum strains for the four cases are .0026 , .0028 , .0033
and .0051 . The intention of this chapter is not to discuss the solution
procedures, but merely to illustrate some effects of more general
interest. The variation of stresses along the hole boundary is very
limited for all the cases, but this is naturally due to the initial choice of
ab= σ1σ
2. However, it tells that the “optimal” hole shape is only
little dependent on the power of the non---linearity.
5.3 Design with orthotropic materials
The problem of shape design for minimum energy concentration is
highly non---linear, and must be solved iteratively. In this section we
shall show some results from Pedersen, Tobiesen and Jensen (1992).
The problem can be converted to a sequence of problems of optimal
redesign, i.e., a given shape can be changed into a better “neighbour-
ing” shape. The solution to this problem involves three steps: finite ele-
ment strain analysis for the given shape, sensitivity analysis with respect
to the parameters describing the design, and optimal decision for rede-
sign.
Orthotropic materials are now frequently used in engineering design.
There is thus a need to understand the effects on optimal shape design
with these materials. Stiffness as well as strength are direction---depen-
dent, and designs that are rather different fromoptimal shapes with iso-
tropicmaterials (shown in section 5.2) will appear. Still we find uniform
energy density along the design boundary.
101
Pauli Pedersen: 5. Stress Concentration Examples
Material # and name EL1011Pa E
T1011Pa ν
LT GLT1011Pa E
LE
T
I Isotropic 1.0 1.0 0.3 0.3846 1.0
II 5% fiber 3.450 1.052 0.3 0.4044 3.281
III 10% fiber 5.900 1.109 0.3 0.4264 5.322
IV 20% fiber 10.80 1.244 0.3 0.4784 8.683
V 30 % fiber 15.70 1.416 0.3 0.5448 11.08
VI Strong orthotropic 60.914 1.4503 0.3 0.13778 42.00
Table 5.1: Applied material data.
Weshall again refer to the specific example in fig. 5.2, but now apply the
material data shown in table 5.1.
The optimal boundary shapes for the materials I---V are shown in fig.
5.6, and the gain in stress concentration relative to the original elliptical
boundary are with material index II---V,
umax initialumax optimal= 1.8II , 1.6III , 1.5IV , 1.2V (5.9)
We conclude that as expected the “degree” of orthotropy has a strong
influence on design.
102
Pauli Pedersen: 5. Stress Concentration Examples
ELET
Fig. 5.6: Optimal boundary shapes for different “degrees” of orthotropy.
Finally, we shall illustrate the influence of combined shape/thick-
ness/orientational design. The specific results are based onmaterial IV
in table5.1.The stressdistributions obtainedwith a 1024 finite element
model are shown in Fig. 5.7.
Fig. 5.7: Stress distributions with a) Shape design only; b) Shape/thickness design;
c) Shape/thickness/orientational design.
103
Pauli Pedersen: 5. Stress Concentration Examples
The resulting energy densities in table 5.2 shows the design influence
on stiffness (mean value of strain energy density) as well as on strength
(maximum value of strain energy density).
Shape optimal for uniform thickness and orientation
uniformthicknessandorientation
optimalthicknessuniformorientation
uniformthicknessoptimalorientation
optimalthicknessandorientation
umean 3209 2900 1425 1299
umax 12370 3320 3887 1436
Table 5.2: Resulting relative strain energy densities.
With optimal thickness distribution, then uniform energy density
should give umean = umax . The differences in table 5.2, 2900≠ 3320
and 1299≠ 1436 , are a result of inforced thickness limits.
104
Pauli Pedersen: 5. Stress Concentration Examples
105
Pauli Pedersen: 6. Stationarity and Extremum Principles of Mechanics
6. STATIONARITY AND EXTREMUM
PRINCIPLES OF MECHANICS
Work equation, virtual work principles,
Clayperon, Castigliano, minimum potential principles
6.1 The work equation, an identity
GOAL OF Energy principles play a central role in mechanics, but surprisingly few
THE CHAPTER books treat the subject in a structured way. It is difficult to get an over-
view of the many different principles, and important questions are not
presented, especially in relation to the necessary conditions for a cer-
tain principle.
The present chapter is written as an alternative to the classical presen-
tations, as by Langhaar (1962) andWashizu (1975). We shall show that
all principles are specific interpretations of the same identity, and nec-
essary conditions will not be introduced until absolutely needed. We
shall refer only to a Cartesian 3---D coordinate system, and traditional
tensor notation for summation and differentiation is applied.
NO PHYSICAL Without physical interpretation of the quantities involved, we shall
INTERPRETATION derive an important identity. The superscripts a and b are only part
of the names (not powers) and their use will be explained later. From
σaijbij= σ
aijb
ij–
12vb
i,j+ vb
j,i+ σ
aij12vb
i,j+ vb
j,i (6.1)
106
Pauli Pedersen: 6. Stationarity and Extremum Principles of Mechanics
and
σaijvbj,i= σ
ajivbi,j= σa
ji– σa
ijvb
i,j+ σ
aijvbi,j
(6.2)
follows
V
σaijbijdV –
V
σaijb
ij–1
2vb
i,j+ vb
j,idV
(6.3)
–V
12σa
ji– σ
aijvb
i,jdV –
V
σaijvbi,jdV= 0
In (6.3) V is the volume of the domain of interest. Now let A be the
FROM surface that bounds this domain and nj the outward normal at a point
THEOREM OF of the surface. Then, using the theorem of divergence, the last part of
DIVERGENCE (6.3) is rewritten to
V
σaijvbi,jdV=
V
σaijvbi,j– σa
ij,jvbidV=
A
σaijvbin jdA –
V
σaij,jvbidV (6.4)
Once more, by adding and subtracting the same quantities, we obtain
THE the identity from which the stationarity principles of mechanics can be
IDENTITY read without further calculations
V
σaijbijdV –
V
σaijb
ij– 12vb
i,j+ v
bj,idV
–V
1
2σa
ji– σ
aijvb
i,jdV –
A
σaijnj – T
aivb
idA (6.5)
–A
TaivbidA+
V
σaij,j+ pa
ivb
idV –
V
paivbidV= 0
107
Pauli Pedersen: 6. Stationarity and Extremum Principles of Mechanics
With the index a and b we have indicated certain relations, and we
shall now assume that all quantities with index a are related and that
all quantities with index b are related. However, no relations exist
between quantities with different index, and the equations therefore
still have no physical interpretations.
Let bij=
bij(x) be a strain field derived from a displacement field
vbi= vb
i(x) with small strain assumption (Cauchy strains)
bij=
12vb
i,j+ vb
j,i , small strain assumption (6.6)
and let σaij= σ
aij(x) be a stress field with moment and force equilib-
rium with the field of volume forces pai= pa
i(x) and surface tractions
Tai= Ta
i(x)
σaji= σ
aij
σaij,j= – pa
i(6.7)
σaijnj= Ta
i
With (6.6) and (6.7) the identity (6.5) reduces to
THE WORK
EQUATIONV
σaijbijdV=
A
TaivbidA+
V
paivbidV (6.8)
which is often called thework equation, although it ismerely an identity
and does not express physical work when the a and b fields are not
related.
108
Pauli Pedersen: 6. Stationarity and Extremum Principles of Mechanics
6.2 Symbols and definitions
In table 6.1 we show the quantities of the energy principles of mechan-
ics.
Symbol Name Definition
Wvi WORK
of external forcesA
v
i
0
Tiv
i
~dvidA
~
+ V
v
i
0
piv
i
~dvidV
~
uij STRAIN ENERGY
DENSITY
ij
0
σijij
~ dij
~
UTotal
STRAIN ENERGYV
udV
ΠTotal
POTENTIAL ENERGYU – W
WCTi, p
i
COMPLEMENTARYWORK
of external forcesA
T
i
0
viT
i
~dTidA
~
+ V
p
i
0
vip
i
~dpidV
~
uCσij STRESS ENERGY
DENSITYσ
ij
0
ijσij
~ dσij
~
UC Total
STRESS ENERGY
V
uCdV
ΠC
TotalCOMPLEMENTARY
POTENTIAL ENERGYUC – WC
Table 6.1: Symbols and definitions for work and energy.
109
Pauli Pedersen: 6. Stationarity and Extremum Principles of Mechanics
For a better overview, the stationarity principles of mechanics will be
REAL divided into four groups covering the possible combinations of real
FIELDS fields (indexed by a superscript 0) and virtual fields (without index).
The virtual fields are assumed to be sufficiently differentiable and
VIRTUAL admissible, but otherwise arbitrary and non---physical. An admissible
FIELDS displacement field must be kinematically admissible, i.e. it must satisfy
the boundary conditions. An admissible stress field must be statically
admissible, i.e. it must satisfy force equilibrium.
6.3 Real stress field and real displacement field
As a first example of use of the work equation (6.8) we insert the real
stress field σ0
ijin equilibrium with the real load field T0
i, p0
i, and the
real displacement field v0ifromwhich the real strain field
0ijis derived.
We get
V
σ0ij0ijdV=
A
T0iv0idA+
V
p0iv0idV (6.9)
For arbitrary constitutive relations we have
σ0ij0ij=
σ0
ij0ij
0
dσijij
~= 0
ij
0
σij ~
ijd ~
ij+σ0
ij
0
ijσ~
ijdσ~ ij (6.10)
which, together with the definitions in table 6.1, gives
σ0ij0ij= u0
+ u0C
(6.11)
V
σ0ij0ijdV= U0
+U0C
110
Pauli Pedersen: 6. Stationarity and Extremum Principles of Mechanics
Analogously for the external loads we get
A
T0iv0idA+
V
p0iv0idV=W0
+W0C (6.12)
and (6.9) can thus be written
ZERO SUM
OF TOTAL U0+U0C – W0
+W0C= Π0+Π
0C= 0 (6.13)
POTENTIALS
i.e. the sum of the real total potentials is zero.
Especially for a linear elastic material the definitions in table 6.1 give
U0= U0C
=
12V
σ0ij0ijdV (6.14)
In relation to the nature of the external forces, the concept of dead load
is important. For dead loads the forces are independent of the displace-
ment of their point of action, say a gravity load. For a dead load we get
no complementary work W0C= 0 , and the work is thus
W0=
A
T0iv0idA+
V
p0iv0idV (6.15)
For a system with both linear elasticity and dead loads (6.9) gives
CLAYPERON’S
THEOREM U0=W02 (6.16)
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Pauli Pedersen: 6. Stationarity and Extremum Principles of Mechanics
and thus
Π0 := U0 – W0= – W02= – U0 (6.17)
which is often called Clayperon’s theorem for linear elasticity. The
“missing” energy W02 is assumed to be dissipated before the static
equilibrium with which we are concerned.
Note that if we by definition take –W0 as given by (6.15) to be the
external potential, then the assumption of dead load is not necessary,
but then again external potential is hardly a physical quantity.
6.4 Real stress field and virtual displacement field
Assume that vi is a kinematically admissible displacement field and
that ij is the strain field derived from vi . Furthermore, as before,
σ0i, T0
i, p0
iare the real stress, surface traction and volume force fields.
Then the work equation (6.8) reads
V
σ0ijijdV=
A
T0ividA+
V
p0ividV (6.18)
To distinguish the work by the external loads from the work of the reac-
tions we divide the surface area A into
A= AT+Av (6.19)
where AT is the surface area without displacement control and Av is
the surface area with given kinematical conditions. Furthermore, we
describe the virtual field vi by a variation δvi relative to the real field
v0i
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Pauli Pedersen: 6. Stationarity and Extremum Principles of Mechanics
vi= v0i+ δvi , ij=
0ij+ δij (6.20)
which with ij=v
i,j+ vj,i2 gives
δij=12δvi,j+ δvj,i (6.21)
Now, as vi is assumed to be kinematically admissible, we have δvi≡ 0
on the surface Av (but not necessarily vi= 0) , and thus (6.18) reduces
to
(6.22)V
σ0ijδijdV=
AT
T0
iδv
idA+
V
p0iδv
idV
VIRTUAL
WORKPRINCIPLE
This is called the virtual work principle or the principle of virtual dis-
placements. Note that the virtual displacements and strains in (6.22)
are infinitesimal and express energy and work variations without
assumptions of linearity. Note also that the virtual work principle is a
principle about a state, not a process.
Often (6.18) and even (6.8) are also called the virtual work principle,
but in this book we shall assume the virtual displacements and the
virtual strains to be infinitesimal.
Because stresses are fixed in the virtual work principle, a direct physical
interpretation is not clear.However, it can be read as an energy balance
which is valid for any kinematically admissible disturbance of the dis-
placement field.
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Pauli Pedersen: 6. Stationarity and Extremum Principles of Mechanics
Some specific cases of use of the virtual workprinciple leadus to specia-
lised principles. Let us choose the very specific virtual displacement
field
δvi= ∆v corresponding to the single load Q
δvi= 0 corresponding to all other external loads (6.23)
∆ij derived from this field
then (6.22) reduces to
CASTIGLIANO’S1ST THEOREM
σ0ij∆ijdV= Q ∆v or Q= ∆U
∆v=
∂U∂v (6.24)
which is the first theoremofCastigliano. It is useful in determining stiff-
nesses. Note that this theorem is valid independent of the specific
constitutive behaviour σ = σ() . The force Q should be interpreted
UNIT as a generalised force; thus, if Q is an external moment, then v is the
DISPLACEMENT corresponding rotation.
THEOREM FOR
LINEAR Now a much used theorem is obtained from (6.24) if we assume linear
ELASTICITY elasticity, because then we can set the displacement to ∆v= 1 and
∆ij= 1
ijfor the resulting strains from this unit displacement field and
get
Q= V
σ0
ij1ijdV (6.25)
Returning to general non---linear elasticmaterials, we can interpret the
virtual work principle as stationary potential energy. A potential is a
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Pauli Pedersen: 6. Stationarity and Extremum Principles of Mechanics
scalar from which work can be derived. Let us assume that a material
has a potential and the external loads has as well, then (6.22) states
STATIONARY δU= δW or δΠ= 0 (6.26)
TOTALPOTENTIALENERGY
Note that the principle of stationary potential energy is only valid for
a limited class of systems (potential systems), while the virtual work
principle is valid in general.
6.5 Virtual stress field and real displacement field
Avirtual stress field σij is a statically admissible field, i.e. in equilibrium
with the given external loads. Now, inserting also the real displacement
field v0i, 0
ijin (6.8), we get
V
σij0ijdV=
AT
T0iv0idA+
Av
Tiv0idA+
V
p0iv0idV (6.27)
On the surface Av , the surface tractions Ti are the unknownreactions.
Taking the virtual stress field as
σij= σ0ij+ δσij (6.28)
where σ0ijis the real stress fieldand δσij is an infinitesimal virtual stress
field satisfying
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Pauli Pedersen: 6. Stationarity and Extremum Principles of Mechanics
δσij= δσji , δσij,j= 0
(6.29)δσijnj= δTi where δTi = 0 on AT
then using (6.9) we get
V
δσij0ijdV=
Av
δTiv0idA
COMPLEMENTARY
VIRTUAL WORK
PRINCIPLE
(6.30)
which expresses the principle of complementary virtual work, also
called the principle of virtual stresses.
Choosing a specific virtual field
δσij= ∆σij where ∆σijn j= ∆Q
corresponding to displacement v(6.31)
∆T= ∆σijnj= 0 all other places with prescribed vi≠ 0
we get from (6.30)
CASTIGLIANO’S
2ND THEOREM
V
∆σij0ijdV= ∆Qv or v= ∆UC
∆Q=
∂UC
∂Q(6.32)
which is the second theorem named after Castigliano. It is valuable in
determining compliances (flexibilities).
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Pauli Pedersen: 6. Stationarity and Extremum Principles of Mechanics
UNIT LOAD
THEOREM FOR Also, a unit theorem is obtained in complementary energies, read
LINEAR directly from (6.32) when linear elasticity is assumed, i.e. ∆P= 1 and
ELASTICITY ∆σij= σ1ij
v= V
σ1ij0ijdV (6.33)
Finally, the parallel to stationary potential energy is the principle of sta-
tionary complementary potential energy
STATIONARY
TOTAL
COMPLEMENTARY
POTENTIAL
ENERGY
δUC= δWC or δΠ
C= 0 (6.34)
valid for potential systems and then just an alternative statement of the
virtual complementary work principle.
6.6 Minimum total potential
The stationarity principles ofmechanics are based on very few assump-
tions. The principles of virtual work hold for any constitutivemodel and
for any type of load, and for potential systems these virtual principles
give stationary potential energies.
Many approximation methods (like the finite element method) are
MOTIVATION based on and uniquely specified by these stationarity principles. How---
FOR METHODS ever, this does not give us sufficient reason to choose an approximate
solution that satisfies the same stationarity as the unknown real solu-
tion. Energy principles that in addition to stationarity give extremum
117
Pauli Pedersen: 6. Stationarity and Extremum Principles of Mechanics
can justify our choice. We choose the approximation for which the
energy is closest to the real unknown energy. Furthermore, consistent
approximation methods are a reasonable choice also for problems
where an extremum cannot be proven.
We shall firstly prove the principle of minimum total potential
ASSUMPTIONS energy δ2Π> 0 , and for this we need assumptions concerning the
constitutive model as well as for the load behaviour. Let us start with
a single load Q (force ormoment) and the correspondingdisplacement
v (translation or rotation). For this force Q as a function of v , we will
assume
SINGLE LOAD ∂Q∂v≥ 0 , ∂v∂Q> 0 , Q(v= 0)= 0 (6.35)
BEHAVIOUR
as illustrated in fig. 6.1
Q
Q
v
v
WC
W
Fig. 6.1: Illustration of a possible relation between load and corresponding dis-
placement. The work W and the complementary work WC are the shown areas.
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Pauli Pedersen: 6. Stationarity and Extremum Principles of Mechanics
Wenote that Q(v) is a function but, as ∂Q∂v= 0 is apossibility, v(Q)
is not strictly a function.Non---linearity and change of sign for curvature
is possible. From the definition of W and WC in table 6.1 follows
W := v
0
Q v~d v~ =>∂W∂v= Q= Q(v)
(6.36)
WC := Q
0
vQ~dQ~ =>∂WC
∂Q= v= v(Q)
As by (6.12) we have for this case of a simple load
W+WC= Qv (6.37)
Also the sign of the curvature of W=W(v) and WC=WC(Q) is
known from (6.36) and (6.35)
∂2W∂v2
=∂Q∂v≥ 0 , ∂
2WC
∂Q2=∂v∂Q> 0 (6.38)
WORK From this it follows that we get a work function W=W(v) , as illu---
FUNCTION strated in fig. 6.2.
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Pauli Pedersen: 6. Stationarity and Extremum Principles of Mechanics
W
v
b
a
Wb
Wa
vbva
∂W∂v= Qb
∂W∂v= Qa
Fig. 6.2: Work W of the load Q as a function of the displacement v
that corresponds to the load.
From the tangents shown in fig. 6.2 we read the inequalities
Wa+Qavb − va≤Wb
(6.39)
Wb − Qbvb − va≤Wa
which together give us an inequality valid for vb> va as well as
vb< va (convexity)
Qavb − va≤Wb − Wa (6.40)
where the equality only holds for ∂Q∂v≡ 0 in the actual interval
va – vb and naturally for va= vb .
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Pauli Pedersen: 6. Stationarity and Extremum Principles of Mechanics
Same arguments hold for the complementary quantities and thus we
also have
vaQb − Qa≤WCb − WCa (6.41)
with equality only for Qb= Qa because ∂v∂Q> 0 is assumed in
(6.35). Inserting Qava=Wa+WCa from (6.37) in (6.40) or (6.41)we
get the inequality for the mixed products
INEQUALITY
FOR MIXED Qavb≤Wb+WCa (6.42)
PRODUCT
By summation and/or integration we can extend the above results to a
load system.
UNIAXIAL For a uniaxial stress/strain in terms of pure normal σ, or, alterna---
CONSTITUTIVE tively, pure shear τ, γ , we assume a function very parallel to the load
MODEL displacement function (6.35), i.e.
∂σ∂> 0 , ∂∂σ> 0 , σ(= 0)= 0 (6.43)
i.e. well---defined functions for σ= σ() as well as for = (σ)
because strict inequalities hold in (6.43). Such a relation has been
shown earlier in fig. 4.1. From the assumptions (6.43) follows in direct
analogy to load---work arguments which lead to (6.40)---(6.42)
σab –
a< ub – ua for b≠
a
aσb – σ
a< uCb – uCa for σb≠ σ
a (6.44)
σab< ub+ uCa for a≠ b
with the definitions of energy densities from table 6.1 and the pre-
viously discussed relation u+ uC = σ as stated in (4.65).
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Pauli Pedersen: 6. Stationarity and Extremum Principles of Mechanics
For a multidimensional stress/strain state a direct generalisation is not
easy, and the assumption is therefore often stated directly as convexity
of the energy density in the six---dimensional strain/stress spaces. With
tensor symbols this is written
σaijb
ij–
aij< ub – ua for
bij≠
aij
aijσb
ij– σ
aij< uCb – uCa for σ
bij≠ σ
aij
σijij= u+ uC⇒ (6.45)
σaijbij< ub
+ uCa for a≠ b
In matrix symbols the last inequality is
σaTb< ub+ uCa (6.46)
We now have the necessary inequalities to prove the extremum prin-
ciples, and again we start from the work equation (6.8). With real
stresses σ0
ijand virtual displacements, strains vi – v0
i , ij –
0ij , we
get
σ0ijij –
0ijdV=
A
T0ivi – v0
idA+
V
p0ivi – v0
idV (6.47)
From (6.45) follows that the left---hand side satisfies
σ0ij
ij – 0ijdV< U – U0 for ij≠
0ij
(6.48)
and the right---hand side for dead load ∂T0i∂vi = 0 , ∂p0
i∂vi = 0
gives
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Pauli Pedersen: 6. Stationarity and Extremum Principles of Mechanics
A
T0
ivi– v0
idA+
V
p0ivi– v0
idV=W – W0 (6.49)
Using (6.48) as well as (6.49) in (6.47) we get the result
(6.50)
MINIMUM
TOTALPOTENTIALENERGY
U – U0>W – W0 or
Π> Π0 for ij≠
0ij
i.e. the extremum principle for total potential energy. We note that the
dead load assumption (6.49) is a necessary condition if we do not decide
by definition to term the right---hand side of (6.47) as the negativeexter-
nal potential energy.
6.7 Minimum complementary potential and two sided bounds
We can quickly establish the complementary principle because all the
necessary inequalities were derived in the preceding section. In the
work equation (6.8) we now insert v0i, 0
ijand the virtual stresses
σij – σ
0ij and get
V
0ijσij – σ
0ijdV=
A
Ti– T0
iv0
idA+pi – p0
iv0
idV(6.51)
The left---hand side satisfies
V
0ijσ
ij – σ0ijdV< UC – UC0 for σij≠ σ
0ij
(6.52)
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Pauli Pedersen: 6. Stationarity and Extremum Principles of Mechanics
The main part of the right---hand side is zero when σij is statically
admissible because pi – p0i≡ 0 and Ti – T0
ican only be different
from zero at the reactions. For dead loads this part will be WC – WC0 ,
and with (6.52) in (6.51) we get
(6.53)
MINIMUMTOTAL
POTENTIALENERGY
UC – UC0>WC – WC0 or
ΠC> Π
C0 for σij≠ σ0ij
COMPLEMENTARY
i.e. the extremum principle for total complementary potential energy.
Using also the early result (6.13) of Π0+Π
C0= 0 for only real fields
we can with (6.50) and (6.53) set up two---sided bounds on approximate
solutions. For the real solution we have (6.13), and by the sum of (6.50)
and (6.53) we for an approximate solution get
Π+ΠC> 0 or Π> – ΠC (6.54)
Furthermore, substitution of Π0= – ΠC0 in (6.50) and (6.53) then
gives the bounds
(6.55)TWO---SIDED Π> Π0> –ΠC
ΠC> Π
C0> – ΠBOUNDS
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Pauli Pedersen: 6. Stationarity and Extremum Principles of Mechanics
6.8 Table of all principles
In table 6.2 we have illustrated the connections between the many dif-
ferent energy principles.
σaij≡ σ
aji ,
bij≡ ½vbi,j+ vb
j,i
Tai = σ
aijnj , p
ai = – σa
ij,j
V
σaij
bijdV=
A
Taiv
bi dA+
V
pai v
bidV
VIRTUALWORK
PRINCIPLE
VIRTUAL
COMPLEMENTARY
WORK PRINCIPLE
MINIMUMOF TOTAL
ELASTIC POTENTIAL
MINIMUMOF TOTAL
COMPLEMENTARY
ELASTIC POTENTIAL
ASSUMPTIONS ON
LOADS--- AND CONSTITUTIVE
BEHAVIOUR
TWO---SIDED
BOUNDS
MINIMUMOF
INTERNAL
ELASTIC ENERGY
MINIMUMOF INTERNAL
COMPLEMENTARY
ELASTIC ENERGY
CASTIGLIANO I CASTIGLIANO II
LINEARITY LINEARITY
UNIT---LOAD
PRINCIPLE
UNIT---DISPLACEMENT
PRINCIPLE
Table 6.2: Overview of energy principles.
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Pauli Pedersen: 7. Application of Energy Principles
7. APPLICATION OF ENERGY PRINCIPLES
Energy in beams, exact solutions,
approximate solutions, different principles
7.1 Elastic energy in straight beams
In handbooks of strength of materials, like in Hedner ed. (1992), we
find the formula for elastic energy in a straight beam of length (b – a)
U= UC= UC
N+UC
T+UC
M+UC
Mx
(7.1)
= b
a
N2
2EA+ β T2
2GA+ M2
2EI+
M2x
2GKdx
where the cross---sectional forces/moments are
N = normal force
M = bending moment
T = transverse (shear) force
Mx = torsional moment
The material parameters of the assumed isotropic linear elastic beha-
viour are
E = Youngs modulus ν = Poissons ratioG = shear modulus = E2(1+ ν)
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Pauli Pedersen: 7. Application of Energy Principles
and the cross---sectional constants are
A= area (tensioncompression stiffness factor)
I = moment of inertia (bending stiffness factor)
K= torsional stiffness factor
β = factor from the shear stress distribution
NORMAL Letusprimarily prove the individual terms in (7.1) basedon thedefini---
FORCE tions in section six. With only a normal force N and stresses uniformly
distributed over the cross---section we get
σ= NA = σE u= uC= 12σ= 1
2σ2E= 1
22E
(7.2)
i.e. u= uC = 12N2A2E
With uniformly distributed energy density, the energy per length is
uCA= N2(2AE) as stated in (7.1).
BENDING With only a bending moment M the stresses vary linearly through the
MOMENT height h of the beam
σ= (MI)z for –h2≤ z≤ h2
(7.3)
i.e. u= uC = 12z2M2I2E
andwith themoment of inertia defined by I := z2dA , the energy per
length is uCdA=M2(2IE) as stated in (7.1).
