ELASTICITY --- ANISOTROPY --- LAMINATES SOLID MECHANICS

334
ELASTICITY --- ANISOTROPY --- LAMINATES SOLID MECHANICS

Transcript of ELASTICITY --- ANISOTROPY --- LAMINATES SOLID MECHANICS

Page 1: ELASTICITY --- ANISOTROPY --- LAMINATES SOLID MECHANICS

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

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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

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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

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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

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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

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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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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

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Pauli Pedersen: 1. Contents and Introduction

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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.

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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,

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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.

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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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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.

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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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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: 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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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

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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.

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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).

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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.

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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

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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.

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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

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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

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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

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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

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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-

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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

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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)

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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)

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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

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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).

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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

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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

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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 .

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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)

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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

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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.

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Pauli Pedersen: 2. Geometry --- Kinematics

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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.

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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

qq

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.

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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.

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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

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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.

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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.

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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.

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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)

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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

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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

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

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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–

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)

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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.

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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

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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

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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

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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

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)

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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).

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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.

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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)

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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)

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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

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 .

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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

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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

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Pauli Pedersen: 4. Physics --- Constitutive Models

α2:=

1

1111– α

2222x

α3:=

1

1111+ α

2222– 2α

1122+ 2α

1212x

(4.13)

α6:= 1

1112+ α

2212x

α7:= 1

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)

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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

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)

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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)).

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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

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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

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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

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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:

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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.

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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.

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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

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

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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

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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)

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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

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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)

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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)

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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.

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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)

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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

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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

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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

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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.

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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)

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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

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)

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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)

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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)

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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.

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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.

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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

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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

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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.

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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)

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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)

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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

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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

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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)

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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.

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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)

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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.

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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)

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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 .

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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).

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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).

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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.

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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.

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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.

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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.

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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)

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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

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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.

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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.

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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

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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

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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

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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.

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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)

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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)

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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.

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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

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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)

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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)

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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 .

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Pauli Pedersen: 7. Application of Energy Principles

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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.

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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.

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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.

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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

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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

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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.

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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 .

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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

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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.

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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 .

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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.

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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.

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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.

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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).

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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)

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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)

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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)

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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.

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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

=

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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

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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

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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)

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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

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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

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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

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)

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Pauli Pedersen: 9. Bending of Rectangular, Orthotropic Plates

Linearly changing pressure: Q= pα2– α

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)

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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

QQ

α1a

α2a

β1b

β2b

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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 .

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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)

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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 .

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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= .

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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

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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

~

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)

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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)

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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

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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

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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

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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).

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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

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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.

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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

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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 ,

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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

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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

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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.

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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)

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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)

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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.

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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)

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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

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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).

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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.

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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.

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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.

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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.

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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

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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)

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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)

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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)

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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

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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

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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

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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

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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

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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)

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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.

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Pauli Pedersen: 10. Torsion of Cylindrical Bars

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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.

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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

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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

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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

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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

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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)

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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

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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

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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

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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

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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

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

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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)

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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

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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.

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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).

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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

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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).

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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).

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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

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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)

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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

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Pauli Pedersen: 11. Finite element analysis

2---D disc elements (t is disc thickness)

LDTR: [T11]= t

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

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

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Pauli Pedersen: 11. Finite element analysis

[T12]= t

6

3(1 – β)

(1 – β)

00

0

–4(1 – β)

–β

–3β

00

0

1

–1

04

–4

0

0

4

04(1 – 2β)

–4(1 – 2β)

–4

–4

0

0–4(1 – 2β)

4(1 – 2β)

4

–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]=

1

symm.

–1

1

0

00

0

000

[T

22]= hα

(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]=

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 .

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Pauli Pedersen: 11. Finite element analysis

QDTE:

[T11]=

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

− 4β2

8β1− β

− 4β1− β

8β2

1− β

β

− 1

0

− 4

− 81− β

41− β

− 8β

8

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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)

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

0

1

–1

000

–4

–8

4

8

0

T13=

30

3

0

0

–4

0

η

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

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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)

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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

3

symm.

13

000

0008

000− 88

− 4− 40008

3

symm.

13

000

2− 2024

− 220− 2424

− 4− 40008

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Pauli Pedersen: 11. Finite element analysis

Trr

=

πhα15

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

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

31 – β

1 – β

0

0

0

–41 – β

–β

–3β

0

0

0

1

–1

0

4

–4

0

0

4

0

41 – 2β

–41 – 2β

–4

–4

0

0

–41 – 2β

41 – 2β

4

–41 – β

0

–4

4

41 – 2β

31 – β

1 – β

0

21 – β

–21 – β

–41 – β

–β

–3β

0

–2β

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

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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

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.

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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)

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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

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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

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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)

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

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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

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

0dV (11.50)

and the nodal loads equivalent to initial strains are

A0

i=

V

[B]T

i[L]

0dV (11.51)

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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

–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.