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Pauli Pedersen: 7. Application of Energy Principles
TRANSVERSE After the two cases of pure normal stress let us analyse the case of only
FORCE a transverse force T . The distribution of shear stresses τ= τ(z)
(τ= σ12) will depend on the specific cross---sectional shape, and is
here stated as
τ= τmax f(z) with f(z)= 0 for z= h2 (7.4)
Thenwith engineering shear strain γ γ= 212
and as a cross---sec-
tional constant determined by f(z) the analog to (7.2) is
τmax= TA τ= τmax f(z) γ= τG
u= uC = 12τγ= 1
2τ2G= 1
2γ2G (7.5)
i.e. u= uC = 122T2(A2G)f2(z)
Integrating to energy per length uCdA , we see that the constant β
in (7.1) must be defined by
β= f2(z)dA2A (7.6)
For specific values of , β see Hedner ed. (1992).
TORSIONAL Finally the case of only a torsional moment Mx , here restricted to
MOMENT circular cross---sections for which the stress distribution with outer
radius R is
128
Pauli Pedersen: 7. Application of Energy Principles
τmax=MxRK τ= τmaxRr = MxKr
(7.7)for r
min≤ r ≤ R
In parallel to (7.5) we then get
u = uC = 12τ2G= 1
2M2
x(K2G)r2 (7.8)
which with K = r2dA for circular cross---sections gives the energy
per length as stated in (7.1). The non---circular cross---sections will be
covered in chapter ten.
DECOUPLED Energy is not linear in stresses as seen from (7.1), where N , T , M
ENERGIES and Mx are often termed generalized stresses. We therefore need to
prove that the simple addition of the four energies are correct. For the
normal stresses with both N and M we get
σ=NA+
MIz⇒ σ
2=
N2
A2+
M2
I2z2+ 2NM
AIz (7.9)
and the last termwill not give rise to energy because by definition of the
beam axis we have zdA= 0 .
For the shear stresses with both T and Mx it is more simple to look
at the work of T and Mx instead of the elastic energy and then base
the proof on the energy principles of section six. The beam displace-
ments from T give no rotation in the length direction and therefore no
work by Mx . Similar the beam cross---sectional rotation from Mx give
no transverse displacement and therefore no work by T . Thus the
decouplings in (7.1) are correct.
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Pauli Pedersen: 7. Application of Energy Principles
7.2 Simplified beam results
BENDING In fig. 7.1 is shown slender beams and although N≠ 0 and T≠ 0 the
ENERGY elastic energy from bending is often so dominating that we can simplify
ONLY (7.1) to
UC = b
a
M2(2EI)dx (7.10)
or expressed in displacements v from M= EId2vdx2 by
U= b
a
12EId2vdx2
2
dx (7.11)
Q
x
M
x
MA
MB
T
a) b)
Fig. 7.1: Slender beam examples. Case a) cantilever with concentrated load at thefree end and case b) a beam subjected to two end moments only.
NEGLECTED Let us in relation to the example in fig. 7.1a discuss the error in
SHEAR neglecting the term with T2 in (7.1). With load Q we have T= –Q
ENERGY and M= Qx giving
UC
T=
L
0
βQ2
2GAdx= β
Q2L2GA
UC
M=
L
0
Q2x2
2EIdx=
Q2L3
6EI(7.12)
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Pauli Pedersen: 7. Application of Energy Principles
From this follows
UT
UM
=
3βEI
GAL2= 6β(1+ ν)Γ2
(7.13)
with the slenderness ratio defined by Γ := L AI
With β values of the order 1 and slenderness ratio mostly of the order
10---100 we see, how dominating the bending energy is.
ELEMENTARY The important case of linearly varying moment shown in fig. 7.1b give
CASE M(x)=MA+ M
B– M
AxL and then from (7.10) with constant
bending stiffness EI we have
UC = 12EIL
0
MA+ M
B– M
A xL2dx= L
6EIM2
A+M2
B+M
AM
B (7.14)
which could have given (7.12) directly for MA= QL and M
B= 0 .
7.3 Complementary principles to solve a beam problem
Thecantilever, slenderbeamproblem is repeated in fig. 7.2with coordi-
nate axis and the displacement v0= v (x= 0) corresponding to the
load Q .Weshall first determine v0by solving the differential equation
v0
x= 0
Q
x= L
x
Fig. 7.2: Slender cantilever beam problem.
131
Pauli Pedersen: 7. Application of Energy Principles
DIFFERENTIAL d2vdx2=M(EI) ; M= Qx
EQUATION
dvdx=QEIx
0
x~dx~+ C1= 1
2Qx2
EI+ C
1= 1
2QEI
x2 – L2
with C1
from dvdxx=L= 0
v=1
2
Q
EIx
0
x~2 − L2dx~+ C2=
12QEI
13x3 − L2x+ C
2=
12QEI
13x3 − L2x+ 2
3L3
with C2from v
x=L= 0
and the result is
v0= v (x= 0)=
QL3
3EI
How can we obtain this result with an energy principle?
FOUR The complementary virtual work principle states δUC= δWC ,
ENERGY which with δWC= v
0δQ and from (7.12) or (7.14) δUC
=
SOLUTIONS 2QL3(6EI)δQ directly gives (7.14) because v0δQ=
2QL3(6EI)δQ must hold independently of δQ .
TheCastigliano’s 2nd theorem states v0= ∂UC∂Q and thus directly
from (7.12) v0= 2QL3(6EI) .
The unit load theorem (6.33) for linear elasticity as here gives with σ1
from Q= 1 and 0
from Q :
v0=
L
0
A
σ10dAdx =
L
0
A
1xIzQx
IEzdAdx=
L
0
QEI
x2dx=QL3
3EI(7.15)
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Pauli Pedersen: 7. Application of Energy Principles
The complementary total potential energy is ΠC= UC – WC , i.e.
from (7.12) ΠC= L
0
Q2x2(2EI)dx − Q
0
v(Q
)dQ
and thus stationar-
ity of ΠC with respect to variationof Q ∂ΠC∂Q= 0 gives the resultwhen the following differentiation is applied
d
Q
0
v(Q
)dQ
dQ= v(Q)= v
0
Minimumof ΠC ∂2ΠC∂Q2 > 0 gives L3(3EI)> 0 , clearly satis-
fied for EI> 0 .
7.4 Approximate solution examples
The following examples will in fact be closely related to the finite ele-
ment concept, but will be presented without that reference.
Let us return to the example in fig. 2.6, and subject the side x1= a to
a uniform line load of intensity q applied in the x1---direction. The
x2
d2
d1
x1
x3
a
a
q
approximate displacement and strain field are given in (2.55) and
(2.56).Assuming a case of isotropic, plane strain the constitutivematrix
(4.60) then gives the stresses and the variation of the strain energy den-
sity
σ= [C] , δu= σTδ=
T[C]δ (7.16)
Our goal is to determine the parameters d1, d
2from the principle of
virtualwork, i.e. by variations δd1, δd
2.With only the line load thevari-
ation of external work is at x1= a obtained from
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Pauli Pedersen: 7. Application of Energy Principles
δW= a
0
x2aa2δd
1qdx
2=
12qa(δd
1) (7.17)
i.e. independent of δd2.
The variation of strain energy δU= δudV= L δudA is expressed
in δd1, δd
2from (7.16), (4.60) and (2.56). We get
δU= LT[C]∂
∂di
δdidA= LT[C]
x2
a2
0x1
2 a2
δd1+
0x1
a2x2
2 a2
δd2
dA(7.18)
Then δW= δU for any δd1, δd
2give two equations to determine
d1, d
2with the result
d1
d2
= 6(1 – 2ν)qa
(15 – 16ν)(9 – 16ν)GL4(3 – 4ν)
–3 (7.19)
The second example from section two has approximate displacements
and strains from (2.58) and (2.59). The load for this example is a pres-x3
x2
x1
p
h
2R
sure p acting on the free rubber surface. With only one parameter d ,
a variation δd and the virtual work principle will give us the approxi-
mate solution. The variation of external work at x3= h is
δW= R
0
–δv3p2πrdr
(7.20)
= R
0
1 – r2R2hh(δd)p2πrdr = p2π(δd)12R2 – 1
4R2= 1
2πR2p(δd)
134
Pauli Pedersen: 7. Application of Energy Principles
With the six by six isotropic, constitutive matrix, say as specified in
(4.99), we again with variation of strain energy density
δu= T[L]δ and strains in (2.59) get
δU= T[L] ∂∂d
δddVol.
(7.21)
and∂
T
∂d= 0 0 −
1h1 − r2
R 0
2 x1x3
hR2
2 x2x3
hR2
which is worked out to
δU=π(1 – 2ν)h2+ (1 – ν)R2Ed
3(1+ ν)(1 – 2ν)hδd (7.22)
From δW= δU independent of δd then with (7.20) and (7.22) we get
the result
d=3(1+ ν)(1 – 2ν)R2hp
2(1 – 2ν)h2+ (1 – ν)R2E(7.23)
Displacements and strains then follow by inserting d in (2.58) and
(2.59) and stresses can be evaluated from σ= [L] .
The third example from section three will then also be completed. The
2R
x3
x1α
variation of the external work, done by the torsional moment M , is
directly related to the single parameter ω and we have
δW=Mδω (7.24)
135
Pauli Pedersen: 7. Application of Energy Principles
Again the variation of elastic energy density is given by (7.16) and the
isotropic constitutive matrix (4.99). From the strains (2.64) we get
δU= 2πGR2
3 tanα1+ 1
3tan2 αωδω (7.25)
and finally then from δW= δU the result
ω=9 tanαM
2πGR33+ tan2α(7.26)
ACCURACY? From these three examples we see the effectiveness and the generality
in using the virtual work principle. Naturally the question follows: how
accurate are the approximate solutions?No simple answer can be given
but it mainly depends on the chosen displacement assumption, relative
to the actual, unknown real displacement field. Engineering judgment
and experience is important, and for more numerically based applica-
tionwe can perform convergence tests. Theapplication of the finite ele-
ment method, FEM, is a good reference for many more details.
7.5 Complementary approximations
Let us for illustration solve a simple problem first with a stress field
approximation and then with a strain field approximation.
A squared domain with side lengths equal to 2a is totally fixed at one
side and fixed with a possible translation at the opposite side as shown
in fig. 7.3. The remaining two sides are free.
136
Pauli Pedersen: 7. Application of Energy Principles
The shown coordinate system has origo in the middle of the domain.
The thickness t in the x3---direction is constant andmuch smaller than
a (t<< a) .With inspiration from simple beam theory we assume the
following stress field
σT=
σ11
σ22
σ33
2 σ12
2 σ13
2 σ23
(7.27)
= cx1x20 0
2
2a2 – x2
2 0 0
Fig. 7.3: Squared domain with a total force Q acting on the side that can translate.
APPROXIMATE For this stress field to be statically admissible we must have
STRESS
FIELD
Q=
a
–a
σ12tdx
2
x1=a
=
12ct
a
–a
a2 – x22dx
2=
12ct 4
3a3 ,
(7.28)
i.e. c= 3Q2ta3
137
Pauli Pedersen: 7. Application of Energy Principles
Furthermore, for moment equilibrium we must test
Q·a=
a
–a
–σ11tx
2dx
2
x1=–a
+
a
–a
σ11tx
2dx
2
x1=a
(7.29)
= c a
–a
ax22+ ax2
2tdx
2= 2
3ca4t
The variation of complementary external work with Q as a reaction is
δWC = vδQ (7.30)
and with variation of stresses only by variation of the factor c we have
δσ = σcδc . Therefore δUC is obtained from
δUC = tc δc
a
–a
a
–a
σT[L]
–1σdx1dx
2(7.31)
The principle δUC = δWC in matrix notation with δc = 3δQ2ta3gives
v = tQa
–a
a
–a
σT[L]
–1σdx1dx
2(7.32)
Expanding this we get from (4.106) (cubic symmetry)
v = Qt
1+ λ
12
1+ 3λ
+65
12
(7.33)
138
Pauli Pedersen: 7. Application of Energy Principles
which for the isotropic case of 2=
1= gives
v=Qt
17+ 23λ
5(2+ 3λ)(7.34)
APPROXIMATE An alternative approximation could be by the kinematically admissible
STRAIN FIELD displacement assumption
v1= v
3= 0 , v
2=
(x1+ a)
2av (7.35)
which returns the pure shear strain case of only 12=
21≠ 0
12=
21= v(4a) (7.36)
The variation of external work is directly in terms of δv
δW= Qδv (7.37)
and the variation of strain energy follows from
δU= t δudA = t a
−a
a
−a
2σ12δ
12dx
1dx
2(7.38)
which with σ12= 2
212
and δ12= δv(4a) gives
δU= t42v(4a)δv(4a)(2a2a)= t2
2vδv (7.39)
From the principle δU= δW follows the result
v=Qt12
(7.40)
139
Pauli Pedersen: 7. Application of Energy Principles
TWO SIDED The approximation (7.40)will be a too stiff approximation (v too small)
BOUNDS and theapproximation (7.33)will bea too flexible approximation (v too
large). This follows from the two sided inequalities (6.55) because with
only a single load the total potential energy is proportional to the dis-
placement v .
140
Pauli Pedersen: 7. Application of Energy Principles
141
Pauli Pedersen: 8. Laminate Analysis
8. LAMINATE ANALYSIS
plies and stacking, classical plate theory, stiffnesses,
special laminates, advanced modelling
8.1 Introduction to laminates
PLY STACKING A laminate consists of two or more plies and acts as an integral plate.
The plies may be made of different materials, although the same ply
material is often used throughout the laminate. The main difference
between the individual plies is thus their orientation in the laminate,
and these orientations are very important because the plies are often
strongly anisotropic, with modulus ratios of the order 5---40 as illus-
trated in table 4.2.
TEXTBOOKS Laminate analysis is an important subject within themechanics of com-
posite materials. Among themany good textbooks on laminates we can
mention Jones (1975), Tsai & Hahn (1980), Vinson & Sierakowski
(1986) and Whitney (1987). In this chapter we shall present the lami-
nate analysis, consistently with the previous chapters, i.e. with the
2 ---contracted notation that simplifies the rotational transformations.
THIN PLATE Classical laminateanalysis is also basedon classical plate theory; hence,
ASSUMPTION it is restricted to thin laminates. Analysis shows that the effect of shear
deformations is more profound for laminates than for isotropic plates.
142
Pauli Pedersen: 8. Laminate Analysis
The assumptions of thin plate theory should therefore be carefully
examined. More advanced models can be treated numerically.
CONTENTS OF The chapter starts with the laminate strains, curvatures and stresses and
THE CHAPTER the material constitutive relations. Then, integrating through the plate
thickness, the laminate stiffnesses are introduced and the most impor-
tant formulas are derived, after which specific laminate layouts are dis-
cussed. Lastly, advanced laminate models are covered, but without
going into detail.
8.2 Basic assumptions
LAMINATE A laminate is built up (stacked) of a number of layers/plies/laminas,
BUILD UP that often are very thin (0.1 mm). We choose here to call them plies to
OF PLIES indicate also by name the difference from the laminate itself. The lami-
nate is often thin compared with its deformation waves, and thus this
introductory analysis is based on classical thin plate theory. In fig. 8.1
we show the reference coordinate system and a single ply at themidsur-
face x3= z= 0 of the laminate. Note that to be consistent with chap-
ter four, the angle γ is defined from the material principal direction
(fiber direction) to the x1
laminate direction, positive anticlockwise,
so a minus sign appears in fig. 8.1.
143
Pauli Pedersen: 8. Laminate Analysis
x3= z
x2
x1 σ
1111
σ12
12
σ21
21
σ22
22
–γ
Fig. 8.1: Laminate mid---surface (z= 0) with definition of the involved strain and
stress components.
STRAINS The contracted membrane strain vector is defined by
T:=
1122
2 12
= v1,1 v2,2 2 v1,2+ v2,1
2 (8.1)
SMALL STRAIN where vi,j are the in---plane displacements gradients and small strains
ASSUMPTION are assumed. Specifically, the middle surface strains 0 are given the
index 0 . The contracted curvature and twist vector is analogously
defined by
CURVATURES T:=
1122 2 12
= – w,11 w,22 2 w,12 (8.2)
where w, ij are secondorder derivatives of the out---of---plane displace-
ment w .
With the classical assumption that normals to the originally plane mid-
surface remain plane and perpendicular to the deformedmid---surface,
the strains are determined only by mid---surface deformations, i.e.
144
Pauli Pedersen: 8. Laminate Analysis
= 0+ z (8.3)
where superscript 0 refers to z= 0 , i.e. to the mid---surface, and
where z is the coordinate perpendicular to the plane of the laminate
DISPLACEMENT and the plies. This follows directly from the displacement assumption
ASSUMPTION
v1x
1, x
2, z= v0
1x
1, x
2+ z f
1x
1, x
2
v2x
1, x
2, z= v0
2x
1, x
2+ z f
2x
1, x
2 (8.4)
v3= w x
1, x
2
i.e. out---of---plane displacement w independent of z and membrane
displacements v1, v
2linear in z . With the shear strains
13,
23equal
to zero, we get
213= v
1,3+ v
3,1= f
1+ w
,1= 0 ⇒ f
1= – w
,1
(8.5)223= v2,3+ v3,2= f2+ w,2= 0 ⇒ f2= – w,2
i.e.
11= v
1,1= v0
1,1– z w
,11=
011
– z w,11
22= v
2,2= v0
2,2– z w
,22=
022
– z w,22
(8.6)
212= v
1,2+ v
2,1= v0
1,2– z w
,12+ v0
2,1– z w
,21
= 2012
– z w,12
CONSTITUTIVE In a local ply (material) coordinate system y coinciding with fiber
RELATIONS direction we shall assume an orthotropic material, which gives
145
Pauli Pedersen: 8. Laminate Analysis
σy=[C]
yy with
(8.7)
[C]y= C
α1111
α1122
0
α1122
α2222
0
0
0
2α1212
y
as described in detail in chapter four.
In the reference coordinate system x the relations are
σx= [C]xx (8.8)
and the matrix [C]x is obtained from [C]
y and the angle γ according
to the rotational transformations (4.17) or (4.16). The matrix [C]x will
normally include coupling between shear and normal components, i.e.
the zeros of the [C]y matrix are not present in the [C]
x matrix.
MATERIAL The elements of the matrices [C]y and [C]
x are often termed the
STIFFNESSES material stiffnesses with physical dimension equal to stress.
8.3 Laminate stiffnesses
FORCES AND In laminate analysis the term stiffness is used for the coefficients which
MOMENTS multiplied with strain and curvature, give forces and moments. To be
more specific, we define the force vector N and the moment vector
M by
NT:= N
11N22
2 N12
(8.9)
MT := M11M
222 M
12
146
Pauli Pedersen: 8. Laminate Analysis
where Nij are the in---plane forces per unit length (Nii normal forces
and N12 shear force), and Mij are themoments per unit length (Mii
bending moments and M12 torsional moment).
In fig. 8.2 we show the sign definitions of themembrane forces (per unit
of length) and of the mid---surface moments (per unit of length).
a) b)x3= z
v3= w
x1
x2
v1
v2N22
N12
N12
N11
h
x3= z
x2
x1 M12
M11
M22
M12
Fig. 8.2: Laminate with thickness h –h2 ≤ z ≤ h2 with definition of signs for forces in a)
and signs for moments in b).
STIFFNESS Laminate “stiffnesses” [A] , [B] , [D] (all symmetric matrices of
DEFINITION order three) are defined by the relations
NM=
[A]
[B]
[B]
[D]
0
, (8.10)
and it should be remembered that we are dealing with a specific point
of the laminate plate, in contrast to the volume/area---dependent stiff-
nesses in a finite element formulation. The laminate stiffnesses have
physical dimensions of force per length for the [A] matrix, force for the
[B] matrix and force times length for the [D] matrix.
147
Pauli Pedersen: 8. Laminate Analysis
FORCE To determine the [A] matrix of membrane stiffnesses and the [B]
EQUILIBRIUM matrix of coupling stiffnesses, we express force equilibrium through
laminate thickness h by
N= h2
–h2
σdz (8.11)
which, using the constitutive relations, gives
N= h2
–h2
[C] 0+ zdz (8.12)
and thus from the definition (8.10)
[A]= h2
−h2
[C]dz= K
k=1
[C]kzk− z
k− 1=
K
k=1
[C]ktk
(8.13)
where tkis the thickness of ply number k , and
[B]= h2
−h2
[C]zdz=K
k=1
[C]k
12z2k− z2
k− 1=
K
k=1
[C]ktkzk
(8.14)
where zk=
12z
k+ z
k−1 is the position of themiddle of the ply num-
ber k , and where, in the accumulated expressions, it is assumed that
[C]kis constant from z
k–1to z
k, which is the thickness domain of ply
number k . The total number of plies is K .
148
Pauli Pedersen: 8. Laminate Analysis
MOMENT The [D] matrix of bending stiffnesses is obtained from moment
EQUILIBRIUM equilibrium
M= h2
–h2
σzdz= h2
–h2
[C] 0+ zzdz (8.15)
which again gives (8.14) and
[D]= h2
−h2
[C]z2dz=K
k=1
[C]k
13z3
k− z3
k− 1
= K
k=1
[C]ktk
13z2
k+ z2
k−1+ z
kzk−1
≈ K
k=1
[C]ktkz2k
(8.16)
with its approximation for thin plies.
The integrations and summations in (8.12)---(8.16) only have meaning
if all the constitutivematrices [C]kfor k= 1, 2, ..., K , are given in the
same coordinate system, say the laminate system x .
INVERTED Formulas expressing strains and curvatures in terms of forces and
FORMULAS FOR moments are also valuable; therefore, from the matrix index in chapter
FLEXIBILITIES 12, we list
0
=
[E]
[L]T
[L]
[H]
NM (8.17)
with
[H]= [D] – [B][A] –1[B]–1
[L]= – [A]–1[B][H]
[E]= [A]–1[I] – [B][L]T
149
Pauli Pedersen: 8. Laminate Analysis
8.4 Special laminate layouts
A number of specific laminate layouts deserve special attention
because they are often used and give rise to certain simplification in the
analysis. We shall deal with symmetric laminates, skew---symmetric
laminates, cross---ply laminates, angle---ply laminates, specially ortho-
tropic laminates and laminates of the same plies.
SYMMETRIC For the symmetric laminates the total number of plies is (or can be dealt
with as) even, and the integrations/summations of eq. (8.12)---(8.16)
need only be extended over, say the negative z---domain –h2≤ z≤
0 . The coupling stiffnesses vanish and we get
[A]= 2K2
k=1
[C]kz
k– z
k – 1
[B]= [0] (8.18)
[D]= 2K2
k=1
[C]k
13z3
k– z3
k – 1
Therefore, for symmetric laminates, the analysis is very much simpli-
fied, and experiments are better defined. The in---plane and out---of---
plane expressions decouple and we have
DECOUPLED
PROBLEMS N= [A]0 , M= [D] (8.19)
Coupling between shear and normal and between bending and torsion,
may or may not be active.
150
Pauli Pedersen: 8. Laminate Analysis
NON---LAMINATED Non---laminated plates, but not necessarily isotropic plates, can be
PLATES treated as a single ply laminate and thus as a symmetric laminate. The
important decoupling of (8.19) is thus valid for these plates.
SKEW--- If we take a symmetric laminate and change sign of all the angles for the
SYMMETRIC plies in the positive z---domain, then we have a skew---symmetric lami-
LAMINATES nate. For skew---symmetric laminates the total number of plies is also
even, and we need only extend the integrations/summations to half the
thickness.
We need the details of the matrix [C]k, as given by (4.17), to see the
simplifications. We note that the constitutive quantities
α1111
, α2222
, α1122
and α1212
are independent of the sign of γ , while
the quantities α1112
and α2212
directly change signwith γ . Therefore,
the total result for the skew---symmetric laminate stiffnesses is
[A]= 2K2
k=1
[C]kz
k– z
k–1 with A
1112= A
2212= 0
[B]= 2K2
k=1
[C]k
12z2
k– z2
k–1 with B
1111= B
2222= 0
(8.20)B1122= B
1212= 0
[D]= 2K2
k=1
[C]k
13z3
k– z3
k–1 with D
1112= D
2212= 0
There is thus no coupling between shear and normal or between bend-
ing and torsion, but we can have coupling from torsion to normal forces
B1112
andor B2212
≠ 0 .
151
Pauli Pedersen: 8. Laminate Analysis
ANGLE---PLY An angle---ply laminate has alternating fiber directions θ, –θ and only
LAMINATES these two directions are involved. It may be symmetric (odd number of
plies) or skew---symmetric (even number of plies). Then if it has many
plies both the simplicities of the symmetric laminate (8.18) and of the
skew---symmetric laminate (8.20) are approximately valid, and we have
a totally decoupled orthotropic laminate.
For angle---ply laminates we see from eq. (4.17) that the only orienta-
tional factorswill be cos 2θ , cos 4θ , sin 2θ , and sin 4θ . Specific
simple results are obtained if all the plies are of the same material and
of the same thickness.
CROSS---PLY A cross---ply laminate has alternating fiber directions oriented at 0o
LAMINATES and 90o . We see from (4.17), (4.10) how much this simplifies the
constitutive matrix when only θ values of 0 and π2 are active. For
0o we have
[C]k,0o= C
α1+ α
2+ α
3
symm.
α4− α
3
α1− α
2+ α
3
0
0
α1− α4− 2α3
k
(8.21)
and for 90o we have
[C]k,90o= C
α1− α
2+ α
3
symm.
α4− α
3
α1+ α
2+ α
3
0
0
α1− α
4− 2α
3
k
(8.22)
i.e. orthotropic behaviours.
The cross---ply laminate may also be symmetric or skew---symmetric
and furthermore of only one plymaterial, which then naturally give fur-
ther simplifications.
152
Pauli Pedersen: 8. Laminate Analysis
SPECIALLY An in---plane specially orthotropic laminate is defined as a laminate
ORTHOTROPIC where A1112= A
2212= 0 , and this can be obtained also for non---
LAMINATES skew---symmetric laminates. The notion of a balanced laminate is also
often used.
An out---of---plane specially orthotropic laminate is defined as a lami-
natewhere D1112= D
2212= 0 , and again skew---symmetric property
is not always necessary. Again the notion balanced is often used.
A laminate may be in---plane orthotropic without being out---of---plane
orthotropic and vice versa. However, it may also be designed to have
both these properties, like the example of a skew---symmetric laminate.
LAMINATES OF When the laminate is build up of plies of equal thickness, the z---factors
THE SAME PLIES in the laminate stiffnesses (8.13), (8.14), (8.16) are conveniently
expressed by laminate thickness h and the total number K of plies
zk− z
k–1= hK
z2k− z2
k–1= h2K2(2k− 1−K) (8.23)
z3k− z3
k–1= h3K33k(k− 1)+ 1+ 3
4K(K+ 2− 4k)
and if furthermore all plies are of the samematerial simple formulas for
the stiffness can then be derived.
For laminates of only one material the description by lamination
parameters may be an effective alternative. See Hammer et al. (1997),
which also shows the possibility of design for maximum laminate stiff-
ness.
153
Pauli Pedersen: 8. Laminate Analysis
8.5 Advanced laminate models
In laminates the ratio of elastic modulus to shear modulus is often of
the order 10 to 40 , and this magnifies the influence of transverse shear
strains. Therefore, the limitations of classical thin plate theory are met
at an earlier stage than for plates of isotropic materials.
HIGHER ORDER Different higher order theories are available, and many papers have
PLATE THEORY beenpublished after 1980.We shall here only give a short overviewwith
formulas, mainly taken from Reddy (1984). Let the assumed displace-
ment fields be restricted to
v1x
1, x
2, z= v0
1x
1, x
2+ zf
1x
1, x
2+ z2f
3x
1, x
2+ z3f
5x
1, x
2
v2x
1, x
2, z= v0
2x
1, x
2+ zf
2x
1, x
2+ z2f
4x
1, x
2+ z3f
6x
1, x
2 (8.24)
v3x
1, x
2= wx
1, x
2
i.e., the out---of---plane displacement v3= w is constant through the
laminate thickness, while the in---plane displacements v1, v
2are cubic
functions of the thickness parameter z . The functions fn are primarily
independent functions.
A simple higher order theory --- often termed as first---order theory ---
is obtained for f3= f
4= f
5= f
6= 0 . This theory is the parallel to
the Timoshenko beam theory, i.e. a Mindlin type theory.
154
Pauli Pedersen: 8. Laminate Analysis
REDDY, In Reddy (1984) the functions f3, f
4, f
5and f
6are determined by the
LEVINSON conditions that the transverse shear stresses should be zero at the
THEORY laminate surfaces
σx1z= σx
2z= 0 for z= h2 (8.25)
which conditions give the simplified displacement fields
vi= v0
i+ zψ
i− 4
3zh2ψ
i+ w
,i (8.26)
for i= 1 and 2
where the traditional notation for rotation f1= ψ
1, f
2= ψ
2is ap-
plied.
This correspond to the plate theory of Levinson (1981). Finite element
models based on this displacement assumption can be found in Phan
and Reddy (1985) with comparisons of results from classical plate
theory, first---order theory and this higher---order theory.
An even more extended model is used by Frederiksen (1996).
155
Pauli Pedersen: 9. Bending of Rectangular, Orthotropic Plates
9. BENDING OF RECTANGULAR,
ORTHOTROPIC PLATES
equilibrium, differential equation, boundary
conditions, Navier solutions, Levy solutions
9.1 Restricted class of problems
The material stiffnesses of chapter four describe space point relations
and the laminate stiffnesses of chapter eight describe plate point rela-
tions, both without the integrated response through the plate.
Restricted to rectangular, orthotropic (parallel to plate boundaries)
PLATE plates we shall in this section solve the “structure” problem, i.e. find
STRUCTURE the displacements, strains and stresses everywhere in the plate. The
loads will cover any plate load, i.e. any pressure distribution and single
forces, but only in the transverse direction of the plate.
SIGN With the laminate stiffnesses from chapter eight as our basis, the sign
DEFINITION definition will agree with classical lamination theory. In more tradi-
tional theory for plates we often find an alternative sign definition for
moments, which agrees with classical theory for beams.
PURE According to (8.3) and (8.2), for the case of pure bending, i.e.
BENDING 0= 0 we have the following strain field
STRAINS
T= z
1122 2
12=− zw
,11w,22 2 w
,12 (9.1)
156
Pauli Pedersen: 9. Bending of Rectangular, Orthotropic Plates
with curvatures w,11, w,22 and twist w,12 determined as second order
derivatives of the transverse displacement w .
The constitutivematrix will have the form (8.7) everywhere in the plate
butmay change through the thickness αijkl= αijkl(z) . The stresses are
thus determined by (assumed α1112= α2212= 0)
STRESSES σ11= – zCα1111 w,11+ α1122 w,22
σ22= – zCα1122 w,11+ α2222 w,22 (9.2)
σ12= – zC2α1212 w,12
With only orthotropic directions the bending stiffnesses (8.16) will also
STRESS/ be “orthotropic” and we have
MOMENT
EQUILIBRIUM
M11M22
2 M12
= –
D1111
D1122
0
D1122
D2222
0
0
0
2D1212
w,11w,22
2 w,12
(9.3)
The specific case of constant [C] through the plate thickness h gives
[D]= 112
h3[C] (9.4)
and then, from (9.2) and (9.3), the stress---moment relations are
σ11=
12h3
M11z ; σ
22=
12h3
M22z ; σ
12=
12h3
M12z (9.5)
157
Pauli Pedersen: 9. Bending of Rectangular, Orthotropic Plates
9.2 Structural equilibrium
We shall now find the results of moment---force equilibriums for an
infinitesimal plate area as shown in fig. 9.1.
a) b)x3= z
x3= zT
1dx
2
M12dx
2
M11dx
2M12dx
1
T2dx
1
M22dx
1
x1 x
1
x2 x
2
T1+∂T
1
∂x1
dx1dx2
T2+∂T
2
∂x2
dx2dx1M11+∂M
11
∂x1
dx1dx2M12+
∂M12
∂x2
dx2dx1M22+
∂M22
∂x2
dx2dx1
M12+∂M
12
∂x1
dx1dx2
Fig. 9.1: Moments and forces active in the two moment equilibria.
From fig. 9.1 a) we read directly
∂M11
∂x1
dx1dx
2+∂M
12
∂x2
dx2dx
1
(9.6)
– T1dx
2dx
1–∂T
1
∂x1
dx1dx
212dx
1= 0
and for dx1, dx
2→ 0, 0 we get the shear force per unit length T
1deter-
mined by gradients of moments
T1=M
11,1+M
12,2(9.7)
158
Pauli Pedersen: 9. Bending of Rectangular, Orthotropic Plates
SHEAR From fig 9.1 b) moment equilibrium around an axis parallel to the x1---
FORCES axis gives
–∂M
22
∂x2
dx2dx
1–∂M
12
∂x1
dx1dx
2
(9.8)
+ T2dx1dx2+∂T2
∂x2
dx2dx112dx2= 0
and then the second shear force T2
T2=M22,2+M12,1
(9.9)
The next equilibrium is for the shear forces and the assumed pressure
p , with directions as shown in fig. 9.2a.
x3= z
x1
x2
T1dx
2
T2dx
1
pdx1dx
2
T2+∂T
2
∂x2
dx2dx1T1+
∂T1
∂x1
dx1dx2
a) b)x3= z
x1
x2
h#1 #2
#3
#4
a
b
Fig. 9.2: In a) force equilibrium between internal forces Ti(per unit length) and external plate pressure,
b) dimensions of the plate with boundary numbering.
159
Pauli Pedersen: 9. Bending of Rectangular, Orthotropic Plates
From fig. 9.2a we read the equilibrium
∂T2
∂x2
dx2dx
1+∂T
1
∂x1
dx1dx
2+ pdx
1dx
2= 0 (9.10)
and thus for dx1, dx
2→ 0, 0 get
p= – T1,1
– T2,2 (9.11)
9.3 Plate differential equations
Substituting (9.7) and (9.9) in (9.11) we now finally obtain the plate dif-
ferential equation
p= – M11,11
– M22,22
– 2M12,12
(9.12)
or with (9.3)
ORTHOTROPIC
THIN PLATE D1111 w,11,11+ D2222 w,22
,22
DIFFERENTIAL (9.13)
EQUATION + D1122 w,22,11+ D1122 w,11
,22+ 4D1212 w,12
,12= p
For constant bending stiffness over the plate area (9.13) simplifies to
(9.14)D1111
w,1111+D
2222w
,2222+ 2D
1122+ 2D
1212w
,1212= p
IMPORTANT
CASE
and this is the case on which we shall work more extensively. For
isotropic plates we have D1111
= D2222 and D1212
=
160
Pauli Pedersen: 9. Bending of Rectangular, Orthotropic Plates
D1111
– D1122
2 , which, with plane stress assumption, gives the well
known plate equation
ISOTROPIC
CASE Eh3
121 – ν2
w,1111+ 2w
,1212+ w
,2222= p (9.15)
which in the literature is often written in compact form as
∆∆w= pD with D= Eh3
121− ν2(9.16)
where ∆ is the harmonic operator and ∆∆ the biharmonic operator.
9.4 Boundary conditions
In fig. 9.2 b we have shown the plate dimensions and the position in the
coordinate system. Furthermore, we have numbered the four bound-
aries as
#1 : x1, x
2=
0≤ x1≤ a , 0
#2 : x1, x
2= 0≤ x
1≤ a , b
(9.17)
#3 : x1, x
2= 0 , 0≤ x
2≤ b
#4 : x1, x
2= a , 0≤ x
2≤ b
KINEMATICS Kinematic boundary conditions, i.e. known translations and rotations,
can be
w= w– along boundaries 1234
w,2= w
,2
–along boundaries 12 (9.18)
w,1= w
,1
–along boundaries 34
161
Pauli Pedersen: 9. Bending of Rectangular, Orthotropic Plates
where the known quantities are given the superscript --- . We note that
constant boundary condition along a certain axis gives information also
about the gradients, i.e.
w= w– along boundaries 12 gives
w,1 = w,11 = ...= 0 along these boundaries
(9.19)
w= w– along boundaries 34 gives
w,2 = w,22 = ...= 0 along these boundaries
STATICS Static boundary conditions, i.e. known forces and moments along a
boundary, may be given as
M22=M
–
22along boundaries 12
M11=M
–
11along boundaries 34
M12
T1
x2
M12,2> 0
(9.20)
T2+M
12,1= T–
2effective
along boundaries 12
T1+M
12,2= T–
1effective
along boundaries 34
Note that shear force and gradient of torsionalmoment cannot be sepa-
rated, again reminding ourselves that moments are conceptual quanti-
ties. By means of (9.3), boundary conditions for bending moments can
be expressed in curvatures
– D1122
w,11
– D2222
w,22=M
–
22along 12
(9.21)
– D1111
w,11
– D1122
w,22=M
–
11along 34
162
Pauli Pedersen: 9. Bending of Rectangular, Orthotropic Plates
The simply supported case, i.e. no translation and no moment load, is
particularly interesting because of the possibility for analytical solution.
For boundaries 1/2 follows from w≡ 0 also w,11≡ 0 (9.19) and thus
from M–
22≡ 0 also w
,22≡ 0 (9.21), assuming D
2222≠ 0 . Forbound-
aries 3/4 follows w≡ 0⇒ w,22≡ 0 and thus from M
–
11≡ 0⇒
w,11≡ 0 for D
1111≠ 0 . In all, with all boundaries simply supported,
we have
FOURSIMPLY w≡ 0 and w
,22≡ 0 along 12
SUPPORTED (9.22)BOUNDARIES w≡ 0 and w
,11≡ 0 along 34
9.5 Navier analytical solution
The case of arbitrarily loaded, rectangular, orthotropic plates that are
simply supported along all four boundaries can be solved analytically
in a series expansion. This is called a Navier solution.
Any function that satisfies the boundary conditions (9.22) can be writ-
ten in a double sine expansion
wx1, x
2=
m, n
wmn sinπmx
1
a sinπnx
2
b(9.23)
where we have introduced the short notation
m,n
:= ∞
m=1
∞
n=1
(9.24)
163
Pauli Pedersen: 9. Bending of Rectangular, Orthotropic Plates
With a limited number of terms in the expansion, the solution will be
an approximation. The unknowns are the coefficients wmn in (9.23),
and before (9.23) can be inserted in the differential equation (9.14) we
need the derivatives
w,1111=
m, n
wmnπ4m4
a4sin
πmx1
a sinπnx
2
b
w,2222=
m, n
wmnπ4n4
b4sin
πmx1
a sinπnx
2
b(9.25)
w,1212=
m, n
wmnπ4m2n2
a2b2sin
πmx1
a sinπnx
2
b
MODE Introducing the mode parameter ηmn defined by
PARAMETER
ηmn := (mb)(na) (9.26)
we obtain amore simple description. Themode parameter is physically
the ratio between the two half---wavelengths (b/n and a/m) of the
actual deformation and thus couples the information from plate size
a,b with the actual deformation pattern m,n .
With (9.25) and using (9.26) inserted in (9.14), we get
m, n
wmnπ4n4
b4sin
πmx1
a sinπnx
2
b.
(9.27)
η4mnD1111+D
2222+ η
2mn2
D1122+ 2D
1212= p
164
Pauli Pedersen: 9. Bending of Rectangular, Orthotropic Plates
Assuming now that the plate pressure p= px1, x
2 is also given by a
double sine expansion
p= m , n
pmn sinπmx
1
a sinπnx
2
b(9.28)
then with an assumption of linearity we get the Navier solution
wmn=pmnb4
π4n4Φmn
(9.29)
with the Φ definition
Φmn := η4mn D1111+D
2222+ η2mn 2D
1122+ 2D
1212 (9.30)
For the isotropic case we have D2222= D
1111and D
1122+2D
1212=
D1111
and thus get Φmn =η2mn+ 1
2
D1111
.
A simple case of a single “harmonic” load with amplitude A
p11= A and all other pmn= 0 gives
(9.31)
w=Ab4 sin
πx1
a sinπx2
b
π4ba4D1111
+D2222+ b
a22D
1122+ 2D
1212
165
Pauli Pedersen: 9. Bending of Rectangular, Orthotropic Plates
9.6 Fourier expansions of loads
More complicated load fields p= px1, x
2 can also be described by
Fourier coefficients pmn determined by
pmn =4abb
0
a
0
px1, x
2 sin
mπx1
a sinnπx
2
bdx
1dx
2
(9.32)
= 4 β2
β1
α2
α1
pα, β sinmπα sin nπβdαdβ
where α,β aredimensionless variables α := x1a , β := x
2b and the
domain of pα, β ≠ 0 is limited by (0 ≤)α1≤ α≤ α
2(≤ 1) and
(0 ≤)β1≤ β≤ β
2(≤ 1) .
Some practical cases with the corresponding Fourier coefficients
(Q is the total force) are shown below.
pα2a pmn =
4p
π2mncosmπα
2– cosmπα
1cos nπβ
2– cos nπβ
1
Uniform pressure: Q= pα2– α
1β
2– β
1ab
(9.33)
and for the whole plate α2= β
2= 1,α
1= β
1= 0β
2b
β1b
α1a
pmn =16p
π2mnfor
m= 1, 3, 5, ...
n = 1, 3, 5, ...(9.34)
166
Pauli Pedersen: 9. Bending of Rectangular, Orthotropic Plates
Linearly changing pressure: Q= pα2– α
1β
2– β
1ab2
pmn =4p
π2mncos nπβ
2– cos nπβ
1cosmπα
2–
sinmπα2– sinmπα
1
πmα2– α
1
(9.35)
α1a α
2a
p
β2b
β1b
when changing in the x1---direction
pmn =4p
π2mncosmπα
2– cosmπα
1cos nπβ
2–
sin nπβ2– sin nπβ
1
πnβ2– β
1
(9.36)
when changing in the x2---direction as shown.
For loads on the whole plate the two results are
pmn = (– 1)m+1 8p
π2mn
m= 1, 2, 3, 4, ...
n = 1, 3, 5, ...(9.37)
pmn = (– 1)n+1 8p
π2mn
m= 1, 3, 5, ...
n = 1, 2, 3, 4, ...(9.38)
Uniform line load: Q= q
= α2– α
12a2+ β
2– β
12b2 (9.39)
α2a
α1a
β1b
β2b
q
pmn =4q
ab
sinmπα2– nπβ
2 – sinmπα
1– nπβ
1
2π mα2– α
1 – nβ
2– β
1
(9.40)
167
Pauli Pedersen: 9. Bending of Rectangular, Orthotropic Plates
+
sinmπα1+ nπβ
1 – sinmπα
2– nπβ
2
2π mα2– α
1+ nβ
2– β
1
specifically parallel with the x2---direction
pmn =–4qπan sinmπαcos nπβ
2– cos nπβ
1 (9.41)
and parallel with the x1---direction we get
pmn =–4q
πbmsin nπβcosmπα
2– cosmπα
1 (9.42)
Rectangular line load: Q= 2qα2– α
1a+ β
2– β
1b
pmn =–4qπ
1an
sinmπα2+ sinmπα
1 · cos nπβ
2– cos nπβ
1+
(9.43)1bm
sin nπβ2+ sin nπβ
1 · cosmπα
2– cosmπα
1
α2a
β2b
α1a
β1b
q
Point load: Q
pmn =4Qab
sinmπα sin nπβ (9.44)Q α a
β b
Four rectangular point loads: 4Q
pmn =4Qab
sinmπα2+ sinmπα
1 · sin nπβ
2+ sin nπβ
1 (9.45)
Q Q
α1a
α2a
β1b
β2b
168
Pauli Pedersen: 9. Bending of Rectangular, Orthotropic Plates
ANALYTICAL Further analytical results can be found from Mathematica (1992), or
OR NUMERICAL numerical integration can be applied to determine the Fourier coeffi---
INTEGRATION cients pmn .
169
Pauli Pedersen: 9. Bending of Rectangular, Orthotropic Plates
9.7 Levy analytical solution
Another classical solution method, the Levy solution, relates to cases
with only two simply supported boundaries. These boundaries must be
opposite and let us here choose the boundaries number 3 and 4 , i.e.
we have
w≡ 0 and w,11≡ 0 along 34 (9.46)
as stated in (9.22). The boundary conditions along 1/2 can be arbitrary.
The solution expansion now includes unknown functions wmx
2 and
not just constants wmn as in (9.23) and we get
wx1, x
2=
m
wmx
2 sin
πmx1
a
w,1111=
m
wmx
2 π4m4
a4sin
πmx1
a
(9.47)
w,2222=
m
wmx
2,2222
sinπmx
1
a
w,1212= –
m
wmx
2,22
π2m2
a2sin
πmx1
a
which, inserted in the differential equation (9.14), gives
m
sinπmx
1
a π4m4
a4D
1111wm
x2+D2222wmx2 ,2222–
π2m2
a22D
1122+ 2D
1212wm
x2,22= p
(9.48)
Assuming a load p= px1, x
2 described by
p=m
pmx2 sin
πmx1
a (9.49)
170
Pauli Pedersen: 9. Bending of Rectangular, Orthotropic Plates
we then have to solve a number of fourth order differential equations
wmx
2,2222
– π2m2
a2
2D1122+ 2D
1212
D2222
wmx
2,22+
π4m4
a4D
1111
D2222
wmx
2=
pmx2
D2222
(9.50)
which, however, have constant coefficients and can thus be solved ana-
lytically for any boundary conditions.
The differential equation (9.50) is well known from beam theory and
has been discussed in general terms, see Pedersen (1986).
SOLUTION TO For the homogeneous part pm = 0 we get
HOMOGENEOUS
PART
AB = πm
a D2222
D
1122+ 2D
1212 D
1122+ 2D
12122 – D
1111D
2222 (9.51)
In general, A and B can be complex values. For the isotropic case of
D1122+ 2D
1212= D
2222= D
1111, we get A= B and then
wx2= C
1+ C
3x2 coshAx
2+ C
2+ C
4x2 sinhAx
2 (9.52)
with constants C1– C
4determined by the specific boundary condi-
tions along the boundaries 1/2 .
171
Pauli Pedersen: 9. Bending of Rectangular, Orthotropic Plates
171
Pauli Pedersen: 10. Torsion of Cylindrical Bars
10. TORSION OF CYLINDRICAL BARS
10.1 A classic problem of elasticity
Torsion of cylindrical bars includes a class of problems which have
CYLINDRICAL/ analytical or semianalytical solutions and which are therefore tradi---
PRISMATIC tionally included in courses on elasticity. A cylindrical bar is straight
and with constant cross---section. It is also called a prismatic bar, and
BAR/ROD sometimes the name torsional rod is used.
In this sectionwe shall treat four classes of problems that are of practi---
FOUR cal importance. Firstly, circular cross---sections for which the cross---
CLASSES section does not warp during torsion. Secondly, non---circular cross---
OF PROBLEMS sections for which the warping modelling is a problem in elasticity with
an interesting classical solution. Analytical solutions will then be dis-
cussed in detail. The third class is a specialization to open thin---walled
cross---sections and lastly the fourth class is closed thin---walled cross---
sections.
Fig. 10.1 shows a bar of length in the x3---direction, and with the
cross---section in the x1, x
2plane. We assume the bar to be fixed at
x3= 0 and measure the deformation either by means of the angle γ
on the cylinder surface or bymeans of the angle , which is the cross---
sectional rotation at x3= .
172
Pauli Pedersen: 10. Torsion of Cylindrical Bars
x1
x2
x3
MT
MT
R
γ
Fig 10.1: Chosen coordinate system and angle definitions.
We assume small deformations γ<< 1 and from fig. 10.1 directly
read the angle relation
ANGLE
RELATIONS γ = R or
=
γ
R(10.1)
with R being the outer radius of the circular cross---section. (In the litera-
ture an angle per unit length θ := is often introduced. It can give more elegant formulas
but we shall here keep the physical angle as our parameter).
10.2 Circular cross---section
The theory of Coulomb for circular cross---section dates back to 1784.
As we shall prove later, plane cross---sections remain plane after
deformation and are only subjected to pure rotation around the center
of torsion, which for double symmetric cross---sections equals the cen-
ter of gravity of the cross---section.
A displacement field that satisfies these conditions is
173
Pauli Pedersen: 10. Torsion of Cylindrical Bars
v1= x
2x3– = x
2x3– γR
DISPLACEMENT
FIELDv2= x
1x3 = x
1x3γR (10.2)
v3≡ 0
x2
x1x
1
x2
v2
–v1
rα
~
This follows directly from the sketch, from which we read
x1= r cosα , v
1= ~
r( – sinα)
x2= r sinα , v2= ~
r cosα
~
=x3
(10.3)
noting that the cross---sectional rotation
is at position x3 .
The displacement gradients vi,j are then known, and with Cauchy
strains we get
11 = v1,1 = 0 , 22 = v2,2 = 0 , 33 = v3,3 = 0
212 = 221 = v1,2 + v2,1 = 0
STRAIN
FIELD(10.4)
213= 2
31= v
1,3+ v
3,1= x
2 –
223= 2
32= v
2,3+ v
3,2= x
1
i.e., a state of pure shear strain in the cross---sectional plane. The
resulting shear strain γ in the tangential direction is
TANGENTIAL
SHEAR
STRAIN
γ
=2
132+ 2
232= r (10.5)
174
Pauli Pedersen: 10. Torsion of Cylindrical Bars
and thus, at the surface r = R , we find in agreement with (10.1),
γ(r= R)= γ .
Assuming an isotropic Hooke’s law with σij= G2ij for i≠ j , we get
a state of pure shear stress
STRESS
FIELDσ13= σ
31= x
2G – , σ
23= σ
32= x
1G (10.6)
with the resulting tangential shear stress τ being
τ= σ231+ σ2
32 = r G (10.7)
as also followsdirectly from (10.5). Themaximumshear stress τmax will
be at the outer cross---sectional boundary r= R , i.e.
τmax= RG = γG(10.8)
τ = rR τmax
We shall then set up the equilibrium with the external torsional
moment MT
and have
MT=
A
r τdA = R
0
r τmaxrR 2πrdrEXTERNAL
MOMENT
INTERNAL STRESS
(10.9)
⇒MT=
π
2R3
τmax or τmax =
2MT
πR3
where A is the cross---sectional area. Introducing the polar moment of
inertia J
J := A
r2dA= R
0
r22πrdr= π
2R4 (10.10)
175
Pauli Pedersen: 10. Torsion of Cylindrical Bars
and using (10.7), as an alternative to (10.9), we write
τ= MTJ r = r G ⇒ M
T= GJ
(10.11)
τmax = MTJ R ⇒ M
T= τmaxJR
In general for the torsional problem we use two important quantities.
Firstly, the torsional stiffness GK defined by
KDEFINITION
MT= (GK) (10.12)
and thus, in analogy with bending stiffness EI from MB= (EI) and
axial stiffness EA from N = (EA) , a quantity thatgives thedeforma-
tion = MT(GK) . The torsional stiffness has a material factor
G and a cross---sectional factor K , called the cross---sectional torsion
factor. Only for circular cross---sections we have, from (10.11), K = J .
The second important quantity is the torsional resistance WT, defined
by
WT
DEFINITION|M
T| = W
T|τmax| (10.13)
and thus, in analogy with Izmax from bending and A for tension/com-
pression, a quantity that gives the stress level |τmax| = |MT|W
T.
For circular cross---sections we read from (10.11) WT= JR =
πR32 .
10.3 Non---circular cross---sections
Displacements in the x3---direction are important for non---circular
cross---sections, and the assumption v3≡ 0 in (10.2) must therefore
176
Pauli Pedersen: 10. Torsion of Cylindrical Bars
be changed. In 1855 Saint---Venant introduced the warping function
Ψ= Ψx1, x
2 and based a torsional theory on a displacement field,
expressed by
v1= x
2x3– , v
2= x
1x3
DISPLACEMENT
FIELD(10.14)
v3= Ψx
1, x
2
Now from this follows displacement gradients, and the only non---zero
strains will again be 13
and 23
213
= 231
= Ψ,1 – x2STRAINS (10.15)
223 = 232 = Ψ,2+ x1
with the resulting stresses
σ13 = σ31 = GΨ,1 – x2STRESSES (10.16)
σ23 = σ32 = GΨ,2+ x1
The stress equilibrium with volume forces and surface tractions deter-
mine the unknown warping function. Without volume forces, i.e.,
pi≡ 0 , wemust have σji,i≡ 0 , which, with (10.16) and G ≠ 0 ,
gives
(10.17)∆Ψ := Ψ ,11+Ψ,22 ≡ 0
LAPLACE
DIFFERENTIAL
EQUATION
177
Pauli Pedersen: 10. Torsion of Cylindrical Bars
dsdx2
–dx1
α
which is known as the Laplace differential equation. Although this
equation is simple, the boundary conditions will complicate the
solution. These boundary conditions follow from no surface trac---
boundary we must have σijnj≡ 0 . From the
n1= sinα , n
2= cosα , n
3= 0
cosα= – dx1ds , sinα= dx
2ds
(10.18)
and thus, σ31
n1+ σ
32n2= 0from (10.16) with G ≠ 0,
Ψ,1– x
2dx2ds
– Ψ,2+ x
1dx1ds
= 0
BOUNDARY
or
(10.19)Ψ,1
dx2
dx1
–Ψ,2= x
2
dx2
dx1
+ x1
CONDITIONS
n
tions, i.e. at the
sketches we read
gives
EQUILIBRIUM If we are able to determine a Ψ function that satisfies (10.17) and
(10.19), then the torsional problem is solved. The external moment
MT, resulting from the stress distribution, can be determined from the
sketch
178
Pauli Pedersen: 10. Torsion of Cylindrical Bars
x2
x1
x3
MT
σ32dx
1dx
2
σ31dx
1dx
2
= G
A
− Ψ,1x2+ x2
2+Ψ,2x1+ x2
1dA
= G
A
Ψ,2x1 − Ψ,1x2dA+G
J
(10.20)
MT = A
– σ
31x2+ σ
32x1dA
where we have used x21+ x2
2= r2 and J := r2dA
Amore practical but alsomore abstract formulation of the problemwas
suggested by Prandtl in 1903. This formulation is given in terms of a
stress function Φ , defined by the stress relations
STRESS
FUNCTION
DEFINITIONσ31= Φ ,2 , σ32= – Φ,1 (10.21)
which then simplifies the boundary condition σ31 n1+ σ32 n2= 0 to
Φ,2
dx2ds+Φ ,1
dx1ds=
dΦds= 0
BOUNDARY
CONDITIONSor (10.22)
Φ= constant
at the boundaries.Having only a single boundarywe can choose Φ= 0
at this boundary. (In the literature there are different definitions of the stress function.