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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.

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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

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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)

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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

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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.

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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.

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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.

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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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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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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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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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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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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

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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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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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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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Pauli Pedersen: 12. An index to matrices

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

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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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

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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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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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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A11A

22– A2

12> 0 , A

22A33

– A2

23> 0 , A

11A

33– A2

13> 0 , |[A]|> 0

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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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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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see eigenvalue problem.

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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

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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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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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

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.

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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

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Pauli Pedersen: 12. An index to matrices

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Pauli Pedersen: References, List of Symbols and Index

REFERENCES

Arnold, V.J. (1977---1988):GeometricalMethods in theTheory ofOrdi-

nary Differential Equations, Springer, Second ed., 351 p.

Bathe,K.J. (1996):FiniteElement Procedures, Prentice---Hall, 1037 p.

Cheng, G. and Pedersen, P. (1996): On Sufficiency Conditions forOpti-

mal Design based on Extremum Principles of Mechanics, J. of the

Mechanics and Physics of Solids (in press).

Cook, R.D., Malkus, D.S. and Plesha, M.E. (1989): Concepts and

Applications of Finite Element Analysis, Wiley, 3. ed., 630 p.

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-

ics, A/Solids, Vol. 16, 117---140.

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.

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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-

holm (in Swedish), 384 p.

Jones, R.M (1975): Mechanics of Composite Materials, McGraw---

Hill, 365 p.

Ladefoged, T. (1988): Triangular Ring Element withAnalytical Expres-

sions for Stiffness and Mass Matrix, Comp. Meth. in Appl. Mech. and

Eng., Vol. 67, 171---187.

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-

ity, Cambridge University Press, Fourth ed., 643 p.

Lur’e, A.I. (1964): Three---Dimensional Problems of the Theory of

Elasticity, Interscience Publ., 493 p.

Mathematica (S. Wolfram) (1992): A System for Doing Mathematics

by Computer, Addition---Wesley, 2. ed., 961 p.

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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,

Dokl. Akad. Nauk SSSR, III, No. 1, 7.

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

Sound and Vibration, Vol. 105, No. 1, 143---150.

Pedersen, P. (1988): Notes for Lectures on Laminates, Department of

Solid Mechanics, Technical University of Denmark, 140 p.

Pedersen, P. (1992): Noter til Elasticitetslære, Department of Solid

Mechanics, Technical University of Denmark (in Danish), 140 p.

Pedersen, P. (1995a):Materials Optimization--- and EngineeringView,

in J. Herskovits (ed.): Advances in Structural Optimization, Kluwer

Acad. Publ., 223---262.

Pedersen, P. (1995b): Simple Transformations by Prober Contracted

Forms: Can we Change the Usual Practice?, Comm. in Num. Meth. in

Eng., Vol. 11, 821---829.

Pedersen, P. and Frederiksen, P.S. (1992): Identification ofOrthotropic

Material Moduli by a Combined Experimental/Numerical Method,

Measurement, Vol. 10, No. 3, 113---118.

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Pauli Pedersen: References, List of Symbols and Index

Pedersen, P. andMegahed,M.M. (1975): Axisymmetric Element Anal-

ysis using Analytical Computing, Comp. & Struct., Vol. 5, 241---247.

Pedersen, P., Tobiesen, L. and Jensen, S.H. (1992): Shapes of Ortho-

tropic Plates forMinimumEnergy Concentration, Mechanics of Struc-

tures and Machines, Vol. 20, No. 4, 499---514.

Phan, N.D. and Reddy, J.N. (1985): Analysis of Laminated Composite

Plates using a Higher---Order Shear Deformation Theory, Int. J. Num.

Meth. Engrg., Vol. 21, 2201---2219.

Reddy, J.N. (1984): A Simple Higher Order Theory for Laminated

Composite Plates, J. Appl. Mech., Vol. 51, 745---752.

Savin, G.N. (1961): Stress Concentration around Holes, Pergamon

Press, (translation of original from 1951), 430 p.

Sokolnikoff, I.S. (1956):MathematicalTheoryofElasticity, McGraw---

Hill, 476 p.

Tsai, S.W. and Hahn, H.T. (1980): Introduction to Composite Materi-

als, Technomic Publ. Co, 457 p.

Vinson, J.R. and Sierakowski, R.L. (1986): The Behaviour of Struc-

tures Composed of Composite Materials, Nijhoff, 323 p.

Washizu, K. (1975): Variational Methods in Elasticity and Plasticity,

Pergamon Press, 2. ed., 412 p.

Whitney, J.M. (1987): Structural Analysis of Laminated Anisotropic

Plates, Technomic, 342 p.

Zienkiewics, O.C. and Taylor, R.L. (1989---1991): The Finite Element

Method, Vol. 1 & 2, McGraw---Hill, 648 + 807 p.

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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

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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)

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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Pauli Pedersen: References, List of Symbols and Index