If the factor G is included in σ31 = Φ ,2 G and σ32 =−Φ ,1 G more elegant for-
mulas are possible, but we shall keep the definition (10.21) where Φ has the dimension of force
per unit length).
179
Pauli Pedersen: 10. Torsion of Cylindrical Bars
The equilibrium σji,i≡ 0 is directly satisfied by the definition (10.21),
and we shall now derive the relations to thewarping function Ψ . From
(10.21) and (10.16) we get
STRESS
FUNCTION/ Φ,1= – GΨ,2 + x1 , Φ ,2 = GΨ,1 – x2WARPING (10.23)
FUNCTION Ψ,1 =
GΦ ,2 + x2 , Ψ,2 = –
GΦ,1 – x1
so when the stress function Φ is known, the warping function Ψ is also
known and thus the displacements of the torsional bar. Differentiating
and adding from (10.23a), we get
(10.24)∆Φ := Φ,11+Φ
,22≡ –2G
POISSON
DIFFERENTIAL
EQUATION
known as a Poisson differential equation. Determining a function Φ
which satisfies (10.24) and is constant at the cross---sectional bound-
aries (10.22) corresponds to solving the torsional problem.
Equilibrium with external torsional moment as in (10.20) gives, with
integration by parts,
MT = – Φ,2x2 – Φ,1
x1dx
1dx2
= – Φx
2x–
2
x2
+Φdx2dx
1+– Φx
1x–
1
x1
+Φdx1dx
2(10.25)
⇒ MT= 2ΦdA
180
Pauli Pedersen: 10. Torsion of Cylindrical Bars
The fact that Φ is constant along the boundaries makes Φx2x2–
x–2
=
Φx1x1–
x–1
= 0 . We note that the terms σ31
and σ32
give equal contribu-
tions to the torsional moment.
Using (10.25) and the definition (10.12), the torsional stiffness factor
K expressed in the stress function will be
K= 2GΦdA (10.26)
and the resulting shear stress τ from (10.7) will be
|τ|= Φ2
,1+Φ
2,2
= |grad Φ| (10.27)
so that the torsional resistance WT from the definition (10.13) is
WT =
2ΦdA|grad Φ|max
(10.28)
10.4 Analytical solution for hollow elliptic cross---sections
On the basis of the results of the previous sectionwehave the possibility
of obtaining analytical results for certain cross---sectional shapes. A
most important class of such cross---sections is shown in fig. 10.2. For
λ= 0 this covers solid cross---sections, for λ≠ 0 hollow cross---sec-
tions and as a special case, circular cross---sections for a= b , i.e.
γ= 1 . For this last case we shall again obtain the results of section
10.2.
181
Pauli Pedersen: 10. Torsion of Cylindrical Bars
a
b
x2
x1λb
λa
b= γa
0< γ≤ 1
0≤ λ< 1
Fig. 10.2: Hollow cross---section with elliptical boundaries.
In the Cartesian coordinate system the boundaries are given by
outer : x1a
2
+ x2b2= 1
(10.29)
inner : x1a2+ x
2b2= λ
2
and thus a stress function Φ with an unknown coefficient C
Φ= Cx1a2+ x
2b2 – 1 (10.30)
will be constant at both these boundaries
Φouter = 0
(10.31)
Φinner
= Cλ2 – 1
The boundary condition (10.22) is satisfied, and to satisfy also the Pois-
son equation (10.24) we determine C
182
Pauli Pedersen: 10. Torsion of Cylindrical Bars
Φ,11+Φ,22 = C2a2+ 2b2 = – 2G
⇒ C= – a2b2
a2+ b2G
(10.32)
⇒ Φ= – a2b2
a2+ b2G
x1a
2
+ x2b2 – 1
From the definition (10.21) then follows the stresses
STRESSES σ31=
– 2a2
a2+ b2G
x2
, σ32=
2b2
a2+ b2G
x1
(10.33)
|τ|= 2a2+ b2
G
a4x2
2+ b4x2
1
which, for a= b , give the results (10.6) for a circular cross---section.
The external moment MT
in equilibrium with these stresses is given
by (10.25)
MT= 2
A
ΦdA= − 2a2b2
a2+ b2G
1a2 x2
1dA+ 1
b2 x2
2dA –dA
(10.34)
⇒ MT=
π a3b3
a2+ b21 – λ
4G
which follows from the well known results for elliptical shapes
x21dA= I
2, x2
2dA= I
1, dA= A , where I
2, I
1are cross---sec-
tional moments of inertia. The cross---sectional constant K of the tor-
sional stiffness, is thus, by its definition and introducing b= γa ,
183
Pauli Pedersen: 10. Torsion of Cylindrical Bars
K=π1 – λ4 γ3a4
1+ γ2STIFFNESS
with the specific cases
K (solid circle) = πa42
K (hollow circle) = πa41 – λ42 (10.35)
K (solid ellipse) = πa4γ31+ γ2
τ= τmax
τ= constantx2
x1 τ2
a2+ b22
4
G2
= a2x22+ b2x
12 (10.36)
The variation of stresses (10.33) is illustrated by the
isolines of τ2
which are seen to be ellipses with half---axes equal to a2 and b2 . The
|τ|max is therefore found at x1= 0 , x
2= b , i.e. from (10.33)
|τ|max =2a2b
a2+ b2G
=2γ
1+ γ2aG
(10.37)
which, with |τ|max WT= GK , gives
RESISTANCE WT= π
21 – λ
4γ2a3
with the specific cases
184
Pauli Pedersen: 10. Torsion of Cylindrical Bars
WT(solid circle) = πa32
WT(hollow circle) = πa31 – λ42 (10.38)
WT(solid ellipse) = πa3γ22
Fig. 10.3 shows a thin---walled cross---section with an outer elliptical
shape, but due to constant thickness h , the inner shape is not elliptical.
We shall return to thin---walled closed cross---sections more generally
later and here just give the approximate results from (10.35) and
(10.38) for h= (1 – λ)b= (1 – λ)γa and 1 – λ4≈ 4(1 – λ)
K≈ 4πγ2
1+ γ2a3h , W
T≈ 2πγa2h (10.39)
h
b
a
x1
x2
Fig. 10.3: Cross---section with elliptical outer boundary and constant thickness h .
WARPING Inserting the stress function (10.32) or its derivatives by (10.33) in the
relations between the stress function and warping function Ψ (10.23)
we can find the warping displacements. For this case the integration
from Ψ,1
and Ψ,2
to Ψ can be solved analytically and we obtain
185
Pauli Pedersen: 10. Torsion of Cylindrical Bars
Ψ= − a2 − b2
a2+ b2x1x2
0, ba, 0
(10.40)
illustratedby isolines in the sketch.The result (10.40) ismost easily veri-
fied by differentiation and substitution into (10.23). From (10.40) we
find the result for a circular cross---section a= b to be Ψ≡ 0 , and
thus we have verified the assumption of section 10.2.
10.5 Other analytical solutions
Elliptical cross---sections are not the onlyones forwhich analytical solu-
tions can be obtained. The condition for obtaining a solution is that the
boundary description, like the functions (10.29), satisfies Poisson’s
equation, as shown in (10.32). Furthermore, if there is more than one
boundary, the difference between the boundary functions should be a
constant, as in (10.29).This last condition is seldom satisfied, sowe shall
here only show examples of solid cross---sections.
h
h
x2
x1
60o
2 3
Fig. 10.4: Equilateral triangular cross---section in a Cartesian coordinate system.
186
Pauli Pedersen: 10. Torsion of Cylindrical Bars
EQUILATERAL A classical example is the equilateral triangle, shown in fig. 10.4 in a
TRIANGLE Cartesian coordinate system, where the boundaries are described by
x2+ 3 x
1x2 – 3 x
1x2 – h = 0 (10.41)
so that a stress function Φ with this factor will be 0 at all the bound-
aries. A constant C is determined to satisfy Poisson’s eq. (10.24)
Φ,11+Φ
,22= C4h= – 2G (10.42)
giving C = – G(2h) and thus the result
Φ=–G
2hx
2+ 3 x
1x
2– 3 x
1x
2– h
σ31= Φ
,2=
− G
2h3x2
2− 2x
2h− 3x2
1
(10.43)
σ32= – Φ
,1=
–G
2h6x
1x2– 6x
1h
τ2 = σ231+ σ2
32= G
2h2
36x21h – x
22+
–3x21– 2hx
2+ 3x2
22
with stress---isolines for the upper third part (symmetry from each side)
illustrated to the left.
Integrating Φ over the triangular area, we get
MT= 2
G
2hh
0
x2 3
–x2 3
x22– 3x
2
1x
2– hdx
1dx
2(10.44)
187
Pauli Pedersen: 10. Torsion of Cylindrical Bars
and thereby the torsional stiffness factor
STIFFNESS K= h4 3 45 (10.45)
With |τ|max= G
h2
at x1, x
2= 0, h , we can also determine the tor-
sional resistance
RESISTANCE WT= h32 3 45 (10.46)
R R
a
x2
x1
Fig. 10.5: A cross---section which is an outer circle modified by a circular notch.
SHAFT Another example is shown in fig. 10.5, which models a shaft with a
WITH A NOTCH notch. The boundaries are described by
a2 – x21– x2
21 – 2Rx1x21+ x2
2 = 0 (10.47)
and, as before, we adjust a constant C to satisfy Poisson’s equation
Φ,11
+Φ,22
= C(–4) = – 2G (10.48)
188
Pauli Pedersen: 10. Torsion of Cylindrical Bars
and we get the result
Φ=G
2a2 – x2
1– x2
21 – 2Rx
1x2
1+ x2
2
σ31= Φ
,2=
G
− x
2+
2Ra2x1x2
x21+ x2
22
(10.49)
σ32=−Φ
,1=−
G
R− x
1+
Ra2x21− x2
2
x21+ x2
22
The maximum shear stress appears at x
1= a, x
2= 0 , where we get
τmax= G(2R – a) . For a shaft without a notch the maximum
shear will be τmax = GR , as derived in (10.8). The notch thus
gives a stress concentration factor of (2R – a)R = 2 – aR . (An ex-
ercise could be to plot the stress---isolines of τ= σ231+ σ2
32 ) .
10.6 Analytical solution for rectangular solid cross---sections
Rectangular cross---sections are often used as torsional bars and there-
fore deserve special attention.An analytical solution in termsof a series
expansion is possible for this cross---section, shown in fig. 10.6 with the
origo of the Cartesian coordinate system in the center of the rectangu-
lar domain and with the axes parallel to the boundaries of the domain.
189
Pauli Pedersen: 10. Torsion of Cylindrical Bars
x2
x1
a2 a2
b2
b2
Fig. 10.6: Rectangular cross---section with length of the longer side a
and length of the shorter side b .
DOUBLE
FOURIER
EXPANSION
The solution is based on a double Fourier expansion for the stress
function
Φ= m , n
odd
Cmn cosmπx
1
a cosnπx
2
b(10.50)
which directly fulfils the boundary condition Φ= 0 because each indi-
vidual term does. The summation symbol is as used in plate theory
(9.24), and the difference between (10.50) and (9.23) just corresponds
to differently chosen origos.
Inserting (10.50) in Poisson’s equation (10.24), we get
m , n
odd
Cmn cosmπx
1
acos
nπx2
b– m2
π2
a2–
n2π2
b2= – 2G
(10.51)
190
Pauli Pedersen: 10. Torsion of Cylindrical Bars
Now the constant term – 2G on the right---hand side can also be
described by a Fourier expansion (just as in (9.34)) and we have
− 2G
=m , nodd
Amn cosmπx
1
a cosnπx2b
(10.52)
Amn= –2G
–1
n+m
2 16π2mn
From (10.52) and (10.51) then follow the Fourier coefficients Cmn
Cmn = –
–1
n+m
232a2b2G
π4mnb2m2+ a2n2
(10.53)
Having determined the stress function Φ we have in reality solved the
problem and only the practical evaluation of the torsional stiffness fac-
tor K from (10.26) and the torsional resistance WT
from (10.28)
remain. These results are in most cases presented in tabular form
introducing the two functions f1= f
1ab and f
2= f
2ab , here
normalised so that for ab→∞ we have f1→ 1 and f
2→ 1
K= 13
a3b3
a2+ b2f1ab= A4
36Jf1ab
(10.54)
WT= 1
3ab2f
2ab= 2Ar
min
3f2ab
191
Pauli Pedersen: 10. Torsion of Cylindrical Bars
abf1ab
f2ab
1
0.844
0, 63
1, 25
0, 845
0, 66
1, 5
0, 848
0, 69
2
0, 858
0, 74
3
0, 878
0, 80
4
0, 895
0, 84
5
0, 909
0, 87
10
0, 946
0, 94
∞1, 000
1, 00
Table 10.1: Correction factors for determining K and W according to the defini-
tions (10.54), (10.55).
The results of table 10.1 with r := ab follows from
f1(r) := 768
π6
1+ r2
m,n odd
1
m2n2m2+ n2r2
(10.55)
f2(r) := f
1(r) π
3
321+ r2 m,n odd
(–1)2n+m+1
2 mm2+ n2r2
and fig. 10.7 shows agraph basedon (10.55) and inagreementwith table
10.1.
f2
f1
f1
f2
r= ab
Fig. 10.7: Graphical representation of the correctional factors f1and f
2as defined in (10.55).
192
Pauli Pedersen: 10. Torsion of Cylindrical Bars
The right most formulas in (10.54) are often used for cross---sections
which looks like rectangular or elliptical shapes without being so
exactly. The quantity rmin is the shortest distance from the torsional
center to the boundary. This approximation for WT is less safe than
the approximation for K .
For a3b3a2+ b2⇒ ab3 the asymptotic values of K and WT
for
a>> b with f1= f
2= 1 will then be
K⇒ 13ab3 for a
b→∞
(10.56)
WT⇒
13ab2 for a
b→∞
In the next section we shall derive this result in another way and then
use it extensively for the important thin---walled cross---sections.
193
Pauli Pedersen: 10. Torsion of Cylindrical Bars
Fig. 10.8a: Resulting warping displacement field for the first quadrant quarter of rectangular cross---sections in
torsion. Surface plot to the left and corresponding isolines to the right.
194
Pauli Pedersen: 10. Torsion of Cylindrical Bars
Fig. 10.8b: Resulting shear stress field for the first quadrant quarter of rectangular cross---sections in torsion. Sur-
face plot to the left and corresponding isolines to the right.
195
Pauli Pedersen: 10. Torsion of Cylindrical Bars
Tables and figures give a general understanding of the behaviour. For
actual calculations, a small computer program will be more appropri-
ate. Then the numerical evaluation of the warping function Ψ , which
follows from (10.23) when Φ is known, can also be carried out. The
results of such a program are shown in fig. 10.8 for cross---sections with
ab ratios= 1, 2 and 4 .Resulting shear stresses aswell aswarpingdis-
placements are shown, both with surface plots and with contour plots
(isolines). Because of symmetries only the results in the first quadrant
are shown.
10.7 Thin---walled open cross---sections
A thin rectangular cross---section, also named a rectangular strip, as
described in (10.56) with a>> b is shown in fig. 10.9.
Only at the ends x1= a2 will the stress function Φ change in the
x1---direction. Thus, a reasonable assumption is
Φ,1= – σ
32= 0 (10.57)
MT
x2x1
Fig. 10.9: Single thin---walled rectangular cross---section
with stress function Φ illustrated.
196
Pauli Pedersen: 10. Torsion of Cylindrical Bars
so to satisfy Poisson’s equation (10.24) we must have
Φ,22= – 2G ⇒
(10.58)
Φ,2= – 2Gx
2+ C
1
which, with the boundary condition Φx2= b2 = 0 , gives
Φ= G
b24
– x22 (10.59)
and then the resulting stresses
σ31= Φ
,2= – 2Gx
2⇒
(10.60)
|τ|max = Gb
The external torsionalmoment MT in equilibriumwith this stress field
is
MT = 2ΦdA= G
ab3
3(10.61)
It follows that we have again obtained the results (10.56), i.e.
SINGLE STRIP
STIFFNESS AND K= 13ab3 , W
T=
13ab2 (10.62)
RESISTANCE
197
Pauli Pedersen: 10. Torsion of Cylindrical Bars
We can then add thin---walled rectangular strips to obtain any open
cross---section, like those shown in fig. 10.10.
Fig. 10.10: Examples of thin---walled open cross---sections.
MULTIPLE We number each of the strips n and have a total number of N strips.
STRIPS The basic assumption is that all the strips are subjected to the same
cross---sectional rotation
n = for n = 1, 2, , N (10.63)
We can therefore add the torsional moments from the strips, which,
using (10.61), gives
MT=
N
n=1
MTn=
n
G
anb3n3
(10.64)
The combined torsional stiffness constant K and the torsional resis-
tance WTthen follow directly from definitions:
RESISTANCE
STIFFNESS AND
K=n
anb3n3
WT= K(b)
max=n
anb3n3(bn)max
(10.65)
198
Pauli Pedersen: 10. Torsion of Cylindrical Bars
We have assumed that the warping which follows from the stress func-
tion is not influenced by boundary conditions. However, particularly
ADVANCED for open thin---walled cross---section the boundary conditions do have
THEORIES amajor influence,which calls formoreadvanced theory, like theVlasov
theory.
10.8 Thin---walled closed cross---sections
Lastly, we shall analyse thin---walled closed cross---sections, starting
with only a single hole as in fig. 10.3 but not restricted to elliptical shape.
If we cut out an infinitesimal brick as shown in the sketch, force equilib-
rium gives
dx3
dx3
hB
hA
τBhBdx
3= τ
AhAdx
3
=> τBhB= τ
AhA= τh
(10.66)
i.e. constant force per unit length. We say that the shear force flow τh
is constant. For thin---walled strips we can assume τ constant through
the thickness, or we can interpret τ as the mean shear stress.
CONSTANT Integrating to equilibrium with the external moment MT
gives
SHEAR FLOW
MT= (τhds)a= τh ads (10.67)
199
Pauli Pedersen: 10. Torsion of Cylindrical Bars
which, with geometry as shown in the sketch: ads= 2Ao , gives
ds a
MT= τh2Ao
τ=
MT
2hAo
; |τ|max =
|MT|
2Aohmin
(10.68)
Note that Ao is the circled area, not the solid area.
Then the cross---sectional torsional resistance WTwill be
RESISTANCE
SINGLE HOLE(10.69)W
T= 2Aohmin
while the cross---sectional stiffness constant cannot be determined so
directly. We shall here derive it by means of the complementary virtual
work principle (6.30), thereby illustrating another practical use of this
principle.
The variation of the complementary work is by definition
δWC= δMT (10.70)
and the variation of the stress energy (complementary elastic energy)
δUC is determined by
δUC= δuChds (10.71)
200
Pauli Pedersen: 10. Torsion of Cylindrical Bars
where the variation of stress energy density δuC is, by definition,
δuC= ijδσij= γδτ= τGδτ (10.72)
with γ being the resulting engineering shear strain (10.5) correspond-
ing to τ . Eliminating τ from (10.68) we get
δuC =MTδMT
4Gh2A2o
(10.73)
Equations (10.70)---(10.73) used in the principle δWC = δUC for
arbitrary δMT then give
=MT
4GA2o
dsh
(10.74)
SINGLE HOLE which, expressed by the torsional stiffness factor K , is called Bredt’s
STIFFNESS, formula
(10.75)
BREDT’S
FORMULA K = 4A2o ds
h
201
Pauli Pedersen: 10. Torsion of Cylindrical Bars
Fig. 10.11: Examples of thin---walled closed cross---sections with several holes.
SEVERAL As shown in fig. 10.11 thin---walled cross---sections often have several
HOLES holes, and for these cross---sections a computational procedure is
needed.
Let the holes be numbered n= 1, 2, , N , where N is the total num-
ber of holes. Still keeping the assumption of constant rotation for the
202
Pauli Pedersen: 10. Torsion of Cylindrical Bars
whole cross---section we have, from (10.74) (omitting index T on the
torsional moments),
n
=
=
Mn
4GA2on
n
dsh
for n= 1, 2, , N
(10.76)
⇒ n
dsh=
4GA2on
Mn
For a specific hole n , the integrated shear stresses with (10.68) and
(10.76) are
A
on
τds= A
on
Mn
2hAon
ds=Mn
2Aon
dsh=
2GAon (10.77)
COMPUTATIONAL We now have the necessary formulas to specify the procedure. At the
PROCEDURE inner boundary of hole n the constant value of stress function is termed
Φn , and at the outer contour of the total cross---section the stress func-
tion constant is, by definition, set at zero, i.e. Φo = 0 . For a two hole
case, fig. 10.12 illustrates the variation of the stress function with the
three boundary values Φo = 0 , Φ1and Φ
2. For thin---walled cross---
sections it is reasonable to assume the shown linear changes from Φo
to Φ1, from Φo to Φ
2and from Φ
1to Φ
2. With this assumption
and the result (10.27), i.e. τ= gradΦ , we get the shear stress τnm in
the wall between hole n and hole m
203
Pauli Pedersen: 10. Torsion of Cylindrical Bars
n=0
n=1
n=2
Φo = 0
Φ1
Φ2
h10
h12= h
21h20
Fig. 10.12: Thin---walled cross---section with two holes, and a cut
to illustrate the model for the stress function.
τnm =Φn – Φm
hnm(10.78)
where hnm is the actual wall thickness. Note that seen from hole m
the sign changes, i.e. τmn = –τnm .
The approximation (10.78) is then inserted in the integrated formula
(10.77), which for convenience, we shall write as a summation over
constant wall thickness (integration or multiple summation along the
same wall is normally not needed).We assume hnm to be constant and
term the length of this wall Snm . Then (10.77) with (10.78) gives
204
Pauli Pedersen: 10. Torsion of Cylindrical Bars
A
on
Φn – Φm
hnmds=
m
Φn – Φm
hnmSnm =
2GAon
(10.79)
for n = 1, 2, , N
where m
is a summation over the m holes connected to hole n , possi-
bly including the outer contour of m= 0 .We note that this constitutes
N linear equations to determine the N unknown values Φn .
With all Φn known, themaximum shear stress can be determined from
|τ|max= Φn – Φm
hnmmax
(10.80)
and the total torsional moment MT
is determined from the general
result (10.25), i.e.
MT= 2ΦdA= 2
n
ΦnAon (10.81)
whichmeans that thepractical parameters K and WTcan beevaluated
from
STIFFNESS K=
2n
ΦnAon
GAND (10.82)
RESISTANCE
WT=
2n
ΦnAon
|(Φn – Φm)hnm|max
205
Pauli Pedersen: 10. Torsion of Cylindrical Bars
10.9 Examples of thin---walled cross---sections
I II IIIh hh
2a
2b
b
a
1
2
Fig. 10.13: Three examples of thin---walled cross---sections for which torsional stiffness and
resistance are determined.
In fig. 10.13 we show three different thin---walled cross---sections, of
which example III is a combination of examples I and II. Example II is
identical to the elliptical case in fig. 10.3. The dimensions given by a , b
and h satisfy
a≥ b>> h (10.83)
and the actual areas can thus be determined without discussing inner,
outer or medium dimensions. We get
AoI= 4ab , A
oII= πab , A
oI−A
oII4= 1− π4ab (10.84)
with the last area referring to the corners of the combined cross---sec-
tion III.
For example I, Bredt’s formula (10.75) gives the stiffness factor
K = 4(4ab)24(a+ b)h = 16h a2b2
(a+ b)(10.85)
206
Pauli Pedersen: 10. Torsion of Cylindrical Bars
and the resistance follows from (10.69), i.e.
WT= 2(4ab)h= 8abh (10.86)
For example II, Bredt’s formula gives, using the approximated circum-
ference formula: 2π a2+ b22 ,
K= 4π2a2b2h2π a2+ b22 = 2 2 πh a2b2
a2+ b2 (10.87)
and the resistance (10.69) is
WT= 2(πab)h= 2πabh (10.88)
Comparing the two examples I and II we conclude that the rectangular
shape has higher strength (factor 4π) and also higher stiffness (factor
between 4π and 4 2 π) . Itmay also be interesting to compare exam-
ple IIwith the asymptotic results in (10.39), with full agreement in resis-
tance and close agreement in stiffness.
Now the example III is intended to illustrate the use of the procedure
that follows from (10.79) --- (10.81). The central hole we number 1 and
the corner holes are all numbered 2 because of symmetry. Then (10.79)
gives
Φ1– Φ
2
h2π a2+ b
22 =
2Gπab (10.89)
and for one of the corner holes we get from (10.79)
Φ2– Φ
1
hπ2 a2+ b
22 +Φ2– Φ
0
h(a+ b)=
2G1 – π
4ab (10.90)
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Pauli Pedersen: 10. Torsion of Cylindrical Bars
From these two linear equations it follows that, with Φ0= 0 , we for
the circular case of a= b get
Φ1= 2ahG
(10.91)
Φ2= ahG
and thus (remembering the four corner holes) from (10.82)
K= (8+ 2π)a3h
and (10.92)
WT= (8+ 2π)a2h
assuming the ellipse completely inside the rectangle. If (Φ1−Φ
0) is
involved the resistance will only be half the value.
208
Pauli Pedersen: 10. Torsion of Cylindrical Bars
209
Pauli Pedersen: 11. Finite element analysis
11. FINITE ELEMENT ANALYSIS
displacement assumptions, nodal reference, stiffness
matrices, basic matrices, equivalent loads, mass matrices
11.1 Contents of this section
NON--- In this sectionwe shall give a short and rather non traditional introduc---
TRADITIONAL tion to the element analysis which is central for the finite element
method, FEM. We do not cover the system analysis of FEM, and only
elements with translational degrees of freedom (not beams and plates)
will be covered.
SCALAR With these restrictions we can work at the level of scalar displacement,
DISPLACEMENT without involving the total displacement vector. For three dimensional
LEVEL problems this means that the order of our matrices is only one third the
order in a coupled presentation.
BASIC The approach with basic matrices gives a possibility for a description
MATRICES independent of the constitutive relation, and thereby focusing on the
consequences of the displacement assumptions.
2 ---CONTRACTED A further non traditional aspect of the presentation is the use of the
NOTATION 2 ---contracted notation used throughout this book. Detailed results
on a number of practical used element will be given, results which can-
not be found in more extended books on FEM.
210
Pauli Pedersen: 11. Finite element analysis
Thus this section can be read within the contents of this book, and can
be read as a supplement to the many good books on FEM, say Zien-
kiewics & Taylor (1989---1991), Cook, Malcus & Plesha (1989), Bathe
(1996), Crisfield (1994---1997) and Ottosen and Petersson (1992).
11.2 Element geometry and nodal positions
LOCAL The elements will be described in a conveniently placed Cartesian
CARTESIAN coordinatesystem, and rotational transformations will then be treated
separately. For axisymmetric elements where also element translation
is important, alternative simplifications are aimed at.
SIZE & The size of the elements is described by the
FORM
size parameter h (11.1)
and often we shall see a very simple influence from this parameter. The
form of the elements (the word shape has an alternative use in FEM)
are described by the non---dimensional
form parameters α , β , γ , δ , (11.2)
This strategy of element description is best illustrated by examples,
where we also show the position of the nodal points
211
Pauli Pedersen: 11. Finite element analysis
1---D bar elements
LDBA: Linear Displacement BAr(two nodes at the ends, A cross---sectional area)(1)h
x1(0)
AI II
QDBA: Quadratic Displacement BAr(two nodes at the ends and one mid point node)
I IIIIIA
(1)h(½)h(0)x1
2---D disc elements (membrane plates)
LDTR: Linear Displacement TRiangle(three nodes at the corners, t disc thickness)
III
I
II
(α, 0)hx1
x2
(βα , 1)h
t
QDTR: Quadratic Displacement TRiangle(three nodes at the corners andthree nodes at the edge midpoints)
I
III
II
IVV
VI
x2
x1
(0, 0)
(βα , 1)h
(α, 0)h
(½(1+ β)α,½)h
(½α, 0)h
t
(0, 0)
(½βα,½)h
212
Pauli Pedersen: 11. Finite element analysis
3---D volume elements
LDTE: Linear Displacement TEtrahedron(four nodes at the corners)III
III
IVx2
x1
x3
(0, 0, 0) (α, 0, 0)h
βα, γ, 0h
δα, γ, 1h
QDTE: Quadratic Displacement TEtrahedron(four nodes at the corner andsix nodes at the edge midpoints)
I
VII
V
VIII
VIII
III
X
IV
IX
x3
(0, 0, 0)x1
(α, 0, 0)h
x2
βα, γ, 0h
δα, γ, 1h
3---D axisymmetric ringelements
LDRI: Linear Displacement RIng
(three nodal circels at the edges,
r
z
z
r
1
2II
III
I II
III
βα, γ+ 1h
− βα, γ− 1h
0, γh α, γh
− α, γh
QDRI: Quadratic Displacement RIng(three nodal circels at the edges and
three nodal circels in the middle of
the border surfaces)
r
z
− α, γh
II
− βα, γ− 1h
III
r βα, γ+ 1h
III
α, γh
II
0, γh
I
1
2IV
IV
V
VVI
VI
two cases 1 and 2)
z
213
Pauli Pedersen: 11. Finite element analysis
11.3 Displacement assumptions
Already in the geometry examples we have indicated the displacement
assumptions, associated with the names of the elements. For descrip-
tion of a displacement fieldwe at every point in space have the displace-
ment vector
DISPLACEMENT
FIELDv := v
1v2v3 (11.3)
with its components as a function of space coordinates
vi= v
i(x)= v
ix
1, x
2, x
3 (11.4)
The function vi(x) symbolizes v
ix
1 in 1---D, v
ix
1, x
2 in 2---D and
as shown for 3---D in (11.4).
BASIC IDEA The basic idea of the FEM is the approximation of the displacement
OF FEM/ fields by a linear combination of preselected functions. These functions
DISPLACEMENT are in most cases chosen as polynomial terms. Only when the total
APPROXIMATION approximation “function space” is close to the real solution will we get
a good solution. The specific solution chosen among the many possible
ones ina given function spacewill be the “best”one,meaning consistent
with the virtual work principle or to state it differently, with errors that
are in equilibrium.
POLYNOMIAL Choosing a polynomial description we have
DESCRIPTION
vi=
a1i+ a
2ix1+ a
3ix2+ a
4ix21+ ... (11.5)
with constants a to be eliminated by nodal compatibility. In matrix
notation (11.5) is written
214
Pauli Pedersen: 11. Finite element analysis
vi= HT
i
HT := 1 x1h x
2h x
3h x2
1h2 (11.6)
Ti := a
1i
a2i
Note that the vector H of displacement functions do not have the
directional index i . Choosing the same displacement assumption in all
direction simplifies the description and we shall therefore restrict us to
this. Note also, that we have chosen to work with non---dimensional
components in the H vector.
For the elements of this book the displacement assumptions are:
ACTUAL
CASES1---D bar elements
LDBA: HT := 1 x1
h (11.7)
QDBA: HT := 1 x1
h
x21
h2 (11.8)
2---D disc elements
LDTR: HT := 1 x1
h
x2
h (11.9)
QDTR: HT := 1 x1
h
x2
h
x21
h2
x22
h2x1x2
h2 (11.10)
215
Pauli Pedersen: 11. Finite element analysis
3---D volume elements
LDTE: HT := 1 x1
h
x2
h
x3
h (11.11)
QDTE: HT := 1 x1
h
x2
h
x3
h
x21
h2
x22
h2
x23
h2x1x2
h2x1x3
h2x2x3
h2(11.12)
3---D axisymmetric elements
LDRI: HT := 1 zh
rh (11.13)
QDTR: HT := 1 zh
rh
z2
h2r2
h2zrh2 (11.14)
11.4 Node displacements and the configuration matrix
The approximation used in the finite element method is characterized
by a discretization from infinite degrees of freedom to a finite number
of degrees of freedom, chosen to be the nodal degrees of freedom. The
position of the nodes in an element we call the nodal configuration.
ELIMINATION When the number of functions in H (equal to the number of
OF CONSTANTS constants in ) equals the number of nodes in an element, we can
express the unknown constants by the displacements of the nodes.
COMPATIBILITY Let the displacement of node k in direction i be called Dkiand this
OF NODAL must be compatible with the description (11.6). Inserting (x)k=
DISPLACEMENTS x1, x
2, x
3kin H we get
216
Pauli Pedersen: 11. Finite element analysis
Dki= v
i(x)
k=
H(x)kT
i(11.15)
Collecting all nodal displacements in direction i in the vector Di
we get
Di=
H(x)kT
H(x)lT
.
.
.
i= [K]
i(11.16)
from which follows the name configuration matrix for the [K] matrix
because it mainly depends on the node configuration in the element.
Note, that we use index k, l, for nodes and indices i and j for direc-
tion.
INVERTED When the number of nodes is the same as the number of components
CONFIGURATION (functions) in the H vector, then the matrix [K] is a square matrix,
MATRIX which we with a symbolic computer language like Mathematica (1992)
can invert analytically, and solve (11.16) for the constants
i= [K]−1D
i(11.17)
With (11.17) we eliminate the constants and get from (11.6) a most
important equation of the finite element approximation
(11.18)vi= HT[K]–1D
i= NTD
i
The vector N in (11.18) contains the functions, termed in FEM as
shape functions and in this book we keep the factorization
217
Pauli Pedersen: 11. Finite element analysis
NT= HT[K]–1 (11.19)
because the [K]–1
matrix does not depend on space x . However, the
presentation of FEMfrom the shape function point of view is verymuch
INTERPOLATION used and the relation to the present presentation should therefore be
INTERPRETATION clear. We note that vi= NTD
ican be viewed as an interpolation
from nodal displacements Dito displacement v
ieverywhere in the
element.
ACTUAL
CASES1---D bar elements
LDBA: [K]–1= 1
–1
0
1
QDBA: [K]–1=
1–32
0–12
04
–4
2---D disc elements
LDTR: [K]–1=
z0T
z1Tα
z2T – z
1Tβ
where
z0T= 1 0 0
z1T= –1 1 0
z2T= –1 0 1
218
Pauli Pedersen: 11. Finite element analysis
QDTR: [K]–1=
z0T
z1Tα
z2T – βz
1T
z3Tα2
z4T – βz
5T+ β2z
3T
z5T – 2βz
3Tα
where
z0T= 1
z1T= –3
z2T= –3
z3T= 2
z4T= 2
z5T= 4
0
–1
0
2
0
0
0
0
–1
0
2
0
0
0
0
0
0
4
0
0
4
0
–4
–4
0
4
0
–4
0
–4
3---D volume elements
LDTE: [K]–1=
z0T
z1Tα
z2T – βz
1Tγ
z3T – z
2T+ ηz
1T
where
z0T= 1
z1T= –1
z2T= –1
z3T= –1
0
1
0
0
0
0
1
0
0
0
0
1
219
Pauli Pedersen: 11. Finite element analysis
QDTE:
[K]−1= where
z0T
z1Tα
z2T − βz1Tγ
z3T − z
2T+ ηz
1T
z4Tα2
z5T − βz7
T+ β2z4Tγ2
z6T − z9
T+ 2z5T+ ηz8
T+ η2z4T − ηz7
T
z7T − 2βz4Tαγ
z8T − z7
T+ 2ηz4Tα
z9T − βz8T+ β+ η z7
T − 2z5T − 2βηz4
Tγ
z0T=1
z1T=− 3
z2T=− 3
z3T=− 3
z4T=2
z5T=2
z6T=2
z7T=4
z8T=4
z9T=4
0
− 1
0
0
2
0
0
0
0
0
0
0
− 1
0
0
2
0
0
0
0
0
0
0
− 1
0
0
2
0
0
0
0
4
0
0
− 4
0
0
− 4
− 4
0
0
0
4
0
0
− 4
0
− 4
0
− 4
0
0
0
4
0
0
− 4
0
− 4
− 4
0
0
0
0
0
0
0
4
0
0
0
0
0
0
0
0
0
0
4
0
0
0
0
0
0
0
0
0
0
4
3---D axisymmetric elements
LDRI: [K]–1=
z0T –γ z
2T – βz
1T
z1Tα
z2T – βz
1T
where
z0T= 1
z1T= –1
z2T= –1
0
1
0
0
0
1
220
Pauli Pedersen: 11. Finite element analysis
QDRI:
[K]−1= where
z0T − γ z
2T− βz
1T+ γ
2 z4T − βz
5T+ β2z
3T
z1T− γz
5T− 2βz
3Tα
z2T− βz
1T− 2γ z
4T− βz
5T+ β2z
3T
z3Tα2
z4T− βz
5T+ β2z
3T
z5T− 2βz
3Tα
z0T=
1
z1T=
− 3
z2T=
− 3
z3T=
2
z4T=
2
z5T=
4
0
− 3
0
2
0
0
0
0
− 1
0
2
0
0
0
0
0
0
4
0
0
4
0
− 4
− 4
0
4
0
− 4
0
− 4
11.5 The strain/displacement matrix and general equlibrium
The general equilibrium for a displacement based finite element
approximation can be derived from the principles of virtual work, from
stationary potential energy or fromminimumpotential energy. It is nat-
ural to choose the principle with least assumptions, and thus we choose
the virtual work principle (6.22).
With only nodal forces A we in matrix notation have
VIRTUAL
WORK
PRINCIPLE
V
δTσdV= δDTA (11.20)
221
Pauli Pedersen: 11. Finite element analysis
where δD are kinematical admissible variation of nodal displace-
ments, and δ are the variational strains that follows from δD .
This brings us to the definition of the strain/displacement matrix [B]
one of the most important matrices in finite element analysis.
STRAIN
DISPLACEMENT
MATRIX
δ= [B]δD (11.21)
Normally [B] will be a rectangularmatrix and as stated here in thevari-
ational form it is not based on small strain assumption. Inserting
(11.21) in (11.20) we with δD arbitrary directly get the equation of
general equilibrium
V
[B]TσdV= A (11.22)
which we shall give as an equation for each direction xiby separating
also the strains in parts from each directional variation δDi, i.e.
δ= [B]iδD
i(11.23)
using the summation convention
[B]iδD
i:=
2 or 3
i=1
[B]iδD
i
From also δDTA= δDTiA
iand the arbitrariness of δD
follows the directional general equilibrium
222
Pauli Pedersen: 11. Finite element analysis
GENERAL
EQUILIBRIUMV
[B]TiσdV= A
i (11.24)
Independent of our chosen strain measure the displacement gradients
vi,j constitute the origin of the matrices [B]i . From (11.18) with [K]
and Di independent of xj we have
DISPLACEMENT
GRADIENTSvi,j= HT
,j [K]–1Di (11.25)
thus the essential part of [B]i will be the vector of derivatives HT
,j
which for the specific elements are directly read from (11.7)---(11.14).
In general the matrix [B] that relates absolute displacements D to
final strains
= [B]D (11.26)
will be different from the [B] matrix of (11.23), but with a linear strain
measure like the Cauchy strains, the matrix [B] is constant and then
[B]= [B] (11.27)
For thismost important casewe shall give the [B]i matricesof the listed
elements.
223
Pauli Pedersen: 11. Finite element analysis
WITH
CAUCHY
STRAINS
for 1---D bar elements
BT
1 =H
T
,1[K]
–1
for 2---D disc elements
[B]1=
HT,1
0T
HT,2 2
[K]–1
, [B]2 =
0T
HT,2
HT,1 2
[K]–1
for 3---D volume elements
[B]1 =
HT,1
0T
0T
HT,2 2
HT,3 2
0T
[K]–1
, [B]2 =
0T
HT,2
0T
HT,1 2
0T
HT,3 2
[K]–1
, [B]3 =
0T
0T
HT,3
0T
HT,1 2
HT,2 2
[K]–1
3---D axisymmetric elements
[B]z =
HT,z
0
0
HT,r 2
[K]–1
, [B]r =
0T
HT,r
HTr
HT,z 2
[K]–1
With the specific H vectors and the [K]–1
matrices listed previously,
the [B]ican be derived, say with Mathematica (1992).
224
Pauli Pedersen: 11. Finite element analysis
11.6 Stiffness submatrices and basic matrices
For non---linear problems the equilibriums (11.24) must be solved
iteratively and/or incrementally. With large strains [B] will depend on
the resulting displacements D , with non---linear material σ will
also be a non---linear function of D and even the load A can
depend on D and thus not be known in advance.
However, for linear problem we have given A and
σ= [L]= [L][B]D= [L][B]jDj (11.28)
and then (11.24) can be written in a practical directly solvable form
STIFFNESS
SUBMATRICES
V
[B]Ti[L][B]jdV D
j=Ai
or [S]ijDj=
Ai
with [S]ij := V
[B]Ti[L][B]jdV
(11.29)
This is a most important equation in finite element analysis and it is
here written in terms of stiffness submatrices.
From the form of the strain/displacement matrices (see after (11.27))
it follows that with constant constitutive matrix [L] in an element, the
stiffness submatriceswill be linear combinations of basicmatrices [Tij]
defined by
225
Pauli Pedersen: 11. Finite element analysis
(11.30)[Tij] := [K]
–TV
H,iHT,jdV[K]
–1DEFINITION
OF BASIC
MATRICES
The linear combinations are
for 1---D bar elements
[S]11= E
1111[T
11] (11.31)
with E1111
being Youngs modulus.
for 2---D disc elements
[S]11= C
1111[T
11]+ C
1212[T22
]+ C1112[T12
]+ [T12]T
[S]22= C1212[T11
]+ C2222[T22
]+ C2212[T12
]+ [T12]T (11.32)
[S]12= C1122[T12
]+ C1212[T12
]T+ C1112
[T11]+ C2212
[T22]
with Cijkl as defined in (4.3) or (4.5).
226
Pauli Pedersen: 11. Finite element analysis
for 3---D volume elements
[S]11= L
1111[T
11]+ L
1212[T22
]+ L1313[T33
]
[S]22= L1212[T11
]+ L2222[T22
]+ L2323[T33
]
[S]33= L1313[T11
]+ L2323[T22
]+ L3333[T33
]
(11.33)
[S]12= L1122[T12
]+ L1212[T12
]T
[S]13= L1133[T13
]+ L1313[T13
]T
[S]23= L2233[T23
]+ L2323[T23
]T
with Lijkl as defined in (4.1) or (4.6) for an orthotropic material in the
orthotropic directions.
for 3---D axisymmetric elements
[S]zz= Lzzzz[Tzz]+ Lzrzr[Trr]
[S]rr = Lzrzr[Tzz]+ Lrrrr[Trr]+ Lrrθθ[Tr]+ [Tr]
T+ Lθθθθ
[T] (11.34)
[S]zr= Lzzrr[Tzr]+ Lzrzr[Tzr]
T+ L
zzθθ[Tz]
again restricted to orthotropic material in the orthotropic directions.
For non---axisymmetric deformations of axisymmetric models more
extended results can be found in Ladefoged (1988).
227
Pauli Pedersen: 11. Finite element analysis
We note, that with this basicmatrix approach the influence fromconsti-
tutive parameters are shown directly.
11.7 Analytical integration and resulting basic matrices
Polynomial integration are necessary to derive the basic matrices
(11.30). This is not a simple task even for triangular and tetrahedronal
domains. Traditionally substitutions to area and volume coordinate are
often preferred. However, formulas are available also in Castesian
coordinates and we shall keep this description. With the geometry
description of section 11.2 we have for integration over a triangle of
shape α,β and size h
A
xm2xn1dA= hm+n+2
αn+1n!
(m+ n+ 2)!n
p=0
(m+ p)!
p!βp (11.35)
and for integration over a tetrahedron of shape α,β, γ, δ, and size
h
V
x3xm2xn1dV=
h+m+n+3αn+1γm+1n!m!
(+m+ n+ 3)!×
n
p=0
δp
p!n–p
r=0
βr
r!m
q=0
q
q!
(n – p+m – q+ 2)! (+ p+ q)!
(m – q)!× (11.36)
n–p–r+1
s=0
(–1)s
s! (n – p – r – s+ 1)! (m – q+ r+ s+ 1)
For our practical need the following specific cases is listed
228
Pauli Pedersen: 11. Finite element analysis
Triangular disc element
dV= V
x1hdV
x2hdV
x2
1h2
dV
x2
2h2
dV
x1x2h2
dV
=
=
=
=
=
=
tαh22
Vα(1+ β)3
V3
Vα2(1+ β+ β2)6
V6
Vα(1+ 2β)12
x31h3dV
x32h3dV
x21x2h3dV
x1x22h3dV
x41h4dV
x42h4dV
=
=
=
=
=
=
Vα3(1+ β+ β2+ β3)10
V10
Vα2(1+ 2β+ 3β2)30
Vα(1+ 3β)30
Vα4(1+ β+ β2+ β3+ β4)15
V15
(11.37)
Tetrahedron volume element
dV= V
x1hdV
x2hdV
x3hdV
=
=
=
=
αγh36
Vα(1+ β+ δ)4
Vγ(1+ )4
V4
x21h2dV
x22h2dV
x23h2dV
x1x2h2dV
x1x3h2dV
x2x3h2dV
=
=
=
=
=
=
Vα2(1+ β+ δ+ β2+ δ2+ βδ)10
Vγ2(1+ + 2)10
V10
Vαγ(1+ 2β+ + δ+ β+ 2δ)20
Vα(1+ β+ 2δ)20
Vγ(1+ 2)20
(11.38)
229
Pauli Pedersen: 11. Finite element analysis
Axissymmetric ring element
dV= V
zhdV
rhdV
z2h2dV
r2h2dV
zrh2dV
=
=
=
=
=
=
(παh3)(γ 13)
(παh3)α γ(1+ β)3+ (1+ 2β)12
(παh3)(γ2 2γ3+ 16)
(παh3)α2γ(1+ β+ β2)6 (1+ 2β+ 3β2)30
(παh3)(γ3 γ2+ γ2 110)
(παh3)α γ2(1+ β)3+ γ(1+ 2β)6 (1+ 3β)30
(11.39)
Now, with integration performed the basic matrices (11.30) can be
derived by matrix multiplication and often the final results are rela-
tively simple. Thus we shall list these matrices from which a numerical
finite element program can directly be based.
1---D bar elements (A is bar area, h is bar length)
LDBA: [T11]= A
h 1
–1
–1
1
QDBA: [T11]= A
3h
7
symm.
17
–8–816
230
Pauli Pedersen: 11. Finite element analysis
2---D disc elements (t is disc thickness)
LDTR: [T11]= t
2α
1
symm.
–11
000
[T22]= tα
2
(1 – β)2
symm.
β(1 – β)
β2–(1 – β)
– β
1
[T
12]= t
2
(1 – β)
–(1 – β)
0
β
– β
0
–1
1
0
QDTR: [T11]= t
6α
3
symm.
13
000
0008
000
–88
–4–40008
[T22]= tα
6
3(1 – β)2
symm.
–β(1 – β)
3β2(1 – β)
β
3
0
–4β
–4β
81 – β+ β2
–4(1 – β)
0
–4(1 – β)
8β(1 – β)
81 – β+ β2
4β(1 – β)
4β(1 – β)
0
–8(1 – β)
–8β
81 – β+ β2
and
231
Pauli Pedersen: 11. Finite element analysis
[T12]= t
6
3(1 – β)
(1 – β)
00
0
–4(1 – β)
–β
–3β
00
0
4β
1
–1
04
–4
0
0
4
04(1 – 2β)
–4(1 – 2β)
–4
–4
0
0–4(1 – 2β)
4(1 – 2β)
4
4β
–4(1 – β)
0–4
4
4(1 – 2β)
Note, the independence of the triangle size h .
3---D volume elements (h is tetrahedron size)
LDTE: [T11]=
hγ
6α
1
symm.
–1
1
0
00
0
000
[T
22]= hα
6γ
(1 – β)2
symm.
β(1 – β)
β2–(1 – β)
–β
1
0
0
00
[T33]=
hαγ
6
(1 – + η)2
symm.
–η(1 – + η)
η2(1 – + η)
–η
2
–(1 – + η)
η
–
1
[T12]= h
6
(1 – β)
–(1 – β)
00
β
–β
00
–1
1
00
0
0
00
[T
13]=
hγ
6
(1 – + η)
–(1 – + η)
00
–η
η
00
–
00
–1
1
00
[T23]= hα
6
(1 − + η)(1 − β)
β(1 − + η)
− (1 − + η)
0
− η(1 − β)
− βη
η
0
(1 − β)
β
−
0
− (1 − β)
− β
1
0
Note, the linear dependence of the tetrahedron size h .
232
Pauli Pedersen: 11. Finite element analysis
QDTE:
[T11]=
hγ
30α
3
symm.
1
3
0
0
0
0
0
0
0
–4
–4
0
0
8
–1
1
0
0
0
8
–1
1
0
0
0
4
8
1
–1
0
0
0
–8
–4
8
1
–1
0
0
0
–4
–8
4
8
0
0
0
0
0
0
0
0
0
0
[T22
] = hα
30γ·
31− β2
symm.
− β1− β
3β2
1− β
β
3
0
0
0
0
− 1+ 4β1− β
3− 4ββ
1
0
81− β+ β2
− 4+ β1− β
β2
− 4+ 3β
0
− 8β
81− β+ β2
− 1− β2
− 1− ββ1− β
0
41− β
− 41− ββ
81− β2
1− β2
− 3+ ββ
− 1+ 3β
0
− 81− β
81− ββ
− 41− β2
81− β+ β2
− β1− β
− β2
β
0
4β
− 4β2
8β1− β
− 4β1− β
8β2
1− β
β
− 1
0
− 4
4β
− 81− β
41− β
− 8β
8
233
Pauli Pedersen: 11. Finite element analysis
[T12]= h
30
3(1–β)
(1–β)
00
–4(1–β)
–(1–β)
–(1–β)
(1–β)
(1–β)
0
–β
–3β
004β
–β
–β
β
β
0
1
–1
000
–3
1
3
–1
0
0
0
000
0
0
0
0
0
–1+ 4β
–3+ 4
00
4(1–2β)
4
4
–4
–4
0
–4+ β
–β
004
4(1–2β)
–4β
–4(1–2β)
4β
0
–(1–β)
(1–β)
000
4(1–β)
8(1–β)
–4(1–β)
–8(1–β)
0
(1–β)
3+ β
00–4
–4(1–2β)
–4(1–β)
4(1–2β)
4(1–β)
0
–β
β
000
4β
8β
–4β
–8β
0
1
–1
000
–4
–8
4
8
0
T13=
hγ
30
3
0
0
–4
–
–
0
η
3η
0
0
–4η
η
η
–η
–η
0
–
0
0
0
3
–
–3
0
1
–1
0
0
0
1
–3
–1
3
0
–+ 3η
–3+ η
0
0
4+ η
41 –
41 –
–41 –
–41 –
0
– – 3
1+ η
0
0
–4
42+
41+ η
4–2+ – 2η
–41+ η
0
–+ 3
η –
0
0
4
4η –
42 – 1
–4η –
4–1+ 2 – 2η
0
η –
–3+ η
0
0
4
–42η –
–4η –
42η –
4η –
0
1+ η
3 – η
0
0
–4
–41+ η
–41+ 2η
41+ η
41+ 2η
0
1 –
–1 –
0
0
0
4 – 2
42 – 1
42 –
41 – 2
0
where = 1 – + η
234
Pauli Pedersen: 11. Finite element analysis
T33
= hαγ
30
32
symm.
η
3η2−
η
32
− η
3
− + 3η
− 3+ ηη
− (1 − )
(1 − )
82+ η2 − η
− ( − 3)
1+ ηη
(3 − )
1+ η
4 − 2η
82+ 2+
− (+ 3)
η − η
− η −
− (3+ 1)
− 4 − 2η
4η − 2
81 − + 2
η −
− 3+ ηη
− 3η+
− η −
42 − η
− 42η −
− 4η − 2
8η2+ 2 − η
1+ η
3 − ηη1+ η
3η − 1
− 42 − η
− 41+ η2
− 42η −
4η − 2
81+ η+ η2
(1 − )
− (1 − )η
− (3+ )
− (1+ 3)
− 4(1 − )2
− 42 − η
42 − η
− 4 − 2η
4 − 2η
81 − + 2
[T23]= hα
30
31 − β
− β
0
− 1 − 4β
− 4 − β
− 1 − β
1 − β
− β
η1 − β
− 3ηβ
− η
0
η− 3+ 4β
− ηβ
η1 − β
η3+ β
ηβ
− η
− 1 − β
− β
− 3
0
−
4 − 3β
− 1 − β
1+ 3β
− β
1 − β
β
− 1
0
1
− β
− 31 − β
− 1 − β
− 3β
3
− + 3η1 − β
3+ ηβ
(1 − )
0
4 2 − η − + ηβ
4η− β+ [β
4(1 − )1 − β
− 42 − η − (1 − )β
4(1 − )β
− 4(1 − )
− ( − 3)1 − β
− 1+ ηβ
− 3+
0
41+ η+ β
4 − − (2+ )β
41+ η 1 − β
− 41+ η − (2+ )β
41+ ηβ
− 41+ η
− ( − 3)1 − β
− 1+ ηβ
− 3+
0
41+ η+ β
4 − − (2+ )β
41+ η 1 − β
− 41+ η − (2+ )β
41+ ηβ
− 41+ η
− (+ 3) 1 − β
− η − βη −
0
4η − − β
41 − ηβ+ β
4(2 − 1)1 − β
− 4η − 1 − β
41 − 2+ 2ηβ
− 41 − 2+ 2η
η − 1 − β
η+ 3 β3η+
0
− 4η − 2+ β
− 4η − 2ηβ+ β
− 4η − 1 − β
4 η − 2+ − 2η β
− 4η − β
4η −
1+ η 1 − β
η − 3β
− 1+ η
0
− 42 − β+ η
41+ ηβ
− 41+ 2η 1 − β
4 2+ η− 1+ ηβ
− 41+ 2ηβ
41+ 2η
(1 − )1 − β
(1 − )β
3+
0
− 4(1 − )
− 41 − 2β+ β
− 4(1 − 2) 1 − β
41 − − (2 − )β
− 4(1 − 2)β
4(1 − 2)
235
Pauli Pedersen: 11. Finite element analysis
3---D axisymmetric elements (h is ring thickness in radial direction)
LDRI:
Tzz
= πh3α
3γ 1
1
symm.
–11
000
Trr
= πhα
33γ 1
1 – β2
symm.
β1 – β
β2
–1 – β
– β
1
Tzr=
πh3
3γ 1
1 – β
–1 – β
0
β
– β
0
– 1
1
0
T
z
=
πh3
–1–1–1
111
000
Tr= πhα
3
–1 – β
–1 – β
–1 – β
– β
– β
– β
1
1
1
T= πhα
6
2δ1
symm.
δ1
2δ1
δ2
δ2
δ3
The components of the [T] matrix can be found by series expansions,
and are listed in Pedersen and Megahed (1975).
QDRI:
Tzz
=
πh
15α
5γ
3
symm.
13
000
0008
000− 88
− 4− 40008
3
symm.
13
000
2− 2024
− 220− 2424
− 4− 40008
236
Pauli Pedersen: 11. Finite element analysis
Trr
=
πhα15
5γ
31 – β2
symm.
–β1 – β
3β2
1 – β
β
3
0
–4β
–4β
81 – β+ β2
–41 – β
0
–41 – β
8β1 – β
81 – β+ β2
4β1 – β
4β1 – β
0
–81 – β
–8β
81 – β+ β2
31 – β2
symm.
–β1 – β
3β2
21 – β
2β
9
1 – 3β+ 2β2
–3β – 2β2
3 – 14β
81 – 2β+ 3β2
–5+ 7β – 2β2
– β+ 2β2
–11+ 14β
4–1+ 6β – 6β2
82 – 4β+ 3β2
–1+ 5β – 4β2
3β – 4β2
–3
–42 – 3β
41 – 3β
81 – β+ β2
Tzr
=
πh15
5γ
31 – β
1 – β
0
0
0
–41 – β
–β
–3β
0
0
0
4β
1
–1
0
4
–4
0
0
4
0
41 – 2β
–41 – 2β
–4
–4
0
0
–41 – 2β
41 – 2β
4
4β
–41 – β
0
–4
4
41 – 2β
31 – β
1 – β
0
21 – β
–21 – β
–41 – β
–β
–3β
0
2β
–2β
4β
2
–2
0
14
–14
0
1 – 2β
3+ 2β
0
81 – 3β
–81 – 3β
–4
–5+ 2β
1 – 2β
0
–82 – 3β
82 – 3β
4
–1+ 4β
– 3+ 4β
0
– 8
8
41 – 2β
237
Pauli Pedersen: 11. Finite element analysis
Tz
=
πh15
–2111–3–3
–12–13–13
000000
–1–12884
11–2–8–8–4
3–30–440
T=
πhα30
δ1
symm.
δ2
δ1
δ3
δ3
δ4
δ5
δ6
δ8
− 8δ5
δ6
δ5
δ8
− 4δ5
− 8δ5
δ7
δ7
δ9
δ10
δ10
δ11
Tr= πhα
15
–21 – β
1 – β
1 – β
1 – β
– 31 – β
– 31 – β
– β
– 2β
β
–3β
β
–3β
–1
– 1
2
3
3
–1
–1 – β
2+ β
–1 – 2β
81 – β
41 – 2β
42 – β
3 – β
–β
–3+ 2β
–41 – 2β
8β
41+ β
1 – 3β
– 2+ 3β
1
– 42 – β
– 41+ β
–8
The components of the [T] matrix can be found by series expansions,
and are listed in Pedersen andMegahed (1975).More extended results
can be found in Ladefoged (1988).
How to control these basis matrices? Several possibilities exist, saywith
Mathematica (1992) or an alternative symbolic computer program. A
simple test is performed by equilibrium tests (except for axisymmetric
r---direction). The stiffness submatrices must by physical arguments
have sum of columns equal to zero. Because this must hold for any
material it must hold for the individual basis matrices. Finally because
[S]ij=
[S]Tji it must also holds in relations to row sums. Thus at least
two errors must be involved if this simple test will not locate an error.
238
Pauli Pedersen: 11. Finite element analysis
11.8 Equivalent nodal loads and consistent mass matrices
The application of virtual work principle in (11.20) was based on only
nodal forces A . If volume forces pi in direction xi is also actual
we have to add the following virtual work
V
δvipidV= δDT
i[K]−T
V
HpidV (11.40)
where δvi= HT[K]
–1δDi= δD
T
i[K]
–TH follows from
(11.15). To the general equilibrium (11.24) or the linear case (11.29)we
must therefore add the following equivalent nodal forces:
EQUIVALENCE
OF DISTRIBUTED
FORCES
Api= [K]
–TV
HpidV (11.41)
Forces derived from (11.41) are consistent with virtual work principle
in the sence that the work of these nodal loads equals the work of the
distributed loads. Surface tractions and line loads can be treated as spe-
cial cases of (11.41) with integration over area and line as alternative to
the volume integration in (11.41).
The field of volume forces pi = p i(x) can in (11.41) be arbitrary.How-
ever, it is natural to restrict the complexity of the force field to that of
the assumed displacement field. In this light we describe pi = p i(x) by
pi = HTi (11.42)
239
Pauli Pedersen: 11. Finite element analysis
where the vector of constants iare the given specification of the
force field. Introducing (11.42) to (11.41) we get
Api= [K]
–TV
HHTdVi= [P]
i(11.43)
with the load matrix [P] defined by
[P] := [K]–T
V
HHTdV (11.44)
Note that [P][K]–1 is a basic matrix, which with constant mass density
gives themassmatrix [M]= [P][K]–1 . The resultingmassmatrices
are more simple than the load matrices, and we shall therefore not list
the [P] matrices. They can simply be obtained from [P]= [M][K] ,
or they can be found in Pedersen (1984).
INERTIA Although this book is not dealing with dynamics, it is natural here to
FORCES treat the case of inertia loads. The volume forces from inertia are
pi=− ω
2vi=− ω
2HT[K]–1D
i(11.45)
Inserting this in (11.41) we get the nodal loads equivalent to inertia
Api= ω2[K]
–TV
HHTdV[K]–1D
i= ω2[M]D
i (11.46)
by which the consistent mass matrix [M] is defined
240
Pauli Pedersen: 11. Finite element analysis
MASS
MATRIX[M] := [K]
–TV
HHTdV[K]–1= [P][K]
–1(11.47)
and thenwith the last equality equal for constant a basicmatrix [T00]
similar to the definitions of [Tij] in (11.30). The resultingmassmatrices
are more simple than the load matrices [P] and we shall therefore list
them for the treated elements.
1---D bar elements mass == Ah
LDBA:
[M]=621
12
QDBA:
[M]=30
4–12
–142
2216
2---D disc elements = th2α2
LDTR:
[M]=12
211
121
112
QDTR:
[M]= 180
6–1–1–400
–16–10–40
–1–1600–4
–400321616
0–40163216
00–4161632
241
Pauli Pedersen: 11. Finite element analysis
3---D volume elements = h3αγ6
LDTE:
[M]=20
2111
1211
1121
1112
QDTE:
[M]= 420
6111–4–4–4–6–6–6
1611–4–6–6–4–4–6
1161–6–4–6–4–6–4
1116–6–6–4–6–4–4
–4–4–6–632161616168
–4–6–4–616321616816
–4–6–6–416163281616
–6–4–4–616168321616
–6–4–6–416816163216
–6–6–4–481616161632
3---D axisymmetric elements = πh3α(3γ 1)3
LDRI:
[M]=
20(3γ 1)
5γ
211
121
112
212
122
226
QDRI:
[M]=
(3γ 1)
γ
60
6–1–1–400
–16–10–40
–1–1600–4
–400321616
0–40163216
00
–4161632
1
420
61–4–12–8–4
16–4–8–12–4
–4–4301212–4
–12–812964832
–8–1212489632
–4–4–4323232
242
Pauli Pedersen: 11. Finite element analysis
INITIAL The constitutive relation (11.28) assumes that the element before load
STRESSES has no stresses and strains. If we want to include initial stresses σ0
AND STRAINS and/or initial strains 0 the generalized Hooke’s law is
σ= σ0+ [L]−
0 (11.48)
Inserting this in the general equilibrium (11.24) we with known terms
on the right hand side get
V
[B]T
i[L]dV= A
i−
V
[B]T
iσ
0dV+
V
[B]T
i[L]
0dV (11.49)
EQUIVALENT Therefore the nodal loads equivalent to initial stresses are
NODAL LOADS
Aσ0
i=−
V
[B]T
iσ
0dV (11.50)
and the nodal loads equivalent to initial strains are
A0
i=
V
[B]T
i[L]
0dV (11.51)
243
Pauli Pedersen: 11. Finite element analysis
11.9 Note on strain evaluation
In this section we shall see the importance of consistency, when finally
evaluating strains and stresses.We shall limit the description to the sim-
ple LDTR---element, although the consistency problem is general.
DISPLACEMENTS By (11.18), (11.9) and the actual [K]–1 matrix we have for this element
vi= 1 x
1
h
x2
h
1
–1α
–(1− β)
0
1α
–β
0
0
1
dI
dII
dIII
i
(11.52)
with definition of the nodal displacements and the geometry parame-
ters as shown in fig. 11.1.
Fig. 11.1: LDTR---element with parameter definitions.
244
Pauli Pedersen: 11. Finite element analysis
DISPLACEMENT From (11.52) follows directly the displacement gradients
GRADIENTS
v1,1= ∆a(αh)
v1,2= (∆b− β∆a)h
v2,1= ∆c(αh)
v2,2= (∆d− β∆c)h
,
,
,
,
∆a := (dII− dI)1
∆b := (dIII− dI)1
∆c := (dII− dI)2
∆d := (dIII− dI)2
(11.53)
with increments in nodal displacements ∆a , ∆b , ∆c , ∆d as defined.
We can then express the Cauchy strains in the increments of the nodal
displacements and get
CAUCHY
STRAINS
ij
11
22
212
:=
=
=
=
12(vi,j+ vj,i)⇒
∆a(αh)
(∆d− β∆c)h
(α∆b− βα∆a+ ∆c)(αh)
(11.54)
It follows that as a result of a rigidbody translationwewill get no strains
∆a= ∆b= ∆c = ∆
d= 0 ⇒
11=
22=
12= 0 (11.55)
However, this will not be the case for a rigid body rotation, and we shall
analyse this in details. In fig. 11.2 is shown the necessary geometry to
evaluate the nodal displacements, resulting from a rigid body rotation.
245
Pauli Pedersen: 11. Finite element analysis
Fig. 11.2: Geometry of a rigid body rotation of the triangular element.
Measured in the x---coordinate system the increments of the nodal dis-
placements are
RIGID BODY
ROTATION
∆a
∆b
∆c
∆d
=
=
=
=
(r2− r
1)(cos− 1)
r3(cos θ cos− sinθ sin− cos θ)− r
1(cos− 1)
(r2− r
1) sin
r3(sin θ cos+ cosθ sin− sinθ)− r
1sin
(11.56)
and from the further geometry relations then follow
246
Pauli Pedersen: 11. Finite element analysis
(r2− r
1)= αh
(r3cos θ− r
1)= βαh
r3sin θ= h
⇒
∆a
∆b
∆c
∆d
=
=
=
=
αh(cos− 1)
βαh(cos− 1)− h sin
αh sin
βαh sin+ h(cos–1)
(11.57)
STRAIN ERRORS
BY RIGID BODY Inserting the result (11.57) in the Cauchy strain measure (11.54) we get
ROTATION
11= cos− 1 , 22= cos− 1 , 12= 0 (11.58)
COMPRESSIVE i.e. an error corresponding to compressive dilatation. This error is by
DILATATION no means small; for just a rotation angle of 3o we get from (11.58)
11= 22=− 0.0014 , i.e. close to 0.2% strain. The error is indepen-
dent of the geometry parameters α , β and h of the triangle, and with
an alternative element we will get similar results.
In fig. 11.2 is also shown the nodal displacements evaluated in the
y---coordinate system (rotated system). Using these displacements for
EXPANSIVE the strain evaluation wewill get 11= 22= 1− cos , 12= 0 , i.e.
DILATATION an expansive dilatation of the same size.
The conclusion at this pointmay easily be that a Cauchy strainmeasure
should not be used. Before explaining why reality is not so bad as
expected from (11.58), we shall analyse rigid body rotations relative to
GREEN--- the Green---Lagrange strain definition
LAGRANGE
STRAINS ηij :=12v
i,j+ vj,i+ vk,ivk,j (11.59)
247
Pauli Pedersen: 11. Finite element analysis
Inserting the displacement gradients (11.53) in (11.59) we get
η11
η22
2η12
=
=
=
∆a(αh)+ (∆2a+ ∆2
c)(2α2h2)
(∆d− β∆c)h+ ∆2
b+ ∆2
d+ β2(∆2
a+ ∆2c)− 2β(∆a∆b
+ ∆c∆d)(2h2)
(α∆b− βα∆a+ ∆c)(αh)+ ∆a∆b
+ ∆c∆d− β(∆2
a+ ∆2c)(αh2)
(11.60)
and we will see that also the rigid body rotation (11.57) will give zero
strains η11= η
22= η
12= 0 , as expected from the quadratic strain
definition. However, a non---linear strain definition implies that finite
element solutions can only be obtained iteratively and/or incremen-
tally.
Fig. 11.3: Cantilever test example.
TEST EXAMPLE A test example based on Cauchy strains like the cantilever problem in
fig. 11.3will not give the errors, indicated in (11.58). Independent of the
rotation at the free end of the beam, there will only be strains from the
fixed end to the applied force. This could also be argued from the linear
nature of the solution. The rotation in the free end part can therefore
not be a pure rigid body rotation. From the inverse point of view we can
determine the increments of nodal displacements that will give zero
strains
248
Pauli Pedersen: 11. Finite element analysis
11
22
12
=
=
=
0
0
0
⇒
⇒
⇒
∆a
∆d
∆c
=
=
=
0
β∆c
− α∆b
(11.61)
which means that ∆bcan be treated as a free parameter. A closer look
at the result of the linear finite element solution will show thatwe in the
strain free part have a rigid body rotation superposedwith an expansive
dilatation that compensates the error (11.58). A simple displacement
plot can show this.
A suggestion to evaluate strains by the Green---Lagrange definition
when Cauchy strains are the basis for the obtained displacements, will
by (11.61) return the errors
η11
η22
2η12
=
=
=
∆2c(2α
2h2)= ∆2
b(2h2)
(∆2
b+ ∆2
d+ β2∆2
c− 2β∆c∆d)(2h2)= ∆2
b(2h2)
(∆c∆d− β∆2
c)(αh2)= 0
(11.62)
CONCLUSION Thus the conclusion of this note is that evaluation of resulting strains
must be consistent with the strain definition behind the element stiff---
WORK/ENERGY ness matrix. Balance of work by external forces and elastic energy
BALANCE should always be established. With a linear formulation the displace-
ment field includes an artificial dilatation field that preserves this bal-
ance.
249
Pauli Pedersen: 12. An index to matrices
12. AN INDEX TO MATRICES
--- definitions, facts and rules ---
This index is based on the following goals and observations:
¯Togive the userquick reference to anactualmatrix definition or rule,
the index form is preferred. However, the index should to a large
extent be self-explaining.
¯The contents is selected in relation to the importance for matrix for-
mulations in solid mechanics.
¯The existence of good computer software for the numerical calcula-
tions, diminishes the need for details on specific procedures.
¯The existence of good computer software for the formulamanipula-
tions means that extended analytical work is possible.
¯The index is written by a non---mathematician (but hopefully with-
out errors), and is written for readers with a primary interest in
applying the matrix formulation without studying the matrix theory
itself.
¯Available chapters or appendices in books on solid mechanics are
not found extensive enough, and good classic books on linear alge-
bra are found too extensive. For further reference, see e.g.
250
Pauli Pedersen: 12. An index to matrices
Gantmacher, F.R. (1959) ‘The Theory of Matrices’,
Chelsea Publ. Co., Vol. I, 374 p., Vol. II, 276 p.
Gel’fand, I.M. (1961) ‘Lectures on Linear Algebra’,
Interscience Publ. Inc., 185 p.
Muir, T. (1928) ‘A Treatise on the Theory of Determinants’,
Dover Publ. Inc., 766 p.
Noble, B. and Daniel, I.W. (1988) ‘Applied Linear Algebra’,
Prentice---Hall, third ed., 521 p.
Strang, G. (1988) ‘Linear Algebra and its Applications’,
Harcourt Brace Jovanovich, 505 p.
Strang, G. (1986) ‘Introduction to Applied Mathematics’,
Wellesley---Cambridge Press, 758 p.
It will be noticed that the rather lengthy notation with [ ] for matrices
and for vectors (columnmatrices) is preferred for the more simple
boldface or underscore notations. The reason for this is that the reader
by the brackets is constantly remindedabout the fact thatwe aredealing
with a blockof quantities.Tomiss this point is catastrophic inmatrix cal-
culations. Furthermore, the lengthy notation adds to the possibilities
for direct graphical interpretation of the formulas.
Cross-reference in the index is symbolized by boldface writings. The
preliminary advices from colleagues and students are verymuch appre-
ciated, and I shall be grateful for further critics and comments that can
improve the index.
251
Pauli Pedersen: 12. An index to matrices
ADDITION Matrices are added by adding the corresponding elements
of matrices
[C]= [A]+ [B] with Cij= Aij+ Bij
The matrices must have the same order.
ANTI--METRIC or See skew---symmetric matrix.
ANTI--SYMMETRIC
matrix
BILINEAR FORM For a matrix [A] we define the bilinear form by
XT[A]Y
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Pauli Pedersen: 12. An index to matrices
BILINEAR For a symmetric, positive definitematrix [A] we have by definition for
INEQUALITY the following two quadratic forms:
XaT[A]Xa
= ua> 0 for Xa≠ 0
XbT[A]X
b= u
b> 0 for X
b≠ 0
The bilinear form fulfills the inequality
XaT[A]X
b≤ 1
2(ua+ u
b)
i.e. less than or equal to the mean value of the values of the quadratic
forms.
This follows directly from
XaT–X
bT[A]Xa–Xb
≥ 0
and only equality for Xa= Xb . Expanding we get with the defini-
tions
ua+ ub–2Xa
T[A]Xb≥ 0
because [A]T= [A] .
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Pauli Pedersen: 12. An index to matrices
BIORTHOGONALITY From the description of the generalized eigenvalue problem (see this)
conditions with right and left eigenvectors Φiand Ψ
iwe have
ΨTj [A] – λ
i[B]Φ
i= 0
and
ΨTj[A] – λ
j[B]Φi= 0
which by subtraction gives
(λi– λj)ΨT
j [B]Φi= 0
For different eigenvalues
λi≠ λj
this implies
ΨTj [B]Φ
i= 0
and thus also
ΨTj [A]Φ
i = 0
which is termed the biorthogonality conditions.
For a symmetric eigenvalue problem Ψi=
Φi (see orthogonality
conditions).
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Pauli Pedersen: 12. An index to matrices
CHARACTERISTIC From the determinant condition
POLYNOMIUM
(generalized) |[A]λ2+ [B]λ+ [C]|= 0
with the square matrices [A] , [B] and [C] all of order n we obtain
a polynomium of order 2n in λ . This polynomium is termed the char-
acteristic polynomium of the triple ([A] , [B] , [C]).
Specific cases as
|[A]λ2+ [C]|= 0
|[I]λ+ [C]|= 0
are often encountered.
CHOLESKI See factorization of a matrix.
factorization /
triangularization
COEFFICIENTS See elements of a matrix.
of a matrix
COFACTOR The cofactor of a matrix element is the corresponding minor with an
of a matrix element appropriate sign. If the sum of row and column indices for the matrix
element is even, the cofactor is equal to theminor. If this sum is odd the
cofactor is the minor with reversed sign, i.e.
Cofactor (Aij)= (–1)i+j Minor (Aij)
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Pauli Pedersen: 12. An index to matrices
COLUMN A column matrix is a matrix with only one column, i.e. order m× 1 .
matrix The notation is used for a columnmatrix. The name columnvector
or just vector is also used.
CONGRUENCE Acongruence transformation of a squarematrix [A] to a squarematrix
transformation [B] of the same order is by the regular transformation matrix [T] of
the same order
[B]= [T]T[A][T]
Matrices [A] and [B] are said to be congruent matrices, they have the
same rank and the same definiteness, but not necessarily same eigenva-
lues. A congruence transformation is also an equivalence transforma-
tion.
CONJUGATE The conjugate transpose is a transformation of matrices with complex
TRANSPOSE elements. Complex conjugate is denoted by a bar and transpose by a
superscript T . With a short notation (from the name Hermitian) we
denote the combined transformation as
[A]H= [A]
T
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Pauli Pedersen: 12. An index to matrices
CONTRACTED For a symmetricmatrix, a simpler contracted notation in terms of a row
NOTATION or columnmatrix is possible.Of the notations which keep the orthogo---
for a symmetric matrix nal transformation, we choose the form with 2 ---factors multiplied to
the off diagonal elements in the matrix, i.e.
B from [A] with
Bi= A
iifor i= 1, 2, ..., n
Bn+...= 2 Aij for j> i
(The ordering within B symbolized by n+... is not specified).
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Pauli Pedersen: 12. An index to matrices
CONVEX SPACE For a symmetric, positive definitematrix [A] we have by definition for
by positive the following two quadratic forms:
definite matrix
XaT[A]Xa
= ua ; 0< ua
XbT[A]X
b= u
b; 0< u
b≤ ua
The matrix [A] describes a convex space such that for
Xα= αXa+ (1 – α)X
b ; 0≤ α≤ 1
we have for all values of α
XαT[A]Xα
= uα≤ ua
Inserting directly we have with [A]T
= [A]
αXaT+ (1 – α)X
bT[A]αXa
+ (1 – α)Xb
= α2Xa
T[A]Xa+ (1 – α)2X
bT[A]X
b+ 2α(1 – α)Xa
T[A]Xb
= α2ua+ (1 – α)2u
b+ 2α(1 – α)Xa
T[A]Xb
From the bilinear inequality we have
XaT[A]X
b≤ 1
2(ua+ u
b)
and thus with ub≤ ua we can substitutive greater values and obtain
XαT[A]Xα
≤ α2ua+ (1 – α)2ua+ 2α(1 – α)ua= ua
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Pauli Pedersen: 12. An index to matrices
DEFINITENESS
For a symmetric matrix the notions of: are used if, for the matrix:
¯ positive definite ¯ all eigenvalues are positive
¯ positive semi---definite ¯ eigenvalues non---negative
¯ negative definite ¯ all eigenvalues are negative
¯ negative semi---definite ¯ eigenvalues non---positive
¯ indefinite ¯ both positive and negative eigenvalues
See specifically positive definite, negative definite and indefinite for
alternative statement of these conditions.
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Pauli Pedersen: 12. An index to matrices
DETERMINANT The determinant of a squarematrix is a scalar, calculated as the sum of
of a matrix products of elements from the matrix. The symbol of two vertical lines
det ([A])= |[A]|
is used for this quantity.
For a square matrix of order two the determinant is
|[A]|=
A11
A21
A12
A22
=A
11A
22– A
12A
21
For a square matrix of order three the determinant is
|[A]|=
A11
A21
A31
A12
A22
A32
A13
A23
A33
=
A11A
22A
33+A
12A
23A
31+A
13A
21A
32– A
31A
22A
13– A
32A
23A
11– A
33A
21A
12
We note that for each product the number of elements is equal to the
order of the matrix, and that in each product a row or a column is only
represented by one element. Totally for a matrix of order n there are
n! terms to be summed.
For further calculation procedures see determinants by minors/cofac-
tors.
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Pauli Pedersen: 12. An index to matrices
DETERMINANTS A determinant can be calculated in terms of cofactors (or minors), by
BY MINORS / expansion in terms of an arbitrary row or column.
COFACTORS
As an example, for a matrix of order three expansion of the third col-
umn yields:
A11
A21
A31
A12
A22
A32
A13
A23
A33
= A
13Minor(A
13) – A
23Minor(A
23)+A
33Minor(A
33)
See determinant of a matrix for direct comparison.
DETERMINANT The product of the determinants for a regular matrix [A] and its
OF AN INVERSE inverse [A]–1 is equal to 1
matrix
|[A]–1|= 1|[A]|
DETERMINANT The determinant of a product of squarematrices is equal to the product
OF A PRODUCT of the individual determinants, i.e.
of matrices
|[A][B]| = |[A]||[B]|
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Pauli Pedersen: 12. An index to matrices
DETERMINANT The determinant of transposed square matrix is equal to the deter---
OF A TRANSPOSED minant of the matrix itself, i.e.
matrix
|[A]T| = |[A]|
DIAGONAL A diagonal matrix is a matrix where all off diagonal elements have the
matrix value zero
[A] a diagonal matrix when Aij = 0 for i ≠ j
and at least one diagonal element is non---zero. This definition also
holds for non---square matrices, as by singular value decomposition.
DIFFERENTIAL See functional matrix.
matrix
DIFFERENTIATION Differentiation of a matrix is carried out by differentiation of each
of a matrix element
[C]= d([A])db with Cij= d(Aij)db
DIMENSIONS See order of a matrix.
of a matrix
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Pauli Pedersen: 12. An index to matrices
DOT PRODUCT See scalar product of two vectors.
of two vectors
DYADIC PRODUCT The dyadic product of two vectors A and B of the same order
of two vectors n results in a square matrix [C] of order n× n , but only with rank
1
[C]= ABT with Cij= A iBj
Dyadic products of vectors of different order can also be defined,
resulting in a matrix of order m× n .
EIGENPAIR The eigenpair λi , Φi is a solution to an eigenvalue problem. The
eigenvector Φicorresponds to the eigenvalue λ
i.
EIGENVALUES The eigenvalues λi
of a square matrix [A] are the solutions to the
of a matrix standard form for the eigenvalue problem, with
([A] – λi[I])Φ
i= 0⇒ |[A] – λ
i[I]|= 0
which gives a characteristic polynomium.
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Pauli Pedersen: 12. An index to matrices
EIGENVALUE With [A] and [B] being two squarematrices of order n , the general---
PROBLEM ized eigenvalue problem is defined by
[A] – λi[B]Φ
i= 0 for i= 1, 2, ..., n
or by
ΨTi[A] – λ
i[B]= 0
Tfor i= 1, 2, ..., n
The pairs of eigenvalue, eigenvectors are λi, Φ
iand λ
i, ΨT
i with
Φias right eigenvector and Ψ
ias lefteigenvector. Theeigenvalue
problem has n solutions with possibility for multiplicity.
With [B] being an identity matrix we have the standard form for an
eigenvalue problem, while for [B] not being an identity matrix the
name generalized eigenvalue problem is used.
EIGENVECTOR An eigenvector Φiis the vector---part of a solution to an eigenvalue
problem. The word eigen reflects the fact that the vector is transformed
into itself except for a factor, the eigenvalue λi.
ELEMENTS The elements of amatrix [A] are the individual entries Aij . In amatrix
of a matrix of order m× n there are mn elements Aij , for i= 1, 2, ...,m ,
j= 1, 2, ..., n .Elements are also called themembersor the coefficients
of the matrix.
EQUALITY Twomatrices of the sameorder are equal if the corresponding elements
of matrices of each of the matrices are equal, i.e.
[A]= [B] if Aij= B ij for all ij
264
Pauli Pedersen: 12. An index to matrices
EQUIVALENCE An equivalence transformation of a matrix [A] to a matrix [B] (not
transformations necessarily square matrices) by the two square, regular transformation
matrices [T1] and [T
2] is
[B]= [T1][A][T2
]
Matrices [A] and [B] are said to be equivalent matrices and have the
same rank.
EXPONENTIAL The exponential of a square matrix [A] is defined by its power series
of a matrix expansion
e[A]t := [I]+ [A]t+ [A]2 t2
2!+ [A]
3 t3
3!+
The series always converges, and the exponential properties are kept,
i.e.
e[A]te[A]s= e[A](t+s) , e[A]te[A](–t)
= [I] , de[A]tdt= [A]e[A]t
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Pauli Pedersen: 12. An index to matrices
FACTORIZATION A symmetric, regularmatrix [A] of order n can be factorized into the
of a matrix product of a lower triangular matrix [L] , a diagonalmatrix [B] and
the upper triangular matrix [L]T
all of the order n
[A]= [L][B][L]T
In a Gauss factorization the diagonal elements of [L] are all 1 .
A Choleski factorization is only possible for positive semi---definite
matrices, and then [B]= [I] and we get
[A]= [L][L]T
with Lii
not necessarily equal to 1 .
FROBENIUS The Frobenius norm of a matrix [A] is defined as the square root of
norm of a matrix the sum of the squares of all the elements of [A] .
For a square matrix of order 2 we get
Frobenius= A2
11+ A2
22+A2
12+ A2
21
and thus for a symmetricmatrix equal to the squareroot of the invariant
I3.
For a square matrix of order 3 we get
Frobenius= (A2
11+A2
21+A2
31)+ (A2
22+A2
12+A2
32)+ (A2
33+A2
13+ A2
23)½
and thus for a symmetricmatrix equal to the squareroot of the invariant
I4.
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Pauli Pedersen: 12. An index to matrices
FULL RANK See rank of a matrix.
FUNCTIONAL The functional matrix [G] consists of partial derivatives --- the partial
MATRIX derivatives of the elements of a vector A of order m with respect
to the elements of a vector B of order n . Thus the functionalmatrix
is of the order m× n
[G]=∂A
∂Bwith G
ij=∂Ai
∂Bj
The name gradient matrix is also used. A square functional matrix is
named a Jacobimatrix, and the determinant of this matrix as the Jaco-
bian.
GAUSS See factorization of a matrix.
factorization /
triangularization
GENERALIZED See eigenvalue problem.
EIGENVALUE
PROBLEM
GEOMETRIC A vector of order two or three in an Euclidian plane or space. See vec---
vector tors. By a geometric vector we mean a oriented piece of a line (an
“arrow”).
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Pauli Pedersen: 12. An index to matrices
GRADIENT See functional matrix.
matrix
HERMITIAN A square matrix [A] is termed Hermitian if it is not changed by the
matrix conjugate transpose transformation, i.e.
[A]H= [A]
Every eigenvalue of aHermitianmatrix is real, and the eigenvectors are
mutually orthogonal, as for symmetric real matrices.
HESSIAN AHessian matrix [H] is a square, symmetricmatrix containing second
matrix order derivatives of a scalar F with respect to the vector A
[H]= ∂2F∂A∂A
with Hij =∂2F
∂A i∂A j
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Pauli Pedersen: 12. An index to matrices
HURWITZ The Hurwitz determinants up to order eight are defined by
determinants
Hi:=
a1
a0
a3
a2
a1
a0
a5
a4
a3
a2
a1
a0
a7
a6
a5
a4
a3
a2
a1
a0
a8
a7
a6
a5
a4
a3
a2
a8
a7
a6
a5
a4
a8
a7
a6
a8
to be read in the sense that Hiis the determinant of order i defined
in the upper left corner (principal submatrix). More specifically,
H1= a
1
H2= a
1a2– a
0a3
H3= H
2a3– (a
1a4– a
0a5)a
1
··
If the highest order is n , then am = 0 for m> n , and therefore the
highest Hurwitz determinant is given by
Hn= Hn–1
an
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Pauli Pedersen: 12. An index to matrices
IDENTITY An identity matrix [I] is a square matrix where all diagonal elements
matrix have the value one and all off diagonal elements have the value zero
[I] := [A] with Aii= 1, A
ij= 0 for i≠ j
The name unit matrix is also used for the identity matrix.
INDEFINITE Asquare, realmatrix [A] is called indefinite if positive aswell asnega---
matrix tive values of XT[A]X exist, i.e.
XT[A]X ><
0
depending on the actual vector (column matrix) X .
INTEGRATION The integral of a matrix is the integral of each element
of a matrix
[C]= [A]dx with Cij=Aijdx
INVARIANTS For matrices which transforms by similarity transformations we can
of similar matrices determine a number of invariants, i.e. scalars which do not change by
the transformation. The number of independent invariants are equal to
the order of the matrix, and as any combination is also an invariant
many different forms are possible. Tomention some important invaria-
nts we have eigenvalues, trace, determinant, and Frobenius norm. The
principal invariants are the coefficients of the characteristic polyno-
mium.
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Pauli Pedersen: 12. An index to matrices
INVARIANTS For the square, symmetric matrix [A] of order 2 we have
of symmetric, similar
matrices of order 2
[A]= A11
A12
A12
A22
with invariants being the trace I
1by
I1= A
11+A
22
and the determinant I2by
I2= A
11A22
– A2
12
Taking as an alternative invariant I3by
I3= (I
1)2 – 2I
2= A2
11+A2
22+ 2A2
12
we get the squared length of the vector A contracted from [A] by
AT= A11, A
22, 2 A
12
Setting up the polynomium to find the eigenvalues of [A] we find
λ2 – I
1λ+ I
2= 0
and again see the importance of the invariants I1and I
2, termed the
principal invariants.
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Pauli Pedersen: 12. An index to matrices
INVARIANTS For the square, symmetric matrix [A] of order 3 we have
of symmetric, similar
matrices of order 3
[A]=
A11
A12
A13
A12
A22
A23
A13
A23
A33
with invariants being the trace I1by
I1= A
11+A
22+A
33
the norm I2by
I2= A
11A
22– A2
12+ A
22A
33– A2
23+ A
11A
33– A2
13
and the determinant I3by
I3= |[A]|
These three invariants are the principal invariants and they give the
characteristic polynomium by
λ3– I
1λ2+ I
2λ – I
3= 0
The squared length of the vector A contracted from [A] by
AT= A11, A
22, A
33, 2 A
12, 2 A
13, 2 A
23
isI4= A2
11+A2
22+A2
33+ 2A2
12+ 2A2
13+ 2A2
23
related to the principal invariants by
I4= (I
1)2 – 2I
2
and therefore another invariant, equal to the squared Frobenius norm.
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Pauli Pedersen: 12. An index to matrices
INVERSE The inverse of a square, regularmatrix is the square matrix, where the
of a matrix product of the two matrices is the identity matrix. The notation [ ]–1
is used for the inverse
[A]–1[A]= [A][A]
–1= [I]
INVERSE OF A From the matrix product in partitioned form
PARTITIONED
matrix
[A]
[C]
[B]
[D]
[E]
[G]
[F]
[H]=
[I]
[0]
[0]
[I]
follows the four matrix equations
[A][E]+ [B][G]= [I] ; [A][F]+ [B][H]= [0]
[C][E]+ [D][G]= [0] ; [C][F]+ [D][H]= [I]
Solving these we obtain (in two alternative forms)
[H]= [D]–1
– [D]–1
[C][F]
[G]= – [D]–1
[C][E]
[E]= [A] – [B][D]–1[C]–1
[F]= – [E][B][D]–1
[E]= [A]–1
– [A]–1
[B][G]
[F]= – [A]–1
[B][H]
[H]= [D] – [C][A]–1[B]–1
[G]= – [H][C][A]–1
The special case of an upper triangular matrix, i.e. [C]= [0] gives
[H]= [D]–1
[G]= [0]
[E]= [A]–1
[F]= – [A]–1
[B][D]–1
[E]= [A]–1
[F]= – [A]–1
[B][D]–1
[H]= [D]–1
[G]= [0]
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Pauli Pedersen: 12. An index to matrices
The special case of a symmetric matrix, i.e. [C]= [B]T
gives
[H]= [D]–1
– [D]–1
[B]T
[F]
[G]= – [D]–1
[B]T
[E]
[E]= [A] – [B][D]–1[B]T–1
[F]= – [E][B][D]–1
= [G]T
[E]= [A]–1
– [A]–1
[B][G]
[F]= – [A]–1
[B][H]
[H]= [D] – [B]T[A]–1[B]–1
[G]= – [H][B]T
[A]–1
= [F]T
The matrices to be inverted, are assumed to be regular.
INVERSE OF The inverse of a product of square, regular matrices is the product of
A PRODUCT the inverse of the individual multipliers, but in reverse sequence
([A][B])–1= [B]
–1[A]
–1
It follows directly from
([B]–1[A]
–1)([A][B])= [I]
INVERSE OF The inverse of a matrix of order two is given by
ORDER TWO
A11A
12
A21
A22
–1
= A22
–A12
–A21
A11
1|[A]|
with the determinant given by
|[A]|= A11A
22– A
21A
12
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INVERSE OF The inverse of a matrix of order three is given by
ORDER THREE
A11
A12
A13
A21
A22
A23
A31
A32
A33
–1
=
(A22A
33– A
32A
23) , (A
32A
13– A
12A
33) , (A
12A
23– A
22A
13)
(A31A
23– A
21A
33) , (A
11A
33– A
31A
13) , (A
21A
13– A
11A
23)
(A21A
32– A
31A
22) , (A
31A
12– A
11A
32) , (A
11A
22– A
21A
12)
1|[A]|
With the determinant given by
|[A]|=
A11A
22A
33+A
12A
23A
31+A
13A
21A
32– A
31A
22A
13– A
32A
23A
11– A
33A
21A
12
INVERSE OF The inverse and the transpose transformations can be interchanged
TRANSPOSED
matrix
([A]T)–1= ([A]–1)
T= [A]
–T
from which follows the definition of the symbol [ ]–T
.
JACOBI The Jacobi matrix [J] is a square functional matrix. We define it here
matrix as thematrix containing the derivatives of the elements of a vector A
with respect to the elements of a vector B , both of order n
[J]=∂A
∂Bwith J
ij =∂A i
∂Bj
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JACOBIAN The Jacobian J is the determinant of the Jacobi matrix, i.e.
determinant
J= |[J]|
and thus a scalar.
JORDAN BLOCKS A Jordan block is a square upper---triangular matrix of order equal to
themultiplicity of an eigenvalue with a single corresponding eigenvec-
tor. All diagonal elements are theeigenvalue andall theelements of the
first upper codiagonal are 1 . Remaining elements are zero. Thus the
Jordan block [Jλ] of order 3 corresponding to the eigenvalue λ is
[Jλ]=
λ
00
1λ
0
01λ
Multiple eigenvalues with linear independent eigenvectors belongs to
different Jordan blocks.
Jordan blocks or order 1 aremost common, as this results for eigenva-
lue problems described by symmetric matrices.
JORDAN FORM The Jordan form of a square matrix [A] is the similar matrix [J] con-
sisting of Jordan blocks along the diagonal (block diagonal), and with
remaining elements equal to zero.
Only when we havemultiple eigenvalues with a single eigenvector will
the Jordan form be different from pure diagonal form. Jordan forms
represent the closest---to---diagonal outcome of a similarity transfor-
mation.
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LAPLACIAN See determinants by minors/cofactors.
EXPANSION
of determinants
LEFT The left eigenvector ΨT (row matrix) corresponding to eigenvalue
eigenvector λiis defined by
ΨT
i ([A] – λi[B])= 0
T
see eigenvalue problem.
LENGTH The length |A| of a vector is the square---root of the scalar product
of a vector of the vector with itself
|A|= ATA
A geometric vector has an invariant length, but this do not hold for all
algebraic vector definitions.
LINEAR Consider a matrix [A] of order m× n , constituting the n vectors
DEPENDENCE / Aifor i= 1, 2, ..., n . Then if there exist a non---zero vector B of
LINEAR order n such that
INDEPENDENCE
[A]B= [A1A
2A
n]B= 0
then the vectors Aiare said to be linear dependent. The vector B
contains a set of linear combination factors.
If on the other hand
[A]B= 0 only for B= 0
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Pauli Pedersen: 12. An index to matrices
then the vectors Aiare said to be linear independent.
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MEMBERS See elements of a matrix.
of a matrix
MINOR The minor of a matrix element is a determinant, i.e. a scalar.
of a matrix element
The actual squarematrix corresponding to this determinant is obtained
by omitting the row and column corresponding to the actual element.
Thus, for a matrix of order 3, the minor corresponding to element A12
become
Minor(A12)=
A21
A23
A31
A33
= A21A
33– A
31A
23
MODAL The modal matrix corresponding to an eigenvalue problem is a square
matrix matrix constituting all the linear independent eigenvectors
[Φ]= [Φ1Φ
2Φ
n]
and the generalized eigenvalue problem can then be stated as
[A][Φ] – [B][Φ][Γ] = [0]
Note that the diagonalmatrix [Γ] of eigenvaluesmust be post---multi-
plied.
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MULTIPLICATION The product of two matrices is a matrix, where the resulting element
of two matrices ij is the scalar productof the i---th rowof the first matrixwith the j---th
column of the second matrix
[C] = [A][B] with Cij=K
k=1
AikBkj
The number of columns in the first matrix must be equal to the number
of rows in the second matrix (here K) .
MULTIPLICATION A matrix is multiplied by a scalar by multiplying each element by the
BY SCALAR scalar
[C]= b[A] with Cij= bAij
MULTIPLICITY In eigenvalue problems the same eigenvalue may be a multiple solu---
OF EIGENVALUES tion, mostly (but not always) corresponding to linear independent
eigenvectors.As an example a bimodal solution is a solution, where two
eigenvectors correspond to the same eigenvalue. Multiplicity of eigen-
values is also named algebraic multiplicity.
For non---symmetric eigenvalue problems multiple eigenvalues may
correspond to the same eigenvector. We then talk about, e.g., a double
eigenvalue/eigenvector solution (by contrast to a bimodal solution,
where only the eigenvalue is the same). Thismultiplicity is described by
the geometric multiplicity of the eigenvalue. For a specific eigenvalue
we have
1≤ geometric multiplicity≤ algebraic multiplicity
Note that the geometric multiplicity of an eigenvalue counts the number of linear independent
eigenvectors for this eigenvalue, and not the number of times that the eigenvector is a solution.
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Pauli Pedersen: 12. An index to matrices
NEGATIVE DEFINITE A square, real matrix [A] is called negative or negative definite if for
matrix any non---zero vector (column matrix) X we have
XT[A]X< 0
The matrix is called negative semi---definite if
XT[A]X≤ 0
NORMALIZATION Eigenvectors can be multiplied with an arbitrary constant (even a
of a vector complex constant). Thus we have the possibility for a convenient scal-
ing, and often we choose the weighted norm. Here we scale the vector
Aito the normalized vector Φ
i
Φi= A
i AT
i [B]Ai
by which we obtain
ΦTi [B]Φ
i= 1
Alternative normalizations are by other norms, such as the 2---norm
Φi = A
i ATiA
i
or by the ∞ ---norm
Φi = A
i(Max|A j|)
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NULL A null matrix (symbolized [0]) is a matrix where all elements have the
matrix value zero
[0] := [A] with Aij= 0 for all ij
A null matrix is also called a zero matrix. The null vector is a special
case.
ONE A one matrix (symbolized [1]) is a matrix where all elements have the
matrix value one
[1] := [A] with Aij= 1 for all ij
The one vector is a special case. Note the contrast to the identity (unit)
matrix [I] , which is a diagonal matrix.
ORDER The order of a matrix is the (number of rows)×(number of columns) .
of a matrix Usually the letters m× n are used, and a rowmatrix thenhas theorder
1× n while a column matrix has the order m× 1 . For square
matrices a single number gives the order. The order of a matrix is also
called the dimensions or the size of the matrix.
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ORTHOGONALITY For an eigenvalue problem ([A] – λi[B])Φ
i= 0 with symmetric
conditions matrices [A] and [B] the biorthogonality conditions simplifies to
ΦTj [B]Φi= 0 , Φ
Tj [A]Φi= 0
for non---equal eigenvalues, i.e. λi ≠ λj .
For standard form eigenvalue problems with [A] symmetric this fur-
ther simplifies to
ΦTj Φ
i= 0 , ΦTj [A]Φ
i= 0 for λi≠ λj
Using normalization of the eigenvectors we can obtain
ΦTi [B]Φ
i= 1 or ΦTi Φ
i= 1
and thus
ΦTi [A]Φ
i= λi
Orthogonal, normalized eigenvectors are termed orthonormal.
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ORTHOGONAL An orthogonal transformation of a square matrix [A] to a square
transformations matrix [B] of the same order is by the orthogonal transformation
matrix
[T]–1= [T]
T
and thus the transformation is both a congruence transformation and
a similarity transformation
[B]= [T]T[A][T]= [T]–1[A][T]
Matrices [A] and [B] are said to be orthogonal similar, and have same
rank, same eigenvalues, same trace and same determinant (same
invariants).
If matrix [A] is symmetric, matrix [B] is also symmetric, which do not
hold generally for similar matrices.
ORTHONORMAL A orthonormal set of vectors Xi fulfill the conditions
XTi [A]X
j= 01 for
for
i≠ j
i= j
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PARTITIONING Partitioning ofmatrices is a very important tool to get closer insight and
of matrices overview. By the example
[A]=[A]
11[A]
12
[A]21
[A]22
we see that the submatrices are given indices exactly like thematrix ele-
ments themselves.
Multiplication on submatrix level is identical to multiplication on ele-
ment level. For example see inverse of a partitioned matrix.
POSITIVE DEFINITE Asquare, realmatrix [A] is called positive or positive definite if for any
matrix non---zero vector (column matrix) X we have
XT[A]X> 0
The matrix is called positive semi---definite if
XT[A]X≥ 0
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Pauli Pedersen: 12. An index to matrices
POSITIVE DEFINITE The conditions for a square matrix [A] to be positive definite can be
matrix conditions stated in many alternative forms. From the Routh---Hurwitz---Lie-
nard---Chipart teoremwe can directly in termsofHurwitzdeterminants
obtain the necessary and sufficient conditions for eigenvalues with pos-
itive real part.
For a matrix of order 2 we get that
[A]= A11
A21
A12
A22
has positive real part of all eigenvalues if and only if
(A11+ A
22)> 0 and A
11A
22– A
12A
21> 0
and the conditions for a symmetricmatrix (A21= A
12) to be positive
definite is then
A11> 0 , A
22> 0 and A
11A
22– A2
12> 0
For a matrix of order 3 we get that
[A]=
A11
A21
A31
A12
A22
A32
A13
A23
A33
has positive real part of all eigenvalues if and only if
I1= (A
11+A
22+A
33)> 0
I2= (A
11A
22– A
21A
12)+ (A
22A
33– A
32A
23)+ (A
11A
33– A
31A
13)> 0
I3= |[A]|> 0 and I
1I2– I
3> 0
and the conditions for a symmetric matrix to be positive definite will
then be
A11> 0 , A
22> 0 , A
33> 0
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Pauli Pedersen: 12. An index to matrices
A11A
22– A2
12> 0 , A
22A33
– A2
23> 0 , A
11A
33– A2
13> 0 , |[A]|> 0
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Pauli Pedersen: 12. An index to matrices
POSITIVE DEFINITE Assume that the two square, real matrices [A] and [B] of the same
SUM order are positive definite, then their sum is also positive definite.
of matrices Using the symbol for positive definite, we have
[A] 0 , [B] 0 ⇒ ([A]+ [B]) 0
It follows directly from the definition
XT([A]+ [B])X= XT[A]X+ XT[B]X> 0
because both terms are positive for X≠ 0 .
From this also follows directly that
α[A]+ (1 – α)[B] 0 for 0≤ α≤ 1
which implies that [A] 0 is a convex condition.
Identical relations hold for negative definite matrices.
POWER The power of a square matrix [A] is symbolized by
of a matrix
[A]p= [A][A] [A] (p times)
[A]–p = [A]–1[A]–1 [A]
–1(p times)
[A]0= [I] ; [A]
p[A]
r= [A]
(p+r); [A]
pr = [A]pr
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Pauli Pedersen: 12. An index to matrices
PRINCIPAL Theprincipal invariants are the coefficients of the characteristic poly---
INVARIANTS nomium for similar matrices.
PRINCIPAL The principal submatrices of the squarematrix [A] of order n , are the
SUBMATRIX n squared matrices of order k (1≤ k≤ n) found in the upper left
corner of [A] .
PRODUCT See multiplication of two matrices.
of two matrices
PRODUCTS Three different products of vectors are defined. The scalar product or
of two vectors dot product resulting in a scalar. The vector product or cross product
resulting in a vector, and especially used for vectors of order three.
Finally, the dyadic product resulting in a matrix.
PROJECTION A projection matrix different from the identity matrix [I] is a square
matrix singular matrix that is unchanged when multiplied by itself
[P][P]= [P] , [P]–1
non–existent
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PSEUDOINVERSE The pseudoinverse [A+] of a rectangular matrix [A] of order m× n
of a matrix always exists. When [A] is a regular matrix the pseudoinverse is the
same as the inverse. Given the singular value decomposition of [A] by
[A]= [T1][B][T2
]T
then with the diagonal matrix [C] of order n×m defined from the
diagonal matrix [B] of order m× n by
[C] from C ii= 1Bii for B ii≠ 0 (other C ij= 0)
the pseudoinverse [A+] is given by the product
[A+]= [T2][C][T1
]T
Case 1: [A] is a n×m matrix where n> m . The solution to
[A]X= B with the objective of minimizing the error
eTe , e= [A]X− B , is given by
X= [A]T[A]
−1
[A]TB
Case 2: [A] is a n×m matrix where n< m . The solution to
[A]X= B with the objective of minimizing the length of the solu-
tion XTX , is given by
X= [A]
T[A][A]T−1
B
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QUADRATIC By a symmetricmatrix [A] of order n we define the associated qua---
FORM dratic form
XT[A]X
that gives a homogeneous, second order polynomial in the n parame-
ters constituting the vector X . The quadratic form is used in many
applications, and thus knowledge about its transformations, definite-
ness etc. is of vital importance.
RANK The rank of a matrix is equal to the number of linearly independent
of a matrix rows (or columns) of the matrix. The rank is not changed by the trans-
pose transformation.
From a matrix [A] of order (m× n) we can, by omitting a number
of rows and/or a number of columns, get square matrices of any order
from 1 to theminimum of m,n . Normally there will be several differ-
ent matrices of each order.
The rank r is defined by the largest order of these square matrices, for
which the determinant is non---zero, i.e. the order of the “largest” regu-
lar matrix we can extract from [A] .
Only a zero matrix has the rank 0 .
The rank of any other matrix will be
1≤ r≤ min (m, n)
If r = min(m,n) we say that the matrix has full rank.
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REAL With [A] and [B] being two real and symmetricmatrices, then for the
EIGENVALUES eigenvalue problem
([A] – λi[B])Φ
i= 0
¯ if λiis complex, then Φ
iis also complex ([A] and [B] regular)
¯ if λi,Φ
iis a complex pair of solution, then the complex conju-
gated pair λi,Φ
iis also a solution.
The condition derived under biorthogonality conditions for these two
pairs is
(λi– λ
i)(ΦT
i[B]Φ
i)= 0
which expressed in real and imaginary parts are
2 Im(λi)Re(ΦTi )[B] Re(Φi)+ Im(Φ
T
i )[B] Im(Φi)= 0
It now follows that if [B] is a positive definitematrix, then Im(λi)= 0
and we have real eigenvalues.
REGULAR A non---singular matrix, see singular matrix.
matrix
RIGHT The right eigenvector Φi(column matrix) corresponding to eigen---
eigenvector values λi
is defined by
([A] – λi[B])Φ
i= 0
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Pauli Pedersen: 12. An index to matrices
see eigenvalue problem.
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Pauli Pedersen: 12. An index to matrices
ROTATIONAL For two dimensional problems we shall list some important orthogonal
transformation transformation matrices. The elements of these matrices involves
matrices trigonometric functions of the angle θ defined in the figure. For short
notation we also define
θ
c1= cosθ s
1= sin θ
c2= cos 2θ s
2= sin 2θ
c4= cos 4θ s
4= sin 4θ
The two Cartesian coordinate systems with the definition of the angle θ .
We then have for rotation of a geometric vector V of order 2
Vy =
[Γ]Vx
with [Γ]= c1–s1
,,s1c1 ; [Γ]
–1= [Γ]
T
For a symmetric matrix [A] of order 2× 2 , contracted with the
2 ---factor to the vector AT= A11, A22, 2 A12 we have
Ay= [T]Ax
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Pauli Pedersen: 12. An index to matrices
with [T]= 12
1+ c2
1 – c2
– 2 s2
,
,
,
1 – c2
1+ c2
2 s2
,
,
,
2 s2
– 2 s2
2c2
; [T]
–1= [T]
T
For a symmetricmatrix [B] of order 3× 3 , contractedwith the 2 ---
factor to the vector BT = B11, B22, B33, 2 B12, 2 B13, 2 B23 we
have
By = [R]Bx
with [R]–1= [R]
Tand [R]= 1
8·
3+ 4c2+ c
4
3 – 4c2+ c
4
2 – 2c4
2 – 2 c4
– 4s2– 2s
4
– 4s2+ 2s
4
,
,
,
,
,
,
3 – 4c2+ c
4
3+ 4c2+ c
4
2 – 2c4
2 – 2 c4
4s2– 2s
4
4s2+ 2s
4
,
,
,
,
,
,
2 – 2c4
2 – 2c4
4+ 4c4
– 2 2 + 2 2 c4
4s4
– 4s4
,
,
,
,
,
,
2 – 2 c4
2 – 2 c4
– 2 2 + 2 2 c4
6+ 2c4
2 2 s4
– 2 2 s4
,
,
,
,
,
,
4s2+ 2s
4
– 4s2+ 2s
4
– 4s4
– 2 2 s4
4c2+ 4c
4
4c2– 4c
4
,
,
,
,
,
,
4s2– 2s
4
– 4s2– 2s
4
4s4
2 2 s4
4c2– 4c
4
4c2+ 4c
4
Note that the listed orthogonal transformation matrices [Γ] , [T] and
[R] only refer to two dimensional problems, where the rotation is spe-
cified by a single parameter (the angle θ) .
ROW Arowmatrix is amatrixwith only one row, i.e.order 1× n . Thenota---
matrix tion T is used for a row matrix ( for column matrix and T for
transposed). The name row---vector or just vector is also used.
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SCALAR PRODUCT The scalar product of two vectors A and B of the same order n
of two vectors results in a scalar C
(standard
Euclidean norm) C= ATB=n
i=1
AiB
i
The scalar product is also called the dot product.
SCALAR PRODUCT The scalar product of two complex vectors A and B of the same
of two complex vectors order n involves the conjugate transpose transformation
(standard norm)
C= AHB=n
i=1
Re(Ai) – i Im(A
i)Re(B
i)+ i Im(B
i)
With this definition the length of a complex vector A is obtained by
|A|2= AHA=
n
i=1
Re(Ai)
2
+ Im(Ai)
2
SIMILARITY A similarity transformation of a square matrix [A] to a square matrix
transformations [B] of the same order is by the regular transformation matrix [T] of
the same order
[B]= [T]–1[A][T]
Matrices [A] and [B] are said to be similar matrices, they have the
same rank and the same eigenvalues, i.e. the same invariants, but dif-
ferent eigenvectors, related by [T] . A similarity transformation is also
an equivalence transformation.
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Pauli Pedersen: 12. An index to matrices
SINGULAR A singular matrix is a square matrix for which the corresponding
matrix determinant has the value zero, i.e.
[A] is singular if |[A]|= 0 , i.e. [A]–1 does not exist
If not singular, the matrix is called regular or non---singular.
SINGULAR VALUE Any matrix [A] of order m× n can be factorized into the product of
DECOMPOSITION an orthogonal matrix [T1] of order m , a rectangular, diagonalmatrix
[B] of order m× n and an orthogonal matrix [T2]T
of order n
[A]= [T1][B][T
2]T
The r singular values (positive values) on the diagonal of [B] are the
square roots of the non---zero eigenvalues of both [A][A]T
and
[A]T[A] ,and the columns of [T
1] are the eigenvectors of [A][A]
Tand
the columns of [T2] are the eigenvectors of [A]
T[A] .
SIZE See order of a matrix.
of a matrix
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SKEW A skew matrix is a specific skew symmetric matrix of order 3, defined
matrix to have a more workable notation for the vector product of two vectors
of order 3 . From the vector A the corresponding skew matrix is
defined by
[A~
]=
0
A3
–A2
–A3
0
A1
A2
–A1
0
by which A× B= [A~
]B .
The tilde superscript is normally used to indicate this specific matrix.
From B× A= – A× B follows
[B~
]A= – [A~
]B
SKEW SYMMETRIC A square matrix is termed skew---symmetric if the transposed trans---
matrix formation only changes the sign of the matrix
[A]T= – [A] , i.e. Aji= – Aij for all ij (Aii= 0)
The skew symmetric part of a square matrix [B] is obtained by the dif-
ference 12([B]–[B]
T) .
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Pauli Pedersen: 12. An index to matrices
SPECTRAL For a symmetric matrix a spectral decomposition is possible. The
DECOMPOSITION eigenvalues λiof the matrix [A] are factors in this decomposition
of a symmetric matrix
[A]=n
i=1
λi[B]
i=
n
i=1
λiΦ
iΦT
i
where Φi
is the eigenvector corresponding to λi
(orthonormal
eigenvectors).
SQUARE A square matrix is a matrix where the number of rows equals to the
matrix number of columns, thus the order of the matrix is n× n or simply
n .
STANDARD FORM The standard form for an eigenvalue problem is
for eigenvalue problem
[A]Φi= λ
iΦ
i
or
ΨTi[A]= λ
iΨT
i
see eigenvalue problem.
SUBTRACTION Matrices are subtracted by subtracting the corresponding elements
of matrices
[C]= [A] – [B] with Cij= A ij – Bij
The matrices must have the same order.
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SYMMETRIC With [A] and [B] being two symmetric matrices of order n , the left
EIGENVALUE eigenvectors will be equal to the right eigenvectors. From the descrip---
PROBLEM tion of eigenvalue problem this means
Ψi= Φ
i
and thus the biorthogonality conditions simplifies to the orthogonality
conditions. The symmetric eigenvalue problem have only real eigenva-
lues and real eigenvectors.
SYMMETRIC A square matrix is termed symmetric if the transposed transformation
matrix does not change the matrix
[A]T= [A] , i.e. A ji= Aij for all ij
The symmetric part of a square matrix [B] is obtained by the sum12([B]+ [B]
T) .
TRACE The trace of a square matrix [A] of order n is the sum of the diagonal
of a square matrix elements
trace([A])=n
i=1
Aii
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TRANSFORMATION The different transformations like equivalence, congruence, similarity
matrices and orthogonal are characterized by the involved square, regular trans-
formation matrices. The equivalence transformation of
[B]= [T1][A][T2
]
is a congruence transformation if [T1]= [T2
]T
and it is a similarity
transformation if [T1]= [T2
]–1
. The orthogonal transformation,
which at the same time is a congruence and a similarity transformation,
thus assumes [T1]= [T2
]T= [T2
]–1
.
TRANSPOSE The transposed of a matrix is the matrix with interchanged rows/
of a matrix columns. The superscript T is used as notation for this transformation
[B]= [A]T
with Bij= Aji for all ij
The transposed of a row matrix is a column matrix, and vise versa.
The transposed matrix of a transposed matrix is the matrix itself
([AT])T= [A]
TRANSPOSE The transposed of a product of matrices is the product of the trans---
OF A PRODUCT posed of the individual multipliers, but in reverse sequence
([A] [B])T = [B]T[A]
T
It follows directly from
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Pauli Pedersen: 12. An index to matrices
Cij=K
k=1
AikBkj and Cji=K
k=1
AjkBki =K
k=1
BkiAjk
TRIANGULAR A triangularmatrix is a squarematrix with only zeros above the diago---
matrix nal (lower triangular matrix)
[L] with Lij= 0 for j> i
or below the diagonal (upper triangular matrix)
[U] with Uij= 0 for j< i
TRIANGULARIZA-- See factorization of a matrix.
TION
of a matrix
UNIT See identity matrix.
matrix
VECTORS As a common name for row matrices and column matrices, the name
vector is used.
Some authors distinguish between geometric vectors (oriented piece of
a line) of order twoor three andalgebraic vectors.Algebraic vectors are
column matrices and row matrices of any order.
302
Pauli Pedersen: 12. An index to matrices
VECTOR PRODUCT The vector product of two vectors A and B , both of the order 3
of two vectors is a vector C defined by
C= A× B with
C1
C2
C3
=
A2B3
A3B1
A1B2
–
–
–
A3B2
A1B3
A2B1
The vector product is also called the cross product. See skewmatrix for
an easier notation.
ZERO See null matrix.
matrix
303
Pauli Pedersen: 12. An index to matrices
301
Pauli Pedersen: References, List of Symbols and Index
REFERENCES
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Bathe,K.J. (1996):FiniteElement Procedures, Prentice---Hall, 1037 p.
Cheng, G. and Pedersen, P. (1996): On Sufficiency Conditions forOpti-
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Cook, R.D., Malkus, D.S. and Plesha, M.E. (1989): Concepts and
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Crisfield, M.A. (1994): Non---linear Finite Element Analysis of Solids
and Structures, Vol. 1 + 2, Wiley, 345 + 494 p.
Frederiksen, P.S. (1992): Identification of Material Parameters in Ani-
sotropic Plates --- a Combined Numerical/Experimental Method,
Ph.D. Thesis, Department of Solid Mechanics, Technical University of
Denmark, DCAMM #S60.
Frederiksen, P.S. (1996): Application of an Improved Model for the
Identification of Material Parameters, Technical University of Den-
mark, DCAMM #531.
Frederiksen, P.S. (1997a): Numerical Studies for the Identification of
OrthotropicElasticConstants of ThickPlates, European J. ofMechan-
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Frederiksen, P.S. (1997b): Experimental Procedure andResults for the
Identification of Elastic Constants of Thick Orthotropic Plates, J. of
Composite Materials, Vol. 31, 360---382.
302
Pauli Pedersen: References, List of Symbols and Index
Green, A.E. and Zerna,W. (1954): Theoretical Elasticity, Oxford, Cla-
redon Press, 442 p.
Hammer, V.B., Bendsøe, M., Lipton, R. and Pedersen, P. (1997): Para-
metrization in LaminateDesign forOptimalCompliance, Int. J. Solids
Structures, Vol. 34, No. 4, 415---434.
Hedner, G. ed. (1992): Formelsamling i Hållfasthetslära, KTH, Stock-
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Jones, R.M (1975): Mechanics of Composite Materials, McGraw---
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Ladefoged, T. (1988): Triangular Ring Element withAnalytical Expres-
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Lanczos, C. (1949): The Variational Principles of Mechanics, Univer-
sity of Toronto Press, Third ed., 375 p.
Langhaar, H.L. (1962): EnergyMethods inAppliedMechanics, Wiley,
350 p.
Lekhnitskii, S.G (1981): Theory of Elasticity of an Anisotropic Body,
Mir Publ., Moscow, 430 p.
Levinson, M. (1981): An Accurate, Simple Theory of the Statics and
Dynamics of Elastic Plates, Mech. Res. Commun., Vol. 7, 81---87.
Love, A.E.H. (1927):ATreatise on theMathematicalTheory ofElastic-
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Elasticity, Interscience Publ., 493 p.
Mathematica (S. Wolfram) (1992): A System for Doing Mathematics
by Computer, Addition---Wesley, 2. ed., 961 p.
303
Pauli Pedersen: References, List of Symbols and Index
Meldahl, A. (1960): Plejlstangsmekanismens Kinematik og Dynamik,
Akademisk Forlag, København (in Danish), 34 p.
Muskhelishvili, N.I. (1934): A NewGeneral Method of Solution of the
Fundamental Boundary Value Problems in Plane Theory of Elasticity,
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Novozhilov, V.V. (1961): Theory of Elasticity, Pergamon Press, 448 p.
Ottosen, N. and Petersson, H. (1992): Introduction to the Finite Ele-
ment Method, Prentice---Hall, 410 p.
Pedersen, P. (1984): Et Notat om Elementanalyse for FEM, Depart-
ment of Solid Mechanics, Technical University of Denmark (in
Danish), 99 p.
Pedersen, P. (1986): A Note on Vibration of Beam---Columns, J. of
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Pedersen, P. (1988): Notes for Lectures on Laminates, Department of
Solid Mechanics, Technical University of Denmark, 140 p.
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Mechanics, Technical University of Denmark (in Danish), 140 p.
Pedersen, P. (1995a):Materials Optimization--- and EngineeringView,
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304
Pauli Pedersen: References, List of Symbols and Index
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Washizu, K. (1975): Variational Methods in Elasticity and Plasticity,
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Method, Vol. 1 & 2, McGraw---Hill, 648 + 807 p.
305
Pauli Pedersen: References, List of Symbols and Index
LIST OF SYMBOLS
(excluding some symbols in Chapter 11)
LATIN SYMBOLS
a length of a plate, distance, radius, ellipse half---axis
A area, constant, index for load case
A0
area covered by a hole
Amn Fourier coefficient
A ijkl laminate membrane stiffness
b width of a plate, radius
B constant, index for load case
Bijkl laminate coupling stiffness
c2 c4 cos(2θ), cos(4θ)
C material stiffness factor
C constant, index for load case, index for complementary
Cmn Fourier coefficient for the stress function
Cn practical parameter defined from Cijkl
Cijkl constitutive elasticity tensor (2D)
d differential prefix
dn displacement number n
D bending stiffness factor
Dijkl laminate bending stiffness
306
Pauli Pedersen: References, List of Symbols and Index
eij deviatoric strain tensor
E E100 E110 E111 modulus of elasticity
EL modulus of elasticity in the fiber length direction
ET modulus of elasticity in the transverse direction
Es secant modulus of elasticity
Et tangent modulus of elasticity
E0 reference modulus of elasticity
fn function in displacement assumption, correctional factors in torsion
G shear modulus of elasticity
GLT shear modulus of elasticity for anisotropic material
h wall thickness in a torsional rod
h thickness of a plate, height of beam
h general design parameter
I cross---sectional moment of inertia
In invariant (Inα material invariant, In strain invariant,
Inσ stress invariant)
J cross---sectional polar moment of inertia
k number of a specific ply
K cross---sectional torsional stiffness factor, total number of plies, bulk
modulus
length of torsional rod
i direction cosine
L length after deformation
L0 length before deformation
Lijkl constitutive elasticity tensor (3D)
307
Pauli Pedersen: References, List of Symbols and Index
m number index, ratio of axis
mi direction cosine, compliance parameter
MT Mx torsional moment
Mij bending moment in a plate
(Mii bending moment, Mij (i≠ j) torsional moment)
Mijkl material compliance tensor
n exponent in a power law
n number index
ni direction cosine
N normal force, total number of n, index for normal
Nαβ membrane force in a plate
(Nii normal force, Nij (i≠ j) shear force)
p exponent in a power law
p pressure (force per area)
p p i volume force, volume force in the i---direction
pm function for a pressure on a plate
pmn Fourier coefficient for a pressure on a plate
P force
P name of a material point, index for principal, index for plane
q qi line load, line load in the i---direction
Q Qi force, force in the i---direction
r radius
R outer radius
s natural parameter
s2 s4 sin(2θ), sin(4θ)
sij deviatoric stress tensor
S index for shear
308
Pauli Pedersen: References, List of Symbols and Index
t thickness
T index for transverse
T Ti superscript for transpose, transverse force, transverse force in the
i---direction
T Ti surface traction, surface traction in the i---direction
u strain energy density
uC stress energy density (complementary energy density)
U strain energy
UC stress energy (complementary energy)
v vi displacement, displacement in the i---direction
vi,j displacement gradient
V volume
w displacement in the transverse direction
wm function for transverse displacement
wmn Fourier coefficient for transverse displacement
W work of external forces, index for out---of---plane
WC complementary work of external forces
WT torsional resistance
x coordinate system and index referring to this
xi coordinate axis in the i---direction
y coordinate system and index referring to this
y i coordinate axis in the i---direction
z transverse coordinate for a plate (= x3) or a beam
z zi specific moment, specific moment in the i---direction
Z Zi moment, moment in the i---direction
zk− zk–1 domain for ply k in a laminate
309
Pauli Pedersen: References, List of Symbols and Index
GREEK SYMBOLS
α angle
α α1 α2 dimensionless distance in the plate length direction
αm dimensionless material parameter
αijkl dimensionless constitutive tensor
β factor from the shear distribution
β β1 β2 dimensionless distance in the plate width direction
βm dimensionless material parameter
γ angle from axis of material reference (positive counterclockwise)
γ angle of rotation on a cylinder surface, ratio
γij γ for i≠ j engineering shear strain = 2ij , resulting shear strain
Γ slenderness ratio
∂ partial differentiation prefix
δ variational prefix
δij Kronecker delta (1 for i= j , 0 for i≠ j)
∆ incremental prefix
normal strain (Cauchy C ; Hencky H ; Green---Lagrange G ;
Almansi A ; Swaiger S ; Kuhn K)
mean normal strain
e, 0 effective strain and reference strain
the strain field as a whole
n principal strain
ij strain tensor (ii normal strain; ij (i≠ j) shear strain)
η ratio
ηij Green---Lagrange strain tensor
ηmn mode parameter
310
Pauli Pedersen: References, List of Symbols and Index
θ angle from the x1--- to the y1---axis (positive counterclockwise)
θ angle rotation per length for torsion, angle of stress point
λ stretch (extension ratio), Lame constitutive parameter, ratio
λn eigenvalue of constitutive matrix
ratio, factor from the shear distribution
n Lame constitutive parameter
ν Poisson’s ratio
νLT Poisson’s ratio for anisotropic material
Π total potential
ΠC total complementary potential
σ normal stress
σ mean normal stress = hydrostatic stress
σe, σ0 effective stress and reference stress
σ stress field as a whole
σn principal stress
σij stress tensor (σii normal stress; σij (i≠ j) shear stress)
σij,j sum of stress gradients
σ∞
stress at infinity
τ resulting shear stress
τnm shear stress in the wall between hole number n and m
angle rotation of torsional cross---section, angle of load
a bc angle of gauge direction
Φ stress function, function of bending stiffnesses
ψ angle from the 1--- to the y
1---axis (positive counterclockwise)
Ψ warping function
311
Pauli Pedersen: References, List of Symbols and Index
ω rotation angle
Ω Ωn domain, subdomain
MATRICES AND VECTORS
[A] plate membrane stiffnesses
[B] plate coupling stiffnesses
[C] [Ck] constitutive matrix (2D), [C
k] constitutive matrix for ply number k
[D] plate bending stiffnesses
e vector of deviatoric strains
[E] plate membrane flexibilities
[F] transformation matrix for gauge strains
[H] plate bending flexibilities
[I] identity matrix
vector of direction cosines
[L] constitutive matrix (3D)
[L] plate coupling flexibilities
m vector of direction cosines
M vector of moments (per unit length) in a plate
n vector of direction cosines
N vector of membrane forces (per unit length) in a plate
p vector of volume forces
312
Pauli Pedersen: References, List of Symbols and Index
[P] projection matrix
Q force vector
[R] orthogonal transformation matrix
s vector of deviatoric stresses
[T] orthogonal transformation matrix
v displacement vector
Z vector of moments
[α] dimensionless constitutive matrix
α vector of dimensionless constitutive components
[] strain matrix
strain vector
0 strain vector for middle surface
bending vector
vector of direction cosines
[σ] stress matrix
σ stress vector
n eigenvector corresponding to material eigenvalue
[Γ] orthogonal transformation matrix
313
Pauli Pedersen: References, List of Symbols and Index
INDEX(see also the matrix index in chapter 12)
A
accuracy?, 135
advanced, laminate, models, 153
Almansi, strain, 13
angle---ply, laminate, 151
anisotropic, elasticity, 61
approximate
solution, 132
stress, field, 136
area, force, 35
artificial, dilatation, field, 251
axisymmetric, ringelement, 213
B
balanced, laminate, 152
bar, element, 211
basic, matrice, 225
beam, result, 129
bending, stiffness, 148
books, on, FEM, 210
bound, 123
boundary, condition, 160
Bredt’s, formula, 200, 205
bulk, modulus, 86
C
Castigliano’s1st, theorem, 1132nd, theorem, 115
Cauchy, strain, 12, 15
center, of, torsion, 172
characteristic, polynomium, 22
circular
cross---section, 172hole, 94
classical, laminate, analysis, 141
compatibility, condition, 28
complementaryapproximation, 135energy, density, 72
potential, energy, 108virtual, work, principle, 115, 199work, 108
compliance, matrix, 63
compressive, dilatation, 248
condition, for orthotropy, 60
configuration, matrix, 217
consistent, mass, matrice, 240
constant, tangential, stress, 95
constitutivematrix, 63, 85
model, 53, 120relation, 54secant, modulus, 76tangent, matrix, 77
coupling, stiffness, 147
cross---ply, laminate, 151
314
Pauli Pedersen: References, List of Symbols and Index
cross---sectional, torsion, factor, 175
cubic
space, centered, 87
surface, centered, 87
symmetry, 87
curvature, and, twist, vector, 143
cylindrical, bar, 171
D
dead, load, 110
degree, of, freedom, 216
design, 100
determinant, norm, 21, 44
deviatoric
strain, 27
stress, 45
dilatation, 26
directional
general, equilibrium, 223
variation, 222
disc, element, 211
displacement, 9
assumption, 213, 215
field, 10, 107, 154
function, 214
gradient, 10, 246
vector, 9, 213
distortion, 26, 27
double
Fourier, expansion, 189
sine, expansion, 162
E
effective
strain, 75
stress, 75
eigenmode, 85
eigenvalue, 85
elastic, energy, in, straight, beam, 125
element, geometry, 210
elliptical
boundary, 101
hole, 93
energy
per, length, 126principle, 105
engineering
modulus, 64, 79
parameter, 84
strain, 12
equilateral, triangle, 186
equilibrium, 39
equivalent, nodal, load, 240
error, functional, 83
expansive, dilatation, 249
experimental, strategy, 83
extension, ratio, 12
external
force, 36
moment, 37
potential, 111torsional, moment, 174
extreme, normal, strain, 22
extremum, principle, for, total, potential, energy,122
315
Pauli Pedersen: References, List of Symbols and Index
F
FEM, 209
finite, element, method, 209
force, 35
equilibrium, 40, 158
per area, 35per length, 35
per volume, 35
four, rectangular, point, load, 167
Fourier
coefficient, 165, 190
expansion, 164
Frobenius, norm, 21, 44
from, 3---D to 2---D, 67
function, space, 214
G
general, equilibrium, 222
generalised, force, 113
generalized, stress, 128
Green---Lagrange, strain, 13, 15
H
Hencky, strain, 13
high, shearcompliance, 64
stiffness, 63
higher, order, plate, theory, 153
hollow, elliptic, cross---section, 180
hydrostatic, stress, 45
I
identification, 83
incompressible, material, 27
initial
strain, 244stress, 244
integration, over, atetrahedron, 228
triangle, 228
internalforce, 36moment, 37
pressure, 46
interpolation, 217
inverse, formulas, 84
invert, analytically, 217
isotropic, case, 62
K
kinematic, boundary, condition, 160
Kuhn, strain, 14
L
Lame, parameter, 88
lamina, 142
laminate, 141
ply, 80stiffness, 145
lamination, parameter, 153
Laplace, differential, equation, 177
layer, 142
Levy, solution, 168
316
Pauli Pedersen: References, List of Symbols and Index
line, force, 35
linear
elastic, 110
elasticity, 53strain, 12
linearly, changing, pressure, 165
loadbehaviour, 117
matrice, 241
logarithmic, strain, 13
longitudinalstrain, 15
stress, 38
low, shearcompliance, 64
stiffness, 63
M
mass, matrix, 241
material
compliance, tensor, 54point, 10, 38
stiffness, 145tensor, 53
maximum, laminate, stiffness, 153
meannormal, stress, 45
value, of normal strain, 28
measured, gauge, normal, strain, 82
measuring, 79
membranestiffness, 147
strain, vector, 143
minimumcomplementary, potential, 122
energy, concentration, 100stress, concentration, 95total, potential, 116
mode, parameter, 163
model, for
analysis, 81experiment, 81
modulus, matrix, 80
moment, 37equilibrium, 41, 158per area, 37
per length, 37per volume, 37
N
Navier, solution, 162
nodalcompatibility, 214degree, of, freedom, 216point, 210
non---circular, cross---section, 175
non--- linear, elasticity, 72
non---orthotropic, case, 62
norm, 21, 44
normalstrain, 15
in a direction, 24stress, 38
O
optimalboundary, shape, 101decision, 100
redesign, 100shape, design, 95
317
Pauli Pedersen: References, List of Symbols and Index
optimization, problem, 83
orientational, design, 102
orthogonal, transformation, 18, 25
orthotropiccase, 55, 62
direction, 59
material, 59, 144
P
plane
strain, 68assumption, 70
case, 71
stress, 69
assumption, 70
case, 71
plate
differential, equation, 159
equation, 160
theory, 154
ply, 141, 142
point, load, 167
Poisson, differential, equation, 179
Poisson’s, ratio, 79
polar, moment, of, inertia, 174
polynomial
description, 214integration, 228
positive, invariant, 61
potential, energy, 108
power, law, non--- linear, elasticity, 72
practical, parameter, 80
pressure, 35
principalmaterial, direction, 60strain, 19stress, 44
principle, ofcomplementary, virtual, work, 115minimum, total, potential, energy, 117stationary
complementary, potential, energy, 116potential, energy, 114
virtualdisplacement, 112stress, 115
prismatic, bar, 171
projection, matrix, 27, 46
purebending, 155shear, 20
strain, 173stress, 174
R
realdisplacement, field, 109, 114stress, field, 109, 111
rectangularline, load, 167strip, 195
referencemodulus, 76strain, 76stress, 78value, 78
resultingshear, stress, 180
field, 194tangential, shear, stress, 174
rigid, bodyrotation, 247
318
Pauli Pedersen: References, List of Symbols and Index
translation, 247
rotational, transformation, 18, 44, 56
S
secantmodulus, 74
stiffness, tensor, 53
sensitivity, analysis, 100
shapedesign, 100
function, 217
sheardeformation, 141force, flow, 198
strain, 15stress, 39
sign
definition, 155for the stress, 39
simply, supported, 162
skew---symmetric, laminate, 150
slenderness, ratio, 130
specially, orthotropic, laminate, 152
spectral, decomposition, 24, 25, 44, 45
spherical, shell, 46
square, 44
static, boundary, condition, 161
stiffness, 100submatrix, 225
strain, 11
calculation, 29
concept, 11decomposition, 28energy, 108
density, 72, 108
evaluation, 245field, 11, 107gauge, 11
invariant, 21matrix, 17tensor, 15, 17transformation, 20
vector, 17
strain/displacement, matrix, 221
strength, 100
stresscomponent, 39concentration, 93
factor, 95, 188
concept, 38energy, 108
density, 72, 108
field, 107function, 178invariant, 44matrix, 38
tensor, 38vector, 38
stretch, 11
structural, equilibrium, 157
submatrix, description, 67
surfaceorientation, 38traction, 38, 107, 177
equilibrium, 42
Swaiger, strain, 14
symmetric, laminate, 149
319
Pauli Pedersen: References, List of Symbols and Index
T
tangentmodulus, 74stiffness, tensor, 53
tangential
strain, 15stress, 39
tensor, transformation, 56
thin---walledclosed, cross---section, 198open, cross---section, 195
torsional
resistance, 175, 180, 197rod, 171stiffness, 175
constant, 197
factor, 180
trace, norm, 21, 44
transversal, isotropy, 90
transverse, displacement, 156
two, sided, bound, 122, 139
U
uniformline, load, 166
pressure, 165
unit
displacement
field, 113
theorem, 113
load, theorem, 116
V
virtual
displacement, field, 111
stress, field, 114
work, principle, 112
volume
element, 212
force, 35, 107
equilibrium, 41
W
warping, 184
displacement, field, 193
function, 176
work, 108
equation, 105
function, 118
320
Pauli Pedersen: References, List of Symbols and Index